ERWIN KREYSZIG

ERWIN KREYSZIG ADVANCED ENGINEERING •••• MATHEMATICS •••• n .,: I . ' . , ~.;... <,. .•• .. '. • .. ,~. .'~~~ .,.L......· -_., . ::: .' L • ." ~.::::----.... • , "" .... " '" :' '.. " .. - I' 1,1 11 't. "'11 ~ r . i :,t::a'. .. '.. :,., ,-" ""', ~;. ;' ,-:~.;;. ~.~: ,.". "':.'. '-,".~.. ; ,..... .... f '~":~"_ ;11,",; . • - _.. ." "'-;:-"'" '_: . . ,;~" ",. . . ',-' , :":~:~,:.. ... . ,, , . -... ~~ 4' ~:-~"'" . ~,'~.. fif I •• ;;~:, '. " __~~ ,INTERNATIONAL ~ '; l\. . .. ... • ''; 'I It. . la.· '''.:'" '-....- ... ...... "-.:lJ." . . '~ \; i, ." • " "..-:. ;'1 ... l- ' \ .~. 9TH EDITION • o •• e. • ~;. .. ,.... I •• .. RESTRICTED! Not for Sale in .~ the United States .. "'" ....\t'~ ,.~. . ~ '.. :. \ . Advanced Engineering Mathematics EDITION Advanced Engineering Mathematics ERWIN KREYSZIG Professor of Mathematics Ohio State University Columbus, Ohio @ WILEY JOHN WILEY & SONS, INC. Vice President and Publisher: Laurie Rosatone Editorial Assistant: Daniel Grace Associate Production Director: Lucille Buonocore Senior Production Editor: Ken Santor Media Editor: Stefanie Liebman Cover Designer: Madelyn Lesure Cover Photo: © John Sohm/ChromosohmJPhoto Researchers This book was set in Times Roman by GGS Information Services Copyright © 2006 John Wiley & Sons. Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act. without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (508) 750-8400, fax (508) 750-4470. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, E-Mail: PERMREQ@WILEY.COM. Kreyszig, Erwin. Advanced engineering mathematics I Erwin Kreyszig.~9th ed. p. cm. Accompanied by instructor's manual. Includes bibliographical references and index. 1. Mathematical physics. 2. Engineering mathematics. 1. Title. ISBN-13: 978-0-471-72897-9 ISBN-10: 0-471-72897-7 Printed in Singapore 10 9 8 7 6 5 4 3 2 PREFACE See also http://www.wiley.com/coUege/kreyszig/ Goal of the Book. Arrangement of Material This new edition continues the tradition of providing instmctors and students with a comprehensive and up-to-date resource for teaching and learning engineering mathematics, that is, applied mathematics for engineers and physicists, mathematicians and computer scientists, as well as members of other disciplines. A course in elementary calculus is the sole prerequisite. The subject matter is arranged into seven parts A-G: A Ordinary Differential Equations (ODEs) (Chaps. 1-6) B Linear Algebra. Vector Calculus (Chaps. 7-9) C Fourier Analysis. Partial Differential Equations (PDEs) (Chaps. 11-12) D Complex Analysis (Chaps. 13-18) E Numeric Analysis (Chaps. 19-21) F Optimization, Graphs (Chaps. 22-23) G Probability, Statistics (Chaps. 24-25). This is followed by five appendices: App. App. App. App. App. I References (ordered by parts) 2 3 4 5 Answers to Odd-Numbered Problems Auxiliary Material (see also inside covers) Additional Proofs Tables of Functions. This book has helped to pave the way for the present development of engineering mathematics. By a modern approach to those areas A-G, this new edition will prepare the student for the tasks of the present and of the future. The latter can be predicted to some extent by a judicious look at the present trend. Among other features, this trend shows the appearance of more complex production processes, more extreme physical conditions (in space travel, high-speed communication, etc.), and new tasks in robotics and communication systems (e.g., fiber optics and scan statistics on random graphs) and elsewhere. This requires the refinement of existing methods and the creation of new ones. It follows that students need solid knowledge of basic principles. methods, and results, and a clear view of what engineering mathematics is all about, and that it requires proficiency in all three phases of problem solving: • Modeling, that is, translating a physical or other problem into a mathematical form, into a mathematical model; this can be an algebraic equation, a differential equation, a graph, or some other mathemalical expression. • Solving the model by selecting and applying a suitable mathematical method, often requiring numeric work on a computer. • Interpreting the mathematical result in physical or other terms to see what it practically means and implies. It would make no sense to overload students with all kinds of little things that might be of occasional use. Instead they should recognize that mathematics rests on relatively few basic concepts and involves powerful unifying principles. This should give them a firm grasp on the illterrelations amollg theory, computing, and (physical or other) experimentation. v Preface vi PART A PART B Chaps. 1-6 Ordinary Differential Equations (ODEs) Chaps. 7-10 Linear Algebra. Vector Calculus Chaps. 1-4 Basic Material I I t l Chap. 5 Series Solutions Chap. 6 t Laplace Transforms I t Chap. 7 Matrices, Linear Systems Chap. 9 Vector Differential Calculus t Chap. 8 Eigenvalue Problems Chap. 10 Vector Integral Calculus t PARTe PARTD Chaps. 11-12 Fourier Analysis. Partial Differential Equations (PDEs) Chaps. 13-18 Complex Analysis, Potential Theory Chap. 11 Fourier Analysis Chaps. 13-17 Basic Material Chap. 12 Partial Differential Equations Chap. 19 Numerics in General , Chap. 18 Potential Theory PART E PART F Chaps. 19-21 Numeric Analysis Chaps. 22-23 Optimization, Graphs Chap. 20 Numeric Linear Algebra Chap. 21 Numerics for ODEs and PDEs Chap. 22 Linear Programming Chap. 23 Graphs, Optimization PARTG GUIDES AND MANUALS Chaps. 24-25 Probability, Statistics Maple Computer Guide Mathematica Computer Guide Chap. 24 Data Analysis. Probability Theory Student Solutions Manual 1- - - 1 - - - - - - - - - - - - - - - - - / Chap. 25 Mathematical Statistics Instructor's Manual Preface vii General Features of the Book Include: • Simplicity of examples, to make the book teachable-why choose complicated examples when simple ones are as instructive or even better? • Independence of chapters, to provide flexibility in tailoring courses to special needs. • Self-contained presentation, except for a few clearly marked places where a proof would exceed the level of the book and a reference is given instead. • Modern standard notation, to help students with other courses, modern books, and mathematical and engineering journals. Many sections were rewritten in a more detailed fashion. to make it a simpler book. This also resulted in a better balance between theory and applicatio1ls. Use of Computers The presentation is adaptable to various levels of technology and use of a computer or graphing calculator: very little or no use, medium u~e, or intensive use of a graphing calculator or of an unspecified CAS (Computer Algebra System, Maple, Mathematica, or Matlab being popular examples). In either case texts and problem sets form an entity without gaps or jumps. And many problems can be solved by hand or with a computer or both ways. (For software, see the beginnings of Part E on Numeric Analysis and Part G on Probability and Statistics.) More specifically, this new edition on the one hand gives more prominence to tasks the computer cannot do, notably, modeling and interpreting results. On the other hand, it includes CAS projects. CAS problems. and CAS experimellts, which do require a computer and show its power in solving problems that are difficult or impossible to access otherwise. Here our goal is the combination of intelligent computer use with high-quality mathematics. This has resulted in a change from a formula-centered teaching and learning of engineering mathematics to a more quantitative, project-oriented, and visual approach. CAS experiments also exhibit the computer as an instrument for observations and experimentations that may become the beginnings of new research, for "proving" or disproving conjectures, or for formalizing empirical relationships that are often quite useful to the engineer as working guidelines. These changes will also help the student in discovering the experimental aspect of modern applied mathematics. Some routille and drill work is retained as a necessity for keeping firm contact with the subject matter. In some of it the computer can (but must not) give the student a hand, but there are plenty of problems that are more suitable for pencil-and-paper work. Major Changes 1. New Problem Sets. Modem engineering mathematics is mostly teamwork. [t usually combines analytic work in the process of modeling and the use of computer algebra and numeric~ in the process of solution, followed by critical evaluation of results. Our problems-some straightforward. some more challenging, some "thinking problems" not acce~sible by a CAS, some open-ended-reflect this modem situation with its increased emphasis on qualitative methods and applications, and the problem sets take care of this novel situation by including team projects, CAS projects, and writing projects. The latter will also help the student in writing general reports, as they are required in engineering work quite frequently. 2. Computer Experiments. using the computer as an instrument of "experimental mathematics" for exploration and research (see also above). These are mostly open-ended viii Preface experiments, demonstrating the use of computers in experimentally finding results. which may be provable afterward or may be valuable heuristic qualitative guidelines to the engineer, in particular in complicated problems. 3. More on modeling and selecting methods, tasks that usually cannot be automated. 4. Student Solutions Manual and Study Guide enlarged, upon explicit requests of the users. This Manual contains worked-out solutions to carefully selected odd-numbered problems (to which App. I gives only the final answers) as well as general comments and hints on studying the text and working further problems, including explanations on the significance and character of concepts and methods in the various sections of the book. Further Changes, New Features • Electric circuits moved entirely to Chap. 2, to avoid duplication and repetition • Second-order ODEs and Higher Order ODEs placed into two separate chapters (2 and 3) • In Chap. 2, applications presented before variation of parameters • Series solutions somewhat shortened, without changing the order of sections • Material on Laplace transforms brought into a better logical order: partial fractions used earlier in a more practical approach, unit step and Dirac's delta put into separate subsequent sections, differentiation and integration of transforms (not of functions!) moved to a later section in favor of practically more important topics • Second- and third-order determinants made into a separate section for reference throughout the book • Complex matrices made optional • Three sections on curves and their application in mechanics combined in a single section • First two sections on Fourier series combined to provide a better, more direct start • Discrete and Fast Fourier Transforms included • Conformal mapping presented in a separate chapter and enlarged • Numeric analysis updated • Backward Euler method included • Stiffness of ODEs and systems discussed • List of software (in Part E) updated; another list for statistics software added (in Part G) • References updated, now including about 75 books published or reprinted after 1990 Suggestions for Courses: A Four-Semester Sequence The material, when taken in sequence, is suitable for four consecutive semester courses, meeting 3-4 hours a week: 1st Semester. 2nd Semester. 3rd Semester. 4th Semester. ODEs (Chaps. 1-5 or 6) Linear Algebra. Vector Analysis (Chaps. 7-10) Complex Analysis (Chaps. 13-18) Numeric Methods (Chaps. 19-21) ix Preface Suggestions for Independent One-Semester Courses The book is also suitable for various independent one-semester courses meeting 3 hours a week. For instance: Introduction to ODEs (Chaps. 1-2, Sec. 21.1) Laplace Transforms (Chap. 6) Matrices and Linear Systems (Chaps. 7-8) Vector Algebra and Calculus (Chaps. 9-10) Fourier Series and PDEs (Chaps. 11-12, Secs. 21.4-21.7) Introduction to Complex Analysis (Chaps. 13-17) Numeric Analysis (Chaps. 19, 21) Numeric Linear Algebra (Chap. 20) Optimization (Chaps. 22-23) Graphs and Combinatorial Optimization (Chap. 23) Probability and Statistics (Chaps. 24-25) Acknowledgments I am indebted to many of my former teachers, colleagues, and students who helped me directly or indirectly in preparing this book, in particular, the present edition. I profited greatly from discussions with engineers, physicists, mathematicians, and computer scientists, and from their written comments. I want to mention particularly Y. Antipov, D. N. Buechler, S. L. Campbell, R. CalT, P. L. Chambre, V. F. Connolly, Z. Davis, J. Delany, J. W. Dettman, D. Dicker, L. D. Drager, D. Ellis, W. Fox, A. Goriely, R. B. Guenther, J. B. Handley, N. Harbertson, A. Hassen, V. W. Howe, H. Kuhn, G. Lamb, M. T. Lusk, H. B. Mann, I. Marx, K. Millet, J. D. Moore, W. D. Munroe, A. Nadim, B. S. Ng, J. N. Ong. Jr.. D. Panagiotis, A. Plotkin. P. J. Pritchard, W. O. Ray, J. T. Scheick. L. F. Shampine. H. A. Smith, J. Todd, H. Unz, A. L. Villone, H. J. Weiss, A. Wilansky, C. H. Wilcox, H. Ya Fan, and A. D. Ziebur, all from the United States, Professors E. J. Norminton and R. Vaillancourt from Canada, and Professors H. Florian and H. Unger from Europe. I can offer here only an inadequate acknowledgment of my gratitude and appreciation. Special cordial thanks go to Privatdozent Dr. M. Kracht and to Mr. Herbert Kreyszig, MBA, the coauthor of the Student Solutions Manual, who both checked the manuscript in all details and made numerous suggestions for improvements and helped me proofread the galley and page proofs. Furthermore. I wish to thank John Wiley and Sons (see the list on p. iv) as well as GGS InfolTllation Services, in particular Mr. K. Bradley and Mr. J. Nystrom. for their effective cooperation and great care in preparing this new edition. Suggestions of many readers worldwide were evaluated in preparing this edition. Further comments and suggestions for improving the book will be gratefully received. ERWIN KREYSZIG CONTENTS PART A Ordinary Differential Equations (ODEs) 1 CHAPTER 1 First-Order ODEs 2 1.1 Basic Concepts. Modeling 2 1.2 Geometric Meaning of y' = f(x, y). Direction Fields 9 1.3 Separable ODEs. Modeling 12 1.4 Exact ODEs. Integrating Factors 19 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 26 1.6 Orthogonal Trajectories. Optional 35 1.7 Existence and Uniqueness of Solutions 37 Chapter 1 Review Questions and Problems 42 Summary of Chapter 1 43 CHAPTER 2 Second-Order Linear ODEs 45 2.1 Homogeneous Linear ODEs of Second Order 45 2.2 Homogeneous Linear ODEs with Constant Coefficients 53 2.3 Differential Operators. Optional 59 2.4 Modeling: Free Oscillations. (Mass-Spring System) 61 2.5 Euler-Cauchy Equations 69 2.6 Existence and Uniqueness of Solutions. Wronskian 73 2.7 Nonhomogeneous ODEs 78 2.8 Modeling: Forced Oscillations. Resonance 84 2.9 Modeling: Electric Circuits 91 2.10 Solution by Variation of Parameters 98 Chapter 2 Review Questions and Problems 102 Summary of Chapter 2 103 CHAPTER 3 Higher Order Linear ODEs 105 3.1 Homogeneous Linear ODEs 105 3.2 Homogeneous Linear ODEs with Constant Coefficients 111 3.3 Nonhomogeneous Linear ODEs 116 Chapter 3 Review Questions and Problems 122 Summary of Chapter 3 123 CHAPTER 4 5v stems of ODEs. Phase Plane. Qualitative Methods 4.0 Basics of Matrices and Vectors 124 4.1 Systems of ODEs as Models 130 4.2 Basic Theory of Systems of ODEs 136 4.3 Constant-Coefficient Systems. Phase Plane Method 139 4.4 Criteria for Critical Points. Stability 147 4.5 Qualitative Methods for Nonlinear Systems 151 4.6 Nonhomogeneous Linear Systems of ODEs 159 Chapter 4 Review Questions and Problems 163 Summary of Chapter 4 164 CHAPTER 5 Series Solutions of ODEs. Special Functions 5.1 Power Series Method 167 5.2 Theory of the Power Series Method 170 124 166 xi xii Contents 5.3 Legendre's Equation. Legendre Polynomials Pnex) 177 5.4 Frobenius Method 182 5.5 Bessel's Equation. Bessel Functions lvCx) 189 5.6 Bessel Functions of the Second Kind YvCx) 198 5.7 Sturm-Liouville Problems. Orthogonal Functions 203 5.8 Orthogonal Eigenfunction Expansions 210 Chapter 5 Review Questions and Problems 217 Summary of Chapter 5 218 CHAPTER 6 Laplace Transforms 220 6.1 Laplace Transform. Inverse Transform. Linearity. s-Shifting 221 6.2 Transforms of Derivatives and Integrals. ODEs 227 6.3 Unit Step Function. t-Shifting 233 6.4 Short Impulses. Dirac's Delta Function. Pm1ial Fractions 241 6.5 Convolution. Integral Equations 248 6.6 Differentiation and Integration of Transforms. 254 6.7 Systems of ODEs 258 6.8 Laplace Transform: General Formulas 264 6.9 Table of Laplace Transforms 265 Chapter 6 Review Questions and Problems 267 Summary of Chapter 6 269 PART B Linear Algebra. Vector Calculus CHAPTER 7 271 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems 272 7.1 Matrices, Vectors: Addition and Scalar Multiplication 272 7.2 Matrix Multiplication 278 7.3 Linear Systems of Equations. Gauss Elimination 287 7.4 Linear Independence. Rank of a Matrix. Vector Space 296 7.5 Solutions of Linear Systems: Existence, Uniqueness 302 7.6 For Reference: Second- and Third-Order Determinants 306 7.7 Determinants. Cramer's Rule 308 7.8 Inverse of a Matrix. Gauss-Jordan Elimination 315 7.9 Vector Spaces, Inner Product Spaces. Linear Transformations. Optional 323 Chapter 7 Review Questions and Problems 330 Summary of Chapter 7 331 CHAPTER 8 Linear Algebra: Matrix Eigenvalue Problems 8.1 Eigenvalues, Eigenvectors 334 8.2 Some Applications of Eigenvalue Problems 340 8.3 Symmetric, Skew-Symmetric, and Orthogonal Matrices 345 8.4 Eigenbases. Diagonalization. Quadratic Forms 349 8.5 Complex Matrices and Forms. Optional 356 Chapter 8 Review Questions and Problems 362 Summary of Chapter 8 363 333 Contents xiii CHAPTER 9 Vector Differential Calculus. 9.1 Vectors in 2-Space and 3-Space 364 9.2 Inner Product (Dot Product) 371 Grad, Div, Curl 364 9.3 Vector Product (Cross Product) 377 Vector and Scalar Functions and Fields. Derivatives 384 Curves. Arc Length. Curvature. Torsion 389 Calculus Review: Functions of Several Variables. Optional 400 Gradient of a Scalar Field. Directional Derivative 403 Divergence of a Vector Field 410 Curl of a Vector Field 414 Chapter 9 Review Questions and Problems 416 Summary of Chapter 9 417 9.4 9.5 9.6 9.7 9.8 9.9 CHAPTER 10 Vettor Integral Calculus. Integral Theorems 10.1 Line Integrals 420 10.2 Path Independence of Line Integrals 426 10.3 Calculus Review: Double Integrals. Optional 433 10.4 Green's Theorem in the Plane 439 10.5 Surfaces for Surface Integrals 445 10.6 Surface Integrals 449 10.7 Triple Integrals. Divergence Theorem of Gauss 458 10.8 Further Applications of the Divergence Theorem 463 10.9 Stokes's Theorem 468 Chapter 10 Review Questions and Problems 473 420 Summary of Chapter 10 474 PART C Fourier Analysis. Partial Differential Equations (PDEs) CHAPTER 11 Fourier Series, Integrals, and Transforms 11.1 Fourier Series 478 11.2 Functions of Any Period p = 2L 487 11.3 Even and Odd Functions. Half-Range Expansions 490 11.4 Complex Fourier Series. Optiollal 496 11.5 Forced Oscillations 499 11.6 Approximation by Trigonometric Polynomials 502 11.7 Fourier Integral 506 11.8 Fourier Cosine and Sine Transforms 513 11.9 Fourier Transform. Discrete and Fast Fourier Transforms 11.10 Tables of Transforms 529 Chapter 11 Review Questions and Problems 532 Summary of Chapter II 533 478 518 CHAPTER 12 Partial Differential Equations (PDEs) 535 12.1 Basic Concepts 535 12.2 Modeling: Vibrating String, Wave Equation 538 12.3 Solution by Separating Vatiables. Use of Fourier Series 540 12.4 D' Alembert's Solution of the Wave Equation. Characteristics 548 12.5 Heat Equation: Solution by Fourier Series 552 477 xiv Contents Heat Equation: Solution by Fourier Integrals and Transforms 562 Modeling: Membrane, Two-Dimensional Wave Equation 569 Rectangular Membrane. Double Fourier Series 571 Laplacian in Polar Coordinates. Circular Membrane. Fourier-Bessel Series 579 Laplace's Equation in Cylindrical and Spherical Coordinates. Potential 587 Solution of PDEs by Laplace Transforms 594 Chapter 12 Review Questions and Problems 597 Summary of Chapter 12 598 12.6 12.7 12.8 12.9 12.10 12.11 PART D Complex Analysis 601 CHAPTER 13 Complex Numbers and Functions 602 13.1 Complex Numbers. Complex Plane 602 13.2 Polar Form of Complex Numbers. Powers and Roots 607 13.3 Derivative. Analytic Function 612 13.4 Cauchy-Riemann Equations. Laplace's Equation 618 13.5 Exponential Function 623 13.6 Trigonometric and Hyperbolic Functions 626 13.7 Logarithm. General Power 630 Chapter 13 Review Questions and Problems 634 Summary of Chapter 13 635 CHAPTER 14 Complex Integration 637 14.1 Line Integral in the Complex Plane 637 14.2 Cauchy's Integral Theorem 646 14.3 Cauchy's Integral Formula 654 14.4 Derivatives of Analytic Functions 658 Chapter 14 Review Questions and Problems 662 Summary of Chapter 14 663 CHAPTER 15 Power Series, Taylor Series 15.1 Sequences, Series, Convergence Tests 664 15.2 Power Series 673 15.3 Functions Given by Power Series 678 15.4 Taylor and Maclaurin Series 683 15.5 Uniform Convergence. Optional 691 Chapter 15 Review Questions and Problems 698 Summary of Chapter 15 699 664 CHAPTER 16 Laurent Series. Residue Integration 16.1 Laurent Series 701 16.2 Singularities and Zeros. Infinity 707 16.3 Residue Integration Method 712 16.4 Residue Integration of Real Integrals 718 Chapter 16 Review Questions and Problems 726 Summary of Chapter 16 727 CHAPTER 17 Conformal Mapning 728 17.1 Geometry of Analytic Functions: Conformal Mapping 17.2 Linear Fractional Transformations 734 17.3 Special Linear Fractional Transformations 737 701 729 Contents xv 17.4 Conformal Mapping by Other Functions 742 17.5 Riemann Surfaces. Optional 746 Chapter 17 Review Questions and Prohlems 747 Summary of Chapter 17 748 CHAPTER 18 Coml')lex Analysis and Potential Theory 18.1 Electrostatic Fields 750 18.2 Use of Conformal Mapping. Modeling 754 18.3 Heat Problems 757 18.4 Ruid Flow 761 18.5 Poisson's Integral Formula for Potentials 768 18.6 General Properties of Harmonic Function.. 771 Chapter 18 Review Questions and Problems 775 Summary of Chapter 18 776 PART E Numeric Analysis 777 Software 778 CHAPTER 19 Numerics in General 780 19.1 [ntroduction 780 19.2 Solution of Equations by Iteration 787 19.3 [nterpolation 797 19.4 Spline Interpolation 810 19.5 Numeric Integration and Differentiation 817 Chapter 19 Review Questions and Problems 830 Summary of Chapter 19 831 CHAPTER 20 Numeric Linear Algebra 833 20.1 Linear Systems: Gauss EliminatIOn 833 20.2 Linear Systems: LU-Factorization. Matrix Inversion 840 20.3 Linear Systems: Solution by Iteration 845 20.4 Linear Systems: III-Conditioning. Norms 851 20.5 Least Squares Method 859 20.6 Matrix Eigenvalue Problems: Introduction 863 20.7 [ncIusion of Matrix Eigenvalues 866 20.8 Power Method for Eigenvalues 872 20.9 Tridiagonalization and QR-Factorization 875 Chapter 20 Review Questions and Problems 883 Summary of Chapter 20 884 CHAPTER 21 Numerics for ODEs and PDEs 886 21.1 Methods for First-Order ODEs 886 21.2 Multistep Methods 898 21.3 Methods for Systems and Higher Order ODEs 902 21.4 Methods for Elliptic PDEs 909 21.5 Neumann and Mixed Problems. Inegular Boundary 917 21.6 Methods for Parabolic PDEs 922 21.7 Method for Hyperbolic PDEs 928 Chapter 21 Review Questions and Problems 930 Summary of Chapter 21 932 749 xvi Contents PART F Optimization, Graphs 935 CHAPTER 22 Unconstrained Optimization. Linear 22.1 Basic Concepts. Unconstrained Optimization 936 22.2 Linear Programming 939 22.3 Simplex Method 944 22.4 Simplex Method: Difficulties 947 Chapter 22 Review Questions and Problems 952 Summary of Chapter 22 953 CHAPTER 23 Graphs. Combinatorial Optimization 23.1 Graphs and Digraphs 954 23.2 Shortest Path Problems. Complexity 959 23.3 Bellman's Principle. Dijkstra's Algorithm 963 23.4 Shortest Spanning Trees. Greedy Algorithm 966 23.5 Shortest Spanning Trees. Prim's Algorithm 970 23.6 Flows in Networks 973 23.7 Maximum Flow: Ford-Fulkerson Algorithm 979 23.8 Bipartite Graphs. Assignment Problem~ 982 Chapter 23 Review Questions and Problems 987 Summary of Chapter 23 989 PART G Programming 954 Probability, Statistics 991 CHAPTER 24 Data Analysis. Probability Theory 993 24.1 Data Representation. Average. Spread 993 24.2 Experiments, Outcomes, Events 997 24.3 Probability 1000 24.4 Permutations and Combinations 1006 24.5 Random Variables. Probability Distributions 1010 24.6 Mean and Variance of a Distribution 1016 24.7 Binomial. Poisson, and Hypergeometric Distributions 1020 24.8 Normal Distribution 1026 24.9 Distributions of Several Random Variables 1032 Chapter 24 Review Questions and Problems 1041 Summary of Chapter 24 1042 CHAPTER 25 Mathematical Statistics 1044 25.1 Introduction. Random Sampling 1044 25.2 Point Estimation of Parameters 1046 25.3 Confidence Intervals 1049 25.4 Testing Hypotheses. Decisions 1058 25.5 Quality Control 1068 25.6 Acceptance Sampling 1073 25.7 Goodness of Fit. x2-Test 1076 25.8 Nonparametric Tests 1080 25.9 Regression. Fitting Straight Lines. Correlation 1083 Chapter 25 Review Questions and Problems 1092 Summary of Chapter 25 1093 936 xvii Contents APPENDIX 1 References APPENDIX 2 Answers to Odd-Numbered Problems Al APPENDIX 3 Auxiliary Material A60 A3.1 Formulas for Special Functions A60 A3.2 Partial Derivatives A66 A3.3 Sequences and Series A69 A3.4 Grad, Div, Curl, V 2 in Curvilinear Coordinates APPENDIX 4 Additional Proofs APPENDIX 5 Tables PHOTO CREDITS INDEX 11 A94 Pl A74 A7l A4 •••• PA R T . ......... .,It ." f • • -oj 'IIH'II I .. .. .......... 0.'- , ..... e• I,: " A Ordinary Differential Equations (ODEs) C HAP T E R 1 First-Order ODEs C HAP T E R 2 Second-Order Linear ODEs C HAP T E R 3 Higher Order Linear ODEs C HAP T E R 4 Systems of ODEs. Phase Plane. Qualitative Methods C HAP T E R 5 Series Solutions of ODEs. Special Functions C HAP T E R 6 Laplace Transforms Differential equations are of basic importance in engineering mathematics because many physical laws and relations appear mathematically in the form of a differential equation. In Part A we shall consider various physical and geometric problems that lead to differential equations, with emphasis on modeling, that is, the transition from the physical situation to a "mathematical model." In this chapter the model will be a differential equation, and as we proceed we shall explain the most important standard methods for solving such equations. Part A concerns ordinary differential equations (ODEs), whose unknown functions depend on a single variable. Partial differential equations (PDEs), involving unknown functions of several variables, follow in Part C. ODEs are very well suited for computers. Numeric methods for ODEs call be studied directly after Chaps. 1 or 2. See Sees. 21.1-21.3, which are independent of the other sections on numerics. I 9 t~ •••• I CHAPTER ~~ " I .. :;....~.,.. .-- ~ il;;ilil• ~ lillI, --oj... -I' -.~ " 1 I :. First-Order ODEs In this chapter we begin our program of studying ordinary differential equations (ODEs) by deriving them from physical or other problems (modeling), solving them by standard methods, and interpreting solutions and their graphs in terms of a given problem. Questions of existence and uniqueness of solutions will also be discussed (in Sec. l.7). We begin with the simplest ODEs, called ODEs of the first order because they invol ve only the first derivative of the unknown function, no higher derivatives. Our usual notation for the unknown function will be y(x). or yet) if the independent variable is time t. If you wish, use your computer algebra system (CAS) for checking solutions, but make sure that you gain a conceptual understanding of the basic terms, such as ODE, direction field, and initial value problem. COMMENT. Numerics for first-order ODEs call be studied immediately after this chapter. See Secs. 2l.1-2l.2, which are independent of other sections on numerics. Prerequisite: Integral calculus. Sections that may be omitted in a shorter course: l.6, 1.7. References and Answers to Problems: App. I Part A. and App. 2 1.1 Basic Concepts. Modeling If we want to solve an engineering problem (usually of a physical nature). we first have to formulate the problem as a mathematical expression in terms of variables, functions, equations, and so forth. Such an expression is known as a mathematical model of the given problem. The process of setting up a model, solving it mathematically, and interpreting the result in physical or other terms is called mathematical modeling or, briefly, modeling. We shall illustrate this process by various examples and problems because modeling requires experience. (Your computer may help you in solving but hardly in setting up models.) Since many physical concepts, such as velocity and acceleration. are derivatives. a model is very often an equation containing derivatives of an unknown function. Such a model is called a differential equation. Of course, we then want to find a solution (a function that satisfies the equation), explore its properties, graph it, find values of it, and interpret it in physical terms so that we can understand the behavior of the physical system in our given problem. However, before we can tum to methods of solution we must first define basic concepts needed throughout this chapter. SEC. 1.1 3 Basic Concepts. Modeling Velocity v Water level h Falling stone Parachutist y" = g = canst. (Sec. 1.1) mv'=mg-bv Outflowing water 2 h'=-Il'fFt (Sec. 1.2) (Sec. 1.3) y " - ,\, //i/ltl/ :\I \ 1\1 I;J \ \ Displacement y Vibrating mass on a spring my"+ky= 0 (Secs. 2.4, 2.8) \ -'./ I / I I I' I t I \. \ I 'l.. -- . / Current I in an RLC circuit Beats of a vi brati ng system y" + liJ~y = cos rot, roo = (Sec. 2.8) Q) LI" +RI' +lI=E' C (Sec. 2.9) Lotka-Volterra predator-prey model Deformation of a beam V Pendulum EI/ = f(x) L8"+gsinB=O (Sec. 3.3) (Sec. 4.5) Fig. 1. y;= aY I - bY IY2 Y~ = kY 1Y 2 -IY 2 (Sec. 4.5) Some applications of differential equations 4 CHAP. 1 First-Order ODEs An ordinary differential equation (ODE) is an equation that contains one or several derivatives of an unknown function, which we usually call y(x) (or sometimes yet) if the independent variable is time t). The equation may also contain y itself, known functions of x (or t), and constants. For example, I Y = cos x, (1) (2) y" + 9)' 0, = (3) are ordinary differential equations (ODEs). The term ordinary distinguishes them from partial differelltial equations (PDEs), which involve partial derivatives of an unknown function of two or more variables. For instance, a PDE with unknown function u of two variables x and y is PDEs are more complicated than ODEs; they will be considered in Chap. 12. An ODE is said to be of order n if the 11th derivative of the unknown function \" is the highest derivative of y in the equation. The concept of order gives a useful classification into ODEs of first order, second order, and so on. Thus, (1) is of first order, (2) of second order, and (3) of third order. In this chapter we shall consider first-order ODEs. Such equations contain only the first derivative y' and may contain y and any given functions of x. Hence we can write them as (4) F(x, y, y') = 0 or often in the form y' = f(x. \'). This is called the explicit fonn. in contrast with the implicit form (4). For instance, the implicit ODE x- 3 y' - 4y2 = 0 (where x *- 0) can be written explicitly as y' = 4x 3 )'2. Concept of Solution A function y = hex) is called a solution of a given ODE (4) on some open interval a < x < b if h(-r) is defined and differentiable throughout the interval and is such that the equation becomes an identity if y and)' are replaced with hand h', respectively. The curve (the graph) of h is called a solution curve. Here, open interval a < x < b means that the endpoints a and b are not regarded as points belonging to the interval. Also, a < x < b includes infinite intervals -00 < x < b, a < x < 00, -00 < x < 00 (the real line) as special cases. I SEC. 1.1 5 Basic Concepts. Modeling E X AMP L E 1 Verification of Solution = y' = y E X AMP L E 2 *" hex) = c/x tc an arbitrary constant, x 0) is a solution of Xl" = -y. To verify this. differentiate, h' (x) = -c/x2 , and multiply by x to get xy' = -c/x = -y. Thus, xy' = -y, the given ODE. • Solution Curves The ODE y' = dyldx = cos x can be solved directly by integration on both sides. Indeed. using calculus. we obtain y = f cos x dx = sin x + c, where c is an arbitrary constant. This is afamily of solutions. Each value of c, for instance. 2.75 or 0 or -8. gives one of these curves. Figure 2 shows some of them. for c = -3. -2, -1,0, 1,2.3.4. • y Solutions y = sin x Fig. 2. EXAMPLE 3 + c of the ODE y' cos x = Exponential Growth, Exponential Decay From calculu~ we know that y = ce 3t (c any constant) has the derivative (chain rule!) , y = dv dl = 3ce 3t = 3y. This shows that y is a solution of y' = 3)'. Hence this ODE can model exponential growth, for instance. of animal populations or colonies of bacteria. It also applies to humans for small population~ in a large country (e.g .. the United States in early times) and is then known as Malt/illS's law. I We shall say more about this topic in Sec. 1.5. Similarly. y' = -O.2y (with a minus on the right!) has the solution y = ce- O.2t . Hence this ODE models exponential decay, for instance. of a radioactive substance (see Example 5). Figure 3 shows solutions for some positive c. Can you find what the solutions look like for negative c? • y 2.5 12 Fig. 3. Solutions of y' = 14 t -O.2y in Example 3 INamed after the English pioneer in classic economics, THOMAS ROBERT MALTHUS (1766-1834) 6 CHAP. 1 First-Order ODEs We see that each ODE in these examples has a solution that contains an arbitrary constant c. Such a solution containing an arbitrary constant c is called a general solution of the ODE. (We shall see that c is sometimes not completely arbitrary but must be restricted to some interval to avoid complex expressions in the solution.) We shall develop methods that will give general solutions uniquely (perhaps except for notation). Hence we shall say the general solution of a given ODE (instead of a general solution). Geometrically, the general solution of an ODE is a family of infinitely many solution curves, one for each value of the constant c. If we choose a specific c (e.g., c = 6.45 or o or -2.01) we obtain what is called a particular solution of the ODE. A particular solution does not contain any arbitrary constants. In most cases, general solutions exist, and every solution not containing an arbitrary constant is obtained as a particular solution by assigning a suitable value to c. Exceptions to these rules occur but are of minor interest in applications: see Frob. 16 in Problem Set 1.1. Initial Value Problem In most cases the unique solution of a given problem, hence a particular solution, is obtained from a general solution by an initial condition y(xo) = Yo, with given values Xo and Yo, that is used to determine a value of the arbitrary constant c. Geometrically this condition means that the solution curve should pass through the point (xo, Yo) in the .C\:.v-plane. An ODE together with an initial condition is called an initial value problem. Thus, if the ODE is explicit, y' = f(x, y), the initial value problem is of the form (5) E X AMP L E 4 y' = f(x, y), Y(Xo) = Yo· Initial Value Problem Solve the initial value problem , y = dy dx = 3)" yeO) = 5.7. The general solution is y(x) = ce 3x ; see Example 3. From this solution and the inttial condition we obtain yeO) = ceo = c = 5.7. Hence the initial value problem ha~ the solution )'(x) = 5.7e 3x . This is a particular solution. • Solution. Modeling The general importance of modeling to the engineer and physicist was emphasized at the beginning of this section. We shall now consider a basic physical problem that will show the typical steps of modeling in detail: Step I the transition from the physical situation (the physical system) to its mathematical formulation (its mathematical model); Step 2 the solution by a mathematical method; and Step 3 the physical interpretation of the result, This may be the easiest way to obtain a first idea of the nature and purpose of differential equations and their applications. Realize at the outset that your computer (your CAS) may perhaps give you a hand in Step 2, but Steps 1 and 3 are basically your work. And Step 2 SEC. 1.1 7 Basic Concepts. Modeling requires a solid knowledge and good understanding of solution methods available to youyou have to choose the method for your work by hand or by the computer. Keep this in mind, and always check computer results for enors (which may result, for instance, from false inputs). EXAMPLE 5 Radioactivity. Exponential Decay Given an amount of a radioactive substance, say, 0.5 g (gram), find the amount present at any later time. Physical Inj"o171wtio11. Experiments show that at each instant a radioactive substance decomposes at a rate proportIOnal to the the amount present. Step 1. Setting lip a mathematical model (a differential equation) of the physical process. Denote by yet) the amount of substance still present at any time t. By the physical law, the time rate of change y' (t) = dyldt is proporhonal to yet). Denote the constant of proportionality by k. Then dy dt (6) = kyo The value of k is known from experiments for various radioactive substances (e.g.. k = -1.4· lO-llsec -1. approximately, for radium ssRa226). k is negative because ylt) decreases with time. The given initial amount is 0.5 g. Denote the corresponding time by t = O. Then the initial condition is y(O) = 0.5. This is the instant at which the process begins; this motivates the term initial condition (which, however, is also used more generally when the independent variable is not time or when you choose a t other than t = 0). Hence the model of the process is the initial value problem (7) yeO) dt = ky, = 0.5. Step 2. Mathematical soilltion. As in Example 3 we conclude thaI the ODE (6) models exponemial decay and has the general solution (with arbitrary constant c but definite given k) (8) We now use the initial condition to detelwine particular solution governing this process is C. Since yeO) = c from (8), this gives .1'(0) = c = 0.5. Hence the y(t) = O.Se kt (9) (Fig. 4). Always check YOllr reslllt-it may involve human or computer errors! Verify by differentiation (chain rule!) that your solution (9) satisfies (7) as well as yeO) = 0.5: dv --=-dt = O.Ske kt = k' O.Se kt = ky. yCO) = O.Se o = 0.5. Step 3. liltelpretation of reslllt. Formula (9) gives the amount of radioactive substance at tIme I. It starts from the correct given initial amount and decreases with time because k (the constant of proportionality, depending on the kind of substance) is negative. The limit of y as t -> x is zero. • JI~ o 0.5 I 1.5 2 2.5 3 Fig. 4. Radioactivity (Exponential decay, y = 0.5 e kt, with k = -1.5 as an example) CHAP. 1 8 E X AMP L E 6 First-Order ODEs A Geometric Application Geometric problems may also lead to initial value problems. For instance, find the curve through the point (I. I) in the .l,)·-plane having at cach of its points the slope -)1x. Solution. The slope y' should equal -)Jx. This gives the ODE y' = -)1x. Its general solution is y = elx (see Example 1). This is a family of hyperbolas with the coordinate axes as asymptotes. Now, for the curve to pass through (1, I), we must have y = 1 when x = I. Hence the initial condition is y(l) = 1. From this condition and y = elx we get yO) = ell = I; that is, c = 1. This gives the particular • solution y = lIx (drawn somewhat thicker in Fig. 5). y y Solutions of y' Fig. 5. ----- •• = ---- = 15. (Existence) (A) Does the ODE y'2 CALCULUS Solve the ODE by integration. 15-91 2. y' 4. y' = -sin TTX x2/2 = xe (B) Does the ODE solution? cosh 4x VERIFICATION OF SOLUTION + 5. y' = 6. y" + 7T y = 0, + 2,,' + lOy + 2y = 4(x + 1 y2, 2 9. y'" = cos x, 110-141 + y = tan (x Y = = O. 1)2, c) + b sin 7TX 4e- x sin 3x Y = 5e- 2x + 2.1'2 + 2T a cos rrx Y y = -sin x = + ax 2 + bx + + (' INITIAL VALUE PROBLEMS Verify that y is a solution of the ODE. Determine from y the particular solution satisfying the given initial condition. Sketch or graph this solution. 10. y' 0.5y. Y = eeO. 5,,:. y(2) = 2 11. y' = I + 4y2, Y = ~ tan (2x + e), yeO) = 0 12. y' = y - x, y = ce x + x + 1, yeO) = 3 13. y' + 2xy = 0, y = ee-~:2. y( 1) = 1Ie 14. y' = y tan x, = -] have a (real) solution? State the order of the ODE. Verify that the given function is a solution. (a, 17, e are arbitrary constants.) 7. y" 8. y' Particular solutions and Singular solution in Problem 16 u 11-41 1. y' 3. y' Fig. 6. -y/x (hyperbolas) y = c sec x, yeO) = ~7T 1/1 + Iyl = 0 have a general 16. (Singular solution) An ODE may sometimes have an additional solution that cannot be obtained from the general solution and is then called a singular solution. The ODE /2 - XV' + Y = 0 is of the kind. Show by differentiation and substitution that it has the general solution y = ex - e 2 and the singular solution y = x 2/4. Explain Fig. 6. 117-221 MODELING, APPLICATIONS The following problems will give you a first impression of modeling. Many more problems on modeling follow throughout this chapter. 17. (Falling body) If we drop a stone, we can assume air resistance ("drag") to be negligible. Experiments show that under that assumption the acceleration y" = d 2Yldt 2 of this motion is constant (equal to the so-called acceleration of gravity g = 9.80 mlsec 2 = 32 ftlsec 2 ). State this as an ODE for yet), the distance fallen a~ a function of time t. Solve the ODE to get the familiar law of free fall, y = gt 2 /2. SEC. 1.2 18. (Falling body) If in Prob. 17 the stone starts at t = 0 from initial position Yo with initial velocity u = uo, show that the solution is y = gt 2 /2 + uot + Yo. Hov. long does a fall of 100 m take if the body falls from rest? A fall of 200 m? (Guess first.) is the half-life of radium 88Ra226 (in years) in Example 5? 22. (Interest rates) Show by algebra that the investment y(t) from a deposit Yo after t years at an imerest rate r is 19. (Airplane takeoff) If an airplane has a run of 3 km, statts with a speed 6 mlsec, moves wIth constant acceleration, and makes the run in I min, with what speed does it take off? 20. (Subsonic flight) The efficiency of the engines of subsonic airplanes depends on air pressure and usually is maximum near about 36 000 f1. Find the air pressure rex) at this height without calculation. Ph\'Sical i'!formGtioll. The ;ate of change y' (x) is propOliionai to the pressure, and at 18 000 ft the pressure has decreased to half its value )'0 at sea level. 21. (Half-life) The half-life of a radioactive substance is the time in which half of the given amount disappears. Hence it measures the rapidity of the decay. What 1.2 9 Geometric Meaning of y' = f(x, y). Direction Fields + + Ya(t) = yo[1 -"d(t) = .ro[l r]t (Interest compounded annually) (rl365)]365t (Interest compounded daily). Recall from calculus that [1 hent:e [I + + (llll)r - 7 (rln)]nt --+ e as 11 --+ x; e't: thu~ (Interest compounded continuously). What ODE does the last function satisfy? Let the initial investment be $1000 and r = 6%. Compute the value of the investment after I year and after 5 years using each of the three formulas. [s there much difference? Geometric Meaning of y' Direction Fields t(x, y). A first-order ODE (1) y' = f(x, y) has a simple geometric interpretation. From calculus you know that the derivative y' (x) of y(x) is the slope of y(x). Hence a solution curve of (1) that passes through a point (xo, )'0) must have at that point the slope y' (xo) equal to the value of f at that point; that is, Read this paragraph again before you go on, and think about it. It follows that you can indicate directions of solution curves of (I) by drawing short straight-line segments (lineal elements) in the ,,=,·-plane (as in Fig. 7a) and then fitting (approximate) solution curves through the direction field (or slope field) thus obtained. This method is important for two reasons. 1. You need not solve (I). This is essential because many ODEs have complicated solution formulas or none at all. 2. The method shows, in graphical form, the whole family of solutions and their typical properties. The accuracy is somewhat limited, but in most cases this does not matter. Let us illustrate this method for the ODE (2) y , = .X)'. 10 CHAP. 1 First-Order ODEs Direction Fields by a CAS (Computer Algebra System). A CAS plots lineal elements at the points of a square grid. as in Fig. 7a for (2), into which you can fit solution curves. Decrease the mesh size of the grid in regions where I(x, y) varies rapidly. Direction Fields by Using Isoclines (the Older Method). Graph the curves I(x, y) = k = const, called isoclines (meaning curves of eqllal inclination). For (2) these are the hyperbolas I(x, y) = xy = k = const (and the coordinate axes) in Fig. 7b. By (1), these are the curves along which the derivative y' is constant. These are not yet solution curves--don't get confused. Along each isocline draw many parallel line elements of the corresponding slope k. This gives the direction field. into which you can now graph approximate solution curves. We mention that for the ODE (2) in Fig. 7 we would not need the method, because we shall see in the next section that ODEs such as (2) can easily be solved exactly. For the time being, let us verify by substitution that (2) has the general solution y(x) = (c arbitrary). ce"2/2 Indeed, by differentiation (chain rule!) we get y' = x(cex2/2 ) = xy. Of course. knowing the solution, we now have the advantage of obtaining a feel for the accuracy of the method by comparing with the exact solution. The particular solution in Fig. 7 through (x, y) = (1,2) must satisfy y(l) = 2. Thus. 2 = ce 1l2 , c = 21Ve = 1.213, and the particular solution is y(x) = 1.213ex2/2 . A famous ODE for which we do need direction fields is (3) y (It is related to the van der Pol equation of electronics. which we shall discuss in Sec. 4.5.) The direction field in Fig. 8 shows lineal elements generated by the computer. We have also added the isoclines for k = - 5, - 3,~, I as well as three typical solution curves, one that is (almost) a circle and two spirals approaching it from inside and outside. y \ \ \ \ \ \ \( \ \ \ \ \ , , 2 "" ...... 1 I / I / I J I I I I J J / / / I J / -1 , / / ." I / / ! I I / I I I I / I I '" '" / I , / I I I I / / I I / / / I \ \ \ \ J I / -1. / / / y / .) ... / "" \ x x \ " \ \ \ \ \ \ (aj Bya CAS (b) By isoclines Fig. 7. Direction field of y ,= xy SEC. 1.2 Geometric Meaning of y' = f(x, y). Direction Fields 11 y -4 '- " \ \ '- \ \ \ \ \ k =! 4/ \ / k =-3 k = 1 '---~"7"--'--­ / I 4, \ \ \ Fig. S. " '" \ " '- x ~ -4~~C/ Direction field of y' = 0.1 (1 - x 2 ) x - - Y On Numerics Direction fields gi ve "all" solutions, but with limited accuracy. If we need accurate numeric values of a solution (or of several solutions) for which we have no formula, we can use a numeric method. If you want to get an idea of how these methods work, go to Sec. 21.1 and study the first two pages on the Euler-Cauchy method, which is typical of more accurate methods later in that section, notably of the classical Runge-Kutta method. It would make little sense to interrupt the present flow of ideas by including such methods here; indeed, it would be a duplication of the material in Sec. 21.1. For an excursion to that section you need no exn'a prerequisites; Sec. 1.1 just discussed is sufficient. 11-101 DIRECTION FIELDS, SOLUTION CURVES Graph a direction field (by a CAS or by hand). In the field graph approximate solution curves through the given point or points (x, y) by hand. 1. y' = eX - y. (0. 0), (0. I) 2. 4yy' 3. y' = = -9x, (2, 2) 1 + y2, (~1T, I) 111-151 ACCURACY Direction fields are very useful because you can see solutions (as many as you want) without solving the ODE, which may be difficult or impossible in terms of a formula. To get a feel for the accuracy of the method, graph a field, sketch solution curves in it, and compare them with the exact solutions. Ll.y' 4. y' = y - 2y2. (0. 0). (0. 0.25). (0. 0.5). (0. I) 13. y' 5. y' = x2 6. y' = 14. .r' 15. y' 7. V' = 8. y' = - IIy, (I, -2) + siny, (-1, 0), (1,4) 3 y3 + x , (0, 1) 2xy + I, (-I, 2), (0, 0), (1, I -2) 9. y' = y tanh x - 2, (-1, -2), (1,0), (1, 2) 10. y' = eY/x, (I, I), (2, 2), (3, 3) sin ~1TX 12. y' = I/x2 -2y (SoL y = ce- 2x ) 3ylx (Sol. y = cx 3 ) -In x MOTIONS A body moves on a straight line, with velocity as given. and yet) is its distance from a fixed point 0 and t time. Find a model of the motion (an ODE). Graph a direction field. CHAP. 1 12 First-Order ODEs In it sketch a solution curve corresponding to the given initial condition. 16. Velocity equal to the reciprocal ofthe distance, y(l) = 20. CAS PROJECT. Direction Fields. Discuss direction fields as follows. I (a) Graph a direction field for the ODE y' = I - Y and in it the solution satisfying yeO) = 5 showing exponential approach. Can you see the limit of any solution directly from the ODE? For what initial condition will the solution be increasing? Constant? Decreasing? 17. Product of velocity and distance equal to -t, y(3) = -3 18. Velocity plus distance equal to the square of time, yeO) = 6 19. (Skydiver) Two forces act on a parachutist, the attraction by the earth mg (/1/ = mass of person plus equipment. g = 9.8 m/sec 2 the acceleration of gravity) and the air resistance, assumed to be proportional to the square of the velocity vet). Using Newton's second law of motion (mass X acceleration = resultant of the forces), set up a model (an ODE for v(t». Graph a direction field (choosing III and the constant of proportionality equal to 1). Assume that the parachute opens when v = 10m/sec. Graph the corresponding solution in the field. What is the limiting velocity? 1.3 (b) What do the solution curves of y' = _X 3/y3 look like, as concluded from a direction field. How do they seem to differ from circles? What are the isoclines? What happens to those curves when you drop the minus on the right? Do they look similar to familiar curves? First. guess. (c) Compare. as best as you can, the old and the computer methods, their advantages and disadvantages. Write a short report. Separable ODEs. Modeling Many practically useful ODEs can be reduced to the form (1) g(y)y' = f(x) by purely algebraic manipUlations. Then we can integrate on hoth sides with respect to x, obtaining (2) Ig(y) y' £Ix = If(x) eLr + c. On the left we can switch to y as the variable of integration. By calculus, y' d" = dy. so that (3) I g(y) £I)' = I f(x) £Ix + c. If f and g are continuous functions, the integrals in (3) exist, and by evaluating them we obtain a general solution of (1). This method of solving ODEs is called the method of separating variables, and (1) is called a separable equation, because in (3) the variables are now separated: x appears only on the right and y only on the left. E X AMP L E 1 A Separable ODE The ODE y' = 1 dv --·-2 1 +Y + y2 is = d.1:_ separable because it can be written By inlegration, arctany =x+c or y = tan (x + c). SEC. 1.3 13 Separable ODEs. Modeling It is very impOltallt to illtroduce the COl/stallt 0/ illtegratioll immediately whell the integratioll is perjonlled. + c, which is not a solution (\~hen c Verify this. ' . If we wrote arctan ,. = x, then v = tan x. and thell introduced c. we would have obtained ,. = tan x '* 0): Modeling The importance of modeling was emphasized in Sec. 1.1, and separable equations yield various useful models. Let us discuss this in terms of some typical examples. E X AMP L E 2 Radiocarbon Dating2 In September 11)1) I the famous Iceman (OetLi). a mummy from the Neolithic period of the Stone Age found in the ice of the Oetftal Alps (hence the name "Oetzi in Southcrn Tyrolia near the Austrian-Italian border. caused a scientific sensation. When did Oetzi approximately live and die if the ratio of carbon 6Cl4 to carbon 6Cl2 in this mummy is 52.5% of that of a living organism? Physical/II/ormatiol!. In the atmosphere and in living organisms, the ratio of radioactive carbon 6C14 (made radioactive by cosmic rays) to ordinary carbon 6C12 is con~tant. When an orgamsm dies, its absorption of 6C14 by breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the radioactive carbon ratio in the fossil with that in the atmosphere. To do this. one needs to know the half-life of 6C14. which is 5715 years (CRC Halldbook a/Chemistry alld Physics, R3rd ed.. Boca Raton: CRC Press. 2002, page II-52. line 9). OO ) Solutioll. Modelillg. Radioactive decay is governed by the ODE y' = ky hee Sec. 1.1. Example 51. By separation and integration (where t is time and Yo is the initial ratio of ~14 to 6C12) dy = In 13,1 = kt + c. kdt. )' Next we use the half-life H = 5715 to determine k. When t = H. half of the original ~ubstance is still present. Thus. In 0.5 0.693 -0.0001213. k= 5715 H Finally. we use the ratio 52.5% for determining the time t when OetLi died (actually, was kiUed). ekt = e-0 OOOl213t = 0.525, r = In 0.525 -~---=- = -0.0001213 5312. Allswer: About 5300 years ago. Other methods show that mdiocarbon dating values are usually too small. According to recent research. this is due to a variation in that carbon ratio because of industrial pollution and other factors. such as nuclear testing. • E X AMP L E 3 Mixing Problem Mixing problems occur quite frequently in chemical industry. We explain here hov. to solve the basic model involving a single tank. The tank in Fig. I) contains 1000 gal of water in which initially 100 Ib of salt is dissolvcd. Brine runs in at a rate of 10 gal/min. and each gallon contains 51b of dissoved salt. The mixture in the tank is kept uniform by ~tirring. Brine ['Uns our at 10 gal/min. Find the amount of salt in the tank at any time t. Solution. Step 1. Settillg up a model. Let ,.( r) denote the amount of salt in the tank at time r. Its time rate of change is y' = Salt inflow rate - Salt outflow rate "Balance law". 51b times 10 gal gives an inflow of 50 Ib of salt. Now. the outflow is IO gal of brine. This is 1011000 = 0.01 (= 1%) of the total brine content in the tank, hence 0.01 of the salt content yet), that is. 0.01.1'(0. Thus the model is the ODE (4) y' = 50 - O.OJy = -O.Ol(y - 5000). 2Method by WILLARD FRANK UBBY (1908-1980), American chemist, who was awarded for this work the 1960 Nobel Prize in chemistry. CHAP. 1 14 First-Order ODEs Step 2. Sollltioll of the model. The ODE (4) is separable. Separation, integration, and taking exponents on both sides gives dy = -0.01 dl, Y - 5000 In L\' - 50001 = -O.Ol t J' - 5000 = ce- o.Olt . + c*, Initially the tank contains 100 Ib of salt. Hence yeO) = 100 is the initial condition that will give the unique solution. Substituting), = 100 and t = 0 in the last equation give~ 100 - 5000 = ceo = c. Hence c = -4900. Hence the amount of salt in the tank at time t is yet) (5) = 5000 - 4900e -o.Olt. This function shows an exponential approach to the limit 5000 Ib: see Fig. 9. Can you explain physically that yet) should increase with time? That its limit is 5000 Ib? Can you see the limit directly from the ODE? The model di~cllssed becomes more realistic in problems on pollutants in lakes (sec Problem Set 1.5. Prob. 27) or drugs in organs. These types of problems are more difficult because the mixing may be imperfect and the flow rates (in and out) may be different and known only vel)' roughly. • Y -----------_.-=-=--=-=-- 5000 4000 3000 2000 1000 100~ o __ __- L_ _ 100 200 300 ~ Tank __~__- L_ _ _ 400 500 Salt contenty(t) Fig. 9. E X AMP L E 4 ~L- Mixing problem in Example 3 Heating an Office Building (Newton's Law of Cooling}) Suppo,e that in Winter the daytime temperature in a certain office building is maintained at 70°F, The heating is shUl off at 10 P.M. and tumed on again at 6 A.M. On a certain day the temperature inside the building at 2 A.M. was found to be 65°F. The outside temperature was 50°F at 10 P.M. and had dropped to 40°F by 6 A.M. What was the temperature inside the building when the heat was turned on at 6 A.M.? Physical information. Experiments show that the time rate of change of the temperature T of a body B (which conducts heat well, as, for example, a copper hall does) is proportional to the difference hetween T and the temperature of the sUlTounding medium (Newton's law of cooling). SOllitioll. Step L Settillg lip a model. Let T(t) be the temperature inside the building and TA the outside temperature (assumed to be constant in Newton's law). Then by Newton's law, (6) Such experimental laws are derived under idealized assumptions that rarely hold exactly. However. even if a model seems to fit the reality only poorly (as in the present case). it may still give valuable qualitative information. To see how good a model is, the engineer will collect experimental data and compare them with calculations from the modeL 3 Sir ISAAC NEWTON (1642-1727), great English physicist and mathematician. became a professor at Camblidge in 1669 and Master of the Mint in 1699. He and the German mathematician and philosopher GOTTFRlED WILHELM LEIBNIZ (1646--1716) invented (independently) the differential and integral calculus. Newton discovered many basic physical laws and created the method of investigating physical problems by means of calculus. His Philosophiae namralis principia mathematica (Mathematical Principle), of Natural Philosophy, 1687) contains the development of classical mechanic~. His work is of greatest importance to both mathematics and physics. SEC. 1.3 15 Separable ODEs. Modeling Step 2. General solution. We cannot solve (6) because we do not know TA , just that it varied between 50°F and 40°F, so we follow the Goldell Rule: If you cannot solve your problem, try to solve a simpler one. We solve (6) with the unknown function TA replaced with the average of the two known values, or 45°F. For physical reasons we may expect that this will give us a reasonable approximate value of T in the building at 6 A.M. For constant TA = 45 (or any other constant value) the ODE (6) is separable. Separation, integration, and taking exponents gives the general solution <IT T _ 45 = k dt, In IT - 451 = kt T(t) = 45 .,. ce kt + c*, (c = e C \ Step 3. PQlticular solutioll. We choose IO P.M. to be t = O. Then the given initial condition is T(Ol = 70 and yields a particular solution, call it Tp- By substitution. T(O) = 45 + ceo = 70, c = 70 - 45 = 25, Step 4. Detel7nillatioll of k. We llse T(4) = 65, where t = 4 is 2 k into Tp(t) gives (Fig. A.M. Solving algebraically for k and inserting 10l e 4k = 0.8, Step 5. Answer alld illterpretation. 6 A.M. k = ! In 0.8 = -0.056, Tp(t) = 45 + 25e -0.056t. is t = 8 (namely, 8 hours after IO P.M.), and Tp(8) = 45 + 25e -0.056·8 = 6] rOF]. Hence the temperature in the building dropped 9°F, a result that looks reasonable. • y 70 68 66 65 64 -------1' I I I ~i -------~-------_" ~ ~ o 2 4 6 8 t 60 '-------'----'--- Fig. 10. E X AMP L E 5 Particular solution (temperature) in Example 4 Leaking Tank. Outflow of Water Through a Hole (Torricelli's Law) This is another prototype engineering problem that leads to an ODE. It concems the outflow of water from a cylindrical tank with a hole at the bottom (Fig. 11). You are asked to find the height of the water in the tank at any time if the tank has diameter 2 m, the hole has diameter 1 cm. and the initial height of the water when the hole is opened is 2.25 m. When will the tank be empty? Physical information. Under the intluence of gravity the outflowing water has velocity (7) vet) = 0.600V2gh(t) where h(t) is the height of the water above the hole at lime acceleration of gravity at the surface of the earth. t, (TorricelIi's law4 ), and q = 980 cm/sec2 = 32.17 ft/sec 2 is the Solutioll. Step 1. Setti1lg up the model. To get an equation, we relate the decrease in water level h(t) to the outflow. The volume ti V of the outflow during a short time tit is tiV=Av!::.t (A = Area of hole). 4 EV ANGELIST A TORRICELLI (1608-1647), Italian physicist, pupil and successor of GALILEO GALl LEI (1564-1642) at Florence. The "contraction factor" 0.600 was introduced by 1. C. BORDA in 1766 because the stream has a smaller cross section than the area of the hole. 16 CHAP. 1 First-Order ODEs il V must equal the change il V* of the volume of the water in the tank. Now il.V* (B = Cross-sectional area of tank) = -B.lh where illz (> 0) is the decrease of the height h(t) of the water. The minus sign appears because the volume of the water in the tank decreases. Equating il. V and il V* gives -B il.il = Au ilt. We now express u according to Torricelli's law and then let ilt (the length of the time interval considered) approach O--this is a stalldard way of obtaining an ODE as a model. That is. we have llh - ilt and by letting .It -+ A =- - B u A =- - B 0.600Y2gh(t) ' 0we obtain the ODE dh A = -2656 - Yh dt . B ' - where 26.56 = 0.600 Y2' 980. This is our model, a first-order ODE. Step 2. General solution. Our ODE is separable. AlB is constant. Separation and integration gives - dh Yh = A -26.56 - dt B Dividing by 2 and squaring gives il yields the general sol ution = (c - A 2Yh = c* - 26.56 - t. B and 13.28AtIB)2. Inserting 13.28AIB = 13.28' 0.5 2 '/7/1002 '/7 = 0.000332 h(t) = (c - 0.000332t)2. Step 3. Particular solution. The initial height (the initial condition) is h(O) = 225 cm. Substitution of t = 0 and Iz = 225 gives from the general solution c 2 = 225, c = 15.00 and thus the particular solution (Fig. I)) hp(t) = (15.00 - 0.000332t)2. Step 4. Tallk empty. hp(t) = 0 if t = 15.0010.000332 = 45 181 [sec] = 12.6 [hOUTS]. Here you see distinctly the importal/ce of the choice of ul/its-we have been working with the Cgs system, in which time i~ measured in seconds! We used g = 980 cmlsec2 . • Step 5. Checking. Check the result. 1:2.00mj r h 250 ~-f 200 '- 150 2.25 m h(t) L ~ • Outflowing t water Tank Fig. 11. " 100 50 0 0 10000 30000 50000 t Water level h(tl in tank Example 5. Outflow from a cylindrical tank ("leaking tank"). Torricelli's law Extended Method: Reduction to Separable Form Certain nonseparable ODEs can be made separable by transformations that introduce for y a new unknown function. We discuss this technique for a class of ODEs of practical SEC. 1.3 17 Separable ODEs. Modeling importance, namely, for equations (8) Here, f is any (differentiable) function of y/x, such as sin (y/x), (Y/X)4, and so on. (Such an ODE is sometimes called a homogeneous ODE. a term we shall not use but reserve for a more important purpose in Sec. 1.5.) The form of such an ODE suggests that we set y/x = u; thus, (9) y = ux Substitution into),' = fey/x) then gives u' x this can be separated: (10) E X AMP L E 6 y' = u'x and by product differentiation + = u du dx feu) - u x feu) or u' x = + u. feu) - u. We see that Reduction to Separable Form Solve Solution. To gel the usual explicit form, divide the given equarion by 2n·. Now substitute y and y' from (9) and then simplify by subtracting , 1/ I/X+ 1/="2 , 1/ on both sides. 1/ ux=---2 211 211 ' You see that in the last equation you can now separate the variables, 211 dl/ I + u 2 dx By integration, x Take exponents on both sides to get I obtai n (Fig. 12) + 1/ 2 In(l = clx or I + + 2 1/ ) (y/x)2 = -In = clx. Ixl + c* = In I~ I + c*. Multiply the last equation by x 2 to Thus This general solution represents a family of circles passing through the origin with centers on the x-axis y -8 \-4 -.J/'j 8 \~~ Fig. 12. x General solution (family of circles) in Example 6 • CHAP. 1 18 First-Order ODEs 1. (Constant of integl'3tion) An arbitrary constant of integration must be introduced inunediately when the integration is performed. Why is this important? Give an example of your own. [2-91 GENERAL SOLUTION Find a general solution. Show the steps of derivation. Check your answer by substitution. 2. y' + (x + 2»)"2 = 0 3. y' + (y 9X)2 + + 36x = 0 6. y' = (4x 2 + y2)/(xy) 5. )'y' 7. y' sin Y cos TTX = = + )'2 26. (Gompel'tz gl'Owth in tumors) TIle Gompertz model is r' = -Av In \' (A > 0), where yet) is the mass of clinical observations. The declining growth rate with increasing y > 1 corresponds to the fact that cells in the interior of a tumor may die because of insufficient oxygen and nutrients. Use the ODE to discuss the growth and decline of solutions (tumors) and to find constant solutions. Then solve the ODE. 9\" = v) 7TX i)"2 + Y 8. xy' = 9. y' e""x 25. (Radiocal'bon dating) If a fossilized tree is claimed to be 4000 years old. what should be its 6C14 content expressed as a percent of the ratio of 6C14 to 6C12 in a living organism? tu~or cells' at time t. The model agrees well with 2 sec 2y = 4. y' = {y of individuals present, what is the popUlation as a function of time? Figure out the limiting situation for increasing time and interpret it. I 27. (Dl-yel') If wet laundry loses half of its moisture 3 during the first 5 minutes of drying in a dryer and if the rate of loss of moisture is proportional to the moisture content, when will the laundry be practically dry, say, when will it have lost 95% of its moisture? First guess. 11. dr/dt = -2tr, r(O) = ro 28. (Alibi?) Jack, arrested when leaving a bar, claims that L10-191 INITIAL VALUE PROBLEMS Find the particular solution. Show the steps of derivation, beginning with the general solution. (L, R, b are constants.) 10. yy' + 12. 2xyy' 4x = = 3)"2 0, y(O) + = he has been inside for at least half an hour (which would provide him with an alibi). The police check the water temperature of his car (parked near the entrance of the bar) at the instant of arrest and again 30 minutes later, obtaining the values 190°F and 110°F, respectively. Do these results give Jack an alibi? (Solve by inspection.) x 2 , )"(1) = 2 13. L d/ldt + RI = 0, 1(0) = 10 14. v' = vlx + (2x 3 /v) COS(X2), y(v.;;:t2) = v;IS. >xy"= 2(x + 2i y 3, yeO) = l/Vs = 0.45 16. x.v' = y + 4x 5 cos 2(y/x), H2) = 0 17. y'x Inx = y, ."(3) = In 81 18. dr/dO = b[(dr/dO) cos 0 + r sin 0], r(i7T) 0< b < I y2 19. yy' = (x - l)e- , y(O) = 1 7T. 20. (Particulal' solution) Introduce limits of integration in (3) such that y obtained from (3) satisfies the initial condition y(xo) = )'0' Try the formula out on Prob. 19. =1-36J APPLICATIONS, MODELING 21. (Curves) Find all curves in the xy-plane whose tangents all pass through a given point (a, b). 22. (Cm'ves) Show that any (nonverticaD straight line through the origin of the xy-plane intersects all solution curves of y' = g()'/x) at the same angle. 23. (Exponential growth) If the growth rate of the amount of yeast at any time t is proportional to the amount present at that time and doubles in I week, how much yeast can be expected after 2 weeks? After 4 weeks? 24. (Population model) If in a population of bacteria the birth rate and death rate are proportional to the number 29. (Law of cooling) A thermometer, reading 10°C, is brought into a room whose temperature is 23°C, Two minutes later the thermometer reading is 18°C, How long will it take until the reading is practically 23°C, say, 22.8°C? First guess. 30. (TonicelIi's law) How does the answer in Example 5 (the time when the tank is empty) change if the diameter of the hole is doubled? First guess. 31. (TolTicelli's law) Show that (7) looks reasonable inasmuch as V2gh(t) is the speed a body gains if it falls a distance h (and air resistance is neglected). 32. (Rope) To tie a boat in a harbor. how many times must a rope be wound around a bollard (a vertical rough cylindrical post fixed on the ground) so that a man holding one end of the rope can resist a force exerted by the boat one thousand times greater than the man can exert? First guess. Experiments show that the change /1S of the force S in a small portion of the rope is proportional to S and to the small angle /1¢ in Fig. 13, Take the proportionality constant 0.15, SEC. 1.4 Exact ODEs. Integrating Factors 19 Small portion of rope (A) Graph the curves for the seven initial value x2 problems y' = e- / 2 , yeO) = 0, ± I, ±2, ±3, common axes. Are these curves congment? Why? (B) Experiment with approximate curves of nth partial sums of the Maclaurin series obtained by term wise integration of that of y in (A); graph them and describe qualitatively the accuracy for a fixed interval o ~ x ~ b and increasing n, and then for fixed nand increasing b. S+l1S Fig. 13. Problem 32 (C) Experiment with 33. (Mixing) A tank contains 800 gal of water in which 200 Ib of salt is dissolved. Two gallons of fresh water rLms in per minute, and 2 gal of the mixture in the tank. kept uniform by stirring, runs out per minute. How much salt is left in the tank after 5 hours? = cos (x 2 ) as in (8). (D) Find an initial value problem with solution t2 dt and experiment with it as in (8). o 36. TEAM PROJECT. Tonicelli's Law. Suppose that the tank in Example 5 is hemispherical, of radius R, initially full of water, and has an outlet of 5 cm2 crosssectional area at the bottom. (Make a sketch.) Set up the model for outflow. Indicate what portion of your work in Example 5 you can use (so that it can become part of the general method independent of the shape of the tank). Find the time t to empty the tank (a) for any R, (b) for R = I m. Plot t as function of R. Find the time when h = R/2 (a) for any R, (b) for R = 1 m. y = e·-2 L"e- 34. WRITING PROJECT. Exponential Increase, Decay, Approach. Collect. order, and present all the information on the ODE y' = ky and its applications fi'om the text and the problems. Add examples of your own. 35. CAS EXPERIMENT. Graphing Solutions. A CAS can usually graph solutions even if they are given by integrals that cannot be evaluated by the usual methods of calculus. Show this as follows. 1.4 y' Exact ODEs. Integrating Factors We remember from calculus that if a function u(x, y) has continuous partial derivatives, its differential (also called its total differential) is du = au - ax dx au +- dv. ay' From this it follows that if lI(X, y) = C = conST, then du = O. For example, if u = x + x2.\'3 = c, then or y , dy dx an ODE that we can solve by going backward. This idea leads to a powerful solution method as follows. A first-order ODE Mex, y) + N(x, y)y' = 0, written as (use dy = y' d-,;; as in Sec. 1.3) (1) M(x, y) dx + N(x, y) dy = 0 20 CHAP. 1 First-Order ODEs is called an exact differential equation if the differential form M(x, y) dx is exact, that is, this form is the differential au au ax ay + N(x, y) dy du= -dx+ -dy (2) of some function u(x, y). Then (1) can be written du = o. By integration we immediately obtain the general solution of (I) in the form (3) u(x, y) = c. This is called an implicit solution, in contrast with a solution y = hex) as defined in Sec. 1.1, which is also called an explicit solution, for distinction. Sometimes an implicit solution can be converted to explicit form. (Do this for x 2 + )'2 = 1.) If this is not possible, your CAS may graph a figure of the contour lines (3) of the function u(x, y) and help you in understanding the solution. Comparing (I) and (2), we see that (1) is an exact differential equation if there is some function u(x, y) such that (4) (a) au ax =M, (b) au ay =N. From this we can derive a formula for checking whether (1) is exact or not, as follows. Let M and N be continuous and have continuous first partial derivatives in a region in the xy-plane whose boundary is a closed curve without self-intersections. Then by partial differentiation of (4) (see App. 3.2 for notation), aM ay a2 u ayax ' aN a2 u ax ax ay By the assumption of continuity the two second partial derivatives are equal. Thus (5) aM aN ay ax This condition is not only necessary but also sufficient for (1) to be an exact differential equation. (We shall prove this in Sec. 10.2 in another context. Some calculus books (e.g., Ref. [GRIll also contain a proof.) If (I) . is exact, the function u(x, y) can be found by inspection or in the followino 0 systematic way. From (4a) we have by integration with respect to x SEC. 1.4 21 Exact ODEs. Integrating Factors (6) J = M dx + k(y); u in this integration, y is to be regarded as a constant, and k(y) plays the role of a "constant" of integration. To detennine key), we derive aulay from (6), use (4b) to get dkldy, and integrate dkldy to get k. Formula (6) was obtained from (4a). Instead of (4a) we may equally well use (4b). Then instead of (6) we first have by integration with respect to y (6*) u J = N dy + lex). To determine lex), we derive aulax from (6*), use (4a) to get dlldx, and integrate. We illustrate all this by the following typical examples. E X AMP L ElAn Exact ODE Solve cos (x + v) dx + {3y2 + 2y + cos (x + y» dy = O. (7) Solution. Step 1. Test/or exactness. Our equation is of the form (1) with M = cos (x + y), + y). N = 3y2 + 2y + cos (x Thus aM - iiy = aN -.- = dx -sm(x + v), - -sin (x + y). From this and (5) we see that (7) is exact. Step 2. Implicit general solution. From (Ii) we obtain by integration (8) u = f M dx + k(y) = f cos (x + y) dx + k(y) = sin (x + y) + k(y). To find k(y), we differentiate this fonTIula with respect to y and use fonnula (4b), obtaining au ay = cos (x + dk y) + dy Hence dk/dy = 3y2 + 2y. By integration. k we obtain the answer 2 = N = 3y + = l + y2 + lI(X. y) = sin (x + y) 2y + cos (x + y). c*. Inserting this result into (8) and observing 0), + y3 + y2 = c. Step 3. Checldng all implicit solution. We can check by differentiating the implicit solution u(x, y) = c implicitly and see whether this leads to the given ODE (7): all (9) This dlt complete~ all = -a dx + x ay the check. dy = cos (.l. + -y) dx + (cos (x + ),) + 3),2 + 21') d), • = O. • CHAP. 1 First-Order ODEs 22 E X AMP L E 2 An Initial Value Problem Solve the initial value problem (co~ (10) Solutioll. y sinh x + I) dx - sin y cosh \" d,' = You may verify that the given ODE is exact. We find II =- ylll = 2. O. II. For a change. let us use (6*), Jsiny cosh x dy + I(x) = cosy cosh x + I(x). From this. all/ax = cos y sinh x + dlldx = M = cos y sinh x + I. Hence dlldx = 1. By integration, I(x) = x + c*. This gives the general solution II(X. y) = cos y cosh x + x = c. From the initial condition. cos 2 cosh I + I = 0.358 = c. Hence the answer is cos y cosh x + x = 0.358. Figure 14 ~how~ the particular solutions for c = O. 0.358 (thicker curve). 1. 2. 3. Check that the answer satisfies the ODE. (Proceed as in Example I.) Also check thar the initial condition is satisfied. • y 2.5 2.0 0 I II 1.5/ 1.0 0.5 t o 0.5 Fig. 14. E X AMP L E 3 1.0 1.5 2.0 2.5 3.0 x Particular solu.ions in Example 2 WARNING! Breakdown in the Case of Nonexactness The equation -y dx + x dy = 0 is not exact because M = -y and N = x. so that in (5), aMIi)y = -I but aN/ax = I. Let us show that in such a case the present method does not work. From (6), 1I = J M dx + key) = -xy + key), iJl/ hence ay = -x + dk {(I' Now, all/ay should equal N = x, by (4b). However, this is impossible because key) can depend only on y. Try • (6*): it will also fail. Sohe the equation by another method that we have discussed. Reduction to Exact Form. Integrating Factors The ODE in Example 3 is -y dt + x dy = O. It is not exact. However, if we mUltiply it by 1Ix2 , we get an exact equation [check exactness by (5)!], (II) y -ydx + xdy 2 = - 2" dx x x 1 dy x +- Integration of (11) then gives the general solution )f.t (y) =d - = x C = = O. COllst. SEC. 1.4 23 Exact ODEs. Integrating Factors This example gives the idea. All we did was multiply a given nonexact equation, say, (12) + P(x, y) dx Q(x, y) dy = 0, by a function F that, in general, will be a function of both x and y. The result was an equation FP dx (13) that i1S exact, so we can 1Solve it an integrating factor of (12). E X AMP L E 4 a~ + FQ dy = 0 just discussed. Such a function F(x, y) i1S then called Integrating Factor The imegrating factor in (] II is F = L/x 2 . Hence in this case the exact equation (13) is -v dx + 2 x FP dx + FQ d)' = .t d)' . = d ( V ) -'- = x o. y - Solution x These are straight lines y = ex through the origin. It is remarkable that we can readily find other mtegrating factors for the equation -y dx 2 1/)'2, lI(xy). and lI(x + )'2), because (14) -y,h + xdy _ (~) 2 - d . y y -y,h+xdy ----'----'- = x)' -d (Inx- ) . y -ydx+xdy 2 x +Y 2 = c. + x dy = O. namely, Y) . ( =d arctanx • How to Find Integrating Factors In simpler cases we may find integrating factors by inspection or perhaps after some trials, keeping (14) in mind. In the general case, the idea is the following. For M dx + N dy = 0 the exactness condition (4) is aM/a)' = aN/ax. Hence for (13), FP dx + FQ d)' = 0, the exactness condition is a (15) - . (FP) ay = a (FQ). -. ax By the product rule, with subscripts denoting partial derivatives, this gives In the general case, this would be complicated and useless. So we follow the Golden Rule: cannot solve your problem, try to solve a simpler one-the result may be useful (and may also help you later on). Hence we look for an integrating factor depending only on one variable; fortunately, in many practical cases, there are such factors, as we shall see. Thus, let F = F(x). Then Fy = 0, and Fx = F' = dFld'(, so that (15) becomes If you FP y = F'Q + FQx. Dividing by FQ and reshuffling terms, we have (16) 1 dF F d, --=R ' This proves the following theorem. where R = 1 Q (ap _ aQ ) . ay ax CHAP. 1 24 THEOREM 1 First-Order ODEs Integrating Factor F(x) If (12) is sllch that the I ight side R of (16), depe1lds o1lly on x. then (12) has an integrating factor F = F(x), which is obtained by i1lfegrating (16) ([nd taking exponents on both sides, (17) F(x) = exp I R(x) dx. Similarly, if F* = F*(y), then i.nstead of (16) we get 1 dF* - - =R* F* dy , (18) where R* ap) I = p ( ~Q _ iJx av and we have the companion THEOREM 2 Integrating Factor F*(y) If (12) is such that the right side R* of (I8) depends only 011 y, then (12) has an integrating factor F* = F*(y), which is obtained from (18) in the fonn F*(y) (19) E X AMP L E 5 I = exp R*(y) dy. Application of Theorems 1 and 2. Initial Value Problem Using Theorem 1 or 2, find an integrating factor and solve the initial value problem (e x +y + "eY) dx + (xe Y - 1) dy = 0, (20) Solutioll. yeO) = -1 Step 1. NOllexactlless. The exactnes, check fails: -oP = -0 oy ay (e x+ Y + Y ye ) = eX+Y + eY + yeY but aQ 0 - =ax ax (xe Y - I) = eY . Step 2. Integratillg factor. General solutio1l. Theorem 1 fails because R [the right side of (16)] depends on both x and y, R = -I Q (oP - oQ) = - iJy 1 - - - (e - xeY - ax I x+ Y + e Y + yeY - eY ). Try Theorem 2. The right side of (18) b R* - -1 - P (a- Q ax ap) - - - ay - -X, - -I - - - (e Y - eX +Y - eY - veY ) e + Y + yeY • = -I. Hence (19) give, the integrating factor F*(y) = e -Yo From this result and (20) you get the exact equation (eX + y) £Ix + (x - e -Y) dy = O. Test for exactness; you will get I on both sides of the exactne" condition. By imegration, using (4a), u = f (ex + y) £Ix = eX + xy + key). SEC. 1.4 25 Exact ODEs. Integrating Factors Differentiate this with respect to), and use (4b) to get all - ily = dk dk -dy = x + = N = x - e- Y , dy -e-Y + c*. k = e- Y , Hence the general solution is u(x. y) = eX + xy + e-Y = C. Step 3. Particular solution. The initial condition .1'(0) = I gives lI(O. - I) = 1 + 0 + e = 3.72. Hence the answer is eX + xy + e -Y = I + e = 3.72. Figure 15 shows several particular solutions obtained as level curves of u(x, y) = c, obtained by a CAS, a convenient way in cases in which it is impossible or difficult to cast a solution into explicit form. Note the curve that (nearly) ~atisfies the initial condition. Step 4. Checking. Check by substitution that the answer satisfies the given equation as well as the initial condition. • y Fig. 15. Particular solutions in Example 5 ===== --.•........---.--.... -......... - ........ . . ----.~ ~-- ....: 11-20 I EXACT ODEs. INTEGRATING FACTORS Test for exactness. If exact, solve. If not, use an integrating factor as given or find it by inspection or from the theorems in the text. Also, if an initial condition is given, determine the corresponding particular solu[ion. 1. x 3. 3 dx -71' 4. (e Y + sin - 5. 9x dx )'3 71'X dy = 0 sinh)' dx ye X ) dx + 4y dy + (xe Y = 2. (x - y)(dx - dy) + - cos 71'X = 0 11. - y dx + x dy = 12. (e x + y + (xe x + y + - 13. -3\" dx y) dx + 2x dy = O. 17. (cos 0 wX + + 9. 10. - 2xy sin + (2y + 1Ix - xly2) dy = 0 cos 2x) dx + (lIx - 2 sin 2y) dy = 0 ylx2) dx 2 (X ) dx + cos (x 2 ) dy = 0 y(l) = 'iT + (1 + 1) dy 20. (sin y cos y + 11)' (-ylx 2 + 2 x dy) = 0, 0 e-X(-e- Y 7. e- 28 dr - 2re- 28 d() 8. (2x + sin wx) 18. (cos xy + xly) dx 19. e- Y dx 0 F(x. y) = ylx 4 + eX dy = O. y(Q) = + (xly) cos xy) dy = w 6. eX(cos y dx - sin y dy) = 0 = 1) dy = 0 14. (x 4 + )'2) dx - xv dy = 0, ),(2) = I 15. e 2X (2 cos y dx - sin y dy) = 0, y(O) 16. -sinxy (y dx cosh y dy = 0 eX) dy = 0 0 + X dx = 0, 1 0 F = e X+ Y cos 2 y) dx + x dy = 0 21. Under what conditions for the constants A, B. C, D is (Ax + By) dx + (ex + Dy) dy = 0 exact? Solve the exact equation. 26 CHAP. 1 First-Order ODEs (d) [n another graph show the solution curves satisfying y(O) = ::':::1. ::':::2, ::':::3. ::':::4. Compare the quality of (c) and (d) and comment. 22. CAS PROJECT. Graphing Particular Solutions Graph paI1icuiar solutions of the following ODE. proceeding as explained. Y cos x dx (21) I + - y (a) Test for exactness. If nece~sary, find an integrating factor. Find the general solution II(X. y) = c. (b) Solve (21) by separating variables. Is this simpler than (a)? (c) Graph contours II(X, y) = c by your CAS. (Cf. Fig. 16.) Fig. 16. 1.5 (e) Do the same steps for another nonexact ODE of your choice. dy = 0 Particular solutions in CAS Project 22 23. WRITING PROJECT. Working Backward. Start from solutions u(x, v) = c of your choice. find a corresponding exact ODE, destroy exactness by a multiplication or division. This should give you a feel for the form of ODEs you can reach by the method of integrating factors. (Working backward is useful in other areas, too: Euler and other great masters frequently did it.l 24. TEAM PROJECT. Solution by Several Methods. Show this as indicated. Compare the amount of work. (A) eY(sinh x dx + cosh x dy) = 0 as an exact ODE and by separation. (B) (I + 2x) cos y dx + dy!cos y = 0 by Theorem 2 and by separation. (C) (x 2 + y2) eLI: - 2xy dy = 0 by Theorem I or 2 and by separation with v = ylx. (D) 3x 2 .v dx + 4x 3 dy = 0 by Theorems I and 2 and by separation. (E) Search the text and the problems for further ODEs that can be solved by more than one of the methods discussed so far. Make a list of these ODEs. Find further cases of your own. Linear ODEs. Bernoulli Equation. Population Dynamics Linear ODEs or ODEs that can be transformed to linear form are models of various phenomena, for instance, in physics, biology, population dynamics, and ecology, as we shall see. A first-order ODE is said to be linear if it can be written (1) y' + p(x)y = rex). The defining feature of this equation is that it is linear in both the unknown function y and its derivative y' = dyJdJC, whereas p and r may be any given functions of x. If in an application the independent variable is time, we write t instead of x. If the first term is f(x)y' (instead ofy'), divide the equation by j(x) to get the "standard form" (I), with y' as the first term. which is practical. For instance. y' cos x + y sin x = x is a linear ODE, and its standard form is y' + Y tan x = x secx. The function rex) on the right may be a force, and the solution y(x) a displacement in a motion or an electrical cunent or some other physical quantity. In engineering, rex) is frequently called the input, and y(x) is called the output or the response to the input (and, if given, to the initial condition). SEC. 1.5 Linear ODEs. 27 Bernoulli Equation. Population Dynamics Homogeneous Linear ODE. We want to solve (1) in some interval a < x < b, call it J, and we begin with the simpler special case that rex) is zero for all x in J. (This is sometimes written rex) ~ 0.) Then the ODE (l) becomes )"' + p(x)y = 0 (2) and is called homogeneous. By separating variables and integrating we then obtain dy - y = In 1)'1 = - thus -p(x) dx, f p(x) dx + c*. Taking exponents on both sides, we obtain the general solution of the homogeneous ODE (2), .v(x) (3) here we may also choose c interval. = = ce-Ip(X) (c = ±ec * dx 0 and obtain the trivial solution y(x) = when .v ~ 0); 0 for all x in that Nonhomogeneous Linear ODE. We now solve (I) in the case that rex) in (I) is not everywhere zero in the interval J considered. Then the ODE (1) is called nonhomogeneous. It turns out that in this case, (1) has a pleasant property; namely, it has an integrating factor depending only on x. We can find this factor F(x) by Theorem I in the last section. For this purpose we write (1) as + (py - r) dx dy = O. This is P dx + Q dy = 0, where P = py - rand Q = 1. Hence the right side of (16) in Sec. 1.4 is simply l(p - 0) = p, so that (16) becomes 1 dF - F - = p(:r). dx Separation and integration gives dF =pd'l and F Taking exponents on both sides, we obtain the desired integrating factor F(x), F(x) = eIP d:x'. We now multiply (1) on both sides by this F. Then by the product rule, eIp d:l'(y' + py) = (eIp d:t: y )' = eIp d:t: r . By integrating the second and third of these three expressions with respect to x we get e Ip Dividing this equation by e Ip (4) dx dXy = feIP dX r dx + c. and denoting the exponent fp dx by h, we obtain II = fp(x) dx. 28 CHAP. 1 First-Order ODEs (The constant of integration in h does not matter; see Prob. 2.) Formula (4) is the general solution of (l) in the form of an integral. Solving (l) is now reduced to the evaluation of an integral. In cases in which this cannot be done by the usual methods of calculus, one may have to use a numeric method for integrals (Sec. 19.5) or for the ODE itself (Sec. 21.1). The structure of (4) is interesting. The only quantity depending on a given initial condition is c. Accordingly, writing (4) as a sum of two terms, y(x) = e- h fehr dx (4*) + ce- h, we see the following: (5) E X AMP LEI Total Output = Response to the Input r + Response to the Initial Data. First-Order ODE, General Solution Solve the linear ODE y' _ )' = Solutioll. e 2x Here, p = -1. h=fpdx=-x and from (4) we obtain the general solution From (4*) and t5) we see that the response to the input is e 2x . In simpler cases, such as the present. we may not need the general formula (4). but may wish to proceed directly. multiplying the given equation by e h = e -x. This gives Integrating on both sides. we obtain the same result a, before: ye- X = eX E X AMP L E 2 + c. • hence First-Order ODE, Initial Value Problem Solve the initial value problem y' + Y tan x Solutioll. = .1'(0) = I. sin 2x, Here p = tan x, r = sin 2x = 2 sin x cos x, and fp dx = ftan x dx = In Isec xl· From this we see that in (4), e h = sec x, e -h cos x, = ehr = (sec x)(2 sin x cos x) = 2 sin x, and the general solutIOn of our equation is y(x) = cos x (2f sin x d.1 + c) = c cos x - 2cos x. 2 ~rom this and the initial condition, I = c . I - 2· 12; thus c = 3 and the solution of Our initial value problem 2 y = 3 cos x - 2 cos x. Here 3 cos x is the response to the initial data, and -2 cos 2 x is the response to the input sin 2x. • IS SEC. 1.5 Linear ODEs. E X AMP L E 3 29 Bernoulli Equation. Population Dynamics Hormone Level Assume that rhe level of a certain hormone in the blood of a patient varie, with time. Suppose that rhe time rate of change is the difference between a sinusoidal input of a 24-hour period from the thyroid gland and a continuous removal rate proportional to the level present. Set up a model for the hormone level in the blood and find its general solution. Find the particular solution satisfying a suitable initial condition. Solutioll. Step 1. Setting lip a model. Let yet) be the hormone level at time t. Then the removal rate is Ky(t). The input rate is A + B cos (2 7ft124). where A is the average input rate. and A ~ B to make the input nonnegative. (The constants A. B. and K can be determined by measurements.) Hence the model is y' (t) = + B cos (127ft) [n - Out = A y' + or - Ky(t) Ky = A + B cos (127ft). The initial condition for a particular solution Ypart is Ypart(O) = Yo with t = 0 suitably chosen. e.g .. 6:00 Step 2. General solutio". In (4) we have p = K = COllst, h the general solution y(t) = B A K + 144K2 + e- = Kt. and r = ~ ( 144K cos + B cos (127ft). Hence (4) gives A KtJeKt(A + B cos 12 7ft) dt + ce- A.M. Kt 7ft 7ft) Kt 12 + 127f sin 12 + ce- . The last term decreases to 0 as t increases, practically after a short time and regardles~ of c (that is. of the initial condition). The orher part of y(t) is called the stead~'-state solution because it consists of constant and periodic terms. The entire solution is called the transient-state solution because it models the transition from rest to rhe steady state. These terms are used quite generally for physical and other systems whose behavior depends on time. Step 3. Particular soilition. Setting r = 0 in )(t) and choosing Yo = 0, we have A ),(0) = -K + B 2 ? ·144K + c = 0, 144K + '" thus c= A K B 2 2 ·144K. 144K + '" Inserting this result into y(t). we obtain the particular solution A Vpart(t) = -K + B 2 144K 2 + '" ( 7ft 144K cos -2 I 7ft) + 127f sin -2 I I 44KB ) A ( -K + 144K2 + 7f2 e -Kt with the steady-state part as before. To plot Yp",~ we must specify values for the constants, say, A = B = I and K = 0.05. Figure 17 shows this solution. Notice that the transition period is relatively short (although K is small), • and the curve soon looks sinusoidal; this is the response to the input A + B cos (127ft) = I + cos (f27ft). y 25 5 Fig. 17. Particular solution in Example 3 CHAP. 1 30 First-Order ODEs Reduction to Linear Form. Bernoulli Equation Numerous applications can be modeled by ODEs that are nonlinear but can be transformed to linear ODEs. One of the most useful ones of these is the Bernoulli equation5 y' (6) If a + p(x)y = g(x)ya (a any real number). = 0 or a = I, Equation (6) is linear. Otherwise it is nonlinear. Then we set U(x) = [y(X)]l-a. We differentiate this and substitute y' from (6). obtaining Simplification gives u' where yl-a = = (l - al(g - pyl-a), II on the right, so that we get the linear ODE (7) tI' + (l - a)pu = (1 - a)g. For further ODEs reducible to linear from, see Ince's classic [A 111 listed in App. I. See also Team Project 44 in Problem Set 1.5. E X AMP L E 4 Logistic Equation Solve thc following Bemoulli equation. known as the logistic equation (or Verhulst equation6 ): y' (8) Solutioll. to see thaI {l Write (8) in the form = 2, so that It = (6). /-a = = Ay - By2 that is. y -1. Ditferentiate this II and substitute y' from (8). The last term is _Ay-l = -All. Hence we have obtained the linear ODE 5JAKOB BERNOULLI (1654-1705), Swiss mathematician, professor at Basel. also known for his contribution to elasticity theory and mathematical probability. TIle method for solving Bernoulli's equation was discovered by the Leibniz in 1696. Jakob Bernoulli's students included his nephew NJKLACS BERNOULLI (1687-1759). who contributed to probability theory and infinite series. and his youngest brother JOHANN BERNOULLI (1667-1748). who had profound influence on the development of calculus. became Jakob's successor at Basel. and had among his students GABRIEL CRAMER (see Sec. 7.7) and LEONHARD EULER (see Sec. 2.5). His son DANIEL BERNOULLI (I700-1782) is "nown for his basic work in fluid flow and the kinetic theory of gases. 6PIERRE-FRAN<;:OJS VERHLLST, Belgian statistician, who introduced Eq. (8) a:, a model for human population growth in 1838. SEC. 1.5 Linear ODEs. 31 Bernoulli Equation. Population Dynamics 1/' + All 11= ce- At = B. The general solution is [by (4)1 Since II = lIy, this gives the general solution of (8). (9) Directly from + BfA. (8) we see that y 0= 0 y = -;; = (y( t) = 0 for all t) is also a solution. o Fig. 18. (Fig. 18). ce-At + BfA • Timet Logistic population model. Curves (9) in Example 4 with A/B = 4 Population Dynamics The logistic equation (8) plays an important role in population dynamics, a field that models the evolution of populations of plants. animals, or humans over time t. If B = 0, then (8) is y' = dyldt = Ay. In this case its solution (9) is y = (l/c)e At and gives exponential growth, as for a small population in a large counn)' (the United States in early times!). This is called Malthus's law. (See also Example 3 in Sec. 1.1.) The term -By2 in (8) is a "braking term" that prevents the population from growing without bound. Indeed, if we write y' = Ay[l - (BIA)y]. we see that if y < AlB, then y' > O. so that an initially small population keeps growing as long as y < AlB. But if y > AlB. then y' < 0 and the population is decrea-;ing as long as y > AlB. The limit is the same in both cases, namely, AlB. See Fig. 18. We see that in the logistic equation (8) the independent variable t does not occur explicitly. An ODE y' = f(t, y) in which t does not occur explicitly is of the form (10) y' == f(y) and is called an autonomous ODE. Thus the logistic equation (8) is autonomous. Equation (10) has constant solutions, called equilibrium solutions or equilibrium points. These are determined by the zeros of fey), because fey) = 0 gives y' = 0 by (10); hence y = const. These zeros are known as critical points of (10). An equilibIium solution is called stable if solutions close to it for some t remain close to it for all further t. It is called unstable if solutions initially close to it do not remain close to it as t increases. For instance, \" = 0 in Fig. 18 is an unstable equilibrium solution, and \' = 4 is a stable one. . 32 CHAP. 1 E X AMP L E 5 First-Order ODEs Stable and Unstable Equilibrium Solutions. "Phase Line Plot" The ODE y' = (y - l)(y - 2) has the stable equilibrium solution Yl = I and the unstable Y2 = 2, as the direction field in Fig. 19 suggests. The values )'1 and Y2 are the zeros of the parabola fey) = (y - l)ly - 2) in the figure. Now, since the ODE is autonomous, we can "condense" the direction field to a "phase line plot" giving)'1 and Y2, and the direction (upward or downward) of the arrows in the field, and thus giving information about the stability or instability of the equilibrium solutions. • y y(xl /////// ~O. / / / / / / / 2.5' "/ / / / / / / / ///////// 1111111111 '111111111 1I1I1I1,i 11111111' / / / / / / / / / '" ~~~~~~~~/ .///////// -~~~~~~~~~ Y2===-=-~--::::--=-W .==-~--=.-:.-:::. ---------~---------------~r------------~-- -----------------------------------------Yl ------_ ±oe ._------- --------- - ~~~~~~~~/ ~~~~~~~~~ / / / / / / / / ~/ / / / / / / / "'///////// '0:5- / / / / / / / / / 0 2 I: r ,. 111111111 ',111111111 I f f f f 1 I I 1 ~ ,. ,. ,. ,. t I ,. -2 -1 (Al x x (el (Bl Fig. 19. Example 5. (A) Direction field. (B) "Phase line". (C) Parabola f(y) A few further population models will be discussed in the problem set. For some more details of population dynamics, see C. W. Clark, Mathematical Bioecnnvmics, New York, Wiley, 1976. Further important applications of linear ODEs follow in the next section. 1. (CAUTION!) Show that e-1n(sec x) = cos x. e- 1n x = l/x (not -x) and 6. x 2 y' 7. y' 2. (Integration constant) Give a reason why in (4) you may choose the constant of integration in fp dx to be zero. + -t- 3xy = lIx, ky = e 8. y' + 2y = 4 cos 2x, 9. y' y(l) = -1 2kx y(!7T) = 2 6(y - 2.5) tanh 1.5x + 4x 2 y = (4x 2 - x)e- x2/ 2 11. )" + 2y sin 2x = 2e cos 2X, yeO) 10. y' 13-171 GENERAL SOLUTION. INITIAL VALUE PROBLEMS Find the general solution. If an initial condition is given, find also the corresponding particular solution and graph or sketch it. (Show the details of your work.) 3. y' + 3.5y + = 2.8 4. y' = 4y 5. )" + 1.25y = 5, x yeO) 6.6 12. )" tan x = 2y - 8, 0 Y(~7T) = 0 + 4y cot 2x = 6 cos 2x, y(!7T) = 2 14. y' + )' tan x = e- O. Ob cos x, 1'(0) = 0 15. y' + Y/X2 = 2xe 1/ x , y(l) = 13.86 16. y' cos 2 x + 3y = I, y(!7T) = ~ 17. x 3 y' + 3x 2 y = 5 sinh lOx 13. y' SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 118-241 NONLINEAR ODEs Using a method of this section or separating variables, find the general solution. If an initial condition is given, find also the particular solution and sketch or graph it. 18. y' + Y = y2, yeO) = -I y' + = -tan y. y(O) = ~7T + I)y = e 'y3, y(O) = 0.5 22. y' sin 2y + x cos 2y = 2x 23. 2yy' + .\'2 sin x = sin x. yeO) = V2 24. y' + x 2y = (e- sinh X)/(3y2) 21. X (x X3 125-361 FURTHER APPLICATIONS 25. (Investment programs) Bill opens a retirement savings account with an initial amount Yo and then adds $k to the account at the beginning of every year until retirement at age 65. Assume that the interest is compounded continuously at the same rate R over the years. Set up a model for the balance in the account and find the general solution as well as the particular solution, letting t = 0 be the instant when the account is opened. How much money will Bill have in (he account at age 65 if he starts at 25 and invests $1000 initially as well as annually, and the interest rate R is 6%? How much should he invest initially and annually (same amounts) to obtain the same final balance as before if he starts at age 45? First, guess. 26. (Mixing problem) A tank (as in Fig. 9 in Sec. 1.3) contains 1000 gal of water in which 200 Ib of salt is dissolved. 50 gal of brine. each uallon containin u (I + cos t) Ib of dissolved salt, run: into the tank pe~ minute. The mixture. kept unifonn by stirring. nms out at the same rate. Find the amount of salt in the tank at any time t (Fig. 20). 1000 500 200 Fig. 20. 28. (Heating and cooling of a building) Heating and cooling of a building can be modeled by the ODE = k 1 (T - Ta) + k 2 (T - Tw) + P, where T = T(t) is the temperature in the building at time t, Ta the outside temperature, T w the temperature wanted in the building. and P the rate of increase of T due to machines and people in the building, and kl and ~ are (negative) constants. Solve this ODE, assuming P = canst, T w = canst, and To varying sinusoidally over 24 hours, say, Ta = A - C cos (27T/2A)t. Discuss the effect of each tenn of the equation on the solution. 29. (Drug injection) Find and solve the model for drug injection into the bloodstream if, beginning at t = 0, a constant amount A g/min is injected and the drug is simultaneously removed at a rate proportional to the amount of the drug present at time t. 30. (Epidemics) A model for the spread of contagious diseases is obtained by assuming that the rate of spread is proportional to the number of contacts between infected and noninfected persons, who are assumed to move freely among each other. Set up the model. Find the equilibrium solutions and indicate their stability or instability. Solve the ODE. Find the limit of the proportion of infected persons as t -+ x and explain what it means. 31. (Extinction vs. unlimited growth) If in a population y(t) the death rate is proportional to the population, and the birth rate is proportional to the chance encounters of meeting mates for reproduction. what will the model be? Without solving. find out what will eventually happen to a small initial population. To a large one. Then solve the model. 32. (Harvesting renewable resources. Fishing) Suppose that the population yU) of a certain kind of fish is given by the logistic equation (8), and fish are caught at a rate Hy proportional to y. Solve this so-called Schaefer model. Find the equilibrium solutions Yl and Y2 (> 0) when H < A. The expression Y = HY2 is called the equilibrium harvest or sustainable yield corresponding to H. Why? y o concentration p/4. and the mixture is unifonn (an assumption that is only very imperfectly true)? First, guess. T' 19. y' = 5.7y - 6.5 y 2 20. (x 2 + I)y' 33 50 100 Amount of salt y(t) in the tank in Problem 26 27. (Lake Erie) Lake Erie has a water volume of about 450 km3 and a flow rate (in and out) of about 175 km 3 per year. If at some instant the lake has pollution c~nc~ntration p = 0.04%, how long, approximately. wIll It take to decrease it to pl2. assuming that the inflow is much cleaner, say, it has pollution 33. (Harvesting) In Prob. 32 find and graph the solution satisfying yeO) = 2 when (for simplicity) A = B = I and H = 0.2. What is the limit? What does it mean? What if there were no fishing? 34. (Intermittent harvesting) In Prob. 32 assume that you fish for 3 years, then fishing is banned for the next 3 years. Thereafter you start again. And so on. This is called illtermittent /Uln'estillg. Describe qualitatively how the population will develop if intermitting is CHAP. 1 First-Order ODEs 34 continued periodically. Find and graph the solution for the first 9 years, assuming that A = B = I, H = 0.2, and yeO) = 2. y 2 1.8 1.6 43. CAS EXPERIMENT. (a) Solve the ODE y' - ylx = -x- 1 cos (l/x). Find an initial condition for which the arbitrary constanl is zero. Graph the resulting particular solution, experimenting to obtain a good figure near x = O. (b) Generalizing (a) from 11 = I to arbitrary 11, solve the ODE y' - nylx = _xn - 2 cos (llx). Find an initial condition as in (a). and experiment with the graph. 44. TEAM PROJECT. Riccati Equation, Clairaut Equation. A Riccati equation is of the form 1.4 (1 I) 1.2 y' + p(x)y = gp:)y 2 + hex). A Clairaut equation is of the form 0.8 L - _ - - L_ _" - - _ - - ' -_ _->--_ 4 o 2 6 8 8. 21. Fish population in Problem 34 35. (Harvesting) If a population of mice (in multiples of 1000) follows the logistic law with A = I and B = 0.25. and if owls catch at a time rate of 10% of the population present, what is the model, its equilibrium harvest for that catch. and its solution? 36. (Harvesting) Do you save work in Prob. 34 if you first transform the ODE to a linear ODE? Do this transformation. Solve the resulting ODE. Does the resulting yet) agree with that in Prob. 34? I - 7-40 GENERAL PROPERTIES OF LINEAR ODEs These properties are of practical and theoretical importance because they enable us to obtain new solutions from given ones. Thus in modeling, whenever possible, we prefer linear ODEs over nonlinear ones, which have no similar properties. Show that nonhomogeneous linear ODEs (1) and homogeneous linear ODEs (2) have the following properties. Illustrate each property by a calculation for two or three equations of your choice. Give proofs. 37. The sum YI + Y2 of two solutions YI and Y2 of the homogeneous equation (2) is a solution of (2), and so is a scalar mUltiple aYI for any constant a. These properties are not true for (1)1 38. Y = 0 (that is, .v(x) = 0 for all x, also written y(x) "'" 0) is a solution of (2) [not of (I) if rex) =1= 01], called the trivial solution. 39. The sum of a solution of (I) and a solution of (2) is a solution of (1). 40. The difference of two solutions of (l) is a solution of (2). 41. If Yl is a sulution of (I), what can you say about eYl? 42. If YI and Y2 are solutions of y~ + PYI = rl and Y~ + PY2 = r2, respectively (with the same p!), what can you say about the sum.vl + Y2? (12) y = xy' + g(y'). (a) Apply the transformation y = Y + lilt to the Riccati equation (1 I ), where Y is a solution of (11), and oblain for u the linear ODE u' + (2Yg - P)U = -g. Explain the effect of the transformation by writing it as y = Y + v, v = lilt. (b) Show that Y = Y y' - (2x 3 + l)y = x is a solution of = _x 2 y2 - X4 - X + and solve this Riccati equation. showing the details. (c) Solve y' + (3 - 2X2 sin x)y = _y2 sin x + 2x + 3x 2 - X4 sin x, using (and Verifying) that y = x 2 is a solution. (d) By working "backward" from the L1-equation find further Riccati equations that have relatively simple solutions. (e) Solve the Clairautequationy = xy' + 1/y'.Hillt. Differentiate this ODE with respect to x. (f) Solve the Clairaut equation /2 - xy' + Y = 0 in Prob. 16 of Problem Set 1.1. (g) Show that the Clairaut equation (12) has as solutions a family of straight lines Y = ex + gee) and a singular solution determined by g' (s) = -x, where s = y', that forms the envelope of that family. 45. (Variation of parameter) Another method of obtaining (4) results from the following idea. Write (3) as ey*, where y* is the exponential function. which is a solution of the homogeneous linear ODE y*' + py* = O. Replace the arbitrary constant e in (3) with a function II to be determined so that the resulting function y = IIY* is a solution of the nonhomogeneous linear ODE y' + PY = r. 46. TEAM PROJECT. Transformations of ODEs. We have transformed ODEs to separable form, to exact form, and to linear form. The purpose of such transfonnations is an extension of solution methods to larger classes of ODEs. Describe the key idea of each of these transformations and give three typical examples of your choice for each transformation, shOwing each step (not just the transformed ODE). 35 Optional SEC. 1.6 Orthogonal Trajectories. 1.6 Orthogonal Trajectories. Optional An important type of problem in physics or geometry is to find a family of curves that intersect a given family of curves at right angles. The new curves are called orthogonal trajectories of the given curves (and conversely). Examples are curves of equal temperamre (isothel7lls) and curves of heat flow, curves of equal altitude (contour lines) on a map and curves of steepest descent on that map, curves of equal potential (equipotential curves, curves of equal voltage-the concentric circles in Fig. 22). and curves of electric force (the straight radial segments in Fig. 22). Fig. 22. Equipotential lines and curves of electric force (dashed) between two concentric (black) circles (cylinders in space) Here the angle of intersection between two curves is defined to be the angle between the tangents of the curves at the intersection point. Orthogonal is another word for pe rpendicular. In many cases orthogonal trajectories can be found by using ODEs. as follows. Let (1) G(x, y, c) = 0 be a given family of curves in the xy-plane, where each curve is specified by some value of c. This is called a one-parameter family of curves, and c is called the parameter of the family. For instance, a one-parameter family of quadratic parabolas is given by (Fig. 23) or, written as in (1), G(x, y, c) = y - cx 2 = o. Step 1. Find an ODE for which the given family is a general solution. Of course, this ODE must no longer contain the parameter c. In our example we solve algebraically for c and then differentiate and simplify; thus, hence y , 2y x CHAP. 1 36 First-Order ODEs The last of these equations is the ODE of the given family of curves. It is of the form y' = f(x, y). (2) Step 2. Write down the ODE of the orthogonal trajectories. that is. the ODE whose general solution gives the orthogonal trajectOlies of the given curves. This ODE is _, 1 y =--f(x, y) (3) with the same f as in (2). Why? Well, a given curve passing through a point (xo, Yo) has slope f(xo, Yo) at that point, by (2). The trajectory through (xo, Yo) has slope -lIf(xo, Yo) by (3). The product of these slopes is -1, as we see. From calculus it is known that this is the condition for orthogonality (perpendicularity) of two straight lines (the tangents at (xo, Yo», hence of the curve and its orthogonal trajectory at (xo, Yo). Step 3. Solve (3). For our parabolas y = cx 2 we have y' = 2ylx. Hence their orthogonal trajectories are obtained from y' = - xl2y or 2yy' + x = O. By integration, y2 + !x 2 = c*. These are the ellipses in Fig. 23 with semi-axes Vk* and VC*. Here, c* > 0 because c* = 0 gives just the origin, and c * < 0 gives no real solution at all. y Fig. 23. 11-121 Parabolas and orthogonal trajectories (ellipses) in the text ORTHOGONAL TRAJECTORIES Sketch or graph some of the given curves. Guess what their orthogonal trajectories may look like. Find these trajectories. (Show the details of your work.) 1. Y = 4x + c 2. y = clx 3. y = ex 4. y2 = 2X2 + c 2 5. x y = C 6. y = ce- 3x 7. Y = ce x2/2 9. 4x 2 + y2 = e 11. x = ce yl4 113-15/ 8. x 2 10. x 12. x 2 - y2 = c cVy = + (y - c)2 = c 2 OTHER FORMS OF THE ODEs (2) AND (3) 13. (y as independent variable) Show that (3) may be written dx/dy = -f(x, y). Use this form to find the orthogonal trajectories of y = 2x + ce-x . SEC. 1.7 37 Existence and Uniqueness of Solutions = 18. (Electric field) The lines of electric force of two opposite charges of the same strength at (-1. 0) and (1, 0) are the circles through (-1. 0) and (l, 0). Show that these circles are given by x 2 + (y - c)2 = 1 -+ c 2. Show that the equipotential lines (orthogonal trajectories of those circles) are the circles given by lx + C*)2 + )'2 = C*2 - I (dashed in Fig. 25). 14. (Family g(x,y) c) Show that if a family is given as g(x, y) = c, then the orthogonal trajectories can be obtained from the following ODE, and use the latter to solve Prob. 6 written in the form g(x, y) = c. dy ag/ay dx ag/iJx 15. (Cauchy-Riemann equations) Show that for a family u(x, y) = c = const the orthogonal trajectories vex, y) = c* = const can be obtained from the following Cauchy-Riemann equations (which are basic in complex analysis in Chap. 13) and use them to find the orthogonal trajectories of eX sin y = const. (Here, subscripts denote partial derivatives.) 116-20 1 APPLICATIONS 16. (Fluid flow) Suppose that the streamlines of the flow lpaths of the particles of the fluid) in Fig. 24 are 'ltlx, y) = xy = COllst. Find their orthogonal trajectories (called equipotential lines, for reasons given in Sec. 18.4). Fig. 25. Electric field in Problem 18 19. (Temperature field) Let the isotherms (curves of constant temperature) in a body in the upper half-plane y > 0 be given by 4X2 + 9)"2 = c. Find the orthogonal trajectories lthe curves along which heat will flow in regions filled with heat-conducting material and free of heat sources or heat sinks). 20. TEAM PROJECT. Conic Sections. (A) State the main steps of the present method of obtaining orthogonal trajectorics. (B) Find conditions under which the orthogonal trajectories of families of ellipses x 2/a 2 + y2/b 2 = C are again conic sections. Illustrate your result graphically by sketches or by using your CAS. What happens if a~ O?If b~ O? x Fig. 24. Flow in a channel in Problem 16 17. (Electric field) Let the electric equipotential lines (curves of constant potential) between two concentric cylinders (Fig. 22) be given by u(x, y) = x 2 + y2 = c. Use the method in the text to find their orthogonal trajectories (the curves of electric force). 1.7 (C) Investigate families of hyperbolas x 2/a 2 - y2/b 2 = c in a similar fashion. (D) Can you find more complicated curves for which you get ODEs that you can solve? Give it a try. Existence and Uniqueness of Solutions The initial value problem /,1"/ + /y/ = 0, .1'(0) = 1 has no solution because y = 0 (that is, y(x) = 0 for all x) is the only solution of the ODE. The initial value problem y' == 2x, .1'(0) = I 38 CHAP. 1 First-Order ODEs has precisely one solution, namely, y = x 2 xy' = y - I, + 1. The initial value problem yeO) = I has infinitely many solutions, namely, y = 1 + cx, where c is an arbitrary con<;tant because = 1 for all c. From these examples we see that an initial value problem y(O) (1) y' = f(x, y), may have no solution, precisely one solution, or more than one solution. This fact leads to the following two fundamental questions. Problem of Existence Under what conditions does an initial mlue problem nne solutinn (hence one or several solutions)? ~f the form (I) have at least Problem of Uniqueness Under what conditiollS dnes that problem have at 17l0H one solution (hence excluding the case that is has more than one solution)? Theorems that state such conditions are called existence theorems and uniqueness theorems, respectively. Of course, for our simple examples we need no theorems because we can solve these examples by inspection; however, for complicated ODEs such theorems may be of considerable practical importance. Even when you are sure that your physical or other system behaves uniquely, occasionally your model may be oversimplified and may not give a faithful picture of the reality. THEOREM 1 Existence Theorem Let the right side f(x, y) of the ODE ill the initial value problem y' (1) = f(x, y), y(xo) = Yo be continuous at all points (x. y) in some rectangle R: Ix - xol < a. IY - Yol < b (Fig. 26) alld bounded ill R; that is, there is a number K sllch that (2) If(x, y)1 ~ K for all (x, y) ill R. Then the initial value problem (1) has at least one solution y(x). This solution exists at least for all x in the subinterl'al Ix - xol < a of the illfervallx - xol < a; here, a is the smaller ~f the two numbers (/ and b/K. SEC 1.7 39 Existence and Uniqueness of Solutions Y -----<j' Yo R x Fig. 26. Rectangle R in the existence and uniqueness theorems (Example of BOllndedlless. The function f(x, .1') = x 2 + y2 is bounded (with K = 2) in the square Ixl < I, Iyl < 1. The function f(x, y) = tan (x + y) is not bounded for Ix + yl < 7T/2. Explain!) THEOREM 2 Uniqueness Theorem Let f alld its partial derivative fy = rJf/rJy be collfinllOllS for aI/ (x, y) ill the rectallgle R (Fig. 26) alld bOllnded, say, (3) (a) If(x, y)1 ~ K, for all (x, y) ill R. Theil the illitial mille problem (1) has at most aile solutioll y(xJ. TIllis. by TIleorem 1, the problem has precisely aile soilltioll. This solution exists at leastfor aI/ x in that subinterval Ix - xol < a. Understanding These Theorems These two theorems take care of almost all practical cases. Theorem 1 says that if f(x, y) is continuous in some region in the xy-plane containing the point (xo, Yo). then the initial value problem (I) hali at least one solution. Theorem 2 says that if, moreover, the partial derivative aflay of f with respect to y exists and is continuous in that region, then (I) can have at most one solution; hence, by Theorem I, it has precisely one solution. Read again what you have just read-these are entirely new ideas in our discussion. Proofs of these theorems are beyond the level of this book (see Ref. [A II] in App. I); however, the following remarks and examples may help you to a good understanding of the theorems. ~ K; that is, the slope of any Since / = f(x, y), the condition (2) implies that solution curve y(x) in R is at least - K and at most K. Hence a solution curve that passes through the point (xo, Yo) must lie in the colored region in Fig. 27 on the next page bounded by the lines 11 and 12 whose slopes are -K and K, respectively. Depending on the form of R, two different cases may arise. In the first case, shown in Fig. 27a, we have blK ~ a and therefore a = a in the existence theorem, which then asserts that the solution exists for all x between Xo - a and Xo + ll. In the second case, shown in Fig. 27b, we have blK < ll. Therefore, ll' = blK < a, and all we can conclude from the theorems is that the solution exists for all x between Xo - blK and Xo + blK. For larger or smaller x's the solution curve may leave the rectangle R, and since nothing is assumed about f outside R, nothing can be concluded about the solution for those larger or smaller x's: that is, for <;uch x's the solution mayor may not exist-we don't know. 1/1 First-Order ODEs CHAP. 1 40 y Yo + b Yo yo-b x Xo (b) (a) Fig. 27. The condition (2) of the existence theorem. (a) First case. (b) Second case Let us illustrate our discussion with a simple example. We shall see that our choice of a rectangle R with a large base (a long x-interval) will lead to the ca<;e in Fig. 27b. E X AMP L E 1 Choice of a Rectangle Consider the initial valne problem "(0) = 0 and take the rectangle R; I.\i < 5, 1)'1 < 3. Then a I I ilf -:(Iv a = = 5, b = = 21yl ~ M = 6. b K= 3. and 0.3 < a. Indeed. the solntion of the problem is y = tan x (see Sec. 1.3, Example I). This solntion is discontinuous at ± 70/2, and there is no callTilllwl/s solution valid in (he entire imerval Ixl < 5 from which we starred. • The conditions in the two theorems are sufficient conditions rather than necessary ones, and can be lessened. [n particular. by the mean value theorem of differential calculus we have where (x, Yl) and (x, )"2) are assumed to be in R, and:V is a suitable value between)"l and Y2' From this and (3b) it follows that (4) It can be shown that (3b) may be replaced by the weaker condition (4), which is known as a Lipschitz condition.7 However, continuity of f(x, y) is not enough to guarantee the uniquelless of the solution. This may be illustrated by the following example. SEC. 1.7 41 Existence and Uniqueness of Solutions E X AMP L E 2 Nonuniqueness The initial value problem y' ViYj, = yeO) = 0 has the two solutions y=o v* and • = { x2/4 if x ~ 0 2 if x < 0 -x 14 although f(x, ,1') = Vlyl is continuous for all y. The Lipschitz condition (4) is violated in any region that includes the line y = 0, because for h = 0 and positive Y2 we have (5) II(x, )'2) - I(x, Yl)1 IY2 - hi I (~>Ol vy; . and this can be made as large as we plea,e by choosing ,1'2 sufficiently ,mall. whereas (4) requires that the quotient on the left side of (5) should not exceed a fixed constant M. • _ _._-.= _ --. . . . =..._ u .•. .- ... -............. 1. (Vertical strip) If the a~sumptions of Theorems I and 2 are satisfied not merely in a rectangle but in a vertical infinite strip Ix - xol < a, in what interval will the solution of (1) exist? 2. (Existence?) Does the initial value problem (x - l)y' = 2y, y(l) = I have a solution? Does your result contradict our present theorems? 3. (Common points) Can two solution curves of the same ODE have a common point in a rectangle in which the assumptions of the present theorems are satisfied? 4. (Change of initial condition) What happens in Prob. 2 if you replace yO) = 1 with yO) = k? S. (Linear ODE) If p and I' in y' + p(x)y = rex) are continuous for all x in an interval Ix - xol ~ a, show that f(x, y) in this ODE satisfies the conditions of our present theorems, so that a cOlTesponding initial value problem has a unique solution. Do you actually need these theorems for this ODE? 6. (Three possible cases) Find all initial conditions such that lx 2 - 4x)y' = (2x - 4)y has no solution, precisely one solution, and more than one solution. 7. (Length of x-interval) In most cases the solution of an initial value problem (1) exists in an x-interval larger than that guaranteed by the present theorems. Show this fact for y' = 2y2, yO) = I by finding the best possible a (choosing b optimally) and comparing the result with the actual solution. 8. PROJECT. Lipschitz Condition. (A) State the definItion of a Lipschitz condition. Explain its relation to the existence of a partial derivative. Explain its significance in our present context. Illustrate your statements by examples of your own. (B) Show that for a lillear ODE y' + p(:.:)y = rex) with continuous p and r in Ix - xol ~ a a Lipschitz condition holds. This is remarkable because it means that for a linear ODE the continuity of f(x, y) guarantees not only the existence but also the uniqueness of the solution of an initial value problem. (Of course, this also follows directly from (4) in Sec. 1.5.) (C) Discuss the uniqueness of solution for a few simple ODEs that you can solve by one of the methods considered, and find whether a Lipschitz condition is satisfied. 9. (Maximum a) What Example I in the text? IS the largest possible a In 10. CAS PROJECT. Picard Iteration. (A) Show that by integrating the ODE in (I) and observing the initial condition you obtain (6) y(x) = Yo + f f(t. \'(t» dt. Xo 7RUDOLF LIPSCHITZ (1832-1903), Gennan mathematician. Lipschitz and similar conditions are important in modern theories, for instance, in partial differential equations. CHAP. 1 First-Order ODEs 42 (B) Apply the iteration to y' = x + y, yeO) = O. Also solve the problem exactly. (0 Apply the iteration to y' = 2y2. yeO) = 1. Also solve the problem exactly. This fonn (6) of (I) suggests Picard's iteration methodS, which is defined by (7) y,,(x) = )'0 + f' fU. y,.-IU)) dt. n = 1. 2..... Xo 2VY. = y( I) = O. Which of them does Picard's iteration approximate? (D) Find all solutions of)" It gives approximations )'1 • .'"2. )'3' . . . of the unknown solution), of (I). Indeed, you obtain ."1 by substituting y = )'0 on the right and integrating-this is the first step-, then Y2 by substituting y = YI on the right and integrating-this is the second step-. and so on. Write a program of the iteration that gives a printout of the first approximations Yo. )'1' ...• YN as well as their graphs on common axes. Try your program on two initial value problems of your own choice. .. (E) Experiment with the conjecture that Picard's iteration converges to the solution of the problem for any initial choice of y in the integrand in (7) (leaving )'0 outside the integral as it is). Begin with a simple ODE and see what happens. When you are reasonably sure. take a slightly more complicated ODE and give it a try. Ew=.Q U £5 T ION SAN D PRO B L EMS , 14. y' 1. Explain the tenns ordinary d~fferellfial equatiol/ (ODE). partial d~fferellfial equation (PDE). order. gel/eral solution. and particular solutioll. Give examples. Why are these concepts of imp0l1ance? 2. What is an initial condition? How is this condition used in an initial value problem? 16xly 13. Y GENERAL SOLUTION , 15-261 Find the general solution. Indicate which method in this chapter you are using. Show the details of your work. 15. y' = x 2 (1 + .1'2) x 2 + I) 17. yy' + xy2 = x IS. -7T sin TTX cosh 3y dx 3. What is a homogeneous linear ODE? A nonhomogeneous linear ODE? Why are these equations simpler than nonlinear ODEs? 16. y' 4. What do you know about direction fields and their practical importance? 5. Give examples of mechanical problems that lead to ODEs. 19. y' + 6. Why do electric circuits lead to ODEs? 7. Make a list of the solution methods considered. Explain each method with a few short sentences and illustrate it by a typical example. S. Can certain ODEs be solved by more than one method? Give three examples. 9. What are integrating factors? Explain the idea. Give examples. 10. Does every first-order ODE have a solution? A general solution? What do you know about uniqueness of solutions? 111-141 DIRECTION FIELDS Graph a direction field (by a CAS or by hand) and sketch some of the solution curves. Solve the ODE exactly and compare. 11. ,.' = 1 + 4)'2 12. y' 3y - 2~ = x(y - + 21. 3 sin 2y dx 22. x/ 3 cos TTX sinh 3y dy = 0 20. y' - y = 1Iy 2x cos 2y dy = 0 x tan (ylx) + Y = 23. (y cos xy - 2x) dx = 24. xy' + Y sin x = sin x (y - 2X)2 25. sin (y - x) dx 26. xy' = (ylx)3 + +)' lx cos x)' + 2y) dy = 0 (Set.\· - 2x = z.) + [cos (y +y - x) - sin (y - x)] dy = 0 127-321 INITIAL VALUE PROBLEMS Solve the following initial value problems. Indicate the method used. Show the details of your work. 27. yy' + x = O. y(3) = 4 2S.y' 3y=-12 y 2. y(O)=2 29. .v' 30. y' + = I + 7T)' = )'2, + y 21x yeO) = 0 + (2 + 2X2)') dy = o. yeO) = + eX(l + llx)] dx + lx + 2y) dy = 31. (2 xy 2 - sin 32. [2y Y(~7T) = 0 2b cos 7TX, x) dx I 0, y(l) = I ~MILE PICARD (1856-1941). French mathematician. also known for his important contributions to complex analysis (see Sec. 16.2 for his famous theorem). Picard u~ed his method to prove Theorems I and 2 as well as the convergence ofthe sequence (7) to the solution of (I). In precomputer times the iteration was oflittle practical value because of the integrations. 43 Summary of Chapter 1 133-431 APPLICATIONS, MODELING 33. ~Heat flow) If the isothelms in a region are x 2 - )'2 = c, what are the curves of heat flow (assuming orthogonality)? 34. (Law of cooling) A thennometer showing WaC is brought into a room whose temperature is 25°C. After 5 minutes it shows 20°C. When will the thennometer practically reach the room temperature, say, 24.9°C? 35. (Half-life) If 10o/c of a radioactive substance disintegrates in 4 days, what is it~ half-life? 36. (HaIf-life) What is the half-life of a substance if after 5 days, 0.020 g is present and after 10 days, 0.015 g? 37. (HaIf-life) When will 99% of the substance in Prob. 35 have disintegrated? 38. (Air circulation) In a room containing 20000 ft3 of air, 600 ft3 of fresh air tlows in per minute, and the mixture (made practically unifonn by circulating fans) is exhausted at a rate of 600 cubic feet per minute (cfm). What is the amount of fresh air Yl1) at any time if yeO) = 0'1 After what time will 90% of the air be fresh? 39. (Electric field) If the equipotential lines in a region of the x)"-plane are 4x 2 + y2 = c, what are the curves of the electIical force? Sketch both families of curves. 40. (Chemistry) In a bimolecular reaction A + B - ? M, a moles per liter of a substance A and b moles per liter of a substance B are combined. Under constant temperature the rate of reaction is y' = (Law of mass action); k(a - y)(b - .1') that is, y' is proportional to the product of the concentrations of the substances that are reacting. where )'(t) is the number of moles per liter which have reacted after time t. Solve this ODE, assuming that a *- b. 41. (Population) Find the population y(1) if the birth rate is proportional to y(r) and the death rate is proportional to the square of y( f). 42. (Curves) Find all curve~ in the first quadrant of the Ayplane such that for every tangent. the segment between the coordinate axes is bisected by the point of tangency. (Make a sketch.) 43. (Optics) Lambert's law of absorption9 states that the absorption of light in a thin transparent layer is proportional to the thickness of the layer and to the amount of light incident on that layer. Formulate this law as an ODE and solve it. --- .......... ...... -.-.- . _._ ......--- ...... - . . . . . . . ".'11..-. • _ ......... _ - Fi rst-Order ODEs This chapter concerns ordinary differential equations (ODEs) of first order and their applications. These are equations of the form F(x, y, /) = 0 or in explicit form y' = I(x, y) involving the derivative y' = dy/dx of an unknown function y, given functions of x, and, perhaps, y itself. If the independent variable x is time, we denote it by t. In Sec. 1.1 we explained the basic concepts and the process of modeling, that is, of expressing a physical or other problem in some mathematical form and solving it. Then we discussed the method of direction fields (Sec. 1.2), solution methods and models (Secs. 1.3-1.6), and, finally, ideas on existence and uniqueness of solutions (Sec. 1.7). 9]OHANN HEINRICH LAMBERT (1728-1777), German physicist and mathematician. 44 CHAP. 1 First-Order ODEs A first-order ODE usually has a general solution, that is. a solution involving an arbitrary constant, which we denote by c. In applications we usually have to find a unique solution by determining a value of c from an initial condition y(xo) = Yo. Together with the ODE this is called an initial value problem y' = (2) f(x, y), y(xo,) = Yo (xo, Yo given numbers) and its solution is a particular solution of the ODE. Geometrically. a general solution represents a family of curves, which can be graphed by using direction fields (Sec. L.2). And each particular ~ulution corresponds to one of these curves. A separable ODE is one that we can put into the form (3) g(y) dy = f(x) tit (Sec. 1.3) by algebraic manipulations (possibly combined with transformations, such as ylx and solve by integrating on both sides. An exact ODE is of the form (4) where M dx M(x. y) dx + + N(x. y) dy = 0 = u) (Sec. 1.4) N dy is the differential du = Ux dx + uy dy of a function u(x, .v), so that from du = 0 we immediately get the implicit general solution u(x, y) = c. This method extends to nonexact ODEs that can be made exact by mUltiplying them by some function F(x, y), called an integrating factor (Sec. 1.4). Linear ODEs (5) y' + p(x)y = rex) are very important. Their solutions are given by the integral formula (4), Sec. 1.5. Certain nonlinear ODEs can be transformed to linear form in terms of new variables. This holds for the Bernoulli equation y' + p(x)y = g(x)yU (Sec. 1.5). Applications and modeling are discussed throughout the chapter. in particular in Secs. 1.1. 1.3. 1.5 (population dynamics, etc.). and 1.6 (trajectories). Picard's existence and uniqueness theorems are explained in Sec. 1.7 (and Picard's iterati()l1 in Problem Set l.7). Numeric methods for first-order ODEs can be studied in Secs. 21.1 and 21.2 immediately after this chapter, as indicated in the chapter opening. ..... 2 CHAPTER • • -~~ I I :.. Second-Order Linear ODEs Ordinary differential equations (ODEs) may be divided into two large classes, linear ODEs and nonlinear ODEs. Whereas nonlinear ODEs of second (and higher) order generally are difficult to solve, linear ODEs are much simpler because various properties of their solutions can be characterized in a general way, and there are standard methods for solving many of these equations. Linear ODEs of the second order are the most important ones because of their applications in mechanical and electrical engineering (Secs. 2.4. 2.8. 2.9). And their theory is typical of that of all linear ODEs, but the formulas are simpler than for higher order equations. Also the transition to higher order (in Chap. 3) will be almost immediate. This chapter includes the derivation of general and particular solutions, the latter in connection with initial value problems. (Boundary value problems follow in Chap. 5, which also contains solution methods for Legendre's, Bessel's, and the hypergeometric equations.) COM M ENT. NllInerics for second-order ODEs can be studied immediately after this chapter. See Sec. 21.3, which is independent of other sections in Chaps. 19-21. Prerequisite: Chap. 1, in particular. Sec. IS Sections that may be omitted in a shorter course: 2.3. 2.9, 2.10. References and Answers to Problems: App. 1 Part A, and App. 2. 2.1 Homogeneous Linear ODEs of Second Order We have already considered first-order linear ODEs (Sec. 1.5) and shall now define and discuss linear ODEs of second order. These equations have important engineering applications, especiaUy in connection with mechanical and electrical vibrations (Secs. 2.4, 2.8, 2.9) as well as in wave motion, heat conduction, and other parts of physics, as we shall see in Chap. 12. A second-order ODE is called linear if it can be wriuen (1) y" + p(x)y' + q(x)y = r(x) and nonlinear if it cannot be written in this form. The distinctive feature of this equation is that it is linear in y alld its derivatives, whereas the functions p, q, and r on the right may be any given functions of x. If the equation begins with, say, f(x)y", then divide by f(x) to have the standard form (1) with y" as the first term, which is practical. 45 CHAP. 2 46 Second-Order Linear ODEs If rex) == 0 (that is, rex) reduces to = 0 for all x considered; read "r(x) is identically zero"), then ( I) (2) y" + p(x)y' + q(x)y = 0 and is called homogeneous. If rex) =/= 0, then (1) is called nonhomogeneous. This is similar to Sec. 1.5. For instance, a nonhomogeneous linear ODE is y" + 25y = e- x cos x, and a homogeneous linear ODE is xy" + y' + A}' = 0, .v" + -x'v' + Y = O. in standard form An example of a nonlinear ODE is y"y + /2 = O. The functions p and q in (l) and (2) are called the coefficients of the ODEs. Solutions are defined similarly as for first-order ODEs in Chap. 1. A function )' = Iz(x) is called a solution of a (linear or nonlinear) second-order ODE on some open interval I if h is defined and twice differentiable throughout that interval and is such that the ODE becomes an identity if we replace the unknown y by 11, the derivative y' by Iz', and the second derivative y" by It. Examples are given below. Homogeneous Linear ODEs: Superposition Principle Sections 2.1-2.6 will be devoted to homogeneous linear ODEs (2) and the remaining sections of the chapter to nonhomogeneous linear ODEs. Linear ODEs have a rich solution structure. For the homogeneous equation the backbone of this structure is the SUPCI7Jositio11 principle or lillearit), principle, which says that we can obtain further solutions from given ones by adding them or by multiplying them with any constants. Of course, this is a great advantage of homogeneous linear ODEs. Let us first discuss an example. E X AMP L E 1 Homogeneous Linear ODEs: Superposition of Solutions The functions y = cos x and y = sin x are solutions of the homogeneous linear ODE y" + Y = 0 for all x. We verify this by differentiation and ,ubstitution. We obtain (cos r)" = -cos x; hence y" +Y = (cos.r)" + cosx = -cos .. + cos x = O. Similarly for y = sin x (verify!). We can go an impollant step fUrlher. We multiply cos x by any constant. for instance. 4.7. and sin.\" by. say. -2. and take the sum of the results. claiming thm it is a solution. Indeed. differentiation and substitution gives (4.7 cos x - 2 sin r)" + (4.7 cos X - 2 sin x) = -4.7 cos X + 2 sin X + 4.7 cos I - 2 sinx = D. • SEC. 2.1 47 Homogeneous Linear ODEs of Second Order In this example we have obtained from)'1 (= cos x) and)'2 (= sin x) a function of the fonn (CI, C2 arbitrary constants). (3) This is called a linear combination of YI and .1'2' In terms of this concept we can now formulate the result suggested by our example. often called the superposition principle or linearity principle. THEOREM 1 Fundamental Theorem for the Homogeneous Linear ODE (2) For a homogeneous linear ODE (2), an)' linear combination of two solutions on an open interl'Ol I is again a solution of (2) 011 I. In pm1icular, for sucb an equation. sums and cOllSta1l117luitipies of solutions are again solutions. PROOF Let YI and Y2 be solutions of (2) on I. Then by substituting Y = CI)'I + C2Y2 and its derivatives into (2), and using the familiar rule (Cd'i + (2)'2)' = CIY~ + C2Y~' etc., we get )''' + py' + q)' = (CIYI + C2Y2)" + P(CIYI + C2Y2)' + q(CI)'I + = CI)'~ + C2."~ + P(CIY~ + C2Y~) + = CI()'~ + py~ + q(CI)'I + C2Y2) + C2()'~ + PY~ + qY2) = q)'I) C2Y2) 0, since in the last line, ( ... ) = 0 because )'1 and Y2 are solutions, by assumption. This that Y is a solution of (2) on I. show~ • CAUTION! Don't forget that this highly important theorem hold~ for homogeneo/ls linear ODEs only but does not hold for nonhomogeneous linear or nonlinear ODEs, as the following t~o examples illustrate. E X AMP L E 2 A Nonhomogeneous Linear ODE Verify by substitution that the functions y linear ODE ~ I + cos t and y y" but their sum E X AMP L E 3 j, +Y = = I + sin \. are solutions of the nonhomogeneou, I. not a solution. Neither is, for instance, 2( I + cos x) Or 5(1 + sin x). • A Nonlinear ODE Verify by sub~titution that the runctions y = x 2 and y = I are solutions of the nonlinear ODE y")" - xy' = O. but their sum is not a solution. Neither is -x2 , so you cannot even mUltiply by -I! • Initial Value Problem. Basis. General Solution Recall from Chap. I that for a first-order ODE, an initial value problem consists of the ODE and one initial condition y(xo) = Yo. The initial condition is used to determine the arbitrary constant c in the general solution of the ODE. This results in a unique solution, as we need it in most applications. That solution is called a particular solution of the ODE. These ideas extend to second-order equations as follows. CHAP. 2 48 Second-Order Linear ODEs For a second-order homogeneous linear ODE (2) an initial value problem consists of (2) and two initial conditions (4) These conditions prescribe given values Ko and Kl of the solution and its first derivative (the slope of its curve) at the same given x = Xo in the open interval considered. The conditions (4) are used to determine the two arbitrary constants CI and C2 in a general solution (5) of the ODE; here. )'1 and )'2 are suitable solutions of the ODE, with "suitable" to be explained after the next example. This results in a unique solution, passing through the point (xo, Ko) with KI as the tangent direction (the slope) at that point. That solution is called a particular solution of the ODE (2). E X AMP L E 4 Initial Value Problem Solve the initial value problem y" + Y = 0, ,,(0) = y' (0) 3.0, = -0.5. Solution. Step 1. General so/Iltioll. The Functions cos x and sin x are solutions of the ODE (by Example I), and we take y = cl cos x + c2 sinx. This will turn out to be a general solution as defined below. Step 2. ParticlI/ar SO/lItiOIi. We need the derivative y' = we obtain, since cos 0 = I and sin 0 = O. yeO) = cl = 3.0 -cl and sin x + y' (0) = C2 = c2 cos x. From this and the initial values -0.5. This gives as the solution of our initial value problem the particular solution y = 3.0 cos x - 0.5 sin x. Figure 28 shows that at x = 0 it ha, the value 3.0 and the slope -0.5, so that its tangent intersects the x-axis • at x = 3.010.5 = 6.0. (The scales on the axes differ!) y 3 2 0 -1 -2 -3 Fig. 28. ~ I V6'VJ' Particular solution and initial tangent in Example 4 Observation. Our choice of )'1 and )'2 was general enough to satisfy both initial conditions. Now let us take instead two proportional solutions )'1 = cos x and SEC. 2.1 49 Homogeneous Linear ODEs of Second Order Y2 = k cos x, so that y I /.\'2 = 11k = COilS/' Then we can write y = CIY! + C2Y2 in the form )' = c i cos X + c2(k cos x) = C cos x where Hence we are no longer able to satisfy two initial conditions with only one arbitrary constant C. Consequently, in defining the concept of a general solution, we must exclude proportionality. And we see at the same time why the concept of a general solution is of importance in connection with initial value problems. D E FIN I T ION I General Solution, Basis, Particular Solution A general solution of an ODE (2) on an open interval I is a solution (5) in which Yt and Y2 are solutions of (2) on I that are not proportional, and C I and C2 are arbitrary constants. These YI, Y2 are called a basis (or a fundamental system) of solutions of (2) on 1. A particular solution of (2) on I is obtained if we assign specific values to CI and C2 in (5). For the definition of an interval see Sec. 1.1. Also, C I and C2 must sometimes be restIicted to some interval in order to avoid complex expressions in the solution. Furthermore, as usual, Yl and )'2 are called proportional on I if for all x on I, (6) or (b) Y2 = l.vI where k and I are numbers, zero or not. (Note that (a) implies (b) if and only if k =1= 0). Actually, we can reformulate our definition of a basis by using a concept of general importance. Namely, two functions )'1 and Y2 are called linearly independent on an interval I where they are defined if everywhere on I implies (7) And Yl and Y2 are called linearly dependent on I if (7) also holds for some constants k I , k2 not both zero. Then if kl =t= 0 or k2 =1= 0, we can divide and see that YI and Y2 are proportional, or In contrast, in the case of linear independence these functions are not proportional because then we cannot divide in (7). This gives the following DEFINITION Basis (Reformulated) A basis of solutions of (2) on an open interval I is a pair of linearly independent solutions of (2) on I. If the coefficients p and q of (2) are continuous on some open interval I, then (2) has a general solution. It yields the unique solution of any initial value problem (2), (4). It CHAP. 2 ) Second-Order Linear ODEs includes all solutions of (2) on J; hence (2) has no singular solutions (solutions not obtainable from of a general solution; see also Problem Set 1.1). All this will be shown in Sec. 2.6. E X AMP L E 5 Basis, General Solution, Particular Solution and sin x in Example 4 form a basis of solutions of the ODE)"" + )" = 0 for all \" because their quotient is cot x COllst (or tan x COllSt). Hence y = Cl cos x + c2 sin x is a general solution. The solution )" = 3.0 cos x - 0.5 sin x of the initial value problem is a particular solution. • COS \" E X AMP L E 6 "* "* Basis, General Solution, Particular Solution Verify by substitution that)"1 = eX and)"2 = e -x are solutions of the ODE ,," - y = O. Then solve the initial value problem y" - y = 0, y(O) = 6, y'(O) = -2. (ex)" - eX = 0 and (e- x )" - e- x = 0 shows that eX and e- x are solutions. They are not proportional. eXle- x = e 2x COllst. Hence eX, e- x form a basis for aUx. We now write down the corresponding general solution and its derivative and equate their values at 0 to the given initial conditions, Solution. "* = cl + c2 = 6. \'(0) By addition and subtraction. cl = 2. c2 = 4, so that the allswer is y = 2e x + 4e -.<. This is the particular solution satisfying thc two initial conditions. • Find a Basis if One Solution Is Known. Reduction of Order It happens quite often that one solution can be found by inspection or in some other way. Then a second linearly independent solution can be obtained by solving a first-order ODE. This is called the method of reduction of order. l We first show this method for an example and then in general. E X AMP - 7 Reduction of Order if a Solution Is Known. Basis Find a basis of solutions of the ODE (x 2 - X)y" - xy' + Y = O. yr Inspection shows that Yt = x is a solution because )'~ = 1 and = O. so that the first term vanbhes identically and the second and third terms cancel. The idea of the method is to substitute Solution. y = ll)"l = llX, y' = U'( + ,," = tt"X Lt. + 2,,' into the ODE. This gives (x ux and -XII 2 - x)(u"x + 2/1') - r(u'x + ll) + U.1: = O. cancel and we are left with the following ODE. which we divide by x. order. and simplify, {X 2 - xlu" + (x - 2)11' = O. ICredited to the great mathematician JOSEPH LOUIS LAGRANGE (1736-1813). who was born in Turin. of French extraction. got his first professorship when he was 19 (al the Military Academy of Turin). became director of the mathematical section of the Berlin Academy in 1766. and moved to Paris in 1787. His important major work was in the calculus of variations. celestial mechanics, general mechanics (Mecallique a/Ja/ytique, Paris, 1788), differential equations, approximation theory, algebra, and number theory. SEC. 2.1 51 Homogeneous Linear ODEs of Second Order This ODE is of first order in v gives dv x - 2 = - v -2-- x - x = dx = u', namely, (x 2 ( 1 2) ---~ x x-I x) v' - dx + (x - 2) v In ' Ivl = = O. Separation of variables and integration In Ix - II - 2 In Ix - 11 Ixl = In -~2~ x We need no con,tant of integration because we want to obtain a particular solution: similarly in the next integration. Taking exponents and integrating again, we obtain v x-I = --2- = x ~ ""2. - x x u = Iv dx = Ixl + ~x . In hence .1'2 = lIX = x In Ixl + I. Since Y1 = x and Y2 = x In Ixl + 1 are linearly independent (their quotient is not constant), we have obtained a basis of solutions. valid for all positive x. • In this example we applied reduction of order to a homogeneous linear ODE [see (2)] y" + p(x)y' + q(x)y = O. Note that we now take the ODE in standard form, with y", not f(x)y"-this is essential in applying our subsequent formulas. We assume a solution .VI of (2) on an open interval I to be known and want to find a basis. For this we need a second linearly independent solution .\"2 of (2) on 1. To get Y2, we substitute y = Y2 = y' = UY1, y~ = U'YI + uy~, into (2). This gives (8) Collecting terms in u", u', and u, we have Now comes the main point. Since )'1 is a solution of (2), the expression in the last parentheses is zero. Hence u is gone, and we are left with an ODE in u' and u". We divide this remaining ODE by )'1 and set u' = U, u" = V', U" + u' 2y~ + PYI = 0, , V + thus Y1 (2V~ -'- + p )V = O. Y1 This is the desired first-order ODE, the reduced ODE. Separation of variables and integration gives dV V -=~ (2)'; ) - + p dx and .v 1 In By taking exponents we finally obtain (9) V = - 1 Yl 2 e -fpdx. Ivi 52 CHAP. 2 Second-Order Linear ODEs Here U = u', so that u = IV dx. Hence the desired second solution is Y2 The quotient )'21Yl = u = a basis of solutions . _... .•. -........ . ........ ,--.... 11-61 = Yl JV dx. -. ..,. GENERAL SOLUTION. INITIAL VALUE PROBLEM 2),' + 2y = 0, e- x cos x, e- x sin x, yeO) = I. /(0) = -1 4. y" - 6y' + 9y = 0, e 3x , xe 3x , -,,(0) = -1.4, y' (0) = 4.6 5. x 2y" + .n-' - 4v = 0, x 2, x- 2, v(l) = II, y'(1)=-=-6 . . 6. Yl 11 IV dx cannot be constant (since v> 0), so that Yl and Y2 fonn (More in the next problem set.) Verify by substitution that the given functions fonn a basis. Solve the given initial value problem. (Show the details of your work.) 1. y" - 16y = 0, e 4x , e- 4x yeO) = 3, /(0) = 8 2. y" + 25y = O. cos 5x. sin 5x. yeO) = 0.8, ),'(0) = -6.5 3. y" = + x\" - 7rr' + 15)' = 0, x3 , x5 , y(l) = O.~. /(1) = 1.0 [7-141 LINEAR INDEPENDENCE AND DEPENDENCE Are the following functions linearly independent on the given interval? 7. x, x In x (0 < r < 10) 8. 3x 2 , 2x n (0 < X < I) 9. e ax , e- ax (any interval) 10. cos 2 x, sin2 \. (any interval) 11. In x, In x 2 (x > 0) 12. x - 2, x + 2 (-2 < x < 2) 13. 5 sin x co~ x. 3 sin 2x (x > 0) 14. 0, sinh TTX (x > 0) REDUCTION OF ORDER is important because it gives a simpler ODE. A second-order ODE F(x, y, y', y") = 0, linear or not, can be reduced to first order if y does not occur explicitly (Prob. 15) or if x does not occur explicitly (Prob. 16) or if the ODE is homogeneoll~ linear and we know a solution (see the text). 15. (Reduction) Show that F(x, y' , y") = 0 can be reduced to first order in ;: = y' \from which y follows by integration). Give two examples of your own. 16. (Reduction) Show that F(y, y'. y") = 0 can be reduced to a first-order ODE with y as the independent variable and y" = (d;:idr):. where z = y'; derive this by the chain rule. Give two examples. [17-221 Reduce to first order and solve (showing each step in detail). 17. y" = ky' 18. / ' = I + /2 19. -"y" = 4y'2 20. xy" 21. -,," + 2y' + xy + /3 siny = = 0, YI = X-I cos x 0 22. (I - x 2 )y" - 2xy' + 2.\' = 0, )'1 = X 23. (Motion) A small body moves on a straight line. Its velocity equals twice the reciprocal of its acceleration. If at t = 0 the body has distance I m from the origin and velocity 2 m/sec, what are its distance and velocity after 3 sec? 24. (Hanging cable) It can be shown that the curve y(x) of an inextensible flexible homogeneous cable hanging between two fixed points is obtained by solving y" = k~, where thc constant k depends on the weight. This curve is called a catellary (from Latin catella = the chain). Find and graph y(x). assuming k = I and those fixed points are (-1,0) and (1, 0) in a vertical .l}'-plane. 25. (Curves) Find and sketch or graph the curves passing through the origin with slope I for which the second derivative is proportional to the first. 26. WRITING PROJECT. General Properties of Solutions of Linear ODEs. Write a short essay (with proofs and simple examples of your own) that includes the following. (a) The superposition principle. (b) y;: 0 is a solmion of the homogeneous equarion (2) (called the trivial solution). (cl The sum y = YI + )'2 of a solution Y2 of (2) is a solmion of (1). )'1 of (1) and (d) Explore possibilities of making further general statements on solutions of (1) and (2) (sums. differences, multiples). 27. CAS PROJECT. Linear Independence. Write a program for testing linear independence and dependence. Try it out on some of the problems in this problem set and on examples of your own. SEC. 2.2 2.2 51 Homogeneous Linear ODEs with Constant Coefficients Homogeneous Linear ODEs with Constant Coefficients We shall now consider second-order homogeneous linear ODEs whose coefficients a and b are constant, (1) y" + ay' + by = O. These equations have imp0l1ant applications, especially in connection with mechanical and electrical vibrations, as we shall see in Secs. 2.4, 2.8, and 2.9. How to solve (I)? We remember from Sec. 1.5 that the solution of the first-order linear ODE with a constant coefficient k y'+ky=O is an exponential function y function = ce- kx• This gives us the idea to try as a solution of (1) the (2) Substituting (2) and its derivatives and into our equation (1), we obtain (A2 + aA + b)eAX = o. Hence if A is a solution of the important characteristic equation (or auxiliary equation) (3) A2 + aA + b = 0 then the exponential function (2) is a solution of the ODE (1). Now from elementary algebra we recall that the roots of this quadratic equation (3) are (4) A1 = I(-a 2 + Va 2 - 4b) ' (3) and (4) will be basic because our derivation shows that the functions and (5) are solutions of (1). Verify this by substituting (5) into (1). From algebra we f1l11her know that the quadratic equation (3) may have three kinds of roots, depending on the sign of the discriminant a 2 - 4b, namely, (Case I) (Case II) (Case III) Two real roots if a 2 - 4b > 0, A real double roOT if a 2 - 4b = 0, Complex conjugate roots if a 2 - 4b < O. 54 CHAP. 2 Second-Order Linear ODEs Case I. Two Distinct Real Roots "-1 and "-2 In this case, a basis of solutions of (I) on any interval is and because )'1 and )'2 are defined (and real) for all x and their quotient is not constant. The corresponding general solution is (6) E X AMP L E 1 General Solution in the Case of Distinct Real Roots We can now solve -,," - y = 0 in Example 6 of Sec. 2.1 systematically. The characteristic equation is A2 - 1 = O. Its roob are A1 = I and A2 = - I. Hence a basis of solutions is e'" and e -x and gi ves the same general solution as before, • E X AMP L E 2 Initial Value Problem in the Case of Distinct Real Roots Sol ve the initial value problem /' + y' - 2)' Solution. = 0, y'(O) = -5. )'(0) = 4, Step 1. Gel1eral SOlllliol1. The characteristic equation is A2 + A - 2 = O. Its roots are and "-2 =l(-I - V9) =-2 so that we obtain the general solution Slep 2. Particlilar SOllllioll. Since /(x) conditions = cle x - 2c2e-2x. we obtain from the general solution and the initial y(O) y' (0) = Cj = cl + c2 = 4, 2C2 = -5. Hence c 1 = I and c2 = 3. This gives the lI11SII'e1' Y = eX + 3e -2x. Figure 29 shows that the curve begins ar y = 4 with a negative slope (-5, but note that the axes have different scales!), in agreement with the initial conditions, • y 8 :~ 2 °O~~O~,~5---7---1~.~5---=2---x Fig. 29. Solution in Example 2 SEC. 2.2 55 Homogeneous Linear ODEs with Constant Coefficients Case II. Real Double Root A = - 0/2 If the discriminant a 2 - 4b is zero, we see directly from (4) that we get only one root, A = Al = 11.2 = -al2. hence only one solution. To obtain a second independent solution )'2 (needed for a basis), we use the method of reduction of order discussed in the last section, setting Y2 = UY1' Substituting this and its derivatives y~ = U')'l + uy~ and Y~ into (1), we first have (/1'\1 + 2u' Y~ + uy~) + a(u' Y1 + It:r~) + bUYl = o. Collecting terms in u", u', and u, as in the last section, we obtain The expression in the last parentheses is zero, since .\'1 is a solution of (1). The expression in the first parentheses is zero. too. since We are thus left with U"Y1 = O. Hence u" = O. By two integrations, u = C1X + C 2 . To get a second independent solution Y2 = UY1, we can simply choose C1 = 1, C2 = 0 and take u = x. Then Y2 = .lYl' Since these solutions are not proportional, they form a basis. Hence in the case of a double root of (3) a basis of solutions of (l) on any interval is The corresponding general solution is Warning. of (l). E X AMP L E 1 If A is a simple root of (4), then (C1 + c2x)eAcC with C2 * 0 is not a solution General Solution in the Case of a Double Root The characteristic equalion of the ODE yo" + 6y' + 9y = 0 is A2 + 6A + 9 = (A + 3)2 = O. It has the double root A = -3. Hence a basis is e- 3", and xe- 3x . The conesponding general solution is y = (cl + c2x)e-3x. • E X AMP L E 4 Initial Value Problem in the Case of a Double Root Solve the initial value problem /' + y' + 0.25y = 0, yeo) = 3.0, Solution. The characteristic equation is A2 + A + 0.25 = (A This gives the general solution We need its derivative y' (0) = - + 0.5)2 = 3.5. O. It has the double root A = -0.5. 56 CHAP. 2 Second-Order Linear ODEs From this and the initial yeO) = condition~ we obtain 3.0. Cl = /(0) = 0.5cl = -3.5: C2 - hence C2 = -2. • The particular solution of the initial value problem is " = (3 - 2x)e -O.5x. See Fig. 30. y 3 2 \ o 1--"-;:-:zL____ - -.L_-~-L--.L8-=="----'----'L.-4 -1 Solution in Example 4 Fig. 30. Case III. Complex Roots -10 + iwand -10 - iw This case occurs if the discriminant a 2 - 4b of the characteristic equation (3) is negative. In this case, the roots of (3) and thus the solutions of the ODE (I) come at first out complex. However, we show that from them we can obtain a basis of real solutions Yl = e- ax/ 2 cos wx. (8) Y2 = e- ax/ 2 sin wx (w> 0) !a 2 . It can be verified by substitution that these are solutions in the where ~ = b present case. We shall derive them systematically after the two examples by using the complex exponential function. They form a basis on any interval since their quotient cot wx is not constant. Hence a real general solution in Case II] is (9) E X AMP L E 5 = y e- ax/ 2 (A cos wx + B sin wx) (A, B arbitrary). Complex Roots. Initial Value Problem Solve the initial value problem y" + DAy' + 9.04y = D. yeO) = Solution. o. Step 1. General SOllltioli. The characteristic equation is -0.2 ± 3i. Hence w = 3. and a general solution (9) is )' = y' (0) }..2 = + 0.4}" + 9.04 = O. It has the roots e- o.2x(A co~ 3x + B sin 3x). Step 2. Pm1iclIiar soilltioll. The first initial condition gives y(O) = A O 2x )' = Be- . sin 3x. We need the derivative (chain rule!) y' = 3. = O. The remaining expression is B(-O.2e- O. 2 1: sin 3x + 3e-O.2x cos 3x). From this and the second imtial condition we obtain y' (0) = 3B = 3. Hence B = I. OUf solution is y = e -O.2x sin 3x. Figure 31 shows \' and the curves of e- O .2x and _e- O.2x (dashed), between which the curve of)' oscillates. Such "damped vibrations" (with x = t being time) have important mechanical and electrical applications. as we shall soon see (in Sec. 2.4). • SEC. 2.2 57 Homogeneous Linear ODEs with Constant Coefficients y 1.0 IT" 0.5" I n', , /1 ~ I\'~_ o 15·· 20 30 25 x -0.5 -1.0 Fig. 31. E X AMP L E 6 Solution in Example 5 Complex Roots A general solution of the ODE (lU constant, not zero) is y = A cos With lU ld, + B sin lUX. • = 1 this confirms Example 4 in Sec. 2.1. Summary of Cases I-III [ Case Basis of (1) I Distinct real A10 A2 II Real double root A = -~a e- a;</2, III Complex conjugate Al = -~a + iw, A2 = -~a - iw e -a~'/2 e- ax/ 2 I L I i Roots of (2) eA1X, eA2X xe-a~'/2 General Solution of (1) y = y = CIe'''l~' (Cl + + C2C"'2X c2x )e- ax / 2 - [ cos wx sin wx y = e- a :li2 (A cos wx + B sin wx) It is very interesting that in applications to mechanical systems or electrical circuits, these three cases correspond to three different forms of motion or flows of current, respectively. We shall discuss this basic relation between theory and practice in detail in Sec. 2.4 (and again in Sec. 2.8). Derivation in Case III. Complex Exponential Function If verification of the solutions in (8) satisfies you, skip the systematic derivation of these real solutions from the complex solutions by means of the complex exponential function e2 of a complex variable z = r + it. We write r + it, not x + iy because x and y occur in the ODE. The definition of e Z in terms of the real functions eT , cos t, and sin t is (10) 58 CHAP. 2 Second-Order Linear ODEs This is motivated as follows. For real z = r, hence t = 0, cos 0 = I. sin 0 = 0, we get the real exponential function eT • It can be shown that eZ , +Z2 = eZ1 eZ2 , just as in real. (Proof in Sec. 13.5.) Finally, if we use the Maclaurin series of e Z with z = it as well as ;2 = -1, ;3 = -;, i4 = 1, etc., and reorder the terms as shown (this is permissible, as can he proved), we obtain the series (it)2 (it)3 (it)4 (it)5 2! 3! 4! 5! + it + - - + - - + - - + - - + ... = ;~ I - + :~ - + ... + i (t - ~~ + ~~ - + ... ) = cos t + i sin t. (Look up these real series in your calculus book if necessary.) We see that we have obtained the fonnula eit = cos t + i sin t, (11) called the Euler fonnula. Multiplication by eT gives (10). For later use we note that e- it = cos (-t) + i sin (-t) addition and subtraction of this and (II), (12) cos t 1. (e,t = "2 + . e-'t), sin t = cos t 1. =- 2i - i sin t. so that by . (e't - e- tt ), After these comments on the definition (10), let us now turn to Ca~e Ill. In Case III the radicand {/2 - 4b in (4) is negative. Hence 4b - {/2 is positive and, using -v=T = i, we obtain in (4) with w defined as in (8), Hence in (4), A} = ~a Using (10) with r = + iw -~ax and t e~'X = and. similarly, = A2 = ~a - iw. wx, we thus obtain e- CaI2 )x+iwx = e-taI2h(cos wx + i sin wx) We now add these two lines and multiply the result by ~. This gives .\', as in (8). Then we subtract the second line from the first and multiply the result by I1(2i). This gives .\'2 as in (8). These results obtained by addition and multiplication by constants are again solutions, as follows from the superposition principle in Sec. 2.1. This concludes the derivation of these real solutions in Case 111. SEC. 2.3 Differential Operators. Optional 59 - .... - =---I_l~ GENERAL SOLUTION Find a general solution. Check your answer by substitution. 1. y" - 6 .. ' - 7)' = 0 2. lOy" - 7/ + 1.2y = 0 3. 4y" - 20y' + 2Sy = 0 4. y" + + 417y' 417 2 y = 0 7. y" 8. y" 10. y" - 2.1' + 13. y" - 12. y" = 0 144y = 0 30. y" 14. y" 19. + hy + 2.4y' + 4.0)' = + y' - 0.96y = 0 0 0 for the given basis. 16. e O. 5x , e- 3 . 5x 18. 1, e- 3x e<-l+i)X, e-(1+i)x INITIAL VALUE PROBLEMS Solve the initial value problem. Check that your answer satisfies the ODE as well as the initial conditions. (Show the details of your work.) 21. y" - 2y' - 3y = 0, yeO) = 2, /(0) = 14 y" y" + + lOy" y" + lOy" 2.3 = 2Sy = O. y(O) = = O. y' (0) = 40 0, yeO) = 0, y' (0) = 20 l.S S.6)' = 0, y(O) = 4, / (0) = - 3.8 35. (Verification) Show by substitution that.h in (8) is a solution of (1 ). 0, yeO) + l8y' + 0 (D) Limits. Double roots should be limiting cases of distinct roots AI, A2 as, say, A2 ~ AI' Experiment with this idea. (Remember I'H6pital's rule from calculus.) Can you aITive at xe A1X ? Give it a try. + Y = 0, yeO) = 4, /(0) = -6 4/ + S)" = 0, yeO) = 2, /(0) = -S - SOy' + 6Sy = 0, yeO) = l.S, y' (0) = 2y' 17y' -2, /(0) = -12 + Y = 0, yeO) = 3.2, y' (0) = + (k 2 + w 2 )y = 0, yeO) = 1, (A) Coefficient fonnulas. Show how a and b in (I) can be expressed in tenns of Al and A2' Explain how these formulas can be used in constructing equations for given bases. (B) Root zero. Solve y" + 4y' = 0 (i) by the present method, and (ii) by reduction to first order. Can you explain why the result must be the same in both cases? Can you do the same for a general ODE y" + aJ' = O? (C) Double root. Verify directly that xe AX with A = -a/2 is a solution of (1) in the case of a double root. Verify and explain why y = e- 2x is a solution of y" - v' - 6y = 0 but xe- 2x is not. 1_ 1-321 22. 23. 24. 25. 26. + -4.S 34. TEAM PROJECT. General Properties of Solutions = 20. e 4 ", e- 4 ." = 0, yeO) = 4y' y' (0) = 2ky' y'(O) = -k ['5-201 FIND ODE Find an ODE)''' + ay' 15. e 2x , eX 17. e Xv3 , xe x v'3 + 0, yeO) = 2, 33. (Instability) Solve y" - y = 0 for the initial conditions y(O) = 1, y' (0) = -1. Then change the initial conditions to yeO) = 1.001, y' (0) = -0.999 and explain why this small change of 0.001 at x = 0 causes a large change later, e.g., 22 at x = 10. 0 = 917 2 y 29. 20y" 32. y" - 2y' - 24y 9. y" - 2y' - S.2S}' = 0 11. y" + s / + 0.62S)' = 28. y" - 9y 31. y" + 20y' - 99," = 0 + 2),' + S)' = 0 - y' + 2.Sy = 0 + 2.6/ + 1.69)' = 0 5. 100y" 6. y" 27. lOy" = 3, y'(o) = -17 Optional Differential Operators. This short section can be omitted without interrupting the flow of ideas; it will not be used in the sequel (except for the notations D.\', D2y, etc., for y', y", etc.). Operational calculus means the technique and application of operators. Here, an operator is a transformation that transforms a function into another function. Hence differential calculus involves an operator, the differential operator D, which transforms a (differentiable) function into its derivative. In operator notation we write (1) Dy = y' dy dx CHAP. 2 60 Second-Order Linear ODEs Similarly, for the higher derivatives we write D2y = D(Dy) = / ' , and so on. For example, D sin = cos, D2 sin = -sin, etc. For a homogeneous linear ODE y" + ay' + by = 0 with constant coefficients we can now introduce the second-order differential operator L = P(D) = D2 + + aD bl. where I is the identity operator defined by Iy = y. Then we can write that ODE as Ly = PW)y = (D 2 (2) + aD + b/)y = O. P suggests "polynomial." L is a linear operator. By definition this means that if Ly and Lw exist (this is the case if y and ware twice differentiable), then L(e)' + kw) exists for any constants e and k, and L(e}' + kw) = + eLy kLw. Let us show that from (2) we reach agreement with the results in Sec. 2.2. Since (DeA)(.x) = Ae Ax and (D 2e A)(x) = )-..2e A"', we obtain LeA(x) = PW)eA(x) = (D 2 + aD + bf)eA(x) (3) = (11. 2 + all. + b)eA"C = P(A)eA"C = O. This confirms our result of Sec. 2.2 that e AX is a solution of the ODE (2) if and only ~f A is a solution of the characteristic equation peA) = O. peA) is a polynomial in the usual sense of algebra. If we replace A by the operator D, we obtain the "operator polynomial"' P(D). The point of this operatiollal calculus is that P(D) call be treated just like all algebraic quantity. In particular. we can factor it. E X AMP L E 1 Factorization, Solution of an ODE Factor P(Dl = if - 3D - 401 and solve P{D»)" = O. D2 - 3D - 401 = (D - 8l)(D + 51) because P = I. Now (D - 8l)y = y' - 8y = 0 has the solution)'1 = e 8x. Similarly. the solution of (D + 5I)y = 0 is)'2 = e -5x. This is a basis of P(D)y = 0 on any interval. From the factorization we obtain the ODE, as expected. Solutioll. (D - 8[)(D + 5l)y = (D - 8[)(y' + 5y) = D(y' + = y" 5y) - 8(/ + 5.\') + 5/ - 8/ - 40,· = y" - 3y' - 40y = O. Verify that this agrees with the result ot our method in Sec. 2.2. This is not unexpected because we factored P(D) in the same way as the characteristic polynomial PIA) = A2 - 3A - 40. • It was essential that L in (2) has constant coefficients. Extension of operator methods to variable-coefficient ODEs is more difficult and will not be considered here. If operational methods were limited to the simple situations illustrated in this section, it would perhaps not be worth mentioning. Actually, the power of the operator approach appears in more complicated engineering problems, as we shall see in Chap. 6. SEC. 2.4 Modeling: Free Oscillations (Mass-Spring System) --_.... -.-- _... ....-............ -... - ~ ----. -~~ 12. (4D 2 13. (D 2 Apply the given operator to the given functions. (Show all steps in detail.) 1. (D - 1)2~ eX.. xe x .. sin x + cosh - I, ix, sinh eO. 4 ."", e- 5x sin x. 5. (D - 4I)(D + 3l); x 3 - x 2, 4. (D [ ~ 5/)(D -]); ix, ex / e 5x • x 2 e- 3x GENERAL SOLUTION Factor as in the text and ~olve. (Show the details.) 6. (D 2 - 5.5D + 6.66l)y = 0 7. (D + 2l)2y = 0 8. (D 2 - 0.49])y 2 9. (D + 6D + 131))' = 0 10. (lOD 2 + 2D + = 0 0 14. (Double root) If D2 + aD + hI has distinct roots JL and A, show that a particular solution is y = (elL" - eAX)/{JL - A). Obtain from this a solution xeAx by letting JL ~ A and applying I"H6pital's rule. 15. (Linear operator) Illustrate the linearity of L in (2) by taking e = 4, k = -6, y = e 2X , and 11' = cos 2x. Prove that L is linear. 16. (Definition of linearity) Show that the definitIOn of linearity in the text is equivalent to the following. If Lly] and L[w] exist. then L[y + w] exists and L[e)'J and L[kw] exist for all constants e and k, and Ll\" + w] = Lb'] + L[w] as well as L[ey] = eLl\"l and 2 xeO. 4x sin 4x, + 4.1D + 3.II)y = + 47TD + 7T2l)y = + 17.64w2l)y = 0 11. (D 2 APPLICATION OF DIFFERENTIAL OPERATORS 2. 8D 2 + 2D - I; 3. D - 0.41; 2x 3 61 0 L[kw] = kL[w]. 1.7/))' = 0 2.4 Modeling: Free Oscillations (Mass-Spring System) Linear ODEs with constant coefficients have important applications in mechanics, as we show now (and in Sec. 2.8), and in electric circuits (to be shown in Sec. 2.9). In this section we consider a basic mechanical system, a mass on an elastic spring ("mass-spring system," Fig. 32). which moves up and down. Its model will be a homogeneous linear ODE. Setting Up the Model We take an ordinary spring that resists compression as well extension and suspend it vertically from a fixed support, as shown in Fig. 32. At the lower end of the spring we --1----1~--- U nstretched spring - - -(Y=O)l---~ System In static equilibrium (a) Fig. 32. I (b) Y ---- System in motion (c) Mechanical mass-spring system 62 CHAP. 2 Second-Order Linear ODEs attach a body of mass 111. We assume /11 to be so large that we can neglect the mass of the spring. If we pull the body down a certain distance and then release it, it starts moving. We assume that it moves strictly vertically. How can we obtain the motion of the body, say, the displacement y(t) as function of time t? Now this motion is detelmined by Newton's second law (1) Mass X Acceleration = nH''' = Force where y" = d 2y/dr2 and "Force" is the resultant of all the forces acting on the body. (For systems of units and conversion factors, see the inside of the front cover.) We choose the dowllward directioll as the positive direction, thus regarding downward forces as positive and upward forces as negative. Consider Fig. 32. The spring is first unstretched. We now attach the body. This stretches the spring by an amount So shown in the figure. It causes an upward force Fo in the spring. Experiments show that Fo is proportional to the stretch So' say, (Hooke's law2 ). (2) k (> 0) is called the spring constant (or spring modulus). The minus sign indicates that Fo points upward, in our negative direction. Stiff springs have large k. (Explain!) The extension So is such that Fo in the spring balances the weight W = mg of the body (where g = 980 cm/sec 2 = 32.17 ftlsec 2 is the gravitational constant). Hence F0 + W = - kso + 17lg = O. These forces will not affect the motion. Spring and body are again at rest. This is called the static eqUilibrium of the system (Fig. 32b). We measure the displacement yet) of the body from this 'equilibrium point' as the origin y = 0, downward positive and upward negative. From the position y = 0 we pull the body downward. This further stretches the spring by some amount y > 0 (the distance we pull it down). By Hooke's law this causes an (additional) upward force FI in the spring, FI = -kyo F 1 is a restoring force. It has the tendency to restore the system, that is, to pull the body back to y = O. Undamped System: ODE and Solution Every system has damping--otherwise it would keep moving forever. But practically, the effect of damping may often be negligible, for example, for the motion of an iron ball on a spring during a few minutes. Then F 1 is the only force in (I) causing the motion. Hence (1) gives the model 111Y" = -kyor (3) my" + ky = O. 2ROBERT HOOKE (1635-1703), English physicist, a forerunner of Newton with respect to the law of gravitation. SEC 2.4 Modeling: Free Oscillations (Mass-Spring System) 63 By the method in Sec. 2.2 (see Example 6) we obtain as a general solution (4) y(t) = A cos Wot +B sin wof. The corresponding motion is called a harmonic oscillation. Since the trigonometric functions in (4) have the period 27T/WO' the body executes wo!27T cycles per second. This is the frequency of the oscillation, which is also called the natural frequency of the system. It is measured in cycles per second. Another name for cycles/sec is hertz (Hz).3 2 The sum in (4) can be combined into a phase-shifted cosine with amplitude C = + B2 and phase angle 8 = arctan (B/A), VA (4*) y(t) = C cos (wof - 8). To verify this, apply the addition formula for the cosine [(6) in App. 3.1] to (4*) and then compare with (4). Equation (4) is simpler in connection with initial value problems, whereas (4*) is physically more informative because it exhibits the amplitude and phase of the oscillation. Figure 33 shows typical forms of (4) and (4*), all corresponding to some positive initial displacement .\'(0) [which determines A = y(O) in (4)] and different initial velocities.r' (0) [which determine B = y' (O)/wol 0 / " / \ @/ /' /' --- /' /' / / / / /' / / / / ,/ CD Positive } @Zero Initial velocity ® Negative Fig. 33. E X AMP L E 1 Harmonic oscillations Undamped Motion. Harmonic Oscillation If an iron ball of weight W = 98 nt (about 22 Ib) stretches a spring 1.09 m (about 43 in.), how many cycles per minme will this mass-spring system execute? What will its motion be if we pull down the weight an additional 16 cm (abom 6 in.) and let it start with zero initial velocity? Solutio1l. Hooke's law (2) with W a~ the force and 1.09 meter as the stretch gives W = 1.09k: thus 2 k = WII.09 = 98/1.09 = 90 [kg/sec j = 90 [nt/meter]. The mass h III = WIg = 98/9.8 = 10 [kg]. This gives the frequency wo/(27i) = v klml(27T) = 3/(27T) = 0.48 [Hz] = 29 [cycles/min]. 3HEINRlCH HERTZ (1857-1894). German physicist. who discovered electromagnetic waves. as the basis of wireless communication developed by GUGLIELMO MARCONI (1874-1937), Italian physicist (Nobel prize in 1909). 64 CHAP. 2 Second-Order Linear ODEs From (4) and the initial conditions, y(O) = A = 0.16 [meter] y(t) = 0.16 cos 31 [meter] and y' (0) = woB = O. Hence the motion is 0.52 cos 31 [ft] or (Fig. 34). If you have a chance of experimenting with a mass-spring system. don't miss it. You will be surprised about the good agreement between theory and experiment. usuall) within a fraction of one percent if you measure carefully. • y 0.2 0.1 0~--,-L-~~L-~--L-4-~L--r--~--- -0.1 -0.2 Fig. 34. Harmonic oscillation in Example 1 Damped System: ODE and Solutions We now add a damping force F2 to our model (5) = -cy' my" = -ky, so that we have my" = -ky - cy' or my" + cy' + Ie)' = O. Physically this can be done by connecting the body to a dashpot; see Fig. 35. We assume this new force to be proportional to the velocity y' = dyldt, as shown. This is generally a good approximation, at least for small velocities. c is called the damping constant. We show that c is positive. If at some instant, y' is positive. the body is moving downward (which is the positive direction). Hence the damping force F2 = -cy'. always acting against the direction of motion. must be an upward force. which means that it must be negative, F2 = -cy' < 0, so that -c < 0 and c > O. For an upward motion, y' < 0 and we have a downward F2 = -cy > 0; hence -c < 0 and c > 0, as before. The ODE (5) is homogeneous linear and has constant coefficients. Hence we can solve it by the method in Sec. 2.2. The characteristic equation is (divide (5) by m) A2 c A m + -- Fig. 35. + k = O. 111 Damped system 65 SEC. 2.4 Modeling: Free Oscillations (Mass-Spring System) By the usual fonnula for the roots of a quadratic equation we obtain, as in Sec. 2.2, (6) A2 = -a - {3, where a =- C 2m and {3 = _1_ 2m Vc 2 - 4mk. It is now most interesting that depending on the amount of damping (much, medium, or linle) there will be three types of motion cOlTesponding to the three Cases I, II, n in Sec. 2.2: Case [. Case II. Case [II. c 2 > 411lk. c 2 = 4mk. c 2 < 4mk. Distinct real roots AI, A2 . A real double root. Complex conjugate roots. (Overdamping) (Critical damping) (Underdamping) Discussion of the Three Cases Case I. Overdamping If the damping constant c is so large that c 2 > 4mk, then Al and A2 are distinct real roots. In this case the cOlTesponding general solution of (5) is (7) We see that in this case, damping takes out energy so quickly that the body does not oscillate. For t > 0 both exponents in (7) are negative because a > 0, {3 > 0, and 132 = if - kim < if. Hence both terms in (7) approach zero as t ~ 00. Practically speaking, after a sufficiently long time the mass will be at rest at the static equilibrium position (y = 0). Figure 36 shows (7) for some typical initial conditions. Case II. Critical Damping Critical damping is the border case between nonoscillatory motions (Case I) and oscillations (Case III). It occurs if the characteristic equation has a double root, that is, if c 2 = 4mk, y y \ (b) (aJ CD Positive ® Zero } Initial velocity @Negative Fig. 36. Typical motions (7) in the overdamped case (a) Positive initial displacement (b) Negative initial displacement 66 CHAP. 2 Second-Order Linear ODEs so that {3 = 0, Al = A2 = -a. Then the corresponding general solution of (5) i" (8) This solution can pass through the equilibrium position y = 0 at most once because e- at is never zero and Cl + C2t can have at most one positive zero. If both Cl and C2 are positive (or both negative), it has no positive zero, so that y does not pass through 0 at all. Figure 37 shows typical forms of (8). Note that they look almost like those in the previous figure. Case III. Underdamping This is the most interesting case. It occurs if the damping constant c 2 < 4mk. Then {3 in (6) is no longer real but pure imaginary, say, (9) {3 = where iw* w* = _12m V4111k - C is so small that k c2 = (> 0). 171 (We write w* to reserve w for driving and electromotive forces in Secs. 2.8 and 2.9.) The roots of the characteristic equation are now complex conjugate. Al = -a + iw*, A2 = -a - iw* with a = C/(2m), as given in (6). Hence the corresponding general solution is J(t) (10) = e-at(A cos w*t + B sin w*t) = Ce- at cos (w*t - 8) where C2 = A2 + B2 and tan 8 = B/A, as in (4*). This represents damped oscillations. Their curve lies between the dashed curves y = Ce- at and y = -Ce- al in Fig. 38, touching them when w*t - 8 is an integer multiple of 7T because these are the points at which cos (w*t - 8) equals lor-I. The frequency is W*/(27T) Hz (hertz, cycles/sec). From (9) we see that the smaller c (> 0) is, the larger is w* and the more rapid the oscillations become. If c approaches 0, then w* approaches Wo = ~. giving the harmonic oscillation (4), whose frequency WO/(27T) is the natural frequency of the system. y y , \ .......................... ce-at /'''"---- --::------- / CD Positive ® Zero ...-"';"- -- -at -Ce } I nitial velocity @Negative ig. 37. ..,..-- Critical damping [see (8)] Fig. 38. Damped oscillation in Case /1/ [see (10)] SEC 2.4 67 Modeling: Free Oscillations (Mass-Spring System) EX AMP L E 2 The Three Cases of Damped Motion How doe~ the motion in Example 1 change if we change the damping constant c to one of the following three values. with y(O) = 0.16 and /(0) = 0 as before? (1) c = 100 kg/sec. (Ill c = 60 kg/sec. (III) c = 10 kg/sec. Soilltioll. It is interesting to see how the behavior of the system changes due to the effect of the damping, which takes energy from the syslem. so that the oscillations decrease in amplitude (Case III) or even disappear (Cases II and I). (I) With m = 10 and k = 90, as in Example I, the model is the initial value problem lOy" + 100y' + 90y = O. /(0) ~ O. yeO) = 0.16 [meter]. The characteristic equation is IOA + 100A + 90 = IO{A + 9)(A + 1) ~ O. It has the roots -9 and -1. This gives the general solution 2 We also need The initial conditions give cl + c2 = 0.16, -9'"1 the overdamped case the solution is = o. The solution is cl c2 = -0.02, c2 = O.IS. Hence in y = -0.02e- 9t + O.ISe- t . 11 approaches 0 as t - 4 x. The approach is rapid: after a few seconds the solution is practically 0, that is. the iron ball is at res\. (III The model is as before. with c = 60 instead of 100. The characteristic equation now has the form IOA2 + 60A + 90 = IO(A + 3)2 = O. It has the double root -3. Hence the corresponding general solution is We also need The initial conditions give y(Ol ~ solution is cl = 0.16. / '"2 - 3,"[ = O. C2 = 0.4S. Hence in the critical case the (0) = y = (0.16 + OASt)e- 3t . It is always positive and decrea~es 10 0 in a monotone fashion. (III) The model now is lOy" + lOy' + 90)' = o. Since c = 10 is smaller than the critical c, we shall get oscillations. The characteristic equation is IOA2 + lOA + 90 = IO[(A + ~)2 + 9 - ~] = o. It has the complex roots [see (4) in Sec. 2.2 with 1I = 1 and b ~ 9] A ~ -0.5 ::': YO.52 - 9 = -0.5 ::': 2.96i. This gives the general solution }' = e -o.5t(A cos 2.96t + 8 sin 2.96t). Thus ."(0) = A = 0.16. We also need the derivative y' = e -O.5t( -0.5A cos 2.Y6t - 0.58 sin 2.Y6t - 2.YM sin 2.96t Hence /(0) = -0.5A + 2.968 ~ O. + 2.968 cos 2.Y6tl. 8 = 0.5A/2.96 = 0.027. This gives the solution y = e -o.5t(O.16 cos 2. libt + 0.027 sin 2.961) = o. 162e -O.5t cos (2.96t - 0.17). We see that these damped oscillations have a smaller frequency than the harmonic oscillations in Example 1 by about lo/c (since 2.96 is smaller than 3.00 by abom 1o/c I. Their amplitude goes 10 zero. See Fig. 39. • y 0.15 0.1 0.05 --0.05 --0.1 Fig. 39. The three solutions in Example 2 CHAP. 2 68 Second-Order Linear ODEs This section concerned free motions of mass-spring systems. Their models are homogeneous linear ODEs. Nonhomogeneous linear ODEs will arise as models of forced motions, that is, motions under the influence of a "driving force". We shall study them in Sec. 2.8, after we have learned how to solve those ODEs. " "::=» -B"£EM_5:U-F::.4:: 11-81 MOTION WITHOUT DAMPING (HARMONIC OSCILLATIONS) 1. (Initial value problem) Find the harmonic motion (4) 2. 3. that starts from Yo with initial velocity vo. Graph or sketch the solutions for Wo = 71", Yo = I, and various Vo of your choice on common axes. At what t-values do all these curves intersect? Why? (Spring combinations) Find the frequency of vibration of a ball of mass 111 = 3 kg on a spring of modulus (i) kl = 27 nt/m, (ii) k2 = 75 nt/m, (iii) on these springs in parallel (see Fig. 40), (iv) in series, that is, the ball hangs on one spling, which in tum hangs on the other spring. (Pendulum) Find the frequency of oscillation of a pendulum of length L Wig. 41), neglecting air resistance and the weight of the rod, and assuming to be so small that sin practically equals (Frequency) What is the frequency of a harmonic oscillation if the static equilibrium position of the ball is 10 cm lower than the lower end of the spring before the ball is attached? (Initial velocity) Could you make a hannonic oscillation move faster by gi\'ing the body a greater initial push? (Archimedian principle) This principle states that the buoyancy force equals the weight of the water displaced by the body (partly or totally submerged). The cylindrical buoy of diameter 60 cm in Fig. 42 is floating in water with its axis vertical. Wben depressed downward in the water and released, it vibrates with period 2 sec. Wbat is its weight? e 4. 5. 6. e e. 7. (Frequency) How does the frequency of a hannonic motion change if we take (i) a spring of three times the modulus, (ii) a heavier ball? 8. TEAM PROJECT. Harmonic Motions in Different Physical Systems. Different physical or other systems may have the same or similar models, thus showing the ullifyillg power of mathematical methods. Illustrate this for the systems in Figs. 43-45. (a) Flat spring (Fig. 43). The spring is horizontally clamped at one end, and a body of weight 25 nt (about 5.6Ib) is attached at the other end. Find the motion of the system, assuming that its static equilibrium is 2 cm below the horizontal line, we let the system start from this position with initial velocity 15 cm/sec, and damping is negligible. (b) Torsional vibrations (Fig. 44). Undamped torsional vibrations (rotations back and forth) of a wheel attached to an elastic thin rod are modeled bv the ODE loe " + Ke = 0, where e is the angle measured from the state of equilibrium, 10 is the polar moment of inertia of the wheel about its center, and K is the torsional stiffness of the rod. Solve this ODE for Kilo = 17.64 sec- 2 , initial angle 45°, and initial angular velocity 15° sec-I. (c) Water in a tube (Fig. 45). What is the frequency of vibration of 5 liters of water (about 1.3 gal) in a U-shaped tube of diameter 4 cm, neglecting friction? . _____ ~ n -r~Fig. 43. t Flat spring (Project 8a) (y=o) Body of massm F·li!. 40. Parallel springs (Problem 2) Fig. 41. Pendulum (Problem 3) Fig. 44. Torsional vibrations (Project 8b) Water level Fig. 42. Buoy (Problem 6) 19-171 Fig. 45. Tube (Project Be) DAMPED MOTION 9. (Frequency) Find an approximation formula for w* in terms of Wo by applying the binomial theorem in (9) and retaining only the first two terms. How good is the approximation in Example 2, III? SEC. 2.5 69 Euler-Cauchy Equations 10. (Extrema) Find the location of the maxima and minima of)' = e- 2t cos ( obtained approximately from a graph of .1' and compare it with the exact values obtained by calculation. (a) Avoiding ullllecessary generality is part of good modeli1lg. Decide that the initial value problems (A) and (B), (A) 11. (Maxima) Show that the maxima of an underdamped motion occur at equidistant (-values and find the distance. y" + cy' +Y = 0, yeO) = 1, y'(O) = 0 (B) the same with different c and y' (0) = -2 (instead of 0), will give practically as much information as a problem with other m, k, yeO), y' (0). 12. (Logarithmic decrement) Show that the ratio of two consecutive maximum amplitndes of a damped oscillation ( 10) is constant, and the natnral logarithm of this ratio, called the logarithmic decrel1lellt. equals j. = 27Texlw*. Find .1 for the solutions of .1''' + 2y' + 5)' = O. (b) COllsider (A). Choose suitable values of c, perhaps better ones than in Fig. 46 for the transition from Case III to II and I. Guess c for the curves in the figure. (c) Time to go to rest. Theoretically, this time is infinite (why?). Practically, the system is at rest when its motion has become very small, say, less than 0.1 % of the initial displacement (this choice being up to us), that is in our case. 13. (Shock absorber) What is the smallest value of the damping constant of a shock absorber in the suspension of a wheel of a car (consisting of a spring and an absorber) that will provide (theoretically) an oscillation-free ride if the mass of the car is 2000 kg and the spring constant equals 4500 kg/sec 2 ? (1 I) ly(t)1 14. (Damping constant) Consider an underdamped < 0.001 for all ( greater than some tl' In engineering constructions, damping can often be varied without too much trouble. Experimenting with your graphs. frnd empirically a relation between tl and c. motion of a body of mass III = 2 kg. If the time between two consecutive maxima is 2 sec and the maximum amplitude decreases to of its initial value after 15 cycles. what is the damping constant of the system? ! (d) Solve (A) a1lalytically. Give a rea~on why the solution c of Y«(2) = -0.001, with t2 the solution of y' (t) = O. will give you the best possible c satisfying (11). (e) Consider (B) empirically as in (a) and (b). What is the main difference between (B) and (A)? 15. (Initial value problem) Find the critical motion (8) that starts from Yo with initial velocity vo. Graph solution curves for ex = 1, Yo = I and several Vo such that (i) the curve does not intersect the t-axis, (ii) it intersects it at ( = L, 2, ... ,5. respectively. 16. (Initial value problem) Find the overdamped motion (7) that starts from Yo with initial velocity Vo. 10 17. (Overdamping) Show that in the overdamped case, the body can pass through y = 0 at most once. 18. CAS PROJECT. Transition Between Cases I. II, Ill. Study this transition in terms of graphs of typical solutions. (Cf. Fig. 46.) 2.5 Fig. 46. CAS Project 18 Euler-Cauchy Equations Euler-Cauchy equations4 are ODEs of the form (1) x 2 }''' + ax}" + by = 0 4LEONHARD EULER (1707-1783) was an enormously creative Swiss mathematician. He made fundamental contributions to almost all branches of mathematics and its application to physics. His important books on algebra and calculus contain numerous basic results of his own research. The great French mathematician AUGUSTIN LOUIS CAUCHY (1789-1857) is the father of modem analysis. He is the creator of complex analysis and had great influence on ODEs, PDEs, infinite series, elasticity theory, and optics. 70 CHAP. 2 Second-Order Linear ODEs with given constants a and b and unknown -"(Jo). We substitute "= xfn (2) and its derivatives y' = m 111X - l and -,," = 1)x",-2 into (1). This gives /11(111 - We now see that (2) was a rather natural choice because we have obtained a common factor xm. Dropping it, we have the auxiliary equation 11l(m - I) + am + b = 0 or (3) 111 2 + (a - 1)11l + b = O. tNote: a - I, not a.) Hence y = xnt is a solution of (1) if and only if I1l is a root of (3). The roots of (3) are (4) Case I. 1112 = !(l - a) - If the roots 1111 and 1112 V~(l - a)2 - b. are real and different. then solutions are and They are linearly independent since their quotient is not constant. Hence they constitute a basis of solutions of (I) for all x for which they are real. The corresponding general solution for all these x is (5) E X AMP L E 1 (Cl, C2 arbitrary). General Solution in the Case of Different Real Roots The Euler-Cauchy equation x 2 y" + 1.5xy' - 0.5.1' = 0 has the auxiliary equlltion 1112 + 0.5111 - 0.5 = O. The roots are 0.5 and -\. Hence a basis of solutions for all positive x is Yl general solution (Note: 0.5. not 1.5!) = xO. 5 and Y2 = IIx and gives the (x> 0). • Case II. Equation (4) shows that the auxiliary equation (3) has a double root =!O - a) if and only if (I - a)2 - 4b = O. The Euler-Cauchy equation (I) then has the form 1111 (6) A solution is h = x O - a )f2. To obtain a second linearly independent solution, we apply the method of reduction of order from Sec. 2.1 as follows. Starting from )'2 = uy!, we obtain for u the expression (9), Sec. 2.1, namely, u=JUdx where U = ~ exp (- Jp dX) . )'1 SEC. 2.5 71 Euler-Cauchy Equations Here it is crucial that p is taken from the ODE written in standard form. in our case. a y " + -x .y (6*) , + This shows that p = alx (not ax). Hence its integral is a in x = In (xa), the exponential function in U is Ih: a , and division by YI 2 = x 1 - a gives U = l/x, and u = In x by integration. Thus, in this "critical case," a basis of solutions for positive x is Yl = xm and Y2 = X 1n In x, where 111 =!O - a). Linear independence follows from the fact that the quotient of these solutions is not constant. Hence, for all x for which )'1 and Yz are defined and real, a general solution is (7) E X AMP L E 2 y (e1 = + c21nx)xm, 111 = ~(l - a). General Solution in the Case of a Double Root The Euler-Cauchy equation x 2v" - 5xy' + 9y = 0 has the auxiliary equation double root III = 3, so that a general solution for all positive x is nz2 - 6m +9 = O. It has the • Case III. The case of complex roots is of minor practical importance, and it suffices to present an example that explains the derivation of real solutions from complex ones. E X AMP L E 3 Real General Solution in the Case of Complex Roots The Euler-Cauchy equation hm; the auxiliary equation /11 2 - 0.4111 + 16.04 = O. The root" are complex conjugate. /Ill = 0.2 + 4i and 0.2 - 4i, where i = v'=T. (We know from algebra that if a polynomial with real coefficients has complex roots. these are always conjugate.) Now use the trick of writing x = e ln ,. and obtain nz2 = xm1 = xO.2+4i = xO.2(eln X)4i = xO.2/4In Xli, xm2 = xO.2-4i = xO.2(eln X)-4i = xO.2e -(4 In xl i. Next apply Euler's formula (11) in Sec. 2.2 with x m1 X"'2 I = 4 In x to these two formulas. This gives = 2 xO. [cos (4 In x) + i sin (4 In x)], = xO. 2 rcos (4 In x) - ; sin (4 In x)]. Add these two formulas. so that the sine drops uut. and divide the result by 2. Then subtract the second formula from the first, so that the cosine drops out, and divide the result by 2i. This yields XO. 2 cos (4 In x) and XO. 2 sin (4 In x) respectively. By the superposition principle in Sec. 2.2 these are solutions of the Euler-Cauchy equation (I). Since their quotient cot (4 In x) is not constant, they are linearly independent. Hence they form a basis of solutions, and the corresponding real gencral solution for all positive x is (8) y 2 = xo. [A cos (4 In x) + B sin (4Inx)l. Figure 47 shows typical solution curves in the three cases discussed, in particular the basis functions in • Examples I and 3. 72 CHAP. 2 Second-Order Linear ODEs Y, 3.0 xl. 5 Y 1.5 1.0 0.5 xl 2.0 xO. 5 0 2 x Case I: Real roots 0 -0.5 -1.0 -1.5 x O.2 sin (4Inx) \I \ ()\'\ O.ll VV 1 1.4, 2 x '\ '\ "- x O.2 cos (4 In x) Case III: Complex roots Case II: Double root Fig. 47. E X AMP L E 4 1.5 1.0 0.5 x-1. 5 lnx 2 x 0 -0.5 -1.0 -1.5 1.0 Y xlnx xO.5lnx~ Euler-Cauchy equations Boundary Value Problem. Electric Potential Field Between Two Concentric Spheres Find the electrostatic potential v = vCr) between two concentric spheres of radii rl = 5 cm and r2 = 10 cm kept at potentials VI = 110 Y and v2 = 0, respectively. Physicallnjorlll{[tion. vCr) is a solution of the Euler-Cauchy equation rv" + 2v' = O. where v' = dvldr. Solution. vCr) = Cl The auxiliary equation is 1112 + III = O. It has the roots 0 and - 1. This gives the general solution + c2fr. From the "boundary conditions" (the potentials on the spheres) we obtain C2 + 5" v(5) = cl C2 = 110. v(i 0) = Cl + 10 = O. = 110. C2 = llOO. From the second equation. Cl = -c2f10 = -llO. Allswer: + llOOIr Y. Figure 48 shows that the potential is not a straight line. as it would be for a potential By subtraction. c2flO vCr) = -llO between two parallel plates. For example, on the sphere ofradius 7.5 cm it is not 11012 = 55 Y, but considerably less. (Whm is it?) • v 100 80 , "' 60 40 20 0 5 6 Fig. 48. 11-101 8 111-151 GENERAL SOLUTION Find a real general solution, showing the details of your work. 2. 4x 2y" + 4xy' - y = 0 3. x 2 y" - 7xy' + 16y = 0 4. x 2 y" + 3x-,,' + y = 0 5. x 2 )''' 6. 2x 2 )"" + 4x)" + 5y = 0 7 10 9 r Potential v(r) in Example 4 INITIAL VALUE PROBLEM Solve and graph the solution, showing the details of your work. 11. x 2y" - 4xy' + 6)' = 0, y(l) = I, y'(l) = 0 12. x .\''' + 3xy' + y = 0, y(\) = 4, y' (1) = -2 13. (x 2D2 + 2xD + 100.251)y = 0, y(1) = 2. 2 - xy' + 2y 7. (lOx 2D2 - 20xD + 22.4l)y = 0 8. (4x 2D2 + l)y = 0 9. (100x 2D2 10. (I Ox 2D2 + 6xD + 0.5/)y = 0 = 0 y'(1) + 9l)y = 0 = -11 14. (x 2D2 - 2xD y' (1) = 2.5 15. (xD 2 + 4D)y + 2.251)y = 0, y(l) = 2.2, = 0, y(l) = 12, y' (1) = -6 SEC. 2.6 16. TEAM PROJECT. Double Root (A) Derive a second linearly independent solution of by reduction of order; but instead of using (9), Sec. 2.1, perform all steps directly for the present ODE (I). (B) Obtain x In In xby considering the solutionsx m and x m + s of a suitable Euler-Cauchy equation and letting (I) s~O. 2.6 73 Existence and Uniqueness of Solutions. Wronskian (C) Verify by substitution thatx m In x, 111 = (1 - a)/2, is a solution in the critical case. (D) TransfoI1n the Euler-Cauchy equation (1) into an ODE with constant coefficients by setting x = et (x > 0). (E) Obtain a second linearly independent solution of the Euler-Cauchy equation in the "critical case" from that of a constant-coefficient ODE. Existence and Uniqueness of Solutions. Wronskian In this section we shall discuss the general theory of homogeneous linear ODEs y" + p(x)y' (1) + q(x)y = 0 with continuous, but otherwise arbitrary variable coefficients p and q. This will concern the existence and form of a general solution of (1) as well as the uniqueness of the solution of initial value problems consisting of such an ODE and two initial conditions (2) with given xo, Ko, and K 1 . The two main results will be Theorem 1, stating that such an initial value problem always has a solution which is unique, and Theorem 4, stating that a general solution (3) (Cl, Cz arbitrary) includes all solutions. Hence linear ODEs with continuous coefficients have no "singular solutions" (solutions not obtainable from a general solution). Clearly, no such theory was needed for constant-coefficient or Euler-Cauchy equations because everything resulted explicitly from our calculations. Central to our present discussion is the following theorem. THEOREM 1 Existence and Uniqueness Theorem for Initial Value Problems If p(x) and q(x) are continuous functions on some open interval J (see Sec. 1.1) and is in J, then the initial value problem consisting of (1) and (2) has a unique solution y(x) on the interval 1. Xo The proof of existence uses the same prerequisites as the existence proof in Sec. 1.7 and will not be presented here; it can be found in Ref. [All] listed in App. 1. Uniqueness proofs are usually simpler than existence proofs. But for Theorem 1, even the uniqueness proof is long, and we give it as an additional proof in App. 4. CHAP. 2 74 Second-Order Linear ODEs Linear Independence of Solutions Remember from Sec. 2.1 that a general solution on an open interval 1 is made up from a basis ."1> Y2 on I, that is, from a pair of linearly independent solutions on I. Here we call )'t, )'2 linearly independent on 1 if the equation (4) k) = 0, implies k2 = O. We call y) • ."2 linearly dependent on 1 if this equation also holds for constants kh k2 not both O. In this case, and only in this case. YI and Y2 are proportional on I. that is (see Sec. 2.1), (5) (a) or YI = /...)'2 (b) )'2 = [YI for all x on I. For our discussion the following criterion of linear independence and dependence of solutions will be helpful. THEOREM 2 Linear Dependence and Independence of Solutions LeT the ODE (I) Izave cOllfinuous coefficients p(x) and q(x) all an open interval I. Then two solutions YI and .1'2 of (1) on T are linearly dependent on I if and only if their "Wronskian" (6) is 0 lit some Xo in 1. FlIrthermore, if W = 0 at an x = Xo in I. then W == 0 on I: hence if there is all Xl ill 1 at wlzich W is IIOt 0, thell ."1. Y2 are linearly independent on I. PROOF (a) Let."1 and Y2 be linearly dependent on I. Then (5a) or (5b) holds on I. If (5a) holds, then W(YI, .1'2) = YI.\'~ - Y2Y~ = k)'2Y~ - Y2k.1'~ = O. Similarly if (5b) holds. (b) Conversely, we let W()'I' )'2) = 0 for some x = Xo and show thar this implies linear dependence of YI and .\'2 on T. We consider the linear system of equations in the unknowns k I , k2 (7) kIYI(XO) + kl.)'~(xo) + k2)'~(XO) k2 Y2(XO) = 0 = O. To eliminate k 2 • multiply the first equation by Y~ and the second by -Y2 and add the resulting equations. This gives Similarly, to eliminate kI' multiply the first equation by -Y~ and the second by YI and add the resulting equations. This gives SEC. 2.6 75 Existence and Uniqueness of Solutions. Wronskian If W were not 0 at xo, we could divide by Wand conclude that kl = k2 = O. Since W is 0, division is not possible, and the system has a solution for which kl and k2 are not both O. Using these numbers k1> k2 , we introduce the function y(X) = k 1Yl(X) + k 2Y2(X), Since (1) is homogeneous linear, Fundamental Theorem I in Sec. 2.1 (the superposition principle) implies that this function is a solution of (I) on I. From (7) we see that it satisfies the initial conditions )"(xo) = 0, y' (xo) = O. Now another solution of (I) satisfying the same initial conditions is y* == O. Since the coefficients p and q of (I) are continuous. Theorem I applies and gives uniqueness, that is, y == y*, written out on 1. Now since kl and k2 are not both zero, this means linear dependence of )'1> )"2 on l. (e) We prove the last statement of the theorem. If W(xo ) = 0 at an Xo in I, we have linear dependence of .h, Y2 on I by part (b), hence W == 0 by part (a) of this proof. Hence in the case of linear dependence it cannot happen that W(x l ) =f. 0 at an XI in 1. If it does happen, it thus implies linear independence as claimed. • Remark. Determinants. Students familiar with second-order determinants may have noticed that Y~I Y2 I = YIY2 I - Y2Yl' This determinant is called the Wronski deTennina1lt 5 or, briefly, the Wronskian, of two solutions)"1 and)"2 of U), a<; has already been mentioned in (6). Note that its four entries occupy the same positions as in the linear system 0). E X AMP L E 1 Illustration of Theorem 2 The functions)"1 = cos wx and W(cos wx. sin wx) = )"2 = sin wx are solutions of)"" + w2 y = O. Their Wronskian is cos.wx I -WSlfl wX sin wx wcO~ I= YIY~ - Y2Y~ = w cos 2 lUX + w sin 2 wx = w. wX '* Theorem 2 shows that the,e solutions are linearly independent if and only if w O. Of course, we can see this directly from the quotient \'2IYl = tan wx. For w = 0 we have .\"2 == 0, which implies linear dependence (why?). • E X AMP L E 2 Illustration of Theorem 2 for a Double Root A general solution of y" - 2y' + Y = 0 on any interval is y = lCI + C2X)ex. (VeIify!). The corresponding Wronskian is not O. which shows linear independence of eX and xi'" on any interval. Namely. • 5Introduced by WRONSKI (JOSEF MARIA HONE. 1776-1853). Polish mathematician. CHAP. 2 76 Second-Order Linear ODEs A General Solution of (1) Includes All Solutions This will be our second main result, as announced at the beginning. Let us start with existence. Existence of a General Solution THEOREM 3 /fp(x) and q(x) are continuous on an open interval I, then (1) has a general solution on T. PROOF By Theorem 1, the ODE (1) has a solution heX) on T satisfying the initial conditions and a solution Y2(X) on T satisfying the initial conditions The Wronskian of these two solutions has at x = Xo the value Hence, by Theorem 2. these solutions are linearly independent on l. They fonn a basis of solutions of (1) on T, and y = ("1)'1 + C2Y2 with arbitrary c 1 - C2 is a general solution of (1) on T, whose existence we wanted to prove. • We finally show that a general solution is as general as it can possibly be. A General Solution Includes All Solutions THEOREM 4 /f the ODE (1) has cnntinuuus cnefficients p(x) and q(x) on some open interval I, then every solution Y = Y(x) of (1) on 1 is of the form (8) where Yv Y2 is any basis of solutions of (l) on 1 and C v C2 are suitable constants. Hence (1) does not have singular solutions (that is, solutions not obtainablefrom a general solution). PROOF Let y = Y(x) be any solution of (1) on I. Now, by Theorem 3 the ODE (I) has a general solution (9) on 1. We have to find suitable values of C1> C2 such that y(x) = Y(x) on I. We choose any in 1 and show first that we can find values of Cl' C2 such that we reach agreement at xo, that is, y(xo) = Y(xo) and y' (xo) = Y' (xo). Written out in terms of (9), this becomes Xo (10) (a) Clh(Xo) + C2.'"2(XO) = Y(xo) (b) CIY~(XO) + C2Y~(XO) = Y' (xo). SEC. 2.6 Existence and Uniqueness of Solutions. Wronskian 77 We detennine the unknowns Cl and C2' To eliminate C2, we multiply (lOa) by J~(xo) and (lOb) by -Y2(XO) and add the resulting equations. This gives an equation for Cl' Then we multiply (lOa) by -J~(xo) and (lOb) by Yl(XO) and add the resulting equations. This gives an equation for C2' These new equations are as follows, where we take the values of Jb y~. )'2' Y~' Y. y' at Xo, Cl()'IY~ - yzY~) = Cl W(y!> .\'2) = yy~ - Y2 Y ' C2(YIJ~ - J2Y~) = C2 W(Yb )'2) = h Y' - Yy~. Since)'b Yz is a basis, the Wronskian W in these equations is not 0, and we can solve for C2' We call the (unique) solution Cl = C b C2 = C2 . By substituting it into (9) we obtain from (9) the particular solution Cl and Now since C b C2 is a solution of (10), we see from (10) that From the uniqueness stated in Theorem this implies that y* and Y must be equal everywhere on /, and the proof is complete. • Looking back at he content of this section, we see that homogeneous linear ODEs with continuous variable coefficients have a conceptually and structurally rather transparent existence and uniqueness theory of solutions. Important in itself, this theory will also provide the foundation of an investigation of nonhomogeneous linear ODEs, whose theory and engineering applicatiuns we shall study in the remaining four sections of this chapter. -_ _-.. .-i. __ ._.__ .. .. 11-171 ~ -- ..... ..-. ~-- ....... BASES OF SOLUTIONS. CORRESPONDING ODEs. WRONSKIANS Find an ODE (1) for which the given functions are solutions. Show linear independence (a) by considering quotients, (b) by Theorem 2. 1. eO. 5x , e-O. 5x 2. cos 7rX, sin 7rX 3. e kx , xe kx 5. XO. 25 , xO. 25 In 4. x 3 , x- 2 6. e 3 .4x , e- 2 . 5X x 7. cos (2 In X), sin (2 In 8. e- 2x , xe- 2x 10. x- 3 • x- 3 In X) 11. cosh 2.5x, sinh 2.5x x 12. e- 2x cos wx, e- 2x sin wx 13. e- x cos 0.8x, e- X sin 0.8x 14. X-I cos (In x), X-I sin (In x) 15. e- 2 . 5x cos 0.3x. e- 2 . 5x sin 0.3x 16. e- kx cos 7rX, e- kx sin 17. e- 3 . 8 "ITx, xe- 3 .8 "ITx 7rX 18. TEAM PROJECT. Consequences of the Present Theory. This concems some noteworthy general properties of solutions. Assume that the coefficients p and q of the ODE (l) are continuous on some open interval T. to which the subsequent statements refer. (A) Solve y" - Y = 0 (a) by exponential functions, (b) by hyperbolic functions. How are the constants in the corresponding general solutions related? (8) Prove that the solutions of a basis cannot be 0 at the same point. (C) Prove that the solutions of a basis cannot have a maximum or minimum at the same point. (D) Express (Y2/Yl) , by a fOimula involving the Wronskian W. Why is it likely that such a formula should exist? Use it to find Win Prob. 10. (E) Sketch YI(X) = x 3 if X ~ 0 and 0 if x < 0, Y2(X) = 0 if x ~ 0 and x 3 if x < O. Show linear independence on - I < x < 1. What is their Wronskian? What Euler-Cauchy equation do Y10 Y2 satisfy? Is there a contradiction to Theorem 2? CHAP. 2 78 Second-Order Linear ODEs (F) Prove Abel's formula 6 where c = W(Yl (xo), Y2(xo». Apply it to Prob. 12. Him: Write (1) for Y1 and for )'2' Eliminate q algebraically from these two ODEs. obtaining a first-order linear ODE. Solve it. W(.vrlX), Y2lx)) = c exp [ - fXp(t) dt] Xo 2.7 Nonhomogeneous ODEs Method of Undetermined Coefficients In this section we proceed from homogeneous to nonhomogeneous linear ODEs y" (1) + p(x)y' + q(x)y = rex) where rex) =t= O. We shall see that a "general solution" of (1) is the sum of a general solution of the corresponding homogeneous ODE (2) y" + p(x)y' + q(x)y = 0 and a "particular solution" of 0). These two new terms "general solution of (\)" and "particular solution of 0)" are defined as follows. DEFINITION General Solution, Particular Solution A general solution of the nonhomogeneous ODE (I) on an open interval I is a solution of the form (3) here. Yh = ClYl + C2Y2 is a general solution of the homogeneous ODE (2) on I and Yp is any solution of ( 1) on I containing no arbitrary constants. A particular solution of (I) on I is a solution obtained from (3) by assigning specific values to the arbitrary constants Cl and C2 in .rh' Our task is now twofold, first to justify these definitions and then to develop a method for finding a solution yp of (I). Accordingly, we first show that a general solution as just defined satisfies (I) and that the solutions of 0) and (2) are related in a very simple way. THEOREM 1 Relations of Solutions of (1) to Those of (2) (a) The sum of a solution y of (I) 0/1 some open inten>al I alld a solution yof (2) on I is a solution of (1) 0/1 l. In particular. (3) is a solution of (1) on l. (b) The differellce oftH'o solutions of (1) on I is a solution of(2) on I 6 NIELS HENRIK ABEL (1802-1829). Norwegian mathematician. SEC 2.7 79 Nonhomogeneous ODEs PROOF (a) Let L[y] denote the left side of (1). Then for any solutions y of (1) and yof (2) on I, L[y + y] = L[y] + L[y] = r + 0 = (b) For any solutions \" and y';' of (I) on I we have L[y - y*] = r. L[ \"] - L[y*] = r - r = O. • Now for homogeneous ODEs (2) we know that general solutions include all solutions. We show that the same is true for nonhomogeneous ODEs (1). THEOREM 2 A General Solution of a Nonhomogeneous ODE Includes All Solutions If the coefficients p(x), q(x), and the function r(x) in (1) are continuous on some open interval I, then ever), solution of (I) on T is obtained by assigning suitable values to the arbitrary constants CI and C2 in a general solution (3) of (I) on I. PROOF Let y* be any solution of (\) on T and Xo any x in I. Let (3) be any general solution of (1) on T. This solution exists. Indeed, Yh = CIYI + C2Y2 exists by Theorem 3 in Sec. 2.6 because of the continuity assumption, and Yp exists according to a construction to be shown in Sec. 2.10. Now, by Theorem I (b) just proved, the difference Y = y* - Yp is a solution of (2) on I. At Xo we have Theorem I in Sec. 2.6 implies that for these conditions, as for any other initial conditions in I, there exists a unique particular solution of (2) obtained by assigning suitable values to c l , C2 in Yh. From this and y* = Y + YP the statement follows. • Method of Undetermined Coefficients Our discussion suggests the following. To solve the nonhomogeneous ODE (I) or an initial value problel1lfor (1), we have to solve the homogeneolls ODE (2) and find any solution yp of (1), so that we obtain a general solution (3) of (1). How can we fmd a solution Yp of (1)? One method is the so-called method of undetermined coefficients. It is much simpler than another, more general method (to be discussed in Sec. 2.10). Since it applies to models of vibrational systems and electric circuits to be shown in the next two sections, it is frequently used in engineering. More precisely, the method of undetermined coefficients is suitable for linear ODEs with constant coefficients a and b (4) y" + aJ' + by = rex) when rex) is an exponential function, a power of x, a cosine or sine, or sums or products of such functions. These functions have derivatives similar to rex) itself. This gives the idea. We choose a form for Yp similar to rex). but with unknown coefficients to be determined by substituting that yp and its derivatives into the ODE. Table 2.1 on p. 80 shows the choice of Yp for practically important fonns of rex). Corresponding ruIes are as follows. CHAP. 2 80 Second-Order Linear ODEs Choice Rules for the Method of Undetermined Coefficients (a) Basic Rule. If rex) in (4) is one of the fUllctions in the first colU111n in Table 2.1. choose )'p ill the same line and determine its undetermined coefficients hy suhstituting .\"p and its deril'atil'es i1lfo (4). (b) Modification Rule. rl a tenn in your chnice for .\"p happens to be a solution of the homogeneous ODE corresponding to (4). lIlultiply your choice of.\'1' by x (or by x 2 !l this solution c()rre~pollds to a double root of the characteristic equation of the h01l1ogeneous ODE). (C) Sum Rule. rl rex) is a sum of functiolls ill the first column of Table 2.1. choose for .\"p the sum of the functions ill the corresponding lines of the second COllllll1l. The Basic Rule applies when rex) is a single tenn. The Modification Rule helps in the indicated case, and to recognize such a case. we have to solve the homogeneous ODE first. The Sum Rule follows by noting that the sum of two solutions of (I) with r = rl and r = r2 (and the same left side!) is a solution of (1) with r = 1"1 + r2' (Verify!) The method is self-correcting. A false choice for .\"p or one with too few tenns will lead to a contradiction. A choice with too many terms will give a correct result. with superfluous coefficients coming out zero. Let us illustrate Rules (a)-( c) by the typical Examples 1-3. Table 2.1 Method of Undetermined Coefficients Term in rex") E X AMP L E 1 Choice for .\'p(.t) key:r Cey:r kt" (n = O. L· .. ) kcos wx k sin wx ke"'" cos wx ke,,:r sin wx Knxn + Kn_1xn - 1 + } K cos wx + K1x + Ko + M sin wx } e"X(K cos wx + M sin wx) Application of the Basic Rule (a) Sol"e the initial value problem .1'(0) = O. (5) Solutioll. y' (0) = 1.5 . Step 1. Gelleral solutioll of the homogelleolls ODE. The ODEy" + Y = y" = A cos ~ a has the general solution + B sin X. v; Step 2. SoluRon yp Of the nonhomogelleous ODE. We first try .1'1' = Kx 2 . Then = 2K. By substitution. 2K + K\-2 = 0.00Ix 2. For this to hold for all X. the coefficient of each power of x (x 2 and ,0) must be the same on both sides; thus K = 0.001 amI 2K = O. a contradiction. The second line in Table 2.1 suggests (he choice Then 2 Equating the coefficients of x . X, .\ 0 on both sides. we have K2 = 0.001. K] Ko = -2K2 = -0.002. Tills givesyp = 0.001x 2 - 0.002, and y = -,"/z + -'"1' = A cos x + B sin x + O.OOh 2 - = 0.002. 0, 2K2 + Ko = O. Hence SEC. 2.7 81 Nonhomogeneous ODEs Step 3. Solutioll of the il/itial value problem. Setting x = 0 and using the first initial condition gives yeO) = A - 0.002 = O. hence A = 0.002. By differentiation and from the second initial condition, .1" = y;, + Y~ = -A sin x + B cos x + 0.002x and /(0) = B = 1.5. This gives the answer (Fig. 49) y = 0.002 cos x + 1.5 sinx + 0.001x2 - 0.002. Figure 49 shows y as well as the quadratic parabola)')) about which y is oscillating, practically like a sine curve • since the cosine term is smaller by a factor of about 111000. x Fig. 49. E X AMP L E 2 Solution in Example 1 Application of the Modification Rule (b) Solve the initial value problem (6) y" + 3/ + 2.25y = -10 e-1. 5x, y'(O) = O. yW) = I, Solution. Step 1. Gel/eral solutioll of the homogelleous ODE. The characteristic equation of the homogeneous ODE is A2 + 3A + 2.25 = (A + 1.5)2 = O. Hence the homogeneou~ ODE has the general solution Step 2. Solutio" Yp of the "ollhomogelleous ODE. The function e-1.5x on the light would normally require the choice Ce-1. 5x. But we see from .I'h that this function is a solution of the homogeneous ODE. which corresponds to a double root of the characteristic equation. Hence, according to the Modification Rule we have to multIply our choice function by x 2 . That is, we choo~ Then We substitute these expressions into the given ODE and omit the factor e- 1 .5x . This yields C(2 - 6x + 2.25x 2 ) + 3C(2x - 1.5x 2 ) + 2.25Cx 2 = -10. Comparing the coefficients of x 2 • x. xU gives 0 = 0.0 = O. 2C = -10. hence C = -5. This gives the solution Yp = _5x 2 e- 1 .5x . Hence the given ODE has the general solution Step 3. Solutioll of the initial value problem. Setting x = 0 in y and using the first initial condition. we obtain .1'(0) = CI = I. Differentiation of y gives From this and the second initial condition we have gives the answer (Fig. 50) y' (0) = c2 - 1.5cl = O. Hence c2 = 1.5cl = 1.5. This The curve begins with a horizontal tangent. crosses the x-axis at x = 0.6217 (where 1 + 1.5x - 5x 2 = 0) and • approaches the axis ti'om below as x increases. CHAP. 2 82 Second-Order Linear ODEs y 1.0 0.5 Oc---';,--'------''------''------'--==-'-x ~l 2 ~---5 -D.5 ~ -1.0 Fig. 50. E X AMP L E 3 Solution in Example 2 Application of the Sum Rule (c) Solve the initial value problem (7) y" Solution. + 2y' + 5)" = (,0.5" yeo) 40 cos lOx - 190 sin lOx. T = 0.16. y' (0) = 40.08. Step 1. General solutioll of the homogeneous ODE. The characteristic equation A2 + 2A + 5 = (A + 1 + 2i){A + 1 - 20 = 0 shows that a real general solution of the homogeneous ODE is J'h = e -x (A cos 2, +B sin 2x). Step 2. Solution of the Ilonhomogeneous ODE. We write Yp = )"1'1 + exponential term and .\1'2 to the sum of the other twO terms. We set Then .1"1'2, where J'pl corresponds to the and Substitution into the given ODE and omission of the exponemial factor gives (0.25 + 2,0.5 C = 116.25 = 0.16. and )"1'1 = 0.16eo. 5". We now set )"1'2 = K cos lOx + M sin lOx. as in Table 2.1. and obtain )'~2 = -10K sin lOx + 10M cos lOx. \";2 = + 5)C = 1, hence -lOOK cos lOx - 100M sin lOx. Substitution into the given ODE gives for the cosine terms and for the sine tenTIS - lOOK + 2· 10M + 5K = 40, -100M - 2' 10K + 5M = -190 or, by simplification. -95K The solution is K + 20M = 40, -10K - 95M = -190. = O. M = 2. Hence .1'1'2 = 2 ,in lOx. Together, Y = Yh + Ypl + .1'1'2 = e- x (A co~ 2x + B SIl1 2<) + 0.16,,0.5x + 2 sin lOx. Step 3. Sollllion of the initial value problem. From y and the first initial condition. y{O) = A hence A = O. Differentiation gives y' = e -xC -A cos 2x - B sin 2, - 2A sin 2x + 0.16 = 0.16, + 2B cos 2,) + 0.08eO. 5 :< + 20 cos lOx. From this and the second initial condition we have /(0) = -A + 2B + 0.08 + 20 = 40.08, hence B This gives the solution (Fig. 51) Y = lOe- x sin 2x + 0.16,,°·5.< + 2 sin lOx. = 10. The firsllerrn goes to 0 relatively fas!. When x = 4. it is practically O. as the dashed curves::': lOe -x + 0.16eo. 5r show. From then on, the last term, 2 sin lOx, gives an oscillation about 0.16eo. 5 ,", the monotone increasing dashed curve. • SEC 2.7 83 Nonhomogeneous ODEs y 10 ,, -4 Fig. 51. Solution in Example 3 Stability. The following is important. If (and only if) all the roots of the characteristic equation of the homogeneous ODE y" + ay' + by = 0 in (4) are negative, or have a negative real part, then a general solution.Vl, of this ODE goes to 0 as x ~ (Xl, so that the "transient solution" Y = Yh + yp of (4) approaches the "steady-state solution" yp' [n this case the nonhomogeneous ODE and the physical or other system modeled by the ODE are called stable; otherwise they are called unstable. For instance, the ODE in Example 1 is unstable. Basic applications follow in the next two sections. [1-141 GENERAL SOLUTIONS OF NONHOMOGENEOUS ODEs Find a (real) general solution. Which rule are you using? (Show each step of your calculation.) 1. 2. 3. 4. 5. 6. 7. 8. + 3/ + + 4/ + 2y = 30e 2x 3.75y = 109 cos 5x y" - 16y = 19.2e 4 ,' + 60e x -,," y" + 9y = cos x + 4cos 3x + y' - 6)' = 6x 3 - 3x 2 + 12x y" + 4y' + 4)' = e- 2x sin 2x y" + 6/ + 73y = 80e x cos 4x y" + lOy' + 25y = 100 sinh 5x y" )''' 9. y" - 0.16y = 32 cosh O.4x 10. y" 11. y" + + + 4/ + 6.25y = 3.125(x 1.44y = 24 cos 1.2x + 1)2 12. y" 9y = 18x + 36 sin 3x 13. y" + 4v' + 5)' = 25x 2 + 13 sin 2x 14. y" + 2y' + Y = 2x sin x 115-201 INITIAL VALUE PROBLEMS FOR NONHOMOGENEOUS ODEs Solve the initial value problem. State which mles you are using. Show each step of your calculation in detail. 15. y" + 4y = 16 cos 2x, y(O) = 0, y' (0) = 0 16. y" - 3)" + 2.25)' = 27(x 2 - x). yeO) = 20, y' (0) = 30 17. y" + 0.2y' + 0.26)' = 1.22eo. 5x , y(O) = 3.5. y' (0) = 0.35 18. y" - 2/ = 12e 2x - 8e- 2x , yeO) = -2, /(0) = 12 19. y" - v' - 12\" = 144x 3 + 12.5, ~·(O) ~ 5, . y' (0) = -0.5 20. y" + 2y' + y(O) = 6.6, lOy = 17 sin x - 37 sin 3x, y' (0) = -2.2 21. WRITING PROJECT. Initial Value Problem. Write out all [he details of Example 3 in your own words. Discuss Fig. 51 in more detail. Why is it that some of the "half-waves" do not reach the dashed curves. whereas others preceding them (and, of course, all later ones) excede the dashed curves? 22. TEAM PROJECT. Extensions of the Method of Undetermined Coefficients. (a) Extend the method to products of the function in Table 2.1. (b) Extend the method to Euler-Cauchy equations. Comment on the practical significance of such extensions. 23. CAS PROJECT. Structure of Solutions of Initial Value Problems. Using the present method. fmd, graph, and discuss the solutions y of initial value problems of your own choice. Explore effects on solutions caused by 84 CHAP. 2 Second-Order Linear ODEs problem with .1'(0) = 0, y' (0) = O. Consider a problem in which you need the Modification Rule (a) for a simple root, (b) for a double root. Make sure that your problems cover all three Cases I. II. III (see Sec. 2.2). changes of initial conditions. Graph yp' y, .I' - Yp separately, to see the separate effects. Find a problem in which (a) the pan of y resulting from Yh decreases to zero, (b) increases. (c) is not present in the answer y. Study a 2.8 Modeling: Forced Oscillations. Resonance In Sec. 2.4 we considered vertical motions of a mass-spring system (vibration of a mass on an elastic spring, as in Figs. 32 and 52) and modeled it by the homogeneolls linear ODE 111 my" + cy' (1) + ky = O. Here yet) as a function of time t is the displacement of the body of mass 111 from rest. These were free motions, that is, motions in the absence of extemalforces (outside forces) caused solely by internal forces. forces within the system. These are the force of inertia my", the damping force c/ (if c > 0). and the spring force ky acting as a restoring force. We now extend our model by including an external force, call it ret), on the right. Then we have my" (2*) + cy' + ky = ret). Mechanically this means that at each instant t the resultant of the internal forces is in equilibrium with r(t). The resulting motion is called a forced motion with forcing function ret), which is also known as input or driving force, and the solution yet) to be obtained is called the output or the response of the system to the driving force. Of special interest are periodic external forces. and we shall consider a driving force of the form ret) = Fo cos wt (Fo > 0, w > 0). Then we have the nonhomogeneous ODE (2) my" + cy' + ky = Fo cos wt. Its solution will familiarize us with further interesting facts fundamental in engineering mathematics, in particular with resonance. c Dashpot Fig. 52. Mass on a spring SEC. 2.8 85 Modeling: Forced Oscillations. Resonance Solving the Nonhomogeneous ODE (2) From Sec. 2.7 we know that a general solution of (2) is the sum of a general solution Yh of the homogeneous ODE (1) plus any solution yp of (2). To find yp' we use the method of undetermined coefficients (Sec. 2.7), starting from (3) yp(t) = a cos wI + b sin wI. By differentiating this function (chain rule!) we obtain , = -wa Sill . wt Yp + y; = -w2a cos wt - b w cos wt. w2b sin wt. Substituting Yp' y~, and y; into (2) and collecting the cosine and the sine terms, we get [(k - 11lw2)a + web] cos wt + [ -wca + (k - 11lw2 )b] sin wt = Fo cos wt. The cosine terms on both sides must be equaL and the coefficient of the sine term on the left must be zero since there is no sine term on the right. This gives the two equations = Fo web (4) -well for determining the unknown coefficients a and b. This is a linear system. We can solve it by elimination. To eliminate b, multiply the first equation by k - 11lW2 and the second by - we and add the results, obtaining Similarly. to eliminate a. multiply the first equation by we and the second by k and add to get If the factor (k - 11lW2)2 and b, If we set ~ = Wo + w2e 2 is not zero, we can divide by this factor and solve for a (> 0) as in Sec. 2.4, then k = III Wo 2 and we obtain (5) We thus obtain the general solution of the nonhomogeneous ODE (2) in the fonn (6) 2 11lW yet) = y,,(1) + Yp(t). 86 CHAP. 2 Second-Order Linear ODEs Here Yh is a general solution of the homogeneous ODE (1) and Yp is given by (3) with coefficients (5). We shall now discuss the behavior of the mechanical system, distinguishing between the two cases c = 0 (no damping) and c > 0 (damping). These cases will correspond to two basically different types of output. Case 1. Undamped Forced Oscillations. Resonance If the damping of the physical system is so small that its effect can be neglected over the time interval considered, we can set c = O. Then (5) reduces to a = F o/[m(wo 2 - w 2 )] and b = O. Hence (3) becomes (use wo 2 = kim) (7) *" Here we must assume that w2 wo2; physically, the frequency wl(27T) [cycles/sec] of the driving force is different from the natural frequency w o/(27T) of the system, which is the frequency of the free undamped motion [see (4) in Sec. 2.4]. From (7) and from (4*) in Sec. 2.4 we have the general solution of the "undamped system" (8) We see that this output is a superpositioll of two harmollic oscillations of the frequencies just mentioned. Resonance. cos wt = I) We discuss (7). We see that the maximum amplitude of Yp is (put (9) I p= ------::- where I - (wlwO)2 a o depends on w and woo If w ~ wo, then p and a o tend to infinity. This excitation of large oscillations by matching input and natural frequencies (w = w o) is called resonance. p is called the resonance factor (Fig. 53), and from (9) we see that plk = aolFo is the ratio of the amplitudes of the particular solution Yp and of the input Fo cos wt. We shall see later in this section that resonance is of basic importance in the study of vibrating systems. In the case of resonance the nonhomogeneous ODE (2) becomes (10) Then (7) is no longer valid. and from the Modification Rule in Sec. 2.7 we conclude that a particular solution of (10) is of the form Yp(t) = tea cos wot + b sin Wol). SEC. 2.8 Modeling: Forced Oscillations. Resonance 87 p co Fig. 53. Resonance factor p(co) By substituting this into (10) we find a = 0 and b = Fo/(2I1lwo). Hence (Fig. 54) (11) yp(t) Fo 2mwo . = - - - t sm wot. We see that because of the factor t the amplitude of the vibration becomes larger and larger. Practically speaking, systems with very little damping may undergo large vibrations that can destroy the system. We shall return to this practical aspect of resonance later in this section. Fig. 54. Particular solution in the case of resonance Beats. Another interesting and highly important type of oscillation is obtained if w is close to woo Take, for example, the particular solution [see (8)] (12) yet) = Fo 2 2 (cos wt - cos wot) m(wo - w) Using (12) in App. 3.1, we may write this as yet) = 2Fo 2 lIl(wo - . (wo sm 2 W ) + 2 w) t sin ( Wo 2- w t) . Since w is close to wo, the difference Wo - w is small. Hence the period of the last sine function is large, and we obtain an oscillation of the type shown in Fig. 55, the dashed curve resulting from the first sine factor. This is what musicians are listening to when they tune their instruments. CHAP. 2 88 Second-Order Linear ODEs y Fig. 55. Forced undamped oscillation when the difference of the input and natural frequencies is small ("beats") Case 2. Damped Forced Oscillations If the damping of the mass-spring system is not negligibly small, we have e > 0 and a damping term cy' in (1) and (2). Then the general solution y" of the homogeneous ODE (I) approaches zero as t goes to infinity, as we know from Sec. 2.4. Practically, it is zero after a sufficiently long time. Hence the "transient solution" (6) of (2), given by Y = Yh + Yp' approaches the "steady-state solution" yp' This proves the following. THEOREM 1 Steady-State Solution After a sufficiently long time the output of a damped vibrating system under a purely sinusoidal dril'ing force [see (2)1 will practically be a harmonic oscillation whose .!i"eqllency is that of the input. Amplitude of the Steady-State Solution. Practical Resonance Whereas in the undamped case the amplitude of yp approaches infinity as w approaches woo this will not happen in the damped case. In this case the amplitude will always be finite. But it may have a maximum for some w depending on the damping constant c. This may be called practical resonance. It is of great importance because if c is not too large, then some input may excite oscillations large enough to damage or even destroy the system. Such cases happened. in particular in earlier times when less was known about resonance. Machines, cars, ships, airplanes, bridges, and high-rising buildings are vibrating mechanical systems. and it is sometimes rather difficult to find constructions that are completely free of undesired resonance effects, caused, for instance, by an engine or by strong winds. To study the amplitude of yp as a function of w, we write (3) in the form (13) yp(t) = C* cos (wt - 1]). C* is called the amplitude of yp and 1] the phase angIe or phase lag because it measures the lag of the output behind the input. According to (5). these quantities are (14) tan 1](w) = b a we SEC. 2.8 89 Modeling: Forced Oscillations. Resonance Let us see whether C*(w) has a maximum and, if so. find its location and then its size. We denote the radicand in the second root in C* by R. Equating the derivative of C* to zero, we obtain The expression in the brackets [... J is zero if (15) By reshuffling terms we have The right side of this equation becomes negative if c 2 > 2mk, so that then (15) has no real solution and C* decreases monotone as w increases, as the lowest curve in Fig. 56 on p. 90 shows. If c is smaller, c 2 < 2mk, then (15) has a real solution w = W max , where (15*) From (15*) we see that this solution increases as c decreases and approaches Wo as c approaches zero. See also Fig. 56. The size of C*(wrnax ) is obtained from (14), with w2 = w~ax given by (15*). For this 2 w we obtain in the second radicand in (14) from (15*) ( and Wo 2 - 2) c. C --2 2 2m The sum of the right sides of these two formulas is Substitution into (14) gives (16) We see that C*(wrnax ) is always finite when c > O. Furthermore, since the expression in the denominator of (16) decreases monotone to zero as c 2 « 2mk) goes to zero, the maximum amplitude (16) increases monotone to infinity, in agreement with our result in Case 1. Figure 56 shows the amplification C*IFo (ratio of the amplitudes of output and input) as a function of W for m = 1, k = I, hence Wo = 1, and various values of the damping constant c. CHAP. 2 90 Second-Order Linear ODEs Figure 57 shows the phase angle (the lag of the outpllt behind the input), which is less than 7r/2 when w < wo, and greater than 7r/2 for w > woo C' Po '1 TC 4 3 ,----c=o c = 112 c=l c=2 2 c= 2 OO---~--~--~--~-- o Fir- 56. Amplification C*/Fo as a function of w for m = 1, k = 1, and various values of the damping constant c 11-81 STEADY-STATE SOLUTIONS Find the steady-state oscillation of the mass-spring system modeled by the given ODE. Show the details of your calculations. 1. / ' + 6/ + 8)" = 130 cos 3t 2.4.'"" + 8y' + 13." = 8 sin 1.5t 3. y" + y' + 4.25y = 221 cos 4.5t 4. y" + 4/ + 5y = cos t - sin t 5. (D2 + 2D + I)y = -sin 2t 6. (D 2 + 4D + 31)y = cos r + l cos 3r 7. (D 2 + 6D + 181»), = cos 3t - 3 sin 3t 8. (D 2 + 2D + lOl)y = -25 sin 4t 19-141 TRANSIENT SOLUTIONS Find the transient motion of the mass-spring system modeled by the given ODE. (Show the details of your work.) y" + 2y' + 0.75.'" = 13 sin t + 4/ + 4)' = cos 4t 11. 4)''' + 12)"' + 9.,' = 75 sin 3t 12. (D2 + 5D + 41)), = sin 2t 9. 10. y" 13. (D 2 14. (D 2 + 3D + 3.251))' = 13 - 39 cos 2t + 2D + 5l)y = I + sin t 115-201 INITIAL VALUE PROBLEMS Find the motion of the mass-spring system modeled by the ODE and initial conditions. Sketch or graph the sol urian curve. In addition, sketch or graph the curve of w 2 Fig. 57. Phase lag 1) as a function of w for m = 1, k = I, thus Wo = 1, and various values of the damping constant c y - Yp to see when the system practically reaches the steady state. 15. y" -t 2.'" + 26y = 13 cos 3t. ),(0) = 1. y' (0) = 0.4 16. y" + 64y = cos t. y(O) = O. /(0) = 1 y(Q) = 0.7, 17. y" + 6y' + 8." = 4 sin 2t. y' (0) = - 11.8 + 2D + I)y = 75(sin t - ~ sin 21 + l sin 3t), = O. y'(O) = I (4D 2 + 12D + 13l)r = 12 cos t - 6 sin t, yeO) = 1. y' (0) = -] y" + 25.\' = 99 cos 4.9t, .1'(0) = 2, y' (0) = 0 18. (D2 y(O) 19. 20. 21. (Beats) Derive the formula after (12) from (12). Can there be beats if the system has damping? 22. (Beats) How does the graph ofthe solution in Prob. 20 change if you change (a) yeO). (b) the frequency of the driving force? 23. WRITING PROJECT. Free and Forced Vibrations. Write a condensed report of 2-3 pages on the most important facts about free and forced vibrations. 24. CAS EXPERIMENT. Undamped Vibrations. (a) Solve the initial value problem y" + Y = cos wt, w 2 *- I, yeO) = o. y' (0) = O. Show that the solution can be written y(t) = (17) - -2 -2 I-w sin [ -1 (1 2 + sin w)t ] X [~(l - W)t]. SEC. 2.9 91 Modeling: Electric Circuits (b) Experiment with (17) by changing w to see the change of the curves from those for small w (> 0) to beats, to resonance and to large values of w (see Fig. 58). m=0.2 20n 25. TEAM PROJECT. Practical Resonance. (a) Give a detailed derivation of the crucial formula (16). (b) By considering dC*/dc show that C*( w max) increases as c (~ V2111k) decreases. (c) lIIustrate practical resonance with an ODE of your own in which you vary c. and sketch or graph corresponding curves as in Fig. 56. (d) Take your ODE with c fixed and an input of two terms, one with frequency close to the practical resonance frequency and the other not. Discuss and sketch or graph the output. (e) Give other applications (not in the book) in which resonance is important. 26. (Gun barrel) Solve y" + Y ifO~t~7T I = { o ift>7T, .1'(0) = y' (0) = O. -10 m= 0.9 I 0.04 I~ I~ ~11~l 'IOn I -0.04 This models an undamped system on which a force F acts during some interval of time (see Fig. 59), for instance, the force on a gun banel when a shell is fired, the barrel being braked by heavy springs (and then damped by a dashpot, which we disregard for simplicity). Hint. At 7T both y and yf must be continuous. F m=1 ~ k=1 ~ n m=6 Fig. 58. Typical solution curves in CAS Experiment 24 Fig. 59. Problem 26 2.9 Modeling: Electric Circuits Designing good models is a task the computer cannot do. Hence setting up models has become an important task in modern applied mathematics. The best way to gain experience is to consider models from various fields. Accordingly, modeling electric circuits to be discussed will be profitable for all students, not just for electrical engineers and computer scientists. We have just seen that linear ODEs have important applications in mechanics (see also Sec. 2.4). Similarly, they are models of electric circuits, as they occur as portions of large networks in computers and elsewhere. The circuits we shall consider here are basic building blocks of such networks. They contain three kinds of components, namely, resistors, inductors, and capacitors. Figure 60 on p. 92 shows such an RLC-circuit, as they are called. In it a resistor of resistance R n (ohms), an inductor of inductance L H (henrys), and a capacitor of capacitance C F (farads) are wired in series as shown, and connected to an electromotive force E(t) V (volts) (a generator, for instance), sinusoidal as in Fig. 60, or of some other kind. R, L, C, and E are given and we want to find the current I(t) A (amperes) in the circuit. 92 CHAP. 2 Second-Order Linear ODEs Fig. 60. RLC-circuit An ODE for the cunent I(t) in the RLC-circuit in Fig. 60 is obtained from the following law (which is the analog of Newton's second law, as we shall see later). Kirchhoff's Voltage Law (KVL).7 The l'oltage (the electromotive force) impressed all a closed loop is equal to the Slllll of the voltage drops across the other elements of tlze loop. In Fig. 60 the circuit is a closed loop. and the impressed voltage E(t) equals the sum of the voltage drops across the three elements R, L, C of the loop. Voltage Drops. Experiments show that a current 1 flowing through a resistor. inductor or capacitor causes a voltage drop (voltage difference, measured in volts) at the two ends; these drops are (Ohm's law) Voltage drop for a resistor of resistance R ohms (D), RI dI LJ' = L - Voltage drop for an inductor of inductance L henrys (H), dt Q Voltage drop for a capacitor of capacitance C farads (F). C Here Q coulombs is the charge on the capacitor, related to the current by I(t) = dQ dt ' equivalently, Q(t) = fI(t) dr. This is summarized in Fig. 61. According to KVL we thus have in Fig. 60 for an RLC-circuit with electromotive force E(t) = Eo sin wt (Eo constant) as a model the "integro-differential equation" 0') I LJ, + RI + C f I lit = E(t) = Eo sin wt. 7GUSTAV ROBERT KIRCHHOFF (1824--1887). German physicist. Later we shall also need Kirchholrs current law (KCL): At allY poillf of a circlIit, the Slim of the illf/owillg currents is equal to the slim of the outflowil1g currents. The units of measurement of electrical quantities are named after ANDRE MARIE AMPERE (\ 775-1836), French physicist. CHARLES AUGUSTIN DE COULOMB (1736-1806), French physicist and engineer, MICHAEL FARADAY (1791-1867), Engli~h physicist, JOSEPH HENRY (I 797-1 878}, American physicist. GEORG SIMON OHM (1789-1854), Gemmn physicist, and ALESSANDRO VOLTA (1745-1827), Italian physicist. SEC. 2.9 Modeling: Electric Circuits 93 - Notation Symbol Ohm's resistor ---ANVV- R Ohm's resistance ohmsC!l) RI Inductor ...ro0OO"L L Inductance henrys (H) Capacitor ---11- C Capacitance farads CF) L dl dt Q/C Fig. 61. Unit Voltage Drop Name Elements in an RLC-circuit To get rid of the integral, we differentiate (1') with respect to t, obtaining (1) LI" + RI' + ~ I = C E' (t) = Eow cos wt. This shows that the current in an RLC-circuit is obtained as the solution of this nonhomogeneous second-order ODE (1) with constant coefficients. From (l '), using I = Q', hence l' = Q", we also have directly (1") LQ" + RQ , + -I C Q = Eo sin wt. But in most practical problems the current I(t) is more important than the charge Q(t), and for this reason we shall concentrate on (1) rather than on (1"). Solving the ODE (1) for the Current. Discussion of Solution A general solution of (l) is the sum I = Ih + Ip, where Ih is a general solution of the homogeneous ODE corresponding to (1) and lp is a particular solution of (1). We first determine Ip by the method of undetermined coefficients. proceeding as in the previous section. We substitute (2) + Ip = a cos I~ = w( -a sin wt + b cos wt) wt b sin wt I; = w 2 ( -a cos wt - b sin wt) into (I). Then we collect the cosine terms and equate them to Eow cos wt on the right. and we equate the sine terms to zero because there is no sine term on the right. Lw2( -a) + Rwb + alC = Eow (Cosine tenus) LW2(-b) + Rw(-a) + =0 (Sine terms). blC To solve this system for a and b, we first introduce a combination of Land C, called the reactance (3) s= wL - 1 wC 94 CHAP. 2 Second-Order Linear ODEs Dividing the previous two equations by w, ordering them, and substituting S gives -Sa + Rb = Eo = O. -Ra - Sb We now eliminate b by multiplying the first equation by S and the second by R, and adding. Then we eliminate a by mUltiplying the first equation by R and the second by -So and adding. This gives In any practical case the resistance R is different from zero. so that we can solve for a and b, (4) Equation (2) with coefficients a and b given by (4) is the desired pm1icular solution Ip of the nonhomogeneous ODE (I) governing the current I in an RLC-circuit with sinusoidal electromotive force. Using (4). we can write Ip in terms of "physically visible" quantities. namely. amplitude 10 and phase lag () of the cun'ent behind the electromotive force, that is. (5) where [see (4) in App. A3.1] tan () = a S b R V The quantity R2 + 52 is called the impedance. Our formula shows that the impedance equals the ratio Eollo. This is somewhat analogous to Ell = R (Ohm's law). A general solution of the homogeneous equation con'esponding to (I) is where Al and A2 are the roots of the characteristic equation A2 R +- L A+ - We can write these roots in the form Al = -a a= 1 LC = O. + {3 and A2 = -a - {3, where R 2L ' Now in an actual circuit, R is never zero (hence R > 0). From this it follows that Ih approaches zero, theoretically as t ~ x, but practically after a relatively short time. (This is as for the motion in the previous section.) Hence the transient current 1= Ih + Ip tends SEC. 2.9 95 Modeling: Electric Circuits to the steady-state current [p, and after some time the output will practically be a harmonic oscillation, which is given by (5) and whose frequency is that of the input (of the electromotive force). E X AMP L E 1 RLC-Circuit Find the cunent l(f) in an RLC-circuit with R = II n (ohms), L = 0.1 H (henry), C = 1O-2 F (farad), which is connected to a source of voltage E(f) = 100 sin 400f (hence 63~ HL = 63~ cycles/sec, because 400 = 63~' 21T). A,sume that cunent and charge are zero when f = O. Solution. Step I. General solution of the homogeneous ODE. Substituting R, L, C, and the derivative E' (f) into (I), we obtain + 0.11" Ill' + JOOI = 100' 400 cos 400f. Hence the homogeneous ODE is 0.11" + III' + 1001 0.1,1.2 The roob are ,1.1 = + 11,1. = + O. Its characteristic equation is 100 = O. ·10 and ,1.2 = -100. The corresponding general solution of the homogeneous ODE is Step 2. Particular solution Ip of (1). We calculate the reactance S = 40 - 114 = 39.75 and the steady-state current 11'(1) = a cos 400f ~ /J sin 4001 with coefficients obtained from (4) -100· 39.75 ]]2 + 39.752 = -2.3368, [[= b= II 100·11 2 2 = 0.6467. + 39.75 Hence in our present case, a general solution of the nonhomogeneous ODE (1) is + 1(1) = C1e-lOt (6) C2e-lOOt - 2.3368 cos 400f + 0.6467 sin 4(Xlf. Step 3. Particular solution satisfying the initial conditions. How to use Q(O) = O? We finally detennine and C2 from the initial conditions 1(0) = 0 and Q(O) = O. From the first condition and (6) we have (7) 1(0) = ("1 Furthermore, using (1 ') with obtain + L1'(0) Differentiating (6) and setting l' (0) = - IOc1 - 100c2 /(t) = + R'O + f = 0, I C'O c1 = hence 0, 1'(0) = (1'», we o. we thus obtain + 0 + 0.6467' 4(Xl The solution of this and (7) is hence c2 - 2.3368 = 0, = 0 and noting that the integral equals QU) (see the formula before f C1 = 0, hence - Hlc 1 = HXJ(2.3368 - C1) - 258.68. = -0.2776, C2 = 2.6144. Hence the answer is -0.:!.776e- lOt + 2.61..J4e- lOOt - 2.3368 cos 400f + 0.6467 sin 400f. Figure 62 on p. 96 shows I(t) as well as 11'(1), which practically coincide, except for a very short time near f = 0 because the exponential terms go to zero very rapidly. Thus after a velY short time the current will practically execute harmonic oscillations of the input frequency 63~ Hz = 63~ cycles/sec. Its maximum amplitude and phase lag can be ,een from (5), which here takes the form 11'(1) = 2.4246 sin (400f - 1.3(08). • 96 CHAP. 2 Second-Order Linear ODEs y 3 , A . ,, ,, , , , ,,, ,,, ,, \ \ \ \ 2 , o " -1 -2 -3 Fig. 62. Transient and steady-state currents in Example 1 Analogy of Electrical and Mechanical Quantities Entirely different physical or other systems may have the same mathematical model. For instance, we have seen this from the various applications of the ODE y' = Icy in Chap. I. Another impressive demonstration of this unifyi1lg power of mathematics is given by the ODE (I) for an electric RLC-circuit and the ODE (2) in the last section for a mass-spring system. Both equations L/" I + RI' + C I = EOW cos wt and my" + cy' + ky = Fo cos wt are of the same form. Table 2.2 shows the analogy between the various quantities involved. The inductance L corresponds to the mass 111 and, indeed, an inductor opposes a change in current, having an "inertia effect" similar to that of a mass. The resistance R corresponds to the damping constant c, and a resistor causes loss of energy, just as a damping dashpot does. And so on. This analogy is strictly quantitative in the sense that to a given mechanical system we can construct an electric circuit whose current will give the exact values of the displacement in the mechanical system when suitable scale factors are introduced. The practical impOltallce of this analogy is almost obvious. The analogy may be used for constructing an "'electrical moder· of a given mechanical model, resulting in substantial savings of time and money because electric circuits are easy to assemble, and electric quantities can be measured much more quickly and accurately than mechanical ones. Table 2.2 Analogy of Electrical and Mechanical Quantities Electncal System Inductance L Resistance R Reciprocal lIC of capacitance Derivative Eow cos wt of } electromotive force Current I(t) Mechanical System Mass 111 Damping constant c Spring modulus k Driving force Fo cos wt Displacement yet) SEC. 2.9 97 Modeling: Electric Circuits -. 1. (RL-circuit) Model the RL-circuit in Fig. 63. Find the general solution when R. L. E are any constants. Graph or sketch solutions when L = 0.1 H. R = 5 D. E = 12V. Current l(t) c 2. (RL-circuit) Solve Pmb. 1 when E = Eo sin wt and R, L, Eo, ware arbitrary. Sketch a typical solution. 3. (RC-circuit) Model the RC-circuit in Fig. 66. Find the current due to a constant E. 4. (RC-circuit) Find the current in the RC-circuit in Fig. 66 with E = Eo sin wt and arbitrary R. C. Eo> and w. Fig. 67. 5. (LC-circuit) This is an RLC-circuit with negligibly small R (analog of an undamped mass-spring system). Find the current when L = 0.2 H. C = 0.05 F, and E = sin r V, assuming zero initial current and charge. 6. (LC-circuit) Find the current when L = 0.5 H. C = 8 . 10-4 F, E = [2 V and initial current and charge zero. 17-91 RL-circuit Current Btl 5 4 * 3 2t---~~~====----1 ' o 0.02 0.04 Fig. 64. 0.06 0.08 RLC-CIRCUITS (FIG. 60, P. 92) 7. (Tuning) In runing a stereo system to a radio station, we adjust the tuning control (tum a knob) that changes C (or perhaps L) in an RLC-circuit so that the amplitude of the steady-state current (5) becomes maximum. For what C will this happen? 8. (Transient current) Prove the claim in the text that if R 0 (hence R > 0). then the tr~msient cun'ent appmaches Ip as r -'; x. L Fig. 63. Current 1 in Problem 3 0.1 Currents in Problem 1 9. (Cases of damping) What are the conditions for an RLC-circuit to be (I) overdamped. (Il) critically damped. (III) underdamped? What is the critical resistance Rcrit (the analog of the critical damping constant 2v;;;i)? 110-121 Find the steady-state current in the RLC-circuit in Fig. 60 on p. 92 for the given data. (Show the details of your work.) 0.5 H. C = 0.1 F. E = 100 sin 2t V 0.25 H, C = 5' 10-5 F, E = 1I0 V 12. R = 2 D, L = I H, C = 0.05 F, E = 1~7 sin 3r V lO. R 11. R 121t = = 8 D, L 1 D, L = = @-g; I Fig. 65. Typical current I = e- o.lt in Problem 2 + sin (t - ~7T) Find the transient current (a general solution) in the RLC-circuit in Fig. 60 for the given data. (Show the details of your work.) 13. R = 6 D, L = 0.2 H. C = 0.025 F. E = 110 sin lOr V R 14. R = 0.2 D, L = 0.1 H, C = 2 F. E = 754 sin 0.51 V 15. R = 1110 D, L = 112 H. C = 100113 F, E = e- 4t (1.932 cos ~r + 0.246 sin ~r) V I~~ c Fig. 66. RC-circuit Solve the initial l'alue problem for the RLC-circuit in Fig. 60 with the given data. assuming zero initial curren! and charge. Graph or sketch the solution. (Sho\\ the details of your worL) 98 CHAP. 2 16. R = 17. R = E = 18. R = E = Second-Order Linear ODEs 4 fl, L = 0.1 H, C = 0.025 F, E = 10 sin lOt V (b) The complex impedance Z is defined by 6 fl, L = 1 H. C = 0.04 F. 600 (cos t + 4 sin t) V Z = R 3.6 n. L = 0.2 H, C = 0.0625 F, 164 cos lOt V + is = R + ;(WL - ~c). Show that K obtained in (a) can be written as 19. WRITING PROJECT. Analogy of RLe-Circuits and Damped Mass-Spring Systems. (a) Write an essay of 2-3 pages based on Table 2.2. Describe the analogy in more detail and indicate its practical significance. (b) What RLC-circuit with L = I H is the analog of the mass-spring system with mass 5 kg, damping constant 10 kg/sec, spring constant 60 kglsec2 , and driving force 220 cos lOt? (c) Illustrate the analogy with another example of your own choice. 20. TEAM PROJECT. Complex Method for Particular Solutions. (a) Find a particular solution of the complex ODE K= Eo iZ Note that the real part of Z is R. the imaginary part is the reactance S, and the absolute value is the impedance Izi = V R2 + S2 as defined before. See Fig. 68. (c) Find the steady-state solutLm of the ODE [" + 2/' + 31 = 20 cos t, first by the real method and then by the complex method, and compare. (Show the details of your work.) (d) Apply the complex method to an RLC-circuit of your choice. (8) by substituting Ip = Ke'wt (K unknown) and its derivatives into (8), and then take the real part Ip of Ip. showing that Ip agrees with (2), (4). Hint. Use the Euler formula e iwt = cos wt + ; sin wt [(11) in Sec. 2.2 with wt instead of tl Note that Eow cos wt in (1) is the real part of Eowe'wt in (8). Use ;2 = -1. 2.10 Solution R Fig. 68. Real axis Complex impedance Z by Variation of Parameters We continue our discussion of nonhomogeneous linear ODEs (1) y" + p(x»),' + q(x)y = rex). In Sec. 2.6 we have seen that a general solution of (1) is the sum of a general solution )'1, of the corresponding homogeneous ODE and any particular solution yp of (1). To obtain),p when r(x) is not too complicated, we can often use the method of 1I1ldetenllined coefficie11fs. as we have shown in Sec. 2.7 and applied to basic engineering models in Secs. 2.8 and 2.9. However, since this method is restricted to functions r(x) whose derivatives are of a form similar to r(x) itself (powers. exponential functions. etc.). it is desirable to have a method valid for more general ODEs (I)' which we shall now develop. It is called the method of variation of parameters and is credited to Lagrange (Sec. 2.1). Here p, q. r in (1) may be v31iable (given functions of x). but we assume that they are continuous on some open interval I. Lagrange's method gives a particular solution Yp of (1) on I in the form (2) SEC. 2.10 99 Solution by Variation of Parameters where )'1> .\'2 form a basis of solutions of the corresponding homogeneous ODE .\''' + p(x)y' + q(x)y (3) 0 = on I. and W is the Wronskian of YI • .\'2. (4) (see Sec. 2.6). CAUTION! The solution formula (2) is obtained under the assumption that the ODE is written in standard form. with y" as the flfst term as shown in 0). If it starts with f(x)y". divide first by f(x). The integration in (2) may often cause difficulties, and so may the determination of YI . .\'2 if (I) has variable coefficients. If you have a choice. use the previous method. It is simpler. Before deriving (2) let us work an example for which you do need the new method. (Try otherwise.) E X AMP L E 1 Method of Variation of Parameters Solve the nonhomogeneous ODE V" +v = sec x = cos x Solulion. A basis of solutions of the homogeneous ODE on any interval is .1'1 gives the Wronskian lIl(YI' .1'2) = cos x cos = cos x. 1'2 = sin x. This sin x (-sin x) = I. T - From (2). choosing zero constants of integration. we get the particular solution of the given ODE Yp = -cos xfSin x sec x dx + sin xfcos x sec x dr (Fig. 69). = cos x In Icos xl + x sin x Figure 69 shows Yp and its first term, which is small, so that x sin x essentially determines the shape of the curve of )'p. (Recall from Sec. 2.8 that we have seen x sin x in connection with resonance. except for notation.) From yp and the general solution Yh = eIYI + C2.1'2 of the homogeneous ODE we obtain the answer .I' = Yh + Yp = (el + In Icosxl) cosx + (e2 + x) sinx. Had we included integration constants -Cl' c2 in (2), then (2) would have given the additional + e2 sin x = eLYI + e2Y2, that is, a general solution of the given ODE directly from (2). This will always be the case. • ci cos x y 10 5 0 -5 -10 Fig. 69. D V I 2 8 10 12 \J x Particular solution yp and its first term in Example 1 100 CHAP. 2 Second-Order Linear ODEs Idea of the Method. Derivation of (2) What idea did Lagrange have? What gave the method the name? Where do we use the continuity assumptions? The idea is to start from a general solution of the homogeneous ODE (3) on an open interval I and to replace the constants ("the parameters") Cl and C2 by functions u(x) and vex): this suggests the name of the method. We shall determine u and v so that the resulting function (5) is a particular solution of the nonhomogeneous ODE (1). Note that Yh exists by Theorem 3 in Sec. 2.6 because of the continuity of p and q on l. (The continuity of T will be used later.) We determine u and v by substituting (5) and its derivatives into (1). Differentiating (5). we obtain v' = u'v.J + uv' .1 + v\ .2 + vv'. .2 _p Now yp must satisfy (I). This is one condition for lYvo functions u and v. It seems plausible that we may impose a second condition. Indeed, our calculation will show that we can detelmine u and v such that Yp satisfies (1) and u and v satisfy as a second condition the equation (6) This reduces the first derivative Y; to the simpler form , (7) Yp = , UYI , + VY2' Differentiating (7), we obtain " (8) Yp = u "YJ + UYII, + V "Y2 + VY2'" We now substitute yp and its derivatives according to (5), (7), (8) into (1). Collecting terms in u and terms in v, we obtain Since Yl and )'2 (9a) Equation (6) is (9b) are solutions of the homogeneous ODE (3), thIs reduces to u'y{ + v'y~ = T. SEC. 2.10 Solution by Variation of Parameters 101 This is a linear system of two algebraic equations for the unknown functions u' and v' We can solve it by elimination as follows (or by Cramer's rule in Sec. 7.6). To eliminate v', we multiply (9a) by -Y2 and (9b) by y~ and add, obtaining thus Here, W is the Wronskian (4) of Yl' Y2' To eliminate u' we multiply (9a) by Yl' and (9b) by -)'~ and add. obtaining thus Since h, )'2 form a basis, we have W,* 0 (by Theorem 2 in Sec. 2.6) and can divide by W, u (10) , v' By integration, u= - I yr .~ dx, v = I Wdx. "Ir These integrals exist because rex) is continuous. Inserting them into (5) gives (2) and completes the derivation. • 11-171 - .. -- lA _ _ . •• .... GENERAL SOLUTION Solve the given nonhomogeneous ODE by variation of parameters or undetennined coefficients. Give a geneml solution. (Show the details of your work.) 1. Y" + Y = csc x 2. y" - 4y' + 4)' 3. x 2 )''' - 2x),' + 2y = x 3 cos x 4. ,," - 2y' + Y = 5. y" + Y = tan x 6. x 2)''' - xy' 7. yo" +Y x 2ex = = +Y cos x eX sin x In sec x = x + Ixl 8. y" - 4/ + 4-," = 12e 2xlx 4 9. (D2 - 2D + l)y = x 2 + x- 2e x 10. (D 2 - l)y = I1cosh x 11. (D 2 + 4l)y = cosh 2x 12. (x 2D2 + xD - al)y = 3x- 1 + 3x 13. (x 2D2 - 2xD + 2l)y = x 3 sin x 14. (x 2D2 + xD - 4l)y = 11.>:2 15. (D 2 + l)y = sec x - 10 sin 5x 16. (x 2D2 + xD + (x 2 - a)l)y = X 3/2 cos x. Hint. To find)'1> )'2 set Y = UX- 1I2 . 17. (x 2D2 + xD + (x 2 - !)l)y = x 3/2 sin x. Him: As in Prob. 16. 18. TEAM PROJECT. Comparison of Methods. The undetermined-coefficient method should be used whenever possible because it is simpler. Compare it with the present method as follows. (a) Solve y" + 2/ - 15)' = 17 sin 5x by both methods, showing all details, and compare. (h) Solve y" + 9)" = r 1 + /2, rl = sec 3x, r2 = sin 3x by applying each method to a suitable function on the right. (e) Invent an undetermined-coefficient method for by nonhomogeneous Euler-Cauchy equations experimenting. 102 CHAP. 2 Second-Order Linear ODEs :a.' • II-$. 1. What general properties make linear ODEs particularly attractive? 2. What is a general solution of a linear ODE? A basis of solutions? 3. How would you obtain a general solution of a nonhomogeneous linear ODE if you knew a general solution of the corresponding homogeneous ODE? 4. What does an initial value problem for a second-order ODE look like? 5. What is a paI1icuiar solution and why is it more common than a general solution as the answer to practical problems? 6. Why are second-order ODEs more important in modeling than ODEs of higher order? 7. Describe the applications of ODEs in mechanical vibrating systems. What are the elecuical analogs of those systems? 8. If a construction, such as a bridge, shows undesirable resonance. what could you do? 19-181 GENERAL SOLUTION Find a general solution. Indicate the method you are using and sho" the details of your calculation. 9. y" 2/ Sy = 52 cos 6x 10. y" + 6/ + 9y = e- 3x - 27x 2 11. y" + S/ + 25y = 26 sin 3x 12. yy" = 2/ 2 13. (x 2D2 + 2xD - 121)y = Jfx 3 14. (x 2D2 + 61:D + 6/).,· = x 2 15. (D 2 - 2D + I)y = x- 3 e x 16. (D 2 - 4D + 5l)y = e 2x csc x 17. (D 2 - 2D + 21)y = e 7 esc x 18. (4x 2D2 - 24xD + 49/)y = 36x 5 119-251 TIONS AND PROBLEMS 126--341 APPLICATIONS 26. Find the steady-state, solution of the system in Fig. 70 when 111 = 4. c = 4. k = 17 and the driving force is 102 cos 3f. 27. Find the motion of the system in Fig. 70 with mass 0.25 kg. no damping. spring constant I kg/sec2 , and driving force 15 cos O.5f - 7 sin l.5f nt. assuming zero initial displacement and velocity. For what frequency of the driving force would you get resonance? 28. In Prob. 26 find the solution corresponding to initial displacement 10 and initial velocity O. 29. Show that the system in Fig. 70 with 111 = 4, c = O. k = 36, and driving force 61 cos 3.1 T exhibits beats. Him: Choose zero initial conditions. 30. In Fig. 70 let 111 = 2. c = 6, k = 27, and r(t) = 10 cos wf. For what w will you obtain the steadystate vibration of maximum possible amplitude? Determine this amplitude. Then use this wand the undetelmined-coefficient method to see whether you obtain the same amplitude. 31. Find an electrical analog of the mass-spring system in Fig. 70 with mass 0.5 kg, spling constant 40 kg/sec2 , damping constant 9 kg/sec. and driving force 102 cos 6f nL Solve the analog. assuming Lew initial current and charge. 32. Find the current in the RLC-circuit in Fig. 71 when L = 0.1 H. R = 20 n, C = 2· 10-4 F. and E(t) = 11 0 sin 415t V ( 66 cycles/sec). 33. Find the current in the RLC-circuit when L = 0.4 H, R = 40 n. C = 10- 4 F, and E(t) = 220 sin 314t V (50 cycles/sec). 34. Find a pat1icular solution in Prob. 33 by the complex method. (See Team Project 20 in Sec. 2.9.) INITIAL VALUE PROBLEMS Solve the following initial value problems. Sketch or graph the solution. (Show the details of your wOIk) 19. y" + 5y' 14y = 0, yeO) = 6, y' (0) = -·6 20. y" + 6y' + ISy = 0, yeO) = 5. /(0) = -21 21. x 2 y" - xy' - 24y = 0, y(l) = 15, y'(1) = 0 22. x 2 y" + 15x/ + .J.9\" = o. yO) = 2. y'(1) = -II 23. y" + 5/ + 6y = IOSx 2 , yeO) = IS, y' (0) = -26 24. -,," + y' + 2.5y = 13 cos x, yeO) = S.O, y' (0) = 4.5 25. (x 2D2 + xD - 4l)y = x 3 , yO) = -4/5, y' (I) = 93/5 E(t) Fig. 70. Mass-spring system Fig. 71. RLC-circuit 103 Summary of Chapter 2 Second-Order Linear ODEs Second-order linear ODEs are particularly important in applications. for instance. in mechanics (Sees. 2A. 2.8) and electrical engineering (Sec. 2.9). A second-order ODE is called linear if it can be written (1) + y" p(x)y' + = q(x)y rex) (Sec. 2.1). (If the first term is, say. f(x)y", divide by f(x) to get the "standard form" (1) with -,," as the first term.) Equation (1) is called homogeneous if r(x) is zero for all x considered, usually in some open interval; this is written rex) "" O. Then (2) .v" + p(x»)" + q(x)y = O. Equation (I) is called nonhomogeneous if rex) =1= 0 (meaning rex) is not zero for some x considered). For the homogeneous ODE (2) we have the imp0l1ant superposition principle (Sec. 2.1) that a linear combination y = kY1 + 1)'2 of two solutions .'"1, Y2 is again a solution. Two linearly independent solutions ."1,)'2 of (2) on an open interval I form a basis (or fundamental system) of solutions on I. and y = C1Y1 + C2Y2 with arbitrary constants (\, C2 is a general solution of (2) on I. From it we obtain a particular solution if we specify numeric values (numbers) for C1 and C2. usually by prescribing two initial conditions (xo. Ko. K1 given numbers; Sec. 2.1). (3) (2) and (3) together fonn an initial value problem. Similarly for (I) and (3). For a nonhomogeneous ODE (1) a general solution is of the form (4) .v = •vh (Sec. 2.7). +". p Here Yh is a general ~olution of (2) and Yp is a particular solution of (I). Such a yp can be determined by a general method (variation of parameters, Sec. 2.10) or in many practical cases by the method of undetermined coefficients. The latter applies when (I) has constant coefficients p and q, and r{x) is a power of x. sine, cosine, etc. (Sec. 2.7). Then we write (1) as (5) y" + a/ + The corresponding homogeneous ODE y' where ,\. is a root of (6) ,\.2 by = rex) + ay' + + a'\' + b = O. by (Sec. 2.7). = 0 has solutions y = eAX• 104 CHAP. 2 Second-Order Linear ODEs Hence there are three cases (Sec. 2.2): Case Type ot Roots - I 11 III General Solution -Y = cleA1X + C2 eA2X Y = (Cl + c 2 x)e- ax/ 2 y = e- ux/ 2 (A eos w*x + B sin w*x) -- - Distinct real Ab A2 1 Double -"2a Complex -~(/ ± iw* Important applications of (5) in mechanical and electIical engineering in connection with vibrations and resonance are discussed in Secs. 2.4. 2.7. and 2.8. Another large class of ODEs solvable "algebraically" consists of the Euler-Cauchy equations (Sec. 2.5). (7) These have solutions of the form y equation (~) 1112 + = (a - x1n, where l)m 111 +b= is a solution of the auxiliary O. Existence and uniqueness of solutions of (1) and (2) is discussed in Sees. 2.6 and 2.7, and reduction of order in Sec. 2.1. •••• CHAPTER 3 Higher Order Linear ODEs In this chapter we extend the concepts and methods of Chap. 2 for linear ODEs from order n = 2 to arbitrary order n. This will be straightforward and needs no new ideas. However, the formulas become more involved, the variety of roots of the characteristic equation (in Sec. 3.2) becomes much larger with increasing n, and the Wronskian plays a more prominent role. Prerequisite: Secs. 2.1, 2.2. 2.6. 2.7, 2.10. References Gnd Answers to Problems: App. I Pm1 A. and App. 2. 3.1 Homogeneous Linear ODEs Recall from Sec. 1.1 that an ODE is of Ilth order if the nth derivative yen> = dnyldx rt of the unknown function y(x) is the highest occurring delivative. Thus the ODE is of the form F(x, y, y', ... . /n» = 0 where lower order derivatives and y itself mayor may not occur. Such an ODE is called linear if it can be written (1) In> + Pn_l(X)y<n-D + ... + Pl(X)y' + Po(x)y = rex). (For n = 2 this is 0) in Sec. 2.1 with 17] = P and Po = q). The coefficients Po, ... , Pn-l and the function r on the right are any given functions of x, and y is unknown./n > has coefficient I. This is practical. We call this the standard form. (If you have Pn(x)/n>, divide by Pn(x) to get this form.) An nth-order ODE that cannot be written in the form (1) is called nonlinear. If r(x) is identically zero, r(x) == 0 (zero for all x considered. usually in some open interval I). then (1) becomes (2) In) + Pn_l(X)y<n-D + ... + Pl(X)y' + Po(x)y = 0 and is called homogeneous. If rex) is not identically zero. then the ODE is called nonhomogeneous. This is as in Sec. 2.1. A solution of an nth-order (linear or nonlinear) ODE on some open interval/is a function), = hex) that is defined and 11 times differentiable on I and is such that the ODE becomes an identity if we replace the unknown function y and its derivatives by h and its corresponding derivatives. 105 106 CHAP_ 3 Higher Order Linear ODEs Homogeneous Linear ODE: Superposition Principle, General Solution Sections 3_1-3_2 will be devoted to homogeneous linear ODEs and Sec. 3.3 to nonhomogeneous linear ODEs_ The basic superposition or linearity principle in Sec_ 2_1 extends to nth order homogeneous linear ODEs as follows_ THEOREM 1 Fundamental Theorem for the Homogeneous Linear ODE (2) For a homogeneous linear ODE (2), sums and constant multiples of solutions on some open i1lferval 1 are again solutions 011 l. (This does not hold for a nonhomogeneous or nonlinear ODE!) The proof is a simple generalization of that in Sec. 2_1 and we leave it to the student_ Our further discussion parallels and extends that for second-order ODEs in Sec. 2_1_ So we define next a general solution of (2), which will require an extension of linear independence from 2 to n functions_ DEFINITION General Solution, Basis, Particular Solution A general solution of (2) on an open intervall is a solution of (2) on 1 of the form (3) (CI, - - - , C n arbitrary) where Y1- - - - , Yn is a basis (or fundamental system) of solutions of (2) on l: that is, these solutions are linearly independent on l, as defined below_ A particular solution of (2) on 1 is obtained if we assign specific values to the n constants CI' - - - , cn in (3)_ DEFINITION Linear Independence and Dependence functions YI(X), - - - • Yn(x) are called linearly independent on some interval 1 where they are defined if the equation 11 (4) on 1 implies that all kI , - - - , kn are zero_ These functions are called linearly dependent on I if this equation also holds on I for some kb - - - , k n not all zero_ (As in Secs_ 1.1 and 2_1, the arbitrary constants CI, ••• , Cn must sometimes be restricted to some interval.) If and only if .\'1, - - - , Yn are linearly dependent on I, we can express (at least) one of these functions on I as a "linear combination" of the other n - I functions. that is, as a sum of those functions. each multiplied by a constant (zero or not)_ This motivates the term "linearly dependent." For instance, if (4) holds with k1 -=1= 0, we can divide by ki and express YI as the linear combination SEC. 3.1 107 Homogeneous Linear ODEs Note that when n = 2, these concepts reduce to those defined in Sec. 2.1. E X AMP L E 1 Linear Dependence Show that the functions .1'1 Solution. E X AMP L E 2 • Y2 = 0.1'1 + 2.5)'3· This proves linear dependence on any interval. Linear Independence = ..2'.\"3 = x 3 are linearly independent on any interval. for instance, on -I ::'" x ::'" 2. Show that.\"1 = Solution. Equation (4) is k2 E X AMP L E 3 = x 2 , Y2 = 5x, .1'3 = 2x are linearly dependent on any interval. X, .1'2 k 1x + k2X2 + k3X3 = O. Taking (aL~ = -1. (b) x = I. (c) x = 2. we get = 0 from (a) + (b). Then k3 = 0 from (c) -2(b). Then kl = 0 from (b). This proves linear independence. • A better method for testing linear independence of solutions of ODEs will soon be explained. General Solution. Basis Solve the fourth-order ODE /v Solution. 5y" + 4.'" = 0 As in Sec. 2.2 we try and substitute .I' = e Ax . Omitting the common factor eA.'", we obtain the characteristic equation This is a quadratic equation in J.L ~ A2, namely, The roots are J.L interval is = 1 and 4. Hence A = -2. -I. I. 2. This give, four ,olutions. A general solution on any provided those four solutions are linearly independent. This is true but will be shown later. • Initial Value Problem. Existence and Uniqueness An initial value problem for the ODE (2) consists of (2) and 11 initial conditions (5) with given Xo in the open interval I considered. and given Ko, ... , KIl - 1. In extension ofthe existence and uniqueness theorem in Sec. 2.6 we now have the following. THEOREM 2 Existence and Uniqueness Theorem for Initial Value Problems If the coefficients Po(x), ... , Pn-l(x) of (2) are continuous on some open interred I and Xo is in I. then the initial value problem (2), (5) has a unique solution y(x) on 1. Existence is proved in Ref. [All] in App. l. Uniqueness can be proved by a slight generalization of the uniqueness proof at the beginning of App. 4. 108 E X AMP L E 4 CHAP. 3 Higher Order Linear ODEs Initial Value Problem for a Third-Order Euler-Cauchy Equation Solve the following initial value problem on any open intervall on the positive x-axis containing .T = 1. Solution. Step 1. General solution. As in Sec. 2.5 we try /11(111 - 1)(111 - 2)~m y = x"'. - 3111(111 - Ih m /'(1) = -4. /(1)= I, y(1) = 2, + 611n By differentiation and 'Llb,titution. 7n - 6xm = O. Dropping xm and ordering gives 111 3 - 6n? + 11m - 6 = O. If we can guess the root III = I. we can divide by III - I and find the other roots 2 and 3, thus obtaining the solutions x. x 2 , x 3 . which are linearly independent on 1 (see Example 2). [Tn general one shall need a root-finding method, such as Newton's (Sec. 19.2), also available in a CAS (Computer Algebra System).] Hence a general solution is 3 valid on any interval I. even when it includes x = 0 where the coefficients of the ODE divided by x (to have the standard form) are not continuous. Step 2. Pwticular solution. The derivatives are.v' = Cl + 2c2x + 3C3X2 and ,," = 2C2 + 6C3X, From this and y and the initial conditions we get by setting X = I (a) yO) = c1 + C2 + C3 + 2("2 + 3("3 = 2C2 + 6c3 (b) l(1) = cl (c) /'(1) = = = 2 -4. This is solved by Cramer's rule (Sec. 7.6), or by elimination. which is simple, as follows. (b) - (a) gives (d) C2 + 2C3 = -I. Then (e) - 2(d) gives C3 = -l. Then (e) gives C2 = I. Finally Cl = 2 from (a). • Answer: y = 2, + x 2 - x 3 . Linear Independence of Solutions. Wronskian Linear independence of solutions is crucial for obtaining general solutions. Although it can often be seen by inspection. it would be good to have a criterion for it. Now Theorem 2 in Sec. 2.6 extends from order n = 2 to any n. This extended criterion uses the Wronskian W of n solutions )'1, . . • , Yn defined as the nth order determinant Note that W depends on x since )'1, . . . , )'n does. The criteIion states that these solutions form a basis if and only if W is not zero: more precisely: THEOREM 3 Linear Dependence and Independence of Solutions Let the ODE (2) have continuous coefficients Po(x) . .. '. Pn-l(x) all all open interval l. Then n solutions Yb ... , Yn of (2) on 1 are linearly depelldent on 1 if and ollly if their Wronskian is zero for some x = Xo in T. Furtlzenl1ore, if W is zero jar x = X o, thell W is identically zero Oil I. Hence if there is an Xl in 1 at which W is Ilot ::,ero. thell Y1> ... , Yn are linearly indepel1dellf all I. so that they f01111 a basis of solutions of (2) all T. SEC 3.1 109 Homogeneous Linear ODEs PROOF (a) Let.\"1 ..... Yn be linearly dependent solutions of (2) on I. Then. by definition. there are constants kI , . . . , k n not all zero, such that for all x in I. (7) By n - I differentiations of (7) we obtain for all x in I =0 (8) (7). (8) is a homogeneous linear system of algebraic equations with a nontlivial solution kb ... , kn . Hence its coefficient determinant must be zero for every x on I, by Cramer's theorem (Sec. 7.7). But that determinant is the Wronskian W, as we see from (6). Hence W is zero for every x on I. (b) Conversely, if W is zero at an Xo in I, then the system (7), (8) with x = Xo has a solution kI *, ... , kn *, not all zero, by the same theorem. With these constants we define the solution y* = kI*YI + .,. + kn*Yn of (2) on I. By (7), (8) this solution satisfies the initial conditions y*(x o) = 0, ... , y*(n-l>(xo) = O. But another solution satisfying the same conditions is y == O. Hence y* ==)' by Theorem 2, which applies since the coefficients of (2) are continuous. Together, y* = kI*YI + ... + k n *)'n == 0 on I. This means linear dependence of YI' .... )In on I. (c) If W is zero at an Xo in T, we have linear dependence by (b) and then W == 0 by (a). Hence if W is not zero at an Xl in I, the solutions YI, ... , Yn must be linearly independent 001. • E X AMP L E 5 Basis, Wronskian We can now prove that in Example 3 we do have a basis. In evaluating W. pull out the exponential functions columnwise. In the result. subtract Column I from Columns 2. 3. 4 lwithout changing Column I). Then expand by Row 1. In the resulting third-order determinant. subtract Column I from Column 2 and expand the result by Row 2: e- 2 .-.:: e- x eX e -2e -2x -e -x eX 2e 2x -2 -2x e- x e·x 4e 2x 4 W= 4e -8e -2x -e-x eX 2x 8e2x 3 -8 -I 4 2 -3 -3 7 9 0 = 72. 4 -I 16 • 8 A General Solution of (2) Includes All Solutions Let us first show that general solutions always exist. Indeed. Theorem 3 in Sec. 2.6 extends as follows. THEOREM 4 Existence of a General Solution If the coefficients Po(x), .. '. Pn-I(x) 0/(2) are continuous on some opell interval I, then (2) lzas a general solution Oil I. CHAP. 3 110 PROOF Higher Order Linear ODEs We choose any fixed Xo in I. By Theorem 2 the ODE (2) has n solutions )"1' •••• y", where)J satisfies initial conditions (5) with K j - 1 = 1 and all other K"s equal to zero. Their Wronskian at Xo equals 1. For instance, when 11 = 3, then Yl(XO) = 1, y~(xo) = 1, y~(xo) = I, and the other initial values are zero. Thus, as claimed, W(."ICt O)' .\'2(XO), Y3(XO») = 0 )'1(XO) Y2(XO) ."3(XO) Y~(xo) y~(xo) .\'~(xo) 0 y~(xo) Y~Cto) y~(xo) 0 0 0 1. 0 Hence for any n those solutions YI- .... Yn are linearly independent on I, by Theorem 3. They form a basis on I. and y = CIY! + ... + CnY" is a general solution of (2) on I. • We can now prove the basic prope11y that from a general solution of (2) every solution of (2) can be obtained by choosing suitable values of the arbitrary constants. Hence an nth order linear ODE has no singular solutions, that is, solutions that cannot be obtained from a general solution. THEOREM 5 General Solution Includes All Solutions If the ODE (2) has continuolls coefficients Po(x). ... , P,,-1 (x) on some open interval T, then e\'ery solution." = Y(x) of (2) 011 T is of the foml (9) where ."1' ... , y" is a basis of solutions of (2) on T al/d C 1, ... , C n are suitable cOl/sta1/ts. PROOF Let Y be a given solution and y = ('1)'1 + ... + c"."n a general solution of (2) on I. We choose any fixed Xo in I and show that we can find constants CI> •• '. Cn for which y and its first 1/ - 1 derivatives agree with Y and its corresponding derivatives at xo. That is, we should have at x = .\'0 + =Y , + cn )' n = y' (10) But this is a linear system of equations in the unknowns Cl, cn . Its coefficient determinant is the Wronskian W of Yl, ... , Yn at xo. Since .\'1 •... , y" form a basis. they are linearly independent, so that W is not zero by Theorem 3. Hence (10) has a unique solution ('1 = CI> ... , Cn = Cn (by Cramer's theorem in Sec. 7.7). With these values we obtain the particular solution on I. Equation (10) shows that y* and its first n - I derivatives agree at xo with Yand its corresponding derivatives. That is, y* and Y satisfy at xo the same initial conditions. SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 111 The uniqueness theorem (Theorem 2) now implies that y* theorem. Y on T. This proves the • This completes our theory of the homogeneous linear ODE (2). Note that for identical with that in Sec. 2.6. This had to be expected. - .•...--.......-. l ]~ - .. - 11. x To get a feel for higher order ODEs. show that the given functions are solutions and form a basis on any interval. Use Wronskians. (In Prob. 2. x> 0.) .1'iv = 0 l. I, x, x 2 , x 3 , 2 2. 1, x , X4, x2ylll - 3x.1''' + 3.1" = 0 3. eX, xe x . x 2e x . ylll - 3.1''' + 3y' - .1' = 0 4. e2x cos x. e2x sin x, e- 2x cos x, e- 2x sin x, yiv - 6y" + 25y = O. S111 3x, yiv + 9y" = 0 6. TEAM PROJECT. General Properties of Solutions of Linear ODEs. These properties are important in obtaining new solutions ii'om given ones. Therefore extend Team Project 34 in Sec. 2.2 to 11th-order ODEs. Explore statements on sums and multiples of solutions of (1) and (2) systematically and with proofs. Recognize clearly that no new ideas are needed in this extension from 11 = 2 to general 11. 7-191 LINEAR INDEPENDENCE AND DEPENDENCE Are the given functions linearly independent or dependent on the positive x-axis? (Give a reason.) 7. I, eX, e- x 8. x + I. x + 2. x 9. In x, In x 2 • (In X)2 10. e", e- x , sinh 2x 3.2 = 2 it is -- TYPICAL EXAMPLES OF BASES 5. I, x, cos 3x, 11 2 • xlxl. x 12. x. I/x. 0 13. sin 2x. sin x. cos x 14. cos 2 x, sin 2 x, cos 2x 15. tan x. cot x. I 16. (x - 17. sin x, sin ~x 18. cosh x, sinh x, cosh 2 x 2 1)2. (x + 1)2. x 2 19. cos x, sin x. 27T 20. TEAM PROJECT. Linear Independence and Dependence. (a) Investigate the given question about a set 5 of functions on an intervall. Give an example. Prove your answer. (I) If 5 contains the zero function, can 5 be linearly independent? (2) If 5 is linearly independent on a subinterval J of I. is it linearly independent on l? (3) If 5 is linearly dependent on a subinterval J of I. is it linearly dependent on n (4) If 5 is linearly independent on I, is it linearly independent on a subinterval J? (5) If 5 is linearly dependent on 1. is it linearly independent on a subinterval J? (6) If 5 is linearly dependent on I, and if T contains 5, is T linearly dependent on l? (b) In what cases can you use the Wronskian for testing linear independence? By what other means can you perform such a tcst'? Homogeneous Linear ODEs with Constant Coefficients In this section we consider nth-order homogeneous linear ODEs with constant coefficients, which we write in the form (1) where lnJ d'\/dr: n • etc. We shall see that this extends the case 11 = 2 discussed in Sec. 2.2. Substituting y = e AX (as in Sec. 2.2), we obtain the characteIistic equation (2) CHAP. 3 112 Higher Order Linear ODEs of (1). If A is a root of (2), then y = e AX is a solution of (1). To find these roots, you may need a numeric method, such as Newton's in Sec. 19.2, also available on the usual CASso For general 11 there are more cases than for 11 = 2. We shall discuss all of them and illustrate them with typical examples. Distinct Real Roots If all the 11 roots AI' .... An of (2) are real and different, then the 11 solutions (3) Yn =e An X constitute a basis for all x. The corresponding general sulution of (1) is (4) Indeed, the solutions in (3) are linearly independent, as we shall see after the example. E X AMP L E 1 Distinct Real Roots Solve the ODE y'" - 2y" - y' + 2y = O. 2A2 - A + 2 = O. II has the roots - I, I, 2; if you find one of them by inspection. you can obtain the other two roots by solving a quadratic equation (explain!). The corresponding general solution (4) is y = Cle -.< + C2ex + C3e2x. • Soluti01l. The characteristic equation is A3 - Linear Independence of (3). Students familiar with 11th-order determinants may verify that by pulling out all exponential functions from the columns and denoting their product by E, thus E = exp [(AI + ... + An)x], the Wronskian of the solutions in (3) becomes eA1X AleA1X e AnX e A2X A2 e A2X An eAnX AI2eAIX 2 A2X A2 e An2eAnx AI'-leA1X A2:-leA2X A;:-le AnX W= (5) 1 =E Al A2 An AI2 A22 A,,2 A;"-1 A2:- 1 Ann - I The exponential function E is never zero. Hence W = 0 if and only if the determinant on the right is zero. This is a so-called Vandermonde or Cauchy determinantl It can be shown that it equals 'ALEXANDRE THEOPHILE VANDERMONDE (1735-1796). French mathematician, who worked on solution of equations by determinants. For CAUCHY see footnote 4, in Sec. 2.5. SEC. 3.2 113 Homogeneous Linear ODEs with Constant Coefficients (6) where V is the product of all factors Aj - Ak withj < k (~ n); for instance, when II = 3 we get - V = -(AI - A2)(A 1 - A3)(A2 - A3). This shows that the Wronskian is not zero if and only if all the n roots of (2) are different and thus gives the following. THEOREM 1 Basis x Solutiolls YI = e AIX, • • • , )"n = /n of (1) ("with allY real or complex A/ s) f0l711 a basis of solutiollS of (I) on any opell interml if and only if all /l roots of (2) are different. Actually, Theorem I is an important special case of our more general result obtained from (5) and (6): THEOREM 2 Linear Independence Ally !lumber of solutions of (1) of the f0I711 e AX are linearly independent on an open interl'ill I if and only if the correspondillg A are all differe11l. Simple Complex Roots If complex roots occur, they must occur in conjugate pairs since the c~efficients of (1) are real. Thus, if A = 'Y + iw is a simple root of (2), so is the conjugate A = 'Y - iw, and two corresponding linearly independent solutions are (as in Sec. 2.2, except for notation) YI E X AMP L E 2 = Y2 = e1'X sin wx. e1'X cos wx, Simple Complex Roots. Initial Value Problem Solve the initial value problem ,"," - y" + 100,", - 100y = yeo) 0, = y' (0) 4, = 11, .1'''(0) = -299. The characteristic equation is A3 - A2 + 100A - 100 = O. It has the root 1, as can perhaps be seen by in~peclion. Then division by A-I ~hows that the other roots are:+: lOi. Hence a general solution and its derivatives (obtained by differentiation) are Solutioll. Y = cle x + A cos lOx + B sin lOx. y' = ('leX - lOA sin lOx + lOB cos lOx. y" = C1ex - 100A cos IOle - 100B sin lOx. From this and the initial conditions we obtain by setting x = 0 (a) C1 +A = 4. (b) c] + lOB = 11, (c) c1 - 100A = -299. We solve this system for the unknowns A. B. ('1' Equation (a) minus Equation (c) gives lOlA Then ('1 = I [mm (a) and B = I from lb). The solution is (Fig. 72) y = eX + 3 cos This gi\e~ the ~olution curve. which o_cillate, ahout eX lOx = 303, A = 3. + sin lOx. (dashed in Fig. 72 on p. ll-l). • CHAP. 3 114 Higher Order Linear ODEs y 20 10 4 °0~--~----~2----~3--X Fig. 72. Solution in Example 2 Multiple Real Roots If a real double root occurs, say, Al = A2 , then)'1 = )'2 in (3), and we take)'1 and .\)'1 as corresponding linearly independent solutions. This is as in Sec. 2.2. More generally, if A is a real root of order 111, then 111 corresponding linearly independent solutions are (7) We derive these solutions after the next example and indicate how to prove their linear independence. E X AMP L E 3 Real Double and Triple Roots Solve the ODE yV - 3iv Solution. + 3y'" - / ' = O. The characteristic equation ,,5 - "3 = 4 = "5 = I, and the answer is 3A4 + 3A3 - A2 = 0 has the roots Al = A2 = 0 and • (8) Derivation of (7). We write the left side of (\) as L[yJ = y<n) + an _ 1y<n-D + ... + aoy. Let y = e A '. Then by performing the differentiations we have Now let Al be a root of mth order of the polynomial on the right, where In ~ 11. For < 11 let A""+1> ... , An be the other roots, all different from AI' Writing the polynomial in product form. we then have 111 with h(A) = I if 111 = 11, and h(A) = (A - A71<+I) ..• (A - An) if III key idea: We differentiate on both sides with respect to A. (9) < II. Now comes the SEC 3.2 Homogeneous linear ODEs with Constant Coefficients 115 The differentiations with respect to x and A are independent and the occurring derivatives are continuous, so that we can interchange their order on the left: (10) The right side of (9) is zero for A = Al because of the factors A - A} (and m ~ 2 since we have a multiple root!). Hence L[x/'X] = 0 by (9) and (10). This proves that X/IX is a solution of (I). We can repeat this step and produce x2/"'X, ... , X"'-I/IX by another 111 - 2 such differentiations with respect to A. Going one step further would no longer give zero on the right because the lowest power of A - Al would then be (A - A})o, multiplied by m!h(A) and heAl) 0 because h(lI.) has no factors A - AI; so we get precisely the solutions in (7). We finally show that the solutions (7) are linearly independent. For a specific n this can be seen by calculating their Wronskian, which turns out to be nonzero. For arbitrary 111 we can pull out the exponential functions from the Wronskian. This gives (eAx)m = e Amx times a determinant which by "row operations" can be reduced to the Wronskian of 1. x . ... , X",-l. The latter is constant and different from zero (equal to 1 !2! ... (111 - I)!). These functions are solutions of the ODE /mJ = 0, so that linear independence follows from Theroem 3 in Sec. 3.1. • * Multiple Complex Roots In this case, real solutions are obtained as for complex simple-.!oots above. Consequently, if A = 'Y + iw is a complex double root, so is the conjugate A = 'Y - iw. Corresponding linearly independent solutions are (11) e-yX cos wx, e YX sin wx, xeYX cos wx, xeYX sin wx. The fi~t two of these result from eAX and e Ax as before, and the second two from xe AT and xe AX in the same fashion. Obviously, the corresponding general solution is (12) For complex triple roots (which hardly ever occur in applications), one would obtain two more solutions x 2e AX cos wx, x 2e Yx sin wx, and so on. 11-61 ODE FOR GIVEN BASIS Find an ODE (1) for which the given functions fonn a basis of solutions. 1. eX, e 2X , e 3x 3. e:c , e- x • cos x. sin x 17-121 GENERAL SOLUTION Solve the given ODE. (Show the details of your work.) 7. y'" + y' = 0 8. yiv - 29y" + LOOy = 0 4. cos x, sin x, x cos x, x sin x 9. y'" + y" - y' - y = 0 10. 16yiV - 8y" + Y = 0 5. I, x, cos 2x. sin 2x 6. e- 2x , e- x , eX, e 2x , I 11. ylll - 3)''' - 4y' + 6y 12. yiv + 3y" - 4y = 0 = 0 CHAP. 3 116 L13-18! Higher-Order Linear ODEs INITIAL VALUE PROBLEMS Solve by a CAS, giving a general solution and the particular solution and its graph. 13. lV + 0.45.\"'" - 0.165y" + 0.0045y' - 0.00175y = 0, )"(0) = 17.4, y' (0) = -2.82. y"(O) = 2.0485. y'''(0) = -1.458675 14. 4,,'" + 8,," + 41 v' + 37v = 0, 1'(0) = 9, y"(O) = ~6.5, /i(O) = ....:39.75 . + 3.2y" + 4.81/ = 0, ,,(0) = 3.4, /(0) = -4.6.y"(O) = 9.91 15. y'" 16. yiv + 4y = o. .\'(0) = -,,"'(0) = -~ !, y' (0) = -!. y" (0) = ~, 17. 1'iv - 91''' - 400" = O. ,,(0) = O. /(0) = O. ~"(O) ~ 4\. y'''(0) = 0 . . 18. y''' + 7.5.\"" + 14.25/ - 9.125-" .\"(0) = = 0, 10.05, y' (OJ = -54.975, y"(O) = 257.5125 11 program for calculating WronsJ..ians. (b) Apply the program to some bases of third-order and fourth-order constant-coefficient ODEs. Compare 3.3 Extend the solution method in Sec. 2.5 to any order Solve X3y '" + 2x2 y" - 4xy' + 4y = 0 and another ODE of your choice. In each case calculate the Wronskian. 20. PROJECT. Reduction of Order. This is of practical interest since a single solution of an ODE can often be guessed. For second order. see Example 7 in Sec. 2.1. (C) 11. (a) How could you reduce the order of a linear constant-coefficient ODE if a solution is known? (b) Extend the method to a variable-coefficient ODE .1'''' + P2(xly" + PI(X)Y' + Po(x)y = o. Assuming a solution YI to be known, show that another solution is Y2(X) = U(X)YI(X) with u(x) = J z(x) dx and .:: obtained by solving )"1'::" 19. CAS PROJECT. Wronskians. Euler-Cauchy Equations of Higher Order. Although Euler-Cauchy equations have mriable coefficients (powers of x). we include them here because they fit quite well into the present methods. (a) Write the results with those obtained by the program most likely available for Wronskians in your CAS. + (3y; + P2YI)'::' + (3y~ + 2P2 Y ; + PIYI)':: = O. (e) Reduce x 3)"'" - 3x\" + (6 - X2)X/ - (6 - X2»)" = O. using Yl = x (perhaps obtainable by inspection). 21. CAS EXPERIMENT. Reduction of Order. Starting with a basis, find third-order ODEs with variable coefficients for which the reduction to second order turns out to be relatively simple. Nonhomogeneous Linear ODEs We now turn from homogeneous to nonhomogeneous linear ODEs of nth order. We write them in standard form (1) /n) + Pn_I(X)y<n-D + ... + PI(X)'" + Po(x)y = rex) with /n) = d'\ldx n as the first term, which is practical, and r(x) 'i= O. As for second-order ODEs, a general solution of (I) on an open interval I of the x-axis is of the form (2) Here Yh(X) = CIYt(X) homogeneous ODE (3) + ... + cny,,(x) is a general solution of the corresponding /n) + Pn_I(X)/n-D + ... + PI(X)y' + Po(x)y = 0 on I. Also, Yp is any sulution of (l) on I containing no arbitrary constants. If (I) has continuous coefficients and a continuous rex) on J, then a general solution of (1) exists and includes all solutions. Thus (1) has no singular solutions. SEC 3.3 Nonhomogeneous Linear ODEs 117 An initial value problem for (I) consists of (l) and 11 initial conditions (4) with Xo in I. Under those continuity assumptions it has a unique solution. The ideas of proof are the same as those for n = 2 in Sec. 2.7. Method of Undetermined Coefficients Equation (2) shows that for solving (I) we have to determine a particular solution of (1). For a constant-coefficient equation (5) (ao, ... , an - 1 constant) and special r(x) as in Sec. 2.7, such a )'p(x) can be detennined by the method of undetermined coefficients. as in Sec. 2.7, using the following rules. (A) Basic Rule as in Sec. 2.7. (B) Modification Rule. If a term il1 your choice for )'p(x) is a solution of the homogeneous equation (3), thel111l111tiply yp(x) by xk, where k is the smallest positive integer such that no tenn of xkyp(X) is a solution of (3). (C) Sum Rule as in Sec. 2.7. The practical application of the method is the same as that in Sec. 2.7. It suffices to illustrate the typical steps of solving an initial value problem and, in particular, the new Modification Rule, which includes the old Modification Rule as a particular case (with k = 1 or 2). We shall see that the technicalities are the same as for 11 = 2. perhaps except for the more involved detennination of the constants. E X AMP L E 1 Initial Value Problem. Modification Rule Solve the initial value problem (6) y'" + 3y" + 3y' + Y = 30e- x , yeo) = 3, /(0) = Step 1. The characteristic equation is A3 + 3A2 + 3'\ A = -I. Hence a general solution of the homogeneous ODE is Solution. Step 2. If we try Yp = Ce -x, we get -C + 3C - 3C The Modification Rule calls for + 1 = (A 3 )'~ = C(3x 2 - x3)e -x. ).; = C(6x - 6x y;' = C(6 - y"(0) = + -47. 1)3 = O. It has the triple root + C = 30, which has no solution. Try Cxe -x and Cx2e -x. Yp = Cx e- x . Then -3, I Sx 2 + x 3 )e- x , + 9x 2 - x 3 )e -x. 118 CHAP. 3 Higher-Order linear ODEs Substitlllion of these expressions into (6) and omission of the common factor e -x gives The linear, quadratic. and cubic terms drop out. and 6C = 30. Hence C = 5. This gives yp = 5x 3 e- x . Step 3. We now write down y = Jh + yp' the general solution of the given ODE. From it we find C1 by the first initial condition. We insert the value. ditTerenliate, and determine c2 from the second initial condition. insert the value, and finally determine ("3 from /'(0) and the third initial condition: )"(0) = C1 = / y" + C2 + = [-3 = [3 Hence the + 2c3 (-c2 + 2C3}X + (15 - C3}X 2 2 5x je- - + (30 - 4c3)x + (-30 + (3)X + allswer 10 3 X , 5x 3 je- x . 3 /(0) = -3 + ("2 = -3. /'(0) = 3 + 2c3 = -47. C3 = -25. our problem is (Fig. 73) The curve of y begins at (0, 3) with a negative slope. as expected from the initial values. and approaches zero • as x --'> ce. The da~hed curve in Fig. 73 is ypy 5 0 -5 5 \ 10 x Fig. 73. Y and Yp (dashed) in Example 1 Method of Variation of Parameters The method of variation of parameters (see Sec. 2.10) also extends to arbitrary order 11. It gives a particular solution Yp for the nonhomogeneous equation (1) (in standard foml with y<n) as the first term!) by the formula Yp(X) = ~ ..c.. Yk(X) k = 1 (7) = Yl(X) I WI (x) W(x) rex) dx I Wk(x) - - rex) dx W(x) + ... + Yn(X) I Wn(x) W(x) rex) dx on an open interval I on which the coefficients of (I) and rex) are continuous. [n (7) the functions .1'1> •••• )'n form a basis of the homogeneous ODE (3), with Wronskian W. and l1j (j = I, ... , 11) is obtained from W by replacing the jtb culumn of W by the column [0 0 0 J]T. Thus, when 11 = 2. this becomes identical with (2) in Sec. 2.10, .r 1 W= /. , " 1 .2 "/, . )'2 ~I = ."1' The proof of (7) uses an extension of the idea of the proof of (2) in Sec. 2.10 and can be found in Ref [All] listed in App. I. SEC. 3.3 Nonhomogeneous Linear ODEs E X AMP L E 2 119 Variation of Parameters. Nonhomogeneous Euler-Cauchy Equation Solve the nonhomogeneous Euler-Cauchy equation (x> 0). Solution. Step 1. General solution of the homogeneous ODE. Substitution of)' = xm and the derivatives into the homogeneous ODE and deletion of the factor xm give~ /11(/11 - 1)(1/1 - 2) - 3m(m - I) + 6111 - 6 = O. The roots are L 2, 3 and give as a basis x, )'1 = Hence the corresponding general solution of the homogeneous ODE is )'h Step 2. Determinants needed in (7). These = Cl x + C2 x 2 + C3~ 3 fiT x 2x 3x2 = 2x 3 0 2 6x 0 x2 x 0 2, 3x 2 6x W= Wj 3 x2 x x W2 = 3 0 x3 0 3x 2 0 2 =x4 -2x 3 6x 2 0 2x 0 x x W3 = 0 = x2. 2 Step 3. Integration. In (7) we also need the right side rex) of our ODE in standard fonn. obtained by division of the given equation by the coefficient x 3 of /"; thus, rex) = (x 4 In x)/.1' 3 = x In .1'. In (7) we have the simple quotients W 1/W = x/2, W2 /W = -I, W3/W = 11(2,). Hence (7) becomes Yp = x I2 x x In x dx - x ( 3In ~ ~ x - Simplification gives yp = ~x4 (in x 2I 3) ~ x In x dx + x - x 2 (2~ 3I In x - 1 2x x In x dt X2) 4 + .1'3 2 (x In X - x). -11'). Hence the answer is . Figure 74 shows Yv Can you explain the shape of this curve? Its behavior near x = O? The occurrence of a minimum? Its rapid increase? Why would the method of undetermmed coefficients not have given the ~~~ 120 CHAP. 3 Higher-Order Linear ODEs y 30 20 10 0 -10 -20 x Particular solution Yp of the nonhomogeneous Euler-Cauchy equation in Example 2 Fig. 74. Application: 10 J -------"- Elastic Beams Whereas second-order ODEs have various applications, some of the more important ones we have seen, higher order ODEs occur much more rarely in engineering work. An important fourth-order ODE governs the bending of elastic beams, such as wooden or iron girders in a building or a bridge. Vibrations of beams will be considered in Sec. 12.3. E X AMP L E 3 Bending of an Elastic Beam under a Load We consider a beam B of length L and constant (e.g .. rectangular) cross section and homogeneous elastic material (e.g .. ~teel): see Fig. 75. We assume that under its own weight the beam is bent so little that it is practically straight. If we apply a load to B in a vertical plane through the axis of symmetry (the x-axis in Fig. 75). B is bent. Its axis is curved into the so-called elastic curve C (or deflection curw). It is shown in elasticity theory that the bending moment M(x) is proportional to the curvarure k(x) of C. We assume the bending to be small, ~o that the deflection )"(x) and its derivative y' (X) (determining the tangent direction of C) are small. Then. by calculus. k = y"I(1 + /2)312 = /'. Hence M(x) = Ely"(x). El is the constant of proportionality. E is Young's lIlodulus of elasticity of the material of the beam. 1 is the moment of inertia of the cross section about the (horizontal) ~-axis in Fig. 75. Elasticity theory shows further that M"(x) = f(x). where f(x) is the load per unit length. Together, Elyiv = f(x). (8) ~--; --L y Undeformed beam Z Z Fig. 75. Deformed beam under uniform load (simply supported) Elastic Beam SEC. 3.3 Nonhomogeneous linear ODEs 121 The practically most important supports and corresponding boundary conditions are as follows (see Fig. 76). (Al Simply supported y = y" = 0 at x = 0 and L (B) Clamped at both ends y = (C) Clamped at x = 0, free at x = L y' = 0 at x = 0 and L .1'(0) = y' (0) = 0, y"(L) = y"'(L) = o. The boundary condition y = 0 means no displacement at that point, y' = 0 means a horizontal tangent, v" = 0 means no bending moment. and y'" = 0 means no shear force. Let us apply this to the uniformly loaded simply supported beam in Fig. 75. The load is i(x) "" io = const. Then (8) is k (9) io . = EI This can be solved simply by calculus. Two integrations give y"(0) = 0 gives c2 = O. Then y"(L) = L(~kL + Y" cl) = = 0, = Cl -kLl2 (since L '* 0). Hence k 2 - Lx). "2(X Integrating this twice. we obtain with C4 = 0 from yeO) = O. Then 3 yeLl = kL 2 (L12 _ L3 6 + c ) = 0, 3 Inserting the expression for k, we obtain as our solution Y = io 24EI (x 4 - 2Lx 3 3 + Lx). Since the boundary conditions at both ends are the same. we expect the deflection y(x) to be "symmetric" with respect to L12, that is, y(x) = y(L - x). Verify this directly or set.r = u + L12 and show that y becomes an even function of u, From this we can see that the maximum deflection in the middle at II that the positive direction points downward. = 0 (x = CA) Simply supported x=O x=L (B) Clamped at both ends x=L x=0 x =L Fig. 76. (el Clamped at the left end, free at the right end Supports of a Beam 4 L12) is 5i o L /(16 . 24EI). Recall • 122 CHAP. 3 Higher-Order Linear ODEs . . 11 HIE -M-=:3 E T 1-;:-3=______ 11.:-iil GENERAL SOLUTION Solve the following ODEs. (Show the details of your work.) 1. y'" - 2y" - 4/ + 8y = e- 3 ,' + 8\"2 + 3y" - 5/ - 39y = 30 cos x + 0.5y" + 0.0625y = e- x cos 0.5x 4. ,,'" + 2,," - 5y' - 6y = 100e- 3x + 18e- x 5. x 3y'" + 0.75x),· - 0.75.\" = 9X 5 . 5 2. y'" 3. yiv 6. (x0 3 + 4D2)y = 8e x 7. (D 4 + IOD 2 + 9/)y = 13 cosh 2x + 18/))' = e 2x 8. (0 3 - 2D2 - 9D 19-141 INITIAL VALUE PROBLEMS Solve the following initial value problems. (Show the details.) 9. rIll - 9\"" + 27/ - 27\" = 54 sin 3x. :\"' (0) ~ U.S, . y" (0) ;", 38.5 yeO) = 3.5, = 128 cosh 2x, yeO) = 1, ),'(0) = 24. /"(0) = -HiO 11. (x 3 D3 - x 2 D2 - 7xD + 16/)y = 9x In x. yO) = 6. Oy(!) = 18, D2y(l) = 65 12. (0 4 - 26D2 + 25/)y = 50(x + 1)2, yeO) = 12.16, Dy(O) = -6. D2y(0) = 34. D 3 \"lO) = -130 10. )'iv - 16y y"(O) = 20, = • 1. What is the superposition or linearity principle? For what 11th-order ODEs does it hold? 2. List some other basic theorems that extend from second-order to 11th-order ODEs. 3. If you know a general solution of a homogeneous linear ODE. what do you need to obtain from it a general solution of a corresponding nonhomogeneous linear ODE? 4. What is an initial value problem for an 11th-order linear ODE? 5. What is the Wronskian? What is it used for? 16-151 GENERAL SOLUTION Solve the given ODE. (Show the details of your work.) 6. ylll + 6y" + 18y' + 40y = 0 7. 4x 2-,,'" + 12x-,," + 3-,,' = 0 8. yiv + lOy" + 9y = 0 9. 8y'" + 12y" - 2)" - 3)' = 0 10. (D 3 + 3D 2 + 3D + I)' = x 2 13. (D 3 + 40 2 + 850)y = 135xe x , yeO) = 10.4. Dy(O) = -18.1, D2y(0) = -691.6 14. (2D 3 - 0 2 - 8D + 4/)y = sin x. yfO) = I, Dy(O) = O. D2yfO) = 0 15. WRITING PROJECT. Comparison of Methods. Write a report on the method of undetermined coefficients and the method of variation of parameters. discussing and comparing the advantages and disadvantages of each method. Illustrate your findings with typical examples. Try to show that the method of undetermined coefficients. say. for a third-order ODE with constant coefficients and an exponential function on the right, can be derived from the method of vmlation of parameters. 16. CAS EXPERIMENT. Undetermined Coefficients. Since variation of parameters is generally complicated, it seems worthwhile to try to extend the other method. Find out experimentally for what ODEs this is possible and for what not. Hint: Work backward. solving ODEs with a CAS and then looking whether the solution could be obtained by undetermined coefficients. For example. consider y'" 3 x y'" 12,r" + 48/ - 64y = x l12 e 4x +x 2 -,," - 6xy' + and 6,v = x In x. TIONS AND PROBLEMS 11. (xD 4 12. (D 4 = 150x 4 3 2D - SD2)y = 16 cos 2x l)y = ge xl2 + 03)y - 13. lD3 + 14. (x 3D3 - 3x 202 + 6xD - 61)y = 30x- 2 15. (D 3 - D2 - D + /)' = eX 116-201 INITIAL VALUE PROBLEMS Solve the given problem. (Show the details.) 16. y'" - 2-,," + 4/ - 8y = O. yeO) = -I, y' (0) = 30. y" (0) = 28 17. x 3,.,,'" + 7x 2"',," -I f2xv' - 10" = O. yO) l. · Y (I) = - 7 • ." (l) = 44 18. (D 3 + 25D)y = 32 cos 2 4x, yeO) = 0, Dy(O) = 0, D2y(0) = 0 19. (D4 + 40D 2 - 441I)y = 8 cosh x. yeO) = 1.98, Oy(O) = 3, 02y (0) = -40.02. D 3 y(0) = 27 20. (x 3D3 + 5x 2D2 + 2xD - 2/)y = 7x 3/2 , y(l) = 10.6, Dy(l) = -3.6, D2y(l) = 31.2 Summary of Chapter 3 123 Higher Order Linear ODEs 11 = 2). Chapter 3 extends Chap. 2 from order 11 = 2 to arbitrary order 11. An nth-order linear ODE is an ODE that can be written Compare with the similar Summary of Chap. 2 (the case (1) In) + Pn_1(X)/n-1) + ... + P1(X)/ + Po(x)y = r(x) with y(n) = dny/dxn as the first term; we again call this the standard form. Equation (I) is called homogeneous if r(x) == 0 on a given open interval 1 considered, nonhomogeneous if r(x) =1= 0 on 1. For the homogeneous ODE /n) (2) + Pn_1(X)/n-ll + ... -, P1(X)y' + Po(.x)y = 0 the superposition principle (Sec. 3.1) holds, just as in the case 11 = 2. A basis or fundamental system of solutions of (2) on I consists of 11 linearly independent solutions Yi, ... ,.\"n of(2) on I. A general solution of(~) on lis a linear combination of these, (3) r=cr 1. 1 +"'+cr n. n (Cb . . . , C n . arbitrary constants). A general solution of the nonhomogeneous ODE (1) on I is of the form (4) Y = Yh + Yp (Sec. 3.3). Here, Yp is a particular solution of (1) and is obtained by two methods (undetermined coefficients or variation of parameters) explained in Sec. 3.3. An initial value problem for (I) or (2) consists of one of these ODEs and 11 initial conditions (Secs. 3.1, 3.3) (5) with given Xo in I and given Ko, .... K Il - l . If Po • .... Pn-1o r are continuous on I. then general solutions of (I) and (2) on J exist. and initial value problems (I). (5) or (2). (5) have a unique solution. •••• ti "\ ~/ CHAPTER ~ ./ 4 ~1 , V" " ... .:. ;-.~ ~ ... .. ~ +-. ~- .. ...:c.. Systems of ODEs. Phase Plane. Qualitative Methods Systems of ODEs have various applications (see, for instance, Secs. 4.1 and 4.5). Their theory is outlined in Sec. 4.2 and includes that of a single ODE. The practically important conversion of a single nth-order ODE to a system is shown in Sec. 4.1. Linear systems (Secs. 4.3, 4.4, 4.6) are best treated by the use of vectors and matrices, of which, however, only a few elementary facts will be needed here, as given in Sec. 4.0 and probably familiar to most students. Qualitative methods. In addition to actually solving systems (Sec. 4.3, 4.6), which is often difficult or even impossible, we shall explain a totally different method, namely, the powerful method of investigating the general behavior of Whole families of solutions in the phase plane (Sec. 4.3). This approach to systems of ODEs is called a qualitative method because it does not need actual solutions (in contrats to a "quantitative method" of actually solving a system). This phase plane method, as it is called, also gives information on stability of ~olutions. which is of general importance in control theory, circuit theory, population dynamics, and so on. Here, stability of a physical system means that, roughly speaking, a small change at some instant causes only small changes in the behavior of the system at all later times. Phase plane methods can be extended to nonlinear systems, for which they are particularly useful. We will show this in Sec. 4.5, which includes a discussion of the pendulum equation and the Lotka-Volterra population model. We finally discuss nonhomogeneous linear systems in Sec. 4.6. NOTATION. Analogous to Chaps. 1-3, we continue to denote unknown functions by x for functions, Xl (t), X2(t), as is sometimes done in systems of ODEs. y; thus, YI (I), h(!)· This seems preferable to suddenly using Prerequisite: Chap. 2. References and Ansll'ers 4.0 10 Problems: App. 1 Part A. and App. 2. Basics of Matrices and Vectors In discussing li1/ear systems of ODEs we shall use matrices and vectors. This simplifies formulas and clarifies ideas. But we shall need only a few elementary facts (by no means the bulk of material in Chaps. 7 and 8). These facts will very likely be at the disposal of most students. Hence this sectio1l is for reference only. Begin with Sec. 4.1 and consult 4.0 as needed. 124 SEC 4.0 Basics of Matrices and Vectors 125 Most of our linear systems will consist of two ODEs in two unknown functions .' let). )'2(t), for example, (1) (perhaps with additional given functions .Rl(t), g2(t) in the two ODEs on the right), Similarly, a linear system of n first-order ODEs in n unknown functions YI(t). )'n(t) is of the fonn (2) (perhaps with an additional given function in each ODE on the right). Some Definitions and Terms Matrices. In (I) the (constant or variable) coefficients form a 2 x 2 matrix A, that is. an anay A= [-5 2] . for example, 13 ~ Similarly, the coefficients in (2) form an n x n matrix (4) The (lIb (/12' . . . are called entries, the horizontal lines rows, and the vertical lines columns. Thus, in (3) the first row is [al1 a12]' the second row is [a2l a22], and the first and second columns are and a 12 ] . [ a22 In the "double subscript notation" for entries, the first subscript denotes the row and the second the column in which the entry stands. Similarly in (4). The main diagonal is the diagonal an a22 ann in (4), hence all {/22 in (3). We shall need only square matrices, that is, matrices with the same number of rows and columns, as in (3) and (4). 126 CHAP. 4 Vectors. Systems of ODEs. Phase plane. Qualitative Methods A column vector x with n components Xl, x= . . . , Xn is of the form thus if 11 = 2. Similarly. a row vector v is of the form thus if 11 = 1. then Calculations with Matrices and Vectors Equality. Two 11 X Thus for 1l = 2. let A= 11 matrices are equal if and only if corresponding entries are equal. a12 ] [ all a21 and [b B= ll b 21 a22 12 b ] b 22 Then A = B if and only if all = bu. a12 = bI2 a21 = b2I , a22 = b 22 · Two column vectors (or two row vectors) are equal if and only if they both have components and corresponding components are equal. Thus. let Then v =x 11 if and only if Addition is performed by adding corresponding entries (or components); here, matrices must both be II X 11, and vectors must both have the same number of components. Thus for n = 2, (5) Scalar multiplication (multiplication by a number c) is performed by multiplying each entry (or component) by c. For example, if [-: :]. A= then -7A = [-63 14 If v ~ [04]. -13 then IOv = [-:30} -2~J . SEC. 4.0 Basics of Matrices and Vectors 127 Matrix Multiplication_ The product C = AB (in this order) of two n X n matrices A = [ajk] and B = [bjk ] is the n X n matrix C = [Cjk] with entries n (6) Cjk j 2: = aj'mb'mk = 1, ... , n k = 1, ... , n, 'm~1 that is, multiply each entry in the jth row of A by the corresponding entry in the kth column of B and then add these n products. One says briefly that this is a "multiplication of rows into columns." For example, 3J[I-4J 0 2 5 9 [ -2 [9-1+3-2 -2 - 1 + 0 - 2 ~ [~: (-2)-(-4) I -4J [ 9 2 5 - 2 3J 0 =1= BA in general. In our 1 -3 [1-9 + (-4)-(-2) = + 0-5 -2:J Matrix multiplication is not commutative, AB CAUTION! example, [ 9-(-4) + 3-5J + (-4) - OJ 2-3+5-0 2 - 9 + 5 - (- 2) Multiplication of an n X n matrix A by a vector x with n components is defined by the same rule: v = Ax is the vector with the n components n Vj 2: = j = I, - - -, n. ajmx'm 'm~1 For example, Systems of ODEs as Vector Equations Differentiation_ The derivative of a matrix (or vector) with variable entries (or components) is obtained by differentiating each entry (or component). Thus, if yet) = Yl(t)] [ Y2(t) = [e-2t] , I Y (t) = then sin t .Vl(t)] 1 [-2e-2t] = . [ Y~(t) cos t Using matrix multiplication and differentiation, we can now write (1) as I (7) y = [ '] Yl I Y2 = Ay = [ all a 21 12 a ] a22 [VI] . Y2 , e.g., y' = [-5 13 128 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods Similarly for (2) by means of an n X n matrix A and a column vector y with II components, namely, y' = Ay. The vector equation (7) is equivalent to two equations for the components. and these are precisely the two ODEs in (1). Some Further Operations and Terms Transposition is the operation of writing columns as rows and conversely and is indicated by T. Thus the transpose AT of the 2 X 2 matrix a 12 a22 J [-5 2J! - is 13 The transpose of a column vector, say, is a row vector, and conversely. Inverse of a Matrix. The n X n unit matrix I is the 11 X 11 matrix with main diagonal 1, 1, ... ,land all other entries zero. If for a given n X n matrix A there is an n X 11 matrix B such that AB = BA = I, then A is called nonsingular and B is called the inverse of A and is denoted by A -1; thus (8) If A has no inverse, it is called singular. For (9) 11 = 2, A-I = det A where the determinant of A is (10) detA = a 11 l 1I2I (For generaln, see Sec. 7.7, but this will not be needed in this chapteL) Linear Independence. r given vectors v(1), ... , VCT) with n components are called a linearly independent set or, more briefly, linearly independent, if (11) C1 VCl) + ... + c,.vCr) = 0 implies that all scalars c 1 , • • . , c,. must be zero; here, 0 denotes the zero vector, whose n components are all zero. If (II) also holds for scalars not all zero (so that at least one of these scalars is not zero), then these vectors are called a linearly depel1dent set or, brietly, linearly dependent, because then at least one of them can be expressed as SEC. 4.0 129 Basics of Matrices and Vectors a linear combination of the others; that is, if, for instance, can obtain v(1) = - ~ (C2 VC2 ) C1 * 0 in (ll), then we + ... + crvc'·)). C] Eigenvalues, Eigenvectors Eigenvalues and eigenvectors will be very important in this chapter (and, as a matter of fact, throughout mathematics). Let A = [Ujk] be an n X n matrix. Consider the equation Ax = AX (12) where A is a scalar (a real or complex number) to be determined and x is a vector to be determined. Now for every A a solution is x = O. A scalar A such that (12) holds for some vector x 0 is called an eigenvalue of A, and this vector is called an eigenvector of A corresponding to this eigenvalue A. We can write (12) as Ax - AX = 0 or * (A - AI)X (13) = O. These are n linear algebraic equations in the n unknowns Xl> ••• , Xn (the components of x). For these equations to have a solution x 0, the determinant of the coefficient matrix A - AI must be zero. This is proved as a basic fact in linear algebra (Theorem 4 in Sec. 7.7). In this chapter we need this only for n = 2. Then (13) is * (14) in components, =0 (14*) Now A - AI is singular if and only if its determinantdet (A - AI), called the characteristic determinant of A (also for general n), is zero. This gives det (A _ AI) = I Uu - U21 A I U12 U22 - A (15) This quadratic equation in A is called the characteristic equation of A. Its solutions are the eigenvalues Al and A2 of A. First determine these. Then use (14*) with A = Al to determine an eigenvector xCi) of A cOlTesponding to A1' FinaIly use (14*) with A = A2 to find an eigenvector X(2) of A cOlTesponding to A2' Note that if x is an eigenvector of A, so is h for any k O. * CHAP. 4 130 E X AMP L E 1 Systems of ODEs. Phase plane. Qualitative Methods Eigenvalue Problem Find the eigenvalues and eigenvectors of the maliix -4.0 A= [ (16) Soluti01l. 4.0J. -1.6 1.2 The characteristic equation is the quadratic equation -4 - A det [A - All = 1 -1.6 4 1 = A2 1.2 - A + 2.8A + 1.6 = O. It has the solutions Al = -2 and A2 = -0.8. These are the eigenvalues of A. Eigenvectors are obtained from (14*). For A = Al = -2 we have from (14*) (-4.0 + 2.0)x1 + + =0 (1.2 + 2.0)x2 = O. A solution of the fir~t equation is Xl = 2, X2 = I. This also satisfies the second equation. (Why?). Hence an eigenvector of A corresponding to Al = -2.0 is (17) X(2)=[I] Similarly. 0.8 is an eigenvector of A corresponding to A2 = -0.8. as obtained from (14") with A = A2. Verify this. • 4.1 Systems of ODEs as Models We first illustrate with a few typical examples that systems of ODEs can serve as models in various applications. We further show that a higher order ODE (with the highest derivative standing alone on one side) can be reduced to a first-order system. Both facts account for the practical importance of these systems. E X AMP L E 1 Mixing Problem Involving Two Tanks A mixing problem involving a single tank is modeled by a single ODE. and you may first review the corresponding Example 3 in Sec. 1.3 because the principle of modeling will be the same for two taoks. The model will be a system of two first-order ODEs. Taok T1 aod T2 in Fig. 77 contain initially 100 gal of water each. In T1 the water is pure, whereas 150 I b of fertilizer are dissolved io T2. By circulating liquid at a rate of 2 gal/min and stirring (to keep the mixture uniform) the amounts of fertilizer ,vI(t) III T1 and Y2(t) in T2 change with time t. How long should we let the liqUid circulate so that T1 will contain at least half as much fertilizer as there will be left in T2? Setting up the model. As for a single tank. the time rate of change \.~ (t) inflow minus outflow. Similarly for tank T 2 . From Fig. 77 we see that Soluti01l. Step I. y~ = Inflow/min - Outflow/min = 2 100.1'2 - 2 }'~ = Inflow/min - Outflow/min = 100 Of)'l (t) equal& 2 IOOY1 2 \'1 - 100 Y2 Hence the mathematical model of our mixture problem i, the system of first-order ODEs )'~ = -0.02Y1 + 0.02)'2 (Tank T1 ) (Tank T2 ). SEC. 4.1 131 Systems of ODEs as Models yet) 150 - r 2 gal/min - 2 gal/min T] '-- 100 ----T2 System of tanks Fig. 77. Fertilizer content in Tanks T, (lower curve) and T2 As a vector equation with column vector y = [YIJ and matrix A this becomes Y2 -0.02 y' = Ay, 0.02J. A= [ 0.02 -0.02 where Step 2. General solution. As for a single equation, we try an exponential function of t, Then (I) Dividing the last equation AxeAt = Axe).' by eAt and interchanging the left and right sides, we obtam Ax = Ax. We need nontIivial solutions (solutions that are not identically zero). Hence we have to look for eigenvalues and eigenvectors of A. The eigenvalues are the solutions of the characteristic equation (2) 1 I= -0.02 - A 0.02 0.02 -O.oz - A det (A - AI) = (-O.oz - A)2 - 0.022 = A(A + 0.04) = O. We see that Al = 0 (which can very well happen--don't get mixed up-it is eigenvectors that must not be zero) and A2 = -0.04. Eigenvectors are obtained from (14*) in Sec. 4.0 with A = 0 and A = 0.04. For our present A this gives [we need only the fIrst equation in (14*)J -0.02'1 + 0.02'2 = 0 (-0.02 + 0.04)Xl + 0.02.\2 and = 0, respectively. Hence Xl = X2 and Xl = -x2, respectively, and we can take Xl = x2 = 1 and Xl This gives two eigenvectors corresponding to Al = 0 and A2 = -0.04, respectively, namely, = -x2 = 1. and From (I) and the superposition principle (which continues to hold for systems of homogeneous linear ODEs) we thus obtain a solution (3) where Cl and C2 are arbitrary constants. Later we shall call this a general solution. Step 3. Use of initial conditions. The initial conditions are yt(O) = 0 (no fertilizer in tank T 1) and Y2(0) = 150. From this and (3) with t = 0 we obtain [:: : ::J 132 CHAP. 4 Systems of ODEs. phase plane. Qualitative Methods In components this is In cl + c2 = 0, CI - c2 = I SO. The solution is Cl = 7S. ("2 = -7S. This gives the answer component~, )'1 = 75 - 7Se- O.04t (Tank T I , lower curve) + 75e- o.04t (Tank T2 • upper curve>. )'2 = 75 Figure 77 shows the exponential increase of \'1 and the exponential decrease of .\'2 to the common limit 75 lb. Did you expect this for physical reasons? Can you physically explain why the curves look "symmetric"? Would the limit change if TI initially contained 100 Ib of fertilizer and T2 contained 50 Ib? Step 4. Answer. T1 contains half the fertilizer amount of T2 if it contains 113 of the total amount, that is, SO lb. Thus YI = 75 - 75e -O,Mt = SO, e -O.Mt 1 t = 3' = (In 3)/0.04 = 27,5, • Hence the fluid should circulate for at lea,t about half an hour. EXAMPLE 2 Electrical Network Find the CUlTents ft (t) and 12 (1) in the network in Fig. 78. Assume all the instant when the switch is closed. L = 1 henry current~ and charges to be zero at t = 0, C = 0.25 farad SWitchlrI; t=O E = 12 volts-=- R2 = 6 ohms Fig. 78. Electrical network in Example 2 Solution. Step I. Setting up the mathematical model. The model of this network is obtained from Kirchhoff's voiLage law, as in Sec, 2.9 (where we considered single circuits). Let II(t) and 12(t) be the CUiTents in the left and right loops, respectively, In the left loop the voltage drops are Ll~ = I; [V] over the inductor and RI(ft - 12) = 4(h - 12 ) [V] over the resistor, the difference because 11 and 12 flow through the resistor in opposite directions. By Kirchhoff's voltage law the sum of these drops equals the voltage of the battery; that is, I; + 4(/1 - 12 ) = 12. hence (4a) I; = -4ft + 4/2 + 12. In the right loop the voltage drops are R2/2 = 612 [V] and R 1 (l2 - 'I) (I/e)f 12 dt = 4 f 12 dt [V] over the capacitor. and their sum is zero. or = 4(12 - 11) [V] over the resistors and 1012 - 4ft +4 f '2 dt = O. Division by 10 and differentiation gives I~ - O.4/~ + 0.4/2 = O. To simplify the solution process. we first get rid of 0.4/~, which by (4a) equals 0.4(-4/1 Substitution into the present ODE gives I~ = OAli - 0.4/2 = OA( -4ft + 4/2 + 12) - 0.4/2 + 4/2 + 12). SEC. 4.1 131 Systems of ODEs as Models and by simplification I~ = -1.6/ 1 (4b) + 1.212 + 4.8. In matrix form. (4) is (we write J since I is the unit matrix) J' (5) = AJ + g, A= where -4.0 4.0J, [-1.6 l.2 g = [12.0J . 4.8 Step 2. Solving (5). Because of the vector g thi~ is a nonhomogeneous system. and we try to proceed as for a single ODE. solving lim the homogeneous system J' = AJ (thu~.J' - A.J = 0) by substituting J = xeAl. This gives hence Ax = Ax. Hence to obtain a nontrivial solution, we again need the eigenvalues and eigenvectors. For the present matrix A [hey are derived in Example I in Sec. 4.0: X(2) = [ I ] 0.8 Hence a "general solution" of [he homogeneous system is For a particular ~olution of the nonhomogeneous sy~tem (5). since g is constant. we try a constant column vector J p = a with components "I' 112' Then J~ = 0, and sub~titution into (5) gives Aa + g = 0; in components. + 4.0112 + 12.0 = 0 -1.6aI + 1.2l12 + 4.8 = O. -4.0111 The solution is al = 3, {l2 = 0: thus a = [~J . Hence (6) in components. The initial conditions give ('2+ Hence ('1 = -4 and ('2 3 =0 = 5. As the solution of our problem we thus obtain (7) In components (Fig. 79b), 11 = -8e- 2t + 5e- o.8t + 3 12 = _4e- 2t + 4e- O.8t . Now collles an important idea, un which we ~hall elaborate further. beginning in Sec. 4.3. Figure 79a shows and 12 (t) as two separate curves. Figure 79b shows these two currents as a single curve [ft(l), 12 (t)] in the 1I/2-plane. Thi~ is a parametric representation with time t as the parameter. It is often important to know in which sense such a curve is traced. This can be indicated by an arrow in the sense of increasing t. as is shown. The 1I/2-plane is called the phase plane of our system (5), and the curve in Fig. 79b is called a trajectory. We shall see [hat such "phase plane representations" are far more important than graphs as in Fig. 79a because they will give a much better qualitative overall impression of the general behavior of whole familie~ of solutions, • not merely of one solution as in the present case. II(t) CHAP. 4 134 Systems of ODEs. Phase plane. Qualitative Methods 1(t) i'--~ 4 3 ~--------------=== 2 0.5 / OL-~L- o OL-__L -_ _L -_ _L -_ _~_ _~_ __ _ o 2 3 4 5 _ _L -_ _L -_ _ _ _~~_ 2 3 4 5 (a) Currents 11 (b) (upper curve) and 12 Fig. 79. Trajectory 1I1(t), 12(t)]T in the 1/2 -plane (the "phase plane") Currents in Example 2 Conversion of an nth-Order ODE to a System We show that an nth-order ODE ofthe general form (8) (see Theorem 1) can be converted to a system of n first-order ODEs. This is practically and theoretically important-practically because it permits the study and solution of single ODEs by methods for systems. and theoretically because it opens a way of including the theory of higher order ODEs into that of first-order systems. This conversion is another reason for the importance of systems, in addition to their use as models in various basic applications. The idea of the conversion is simple and straightforward, as follows. THEOREM 1 Conversion of an ODE An nth-order ODE y<n) (8) = F(t, y, y', ... , y<n-ll) call be converted to a system of n first-order ODEs by setting (9) Yl = y, Y2 = y', )'3 = y",' .. , Yn = y<n-ll. This system is of the form , Yl , = Y2 )'2 =)'3 (10) , Yn-l y~ PROOF = = Yn F(t, Y10 Y2, ... , Yn)· The first n - 1 of these n ODEs follow immediately from (9) by differentiation. Also, y~ = y<n) by (9), so that the last equation in (10) results from the given ODE (8). • SEC. 4.1 135 Systems of ODEs as Models E X AMP L E 3 Mass on a Spring To gain confidence in the conversion method, let us apply it to an old friend of ours. modeling the free motions of a mass on a spring (see Sec. 2.4) my" + + ky cy' 0 = k -yo y, y " = - -c m or III For this ODE (8) the system (10) is linear and homogeneous, , Y1 = .1'2 , k c Y2 = - - Y1 - m III Setting y = )"2' [Y1J .we get in matrix form Y2 O y' = Ay k = [_ cll [::J . _ m III The characteristic equation is -A det (A - Ali = = A2 c m k --- A 111 III It agrees with that in Sec. 2.4. For an illustrative computation, let A2 + 2A + 0.75 = (A + .!:... A + ~ III + 0.5)(A + = O. 111 = I, c = 2, and k = 0.75. Then 1.5) = O. This gives the eigenvalues Al = -0.5 and A2 = -1.5. Eigenvectors follow from the first equation Lll A - AI = 0, which is -A,y] +.x2 = O. For A] this gives 0.5x1 + x2 = O. say. xl = 2. '\2 = -1. For A2 = -1.5 it gives l.5XI + -'"2 = 0, say, Xl = I, X2 = -1.5. These eigenvectors X<2l=[ I -1.5 J give e Y- [2J -1 - c1 -0.5t + (2. [ 1 Je. -1.5t -1.5 This vector solution has the first component which is the expected solution. The second componenl is its derivative .1'2 ----_ = yi = y' = _c]e- 0 .5t - • 1. 5c2 e -1.5t. .... - 11-61 MIXING PROBLEMS 1. Find out without calculation whether doubling the flow rate in Example 1 has the same effect as halfing the tank sizes. (Give a reason.) 2. What happens in Example 1 if we replace T2 by a tank containing 500 gal of water and ISO Ib of fertilizer dissolved in it? 3. Derive the eigenvectors consulting this book. III Example 1 without 4. In Example 1 find a "general solution" for any ratio a = (flow rate)/(tank si:;;e), tank sizes being equal. Comment on the result. 5. [f you extend Example I by a tank T3 of the same size as the others and connected to T2 by two tubes with CHAP. 4 136 Systems of ODEs. Phase Plane. Qualitative Methods 16. TEAM PROJECT. Two Masses on Springs. (a) Set up the model for the (undamped) system in Fig. 80. flow rates a~ between T1 and T2 , what system of ODEs will you get? 6. Find a "general solution" of the system in Prob. 5. 17-10 I (b) Solve the ~ystem of ODEs obtained. Him. Try 2 = xe'"' and set w = A. Proceed as in Example I or 2. y ELECTRICAL NETWORKS (e) Describe the influence of initial conditions on the possible kind of motions. 7. Find the currents in Example 2 if the initial cun-ents are 0 and - 3 A (minus meaning that 12 (0) flows against the direction of the an-ow). 8. Find the cun-ents in Example 2 if the resistance of R1 and R2 is doubled (general solution only). First, guess. 9. What are the limits of the CUlTents in Example 27 Explain them in terms of physics. 10. Find the cun-ems in Example 2 if the capacitance is changed to C = 115.4 F (farad). 111-151 (Net change in spring length CONVERSION TO SYSTEMS =Y2- Y l) Find a general solution of the given ODE (a) by first converting it to a system. (b). as given. (Show the detaib of your work.) II. y" - 4y = 0 12. y" + 2y' - 24y = 0 13. y" - y' = 0 14. y" + 15y' + SOy = 0 15. 64y" - 48/ - 7." = System in static equilibrium 0 Fig. 80. System in motion Mechanical system in Team Project 16 4.2 Basic Theory of Systems of ODEs In this section we discuss some basic concepts and facts about systems of ODEs that are quite similar to those for single ODEs. The first-order systems in the last section were special cases ofthe more general system y~ = flU. Y1' ...• )'n) )'~ = f2(t· ."1, .... ),,,) (1) We can write the system (I) as a vector equation by introducing the column vectors Yn]T and f = [fl fn]T (where T means transposition and saves us the space that would be needed for writing y and f as columns). This gives y = [."1 (1) y' = fU. y). This system (1) includes almost all cases of practical interest. For Il = I it becomes )'~ = flU. )'1) or. simply, y' = f(t, y). well known to us from Chap. I. A solution of (I) on some interval (/ < t < b is a set of n differentiable functions ),,, = 1I,,(t) SEC. 4.2 137 Basic Theory of Systems of ODEs on a < t < b that satisfy (1) throughout this interval. In vector form, introducing the hnr (a column vector!) we can write "solution I'ector" h = [hI = y h(t). An initial value problem for (I) consists of (1) and 11 given initial conditions (2) in vector form, y(to) = K, where to is a specified value of t in the interval considered and Knr are given numbers. Sufficient conditions for the the components of K = [Kl existence and uniqueness of a solution of an initial value problem (I), (2) are stated in the following theorem, which extends the theorems in Sec. 1.7 for a single equation. (For a proof, see Ref. [A 7].) Existence and Uniqueness Theorem THEOREM 1 Let f 1, • . . , f n in (1) be continuousfwlctiolls havi1lg collfil1UOUS pm1illl derivatil'es afl/aYI, ... , afl/aYn' ... , af,/iJYn in some domain R of f)·1.\"2 ••• Yn-space containing tlte point (to, K I , . • . , K,,). Theil (I) has a solutioll 011 some illfen'al to - a < t < to + a satisfying (2). and this solution is unique. Linear Systems Extending the notion of a linear ODE. we call (1) a linear system if it is linear in Yl ... , Yn; that is, if it can be written (3) In vector form. this becomes (3) = Ay + g y' where a~] ., y ann ., g = [~]. . gn = 0, so that it is y' * [h] Yn This system is called homogeneous if g (4) = = Ay. If g 0, then (3) is called nonhomogeneous. The system in Example I in the last section is homogeneous and in Example 2 nonhomogeneous. The system in Example 3 is homogeneous CHAP. 4 138 Systems of ODEs. Phase plane. Qualitative Methods For a linear system (3) we have atl/aYI = an(t), ... , at nlaYn = ann(t) in Theorem l. Hence for a linear system we simply obtain the following. THEOREM 2 Existence and Uniqueness in the Linear Case Let the ajk' sand g/ s in (3) be continuous functions of t on an open interval a < t < f3 containing the point t = to. Then (3) has a solution y(t} on this inten'al satisfying (2), and this solution is unique. As for a single homogeneous linear ODE we have THEOREM 3 Superposition Principle or Linearity Principle lfy(1) and y(2) are solutions of the homogeneous linear system (4) on some interval, so is any linear combination y = c l y(1) + C2y(2). PROOF Differentiating and using (4), we obtain • The general theory of linear systems of ODEs is quite similar to that of a single linear ODE in Secs. 2.6 and 2.7. To see this, we explain the most basic concepts and facts. For proofs we refer to more advanced texts, such as [A7J. Basis. General Solution. Wronskian By a basis or a fundamental system of solutions ofthe homogeneous system (4) on some interval J we mean a linearly independent set of n solutions y(1}, ... , yCn) of (4) on that interval. (We write J because we need I to denote the unit matrix.) We call a conesponding linear combination (5) (CI, ... , Cn arbitrary) a general solution of (4) on J. It can be shown that if the ajk(t) in (4) are continuous on J, then (4) has a basis of solutions on J. hence a general solution. which includes every solution of (4) on J. We can write n solutions ym, ... , yCn) of (4) on some interval J as columns of an 11 X 11 matrix (6) SEC. 4.3 Constant-Coefficient Systems. Phase plane Method 139 The determinant of Y is called the Wronskian of ym, ... , yen>, written . I . I V(2) _yCn) 1 v(1) y~2) y~n) " (1) .n . n ,,(1) (7) W(yCl), ... , yCn») = .2 V Cn ) \' (2) . n The columns are these solutions, each in terms of components. These solutions form a basis on 1 if and only if W is not zero at any 11 in this interval. W either is identically zero or is nowhere zero in 1. (This is similar to Sees. 2.6 and 3.l.) If the solutions y(1), . . . , yen) in (5) form a basis (a fundamental system), then (6) is cnl T , often called a fundamental matrix. Introducing a column vector e = [CI C2 we can now write (5) simply as (8) y = Ye. Furthermore, we can relate (7) to Sec. 2.6, as follows. If y and second-order homogeneous linear ODE, their Wronskian is W(y,.:) = v y' I To write this ODE as a system, we have to set y = h, Y' = y~ = (see Sec. 4.1). But then W(y, z) becomes (7), except for notation. 4.3 z are solutions of a )'2 and similarly for z Constant-Coefficient Systems. Phase Plane Method Continuing, we now assume that our homogeneous linear system y' (1) = Ay under discussion has constant coefficients, so that the n X n matrix A = [OjkJ has entries not depending on t. We want to solve (I). Now a single ODE y' = ky has the solution y = Ce kt • So let us try (2) Substitution into (I) gives y' eigenvalue problem (3) Axe>" = Ay = Ax = AX. AxeAl. Dividing by eAt, we obtain the 140 CHAP. 4 Systems of ODEs. phase Plane. Qualitative Methods Thus the nontrivial solutions of (1) (solutions that are not zero vectors) are of the form (2), where A is an eigenvalue of A and x is a con·esponding eigenvector. We assume that A has a linearly independent set of Il eigenvectors. This holds in most applications, in particular if A is symmetric (okj = Ojk) or skew-symmetric (okj = -Ojk) or has Il differellt eigenvalue~. Let those eigenvectors be XCll, .... x(n) and let them correspond to eigenvalues AI> ... , An (which may be all different, or some--or even all-may be equal). Then the corresponding solutions (2) are (4) Their Wronskian W = W(yCll. . . . . yen»~ [(7) in Sec. 4.2] is given by e Xl = (y(1), ... , y(n» = X2 e = \: (n)e Ant (1) Alt Xn e (n) Ant X2 e (1) Alt W xil) (n) Ant (1) Alt Xl e - n eAlt + .. +An t X~l) XU) -n yen) -'n On the right, the exponential function is never zero, and the determinant is not zero either because its columns are the n linearly independent eigenvectors. This proves the following theorem, whose assumption is true if the matrix A is symmetric or skew-symmetric, or if the 11 eigenvalues of A are all different. THEOREM 1 General Solution If the constant matrix A in the system (I) has a linearly indepelldent set of Il eigenvectors, then the corresponding solutions y(1), .•. ,y(n) in (4)for711 a basis of solutiol1s of (l). olld the con'espollding general solution is (5) l __________________~ How to Graph Solutions in the Phase Plane We shall now concentrate on systems (I) with constant coefficients con ~isting of two ODEs (6) y' = Ay; in components, Of course, we can graph solutions of (6). (7) y(t) = YI(t)] , [Y2(t) SEC. 4.3 141 Constant-Coefficient Systems. phase Plane Method as two curves over the t-axis, one for each component of ytf). (Figure 79a in Sec. 4.1 shows an example.) But we can also graph (7) as a single curve in the )'lY2-plane. This is aporamefric representation (parametric equation) with parameter t. (See Fig. 79b for an example. Many more follow. Parametric equations al~o occur in calculus.) Such a CUI ve is called a trajectory (or sometimes an orbit or path) of (6). The YIY2-plane is called the phase plane. 1 If we fill the phase plane with trajectories of (6), we obtain the so-called phase portrait of (6). E X AMP L E 1 Trajectories in the Phase Plane (Phase Portrait) In order to see what is going on, let u, find and graph solutions of the system I] -3 (8) y' = Ay = Solution. By substituting y = The characteristic equation is I [ xeAt y. thus -3 yi = y~ = + -3)'1 )'1 - ,1'2 3.1'2' and y' = lLxe At and dropping the exponential function we get Ax = Ax. det (A - AI) = 1- 3 - A I 1= I ,1.2 + 6,1. + 8 = O. -3 - A Thi. gives the eigenvalue, Al = -2 and ,1.2 = -4. Eigenvectors are then obtained from For Al = -2 this is -xl + t2 = O. Hence we can take x(1) = 1I and an eigenvector is x(2) = [I JlT. For ,1.2 = -4 this becomes Xl + x2 = O. _I]T. Tins gives the general solution . Figure 81 on p. 142 shows a phase pomait of some of the trajectories (to which more trajectories could be added if so desired). The two straight trajectories correspond to Cl = 0 and C2 = 0 and the olhers to other choices of ~~ Studies of solutions in the phase plane have recently become quite important, along with advances in computer graphics, because a phase portrait gives a good general qualitative impression of the entire family of solutions. This method becomes particularly valuable in the frequent cases when solving an ODE or a system is inconvenient or impossible. Critical Points of the System (6) The point y = 0 in Fig. 81 seems to be a common point of all trajectories, and we want to explore the reason for this remarkable observation. The answer will follow by calculus. Indeed, from (6) we obtain , (9) ."2 (21)'1 Y1 0U)'l , + 022)'2 + 012)'2 1A name that come, from physio. where il is the Y-(/IIv)-plane. used to plot a motion in terms of po,ition = v (m = mass): but the name is now used quite generally for the YlY2-plane. The use of the phase plane is a qualitatin method. a method of obtaining general qualitative information on solutions without actually solving an ODE 01' a system. This method was created by HENRI POINCARE (1854-1912). a great French mathematician, whose work was also fundamental in complex analysis, divergent series, topology, and astronomy. y and velocity / CHAP. 4 142 Systems of ODEs. phase Plane. Qualitative Methods This associates with every point P: (.vI' )'2) a unique tangent direction d.v2ldYl of the trajectory passing through P, except for the point P = Po: (0.0), where the right side of (9) becomes 0/0. This point Po, at which dY2idYl becomes undetermined. is called a critical point of (6). Five Types of Critical Points There are five types of critical points depending on the geometric shape of the trajectories near them. They are called improper nodes, proper nodes, saddle points, centers, and spiral points. We define and illustrate them in Examples 1-5. E X AMP L E 1 (Continued) Improper Node (Fig. 81) An improper node is a critical point Po at which all the trajectories. except for two of them, have the same limiting direction of the tangent. The two exceptional trajectories also have a limiting direction of the tangent at Po which, however, is different. The system (8) has an improper node at 0, as its phase portrait Fig. 81 shows. The common limiting direction at 0 is that of the eigenvector xU) = [lIlT because e -4t goes to zero faster than e -2t as r increases. The two exceptional limiting tangent directions are those of x(2) = [1 _l]T and -x(2) = [-I UT. • E X AMP L E 2 Proper Node (Fig. 82) A proper node is a critical point Po at which every trajectory has a definite limiting direction and for any given direction d at Po there is a trajectory having d as its limiting direction. The system (10) y' = , [~ .\"1 =)'1 , thus .\"2 =)'2 has a proper node at the origin (see Fig. 82). Indeed, the matrix is the unit matrix. Its chmacteristic equation 0 is an eigenvector, and we can take [1 ol and [0 l]T. Hence (I - ),)2 = 0 has the root)' = 1. Any x a general solution is '* y = cl [~J t e + c2 [~J )"1 = cle t t or e or )'2 = Y2 ," "- - ---1' '" '" / /' ~ \ / \ Fig. 81. • Y21 ,\j~ ~\/~ }!:-' "- c!Y2 = c2Vl· C2 et y t2l CtJ Trajectories of the system (8) (Improper node) EO Yj ;)' ~ ~ \' Fig. 82. Trajectories of the system (10) (Proper node) Yj SEC. 4.3 Constant-Coefficient Systems. phase plane Method E X AMP L E 3 143 Saddle Point (Fig. 83) A saddle point is a critical point Po at which there are two incoming trajeclOries. all the other trajectories in a neighborhood of Po bypass Po. The system 1 OJ Y, = [0 -1 (11) , y. Yl = thus TWO outgoing trajeclOries. and Yl has a saddle point at the origin. Its characteristic equation (I - ,1.)( - I - A) = 0 has the roots ,1.1 = I and ,1.2 = -I. For A = I an eigenvector [l OlT is obtained from the second row of (A - AI)x = 0, that is. OXI + (-I - I)x2 = O. For ,1.2 = -1 the fIrst row gives [0 Hence a general solution is nT. or or • This is a family of hyperbolas (and the coordinate axes); see Fig. 83. E X AMP L E 4 YIY2 = c{)nst. Center (Fig. 84) A center is a critical point that is enclosed by infinitely many closed trajectories. The system , y' (12) = [ ,vI = Y2 0 thus -4 Y~ = -4Y1 has a center at the origin. The characteristic equation ,1.2 + 4 = 0 gives the eigenvalues 2i and -2i. For 2i an eigenvector follows from the fIrst equation -2ixl + X2 = 0 of (A - AI)x = 0, say. [l 2ilT. For A = -2i that equation is -(-2i)xl + X2 = 0 and gives. say. [I -2ilT. Hence a complex general solution is (12*) thus The next step would be the transfOlmation of this solution 10 real form by the Euler formula (Sec. 2.2). But we were just curious to see what kind of eigenvalues we obtain in the case of a center. Accordingly, we do not continue. but start again from the beginning and use a shortcut. We rewrite the given equations in the form yi. = )'2. 4"1 = -y~; then the product of the left sides must equal the product of the right sides. By integration, This is a family of ellipses (see Fig. 84) enclosing the center at the origin. Fig. 83. Trajectories of the system (11) (Saddle point) Fig. 84. • Trajectories of the system (12) (Center) 144 E X AMP L E 5 CHAP. 4 Systems of ODEs. phase Plane. Qualitative Methods Spiral Point (Fig. 85) A spiral point is a critical point Po about which the trajectories spiral. approaching Po as t ---'> rx (or tracing these spirals in the opposite sense. away from Po). The system [-I IJ y' = (13) thus y, -I -\ has a spiral point at the origin, as we shall see. The characteristic equation is A2 + 2A + 2 = O. It gives the eigenvalues -\ + i and -\ - i. Corresponding eigenvectors are obtained from (-\ - A)XI + -"2 = O. For A = -I + ; this becomes -if1 + X2 = 0 ami we can take [I ;]T as an eigenvector. Similarly. an eigenvector corresponding to -1 - ; is [I _;]T. This gives the complex general solution Y= ci [IJ e(-I+i)t + c2 i [IJ e(-I-i)t . -; The next step would be the transformation of this complex solution to a real general solution by the Euler fOlmula. But. as in the last example. we ju~t wanted to see what eigenvalues to expect in the ca,e of a spiIaI point. Accordingly. we start again from the beginning and instead of that rather lengthy systematic calculation we use a shortcut. We multiply the f,rst equation in (13) by YI. the second by Y2. and add. obtaining We now introduce polar coordinates r. t. where r2 = ."12 + .'"22. Differentiating this with respect to t gives 21T' = 2YIY~ + 2Y2Y~' Hence the previous equation can be written , rr = 2 -r . For each real c drlr = -dt. Thus. thi~ is a spiral. a~ Fig. 85. E X AMP L E 6 In Irl = -t + c"'. claimed. (see Fig. 85). r = ce -t . • Trajectories of the system (13) (Spiral point) No Basis of Eigenvectors Available. Degenerate Node (Fig. 86) This cannot happen if A in (1) is symmetric (alQ = ajk. as in Examples 1-3) or skew-symmetnc (akj = -ajk. thus ajj = 0). And ,t does not happen in many other cases (see Examples 4 and 5). Hence it suffices to explain the method to be used by an example. SEC 4.3 145 Constant-Coefficient Systems. phase plane Method Find and graph a general ~olution of y' (14) Solution. = Ay = [ 4 -I IJ y. 2 A is not skew-symmetric! Its characteristic equation is det (A - AI) = 14 - A -I I 1= ,1.2 - 6,1. + 9 2 - A = (A - 3)2 = O. It has a double root A = 3. Hence eigenvectors are obtained from (4 - ,1.).\"1 + x2 = O. thus from Xl + -'"2 = O. say. x(l) = [I - J]T and nonzero multiples of it (which do not help). The method now is to substitute with constant u = [Ill 112{ into (14). (TIle xT-term alone, the analog of what we did in Sec. 2.2 in the case of a double rool. would nol be enough. Try iLl This gives On the right. Ax = Ax. Hence the terms AXTe At cancel, and then division by x+Au=Au, Here A = 3 and x = [l thus eAt gives (A - Anu = x. -nT. so that 4 - 3 (A - 31)u = thus [ -I A solution. linearly independent of x = [l _lJT. is u = [0 lJT. This yields the answer (Fig. 86) The critical point at the origin is often called a degenerate node. c1y(l) gives the heavy straight line, with > 0 the lower part and c1 < 0 the upper part of it.l2J give~ the right part of the heavy curve from 0 through • the second, first. and-finally-fourth quadrants. -l2) gives the other part of that curve. ("1 y (2) Y Fig. 86. Degenerate node in Example 6 CHAP. 4 146 Systems of ODEs. Phase plane. Qualitative Methods We mention that for a system (1) with three or more equations and a triple eigenvalue with only one linearly independent eigenvector, one will get two solutions. as just discussed, and a third linearly independent one from with v from 11-91 GENERAL SOLUTION Find a real general solutiun of the following systems. (Show the details.) 2. y~ , Y2 3. Y ~ = Yl + 4. Y~ )"2 9Yl , 6. y~ 2Yl .v 2 2Yl , Y2 + 1.5)'1 Y2 , + 13.5Y2 9Y2 2)"2 + 2."2 !16-171 Y~ = -4Yl - 10Y2 + CONVERSION 16. The system in Prob. 8. 17. The system in Example 5. 18. (Mixing problem, Fig. 87) Each of the two tanks contains 400 gal of water. in which initially 100 lb (Tank T l ) and 40 Ib (Tank T2 ) of fertilizer are dissolved. The intlow, circulation, and outflow are shown in Fig. 87. The mixture is kept uniform by stirring. Find the fertilizer contents Yl(t) in T] and Y2(t) in T2 . (P 2Y3 + Av = Av. Find a general solution by conversion to a single ODE. 4S . . .a lJ...... }, ----;- 7. y~ u Nat ..........TJ y~ = -4.\"1 - 4.\"2 - 4)"3 '8 --" T2 non ----;- I. ----;- Fig. 87. Tanks in Problem 18 --- 19. (Network) Show that a model for the currents I] (1) and 12 (t) in Fig. 88 is y~ 0.IY2 =)"1 - + 1.4Y3 110-151 INITIAL VALUE PROBLEMS Solve the following initial value problems. (Show the details.) 10. Y ~ = Y1 + )"2 y~ y~ = 4Yl + Y2 6 = ~Yl + .\'2 ."1(0) = 16, )"2(0) = -2 Find a general solution, assuming that R = 20 .0, L = 0.5 H, C = 2' 10-4 F. 20. CAS PROJECT. Phase Portraits. Graph some of the figures in this section, in particular Fig. 86 on the degenerate node. in which the vector y(2J depends on t. In each figure highlight a trajectory that satisfies an initial condition of your choice. c R 14. )"~ = -Yl + 5Y2 y~ = -Yl + 3Y2 15. Y~ = Y; = 2Yl 5Yt + + 5Y2 L l2.5Y2 Fig. 88. LI -Y Network in Problem 19 SEC. 4.4 Criteria for Critical Points. Stability 147 4.4 Criteria for Critical Points. Stability We continue our discussion of homogeneous linear systems with constant coefficients (1) y' (lU = Ay = in components, [ (121 From the examples in the last section we have seen that we can obtain an overview of families of solution curves if we represent them parametrically as yet) = lVI(t) Y2(t)]T and graph them as curves in the YLv2-plane, called the phase plane. Such a curve is called a trajectory of (I), and their totality is known as the phase portrait of (I). Now we have seen that solutions are of the form Substitution into (1) gives Dropping the common factor eAt, we have Ax = Ax. (2) Hence yet) is a (nonzero) solution of (1) if A is an eigenvalue of A and x a corresponding eigenvector. Our examples in the last section show that the general form of the phase portrait is determined to a large extent by the type of critical point of the system (1) defined as a point at which dY2/dYI becomes undetermined, DID; here [see (9) in Sec. 4.3] (l21YI (3) (lUYI + + (l22Y2 1112Y2 We also recall from Sec. 4.3 that there are various types of critical points, and we shall now see how these types are related to the eigenvalues. The latter are solutions A = Al and A2 of the characteristic equation (4) det (A - AI) = I(lU - A al2 U21 (122 - A This is a quadratic equation A2 - pA given by + I = A2 - (au + a22)A + det A = O. q = 0 with coefficients p. q and discriminant D. From calculus we know that the solutions of this equation are (6) Furthermore, the product representation of the equation gives 148 CHAP. 4 Systems of ODEs. Phase plane. Qualitative Methods Hence]J is the sum and q the product of the eigenvalues. Also Al - A2 Together. = VA from (6). (7) This gives the criteria in Table 4.1 for classifying critical points. A derivation will be indicated later in this section. Table 4.1 I Eigenvalue Criteria for Critical Points (Derivation after Table 4.2) Name - (a) Node (b) Saddle point I (c) Center (d) Spiral point p = Al + p=O p=l=O A2 q = AIA2 q>O q<O q>O !:J. = (AI ~~O ~ <0 A2)2 Comments on A1 • A2 Real, same sign Real, opposite sign Pure imaginary Complex. not pure imaginary Stability Critical points may also be classified in terms of their stability. Stability concepts are basic in engineering and other applications. They are suggested by physics. where stability means, roughly speaking, that a small change (a small disturbance) of a physical system at some instant changes the hehavior of the system only slightly at all future times t. For critical points, the following concepts are appropriate. DEFINITIONS Stable, Unstable, Stable and Attractive A critical point Po of (\) is called stable 2 if, roughly, all trajectories of (\) that at some instant are close to Po remain close to Po at all future times: precisely: if for every disk D" of radius E > 0 with center Po there is a disk D t; of radius 8 > 0 with center Po such that every trajectory of (l) that has a point PI (corresponding to t = t 1 , say) in Dt; has all its points corresponding to t ~ t1 in DE" See Fig. 89. Po is called unstable if Po is not stable. Po is called stable and attractive (or asymptotically stable) if Po is stable and every trajectory that has a point in DB approaches Po as t ~ x. See Fig. 90. Classification criteria for critical points in terms of stability are given in Table 4.2. Both tables are summarized in the stability chart in Fig. 91. In this chart the region of instability is dark blue. 21n the sense of the Russian mathematician ALEXANDER MICHAILOVICH LJAPUNOV (1857-1918), whose work was fundamental in stability theory for ODEs. This is perhaps the most appropriate defmition of stability (and the only we shall lise), but there are others, too. SEC 4.4 Criteria for Critical Points. Stability 149 Fig. 89. Stable critical point Po of (1) (The trajectory initiating at P1 stays in the disk of radius E.) Table 4.2 Fig. 90. Stable and attractive critical point Po of (1) Stability Criteria for Critical Points Type of Stability p (a) Stable and attractive (b) Stable (e) Unstahle Al = + q A2 p<O AIA2 -- q>O q>O q<O p~O p>O = OR - q Saddle point p Fig. 91. Stability chart of the system (1) with p, q, ~ defined in (5). Stable and attractive: The second quadrant without the q-axis. Stability also on the positive q-axis (which corresponds to centers). Unstable: Dark blue region We indicate how the criteria in Tables 4.1 and 4.2 are obtained. If q = Al A2 > 0, both eigenvalues are positive or both are negative or complex conjugates. If also p = Al + A2 < O. both are negative or have a negative rea] part. Hence Po is stable and attractive. The reasoning for the other two lines in Table 4.2 is similar. If fj, < 0, the eigenvalues are complex conjugates, say, Al = a + i{3 and A2 = a - i{3. If also p = Al + A2 = 20' < 0, this gives a spiral point that is stable and attractive. If p = 20' > 0, this gives an unstable spiral point. If p = 0, then A2 = -AI and q = AIA2 = -AI2. If also q > 0, then A12 = -q < 0, so that AI, and thus A2, must be pure imaginary. This gives periodic solutions, their trajectories being closed curves around Po, which is a center. E X AMP L E 1 Application ofthe Criteria in Tables 4.1 and 4.2 In Example I, Sec. 4.3, we have y' = [-: lJ y,p = -6, q = 8, -3 is stable and attractive by Table 4.2(a). ~ = 4, a node by Table 4.1(a), which • CHAP. 4 150 E X AMP L E 2 Systems of ODEs. phase plane. Qualitative Methods Free Motions of a Mass on a Spring What kind of critical point does my" + 0" + ky = Solution. Division by m gives y" = -(kIm))" 4.1). Then)'~ = y" = -(klm)Yl - (elm)Y2' Hence y' = 0 [ I ] y -kIm -elm det ' (A - 0 in Sec. 2.4 have? (elm)y'. To get a system, set AI) = I )'1 = y, Y2 = Y' (see Sec. -A -kIm We see thatp = -elm. q = kIm, l:!. = (elm)2 - 4klm. From this and Tables 4.1 and 4.2 we obtain the following results. Note that in the last three cases the discriminant l:!. plays an essential role. No damping. c = 0, p = 0, q > 0, a center. UnderdaJllping. c 2 < 4mk, p < O. q > O. ~ < 0, a stable and attractive spiral point. Critical damping. c 2 = 4mk. p < O. q > O. l:!. = O. a stable and attractive node. Overdamping. c 2 > 4mk, p < 0, q > 0, l:!. > 0, a stable and attractive node. -- .•.... -.-.... 11-91 ... .. .. ... .-..---....... --.---~ TYPE AND STABILITY OF CRITICAL POINT Determine the type and stability of the critical point. Then find a real general solution and sketch or graph some of the trajectories in the phase plane. (Show the details of your work.) 1. , , Yl 2. 2)'2 Y2 3. Yl = 2Yl + Y2 .y 1 + 2)'2 , Y2 = , , 5. Yl Y2 )'1 - , , 4Y2 Y2 = 9. Y~ = , Y2 6. Yl , -5)'1 Yl + - 2)'2 IOY2 Y2 = 7)'1 , 3)'1 8Y2 + Y2 = -5)'1 - 8Y1 + 2)'2 y~ = 2Yl + Y2 5Y2 3Y2 FORM OF TRAJECTORIES What kind of curves are the trajectories of the following ODEs in the phase plane? 10. y" + 11. y" - 12. Y" + 5)" = 0 k 2y = 0 14. (Transformation of variable) What happens to the system (1) and its critical point if you introduce as a new independent variable? T = -t 15. (Types of critical points) Discuss the critical points in (10)-( 1-1-) in Sec. 4.3 by applying the criteria in Tables 4.1 and 4.2 in tlus section. 16. (Perturbation of center) If a system has a center as 3Y2 4. Y~ = Y2 , )'1 110-121 = 8. Yl -2Y2 7. Yl 4Yl , + -4Yl Y2 , )'1 , Y2 = 8y! , • tBY = 0 13. (Damped oscillation) Solve y" + 4y' + 5y = O. What kind of curves do you get as trajectories? its critical point, whal happens if you replace the matrix A by A = A + kI with any real number k =1= 0 (representing measurement errors in the diagonal entries)? 17. (Perturbation) The system in Example 4 in Sec. 4.3 has a center as its critical point. Replace each Gjk in Example 4, Sec. 4.3, by ajk + b. Find values of b such that you get (a) a saddle point. (b) a stable and attractive node, (c) a stable and attractive spiral. (d) an unstable spiral, (e) an unstable node. 18. CAS EXPERIMENT. Phase Portraits. Graph phase portraits for the systems in Prob. 17 with the values of b suggested in the answer. Try to illustrate how the phase portrait changes "continuously" under a continuous change of b. 19. WRITING EXPERIMENT. Stability. Stability concepts are basic in physics and engineering. Write a two-part report of 3 pages each (A) on general applications in which stability plays a role (be as precise as you can), and (B) on material related to stability in this section. Use your own formulations and examples: do not copy. 20. (Stability chart) Locate the critical points of the systems (0)-(14) in Sec. 4.3 and of Probs. 1,3,5 in this problem set on the stability chart. 151 SEC. 4.5 Qualitative Methods for Nonlinear Systems 4.5 Qualitative Methods for Nonlinear Systems Qualitative methods are methods of obtaining qualitative information on solutions without actually solving a system. These methods are particularly valuable for systems whose solution by analytic methods is difficult or impossible. This is the case for many practically important nonlinear systems y' (1) = fey), thus Y~ = fl(Yl, Y2) Y~ = f 2(Yb Y2)· In this section we extend phase plane methods, as just discussed, from linear systems to nonlinear systems (1). We assume that (1) is autonomous, that is, the independent variable t does not occur explicitly. (All examples in the last section are autonomous.) We shall again exhibit entire families of solutions. This is an advantage over numeric methods, which give only one (approximate) solution at a time. Concepts needed from the last section are the phase plane (the YIY2-plane), trajectories (solution curves of (1) in the phase plane), the phase portrait of (1) (the totality of these trajectories), and critical points of (1) (points (Yb .\'2) at which both fl()'b )'2) and f2(Yb )'2) are zero). Now (1) may have several critical points. Then we discuss one after another. As a technical convenience, each time we first move the critical point Po: (a, b) to be considered to the origin (0, 0). This can be done by a translation which moves Po to (0, 0). Thus we can assume Po to be the origin (0, 0), and for simplicity we continue to write .\'1' Y2 (instead of 511, )'2). We also assume that Po is isolated, that is, it is the only critical point of (1) within a (sufficiently small) disk with center at the origin. If (1) has only finitely many critical points, this is automatically true. (Explain!) Linearization of Nonlinear Systems How can we determine the kind and stability property of a critical point Po: (0, 0) of (1)? In most cases this can be done by linearization of (1) near Po. writing (1) as y' = fey) = Ay + hey) and dropping hey), as follows. Since Po is critical, fl(O. 0) = 0, f2(0, 0) = O. so that fl and f2 have no constant terms and we can write (2) y' = Ay + hey), thus A is constant (independent of t) since (1) is autonomous. One can prove the following (proof in Ref. [A7], pp. 375-388, listed in App. 1). CHAP. 4 152 Systems of ODEs. Phase Plane. Qualitative Methods Linearization THEOREM 1 If f 1 and f 2 ill (1) are comillllolls alld have contilluollS partial derivatives ill a neighborhood of the critical point Po: (0, 0). alld !f det A =1= 0 in (2), then the kind and stability of the critical poillt of (1) {Ire the same as those of the linearized system (3) , , Ay, Y .r I = aIL"l + (/12Y2 Y2 = a21Yl + a22Y2' , thus Exceptions occllr!f A has equal or pure i111agilwry eigellvalues; then (1) may have the same kind of critical point as (3) or a spiral point. E X AMP L E 1 Free Undamped Pendulum. Linearization Figure 91a shows a pendulum comisting of a body of mass 111 (the bob) and a rod of length L. Detenmne the locations and types of the critical points. Assume that the mass of the rod and air resistance are negligible. Solutioll. Step 1. Settillg lip the mathematical model. Let () denote the angular displacement, measured counterclockwise from the equilibrium position. The weight of the bob is mg (g the acceleration of gravity). It causes a restoring force IIlg sin () tangent to the curve of motion (circular arc) of the bob. By Newton's second law. at each instant this force is balanced by the force of acceleration mL()", where L()" is the acceleration: hence the resultant of these two forces is zero. and we obtain as the mathematical model IIlL()" + IIlg sin () = O. Dividing this by mL. we have ()" + k sin () (4) = 0 When () is very small. we can approximate sin () rather accurately by () and obtain as an approximate solution A cos V kt + B sin Vkt. but the exact solution for any () is not an elementary function. Step 2. Critical po;"ts (0, 0), ±(2rr, 0), ±l4rr, 0), ... , Lilleari;;.atioll. To obtain a system of ODEs. we set () = .1"1' ()' = )"2' Then from (4) we obtain a nonlinear system (I) of the form Y~ = hlYl, Y2) = Y2 y~ = .12(.1"1 . .1"2) = -k sinY1· The right sides arc both zero when .1'2 = 0 and sin.\"1 = O. This gives infinitely many critical points where Il = O. ± I. ±2, .... We consider (0, 0). Since the Maclaurin series is (111T. 0). sin."1 = .\'1 - ~Y13 + - ... = .\'1' the linearized system at (0. 0) i, y' = Ay = [ 0 thus -k To apply our criteria in Sec. -lA we calculate p = all + a22 = 0, q = det A = k = gIL (> 0), and j. = p2 - 4q = -4k. From this and Table 4.1 (c) in Sec. 4.4 we conclude that (0. 0) is a center. which is always stable. Since sin (j = sinYI is periodic with period 11T. the critical points (/l1T. 0), /I = ±1. ±4..... are all centers. Step 3. Critical poillts ±(rr. 0). ±(3rr. 0), ±(5rr. 0) •.. '. Lilleari:.atioll. We now consider the critical point (1T. 0). setting () - 1T = Yl and «() - 1T)' = ()' = .\'2' Then in (4). sin () = sin 1.\'1 + 1Tl = -sin Yl = -Yl + ~YI 3 - + ... = -\'1 SEC. 4.5 153 Qualitative Methods for Nonlinear Systems and the linearized system at (7T, 0) is now thus We see that p = 0, q = - k « 0), and D. = -4q = 4k. Hence, by Table 4.1(b), this gives a saddle point, which is always unstable. Because of periodicity, the critical points (/17T, 0), /1 = ::'::1, ::'::3, .. " are all saddle points. • These results agree with the impression we get from Fig. Y2b. mg (a) Pendulum (b) Solution curvesY2(Yj) of (4) in the phase plane Fig. 92. EXAMPLE 2 Example 1 (C will be explained in Example 4.) Linearization of the Damped Pendulum Equation To gain further experience in investigating critical points, as another practically imponant case. let us see how Example I changes when we add a damping term ce' (damping proportional to the angular velocity) to equation (4), so that it becomes e" + ce' (5) + k sin e= 0 where k > 0 and c ::::; 0 (which includes Ollr previous casc of no damping, c = o1. Setting before, we obtain the nonlinear system (use e" = )'~) e= -"1, e' = Y2' as , Y1 =)'2 y~ = -k sinYI - cY2. We see that the critical poinrs have the same locations as before. namely. (0, 0). (::'::7T. 0), (::'::27T. 0), .... We consider (0, 0). Linearizing sin Yl = )'1 as in Exan1ple 1, we get the linearized system at (0, 0) , (6) y' = Ay = [0 -k )'1 = lJ Y -c Y2 thus ' This is identical with the system in Example 2 of Sec 4.4, except for the (positive!) facror I1l (and except for the physical meaning of Yl)' Hence for c = 0 (no damping) we have a center (see Fig. 92b). for small damping we have a spiral point (see Fig. 93), and so on. We now consider the critical point (7T, 0). We set e - 7T = Yl, (e - 7T/ = e' = Y2 and linearize sin e ~ sin(YI + 7T) = - sinYl = -y!. This gives the new linemized system at (7T, 0) , (6*) y'=AY=[O k IJy, -c thus Yj = Y2 154 CHAP. 4 Systems of ODEs. Phase plane. Qualitative Methods For our criteria in Sec 4.4 we calculate p = au + a22 = -c. q = det A This gives the following results for the critical point at (1T, 0). = -k, and D. = p2 - 4q = c 2 + 4k. No damping. c = 0, p = O. q < 0, ~ > O. a saddle point. See Fig. 92b. Damping. c > 0, p < O. q < 0, D. > O. a saddle point. See Fig. 93. Since sin ,1'1 is periodic with period 21T, the critical points (:+:21T, 0), (:+:41T, 0), ... are of the same type a~ (0.0). and the critical points (-1T, 0), (:+:31T. 0), ... are of the same type as (1T. 0), so that our task is finished. Figure 93 shows the trajectories in the case of damping. What we see agrees with our physical intuition. Indeed. damping means loss of energy. Hence instead of the closed trajectories of periodic solutions in Fig. 92b we now have trajectories spiraling around one of the critical points (0, 0), (:!:21T, 0), .... Even the wavy trajectorie~ corresponding to whirly motions eventually spiral around one of these points. Furthermore. there are no more trajectories that connect critical points (as there were in the undamped case for the saddle points). • Fig. 93. Trajectories in the phase plane for the damped pendulum in Example 2 Lotka-Volterra Population Model E X AMP L E 3 Predator-Prey Population Model3 This model concerns two species, say, rabbits and foxes, and the foxes prey on the rabbits. Step 1. Setting lip the model. We assume the following. 1. Rabbits have unlimited food supply. Hence if there were no foxes. their number Yl(t) would grow exponentially. \'~ = ay!. 2. Actually, Yl is decrea,ed because of the kill by foxes. say. at a rate proportional to YIY2' where the number of foxes. Hence y~ = aYl - bYIY2, where a > 0 and b > O. )'2(t) is 3. If there were no rabbits, then Y2(f) would exponentially decreaSe to zero, y~ = -1.\'2' However, Y2 is increased by a rate proportional to the number of encounters between predator and prey; together we have Y~ = - IY2 + kyl)'2' where k > 0 and 1 > O. This gives the (nonlinear!) Lotka-Volterra system (7) Y; = h(Yl, )'2) = aYl - bylY2 ,1'2 = f 2(Yl,.\'2) = /"'}'1.'·2 - lY2 . 3lntroduced by ALFRED 1. LOTKA (1880-1949), Amelican biophySicist. and VITO VOLTERRA (1860-1940), Italian mathematician, the initiator of functional analysis (see [GR7] in App. 1). SEC. 4.5 Qualitative Methods for Nonlinear Systems 155 Step 2. Critical point (0, 0), Linearization. We see from (7) that the critical pomts are the solutIOns of (7*) I a The solutions are (Yl, Y2) = (0,0) and (k' b)' We consider (0, 0). Dropping -bYIY2 and kYIY2 from 0) gives the linearized system OJ y. -I Its eigenvalues are Al = a > 0 and ,1,2 = -I < O. They have opposite signs, so that we get a saddle point. Step 3. Critical point (lIk, alb), Linearization. We set Yl = Yl + Ilk, Y2 = 3'2 + alb. Then the critical point (Ilk, alb) corresponds to (Yl, Y2) = (0,0). Since Y; = y;, y~ = y~, we obtain from 0) [factorized as in (8)] yi = (Yl + ±) y~ (Y2 + i) [k(S\ + ±) - lJ = [a - b(Y2 + ~) J = (Yl ±) + (3'2 + (-byv i )k)\. Dropping the two nonlinear terms -bYIY2 and "5'IY2, we have the linearized system ~, (a) Yl = - (b) Y2= Ib T ~ Y2 (7**) ~, ak b ''t. The left side of (a) times the right side of (b) must equal the right side of (a) times the left side of (b), ak ~ By integration. bYI 2 lb ~ + T)'2 2 = const. This is a family ellipses, so that the critical point (Ilk, alb) of the linearized system 0**) is a center (Fig. 94). It can be shown by a complicated analysis that the nonlinear system (7) also has a center (rather than a spiral point) at (Ilk, alb) surrounded by closed trajectories (not ellipses). We see that the predators and prey have a cyclic variation about the critical point. Let us move counterclockwise around the ellipse, beginning at the right vertex, where the rabbits have a m1L'dmum number. Foxes are sharply increasing in number until they reach a maximum at the upper vertex, and the number of rabbits is then sharply decreasing until it reaches a minimum at the left vertex, and so on. Cyclic variations of this kind have been observed in nature, for example, for lynx and snowshoe hare near the Hudson Bay, with a cycle of about 10 years. For models of more complicated situations and a systematic discussion, see C. W. Clark, Mathematical Bioeconolllics (Wiley, 1976), • Y2 ~ b ~ ------~ I k Ecological equilibrium pOint and trajectory of the linearized Latka-Volterra system (7**) Fig. 94. 156 CHAP. 4 Systems of ODEs. phase plane. Qualitative Methods Transformation to a First-Order Equation in the Phase Plane Another phase plane method is based on the idea of transforming a second-order autonomous ODE (an ODE in which t does not occur explicitly) F()'. to first order by taking), y" by the chain rule, y'. -,"") = 0 = )"1 as the independent variable, setting y' = )'2 and transforming )' dY2 dy! " =)'2, = dy! dt Then the ODE becomes of first order, (8) and can sometimes be solved or treated by direction fields. We illustrate this for the equation in Example I and shall gain much more insight into the behavior of solutions. E X AMP L E 4 An ODE (8) for the Free Undamped Pendulum If in (4) 6" +k sin 6 = 0 we set 6 = .\"1. 6' = .1"2 (the angular velocity) and use 6" dY2 dt =- dY2 ((VI dYl dt we get Separation of variables gives.l'2 dY2 = -k sin Yl elYl' By integration. (9) ll'22 = k cosYl + e (e constant). Multiplying this by mL2. we get We see that these three terms are energies. Indeed. Y2 is the angular velocity. so that LY2 is the velocity and the tirst term b the kinetic energy. The ~ecoml term (including the minus sign) is the potential energy of the pendulum. 2 and mL e is its total energy, which is constant, as expected from the law of conservation of energy, because there is no damping (no loss of energy). The type of motion depends on the total energy. hence on C. as follows. Figure 92b on p. 153 shows trajeclOries for various values of C. These graphs continue periodically with period 27TtO the left and to the right. We see that some of them are ellipse-like and closed. others are wavy, and there are two trajectories (passing through the saddle points (1/7T. 0). n = ::':: I. ::'::3.... I that ~eparate those two types of trajectories. From (9) we see that the smallest possible e is e = -k; then.l"2 = 0, and cos VI = I. so that the pendulum is at rest. The pendulum will change its direction of motion if there are points at which Y2 = e' = O. Then k cos Yl + e = 0 by (9). If,vl = 7T, then cos .1'1 = - I and e = k. Hence if -J.. < e < k, then the pendulum reverses its direction for a IYll = lei < 7T. and for these values of e with lei < Ii: the pendulum oSl:iIlates. This corresponds to the closed trajectories in the figure. However. if e > k, then Y2 = 0 is impossible and the pendulum makes a whirly motion that appears as a wavy trajectory in the YIY2-plane. Finally. the value e = k correspond, to the two "separating trajectories" in Fig. 92b connecting the saddle points. • The phase plane method of deriving a single first-order equation (8) may be of practical interest not only when (8) can be solved (as in Example 4) but also when solution is not possible and we have to utilize direction fields (Sec. 1.2). We illustrate this with a very famous example: SEC. 4.5 Qualitative Methods for Nonlinear Systems E X AMP L E 5 157 Self-Sustained Oscillations. Van der Pol Equation There are physical systems such that for small oscillations, energy is fed into the system, whereas for large oscillations. energy is taken from the ~y~tem. In o!her words, large oscillations will be damped, whereas for small oscillations there is "negative damping" (feeding of energy into the system). For physical reason~ we expect such a system to approach a periodic behavior, which will thus appear as a closed trajectory in the phase plane. called a limit cycle. A differential equation describing such vibrations is the famous van der Pol 4 equation (M > 0, constant). (10) It first occurred in the study of electrical circuits containing vacuum tubes. For M = 0 this equation becomes Y" + ." = 0 and we obtain harmonic oscillations. Let M > O. The damping term has !he factor -M(I _ y2). This is negative for small oscillation~, when y2 < I. so that we have "negative damping," is .lero for y2 = I (no damping), and is positive if)'2 > 1 (positive damping, loss of energy). If M is small, we expect a limit cycle that is almost a circle because then our equation differs bm little from y" + )' = O. If M is large. the limit cycle will probably look different. Setting y = ."1, y' = )'2 and using y" = (dY2/dYl)Y2 as in (8), we have from (10) (II) The isoclines in the YIY2-plane (the phase plane) are the curves dY2/dYl ~ K = consf, thaI is, Solving algebraically for Y2, we see that the isoclines are given by Yl (Figs. 95, 96). Y2 5 K=O K=-l K= l 4 K= 1 K=-5 / K=l K=-5 K=-l -5 Fig. 95. Direction field for the van der Pol equation with fL = 0.1 in the phase plane, showing also the limit cycle and two trajectories. See also Fig. 8 in Sec. 1.2. 4BALTHASAR VAN DER POL (I 88<j-I<jS9), Dutch physicist and engineer. CHAP. 4 158 Systems of ODEs. Phase plane. Qualitative Methods Figure 9S shows some isoclines when fL is small, f.L = 0.1. the limit cycle (almost a circle), and two (blue) trajectories approaching it. one from the outside and the other from the inside. of which only the initial portion, a small spiral, is shown. Due to this approach by trajectories. a limit cycle differs conceprually from a closed curve (a trajectory) surrounding a center, which is not approached by trajectories. For larger fL the limit cycle no longer resembles a circle, and the trajectories approach it more rapidly than for smaller fL. Figure 96 illustrates this for fL = I. • K=D K=-l K=-l K= 1 \ K=O ,,/'" K=-5 -3 K=O K=-5 K=O K= 1 K=-l Fig. 96. . CRITICAL POINTS, LINEARIZATION 9 Detennine the location and type of all critical points by linearization. In Probs. 7-12 first transform the ODE to a system. (Show the details of your work.) 1. y~ = Y2 3. y~ 2. )"22 5. )' ~ -YI , + Y . 2 \' .2 - + .\' 1 2 + Y2 - Y2 y~ = YI - 3Y2 2 Y2 )'2 - 2 6. y~ = Y2 - Y2 2 y~ = Yl - Y1 2 -Yl - Y2 Y2 4\' . 1 4. y~ = -3Yl 2Yl - h .\'2 , •\' 1 ,. , 4Y2 , 7. y" + o K=-l Direction field for the van der Pol equation with IL = 1 in the phase plane. showing also the limit cycle and two trajectories approaching it 8. y" + 9)' + y2 o 2 \' " + cos y Ll. Y" + 4)' - 0 10. y" y3 = 0 12. Y" = + sin y = + Y' + 2}' - 0 y2 = 0 13. (Trajectories) What kind of curves are the trajectories of -,~y" + 2/ 2 = O? 14. (Trajectories) Write the ODE y" - 4y + )'3 = 0 as a system. ~llive it for Y2 as a function of ."1. and sketch or graph some of the trajectories in the phase plane. 15. (Trajectories) What is the radius of a real general solution of y" + Y = 0 in the phase plane? 16. (Trajectories) In Prob. 14 add a linear damping tenn to get y" + 2y' - 4y + y3 = O. Using arguments from mechanics and a comparison with Prob. 14, as well as with Examples I and 2. guess the type of each critical point. Then determine these types by linearization. (Show all details of your work.) SEC. 4.6 17. (Pendnlum) To what state (position, speed, direction of motion) do the four points of intersection of a closed trajectory with the axes in Fig. 92b correspond? The point of intersection of a wavy curve with the Y2- axis ? 18. (Limit cycle) What is the essential difference between a limit cycle and a closed trajectory surrounding a center? 19. CAS EXPERIMENT. Deformation of Limit Cycle. Convert the van der Pol equation to a system. Graph the limit cycle and some approaching trajectories for fL = 0.2,0.4,0.6, 0.8, 1.0, l.5, 2.0. Try to observe how the limit cycle changes its form continuously if you vary IL continuously. Describe in words how the limit cycle is deformed with growing fL. 20. TEAM PROJECT. Self-sustained oscillations. (a) Van der Pol Equation. Determine the type of the critical point at (0, 0) when IL > 0, IL = 0, IL < O. 4.6 159 Nonhomogeneous Linear Systems of ODEs Show that if IL -) 0, the isoclines approach straight lines through the origin. Why is this to be expected? (b) Rayleigh equation. Show that the so-called Rayleigh equation5 y" - IL(I - §y'2)y' + Y = 0 (IL> 0) also describes self-sustained oscillations and that by differentiating it and setting y = y' one obtains the van der Pol equation. (c) Duffing equation. The Duffing equation is y" + wo2 y + f3y 3 = 0 where usually 1f31 i~ small. thus characterizing a small deviation of the restoring force from linearity. f3 > 0 and f3 < 0 are called the cases of a hard spring and a soft spring, respectively. Find the equation of the trajectories in the phase plane. (Note that for f3 > 0 all these curves are closed.) Nonhomogeneous Linear Systems of ODEs In this last section of Chap. 4 we discuss methods for solving nonhomogeneous linear systems of ODEs (1) y' = Ay + g (see Sec. 4.2) where the vector g(t) is not identically zero. We assume g(t) and the entries of the 11 X II matrix A(t) to be continuous on some interval 1 of the t-axis. From a general solution y(h)(t) of the homogeneous system y' = Ay on J and a particular solution y(P)(t) of (1) on J [i.e., a solution of (1) containing no arbitrary constants], we get a solution of (l), (2) y is called a general solution of (I) on 1 because it includes every solution of (l) on 1. This follows from Theorem 2 in Sec. 4.2 (see Prob. 1 of this section). Having studied homogeneous linear systems in Secs. 4.1-4.4. our present task will be to explain methods for obtaining particular solutions of (I). We discuss the method of undetermined coefficients and the method of the variation of parameters; these have counterparts for a single ODE, as we know from Secs. 2.7 and 2.10. 5 LORD RAYLEIGH (JOHN WILLIAM STRUTI) (1842-1919). great English physicist and mathematician. professor at Cambridge and London. known by his important contributions to the theory ot waves, elasticity theory. hydrodynamics. and various other branches of applied mathematics and theoretical physics. In 1904 he received the Nobel Prize in physics. 160 CHAP. 4 Systems of ODEs. Phase plane. Qualitative Methods j Method of Undetermined Coefficients As for a single ODE, this method is suitable if the entries of A are constants and the components of g are constants, positive integer powers of t, exponential functions, or cosines and sines. In such a case a particular solution yep) is assumed in a fonn similar to g; for instance, y(P) = U + vt + wt2 if g has components quadratic in t, with u, v, w to be determined by substitution into (I). This is similar to Sec. 2.7, except for the Modification Rule. It suffices to show this by an example. E X AMP L E 1 Method of Undetermined Coefficients. Modification Rule Find a general solution of Solution. A general equation of the homogeneous system is (see Example I in Sec. 4.3) Since A = -2 is an eigenvalue of A, the function e- 2t on the right also appears in yChl, and we must apply the Modification Rule by setting (rather than ue -2t). Note that the first of these two terms is the analog of the modification in Sec. 2.7. but it would not be sufficient here. (Try It.) By substitution, Equating the Ie -2t-terms on both sides, we have - 2u = Au. Hence u is an eigenvector of A corresponding to A = -2; thus [see (5)] u = all lIT with any a 'F O. Equating the other terms gives thus Collecting terms and reshuffling gives -v 1 + V2 = By addition, 0 = -2a - 4, a = -2, and then v2 = VI We can simply choose k = O. This gives the answer -a + 2. + 4, say, VI For other k we get other v; for instance, k = -2 gives v = [-2 = k, v2 = k + 4, thus, v = [k + 4]T. 2]T, so that the answer becomes (5*) etc. • Method of Variation of Parameters This method can be applied to nonhomogeneous linear systems (6) k y' = A(t)y + get) SEC. 4.6 161 Nonhomogeneous Linear Systems of ODEs with variable A = ACt) and general get). It yields a particular solution y(p) of (6) on some open interval J on the t-axis if a general solution of the homogeneous system y' = A(t)y on J is known. We explain the method in terms of the previous example. EXAMPLE 2 Solution by the Method of Variation of Parameters Solve (3) in Example I. Solutioll. A basis of solutions of the homogeneous, ystem is [e -2t the general solution (4) of the homogenous system may be written -2t y(hl = (7) [ e e- 2t -e -4t]T. Hence e '"I -4t] [ ] _e-4t C2 = YU)e. r Here, Y(n = [y(!) y(2J is the fundamental matrix (see Sec. 4.2). As in Sec 2.10 we replace the constant vector e by a variable vector u(t) to obtain a particular solution yep) = Y(t)u(t). Substitution into (3) y' = Ay + g gives Y'u+Yu'=AYu+g. (8) Now since y(lJ and y(2J are solutions of (he homogeneous system. we have Y' = AY. thus Hence Y' u = AYu, so that (8) reduces to Yu' = g. The solution is here we use that the inverse y- 1 of Y (Sec. 4.0) exists because the detenninant of Y is the Wronskian W, which is not zero for a basis. Equation (9) in Sec. 4.0 gives the form of y-l. y-l = -e -4t - 2e -6t [ -e -2t We multiply this by g, obtaining Integration is done componentwi,e (just a~ differentiation) and gives L t [ u(t) = -2 - ] dt = [ - 2 t ] -4e2t _2e2t + 2 (where + 2 comes from the lower limit of integration). From this and Y in (7) we obtain e- 2t Yu = [ e -2t 4t e- ] [ -e- 4t -2t -2e2t + 2 ] [-2Ie= 2t - 2e- 2t + 2e- 4t -2te- 2t + 2e- 2t _ 2e- 4t [-2t - ] = 2J -21 + 2 2t e- + The last term on the right is a solution of the homogeneous system. Hence we can absorb it into lh). We thus obtain as a general solution of the system (3). in agreement with (5*). (9) • 162 CHAP. 4 Systems of ODEs. Phase plane. Qualitative Methods - ----_ Z.Q.i= ~CI==~ 1. (General solution) Prove that (2) includes every solution of (I). 12-91 14. )' ~ = 3YI - 4Y2 + 20 )'~ co~ t GENERAL SOLUTION Find a general solution. (Show the details of your work.) 2'Y~=Y2+t y~ = 4. Y; = + + 5 cos t 5. J~ )'2 "2 - 5 sin t )"~ = 3Y1 - 7. Y~ = -14)"1 y~ = -5Yl 8. y; = + + = + = = y~ = -4Y1 2)'1 + 5)'1 - 15. + 5 2Y2 + 12 )"2 - 30 + 10(1 - t - t 2 ) IOY2 + 4 - 20t - 6t 2 4)'2 + + lit + 3e- t 6.\'2 16. 4Yl + 8Y2 + 2 cos t y~ = 6YI + 2Y2 + = - cust - 16 sin t 14 sint + 162 IOY2 Y2 - 3241 -3Ji - )'~ = 5Y1 + 9t = 4Y2 6)'2 IOY1 - y~ = 6YI - 9 . .\'~ Y~ 3t Y1 )"1 3. Y~ - 17. (Network) Find the currents in Fig. 97 when R = 2.5 D. L = 1 H, C = 0.04 F, E(t) = 845 sin t Y, and 11(0} = 0, [2(0) = O. (Show the details.) 18. (Network) Find the currents in Fig. 97 when R = I D. L = 10 H, C = 1.25 F, E(t) = 10 kY, and 11 (0) = 0, [2(0) = O. (Show the details.) 15 15T - 20 c E 10. CAS EXPERIMENT. Undetermined Coefficients. Find out experimentally how general you mLLst choose y(jJ). in particular when the components of g have a different form (e.g., as in Prob. 9). Write a short report, covering also the situation in the case of the modification rule. = 1-161 INITIAL VALUE PROBLEM Solve {showing details): II. y~ = -2Y2 y~ = + ." ~ = Network in Probs. 17, 18 19. (Network) Find the CUiTents in Fig. 98 when R1 = 2 D, R2 = 8 n. L = 1 H. C = 0.5 F. E = 200 Y. (Show the details.) L 4t 2YI - 2t ! )'1(0) = 4, )'2 (0) = 12. Fig. 97. 4Y2 + 5e t Switch 20e- t y~ = -Yl - Fig. 98. 13. Y; = YI y~ = )'1(0) = 1, + )'2 2.\"2 + )"2(0) 1 + e 2t + t = -4 - 2t c Network in Prob. 19 20. WRITING PROJECT. Undetermined Coefficients. Write a short report in which YOLL compare the appl ication of the method of undetennined coeflicients to a single ODE and to a system of two ODEs, using ODEs and systems of your choice. 163 Chapter 4 Review Questions and Problems , .• TIONS AND PROBLEMS 1. State some applications that can be modeled by systems of ODEs. 23. Y ~ = 24. y~ = Y1 - 4.h + 3Y2 + 2 2. What is population dynamics? Give examples. 3. How can you transform an ODE into a system of ODEs? 4. What are qualitative methods for systems? Why are they important? y~ = 2Y2 - sin t 3Y1 - 4Y2 - cos t 5. What is the phase plane? The phase plane method? The phase portrait of a system of ODEs? 6. What is a critical point of a system of ODEs? How did we classify these points? 7. What are eigenvalues? What role did they play in this chapter? 8. What does stability mean in general? In connection with critical points? 9. What does linearization of a system mean? Give an example. 10. What is a limit cycle? When may it occur in mechanics? 26. (Mixing problem) Tank Tl in Fig. 99 contains initially 200 gal of water in which 160 lb of salt are dissolved. Tank T2 contains initially 100 gal of pure water. Liquid is pumped through the system as indicated. and the mixtures are kept uniform by stirring. Find the amounts of salt Y1(t) and Y2(t) in Tl and T2 , respectively. Water, -- 10< Il~-!2J GENERAL SOLUTION. CRITICAL POINTS Find a general solution. Determine the kind and stability of the critical point. (Show the details of your work.) 11. Y~ 12. Y~ = 9Y1 = 4.\'2 I Y2 13. Y~ I 14. Y1 = Y2 I Y2 15. y~ = 16 gal/min 1.5.h - 6Y2 I 16. Y1 I .\'2 I 18. Y1 = Fig. 99. Y2 3Y2 3)'1 3)'1 + 3)'2 -3Y1 2Y2 -2Y1 3Y2 3Y1 + I Y2 = -5Yl 3Y2 NONHOMOGENEOUS SYSTEMS Find a general solution. (Show the details.) y~ = 12Y1 22. y~ = )'1 + + 6t + 1 Y2 + sin t Tanks in Problem 26 5Y2 - [u.!.-~ = 3)'2 --- 27. (Critical point) What kind of critical point does y' = Ay have if A has the eigenvalues -6 and I? 28. (Network) Find the currents in Fig. 100. where R1 = 0.5 fl, R2 = 0.7 fl, Ll = 0.4 H, L2 = 0.5 H, E = 1 kV = 1000 V, and ll(O) = 0,/2(0) = O. Fig. 100. 20. y~ Mixture, o gal/min Network in Problem 28 29. (Network) Find the currents in Fig. 10 1 when R = 10 fl, L = 1.25 H. C = 0.002 F. and 11 (0) = liG) = 3 A. 21. )'~ = )'1 + 2.\'2 + e 2t Fig. 101. Network in Problem 29 CHAP. 4 164 Systems of ODEs. Phase plane. Qualitative Methods 130-331 LINEARIZATION Detelmine the location and kind of all critical points of the given nonlinear system b) linearization. 30. y~ 31. )'~ = )'2 , Y2 ==-~::.".::'.I' -;:~==]= .. :: = -9Y2 = smYI 32. )' ~ 33. y~ = COS)'2 = Y2 - 2Y2 Y; = Yl - 2Y1 2 2 . :.:==: Systems of ODEs. Phase Plane. Qualitative Methods Whereas single electric circuits or single mass-spring systems are modeled by single ODEs (Chap. 2). networks of several circuits. systems of several masses and springs. and other engineering problems lead to systems of ODEs, involving several unknown functions ."1(1), ... , YI1(1)· Of central interest are first-order systems (Sec. 4.2): y' = f(t, y), in components, to which higher order ODEs and systems of ODEs can be reduced (Sec. 4.1). In this summary we let 11 = 2. so that y' (1) = f(t, y), Y; = fl(t, Yh Y2) in components. .\'~ = f2(1, .\'1, .\'2) Then we can represent solution curves as trajectories in the phase plane (the YIY2-plane), investigate their totality [the "phase portrait" of (1 )J, and study the kind and stability of the critical points (points at which both f 1 and f 2 are zero), and classify them as nodes, saddle points, centers, or spiral points (Secs. 4.3, 4.4). These phase plane methods are qualitative; with their use we can discover various general properties of solutions without actually solving the system. They are primarily used for autonomous systems, that is, systems in which t does not occur explicitly. A linear system is of the fonn (2) y' = Ay + g, where A = [(/11 °21 If g (3) L :~:J' y = [:J, [:J . = 0, the system is called homogeneous and is of the form y' = Ay. g = 165 Summary of Chapter 4 If all, . . . , a22 are constants, it has solutions Y = quadratic equation and x -:f- 0 has components Xl' X2 xeAt, where A is a solution of the determined up to a multiplicative constant by (These A's are called the eigenvalues and these vectors x eigenvectors of the matrix A. Further explanation is given in Sec. 4.0.) A system (2) with g -:f- 0 is called nonhomogeneous. Its general solution is of the form Y = Yh + Yp, where Yh is a general solution of (3) and Yp a particular solution of (2). Methods of determining the latter are discussed in Sec. 4.6. The discussion of critical points of linear systems based on eigenvalues is summarized in Tables 4.1 and 4.2 in Sec. 4.4. It also applies to nonlinear systems if the latter are first linearized. The key theorem for this is Theorem L in Sec. 4.5, which also includes three famous applications, namely the pendulum and van der Pol equations and the Lotka-Volterra predator-prey population model. • ••• n .~ ~ ~, CHAPTER ~I\ ,,1 5 ........... ---r • I 'ill;. 1 1 1/'1\ ' I', ... Series Solutions of ODEs. Special Functions In Chaps. 2 and 3 we have seen that linear ODEs with constant coefficients can be solved by functions known from calculus. However. if a linear ODE has variable coefficients (functions of x). it must usually be solved by other methods. as we shall see in this chapter. Legendre polynomials, Bessel functions, and eigenfunction expansions are the three main topics in this chapter. These are of greatest importance to the applied mathematician. Legendre's ODE and Legendre polynomials (Sec. 5.3) are likely to occur in problems showing spherical symmetry. They are obtained by the power series method (Secs. 5.1, 5.2). which gives solutions of ODEs in power series. Bessel's ODE and Bessel functions (Secs. 5.5, 5.6) are likely to occur in problems showing cylindrical symmetry. They are obtained by the Frobenius method (Sec. 5.4), an extension of the power series method which gives solutions of ODEs in power series, possibly multiplied by a logarithmic tenn or by a fractional power. Eigenfunction expansions (Sec. 5.8) are infinite series obtained by the SturmLiouville theory (Sec. 5.7). The terms of these series may be Legendre polynomials or other functions, and their coefficients are obtained by the orthogonality of those functions. These expansions include Fourier series in terms of cosine and sine, which are so in1portant that we shall devote a whole chapter (Chap. II) to them. Special functions (also called higher functions) is a name for more advanced functions not considered in calculus. If a function occurs in many applications, it gets a name, and its properties and values are investigated in all details, resulting in hundreds of formulas which together with the underlying theory often fill whole books. This is what has happened to the gamma, Legendre, Bessel, and several other functions (take a look into Refs. [GRI], [GRIO], [All] in App. 1). Your CAS knows most of the special functions and corresponding formulas that you will ever need in your later work in industry, and this chapter will give you a feel for the basics of their theory and their application in modeling. COMMENT You can study this chapter directly after Chap. 2 because it needs no material from Chaps. 3 or 4. Prerequisite: Chap. 2. Sections that may be omitted il1 a shorter course: 5.2, 5.6-5.8. References and Answers to Problems: App. I Part A, and App. 2. 166 167 SEC. 5.1 Power Series Method 5.1 Power Series Method The power series method is the standard method for solving linear ODEs with variable coefficients. It gives solutions in the form of power series. These series can be used for computing values, graphing curves, proving formulas, and exploring properties of solutions, as we shall see. In this section we begin by explaining the idea of the power series method. Power Series From calculus we recall that a power series (in powers of x - xo) is an infinite series of the form (1) 2: a",(x - xo)m = ao + al (x - xo) + + a2(x - XO)2 m=O Here, x is a variable. ao, at. a 2, ... are constants, called the coefficients of the series. is a constant, called the center of the series. In particular, if Xo = 0, we obtain a power series in powers of x Xo 2: (2) amx ln = ao + alx a3 x3 + + a2x2 + m=O We shall assume that all variables and constants are real. Familiar examples of power series are the Maclaurin series I 00 --- = 1- x eX 2: (Ixl < = 1 + x + x 2 + ... xm 1, geometric series) m=O = 2: m=O cosx = 2: '1'11=0 sin x = 2: m=O x2 xm 111! = I +x+ 2! (_l)mx2m (2111) ! (_I)mx 2m+l (2m + I)! I + - x2 2! =x- x3 3! + ... X4 + x3 + ... - 4! x5 + 5! 3! - + .... We note that the term "power series" usually refers to a series of the form (1) lor (2)] but does not include series of negative or fractional powers of x. We use 111 as the summation letter, reserving n as a standard notation in the Legendre and Bessel equations for integer values of the parameter. Idea of the Power Series Method The idea of the power series method for solving ODEs is simple and natural. We describe the practical procedure and illustrate it for two ODEs whose solution we know, so that 168 CHAP. 5 Series Solutions of ODEs. Special Functions we can see what is going on. The mathematical justification of the method follows in the next section. For a given ODE y" + p(x)y' + q(x)y =0 we first represent p(x) and q(x) by power series in powers of x (or of x - Xo if solutions in powers of x - xo are wanted). Often p(x) and q(x) are polynomials, and then nothing needs to be done in this first step. Next we assume a solution in dle form of a power series with unknown coefficients, (3) y L = arnxTn = ao + (/tX + a2x2 + a3x3 + m=O and insert this series and the series obtained by term wise differentiation, (a) )" =L mamxrn - 1 = a] + 2a2x + 3a3x2 + ... m~] (4) (b) )''' = L m(m - l)a m x m - 2 = 2(/2 + 3· 2a3x + 4· 3a4x2 + .. m=2 into the ODE. Then we collect like powers of x and equate the sum of the coefficients of each occuning power of x to zero, starting with the constant terms, then taking the terms containing x, then the terms in x 2 , and so on. This gives equations from which we can determine the unknown coefficients of (3) successively. Let us show this for two simple ODEs that can also be solved by elementary methods, so that we would not need power series. E X AMP L E 1 Solve the following ODE by power series. To grasp the idea. do this by hand: do not use your CAS (for which you could program the Whole process). y' = 2xy. Soluti01l. We insert (3) and (4a) into the given ODE. obtaining We must perform the multiplication by a] + 211 2 X 2110X 2~ + + on the right and can write the resulting equation conveniently as 3a3x2 2111X2 + 4a4x3 + 3 2112 X + 5a5-\·4 + 6a&y5 + ... + 2114X5 + .. 2a3x4 + For this equation to hold, the two coefficients of every power of x on both sides must be equal. that is. Hence a3 = 0, 1I5 = 0, ... and for the coefficients with even SUbscripts. ao 3! ' SEC. 5.1 Power Series Method 169 ao remain~ arbitrary. With these coefficients the series (3) gives the following solution. which you should confirm by the method of separating variables. More rapidly, (3) and (4) give for the ODE y' = 2\:" x ] ·UIXo + x x L L m 1 InunzX = 2\- m=O 1n=2 L = DmXTn 2am x'11 + 1 m=O Now, to get the same general power on both sides, we make a "shift of index" on the left by sening III = S + 2, thus 111 - I = s + I. Then am becomes lIs+2 and x",-I becomes i'+I. Also the summation. which started with m = 2. now starts with s = 0 because s = /11 - 2. On the right we simply make a change of notation /11 = S, hence lim = as and X"H I = x s + 1: abo the summation now starts with s = O. This altogether gives <Xl L (s + 2)aS+2xs+I = L 2llsXS+I. + al s=o s~o Every occurring power of x must have the same coefficient on both sides: hence and (s + 2 or 2kls+2 = 2l1s a s +2 = s + 2 as' For s = 0, I. 2.... we thus have a2 = (2/2)lIo, a3 = (2/3)aI = O. a 4 = (2/4)a2' ... as before. EXAMPLE 2 • Solve y" + Solutioll. O. y = By in,erting (3) and (4b) into the ODE we have x L /11(111 - l)llmxm-2 m~2 L + 7n am x = O. m~O To obtain the same general power on both selies. we set 11/ = and then we take the laner to the right side. This gives S + 2 in the first series and 111 = s in the second, "" L (s + 2)(.\' + ex; I)lI s +2"'s = 5=0 L {{sXS. 5=0 S Each power X must have the same coefficient on both sides. Hence recursion formula (s + 2)(s + I )lls+2 = -as' This gives the {{s a s +2 = - -,----,--"----(s + 2)(s + I) (s = 0, 1, .. '). We thu, obtain successively 112 = lI4 = and so on. ao and {{I 110 lIo 2' I 2! a2 110 4'3 4! a3 = a5 = al al 3·2 3! a3 al 5'4 5! remain arbitrary. With these coefficients the series (3) becomes '" = ao + {{IX - ao 2! x 2 - ~ \.3 + {{o \.4 +!!.!. 3! . 4! . 5! X 5 + CHAP. 5 170 Series Solutions of ODEs. Special Functions Reordering terms lwhich is permissible for a power series), we can write this in the form + X3 a1 ( x - 3! + X5 5! - + ... ) and we recognize the familiar general solution • y = Go cosx + G1 sinx. Do we need the power series method for these or similar ODEs? Of course not; we used them just for explaining the idea of the method. What happens if we apply the method to an ODE not of the kind considered so far, even to an innocent-looking one such as y" + xy = 0 ("Airy's equation")? We most likely end up with new special functions given by power series. And if such an ODE and its solutions are of practical (or theoretical) interest, we name and investigate them in terms of formulas and graphs and by numeric methods. We shall discuss Legendre's, Bessel's, and the hypergeometric equations and their solutions, to mention just the most prominent of these ODEs. To do this with a good understanding, also in the light of your CAS. we first explain the power series method (and later an extension, the Frobenius method) in more detail. 11-10 I POWER SERIES METHOD: TECHNIQUE, FEATURES Apply the power series method. Do this by hand, not by a CAS, so that you get a feel for the method, e.g., why a series may terminate, or has even powers only, or has no constant or linear terms, etc. Show the details of your work. 1. y' - y = 0 3. y" + 4y = 0 5. (2 + x)y' = y 7. y' = Y 9. y" - )" 111-161 + x = 0 2. y' + xy = 0 + 3(1 + X2)y = 0 + 12x2»)' 8. (x 5 + 4x 3 )y' = (5x 4 10. y" - xy' + y = 0 CAS PROBLEMS. INITIAL VALUE PROBLEMS Solve the initial value problems by a power series. Graph the partial sum s of the powers up to and including x 5 . Find the value of s (5 digits) at Xl' 5.2 + 4y = 1. = 1 + = y - yeO) = 1.25. 15. y" + yeO) y2, yeO) = 3xy' = = y2, 14. (x - 2)y' = xy, /(0) 4. y" - y = 0 6. y' 11. y' 12. y' 13. / 1, + + !, Xl = 1 yeO) = 30y = 0, Xl = !1T Xl yeO) = 4, 2y = 0, Xl = 0.5 16. (1 - X2)y" - 2xy' / (0) = l.875, Xl = 0.2 = 0, Xl = 2 1, yeO) 0, 0.5 17. WRITING PROJECT. Power Series. Write a review (2-3 pages) on power series as they are discussed in calculus, using your own formulation and examplesdo not just copy passages from calculus texts. 18. LITERATURE PROJECT. Maclaurin Series. Collect Maclaurin series of the functions known from calculus and arrange them systematically in a list that you can use for your work. Theory of the Power Series Method In the last section we saw that the power series method gives solutions of ODEs in the form of power series. In this section we justify the method mathematically as follows. We first review relevant facts on power series from calculus. Then we list the operations on power series needed in the method (differentiation, addition, multiplication, etc.). Near the end we state the basic existence theorem for power series solutions of ODEs. SEC. 5.2 Theory of the Power Series Method 171 Basic Concepts Recall from calculus that a power series is an infinite series of the form oc (1) ~ am(x - xoyn = ao + al (x - Xo) + a2(X - XO)2 + m~O As before, we assume the variable x, the center .\"0' and the coefficients ao, real. The nth partial sum of (1) is aI, • . • to be where n = 0, 1, .... Clearly, if we omit the terms of s" from (I), the remaining expression is (3) This expression is called the remainder of (1) after the tenn a,/x - xo)n. For example, in the case of the geometric series I + x + X2 + ... + xn + ... we have So = 1, Sl = + x. etc. In this way we have now associated with (1) the sequence of the partial sums so(x), SI(X), S2(X), .... If for some x = Xl this sequence converges, say, lim sn(x I ) = S(XI)' 11.-----"'00 then the series (I) is called convergent at X sum of (I) at Xl, and we write = Xl, the number S(XI) is called the value or 00 S(XI) = ~ am(XI - XOrn. m~O Then we have for every n, (4) If that sequence diverges at X = Xl> the series (I) is called divergent at X = Xl. In the case of convergence, for any positive E there is an N (depending on E) such that, by (4), (5) for all n > N. 172 CHAP. S Series Solutions of ODEs. Special Functions Geometrically, this means that all Sn(Xl) with n > N lie between s(x l ) - E and s(x l ) + E (Fig. 102). Practically, this means that in the case of convergence we can approximate the sum S(Xl) of (I) at Xl by Sn(Xl) as accurately as we please, by taking 11 large enough. Convergence Interval. Radius of Convergence With respect to the convergence of the power series (I) there are three cases, the useless Case I, the usual Case 2, and the best Case 3, as follows. Case 1. The series (1) always converges at x = xo, because for x = Xo all its terms are zero, perhaps except for the first one, ao. In exceptional cases x = Xo may be the only x for which (l) converges. Such a series is of no practical interest. Case 2. If there are further values of x for which the series converges, these values form an interval, called the convergence interval. If this interval is finite, it has the midpoint xo, so that it is of the form Ix - xol < (6) (Fig. 103) R and the series (1) converges for all x such that Ix - xol < R and diverges for all x such that Ix - xol > R. (No general statement about convergence or divergence can be made for x - Xo = R or -R.) The number R is called the radius of convergence of 0). (R is caned "radius" because for a complex power series it is the radius of a disk of convergence.) R can be obtained from either of the formulas (7) (a) R = l/lim 111~'JC Vfa:f (b) = R 1 him / m_x, I am+l lint I provided these limits exist and are not zero. [If these limits are infinite, then only at the center xo.] (1) converges Case 3. The convergence interval may sometimes be infinite, that is, (l) converges for all x. For instance, if the limit in (7a) or (7b) is zero, this case occurs. One then writes R = x, for convenience. (Proofs of all these facts can be found in Sec. 15.2.) For each x for which (1) converges. it has a certain value sex). We say that (1) represents the function sex) in the convergence interval and write 00 seX) = L (Ix - xol < {/m(X - Xo)m R). m~O Let us illustrate these three possible cases with typical examples. Divergence iconvergence ~E--+-E_I I I I Fig. 102. I Inequality (S) I Fig. 103. -R· ~I' I ------j Divergence R-I I Convergence interval (6) of a power series with center Xo SEC. 5.2 173 Theory of the Power Series Method E X AMP LEI The Useless Case 1 of Convergence Only at the Center In the case of the series ~ m!x'" = 1 + x + 2x2 + 6x 3 + ... m=O we have am. = Ill!, and in (7b). (1/1 a",+1 --= am. + I)! =m+1-,,<o as ,n ----7 • Thus this series converges only at the center x = O. Such a series is useless. E X AMP L E 2 ro. In! The Usual Case 2 of Convergence in a Finite Interval. Geometric Series For the geometric series we have 1 x --=~x I-x m =I+x+x 2 (Ixl + ... In fact, am = 1 for all m, and from (7) we obtain R = I, that is. the geometric series converges and 1/(1 - x) when Ixl < L E X AMP L E 3 < I). m=O represent~ • The Best Case 3 of Convergence for All x In the case of the series + 1+x x2 + ... 2! we have a", = 11m!. Hence in (7b), l/{m + I)! 11m! 111 + 1 -,,0 as ~ co, • so that the series converges for all x. E X AMP L E 4 111 Hint for Some of the Problems Find the radius of convergence of the series OJ ~ L.J ( __I)'" _ 8'" x3 .3m_ .\ - I - 8 + x6 x9 ill + - .... 64 - 'tn=O Solution. This is a senes in powers of t = x 3 with coefficients am = (-1)"'/8"', so that in (7b), I a"'+1 Thus R = 8. Hence the series converges for am Itl = I= ~ 8",+1 Ix 3 1< = .!. 8' 8, that is, Ixl < 2. • Operations on Power Series In the power series method we differentiate, add, and multiply power series. These three operations are permissible, in the sense explained in what follows. We also list a condition about the vanishing of all coefficients of a power series, which is a basic tool of the power series method. (Proofs can be found in Sec. 15.3.) 174 CHAP. 5 Series Solutions of ODEs. Special Functions Termwise Differentiation A power series may be d(fferenlialed Term by Term. More precisely: if "L y(x) = am(x - X O)111 m~O converges for Ix - xol < R, where R > 0, then the series obtained by differentiating term by term also converges for those x and represents the derivative y' of y for those x, that is, x y' (x) = "L 17Ul m {.X - xo)'n-l (Ix - xol < R). m~l Similarly, y"(x) = "L m(m - l)am(x - xo)m-2 (Ix - xol < R), etc. m~2 Termwise Addition Two power series lIlay be added term by term. More precisely: if the series GC "L and (8) bm(x - xo)m m~O have positive radii of convergence and their sums are f(x) and g(x). then the series CXJ "L (am + bm)(x - xo)m m~O converges and represents f(x) + g{x) for each x that lies in the interior of the convergence interval of each of the two given series. Termwise Multiplication Two power series may be multiplied Tel7ll by Term. More precisely: Suppose that the series (8) have positive radii of convergence and let f(x) and g(x) be their sums. Then the series obtained by multiplying each term of the first series by each term of the second series and collecting like powers of x - Xo, that is, GC "L (aob m + a1b m- 1 + ... + ambo)(x - xo)m m~O converges and represents f(x)g(x) for each x in the interior of the convergence interval of each of the two given series. SEC. S.2 Theory of the Power Series Method 175 Vanishing of All Coefficients If a power series has a positive radius of convergence and a sum that is identically zero throughout its illterval of convergence, then each coeffIcient of the series must be zero. Existence of Power Series Solutions of ODEs. Real Analytic Functions The properties of power series just discussed form the foundation of the power series method. The remaining question is whether an ODE has power series solutions at all. An answer is simple: If the coefficients p and lj and the function r on the right side of (9) y" + p(x)y' + q(x)y = r(x) have power series representations, then (9) has power series solutions. The same is true if h, p, q, and r in (10) h(x)y" + p(x)y' + q(x»)' = r(x) have power series representations and h(xo) *- 0 (xo the center of the series). Almost all ODEs in practice have polynomials as coefficients (thus te1l11inating power series), so that (when r(x) == 0 or is a power series, too) those conditions are satisfied, except perhaps the condition h(xo) *- O. If h(xo) *- 0, division of (10) by h(x) gives (9) with p = pIli, q = qlh, r = 'ilh. This motivates our notation in (0). To formulate all this in a precise and simple way, we use the following concept (which is of general interest). DEFINITION Real Analytic Function A real function f(x) is called analytic at a point x = Xo if it can be represented by a power series in powers of x - Xo with radius of convergence R > O. Using this concept, we can state the following basic theorem. THEOREM 1 Existence of Power Series Solutions If p, x = q, and r in (9) are analytic at x = x o, then every SoluTion of (9) is analYTic aT Xo and can thus be represenTed by a power series in powers of x - Xo with radius of convergence R > O. Hence the same is true if h, p, q, and r in (10) are analytic at x = Xo and h(xo) *- O. The proof of this theorem requires advanced methods of complex analysis and can be found in Ref. [All] listed in App. 1. We mention that the radius of convergence R in Theorem I is at least equal to the distance from the point x = Xo to the point (or points) closest to Xo at which one of the functions p, q, r, as functions of a complex variable, is not analytic. (Note that that point may not lie on the x-axis but somewhere in the complex plane.) 176 CHAP. 5 .. = /1-12/ Series Solutions of ODEs. Special Functions RADIUS OF CONVERGENCE Determine the radius of convergence. (Show the details.) 0:: 1. L y7n ~ 1= 0) (c m~O C 15. 2 L p~l (p /16-23/ (m + L I)m (x - 3)271> (-I )7l1 x 4m "n1=O 5. L m~O 00 4 xm (2111 + 2 )(2m + ) x (m!) X 2m" 10 (-l)m ~ (1)2m 7 'L..~xm~2 ~ 8. L.. (4m)! --4 71>~1 (m!) (Ill m=4 10. + xnt 3)2 x m. (111 - 3)4 = (7_111. )' L.. - - 2 - ~ 11. L.. Xm 111 1 "r" 1m (x - 2 1T) 7n=1 12• (m + 1)111 ~ L.. 71l~1 (2m + \)! /13-15/ 16. y " + xy = 0 , 17. .I' " - Y + x 2 y = 0 , 18. y " - y + xy = 0 , 19. y " + 4xy = 0 20. y " + lxy + y = 0 21. y" + (I + X2»)' = 23. (2x 2 - 2 0 2).1' = 0 3x + I)y" + 2xv' - 2." = 0 24. TEAM PROJECT. Properties from Power Series. In the next sections we shall define new functions (Legendre functions. etc.) by power series. deriving properties of the functions directly from the series. To understand this idea. do the same for functions familiar from calculus. using Maclaurin series. (a) Show that cosh x + sinh x = eX. Show that cosh x > 0 for all x. Show that eX ;:;:; e- x for all x;:;:; O. (b) Derive the differentiation formulas for eX. cos x, sinx. 11(1 - x) and other functions of your choice. Show that (cos xl" = -cos x. (cosh :d' = cosh x. Consider integration similarly. ~ ",~1 POWER SERIES SOLUTIONS 22. y" - 4xy' + (4x I)m --6. ~ L.. 2 m~O l)! , (2m)! ( + Find a power series solution in powers of x. (Show the details of your work.) m~l 4. xp + 4 p x21n+l SHIFTING SUMMATION INDICES (CF. SEC. 5.1) Thi~ is often convenient or nece~sary in the power series method. Shift the index so that the power under the summation sign is xS. Check by writing the first few terms explicitly. Also determine the radius of convergence R. (c) What can you conclude if a series contains only odd powers? Only even powers? No constant tenn? If all its coefficients are positive? Give examples. (d) What properties of cos x and sin x are lIot obvious from the Maclaurin series? What properties of other functions? 25. CAS EXPERIMENT. Information from Graphs of Partial Sums. In connection with power series in numerics we use partial sums. To get a feel for the accuracy for various x. experiment with sin x and graphs of partial sums of the Maclaurin series of an increasing number of terIllS, describing qualitatively the "breakaway points" of these graphs from the graph of sin x. Consider other examples of your own choice. SEC. 5.3 5.3 Legendre's Equation. Legendre Polynomials Pn(x) Legendre's Equation. Legendre Polynomials Pn{x) In order to first gain skill, we have applied the power series method to ODEs that can also be solved by other methods. We now turn to the first "big" equation of physics, for which we do need the power series method. This is Legendre's equationl (l - x 2 )y" - 2AY' + n(n (1) + I)y = 0 where n is a given constant. Legendre's equation arises in numerous problems, particularly in boundary value problems for spheres (take a quick look at Example I in Sec. 12.10). The parameter n in (1) is a given real number. Any solution of (1) is called a Legendre function. The study of these and other "higher" functions not occurring in calculus is called the theory of special functions. Further special functions will occur in the next sections. Dividing 0) by the coefficient 1 - x 2 of y". we see that the coefficients -2x/(1 - x 2 ) and n(n + 1)/(1 - x 2 ) of the new equation are analytic at x = O. Hence by Theorem I, in Sec. 5.2. Legendre's equation has power series solutions of the form (2) Substituting (2) and its derivatives into (1), and denoting the constant n(n k, we obtain ex:; l) amxm.-2 00 2x.L mamxm - l - + 1) simply by + k"L amx m = O. m=1 "m,=2 By writing the first expression as two separate series we have the equation 00 "L 00 X) m(m l)amx m - 2 - "L m(1Il - l)am x m - "L X) 2mam x m + "L kamx m = O. m=2 To obtain the same general power X S in all four series, we set m - 2 = s (thus m = s + 2) in the first series and simply write s instead of III in the other three series. This gives 00 "L (s + 2)(s + 00 l)as +2-C\:s - "L s(s - 00 I)asx s - 00 "L 2sasxs + "L kasxs = O. lADRIEN-MARIE LEGENDRE (1752-1833). French mathematician. who became a professor in Paris in 1775 and made important contributions to special functions, elliptic integrals, number theory, and the calculus of variations. His book Elements de geollletrie (1794) became very famous and had 12 editions in less than 30 years. Fonnulas On Legendre functions may be found in Refs. [GRJ] and [GRIO]. 178 CHAP. 5 Series Solutions of ODEs. Special Functions (Note that in the first series the summation begins with s = 0.) Since this equation with right side 0 must be an identity in x if (2) is to be a solution of (1), the sum of the coefficients of each power of x on the left must be zero. Now X O occurs in the first and fourth series and gives [remember that k = n(n + 1)] (3a) "\"1 2 . la2 + l1(n + = O. 1) ao occurs in the first, third, and fourth series and gives (3b) 3' 2a3 +[-2 + The higher powers x 2 , x 3 , ••• (3c) (s + + 2)(s + n(n = O. l)]aI occur in all four series and give l)aS +2 + [-s(s - I) - 2s + + n(n l)]a s = O. The expression in the brackets [ .. ·1 can be written (n - s)(n + s + I), as you may readily verify. Solving (3a) for a2 and (3b) for a 3 as well as (3c) for a s +2' we obtain the general formula (n - s)(n (4) (s +s + + 2)(s + 1) (s 1) = 0, 1, ... ). This is called a recurrence relation or recursion formula. (Its derivation you may verify with your CAS.) It gives each coefficient in terms of the second one preceding it except for a o and aI, which are left as arbitrary constants. We find successively n(n a2 = - + (n - I) 2! I)(n + 2) + 4) 3! (II - 2)(11 + 3) (n - 5·4 4·3 (n - 2)1l(11 3)(n + 1)(/1 + (11 - 3)tll - 3) I )(Il 4! + 2)(11 + 4) 5! and so on. By inserting these expressions for the coefficients into (2) we obtain (5) y(x) = aoY! (x) (n - 2)11(11 + aIY2(x) where + 11(11 1) 2! + (6) Yl(X) = 1- (7) )'2(X) = x - - - - - - - x3 (n - X2 l)(ll 3! + 2) + 1)(11 + 3) 4! + (n - 3)(11 - x4 - 1)(11 5! + + ... 2)(11 + 4) x5 - + .... These series converge for Ixl < (see Prob. 4; or they may terminate, See below). Since (6) contains even powers of x only, while (7) contains odd powers of x only, the ratio YtiY2 is not a constant, so that Yl and Y2 are not proportional and are thus linearly independent solutions. Hence (5) is a general solution of (I) on the interval - I < x < I. SEC. 5.3 179 Legendre's Equation. Legendre Polynomials Pn(x) Legendre Polynomials Pn{x) In various applications. power series solutions of ODEs reduce to polynomials. that i~. they terminate after finitely many terms. This is a great advantage and is quite common for special functions. leading to various important families of polynomials (see Refs. [GR I] or [GRIO] in App. 1). For Legendre's equation this happens when the parameter n is a nonnegative integer because then the right side of (4) is zero for s = n, so that an +2 = 0, a n +4 = 0, (In+6 = 0, .... Hence if n is even, hex) reduces to a polynomial of degree n. If II is odd, the same is true for Y2(X). These polynomials, multiplied by some constants. are called Legendre polynomials and are denoted by Pn(x). The standard choice of a constant is done as follows. We choose the coefficient an of the highest power xn as I . 3 . 5 . . . (2n - I) an = (8) (n n! a positive integer) (and an = 1 if n = 0). Then we calculate the other coefficients from (4). solved for as in terms of as +2, that is, (9) (Is (s + 2)(s + 1) = - ------(n - s)(n + s + 1) (ls+2 (s ~ n - 2). The choice (8) makes P ,,(I) = 1 for every n (see Fig. 104 on p. 180); this motivates (8). From (9) with s = 11 - 2 and (8) we obtain 11(11 - 1)(2n)! 1) n(n - l) 2(211 - Using (2n)! = 2n(211 - 1)(211 - 2)!, obtain II! (In = - = n(1I - 2(211 - 1)2n( 11 !)2 1)!, and n! = n(1I - n(n - 1)2n(2n - 1)(2n - 2)! an -2 = - ----'----'-----'---_..:...:.._---'--2(211 - 1)2nn(1I - I)! n(n - 1)(n - 2)! n(11 - 1)211(211 - 1) cancels. so that we get (211 - 2)! Similarly, (11 - 2)(n - 3) 4(211 - 3) (211 - 4)! 2 n 2! (11 - and so on, and in general, when (10) (In-2m = 11 - 2111 ~ 2)! (n - 4)! 0, (211 - 2111)! (-I)m - - - - - - - - - 2nm! (n - Ill)! (n - 2m)! 1)(n - 2)!, we 180 CHAP. 5 Series Solutions of ODEs. Special Functions The resulting solution of Legendre's differential equation (l) is called the Legendre polynomial of degree n and is denoted by P n(x). From (10) we obtain M P n(x) = 'L (2n - 2m)! (-1)'m - - - ' - - - - - - ' - - - - x n - 2'm 2nm! (n - m)! (n - 2m)! (11) where M = nl2 or (11 - I )/2, whichever is an integer. The first few of these functions are (Fig. 104) (11') Po(X) = 1, P 2 (x) = P 4 (x) = ~(35x4 - 30x 2 + 3), !(3x 2 - 1), PI(X) = P 3 (x) = !(5x 3 P 5(x) = x - 3x) ~(63x5 - 70x 3 + I5x) and so on. You may now program (11) on your CAS and calculate Pn(x) as needed. The so-called orthogonality of the Legendre polynomials will be considered in Sees. 5.7 and 5.8. x Fig. 104. ======= =" Legendre polynomials - 1. Verify that the polynomials in (11') satisfy Legendre's equation. 2. Derive (11 ') from (11). 3. Obtain P6 and P 7 from (11). 4. (Convergence) Show that for any 11 for which (6) or (7) does nol reduce to a polynomial, the series has radius of convergence 1. 5. (Legendre function Qo(x) for n = 0) Show that (6) with 11 = 0 gives Yl(X) = Po(x) = I and (7) gives Y2(X) = x 2 + - =x+ 3! x3 3 x3 + (-3)(-1)·2·4 5! x5 + ... X5 I I +x +-+···=-In--. 5 2 I-x Legendre's Equation. Legendre Polynomials Pn{x) SEC. 5.3 Verify this by solving (\) with and separating variables. z 0, setting /I = = y' = 6. (Legendre function -Ql(X) for II 1) Show that (7) with 11 = I gives )"2(X) = PI(x) = x and (6) gives )"I(X) = -Ql(X) (the minus sign in the notation being conventional), )"I(X) = I - 3 I - x 5 1 7. (ODE) Find a solution of 2 + + l)y = 0. a by reduction to the Legendre equation. (a - x )y" - 2xy' n(/1 *' r 0. 8. [Rodrigues's formula (12)]2 Applying the binomial theorem to (X2 - I)n, differentiating it 11 times term by term. and comparing the result with (II), show that (12) 9. (Rodrigues's formula) Obtain (II ') from (12). 110-131 (13) (b) Potential theory. Let Al and A2 be two points in space (Fig. 105, r2 > 0). Using (13), show that I +x I - x = 1- - x l n - - 2 (a) Legendre polynomials. Show that is a generating function of the Legendre polynomials. Hint: Start from the binomial expansion of 11"\ 1 - v. then set v = 2xlI - u 2 • multiply the powers of 2m - u 2 out. collect all the terms involving un, and verify that the slim of these terms is Pn(x)u n . (r + .~ + .~5 + ...) I 2 181 Vr12 + r22 2rlr2 cos - e This formula has applications in potential theory. (Qlr is the electrostatic potential at A2 due to a charge Q located at Al . And the series expresses I1r in terms of the distances of Al and A2 from any origin o and the angle e between the segments OA I and OA 2 ·) CAS PROBLEMS 10. Graph P 2 (x) • ...• P IO (.\) on common axes. For what \" (approximately) and II = 2... " 10 is Ipn(x)1 <!? It. From what /I on will your CAS no longer produce faithful graphs of P n(x)? Why? 12. Graph Qo(x), QI (x), and some further Legendre functions. 13. Substitute asxs + a s + IX s + 1 + as+2xs+2 into Legendre's Fig. 105. Tearn Project 14 (c) Further applications of (13). Show that Pn(l) = I, P n ( -I) = (-It'. P 2n + 1 (0) = 0, and equation and obtain the coefficient recursion (4). P 2n(O) 14. TEAM PROJECT. Generating Functions. Generating functions playa significant role in modem applied mathematics (see [GR5]). The idea is simple. If we want to study a certain sequence (fn(x» and can find a function C(u, x) = L"" fn(x)u n , n=O we may obtain properties of (f ,,(x» from those of C. which ""generates" this sequence and is called a generating function of the sequence. = (-I)n'I'3'" (211 - 1)/[2,4", (2n)]. (d) Bonnet's recursion. 3 Differentiating (\3) with respect to u, using (13) in the resulting formula, and comparing coefficients of un, obtain the Bonnet recursion (14) (II + l)Pn+l(x) = (211 + I)xP,/:r) - IlPn_I(X), where Il = I, 2, .... This formula is useful for computations, the loss of significant digits being small (except near zeros). Try ( 14) out for a few computations of your own choice. 20UNDE RODRIGUES (1794-1851). French mathematician and economist. 30SSIAN BONNET (1819-1892), French mathematician. whose main work was in differential geometry. CHAP. 5 Series Solutions of ODEs. Special Functions 182 15. (Associated Legendre functions) The associated and are solutions of the ODE Legendre functions Pnk(x) play a role in quantum physics. They are defined by (I - x 2 )y" (16) + (15) 2\)" [n(n + I) - ~ ] 1- x 2 y = O. Find P 11(X), P 21(X), P 22(X), and P42(X) and verify that they satisfy (16). 5.4 Frobenius Method Several second-order ODEs of considerable practical importance-the famous Bessel equation among them-have coefficients that are not analytic (definition in Sec. 5.2), but are "not too bad," so that these ODEs can still be solved by series (power series times a logarithm or times a fractional power of x, etc.). Indeed, the following theorem permits an extension of the power series method that is called the Frobenius method. The latteras well as the power series method itself-has gained in significance due to the use of software in the actual calculations. THEOREM 1 Frobenius Method Let b(x) and c(x) be any Junctions that are analytic at x y" (1) b(x), c(x) x + -- y + -v 2 X = = O. Then the ODE 0 - has at least one solution that can be represented in the JOI7I1 co y(x) = xT.L a",x'in = xT(ao (2) + alx + a 2 x 2 + ...) (ao *" 0) 7n~0 where the exponent r may be any (real or complex) number (and r is chosen so that ao 0). The ODE (1) also has a second solution (such that these two solutions are linearly independent) that may be similar to (2) (with a different r and different coefficients) or m£ly contain a logarithmic tenn. (Details in Theorem 2 below.)4 *" For example, Bessel's equation (to be discussed in the next section) y" + 1 y' + X (X2 - V2) x2 V = 0 (va parameter) - 4GEORG FROBENIUS (1849-1917), German mathematician, also known for his work on matrices and in group theory. 0 is no restriction: it In this theorem we may replace x by x - Xo with any number xo. The condition ao simply means that we factor out the highest possible power of x. The singular point of (1) at x = 0 is sometimes called a regular singular point, a term confusing to the student, which we shall not use. * SEC. 5.4 183 Frobenius Method = = = is of the form (I) with b(x) 1 and c(x) x 2 - v 2 analytic at x 0, so that the theorem applies. This ODE could not be handled in full generality by the power series method. Similarly, the so-called hypergeometric differential equation (see Problem Set 5.4) also requires the Frobenius method. The point is that in (2) we have a power series times a single power of x whose exponent r is not restricted to be a nonnegative integer. (The latter restriction would make the whole expression a power series, by definition; see Sec. 5.1.) The proof of the theorem requires advanced methods of complex analysis and can be found in Ref. [A 11] listed in App. I. Regular and Singular Points The fonowing commonly used terms are practical. A regular point of y" + p(x)y' + q(x)y = 0 is a point Xo at which the coefficients p and q are analytic. Then the power series method can be applied. If Xo is not regular, it is called singular. Similarly, a regular point of the ODE h(x)y" + p(x)y' (x) + q(x)y = 0 is an Xo at which h. p. q are analytic and h(xo) -=I=- 0 (so what we can divide by h and get the previous standard form). If Xo is not regular. it is called singular. Indicial Equation, Indicating the Form of Solutions We shall now explain the Frobenius method for solving (1). Multiplication of (1) by x 2 gives the more convenient form 2 x y" (1') + xb(x)y' + c(x)y = O. We first expand b(x) and c(x) in power series. or we do nothing if b(x) and c(x) are polynomials. Then we differentiate (2) term by term, finding GC y' (x) =.L (/11 + r)amx'm+r-l (m + r)(m = xr - 1 [rao 'm=o co (2*) -,",'(x) = .L + r - I ) amx'm+r-2 m~O By inserting all these series into (1') we readily obtain (3) + (r + l)alx + ... ] 184 CHAP. 5 Series Solutions of ODEs. Special Functions We now equate the sum of the coefficients of each power XT, XT+l, XT+2, ••• to zero. This yields a system of equations involving the unknown coefficients (1m- The equation cOlTesponding to the power x" is [r(r - I) Since by assumption ao + bor + colao = o. *- 0, the expression in the brackets [ ... ] must be zero. This gives r(r - I) + bor + Co = O. (4) This important quadratic equation is called the indicial equation of the ODE (I ). Its role is as follows. The Frobenius method yields a basis of solutions. One of the two solutions will alway~ be of the form (2), where r is a root of (4). The other solution will be of a form indicated by the indicial equation. There are three cases: Case 1. Distinct roots not differing by an integer I. 2. 3..... Case 2. A double root. Case 3. Roots differing by an integer I. 2, 3..... Cases I and 2 are not unexpected because of the Euler-Cauchy equation (Sec. 2.5), the simplest ODE of the form (1). Case I includes complex conjugate roots r 1 and r2 = rl because rl - r2 = rl - rl = 2i 1m /"1 is imaginary. so it cannot be a real integer. The form of a basis will be given in Theorem 2 (which is proved in App. 4). without a general theory of convergence, but convergence of the occurring series can he tested in each individual case as usuaL Note that in Case 2 we must have a logarithm, whereas in Ca'>e 3 we mayor may 110t. THEOREM 2 Frobenius Method. Basis of Solutions. Three Cases Suppose that the ODE (1) satisfies the assumptions in Theorem I. LRt /"1 and r2 be the roots of the indicial equation (4). Then we have the following three cases. Case 1. Distinct Roots Not Differing by all Integer. A basis is (5) al1d (6) with coefficients obtained successivelyfrom (3) with r = rl and r = r2, respectivel)'. Case 2. Double Root rl = r2 = r. A basis is (7) [r = ~(1 - bo)] (of the .\£lme general form as before) and (8) (x> 0). SEC. 5.4 Frobenius Method 185 Case 3. Roots Differing by an Integer. A basis is (9) (of the same generalfoml as before) and (10) > where the roots are so denoted that rl - r2 0 and k may tum out to be zero. Typical Applications Technically, the Frobenius method is similar to the power series method, once the roots of the indicial equation have been determined. However, (5)-00) merely indicate the general form of a basis, and a second solution can often be obtained more rapidly by reduction of order (Sec. 2.1). E X AMP L E 1 Euler-Cauchy Equation, Illustrating Cases 1 and 2 and Case 3 without a Logarithm For the Euler-Cauchy equation (Sec. 2.5) (b o, Co constant) substitution of y ~ x T gives the auxiliary equation r(r - which is the indicial equation [and y =x T 1) + bor + Co = 0, is a very special form of (2)!]. For different roots rI, r2 we get a basis YI = XTt,.I'2 = XT2, and for a double root r we get a basis XT, x T lnx. Accordingly, for this simple ODE, Case 3 plays no extra role. • E X AMP L E 2 Illustration of Case 2 (Double Root) Solve the ODE x(x - (11) I)y" + (3x - I)y' + Y= O. (This is a special hypergeometric equation, as we shall see in the problem set.) Solution. Writing (11) in the standard form (1), we see that it satisfies the assumptions in Theorem I. [What are b(x) and c(x) in (II )?] By inserting (2) and its derivatives (2*) into (11) we obtain co L (m + r)(m + r - I)a",xm + r - m=O L (m + r)(m + r - l)a'mx'm+T-l 7n=O (12) m=O 7n=O 7n=O The smallest power is xT-t, occurring in the second and the fourth series; by equating the sum of its coefficients to zerO we have [-r(r - 1) - r]ao = 0, Hence this indicial equation has the double root r = O. thus 186 CHAP. 5 Series Solutions of ODEs. Special Functions First Solutioll. X S We insert this value 0 into (12) and equate the ~um of the coefficients of the power I" = to zero. obtaining s(s - thus {/s+1 = (/s. l)0s - (s = Hence 00 01 = 02 + + llSOs+1 3sos - + I )os+1 + (s Os = 0 = .... and by choosmg 00 = I we obtain the solution I m L 0::: \'I(X) = x (Ixl = --- I-x 1n=O < I). Second Solution. We get a second independent solution)"2 by the method of reduction of order (Sec. 2.1). substituting)"2 = 11.\"1 and its derivatives into the equation. This leads to (9). Sec. 2.1. which we shall use in this example. instead of starting reduction of order from scratch (as we shall do in the next example). In (9) of Sec. 2.1 we have p = (3.1' - I )/(x2 - x). the coefficient of y' in (11) ill stalldard form. By partial fractions. -J pdt = -J 3x - I .1'(.l-1) dx = -J (_2_ + -.:.) dx = x-I, -2 In (x - I) - In x . Hence (9), Sec. 2.1, becomes )"1 Inx In x, 11= x )"2 = 11.\'1 = I -x . and )"2 are shown in Fig. 106. These functions are linearly independent and thus form a basis on the interval 1 (as well as on I < x < X). • o< x < Fig. 106. E X AMP L E 3 Solutions in Example 2 Case 3, Second Solution with Logarithmic Term Solve the ODE (x 2 - (13) Solution. t)y" - t)-' 2 - o. = Substituting (2) and (2*) into (13), we have cc 00 (x + )" x) L (111 + r)(111 + r - 1){/ m x 'llt+T-2 - t m 0 x L (111 + rJo",xm +,'- l + 0 7ft = L omx7lt+T = O. TTL=O We now take x 2 , x. and x inside the summation~ and collect all tenns with power x"'+r and simplify algebraically, 'XC L + (m r - 1n=O (/II + 1")(/11 + r - l)lI m x",+r-l = O. nz,=O In the first scnes We set /II = S and in the second oc (14) L l)2omxm+r - L s=o III = S + L thus s = /II - + r)lI S +1.t'H" = 1. Then x (s + I" - 1)2{/sxs+r - L s=-l (s r + I)(s + O. SEC. 5.4 Frobenius Method 187 The lowest power is x r - 1 = (take s -I in the second series) and gives the indicial equation r(r - The roots are /"1 o. 1) = = 1 and r2 = U. They differ by an integer. This is Case 3. First Solution. From (14) with r = rl = I we have L 1·,211s - + 21(s + IllIs +ljxS + 1 = O. (s s~o This gives the recurrence relation as+l = Hence O. "1 = {/2 + (s = 0, ... successively. Taking 2)(s + ~ 1, we get "0 (s = 0, I, ., '). 1) as a." a first solution Second Solution. Y~ = xu" + Applying reduction of order (Sec. 2.1), we substitute Y2 = 2u' into the ODE, obtaining (x 2 xu drops Olll. - XJlXll" + 2u') - x(x,,' + ul + XII = 1'1 Ylll = ./lao = x. = Xli, y~ = Xll' + u and O. Division by x and simplification give (x 2 - x)u" + (x - 2)11' = O. From this. using partial fractions and integrating (taking the integration constant zero). we get u" 2 1/ x , + ~ I x - I Inu'=ln ~. I - I Taking exponents and integrating (again taking the integration constant zero), we obtain 1/ , = x X 1/ = 2 • Inx + )'2 = XI/ = x x Inx + 1. Yl and ."2 are linearly independent. and."2 has a logarithmic term. Hence."l and."2 constitute a basis of solutions • for positive .t. The Frobenius method solves the hypergeometric equation. whose solutions include many known functions as special cases (see the problem set). In the next section we use the method for solving Bessel's equation. BASIS OF SOLUTIONS BY THE FROBENIUS METHOD Find a basis of solutions. Try to identify the series as expansions of known functions. (Show the details of your work.) 1. xy" + 2y' - xy = 0 2. (x + 2)2)''' - 2)' = 0 11-171 3. xv" + 51" + xy = 0 4. 2xy" + (3 - 4.1.'lY' + (2x - 3)y 2 5. x )"" + -1-.1:\" + (x + 2»)" 6. 4.1.')," + 2/ + y = 0 7. (x 2 = 0 x/' + = 0 (x + 1»), = 0 + 2x 3 ),' + (x 2 - 2)y = 0 11. (x 2 + f)Y" + (4x + 2»),' + 2)" = 0 12. x 2 y" + 6xy' + (4x 2 + 6)y = 0 13. 2x)"" - (8x - 1)y' + (8x - 2).\' 0 14• .1.'y" + y' - xy = 0 10. (2x + 1)/ + .1'2)''' 15. (x - 4)2)''' - (x - 4)y' - 35y 0 + 3)2)''' - 9(x + 3»),' + 25y 8. xy" - }' 9. 16. x 2 y" o 17. v" + + 4xy' - (x 2 (x - 6»)' = 0 - 2)y = 0 o CHAP. 5 188 Series Solutions of ODEs. Special Functions 18. TEAM PROJECT. Hypergeometric Equation, Series, and Function. Gauss' 5 hypergeometric ODE5 is (15) + TO - x)}"" [e - (0 + b + l)x]/ - aby = O. Here. a. b, e are constants. This ODE is of the form P2y" + Ply' + Po)" = 0, where P2' PI' Po are polynomials of degree 2, 1, 0, respectively. These polynomials are written so that the series solution takes a most practical form. namely. In (1 + -- heX) = ] I! e x + ala + + I)b(b I) 2! e(e + I) I, 2; = xF(l. 1 + T In - - ' = 2tF(.12' I , I -x -x), .3. 2' .T2) . Find more such relations from the literature on special functions. (d) Second solution. Show that for /"2 = I - c the Frobenius method yields the following solution (where e =fo 2.3.4.... ): \'2(.\") = x ] -c I ( a - c + x2 + I)(h - c + I) I! (-c + 2) 07} + Ca - c + ab + x) l)(a - c + 2}Ch - c + I)Ch - c 2! (-c + 2)(-c + 3) x + 2) x2 + .. -). (16) + a(a + I)(a + 3! e(e + 1)(b + + 1)(e + 2) 2)b(b 2) T3 + .... This series is called the hypergeometric series. Its sum By choosing specific values of 0, b. e we can obtain an incredibly large number of special functions as solutions of (15) [see the small sample of elementary functions in part (c)]. This accounts for the importance of (15). (a) Hypergeometric series and function. Show that the indicial equation of (15) has the roots /"1 = 0 and /"2 = I-c. Show that for /"1 = 0 the Frobenius method gives (16). Motivate the name for (16) by showing that 1 F(1, I. I; x) = HI. b, b; x) = F(a. 1. a; x) = - - I - x (b) Convergence. For what 0 or b will (16) reduce to a polynomial? Show that for any other a, b. e (e =fo 0, - I, - 2, ... ) the series (16) converges when Ixl < 1. (I -I- (I - x)n = I - 1IxF(1 - 11, 1.2: x). arctan x = x F(!, 1.~; -x2 ). arcsin x = x F(!, !, ~: x 2 ). I, b - e + 1,2 - e;x). (e) On the generality of the hypergeometric equation. Show that (18) (12 + At + B)y + (Ct + D»)' + K)' = 0 with .,. = dyldt. etc.. constant A, B. C. D. K. and + At + B = (t - tl)(t - t 2), tl =fo t 2, can be reduced to the hypergeometric equation with independent variable t2 x= and parameters related by Ct] + D = -e(t2 - tl), e = a + b + I, K = abo From this you see that (15) is a "normalized fonn" of the more general (18) and that various cases of (18) can thus be solved in terms of hypergeometric functions. 119-241 HYPERGEOMETRIC EQUATIONS Find a general solution in terms of hypergeometric functions. 19. x(\ - x»)''' x)n = F( -11, b. b; -x), + )'2(X) = TI-CF(a - e .\'I(X) is called the hypergeometric function and is denoted by F(a, b, e; x). Here. e =fo 0, -I, -2..... (c) Special cases. Show that Show that + (! - 2x)), , - + + h' + !y = 0 20. 2x(l - x»," - (1 6x)y' - 2y = 0 21. x(1 - x),," 2.\' = 0 + t)y 23. 2(£2 - + t)' - )' = 0 5t + 6)5; + (2t - 3).}· 24. 4(t 2 3t 22. 3[( I - + 2)5: - 2." + Y 8y = 0 o 5 CARL FRIEDRICH GAUSS (1777-1855 J. great German mathemmician. He already made the first of his great discoveries as a student at Helmstedt and Gottingen. In 1807 he became a professor and director of the Observatory at Giittingen. His work was of basic importance in algebra. number theory, differential equations. differential geometry. non-Euclidean geometry. complex analysis. numeric analysis. a~trollomy. geodesy. electromagnetism. and theoretical mechanics. He also paved the way for a general and systematic use of complex numbers. SEC. 5.5 Bessel's Equation. Bessel Functions},,(x) 5.5 Bessel's Equation. Bessel Functions Jv (x) 189 One of the most important ODEs in applied mathematics in Bessel's equation,6 (1) Its diverse applications range from electric fields to heat conduction and vibrations (see Sec. 12.9). It often appears when a problem shows cylindrical symmetry (just as Legendre's equation may appear in cases of spherical symmetry). The parameter v in (1) is a given number. We assume that v is real and nonnegative. Bessel's equation can be solved by the Frobenius method, as we mentioned at the beginning of the preceding section, where the equation is written in standard form (obtained by dividing 0) by x 2 ). Accordingly, we substitute the series (2) y(x) ~ amxm + r = (ao -=/=- 0) 7n~O with undetermined coefficients and its derivatives into (1). This gives :x: 00 We equate the sum of the coefficients of x s + r to zero. Note that this power X S + T corresponds to 111 = s in the first, second, and fourth series. and to 111 = S - 2 in the third series. Hence for s = 0 and s = I, the third series does not contribute since 111 ~ O. For s = 2, 3, ... all four series contribute, so that we get a general formula for all these s. We find (3) (a) r(r - l)ao + rao - (b) + (r l)ral + + (r = 0 V2ao 1)(/1 - 2 V (/1 = 0 (s = 0) (8 = 1) (8 = 2,3, ... ). From (3a) we obtain the indicial equation by dropping ao, (4) The roots are r I (r v(~ 0) and 1"2 + v)(r - v) = O. = -v. 6FRIEDRICH WILHELM BESSEL (1784-1846l. German astronomer and mathematician. studied astronomy on his own in his spare time as an apprentice of a trade company and finally became director of the new Konigsberg Observatory. Formulas on Bessel functions are contained in Ref. [GRI] and the standard treatise [AB]. 190 CHAP. 5 Series Solutions of ODEs. Special Functions Coefficient Recursion for r = rl = v. For r = v, Eq. (3b) reduces to (2v + I)al = 0. Hence al = since v ~ 0. Substituring r = v in (3c) and combining the three terms containing as gives simply ° (5) (s + 2v)sas + lIs-2 = 0. ° Since al = and v ~ 0, it follows from (5) that a3 = 0, a5 = 0, .... Hence we have to deal only with even-numbered coefficients as with s = 2m. For s = 2m. Eq. (5) becomes + (2m + 2v)2ma2rn lI2m-2 = O. Solving for a2m gives the recursion formula (6) - - ; ; : 2 - - - - a2m-2, 2 m(v From (6) we can now determine a2' £14' • • • + m = I, 2, .... III = m) successively. This gives and so on, and in general (7) II 2m = ( -l)'mao --;;:-------~------ 2211117l! (V + l)(V + 2) ... (v + Ill) 1, 2, .... Bessel Functions In(x) For Integer v = n Illteger vailles oIv are denoted by 11. This is standard. For v = (8) = (/2 111 II the relation 0) becomes (-l)'mao --;;:-------------- 22 'mm! (11 + l)(n + 2) . . . (n + m) m = 1,2,···. ao is still arbitrary, so that the series (2) with these coefficients would contain this arbitrary factor ao. This would be a highly impractical situation for developing formulas or computing values of this new function. Accordingly, we have to make a choice. ao = 1 would be possible, but more practical turns out to be (9) because then n!(11 (10) lIo + 1) ... (n = 2"n! . + m) = (Ill + a2m = n)! in (8), so that (8) simply becomes (-l)m """"="2----'-----2 1n+n m! (n 1Il)! + m = 1,2,···. SEC. 5.5 Bessel's Equation. Bessel Functions },,(x) 191 This simplicity of the denominator of (10) partially motivates the choice (9). With these coefficients and rl = v = II we get from (2) a particular solution of (I), denoted by in(x) and given by (11) i,,(x) is called the Bessel function of the first kind of order 11. The series (II) converges for all x. as the ratio test shows. In fact. it converges very rapidly because of the factorials in the denominator. E X AMP L E 1 Bessel Functions lo(x) and l,(x) For 11 = 0 we obtain from (I I) the Bessel function of order 0 (12) ./o(x) = L- x6 26(3!)2 (_I)'nJ'x 2m 2211t(m!)2 + - ." 1l1,-O which looks similar to a cosine (Fig. 107). For hex) (131 L- = 1n=O (-I ),,'x2111 + 1 2m 1 + 11ll! (Ill + l)! II = I we obtain the Bessel function of order I X 2 :3 2:3 1!2! + x5 5 2 1!3! x7 7 2 3!4! + - ... which looks similar to a sine (Fig. 107). But the zeros of these functions are not completely regularly spaced 2 2 (see also Table Al in App. 5) and the height of the "waves" decreases with increasing x. Heuristically. n /x in (I) in standard form [( I) divided by .\'21 is zero (if It = 0) or small in absolute \alue for large x. and so is \·'/x. so that then Besser s equation comes close to / ' + Y = O. the equation of cos y and sin y; also / Ix acts as a "damping term."' in part responsible for the decrea~e in height. One can show that for large x. (141 ./,,(x) - [2 cos (x ~ -:;;:;:- - 2itT. - 4T.) where - is read "asymptotically equal"' and means thatfor.fhed II the quotient of the two aSX---i> ~ide~ appruache~ I x. Formula (14) is surprisingly accurate even for smaller x (> 0). For instance. it will give you good starting values in a computer program for the basic task of computing Leros. For example. for the first three zeros of ./0 you obtain the values 2.356 (2.405 exact to 3 decimals. error 0.049). 5.498 (5.520. error 0.022), 8.639 (8.654. error 0.015), etc. • 0.5 / "" / / / "" / O~---L----~~-L----~--~~L-~--,,~/~--~--~~--~lO~--~--~/~/~--x ..... Fig. 107. _--/ ,,/ Bessel functions of the first kind Jo and JI 192 CHAP. 5 Series Solutions of ODEs. Special Functions Bessel Functions Jv{x) for any JJ ::> o. Gamma Function We now extend our discussion from integer v = 11 to any v ~ O. All we need is an extension of the factorials in (9) and (11) to any v. This is done by the gamma function [( v) defined by the integral (15) (v> 0). By integration by parts we obtain The first expression on the right is zero. The integral on the right is f( v). This yields the basic functional relation (16) rev + l) = v rev). Now by (I5) From this and (16) we obtain successively r(2) = f(J) and in general nil + (17) 1) = I!, [(3) = 2f(2) = n! (n = 2!, ... = O. L .. '). This shows the the gamma function does in fact generalize the factorial function. Now in (9) we had ao = 1I(2nn!). This is 1!(271 + 1)) by (17). It suggests to choose, for any v, rell (18) Then (7) becomes 22mm! (v + l)(v + 2) ... (v + m)2'T(v + 1) . But (16) gives in the denominator (v + l)[(v + IJ = rev + 2), (v + 2)nv + 2) = rev and so on, so that (v + 1)(11 + 2) ... (v + 171) rev + I) = rev + 11l + 1). + 3) SEC. 5.5 Bessel's Equation. Bessel Functions JAx) 193 Hence because of our (standard!) choice (18) of ao the coefficients (7) simply are (-1)= (19) a2nz = 2 2 m+"m! With these coefficients and r by i,.(x) and given by = r) = (20) = rev + m + 1) v we get from (2) a particular solution of (1), denoted 00 i,.(x) x"L m=O 22m +"I1l! rev + m + 1) i,,(x) is called the Bessel function of the first kind of order v. The series (20) converges for all x, as one can verify by the ratio test. General Solution for Noninteger v. Solution )-'/ For a general solution, in addition to I,. we need a ~econd linearly independent solution. For v not an integer this is easy. Replacing v by - v in (20), we have (21) Since Bessel's equation involves v 2 , the functions i" and i_,. are solutions of the equation for the same v. If v is not an integer, they are linearly independent, because the first term in (20) and the first term in (21) are finite nonzero multiples of x" and x-", respectively. x = 0 must be excluded in (21) because of the factor x-v (with v> 0). This gives THEOREM 1 General Solution of Bessel's Equation If v is not an integer. a general solution of Bessel's equation for all x -=I=- 0 is (22) But if v is an integer, then (22) is not a general solution because of linear dependence: THEOREM 2 Linear Dependence of Bessel Functionsln andl_ n For integer v = n the Bessel functions in(x) and i_n(x) are linearly dependent, because (23) (n = 1,2, .. '). CHAP. 5 Series Solutions of ODEs. Special Functions 194 PROOF We use (21) and let v approach a positive integer n. Then the gamma functions in the coefficients of the first n terms become infinite (see Fig. 552 in App. A3.1). the coefficients become zero. and the summation starts with rn = II. Since in this case rem - n + 1) = (m - Il)! by (17). we obtain (m = 11 + s). The last series represents (-I)nIn{x), as you can see from (11) with m replaced by s. This completes the proof. • A general solution for integer interesting ideas. 11 will be given in the next section, based on some further Discovery of Properties From Series Bessel functions are a model case for showing how to discover properties and relations of functions from series by which they are defined. Bessel functions satisfy an incredibly large number of relationships-look at Ref. [AI3] in App. I; also, find out what your CAS knows. In Theorem 3 we shall discuss four formulas that are backbones in applications. THEOREM 3 Derivatives, Recursions The derivative of l,,(x) with respect to x can be expressed by lv_lex) or Iv+I(X) by the fOl1llu/lls (a) [xVI,,(x)]' (b) [x-vI,,(x)]' (24) = xVJ,,_I(X) = -X-vI,,+I(X). Furthermore. J,,(x) and its derivative satisfy the recurrence relations lv+l(x) 2v = -lv(x) lv_leX) (d) lv_leX) - Iv+I(X) = 2l~(x). (24) PROOF + (c) x (a) We multiply (20) by xl' and take X2v under the summation sign. Then we have We now differentiate this, cancel a factor 2, pull X 2v- 1 out, and use the functional relationship n v + III + 1) = (v + l7l)re v + m ) [see (16)]. Then (20) with v-I instead of v shows that we obtain the right side of (24a). Indeed, SEC. 5.5 195 Bessel's Equation. Bessel Functions J Jx) (b) Similarly, we multiply (20) by x-", so that x" in (20) cancels. Then we differentiate, cancel 2111, and use Ill! = m(m - I)!. This gives, with III = s + I, Equation (20) with v + I instead of v and s instead of m shows that the expression on the right is -x- vJ v + 1 (x). This proves (24b). (e), (d) We perform the differentiation in (24a). Then we do the same in (24b) and multiply the result on both sides by x2v. This gives (a*) (b*) vx,.-lJ .. -vX,·-IJ v + x''''~ = x"J,,_1 + xVJ:, = -x"J v + 1' Substracting (b*) from (a*) and dividing the result by x" gives (24c). Adding (a*) and (b~') and dividing the result by xl' gives (24d). • E X AMP L E 2 Application of Theorem 3 in Evaluation and Integration Formula (24c) can be used recursively in the form for calculating Bessel functions of higher order from those of lower order. For instance, J 2 (x) = 2J1(.1')/.1' - JoC>:), so that J 2 can be obtained from tables of J o and h (in App. 5 or, more accurately, in Ref. [GRI] in App. 1). To illustrate how Theorem 3 helps in integration. we use (24b) with v = 3 integrated on both sides. This evaluates. for instance. the integral A table of J 3 (on p. 398 of Ref. [GR I]) or yom' CAS will give you - A· 0.128943 + 0.019563 = 0.003445. Yom CAS (or a hnman computer in precomputer times) obtains h from (24), first u~ing (24e) with v = 2, that is, J 3 = 4.1'- 1J 2 - J 1 • then (24c) with v = I. thal is. J 2 = 2r:- 1h - J o. Together. 1= x- 3 (4r:- 1 (2r- 1h - J o) - hI I: = -A[2h(2) - 2Jo(2) = -AJ1 (2) + !Jo(2) + 7h(l) - 4JoO). - h(2)] + [8h(l) - 4Jo(l) - hill] This is what you get, for instance. with Maple if you type int(·· '). And if you type evalf(int(·· 0.00344544K in agreement with the result near the beginning of the example. .», you obtain • In the theory of special functions it often happens that for certain values of a parameter a higher function becomes elementary. We have seen this in the last problem set, and we now show this for J r,. CHAP. 5 196 THEOREM 4 Series Solutions of ODEs. Special Functions Elementary}" for Half-Integer Order v Besselfunctions J" of orders ±!, ±~, ±~, ... are elementary; the}' call be expressed by fillitely many cosines and sines and powers of x. In particular, (25) PROOF When lJ 11/2(X) (a) = = J 2 sin x. (b) Ll/2(x) 7T"X = J 2 cos X. 7T"X !, then (20) is To simplify the denominator. we first write it out as a product AB. where A = 2mm! = 2m(2111 - 2)C2m - 4) ... 4·2 and [use (16)J = (2m + 1)(2111 - I) ... 3· 1 • v.r;; here we used (26) We see that the product of the two right sides of A and B is simply (2111 J 1/2 becomes 1 1 / 2 (x) = [f - x ~ 7T"X m=O (_I}lnx 2m+1 (2m + = I)! as claimed. Differentiation and the use of (24a) with + I )!v.r;, so that [f. - sin x, 7T"X lJ = ! now gives This proves (25b). From (25) follow further formulas successively by (24c), used as in Example 2. This completes the proof. • E X AMP L E 3 Further Elementary Bessel Functions From (24c) with v = ~ and v = -~ and (25) we obtain L 3/ 2(X) respectively, and so on. = - -I x L 1/ 2(X) - h/2(X) =- [f TTX (cosx -- x + sinx) • SEC 5.5 Bessel's Equation. Bessel Functions JJx) 197 We hope that our study has not only helped you to become acquainted with Bessel functions but ha<; also convinced you that series can he quite useful in obtaining various properties of the corresponding functions. PROBL~ME5EE~E3~.35L- _____ 1. (Convergence) Show that the series in (I I) converges 121-281 for all x. Why is the convergence very rapid? 2. (Approximation) Show that for small Ixl we have 10 = I - 0.25x2 . From this compute 10(.'r) for x = O. 0.1. 0.2..... 1.0 and determine the error by using Table Al in App. 5 or your CAS. Use the powerful formulas (24) to do Probs. 21-28. (Show the details of your work.) 21. (Derivatives) Show that l~(x) = - 1 1 (x), l~(x) = 10(x) - 11(X)/X, l~(x) 3. ("'Large" values) Using 04), compute 10lx) for x = 1.0, 2.0. 3.0..... 8.0, determine the error by Table Al or your CAS. and comment. 4. (Zeros) Compute the fIrst four positive zeros of 10 (.r) and 1 1 (x) from (14). Determine the error and comment. 15-201 ODEs REDUCIBLE TO BESSEL'S EQUATION 6. x 2 / ' + xy' + (x 2 - -W)Y 7. x 2)''' + xy' + !(x - JJ2)y = = = 0 19. x 2 )"" 20. = x "1" (x) + C, 0 CVx = z) = 0 26. (Integration) Evaluate integrate by parts.) I x- 11 4 (x) dx. (Use Prob. 25; 27. (Integration) Show that Ix 2 10 (x) dx = x21 1 (x) + x10(x) - Il o(X) dx. (The last integral is nonelementary; tables exist, e.g. in Ref. [A13J in App. L) 28. (Integration) Evaluate I1 5 (X) dx. 29. (Elimination of first derivative) Show that y = ltV with vex) = exp (-~ J p(x) dx) gives from the ODE y" + p(x)y' + q(x)y = 0 the ODE xl/4 u. X1l4 = ::.;) 17. 36x2,," + 18x\" (y =. x 1l4 u, ix i/4 18. x 2 )''' IX"l"-l(X) dx lAx = ;::) + 1 )2y" + 2(2x + 1)/ + 16x( t + l)y (2x + I = ::.;) 9. x/' - / + 4ry = 0 (\. = XII, 2t = z) 10. x 2 )"" + x.v' + !(x 2 - I)), = 0 lx = 2::.;) 11. xy" + (2v + l)y' + xy = 0 (y = x-Vu) 12. x 2)"" + xy' + 4(x 4 - JJ2)y = 0 (x 2 = z) 13 .... 2y" + xy' + 9(x 6 - v 2 »' = 0 (x 3 = z) I-I. y" + (e 2x - ~)y = 0 (eX = .::.) 15. xy" + y = 0 l)' = Vx u. 2Vx = z) 16. 16x2 )''' + 8xy' + (x1/ 2 + :a)y = 0 = 13(x)l. 0 8. (2x (y = Ml 1(x) - 22. (Interlacing ofzeros) Using (24) and Rolle's theorem, show that between two consecutive zeros of 10(x) there is precisely one zero of 11 (x). 23. (Interlacing ofzeros) Using (24) and Rolle's theorem. show that between any two consecutive positive zeros of In(x) there i~ precisely one J:ero of l,,+I(X). 24. (Bessel's equation) Derive (I) from (24). 25. (Basic integral formulas) Show that Using the indicated substitution~. find a general solution in temlS of 1 v and 1 - v or indicate when this is not possible. (This is just a sample of various ODEs reducible to Bessel's equation. Some more follow in the next problem set. Show the details of your work.) 5. (ODE "ith two parameters) x 2)"" + xy' + (A 2x 2 - JJ2)y APPLICATION OF (24): DERIVATIVES, INTEGRALS + Vx V = z) . = 0 + xy' + Vx y = 0 (4t 1l4 = z) + !xy' + Vx y = 0 ly = x 2/5 u. 4X1l4 = + (I - 2v)xy' + v 2(x 2v + 1 - v 2 )y = x 2/ ' (y = t"u, XV = ::.;) .::.) 0 no longer containing the first derivative. Show that for the Bessel equation the substitution is y = UX- 1I2 and gives (27) 198 CHAP. 5 Series Solutions of ODEs. Special Functions 30. (Elementary Bessel functions) Derive (25) in Theorem 4 from (27). 31. CAS EXPERIMENT. Change of Coefficient. Find and graph (on common axes) the solutions of .v" + h- 1y' + Y = O• .1'(0) = I. y' (0) = (c) Conclude that possible frequencies wl27f are those for which s = 2wv Llg is a zero of 10 , The con-esponding solutions are called normal modes. Figure 108 shows the first of them. What does the second normal mode look like? The third? What is the frequency (cycles/min) of a cable of length 2 m? Of length 10 m? O. for k = 0, 1, 2... '. 10 (or as far as you get useful graphs). For what k do you get elementary functions? Why? Try for noninteger k. particularly between 0 and 2. to see the continuous change of the curve. Describe the change of the location of the zeros and of the extrema as k increases from O. Can you interpret the ODE as a model in mechanics, thereby explaining your observations? 32. TEAM PROJECT. Modeling a Vibrating Cable (Fig. 108). A flexible cable, chain, or rope of length L and density (mass per unit length) p is fixed at the upper end (x = 0) and allowed to make small vibrations (small angles a in the horizontal displacement u(x. t), t = time) in a vertical plane. (a) Show the following. The weight of the cable below a point x is W(x) = pg(L - x). The restoring force is F(x) = W sin a = Wu.". U x = (JuliJx. The difference in force between x and x + D.X is D.x (Wu~')x' Newton's second law now gives Equilibrium position Fig. 108. 33. CAS EXPERIMENT. Bessel Functions for Large x. (a) Graph l,,(x) for 11 = 0, ... , 5 on common axes. (b) Experiment with (14) for integer II. Using graphs. find out from which x = Xn on the curves of (II) and (14) practically coincide. How does Xn change with n? (c) What happens in (b) if II = ::!::~? (Our usual notation in this case would be v.) (d) How does the en-or of (14) behave as function of x for fixed II? [En-or = exact value minus approximation (14).1 (e) Show from the graphs that 10 (x) has extrema where 1 1 (x) = O. Which formula proves this? Find further relations between zeros and extrema . (f) Raise and answer questions of your own. for instance. on the zeros of 10 and 1 1 , How accurately are they obtained from (14)? p .1x II tt = .lx pg[(L - X)lI"Jx. For the expected periodic motion = .r(x) cos (wt + 8) the model of the problem is the ODE u(x, t) (L - x).r" - y' + A\ = O. (b) Transform this ODE to~; + ,1'-1" + Y = O. dylds. s = 2A::,1I2. ::. = L - x. so that the solution is ." = y(x) = 10 (2wV(L - >;)Ig). 5.6 Vibrating cable in Team Project 32 Bessel Functions of the Second Kind Yv(x) From the last section we know that I" and I_v fonn a basis of ~olution<; of Bessel's equation, provided v is not an integer. But when v is an integer, these two solutions are linearly dependent on any interval (see Theorem 2 in Sec. 5.5). Hence to have a general solution also when v = 11 is an integer, we need a second linearly independent solution besides In. This solution is called a Bessel function of the second kind and is denoted by Yn . We shall now derive such a solution. beginning with the case II = O. n = 0: When (1) II Bessel Function of the Second Kind Yo(x) = 0, Bessel's equation can be written xy" + y' + .\}' = O. SEC 5.6 199 Bessel Functions of the Second Kind Y,.(x) Then the indicial equation (4) in Sec. 5.5 has a double root r = O. This is Ca<;e 2 in Sec. 5.4. In this case we first have only one solution. 10(x). From (8) in Sec. 5.4 we see that the desired second solution must be of the form 00 (2) = Y2(X) + "L 10(x) In x Amx7n. m=1 We substitute .'"2 and its derivatives I I .'"2 = 10 In x + "" 21~ 10 Inx + -x x 111A",x"'- 1 m=1 10 )'2 = + L.. ~ 10 x2 + "L 'Xc 111 (m I )A m x m - - 2 Jll,=l into (1). Then the sum of the three logarithmic terms x 1~ In x, 1 ~ In x, and xl0 In x is Lero because 10 is a solution of (l). The terms - 10 Ix and 10 lx (from xy" and :v') cancel. Hence we are left with x 21~ x + "L m(m - I)Am x m - 1 + "L x mAm·\·m-I m=l 111=1 "L + Amxm+I = O. ==1 Addition of the first and second series gives Lm2Amxm-l. The power series of 1~(x) is obtained from 1I2) in Sec. 5.5 and the use of m!/m = (111 - I)! in the form 00 ="L In=l Together with Lm 2A",xm - 1 and LAmxm+l this gives (3*) "L.. m=1 ( -1 )nlx2m-l 2 ,-2 2 n Ill! (m - I)! +" L.. 'Xc 111 2A mX"In-I ==1 +" L..A .,..,..':.=+1 = O. m=1 First, we show that the Am with odd subscripts are all zero. The power X O occurs only in the second series. with coefficient AI' Hence Al = O. Next, we consider the even powers X2s. The first series contains none. In the second serie", m - 1 = 25 gives the term (25 + 1)2A2s+1X2s. In the third series. m + 1 = 25. Hence by equating the sum of the coefficients of x 2s to zero we have 5 Since Al = 0, we thus obtain A3 = O. A5 = O.... , successively. We now equate the sum of the coefficients of X 2s + I to zero. For s -1 + 4A2 = 0, = L. 2 ..... = 0 this gives thus For the other values of s we have in the first series in (3*) 2111 - I = 2s + 1, hence m = s + 1, in the second m - 1 = 2s + 1, and in the third 111 + 1 = 2s + I. We thus obtain (_l)s+1 2s 1 2 (s + )! s! + (2s + 2)2A 2s + 2 + A2s = O. 200 CHAP. 5 For s Series Solutions of ODEs. Special Functions = J tills yields 3 128 thus and in general (3) A 2m = (-I )m-l ( 22m(m!)2 1 2 + = ]+ 2 1 + 1 3 J ) + ... + - III . m = 1,2,···. Using the short notations (4) and inserting hm (4) and Al = A3 = . . . = 0 + ... + into (2). m m = 2.3.··· we obtain the result (5) = Jo(x) lnx + 1 2 4x _3_ 128 X4 + _1_1_ x 6 13824 + Since 10 and )"2 are linearly independent functions, they fonn a basis of (I) for x > o. Of course, another basis is obtained if we replace )"2 by an independent particular solution of the form a(Y2 + b10), where a (of:. 0) and b are constants. It is customary to choose a = 2/7T and b = y - ]n 2. where the number y = 0.577 215 664 90 ... is the so-called Euler constant, which is defined as the limit of 1 + 2 + ... + s - In s as s approaches infinity. The standard particular solution thus obtained is called the Bessel function of the second kind of order ~ero (Fig. 109) or Neumann's function of order zero and is denoted by Yo(x). Thus [see (4)] (6) For small x > 0 the function Yo(x) behaves about like In x (see Fig. 109, why?), and ~ - 'XJ as x ~ o. Yo(x) Bessel Functions of the Second Kind Yn(x) For v = 11 = 1, 2, ... a second solution can be obtained by manipulations similar to those for 11 = 0, st<uting from (10), Sec 5.4. It turns out that in these cases the solution also contains a logarithmic term. The situation is not yet completely satisfactory, because the second solution is defined differently, depending on whether the order v is an integer or not. To provide uniformity SEC. 5.6 Bessel Functions of the Second Kind Y•. (x) 201 of fonnalism, it is desirable to adopt a form of the second solution that is valid for all values of the order. For this reason we introduce a standard second solution Y,,(x) defined for all v by the formula (a) Y,.(X) = Sill (7) V7T [1,.(x) cos V7T - i_.,(x)] Yn(X) = lim VAx). (b) ,,~n This function is called the Bessel function of the second kind of order vor Neumann's function 7 of order v. Figure 109 shows Yo(x) and YI(X), Let LIS show that i,. and YI' are indeed linearly independent for all v (and x > 0). For non integer order v. the function Y,lx) is evidently a solution of Bessel's equation because i,.(x) and 1 -,,(x) are solutions of that equation. Since for those v the solutions 1" and i_,. are linearly independent and Y,. involves I_v, the functions i,. and Y" are linearly independent. Furthermore, it can be shown that the limit in (7b) exists ,md Yn is a solution of Bessel's equation for integer order; see Ref. [A13] in App. 1. We shall see that the series development of Yn(x) contains a logarithmic term. Hence i,lr) and Yn(x) are linearly independent solutions of Bessel's equation. The series development of Yn(x) can be obtained if we inseI1 the series (20) and (21), Sec. 5.5. for i,.(x) and L,.(x) into (7a) and then let v approach 11; for details see Ref. [A13]. The result is Y,,(x) (x 2 ) xn ~ = - 1n(x) In -2 + 'Y + 7T (8) _x -n n - l '" ~ 7T where x > 0, h m 11 m~O (n - m - 1 2 Fig. 109. 111 (-I)m-l(h m 2 2 I)! m+1l-17l ! -----;::----- X 2m n 2 - m! = O. 1. .... and [as in (4)] =1+-+ .. ·+ ~ 7T m~O ho hm+n + h m+ n ) (Ill + n)! 2 1ll = O. hI = L = 1+ 1 2. + ... + 111 + 11 Bessel functions of the second kind Yo and Y,. (For a small table, see App. 5.) 7CARL NEUMANN (1832-\9251. German mathematician and physicist. His work on potential theory sparked the development in the field of integral equations by VITO VOLTERRA (1860-1940) of Rome. ERIC IVAR FREDHOLM (I 866--19D) of Stod.holm. and DAVID HILBERT (1862-1943) of Giittingen (see the footnote in Sec. 7.91. The solutions Y,.(X) are sometimes denoted by N,.(x); in Ref. [A13] they are called Weber's functions; Euler', constant in (6) is often denoted by C or In 1'. 101 CHAP. 5 Series Solutions of ODEs. Special Functions For n = 0 the last sum in (8) is to be replaced by 0 [giving agreement with (6)]. Furthermore, it can be shown that Our main result may now be formulated as follows. THEOREM 1 General Solution of Bessel's Equation A general solution of Bessel's equatimz for all values of v (and x > 0) is (9) We finally mention that there is a practical need for solutions of Bessel's equation that are complex for real values of x. For this purpose the solutions (10) H~~\>;;) = J v(X) + iY,,(x) H~~)(x) = iY,,(x) Jv(x) - are frequently used. These linearly independent functions are called Bessel functions of the third kind of order v or first alld second Hankel functions B of order v. This finishes our discussion on Bessel functions, except for their "orthogonality," which we explain in Sec. 5.7. Applications to vibrations follow in Sec. 12.9. 11-10 1 SOME FURTHER ODEs REDUCIBLE TO BESSEL'S EQUATIONS (See also Sec. 5.5.) Using the indicated substinl!ions, find a general solution in terms of J v and Y •. Indicate whether you could also use J- v instead of Y v ' (Show the details of your work.) 1. x2y" + x)" + (x 2 - 25)y = 0 2 2 2. x -,," + x/ + (9x - ~)y = 0 (3x = .:::) 3. 4xy" + 4/ + y = 0 (~=.:::) 4. xy" + y' + 36)" = 0 (\ 2~ = z) 5. x 2 y" + xy' + (4 X 4 - 16)y = 0 (x 2 = z) 6. x 2 -,," + x/ + (x 6 - I)." = 0 (~x3 = z) 7. xy" + 11/ + xy = 0 (y = x- 5 ,,) 8. y" + -1-x 2 y = 0 (y = u~. x 2 = z) 9. x 2 y" - 5xy' + 9(x 6 - 8)y = 0 (y = x 3 u, x 3 = z) 10. xy" + 7/ + 41·Y = 0 (y = x- 3 u. 2x = z) 11. (Hankel functions) Show that the Hankel functions ( 10) form a basis of solutions of Bessel's equation for any v. 12. CAS EXPERIMENT. Bessel Functions for Large x. It can be shown that for large x. (11) Yn(x) - v'2/( 7TX) sin (x - ! 1l7T - ~7T) with - defined as in (14) of Sec. 5.5. (a) Graph Yn(x) for 11 = O.... , 5 on common axes. Are there relations between zeros of one function and extrema of another? For what functions? (b) Find out from graphs from which x = Xn on the curves of (8) and (11) (both obtained from your CAS) practically coincide. How does Xn change with 11? (c) Calculate the first ten Leros X m ' In = I, ... , 10, of Yo(x) from your CAS and from (II). How does the error behave as 171 increases? (d) Do (c) for Yl(X) and Y2(x). How do the errors compare to those in (c)? BHERMANN HANKEL (1839-1873). German mathematician. SEC. 5.7 Sturm-Liouville Problems. Orthogonal Functions 203 13. Modified Bessel functions of the first kind of order P are defined by Iv(x) = i-vJ,,(ix), i = \1"="1. Show that Iv satisfies the ODE 14. (Modified Bessel functions I.,) Show that I,,(x) is real for all real x (and real v), 1v (x) "* 0 for all real x "* 0, and Ln(x) = In(x). where n is any integer. x 2 y" + xy' - (x 2 + v 2 )y = 0 15. Modified Bessel functions of the third kind (sometimes called of the second kind) are defined by the formula (14) below. Show that they satisfy the ODE (12). (12) and has the representation cc (13) [,Jx) = x2m.+v L 22m+vm! rem + v + (14) I) KJI:) = 711=-0 5.7 7r [Lv(.\) - IJr)] . 2 sm V7r Sturm-Liouville Problems. Orthogonal Functions So far we have considered initial value problems. We recall from Sec. 2.1 that such a problem consists of an ODE, say, of second order, and initial conditions .1'(xo) = Ko, y' (xo) = KI referring to the same point (initial point) x = Xo. We now turn to boundary value problems. A boundary value problem consists of an ODE and given boundary conditions referring to the two boundary points (endpoints) x = a and x = b of a given interval a ~ x ~ b. To solve such a problem means to find a solution of the ODE on the interval a ~ x ~ b satisfying the boundary conditions. We shall see that Legendre's, Bessel's, and other ODEs of importance in engineering can be written as a Sturm-Liouville equation (1) [p(x)y']' + [q(x) + Ar(x)]y = 0 involving a parameter A. The boundary value problem consisting of an ODE (1) and given Sturm-Liouville boundary conditions + (a) kIy(a) (b) IIy(b) (2) k2 y' (a) = 0 + 12y'(b) = 0 is called a Sturm-Liouville problem. 9 We shall see further that these problems lead to useful series developments in terms of particular solutions of (1), (2). Crucial in this connection is orthogonality to be discussed later in this section. In (1) we make the assumptions thatp, q, r, andp' are continuous on a ~ x ~ b, and rex) >0 (a ~ x ~ b). In (2) we assume that kI , k2 are given constants, not both zero, and so are II, 12, not both zero. 9JACQUES CHARLES FRAN<;:OIS STURM (1803-1855), was born and studied in Switzerland and then moved to Paris, where he later became the successor of Poisson in the chair of mechanics at the Sorbonne (the University of Paris). JOSEPH LIOUVILLE (1809-1882), French mathematician and professor in Paris, contributed to various fields in mathematics and is particularly known by his important work in complex analysis (Liouville's theorem; Sec. 14.4), special functions, differential geometry, and number theory. 204 E X AMP L E 1 CHAP. 5 Series Solutions of ODEs. Special Functions Legendre's and Bessel's Equations are Sturm-Liouville Equations Legendre's equation (I - x2)y" - 2.\)"' + "(,, + I)y = 0 may be written A= 11(11 + I). This is (1) With P = I - x 2 , q = O. and r = 1. In Bessel's equation .,. = dyldx. etc. as a model in physics or elsewhere. one often likes to have another parameter k in addition to II. For this reason we set x = h. Then by the chain rule .,. = dyld.y = «(~,"ldx) drldx = y'lk. ;.' = y"lk 2 . In the first two lerms. k 2 and k drop out and we get Division by x gives the Sturm-Liouville equation [xy'l' + This is (I) with p = x. q = _/12/x, + A.\}' 0 = • and r = x. Eigenfunctions, Eigenvalues Clearly, y == 0 is a solution-the "trivial solution"-for any A because (I) is homogeneous and (2) has zeros on the right. This is of no interest. We want to find eigenfunctions y(x), that is, solutions of (l) satisfying (2) without being identically zero. We call a number A for which an eigenfunction exists an eigenvalue of the Sturm-Liouville problem (1), (2). E X AMP L E 2 Trigonometric Functions as Eigenfunctions. Vibrating String Find the eigenvalues and eigenfunctions of the Sturm-Liouville problem y" + Ay = O. (3) .1'(0) = 0, .1'( 77) = O. This problem arises. for instance. if an elastic SIring (a violin sIring, for example) is slrelched a little and then fixed at its ends x = 0 and x = 77 and allowed to vibrate. Then y(x) is the "space function'" of the deflection II(X. 1) of the string. assumed in the fom1 II(X, 1) = y(x)II'(1). where t is time. (This model will be discussed in great det:Jil in Secs. 12.2-12.4.) Sollltion. k2 = 12 From (I) and (2) we see that p = I, if = O. r = I in (I I. and a = O. b = 77. kl = 11 = I. = 0 in (2). For negative A = - v 2 a general solution of the ODE in (3) is y(x) = ("levx + c2e -VX. From the boundary conditions we obtain ("1 = ("2 = O. so that y == O. which is not an eigenfunction. For A = 0 the situation is similar. For positive A = v 2 a general solution is y(x) = A cos vx -t- B sin vx. From the first boundary condition we obtain \'(0) = A = O. The second boundary condition then yields )'(77) = B sin V77 = 0, For v = 0 we have y == O. For A = v 2 thus = 1. 4. 9. 16..... laking B y(X) = sin vr Hence the eigenvalues of the problem are A = v = sin VX, where" = 1. 2. . . . . y(x) v = O. :!:I, :+:2,···. 2 , = I. we obtain (v = 1,2, .. '). where v = I, 2, ... , and corresponding eigenfunctions are • Existence of Eigenvalues Eigenvalues of a Sturm-Liouville problem (I), (2), even infinitely many, exist under rather general conditions on p. q. r in (1). (Sufficient are the conditions in Theorem I, below, together with p(x) > 0 and r(x) > 0 on a < x < b. Proofs are complicated; see Ref. LA3] or [All] listed in App. 1.) SEC. 5.7 205 Sturm-Liouville Problems. Orthogonal Functions Reality of Eigenvalues Furthermore, if p, q, r, and p' in (l) are real-valued and continuous on the interval a ~ x ~ band r is positive throughout that interval (or negative throughout that interval). then all the eigenvalues of the Sturm-Liouville problem (l), (2) are real. (Proof in App. 4.) This is what the engineer would expect since eigenvalues are often related to frequencies, energies. or other physical quantities that must be real. Orthogonality The most remarkable and important property of eigenfunctions of Sturm-Liouville problems is their 0I1hogonality, which will be crucial in series developments in terms of eigenfunctions. DEFINITION Orthogonality Functions Yl(X), Y2(X), ... defined on some interval a ~ x ~ b are called orthogonal on this interval with respect to the weight function rex) > 0 if for all 111 and all n different from tn, I (4) b rex) Ym(X) Yn(X) dx = 0 (m *- n). a II Ym II The norm of Ym is defined by I IIYmll (5) b r(x)Ym2(x) dr: . a Nore thar this is the "quare roor of the integral in (4) with Il = Ill. The functions .'"1> )"2, •.• are called orthonormal on a ~ x ~ b if they are orthogonal on this interval and all have norm I. If r{x) = I, we more briefly call the functions orthogonal instead of orthogonal with respect to rex) = I; similarly for orthonormality. Then I b Ym(X) Yn(X) dx = 0 *- (m IIYmll = 11), a E X AMP L E 3 I b 2 Ym (x) dx. a Orthogonal Functions. Orthonormal Functions The functions y",(x) = sin mx, m = I. 2.... form an orthogonal set on the interval m 1= 11 we obtain by integration [see (II) in App. A3. II IT>' J'm(X)Yn(X) d,' = IT>' sin Ill\' sin In dx = -77' The norm 1I" IiYmli cos (111 - II)T tiT -1 IT>' -71 -r, 7T ;§; X ;§; 7T, cos (/II because for + II)X dx = O. -17 equals V;. because IIYmli 2 = I:"sin2 /11x dx = 7T tm = 1,2," 'J. Hence the corre'ponding orthonormal set, obtained by division by the norm, is sinx v;,. , sin2x v;,. , sin 3x V; , • CHAP. 5 206 Series Solutions of ODEs. Special Functions Orthogonality of Eigenfunctions Orthogonality of Eigenfunctions THEOREM 1 Suppose that the functions p, q, r, and p' in the Snl17ll-Liouville equation (l) are real-valued alld contilluous alld r(x) > 0 Oil the interval a ~ x ~ b. Let Ym(x) and Yn(x) be eiRellfilllctiolls of the Sturm-Lioul'iIIe problem (I), (2) that correspond to different eigelll'alues Am alld An' respectively. Theil .1'm, Yn are orthogollal on that interval with respect to the weight jilllctioll r. that is. J b (6) r(x)ym(x)yn(x) dl: =0 (111 =1= 11). a fr pea) = 0, thell (2a) can be dropped from the problem. fr pCb) = 0, thell (2b) can be dropped. [It is then required that y and .1" remain bounded at such a point, and the problem is called singular, as opposed to a regular problem in which (2) is used.] ljp(a) = pCb), thell (2) call be replaced by the "periodic boundary conditions" (7) yea) = y(b), )" (a) = y' (b). The boundary value problem consisting of the Sturm-Liouville equation (I) and the periodic boundary conditions (7) is called a periodic Sturm-Liouville problem. PROOF By assumption. Ym and )'n satisfy the Sturm-Liouville equations (PY;nJ' + (q + Amr)Ym = 0 (py;J' + (q + An r »)'" = 0 respectively. We multiply the first equation by .1'n, the second by -Ym' and add. where the last equality can be readily verified by performing the indicated differentiation of the last expression in brackets. This expression is continuous on a ~ x ~ b since p and p' are continuous by assumption and Y>n' Yn are solutions of (I). Integrating over x from a to b, we thus obtain (8) (a < b). The expression on the right equals the sum of the subsequent Lines I and 2, (9) p(b)L,,:,(iJ)Ym(b) - y';"(b)Yn(b)] (Line I) -p(a)[y~(a)Ym(a) - Y:n(a)Yn(a)] (Line 2). Hence if (9) is zero. (8) with Am - An =1= 0 implies the orthogonality (6). Accordingly, we have to show that (9) is zero, using the boundary conditions (2) as needed. SEC. 5.7 207 Sturm-Liouville Problems. Orthogonal Functions Case 1. pea) Case 2. pea) = pCb) = O. Clearly. (9) is zero, and (2) is not needed. '* 0, pCb) = O. Line I of (9) is zero. Consider Line 2. From (2a) we have k2y~(a) k1Yn(a) + k1Ym(a) + k 2Y;n(a) = O. 0, = Let k2 =F O. We multiply the first equation by Ym(a). the last by -y,,(a) and add. k2[Y;t(a)Ym(a) - y,',.,(a))",,(a)] = O. This is k2 times Line 2 of (9), which thus is zero since k2 =F O. If k2 assumption, and the argument of proof is similar. = O. then kl =F 0 by '* '* O. We use both (2a) and (2b) and proceed as in Cases 2 and 3. Case 3. pea) = O,p(b) O. Line 2 of (9) is zero. From (2b) it follows that Line 1 of (9) is zero; this is similar to Case 2. Case 4. pea) '* Case 5. pea) = pCb). Then (9) becomes O,p(b) The expression in brackets [.. '1 is zero, either by (2) used as before, or more directly by (7). Hence in this case, (7) can be used instead of (2), as claimed. This completes the proof of Theorem 1. • E X AMP L E 4 Application of Theorem 1. Vibrating Elastic String The ODE in Example 2 is a Sturm~Liouville equation with p = 1. q = O. and r = I. From Theorem I it follows • that the eigenfunctions Yilt = sin m~ (111 = 1.2.... ) are orthogonal on the interval 0 ~ x ~ 7T. E X AMP L E 5 Application of Theorem 1. Orthogonality of the Legendre Polynomials Legendre's equation is a Sturm~Liouville equation (see Example I) [ (I-x).'" 2 'l' +A.I'=O. A = nen + I) with I' = I - x 2 • q ~ 0, and r = I. Since p( -1) = p(l) = 0, we need no boundary conditions. but have a ~illglliar Sturm-Liouville problem on the interval -1 ~ x ~ 1. We know that for 11 = 0, I, ... , hence A = O. I . 2, 1 . 3....• the Legendre polynomials P n(x) are solution, of the problem. Hence these are the eigenfunctions. From Theorem I it follows that they are orthogonal on that interval. that is, f (10) 1 Pm(x)Pn(x) dx = 0 (Ill 1= n). • ~1 E X AMP L E 6 Application of Theorem 1. Orthogonality of the Bessel Functions In(x) The Bessel function in(x) with fixed integer II ~ 0 satisfies Bessel's equation (Sec. 5.5) where j" = dJnldx. j~ = d 2 i,/dx2 . In Example 1 we transformed this equation. by setting equation x= h. into a Stuml~Liouville with pIx) = ~,q(x) = -1l /x, rex) = x, and parameter A = k 2 . Since 1'(0) = O. Theorem 1 implies orthogonality on an interval 0 ~ x ~ R (R given. fixed) of those solutions JnVer) that are zero at x = R. that is. 2 (11) in(kR) = 0 (II fixed). 208 CHAP. 5 Series Solutions of ODEs. Special Functions [Note that q(x) = -1l 2 /x is discontinuous at O. but this does not affect the proof of Theorem 1.1 It can be shown (see Ref. [A 13]) that In(Ji) ha~ infinitely many zeros, say, = an.l < 0'n.2 < ... (see Fig. 107 in Sec. 5.5 for II = 0 and I). Hence we must have x (12) This THEOREM 2 thus kR = O'n,711 prove~ (Ill = 1,2." '). the following orthogonality property. Orthogonality of Bessel Functions For each fixed nonnegative integer n the sequence of Bessel functions of the first ki1ld In(k,,.lX), I n (kn.2 x), ... with ~.m as in (12) forms an orthogonal set on the imell'al 0 ~ x ~ R with respect to the weight function r(x) = x, that is. R f xJn(kn,mx)Jn(kn,jx) dx (13) o = (j 0 *- 111, II fixed). Hence we have obtained illfinitely lIIallY orthogollal sets. each conesponding to one of the fixed values Il. This also illustrates the importance of the zeros of the Bessel functIons. • E X AMP L E 7 Eigenvalues from Graphs Solve the Sturm-Liouville problem y" + Ay = O. .1'(0) + y' (0) = O. y( 17) - y' (17) = o. Solution. A general solution and its derivative are y=Acosb:+Bsinb: y' and The fust boundary condition gives yeO) + and substitution of A ~ -B/.. give = . Ak sin b: + Bk cos b:. k= VA. y' (0) = A + Bk = O. hence A = - Bk. The second boundary condilion )'(17) - y' (17) = A cos 17/... + B sin 17k + Ak sin 17k - Bk cos 17k = -Bk "* We must have B B cos 17k gives cos 17k +B sin 17k - Bk 2 sin 17k - Bk cos 17k = D. 0 since otherwise B = A = O. hence y = O. which is not an eigenfunction. Division by -k + tan 17k - k 2 tan 17k - k = 0, thus tan 17k = -2k k2 - I . The graph in Fig. I IO now shows u, where to look for eigenvalues. These conespond to the k-values of the points of intersection of tan 17k and the right side - 2k/(k 2 - I) of the last equation. The eigenvalues are Am = k",2, where 1..0 = 0 with eigenfunction Yo = I and (he other An, are located near 22 , 32 ,42 , . . . , with eigenfunctions cos k",x and sin k",x, 111 = 1,2, .... The precise numeric determination of the eigenvalues would require a root-finding method (such as those given in Sec. 19.2). • y 1 1/: Or-~-+--r-7--+~~+--r~--+--- k -1 -2 -3 Fig. 110. Example 7. Circles mark the intersections of tan 1Tk and - 2k/(e - 1) SEC. 5] 209 Sturm-Liouville Problems. Orthogonal Functions .......... _..... ---. _....... .......... -_ .............. ..- 1. (Proof of Theorem 1) CalTY out the details in Cases 3 and 4. 2. Normalization of eigenfunctions Ym of (I), (2) means that we multiply Ym by a nonzero constant em such that emYm has norm I. Show that 2m = ey", with any e 0 is an eigenfunction for the eigenvalue corresponding to vm 3. (Change of x) Show that if the functions Yo(x), YI(X), ... form an orthogonal set on an interval a ~ x ~ b (with rex) = I), then the functions )'o(er + k), )'l(er + k), ... , e > 0, form an orthogonal set on the interval * ~ I ~ (a - k)/e STURM-LIOUVILLE PROBLEMS Write the given ODE in the form (I) if it is in a different form. (Use Prob. 6.) Find the eigenvalues and eigenfunctions. Verify orthogonality. (Show the details of your work.) 7. y" + Ay = o. yeO) = 0, )'(5) = 0 8. y" + Ay = 0, ),'(0) = 0, y'(w) = 0 9. y" + Ay = 0, yeO) = 0, )" (L) = 0 10. y" + Ay = 0, yeo) = yO), y' (0) = y' 0) 11. y" + Ay = O. yeO) = y(2w), y' (0) = y' (2rr) 12. .'v" + yO) , AV . = O. + \,(0) .,. + -v'(O) = o. o. yO) = 0, y(2) = O. (a) Chebyshev pol)nomiaIs lO of the first and second kind are defined by Tn(x) = cos (n arccos x) Un(x) = respectively, where + (A (Set x = e t .) + I)y = 0, + (A + 16)y = 0, + I) arccos x] ~ sin [(n 11 = O. 1, .. '. Shuw that To = 1, Uo = 1, Show that the Chebyshev polynomials Tn(x) are orthogonal on the interval - I ~ x ~ I with respect to the weight function rex) = 1I~. (Hint. To evaluate the integral, set arccos x = e.) Verify that T1l (x), 11 = 0, I, 2. 3, satisfy the Chebyshev equation (1 - x 2 )y" - xy' + n2 y = O. (b) Orthogonality on an infinite interval: Laguerre polynomials l l are defined by Lo = I, and 0, 13. y" + Ay = 0, yeO) =0, y(\)+y'(1)=O 14. (xY')' + Ax-Iy = 0, yO) = 0, y'(e) = O. (Set x = e t .) 15. (x-1y')' + (A + L)x- 3y = 0, y(l) = O. y(e"") = = 20. TEAM PROJECT. Special Functions. Orthogonal polynomials playa great role in applications. For this reason, Legendre polynomials and various other orthogonal polynomials have been studied extensively; see Refs. [GR1], [GRIO] in App. 1. Consider some of the most important ones as follows. y (I) = 0 16. y" - 2/ yO) = 0 17. y" + 8y' y( rr) = 0 + (k 2 + 2x- 2 )y (Use a CAS or set y = xu.) (b - k)!c. 4. (Change of x) Using Prob. 3, derive the orthogonality of I, cos wx, sin wx, cos 2wx. sin 2wx. ... on -1 ~ x ~ I (r(x) = 1) from that of 1, cos x, sin x, cos 2x, sin 2x, ... on -w ~ x ~ rr. 5. (Legendre polynomials) Show that the functions P,,{cos 6), n = 0, I, ... , form an orthogonal set on the interval 0 ~ e ~ rr with respect to the weight function sin e. 6. (Tranformation to Sturm-Liom iIIe form) Show that Y" + fy' + (g + Ah»)' = 0 takes the form (I) if you set p = exp (If dx), q = pg, r = hp. Why would you do such a transformation? 17-191 19. y" - 2x- 1 y' n = 1,2,'" Show that yeO) = 0, yeO) 18. xY" + 2y' + Axy = 0, yew) = 0, (Use a CAS or set y = x-1u.) = 0, n2rr) = O. Prove that the Laguerre polynomials are orthogonal on the positive axis 0 ~ x < :x; with respect to the weight function /"p:) = e- x . Hint. Since the highest power in Lm is x"', it suffices to show that f e-xxkLn dx = 0 for k < n. Do this by k integrations by parts. IOpAFNUTI CHEBYSHEV (1821-1894), Russian mathematician. is known for his work in approximation theory and the theory of numbers. Another transliteration of the name is TCHEBICHEF. llEDMOND LAGUERRE (1834-1886). French mathematician. who did research work in geometry and in the theory of infinite series. 210 CHAP. 5 Series Solutions of ODEs. Special Functions 5.8 Orthogonal Eigenfunction Expansions Orthogonal functions (obtained from Sturm-Liouville problems or otherwise) yield important series developments of given functions. as we shall see. This includes the famous FOllrier series (to which we devote Chaps. 11 and 12), the daily bread of the physicist and engineer for solving problems in heat conduction. mechanical and electrical vibrations, etc. Indeed, orthogonality is one of the most useful ideas ever introduced in applied mathematics. Standard Notation for Orthogonality and Orthonormality The integral (4) in Sec. 5.7 defining orthogonality is denoted by (Ym. Yn). This is standard. Also. Kronecker's deJta 12 omn is defined by omn = 0 if 111 *- II and omn = 1 if 111 = 11 (thus on" = I). Hence for orthonormal functions Yo, ."1' Y2' ... with respect to weight rex) (> 0) on [I ;:::; x ;:::; b we can now simply write (Ym' Yn) = om11,' written out if /11 *- if 11l = n. 11 (1) Also. for the norm we can now write lIyll = \I(y",. Ym) (2) = f b 2 r(x)Ym (x) d.r. a Write down a few examples of your own, to get used to this practical ~hort notation. Orthogonal Series Now comes the instant that shows why orthogonality is a fundamental concept. Let Yo, .1'1' .1'2• .•. be an orthogonal set with respect to weight r(x) on an interval [I ;:::; x;:::; b. Let J(x) be a function that can be represented by a convergent series x J(x) = ~ [lmY",(x) = [loYo(x) (3) + [lIYl(X) + m~O This is called an orthogonal expansion or generalized Fourier series. If the Ym. are eigenfunctions of a Sturm-Liouville problem. we call (3) an eigenfunction expansion. In (3) we use again 111 for summation since 11 will be used as a fixed order of Bessel functions. Given J(x). we have to determine the coefficients in (3), called the Fourier constants of J(x) with re:,pect to Yo, )'1, . . . . Because of the orthogonality this is simple. All we have to do is to multiply both sides of (3) by r(x)y,,(x) (nfixed) and then integrate on both sides from a to b. We assume that term-by-term integration is permissible. (This is justified, for instance, in the case of "uniform convergence," as is shown in Sec. 15.5.) Then we obtain (J, )'n) = f a b rJ)'n dr = f r (x~ b a m=O [I",Ym ) .1'" dx = ~ ex: [lm(Y",., Yn)' m~O 12LEOPOLD KRONECKER (1823-1891 l. German mathematician at Berlin University. who made important to algebra. group theory. and number theory. contribution~ SEC. 5.8 211 Orthogonal Eigenfunction Expansions Because of the orthogonality all the integrals on the right are zero. except when Hence the whole infinite series reduces to the single term 111 n. Assuming that all the functions Yn have nonzero norm, we can divide by II.vn 112; writing again 111 for n, to be in agreement with (3), we get the desired formula for the Fourier constants (f, (4) EXAMPLE 1 1 YIll) Ily",11 2 f b (m r(x)f(x)Ym(x) dx 0, 1, .. '). a Fourier Series A mo,t important c\as, of eigenfunction expansions is obtained from the periodic Sturm-Liouville problem y" + Ay = 0, y'('iT) A general solution of the ODE is y = A cos kx + B sin kx, where k into the boundary conditions, we obtain A cos k'iT -kA sin k7T + + B ,in k7T = A cos (-k'iT) + kB cos k'iT = -kA sin (-k'iT) = y'(-'iT). = VA. Substituting y and its derivative B sin (-k7T) + kB cos (-k7T). Since cos l-a) = cos a, the cosine terms cancel, so that these equations give no conditlOn for these terms. Since sin (-a) = -sin a, thc equations gives the condition sin k7T = 0, hence k'iT = 1II'iT, k = 11/ = 0, 1,2, ... , so that the eigenfunctions are cos 0 = 1, sin x, cos X, sin 2x, .. " cos 2x, cos IIIX, sinlllx, ... corresponding pairwise to the eigenvalues A = k 2 = 0, 1,4, ... , m 2 , . . . . lsin 0 = 0 is not an eigenfunction.) By Theorem I in Sec. 5.7, any two of these belonging to different eigenvalues are orthogonal on the interval -7T ~ X ~ 7T (note that rex) = 1 for the present ODE). The orthogonality of cos I1lX and sin 111X for the same 111 follows by integration. III I I, For the 1l0mlS we get = \1'2;, and v:;;: for all the others, as you may verify by integrating cos2 'y, 2 sin x. etc .. from -'iTto 'iT. This gives the series (with a slight extension of notation sincc we have two functions for each eigenvalue I, 4, 9, ... ) (5) f(x) = ao + L (am cos IIIX + b m sin fI1x). 1#l.=1 According to (4) the coefficients (with 111 = 1,2, ... ) are b1ll. = (6) ..!.. 7T f'" J(x) sin 111X £lx. -71 The series (5) is called the Fourier series of f(x). Its coefficients are called the Fourier coefficients of f(x), as given by the so-called Euler formulas l6) lnot to be confused with the Euler formula (11) in Sec. 2.2). For instance, for the "periodic rectangular wave" in Fig. III, given by f(x) = { -I 1 if -7T<X<O if O<x<'iT and f(x + 27T) = ./(x), 212 CHAP. 5 Series Solutions of ODEs. Special Functions ° we get from (6) the values ao = and [I~,,-(-I)COSIIIXdX+ L"-I'COS11lXdT] =0, 7r bm = [I~,,-(-I)SinIllXdT+ Io"-l,sinIllXdT] 7r 4f( mil) 7Tln ifm ° [I - 2 cos 1117r + I] = { = 1.3.···. if m = 2,4···. Hence the Fourier senes of the periodic rectangular wave is f (x) = ~ 7r ~ ~ 1C 0 1C x 21C -~-l ~ Fig. 111. • (Sin x + 3 sin 3x + 5 sin 5, - + ... ) . Periodic rectangular wave in Example 1 Fourier series are by far the most important eigenfunction expansions. so important to the engineer that we shall devote two chapters (11 and 12) to them and their applications, and discuss numerous examples. Did it surprise you that a series of continuous functions (sine functions) can represent a discontinuous function? More on this in Chap. 11. E X AMP L E 2 Fourier-Legendre Series A Fourier-Legendre series is an eigenfunction expansion .f(x) = L a'mP'm(x) = aoPo + a1 P1(x) + (l2 P 2(x) + - - - = ao + a1x + a2(ix2 - i) -'- - m=O in terms of Legendre polynomials (Sec_ 5.3). The latter are the eigenfunctions of the Sturm-Liouville problem in Example 5 of Sec. 5.7 on the interval -I ~ x ~ I. We have rex) = I for Legendre'S equation, and (4) gives 2111+ I 2 (7) becau~e (8) am = - - - I 1 f(x)Pm(x) d>:. 111 = 0, I,'" -1 the norm is I 1 -1 Pm(x)2 dx = 1_2_ + 1 V (Ill = 0, I, ... ) 2111 as we state without proof (The proof is tricky; it uses Rodrigues's formula in Problem Set 5 3 and a reduction of the resulting integral to a quotient of gamma functions_) SEC. 5.8 213 Orthogonal Eigenfunction Expansions For instance, let j(x) = sin 7TX. Then we obtain the coefficients 1 1 2nz+II (sin 7TX) P'In(x) dx, a'ln = --2-- 3 2 a1 = thus -1 J x sin TTX tIT: = 3 = 0.95493, -1 etc. 7T Hence the Fourier-Legendre series of sin TTX is sin TTX = 0.95493P 1 (x) - 1.15824P3(x) + 0.21429P 5 (x) - 0.OJ664P7 (x) + 0.00068P9 (x) - 0.OO002P l l(x) + .... The coefficient of P 13 is about 3 . 10- 7 . The sum of the first three nonzero terms gives a curve that practically coincides with the sine curve. Can you see why the even-numbered coefficients are 7ero? Why a3 is the absolutely biggest coefficient? • E X AMP L E 3 Fourier-Bessel Series In Example 6 of Sec. 5.7 we obtained infinitely many orthogonal sets of Bessel functions, one for each of Jo, ft, J2 , • • • . Each set is orthogonal on an interval 0 ~ x ~ R with a fixed positive R of our choice and with respect to the weight x. The orthogonal set for I n is In(kn,lX), In(kn,2X), In(kn,3x), ... , where n is fixed and kn,'In is given in (12), Sec. 5.7. The corresponding Fourier-Bessel series is (9) f(x) = L a",Jn(kn,'ln x ) = a1Jn(kn ,1 x) + a2Jn(kn,2X) + a3Jn(kll,3X) + (n fixed). 1n=1 The coefficients are (with O'n,m = kn,mR) a'In = (10) m = 1,2," because the square of the norm is (11) as we state without proof (which is tricky; see the discussion beginning on p. 576 of [A13]). For instance, let us consider f(x) = I - x 2 and take R = I and n = 0 in the series (9), simply writing A for 0'0,'In' Then kn,m = O'O,m = A = 2.405, 5.520, 8.654, 11.792, etc. luse a CAS or Table Al in App. 5). Next we calculate the coefficients ~ by (10), am = 2 J 1 (A) -2-- II 2 x(l - x )Jo(Ax) dx. 0 This can be integrated by a CAS or by formulas as follows. First use [xftlAx)], = AxJolAx) from Theorem 3 in Sec. 5.5 and then integration by parts, am = 2 -2J 1 (Al I 1 2 x(1 - x )Jo(Ax) dx 2 = -2- [ ft (A) 0 I - A 2 (l - x )xft(Ax) 11 0 The integral-free part is zero. The remaining integral can be evaluated by [x2lz(Ax) 3 in Sec. 5.5. This gives I A l' = I 1 xft(Ax)(-2x) d, ] 0 Ax2ft (Ax) from Theorem a'In = (A = 0'0,,,,)' Numenc values can be obtained from a CAS (or from the table on p. 409 of Ref. [GRI] in App. I, together with the formula J2 = 2,-Ift - Jo in Theorem 3 of Sec. 5.5). This gives the eigenfunction expansion of I - x 2 in terms of Bessel functions J0, that is, I - x 2 = 1.1081Jo(2.405x) - 0.1398Jo(5.520x) 2 + 0.0455Jo(8.654x) 0.02IOJo(11.792,) + .... A graph would show that the curve of I - x and that of the sum of the first three terms practically coincide. • 214 CHAP. 5 Series Solutions of ODEs. Special Functions Mean Square Convergence. Completeness of Orthonormal Sets The remaining part of this section will give an introduction to a convergence suitable in connection with orthogonal series and quite different from the convergence used in calculus for Taylor series. In practice, one uses only orthonormal sets that consist of "sufficiently many" functions, so that one can represent large classes of functions by a generalized FOUlier series (3)certainly all continuous functions on an interval a ~ x ~ b, but also functions that do "not have too many" discontinuities (see Example 1). Such orthonormal sets are called "complete" (in the set of functions considered; definition below). For instance, the orthonormal sets corresponding to Examples 1-3 are complete in the set of functions continuous on the intervals considered (or even in more general sets of functions; see Ref. [OR7], Secs. 3.4-3.7, listed in App. 1. where "complete sets" bear the more modem name "total sets"). In this connection, convergence is convergence in the norm, also called mean-square convergence; that is. a sequence of functions fk is called convergent tvith the limit f if (12*) lim k~x Ilfk - fll = 0; written out by (2) (where we can drop the square root, as this does not affect the limit) lim (12) I b r(x)[fk(X) - f(X)]2 dx k_x a Accordingly, the series (3) converges and represents (13) I lim = O. f if b k_x a r(x)[Sk(X) - f(X)]2 dx = 0 where Sk is the hh partial sum of (3), k (14) sk(x) = L (lmYm(x). 1n=0 By definition, an 0l1honormal set )'0' YI' . . . on an interval {l ~ x ~ b is complete ill set of fimctiolls S defined on (/ ~ x ~ b if we can approximate every f belonging to S arbitrarily closely by a linear combination ao)'o + al)'1 + ... + akYk, that is, technically, if for every E > 0 we can find constants (/0, . . . , {lk (with k large enough) such that {l (15) An interesting and basic consequence of the integral in (13) is obtained as follows. Performing the square and using (14), we first have I b r(x)[Sk(X) - f(X)]2 dr = a I b a = I b a I J2 b 2 rS k d-r - [ I' k L 1n.=O 2 amY", rfSk d-r + a I b rf2 dr a k dx - 2 L Ul,=o ! am I b a rfYm dx + I b rp dx. a The, f~rst in~egral on the right equals L a",2 ?ecause rYmYz dr = 0 for III =1= I, and Ir)m dx - 1. In the second sum on the fIght, the mtegral equals am' by (4) with SEC. 5.B Orthogonal Eigenfunction Expansions 215 II Ym II 2 = 1. Hence the first term on the right cancels half of the second term, so that the right side reduces to k - 'L am 2 + I b rp dx. a 1lZ.=O This is nonnegative because in the previous formula the integrand on the left is nonnegative (recall that the \Veight r(x) is positive!) and so is the integral on the left. This proves the important Bessel's inequality 11/112 = (16) I b (k r(x)/(x)2 dx = 1,2, .. '). a Here we can let k ---7 YJ, because the left sides foml a monotone increasing sequence that is bounded by the right side, so that we have convergence by the familiar Theorem 1 in App. A3.3. Hence (17) m=O Furthermore, if )'0, )'1 • . . . is complete in a set of function~ S. then (13) holds for every By (15) this implies equality in (16) with k ---7 :>C. Hence in the case of completeness every 1 in S satisfies the so-called Parseval's equality 1 belonging to S. II 1 112 = (18) I b r(x)/(x)2 dr. a As a consequence of (18) we prove that in the case of complete1less there is no function orthogonal to every function of the orthonormal set. with the trivial exception of a function of zero norm: THEOREM 1 Completeness Let Yo, )'1 •... be a complete ort/101101711al set on a ~ x ~ b in (I set offunctions S. Then if a .function 1 belongs to S alld is orthogonal to every )'m, it must have norm zero. In particular, if 1 is continuous, then 1 must be identical!.v zero. PROOF E X AMP L E 4 Since 1 is orthogonal to every )'m' the left side of (18) must be zero. If 1 is continuous, then II 1 II = 0 implies I(x) == 0, as can be seen directly from (2) with 1 instead of )'m because r(x) > O. • Fourier Series The orthonormal set in Example I is complete in the set of continuous functions on -7T~ x that fIx) == 0 is the only continuous function orthogonal to all the functions of that set. Solutioll. Lef f be any continuou~ function. By the orthogonality (we can unlit J" ~ 7T. Vh and \ J(x) sin IIIX d," Verify directly S), ~ O. -r. Hence am ~ 0 and b m ~ 0 in (6) for all III. su thaI (3) reduces to J(X) == O. • CHAP. 5 216 Series Solutions of ODEs. Special Functions This is the end of Chap. 5 on the power series method and the Frobenius method, which are indispensable in solving linear ODEs with variable coefficients, some of the most important of which we have discussed and solved. We have also seen that the latter are important sources of special functions having orthogonality properties that make them suitable for orthogonal series representations of given functions. 11-41 - : FOURIER-LEGENDRE SERIES Showing the details of your calculations, develop: 1. 7x 4 - 6x 2 2. (x + 1)2 5. Prove that if f(x) in Example 2 is even [that is, f(x) = f( - x)], its series contains only P m(x) with even 111. 16-161 CAS EXPERIMENTS. FOURIER-LEGENDRE SERIES Find and graph (on common axes) the partial sums up to that S"'o whose graph practically coincides with that of/ex) within graphical accuracy. State what 1Il0 is. On what does the size of IIlO seem to depend? 6. f(x) sin TTX 7. f(x) sin 27TX 8. f(x) cos TTX 9. f(x) cos 27TX 10. f(x) cos 37TX n. f(x) eX on the speed of convergence by observing the decrease of the coefficients. {c) Take f(x) = 1 in (19) and evaluate the integrals for the coefficients analytically by (24a), Sec. 5.5, with JJ = I. Graph the first few partial sums on common axes. 18. TEAM PROJECT. Orthogonality on the Entire Real Axis. Hermite Polynomials. 13 These orthogonal polynomials are defined by HeoO) = 1 and REMARK. As is true for many special functions, the literature contains more than one notation, and one sometimes defines as Hermite polynomials the functions H *(x) x2 12. f(x) e13. f(x) = 1/(1 + x 2) 14. f(x) = 10(aO,1x). where aO,1 is the first positive zero of 10 15. f(x) = 10(aO,2x), where aO,2 is the second positive zero of 10 16. f(x) = 11ta1,1x), where a1,1 is the first positive zero of 11 17. CAS EXPERIMENT. Fourier-Bessel Series. Use Example 3 and again take n = 10 and R = 1. so that you get the series (19) flx) = al10lao,lX) + a210(aO,2x) + a310lao,3x) + ... with the zeros 0:0,1 0:0,2' Table AI in App. 5). .•. from your CAS (see also (a) Graph the terms 10(aO,lx), ... , 10(aO,lOx) for ~ l on common axes. o~ x (b) Write a program for calculating partial sums of (9). Find out for what f(x) your CAS can evaluate the integrals. Take two such f(x) and comment empirically n 2 d ne _X2 = (-l)ne" - - dxn This differs from our definition, which is preferred in applications. (a) Small Values of n. Show that He1(X) = x, He3(X) = x 3 - 3x, He2(X) = x 2 - 1, He4(X) = X4 - 6x 2 + 3. (b) Generating Function. A generating function of the Hermite polynomials is -n (20) etx-t2/2 = L anlx)f n n=O because He.,(x) = n!anlx), Prove this. Hint: Use the formula for the coefficients of a Maclaurin series and note that tx - ~f2 = ~X2 - ~(x - t)2. (C) Derivative. Differentiating the generating function with respect to x, show that (21) 13CHARLES HERMITE (1822-1901), French mathematician, is known for his work in algebm and number theory. The great HENRI POINCARE 11854-1912) was one of his students, 217 Chapter 5 Review Questions and Problems (d) Orthogonality on the x-Axis needs a weight function that goes to zero sufficiently fast as x ---? ::,:::cc. (Why?) Show that the Hermite polynomials are orthogonal on -<Xl < X < cc with respect to the weight function rex) = e- x2/2 . Hint. Use integration by parts and (21). (e) ODEs. Show that (22) He~(x) = xHen(x) - Hen+l(x). Using this with n - 1 instead of nand (21), show that y = Hen(x) satisfies the ODE (23) y" - xy' Show that w equation14 = (24) w" + O. = e- X2/4y is a solution of Weber's +~- (n + ny !x 2 )w = 0 (n=O,l,···). 19. WRITING PROJECT. Orthogonality. Write a short report (2-3 pages) about the most important ideas and facts related to orthogonality and orthogonal series and their applications. TIONS AND PROBLEMS 1. What is a power series? Can it contain negative or fractional powers? How would you test for convergence? 18. (x 2 - 2. Why could we use the power series method for Legendre's equation but needed the Frobenius method for Bessel's equation? 3. Why did we introduce twO kinds of Bessel functions, J and Y? 4. What is the hypergeometric equation and why did Gauss introduce it? 5. List the three cases of the Frobenius method, giving examples of your own. 6. What is the difference between an initial value problem and a boundary value problem? 7. What does orthogonality of functions mean and how is it used in series expansions? Give examples. 8. What is the Sturm-Liouville theory and its practical importance? 9. What do you remember about the orthogonality of the Legendre polynomials? Of Bessel functions? 10. What is completeness of orthogonal sets? Why is it important? 20. x 2y" + ~1-251 BESSEL'S EQUATION [I ~-~ SERIES SOLUTIONS Find a basis of solutions. Try to identify the series as expansions of known functions. (Show the details of your work.) 11. y" - 9y = 0 12. (1 - X)2y" + (1 - x)y' - 3y = 0 13. xy" - (x + l)y' + y = 0 14. x 2 y" - 3xy' + 4y = 0 15. y" + 4xy' + (4x 2 + 2)y = 0 16. x 2 y" - 4xy' + (x 2 + 6)y = 0 17. xy" + (2x + I)Y' + (x + l)y 19. (x 2 0 l)y" + - + xy' (4x 4 0 0 = = 1)y = 0 - Find a general solution in terms of Bessel the indicated transformations and show the 21. x 2y" + xy' ., (36x 2 - 2))" = 0 22. x 2y" + 5xy' + (x 2 - 12)y = 0 23. x 2y" + xy' + 4(x 4 - l)y = 0 functions. (Use details.) (6x = z) (y = u/x 2) (x 2 = z) 24. 4x 2y" - 20xy' + (4x 2 + 35)y = 0 25. y" + k 2 x 2y = 0 (y = uVx, ~kX2 y' (0) = z) 30. y" 0, + Ay 131-~ = 0, y(l) + xy' + (Ax 2 - J)y = 0 y' (1) = 28. (xy')' + Ax- 1 y (Set x = e t .) 29. x 2y" yeO) x 3 u) (y = 126-301 BOUNDARY VALUE PROBLEMS Find the eigenvalues and eigenfunctions. 26. y" + Ay = 0, yeO) = 0, y' (7T) 27. y" + Ay = 0, yeO) = y(I). = y(1) 0, = = 0, = y(e) = O. 0, 0 yeO) + y'(O) = 0, y(27T) = 0 CAS PROBLEMS Write a program, develop in a Fourier-Legendre series, and graph the first five partial sums on common axes, together with the given function. Comment on accuracy. 31. e 2x ( - 1 ~ x ~ I) 32. sin ( 7TX 2) ( - 1 ~ x ~ I) 33. 11(1 = + 2y 4xy' + 2y l)y" - 2xy' + Ixl) (-1 ~ x ~ 1) 34. Icos 7Txl (-1 ~ x ~ 1) 35. x if 0 ~ x ~ 1,0 if -1 14HEINRICH WEBER (1842-1913), German mathematician. ~ x < 0 218 CHAP. 5 Series Solutions of ODEs. Special Functions Series Solution of ODEs. Special Functions The power series method gives solutions of linear ODEs y" (1) + p(.'l)y' + q(.'l)Y = 0 with variable coefficients p and q in the form of a power series (with any center e.g., .'lo = 0) .'lo, (2) Y(.'l) = L am(.'l - .'lo)m = (10 + (l1(.'l - .'lo) + (l2(X - XO)2 + .... In=O Such a solution is obtained by substituting (2) and its derivatives into (I). This gives a recurrence formula for the coefficients. You may program this formula (or even obtain and graph the whole solution) on your CAS. If p and q are analytic at .'lo (that is. representable by a power series in powers of x - .'lo with positive radius of convergence: Sec. 5.2). then (I) has solutions of this form (2). The same holds if h. p. q in h(x»)"" + p(.'l)y' + q(.'l)Y = 0 are analytic at .'lo and h(.'lo) oF O. so that we can divide by h and obtain the standard form 0). Legendre's equation is solved by the power series method in Sec. 5.3. The Frobenius method (Sec. 5.4) extends the power selies method to ODEs y" (3) (I(X), + - - - \" + X - .'lo . b(x) (x - \" xO)2 . = 0 whose coefficients are singular (i.e., not analytic) at xo. but are "not too bad," namely, such that a and b are analytic at Xo. Then (3) has at least one solution of the form (4) y(x) = (x - xor L 111, (I",(x - .'loY'" = (lo(x - xor + (/l(X - XO)'"+I + ... 0 where r can be any real (or even complex) number and is determined by substituting (4) into (3) from the indicial equation (Sec. 5.4), along with the coefficients of (4). A second linearly independent solution of (3) may be of a similar form (with different rand (l11/S) or may involve a logarithmic term. Bessel's equation is solved by the Frobenius method in Secs. 5.5 and 5.6. "Special functions" is a common name for higher functions. as opposed to the usual functions of calculus. Most of them arise either as nonelementary integrals fsee (24)-(44) in App. 3.11 or as solutions of (1) or (3). They get a name and notation and are included in the usual CASs if they are important in application or in theory. Summary of Chapter 5 219 Of this kind, and particularly useful to the engineer and physicist, are Legendre's equation and polynomials Po, PH ... (Sec. 5.3), Gauss's hypergeometric equation and functions F(a, b, c; x) (Sec. 5.4), and Bessel's equation and functions J v and Y v (Secs. 5.5, 5.6). Modeling involving ODEs usually leads to initial value problems (Chaps. 1-3) or boundary value problems. Many of the latter can be written in the form of Sturm-Liouville problems (Sec. 5.7). These are eigenvalue problems involving a parameter A that is often related to frequencies, energies, or other physical quantities. Solutions of Sturm-Liouville problems, called eigenfunctions, have many general properties in common, notably the highly important orthogonality (Sec. 5.7), which is useful in eigenfunction expansions (Sec. 5.8) in terms of cosine and sine (··Fourier series", the topic of Chap. 11), Legendre polynomials, Bessel functions (Sec. 5.8), and other eigenfunctions. •••• CHAPTER 6 Laplace Transforms The Laplace transform method is a powerful method for solving linear ODEs and corresponding initial value problems, as well as system~ of ODEs arising in engineering. The process of solution consists of three steps (see Fig. 112). Step 1. The given ODE is transformed into an algebraic equation ("subsidiary equation"). Step 2. The subsidiary equation is solved by purely algebraic manipulations. Step 3. The solution in Step 2 is transformed back, resulting in the solution of the given problem. IVP IInitial Value f---~ Problem Fig. 112. 1--- Solution of the ® IVI" [ I ~ Solving an IVP by Laplace transforms Thus solving an ODE is reduced to an algebraic problem (plus tho~e transformations). This switching from calculus to algebra is called operational calculus. The Laplace transform method is the most important operational method to the engineer. This method has two main advantages over the usual methods of Chaps. 1-4: A. Problems are solved more directly, initial value problems without first determining a general solution. and nonhomogeneous ODEs without first solving the corresponding homogeneous ODE. B. More importantly, [he use of the unit step function (Heaviside function in Sec. 6.3) and Dirac's delta (in Sec. 6.4) make the method particularly powerful for problems with inputs (driving forces) that have discontinuities or represent short impulses or complicated periodic functions. In this chapter we consider the Laplace transform and its application to engineering problems involving ODEs. PDEs will be solved by the Laplace transform in Sec. 12.11. General formulas are listed in Sec. 6.8, transforms and inverses in Sec. 6.9. The usual CASs can handle most Laplace transforms. Prerequisite: Chap. 2 Sections that lIlay be omitted in a shorter course: 6.5, 6.7 References and Answers to Problems: App. 1 Part A, App. 2. 220 SEC. 6.1 6.1 Laplace Transform. Inverse Transform. Linearity. s-Shifting 221 Laplace Transform. Inverse Transform. Linearity. s-Shifting If f(t) is a function defined for all t ~ 0, its Laplace transform l is the integral of f(t) times e- st from t = 0 to x. It is a function of s, say, F(s), and is denoted by ;£(f); thus F(s) = 9::(f) = fCe-stf(t) dt. (1) o Here we must assume that f(t) is such that the integral exists (that is, has some finite value). This assumption is usually satisfied in applications-we shall discuss this near the end of the section. Not only is the result F(s) called the Laplace transform, but the operation just described, which yields F(s) from a given f(t), is also called the Laplace transform. It is an "integral transform" F(s) = fC k(s, t)f(t) dt o with '·kernel" k(s, t) = e- st . Furthermore, the given function f(t) in (1) is called the inverse transform of F(s) and is denoted by 9::- I (F); that is, we shall write f(t) (1*) = 9::- l (F). Note that (1) and (1 *) together imply 9::-\9::(f) = f and 9::(9::- l (F» = F. Notation Original functions depend on t and their transforms on s-keep this in mind! Original functions are denoted by lowercase letters and their transforms by the same letters in capital, so that F(s) denotes the transform of f(t), and Y(s) denotes the transform of y(t), and so on. E X AMP L E 1 Laplace Transform Let l(t) = 1 when t Solution. ~ O. Find F(s). From (1) we obtain by integration 5£(f) = 5£(1) = LOG e- st dt = o _ ~s e- st I"" 0 s (s> 0). IPIERRE SIMON MARQUIS DE LAPLACE (1749-1827), great French mathematician, was a professor in Paris. He developed the foundation of potential theory and made important contributions to celestial mechanics, astronomy in general, special functions. and probability theory. Napoleon Bonaparte was his student for a year. For Laplace's interesting political involvements. see Ref. [GR2], listed in App. I. The powerful practical Laplace transform techniques were developed over a century later by the English electrical engineer OLIVER HEAVISIDE (1850-1925) and were often called "Heaviside calculus." We shall drop variables when this simplifies formulas without causing confusion. For instance, in (1) we wrote 5£(f) instead of 5£(f)(s) and in (I *) 5£-l(F) instead of 5£-\F)(t). CHAP. 6 222 Laplace Transforms Our notation is convenient, but we should say a word about it. The interval of integration in (1) is infinite. Such an integral is called an improper integral and, by definition, is evaluated according to the rule oc e -stfU) dt = Jim ( 1~cx:; Jo f T e -Si.f(t) dt. 0 Hence our convenient notation means (s> 0) . • We shall use thi, notation throughout this chapter. E X AMP L E 2 Laplace Transform .:£(eat ) of the Exponential Function eat Let f(t) = eat when t :0;; 0, where a is a constant. Find :.f(f). Solution. Again by (1), hence, when s - a > 0, • Must we go on in this fashion and obtain the transform of one function after another directly from the definition? The answer is no. And the reason is that new transforms can be found from known ones by the use of the many general properties of the Laplace transform. Above all, the Laplace transform is a "linear operation," just as differentiation and integration. By this we mean the following. THEOREM 1 Linearity of the Laplace Transform The Laplace transform is a linear operation; that is, for anyfunctions f(t) a17d g(t) whose transjol7lls exist and any constants a and b the tran,~for11l of afft) + bg(t) exists, and .:£{af(t) PROOF a.:£{f(t)} + b.:£{g(t)}. By the definition in (1), .:£{af(t) + bg(t)} = L=e-st[af(t) o =a E X AMP L E 3 + bgU)} = + bg(t)l dt f'e-stf(t) dt o + b ICCe-stg(t) dt = a.:£{f(t)} + b.:£{gU)}. • 0 Application of Theorem 1: Hyperbolic Functions Find the transforms of cosh at and sinh aI. Solution. Since cosh at ~(eat = + e -at) and sinh at = ~(eat - e -at), we obtain from Example 2 and Theorem 1 1 . =1 .'£(coshat) = -2 (:£(e at ) !f(smhat) + -(!f(e a t ) - !f(e-at» 2 I( I + -+1) - = -1( - - _ _1) _ + 5£'(e- at» = - 2 -- s - a s a J 2 s - a s a - s -2 s2 - a a_ = __ .1'2 - a 2 • • SEC. 6.1 Laplace Transform. Inverse Transform. Linearity. s-Shifting EXAMPLE 4 223 Cosine and Sine Derive the formulas .:f(cos wt) s = s2 w 9'(sin wt) = + w2 ' 2 S + 2' w Solution by Calculus. We wlite Lc = !£(cos wt) and Ls = .'t:(sin wt). Integrating by pmts and noting that the integral-free parts give no contribution from the upper limit 00. we obtain e~:t coswtl~- Lc= lCOe-stcoswtdt= Ls = fooo e -st sin wt dt = -; lCOe-stsin wtdt = ~ e -st sin wtl = + -s 0 s (= e -st cos wt dt = Jo s ~ Lc' s By substituting Ls into the formula for Lc on the light and then by substituting Lc into the formula for Ls on the right. we obtain L = C L s = ..!..s - ~s(!.'!..L) se' ~S (..!..S - ~L) s S s w2) 1 Lc ( 1+2 = - , s s w ' Solution by Transforms Using Derivatives. S 2 Ls ' See next section. In Example 2, if we set a = iw with i = Soilltion by Complex Methods. w = \1=1, we obtain s + iw s w . t I s + iw :i(e'W ) = - - - = - - - - - - = - - - = - - - + i - - s2 + w2 s2 + w 2 ,,2 + w2 < - iw (s - iw)(s + iw) Now by Theorem I and e iwt = . cos wt + i sin wt [see (11) in Sec. 2.2 with wt instead of t] we have 5£(eiwt ) = 5£(cos wt + i sin wt) = ':ECcos wt) t- i.'£(sin wt). If we equate the real and imaginary parts of this and the previous equation, the result follows. (This formal calculation can be justified in the theory of complex integration.) • Basic transforms are listed in Table 6.1. We shall see that from these almost all the others can be obtained by the use of the general propelties of the Laplace transform. Formulas 1-3 are special cases of formula 4, which is proved by induction. Indeed, it is true for n = 0 because of Example 1 and O! = 1. We make the induction hypothesis that it holds for any integer n ~ 0 and then get it for /1 + 1 directly from (1). Indeed, integration by parts first gives Now the integral-free part is zero and the lasl part is (n and the induction hypothesis, n + 1 s This proves formula 4. n + s 1 n! + 1)/s times :£(tn). From this (n + I)! 224 CHAP. 6 Table 6.1 I Laplace Transforms Some Functions f(t) and Their Laplace Transforms ~(f) f(t) ~(f) 1 1 lis 7 cos wt 2 t lIs2 8 sin wt 3 t2 2 !Is 3 9 cosh ar s S2 - a 2 4 tn (n = 0, 1, ... ) sn+l 10 sinh at a S2 - a 2 5 ta (a positive) sa+l 11 eat cos wt s-a (s - a)2 + uJ 6 eat L -s-a 12 eat sin wt w (s - a)2 ~(f) J(t) n! rCa + 1) I s S2 + w2 W S2 + uJ + uJ I f(a + I) in formula 5 is the so-called gamma function [(15) in Sec. 5.5 or (24) in App. A3.1]. We get formula 5 from (1), setting st = x: where s > O. The last integral is precisely that defining f(a + 1), so we have f(a + I)/sa+t, as claimed. (CAUTION! f(a + 1) has x a in the integral, not x a + 1 .) Note the formula 4 also follows from 5 because f(n + I) = n! for integer n ;::::; O. Formulas 6-10 were proved in Examples 2-4. Fonllulas 11 and 12 will follow from 7 and 8 by "shifting," to which we tum next. s-Shifting: Replacing 5 by 5 - a in the Transform The Laplace transform has the very useful property that if we know the transform of f(t), we can immediately get that of eatf(t), as follows. THEOREM 2 First Shifting Theorem, s-Shifting Iff(f) has the transfonn F(s) (where s > kfor some k), thell eatf(t) has the transform F(s - a) (where s - a > k). In fonnuias, or, (f we take the inverse on both sides, SEC. 6.1 225 Laplace Transform. Inverse Transform. Linearity. s-Shifting PROOF We obtain F(s - a) by replacing s with s - a in the integral in (1), so that F(s - a) = {'e-Cs-a)tf(t) dt = ["e-st[eatf(t)] dt = .:£{eatf(t)}. o 0 If F(s) exists (i.e., is finite) for s greater than some k, then our first integral exists for s - a > k. Now take the inverse on both sides of this formula to obtain the second formula • in the theorem. ( ~AUTION! -a in F(s - a) but +a in eatf(t).) E X AMP L E 5 s-Shifting: Damped Vibrations. Completing the Square From Example 4 and the first shifting theorem we immediately obtain formulas II and 12 in Table 6.1, ~{eat s-a cos wt} = ------;;------;;(s a)2 + u} , For instance, use these formulas to find the inverse of the transform 3s - 137 5£(f) Solution. I = s 2 + 2s + 401 . Applying the inverse transform. using its linearity (Prob. 28). and completing the square. we obtain = ~-1{ 3(s + 1) - 140} = 3~-1{ (s + 1)2 + 400 (s s + I } _ 7:J:- 1 { + 1)2 + 202 - (s + 20 }. 1)2 + 202 We now see that the inverse of the right side is the damped vibration (Fig. 113) I(t) 6 4 = e -t(3 cos 20 t - 7 sin 20 t). • ~ 2 A 0 05 .0 -2 -4 ' -6 Fig. 113. Vibrations in Example 5 Existence and Uniqueness of Laplace Transforms This is not a big practical problem because in most cases we can check the solution of an ODE without too much trouble. Nevertheless we should be aware of some basic facts. A function f(t) has a Laplace transform if it does not grow too fast, say, if for all t ~ 0 and some constants M and k it satisfies the "growth restriction" (2) 226 CHAP. 6 Laplace Transforms (The growth restriction (2) is sometimes called "growth of exponential order," which may be misleading since it hides that the exponent must be kt, not kt 2 or similar.) f(t) need not be continuous, but it should not be too bad. The technical term (generally used in mathematics) is piecewise continuity. f(t) is piecewise continuous on a finite interval a ~ t ~ b where f is defined, if this interval can be divided into finitely many subintervals in each of which f is continuous and has finite limits as t approaches either endpoint of such a subinterval from the interior. This then gives finite jumps as in Fig. 114 as the only possible discontinuities, but this suffices in most applications, and so does the following theorem. a \", b Fig. 114. Example of a piecewise continuous function fIt). (The dots mark the function values at the jumps.) THEOREM 3 Existence Theorem for Laplace Transforms If f(t) is defined alld piecewise colltinuolls on every finite imerval on the semi-alCis t ~ 0 and satisfies (2) for all t ~ 0 and some constants M and k, then the Laplace transform ;t(n exists for all s > k. PROOF Since f(t) is piecewise continuou~, e-stf(t) is integrable over any finite interval on the t-axis. From (2), assuming that s > k (to be needed for the existence of the last of the following integrals), we obtain the proof of the existence of ;t(n from Note that (2) can be readily checked. For instance, cosh t < e t , t n < n!e t (because t n /ll! is a single term of the Maclaurin series), and so on. A function that does not satisfy (2) t2 for any M and k is e (take logarithms to see it). We mention that the conditions in Theorem 3 are sufficient rather than necessary (see Prob. 22). Uniqueness. If the Laplace transform of a given function exists, it is uniquely determined. Conversely, it can be shown that if two functions (both defined on the positive real axis) have the same transform. these functions cannot differ over an interval of positive length, although they may differ at isolated points (see Ref. [AI4] in App. 1). Hence we may say that the inverse of a given transform is essentially unique. In particular, if two cominuolls functions have the same transform, they are completely identical. --. :.:- : 11-20 I LAPLACE TRANSFORMS Find the Lapial:e transforms of the following functions. Show the details of your work. (a. b, k, w, B are constants.) 1. t 2 - 2t 2. (t2 - 3 f 4. sin 2 4r 6. e- t sinh 5t 3. cos 2,Trt 5. e 2t cosh t 7. cos (wt 9. e3a-2bt + B) 8. sin (3t - ~) 10. -8 sin O.2t SEC. 6.2 Transforms of Derivatives and Integrals. ODEs kD _ 11. sin t cos t 13. 12. (t 14. 1~ ll 16. I I I I a b k~ 28. (Inverse transform) Prove that :£-1 is linear. Hint. Use the fact that :£ is linear. 129-401 INVERSE LAPLACE TRANSFORMS Given F(s) = :£(f), find f(t). Show the details. (L, n, k, a, 17 are constants.) 29. 31. 18. 4.1 .1 b 2 170 b 1)3 klI=L b 15. + 227 k~ .1 2 4 31T + ~ - 3.1 .1 33. 2 30. + 12 5 n1TL L 2 s2 + 32. 34. n2~ b b 19. '~ -1 I I 20. 1~ I I 37. 36. + 4.1 1 v3)(s (.I - 39. 22. (Existence) Show that :£(llVt) = ~. [Use (30) r@ = V; in App. 3.1.J Conclude from this that the conditions in Theorem 3 are sufficient but not necessary for the existence of a Laplace transform. 23. (Change of scale) If :£(f(t» = F(s) and c is any positive constant, show that .'£(f(ct» = F(s/c)/c. (Hint: Use (1).) Use this to obtain :£(cos wt) from :£(cos t). 24. (Nonexistence) Show that e condition of the form (2). t2 does not satisfy a 25. (Nonexistence) Give simple examples of functions (defined for all x ~ 0) that have no Laplace transform. 26. (Table 6.1) Derive formula 6 from formulas 9 and lO. 27. (Table 6.1) Convert Table 6.1 from a table for finding transforms to a table for finding inverse transforms (with obvious changes. e.g .. :£-l(lIs n ) = t n - 1 /(n - I)!. etc.). + 16 - 16 10 2.1 +Yz 20 (.I - (k L + Vs) 1 -----2 .1 5 .I 5 + + 38. 40. + 4) 1)(.1 .I k~l 2 -----' 21. Using :£(f) in Prob. 13, find :£(f1), where fN) = 0 if t ~ 2 and f1(r) = 1 if t > 2. 6.2 .1 2 .1 2 4 8 35. 2.1 + 1)2 + k2 18.1 - 12 2 9.1 - 1 (.I + + b) a)(s APPLICATIONS OF THE FIRST SHIFTING THEOREM (s-SHIFTING) 141-541 In Probs. 41--46 find the transform. In Probs. 47-54 find the 1l1verse transform. Show the details. 41. 3.Ste 2 . 4t 42. - 3t 4 e- O . 5f at 43. 5e- sin wt 44. e- 3t cos 1Tt 45. e-kt(o cos t + 17 sin t) 46. e-t(ao + a1t + ... + o.J n ) 47. 7 48. (.I - 1)3 Vs 49. + Yz)3 2 + 4.1 + 2 + 50. 15 .1 .1 1T)2 1)2 + 4 (.I - 4.1 - 52. 29 1T 53. + .1-6 (.I 51. 1T (.I 10m + 24~ 54. .1 2 - 6.1 2 + 18 2.1 - 56 .1 2 - 4.1 - 12 Transforms of Derivatives and Integrals. ODEs The Laplace transform is a method of solving ODEs and initial value problems. The crucial idea is that operations of calculus on functions are replaced by operations of algebra on transfonns. Roughly, differentiation of f(t) will correspond to multiplication of 5£(f) by s (see Theorems 1 and 2) and integration of f(t) to division of 5£(f) by s. To solve ODEs, we must first consider the Laplace transform of derivatives CHAP.6 228 THE 0 REM 1 Laplace Transforms Laplace Transform of Derivatives The transforms of the first and second derivatives of f(t) satisfy 5£(j') = s5£(f) - f(O) (1) (2) 5£(f") = s25£(f) - sf(O) - ff (0). F0I711ula (1) holds if f(t) is contilluousforall t ~ 0 and satisfies the growth restriction (2) ill Sec. 6.1 and f' (t) is piecewise continuous on every finite imen'al on the semiaxis t ~ O. Similarly, (2) holds if f and f' are continuous for all t ~ 0 and satisfy the growth restriction and f" is piecewise continuous on every finite interval on the semi-ar:is t ~ O. PROOF We prove (l) first under the additional assumption that j' is continuous. Then by the definition and integration by parts, Since f satisfies (2) in Sec. 6.1, the integrated part on the right is zero at the upper limit when s > k, and at the lower limit it contributes - f(O). The last integral is 5£(f). It exists for s > k because of Theorem 3 in Sec. 6.1. Hence .c£(j' ) exists when s > k and (1) holds. If j' is merely piecewise continuous, the proof is similar. In this case the interval of integration of f' must be broken up into parts such that j' is continuous in each such part. The proof of (2) now follows by applying (1) to f" and then substituting (1), that is .c£(f") = s.c£(f') - reO) = s[s.c£(f) - f(O)] = s 2.c£(f) - sf CO) - rCO). • Continuing by substitution as in the proof of (2) and using induction, we obtain the following extension of Theorem 1. THEOREM 2 Laplace Transform of the Derivative f (n) of Any Order Let f, j', . .. , In-ll be continuous for all t ~ 0 and satisfy the growth restriction (2) in Sec. 6.1. Furthermore, let In} be piecewise continuous on every finite interval on the semi-axis t ~ O. Then the transform of In} satisfies E X AMP L E 1 Transform of a Resonance Term (Sec. 2.8) Let f(t) = by (2), ! sin Cd!. Then f{O) = 0, f' (t) = sin Cdr s - 2 - Cd2 f£(f) = s 2 f£(f), f£(f" ) = 2Cd - 2 - s + Cd + Cdr cos Cdr, f' (0) thus = 0, I' = 2Cd cos Cdr - Cd2 r sin Cdr. Hence • SEC. 6.2 229 Transforms of Derivatives and Integrals. ODEs E X AMP L E 2 Formulas 7 and 8 in Table 6.1, Sec. 6.1 This is a third derivation of ;£(cos wt) and ;£(sin wt); cf. Example 4 in Sec. 6.1. Let f(t) = cos wt. Then f(O) = I, f' (0) = 0, f"(t) = _w2 cos wt. From this and (2) we obtain XU") = s2:£(f) - .I .I -w2X(f). = P( cos wt) = - 2 - - 2 .I + w By algebra, Similarly, let g = sin wt. Then g(O) = 0, g' = w cos wf. From this and (I) we obtain X(g') = s'£(g) = w:£(cos w .:r(sin wt) = -:£(cos wt) = Hence wt). .I • w -2--2 . .I + w Laplace Transform of the Integral of a Function Differentiation and integration are inverse operations, and so are multiplication and division. Since differentiation of a function J(t) (roughly) corresponds to multiplication of its transform ::£(f) by s, we expect integration of J(t) to correspond to division of ::£(f) by s: THEOREM 3 Laplace Transform of Integral Let F(s) denote the transfonn of a function J(t) which is piecewise continuous for t ~ 0 and satisfies a growth restriction (2), Sec. 6.1. Then, for s > 0, s > k, and t> 0, (4) PROOF thus Denote the integral in (4) by get). Since J(t) is piecewise continuous, get) is continuous, and (2), Sec. 6.1, gives Ig(t)1 = I I tt ~ (IJ(T)I dT ~ M ( e t (J(T) dT 10 10 kT 10 M M = _(ekt dT - k L):S _ekt - k (k> 0). This shows that get) also satisfies a growth restriction. Also, g' (t) = J(t), except at points at which J(t) is discontinuous. Hence g' (t) is piecewise continuous on each finite interval and, by Theorem 1, since g(O) = 0 (the integral from 0 to 0 is zero) ::£{f(t)} = ::£{g'(t)} = s::£{g(t)} - g(O) = s::£{g(t)}. Division by s and interchange of the left and right sides gives the first formula in (4), • from which the second follows by taking the inverse transform on both sides. E X AMP L E 3 Application of Theorem 3: Formulas 19 and 20 in the Table of Sec. 6.9 I Using Theorem 3. find the inverse of 2 .1(.1 I 2 + w ) and 2 2 .I (.I + 2 W ) Solution. From Table 6.1 in Sec. 6.1 and the integration in (4) (second formula with the sides interchanged) we obtain C :£ _I{ __I_} s 2 + w2 = sin wt w ' ;£ -I{ 2 .1(.1 I 2 + w) } = It 0 sin WO' -- w dO' = I (l - cos wt). w 2 230 CHAP. 6 Laplace Transforms This is formula 19 in Sec. 6.9. lmegraring this result again and using (4) as before, we obtain formula 20 in Sec. 6.9: ;;e-l{ 2 21+ 2} = ~ ft(1 - cos wr) dr = [-;. s (s w ) W O W JI sin :r sin wt w 0 W w3 2 It is typical that results such as these can be found in several ways. In this example. try partial fraction reduction. • Differential Equations, Initial Value Problems We shall now discuss how the Laplace transfonn method solves ODEs and initial value problems. We consider an initial value problem (5) y" + ay' + by = ret), yeO) = Ko, y'(O) = Kl where a and b are constant. Here ret) is the given input (driving force) applied to the mechanical or electrical system and yet) is the output (response to the input) to be obtained. In Laplace's method we do three steps: Step 1. Setting up the subsidiary equation. This is an algebraic equation for the transfonn Y = .;£(y) obtained by transforming (5) by means of (J) and (2), namely, [S2y - sy(O) - ),'(0)] + a[sY - yeO)] + bY = R(s) where R(s) = ;£(r). Collecting the Y-tenns, we have the subsidiary equation (S2 + as + b)Y = (s + a)y(O) + y' (0) + R(s). Step 2. Solution of the subsidiary equation by algebra. We divide by use the so-called transfer function (6) s2 + as + band 1 Q(s) = s2 + as + b (Q is often denoted by H. but we need H much more frequently for other purposes.) This gives the solution (7) Yes) = [(s + a)y(O) + y' (O)]Q(s) + R(s)Q(s). If yeO) = y' (0) = 0, this is simply Y = RQ; hence Q= Y ;£(output) R ;£(input) and this explains the name of Q. Note that Q depends neither on ret) nor on the initial conditions (but only on a and b). Step 3. III version ofY to obtain y = ;£-1(1'). We reduce (7) (usually by partialfractiolls as in calculus) to a sum of tenns whose inverses can be found from the tables (e.g .• in Sec. 6.1 or Sec. 6.9) or by a CAS, so that we obtain the solution yet) = ?l(y) of (5). SEC. 6.2 231 Transforms of Derivatives and Integrals. ODEs E X AMP L E 4 Initial Value Problem: The Basic Laplace Steps Solve Solution. y' (0) y(o) = 1, y" - y = t, I. = Step 1. Prom (2) and Table 6.1 we get the subsidiary equation [with Y = .P(yl] thus Step 2. The transfer function is Q = 1/(.1'2 - 1), and (7) becomes Y = (.I' + + I)Q I .1'2 Q .I' + I .1'2 _ 1 = + .1'2(.1'2 - 1) Simplification and partial fraction expansion gives 1 (1 --- Y=--+ s-] .1'2-1 ~). s - Step 3. From this expression for Y and Table 6.1 we obtain the solution _,-1 _ _1{_I} '_I{_1 } y(t) -:£ (Y) - :£ +Y .1'-1 2 .1'-1 - c -1{~} Y 2 .I' _ - . et + smh t _ t. • The diagram in Pig. liS summarizes our approach. s-space t-space Subsidiary equation Given problem y" -y = t (s2 - I)Y = s + 1 + 1Is2 ~ y(O) =1 y'(O) =1 t Solution of subsidiary equation Solution of given problem yet) = e' + sinh t - -E- t Y= _1_ +_I__ ...l s-1 Fig. 115. E X AMP L E 5 s2-1 s2 Laplace transform method Comparison with the Usual Method Solve the initial value problem y" Solution. + y' + 9y y(O) = 0.16, = 0, )"(0) = o. From (1) and (2) we see that the subsidiary equation is s2y - 0.16.1' + sY - 0.16 + 9Y = 0, ts 2 thus + .I' + 9)Y = 0.16(.1' + I). The solution is 0.16(.1' + 1) + ~) + 0.08 + ~)2 + ~ 0.16(.1' (.I' Hence by the first shifting theorem and the formulas for cos and sin in Table 6.1 we obtain yet) = Y-\y) = e- tI2 (0.16 cos = e -0.5\0.16 08 f35 t + 1°. V4 2 V35 cos 2.96t sin f35 t) V4 + 0.027 sin 2.96tl. This agrees with Example 2, Case (TIl) in Sec. 2.4. The work was less. • CHAP. 6 232 Laplace Transforms Advantages of the Laplace Method 1. Solving a nonhomogeneous ODE does not require first solving the homogeneous ODE. See Example 4. 2. Initial values are automatically taken care of See Examples 4 and 5. 3. E X AMP L E 6 Complicated inputs ret) (right sides of linear ODEs) can be handled very efficiently, as we show in the next sections. Shifted Data Problems This means initial value problems with initial conditions given at some 1 = 10 > 0 instead of t = O. For such a problem set 1 = 7 + to, so that t = to gives 7 = 0 and the Laplace transform can be applied. For instance, solve y" + Y = 2t, Solution. We have to = !7T y" + Y = and we set r = 7 + !7T. Then the problem is 2(1 + !7T), vi y'(O) = 2 - yeO) = !7T, where y(7) = y(t). Using (2) and Table 6.1 and denoting the transform of y by Y, we see that the subsidiary equation of the "shifted·' initial value problem is 2~ 2 17T S Y - S·!7T - (2 - V 2) + Y = 2" + ~ S Solving this algebraically for Y, (s 2 + - I)Y = !7T 2 (s2 + l)s2 + (s2 2- !7TS + l)s + s2 = .;e-l(y) = 2(7 - sin 7) + = 1 - Now 7 = t - 47T, sin 1 = 27 + 1 Yz (sin t - Using (l) or (2), find ;£(f) if fU) equals: 2. 3. sin2 wt 2 5. sinh at 7. 1 !7T(I - sin ~'1Tr 1 7rt 6. cosh 2 8. sin4 t cos 7) + = 1), and the last two terms give cos !7T cos 7 + (2 - Yz) sin 7 !t (Use Prob. 3.) 9. (Derivation by different methods) It is typical that various transforms can be obtained by several methods. Show this for Prob. 1. Show it for ;£(cos 2 !t) (a) by • = 21 - sin t + cos t. expressing cos2 !t in terms of cos t, (b) by using Prob.3. 110-241 cos 51 4. cos 2 Yz cos 1), so that the answer (the solution) is OBTAINING TRANSFORMS BY DIFFERENTIATION 1. te kt ~ 2 - V 2. !7T - Yz sin 7. y 11-81 2"1 7TS + + S + 1 + ~. The inverse of the first two terms can be seen from Example 3 (with w and sin, y 2 s 2" + we obtam _ Y = thus S INITIAL VALUE PROBLEMS Solve the following initial value problems by the Laplace transform. (If necessary, use partial fraction expansion as in Example 4. Show all details.) + y' + 10. y' 4y = O. 11. h = 17 yeO) = 2.8 sin 2t, 12. y" - y' - 6y = 0, y'(O) = 13 yeO) = -} yeO) = 6, SEC. 6.3 Unit Step Function. t-Shifting 13. y" - h 233 yeO) = 4, = 0, 14. y" - 4y' + 4y = 0, y' (0) = 3.9 yeO) = 2.1, 15. y" + 2y' + 2y yeO) = 1, f(t) 0, = own to illustrate the advantages of the present method (to the extent we have seen them so far). y' (0) = 0 I :/f(a-O) 1 :.....--f(a + Ol y'(O) = -3 16. y" + ky' - 2k 2 y = O. y' (0) = 2k 17. y" + 7y' + 12y y'(0) = -10 18. y" = + 9y = lOe-t, + 3y' + 2.2Sv 19. y" y'(O) = 31.S 21e 3t o yeO) = 3.S, , = 9t 3 + 64. /(0) yeo) = = 0 1, y(O) = 3.2. (a) .;£(t cos wt) = + 2y' + Sy = SOt - ISO, y'(3) = 14 + 0) (s (e) .;£(t cosh at) = (S2 _ 2 2 w) a 2 )2 2as (f) OBTAINING TRANSFORMS BY INTEGRATION 127-341 Using Theorem 3, find f(t) if .;£(f) equals: 1 10 28. 27. S3 - "'S2 S2 + s/2 29. - f(a - O)]e-a.s. (c) Verify (1 *) for f(t) = e- t if 0 < t < 1 and 0 if t> 1. (d) Verify (l *) for two more complicated functions of your choice. (e) Compare the Laplace transform of solving ODEs with the method in Chap. 2. Give examples of your w2 + y(3) = -4, 25. PROJECT. Comments on Sec. 6.2. (a) Give reasons why Theorems 1 and 2 are more important than Theorem 3. (b) Extend Theorem 1 by showing that if f(t) is continuous, except for an ordinary discontinuity (finite jump) at some t = a (> 0), the other conditions remaining as in Theorem 1, then (see Fig. 116) (1*) .;£(f') = s.;£(f) - f(O) - [f(a S2 2 and from this and Example 1: (b) formula 21, (c) 22, (d) 23 in Sec. 6.9, S 24. y" 6.3 Formula (1*) 26. PROJECT. Further Results by DifferentiatiolL Proceeding as in Example 1, obtain y(2) = 4 21. (Shifted data) y' - 6y = 0, 22. y" - 2y' - 3y = 0, yO) = -3, y'(l) = -17 y(l) = 4, 23. y" + 3y' - 4y = 6e 2t - 2 • = a Fig. 116. y(O) = 0, 20. y" - 6y' + Sy = 29 cos 2t. y' (0) = 6.2 y'(l) i~ yeO) = 2. 31. 33. S3 - ks 2 S S3 - Ss S4 - 4s 2 30. 32. 34. S4 + S3 + 9s S4 + S2 2 7f2s 2 35. (Partial fractions) Solve Probs. 27, 29, and 31 by using partial fractions. Unit Step Function. f-Shifting This section and the next One are extremely important because we shall now reach the point where the Laplace transform method shows its real power in applications and its superiority over the classical approach of Chap. 2. The reason is that we shall introduce two auxiliary functions, the unit step function or Heaviside function u(t - a) (below) and Dirac's delta (jet - a) (in Sec. 6.4). These functions are suitable for solving ODEs with complicated right sides of considerable engineering interest, such as single waves, inputs (driving forces) that are discontinuous or act for some time only, periodic inputs more general than just cosine and sine, or impUlsive forces acting for an instant (hammerblows, for example). 234 CHAP. 6 Laplace Transforms Unit Step Function (Heaviside Function) u(t - a) The unit step function or Heaviside function u(t - a) is 0 for t < a, has a jump of size I at t = a (where we can leave it undefined), and is I for t > a, in a formula: (1) u(t - a) = if t < a {~ (a ~ 0). if t > a Figure 117 shows the special case u(t), which has its jump at zero, and Fig. 118 the general case u(t - a) for an arbitrary positive a. (For Heaviside see Sec. 6.1.) The transform of u(t - a) follows directly from the defining integral in Sec. 6.1, ~{u(t - a)} = lXo e - stU(l - = JX e - st • 1 dt = st _ e- a here the integration begins at t = a (~ ~{u(t (2) a) dt I"" ; S t=a 0) because u(t - a) is 0 for 1 < a. Hence - a)} (S > 0). S The unit step function is a typical "engineering function" made to measure for engineering applications. which often involve functions (mechanical or electrical driving forces) that are either "off' or "on." Multiplying functions f(t) with U(l - a). we can produce all sorts of effects. The simple basic idea is illustrated in Figs. 119 and 120. In Fig. 119 the given function is shown in (A). In (B) it is switched off between t = 0 and t = 2 (because u(t - 2) = 0 when t < 2) and is switched on beginning at t = 2. In (C) it is shifted to the righl by 2 units, say, for instance, by 2 secs, so that it begins 2 secs later in the same fashion as before. More generally we have the following. Let f(t) = 0 for all negative t. Then f(t - a)u(t - a) with a (translated) to the right by the amount a. > 0 is f(t) shifted Figure 120 shows the effect of many unit step functions, three of them in (A) and infinitely many in (B) when continued periodically to the right: this is the effect of a rectifier that clips off the negative half-waves of a sinuosidal voltage. CAUTION! Make sure that you fully understand these figures, in particular the difference between parts (B) and (C) of Figure 119. Figure 119(C) will be applied next. o Fig. 117. o t Unit step function u(tJ Fig. 118. a Unit step function u(t - oj SEC. 6.3 235 Unit Step Function. t-Shifting 5'L 5f\ -5 V -5 V (tt) I o (B) (A) ((t) = 5 sin t Fig. 119. I 2 11: 211: {(t)u(t - t 2) 0 (el 2 11:-1-2211:+2 {(t - 2)u(t - 2) Effects of the unit step function: (A) Given function. (B) Switching off and on. (e) Shift. k~'-------'4 I I , , 1. 6 ---t ~ -k 02468lO (B) 4 sin (~11:t)[u(t) - u(t - 2) + u(t - 4) - + ... ] (A) k[u(t - 1) - 2u(t - 4) + u(t - 6)] Fig. 120. Use of many unit step functions. Time Shifting (t-Shifting): Replacing t by t - a in f{t) The first shifting theorem ("s-shifting") in Sec. 6.1 concerned transforms pes) = ~{f(t)} and F(s - a) = ~{eatJ(t)}. The second shifting theorem will concern functions J(t) and J(t - a). Unit step functions are just tools, and the theorem will be needed to apply them in connection with any other functions. THEOREM 1 Second Shifting Theorem; Time Shifting If J(t) has the tran~fonl! let) (3) F(s), then the "shifted function" if t = J(t - a)u(t - a) = { J(t - a) has the (4) Or, (4*) tran~fonn e-aSP(s). That is, ~{f(t < a 0 if t > a if ~{f(t)} = pes), then - a)uCt - a)} = e-asF(s). !f we take the inverse on both sides, we can write J(t - a)u(t - a) = ~-l{e-a.sp(s)}. Practically speaking, if we know pes), we can obtain the transform of (3) by multiplying pes) bye-as. In Fig. 119, the transform of 5 sin tis F(s) = 5/(S2 + 1), hence the shifted function 5 sin (t - 2) u(t - 2) shown in Fig. 119(C) has the transform CHAP. 6 236 PROOF Laplace Transforms We prove Theorem 1. In (4) on the right we use the definition of the Laplace transform, writing 'T for t (to have t available later). Then, taking e- as inside the integral, we have x :x e-asp(s) Lo e- as = e- S7J('T) d'T = L e-scT+a>f('T) d'T. 0 Substituting 'T + a = t, thus 'T = t - a, d'T = dt, in the integral ( 'AUTION, the lower limit changes!), we obtain J 00 e-asp(s) = e-stf(t - a) dt. a To make the right side into a Laplace transform, we must have an integral from 0 to 00, not from a to IX. But this is easy. We multiply the integrand by u(l - a). Then for t from o to a the integrand is 0, and we can write, with f as in (3), (Do you now see why u(t - a) appears?) This integral i., the left side of (4), the Laplace transform of f(t) in (3). This completes the proof. • E X AMP L E 1 Application of Theorem 1. Use of Unit Step Functions Write the following function using unit step functions and find its rransform. iro < if I if Solulioll. I < I < I <!1' I> !1'. (Fig. 121) Step 1. In terms of unit step functions, f(t) = 2(1 - u(t - I)) + !t2(1l(1 - !7T» I) - lI(t - + (cos I)U(I - !1'). Indeed. 2(1 - 11(1 - I)) gives I(t) for 0 < I < 1, and so on. I(tl in the form 1(1 - a)u(1 - a). Thus, 2(\ - 1I(t - I» remains as it is and gives the transform 2(1 - e -')1.1'. Then Step 2. To apply Theorem \, we must write each term in {12 ;e 21 11(1 - } (I I) ~ 9., "2(t - I) 2+ (t - 1) + I)} (17i + --;;:1 + 1) 2 lI(t - 1) ~ 2s e- S Together, I "3" ( s I + -I + ""2 s 2.1' ) e- s - ( -I .1'3 + - 7T + -1'2 ) e- ws/2 2s2 8s - I _ e- ws12 __ s2 + I . SEC. 6.3 237 Unit Step Function. t-Shifting If the conversion of f( t) to f( t - a) is inconvenient. replace it by ~{f(t)u(t (4**) - a)} = e-as~{f(t (4**) follows from (4) by writing f(t - a) = get). hence f(t) = get + a)}. + a) and then again writing f for g. Thus. as before. Similarly for .'£{~t2u(t - ~'17)}. Finally, by (4**). fIt) 2 Or--L~--~--~r----L--~~--~---T----L---~--~ 1T 21T 41T -1 Fig. 121. t(t} in Example 1 E X AMP L E 2 Application of Both Shifting Theorems. Inverse Transform Find the inverse transform f(t) of Solution. Without the exponential functions in the numerator the three terms of F(s) would have the inverses (sin '17t)/'17, (sin '17t)/'17, and te -21; because IIs 2 has the inverse t, so that 1/(s + 2)2 has the inverse te -2t by the first shifting theorem in Sec. 6.1. Hence by the second shifting theorem (t-shifting), 1 f(t) = - sin ('17(t - 1» u(t - 1) '17 1 + - '17 sin ('17(t - 2» u(t - 2) + (t - 3)e -2(t-3) u(t - 3). Now sin ('17t - '17) = -sin '17t and sin ('17t - 2'17) = sin '17t, so that the second and third terms cancel each other when t > 2. Hence we obtain f(t) = 0 if 0 < t < 1, -(sin '17t)/'17ifl < t < 2,0 if2 < t < 3, and (t - 3)e -2(t-3) if t > 3. See Fig. 122. • 0.3 0.2 0.1 OL------L______L-____~______~____~~~==~___ o 2 3 4 5 6 Fig. 122. t(t} in Example 2 E X AMP L E 3 Response of an RC-Circuit to a Single Rectangular Wave Find the current i(t) in the RC-circuit in Fig. 123 if a single rectangular wave with voltage Vo is applied. The circuit is assumed to be quiescent before the wave is applied. 238 CHAP. 6 Laplace Transforms c vet) v(t) a R Fig. 123. RC-circuit, electromotive force v(t), and current in Example 3 Solutioll. The input is Vo[lI(t equation (see Sec. 2.9 and Fig. 123) + Ri(t) a) - = Ri(t) + ~ q(t) C b)]. Hence the circuit is modeled by the integro-differential 11([ - C f t i(T) dT 0 = vet) = Vo [lI(t - a) - tI(t - b)l. Using Theorem 3 in Sec. 6.2 and formula (I) in this section, we obtain the subsidiary equation RI(s) + Irs) = Vo [e -as _ e -bsj. sC s Solving this equation algebmically for I(s). we get where VoIR F(s) = - - - " - - s + J/(RC) and the last expression being obtained fi'om Table 6.1 in Sec. b.l. Hence Theorem 1 yields the solution (Fig. 123) that is. i(l) = 0 if t < a. and ifa<l<b if a > b • E X AMP L E 4 Response of an RLC-Circuit to a Sinusoidal Input Acting Over a Time Interval Find the response (the current) of the RLC-circuit in Fig. 124, where E(t) is sinusoidal. acting for a short time interval only. say. E(t) = 100 sin 400t if 0 < t < 27T and E(t) = 0 if t > 27T and current and charge are initially zero. Solution. The electromotive force E(t) can be represented by (100 sin 400t){l - u(t - 27T)). Hence the model for the current i(t) in the circuit is the integro-differential equation (see Sec. 2.9) O.li' + I Ii + 100 f t i(T) dT = (100 sin 400t)(1 - tI(t - 27T», ;(0) = O. o From Theorems 2 and 3 in Sec. 6.2 we obtain the subsidiary equation for Irs) = 5£(i) / O.ls/ + 111 + 100- s 2 100' 400s (-sl _ e2 s2 + 400 s r.S). /(0) = O. SEC. 6.3 239 Unit Step Function. t-Shifting Solving it algebraically and noting that .1 2 + 110.1 + 1000 = (s + 10)(.1 + 100), we obtain I(s) = (.I 2 se- .,,-s s2 + 4002 (s 1000·400 + 10)(.1 + 100) .1 2 + 4002 - ) . For the first term in the parentheses ( ... ) times the factor in front of them we use the partial fraction expansion 400000.1 A (s + 10)(.1 + 100)(s2 + 4002) .1+10 + B + s+100 Ds +K ~-----= s2+4002 Now determine A, B, D, K by your favorite method or by a CAS or as follows. Multiplication by the common denominator gives 400000.1 = A(s + 100)(.1 2 + 4002) + B(s + 1O)(s2 + 400 2) + (Ds + K)(s + 10)(.1 + 100). We sets = -10 and -100 and then equate the sums of the s3 and .1 2 terms to zero, obtaining (all values rounded) -4000000 = 90(102 + 4002)A, (s = -10) (s Since K = A = -0.27760 -40000000 = -90(1002 + 4002)B, -100) B = 2.6144 + D, (s3- terms ) 0= A + B (s2-terms) 0= 100A + lOB + llOD + K, D = -2.3368 = 258.66 = 0.6467' 400, we thus obtain for the first term I1 in I = 2.6144 2.3368.1 s + 100 .1 2 + 4002 0.2776 h = - ----- + -----.I + 10 K = 258.66. II - 12 0.6467 . 400 + ---,;;--------;:2 s2 + 400 From Table 6.1 in Sec. 6.1 we see that its inverse is i1 U) = -0.2776e- lOt + 2.6144e- lOOt - 2.3368 cos 400t + 0.6467 sin 400/. This is the cunent i(t) when 0 < t < 27T. It agrees for 0 < r < 27T with that in Example 1 of Sec. 2.9 (except for notation), which concerned the same RLC-circuit. Its graph in Fig. 62 in Sec. 2.9 shows that the exponential terms decrease very rapidly. Note that the present amount of work wa, substantially less. The second term h of 1 differs from the first term by the factor e -2.,,-s. Since cos 400(1 - 27T) = cos 400t and sin 400(1 - 27T) = sin 400t, the second shifting theorem (Theorem I) gives the inverse i2(t) = 0 if o < t < 27T. and for > 27T it gives i2(t) = -0.2776e- lOCt - 2.,,-) + 2.6144e- lOO \t-2r.) - 2.3368 cos 400t + 0.6467 sin 400t. Hence in i(t) the cosine and sine terms cancel, and the current for t > 27T is i(t) = -0.2776(e- lOt - e- lOCt - 2.,,-) + 2.6144(e- lOOt It goes to zero very rapidly, practically within 0.5 sec. E(t) Fig. 124. RLC-circuit in Example 4 _ e-lOOCt-2.,,-). • 240 CHAP. 6 Laplace Transforms 1. WRITING PROJECT. Shifting Theorem. Explain and compare the different roles of the two shifting theorems, using your own formulations and examples. \2-13\ UNIT STEP FUNCTION AND SECOND SHIFTING THEOREM Sketch or graph the given function (which is assumed to be zero outside the given interval). Represent it using unit step functions. Find its transform. Show the details of your work. 2. 4. 6. 8. 10. 12. < t < 1) 3. e t (0 < t < 2) 5. t 2 (I < t < 2) sin 3t (0 < t < 'IT) 7. cos 'lTf (1 < t < 4) t 2 It > 3) t I - e- (0 < t < 'IT) 9. t (5 < t < 10) 11. 20 cos 7ft (3 < f < 6) sin wt (t > 6 'IT/ w) 13. e'Ut (2 < t < 4) sinh t (0 < t < 2) t (0 \14-22\ INVERSE TRANSFORMS BY THE SECOND SHIFTING THEOREM 30. y" - 16)' = r(t), o if t > 4: 31. ret) = 48e y(O) 21 if 0 = 3. < t < 4 and y'CO) = -4 y" + y' - 2)" = 1'(1), r(t) = 3 sin t - cos t if o < t < 2'IT and 3 sin 2t - cos 2t if t > 2 'IT; yeO) = 1, y'(O) = 0 8y' + 15)' = ret), r(t) = 35e 21 if t < 2 and 0 if t > 2; yeO) = 3. y'(O) = -8 32. y" + o< 33. (Shifted data) v" + 4v = 8t 2 if 0 < t < 5 and 0 if t > 5; y{ 1) ~ I + 'cos 2, y' (1) = 4 - 2 sin 2 + 2)"' + 5)' = 10 sin t if 0 < t < 2'IT and 0 if t > 2'IT; Y('IT) = I, y'('IT) = 2e-r. - 2 34. y" MODELS OF ELECTRIC CIRCUITS 35. (Discharge) Using the Laplace transform, find the charge q(t) on the capacitor of capacitance C in Fig. 125 if the capacitor is charged so that its potential is Vo and the switch is closed at f = O. Find and sketch or graph f(t) if ;e(n equals: 14. se- s /(s2 4s 15. e- /s 16. S-2 - w 2) + R 2 + (s-2 17. (e- 27TS - s-l)e- S e- Br.s)/(s2 + Fig. 125. I) + 2s + 2) 19. e- 2s /s 5 e-s+k)/(s - k) 21. se- 3s /(s2 - 18. e- 7Ts /{s2 20. (1 - 22. 2.5(e- 3 . Bs \23-34\ 4) e- 2 . 6S )/s - INITIAL VALUE PROBLEMS, SOME WITH DISCONTINUOUS INPUTS Using the Laplace transform and showing the details, solve: 23. y" + 2y' + 2v y' (0) = = O. 0, yeO) = + Y = 0, \36-38\ RC-CIRCUIT Using the Laplace transform and showing the details. rmd the current i(f) in the circuit in Fig. 126 with R = 10 fl and C = 10- 2 F, where the current at t = 0 is assumed to be zero. and: 36. 100 V if 0.5 < f Why does i(t) have jumps? v(t) = < 0.6 and 0 otherwise. 37. v = 0 if t < 2 and 100 (t - 2) V if t > 2 38. v = 0 if t < 4 and 14' 106 e -3t V if t > 4 1 24. 9)''' - 6)"' Problem 35 yeO) = 3, y'(O) = 1 25. y" + 4y' + 13y = 145 cos 2t, yeO) = 10, C y'(O) = 14 yeO) = ~, 26. y" + lOy' + 24y = 144t 2 , y'(0) = -5 + 9y = if t > 'IT; 27. y" r(r), 1'(1) = 8 sin t if 0 yeO) = 0, y'(O) = 4 < t < 'IT and 0 28. y" + 3;-' + 2y = r(t), r(t) = 1 if 0 < t < 1 and o if t > 1; yeO) = 0, y' (0) = 0 29. y" + t> 1; y r(t), yeO) r(t) = t if 0 = y'(O) = 0 < t < 1 and 0 if L Fig. 126. /39-411 R vet) Problems 36-38 RL-CIRCUIT Using the Laplace transform and showing the details, find the current i(t) in the circuit in Fig. 127, assuming i(O) = 0 and: SEC. 6.4 241 Short Impulses. Dirac's Delta Function. Partial Fractions 39. R = 10 fl, L = 0.5 H, v = 200t V if 0 < t < 2 and Oift>2 C 40. R = 1 kfl (= 1000 fl), L = 1 H, v = 0 if o < t < r., and 40 sin t V if t > 'iT' 41. R = 25 fl, L = 0.1 H, v = 490e- 5t V if o < t < 1 and 0 if t > 1 L Fig. 128. L u(t) Problems 42-44 RLC-CIRCUIT R L u(t) Fig. 127. 142-441 Problems 39-41 LC-CIRCUIT Usmg the Laplace transform and showing the details, find the current i(t) in the circuit in Fig. 128, assuming zero initial current and charge on the capacitor and: 42. L = o< 1 H, C = 0.25 F, v t < 1 and 0 if t > 1 = 200(t - tt 3 ) Using the Laplace transform and showing the details, find the current i(t) in the circuit in Fig. 129. assuming zero initial current and charge and: 45. R = 2 n, L = I H. C = 0.5 F. vet) = 1 kV if o < t < 2 and 0 if t > 2 46. R = 4 n, L = I H, C = 0.05 F, v = 34e- t V if 0 < t < 4 and 0 if t > 4 47. R = 2 n, L = I H, C = 0.1 F, v = 255 sin t V if 0 < t < 2r. and 0 if t > 2r. V if 43. L = I H, C = 10- 2 F, v = -9900 cos t V if 'iT' < t < 3 r. and 0 otherwise 44. L = 0.5 H, C = 0.05 F, v = 78 sin t V if o < t < r. and 0 if t > r. 6.4 u(t) Fig. 129. Problems 45-47 Short Impulses. Dirac's Delta Function. Partial Fractions Phenomena of an impulsive nature. such as the action of forces or voltages over short intervals of time, arise in various applications, for instance, if a mechanical system is hit by a hammerblow, an airplane makes a "hard" landing, a ship is hit by a single high wave, or we hit a tennisball by a racket, and so on. Our goal is to show how such problems are modeled by "Dirac's delta function" and can be solved very efficiently by the Laplace transform. To model situations of that type, we consider the function (1) Ilk h(t - a) = { 0 ifa~t~a+k (Fig. 130) otherwise (and later its limit as k --'.> 0). This function represents. for instance. a force of magnitude 11k acting from t = a to t = a + k, where k is positive and small. In mechanics, the integral of a force acting over a time intervalll ~ t ~ a + k is called the impulse of the 242 CHAP.6 Laplace Transforms force; similarly for electromotive force!. E(t) acting on circuits. Since the blue rectangle in Fig. 130 has area 1, the impulse of fk in (l) is a+k I a) dt = dt = 1. oak J - -: :x: h (2) = L fk(t - To find out what will happen if k becomes smaller and smaller, we take the limit of fk as k ~ 0 (k > 0). This limit is denoted by 8(t - a), that is, 8(t - a) = k-O lim h(t - a). 8(t - a) is called the Dirac delta function 2 or the unit impulse function. 8(t - a) is not a function in the ordinary sense as used in calculus, but a so-called generalizedjullction. 2 To see this, we note that the impulse lk of fk is I, so that from (1) and (2) by taking the limit as k ~ 0 we obtain if t (3) 8(t - a) = {: = a 100 8(t - and a) dt = 1, o otherwise but from calculus we know that a function which is everywhere 0 except at a single point must have the integral equal to O. Nevertheless, in impulse problems it is convenient to operate on 8(t - a) as though it were an ordinary function. In particular, for a continuous function get) one uses the property [often called the sifting property of B(t - a), not to be confused with shifting 1 IXo g(t) B(t - (4) a) dt = g(a) which is plausible by (2). To obtain the Laplace transform of 8(t - a), we write fk(t - a) I = - k [u(t - a) - u(t - (a + k»] r~Area=l 11k t Fig. 130. a a+k t The function fk(t - 0) in (1) 2 PAUL DIRAC (1902-1984), English physicist, was awarded the Nobel Plize [jointly with the Austrian ERWIN SCHRODINGER (1887-1961)] in 1933 for his work in quantum mechanics. GeneralIzed functions are also called distributions. Their theory was created in 1936 by the Russian mathematician SERGEI L'VOVICH SOBOLEV (1908-1989). and in 1945. under wider aspects, by the French mathematician LAURENT SCHWARTZ (1915-2002). SEC. 6.4 243 Short Impulses. Dirac's Delta Function. Partial Fractions and take the transform [see (2)] 1 _ [e- as ks :£{h(t - a)} e-Ca+k)s] _ = e- as 1 - e- ks ---- ks We now take the limit as k----7 O. By l'H6pital's rule the quotient on the right has the limit 1 (differentiate the numerator and the denominator separately with respect to k, obtaining se- ks and s, respectively, and use se-ks/s ----7 1 as k ----7 0). Hence the right side has the limit e- as • This suggests defining the transform of 8(t - a) by this limit, that is, :£(8(t - a)} = e- as • (5) The unit step and unit impulse functions can now be used on the right side of ODEs modeling mechanical or electrical systems, as we illustrate next. E X AMP L E 1 Mass-Spring System Under a Square Wave Determine the response of the damped mass-spring system (see Sec. 2.8) under a square wave, modeled by (see Fig. 131) y" + 3y' + 2y = r(t) = u(r - 1) - u(t - 2). /(0) = o. .1'(0) = O. Solution. From (1) and (2) in Sec. 6.2 and (2) and l4) in this section we obtain the subsidiary equation s2y -t 3sY + 2Y = - 1 s e- 2S ). (e- s 1 Yes) = Solution 2 s(s + 3s + 2) (e- s - e- 2S ). Using the notation F(,,) and partial fractions, we obtain F(s) = 1 s(s2 1 + 3s + 2) = ----s(s + 1)(s + 2) 112 s --+ s+1 112 s+2 From Table 6.1 in Sec. 6.1. we see that the inverse is Therefore, by Theorem 1 in Sec. 6.3 (t-shifting) we obtain the square-wave response shown in Fig. 131, y = :g-\F(s)e- s - F(s)e- 2s ) = f(t - I)u(t - I) - f(t - 2)u(t - 2) o = ~ _ e-Ct-1) + ~e-2Ct-l) { -e -Ct-l) + e -Ct-2) + "2e 1 -2(t-1) - 1 "2e -2(t-2) yet) 0.5 2 3 4 Fig. 131. Square wave and response in Example 1 (0 < t< J) (1 < t< 2) (t> 2). • 244 E X AMP L E 2 CHAP.6 laplace Transforms Hammerblow Response of a Mass-Spring System Find the response of the system in Example I with the square wave replaced by a unit impulse at time t = L Solutioll. We now have the ODE and the subsidiary equation y" + 3/ + 2y I). = S(t - and (S2 + 3s + 2)Y = e- s . Solving algebraically gives ( 1 e -s Yes) = (s + + 1)(5 I) e . ~-s+2 2) -s By Theorem 1 the inverse is y(t) = ~C\Y) = ifO<t<1 0 { e -(t-l) - e -2(t-ll if t> 1. y(t) is shown in Fig. 132. Can you imagine how Fig. 131 approaches Fig. 132 as the wave becomes shorter and shorter. the area of the rectangle remaining I? • yet) 0.2 0.1 ,3 Fig. 132. E X AMP L E 3 5 Response to a hammerblow in Example 2 Four-Terminal RLe-Network n. Find the output voltage response in Fig. 133 if R = 20 L = I H, C = 10- 4 F, the input is S(t) (a umt impulse at time t = 0). and current and charge are zero at time t = O. Solutioll. To understand what is going on, note that the network is an RLC-cucuit to which two wires at A and B are attached for recording the voltage v(r) on the capacitor. Recalling from Sec. 2.9 that current i(t) and charge q(t) are related by i = q' = dqldt, we obtain the model Li' + Ri + !!.. C = Lq" + Rq' + q C = q" + 20q' + 10000q = Set). From (1) and (2) in Sec. 6.2 and (5) in this section we obtain the subsidiary equation for Q(s) = '£(q) (S2 + 20s + 10OOO)Q = 1. Solution Q=------::--(s + 10)2 + 9900 By the first shifting theorem in Sec. 6.1 we obtain from Q damped oscillations for q and v; rounding 9900 = 99.502 , we get (Fig. 133) I lOt q = ~C\Q) = 99.50 esin 99.50t and v = q = lOO.5e- 1ot sin 99.50t. C • SEC. 6.4 245 Short Impulses. Dirac's Delta Function. Partial Fractions 8(t) v 80 R /\ 40 L C 0 B A V -40 v(t); -80 ? f\ O.O~ 0- \ dlj O.l!'L Network "" 0.2 0.3 :25 Voltage on the capacitor Fig. 133. Network and output voltage in Example 3 More on Partial Fractions We have seen that the solution Y of a subsidiary equation usually appears as a quotient of polynomials Y(s) = F(s)JG(s), so that a partial fraction representation leads to a sum of expressions whose inverses we can obtain from a table, aided by the first shifting theorem (Sec. 6.1). These representations are sometimes called Heaviside expansions. An un repeated factor s - a in G(s) requires a single partial fraction AJ(s - a). See Examples I and 2 on pp. 243, 244. Repeated real factors (s - a)2, (s - a)3, etc., require partial fractions etc., The inverses are (A2t + AI)e at , (iA 3t 2 + A2t + Al)e at , etc. Unrepeated complex factors (s - a)(s - a), a = 0: + i{3, a = 0: - i{3, require a partial fraction (As + B)J[(s - 0:)2 + 132]. For an application, see Example 4 in Sec. 6.3. A further one is the following. E X AMP L E 4 Unrepeated Complex Factors. Damped Forced Vibrations Solve the initial value problem for a damped mass-spring system acted upon by a sinusoidal force for some time interval (Fig. 134), y" + 2y' + 2y = rlt), r(t) = 10 sin 2t if 0 < t < 7T > and 0 if t 7T; y(O) = I, y' (0) Solution. = ~5. From Table 6.1, (1), (2) in Sec. 6.2, and the second shifting theorem in Sec. 6.3, we obtain the subsidiary equation (s2y ~ S + 5) + 2(sY ~ I) + 2Y = 10 2 ~2-- (1 ~ e- 7TS ). s +4 We collect the Y-terms, (s2 + 2s + 2)Y, take ~s + 5 - 2 = ~s + 3 to the right, and solve, (6) 20 y = ----;;----;:---(s2 + 4)(s2 + 2s + 2) s (S2 + 4)(s2 + 2s + For the last fraction we get from Table 6.1 and the first Shifting theorem (7) ';£- I{ S+I~4} 2 + I) + I (s = t e- (cost ~ ~ 3 2) + s2 + 2s + 2 4sint). 246 CHAP. 6 Laplace Transforms In the first fraction in (6) we have unrepeated complex roots, hence a partial fraction representation 20 As + Ms+N B (S2 + 4)(s2 + 2s + 2) Multiplication by the common denominator gives 20 = (As + 8)(s2 + 2s + 2) + (Ms + M(s2 + 4). We determine A, B, M, N. Equating the coefficients of each power of s on both sides gives the four equations = A +M 0 = 2A + (a) [s3]: 0 (c) [sl: 2B + 4M (b) [s2]: 0 = 2A (d) [sol: 20 = 2B + B +N + 4N. We can solve this, for instance, obtaining M = -A from (a), then A = B from (c), then N = - 3A from (b), and finally A = -2 from (d). Heuce A = -2, B = -2, M = 2. N = 6. and the fust fraction in (6) has the representation -2s - 2 (8) s2 +4 + 2(s + I) + 6 - 2 + 1)2 + I -2 cos 2f - sin 2f + e- t (2 cos f + 4 sin f). Inverse transform: (s The ,urn of this and (7) is the ,olution of the problem for 0 < f < 1'. namely (the sines cancel). y(t) = 3e- t cos f - 2 cos 2f - sin 21 (9) if 0<1<1'. In the second fraction in (6) taken with the minus sign we have the factor e--rrs, so that from (8) and the second shifting theorem (Sec. 6.3) we get the inverse transform +2 cos (2f - 21') + sin (2, - 217) - e-(t--rr) [2 cos (I - 17) + 4 sin (I - 1T)] = 2 cos 2f + sin 21 + e -(t-.,,-) (2 cos, + 4 sin f). The sum of this and (9) is the solution for y(l) (iO) = f > 17, e- t [(3 + 2e"') cos f + 4e 7T sin I] if'> 17. Figure 134 shows (9) (for 0 < , < 17) and (10) (for' > 17). a beginning vibration, which goes to zero rapidly • because of the damping and the absence of a driving force after r = 1'. yet) 2 -----l y = 0 (Equilibrium position) y 2rr 3rr -1 Driving force l----II'=~ Dashpot (damping) -2 Mechanical system Output (solution) Fig. 134. Example 4 The case of repeated complex factors [(s - a)(s - a)]2, which is important in connection with resonance, will be handled by "convolution" in the next section. SEC. 6.4 247 Short Impulses. Dirac's Delta Function. Partial Fractions PROBLEM SH 6:;:4 11-121 Showing the details. find. graph. and discuss the + Y = OCt y' (0) = 0 v" + 2v' + 2v 1. y" 2. yeO) 217), = = ~olution. 10. e- t + Sfj(t - 2). 'y' (0) = I :,,(0) = '0. !) - .v" - 3. for the differential equation. involving k, take specific k's from the beginning. EFFECT OF DELTA FUNCTION ON VIBRATING SYSTEMS y = 100(t 1000U yeO) = 10. y' (0) = 1 1). + 3/ + 2)' = 10(sin t + oCt - I». = 1. y' (0) = - I 5. y" + 4y' + Sy = [I - ult - lO)]et - elOo(t yeO) = 0, y' (0) = I 6. -,," + 2\"' - 3y = 1000(t - 2) + 1000(t 4. y" yeO) I. yeO) = 7. y" . + = I. 8. y" + S/ + 6)' = I y (0) yeo) = 0, = o(t - !17) -,,'(0) = 0 + 9. y" + 2/ + Sy = 2St - 1000(1 yeo) = -2, y' (0) = S u(t - 17) 3," 12. v" + y cos t, Y' (0)'= -2 sin t + 100(t - 17), • (.I' - a)F(s) hm -----'--'- s~a G(s) F(s) G(s) = -2. - 41' = 2e t - 8e 2 0(t - 2), ~'(0)='2 . . /(0)=0 = = (b) Similarly, show that for a root a of order m and fractions In 17), 10. y" + S)' = 2St - 1000(t - 17). yeo) y' (0) = S. (Compare with Prob. 9.) + A lOy = 10[1 - lilt - 4)] - 100(t - S). . I y(O) 11. v" 3L 15. PROJECT. Heaviside Formulas. (a) Show that for a simple root a and fraction AI(s - a) in F(s)/G(s) we have the Heal'iside formllia y'(O) = 0 + 2\"' - 10), (b) Experiment on the response of the ODE in Example 1 (or of another ODE of your choice) to an impulse OCt - a) for various systematically chosen a (> 0); choose initial conditions yeO) =1= 0, y' (0) = O. Also consider the solution if no impulse is applied. Is there a dependence of the response on a? On b if you choose bo(t - a)? Would -o(t - Ii) with a > a annihilate the effect of o(t - a)? Can you think of other questions that one could consider experimentally by inspecting graphs? Al s-a + - - + further fractions we have the Heaviside formulas for the first coefficient yeO) = 0, . (.I' - a)mF(s) Am = hm - - - - - I G(s) s-.a 13. CAS PROJECT. Effect of Damping. Consider a vibrating system of your choice modeled by y" + cy' + ky = I'lt) with r(t) involving a B-function. (a) Using graphs of the solution. describe the etJect of continuously decreasing the damping to O. keeping k constant. (b) What happens if c is kept constant and k is continuously increased, starting from O? (c) Extend your results to a system with two o-functions on the right. acting at different times. 14. CAS PROJECT. Limit of a Rectangular Wave. Effects of ImpUlse. (a) In Example I. take a rectangular wave of area 1 from 1 to I + k. Graph the responses for a sequence of values of k approaching zero. illustrating that for smaller and smaller" those curves approach the curve shown in Fig. 132. Hint: If your CAS gives no solution and for the other coefficients Ak = I (m - k)! m k d ---k s~a d.~m- lim [(.I' - ai"'F(s) ] G(s) k= 1."'.111-\. 16. TEAM PROJECT. Laplace Transform of Periodic Functions (a) Theorem. The Laplace trallSform of a piecewise COlltilluOllS fUllctioll f(t) n'ith period p is (I \) ~(f) = I 1 _ e-Ps JVe-stf(t) dt (.I' 0 P tv > 0). Prove this theorem. Hint: Write {o'" = I o + v + ... . Set t = (n - I)p in the nth integral. Take out e -(n-l)p from under the integral sign. Use the sum formula for the geometric series. 248 CHAP. 6 Laplace Transforms (b) Half-wave rectifier. Using (11), show that the half-wave rectification of sin wt in Fig. 135 has the Laplace transform (c) Full-wave rectifier. Show that the Laplace transform of the full-wave rectification of sin wt is W 2 S + 7TS w 2 coth - . 2w (d) Saw-tooth wave. Find the Laplace transform of the saw-tooth wave in Fig. 137. w fit) k (A half-wave rectifier clips the negative portions of the I i/ curve. Afull-wave rectifier converts them to positive; see Fig. 136.) o p Fig. 137. v~_ 2rrlm Fig. 135. f(t) 3rrlm Half-wave rectification Fig. 136. 2rrlm Saw-tooth wave ~':~ lr",~~Y~ TrIm 3p (e) Staircase function. Find the Laplace transform of the staircase function in Fig. 138 by noting that it is the difference of ktlp and the function in (d). I o 2p O~-----pL------2~p----~3pL------ 3rrlm Full-wave rectification Fig. 138. Staircase function 6.5 Convolution. Integral Equations Convolution has to do with the multiplication of transforms. The situation is as follows. Addition of transforms provides no problem; we know that :£(f + g) = :£(f) + :£(g). Now multiplication of transforms occurs frequently in connection with ODEs, integral equations, and elsewhere. Then we usually know :£(f) and :£(g) and would like to know the function whose transform is the product :£(f):£(g). We might perhaps guess that it is fg, but this is false. The transform of a product is generally differentfrom the product of the transforms of the factors, :£(fg) =1= :£(f):£(g) in general. To see this take f = et and g = 1. Then fg = et , :£(fg) = lI(s - 1), but :£(f) = lI(s - 1) and :£(1) = lis give :£(/):£(g) = lI(s2 - s). According to the next theorem, the correct answer is that :£(f):£(g) is the transform of the convolution of f and g, denoted by the standard notation f * g and defined by the integral fof( t (1) h(t) = (f * g)(t) = 'T)g(t - 'T) d'T. SEC 6.5 249 Convolution. Integral Equations THEOREM 1 Convolution Theorem If two functions f and g satisfy the assumption in the existence theorem in Sec. 6.1, so that their transfonns F and G exist, the product H = FG is the transform of h given by (1). (Proof after Example 2.) E X AMP L E 1 Convolution Let H(s) 1/[(s - a)s]. Find h(t). = Solution. g(t - 7) 1/(s - a) has the inverse fft) = eat. and lis has the inverse get) = 1. With f( 7) = eM and == I we thus obtain from (I) the answer h(t) = eat * I = I t eaT. I d7 = .!.. (eat I). - a o To check. calculate H(s) = :£(h)(s) = E X AMP L E 2 (_1- a a .!..) s s-a - ~ • --- = s2-as I 1 s-a s -- . - = :£(eat ) :£(1). • Convolution Let H(s) = 1I(s2 + w2)2. Find h(t). The inverse of l/(s2 t- w2 ) is (sin wt)lw. Hence from (I) and the trigonometric formula (11) in App. 3.1 with x = ~(wt + W7) and y = ~(wt - W7) we obtam Solution. sin wt h(t) = - - w sin wt *-w = 2 1 W I = --2 It It sin W7 sin w(t - 7)d7 0 2w [-cos wt + cos W7] d7 0 [ -7COS sin W7 wt + - w - Jt 7~O 2w2 [ -t cos wt + sinwwtJ • in agreement with formula 21 in the table in Sec. 6.9. PROOF We prove the Convolution Theorem 1. CAUTION! Note which ones are the variables of integration! We can denote them as we want, for instance, by T and p, and write and We now set t to 00. Thus = p + T, G(s) = where T is at first constant. Then p t)Oe-sCt-'r)g(t T G(s) = T) dt = eST L=o e-SPg(p) dp. = t - t)Oe-stg(t T T, and t varies from T) dt, T 250 CHAP.6 Laplace Transforms T in F and t in G vary independently. Hence we can insert the G-integral into the F-integral. Cancellation of e - ST and eST then gives F(s)G(s) = I x e-STf(T)eST o I I cc = e-stg(t - T) dulT I X:xl f(T) e-stg(t - T) dtdT. T O T Here we inregrate for fixed T over T from T to rye and then over T from 0 to co. This is the blue region in Fig. 139. Under the assumption on f and g the order of integration can be reversed (see Ref. [A5] for a proof using uniform convergence). We then integrate first over T from 0 to t and then over t from 0 to x, that is, F(s)G(s) = I x e- st o f t f(T)g(t - T) cIT dt = 0 I oc e-sth(t) dt = ~(h) = H(s). 0 • This completes the proof. r Fig. 139. Region of integration in the tT-plane in the proof of Theorem 1 From the definition it follows almost immediately that convolution has the properties (commutative law) f*g=g*f f * (gl + g2) = (f * g] + f * g2 f * g) * v = .f * (g * v) (distributive law) (associative law) f*O=O*f=O similar to those of the multiplication of numbers. Unusual are the following two properties. E X AMP L E 3 Unusual Properties of Convolution f *I "* f in general. For instance. t (* (f * j)(t) ~ I = f .. I d. o = ~ (2 "* t. 0 may not hold. For instance. Example 2 with w = I gives sin t * sin t = -~ t cos t + ~ sin t (Fig. 140). • SEC. 6.5 251 Convolution. Integral Equations 4 2 -2 -4 Fig. 140. Example 3 We shall now take up the case of a complex double root (left aside in the last section in connection with partial fractions) and find the solution (the inverse transform) directly by convolution. E X AMP L E 4 Repeated Complex Factors. Resonance In an undamped mass-spring system, resonance occurs if the frequency of the driving force equals the natural frequency of the system. Then [he model is lsee Sec. 2.8) where "'02 = kIm, k is the spring constant. and 111 is the mass of the body attached to the spring. We assume yeO) = and y' (0) = 0, for simplicity. Then the subsidiary equation is ° 2 s Y+ 2 "'0 K",o Y= 2 S + Its solution is 2 . "'0 This is a transform as in Example 2 with", = directly that the solution of our problem is y(t) = K",o ( --2 -tcoS "'ot + 2"'0 "'0 and multiplied by K",o' Hence from Example 2 we can see sin "'ot ) "'0 = K - - 2 (-"'olCOS "'ol + sin "'ot)· 2"'0 We see that the first term grows without bound. Clearly, in the case of resonance such a tenn must occur. (See • also a similar kind of solution in Fig. 54 in Sec. 2.8.) Application to Nonhomogeneous Linear ODEs Nonhomogeneous linear ODEs can now be solved by a general method based on convolution by which the solution is obtained in the form of an integral. To see this, recall from Sec. 6.2 that the subsidiary equation of the ODE y" (2) + ay' + by = ret) (a. b constant) has the solution [(7) in Sec. 6.2] Yes) = [(s + a)y(O) + y' (O)jQ(s) + R(s)Q(s) with R(s) = :£(r) and Q(s) = 1/(S2 + as + b) the transfer function. Inversion of the first term [... ] provides no difficulty; depending on whether 2 - b is positive, zero, or negative, its inverse will be a linear combination of two exponential functions, or of the !a 252 CHAP. 6 Laplace Transforms form (Cl + c2t)e-at/2, or a damped oscillation, respectively. The interesting term is R(s)Q(s) because ret) can have various forms of practical importance, as we shall see. If yeo) = 0 and y' (0) = 0, then Y = RQ, and the convolution theorem gives the solution f qV t (3) E X AMP L E 5 y(t) = T)r( T) dT. o Response of a Damped Vibrating System to a Single Square Wave Using convolution. determine the response of the damped mass-spring system modeled by y" + 3/ + 2)' = r(t), r(t) = 1 if I < t < 2 and 0 otherwise, yeO} =/ (O) = o. This system with an input (a driving force) that acts for sOllie lillie ollly (Fig. 141) has been solved by partial fraction reduction in Sec. 6.4 (Example I). Solution by Convolution. Q(s} = s2 + 3s + 2 The transfer function and its inverse are (s + I}(s + 2) s + I hence s+2' Hence the convolution integral (3) is (except for the limits of integration) () I ( ) Yt = q t - T' I dT = I[ e -(t-T) - e -2(t-T)] d T = e-(t-T) - 2e 1 -2(t-T) . Now comes an important point in handling convolution. reT} = 1 if I < T < 2 only. Hence if t < I. the integral is zero. If I < t < 2. we have to integrate from T = I (not 0) to t. This gives (with the first two terms from the upper limit) If t > 2, we have to integrate from T = I to 2 (not to t). This gives Figure 141 shows the input (the square wave) and the interesting output, which is zero from 0 to I. then increases, reaches a maximum (near 2.6) after the input has become zero (why?), and finally decreases to zero in a monotone fashion. • yet) 1 /output (response) 0.5 2 Fig. 141. 3 4 Square wave and response in Example 5 Integral Equations Convolution also helps in solving certain integral equations. that is, equations in which the unknown function yet) appears in an integral (and perhaps also outside of it). This concerns equations with an integral of the form of a convolution. Hence these are special and it suffices to explain the idea in terms of two examples and add a few problems in the problem ~et. SEC. 6.5 Convolution. Integral Equations E X AMP L E 6 253 A Volterra Integral Equation of the Second Kind Solve the Volterra integral equation of the second kind3 f yet) - t yeT) sin (t - T) dT = t. o Solutioll. From (I) we see that the given equation can be written as a convolution. y - y * sin t = Y = !ley) and applying the convolution theorem, we obtain t. Writing I s2 yes) - Yes) - 2 - - = yes) - 2 - s+1 s+l The solution is and give~ t3 the answer yet) = t + (5 . Check the result by a CAS or by substitution and repeated integration by parts (which will need patience) . • E X AMP L E 7 Another Volterra Integral Equation of the Second Kind Solve the Volterra integral equation y(t) - t J (1 + T) ylt - T) dT = I - sinh r. o Solution. By (1) we can write y - (I + t) * Y = 1 - sinh t. Writing Y convolution theorem and then taking common denominators yeS) [I - (.!.. S (S2 - S - + ~)J s2 = .!..s s CONVOLUTIONS BY INTEGRATION Find by integration: 1. 1 * 1 yet) = cosh t. • Y(s) • * et 4. 5. 1 * cos wt 6. 1 8. s2 _ I and the solution is 13. S2(S2 + 14. 1) s (S2 + 16)2 (S2 + 1)(S2 *t 2. t 3. t 7. ekt * e- kt hence s2 - I we obtain by using the l)ls cancels on both sides, so that solving for Y simply gives Yes) = 11-81 _1_. = !ley), eat * e bt (a * f(t) sin t * cos t i= b) 15. 16. S(S2 - 9) 5 + 25) 17. (Partial fractions) Solve Probs. 9, 11, and 13 by using partial fractions. Comment on the amount of work. 19-161 INVERSE TRANSFORMS BY CONVOLUTION 9. 11. 1 (s - 3)(s --=-2-- s(s + 4) + 5) SOLVING INITIAL VALUE PROBLEMS 118-251 Find f(t) if 5£(f) equals: USing the convolution theorem, solve: 10. 12. 18. y" s(s - 1) -::-2- - S (s - 2) 19. y" 20. y" + + + y' (0) y = 4)' 5/ yeO) sin t. sin 3t, = + 4)' = 2e- = O. yeO) = 2t , /(0) = 0 o. /(0) yeO) = 0, 0 = 0 31f the upper limit of integration is variable, the equation is named after the Italian mathematician VITO VOLTERRA (1860-1940), and if that limit is collsta1l1, the equation is named after the Swedish mathematician IVAR FREDHOLM (1866-1927). "Of the second kind (fust kind)" indicates that y occurs (does not occur) outside of the integral. CHAP.6 254 Laplace Transforms + 9y = 8 sin t if 0 < t < IT and 0 if t > 7.; y'(O) = 4 22. y" + 3y' + 2y = 1 if 0 < t < a and 0 if t > a; yeO) = 0, y' (0) = 0 21. y" yeO) = 0, 23. y" + 4)' = 5u(t - I); y(O) 24. y" + 5/ + 6y = 8(t y'(O) = 0 25. y" + 6/ y(o) = L + 3); INTEGRAL EQUATIONS Using Laplace transforms and showing the details, solve: t 27. y(t) - = 0, y' (0) = 0 yeo) = 1, Jo Y( T) (IT = I f f t 28. y(t) + cosh (t - T) dT = t y( T) + e' o 8y = 28(t /(0) 127-341 I) + 28(t - 2); 29. = 0 y(T) - t sin (t - T) dT = cos t y( T) o t 26. TEAM PROJECT. Properties of Convolution. Prove: (a) Commutativity. f * g = g * f * g) 7- v = f * (g * v) Distrihutivity, f * (gl + g2) = f * gl + f * g2 30. y(t) 2 Y(T) cos (t - T) d. = cos t t 31. y(t) (b) Associativity, (f (c) (d) Dirac's delta. Derive the sifting formula (4) in Sec. 6.-1- by using h· with a = 0 [(1), Sec. 6.4] and applying the mean value theorem for integrals. Jo + J(t o Jo + Jo + T)Y(T) dT = 1 t 32. y(t) - y( T)(t - T) dT = 2 - 4t 2 t 33. y(t) 2e t e-TY(T) ciT = te' (e) Unspecified drhing force. Show that forced vibrations governed by ),'(0) = K2 with w =1= 0 and an unspecified driving force r(t) can be written in convolution form. I Y = - sin wt w 6.6 * ret) + Kl cos wt K2 + - w sin wt. 35. CAS EXPERIJ\iIENT. Variation of a Parameter. (a) Replace 2 in Prob. 33 by a parameter k and investigate graphically how the solution curve changes if you vary k, in particular near k = - 2. (b) Make similar experiments with an integral equation of your choice whose solution is oscillating. Differentiation and Integration of Transforms. ODEs with Variable Coefficients The variety of methods for obtaining transforms and inverse transforms and their application in solving ODEs is surprisingly large. We have seen that they include direct integration. the use oflinearity (Sec. 6.1), shifting (Secs. 6.1, 6.3), convolution (Sec. 6.5). and differentiation and integration of functions f(t) (Sec. 6.2). But this is not all. In this section we shall consider operations of somewhat lesser importance. namely. differentiation and integration of transforms F(s) and corresponding operations for functions f(t), with applications to ODEs with variable coefficients. Differentiation of Transforms It can be shown that if a function f(t) satisfies the conditions of the existence theorem in Sec. 6.1, then the derivative F' (s) = dF/ds of the transform F(s) = ::£(f) can be obtained by differentiating F(s) under the integral sign with respect to s (proof in Ref. LGR4] listed in App. 1). Thus, if F(s) = {Oo e-stf(t) dt, then F'(s) = - fOe-sttf(t) dt. o SEC. 6.6 255 Differentiation and Integration of Transforms. ODEs with Variable Coefficients Consequently, if SE(f) = F(s), then (1) = -F'(s), SE{tf(t)} SE-I{F'(s)} hence = -rf(t) where the second formula is obtained by applying SE- I on both sides of the first formula. In this way, differentiation of the transform of a function corresponds to the multiplication of the function by -to E X AMP L E 1 Differentiation of Transforms. Formulas 21-23 in Sec. 6.9 We shall derive the following three formulas. f(t) SE(f) 1 (2) (S2 + ,I (S2 + I (S2 + 1 --3 (32)2 2f3 s (3) (32)2 t sin f3t 2f3 (32)2 (sin f3t 2f3 S2 (4) (sin f3t - f3t cos f3t) 1 + f3t cos f3t) Solutioll. From (I) and formula 8 (with w = (3) in Table 6.1 of Sec. 6.1 we obtain by differentiation (CAUTION! Chain rule') Dividing by 2f3 and using the linearity of 5£. we obtain (3). Formulas (2) and (4) are obtained as follows. From (I) and formula 7 (with w = (3) in Table 6.1 we find (S2 +~) - 2s2 (5) f(t co~ f3t) = - From this and formula 8 (with w = 2 (s r:2 2 + p ) (3) in Table 6.1 we have ( I) 5£ t cos f3t ::':: - sin f3t f3 s2_f32 = (s 2 2 2 ::':: _,,2 + f3 ) , + f32 On the right we now take the common denominator. Then we see that for the plus sign the numerator becomes s2 - ~ + s2 + f32 = 2.. 2 , so that (4) follows by division by 2. Similarly. for the minus sign the numerator takes the form s2 - f32 - s2 - ~ = -2~. and we obtain (2). This agrees with Example 2 in Sec. 6.5. • Integration of Transforms Similarly, if f(t) satisfies the conditions of the existence theorem in Sec. 6.1 and the limit of f(t)/t, as t approaches 0 from the tight, exists, then for s > k, (6) 5£{ f;t) } = ["F('S) df hence In this way, illfegration of the tmllsform of a function f(t) corresponds f(t) by t. 10 the division of 256 CHAP. 6 laplace Transforms We indicate how (6) is obtained. From the definition it follows that and it can be shown (see Ref. [GR4] in App. 1) that under the above a:.:.umptions we may reverse the order of integration, that is, Integration of e-st with respect to equals e-st/t. Therefore, s gives e-st/( -t)o Here the integral over s on the right to fO 5£{ J(t) } F(s) d'S = e- st J(t) dr = s o t E X AMP l E 2 (s> k). t • Differentiation and Integration of Transforms W2) (I + 7 Find the inverse transform of In Solution. = In s2 + w2 --S-2- Denote the given transfonn by F(s). Its derivative is 2 2 2) 2s d ( In (s + w ) - In s = -2--2 s s + w , F (s) = -d - 2s "2 . s Taking the inverse transform and using (I), we obtain ;e- II F '} (5) =;e- I{ - 22s --2 s + w Hence the inverse fell of H,I') is fO) Alternatively, if we let G(s) 2s = - s + w 2} = - s 2 cos wt - 2 = -tf(t). 2(1 - cos wt)/t. This agrees with formula 42 in Sec, 6.9. = 2 -2---2 - - s g(t) = ;e-1(G) = 2(cos wr - then • 1). From this and (6) we get. in agreement with the answer just obtained, In s 2 +w 2 J 00 glt) G(s) ds = - s s t --2- = = 2 - (1 - cos wt). t the minus occurring since s is the lower limit of integration. In a similar way we obtain formula 43 in Sec. 6.9, ;e-1 fin (I - ::)} = ~ (I - cosh at). • Special Linear ODEs with Variable Coefficients Formula (I) can be used to solve certain ODEs with variable coefficients. The idea is this, Let 5£(y) = Y. Then 5£(/) = sY - yeO) (see Sec. 6.2). Hence by (I), (7) , d 5£(ty ) = - - ds [sY - yeO)] dY -Y- s - . ds SEC. 6.6 257 Differentiation and Integration of Transforms. ODEs with Variable Coefficients Similarly, :iCy") = s2y - sy(O) - y' (0) and by (1) (8) dY :i(ty" ) =d - [ - 2 s Y - sy(O) - y ' (0) ] = -2sY - s 2 ~ ~ + y(O). Hence if an ODE has coefficients such as at + b, the subsidiary equation is a first-order ODE for Y, which is sometimes simpler than the given second-order ODE. But if the latter has coefficients at2 + bt + c, then two applications of (1) would give a second-order ODE for Y, and this shows that the present method works well only for rather special ODEs with variable coefficients. An important ODE for which the method is advantageous is the following. E X AMP L E 3 Laguerre's Equation. Laguerre Polynomials Laguerre's ODE is + (I ty" (9) We determine a ,olution of (9) with [ 11 = O. t)y' + ny - = O. I. 2..... From (7)-(9) we get the subsidiary equation dY . 2sY_s2dY +y(O)] +sY-y(O)- (_Y_S ) +IIY=O. ds ds Simplification gives 2 dY (s - s ) ds + + I - slY = O. (n Separating variables, using partial fractions, integrating (with the constant of integration taken zero), and taking exponentials. we get (10*) dY Y =_ II + I s - s ds (_'_1__ ~) ds I = s2 S - and s We write In = Y-\Y) and prove Rodrigues's formula 10 = I, (10) II = 1,2,···. These are polynomials because the exponential terms cancel if we perform the indicated differentiations. They are called Laguerre polynomials and are usually denoted by Ln (see Problem Set 5.7, but we continue to reserve capital letters for transforms). We prove (10). By Table 6.1 and the first shifting theorem (s-shifting), hence by (3) in Sec. 6.2 because the derivatives up to the order II (10) and then (10*)] - I are zero at O. Now make another shift and divide by n! to get [see ;£(1,,) 11-121 TRANSFORMS BY DIFFERENTIATION Showing the details of your work. find 5£(f) if f(t) equals: 1. 4te t 3. t sin wt 2. - t cosh 2t 4. t cos (t + = s = 5. te- 2t ~in t 7. t 2 sinh 4t 2 sin wt 11. t sin (t + k) 9. t k) (s _ I)n ~ • Y. 6. 8. 10. t 2 sin 3t tne kt t cos WI 12. te -kt sin I 258 CHAP.6 Laplace Transforms I 13-20 INVERSE TRANSFORMS Using differentiation, integration. s-shifting. or convolution (and showing the details). find f(t) if 5£(f) equals: 6 s 13. 14. (s 15. 17. + 1)2 2(s [(s + (S2 2) + 2)2 + 16. 1]2 16)2 s (S2 _ 1)2 and calculate 10 , (c) Calculate 10 , ' II=I-tby s+a 18. I n - s+b 2 (s - k)3 s s 19. I n - S - + (b) Show that 20. arccot - 1 w 21. WRITING PROJECT. Differentiation and Integration of Functions and Transfonns. Make a shOlt draft of these four operations from memory. Then compare your notes with the text and write a report of 2-3 pages on these operations and their significance in applications. 22. CAS PROJECT. Laguerre Polynomials. (a) Write a CAS program for finding In(t) in explicit form from (10). Apply it to calculate 10 , ••• , /10 , Verify that 10 , • • . ,110 satisfy Laguerre's differential equation (9). (11 + ••• , ••• 110 from this formula. 110 recursively from 10 = 1, 1)ln+l = (2/l + I - t)ln - /lIn_I' (d) Experiment with the graphs of 10 , •••• 110 , finding out empirically how the first maximum. first minimum. ... is moving with respect to its location as a function of II. Write a short report on this. (e) A generating function (definition in Problem Set 5.3) for the Laguerre polynomials is L I,,(t)x n = (1 - X)-IetX/(X-ll. n=O Obtain 10 , • • . ,110 trom the corresponding partial sum of this power series in x and compare the In with those in (a), (b), or (e) . f,.7 Systems of ODEs The Laplace transform method may also be used for solving systems of ODEs. as we shall explain in terms of typical applications. We consider a first-order linear system with constant coefficients (as discussed in Sec. 4.1) (1) Writing Y 1 = ~(Yl)' Yz = ':£(yz)· G 1 Sec. 6.2 the suhsidiary system = ':£(gl)' Gz = .:£(gz), we obtain from (I) in By collecting the Y r and Yz-tenns we have (2) By solving this system algebraically for Y1 (s), Yz(s) and taking the inverse transform we obtain the solution h = ~-l(Yl)' yz = ~-l(yZ) of the given system (I). SEC. 6.7 259 Systems of ODEs Note that (1) and (2) may be written in vector form (and similarly for the systems in the examples); thus, setting y = [h YZ]T, A = [ajk], g = [gl g2]T, Y = [Y1 Yz]T, G = [G 1 GZ]T we have y' = Ay E X AMP L E 1 +g (A - sI)Y and = -yeO) - G. Mixing Problem Involving Two Tanks Tank Tl in Fig. 142 contains initially 100 gal of pure water. Tank T2 contains initially 100 gal of water in which 150 Ib of salt are dissolved. The inflow into Tl is 2 gal/min from T2 and 6 gal/min containing 6 Ib of salt trom the outside. The inflow into T2 is 8 gal/min from T 1. The outflo\>; from T2 is 2 + 6 = 8 gal/min. as shown in the figure. The mixtures are kept unifonn by stirring. Find and plot the salt contents )"1(1) and Y2(t) in Tl and T2 , respectively. Solutioll. The model is obtained in the form of two equations Time rate of change = Inflow/min - Outflow/min for the two tanks (see Sec. 4.1). Thus, , )'1 8 = - Wo , 2 Y1 + 100)'2 + 6, 8 )'2 = 8 Wo ."1 - Wo ."2' The initial conditions are )'1(0) = 0, )'2(0) = 150. From this we see that the subsidiary system (2) is 6 (-0.08 - s)Y1 + + s (-0.08 - S)Y2 = -150. We solve this algebraically for YI and Y2 by elimination (or by Cramer's rule in Sec. 7.7), and we write the solutions in terms of partial tractions, 9s + 0.48 YI = - - - - - - - s(s Y2 = + 0.12)(s 150s2 s(s + + 12s 0.12)(s 100 + 0.(4) + 0.48 + 0.04) 62.5 s s + 0.12 37.5 s + 0.04 100 125 75 s s + 0.12 s + 0.04 --+ By taking the inverse transfonn \>;e arrive at the solution )'1 = 100 - 62.5e- O. 12t - 37.5e-O.04t )'2 = 100 + 125e- O. 12t - 75e- o.04t . Figure 142 shows the interesting plot of these functions. Can you give physical explanations for their main features? Why do they have the limit 100? Why is ."2 not monotone, whereas Yl is? Why is )'1 from some time on suddenly larger than Y2? Etc. • y(tl 2 gal/min 5u Fig. 142. l/min 50 Mixing problem in Example I 100 150 200 260 CHAP.6 laplace Transforms Other systems of ODEs of practical importance can be solved by the Laplace transform method in a similar way, and eigenvalues and eigenvectors as we had to determine them in Chap. 4 will come out automatically, as we have seen in Example 1. E X AMP l E 2 Electrical Network Find the currents ;1(1) and ;2(1) in the network in Fig. 143 with Land R measured in terms of the usual units (see Sec. 2.9). U(I) = 100 volts if 0 ~ I ~ 0.5 sec and 0 thereafter, and ;(0) = 0, ;' (0) = o. L J =0.8H 0.5 1.5 2 2.5 3 Currents Network Fig. 143. Electrical network in Example 2 Solution. The model of the network is obtained from Kirchhoff's voltage law as in Sec. 2.9. For the lower circuit we obtain 0.8;~ + l(i} - ;2) + 1.4i} = 100[1 - U(I - i)l and for the upper O. = Division by 0.8 and ordering gives for the lower circuit ;~ + 3;1 - 1.25;2 = 125[1 - i)l 11(1 - and for the upper With i}(O) = O. ;2(0) = 0 we obtain from (I) in Sec. 6.2 and the second shifting theorem the subsidiary system (s 1_ 2 + 3)/ 1.25/ = 125 (+ _ -:/2) e -I} + (s + 1)12 = O. Solving algebraically for I} and 12 gives I} = 12 = 125(s + I) s(s + !)(s + ~) s(s + i)(s + ~) 125 e (1 - -s/2 ), (l - e -s/2). The right sides without the factor I - e -S/2 have the partial fraction expansions 500 h and 125 - 3(s + 625 i) - 21(s + ~) SEC. 6.7 261 Systems of ODEs 500 250 250 7s 3(s + !) + ------,,21(s + ~) respectively. The inverse transform of this gives the solution for 0 ~ t ~ ;1(t) = - -- 125 3 e- t12 ;2(t) = - -- 250 3 e- tJ2 625 --e -7t/2 21 + !, 500 7 (0 ~ t ~ According to the second shifting theorem the solution for t > I is ;1(t) - 125 3 ;1(t - !) and ;2(t) - ;2(t -I), that is, 625 21 ;1(t) = - - - (I - e 1l4 )e- tl2 - - - (I - e7/4)e-7t12 (t 250 i). 500 -7t/2 + -250 -e + -21 7 ;2(t) = - -3- (1 - e 1l4 )e-t/2 250 + ~ > i) (I - e7/4 )e-7t/2 Can you explain physically why both currents eventually go to zero, and why i 1(t) has a sharp cusp whereas ;2(t) has a continuous tangent direction at t = I? • Systems of ODEs of higher order can be solved by the Laplace transform method in a similar fashion. As an important application, typical of many similar mechanical systems, we consider coupled vibrating masses on springs. l' o~ mj ~, Yj Q °l 1, Y2 Fig. 144. E X AMP L E 3 =I m 2 =1 Example 3 Model of Two Masses on Springs (Fig. 144) The mechanical system in Fig. 144 consists of two bodies of mass I on three springs of the same spring constant k and of negligibly small masses of the springs. Also damping is assumed to be practically zero. Then the model of the physical system is the system of ODEs Y~ = -kYI + k(Y2 - Yl) (3) Y; = -k(Y2 - Yl) - kY2· Here Yl and Y2 are the displacements of the bodies from their positions of static equilibrium. These ODEs follow from Newton's second law, Mass X Acceleration = Force, as in Sec. 2.4 for a single body. We again regard downward forces as positive and upward as negative. On the upper body, -kyl is the force of the upper spring and k(Y2 - Yl) that of the middle spring, Y2 - Yl being the net change in spring length-think this over before going on. On the lower body, -k(Y2 - Yl) is the force of the middle spring and -ky2 that of the lower spring. 262 CHAP. 6 Laplace Transforms We shall determine the solution corresponding to the initial conditions )"1(0) = I, Y2(0) = I. y~(O) = V3k, y~(O) = -V3k. Let Y 1 = ~(Yl) and Y2 = ~(Y2)' Then from (2) in Sec. 6.2 and the initial conditions we obtain the subsidiary system s2Yl - s - V3k = -kYl s2Y2 - s + "\'3i: + k(Y2 = -k(Y2 - Y1 ) - Y1) kY2. - This system of linear algebraic equations in the unknowns Y1 and Y2 may be written (S2 + 2k)Yl -kYI kY2 + (,,2 + S = + V3i: 2k)Y2 = s - V3k. Elimination (or Cramer's rule in Sec. 7.7) yields the solution, which we can expand in terms of partial fractions. (s + V3k)(S2 + 2k) + k(s - V3k) (s2 (S2 + 2k)2 s - k2 + k + s2 + 3k S2 + 2k)(s - V3k) + k(s + V3k) (s2 + 2k)2 - k 2 s Hence the solution of our initial value problem is (Fig. 145) Yl(t) = ~-l(Yl) = cos Ykt + sin V3kt Y2(t) = ~-I(Y2) = cos Ykt - sin V3kt. We see that the motion of each ma,s is harmonic (the system is undamped !), being the superposition of a "slow" oscillation and a "rapid" oscillation. • 0, "1jJ 2, -2 Fig. 145. Solutions in Example 3 ..-=.. lE5-H 6 7 ~ 11-201 SYSTEMS OF ODES Usmg the Laplace transform and showing the details of YOUT work, solve the initial value problem: 1. )'~ = -)'1 - Y1(0) = 0, )'2, , )'2 = )'1 - )'2, Y2(0) = I 2. Y~ = 5Yl + Y2' y~ = Y1 + 5Y2' Y2(0) = -3 Yl(O) = 1, 3. Y~ = -6)'1 + 4Y2, )'1(0) = -2, 4. y~ + Y2 = )"1(0) = 1, 0, Y~ = -4Yl )'2(0) = -7 4V2, Yl + y~ = 2 cos T, h(O) = 0 5. y~ = -4Y1 - 2Y2 Yl(O) = 5.75, + t, Y; = 3\'1 + Y2(0) = -6.75 6. y~ = 4Y2 - 8 cos 4t, )'1(0) = 0, + y; = -3Yl - Y2(0) = 3 )'2 - t, 9 sin 4t, SEC 6.7 7. y~ 263 Systems of ODEs 5)'1 - = 17t 2 - )'~ = lOy! - 7Y2 - Yl (0) = 2, 8. y~ = 6Yl )'1(0) = Y2(0) + FURTHER APPLICATIONS 9t 2 + 2t, 4)'2 - 2t, 22. (Forced vibrations of two masses) Solve the model in Example 3 with k = 4 and initial conditions Yl(O) = 1, )'~(O) = 1, Y2(0) = 1, y~(O) = -1 under the assumption that the force II sin t is acting on the first body and the force -11 sin t on the second. Graph the two curves on common axes and explain the motion physically. = 0 y~ = 9Yl + 6)'2' Y2(0) = -3 Y2, -3, 9. y~ = 5Yl + 5)'2 - 15 cost + 27 sint, y~ = -lOVI - 5Y2 - 150 sin t. Yl(O) = 2, Y2(0) = 2 10. y~ = . 2Yl + 3Y2, y~ = 4)'1 Yl(O) = 4, Y2(0) = 3 11. y~ = Y2 y~ = )'2(0) + -)'1 = 1 - u(t - + I - 13. y~ = Yl + = 0, 14. y~ = Y~ = -3)'1 -4Yl = 16. )": = 0, 2)e 4t , )'2(0) Y; -)'2, Yl(O) = YI(O) = 0, Y; = y~(O) = 1)e t , t l)e , Y; = 2.\'1 - 5Y2, 0, Y2(0) = 3, y;(O) 17. y~ = 4Yl + 8)'2, Y; = 5Yl + .1'2' Yl(O) = 8, )'~(O) = -18, ),;(0) = -21 18. y~ +)'2 = -101 sin lOt, Yl(O) = 0, y~(O) = -6 19. y~ + y~ y~ + y~ = = Y1(0) = 0, 20. 4y~ 2.\'2, = 1 + Y2 + u(t + 2Y2 + u(t Y2(0) = 3 = 1, + )'1 I), -Yl + 2[1 - u(t - 217)] cos t, Y2(0) = 0 = -2YI + 2)'2, .\'1(0) I), 1), y~ = 4Yl + 2Y2 + 64tu(t Y2(0) = 0 6u(t - Yl(O) = 1, 15. y~ Y2, - 0 12. y~ = 2Yl + Y2, Yl(O) = 2, .\'1(0) u(t - 23. CAS Experiment. Effect of Initial Conditions. In Prob. 22, vary the initial conditions systematically, describe and explain the graphs physically. The great variety of curves will surprise you. Are they always periodic? Can you find empirical laws for the changes in terms of continuous changes of those conditions? Y; Y~(O) = 6, )'2(0) = 1, 0 Y2(0) = 5, 2 sinh t, = 2H = 4H Network ."3(0) = 1 -2y~ + y~ + 25. (Electrical network) Using Laplace transforms, find the currents i 1 (t) and i2 (t) in Fig. 146, where u(t) = 390 cos l and i 1(0) = 0, i 2(0) = O. How soon will the currents practically reach their steady state? Yl = 101 sin lOt, Y2(0) = 8, y~ + Y~ 2e t + e- t , et y~ - 2y~ = O. 2y~ - 4y~ = -I6t y](O) = 2, Y2(O) + = 24. (Mixing problem) What will happen in Example I if you double all flows (in particular, an increase to 12 gal/min containing 12 Ib of salt from the outside), leaving the size of the tanks and the initial conditions as before? First guess, then calculate. Can you relate the new solution to the old one? i(t) l. 40 = 0, Y3(0) = 0 20 21. TEAM PROJECT. Comparison of Methods for Linear Systems of ODEs. -20 (a) Models. Solve the models in Examples 1 and 2 of Sec. 4.1 by Laplace transfonns and compare the amount of work with that in Sec. 4.1. (Show the details of your work.) -40 (b) Homogeneous Systems. Solve the systems (8), (11)-(13) in Sec. 4.3 by Laplace transfonns. (Show the details.) (c) Nonhomogeneous System. Solve the syslem (3) in Sec. 4.6 by Laplace transfonns. (Show the details.) Currents Electrical network and currents in Problem 25 Fig. 146. 26. (Single cosine wave) Solve Prob. 25 when the EMF (electromotive force) is acting from 0 to 217 only. Can you do this just by looking at Prob. 25, praclically without calculation? 264 6.8 CHAP. 6 Laplace Transforms Laplace Transform: General Formulas Fonnula F(s) = ~(f(t)} = Name, Comments x L e-stf(t) dl Definition of Transfonn 0 6.1 f(t) = ~-lIF(s)} ~(af(t) + hg(t)} = a~{f(t)} Inverse Transform + h~(g(t)} = F(s - a) ~{eO'tf(t)} :;P-l{F(s - a)} Sec. = eatf(t) Linearity 6.1 s-Shifting (First Shifting Theorem) 6.1 ;£(f') = s~(f) - f(O) ~(f") = s2~(f) - sf(O) - 'f(t n ) = sn~(f) t' (0) - s<n-l)f(O) - ... Differentiation of Function 6.2 ... - tn-0(O) ~ {{f(T) dT} = ~ ~(f) * g)tt) = (f I I Integration of Function t f( T)g(t - T) dT 0 = t f(t - T)R(T) dT Convolution 6.5 t-Shifting (Second Shifting Theorem) 6.3 0 ~(f ~(f(t * g) = ~(f)~(g) - a) u(t - a)} = e-asF(s) ~-l(e-asF(s)} = f(t - a) u(t - (I) 'f(tf(t)} = -F'(s) ~ { f~t) } ~(f) = = 1 _ 1e-Ps (0 I 6.6 F{S) di' r e-stf(t) dt 0 Differentiation of Transform Integration of Transform f Periodic with Period p 6.4 Project 16 265 SEC. 6.9 Table of Laplace Transforms 6.9 Table of Laplace Transforms For more extensive tables, see Ref. [A9] in Appendix 1. F(s) = ~(f(t)} 1 lis 2 lIs2 3 lIs n 4 1/V:;: 5 lIs3/2 (n = 1,2, ... ) 8 9 10 11 12 Sec. t tn-l/(n - 1)! 6.1 lIYm 2vthr 6 7 f(t) ta-l/f(a) (a> 0) - s - a (s - a)2 6.1 1 (s - a)n 1 (n=I,2,···) (k> 0) (s - a)(s - b) (a * b) (a * b) s (s - a)(s - b) 1 - - - (ae at - (a - b) bebt ) --+------------~-------------+---- 13 14 1 s sin wt cos wt 1 - sinh at a s 17 18 1 w 15 16 - 6.1 cosh al - 1 eat w s-a eat sin wI cos wt 1 19 2 w 20 3 21 --3 I w (l - cos wt) (wt - sin wt) 1 2w (sin wt - wt cos wt) 6.6 ( continued) 266 CHAP. 6 Laplace Transforms Table of Laplace Transforms (continued) F(s) = !f{f(t)} 22 W 2)2 t sinwt 2w 1 (sin wt 2w s + (S2 S2 23 24 25 26 27 28 29 30 (S2 + W 2)2 (S2 + £l2)(S2 s + b 2) (a 2 =t- b 2) 1 2 b - a 1 1 S4 + 4k4 S4 + Sec. J(t) -3 4k 2 } 66 + wt cos wt) (cos at - cos bt) (sin kt cos kt - cos kt sinh kt) 1 2k2 sin kt sinh kt s 4k4 1 1 -3 s4 - k4 2k s 1 -2 S4 - k4 2k ~-Vs=b 1 ~~ (sinh kt - sin kt) (cosh kt - cos kt) 1 _ (e bt _ eat) __ 2v:;;(i - b l) e-(a+b)t/21o (a -2 5.6 Jo(at) 5.5 1 31 + VS2 a2 I 32 s (s - a)3/2 33 1 (S2 _ a 2)k v:;;t eat(l V; r(k) ( -t- 7ft (k> 0) 2a + r- 2at) 1 2 / 1 at k-1/2( ) 5.6 e-as/s u(t - a) e- as ti(t - a) 6.3 6.4 36 -1 e -k/s fo(2Vkt) 5.5 37 -1e -kls 38 _ 39 e- kVs 40 1 -Ins S 34 35 s 1 v:;;t 7ft Vs 1 ~/2 cos 2Vkt 1 e kls v:;;:k sinh 2Vkt 7fk (k> 0) - -k - e _k2/4t 2v:;;(i -int-I' (1' = 0.5772) 5.6 (continued) 267 Chapter 6 Review Questions and Problems Table of Laplace Transforms (continued) F(s) = ~ (f(t)} f(t) 1 .\'-a 41 In-- 42 In 43 In 44 w arctan - _ (e bt _ eat) s-h .1'2 + t w2 2 -(1 - cos wt) .\'2 2 -(1 - cosh at) .\'2 t I - sin wt t .I' 45 - I 6.6 t a2 .1'2 - Sec arccot s App. A3.1 Si(t) .I' ======- = : .:.'= -= === :_:=. -£¥IE:--w.::::Q--u::E-"S T ION 5 1. What do we mean by operational calculus? 2. What are the steps needed in solving an ODE by Laplace transform? What is the subsidiary equation? 3. The Laplace transform is a linear operation. What does this mean? Why is it important? AND PRO B L EMS 13. sin 2 t 15. tu(t - '17) 17. e t * cos 2t 19. sin t + sinh t 21. eat _ e bt (a 14. cos 2 4t 2'17) sin t 16. l/(t - "* h) 18. (sin wt) * (cos wt) 20. cosh 1 - cos t 22. cosh 2t - cosh t 4. For what problems is the Laplace transform preferable over the usual method? Explain. INVERSE LAPLACE TRANSFORMS 5. What are the unit step and Dirac's delta functions"? Give examples. Find the inverse transform (showing the details of your work and indicating the method or formula used): 6. What is the difference between the two shifting theorems? When do they apply? 23. 10.1' +2 .1'2 + 4.\' + 20 7. Is .'£[f(t)g(t») = .'£ [f(t)}.'£{g(t))? Explain. 8. Can a discontinuous function have a Laplace transform? Does every continuous function have a Laplace transform? Give reasons. 9. State the transforms of a few simple functions from memory. 25. 29. 111-221 31. Find the transform (showing the details of your work and indicating the method or formula you are Using): 11. te 3t 12. e- t sin 2t 12 5.1' + 4 - e- 2s 27. - .1'2 10. If two different cuntinuous functions have transforms, the latter are different. Why is this practically important? LAPLACE TRANSFORMS 24. .1'2 33. (S2 2s + 4 + 4.1' + 5)2 (.~ +2 .1'3 ) e-s 26. 15 3.1' .1'2-2.1'+2 2.1' - 10 28. .1'3 32. (.1'2 180 + w2 ) e- 5s 16 .1'2 - 30. + 16)2 + 18.1'2 + .1'7 2 IT .1'2(.1'2 4 .1'2 - 34. 2s2 + 2.1' + I 3.1'4 268 CHAP.6 135-501 Laplace Transforms o< t < SINGLE ODEs AND SYSTEMS OF ODEs Solve by Laplace transforms. showing the details and graphing the solution: 35. y" + Y = uCt y' (0) = 20 - yeO) = 1). 36. y" + L6y = 48(r - 'iT), y'(O) = 0 37. y" + 4)" = 88(r - 5), y' (0) 38. y" + 'iT, v(t) = 0 if t 'L yeO) = -1, /(0) = 0 39. y" + 2y' + vet) yeO) = 10, 2). \'(0) = 0, lOy = 0, + 4/ + y'(o) = -5 LC-circuit 53. (RLC-circuit) Find and graph the current i(t) in the RLC-circuit in Fig. 149, where R = 1 n. L = 0.25 H, C = 0.2 F, v(t) = 377 sin 20t V, and CUlTent and charge yeo) = 7, / (0) = -1 40. y" L Fig. 148. Y = 1I(t - and CUlTent and charge at 'iT, o. -1 = > t = 0 are O. at t = 0 are O. yeo) = 5. 5y = SOt. 41. y" - y' - 2y = 12u(t - 'iT) sin t. y(O) = 1, y' (0) = -1 42. y" - 2/ yeO) = + y = t8(t - 1), o. ),'(0) = 0 43. y" - 4/ + 4y = 8(t - 1) - 8(r - 2). yeO) = o. y' (0) = 0 44. y" + 4y = 8(t - iT) - 8(t - 2 'iT), yeO) = I, /(0) = 0 45. Y~ + Y2 = sin t, y~ + Yl = -sint, vet) Fig. 149. 54. (Network) Show [har by Kirchhoff's voltage law (Sec. 2.9), the CUlTents in the network in Fig. L50 are obtained from the system Y2(0) = 0 h(O) = 1, 46. y~ = -3Yl + Y2 - 12t, y~ = -4YI Yl (0) = 0, Y2(0) = 0 + 2Y2 + 12t, )'~ = J2, h(O) = = 0, -4Yl + 8Ct - )"2' = 2. = L6Y2, }'~ = 16Y1, h(O) = 2. y~(O) = 12, R(i l .f 11) i 2 ) = v(t) - 1. + C 12 = O. Solve this system. where R = 1 n, L = 2 H. C = 0.5 F. v(t) = 90e- t /4 V. i 1 (Q) = O. i 2 (Q) = 2 A. 'iT). y~(O) = 3 50. y~ y~(O) = 4 MODELS OF CIRCUITS AND NETWORKS 51. (RC-circuit) Find and graph the CUlTent i(t) in the RCcircuit in Fig. 147, where R = 100 n, C = 10- 3 F, v(t) = 100rV if 0 < t < 2, v(t) = 200 V if t > 2 and the initial charge on the capacitor is O. ~c R + R(12 - )'2(0) = 0 49. y~ = 4Y2 - 4e t . y~ = 3Yl + )'1 (0) = 1. Y ~ (0) = 2. Y2(O) Li~ ., 47. y~ = Y2, y~ = -5Yl - 2Y2, h (0) = 0, Y2(0) = 1 48. y~ RLC-circuit L w)c:~r oJ Fig. 150. Network in Problem 54 55. (Network) Set up the model of the network in Fig. 151 and find and graph the CUlTents. assuming that [he currents and the charge on the capacitor are 0 when the switch is closed at t = O. L=lH vet) Fig. 147. c= 0.01 RC-circuit 52. (LC-circuit) Find and graph the charge q(t) and the current i(t) in the LC-circuit in Fig. 148, where L = 0.5 H, C = 0.02 F, v(t) = 1425 sin 51 V if Switch Fig. 151. R2 = 30 n Network in Problem 55 F 269 Summary of Chapter 6 Laplace Transforms The main purpose of Laplace transforms is the solution of differential equations and systems of such equations, as well as corresponding initial value problems. The Laplace transform Hs) = :£(f) of a function f(t) is defined by F(s) (1) = :£(f) = L'X)e-stf(t) dt (Sec. 6.1). o This definition is motivated by the property that the differentiation of f with respect to t corresponds to the multiplication of the transform F by s; more precisely, 5£(f') = s:£(f) - f(O) (2) :£(f") = s2:£(f) - sf(O) - (Sec. 6.2) f' (0) etc. Hence by taking the transform of a given differential equation y" + ay' + by = ret) (a, b constant) and writing :£(y) = Yes), we obtain the subsidiary equation (4) (S2 + as + b)Y = :£(r) + sf(O) + t' (0) + afCO). Here, in obtaining the transform :£(r) we can get help from the small table in Sec. 6.1 or the larger table in Sec. 6.9. This is the first step. In the second step we solve the subsidiary equation algebraically for Y(s). In the third step we determine the inverse transform yet) = :;e-l(y). that is, the solution of the problem. This is generally the hardest step. and in it we may again use one of those two tables. Yes) will often be a rational function, so that we can obtain the inverse :£-1(Y) by partial fraction reduction (Sec. 6.4) if we see no simpler way. The Laplace method avoids the determination of a general solution of the homogeneous ODE. and we also need not determine values of arbitrary constants in a general solution from initial conditions; instead, we can insert the latter directly into (4). Two further facts account for the practical importance of the Laplace transform. First, it has some basic properties and resulting techniques that simplify the determination of transforms and inverses. The most important of these properties are listed in Sec. 6.8, together with references to the corresponding sections. More on the use of unit step functions and Dirac's delta can be found in Secs. 6.3 and 6.4, and more on convolution in Sec. 6.5. Second, due to these properties, the present method is particularly suitable for handling right sides r(t) given by different expressions over different intervals of time, for instance, when ret) is a square wave or an impulse or of a form such as ret) = cos t if 0 ~ t ~ 47T and 0 elsewhere. The application of the Laplace transform to systems of ODEs is shown in Sec. 6.7. (The application to PDEs follows in Sec. 12.11.) PA RT _.1 B Linear Algebra. Vector Calculus .~ '.' • CHAPTER 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems CHAPTER 8 Linear Algebra: Matrix Eigenvalue Problems CHAPTER 9 Vector Differential Calculus. Grad, Div, Curl CHAPTER 10 Vector Integral Calculus. Integral Theorems Linear algebra in Chaps. 7 and 8 consists of the theory and application of vectors and matrices, mainly related to linear systems of equations, eigenvalue problems, and linear transformations. Linear algebra is of growing importance in engineering research and teaching because it forms a foundation of numeric methods (see Chaps. 20-22), and its main instruments, matrices, can hold enormous amounts of data-think of a net of millions of telephone connections-in a form readily accessible by the computer. Linear analysis in Chaps. 9 and 10. usually called vector calculus, extends differentiation of functions of one variable to functions of several variables-this includes the vector differential operations grad, div, and curl. And it generalizes integration to integrals over curves, surfaces, and solids, with transformations of these integrals into one another, by the basic theorems of Gauss, Green, and Stokes (Chap. 10). Software suitable for linear algebra (Lapack, Maple, Mathematica, Matlab) can be found in the list at the opening of Part E of the book if needed. Numeric linear algebra (Chap. 20) can be studied directly after Chap. 7 or 8 because Chap. 20 is independent of the other chapters in Part E on numerics. 271 CHAPTER .. ., ,-, • '\ 7 -~ Linear Algebra: Matrices, Vectors, Determinants. Linear Systems This is the first of two chapters on linear algebra, which concerns mainly systems of linear equations and linear transformations (to be discussed in this chapter) and eigenvalue problems (to follow in Chap. 8). Systems of linear equations, briefly called linear systems, arise in electrical networks, mechanical frameworks. economic models_ optimization problems, numerics for differential equations, as we shall see in Chaps. 21-23, and so on. As main tools. linear algebra uses matrices (rectangular arrays of numbers or functions) and vectors. Calculations with matrices handle matrices as single objects, denote them by single letters, and calculate with them in a very compact form, almost as with numbers, so that matrix calculations constitute a powerful "mathematical shorthand". Calculations with matrices and vectors are defined and explained in Secs. 7.1-7.2. Sections 7.3-7.8 center around linear systems, with a thorough discussion of Gauss elimination, the role of rank. the existence and uniqueness problem for solutions (Sec. 7.5), and matrix inversion. This also includes determinants (Cramer's rule) in Sec. 7.6 (for quick reference) and Sec. 7.7. Applications are considered throughout this chapter. The last section (Sec. 7.9) on vector spaces, inner product spaces, and linear transformations is more abstract. Eigenvalue problems follow in Chap. 8. COMMENT. Numeric linear algebra (Sees. 20.1-20.5) call be studied immediately after this chapter. Prerequisite: None. Sections thatma)" be omitted in a short course: 7.5, 7.9. References lind Answers to Problems: App. I Part B, and App. 2. 7.1 Matrices, Vectors: Addition and Scalar Multiplication In this ~ection and the next one we introduce the basic concepts and rules of matrix and vector algebra. The main application to linear systems (systems of linear equations) begins in Sec. 7.3. 272 SEC. 7.1 Matrices, Vectors: Addition and Scalar Multiplication 273 A matrix is a rectangular array of numbers (or functions) enclosed in brackets. These numbers (or fUnctions) are called the entries (or sometimes the elements) of the matrix. For example, -0.2 ~:J ' [0~3 (1) x [ee 6x 2 2X 4x J, a,,] ran a12 a2l a 22 °23 a3l a32 a33 [al a2 a3]' [:J are matrices. The first matrix has two rows (horizontal lines of entries) and three columns (vertical lines). The second and third matrices are square matrices, that is, each has as many rows as columns (3 and 2, respectively). The entries of the second matrix have two indices giving the location of the entry. The first index is the number of the row and the second is the number of the column in which the entry stands. Thus, a23 (read a 111'0 three) is in Row 2 and Column 3, etc. This notation is standard, regardless of whether a matrix is square or not. Matrices having just a single row or column are called vectors. Thus the fourth matrix in (l) has just one row and is called a row vector. The last matrix in (1) has just one column and is called a column vector. We shall see that matrices are practical in various applications for storing and processing data. As a first illustration let us consider two simple but typical examples. E X AMP L E 1 Linear Systems, a Major Application of Matrices In a system of linear equations, briefly called a linear system, such as the coefficients of the unknowns Xl, X2, X3 are the entries of the coefficient matrix, call it A, The matrix 6 9 o -2 -8 is obtained by augmenting A by the right sides of the linear system and is called the augmented matrix of the system. In A the coefficients of the system are displayed in the pattern of the equations. That is, their position in A corresponds to that in the system when written as shown. The same is true for A. We shall see that the augmented matrix A contains all the informatIon about the solutions of a system, so that we can solve a system just by calculations on its augmented matrix. We shall discuss this in great detail, beginning in Sec. 7.3. Meanwhile you may verify by substitution that the solution is xl = 3, x2 = !, X3 = -1. The notation letters. Xl, X2, X3 for the unknowns is practical but not essential; we could choose x, y, Z or some other • 274 E X AMP L E 2 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Sales Figures in Matrix Form Sales figures for three products I. II. !II in a store on Monday (M). Tuesday (T). ... may for each week be arranged in a matrix M A= [7 100 T Vv Th F 330 810 0 210 470] 120 780 SOO SOO 960 II 0 0 270 430 780 III S If the company has ten stoTes. we can set up ten such matrices, one for each store. Then by adding corresponding entries of these mmrices we can get a mmrix ,howing the IOtal sale~ of each product on each day. Can you think of other data for which matrices are feasible? FOT instance. in transportation or storage problems? Or in recoTding • phone calls. or in li,ting distances in a network of roads? General Concepts and Notations We shall denote matrices by capital boldface letters A, B, C. ... ,or by writing the general entry in brackets; thus A = [ajk], and so on. By an m x 11 matrix (read 171 by n matrix) we mean a matrix with m rows and n columns-rows come always first! m X 11 is called the size of the matrix. Thus an 17l X 11 matrix is of the form (2) The matrices in (I) are of sizes 2 X 3.3 X 3,2 X 2, I X 3. and 2 X l. respectively. Each entry in (2) has two subscripts. The first is the row number and the second is the column number. Thus (/21 is the entry in Row 2 and Column I. If m = n, we call A an n X n square matrix. Then its diagonal containing the emries a11, a22, . . . , ann is called the main diagonal of A. Thus the main diagonals of the two square matrices in (1) are an, (/22' a33 and e- x , 4x, respectively. Square matrices are particularly important. as we shall see. A matrix that is not square is called a rectangular matrix. Vectors A vector is a matrix with only one row or column. Its entries are called the components of the vector. We shall denote veCIOrs by lowercase boldface letters a, b, ... or by its general component in brackets, a = [OJ], and so on. Our special vectors in (I) suggest that a (general) row vector is of the form For instance, a = [-2 5 0.8 0 1]. SEC. 7.1 275 Matrices, Vectors: Addition and Scalar Multiplication A column vector is of the form b= For instance, Matrix Addition and Scalar Multiplication What makes matrices and vectors really useful and particularly suitable for computers is the fact that we can calculate with them almost as easily as with numbers. Indeed, we now introduce rules for addition and for scalar multiplication (multiplication by numbers) that were suggested by practical applications. (Multiplication of matrices by matrices follows in the next section.) We first need the concept of equality. D E FIN I T ION Equality of Matrices Two matrices A = [ajk] and B = [bjk ] are equal, written A = B, if and only if they have the same size and the corresponding entries are equal, that is. a11 = b 11 , a12 = b 12 , and so on. Matrices that are not equal are called different. Thus, matrices of different sizes are always different. E X AMP L E 3 Equality of Matrices Let and B= [43 -1OJ = 4. a12 Then A=B all = if and only if o. The following matrices aTe all different. Explain! [~ ~J DEFINITION [~ 3 2 ~J [~ 4 Addition of Matrices The sum of two matrices A = [ajk] and B = [bjkJ of the same size is written A + B and has the entries ajk + bjk obtained by adding the corresponding entries of A and B. Matrices of different sizes cannot be added. As a special case, the sum a + b of two row vectors or two column vectors, which must have the same number of components, is obtained by adding the corresponding components. 276 E X AMP L E 4 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Addition of Matrices and Vectors -4 If .\ = [ 6 o ~J and B = [: -I °oJ· then A +B = [~ 5 2 A in Example 3 and our present A cannot be added. If a = [5 7 21 and b = [-6 a+b=[-I 9 21. An application of matrix addition was suggested in Example 2. Many others will follow. 2 OJ. then • Scalar Multiplication (Multiplication by a Number) DEFINITION The product of any /1l X 1l matrix A = [ajk] and any scalar c (number c) is written cA and is the I1l X 11 matrix cA = [cajk] obtained by mUltiplying each entry of A by c. Here (-I)A is simply written -A and is called the negative of A. Similarly, (- k)A is written - kA. Also, A + (- B) is written A - B and is called the difference of A and B (which must have the same size!). E X AMP L E 5 Scalar Multiplication If A = 2.7 -1.8] [ 0 0.9 9.0 then -A = -2.7 1.8] [ 0 -0.9 -9.0 -4.5 ~OA= 4.5 [ 3-2] 0 10 [0 0] 1,OA=0 -5 0 O. 0 If a matrix B shows the distances between some cities in miles. 1.60,)B gives these distances in kilometers. • Rules for Matrix Addition and Scalar Multiplication. From the familiar laws for the addition of numbers we obtain similar laws for the addition of matrices of the same size 111 X 11, namely, (a) A+B=B+A (b) (A + B) + C = A + (B + C) (3) A (c) (d) (written A + B + C) +0=A A+(-A)=O. Here 0 denotes the zero matrix (of size 111 X 11), that is. the III X 11 matrix with all entries zero. (The last matrix in Example 5 is a zero matrix.) Hence matrix addition is commutative and associative [by (3a) and (3b)]. Similarly, for scalar multiplication we obtain the rules (4) (a) c(A + B) = cA + cB (b) (c + k)A = cA + kA (c) c(kA) = (ck)A (d) IA = A. (written ckA) SEC 7.1 . 1fD=B""l £:M::::S E T 7 11-81 ~] ADDITION AND SCALAR MULTIPLICATION OF MATRICES AND VECTORS Let A = [-~ ~ :l, 6 C 5 B = -4J ~l. = [: 1 3J u~ []. [-~ -: -3 D = F 4 -:l, oj [-~ ~ l -8 3J [-~1 Find the following expressions or give reasons why they are undefined. 1. C 277 Matrices, Vectors: Addition and Scalar Multiplication + D, D + C, 6(D - C), 6C - 6D 2. 4C, 2D, 4C + 2D, 8C - OD 3. A + C - D, C - D, D - C, B + 2C + 4D 4. 2(A + B), 2A + 2B, 5A - ~B, A + B + C 5. 3C 8D, 4(3A). (4' 3)A, B - fDA 6.5A 3C, A - B + D, 4(B - 6A), 4B - 24A 7. 33u, 4v + 15. (General rules) Prove (3) and (4) for general 3 X 2 matrices and scalars c and k. 16. TEAM PROJECT. Matrices in Modeling Networks. Matrices have various applications, as we shall see, m a form that the~e problems can be efficiently handled on the computer. For instance, they can be used to characterize connections in electrical networks, in nets of roads, in production processes, etc., as follows. (a) Nodal incidence matrix. The network in Fig. 152 consists of 5 branches or edges (connections, numbered 1, 2, .. ·,5) and 4 nodes (points where two or more branches come together), with one node being grounded. We number the nodes and branches and give each branch a direction shown by an arrow. This we do arbitrarily. The network can now be described by a "nodal incidence matrix" A = [ajk], where (J) - 1 if branch k enters node (j) { o if branch k does not touch (J). + 1 if branch k leaves node Gjk = Show that for the network in Fig. 152 the matrix A has the given form 9u, 4(v + 2.25u), u - v 8. A + u, 12u + lOy, O(B - v), OB + u 9. (Linear system) Write down a linear system (as in Example I) whose augmented matrix is the matrix B in this problem set. 10. (Scalar multiplication) The matrix A in Example 2 shows the numbers of items sold. Find the matrix showing the number of units sold if a unit consists of (a) 5 items, (b) 10 items? 11. (Double subscript notation) Write the entries of A in Example 2 in the general notation shown in (2). 12. (Sizes, diagonal) What sizes do A, B, C, D, u, v in this problem set have? What are the main diagonals of A and B, and what about C? 13. (Equality) Give reasons why the five matrices in Example 3 are different. 14. (Addition of vectors) Can you add (a) row vectors whose numbers of components are different, (b) a row and a column vectOr with the same number of components, (c) a vector and a scalar? Branch Node CD ® Node ® Node Node@ 1 [-~ 2 3 4 -1 -1 0 1 0 1 0 1 0 0 0 -1 5 -~l Network and nodal incidence matrix in Team Project 16(a) Fig. 152. (b) Find the nodal incidence matrices of the networks in Fig. 153. CHAP. 7 278 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems + 1 if branch k is in mesh [2J and has the same orientation 11Ijk ~ - 1 if branch k is in mesh [2J and has the opposite orientation o if branch k is not in mesh [2J and a mesh is a loop with no branch in its interior (or in its exterior). Here, the meshes are numbered and directed (oriented) in an arbitrary fashion. Show that in Fig. 154 the matrix M corresponds to the given figure, where Row I corresponds to mesh I, etc. Fig. 153. Networks in Team Project 16{b) (c) Graph the three networks corresponding to the nodal incidence matrices -I] -I o -I I o o -1 o -1 o o o o o o -1 -I , o -I o o o 0 -1 o -I o Fig. 154. -1 (d) Mesh incidence matrix. A network can also be characterized by the mesh incidence matrix M = [mjkl. where 7.2 M~ [~ o 0-1 o o 1 0 -1 0 0 0 1 -1 -1 1 0 1 0 1 0 0 ~l Network and matrix M in Team Project 16{d) Number the nodes in Fig. 154 from left to right I, 2, 3 and the low node by 4. Find the corresponding nodal incidence matrix. (e) Matrix Multiplication Matrix mUltiplication means multiplication of matrices by matrices. This is the last algebraic operation to be defined (except for transposition, which is of lesser importance). Now matrices are added by adding corresponding entries. In multiplication, do we multiply corresponding entries? The answer is no. Why not? Such an operation would not be of much use in applications. The standard definition of multiplication looks artificial, but will be fully motivated later in this section by the use of matrices in "linear transformations," by which this multiplication is suggested. SEC. 7.2 Matrix Multiplication 279 Multiplication of a Matrix by a Matrix DEFINITION The product C = AB (in this order) of an 111 X 11 matrix A = [Gjk] times an r X p matrix B = [bjk ] is defined if and only if r = 11 and is then the 111 X P matrix C = [Cjk] with entries n (1) Cjk = L Gjtb/k = + Gjl b lk Gj2 b 2k + ... + Gjnbnk t~l j = 1.···. m k = L··· .p. The condition r = n means that the second factor, B, must have as many rows as the first factor has columns, namely n. As a diagram of sizes (denoted as shown): A C B [m X 11] [n X r] = [111 X r]. in (1) is obtained by multiplying each entry in thejth row of A by the corresponding entry in the kth column of B and then adding these 11 products. For instance, C21 = G21bl1 + G22b21 + ... + G2nbnl, and so on. One calls this briefly a "multiplicatioll of rows into columlls." See the illustration in Fig. 155, where 11 = 3. Cjk Fig. 155. E X AMP L E 1 Notations in a product AB = C Matrix Multiplication 3 AB = [ : : -6 -3 -:] [: 2 '0 = 1 -2 43 -16 14 4 -37 -9 on. The entry in the box is ("23 = 42] 6 - 28 4' 3 + O· 7 + 2 . 1 = 14. • Multiplication of a Matrix and a Vector 2J [3J = [4'3 + 2'5J = [22J 8 5 I . 3 + 8· 5 43 E X AMP L E 3 7 9-4 Here ell = 3 . 2 + 5 . 5 + (- I) . 9 = 22, and The product BA is not defined. E X AMP L E 2 -: 1] [2226 8 whereas is undefined. • Products of Rowand Column Vectors 6 [3 6 12 24 • 280 E X AMP L E 4 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems CAUTION! Matrix Multiplication Is Not Commutative, AB '* BA in General This is illustrated by Examples I and 2. where one of the two products is not even defined. and by Example 3. where the two products have different sizes. But it also holds for square matrices. For instance. I] [99 -9999J . but 100 = -<)<) It is interesting that this also shows that AB = 0 does 1101 necessarily imply BA = 0 or A = 0 or B = O. We shall discuss this further in Sec. 7.8. along with reasons when this happens. • Our examples show that the order offactors in matrix products must always be obse",ed vel)' carefully. Otherwise matrix multiplication satisfies rules similar to those for numbers, namely. (kA)B = k(AB) = A(kB) (a) written kAB or AkB written ABC A(BC) = (AB)C (b) (2) + mc = AC + BC (c) (A (d) C(A + B) = CA + CB provided A, B, and C are such that the expressions on the left are defined; here, k is any scalar. (2b) is called the associative law. l2c) and (2d) are called the distributive laws. Since matrix mUltiplication is a multiplication of rows into columns. we can write the defining formula (1) more compactly as j = 1. .... 111: (3) k = 1. .... p. where aJ is the jth row vector of A and b k is the hh column vector of B, so that in agreement with (1), ll b: a Jn ] '] = aj1b1k + aj2 b2k + ... + aj"bnk . [ bnk E X AMP L E 5 Product in Terms of Rowand Column Vectors If A = [ajkl is of si/e 3 x 3 and B = [bjkl is of size 3 x 4. then alb l AB = (4) a2bl [ a3 b l Taking al = [3 5 -11. a2 = [4 0 2]. etc .. verify (4) for the product in Example I. • Parallel processing of products on the computer is facilitated by a variant of (3) for computing C = AB, which is used by standard algorithms (such as in Lapack). In this method, A is used as given. B is taken in terms of its column vectors, and the product is compUled columnwise: thus, (5) SEC 7.2 281 Matrix Multiplication Columns of B are then assigned to different processors (individually or several IO each processor), which simultaneously compute the columns of the product matrix Ab l , Ab2, etc. E X AMP L E 6 Computing Products Columnwise by (5) To obtain o 4 AB = [ -5 4 7J - [11 -17 6 4 8 34J -23 from (5), calculate the columns of AB and theu wnte them as a single matrix, as shown in the first formula ou the right. Motivation of Multiplication • by Linear Transformations LeL us now motivate the "unnatural" matrix multiplication by its use in linear transformations. For II = 2 variables these transformations are of the form (6*) and suffice to explain the idea. (For general n they will be discussed in Sec. 7.9.) For instance, (6*) may relate an xlx2-coordinate system to a YIY2-coordinate system in the plane. In vectorial form we can write (6*) as (6) y- [ YI] -Ax.\'2 [au 1121 Now suppose further that the xlx2-system is related to a wlw2-system by another linear transformation, say, (7) Then the )'IY2-system is related to the ~1'lw2-system indirectly via the x1x2-system, and we wish to express this relation directly. Substitution will show that this direct relation is a linear transformation, too, say, (8) e 11 y = Cw = [ C21 Indeed, substituting (7) into (6), we obtain Y1 = all(b 11 H."1 = Y2 + b 12 W 2) (allhll = a21(b ll w l + + b 12 W 2) = (a2I b 11 + a12(b21 W I + a12 h 21)wI + (a11hI2 + a22(b21 W I + + a22 b21)WI + b 22 W 2) + a12b22)W2 b22W2) (a21 b I2 + a22 b 22)W2' 282 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Comparing this with (8), we see that C12 = an b 12 + a12 b 22 C2 2 = a21 b 12 + a22 b 22' This proves that C = AB with the product defined as in (I). For larger matrix sizes the idea and result are exactly the same. Only the number of variables changes. We then have III variables y and n variables x and p variables w. The matrices A, B, and C = AB then have sizes III X Il, 11 X p. and m X p. respectively. And the requirement that C be the product AB leads to formula (1) in its general form. This motivates matrix multiplication completely. Transposition Transposition provides a transition from row vectors to column vectors and conversely. More generally, it gives us a choice to work either with a matrix or with its transpose. whatever will be more practical in a specific situation. DEFINITION Transposition of Matrices and Vectors The transpose of an III X /I matrix A = [ajk] is the 11 X m matrix AT (read A transpose) that has the rust row of A as its first column, the second row of A as its second column. and so on. Thus the transpose of A in (2) is AT = [a/<J], written out (9) As a special case, transposition converts row vectors to column vectors and conversely. E X AMP L E 7 Transposition of Matrices and Vectors If 5 -8 A= [ 4 0 then A little more compactly, we can write J+: :l -8 0 [: [6 2 3]T ~ or [38 -1 [:l [:r" = 2 [30 -18J 3]. Note that for a square matrix. the transpose is obtained by interchanging entries that are symmetrically positioned • with respect to the main diagonal, e.g., a12 and a21. and so on. SEC. 7.2 283 Matrix Multiplication Rules for transposition are (AT)T (a) (A (b) = A + B)T = AT + BT (10) (c) (CA)T = CAT (d) (AB)T = BTAT. CAUTION! Note that in (lOd) the transposed matrices are ill reversed order. We leave the proofs to the student. (See Prob. 22.) Special Matrices Certain kinds of matrices will occur quite frequently in our work, and we now list the most important ones of them. Symmetric and Skew-Symmetric Matrices. Transposition gives rise to two useful classes of matrices, as follows. Symmetric matrices and skew-symmetric matrices are square matrices whose transpose equals the matrix itself or minus the matrix, respectively: (thus akj = -ajk, hence ajj = 0). (11) Svmllletric Matri ... E X AMP L E 8 Ske\\ oS) mmetric M.ltrix Symmetric and Skew-Symmetric Matrices 120 200] 10 150 150 30 is symmetric. and is skew-symmetric. For instance, if a company has three building supply centers C1 , C2 , C3 , then A could show costs, say, ajj for k) the cost of shipping 1000 bags from Cj to C k . handling 1000 bags of cement on ceoter Cj , and ajl, (j Clearly. ajk = lI~j because shipping in the opposite direction will usually cost the same. Symmetric matrices have several general pmperties which make them importaot. This will be seen as we proceed. • "* Triangular Matrices. Upper triangular matrices are square matrices that can have nonzero entries only on and above the main diagonal, whereas any entry below the diagonal must be zero. Similarly, lower triangular matrices can have nonzero entries only on and below the main diagonaL Any entry on the main diagonal of a triangular matrix may be zero or not. EXAMPLE 9 Upper and Lower Triangular Matrices [~ :J. 4 3 [: 0 LIpper triangular :l () -I [: 6 :l [~ ~l 0 0 -3 0 0 2 9 3 Lo\\ er Irian!!ul"r • 284 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Diagonal Matrices. These are square matrices that can have nonzero entries only on the main diagonal. Any entry above or below the main diagonal must be zero. If all the diagonal entries of a diagonal matrix S are equal. say, c, we call S a scalar matrix because mUltiplication of any square matrix A of the same size by S has the same effect as the multiplication by a scalar, that is, AS (12) = SA = cA. In particular, a scalar matrix whose entries on the main diagonal are all 1 is called a unit matrix (or identity matrix) and is denoted by In or simply by I. For I, formula (12) becomes AI = IA = A. (13) E X AMP L E 1 0 Diagonal Matrix D. Scalar Matrix S. Unit Matrix I • Applications of Matrix Multiplication Matrix multiplication will play a crucial role in connection with linear systems of equations, beginning in the next section. For the time being we mention some other simple applications that need no lengthy explanations. E X AMP L E 11 Computer Production. Matrix Times Matrix Supercomp Ltd produces two computer models PC I086 and PC 1186. The matrix A shows the cost per computer (in thollsands of dollars) and B the production figures for the year 2005 (in multiples of 10000 units.) Find a mutrix C that shows the shareholders the cost per quarter (in millions of dollars) for raw muterial. labor. and miscellaneous. Quarter PCIOH6 A = PCIIH6 1.2 1.6] Raw Components 0.3 0.4 Labor 0.6 Miscellaneous [0.5 2 3 8 6 PCI086 2 4 PCI186 4 Solutioll. Quarter 2 C =AB = 3 [132 12.8 13.6 3.3 3.2 3.4 5.1 5.2 5.-l- 4 156] Raw Components 3.9 Labor 6.3 Miscellaneous Since cost b given io multiples of $1000 aod production in multiples of 10 000 units the eotries of Care multiples of $10 millioos; thus ell = 13.2 means $132 miUion. etc. • SEC. 7.2 285 Matrix Multiplication E X AMP L E 12 Weight Watching. Matrix Times Vector Suppose that in a weight-watching program. a person of 1851b burns 350 callhr in walking (3 mph). 500 in bycycling (13 mph) and 950 in jogging (5.5 mph). Bill. v.eighing 185 lb. plans to exercise according to lhe matrix shown. Verify the calculmions (W = Walking. B = Bicycling. J = Jogging). W B MON 0 LO 1.0 WED [ l.0 0 J ::] [=] ~ [I::] FRI 1.5 SAT 2.0 1.5 1.0 U.5 MON WED 1000 FRl 2400 SAT • 950 EXAMPLE 13 Markov Process. Powers of a Matrix. Stochastic Matrix Suppose that the 2004 state of land use in a city of 60 mi 2 of built-up area i~ C: Commercially Used 25<lc I: Industrially Used 20% R: Residentially Used 55%. Find the stales in 2009, 2014. and 2019, assuming that the transition probabilitie~ for 5-year intervals are given by the matrix A and remain practically the same over the time considered. From C A = From [ FTomR 0.[ [0' 0.2 0.9 0.[ 0 ToC 0°, ] To I 0.8 ToR A is a stochastic matrix, that is, a square matrix with all entries nonnegative and all column sums equal to I. Our example concerns a Markov process1 , that is. a process for which thc probability of entering a certain state depends only 00 the last state occupied (and the matrix A), not on any earlier state. Solutioll. From the matrix A and the 2004 state we can compute the 2009 state. C + 0.[ ·20 + 0,55] 0.2 . 25 + 0.9' 20 + 0.2' 55 0.7'25 [ R 0.1,25 + 0·20 = + 0.8,55 [0.7 U.I 0.2 0.9 0.1 o 0.20] 0.8 [25] 20 55 [19.5] 34.0. 46.5 To explain: The 2009 figure for C equals 25o/c times the probability 0.7 that C goes into C, plus 20'7< probability 0.1 that I goes into C, plus 55% times the probability U that R goes into C. Together, 25· 0.7 + 20· 0.1 + 55' 0 = 19.5 ['k]. Also time~ the 25' 0.2 + 20' 0.9 + 55· 0.2 = 34 [%]. Similarly. the new R is 46.5%. We see that the 200!) state vector is the column vector y = [19.5 34.0 46.5]T = Ax = A [25 20 55]T where the column ~ector X = [25 20 55] T is the given 2004 state vector. Note that the ~um of the entries of y is 100 ['7<]. Similarly. you may verify that for 2014 and 20[9 we get the state vectors z = Ay = A(Ax) = A 2 x = [17.05 u = Az = A~' = A3x = [16.315 43.80 50.660 39.15]T 33.025]T. lANDREI ANDREJEVITCH MARKOV (1856-1922), Russian mathematician, known for his work in probability theory. CHAP. 7 286 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems 2 In 2009 the commercial area will be 19.5% (11.7 mi 2 ). the industrial 34% (20.4 mi ) and the residential 46.5% (27.9 mi 2 ). For 2014 the conesponding figures are 17.05%.43.80%. 39.15o/r. For 2019 they are 16.315%. 50.660%. 33.0:!5o/c. (In Sec. 8.2 we shall see what happens in the limit. assuming that those probabilities remain the same. In the meantime. can you experiment or guess?) • Answer. - __ .............. ___ ..., •... -.•.-- .....,J __....• ::-~ 11-141 Let A = MULTIPLICATION, ADDITION, AND TRANSPOSITION OF MATRICES AND VECTORS l L~ =: -~J. l: : -~J' B -10 C l: 5 = 1 -4 ~ -~l· ~ [l ~ b 0 [3 11 0 81· Calculate the following products and sums or give reasons why they are not defined. (Show all intermediate results.) 1. Aa. Ab, Ab T, AB 2. Ab T + Bb T• (A + B)bT. bA. B - BT 3. AB, BA, AAT, ATA 4. A2. B2, (AT)2, (A2)T 5. aT A, bA, 5B(3a + 2b T). 15Ba + lOBb T 6. ATb, bTB, (3A - 2B)Ta, a T(3A - 28) 7. ab, ba, (ab)A, a(bA) 8. ab - ba, -(4b)(7a), -28ba, 5abB 9. (A + B)2, A2 + AB + BA + B2, A2 + 2AB + B2 10. (A + Bj(A - B). A2 - AB + BA - B2, A2 - B2 11. A2B, A3. (AB)2. A2B2 12. B 3 , BC, (BC)2. (BC)(BC)T 13. aTAa, aT(A + AT)a, bBb T. b(B - BT)b r 14. aTCCTa, a TC 2a. bCTCb T. bCCTb T 15. (General rules) Prove (2) for 2 X 2 matrices A = [ajk]. B = [bjk ]. C = [Cjk] and a general scalar. 16. (Corrunutativity) Find all 2 x 2 matrices A = [ajk] that commute with B = [bjk ]. where bjk = j + k. 17. (Product) Write AB in Probs. 1-14 in terms of row and column vectors. 18. (Product) Calculate AB in Prob. 1 column wise. (See Example 6.) 19. TEAM PROJECT. Symmetric and SkewSymmetric J\;latrices. These matrices occur quite frequently in applications. so it is worthwhile to study some of their most important properties. (a) Verify the claims in (11) that (/kj = ajk for a symmetric matrix. and akj = -ajk for a skew-symmetric matrix. Give examples. (b) Show that for every square matrix C the matrix C + C T is symmetric and C - C T is skew-symmetric. Write C in the form C = S + T, where S is symmetric and T is skew-symmetric and find Sand T in terms of C. Represent A and B in Probs. 1-14 in this form. (c) A linear combination of matrices A, B, C, ... , M of the same size is an expression of the form (14) aA + bB + cC + ... + 111M. where a . ... , III are any scalars. Shuw that if these matrices are square and symmetric, so is (14): similarly, if they are skew-symmetric. so is (14). (d) Shuw that AB with symmetric A and B is symmetric if and only if A and B commute, that is. AB = BA. (e) Under what condition is the product of skewsymmetric matrices skew-symmetric? 20. (Idempotent and nilpotent matrices) By definition, A is idempotent if A2 = A, and B is nilpotent if Bm = 0 for some positive integer 111. Give examples (different from 0 or I). Also give examples such that A2 = I (the unit matrix). 21. (Triangular matrices) Let VI' V 2 be upper triangular and L 1. L2 lower triangular. Which of the following are triangular? Give examples. How can you save half of your work by transposition? U1 + U2, V\V2 , V12. VI + L 1. U1L 1, L1 + L 2. L 1L 2, L12 22. (Transposition of products) Prove (lOa)-(lOc). lllustrate the basic formula (lOd) by examples of your own. Then prove it. APPLICATIONS 23. (Markov process) If the transition matrix A has the entries all = 0.5, a12 = 0.3, a21 = 0.5, (/22 = 0.7 and the initial state is [1 1] T, what will the next three states be? 24. (Concert subscription) In a community of 300000 adults, subscribers to a concert se1ies tend to renew their SUbSCliption with probability 90% and persons presently not SUbsClibing will subscribe for the next season with probability 0.1 %. If the present number of subscribers is 2000, can one predict an increase, denease, or no change over each of the next three seasons? SEC. 7.3 Linear Systems of Equations. Gauss Elimination 25. CAS Experiment. Markov Process. Write a program for a Markov process. Use it to calculate further steps in Example 13 of the text. Experiment with other stochastic 3 X 3 matlices, also using different starting values. 26. (Production) In a production process, let N mean "no trouble" and r'trouble." Let the transition probabilities from one day to the next be 0.9 for N --> N, hence 0.1 for N --> T, and 0.5 for T --> N, hence 0.5 for T --> T. If today there is no trouble, what is the probability of N two days after today? Three days after today? 27. (Profit vector) Two factory outlets Fl and F2 in New York and Los Angeles sell sofas (S), chairs (C). and tables (T) with a profit of $110, $45, and $80, respectively. Let the sales in a certain week be given by the matrix T S c A = 400 [600 300 100J 205 820 Introduce a "profit vector" p such that the components of v = Ap give the total profits of Fl and F 2 . 28. TEAM PROJECT. Special Linear Transformations. Rotations have various applications. We show in this project how they can be handled by matrices. (a) Rotation in the plane. Show that the linear transformation y = Ax with matrix A = [COS 8 sin 8 -sin 8] and (c) Addition formulas for cosine and sine. By geometry we should have [ c~s a sm a -sin a] cos a COS f3 [ sin f3 = [cos (a sin (a -sin f3] cos f3 m + + (3) -sin (a + m] . cos (a + (3) Del;ve from this the addition formulas (6) in App. A3.1. (d) Computer graphics. To visualize a threedimensional object with plane faces (e.g., a cube), we may store the position vectors of the vertices with respect to a suitable XIX2x3-coordinate system (and a list of the connecting edges) and then obtain a twodimensional image on a video screen by projecting the object onto a coordinate plane, for instance, onto the xlx2-plane by setting -'"3 = O. To change the appearance of the image. we can impose a linear transformation on the position vectors stored. Show that a diagonal matrix D with main diagonal entries 3, 1, ~ gives from an x = [Xj] the new position vector y = Dx, where Yl = 3Xl (stretch in the Xl-direction by a factor 3), Y2 = X2 (unchanged), }"3 = ~X3 (contraction in the x3-direction). What effect would a scalar matrix have? (e) Rotations in space. Explain y = Ax geometrically when A is one of the three matrices cos 8 y= An = [ 0 [J is a counterclockwi~e rotation of the Cartesian XIX2coordinate system in the plane about the origin. where 8 is the angle of rotation. (b) Rotation through nO. Show that in (a) COS 11f1 - sin I1f1J sin 1If1 cos 118 Is this plausible? Explain this in words. 7.3 287 cos fI l: 0 lOO' · Si~ cp sin 8 cos fI -,m.] l'~· o 0 -'~n' 1 cos cp . . -sin '" sin '" cos '" 0 0 ] What effect would these transformations have in situations such as that described in (d)? Linear Systems of Equations. Gauss Elimination The most important use of matrices occurs in the solution of systems of linear equations, briefly called linear systems. Such systems model various problems, for instance, in frameworks, electrical networks, traffic flow, economics, statistics, and many others. In this section we show an important solution method, the Gauss elimination. General properties of solutions will be discllssed in the next sections. 288 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Linear System, Coefficient Matrix, Augmented Matrix A linear system of m equations in 11 unknowns"\ b . . . 'Xn is a set of equations of the form (1) The system is called linear because each variable Xj appears in the first power only, just as in the equation of a straight line. alb"', a mn are given numbers, called the coefficients of the system. b I , . . . , bm on the right are also given numbers. [f all the bj are zero, then (1) is called a homogeneous system. If at least one bj is not zero, then (1) is called a nonhomogeneous system. A solution of (1) is a set of numbers Xl' • . • • Xn that satisfies all the m equations. A solution vector of (1) is a vector x whose components form a solution of (1). If the system (1) is homogeneous. it has at least the trivial solution Xl = 0, .... Xn = O. Matrix Form of the Linear System (1). From the definition of matrix multiplication we see that the m equations of (1) may be written as a single vector equation (2) Ax =b where the coefficient matrix A = [ajk] is the au ([12 a1n (/21 a22 a2n In and A= amI a m2 x n matrix x= and b= Q·'tnn Xn are column vectors. We assume that the coefficients (/jk are not all zero, so that A is not a zero matrix. Note that x has 11 components, whereas b has III components. The matrix A= is called the augmented matrix of the system (1). The dashed vertical line could be omitted (as we shall do later); it is merely a reminder that the last column of A does not belong to A. The uugmellted mutrix A determines the system (1) completely becam,e it contains all the given numbers appearing in (1). SEC. 7.3 289 Linear Systems of Equations. Gauss Elimination E X AMP L E 1 Geometric Interpretation. Existence and Uniqueness of Solutions If m Unique solution = 11 = 2. we have two equations in two unknowns Xl, X2 If we interpret Xl, X2 as coordinates in the xlx2-plane. then each of the two equations represems a slraight line. and (Xl. -'"2) is a solution if and only if the point P with coordinates Xl' X2 lies on both lines. Hence there are three possible cases: (aJ Precisely one solution if the lines intersect. (b) Infinitely many solutions if the lines coincide. (c) No solution if the lines are parallel For instance, Xl +X2 2xl-x2 =1 =0 Xl +X2 2xl + = 1 2x2 = Case (b) Case (a) = 1 xl +X2 = 0 Xl +X2 2 Case (c) x2 / Infinitely many solutions ~ .p / I / :I xl If the system is homogenous, Case (c) cannot happen. because then those two straight lines pass through the origin. whose coordinates O. 0 constitute the trivial solution. If you wish, consider three equations in three unknowns as representations of three planes in space and discuss the various possible cases in a similar fashion. See Fig. 156. • Our simple example illustrates that a system (I) may perhaps have no solution. This poses the following problem. Does a given system (1) have a solution? Under what conditions does it have precisely one solution? If it has more than one solution, how can we characterize the set of all solutions? How can we actually obtain the solutions? Perhaps the last question is the most immediate one from a practical viewpoint. We shall answer it first and discuss the other questions in Sec. 7.5. Gauss Elimination and Back Substitution No solution Fig. 156. Three equations in three unknowns interpreted as planes in space This is a standard elimination method for solving linear systems that proceeds systematically irrespective of particular features of the coefficients. It is a method of great practical importance and is reasonable with respect to computing time and storage demand (two aspects we shall consider in Sec. 20.1 in the chapter on numeric linear algebra). We begin by motivating the method. If a system is in "triangular form," say, 2Xl + 5X2 = 13x2 2 = -26 we can solve it by "back substitution," that is, solve the last equation for the variable. = -26113 = -2, and then work backward, substituting X2 = -2 into the fIrst equation X2 290 CHAP. 7 Linear Algebra: Matrices. Vectors. Determinants. Linear Systems and solve it for Xl' obtaining Xl = ~(2 - 5x2 ) = ~(2 - 5· (-2» = 6. This gives us the idea of fIrst reducing a general system to triangular form. For instance, let the given system be -3~J . 5 -4Xl + 3x2 = - 30. Its augmented matrix is 3 We leave the fust equation as it is. We eliminate Xl from the second equation. to get a triangular system. For this we add twice the fIrst equation to the second, and we do the same operation on the rows of the augmented matrix. This gives -4Xl + 4Xl + 3X2 + 10x2 = -30 + 2· 2, that is, 2 -26 Row :2 + 2 Row I [~ 5 13 where Row :2 + :2 Row I means "Add twice Row 1 to Row T in the original matrix. This is the Gauss elimination (for 2 equations in 2 unknowns) giving the triangular form, from which back substitution now yields X2 = - 2 and Xl = 6, as before. Since a linear system is completely determined by its augmented matrix, Gauss elimination call be dOlle by merely considering the matrices, as we have just indicated. We do this again in the next example. emphasizing the matrices by writing them first and the equations behind them. just as a help in order not to lose track. E X AMP L E 2 Gauss Elimination. Electrical Network Solve the linear system This is the ~ystem for the unknown currellIs in the electrical network in Fig. 157. To obtain it. we label the currents as shown. choosing directions arbitrarily: if a current will come out negative. this will simply mean that the current flows against the direction of our arrow. The current entering each battery will be the same as the current leaving it. The equations for the CUlTents result from Kirchhoff's laws: Derivation from the circuit ill Fig. 157 (Optional). Xl = i l . x2 = i 2 • x3 = i3 Kirchhoff's currellt law (KCL). At allY poim of a circuit. rhe sum of the illf/owillg Cl/rrems equals the of fhe olltf/oll"illg ("lIrrellTs. Sll111 Kirclzhoff's I'oltage law (KVL). 111 allY closed loop. the slim of all I'Olrage drops eqllals rhe impressed electromotil'e force. Node P gives the first equation, node Q the second, the right loop the third. and the left loop the fourth, as indicated in the figure. 80 v~ rtJ P Fig. 157. 15Q fWV NodeP: i1 - i2 + i3 = NodeQ: -ir + i2 - 13 Right loop: Left loop: 0 = 0 1Oi2 + 25i 3 = 90 20;1 + lOi2 Network in Example 2 and equations relating the currents =80 SEC. 7.3 291 Linear Systems of Equations. Gauss Elimination Solution by Gauss Elimination. This system could be solved rather quickly by noticing its particular form. But this is not the point. The point is that the Gauss elimination is systematic and will work in general, also for large systems. We apply it to our system and then do back substitution. As indicated let us write the augmented matrix of the system first and then the system itself: '-'fI;I Augmented Matrix CD-I Flimi",'e~ Pi"" A Equations Pivotl~~- X2 -I 10 25 10 0 I I I I I 9:] Elimlllate ~ 80 Cl 20xl + X3 = 0 x3 = 0 + 25x3 = 90 '\2 lOx2 + IOx2 = 80 Step 1. Elimination of Xl Call the first row of A the pivot row and the first equation the pivot equation. Call the coefficient I of its xrterm the pivot in this step. Use this equation to eliminate Xl (get rid ot xl) in the other equations. For this, do: Add I times the pivot equation to the second equation. Add -20 times the pivot equation to the fourth equation. This corresponds to row operations on the augmented matrix as indicated in BLUI behind the new matrix in (3). So the operations are performed on the preceding matrix. The result is -I (3) [; X2 Xl - 0 0 to 25 30 -20 + X3 = Row 2.L Row I 0= 0 i] 80 0 IOx2 Row 4 - 20 Row I + 25x3 30x2 - 20x3 = 90 = 80. Step 2. Elimination of X2 The first equation remains as it is. We want the new second equation to serve as the next pivot equation. But since it has no x2-term (in fact, it is 0 = 0), we mllst first change the order of the equations and the corresponding rows of the new mauix. We put 0 = 0 at the end and move the third equation and rhe fourth equation one place up. This is called partial pivoting (as opposed to the rarely used total pivoting, in which also the order of the unknowns is changed). It gives l -I Pivotlt~ 0 @ 25 Eliminate 3,,~ : ~ -20 r 0 Xl - ;] Pivot It X3 = 0 25x3 = 90 Eliminate 30-'2 ~ 130x21- 2o.r3 = 80 80 0 + ,@;)+ X2 0 0 = 0 To eliminate X2' do: Add -3 times the pivot equation to the third equation. The result is (4) r~ -I to 0 0 I' 0] 25: -95 i- 01 90 190 X2 Xl - I IOx2 Row 3 - 3 Row 2 + X3 = + 25x3 = 0 90 - 9SX3 = -190 0 0= 0 Back Substitution. Determination ofx3' x2' Xl (in this order) Working backward from the last to the first equation of this "triangular" system (4), we can now readily find x3, then .\'2, and then xl: -95x3 = -190 90 X3 =;3 = 2lAJ X2 = fo(90 - 25x3) = i2 = 4 [AJ o where A stands for "amperes." This is the answer to our problem. The solution is unique. • 292 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Elementary Row Operations. Row-Equivalent Systems Example 2 illustrates the operations of the Gauss elimination. These are the first two of three operations. which are called Elementary Row Operations for Matrices: Interchange (~f two rows Addition of a constant multiple of one row to another row Multiplication of a row by a nonzero constant c. CAUTION! following These operations are for rows, not for columns! They correspond to the Elementary Operations for Equations: Interchange of two equations Addition of a constant multiple of one equation to another equation Multiplication of an equation by a nonzero constant c. Clearly, the interchange of two equations does not alter the solution set. Neither does that addition because we can undo it by a corresponding subtraction. Similarly for that multiplication, which we can undo by multiplying the new equation by lIc (since c =1= 0), producing the original equation. We now call a linear system SI row-equivalent to a linear system S2 if SI can be obtained from S2 by (finitely many!) row operations. Thus we have proved the following result, which also justifies the Gauss elimination. THEOREM 1 Row-Equivalent Systems Row-equivalent linear systems have the same set of solutions. Because of this theorem, systems having the same solution sets are often called equivalent systems. But note well that we are dealing with row operations. No column operations on the augmented matrix are pennitted in this context because they would generally alter the solution set. A linear system (1) is called overdetermined if it has more equations than unknowns. as in Example 2. determined if m = n. as in Example I. and underdetermined if it has fewer equations than unknowns. Furthermore, a system (1) is called consistent if it has at least one solution (thUS, one solution or infinitely many solutions), but inconsistent if it has no solutions at all, as Xl + X2 = I, Xl + X2 = 0 in Example l. Gauss Elimination: The Three Possible Cases of Systems The Gauss elimination can take care of linear systems with a unique solution (see Example 2), with infinitely many solutions (Example 3, below), and without solutions (inconsistent systems; see Example 4). SEC 7.3 293 Linear Systems of Equations. Gauss Elimination E X AMP L E 3 Gauss Elimination if Infinitely Many Solutions Exist Solve the following linear systems of three equatIons in four unknowns whose augmented matrix is (S) [3U 2.0 2.0 -S.O 0.6 I.S I.S -S.4 1.2 -0.3 -0.3 2.4 I I I I I I 'U] 2.7 ~ + 2.0X2 + 2·(l~3 - S.OX4 . Thus. 2.1 = 10.6X11: I.SX2: I.Sx3 - S.4x4 : 1.2Tl 0.3'\2 0.3X3 + 2.4x4 - 8.0 2.7 2.1. Solutioll. As in the previous example. we circle pivots and box terms of equations and corresponding entries to be eliminated. We indicate the operations in terms of equations and operate on both equations and matrices. Step 1. Elimillation O/Xl from the second and third equations by adding - 0.6/3.0 = -0.2 times the first equation to the second equation, - 1.2/3.0 ~ -0.4 times the first equation to the third equation. This gives the following, in which the pivot of the next step is circled. 2.0 l.l 1.1 -4.4 -1.1 -1.1 4.4 (6) Step 2. Elimillatioll 3.0Xl + 2.0X2 + 2.Ox3 - S.Ox4 = !l.0 Row 2 - 0.2 Row I ~+ 1.1x3 - 4.4x4 = l.l Row 3 - OA Row I 1-l.1x21- 1.1x3 + 4.4x4 = - l.l -S.O 2.0 0/ x2 from 8.0] 1.1 -1.1 the third equation of (6) by adding 1.111.1 = I times the second equation to the third equation. This gives (7) 2.0 2.0 1.1 1.1 o -S.O i -4.4 I 8.0 8.0] l.l 1.1 I o 010 RO\\ 3 ;- RO\\ 2 0= o Back Substitution. From the second equation. X2 = 1 - X3 + 4x4' From this and the fIrst equation. Xl = 2 - X4' Since x3 and x4 remain arbitrary. we have infinitely many solutions. If we choose a value of x3 and a value of X4. then the corresponding values of Xl and x2 are uniquely determined. If unknowns remain arbitrary. it is al~o customary to denote them by other letters 11, 12•.... In this example we may thus write Xl = 2 - X4 = 2 - 12. x2 = I - x3 + 4X4 = I - '1 + 412. x3 = 11 (flrst arbitrary unknown), X4 = 12 (second arbitrary unknown). • Oil Notation. E X AMP L E 4 Gauss Elimination if no Solution Exists What will happen if we apply the Gauss elimination to a linear system that has no solution? The answer is that in this case the method will show this fact by producing a contradiction. For instance. consider 3 2 1: 3] [ 2 1 : 0 4 I 6 I 6 2 @+ 2~2 + X3 ~+ X2 + X3 = 0 ~+ 2X2 + 4x3 = 6. = 3 Step 1. Eliminatioll o/x] from the second and third equations by adding -~ time, the fIrst equation to the second equation. -i = -2 times the first equation to the third equation. This give, 2 1 [: -3 -2 3xl + 2~2 + x3 = 3 : 3J :-2 Row ]. - ~ Ron 1 (B+ - 3x 2 2 I I RO\I J - 2 Row I 1- U2/+ U3 = 1 0 3 -2 31 x 3- O. 294 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Step 2. Elimillatioll of X2 from the third equation gives 2 1! ::-23] o o The false statement 0 ~ I I 12 Rm" 3 6 Ro\\ 2 12 shows that the system has no 0= 12. • ~olution. Row Echelon Form and Information From It At the end of the Gauss elimination the form of the coefficient matrix, the augmented matrix, and the system itself are called the row echelon form. In it. rows of zeros. if present. are the la"t rows. and in each nonzero row the leftmost nonzero entry is farther to the right than in the previous row. For instance. in Example 4 the coefficient matrix and its augmented in row echelon fonn are 2 1 [: -3 0 2 :] and 1 [: 1 -3 3" 0 0 -+ 12 Note that we do not require that the leftmost nonzero entries be I since this would have no theoretic or numeric advantage. (The so-called reduced echelon form, in which those entries are I, will be discussed in Sec. 7.8.) At the end of the Gauss elimination (before the back substitution) the row echelon form of the augmented matrix will be (8) Here, r ~ 111 and (/11 =1= 0, C22 =1= 0, ... , k1T =1= 0, and all the entries in the blue triangle as well as in the blue rectangle are zero. From this we see that with respect to solutions of the system with augmented matrix (8) (and thus with respect to the originally given system) there are three possible cases: (a) Exactly one solution if r = 1l and b,,+ .. .... bm' if present. are zero. To get the solution. solve the nth equation corresponding to (8) (which is knnxn = bn) for X n ' then the (n - l)st equation for Xn-l, and so on up the line. See Example 2, where r = n = :3 and I7l = 4. (b) Infinitely many solutions if r < 11 and b,.+!, .... bm' if present, are zero. To obtain any of these solutions, choose values of X r + l , ••• 'Xl1 arbitrarily. Then solve the 7th equation for x,., then the (1' - l)st equation for X,._!, and so on up the line. See Example 3. (c) No solution if r < 111 and one of the entries br + I' 4, where r = 2 < m = 3 and br + 1 = b3 = 12. •.. , bm is not zero. See Example SEC. 7.3 11-161 GAUSS ELIMINATION AND BACK SUBSTITUTION 20.9 2. 3.0x + 6.2y = 0.2 -x + 4y = -11).3 2.lx + 8.5y = 4.3 1. 5x - 2y + 3. 0.5x = 3.5)' = 5.7 -x + 5.0.1' = 7.8 4. 4x 5. 0.8x + 1.2.1' - 0.6::: = -7.8 + 1.7z = 15.3 4.0x - 7.3y - 1.5::: = l.l 2.6x 6. 14x - 2y - 47 = 0 18x - 2}' - 6;: = 0 + 8)' - J4z = 0 4x 8. 2x + y y + 3z = -1 -4x + 2y - 6z = 2 - 2y - 2: 4y - 5z = -4w -2w 14. + IV - - 31-1' + 7 81<' + 4y - 2::: = 0 + = 13 + 29 8y - 4::: = 24 - 6x 2:: 17x + 4y + 3: = 7w 2x 0 + 3y + 8y - 6: = -20 511' - 13x - y - + 2- = 5.: = 0 16 -= x + y + <. = 4y + 4::: = 24 27 = -6 = 18 11)' - 17y + -2 = -12 z 2 f3. A:2A I ~ 1/2 ViC I 'L~ _Eo x 3x + 19·~1!;OQ 5Q ~tJ--~3_5_V___- , 13 = y -1.3 8 - x + y - 2: 13. = 0.3)' - 0.4: = -1.9 = + = 6;: 9. -4.6x + 0.5)' + 1.2z 3x z + - 12. 2 4y 6x 2x - = + 8x - y + 7z =0 11. 8y + : 16. -211' Y 3x + + 17.~ + 2::: = 3 0.6x 511' + 4x MODELS OF ELECTRICAL NETWORKS Using Kirchhoff's laws (see Example 2), find the currents. (Show the details of your work.) - 5x 7y - 4: = -46 117-191 7. 3: = 8 - + + -lI' 4y - 2;:: 6x - 2)' + 3x 15. Solve the following systems or indicate the nonexistence of solutions. (Show the details of your work.) 10. 295 Linear Systems of Equations. Gauss Elimination + 0 = 2:: = -4 3y - 62 = -2 2x + 5y - 3z = 0 6x + v + : 2w - 4x + 3y - z = 0 = 3 Wheatstone bridge Net of one-way streets (Prob. 20, next page) (Prob. 21, next page) 296 CHAP. 7 Linear Algebra: Matrices, Vectors. Determinants. Linear Systems 20. (Wheatstone bridge) Show that if RxlR3 = Rl/R2 in the figure. then T = O. (Ro is the resistance of the instrument by which I is measured.) This bridge is a method for determining R.r . R I • R 2 • R3 are known. R3 is variable. To get Rx. make I = 0 by varing R 3 . Then calculate Rx = R 3R I /R z . to do row operations directly, mUltiplication by E.) (a) Show that the following are elementary matrices, for interchanging Rows 2 and 3. for adding -5 times the first row to the third, and for mUltiplying the fourth row by 8. 21. (Traffic flow) Methods of electrical circuit analysis have applications to other fields. For instance, applying the analog of Kirchhoff's current law, find the traffic flow (cars per hour) in the net of one-way streets (in the directions indicated by the an'ows) shown in the figure. Is the solution unique? 22. (Model., of markets) Determine the equilibrium solution (D 1 = SI, D2 = S2) of the two-commodity market with linear model (D, S, P = demand, supply, price: index I = first commodity. index 2 = second commodity) -5 + 14 o Dl = 60 - 2P I D2 = 4Pl - P 2 - P2• 4P l - 2P 2 + 10. 5P 2 - 2. 23. (Equiyalence relation) By definition, an equivalence relation on a set is a relation satisfying three conditions (named as indicated): (i) Each element A of the set is equivalent to itself ( "R~f7exivity"). (iil If A is equi\'alent to B. then B is equivalent to A ("SYlIlllletn- "). (iii) If A is equivalent to B and B is equivalent to C, then A is equivalent to C ("Transitivity"). Show that row equivalence of matrices satisfies these three conditions. Him. Show that for each of the three elementary row operations these conditions hold. rather than by o o o o o o 1 o o o o o o o o o o o o o o o o o o o o o o o o o o o o 8 Apply E10 E 2 , E3 to a vector and to a 4 X 3 matrix of your choice. Find B = E3E2EIA, where A = [ajk] is the general 4 X 2 matrix. Is B equal to C = EIE2E3A? (b) Conclude that Eb E 2 , E3 are obtained by doing the corresponding elementary operations on the 4 X 4 unit matlix. Prove that if lVl is obTained from A hy an elementary rOil' operation. then M=EA, 24. PROJECT. Elementary Matrices. The idea is that elementary operations can be accomplished by matrix multiplication. If A is an 111 X Il matrix on which we want to do an elementary operation, then there is a matrix E such that EA is the new matrix after the operation. Such an E is called an elementary matrix. This idea can be helpful, for instance. in the design of algorithms. (Computationally, it is generally preferable 7.4 where E is obtained from the the same row operation. 11 X Il unit matrix In by 25. CAS PROJECT. Gauss Elimination and Back Substitution. Write a program for Gauss elimination and back substitution (a) that does not include pivoting, (b) that does include pivoting. Apply the programs to Probs. 13-16 and to some larger systems of your choice. Linear Independence. Rank of a Matrix. Vector Space In the last section we explained the Gauss elimination with back substitution, the most important numeric solution method for linear systems of equations. It appeared that such a system may have a unique solution or infinitely many solutions. or it may be inconsistent, that is, have no solution at alL Hence we are confronted with the questions of existence and uniqueness of solutions. We shall answer these questions in the next section. As the SEC. 7.4 297 Linear Independence. Rank of a Matrix. Vector Space key concept for this (and other questions) we introduce the rallk of a matrix. To define rank, we first need the following concepts, which are of general importance. Linear Independence and Dependence of Vectors Given any set of 111 vectors ~1J' . • • , ~m) (with the same number of components), a linear combination of these vectors is an expression of the form where Cl' C2, ••• , em are any scalars. Now consider the equation (1) Clearly, this vector equation (I) holds if we choose all c/s zero, because then it becomes O. [f this is the only m-tuple of scalars for which (1) holds, then our vectors a(1), ... , a('m) are said to fOlm a linearly independent set or, more briefly, we call them linearly independent. Otherwise, if (1) also holds with scalars not all zero, we call these vectors linearly dependent, because then we can express (at least) one of them as a linear combination of the others. For instance, if (l) holds with, say, Cl =1= 0, we can solve (I) for a(1): o= (Some k/s may be zero. Or even all of them, namely, if a(1) = 0.) Why is this important? Well, in the case of linear dependence we can get rid of some of the vectors until we anive at a linearly independent set that is optimal to work with because it is smallest possible in the sense that it consists only of the "really essential" vectors, which can no longer be expressed linearly in terms of each other. This motivates the idea of a "basis" used in various contexts, notably later in our present section. E X AMP L E 1 Linear Independence and Dependence The three vectors 3 0 2 2] [-6 42 24 54J = [21 -21 o -15] 3{l)=[ 3(2) = 3(3) are linearly dependent because Although this is easily checked (do it!), it is not so ea~y to discover. However. a systematic method for finding out about linear independence and dependence follows below. The first two of the three vectors are linearly independent because c13m + c23c2) = 0 implies c2 = 0 (from • the second components) and then C1 = 0 (from any other component ot 3(U)' Rank of a Matrix DEFINITION The rank of a matrix A is the maximum number of linearly independent row vectors of A. It is denoted by rank A. 298 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Our further discussion will show that the rank of a matrix is an important key concept for understanding general properties of matrices and linear systems of equations. EXAMPLE 2 Rank The matnx [ (2) 2 42 24 54 -21 0 -15 J ~ = ~~ '] 0 has rank 2. because Example 1 shows that the first two TO'" vectors are linearly independent. whereas all three row vectors are linearly dependent. Note further that rank A = 0 if and only if A - O. This follows directly from the definition. • We call a matrix Al row-equivalent to a matrix A2 if Al can be obtained from A2 by (finitely many!) elementary row operations. Now the maximum number of linearly independent row vectors of a matrix does not change if we change the order of rows or multiply a row by an nonzero c or take a linear combination by adding a multiple of a row to another row. This proves that rank is invariant under elementary row operations: THEOREM 1 Row-Equivalent Matrices Row-equivalent matrices hal'e the slime rank. Hence we can determine the rank of a matrix by reduction to row-echelon form (Sec. 7.3) and then see the rank directly. E X AMP L E 3 Determination of Rank For the matrix in Example 2 we obtain successively '] '] A+: 42 24 54 21 -21 0 -IS 3 0 2 0 42 28 58 0 -21 14 -29 3 0 2 0 42 28 0 0 0 [ [ 0 2 ':] (given) Row 2 + 2 Row I Row 3 - 7 Row I Row 3 +! Row 2 Since rank is defined in terms of two vectors, we immediately have the useful THEOREM 2 Linear Independence and Dependence of Vectors p vectors with 11 components each are linearly i1ldependent if the matrix with these vectors as row vectors has rank p, but they are linearly dependent if that rank is less than p. • SEC. 7.4 Linear Independence. Rank of a Matrix. Vector Space 299 Further impOltant properties will result from the basic Rank in Terms of Column Vectors THEOREM 3 The rank r of a matrix A equals the maximum number of linearly independent column vectors of A. Hence A alld its transpose AT have the same rallk. PROOF In this proof we write simply "rows" and "columns" for row and column vectors. Let A be an 171 X n malIu of rank A = r. Then by definition of rank, A has r Linearly independent rows which we denote by v(1), ... , V(T) (regardless of their position in A), and all the rows a(l), • . • , a(m) of A are linear combinations of those, say, (3) These are vector equations for rows. To switch to columns, we write (3) in terms of components as n such systems, with k = ], . . . , n, alk = cnulk + Cl2 U 2k + ... + CITUTk a2k = C21 U lIc + C22 U 2k + ... + C2T U Tk ({mk = CmlUlk + c'm2 u 2k + ... + CmTUTk (4) and collect components in columns. Indeed. we can write (4) as cn ({Ik ({2k (5) C12 C2 1 = C22 + U 1k U2k C",I ({mk cIT + ... + C2T UTk C.n~T Cm.2 where k = I,· .. , n. Now the vector on the left is the hh column vector of A. We see that each of these n columns is a linear combination of the same r columns on the right. Hence A cannot have more Linearly independent columns than rows, whose number is rank A = r. Now rows of A are columns of the transpose AT. For AT our conclusion is that AT cannot have more linearly independent columns than rows, so that A cannot have more linearly independent rows than columns. Together, the number of Linearly independent columns of A must be r, the rank of A. This completes the proof. • E X AMP L E 4 Illustration of Theorem 3 The matrix in (2) has rank 2. From Example 3 we see that the first two row vectors are linearly independent and by "working backward" we can verify that Row 3 = 6 Row I Row 2. Similarly, the first two columns are linearly independem. and by reducing the last matnx in Example 3 by columns we find that -i Column 3 = ~ Column I + ~ Column 2 and Column 4 = ~ Column I + ~ Column 2. • CHAP. 7 300 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Combining Theorems 2 and 3 we obtain THEOREM 4 Linear Dependence of Vectors p vectors witll n < p components are always linearly dependent. PROOF The matrix A with those p vectors as row vectors has p rows and 11 < P columns; hence by • Theorem 3 it has rank A ~ II < p, which implies linear dependence by Theorem 2. Vector Space The following related concepts are of general interest in linear algebra. In the present context they provide a clarification of essential properties of matrices and their role in connection with linear systems. A vector space is a (nonempty) set V of vectors such that with any two vectors a and b in Vall their linear combinations aa + f3b (a, f3 any real numbers) are elements of V, and these vectors satisfy the laws (3) and (4) in Sec. 7.1 (written in lowercase letters a, b, u, ... , which is our notation for vectors). (This definition is pre~ently sufficient. General vector spaces will be discussed in Sec. 7.9.) The maximum number of linearly independent vectors in V is called the dimension of Vand is denoted by dim V. Here we assume the dimension to be finite; infinite dimension will be defined in Sec. 7.9. A linearly independent set in V consisting of a maximum possible number of vectors in V is called a basis for V. Thus the number of vectors of a basis for V equals dim V. The set of all linear combinations of given vectors a(l), . . . , alP) with the same number of components is called the span of these vectors. Obviously, a span is a vector space. By a subspace of a vector space V we mean a nonempty subset of V (including V itself) that forms itself a vector space with respect to the two algebraic operations (addition and scalar multiplication) defined for the vectors of V. E X AMP L E 5 Vector Space, Dimension, Basis The span of the three vecrors in Example I is a vector space of dimension 2, and a basis is 3(1), 3(2), for instance, or 3(l). 3(3), etc. • We further note the simple THEOREM 5 Vector Space R" Tile vector space Rn consisting of all vectors with n cOlllpOnel1lS has dimension 11. PROOF A basis of 3cn) = [0 11 vectors is aU) 0 1]. [1 0 [0 (11 real numbers) o 0], ... , • In the case of a matrix A we call the span of the row vectors the row space of A and the span of the column vectors the column space of A. SEC. 7.4 Linear Independence. Rank of a Matrix. Vector Space 301 Now, Theorem 3 shows that a matrix A has as many linearly independent rows as columns. By the definition of dimension, their number is the dimension of the row space or the column space of A. This proves Row Space and Column Space THEOREM 6 The row space and the column space ofa matrix A have the same dimension, equal to rank A. Finally, for a given matrix A the solution set of the homogeneous system Ax = 0 is a vector space, called the null space of A, and its dimension is called the nullity of A. In the next section we motivate and prove the basic relation rank A (6) 11-121 + nullity A = Number of columns of A. 2 3 4 2 3 4 5 3 4 5 6 4 5 6 7 RANK, ROW SPACE, COLUMN SPACE Find the rank and a basis for the row space and for the column space. Hint. Row-reduce the matrix and its transpose. (You may omit obvious factors from the vectors of these bases.) 10. 2 4 16 8 842 16 11. ~J b 4. [: a 4 8 16 2 2 16 8 4 o o o o 7 5 0 -7 5 0 :2 o :2 0 12. o 4 o 2 o 4 0 o 4 8 -2 3 -4 113--20 I 2 -3 4 -1 Are the following sets of vectors linearly independent? (Show the details.) 2 3 -4 o 4 -[ 8. 7. o 3 0 o 5 8 -37 3 8 7 0 o -37 o 37 9. -2 2 -3 13. [3 [2 14. [1 LINEAR INDEPENDENCE -2 0 4], [5 0 0 0 0 3] 0], [1 0 0]. [1 1], L-6 [ 0 1] 15. [6 0 3 1 4 2], [0 -1 [12 3 0 -19 8 -11] 2 7 16. [3 4 7], [2 0 3], [8 2 17. [0.2 1.2 5.3 2.8 1.6], [4.3 3.4 0.9 2.0 -4.3] 3], [5 0 5 5], 6] I], CHAP. 7 302 18. [3 2 19. [ I I [! 20. [I 2 ~ 5 2 I]. [0 0 0]. [4 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems 3 25. If the row vectors of a square matrix are linearly independent. so are the column vectors. and conversely. 61 ! !]. [~ ! ! H [! ! ! !]. ! t] 3 4], [2 3 4 5]. [3 4 5 26. Give examples showing that the rank of a product of matrices cannot exceed the rank of either factor 6], [4 5 6 7] 21. CAS Experiment. Rank. (a) Show experimentally that the 11 X II matrix A = [ajk] with lIjk = j + k - I has rank 2 for any 11. (Problem 20 shows 11 = 4.) Try to prove it. (b) Do the same when lIjk = j + k + c. where c is any positi ve integer. (c) What is rank A if ajk = 2 j + k - 2 ? Try to find other large matrices of low rank independent of 11. 122-261 PROPERTIES OF RANK AND CONSEQUENCES Show the following. 22. rank BTAT = rank AB. (Note the order!) 23. rank A = rank B does lIot imply rank A2 (Give a counterexample.) VECTOR SPACES Is the given set of vectors a vector space? (Give reason.) If your answer is yes, determine the dimension and find a basis. (Vb V2, • . . denote components.) 27. All vectors in such that VI 2V2 - 28. All vectors in R4 such that 29. All vectors in R3 with 30. All vectors in R2 31. All vecrors in R3 rank B2. = 0 3v 4 = k VI ~ O. V 2 = -4V3 with VI ~ V2 with 4VI 32. All vectors in R4 with VI - 33. All vectors in R with = + V2 R3 n 24. If A is not square, either the row vectors or the column vectors of A are linearly dependent. 7.5 127-361 + V3 = V2 = IvA ~ O. 0, 3v 2 = V3 V3 = 5v I • v 4 = I for j = I, ... 0 ,11 34. All ordered quadruples of positive real numbers 35. All vectors in R 5 with VI = 2V2 = 3V3 = 4V4 = 5v5 36. All vectors in R4 with 3VI - V3 = O. 2VI + 3v 2 - 4V4 = 0 Solutions of Linear Systems: Existence, Uniqueness Rank as just defined gives complete information about existence, uniqueness, and general structure of the solution set of linear systems as follows. A linear system of equations in 11 unknowns has a unique solution if the coefficient matrix and the augmented matrix have the same rank 11, and infinitely many solution ifthat common rank is less than 11. The system has no solution if those two matrices have different rank. To state this precisely and prove it. we shall use the (generally important) concept of a submatrix of A. By this we mean any matrix obtained from A by omitting some rows or columns (or both). By definition this inclUdes A itself (as the matrix obtained by omitting no rows or columns); this is practical. THEOREM 1 Fundamental Theorem for Linear Systems (a) Existence. A linear SYSTem of m equaTions ill n unknowlls (1) Xl' . . . , Xn SEC. 7.5 Solutions of Linear Systems: Existence. Uniqueness 303 is consistent, that is, has solutions, !f and only augmented matrix A have the same rallk. Here, A= and (f the coefficient matrix A and the A= Q rnn (b) Uniqueness. The system (l) has precisely one solution ~f and only !f this common rank r of A and A equals n. (c) Infinitely many solutions. {f this commOn rank r is less thann, the system (l) has infinitely mallY solutions. All of these solutions are obtained by determining r suitable unlmowns (whose submatrix of coefficients must have rank r) in tenl1S of the remaining n - r unknowns, to which arbitrary values can be assigned. (See Example 3 in Sec. 7.3.) (d) Gauss elimination (Sec. 7.3). If solutions exist, they can all be obtained by the Gauss elimination. (This method will automatically reveal whether or not solutions exist; see Sec. 7.3.) PROOF (a) We can write the system (I) in vector form Ax = b or in terms of column vectors c(l), • • • , c(n) of A: (2) A is obtained by augmenting A by a single column b. Hence, by Theorem 3 in Sec. 7.4, rank A equals rank A or rank A + 1. Now if (1) has a solution x, then (2) shows that b must be a linear combination of those column vectors, so that A and A have the same maximum number of linearly independent column vectors and thus the same rank. Conversely, if rank A = rank A, then b must be a linear combination of the column vectors of A, say, (2*) since otherwise rank Xl = 0'1' . . . • Xn = A= an, rank A + 1. But (2*) means that (1) ha<; a solution. namely, as can be seen by comparing (2*) and (2). (b) If rank A = n. the n column vectors in (2) are linearly independent by Theorem 3 in Sec. 7.4. We claim that then the representation (2) of b is unique because otherwise This would imply (take all terms to the left, with a minus sign) and Xl scalars - Xl 0, ... , Xn - xn = 0 by linear independence. But this means that the in (2) are uniquely determined, that is, the solution of (l) is unique. Xl, . . . , Xn 304 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems (c) If rank A = rank A = I' < Il, then by Theorem 3 in Sec. 7.4 there is a linearly independent set K of I' column vectors of A such that the other n - I' column vectors of A are linear combinations of those vectors. We renumber the columns and unknowns, denoting the renumbered quantities by A, so that (C(1), ... , c(r)} is that linearly independent set K. Then (2) becomes are linear combinations of the vectors of K, and so are the vectors Expressing these vectors in terms of the vectors of K and collecting terms, we can thus write the system in the form CCr+l)' . . . , c(n) Xr+IC(r+U' • . . • xnc(n)' (3) with Xi = Xj + {3j, where {3j resulls from the 11 - I' terms c(r+UXr+b ••. , c(n)xn ; here, = 1, ... , r. Since the system has a solution, there are Yt> ... , Yr satisfying (3). These scalars are unique since K is linearly independent. Choosing xr+l> ... , xn fixes the {3j and corresponding Xj = )J - {3j, where j = I,' .. , r. j (d) This was discussed in Sec. 7.3 and is restated here as a reminder. • The theorem is illustrated in Sec. 7.3. In Example 2 there is a unique solution since rank A = rank A = n = 3 (as can be seen from the last matrix in the example). In Example 3 we have rank A = rank A = 2 < n = 4 and can choose X3 and X4 arbitrarily. In Example 4 there is no solution because rank A = 2 < rank A = 3. Homogeneous Linear System Recall from Sec. 7.3 that a linear system (I) is called homogeneous if all the b/ s are zero, and nonhomogeneous if one or several b/ s are not zero. For the homogeneous system we obtain from the Fundamental Theorem the following results. THEOREM 2 Homogeneous Linear System A homogeneolls linear system (4) always hm the trivial solution Xl = 0, ... , Xn = O. Nontrivial solutions exist ~f and ollly if rallk A < 11. ff rank A = I' < n, these solutions. together with x = 0, form a vector :;pace (5ee Sec. 7.4) of dimension n - 1', called the solution space of (4). III particular, !fXcl) and x(2) are solution vectors qf(4), then x = c l x(1) + C2Xc2) with any sC(llars CI and C2 is a solution vector qf (4). (This does not hold for nonhomogeneous systems. Also, the term solution space is used for homogeneous systems only.) SEC. 7.5 305 Solutions of Linear Systems: Existence, Uniqueness PROOF The first proposition can be seen directly from the system. It agrees with the fact that b = 0 implies that rank A = rank A, so that a homogeneous system is always consistent. If rank A = n, the trivial solution is the unique solution according to (b) in Theorem l. If rank A < n, there are nontrivial solutions according to (c) in Theorem 1. The solutions form a vector space because if x(l) and Xt.2) are any of them, then AX(1) = 0, AXt.2) = 0, and this implies A(x(1) + X(2) = AX(l) + AX(2) = 0 as well as A(cx(1) = cAx(1) = 0, where c is arbitrary. If rank A = r < n, Theorem I (c) implies that we can choose n - r suitable unknowns. call them Xr+ 10 ••• , X n , in an arbitrary fashion, and every solution is obtained in this way. Hence a basis for the solution space, briefly called a basis of solutions of (4), is Y(1), •.• , Y(n-r), where the basis vector Y(j) is obtained by choosing xr+j = 1 and the other xr+ 1, . . . , xn zero; the corresponding first I' components of this solution vector are then determined. Thus the solution space of (4) has dimension n - r. This proves Theorem 2. • ° The solution space of (4) is also called the null space of A because Ax = for every x in the solution space of (4). Its dimension is called the nullity of A. Hence Theorem 2 states that rank A + nullity A (5) =n where n is the number of unknowns (number of columns of A). Furthermore, by the definition of rank we have rank A ~ min (4). Hence if m < n, then rank A < 11. By Theorem 2 this gives the practically important THEOREM 3 Homogeneous Linear System with Fewer Equations Than Unknowns A homogeneous linear system with fewer equations than unknowns has always nontrivial solutions. Nonhomogeneous Linear Systems The characterization of all solutions of the linear system (I) is now quite simple. as follows. THEOREM 4 Nonhomogeneous Linear System !f a nonhomogeneous linear system (l) is consistent. then all of its solutions are obtained as (6) where Xo is any (fixed) solution Qf (l) and Xh runs through all the solutions Qf the corresponding homogeneous system (4). PROOF The difference Xh = x - Xo of any two solutions of (1) is a solution of (4) because xo) = Ax - Axo = b - b = 0. Since x is any solution of (1), we get all the solutions of (l) if in (6) we take any solution Xo of (l) and let Xh vary throughout the solution space of (4). • AXh = A(x - 306 7.6 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems For Reference: Second- and Third-Order Determinants We explain these determinants separately from the general theory in Sec. 7.7 because they will be sufficient for many of our examples and problems. Since this section is for reference, go on to the Ilext sectioll, consulting this material ollly when needed. A determinant of second order is denoted and defined by {/ll (1) D = det A = I a21 So here we have bars (whereas a matrix has brackets). Cramer's rule for solving linear systems of two equations in two unknowns (2) is Xl = Ihl {/121 b2 {/22 b 1{/22 - (/12 b2 D D (3) la X2 = ll bll b2 a21 l/llb2 D - b 1{/21 D with D as in (l), provided D"*O. ° The value D = appears for inconsistent nonhomogeneous systems and for homogeneous systems with nontrivial solutions. PRO 0 F We prove (3). To eliminate X2' multiply (2a) by Similarly, to eliminate Xl' multiply (2a) by "* {/22 -a21 and (2b) by and (2b) by all -a12 and add, and add. Assuming that D = all{/22 - {/12{/21 0, dividing, and writing the right sides of these two equations as detelminants, we obtain (3). • SEC. 7.6 307 For Reference: Second- and Third-Order Determinants E X AMP L E 1 Cramer's Rule for Two Equations 4XI + 3.\"2 = 12 2~1 + = -8 If then xl = 5x2 112 -8 :1 4 1 2 :1 = 84 14 =6 ' X2 = I: I: 121 8 --- -56 14 -4. • :1 Third-Order Determinants A determinant of third order can be defined by 12 a231_ a21 la a33 a32 (4) a131 a33 + a31 12 la a22 Note the following. The signs on the right are + - +. Each of the three terms on the right is an entry in the first column of D times its minor, that is, the second-order determinant obtained from D by deleting the row and column of that entry; thus. for all delete the first row and first column, and so on. If we write out the minors in (4), we obtain Cramer's Rule for Linear Systems of Three Equations (5) IS (6) (D *- 0) with the determinant D afthe system given by (4) and DI hi a12 al3 = h2 a22 a23 , h3 a32 a33 D2 all hI al3 = a2i h2 a23 , a 31 h3 a33 D3 all al2 hI = a21 a22 h2 a3i a32 h3 Note that D], D 2 , D3 are obtained by replacing Columns 1, 2, 3, respectively, by the column of the right sides of (5). Cramer's rule (6) can be derived by eliminations similar to those for (3), but it also follows from the general case (Theorem 4) in the next section. 308 7.7 CHAP. 7 Linear Algebra: Matrices, Vectors. Determinants. Linear Systems Determinants. Cramer's Rule Determinants were originally introduced for solving linear systems. Although impractical in computations, they have important engineering applications in eigenvalue problems (Sec. 8.1), differential equations, vector algebra (Sec. 9.3), and so on. They can be introduced in several equivalent ways. Our definition is particularly practical in connection with linear systems. A determinant of order n is a scalar associated with an 11 X 11 (hence square!) matrix A = [ajk]' which is written (1) D=detA= and is defined for n = (2) and for n (3a) I by D ~ = au 2 by (j = 1. 2..... or n) or (3b) Here, and Mjk is a determinant of order n - I. namely, the determinant of the submatrix of A obtained from A by omitting the row and column of the entry ajb that is, the jth row and the kth column. In this way, D is defined in terms of n determinants of order n - 1, each of which is, in turn, defined in terms of n - I determinants of order n - 2, and so on; we finally arrive at second-order determinants, in which those submatrices consist of single entries whose determinant is defined to be the entry itself. From the definition it follows that we may expand D by any row or column, that is, choose in (3) the entries in any row or column, similarly when expanding the Cjk's in (3), and so on. This definition is unambiguous, that is, yields the same value for D no matter which columns or rows we choose in expanding. A proof is given in App. 4. SEC. 7.7 309 Determinants. Cramer's Rule Terms used in connection with determinants are taken from matrices. In D we have n 2 entries ajk, also n rows and n columns, and a main diagonal on which all, a22, . . . , ann stand. Two terms are new: is called the minor of ajk in D, and Cjk the cofactor of ajk in D. For later use we note that (3) may also be written in terms of minors Mjk n D = (4a) 2: (- L)j+kajkMjk (j = 1, 2, ... , or n) k~l n D = (4b) 2: (-l)j+kajkMjk (k = 1, 2, ... , or n). j~l E X AMP L E 1 Minors and Cofactors of a Third-Order Determinant In (4) of the previous section the minors and cofactors of the entries in the first column can be seen directly. For the entries in the second row the minors are and the cofactors are C21 = -M2b C22 = +M22 , and C23 = -M23. Similarly for the third row-write these down yourself. And verify that the signs in Cjk fonn a checkerboard pattern + + + + E X AMP L E 2 • + Expansions of a Third-Order Determinant 3 D= 2 6 -1 o 0 = + 4) + 0(0 + 6) 1(12 - 0) - 3(4 = -12. This is the expansion by the first row. The expansion by the third colmlll is D = 0 I :I = 0 - 2 -I 12 + 0 = -12, • Verify that the other four expansions also give the value -12. E X AMP L E 3 Determinant of a Triangular Matrix -3 o 6 4 -1 2 :I = - 3· 4 . 5 = -60. Inspired by this, can you formulate a little theorem on determinants of triangular matrices? Of diagonal matrices? • CHAP. 7 310 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems General Properties of Determinants To obtain the value of a determinant (1), we can first simplify it systematically by elementary row operations. similar to those for matrices in Sec. 7.3. as follows. THEOREM 1 Behavior of an nth-Order Determinant under Elementary Row Operations (a) Interchange of two rows multiplies the value (~f the determinant by -1. (b) Addition of a multiple of a row to another roH' does not alter the value of the determinant. (e) Multiplication of a row by a IlOn;:.ero constant c multiplies the I'aille of the detenninant by c. (This holds also when c = 0, but gives no longer an elementary row operation.) PROOF (a) By induction. The statement holds for n = 2 because I: bld = ad - dlb = bc _ ad. but bc, We now make the induction hypothesis that (a) holds for detenninants of order n - I ~ 2 and show that it then holds for determinants of order 11. Let D be of order n. Let E be obtained from D by the interchange of two rows. Expand D and E by a row that is not one of those interchanged. call it the jth row. Then by (4a). n (5) D = L E = (-I)j+kajkMjk' L (-l)j+kajkNjk k=l k=l where Njk is obtained ti'om the minor Mjk of ajk in D by the interchange of those two rows which have been interchanged in D (and which Njk must both contain because we expand by another row!). Now these minors are of order 11 - I. Hence the induction hypothesis applies and gives Njk = -Mjk . Thus E = -D by (5). (b) Add c times Row i to Row j. Let i5 be the new determinant. Its entries in Row j are + CGik- If we expand i5 by this Row j, we see that we can write it as i5 = DI + cD2 , where DI = D has in Row j the ajk, whereas D2 has in that Row j the (/ik from the addition. Hence D2 has aik in both Row i and Row j. Interchanging these two rows gives D2 back. but on the other hand it gives -D2 by (a). Together D2 = -D2 = O. so that i5 = DI = D. ajk (e) Expand the determinant by the row that has been multiplied. CAUTION! E X AMP L E 4 det (cA) = c n det A (not c det A). Explain why. • Evaluation of Determinants by Reduction to Triangular Form Because of Theorem 1 we may evaluate determinants by reduction to triangular form. as in the Gauss elimination for a matrix. For instance (with the blue explanations always referring to the precedillg determinallt) D= 2 o 6 4 5 o o 2 6 8 9 -3 -I SEC 7.7 311 Determinants. Cramer's Rule 2 0 -4 6 0 5 9 -12 0 2 6 -1 0 8 3 10 2 0 -4 6 0 5 9 -12 0 0 2.4 3.8 R"v. 3 - 004 Row 2 0 0 -11.4 29.2 R"v. 4 - 1.6 Rov. 2 2 0 -4 6 0 5 9 12 0 0 2.4 0 0 -0 Row 2 2 Row I Rov. -l 1.5 Rov. I 3.8 47.25 Row 4 + 4.75 Row 3 • = 2·5·2.4· 47.25 = 1134. THEOREM 2 Further Properties of nth-Order Determinants (a)-(c) ill Theorem I hoLd also for coLumlls. (d) Trallsposition leaves the value of a detenninanl unaLtered. (e) A zero row or columll renders the value of a detennillant ~ero. (f) Proportional rows or columlls render the value of a determinant ::.ero. In particular, a detemlil1ant with two identical rows or columlls has the I'aille ~ero. PROOF (a)-(e) follow directly from the fact that a determinant can be expanded by any row column. In (d), transposition is defined as for matrices, that is, the jth row becomes the jth column of the transpose. (f) If Row j = c times Row i, then D = CDb where Dl has Row j = Row i. Hence an interchange of these rows reproduces Db but it also gives -D 1 by Theorem l(a). Hence Dl = 0 and D = cD l = O. Similarly for columns. • It is quite remarkable that the important L:oncept of the rank of a matrix A, which is the maximum number of linearly independent row or column vectors of A (see Sec. 7.4), can be related to determinants. Here we may assume that rank A > 0 because the only matrices with rank 0 are the zero matrices (see Sec. 7.4). THEOREM 3 Rank in Terms of Determinants = [ajk] has rank I' ~ I ~f and only if A has all I' X rSlliJ111atrix with non::.ero detel71zinant, ~l'hereas eve0' square suiJmatrix with more than I' rows that A has (or does IlOt have!) has determinant equal to zero. In particular, if A is square, n X n, it has rank 11 if and ol1ly if All 111 X n matrix A detA "* O. 312 CHAP. 7 PROOF Linear Algebra: Matrices, Vectors, Determinants. Linear Systems The key idea is that elementary row operations (Sec. 7.3) alter neither rank (by Theorem 1 in Sec. 7.4) nor the property of a determinant being nonzero (by Theorem 1 in this section). The echelon fonn A of A (see Sec. 7.3) has r nonzero row vectors (which are the first r row vectors) if and only if rank A = r. Let R be the r X r submatrix in the left upper corner of A (so that the entries of R are in both the first r rows and r columns of A). Now R is triangular, with all diagonal entries l'Jj nonzero. Thus, det R = r11 ... Ir,. =I=- O. Also det R =I=- 0 for the corresponding r X r submatrix R of A because R results from R by elementary row operations. Similarly, det S = 0 for any square submatrix S of r + I or more rows perhaps contained in A because the corresponding submatrix S of A must contain a row of zeros (otherwise we would have rank A ~ r + I), so that det S = 0 by Theorem 2. This proves the theorem for an m X n matrix. In particular. if A is square. n X n. then rank A = n if and only if A contains an 11 X n submatrix with nonzero determinant. But the only such submatrix can be A itself. hence detA =I=- O. • Cramer's Rule Theorem 3 opens the way to the classical solution formula for linear systems known as Cramer's rule 2 , which gives solutions as quotients of determinants. Cramer's rule is not practical ill computations (for which the methods in Secs. 7.3 and 20.1-20.3 are suitable), but is of theoretical interest in differential equations (Secs. 2.10, 3.3) and other theories that have engineering applications. THEOREM 4 Cramer's Theorem (Solution of Linear Systems by Determinants) (a) If a linear system qf n equatio/lS in the same /lumber of unknow/lS x I, • . . , Xn (6) has a nonzero coefficient determinant D = det A, the system has precisely one solution. This solution is given by the f017nulas (7) xn = (Cramer's rule) where Dk is the determinant obtained from D by replacing in D the kth columll by the column with the entries b I , . . . ,bn(b) Hence if the SYSTem (6) is homogeneous and D =I=- 0, it has only The Trivial soluTion Xl = 0, X2 = 0, ... , Xn = O. If D = 0, the homogeneous system also has nontrivial solutions. 20ABRIEL CRAMER (1704--1752), Swiss mathematician. SEC. 7.7 Determinants. Cramer's Rule PROOF The augmented matrix at most n. Now if (8) A of the system (6) is of size n X (n + 1). Hence its rank can be D=detA= then rank A = n by Theorem 3. Thus rank A = rank A. Hence. by the Fundamental Theorem in Sec. 7.5, the system (6) has a unique solution. Let us now prove (7). Expanding D by its kth column, we obtain (9) where Cik is the cofactor of entry (lik in D. If we replace the entries in the kth column of D by any other numbers, we obtain a new determinant, say, D. Clearly, its expansion by the kth column will be of the form (9), with alk, . . . , (Ink replaced by those new numbers and the cofactors Cik as before. In particular, if we choose as new numbers the entries (Ill, • . • , (lnl of the lth column of D (where I k), we have a new determinant D which T has twice the column [all (lnzl • once as its lth column. and once as its kth because of the replacement. Hence D = 0 by Theorem 2(f). [f we now expand b by the column that has been replaced (the kth column). we thus obtain *' (10) (l We now multiply the first equation in (6) by Clk on both sides. the second by the last by Cnk , and add the resulting equations. This gives *' k). C 2k , • • . • (11) Collecting terms with the same Xj' we can write the left side as From this we see that Xk is multiplied by Equation (9) shows that this equals D. Similarly, Equation (10) shows that this is zero when I simply xkD, so that (11) becomes Xl is multiplied by *' k. Accordingly, the left side of (11) equals CHAP. 7 314 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Now the right side of this is Dk as defined in the theorem. expanded by its kth column. so that division by D gives 0). This proves Cramer's rule. If (6) is homogeneous and D 0, then each Dk has a column of zeros, so that Dk = 0 by Theorem 2(e). and (7) gives the trivial solution. Finally, if (6) is homogeneous and D = 0, then rank A < 11 by Theorem 3, so that • nontrivial solutions exist by Theorem 2 in Sec. 7.5. "* Illustrations of Theorem ..J. for 11 = 2 and 3 are given in Sec. 7.6. and an important application of the present formulas will follow in the next section. --PJlOBLEM3£T~ 1. (Second-order detenninant) Expand a general secondorder determinant in four possible ways and show that the results agree. 2. (Minors, cofactors) Complete the list of minors and cofactors in Example 1. 3. (Third-order detenninant) Do the task indicated in Example 2. Also evaluate D by reduction to triangular form. 4. (Scalar multiplication) Show that det (kA) = k n det A (not k det A), where A is any 11 X 11 matrix. Give an example. 3 15. 81 -2 7. 6. 7 Icos a sin al sin f3 cos f3 0.3 0.8 0 0.5 2.6 0 0 -1.9 118 2 5 2 o 8 5 8-2 2 10. o o 3 -1 11. -3 0 -4 4 0 l/ 13. U 12. l/ U U W l/ 6 7 8 -I o 2 0 -4 -1 o 10 15 20 25 Time 0.004 sec 22 min 77 years 0.5' 109 years 18. 2x - 5y = 23 4x + 6y = -2 2 -2 19. 2 2 -2 0 a b -a 0 c -b -c 0 3y 4x 20. + 0 0 4 3 5 0 0 2 7 5 3w 14. 0 2 4 + 4::: 2y - y + = 14.8 = -6.3 5z = 13.5 7 w +2x 2w -2 0 5 118-201 CRAMER'S RULE Solve by Cramer's rule and check by Gauss elimination and back substitution. (Show details.) w w 4 11 x - 0 0-2 16. cos 118 14 8. 70.4 9. I -sin o o 2 o -2 17. (Expansion numericallJ impractical) Show that the computation of an nth-order determinant by expansion involves n! multiplications, which if a multiplication takes 10- 9 sec would take these times: 15-161 EVALUATION OF DETERMINANTS Evaluate, showing the details of your work. cos 118 sin 11 81 5.113 4 o o o o o 2 121-231 + - 3::: = 30 4x - 5)" + 8x - 4y + 2::: = 13 z = 42 + y - 5;: = 35 RANK BY DETERMINANTS Find the rank by Theorem 3 (which is not a very practical way) and check by row reduction. (Show details.) SEC 7.8 21. [-: (a) Line through two points. Derive from D = 0 in (12) the familiar fonnula -:J y - Yl 22.ll~ -13 1:] l-3 23. [ 5 -4 0.4 o -2.4 1.2 0.6 o 3.0] 0.3 o 1.2 1.2 o 24. TEAM PROJECT. Geometrical Applications: Curves and Surfaces Through Given Points. The idea is to get an equation from the vanishing of the detenninant of a homogeneous linear system as the condition for a nontrivial solution in Cramer's theorem. We explain the trick for obtaining such a system for the case of a line L through two given points PI: (x 1> Y 1) and P 2 : (X2, )'2)· The unknown line is ax + by = -c, say. We write it as ax + by + c· I = O. To get a nontrivial solution a, b, c, the determinant of the "coefficients" x, y, I must be zero. The system is (2) 7.8 315 Inverse of a Matrix. Gauss-Jordan Elimination ax + by + c· aXI + bYI + c· I o o o (Line L) (PIon L) (P 2 on L). (b) Plane. Find the analog of (12) for a plane through three given points. Apply it when the points are (I, I, I), (3, 2, 6), (5, 0, 5). (c) Circle. Find a similar formula for a circle in the plane through three given points. Find and sketch the circle through (2. 6). (6. 4). (7. I). (d) Sphere. Find the analog of the formula in (c) for a sphere through four given points. Find the sphere through (0, 0, 5), (4, 0, I), (0,4, I), (0, 0, 3) by this formula or by inspection. (e) General conic section. Find a fonnula for a general conic section (the vanishing of a detenninant of 6th order). Try it out for a quadratic parabola and for a more general conic section of your own choice. 25. WRITING PROJECT. General Properties of Determinants. Illustrate each statement in Theorems I and :2 with an example of your choice. 26. CAS EXPERIMENT. Determinant of Zeros and Ones. Find the value of the determinant of the n X n matrix An with main diagonal entries all 0 and all others I. Try to find a formula for this. Try to prove it by induction. Interpret A3 and ~ as "incidence lI1i1frices" (as in Problem Set 7.1 but without the minuses) of a triangle and a tetrahedron, respectively; similarly for an un-simplex". havingn vertices andn(n- l)/2edges (and spanning R"-I, 11 = 5,6, ... ). Inverse of a Matrix. Gauss-Jordan Elimination In this section we consider square matrices exclusively. The inverse of an n X n matrix A = [ajk] is denoted by A -1 and is an 11 such that X n matrix (1) where I is the n X 11 unit matrix (see Sec. 7.2). If A has an inverse, then A is called a nonsingular matrix. If A has no inverse, then A is called a singular matrix. If A has an inverse, the inverse is unique. Indeed, if both Band C are inverses of A, then AB = I and CA = I, so that we obtain the uniqueness from B = IE = (CA)B = CCAB) = CI = C. CHAP. 7 316 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems We prove next that A has an inverse (is nonsingular) if and only if it has maximum possible rank n. The proof will also show that Ax = b implies x = A -lb provided A-] exists. and will thus give a motivation for the inverse as well as a relation to linear systems (But this willilot give a good method of solving Ax = b Illlmerically because the Gauss elimination in Sec. 7.3 requires fewer computations.) THEOREM 1 Existence of the Inverse The inverse A-I of an n X n matrix A exists if and only if rank A = n, thus (by Theorem 3, Sec. 7.7) if and onZy if det A O. Hence A is nonsingular if rank A = n, and is singular if rank A < n. "* PROOF Let A be a given /1 X n matrix and consider the linear system (2) Ax = h. If the inverse A-I exists, then multiplication from the left on both sides and use of (1) gives A-lAx = x = A-lb. This shows that (2) has a unique solution x. Hence A must have rank /1 by the Fundanlental Theorem in Sec. 7.5. Conversely, let rank A = n. Then by the same theorem, the system (2) has a unique solution x for any b. Now the back substitution following the Gauss elimination (Sec. 7.3) shows that the components Xj of x are linear combinations of those of b. Hence we can write x (3) = Bb with B to be determined. Substitution into (2) gives Ax for any b. Hence C get = A(Bb) = (AB)b = Cb = b AB) = AB = I, the unit matrix. Similarly, if we substitute (2) into (3) we x for any x (and b (C = = Bb = B(Ax) = (BA)x = Ax). Hence BA = I. Together, B = A-I exists. • 3 WILHELM JORDAN (IR42-1899), German mathematician and geodesist. [See American Mathematical Monthly 94 (1987). 130-142.] We do not recommend it as a method for solving sy~tems of linear equations, since the number of operations in addition to those of the Gauss elimination is larger than that for back substitution, which the Gauss-Jordan elimination aVOlds. See also Sec. 20.1. SEC. 7.8 Inverse of a Matrix. Gauss-Jordan Elimination 317 Determination of the Inverse by the Gauss-Jordan Method For the practical determination of the inverse A-I of a nonsingular n X 11 matrix A we can use the Gauss elimination (Sec. 7.3), actually a variant of it, called the Gauss-Jordan elimination3 (footnote of p. 316). The idea of the method is as follows. Using A, we form n linear systems e(n) are the columns of the 11 X n unit matrix I; thus, where eel), em = [\ 0 O]T, e(2) = [0 I 0 OlT, etc. These are 11 vector equations in the unknown vectors xm, ... , x(n)' We combine them into a single matrix equation AX = I, with the unknown matrix X having the columns xm'···. x(n)' Correspondingly, we combine the n augmented matrices [A em],"', [A e(n)] into one n X 2n "augmented matrix" A = [A I]. Now multiplication of AX = I by A- 1 from the left gives X = A -II = A -1. Hence, to solve AX = I for X, we can apply the Gauss elimination to A = [A I]. This gives a matrix of the form [U H] with upper triangular U because the Gauss elimination triangularizes systems. The Gauss-Jordan method reduces U by further elementary row operations to diagonal form, in fact to the unit matrix I. This is done by eliminating the ennies of U above the main diagonal and making the diagonal entries all 1 by multiplication (see the example below). Of course, the method operates on the entire matrix rU Hl, transforming H into some matrix K, hence the entire [U H] to [I K]. This is the "augmented matrix" of IX = K. Now IX = X = A -t, as shown before. By comparison. K = A -t, so that we can read A- J directly from [I K]. The following example illustrates the practical details of the method. E X AMP L E 1 Inverse of a Matrix. Gauss-Jordan Elimination Determine the inverse A-I of A = [-~ ~l -1 -1 3 4 Solution. We apply the Gauss elimination (Sec. 7.3) to the following n X 2n always refers to the previous matrix. 3 X n matrix, where BLUE o 2 4 = o o 0 o 2 7 3 2 -] Row 2 + 3 Row] 0 Row 3 - Row] o 2 7 3 -5 -4 -I Row 3 - Row 2 318 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems This is IV H] as produced by the Gauss elimination. Now follow the additional Gauss-Jordan steps. reducing U to I, that is. to diagonal form with entries I on the main diagonal. [~ -I -2 -I 1.5 0.5 0.8 O.:! 0 0.6 0.4 0 1.3 -0.2 0.8 0.2 0 -0.7 0.2 0 -1.3 -0.2 0.8 0.2 3.5 0 -I [: 0 0 [: Row I 0 0 -0:] -0'] 0.7 0.5 Row 2 -0.2 Row 3 Rov, I 2 Rov, 3 Rov, 2 - 3.5 Row .3 -0.2 03] Rov, I + Row 2 0.7 -0.2 The last three columns constitute A -1. Check: [-; -I -) .3 2] [-07 0.2 L -1.3 -0.2 4 0.8 0.2 03] [' 0.7 = -0.2 0 0 0 0 :l • Hence AA- 1 = I. Similarly, A- 1 A = I. Useful Formulas for Inverses The explicit formula (4) in the following theorem is often useful in theoretical studies (as opposed to computing inverses). In fact, the special case 11 = 2 occurs quite frequently in geometrical and other applications. THEOREM 2 Inverse of a Matrix The inverse of a 110nsi11gular n X (4) A-I 11 matrix A = [ajk] is given by _ I = -I- [Cd T -- det A J Cl l C21 Cnl Cl2 C22 Cn2 C1n C2n Cnn detA where Cjk is the cofactor of ajk in det A (see Sec. 7.7). (CAUTION! Note well that in A -\ the cofactor Cjk occupies the same place as alrj (not ajk) does in A.) III particular. the inverse of (4*) is A-I = detA SEC. 7.8 Inverse of a Matrix. Gauss-Jordan Elimination PROOF 319 We denote the right side of (4) by B and show that BA = I. We first write (5) and then show that G = I. Now by the definition of matrix multiplication and because of the form of B in (4), we obtain (r AUTION! Csb not C ks ) (6) Now (9) and (l0) in Sec. 7.7 show that the sum ( ... ) on the right is D = det A when I = k, and is zero when I =1= k. Hence 1 detA = - - detA = 1, gkk = 0 gkZ (I k), =I=- In particular, for n = 2 we have in (4) in the first row C ll = the second row C 12 = -a2b C 22 = all' This gives (4*). E X AMP L E 2 Inverse of a 2 x = -a 12 and in • 2 Matrix A-I _ ~ [ 10 E X AMP L E 3 C 21 a22, -IJ 4 -2 = [ 3 0.4 -0.2 -O.IJ • 0.3 Further Illustration of Theorem 2 Using (4), find the inverse of Solution. Cl l = -1 -1 3 :] 4 We obtain detA = -1(-7) - 1'13 + 2·8 = 10, and in (4), I-I :1 3 CI2 = _I -1 CI3 I-1 -'I = A = [-: 3 3 3 = -7, :1 = = 8, -13, C2I = -I~ :1 C22 = = 1-1 :1 -1 1-1 ~I -1 = C23 = - I-11 2, C31 = -2, C32 = - = 2, C33 = :1 = 1-1 :1 3, 3 = 1-1 11 -1 = -2, 3 7, so that by (4), in agreement with Example 1, -0.7 A-I = -1.3 0.2 -0.2 0.8 0.2 0.3] 0.7. [ -0.2 • CHAP. 7 320 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Diagonal matrices A = [ajk]' (/jk = 0 when j Ojj =I=- O. Then A-I is diagonal, too, with entries PROOF =I=- k. have an inverse if and only if all l/onn. 1/(/11' • . . , For a diagonal matrix we have in (4) ell etc. D E X AMP L E 4 • Inverse of a Diagonal Matrix Let A= [ -0.5 0 OJ 0 4 O. ° 0 I Then the inverse is o 0.25 o • Products can be inverted by taking the inverse of each factor and mUltiplying these inverses in reverse order, (7) Hence for more than two factors, (8) PROOF The idea is to start from (I) for AC instead of A, that is, AC(Aq-1 = I, and mUltiply it on both sides from the left, first by A -t, which because of A -IA = I gives A-1AC(Aq-1 = C(Aq-1 = A-II = A-I, and then multiplying this on both sides from the left, this time by C- l and by using C-1C = I, This proves (7). and from it. (8) follows by induction. We also note that the inverse of the inverse is the given matrix, as you may prove, (9) • SEC. 7.8 :m Inverse of a Matrix. Gauss-Jordan Elimination Unusual Properties of Matrix Multiplication. Cancellation Laws Section 7.2 contains warnings that some properties of matrix multiplication deviate from those for numbers, and we are now able to explain the restricted validity of the so-called cancellation laws [2.] and [3.] below, using rank and inverse, concepts that were not yet available in Sec. 7.2. The deviations from the usual are of great practical importance and must be carefully observed. They are as follows. [1.] Matrix multiplication is not commutative, that is, in general we have AB =1= BA. [2.] AB = 0 does not generally imply A = 0 or B = 0 (or BA = 0); for example, [~ [3.] AC = AD does not generally imply C = D (even when A =1= 0). Complete answers to [2.] and [3.] are contained in the following theorem. THEOREM 3 Cancellation Laws Let A, B, C be n X n matrices. Then: (a) If rank A = nand AB = AC, then B = C. (b) lfrank A = n, then AB = 0 implies B = O. Hence if AB = 0, but A as well as B =1= 0, then rank A < n and rank B < n. =1= 0 (c) If A is singular, so are BA and AB. PROOF (a) The inverse of A exists by Theorem 1. Multiplication by A-I from the left gives A -lAB = A -lAC, hence B = C. (b) Let rank A = n. Then A -1 exists, and AB = 0 implies A -lAB = B = O. Similarly when rank B = n. This implies the second statement in (b). (c l ) Rank A < n by Theorem 1. Hence Ax = 0 has nontrivial solutions by Theorem 2 in Sec. 7.5. Multiplication by B shows that these solutions are also solutions of BAx = 0, so thaI rank (BA) < n by Theorem 2 in Sec. 7.5 and BA is singular by Theorem 1. (c 2 ) AT is singular by Theorem 2(d) in Sec. 7.7. Hence B TAT is singular by part (c 1 ), and is equal to (AB)T by (lOd) in Sec. 7.2. Hence AB is singular by Theorem 2(d) in Sec. 7.7. • Determinants of Matrix Products The detelminant of a matrix product AB or BA can be written as the product of the determinants of the factors, and it is interesting that det AB = det BA, although AB =1= BA in general. The corresponding formula (10) is needed occasionally and can be obtained by Gauss-Jordan elimination (see Example 1) and from the theorem just proved. 311 CHAP. 7 THE 0 REM 4 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Determinant of a Product of Matrices For am: n X n matrices A and B, (10) det (AB) = del (BA) = det A det B. If A or B is singular. so are AB and BA by Theorem 3(c), and (10) reduces to 0 = 0 by Theorem 3 in Sec. 7.7. Now let A and B be nonsingular. Then we can reduce A to a diagonal matrix A = [ajk] by Gauss-Jordan steps. Under these operations, det A retains its value, by Theorem I in Sec. 7.7, (a) and (b) [not (c)] except perhaps for a sign reversal in row interchanging when pivoting. But the same operations reduce AB to AB with the same effect on det (AB). Hence it remains to prove (10) for AB; written out, PROOF all 0 0 b ll b I2 bIn 0 a22 0 b 2I b 22 b 2n 0 0 ann b nl b n2 b nn AB= al1 b l1 al1 b I2 allb in a22 b 2I a22 b 22 a22 b 2n annbnl a nn b n2 annbnn all ... We now take the determinant det (AB). On the right we can take out a factor from the first row, a22 from the second, ... , ann from the nth. But this product ~2 ann equals det A because A is diagonal. The remaining determinant is det B. This proves (10) for det (AB), and the proof for det (BA) follows by the same idea. • all This completes our discussion of linear systems (Secs. 7.3-7.8). Section 7.9 on vector spaces and linear transformations is optional. Numeric methods are discussed in Secs. 20.1-20.4, which are independent of other sections on numerics . .• ~ ~-=-12! -.-.. ....... ... ..-._.. INVERSE Fmd the inverse by Gauss-Jordan [or by (4*) if 11 = 2] or state that it does not exist. Check by using (1). 1. [ 1.20 4.64J 0.50 3.60 0.6 2. [ 0.8 3. [ cos 28 0.8J -0.6 sin 28] -sin 28 cos 28 SEC. 7.9 Vector Spaces. Inner Product Spaces. Linear Transformations -11 -I S. [-I; -:] [: ] 29 6. [ -160 61 -55 -2 55 -21 19 0 7. [: [ J [ 8. 4 9. 10. 0 0 2 11. [: -1 4 7.9 10] 6 1:] 12. [-; 13. (Triangular matrix) Is the inver~e of a triangular matrix always triangular (as in Prob. 7)? Give reason. 14. (Rotation) Give an application of the matrix in Prob. 3 that makes the form of its inverse obvious. 15. (Inverse of the square) Verify (A2 r A in Prob. 5. 2 1] -I 4 :] 0 = (A-If for 16. Prove the formula in Prob. 15. 17. (Inverse of the transpose) Verify (AT) -1 for A in Prob. 5. [ 1-231 -4 -9] -1 2 2 1 = (A _1)T 18. Prove the formula in Prob. 17. 19. (Inverse of the inverse) Prove that (A -1)-1 = A. 20. (Row interchange) Same question as in Prob. 14 for the matrix in Prob. 9. 8 0 323 Optional 19 EXPLICIT FORMULA (4) FOR THE INVERSE Formula (4) is generally not very practical. To understand its use, apply it: 21. To Prob. 9. 22. To Prob. 4. 23. To Prob. 7. Vector Spaces, Inner Product Spaces, Linear Transformations Optional In Sec. 7.4 we have Seen that special vector spaces arise quite naturally in connection with matrices and linear systems, that their elements, called vectors, satisfy rules quite similar to those for numbers [(3) and (4) in Sec. 7.1], and that they are often obtained as spans (sets of linear combinations) of finitely many given vectors. Each such vector has n real numbers as its compollents. Look this up before going on. Now if we take all vectors with II real numbers as components ("real vectors"), we obtain the very important realll-dimensional vector space Rn. This is a standard name and notation. Thus, each vector in R n is an ordered n-tuple of real numbers. Pat1icular cases are R2, the space of all ordered pairs (""vectors in the plane") and R 3, the space of all ordered triples ("vectors in 3-space"). These vectors have wide applications in mechanics, geometry, and calculus that are basic to the engineer and physicist. Similarly, if we take all ordered n-tuples of complex numbers as vectors and complex numbers as scalars, we obtain the compleJ!: vector space en, which we shall consider in Sec. 8.5. This is not alL There are other sets of practical interest (sets of matrices, functions, transformations, etc.) for which addition and scalar multiplication can be defined in a natural way so that they foml a "vector space". This suggests to create from the "COil crete model" R n the "abstract cOllcept" of a "real vector space" V by taking the basic properties (3) and (4) in Sec. 7.1 as axioms. These axioms guarantee that one obtains a useful and applicable theory of those more general situations. Note that each axiom expresses a simple property of R n or, as a matter of fact. of R3. Selecting good axioms needs experience and is a process of trial and error that often extends over a long period of time. 324 DEFINITION CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Real Vector Space A nonempty set V of elements a, b, ... is called a real vector space (or real linear space), and these elements are called vectors (regardless of their nature, which will come out from the context or will be left arbitrary) if in V there are defined two algebraic operations (called vector addition and scalar multiplication) as follows. I. Vector addition associates with every pair of vectors a and b of V a unique vector of V, called the sum of a and b and denoted by a + b, such that the following axioms are satisfied. 1.1 Commutativity. For any two vectors a and b of V, a+b=b+a. 1.2 Associativity. For any three vectors u. v. w of V, (u + v) + w = u + (v + w) (written u + v + w). 1.3 There is a unique vector in V, called the zero vector and denoted by 0, such that for every a in V, a+O=a. 1.4 For every a in V there is a unique vector in V that is denoted by -a and is such that a + (-a) = O. II. Scalar multiplication. The real numbers are called scalars. Scalar multiplication associates with every a in V and every scalar c a unique vector of V, called the product of c and a and denoted by ca (or ac) such that the following axioms are satisfied. 11.1 Distributivity. For every scalar c and vectors a and b in V, c(a + b) = ca + cb. 11.2 Distributivity. For all scalars c and k and every a in V, (c + k)a = ca + ka. 11.3 Associativity. For all scalars c and k and every a in V, c(ka) = (ck)a (written cka). 11.4 For every a in V, la = a. A complex vector space is obtained if, instead of real numbers, we take complex numbers as scalars. SEC. 7.9 325 Optional Vector Spaces, Inner Product Spaces, Linear Transformations Basic concepts related to the concept of a vector space are defined as in Sec. 7.4. A linear combination of vectors a(l),"', a(m) III a vector space V is an expression (C1, .•. , C m any scalars). These vectors form a linearly independent set (briefly, they are called linearly independent) if (1) implies that C1 = 0, ... , Cm = O. Otherwise, if (1) also holds with scalars not all zero, the vectors are called linearly dependent. Note that (1) with 11l = I is ca = 0 and shows that a single vector a is linearly independent if and only if a =F O. V has dimension n, or is n-dimensional, if it contains a linearly independent set of n vectors, whereas any set of more than n vectors in V is linearly dependent. That set of n linearly independent vectors is called a basis for V. Then every vector in V can be written as a linear combination of the basis vectors; for a given basis, this representation is unique (see Prob. 14). E X AMP L E 1 Vector Space of Matrices The real 2 X 2 matrice, form a four-dimensional real vector space. A ba~is is ~J ~J because any 2 X 2 matrix A = [ajkJ has a unique representation A = allB11 + 012B12 + 021B21 + 022B22' Similarly. the real 111 X II matrices with fixed 111 and n form an mil-dimensional vector space. What is the • dimension of the vector space of all 3 X 3 skew-symmetric matrices'! Can you find a basis? E X AMP L E 2 Vector Space of Polynomials The set of all constant, linear, and quadratic polynomials in x together is a vector space of dimension 3 with basis {I. x, .r 2 } under the usual addition and multiplication by real numbers because these two operations give polynomials not exceeding degree 2. What is the dimension of the vector space of all polynomials of degree not exceeding a given fixed n'! Can you find a basis? • If a vector space V contains a linearly independent set of 11 vectors for every n, no matter how large, then V is called infinite dimensional, as opposed to a finite dimensional (n-dimensional) vector space just defined. An example of an infinite dimensional vector space is the space of all continuous functions on some interval [ll, b J of the x-axis, as we mention without proof. Inner Product Spaces If a and b are vectors in Rn, regarded as column vectors, we can form the product a Tb. This is a 1 X 1 matrix, which we can identify with its single entry, that is, with a number. This product is called the inner product or dot product of a and b. Other notations for it are (a, b) and a·b. Thus aTb = (a, b) = a·b = [al' .. an] [~1] : n = ~ alb l = alb l l=l bn + ... + anbn- 326 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems We now extend this concept to general real vector spaces by taking basic properties of (a, b) as axioms for an "abstract inner product'" (a, b) as follows. Real Inner Product Space DEFINITION A real vector space V is called a real inner product space (or real pre-Hilbert4 space) if it has the following property. With every pair of vectors a and b in V there is associated a real number, which is denoted by (a, b) and is called the inner product of a and b, such that the following axioms are satisfied. I. For all scalars ql and q2 and all vectors a, b, c in V, (Linearity). II. For all vectors a and b in V. (a, b) = (b, (Symmetry). a) III. For every a in V, (a, a) ~ 0, (Positive-definiteness). (a, a) =0 if and only if Vectors whose inner product is zero are called orthogonal. The length or norm of a vector in V is defined by lIall = Yea, a) (2) (~ 0). A vector of norm 1 is called a unit vector. From these axioms and from (2) one can derive the basic inequality (3) (Callchy-Schwarz5 inequality). From this follows (Triangle inequality). (4) A simple direct calculation gives (5) lIa + bll 2 + lIa - bll 2 = 2( lIall 2 + lib II 2) (Parallelogram equality). 4DAVID HILBERT (1862-1943), great Gennan mathematician, taught at Konigsberg and Gottingen and was the creator of the famous Gottingen mathematical schooL He is known for his basic work in algebra. the calculus of variations. integral equations, functional analysis, and mathematical logic. His "Foundations of Geometry" helped the axiomatic method to gain general recognition. His famous 23 problems (presented in 1900 at the International Congress of Mathematicians in Paris) considerably influenced the development of modem mathematics. If V b finite dimensional. it is actually a so-called Hilbert :lpace; see Ref. [GR7], p. 73, listed in App. L 5 HERMANN AMANDUS SCHWARZ (1843-1921). Gennan mathematician, known by his work in complex analysis (confonnal mapping) and differential geometry. For Cauchy see Sec. 2.5. SEC. 7.9 Optional Vector Spaces, Inner Product Spaces, Linear Transformations E X AMP L E 3 327 n-Dimensional Euclidean Space R n with the inner product (6) (where both a and b are column vectors) is called the n-dimensional Euclidean space and is denoted by En or again simply by Rn. Axioms I-III hold, as direct calculation shows. Equation (2) gives the "Euclidean norm" • (7) E X AMP L E 4 An Inner Product for Functions. Function Space The set of all reaT-valued continuous functions I(x), g(x), ... on a given interval a ::'" x ::'" f3 is a real vector space under the usual addition of functions and multiplication by scalars (real numbers). On this "function space" we can define an inner product by the integral {3 (f, g) = (8) {I(X) g(x) ,ll-. Axioms I-ITT can be verified by direct calculation. Equation (2) gives the norm (3 IIIII (9) Y(f, I) = = • {f"(X)2 d". Our examples give a first impression of the great generality of the abstract concepts of vector spaces and inner product spaces. Further details belong to more advanced courses (on functional analysis. meaning abstract modern analysis; see Ref. [OR7] listed in App. 1) and cannot be discussed here. Instead we now take up a related topic where matrices play a central role. Linear Transformations Let X and Y be any vector spaces. To each vector x in X we assign a unique vector y in y. Then we say that a mapping (or transformation or operator) of X into Y is given. Such a mapping is denoted by a capital letter, say F. The vector y in Yassigned to a vector x in X is called the image of x under F and is denoted by F(x) [or Fx, without parentheses J. F is called a linear mapping or linear transformation if for all vectors v and x in X and scalars c, F(v + x) = F(v) + F(x) (10) F(cx) = cF(x). Linear Transformation of Space R n into Space R m From now on we let X = Rri and Y = RIn. Then any real m X n matrix A = [ajk] gives a transformation of R n into Rnl, y (11) Since A(u + x) = Ax. = Au + Ax and A(cx) = cAx, this transformation is linear. We show that, conversely, every linear transformation F of R" into R'm can be given in terms of an 111 X n matrix A, after a basis for R n and a basis for R m have been chosen. This can be proved as follows. 328 CHAP. 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Let em, . . . . e(n) be any basis for Rn. Then every x in R n has a unique representation Since F is linear, this representation implies for the image F(x): Hence F is uniquely determined by the images of the vectors of a basis for R". We now choose for R n the "standard basis" o o o o (12) o o where e(j) has its jth component equal to 1 and all others O. We show that we can now determine an I1l X n matrix A = [~jk] such that for every x in R n and image y = F(x) in R m , y = F(x) = Ax. Indeed, from the image y(1) = F(e(1) of e(n we get the condition o o \" 0) .m from which we can determine the first column of A, namely lIll = yill. (/21 = y~l), ... , = Y1~1)· Similarly, from the image of e(2) we get the second column of A, and so on. This completes the proof. • amI We say that A represents F, or is a representation of F, with respect to the bases for R'n and Rm. Quite generally, the purpose of a "representation" is the replacement of one object of study by another object whose properties are more readily apparent. In three-dimensional Euclidean space £3 the standard basis is usually written eO) = i, e(2) = j, e(3) = k. Thus. (13) j SEC. 7.9 Optional Vector Spaces, Inner Product Spaces, Linear Transformations 329 These are the three unit vectors in the positive directions of the axes of the Cartesian coordinate system in space, that is, the usual coordinate system with the same scale of measurement on the three mutually perpendicular coordinate axes. E X AMP L E 5 Linear Transformations Interpreted as transformations of Cartesian coordinates in the plane, the matrices [1 OJ 1J [0I 0 ' ° represent a reflection in the line x2 = (when a> 1, or a contraction when °< E X AMP L E 6 [a OJ [ . 1 OJ ° -I' ° -I' I a reflection in the xraxis, a reflection in the origin. and a stretch a < 1) in the xrdirection, respectively. • Xl, Linear Transformations Our discussion preceding Example 5 is simpler than it may look at first sight. To see this, find A representing the linear transfonnation that maps (Xl, X2) onto (~XI - 5X2' 3XI + 4X2)' Solution. Obviously, the transfonnation is From this we can directly see that the matrix is [ YIJ [23 -5J4 [XIJ Check: • = Y2 X2 If A in (11) is square, 11 X 11, then (11) maps R'n into Rn. If this A is nonsingular, so that A -1 exists (see Sec. 7.8), then multiplication of (11) by A - I from the left and use of A -lA = I gives the inverse transformation x (14) = A-1y. It maps every y = Yo onto that x, which by (11) is mapped onto Yo. The inverse of a linear transformation is itself linear. because it is given by a matrix, as (14) shows . . .R 0 B L E M. r=uJ 5.£"E7.:~'F VECTOR SPACES 6. All vectors in (Additional problems in Problem Set 7.4.) Is the given set (taken with the usual addition and scalar multiplication) a vector space? (Give a reason.) If your answer is yes. find the dimension and a basis. 1. All vectors in R3 2. All vectors in VI - 4V2 + V3 satisfying 5VI - satisfying = 0 R3 2VI 3v 2 + + 2V3 = 3V2 - 0 V3 3. All 2 X 3 matrices with all entries nonnegative 4. All symmetric 3 x 3 matrices S. All vectors in R 5 with the first three components 0 0, R4 with VI + V2 = 0, V3 - V4 = I 7. All skew-symmetric 2 X 2 matrices 8. All n X n matrices A with fixed n and det A = 0 9. All polynomials with positive coefficients and degree 3 or less 10. All functions f(x) constants a and b = a cos x + h sin x with any 11. All functions I(x) = (ax + b)e-X with any constants a and b 12. All 2 X 3 matrices with the second row any multiple of [4 0 -9] CHAP. 7 330 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems 13. (Different bases) Find three bases for R2. 14. (Uniqueness) 20. Y1 = the representation v = c1a(1) + ... + cna(n) of any given vector in an n-dimensional vector space V in terms of a given basis a(1)' ... , a(n) for V is unique. [IS-20 1 Show that Find the inverse transformation. (Show the details of your work.) X2 Y2 = 3X1 - x 2 17. Y1 = 3X1 - X2 18. Y1 = 0.25x1 Y2 = 19. h = .. ---. ..... --. INNER PRODUCT. ORTHOGONALITY 121-261 LINEAR TRANSFORMATIONS 16. Y1 = 5X1 - )'2 = ...-. Find the Euclidean nonn of the vectors 21. [4 2 -6]T 22. [0 -3 3 0 5 I]T 23. [16 -32 O]T 24. [~ 2S. [0 I ! 26. [~ -~ 1 2f 0 -1 0 if 27. (Orthogonality) Show that the vectors in Probs. 21 and 23 are orthogonaL 28. Find all vectors v in R3 orthogonal to [2 0 I]T. 29. (Unit vectors) Find all unit vectors orthogonal to [4 -3]T. Make a ~ketch. 30. (Triangle inequality) Verify (4) for the vectors in Probs. 21 and 23. TIONS AND PROBLEMS 1. What properties of matrix multiplication differ from those of the multiplication of numbers? What about division of matrices? 11. 9x - 3y = 15 2. Let A be a 50 x 50 matrix and B a 50 X 20 matrix. Are the following expressions defined or not? A + B, A 2 , B2, AB. BA. AAT. BTA. BTB, BBT, BTAB. (Give reasons.) 3. How is matrix mUltiplication motivated? 12. -2x - 4y 4. Are there any linear systems without solutions? With one solution? With more than one solution? Give simple examples. S. How can you give the rank of a matrix in tenns of row vectors? Of column vectors? Of determinants? 6. What is the role of rank in connection with solving linear systems? 7. What is the row space of a matrix? The column space? The null space? 8. What is the idea of Gauss elimination and back substitution? 9. What is the inverse of a matrix? When does it exist? How would you determine it? 10. What is Cramer's rule? When would you apply it? IU-191 LINEAR SYSTEMS Find all solution~ or indicate that no solution exists. (Show the details of your work.) -l]T 5x + 4y x 13. 3x x + = 2y 48 + + 7z = -6 16:;: = 3 + 5y - 8z = ] R + 3z = 2}' - + 15. - 8 x 6y 12x + 14. 5x - 6 = -x + 2 v = 13 6 6y = 2y + z 5x - 4)" + = 3z = 0 18. -x + 4y - 2z = 5x - 4)' = 47 3x + 4)" 17. 3x 6x + + 7y = 9y = 15 19. 7 x + 9)' - 14z = -x - 3y + 2x + Y - x - 2y 36 2z = -12 4z = 4 -] z=-12 2x+3y- 3 =2 2y + 16. 2z = + 4z 3x lOy = + 6z = + 2z = 32 331 Summary of Chapter 7 120-301 CALCULATIONS WITH MATRICES AND VECTORS Calculate the following expressions (showing the details of your work) or indicate why they do not exist, when A ~ r: 2 IT r-: [l b~ B= 18 -6 15 10 a= 2 0 3 -J 21. A - AT + B2 T 24. AA , ATA 23. det A, det B, det AB 26. Aa, aTA, aTAa 27. aTb, bTa, ab T 28. bTBb 29. aTB, ETa 22. A2 + Find the inverse or state why it does not exist. (Show details.) 37. 38. 39. 40. 41. 42. Of the coefficient matrix in Prob. 11 Of the coefficient matrix in Prob. 15 Of the coefficient matrix in Prob. 16 Of the coefficient matrix in Prob. 18 Of the augmented matrix in Prob. 14 Of the diagonal matrix with entries 3, -1, 5 143--451 NETWORKS Find the currents in the following networks. [] 20. AB, BA 30. O.I(A INVERSE 137-@ 12 II 32. In Frob. 12 33. In Prob. 17 35. In Prob. 19 36. In Prob. 18 --- 3400 V 40 Q 100Q 131-361 RANK Determine the ranks of the coefficient matrix and the augmented matrix and state how many solutions the linear system will have. 31. In Frob. 13 - 20U 220V 45. 34. In Prob. 14 3800 V 44. 13 25. O.2BBT AT)(B - BT) lOQ 43. ~1020 V ~540V 20Q Linear Algebra: Matrices, Vectors, Determinants Linear Systems of Equations An m X n matrix A = [ajk] is a rectangular array of numbers or functions ("entries", "elements") arranged in 111 horizontal rows and n vertical columns. If 111 = n, the matrix is called square. A 1 X 11 matrix is called a row vector and an m X 1 matrix a column vector (Sec. 7.1). The sum A + B of matrices of the same size (i.e., both m X n) is obtained by adding corresponding entries. The product of A by a scalar c is obtained by multiplying each ajk by c (Sec. 7.1). The product C = AB of an m X n matrix A by an r X p matrix B = [bjk ] is defined only when r = n, and is the 111 X P matrix C = [Cjk] with entries (1) (row j of A times column k of B). 332 CHAP. 7 Linear Algebra: Matrices. Vectors. Determinants. Linear Systems This multiplication is motivated by the composItIOn of linear transfonnations (Secs. 7.2, 7.9). It is associative, but is 1I0t commutative: if AB is defined, BA may not be defined, but even if BA is defined, AB =t- BA in general. Also AB = 0 may not imply A = 0 or B = 0 or BA = 0 (Secs. 7.2, 7.8). Illustrations: [~ ~J [-: = [~ ~J [-1 IJ [I2 ~J = [-: -:J 2] [:J = [11], [:J [I 2] = [: J [l -:J -1 :J. The transpose AT of a matrix A = [ajk] is AT = [akj]; rows become columns and conversely (Sec. 7.2t Here. A need not be square. If it is and A = AT, then A is called symmetric; if A = _AT, it is called skew-symmetric. For a product. (AB)T = BTAT (Sec. 7.2). A main application of matrices concerns linear systems of equations (2) Ax = b (Sec. 7.3) (m equations in n unknowns Xl' . . . ,xn ; A and b given). The most important method of solution is the Gauss elimination (Sec. 7.3), which reduces the system to "triangular" form by elementary row operations. which leave the set of solutions unchanged. (Numeric aspects and variants. such as Doolittle's and Cholesky's methods. are discussed in Secs. 20.1 and 20.2) Cramer's rule (Sees. 7.6, 7.7) represents the unknowns in a system (2) of n equations in Il unknowns as quotients of determinants; for numeric work it is impractical. Determinants (Sec. 7.7) have decreased in importance, but will retain their place in eigenvalue problems, elementary geometry, etc. The inverse A -I of a square matrix satisfies AA -I = A -IA = I. It exists if and only if det A =t- O. It can be computed by the Gauss-Jordan elimination (Sec. 7.8). The rank r of a matrix A is the maximum number of linearly independent rows or columns of A or, equivalently, the number of rows of the largest square submatrix of A with nonzero determinant (Secs. 7.4. 7.7). The system (2) has solutions if and only if rank A = rank [A b], where [A b] is the augmented matrix (Fundamental Theorem, Sec. 7.5). The homogeneous system (3) Ax = 0 has solutions x =t- 0 ("nontrivial solutions") if and only if rank A < m = n equivalently if and only if det A = 0 (Secs. 7.6. 7.7). 11, in the case Vector spaces, inner product spaces, and linear transformations are discus'ied in Sec. 7.9. See also Sec. 7.4. .,J -------.j' :.;-" . . ·f . - ' ! C HAP T E R ~ ~;/ 8 " Linear Algebra: Matrix Eigenvalue Problems Matrix eigenvalue problems concern the solutions of vector equations (1) Ax = AX where A is a given square matrix and vector x and scalar A are unknown. Clearly, x = 0 is a solution of (I), giving 0 = O. But this of no interest, and we want to find solution vectors x*-O of (l), called eigenvectors of A. We shall see that eigenvectors can be found only for certain values of the scalar A: these values A for which an eigenvector exists are called the eigenvalues of A. Geometrically, solving (1) in this way means that we are looking for vectors x for which the multiplication of x by the matrix A has the same effect as the multiplication of x by a scalar A, giving a vector Ax with components proportional to those of x, and A as the factor of proportionality. Eigenvalue problems are of greatest practical interest to the engineer, physicist, and mathematician, and we shall see that their theory makes up a beautiful chapter in linear algebra that has found numerous applications. We shall explain how to solve that vector equation (1) in Sec. 8.1, show a few typical applications in Sec. 8.2, and then discuss eigenvalue problems for symmetric, skew-symmetric. and orthogonal matrices in Sec. 8.3. In Sec. 8.4 we show how to obtain eigenvalues by diagonalization of a matrix. We also consider the complex counterparts of those matrices (Hermitian. skew-Hermitian. and unitary matrices, Sec. 8.5). which playa role in modern physics. COMMENT. Numerics for eigenvalues (Sees. 20.6-20.9) can be studied immediately after this chapter. Prerequisite: Chap. 7. Sections that may be omitted il1 a shorter course: 8.4, 8.5 References and Answers to Problems: App. I Part B, App. 2. 333 CHAP. 8 334 8.1 Linear Algebra: Matrix Eigenvalue Problems Eigenvalues, Eigenvectors From the viewpoint of engineering applications, eigenvalue problems are among the most important problems in connection with matrices, and the student should follow the present discussion with particular attention. We begin by defining the basic concepts and show how to solve these problems, by examples as well as in general. Then we shall turn to applications. Let A = [ajk] be a given Il X 11 matrix and consider the vector equation (1) Ax = AX. Here x is an unknown vector and A an unknown scalar. Our task is to determine x's and A's that satisfy (I). Geometrically, we are looking for vectors x for which the multiplication by A has the same effect as the multiplication by a scalar A; in other words, Ax should be proportional to x. Clearly. the zero vector x = 0 is a solution of (I) for any value of A. because AO = O. This is of no interest. A value of A for which (I) has a solution x =1= 0 is called an eigenvalue or characteristic value (or latent root) of the matrix A. ("Eigen" is German and means "proper" or "characteristic.") The corresponding solutions x =1= 0 of (l) are called the eigenvectors or characteristic vectors of A corresponding to that eigenvalue A. The set of all the eigenvalues of A is called the spectrum of A. We shall see that the spectrum consists of at least one eigenvalue and at most of n numerically different eigenvalues. The largest of the absolute values of the eigenvalues of A is called the spectral radius of A, a name to be motivated later. How to Find Eigenvalues and Eigenvectors The problem of determining the eigenvalues and eigenvectors of a matrix is called an eigenvalue problem. (More precisely: an algebraic eigenvalue problem, as opposed to an eigenvalue problem involving an ODE, POE (see Sees. 5.7 and 12.3) or integral equation.) Such problems occur in physical, technical, geometric, and other applications, as we shall see. We show how to solve them, first by an example and then in general. Some typical applications will follow afterwards. E X AMP L E 1 Determination of Eigenvalues and Eigenvectors We illustrate all the steps in terms of the matrix A [-5 2J. = 2 Soluti01l. -2 tal Eige1lvalues. These must be determined first. Equation (1) is Ax = [52 -22J [X1J = A [X1J ; -"2 Transferring the term~ in components. X2 on the right to the left. we get (-5 - A)xl + =0 (2*) 2X1 This can be written in matrix notation + (-2 - A)X2 = O. SEC. 8.1 335 Eigenvalues, Eigenvectors (A - AI)x = 0 (3*) because (1) is Ax - Ax = Ax - Alx = (A - M)x = 0, which gives (3*). We see that this is a homogeneous linear system. By Cramer's theorem in Sec. 7.7 it has a nontrivial solution x 0 (an eigenvector of A we are looking for) if and only if its coefficient determinant is zero, that is, '* (4*) 2 1=(-5-,\)(-2-A)-4=A2+7,\+6=0. -2 - A D(A)=detCA-AI)=I-S-A 2 We call D(A) the characteristic detenninant or, if expanded, the characteristic polynomial, and D(A) = 0 the characteristic equation of A. The solutions of this quadratic equation are Al = -I and A2 = -6. These are the eigenvalues of A. (hI) Eigellvector of A correspollding to AI' This vector is obtained from (2*) with A = Al = -I, that is, A solution is x2 = 2~'1> as we see from either of the two equations, so that we need only one of them. ThIs detennines an eigenvector corresponding to Al = -I up to a scalar multiple. If we choose Xl = 1. we obtain the eigenvector Check: AXI = [-5 2J [IJ [-IJ 2 -2 2 -2 = (-I1xI = Alxl' (b2 ) Eigenvector of A correspollding to A2 • For A = A2 = -6. equation (2*) becomes Xl 2xI A solution is x2 = - xI I2 with arbitrary corresponding to A2 = - 6 is Xl' + 2x2 = 0 + 4x2 = O. If we choose Xl = 2, we get X2 = -1. Thus an eigenvector of A Check: This example illustrates the general ca'ie as follows. Equation (I) written in components is Transferring the tenns on the right side to the left side, we have (2) + + (a22 - A)X2 + ... + =0 + ... + =0 + ... + In matrix notation, (3) (A - AI)x = o. (ann - A)Xn = O. CHAP. 8 336 Linear Algebra: Matrix Eigenvalue Problems By Cramer's theorem in Sec. 7.7, this homogeneous linear system of equations has a nontrivial solution if and only if the corresponding determinant of the coefficients is zero: (4) D(A) = det (A - AI) = A a l2 al n a21 a22 - A a2n anI a n2 all - = O. ann - A A - AI is called the characteristic matrix and D( A) the characteristic determinant of A. Equation (4) is called the characteristic equation of A. By developing D(A) we obtain a polynomial of nth degree in A. This is called the characteristic polynomial of A. This proves the following important theorem. THEOREM 1 Eigenvalues The eif;envalues of a square matrix A are the roots of the characteristic equatioll of A. Hence an Il X n matrix has at least one eigellvalue and at most Il numerically different eigellvalues. (4) For larger Il. the actual computation of eigenvalues will in general require the use of Newton' s method (Sec. 19.2) or another numeric approximation method in Secs. 20.7-20.9. The eigenValues must be determined first. Once these are known, corresponding eigenvectors are obtained from the system (2), for instance, by the Gauss elimination. where A is the eigenvalue for which an eigenvector is wanted. This is what we did in Example I and shall do again in the examples below. (To prevent misunderstandings: numeric approximation methods (Sec. 20.8) may determine eigenvectors first.) Eigenvectors have the following properties. THEOREM 2 Eigenvectors, Eigenspace Jfw and x are eigenvectors of a matrix A corresponding to the same eigenvalue A, so are w + x (provided x -w) and kxfor allY k O. * * Hence the eigenvectors correspollding to one and the same eigenvalue A of A, together with 0, fOl1n a !'ector space (cf. Sec. 7.4), called the eigenspace of A corresponding to that A. PROOF Aw = Aw and Ax = Ax imply A(w + x) = Aw + Ax = Aw = k(Aw) = k(Aw) = A(kw); hence A(kw + ex) = A(kw A(kw) + Ax = A(w + .fx). + x) and • In particular. an eigenvector x is detel1nined only up to a constant factor. Hence we can normalize x, that is. multiply it by a scalar to get a unit vector (see Sec. 7.9). For instance, Xl = [I 2JT in Example I has the length "Xl" = V 12 + 22 [lIvS 21vSf is a normalized eigenvector (a unit eigenvector). = vS; hence SEC. 8.1 337 Eigenvalues. Eigenvectors Examples 2 and 3 will illustrate that an n X n matrix may have n linearly independent eigenvectors. or it may have fewer than n. In Example 4 we shall see that a real matrix may have complex eigenvalues and eigenvectors. E X AMP L E 2 Multiple Eigenvalues Find the eigenvalues and eigenvectors of A = [-~ -1 Solutioll. =:]. 2 -2 0 For our matrix, the characteristic determinant gives the characteristic equation g -A - The roots (eigenvalues of A) are Al (Sec. 7.3) to the system (A - Al)x matrix is -7 A - AI = A-51 = A2 + 21A + 45 = O. = 5, A2 = Ag = -3. To find eigenvectors, we apply the Gauss e1immation = O. first with A = 5 and then with A = -3. For A = 5 the characteristic 2 It row-reduces to [ -1 Hence it has rank 2. Choosing Xg = -1 we have x2 = 2 from -¥X2 - ~Xg -7xI + 2X2 - 3xg = O. Hence an eigenvector of A coresponding to A = 5 is For A = - 3 the characteristic matrix A - AI = A + 31 = [ 2 -I : =:] -2 Xl = 0 and then xl = I from = [I 2 _I]T. row-reduce, to 3 Hence it has rank I. From Xl + 2Y2 - 3xg = 0 we have Xl = - i l2 + 3xg. Choosing x2 = 1. Xg = 0 and = 0, Xg = 1, we obtain two linearly independent eigenvectors of A corresponding to A = -3 [as they must exist by (5), Sec. 7.5. with rank = 1 and /I = 3], x2 ~ ~ [:] and x,~ [:1 • The order M). of an eigenvalue A as a root of the characteristic polynomial is called the algebraic multiplicity of A. The number 11l}. of linearly independent eigenvectors corresponding to A is called the geometric multiplicity of A. Thus In}. is the dimension of the eigenspace corresponding to this A. Since the characteristic polynomial has degree n, the sum of all the algebraic multiplicities must equal n. In Example 2 for A = - 3 we have 111)0. = M)o. = 2. In general, 111)0. ~ MAo as can be shown. The difference 6..}. = M)o. - 11l}. is called the defect of A. Thus 6..-3 = 0 in Example 2, but positive defects 6..}. can easily occur: CHAP. 8 338 E X AMP L E 3 Linear Algebra: Matrix Eigenvalue Problems Algebraic Multiplicity, Geometric Multiplicity. Positive Defect The characteristic equation of the matrix is Hence A = 0 is an eigenvalue of algebraic multiplicity Mo = 2. But its geometric multiplicity is only tno = I, since eigenvectors result from -Oxl + x2 = 0, hence x2 = 0, in the form [Xl 0] T. Hence for A = 0 the defect is 6. 0 = 1. Similarly, the characteristic equation of the matrix 3 det(A _ AI) = 1 - A is o 2 1 = (3 - A)2 = O. 3 - A Hence A = 3 is an eigenvalue of algebraic multiplicity M3 = 2. but its geometric multiplicity is only tn3 = I, since eigenvectors result from OXI + 2x2 = 0 in the form [Xl OlT. • E X AMP L E 4 Real Matrices with Complex Eigenvalues and Eigenvectors Since real polynomials may have complex roots (which then occur in conjugate pairs), a real matrix may have complex eigenvalues and eigenvectors. For instance. the characteristic equation of the skew-symmetric matrix is det(A-AI) = I-A -I II=A2 +1=0. -A It gives the eigenvalues Al = i (=v=I), A2 = -i. Eigenvectors are obtained from -ixl iXl + x2 = + X2 = 0 and 0, respectively, and we can choose Xl = I to get [J • and In the next section we shall need the following simple theorem. THEOREM 3 Eigenvalues of the Transpose The transpose AT of a square matrix A has the same eigenvalues as A. PROOF Transposition does not change the value of the characteristic determinant, as follows from Theorem 2d in Sec. 7.7. • Having gained a first impression of matrix eigenvalue problems, in the next section we illustrate their importance with some typical applications. 11-251 EIGENVALUES AND EIGENVECTORS Fmd the eigenvalues and eigenvectors of the following matrices. (Use the given A or factors.) 1. [-2o 0] 0.4 4. S. [5 -2J 9 -6 [~ ~J SEC. 8.1 7. o.s [ 0.6 Eigenvalues, Eigenvectors -0.6J 8. O.S 10. 339 [~ ~J e sin e COS [ -sin eJ cos e 20. 21. :[ : -~ -:] o 0 19 -1 o 0 -1 19 o -2 2 -4 2 -2 4 o o 2 2 -4 2 -6 4 o ,A = 4 o 4 -2 22. ~] 4 -1 -2 2 -2 3 , (A - 3)2 23. 24. 0.2 0.1] 1.0 1.5 o 3.5 0] 12 25. , (A + 1)2 -4 -1 17. 26. (Multiple eigenvalues) Find further 2 X 2 and 3 X 3 matrices with multiple eigenvalues. (See Example 2.) o 18. 3 6 27. (Nonzero defect) Find further 2 X 2 and 3 X 3 matrices with positive defect. (See Example 3.) 12] : ,A = 9 28. (Transpose) Illustrate Theorem 3 with examples of your own. 29. (Complex eigenvalues) Show that the eigenvalues of a real matrix are real or complex conjugate in pairs. 19. 30. (Inverse) Show that the inverse A -1 exists if and only if none of the eigenvalues AI, ... , An of A is zero, and then A-I has the eigenvalues lIAl>' .. , lIAn- CHAP. 8 340 8.2 Linear Algebra: Matrix Eigenvalue Problems Some Applications of Eigenvalue Problems In this section we discuss a few typical examples from the range of applications of matrix eigenvalue problems, which is incredibly large. Chapter 4 shows matrix eigenvalue problems related to ODEs governing mechanical systems and electrical networks. To keep our present discussion independent of Chap. 4, we include a typical application of that kind as our last example. E X AMP L E 1 Stretching of an Elastic Membrane An elastic membrane in the Xlx2-plane with boundary circle X1 2 point P: (Xl, X2) goes over into the point Q: (Y1, Y2) given by y (I) = [Y1J = Ax = [5 3J [XIJ ; 3 Y2 5 + X2 2 I (Fig. 158) is stretched so that a in components. x2 Find the prinCipal directions, that is, the directions of the position vector x of P for which the direction of the position vector y of Q is the same or exactly opposite. What shape does the boundary circle take under this deformation? We are looking for vectors x ~uch that y of an eigenvalue problem. In components, Ax = Ax IS Solutioll. 5X1 + = (5 - A)X1 3x2 = '\'\:1 3x 1 + Ax, this gives Ax + or (2) = Ax. Since y Ax, the equation =0 + (5 - 5x2 = A.\:2 = A)X2 = O. The characteristic equation is 3 (3) 5-A I = (5 - A)2 - 9 = O. Its solutions are Al = 8 and '\'2 = 2. These are the eigenvalues of our problem. For A = Al = 8, our system (2) becomes Solution.\"2 = Xl' for instance, Xl = Xl arbitrary, X2 = I. For A2 = 2, our system (2) becomes Solution -Xl. X2 = for instance. Xl = 1, Xl arbitrary, x2 = -1. We thus obtain as eigenvectors of A, for instance, [1 I]T corresponding to Al and [I I]T corresponding to A2 (or a nonzero scalar multiple of these). These vectors make 45° and 135 0 angles with the positive Xl-direction. They give the principal directions. the answer to our problem. The eigenvalues show that in the principal directions the membrane is stretched by factors 8 and 2, respectively; see Fig. 158. Accordingly. if we choose the principal directions as directions of a new Cartesian "1"2-coordinate system. say, with the positive lll-semi-axis in the first quadrant and the positive 112-senu-axis in the second quadrant of the xlx2-system. and if we set III = rcos cP. "2 = rsin cP. then a boundary point of the unstretched circular membrane has coordinates cos cP, sin cP. Hence. after the stretch we have 21 Since cos (4) 2 cP + sin 2 cP = = 8 cos cP. :2 = 2 sin cb. I. this shows that the deformed boundary is an ellipse (Fig. 158) 1. • SEC. 8.2 Some Applications of Eigenvalue Problems 341 / / / / Fig. 158. E X AMP L E 2 Undeformed and deformed membrane in Example 1 Eigenvalue Problems Arising from Markov Processes Markov processes as considered in Example 13 of Sec. 7.2 lead to eigenvalue problems if we ask for the limit state of the precess in which the stale vecmr x is reproduced under the multiplication by the smchastic marrix A governing the process. that is, Ax = x. Hence A should have the eigenvalue I, and x should be a corresponding eigenvector. This is of practical interest because it shows the long-term tendency of the development modeled by the process. In that example, A = 0.7 0.1 0.2 0.9 For the transpose, [ 0.7 0.2 0.1 0.9 ° 0.2 [ 0.1 ° Hence AT has the eigenvalue I, and the same is true for A by Theorem 3 in Sec. 8.1. An eigenvector x of A for A = I is obtained from A ~ [~::: I = U.l 0.1 ~O.I o :.2] , 1110 row-reduced to ~0.2 ~1I30 ° Taking x3 = 1, we get x2 = 6 from ~x2/30 + x3/5 = 0 and then Xl = 2 frem ~3XI/1O + x2/1O = O. This gives x = [2 6 I]T It means that in the long run. the ratio Commercial: Industrial: Residential will approach 2:6: I, provided that the probabilities given by A remain (about) the same. (We switched to ordinary fractions to avoid rounding errors.) • E X AMP L E 3 Eigenvalue Problems Arising from Population Models. Leslie Model The Leslie model describes age-specified population growth, as follows. Let the oldest age attained by the females in some animal population be 9 years. Divide the population into three age classes of 3 years each. Let the "Leslie matrix" be 2.3 (5) ° 0.3 where i llc is the average number of daughters born to a single female during the time she is in age class k, and (j = 2, 3) is the fraction of females in age class j ~ I that will survive and pass into class j. (a) What is the number of females in each cia" after 3, 6, 9 years if each class initially consists of 400 females? (b) For what initial distribution will the number of females in each class change by the same proportion? What is this rate of change? lj,j_l 342 CHAP. 8 Solution. Linear Algebra: Matrix Eigenvalue Problems (a) Initially, x;o) = [400 400 2.3 ~3) = L,,<O) = n 400]. After 3 years, [0:' 0 04] o 0.3 o 400 = [1000] 240. 400 120 Similarly. after 6 years the number of females in each class is given by X;6) = (LX(3»T = [600 648 72]. and after 9 years we have X;9) = (LX(6»T = [1519.2 360 194.4]. (b) Proportional change means that we are looking for a distribution vector x such thai Lx = Ax, where A is the rate of change (growth if A > I, decrease if A< I). The characteristic equation is (develop the characteristic determinant by the first column) det (L - AI) = - A3 - 0.6( -2.3A - 0.3·0.4) = - A3 + 1.38A + 0.072 = O. A positive root is found to be (for instance. by Newton's method. Sec. 19.2) A = 1.2. A corresponding eigenvector x can be detennined from the characteristic matrix 2.3 0.4] -1.2 0.3 o . say, -\.2 X=[0.5] 0.125 where x3 = 0.125 is chosen, X2 0.5 then follows from 0.3x2 - 1.2<3 = 0, and Xl = 1 from -1.2<1 + 2.3x2 + 0.4x3 = O. To get an initial population of 1200 as before, we multiply x by 1200/(1 + 0.5 + 0.125) = 738. Answer: Proportional growth of the numbers of females in the three classes will occur if the initial values are 738, 369, 92 in classes I, 2, 3, respectively. The growth rate will be 1.2 per 3~ • E X AMP L E 4 Vibrating System of Two Masses on Two Springs (Fig. 159) Mass-spring systems involving several masses and springs can be treated as eigenvalue problems. For instance, the mechanical system in Fig. 159 is governed by the system of ODEs (6) where Yl and Y2 are the displacements of the masses from rest. as shown in the figure, and primes denote derivatives with respect to time t. In vector form, this becomes (7) y" = [Y~J = Ay = [-5 Y2 2 2J [YlJ. -2 Y2 (Net change in spring length =Y 2 -Y 1 ) Y2 System in static equilibrium Fig. 159. System in motion Masses on springs in Example 4 SEC 8.2 Some Applications of Eigenvalue Problems 343 We try a vector solution of the form (8) This is suggested by a mechanical system of a single mass on a spring (Sec. 2.4), whose motion exponential functions (and sines and cosines). Substitution into (7) gives IS given by Dividing by e wt and writing w2 = A, we see that our mechanical system leads to the eigenvalue problem Ax = Ax (9) From Example I in Sec. 8.1 we see that A has the eigenvalues Al = -I and A2 w = V=! = -::'::.i and Y=6 = -::'::.iV6, respectively. COlTesponding eigenvectors are -6. Consequently, and (10) From (8) we thus obtain the four complex solutions [see (10), Sec. 2.21 Xle±it = Xl(cost -::': . isint), X2e±iV6t = x2 (cos V6 t -::'::. i sin V6 t). By addition and subtraction (see Sec. 2.2) we get the four real solutions Xl co~ t. Xl sin t. X2 cos V6 t, x2 sin V6 t. A general solution is obtained by taking a linear combination of these. with arbitrary constants aI, b l , a2. b 2 (to which values can be assigned by prescribing initial displacement and initial velocity of each of the two masses). By (10), the components of yare Yl = al cos t + b l sin t + 2a2 cos Y2 = 2al cos t + 2b l V6 t + sin t - a2 cos V6 t 2b2 sin V6 t - b 2 sm V6 t. These functions describe harmonic oscillations of the two masses. Physically, this had to be expected because we have neglected damping. • ••• 11-61 - LINEAR TRANSFORMATIONS Find the matrix A in the indicated linear transformation y = Ax. Explain the geometric significance of the eigenvalues and eigenvectors of A. Show the details. 1. Reflection about the y-axis in R2 2. Reflection about the xy-plane in R3 17-141 7. 3. Orthogonal projection (perpendicular projection) of R2 onto the x-axis 4. Orthogonal projection of R3 onto the plane y = x 9. the origin in R2 [: [2.S I.S 5. Dilatation (uniform stretching) in R2 by a factor S 6. Counterclockwise rotation through the angle 7r12 about ELASTIC DEFORMATIONS Given A in a deformation y = Ax, find the principal directions and corresponding factors of extension or contraction. Show the details. 11. :J 1.SJ 6.S [~ V:J 8. [0.4 0.8 0.8J 0.4 [: I;J 12. [S 2 l~J 10. 344 CHAP. 8 13. [-2 3 3J -2 Linear Algebra: Matrix Eigenvalue Problems 14. [ 10.5 1IY2] lfY2 10.0 20. [ 0.6 0.1 0~4 0.1 15. (Leontief1 input-output model) Suppose that three industries are interrelated so that their outputs are used as inputs by themselves, according to the 3 X 3 consumption matrix 0.5 0.2 A = [ajk] = ° 0.6 [ 0.2 0.5 where ajk is the fraction of the output of industry k consumed (purchased) by industry j. Let Pj be the price charged by industry.i for its total output. A problem is to find prices so that for each industry, total expenditures equal total income. Show that this leads to Ap = p, where p = [PI PZ P3]T, and find a solution p with nonnegative PI, Pz, P3' 16. Show that a consumption matrix as considered in Prob. 15 must have column sums 1 and always has the eigenvalue I. 17. (Open Leontief input-output model) If not the whole output but only a portion of it is consumed by the industries themselves, then instead of Ax = x (as in Prob. 15), we have x - Ax = y, where x = [Xl Xz X3]T is produced, Ax is consumed by the industries, and, thus, y is the net production available for other consumers. Find for what production x a given demand vector y = [0.136 0.272 0.136]T can be achieved if the consumption matrix is 0.2 A = 0.3 [ 0.2 0.4 ° U.4 0.2] 0.1 . 0.5 118-201 MARKOV PROCESSES Find limit states of the Markov processes modeled by the following matrices. (Show the details.) 18. [0.1 0.9 19. [05 0.3 U.5 0.2 0.2 0.4 0.4 0.8 POPULATION MODEL WITH AGE SPECIFICATION Find the growth rate in the Leslie model (see Example 3) with the matrix as given. (Show details.) 3.45 21. [0: ° 0.45 12.0 n. [0:5 ° 0.30 7.280 U [0:60 ° 0.420 24. TEAM PROJECT. General Properties of Eigenvalues and Eigenvectors. Prove the following statements and illustrate them with examples of your own choice. Here, A]o ... , An are the (not necessarily distinct) eigenvalues of a given /J X 11 matrix A = [ajk]. (a) Trace. The sum of the main diagonal entries is called the trace of A. It equals the sum of the eigenvalues. (b) "Spectral shift." A - kI has the eigenvalues Al - k, ... , An - k and the same eigenvectors as A. (c) Scalar multiples, powers. kA has the eigenvalues I..A1 , . . . • kA.,. Am (Ill = I. 2.... ) has the eigenvalues At'. .... k,,"'. The eigenvectors are those of A. (d) Spectral mapping theorem. The ''polynomial matrix" has the eigenvalues O.4J 0.6 0.3 0.2] p(Aj ) = kmA/" 02] 0.2 0.6 + k",_IA/n - 1 + ... + k1Aj + ko where.i = I,' .. , 11, and the same eigenvectors as A. (e) Perron's theorem. Show that a Leslie matrix L with positive lIZ' 113, Iz]o 13z has a positive eigenvalue. (This is a special case of the famous Perron-Frobenius theorem in Sec. 20.7, which is difficult to prove in its general form.) lWASSILY LEONTIEF (1906-1999). American economist at New York University. For his input-output analysis he was awarded the Nobel Prize in 1973. SEC. 8.3 8.3 Symmetric, Skew-Symmetric, and Orthogonal Matrices 345 Symmetric, Skew-Symmetric, and Orthogonal Matrices We consider three classes of real square matrices that occur quite frequently in applications because they have several remarkable properties which we shall now discuss. The first two of these classes have already been mentioned in Sec. 7.2. DEFINITIONS Symmetric, Skew-Symmetric, and Orthogonal Matrices A real square matrix A = [ajk] is called symmetric if transposition leaves it unchanged, AT = A, (1) thus skew-symmetric if transposition gives the negative of A, AT = -A, (2) thus orthogonal if transposition gives the inverse of A, (3) E X AMP L E 1 Symmetric, Skew-Symmetric, and Orthogonal Matrices The matrices ~ -::]. -20 0 are symmetric, skew-symmetric, and orthogonal, respectively, as you should verify. Every skew-synuuetric matrix has all main diagonal entries zero. (Can you prove this?) • Any real square matrix A may be written as the sum of a symmetric matrix R and a skew-symmetric matrix S, where (4) EXAMPLE 1 and Illustration of Formula (4) A ~ 5 [; '] 3 -8 4 3 = R + S = 3.5 [9C 3.0 35] + [ 0 1.5 3.5 -~.O -1.5 3.5 -2.0 3.0 1.5 6.0 0 -15] -6.0 0 • 346 CHAP. 8 Linear Algebra: Matrix Eigenvalue Problems Eigenvalues of Symmetric and Skew-Symmetric Matrices THEOREM 1 (a) The eigenvalues of a symmetric matrix are real. (b) The eigenvalues of a skew-symmetric matrix are pure imaginal}' or zero. This basic theorem (and an extension of it) will be proved in Sec. 8.5. E X AMP L E 3 Eigenvalues of Symmetric and Skew-Symmetric Matrices The matrices in (1) and (7) of Sec. 8.2 are symmetric and have real eigenvalues. The skew-symmetric matrix in Example 1 has the eigenvalues O. -25 i, and 25 i. (Verify this.) The following matrix has the real eigenvalues 1 and 5 but is not symmetric. Does this contradict Theorem I? • Orthogonal Transformations and Orthogonal Matrices OrthogonaJ transformations are transformations (5) y = where A is an orthogonal matrix. Ax With each vector x in R n such a transformation assigns a vector y in Rn. For instance, the plane rotation through an angle 0 y = (6) [Yl] Y2 = [C~s 0 sm e -sin cos 0] [Xl] e X2 is an orthogonal transformation. It can be shown that any orthogonal transformation in the plane or in three-dimensional space is a rotation (possibly combined with a reflection in a straight line or a plane. respectively). The main reason for the importance of orthogonal matrices is as follows. THEOREM 2 In variance of Inner Product An orthogonal transfonnation preserves the value of the inner product of vectors a and bin Rn. defined by (7) That is, for any a and b in R n , orthogonaln X 11 matrix A, and u = Aa, v = Ab we have u·v = a·b. Hence the tra11Sfonnation also preserves the length or norm of any vector a in Rn given by (8) II a II = v'a.3 = Wa. SEC 8.3 347 Symmetric, Skew-Symmetric, and Orthogonal Matrices PROOF Let A be orthogonal. Let u = Aa and v = Ab. We must show that u· v = a· b. Now (Aa)T = a TAT by (lOd) in Sec. 7.2 and ATA = A-I A = [by (3). Hence (9) From this the invariance of II a II follows if we set b = • a. Orthogonal matrices have further interesting properties as follows. THEOREM 3 Orthonormality of Column and Row Vectors A real square matrix is orthogonal if and only if its column vectors a1> ... , an (and also its row vectors) form an orthonormal system, that is, (10) if j = k. PROOF (a) Let A be orthogonal. Then A-I A = ATA = I, in tenns of colullUl vectors a1> ... , an' (11) T 1 I=A- A=A A= 1T a ] lalTal :T [al···a,.,]= ~ l an an~ 1 al T. a2 : : : a T.an ] . a n Ta2' .. anTan The last equality implies (0), by the definition of the n X n unit matrix I. From (3) it follows that the inverse of an orthogonal matrix is orthogonal (see CAS Experiment 20). Now the column vectors of A-I (= AT) are the row vectors of A. Hence the row vectors of A also form an orthonormal system. (b) Conversely, if the column vectors of A satisfy (10), the off-diagonal entries in (11) must be 0 and the diagonal entries 1. Hence ATA = I, as (11) shows. Similarly, AAT = I. This implies AT = A-I because also A- 1A = AA -1 = I and the inverse is unique. Hence A is orthogonal. Similarly when the row vectors of A form an orthonormal system, by • what has been said at the end of part (a). THEOREM 4 Determinant of an Orthogonal Matrix The detenninant of an orthogonal matrix has the value PROOF + 1 or -1. From det AB = det A det B (Sec. 7.8, Theorem 4) and det AT = det A (Sec. 7.7, Theorem 2d), we get for an orthogonal matrix I = det I = det (AA -1) = det (AA T) = det A det AT = (det A)2. E X AMP L E 4 • Illustration of Theorems 3 and 4 The last matrix in Example I and the matrix in (6) illustrare Theorems 3 and 4 because their determinants are -1 and + 1. as you should verify. • 348 CHAP. 8 THEOREM 5 Linear Algebra: Matrix Eigenvalue Problems Eigenvalues of an Orthogonal Matrix The eigenvalues of an orthogonal matrix A are real or complex conjugates in pairs and have absolute value I. PROOF E X AMP L E 5 The first part of the statement holds for any real matrix A because its characteristic polynomial has real coefficients. so that its zeros (the eigenvalues of A) must be as indicated. The claim that IAI = 1 will be proved in Sec. 8.5. • Eigenvalues of an Orthogonal Matrix The orthogonal matrix in Example 1 has the characteristic equation Now one of the eigenvalues must be real (why?). hence + I or -1. Trying. we find -1. Division by A + I gives _(A2 - 5Al3 + 1) = 0 and the two eigenvalues (5 + iVll)/6 and (5 - iVll)/6. which have absolute value I. Verify all of this. • Looking back at this section. you will find that the numerous basic results it contains have relatively short, straightforward proofs. This is typical of large portions of matrix eigenvalue theory . . . 08 LE~--S-EI-83" e sin e 1. (Verification) Verify the statements in Example 1. COS 2. Verify the statements in Examples 3 and 4. 3. Are the eigenvalues of A + B of the form Aj + Mj. where A.i and p; are the eigenvalues of A and B, respecti vely? 4. (Orthogonality) Prove that eigenvectors of a symmetric matrix corresponding to different eigenvalues are orthogonal. Give an example. 11. 12. e cos e -sin [ ° ° 13. 5. (Skew-symmetric matrix) Show that the inverse of a skew-symmetric matrix is skew-symmetric. 6. Do there exist nonsingular skew-symmetric matrices with odd II? 11 X 11 15. 7. (Orthogonal matrix) Do there exist skew-symmetric orthogonal 3 X 3 matrices? 8. (Symmetric matrix) Do there exist nondiagonal symmetric 3 X 3 matrices that are orthogonal? 19-171 17. EIGENVALUES OF SYMMETRIC, SKEWSYMMETRIC, AND ORTHOGONAL MATRICES Are the following matrices symmetric, skew-~ymmetric, or orthogonal? Find their spectrum (thereby illustrating Theorems 1 and 5). (Show the details of your work.) 9. [0.96 0.28 -0.28J 0.96 10. [a -b bJ a 18. (Rotation in space) Give a geometric interpretation of the transformation y = Ax with A as in Prob. 12 and x and y referred to a Cartesian coordinate system. 19. WRITING PROJECT. Section Summary. Summarize the main concepts and facts in this section, with illustrative examples of your own. SEC 8.4 Eigenbases. Diagonalization. Quadratic Forms 20. CAS EXPERIMENT. Orthogonal Matrices. (a) Products. Inverse. Prove that the product of two orthogonal matrices is orthogonaL and so is the inverse of an orthogonal matrix. What does this mean in terms of rotations? (b) Rotation. Show that (6) is an orthogonal transformation. Verify that it satisfies Theorem 3. Find the inverse transformation. (e) Powers. Write a program for computing powers Am (17l = 1. 2.... ) of a 2 X 2 matrix A and their 8.4 349 spectra. Apply it to the matrix in Prob. 9 (call it A). To what rotation does A correspond? Do the eigenvalues of Am have a limit as 111 _ x? (d) Compute the eigenvalues of (O.9A)"'. where A is the matrix in Prob. 9. Plot them as points. What is their limit? Along what kind of curve do these points approach the limit? (e) Find A such that y = Ax is a counterclockwise rotation through 30° in the plane. Eigenbases. Diagonalization. Quadratic Forms So far we have emphasized properties of eigenvalues. We now turn to general properties of eigenl'ectors. Eigenvectors of an n X n matrix A may (or may not!) form a basis for Rn. If we are interested in a transformation y = Ax, such an "eigenbasis" (basis of eigenvectors)-if it exists-is of great advantage because then we can represent any x in R n uniquely as a linear combination of the eigenvectors Xl' • . . , Xn , say, And, denoting the corresponding (not necessarily distinct) eigenvalues of the matrix A by AI' ... , An. we have AXj = A:iXj, so that we simply obtain y (I) = Ax = A(clX I + ... + cnxn) = ciAx i + ... + cnAx" = cIAlx I + ... + cnAnxn· This shows that we have decomposed the complicated action of A on an arbitrary vector x into a sum of simple actions (multiplication by scalars) on the eigenvectors of A. This is the point of an eigenbasis. Now if the n eigenvalues are all different, we do ohtain a hasis: Basis of Eigenvectors THEOREM 1 If an n X Xl • . . . • PROOF n matrix A has 11 distillct eigenvalues, then A has a basis of eigenvectors xnfor Rn. All we have to show is that Xl' . • • • Xn are linearly independent. Suppose they are not. Let r be the largest integer such that {Xl> . • • • x,.} is a linearly independent set. Then r < n and the set {Xl.' ••• Xn xr+d is linearly dependent. Thus there are scalars CI> ••• , Cr+I, not all zero, such that (2) (see Sec. 7.4). Multiplying both (3) side~ by A and using AXj = AjXj. we obtain 350 CHAP. 8 Linear Algebra: Matrix Eigenvalue Problems To get rid of the last term, we subtract Ar+l times (2) from this, obtaining Here cl(A l - Ar+l) = O..... c,·(Ar - A,-+l) = 0 since {Xl' ... ,xrl is linearly independent. Hence Cl = ... = cr = 0, since all the eigenvalues are distinct. But with this, (2) reduces to c r +lxr +l = 0, hence Cr +l = 0, since xr+l 1:- 0 (an eigenvector!). This contradicts the fact that not all scalars in (2) are zero. Hence the conclusion of the theorem must hold. • E X AMP L E 1 Eigenbasis. Nondistinct Eigenvalues. Nonexistence The matrix A = [ : :J has a ba~is of eigenvectors [: J' [-:J corresponding to the eigenvalues = 8. A2 = 2. (See Example I in Sec. 8.2.) Even if not all n eigenvalues are different, a matrix A may still provide an eigenbasis for RTl. See Example 2 in Sec. 8.1, where n = 3. On the other hand, A may not have enough linearly independent eigenvectors to make up a basis. For instance, A in Example 3 of Sec. 8.1 is Al ~J and has only one eigenvector (k *- O. arbitrary). • Actually, eigenbases exist under much more general cunditions than those in Theorem L. An important case is the following. THEOREM 2 Symmetric Matrices A symmetric matrix has all ortllOno1711al basis of eigellvectors for Rn. For a proof (which is involved) see Ref. [B3], vol. I, pp. 270-272. E X AMP L E 2 Orthonormal Basis of Eigenvectors The first matrix in Example I i~ symmetric, and an orthonormal basis of eigenvectors is [IIV:>: [1IV2 -ltif2:r Diagonalization of Matrices Eigenbases also playa role in reducing a matrix A to a diagonal matrix whose entries are the eigenvalues of A. This is done by a "similarity transformation." which is defined as follow,> (and will have various applications in numerics in Chap. 20). DEFINITION Similar Matrices. Similarity Transformation An n X 11 matrix A is called similar to an 11 X 11 matrix A if (4) for some (nonsinguiar!) 11 X n matrix P. Thi~ transformation. which gives A. is called a similarity transformation. A from SEC 8.4 )51 Eigenbases. Diagonalization. Quadratic Forms The key property of this transformation is that it preserves the eigenvalues of A: THEOREM) Eigenvalues and Eigenvectors of Similar Matrices If A is similar to A, then A !las the same eigenvalues as A. Furthennore, {f x is all eigenvector of A, then y = p-1x is an eigenvector of A corresponding to the same eigenvalue. PROOF From Ax = Ax (A an eigenvalue, x *- 0) we get P-1Ax = AP-1x. Now 1= pp-l. By this "identity trick" the previous equation gives Hence A is an eigenvalue of A and p-1x a corresponding eigenvector. Indeed, p-1x = 0 would give x = Ix = pp-1x = PO = 0, contradicting x *- O. • E X AMP L E) Eigenvalues and Vectors of Similar Matrices A= [6 -3J Let 4 A= [ Then and -I p= [I 3J I 4 4 -3JI [Ii4 -3J [I1 43J [30 OJ. -I 2 -I Here p-l was obtained from (4*) in Sec. 7.8 with det P = 1. We see that A has the eigenvalues Al = 3, A2 The characteristic equation of is A)( A) + A2 - 5A + It has the roots (the eigenvalues of A) Al = 3, A2 = 2, confirming the first part of Theorem 3. We confirm the second part. From the first component of (A - AI)x = 0 we have (6 - A)XI - 3"2 = O. For A = 3 this gives 3.'1 - 3X2 = O. say. Xl = [1 l]T. For A = 2 it gives 4xl - 3X2 = O. say. X2 = [3 4]T. In Theorem 3 we thus have =2. A l6 - -1 - 12 = 6=O. Indeed. these are eigenvectors of the diagonal matrix A. Perhaps we see that Xl and x2 are the columns of P. This suggests the general method of transforming a matrix A to diagonal form D by using P = X, the matrix with eigenvectors as columns: • THEOREM 4 Diagonalization of a Matrix If an n X 11 matrix A has a basis of eigenvectors, then (5) is diagonal, with the eigenvalues of A as the entries on the main diagonal. Here X is the matrix H·ith these eigenvectors as colu11ln vectors. Also, (5*) D m = X-1A"'X (/11 = 2,3,' . '). CHAP. 8 352 PROOF Linear Algebra: Matrix Eigenvalue Problems Let Xb ... , Xn constitute a basis of eigenvectors of A for Rn. Let the corresponding eigenvalues of A be Ab ... , An' respectively, so that Ax] = A1Xb· .. ,AXn = Anx". Then X = [Xl x.,] has rank n, by Theorem 3 in Sec. 7.4. Hence X-I exists by Theorem 1 in Sec. 7.8. We claim that (6) AX = A[X1 where D is the diagonal matrix as in (5). The fourth equality in (6) follows by direct calculation. (Try it for n = 2 and then for general n.) The third equality uses AXk = Akxk. The second equality results if we note that the first column of AX is A times the first column of X, and so on. For instance. when n = 2 and we write Xl = [XU X21]T, X2 = [X12 X22]T, we have AX = A[X1 X2] [(/11 (/12] (/21 (/22 [(/l1 X 11 (/21X 11 X12 ] [X11 X 21 X22 + (/12 X 21 + (/22 X 21 (/l1X 12 + (/21 X 12 + (/22 X 22 Column I (/12 X 22 ] = [AX1 Ax2]. Column 2 If we multiply (6) by X-I from the left, we obtain (5). Since (5) is a similarity transformation, Theorem 3 implies that D has the same eigenvalues as A. Equation (5*) follows if we note that • etc. E X AMP L E 4 Diagonalization Diagonalize 7.3 0.2 -3.7] A = -11.5 1.0 5.5 17.7 1.8 -9.3 [ . The characteristic determinant gives the characteristic equation -A3 - A2 + 12A = O. The roots (eigenvalues of A) are Al = 3, A2 = -4, A3 = O. By the Gauss elimination applied to (A - AI)x = 0 with A = A]. A2• A3 we find eigenvectors and then X-I by the Gauss-Jordan elimination (Sec. 7.8. Example 1). The results are Solution. 0.2 0.3] -0.2 0.2 0.7. -0.2 Calculating AX and multiplying by X-I from the left, we thus obtain D = X-lAX = [-0' -1.3 -0.2 0.8 0.2 0.2 0'][-3 -4 0.7 9 4 -0.2 -3 -12 :] ~ [: 0 -4 0 :1 • SEC. 8.4 353 Eigenbases. Diagonalization. Quadratic Forms Quadratic Forms. Transformation to Principal Axes By definition, a quadratic form Q in the components Xl, ... , Xn of a vector x is a sum of n 2 terms, namely, n T Q = x Ax = n 2. 2. ajkxjxk j~l k~l (7) + ........................ . A = [ajk] is called the coefficient matrix of the fmID. We may assume that A is symmetric, because we can take off-diagonal telIDS together in pairs and write the result as a sum of two equal terms; see the following example. E X AMP L E 5 Quadratic Form. Symmetric Coefficient Matrix Let Here 4 t- 6 = 10 = 5 -t 5. From the corresponding symmetric matrix C thus Cll = 3, C12 = C21 = 5, C22 = 2, we get the same result; indeed, = [Cjk], where Cjk = ~(ajk + aid)' • Quadratic forms occur in physics and geometry. for instance. in connection with conic sections (ellipses X12/a 2 + X22/b 2 = 1, etc.) and quadratic surfaces (cones, etc.). Their transformation to principal axes is an important practical task related to the diagonalization of matrices, as follows. By Theorem 2 the symmetric coefficient matrix A of (7) has an orthonormal basis of eigenvectors. Hence if we take these as column vectors, we obtain a matrix X that is orthogonal, so that X-l = XT. From (5) we thus have A = XDX- 1 = XDXT. Substitution into (7) gives (8) If we set XTx (9) = y, then, since XT = x-I, we get x = Xy. Furthermore, in (8) we have xTX = (XTX)T = yT and XTx = y, so that Q becomes simply (10) 354 CHAP. 8 Linear Algebra: Matrix Eigenvalue Problems This proves the following basic theorem. THEOREM 5 Principal Axes Theorem The substitution (9) transforms a quadratic form n Q = xTAx = n 2. 2. ajkXjXk j=l k=I to the principal axes form or canonical form (10), where Ab ...• An are the (not necessarily distinct) eigenvalues of the (s)"lnmetric!) matrix A, and X is an orthogonal matrix with corresponding eigenvectors Xl' . . . , xn , respectively, as colu11ln vectors. E X AMP L E 6 Transformation to Principal Axes. Conic Sections Find out what type of conic section the following quadratic form represents and transform it to principal axes: Solution. We have Q = xTAx. where A=[ I7 -15 -15J. I7 This gives the characteristic equation 07 - A)2 - 15 2 = O. It has the roots A1 = 2. A2 = 32. Hence (10) becomes We see that Q = 128 represents the ellipse 2.1'/ + 32yl = 128. that is. If we want to kno" the direction of the principal axes in the Xlx2-coordinates. we have to determine normalized eigenvectors from (A - AI)x = 0 with A = Al = 2 and A = A2 = 32 and then use (9). We get [ I/V~. ] and II\., 2 [-IIVzJ. IIVz hence x = XJ' = IIVz -IIVz J [YIJ [IIVz 1IV2 .1'2' Xl = yI/Vz - Y2 /Vz X2 = y 1 /V2 + Y2/V2. This is a 45° rotation. Our results agree with those in Sec. 8,2, Example I, except for the notations. See also • Fig. 158 in that example. SEC 8.4 .... __ r------." ;...= ;...• ... r - - - - - .. 11-91 3. 5. [: ====-_ ~-- ~ (d) Diagonalization. What can you do in (5) if you want to change the order of the eigenvalues in D, for instance. interchange d l l = Aland d22 = A2? DIAGONALIZATION OF MATRICES Find an eigenbasis (a basis diagonalize. (Show the details.) 1. 355 Eigenbases. Diagonalization. Quadratic Forms of eigenvectors) 2. :J [~ ~J 4. [1.0 6.0J 1.5 1.0 6. and [~ I~J [-: -~J SIMILAR MATRICES HAVE EQUAL SPECTRA 113-181 Verify this for A and A = P-lAP. Find eigenvectors y of A. Show that x = Py are eigenvectors of A. (Show the details of your work.) 13. A [-5o OJ2 ' P = = [: -:J 4 -2JI [ -3 3 14. A = [ 3 0 :1 10 -1:1 9. l-1: 39 -24 40 -15 7. [4 -6 0 8. -5 l-' -5 1:1 -9 -9 13 10. (Orthonormal basis) Illustrate Theorem 2 with further examples. 11. (No basis) Find further 2 X 2 and 3 X 3 matrices without eigenbases. 12. PROJECT. Similarity of Matrices. Similarity is basic, for instance in designing numeric methods. (a) Trace. By definition, the trace of an 11 X 11 matrix A = [ajk] is the sum of the diagonal entries, trace A = all + a22 + ... + ann' Show that the trace equals the sum of the eigenvalues, each counted as often as its algebraic multiplicity indicates. Illustrate this with the matrices in Probs. 1. 3,5.7,9. (b) Trace of product. Let B = [bjk } be 11 X n. Show that similar matrices have equal traces, by first proving n n trace AB = ~ ~ 15. A = [ = trace BA. i~ll~l [1 3 -6 1 6 l-~ ~0 ~:1· l~0 = P = -5 15 ~ ~] 0 10 0 0 ~1 1 TRANSFORMATION TO PRINCIPAL AXES. CONIC SECTIONS 119-281 What kind of conic section (or pair of straightlines) is given by the quadratic form? Transform it to principal axes. Express xT = [Xl X2] in terms of the new coordinate vector yT = [Yl ."2]' as in Example 6. 19. X1 2 + 24xlX2 - 6X22 = 5 20. 3X12 + 4V3x l X2 + 7X 2 2 = 9 21. 3-'"1 2 - 8XlX2 - 3-'"22 = 0 23. 4X12 + + 24. 7X12 (c) Find a reiationship between A in (4) and A = PAP-I. P= P = 2] 18. A 2J -2 l : -~ ~] , l~ 17. A = 22. 6X12 ailbli 4 -4 25. X1 2 - 16-'"1-'"2 - 6X22 = 24xlX2 12xl-'"2 20 + 2X22 = 10 = 144 2V3xlX2 + X2 2 = 35 CHAP. 8 356 Linear Algebra: Matrix Eigenvalue Problems + 22xIX2 + 3X22 = 0 12xI2 + 32xIX2 + 12x22 = 6.5xI 2 + 5.0X]x2 + 6.5x22 26. 3X]2 27. 28. Q(x) 112 36 = 29. (Definiteness) A quadratic fonn Q(x) = xT Ax and its (symmetric!) matrix A are called (a) positive definite if Q(x) > 0 for all x 0, (b) negative definite if Q(x) < 0 for all x O. (e) indefinite if Q(x) takes both positive and negative values. (See Fig. 160.) [Q(x) and A are called positive semidefinite (negative semidefinite) if Q(x) ~ 0 (Q(x) ~ 0) for all x.J A necessary and sufficient condition for positive definiteness is that all the "principal minors" are positive (see Ref. [B3]. vol. 1. p. 306), that is. "* "* lall (/]2 all> 0, .1:2 (a) PosItIve defimte form Q(x) aI21 >0, a22 (b) Negative defm Ite form all aI2 a]3 a I2 a22 a23 >0, a]3 a 23 a33 detA > O. Q(x) Show that the form in Prob. 23 is positive definite, whereas that in Prob. 19 is indefinite. 30. (Definiteness) Show that necessary and sufficient for (a), (b), (c) in Prob. 29 is that the eigenvalues of A are (a) all positive, (b) all negative, tc) both positive and negative. Hint. Use Theorem 5. 8.5 (c) Indefinite form Fig. 160. QuadratiC forms in two variables Optional Complex Matrices and Forms. The three classes of real matrices in Sec. 8.3 have complex counterparts that are of practical interest in certain applications, mainly because of their spectra (see Theorem 1 in this section), for instance, in quantum mechanics. To define these classes, we need the following standard Notations A = [ajk] is obtained from A = [lljk] (a,.!! real) with its complex conjugate ajk = a + i{3 = [al0] is the transpose by replacing each entry lljk = 0'- i{3. Also, AT of A, hence the conjugate transpose of A. E X AMP L E 1 Notations 3 + 4i If A = [ 6 J, 1- i :>. - 5i then A = [3 - 4i 6 1+ i ] 2+5; and -T A = [3 - 4; 1+; 6 ] 2 + 5; •• SEC. 8.5 Complex Matrices and Forms. DEFINITION Optional 357 Hermitian, Skew-Hermitian, and Unitary Matrices A square matrix A = [aid] is called Hermitian if r=A, skew-Hermitian if unitary if AT = A-I. AT =-A, that is, akj = ajk that is, akj = -ajk The first two classes are named after Hermite (see footnote 13 in Problem Set 5.8). From the definitions we see the following. If A is Hermitian. the entries on the main diagonal must satisfy ajj = ajj; that is, they are rea1. Similarly, if A is skew-Hermitian, then ajj = -aii' If we set aij = 0' + i{3, this becomes 0' - i{3 = -(0' + i(3). Hence 0' = 0, so that aji must be pure imaginary or O. E X AMP L E 2 Hermitian, Skew-Hermitian, and Unitary Matrices A= 4 [ 1+ 3i 1, 3i B= [ -2 c= t- i 2' [ !V3 are Hennitian, skew-Hennitian, and unitary matrices, respectively, as you may verify by using the definitions. If a Hermitian matrix is real, then AT = AT = A. Hence a real Hermitian matrix is a symmetric matrix (Sec. 8.3.). Similarly, if a skew-Hermitian matrix is real, then AT = AT = -A. Hence a real skew-Hermitian matrix is a skew-symmetric matrix. Finally, if a unitary matrix is real, then AT = AT = A -1. Hence a real unitary matrix is an orthogonal matrix. This shows that Hennitian, skew-HeI7l1itian, and unitary matrices generalize symmetric, skew-symmetric, and orthogonal matrices, respectively. Eigenvalues It is quite remarkable that the matrices under consideration have spectra (sets of eigenvalues; see Sec. 8.1) that can be characterized in a general way as follows (see Fig. 161). 1m A I Skew-Hermitian (skew-symmetric) Unitary (orthogonal) Hermitian (symmetric) ReA Fig. 161. Location of the eigenvalues of Hermitian, skew-Hermitian, and unitary matrices in the complex A-plane CHAP. 8 358 Linear Algebra: Matrix Eigenvalue Problems Eigenvalues THEOREM 1 (a) The eigenvalues qf a Hermitian matrix tand tllUS of a symmetric matrix) are real. (b) The eigenvalues (~f a skew-Hermitial1 matrix (and thus matrix) are plIre imaginwy or ::,ero. qf a skew-symmetric (e) The eigenvalues of a unitary matrix (and thus of an orthogonal matrix) have absolute FaIlle L E X AMP L E 3 Illustration of Theorem 1 For the matrices in Example 2 we find by direct calculation Matrix A B C Hermitian Skew-Hennitian Unitary Charactenstic Equation A2 - llA + I~ = 0 A2 - liA + 8 = 0 A2 - iA - 1 = 0 Eigenvalues 9, 4i, 2 -2i ~V3 + ii. -iV3 + ~i • PROOF We prove Theorem L Let A be an eigenvalue and x an eigenvector of A. Multiply Ax = Ax from the left by xT. thus xTAx = AXTX. and divide by xTx = XIXI + ... + xnxn = IXll2 + ... + IXnI2, which is real and not 0 because x *- O. This gives A= (1) (a) If A is Hennitian, AT = A or AT = A and we show that then the numerator in (l) is real, which makes A reaL xT Ax is a scalar; hence taking the transpose has no effect. Thus (2) Hence, xT Ax equals its complex conjugate, so that it must be reaL (a implies b = 0.) (b) If A is skew-Hermitian, AT = -A and instead of (2) we obtain + ib a - ib (3) so that xT Ax equals mmus its complex conjugate and is pure imaginary or O. (a + ib = -(a - ib) implies a = 0.) (e) Let A be unitary. We take Ax = Ax and its conjugate transpose - (AX) T - T - = (Ax) = AXT and multiply the two left sides and the two right sides, SEC. 8.5 Complex Matrices and Forms. But A is unitary. Optional AT = A -I, 359 so that on the left we obtain Together, "TX = IAI2"Tx. We now divide by "TX (* 0) to get IAI2 = 1. Hence IAI This proves Theorem 1 as well as Theorems I and 5 in Sec. 8.3. 1. • Key properties of orthogonal matrices (invariance of the inner product, orthonormality of rows and columns; see Sec. 8.3) generalize to unitary matrices in a remarkable way. To see this, instead of R n we now use the complex vector space of all complex vectors with 11 complex numbers as ~omponents, and complex numbers as scalars. For such ,complex vectors the inner product is defined by (note the overbar for the complex conjugate) en (4) The length or norm of such a complex vector is a real number defined by THE 0 REM 2 Invariance of Inner Product A unitary transformation, that is, y = Ax with a unitaJ:': matrix A, preserves the value of the inner product (4), hence also the norm (5). PROOF The proof is the same as that of Theorem 2 in Sec. 8.3, which the theorem generalizes. In the analog of (9), Sec. 8.3, we now have bars, T u·v = fi v = -- T (Aa) Ab = _T-T _ _ a A Ab = aTlb = aTb = a·b. • The complex analog of an orthonormal systems of real vectors (see Sec. 8.3) is defined as follows. DEFINITION Unitary System A unitary system is a set of complex vectors satisfying the relationships if j (6) *k if j = k. Theorem 3 in Sec. 8.3 extends to complex as follows. THEOREM 3 Unitary Systems of Column and Row Vectors A complex square matrix is unitary row vectors) fOl71l a unitary system. if and only if its column vectors (and also its CHAP. 8 360 PROOF Linear Algebra: Matrix Eigenvalue Problems The proof is the same as that of Theorem 3 in Sec. 8.3. except for the bars required in AT = A -1 and in (4) and (6) of the present section. • THE 0 REM 4 Determinant of a Unitary Matrix Let A be a unitary Inarrix. Then iTS determinanT has absolute m/ue one, that is, AI Idet PROOF = 1. Similarly as in Sec. 8.3 we obtain I = det (AA -1) = det (AAT) = det A det AT = det A det A = det A det A = Idet A12. Hence Idet AI E X AMP L E 4 = 1 (where det • A may now be complex). Unitary Matrix Illustrating Theorems lc and 2-4 For the vectors and with 3 T = [2 -il and bT = [I + i 4i] we get aT 0.8i 0.6 A= [ 0.6 J also 0.8i A3 = as one can readily verify. This gives (Aa)TAb = -2 columns form a unitary system. a}T3} = [:J + = -0.8i· 0.8i + 0.62 = I. [2 i]T amlaTb = 2(1 + i) - 4 = -2 + 2i -0.8 + 3.2;J <\b = and [ -2.6 + 0.6i , 2i. illustrating Theorem 2. The matrix is unitary. Its a}T 32 = -0.8i· 0.6 + 0.6' 0.8i = O. ~T 32 = 0.6 + (-0.8i)0.8i = I 2 and so do its rows. Also. det A = -1. The eigenvalues are 0.6 + O.Si and -U.6 [I I]T and [I _I]T. respectively. + O.Si, with eigenvectors • Theorem 2 in Sec. 8.4 on the existence of an eigenbasis extends to complex mahices as follows. THEOREM 5 Basis of Eigenvectors A Hen71itian, skew-Hemzitian, or unitGl)' matrix has a basis of eigenvectors for that is a unitary system. en For a proof see Ref. [B3], vol. 1, pp. 270-272 and p. 244 (Definition 2). E X AMP L E 5 Unitary Eigenbases The matrices A, E, C in Example 2 have the following unitary systems of eigenvectors, as you should verify. I A: ---r= [I - 3i E: --[1-2; C: -[I 5]T \. 35 I (A -5]T vTo I V2 I]T (A = ~(i = (A I 9), - - f l - 3i = -2i), I vTo[5 + V3», -2]T Vi4 I -[I V2 1 + 2i]T -I]T (A (A (A = 2) = 4i) = !(i - V3». • SEC. 8.5 Optional Complex Matrices and Forms. 361 Hermitian and Skew-Hermitian Forms The concept of a quadratic form (Sec. 8.4) can be extended to complex. We call the numerator "TAx in (1) a form in the components Xl> • . • , Xn of x, which may now be complex. This form is again a sum of n 2 terms n "TAx = n 2. 2. ajkXjXk j~l k~l (7) + ................. . A is called its coefficient matrix. The fOlTn is called a Hermitian or skew-Hermitian form if A is Hermitian or skew-Hermitian, respectively. The value qf a Hermitianfonn is real. and that of a skew-Hennitiall form is pllre imaginw}' or z.ero. This can be seen directly from (2) and (3) and accounts for the importance of these forms in physics. Note that (2) and (3) are valid for any vectors because in the proof of (2) and (3) we did not use that x is an eigenvector but only that "TX is real and not O. E X AMP L E 6 Hermitian Form For A in Example 2 and, say, x = [I xTAx=[l-i -SiJ [ 4 1 + 3; + i SilT we get 1- 3iJ [1 + iJ 7 = [I - . . [4(1 + i) + (I - 3i)' Si] _ 223. -SI] I Si - (l + 3;)(1 t- i) • + 7· S; Clearly, if A and x in (4) are real, then (7) reduces to a quadratic form, as discussed in the last section . ....... 1. (Verification) Verify the statements in Examples 2 and 3. 2. (Product) Show (BA{ = - AB for A and B in Example 2. For any n X n Hermitian A and skew-Hermitian B. 3. Show that (ABC{ = -C-1BA for any n X n Hermitian A, skew-Hermitian B, and unitary C. 4. (Eigenvectors) Find eigenvectors of A, B, C in Examples 2 and 3. 15-111 7. 0 9. [5: EIGENVALUES AND EIGENVECTORS -i 2 6. [0 2i 2~J s] 0 5i Are the matrices in Probs. 5-11 Hermitian? SkewHermitian? Unitary? Find their eigenvalues (thereby verifying Theorem 1) and eigenvectors. 5. [4 iJ -~:J r-: 8. 10. [I :i I + i 0 1- i : i] [~ ~J 362 CHAP. 8 Linear Algebra: Matrix Eigenvalue Problems COMPLEX FORMS 113-151 Is the given matrix lcall itA) Hermitian or skew-Hermitian? Find x:TAx. (Show all (he details.) a, b, e, k are reaL 13. [ O. 12. PROJECT. Complex Matrices (a) Decomposition. Show (hat any square matrix may be written as the sum of a Hermitian and a skew-Hermitian matrix. Give examples. (b) Normal matrix. This important concept denotes a matrix that commutes with its conjugate transpose, AAT = ATA. Prove that Hermitian, skew-Hermitian. and unitary matrices are normal. Give corresponding examples of your own. (c) Normality criterion. Prove (hat A is normal if and only if the Hermitian and skew-Hermitian matrices in (a) commute. (d) Find a simple matrix that is nol normal. Find a nOimal matrix that is not Hermitian, skew-Hermitian. or unitary. (e) Unitary matrices. Prove that the product of two unitary 11 X 11 mau'ices and the inverse of a unitary matrix are unitary. Give examples. (f) Powers of unitary matrices in applications may sometimes be very simple. Show that C 12 = I in Example 2. Find further examples. - 3; -31 14. [ 0 J. x = [4 h a. + ;CJ b - Ie + 3 - ' ~J I X = k [Xl] X2 15. 16. (Pauli spin matrices) Find the eigenvalues and eigenvectors of the so-called Pallli Spill111afriees and show that SxSy = is,, SySX = -iSz, Sx2 = Sy2 = S/ = I, where Sy S z = = 0 [ i -iJ 0' I OJ [0-1 C H A-P T E R-8::: R £ V-I E W-=Q U EST ION 5 AND PRO B L EMS 1. In solving an eigenvalue problem. what is given and what is sought? 2. Do there exist square matrices without eigenvalues? Eigenvectors corresponding to more than one eigenvalue of a given matrix? 14.4 10. [ -11.2 11. [-14 -10 3. What is the defect? Why is it important? Give examples. 4. Can a complex matrix have real eigenvalues? Real eigenvectors? Gi\'e reasons. 5. What is diagonalization of a matrix? Transformation of a form to principal axes? 12. r 7. Does a 3 X 3 matrix always have a real eigenvalue? 8. Give a few typical applications in which eigenvalue problems occur. ~-:BJ 13. Find an eigenbasis and diagonalize. (Show the details.) 9. 101 [ -144 72J -103 r 14. 10J 11 11. = -2 -7 : : -:] -4 DIAGONALIZATION 102.6 I: I: -:], 18 -12 6. What is an eigenbasis'? When does it exist? Why is it important? -11.2J -: r -4 -~ -~ 11 10 Summary of Chapter 8 ~ ~-iil SIMILARITY Verify that A and Here, A, Pare: A = P- 1AP have the same spectrum. [-~: ~~ ~~], [~ 28 -5l:r ,. -14 29 2 17. [~ 2 I ~J 3.8 15. [ 2.4 16. 363 : 8 ~] 0 -1 =~], [~ ~ :] -1 4 3 2 Transformation to Canonical Form. Reduce the quadratic form to principal axes. 18. 11.56x12 + 20.16x1-'2 + 17 .44x22 = 100 19. 1.09x/ - 0.06X1X2 + \.0\ xl = 1 20. 14x12 + 24xIX2 - 4X22 = 20 1FEIJF~-HA-p..T£R- 8:= Linear Algebra: Matrix Eigenvalue Problems The practical importance of matrix eigenvalue problems can hardly be overrated. The problems are defined by the vector equation Ax (I) = Ax. A is a given square matrix. All matrices in this chapter are square. 11. is a scalar. To solve the problem (1) means to determine values of A, called eigenvalues (or characteristic values) of A, such that (I) has a nontrivial solution x (that is, x =1= 0), called an eigenvector of A corresponding to that A. An 11 X 11 matrix has at least one and at most 11 numerically different eigenvalues. These are the solutions of the characteristic equation (Sec. 8.1) all - (2) D(A) = det (A - AI) = a21 anI A a12 a22 - an2 ain A a2n ann - = O. A D(A) is called the characteristic determinant of A. By expanding it we get the characteristic polynomial of A, which is of degree n in A. Some typical applications are shown in Sec. 8.2. Section 8.3 is devoted to eigenvalue problems for symmetric (AT = A), skew-symmetric (AT = -A), and orthogonal matrices (AT = A-I). Section 8.4 concerns the diagonalization of matrices and the transformation of quadratic forms to principal axes and its relation to eigenvalues. Section 8.5 extends Sec. 8.3 to the complex analogs of those real matrices, called Hermitian (AT = A). skew-Hermitian (AT = -At and unitary matrices (AT = A-I). All the eigenvalues of a Hermitian matrix (and a symmetric one) are real. For a skew-Hermitian (and a skew-symmetric) matrix they are pure imaginary or zero. For a unitary (and an orthogonal) matrix they have absolute value l. CHAPTER · . 9 Vector Differential Calculus. Grad, Div, Curl This chapter deals with vectors and vector functions in 3-space, the space of three dimensions with the usual measurement of distance (given by the Pythagorean theorem). This includes 2-space (the plane) as a special case. It extends the differential calculus to those vector functions and the vector fields they represent. Forces, velocities, and various other quantities are vectors. This makes the algebra, geometry, and calculus of these vector functions the natural instrument for the engineer and physicist in solid mechanics. fluid flow, heat flow, electrostatics, and so on. The engineer must understand these vector functions and fields as the basis of the design and consuuction of systems, such as airplanes, laser generators, and robots. In Secs. 9.1-9.3 we explain the basic algebraic operations with vectors in 3-space. Calculus begins in Sec. 9.4 with the extension of differentiation to vector functions in a simple and natural fashion. Application to curves and their use in mechanics follows in Sec. 9.5. We finally discuss three physically important concepts related to scalar and vector fields, namely, the gradient (Sec. 9.7), divergence (Sec. 9.8), and curl (Sec. 9.9). (The use of these concepts in integral theorems follows in the next chapter. Their form in cunilinear coordinates is given in App. A3.4.) We shall keep this chapter independe1lt of Chaps. 7 alld 8. Our present approach is in harmony with Chap. 7, with the restriction to two and three dimensions providing for a richer theory with basic physical, engineering, and geometric applications. Prerequisite: Elementary use of second- and third-order determinants in Sec. 9.3. Sections that may be omitted in a shorter course: 9.5, 9.6. References and Answers to Problems: App. I Part B, App. 2. 9.1 Vectors in 2-Space and 3-Space In physics and geometry and its engineering applications we use two kinds of quantities: scalars and vectors. A scalar is a quantity that is determined by its magnitude; this is the number of units measured on a suitable scale. For instance, length. voltage. and temperature are scalars. A vector is a quantity that is determined by both its magnitude and its direction. Thus it is an arrow or directed line segment. For instance, a force is a vector, and so is a velocity, giving the speed and direction of motion (Fig. 162). 364 SEC 9.1 365 Vectors in 2-Space and 3-Space We denote vectors by lowercase boldface letters a, b. v, etc. In handwriting you may use arrows, for instance ii (in place of a), b, etc. A vecror (arrow) has a tail, called its initial point, and a tip, called its terminal point. This is motivated in the translation (displacement without rotation) of the triangle in Fig. 163, where the initial point P of the vector a is the original position of a point, and the terminal point Q is the terminal position of that point, its position after the translation. The length of the arrow equals the distance between P and Q This is called the length (or magnitude) of the vector a and is denoted by lal. Another name for length is norm (or Euclidean nonn). A vector of length 1 is called a unit vector. Velocity --~Earth I I I I " I " Force \ \ \ I I I Fig. 162. Sun Fig. 163. Translation Force and velocity Of course, we would like to calculate with vectors. For instance, we want to find the resultant of forces or compare parallel forces of different magnitude. This motivates our next ideas: to define compollents of a vector. and then the two basic algebraic operations of vector addition and scalar multiplication. For this we must first define equality of vectors in a way that is practical in connection with forces and other applications. DEFINITION Equality of Vectors Two vectors a and b are equal, written a = b, if they have the same length and the same direction [as explained in Fig. 164; in particular, note (B)l- Hence a vector can be arbitrarily translated; that is, its initial point can be chosen arbitrarily_ \~ //: Equal vectors, a=b (Al 7/ ~ b Vectors having the same length but different direction Vectors having the same direction but different length Vectors having different length and different direction eE) (e) (D) Fig. 164. (A) Equal vectors. (B)-(D) Different vectors 366 CHAP.9 Vector Differential Calculus. Grad, Div, Curl Components of a Vector We choose an xyz Cartesian coordinate system l in space (Fig. 165). that is, a usual rectangular coordinate system with the same scale of measurement on the three mutually perpendicular coordinate axes. Let a be a given vector with initial point P: (xl' YI, ZI) and tenninal point Q: (X2' Y2, Z2)' Then the three coordinate differences (1) are called the components of the vector a with respect to that coordinate system, and we write simply a = [at> a2, a 3]. See Fig. 166. The length lal of a can now readily be expressed in tenns of components because from (l) and the Pythagorean theorem we have (2) E X AMP L E 1 Components and Length of a Vector The vector a with initial point P: (4, 0, 2) and terminal point Q: (6, -1. 2) ha, the "omponents al = 6 - 4 = 2, az = -1 - 0 = - I, {l3 = 2 - 2 = O. Hence a = [2. -I. OJ. (Can you sketch a, as in Fig. 166'!) Equation (2) gives the length If we "hoose (-I, 5, 8) as the initial point of a, the corresponding terminal point is (I, 4, 8). If we choose the origin (0. O. 0) as the initial point of a, the conesponding terminal point is (2, - I, 0); its coordinate, equal the components of a. This suggests that we can determine each point in space by a vector, • called the positiol! I'ector of the point. as follows. A Cartesian coordinate system being given, the position vector r of a point A: (x. y, z) is the vector with the origin (0, 0, 0) as the initial point and A as the terminal point (see Fig. 167). Thus in components, r = [x, y, z]. This can be seen directly from (1) with Xl = .\'1 = ;::1 = O. z ,, ,, I( r /" ---x y x Fig. 165. Cartesian coordinate system -_ 1 1 1 1 \ \ 11""'-----::> ___ --_ -" \1" -"""y .... Fig. 166. Components of a vector Fig. 167. Position vector r of a point A: (x, y, z) INamed after the French philosopher and mathematician RENA TUS CARTESIUS. latinized for RENE DESCARTES (1596--1650), who invented analytic geometry. His basic work Geometrie appeared in 1637. as an appendix to his Discours de fa mitftode. SEC 9.1 367 Vectors in 2-Space and 3-Space Furthennore, if we translate a vector a, with initial point P and terminal point Q, then corresponding coordinates of P and Q change by the same amount, so that the differences in (1) remain unchanged. This proves THEOREM 1 Vectors as Ordered Triples of Real Numbers A fixed Cartesian coordinate system being given, each vector is uniquely determined by its ordered triple of corresponding components. Conversely, to each ordered triple of real numbers (ab a2, a3) there corresponds precisely one vector a = [aI' a2, a3], with (0, 0, 0) corresponding to the zero vector 0, which has length 0 and no direction. Hence a vector equation a = b is equivalent to the three equations al = bl , a2 = b2, a3 = b3 for the components. We now see that from our "geometric" definition of a vector as an arrow we have arrived at an "algebraic" characterization of a vector by Theorem 1. We could have started from the latter and reversed our process. This shows that the two approaches are equivalent. Vector Addition, Scalar Multiplication Applications suggest calculation with vectors that are practically useful and are almost as simple as the arithmetics for real numbers. The first is addition and the second is multiplication by a number. DEFINITION Addition of Vectors The sum a + b of two vectors a = [ab a2, adding the corresponding components, a3J and b = [bI> b2, b3J is obtained by (3) Fig. 168. Vector addition Geometrically, place the vectors as in Fig. 168 (the initial point of b at the terminal point of a); then a + b is the vector drawn from the initial point of a to the terminal point of b. For forces, this addition is the parallelogram law by which we obtain the resultant of two forces in mechanics. See Fig. 169. Figure 170 shows (for the plane) that the "algebraic" way and the "geometric way" of vector addition give the same vector. Fig. 169. Resultant of two forces (parallelogram law) CHAP.9 368 Vector Differential Calculus. Grad, Div, Curl Basic Properties of Vector Addition. (see also Figs. 171 and 172) Familiar laws for real numbers give immediately (a) a+b=b+a (b) tu + v) + w = u + (v + w) (C) a+O=O+a=a (d) a + (-a) = O. ( COllllllutativity) (Associativity) (4) Here -a denotes the vector having the length lal and the direction opposite to that of a. In (4b) we may simply write u + v + w, and similarly for sums of more than three vectors. Instead of a + a we also write 2a, and so on. This (and the notation -a used just before) motivates defining the second algebraic operation for vectors as follows. YI r~-----------­ 2 C r u: - - : - : ---- I L:~ ___ b!...l_ _ x Fig. 170. Vector addition Fig. 171. Cummutativity of vector addition Fig. 172. Associativity of vector addition Scalar Multiplication (Multiplication by a Number) DEFINITION The product ca of any vector a = [aI, {/2' a3] and any scalar c (real number c) is the vector obtained by multiplying each component of a by c, I a 2a -a (5) -.! a 2 Fig. 173. Scalar multiplication [multiplication of vectors by scalars (numbers)] *" Geometrically, if a 0, then ca with c > 0 has the direction of a and with c < 0 the direction opposite to a. In any case, the length of ca is leal = lellal, and ca = 0 if a = 0 or c = 0 (or both). (See Fig. 173.) Basic Properties of Scalar Multiplication. From the definitions we obtain directly (a) c(a + b) = ca + cb (b) (c + k)a = ca + ka (c) c(ka) = (ck)a (d) la = a. (6) (written cka) SEC. 9.1 369 Vectors in 2-Space and 3-Space You may prove that (4) and (6) imply for any vector a (a) Oa = 0 (b) (-I)a = -a. (7) + Instead of b E X AMP L E 2 (-a) we simply write b - a (Fig. 174). Vector Addition. Multiplication by Scalars With respect to a given coordinate system. let a = [4. O. I] Then -a = [-4, o. -I]. 7a and = [28,0,71, 2(a - b) a + b b = [6, = 2[2, 5, ~l = [4, = [2. 5. n -5. nand 10. ~l = 2a • - 2b. Besides a = lab a2, a3] another popular way of writing vectors is Unit Vectors i, j, k. In this representation, i, j, k are the unit vectors in the positive directions of the axes of a Cartesian coordinate system (Fig. 175). Hence, in components, i = [1, (9) 0, 0], j = [0, l. 0], k = [0, o. 1] and the right side of (8) is a sum of three vectors parallel to the three axes. E X AMP L E 3 i j k Notation for Vectors • In Example 2 we have a = 4i + k, b = 2i - 5j + ~k, and so on. All the vectors a = [aI, lI2, a3] = ali + a2j + a3k (with real numbers as components) form the real vector space R3 with the two algebraic operations of vector addition and scalar multiplication as just defined. R3 has dimension 3. The triple of vectors i, j, k is called a standard basis of R3. A Cartesian coordinate system being given. the representation (8) of a given vector is unique. Vector space R3 is a model of a general vector space, as discussed in Sec. 7.9, but is not needed in this chapter. ZI ,kl , a/ -a/< ,r b ".... ''l~ ~ ",-a Fig. 174. Difference of vectors /~ x y Fig. 175. The unit vectors i, j, k and the representation (8) CHAP. 9 370 :u 11-61 Vector Differential Calculus. Grad, Div, Curl --.- COMPONENTS AND LENGTH Find the components of the vector v with given initial point P and terminal point Q. Find Ivl. Sketch Ivl. Find the unit vector in the direction of v. Q: (5, -2,0) 1. P: (3, 2, 0), Q: (-4, -4. -4) 2. P: (1. I, 1). 3. 4. 5. 6. P: P: P: P: (1. 0, 1.2), (2, -2,0), (4, 3, 2), (0, O. 0). Q: (0, 0, 6.2) Q: (0,4,6) Q: (-4, -3,2) Q: (6, 8, 10) -~2J Given the components vI> V2, V3 of a vector v and a particular initial point P, find the corresponding terminal point Q and the length of v. 7. 3, -1,0; P: (4.6,0) 8. 8,4, -2; P: (-8, -4,2) L 9. !, 2,~; 10. 3,2.6; 11. 4,~, -~; 12. 3, -3,3; ~ 3-20 I P: (0. -~. ~); P: (0. O. 0) P: (-4,~, 2) P: (1,3, -3) VECTOR ADDITION AND SCALAR MULTIPLICATION Let a = [2, - I, 0] = 2i b = [-4, 2, 5] = -4i + 2j Find: 13. 2a, -a, -~a 15. 5(a - c). 5a - 5c 16. (3a - 5b) + 2c, 3a + 17. 6a - 4b + 2c, 2(3a 18. (lIla l)a, (1IIcl)c j, + 5k, c = [0,0, 3J = 3k. 14. a + 2b, 2b + a (-5b + 2c) 2b + c) 19. a + b + c, -3a - 3b - 3e 20. la + hi, lal + lb: 21. What laws do Probs. 14-17 illustrate? 22. Prove (4) and (6). 23. Find the midpoint of the segment PQ in Probs. 7 and 9. =-1-281 Find the 24. p = v = 25. P = 26. P = 27. P = 28. P = FORCES resultant (in components) and its magnitude. [1,2,0]. q = [0,4, -I], U = [4.0, -3], [6,2,4] [2,2,2], q = [-4, -4,0], U = [2,2,7] [-I, -3, -5]. q = [6.4, 2J, u = [-5, -1. 3] [8,2. -4], q = 3p, U = -5p [3.0, -2], q = [2,5, I], u = 4q 29. Find v so that v, p, q. U in Prob. 25 are in equilibrium. 30. For what c is the resultant of [3, I, 7], [4, 4, 5], and [3. 2. c] parallel to the x\"-plane? 31. Find forces P. q, U in the direction of the coordinate axes such that p, q, U, v = [2.3.0], w = [7. -I, 11] are in equilibrium. Are p. q, U uniquely determined? 32. If Ipi = I and Iql = 2, what can be said about the magnitude and direction of the resultant? Can you think of an application where this matters? 33. Same question as in Prob. 32 if IPI = 3. Iql = 2, :ul = I. 34. (Relative velocity) If airplanes A and B are moving southwest with speed IVAI = 500 mph and northwest with speed IVBI = 400 mph. respectively, what is the relative velocity v = VB - VA of B with respect to A? 35. (Relative velocity) Same question as in Prob. 34 for two ships moving northwest with speed IVAI = 20 knots and northeast with speed IVBI = 25 knots. 36. (Reflection) If a ray of light is reflected once in each of two mutually perpendicular mirrors, what can you say about the reflected ray? 37. (Rope) Find the magnitude of the force in each rope in the figure for any weight wand angle a. 38. TEAM PROJECT. Geometric Applications. To increase your skill in dealing with vectors, use vectors to prove the following (see the figures). (a) The diagonals of a parallelogram bisect each other. (b) The line through the midpoints of adjacent sides of a parallelogram bisects one of the diagonals in the ratio I: 3. (e) Obtain (b) from (a). (d) The three medians of a triangle (the segments from a vertex to the midpoint of the opposite side) meet at a single point, which divides the medians in the ratio 2:1. (e) The quadrilateral whose vertices are the midpoints of the sides of an arbitrary quadrilateral is a parallelogram. (n The four space diagonals of a parallelepiped meet and bisect each other. (g) The sum of the vectors drawn from the center of a regular polygon to its vertices is the zero vector. ~ a w Problem 37 Team Project 38(a) o~ a Team Project 38{d) c C _ b D~B ~ A a Team Project 38(e) 371 SEC. 9.1 Inner Product (Dot Product) 9.2 Inner Product (Dot Product) We shall now define a multiplication of two vectors that gives a scalar as the product and is suggested by various applications. in particular when angles between vectors and lengths of vectors are involved. DEFINITION Inner Product (Dot Product) of Vectors The inner product or dot product a" b (read "a dot b") of two vectors a and b is the product of their lengths times the cosine of their angle (see Fig. 176), a" b (1) = lallbl cos '}' if a*O,b*O if a = 0 or b = O. a"b = 0 The angle '}', 0 ~ '}' ~ 7T, between a and b is measured when the initial points of the vectors coincide, a~ in Fig. 176. In components, a = [a!> a2, ag], b = [bI> b 2, bg], and (2) o or b The second line in (l) is needed because '}' is undefined when a derivation of (2) from (1) is shown below. 2J(_ b a.b>O Fig. 176. t a[l b - a.b=O O. The ~b a·b<O Angle between vectors and value of inner product Orthogonality. Since the cosine in (1) may be positive, 0, or negative. so may be the inner product (Fig. 176). The case that the inner product is zero is of particular practical interest and suggests the following concept. A vector a is called orthogonal to a vector b if a" b = O. Then b is also orthogonal to a, and we call a and b orthogonal vectors. Clearly, this happens for nonzero vectors if and only if cos '}' = 0; thus '}' = rr/2 (90°). This proves the important THEOREM 1 Orthogonality The inner product of two nonzero vectors is 0 if and only if these vectors are perpendicular. 372 CHAP 9 Vector Differential Calculus. Grad, Div, Curl Length and Angle. Equation (1) with b (3) lal = = a gives a'a = lal 2. Hence v;;a. From (3) and (1) we obtain for the angle 'Y between two nonzero vectors (4) E X AMP L E 1 = cos 'Y Inner Product. Angle Between Vectors Find the inner product and the lengths of a vectors. Solution. a'b = J ·3 = [I. + 2 'l-2) + O· I = 2, 0] and b -I, lal = = [3, - 2, I] as well as the angle between these Va~ = "\ '5, Ibl = v'b-b = v'J4, and l4) gives the angle a'b y= arccos lallbl = arccos (-0.1 1952) = 1.69061 = 96.865°. • From the definition we see that the inner product has the following properties. For any vectors a, b, C and scalars ql, q2, (a) (5) (qla + q2b)'C = qla'c + q2b ' c (b) a'b = b'a a'a ~ (Linearity) (Symmetry) 0 (c) } (Positive-definitelless). a'a = 0 if and only if a=O Hence dot multiplication is commutative [see (5b)] alld is distributive 'with re.\pect to vector addition; in fact, from (Sa) with ql = I and q2 = I we have (5a*) (a Furthermore, from (1) and Icos + b)'c = a'c + b'c 'YI (6) (Distributivity). 3 I we see that la· bl 3 lallbl (Cauchy-Schwarz inequality). Using this and (3). you may prove (see Prob. 18) (7) la + bl 3 lal + Ibl (Triangle inequality). Geometrically, (7) with < says that one side of a triangle must be shorter than the other two sides together; this motivates the name of (7). A simple direct calculation with inner products shows that (8) la + bl 2 + la - bl 2 = 2(la1 2 + Ib1 2) (Parallelogram equality). Equations (6)-(8) play a basic role in so-called Hilbert .\paces (abstract inner product spaces), which form the basis of quantum mechanics (see Ref. [GR7] listed in App. I). SE.c. 9.2 373 Inner Product (Dot Product) Derivation of (2) from (1). We write a = ali + a2j + ({3k and b = bli + b2 j as in (8) of Sec. 9.1. If we substitute thi~ into a-b and use (5a*), we first have a 3 X 3 = 9 products + b3k, ~um of Now i, j, k are unit vectors, so that i- i = j -j = k - k = I by (3). Since the coordinate axes are perpendicular, so are j, j, k, and Theorem I implies that the other six of those nine products are 0, namely, j-j = j-j = j-k = k-j = k-j = j-k = O. But this reduces our sum for a-b to (2). • Applications of Inner Products Typical applications of inner products are shown in the following examples and in Problem Set 9.2. E X AMP L E 2 Work Done by a Force Expressed as an Inner Product This is a major application. It concerns a body on which a cO/lSlOl/l force p acts. (For a l'C/r;able force. see Sec. 10.1.) Let the body be given a displacement d. Then the work done by p in the displacement is defined as W = (9) ipiidi cos a = pod, that is. magnitude ipi of the force times length idi of the displacement times the cosine of the angle a between p and d (Fig. 177). If a < 90°. as in Fig. 177. then W> O. If P and d arc orthogonal, then the work is zero (why'!). If a> 90°. then W < O. which means that in the displacement one has to do work against the force. • (Think of swimming across a river at some angle a against the current.) y ~/~ I I • I '",8 d Fig. 177. E X AMP L E 3 Work done by a force Fig. 178. Example 3 Component of a Force in a Given Direction What force in the rope in Fig. 178 will hold a car of 5000 lb in equilibrium if the ramp makes an angle of 25° with the hori70ntal? Solutioll. Introducing coordinates as shown. the weight is a = [0. -5000] because this force points downward. in the negative .,·-direction. We have to represent a as a sum (resultant) of two forces. a = c + p, where c is the force the car exerts on the ramp. which is of no interest to us. and p is parallel to the rope. of magnimde (see Fig. 178) ipl = lal cos l' = 5000 cos 65° = 2113 [Ib) and direction of the unit vector U opposite to the direction of the rope; here l' between a and p. Now a vector in the direction of the rope is b = [-I. tan 25°] = [-I. 0.466311, thus ,bl = 90° - 25° = 1.10338. = 65° is the angle 374 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl so that U = Since lUI = - 1 Tbf b = [0.90631, -0.42262]. I and cos y > 0, we see that we can also write our result as Ipi = (Ial co, y)lul = a'u = a' b -lbi = 5000·0.46631 1.l0338 = 2113 [Ib]. Answer: AbDUl 2100 lb. • Example 3 is typical of applications in which one uses the concept of the component or projection of a vector a in the direction of a vector b (*- 0), defined by (see Fig. 179) (10) p = lal cos y. Thus p is the length of the orthogonal projection of a on a straight line I parallel to b, taken with the plus sign if pb has the direction of b and with the minus sign if pb has the direction opposite to b; see Fig. 179. a~ l~~-: b '-----v-------- P (p>O) Fig. 179. Multiplying (10) by (p=O) (p<O) Component of a vector a in the direction of a vector b Ibi/lbi = I, we have a-b in the numerator and thus p= (11) (b *- 0). If b is a unit vector, as it is often used for fixing a direction, then (11) simply gives (12) p = a-b (Ibl = 1). Figure 180 shows the projection p of a in the direction of b las in Fig. 179) and the projection q = Ibl cos 'Y of b in the direction of a. a q ..-/"i rZ\ : ~ p Fig. 180. Projections p of a on band q of b on a SEC. 9.2 375 Inner product (Dot Product) E X AMP L E 4 Orthonormal Basis By definition. an orthonormal basis for 3-space is a basis (a. b. c) consisting of orthogonal unit vectors. It has the great advantage that the determination of the coefficients in representations v = 11 a + 12 b + '3c of a given vector v is very simple. We claim that '1 = a v. 12 = b V, 13 = Co v. Indeed. this follows simply by taking the inner products of the representation with a, b, c, respectively, and using the orthononnality of the basis, aov = Ilaoa + ' 2a o b + '3aoc = II, etc. For example, the unit vectors i. j. k in (8), Sec. 9.1, associated with a Cartesian coordinate system form an orthonormal basis. called the standard basis with respect to the given coordinate system. • 0 E X AMP L E 5 0 Orthogonal Straight Lines in the Plane Find the straight line LI through the point P: O. 3) in the -":I'-plane and perpendicular to the straight line - 2y + 2 = 0; see Fig. 181. Lz: x Solution. The idea is to write a general straight line L I : alx + a2.\" = c as a r = c with a = [aI, a2] "" 0 and r = [x. y]. according to (2). Now the line Ll * through the origin and parallel to Ll is a r = O. Hence, by Theorem I, the vector a is perpendicular to r. Hence it is perpendicular to L J * and also to LI because LI and Ll l' are parallel. a is called a nOl'mal vector of LI (and of Ll *). Now a nonnal vector of the given line x - 2y + 2 = 0 is b = [I, -2]. Thus L J is perpendicular (0 L2 if boa = al - 2a2 = 0, for instance, if a = [2, I]. Hence L J is given by 2x + Y = c. It passes through P: (I, 3) when 2· I + 3 = c = 5. Answer: y = -2x + 5. Show that the point of intersection is (x • .1') = (1.6, 1.8). • 0 0 E X AMP L E 6 Normal Vector to a Plane Find a unit vector perpendicular to the plane 4x Solutioll. + 2y + 4;:: = -7. Using (2). we may write an) plane in space as where a = [al' a2. a31 "" 0 and r = [x. y, z]. The unit vector in the direction of a is (Fig. IS2) I n = Iaf a. Dividing by lal. we obtain from (13) nor (14) = p where From (12) we see that p is the projection of r in the direction of n. This projection ha~ the same constant value ellal for the position vec(Or I' of any point in the plane. Clearly this holds if and only if n is perpendicular to the plane. n is called a unit nonnal vector of the plane (the other bcing - nl. Furthermore. from this and the definition of projection it follows that Ipl is the distance of the plane from the origin. Representation 04) is called Hesse's2 nonnal form of a plane. In our case, a = [4, 2, 41. c = -7, lal = 6. n = ~a = [~. !. and the plane has the distance 7/6 frum the origin. • n n 2 Fig. 181. 2 3 Example 5 x Fig. 182. Normal vector to a plane LUDWIG OTTO HESSE (1811-1874), Gennan mathematician who contributed to the theory of curves and surfaces. CHAP. 9 376 INNER PRODUCT 11-121 Let a = [2. I. 4]. b = [-4, 0.3], c = [3, -2, 1]. Find 1. aob,boa 3. 5. 7. 9. Vector Differential Calculus. Grad, Div, Curl 13a - 2bl, 12b - 3al (aoblc, a(b·c) (a - bloc, a·c - boc ao(b - c), ao(c - b) 11. 6(a + b) • (a - b) 2. lal, Ibl, Icl 4. ao(b + c), aob + aoc 6. aob + boc + coa 8. 4a 3c, 12a c 0 10. Ib + cl, Ibl 12. la cl, lallcl + Icl 0 = uow with u 0/= 0 imply that v = w? inequality. and the parallelogram equality for the above a and b. 17. Prove the parallelogram equality. 18. (Triangle inequality) Prove (7). Hint. Use (3) for la + bl and (6) to prove the square of (7). then take roots. 119-221 WORK Find the work done by a force p acting on a body if the body is displaced from a point A to a point B along the straight segment AB. Sketch p and AB. (Show the details of your work.) 19. p = [8. -4. 11], A: (I. 2. 0). B: (3, 6. 0) 20. p = [2. 7. -4], A: (3. I. m. B: (0. 2, 0) 21. p = [5, -2, I], A: (4, 0, 3), B: (6, 0, 8) 22. p = [4.3. 6J, A: (5. 2. 10). B: (1, 3. l) 23. Why is the work in Prob. 19 zero? Can work be negative? Explain. 24. Show that the work done by the resultant of p and q in a displacement from A to B is the sum of the work done by each force in that displacement. 25. Find the work W = pod if d = 2i and p = i, i + j, j, -i + j and sketch a figure similar to Fig. 177. I ANGLE BETWEEN VECTORS. ORTHOGONALITY Let a = ll, I, I], b angle between: = [2.3.1], c = 26. a, b 27. b, c 29. a + b, c [-I, 1,0]. Find the 18. a-c,b-c 30. a, h + C 31. (Planes) Find the angle between the planes x + y + .;; = 1 and 2x - J + 2;: = O. 34. (Addition law) Obtain cos (a - (3) = cos a cos {3 + sin a sin {3 by using a = [cos a, sin a]. b = [cos {3, sin {3]. where o ~ a ~ (3 = 27f. 35. (Parallelogram) Find the angles if the sides are [5, 0] and [I. 21- IS. Prove the Cauchy-Schwarz inequality. 16. Verify the Cauchy-Schwarz inequality, the triangle 126--30 33. (Triangle) Find the angles of the triangle with vertices [0, 0, 0], [1, 2, 3], [4, -1, 3]. 0 13. What laws do Probs. I. 3.4, 7, 8 illustrate? 14. Does uov 32. (Cosine law) Deduce the law of cosines by using vectors a, b, and a-b. 36. (Distance) Find the distance of the plane 5x + 2y + z = 10 from the origin. 13740 I COMPONENTS IN THE DIRECTION OF A VECTOR Find the component of a in the direction of b. 37. a = [I. 1,3]. b = [0, O. 5] 38. a = [2. O. 6], b = [3. 4, - 11 39. a = [0.4, -3]. b = [0.4,3] 40. a = [-1,2,0]. b = [1, -2,0] 41. Cnder what condition will the projection of a in the direction of b equal the projection of b in the direction of a? 42. TEAM PROJECT. Orthogonality is particularly important. mainly because of the use of orthogonal coordinates. such as Cartesian coordinates, whose "natural basis" (9). Sec. 9.1. consists of three 0l1hogonal unit vectors. (a) Show that a = [2. -2.4]. b = [0.8.4], c = [-20. -4. 8] are orthogonaL (b) For what values of al are a b = [3.4, -IJ 0l1hogonal? = (c) Show that the straight lines 4x 5x - 10)" = 7 are orthogonaL [aI- 2, 0] and + 2y = 1 and (d) Find all unit vectors a = [a lo a2] in the plane orthogonal to [4, 31. (e) Find all vectors orthogonal to a = [2. l. 0]. Do they fonn a vector space? (I) For what c are the plane!> 4x - 2)" + 3.;; = 6 and 2x - cy + 5::: = 1 orthogonal? (g) Under what condition will the diagonals of a parallelogram be orthogonal? (Prove your answer.) (h) What is the angle between a light ray and its reflection in three orthogonal plane minors (known as a "corner reflector")"? (i) Discuss further applications in physics and geometry in which orthogonality plays a role. 377 SEC. 9.3 Vector Product (Cross Product) 9.3 Vector Product (Cross Product) The dot product in Sec. 9.2 is a scalar. We shall see that in some applications, for instance, in connection with rotations, we shall need a product that is again a vector: DEFINITION Vector Product (Cross Product, Outer Product) of Vectors The vector product (also called cross product or outer product) a x b (read "a crOss b"') of two vectors a and b is the vector v=axb as follows. If a and b have the same or opposite direction, or if a = 0 or b = 0, then v = a x b = O. In any other case v = a x b ha~ the length (1) Ivl = la x bl = lallbl sin 'Y. This is the area of the blue parallelogram in Fig. 183. 'Y is the angle between a and b (as in Sec. 9.2). The direction of v = a x b is perpendicular to both a and band such that a, b, v, in this order, form a right-hallded triple as in Figs. 183-185 (explanation below). In components, let a = [aI' a2' a3] and b = [b 1, b 2, b3]. Then v has the components = [Vb V2, V3] = a x b (2) Here the Cartesian coordinate system is right-handed, as explained below (see also Fig. 186). (For a left-handed system, each component of v must be multiplied by -1. Derivation of (2) in App. 4.) Right-Handed Triple. A triple of vectors a, b, v is right-handed if the vectors in the given order assume the same sort of orientation as the thumb, index finger, and middle finger of the right hand when these are held as in Fig. 184. We may also say that if a is rotated into the direction of b through the angle 'Y « 'IT), then v advances in the same direction as a right-handed screw would if turned in the same way (Fig. 185). a Fig. 183. a Vector product Right-handed triple of vectors a, b, v Fig. 184. Fig. 185. Right-handed screw CHAP. 9 378 Vector Differential Calculus. Grad, Div, Curl z k ~j x / ~ :k x y y z (b) Left-handed (a) Right-handed Fig. 186. The two types of Cartesian coordinate systems Right-Handed Cartesian Coordinate System. The system is called right-handed if the corresponding unit vectors i, j, k in the positive directions of the axes (see Sec. 9.1) form a right-handed triple as in Fig. 186a. The system is called left-handed if the sense of k is reversed, as in Fig. 186b. In applications, we prefer right-handed systems. How to Memorize (2). (2) can be written If you know second- and third-order determinants, you see that (2*) and v = [VI, V 2 , V3] = VIi + V~ + V3k is the expansion of the following symbolic determinant by its first row. (We call the determinant "symbolic" because the first row consists of vectors rather than of numbers.) j (2**) v= a x b = k a} For a left-handed system the determinant has a minus sign in front. E X AMP L E 1 Vector Product For the vector product v = a x b of a = [I, I, OJ and b = [3, 0, 0] in right-handed coordinates we obtain from (2) VI = 0, V3 = I ·0 - 1·3 = -3. We confirm this by (2**): j k ~I k v=axb= 3 o = -3k = [0,0, -3]. o To check the result in this simple case, sketch a, b, and v. Can you see that two vectors in the .xy-plane must always have their vector product parallel to the z-axis (or equal to the zero vector)? • SEC 9.3 379 Vector Product (Cross Product) E X AMP l E 2 Vector Products of the Standard Basis Vectors i x j = k, jxk=i, k x i = j k x j = -i, i x k = -j. (3) j x i = - k, • We shall use this in the next proof. THEOREM 1 General Properties of Vector Products (a) For every scalar I, (4) (fa) x b = lea x b) = a x (fb). (b) Cross multiplication is distributive with respect to vector addition; that is. x b) + (a x c), b) x c = (a x c) + (b x c). (0') a x (b ({3) (a (5) + + c) = (a (c) Cross multiplication is not cOllllllutative bllt alltico11l11lutative; that is, b x a (6) a I bXRf Fig. 187. Anticommutativity of cross multiplication PROOF = -l3 x b) (Fig. 187). (d) Cross multiplication is not associative; that is, in general, (7) a x (b x C) "* (a x b) x c so that the parentheses cannot be omitted. (4) follows directly from the definition. In (50'), formula (2*) gives for the first component on the left I= a3 b3 + a2(b 3 + C3) - a3(b 2 + C2) C3 By (2*) the sum of the two determinants is the fIrst component of (a x b) + (a x C), the right side of (Sa). For the other components in (50') and in (5{3), equality follows by the same idea. Anticommutativity (6) follows from (2**) by noting that the interchange of Rows 2 and 3 multiplies the determinant by -I. We can confirm this geometrically if we set a x b = v and b x a = w; then Ivl = Iwl by (1), and for b, a, w to form a right-handed triple, we must have w = -v. Finally. i X (i x j) = i x k = -j, whereas (i x i) x j = 0 X j = 0 (see Example .. 2). This proves (7). 380 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl Typical Applications of Vector Products E X AMP L E 3 Moment of a Force In mechanics the moment III of a force p about a point Q is defined as the product III = Ipld, where d is the (perpendicular) distance between Q and the line of action L of p (Fig. 188). If r is the vector from Q to any point A on L, then d = 11'1 sin l' (Fig. 188) and 111 Since 1'is the angle between = Irllpl sin 1'. and p, we see from (I) that I' m = (8) I' X III = Ir x pI. The vector P * is called the moment \'ector or "ector moment of p about Q. Its magnitude is 111. If m 0, its direction is that of the axis of the rotation about Q that p has the tendency to produce. This axis i~ perpendicular to both I' and p. • L Qy-_ _ p~ l' y \ d \ \ Moment of a force p Fig. 188. E X AMP L E 4 Moment of a Force Find the moment of the force p in Fig. 189 about the center Q of the wheel. Solutioll. Introducing coordinates as shown in Fig. 18<), we have p = [1000 cos 30°, 1000 sin 30°, 0] = [866, 500, r = [0, 0], 1.5, 0]. (Note that the center of the wheel is at y = -1.5 on the y-axis.) Hence (8) and (2**) give j m=rxp= k 0 1.5 0 = Oi - OJ + 866 SOU 0 18~6 5 1. k = [0,0, -1299]. 500 1 This moment vector is normal (perpendicular) to the plane of the wheel; hence it has the direction of the axis of rotation about the center of the wheel that the fon;e has the tendency to produce. m points in the negative ;:-direction, the direction in which a right-handed screw would advance if turned in that way. • y Ipl = 1000 Ib x Q Fig. 189. Moment of a force p SEC. 9.3 Vector Product (Cross Product) E X AMP L E 5 381 Velocity of a Rotating Body A rotation of a rigid body B In space can be simply and uniquely described by a vector w as follows. The direction of w i, that of the axis of rotation and such that the rotation appears clockwise if one looks from the initial point of w to its terminal point. The length of w is equal to the angular speed w (> 0) of the rotation. thai is. the linear (or tangential) speed of a point of B divided by its dislance from the axis of rotation. Let P be any point of Band d its distance from the axis. Then P has the speed wd. Let r be the position vector of P referred to a coordinate system with origin 0 on the axis of rotation. Then d = Irl sin 'Y. where 'Y is the angle between wand r. Therefore. wd = Iwllrl sin 'Y = Iw x rl. From this and the definition of vector product we see that the velocity vector v of P can be represented in the form (Fig. 190) v = w x r. (9) • This simple formula is useful for determining v at any point of B. d Fig. 190. Rotation of a rigid body Scalar Triple Product The most important product of vectors with more than two factors is the scalar triple product or mixed triple product of three vectors a. b, c. It is denoted by (a b c) and defined by \a (10*) b c) = ao(b x c). Because of the dot product it is a scalar. In terms of components a = [at. {/2, l/3]. b = [b 1 • b 2 , b3 ]. C = [Cb c2, C3] we can write it as a third-order detenninant. For this we set b x c = v = [VI. V 2 , V3]' Then from the dot product in components [formula (2) in Sec. 9.2] and from (2*) with band c instead of a and b we first obtain The sum on the right is the expansion of a third-order determinant by its first row. Thus (10) (a b c) = ao(b x c) = b1 CHAP. 9 382 Vector Differential Calculus. Grad, Div, Curl The most important properties of the scalar triple product are as follows. THEOREM 2 Properties and Applications of Scalar Triple Products (a) In (10) the dot and cross can be interchanged: (a (11) b = ao(b x c) = (a x b)oc. c) (b) Geometric interpretation. The absolute value I(a b c)1 of (10) is the volume of the parallelepiped (oblique box) with a, b. c as edge vectors (Fig. 191). (c) Linear independence. Three vectors in R3 are linearly independent only ~f their scalar triple product is not zero. PROOF if and (a) Dot multiplication is commutative. so that by (10) (a x b)oc = coCa x b) = al a2 a3 b1 b2 b3 From this we obtain the determinant in (10) by interchanging Rows 1 and 2 and in the result Rows 2 and 3. But this does not change the value of the determinant because each interchange produces a factor - I, and (- 1)( -}) = 1. This proves (11). (b) The volume of that box equals the height h = lallcos 'YI (Fig. 191) times the area of the base, which is the area Ib x cl of the parallelogram with sides b and c. Hence the volume is lallb cllcos 'YI X = lao (b x c)1 (Fig. 191) as given by the absolute value of (II). (c) Three nonzero vectors, if we let their initial points coincide, are linearly independent if and only if they do not lie in the same plane (or do not lie on the same straight line). This happens if and only if the triple product in (b) is not zero, so that the independence criterion follows. (The case that one of the vectors is the zero vector is trivial.) • I I bxc I I 1 I :h/~-----/ /1 // 1 Fig. 191. E X AMP L E 6 b Geometric interpretation of a scalar triple product Tetrahedron A tetrahedron is determined by three edge vectors a, b, c, as indicated in Fig. 192. Find its volume when a = [2. O. 3]. b Soilltioll. = [0.4. I]. c = [5.6. OJ. The volume Vof the parallelepiped with these vectors as edge vectors is the absolute value of the scular triple product SEC. 9.3 383 Vector Product (Cross Product) (a b b c) = 2 o 0 4 5 6 3 :1 = -12 - 60 = -72. o Hence V = 72. The minus sign indicates that if the coordinates are right-handed, the triple a, b, c is left-handed. The volume of a tetrahedron is ~ of that of the parallelepiped (can you prove it?). hence 12. Can you sketch the tetrahedron, choosing the origin as the common initial point of the vectors? What are the coordinates of the four vertices? • Fig. 192. Tetrahedron This is the end of vector algebra (in space R3 and in the plane). Vector calculus (differentiation) begins in the next section. [ 1-20 VECTOR PRODUCT, SCALAR TRIPLE PRODUCT 1 With respect to right-handed Cartesian coordinates, let a = [1. 2. 0]. b = [3. -4,0], c = [3.5.2]. d = [6,2, -3]. Showing details. find: 1. a x b, b x a 2. a x c, la xci, aoc 3. (a + b) x c, a x c + b x c 4. (c + d) x d, c x d 5. 2a x 3b, 3a x 2b, 6a x b 6.bxc+cxb 7. ao(b x c), (a x bloc 8. (a 9. (a 10. (a + x x b) x (b 18. (a + b 19. (a - c 20. (4a 3b (12) la x bl = Y(aoa)(bob) - (a ob)2 (13) b x (c x d) = (bod)c - (boc)d (14) (a x b) x (c x d) = (15) (a b d)c - (a b c)d (a x b)o(c x d) = (aoc)(bod) - (aod)(boc) (a b c) = (b (16) = -(c b) x c, a x (b x c) c a) b = (c a) = -(a a b) c b) 11. d x c, Id x cI. IC x dl 12. (a + b) x (c + d) 13. a x (b + c - dl [25-281 MOMENT OF A FORCE Find the moment vector m and the moment 111 of a force p about a point Q when p ads on a line through A. 14. (i 25. P = L4, 4, 0], Q: (2, 1,0), A: (0.3,0) j k), (i k j) ..:1'\1 U).ltii U Formula (15) is called Lagrange's identity. + a) b)o(c x d). (b x a)o(d x c) 15. (i + j j + k k + i) 16. (b x Clod, bo(c x d) .,. til each side of (13) then equals [-b2C2dl' b1C2dl' 0]. and give reasons why the two sides are then equal in any Cartesian coordinate system. For (14) and (15) use (13). 27. P = [1,2,3]. Q: (0, I, 1), A: (1. 0, 3) fL ~ .n b + c c + d) b - c c). (a b lc). 24(b 26. P = [0. O. 5]. Q: (3. 3. 0), A: (0, O. 0) c c) a) 21. What properties of cross multiplication do Probs. I, 3, 8, 10 illustrate? 22. Give the details of the proofs of (4) and (5). 23. Give the details of the proofs of (6) and (11). 24. TEAM PROJECT. Useful Fonnulas for Two and More Vectors. Prove (12)-06). which are often useful in practical work. and illustrate each formula with two examples. Hillts. For (13) choose Cartesian coordinates such that d = [c11 , 0, 0] and c = [Cl, C2. 0]. Show that 28. p = [4. 12.8]. Q: (3.0,5). A: (4. 3. 7) 29. (Rotation) A wheel is rotating about the y-axis with angular speed w = 10 sec-I. The rotation appears clockwise if one looks from the origin in the positive y-direction. Find the velocity and speed at the point (4, 3, 0). 30. (Rotation) What are the velocity and speed in Prob. 29 at the point (4. 2. -2) if the wheel rotates about the line y = x, Z = 0 with w = 5 sec-I. GEOMETRIC APPLICATIONS 31. (Parallelogram) Find the area if the vertices are (2, 2), (9. 2), (10, 3), (3, 3). CHAP. 9 384 Vector Differential Calculus. Grad, Div, Curl 32. (Parallelogram) Find the area if the vertices are (3, 9, 8), (0, 5, 1), (-1, -3, -3), (2, 1, 4). 33. (Triangle) Find the area if the vertices are (1, 0, 0). (0. 1. 0), (0. O. 1). 34. (Triangle) Find the area if the vertices are (4. 6. 5). (4. 9, 5), (8.6. 7). 35. (Plane) Find a nonnal vector and a representation of the plane through the points (4, 8, 0), (0, 2, 6), (3, 0, 5). 36. (Plane) Find the plane through (2, I, 3), (4, 4. 5), (I, 6, 0). 9.4 37. (Parallelepiped) Find the volume of the parallelepiped detennined by the vertices (1, 1, 1), (4, 7, 2). (3, 2, 1), (5, 4, 3). 38. (Tetrahedron) Find the volume of the tetrahedron with vertices lO, 2, 1), (4, 3, 0), (6, 6, 5), (4, 7, 8). 39. (Linear dependence) For what c are the vectors [9, 1, 2J. [-I, c, 5]. [4, c. 5] linearly dependent? 40. WRITING PROJECT. Applications of Cross Products. Summarize the most important applications we have discussed in this section and give a few simple examples. No proofs. Vector and Scalar Functions and Fields. Derivatives We now begin with vector calculus. This calculus concerns two kinds offunctions, namely, vector functions, whose values are vectors depending on the points P in space, and scalar functions, whose values are scalars f = f(P) depending on P. Here, P is a point in the domain of definition, which in applications is a (three-dimensional) domain or a surface or a curve in space. We say that a vector function defines a vector field, and a scalar function defines a scalar field in that domain or on that surface or curve. Examples of vector functions are shown in Figs. 193-196. Examples of scalar fields are the temperature field in a body or the pressure field of the air in the earth's atmosphere. Vector and scalar functions may also depend on time t or on some other parameters. Notation. write If we introduce Cartesian coordinates x, y. z, then instead of v(P) we can also "Fig. 193. Field of tangent vectors of a curve Fig. 194. Field of normal vectors of a surface SEC. 9.4 Vector and Scalar Functions and Fields. Derivatives 385 but we keep in mind that components depend on the choice of a coordinate system, whereas a vector field that has a physical or a geometric meaning should have magnitude and direction depending only on P, not on that choice. Similarly for the value of a scalar field f(P) = f(x, y, z). E X AMP L E 1 Scalar Function (Euclidean Distance in Space) The distance f(P) of any point P from a fixed point Po in space is a scalar function whose domain of definition is the whole space. f(P) defines a scalar field in space. If we introduce a Cartesian coordinate system and Po has the coordinates xo, Yo, Zo, then f is given by the well-known formula f(P) = f(x, y, z) = Vex - .\'0)2 + (y - YO)2 + (z - -::.0)2 where x, y, z are the coordinates of P. If we replace the given Cartesian coordinate system with another such system by translating and rotating the given system, then the values of the coordinates of P and Po will in general change, but J(P) will have the same value as before. Hence f(P) is a scalar function. The direction cosines of the straight line through P and Po are not scalars because their values depend on the choice of the coordinate system. • E X AMP L E 2 Vector Field (Velocity Field) At any instant the velocity vectors v(P) of a rotating body B constitute a vector field, called the velocity field of the rotation. If we introduce a Cartesian coordinate system having the origin on the axis of rotation, then (see Example 5 in Sec. 9.3) (1) vex, y, z) = w xr = w X [x, y, zl = w x (xi + yj + zk) where x, y, z are the coordinates of any point P of B at the instant under consideration. If the coordinates are such that the z-axis is the axis of rotation and w points in the positive z-direction, then w = wk and k v = 0 o w x y z = w[ -y, x, 0] = w(-yi + xj). An example of a rotating body and the corresponding velocity field are shown in Fig. 195. • I I I I I --~-=~-1~ K =-=~~ I I c0 Fig. 195. E X AMP L E 3 Velocity field of a rotating body Vector Field (Field of Force, Gravitational Field) Let a particle A of mass M be fixed at a point Po and let a particle B of mass m be free to take up various positions P in space. Then A attracts B. According to Newton's law of gravitation the corresponding gravitational force p is directed from P to Po, and its magnitude is proportional to IIr 2, where r is the distance between P and Po, say, (2) c= GMIIl. 386 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl Here G = 6.67' 10-8 cm3 /(gm· sec2 ) is the gravitational constant. Hence p defines a vector field in space. If we introduce Cartesian coordinates such that Po has the coordinates xo. Yo. Zo and P has the coordinate~ x. y. z. then by the Pythagorean theorem. (~ Assuming that r 0). > 0 and introducing the vector r = [x - = Ix - xo, Y - .1'0' :: - ::01 xo)i + {y - yo)j + {z - :::o)k. we have Irl = ,.. and {- IIrjr is a unit vector in the direction of p; the minus sign indicates that p is directed from P to Po (Fig. 196). From this and (2) we obtain p = Ipl ( I) - - r ,. = - - c r = -c --3-) - r3 = [ -c x-xo -- y - Yo ,. -c r3 --3-' (3) x - xo . r c y - Yo . c --3-J - r ::: r :::0 --3- b.. • This vector function describes the gravitational force acting on B ~p t --- 00.....-. t Fig. 196. Gravitational field in Example 3 Vector Calculus We show next that the basic concepts of calculus, namely. convergence. continuity. and differentiability, can be defined for vector functions in a simple and natural way. Most imp0l1ant here is the derivative. Convergence. An infinite sequence of vectors if there is a vector a such that (4) lim la(n) - al n_x 3(n)' n = L 2..... is said to converge = O. a is called the limit vector of that sequence. and we write (5) lim a(n) = a. n~oo Cartesian coordinates being given, this sequence of vectors converges to a if and only if the three sequences of components of the vectors converge to the corresponding components of a. We leave the simple proof to the student. SEC. 9.4 387 Vector and Scalar Functions and Fields. Derivatives Similarly, a vector function v(t) of a real variable t is said to have the limit 1 as t approaches to, if vet) is defined in some neighborhood of to (possibly except at to) and (6) o. 11 = lim Iv(t) - t--.+to Then we write lim v(t) (7) t---+to =L Here, a neighborhood of to is an interval (segment) on the t-axis containing to as an interior point (not as an endpoint). Continuity. A vector function v(t) is said to be continuous at t some neighborhood of to (including at to itself!) and (8) lim vet) t---+to = to if it is defined in = veto). If we introduce a Cartesian coordinate system, we may write Then v(t) is continuous at to if and only if its three components are continuous at to. We now state the most important of these definitions. DEFINITION Derivative of a Vector Function A vector function v(t) is said to be differentiable at a point t if the following limit exists: (9) , . v (t) = hm At~O v(t + t1t) - v(t) A ut This vector v'(t) is called the derivative of v(t). See Fig. 197. Fig. 197. Derivative of a vector function In components with respect to a given Cartesian coordinate system. (10) v' (t) = [v~(t), v~(t), v~(t)]. Hence the derivative v' (t) is obtained by differentiating each component separately. For instance, if v = [t, t 2 , 0], then v' = [1, 2t, 0]. 388 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl Equation (10) follows from (9) and conversely because (9) is a "vector form" of the usual formula of calculus by which the derivative of a function of a single variable is defined. [The curve in Fig. 197 is the locus of the terminal points representing v(t) for values of the independent variable in some interval containing 1 and 1 + At in (9)]. It follows that the familiar differentiation rules continue to hold for differentiating vector functions, for instance, = (cv)' (0 + v)' = cv' (c + 0' constant). v' and in particular (O'v)' = u'·v (11) (12) (13) (0 X (0 v w)' = (0' + v)' = u' x v v w) + (0 u'v' +0 v' X v' w) + (0 v w'). The simple proofs are left to the student. In (12), note the order of the vectors carefully because cross multiplication is not commutative. E X AMP L E 4 Derivative of a Vector Function of Constant Length Let vet) be a vector function whose length is constant. say, Iv(t)1 = c. Then Ivl 2 = v·v = c 2 , and (v· v)' = 2v· v' = 0, by differentiation [see (11)]. This yields the following result. The derivative of a vector • function vet) of constant length is either the zero vector or is perpendicular to vet). Partial Derivatives of a Vector Function Our present discussion shows that partial differentiation of vector functions of two or more variables can be introduced as follows. Suppose that the components of a vector function are differentiable functions of n variables tlo . . . , tn' Then the partial derivative of v with respect to 1m is denoted by av/atm and is defined as the vector function av Similarly. second partial derivatives are and so on. E X AMP L E 5 Partial Derivatives ilr ilt1 = -a sin t1 i + a cos t1 j and ilr -=k. ilt2 • Various physical and geometric applications of derivatives of vector functions will be discussed in the next sections and in Chap. 10. SEC. 9.5 ~1~ SCALAR FIELDS 7. (Isobars) For the pressure field f(x. y) = 9x2 + 16y2 find the isobars f(x, y) = const, the pressure at (4, 3), (- 2, 2), (I, 5), and the regIOn in which the pressure is between 4 and 16. S. CAS PROJECT. Scalar Fields in the Plane. Sketch or graph isotherms of the following fields and describe what they look like. (a) x 2 - 4x - (b) x 2y - y3/3 y2 (c) cos x sinh) (e) eX sin y (d) sin x sinh y (f) e 2x cos 2)' (g) x4 _ 6x2y2 (h) x 2 - + )'4 f = z; - 13. f = 4x 11. Determine the isotherms (curves of constant temperature T) of the temperature fields in the plane given by the following scalar functions. Sketch some isotherms. 1. T = xy 2. T = 4x - 3) 3. T = y2 - x 2 4. T = x/(x 2 + ."2) 2 5. T = y/(x + y2) 6. T = x 2 - y2 + 8y 115-201 Vx 2 + y2 + 12. f + x 2j = yi - xj = Z VECTOR FIELDS Sketch figures similar to Fig. 196. 15. v = i - j 16. v 17. v 19. v = 4.\'2 - 3)' - 5z i IS. v = yi = xi + xj + yj 20. v = (x - y)i + (x + v)j 121-251 DIFFERENTIATION 21. Prove (11)-(13). Give two examples for each formula. 22. Find the first and second derivatives of [4 cos t, 4 sin t, 2tl 23. Find the first partial derivatives of [4x 2, 9z 2, xyz] and [yz, zx, .I.}']. 2x _ y2 I 9-1.::' SCALAR FIELDS IN SPACE What kind of surfaces are the level surfaces f(x,)" z) = cOllst? 9. f = x 2 + )"2 + 4~2 10. f = x 2 + 4y2 9.5 389 Curves. Arc Length. Curvature. Torsion 24. Find the first partial derivatives of [sin x cosh y, cos x sinh yJ and [eX cos)" eX sin y]. 25. WRITING PROJECT. Differentiation of Vector Functions. Summarize the essential ideas and facts and gi ve examples of your own. Curves. Arc Length. Curvature. Torsion A major application of vector calculus concerns curves (this section) and surfaces (Sec. 10.5) and their use in physics and geometry. This field is called differential geometry. It plays a role in mechanics, computer-aided and traditional engineering design, geodesy and geography, space travel, and relativity theory (see Refs. [GR8], [GR9] in App. I). Curves C in space may occur as paths of moving bodies. This and other applications motivate parametric representations with parameter t, which may be time or something else (see Fig. 198) (1) r(t) = [x(t), Fig. 198. y(t), z(t)] = x(t)i + y(t)j + z(t)k. Parametric representation of a curve 390 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl Here x, y, z are Cartesian coordinates (the usual rectangular coordinates; see Sec. 9.1). To each value t = to there corresponds a point of C with position vector r(to), that is, with coordinates x(to), y(to). :(to). Parametric representations (1) have a key advantage over representations of a curve C in terms of its projections into the x:v-plane and into the xz-plane, that is, y = f(x). (2) z = g(x) (or by a pair of equations with y or with z as the independent variable). The advantage is that in (I) the coordinates x, y, : play the same role: all three are dependent variables. Moreover, the sense of increasing t, called the positive sense on C, induces an orientation of C, a direction of travel along C. The sense of decreasing t is then called the negative sense on C, given by (I). EXAMPLE 1 Circle The circle.£2 + ... 2 = 4, ;: = 0 in the :\~\'-plane with center 0 and radius 2 can be represented parametrically by r(t) = [2 cos t. 2 sin t, 0] or simply by r(t) = [2 cos t. 2 sin t] (Fig. 199) where 0 ;;; t ;;; 21T. Indeed. x 2 + y2 = (2 co, 1)2 + (2 sin t)2 = 4(cos2 t + sin 2 t) = 4. For t = 0 we have r(O) = [2, 0], for t = ~1T we get r(~1T) = [0. 2]. and so on. The positive sense induced by this representation is the counterclockwise sense. if we replace t with t* = t, we have t = -t* and get r*(t*) = [2 cos (-t*). 2 sin (-t*)) = [2 cos t*, -2 sin t*]. • This has reversed the orientation. and the circle is now oriented clockwise. E X AMP L E 2 Ellipse The vector function r(t) = [a cos t. (3) bsint, 0] = acost i + bsint j (Fig. 200) represents an ellipse in the \}'-plane with center at the origin and principal axes in the direction of the x and y axes. In fact, since cos2 t + sin 2 t = 1, we obtain from (3) z = O. If b = • a, then (3) represents a circle of radius a. "~\~~ -~y\ (t= Fig. 199. ~1I)T Circle in Example 1 (t Fig. 200. = ~1I)1 (t = 0) Ellipse in Example 2 SEC. 9.5 391 Curves. Arc Length. Curvature. Torsion E X AMP L E 3 Straight Line A straight line L through a point A with position vector a in the direction of a constant vector b (,ee Fig. 201) can be represented parametrically in the form (4) If b is a unit vector. its components are the direction cosines of L. In this case. It I mea,ures the distance of the points of L from A. For instance. the straight line in the xv-plane through A: (3, 2) having slope l is (sketch it) r(t) ~ [3, 2, + 0] 1, t[l, 0] ~ [3 + t, 2 + t, • 0]. A z / ------ y X Fig. 201. a Parametric representation of a straight line A plane curve is a curve that lies in a plane in space. A curve that is not plane is called a twisted curve. A standard example of a twisted curve is the following. E X AMP L E 4 Circular Helix The twisted curve C represented by the vector function (5) r(t) ~ [a cos t. a sin t. etl ~ a cos t i + a sin t j + et k (c'* 0) is called a circlilar helix. It lies on the cylinder x 2 + y2 = a 2 . If c > O. the helix is shaped like a right-handed screw (Fig. 202). If c < 0, it looks like a left-handed screw (Fig. 203). If c = 0, then (5) is a circle. • y I I I / .P-- ../ / y x Fig. 201. Right-handed circular helix Fig. 203. Left-handed Circular helix A simple curve is a curve without multiple points, that is, without points at which the curve intersects or touches itself. Circle and helix are simple. Figure 204 shows curves that are not simple. An example is [sin 2t, cos t, 0]. Can you sketch it? An arc of a curve is the portion between any two points of the curve. For simplicity, we say "curve" for curves as well as for arcs. 392 CHAP. 9 Vector Differential Calculus. Grad. Div. Curl Fig. 204. Curves with multiple points Tangent to a Curve The next idea is the approximation of a curve by straight lines, leading to tangents and to a definition of length. Tangents are straight lines touching a curve. The tangent to a simple curve C at a point P of C is the limiting position of a straight line L through P and a point Q of C as Q approaches P along C. See Fig. 205. If C is given by ret), and P and Q cOlTespond to T and t + b..t, then a vector in the direction of L is I (6) [ret .:1t + Ilt) - r(t)]. In the limit this vector becomes the derivative , r (t) (7) = lim :,t~O I A ul Ir(t + b..t) - r(t)l, provided r(t) is differentiable, as we shall assume from now on. If r' (t) =F 0, we call r' (t) a tangent vector of C at P because it has the direction of the tangent. The cOlTesponding unit vector is the unit tangent vector (see Fig. 205) (8) u= 1 , !r'! r. Note that both r' and u point in the direction of increasing t. Hence their sense depends on the orientation of C. It is reversed if we reverse the orientation. It is now easy to see that the tangent to C at P is given by (9) q(w) = r + wr' (Fig. 206). This is the sum of the position vector r of P and a multiple of the tangent vector r' of C at P. Both vectors depend on P. The variable w is the parameter in (9). L o Fig. 205. Tangent to a curve Fig. 206. Formula (9) for the tangent to a curve SEC. 9.5 393 Curves. Arc Length. Curvature. Torsion E X AMP L E 5 Tangent to an Ellipse Find the tangent to the ellipse ~x2 + y2 = 1 at P: CV2, 11V2). Solution. r' (t) = Equation (3) with semi-axes a = 2 and b = 1 gives r(t) = [2 cos t, sin t]. The derivative [-2 sin t. cos t]. Now P corresponds to t = 7T/4 because r(7T/4) Hence r' (7T/4) = [ - V2, q(w) = = = [V2. [2 cos (7T/4). sin (7714)] IS 11V2]. I/V2]. From (9) we thus get the answer [V2, 11V2] + 1\'[-V2, 11V2] = [V2(1 - 11'), (lNi)(l t- 11')]. • To check the result, sketch or graph the ellipse and the tangent. Length of a Curve We are now broken lines let ret), a ~ interval a ~ ready to define the length I of a curve. I will be the limit of the lengths of of n chords (see Fig. 207, where n = 5) with larger and larger n. For this, t ~ b, represent C. For each n = I, 2, ... we subdivide ("partition") the t ~ b by points where This gives a broken line of chords with endpoints r(to), ... , r(tn). We do this arbitrarily but so that the greatest ILlt.,nl = Itm - t m-ll approaches 0 as n ~ co. The lengths II' 12 , • • • of these chords can be obtained from the Pythagorean theorem. If ret) has a continuous derivative r' (t), it can be shown that the sequence II' 12 , ••• has a limit, which is independent of the particular choice of the representation of C and of the choice of subdivisions. This limit is given by the integral (10) I is called the length of C, and C is called rectifiable. Formula (10) is made plausible in calculus for plane curves and is proved for curves in space in [GR8] listed in App. 1. The practical evaluation of the integral (10) will be difficult in general. Some simple cases are given in the problem set. Arc Length 5 of a Curve The length (10) of a curve C is a constant, a positive number. But if we replace the fixed b in (10) with a variable t, the integral becomes a function of t, denoted by s(t) and called the arc length function or simply the arc length of C. Thus t (11) s(t) = I Vr or ~ rtI dt- a Fig. 207. Length of a curve ~n· 394 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl Here the variable of integration is denoted by t because t is now used in the upper limit. Geometrically, s(to) with some to > a is the length of the arc of C between the points with parametric values a and to. The choice of a (the point s = 0) is arbitrary; changing a means changing s by a constant. Linear Element ds. (12) ds ( -dt If we differentiate )2 = -dr • -dr dt dt = Ir U 1) and square, we have '2 (t)1 = (-dx )2 + (-dy )2 + (-d::. )2. dt dt dt It is customary to write dr (13*) [dx, dy, d;::] = dxi = + dyj + d;::k and (13) ds is called the linear element of C. Arc Length as Parameter. The use of sin (1) instead of an arbitrary t simplifies various formulas. For the unit tangent vector (8) we simply obtain (14) U(s) = r' (s). Indeed, Ir' (s)1 = (ds/ds) = I in (12) shows that r' (s) is a unit vector. Even greater simplifications due to the use of s will occur in curvature and torsion (below). E X AMP L E 6 Circular Helix. Circle. Arc Length as Parameter The helix r(l) = [a cos I. a sin I. el] in (5) has the derivative r' (I) = [-a sin t. a cos t. d. Hence r' • r' = a 2 + e 2• a constant. which we denote by K2. Hence the integrand in ( II) is constant. equal to K. and the integral is s = Kt. Thus I = 51K. so that a representation of the helix with the arc length s as parameter is (15) r*(s) = s ) r( K = [ a cos K5 asin 5 cSJ K , K' K = Va 2 + c2 . A circle is obtained if we seI c = O. Then K = a. t = sla. and a representation with arc length s as parameter is r*(s) = r( ~) = [a cos ~ a sin ~ J. • Curves in Mechanics. Velocity. Acceleration Curves playa basic role in mechanics, where they may serve as paths of moving bodies. Then such a curve C should be represented by a parametric representation rV) with time t as parameter. The tangent vector (7) of C is then called the velocity vector v because, being tangent, it points in the instantaneous direction of motion and its length oives the speed Ivl = Ir'l = ~ = dsldt; see (2). The second derivative of r(t)~is c':uled the SEC. 9.5 395 Curves. Arc Length. Curvature. Torsion acceleration vector and is denoted by a. Its length lal i<; called the acceleration of the motion. Thus (16) v(t) = r' (t), aCt) = v' (t) = r"(t). Tangential and Normal Acceleration. Whereas the velocity vector is always tangent to the path of motion, the acceleration vector will generally have another direction, so that it will be of the form (17) where the tangential acceleration vector atan is tangent to the path (or, sometimes, 0) and the normal acceleration vector a norm is normal (perpendicular) to the path (or, sometimes, 0). Expressions for the vectors in (17) are obtained from (16) by the chain rule. We first have dr dr ds dt ds vet) = - - dt = ds u(s)- dt where u(s) is the unit tangent vector (4). Another differentiation gives 2 (18) aCt) = -dv = -d ( U(s) -ds ) = -du (ds)2 + u(s) -d 2s dt dt dt ds dt dt Since the tangent vector u(s) has constant length (one), its derivative du/ds is perpendicular to u(s) (by Example 4 in Sec. 9.4). Hence the first term on the right of (18) is the normal acceleration vector, and the second term on the right is the tangential acceleration vector, so that (18) is of the form (17). Now the length of a tan is the projection of a in the direction of v, given by (II) in Sec. 9.2 with b = v; that is, latanl = a·v/lvl. Hence atan is this expression times the unit vector (1IIvl)v in the direction of v; that is, a·v atan = - - v. (18*) Also. v·V a norm = a - a tan . Let us consider two basic examples, involving centripetal and centrifugal accelerations and Corio lis acceleration, as it occurs. for instance. in space travel. E X AMP L E 7 Centripetal Acceleration. Centrifugal Force The vector function ret) = [R cos wt. R sin wtj = R cos wt i + R sin wt j (Fig. 208) (with fixed i and j) represents a circle C of radIUS R with center at the origm of the .\)"-plane and describes the motion of a small body B counterclockwise around the circle. Differenriarion gives the velocity vector v = r' = [- Rw sin wt. v is tangent to C. It~ magnitude, the speed. Rw cos wt] = - Rw sin wt i i~ Ivl = Ir'l = w-:-;:' = Rw. T Rw cm, wt .i (Fig. 208). 396 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl y x Fig. 208. Centripetal acceleration a Hence it is constant. The speed divided by the distance R from the center is called the angular speed. It equals w, so that it is constant. too. Differentiating the velocity vector, we obtatn the acceleration vector (19) a = v' = [-Rw2 cos wt, -Rw2 sin wt] = -Rw2 cos wt i - Rw2 sin wt j. Thi~ shows that a = _w2 r (Fig. 208). so that there is an acceleration IOwatd the center. called the centripetal acceleration of the motion. It occurs because the velocity vector is changing direction at a constant rate. Its magnitude is constant, lal = w2 1rl = w2 R. Multiplying a by the mass m of B, we get the centripetal force ma. The oppo,ite vector -ilia is called the centrifugal force. At each instant these two forces are in equilibrium. We see that in this motion the acceleration vector is normal (perpendicular) to C; hence there is no tangential acceleration. • E X AMP L E 8 Superposition of Rotations. Coriolis Acceleration A projectile IS moving with constant speed along a meridian of the rotating eatth in Fig. 209. Find its acceleration. a ~ ~"\ -----p --- , , --\. I', 1 J 1 Fig. 209. Example 8. Superposition of two rotations Solution. Let x. y, :: be a tixed Cartesian coordinate system in space. with unit vectors i, j, k in the directiuns of the axes. Let the earth, together with a unit vector b, be rotating about the z-axis with angulat· speed w > 0 (see Example 7). Since b is rotaing together with the earth. it is of the form b(t) = cos wt i + sin WI j. Let the projectile be moving on the meridian whose plane is spanned by band k (Fig. 209) with constant angular speed y > O. Then its position vector in terms of band k IS r(l) = R cos yt btl) + R sin yl k (R = Radius of the earth). SEC. 9.5 Curves. Arc Length. Curvature. Torsion 397 This is the modeL The rest is calculation. The result will be unexpected and highly relevant for air and space travel. The first and second derivatives of b with respect to 1 are b ' (tl = -w sin wt i + w cos wt j (20) The first and second derivatives of rtt) with re<;pect to tare v = r'(t) = R cos ')'t b ' - ')'R sin (21) ')'t b + ')'R cos ')'t k a = v' = R cos ')'t b" - 2')'R sin ')'t b ' - ')'2R cos ')'t b - ')'2R sin ')'t k = R cos ')'t bIt - 2')'R SIn ')'t bI - ')'2r. By analogy with Example 7 and because of bIt = - w2 b in (20) we conclude that the first term in a (involving win bIt!) is the centripetal acceleration due to the rotation of the earth. Similarly, the third term in the last line (involving ')'!) i, the centripetal acceleratiun due to the motion of the projectile un the meridian M of the rotating earth. The second, unexpected term -2')'R sin ')'t b ' in a is called the Coriolis acceleration3 (Fig. 209) and is due to the interaction of the two rotations. On the Northern Hemisphere, sin ')'t > 0 (for 1 > 0; also ')' > 0 by assumption), so that a cor has the direction of - b I. that is, opposite to the rotation ufthe earth. lacorl is maximum at the North Pole and zero at the equator. The projectile B of mass 1Il0 experiences a force -Illoa cor opposite to 1I10 a co l"' which tends to let B deviate from M to the right (and in the Southern Hemisphere, where sin ')'1 < O. to the left). This deviation has been observed for missiles. rockets. shells. and atmospheric air flow. • Curvature and Torsion. Optional This optional portion of the section completes our discussion of curves from the viewpoint of vector calculus. The curvature K(S) of a curve C: rts) (s the arc length) at a point P of C measures the rate of change lu' (s)1 of the unit tangent vector u(s) at P. Hence K(S) measures the deviation of C at P from a straight line (its tangent at P). Since u(s) = r' (s). the definition is (22) K(S) (' = d/ds). = lu' (s)1 = Ir"(s)1 The torsion res) of C at P measures the rate of change of the osculating plane 0 (the plane spanned by u and u'. see Fig. 210) of C at P. Hence res) measures the deviation rn E 5c co Rectifying plane b Normal plane PrinCipal - p _n()fl7Jal Osculating plane Fig. 210. Trihedron. Unit vectors u, p, b and planes 3GUSTAVE GASPARD CORIOLIS (1792-1843), French engineer who did research in mechanics. CHAP. 9 398 Vector Differential Calculus. Grad, Div, Curl of C at P from a plane (from 0 at Pl. Now the rate of change is also measured by the derivative b' of a normal vector bat 0. By the definition of vector product, a unit normal vector of 0 is b = u X (I/K)U' = U x p, where p = (IIK)U' is called the unit principal normal vector and b is called the unit binormal vector of C at P; see Fig. 210. Here we must assume that K =1= 0; hence K > O. The absolute value of the torsion is now defined by (23*) Whereas K(S) is nonnegative. It IS practical to give the torsion a sign. motivated by "right-handed" and "left-handed" (see Figs. 202. 203). This needs a little further calculation. Since b is a unit vector, it has constant length. Hence b' is perpendicular to b (see Example 4 in Sec. 9.4). Now b' is also perpendicular to U because by the definition of vector product we have bou = 0, bou' = O. This implies (bou)' = 0; that is, b'ou + bou' = b ' °U + 0 = O. "* Hence if b' 0 at P, it must have the direction of p or -p, so that it must be of the form b' = -7p. Taking the dot product of this by p and using pop = I gives (23) 7(S) = -p(s)ob'(s). The minus sign is chosen to make the torsion of a right-handed helix positive and that of a left-handed helix negative (Figs. 202, 203). The orthonormal vector triple u, p, b is called the trihedron of C. Figure 210 also shows the names of the three straight lines in the directions of u, p, b, which are the intersections of the osculating plane, the normal plane. and the rectifying plane. = 11-101 PARAMETRIC REPRESENTATIONS Find a parametric representation of the following curves. 1. Circle of radius 3, center (4, 6) 2. Straight line through (5. 1. 2) and (11, 3. 0) 15. [\!CoSt, Vsin t, oj ("Lame 16. [cosh t, sinh t. 0] 17. [t, lit, 0] 18. [1,5 + t, -5 + lit] ClI/Te n ) 3. Straight line through (2, O. 4) and (- 3. O. 9) 4. Straight line y 5. Circle y2 6. Ellipse x 2 = 2x + 4y + + y2 = + 3, :;; Z2 = 5, x = 3 I. z = y 7. Straight line through ta, b, c) and (a 8. Intersection of x 19. Show that setting t = -t* reverses the orientation of [a cos t. a sin t. 0]. = 7x + y - ::: = = 1.::: = y + 3, b - 2. C + 5) +z= 3 2, 1x - 5y 20. If we set t Explain. = et in Prob. 12, do we get the entire line? 21. CAS PROJECT. Curves. Graph the following more complicated curves. 9. Circle ~x2 + y2 10. Helix x 2 + y2 = 9, ::: = 4 arctan tylx) (a) r(t) = [2 cos t + cos 2t, (Steiner's hypocycloid) 111-181 (b) r(t) = [cos t + k cos 2t. k = 10,2. 1,~, O. -~, -) What curves are represented as tollows? 11. [2 + r cos 4t. 6 + r sin 4t, 2t] 12. [4 - 2t, 8t, -3 + 5t] 13. [2 + cos 3t, - 2 14. [t, t 2 , t 3 ] + sin 3t, 5] (c) r(t) = [cos t. (d) r(t) = [cos t, closed? 2 sin t - sin 2t] sin t - k sin 2t] with sin 5t] (a Lissajolls cline) sin kt]. For what k's will it be (e) r(t) = [R sin wt + wRt, R cos wt + R] (cycloid). SEC. 9.5 Curves. Arc Length. Curvature. Torsion 122-251 TANGENT 399 23. ret) = [5 cos t, 5 sin t, 0], 24. ret) = [3 cos t, 3 sin t, 4t], 25. r(t) = [cosh t, CURVES IN MECHANICS 132-341 Given a curve C: r(t), find a tangent vector r' (t), a unit tangent vector u' (t), and the tangent of C at P. Sketch the curve and the tangent. 22. ret) = [t, t 2, 0], P: (2,4,0) P: (4, 3, 0) 32. r(t) 33. ret) P: (3, 0, 87T) P: (~, ~) sinh t], Velocity and Acceleration. Forces on moving objects (cars, airplanes, etc.) require that the engineer knows corresponding tangential and normal accelerations. Find them, along with the velocity and speed, for the following motions. Sketch the path. = [4t, = [1. -3t, t 2 34. ret) = [cos t, 0] 0] , 2 sin t, 0] LENGTH 126-281 35. (Cycloid) Given Find the length and sketch the curve. 26. Circular helix r(t) = [2 cos t, 2 sin t, 6t] from (2, 0, 0) to (2, 0, 247T) 27. Catenary ret) = [t, 28. Hypocycloid ret) = cosh t] from t = 0 to t = 1 la cos 29. Show that (10) implies € 3 = 3 a sin t]. total length t. I b ~ cir for the a length of a plane curve C: y = f(x), z = 0, a 30. Polar coordinates p = give€ = I 13 Yr + y2, ~ x ~ b. e = arctan (ylx) Vp2 + p'2 de, where p' = dplde. Derive ex this. Use it to find the total length of the cardioid p = a(l - cos e). Sketch this curve. Hint. Use (10) in App. 3.l. 31. CAS PROJECT. Polar Representations. Use your 4 p = ae p= p p = -- e 2 e e Cissoid of Diocles + b Conchoid of Nic011ledes Hyperbolic spiral 3a sin 2e cos3 + (R cos wt + R)j. This cycloid is the path of a point on the rim of a wheel of radius R that rolls without slipping along the x-axis. Find v and a at the maximum y-values of the curve. 36. CAS PROJECT. Paths of Motions. Gear transmissions and other engineering constructions often involve complicated paths whose study is greatly facilitated by the use of a CAS. To grasp the idea, graph the following paths and find the velocity, the speed, and the tangential and normal accelerations. (a) ret) = [2 cos t + cos 2t, (Steiner's hypocycloid) (b) ret) = [cos t (c) ret) = [cos t, + cos 2t, sin t sin 2t, (d) r(t) = [ct cos t, 2 sin t - sin 2t] cos 2t] ct sin t, sin 2t] ct] (c * 0) 38. (Earth and moon) Find the centripetal acceleration of cos p = ale wRt) i Logarithmic spiral 2a sin cos + (R sin wt Spiral of Archimedes be a = = 37. (Sun and earth) Find the acceleration of the earth toward the sun from (19) and the fact that the earth revolves about the sun in a nearly circular orbit with an almost constant speed of 30 kmIsec. CAS to graph the following famous curves and investigate their form depending on parameters a and b. p = ae r{t) e + sin3 e sin 3e p = 2a - - sin 2e p = 2a cos e + b Folium of Descartes Maclaurin's trisectrix Pascal's snail the moon toward the earth, assuming that the orbit of the moon is a circle of radius 239,000 miles = 3.85· 108 m, and the time for one complete revolution is 27.3 days = 2.36· L06 sec. 39. (Satellite) Find the speed of an artificial earth satellite traveling at an altitude of 80 miles above the earth's surface, where g = 31 ft/sec 2 . (The radius of the earth is 3960 miles.) 40. (Satellite) A satellite moves in a circular orbit 450 miles above the earth's surface and completes I revolution in 100 min. Find the acceleration of gravity at the orbit from these data and from the radius of the earth (3960 miles). 4Named after ARCHIMEDES (c. 287-212 B.C.), DESCARTES (Sec. 9.1), DlOCLES (200 B.C.), MACLAURIN (Sec. 15.4), NICOMEDES (250? B.C.) ETIENNE PASCAL (1588-1651), father of BLAISE PASCAL (1623-1662). CHAP. 9 400 141-501 Vector Differential Calculus. Grad, Div, Curl CURVATURE AND TORSION 41. Show that a circle of radius a has curvature lIa. 42. Using (22), show that if C is represented by ret) with arbitrary t, then (23***) VCr' r' )(r" r") - (r' r")2 0 0 45. Show that the torsion of a plane curve (with K > 0) is identically zero. 46. Show that if C is represented by r(t) with arbitrary parameter t. then. assuming K > 0 as before. 0 (r' T(t) r" rIll) = ---'-----'---- (r' or' )(r" or") - (r' or")2 (22*) K(t) = - ' - - - - - - - ' - - - - - - ' - (r' r')3/2 0 43. Using (22*), show that for a curve y = i{x) in the xy-plane. dr , etc. ) . ( y' = -'dx (22**) 44. Using b = u x p and (23), show that (23**) T(S) = (u p p') = (r' r" r"')/K'(K 9.6 > 0). 47. Find the torsion of C: r(t) = [t. t 2 , t 3 ] (which looks similar to the curve in Fig. 2ID). 48. (Helix) Show that the helix [Cl cos t. CI sin t, ctl can be represented by [a cos (sIK), a sin (sIK), cslKl, where K = VCl 2 + c2 and .I" is the arc length. Show that it has constant curvature K = cdK2 and torsion T= dK2. 49. Obtain K and Tin Prob. 48 from (22*) and (23***) and the Oliginal representation in Prob. 48 with parameter t. 50. (Frenet5 formulas) Show that u' = KP, p' = -KU + Tb, b' = -TP. Calculus Review: Functions of Several Variableso Optional Curves required vector functions of a single variable x or s, and we now proceed to vector functions of several variables, beginning with a review from calculus. Go on to the next section, consulting this material only when needed. (We include this short section to keep the book reasonably ~elf-contained. For partial derivatives see App. A3.2.) Chain Rules Figure 211 shows the notations in the following basic theorem. " 1 D ~[X(U'V).Y(u.L').z(u.V)l B u Fig. 211. Notations in Theorem 1 5JEAN-FREDERIC FRENET (l816-1900), French mathematician. SEC. 9.6 Calculus Review: Functions of Several Variables. Optional 401 Chain Rule THEOREM 1 Let w = f(x, )', z) be continuous and have continuous first partial derivarives in a domain D in xy:;:-space. Let x = x(u, v), y = y(u, v), :;: = z(u, v) be funcTions that are colltinuous and hal'e first partial derivatives in a domain B in the uv-plane, where B is such that for every point (u, v) ill B, the corresponding point Ix(u, v), y(u, v), :;:(ll, v)] lies in D. See Fig. 21l. Then the function w = f(x(u. v), y(u. v). z(u. v» is defined in B, has first partial deril'Otil'es lI'ith respect to u and v in B, and a" aw a- aw aw ax aw Au Ax au ay Au az Au aw aw ax away aw az av ax av ay av az av -=--+--~-+-~ (1) -=--+--+-- In this theorem, a domain D is an open connected point set in xyz-space, where "connected" means that any two points of D can be joined by a broken line of finitely many linear segments all of whose points belong to D. "Open" means that every point P of D has a neighborhood (a little ball with center P) all of whose points belong to D. For example. the interior of a cube or of an ellipsoid (the solid without the boundary surface) is a domain. In calculus, x, y, Z are often called the intermediate variables, in contrast with the independent variables u, v and the dependent variable w. Special Cases of Practical Interest If w = f(x, y) and x = x(u, v), y = y(u, v) as before, then (1) becomes av aw aw ax aw au ax au ay au aw aw ax away av ax av ay av -=--+---- (2) -=--+-- If no = f(x, y, .:::) and x = xU), y = yet), .::: = (3) z(t), then dw aw dx aw dt ax dt ay dt dy 1I) gives all' az dt If w = f(x, y) and x = x(t), y = y(t), then (3) reduces to (4) dz -=--+--+-- dw aw dt ax dt dx aw dr -=--+-_.- ay dt· 402 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl Finally, the simplest case w = f(x), x = x(t) gives (5) E X AMP L E 1 dw dw dx dt dx dt Chain Rule If w = x 2 i and we define polar coordinates r, 8 by x = r cos 8, y = r sin 8, then (2) gives ~ a; = 2xcos 8 - 2ysin 8 aw 2x(-r sin 8) - 2y(r cos lJ) = -2r 2 cos 8 sin lJ - 2r2 sin lJcos 8 = -2r 2 sin 28. a8 = 2 2 2rcos 8 - 2rsin 8 = 2rcos28 = • Mean Value Theorems THEOREM 2 Mean Value Theorem Let f(x, y, z) be continuous and have continuous first partial derivatives in a domain D in xyz-space. Let Po: (xo, Yo, zo) and P: (xo + h, Yo + k, Zo + l) be points in D such that the straight line segment PoP joining these points lies entirely in D. Then (6) f(xo + h, Yo + k, Zo + l) - f(xo, Yo, af Zo) = h- ax + af kay + af l-, az the partial derivatives being evaluated at a suitable point of that segment. Fig. 212. Mean value theorem for a function of two variables [Formula (7)] Special Cases For a function f(x, y) of two variables (satisfying assumptions as in the theorem), formula (6) reduces to (Fig. 212) (7) f(xo + h, Yo + k) at - f(xo, Yo) = h ax +k at , a; SEC. 9.7 Gradient of a Scalar Field. Directional Derivative 403 and for a function f(x) of a single variable, (6) becomes f(xo + 11) - f(xo) (8) = df 11-, dx where in (8), the domain D is a segment of the x-axis and the derivative is taken at a suitable point between Xo and Xo + h. [1-51 DERIVATIVE Find dwldt by (3) or (4). Check the result by substitution and differentiation. (Show the details.) 1. w = + y2, X = e 2t , y = e- 2t V:>? = ylx, x = g(t), y = h(t) 3. w = xY, x = cosh t. y = sinh t 4. w = xy + yz + zx, x = 2 cos t, Y = 2 sin T, z = 5t 2. w 5. w = (x 2 + y2 + Z2)3, X = (2, Y = (4, Z = (2 /6-91 PARTIAL DERIVATIVES Find iJwliJu and iJwlav by (1) and (2). Check the result by substitution and differentiation. (Show the details.) 6. w = 4x 2 - 4y2, X = U + 2v, y = 2u - v 7. W =x 2y2.x= eUcosv.y = eUsinv 8. w = 9.7 X4 - 4x 2y2 = 1/(x 2 Z = 2uv 9. w + )'4, X = uv, y = ulv + y2 + Z2), X = u 2 + v 2, Y = u 2 - v 2, 10. (Partial derivatives on a surface) Let w = f(x, y, z), and let z = g(x, y) represent a surface S in space. Then on S, the function becomes w(x, y) = f[x, y, g(x, y)]. Show that its partial derivatives are obtained from aw af af iJx ax az ax' iJg -=-+-- aw af af ay ay az ag a)' -=-+-[;: = g(x. y)]. Apply this to f = x 3 + )'3 + Z2, g = x 2 + y2 and check by substitution and direct differentiation. (The general formula will be needed in Sec. 10.9.) Gradient of a Scalar Field. Directional Derivative We shall see that some of the vector fields in applications-not all of them!---can be obtained from scalar fields. This is a considerable advantage because scalar fields can be handled more easily. The relation between these two kinds of fields is obtained by the "gradient," which is thus of great practical importance. DEFINITION 1 Gradient The gradient of a given scalar function f(x, y, z) is denoted by grad f or Vf (read nabla f) and is the vector function defined by ll) gradf at, at, -at] = at). + -.-J at. +-k. at = Vt = [ ax ay az ax dy az Here x. y, z are Cartesian coordinates in a domain in 3-space in which f is defined and differentiable. (For curvilinear coordinates see App. 3.4.) CHAP. 9 404 Vector Differential Calculus. Grad, Div, Curl For instance, if f(x, y, z) = 2)'3 + 4xz + 3x, then grad f = [4z + 3, 6)'2, 4x]. The notation \' f is suggested by the differential operator V (read nabla) defined by V = (1*) a -j ax a a ay iJ;:. + - j + -k. Gradients are useful in several ways, notably in giving the rate of change of f(x. y. ;:.) in any direction in space, in obtaining surface normal vectors, and in deriving vector fields from scalar fields, as we are going to show in this section. Directional Derivative From calculus we know that the partial derivatives in (1) give the rates of change of f(x. y. z) in the directions of the three coordinate axes. It seems natural to extend this and ask for the rate of change of f in an arbitrw:v direction in space. This leads to the following concept. Dr INITION 2 Directional Derivative The directional derivative Dbf or dflds of a function f(x, y, z) at a point P in the direction of a vector b is defined by (see Fig. 213) (2) 1_ df . f(Q) - f(P) Dbf = - = hm . s->O S ds Here Q is a variable point on the straight line L in the direction of b, and lsi is the distance between P and Q. Also, s > 0 if Q lies in the direction of b (as in Fig. 213), s < 0 if Q lies in the direction of -b, and s = 0 if Q = P. Fig. 213. Directional derivative The next idea is to use Cartesian .x),z-coordinates and for b a unit vector. Then the line L is given by (3) res) = x(s)i + y(s)j + z(s)k = Po + sb where Po the position vector of P. Equation (2) now shows that Dbf dflds is the derivative of the function f(x(s), yes), z(s)) with respect to the arc length s of L. Hence. assuming that f has continuous partial derivatives and applying the chain rule [formula (3) in the previous section], we obtain (4) df af, af, af, Dbf=-=-x + - y + - z ds ax ay az SEC 9.7 405 Gradient of a Scalar Field. Directional Derivative where primes denote derivatives with respect to s (which are taken at s = 0). But here, differentiating (3) gives r' = x'i + y'j + z'k = b. Hence (4) is simply the inner product of grad f and b [see (2), Sec. 9.2]; that is, (5) Dbf ATTENTION! ds = b·grad f (Ibl = 1). If the direction is given by a vector a of any length (oF 0), then 1 df (5*) E X AMP L E 1 df =- Daf = -ds = -I I a·gradf· a Gradient. Directional Derivative Find the directional derivative of f(x. y, .:) = 2x2 + 3.1'2 + Z2 at P: (2, L 3) in the direction of a = [1, 0, -2]. Solution. since lal = grad J = [4x. fl\,. 2.:] gives at P the vector grad J(p) = [8. fl. 6]. From this and (5*) we obtain, Vs. DaJ(PI= 1 1 4 V5 [1.0.-2]"[8.6.61= Vs (8+0-12)=- Vs =-1.789. The minus sign indicates that at P the function f i~ decreasing in the direction of a. • Gradient Is a Vector. Maximum Increase grad f in (I) looks like a vector-after all, it has three components! But to prove that it actually is a vector. since it is defined in telms of components depending on the Cartesian coordinates, we must show that grad f has a length amI direction independent of the choice of those coordinates. In contrast, raflax, 2aflay, afli'J:;::] also looks like a vector but does not have a length and direction independent of the choice of Cartesian coordinates. Incidentally, the direction makes the gradient eminently useful: grad f points in the direction of maximum increase of f. Vector Character of Gradient. Maximum Increase THEOREM 1 Let f(P) = f(x. y. :;::) be a scalar function having continuous first partial derivatives in some domain B in space. Then grad f exists in B and is a vector, that is, its lellgth and direction are independent of the particular choice of Cartesian coordinates. {f grad f(P) oF 0 at some point P, it has the direction of maximum illcrease of f at P. PROOF From (5) and the definition of inner product [(1) in Sec. 9.2] we have (6) Dbf = Ibllgrad fl cos l' = Igrad fl cos l' where l' is the angle between b and grad f. Now f is a scalar function. Hence its value at a point P depends on P but not on the particular choice of coordinates. The same holds for the arc length s of the line L in Fig. 213, hence also for Dbf. Now (6) shows that Dbf is maximum when cos l' = \, l' = 0, and then Dbf = Igrad fl. It follows that the length and direction of grad f are independent of the choice of coordinates. Since l' = 0 if and only if b has the direction of grad f, the latter is the direction of maximum increase of f at P, provided grad f oF 0 at P. • 406 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl Gradient as Surface Normal Vector Gradients have an important application in connection with surlaces, namely, as surlace normal vectors, as follows. Let S be a surlace represented by f(x, y, z) = C = COllst, where f is differentiable. Such a surface is called a level surface of f, and for different c we get different level surlaces. Now let C be a curve on S through a point P of S. As a curve in space, C has a representation ret) = [x(t), yet), z(t)]. For C to lie on the surlace S, the components of r(1) must satisfy f(x, y, z) = c, that is, f(x(t), y(1), z(t» = c. (7) , [ , , '] Now a tangent vector of C . IS r (1) = x (1), Y (f), z (f) . And the tangent vectors of all curves on S passing through P will generally form a plane, called the tangent plane of S at P. (Exceptions occur at edges or cusps of S, for instance, for the cone in Fig. 215 at the apex.) The normal of this plane (the straight line through P perpendicular to the tangent plane) is called the surface normal to S at P. A vector in the direction of the surface normal is called a surface normal vector of Sat P. We can obtain such a vector quite simply by differentiating (7) with respect to t. By the chain rule, af, af, af, ax ay' iJz -x + -v + -z = , o. (gradf)or = Hence grad f is orthogonal to all the vectors r' in the tangent plane, so that it is a normal vector of Sat P. Our result is as follows (see Fig. 214). grad~Tangent plane f= cons) ~ /p / Fig. 214. THEOREM 2 Gradient as surface normal vector Gradient as Surface Normal Vector Let f be a differentiable scalar function ill space. Let f(x, y, z) = c = COllst represent a surface S. Tlzell if tlze gradient of f at a poim P of 5 is /lOT the zero vector, if is a normal vector of 5 at P. E X AMP L E 2 Gradient as Surface Normal Vector. Cone Find a unit nonnal vector n of the cone of revolution Solution. The cone is the level surface I grad I n = ~ [8x, 8y, = ;:.2 0 of I(x, = 4(x 2 + y2) at the point P: (I, U, 2). y, 4(x 2 + y2) - z2. Thus (Fig. 215), - 22], I Igrad I(P)I grad I(P) = [ z) = grad I(P) = [8, 2 V5' 0, - U, -4] I ] V5 n points downward since it has a negalJve z-component. The other unit normal vector of the cone at P is -no • SEC. 9.7 407 Gradient of a Scalar Field Directional Derivative n/: p I I I I I I ~ Fig. 215. Cone and unit normal vector n Vector Fields That Are Gradients of Scalar Fields ("Potentials") At the beginning of this section we mentioned that some vector fields have the advantage that they can be obtained from scalar fields, which can be handled more easily. Such a vector field is given by a vector function yep), which is obtained as the gradient of a scalar function. say, vW) = grad f(P). The function f(P) is called a potential function or a potential of yep). Such a v{P) and the conesponding vector field are called conservative because in such a vector field, energy is conserved; that is, no energy is lost (or gained) in displacing a body (or a charge in the case of an electrical field) from a point P to another point in the field and back to P. We show this in Sec. 10.2. Conservative fields playa central role in physics and engineering. A basic application concerns the gravitational force (see Example 3 in Sec. 9.4) and we show that it has a potential which satisfies Laplace's equation. the most important partial differential equation in physics and its applications. THEOREM 3 Gravitational Field. Laplace's Equation The force of attraction (8) p = c -r r3 = _c[x - Xo 1'3 Y - Yo . r3 z - zoJ . r3 between two particles at points Po: (Xo, Yo, zo) and P: (x. y, z) (as given by Newton's law of gravitation) has the potellfial f(x. y. z) = clr. where r (> 0) is the distance between Po alld P. TllllS P = grad f = grad (elr). This potential f is a solution o/Laplace's equation (9) [v 2 f (read nabla squared f) is called the Laplacian of f.] CHAP. 9 408 PROOF Vector Differential Calculus. Grad, Div, Curl That distance is r = «x - XO)2 + (Y - .\'0)2 + (z - <:2)2)1/ 2 . The key observation now is that for the components of p = [PI' P2. P3] we obtain by partial differentiation x - Xo (lOa) and similarly ;" (~) (lOb) :<: Y - Yo ---- r3 (~) = z- '::0 ---- ,-3 From this we see that, indeed. p is the gradient of the scalar function f = eI,-. The second statement of the theorem follows by partially differentiating (10), that is. a~2 (~) --+ r3 a iJy2 C) --+ r3 :Z22 (~) 2 I I r I = --+ r3 3(x - xO)2 r5 3(y - )'0)2 r5 3(.:: - ZO)2 ,-5 and then adding these three expressions. Their common denominator is r5. Hence the three terms -1/,-3 contribute - 3r 2 to the numerator, and the three other terms give the sum so that the numerator is 0, and we obtain (9). • V2 f is also denoted by I:::.f. The differential operator (11) (read "nabla squared" or "delta") is called the Laplace operator. It can be shown that the field of force produced by any distribution of masses is given by a vector function that is the gradient of a scalar function f. and f satisfies (9) in any region that is free of matter. The great importance of the Laplace equation also results from the fact that there are other laws in physics that are of the same form as Newton's law of gravitation. For instance, in electrostatics the force of attraction (or repulsion) between two particles of opposite (or SEC. 9.7 Gradient of a Scalar Field Directional Derivative 409 like) charge QI and Q2 is k (Coulomb's law6 ) p=-r r3 (12) Laplace's equation will be discussed in detail in Chaps. 12 and 18. A method for finding out whether a given vector field has a potential will be explained in Sec. 9.9. 11-61 CALCULATION OF GRADIENTS Find Vf. Graph some level curves f = const. lndicate Vf by arrows at some points of these curves. 2. f = x 2 + ty2 x 3. f = - 4. I = X4 + )'4 Y (x - 2)(y f = 6. f = (x 5. 17-121 3)2 + 2) + Cr - 1)2 8. 9. 10. 11. 12. = x2 + )'2 = In (x 2 + y2), + ;::2, P: (3, 2, 2) P: (4. 3) = cos x cosh y. P: (!7T. In 2) = x 2 + 4y2 + 9;::2, P: (3, 2. I) = eX sin y. P: (I. 7T) = (x 2 + )'2 + Z2)-I/2, P: (2, 1, [13-18] - y2, P: (2, I) \. 15. T = x 3 - 16. T = xl(x 2 17. T = 3x 2 )' =-- , P: t2, 2) x 3X)'2, P: ('VB, V2) + )'2), P: (4.0) - )'3. 1)2 - (y = yl(x 2 + )'2), = x2 2x - = 1)2. P: (4, - 3) )'2, P: t-2, 6) In (x 2 = (x 2 = - + P: (5, 3) + y2), P: (3, 3) + y2 + ~2)-1/2, P: x 2y - (12,0, 16) h 3 , P: (2, 3) 25. (Gradient) What does it mean if Igrad I(p)1 < Igrad I( QJI at two points P and Q in a scalar field? (Landscape) If ;::(x. yl = 2000 - 4x 2 - y2 [meters] gi ves the elevation of a mountain above sea level. what is the direction of steepest ascent at P: (3, -6)? What does the mountain look like? P: (4, -2) 18. T = sin x cosh y. P: (~7T. In 5) SURFACE NORMAL Find a normal vector of the surface at the given point P. 2) HEAT FLOW 14. T = arctan 24. = (x - ~7-321 Experiments show that in a temperature field, heat flows in the direction of maximum decrease of temperature T. Find this direction in general and at a given point P. Sketch that direction at P as an arrow. 13. T = x 2 21. I I I f f I 26. v(P). f f f f f f 19. 20. 23. USE OF GRADIENTS. VELOCITY FIELDS ELECTRIC FORCE The force in an electrostatic field I(x, y, z) has the direction of the gradient of f. Find VI and its value at P. 22. Given the velocity potential f of a flow. find the velocity v = vI of the flow and its value at P. Make a sketch of 7. 119-241 29. x 2 + by + cz = d. any P + 3y2 + ;::2 = 28, P: (4, 1. 3) + y2 = 25, P: (4, 3, 8) 30. x 2 - 27. ax 28. x 2 31. X4 y2 + 4;::2 = 67. P: (-2. 1, 4) + y4 + Z4 = + y2, P: 32. z = x 2 133-381 243, P: (3, 3, 3) (3, 4. 25) DIRECTIONAL DERIVATIVE Find the directional derivative of of a. I at P in the direction I = x 2 + )'2 - z, P: 0, l. -2). a = [I, 1. 2] 34. I = x 2 +)'2 + .;:2. P: (2, -2, 1), a = [-1, -1. 0] 35. I = xy.;:, P: (-I, 1,3), a = [I, -2.2] 33. 6CHARLES AUGUSTIN DE COULOMB (1736--1806), French phYSicist and engineer. Coulomb's law was derived by him from his own very precise measurements. CHAP. 9 410 36. f = (x 2 + y2 + :;,2)-112, P: Vector Differential Calculus. Grad, Div, Curl (4, 2, -4), a each of them two examples showing when they are advantageous. = [1,2, -2] 37. f = eX sin y, P: (2, ~'7T, 0), a = [2, 3, 0] = 4x 2 + y2 + 9:;,2, P: (2.4. 0). a = [-2. -4, 3] 38. f v(fg) = fvg v(f") = POTENTIALS for a given vector field-if they exist!--can be obtained by a method to be discussed in Sec. 9.9. In simpler cases. use inspection. Find a potential f = grad v for given v(x, y, ;:). 39. v = [3x, 5y, -4z] 40. v 41. v = [ye X, eX, = nf"-lvf v{flg) = (Ilg2)(gY'f - f'\g) V2(fg) = gV 2f + 2vf o vg + fY'2g 43. CAS PROJECT. Equipotential Curves. Graph some isotherms (curves of constant temperature) and indicate directions of heat flow by arrows when the temperature T(x. y) equals: 2;:J [4x 3 • 3y2, -6;:] 42. Project. Useful Formulas for Gradients and Laplacians. Prove the following formulas and give for 9.8 + gY'f (a) x 3 (b) sin x sinh y 3.\),2 - (c) eX sin y. Divergence of a Vector Field Vector calculus owes much of its importance in engineering and physics to the gradient, divergence, and curL From a scalar field we can obtain a vector field by the gradient (Sec. 9.7). Conversely, from a vector field we can obtain a scalar field by the divergence, or another vector field by the curl (to be discussed in Sec. 9.9). These concepts were suggested by basic physical applications, as we shall see. To begin, let Vlx, y, z) be a differentiable vector function, where x, y, z are Cartesian coordinates, and let vI> V2, V3 be the components of v. Then the function . (1) dlV v aVl aV2 aV3 ax ay az = -- + -- + -- is called the divergence of v or the divergence of the vector field defined by v. For example. if v = [3xz , 2n', _)'Z2] = 3x.:i ~ + 2.ni - r.:::2k .... d ~ then , div v = 3z + 2x - 2yz. Another common notation for the divergence is div v [a a] • = V· v = - . -a . ax ay az [Vb V2' V3] with [he understanding [hat the "product" (alax)v 1 in the dot product means the partial derivative av1lax. etc. Thi~ is a convenient notation, but nothing more. Note that V· v means the scalar div v, whereas V! means the vector grad! defined in Sec. 9.7. SEC. 9.8 411 Divergence of a Vector Field In Example 2 we shall see that the divergence has an important physical meaning. Clearly, the values of a function that characterizes a physical or geometric property must be independent of the particular choice of coordinates: that is, those values must be invariant with respect to coordinate transformations. Accordingly, the following theorem should hold. THEOREM 1 Invariance of the Divergence The divergence div v is a scalar jimctioll. that is, its mlues depend only on the points ill space (and. of course, on v) bllt not on the choice of the coordinates in (I). sO that with respect to other Cartesian coordinates x*, y*, z* and corre~ponding components Vi *, V2*' V3* of v, (2) We shall prove this theorem in Sec. 10.7, using integrals. Presently, let us mm [0 the more immediare practical task of gaining a feel for the significance of the divergence as follows. Let f(x, y, z) be a twice differentiable scalar function. Then its gradient exists, v af, -.af , -af] = -.-1 af. + -at.J + -at k = grad t = [ ax iI) az ilx a)' az and we can differentiate once more, the first component with respect to x, the second with respect to y. the third with respect to z, and then form the divergence, Hence we have the basic result thal the divergence of the gradient is the Laplacian (Sec. 9.7). (3) E X AMP L E 1 div (grad f) = ",2t. Gravitational Force. Laplace's Equation The gravitational force p in Theorem 3 of the last section is the gradient of the scalar function f(x, y, z) = clr, which satisfies Laplaces equation V2 f = U. According to (3) this implies that div p = 0 (r > 0). • The following example from hydrodynamics shows the physical significance of the divergence of a vector field. (More physical details on this significance will be added in Sec. 10.8.) 412 E X AMP L E 2 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl Flow of a Compressible Fluid. Physical Meaning of the Divergence We consider the motion of a fluid in a region R having no sources or sinks in R, that is, no points at which fluid is produced or disappears. The concept of fluid state is meant to cover also gases and vapors. Fluids in the restricted sense, or liquids l water or oil, for instance), have very small compressibility, which can be neglected in many problems. Gases and vapors have large compressibility; that is, their density p (= mass per unit volume) depends on the coordinates x, y, z in space (and may depend on time t). We assume that our t1uid is compressible. We consider the flow through a rectangular box B of small edges ax. /:J.y . ..k parallel to the coordinate axes (Fig. 216), (/:J. is a standard notation for small quantities; of course, it ha;, nothing to do with the notation for the Laplacian in (11) of Sec. 9.7.) The box B has the volume.1V = !:J.x /:J.y.1z. Let v = [VI, V2, V3] = VIi + V2j + V3k be the velocity vector of the motion. We set (4) and assume that u and v are continuously differentiable vector functions of x, y, z, and t (that is, they have first partial derivatives which are continuous). Let us calculate the change in the mass included in B by considering the flux across the boundary, that is_ the lotal loss of mass leaving B per unit time. Consider the flow through the left of the three faces of B that are visible in Fig_ 216, whose area is .1x j,z.. Since the vectors VI i and V3 k are parallel to that face, the components VI and V3 of v contribute nothing to this flow. Hence the mass of fluid entering through that face during a short time interval 0.t is given approximately by where the subscnpt y indicates that this expre%ion refers to the left face_ The mass of fluid leaving the box B through the opposite face during the same time interval is approximately (U2)y+.'l.Y /:J.x /:J.z /:J.t_ where the subscript y + ~y indicates that this expression refers to the right face (which is not visible in Fig. 216)_ The difference is the approximate loss of mass. Two similar expressions are obtained by considering the other two pairs of parallel faces of B.. If we add these three expressions, we find that the total loss of mass in B during the time interval /:J.l is approximately where and This loss of mass in B is caused by the time rate of change of the density and is thus equal to up ~-Ll.VLl.l. at Box B !1X Fig. 216. Physical interpretation of the divergence SEC 9.8 413 Divergence of a Vector Field If we equate both expressions. divide the resulting equation by ~V ::J.t, and let ..h. ~Y • .1::. and .it approach zero. then we obtain di, u = div (pv) fJp ill = - or ap a, + div (pv) = (5) O. This important relation is called the condition for the collsermtiolJ of lIIasS or the continuity equation of a cOlllpre.u1bie fluid flow. If the flow is steady, that b. independent of time. then aplat = 0 and the continuity eljuation is div (pv) (6) = o. If the density p is constant. so that the t1uid is incompressible, then equation (6) becomes divv (7) O. = This relation is known as the condition of incompressibility. It expresses the fact that the balance of outtlow and inflow for a given volume element is zero at any time. Clearly. the assumption that the tlow has no sources or sinks in R is essential to our argument. From this discussion you should conclude and remember that. roughly speaking. tile dh'ergellce measures outflow millus ;'l!1oW. • Comment. The divergence theorem of Gauss, an integral theorem involving the divergence, follows in the next chapter (Sec. 10.7). P R Olil;£M -S E~~ lf7] CALCULATION OF THE DIVERGENCE Find the divergence of the following vector functions. 1. [x 3 + y3, 3xy2, 3<:.\·2] [e 2x cos 2.\". e2x sin 2y. 5e 2z ] 3. [x 2 + y2, 2~yz, Z2 + x 2] 4. (x 2 + y2 + ::2)-3/2rx, v, zl 2. 5. [sin xy. 6. 7. [VI(Y, z), X 2 y 2 Z 2[X, sin xy, Z co~ V2(Z, x), y. xyl v 3 (x, y)l zl = [x, y. V 3 ]. Find a V3 such that (a) div v > 0 everywhere. (b) div v > 0 if Izl < I and div v < 0 if 1:::1 > l. 9. (Incompressible flow) Show that the flow with velocity vector v = yi is incompressible. Show that the particles that at time t = 0 are in the cube whose faces are portions of the planes x = 0, x = I, y = O. Y = I, Z = 0, Z = I occupy at t = I the volume 1. 8. Let v 10. (Compressible flow) Consider the flow with velOCIty vector v = xi. Show thm the individual particles have the position vectors r( t) = C I e t i + c 2j + C3k with constant C1 , ('2, ('3' Show that the particles that at I = 0 are in the cube of Prob. 9 at t = I occupy the volume e. 11. (Rotational flow) The velocity vector vex, y. <:) of an incompressible fluid rotating in a cylindrical vessel is of the form v = w X r, where w is the (constant) rotation vector; see Example 5 in Sec. 9.3. Show that div v = O. Is this plausible because of our present Example 27 12. CAS PROJECT. Visualizing the Divergence. Graph the given velocity field v of a fluid flow in a square centered at the origin with sides parallel to the coordinate axes. Recall that the divergence measures outflow minus inflow. By looking at the flow near the sides of the square, can you see whether div v must be positive or negative or may perhaps be zero? Then calculate div v. First do the given flows and then do some of your own. Enjoy it. (a) v = i (b) v = xi (c) v = xi - yj (d) v = xi + yj (e) v = - ri - yj (0 v = (x 2 + y2)-I(_yi + xj) CHAP. 9 414 Vector Differential Calculus. Grad, Div, Curl ~4.=-20 I 13. PROJECT. Useful Formulas for the Divergence. Prove (a) div (kv) = k div v (k constant) (b) div(fv) = fdi\'v + vo"\f (c) div (f\g) = f\2g + '\fo'\g (d) div (f'\ g) - div (gV f) = fV 2g - g,\2f. CALCULATION OF THE LAPLACIAN BY (3) Find "\2f by (3). Check by ditlerentiation. Indicate when (3) is simpler. (Show the details of your work.) 14. 15. Verify (b) for f = e and v = ad + byj + c:::k. Obtain the answer to Prob. 4 from (b). Verify (c) for f = x 2 - y2 and g = eX + Y . Give examples of your own for which (a)-(d) are advantageous. X1JZ 9.9 f f = xyl:::2 = (y + x)/(y - x) f =::: - 4Vx + )'2 18. f = arctan (ylx) 20. f = cos 2 X - sin2 )' 2 16. 17. f 19. f = ~2_y2 = cos 2xy eXYz Curl of a Vector Field Gradient (Sec. 9.7), divergence (Sec. 9.8), and curl are basic in connection with fields, and we now define and discuss the curl. Let vex, y, z) = [Vb V2' V3] = VIi + V2j + vsk be a differentiable vector function of the Cartesian coordinates x, y, z. Then the curl of the vector fUllction v or of the vector field gil'en by v is defined by the "symbolic" determinant curl v = "\ x v = j k a a a ax ay az VI V2 V3 (1) avs _ aV2)i + (aVI _ avs)j + (aV2 _ aVl)k. ( ay a::; az ax ax ay This is the formula when x. J, z are right-handed. If they are left-handed. the determinant has a minus sign in front (just as in (2**) in Sec. 9.3). Instead of curl v one also uses the notation rot v (suggested by "rotation"; see Example 2). E X AMP L E 1 Curl of a Vector Function Let v = [yz. zl = 3;:x. yzi + 3zxj + zk with right-handed x, y, z. Then (1) gives k curl, = alax alay iJIiJ~ = -3xi + yj + (3;: 3::x - :)k = -3d + yj + 2zk. • The curl plays an important role in many applications. Let us illustrate this with a typical basic example. More about the nature and significance of the curl will be said in Sec. 10.9. E X AMP L E 2 Rotation of a Rigid Body. Relation to the Curl We have seen in Example 5. Sec. 9.3, thar a rotation of a rigid body B about a fixed axis in space can be described by a vector w of magnitude w in the direction of the axis of rotation, where w (> 0) is the angular speed of the rotation, and w is directed so that the rotation appears clockwise if we look in the direction of w. According to (9), Sec. 9.3, the velocity field of the rotation can be represented in the form v = w X r SEC. 9.9 415 Curl of a Vector Field where r is the position vector of a moving point with respect to a Cartesian coordinate system harillg the origill on the axis of rotation. Let us choose right-handed Cartesian coordinates such that the axis of rotation is the ::-axis. Then (see Example 2 in Sec. 9.4) w = [0. 0, wI v = w X r = [-iVY, = ivk, WX, 0] = -Wl"i + ivXj. Hence k j curl v = This THEOREM 1 prove~ iJ a a ax ay iJ:: -wy WX o = [0, O. 2wJ = 2wk = 2w. • the following theorem. Rotating Body and Curl The curl of the velocity field of a mtating rigid hody has the direction of the axis of the rotation, and its magnitude equals twice the angular ~peed of the rotation The following two relations among grad, div, and curl are basic and shed further light on the nature of the curl. THEOREM 2 Grad. Div, Curl Gradient fields are irrotational. That is, if a continllol/sly d{fferentiable vector function is the gradient of a scalar function f, then its cllrl is the zero vector, (2) curl (grad f) = O. Furfhel71lOre, the divergence of the cllrl of a t'rvice continllously dijferentiable vector function v is :ero, (3) PROOF E X AMP L E 3 div (curl v) = O. Both (2) and (3) follow directly from the definitions by straightforward calculation. In the proof of (3) the six terms cancel in pairs. • Rotational and Irrotational Fields The field in Example 2 is not motatlOnal. A similar velocity field is obtained by stirring tea or coffee in a cur The gravitational field in Theorem 3 of Sec. 9.7 has curl p = O. It is an irrotational gradient field. • The term "irrotationar' for curl v = 0 is suggested by the use of the curl for characterizing the rotation in a field. If a gradient field occurs elsewhere, not as a velocity field, it is usually called conservative (see Sec. 9.7). Relation (3) is plausible because of the interpretation of the curl as a rotation and of the divergence as a flux (see Example 2 in Sec. 9.8). Finally, since the curl is defined in terms of coordinates. we should do what we did for the gradient in Sec. 9.7, namely, to find out whether the curl is a vector. This is true, as follows. CHAP. 9 416 THEOREM 3 Vector Differential Calculus. Grad, Div, Curl Invariance of the Curl curl v is a vector. That is, it has a length and direction that are independent of the particular choice of a Cartesian coordinate system in space. (Proof in App. 4.) 11-61 CALCULATION OF CURL Find curl v for v given with respect to right-handed Cartesian coordinates. Show the details of your work. 2 2x , 2. [yn, z n , 0] 1. [yo xn] eX siny, 3. [ex cos y, 4. (x 2 + (n + y2 5. [In (x 2 + 6. [sin y. cos > 0, integer) 0] Z2)-3/2[X, y, y2), 2 arctan (y/x), Z, z] 0] -tan x] 7. What direction does curl v have if v is a vector parallel to the xz-plane? S. Prove Theorem 2. Give two examples for (2) and (3) each. 19-141 FLUID FLOW Let v be the velocity vector of a steady fluid flow. Is the flow irrotational? Incompressible? Find the streamlines (the paths of the particles). Hint. See the answers to Probs. 9 and 11 for a determination of a path. 9. v = [0, 10. v = [_y2, 11. v = [y, 12. v = [csc x, 0] Z2, 4, -x, 13. v L4. v = [x, = [y3, L5. WRITING PROJECT. Summary on Grad, Div, Curl. List the definition and most important facts and formulas for grad, div, curl, and '17 2 • Use your list to write a corresponding essay of 3-4 pages. Include typical examples of your own. L6. PROJECT. Useful Formulas for the Curl. Assuming sufficient differentiability, show that (a) curl (u + v) = curl u + curl v (b) div (curl v) = 0 (c) curl (fv) = (grad f) x v + f curl v (d) curl (grad f) = 0 (e) div (u x v) 117-~ = v-curl u - u-curl v. EXPRESSIONS INVOLVING THE CURL With respect to right-handed coordinates, let u = [y2, .;:2, x 2], v = [YZ, ;:x, .\)'], f = xyz, and g = x + Y + z. Find the following expressions. If one of the formulas in Project 16 applies. use it to check your result. (Show the details of your work.) 17. curl v, curl (fv), curl (gv) 0] LS. curl (fu), curl (gu) L9. u x curl v, v x curl v, u-curl v, v-curl u 0] sec x, 0] .• 20. curl (u x v), curl (v x u) TIONS AND PROBLEMS 1. Why did we discuss vectors in R2 and ~ in a separate chapter, in addition to Chap. 7 on R n ? 6. Explain "right-handed coordinates," "orthonormal basis," "tangential acceleration." 2. What are applications that motivate inner products, vector products, scalar triple products? 7. What is the definition of the divergence? Its physical meaning? Its relation to the Laplacian? 3. What is wrong with the expression a x b x c? With a-b-c? With (a-b) x c? 4. What are scalar fields? Vector fields? Potentials? Give examples. 8. Granted sufficient differentiability of a scalar function f and a vector function v, which of the following make sense? gradf, f gradf, v gradf, v-gradf, divf, div v, div (fV), curl (fv), curl f, .f curl v, v curl f. 5. What is the gradient? How is it related to directional derivatives? 9. If ret) represents a motion, what is r' (t), Ir' (01, r"(t), Ir"(t)I? 417 Summary of Chapter 9 10. How do you express the resultant of forces, the moment of a force, and the work done by a force in terms of vectors? L1-201 VECTOR ADDITION, SCALAR MULTIPLICATION, PRODUCTS In right-handed coordinates let a = [3, 2, 7], b = [6, 5, -4], c = [1, 8, 0], d = [9, -2, Find 11. 4a + b - c - 2d 0]. 3a· 4a, I 2a • a, I21a1 2 , Ibl 2 2c x 5d, 10c x d (a x b)·c, a·(b x c), (a b 20. Iial - Ibll, la + c) b) bl, lal 31. (Moment) In what cases is the moment of a force p "" 0 zero? 32. (Velocity, acceleration) Find the velocity, speed, and acceleration of the motion given by at the point P: [5/'\1'2, curve is the path? (a x b) x c, a x (b x c) 18. llllal)a, (lIlcl)c 19. (a b d), (d a 30. (Moment) Find the moment vector m of p = [4, 2, 01 about P: (5, 1, 0) if p acts on a line through (1, 4, 0). Make a sketch. ret) = [5 cos t, 12. a·b, a·c, a x c 13. b x b, a x b, b x a 14. 15. 16. 17. 29. (Component) When is the component of a in the direction of b negative? Zero? + Ibl 21. (Angle) Find the angle between a and b. Between c and d. Sketch c and d. 22. (Angle) Find the angle between the planes 4x + 3y - z = 2 and x + y + Z = 1. 23. In what case is u x v = v x u? u·v = v·u·! 24. (Resultant) Find u such that a, b, c, d above, and u are in equilibrium. 25. (Resultant) Find the most general v such that the resultant of a, b, c, d above, and v is parallel to the .1y-plane. 26. (Work) Find the work done by q = [5, 1, 0] in the displacement from (4, 4, 0) to (6, -1, 0). 27. (Component) Find thecomponentofu= [-1, 5, 0] in the direction of v = [3, 4, 0). 28. (Component) In what cases is the component of a in the direction of b equal to the component of b in the direction of a? sin t, 2t] 1/'\1'2, 7T121 What kind of 33. (Tetrahedron) Find the volume of the tetrahedron with vertices (0, 0, 0), (I, 2, 0), (3, -3,0), (I, 1,5). 34. (Plane) Find an equation of the plane through (1, 0, 2), (2, 3, 5), (3, 5, 7). 35. (Linear dependence) Are [2, -1, 3], [4, 2, -5], [-1, 6, 0] linearly dependent? (Give reason.) 136-451 GRAD, DIV, CURL, V 2 , DIRECTIONAL DERIVATIVE Let f = zy Find + 36. grad f and 37. (grad f) X yx, v = [y, z, 4~ - x], w = b· 2 , Z2, x 2 ]. f grad f at (3, 4, 0) grad f, (grad f). grad f 38. div v, div w 39. curl v, curl w 40. curl (grad f), div (grad f), div v 41. V2(f), V2(f2) 42. Dwf at (1. 2, 0) 43. Dvf at (3, 7, 5) 44. div (v x w) 45. curl (v x w) + curl (w x v) Vector Differential Calculus. Grad, Div, Curl All vectors of the form a = [aI' (/2, (13] = (IIi + a2j vector space R3 with componentwise vector addition and componentwise scalar multiplication (2) (c + (/3k constitute the real a scalar, a real number) (Sec. 9,1). 418 CHAP. 9 Vector Differential Calculus. Grad, Div, Curl For instance, the resultant of forces a and b is the sum a + b. The inner product or dot product of two vectors is defmed by (Sec. 9.2) (3) where'}' is the angle between a and b. This gives for the norm or length lal of a (4) as well as a formula for '}'. If a- b = O. we call a and b orthogonal. The dot product is suggested by the work W = p - d done by a force p in a displacement d. The vector product or cross product v = a x b is a vector of length (5) la x hi = (Sec. 9.3) lallbl sin '}' and perpendicular to both a and b such that a, b, v form a right-handed triple. In terms of components with respect to right-handed coordinates, j k (Sec. 9.3). (6) The vector product is suggested, for instance, by moments of forces or by rotations. CAUTION! This multiplication is anticommutative, a x b = -b x a, and is not associative. An (oblique) box with edges a, b, c has volume equal to the absolute value of the scalar triple product (7) (a b c) = a-(b x c) = (a x b)-c. Sections 9.4-9.9 extend differential calculus to vector functions and to vector functions of more than one variable (see below). The derivative of v(t) is (8) ,dv v(t v = - = lim dt .It_O + !1t) !1t - v(t) = [' "] Vb V2, V3 = ,. VII + ,. V2J + 'k V3 • Differentiation rules are as in calculus. They imply (Sec. 9.4) (u-v)' = u' -v + u-v', (u x v)' = u' x v + u X v'. Curves C in space represented by the position vector r(t) have r' (t) as a tangent vector (the velocity in mechanics when t is time), r' (s) (s arc length, Sec. 9.5) as the unit tangent vector, and Ir"(s)/ = K as the curvature (the acceleration in mechanics). 419 Summary of Chapter 9 Vector functions vex. y. z) = [UI(X. y. z), U2(x, y. z), U3(X, y, z)] represent vector fields in space. Partial derivatives with respect to the Cartesian coordinates x. y. Z are obtained componentwise. for instance, ~ -_ [aUI ~ aU2 aU3J fu'~'ili The gradient of a scalar function ~ fu ~ (Sec. 9.6). is af] = V f = [ -af , -af , -.ax aJ Clz grad f (9) f aU2 • =aU-Il.+ J +aU3 -k (Sec. 9.7). The directional derivative of f in the direction of a vector a is df I D f = = -a-vf (10) (Sec. 9.7). lal ds a The divergence of a vector function v is . dlv (11) aU ax aU2 ay aU3 az' = - -I + - - +-- v = v-v (Sec. 9.8). The curl of v is k j (12) curl v = \" x v = a a iJ ax ay az UI U2 u3 (Sec. 9.9) or minus the determinant if the coordinates are left-handed. Some basic formulas for grad, div. curl are (Secs. 9.7-9.9) Wfg) = fvg + gVf (13) v(f/g) O/g2)(gVf - fVg) = div(fv) = fdivv + v-vf div (fVg) = fv 2g + vf-vg (14) v2f (15) = div (\"f) \"2(fg) = gv 2f curl (fV) + 2vf-vg + fv 2g = V f x v + f curl v (16) div (u x v) = v-curl u - u-curl v curl (V'f) (17) div (curl v) = 0 = o. For grad, diy, curl, and v 2 in curvilinear coordinates see App. A3.4. CHAPTER ., -, 10 Vector Integral Calculus. Integral Theorems This chapter is the companion to Chap. 9. Whereas the previous chapter dealt with differentiation in vector calculus, this chapter concerns integration. This vector integral calculus extends integrals as known from calculus to integrals over curves ("line integrals"). surfaces ("surface integrals"). and solids. We shall see that these integrals have basic engineering applications in solid mechanics, in fluid flow. and in heat problems. These different kinds of integrals can be transformed into one another. This is done to simplify evaluations or to gain useful general formulas, for instance, in potential theory (see Sec. 10.8). Such transformations are done by the powerful formulas of Green (line integrals into double integrals or conversely, Sec. 10.4), Gauss (surface integrals into triple integrals or conversely. Sec. 10.7), and Stokes (line integrals into surface integrals or conversely, Sec. 10.9). The root of these transformations was largely physical intuition. The corresponding formulas involve the divergence and the curl and will thus lead to a deeper physical understanding of these two operations. Prerequisite: Elementary integral calculus, Sees. 9.7-9.9 Sections that may be omitted in a shorter course: 10.3. 10.5. 10.8 References and Answers to Problems: App. I Part B. App. 2 10.1 Line Integrals The concept of a line integral is a simple and natural generalization of a definite integral J b (1) f(x) dx a known from calculus. [n (I) we integrate the integrand f(x) from x = a along the x-axis to x = b. [n a line integral we shall integrate a given function, also called the integrand, along a curve C in space (or in the plane). Hence curve integral would be a better name, but line integral is standard. We represent the curve C by a parametric representation (as in Sec. 9.5) (2) 420 ret) = [x(t), yet), z(t)] = x(t)i + y(t)j + z(t)k (a ~ t ~ b). SEC. 10.1 Line Integrals 421 ) B (C (a) (b) Fig. 217. Oriented curve The curve C is called the path of integration, A: rea) its initial point, and B: reb) its terminal point. C is now oriented. The direction from A to B, in which t increases, is called the positive direction On C and can be marked by an arrow (as in Fig. 217a). The points A and B may coincide (as in Fig. 217b). Then C is called a closed path. C is called a smooth curve if it has at each point a unique tangent whose direction varies continuously as we move along C. Technically: r(t) in (2) is differentiable and the derivative r' (t) = drldt is continuous and different from the zero vector at every point of C. General Assumption In this book, every path of integration of a line integral is assumed to be piecewise smooth; that is, it consists of finitely many smooth curves. For example, the boundary curve of a square is piecewise smooth, consisting of four smooth curves (segments, the four sides). Definition and Evaluation of Line Integrals A line integral of a vector function F(r) over a curve C: r(t) [as in (2)] is defined by JF(r)edr J b (3) C r F(r(t)er'(t)dt = a (see Sec. 9.2 for the dot product). In terms of components, with dr in Sec. 9.5 and ' = dldt, formula (3) becomes = [dx, dy, , dr dt dz} as JF(r)edr = J(FI dx + F2 dy + F3 dz) C C (3') b = J (FIX' + F 2 y' + F 3 z') dt. a If the path of integration C in (3) is a closed curve, then instead of we also write f· c Note that the integrand in (3) is a scalar, not a vector, because we take the dot product. Indeed, Fer'/lr'l is the tangential component of F. (For "component" see (11) in Sec. 9.2.) CHAP. 10 422 Vector Integral Calculus. Integral Theorems We see that the integral in (3) on the right is a definite integral of a function of t taken over the interval a ~ t ~ b on the t-axis in the positive direction (the direction of increasing t). This definite integral exists for continuous F and piecewise smooth C, because this makes For' piecewise continuous. Line integrals (3) arise naturally in mechanics. where they give the work done by a force F in a displacement along C (details and examples below). We may thus call the line integral (3) the work integral. Other forms of the line integral will be discussed later in this section. E X AMP L E 1 Evaluation of a Line Integral in the Plane Find the value of the line integral (3) when F(r) = [-y, Fig. 218 from A to B. Solution. YL R tit) = em. t. I) We may represent C by ret) = [cos t. sin t] = cos t i yet) = sin t. and = F(r(t» -y(tH - x(t)y(t)j By differentiation. r' (t) = [-sin t, _cos t = II in the second term] A_ 1 Fig. 218. f = f = f~"2 ~ (l - cos t] = ~in t :'§ 71"/2. Then -cos t sin t] = -sin t i-cos t sin t j. t, -sin t i + cos t j, so that by (3) [use (10) in App. 3.1; set F(r) - dr f rr/2 rr/2 f OI/2(-dl/) I cos 2t) dt - f- i 0 - = • = 0.4521. Line Integral in Space The evaluation of line integrals in space is practically the same as it is in the plane. To see this. find the value of (3) when F(r) = [~, x. y] = :::i + xj + yk and C is the helix (Fig. 219) z r(t) = [cos t. sin t. 3tJ = cos t i (4) Solution. From (4) we have lett) Example 2 The dot product is 3t( -sin t) + f 2,,- c F(r)-dr = f° + sin t j + 3tk = cos t. y(t) = sin t, :::(t) = + costj + F(r(t))-r'(t) = (3ti Fig. 219. = [- + sin t j, where U :'§ [-sin t. -cos t sin t[ - [-sin t, cos t] dt = (sin2 t - cos2 t sin t) dt c o o x Example 1 EXAMPLE 2 = -yi - xyj and C is the circular arc in -xy] cos 2 t + 3 sin t. (-3tsint + 3t. Thus sintk)-(-sinti + costj + 3k). Hence (3) gives cos 2 t + 3sint)dt = 671" + 71" +0 = 771"= 21.99. • Simple general properties of the line integral (3) follow directly from corresponding properties of the definite integral in calculus, namely, f kFodr = k f (Sa) c (5b) f (F c + G)°dr = C Fig. 220. Formula (Sc) (5c) (k constant) Fodr f Fodr + f Godr c C f Fodr = f Fodr + f Fodr C c, C2 (Fig. 220) SEC. 10.1 423 Line Integrals where in (Sc) the path C is subdivided into two arcs C 1 and C2 that have the same orientation as C (Fig. 220). In (Sb) the orientation of C is the same in all three integrals. If the sense of integration along C is reversed, the value of the integral is multiplied by -1. However, we note the following independence if the sense is preserved. THEOREM 1 Direction-Preserving Parametric Transformations Any representations of C that give the same positive direction on C also yield the same value of the line integral (3). PROOF A proof follows by the chain rule, where ret) is the given representation, t = cp(t*) with a positive derivative dtldt* is the transformation, with a* ~ t* ~ b* corresponding to a ~ t ~ bin (3), and we write ret) = r(cp(t*» = r*(t*). Then dt = (dtldt*) dt* and f c F(r*)odr* dr dt F(r(cp(t*») ° ~~ dt* u" dt dt* f =f = b* dr F(r(t»° - dt dt b u = f c F(r) ° dr. • Motivation of the Line Integral (3): Work Done by a Force The work W done by a constant force F in the displacement along a straight segment d is W = Fod; see Example 2 in Sec. 9.2. This suggests that we define the work W done by a variable force F in the displacement along a curve C: ret) a~ the limit of sums of works done in displacements along small chords of C. We show that this definition amounts to defining W by the line integral (3). For this we choose points to (= a) < tl < ... < tn (= b). Then the work LlWm done by F(r(tm in the straight displacement from r(tm} to r(tm.+ 1) is » The sum ofthese n works is Wn = LlWo + ... + LlWn_1.lfwe choose points and consider Wn for every II arbitrarily but so that the greatest Lltl11 approaches zero as n ---? 00, then the limit of Wn as n ---700 is the line integral (3). This integral exists because of our general assumption that F is continuous and C is piecewise smooth: this makes r' (t) continuous, except at finitely many points where C may have comers or cusps. • E X AMP L E 3 Work Done by a Variable Force IfF in Example I is a force. the work done by F in the displacement along the quarter-circle is 0.4521, measured in snitable nnits, say, newton-meters (nt'm, also called joules, abbreviation J; see also front cover). Similarly in Example 2. • 424 E X AMP L E 4 CHAP. 10 Vector Integral Calculus. Integral Theorems Work Done Equals the Gain in Kinetic Energy Let F be a force, so that (3) is work. Let t be time, so that dr/dt (6) W= f v, = velocity. Then we can write (3) as b F·dr J = F(r(t))·v(t)dt. a C Now by Newton's second law (force = mass X acceleration), F = mr"(t) where III mv' (t), = is the mass of the body displaced. Substitution into (5) gives [see (11), Sec. 9.4] b J W = IIlV' • Jb (v.v)' dt vdt = III a -- Ivl It~b m 2 = 2 - 2 a t~a . On the right, mlvl 2 /2 is the kinetic energy. Hence the work done equals the gain in kinetic energy. This is a basic law in mechanics. • Other Forms of Line Integrals The line integrals (7) are special cases of (3) when F = F1i or F 2 j or F3k, respectively. Furthermore, without taking a dot product as in (3) we can obtain a line integral whose value is a vector rather than a scalar, namely, b (8) J F(r) dt = J F(r(t» C dt = a f b [F1(r(t», F 2 (r(t», Obviously, a special case of (7) is obtained by taking Fl f fer) (8*) F 3 (r(t»] dt. a C dt = f = f, F2 = F3 = O. Then b f(r(t» dt a with C as in (2). The evaluation is similar to that before. E X AMP L E 5 A Line Integral of the Form (8) Integrate F(r) = [xy. yz, zJ along the helix in Example 2. Solution. F(r(t)) = [cos t sin t, 3t sin t, 3t] integrated with respect to t from 0 to 271" gives 2-,,- fo F(rVJJ dt [ = - ~ cos2 t, 3 sin t - 3t cos t, "23 2 t ] 127T 0 = [0, • Path Dependence Path dependence of line integrals is practically and theoretically so important that we formulate it as a theorem. And a whole section (Sec. 10.2) will be devoted to conditions under which path dependence does not occur. SEC 10.1 425 Line Integrals THEOREM 2 r Path Dependence The line integral (3) generally depends not ollly all F alld all the endpoints A and B of the path, but also on the path Use?! along which the integral is taken. PROOF Almost any example will show this. Take, for instance. the straight segment C1: rl(t) = [t, t, 0] and the parabola C2: r 2(t) = [t, t 2, 0] with 0 ~ t ~ 1 (Fig. 22]) and integrate F = [0, xy, 0]. Then F(r 1 (t»· rl (t) = t 2, F(r2(t»· r2(t) = 2t 4 , so that integration gives L/3 and 2/5, respectively. • l~B 1 Fig. 221. Proof of Theorem 2 ... 11-121 Calculate LINE INTEGRAL. WORK DONE BY A FORCE 11. F = [ex, e Y , e Z ], r = (2, 4, 4). Sketch C. f 12. F = [y2, 2. F as in Prob. 1, C the shortest path from A to B. Is the integral smaller? Give reason. 3. F as in Prob. 1, C from A straighL to (2. 0). then vertically up to B = cos 2 :::], C as in Prob. 7. Sketch C. F(r)· dr for the following data. If F is a force. c this gives the work done in the displacement along C. (Show the details.) 1. F = [y3, x 3], C the parabola y = 5x 2 from A: (0, 0) to B: (2,20) 4. F x 2, [t, P, t2 ] from (0, 0, 0) to [x 2, y2, (-2.0), y ~ 0], C the semicircle from (2, 0) to 0 5. F = [xy2, x~], C: r = [cosh t, sinh t, 0], o ~ t ~ 2. Sketch C. 6. F = [ex, e Y ] clockwise along the circle with center (0, 0) from (1, 0) to (0, -1) 7. F = [z, x, y], C: r to (1, 0, 417) = [cos t, sin t, t] from (1, 0, 0) 13. WRITING PROJECT. From Definite Integrals to Line Integrals. Write a short report (1-2 pages) with examples on line integrals as generalizations of definite integrals. The latter give the area under a curve. Explain the corresponding geometric interpretation of a line integral. 14. PROJECT. Independence of Representation. Dependence on Path. Consider the integral where F = [xy, _y2]. f F(r)· dr, C (a) One path, several representations. Find the value of the integral when r = [cos t, sin t], 0 ~ t ~ 1712. Show that the value remains the same if you set t = - p or t = p2 or apply two other parametric transformations of your own choice. (b) Several paths. Evaluate the integral when C: y = x n , thus r = [t, t"l, 0 ~ t ~ 1, where n = 1,2,3, .... Note that these infinitely many paths have the same endpoints. 8. F = [coshx, sinhy. eZ ] . C: r = [t. P, t3 ] from (0, 0, 0) to (!, ~) 9. F as in Prob. 8. C the straight segment from (0. O. 0) to (!, ~) (c) Limit. What is the limit in (b) as n --+ oo? Can you confirm your result by direct integration without referring to (b)? 10. F = [x, -z, 2y] from (0, 0, 0) straight to (1, 1,0), then to (1, 1, 1), back to (0, 0, 0) (d) Show path dependence with a simple example of your choice involving two paths. i, i, CHAP. 10 426 115-181 Vector Integral Calculus. Integral Theorems INTEGRALS OF THE FORMS (8) AND (8*) Evaluate (8) or (8*) with F or f and C as follows. f = x 2 + y2, c: r = [t, 4t, 0], 0 ~ t ~ 1 16. f = 1 - sinh2 x, C the catenary r = [t, cosh t], 15. 0~t32 17. F = [y2, [3 cos t, (L = Length of 0. (9) X2], C the helix 3 sin t, 2t], 0 3 t ~ 817 Z2, 20. Using (9), find a bound for the absolute value of the work W done by the force F = [x 2 , y] in the 18. F = [(xy)1/3, (y/x) 1/3, 0], C the hypocycloid r = [cos 3 t. sin 3 t, 0]. 0 ~ t ~ 17/4 10.2 19. (ML-Inequality, Estimation of Line Integrals) Let F be a vector function defined on a curve C. Let IFI be bounded. say. IFI ~ M on C, where M is some positive number. Show that displacement along the segment from (0. Q) to (3, 4). Path Independence of Line Integrals In this section we consider line integrals (1) Fig. 222. Path independence (dr = [£lx, d.\', dz]) as before, and we shall now find conditions under which (I) is path independent in a domain D in space. By definition this means that for every pair of endpoints A, B in D the integral (1) has the same value for all paths in D that begin at A and end at B. (See Fig. 222. See Sec. 9.6 for "domain.") Path independence is important. For instance, in mechanics it may mean that we have to do the same amount of work regardless of the path to the mountaintop, be it short and steep or long and gentle. Or it may mean that in releasing an elastic spring we get back the work done in expanding it. Not all forces are of this type-think of swimming in a big round pool in which the water is rotating as in a whirlpool. We shall follow up three ideas that will give path independence of (1) in a domain D if and only if: = grad j (see Sec. 9.7 for the gradient). (Theorem]) F (Theorem 2) Integration around closed curves C in D always gives O. (Theorem 3) curl F = 0 (provided D is simply connected, as defined below). Do you see that these theorems can help in understanding the examples and counterexample just mentioned? Let us begin our discussion with the following very practical criterion for path independence. THEOREM 1 Path Independence A line integral (1) with continuous Fl , F2 , F3 ill a domain D in space is path independent in D if and only ifF = [Flo F2 , F 3 ] is the gradient of some function jill D, (2) F = gradj, thus. SEC. 10.2 427 Path Independence of Line Integrals PROOF (a) We assume that (2) holds for some function .f in D and show that this implies path independence. Let C be any path in D from any point A to any point B in D, given by ret) = [x(t), yet), ::(t)], where a ~ t ~ b. Then from (2). the chain rule in Sec. 9.6, and (3') in the last section we obtain Ic (F1dx + F2 dy + Ic (~f F3 d::) = cJx ~f iJy + dx dy + ~f iJz d::) Jb( - - + - - +aZ.- bdf It=b =I = c.t = af dx ax dt a -f' dt af dy ay dt af dZ) dt dt f[x(t), yet), z(t)J a t=a = f(x(b), = y(b), z(b)) - .f(x(a), yea), z(a) feB) - f(A). (b) The more complicated proof of the converse, that path independence implies (2) for some f, is given in App. 4. • The last formula in part (a) of the proof, J B (3) (F] dx + F2 dy + F3 dz) = .f(B) - f(A) [F = grad.f] A is the analog of the usual formula for definite integrals in calculus. J b g(X) dx = C(x) a Ib = G(b) - C(a) [C'(x) = g(x)]. a Formula (3) should be applied whenever a line integral is independent of path. Potential theory relates to our present discussion if we remember from Sec. 9.7 that f is called a potential of F = grad f. Thus the integral (1) is independent of path in D if and only if F is the gradient of a potential in D. E X AMP L E 1 Path Independence Show that the integrdl f F dr = 0 c f c (2x dx + 2)' dy + 4;: dz) is path independent in any domain in space and find its value in the integration from A: (0, O. 0) to B: (2. 2. 2). Solution. F = [2y. 2)" 4;:] = grad i. where i =,\'2 + )'2 + ai/a::. = 4.:: = F3 . Hence the integral is independent of ai/ax = 2y = F I , ai/ay = 2)' = F2 , path according to Theorem I, and (3) gives 2;:2 because I(B) - I(A) = i(l. 1. 1) - ItO. O. 0) = 4 + 4 + 8 = 16. If you want to check this. use the most convenient path C: ret) = [I. I, I]. 0 ~ I ~ 1. on which F(r(l) = [2/, 21, 411, so that F(r(/j) r'(I) = 21 + 21 + 41 = 8/. and integration ti-om 0 to 2 gives 8.22/2 = 16. If you did not see the potential by inspection. use the method in the next example. • 0 E X AMP L E 2 Path Independence. Determination of a Potential Evaluate the integrall = f 2 (3x dy c has a potential and applying (3). + 2)'.:: d)' + y2 d:;;) from A: (0, I, 2) to B: (I, - I, 7) by showing that F 428 CHAP. 10 Vector Integral Calculus. Integral Theorems Solution. If F has a potential f. we should have Iy = F2 = 2yz, We show that we can satisfy these conditions. By integration of fx and differentiation, I = x This gives f(x, y, 3 + g(y, z), z) = x 3 + I y2;: fy = gy = 2y;:, g = y2Z + hi = 0 h = 0, say. I h(;:,), = x3 + y2 Z + h(::.) and by (3), = 1(1, -1, 7) - f(O, 1, 2) = 1 + 7 - (0 + 2) = 6. • Path Independence and Integration Around Closed Curves The simple idea is that two paths with common endpoints (Fig. 223) make up a single closed curve. This gives almost immediately THEOREM 2 Path Independence The integral (1) is path independent in a domain D if and only if its value around ever}' closed path in D is zero. PROOF If we have path independence, then integration from A to B along C1 and along C2 in Fig. 223. Proof of Theorem 2 Fig. 223 gives the same value. Now C1 and C2 together make up a closed curve C, and if we integrate from A along C1 to B as before, but then in the opposite sense along C2 back to A (so that this second integral is multiplied by -]), the sum of the two integrals is zero, but this is the integral around the closed curve C. Conversely, assume that the integral around any closed path C in D is zero. Given any points A and B and any two curves C 1 and C2 from A to B in D, we see that C1 with the orientation reversed and C2 together form a closed path C. By assumption, the integral over C is zero. Hence the integrals over C] and C2 , both taken from A to B, must be equal. This proves the theorem. • B Work. Conservative and Nonconservative (Dissipative) Physical Systems Recall from the last section that in mechanics, the integral (1) gives the work done by a force F in the displacement of a body along the curve C. Then Theorem 2 states that work is path independent in D if and only if its value is zero for displacement around every closed path in D. Furthermore, Theorem] tells us that this happens if and only if F is the gradient of a potential in D. In this case, F and the vector field defined by F are called conservative in D because in this case mechanical energy is conserved; that i!>, no work is done in the displacement from a point A and back to A. Similarly for the displacement of an electrical charge (an electron, for instance) in a conservative electrostatic field. Physically, the kinetic energy of a body can be interpreted as the ability of the body to do work by virtue of its motion, and if the body moves in a conservative field of force, after the completion of a round trip the body will return to its initial position with the same kinetic energy it had originally. For instance, the gravitational force is conservative; if we throw a ball vertically up, it will (if we assume air resistance to be negligible) return to our hand with the same kinetic energy it had when it left our band. SEC. 10.2 429 Path Independence of Line Integrals Friction, air resistance, and water resistance always act against the direction of motion, tending to diminish the total mechanical energy of a system (usually converting it into heat or mechanical energy of the surrounding medium. or both), and if in the motion of a body these forces are so large that they can no longer be neglected, then the resultant F of the forces acting on the body is no longer conservative. Quite generally, a physical system is called conservative if all the forces acting in it are conservati ve; otherwise it is called non conservative or dissipative. Path Independence and Exactness of Differential Forms Theorem I relates path independence of the line integral (I) to the gradient and Theorem 2 to integration around closed curves. A third idea (leading to Theorems 3* and 3, below) relates path independence to the exactness of the differential form (or Pfaff/an f017l11) (4) under the integral sign in (1). This form (4) is called exact in a domain D in space if it is the differential df af af af = - £Ix + - dv + - d::. = (uradf)-dr ax ay' az to of a differentiable function f(x, y, z) everywhere in D. that is, if we have F-dr = df. Comparing these two formulas. we see that the form (4) is exact if and only if there is a differentiable function f(x, y, z) in D such that everywhere in D. (5) F = gradf, thus, Fl = af ax ' Hence Theorem l implies THEOREM 3* Path Independence The integral (1) is path independent in a domain D in :,pace (f and only if the d(fferentialfo171l (4) has continuous coefficient functions Flo F2 , F3 and is exact in D. This theorem is practically important because there is a useful exactness criterion To formulate the criterion, we need the following concept, which is of general interest. A domain D is called simply connected if every closed curve in D can be continuously shrunk to any point in D without leaving D. For example, the interior of a sphere or a cube. the interior of a sphere with finitely many points removed. and the domain between two concentric spheres are simply IJOHANN FRIEDRICH PFAFF (1765-1825), German mathematician. CHAP. 10 430 Vector Integral Calculus. Integral Theorems connected. while the interior of a torus (a doughnut; see Fig. 247 in Sec. 10.6) and the interior of a cube with one space diagonal removed are not simply connected. The criterion for exactness (and path independence by Theorem 3*) is now as follows. THEOREM 3 Criterion for Exactness and Path Independence Let F b F2' F3 in the line integral (I), fc F(r)"dr = fc (PI dx + F2 d. . . + F3 d::.), be contillllOllS and have cominuous first partial derivatives ill a domain D in space. Then: (a) lfthe differel1Tialform (4) is eX({Ci ill D-al1d thus (I) is path independent by Theorem 3*-, then in D, (6) curl F = 0; ill components (see Sec. 9.9) (6') (b) If (6) holds in D and D is simply connected. thell (4) is exact in D-and thus (I) is path independent by Theorem :1*. PROOF (a) If (4) is exact in D, then F = grad f in D by Theorem 3*, and, furthermore, curl F = curl (grad.f) = 0 by (2) in Sec. 9.9, so that (6) holds. (b) The proof needs "Stokes's theorem"" and will he given in Sec. 10.9. • f F( r) "dr = f (F I dx + F2 dy) the curl has only one c c component (the z-component), so that (6') reduces to the single relation Line Integral in the Plane. For (6") (which also occurs in (5) of Sec. 1.4 on exact ODEs). E X AMP L E 3 Exactness and Independence of Path. Determination of a Potential Using (6'), show that the differential form under the integral sign of is exact, so that we have independence of path in any domain, and find the value of I from A: (0, 0, 1) to B: (l, 7r/4, 2). SEC. 10.2 431 Path Independence of Line Integrals Solution. Exactness follows from (6'), which gives 2 (F3 )y = 2x z + = (Fl)z = 4xyz (F2 )x To find and F 3 , J, = 2xz 2 cosyz - yzsinyz = (F2 )z (F3 )x (F1)y' = we integrate F2 (which is "long," so that we save work) and then differentiate to compare with Fl fz = 2 2x zy + Y cos yz + = h' 2 2x zy = F3 + Y cos yz, h' = O. h' = 0 implies h = const and we can take h = 0, so that g = 0 in the first line. This gives, by (3), f(x, y, z) = x 2YZ 2 -t sin yz, 7T 7T 4 . 4 + sin 2 - feB) - f(A) = 1 . 0 = 7T -t • 1. The assumption in Theorem 3 that D is simply connected is essential and cannot be omitted. Perhaps the simplest example to see this is the following. E X AMP L E 4 On the Assumption of Simple Connectedness in Theorem 3 Let x Y F] = - - 2 - - 2 ' X +Y (7) F2 = - 2 - - 2 ' x +Y F3 = o. Differentiation shows that (6') is satisfied in any domain of the xy-plane not containing the origin. for example, in the domain D: ~ < ~ + ; < ~ shown in Fig. 224. Indeed, Fl and F2 do not depend on z, and F3 so that the first two relations in (6') are trivially true. and the third is verified by differentiation: V aF2 ax x 2 +y2-x·2x y + (x2 (x2 x2 aFl ay y2)2 + y2 - y-2y (x 2 + V2 )2 2 = 0, - x2 + y2)2 y 2 (x2 - x2 + y2)2 Clearly, D in Fig. 224 is not simply connected. If the integral 1= f c (Fl dx + F2 dy) = f C -ydx + xdy 2 x + y2 were independent of path in D, then I = 0 on any closed curve in D, for example, on the circle x 2 But setting x = r cos 8, y = r sin e and noting that the circle is represented by r = I, we have x=cose. so that -y dx + x dy = 2 sin e de dx = -sin ed8, + cos2 8 de = de 1= f y=sine. + y2 = 1. dy = cos ede, and counterclockwise integration gives 2.,,- o de = I 27T. Since D is not simply connected. we cannot apply Theorem 3 and cannot conclude that I is independent of path in D. Although F = grad f, where f = arctan (ylx) (verify!), we cannot apply Theorem I either because the polar angle f = 8 = arctan (y Ix) is not single-valued, as it is required for a function in calculus. • CHAP. 10 432 Vector Integral Calculus. Integral Theorems y 3 ;1 Example 4 Fig. 224. .... ,,-- -.-. ....- . - 11-81 PATH-INDEPENDENT INTEGRALS Show that the fonn under the integral sign is exact in the plane (Probs. 1-4) or in space (Probs. 5-8) and evaluate the integral. (Show the details of your work.) 1. f f (4."./8) (y +x cosxJ dx (c) Integrate from (0, 0) along the straight-line segment to (c, I), 0 ~ c ~ I, and then horizontally to (1, I). For c = I, do you get the same value as for b = I in (b)? For which c is I maximum? What is its maximum value? cosxy dy) y (0.0> 2. x (c,l) 1 (0,5) 2 (y e 2x dx ye 2x dy) + (5.0) 3. fO,l) e- X2 _ y2 (x dx + (1, b) y dy) (-1,-1) 4. f <0.0) (6.w) (cos 2 Y dx - 2x cos y sin y dy) (2,0) 5. f f f f 111-191 (O,1,2) (z e Xz dx + dy + xe xz dz) + y dy - d;:.) (0.0.0) 7. (7.8.0) (2xy dx + x 2 dy + sinh z £Iz) 0.0.0) 8. [2X(y3 - Z3) dx + 3x 2 )'2 dy - 3x 2Z 2 dz] 9. Show thar in Example 4 of the text we have F = grad (arctan (ylx». Give examples of domains in which the integral is path independent. 10. PROJECT. Path Dependence. (a) Show that 1 (x\ dx c xy-plane. = + 2xy2 dy) is path dependent in the (b) Integrate from (0. 0) along the straight-line segment to (1. b). 0 ~ b ~ I, and then vertically up to (I, I); see the figure. For which of these paths is I maximum? What is its maximum value? t' dz) 12. (3x 2e 2Y + x) dx + 2x 3e 2Y dy 2 13. 3x y {lY + x 3 dy + Y d: 14. 2x sin J dx + x 2 cosy dy + y2 dz 15. (ze e Y ) dx - xe Y dy + eX dz 16. eX cos 2.1' dx - 2e x sin 2y dy - xz dz 17. xy Z2 d.1. + !X 2Z2 dy + x 2)'z do;:. 18. yz cosh x dx + Z sinh x dy + J sinh x dz 19. Y dt' + (x - 2y) dy + 4x dz X (4.4,0> (2.0,1> I CHECK FOR PATH INDEPENDENCE 11. (cosh x::)(:: dx + O .1 •0 ) ex2+y2-2z (x dx x and, if independent, integrare from (0, 0, 0) to (a, b, c). (2,3.0) 6. 1 Project 10. Path Dependence - 20. WRITING PROJECT. Ideas on Path Independence. Make a list of the main ideas on path independence and dependence in this section. Then work this list into an essay. including explanations of all definitions and on the practical usefulness of the theorems, but no proofs. Include illustrating examples of your own. Explain what happens in Example 4 if you take the domain 0 < Vr + y2 < ~. SEC. 10.3 10.3 Calculus Review: Double Integrals. Optional 433 Calculus Review: Double Integrals. Optional Students familiar with double integrals from calculus should go on to the next section, skipping the present review, which is included to make the book reasonably self-contained. In a definite integral (1), Sec. 10.1, we integrate a function f(x) over an interval (a segment) of the x-axis. In a double integral we integrate a function f(x, y), called the integrand, over a closed bounded region2 R in the xy-plane, whose boundary curve has a unique tangent at each point, but may perhaps have finitely many cusps (such as the vertices of a triangle or rectangle). The definition of the double integral is quite similar to that of the definite integral. We subdivide the region R by drawing parallels to the x- and y-axes (Fig. 225). We number the rectangles that are entirely within R from 1 to n. In each such rectangle we choose a point, say, (Xk, Yk) in the kth rectangle, whose area we denote by LlA k. Then we form the sum n in = 2: f(xk, Yk) LlAk- k~l This we do for larger and larger positive integers II in a completely independent manner, but so that the length of the maximum diagonal of the rectangles approaches zero as n approaches infinity. In this fashion we obtain a sequence of real numbers i n" i n2 , . . . . Assuming that f(x, y) is continuous in Rand R is bounded by finitely many smooth curves (see Sec. 10.1), one can show (see Ref. [GR4] in App. 1) that this sequence converges and its limit is independent of the choice of subdivisions and corresponding points (xk, Yk). This limit is called the double integral of f(x, y) over the region R, and is denoted by f ff(x, y) dxdy R or f ff(x, y) dA. R y x Fig. 225. Subdivision of a region R 2 A region R is a domain (Sec. 9.6) plus, perhaps, some or all of its boundary points. R is closed if its boundary (all its boundary points) are regarded as helonging to R; and R is bounded if it can be enclosed in a circle of sufficiently large radius. A boundary point P of R is a point (of R or not) such that every disk with center P contains points of R and also points not of R. 434 CHAP. 10 Vector Integral Calculus. Integral Theorems Double integrals have properties quite similar to those of definite integrals. Indeed, for any functions f and g of (x, y), defined and continuous in a region R, f f f kf dx dy = k f f R (1) ff(f + (k constant) dx dy R g)dxdy = fffdxdy+ ffgdxdy R R f ffdxdy R f ffdxdy = R + f ffdxdy Rl (Fig. 226). R2 Furthermore, if R is simply connected (see Sec. 10.2), then there exists at lea')t one point (xo, Yo) in R such that we have (2) f f f(x, y) dx dy = f(xo, yo)A· R where A is the area of R. This is called the mean value theorem for double integrals. Fig. n6. Formula (l) Evaluation of Double Integrals by Two Successive Integrations Double integrals over a region R may be evaluated by two successive integrations. We may integrate first over y and then over x. Then the formula is (3) R hex) b [ f f f(x, y) dx dy = f f a ] f(x, y) dy dx (Fig. 227). g(x) Here y = g(x) and y = hex) represent the boundary curve of R (see Fig. 227) and, keeping x constant, we integrate f(x, y) over y from g(x) to hex). The result is a function of x. and we integrate it from x = a to x = b (Fig. 227). Similarly, for integrating first over x and then over y the formula is (4) f ff(x,y)dxdy R d[ = f f C q(y) p(y) f(x,y)dx J dy (Fig. 228). SEC. 10.3 435 Optional Calculus Review: Double Integrals. y y d h(X)~J C I : I I R : I I ,---: I a Fig. 227. < g(x) I b ---------7) P(y)~ R ----c~ / '--"y) x x Evaluation of a double integral Fig. 228. Evaluation of a double integral The boundary curve of R is now represented by x = p(y) and x = q(y). Treating y as a constant, we first integrate f(x. y) over x from p(y) to q(y) (see Fig. 228) and then the resulting function of y from y = c to y = d. In (3) we assumed that R can be given by inequalities a ~ x ~ b and g(x) ~ y ~ hex). Similarly in (4) by c ~ y ~ d and p(y) ~ x ~ q(y). If a region R has no such representation, then in any practical case it will at least be possible to subdivide R into finitely many portions each of which can be given by those inequalities. Then we integrate f(x, y) over each portion and take the sum of the results. This will give the value of the integral of f(x. y) over the entire region R. Applications of Double Integrals Double integrals have various physical and geometric applications. For instance. the area A of a region R in the xy-plane is given by the double integral A = II dxdy. R The volume V beneath the surface is (Fig. 229) z= f(x, y) (> 0) and above a region R in the xy-plane V= f ff(x,y)dxdy R because the term f(Xk, Yk) ilAk in in at the beginning of this section represents the volume of a rectangular box with base of area ilAk and altitude f(xk, Yk)' z x '. Fig. 229. - Double integral as volume y 436 CHAP. 10 Vector Integral Calculus. Integral Theorems As another application. let f(x, y) be the density (= mass per unit area) of a distribution of mass in the \)·-plane. Then the total mass M in R is M II = f(x. y) dx dy: R the center of gravity of the mass in R has the coordinates X, x = ~ ~ y= and I I xf(x, y) dx dy R y, where I I yf(x, y) dx dy; R the moments of inertia Ix and Iy of the mass in R about the x- and y-axes, respectively, are Ix = I I y2f(x, y) dx dy, II Iy = R x 2 f(x. y) dx dy; R and the polar moment of inertia 10 about the origin of the mass in R is 10 = + Iy = Ix I I(x 2 + y2)f(x, y) dx dy. R An example is given below. Change of Variables In Double Integrals. Jacobian Practical problems often require a change of the variables of integration in double integrals. Recall from calculus that for a definite integral the formula for the change from x to u is b (5) Ia f(x) dx = I/3 0' f(x(u») dx duo du _0 Here we assume that x = x(u) is continuous and has a continuous derivative in some interval a ~ II ~ f3 such that x(a) = G, x(f3) = b [or x(a) = b. x(f3) = G] and X(ll) varies between G and b when u varies between a and f3. The formula for a change of variables in double integrals from x, y to ll, U is (6) IRI f(x, y) dx d.\' = IR*I f(x(u, u), y(u, u)) a(x, y) I-a(u,-u)- Idu du; that is, the integrand is expressed in terms of u and u, and dx dv is replaced by du du times the absolute value of the Jacobian 3 (7) J= ax ax B(x, y) au au ax ay ax ay a(u. u) ay ay all au iJu au au au --- 3 Named after the German mathematician CARL GUSTAV JACOB JACOBI (1804-1851), known for his contributions to elliptic functions. partial differential equations, and mechanics. SEC. 10.3 Calculus Review: Double Integrals. Optional 437 Here we assume the fol1owing. The functions x = .Y(u, u), y = y(u, u) effecting the change are continuous and have continuous partial derivatives in some region R* in the uu-plane such that for every (u, u) in R* the corresponding point (.Y, y) lies in R and, conversely, to every (x, y) in R there corresponds one and only one (u, v) in R*; furthermore, the Jacobian J is either positive throughout R* or negative throughout R*. For a proof. see Ref. [GR4] in App. 1. E X AMP L E 1 Change of Variables in a Double Integral Evaluate the following double integral over the square R in Fig. 230. Solution. The shape of R suggest~ the transfonnation x J = !(u - v). The Jacobian is J = R corresponds to the square 0 II ~ u 2 ~ 2, 0 a(x, y) = a(ll, v) ~ v ~ I! !I 1 = _1 2 = v. Then x Y= ll, X - = !lu + v), -!.. . 2 2 2. Therefore, f2f21 2 (x -t Y ) dx dy = - R + y 00 2 (u 2 1 2 + v ) - du dv 8 = - 2 . 3 • y x Fig. 230. Region R in Example 1 Of pmticular practical interest are polar coordinates r and ti, which can be introduced by setting x = r cos ti, y = r sin e. Then J = a(x, y) = a(r, Icos e -r sin el e r cos e e) sin = r and (8) JJf(x, y) dx dy JJfer cos e. r sin e) r dr de = R R* where R* is the region in the re-plane corresponding to R in the xy-plane. CHAP. 10 438 EXAMPLE 2 Vector Integral Calculus. Integral Theorems Double Integrals in Polar Coordinates. Center of Gravity. Moments of Inertia Let f(x, y) == I be the mass density in the region in Fig. 231. Find the total mass, the center of gravity, and the moment' of inertia lx, Iy, 10 , y~ Solution. We use the polar coordinates just defined and formula (8), This gives the total mass x 1 Fig. 231. The center of gravity has the coordinares Example 2 4 x = 71" I "2 f 7 0 4 y=- I 0 1.,,/2 1 cos fI dfl 4 r cos B r dr dB = 71" - 3 0 4 = 3..,- = 0.4244 for reasons of symmetry. 371" The moments of inertia are Ix = II If ."./2 y2 dx dy = 1 2 = r2 sin B r dr dB ROO ."./2 ={ 1T 1=y 16 Why are I ."./2 ~ sin 2 B dB 0 i(1- cos28)dB= i (-i - 0) = I: = 0.1963 IT for reasons of symmetry, 10 = Ix + ly = "8 = 0.3927. • x and y less than~? This is the end of our review on double integrals. These integrals will be needed in this chapter. beginning in the next section. • • w ...... _ n ... w,oc:: ,_IiI_ ___ _._ 1. (Mean value theorem) lllustrate (2) with an example. 12-91 DOUBLE INTEGRALS Describe the region of integration and evaluate. (Show the details.) ff 1 2. o 3. (x + y)2 dy 10 Ix"(1 - 11 o 2xy) dy dx y cosh tx + y) dx dy 0 6. As Prob. 5, order reversed 4 7. 1J o 7T'/2 9. o x 2 )' dy dx I-x sin y eX cos y dx dy 0 X -x e x 2y + dy dx y2 10. Integrate xye _ over the triangular region with vertices (0. 0). O. I). (1. 2). 4. As Prob. 3, order reversed 5. o X2 dx x 3 I-x" ff I f 2x x 1 1 8. 111-131 VOLUME Find the volume of the following regions in space. 11. The region beneath z = x 2 + y2 and above the square with vertices (1, I), (-I, 1), (-1, -I), (1, -1) 12. The tetrahedron cut from the first octant by the plane ~x + 2)' + z = 6. Check by vector methods. 13. The first octant section cut from the region inside the + Z2 = 1 by the planes y = z = 0, x = y. cylinder r 0, SEC. lOA 439 Green's Theorem in the Plane [14-161 CENTER OF GRAVITY Find the center of gravity (x, y) of a mass of density j(x, y) = I in the given region R. 14. R the semi disk x 2 + y2 ~ a 2 , y ~ 0 15. engineer is likely to need. along with other profiles listed in engineering handbooks). 18. R as in Prob. 16. 17. R as in Prob. 15. y 19. '~ h 2+----.. h~ x b x y 20. h+--..... 117-201 MOMENTS OF INERTIA Find the moments of ineltia Ix, Iy , 10 of a mass of density j(x, y) = 1 in the region R shown in the figures (which the 10.4 o a 2 a 2 x Green's Theorem in the Plane Double integrals over a plane region may be transformed into line integrals over the boundary of the region and conversely. This is of practical interest because it may simplify the evaluation of an integral. It also helps in the theory whenever we want to switch from one kind of integral to the other. The transformation can be done by the following theorem. THEOREM 1 Green's Theorem in the Plane4 (Transformation between Double Integrals and Line Integrals) Let R be a closed bounded region tsee Sec. 10.3) in the xy-plane whose boundary C consists offinitely many smooth curves (see Sec. 10.1). Let FI(x, y) and F2(x, y) befunctions that are continuous and have continuous partial derivatives aF I lay and aF2/ax everywhere in some domain containing R. Then (1) II ( R aF2 aFI ) 1 WI dx + F2 dy). - - - - - dx:dy = ax ay c r Here we integrate along the entire boundary C of R il1 such a sense that R is the left as we advance ill the direction of illlegratioll (see Fig. 232 on p. 440). 011 4GEORGE GREEN (1793-1841), English mathematician who was self-educated, started out as a baker, and at his death was fellow of Caius College, Cambridge. His work concemed potential theory in connection with electricity and magnetism, vibrations, waves, and elasticity theory. It remained almost unknown. even in England. until after his death. A "domain containing R" in the theorem guarantees that the assumptions about FI and F2 at boundary poims of R are the same as at other poims of R. CHAP. 10 440 Vector Integral Calculus. Integral Theorems y x Fig. 232. Region R whose boundary C consists of two parts: C1 is traversed counterclockwise, while C2 is traversed clockwise in such a way that R is on the left for both curves Setting F = [Fl' F 2 ] = Fli + F 2 j and using (1) in Sec. 9.9, we obtain (1) in vectorial form, I I (curl F)-k (1') dxdy f F-dr. = R C The proof follows after the first example. For if> see Sec. 10.1. E X AMP L E 1 Verification of Green's Theorem in the Plane Green's theorem in the plane will be quite important in our further work. Before proving it. let us get used to 2 it by verifying it for Fl = -,,2 - ?y, F2 = 2\")' + 2x and C the circle x + .1'2 = l. Solutioll. In (1) on the left we get II( ilF2 ~ R - iJFl) ~ dxdy = II . [(2)' + 2) - (2y - 7)ldxdy = 9 II dxdy = 911 R R since the circular disk R has area 7r. We now show that the line integral in (1) on the right gives the same value, 97r. We must orient C counterclockwise, say. ret) = [cos t, sin tl. Then r' (t) = [-sin t, cos tl. and on C, F2 = 2xy + 2x = 2 cos t sin t + 2 cos t. Hence the line integral in (1) becomes, verifYing Green's theorem, ~ c f f (2.,,- (FiX' + F 2 y') dt = [(sin 2 t - 7 sin t)( -sin t) + 2(cos t sin t + cos t)(cos t)J dt 0 2.,,- = 2 sin3 t + 7 sin2 t + 2 cos t sin t o = PROOF 0 + 2 cos 2 t) dt • + 711 - 0 + 27r = 97r. We prove Green's theorem in the plane. first for a special region R that can be represented in both forms a ~ x ~ b, ~ y ~ vex) (Fig. 233) p(y) ~ x ~ q(y) (Fig. 234). U(x) and c~)'~d, SEC. 10.4 441 Green's Theorem in the Plane y u(x) a Fig. 233. x b x Example of a special region Fig. 234. Example of a special region Using (3) in the last section, we obtain for the second term on the left side of (1) taken without the minus sign JJ-~ dx(~v = Jb[ J aF (2) ay R aF vex) a ] _1 u(x) ay dy lix (see Fig. 233). (The first term will be considered later.) We integrate the inner integral: JvCx) aFt IY=V(Xl dy ay u(x) = = FI(x. y) FI[X, V(X)] - FI[x, u(x)]. y=u(xl By inserting this into (2) we find (changing a direction of integration) I J__ aF 1 ay R dl: dy = I b FI[x. vex)] dx - a = - I b F1[x. u(x)] dx a f a J b F 1 [x, VeX)] lix - F 1 [x, u(x)] lix. a b Since y = vex) represents the curve C** (Fig. 233) and y = u(x) represents C*, the last two integrals may be written as line integrals over C** and C* (oriented as in Fig. 233); therefore, (3) JJ aF]ay dx dy = - f R FI(x. y) dx - C** = -~ c f FI(x. y) dx C* FI(x, y) dx. This proves (I) in Green's theorem if F2 = O. The result remains valid if C has portions parallel to the y-axis (such as C and Cin Fig. 235). Indeed, the integrals over these portions are zero because in (3) on the right we integrate with respect to x. Hence we may add these integrals to the integrals over C* and C** to obtain the integral over the whole boundary C in (3). We now treat the first term in (I) on the left in the same way. Instead of (3) in the last section we use (4), and the second representation of the special region (see Fig. 234). Then (again changing a direction of integration) 442 CHAP. 10 Vector Integral Calculus. Integral Theorems aF I I_2 ax R dx dy Id[ Iq(Y)aFax =I = _2 c dx ] dy p(y) d + F2(q(y), y) dy I c F2(P(Y), y) dy d c y y x Fig. 235. x Proof of Green's theorem Fig. 236. Proof of Green's theorem Together with (3) this gives (1) and proves Green's theorem for special regions. We now prove the theorem for a region R that itself is not a special region but can be subdivided into finitely many special regions (Fig. 236). In this case we apply the theorem to each subregion and then add the results; the left-hand members add up to the integral over R while the right-hand members add up to the line integral over C plus integrals over the curves introduced for subdividing R. The simple key observation now is that each of the latter integrals occurs twice. taken once in each direction. Hence they cancel each other, leaving us with the line integral over C. The proof thus far covers all regions that are of interest in practical problems. To prove the theorem for a most general region R satisfying the conditions in the theorem, we must approximate R by a region of the type just considered and then use a limiting process. For details of this see Ref. [GR4] in App. 1. • Some Applications of Green's Theorem E X AMP L E 2 Area of a plane Region as a Line Integral Over the Boundary In (I) we first choose Fl = 0, F2 = x and then Fl = -y, F2 = O. This gives II R dxdy = fXdY and C respectively. The double integral is the area A of R. By addition we have (4) A = 2.2 fc (xdy - ydr;) where we integrate as indicated in Green's theorem. This interesting formula expresses the area of R in terms of a line integral over the boundary. It is used, for instance, in the theory of certain planimeters (mechanical instruments for measuring area). See also Prob. 17. SEC lOA 443 Green's Theorem in the Plane For an ellipse x 2/0 2 + y2/b 2 = I or x = 0 cos t, Y = b sin t we get x' = - 0 sin t,),' = b cos t; thus from (4) we obtain the familiar fonnula for the area of the region bounded by an ellipse, 2w A E X AMP L E 3 = ' 2I0f(x)' 2'71'" , - yx ) dt = 2"If[ 0 ab cos2 t Area of a Plane Region in Polar Coordinates = r cos e. y = r sin e. Then Let rand e be polar coordinates defined by x dx = cos edr - rsin ede, and (4) dy = sin edr + rCos ede. becomes a formula that is well known from calculm.. namely, A = -I (5) 2 f' c r 2 de. As an application of (5), we consider the cardioid r = a(l - cos 2 A E X AMP L E 4 • - (-ob sin2] t) dt = Trab = -a 2 f e), where 0 ~ e~ 2Tr (Fig. 237). We find 2". (l - cos e)2 de = -3Tr 2 0 • 02 Transformation of a Double Integral of the Laplacian of a Function into a Line Integral of Its Normal Derivative The Laplacian plays an imponanl role in physics and engineering. A first impression of this was obtained in Sec. 9.7, and we shall discuss this further in Chap. 12. At present, let us use Green's theorem for deriving a basic integral formula involving the Laplacian. We take a function w(x. y) that is conti nuous and has continuous first and second partial derivatives in a domain of the xy·plane containing a region R of the type indicated in Green's theorem. We set FI = -aw/ay and F2 = aw/ax. Then aFI/ay and aF2 /ax are continuous in R. and in (I) on the left we obtain (6) the Laplacian ofw (see Sec. 9.7). Furthermore. using those expressions for FI and F2 • we get in (I) on the right (7) f·WI c dx + F2 dy) = f' (d.x + c f' ( dl' ) F2 -d' ds = sSe FI -d - aw d.x ay ds -;- - dV) + -au' --"ax ds ds where s is the arc length of C, and C is oriented as shown in Fig. 238. The integrand of the last integral may be written as the dot product (8) (grad wJon = aw] [dV [ aw ax ay ds - , -;- 0 --"-, - - d.x] aw dy aw d.x ds ax ds CJy ds' y y x x Fig. 237. Cardioid Fig. 238. Example 4 CHAP. 10 444 Vector Integral Calculus. Integral Theorems The vector n IS a untt normal vector to C, because the vector r' (s) = drlds = [dxlds, dvldsl is the unit tangent vector of C. and r' • n = O. so that n is perpendicular to r'. Also, n is directed to the exterior of C because in Fig. 238 the positive x-component dxlds of r' is the negative y-component of n. and similarly at other points. From this and (4) in Sec. 9.7 we see that the left side of (8) is the derivative of u' in the direction of the outward normal of C. This derivative is called the normal derivative of w and is denoted by awlall: that is. au-Iall = (grad w)· n. Because of (6), (7), and (8). Green's theorem gives the desired formula relating the Laplacian to the normal derivative. II V (9) 2 wdxdy = R fc awan ds. 2 For instance. \I" = x - y2 sati~fies Laplace's equation ,2w = O. Hence its nomml derivative integrated over a closed curve must give O. Can you verify this directly by integration. say. for the square 0 -<:: x -<:: 1. 0-<::)" -<:: I? • Green's theorem in the plane may facilitate the evaluation of integrals and can be used in both directions, depending on the kind of integral that is simpler in a concrete case. This is illustrated further in the problem set. Moreover, and perhaps more fundamentally, Green's theorem will be the essential tool in the proof of a very important integral theorem, namely, Stokes's theorem in Sec. 10.9. • ; ITM'=SE 'F-:)-O 11-121 _]1-== ____ EVALUATION OF LINE INTEGRALS BY GREEN'S THEOREM Using Green's theorem, evaluate f F(r)-drcounterclockwise c around the boundary curve C of the region R, where 1. F = l~XV4, ~x~l. R the rectangle with vertice~ (0. 0). (3, 0)' (3, 2), (0. 2) 2. F = [y sin x. 2x cos y]. R the square with vertices (0, 0), (~7T. 0). ~7T, !7T). (0. ~7T) 3. F = [_y3. x 3], C the circle x 2 + )"2 = 25 4. F = [-eY • eX]. R the triangle with vertices (0. 0). (2, 0). (2. l) 5. F = [e x + y • eX 0. 1),0.2) Y ]. R the triangle with vertices (0. 0). 6. F = [x cosh y. x 2 sinh y]. R: x 2 ~ y ~ x. Sketch R. 7. F = [x 2 x 2 - )'2], R: 1 ~ Y ~ 2 - x 2. Sketch + y2. 113--161 INTEGRAL OF THE NORMAL DERIVATIVE Using (9). evaluate 1, ~w ds counterclockwise over the Jc [In boundary curve C of the region R. 13. w = sinh x, R the triangle with vertices (0, 0), (2, 0), (2, 1) 14. w = t 2 + )'2, C: x 2 direct integration. 15. w = 2 In (x 2 + y2) 16. w = x 6y + xy6, + .\'2 = l. Confirm the answer by + xy3, R: 1 :;S )' ~ 5 R: x 2 + y2 ~ 4, " ~ 0 - x 2• X ~ 0 17. CAS EXPERIMENT. Apply (4) to figures of your choice whose area can also be obtained by another method and compare the results. 18. (Laplace's equation) Show that for a solution w(x, y) of Laplace's equation \,2U- = 0 in a region R with boundary curve C and outer unit normal vector n, R. 8. F x2 = [eX cosy. + y2 -ex siny]. R the semidisk ~ a 2• x ~ 0 (10) 9. F = grad (x 3 cos 2 (xv)), R the region in Prob. 7 10. F = [x In y. yeo,,]. R the rectangle with vertices (0. I). (3. 1), (3. 2). (0. 2) 11. F = [2\" - 3y. x + 5y]. R: 16x2 + 25."2 ~ 400. y ~ 0 12. F = [x~,2. -xly2]. R: I ~ x 2 + y2 ~ 4. x ~ 0, y ~ x. Sketch R. 1, = dw Jc w an ds. 19. Show that w = 2e x cos)' satisfies Laplace's equation V2 w = 0 and. using (0), integrate w(ilwldll) counterclockwise around the boundary curve C of the square 0 ~ x ~ 2, 0 ~ y ~ 2. SEC. 10.5 445 Surfaces for Surface Integrals 20. PROJECT. Other Forms of Green's Theorem in the Plane. Let Rand C be as in Green's theorem, r' a unit tangent vector. and n the outer unit normal vector of C (Fig. 238 in Example 4). Show that (1) may be written (11) I I div F dx dy R 10.5 = f or (12) I I(CUrIF)OkdXdy R = f For' ds C where k is a unit vector perpendicular to the xy-plane. Velify (11) and (12) for F = [7x, - 3)'] and C the circle x 2 + )'2 = 4 as well as for an example of your own choice. F n ds 0 C Surfaces for Surface Integrals Having introduced dquble integrals over regions in the plane, we turn next to surface integrals, in which we integrate over surfaces in space, such as a sphere or a portion of a cylinder. For this we must first see how to represent a surface. And we must discuss surface normals, since they are also needed in surface integrals. For simplicity we shall say "surface" also for a portion of a surface. Representation of Surfaces Representations of a surface S in xyz-space are z (1) f(x. y) = g(x, y, z) = or o. For example, z +Va 2 - x 2 - y2 or x 2 + y2 + Z2 - a 2 = 0 (z ~ 0) represents a hemisphere of radius a and center O. Now for cun'es C in line integrals. it was more practical and gave greater flexibility to use a parametric representation r = r(t). where a ~ t ~ b. This is a mapping of the interval a ~ t ~ b, located on the t-axis, onto the curve C (actually a portion of it) in x)'z-space. It maps every t in that interval onto the point of C with position vector ret). See Fig. 239A. yr(t) ~ec ~ x /d in space Z I/ n- y /' r(u,v) SurfaceS in space v -I 1 a b (t-axis) u (A) Curve Fig. 239. (E) Surface Parametric representations of a curve and a surface 446 CHAP. 10 Vector Integral Calculus. Integral Theorems Similarly, for surfaces S in surface integrals, it will often be more practical to use a parametric representation. Surfaces are two-dimensional. Hence we need two parameters, which we call u and v. Thus a parametric representation of a surface S in space is of the form r(u. v) (2) [x(u, v), y(u, v), .-:(u, v)] = x(u, v)i + y(u, v)j + :::(u, v)k where (u, v) varies in some region R of the uv-plane. This mapping (2) maps every point (u, v) in R onto the point of S with position vector r(u, v). See Fig. 239B. E X AMP L E 1 Parametric Representation of a Cylinder The circular cylinder x 2 + y2 = a 2 , - ) ;a Z "" ), has radius a, height 2, and the ~-axis as axis. A parametric representation is (Fig. 240). r(u. v) = [a cos 1/, asinu, vl = acosui + asinuj + vk The components of r are x = a cos u, y = a sin u, Z = v. The parameters u, v vary in the rectangle R: 0 ;a u "" 2'IT, -) ;a v ;a I in the uv-plane. The curves u = COllst are vertical straight lines. The curves v = COllst are parallel circles. The point P in Fig. 240 corresponds to 1/ = 'lT13 = 60°, v = 0.7. • z , p -------q-- (v = 1) \. \. ,P (v=O) ~ x (v =-1) Fig. 240. E X AMP L E 2 Fig. 241. Parametric representation of a cylinder Parametric representation of a sphere Parametric Representation of a Sphere A sphere x 2 + y2 + ~2 = a 2 can be represented in the form r(u, v) = a cos v cos (3) II i + a cos v sin 11 j + a sin v k where the parameters u, v vary in the rectangle R in the uv-plane given by the inequalities 0 ;a u "" 2'IT. - 'IT!2 ;a v ;a 'lT12. The components of r are x = a cos V cos u. y = a cos v sin u. Z = a sin v. The curves u = COllst and v = COllst are the "meridians" and "parallels" on S (see Fig. 241). This represel1lalion is used ill geography for measurillg the latitude and longitude of points 011 the globe. Another parametric representation of the sphere also used in mathematics is (3*) where 0 "" r(u, V) = 1/ "" 2'IT, 0 "" V "" 'IT. a cos 1/ sin v i + a sin u sin v j + a cos V k • SEC. 10.5 447 Surfaces for Surface Integrals EX AMP L E 3 Parametric Representation of a Cone A circular cone z = Yx 2 + r(u, in components x 2 Check that x + i, 0 ~ t ~ H can be represented by V) = [u cos V, u sin V, u1 = u cos V i + u sin V j + uk, = it cos V, y = u sin v,::: = u. The parameters vary in the rectangle R: 0 ~ u ~ )'2 = Z2, as it should be. What are the curves u = const and V = COllst? H. 0 ~ V :0; 2n. • Tangent Plane and Surface Normal Recall from Sec. 9.7 that the tangent vectors of all the curves on a surface S through a point P of S form a plane, called the tangent plane of S at P (Fig. 242). Exceptions are points where S has an edge or a cusp (like a cone), so that S cannot have a tangent plane at such a point. Furthermore, a vector perpendicular to the tangent plane is called a normal vector of S at P. Now since S can be given by r = r(u, v) in (2), the new idea is that we get a curve C on S by taking a pair of differentiable functions u = u(t), v = v(t) whose derivatives u' = dll/dt and v' = dv/dt are continuous. Then C has the position vector i(t) = r(u(t), vet)). By differentiation and the use of the chain rule (Sec. 9.6) we obtain a tangent vector of C on S di I i (t) = dl ar -u iJu I ar av + -v I Hence the partial derivatives ru and rv at P are tangential to Sat P. We assume that they are linearly independent, which geometrically means that the curves u = canst and v = canst on S intersect at P at a nonzero angle. Then r u and r v span the tangent plane of S at P. Hence their cross product gives a normal vector N of Sat P. (4) The corresponding unit normal vector n of S at P is (Fig. 242) (5) n= I lNI N = 1 Iru X rvl n s Fig. 242. Tangent plane and normal vector CHAP. 10 448 Vector Integral Calculus. Integral Theorems Also, if S is represented by g(x, y, z) = 0, then, by Theorem 2 in Sec. 9.7, I n= - - - gradg. Igrad gl (5*) A surface S is called a smooth surface if its surface normal depends continuously on the points of S. S is called piecewise smooth if it consists of finitely many smooth portions. For instance, a sphere is smooth, and the surface of a cube is piecewise smooth (explain!). We can now summarize our discussion as follows. Tangent Plane and Surface Normal THEOREM 1 If a suiface S is given by (2) with continuous ru = Br/rJu LInd rv = Br/CJv satisfying (4) at every point of S, then S has at every point P a unique tangent plane passing through P and spanned by ru and r v ' and a unique normal whose direction depends continuously on the points of S. A /lonnal vector is given by (4) and the corresponding unit /lonnal vector by (5). (See Fig. 242.) E X AMP L E 4 Unit Normal Vector of a Sphere From (5*) we find that the sphere g(x. y. n(x. y. z) = z) = x 2 [ X )' a a -, - + y2 z] . - a We see that n has the direction of the position vector [x, must be the case? E X AMP L E 5 + Z2 - x = - a a2 i 0 ha~ the unit normal vector , + -y a j z + - a k. y, z] of the corresponding point. Is it obvious that this • = 0 in Example 3, the unit nonnal vector n becomes Unit Normal Vector of a Cone At the apex of the cone g(x. y, z) = -z + undetermined because from (5*) we get Vf + i We are now ready to discuss surface integrals and their applications. beginning in the next section . --... ..-. ... 11-lOl PREPARATION FOR SURFACE INTEGRALS: PARAMETRIC REPRESENTATION, NORMAL Familiarize yourself with parametric representations of important surfaces by deriving a representation (1), by finding the parameter curves (curves II = eonst and u = eonst) of the surface and a nonmal vector N = r ll x rv of the surface. (Show the details of your work.) 1. xy-plane r(ll, u) = [u, Probs. 2-10) uJ (thus ui + uj; similarly in 2. xy-plane in polar coordinates r(u, u) = [u cos u, u sin u] (thus u = 3. Elliptic cylinder r(u, u) = [a cos u, b sin u, 4. Paraboloid of revolution r(u, u) = [u cos u, u sin u, 5. Cone r(u, U) = [au cos u, r, u = 0) u] u2] au sin u. 6. Hyperbolic paraboloid r(u, u) = [4u cosh u. u sinh u, u2 eu] J 7. Elliptic paraboloid r(u, u) = [3u cos u, 4u sin u, u2] SEC 10.6 8. 449 Surface Integrals Helicoid r(u, v) = [1/ cos v, v J. Explain the u sin v, 20. (Representation z = f(x,y)) Show that z = f(x, y) or g = z - flx, )'J = 0 can be written (fu = ilf/ilu. etc.) nanl.e. 9. Ellipsoid r(u. v) = [2 cos v cos u. 10. Ellipsoid r(Lt, v) = [a cos v cos Lt, 3 co~ v sin Lt. b cos v sin Lt, 4 sin vI (6) r(Lt, v) = [Lt, feLt. v)j v, N = gradg = and I-f,,, -fv, 1]. c sin vI 21. (Orthogonal parameters) Show that the parameter 11. CAS EXPERIMENT. Graphing Surfaces, Dependence on a, b, c. Graph the surfaces in Probs. 1-10. In Probs. 6--9 generalize the surfaces by introducing parameters a. b. c and then find out in Probs. 3-10 how the shape of the surfaces depends on a. h. c. curves II = const and v = COllst on a surface r(lI, v) are orthogonal (intersect at right angles) if and only if rU.-rv = o. 22. (Condition (4)) Find the points in Probs. 2-7 at which (4) N 0 does not hold and state whether this is owing to the shape of the surface or to the choice of the representation. 23. (Change of representation) Represent the paraboloid in Proh. 4 so that N(O, 0) 0, and show N. 24. PROJECT. Tangent Planes T(P) will be less important in our work, bur you should know how to represent them. ru rv) = 0 (a) If S: rILl, V), then T(P): (r* - r (a scalar triple product) or r*(p. q) = rIP) + pr,,(P) + qrvCP). lb) If S: g(x, y, z) = 0, then T(P): (r* - rtp) - vg = O. * 112-191 DERIVATION OF PARAMETRIC REPRESENTATIONS Find a parametric representation and a normal vector. (The answer gives one of them. There are many.) + y - 3z = 30 Plane 4x - 2y + 10.;: = 16 Sphere (x - 1)2 + (y + 2)2 + Z2 = Sphere Ix + 2)2 + y2 + (z - 2)2 = Elliptic paraboloid.;: = 4x2 + \.2 Parabolic cylinder z = 3)'2 * 12. Plane 5x 13. 14. 15. 16. 17. 18. Hyperbolic cylinder 9x 2 19. Elliptic cone z = - 25 I (c) If S: z = f(x, y), then T(P): z* - z = (x* - x)fx(P) 4.\'2 = 36 + (y* - y)fY(P)). Interpret (a)-(c) geometrically. Give two examples for (a), two for (b), and two for (c). V9x2 + y2 10.6 Surface Integrals To define a surface integral, we take a surface 5, given by a parametric representation as just discussed, (1) r(u, v) = Ix(u, v), y(u, v), z(u, v)] = x(u, v)i + y(u, v)j + z(u, v)k where (u. v) varies over a region R in the uv-plane. We assume 5 to be piecewise smooth (Sec. 10.5). so that 5 has a normal vector (2) and unit normal vector n= 1 TNfN at every point (except perhaps for some edges or cusps, as for a cube or cone). For a given vector function F we can now define the surface integral over 5 by (3) JJF-n dA JJFCrCu, v))-N(u, v) du dv. = S R 450 CHAP. 10 Vector Integral Calculus. Integral Theorems Here N = INln by (2), and INI = Iru x rvl is the area of the parallelogram with :-ides ru and r by the definition of cross product. Hence L " n dA = n INI dll du = N dll du. (3*) And we see that dA = INI du du is the element of area of S. Also F-n is the normal component of F. This integral arises naturally in flow problems, where it gives the flux across S (= mass of fluid crossing S per unit time; see Sec. 9.8) when F = pv. Here. p is the density of the fluid and v the velocity vector of the flow (example below). We may thus call the surface integral (3) the flux integral. We can write (3) in components, using F = [Fb F2 , F3]' N = [Nb N2 , N3]' and n = [cos a, cos f3, cos '}']. Here, a, f3, I' are the angles between n and the coordinate axes; indeed, for the angle between nand i, formula (4) in Sec. 9.2 gives cos a = noj/lnllil = noi, and so on. We thus obtain from (3) II (4) II = II Fon dA = s + (FI cos a F2 cos f3 + F3 cos '}') dA s (FIN] + F2N2 + F3 N 3) du du. R f3 dA = dz. dx. In (4) we can write cos adA = dy dz, cos becomes the following integral for the flux: (5) II Fon dA = s II (F] dy dz. + F2 d::. dx cos '}'dA = dx dy. Then (4) + F3 dx dy). s We can use this formula to evaluate surface integrals by converting them to double integrals over regions in the coordinate planes of the xy::.-coordinate system. But we must carefull) take into account the orientation of S (the choice of n). We explain this for the integrals of the F3 -terms, II (5') F3 cos '}'dA = s II F3 dx dy. s If the surface S is given by z = hex, y) with (x, y) varying in a region R in the xv-plane, and if S is oriented so that cos I' > 0, then (5') gives (5") II S F3 cos '}'dA = + II F3(x. y. hex. y) dxdy. R But if cos I' < 0. the integral on the right of (5") gets a minus sign in front. This follows if we note that the element of area dx d..v in the xy-plane is the projection Icos '}'I dA of the element of area dA of S; and we have cos I' = + Icos '}'I when cos I' > 0, but cos I' = -leos '}'I when cos I' < O. Similarly for the other two terms in (5). At the same time, this justifies the notations in (5). Other forms of surface integrals will be discussed later in this section. SEC. 10.6 451 Surface Integrals E X AMP L E 1 Flux Through a Surface 2 Compute the flux of water through the parabolic cylinder S: y = x , 0 "'" X "'" 2, 0 "'" ::: "'" 3 (Fig. 243) if the velocity vector is v = F = [3,=2, 6, 6x:::], speed being measured in meters/sec. (Generally, F = pv, but water 3 3 has the density p = I gmlcm = I ton/m .) Fig. 243. Solution. Writing x = [/ and z = v, Surface 5 in Example 1 we have y = x2 = S: [/2. Hence a representation of S is [u, r-?, v] r = (0 "'" [/ "'" 2, 0 "'" v "'" 3). By differentiation and by the definition of the cross product, N = r-u x rv = [1. 2u. x [0. O. 0] II ff f 3 = F'n dA s 0 = 2 (6uv 2 f = - 6) du dv 0 = 3 (3[/2V 2 (12v 2 - 12) dv = (4v 3 - 0]. = 6uv 2 - 6. By 12 - 6[/) dv U~O 0 3 -1. [2u. 2 [3v . 6. 6uv]. Hence F(S)' N = On S, writing simply F(S) for F[r(u. v)], we have F(S) integration we thus get from (3) the flux I] 12v) o 13 = 10~ - 36 = 72 [m3 /sec] v~O or 72 000 liters/sec. Note that the y-component of F is positive (equal to 6), so that in Fig. 243 the flow goes from left to right. Let us confirm this result by (5). Since N = INln = INllcos a, cos {3, cos 'YI = l2u, -I, 0] = I, [2x, 0] we see that cos a > 0, cos (3 < 0, and cos 'Y ~ O. Hence the second term of (5) on the right gets a minus sign, and the last term is absent. This gives. in agreement with the previous result. IIF' s E X AMP L E 2 ff 34 n dA = 00 ff 23 3,=2 dy dz - 6 dz ll\- 00 = f 3 0 2 4(3z ) dz - f 2 6· 3 dx = 4' 3 3 6' 3 . 2 - = 72. • 0 Surface Integral Evaluate (3) when F (Fig. 244). [x 2, 0, 3y2] and S is the portion of the plane x + x Fig. 244. Portion of a plane in Example 2 y + z 1 In the first octant 452 CHAP. 10 Vector Integral Calculus. Integral Theorems Solution. Writing x = " and y = v. we have z = I - x - y = I - II - v. Hence we can represent the plane x + y + Z = I in the form r(u. v) = [II, V. I - u - v]. We obtain the first-octdnt portion S of this plane by restricting = II and v = v to the projection R of S in the xy-plane. R is the triangle bounded by the two coordinate axes and the straight line x +}' = I, obtained from x + y + Z = I by setting z = O. Thus 0;;;; x;;;; I - y. 0;;;;)';;;; I. By inspection or by differentiation. x N Hence F(S)oN = fl? II FondA S = ru X rv = [I. o. = -11 x [0. I, -I] [I. 1, II. O. 3v2]o[l. I. I] = u 2 + 3v 2 . By (3). = JIc 1 u2 + 3v 2)dlldv = ff I-v 2 (u + 3V2)dudv ROO L[+ 1 = (I - V)3 2 + 3v (l - V)] dv = + • Orientation of Surfaces From (3) or (4) we see that the value of the integral depends on the choice of the unit normal vector D. (Instead of D we could choose -D.) We express this by saying that such an integral is an integral over an oriented surface S, that is, over a surface S on which we have chosen one of the two possible unit normal vectors in a continuous fashion. (For a piecewise smooth surface. this needs some further discussion, which we give below.) If we change the orientation of S, this means that we replace 0 with -D. Then each component of 0 in (4) is multiplied by -I, so that we have THEOREM 1 Change of Orientation in a Surface Integral The replacement ofn by - 0 (hence ofN by -N) corresponds to the lI1u/tipli("(ltion of the integral in (3) or (4) by -1. How do we effect such a change of N in practice if S is given in the form (l)? The simplest way is to interchange u and v, because then ru becomes rv and conversely, so that N = ru X rv becomes rv X ru = -ru X rv = -N, as wanted. Let us illustrate this. E X AMP L E 3 Change of Orientation in a Surface Integral In Example I we now repre~ent S by r = [v. v2 • 11],0;;;; V N = ru x r" = ;;;; 2. 0 ;;;; II ;;;; [0,0, I] x [I, 2v, 0] = [-2v, 1,0]. 2 For F = [3z • 6. 6x.::] we now get FCS) = [3u 2, 6. 6uv]. Hence FCS) the old result times - I, II 3. Then 0 N= -6u 2v + 6 and integration gives 3 2 3 F(S)'Ndvdu = ff 2 C-6u v + 6)dvd" = ROO f 2 (-1211 + 12)du = -72. • 0 Orientation of Smooth Surfaces A smooth surface S (see Sec. 10.5) is called orientable if the positive normal direction, when given at an arbitrary point Po of S, can be continued in a unique and continuous way to the entire surface. For smooth surfaces occurring in applications this is always true. SEC. 10.6 Surface Integrals 453 n s c s c i' I L (a) Smooth surface (b) Piecewise smooth surface Fig. 245. Orientation of a surface Orientation of Piecewise Smooth Surfaces Here the following idea will do it. For a smooth orientable smface S with boundary curve C we may associate with each of the two possible orientations of S an orientation of C, as shown in Fig. 245a. Then a piecewise smooth surface is called orientable if we can orient each smooth piece of S so that along each curve C* which is a common boundary of two pieces Sl and S2 the positive direction of C* relative to Sl is opposite to the direction of C* relative to S2' See Fig. 245b for two adjacent pieces; note the arrows along C*. Theory: Nonorientable Surfaces A sufficiently small piece of a smooth swface is always orientable. This may not hold for entire surfaces. A well-known example is the Mobius strip 5, shown in Fig. 246. To make a model, take the rectangular paper in Fig. 246. make a half-twist, and join the short sides together so that A goes onto A, and B onto B. At Po take a normal vector pointing, say. to the left. Displace it along C to the right (in the lower part of the figure) around the strip until you return to Po and see that you get a normal vector pointing to the right, opposite to the given one. See also Prob. 21. B I A A c Po 1 B Fig. 246. Mobius strip 5AUGUST FERDINAND MOBIUS (1790-1868). German mathemallcian, srudeI1l of Gauss, known for his work in surface theory, geometry, and complex analysis (see Sec. 17.2). 454 CHAP. 10 Vector Integral Calculus. Integral Theorems Surface Integrals Without Regard to Orientation Another type of surface integral is f f G(r) dA = f f G(r(u, v))IN(u, v)1 du £Iv. (6) S R Here dA = INI du dv = Iru x rei du dv is the element of area of the surface S represented by (1) and we disregard the orientation. We shall need later (in Sec. 10.9) the mean value theorem for surface integrals, which state~ that if R in (6) is simply connected (see Sec. 10.2) and G(r) is continuous in a domain containing R, then there is a point (uo, vo) in R such that ff (7) G(r) dA = G(r(uo, vo»)A (A = Area of S). S G As for applications, if G(r) is the mass density of S, then (6) is the total ma'>s of S. If = 1, then (6) gives the area A(S) of S, (8) = f fdA = f fir" x rvl du dv. A(S) S R Examples 4 and 5 show how to apply (8) to a sphere and a torus. The final example, Example 6. explains how to calculate moments of inertia for a surface. E X AMP L E 4 Area of a Sphere For a sphere r(lI. v) = [a cos v cos II, a cos v sin II, in Sec. 10.51 we obtain by direct calculation (verify!) a a Using cus2 l/ 2 sin v), cos 2 0 ~ u ~ . 27T, {l2 V Sin Ii. -7T/2 ~ v ~ 7T/2, [see (3) cos V sin uJ. + sin2 l/ = I and then cos2 v + sm2 V = 1. we obtain With this, (8) gives the familiar formula (note that leos vi = cos v when -7T/2 ~ v ~ 7T/2) f f 11"12 A(S) = a2 -r./2 E X AMP L E 5 211" Icos vi dll dv = 27Ta 2 0 f 11"12 cos v dv = 47Ta 2. -.".12 • Torus Surface (Doughnut Surface): Representation and Area A torus swface S is obtained by rotating a circle C about a straight line L in space so that C does not intersect or touch L but its plane always passes through L. If L is the ~-axis and C has radius b and its center has distance a (> b) from L, as in Fig. 247, then S can be represented by r(lI. v) = (a where 0 :;:: u :;:: 27T. 0 :;:: V ~ fu (a + b cos v) sin uj + b sin v k 27T. Thus = -(a fu = ru X ru + b cos v) cos II i + + bcosv)sinui + (a + hcosv)cosl/j -bSinVCOSlli - bsinvsinllj ~ b(lI + bcosV)(CO~llcosvi + bcosvk + sin llCOS vj + sinvk). SEC. 10.6 Surface Integrals 455 Hence Iru x [vi = b(a + b cos v). and (8) gives JJ 27T (9) A(S) = o the total area of the torus. 27T b(a 0 • + b cos v) dll dv = 4~ab. z c y 1 1 ~b~ 1 1 1 1 1 1 I: 1 1 1 4-ja~ 1 1 1 1 1 1\1 1.1 1 1 I y x Fig. 247. E X AMP L E 6 Torus in Example 5 Moment of Inertia of a Surface Find the moment of inertia / of a spherical lamina S: x 2 mass M about the .:-axis. + )'2 + :2 = a 2 of constant mass density and total Solutioll. If a mass is distributed over a surface S and fLeX, y, ;:) is the density of the mass (= mass per unit area), then the moment of inertia I of the mass with respect to a given axis L is defined by the surface integral (10) /= II fLD 2 dA s where D(x. y. z) is the distance of the point lx. y. ;:) from L. Since. in the present example. fL is constant and S has the area A = -I-7Ta 2 , we have fL = MIA = MI(47Tc?). For S we use the same representation as in Example 4. Then D2 = x 2 + y2 = a 2 cos2 v. Also, as in that example, dA = a 2 cos v dll dv. This gives the following result. [Tn the integration, use cos 3 v = cos v (1 - sin2 v).l /= 2 II S fLD dA = M ~ 47Ta Representations z = f(x,y). = y, r = [l/, v, f] gives I"'/2 J27ra4cos3vdlldv = -",/2 0 M~' 2I"'/2 cos3 vdv --../2 ?Ma 2 = 3 If a surface S is given by z = f(x, y), then setting l/ V INI = Iru x rvl = 1[I,O.f,,] x [0, 1,fv]1 = I[-fw -fv, 1]1 and, since fu (11) = fx, fv = fy, formula (6) becomes IIc(r)dA= I IC(X,Y,f(X,y)Jl S R* + (ilf)2 ax + ( -of )2 dxdy. ay • = x, CHAP. 10 456 Vector Integral Calculus. Integral Theorems R* y Fig. 248. Formula (11) Here R* is the projection of S into the _\y-plane (Fig. 248) and the normal vector N on S points up. If it points down, the integral on the right is preceded by a minus sign. From (11) with G = 1 we obtain for the area A(S) of S: z = j(x, y) the formula (12) A(S) ~ UJI + (:~)' + (~;.)' dxdJ where K:' is the projection of S into the xy-plane, as before. 11-121 FLUX INTEGRALS (3) f Fon dA 5 Evaluate these integrals for the following data. Indicate the kind of surface. (Show the details of your work.) 1. F = [2x, 5-", s: r 0]. 4u = [u, v, ~ 0, z + 3v]. 0~1I~1,-8~v~8 2. F = S: x [x 2, y2, +y +Z= 3. F [x - z. = s: r ,:2]. ~ 4, x y - = [u cos V, 0, y x. Z - y]. 1/ sin v, u], 0 ~ 1]4-20 1 0 ~ II ~ 3. 0 ~ v ~ 71" 4. F = leY, -e Z , eX], s: x 2 + .1'2 = 9, x ~ O. -" ;;:; O. 0 ~ z ~ 2 5. F = y, [x, z], S: r o ~ II ~ 4, = [1/ cos v, 7. F = [1, 8. F = [tanxy. 10. F S: 1, = [y2, z= sin v. 2 11 ], x 2y. -z], S: y2 = I, 1 ~ x ~ 4 0], + x 2, Z2 = a 2, x ~ 0, y ;;:; 0, Z ~ 0 Z4], 4Y~ + )2, 0 ~ z ~ 8, Y ~ 0 x 3, 11. F = [y3, S: x 2 + 4)'2 12. F = [coshy, = Z3], 4, x ~ 0, y ~ O. 0 ~ z ~ h 0, sinh x], ~z=x+~O~y~~O~x~1 G(r) dA Evaluate these integrals for the following data. Indicate the kind of surface. (Show the details.) + sinx, ~x+y+z=~x~~y;;:;~z~O 15. G = 5(x S: z = x + )' + z), + 2y, 0 ~ y ~ x, 0 ~ x ~ 2 + e', 16. G = .vex + S: x 2 + y2 = 16, Y ~ 0, 0 ~ 0 + ~::2 IJ xeY 1], S the sphere of radius 1 and center 0 x, y2 Z ~ SURFACE INTEGRALS (6) 14. G = cosy -71" ~ V ~ 71" 6. F = [cosh yz. O. )'4]. S: -,,2 + Z2 = 1. 0 ~ x ~ 20. 9. F = [0, s: x 2 + II 13. CAS EXPERIMENT. Write a program for evaluating sUiface integrals (3) that prints intennediate results (F, F ° N, the integral over one of the two variables). Can you experimentally obtain rules on functions and surfaces giving integrals that can be evaluated by the usual methods of calculus? Make a list of positive and negative results. 17. G = (x 2 + )'2 + z2f, S: Z ~ 4 Z = Vx 2 + y2, y ~ 0, 0~z~2 18. G = ax + by + cz, S: x 2 + y2 + Z2 = I, y ~ O. z ~ 0 19. G = arctan (v/x), S: z = x 2 + ),2, l ~ z ~ 9, x ~ 0, y ~ 0 20. G = 3x),. S: z = xy. 0 ~ x ~ 1. 0 ~ y ~ 1 21. (Fun with Mobius) Make Mobius strips from long slim rectangles R of grid paper (graph paper) by pasting the short sides together after giving the paper a halftwist. In each case count the number of parts obtained SEC. 10.6 Surface Integrals 457 by cutting along lines parallel to the edge. (a) Make R three squares wide and cut until you reach the beginning. (b) Make R four squares wide. Begin cutting one square away from the edge until you reach the beginning. Then cut the portion that is still two squares wide. (c) Make R five squares wide and cut similarly. (d) Make R six squares wide and cut. Formulate a conjecture about the number of parts obtained. (13) ds 2 = E du 2 + 2F du dv + G dv 2 with coefficients 22. (Center of gravity) Justity the following formulas for is called the first fundamental form of S. (E, F, G are standard notations that have nothing to do with F and G that occur at some other places in this chapter.) The first fundamental form is basic in the theory of surfaces, since with its help we can determine lengths, angles, and areas on S. To show this, prove the following. the mass M and the center of gravity (x, y, Z) of a lamina S of density (mass per unit area) u(x, y, z) in space: (a) For a curve C: u = u(t), v = vet), a ~ t ~ b, on S, formulas (10), Sec. 9.5, and (14) give the length APPLICATIONS II :v = ~ II M= x= udA. s ~ If.rudA , s Z= yudA, ~ s II I = = for the moments of inertia of the lamina in Prob. 22 about the x-, y-. and ::;-axes. respectively: v'r'(t).r'(t) dt b YEu'2 Iz = II (x 2 + y2)udA. 24. Find a fonnula for the moment of inertia of the lamina in Prob. 22 about the line y = x, ::; = O. Find the moment of inertia of a lamina S of density 1 about an axis A, where x2 + y2 = I, 26. S as in Prob. 25. 2 27. S: x + y2 = Z2, 0 ~ + Gv'2dt. cos 'Y = s s 25. S: 2Fu'v' (b) The angle 'Y between two intersecting curves C r : u = gUY, v = h(t) and C 2 : u = p(t), v = q(t) on S: r(u, v) is obtained from (16) s + a s 23. (Moments of inertia) Justify the following formulas b a (15) zudA. I I where a = rug' + rvh' and b tangent vectors of C r and C 2 . (c) The square of the length of the normal vector N can be written so that formula (8) for the area A(S) of S becomes z ~ h, A: the z-axis A: the line::: = h/2 in the xc-plane 0 ~ Z ~ h, A: the z-axis 28. (Steiner's theorem6 ) If IA is the moment of inertia of a mass distribution of total mass M with respect to an axis A through the center of gravity, show that its moment of inertia IB with respect to an axis E, which is parallel to A and has the distance k from it. is = rup' + rvq' are A(S) = II I I INI II Y dA = S (18) = du dv R EG - F2 du dv. R (d) For polar coordinates u (= r) and v (= 8) defined by x = u cos v, y = u sin v we have E = 1, F = O. G = u 2 , so that ds 2 = du 2 + u 2 dv 2 = dr 2 + r2 d8 2. 29. Using Steiner's theorem, find the moment of inertia of S in Prob. 26 about the x-axis. Calculate from this and (18) the area of a disk of radius a. 30. TEAM PROJECT. First Fundamental Form of a Surface. Given a surface S: r(lI, v), the corresponding quadratic differential fonn (e) Find the tirst fundamental fOlm of the torus in Example 5. Use it to calculate the area A of the torus. Show that A call also be obtained by the theorem of 6JACOB STEINER (1796-1863), Swiss geometer, born in a small village, learned to write only at age 14, became a pupil of Pestalozzi at 18. later studied at Heidelberg and Berlin and, finally, because of his outstanding research, was appointed professor at Berlin University. 458 CHAP. 10 Vector Integral Calculus. Integral Theorems Calculate the first fundamental form for the usual representations of important surfaces of your own choice (cylinder, cone, etc.) and apply them to the calculation of length~ and areas on these ~urfaces. Pappos,7 which states that the area of a surface of revolution equals the product of the length of a meridian C and the length of the path of the center of gravity of C when C is rotated through the angle In. 10.7 (I) Triple Integrals. Divergence Theorem of Gauss In this section we discuss another "big" integral theorem, the divergence theorem, which transforms surface integrals into triple integrals. So let us begin with a review of the latter. A triple integral is an integral of a function f(x, y, z) taken over a closed bounded (three-dimensional) region T in space (where "clo!o.ed" and "bounded" are defined as in footnote 2 of Sec. 10.3, with "sphere" substituted for "circle"). We subdivide T by planes parallel to the coordinate planes. Then we consider those boxes of the subdivision (rectangular parallelepipeds) that lie entirely inside T, and number them from I to n. In each such box we choose an arbitrary point, say, tXk, Yk, z,.J in box k. The volume of box k we denote by Ll V k . We now form the sum n in 2: = f(xk, Yk, Zk) .1 Vk · k~l This we do for larger and larger positive integers 11 arbitrarily but so that the maximum length of all the edges of those 11 boxes approaches zero as II approaches infinity. This gives a sequence of real numbers in}' Jn2 , . . . . We assume that f(x, Y, z) is continuous in a domain containing T, and T is bounded by finitely many smooth sU/jaces (see Sec. 10.5). Then it can be shown (see Ref. [GR4] in App. I) that the sequence converges to a limit that is independent of the choice of subdivisions and corresponding points (Xk, Yk, Zk)' This limit is called the triple integral of f(x, y, .;:) orer the region T and is denoted by III f(x, y, z) dx dy d.;: or by T III f(x, y, .;:) dV. T Triple integrals can be evaluated by three successive integrations. This is similar to the evaluation of double integrals by two successive integrations, as discussed in Sec. LO.3. An example is shown below (Example 1). Divergence Theorem of Gauss Triple integrals can be transformed into surface integrals over the boundary surface of a region in space and conversely. Such a transformation is of practical interest because one of the two kinds of integral is often simpler than the other. It also helps in establishing fundamental equations in fluid flow, heat conduction, etc .. as we shaH see. The transformation is done by the divergence theorem. which involves the divergence of a vector function F = [FI, F 2 , F 3 ] = F1i + F 2 j + F3k, namely, 7PAPPUS OF ALEXANDRIA (about A.D. 300), Greek mathematician. The theorem is also called Guldin's theorem. HABAKUK GULDIN (1577-1643) was born in St. Gallen, Switzerland. and later became professor in Graz and Vienna. SEC 10.7 459 Triple Integrals. Divergence Theorem of Gauss (Sec. 9.8). (1) Divergence Theorem of Gauss (Transformation Between Triple and Surface Integrals) THEOREM 1 Let T be a closed bounded region in space whose bOl/ndary is a piecewise smooth orielltable sll1face S. Let F(x, y, ;:) be a vector function that is cOlltinllous and has continllolls first partial derivatives ill some domain containing T. Then f f f div F dV = f f Fen dA. (2) s T = [Fl , F 2 , Fg] and of the (Juter unit 1101711al vector In components ofF n = [cos a, cos f3, cos y] of S (as in Fig. 250), formula (2) becomes f f f( T iJ~2 + aFI + ax rJy = (2*) fI g iJF az (FI cos a ) + eLl: d y d::. F2 cos f3 + Fg cos y) dA s f f(F = 1 dy dz + F2 dzdx + Fg dxdy). s The proof follows after Example 1. "Closed bounded region'" is explained above. "piecewise smooth orientable" in Sec. I 0.5, and "domain containing T" in footnote 4, Sec. 10.4, for the two-dimensional case. EXAMPLE Evaluation of a Surface Integral by the Divergence Theorem Before we prove the theorem. let us show a typical application. Evaluate z b I I I ff I= 3 2 (x dy dz + x 2y d: ell: + x : dt dy) S _----T----, where S is the closed surface in Fig. 249 consisting of the cylinder x 2 + disks::: = 0 and: = b (x 2 + y2 ~ a 2 ). x Solution. - -....l. Fig. 249. Surface 5 in Example 1 a (0 ~ ~ ~ b) and the circular 2 Hence div F = 3x2 + x2 + x 2 = 5x 2 . The form of the surface suggests that we introduce polar coordinates r. edefined by x = r cos e. y = r sin e fthu~ cylindrical coordinates e. :). Then the volume element is = and we obtain F1 = x 3 • F2 = x 2 )". )"2 = r. F3 = .\"2:;:. dl: dy dz rdr dl:J d:. b /= fffSX2dXdYdZ= f T Z~O f 27T 6~O a f (Sr cos e)rdrdl:Jdz 2 2 T~O • 460 CHAP. 10 PROOF Vector Integral Calculus. Integral Theorems We prove the divergence theorem, beginning with the first equation in (2*). This equation is true if and only if the integrals of each component on both sides are equal; that is, (3) IIIo.F1 dxdydz= IIF1 cos adA, T ox s (4) III T (5) III T O~2 f3 dA, dx dy dz = I I F2 cos s 0) aF3 dxdydz = II F3cosydA. s 7 o~ We first prove (5) for a special region T that is bounded by a piecewise smooth orientable surface S and ha~ the property that any straight line parallel to anyone of the coordinate axes and intersecting T has at most aile segment (or a single point) in common with T. This implies that T can be represemed in the form (6) g(x, y) ~ z ~ /z(x, y) where (x, y) varies in the orthogonal projection R of T in the xy-plane. Clearly, R(X, y) represents the "bottom" S2 of S (Fig. 250), whereas z = hex, y) represents the "top" Sl of S, and there may be a remaining vertical portion S3 of S. (The portion S3 may degenerate into a curve, as for a sphere.) To prove (5), we use (6). Since F is continuously differentiable in some domain containing T, we have z= . II1 --:-::- dxdydz of (7) 0<0 T [ = I I R h<x, y) I g(x, y) "JF ~ dz ] OZ dxdy. Integration of the inner integral [ ... ] gives F3 [x, y, hex, y)] - F 3 [x, y, glx, y)]. Hence the triple integral in (7) equals (8) I I F3[x, y, hex, y)] dx dy - I I F3[x, y, g(x, y)] dx dy. n'11 ~/Sl z t Y~~ , 1 fJ n 1 1I~I1 x Fig. 250. 11 11_ _ 1 1 1 Example of a special region Y SEC. 10.7 Triple Integrals. Divergence Theorem of Gauss 461 But the same result is also obtained by evaluating the right side of (5); that is [see also the last line of (2*)], JJFgcOS ydA = JJFgdxdy s s = + JJ Fg[x, y, hex, y)] dx dy - JJ Fg[x, y. g(x, y)] dx dy, where the first integral over R gets a plus sign because cos y> 0 on S1 in Fig. 250 [as in (5"), Sec. 10.6], and the second integral gets a minus sign because cos y < 0 on S2' This proves (5). The relations (3) and (4) now follow by merely relabeling the variables and using the fact that, by assumption, T has representations similar to (6). namely, g(.\', z) ~ x ~ hey, g(z, x) ~ y ~ h(z, x). and z) This proves the first equation in (2*) for special regions. It implies (2) because the left side of (2*) is just the definition of the divergence, and the right sides of (2) and of the first equation in (2*) are equal, as was shown in the first line of (4) in the last section. Finally, equality of the right sides of (2) and (2*), last line, is seen from (5) in the last section. ThIS establishes the divergence theorem for special regions. For any region T that can be subdivided into finitely many special regions by means of auxiliary surfaces. the theorem follows by adding the result for each part separately; this procedure is analogous to that in the proof of Green's theorem in Sec. LO.4. The sUlface integrals over the auxiliary surfaces cancel in pairs, and the sum of the remaining surface integrals is the surface integral over the whole boundary surface S of T; the triple integrab over the parts of T add up to the triple integral over T. The divergence theorem is now proved for any bounded region that is of interest in practical problems. The extension to a most general region T of the type indicated in the theorem would require a certain limit process: this is similar to the situation in the case of Green's theorem in Sec. 10.4. • E X AMP L E 2 Verification of the Divergence Theorem Evaluate JJ(7xi - ;:k)"n dA over the sphere S: x 2 + y2 + Z2 = 4 (a) by (2). (b) directly. s 3 = div [7x, 0, -z] = div [7xi - zk] = 7 - J = 6. Answer: 6' (4J3)1T' 2 = 641T. (b) We can represent S by (3). Sec. 10.5 (with a = 2). and we shall use n dA = N dlt dv [see (3*), Sec. 10.61. Accordingly, Solution. (a) div F S: Then r = [2 cos v cos ru= l-2co$vsinll. rv N Now on S we have x = r ll x r,. = 2 cos v sin II. 2cosvcosu, [-2sinvcoslI, - 2sinvsinu. = [4 cos 2 v cos II, 2 4 cos v sin II. = 2 cos v cos II, Z = 2 sin v, so that F = [7x. F(S) and It. F(S)"N = [14cosvco~lI. O. 2 sin v]. 0] 2 cos v] 4 cos v sin vJ. O. -;oj becomes on S -2 sin v] = (\4 cos v cos 1I)'4cos2 v cos II = 56 cos3 v cos 2 u - 8 cos v sin 2 v. + (-2 sin v)'4 cos v sin v 462 CHAP. 10 Vector Integral Calculus. Integral Theorems On S we have to integrate over II from 0 to 277". This gives 77"' 56 cos 3 v - 277"' 8 cos v sin2 v. The integral of cos v sin2 v equal~ (sin3 v)/3, and that of cos 3 v = cos v (I - sin2 v) equals sin v - (sin 3 v)13. On S we have -77"/2 ;'" v ;'" 77"12, so that by sub~tituting these limits we get 5677"(2 - 2/3) - ) 677"' 2/3 = 6477" a~ • hoped for. To see the point of Gauss's theorem, compare the amounts of work. Coordinate Invariance of the Divergence. The divergence (I) is defined in terms of coordinates, but we can use the divergence theorem to show that div F has a meaning independent of coordinates. For this purpose we first note that triple intgrals have properties l/uite similar to thuse of double integrals in Sec. 10.3. In particular. the mean value theorem for triple integrals asserts that for any continuous function f(x, y, .::) in a bounded and simply connected region T there is a point Q: (xo, )'0, <'0) in T such that III (9) f(x, y, z) dV = f(xo, Yo, '::0)V(T) (V(l) = volume of T). T In this formula we interchange the two sides, divide by veT), and set f = div F. Then by (he divergence theorem we obtain for the divergence an integral over the boundary surface SeT) of T, (10) div F(xo, Yo, '::0) = _1_ veT) III div F dV = _1_ sm I IFon dA. veT) T We now choose a point P: (Xl> ,vI' ZI) in T and let T shrink down onto P so that the maximum distance den of the points of T from P goes to zero. Then Q: (xo. )'0' 20) must approach P. Hence (10) becomes (11) divF(P)=lim d(T)->O -I-IIFOUdA. VeT) SeT) This proves THEOREM 2 Invariance of the Divergence The divergence of a vector function F with cOlltinuolls first partial derivatives in a region T is independent of the particular choice of Cartesian coordinates. For any Pin T it is given by (II). Equation (l1) is sometimes used as a definition of the divergence. Then the representation (1) in Cartesian coordinates can be derived from (11). Further applications of the divergence theorem follow in the problem set and in the next section. The examples in the next section will also shed further light on the nature of the divergence. SEC. 10.8 Further Applications of the Divergence Theorem .------ C .. C_i, _ ••• APPLICATION OF TRIPLE INTEGRALS: MASS DISTRIBUTION 11-81 14. The paraboloid Find the total mass of a mass distribution of density u in a region T in space. (Show the details of your work.) 1. u 2. u Ixl kl x2y2~2, T the box ~ a, iYI ~ b, ~ c 2 = x + y2 + ~2, T the box 0 ~ x ~ 4, 0 ~ J ~ 9, = O~:::~l = 5. u 6. u = ~(X2 7. u = + y2)2, T the cylinder x 2 + )'2 ~ 4. Izl ~ 2 = 30::;. T the region in the first octant bounded by y = 1 - x 2 and z = x. Sketch it. + Y + ~2, 1 T the cylinder ,,2 + ::.2 ~ 9, l~x~9 8. u = x 2 APPLICATION OF TRIPLE INTEGRALS: MOMENT OF INERTIA 19-141 Ix = + y2. T the ball x 2 + y2 + :;2 ~ a 2 JJJ (y2 + z2) dx dy dz of a mass of density 1 in T a region T about the x-axis. Find Ix when T is as follows. 9. 10. 11. 12. 13. The cube 0 ~ x ~ a, 0 ~ y ~ a, 0 ~ z ~ The box 0 ~ x ~ 1I, -bt2 ~)' ~ bt2. -el2 The cylinder y2 + :;2 ~ c?, 0 ~ x ~ Iz + .\'2 + ~2 ~ a 2 The cone y2 + ~2 ~ x 2 • 0 ~ x The ball x 2 10.8 ~ Iz a ~ ~ ~ el2 y2 + Z2 ~ X, 0 ~ x ~ h 1 (h 15. Show that for a solid of revolution, f" = 27T L r\x) dx. U~t: this to solve Probs. 11-14. 0 16. Why is Tx in Prob. 13 for large h larger than I,,, in Prob. 14? Why is it smaller for h = I? Give physical reason. 117-251 sin x cos y, T: 0 ~ x ~ ~7T. ~7T - X ~ Y ~ ~7T, o ~ ~ ~ 12 4. u = e-"'-Y-z, Tthe tetrahedron with vertices (0. O. 0). (2, O. a), (0, 2. a). (0, 0, 2) 3. u 463 APPLICATION OF THE DIVERGENCE THEOREM: F· n dA SURFACE INTEGRALS s JJ Evaluate this integral by the divergence theorem. (Show the details.) ;::], S the sphere x 2 + y2 + ::;2 = 9 18. F = [4x, 3z, 5y]. S the surface of the cone x 2 + y2 ~ :;2. 0 ~ ~ ~ 2 19. F = [z - y y3, 2;::3]. S the surface of y2 + Z2 ~ 4, -3 ~x~ 3 20. F = [3x)' 2, yx 2 - y3, 3zx2], S the surface of 2 t + y2 ~ 25. 0 ~ Z ~ 2 21. F = [sin y, cos x. cos ;::], S tile surface of x 2 + )'2 ~ 4. Izl ~ 2 22. F = [x 3 - .\'3, )'3 - Z3, ;::3 - x3 ]. S the surface of x 2 + y2 + ;::2 ~ 25, z :0=: 0 23. F = [4x 2 • 2x + y2. x 2 + Z2]. S the surface of the tetrahedron in Prob. 4 24. F = [4x 2 • y2. -2 cos m:], S the surface of the tetrahedron with vertices (0, O. a). (l. O. a). (0. l. m. (0,0, 1) 25. F = [5x 3 . 5y 3, 5;::3], S: x 2 + y2 + Z2 = 4 17. F = [x, y, Further Applications of the Divergence Theorem We show in this section that the divergence theorem has basic applications in fluid flow, where it helps characterize sources and sinks of fluid, in /zeat flo~r, where it leads to the basic heat equation, and in potential theory, where it gives basic properties of the solutions of Laplace's equation. Here the region T and its boundary surface S are assumed to be such that the divergence theorem applies. E X AMP L E 1 Fluid Flow. Physical Interpretation of the Divergence From the divergence theorem wc may obtain an intuitive interpretation of the divergence of a vector. For this purpose we consider the flow of an incompressible fluid (see Sec. 9.8) of constant density p = I which is steady. that is. does not vary with time. Such a flow is determined by the field of its velocity vector yep} at any poimP. 464 CHAP. 10 Vector Integral Calculus. Integral Theorems Let S be the boundary surface of a region T in space, and let n be the outer unit normal vector of S. Then von is the normal component of v in the direction of n, and Ivon dAI is the mass of fluid leaving T (if von> 0 at some P) or enterillg T (if von < 0 at P) per unit time at some point P of S through a small portion 6.S of S of area 6.A. Hence the total mass of fluid that flows across S from T to the outside per unit time is given by the surface integral II vondA. s Division by the volume Vof T give, the average flow out of T: (1) Since the flow is steady and the fluid is incompressible. the amount of fluid flowing outward must be continuously supplied. Hence. if the value of the integral (I) is different from zero, there must be sources (positive sources and negat;,'e sources. called sinks) in T. that is, points where fluid is produced or disappears. If we let T shrink down to a fixed point P in T, we obtain from (I) the source intensity at P given by the right side of (11) in the last section with F n replaced by von, that is, 0 (2) div vlP) =d(T)~O lim _1_ V(1) IIvon dA. SeT) Hence the dil'erge1lce of the "e/ocity ,'ector v of a steady incompressible floll' is the source intensit--.- of the flow at the correJoponding point. There are no sources in T if and only if div v is zero everywhere in T. Then for any closed surface S in T we have I IvondA = o. s E X AMP L E 2 • Modeling of Heat Flow. Heat or Diffusion Equation Physical experiments show that in a body, heat flows in the direction of decreasing temperature, and the rate of flow is proportional to the gradient of the temperature. This means that the velocity v of the heat flow in a body is of the form (3) v = -Kgrad V where V(x, y, z. t) is temperature, t is time. and K is called the thermal conductil'ity of the body: in ordinary physical circumstances K is a constant. Using this information, set up the mathematical model of heat flow, the so-called heat equation or diffusion equation. Solution. Let T be a region in the body bounded by a surface S with outer unit normal vector n such that the divergence theorem applies. Then von is the component of v in the direction of n. and the amount of heat leaving T per unit time is IIvondA. s This expression is obtained similarly to the corresponding surface integral in the last example. Using (the Laplacian; see (3) in Sec. 9.8), we have by the divergence theorem and (3) I IvondA = -K I I IdiV(grad U)dxdyd::. S T (4) = -K I I T 2 I V Vdxdyd::.. SEC. 10.8 465 Further Applications of the Divergence Theorem On the other hand, the total amount of heat H in T is H = JJJ apU d:'(dydz T where the constant u is the specific heat of the material of the body and p is the density volume) of the material. Hence the time rate of decrease of H is _ aH at and thi~ = _ JJJu pau T (= mass per unit d.ydvdz at' must be equal to the above amount of heat leaving T. From (4) we thus have or JJJ(up:)~ - K\' 2 U ) dxdydz = O. T Since this holds for any region T in the body, the integrand (if continuous) must be zero everywhere; that is, (5) c2 K =-- up where c 2 is called the thermal diJfusil'ity of the material. This partial differential equation is called the heat equation. It is the fundamental equation for heat conduction. And our derivation is another impressive demonstration of the great importance of the divergence theorem. Methods for solving heat problems will be shown in Chap. 12. The heat equation is also called the diffusion equation because it also models diffusion processes of motIOns of molecules tending to level off differences in den,ity or pressure in gases or liquids. If heat flow does not depend on time, it is called steady-state heat flow. Then aUlat = 0, so that (5) reduces to Laplace's equation 'iJ 2U = O. We met this equation in Secs. 9.7 and 9.8, and we shall now see thaI the divergence theorem adds basic insights into the nature of solutions of this equation. • Potential Theory. Harmonic Functions The theory of solutions of Laplace's equation (6) is called potential theory. A solution of (6) with continuous second-order partial derivatives is called a harmonic function. That continuity is needed for application of the divergence theorem in potential theory, where the theorem plays a key role that we want to explore. Further details of potential theory follow in Chaps. 12 and 18. E X AMP L E 3 A Basic Property of Solutions of Laplace's Equation The integrands in the divergence theorem are div F and F' n (Sec. 10.7). If F is the gradient of a scalar function, say. F = grad f, then div F = div tgrad f) = 'iJ2f ; see (3). Sec. 9.8. Also, F' n = n' F = n' grad f. TIris is the directional derivative of f in the outer normal direction of S. the boundary surface of the region T in the theorem. This derivative is called the (outer) normal derivative of f and is denoted by aflan. Thus the formula in the divergence theorem becomes CHAP. 10 466 Vector Integral Calculus. Integral Theorems (7) This is the three-dimensional analog of (9) in Sec. 10.4. Because of the assllmptions in the divergence theorem this gives the following result. • THEOREM 1 r I X AMP L E 4 A Basic Property of Harmonic Functions Let f(x, y, z) be a harmonic function in some domain D is space. Let S be any piecewise smooth closed orientable st/1jace in D whose entire region it encloses belongs to D. Then the integral of the nonna/ derivative of f taken over S is -;ero. (For "piecewise smooth" see Sec. 10.5.) Green's Theorems Let f and g be scalar functions such that F = f grad g satisfies the assumptions of the divergence theorem in some region T. Then div F = div (f grad g) = iJg iJg iJg div ([ f -;- . f -;- . f -;iJx iJy iJz J) Also, since f is a scalar function, Fon = noF = no(fgradg) = (n grad g)f. 0 Now n° grad g is the direcl10nal derivative ag/iJll of g in the outer normal direction of S. Hence the formula in the divergence theorem becomes "Green's first formula" (8) JJJ(fV2 T g + grad f-grad g) dV = JJf an dA. iJg S Formula (8) together with the assumptions is known as thefirstform of Greel1's theorem. Interchanging f and g we obtain a similar formula. Subtracting this formula from (8) we find (9) This formula is called Green's second formula or (together with the assumptions) the secolldform ofGreell's theorem. • SEC. 10.8 467 Further Applications of the Divergence Theorem E X AMP L E 5 Uniqueness of Solutions of Laplace's Equation Let I be harmonic in a domain D and let I be zero everywhere on a piecewise smooth closed orientable surface S in D whose entire region T it encloses belongs to D. Then V 2 g is zero in T. and the surface integral in (8) is zero, so that (8) with g = I gives IIJ grad I . grad I dV = IJI T Igrad 112 dV = O. T Since I is harmonic, grad I and thus Igrad II are continuous in T and on S, and since Igrad II is nonnegative, to make the integral over T zero. grad I must be the zero vector everywhere in T. Hence Ix = I y = I z = O. and f is constant in T and, because of continuity, it is equal to its value 0 on S. This proves the following theorem. THEOREM 2 Harmonic Functions Let j"(x, y, z) be harmonic in some dOll/ain D and zero at eVel)' point of a piecewise smooth closed orientable suiface S in D whose entire region T it encloses belongs to D. Then f is identically zero in T. This theorem has an important conseq LIenee. Let II and 12 be functions that satisfy the assumptions of Theorem I and take on the same values On S. Then their difference II - 12 satisfies those assumptions and has the value o everywhere on S. Hence, Theorem 2 implies that II -h=O throughout T, and we have the following fundamental result. THEOREM 3 Uniqueness Theorem for laplace's Equation Let T be a region that satisfies the assumptions of the divergence theorem, and let f(.\", y, z) be a hal11lOnic function in a domain D that contains T and its /JoundGl)' surface S. Then f is uniquely detennined in T by its values on S. The problem of determining a solution u of a partial differential equation in a region T such that u assumes given values on the boundary surface S of Tis called the Dirichlet problem.8 We may thus reformulate Theorem 3 as follows. THEOREM 3* Uniqueness Theorem for the Dirichlet Problem if the assumptions in Theorem 3 are satisfied and the Dirichlet problem for the Laplace equation has a solution in T, then this solution is unique. These theorems demonstrate the extreme importance of the divergence theorem in potential theory. • 8PETER GUSTAV LEJEUNE DIRICHLET il805-1859), German mathematician, studied in Paris LInder Cauchy and others and sLlcceeded Gauss at G6ttingen in 1855. He became known by his important research on Fourier series (he knew Fourier personally) and in number theory. 468 CHAP. 10 Vector Integral Calculus. Integral Theorems 1. (Hannonic functions) Verify Theorem 1 for f = 2x2 + 2y2 - 4z 2 and S the surface of the cube o~ x 2. 3. 4. 5. 6. I, 0 ~ y ~ 1, 0 ~ z ~ 1. (Hannonic functions) Verify Theorem 1 for f = y2 - x 2 and the surface of the cylinder x 2 + y2 ~ I, 0 ~ z ~ 5. (Green's first formula) Verify (8) for f = 3y2, g = x 2 , S the surface of the cube in Prob. I. (Green's first formula) Verify (8) for f = x, g = y2 + ;:2. S the surface of the box 0 ~ x ~ 1, o ~ Y ~ 2, 0 ~ z ~ 3. (Green's second formula) Verify (9) for the data in Prob.3. (Green's second formula) Verify (9) for f = x4, g = y2 and the cube in Prob. l. = S ~ = II (a) :! dA = I IIlgrad gl2 dV. T S II (f (c) tP 8. Find the volume of a ball of radius a by means of the formula in Prob. 7. 9. Show that a region T with boundary surface S has the volume S Of) dA = O. og - g on 011 (d) If ofIan = fJglon on S, then f = g c is a constant. + c in T, where (e) The Laplacian can be represented independently of coordinate systems in the form v2 = lim _1_ f d(T)~O VeT) IIXdydz JI of dA an S(T) where d(T) is the maximum distance of the points of a region T bounded by SeT) from the point at which the Laplacian is evaluated and veT) is the volume of T. S = g S (b) If aglan = 0 on S, then g i8 constant in T. where r is the distance of a variable point P: (x, y, z) on S from the origin 0 and is the angle between the directed line OP and the outer normal of Sat P.(Make a sketch.) V= II(XdydZ + ydzdx + zdxdy). 10. TEAM PROJECT. Divergence Theorem and Potential Theory. The importance of the divergence theorem in potential theory is obvious from (7)-(9) and Theorems I - 3. To emphasize it further, consider functions f and g that are harmonic in some domain D containing a region Twith boundary surface S such that T satisfies the assumptions in the divergence theorem. Prove and illustrate by examples that then: ~ IJrcostPdA 3 ~ S 7. (Volume as a surface integral) Show that a region T with boundary surface S has the volume V= IIZdxdy IIyd::dx S 10.9 Stokes's Theorem Having seen the great usefulness of Gauss's divergence theorem, we now tum to the second "big" theorem in this chapter, Stokes's theorem. This theorem transforms line integrals into surface integrals and conversely. Hence it generalizes Green's theorem of Sec. 10.4. Stokes's theorem involves the curl j (1) curl F = a/ax k (see Sec. 9.9). SEC.10.9 469 Stokes's Theorem THEOREM 1 Stokes's Theorem 9 {Transformation Between Surface and Line Integrals} Let S be a piecewise STllooth 9 oriented suiface in space and let the boundary of S be a piecewise smooth simple closed curve C. Let F(x, y, z) be a continuous vector function that has continuous first partial derivatives in a domain in space containing S. Then JsJ(curl F)en dA = f Fer' (s) ds. (2) C Here n is a unit nonnal vector of S and, depending on n, the integration around C is taken in the sense shown in Fig. 251. Furthermore, r' = dr/ds is the unit tangent vector and s the arc length of C. In components, formula (2) becomes (2*) = f_(Fl dx + F2 dy + F3 dz). C Here, F = [Flo F2 , F3]' N = [Nl , N 2, N3]' n dA = N du du, r' ds = [dx. dy, dz]. and R is the region with boundary curve C in the uv-plane corresponding to S represented by r(u. v). z The proof follows after Example 1. (\ ~c r'~ J r' Fig. 251. E X AMP L E 1 x n Stokes's theorem Fig. 252. y Surface 5 in Example 1 Verification of Stokes's Theorem Before we prove Stokes's theorem, let (Fig. 252) LIS first get L1sed to it by verifying it for F = [y, z, xl and S the paraboloid z ~ O. Solution. The curve C, oriented as in Fig. 252, is the circle r(s) = [cos s, sin s, 0]. Its unit tangent vector is r' (s) = I-sin s, cos s, 0]. The function F = [y, z, x] on Cis F(r(s)) = [sin s, 0, cos s]. Hence f J 27T Fodr = F(r(s))or'(s)ds= 2'17" J [(sins)(-sins)+O+O]ds=-'7T. C O O 9 Sir GEORGE GABRIEL STOKES (l819-1903).lrish mathematician and physicist who became a professor in Cambridge in 1849. He is also known for his important contribution to the theory of infinite series and to viscous flow (Navier-Stokes equations), geodesy, and optics. "Piecewise smooth" curves and surfaces are defined in Sees. 10.1 and 10.5. CHAP. 10 470 Vector Integral Calculus. Integral Theorems We now consider the surface integral. We have Fl = y. F2 = :, F3 = X. so that in (2*) we obtain curlF= Cllrl[Fl' F3 ] = cllrl[y. F2, " -I. xj=[-1. -11. A normal vector of Sis N = grad(;: - J(x, y)) = [2.-.2,'. I]. Hence (curl F)oN = -2\' - 2y - I. Now n dA = N dx dy (see (3'') in Sec. 10.6 with x, y instead of II, u). Using polar coordinates r. e defined by x = r cos e, y = r sin e and denoting the projection of S into the x\'-plane by R. we thus obtain f I(curIF)ondA = f I<CLlrlF)ON dxdy = I f<-2X - 2y - I)dxdy S R R I I f (- f 2.". = 1 (-2r(cos8+ sin 8) - I)rdr£le 8=0 7·=0 2.". = (cos e+ sin 8) O~O PROOF -1) d8 = 0 + 0 -1 (21T) = -1T. • We prove Stokes's theorem. Obviously, (2) holds if the integrals of each component on both c;ides of (2*) are equal; that is, {I( (3) rc aaF?l N.2 - -.a FN3 l ) du dv = ,( Fl dx ely (4) (5) We prove this first for a surface S that can be represented simultaneously in the forms (6) (a) ;: = f(x, y), We prove (3), using (6a). Setting u r(u, v) = y (b) = g(x, .:), (c) x = h(y, ;:). = x, v = y, we have from (6a) rex, y) = [x, y, f(x, y)] = xi + yj + fk and in (2), Sec. 10.6. by direct calculation Note that N is an upper normal vector of S, since it has a positive z-component. Also, R = S*, the projection of S into the x,v-plane, with boundary curve E = C* (Fig. 253). Hence the left side of (3) is (7) I I [aF S* l (-fy) - az aFlJ dxdv. iJy' We now consider the light side of (3). We transform this line integral over E = C* into a double integral over S* by applying Green's theorem [formula (1) in Sec, 10.4 with F2 = 0]. This gives ,( Fldx = Jc * ffS* aFl dxd)'. ay SEC. 10.9 471 Stokes's Theorem ~ ::n z 1 1 S 1 1 1 1 1 1 1 1------ ~y ~ C* x Fig. 253. Here, Fl = FI(x, y, Proof of Stokes's theorem f(x, y)). Hence by the chain rule (see also Prob. 10 in Problem Set 9.6), (!Fl(X, y, f(x, y)) iJFl(X, y, z) iJy iJy Cly (Jz [z = f(x. y)]. We see that the right side of this equals the integrand in (7). This proves (3). Relations (4) and (5) follow in the same way if we use (6b) and (6c), respectively. By addition we obtain (2*). This proves Stokes's theorem for a surface S that can be represented simultaneously in the forms (6a), (6b), (6c). As in the proof of the divergence theorem, our result may be immediately extended to a surface S that can be decomposed into finitely many pieces, each of which is of the kind just considered. This covers most of the cases of practical imerest. The proof in the case of a most general surface S satisfying the assumptions of the theorem would require a limit • process; this is similar to the situation in the case of Green's theorem in Sec. lOA. E X AMP L E 2 Green's Theorem in the Plane as a Special Case of Stokes's Theorem Let F = IFl' F21 = F1 i + F2 j be a vector function that is continuously differentiable in a domain in the \"y-plane containing a simply connected bounded closed region S whose boundary C is a piecewise smooth simple closed curve. Then. according to (I), (curIF)on aF aF l = (curlF)ok = ---2 - ---. ax ay Hence the formula in Stokes's theorem now takes the form II( a~2 s rJx - (IFI) dA = Ay J. 'j c (F1 dx -'- F2 dy). This shows that Green's theorem in the plane (Sec. 10.4) is a special case of Stokes's theorem (which we needed • in the proof of the latter!). E X AMP L E 3 Evaluation of a Line Integral by Stokes's Theorem Evaluate f c For' ds, where C is the circle x 2 + y2 = 4, z = - 3, oriented counterclockwise as seen by a person standing at the origin, and. with respect to right -handed Cartesian coordinates. i ~ 4 in the plane;: = - 3. Then n in Stokes's theorem points in the po~itive ;:-direction; thus n = k. Hence (curl F)on is simply the compone~t of curl F in the positive ~-direction. Since F with;:: = -3 has the components F1 = y, F2 = -27x, F3 = 3y , we thus obtain Solutioll. As a surface S bounded by C we can take the plane circular disk x2 + (curl F)on (JF iJF iJx iJy = -.-2 - ---1 = -27 - I = -28. CHAP. 10 472 Vector Integral Calculus. Integral Theorems Hence the integral over S in Stokes· s theorem equals - 28 times the area 47T of the disk S. This yields the answer -28' 47T = -1127T = -352. Confirm this by direct calculation, which involves somewhat more work. • E X AMP L E 4 Physical Meaning of the Curl in Fluid Motion. Circulation Let ST be a circular disk of radius "0 and center P bounded by the circle CTo (Fig. 254), and let O F(Q) == F(x, y, :::) be a continuously differentiable vector function in a domain containing ST • Then by Stokes's O theorem and the mean value theorem for sUiface integrab (see Sec. 10.6), where ATo is the area of S'o and Fig. 254. P~ is a ~uitable point of S"o. This may be written in the form Example 4 In the case of a fluid motion with velocity vector F = v, the integral is called the circulation of the t10w around C ro . It measures the extent to which the corresponding fluid motion is a rotation around the circle CTO • If we now let ro approach zero, we find (8) that is. the component of the curl in the positive normal direction can be regarded (circulation per unit area) of the flow in the sUiface at the corresponding point. E X AMP L E 5 a~ the specific circulation • Work Done in the Displacement around a Closed Curve Find the work done by the force F = 2ry3 sin::: i + 3x\2 sin::: j + x 2 cos::: k in the displacement around the curve of intersection of the paraboloid z = x2 + y2 and the cylinder (r - 1)2 + y2 = l. .l Solutioll. This work is given by the line integml in Stokes's theorem. Now F = grad f, where f = X 2 y3 sin::: and curl(grad f) = 0 (see (2) in Sec. 9.9). so that (cur! F)-n = 0 and the work is 0 by Stokes's theorem. This agrees with the fact that the present field is conservative (definition in Sec. 9.7). • Stokes's Theorem Applied to Path Independence We emphasized in Sec. 10.2 that the value of a line integral generally depends not only on the function to be integrated and on the two endpoints A and B of the path of integration C, but also on the particular choice of a path from A to B. In Theorem 3 of Sec. 10.2 we proved that if a line integral I F(r)odr = I (9) C c (FI dx + F2 dy + F3 d;:;) (involving continuous F], F2 , F3 that have continuous first partial derivatives) is path independent in a domain D, then curl F = 0 in D. And we claimed in Sec. 10.2 that. conversely. curl F = 0 everywhere in D implies path independence of (9) in D provided D is simply connected. A proof of this needs Stokes's theorem and can now be given as follows. Let C be any closed path in D. Since D is simply connected. we can find a surface S in D bounded by C. Stokes's theorem applies and gives fc (Fl dx + F2 dy + F3 d;:;) = fc For' ds = J J(curl F)on dA s 473 Chapter 10 Review Questions and Problems for proper direction on C and nonnal vector n on S. Since curl F = 0 in D, the surface integral and hence the line integral are zero. This and Theorem 2 of Sec. 10.2 imply that the integral (9) is path independent in D. This completes the proof. • - ..-.- . 11-81 DIRECT INTEGRATION OF THE SURFACE INTEGRALS Evaluate the integral F and S. 1. F = [4Z2, 2. F = [0, 16x, 0, s: x 2 + y2 3. F = II = 0 0], s: Z = Y (0 ~ x ~ 1, 0 ~ y ~ I) 5x cos z], 4, Y ~ 0, 0 ~ eZ, [-e Y , (curl F) n dA directly for the given S z~ ~7T eX], s: Z = x + y (0 ~ x ~ 1, 0 ~ y ~ 1) 4. F = [3 cos y, cosh z, x], S the square 0 ~ x ~ 2, 0 ~ y ~ 2, z 2Z Z = 18. F = 9. Verify Stokes's theorem for F and S in Prob. 7. 10. Verify Stokes's theorem for F and S in Prob. 8. EVALUATION OF f For' ds c Calculate this line integral by Stokes's theorem, clockwise as seen by a person standing at the origin, for the following F and C. Assume the Cartesian coordinates to be righthanded. (Show the details.) : - -_11" [z, x, y]. C as in Prob. 13 7 ~ ? .... __ 7. F = [Z2. ~x, 0], S the square 0 ~ x ~ a, 0 ~ y ~ a, Z = 1 8. F = [y3. -x3, 0], S: x 2 + y2 ~ I. Z = 0 111-181 13. F = [y2, x2, -x + z], around the triangle with vertices (0, 0, I). (I. O. I), (1, 1, I) 14. F = [y, xy3, - Zy 3], C the circle x 2 + y2 = a 2, Z = b (> 0) IS. F = [y, Z2, x 3 ], C as in Prob. 12 16. F = [x 2, y2, Z2], C the intersection of x 2 + y2 + Z2 = 4 and z = y2 17. F = [cos 7T)" sin 7TX, 0], around the rectangle with vertices (0, 1,0), (0, 0, I), (1, 0, I), (1, 1. 0) 4 Z S. F = [e , e sin y, e cos y], S: Z = y2 (0 ~ X ~ 4, 0 ~ y ~ I) 2 = x 2 + ),2, ] 2 7 ,,2: .(., X-, _v 2 ] , S'..... _ 0, 0 ~ _ 6• F = [_2 11. F = [-3y. 3x. z], C the circle x 2 + y2 = 4. z = 1 12. F = [4z, -2x, 2x], C the intersection of x 2 + )'2 = I and z = y + 1 20. WRITING PROJECT. Grad, Div, Curl in Connection with Integrals. Make a list of ideas and results on this topic in this chapter. See whether you can rearrange or combine parts of your material. Then subdivide the material into 3-5 portions and work out the details of each portion. Include no proofs but simple typical examples of your own that lead to a better understanding of the material. .. 1. List the kinds of integrals in this chapter and how the integral theorems relate some of them. 2. How can work of a variable force be expressed by an integral? 3. State from memory how you can evaluate a line integral. A double integral. 4. What do you remember about path independence? Why is it important? 5. How did we Use Stokes's theorem in connection with path independence? 6. State the definition of curl. Why is it important in this chapter? 7. How can you transform a double integral or a surface integral into a line integral? f Fo r' ds, c F = (x 2 + y2)-1[ -y,x], C: x 2 + y2 = I, z = 0, oriented clockwise. Why can Stokes's theorem not be applied? What (false) result would it give? 19. (Stokes's theorem not applicable) Evaluate AND PROBLEMS 8. What is orientation of a surface? What is its role in connection with surface integrals? 9. State the divergence theorem and its applications from memory. 10. State Laplace's equation. Where in physics is it important? What properties of its solutions did we discuss? 111-201 LINE INTEGRALS I (WORK INTEGRALS) F(r)odr C Evaluate. with F and C as given, by the method that seems most suitable. Recall that if F is a force, the integral gives the work done in a displacement along C. (Show the details.) 11. F = [x 2• y2, Z2], C the straight-line segment from (4, I, 8) to (0, 2, 3) 474 CHAP. 10 Vector Integral Calculus. Integral Theorems I 25. I 12. F = [cos;::, -sin z, -x sin;:: - y cos ;::]. C the straight-line segment from (-2. 0, ~'iT) to (4. 3. 0) 13. F = [x - y, 0, eZ ], C: y = 3x2 , Z = 2x for x from 0 to 2 24. 14. F = [yz, 2;::x, xy], C the circle x 2 + y2 = 9, 126-35 1 Z = = x2 + )'2, R: x 2 + )'2 = 2x2, ~ I, x ~ 0, y ~ 0 R the region below y = x + 2 and above )' = x 2 SURFACE INTEGRALS ff l, counterclockwise 15.F=[-3v3 , 3x3 +cosy. 0]. C the circle x 2 + )'2 = 16. z = 0, counterclockwise 16. F = [sin 10', cos m:, sin 17X]. C the boundary of 0 ~ x ~ 112, 0 ~ y ~ 2, z = 2x Evaluate this integral directly or. if divergence theorem. (Show the details.) 17. F = [9z, 5x, 3.\'], C the ellipse x 2 + )'2 = 9. z = x + 2 18. F = [cosh x, e 4y , tan z], C: x 2 + )'2 = -1-, (Sketch C.) 19. F = [Z2. x 3• y2], C: x 2 + )'2 = 4, x + Y + 27. F = [yo -x. 0]. S: 3 t' + 2 Y + z = 6, x 2 Z = x . Z = 0 2 = [x • y2, )'2X], C the helix 20. F r = [2 cos I. 2 sin I, 61] from (2. O. 0) to (0. 2, ~... 1-251 317) DOUBLE INTEGRALS, CENTER OF GRAVITY Find the coordinmes .i. y of the center of gravity of a mass of density I(x. y) in the region R. (Sketch R. Shmv the details.) 21. I = 2x)" R the triangle with vertices (0, 0), (1, 0), 22. 23. Fon dA 5 26. F = [2X2, 4.", 0], S: x + y + z = 1, x ~ 0, y ~ ~ 0, y 0, z ~ pos~ible. ~ 0, z by the 0 ~ 0 28. F = [x - y, y - z, z - x], S the sphere of radius 5 and center 0 29. F = [y2, x 2, Z2]. S the surface of x 2 + y2 ~ 4, 0 ~ Z ~ 5 30. F = [-,,3, x 3 , 3z 2 ], S the portion of the paraboloid z = x2 + 31. F = [sin2 x, -y sin 2x, 5;::]. S the sul1'ace of the box Ixl ~ a, 32. F = [1, 33. F = [x, I, xy, a]. S: x 2 z], S: x 2 y2, Iyl ~ b, Izl z~4 ~ c + )'2 + 4;::2 = 4, z ~ 0 + y2 = I, 0 ~ z ~ h (1, I) 34. F as in Prob. 33, S the complete boundary of x2 + )'2 ~ I, 0 ~ z ~ II I I 35. F = leY, 0, ze X ]. Sthe rectangle with vertices (0, O. 0). (1.2,0), (0, O. 5), (1, 2, 5) R: 0 ~ y ~ I - x 2 = 1. R: x 2 + y2 ~ a 2, y ~ 0 = I, . ...... .:..:... :. . ........ _\I... 1ft .... - Vector Integral Calculus. Integral Theorems Chapter 9 extended differential calculus to vectors, that is, to vector functions vex, y, z) or vet). Similarly. Chapter 10 extends integral calculus to vector functions. This involves line integrals (Sec. 10.1), double integrals (Sec. 10.3), swface integrals (Sec. 10.6), and triple integrals (Sec. 10.7) and the three "big" theorems for transforming these integrals into one another, the theorems of Green (Sec. 10.4), Gauss (Sec. 10.7), and Stokes (Sec. lO.9). The analog of the definite integral of calculus is the line integral (Sec. 10.1) (1) where C: r(t) = [x(t), y(t), z(t)] = x(t)i + y(t)j + z(t)k (a ~ t ~ b) is a curve in space (or in the plane). Physically. (I) may represent the work done by a (variable) force in a displacement. Other kinds of line integrals and their applications are also discussed in Sec. 10.1. Summary of Chapter 10 475 Independence of path of a line integral in a domain D means that the integral of a given function over any path C with endpoints P and Q has the same value for all paths from P to Q that lie in D; here P and Q are fixed. An integral (1) is independent of path in D if and only if the differential form Fl dx + F2 dy + F3 dz with continuous F I , F2 • F3 is exact in D (Sec. LO.2). Also, if curl F = 0, where F = [Fl' F2 , F3]' has continuous first partial derivatives in a simp/" connected domain D, then the integral (1) is independent of path in D (Sec. 10.2). Integral Theorems. The formula of Green's theorem in the plane (Sec. 10.4) (2) II( -iJF2 ax R l - -iJF ay ) dr: dy . = T c (F dx I + F dy) 2 . transforms double integrals over a region R in the xy-plane into line integrals over the boundary curve C of R and conversely. For other forms of (2) see Sec. lOA. Similarly, the formula of the divergence theorem of Gauss (Sec. 10.7) (3) I I I div F dV = I I F- n dA T S transforms triple integrals over a region T in space into surface integrals over the boundary surface S of T. and conversely. Formula (3) implies Green's formulas (4) III (f'\P g + Vf-Vg)dV= T IIf S ~g an dA, (5) Finally, the formula of Stokes's theorem (Sec. 10.9) (6) IsI(curl F)-n dA Tc F-r' (s) ds = transforms surface integrals over a surface S into line integrals over the boundary curve C of S and conversely. PA RT c Fourier Analysis. ••• Partial Differential Equations C HAP T E R 11 Fourier Series, Integrals, and Transforms C HAP T E R 1 2 Partial Differential Equations (PDEs) Fourier analysis concerns periodic phenomena, as they occur quite frequently in engineering and elsewhere-think of rotating parts of machines, alternating electric currents, or the motion of planets. Related periodic functions may be complicated. This situation poses the important practical task of representing these complicated functions in terms of simple periodic functions. namely. cosines and sines. These representations will be infinite series, called Fourier series. l The creation of these series was one of the most path-breaking events in applied mathematics, and we mention that it also had considerable influence on matl1ematics as a whole, on the concept of a function. on integration theory, on convergence tl1eory for series. and so on (see Ref. [OR7] in App. 1). Chapter II is concerned mainly with Fourier series. However, the underlying ideas can also be extended to nonperiodic phenomena. This leads to Fourier integrals and fransjonl1s. A common name for the whole area is Fourier analysis. Chapter 12 deals witl1 the most important partial differential equations (PDEs) of physics and engineering. This is the area in which Fourier analysis has its most basic applications, related to boundary and initial value problems of mechanics, heat flow, electrostatics, and other fields. IJEAN-BAPTISTE JOSEPH FOURIER (1768-1830). French physicist and mathematician, lived and taught in Paris. accompanied Napoleon in the Egyptian War. and was later made prefect of Grenoble. The beginnings on Fourier series can be found in works by Euler and by Daniel Bernoulli, but it was Fourier who employed them in a systematic and general manner in his main work, Theorie allalyflque de la chaleur (Analytic Theory of Heat. Paris, 1822). in which he developed the theory of heat conduction (heat equation; see Sec. 12.5), making these series a most important tool in applied mathematics. - 477 · / .' . CHAPTER \ 11 Fourier Series, Integrals, and Transforms Fourier series (Sec. 11.1) are infinite series designed to represent general periodic functions in terms of simple ones, namely. cosines and sines. They constitute a very important tool, in particular in solving problems that involve ODEs and PDEs. In this chapter we discuss Fourier series and their engineering use from a practical point of view, in connection with ODEs and with the approximation of periodic functions. Application to PDEs follows in Chap. 12. The theory of Fourier series is complicated. but we shall see that the application of these series is rather simple. Fourier series are in a certain sense more universal than the familiar Tay lor series in calculus because many discontinuous periodic functions of practical interest can be developed in Fourier series but, of course, do not have Taylor series representations. In the last sections (11.7-11.9) we consider Fourier integrals and Fourier transforms, which extend the ideas dnd techniques of Fourier series to nonperiodic functions and have basic applications to PDEs (to be shown in the next chapter). Prerequisite: Elementary integral calculus (needed for Fourier coefficients) Sections that lIlay be nmitted in a shorter course: 11.4-11.9 References alld Answers to Problems: App. 1 Part C. App. 2. 11.1 Fourier Series Fourier series are the basic tool for representing periodic functions, which play an important role in applications. A function f(x) is called a periodic function if f(x) is defined for all real x (perhaps except at some points, such as x = ±7T!2, ±37T/2, ... for tan x) and if there is some positive number p. called a period of f(x). such that (1) f(x + p) = f(x) for all x. The graph of such a function is obtained by periodic repetition of its graph in any interval of length p (Fig. 255). Familiar periodic functions are the cosine and sine functions. Examples of functions that are not periodic are x, x 2 , x 3 , eX, cosh x, and In x, to mention just a few. If f(x) has period p, it also has the period 2p because (I) implies f(x + 2p) = f([x + p] + p) = f(x + p) = f(x), etc.; thus for any integer 11 = 1,2,3, .. " (2) 478 f(x + np) = f(x) for all x. SEC. 11.1 479 Fourier Series {(x) x Fig. 255. Periodic function Furthermore if f(x) and g(x) have period p, then af(x) + bg(x) with any constants a and b also has the period p. Our problem in the first few sections of this chapter will be the representation of various functions f(x) of period 217 in terms of the simple functions (3) I, cos x, cos 2x, sin x, sin 2x, ... , cos In:, sin /lX, . • • . All these functions have the period 27T. They form the so-called trigonometric system. Figure 256 shows the fIrst few of them (except for the constant 1, which is periodic with any period). The series to be obtained will be a trigonometric series, that is, a series of the form ao + a1 cos x (4) + bi ao = sin x + .L + a2 cos 2\'" + b 2 sin 2x + (an cos + b n sin nx). IlX n~I ao, Lib b l . a2, b2, ... are constants, called the coefficients of the series. We see that each term has the period 27T. Hence if the coefficients are such that the series converges, its sum will be a function of period 27T. It can be shown that if the series on the left side of (4) converges, then inserting parentheses on the right gives a series that converges and has the same sum as the series on the left. This justifIes the equality in (4). Now suppose that f(x) is a given function of period 27T and is such that it can be represented by a series (4), that is, (4) converges and, moreover, has the sum f(x). Then, using the equality sign, we write f(x) (5) = ao + .L (an cos nx + bn sin nx) n~I :\vnv /:\ L o :\ f\,!\ 2n cos 2x cos x V\ V sin x cos 3x 1f!\. 2n V sin 2x Fig. 256. Cosine and sine functions having the period 2IT Sin 3x (, 480 CHAP. 11 Fourier Series, Integrals, and Transforms and call (5) the Fourier series of f(x). We shall prove that in this case the coefficients of (5) are the so-called Fourier coefficients of f(x), given by the Euler formulas (a) ao = - f I 71" 27T -71" -f I (6) an = (b) 7T 7T f(x) cos I1X dx n = 1.2.··· f(x) sin 11-1: dx 11 -71" f I bn = 7T (c) f(x) dx 7T = 1,2, .... -7T The name "Fourier series" is sometimes also used in the exceptional case that (5) with coefficients (6) does not converge or does not have the sum f(x)-this may happen but is merely of theoretical interest. (For Euler see footnote 4 in Sec. 2.5.) A Basic Example Before we derive the Euler formulas (6). let us become familiar with the application of (5) and (6) in the case of an important example. Since your work for other functions will be quite similar, try to fully understand every detail of the integrations, which because of the 11 involved differ somewhat from what you have practiced in calculus. Do not just routinely use your software, but make observations: How are continuous functions (cosines and sines) able to represent a given discontinuous function? How does the quality of the approximation increase if you take more and more terms of the series? Why are the approximating functions, called the partial sums of the series, always zero at 0 and 7T? Why is the factor lin (obtained in the integration) important'? E X AMP L E 1 Periodic Rectangular Wave (Fig. 257a) Find the Fourier coefficients of the periodic function f(x) in Fig. 257a. The formula is (7) f(x) = -k { k if -71"<X<O if O<X<71" and f(x + 271") = f(x). Functions of this kind occur as external forces acting on mechanical systems, electromotive forces in electric circuits, etc. (The value of f(x) at a single point does not affect the integral: hence we can leave f(x) undefined at x = 0 and x = 2:71".) Solution. From (6a) we obtain ao = O. This can also be seen without integration, since the area under the curve of f(x) between -71" and 71" is zero. From (6bl. an = - I 7r f'" f(x) cos nxdx = -'iT I 7T [ f 0 'IT (-k)COSI1Xdx+f kcosl1xdx 0 7r [ ] 0 -'iT nx sin -k n 1 -7T sin-nx +k n I"'] cO 0 because sin nx = 0 at -71", 0, and 71" for all n = 1, 2, .... Similarly, from (6cl we obtain bn = ~ 71" f'"_'" f(x) sin nx dx = 7r [f O(-k) 1 [ kcos - nx n 'IT sin nx dx + f'" k sin 17X dX] 0 _" /0 -'iT - kcos -n.r - /"'] . n 0 SEC. 11.1 481 Fourier Series -n n 0 -----l-k L 2n x 1- J (a) The gIven function {(x) (Periodic rectangular wave) x " '~< n / x 4k sin 3x 3" '-, n ...... _ / x 4k sin 5x 5" (b) The first three partial sums of the corresponding Fourier series Fig. 257. Eample 1 Since cos ( -a) = cos a and cos 0 = 1, this yields bn = k nn [cos 0 - cos (-n7T) - cos n7T + cos 0] = U ~ nn (1 - cos n7T). Now, cos 71" = -1, cos 271" = 1, cos 371" = -1, etc.; in general, cos n71" = { -I I for odd n, e for odd n, and thus for even n, I - cosn71" = for cven n. Hence the Fourier coefficients h n of our function are 4k 4k h5 = 571" ' CHAP. 11 482 Fourier Series, Integrals, and Transforms Since the an are 7ero, the Fourier series of f(x) is + ..!.. sin 3x + ..!.. sin 5x + ... ) 3 5 . 4k (Sin x (8) 7T The partial sums are 4k S = 4k (sin x Sl = ~sinx, 2 7T + ..!.. 3 sin 3X) ' etc., Their graphs in Fig. 257 ,eem to indicate that the series is convergent and has the sum f(x), the given function. We notice that at x = 0 and x = 7T, the points of discontinuity of f(x), all partial sums have the value zero, the arithmetic mean of the limits -k and k of our function, at these points. Furthermore, assuming that f(x) is the sum of the series and setting x = 7TI2, we have thus 1 1 1 7T 1--+---+-···=-. 3 5 7 4 This is a famous result obtained by Leibniz in 1673 from geometric considerations. It illustrates that the value, of various series with constant terms can be obtained by evaluating Fourier series at specific points. • Derivation of the Euler Formulas (6) The key to the Euler formulas (6) is the orthogonality of (3), a concept of basic importance, as follows. THEOREM 1 Orthogonality of the Trigonometric System (3) The trigonometric system (3) is orthogonal on the interval -7T ~ X ~ 7T (hence also on 0 ~ x ~ 27T or any other interval of length 27T because of periodicity): that is, the integral of the product of any two functions in (3) over that interval is 0, so that for any integers nand nz, (a) J7T cos nx cos nIX dx = 0 (n =/=- m) 0 (n =/=- m) 0 (n =/=- m or n -7T (9) J" sin nx sin mx dx = (b) -7T J7T sin nx cos mx dx = (e) = m). -7T PROOF This follows simply by transfonning the integrands trigonometrically from product'> into sums. In (9a) and (9b), by (11) in App. A3.I, I 1"" 7T cos nx cos nIX dx -7T = - 2 1 J cos (n + m)x dx + -2 J_.".cos (n - m)x dx -7T J"" sin nx sin nzx dx = -2 J cos (n -7T 17T 7T -7T J m)x dx - 2 J cos (n + m)x dx. 7T -7T SEC 11.1 483 Fourier Series Since m * n (integer!), the integrals on the right are all O. Similarly, in (9c), for all integer m and n (without exception; do you see why?) I ~ sin nx cos mx dx = - J sin (n _~ 2 _~ ~ J ~ + m)x dr: + 2 J sin (n - lIl)x dr: = 0 _~ + O. • Application of Theorem 1 to the Fourier Series (5) We prove (6a). Integrating on both sides of (5) from ~ ~ L}(X) dx = L~ [ ao + ~l 00 -7T (an cos rue to 7T, we get + bn sin Itt) ] dx. We now assume that termwise integration is allowed. (We shall say in the proof of Theorem 2 when this is true.) Then we obtain The first tenn on the right equals 27Tao. Integration shows that all the other integrals are O. Hence division by 27T gives (6a). We prove (6b). Multiplying (5) on both sides by cos 11/X with any fixed positive integer m and integrating from - 7T to 7T, we have (10) ~ J_~f(X) cos mx dx = ~ J_~ [ ao + ~1 YO (an cos nx ] + hn sin nx) cos mx dx. We now integrate term by term. Then on the right we obtain an integral of ao cos mx. which is 0; an integral of an cos nx cos 17U, which is am 7T for n = 11/ and 0 for n =/=- 111 by (9a); and an integral of bn sin In cos 111X, which is 0 for all nand 111 by (9c). Hence the right side of (10) equals a m 7T. Division by 7T gives (6b) (with 111 instead of n). We finally prove (6c). Multiplying (5) on both sides by sin my with any fixed positive integer 111 and integrating from - 7T to 7T, we get (II) ~ L}(X) sin mx dx = LTi~ [ ao + ~l= (an cos nx + hn sin nx) ] sin mx dr:. Integrating term by term, we obtain on the right an integral of a o sin mx, which is 0; an integral of an cos nx sin mx, which is 0 by (9c); and an integral of h n sin 11.)( sin llU", which is hm 7T if n = 1ll and 0 if 17 =/=- m, by (9b). This implies (6c) (with n denoted by m). This completes the proof of the Euler formulas (6) for the Fourier coefficients. • CHAP. 11 484 Fourier Series, Integrals, and Transforms Convergence and Sum of a Fourier Series The class of functions that can be represented by Fourier series is surprisingly large and general. Sufficient conditions valid in most applications are as follows. Representation THEOREM 2 by a Fourier Series Let f(x) he periodic with period 271" and pieceJ,vise cOlltinuous (see Sec. 6.1) in the interval -71" ~ X ~ 71". Furthermore, let f(x) have a left-hand derivative and a right-hand derivative at each point of that interval. Then the Fourier series (5) of f(x) [with coefficients (6)] conver!!es. Its sum is f(x), except at points Xo where f(x) is discontinuous. There the slim of the series is the average of the left- and right-hand limits 2 of f(x) at Xo. PROOF We prove convergence in Theorem 2. We prove convergence for a continuous function f(x} having continuous first and second derivatives. Integrating (6b) by parts, we obtain. an = -1 I71" f(x) cos nx dr: = f(x) sin IlX 71" -71" - - I I71" f ,(x) sin nx dt. 17T -7T n71" n71"_7T The first teml on the right is zero. Another integration by parts gives an = t' (.x) 2cos nx 171" n 71" - I n 71" -2- -7T I7T f "(x) cos nx dx. -7r The firs I term on the right is zero because of the periodicity and continuity of f' (x). Since f" is continuous in the interval of integration, we have If"(x)1 < M for an appropriate constant M. Furthermore, Icos nxl ~ 1. It follows that lanl = -i-I n 71" I 7T {'ex) cos nx dxl < -7T -iI7T M dx n 71" = -7T 2M n2 . f(x) f(l- 0) j~ o 2 The left-hand limit of f(x) at Xo is defined as the limit of f(x) as x approaches Xo from the left and is commonly denoted by f(xo - 0). Thus Left- and right-hand limits Fig. 258. + 0) =i of the function X2 I(x) = { x/2 if x < 1 0 through positive values. + h) as h ---> 0 through positive values. h~O The right-hand limit is denoted by f(xo f(xo + 0) + 0) and = lim f(xo 11._0 1(1 - O} = 1, 1(1 ~ f(xo - 0) = lim f(xo - Iz) as h x The left- and right-hand derivatives of f(x) at xo are defined as the limits of f(x o - Iz) - f(x o - 0) -Iz and f(xo + Iz) - f(xo + 0) It respectively, as Iz ---> 0 through positive values. Of course if f(x) is continuous at X()o the last tenn in both numerators is simply flxo). SEC. 11.1 Fourier Series 485 Ibnl Similarly, < 2 Mln 2 for alln. Hence the absolute value of each teml of the Fourier series of f(x) is at most equal to the corresponding term of the series I I + 2M ( 1 + 1 + -221 + -22I+ I -32 + -32 + ... ) la o which is convergent. Hence that Fourier series converges and the proof is complete. (Readers already familiar with uniform convergence will see that, by the Weierstrass test in Sec. 15.5, under our present assumptions the Fourier series converges uniformly, and our derivation of (6) by integrating term by term is then justified by Theorem 3 of Sec. 15.5.) The proof of convergence in the case of a piecewise continuous function f(x) and the proof that under the assumptions in the theorem the Fourier series (5) with coefficients (6) represents f(x) are substantially more complicated; see, for instance, Ref. [C121. • E X AMP L E 2 Convergence at a Jump as Indicated in Theorem 2 The rectangular wave in Example I has a jump at x = O. Its left-hand limit there is -k and its right-hand limit is k (Fig. 257). Hence the average of these limits is O. The Fourier series (8) of the wave does indeed converge to this value when x = 0 because then all its terms are O. Similarly for the other jumps. This is in agreement with Theorem 2. • Summary. A Fourier series of a given function f(x) of period 271' is a series of the form (5) with coefficients given by the Euler formulas (6). Theorem 2 gives conditions that are sufficient for this series to converge and at each x to have the value f(x), except at discontinuities of f(x), where the series equals the arithmetic mean of the left-hand and right-hand limits of f(x) at that point. .. =J~ 1. (Calculus review) Review integration techniques for integrals as they are likely to arise from the Euler formulas, for instance, definite integrals of x cos /lX, x 2 sin I1X, e- 2x cos I1X, etc. @-iJ FUNDAMENTAL PERIOD Theful1damental period is the smallest positive period. Find it for sin 2x, 2. cos x, sinx. cos 2x. cos 27TX, sin 27TX 3. cos I1X. 27TI1X sin nx. cos - k - , 27TX cos -k- cos 7TX, sin 7TX. 27TX 6. (Change of scale) If f(x) has period 17, show that f(ax), a *- O. and f(x/b) , b *- O. are periodic functions of x of periods pia and bp, respectively. Give examples. 17-121 GRAPHS OF 21T"PERIODIC FUNCTIONS Sketch or graph f(x), of period 27T, which for -7T < X < is given as follows. 7. f(x) =x 8. f(x) = e- lxl 9. f(x) 11. f(x) sin - k ' 27TI1X sin - - 12. f(x) k 4. Show that f = COl1st is periodic with any period but has no fundamental period. S. If f(x) and g(x) have period p, show that hex) = af(x) + bg(x) (a, b, constant) has the period p. Thus all functions of period 17 form a vector space. 113-241 10. f(x) Ixl 7T - 3 if x3 if {-X Losl ~x -7T 7T Isin 2xI < x < 0 O<X<7T if-7T<x<O if O<x< 7T FOURIER SERIES Showing the details of your work, find the Fourier series of the given f(x)' which is assumed to have the period 27T. Sketch or graph the pattial sums up to that including cos 5x and sin 5x. 486 CHAP. 11 lIn I 13. l _ -IT 0 24. f{x) lIT 0 -IT 15. 16. IT 1 "" "- '. + ~ sin 3x ! sin 5x + ... ) + i sin 4x + i sin 6x + ... ) + ----z (cos x + 9 cos 3x + :l5 cos Sx + ... ) + 4(cos x + - ... ) (c) ~~ i cos 2x + i cos 3x - -h cos 4x 27. CAS EXPERIMENT. Order of Fourier Coefficients. The order seems to be lin if f is discontinous. and 11112 if f is continuous but f' = dfldx is discontinuous. 1In 3 if f and J' are continuous but fff is discontinuous, etc. Try to verify this for examples. Try to prove it by integrating the Euler formulas by parIs. Whal is the practical significance of this? Tr '/ -Tr 7T 7T Tr 0 18. 0 < x < 1 4 1 (b) 2 Tr -Tr if < x < 0 - 2( i sin 2x "2Tr 17. 4x -7T 26. CAS EXPERIMENT. Graphing. Write a program for graphing partial sums of the following series. Guess from the graph what f(x) the series may represent. Confirm or disprove your guess by using the Euler IT ~ ~ 2 if (a) 2{sinx 0 -Tr -4X { formula~. /~ -Tr = 25. (Discontinuities) Verify the last statement in Theorem 2 for the discontinuities of f(x) in Prob. 13. 2 14. Fourier Series, Integrals, and Transforms 28. PROJECT. Euler Formulas in Terms of Jumps Without Integration. Show that for a function whose third derivative is identically zero, IT 0 an 19. = 1l'iT I + n2 1 -n n7T - L is.ff.Sill nxs ] Lj~ sinnxs 2 I ~ _ff ] L..J is cos nxs 11 where n = I, 2, ... and we sum over all the jumps js, J', J'. respectively. located atxs' 20. j~,j; of f, 29. Apply the formulas in Project 28 to the function in Prob. 21 and compare the results. 21. f(x) = x 2 22. f(x) 23. f{x) ( - 7T = x 2 (0 < < X X < 27T) < 7T) 30. CAS EXPERIMENT. Orthogonality. Integrate and graph the integral of the product cos mx cos nx (with various integer m and IJ of your choice) from -a to a as a function of a and conclude orthogonality of cos mx and cos nx (m *- Il) for a = 7T from the graph. For what m and n will you get orthogonality for a = 7T/2, rr/3, 7T14? Other a? Extend the experiment to cos mx sin Il\: and sin 111 ,. sin l1X. SEC. 11.2 11.2 Functions of Any Period p = 2L 487 = 2L Functions of Any Period p The functions considered so far had period l7T, for the simplicity of the formulas. Of course, periodi.c function~ in applications will generally have other periods. However, we now show that the transition from period p = 27T to a period 2L is quite simple. The notation p = 2L is practical because L will be the length of a violin string (Sec. 12.2) or the length of a rod in heat conduction (Sec. 12.5), and so on. The idea is simply to find and use a cha1lge of scale that gives from a function g(v) of period 27T a function of period 2L. Now from (5) and (6) in the last section with g(v) instead of I(x) we have the Fourier series (1) g(v) "" = ao + 2: (an cos flV + bn sin llv) n=l with coefficients 1 ao J 27T = - J 7T = - J 7T 7T = - g(v) dv -71" 1 (2) lin 7T g(v) cos flV dv -7T 1 bn 7T g(v) sin llV dv. -7T We can now write the change of scale as v = kx with k such that the old period v = 27T gives for the new variable x the new period x = 2L. Thus, 27T = k2L. Hence k = 7TIL and v = kx = 7TXIL. (3) This implies dv = (7TIL) dx. which upon substitution into (2) cancels 1I27T and 1I7T and gives instead the factors 1I2L and IlL. Writing (4) g(v) = I(x), we thus obtain from (1) the Fourier series of the function f(x) of period 2L (5) f(x) 1l7T = ao + ~l 00 ( an cos L x + bn 1l7T sin L x) with the Fourier coefficients of f(x) given by the Euler formulas 1 (a) (6) (b) (C) 1 an = L 1 bn = L J L ao = -;;...L f(x) dx -L JL f(x) cos -ll7TX- dx -L JL I(x) sin -ll7TX- dx -L 11 = 1,2, ... n = 1,2, ... L L 488 CHAP. 11 Fourier Series, Integrals, and Transforms Just as in Sec. 11.1, we continue to call (5) with any coefficients a trigonometric series. And we can integrate from 0 to 2L or over any other interval of length p = 2L. E X AMP L E 1 Periodic Rectangular Wave Find the Fourier series of the function (Fig. 259) f(x) = o if -2 < x < -1 Ok if -1 { if Solution. < < 1 1<..1.'< 2 x p = 2L = 4. = 2. L From (6a) we obtain ao = kl2 (verify!). From (6h) we ohtain an I {/(X) = 11;' cos = dx I f/ ,~: sin 1127T . = cos 11;..1.' dx Thus an = 0 if 11 is even and an = 2kln7T if n = 1, 5, 9, .... = 0 for 11 = From (6c) we find that b n k f(x) = - 2 + -2k 7T ( an = -2kll17T if 1,2..... Hence the Fourier series is 7T 1 37T cos - x - - cos - x . 2 3 2 I + -5 cos ' ~b -2 -1 57T + ... ) -..I. - 2 I 1'-----:!~:---' 0 Fig. 259. E X AMP L E 2 = 3,7, 11, .... n x Example 1 Periodic Rectangular Wave Find the Fourier serie~ of the function (Fig. 260) if -2 < x < 0 kif 0<..1.'<2 -k f(x) = { Solution. ao = L = 2. p = 2L = 4, 0 from (6a). From (6bt with IlL = 112, an = 2 [f a Il1iX _2(-k) cos -2- dt: + {2 0 n7TX 0 2 [ _ 2k sin 117TX 1 n7T 2 -2 + ] k cos -2- dx 2 2k sin 1l7TX 1 1l7T 2 0 J = O. so that the Fourier series has no cosine terms. From (6c). 0 bn 2 [2k fl7TX 1l7TX cos - 1 - -2k cos -1 2 1l7T 2 -2 1l7T 2 0 = -I = 117T k (I - cos n7T - cos 1l7T + 1) = J {4klll7T if 0 if Il II = L 3•... = 2, 4, .... . • SEC lU Functions of Any Period p =lL 489 Hence the Fourier series of f(.1:) is ~ f(x) 4k (Sin ~ x + ~ sin 37T x + ~ sin 57T x + ... ) 7T 2 3 2 5 2 . It is interesting that we could have derived this from (8) in Sec. 11.1, namely, by the scale change (3). Indeed. writing v instead of x, we have in (8), Sec. 11.1, : + ~ sin 3v + (sin v + sin 5v + ... ) . Since the period 27T in v corresponds to 2L = 4, we have k = 7TIL = 7T12 and v = kx = TTXI2 in (3); hence we obtain the Fourier series of f(x), as before. • {(x) k x 2 L- '-------j-k ~, -rr/m Example 2 Fig. 260. EXAMPLE 3 ""'6 o rr/m Half-wave rectifier Fig. 261. Half-Wave Rectifier A sinusoidal voltage E sin WT. where T is time. is passed through a half-wave rectifier that clips the negative portion of the wave (Fig. 261). Find the Fourier series of the resulting periodic function if E sin Solution. -L 0 u(t) = { < t < O. p = 2L wi ~ L= W 0< r < L if 7T W Since u = 0 when -L < t < 0, we obtain from (6a), with t instead of x, f7C/W W ao = - 27T 0 E E sin wt dt = 7T and from (6b), by using formula (11) in App. A3.1 with x = wt and y = I1wt, an = !:'!... f7C/W E sin WT cos I1WT dt 7T ~E _7T = 0 If 11 = I, the integral on the right is zero, and if 11 an = = f",IW[Sin (1 + + sin (1 - I1)M] (~ : ~7T + If 11 is odd, thi, is equal to zero, and for even 11 we have cos <I - I1)Wt ] 7C/w (I - l1)w 0 + _-_c_o_s_<_I_-_")_7T_+_1 ) . 1 - 11 2E (11 - In a similar fashion we find from (tiC) that b i = E12 and bn u(t) = ~ + f sin wt - z: dt. 2, 3, ... , we readily obtain wE [_ cos (1 + Il)wt 27T (1 + I1)W 2: (-cos 111M 0 ~ = 0 for 1)(11 11 = ~ + 1)7T (Il = 2,4, .. '). 2,3, .... Consequently, (1 3 cos 2mt + 3 5 cos 4wt + .. -) . • CHAP. 11 490 11-111 Fourier Series, Integrals, and Transforms FOURIER SERIES FOR PERIOD P = lL Fmd the Fourier series of the function f(x), of pedol! p = 2L, and sketch or graph the first three partial sums. (Show the details of your work.) 1. f(x) = -1 (-2 < x < 0). f(x) = 1 (0 < x < 2). p = 4 2. f(x) = 0 (- 2 < x < 0). f(x) = 4 (0 < x < 2). p = 4 3. f(x) = x 2 ( - I < x < 1), p = 2 4. f(x) 7Tx 312 (-I < x < I), p = 2 5. f(x) sin TTX (0 < X < I), P = 1 6. f(x) cos TTX < x < ~), p = 1 7.f(x) Ixl (-l<x<l). p=2 I + x if - I < x < 0 8. f(x) = { 1 _ x if 0 < x < 1. p = 2 (-4 9. f(x) = I - ~2 (-1 < x < I), P = 2 10. f(x) = 0 (-2 < x < 0), f(x) = x (0 < x < 2), p = 4 ll.f(x)=-x (-I<x<O), f(x)=x (O<x<I). f(x) = 1 tl < x < 3), p = 4 12. (Rectifier) Find the Fourier series of the function obtained by passing the voltage v(t) = Vo cos 100m through a half-wave rectifier. 13. Show that the familiar identities cos3 x =! cos x + ~ cos 3x and sin 3x can be interpreted as sin3 x = ~ sin x Fourier series expansions. Develop cos4 x. ! 11.3 14. Obtain the series in Prob. 7 from that in Prob. 8. 15. Obtain the series in Prob. 6 from that in Prob. 5. 16. Obtain the series in Prob. 3 from that in Prob. 21 of Problem Set 11.1. 17. Using Prob. 3, show that I - !+~ 18. Show that I - k + - . . . = fz7T +! + ~ + k + ... 2 = ~7T2. 19. CAS PROJECT. Fourier Series of 2L-Periodic Functions. (a) Write a program for obtaining partial sums of a Fourier series (1). (b) Apply the program to Probs. 2-5. graphing the first few partial sums of each of the four series on common axes. Choose the first five or more partial sums until they approximate the given function reasonably well. Compare and comment. 20. CAS EXPERIMENT. Gibbs Phenomenon. The partial sums ,1'n(X) of a Fourier series show oscillations near a discontinuity point. These oscillations do not disappear as 1l increases but instead become sharp "spikes." They were explained mathematically by 1. W. Gibbs3 • Grdph sn(x) in Prob. 10. When 11 = 50. "ay. you will see those oscillations quite distinctly. Consider other Fourier series of your choice in a similar way. Compare. Even and Odd Functions. Half-Range Expansions The function in Example 1, Sec. 11.2, is even, and its Fourier series has only cosine terms. The function in Example 2, Sec. 11.2, is odd, and its Fourier series has only sine terms. Recall that g is even if g( - x) = g(x), so that its graph is symmetric with respect to the vertical axis (Fig. 262). A function h is odd if h( - x) = - hex) (Fig. 263). Now the cosine terms in the Fourier series (5), Sec. I L.2. are even and the sine terms are odd. So it should not be a surprise that an even function is given by a series of cosine terms and an odd function by a series of sine terms. Indeed, the following holds. 3 JOSIAH WILLARD GIBBS (1839-1903). American mathematician. professor of mathematical physics at Yale from 1871 on. one of the founders of vector calculus [another being O. Heaviside (see Sec. 6.1)], mathematical thermodynamics. and statistical mechanics. His work was of great importance to the development of mathematical physics. SEC. 11.3 491 Even and Odd Functions. Half-Range Expansions y y x Fig. 262. THEOREM 1 Even function Fig. 263. Odd function Fourier Cosine Series, Fourier Sine Series The Fourier series of an even function of period 2L is a "Fourier cosine series" ro (1) f(x) = £/0 + 2: 117T an cos L (f even) X n=l with coefficients (note: integration from 0 to L only!) 1 (2) ao = - L J L f(x) dx, 0 2 JL an = L 0 tl7TX f(x) cos - - dx, L n = 1,2, .. '. The Fourier series of an odd function of period 2L is a "Fourier sine series" ro (3) = f(x) 2: bn sin n7T L x (f odd) n=l with coefficients (4) PROOF 2 JL n7TX f(x) sin - - (h. L 0 L bn = - Since the definite integral of a function gives the area under the curve of the function between the limits of integration. we have L J J L g(x) d>:: =2 -L J g(x) dx for even g 0 L hex) dx = 0 for odd h -L as is obvious from the graphs of g and h. (Give a formal proof.) Now let f be even. Then (6a), Sec. 11.2, gives ao in (2). Also, the integrand in (6b), Sec. 11.2, is even (a product of even functions is even), so that (6b) gives an in (2). Furthermore, the integrand in (6c), Sec. 11.2, is the even f times the odd sine, so that the integrand (the product) is odd, the integral is zero, and there are no sine terms in (1). 492 CHAP.11 Fourier Series, Integrals, and Transforms Similarly, if f is odd. the integrals for ao and an in (6a) and (6b). Sec. 11.2. are zero, f times the sine in (6c) is even. (6c) implies (4), and there are no cosine terms in (3). • oc If L = The Case of Period 27T. coefficients I (2*) ao =- 7f 7f, + then f(x) = ao ~ an cos nx (f even) with n~l 2 f f(x) dx, 'iT an = - 0 7f f f(x) cos nx dx, 'iT n = 1,2, ... 0 co and f(x) = ~ bn sin nx (f odd) with coefficients n=l 2 (4*) bn = 7f f 'iT f(x) sin nx dx, n = 1,2,···. 0 For instance, f(x) in Example I, Sec. ILl, is odd and is represented by a Fourier sine series. Further simplifications result from the following property, whose very simple proof is left to the student. THEOREM 2 Sum and Scalar Multiple The Fourier coefficients of a sum h + f2 are the sums of the corresponding Fourier coefficients of f 1 and f 2· The Fourier coefficients of cf are c times the corresponding Fourier coefficiencs off· E X AMP L E 1 Rectangular Pulse The function f"(x) in Fig. 264 is the sum of the function f(x) in Example I of Sec 11.1 and the constant k. Hence. from that example and Theorem 2 we conclude that f*(x) = k E X AMP L E 2 4k + -:; ( sin x + "31 sin 3x + 5"1 sin 5x • + . .. ) . Half-Wave Rectifier The function u(t) in Example 3 of Sec. 11.2 has a Fourier cosine series plus a single term vCr) = (E/2) sin wi. We conclude from this and Theorem 2 that U(l) - Vel) must be an even function. Verify this graphically. (See Fig. 265.) • y [*(x) 2k -1r o Fig. 264. Example 1 Fig. 265. u(t) - v(t) with E = 1, W = 1 SEC. 11.3 493 Even and Odd Functions. Half-Range Expansions EXAM PLE 3 Sawtooth Wave Find the Fourier series of the function (Fig. 266) f(x) = x + 7T if -7T < x (a) (b) < 7T l(x and + 27T) = f(x). The functionf(x) Partial sums 81> 8 2 , 8 3, 8 20 Example 3 Fig. 266. Solution. We have f = iI + f2' where h = x and f2 = 7T. The Fourier coefficients at f2 are zero, except for the first one (the constant term). which is 7T. Hence, by Theorem 2. the Fourier coefficients an' bn are those of iI, except for ao, which is 7T. Since iI is odd, an = 0 for n = 1,2, ... , and bn = 2.7T J('iIlX)sin ny dx = 2.7T J("x sinllx cL--.:. o o Integrating by parts, we obtain b n 2 7T =- [ --XCOSIlX 11 Hence b i = 2, b 2 = - 2/2, bs = 2/3, b4 f(x) = 7T + I'" + - f'"cosl1xcL>:] 1 0 = 11 = - 0 2 COSl17T. - 11 -214, ... , and the Fourier series of f(x) is 2 (Sin x - ~ sin 2x + ~ sin 3x - + ... ) . • Half-Range Expansions Half-range expansions are Fourier series. The idea is simple and useful. Figure 267 explains it. We want to represent f(x) in Fig. 267a by a Fourier series. where f(x) may be the shape of a distorted violin string or the temperature in a metal bar of length L, for example. (Corresponding problems will be discussed in Chap. 12.) Now comes the idea. 494 CHAP. 11 Fourier Series, Integrals, and Transforms [(x)~ L x (a) The given function [(x) -L (b) [(x) x L extended as an even periodic function of period 2L (e) [(x) extended as an odd periodic function of period 2L Fig. 267. (a) Function fIx) given on an interval 0 ~ x ~ L (b) Even extension to the full "range" (interval) -L ~ x ~ L (heavy curve) and the periodic extension of period 2L to the x-axis (c) Odd extension to -L ~ x ~ L (heavy curve) and the periodic extension of period 2L to the x-axis We could extend I(x) as a function of period L and develop the extended function into a Fourier series. But this series would in general contain both cosine and sine terms. We can do better and get simpler series. Indeed, for our given I we can calculate Fourier coefficients from (2) or from (4) in Theorem l. And we have a choice and can take what seems more practicaL If we use (2). we get (1). This is the even periodic extension II of f in Fig. 267b. If we choose (4) instead. we get (3), the odd periodic extension I2 of I in Fig. 267c. Both extensions have period 2L. This motivates the name half-range expansions: I is given (and of physical interest) only on half the range, half the interval of periodicity of length 2L. Let us illustrate these ideas with an example that we shall also need in Chap. 12. E X AMP L E 4 "Triangle" and Its Half-Range Expansions Find the two half-range expansions of the function (Fig. 268) L 2k L x o Ll2 Fig. 268. The given function in Example 4 T f(x) = X { 2k T(L - Solution. (a) E,'en periodic extell.~ion. an 2 =L [2k X} if O<x<2" if 2"<x<L. L From (2) we obtain rUzxcosT xdx + T2k fLuz(L-x)cosT 117T xdx] . T J o 1l7T SEC. 11.3 495 Even and Odd Functions. Half-Range Expansions We cunsider an' For the first integral we obtain by integration by parts r J 0 ~ sin l2 X cos nTT x (iT = L nTT 2 I L fL/ S1I1 nTT x dx nTT 0 L nTT x L/2 L 0 L2 sin nTT 211TT 2 + 1) . 2L22 (cos 112TT _ n TT Similarly, for the second integral we obtain f L nTT x dx = - (L - x) cos - L/2 L L 11TT (L - x) sin - nTT IL x L + L/2 L 11TT f L nTT sin -xdx L L/2 We insert these two results into the formula for an' The sine terms cancel and so does a factor L2. This gives (2 cos 112TT - cos n TT - 4k n 2 TT2 1) . Thus, and a" = 0 if n * 2.6. 10. 14..... Hence the first half-range expansion of f(x) = ~ 2 - ~ (~ 22 TT2 cos 2TT X L + ~2 cos 6TT X L 6 f(x) is (Fig. 269a) + ... ) . This Fourier cosine series represents the even periodic extension of the given function f(x), of period 2L. (b) Odd periodic exte11sio11. Sunilarly, from (4) we obtain 8k 11TT bn = 2 2 sin - . n 1T 2 (5) Hence the other half-range expansion of f(x) is (Fig. 269b) 8k f(X) = TT2 ( 1 TT 1 12 sin LX - 32 3TT sin LX + 52 5TT sin L X - This series represents the odd periodic extension of f(x), of period 2L. Basic applications of these results will be shown in Sees. 12.3 and 12.5. -L o • x L (a) Even extension x (b) Odd extension Fig. 269. Periodic extensions of [(xl in Example 4 CHAP. 11 496 Fourier Series, Integrals, and Transforms S£E?H:r-~-- [I~ EVEN AND ODD FUNCTIONS 12. fIx) = Are the following functions even. odd. or neither even nor odd? 1. lxi, x 2 sin IIX, x + x 2 • e- 1xl , In x, x cosh x 2. sin (X2), sin 2 x, x sinh x, Ix3 1, e=-, xe x , tan 2x, xlO + x2) Are the following functions, which are assumed to be periodic of period 27T. even. odd, or neither even nor odd? 3. lex) = x 3 ( - 7T < X < 7T) 4. lex) = x 2 (-7T/2 < x < 37T12) 5. f(x) = e- 4x (-7T < x < 7T) lex) = x 3 sin x (-7T < X < 17) 7. lex) = xlxl - x 3 (-17 < X < 7T) 8. lex) = I - x + x 3 - x 5 (-7T < 9. f(x) = 1/(1 + \"2) if -17 < x < o. f(x) X < 7T) = -1/(1 + x2) ifO<x<7T 10. PROJECT. Even and Odd Functions. (a) Are the following expressions even or odd? Sums and products of even functions and of odd functions. Products of even times odd functions. Absolute values of odd functions. f(x) + f( -xl and f(x) - f( -x) for arbitrary f(x). (b) Write ekx , lI(l - x). sin (x + k), cosh (x + k) as sums of an even and an odd function. (c) Find all functions that are both even and odd. (d) Is cos3 \" even or odd? sin 3 x? Find the Fourier series of these functions. Do you recognize familiar identities? 111-161 FOURIER SERIES OF EVEN AND ODD FUNCTIONS 11.4 13. f(x) = { 17-X < 1712 7T12 < x < 31712 if if-17<x<O 14. lex) if 0 e if 15. f(x) -2 if 0 lex) 117-251 =c - !Ixl <X<17 <x<O <x<2 if -2 < x < 2 (p 0 2<x<6 if = 8) HALF-RANGE EXPANSIONS Find (a) the Fourier cosine series, (b) the Fourier sine serie~. Sketch J(x) and its two periodic extensions. (Show the details of your work.) 17. f(x) = I (0 < x < 2) 18. f(x) = x (0 19. lex) = 2 - 20. lex) = 21. f(x) < x < ~) (0 < x < 2) (0 < x < 2) x o {1 (2 1 -- {2 (0 (I 22. f(x) = { x 7T/2 23. f(x) = x < x < < x < 4) I) < x < 2) (0 < x < 7T12) (1712 < x < 7T) < x < L) (0 < \. < L) X (0 < x < 17) (0 24. f(x) = x 2 Is the given function even or odd? Find its Fourier series. Sketch or graph the function and some partial sums. (Show the details of your work.) 11. lex) = 17 - Ixl (-7T < x < 7T) (-I<x<l) if -1712 < x X 16. 6. 2xlxl 25. f(x) = 7T - 26. Illustrate the formulas in the proof of Theorem I with examples. Prove the formulas. Complex Fourier Series. Optional In this optional section we show that the Fourier series (1) f(x) = ao +~ (an cos IIX + bn sin I1X) n~l can be written in complex form, which sometimes simplifies calculations (see Example 1, on page 498). This complex form can be obtained because in complex, the exponential function e it and cos t and sin t are related by the basic Euler formula (see (11) in Sec. 2.2) (2) eit = cos t + i sin T. Thus e- it = cos t - i sin t. SEC. 11.4 Complex Fourier Series. 497 Optional Conversely, by adding and subtracting these two fonnulas, we obtain (b) (3) 1. . sin t = _(e't - e- lt ). 2i From (3), using 1Ii = -i in sin t and setting t = nx in both formulas, we get 1 . 1nx -2 (an - ib.,o}e . We insert this into (1). Writing ao = we get from (l) + "2I !(an - ibn) Co' (a n . -'nx lbn )e • . + = Cn' and !(an + ibn) = kn , 00 (4) f(x) = + Co ~ (cne inx + kne- inx ). n=1 The coefficients Cl' C2, • • • • and klo k2 • then (2) above with t = nx. 1 C n = -2 (an - ibn) = - 1 27T f •.• are obtained from (6b), (6c) in Sec. 11.1 and 7T f(x)(cos nx - i sin llX) dx _" = - 1 27T f 7T • f(x)e- mx dx -TT (5) 1 kn = - (an 2 + ibn) f 27T = - 1 f(x)(cos llX + f 27T 1 7T i sin 1LX) dx = - -7T 7T _ f(x)e 1nx dx. -7T Finally, we can combine (5) into a single formula by the trick of writing kn = (4). (5), and Co = ao in (6a) of Sec. ll.l give (summation from -cx::!) en' Then 00 cnein.r , f(x) = ~ n=-co (6) = - C 11. 1 f 7T f(x)e- tnx dx, 11 27T_-rr = O. ±1, ±2, .. '. This is the so-called complex fOl"l/l of the Fourier series or, more briefly, the complex Fourier series, of f(x). The Cn are called the complex Fourier coefficients of f(x). For a function of period 2L our reasoning gives the complex Fourier series 00 f(x) (7) = ~ Cnein7rxlL, n=-x 11 = 0, ±l, ±2,···. 498 E X AMP L E 1 CHAP. 11 Fourier Series, Integrals, and Transforms Complex Fourier Series Find the complex Fourier series of fex) = eX if -7T < x < 7T and f(x Fourier series. + 27T) = f(x) and obtain from it the usual Solution. Since sin n7T = 0 for integer n, we have e"'in7T = cos n7T ::':: i sin n7T = cos n7T = (-I)n. With this we obtain from (6) by integration en = 27T ITT' eXe- inx dx = eX-inxl"" __ 1_ X~-7T 27T 1 - in -7T On the right, I I - in + in (I - in)(I + I in) I + in + n2 e 7T - e- 7T = 2 sinh 7T. and Hence the complex Fourier ,erie, is sinh 7T (8) 7T 00 L I + in . (_l)n - - - 2 - e1nx n~-oo I + From this let us derive the real Fourier series. Using (2) with t (I + il1)i nx = (I + ill)(cos IU: + Now (8) also has a corresponding term with sin (-In) = -sin IIX, we obtain in this term i sin In) -II ( - 7T < = l1X and i 2 = = (cos nx - X < 7T). n -1, we have in (8) n sin l1x) + ;(n cos nx + sin l1x). instead of n. Since cos (-nx) = cos IU: and (I - in)e- inx = (I - ;n)(cos nx - i sin nx) = (cos l1X - n sin IU:) - i(n cos nx + sin nx). If we add these two expressions, the imaginary parts cancel. Hence their sum is 2(cos nx - n sin nx), n = 1,2,···. For II = 0 we get I (not 2) because there is only one term. Hence the real Fourier series is (9) 2sinh7T[1 I - - - - - (cos x - sin x) 7T 2 I + 12 eX = - - - I + -- (cos il - 2 sin il) - + .. 1 + 22 J In Fig. 270 the poor approximation near the jumps at ::'::7T is a case of the Gibbs phenomenon (see CAS Experiment 20 in Problem Set 11.2). • y 25 ~ 20 15 10· / 5/ -lC, Fig. 270. o lC Partial sum of (9), terms from n = X 0 to 50 SEC. 11.5 499 Forced Oscillations 1. (Calculus review) Review complex numbers. 2. (Even and odd functions) Show that the complex Fourier coefficients of an even function are real and those of an odd function are pure imaginary. 3. (Fourier coefficients) Show that ao = Co, an = C n + C- n , b n = i(c n - c- n )· 4. Verify the calculations in Example 1. S. Find further temlS in (9) and graph partial sums with your CAS. 6. Obtain the real series in Example 1 directly from the Euler formulas in Sec. II. [7-131 10. Convert the series in Prob. 9 to real form. X < 7r) 12. Convert the series in Prob. II to real form. 13. f(x) = x (0 < x < 27r) 14. PROJECT. Complex Fourier Coefficients. It is very interesting that the Cn in (6) can be derived directly by a method sinlllar to that for an and b n in Sec. 11.1. For this, mUltiply the series in (6) by e- imx with fixed integer m, and integrate term wise from -7r to 7r on both sides (allowed, for instance, in the case of uniform convergence) to get COMPLEX FOURIER SERIES I7T f(x)e- imx dx = ~ Find the complex Fourier series of the following functions. (Show the details of your work.) 7. f(x) = -1 if - 7r < X < 0, f(x) = 1 if 0 < x < 7r 8. Convert the series in Prob. 7 to real form. 9. f(x) = x (-7r < X < 7r) 11.5 (-7r < 11. f(x) = x 2 cn I7T ei(n-m)x dx. n=-OO-71" -7r Show that the integral on the right equals 27r when n = m and 0 when n =1= m [use (3b)], so that you get the coefficient formula in (6). Forced Oscillations Fourier series have important applications in connection with ODEs and PDEs. We show this for a basic problem modeled by an ODE. Various applications to PDEs will follow in Chap. 12. This will show the enormous usefulness of Euler's and Fourier's ingenious idea of splitting up periodic functions into the simplest ones possible. From Sec. 2.8 we know that forced oscillations of a body of mass m on a spring of modulus k are governed by the ODE (1) my" + cy' + f....), = ret) where y = yet) is the displacement from rest, c the damping constant, k the spring constant (spring modulus), and r(t) the external force depending on time t. Figure 271 shows the model and Fig. 272 its electrical analog, an RLC-circuit governed by c R L E(t) Fig.271. Vibrating system under consideration Fig. 272. Electrical analog of the system in Fig. 271 (RLC-circuit) 500 CHAP.11 Fourier Series, Integrals, and Transforms Ll" (1*) I + RT' + - C T = E' (t) (Sec. 2.9). We consider (1). If ret) is a sine or cosine function and if there is damping (c > 0), then the steady-state solution is a harmonic oscillation with frequency equal to that of r(t). However, if r(t) is not a pure sine or cosine function but is any other periodic function, then the steady-state solution will be a superposition of harmonic oscillations with frequencies equal to that of r(t) and integer multiples of the latter. And if one of these frequencies is close to the (practical) resonant frequency of the vibrating system (see Sec. 2.8), then the corresponding oscillation may be the dominant patt of the response of the system to the external force. This is what the use of Fourier series will show us. Of course, this is quite surprising to an observer unfamiliar with Fourier series, which are highly important in the study of vibrating systems and resonance. Let us discuss the entire situation in terms of a typical example. E X AMP L E 1 Forced Oscillations under a Nonsinusoidal Periodic Driving Force In (I), let In = 1 (gm), C = 0.05 (gmfsec), and k y" (2) where r(t) is measured in gm • cmfsec t 2 . = 25 (gmfsec 2 ), so that (1) + 0.05/ + 25y = r(t) Let (Fig. 273) +~ if -7T<t<O, if O<t<7T, 2 r(t) = { -( + becomes r(t IT 2" + 27T) = r(t). Find the steady-state solution yet). Fig. 273. Solution. We represent ret) by a Fourier series. finding (3) r(t) = ~ (cos t 7T (take the answer (4) Force in Example 1 [0 + ~ 3 cos 3t + ~ 52 cos 5t + ... ) Prob. 11 in Problem Set 11.3 minus ~7T and write .v " + 0.05y. I + 251'- = -24- cos Ilt 11 t (n for x). Then we consider the ODE = 1. 3.... ) 7T whose right side is a single term of the series (3). From Sec. 2.8 we know that the steady-state solution vn(t) of (4) is of the form (5) Yn = An cos I1f + Bn sin nt. SEC. 11.5 SOl Forced Oscillations By substituting this into (4) we find that (6) 0.2 An = Since the ODE (2) where i~ linear. we may expect the steady-state solution to be (7) Y = .1'1 + )'3 + Y5 + ... where )'n is given by (5) and (6). In fact, this follows readily by substituting (7) into (2) and using the Fourier series of r( t), provided that termwise differentiation of (7) is permissible. (Readers already fami) iar with the notion of uniform convergence [Sec. 15.51 may prove that (7) may be diilerentiated term by term.) From (6) we find that the amplitude of (5) is (a factor Vii;. cancels out) Numeric values are C 1 = 0.0531 C3 = 0.0088 C5 = 0.2037 C7 = 0.0011 C9 = 0.0003. Figure 274 shows the input (multiplied by 0.1) and the output. For n = 5 the quantity Dn is very small. the denominator of C5 is small, and C5 is so large that Y5 is the dominating term in (7). Hence the output is almost a harmonic oscillation of five times the frequency of the driving force, a little distorted due to the term Yl, whose amplitude is about 25% of that of Y5' You could make the situation still more extreme by decreasing the damping • constant c. Try it. y 0.3 Fig. 274. Input and steady-state output in Example 1 1. (Coefficients) Derive the fonnula for en from An and Bn. 2. (Spring constant) What would happen to the amplitudes en in Example 1 (and thus to the fonn of the vibration) if we changed the spring constant to the value 97 If we took a stiffer spring with k = 817 First guess. 3. (Damping) In Example I change c to 0.02 and discuss how this changes the output. 4. (Input) What would happen in Example I if we replaced ret) with its derivative (the rectangular wave)? What is the ratio of the new en to the old ones? 502 CHAP.11 15-111 Fourier Series, Integrals, and Transforms 114-171 GENERAL SOLUTION w 2y = ret) Find a general solution of the ODE y" + r(t) as given. (Show the details of your work.) Find the steady-state oscillation of y" + c/ + Y = r(t) with c > 0 and ret) as given. (Sho\'i the details of your work.) 14. ret) = an cos III 15. r(t) = sin 3t if -7T12 < t < ,,12 7Tt with 5. r(t) = cos wt, w = 0.5, 0.8, U, 1.5, 5.0, 10.0 6. r(t) = cos WIt + cos w2t (w 2 2 =1= W1 , W22) N 7. r(t) = 2 an cos Ilt, Iwl 1, 2, ... , N =1= 16. reT) n=1 8. r(t) sin t + l sin 3t + t+7f if 9. r(t) { and r(t = -t + + { r(t + and 17. ret) = O<t<7T 27T) = ret), Iwl =1= 0, 1,3, - 7T12 < t < 7f( 7T - t) + r(t if 7T12 < t < 3,,/2 27f) = ret) if + 7T "4 7T12 Isin t/ < t < 37T12 Iwl 27T) = reT), =1= 1,3,5, . . . if -7T < t < 7T and 27T) = ret). Iwl =1= o. 2 b n sin nt 18. CAS EXPERIMENT. Maximum Output Graph and discus~ outputs of y" + cy' + /...y with r(t) as in Example I for various c and emphasis on the maximum Cn and its ratio second largest Icni. 7T12 Term. = ret) k with to the ~9_-~ RLC-CIRCUIT Find the steady-state current I(t) in the RLC-circuit in Fig. 272, where R = 100 n, L = 10 H, C = 10- 2 F and E(t) V as follows and periodic with period 27f. Sketch or graph the first four partial sums. Note that the coefficients of the solution decrease rapidly. 19. E(t) = 200t( 7T2 - t 2 ) (- 7T < t < 7f) 2. 4 . . . . 12. (CAS Program) Write a program for solving the ODE just considered and for jointly graphing input and output of an initial value problem involving that ODE. Apply the program to Probs. 5 and 9 with initial values of your choice. 13. (Sign of coefficients) Some An in Example 1 are positive and some negative. Is this physically understandable? 11.6 { n=l ( r(t 11. ret) = t sin 7t -7T<t<O 7f if 7T- and + N if 10. r(t) ! sin 5t STEADY-STATE DAMPED OSCILLATIONS 100 (7Tt 20. E(t) = { + lOO( 7Tt - t 2) if - 7f < t < 0 (2) if 0 < t < 7T Approximation by Trigonometric Polynomials Fourier series playa prominent role in differential equations. Another field in which they have major applications is approximation theory, which concerns the approximation of functions by other (usually simpler) functions. In connection with Fourier series the idea is as follows. Let lex) be a function on the interval -7T" ~ X ~ 7f that can be represented on this interval by a Fourier series. Then the Nth partial sum of the series N (1) f(x) = 00 + 2: (on cos nx + bn sin nx) n=l is an approximation of the given f(x). It is natural to ask whether (l) is the "best"" approximation of f by a trigonometric polynomial of degree N, that is, by a function of the form N (2) F(x) = Ao + 2: (An cos nx + Bn sin nx) (N fixed) n=l where "best" means that the "error" of the approximation is as small as possible. SEC. 11.6 503 Approximation by Trigonometric Polynomials Of course, we must first define what we mean by the error E of such an approximation. We could choose the maximum of If - Fl. But in connection with Fourier series it is better to choose a definition that measures the goodness of agreement between f and F on the whole interval - 7T ~ X ~ 7T. This seems preferable, in particular if f has jumps: F in Fig. 275 is a good overall approximation of f, but the maximum of If - FI (more precisely, the supremum) is large (it equals at least half the jump of fat Xo). We choose E = (3) J'" (f - Fi dx . -'" This is called the square error of F relative to the function f on the interval -7T ~ X ~ 7T. Clearly, E ~ O. N being fixed. we want to determine the coefficients in (2) such that E is minimum. Since (f - Ff = f2 - 2fF + F2, we have E = (4) J'" f2 dx - 2 J'" fF dx + J'" F2 dx. -~ -~ -~ We square (2), insert it into the I&<;t integral in (4), and evaluate the occurring integrals. This gives integrals of cos 2 m: and sin2 1u (n ~ 1), which equal 7T, and integrals of cos nx, sin 1Z:r. and (cos nx)(sin mx). which are zero (just as in Sec. 11.1). Thus N(An cos L7T'" F2 dx = L7T'" [Ao + ~I llX + Bn sin nx) J2 dx We now insert (2) into the integral of fF in (4). This gives integrals of f cos nx as well as f sin IU, just as in Euler's formulas, Sec. 1l.1, for an and bn (each multiplied by An or Bn)' Hence J'" fF dx = 7T( 2A oao + AlaI + ... + ANaN + Bib i + ... + -'" With these expressions, (4) becomes E = J:J 2 [2A oao + d, - 27T 7T [ 2A02 (5) + + ~l (Anan + Bnbn) J ~I (An2 + B n2)J . x Fig. 275. Error of approximation BNbN)· 504 CHAP. 11 Fourier Series, Integrals, and Transforms We now take An = an and Bn = b n in (2). Then in (5) the second line cancels half of the integral-free expression in the first line. Hence for this choice of the coefficients of F the square error, call it E*, is (6) We finally subtract (6) from (5). Then the integrals drop out and we get terms An2 - 2Anan + an 2 = (An - a n )2 and similar terms (Bn - bn )2: Since the sum of squares of real numbers on the right cannot be negative, E - E* ~ 0, thus E~E*, and E = E* if and only if Ao = ao, ... , EN = bN . This proves the following fundamental minimum property of the partial sums of Fourier series. Minimum Square Error THEOREM 1 The square error of Fill (2) (with fixed N) relative to f on the interval -7T ~ X ~ 7T is millimum if alld ollly if the coefficients of Fill (2) are the Foltrier coefficients of f. This millimllm vallle E* is givell by (6). From (6) we see that E* cannot increase as N increases, but may decrease. Hence with increasing N the partial sums of the Fourier series of f yield better and better approximations to f, considered from the viewpoint of the square error. Since E* ~ 0 and (6) holds for every N, we obtain from (6) the important Bessel's inequality (7) for the Fourier coefficients of any function f for which integral on the right exists. (For F. W. Bessel see Sec. 5.5.) It can be shown (see [eI2] in App. 1) that for such a function f, Parseval's theorem holds; that is, formula (7) holds with the equality sign, so that it becomes Parseval's identity4 (8) 4 MARC ANTOINE PARSEV AL (1755-1836), French mathematician. A physical interpretation of the identity follows in the next section. SEC. 11.6 505 Approximation by Trigonometric Polynomials E X AMP L E 1 Minimum Square Error for the Sawtooth Wave Compute the minimum square error E* of F(x) with N = 1, 2, ... , 10, 20, ... , 100 and 1000 relative to f(x) = x on the interval - + 71' 71' ~ X ~ 71'. 1 Solution. + 2 (sin x Sec. 11.3. From this and (6), F(x) = 71' '2 1 sin 2, + '3 (_l)N+l sin 3x - + ... + - ~ sin Nx) by Example 3 in Numeric values are: N E* N E* N E* N E* 1 2 3 4 5 8.1045 4.9629 3.5666 2.7812 2.2786 6 7 8 9 10 1.9295 1.6730 1.4767 l.3216 L.1959 20 30 40 50 0.6129 0,4120 70 80 90 100 1000 0.1782 0.1561 0.1389 0.1250 0.0126 0.3103 0.2488 0.2077 (iO F = S1. S2, S3 are shown in Fig. 266 in Sec. 11.3, and F = S20 is shown in Fig. 276. Although l.r(x) - F(x)1 o -IT N IT X Fig. 276. F with = 20 in Example 1 is large at :+: 71' (how large?), where f is discontinuous, F approximates f quite well on the whole interval, except near :+:71', where "waves" remain owing to the Gibbs phenomenon (see CAS Experiment 20 in Problem Set 11.2). Can you think of functions f for which E* decreases more quickly with increasing N? • This is the end of our discussion of Fourier series, which has emphasized the practical aspects of these series, as needed in applications. In the last three sections of this chapter we show how ideas and techniques in Fourier series can be extended to nonperiodic functions. L:il MINIMUM SQUARE ERROR Find the trigonometric polynomial F(x) of the form (2) for which the square error with respect to the given f(x) on the interval - 7T ~ x ~ 7T is minimum, and compute the minimum value for N = 1, 2.... , 5 (or also for larger values if you have a CAS). 1. f(x) = x (-7T < X < 7T) 2. f(x) = x 2 (-7T < X < 7T) 3. f(x) = Ixl (-7T < x < 7T) 4. f(x) = .\'3 (-7T < X < 7T) 5.f(x) 6. f(x) ISinxl(-7T<x<7T) X < 7T) e- 1xl (- 7T < 8. f(x) = { X if o if < x < !7T !7T < + 7T) if -7T < < x < 7T 9. f(x) = .r(x if 0 -!7T X < ~7T x < 0, f(x) = xC-x + 7T) 10. CAS EXPERIMENT. Size and Decrease of E*. Compare the size of the minimum square error E* for functions of your choice. Find experimentally the factors on which the decrease of E* with N depends. For each function considered find the smaIIest N such that E* < 0.1. if -7T<X<O 11. (Monotonicity) Show that the minimum square error if O<X<7T (6) is a monotone decreasing function of N. How can you use this in practice? 7. f(x) CHAP.11 506 Fourier Series, Integrals, and Transforms [12-161 PARSEVAL'S IDENTITY Usmg Parseval"s identity, prove that the series have the indicated sums. Compute the first fev; partial sums to see that the convergence is rapid. 12. L + 34 + 54 + 4 7T 7 4 + ... = - 96 rr 1 "2 = 0.116850275 16 (Use Prob. 5. this set.) = 1.014678032 15. J + - I (Use Prob. 15 in Sec. 11.1.) 4 I 3 7T + - 4 + .. 24 1.08232 3234 90 (Use Prob. 21 in Sec. 1l.1.) 13. L 1 + -2 + 3 1 52 + ... = ~ 1.23370 0550 8 16. I + - 6 + ... = - 56 7 3 (Use Prob. 9, this set.) (Use Prob. 13 in Sec. 11.1.) 11.7 ILl + -6 + - 6 7T 960 = 1.001447078 Fourier Integral Fourier series are powerful tools for problems involving functions that are periodic or are of interest on a finite interval only. Sections 11.3 and L1.5 first illustrated this, and various further applications follow in Chap. 12. Since, of course, many problems involve functions that are nonperiodic and are of interest on the whole x-axis, we ask what can be done to extend the method of Fourier series to such functions. This idea will lead to "Fourier integrals." In Example I we stan from a special function fL of period 2L and see what happens to its Fourier series if we let L ~ x. Then we do the same for an arbitral}' function fL of period 2L. This will motivate and suggest the main result of this section, which is an integral representation given in Theorem 1 (below). E X AMP L E 1 Rectangular Wave Consider the periodic rectangular wave fdx) of period 2L J,N~ {; > 2 given by if -£<x<-1 if -I<x< I<x< L. if The left part of Fig. 277 shows this function for 2L = 4, 8, 16 as well as the nonperiodic function f(x), which we obtain from fL if we let L ~ x, f(x) = lim L-"" hex) = {I if -I <x 0 II -1 I dx - £ ' I an = £ II -I 2 - dx = L 1171"X cos - I otherwise. We now explore what happens to the Fourier coefficients of fL as L all n. For an the Euler formulas (6), Sec. 11.2. give I a - 0- 2L < L increa~es. II 0 Since fL is even, b n = 0 for Il71"X 2 sin (1171"IL) cos - - dx = - - - - £ L 1171"IL This sequence of Fourier coefficients is called the amplitude spectrum of fL because lanl is the maximum amplitude of the wave an cos (Ilm:lL). Figure 277 shows this spectrum fOf the periods 2L = 4, 8, 16. We see that for increasing L these amplitude, become more and more dense on the positive wn-axis. where Wn = 1l71"1L. Indeed, for 2£ = 4, 8, 16 we have I. 3, 7 amplitudes per "half-wave" of the function (2 sin wn)/(Lw n ) (dashed in the figurel. Hence for 2L = 2k we have 2k - 1 - I amplitudes per half-wave. so that these amplitudes will eventually be everywhere dense on the positive w.,-axis (and will decrease to zero). The outcome of this example gives an intuitive impression of what about to expect if we turn from our special • function to an arbitrary one, as we shall do next. SEC. 11.7 507 Fourier Integral Waveform fL (x) 1 , rno, Amplitude spectrum un(wn) , wn=nn/L fn=5 \ x ~ ,, ~ , wn ,I" n=3/ 2L=4 , n=:} 1 2 1'r\n=2 £n=lO -, x 'L n=6/ r--2L=8~ fL(;;6 1 IE W n= 14 n n=4 [1l;'I',uv 1-- . C r =20 n 1 -8 J ,..J -'=1-_ 1 x 8 0 2L= 16 4 n=12/ n=28/ Wn "'I ___________f_(;;6~_______________________ -101 Fig. 277. x Waveforms and amplitude spectra in Example 1 From Fourier Series to Fourier Integral We now consider any periodic function fL(X) of period 2L that can be represented by a Fourier series 00 n7T wn = fdx) = ao + ~ (an cos WnX + bn sin wnx), L n=l and find out what happens if we let L~ 00. Together with Example I the present calculation will suggest that we should expect an integral (instead of a series) involving cos wx and sin wx with W no longer restricted to integer multiples W = Wn = 117TIL of 7TIL but taking all values. We shall also see what form such an integral might have. If we insert an and bn from the Euler formulas (6), Sec. 1 1.2, and denote the variable of integration by v, the Fourier series of fdx) becomes We now set ll.w = Wn+l - Wn = (n + L 1)7T 1l7T 7T L L 508 CHAP. 11 Fourier Series, Integrals, and Transforms Then lIL = /1W/7T, and we may write the Fourier series in the form (1) fdx) = - 1I 2L L iLlv) dv 1=[ + - ~ (cos wnx) Llll' 7T n=l -L I L iLtv) cos WnV dv -L + (sin wnx) .lw f:/dV) sin wnv dVJ This representation is valid for any fixed L, arbitrarily large, but finite. We now let L ~ x and assume that the resulting nonperiodic function f(x) = lim iLlx) L_x is absolutely integrable on the x-axis: that is, the following (finite!) limits exist: (2) lim a~-x IOlf(x)1 dx a + lim fblf(x)1 dx b~x (written I_oo=lf(X) 1 dX) . 0 Then lIL ~ 0, and the value of the first term on the right side of (l) approaches zero. Also LlW = 7T/L ~ 0 and it seems plausible that the infinite series in (l) becomes an integral from 0 to Xl, which represents f(x), namely, f(x) (3) = - I 7T L°=[ cos wx I = f(v) cos wv dv + sin wx I -x = f(v) sin wv dv ] dw. -00 If we introduce the notations 1 (4) A(w) 7T I co = - I f(v) cos wv dv, -cc B(w) co = - I 7T f(v) sin wv dv -co we can write this in the form (5) f(x) = LX lA(w) cos wx + B(w) sin wx] dw. ° This is called a representation of f(x) by a Fourier integral. It is clear that our naive approach merely suggests the representation (5), but by no means establishes it; in fact. the limit of the series in (I) as Llw approaches zero is not the definition of the integral (3). Sufficient conditions for the validity of (5) are as follows. THEOREM 1 Fourier Integral If f(x) is piecewise continllous (see Sec. 6.1) ill eve/}' finite interml and has a right-hand derimtive alld a left-hand derivative at evel), point (see Sec ILl) and call be represented by a Fourier imegral (5) with A and B given by (4). At a point where f(x) is disconti1lllolis the value of the Fourier integral equals the average of the left- and right-hand limits of f(x) at that point (see Sec. 11.1). (Proof in Ref. [C 12]; see App. 1.) if the integral (2) exists, then f(x) SEC. 11.7 509 Fourier Integral Applications of Fourier Integrals The main application of Fourier integrals is in solving ODEs and PDEs, as we shall see for PDEs in Sec. 12.6. However, we can also use Fourier integrals in integration and in discussing functions defined by integrals, as the next examples (2 and 3) illustrate. E X AMP L E 2 Single Pulse, Sine Integral Find the Fourier integral representation of the function C f(x) = Ixl < loll> if if ---_~~:j Fig. 278. Solutioll. I (Fig. 278). I l--x Example 2 From (4) we obtain A(w) = -I 7T Joe f(v) cos >l'U dv = - -x Jl cos wv dv = -sin- - 11 I II"V 7T B(w) = - mv -1 Jl sin I 7r wU -1 2 sin w 7n1' dv = 0 -1 and (5) gives the answer (6) cos wx sin w f(x) = w dw. The average of the left- and right-hand limits of f(x) at x = I is equal to (I Furthermore. from (6) and Theorem I we obtain (multiply by 7r12) (7) f X o cos I n sin 11' dll" = II" r if + 0)/2. that is. 111. o ~x< I. = 1, 71/4 if x 0 if x> 1. We mention that this integral is called Dirichlet's discontinous factor. (For P. L. Dirichlet see Sec. 10.8.) The case x = 0 is of particular interest. If x = O. then (7) gives co (8*) We see that this integral is the limit of the 1 o sin II" 7r --dw=-. w 2 ~o-called sine integral u (8) Si(lI) = 1 sinw -- o dll' W II ~ x. The graphs of Si(lI) and of the integrand are shown in Fig. 279. Tn the case of a Fourier series the graphs of the partial sums are approximation curves of the curve of the periodic function represented by the series. Similarly, in the case of the Fourier integral (5). approximations are obtained by replacing GO by numbers a. Hence the integral as (9) 21 7r a _c_os_w_x_s_in_'_v 0 approximares the right side in (6) and therefore f(x). W dw 510 CHAP.11 Fourier Series, Integrals, and Transforms y Integrand 1 ~ 1C 2 ,; Fig. 279. Sine integral Situ} and integrand Figure 280 shows o,cillations near the points of discontinuity of f(x). We might expect that these oscillations disappear as a approaches infinity. But this is not tfile; with increasing a, they are shifted closer to the points x = :!: I. Thi~ unexpected behavior. which also occurs in connection with Fourier series. is known as the Gibbs phenomenon. (See also Problem Set 1l.2.) We can explain it by representing (9) in terms of sine integmls as follows. Using (II) in App. A3.1. we have 2 - I a cos wx sin w 7To dll' = U' - I I a sin (w + , "x) 1 + - U' 7To In the fIrst integral on the right we set w + dw wx = wx 7To r. Then dw/w = 0::<; t::<; (x + 1)1I. In the last integral we set w ~ -I. Then dw/w 0::<; t ::<; (x - I)a. Since sin (-t) = -sin t. we thus obtain 2 1T I a .sm w cos wx 0 I dw=- w 1T I a sin (w - wx) dt/t, and 0 ::<; w ::<; a corresponds to = dt/t, and 0::<; w::<; a corresponds to I(x+lla Sin. t --dtt 0 dw. l'\,' 1T I (X-lla smt . 0 - - dt. t From this and (8) we see that our integral (9) equab - I 1T Si(a[x + I]) - - I Si(lI[x - I]) 1T and the oscillations in Fig 280 result from those in Fig. 279. The increa~e of a amounts to a transformation of the scale on the axis and causes the shift of the oscillations (the waves) toward the points of discontinuity -1 and 1. • y a= 16 2x Fig. 280. -2 -1 0 2x -2 -1 0 The integral (9) for a = 8, 16, and 32 I 2x SEC. 11.7 511 Fourier Integral Fourier Cosine Integral and Fourier Sine Integral For an even or odd function the Fourier integral becomes simpler. Just as in the case of Fourier series (Sec. 1l.3), this is of practical interest in saving work and avoiding errors. The simplifications follow immediately from the formulas just obtained. Indeed. if f(x) is an evell function. then B(w) = 0 in (4) and 2 I Alw) = - (10) :x: feu) cos wu du. 0 7r The Fourier integral (5) then reduces to the Fourier cosine integral fO f(x) = (11) A(w) cos (f even). WX {hI o Similarly, if f(x) is odd, then in (4) we have A(w) = 0 and 2 (12) B(w) L GC = - f(u) sin wu du. 7r 0 The Fourier integral (5) then reduces to the Fourier sine integral IXo Blw) sin \1'X dw f(x) = (13) (f odd). Evaluation of Integrals Earlier in this section we pointed out that the main application of the Fourier integral is in differential equations but that Fourier imegral representations also help in evaluating certain integrals. To see this, we show the method for an important case, the Laplace integrals. E X AMP L E 3 Laplace Integrals We shall derive the Fourier cosme and Fourier sine integrals of f(x) = e -kX, where x> 0 and k > 0 (Fig. 2lll). The re~ult will be used to evaluate the so-called Laplace integrals. Solutioll. (a) From tIO) we have A(lI") = ~ IXe7T I Fig. 281. fIx) Example 3 in e -kv cos nov dv = - 2 k kv cos wv dv. Now. by integration by parts, 0 k +w If v = O. the expression on the right equals -kl(k 2 lero because of the exponential factor. Thus 2 e -lw ( , -lI" sm k - II'V + cos wv ) . + w 2 ). If v approaches infinity. that expression approache~ (14) By substituting this into (II) we thus obtain the Fourier cosine integral representation L :x: 2k f(x) = e -k.-.; = - 7T 0 cos 2 k liT +w 2 dll" (x> 0, k> 0), 512 CHAP. 11 Fourier Series, Integrals, and Transforms From this representation we see that f oo coswx (15) ~ dw = o k 2 +w 2 ~ fooe- (b) Similarly, from (12) we have B(w) = 7T f This equals -wl(k 2 . e -kv slllwudu=- + ",2) if u e-kx 2k kv (x> 0, k > 0). (x> 0, k > 0). sin wu du. By integration by parts 0 2 k 11' (k-SIllWU+Coswu. . ) e -ku 2 +w w = 0, and approaches 0 as u ~ ce. Thus (16) From (13) we thus obtain the Fourier sine integral representation L 00 2 k f(x) = e - x = - 7T 11'sinwx 2 2 dw. +w k 0 From this we see that = w sin U'X o k 2 + 11'2 L (17) -kx 7T 2 d11' = e • The integrals (15) and (17) are called the Laplace integrals. 11-61 EVALUATION OF INTEGRALS L 0:: Show that the given integral represents the indicated function. Hint. Use (5), (11), or (13); the integral tells you which one, and its value tells you what function to consider. (Show the details of your work.) if 0 00 1. L + w sin xw dw = 1 + w2 cosxw { wl2 rre- x x< 0 if x= 0 if x> 0 4. o oo 5. L r • smw - cos xw dw H' cos (rrwI2) l-w o 2 if O~x< rr/4 if x= 0 if x> cosxw dw if 0 < if 00 2. L • Slnw-wcosw w o 2 sinxw dw 12 = 00 3. L cosxw - - -2 dw o 1 + w rr { :0/4 - e - x if x 2 > 0 6. if 0 < x < 1 = if x if x> o:: L sin rrw sin xw 1- o 17-121 2 dw = {¥ W if 0 if FOURIER COSINE INTEGRAL REPRESENTATIONS Represent j(x) as an integral (11). 7. f(x) sin x = { I if O<x<a o if x>a Ixl Ixl < rr/2 ~ 7T12 SEC. 11.8 2 8. f(x) 9. f(x) 10. f(x) if =r 0 =e ~f = O<x<a if 16. f(x) = x>a if 0< x < 1 if x > 1 if xl2 17. f(x) = o<x< 1 - x/2 if <x<2 0 if x>2 11. f(x) = rnx 0 12. f(x) 513 Fourier Cosine and Sine Transforms {e~X if O<X<7T if X>7T if O<x<a if x>a {I 0 x 19. f(x) = 0 FOURIER SINE INTEGRAL REPRESENTATIONS Represent f(x) as an integral (13). 14. f(x) = e if O<x<a if x> a SlllX 15. f(x) = 11.8 { if 0 if O<X<7T if x> O<X<7T r X> if O<X<7T if ~x 7T 7T x> if O<x<a if x>a 20. PROJECT. Properties of Fourier Integrals (a) Fourier cosine integral. Show that (11) implies (al) (a2) f(ax) = ~ xf(x) = fOA( :) cos xw dw (Scale change) fOO B*(w) sin xw dw, o replacing co with finite upper limits of your choice. Compare the quality of the approximations. Write a short report on your empirical results and observations. 114-191 O<x< if (a> 0) 13. CAS EXPERIMENT. Approximate Fourier Cosine Integrals. Graph the integrals in Prob. 7, 9, and 11 as functions of x. Graph approximations obtained by if if {7T - x ro~x 18. f(x) = 2 B* (a3) = - dA dw ' A as in (10) x 2 f(x) = f=A*(W) cosxw dw, o A* = - d2A dw 2 . (b) Solve Prob. 8 by applying (a3) to the result of Prob.7. (c) Verify (a2) for f(x) = I if 0 < x < a and f(x) = 0 if x > a. (d) Fourier sine integral Find formulas for the Fourier sine integral similar to those in (a). Fourier Cosine and Sine Transforms An integral transform is a transformation in the form of an integral that produces from given functions new functions depending on a different variable. These transformations are of interest mainly as tools for solving ODEs, PDEs, and integral equations, and they often also help in handling and applying special functions. The Laplace transform (Chap. 6) is of this kind and is by far the most important integral transform in engineering. The next in order of importance are Fourier transforms. We shall see that these transforms can be obtained from the Fourier integral in Sec. 11.7 in a rather simple fashion. In this section we consider two of them, which are real, and in the next section a third one that is complex. 514 CHAP. 11 Fourier Series, Integrals, and Transforms Fourier Cosine Transform For an even function f(x), the Fourier integral is the Fourier cosine integral (1) (a) f(x) = t"'A(W) cos wx dw. where (b) A(w) o 2 = - 7T [see (10), (11), Sec. 11.71. We now set A(w) Then from (1 b), writing v = x, we have L"" f(v) cos wv dv 0 = "'v'ij; icCw), where c suggests "cosine." ac ~- L icCw) = (2) f(x) cos wx dx 0 'iT and from (la), CIJ (3) f(x) = ~L - 7T ie(w) cos WX dll'. 0 ATTENTION! In (2) we integrate with respect to x and in (3) with respect to 11". Formula (2) gives from f(x) a new function ie(w), called the Fourier cosine transform of f(x). Formula (3) gives us back f(x) from ie(w), and we therefore call f(x) the inverse Fourier cosine transform of ie(w). The process of obtaining the transform ie from a given f is also called the Fourier cosine transform or the Fourier cosine transJoml method. Fourier Sine Transform Similarly, for an odd function f(x), the Fourier integral is the Fourier sine integral [see (12), (13), Sec. 11.7] L 2 CIJ (4) (a) f(x) = o where B(w) sin wx dw, (b) B(w) =- 7T L ac f(v) sin wv dv. 0 We now set B(w) = "'v'ij; isCw), where s suggests "sine:' Then from (4b), writing v = x, we have (5) fsCw) A = ~ 7T L CXJ f(x) sin wx dx. 0 This is called the Fourier sine transform of f(x). Similarly, from (-I-a) we have (6) f(x) = ""A L ~ - 7T fs(w) sin wx dw. 0 This is called the inverse Fourier sine transform of is(w). The process of obtaining iAw) from f(x) is also called the Fourier sine transform or the Fourier sine transJoT11111letlzod. Other flotations are and g;;;-l and 9F;! for the inverses of ;!Fe and 9Fs, respectively. SEC. 11.8 Fourier Cosine and Sine Transforms EXAMPLE 515 Fourier Cosine and Fourier Sine Transforms Find the Fourier cosine and Fourier sine tran~forms x=a x fIx) in Solution. ifO<x<a {k j(x) = Fig. 282. of the function (Fig. 282). o if x> a From the definitions (2) and (5) we obtain by integration Example 1 , Ic(w) = fa cos wx dt = -Vf2 aw ) -:;; k (sin -w- -Vf2 -:;; k 0 , If fa Iiw) = k - 7T sin ll'X dx = 0 If - k (1 - cos aw ) . W 7T This agrees with formulas 1 in the first two tables in Sec. 11.10 (where k = 1). Note that for I(x) = k = const (0 < x < co). these transforms do not exist. (Why?) E X AMP L E 2 • Fourier Cosine Transform of the Exponential Function Find ?Fc(e -x). Solution. By integration by parts and recursion. - ?Fc(e x) = ~f2 Leo -7T 0 e -1· . cos wx dx = f; -7T e-·l' ---2 l+w This agrees with formula 3 in Table 1. Sec. 11.10. with 1I = (-cos wx + w sin wx) 1= 0 V2hr = ---2 l+w 1. See also the next example. • What did we do to introduce the two integral transforms under consideration? Actually not much: We changed the notations A and B to get a "symmetric" distribution of the constant 2/7T in the original formulas (10)-(13), Sec. 11.7. This redistribution is a standard convenience, but it is not essential. One could do without it. What have we gained? We show next that these transforms have operational properties that permit them to convert differentiations into algebraic operations (just as the Laplace transform does). This is the key to their application in solving differential equations. Linearity, Transforms of Derivatives If f(x) is absolutely integrable (see Sec. 11.7) on the positive x-axis and piecewise continuous (see Sec. 6.1) on every finite interval, then the Fourier cosine and sine transforms of f exist. Furthermore, if f and g have Fourier cosine and sine transforms, so does af + bg for any constants a and b, and by (2), oo 9F cCaf + bg) ~L cc = ~L = [af(x) - 7T a + bg(x)] cos lIJX dx 0 - 7T f(x) cos wx d-1: + b 0 ~oo gCx) cos wx dx. 7T L 0 The right side is a9Fc(f) + b2Fc(g). Similarly for 2Fs, by (5). This shows that the Fourier cosine and sine transforms are linear operations, (7) (a) '*c(af + bg) = a9Fc(f) + b9FcCg), (b) 9FsCaf + bg) = a9FsCf) + b9Fs(g). CHAP. 11 516 THEOREM 1 I Fourier Series, Integrals, and Transforms Cosine and Sine Transforms of Derivatives t' Let f(x) be continuous and absolutely integrable on the x-axis, let (x) be piecewise continuous on eve'}' finite interval, and let let f(x) ~ 0 as x ~ 00. Then (a) = w9's{f(x)} - 9' e{f' (x)} (8) 9's{f'(x)} (b) PROOF [f f(O), = -w9'e{f(x)}. This follows from the definitions by integration by parts, namely, = [f I 7T = CC - f'(x) cos wx dx 0 ~ [f(X) cos wx = - [f f(O) I: + w f~ f(x) sin wx dxJ + w9's{f(x)}; and similarly, , [fIX' 9'sff (x)} = f (x) sin wx dx - 7T 0 = ~ [f(X) sin wx = 0 - w9'e{f(x)}. I: - LX w f(x) cos wx dx ] • Formula (8a) with t' instead of f gives (when f', f" satisfy the respective assumptions for f, J' in Theorem 1) ~e { Q7; f " (x)} = Q7;' w~s{f (x)} - -V{2 -:;;. f '·0 ( ); hence by (8b) (9a) Similarly, (9b) A basic application of (9) to PDEs will be given in Sec. 12.6. For the time being we show how (9) can be used for deriving transforms. SEC. 11.8 517 Fourier Cosine and Sine Transforms E X AMP L E 3 An Application of the Operational Formula (9) Find the Fourier cosine transform <;IFc(e -ax) of f(x) = e -ax, where a > O. Solution. By differentiation, (e- ax)" = a 2 e- ax ; thus From this, (9a), and the linearity (7 a), Hence The answer is (see Table I, Sec. ll.lO) (a> 0). • Tables of Fourier cosine and sine transfonns are included in Sec. 11.10. I -10 1 12. Find the answer to Prob. 11 from (9b). FOURIER COSINE TRANSFORM < 2. 13. Obtain formula 8 in Table II of Sec. 11.1 I from (8b) and a suitable formula in Table I. 2. Let f(x) = x if 0 < x < k, f(x) = 0 if x > k. Find 14. Let f(x) = sinx if 0 < x < 7T and 0 if x> 7T. Find 9's(f). Compare with Prob. 6 in Sec. 11.7. Comment. 1. Let f(x) = - I if 0 < x < L f(x) f(x) = 0 if x > 2. Find ic(w), = 1 if 1 < x Ic(w), 3. Derive formula 3 in Table 1 of Sec. 11.10 by integration. 4. Find the inverse Fourier cosine transform f(x) from the answer to Prob. 1. Hint. Use Prob. 4 in Sec. 11.7. 5. Obtain 9';:-1(1/(1 + w 2 )) from Prob. 3 in Sec. 11.7. 6. Obtain 9';:-I(e- W ) by integration. 2 7. Find 9'c«(1 - X )-1 cos (7TX/2». Hint. Use Prob. 5 in Sec. 11.7. 8. Let f(x) = x 2 if 0 < x < I and 0 if x> 1. Find 9'cCf). 9. Does the Fourier cosine transform of X-I sin x exist? Of X-I cos x? Give reasons. 10. f(x) = 1 (0 < x < (0) has no Fourier cosine or sine transform. Give reasons. /11-201 FOURIER SINE TRANSFORM 11. Find 9's(e-"'-X) by integration. 15. In Table II of Sec. 11.10 obtain formula 2 from formula 4, using r@ = y.;;: [(30) in App. 3.1]. 16. Show that 9'sCx- 1I2 ) = w- 1I2 by setting wx = t 2 and using S(oo) = y:;;j8 in (38) of App. 3.1. 17. Obtain 9'sCe- ax ) from (8a) and formula 3 in Table I of Sec. 11.10. 18. Show that 9's(x- 3/2 ) = 2w 1/2• Hint. Set wx = t 2 , integrate by parts, and use C(oo) = y:;;j8 in (38) of App.3.1. 19. (Scale change) Using the notation of (5), show that f(ax) has the Fourier sine transform (1/a)IsCw/a). 20. WRITING PROJECT. Obtaining Fourier Cosine and Sine Transforms. Write a short report on ways of obtaining these transforms, giving illustrations with examples of your own. 518 11.9 CHAP. 11 Fourier Series, Integrals, and Transforms Fourier Transform. Discrete and Fast Fourier Transforms The two transforms in the last section are real. We now consider a third one, called the Fourier transform, which is complex. We shall obtain this transform from the complex Fourier integral. which we explain first. Complex Form of the Fourier Integral The (real) Fourier integral is [see (4), (5), Sec. 11.7] LX [A(w) cos wx + B(w) sin wx] o f(x) = (hI' where 1 A(w) = 7T f 1 x f(v) cos wv dv, 7T -x Suhstituting A and B into the integral for 1 f(x) = - B(w) Lf GC = 7T 0 J= f(v) sin wv dv. -0<: f, we have x f(v) Lcos wv cos IVX + sin wv sin wx] dv dlV. -x By the addition formula for the cosine L(6) in App. A3.1] the expression in the brackets [... ] equals cos (wv - wx) or, since the cosine is even, cos (wx - wv). We thus obtain (1 *) f(x) L=[ f"" = -1 7T 0 f(v) cos (wx - wv) dv ] dw. -x The integral in brackets is an even function of w. call it F(w). because cos (wx - wv) is an even function of w, the function f does not depend on IV, and we integrate with respect to v (not w). Hence the integral of F(w) from w = 0 to x is 1/2 times the integral of F(w) from - x to x. Thus (note the change of the integration limit!) (1) f(x) = Lx= [ L::.l(V) cos (wx - 1 = 27T wv) dv ] dw. We claim that the integral of the form (1) with sin instead of cos is zero: (2) -27T f J 1 x -0<: [ 00 f(v) sin (wx - wv) dv ] dw = O. -x This is true since sin (wx - IVV) is an odd function of lV, which makes the integral in brackets an odd function of w, call it G(w). Hence the integral of G(W) from - x to ex; is zero, as claimed. We now take the integrand of (1) plus i (= -v=T) times the integrand of (2) and use the Euler formula l( 11) in Sec. 2.2] (3) e ix = cos x + i sin x. SEC 11.9 519 Fourier Transform. Discrete and Fast Fourier Transforms Taking wx - IrV instead of x in (3) and multiplying by f(v) gives + f(v) cos (~~:\" - wv) if(v) sin (wx - wv) = f(v)ei(WX-Wv) Hence the result of adding (1) plus i times (2), called the complex Fourier integral, is (4) f f 27T = - f(x) I =:xl -oc . -= f(v)e'W(X-v) dv dw (i = v=i). It is now only a very short step to our present goal, the Fourier transform. Fourier Transform and Its Inverse Writing the exponential function in (4) as a product of exponential functions, we have 1 f(x) = - - (5) yI2; f [I -f yI2; x cc -x -cc . f(v)e-'WV dv J. e'wx dw. The expression in brackets is a function of tV, is denoted by 1(».'), and is called the Fourier transform of f; writing v = x, we have A few) (6) = -1~ f= f(x)e-'wx . dx. -x With this, (5) becomes I f(x) = - - (7) yI2; f= f(w)e'WX dw A • -0:; and is called the inverse Fourier transform of jew). Another notation for the Fourier transform is I = ?F(f), so that The process of obtaining the Fourier transform ':!F(f) = I from a given f is also called the Fourier transform or the Fourier transfon1l method. Conditions sufficient for the existence of the Fourier transform (involving concepts defined in Secs. 6.1 and 11.7) are as follows. as we state without proof. THEOREM 1 Existence of the Fourier Transform {tJCx) is absolutely integrable on the x-axis and piecewise continuous on every finite interval. then tile Fourier transform ICw) of f(x) given by (6) exists. 520 E X AMP L E 1 CHAP. 11 Fourier Series, Integrals, and Transforms Fourier Transform Find the Fourier transfonn of f(x) = I if Ixl < 1 and f(x) = 0 otherwise. Solution. Using (6) and integrating, we obtain - few) I = -- v'2; Jl e- iwx 1 e- iwx dx = - - . - - .- v'2; -1 -IW As in (3) we have e iw = cos W + i sin w, e- iw = cos w - i sin w, and by subtraction i w - e- iw = 2; sin w. Substituting this in the previous formula on the right, we see that i drops out and we obtain the answer _ few) E X AMP L E 2 "2 --;-. • ,-:;; sin w = Fourier Transform Find the Fourier transfonn '!F(e -ax) of f(x) = e -ax if X > 0 and f(x) = 0 if x < 0; here a > + iw) o. Solution. From the definition (6) we obtain by integration L oo m( ere -ax)= 1 v27r ,~ e -axe -iwx dx 0 e-ca+iw)X v'2; -(a + iw) v'2;(a • This proves fonnula 5 of Table III in Sec. 11.10. Physical Interpretation: Spectrum The nature of the representation (7) of f(x) becomes clear if we think of it as a superposition of sinusoidal oscillations of all possible frequencies, called a spectral representation. This name is suggested by optics, where light is such a superposition of colors (frequencies). In (7), the "spectral density" jew) measures the intensity of f(x) in the frequency interval between wand w + Aw (Aw small, fixed). We claim that in connection with vibrations, the integral f,o Ij(w)1 2 dw -co can be interpreted as the total energy of the physical system. Hence an integral of Ij(w)1 2 from a to b gives the contribution of the frequencies w between a and b to the total energy. To make this plausible, we begin with a mechanical system giving a single frequency, namely, the harmonic oscillator (mass on a spring, Sec. 2.4) my" + ky = o. Here we denote time t by x. Multiplication by y' gives my' y" !mv 2 + ky' y = O. By integration, + !ky2 = Eo = const where v = y' is the velocity. The first term is the kinetic energy, the second the potential energy, and Eo the total energy of the system. Now a general solution is (use (3) in Sec. 11.4 with t = x) SEC 11.9 521 Fourier Transform. Discrete and Fast Fourier Transforms w0 2 = kIm iwoX )/2, C l = CI = (al + ib l )/2. We write simply A cle , B = = A + B. By differentiation, v = y' = A' + B' = iWo(A - B). Substitution of v and)' on the left side of the equation for Eo gives where = (al - ib I iWoX cle• Then y Here Cl Wo 2 = kim. as just stated: hence mwo2 = k. Also i2 = -I. so that Hence the energy is proponional to the square of the amplitude lell. As the next step, if a more complicated system leads to a periodic solution y = f(x) that can be represented by a Fourier series, then instead of the single energy term IeII2 we get a series of squares Ienl 2 of Fourier coefficients Cn given by (6), Sec. 11.4. In this case we have a "discrete spectrum" (or "point spectrum") consisting of countably many isolated frequencies (infinitely many, in general), the corresponding Ienl 2 being the contributions to the total energy. Finally, a system whose solution can be represented by an integral (7) leads to the above integral for the energy, as is plausible from the cases just discussed. Linearity. Fourier Transform of Derivatives New transforms can be obtained from given ones by THEOREM 2 Linearity of the Fourier Transform The Fourier transform is a linear operation; that is, for any functions f(x) and g(x) whose Fourier transforms exist and any constants a and b, the Fourier transform of af + bg exists, and (8) PROOF g;(af + bg) = a'2F(f) + bgjP(g). This is true because integration is a linear operation, so that (6) gives I gjP{af(x) f v 27T + bg(x)} = ~ ~ co _ [af(x) + bg(x)]e- tWX dx -::>0 f yI2; 1 ::>0. = a -- f(x)e- tWX dx -co + J g(x)e- 1::>0. b -- yI2; tWX dx -::>C • = agjP{f(x)} + b'2F{g(x)}. In applying the Fourier transform to differential equations, the key property differentiation of functions corresponds to multiplication of transforms by iw: IS that 522 CHAP. 11 THEOREM 3 Fourier Series, Integrals, and Transforms Fourier Transform of the Derivative of I(x) Let f(x) be continuous on the x-axis and f(x) ~ 0 as f' (x) be absolutely integrable on the x-axis. Then co. Furthermore, let ~(f'(x)} = iw~{f(x)}. (9) PROOF Ixl ~ From the definition of the Fourier transform we have , ~{f (x)} = 1 -- IX,f (x)e- . ZWX yI2; -= dx. Integrating by parts, we obtain ~{f'(x)} Since f(x) ~ 0 as = ~ yI2; [f(x)e- loc - iWX -= (-iw) I oo -= f(x)e- iwx dX] . Ixl ~ ro, the desired result follows, namely, ~(f'(x)} = + 0 iw~{f(x)}. • Two successive applications of (9) give ~(f") = iw~(f') = (iwf~(f). Since (iW)2 = -w 2, we have for the transform of the second derivative of f ~{f"(x)} (10) -w2~{f(X)}. = Similarly for higher derivatives. An application of (l0) to differential equations will be given in Sec. 12.6. For the time being we show how (9) can be used to derive transforms. E X AMP L E 3 Application of the Operational Formula (9) Find the Fourier transform of xe Solution. -x' from Table III. Sec 11.10. We use (9). By formula 9 in Table III. ~(~e-X2) = ~{ . +(e-X2)'} = - ~ ~{(e-x'n = - - I 2 ill' - 1 v'2 e- w 2} 4 • SEC. 11.9 Fourier Transform. Discrete and Fast Fourier Transforms Convolution The convolution f * g of functions f (11) h(x) = (f * g)(x) = I 523 and g is defined by oo f(P)R(x - p) dp = t; f(x - p)g(p) dp. -x -~ The purpose is the same as in the case of Laplace transforms (Sec. 6.5): taking the convolution of two functions and then taking the transform of the convolution is the same as multiplying the transforms of these functions (and multiplying them by \.1'2;): THEOREM 4 Convolution Theorem Suppose that f(x) and g(x) are piecewise continuous, bounded. and absolutely intes;rable Oil the x-axis. Then '?Jf(f (12) PROOF * g) \.1'2; '?Jf(f)'?Jf(g). = By the definition, '?Jf(f * g) I I I \.1'2; x x = -- -00 . f(p)g(x - p) dp e- ZWX dx. -00 An interchange of the order of integration gives I I \.1'2; I '?Jf(f X:JC * g) = - - -x . f(p)g(x - p)e- ZWX d-..: dp. -x Instead of x we now take x - p = q as a new variable of integration. Then x = p '?Jf(f * g) I I \.1'2; I 00 00 = -- -x + q and . f(p)g(q)e-ZW(p+q) dq dp. -x This double integral can be written as a product of two integrals and gives the desired result lOX:. 00 . '?Jf(f * g) = - f(p)e- ZWP dp g(q)e-·wq dq \.1'2; I I -cxo -00 • By taking the inverse Fourier transform on both sides of (12), writing j = '?Jf(f) and as before, and noting that \.1'2; and l/\.I'2; in (12) and (7) cancel each other, we obtain g = '?Jf(g) (13) (f * g)(X) = {C j(w)g(w)e iwx dw, -x a formula that will help us in solving partial differential equations (Sec. 12.6). 524 CHAP. 11 Fourier Series, Integrals, and Transforms Discrete Fourier Transform (OFT), Fast Fourier Transform (FFT) In using Fourier series, Fourier transforms, and trigonometric approximations (Sec. 11.6) we have to assume that a function f(x), to be developed or transformed, is given on some interval, over which we integrate in the Euler formulas, etc. Now very often a function f(x) is given only in terms of values at finitely many points. and one is interested in extending Fourier analysis to this case. The main application of such a "discrete Fourier analysis" concerns large amounts of equally spaced data, as they occur in telecommunication, time series analysis, and various simulation problems. In these situations. dealing with sampled values rather than with functions. we can replace the Fourier transform by the so-called discrete Fourier transform (DFT) as follows. Let f(x) be periodic, for simplicity of period 27f. We assume that N measurements of f(x) are taken over the interval 0 ~ x ~ 27f at regularly spaced points 27fk (14) k = 0, 1, ... , N - 1. N' We also say that f(x) is being sampled at these points. We now want to determine a complex trigonometric polynomial N-l (15) q(x) = 2: inxk cne n~O that interpolates f(x) at the nodes (14). that is. q(Xk) = f(Xk). written out, with fk denoting f(Xk). N-l fk = (16) f(xk) = q(Xk) = 2: c,/nxk k = 0, 1, ... , N - 1. n~O Hence we must determine the coefficients co, ... , CN - 1 such that (16) holds. We do this by an idea similar to that in Sec. 11.1 for deriving the Fourier coefficients by using the orthogonality ofthe trigonometric system. Instead of integrals we now take sums. Namely, we multiply (16) by e- imxk (note the minus!) and sum over k from 0 to N - 1. Then we interchange the order of the two summations and insert Xk from (14). This gives N-l (17) ~ £..J k~O f N-IN-l ke -imxk _ ~ ~ iCn-m)xk _ - £..J £..J cne N-l N-l ~ ~ n=O k=O iCn-m)27TkIN - £..J Cn £..J e k~O n~O . Now We donote [ ... ] by r. For n = m we have r = eO = 1. The sum of these terms over k equals N, the number of these terms. For n m we have r 1 and by the formula for a geometric sum [(6) in Sec. 15.1 with q = rand n = N - 1] "* 1 - rN N-l 2: k=o "* rk = 1- =0 r SEC. 11.9 525 Fourier Transform. Discrete and Fast Fourier Transforms because r N = 1; indeed, since k, m, and n are integers, rN = eiCn - m )27Tk = cos 27Tk(n - m) + i sin 27Tk(n - m) = I + 0 = 1. This shows that the right side of (17) equals cmN. Writing n for m and dividing by N, we thus obtain the desired coefficient formula (18*) fk = f(Xk), n = 0, 1, ... , N - 1. Since computation of the Cn (by the fast Fourier transform, below) involves successive halfing of the problem size N, it is practical to drop the factor lIN from Cn and define the discrete Fourier transform of the given signal f = [fo fN_I]T to be the vector f = [io iN-I] with components in = (18) N-l NC = ~ fke-inXk, n fk = f(Xk), n = 0, ... , N - 1. k=O This is the frequency spectrum of the signaL In vector notation, f = FNf, where the N X N Fourier matrix FN = [enk] has the entries [given in (18)] (19) where n, k = 0, ... , N - 1. E X AMP L E 4 Discrete Fourier Transform (OFT). Sample of N = 4 Values Let N = 4 measurements (sample values) be given. Then w = e -2r.i/N = e -",i/2 = -i and thus w nk = (_i)nk. Let the sample values be. say f = [0 I 4 9]T. Then by (18) and (19). A (20) [ w' wO f = F4f = wO wO wO wO wI w2 w w 2 3 w w 4 ::1 6 9 w6 w f = [: I -i -I 1 -I :1 [~1 [-41: 8i1 = -[ -I 4 -6 -i 9 -4 - 8i From the first matrix in (20) it is easy to infer what F N looks like for arbitrary N. which in practice may be 1000 or more, for reasons given below. • f = FNf we can recreate the given signal = FIV 1 f, as we shall now prove. Here F N and its complex conjugate FN = ~ [wnk] satisfy From the DFT (the frequency spectrum) f (21a) where I is the N (21b) X N unit matrix; hence FN has the inverse -1 _ FN - 1 N FN • CHAP. 11 526 PROOF Fourier Series. Integrals. and Transforms We pr~ve (21). By the multiplication rule (row times col~n) the product matrix G N = FNFN = [gjk] in (21a) has the entries gjk = Row j ofFN times Column k ofFN . That is, writing W = wjw\ we prove that = W O + WI + ... + WN- 1 = {~ if j=l=-k if j = k. Indeed, when j = k, then w\v k = (wwl = (e2TTiINe-2uiIN)k = 1k = L so that the sum of these N tenns equals N; these are the diagonal entries of G N . Also, when j =f=. k, then W =I=- 1 and we have a geometric sum (whose value is given by (6) in Sec. 15.1 with q=Wandn=N-l) WO + WI + ... + WN-1 = 1 - WN = 0 1- W • We have seen that f is the frequency spectrum of the signal f(x). Thus the components in of f give a resolution of the 2 IT-periodic function f(x) into simple (complex) harmonics. Here one should use only n's that are much smaller than N!2, to avoid aliasing. By this we mean the effect caused by sampling at too few (equally spaced) points, so that. for instance, in a motion picture, rotating wheels appear as rotating too slowly or even in the wrong sense. Hence in applications, N is usually large. But this poses a problem. Eq. (18) requires O(N) operations for any particular n, hence O(N2) operations for, say. alln < N!2. Thus, already for 1000 sample points the straightforward calculation would involve millions of operations. However, this difficulty can be overcome by the so called fast Fourier transform (FFT), for which codes are readily available (e.g. in Maple). The FFT is a computational method for the DFT that needs only O(N) log2 N operations instead of O(N2). It makes the DFT a practical tool for large N. Here one chooses N = 2P (p integer) and uses the special fonn of the Fourier matrix to break down the given problem into smaller problems. For instance. when N = 1000, those operations are reduced by a factor lOOO/log2 1000 = 100. The breakdown produces two problems of size M = N12. This breakdown is possible because for N = 2M we have in (19) The given vector f = [fo fN_I]T is split into two vectors with M components each, namely, fev = [fo f2 fN_2]T containing the even components of f, and fod = [fl!3 fN_d T containing the odd components of f. For fev and fod we determine the DFTs r= r fev = fev.o iev.2 ~ fev.N-2 fOd = [iOd,1 iOd.3 f~od,N-l A [~ and FMfev = F Mfod involving the same M X M matrix F M' From these vectors we obtain the components of the DFT of the given vector f by the formulas (22) (a) (b) in+M = iev,n - wNniod.n 11 = 0,"', M - 1 11 = 0,"', M - 1. SEC. 11.9 527 Fourier Transform. Discrete and Fast Fourier Transforms For N = 2P this breakdown can be repeated p - 1 times in order to finally arrive at NI2 problems of size 2 each, so that the number of multiplications is reduced as indicated above. We show the reduction from N = 4 to M = NI2 = 2 and then prove (22). E X AMP L E 5 Fast Fourier Transform (FFT). Sample of N = 4 Values When N = 4. then Consequently. = WN = W fev -i as in Example 4 and M = [/0] A = F2 f ev = f2 = N12 = 2. hence It' = "'M = e-2m/ 2 = e--rr; = [I I] [foJ I -I -I. [fo + 12J = fo - f2 f2 From this and (22a) we obtain lo II lev.o + wN°lod.O = (fo + f2) + (fl + 13) = = lev,1 + 1 H'N lod.l = = fo + ft +,(2 + 13 efo - f2) - i(fl + f3) = fo - iIt - f2 + i13- Similarly. by (22b). A 0 A f2 = A fev,o - 1 A f3 = A fod,O = II'N A f ev.l - wN f ud.l = efo + f2) - (fl + f3) = fo - (fo - f2) - (-i)(fl - f3) = This agrees with Example 4, as can be seen by replacing O. l. 4, 9 with ft + fo + if1 12 - f3 - f2 - if3' fo. ft, 12, /3, • We prove (22). From (8) and (19) we have for the components of the DFT Splitting into two sums of M fn = = NI2 terms each gives M-I ~ .c.., M-I 2kn WN f2k + ~ .c.., (2k+ 1m WN .f2k+l· k=O We now use WN 2 = WM and pull out n WN from under the second sum, obtaining M-l (23) ~ f- n =.c.., k=O M-I knf WM eV,k + WN n ~ .c.., knf WM od,k' k=O The twO sums are f eV,n and f od.no the components of the "half-size" transforms F fev and Ffod ' Formula (22a) is the same as (23). In (22b) we have 11 + M instead of n, This causes a sign change in (23), namely -wN n before the second sum because This gives the minus in (22b) and completes the proof. • CHAP. 11 528 Fourier Series, Integrals, and Transforms -1. (Review) Show that 1Ii e ix - e -;,,, = 2i sin x. 12-91 = -i, e ix + e- ix = xe- X 2 cos x, 8. f(x) = { o if -1 < x < 0 otherwise FOURIER TRANSFORMS BY INTEGRATION Find the Fourier transform of f(x) (without using Table III in Sec. ILl 0). Show the details. -I 9. f(x) 01 = if-l<x<O if { 2. f(x) = { e kX if x < 0 o ifx>O 0 < x < otherwise (k> 0) OTHER METHODS k ifO<x<b 3. f(x) = { 0 otherwise e2iX 4. f(x) = { if - I < x < o otherwise e if-l<x< 5. f(x) = otherwise if -I < x < 1 6. f(x) = {: otherwise ifO<x< 7. f(x) = {: otherwise 10. Find the Fourier transform of f(x) = xe- x if x> 0 and < 0 from formula 5 in Table III and (9) in the text. Him: Consider xe- x and e- x . o if x 11. Obtain '!-F(e-x"/2) from formula 9 in Table m. 12. Obtain formula 7 in Table III from formula 8. 13. Obtain formula 1 in Table III from formula 2. 14. TEAM PROJECT. Shifting. (a) Show that if f(x) has a Fourier transform, so does f(x - a), and SC{f(x - a)} = e-iwaSC[f(x)}. (b) Using (a), obtain formula 1 in Table III, Sec. l1.10, from formula 2. (e) Shifting on the w-Axis. Show that if j(lI') is the Fourier transform of f(x), then J(w - a) is the Fourier transform of eiaxf(x). (d) Using (c), obtain formula 7 in Table TTTfrom 1 and formula 8 from 2. SEC 11.10 Tables of Transforms 529 11.10 Tables of Transforms Table I. Fourier Cosine Transforms See (2) in Sec. 11.8. f(x) I I {~ X < a otherwise 2 xa- 3 e- ax 4 e- x2/2 5 e-ax" 6 xne- ax (a> 0) ro~, ifO<x<a 7 1 (0 < a< I) (a> 0) H7f H7f H no) cos wa I ?]PeW a7f (na) see App. A3.1.) 2 ((12: W2 ) e- w2/ 2 (0) 0) otherwise _I_ e- w2/(4a> '\~ H n! (a 2 1 (W2 4a - 7f) "4 1 C1,2 4a "4 (a> 0) V2c; cos 9 sin (llX 2) (a> 0) vTc; co~ smax -x (a> 0) e- x sin x Re (a + w2)n+l _1_ [ sin a(l - w) \12; I-w cos «(lX 2) II = sinaw w 8 \0 I if 0 < fet w ) + + + Re = Real part iW)71+1 sin o( 1 + + w I w) ] 7f) H (See Sec. 6.3.) (1 - lI(w - a» I 2 - - arctan2 \12; X w I 12 lo(ax) (a> 0) \ff I Va2 - w 2 (l - lI(w - 0)) (See Secs. 5.5, 6.3.) 530 CHAP. 11 Fourier Series. Integrals. and Transforms Table II. Fourier Sine Transforms See (5) in Sec. 11.8. I J(x) {~ 1 ifO<x<a otherwise is(w) H[ 1 - cosaw ] W 7f 2 1/~ 1/,,;;' 3 1/J3/2 2~ 4 xa-I (O<a<l) = ~s(J) H r(a) sin w 7f (!7f a (rca) see App. A3.1.) 2 I e- ax 5 fI ( (a> 0) V e- ax la > 0) -- 6 x 7 xne- ax 8 xe- x2/2 9 xe- ax2 10 {Si~X (a> 0) H - IT V 7f (a > 0) if 0 < 11 cos (I\" -x 12 2a arctan - X (a x I H.· 2 ) arctan -w a II! (0 2 + ~1,2)n+ 1 1m (a + iw)n+l 1m = Imaginary part we- w2/2 < a otherwise I w + 7f I i a2 7f > 0) (a > 0) _~_v_ e-w2/4a (2a)3/2 _1_ [ sin aU - w) _ sin aO + w) ] yI2; I-w 1+w E V 2 yI2; (See Sec. 6.3.) u(w - a) sinh (/11' w e -aw SEC. 11.10 531 Tables of Transforms Fourier Transforms Table III. See (6) in Sec. 1l.9. --- I i(x) 1 2 3 4 ifb<x<c e- ibw +a {2X: , 2 (0 = ge(f) sinbw w e- icw _ iw\l2; otherwise I x 7T otherwise {~ 2 H if -17 < x < b C jeW) r; > 0) e- \I 2 a1wl a ifO<x<b b + -1 if b < x < 2b ibw 2e - e- \l2;w 2ibw 2 otherwise 5 r~~T if x > 0 1 (a> 0) \I2;(a otherwise 6 7 8 9 10 r~x e: {e: x if b < x < c eCa-iw)c _ e-a:il sin ax x -- eCa-iw)b V2-ii-Ca - iw) otherwise IT \I if -b <.r < b sin b(w - a) if b < x < c eibCa-w) _ i otherwise V2; (a> 0) _1_ I w-a 7T otherwise x + iw) eicCa-w) a-w e-w2/4a V2a (a> 0) -- H if Iwl < a; o iflwl > a 532 CHAP. 11 Fourier Series, Integrals, and Transforms --.................-. .. _ ......... =..... ........ _ • . - , ......... _ _ .. , .... _ _~ _. ........... _a iJ • 1. What is a Fourier series? A Fourier sine series? A half-range expansion? 2. Can a discontinuous function have a Fourier series? A Taylor series? Explain. 3. Why did we start with period 27f? How did we proceed to functions of any period p? 4. What is the trigonometric system? Its main property by which we obtained the Euler formulas? TIONS AND PROBLEMS [21-23J Using the answers to suitable odd-numbered problems, find the sum of 21. I - ~ + 22. + 1·3 ~ ~ - + + 3·5 5·7 + ... 23.1+b+~+ 5. What do you know about the convergence of a Fourier . ? senes. 6. What is the Gibbs phenomenon? 7. What is approximation by trigonometric polynomials? The minimum square error? 8. What is remarkable about the response of a vibrating system to an arbitrary periodic force? 9. What do you know about the Fourier integral? Its applications? I 26. (Half-range expansion) Find the half-range sine series of f(x) = 0 if 0 < x < 7f/2, f(x) = 1 if 7f/2 < x < 7f. Compare with Prob. 12. series of f(x) = x (0 < x < 27f). Compare with Prob.20. FOURIER SERIES Find the Fourier series of f(x) as given over one period. Sketch f(x). (Show the details of your work.) if -1 < x < 0 {-: 11. f(x) = O<x<1 14. f(x) if 29. For f(x) 7f/2 < x < 37f/2 X < 27f) ~0-311 X (-7f < X < 7f). GENERAL SOLUTION Solve y" + (lly = ret). where Iwl is 27f-periodic and: 31. r(t) = {2 - x IS. f(x) 16. f(x) = { - I - x 1 - x '* o. I. 2 . . . . . r(t) = < x < 3 if if -1 < x < 0 if 0 < x < 1 Isin 8m:1 (-118 < x < 1/8) 19. f(x) = x 2 (-7f/2 < x < 7f/2) = (2 if -I < x < I 18. f(x) = eX (-7f < x < 7f) 20. f(x) = (-2<x<2) x 17. f(x) MINIMUM SQUARE ERROR Compute the minimum square errors for the trigonometric polynomials of degree N = I, ... , 8: if -7f/2 < x < 7f/2 (-27f < x ~8-291 28. For f(x) in Prob. 12. if C 12. f(x) 13. f(x) 2S. What are the sum of the cosine terms and the sum of the sine terms in a Fourier series whose sum is f(x)? Give two examples. 27. (Half-range cosine series) Find the half-range cosine 10. What is the Fourier sine transform? Give examples. 11l-20 24. (Parseval's identity) Obtain the result of Prob. 23 by applying Parseval's identity to Prob. 12. x (0 < x < 27f) ~2-371 FOURIER INTEGRALS AND TRANSFORMS Sketch the given function and represent it as indicated. If you have a CAS, graph approximate curves obtained by replacing ::to with finite limits; also look for Gibbs phenomena. 32. f(x) = I if I < x < 2 and 0 otherwise, by a Fourier integral 33. f(x) = x if 0 < x < 1 and 0 otherwise, by a Fourier integral Summary of Chapter 11 533 + x/2 if -2 < x < o. f(x) = I - x/2 if o < x < 2, ((x) = 0 othenvise, by a Fourier cosine integral 35. f(x) = -I - x/2 if -2 < x < O. f(x) = 1 - x/2 if o < x < 2, f(x) = 0 otherwise. by a Fourier sine integral 36. f(x) = -4 + x 2 if -2 < x < 0, f(x) = 4 - x 2 if o < x < 2, f(x) = 0 otherwise, by a Fourier sine integral 34. f(x) .. = I .... .. - - -- -___ .·."4"."...... _.......""''''-''---'-..... -..... ·_._ ...... .... --~ =4 - x 2 if -2 < x < 2. f(x) a Fourier cosine integral 37. f(x) = 0 otherwise. by 38. Find the Fourier transform of f(x) = k if a < x < b. f(x) = 0 otherwise. 39. Find the Fourier cosine transform of f(x) x > 0, f(x) = 0 if x < O. = e- 2x if 40. Find 9' c(e- 2x ) and 9' s(e- 2x ) by formulas involving second derivatives _-_ ~ Fourier Series, Integrals, Transforms Fourier series concern periodic functions f(x) of period p = 2L, that is. by definition f(x + p) = f(x) for all x and some fixed p > 0; thus. f(x + IIp) = f(x) for any integer 11. These series are of the form 117T = + ~ f(x) = ao (I) ( an cos - n~l + bn x L 117T X) sin - (Sec. 11.2) L with coefficients, called the Fourier coefficients of f(x), given by the Euler formulas (Sec. 11.2) - f 1 2L (2) L 11 = 1. (l *) 2. • ••. 1 f(x) = fL 117TX f(x) cos - - dx -L L 117TX f(x) sin - - dx -L L L 27T +~ ao L fL = - For period 1 = - -L bn where an f(x) dx. we simply have (Sec. 11.1) (an cos nx + bn sin I1X) dx, bn 71=1 with the Fourier coefficients of f(x) (Sec. L1.1) I ao = 27T I .,,- L!(X) dx, an = - 7T f 1 .,,- _.,,- f(x) cos I1X = - 7T f .,., f(x) sin nx dx. _.,,- Fourier series are fundamental in connection with periodic phenomena, pm1icularly in models involving differential equations (Sec. 11.5, Chap. 12). If f(x) is even [f( -x) = f(x)] or odd [f( - x) = - f(x)], they reduce to Fourier cosine or Fourier sine series, respectively (Sec. 11.3). If f(x) is given for 0 ~ x ~ L only, it has two half-range expansions of period 2L, namely, a cosine and a sine series (Sec. 11.3). 534 CHAP.11 Fourier Series, Integrals, and Transforms The set of cosine and sine functions in (I) is called the trigonometric system. Its most basic property is its orthogonality on an interval of length 2L; that is, for 111 we have all integers /11 and n * I L 1117TX 117TX cos - - cos - - dx -L L L and for all integers m and = I 0, L 11l7TX n 7[;r sin - - sin - - dx L -L L =0 11. I L ITI7rX WlTX L L cos - - sin - - dx = O. -L This 0l1hogonality was crucial in deriving the Euler formulas (2). Partial sums of Fourier series minimize the square error (Sec. 11.6). Ideas and techniques of Fourier series extend to non periodic functions f(x) defined on the entire real line: this leads to the Fourier integral (3) f(x) = {O + B(w) sin wx] [A(w) cos wx dw (Sec. 11.7) o where 1 (4) A(w) = Tr I_= I x f(v) cos wv dv, = - B(w) 7r I x f(v) sin wv dv -GC or, in complex form (Sec. 11.9), (5) 1 f(x) = - - V2; IX f(w)e A twx • dw -x where (6) jew) I V2; 1 GC . f(x)e-tw:r dx. = -- -x Formula (6) transforms f(x) into its Fourier transform jew), and (5) is the inverse transform. Related to this are the Fourier cosine transform (Sec. 11.8) (7) je(w) = p; I x - 7r f(x) cos H'X dx 0 and the Fourier sine transform (Sec. 11.8) (8) A fs(W) = p; I - 7r 00 f(x) sin wx dx. 0 The discrete Fourier transform (DFT) and a practical method of computing it, called the fast Fourier transform (FFT), are discussed in Sec. I 1.9. CHAPTER 12 'f· Partial Differential Equations (PDEs) PDEs are models of various physical and geometrical problems, arising when the unknown functions (the solutions) depend on two or more variables, usually on time t and one or several space variables. It is fair to say that only the simplest physical systems can be modeled by ODEs, whereas most problems in dynamics, elasticity, heat transfer, electromagnetic theory, and quantum mechanics require PDEs. Indeed, the range of applications of PDEs is enormous, compared to that of ODEs. In this chapter we concentrate on the most important PDEs of applied mathematics, the wave equations governing the vibrating string (Sec. 12.2) and the vibrating membrane (Sec. 12.7), the heat equation (Sec. 12.5), and the Laplace equation (Secs. 12.5, 12.10). We derive these PDEs from physics and consider methods for solving initial and boundary value problems, that is. methods of obtaining solutions satisfying conditions that are given by the physical situation. In Secs. 12.6 and 12.11 we show that PDEs can also be solved by Fourier and Laplace transform methods. COMMENT. Numerics for PDEs is explained in Secs. 21.4-21.7. Prerequisites: Linear ODEs (Chap. 2), Fourier series (Chap. 11) Sections that may be omitted ill a shorter course: 12.6, 12.9-12.11 References and Answers to Problems: App. 1 Part C, App. 2 12.1 Basic Concepts A partial differential equation (PDE) is an equation involving one or more partial derivatives of an (unknown) function, call it u, that depends on two or more variables, often time t and one or several variables in space. The order of the highest derivative is called the order of the PDE. As for ODEs. second-order PDEs will be the most imp0l1ant ones in applications. Just as for ordinary differential equations (ODEs) we say that a PDE is linear if it is of the first degree in the unknown function u and its partial derivatives. Otherwise we call it nonlinear. Thus, all the equations in Example 1 on p. 536 are linear. We call a linear PDE homogeneous if each of its terms contains either u or one of its partial derivatives. Otherwise we call the equation nonhomogeneous. Thus, (4) in Example I (with f not identically zero) is nonhomogeneous, whereas the other equations are homogeneous. 535 CHAP. 12 536 E X AMP L E 1 Partial Differential Equations (PDEs) Important Second-Order POEs (I) Olle-dimellsiollal WUl'e equation (2) One-dimensional heat equation (3) Two-dimensiollal Laplace equatioll (4) Two-dimensional Poissoll equatioll (5) (J211 = ilt2 (6) - a2 u iJx 2 c 2 2 (iJ U ax 2 iJ2 11 + - iJy2 2 + a u) Two-dimellsiollal wave equatioll iJy2 (J2 u + - iJ72 =0 Three-dimellsiollal Laplace equation Here c is a posItive constant, t is time, x. y • .: are CartesIan coordinates, and dimensioll is the number of these coordinates in the equation. • A solution of a PDE in some region R of the space of the independent variables is a function that has all the partial derivatives appearing in the PDE in some domain D (definition in Sec. 9.6) containing R, and satisfies the PDE everywhere in R. Often one merely requires that the function is continuous on the boundary of R. has those derivatives in the interior of R, and satisfies the PDE in the interior of R. Letting R lie in D simplifies the situation regarding derivatives on the boundary of R, which is then the same on the boundary as it is in the interior of R. In general, the totality of solutions of a PDE is very large. For example, the functions u = eX cosy, u = sin x cosh y, which are entirely different from each other, are solutions of (3), as you may verify. We shall see later that the unique solution of a PDE corresponding to a given physical problem will be obtained by the use of additional conditions arising from the problem. For instance, this may be the condition that the solution u assume given values on the boundary of the region R ("boundary conditions"). Or, when time t is one of the variables, u (or U t = uu/ut or both) may be prescribed at t = 0 ("initial conditions"). We know that if an ODE is linear and homogeneous, then from known solutions we can obtain further solutions by superposition. For PDEs the situation is quite similar: THEOREM 1 Fundamental Theorem on Superposition If Ul and U2 are solutions of a homoge1leous linear PDE in some regioll R, then with an.v constants Cl and C2 is also a solution of that PDE in the region R. The simple proof of this imp0l1ant theorem is quite similar to that of Theorem I in Sec. 2.1 and is left to the student. SEC. 12.1 Basic Concepts 537 Verification of solutions in Probs. 14-25 proceeds as for ODEs. Problems 1-12 concern PDEs solvable like ODEs. To help the student with them. we consider two typical examples. E X AMP L E 2 Solving Uxx - Find solutions U = 0 Like an ODE of the PDE II Uxx - It = 0 depending on x and y. Soluti01l. Since no y-derivatives occur. we can solve this PDE like uTI - u = O. In Sec. 2.2 we would have obtained u = Ae x + Be with constant A and B. Here A and B may be functions of y. so that the answer i, -.l' u(x, A(y)e x + B(y)e-x y) = with arbitrary functions A and B. We thus have a great variety ot solution~. Check the result by differentiation. • E X AMP L E 3 Solving u xy Find solutions = -UK Like an ODE II = lI(X. y) of this PDE. Soluti01l. Setting "x = p. we have Py = -p. pylp = -I. y p = c!x)e- and by lnp = -y + ('(x). integration with respect to x, u(x, yl = J(x)e- y + .!,'(y) where J(x) = f c(x) £Ix; • here, J(x) and g(y! are arbitrary. --- -11-121 PDEs SOLVABLE AS ODEs 118-211 This happens if a PDE involves derivatives with respect to one variable only (or can be transformed to such a form), so that the other variable(s) can be treated as parameter(s). Solve for u = u(x. y): 1. U yy + 16u = 0 2. U.l : X = U 4. u y + 2yu 3. Uyy = 0 5. u y + U = e XY 7. u y = (cosh x)yu 9. y 2 u yy + 2yu y - 2u = 0 11. u xy 12. U yy = + 0 6. U xx = 4y 2 u 8. u y = 2xyu 10. Uyy = 4xl/y Ux lOuy + 25u = e- 5y 13. (Fundamental Theorem) Prove Fundamental Theorem I for second-order PDEs in two and three independent variables. 114-251 18. It 2U. It - Heat Equation (2) with suitable c e- 2kt cos 8x 19. II e- w2t sin 4x e- 4w2t sin wx 21. U e-w2c2t 122-251 cos wx Laplace Equation (3) 22. U in (7) in the text 23. U cos 2y sinh 2x 24. U = arctan (ylx) 25. U - e y2 x2 - sin 2xJ 26. TEAM PROJECT. Verification of Solutions (a) Wave equation. Verify that u(.\". t) = vex + ell + w(x - el) with any twice differentiable functions v and w satisfies (I). (b) Poisson equation. Verify that each u satisfies (4) with f(x. y) as indicated. u VERIFICATION OF SOLUTIONS = U = Verify (by substitution) that the given function is a solution of the indicated PDE. Sketch or graph the solution as a surface in space. u X4 + y4 cos.\ sin y = .vlx f = 12(x 2 + y2) f = -2 cos x sin f = y 2)'lx 3 (c) Laplace equation. Verify that [14-171 Wave Equation (1) with suitable c + 14. U - 4x 2 16. U = sin 3x sin 18t (2 15. U sin 8x cos 2l IIVx 2 + y2 + Z2 satisfies (6) and u = In (x 2 + y2) satisfies (3). Is u = 17. U = sin kx cos ket solution of (3)? Of what Poisson equation? u = l/Vx 2 + y2 a 538 CHAP. 12 Partial Differential Equations (PDEs) (d) Verify that II with any (sufficiently often differentiable) v and w satisfies the given PDE. + 1I = vex) !I = v(xhl"(y) 1/ = vex + equation (3) and determine a and b so that u satisfies the boundary conditions 1I = 110 on the circle x 2 + .1'2 = 1 and II = 0 on the circle.\"2 + y2 = 100. II"(Y) 31) 12S-301 + w( \" - 31) 27. (Boundary value problem) Verify that the function u(x, y) = a In (x 2 + ,.2) + b satisfies Laplace's 12.2 SYSTEMS OF PDEs Solve 28. !Ix 29. U"X = 0, !l xy = 0 30. l/xx = n, llyy = = 0, lly = 0 n Modeling: Vibrating String, Wave Equation As a first important POE let us derive the equation modeling small transverse vibrations of an elastic string, such as a violin string. We place the string along the x-axis, stretch it to length L, and fasten it at the ends x = 0 and x = L. We then dist0l1 the string, and at some instant. call it t = 0, we release it and allow it to vibrate. The problem is to determine the vibrations of the string. that is, to find its deflection u(x, 1) at any point x and at any time t > 0; see Fig. 283. u(x, t) will be the solution of a POE that is the model of our physical system to be derived. This POE should not be too complicated. so that we can solve it. Reasonable simplifying assumptions Uust as for ODEs modeling vibration~ in Chap. 2) are as follows. Physical Assumptions 1. The mass of the string per unit length is constant ("homogeneous stl;ng"). The string is perfectly elastic and does not offer any resistance to bending. 2. The tension caused by stretching the string before fastening it at the ends is so large that the action of the gravitational force on the string (trying to pull the string down a little) can be neglected. 3. The string performs small transverse motions in a ve11ical plane: that is, every particle ofthe string moves strictly vel1ically and so that the deflection and the slope at every point of the string always remain small in absolute value. Under these assumptions we may expect solutions lI(X, t) that describe the physical reality sufficiently well. u f3 p Q:....",."'""'~~. T2 -....--', r r r r r r o x Fig. 283. x+ili: L Deflected string at fixed time t. Explanation on p. 539 SEC. 12.2 Modeling: Vibrating String. Wave Equation 539 Derivation of the PDE of the Model (,lWave Equation") from Forces The model of the vibrating string will consist of a PDE (""wave equation") and additional conditions. To obtain the PDE, we consider the forces acting 011 a small portion of the string (Fig. 283). This method is typical of modeling in mechanics and elsewhere. Since the string offers no resistance to bending. the tension is tangential to the curve of the string at each point. Let T] and T2 be the tension at the endpoints P and Q of that portion. Since the points of the string move vertically, there is no motion in the horizontal direction. Hence the horizontal components of the tension must be constant. Using the notation shown in Fig. 283. we thus obtain (1) Tl cos ll' = T2 cos f3 = T = COllst. In the vertical direction we have two forces. namely, the vertical components - T] sin ll' and T2 sin f3 of Tl and T2; here the minus sign appears because the component at P is directed downward. By Newton's second law the resultant of these two forces is equal to the mass p ~x of the portion times the acceleration a2uICJ(2, evaluated at some point between x and x + ~x; here p is the mass of the un deflected string per unit length. and ~x is the length of the portion of the undeflected string. (~ is generally used to denote small quantities; this has nothing to do with the Laplacian V2, which is sometimes also denoted by ~.) Hence Using 0), we can divide this by T2 cos (2) Now tan ll' and tan tan Tl sin ll' Tl cos ll' f3 = Tl cos ll' = T, obtaining = tan f3 - tan ll' = p ~x -T f3 are the slopes of the string at x and x + ll' = (~u) I dx and tanf3= x a2 u 0"(2 ~x: I (-au) ax . x+.lx Here we have to write partial derivatives because u depends also on time (. Dividing (2) b) ~x, we thus have If we let (3) ~x approach zero, we obtain the linear PDE p This is called the one-dimensional wave equation. We see that it is homogeneous and of the second order. The physical constant Tip is denoted by c 2 (instead of c) to indicate 540 CHAP. 12 Partial Differential Equations (PDEs) that this constant is positive. a fact that will be essential to the form of the solutions. "One-dimensional" means that the equation involves only one space variable. x. In the next section we shall complete setting up the model and then show how to solve it by a general method that is probably the most imp0l1ani one for PDEs in engineering mathematics. 12.3 Solution by Separating Variables. Use of Fourier Series The model of a vibrating elastic string (a violin string, for instance) consists of the one-dimensional wave equation (1) p for the unknown deflection tI(x, t) of the string, a PDE that we have just obtained, dnd some additiollal cOllditiolls, which we shall now derive. Since the string is fastened at the ends x = 0 and x = L (see Sec. 12.2). we have the two boundary conditions (2) ~a) ufO, t) = 0, (b) ll(L, t) = 0 for all T. FUl1hermore, the form of the motion of the string will depend on its initial deflection (deflecrion ar time t = 0), call it f(x), and on its iniTiall'eiocity (velocity at t = 0). call it g(x). We thus have rhe rwo initial conditions (3) (a) l/(x. 0) = f(x) , (b) Ut(x. 0) = g(x) (0 ~ x ~ L) where lit = ilu/at. We now have to find a solution of the PDE (I) satisfying the conditions (2) and (3). This will be the suI uti on of our problem. We shall do this in three steps, as follows. Step 1. By the "method of separating variables" or product meThod. setting u(x. t) = F(x)G(t). we obtain from (l) two ODEs. one for F(x) and the other one for G(t). Step 2. We determine solutions of these ODEs that satisfy the boundary conditions (2). Step 3. FinaIl y, using Fourier series, we compose the solutions gained in Step 2 to obtain a solution of (l) satisfying both (2) and (3), that is, the solution of our model of the vibrating string. Step 1. Two ODEs from the Wave Equation (1) In the method of separating variables, or product method, we determine solutions of the wave equation (I) of the form (4) ll(X, 1) = F(x)G(T) SEC 12.3 Solution by Separating Variables. Use of Fourier Series 541 which are a product of two functions. each depending only on one of the variables x and t. This is a powerful general method that has various applications in engineering mathematics. as we shall see in this chapter. Differentiating (4), we obtain and where dots denote derivatives with respect to T and primes derivatives with respect to x. By inserting this into the wave equation (1) we have Dividing by c 2 FG and simplifying gives The variables are now separated, the left side depending only on t and the right side only on x. Hence both sides must be constant because if they were variable. then chdnging t or x would affect only one side. leaving the other unaltered. Thus. say, F" - F =k . Multiplying by the denominators gives immediately two ordinary DEs F" - kF = 0 (5) and (6) Here. the separation constant k is still arbitrary. Step 2. Satisfying the Boundary Conditions (2) We now determine solutions F and G of (5) and (6) so that u = FG satisfies the boundary conditions (2), that is, (7) u(O, t) = We first solve (5). If G and then by (7), (8) F(O)G(t) 0= (a) = 0, 0, then u = FG F(O) = o. u(L, t) =0 = F(L)G(t) = 0 for all T. 0, which is of no interest. Hence G "'" 0 (b) F(L) = o. We show that k must be negative. For k = 0 the general solution of (5) is F = ax + b, and from (8) we obtain a = b = 0, so that F 0= 0 and II = FG 0= 0, which is of no interest. For positive k = J..t2 a general solution of (5) is 542 CHAP. 12 Partial Differential Equations (PDEs) and from (8) we obtain F ~ 0 as before (verify!). Hence we are left with the possibility of choosing k negative, say, k = _p2. Then (5) becomes F" + p2F = 0 and has as a general solution F(x) = +B A cos px sin px. From this and (8) we have F(O) We must take B *- = =0 pL = F(L) and then 0 since otherwise F (9) Setting B A ~ = 117T, = B sinpL = O. O. Hence sinpL = O. Thus 117T p=- so that 1, we thus obtain infinitely many solutions F(x) (10) Fn(x) = sin (ll integer). L = F,l>:), where 117T L In -x = I, 2, .. '). These solutions satisfy (8). [For negative integer 11 we obtain essentially the same solutions, except for a minus sign, because sin (-ex) = -sin ex.] We now solve (6) with k = _p2 = -(Il7TIL)2 resulting from (9), that is. (11 *) An = cp = where C177T L A general solution is Hence solutions of (I) satisfying (2) are unlx, t) = Fn(x)Gn(t) = Gn(t)Fn(X), written out (11) (11 = I. 2... '). These functions are called the eigenfunctions, or characteristic junctions, and the values A" = C177TIL are called the eigenvalues, or characteristic ralues, of the vibrating sHing. The set {AI, A2 , ••• } is called the spectrum. Discussion of Eigenfunctions. We see that each Un represents a harmonic motion having the frequency A,,!27T = cnl2L cycles per unit time. This motion is called the 11th normal mode of the string. The first normal mode is known as the fillldalllelltal mode (17 = 1), and the others are known as overto17es; musically they give the octave, octave plus fifth, etc. Since in (II) . 177TX L Sll1 - - =0 at x= L 2L 11 11 11-1 --L, 11 the nth normal mode has 11 - I nodes, that is, points of the string that do not move (in addition to the fixed endpoints); :-.ee Fig. 284. SEC. 12.3 Solution by Separating Variables. Use of Fourier Series 543 I~'J o L n=l n=2 Fig. 284. n=4 n=3 Normal modes of the vibrating string Figure 285 shows the second normal mode for various values of t. At any instant the string has the form of a sine wave. When the left part of the string is moving down, the other half is moving up, and conversely. For the other modes the situation is similar. Tuning is done by changing the tension T. Our formula for the frequency AJ27T = cll!2L of Un with c = [see (3), Sec. 12.2] confirms that effect because it shows that the frequency is proportional ro the tension. T cannot be increased indefinitely, but can you see what to do to get a string with a high fundamental mode? (Think of both Land p.) Why is a violin smaller than a double-bass? VfiP ........-, ..,.. x ___ -'.0" Fig. 285. Second normal mode for various values of t Step 3. Solution of the Entire Problem. Fourier Series The eigenfunctions (II) satisfy the wave equation (l) and the boundary conditions (2) (string fixed at the ends). A single Un will generally not satisfy [he initial conditions (3). But since the wave equation (I) is linear and homogeneous, it follows from Fundamental Theorem 1 in Sec. 12.\ that the sum of finitely many solutions Un is a solution of (I). To obtain a solution that also satisfies the initial conditions (3), we consider the infinite series (with An = Cl17TIL as before) <Xl (12) u(x, t) = ~ ex: un(x, 1) = ~ (Bn cos Ant n~l 117T + Bn * sin Ant) sin LX. n~l Satisfying Initial Condition (3a) (Given Initial Displacement). we obtain 117T <Xl (\3) From (12) and (3a) u(x, 0) =~l Bn sin LX = f(x). Hence we must choose the Bn's so that u(x. 0) becomes the Fourier sine series of f(x). Thus, by (4) in Sec. 11.3, (14) Bn 2 = - L fL f(x) sin --dx, 117TX 0 L 11 = 1,2,·· '. 544 CHAP. 12 Partial Differential Equations (PDEs) Satisfying Initial Condition (3b) (Given Initial Velocity). (12) with respect to t and using (3b). we obtain 117TX co .L = Similarly, by differentiating Bn*An sin L = g(x). n~l Hence we must choose the Bn*' s so that for t = 0 the derivative aulat becomes the Fourier sine series of g(x). Thus, again by (4) in Sec. 11.3, BrI *ArI Since An = CI117IL, = 2 L t 0 g(x) sin L1117X dx. we obtain by division BrI * (15) = -2- t CIl17 0 1117X g(x) sin - - d\:, 11 L = 1,2,···. Result. Our discussion shows that u(x, t) given by (12) with coefficients (14) and (15) is a solution of (I) that satisfies all the conditions in (2) and (3), provided the series (12) converges and so do the selies obtained by differentiating (12) twice tennwise with respect to x and t and have the sums a2 ulax 2 and a2 ulat 2 . respectively, which are continuous. Solution (12) Established. According to our derivation the solution (12) is at first a purely formal expression, but we shall now establish it. For the sake of simplicity we consider only the case when the initial velocity g(x) is identically zero. Then the Bn* are zero, and (12) reduces to 1117X co u(x, t) = (16) .L C1117 Bn cos Ant sin - L - , L n~l It is possible to sum this series, that is. to wlite the result in a closed or finite form. For this purpose we use the formula [see (l L), App. A3.I] cos 1117 1 LC1117 t sin LX = 2 [ sin { L 1117 (x - ct) } + sin { L 1117 (x + ct) } ] Consequently, we may write (16) in the form l/(x, t) = co 21 .L Bn sin L { 1117 (x - ct) } + 21 n~l co .L Bn sin (x + cO } . n~l These two series are those obtained by substituting x - ct and x the variable x in the Fourier sine series (13) fOf j(x). Thus (17) L { 1117 u(x, t) = Hf*(x - ct) + f*(x + ct)] + ct, respectively, for SEC 123 545 Solution by Separating Variables. Use of Fourier Series where f* is the odd periodic extension of f with the period 2L (Fig. 286). Since the initial deflection f(x) is continuous on the interval 0 ~ x ~ L and zero at the endpoints, it follows from (17) that u(x, t) is a continuous function of both variables x and t for all values of the variables. By differentiating (17) we see that u(x, t) is a solution of (1), provided f(x) is twice differentiable on the interval 0 < x < L, and has one-sided second derivatives at x = 0 and x = L, which are zero. Under these conditions u(x. t) is established as a solution of (1), satisfying (2) and (3) with g(x) == O. • 'J Fig. 286. J..............- Odd periodic extension of {{x) Generalized Solution. 1ft' (x) and f"(x) are merely piecewise continuous (see Sec. 6.1), or if those one-sided derivatives are not zero, then for each t there will be finitely many values of x at which the second delivatives of u appearing in (1) do not exist. Except at these points the wave equation will still be satisfied. We may then regard u(x, t) as a "generalized solution," as it is called, that is, as a solution in a broader sense. For instance, a triangular initial deflection as in Example I (below) leads to a generalized solution. Physical Interpretation of the Solution (17). The graph of f*(x - ct) is obtained from the graph of f*(x) by shifting the latter ct units to the right (Fig. 287). This means thal f*(x - ct) (c > 0) represents a wave that is traveling to the right as t increases. Similarly, f*(x + ct) represents a wave that i:. traveling to the left. and u(x. t) is the superposition of these two waves. x Fig. 287. E X AMP L E 1 Interpretation of (l7) Vibrating String if the Initial Deflection Is Triangular Find the solution of the wave equation (1) cOlTesponding to the triangular initial deflection 2J.. f(x) = ~x { T(L - x) L O<x<- if 2 L -<x<L if 2 and initial velocity zero. (Figure 288 shows f(x) = I/(X, 0) at the top.) Solutioll. Since g(x) == 0, we have Bn * = 0 in (12). and from Example 4 in Sec. 11.3 we see that the Bn are given by (5), Sec. 11.3. Thus (12) takes the fonn I/(x, t) = 81.. [ 7T2 I )2 sin I L7T x cos L7TC t - )2 sin 37T 37TC LX cos L t + - ." J. For graphing the solution we may use u(x, 0) = fer) and the above interpretation of the two functions in the representation (17). This leads to the graph shown in Fig. 2g8. • 546 CHAP. 12 Partial Differential Equations (PDEs) /s: o t=0 L / t = Ll5e v'''--______'''.,. t = 2L15e t = Ll2c -- -t = 4L15c , ,1 t = LIe ~2f*(X-L) =!{"(x +L) 2 Fig. 288. Solution u(x, t) in Example 1 for various values of t (right part of the figure) obtained as the superposition of a wave traveling to the right (dashed) and a wave traveling to the left (left part of the figure) 11-10 I DEFLECTION OF THE STRING 7. Find u(x, t) for the string of length L = 1 and c 2 = 1 when the initial velocity is zero and the initial deflection with small k (say, 0.01) is as follows. Sketch or graph U(X, 1) as in Fig. 288. 1. k sin 2. k(sin TTX -l sin 4. kx(l - X 2 ) 27TX 3. h(l - x) 5. 37TX) 1 4 8. ol t/ 1 4 1 "2 1 4 L /' 1 0.1~._! 6 *b 4 1 -4 0.5 " 3 4 9. " It vA, 1 4 1 "2 3 4 SEC. 12.3 Solution by Separating Variables. Use of Fourier Series 547 10. 2A(1 - cos + 0.8 11. (Frequency) How does the frequency of the fundamental mode of the vibrating string depend on the length of the string? On the mass per unit length? What happens to the string if we double the tension? Why is a contrabass larger than a violin? 12. (Nonzero initial velocity) Find the deflection II(X, t) of the string of length L = 7f and ("2 = I for zero initial displacement and "triangular" initial velocity II t (x, 0) = 0.0 I x if 0 :;;; x ~ ~7f, II t (X, 0) = 0.0 I (7f - x) if ~7f ~ x :;;; 7f. (Initial conditions with lIix, 0) ~ 0 are hard to realize experimentally.) 13. CAS PROJECT. Graphing Normal Modes. Write a program for graphing lin with L = 7f and c 2 of your choice similarly as in Fig. 284. Apply the program to 112, U3, 114' Also graph these solutions as surfaces over thext-plane. Explain the connection between these two kinds of graphs. nTo(An 2 - 1I7f) 2 w ) . sm wt. Determine B" and Bn * so that II satisfies the initial conditions u(x, 0) = f(x), uix, 0) = O. (d) (Resonance) Show that if An = w, then - A - - (1 n7fW cos 1l7f)T cos WT. (e) (Reduction of boundary conditions) Show that a problem (1)-(3) with more complicated boundary conditions 11(0, t) = 0, u(L, t) = h(t), can be reduced to a problem for a new function v satisfying conditions v(O, t) = v(L, f) = 0, vex. 0) = fl(x), Vt(x, 0) = gl(X) but a nonhomogeneous wave equation. Him: Set II = V + I\" and determine w suitably. 14. TEAM PROJECT. Forced Vibrations of an Elastic String. Show the following. (a) Substitution of x (17) II(X, t) = Il7fX L GnU) sin L n=l y u (L = length of the string) into the wave equation (I) governing free vibrations leads to [see •• (18) Gn 2_ + An G - Fig. 189. (lO'~)J Elastic beam SEPARATION OF A FOURTH-ORDER POE. VIBRATING BEAM 0, (b) Forced vibrations of the string under an external force P(x, t) per unit length acting normal to the string are governed by the PDE By the prinCiples used in modeling the string it can be shown that small free vertical vibrations of a uniform elastic beam (Fig. 289) are modeled by the fourth-order PDE (21) (19) (c) For a sinusoidal force P = Ap sin wt we obtain p p n7fX ex = A sin wt = L n~l knCt) sin L ' (20) _ {(4A11l7f) k,,(t) - sin wt o (11 odd) (n even). Substituting (17) and (20) into (19) gives •• Gn + 2 _ An G n - 2A - f/7f (l - ,,' cos 1l7f) sm wt. (Ref. [CII]) where c 2 = EIIpA (E = Young's modulus of elasticity, I = moment of intertia of the cross section with respect to the y-axis in the figure, p = density, A = cross-sectional area). (Bending of a beam under a load is discussed in Sec. 3.3.) 15. Substituting II = F(x)G(t) into (21), show that F(4)/F = -C/c 2 G = 13 4 = conST, + B sin f3x + C cosh f3x + D sinh f3x, F(x) = A cos f3x G(l) CHAP. 12 548 Partial Differential Equations (PDEs) .f =-=:l1.. ~ x=o x=L += (A) Simply supported (E) Clamped at both ends x=L (e) Clamped at the left I I x=L x=o Fig. 290. end, free at the right end Supports of a beam 16. (Simply supported beam in Fig. 290A) Find solutions un = Fn(x)Gn(t) of (21) corresponding to zero initial velocity and satisfying the boundary conditions (see Fig. 290A) u(O, t) = 0, u(L, t) = 0 (ends simply supported for all times t), uxx(O, t) = 0, uxx(L, t) = 0 (zero moments, hence zero curvature, at the ends). 17. Find the solution of (21) that satisfies the conditions in Prob. 16 as well as the initial condition u(x, 0) = f(x) = x(L - x). 12.4 18. Compare the results of Probs. 17 and 3. What is the basic difference between the frequencies of the normal modes of the vibrating string and the vibrating beam? 19. (Clamped beam in Fig. 290B) What are the boundary conditions for the clamped beam in Fig. 290B? Show that F in Prob. 15 satistles these conditions if {3L is a solution of the equation (22) cosh {3L cos {3L = I. Determine approximate solutions of (22), for instance, graphically from the intersections of the curves of cos {3L and lIcosh {3L. 20. (Clamped-free beam in Fig. 290C) If the beam is clamped at the left and free at the right (Fig. 290C), the boundary conditions are u(O, t) = 0, uxx(L, t) = 0, uxxx(L, t) = O. Show that F in Prob. 15 satisfies these conditions if {3L is a solution of the equation cosh {3L cos (3L (23) = - 1. Find approximate solutions of (18). D'Alembert's Solution of the Wave Equation. Characteristics It is interesting that the solution (17), Sec. 12.3, of the wave equation (1) p can be immediately obtained by transforming (1) in a suitable way, namely, by introducing the new independent variables (2) v = x + ct, w = x - ct. Then u becomes a function of v and w. The delivatives in (1) can now be expressed in terms of delivatives with respect to v and w by the use of the chain rule in Sec. 9.6. Denoting partial delivatives by subscripts, we see from (2) that Vx = I and Wx = I. For simplicity let us denote u(x, t), as a function of v and w, by the same letter u. Then SEC 12.4 D'Alembert's Solution of the Wave Equation. Characteristics 549 We now apply the chain rule to the right side of this equation. We assume that all the partial derivatives involved are continuous, so that U wv = uvw ' Since Vx = I and Wx = 1, we obtain Transforming the other derivative in (1) by the same procedure. we find By inserting these two results in (I) we get (see footnote 2 in App. A3.2) (3) = awav o. The point of the present method is that (3) can be readily solved by two successive integrations, first with respect to wand then with respect to v. This gives all av = h(v) and = fh(V) u dv + I/J(w). Here h(v) and I/J(w) are arbitrary functions of v and w, respectively. Since the integral is a function of v, say, cfJ(v). the solution is of the form u = cfJ(v) + I/J(w). In tem1S of x and t, by (2), we thus have (4) u(x, t) = cfJ(x + + ct) I/J(x - ct). Thi~ is known as d' Alembert's solution l of the wave equation (1), Its derivation was much more elegant than the method in Sec. 12.3. but d'Alembert's method is special, whereas the use of Fourier series applies to various equations, as we shall see. D'Alembert's Solution Satisfying the Initial Conditions (5) (a) U(X. 0) = f(x). (b) Ut(X. 0) = g(x). These are the same as (3) in Sec. 12.3. By differentiating (4) we have (6) Ut(x, t) = ccfJ' (x + ct) - cl/J' (x - ct) IJEAN LE RONO O'ALEMBERT (1717-1783). French mathematician, also known for his important work in mechanics. We mention that the general theory of POEs provides a systematic way for rmding the transformation (2) that simplifies (I). See Ref. [e8] in App. I. sso CHAP. 12 Partial Differential Equations (PDEs) where primes denote delivatives with respect to the entire arguments x + ct and x - ct, respectively, and the minus sign comes from the chain rule. From (4}-(6) we have + (7) U(x, 0) = <p(x) (8) Ut(X, 0) = C<p' (x) - crJ/ (x) ",(x) = f(x), = g(x). Dividing (8) by c and integrating with respect to x. we obtain I (9) <p(x) - "'(x) = k(xo) + - c I x g(s) ds, Xo If we add this to (7). then '" drops out and division by 2 gives I (10) = - <p(X) 2 f(x) + - I ~ I I x g(s) ds ~ + - k(xo)· 2 Similarly, subtraction of (9) from (7) and division by 2 gives ",(x) = -1 f(x) - (11) 2 -I 2c r -I k(xo). g(s) ds - 2 Xo In (10) we replace x by x + ct; we then get an integral from Xo to x + ct. In (11) we replace x by x - ct and get minus an integral from Xo to x - ct or plus an integral from x - ct to xo. Hence addition of <p(x + ct) and "'(x - ct) gives u(x, t) [see (4)] in the form 1 (12) u(x. t) = - 2 [J(x + ct) + f(x - ct)] +- 1 2c I x+ct g(s) ds. x-ct If the initial velocity is zero. we see that this reduces to (13) u(x, t) = Uf(x + ct) + f(x - ct)], in agreement with (17) in Sec. 12.3. You may show that because of the boundary conditions (2) in that section the function f must be odd and must have the period 2L. Our result shows that the two initial conditions [the functions f(x) and g(x) in (5)] determine the solution uniquely. The solution of the wave equation by the Laplace transform method will be shown in Sec. 12.11. Characteristics. Types and Normal Forms of PDEs The idea of d' Alembert's solution is just a special instance of the method of characteristics. This concems PDEs of the form (14) SEC. 12.4 sst D'Alembert's Solution of the Wave Equation. Characteristics (as well as PDEs in more than two variables). Equation (14) is called quasilinear because it is linear in the highest delivatives (but may be arbitrary otherwise). There are three types of PDEs (14), depending on the discliminant AC - B2, as follows. Type Defining Condition Hyperbolic AC - B2 < 0 Wave equation (1) Parabolic AC - B2 = 0 Heat equation (2) Elliptic AC - B2 > 0 Laplace equation (3) Example in Sec. 12.1 Note that (I) and (2) in Sec. 12.1 involve t, but to have y as in (14), we set y = ct in (1). obtaining Utt - c 2 u xx = c 2 (u yy - uxx ) = O. And in (2) we set .r = c 2 t. so that Ut c 2 U= -_ C 2(u y - Ux~') . A, B, C may be functions of x, y, so that a PDE may be of mixed type, that is, of different type in different regions of the xy-plane. An important mixed-type PDE is the Tricomi equation (see Prob. 10). Transformation of (14) to Normal Form. The normal forms of (14) and the corresponding transformations depend on the type of the PDE. They are obtained by solving the characteristic equation of (14), which is the ODE Ay'2 - 2By' + C = 0 (15) where y' = dy/dx (note -2B, not +2B). The solutions of (I5) are called the characteristics of (14), and we write them in the form (I\x, y) = const and 'l'(x, y) = const. Then the transformations giving new variables v, winstead of x, y and the normal fonns of (14) are as follows. Type New Variables Normal Form Hyperbolic v=<fJ w='l' uvw = Fl Parabolic v=x w=<fJ='l' U ww Elliptic V = ~(<fJ + 'l') 1 tv = 2i (<fJ - 'l') = F2 uvv + uww = Fg Here, <I) = (I\x, y), 'l' = 'l'(x, y), Fl = F1(v, w, u, u v , uw ), etc., and we denote u as function of v, w again by u, for simplicity. We see that the normal form of a hyperbolic PDE is as in d' Alembert's solution. In the parabolic case we get just one family of solutions <I) = 'l'. In the elliptic case, i = yq, and the characteristics are complex and are of minor interest. For derivation, see Ref. [GR3] in App. 1. EX AMP LEt D'Alembert's Solution Obtained Systematically The theory of characteristics gives d' Alembert's solution in a systematic fashion. To see this, we write the wave 2 equation Utt - c u xx ~ 0 in the form (14) by setting Y ~ ct. By the chain rule, Ut = uyYt = cU y and 2 2 Htt = C Hyy. Division by c gives H.= - Uyy = 0, as stated before. Hence the characteristic equation is y'2 - I = (y' + 1)(/ - 1) = O. The two families of solutions (characteristics) are <1>(x, y) = )' + x = const and 'I'(x, y) = y - x = const. This gives the new variables v = <1> = y + x = ct + x and w = 'I' = y - x = ct - x and d'Alembert's solution u = h(x + ct) + f2(X - ct). • CHAP. 12 552 -- =- .. Partial Differential Equations (PDEs) - -...... .-. ......w................ ___ . _ 1. Show that c is the speed of each of the two waves given by (4). 2. Show that because of the boundaty conditions (2). Sec. 12.3, the function f in (13) of this section must be odd and of period 2L. 3. If a steel wire 2 m in length weighs 0.9 nt (about 0.20 lb) and is stretched by a tensile force of 300 m (about 67.4 Ib). what is the corresponding speed of transverse waves? 4. What are the frequencies of the eigenfunctions in Prob.3? 5. Longitudinal Vibrations of an Elastic Bar or Rod. These vibrations in the direction of the x-axis are modeled by the wave equation Utt = r 2 u xx , c 2 = EI P (see Tolstov [C9]. p. 275). If the rod is fastened at one end. x = 0, and free at the other. x = L. we have 11(0. t) = 0 and II.AL, t) = O. Show that the motion corresponding to initial displacement u(x, 0) = f(x) and initial velocity zero is = II = 2: An sin Pnx cos Pnct, n=O 2 L An = - 12.5 fL f(x) sin 0 (211 PnX dx, Pn = + 2L 1)7T /6-91 GRAPHING SOLUTIONS using (13), st...etch or graph a figure (similar to Fig. 288 in Sec. 12.3) of the deflection u(x. t) of a vibrating string (length L = L ends fixed, c = I) starting with initial velocity 0 and initial deflection (k small, say, k = 0.0 I). 6. f(x) k sin 8. kx(l - x) .f(x) 7. f(x) TTX cos = k(l - 27TX) 10. (Tricomi and Airy equations2 ) Show that the Tricomi equatioll YU xx + lIyy = 0 is of mixed type. Obtain the Airy equation G" - yG = 0 from the Tricomi equation by separation. (For solutions, see p. 446 of Ref. [GR I] listed in App. I.) /11~01 NORMAL FORMS Find the type, transform to normal form. and solve. (Show the details of your work.) 11. llxy - 13. U xx 15. U= 17. u"", - + + 19. XlIxx - Heat Equation: Solution U yy = 12. l"~x 0 9uyy =0 2uxy + 4uxy + Xl/xy = Uyy 211xy lI XY - lI~.x = 0 16. XU"lI - = 0 18. uxx 411yy 0 - + 14. 20. + lixx - + yU yy = 2u xy 4uxy U yy 2u yy = 0 =0 0 + 5uyy + 3l1 yy = 0 = 0 by Fourier Series From the wave equation we now turn to the next "'big" PDE, the heat equation K up which gives the temperature u{x, y, ~, t) in a body of homogeneous material. Here c 2 is the thermal diffusivity, K the thermal conductivity, u the specific heat, and p the density of the material of the body. V2 11 is the Laplacian of u, and with respect to Cartesian coordinates x, y, .:::, The heat equation was derived in Sec. 10.8. It is also called the diffusion equation. As an important application. let us first consider the temperature in a long thin metal bar or wire of constant cross section and homogeneous material, which is oriented along the x-axis (Fig. 291) and is perfectly insulated laterally. so that heat flows in the x-direction 2SIR GEORGE BIDELL AIRY (1801-1892), English mathematician, known for his work in elasticity. FRANCESCO TRICOMI (1897-1978), Italian mathematIcian, who worked in integral equations. SEC. 12.5 Heat Equation: Solution by Fourier Series 553 o x=L Fig. 291. Bar under consideration only. Then u depends only on x and time t. and the heat equation becomes the one-dimensional heat equation (1) This seems to differ only very little from the wave equation, which has a term lttt instead of LIt, but we shall see that this will make the solutions of (1) behave quite differently from those of the wave equation. We shall solve (1) for some important types of boundary and initial conditions. We begin with the case in which the ends x = 0 and x = L of the bar are kept at temperature zero, so that we have the boundary conditions (2) u(O, t) = 0, u{L, t) =0 Furthermore, the initial temperature in the bar at time t have the initial condition (3) u(x, 0) = for all t. = 0 is given. say, f(x). so that we f(.t) [f(x) given]. Here we must have teO) = 0 and ttL) = 0 because of (2). We shall determine a solution /I(x, t) of (I) satisfying (2) and (3)-one initial condition will be enough, as opposed to two initial conditions for the wave equation. Technically. our method will parallel that for the wave equation in Sec. 12.3: a separation of variables. followed by the use of Fourier series. You may find a step-by-step comparison worthwhile. Step 1. Two ODEs from the heat equation (1). Substitution of a product u(x, t) = F{x)G(t) into (I) gives FG = c 2F"G with G = dG/dt and F" = d 2Fldr;2. To separate the variables, we divide by c 2 FG, obtaining (4) G F" F The left side depends only on t and the right side only on x, so that both sides must equal a constant k (as in Sec. 12.3). You may show that for k = 0 or k > 0 the only solution u = FG satisfying (2) is u == O. For negative k = _p2 we have from (4) Multiplication by the denominators gives immediately the two ODEs (5) 554 CHAP. 12 Partial Differential Equations (PDEs) and (6) Step 2. Satisfying the boundary conditions (2). We first solve (5). A general solution is Ex) (7) = A cos px + B sin px. From the boundary conditions (2) it follows that lI(O. t) = F(O)G(t) =0 u(L, t) = F(L)G(t) = O. and Since G == 0 would give LI == O. we require F(O) = O. F(L) = 0 and get F(O) = A = 0 by 0) and then F(L) = B sinpL = 0, with B =1= 0 (to avoid F == 0); thus, sinpL Setting B = = 0, hence II = I. 2, .... 1, we thus obtain the following solutions of (5) satisfying (2): 117TX Fn(x) = sin L' II = 1,2..... (As in Sec. 12.3, we need not consider negative integral values of Il.) All this was literally the same as in Sec. 12.3. From now on it differs since (6) differs from (6) in Sec. 12.3. We now solve (6). For p = Il7TIL, as just obtained, (6) becomes G + An 2G = 0 An = where CIl7T L It has the general solution = I, 2, ... 11 where Bn is a constanl. Hence the funclions (8) L1 n(X, t) = Fn(x)Gn(t) = Bn sin Il7TX L 2 e- An t (n = 1, 2, ... ) are solutions of the heat equation (1), satisfying (2). These are the eigenfunctions of the problem. cOlTesponding to the eigenvalues An = cll7TIL. Step 3. Solution of the entire problem. Fourier series. So far we have solutions (8) satisfying the boundary conditions (2). To oblain a solution that also satisfies the initial condition (3), we consider a series of these eigenfunctions, ex; (9) u(x, t) = 2: n=1 un(x, t) = 2: ex; 117TX Bn sin - - e-J. n 2t n=1 L ( An = CII7T) L . SEC 12.5 555 Heat Equation: Solution by Fourier Series From this and (3) we have n17X x L = u(x, 0) Bn sin L = j(x). n=l Hence for (9) to satisfy (3), the Bn' s must be the coefficients of the Fourier sine series, as given by (4) in Sec. 11.3; thus (10) 2 L = - Bn fL 1117.\ I(x) sin - - dr (11 L 0 = 1. 2... '). The solution of our problem can be established, assuming that I(x) is piecewise continuous (see Sec. 6.1) on the interval 0 ~ x ~ L and has one-sided derivatives (see Sec. 11.1) at all interior points of that interval; that is, under these assumptions the series (9) with coefficients (10) is the solution of our physical problem. A proof requires knowledge of uniform convergence and will be given at a later occasion (Probs. 19, 20 in Problem Set 15.5). Because of the exponential factor, all the terms in (9) approach zero as t approaches infinity. The rate of decay increases with 11. E X AMP L E 1 Sinusoidal Initial Temperature Find the temperature 1I(.I. r} in a laterally insulated copper bar 80 cm long if the initial temperature i~ 100 sin (Tlx/80) °C and the ends are kept at O°c. How long will it take for the maximum temperature in the bar to drop to 50°C? First guess. then calculate. Physical data for copper: density 8.92 gm/cm3 • specific heat 0.092 call(gm °C), thermal conductivity 0.95 cal/(cm sec 0c}. Solution. The initial condition gives u(x. O} = x 2:: 1l7T\" Bn sin 80 = 7TX f(x} = 100 sin 80 . 1'1-1 Hence. by inspection or from (9) we get Bl = 100. B2 = B3 = ... = O. In (9) we need A12 = where c 2 = KI(ap) = 0.95/(0.092' 8.92) = 1.158 [cm2/sec]. Hence we obtain The solution (9) 2 C T121L2, is U(l ., TlX t) = 100 sin e-O.00l785t 80 . Also, 100e-O.001785t = 50 when r = (\nO.5)/(-0.00l785) = 388 rsec] = 6.5 [min1. Does your guess. or at least its order of magnitude. agree with this result'! • E X AMP L E 1 Speed of Decay Solve the problem in Example 1 when the initial temperature is 100 sin (3T11/80) °C and the other data are as before. Solutioll. In (9). instead of 11 = I we now have 11 = 3. and A32 = 32A12 = 9 • 0.001 785 = 0.01607, so that the solution now is 3T1x 001607t !l(x r) = 100 sin - - e- . , 80 . Hence the maximum temperature drops to SO°C in t = (in 0.5)/( -0.01607) = 43 [secondsl, which is much faster (9 times as fast as in Example I; why?). 556 CHAP. 12 Partial Differential Equations (PDEs) Had we chosen a bigger 11. the decay would have been still faster. and in a sum or series of such terms, each term has it~ own rate of decay. and terms with large 11 are practically 0 after a very short time. Our next example is of this type. and the curve in Fig. 292 corresponding to r = 0.5 looks almost like a sine curve; that is, it is practically the graph of the first term of the solution. • ~=O "I L-\ n x x u~ x u I_--_-:t = 2 ~ =:-:--"1 x Fig. 292. Example 3. Decrease of temperature with time t for L = 'if and c = 1 E X AMP L E 3 "Triangular" Initial Temperature in a Bar Find the temperalllre in a laterally insulated bar of length L whose ends are kept at temperature 0, assuming that the initial temperature is if f(x) = { 0<.l<Ll2, x L-x if LI2 < x < L. (The uppermost part of Fig. 292 shows this function for the special L = 17.) Solutioll. From (10) we get 2 (JLl2x sin 1117< L dx + Bn = L o r 1l17X) . (L - x) sin - - dx L L/2 Integration gives Bn = 0 if II is even. 4L Bn = 2 11 7i 2 (n = I, 5, 9, ... ) and Bn = - 4L f1 2 'iT 2 (II = 3,7, 11, .. '). (see also Example 4 in Sec. 11.3 with k = Ll2). Hence the solution is !I(x, t) = Cl7)2 t ] -"9I ~in L317X 4L [ sin L 17" exp [- ( L 17 2 exp [- Figure 292 shows that the temperalllre decreases with increasing t. because of the heat loss due to the cooling of the ends. Compare Fig. 292 and Fig. 288 in Sec. 12.3 and comment. • SEC. 12.5 557 Heat Equation: Solution by Fourier Series E X AMP L E 4 Bar with Insulated Ends. Eigenvalue 0 Find a solution formula of (I). (3) with (2) replaced by the condition that both ends of the bar are insulated. Solution. Physical experiments show that the rate of heat flow is proportional to the gradient of the temperature. Hence if the ends x = 0 and x = L of the bar are in~ulated. so that no heat can t10w through the ends. we have grad u = lIx = iJuliJx and the boundary conditions C2""; " x CO,1) = O. "xCL. Since II(X. t) = F{x)G{t). this gives u,,{O, t) = F'(O)G(t) we have F' (x) = - Ap sin px + Bp cos px. '0 that FiCo) = Bp = 0 = f) for all r. = 0 0 and "x(L, t) and then F'CL) = = F'{L)G(t) -Ap sinpL = O. Differentiating 0). = O. The second of these conditions gives p = Pn = Il'TTIL. (11 = O. l. 2, .. '). From this and (7) with A = I and B = 0 we get Fn{x) = cos (Il'TTxIL). (II = 0, l. 2... '). With G n as before, thi~ yield~ the eigenfunctions (II) UniX. t) 117iX -A 21 Len = Fnl,)GnltJ = An cos (n = 0, I,"'J corresponding to the eigenvalues An = cn7TlL. The latter are as before, but we now have the additional eigenvalue Ao = 0 and eigenfunction 110 = consr, which is the solution of the problem if the initial temperature fIx) is constant. This ,hows the remarkable fact that a separatioll cOllstallt call very well be zero. alld ;;,ero call be all eigellvalue. Furthermore. whereas (8) gave a Fourier sine series. we now get from (II) a Fourier cosine series x (ll) II(X. t) = x L = ll.ix. 1) Il7iX L n=O 11.=0 An co~ - - e -A" L 2 t ( An = C/l7T) L' Its coefficients result from the initial condition (3), x tI{x. 0) = f17TJ..- L An cos L = fIx), n-O in the form (2). Sec. 11.3. that is. (3) E X AMP L E 5 ±f Ao = L o An = fIx) dx. 2 L f L o I17TX fix) cos - - dx. L 11 • = 1.2." "Triangular" Initial Temperature in a Bar with Insulated Ends Find the temperature in the bar in Example 3, assuming that the ends are insulated (instead of being kept at temperature 0). Solutioll. For the triangular initial temperature. (13) gives Ao = Ll4 and (see also Example 4 in Sec. 11.3 with k = U2) An 2 = - L [ f L/2 X 11 'iTt cos - - dx 0 L + f L ] Il1TX 2L (L - t) cos - - dx L/2 ( L 2 cos 117T ""2 - COSIl7r- I) . Hence the solution {I 2) is L 8L { I 2nt [ - cos - - exp 4 ~ 22 L u{x r) = - , ( L2e7T)2 I ] + I 62 cos L67TX exp [- ( L6C7T)2 r] +...} . We see that the terms decrease with increasing t. and II - 4 Ll4 as t -> "'; this is the mean value of the initial temperature. Thi~ is plausible because no heat can escape from this totally insulated bar. In contrast. the cooling of the ends in Example 3 led to heat loss and u -> O. the temperature at which the ends were kept. • 558 CHAP. 12 Partial Differential Equations (PDEs) Steady Two-Dimensional Heat Problems. Laplace's Equation We shall now extend our discussion from one to two space dimensions and consider the two-dimensional heat equation for steady (that is. time-indepelldem) problems. Then all/at = 0 and the heat equation reduces to Laplace's equation (14) (which has already occUlTed in Sec. 10.8 and will be considered fUlther in Secs. 12.7-12.10). A heat problem then consists of this POE to be considered in some region R of the xy-plane and a given boundary condition on the boundary curve C of R This is a boundary value problem (BVP). One calls it: First BVP or Dirichlet Problem if u is prescribed on C ("Dirichlet boundary condition") Second BVP or ~eumann Problem if the normal derivative u., prescribed on C ("Neumann boundary condition") = aU/all is Third BVP, Mixed BVP, or Robin Problem if 1I is prescribed on a portion of C and u" on the rest of C ("Mixed buundary condition"). y u = {(x) b~----------~--------~ u=o u=o R o-r----------u-=-o----------~--------x o Fig. 293. a Rectangle R and given boundary values Dirichlet Problem in a Rectangle R (Fig. 293). We consider a Dirichlet problem for Laplace's equation (14) in a rectangle R. assuming that the temperature u(x, y) equals a given function f(x) on the upper side and 0 on the other three sides of the rectangle. We solve this problem by separating variables. Substituting u(x, y) = F(x)G(y) into (14) written as Uxx = -Uyy , dividing by FG, and equating both sides to a negative constant, we obtain SEC. 12.5 Heat Equation: Solution by Fourier Series 559 -k. From this we get + kF = d'(2 0, and the left and right boundary conditions imply F(O) = 0, This gives k = F(a) = O. and (1l7T/a)2 and corresponding nonzero solutions F(x) (15) The ODE for G with k = = FnlX) = sin 1l7T -x, a 11 = L 2,···. (117T/a)2 then becomes 2 d .G _ (1l7T)2 G d\"2 a = O. Solutions are Now the boundary condition II = 0 on the lower side of R implies that Gn(O) = 0; that is, Gn(O) = An + En = 0 or En = -An. This gives From this and (15), writing 2An (16) un(x, y) = = A~, we obtain as the eigenfunctions of our problem Fn(x)GnCv) x . Il7TX. Il7TY = A':' sm - - smh - - . a a These solutions satisfy the boundary condition u = 0 on the left, right. and lower sides. To get a solution also satisfying the boundary condition u(x, b) = f(x) on the upper side, we consider the infinite series co u(x, y) = ~ un(x, y). 71=1 From this and (16) with y = b we obtain DC * Il7TX 117Tb a a u(x. b) = f(x) = ~ An sin - - sinh - - . We can write this in the form U(x, b) = ~1 (A! sinh n:b) . Il7TX sm--. a 560 CHAP. 12 Partial Differential Equations (PDEs) This shows that the expressions in the parentheses must be the Fourier coefficients bn of f(x); that is, by (4) in Sec. 11.3, bn * fa j(x) sin - - dx. 2 a n7Tb Il7TX = An sinh - - = a a 0 From this and (16) we see that the solution of our problem is " "* sin -a- sinh -a- ' "" u(x, y) (17) x = .c.. un(x, y) = .c.. An n7TX 117TY where A~ (18) = 2 a sinh (l17Tbla) fa f(x) sin - - dx. n7TX 0 a We have obtained this solution formally, neither considering convergence nor showing that the series for u, U:rx' and U yy have the right sums. This can be proved if one assumes that f and f' are continuous and f" is piecewise continuous on the interval 0 ~ x ~ a. The proof is somewhat involved and relies on uniform convergence. It can be found in [C4] listed in App. l. Unifying Power of Methods. Electrostatics, Elasticity The Laplace equation (14) also governs the electrostatic potential of electrical charges in any region that is free of these charges. Thus our steady-state heal problem can also be interpreted as an electrostatic potential problem. Then (17), (18) is the potential in the rectangle R when the upper side of R is at potential f(x) and the other three sides are grounded. Actually, in the steady-state case, the two-dimensional wave equation (to be considered in Secs. 12.7, 12.8) also reduces to (14). Then (17), (18) is the displacement of a rectangular elastic membrane (rubber sheet, drumhead) that is fixed along its boundary, with three sides lying in the x),-plane and the fourth side given the displacement f(x). This is another impressive demonstration of the unifying power of mathematics. It illustrates that entirely different physical systems may have the same mathematical model and can thus be treated by the same mathematical methods . ..,_ ... • ~- 1. WRITING PROJECT. Wave and Heat Equations. Compare the two PDEs with respect to type, general behavior of eigenfunctions. and kind of boundary and initial conditions and resulting practical problems. Also discuss the difference between Figs. 288 in Sec. 12.3 and 292. 2. (Eigenfunctions) Sketch (or graph) and compare the first three eigenfunctions (8) with Bn = 1, c = I. L = 7T for t = 0, 0.2, 0.4, 0.6, 0.8. 1.0. 3. (Decay) How does the rate of decay of (8) with fIxed /J depend on the specific heat, the density, and the thermal conductivity of the material? SEC. 12.5 Heat Equation: Solution 561 by Fourier Series 4. If the first eigenfunction (8) of the bar decreases to half its value within 10 sec, what is the value of the diffusi vity? 15-91 LATERALLY INSULATED BAR A laterally insulated bar of length 10 cm and constant cross-sectional area I cm2 , of density 10.6 gmlcm3 , thermal conductivity 1.04 call( cm sec DC), and specific heat 0.056 cal/(gm DC) (this corresponds to silver, a good heat conductor) has initial temperature f(x) and is kept at ODC at the ends x = 0 and x = 10. Find the temperature u(x, t) at later times. Here, f(x) equals: 5. f(x) sin 0.411X 6. f(x) sin 0.11lx + i sin 0.2m: 7. f(x) 0.2x if 0 < x < 5 and 0 otherwise 8. f(x} I - 0.21x - 51 9. f(x) = x if 0 < x < 2.5, f(x) = 2.5 if2.5 < x < 7.5, f(x) = 10 - x if 7.5 < x < 10 (Arbitrar~ temperatures at ends) If the ends x = 0 and x = L of the bar in the text are kept at constant temperatures VI and V 2 , respectively, what is the temperature UI(X) in the bar after a long time (theoretically, as t -'? "YO)? First guess, then calculate. 11. Tn Prob. 10 find the temperature at any time. 12. (Changing end temperatures) Assume that the ends of the bar in Probs. 5-9 have been kept at 100DC for a long time. Then at some instant. call it t = 0, the temperature at x = L is suddenly changed to ODC and kept at ODC, whereas the temperature at x = 0 is kept at 100De. Find the temperature in the middle of the bar at t = L 2. 3, 10, 50 sec. First guess, then calculate. 10. BAR UNDER ADIABATIC CONDITIONS "Adiabatic" means no heat exchange with the neighborhood. because the bar is completely insulated, also at the ends. PhysicolTnformation: The heat flux at the ends is proportional to the value of aulax there. 13. Show that for the completely insulated bar. ux(O, t) = 0, uAL, t) = 0, /leX, t) = f(x) and separation of variables gives the following solution, with An given by (2) in Sec. 11.3. u(x. t) = Ao 114-19J C = 14. 16. 18. "" An cos ";x e-{cn7T/L)2t + L 21. The boundary condition of heat transfer (19) -u x (1I, t) = k[II('IT. t) - uo] applies when a bar of length 'IT with c = I is laterally insulated, the left end x = 0 is kept at O°C, and at the right end heat is flowing into air of constant temperature uo. Let k = I for simplicity, and 110 = O. Show that a solution is lI(x, t) = sin px e- p2t • where P is a solution of tan p1I = - p. Show graphically that this equation has infinitely many positive solutions PI> P2, P3, ... , where Pn > 11 - i and lim (Pn - n_cc 11 + ~) = O. (Formula (19) is also known as radiation boundary condition, but this is misleading; see Ref. [C3], p. 19.) 22. (Discontinuous f) Solve (\), (2), (3) with L = 11 and f(x) = Vo = const (=1= 0) if 0 < x < 11/2, f(x) = 0 if 1112 < x < 11. 23. (Heat flux) The heat fiLix of a solution II(X, t) across x = 0 is defined by cp(t) = - KlIx(O, t). Find CPU) for the solution (9). Explain the name. Is it physically understandable that cp goes to 0 as t -'? x? OTHER HEAT EQUATIONS 24. (Bar with heat generation) If heat is generated at a constant rate throughout a bar of length L = 'IT with initial temperature f(x) and the ends at x = 0 and 'IT are kept at temperature 0, the heat equation is 2 U t = c u xx + H with constant H > O. Solve this problem. Hint. Set u = v - Hx(x - 'IT)/(2c 2 ). 25. (Convection) If heat in the bar in the text is free to flow through an end into the surrounding medium kept at O°C, thePDEbecomesv t = c 2 v xx - f3v. Show that it can be reduced to the form (I) by setting vex, t) = II(X, t)w(t). 26. Consider v t = c 2 v xx - v (0 < x < L, t > 0), v(O, t) = 0, veL, t) = 0, vex, 0) = f(x), where the term -v models heat transfer to the surrounding medium kept at temperature O. Reduce this PDE by setting vex, t) = u(x, t)w(t) with w such that U is given by (9), (10). 27. (Nonhomogeneous heat equation) Show that the problem modeled by n~l Find the temperature in Prob. 1, and f(x) = t" 15. f(x) 17. f(x) f(x) = 0.5 cos 4x 19. f(x) f(x) = ~11 - Ix - ~1I1 13 with L = 1 = 112 - x 2 = (x - 11. ~11)2 20. Find the temperature of the bar in Prob. 13 if the left end is kept at ODC, the right end is insulated, and the initial temperature is Vo = const. and (2), (3) can be reduced to a problem for the homogeneous heat equation by setting u(x, t) = vex, t) + w(x) and determining w so that v satisfies the homogeneous PDE and the conditions v(O, t) = veL, t) = 0, v(x.O) = f(x) - w(x). (The term Ne- ax may represent heat loss due to radioactive decay in the bar.) CHAP. 12 562 28-351 Partial Differential Equations (PDEs) TWO-DIMENSIONAL PROBLEMS 28. (Laplace equation) Find the potential in the rectangle o ~ x ~ 20, 0 ~ y ~ 40 whose upper side is kept at potential 220 V and whose other sides are grounded. 29. Find the potential in the square 0 ~ x ~ 2, 0 ~ y ~ 2 if the upper side is kept at the potential sin 47TX and the other sides are grounded. 30. CAS PROJECT. Isotherms. Find the steady-state solutions (temperatures) in the square plate in Fig. 294 with a = 2 satisfying the following boundary conditions. Graph isotherms. (a) II = sin TTX on the upper side. 0 on the others. (b) u = 0 on the vertical sides. assuming that the other sides are perfectly insulated. (c) Boundary conditions of your choice (such that the solution is not identically zero). YI a~ -, Fig. 294. 12.6 a x Square plate 31. (Heat flow in a plate) The faces of the thin square plate in Fig. 294 with side a = 24 are perfectly insulated. The upper side is kept at 20°C and tl1e other sides are kept at O°C. Find the steady-state temperature II(X, y) in the plate. 32. Find the steady-state temperature in the plate in Prob. 31 if the lower side is kept at UOdc, the upper side at U I dc, and the other sides are kept at O°c. Hint: Split into two problems in which the boundary temperature is 0 on three sides for each problem. 33. <Mixed boundary value problem) Find the steady~tate temperature in the plate in Prob. 31 with the upper and lower sides perfectly insulated. the left side kept at O°c. and the right side kept at f(y)°C. 34. (Radiation) Find steady-state temperatures in the rectangle in Fig. 293 with the upper and left sides perfectly insulated and the right side radiating into a medium at O°C according to 1I,.(a. y) + hu(a, y) = O. " > 0 constant. (You will get many solutions since no condition on the lower side is given.) 35. Find formulas similar to (\7). (18) for the temperature in the rectangle R of the text when the lower side of R is kept at temperature f(x) and the other side~ are kept at O°c. Heat Equation: Solution by Fourier Integrals and Transforms Our discussion of the heat equation (1) in the last section extends to bars of infinite length, which are good models of very long bars or wires (such a<; a wire of length, say, 300 ft). Then the role of Fourier series in the solution process will be taken by Fourier integrals (Sec. 11.7). Let us illustrate the method by solving (1) for a bar that extends to infinity on both sides (and is laterally insulated as before). Then we do not have boundary conditions. but only the initial condition (2) u(x, 0) = f(x) (-co where f(x) is the given initial temperature of the bar. To solve this problem, we start as in the last section, substituting uex, t) into (1). This gives the two ODEs (3) < x< ro) F(x)G(t) [see (5), Sec. 12.5] SEC. 12.6 563 Heat Equation: Solution by Fourier Integrals and Transforms and [see (6), Sec. 12.5]. (4) Solutions are = A cos px + B sin px F(x) and respectively. where A and 8 are any constants. Hence a solution of (1) is u(x, t; p) = FG = (A cuspx (5) + 8 sinpx)e-C2p2t. Here we had to choose the separation constant k negative. k = _p2, because positive values of k would lead to an increasing exponential function in (5), which has no physical meaning. Use of Fourier Integrals Any series of functions (5), found in the usual manner by taking p as multiples of a fixed number, would lead to a function that is periodic in x when t = O. However, since f(x) in (2) is not assumed to be periodic, it is natural to use Fourier integrals instead of Fourier series. Also, A and B in (5) are arbitrary and we may regard them as functions of p, writing A = A(p) and 8 = 8(p). Now, since the heat equation (I) is linear and homogeneous. the function (6) u(x, t) = {:U(X, t; p) dp = {c[A(P) o COSpl + 8(p) sinpx]e-C2p2t dp 0 is then a solution of (I), provided this integral exists and can be differentiated twice with respect to x and once with respect to t. Determination ofA(p) and R(p) from the Initial Condition. (7) fleX, 0) = From (6) and (2) we get LX [A(p) cos px + 8(p) sin px] dp = f(x). o This gives A(p) and 8(p) in teons of f(x); indeed. from (4) in Sec. 11.7 we have 1 (8) A(p) = 7T According to written (l *). f 1 :x:: feu) cos pu du. 8(p) = 7T -00 f :x:: feu) sinpu du. -GO Sec. 11.9. our Fourier integral (7) with these A(p) and B(p) can be u(x, 0) = I 7T ~X[:X:: icof(U) cos (px - pu) du ] dp. Similarly, (6) in this section becomes u(x, t) = -1 7T iX[J~ 0 -::x; feu) cos (px - pu) e-c2p2t du ] dp. 564 CHAP. 12 Partial Differential Equations (PDEs) Assuming that we may reverse the order of integration. we obtain (9) u(x, t) = -1 7T IW [W L f(v) cos (px - pv) dp e-c2p2t ] dv. 0 -ex. Then we can evaluate the inner integral by using the formula oo L -" e-~- o (10) v;. b2 cos 2bs ds = -2- e- . [A derivation of (10) is given in Problem Set 16.4 (Team Project 28).] This takes the form of our inner integral if we choose p = s/(eVt) as a new variable of integration and set x-v b=-2eVt . Then 2bs = (x - v)p and ds = eVtdp, so that (10) becomes L x o cos (px - pv) dp = e-c2p2t V; ~ r 2evt { ( X - V)2 } 4e 2 t ' exp - By inserting this result into (9) we obtain the representation (11) u(x, t) = I ~ r-: I 2ev 7T1 oo -00 f(v) exp {-( X - 2V)2 } dv. 4c t Taking.:: = (v - x)f(2eVt) as a variable of integration, we get the alternative form 1 (12) /lex. t) = ~~ v 7T I oc f(x + 2ezVt) e- z2 dz. -x If f(x) is bounded for all values of x and integrable in every finite interval, it can be shown (see Ref. [ClOD that the function (11) or (12) satisfies (I) and (2) Hence this function is the required solution in the present case. E X AMP L E 1 Temperature in an Infinite Bar Find the temperature in the infinite bar if the initial temperature is (Fig. 295) I(x) = Vo = cons' if ~TI o if Ixl> I. { < I, {(xli ~ I I I -1 Fig. 295. x Initial temperature in Example 1 SEC. 12.6 Heat Equation: Solution by Fourier Integrals and Transforms Solution. 565 From (11) we have u(x. t) U, •~ = Ii { exp - -1 2CV7Tt (x - v)2 } --24c I dv. If we introduce the above variable of integration ::. then the integration over v from - I to 1 corre~ponds to the integration over:: from (-I - x)/(2cVi) to (l - x)/(2cVi). and (13) !I(X, I y:;;: (1 -xl/(2cVt) Vot) = - e- Z2 dz (t> 0). -0 +xl/(2cVtl We mention that this integral is not an elementary function, but can be expressed in terms of the error function, whose values have been tabulated. (Table A4 in App. 5 contains a few values; larger tables are listed in Ref. [GRI] in App. I. See also CAS Project 10. p. 568.) Figure 296 shows I/(X. t) for Vo = 100°C, c 2 = 1 cm2 /sec. and several values of t. • ulx, -3 -2 -1 t) o Fig. 296. Solution u(x, t) in Example 1 for Uo = lOO·C, c 2 = 1 cm 2 /sec, and several values of t Use of Fourier Transforms The Fourier transform is closely related to the Fourier integral, from which we obtained the transform in Sec. 11.9. And the transition to the Fourier cosine and sine transform in Sec. 11.8 was even simpler. (You may perhaps wish to review this before going on.) Hence it should not surprise you that we can use these transforms for solving our present or similar problems. The Fourier transform applies to problems concerning the entire axis. and the Fourier cosine and sine transforms to problems involving the positive half-axis. Let us explain these transform methods by typical applications that fit our present discussion. E X AMP L E 2 Temperature in the Infinite Bar in Example 1 Solve Example I using the Fourier transfonn. Solutioll. The problem consist, of the hear equation (I) and the initial condition (2), which in this example is f(x) = Vo = COllst if Ixl < I and 0 otherwise. Our strategy is Lo take the Fourier transform with respect to x and then to solve the resulting ordinary DE in t. The details are as follows. 566 CHAP. 12 Partial Differential Equations (PDEs) LeI i; = g;'(1I) denote the Fourier transform of II. regarded as afllllcliOlI ofx. From (l0) in Sec. 11.9 we see that the heat equation (1) gives On the left, assuming that we may interchange the order of differentiation and integmtion. we have ~(Ut) = 1 .~ V217 f""· Ute-tWX dx = -00 fcc. au lie-twX dx = -:;- . I =-a at V217 -00 at Thus au at = -c 2 w 2"u. Since this equation involves only a derivative with respect to t but none with respect to w, this is a tirst-order ordinary DE. with t as the independent variable and was a parameter. By separating variables (Sec. 1.3) we get the general solution 2 2 u(w, t) = C(w)e -C w t with the arbitrary "constant" C(w) depending on the parameter w. The initial condition (2) yields the relationship u(w, 0) ~ C(w) ~ J(w) = ~(f). Our intermediate result is The inversion fonnula (7). Sec. 11.9. now gives the solution (14) In this SolUTion we may insert the Fourier transform " few) ~ I fCC ivw vz; _ccf(u)edu. Assuming that we may inver! the order of integration, we then obtain By the Euler formula (3). Sec. l1.9. the integrand of the inner integral equals We see that it~ imaginary part is an odd function of w, so that its integral is O. (More precisely, this is the principal part of the integral; see Sec. 16.4.) The real part is an even function of w, so that its integral from -:x; to :x; equals twice the integral from 0 to x: This agrees with (9) (withp = w) and leads to the further formulas (il) and (13). E X AMP L E 3 Solution in Example 1 by the Method of Convolution Solve the heat problem in Example I by the method of convolution. Solution. The beginning is as in Example 2 and leads to (14). that is, • SEC 12.6 567 Heat Equation: Solution by Fourier Integrals and Transforms Now comes the crucial idea. We recognize that this is of the form (13) in Sec. 11.9. that (16) uCr. t) = (j '" g)Cr) = IX i~. I(w)g(w)eiwx dw -00 where (17) Since. by the definition of convolution [(II l. Sec. 11.9], (f (18) * g)(x) = {YO f(p)g(x - p) dl', -:lc as our next and last step we must determine the inverse Fourier tmnsform g of g. For this we can use formula 9 in Table III of Sec. 11.10, with a suitable a With c 2 t = 1I(4a) or a = 1I(4c 2 t). using (17) we obtain Hence g has the inverse e Replacing x with x - I' and (19) suh~tituting u(x, t) = (f -XZ/(4c2 t) this into (18) we finally have I * g)(x) = • ~ 2c V7Tl foo f(p) exp {-( X - p)2 } - - 2 - dp. 4c -:>G This solution formula of our problem agrees with (II l. We wrote (f with respect to which we did not integrate. E X AMP L E 4 . I * g)(x). without indicating the parameter I • Fourier Sine Transform Applied to the Heat Equation If a latemlly insulated bar extends from" = 0 to infinity, we can use the Fourier sine transform. We let the initial temperature be u(x, 0) = f(x) and impose the boundary condition 11(0, t) = O. Then from the heat equation and (9b) in Sec. 11.8, since frO) = £1(0, 0) = 0, we obtain This is a first-order ODE aUs/ill + u 2 2 C 1l. s = O. Its solution is u From the initial condition u(x. OJ = f(x) we have s ("', 0) = IS<w) = C(w). Hence "s(w, t) = is(w)e -dlw 2 t. Taking the inverse Fourier sine transform and ,ubstituting Is(w) = IT ~-; {"'"f(P) sin lI'p dp 0 CHAP. 12 568 Partial Differential Equations (PDEs) on the right. we obtain the solution formula u(x. t) ~ -2 (20) 7T L=L"" f(p) sin 1\'1' e - c 0 2 2 w t sin lH dp dll". 0 Figure 297 shows (20) with c = I for f(x) = I if 0 ~ x ~ 1 and 0 otherwise. graphed over the xt-plane for 0;;;; x;;;; 2. 0.01 ;;;; t ~ 1.5. Note that the curves of u(x, t) for constant t resembk- those in Fig. 296 on p. 565 . • Fig. 297. 11-71 SOLUTION IN INTEGRAL FORM Using (6). obtain the solution of (I) in integral form satisfying the initial condition u(x. m = lex). where lex) l if Ixl < a and 0 otherwise 2. f(x) = e- klxl (/.. > m 1. 3. f(x) LI( I + X2). [Use (15) in Sec. 11.7.] 4. I(x) = (sin xl/x. [Use Prob. 4 in Sec. 11.7.] 5. I(x) = (sin 7fx)/x. [Use Prob. 4 in Sec. 11.7.] 6. f(x) = x if Ixl < I and 0 otherwise 7. I(x) = Ixl if Ixl < 1 and 0 otherwise. 8. Verify that II in Prob. 5 satisfies the initial condition. 9. CAS PROJECT. Heat Flow. (a) Graph the basic Fig. 296. (b) In (a) apply animation to "see" the heat flow in terms of the dccrease of temperature. (c) Graph u(x, t) with c xt-half-plane. = I as a surface over the upper 2 erfx = - ~ rX J 0 e- w2 This function is imp0l1ant in applied mathematics and physics (probability theory and statistics. thermodynamics. etc.) and fits our present discussion. Regarding it as a typical case of a special function defined by an integral that cannot be evaluated as in elementary calculus. do the following. (a) Sketch or gmph the bell-shaped curve [the curve of the integrand in (21 )J. Show that erf x is odd. Show that I b e- w2 dw = 2~ (erfb - erfa), a I b e- w2 dw = .y:;;: erf b. -b (b) Obtain the Maclaurin series of erf x from that of the integrand. Use that series to compute a table of erfx for x = OCO.OI)3 (meaning x = O. O.OL 0.02, .... 3). (c) Obtain the values required in (b) by an integration command of your CAS. Compare accuracy. to. CAS PROJECT. Error Function (21) Solution (20) in Example 4 dw (d) [t can be shown that erf (x) = 1. Confirm this experimentally by computing erf x for large x. SEC. 12.7 569 Modeling: Membrane, Two-Dimensional Wave Equation (e) Let J(x) = 1 when x> 0 and 0 when x < O. Using erf(X) = 1, show that (12) then gives 1 u(x, t) = • I v 7f I x (g) Show that ¢(t) = 1 =-"27f IX e- s /2 ds 2 -ex: 2 e- Z dz -x/(2cVtJ (t> 0). (1) Express the temperature (13) in tenns of the error Here. the integral is the definition of the "distribution function of the normal distribution" to be discussed in function. Sec. 24.8. 12.7 Modeling: Membrane, Two-Dimensional Wave Equation The vibrating string in Sec. 12.2 is a basic one-dimensional vibrational problem. Equally important is its two-dimensional analog, namely, the motion of an elastic membrane. such as a drumhead, that is stretched and then fixed along its edge. Indeed. setting up the model will proceed almost as in Sec. 12.2. Physical Assumptions 1. The mass of the membrane per unit area is constant ("homogeneous membrane"). The membrane is perfectly flexible and offers no resistance to bending. 2. The membrane is stretched and then fixed along its entire boundary in the xy-plane. The tension per unit length T caused by stretching the membrane is the same at all points and in all directions and does not change during the motion. 3. The deflection u{x. y. t) of the membrane during the motion is small compared to the size of the membrane, and all angles of inclination are small. Although these assumptions cannot be realized exactly, they hold relatively accurately for small transverse vibrations of a thin elastic membrane, so that we shall obtain a good model, for instance. of a drumhead. Derivation of the POE of the Model ("Two-Dimensional Wave Equation") from Forces. As in Sec. 12.2 the model will consist of a POE and additional conditions. The POE will be obtained by the same method as in Sec. 12.2, namely, by considering the forces acting on a small portion of the physical system, the membrane in Fig. 298 on the next page, as it is moving up and down. Since the deflections of the membrane and the angles of inclination are small. the sides of the portion are approximately equal to Ax and Ay. The tension T is the force per unit length. Hence the forces acting on the sides of the portion are approximately T Ax and T Ay. Since the membrane is perfectly flexible, these forces are tangent to the moving membrane at every instant. Horizontal Components of the Forces. We first consider the horizontal components of the forces. These components are obtained by multiplying the forces by the cosines of the angles of inclination. Since these angles are small, their cosines are close to I. Hence 570 CHAP. 12 Partial Differential Equations (PDEs) Membrane) x u ~ Z·; ___ .P". 4p Tl'.y TI'.X ~~T!'.y - jl--""\-.- a Tl'.y x+~x 1 {3 : 1 1 1 1 1 1 1 1 1 T!'>.x 1 1 1 1 1 1 1 1----.,,1 1 1 : .,,/ -,. ____ 1 -----.:k.......- I 1 ---J..... " / . . . . :J r x+~x Fig. 298. Vibrating membrane the horizontal components of the forces at opposite sides are approximately equaL Therefore, the motion of the particles of the membrane in a horizontal direction will be negligibly small. From this we conclude that we may regard the motion of the membrane as transversal; that is, each particle moves vertically. Vertical Components of the Forces. left side are (Fig. 298), respectively, T ~y sin f3 These components along the righr side and the -T ~v sin a. and Here a and f3 are the values of the angle of inclination (which varies slightly along the edges) in the middle of the edges, and the minus sign appears because the force on the left side is directed downward. Since the angles are small, we may replace their sines by their tangents. Hence the resultant of those two vertical components is T ily (sin (I) f3 - sin a) = T .ly (tan f3 - tan a) = where subscripts x denote partial T~)' [ux(x derivative~ + ~X, )'1) - ux(X, )'2)] and Yl and .\'2 are values between), and )' + .ly. Similarly. the resultant of the vertical components of the forces acting on the other two sides of the portion is (2) where Xl and X2 are values between X and x + .lx. Newton's Second Law Gives the POE of the Model. By Newton's second law (see Sec. 2.4) the sum of the forces given by (I) and (2) is equal to the mass p~A of that small SEC. 12.8 Rectangular Membrane. Double Fourier Series 571 portion times the acceleration a2 u/at 2 ; here p is the mass of the undeflected membrane per unit area, and .lA = Llx Lly is the area of that portion when it is undeflected. Thus a2 u p Llx~)' at 2 + = T Lly [ux(x + T ~.\ [lIiXI' )' .lX• .'"1) - lIx(X, )'2)] + .1y) - Uy (X2, y)] where the derivative on the left is evaluated at some suitable point (x, Y) corresponding to that portion. Division by p .lx LlY gives 2 a u at 2 = + .lx. )'1) - II"J', }'2) p.lx T [lI x (X + lIy(x., y + ~)') - lIiX2. Y) ] . .ly If we let Llx and Lly approach zero, we obtain the PDE of the model (3) p This PDE is called the two-dimensional wave equation. The expression in parentheses is the Laplacian V2 u of u (Sec. 10.8). Hence (3) can be written (3') Solutions of the wave equation (3) will be obtained and discussed in the next section. 12.8 Rectangular Membrane. Double Fourier Series The model of the vibrating membrane for obtaining the displacement u(x, y, t) of a point (x, y) of the membrane from rest (u = 0) at time tis (1) (2) (3a) (3b) u = 0 on the boundary u(x, y. 0) = f(x. y) y. 0) = g(x. v). lit (x, Here (1) is the two-dimensional wave equation with c 2 = TIp just derived, (2) is the boundary condition (membrane fixed along the boundary in the xy-plane for all times t ~ 0), and (3) are the initial conditions at t = O. consisting of the given initial displacement (initial shape) f(x, y) and the given initial velocity g(x, y), where Ut = au/at. We see that these conditions are quite similar to those for the string in Sec. 12.2. 572 CHAP. 12 Partial Differential Equations (PDEs) y bt------, R a Fig. 299. x Rectangular membrane As a first important model, let us consider the rectangular membrane R in Fig. 299, which is simpler than the circular drumhead to follow. Then the boundary in (2) is the rectangle in Fig. 299. We shall solve this problem in three steps: Step 1. By separating variables, setting !leX, y, t) = F(x, y)C(t) and later F(x, y) = H(x)Q(y) we obtain from (I) an ODE (4) for G and later from a PDE (5) for F two ODEs (6) and (7) for Hand Q. Step 2. From the solutions of those ODEs we determine solutions (13) of (1) ("eigenfunctions" Limn) that satisfy the boundary condition (2). Step 3. We compose the Umn into a double series (14) solving the whole model (I), (2), (3). Step 1. Three ODEs From the Wave Equation (1) To obtain ODEs from (I), we apply two successive separations of variables. In the first separation we set u(x, y, t) = Flx, y)G(t). Substitution into (I) gives where subscript'> denote partial derivatives and dots denote derivatives with respect to t. To separate the variables, we divide both sides by c 2 FG: C c2C = 1 F (Fxx + Fyy). Since the left side depends only on t. whereas the right side is independent of t. both sides must equal a constant. By a simple investigation we see that only negative values of that constant will lead to solutions that satisfy (2) without being identically zero: this is similar to Sec. 12.3. Denoting that negative constant by - v 2 , we have This gives two equations: for the "time function" G(t) we have the ODE (4) where 11. = cv. and for the "amplitude function" F(x. y) a PDE. called the two-dimef15iofl({/ Helmholtz3 equation (5) 3HERMANN VON HELMHOLTZ (!821-J894), German physici~t, known for his basic work in thermodynamics, fluid flow. and acoustics. SEC. 12.8 573 Rectangular Membrane. Double Fourier Series Separation of the Helmholtz equation is achieved if we set F(x, y) substitution of this into (5) we obtain = H(x)Q(y). By To separate the variables, we divide both sides by HQ, finding Both sides must equal a constant, by the usual argument. This constant must be negative, say, -k 2 , because only negative values will lead to solutions that satisfy (2) without being identically zero. Thus This yields two ODEs for Hand Q, namely, (6) and (7) Step 2. Satisfying the Boundary Condition General solutions of (6) and (7) are H(x) = A cos kx + B sin kx and Q(y) = C cospy + D sinpy with constant A. B. C, D. From II = FG and (2) it follows that F = HQ must be zero on the boundary, that is, on the edges x = 0, x = a, Y = 0, Y = b; see Fig. 299. This gives the conditions H(O) = 0, H(a) = 0, Q(O) = 0, Q(b) = O. Hence H(O) = A = 0 and then H(n) = B sin ka = O. Here we must take B otherwise H(x) == 0 and F(x, y) == O. Hence sin ka = 0 or ka = 11177, that is, k= 11177 a (m integer). *' 0 since 574 CHAP. 12 Partial Differential Equations (PDEs) In precisely the same fashion we conclude that C = 0 and p must be restricted to the values p = n7Tlb where n is an integer. We thus obtain the solutions H = Hm, Q = Qm where Hm(x) I717TX = sin - a 171 = 117TY and Qn(Y) = sin -:- ' 11 1.2.... , = 1,2, .... As in the case of the vibrating string, it is not necessary to consider 111, n = - I, - 2, ... since the corresponding solutions are essentially the same as for positive m and n, except for a factor - I. Hence the functions (8) F mn(x, y) = Hm(x)Qn(Y) = • 11l7TX . 117TY Sill - - Sill -b- a III = , 11 1,2, ... , = I, 2, ... , are solutions of the Helmholtz equation (5) that are zero on the boundary of our membrane. Eigenfunctions and Eigenvalues. Having taken care of (5), we tum to (4). Since = 1J2 - k 2 in (7) and A = CIJ in (4). we have p2 Hence to k = l717Tla and p = l17Tlb there corresponds the value III = 1,2, ... , Il = 1,2, ... , (9) in the ODE (4). A corresponding general solution of (4) is It follows that the functions (10) umn(x, y, t) = 111Ttn(.\:' y. t) = (Brnn cos Amnt + F mn(x. y)Gmn(tt written out *. Bmn SIn Amnt) . 1117TX . Il7TY SIn - - SIn - a b with Amn according to (9), are solutions of the wave equation (I) that are zero on the boundary of the rectangular membrane in Fig. 299. These functions are called the eigenfunctions or characteristic jilllctiol1S. and the numbers Amn are called the eigenvalues or characteristic values of the vibrating membrane. The frequency of U mn is Amn I27T. Discussion of Eigenfunctions. It is very interesting that, depending on a and b, several functions Fm11 may correspond to the same eigenvalue. Physically this means that there may exist vibrations having the same frequency but entirely different nodal lines (curves of points on the membrane that do not move). Let us illustrate this with the following example. SEC. 12.8 Rectangular Membrane. Double Fourier Series E X AMP L E 1 575 Eigenvalues and Eigenfunctions of the Square Membrane Consider the square membrane with a = b = I. From (9) we obtain its eigenvalues ~ 1""""22 + n"-. Autn = C7fV Ill'" (II) Hence Amn = An",' but for 111 * F mn = sin 11 the correspouding functions II/1n sin 117T)' are certainly different. For example. to F12 = sin 'lTX and A12 = A2l = C'lT\' sin 2'lTV Fum = sin /I'lTT sin 1117T)' '5 there correspond the two functions and F2l = sin 2'lTX sin 'lTy. Hence the corresponding solutions and have the nodal lines J = ~ and x = ~. respectively (see Fig. 300). Taking obtain Bl2 = 1 and B~2 = 8;1 = O~ we (12) which represents another vibration corresponding to the eigenValue C7TVs. The nodal line of this function is the solution of the equation F12 + B21F2l = sin 'lTX sin 2'lTY + B21 sin 2'lTX sin 'IT)" = 0 or, since sin 2a = 2 sin a cos a, sin 1T.r sin (13) 'IT.\' (cos 'lTY + B21 cos 'lTx) = O. This solution depends on the value of B2l (see Fig. 301). From \ I I) we see that even more than two functions may correspond to the same numerical value of Amn. For example, the four functions F 1S• F S1 , F 47 • and F74 correspond to the value because This happens because 65 can be expressed as the sum of two squares of positive integers in several ways. According to a theorem by Gauss, this is the case for every sum of two squares among whose prime factors there are at least two different ones of the form 411 + I where II is a positive integer. In our case we have • 65 = 5·13 = (4 + 1)(12 + I). DOC] 'Tln W U Un. Un. U 21 • U 22• =-10 B2l = -0.5 [J I I I I I Nodal lines of the solutions un. U 31 in the case of the square membrane Fig. 300. B21 , -_ _..L:.._----, B21 =-1 B21 =0 B2l = " - - - - - - - - ' B21 = 0.5 1 Fig. 301. Nodal lines of the solution (12) for some values of B21 576 CHAP. 12 Partial Differential Equations (PDEs) Step 3. Solution of the Model (1), (2), (3). Double Fourier Series So far we have solutions (10) satisfying (I) and (2) only. To obtain the solution that also satisfies (3), we proceed as in Sec. 12.3. We consider the double series x u(x, y, t) = x L L umn~x, y, t) m=ln=l (14) = x x 2: L + (Bmn cos Amn t m=ln=l 11l17X '" 1l17" B;nn sin Amnt) sin - - sin - - ' a b (without discussing convergence and uniqueness). From (14) and (3a), setting t have (15) u(x, y, 0) = 1Il17X x 00 1117" L L Bmn sin - - sin -b' m=ln=l a = = 0, we f(x, y). Suppose that f(x. y) can be represented by (15). (Sufficient for this is the continuity of f, afli)x, BflBy, a2 ftr)xBy in R.) Then (15) is called the double Fourier series of f(x, y) Its coefficients can be determined as follows. Setting :>0 (16) = KmCY) 1l17y • L b Bmn sm n=l we can write (15) in the form 2: f(x, y) = 17l17X Knb) sin - - . a 7n=1 For fixed y this is the Fourier sine series of f(x, y), considered as a function of x. From (4) in Sec. 11.3 we see that the coefficients of this expansion are (17) KmC\') = 2 la a 0 - m17X f(x, y) sin - - dx. a Furthermore, (16) is the Fourier sine series of Km(Y), and from (4) in Sec. 11.3 it follows that the coefficients are B mn 2 = -b lbK (,,) 0 m. 1117,,' sin - - ' d". b' From this and (17) we obtain the generalized Euler formula (18) Bmn = 4 -b a Ib1a f(x, y) sin - - sin - - ' dx dy 0 0 11117X 1117\' a b 17l = 1, 2, ... n = 1,2, .,. SEC. 12.8 Rectangular Membrane. Double Fourier Series 577 for the Fourier coefficients of f(x, y) in the double Fourier series (15). The Bmn in (14) are now determined in terms of f(x, y). To determine the B;;m, we differentiate (14) termwise with respect to t; using (3b), we obtain au at I t=O * :L :L ex; 00 = . IIl7TX n7TY BmnAmn Sin - - sin -b' = g(x, y). a m=l n=l Suppose that g(x. y) can be developed in this double Fourier series. Then. proceeding as before. we find that the coefficients are 4 abAmn B,~n = - - - (19) JbJag(x, 0 m7TX n7TV a b 111 y) sin - - sin --- dx d,' 0 . n = 1, 2, ... = 1, 2, .... Result. If f and g ill (3) are such that u can be represented by (14), then (14) with coefficients (18) and (19) is the solution of the model (1), (2). (3). E X AMP L E 2 Vibration of a Rectangular Membrane Find the vibrations of a rectangular membrane of sides a = 4 ft and b = 2 ft (Fig. 302) if the tension is 12.5 Ib/ft. the dem,ity is 2.5 slugs/fr (as for light rubber). the initial velocity is O. ami the initial displacement is (20) y 2 1--------. R x 4 Membrane Initial displacement Fig. 302. Solution. c 2 = TIp = 12.512.5 = 5 [ft2/sec 4 JJ 2 0 J 2 1. Also. B~tn = 0.1(4x - m7fX 117fl' x 2 )(2y - y 2 ) sin - sin --dx dy 0 4 I 20 0 from (19). From (18) and (2m. 4 Bmn = - 4'2 Example 2 J 4 2 2 In7Tr (4x - x 2 ) sin -4- dr o 2 (2l' - y ) sin 2 117TY dy. 0 Two integrations by parts give for the first integral on the right (m odd) and for the second integral (11 odd). For even m or 11 we get O. Together with the factor 1120 we thus have B.rnn = 0 if 111 or 11 is even and 256· 32 Bmn = 20m 33 6 Il 7f (m and Il both odd) CHAP. 12 578 Partial Differential Equations (PDEs) From this. (9), and (14) we obtain the answer _ nl7TX _ n7Ty SlIl-- SIn-- 4 (21) = 0.426050 ( cos + 1 27 cos Vs1TVs 4 t Vs1Tv'13 4 f sin sin 1TX 4 31TX 4 ~in sin 21Tl' 1Ty 2 + + J 27- Vs1TV37 cos I 729 cos 2 4 I Vs1TV45 4 1 sin sin 1TX 4 3m 4 31TY SIn -2- 31Tl' ) sin ~ + . .. . To discuss this solution, we note that the first term i~ very similar to the initial shape of the membrane. has no nodal lines, and is by far the dominating term because the coefficients of the next terms are much smaller. The second term has two horizontal nodal lines ly = 2/3, 4/3), the third term two vertical ones lx = 4/3, 8/3), the fourth term two horizontal and two vertical ones, and so on. • 1. (Frequency) How does the frequency of the eigenfunctions of the rectangular membrane change if (a) we double the tension, (b) we take a membrane of half the mass of the original one, (c) we double the sides of the membrane? (Give reason.) SQUARE MEMBRANE 2. Determine and sketch the nodal lines of the eigenfunctions of the square membrane for m 3, 4 and n = I, 2, 3, 4. = I, 2, If-~ Double Fourier Series. Represent f(x, y) by a series (15), where 0 < x < I. 0 < Y < I. 3. f(x, y) = Fig. 303. \ 4. f(x, y) = x 5. f(x, y) = y 6. f(x, y) = x + y 7. f(x, y) = xy 8. f(x, y) = xy(1 - x)(l - y) 9. CAS PROJECT. Double Fourier Series. (a) Wlite a program that gives and graphs partial sums of (\5). Apply it to Probs. 4 and 5. Do the graphs show that those partial sums satisfy the boundary condition (3a)? Explain Why. Why is the convergence rapid? (b) Do the tasks in (a) for Prob. 3. Graph a portion, say, 0 < x < ~, 0 < Y < ~, of several partial sums on common axes, so that you can see how they differ. (See Fig. 303.) (c) Do the tasks in (b) for functions of your choice. Partial sums 52 •2 and 510.10 in CAS Project 9b 10. CAS EXPERIMENT. Quadruples of F mn- Write a program that gives you four numerically equal "mn in Example I, so that four different Fmn correspond to it. Sketch the nodal lines of F 18 , F 81 , F 47 , F74 in Example I and similarly for further F mn that you will find. 111-131 Deflection. Find the deflection u(x, y, t) of the square membrane of side 7r and c 2 = 1 if the initial velocity is 0 and the initial deflection is 11. k sin 2x sin 5y 12. 0.1 sin x siny 13. O.lxy( 7r - x)( 7r - y) RECTANGULAR MEMBRANE 14. VerifY the discussion of the terms of (21) in Example 2. 15. Repeat the task of Prob. 2 when a = 4 and b = 1. SEC. 12.9 21. f(x, y) = xy(a 2 16. Verify the calculation of Bmn in Example 2 by integration by parts. 17. Find eigenvalues of the rectangular membrane of sides a = 2 and b = I to which there correspond two or more different (independent) eigenfunctions. 18. (Minimum property) Show that among all rectangular membranes of the same area A = ab and the same c the square membrane is that for which Un [see (10)] has the lowest frequency. - x 2 )(b 2 - y2) 22. j(x. y) = xy(a - x)(b - y) 23. (Deflection) Find the deflection of the membrane of sides a and b with c 2 = I for the initial deflection f (x, y) 3'7TX . 4'7TY . I veI oClty . 0. . sm - sm - an d"mltla a b = 24. Repeat the task in Prob. 23 with c 2 = 1, for f(x, y) as in Prob. 22 and initial velocity O. 25. (Forced vibrations) Show that forced vibrations of a membrane are modeled by the PDE U tt = C 2 V 2 U + PIp, where P(x, y, t) is the external force per unit area acting perpendicular to the xy-plane. 119-221 Double Fourier Series. Represent f(x, y) (0 < x < a, 0 < Y < b) by a double Fourier series (15). 19. f(x, y) = k 20. f(x, y) = 0.25x)" 12.9 579 Laplacian in Polar Coordinates. Circular Membrane. Fourier-Bessel Series Laplacian in Polar Coordinates. Circular Membrane. Fourier-Bessel Series In boundary value problems for PDEs it is a general principle to use coordinates in which the fonTIula for the boundary is as simple as possible. Since we want to discuss circular membranes (drumheads), we first transform the Laplacian in the wave equation (1), Sec. 12.8, (1) (subscripts denoting partial derivatives) into polar coordinates e= Hence x = r cos e, y= r sin v arctan -'-- . x e. By the chain rule (Sec. 9.6) we obtain Differentiating once more with respect to x and using the product rule and then again the chain rule gives (2) Also, by differentiation of rand rx = x -v"~=+=y=2 x r (J we find e = -----;:x 1 + (Y/X)2 y r2 . 580 CHAP. 12 Partial Differential Equations (PDEs) Differentiating these two formulas again, we obtain r xx = exx = r -\"- (-~) 1'3 r x = 2xy 1'4· We substitute all these expressions into (2). Assuming continuity of the first and second partial derivatives, we have U rfJ = lie,., and by simplifying, (3) In a similar fashion it follows that (4) By adding (3) and (4) we see that the Laplacian of II in polar coordinates is (5) Circular Membrane Circular membranes occur in drums, pumps, microphones, telephones, and so on. This accounts for their great importance in engineering. Whenever a circular membrane is plane and its material is elastic, but offers no resistance to bending (this excludes thin metallic membranes!), its vibrations are modeled by the two-dimensional wave equation in polar coordinates obtained from (l) with y 2 u given by (5), that is, (6) Y R x Fig. 304. Circular membrane p We shall consider a membrane of radius R (Fig. 304) and determine solutions u(r. t) that are radially symmetric. (Solutions also depending on the angle e will be discussed in the problem set.) Then um] = 0 in (6) and the model of the problem (the analog of (1). (2), (3) in Sec. 12.8) is (7) (8) (9a) u(R, t) = 0 for all t u(r, 0) = ~ 0 fer) (9b) Here (8) means that the membrane is fixed along the boundary circle deflection fer) and the initial velocity g(r) depend only on 1', not on expect radially symmetric solutions u(r, t). l' e, = R. The initial so that we can SEC. 12.9 581 Laplacian in Polar Coordinates. Circular Membrane. Fourier-Bessel Series Step 1. Two ODEs From the Wave Equation (7). Bessel's Equation Using the method of separation of variables, we first determine solutions u( r, t) = W(r) GCt). (We write W, not F because W depends on r, whereas F, used before, depended on x.) Substituting u = WG and its derivatives into (7) and dividing the result by c 2 WG, we get G W1 (W"+ -1) W' r -2 - = c G where dots denote derivatives with respect to t and primes denote derivatives with respect to r. The expressions on both sides must equal a constant. This constant must be negative, say, -k 2 , in order to obtain solutions that satisfy the boundary condition without being identically zero. Thus, This gives the two linear ODEs where A = ck (10) and w"+ (11) r We can reduce (11) to Bessel's equation (Sec. 5.5) if we set s = kr. Then IIr retaining the notation W for simplicity, we obtain by the chain rule W'= dW dW ds dr ds = kls and, 2 dW =-k dr ds and d W 2 W " =-2-k. ds By substituting this into (11) and omitting the common factor k 2 we have (12) d2W ds 2 1 dW + - S ds + W= O. This is Bessel's equation (I), Sec. 5.5, with parameter v = o. Step 2. Satisfying the Boundary Condition (8) Solutions of (12) are the Bessel functions 10 and Yo of the first and second kind (see Secs. 5.5, 5.6). But Yo becomes infinite at 0, so that we cannot use it because the deflection of the membrane must always remain finite. This leaves us with (13) W(r) = 10(5) = lo(kr) (5 = kr). 582 CHAP. 12 Partial Differential Equations (PDEs) On the boundary r = R we get W(R) = 10(kR) = 0 from (8) (because G == 0 would imply u == 0). We can satisfy this condition because 10 has (infinitely many) positive zeros, S = 0'1' 0'2, ••• (see Fig. 305), with numerical values 0'1 = 2.4048, 0'2 = 5.5201, 0'3 = 8.6537, 0'4 = 11.7915, 0'5 = 14.9309 and so on. (For further values, consult your CAS or Ref. [GRI] in App. l.) These zeros are slightly inegularly spaced. as we see. Equation (13) now implies kR = am (14) 111 thus = I, 2, .... Hence the functions m = 1,2,'" (15) are solutions of (I 1) that are zero on the boundary circle r = R. Eigenfunctions and Eigenvalues. For Wm in (15), a corresponding general solution of (10) with A = Am = ckm = camlR is Hence the functions with III = 1,2, ... are solutions of the wave equation (7) satisfying the boundary condition (8). These are the eigenfunctions of our problem. The corresponding eigenvalues are Am. The vibration of the membrane conesponding to Urn is called the 111th normal mode; it has the frequency Am l27r cycles per unit time. Since the zeros of the Bessel function 10 are not regularly spaced on the axis (in contrast to the zeros of the sine functions appearing in the case of the vibrating string), the sound of a drum is entirely different from that of a violin. The fonTIs of the normal modes can easily be obtained from Fig. 305 and are shown in Fig. 306. For 111 = I, all the points of the membrane move up (or down) at the same time. For 111 = 2, the situation is as follows. The function W2(r) = 10 (a 2r1R) is zero for a2r1R = 0'1' thus r = a1R1a2' The circle r = alR1cx2 is, therefore, nodal line, and when at some instant the central part of the membrane moves up, the outer part (r > a l Rl(2 ) moves down. and conversely. The solution um(r. t) has 111 - I nodal lines, which are circles (Fig. 306). ) -10 / \ 5 ~ 10 ~'~__~~__~~__~~__~__~__-L~__~'~~~_ __ _ -04 -03 01 - - - - / 02 Fig. 305. Bessel function Jo (5) 03 04 s SEC. 12.9 583 Laplacian in Polar Coordinates. Circular Membrane. Fourier-Bessel Series I I I I i~.1 {" i __ . 1 t( I I f 1/ 1 ~_ I 1/ \ I I 1'-. I 1 1 1 \ '''-. --~ m= 1 Fig. 306. -' ,I ',1 I / m=3 m=2 Normal modes of the circular membrane in the case of vibrations mdependent of the angle Step 3. Solution of the Entire Problem To obtain a solution lI(r, t) that also satisfies the initial conditions (9), we may proceed as in the case of the string. That is. we consider the series (17) u(r, t) = ~1 W"lr)Gm(t) = ~1 (Am cos Am! + Bm sin A",I) 10 ( (leaving aside the problems of convergence and uniqueness). Setting! we obtain = ~n 1') 0 and using (9a). (18) Thus for the series (17) to satisfy the condition (9a), the constants Am must be the coefficients of the Fourier-Bessel series (18) that represents fer) in terms of 10 (O'm rlR ); that is [see (10) in Sec. 5.8 with 11 = O. O'O,rn = and x = 1'1, am, (19) Am = 2 R II 2 2 (O'Ul) JR rf(r)1 (am)r dr 0 0 - R (111 = 1, 2, .. '). Differentiability of fer) in the interval 0 ~ r ~ R is sufficient for the existence of the development (18); see Ref. [Al3]. The coefficients Em in (17) can be determined from (9b) in a similar fashion. Numeric values of Am and Em may be obtained from a CAS or by a numeric integration method. using tables of 10 and 11 , However, numeric integration can sometimes be avoided, as the following example shows. 584 E X AMP L E 1 CHAP. 12 Partial Differential Equations (PDEs) Vibrations of a Circular Membrane Find the vibrations of a circular drumhead of radius I ft and density 2 slugs/ft2 if the tension is 8 Iblft, the initial velocity is O. and the initial displacement is f(,.) Solutioll. = I - r2 [ftl. c 2 = TIp = 8/2 = 4 [ft2/sec 2 Also Bm = 0, since the initial velocity is O. From (19) and Example 1. 3 in Sec. 5.8, since R = I, we obtain 4J2 (am ) a",21r 2(a",) 8 where the last equality follows from (24c). Sec. 5.5, with v = I, that b. J2{am ) = - 2 Urn 2 h(crm ) - JO(u,n) = J 1(am )· am Table 9.5 on p. 409 of [GRI] gives lYm dnd J~(a",). From this we get h(am) = -J~(a",) by (24b), Sec. 5.5. with v = 0, and compute the coefficients Am: 171 am. 11 (0'",) 12 (0CyJ Am 2 3 4 5 6 7 8 9 10 2.40483 5.52008 8.65373 11.79153 14.93092 18.07106 2l.21164 24.35247 27.49348 30.63461 0.51915 -0.34026 0.27145 -0.23246 0.20655 -0.18773 0.17327 -0.16170 0.15218 -0.14417 0.43176 -0.12328 0.06274 -0.03943 0.02767 -0.02078 0.01634 -0.0l328 0.01107 -0.00941 1.10801 -0.l3978 0.04548 -0.02099 0.01164 -0.00722 0.00484 -0.00343 0.00253 -0.00193 Thus fer) = 1.108Jo(2.4048,.) - 0.140Jo(5.520Ir) + 0.045Jo(!!.6537r) - .... We see that the coefficients decrease relatively slowly. The sum of the explicitly given coefficients in the table is 0.99915. The sum of all the coefficients should be I. (Why?) Hence by the Leibniz test in App. A3.3 the partial sum of those terms gives about three correct decimals of the amplitude (fr). Since from (17) we tl1U~ obtain the solution (with,. measured in feet and t in seconds) lI{r. t) = 1.1 08Jo(2.4048r) cos 4.8097 t - 0.140J0<5.5201 r) cos 11.0402t + 0.045JO<8.6537r) (;O~ 17.3075t - .... In Fig. 306, m = I gives an idea of the motion of the tlrst term of our series, 111 = 2 of the second term, and = 3 of the third term, so that we can "see" our result about as well as for a violin string in Sec. 12.3. • 111 SEC 12.9 Laplacian in Polar Coordinates. Circular Membrane. Fourier-Bessel Series ========= -. 3:£2:-- SET 1. Why did we use polar coordinates in this section? 2. Work out the details of the calculation leading to with arbitrary Ao and the I Laplacian in polar coordinates. 3. If --n-_--cl e, 7rIlR is independent of then (5) reduces to y 2 u = II'T + uTlr. Derive this directly from the Laplacian in Cartesian coordinates. l/ 585 1 il riJU) 4. An alternative form of (5) is v2 u = r ill' ( ilr iJ211 + r2 ae 2 . Derive this from (5). 7rIlR TI" f(A) cos nA de, -TI" I Bn = f n-l f " fee) sin lie de. _ .. (e) Compatibility condition Show that (9), Sec. 10.4, impo~es on f(O) in (d) the "compatibility condition" 5. (Radial solution) Show that the only solution of y 2 u = 0 depending only on r = V.~ + u = a In r + b with constant a and b. 6. TEAM PROJECT. Nemnann Problems Series for i is Dirichlet and "n (a) Show that lin = 1'71 cos lie. = rn sin ne, II = 0, I, ... , are solutions of Laplace's equation -V2 u = 0 with ,211 given by (5). (What would Un be in Cartesian coordinates'? Experiment with small II.) (b) Dirichlet problem (See Sec. 12.5) Assuming that term wise differentiation is permissible. show that a solution of the Laplace equation in the disk r < R satisfying the boundary condition u(R, e) = I(e) (f given) is u(r, B> = 00 x + ~l (20) + bn [ an (r)n Ii cos lie f (see e (d) Neumann problem Show that the solution of the Neumann problem y211 = 0 if r < R, llN(R, e) = f(B) (where LIN = iJ"/iJN is the directional de11vative in the direction of the outer normal) is 0) = Ao + L n~1 ELECTROSTATIC POTENTIAL. STEADY-STATE HEAT PROBLEMS The electrostatic potential satisfies Laplace's equation 'V 2 11 = 0 in any region free of charges. Also the heat equation lit = C 2 ,211 (Sec. 12.5) reduces to Laplace's equation if the temperature u is tinIe-independent ("steady-state case"). Using (20), find the potential (equivalently: the steady-state temperature) in the disk r < I if the boundary values are (sketch them, to see what is going on). 7. u(l. 01 = 40 cos3 0 8. u( I, e) 800 sin3 0 9. 1I(l, 0) < e < ~7r and 0 otherwise < e < ~7r and 0 otherwise I0 I if - 7r < 0 < 7r 0 2 if - 7r < 0 < 7r 12. lI( I. 0) (c) Dirichlet problem Solve the Dirichlet problem using (20) if R = I and the boundary values are u(O) = -100 volts if -7r < 0 < O. u(O) = 100 volts if 0 < < 7r. (Sketch this disk, indicate the boundary values.) u(r, 17-121 11. u(l, 0) nO where (In' bn are the Fourier coefficients of Sec. 11.I). o. 10. u( I, e) ( r)n sin ] R (f) Neumann problem Solve y 2 u = 0 in the annulus I < r < 3 if liTO, 0) = sin 0, U,(3, e) = rn(An cos IlO + Bn sin lie) I IO if o if -!7r -!7r 13. CAS EXPERIMENT. Equipotential Lines. Guess what the equipotential lines tI(r, e) = const in Probs. 9 and 11 may look like. Then graph some of them, using partial sums of the series. 14. (Semidisk) Find the electrostatic potential in the semidisk r < I, 0 < < 7r which equals I IO O( 7r - B> on the semicircle I' = I and 0 on the segment e -I<x<l. 15. (Semidisk) Find the steady-state temperature in a semicircular thin plate r < a, 0 < < 7r with the semicircle I' = a kept at constant temperature 110 and the segment - ( l < X < a at O. e 16. (Illvariance) Show that y 2 u is invariant under translations x* = x + (l, y* = Y + b and under rotations x* = x cos a - y sin a, y* = x sin a + y cos a. 586 CHAP. 12 Partial Differential Equations (PDEs) CIRCULAR MEMBRANE 17. (Frequency) What happens to the frequency of an eigenfunction of a dfilm if you double the tension? 18. (Size of a drum) A small dfilm should have a higher fundamental frequency than a large one, tension and density being the same. How does this follow from our formulas? 19. (Tension) Find a formula for the tension required to produce a desired fundamental frequency f I of a drum. 20. CAS PROJECT. Normal Modes. (a) Graph the nOimal modes 114' 115' 116 as in Fig. 306. (b) Write a program for calculating the Am's in Example 1 and extend the table to III = 15. Verify numerically that am = (Ill - ~) 7T and compute the error for 111 = 1, . . . , 10. (c) Graph the initial deflection fer) in Example 1 as well as the fIrst three partial sums of the series. Comment on accuracy. (d) Compute the radii of the nodal lines of U2' U3' 114 when R = I. How do these values compare to those of the nodes of the vibrating string of length I? Can you establish any empirical laws by experimentation with further 11m? 21. (Nodal lines) Is it possible that for fixed L" and R two or more II", [see (16)] with ditlerent nodal lines correspond to the same eigenvalue? (Give a reason.) 22. Why is Al + A2 + ... = 1 in Example I? Compute the first few partial sums until you get 3-digit accuracy. What does this problem mean in the field of music? 0, (24) where A = ck, r Show that the PDE can now be separated by substituting F = W(r)Q(O), giving (25) 25. (Periodicity) Show that Q(8) must be periodic with period 27T and. therefore. 11 = 0, 1, 2•••. in (25) and (26). Show that this yields the solutions Qn = cos 110, Qn * = sin nO, Wn = In(kr), 11 = 0, 1, . . . . 26. (Boundary condition) Show that the boundary condition (27) u(R. O. t) = 0 leads to k = k mn = amnlR, where s = a mn is the mth positive zero of In(s). 27. (Solutions depending on both rand 8) Sho\\ that solutions of (22) satisfying (27) are (see Fig. 307) (28) 23. (Nonzero initial velocity) Show that for (17) to satisfy (9b) we must have (21) X f R o rg(r)J o(a 1ll rlR) dr. VIBRATIONS OF A CIRCULAR MEMBRANE DEPENDING ON BOTH rAND (J 24. (Separations) Show that substitution of II = F(r, (})G(t) into the wave equation (6), that is, gives an ODE and a PDE CD Fig. 307. Nodal lines of some of the solutions (28) 28. (Initial condition) Show that Bmn = 0, Bi;,n = 0 in (28). 29. Show that II~,O = 0 and the current section. UmO II t {r, O. 0) = 0 gives is identical with (16) in 30. (Semicircular membrane) Show that Ull represents the fundamental mode of a semicircular membrane and fInd the corresponding frequency when c 2 = I and R = 1. SEC 12.10 Laplace's Equation in Cylindrical and Spherical Coordinates Potential 12.1 0 587 Laplace's Equation in Cylindrical and Spherical Coordinates. Potential Laplace's equation (1) is one of the most important PDEs in physics and its engineering applications. Here, x, y, z are Cartesian coordinates in space (Fig. 165 in Sec. 9.1), /l xx = a2u/ax2, etc. The expression V 2 u is called the Laplacian of u. The theory of the solutions of (1) is called potential theory. Solutions of (I) that have COlltillUOUS second partial derivatives are known as harmonic functions. Laplace's equation occurs mainly in gravitation, electrostatics (see Theorem 3, Sec. 9.7). steady-state heat flow (Sec. 12.5), and fluid flow (to be discussed In Chap. 18.4). Recall from Sec. 9.7 that the gravitational potential u(x. y.::) at a point (x. y. z) resulting from a single mass located at a point (X. Y. Z) is (2) u(x, y, z) = c c V(x - r X)2 + (y - y)2 + (r> 0) (z - Z)2 and u satisfies (1). Similarly, if mass is distributed in a region T in space with density p(X, Y, Z), its potential at a point (x, y, ::) not occupied by mass is (3) u(x, y, z) = k III p(X, Y, Z) T dX dY dZ. r = 0 (Sec. 9.7) and p is not a function of x, y, ::. Practical problems involving Laplace's equation are boundary value problems in a region T in space with boundary surface S. Such a problem is called (see also Sec. 12.5 for the two-dimensional case): It satisfies (I) because V2(\/r) (I) First boundary value problem or Dirichlet problem if u is prescribed on S. (II) Second boundary value problem or Neumann problem if the normal derivative Un = au/an is prescribed on S. (III) Third or mixed boundary value problem or Robin problem if II is prescribed on a portion of S and lin on the remaining portion of S. Laplacian in Cylindrical Coordinates The first step in solving a boundary value problem is generally the introduction of coordinates in which the boundary surface S has a simple representation. Cylindrical related to x, symmetry (a cylinder as a region T) calls for cylindrical coordinates r, y, z by e, :: (4) x = r cos e, y = r sin e, z=z (Fig. 308, p. 588). 588 CHAP. 12 Partial Differential Equations (PDEs) 2 2 'i' (r.e.z) Ir,8,1/» 1 1 1 r Iz :----- " e r" e Y 1 ,I " x Fig. 308. x Cylindrical coordinates Fig. 309. For these we get y 2 U immediately by adding U zz 1-1 1 1 1 1 Y 'I ',I... Spherical coordinates to (5) in Sec. 12.9; thus, (5) Laplacian in Spherical Coordinates Spherical symmetry (a ball as region T bounded by a sphere S) requires spherical z by coordinates r, e, lb related to x, y, (6) x = r cos e sin efy. y= r sin e sin efy. (Fig. 309). z = r cos efy Using the chain rule (as in Sec. 12.9), we obtain V- 2 u in spherical coordinates (7) We leave the details as an exercise. It is sometimes practical to write (7) in the form I (7) 2 I [a Yu=r2 ar (2r -au) ar 2 I -a (.smefyau ) + 1 -au] . +2 2 sin efy aefy defy sin efy ae Remark on Notation. Equation (6) is used in calculus and extends the familiar notation for polar coordinates. Unfortunately, some books use e and efy interchanged, an extension of the notation x = r cos efy, y = r sin efy for polar coordinates (used in some European countries). Boundary Value Problem in Spherical Coordinates We shall solve the following Dirichlet problem in spherical coordinates: (8) (9) (10) a [ -ar (2r -au) + -- -a ar sin efy aefy I u(R, efy) = J(efy) lim u(r, efy) = O. '1"-->00 (. SIll efy -au ) ] = aefy o. SEC. 12.10 Laplace's Equation in Cylindrical and Spherical Coordinates. Potential 589 The PDE (8) follows from (7) by assuming that the solution £I will not depend on ebecause the Dirichlet condition (9) is independent of e. This may be an electrostatic potential (or a temperature) J(ep) at which the sphere S: r = R is kept. Condition (10) means that the potential at infinity will be zero. Separating Variables by substituting u(r. ep) = C(r)H(ep) into (8). MUltiplying (8) by r2, making the substitution and then dividing by CH, we obtain I G (2r dr dC) d dr = - I d (. dH ) H sin lb dlb sm ep dlb . By the usual argument both sides must be equal to a constant k. Thus we get the two ODEs I (11) d C dr ( =k r2 dC) dr or and ~ (12) (Sin cb sin ep dep dH) dep + kH = O. The solutions of (11) will take a simple form if we set k C' = dC/dr, etc., we obtain r 2C" (13) + 2rC' - n(1l + = n(n + 1). Then, writing l) C = O. This is an Euler-Cauchy equation. From Sec. 2.5 we know that it has solutions C Substituting this and dropping the common factor r a gives a(a - I) + 2a - n(n + 1) = O. a = nand The roots are Hence solutions are (14) and We now solve (12). Setting cos ep d Consequently, (12) with k (15) = n(n = w, we have sin2 lb = 1 d dep dw dl!' dep + C~(r) = d = -sin ep - . dl\' 1) takes the form d [ (I - w 2 ) -dH] -d II' dw + n(n + l)H = O. w 2 and -n - = ra I 590 CHAP. 12 Partial Differential Equations (PDEs) This is Legendre's equation (see Sec. 5.3), written out (1 - w 2 ) (15') d 2H dH 2w - -- - d1l'2 dw + n(n + 1)H = O. For integer II = 0, 1, ... the Legendre polynomials 11 = 0,1, " . , are solutions of Legendre's equation (15). We thus obtain the following two sequences of solution II = GH of Laplace's equation (8), with constant An and Bn, where n = 0, 1, ... , (16) (a) (b) Use of Fourier-Legendre Series Interior Problem: Potential Within the Sphere S. We consider a series of terms from (16a), lI(r. ¢) (17) = :L Anrnpn(cos ¢) (r ~ R). n~O Since S is given by r we must have = R, for (17) to satisfy the Dirichlet condition (9) on the sphere S, (18) n=O that is, (18) must be the Fourier-Legendre series of f(¢). From (7) in Sec. 5.8 we get the coefficients AnRn = (19*) 211 + I 2 II - f(w)Pn(w) dw -1 where few) denotes f(¢) as a function of IV = cos ¢. Since dll· = -sin ¢ d¢, and the limits of integration -I and I correspond to ¢ = 7T" and ¢ = 0, respectively, we also obtain (19) An = 2n + lR n I 1 7r 0 f(¢)Pn(cos ¢) sin ¢ d¢, 11 = 0,1, ... If f(¢) and j'(¢) are piecewise continuous on the interval 0 ~ ¢ ~ 7T", then the series (17) with coefficients (19) solves our problem for points inside the sphere because it can be shown that under these continuity assumptions the series (17) with coefficients (19) gives the derivatives occuning in (8) by termwise differentiation, thus justifying our derivation. SEC. 12.10 591 laplace's Equation in Cylindrical and Spherical Coordinates. Potential Exterior Problem: Potential Outside the Sphere S. Outside the sphere we cannot use the functions Un in (16a) because they do not satisfy (to). But we can use the ll~ in (16b). which do satisfy (to) (but could not be used inside S; why?). Proceeding as before leads to the solution of the exterior problem (20) (r ~ R) satisfying (8), (9), (10), with coefficients (21) En = + 2n 1 Rn + 1 2 IT.f(c/J)Pn(cos c/J) sin c/J dd>. 0 The next example illustrates all this for a sphere of radius I consisting of two hemispheres that are separated by a small strip of insulating material along the equator, so that these hemispheres can be kept at different potentials OW V and 0 V). E X AMP L E 1 Spherical Capacitor Find the potential inside and outside a spherical capacitor consisting of two metallic hemispheres of radius I ft separated by a small slit for reasons of insulation, if the upper hemisphere is kept at 110 V J.nd the lower i~ grounded (Fig. 310). Solutioll. The given boundary condition is (recall Fig. 309) IIO { 0 f(<I» = 0""-<1><71"/2 if if 71"/2 < <I> ""- 71". Since R = I, we thus obtain from (19) 2n + I An = - - - . 110 2 I I .,,/2 Pn(COS <1» sin <I> d<l> 0 2n + I - - 2 - . 110 1 Pn(w) dw o where U' = cos <1>. Hence Pn(COS <1» sin <1> d<1> = -p ..(w) tin'. we integrate from I to O. and we finally get rid of the minus by integrating from 0 to I. You can evaluate this integral by your CAS or continue by using (II) in Sec. 5.3, obtaining M L An = 55(211 + 1) m~O where M = nl2 for even nand M = (II - (-I)'" "n 1 (2n - 2m)! 1 ~ m.(n - m).(n - 2)1 II!. II 1)/2 for odd n. The integral equals lI(n - 2m 110 volts x Fig. 310. ".n-2m dll" 0 y Spherical capacitor in Example 1 + I). Thus 592 CHAP. 12 (22) Partial Differential Equations (PDEs) An = 55(211 + 1) ~ m (211 - 2m)! 21Z L.J (-I) m!(11 - m)!(11 - 2m ·m=O Taking 11 = 0, we get Ao = 55 tsince O! = 1). For 11 = I, 2, 3,' .. we get 2! 165 2 165 2 ' 0!1!2! 275 ( 4! A2 = - 0!2!3! 4 385 A3 = 2! ) 1!1!l ! (6! 0!3!4! -- 8 + I)! 4!) 1!2!2! ~ 0, 385 8 --- etc. Hence the potential (17) inside the sphere is (since Po = 1) (23) u(r, </J) = 55 + 165 2 r P1tcos </J) - 385 3 -8- r P 3 (cos </J) + ... (Fig. 311) with PI, P3 , ... given by (Il '), Sec. 5.3. Since R = I, we see from (19) and (2l) in this section that = An' and (20) thus gives the potential outside the sphere En 55 (24) u(r, </J) = r + 165 -2- 2r PI (cos </J) - 385 -4- 8r P 3 (cos </J) + .... Partial sums of these series can now be used for computing approximate values of the inner and outer potential. Also, it is interesting to see that far away from the sphere the potential is approximately that of a point charge, • namely, 55/r. (Compare with Theorem 3 in Sec. 9.7.) y o Jr IT 2 Partial sums of the first 4, 6, and 11 nonzero terms of (23) for r = R = 1 Fig. 311. E X AMP L E 2 Simpler Cases. Help with Problems The technicalities occurring in cases like that of Example I can often be avoided. For instance, find the potential inside the sphere S: r = R = I when S is kept at the potential j(</J) = cos 2</J. (Can you see the potential on S? What is it at the North Pole? The equator? The South Pole? I Solution. 2 w = cos </J, cos 2</J = 2 cos </J - 1 = 2w 2 - I = ~P2(W) - ~ = ~(~w2 - ~) - ~. Hence the potential in the interior of the sphere is • SEC. 12.10 -_. ..........-............... ---. .-.-... 1. Derive (7) from V2 11 in Cartesian coordinates. (Show the details.) 2. Find the surfaces on which the functions zero. 3. Sketch the functions P,,(cos c/J) for (11') in Sec. 5.3). 4. Sketch the functions P3(COS 5. Verify that * in (16) !:!il lin and lin c/J) and II = are l/I' lI2. l/3 O. I. 2 (see c/J). + y2 + :2 is II 13. f(tb) = 100 16. f( c/J) sin 2 c/J 17. f( c/J) 35 cos 4c/J + 20 cos 2tb + 9 = elr + k with constant c sphere is the same as that of a point charge at the origin. Is this physically plausible? 19. Sketch the intersection of the equipotential surfaces in Prob. 14 with the xo-plane. 7. (Dimension 3) Verify that Vx 2 Find the potential in the interior of the sphere S: r = R = I if this interior is free of charges and the potential on Sis: 18. Show that in Prob. 13 the potential exterior to the and k. = BOUNDARY VALUE PROBLEMS IN SPHERICAL COORDINATES r, 8, c/J 15. f( cp) = cos 3c/J are solutions of (8). POTENTIALS DEPENDING ONLY ON r r = Vx 2 113-171 14. f(c/J) = cos c/J P 4 (cos 6. (Dimension 3) Show that the only solution of the Laplace equation depending only on r 593 Laplace's Equation in Cylindrical and Spherical Coordinates. Potential + y2 1I = dr. + :2. satisfies Laplace's equation in spherical coordinates. 8. (Dirichlet problem). Find the electrostatic potential between two concentric spheres of radii rl = 10 cm and r2 = 20 em kept at potentials VI = 260 V amI V 2 = 110 V. respectively. 9. (Dimension 2. logarithmic potential) Show that the onl) solution of the two-dimensional Laplace equation depending only on r = with constant c and k. V>;;2 + .1'2 is l/ = c In r + k 10. (Logarithmic potential) Find the electrostatic potential between two coaxial cylinders of radii r l = 10 cm and r2 = 20 cm kept at potentials VI = 260 V and V 2 = 110 V. respectively. Compare with Prob. 8. Comment. 11. (Heat problem) If the sUiface of the ball 2 r2 = x + y2 + :2 ~ R2 is kept at temperature zero and the initial temperature in the ball is f( 1'). show that the temperature u(r, t) in the ball is a solution of lit = c 2 (u r l" + 2u,./r) satisfying the conditions u(R.1) = O. u(r, 0) = fer). Show that setting v = ru gives v t = C2VJT' vCR. t) = O. vCr. 0) = rf(r). Include the condition v(O. t) = 0 (which holds because II must be bounded at r = 0). and solve the resulting problem by separating variables. 20. Find the potential exterior to the sphere in Example 2 of the text and in Prob. 15. 21. What is the temperature in a ball of radius I and of homogeneous material if its lower boundary hemisphere is kept at O°C and its upper at 100°C? 22. (Renection in a sphere) Let r, 0, c/J be spherical coordinates. If u(r. O. c/J) satisfies V 2 11 = O. show that vCr. O. c/J) = lI( Ilr. O. c/J)lr satisfies V 2 v = O. What does this give for (l6)? 23. (Renection in a circle) Let r, 0 be polar coordinates. If lI(r. 0) satisfies V2 l/ = 0, show that the function v(r. 0) = lI(llr, 0) satisfies V2 v = O. What are l/ = r cos 0 and v in terms of x and y? Answer the same question for u = r2 cos 0 sin 0 and v. 24. TEAM PROJECT. Transmission Line and Related PDEs. Consider a long cable or telephone wire (Fig. 312) that is imperfectly insulated. so that leaks occur along the entire length of the cable. The source S of the current i(x, t) in the cable is at x = 0, the receiving end T at x = I. The current flows from S to T. through the load, and returns to the ground. Let the constants R, L. C. and G denote the resistance, inductance, capacitance to ground. and conductance to ground. respectively. of the cable per unit length. 12. (Two-dimensional potential problems) Show that the functions x 2 - )'2. XY. xl(x 2 + y2). eX cos y. r~ sin r. cos x cosh y. I~ (x 2 ' + y2). and arctan <.,:Ix) satisfy Laplace's equation "xx + l/yy = O. (Two-dimensional potential problems are best solved by complex allalysis, as we shall see in Chap. 18.) x=O Fig. 312. x=l Transmission line CHAP. 12 594 Partial Differential Equations (PDEs) (d) Telegraph equations. For a submarine cable, G is negligible and the frequencies are low. Show that this leads to the so-called submarine cable equations or telegraph equations (a) Show that ("first transmission line equation") au ax ai at =Ri + L - - - where u(x, t) is the potential in the cable. Hint: Apply Kirchhoff's voltage law to a small portion of the cable between x and x + !n (difference of the potentials at x and x + !n = resistive drop + inductive drop). Find the potential in a submarine cable with ends (x = 0, x = l) grounded and initial voltage distribution Va = const. (e) High-frequency line equations. Show that in the case of alternating currents of high frequencies the equations in (c) can be approximated by the so-called high-frequency line equations (b) Show that for the cable in (a) ("second transmission line equation"), - ai ax - = Gu au at + C- Hint: Use Kirchhoff's current law (difference of the currents at x and x + LlX = loss due to leakage to ground + capacitive loss). (c) Second-order PDEs. Show that elimination of i or u from the transmission line equations leads to Solve the first of them, assuming that the initial potential is Vo sin (7fx/l). u xx = ixx = 12.11 LCutt LCi tt + (RC + (RC + + GLhl t + RGu. GL)i t + RGi. and ut(x. 0) = 0 and u = 0 at the ends x = 0 and = I for all t. x Solution of PDEs by Laplace Transforms Readers familiar with Chap. 6 may wonder whether Laplace transforms can also be used for solving partial differential equations. The answer is yes, particularly if one of the independent variables ranges over the positive axis. The steps to obtain a solution are similar to those in Chap. 6. For a PDE in two variables they are as follows. 1. Take the Laplace transform with respect to one of the two variables, usually t. This gives an ODE for the transform of the unknown function. This is so since the derivatives of this function with respect to the other variable slip into the transformed equation. The latter also incorporates the given boundary and initial conditions. 2. Solving that ODE, obtain the transform of the unknown function. 3. Taking the inverse transform, obtain the solution of the given problem. If the coefficients of the given equation do not depend on t, the use of Laplace transforms will simplify the problem. We explain the method in terms of a typical example. E X AMP LEI Semi-Infinite String Find the displacement It'(X. t) of an elastic string subject to the following conditions. (We write u to denote the unit step function.) (i) The suing is initially at rest on the x-axis from x (ii) For t > 0 the left end of the string sine wave (x = 0) w(O, t) = f(t) = = 0 to 00 IV since we need ("semi-infiniTe string"). is moved in a given fashion, namely, according to a single sin t { 0 if 0 ~ t ~ 27T otherwise (Fig. 313). SEC. 12.11 595 Solution of PDEs by Laplace Transforms Fig. 313. Motion of the left end of the string in Example 1 as a function of time t (iii) Furthermore, lim w(x, t) = 0 ~ for t O. X_Xl Of course there is no infinite string, but our model describes a long string or rope (of negligible weight) with its right end fixed far out on the x-axis. Solutioll. We have to solve the wave equation (Sec. 12.2) (1) p for positive x and t, subject to the "boundary conditions" lim w(x. t) = 0 w(O. t) = JU). (2) (t ~ 0) X_!]C with J as given above. and the initial conditions (3) "'(x, 0) = 0, (a) wt(x, 0) = (b) o. We take the Laplace transform with respect to t. By (2) in Sec. 6.2. The expression - sw(x. 0) - "'t(x, 0) drops out because of (3). On the right we assume that we may interchange integration and differentiation. Then 2 ;£ 2 {a[Ix J= J('°e_ st 2 iJ :, dt ax : o = iJ ax 2 2 (Oe-stw(x, t) dt Jo = iJ [I~2 ;£(w(x, t)}, Writing W(x, s) = ;£{w(x, tl}, we thus obtain s2W = c 2 a2 w thus -2-' ax Since this equation contains only a derivative with respect to x, it may be regarded as an ordillary differelltial equatioll for W(x, s) considered as a function of x. A general solution is W(x, s) (4) = A(s)esx1c + B(s)e-sxlc. From (2) we obtain. writing F(s) = ;£{.f(tl}. W(Q. s) = ~'(1I'W. t)) = ;£{J(t») = Frs). Assnming that we can interchange integration and taking the limit, we have 00 lim W(.\", s) = lim x-x x.-oo 00 ( e -stw(x, t) tit = ( e -st lim w(x. t) tit = O. Jo Jo x_oo This implies A(s) = 0 in (4) because c > O. so that for every fixed positive s the function eSx!c increases as x increases. Note that we may assume s > 0 since a Laplace transfonn generally exists for all s greater than some fixed k (Sec. 6.2). Hence we have W(O, s) = B(s) = F(s), 596 CHAP. 12 Partial Differential Equations (PDEs) so that (4) becomes Wtx, s) = F(s)e -sxle. From the second shifting theorem (Sec. 6.3) with a = xle we obtain the inverse transform (5) W(x, t) = +- +~) ~) (Fig. 314) that is, W(x, t) = sin (t - ~) x x c C - < t < - + 27T if ct> x > (t - 27T)C or and zero otherwise. This is a single sine wave traveling to the right with speed c. Note that a point x remains at rest until t = x/c, the time needed to reach that x if one stalts at t = 0 (start of the motion of the left end) and travels with speed c The result agrees with our physical intuition. Since we proceeded formally, we must verify that (5) satisfies the given conditions. We leave this to the student. • (t=O)LI______________________ x (t=2ro~~L-~I---------------~ 2nc x (t = 4n) LI _ _ _~-~LC---"'----. 'C7 x (t = 6n) LI______________-.--_-,/-/'- 'C7 Fig. 314. x Traveling wave in Example 1 This is the end of Chap. 12, in which we concentrated on the most important partial differential equations (PDEs) in physics and engineering. This is also the end of Part C on Fourier analysis and PDEs. We have seen that PDEs have various basic engineering applications, which make them the subject of many ongoing research projects. Numerics for PDEs follows in Secs. 21.4-21.7, which are independent of the other sections in Part E on numerics. In the next part, Part D on complex analysis, we tum to an area of a different nature that is also highly important to the engineer, as our examples and problems will show. This will include another approach to the (two-dimensional) Laplace equation and its applications in Chap. 18. f is "triangular" as in Example 1, Sec. 12.3. 2. How does the speed of the wave in Example I depend on the tension and on the mass of the string? 3. Verify the solution in Example 1. What traveling wave do we obtain in Example I in the case of a 1. Sketch a figure similar to Fig. 314 if c = 1 and (nonterminating) sinusoidal motion of the left end starting at t = O? /4-6/ SOLVE BY LAPLACE TRANSFORMS aw aw + xax at 4. - = x, w(x, 0) = 1, w(O, t) = Chapter 12 Review Questions and Problems aw s.xax + aw at Applying the convolution theorem. show that = Xl, w(x, 0) = 0 if x w(O, t) 6. a2w ax2 597 a2w 1002 ot OW + 100+ at o if t ~ ~ 0, 0 25w, w(x, 0) = 0 if x ~ 0, "'t(x, 0) = 0 if t w(O, t) = sin t if t ~ 0 ~ 2C~;; w(x, t) = 0, L t J(t - T)T-3/2e-x2t(4c?T) dT. 9. Let w(O, t) = J(t) = u(t) (Sec. 6.3). Denote the con-esponding w. W, and F by wo, Wo, and Fo. Show that then in Prob. 8. I 2cV; t wo(x, t) = __l : _ 7. Solve Prob. 5 by another method. 18-101 T-3/2e-x2t(4c?T) dT 0 l-erf(~~) HEAT PROBLEM Find the temperature w(x, t) in a semi-infinite laterally insulated bar extending from x = 0 along the x-axis to infinity, assuming that the initial temperature is 0, w(x, t) -> 0 as x -> 00 for every fixed 1 ~ 0, and w(O. t) = J(t). Proceed as follows. with the en-or function erf as defined in Problem Set 12.6. 10. (Duhamel's formula4 ) Show that in Prob. 9, 8. Set up the model and show that the Laplace transform leads to and the convolution theorem gives Duhamel's formula and W = F(s)e-'- sxte w(x, 1) = (F = .'i{f}). awo J(t - T) - - dT. o OT I t :;:....._= .. =. S T ION SAN D PRO B L EMS 1. Write down the three probably most important PDEs from memory and state their main applications. 2. What is the method of separating variables for PDEs? Give an example from memory. 3. What is the superposition principle? Give a typical application. 4. What role did Fourier series play in this chapter? Fourier integrals? S. What are the eigenfunctions and their frequencies of the vibrating string? Of the heat equation? 6. What additional conditions did we consider for the wave equation? For the heat equation? 7. Name and explain the three kinds of boundary conditions. 8. What do you know about types of PDEs? About transformation to normal forms? 9. What is d' Alembert's method? To what PDE does it apply? 10. When and why did we use polar coordinates? Spherical coordinates? 11. When and why did Legendre's equation occur in this chapter? Bessel's equation? 12. What are the eigenfunctions of the circular membrane? How do their frequencies differ in principle from those of the eigenfunctions of the vibrating string? 13. Explain mathematically (not physically) why we got exponential functions in separating the heat equation, but not for the wave equation. 14. What is the en-or function? Why did it occur and where? 15. Explain the idea of using Laplace transform methods for PDEs. Give an example from memory. 16. For what k and 111 are X4 + kx 2y2 + y4 and sin nIX sinh y solutions of Laplace's equation? 17. Verify that (x 2 - y2)/(x 2 + y2)2 satisfies Laplace's equation. I 1tH-21 18. U yy 19. "xx 20. 21. u xy U yy Solve for + + + + II = lI(X, 1611 = 0 - 2u = 0 uy + x + y + y): Ux uy = O. u(x, 0) ] = 0 = J(x), 22. Find all solution u(x, y) equation in two variables. = 4JEAN-MARlE CONSTANT DUHAMEL (1797-1872), French mathematician. lIyCX, 0) = g(x) F(x)G(y) of Laplace's CHAP. 12 598 Partial Differential Equations (PDEs) li3~261 where Find and sketch or graph (as in Fig. 285 in Sec. 12.3) the deflection u(x, t) of a vibrating string of length 7r, extending from x = 0 to x = 7r, and e 2 = TIp = I. starting with velocity 0 and deflection 4sin 2x 23 . .f(x) = sin x 24 • .f(x) = !7r - Ix - x( 7r - 27 • .f(x) = sin (7rxI50) 28 • .f(x) = x(50 - x) 29 . .f(x) = 25 - 125 - xl 30• .f(x) = 4 sin 3 (7rxIlO) 131-331 Find the temperature IItx, t) in a laterally insulated bar of length 7r. extending from x = 0 to x = 7r. with e 2 = I for adiabatic boundary condition (see Problem Set 12.5) and initial temperature 31. 100 cos 4x 32. 3x 2 21x - ~7r1 34. Using partial sums, graph lI(x, t) in Prob. 33 for several constant f on conunon axes. Do these graphs agree with your physical intuition? 35. Let .f(x, y) = utx, y, 0) be the initial temperature in a thin square plate of side 7r with edges kept at onc and faces perfectly insulated. Separating variables. obtain from U t = C 2 V 2 U the solution x u(x, y, t) = L L Tr 0 7C .f(x, y) sin mx sin ny dx dy. 0 x)y( 7r - y). 137-=,~ x) Find the temperature distribution in a laterally insulated thin copper bar (e 2 = Klpu = 1.158 cm2 /sec), 50 cm long and of constant cross section with endpoints at x = 0 and 50 kept at O°C and initial temperature 7r - ff .f(x, y) = x( 7r - 127-30 I 33. 4 ----:2 7r !7r1 25 . .f(x) = sin x = = 36. Find the temperature in Prob. 35 if 3 26 • .f(x) Bmn Bmn sin mx sin II}" e- C2 (m2+n2)t m=l n=l Transform to normal form and solve (showing the details!) 37. u xy = U xx 38. 39. lIxx U xx + + 411 xy + 4u yy = 0 411yy = 0 40. 2u xx + SU xy + 2u yy = 0 41. U xx + 2l1 xy + U yy = 0 42. U yy + u xy - 2u xx = 0 L43-4~1 Show that the following membranes of area with e 2 = I have the frequencies of the fundamental mode as given (4-decimal values). Compare. 43. Circle: a/(2V:;;:) = 0.6784 44. Square: Iv'l = 0.7071 45. Rectangle (sides 1: 2): ~ = 0.7906 46. Semicircle: 3.832/vs.;;: = 0.7644 47. Quadrant of circle: aI2/(4v'";) = 0.7244 (a 12 = S.13562 = first positive zero of J2 ) ,!8-50 1 Find the electrostatic potential in the following (charge-free) regions: 48. Between two concentric spheres of radii ro and 1"1 kept at the potentials Uo and u I , respectively. 49. Between two coaxial circular cylinders of radii 1"0 and 1"1 kept at the potential 110 and lI., respectively. (Compare with Prob. 48.) 50. In the interior of a sphere of radius 1 kept at the potential .f(c/J) = cos 3c/J + 3 cos c/J (referred to our usual spherical coordinates). 1 Partial Differential Equations (PDEs) Whereas ODEs (Chaps. 1-6) serve as models of problems involving only one independent variable, problems involving (H'O or more independent variables (space variables or time t and one or several space variables) lead to PDEs. This accounts for the enonnous importance of PDEs to the engineer and physicist. Most important are: (1) U tt = (2) Utt = c 2 (u xx + U yy ) ('2U= One-dimensional wave equation (Sees. 12.2-12.4) Two-dimensional wave equation (Sees. 12.7-12.9) 599 Summary of Chapter 12 One-dimensional heat equation (Secs. 12.5. 12.6) (4) (5) + U yy = 0 Two-dimensional Laplace equation (Secs. 12.5, 12.9) = U xx + uYV + U zz = 0 Three-dimensional Laplace equation V 2 u = u"x 2 V u (Sec. 12.10). Equations (I) and (2) are hyperbolic. (3) is parabolic. (4) and (5) are elliptic. In practice, one is interested in obtaining the solution of such an equation in a given region satisfying given additional conditions, such as initial conditions (conditions at time t = 0) or boundary conditions (prescribed values of the solution u or some of its derivatives on the boundary surface S, or boundary curve C, of the region) or both. For (1) and (2) one prescribes two initial conditions (initial displacement and initial velocity). For (3) one prescribes the initial temperature distribution. For (4) and (5) one prescribes a boundary condition and calls the resulting problem a (see Sec. 12.5) Dirichlet problem if u is prescribed on S. Neumann problem if lin = all/an is prescribed on S, Mixed problem if u is prescribed on one part of S and lin on the other. A general method for solving such problems is the method of separating variables or product method, in which one assumes solutions in the form of products of functions each depending on one variable only. Thus equation (1) is solved by setting lItx, t) = F(x)G(t); see Sec. 12.3; similarly for (3) (see Sec. 12.5). Substitution into the given equation yields ordinary differential equations for F and G, and from these one gets infinitely many solutions F = Fn and G = G n such that the corresponding functions are solutions of the PDE satisfying the given boundary conditions. These are the eigenfunctions of the problem. and the corresponding eigenvalues determine the frequency of the vibration (or the rapidity of the decrease of temperature in the case of the heat equation. etc.). To satisfy also the initial condition (or conditions). one must consider infinite series of the Un. whose coefficients tum oul to be the Fourier coefficients of the functions f and g representing the given initial conditions (Secs. 12.3, 12.5). Hence Fourier series (and Fourier integrals) are of basic importance here (Secs. 12.3. 12.5, 12.6, 12.8). Steady-state problems are problems in which the solution does not depend on time I. For these, the heat equation lit = C 2 V 2 U becomes the Laplace equation. Before solving an initial or boundary value problem. one often transforms the PDE into coordinates in which the boundary of the region considered is given by simple formulas. Thus in polar coordinates given by x = r cos e. y = r sin e. the Laplacian becomes (Sec. 12.9) (6) y 2U = 1 li,T + - r U, . + ""2 u fi /!; r for spherical coordinates see Sec. 12.10. If one now separates the variables. one gets Bessel's equation from (2) and (6) (vibrating circular membrane, Sec. 12.9) and Legendre's equation from (5) transf01med into spherical coordinates (Sec. 12.10). .. .. ' , PA R T D Complex Analysis C HAP T E R 13 Complex Numbers and Functions C HAP T E R 14 Complex Integration C HAP T E R 1 5 Power Series, Taylor Series C HAP T E R 1 6 Laurent Series. Residue Integration C HAP T E R 17 Conformal Mapping C HAP T E R 18 Complex Analysis and Potential Theory Many engineering problems can be modeled, investigated, and solved by functions of a complex variable. For simpler problems, some acquaintance with complex numbers will suffice. This is true for simpler electric circuits and mechanical vibrating systems. For more complicated problems in heat conduction, fluid flow, electrostatics, etc., one needs the theory of complex analytic functions, briefly called complex analysis. The importance of the latter in applied mathematics has three main reasons: 1. Most importantly, the real and imaginary parts of an analytic function satisfy Laplace's equation in two real variables. Hence two-dimensional potential problems can be solved by methods for analytic functions, and this is often simpler than working in real. 2. Many complicated real and complex integrals in applications can be evaluated by the elegant methods of complex integration. 3. Most functions in engineering mathematics are analytic functions, and their study as functions of a complex variable leads to a deeper understanding of their properties and to interrelations in complex that have no analog in real calculus. 601 CHAPTER 13 Complex Numbers and Functions Complex numbers and their geometric representation in the complex plane are discussed in Secs. 13.1 and 13.2. Complex analysis i:-. concerned with complex analytic functions as defined in Sec. 13.3. Checking for analyticity is done by the Cauchy-Riemann equations (Sec. 13.4). These equations are of basic importance, also because of their relation to Laplace's equation. The remaining sections of the chapter are devoted to elementary complex functions (exponential, trigonometric, hyperbolic, and logarithmic functions). These generalize the familiar real functions of calculus. Their detailed knowledge is an absolute necessity in practical work. just as that of their real counterparts is in calculus. Prerequisite: Elementary calculus. References and Answers to Problems: App. I Part D. App. 2. 13.1 Complex Numbers. Complex Plane Equations without real solutions, such as x 2 = -1 or x 2 - lOx + 40 = O. were observed early in history and led to the introduction of complex numbers.1 By definition, a complex number z is an ordered pair (x, y) of real numbers x and y, written z = (x, y). x is called the real part and y the imaginary part of z. written x = Re;::, y = [m Z. By definition, two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. (0, 1) is called the imaginary unit and is denoted by i, (1) i = (0, 1). IFirst to use complex numbers for this purpose was the Italian mathematician GIROLAMO CARDANO (1501-1576). who found the formula for solving cubic equations. The term "complex number" was introduced by CARL FRlEDRICH GAUSS (see the footnote in Sec. 5.4). who also paved the way for a general u~e of complex numbers. 602 SEC. 13.1 Complex Numbers. Complex Plane 603 + iy Addition, Multiplication. Notation z = x Addition of two complex numbers ::1 = (Xl> Yl) and ::2 = (.\"2, ."2) is defined by (2) Multiplication is defined by (3) In particular, these two definitions imply that (Xl> 0) + (X2' 0) = (Xl + X2, 0) and (Xl> 0)(X2' 0) as for real numbers can thus write Xl> (x,O) = (XlX2, 0) X2. Hence the complex numbers = x. "extend" the real numbers We (0, y) Similarly, = iy because by (1) and the definition of multiplication we have iy = (0, l)y = (0, l)(y, 0) = (0· y - 1·0, 0·0 + 1 . y) = (0, y). Together we have by addition (x, y) = (x, 0) + (0, y) = x + iy: In practice, complex numbers z = (x, y) are written (4) z= x + iy or z = x + .vi, e.g., 17 + 4i (instead of i4). Electrical engineers often write j instead if i because they need i for the cunent. If .r = 0, then z = iy and is called pure imaginary. Also, (1) and (3) give (5) because by the definition of multiplication, i 2 = ii = (0, 1)(0, 1) For addition the standard notation (4) gives [see (2)] = (-1, 0) = -l. For multiplication the standard notation gives the following very simple recipe. Multiply each term by each other term and use i 2 = -1 when it occurs [see (3)]: This agrees with (3). And it shows that x numbers than (x, y). + iy is a more practical nmation for complex 604 CHAP. 13 Complex Numbers and Functions If you know vectors. you see that (2) is vector addition. whereas the multiplication (3) has no counterpart in the usual vector algebra. E X AMP L E 1 Real Part, Imaginary Part, Sum and Product of Complex Numbers LeI '::1 = 8 + 3i and '::2 = 9 ~ 2i, Then Re <:1 = 8. Im:::l = 3, Re:::2 = 9, [m:::2 = :::1 + :::2 = (8 + 3i) + (9 :::1:::2 = (8 + 3i)(9 ~ 2;) ~ = 72 + 6 + 2i) = ;(~16 ~2 and 17 + i. • + 27) = 78 + IIi. Subtraction, Division Subtraction and division are defined as the inverse operations of addition and multiplication, respectively. Thus the difference z = ZI - ':2 is the complex number.:: for which ZI = .: + ':2' Hence by (2), (6) The quotient z = z1/22 (Z2 *- 0) is the complex number z for which':l = 2Z2' If we equate the real and the imaginary parts on both sides of this equation, setting 2 = x + iy, we obtain Xl = X2X - Y2Y')'I = Y2X + X2)" The solution is (7*) z= 21 =x + X l X2 2 X2 x= iy. + )'1."2 + ."22 y= Xl."2 X2Yl X2 2 + )'22 The practical rule used to get this is by multiplying numerator and denominator of z l /z2 by X2 - iY2 and simplifiying: (7) Xl X2 E X AMP L E 2 + + iYl (Xl iY2 (X2 + i)'l) (X2 + i)'2) (x2 i."2) - - i."2) XI X 2 2 X2 + ."1)'2 + + ."22 i X2Yl 2 X2 X l .\'2 + yl Difference and Quotient of Complex Numbers For :::1 = 8 + 3; and :2 = 9 ZI ~ 2; we get:1 8 + 3i 9 ~ 2; ~ :2 = (8 + 3;) (8 + 3i)(9 + 2i) (9 ~ Check the division by multiplication to get 8 2i)(9 + 2;) + 3i. ~ (9 ~ 2i) 66 + 43i 81 + 4 = ~1 + 5; and 66 43 + -i. 85 85 - • Complex numbers satisfy the same commutative. associative. and distributive laws as real numbers (see the problem set). Complex Plane This was algebra. Now comes geometry: the geometrical representation of complex numbers as points in the plane. This is of great practical importance. The idea is quite simple and natural. We choose two perpendicular coordinate axes, the horizontal x-axis. called the real axis, and the vertical y-axis, called the imaginary axis. On both axes we choose the same unit of length (Fig. 315). This is called a Cartesian coordinate system. SEC. 13.1 605 Complex Numbers. Complex Plane y (Imaginary axis) y 5 p z =X x +iy (Real C F - - - ! - - - - - - - - - x - axis) Fig. 315. -3 The complex plane ---------- 4-3i Fig. 316. The number 4 - 3; in the complex plane We now plot a given complex number z = (x. y) = x + iy as the point P with coordinates x, y. The xy-plane in which the complex numbers are represented in this way is called the complex plane. 2 Figure 31 t1 shows an example. Instead of saying "the point represented by z in the complex plane" we say briefly and simply "the point z in the complex plane." This will cause no misunderstandings. Addition and subtraction can now be visualized as illustrated in Figs. 317 and 318. y y x I I I I I 6---z2 Fig. 317. Addition of complex numbers Fig. 318. Subtraction of complex numbers Complex Conjugate Numbers The complex conjugate z of a complex number z = x z= x + iy is defined by - iy. It is obtained geometrically by reflecting the point this for z = 5 + 2i and its conjugate Z = 5 - 2i. z in the real axis. Figure 319 shows Y 2 Fig. 319. ~- z =x + iy = 5 + 2i Complex conjugate numbers 2Sometimes called the Argand diagram, atter the French mathematician JEAN ROBERT ARGAND (1768-1822). born in Geneva and later librarian in Paris. His paper on the complex plane appeared in 1806. nine years after a similar memoir by the Norwegian mathematician CASPAR WESSEL (1745-1818). a surveyor of the Danish Academy of Science. CHAP. 13 606 Complex Numbers and Functions The complex conjugate is important because it permits us to switch from complex to real. Indeed, by multiplication, = x 2 + )'2 (verify!). By addition and subtraction. z + Z = 2x. z = 2iy. We thus obtain for the real part x and the imaginary part y (not iy!) of::: = .\ + iy the important formulas zz z (8) Re I I 2 (::: + Z), z=x= 1m z = y = --;;: (z - z). _I If z is real, Z = x, then Z = z by the definition of Z, and conversely. Working with conjugates is easy, since we have (9) E X AMP L E 3 Illustration of (8) and (9) Let Zl =4 + 3i and :2 =2+ 5i. Then by (8), 3i + 3i I 1m:1 = 2i [(4 + 3i) - (4 - 3i)] = -2-i- = 3. Also, the multiplication formula in (9) is verified by (':1':2) ~ (4 Zl::2 ===== --.•. ........ :... -.-. to. + 3i)(2 + 5i) = (4 - 3i)(2 - = (-7 • "1-=-13. (4z 1 2. (Rotation) \1ultiplication by i is geometrically a counterclockwise rotation through rr12 (90°). Verify this by graphing <. and iz and the angle of rotarian for z = 2 + 2i, : = - I - 5i, z = 4 - 3i. 3. (Dhision) Verify the calculation in (7). 116-]2] 4. (Multiplication) If the product of two complex numbers is zero, show that at least one factor must be zero. S. Show that: = x + iy is pure imaginary if and onJy if;: = -:. 6. (Laws for conjugates) Verify (9) for ':2 = 4 + 6i. Zl = 24 + 10i. 15. - + (Zl :2)2 z2)/(zl - + iy. Find: 17. Re (lIZ) 18. 1m [0 + i)8;;:2] 19. Re (1/z2) 20. (Laws of addition and multiplication) Derive the following laws for complex numbers from the corresponding laws for real numbers. + ':2) + ':3 = ':1 + (:::2 + ':3)' (Associative laws) Let': l = 2 + 3i and Z2 = 4 - 5i. Showing the details of your work. find (in the form x + iy): 8. ;:1;:2 10. Re (:22), (Re Z2) Let z = x 16. Im:3, (1m Z)3 (::1 COMPLEX ARITHMETIC 7. (5':1 + 3::z f 9. Re (1/: 1 2 ) = -7 - 26i. 5;) = -7 - 26i. 1. (powersofi)Showthari 2 = -I, i 3 = -i, i4 = I, ;5 = i .... and IIi = -i. Ili 2 = -I. lIi 3 = i ..... 17-151 + 26i) (ZlZ2)Z3 = Zl(Z2Z3) (Distributive law) o+ Z = 22)2 Z + (- z) = (- z) + Z + Z = 0 0, = z, Z' 1 z. SEC. 13.2 13.2 607 Polar Form of Complex Numbers. Powers and Roots Polar Form of Complex Numbers. Powers and Roots The complex plane becomes even more useful and gives further insight into the arithmetic operations for complex numbers if besides the xy-coordinates we also employ the usual polar coordinates r. e defined by (1) x We see that then::: = r cos e, y= r sin e. = x + iy takes the so-called polar form : : = r(cos e + (2) i sin 8). r is called the absolute value or modulus of.: and is denoted by 1::1 (3) = r = V.~ + );2 = Izl. Hence V2 . Geometrically, Izl is the distance of the point z from the origin (Fig. 320). Similarly, 1'::1 - :::21 is the distance between Zl and 22 (Fig. 321). e is called the argument of z and is denoted by arg z. Thus (Fig. 320) e= (4) y arg::: = arctan .:.... (z X *" 0). Geometrically, e is the directed angle from the positive x-axis to OP in Fig. 320. Here. as in calculus, all angles are measured in radians and positive in the counterclockwise sense. For z = 0 this angle e is undefined. (Why?) For a given z 0 it is determined only up to integer multiples of 27r since cosine and sine are periodic with period 27r. But one often wants to specify a unique value of arg ::: of a given::: O. For this reason one defines the principal value Arg::: (with capital A!) of arg ::: by the double inequality *" *" -7r < Arg (5) z~ 7r. Then we have Arg z = 0 for positive real.:: = x, which is practical, and Arg z = 7r (not -7r!) for negative real :::, e.g., for z = -4. The principal value (5) will be important in connection with roots, the complex logarithm (Sec. 13.7), and certain integrals. Obviously, for a given z 0 the other values of arg ::: are arg::: = Arg::: ± 21l7r (11 = ± I. ±2... '). *" Imaginary axis Y p Y -----------. z . =X + 'Y I I I 06"'---'---------!~c--- :~~I Fig. 320. Complex plane, polar form of a complex number x Fig. 321. Distance between two points in the complex plane CHAP. 13 608 E X AMP L E 1 y Complex Numbers and Functions Polar Form of Complex Numbers. Principal Value Arg z z= I + ; (Fig. 322) has the polar form z = V2 (COS!7T + i sin !7T). Hence we obtain arg::: =!7T:!: 1+i Similarly. z = 3 + 3V3i = 21l7T(1l = 6 (cos ~7T D, I." .), + i sin ~7T). and Izl = Arg::: =!7T (the principal value). • 6. and Arg::: = ~7T. lfl4 CArTION! [n using (4), we must pay attention to the quadrant in which::: lies, since tan 6 has period 7r, so that the arguments of z and -z have the same tangent. Example: Example 1 for 6 1 = arg (1 + i) and 62 = arg (-] - i) we have tan 61 = tan 62 = 1. x g. 322. Triangle Inequality Inequalities such as Xl < X2 make sense for real numbers, but not in complex because there is 110 lIatural WllY of ordering complex 11 umbers. However, inequalities between absolute values (which are real!), such as IZII < 1:: 21 (meaning that ZI is closer to the origin than Z2) are of great importance. The daily bread of the complex analyst is the triangle inequality (6) (Fig. 323) which we shall use quite frequently. This inequality follows by noting that the three points 0, .(;1' and;::1 + ':2 are the vertices of a triangle (Fig. 323) with sides 1z.1, 1.:21, and 1;::1 + 221. and one side cannot exceed the sum of the other two sides. A formal proof is left to the reader (Prob. 35). (The triangle degenerates if:::l and :::2 lie on the same straight line through the origin.) Y I -,P" ~"." ~ x Fig. 323. Triangle inequality By induction we obtain from (6) the generalized triangle inequality 7 + 1-1 (6*) ~ ~2 + ... + -n 7 I :so; - that is. the absolute value of a terms. A .. 2 SUIIl Iz1 I + 17-2 I + ... + Izn I', callnot exceed the sum of the absolute vailies of The Triangle Inequality If:::l = I + ; and :::2 = -2 1:::1 + + 3;. then (sketch a figure!) :::21 = I-I + 4il = \'17 = 4.123 < \'2 + Vi] = 5.020. Multiplication and Division in Polar Form This will give us a "geometrical"' understanding of multiplication and division. Let and • SEC. 13.2 609 Polar Form of Complex Numbers. Powers and Roots Multiplication. By (3) in Sec. 13.1 the product is at first The addition rules for the sine and cosine [(6) in App. A3.1] now yield (7) Taking absolute values on both sides of (7), we see that the absolute value of a product equals the product of the absolute values of the factors, (8) Taking arguments in (7) shows that the argument of a product equals the sum of the arguments of the factors, (9) (up to multiples of 27T). Division. by 1'<:21 We have ~l = Hence (ZlIz2)z2. IZ11 = I(zI 1z2)z21 = IZ11z211z21 and by division (10) Zl arg - (11) = arg Z1 - arg (up to multiples of 27T). Z2 22 Combining (10) and (II) we also have the analog of (7), (12) To comprehend this formula. note that it is the polar form of a complex number of absolute value r1/r2 and argument (it - 82 . But these are the absolute value and argument of zl lz2 , as we can see from (10). (II), and the polar forms of Zl and Z2. E X AMP L E 3 Illustration of Formulas (8)-(11) Let Zl = -2 + 2; and::2 = 3i. Then ~IZ2 = -6 - 6i, zl fz2 = 213 + (213);. Hence (make a sketch) and for the argumems we obtain Arg::1 = 3m4, Arg;:2 = 7[12, 37[ Arg (::1::2) = - 4 = Arg;:1 + Arg::2 - 27[, Arg (::/<:2) = ; = Arg Z1 - Arg ;:2· • 610 E X AMP L E 4 CHAP. 13 Complex Numbers and Functions Integer Powers of z. De Moivre's Formula From (8) and (9) with ~1 = ~2 = Z we obtain by induction for (13) Z,n Similarly. (l~) with:: 1 = I and::2 De Moivre's formula3 rn (COS = : " gives (cos (13*) = e+ i ne + i O. 1,2.... 11 = sin I/e). (I3)for 11 = - I, -2..... For 1::1 = r = I, tormula (13) becomes sin e)'" = cos I/e + i sin nfl. We can use this to express cos 118 and sin lI8 in terms of powers of cos 8 and sin 8. For instance, for II = 2 we have on the left cos2 0 + 2; cos 0 sin 0 - sin2 O. Taking the real and imaginary parts on both sides of (13"') with /I = 2 gives the familiar formulas cos 28 = cos2 0 - sin 2 8, sin 20 = 2 cos 0 sin O. This shows that complex methods often simplify the derivation of real formulas. Try /I = 3. • Roots If ;: = w" (n = 1. 2, .. '). then to each value of w there corresponds olle value of ;:. We shall immediately see that, conversely, to a given z =1= 0 there correspond precisely 11 distinct values of w. Each of these values is called an nth root of ;:, and we write (14) = W ~nf V z. Hence this symbol is l1lultivalued, namely, n-va/ued. The 11 values of ~ can be obtained as follows. We write;: and w in polar form z= r(cos Then the equation w" e + i sin tJ) and = z becomes. w" = R"(cos Ilc/J w = R(cos c/J + i sin c/J). by De Moivre's formula (with +i sin 11c/J) = .: = r(cos e+ c/J instead of e) i sin e). The absolute values on both sides must be equal: thus. R n = r. so that R = Vr , where ~"f v r is positive real (an absolute value must be nonnegative!) and thus uniquely determined. Equating the arguments 11c/J and e and recalling that e is determined only up to integer multiples of 21T, we obtain 11c/J = e + 2k1T, thus c/J= e 11 + 2k1T n where k is an integer. For k = O. I, .... n - I we get 11 distinct values of w. Further integers of k would give values already obtained. For instance, k = n gives 2k7r1n = 271", 3 ABRAHAM DE MOIVRE (1667-1754), French mathematician. who pioneered the use of complex numbers in trigonometry and also contributed to probability theory (see Sec. 24.8). SEC. 13.2 611 Polar Form of Complex Numbers. Powers and Roots hence the w corresponding to k values = 0, etc. Consequently, ~nl ~nl ( v z = v r cos (15) Vz, for z *- 0, has the 11 distinct + 2k7r + 1.Sin ' () + 2k7r) ---- () 11 11 ~nl where k = 0, I, ... , 11 - 1. These n values lie on a circle of radius v r with center at the origin and constitute the vertices of a regular polygon of 11 sides. The value of Vz obtained by taking the principal value of arg z and k = 0 in (15) is called the principal ~"I value of w = v .:: . Taking.:: = I in (15), we have ~nr. v 1 (16) Izl = r = I and Arg::: = O. Then 2br n 2br n = cos - - + i sin - - , (15) gives k = 0, 1, .... n - 1. These 11 values are called the nth roots of unity. They lie on the circle of radius I and center 0, briefly called the unit circle (and used quite frequently!). Figures 324-326 show ~3r.1 ....V11 -- +- I • +' v I -- I , _12 -+ 1~ 2 V r;;3 .:j I . , _I, an d ~5r.l VI. If w denotes the value corresponding to k written as = I in (6). then the 11 value" of VI can be More generally, if WI is any nth root of an arbitrary complex number II values of Vz in (15) are z (*- 0), then the (17) because multiplying ~~'! by w k corresponds to increasing the argument of WI by 2k7r/n. Formula (17) motivates the introduction of roots of unity and shows their usefulness. y y y OJ OJ OJ x Fig. 324. 11-81 Vl POLAR FORM Do these problems very carefully since polar forms will be needed frequently. Represent in polar form and graph in the complex plane as in Fig. 322 on p. 608. (Show the details of your work.) x :r Fig. 325. ~l Fig. 326. L 3 - 3i 2. 2i. -2; 3. -5 4. ~ 5. 1 + 1 - ; 6. + "\ll ~1Ti 3V2 + 2i -VI - (2/3); CHAP. 13 612 7. + 5; -6 + + 2 8. 5 3i [9-151 Complex Numbers and Functions where sign y = I if y ~ 0, sign y = - I if y < 0, and all square roots of positive numbers are taken with positive sign. Hint: Use (10) in App. A3.1 with x = 012. 3; 4i PRINCIPAL ARGUMENT (e) Find the square roots of 4;, 16 - 30i, and + 8 v7 i by both (18) and (19) and comment on the work involved. Determine the principal value of the argument. 9. - I - i 10. - 20 + ;, - 20 - ; 12. -7T 2 14. (l + 11. 4 ::':: 3; 13. 7 ::':: 7; IS. (9 + 9;)3 9 (d) Do some further examples of your own and apply a method of checking your results. i)12 127-301 116-20 I EQUATIONS CONVERSION TO X + iy Represent in the form x + iy and graph it in the complex Solve and graph all solutions, showing the details: 27. ::2 - (8 - 5i)::; + 40 - 20; = 0 (Use (19).) plane. 28. ::4 + 29. 30. + 16. COS!7T + ; sin (::'::!7T) 18. 4(COS!7T ::':: ; sin !7T) 20. 12(cos ~7T + ; sin ~7T) 121-251 17. 3(cos 0.2 + ; sin 0.2) 19. cos (-I) + ; sin (-I) ROOTS 23. ~ 2S.~ 24. ~ 3 + 4; 26. TEAM PROJECT. Square Root. (a) Show that w = ~ has the values = Vi- (36 - 6i)z + 42 - + Wi) = 0 I Ii = 0 16 = O. Then use the solutions to factor Z4 into quadratic factors with real coefficients. Z4 + 16 31. CAS PROJECT. Roots of Unity and Their Graphs. Find and graph all roots in the complex plane. 21. V-i 22. {Y] }\'1 8::;2 - (5 - 14i)::2 - (24 [cos ~ Write a program for calculating these roots and for graphing them as poims on the unit circle. Apply the program to z n = 1 with n = 2, 3. . . . , 10. Then extend the program to one for arbitrary roots. using an idea near the end of the text, and apply the program to examples of your choice. 132-351 + ; sin ~] INEQUALITIES AND AN EQUATION Verify or prove as indicated. ' 32. (Re and 1m) Prove IRe zl ~ Izl, lIm zl ~ Izl· 33. (parallelogram equality) Prove 1::1 + 2212 + 1.:::1 - ::;21 2 = 2(h1 2 + IZ212). Explain the name. (b) Obtain from (8) the often more practical formula (19) V~ 13.3 = ::,::[v'~ (1.;:1 +x) + (signy)iv'~ (izl +x)j 34. (Triangle inequality) Verify (6) for ::2 = 5 + 1;. 35. (Triangle inequality) Prove (6). ZI = 4 + 7i. Derivative. Analytic Function Our study of complex functions will involve point sets in the complex plane. Most important will be the following ones. Circles and Disks. Half-Planes The unit circle Izl = 1 (Fig. 327) has already occurred in Sec. 13.2. Figure 328 shows a general circle of radius p and center a. Its equation is Iz - al = p SEC. 13.3 613 Derivative. Analytic Function y y y 1 x Fig. 127. Unit circle a x x Fig. 128. Circle in the complex plane Fig. 129. Annulus in the complex plane because it is the set of all : whose distance Iz - al from the center 1I equals p. Accordingly, its interior ("open circular disk") is given by Iz - al < p, its interior plus the circle itself ("closed circular disk") by Iz - al ~ p, and its exterior by Iz - al > p. As an example, sketch this for a = 1 + i and P = 2, to make sure that you understand these inequalities. An open circular disk Iz - 1I1 < P is also called a neighborhood of aor, more precisely, a p-neighborhood of 1I. And 1I has infinitely many of them. one fur each value of P (> 0), and a is a point of each of them, by definition! In modem literature any set containing a p-neighborhood of a is also called a neighborhood of a. Figure 329 shows an open annulus (circular ring) PI < Iz - al < P2, which we shall need later. This is the set of all z whose distance Iz - al from 1I is greater than PI but less than P2. Similarly, the closed annulus PI ~ Iz - al ~ P2 includes the two circles. Half-Planes. By the (open) upper half-plane we mean the set of all points: = x + iy such that y > O. Similarly, the condition y < 0 defines the lower half-plane, x > 0 the right half-plane, and x < 0 the left half-plane. For Reference: Concepts on Sets in the Complex Plane To Our discussion of special sets let us add some general concepts related to sets that we shall need throughout Chaps. 13-18: keep in mind that you can find them here. By a point set in the complex plane we mean any sort of collection of finitely many or infinitely many points. Examples are the solutions of a quadratic equation, the points of a line, the points in the interior of a circle as well as the sets discussed just before. A set S is called open if every point of S has a neighborhood consisting entirely of points that belong to S. For example, the points in the interior of a circle or a square form an open set, and so do the points of the right half-plane Re z = x > O. A set S is called connected if any two of its points can be joined by a broken line of finitely many straight-line segments all of whose points belong to S. An open and connected set is called a domain. Thus an open disk and an open annulus are domains. An open square with a diagonal removed is not a domain since this set is not connected. (Why?) The complement of a set S in the complex plane is the set of all points of the complex plane that do 1I0t belo1lg to S. A set S is called closed if its complement is open. For example, the points on and inside the unit circle form a closed set ("closed unit disk") since its complement Izl > I is open. A boundary point of a set S is a point every neighborhood of which contains both points that belong to S and points that do not belong to S. For example, the boundary 614 CHAP. 13 Complex Numbers and Functions points of an annulus are the points on the two bounding circles. Clearly, if a set S is open. then no boundary point belongs to S; if S is closed, then every boundary point belongs to S. The set of all boundary points of a set S is called the boundary of S. A region is a set consisting of a domain plus, perhaps, some or all of it'> boundary points. WARNING! "Domain" is the modem term for an open connected set. Nevertheless, some authors still call a domain a "region" and others make no distinction between the two terms. Complex Function Complex analysis is concerned with complex functions that are differentiable in some domain. Hence we should first say what we mean by a complex function and then define the concepts of limit and derivative in complex. This discussion will be similar to that in calculus. Nevertheless it needs great attention because it will show interesting basic differences between real and complex calculus. Recall from calculus that a real function f defined on a set S of real numbers (usually an interval) is a rule that assigns to every x in S a real number f(x), called the value of f at x. Now in complex, S is a set of complex numbers. And a function f defined on S is a rule that assigns to every.::: in S a complex number lV, called the vallie of fat.:::. We write w = f(.:::). Here z varies in S and is called a complex variable. The set S is called the domain of definition of f or, briefly, the domain of f. (In most cases S will be open and connected, thus a domain as defined just before.) Example: w = fez) = Z2 + 3.::: is a complex function defined for all z; that is, its domain S is the whole complex plane. The set of all values of a function f is called the range qf f. w is complex, and we write w = u + iv, where u and v are the real and imaginary parts, respectively. Now H' depends on .::: = x + iy. Hence u becomes a real function of x and y. and so does v. We may thus write w = fez) = u(x, y) + iv(x, y). This shows thaI a complex function f(z) is equivalent to a pair of real functions u(x, v) and vex, y), each depending on the two real variables x and y. E X AMP L E 1 Function of a Complex Variable Let w = 1(:) = ;;:2 Solutio1l. /I + 3::. Find II and v and calculate the value of I at :: = I + 3i. = Re 1(:::) = x 2 1(1 + 3i) = .\"2 - (I + 3~ and v = 2~y + 3y. Also. + 3i)2 + 30 + 3i) = I - 9 + 6i + 3 + 9i = - 5 + 15i. This shows that 11(1. 3) = -5 and vO. 3) = 15. Check this by using the expressions for II and v. E X AMP L E 2 • Function of a Complex Variable Let w = f(:;:) = 2iz + 6z. Find u and v and the vallie of f Solution. 1(;::) = 2i(x + iy) I(! + 4i) = Check thIS a~ III Example I. + 6(x - iy) gives Lt(x. y) 2i(~ at z = = ~ + 4i. 6x - 2)" and vex. y) = 2.< - 6.\". Also, + 4i) + 6(! - 4i) = i - 8 + 3 - 24i = -5 - 23;. • SEC 13.3 615 Derivative. Analytic Function Remarks on Notation and Terminology 1. Strictly speaking, fez) denotes the value of f at z, but it is a convenient abuse of language to talk about the junction fez) (instead of the junction f), thereby exhibiting the notation for the independent variable. 2. We assume all functions to be sillgle-valued relatiolls, as usual: to each.: in S there corresponds but one value w = f(.:) (but. of course, several z. may give the same value tv = fez), just as in calculus). Accordingly, we shall not lise the term "multi valued function" (used in some books on complex analysis) for a multivalued relation. in which to a.: there corresponds more than one w. Limit, Continuity A function f(;:.) is said to have the limit I as ;:. approaches a point lim fez) (1) ':0, written I, = z-Z'o if f is defined in a neighborhood of ':0 (except perhaps at Zo itself) and if the values of f are "close" to I for all z. "close" to Zo; in precise terms, if for every positive real E we can find a positive real 0 such that for all z ':0 in the disk Iz - 201 < 0 (Fig. 330) we have * If(z) - (2) II < E; * geometrically. if for every.::: ':0 in that 8-disk the value of f lies in the disk (2). Formally, this definition is similar to that in calculus. but there is a big difference. Whereas in the real case, x can approach an Xo only along the real line. here, by definition. z may approach Zofrolll allY direction in the complex plane. This will be quite es~ential in what follows. If a limit exists, it is unique. (See Team Project 26.) A function fez) is said to be continuous at (3) lim f(.:) z= = ':0 if f(.:o) is defined and f(;:.o)· Z-Zo Note that by definition of a limit this implies that fez) is defined in some neighborhood of ':0' f(;:.) is said to be continuous in a domain if it is continuous at each point of this domain. v y ,,.---- ........, " ---- " --.1_ I , E~l I ,.....------ , --"-0 , I f(z) , I , x I , Fig. :no. Limit .... ' .... _---,,. " " U 616 CHAP. 13 Complex Numbers and Functions Derivative The derivative of a complex function f at a point ':0 is written J' (~o) and is defined by (4) provided this limit exists. Then f is said to be differentiable at zoo If we write b.z we have z = '::0 + .1.: and (4) takes the fonn f' (zo) (4') ':0, fez) - f(zo) lim = = :: - Z - 2-20 20 Now comes an important point. Remember that, by the definition of limit. f(.::) is defined in a neighborhood of Zo and z in (4') may approach Zo from any direction in the complex plane Hence differentiability at '::0 means that. along whatever path.:: approaches ':0' the quotient in (4') always approaches a certain value and all these values are equal. This is important and should be kept in mind. E X AMP L E 1 Differentiability. Derivative The function I(;:;) = ~2 is differentiable for all.: and has the derivative I'(.:) = 2.: because • The differentiation rules are the same as in real calculus, since their proofs are literally the same. Thus for any analytic functions f and g and constants c we have (cf)' = cJ', (f + g)' = J' + g', (fg)' = f'g + fg', J'g - fg' (;)' = If as well as the chain rule and the power rule (:::n)' = 11Zn - 1 (11 integer). Also, if f(.::) is differentiable at zoo it is continuous at '::0' (See Team Project 26.) E X AMP L E 4 i not Differentiable It may come as a surprise that there are many complex functions that do not have a derivative at any point. For instance. II.:) = ;: = f - iy is such a function. To ~ee this. we write .l:: = .l" + ;.ly and obtain I(~ (5) + .l::) - I(::) (z + j.::) Cl.x - iCl.y - ;: Cl.7 Cl.:: Cl.x + iCl.y If .ly = O. thi, i, + I. If j.x = O. this is - I. Thu, (5) approaches + I along path I in Fig. 331 but -I along path H. Hence. by definition. the limit of (5) as .l: -> 0 does not exist at any.:. • y x Fig. 331. Paths in (5) SEC. 13.3 617 Derivative. Analytic Function Surprising as Example 4 may be. it merely illustrates that differentiability of a compler function is a rather severe requirement. The idea of proof (approach of z. from different directions) is basic and will be used again as the crucial argument in the next section. Analytic Functions Complex analysis is concerned with the theory and application of "analytic functions," that is. functions that are differentiable in some domain. so that we can do "calculus in complex." The definition is as follows. DEFINITION Analyticity A function f(::.) is said to be allalytic ill a domaill D if f(~) is defined and differentiable at all points of D. The function f(z.) is said to be analytic at a point Z. = Zo in D if fez) is analytic in a neighborhood of zoo Also, by an analytic function we mean a function that is analytic in some domain. Hence analyticity of fez) at :0 means that fez) has a derivative at every point in some neighborhood of Zo (including Zo itself since, by definition, Zo is a point of all its neighborhoods). This concept is motivated by the fact that it is of no practical interest if a function is differentiable merely at a single point ::'0 but not throughout some neighborhood of zoo Team Project 26 gives an example. A more modem term for analytic in D is bolomorphic in D. E X AMP L E 5 Polynomials, Rational Functions The nonnegative integer powers I, that is, functions of the form z, ::.2•••• are analytic where Co• • • • • C n are complex constants. The quotient of two polynomials g(::.) and in the entire complex plane. and so are polynomials, h(;;:), g(::.) I(:) = he:) , is called a rational function. This I is analytic except at the points where /i(::;) = 0: here we assume that common factors of .Ii and h have been canceled. Many further analytic functions will be considered in the next sections and Chapters. • The concepts discussed in this section extend familiar concepts of calculus. Most important is the concept of an analytic function, the exclusive concern of complex analysis. Although many simple functions are not analytic, the large variety of remaining functions will yield a most beautiful branch of mathematics that is very useful in engineering and physics. 11-101 CURVES AND REGIONS OF PRACTICAL INTEREST 3. 0 < Iz - 3 - 2il = ~ 2. 1 ~ Iz - < 1 5. 1m Z2 = 2 Find and sketch or graph the sets in the complex plane given by 1. Iz - 11 I + 4il ~ 5 7. Iz + 11 = 4. -7r<Re;:<7r 6.Rez>-I Iz - 11 9. Re z 21m.:: 8. IArg zl 10. Re (1/:) ~ ~7r < 1 618 CHAP. 13 Complex Numbers and Functions 11. WRITING PROJECT. Sets in the Complex Plane. Extend the part of the text on sets in the complex plane by fonnulating that part in your own words and including examples of your own and comparing with calculus when applicable. 25. CAS PROJECT. Graphing Functions. Find and graph Re f. 1m f. and IfI as surfaces over the ::-plane. Also graph the two families of curves Re Ie::) = COllSt and 1m if:::) = COllst in the same figure, and the curves If(zli = COIlS! in anoth€r figure, where (a) fez) = ::2, (b) I(z) = liz, (c) fez) = Z4. COMPLEX FUNCTIONS AND DERIVATIVES 26. TEAM PROJECT. Limit, Continuity, Derivative (a) Limit. Prove that (I) is equivalent to the pair of relations 112-151 Function Values. Find Re I and 1m f. Also find their values at the given point :::. f = 3::: 2 - 6::: + 3i, z = 2 + f .:::/(z + I), z = 4 - 5i 14. f 1/( I - :::), ::: = l + !i 15. f 1/:::2, ::: = I + ; 12. 13. lim Re i(z) = Re t, (b) Limit. If lim I(:::) exists, show that this limit is z-zo unique. (e) Continuity. If:::}o ::2' ... are complex numbers for which lim ::" = a, and if i(:) is continuous at 116-191 Continuity. Find out (and give reason) whether .f(z) is continuous at ::: = 0 if I(O) = 0 and for z =1= 0 the function I is equal to: 17. [1m (::2)]/1z1 16. [Re (::2)]/ld 2 19. (1m ::)/(1 18. 1z12 Re (1/::) 1:::1) 120-241 Derivative. Differentiate 20. (.:::2 - 9)/(:::2 + I) 21. (:3 22. (3:: + 4i)/( 1.5;: - 2) 24. ::2/(: + ;)2 13.4 lim 1m Ie::) = 1m l. Z-Zo 2-----;"2'0 'it_CO z = a, show that lim i(::n) = i(a). n-----'""x (d) Continuity. If if:::) is differentiable at :::0' show that if:::) is continuous at :::0' (e) Differentiability. Show that if::) = Re z = x is not differentiable at any z. Can you find other such functions? + ;)2 23. i/(l - ;::)2 Differentiability. Show that if::) = 1:::12 is differentiable only at:: = 0; hence it is nowhere analytic. (l) Cauchy-Riemann Equations. Laplace's Equation Tlte Cauchy-Riemall1l equatiolls are tile most importallt equatiolls ill tltis chapter and one of the pillars on which complex analysis rests. They provide a criterion (a test) for the analyticity of a complex function w = fez) = u(x, y) + iv(x, y). Roughly, f is analytic in a domain D if and only if the first partial derivatives of u and v satisfy the two Cauchy-Riemann equations4 (1) 4 The French mathematician AUGUSTIN-LOUIS CAUCHY (see Sec. 2.5) and the German mathematicians BERNHARD RIEMANN (l1l26-Hl66) and KARL WEIERSTRASS (1815 ·1897: see also Sec. 15.5) are the founders of complex analysis. Riemann received his Ph.D. (in 1851) under Gauss (Sec. 5.4) at Gilttingen. where he also taught until he died, when he was only 39 years old. He introduced the concept of the integral as it is used in basic calculus courses. and made important contributions to differential equations. number theory. and mathematical physics. He also developed the s(}-called Riemannian geometry. which is the mathematical foundation of Einstein's theory of relativity; see Ref. [GR9] in App. I. SEC. 13.4 Cauchy-Riemann Equations. Laplace's Equation 619 everywhere in D; here U x = alliax and u y = aulay (and ~imilarly for v) are the usual notations for partial derivatives. The precise formulation of this statement is given in Theorems I and 2. Example: fez) = ;:,2 = x 2 - ."2 + 2ixy is analytic for all:: (see Example 3 in Sec. 13.3), and II = x 2 - ."2 and v = 2xy satisfy (1), namely, U x = 2x = Vy as well as lIy = -2y = -v x . More examples will follow. Cauchy-Riemann Equations THEOREM 1 Let fez) = lI(X, y) + iv(x, y) be defined and continuous in some neighborhood of a point :: = x + iy and d(fferentiable at :: itself. Then at that point, the first-order partial derimtil'es of u and v exist and satisfy the Cauchy-Riemann equations (I). Hence if ft::) is analytic ill a domain D, those partial deriI'Gtil'es exist and satisfr (l) at all points of D. PROOF By a~~umption. the derivative f' (.:) at .: exists. It is given by f' (z) (2) = lim fez + !>z~O ilz) - fez) ilz The idea of the proof is very simple. By the definition of a limit in complex (Sec. 13.3) we can let S~: approach zero along any path in a neighborhood of ;:.. Thus we may choose the two paths I and II in Fig. 332 and equate the results. By comparing the real parts we shall obtain the fir.;t Cauchy-Riemann equation and by comparing the imaginary parts the second. The technical details are as follows. We write ..k = ~x + i:1y. Then.: + .1.: = x + :1x + iCy + :1.\"), and in terms of /I and v the derivative in (2) becomes (3) f' (;:.) = lim [lI(x + ilx, y + ily) + iv(x ..lz~O + ilx, )' .!lx + + ily)] - [II(X, .1') + iv(x, y)] i.!ly We first choose path I in Fig. 332. Thus we let ily ~ 0 first and then ilx ~ O. After ily is zero, il:: = ilx. Then (3) becomes. if we first write the two u-tenns and then the two v-terms, f'(.:) = lim ..lx~O lI(X + .!lx, .r) - lI(X, .r) .1.\ +i lim .l.x~O y x Fig. 332. Paths in (:2) vex + .lx, r) - vex, r) . . 6..\ 620 CHAP. 13 Complex Numbers and Functions Since f' (z) exists, the two real limits on the right exist. By definition, they are the partial derivatives of u and v with respect to x. Hence the derivative f' (z) of fez) can be written (4) Similarly, if we choose path II in Fig. 332. we let ~x ~ 0 first and then is zero, ~:: = i:1y, so that from (3) we now obtain ~y ~ O. After ~x f' (::) = lim ..ly~O II(X, \' . + .1,') .. I .1y u(x, ,.) - + i lim .ly~O vex, " + .1 \') - _. i.ly vex, ") - Since f' (.:) exists, the limits on the right exist and give the partial derivatives of u and v with respect to y; noting that 1Ii = -i, we thus obtain j'(z) = -illy (5) + Vy . The existence of the derivative f' (z) thus implies the existence of the four partial derivatives in (4) and (5). By equating the real parts liT and Vy in (4) and (5) we obtain the first Cauchy-Riemann equation (1). Equating the imaginary parts gives the other. This proves the first statement of the theorem and implies the second because of the definition of analyticity. • FOlmulas (4) and (5) are also quite practical for calculating derivatives see. E X AMP L E 1 f' (z), as we shall Cauchy-Riemann Equations J(~) = ::2 is analytic for all ~. It follow, that the Cauchy-Riemann equation, mu,t be ,atisfied (as we have verified abuve). For f(::) = :: = x - iy we have /I = X, V = -.1' and see that the second Cauchy-Riemann equation is satisfied. /l y = -v x = O. but the tlrst is not: "x = I Vy = -1. We conclude that f(::) = :: is not analytic. confirming Example 4 of Sec. 13.3. Note the savings in calculation! • * The Cauchy-Riemann equations are fundamental because they are not only necessary but also sufficient for a function to be analytic. More precisely, the following theorem holds. THEOREM 2 Cauchy-Riemann Equations If two real-valued continllolls functions lI(X. y) and vex. y) of two real variables x and y have COlltillUOUS first partial derivatives that satisfy the Cauchy-Riemll1ln equlItions in some domain D, then the complex jilllctioll fez) = lI(X, y) + iv(x, y) is allalytic ill D. The proof is more involved than that of Theorem 1 and we leave it optionallsee App. 4). Theorems I and 2 are of great practical importance, since by using the Cauchy-Riemann equations we can now easily find out whether or not a given complex function is analytic. SEC. 13.4 Cauchy-Riemann Equations. Laplace's Equation E X AMP L E 2 621 Cauchy-Riemann Equations. Exponential Function Is i(:::) = II(X. y) + Solution. iv(x, y) = eX(cos y + i sin y) analytic? We have II = eX cos y, v = eX sin ltx = eX lIy = y and by differentiation vy cosy. sin y. -ex = eX Vx = e x cos .\" . smy. We see that the Cauchy-Riemann equations are satisfied and conclude that be the complex analog of eX known from calculus.) I(~) is analytic for all ~. (f(~) will • E X AMP L ElAn Analytic Function of Constant Absolute Value Is Constant The Cauchy-Riemann equations also help in deriving geneml properties of analytic functions. For instance. show that if I(~) is analytic in a domain D and II(::) I = k = CO/1St in D. then I(:) = D. (We shall make crucial use of thb in Sec. 18.6 in the proof of Theorem 3.1 Solutioll. Now use By assumption. Vx = -lly IJI2 = lu + ivl2 in the first equation and = I? + v2 in 2 = k . By differentiation, IIllX + vVx = lllly + VVy = Vy = llx COIlst o. o. in the second. to gel (a) llllx - (b) lilly Vlly = 0, (6) To get rid of lly. multiply (6a) by (6b I by II and add. l1lis yields II and (6b) by + Vllx = O. v and add. Similarly. to eliminate 2 + V 2 + 2 V )lIy = (11 (11 2 llx. multiply (6a) by -v and )lIx = O. O. * If k 2 = ll2 + v 2 = O. then II = v = 0; hence I = O. If k 2 = ll2 + v 2 O. then IIx = lIy = O. Hence. by the Cauchy-Riemann equations. also Vx = Vy = O. Together this implies II = COllst and v = canst; hence I = canst. • We mention that if we use the polar fom1 z = r(cos 6 + i sin 6) and set fez) = u(r, 6) + iv(r, 6), then the Cauchy-Riemann equations are (Prob. 11) LIT = (7) vT = r v e, (r> 0). r LI/I Laplace's Equation. Harmonic Functions The great importance of complex analysis in engineering mathematics results mainly from the fact that both the real part and the imaginary part of an analytic function satisfy Laplace's equation, the most important PDE of physics. which Occurs in gravitation, electrostatics, fluid flow, heat conduction, and so on (see Chaps. 12 and 18). CHAP. 13 622 THEOREM 3 Complex Numbers and Functions Laplace's Equation If fez) = u(x, y) + iv(x, y) is lInalytic in Laplace's equation II d0111l1in D. then both II and v sati.~f\' (8) (V2 read "nabla squared") and (9) in D and h(lI'e continuous second partial derivatives in D. PROOF Differentiating Ux = Vy with respect to x and = uy -vx with respect to y, we have (10) Now the derivative of an analytic function is itself analytic. as we shall prove later (in Sec. 14.4). This implies that u and v have continuous partial derivatives of all orders: in particular, the mixed second derivatives are equal: vYT = v XY ' By adding (10) we thus obtain (8). Similarly, (9) is obtained by differentiating Ux = Vy with respect to y and lty = -v x with respect to x and subtracting, using uxy = uyx ' • Solutions of Laplace's equation having conti1luous second-order partial derivatives are called harmonic functions and their theory is calIed potential theory (see also Sec. 12.10). Hence the real and imaginary parts of an analytic function are harmonic functions. If two harmonic functions u and v satisfy the Cauchy-Riemann equations in a domain D, they are the real and imaginary parts of an analytic function f in D. Then v is said to be a harmonic conjugate function of u in D. (Of course, this has absolutely nothing to do with the use of "conjugate" for z.) E X AMP L E 4 How to Find a Harmonic Conjugate Function by the Cauchy-Riemann Equations Verify that v of 1/. 1/ = x2 - \,2 - Y is harmonic in the whole complex plane and find a harmonic conjugate function Solution. ,21/ = 0 by direct calculation. Now lIx = 2x and Cauchy-Riemann equations a conjugate v of 1/ must satisfy vx 2x, Vy = lIx = lIy ~ -1/ y ~ = - 2.1' - I. Hence because of the 2,· _ + 1. Integrating the first equation with respect to )' and differentiating the result with respect to .t. we obtain v = 2.\)' + h(x). dh 2y + dx . Vx = A comparison with the second equation shows that dh/dr: = 1. This gives hex) = x + c. Hence v = 2.\)' + X + c (c any real constant) is the most general hannonic conjugate of the given II. The conesponding analytic function is I(::.) = II + iv ~ x 2 - )'2 - )' + ;(2.\)' + X + c) = ~2 + ;: + ;e. • SEC 13.5 Exponential Function 623 Example 4 illustrates that a conjugate of a given harmonic function is uniquelv determilled up to an arbitrary real additive constant. The Cauchy-Riemann equations are the most important equations in this chapter. Their relation to Laplace's equation opens wide ranges of engineering and physical applications, as we shall show in Chap. 18 . . ........ ~ CAUCHY-RIEMANN EQUATIONS 22. U = e 3:]; co~ ay 23. u = sin x cosh cy Are the following functions analytic? [Use (1) or (7).] 1. f(:;.) = 2 2. f(::.) = 1m :;.4 3. e x(cos y 5. e-X(cos + i sin y) y - i sin y) 7. f(z) = Re z + 1m z 9. f(:;.) = i/::. 8 (:;.2) 4. f(:;.) = I/O - 6. fez) = Arg 10. f(:;.) = 7TZ. Izl 8. f(z.) = In ::.2 :;.4) + + i Arg z I/:;.2 11. (Cauchy-Riemann equations in polar form) Derive (7) from (1). 112-21/ HARMONIC FUNCTIONS f(:;.) = u (x, y) + iv(x, y). 13. v = xy - yl(x 2 14. v 16. v = In Izl 18. Lt = I/(x 2 + 20. Lt = + y2) )'2) 15. u = In Izl 17. II = x 3 - 3xy2 19. U = (x 2 _ )'2)2 21. cos x cosh y 122-241 l/ = e- x sin 2)' Determine a, b, C such that the given functions are harmonic and find a harmonic conjugate. 13.5 26. TEAM PROJECT. Conditions for fez) = COllst. Let f(:;.) be analytic. Prove that each of the following conditions is sutIicient for f(:;.) = COllst. (a) Re fez) = comt (b) [m f(:;.) = (c) Are the following functions harmonic? If your answer is yes, find a corresponding analytic function 12. u = x)' 25. (Harmonic conjugate) Show that if II is harmonic and v is a harmonic conjugate of II, then II is a harmonic conjugate of -v. f' (z) = COIUT 0 (d) If(z)1 = COllst (see Example 3) 27. (Two further formulas for the derivative). Formulas (4). (5), and (J I) (below) are needed from time to time. Derive (II) J'(;:;) = Ux - illy, f' (z) = Vy + iv x ' 28. CAS PROJECT. Equipotential Lines. Write a program for graphing equipotential lines II = comt of a harmonic function II and of its conjugate v on the same axes. Apply the program to (a) II = x 2 - )'2, U = 2xy, (b) u = x 3 - 3xy2, U = 3x 2y _ y3, (c) U = eX cos )', v = eX sin y. Exponential Function In the remaining sections of this chapter we discuss the basic elementary complex functions, the exponential function, trigonometric functions. logarithm, and so on. They will be counterparts to the familiar functions of calculus, to which they reduce when z = x is real. They are indispensable throughout applications, and some of them have interesting properties not shared by their real counterparts. We begin with one of the most important analytic functions, the complex exponential function also written exp Z. The definition of e Z in terms of the real functions eX, cos y, and sin y is (1) 624 CHAP. 13 Complex Numbers and Functions This definition is motivated by the fact the eZ extends the real exponential function eX of calculus in a natural fashion. Namely; (A) eZ = eX for real z = x because cos Y = 1 and sin y = 0 when y = o. (B) eZ is analytic for all z. (Proved in Example 2 of Sec. 13.4.) (e) The derivative of eZ is eZ • that is. (2) This follows from (4) in Sec. 13.4. REMARK. This defInition provides for a relatively simple discussion. We could defme eZ by the familiar series I + x + x2/2! + x 3 /3! + ... with x replaced by Z, but we would then have to discuss complex series at this very early stage. (We will show the connection in Sec. 15.4.) Further Properties. A function I(::) that is analytic for all :: is called an entire function. Thus, eZ is entire. Just as in calculus the fUllctional relation (3) holds for any 21 = + Xl iYl and Z2 = X2 + iYz. Indeed, by (1), Since e e = e + for these real functions, by an application of the addition fonnulas for the cosine and sine functions (similar to that in Sec. 13.2) we see that X1 X2 X1 X2 as asserted. An interesting special case of (3) is Zl = X, Z2 = iy; then (4) Furthennore, for Z = iy we have from (5) e iy = (1) the so-called Euler formula cosy + i siny. Hence the polar form of a complex number, ;:: (6) From (5) we obtain (7) as well as the important formulas (verify!) (8) e 7Ti = -1, = r(cos e + i sin 0). may now be written SEC. 13.5 Exponential Function 625 Another consequence of (5) is leiYI = (9) leos y + i sin yl = Vcos 2 y + sin2 y = 1. That is, for pure imaginary exponents the exponential function has absolute value I, a result you should remember. From (9) and (1), (10) argeZ = y ± 2nn (n = 0, 1,2," .), Hence since !ezi = eX shows that (1) is actually ~ in polar form. From lezi = eX *- 0 in (0) we see that (11) for all z. So here we have an entire function that never vanishes, in contrast to (nonconstant) polynomials, which are also entire (Example 5 in Sec. 13.3) but always have a zero, as is proved in algebra. Periodicity of e Z with period 27Ti, (12) for all z is a basic property that follows from (1) and the periodicity of cos y and sin y. Hence all the values that w = e Z can assume are already assumed in the horizontal strip of width 27T (13) (Fig. 333). -n<Y~7T This infinite strip is called a fundamental region of eZ • E X AMP L E 1 Function Values. Solution of Equations. Computation of values from (I) provides no problem. For instance. verify that 4 e1. - O.6i = e1. 4 tcos 0.6 - i sin 0.6) = 4.055(0.8253 - 0.5646i) = 3.347 - 2.289; Arg e1.4 - 0 .6i = -0.6. To illustrate (3), take the product of e2 + i = e2 (cos 1 +i sin I) and y x -Tr: Fig. 333. Fundamental region of the exponential function e in the z-plane Z CHAP. 13 626 Complex Numbers and Functions To solve the equation eZ = 3 solutions. Now, since eX = 5, eX cosy = 3. + 4i, note first that lezi = eX = 5. eX sin v = 4. cosy = 0.6. X = In 5 = 1.609 is the real part of all siny = 0.8. y = 0.927. Am. :: = 1.609 + 0.927i::': 211'11'; (n = O. 1.2, ... ). The~e are infinitely many solutions (due to the periodicity of eZ ). They lie on the vertical line x = 1.609 at a distance 27i" from their neighbors. • To summarize: many properties of eZ = exp z parallel those of eX; an exception is the periodicity of £f with 2ni, which suggested the concept of a fundamental region. Keep in mind that ~ is an entire function. (Do you still remember what that means?) 1. Using the Cauchy-Riemann equations, show that e Z is entire. 12-81 Values of eZ • Compute eZ in the form u + iv and lezl, where ~ equals: 3. I + 2i 2. 3 + 71'i 4. Vz - !71'i 5. 771'il2 7. 0.8 - 5i 6. (l + i)71' 8. 971'i/2 19-12 1 Real and Imaginary Parts. Find Re and 1m of: 9. e- 2z 11. e 113-171 13. 15. 10. e z3 z2 Vi Polar Form. Write in polar form: 14. 1 + V; 17. -9 13.6 16. 3 + 4i 118-21\ Equations. Find all solutions and graph some of them in the complex plane. 18. e 3 • = 4 19. e Z = -2 Z 21. e Z = 4 - 3i 20. e = 0 22. TEAM PROJECT. Further Properties of the Exponential Function. (a) Analyticity. Show that c is entire. What about el/z? e Z? eX(cos ky + i sin ky)'? (Use the Cauchy-Riemann equations.) (b) Special values. Find all ;;: such that (i) e Z is real. (ii) le-zi < 1, (iii) e Z = 'if. (c) Harmonic function. Sho~- that u = e XY cos (x 2 /2 - )'2/2) is harmonic and find a conjugate. (d) Uniqueness. [t is interesting that f(z) = e Z is uniquely determined by the two properties f(x + iO) = eXand!'(;;:) = f(z).wherefisassumed to be entire. Prove this using the Cauchy-Riemann equations. Trigonometric and Hyperbolic Functions Just as we extended the real eX to the complex eZ in Sec. 13.5. we now want to extend the familiar real trigonometric functions to complex trigonometric flillctiollS. We can do this by the use of the Euler formulas (Sec. 13.5) eix = cos x + i sin x, e- ix = cosx - i sinx. By addition and subtraction we obtain for the real cosine and sine This suggest,> the following definitions for complex values z = x + iy: SEC 13.6 627 Trigonometric and Hyperbolic Functions (1) slnz = It is quite remarkable that here in complex. functions come together that are unrelated in real. This is not an isolated incident but is typical of the general situation and shows the advantage of working in complex. Furthermore, as in calculus we define (2) tanz = sin z cos cot;:: z z sin z cos = and (3) sec z= I cos csc z z= sin z Since eZ is entire, cos z and sin z are entire functions. tan z and sec z are not entire; they are analytic except at the points where cos;:. is zero; and cot z and csc z are analytic except where sin z is zero. Formulas for the derivatives follow readily from (~)' = eZ and (1)-(3); dS in calculus, (4) (cos ;:.)' (sin z)' -sin?. (tan z)' = sec 2 = cos z. z, etc. Equation (I) also shows that Euler's formula is valid ill complex: eiz = cos;:. + i sin z (5) for all z. The real and imaginary parts of cos z and sin z are needed in computing values, and they also help in displaying properties of our functions. We illustrate this with a typical example. E X AMP L E 1 Real and Imaginary Parts. Absolute Value. Periodicity Show that (a) cos ~ = cos x cosh Y - i sin x sinh y (b) sin z. = sin x cosh y (6) + i cos x sinh y and Icos :12 = cos2 x (a) + sinh2 y (7) Ib) and give some applications of these Solution. forrnula~. From (1). cos z = ~(ei(x+iYJ + = ~e -Y(cos x = ~(eY + + i sin x) + ~eY(cos X e- Y) This yields (6a) since. as is known fonn calculus, (8) e -i(x+iYJ) cos x - ~i(eY - e- Y) - i sin xl sinx. 628 CHAP. 13 Complex Numbers and Functions (6b) is obtained similarly. From (6a) and cosh2 y = I Icos zl2 ~ (cos2 x) (I + sinh2 y we obtain + sinh2 y) + sin2 x sinh2 y. Since sin2 x + cos2 x = I, this gives (7a). and (7bl is obtained similarly. For instance, cos (2 + 3i) = cos 2 cosh 3 - i sin 2 sinh 3 = -4.190 - 9.109i. From (6) we see that cos z and sin z are periodic with period 2n, just as in real. Periodicity of tan;: and cot z with period 7r now follows. Formula (7) points to an essential difference between the real and the complex cosine and sine; whereas Icos xl ~ I and Isin xl ~ I, the complex co~ine and sine functions are 110 10llger boullded but approach infinity in absolute value as y --'> x, since then sinh y ~ 00 in (7). • E X AMP L E 2 Solutions of Equations. Zeros of cos z and sin z Solve la) cos z = 5 (wluch has no real solution!), (b) cos z = 0, (e) sin z = o. (a) e 2iz - 10iz + I = 0 from (1) by multiplication by e iz . This is a quadratic equation in e iz, with solutions (rounded off to 3 decimals) Solution. i z = e -y+ix = 5 :':: V25=""l ~ 9.899 and 0.1O\. Thus e- Y = 9.899 or 0.\01, e ix = I, Y = :'::2.292, x = 2wTi". AilS. Z ~ ±21l7r ± 2.292i (11 = 0, 1,2, .. '). Can you obtain this from (6a)? (b) cos x = 0, sinh y = 0 by (7a), y = O. Ans. z = ::':~(2n + 1)7r (11 = 0, 1,2, .. '). (C) sin x = 0, sinh y = 0 by (7b). Ans. z = :'::1l7r (11 = 0, I, 2, .. '). Hence the only zeros of cos z and sin;: are those of the real cosine and sine functions. • General formulas for the real trigonometric functions continue to hold for complex values. This follows immediately from the definitions. We mention in particular the addition rules (9) cos (Zl ± Z2) = sin (Zl ± Z2) cos Zl cos Z2 =+= sin Zl sin Z2 = sin Zl cos Z2 ± sin Z2 cos Z] and the fOilliula cos 2 Z (10) + sin2 Z = 1. Some further useful formulas are included in the problem set. Hyperbolic Functions The complex hyperbolic cosine and sine are defined by the formulas (11) This is suggested by the familiar definitions for a real variable [see (8)]. These functions are entire, with derivatives (12) (cosh z)' = sinh z, (sinh z)' = cosh z, as in calculus. The other hyperbolic functions are defined by SEC. 13.6 Trigonometric and Hyperbolic Functions tanh z = 629 sinh.: cosh z coth z = cosh z sinh .: (13) sech z = 1 cosh z , csch z = sinh z If in (11), we replace Complex Trigonometric alld Hyperbolic FUllctions Are Related. z by iz and then use (1), we obtain (14) cosh iz = cos z;, sinh iz = i sin z. Similarly, if in (1) we replace z by i;:. and then use (II), we obtain conversely (15) cos iz = cosh z, sin iz = i sinh z. Here we have another case of unrelated real functions that have related complex analogs. pointing again to the advantage of working in complex in order to get both a more unified formalism and a deeper understanding of special functions. This is one of the main reasons for the importance of complex analysis to the engineer and physicist. 1. Prove that cos z, sin z, cosh z, sinh Z are entire functions. 2. Verify by differentiation that Re cos z and 1m sin z are harmonic. 13-61 FORMULAS FOR HYPERBOLIC FUNCTIONS Show that 14. sinh (4 - 3i) 15. cosh (4 - 67Ti) 16. (Real and imaginary parts) Show that Re tan z 1m tan 3. cosh z sinh z 4. cosh (ZI sinh (ZI = = + i sinh x sin y sinh x cos y + i cosh x sin y. cosh x cos Y + Z2) = cosh ZI cosh Z2 + sinh ZI sinh Z2 + Z2) = sinh Zl cosh Z2 + cosh ZI sinh Z2' 5. cosh2 Z - sinh2 z. 6. cosh2 Z + sinh2 Z = = z= sinhy coshy cos X + sinh y . --=--'---'-=-2 2 117-211 Equations. Find all solutions of the following equations. 17. cosh z = 0 18. sin z = 100 19. cos Z = 2i 20. cosh z = - 1 21. sinh z = 0 22. Find all z for which (a) cos z, (b) sin z has real values. 1 cosh 2z 17-151 Function Values. Compute (in the form u 7. cos(l + i) 8. sin(1 + i) 9. sin 5i, cos 5i 10. cos 37Ti 11. cosh (-2 + 3i), cos (-3 - 2i) 12. - i sinh (- 7T + 2i), sin (2 + 7Ti) 13. cosh (2n + 1)7Tl, n = 1,2, ... sin x cos x = --=-------,=-cos 2 X + sinh2 y , + iv) 123-25] Equations and Inequalities. Using the definitions, prove: 23. cos z is even. cos (-z) = cos z, and sin z is odd, sin (-z) = -sin z. 24. Isinh yl ~ lcos zl ~ cosh y, Isinh yl ~ Isin zl ~ cosh y. Conclude that the complex cosine and sine are not bounded in the whole complex plane. 25. sin ZI cos Z2 = Hsin (ZI + Z2) + sin (Zl - Z2)] 630 13.7 CHAP. 13 Complex Numbers and Functions Logarithm. General Power We finally introduce the complex logarithm, which is more complicated than the real logarithm (which it includes as a special case) and historically puzzled mathematicians for some time (so if you first get puzzled-which need not happen!-be patient and work through this section with extra care). The natural logarithm of z = x + iy is denoted by In z (sometimes also by log z) and is defined as the inverse of the exponential function; that is, W = In z is defined for z =1= 0 by the relation (Note that z = 0 is impossible, since w = u + iv and:: = reifl , this becomes e W 0 for all w; see Sec. 13.5.) [f we set =1= Now from Sec. 13.5 we know that eu + iv has the absolute value eU and the argument v. These must be equal to the absolute value and argument on the right: v = 8. eU = r gives u = In r, where In r is the familiar real natural logarithm of the positive number r = Izi. Hence w = u + iv = In z is given by (1) In:: = In,. + i8 (r = Izl > 8 = arg z). 0, Now comes an important point (without analog in real calculus). Since the argument of z is determined only up to integer mUltiples of 271", the complex 1latural logarithm In z (z 0) is i1lfi1litely many-valued. The value of In:: conesponding to the principal value Arg z (see Sec. 13.2) is denoted by Ln :: (Ln with capital L) and is called the principal value of In::. Thus * (2) Ln z = In Izl + i Arg z (z =1= 0). The uniqueness of Arg z for given z (=1= 0) implies that Ln z is single-valued, that is, a function in the usual sense. Since the other values of arg :: differ by integer multiples of 271", the other values of In:: are given by (3) In z = Ln z ::'::: 2n71"i (n = 1. 2.... ). They all have the same real part, and their imaginary paJ1s differ by integer multiples of 271". If:: is positive real, then Arg z = 0, and Ln z becomes identical with the real natural logarithm known from calculus. If z is negative real (so that the natural logarithm of calculus is not defined!), then Arg z = 71" and Ln z = In Izi + 71"i (z negative real). SEC 13.7 631 Logarithm. General Power From (l) and e1n r = r for positive real r we obtain (4a) as expected, but since arg (e Z ) = y ± 2nn is multi valued. so is In (c) = (4b) E X AMP L E 1 Z ± 21lni, n = 0, I,···. Natural Logarithm. Principal Value In 1 = 0, ±2wi, ±4wi, ... Ln 1= 0 In 4 = 1.386 294 ± 211wi Ln 4 = 1.386294 In (-1) = ± m, ±3w;. ±Swi, .. Ln (-I) = wi In (-4) = 1.386294 ± (211 + I)wi Ln (-4) = 1.386294 In i = wil2. - 3 w/2. S wi12 . .•. + wi Ln i = wi/2 In 4; = 1.386294 + wi/2 ± 21lwi Ln 4; = 1.386 294 + wil2 In (-4i) = 1.386294 - wi/2 ± 21lwi Ln (-4i) = 1.386 294 - wi/2 In (3 - 4i) = In S + i arg (3 - 4i) = Ln (3 - 4i) = 1.609438 - 0.927 29S; (Fig. 334) 1.609438 - 0.927 29Si ± 21171'i v -0.9 + 6n • 1 • , 1 -0.9 + 4n 1 1 -0.9 + 2n + o I---....I....----il- - ' - -0.9 +2 u 1 -0.9 - 2n Fig. 334. • 1 Some values of In (3 - 4;) in Example 1 The familiar relations for the natural logarithm continue to hold for complex values, that is. (5) (a) In (~1::2) = In Zl + In ':2, but these relations are to be understood in the sense that each value of one side is also contained among the values of the other side: see the next example. E X AMP L E 2 Illustration of the Functional Relation (5) in Complex Let ;:1=;:2=e"1Ti=-1. If we take the principal values Ln;::1 = LnZ2 = wi. then (Sa) holds provided we write In (:1:::2) = In I = 2wi; however. it is not true for the principal value, • Ln (ZIZ2) = Ln 1 = O. 632 CHAP. 13 THE 0 REM 1 Complex Numbers and Functions Analyticity of the Logarithm For el'ery n = 0, ::t::: 1, ::t:::2, ... formula (3) defines a function, which is analytic, except at 0 and 011 the Ilegarire real axis, alld has the derivative (6) PROOF (ln~) , I (z not 0 or negative real). =- z. We show that the Cauchy-Riemann equations are satisfied. From (I )-(3) we have In z = In r + ice + c) .!.2 In (x 2 + v ' 2 = ) + i(arctan I. + c) x where the constant c is a multiple of 27r. By differentiation, ux x = 2 X + y 2 = Vy = 1 + (ylx) 2 x Hence the Cauchy-Riemann equations hold. [Confirm this by using these equations in polar form. which we did not use since we proved them only in the problems (to Sec. 13.4).j Formula (4) in Sec. 13.4 now gives (6). (- ;~) x - iy = x2 + y2 z • Each of the infinitely many functions in (3) is called a branch of the logarithm. The negative real axis is known as a branch cut and is usually graphed as shown in Fig. 335. The branch for 11 = 0 is called the principal branch of In z. Fig. 335. Branch cut for In z General Powers General powers of a complex number z (7) Since In z is infinitely many-valued, value is called the principal value of zC. = x + iy are defined by the formula (c complex, z ZC oF 0). will, in general, be multi valued. The particular SEC. 13.7 633 Logarithm. General Power If c = n = 1. 2, ... , then zn is single-valued and identical with the usual nth power of z. If c = -1, -2, ... , the situation is similar. If c = I/n, where n = 2, 3, ... , then (z oF 0), the exponent is determined up to multiples of 27Ti/n and we obtain the n distinct values of the nth root, in agreement with the result in Sec. 13.2. [f c = p/q, the quotient of two positive integers, the situation is similar, and ZC has only finitely many distinct values. However, if c is real irrational or genuinely complex, then ZC is infinitely many-valued. E X AMP L E 3 General Power ji =i In i = exp (i In i) = exp [i ( -i j ± 2n17i) ] = e -(",/2)+2n". All these values are real, and the principal value (n = 0) is e -",/2. Similarly, by direct calculation and multiplying out in the exponent, (I + i)2-i = exp [(2 - i) In (I = 2e",/4"'2n"'[sin + i)] = exp [(2 - i) {In V2 + !17i ± 21l17il] (~In 2) • + i cos (~In 2)]. It is a convention that for real positive z = x the expression ZC mean:" e In x where In x is the elementary real natural logarithm (that is, the principal value Ln z (z = x > 0) in the sense of our definition). Also, if z = e, the base of the natural logarithm, ZC = eC is conventionally regarded as the unique value obtained from (1) in Sec. l3.5. From (7) we see that for any complex number a, C (8) aZ = e" In a. We have now introduced the complex functions needed in practical work. some of them (e", cos z, sin z. cosh z, sinh z) entire (Sec. 13.5), some of them (tan z, cot z, tanh z. coth z) analytic except at certain points, and one of them (In z) splitting up into infinitely many functions, each analytic except at 0 and on the negative real axis. For the inverse trigonometric and hyperbolic functions see the problem set. ~-il Principal Value Ln z. Find Ln z when z equals: 1. - to 2. 2 + 2; 3. 2 - 2i 4. -5 ~ O.li 5. -3 - 4i 6. -100 7. 0.6 + O.Si B. -ei 9. 1 - i 110-161 All Values of In z. Find all values and graph some of them in the complex plane. 10. In 1 n. In (-I) 13. In (-6) 12. In e 14. In (4 + 3i) 15. In (-e-') 16. In (e 3i ) 17. Show that the set of values of In (i2) differs from the set of values of 2 In i. 11B-21/ lB. In Equations. Solve for z: z = (2 - !i)7T 20.lnz=e-7Ti 19. In z = 0.3 + 0.7; 21. In z = 2 + ~7Ti CHAP. 13 634 \22-2S \ 22. 24. 26. 2S. General Powers. Showing the details of your work, find the principal value of: i2i, (2i)i (l - Complex Numbers and Functions i)l+i (-I )1-2i 23. 4 3 + i 25. (l + 27. ;112 Yz2 - (a) arccos.;: = -i In (;: + + ~) (b) arcsin z: = -i In (i.;: (c) arccosh z = In (z: il-' (d) arcsinh;:: = In (z 1) + -w-=-I) + W+l) (3 - 4i)1I3 i i + Z (e) arctan;:: = - In - 2 i - :: 29. How can you find the answer to Prob. 24 from the answer to Prob. 25? 30. TEAM PROJECT. Inverse Trigonometric and Hyperbolic Functions. By definition. the inverse sine w = arcsin z is the relation such that sin w = z. The inverse cosine w = arccos:: is the relation such that cos W = ::. The in,-erse tangent, inverse cotangent, innrse hyperbolic sine, etc .. are defined and denoted in a similar fashion. (Note that all these relations are mllitivailled.) Using sin w = (e i '" - e- iW )/(2i) and similar representations of cos w, etc .. show that <n I arctanh;:: = 2" I +z z In 1 - (g) Show that w = arcsin:: is infinitely many-valued. and if WI is one of these values, the others are of the form WI ~ 11l7T and 7T - WI ~ 21l7T, 11 = 0, I, .... (The principal mlue of w = u + iv = arcsin z is defined to be the value for which -7T!2 ;;; U ;;; 7T!2 if v ~ 0 and - 7T!2 < 1I < 7T!2 if v < 0.) -C-l-FA PT-"E~33_ R E-Vl..E-w=:QlJ EST ION SAN D PRO B L EMS 1. Add. subtract. multiply. and divide 26 3 + 4i as well as their complex conjugates. 7i and 2. Write the two given numbers in Prob. I in polar form. Find the principal value of their arguments. 3. What is the triangle inequality? Its geometric meaning? Its significance? 4. If you know the values of {,fl, how do you get from them the values of ~ for any;:? 5. State the definition of the derivative from memory. It looks similar to that in calculus. But what is the big difference? 6. What is an analytic function? How would you test for anal yticity? 7. Can a function be differentiable at a pomt without being analytic there? If yes, give an example. [16-2D Complex Numbers. Find, in the fonn x showing the details: 16. (1 + ;)12 17. (- 2 + 6;)2 IS. 1/(3 - 7i) 20. \/-5 - 12i 19. (l - ;)/(1 + + iy. i)2 21. (43 - 19i)/(8 + ;) 122-261 Polar Form. Represent in polar form. with the principal argument: 23. -6 + 6i 22. 1 - 3i 25. -12i 24. YW/(4 + 2i) 26. 2 + 2; \27-30\ Roots. Find and graph all values of 27. V& 29.~ 2S. V'256 30. VC"32-:------::-24-i [31-~ 9. State the definitions of eZ , cos z. sin ;::. cosh z. sinh;:: and the relations between these functions. Do these relations have analogs in real? Analytic Functions. Find f(.::) = u(x.y) + ;v(x.y) with 1I or v as given. Check for analyticity. 31. 11 = x/(x 2 + y2) 32. v = e- 3x sin 3y 2 33. u = x - 2xy - y2 34. 1I = cos 1x cosh 2y y2 35. v = e X2 sin 2xy 10. What properties of C are similar to those of eX ? Which one is different? @6-391 Harmonic Functions. Are the following functiuns hannonic? If so, find a hannonic conjugate. II. What is the fundamental region of eZ ? Its significance? 36. x 2 y2 3S. e- x / 2 cos!y S. Are 1::1, .:. Re;::, 1m:: analytic? Give reason. 12. What is an entire function? Give examples. 13. Why is In z much more complicated than In x? Explain from memory. 14. What is the principal value of In z? 15. How is the general power:;c defined? Give examples. 37. xy 39. x 2 + y2 \40-451 Special Function Values. Find the values of 40. sin (3 + 47Ti) 41. sinh 47Ti 42. cos (57T + 2;) 43. Ln CO.8 + 0.6i) 44. tan (I + i) 45. cosh (I + 7Ti) Summary of Chapter 13 635 Complex Numbers and Functions For arithmetic operations with complex numbers (l) = z Izl + x iy = re iIJ = r(cos e + i sin 8). e = arctan (y/x). and for their representation in the complex plane, see Secs. 13.1 and 13.2. A complex function f(:;) = u(x, y) + iv(x. y) is analytic in a domain D if it has a derivative (Sec. 13.3) r= = \lx2 + )'2, t' (z) = lim (2) fez + Llz) - fez) Llz .lz~o everywhere in D. Also, fez) is analytic at a point z = '<:0 if it has a derivative in a neighborhood of Zo (not merely at Zo itself). If fez) is analytic in D. then u(x. y) and v (x. y) satisfy the (very important!) Cauchy-Riemann equations (Sec. 13.4) (3) everywhere in D. Then (4) au au au au ax dy , ay ax II and v also satisfy Laplace's equation + Uyy = U XX 0, everywhere in D. If 1I(x, y) and u(x. y) are continuous and have continuous partial derivatives in D that satisfy (3) in D. then fez) = u(x, y) + iu(x, y) is analytic in D. See Sec. 13.4. (More on Laplace's equation and complex analysis follows in Chap. 18.) The complex exponential function (Sec. 13.5) (5) e Z = exp z = eX (cos y + i sin y) reduces to e T if z = x (y = 0). It is periodic with 27Ti and has the derivative eZ • The trigonometric functions are (Sec. 13.6) I. cos z = - (en 2 + e- . ZZ ) = cos x cosh V - i sin x sinh \' - (6) 1. . sin z = - (eOZ - e- OZ ) = sin x cosh \' + i cos x sinh \' 2i - - and, furthermore, tan z = (sin z)Jcos z, cot Z = lItan:. etc. 636 CHAP. 13 Complex Numbers and Functions The hyperbolic functions are (Sec. 13.6) (7) cosh z = !(eZ + e-Z) = cos iz, etc. The functions (5)-(7) are entire, that is, analytic everywhere in the complex plane. The natural logarithm is (Sec. 13.7) (8) In z = In Izl + i arg z = In Izl +i Arg z ::!: 2117Ti where z =F 0 and II = 0, 1, . . . . Arg z is the principal value of arg z, that is. -7T < Arg z ~ 7T. We see that In z is infinitely many-valued. Taking n = 0 gives the principal value Ln z of In z; thus Ln z = In Izl + i Arg z. General powers are defined by (Sec. 13.7) (9) (c complex. z =F 0). ~CHAPTER ~7 / 14 Complex Integration Two main reasons account for the importance of integration in the complex plane. The practical reason is that complex integration can evaluate certain real integrals appearing in applications that are not accessible by real integral calculus. The theoretical reason is that some basic properties of analytic functions are difficult to prove by other methods. A striking property of this type is the existence of higher derivatives of an analytic function. Complex integration also plays a role in connection with special functions. such as the gamma function (see [GRll. p. 255), the error function. various polynomials (see [GRIOD and others. and the application of these functions in physics. In this chapter we define and explain complex integrals. The most important result in the chapter is Cauchy's integral theorem or the Callchy-Goursat theorem, as it is also called (Sec. 14.2). It implies Cauchy's integral formula (Sec. 14.3), which in tum implies the existence of all higher derivatives of an analytic function. Hence in this respeCl, complex analytic functions behave much more simply than real-valued functions of real variables, which may have derivatives only up to a certain order. A further method of complex integration, known as integration by residues, and its application to real integrals will need complex series and follows in Chap. 16. Prerequisite: Chap. 13 References alld Answers to Prohlems: App. I Part D, App. 2. 14.1 Line Integral in the Complex Plane As in calculus we distinguish between definite integrals and indefinite integrals or antiderivatives. An indefinite integral is a function whose derivative equals a given analytic function in a region. By inverting known differentiation formulas we may find many types of indefinite integrals. Complex definite integrals are called (complex) line integrals. They are wlitten fc fez) dz. Here the integrand fez) is integrated over a given curve C or a portion of it (an arc. but we shall say "curve" in either case, for simplicity). This curve C in the complex plane is called the path of integration. We may represent C by a parametric representation (1) z(t) = x(t) + iy(t) (a ~ t ~ b). 637 638 CHAP. 14 Complex Integration The sense of increasing t is called the positive sense on C, and we say that C is oriented by (1). For instance, 2(t) = t + 3it (0 ~ t ~ 2) gives a portion (a segment) of the line)' = 3x. The function .:(t) = 4 cos t + 4i sin t (- 7T ~ t ~ 7T) represents the circle Izl = 4. and so on. More examples follow below. We assume C to be a smooth curve, that is, C has a continuous and nonzero derivative . d: . dt = x(t) z(t) = - + . .. i,,(t) at each point. Geometrically this means that C has everywhere a continuously turning tangent, as follows directly from the definition • z{t) . = hm z(t + .It) - :(1) (Fig. 336). !1t ~t~O Here we use a dot since a prime' denotes the derivative with respect to z. Definition of the Complex Line Integral This is similar to the method in calculus. Let C be a smooth curve in the complex plane given by (1), and let fez) be a continuous function given (at least) at each point of C. We now subdivide (we "partition") the interval a ~ t ~ b in (1) by points where to < points tl < ... < tn' To this subdivision there corresponds a subdivision of C by Zn-l' Zn (= Z) (Fig. 337). Z / m z z(t) o Fig. 336. Tangent vector i(t) of a curve C in the complex plane given by z(t). The arrowhead on the curve indicates the positive sense (sense of increasing t). Fig. 337. Complex line integral where Zj = :(tj)' On each portion of subdivision of C we choose an arbitrary point, say, a point (1 between Zo and 21 (that is, (1 = z(t) where t satisfies to ~ t ~ t 1), a point (2 between '::1 and 22' etc. Then we form the sum n (2) where m=1 We do this for each 11 = 2, 3, ... in a completely independent manner. but so that the greatest 1!11ml = It.m- tm - 1 1 approaches zero as n ~ 00. This implies that the greatest SEC. 14.1 Line Integral in the Complex plane 639 ILlZmI also approaches zero. Indeed, it cannot exceed the length of the arc of C from Zm-I to Zm and the latter goes to zero since the arc length of the smooth curve C is a continuous function of t. The limit of the sequence of complex numbers S2, S3 • ... thus obtained is called the line integral (or simply the imegral) of f(z) over the path of integration C with the oriemarion given by (l). This line integral is denoted by I (3) c f or by f(::) d::., c f(z) d: if C is a closed path (one whose terminal point Z coincides with its initial point a circle or for a curve shaped like an 8). ::'0, as for General Assumption. All paths of integration for complex line integrals are assllmed to be piecewise smooth, that is. they consist offinitely many smooth curves joined end to end. Basic Properties Directly Implied by the Definition 1. Linearity. Integration is a linear operation, thar is, we can imegrare sums term by term and can take out constant factors from under the imegral sign. This mean~ that if the integrals of f 1 and .f2 over a path C exist, so does the integral of kIf 1 + k2f2 over the same path and 2. Sense reversal in integrating over the same path, from Z.o (right), introduces a minus sign as shown, I (5) z Iz ::'0 to Z (left) and from Z to Zo f(:) dz = Zo - fez) dz. 3. Partitioning of path (see Fig. 338) (6) I f(;:) d:: C Fig. 338. = I f(::.) d::. C1 + I f(z) dz. C2 Partitioning of path [formula (6)] Existence of the Complex Line Integral Our assumptions that f(:) is continuous and C is piecewise smooth imply the exi'itence of the line integral (3). This can be seen as follows. As in the preceding chapter let us write f(z) = H(X. y) -t iv(x, y). We also set and 640 CHAP. 14 Complex Integration Then (2) may be written (7) Sn .L (u + iv)(·.h + i!l.Ym) = m where u = u«(m, 7]",). v = V«(7m 7]",) and we sum over I7l from 1 to n. Performing the multiplication. we may now split up Sn into four sums: These sums are real. Since f is continuous, u and v are continuous. Hence, if we let n approach infinity in the aforementioned way, then the greatest !l.xm and .lYm will approach zero and each sum on the right becomes a real line integral: lim SII = (8) n--+oc f f(z) dz C = f u dx - C f v dy C +i [f C u dy + f v dX] . C This shows that under our assumptions on f and C the line integral (3) exists and its value • is independent of the choice of subdivisions and intermediate poinb (m' First Evaluation Method: Indefinite Integration and Substitution of Limits This method is the analog of the evaluation of definite integrals in calculus by the well-known formula f b f(x) dx = F(b) - F(a) [F' (x) = f(x)]. a It is simpler than the next method. but it is suitable for analytic functions only. To formulate it, we need the following concept of general interest. A domain D is called simply connected if every simple closed curve (closed curve without self-intersections) encloses only points of D. For instance, a circular disk is simply connected, whereas an annulus (Sec. 13.3) is not simply connected. (Explain!) THEOREM 1 Indefinite Integration of Analytic Functions Let f(;:.) be analytic in a simply connected domain D. Then there exists an indefinite integral of f(::.) in the domain D, that is, an analytic function F(::.) such that F' (::.) = f(::.) in D, and for all paths in D joining two poims 20 and ZI in D we have f'f(z) dz (9) = F(ZI) - F(zo) [F' (z) = f(z)]. 20 (Note that we can write those C from ::'0 to 21') 20 and ::'1 instead of C, since we get the same value for all SEC. 14.1 Line Integral in the Complex plane 641 This theorem will be proved in the next section. Simple connectedness is quite essential in Theorem I, as we shall ~ee in Example 5. Since analytic functions are our main concern, and since differentiation formulas will often help in finding F(z) for a given fez) = F' (z), the present method is of great practical interest. If fez) is entire (Sec. 13.5), we can take for D the complex plane (which is certainly simply connected). EXAMPLE 1 EXAMPLE 2 f I f 1+i _2 do 0-'3 Z3 .cos::: d::: = sin::: S-3m 1 (1 3 = - 0 7Ti -m E X AMP L E 3 11 + i 1 = - I' I To • + 2 • • 2 = - - + - i i)3 33 = 2 sin 7ri = 2i sinh r. = 23.097i -m e z/ 2 d::: = S- 3 m 2eZ / 2 8+m = e4 + mt2 ) 2(e4-37Ti/2 - =0 8+m • since eZ is periodic with period 2r.i. E X AMP L E 4 I~i ~ = Ln i - Ln (-i) = i ; - (- i;) = ir.. Here D is the cumplex plane without 0 and the negative real axis (where Ln::: is not analytic). Obviously, D is a simply connected domain. • Second Evaluation Method: Use of a Representation of a Path This method is not restricted to analytic functions but applies to any continuous complex function. THEOREM 2 Integration by the Use of the Path Let C be a piecewise smooth path, represented by Z fez) be a continuous function on C. Then f f(;:;) (10) z(t), where a 3 t 3 b. Let f f[z(t)]Z(t) dt b d: C PROOF = = a The left side of (10) is given by (8) in tenns of real line integrals, and we show that the right side of (10) also equals (8). We have;:; = x + iy, hence = + iY. We simply wnte II for u[x(t), y(t)] and v for v[x(t). y(t)]. We also have d"t = dt and dy = Ydt. Consequently, in (10) z x x b b f f[z(t)]Z(t) dt = f (u + = f [u dx - iv)(.t a + (),) dt a v dy + i(u dy + V dx)] C = fc(u dx - V dy) + ifc (u dy + V dx). • 642 CHAP. 14 Complex Integration COMMENT. In (7) and (8) of the existence proof of the complex line integral we referred to real line integrals. If one wants to avoid this, one can take (to) as a definition of the complex line integraL Steps in Applying Theorem 2 (A) Represent the path C in the form z(t) (a ~ t ~ b). (B) Calculate the derivative z(t) = dz/dt. (C) Substitute zlt) for every z in .«z) (hence x(t) for x and y(t) for y). (D) Integrate .f[z(t)]z(t) over t from a to b. E X AMP L E 5 A Basic Result: Integral of 1/z Around the Unit Circle We show that by integrating Sec. 13.3) we obtain 1I~ counterclockwise around the unit circle (the circle of radius I and center 0; see J, dz = 21Ti (11) (C the unit circle, counterclockwise). Jc z This is a vel)' importalll result that we shall need quite often. Solutioll. (A) We may represent the unit circle C in Fig. 327 of Sec. 13.3 by z(t) = cos T + i sin r = eit (0 ~ r ~ 27T). so that counterclockwise integration corresponds to an increase of t from 0 to 27T. (B) Differentiation gives :(t) = ieit (chain rule!). (C) By substitution. f(~(t) = 1I~(t) = e -it. (D) From (10) we thus obtain the result f dz - = f 2.,,-.. f2"'-dt = 27Ti. e -'tie't dt = i c zoo Check this result by using ~(tl = cos t + i sin t. Simple cOllllectedlless is esselltial ill Tlzeorem 1. Equation (9) in Theorem I gives 0 for any closed path because then ;:1 = ;:0, so that F(;:1) - F(;:o) = O. Now 1/;: is not analytiC at z = O. But any simply connected domain containing the unit circle must contain ~ = 0, so that Theorem I does not apply-it is not enough that liz is analytic in an annulus. say. ~ < Izl < i, because an annulus is not simply connected! • E X AMP L E 6 Integral of 1/z m with Integer Power m Let fez) = (z - :Co)m where m is the integer and of radius p with center at ~o (Fig. 339). ;:0 a constant. Integl'ate counterclockwise around the circle C y x Fig. 339. Path in Example 6 SEC. 14.1 643 Line Integral in the Complex Plane Solutioll. We may repre,ent C in the form = :0 :e(t) = Zo + p(cos t + i sin t) + peit (0 "'" t "'" 2'IT). Then we have t dz = ipi dt and obtain By the Euler formula (5) in Sec. 13.6 the right side equals fo 2w ip>n+l [ 2w ] COS(n1+l)tdt+if sin(m+l)rdt o . '* If 111 = - I. we have pm+l = I, cos 0 = 1, sin 0 = O. We thus obtain 2'ITi. For integer nI 1 each of the two integrals is zero because we integrate over an interval of length 2'IT, equal to a period of sine and cosine. Hence the result is (m = (111 *" -) and integer). (12) -1), • Dependence on path. Now comes a very important fact. If we integrate a given function J(z) from a point Zo to a point Zl along different paths, the integrals will in general have different values. In other words. a complex lille illtegral depellds Ilot ollly Oil the elldpoillts o/the path but ill gelleral also Oil the path itself. The next example gives a first impression of this, and a systematic discussion follows in the next section. E X AMP L E 7 Integral of a Nonanalytic Function. Dependence on Path Integrate f(:) = Re: = " from 0 to I + 2i (a) along C* in Fig. 340, (b) along C consisting of C1 and C 2 · y 2 z=1+2i I I I I C'I z I " C I 1 " C Fig. 340. x Paths in Example 7 Solutioll. (a) C* can be represented by z(t) = r + 2it (0 "'" t"", I). Hence z(t) = I + 2i and f[z(t)] = xCt) = t on C*. We now calculate f Re z d::: = c* f 1 t(l + 2i) dt = ~(I + 2i) =! + i. 0 (b) We now have C1 : z(t) = t. itt) C2 : Zll) = I + it, ~(t) = i, = I, f(::(t)) = x(t) = t f(zv» = X(I) = 1 (0 "'" t "'" 1) (0"", t"", 2). CHAP. 14 644 Complex Integration Using (6) we calculate f Re.: dz = C f Re:: dz + C1 f Re:: dz = C2 f 1 2 t dt + f l . i dt 0 = ~ + 2i. 0 • Note that this Tesult diffeTs from the result in (a). Bounds for Integrals. ML -Inequality There will be a frequent need for estimating the absolute value of complex line integrals. The basic formula is II/(z) dzl ~ (13) ML (ML-inequality); L is the length of C and M a constant such that If(z) 1 ~ M everywhere on C. PROOF Taking the absolute value in (2) and applying the generalized inequality (6*) in Sec. 13.2. we obtain Now ILlZml is the length of the chord whose endpoints are Z",,-l and Zm (see Fig. 337 on p. 638). Hence the sum on the right represents the length L * of the broken line of chords whose endpoints are zo, Zl, •.• , Zn (= Z). If n approaches infinity in such a way that the greatest ILltml and thus ILlZ'>n1 approach zero, then L * approaches the length L of the curve C, by the definition of the length of a curve. From this the inequality (13) follows. • We cannot see from (13) how close to the bound ML the actual absolute value of the integral is, but this will be no handicap in applying (13). For the time being we explain the practical use of (13) by a simple example. EXAMPLE 8 '~ Estimation of an Integral Find an UppeT bound fOT the absolute value of the integral C the straight-line segment from 0 to I + i. Fig. 341. Solution. L = V2 and If(z) 1= Iz21 ~ 2 on C gives by (13) Fig.341. Path in Example 8 IIc I~ 2V2 Z2 I dz I The absolute value of the imegral is - -2 + -2 i = -2 333 = 2.8284. Vz = 0.9428 (see Example I). • Summary on Integration. Line integrals of .Hz) can always be evaluated by (10), using a representation (I) of the path of integration. If f(z) is analytic, indefinite integration by (9) as in calculus will be simpler. SEC. 14.1 645 Line Integral in the Complex plane . ..... .- . . ---- -- ....... PARAMETRIC REPRESENTATIONS 11-91 26. Find and sketch the path and its orientation given by: 1. zU) = (l + 3i)t (1 ~ 2. :.:(1) = 5 - 2it (- 3 3. zU) = 4 4. z(t) 5. = } ~ 4) 7. :.:(1) + 4i + 5e (7T ~ 1 ~ = 1) +i + ib to c x 2 to f sec 2 :.: d:.:, C any path from 7T/4 to 7Ti14 Tm Z2 dz counterclockwise around the triangle with c vertices:.: = 0, I, i :.:e z2/2 d:.:, C from i along the axes to I Z2, where Cis 31. (Path partitioning) Verify (6) for f(:.:) = 11:::: and C 1 and C2 the upper and lower halfs of the unit circle to 4 - 2i 11. Unit circle (c1ocl\.wise) 12. Segment from a l' = i 30. (Sense reversal) Verify (5) for fez) = the segment from -1 - i to I + i. Sketch and represent parametrically: 10. Segment from I i along the parabola c PARAMETRIC REPRESENTATIONS 110-181 29. 27T) + 2t + 8il 2 (-1 ~ 1 ~ t + !it 3 (-1 ~ 1 ~ 2) + c ~ T ~ 7T) :.:(t) = I 9. :.:(t) 28. f f 27. e-.,.it (0 ~ 1 ~ 2) 6 cos 21 + 5i sin 2t (0 = + I 1 ~ 3) 3e it (0 ~ t ~ 27T) it dz, C from -} C z(t) = e it (0 ~ t ~ 7T) 6. :.:(1) = 3 8. +i + +i+ t ~ fz 32. (ML-inequality) Find an upper bound of the absolute value of the integral in Prob. 19. + id + i to 4 + ~i 13. Hyperbola xy = 1 from I 14. Semi-ellipse x 21a 2 + y21b 2 = 1. Y ~ 0 33. (Linearity) Illustrate (4) with an example of your own. Prove (4). 15. Parabola y = 4 - 4x 2 (-I ~ x ~ 1) 34. TEAM PROJECT. Integration. (a) Comparison. Write a short repOit comparing the essential points of the two integration methods. + 3il 16. I:.: - 2 17. Iz + a + ibl = 4 (counterclockwise) = r (clockwise) 18. Ellipse 4(x - 1)2 + 9(y + 2)2 = 36 (b) Comparison. Evaluate Integrdte by the first method or state why it does not apply and then use the second method. (Show the details of your work.) 19. f f f f fc Re:.: d:.:, C the shortest path from 0 to I + i c 20. Re z dz, C the parabola y = x 2 from 0 to I + i c 21. e 2z d:.:, C the shonest path from 7Ti to 27Ti c 22. sin z dz, C any path from 0 to 2i c 23. cos 2 :.: d:.: from -7Ti along Izl = 7T to 7Ti in the right half-plane 24. f f c 25. c (z + l and check the result by Theorem 2, where: INTEGRATION 119-291 I/(:::) d::: by Theorem [1) dz, C the unit circle (counterclockwise) cosh 4:.: d:.:, C any path from - 7Ti/8 to 71i/8 = ::4 and C is the semicircle Izi - 2i to 2i in the right half-plane, (i) f(::.) = 2 from (ii) f(:.:) = e2z and C is the shonest path from 0 to 1 + 2i. (e) Continuous deformation of path. Experiment with a family of paths with common endpoints, say, z(t) = 1 + ia sin t, 0 ~ t ~ 71. with real parameter a. Integrate nonanalytic functions (Re:c, Re (:.:2), etc.) and explore how the result depends on a. Then take analytic functions of your choice. (Show the details of your work.) Compare and comment. (d) Continuuus deformation of path. Choose another family. for example. semi-ellipses z(t) = a cos 1 + i sin I, -7T/2 ~ t ~ 71'/2, and experiment as in (c). 35. CAS PROJECT. Integration. Write programs for the two integration methods. Apply them to problems of your choice. Could you make them into a joint program that also decides which of the two methods to use in a given case? 646 14.2 CHAP. 14 Complex Integration Cauchy's Integral Theorem We have just seen in Sec. 14.1 that a line integral of a function fez) generally depends not merely on the endpoints of the path, but also on the choice of the path itself. This dependence often complicates situations. Hence conditions under which this does not occur are of considerable importance. Namely. if .Hz) is analytic in a domain D and D is simply connected (see Sec. 14.1 and also below), then the integral will not depend on the choice of a path between given points. This result (Theorem 2) follows from Cauchy's integral theorem. along with other basic consequences that make Cauchy's integral theorem the most importallt theorem in this chapter and fundamental throughout complex analysis. Let us begin by repeating and illustrating the definition of simple connectedness (Sec. 14.1) and adding some more details. 1. A simple closed path is a closed path (Sec. 14.1) thaJ does not intersect or touch itself (Fig. 342). For example, a circle is simple, but a curve shaped like an 8 is not simple. ( \ Simple Not simple Not simple Simple Fig. 342. Closed paths 2. A simply connected domain D in the complex plane is a domain (Sec. 13.3) such that every simple closed path in D encloses only points of D. Examples: The interior of a circle ("open disk"). ellipse. or any simple closed curve. A domain that is not simply connected is called mUltiply connected. Examples: An annulus (Sec. 13.3), a disk without the center, for example, 0 < Izl < 1. See also Fig. 343 . .. -, ... , ,, '" ," .../,' , '" ,, , '-' """, " ., : ,'--', \ -, " .,_---"!!Ir I , , ", I I I \ \ ,-, " ' ......... _." , . ,' " ....... _-' " ~ I. \ ,.,-.".-- ..... , , , 1 , I I ,"" '-', " '\ ','" " ' .... _--'--,,' .... ' '\ Simply connected Simply connected Fig. 343. '\ \ I " \ I ".., ' ... _..... ' .... I I _---- " I I ", I " • ,, , ... ---"." ,"'-, I , , . '\ I ,, , ' .. --' ,,--, ,I I I , \ '.. Doubly connected , , " I " ,-_/, I ' ..... _----'" Triply connected Simply and multiply connected domains More precisely, a bounded domain D (thal is, a domain that lies entirely in some circle about the origin) is called p-fold connected if its boundary consists of p closed connected sets without common points. These sets can be curves, segments, or single points (such as z = 0 for 0 < Izl < I. for which p = 2). Thus, D has p - I "holes", where "hole" may also mean a segment or even a single point. Hence an annulus is doubly connected (p = 2). SEC 14.2 Cauchy's Integral Theorem THEOREM 1 647 Cauchy's Integral Theorem Iff(z) is analytic in a simply connected domain D. tllenfor every simple closed path C in D. f (1) c fez) dz = O. See Fig. 344. --- ----- "/'-0 " ... ... \ I ,,-' -' -' ,,-' I / / : D " ..... C _---- ------------' Fig. 344. ,,/ Cauchy's integral theorem Before we prove the theorem. let us consider some examples in order to really understand what is going on. A simple closed path is sometimes called a colltour and an integral over such a path a contour integral. Thus, (1) and our examples involve contour integrals. E X AMP L E 1 No Singularities (Entire Functions) fez dz c = O. f (n = cos zdz = 0, for any closed path, since these functions are entire (analytic for all E X AMP L E 2 0, 1.... ) c • ~). Singularities Outside the Contour f sec c ~d;: = J, O. Jc d::. <.2 +4 = 0 where C is the unit circle, sec z = IIcos <. is not analytic at ;;: = ± 7rf2, ±37Tf2 • ... , but all these points lie outside C; none lies on C or inside C. Similarly for the second integral, whose integrand is not analytic at z = ::':2i outside C. • E X AMP L E 3 Nonanalytic Function f:: = I 2,,- e-itiit dt d;:. C where C: ~(t) analytic. E X AMP L E 4 = = 27ri 0 e it is the unit circle. This does not contradict Cauchy's theorem because f(z) = :: is not • Analyticity Sufficient, Not Necessary J, r dz 7 C~ = U 2 where C is the unit circle. This result does not follow from Cauchy'~ theorem. because f(::;) = 1/;:.2 is not analytic at z = O. Hence the condition that f be analytic ill D is sujficiellf rather thall neces.mr\' for ( I) to be true. • 648 CHAP. 14 E X AMP L E 5 Complex Integration Simple Connectedness Essential d: rJ,c --== 27ri - for counterclockwise imegrarion around rhe unit circle (see Sec. 14.1). C lies in the annulus ~ < 1:1 < ~ where If: is analytic. but this domain is not simply connected. so that Cauchy"s theorem cannot be applied. Hence the condition that tile doma;'1 D be simply connected is essential. In other word,. by Cauchy's theorem. if II:) is analytic on a simple closed path C and everywhere inside C, with no exception. not even a single point. then (I) holds. The point that causes trouble here is : = 0 where If: is not analytic • PROOF Cauchy proved his integral theorem under the additional assumption that the derivative f' (z) is continuous (which is true. but would need an extra proof). His proof proceeds as follows. From (8) in Sec. 14.1 we have fc f(z) dz = fc (u dx - v dy) + i fc (u dy + v dx). Since .f(z) is analytic in D, its derivative .f' (z) exists in D. Since .f' (z) is assumed to be continuous, (4) and (5) in Sec. 13.4 imply that u and v have continuous partial derivatives in D. Hence Green's theorem (Sec. 10.4) (with u and -v instead ofF! and F 2 ) is applicable and gives f (u dx - v dy) = C I I (- ~ax - ~) a) dr dy R where R is the region bounded by C. The second Cauchy-Riemann equation (Sec. 13.4) shows that the integrand on the right is identically zero. Hence the integral on the left is zero. In the same fashion it follows by the use of the first Cauchy-Riemann equation that • the last integral in the above formula is zero. This completes Cauchy's proof. Goursat's proof without the cOllditioll that f' (z) is cOlltillllOUS 1 is much more complicated. We leave it optional and include it in App. 4. Independence of Path We know from the preceding section that the value of a line integral of a given function Z1 to a point Z2 will in general depend on the path C over which we integrate, not merely on Z1 and Z2' It is imp0l1ant to characterize situations in which this difficulty of path dependence does not occur. This task suggests the following concept. We call an integral of .f(z) independent of path in a domain D if for every ::::1, Z2 in D its value depends (besides on f(::::), of course) only on the initial point ::::1 and the terminal point Z2, but not on the choice of the path C in D [so that every path in D from Z1 to ::::2 gives the same value of the integral of f(z)]. .f(z) from a point ItDOUARD GOURSAT (1858-1936). French mathematician. Cauchy published rhe theorem in 1825. The removal of that condition by GourSal (see Transactions Amer. Math. Soc.. vol. I. 1900) is quite important. for instance, in connection with the fact thai derivatives of analytic functions are also analytic. as we shall prove soon. Goursat also made important contributions to PDEs. SEC. 14.2 649 Cauchy's Integral Theorem THEOREM 2 Independence of Path If fez) is analytic in a simply connected domain D, then the integral of .f(z) is independent of path in D. PROOF Let ZI and Z2 be any points in D. Consider two paths C 1 and C 2 in D from Zl to Z2 without further common points, as in Fig. 345. Denote by ci the path C2 with the orientation reserved (Fig. 346). Integrate from :::1 over C 1 to ;:2 and over C~ back to Zl' This is a simple closed path, and Cauchy's theorem applies under our assumptions of the present theorem and gives zero: (2') I f dz + I el c~ f dz I f dz thus = 0, C, = - I c,; f dz. But the minus sign on the right disappears if we integrate in the reverse direction, from to Z2, which shows that the integrals of fez) over C 1 and C2 are equal. ZI Ic, (2) fC:;) dz = Ic, (Fig. 345). .f(z) dz This proves the theorem for paths that have only the endpoints in common. For paths that have finitely many further common points, apply the present argument to each "loop" (portions of C1 and C2 between consecutive common points; four loops in Fig. 347). For paths with infinitely many commOn points we would need additional argumentation not • to be presented here. Fig. 345. Formula (2) Fig. 346. Formula (2') Paths with more common points Fig. 347. Principle of Deformation of Path This idea is related to path independence. We may imagine that the path C2 in (2) was obtained from C 1 by continuously moving C 1 (with ends fixed!) until it coincides with C2 . Figure 348 shows two of the infinitely many intermediate paths for which the integral always retains its value (because of Theorem 2). Hence we may impose a continuous deformation of the path of an integral, keeping the ends fixed. As long as our deforming path always contains only points at which .f(z) is analytic, the integral retains the same value. This is called the principle of deformation of path. 650 CHAP. 14 Complex Integration c) ------, ...." , ... ---~.... ", , \ \ \ ... \ \ \ I \ I . . . . . . . \1 2) Fig. 348. E X AMP L E 6 Continuous deformation of path A Basic Result: Integral of Integer Powers From Example 6 in Sec. 14.1 and the principle of deformation of path it follows that f (3) zor d" (z - = 2m { 0 (m = -I) (m '* - I and integer) for counterclockwise integration around allY simple closed path cOlltaillillg Zo ill its illterior. Indeed. the circle Iz - .::01 = P in Example 6 of Sec. 14.1 can be continuously defonned in two steps into a path as just indicated. namely. by first defomling. say. one semicircle and then the other one. (Make a sketch). • Existence of Indefinite Integral We shall now justify our indefinite integration method in the preceding section [formula (9) in Sec. 14.1]. The proof will need Cauchy's integral theorem. Existence of Indefinite Integral THEOREM 3 If f(::.) is analytic in a simply c01lnected domain D, tben there exists an indefinite integral F(z) of f(z) in D-thus, F'(z) = f(z)-which is analytic in D, and for all paths in D joining any two points ::'0 alld ::'1 in D, the integral of f(z) from ::'0 to ZI call be evaluated by fOl1llula (9) in Sec. 14.1. PROOF The conditions of Cauchy's integral theorem are satisfied. Hence the line integral of f(z) from any Zu in D to any z in D is independent of path in D. We keep Zo fixed. Then this integral becomes a function of z. call if F(z}, (4) = F(z) r f(z*) dz* Zo which is uniquely detennined. We show that this F(z) is analytic in D and F' (z) = .f(z). The idea of doing this is as follows. Using (4) we form the difference quotient F(z (5) + .lz) - F(z) ll.'-. = 1 A LlZ [Z+l1Z I Zo f(z*) dz* - IZ f(::.*) d::.* Zo ] = 1 A _ Ll-<, f z+!lz f(z*) dz*. Z We now ~ubtract f(z) from (5) and show that the resulting expression approaches zero as ll.z ~ O. The details are as follows. SEC. 14.2 651 Cauchy's Integral Theorem We keep z fixed. Then we choose z + fl.z in D so that the whole segment with endpoints z and z + fl.z is in D (Fig. 349). This can be done because D is a domain, hence it contains a neighborhood of z. We use this segment as the path of integration in (5). Now we subtract fez). This is a constant because z is kept fixed. Hence we can write J J Z+.'1z 1 z+.'1z fez) dz* = fez) z = fez) fl.z. dz* Thus fez) = z A tiZ J z+.'1z fez) dz*. z By this trick and from (5) we get a single integral: F(z + fl.z) - F(z) fl. - z 1 f tiZ Since .f(z) is analytic, it is continuous. An such that I.f(z*) - f(z) 1< E when Iz* - zl ML-inequality (Sec. 14.1) yields I F(z + I fl.z) - F(z) fl.z - fez) 1 = Ifl.zl Z = A f(::.) If +.'1Z . [f(z*) - fez)] dz"'. Z > 0 being given, we can thus find a 8 > 0 < 8. Hence. letting Ifl.zl < 8. we see that the E Z +.'1Z z [.f(z*) - fez)] d-;;* I :S By the definition of limit and derivative, this proves that , . F(z F (z) = hm + .'1z->O fl.z) - F(z) = fl.::. .f{z) . Since Z is any point in D, this implies that F(z) is analytic in D and is an indefinite integral or antiderivative of f(z) in D, written F(z) = ff(z) dz. Also, if c' (z) = fez), then F' (z) - c' (z) ~ 0 in D; hence F(z) - C(z) is constant in D (see Team Project 26 in Problem Set 13.4). That is, two indefinite integrals of fez) can differ only by a constant. The latter drops out in (9) of Sec. 14.1, so that we can use any • indefinite integral of fez). This proves Theorem 3. -------- , ..... ",,,,,,,,,,,,,,,,,,, / / , " '" / \ \ D ",/ -'" I '" Zo ",,,,,,,,,,,,,,,, \,----'" Fig. 349. , I I " I z + t.z Path of integration 652 CHAP. 14 Complex Integration Cauchy's Integral Theorem for Multiply Connected Domains Cauchy's theorem applies to multiply connected domains. We first explain this for a doubly connected domain D with outer boundary curve C1 and inner C2 (Fig. 350). If a function fez) is analytic in any domain D* that contains D and its boundary curves, we claim that f (6) fez) d::. = C1 f (Fig. 350) fez) dz C2 both integrals being taken counterclockwise (or both clockwise, and regardless of whether or not the full interior of C2 belongs to D*). Fig. 350. PROOF By two cuts Paths in (5) C\ and C2 (Fig. 351) we cut D into two simply connected domains Dl and .Hz) is analytic. By Cauchy's integral theorem the integral over the entire boundary of Dl (taken in the sense of the arrows in Fig. 351) is zero, and so is the integral over the boundary of D 2 , and thus their sum. [n this sum the integrals over the cuts C1 and C2 cancel because we integrate over them in both directions-this is the key-and we are left with the integrals over C1 (counterclockwise) and C2 (clockwise; see Fig. 351); hence by reversing the integration over C2 (to counterclockwise) we have D2 in which and on whose boundaries f C1 f d z - f fd::;=O C2 • and (6) follows. For domains of higher connectivity the idea remains the same. Thus, for a triply connected domain we use three cuts Cb C2, C3 (Fig. 352). Adding integrals as before, the integrals over the cuts cancel and the sum of the integrals over C1 (counterclockwise) and C2 , C3 (clockwise) is zero. Hence the integral over C1 equals the sum of the integrals over C2 and C3 , all three now taken counterclockwise. Similarly for quadruply connected domains, and so on. - ~ Ldr~~c Cl'~~J.2 D2 Fig. 351. C CI j Doubly connected domain Fig. 352. Triply connected domain SEC. 14.2 Cauchy's Integral Theorem 653 CAUCHY'S INTEGRAL THEOREM APPLICABLE? 2:: f(::) = (i) _2 + 3i +1 - Integrate f(::) counterclockwise around the unit circle. indicating whether Cauchy's integral theorem applies. (Show the details of your work.) 1. f(::) = Re:: 3. f(::) = ez 6. f(::) = sec (::/2) 7. f(::) = 11(::8 - 1.2) 9. f(::) = 1/(21<:13) 11 •.f(z) = .:2 cot .: 112-171 = II: 4. f(::) /2 5. f(::) = tan:: 2 8. f(::) = 1/(4z - 3) 10. f(::) l = 119-301 COMMENTS ON TEXT AND EXAMPLES 12. (Singularities) Can we conclude in Example 2 that the integral of 11(::2 + 4) taken over (a) Iz - 21 = 2, (b) I:: - 21 = 3 is zero? Give reasons. 13. (Cauchy's integral theorem) Velify Theorem 1 for the integral of ::2 over the boundary of the square with vertices I + i, -I + i. -I - i, and I - i (counterclockwise). f(::) = 4 z+I -=---:;:2 + 2:: (c) Deformation of path. Review (c) and (d) of Team Project 34, Sec. 14.\. in the light of the principle of deformation of path. Then consider another family of paths with common endpoints. say, ::(t) = r + ia(r - (2). o~ (~ 1. and experiment with the integration of analytic and nonanalytic functions of your choice over these paths (e.g., ::. 1m::. ::2, Re .:2, 1m Z2, etc). 2. f(::) = 11(3:: - 1Ii) 2 (ii) FURTHER CONTOUR INTEGRALS Evaluate (showing the details and using partial fractions if necessary) 19. ,.( 'f 2-d::_ i . C the circle c - 20. 21. f f c tanh::: d::, C the circle IIz = Iz - 3 (counterclockwise) !7Til = ~ (clockwise) Re 2::: d:;:, C as shown c 14. (Cauchy's integral theorem) For what contours C will it follow from Theorem I that (a) f ,.( d:: = 0, c :: (c) (b) 'f cos ::: _6 _ c- fc:: _2 c d:: = O. - -2-- +9 1 x -1 elfz d:: = O? 15. (Deformation principle) Can we conclude from Example 4 that the integral is also zero over the contour in Problem 13? 22. f 7z - 6 _2 _ c~ 2- d::, C as shown - _--_c 16. (Deformation principle) If the integral of a function fez) over the unit circle equals 3 and over the circle Izl = 2 equals 5, can we conclude that fez) is analytic everywhere in the annulus I < Izl < 2? 17. (Path independence) Verify Theorem 2 for [he integral of cos:: from 0 to (l + i}7T(a) overthe shortest path. (b) over the x-axis to 7T and then straight up to (l + i)7T. x 23.,.( 2 d:: Jcz - I , C as shown y 18. TEAM PROJECT. Cauchy's Integral Theorem. (a) Main Aspects. Each of the problems in Examples 1-5 explains a basic fact in connection with Cauchy's theorem. Find five examples of your own, more complicated ones if possible. each illustrating one of those facts. (b) Partial fractions. Write f(::) in terms of partial fractions and integrate it counterclockwise over the unit circle, where x 24. ,.( e 2z 'f -4- d::. C consists of 1:::/ c (counterclockwise) o. = 2 (clockwise) and /<:/ = f CHAP. 14 654 25. cos rJ:c ~ dz, C consists of Izl = (. 7 and 26. Complex Integration Izi f Ln + ::) d::, C the boundary of the square with c vertices 27. J: 2 d:: Jcz + • C: (a) Izi !. (b) Iz - il 3 2: (counterclockwise) 14.3 -2-- , 29. f 30. f :!: 1, :!: i 1 rJ:c:: d::+ 1 C: (a) Iz + il I, (b) Iz - il (counterclockwise) 3 (clockwise) = (2 28. 1 (counterclockwise) -sin:: - . d:;., C: c:: + 21 tan (::/2) 4 16 veltices :!: 1, cZ - Iz - d::., C ~i 4 - 2i I = 5.5 (clockwise) the boundary of the square with (clockwise) Cauchy's Integral Formula The most important consequence of Cauchy's integral theorem is Cauchy's integral formula. This formula is useful for evaluating integrals, as we show below. Even more important is its key role in proving the surprising fact that analytic functions have derivatives of all orders (Sec. 14.4), in esrablishing Taylor series representations (Sec. 15.4), and so on. Cauchy's integral formula and irs conditions of validity may be stated as follows. Cauchy's Integral Formula THEOREM 1 Let fez) be analytic il1 a simply connected domain D. Then for allY POi11T alld any simple closed path C in D that encloses Zo (Fig. 353), (1) f(::) rJ:C<·-_-_dz = 27Tif(zo) ':0 ill D (Cauchy's integral formula) ---0 the integration being taken cuullterc!ockwise. Alternatively (for representing f(zo) by a contour integral, divide (I) by 27Ti), (1*) PROOF f(zo) 1 27Tl = -. fez) rJ:C --. dz Z - Zo (Cauchy's integral formula). By addition and subtraction, fez) = f(zo) + [fez) - fC2{)]. Inserting this into (l) on the left and taking the constant factor f(.::o) out from under the integral sign, we have (2) The first term on the right equals f(;:.o)· 27Ti (see Example 6 in Sec. 14.2 with 111 = - I). This proves the theorem. provided the second integral on the right is zero. This is what we are now going to show. Its integrand is analytic, except at Zoo Hence by (6) in Sec. 14.2 we can replace Cby a small circle K of radius p and center.::o (Fig. 354), without SEC. 14.3 Cauchy's Integral Formula 655 c Fig. 353. Fig. 354. Cauchy's integral formula o K Proof of Cauchy's integral formula altering the value of the integral. Since f(~) is analytic, it is continuous (Team Project 26, Sec. 13.3). Hence an E > 0 being given, we can find aD> 0 such that 1ft.:) - f(~o)1 < E for all z in the disk Iz - 201 < o. Choosing the radius p of K smaller than 0, we thus have the inequality ~ fez) - f(zo) 1 < P Z - 20 1 at each point of K. The length of K is 27fp. Hence, by the ML-inequality in Sec. 14.1, l f(z~ 1... I K _- f_(2 0 ) d-:I- < -E 27fp = 27fE. P "-.(.0 Since E (> 0) can be chosen arbitrarily small, it follows that the last integral in (2) must have the value zero, and the theorem is proved. • E X AMP L E 1 Cauchy's Integral Formula lJ ~_~ 2 c d::. = 2'ITie Z I z~2 = 2'ITie 2 = 46.4268; for any contour enclosing ::'0 = 2 (since eZ is entire). and zero for any contour for which ::'0 = 2 lies outside (by Cauchy's integral theorem). • E X AMP L E 2 Cauchy's Integral Formula f C Z3 - 6 -2--. dz = Z - f ~Z3 - 3 --1-· dz C Z - 2' I = 2'ITi[~::.3 - 3]1 z~i/2 'IT = E X AMP L E 3 "8 - 6'ITi Integration Around Different Contours Integrate Z2 +1 g(z) = - - Z2 - 1 ::.2 + I ----(::. + 1)(z - 1) counterclockwise around each of the four circles in Fig. 355. (::'0 = li inside C) . • 656 CHAP. 14 Complex Integration Solution. g(::) is not anal)1ic at -I and L These are the points we have to watch for. We consider each circle separately. (a) The circle write I:: - 11 = I encloses the point ::0 = 1 where g(;:J is not analytic. Hence in (1) we have to ;:2 + I g(::) = - Z2 - I <: + z- 1 ' 1 thus fez) = _2 + 1 ~+I and (I) gives + I fzZ2 c -2-- dz = 27Tif(l) = 27Ti I [ Z2 + 1] --- z+ I = 27Ti. z-l lb) gives the same as (a) by the principle of deformation of path. (c) The function glz) is as before, but fez) changes because we must take Zo = -I (instead of 1). This gives a factor z - ~o = z + 1 in (1). Hence we must write z- 1 z+ I ' thus Compare this for a minute with the previous expression and then go on: f c 2+1 --2-- d;: = '27Tif(- I) = 27Ti c:: - I [-2+IJ ~ I z~-l = -hi • (d) gives O. Why? y x Example 3 Multiply connected domains may be handled as in Sec. 14.2. For instance, if fez) is analytic on C 1 and C2 and in the ring-shaped domain bounded by C 1 and C2 (Fig. 356) and ~o is any point in that domain, then (3) 1 -)=f( -0 2· m f I-f(~) -d- + 2· m ~z-~ f -fez) - d --.. ~z-~ where the outer integral (over C 1) is taken counterclockwise and the inner clockwise, as indicated in Fig. 356. SEC. 14.3 Cauchy's Integral Formula 657 c] Fig. 356. Formula (3) Our discussion in this section has illustrated the use of Cauchy's integral formula in integration. In the next section we show that the formula plays the key role in proving the surprising fact that an analytic function has derivatives of all orders, which are thus analytic functions themselves. ::a-III',-- 11-41 CONTOUR INTEGRATION Integrate circle: (Z2 - 1. Iz - il + 4) counterclockwise around the 4)/(Z2 2. Iz - 11 2 = 3. Iz + 3il = 2 l ::!.iI 4. Izl = = 2 71"!2 CONTOUR INTEGRATION Using Cauchy's integral formula (and showing the details), integrate counterclockwise (or as indicated) ,( 7 ,( e- 3 r.z 13. :r - - _ dz. C the boundary of the square with c2z + I vertices ±1, ±i 14. C- Iz - 11 2 = z- I 7. f sinh rr;:; -2-- cZ 3z - ,( 8. :r c Z2 _ dz ,( 9. :r c Z2 - dz, 1 . C: Iz ,( cosz 11. :r - - dz, c 2z ,( tanz 12. :r - - d;:;, z- i 1 IZ - 1I = C: dz = 17. f + 11 C: Izl = 71"/2 = 4 ~ C the boundary of the triangle with vertices 0 and ± 1 + 2i ~. - d:., C: Iz - 41 = 2 Cconsistsoflzl =3 (counterclockwise) 21Z f cosh2 z. c(z-l-i)z 2 dz, C as in Prob. 16 f (z - Z1)-\Z - Z2)-1 dz = 0 for a simple c closed path C enclosing Z1 and Z2, which are arbitrary_ = 1 C- Iz - 2il sm" 2 18. Show that c-Izl = I l' 10. ,( ~ dz, :rc z - 2z c C- Izl + 1) 2 and Izl = 1 (clockwise) C ,. ,( e 3z 6. :r -3- . dz, (z dz, C consists of Iz - 2il = 2 c z + 1 (counterclockwise) and Iz - 2il = ~ (clockwise) cZ 5. :r : _ 2 dz., C Ln ,( Ln (z. - 1) 15. :r d;:;, c z-5 16. +2 f 19. CAS PROJECT. Contour Integration. Experiment to find out to what extent your CAS can do contour integration (a) by using the second method in Sec. 14.1, (b) by Cauchy's integral formula. 20. TEAM PROJECT. Cauchy's Integral Theorem. Gain additional insight into the proof of Cauchy's integral theorem by producing (2) with a contour enclosing ;:;0 (as in Fig. 353) and taking the limit as in the text. Choose (b) ,( sin;:; :r - - 1 - dz, c z - 271" and (c) two other examples of your choice. 658 14.4 CHAP. 14 Complex Integration Derivatives of Analytic Functions In this section we use Cauchy's integral formula to show the basic fact that complex analytic functions have derivatives of all orders. This is very surprising because it differs strikingly from the situation in real calculus. Indeed, if a real function is once differentiable. nothing follows about the existence of second or higher detivatives. Thus. in this respect, complex analytic functions behave much more simply than real functions that are once differentiable. The existence of those derivatives will result from a general integral formula, as follows. THEOREM 1 Derivatives of an Analytic Function If fez) is analytic in a domain D, then it has derivatives of all orders in D. which are then also analytic functions in D. The values of these derivatives at a point Zo in D are given by the fOl7llulas (l ') f , 1 (zo) J. fez) = -2 . 7TI r (7 _ 7 )2 dz n'. . -2 f fez) C ~ ,·0 (I ") lind in general (1) tn)(Zo) = 7TI C (z - zo)n+l d::. (n = 1,2, ... ); here C is any simple closed path i11 D that encloses Zo and whose full interior belongs to D; and we integrate counterclockwise arollnd C (Fig. 357). Fig. 357. Theorem 1 and its proof COI\IMENT. For memorizing (I). it is useful to observe that these formulas are obtamed formally by differentiating the Cauchy formula (l *), Sec. 14.3, under the integral sign with respect to zoo SEC. 14.4 659 Derivatives of Analytic Functions PROOF We prove (1 '), starting from the definition of the delivative f '(~~'0) -On the right we represent f(zo f(zo + + I' f(zo 1m + "'2-->0 .6.z) - f(~) .6.z .6.z) and f(zo) by Cauchy's integral formula: .6.z) - f(zo) = _1_ .6.z 27Ti.6.z [1 Jc z fez) dz - (Zo + .6.z) 1~ Jc z dZ] . - Zo We now write the two integrals as a single integral. Taking the common denominator gives the numerator f(z){z - Zo - [z - (zo + ilz)]} = fez) .6.z, so that a factor ilz drops out and we get f(zo + .6.z) - f(zo) = _1_ .6.z 27Ti 1 fez) dz . - Zo - .6.z)(z - zo) Jc (z Clearly, we can now establish (1') by showing that, as .6.z ---'> 0, the integral on the right approaches the integral in (1 '). To do this, we consider the difference between these two integrals. We can write this difference as a single integral by taking the common denominator and simplifying the numerator (as just before). This gives fc _Zo (z !(z)_ _ dz .6..:.)(z zo) fc (z ~z: 2 dz = ':'0) fc f(z).6.z 2 dz. (z - Zo - .6.z)(z - zo) We show by the ML-inequality (Sec. 14.1) that the integral on the right approaches zero as .6.z ---'> O. Being analytic, the function fez) is continuous on C, hence bounded in absolute value, say, If(z)1 ~ K. Let d be the smallest distance from Zo to the points of C (see Fig. 357). Then for all z on C, 17- 712>= d 2 , ..... .....0 hence Furthermore, by the triangle inequality for all d ~ Iz - zol = Iz - Zo - .6.;:: z on C we then also have + .6.21 ~ Iz - Zo - .6.zl + l.6.zl· We now subtract lilzl on both sides and let l.6.zl ~ dl2, so that -lilzl ~ -d/2. Then id ~ d - l.6.zl ~ Iz - Zo - .6.zl· Hence 2 -:---------,- ::; - Iz - Zo - .6.zl Let L be the length of C. If l.6.zl ~ dl2, then by the ML-inequality d 660 CHAP. 14 Complex Integration This approaches zero as !:::..Z ---,> O. Formula (1 ') is proved. Note that we used Cauchy's integral formula (1 *), Sec. 14.3, but if all we had known about f(zo) is the fact that it can be represented by (1 *), Sec. 14.3, our argument would have established the existence of the derivative t' (zo) of fez). This is essential to the continuation and completion of this proof, because it implies that (1") can be proved by a similar argument, with f replaced by f', and that the general formula (1) follows by induction. • E X AMP L E 1 Evaluation of Line Integrals From (1 '), for any contour enclosing the point 71"i (counterclockwise) f _c_o_s2-.-:0 dz = 271"i(cos c (z - m) 2 E X AMP L E 2 • = -271"i sin 71"i = 271" sinh 71" z~.".i From (1 "), for any contour enclosing the point - i we obtain by counterclockwise integration f. rc E X AMP L E 3 Z)'I 2 Z4 - (Z 3z + +6 ·3 I) dz = 2"1 4 71"i(z - 3z + 6) . = z~-, 71"i[12z 2- 61z~-i = -1871"i. • By (1'), for any contour for which 1 lies inside and ±2i lie out~ide (counterclockwise), f :z c (z - 1) (z 2 + 4) dz = 271"i(-/-)' I + z 4 z . e (z2 = 271"1 + 4) 2 (z + z~l Z - e 2z 2 I z~l 4) 6e71" . . = - - 1 = 2.0501. 25 • Cauchy's Inequality. Liouville's and Morera's Theorems As a new aspect, let us now show that Cauchy's integral theorem is also fundamental in deliving general results on analytic functions. Cauchy's Inequality. Theorem I yields a basic inequality that has many applications. To get it, all we have to do is to choose for C in (1) a circle of radius r and center Zo and apply the ML-inequality (Sec. 14.1); with If(z)1 ~ M on C we obtain from (I) 7 1 -f (n) Co) 1 If. ~ 2 r 7T (7 _ c~ fez) Zo 71 )n+l d_ :5= This gives Cauchy'S inequality (2) To gain a first impression of the importance of this inequality, let us prove a famous theorem on entire functions (definition in Sec. 13.5). (For Liouville, see Sec. 5.7.) SEC. 14.4 661 Derivatives of Analytic Functions THEOREM 2 Liouville's Theorem If an entire function is bounded in absolute value in the whole complex plane, then this function must be a constant. PROOF By assumption, If(z)1 is bounded, say, If(z)1 < K for all z. Using (2), we see that If' (zo)1 < Klr. Since fez) is entire, this holds for every r, so that we can take r as large as we please and conclude that f' (zo) = O. Since Zo is arbitrary, f' (z) = Ux + ivx = 0 for all z (see (4) in Sec. 13.4), hence U x = Vx = 0, and uy = Vy = 0 by the Cauchy-Riemann equations. Thus u = const, v = const, and f = u + iv = const for all z. This completes . ~~ Another very interesting consequence of Theorem 1 is THEOREM 3 Morera's2 Theorem (Converse of Cauchy's Integral Theorem) If fez) is continuous in a simply connected domain D and if f (3) fez) dz = 0 c for every closed path in D, then fez) is analytic in D. PROOF In Sec. 14.2 we showed that if fez) is analytic in a simply connected domain D. then F(z) = r f(z*) dz* Zo is analytic in D and F' (z) = fez). In the proof we used only the continuity of fez) and the property that its integral around every closed path in D is zero; from these assumptions we concluded that F(z) is analytic. By Theorem 1, the derivative of F(z) is analytic, that • is, fez) is analytic in D, and Morera's theorem is proved. 11-81 CONTOUR INTEGRATION Imegrate counterclockwise around the circle Izl = 2. (n is a positive integer, a is arbitrary.) Show the details of your work. 1. cosh 3z 2. sin z (z - 7fil2) 3. 5. e Z cos (z - Z 7f12)2 sinh az Z4 cos z 4. ?n+l 6. Zn 4 7. (z - a)n+l 8. Ln (z + 3) + cos z + 1)2 (z eZ (z - a)n 2GlACINTO MORERA (1856-190Y), Italian mathematician who worked in Genoa and Turin. 662 CHAP. 14 [9-131 Complex Integration INTEGRATION AROUND DIFFERENT CONTOURS 14. TEAM PROJECT. Theory on Growth (a) Growth of entire functions. If fez) is not a constant and is analytic for all (finite) z, and Rand M are any positive real numbers (no matter how large), show that there exist values of z for which Izl > Rand If(z)1 > M. (b) Growth of polynomials. If fez) is a polynomial of degree n > 0 and M is an arbitrary positive real number (no matter how large), show that there exists a positive real number R such that If(z)1 > M for all Izl >R. (c) Exponential function. Show that fez) = e Z has the property characterized in (a) but does not have that characterized in (b). Integrate around C. Show the details. 9. 10. + 2:::) cosz (I 2' (2:: - 1) sin 4z 3 (z - 4 ) and ' I:: - 31 tan 7rZ :: C the unit circle. counterclockwise C consists of 1.:1 = ~ (clockwise) C the ellipse 16 x 2 11. --2-' 12. e 2z , C consists of Iz z(:: - 2i)2 and Izl = + .v2 = - i I= L counterclockwise 3 (counterclockwise) (d) Fundamental theorem of algebra. rf fez) is a po/ynDmial in z, IUlt a constant, then fez) = 0 for at least one value (If z. Prove this, using (a). C the circle I:: - 2 - i I = 3, counterclockwise 15. (Proof of Theorem 1) Complete the proof of Theorem 1 by performing the induction mentioned at the end. 1 (clockwise) ez / 2 13. (__, _ a) 5 (counterclockwise) 4' o .- S T ION SAN D PRO B L EMS J J 1. What is a path of integration? What did we assume about paths? 13. Is Re 2. State the definition of a complex line integral from memory. 14. How did we use integral formulas for derivatives in integration? 3. What do we mean by saying that complex integration is a linear operation? 15. What is Liouville's theorem? Give examples. State consequences. 4. Make a list of integration methods discussed. lllustrate each with a simple example. 116-301 5. Which integration methods apply to analytic functions only? 16. 4z 3 6. What value do you get if you integrate liz counterclockwise around the unit circle? (You should memorize this basic result.) If you integrate liz 2, 1/z3, . . . ? IS. :: + liz counterclockwise around I:: -I 3il = 2 19. e 2z from -2 + 37ri along the straight segment to 7. Which theorem in this chapter do you regard as most important? State it from memory. S. What is independence of path? What is the principle of deformation of path? Why is this important? 9. Do not confuse Cauchy's integral theorem and Cauchy's integral formula. State both. How are they related? 10. How can you extend Cauchy's integral theorem to doubly and triply connected domains? 11. If integrating fez) over the boundary circles of an annulus D gives different values, can fez) be analytic in D? (Give reason.) 12. Is IfJ(Z) I dz = fel f(::)1 dz? How would you find a bound for the integral on the left? fez) dz e = Re fez) dz? Give examples. e INTEGRATION Integrate by a suitable method: + 2z from - i to 2 +i along any path 17. 5z - 3/z counterclockwise around the unit circle -2 20. 21. 22. 23. 24. e z2 + 57ri /(z - 1)2 counterclockwise around Izi = 2 z1(z2 + 1) clockwise around Iz + il = 1 Re:: from 0 to 4 and then vertically up to 4 + 3i cosh 4z from 0 to 2i along the imaginary axis eZ/z over C consisting of 1.::1 = I (counterclockwise) and Izl = ! (clockwise) 25. (sin z)/z clockwise around a circle containing z = 0 in its interior 26. 1m z counterclockwi~e around /:::1 = r 27. (Ln z)/(z - 20 2 counterclockwise around /z - 2i/ = 1 2S. (tan 7r:::)/(z - 1)2 counterclockwise around /z - 1/ = 0.2 29. Izi + z clockwise around the unit circle 30. (z - i)-3(Z3 + sin z) counterclockwise around any circle with center i 663 Summary of Chapter 14 Complex Integration The complex line integral of a function fez) taken over a path C is denoted by (1) J f or. if C is closed. also by fez) dz c (Sec. 14.1). fez) c If fez) is analytic in a simply connected domain D, then we can evaluate (1) as in calculus by indefinite integration and substitution of limits, that is, J (2) fez) dz = [F' (z) = fez)] F(z1) - F(zo) c for every path C in D from a point Zo to a point Z1 (see Sec. 14.1). These assumptions imply independence of path, that is, (2) depends only on Zo and Z1 (and on fez), of course) but not on the choice of C (Sec. 14.2). The existence of an F(z) such:that F' (z) = fez) is proved in Sec. 14.2 by Cauchy's integral theorem (see below).i A general method of integration, not restricted to analytic functions, uses the equation z = z(t) of C, where a ~ t ~ b, J (3) J b fez) dz = c f(z(t))z(t) dr z = dZ) . 0 ( a dl Cauchy's integral theorem is the most important theorem in this chapter. It states that if fez) is analytic in a simply connected domain D, then for every closed path C in D (Sec. 14.2), f (4) fez) dz = c o. Under the same assumptions and for any Zo in D and closed path C in D containing <:0 in its interior we also have Cauchy's integral formula 1 ,( f(zo) = - 2. (5) 7T'1 fez) r- dz. Z - Zo C Furthermore, under these assumptions fez) has derivatives of all orders in D that are themselves analytic functions in D and (Sec. 14.4) (6) (n) f _ n! ~ (zo) - - 2. 7T'1 fez) rC ( __ )n+1 Z dz (n = 1,2.· .. ). '-0 This implies Morera's theorem (the converse of Cauchy's integral theorem) and Cauchy's inequality (Sec. 14.4), which in turn implies Liouville's theorem that an entire function that is bounded in the whole complex plane must be constant. CHAPTER 15 Power Series, Taylor Series Complex power series, in particular, Taylor series, are analogs of real power and Taylor series in calculus. However, they are much more fundamental in complex analysis than their real counterparts in calculus. The reason is that power series represent analytic functions (Sec. 15.3) and, conversely, every analytic function can be represented by power series, called Taylor series (Sec. 15.4). Use Sec. 15.1 for reference if you are familiar with convergence tests for real seriesin complex this is quite similar. The last section (15 .5) on uniform convergence is optional. Prerequisite: Chaps. 13, 14. Sections thar may be omitted in a shorter course: 14.1, 14.5. References and Answers 10 Problems: App. I Part D, App. 2. 15.1 Sequences, Series, Convergence Tests In this section we define the basic concepts for complex sequences and series and discuss tests for convergence and divergence. This is very similar to real sequences and series in calculus. If you feel at home with the latter and want to take for granted that the ratio test also holds ill complex, skip this section and go to Sec. 15.2. Sequences The basic definitions are as in calculus. An ii!finite sequence or, briefly, a sequence, is obtained by assigning to each positive integer 11 a number Zn, called a term of the sequence, and is wlitten or ZI' 22, ••• or briefly We may also write 20, Z1, ••• or :::2, :::3, ••• or start with some other integer if convenient. A real sequence is one whose terms are real. Convergence. A convergent sequence lim Zn = C 21. Z2, ••• or simply n~oo By definition of limit this means that for every (1) 664 Izn - cl E < is one that Zn~ ha~ a limit c, written c. > 0 we can find an N such that E for all 11 > N; SEC. 15.1 665 Sequences, Series, Convergence Tests geometrically, all terms Zn with n > N lie in the open disk of radius E and center c (Fig. 358) and only finitely many terms do not lie in that disk. [For a real sequence, (1) gives an open interval of length 2E and real midpoint c on the real line; see Fig. 359.] A divergent sequence is one that does not converge. y : x Fig. 358. E X AMP L E 1 C-E Convergent complex sequence C +E C Fig. 359. Convergent and Divergent Sequences The sequence {inln} = Ii, -112, -;/3, 114, ... } is convergent with limit O. The sequence {in} = [i. -I. -i. I .... } is divergent. and so is {zn} with zn = (1 E X AMP L E 2 x Convergent real sequence + On. • Sequences of the Real and the Imaginary Parts The sequence [zn) with zn = xn + iYn = I - Iln 2 + ;(2 + 41n) is 6i, 3/4 + 4i, 8/9 + 1Oi/3, 15/16 + 3i, . (Sketch it.) It converges with the limit c = I + 2;. Observe that {x",} has the limit 1 = Re c and {Yn} has the limit 2 = 1m c. This is typical. It illustrates the following theorem by which the convergence of a complex sequence can be referred back to that of the two real sequences of the real parts and the imaginary parts. • THEOREM 1 Sequences of the Real and the Imaginary Parts A sequence Z1, Z2, ... , Zn, . .. of complex numbers Zn = Xn + iYn (where n = 1, 2, ... ) converges to c = a + ib if and only if the sequence of the real parts Xl> X2 , • • • converges to a and the sequence of the imaginary parts Yl> )'2' . . . converges to b. PROOF Convergence zn ~ C = a + ib implies convergence Xn ~ a and Yn ~ b because if IZn - cl < E, then Zn lies within the circle of radius E about c = a + ib, so that (Fig. 360a) IYn - bl < y E. y ": -!l~ ~ b+~ b b-~ b-E I : I a-E a a+E x -@ I : I Eft a \ a-:2 Cal Fig. 360. E a+:2 Chl Proof of Theorem 1 x 666 CHAP. 15 Power Series, Taylor Series Conversely. if Xn ~ a and Yn so large that, for every II > N, Ix n ~ b as n ~ x. E - al <-2' then for a given I.vn - E > 0 we can choose N E bl < -2 . These two inequalities imply that Zn = xn + iYn lies in a square with center c and side E. Hence, zn must lie within a circle of radius E with center c (Fig. 360b). • Series Given a sequence we may form the sequence of the sums Z10 Z2, . . . , ::m, ... , and in general (11 (2) sn is called the nth partial sum of the i/~fillite = 1. 2 ... '). series or series cc 2: (3) Zm = Z1 + Z2 + m~1 The Z10 Z2, • . . are called the terms of the series. (Our usual summation letter is 11, unless we need 11 for another purpose, as here, and we then use m as the summation letter.) A convergent series is one whose sequence of partial sums converges. say, x lim n--.oo Sn = S. Then we write S = 2: ::m = ZI + 22 + m~1 and call s the sum or value of the series. A series that is not convergent is called a divergent series. If we omit the terms of sn from (3), there remains (4) Rn = Zn+1 + Zn+2 + Zn+3 + This is called the remainder o/the series (3) after the term and has the sum s, then Zn' Clearly, if (3) converges thus Now Sn ~ S by the definition of convergence; hence Rn ~ O. In applications, when s is unknown and we compute an approximation Sn of s, then IRnl is the error, and Rn ~ 0 means that we can make /Rn/ as small as we please, by choosing 11 large enough. An application of Theorem I to the partial sums immediately relates the convergence of a complex series to that of the two series of its real parts and of its imaginary parts: SEC. 15.1 667 Sequences, Series, Convergence Tests THEOREM 2 Real and Imaginary Parts A series (3) with Zm = Xm + iYm converges and has the sum s = u + iv if and only ifx] + X2 + ... converges and has the sum u and Yt + Y2 + ... converges and has the sum v. Tests for Convergence and Divergence of Series Convergence tests in complex are practically the same as in calculus. We apply them before we use a series, to make sure that the series converges. Divergence can often be shown very simply as follows. THEOREM 3 Divergence if a series Zl + Z2 + ... converges, then lim Zm. = O. Hence if this does not hold, m,~oo the series diverges. PROOF [f Zl + Z2 + ... lim rrz,--...:..x converges, with the sum s, then, since Zn, = Zm = 71llim (sm __ 00 - Sm-1) = 'H1lim __ (X) Sm - lim Sm-1 -1'11-----"'00 Sm - = S - Sm-1' S = o. • Zm ~ 0 is necessary for convergence but not sufficient, as we see from the harmonic series I + ! + ~ + f + ... , which satisfies this condition but diverges, as is shown in calculus (see, for example, Ref. [GRll] in App. I). CAUTION! The practical difficulty in proving convergence is that in most cases the sum of a series is unknown. Cauchy overcame this by showing that a series converges if and only if its partial sums eventually get close to each other: THEOREM 4 Cauchy's Convergence Principle for Series A series Zl + Z2 + ... is convergent if and only iffor every given E> 0 (no matter how small) we can find an N (which depends on E, in general) such that (5) Izn+1 + Zn+2 + ... + Zn+pl < E for every n > Nand p = 1. 2, ... The somewhat involved proof is left optional (see App. 4). A series Zl + Z2 series of the absolute values of the terms Absolute Convergence. + ... is called absolutely convergent if the co is convergent. If Zl + Z2 + ... converges but 1z11 + IZ21 + ... diverges, then the series is called, more precisely, conditionally convergent. Zl + Z2 + ... CHAP. 15 668 E X AMP L E 3 Power Series, Taylor Series A Conditionally Convergent Series The series I - i + ! - ! + - ... converges. but only conditionally since the harmonic series diverges, as mentioned above (after Theorem 3). • If a series is absolutely convergent, it is convergent. This follows readily from Cauchy's principle (see Team Project 30), This principle also yields the following general convergence test. THEOREM 5 Comparison Test If a series::l + Z2 + ... is given and we C([nfind a convergent series b l + b2 + ... IZII with nonnegative real fenns such that converges, even absolutely. PROOF By Cauchy's principle, since b l an N such that + b2 ~ + ... bI> IZ21 ~ b2 , .•• , converges, for any given for every n > Nand p From this and IZII ;0; bI> 1z21 ~ b2 • .•. Hence, again by Cauchy's principle, is absolutely convergent. then the given series = E > 0 we can find 1,2, .... we conclude that for those n and p, Izil + IZ21 + ... converges, so that Zl + Z2 + • A good comparison series is the geometric series. which behaves as follows. THEOREM 6 Geometric Series The geometric series 00 2, (6*) qm = I + q + q2 + ... m=O converges with the sum I/O - q) PROOF if Iql < I and diverges if Iql ~ 1. If Iql ~ 1. then !qml ~ I and Theorem 3 implies divergence. Now let Iql < l. The nth partial sum is sn = I + q + From this, On subtraction, most terms on the right cancel in pairs, and we are left with SEC. 15.1 669 Sequences, Series, Convergence Tests Now 1 - q =1= 0 since q =1= 1, and we may solve for S/l.' finding 1 - qn+l (6) 1 - q l-q l-q Since Iql < L the last tenn approaches zero as II ~ rx. Hence if convergent and has the sum 11(1 - q). This completes the proof. Iql < L the series is • Ratio Test This is the most important test in our further work. We get it by taking the geometric series as comparison series b 1 + b 2 + ... in Theorem 5: THEOREM 7 Ratio Test <:1 + <:2 + ... with Zn eve,)" n greater than some N, If a series =1= 0 (/1 I, 2, ... ) has the property that for I Z~:1 I ~ q < (7) (n I (where q < I is fixed), this series cunverges absolutely. If for N) every n > N, IZ::1 I ~ 1 (8) > (n > N), the series diverges. PROOF If (8) holds. then IZn+ll ~ Iznl for n > N, so that divergence of the series follows from Theorem 3. If (7) holds, then Izn+ll ~ IZnl q for Il > N, in particular, etc., and in general, IZN+pl ~ IzN+llqP-l. Since q Absolute convergence of Zl + <:2 + ... < 1, we obtain from this and Theorem 6 now follows from Theorem 5. • CAUTION! The inequality (7) implies IZn+llznl < 1, but this does not imply convergence, as we see from the harmonic series, which satisfies ::'n+llzn = n/(n + 1) < I for alln but diverges. If the sequence of the ratios in (7) and (8) converges, we get the more convenient 670 CHAP. 1S THEOREM 8 Power Series, Taylor Series Ratio Test if a series Z1 + Z2 + ... with Zn ~ * 0 (n = 1,2, ... ) is such that 2.! I Zn+1 (a) If L < 1, the series converges absolutely. (b) if L > ~ I= L, 1, the series diverges. (c) {f L = 1, the series may converge or diverge, so that the test fails and permits no conclusion. PROOF (a) We write k n = IZn+1/2nl and let L = I - b < 1. Then by the definition of limit, the kn must eventually get close to 1 - b, say, kn ~ q = 1 - ~b < 1 for alln greater than some N. Convergence of 21 + Z2 + ... now follows from Theorem 7. (b) Similarly, for L = I + c> 1 we have k n ;:::; I + ~c > I for alln > N* (sufficiently large), which implies divergence of Z1 + Z2 + ... by Theorem 7. (c) The harmonic series I ~ + + ~ + ... has Zn+1/Zn = 11/(11 + I), hence L = 1, and diverges. The series 1+ 4 hence also L 1 + + 9 16 + 1 25 + ... Zn+1 has = 1, but it converges. Convergence follows from (Fig. 361) Sn I 4 I = I + - + ... + - 2 n ~ fndX - 1+ 1 =2- - , n 1 X2 so that S10 S2, ••• is a bounded sequence and is monotone increasing (since the terms of the series are all positive); both properties together are sufficient for the convergence of the real sequence S10 S2, • • • . (In calculus this is proved by the so-called integral test. whose idea we have used.) • y \ Area 1 o 2 Convergence of the series 1 + Fig. 361. E X AMP L E 4 4 3 t + i + k + ... Ratio Test Is the following seIies convergent or divergent? (First guess, then calculate.) L n=O (100 + 75i)n n! = 1 + (100 + 75;) + - I 2! (100 + 75;)2 + ... x SEC. 15.1 671 Sequences, Series, Convergence Tests Solution. By Theorem 8, the series is convergent, since Zn+l I E X AMP L E S I ;100 = + 75il n +l/(n + I)! = 1100 1100 + 75il n /11! Zn + 75il + 11 125 1 11 + • L = O. 1 Theorem 7 More General than Theorem 8 Let an = il2 3n and bn ao = lI23n + 1. Is the following selies convergent or divergent? + b o + al + b 1 + ... = i + 1 2 + iIi 64 "8 + 16 + I + 128 + ... Solution. The ratios of the absolute values of successive terms are!,!,!,!, .... Hence convergence follows from Theorem 7. Since the sequence of these ratios has no limit, Theorem 8 is not applicable. • Root Test The ratio test and the root test are the two practically most important tests. The ratio test is usually simpler, but the root test is somewhat more general. THEOREM 9 Root Test If a series ::1 + Z2 + ... is such that for every n greater thall some N, (9) (n (where q < I is .fixed), this series converges absolutely. If for > N) infinitely mall)' n, (10) the series diverges. PROOF If (9) holds, then Iznl ~ qn < I for all n > N. Hence the series 1:::11 + IZ21 + ... converges by comparison with the geometric series, so that the series ZI + Z2 + ... converges absolutely. If (10) holds, then Iznl ~ 1 for infinitely many n. Divergence Of::l +::2 + ... now follows from Theorem 3. • Vfz:J CAUTION! Equation (9) implies < 1, but this does not imply convergence, as we see from the harmonic series, which satisfies ~ < I (for n > 1) but diverges. If the sequence of the roots in (9) and (10) converges, we more conveniently have THEOREM 10 Root Test If a series ZI + Z2 + ... is such that lim r~co (a) The series converges absolute(..' if L > 1. ff L = 1, the test fails; that is, Vfz:J = if L < L, then: 1. (b) The series diverges (c) no conclusion is possible. 672 CHAP. 15 PROOF Power Series, Taylor Series The proof parallels that of Theorem 8. (a) Let L = 1 - a* < 1. Then by the definition of a limit we have ~ < q = 1 - ~a* < 1 for all n greater than some (sufficiently large) N*. Hence IZnl < qn < I for all n > N*. Absolute convergence of the series ZI + Z2 + ... now follows by the comparison with the geometric series. VfzJ (b) If L > 1, then we also have > I for all sufficiently large n. Hence for those n. Theorem 3 now implies that ZI -f;, Z2 + ... diverges. 1 (c) Both the divergent harmonic series and the convergent series t 116 + 2~ + .. give L = 1. This can be seen from (In n)/n ---7 0 and SEQUENCES 11. Illustrate Theorem 1 by an example of your own. 12. (Uniqueness of limit) Show that if a sequence converges. its limit is unique. 13. (Addition) If ZI, Z2, ... converges with the limit [and ZI *, Z2 *, ... converges with the limit [*, show that Zl + Zl *':::2 + Z2*' ... converges with the limit [ + [*. 14. (Multiplication) Show that under the assumptions of Prob. L3 the sequence ZlZ1*' Z2Z2*' ... converges with the limit U*. 15. (Boundedness) Show that a complex sequence is bounded if and onl y if the two corresponding sequences of the real parts and of the imaginary parts are bounded. SERIES Are the following series convergent or divergent? (Give a reason.) (10 - ISi)n 16.2:---n! n=O rye 18. :L n=O ·n 1 -2-- - . n - 21 20.2: n=2 In n o . e(2/nHn n CC Are the following sequences Zl, Z2, ... , Zn> ... bounded? Convergent? Find their limit points. (Show the details of your work.) 2. Zn = e- nwi / 4 1. Zn = (_l)n + il2" 3. Zn = (-1)n/(n + i) 4. Zn = (I + i)n 5. Zn = Ln «2 + i)n) 6. Zn = (3 + 4i)n/n! 7. Zn = sin (n'1T/4) + in 8. Zn = [(1 + 3i)rVioT 9. Zn = (0.9 + 0.li)2n 10. Zn = tS + Si)-n 116-241 > + ~ +! e 11-101 IZnl cc 17. ~o CC 19.2: n~l (-I)n(1 (2n I Vn n + 2i)2n+l + I)! (n 1)3 22. 2: - ' - (1 n~O (3n)! + on e o . • n - i 23.2: n=O 3n + 2i 25. What is the difference between (7) and just stating IZn+l/Znl < I? 26. Illustrate Theorem 2 by an example of your choice. 27. For what n do we obtain the term of greatest absolute value of the series in Example 4? About how big is it? First guess, then calculate it by the Stirling formula in Sec. 24.4. 28. Give another example showing that Theorem 7 is more general than Theorem 8. 29. CAS PROJECT. Sequences and Series. (a) Write a program for graphing complex sequences. Apply it to sequences of your choice that have interesting "geometrical" properties (e.g., lying on an ellipse, spiraling toward its limit, etc.). (b) Write a program for computing and graphing numeric values of the first n partial sums of a series of complex numbers. Use the program to experiment with the rapidity of convergence of series of your choice. 30. TEAM PROJECT. Series. ta) Absolute convergence. Show that if a series converges absolutely, it is convergent. (b) Write a short report on the basic concepts and properties of series of numbers, explaining in each case whether or not they carry over from real series (discussed in calculus) to complex series, with reasons given. SEC. 15.2 673 Power Series (c) Estimate of the remainder. Let Izn+llznl ~ q < 1, so that the series Zl + Z2 + ... converges by the ratio test. Show that the remainder Rn = Zn+l + Zn+2 + ... satisfies the inequality IRnl ~ Izn+ll/(I - q). (d) Using (c), find how many terms suffice for computing the sum s of the series 15.2 CC 2: n=l n + i 2nn with an error not exceeding 0.05 and compute s to this accuracy. (e) Find other applications of the estimate in (c) Power Series Power series are the most important series in complex analysis because we shall see that their sums are analytic functions, and every analytic function can be represented by power series (Theorem 5 in Sec. 15.3 and Theorem 1 in Sec. 15.4). A power series in powers of z - Zo is a series of the form (1) 2: an(z - zo)n = ao + al(z - zo) + a2(Z - zoi + n~O where z is a complex variable, ao, ab . . . are complex (or real) constants, called the coefficients of the series, and zo is a complex (or real) constant, called the center of the series. This generalizes real power series of calculus. If Zo = 0, we obtain as a particular case a power series in powers of z: GO (2) 2: anz n = + ao alZ + a2z2 + n~O Convergence Behavior of Power Series Power series have variable terms (functions of z), but if we fix z, then all the concepts for series with constant terms in the last section apply. Usually a series with variable terms will converge for some z and diverge for others. For a power series the situation is simple. The series (l) may converge in a disk with center Zo or in the whole z-plane or only at zo0 We illustrate this with typical examples and then prove it. E X AMP L E 1 Convergence in a Disk. Geometric Series The geometric series converges absolutely if Izl < I and diverges if Izl E X AMP L E 1 ~ 1 (see Theorem 6 in Sec. 15.1). Convergence for Every z The power series (which will be the Maclaurin series of eZ in Sec. 15.4) 00 zn L -, n. n=O ,2 Z3 =1+z+-+-+'" 2! 3! • 674 CHAP.15 is Power Series, Taylor Series ab~olutely convergent for every ~, In fact. by the ratio test. for any fixed ::. • as E X AMP L E 3 Convergence Only at the Center. (Useless Series) The following power series converges only at z = 0, but diverges for every z *" 0, as we shall show. 00 L n!zn = 1 + Z I) Izl + 2::2 + 6;:3 + ... n=O In fact, from the ratio test we have (II + I)!zn+l ---n-n!z I THEOREM 1 I + = (n x -4 as (;: fixed and *" 0). • Convergence of a Power Series (a) Every power series (1) converges at the center Zo. rr (b) (1) converges at a point Z = Zl *- zo, it converges absolutely for every closer to Zo than Zl, that is, Iz - zol < IZI - zol. See Fig. 362. (C) If (1) diverges at a z than Z2' See Fig. 362. Z Z2, it diverges for every zfarther away from Zo = y , /' I ; ' - - .... I I \ ----- .... " , ,Divergent qz] \ Cony. \ 0 I " ..... _-"" Zo,' \ I ,~Z2 , ' .... _----' Fig. 362. PROOF (a) For z :x: Theroem 1 = Zo the series reduces to the single tenn ao. (b) Convergence at z = Zl gives by Theorem 3 in Sec. 15.1 an(ZI - zo)n ~ 0 as n ~ This implies boundedness in absolute value, for every n Multiplying and dividing an(z - Zo)n by (Zl - = 0, 1..... zo)n we obtain from this 00. SEC. 15.2 675 Power Series Summation over Il gives (3) Now our assumption Iz - 201 < 1:1 - ~ol implies that I(z - z.o)/(Zl - zo)1 < 1. Hence the series on the right side of (3) is a converging geometric series (see Theorem 6 in Sec. 15.1). Absolute convergence of (1) as stated in (b) now follows by the comparison test in Sec. 15.1. (c) If this were false, we would have convergence at a Z3 farther away from Zo than Z2' This would imply convergence at Z2, by (b), a contradiction to our assumption of divergence at Z2' • Radius of Convergence of a Power Series Convergence for every z (the nicest case, Example 2) or for no z *- Zo (the useless case, Example 3) needs no further discussion, and we put these cases aside for a moment. We consider the smallest circle with center Zo that includes all the points at which a given power series (I) converges. Let R denote its radius. The circle Iz - zol =R (Fig. 363) is called the circle of convergence and its radius R the radius of convergence of (l). Theorem I then implies convergence everywhere within that circle, that is, for all z for which (4) Iz - zol < R (the open disk with center:o and radius R). Also. since R is as small as possible. the series (l) diverges for all z for which Iz - ~ol > R. (5) No general statements can be made about the convergence of a power series (1) on the circle of convergence itself. The series (I) may converge at ~ome or all or none of these points. Details will not be essential to us. Hence a simple example may just give us the idea. DiVergent C3 co::;;r Zo Fig. 363. Circle of convergence CHAP.15 676 E X AMP L E 4 Power Series, Taylor Series Behavior on the Circle of Convergence On the circle of convergence (radius R L ~ nln L 2 ~nl" = 1 in all three series), converges everywhere since L 1/,,2 converges. converges at -1 (by Leibniz's test) but diverges at 1, • diverges everywhere. = Notations R oc and R notation, we write R = x = O. To incorporate these two excluded cases in the present z (as in Example 2), converges only at the center z = ~o (as in Example 3). if the series 0) converges for all R = 0 if (1) These are convenient notations, but nothing else. Real Power Series. In this case in which powers, coefficients. and center are real. formula (4) gives the convergence interval Ix - xol < R of length 2R on the real line. Determination of the Radius of Convergence from the Coefficients. important practical task we can use THEOREM 1 For this Radius of Convergence R Suppose that the sequence lan +1lan l, n = 1. 2, ... , converges with limit L *. !f L * = 0, thell R = x; that is, the power series (1) converges for all ~. If L * 0 (hence L * > 0), then *" (6) R= L* I~I = n~CXl lim an+l (Cauchy-Hadamard formula l ). If lan+l/anl ~ x, then R = 0 (convergence only at the center 20). PROOF For (1) the ratio of the terms in the ratio test (Sec. 15.1) is The limit is *" L = L*lz - zol. :01 Let L* 0, thus L* > O. We have convergence if L = L*lz < 1, thus Iz - zol < IIL*, and divergence if k - !ol > IIL*. By (4) and (5) this shows that IIL* is the convergence radius and proves (6). If L * = 0, then L = 0 for every z, which gives convergence for all z by the ratio test. If lan+l/anl ~ co, then lan+l/anllz - zol > I for any z Zo and all sufficiently large n. This implies divergence for all z Zo by the ratio test (Theorem 7, Sec. 15.1). • *" *" INamed after the French mathematicians A. L. CAUCHY (see Sec. 2.5) and JACQUES HADAMARD (1865-1963). Hadamard made basic contributions to the theory of power series and devoted his lifework to partial differential equations. SEC. 15.2 677 Power Series Formula (6) will not help if L * does not exist, but extensions of Theorem 2 are still possible, as we discuss in Example 6 below. E X AMP L E 5 Radius of Convergence By (6) the radIUs of convergence of the power ~eries R = (211)! lim -n~'X [ (1I!)2 I (2n «11 + 2)! ] + 1)!)2 = The series converges in the open disk E X AMP L E 6 (211)! - - 2 (z - ,,~O (n!) 3i)n is [(211)! «n + 1)!)2 ] . (2n + 2)! (1I!)2 lim n~oc Iz - oc L = + 1)2 1 + 2)(211 + 1) 4 (11 lim n~x (211 • 3;1 < ! of radius! and center 3;. Extension of Theorem 1 Find the radius of convergence R of the power series Solution. The sequence of the ratios 1/6. 2(2 2 is of no help. It can be shown that + !), 1/(8(2 + !».... doe~ not converge. so that Theorem R = 1/L, (6*) This still does not help here, since whereas for even II we have (V/i~) does not converge because Vfa:J = V2 + 112n~ 1 n as V);J ---+ = ~ = 112 for odd II. 00, ."r.--: so that V lanl has the two limit points 112 and I. It can further be shown that R (6**) = Tthe greatest limit point of the sequence 1/T, Here T = I. so that R = I. Answer. The series converges for Izl < {Vj:J}. • 1. Summary. Power series converge in an open circular disk or some even for every z (or some only at the center. but they are useless): for the radius of convergence. see (6) or Example 6. Except for the useless ones, power series have sums that are analytic functions (as we show in the next section); this accounts for their importance in complex analysis. .. - .•= 33-3;=-- - - - 1. (Powers missing) Show that if ~ a"z'YI has radius of convergence R (assumed finite), then ~ a,,:!''' has radius of convergence "\'R. Give examples. 2. (Convergence behavior) Illustrate the facts shown by Examples 1-3 by further examples of your own. 13-181 n~O (z + l)n n 6. 2: 2 /1 4. 2: - n~O n! (z + 2i)n 9. 2: x (n - i)"z" n=O 11. 2: n=l 2100., --z" 11! n=O 8.2: RADIUS OF CONVERGENCE 2: n=l Il! n~O Find the center and the radius of convergence of the following power series. (Show the details.) (z + i)n x 11" 3. 00 5. 2: n 10.2: n=O (_1)n+1 zn 11 12. 2: n=O (-I)" 22"(11!)2 _2n - (2:::)2" --- (211)! 4" (1 + i)" (::: - 5)" CHAP.15 678 13. 2: 11(11 - 1)(: - 3 + Power Series. Taylor Series 2i)n n=2 14.2: S co I)n n~O (211). 16. 2: :JC 15. 2: :271 2n (z - i)4" n=O ('>+3·)n ~5 _ i' (z - 7T)n n=O :x; 18. 2: n~O (411)! ~ 2 (II!) * * (: + 7Ti)" 19. CAS PROJECT. Radius of Convergence. Write a program for computing R from (6), (6*), or (6"'*). in this order, depending on the existence of the limits needed. Test the program on series of your choice and 15.3 such that all three formulas (6). (6*), and (6**) will come up. 20. TEAM PROJECT. Radius of Convergence. (a) Formula (6) for R contains iOn/On+li, not iOn+l/Oni. How could you memorize this by using a qualitative argument? (b) Change of coefficients. What happens to R (0 < R < 00) if you (i) multiply all On by k O. (ii) multiply On by k fl O. (iii) replace On by lion? (c) Example 6 extends Theorem 2 to nonconvergent cases of O../On+l' Do you understand the principle of "mixing" by which Example 6 was obtained? Use this principle for making up further examples. (d) Does there exist a power series in powers of z that converges at z = 30 + 10i and diverges at z = 31 - 6i? (Give reason.) Functions Given by Power Series The main goal of this section is to show that power series represent analytic functions (Theorem 5). Along our way we shall see that power series behave nicely under addition, multiplication, differentiation. and integration. which makes these series very useful in complex analysis. To simplify the formulas in this section. we take :0 = 0 and write (1) This is no restriction because a series in powers of reduced to the fonn (I) if we set i - Zo = z. £- Zo with any Zo can always be Terminology and Notation. If any given power selies (1) has a nonzero radius of convergence R (thus R > 0), its sum is a function of z. say fez). Then we write x (2) fez) = 2: a"z''' = ao + 2 alz' + a2z + (izi < R). ,,~o We say that fez) is represented by the potrer series or that it is developed in the power series. For instance. the geometric series represents the function fez) = lI( I - z) in the interior of the unit circle IzI = 1. (See Theorem 6 in Sec. 15.1.) Uniqueness of a Power Series Representation. This is our next goal. It means that a jUllctioll f(:;:;) cannot be represented by two different power series with the same center. We claim that if fez) can at all be developed in a power series with center zoo the development is unique. This important fact is frequently used in complex analysis (as well as in calculus). We shall prove it in Theorem 2. The proof will follow from SEC. 15.3 Functions Given by Power Series THEOREM 1 679 Continuity of the Sum of a Power Series If afunction fez) CUll he represented by a power series (2) with radius of convergence R > 0, then fez) is continuous at ;:: = o. PROOF From (2) with:: = 0 we have f(O) = ao. Hence by the definition of continuity we must show that limz~o fez) = f(O) = ao. That is, we must show that for a given E> 0 there is a 8 > 0 such that k:1 < 8 implies If(z) - aol < E. Now (2) converges absolutely for Izl ;:; r with any r such that 0 < r < R, by Theorem 1 in Sec. 15.2. Hence the series 1 co L n= I lanlrn-l = r co L lanlrn n~l converges. Let S*-O be its sum. (S = 0 is trivial.) Then for 0 < Izl ;:; r. and Izls < E when Izi < 8, where 8 > 0 is less than r and less than EIS. Hence lzls < 8S < (EIS)S = E. This proves the theorem. • From this theorem we can now readily obtain the desired uniqueness theorem (again assuming ':0 = 0 without loss of generality): THEOREM 1 Identity Theorem for Power Series. Uniqueness Let the power series ao + alZ + (/2Z2 + ... and b o + bIZ + b 2z2 + ... both be convergent for l:::l < R, where R is positive, and let them both have the same SUlII for all these z. Then the series are identical, that is, ao = bo, al = bI> a2 = b 2, .... Hence if afullction f(;:;) can be represellted by a power series with any cellfer ZO, this representation is unique. PROOF We proceed by induction. By assumption, (Izl < R). The sums of these two power series are continuous at z = 0, by Theorem 1. Hence if we consider 1::1 > 0 and let z ~ 0 on both sides, we see that a o = bo: the assertion is true for n = O. Now assume that an = bn for n = 0, 1, ... , m. Then on both sides we may omit the terms that are equal and divide the result by zm+l (*- 0); this gives Similarly as before by letting completes the proof. z~ 0 we conclude from this that am+l bm + l . This • CHAP. 15 680 Power Series, Taylor Series Operations on Power Series Interesting in itself, this discussion will serve as a preparation for our main goal, namely, to show that functions represented by power series are analytic. Termwise addition or subtraction of two power series with radii of convergence RI and R2 yields a power series with radius of convergence at least equal to the smaller of RI and R2. Proof Add (or subtract) the partial sums Sn and s:; term by term and use lim (sn ::!: s:;) = lim Sn ::!: lim s:;. Termwise multiplication of two power series f(;::) = L akz k = ao + (lIZ + k~O and g(Z) L = bmz m = b o + bIZ + m~O means the multiplication of each term of the first series by each term of the second series and the collection of like powers of z. This gives a power series, which is called the Cauchy product of the two series and is given by = L (aob n + albn - l + ... + (lnbO)zn. n~O We mention without proof that this power series converges absolutely for each Z within the circle of convergence of each of the two given series and has the sum s(;::) = f(;::)g(z). For a proof. see [D5] listed in App. 1. Termwise differentiation and integration of power series is permissible, as we show next. We call derived series qf the power series (I) the power series obtained from (1) by termwise differentiation, that is, x L (3) nanZ n - 1 = al + 2a 2z + 3a3z2 + n~l THEOREM 3 Termwise Differentiation of a Power Series The derived series of a power series has the same radius of convergence original series. PROOF af the This follows from (6) in Sec. 15.2 because . lim n~x (11 + nlanl 1) lan+ll I I .I I . -n - lim . = hm -an- = hm -{Inn~ n + I ~= an+l n~::>:) a n +l or, if the limit does not exist, from (6**) in Sec. 15.2 by noting that \Yn ~ I as Il ~::xl. • SEC. 15.3 Functions Given by Power Series E X AMP L E 1 681 Application of Theorem 3 Find the mdius of convergence R of the following series by applying Theorem 3. ~ (n) n~2 zn = Z2 t- 3;::3 + 6:;;4 ~ IOz5 + . 2 Solution. Differentiate the geometric series twice term by term and mUltiply the result by z2f2 This yields the given series. Hence R = 1 by Theorem 3. • Termwise Integration of Power Series THEOREM 4 The power series an - - - ~n+ n + I 1 = a <.. 7 o~ + -al z2 + -a2 z3 + ... 2 3 obtained by integrating the series ao + al::' + a2z2 same radius of convergence as the original series. + tenll by term has the The proof is similar to that of Theorem 3. With Theorem 3 as a tool, we are now ready to establish our main result in this section. Power Series Represent Analytic Functions Analytic Functions. Their Derivatives THEOREM 5 A power series with a non:;:,ero radius of convergence R represents an analytic function at eve I)· point interior to its circle of convergence. The derivatives of this function are obtained by differentiating the original series tenn by tenn. All the series thus obtained have the same radius of convergence as the original series. Hence, by the first statement, each of them represents an a7lalytic function. PROOF (a) We consider any power series (1) with positive radius of convergence R. Let fez) be its sum and fl(:) the sum of its derived series; thus 00 (4) and fl(::') = L na n z-n - 1 . n~l We show that fez) is analytic and has the derivative f1(z) in the interior of the circle of convergence. We do this by proving that for any fixed z with Izl < Rand /1;:. ~ 0 the difference quotient [fez + /1;::) - f(::.)]//1z approaches fl(z). By termwise addition we first have from (4) Note that the summation starts with 2, since the constant term drops out in taking the difference fez + /1.;:) - fez), and so does the linear term when we subtract f 1 (z) from the difference quotient. CHAP. 15 682 Power Series, Taylor Series (b) We claim that the series in (5) can be written (0) 2: anLl::[(:: + LlZ)n-2 + + 2z(z .lZ)"-3 + + (n - 2)Zn-3(;:: n=2 + + LlZ) (11 - 1)::n-2]. The somewhat technical proof of this is given in App. 4. (e) We consider (6). The brackets contain 11 - I terms, and the largest coefficient is 1. Since (11 - 1)2 ~ 11(11 - 1), we see that for Izl ~ Ro and Iz + ~;::I ~ Ro, Ro < R. the absolute value of this series (6) cannot exceed 11 - (7) This series with lin instead of lanl is the second derived series of (2) at Z = Ro and converges absolutely by Theorem 3 of this section and Theorem I of Sec. 15.2. Hence our present series (7) converges. Let the sum of (7) (without the factor ILlzl) be K(Ro). Since (6) is the right side of (5), our present result is Letting .lz ~ 0 and noting that Ro « R) is arbitrary, we conclude that f(;::) is analytic at any point interiorto the circle of convergence and its derivative is represented by the derived series. From this the statements about the higher derivatives follow by induction. • Summary. The results in this section show that power series are about as nice as we could hope for: we can differentiate and integrate them term by term (Theorems 3 and 4). Theorem 5 accounts for the great importance of power series in complex analysis: the sum of such a series (with a positive radius of convergence) is an analytic function and has derivatives of all orders, which thus in turn are analytic functions. But this is only part of the story. In the next section we show that, conversely, every given analytic function f(:::') can be represented by power series, called Taylor series and being the complex analog of the real Taylor series of calculus. 11-10 I RADIUS OF CONVERGENCE BY DIFFERENTIATION OR INTEGRATION Find the radius of convergence in two ways: (a) directly by the Cauchy-Hadamard formula in Sec. 15.2. (b) from a series of simpler telms by using Theorem 3 or Theorem 4. cc 1. L 11(11 - :x; (.;: - 2i)n _ LOG !:I . 3n n(11 + 1) (7 - 1)2n 5"- n=l 6. n=2 i; (11) (±)n n=k -l-n 2·L--n=1 n(n + 1) (::)2n+l n=O 1) 3n (-I)n 4'L-211+17r k :x; 7. (-7)" L ---'------'---n= 1 8. 11(11 + 1)(11 211(21l _ I) L -nn- - + 2) ..2n :x; n=l ..2n-2 SEC. 15.4 9. L [(11 + 1e)]-1 zn+k cc n~O 10. 683 Taylor and Maclaurin Series L DC Ie (n n~O + 111) z" 17. (Odd function) If .f(z) in (1) is odd (i.e., .f(-z) = -.f(z», show that an = 0 for even n. Give examples. I1l 11. (Addition and subtraction) Write our the details of the proof on terrnwise addition and subtraction of power series. 18. (Even functions) If .f(z) in (1) is even (i.e., .f( - z) = .f(z», show that an = 0 for odd n. Give examples. 19. Find applications of Theorem 2 in differential equations 12. (Cauchy product) Show that (1 - Z)-2 = L';;~O (n + l)zn tal by using the Cauchy product, (b) by differentiating a suitable series. and elsewhere 20. TEAM PROJECT. Fibonacci nmnbers.2 tal The 13. (Cauchy product) Show that the Cauchy product of L~~O zn/n! multiplied by itself gives L~~O (2zyn/n!. Vn ~ I as n ~ claimed in the proof of Theorem 3). 14. (On Theorem 3) Prove that ex; (as 15. (On Theorems 3 and 4) Find further examples of your own. 116-201 APPLICATIONS OF THE IDENTITY THEOREM State clearly and explicitly where and how you are using Theorem 2. 16. (Bionomial coefficients) Using (1 + z)P(J relation 15.4 + z)q = (1 + z)p+q. obtain the basic Fibonacci numbers a o = al = 1. a n +l are recursively defined by if n = 1. 2 ..... Find the limit of the sequence (an+l/an)' (b) Fibonacci's rabbit problem. Compute a list of a 1. .... a12' Show that a12 = 233 is the munber of pairs of rabbits after l2 months if initially there is 1 pair and each pair generates I pair per month, beginning in the second month of existence (no deaths occuning). (c) Generating function. Show that the generating junction of the Fibonacci numbers is .f(z) = I/(1 - z - Z2); that is, if a power series (l) represents this .f(z), its coefficients must be the Fibonacci numbers and conversely. Hint. Start from .f(z) (1 - z - Z2) = I and use Theorem 2. = an + an-l Taylor and Maclaurin Series The Taylor series 3 of a function fez), the complex analog of the real Taylor series is (1) where or, by (l), Sec. 14.4, (2) 1 a - -n 21Tl' f C f(::;*) dz*. (z* - zd n + 1 In (2) we integrate counterclockwise around a simple closed path C that contains ::'0 in its interior and is such that f(:::) is analytic in a domain containing C and every point inside C. A Maclaurin series 3 is a Taylor series with center zo = O. 2LEONARDO OF PISA, called FIBONACCI (= son of Bonaccio), about 1180-1250, Italian mathematician. credited with the first renaissance of mathematics on Christian soil. 3BROOK TAYLOR (1685-1731), English mathematician who introduced real Taylor series. COLIN MACLAURIN (1698--1746), Scots mathematician, professor at Edinburgh. 684 CHAP. 15 Power Series, Taylor Series The remainder of the Taylor series (1) after the telm an(z - zo)n is (3) (proof below). Writing out the corresponding pmtial sum of (1). we thus have fez) = f(2o) + z - zo, -l-!- f (zo) + (z - ZO)2 " 2! f (zo) + ... (4) This is called Taylor's formula with remainder. We see that Taylor series are power series. From the last section we know that power series represent analytic functions. And we now show that eve I}' analytic function can be represented by power series, namely, by Taylor series (with various centers). This makes Taylor series very important in complex analysis. Indeed. they me more fundamental in complex analysis than their real counterparts me in calculus. Taylor's Theorem THEOREM 1 Let fez) be analytic in a domain D, and let z = 20 be any point in D. Then there exists precisely one Taylor series (1) with center ':0 that represents fez). This representation is mlid in the largest open disk with center.:o in which fez) is analytic. The remainders Rn(z) of (1) can be represented in the f0171l (3). The coefficients satisfy the inequality M lal:::S;n rn (5) Irhere M is the 1I1(n:ill1ll1ll of If(z)1 also in D. PROOF 011 a circle Iz - :01 = r ill D whose interior is The key tool is Cauchy's integral formula in Sec. 14.3; writing z (so that z* is the vmiable of integration), we have (6) fez) = -1. f --- 21Tl f(z*) C z* - z z and z* instead of 20 and dz*. z lies inside C, for which we take a circle of radius r with center Zo and interior in D (Fig. 364). We develop 1/(z* - z) in (6) in powers of z - z{). By a standard algebraic manipulation (worth remembering!) we first have (7) z* - z 1 z* - zo - (z - z{)) SEC. 15.4 685 Taylor and Maclaurin Series For later use we note that since z* is on C while z is inside C, we have z-zol<] z* - 20 I (7*) (Fig. 364). y x Fig. 364. Cauchy formula (6) To (7) we now apply the sum formula for a finite geometric sum qn+l I - qn+l 1 (8*) + q + ... + qn = --'----I - q I - q (q =1= 1), i-q which we use in the form (take the last term to the other side and interchange sides) I (8) I + q + ... + qn I - q qn-t-l + ]-q Applying this with q = (z - zo)/(z* - zo) to the right side of (7), we get I z* - .,. z* - Zo [ ]+ + z - Zo z* - Zo ( Z - Zo Z* - Zo + )2 + + ( Z- 20 z* - Zo )nJ I ( z - Zo )n+l z* - Z Z* - Zo We insert this into (6). Powers of z - Zo do not depend on the variable of integration z*. so that we may take them out from under the integral sign. This yields fez) . I 21Ti = - f f(z*) dz* c z* - zo· Z - <'0 + -21Ti ... + 1 r f(z*) dz* c (z* - 20)2 (z - zo)n 21Ti f + ... f(z*) c (z* - zo)n+l dz* + Rn(z) with Rn(z) given by (3). The integrals are those in (2) related to the derivatives, so that we have proved the Taylor formula (4). Since analytic functions have derivatives of all orders, we can take n in (4) as large as we please. If we let n approach infinity, we obtain (I). Clearly, (I) will converge and represent f(z) if and only if (9) lim Rn(z) = O. n-->oo 686 CHAP. 15 Power Series, Taylor Series We prove (9) as follows. Since .:* lies on C. whereas.: lies inside C (Fig. 364). we have 1.:* - zI > O. Since fez) is analytic inside and on C, it is bounded, and so is the function f(::.*)/(z* - z). say. -f(::.*) -z* - z I I ~M for all z"" on C. Also. C has the radius r = k* ML-ineguality (Sec. 14.1) we obtain from (3) IRnl = (10) ..,:; 1z - '.0 In+l 27T Iz - zoln+l 27T - zol and the length 27Tr. Hence by the If f(z*) c (.:* - .:o)n+l(z* - .:) M 1 r n+l -I d::.*1 z - Zo 27T1" -_ M -- r r+ 1 Now Iz - ':01 < r because 2 lies inside C. Thus Iz - 20111" < L so that the right side approaches 0 as n ~ x. This proves the convergence of the Taylor series. Uniqueness follows from Theorem 2 in the last section. Finally, (5) follows from 0) and the Cauchy • inequality in Sec. 14.4. This proves Taylor's theorem. Accuracy of Approximation. We can achieve any preassinged accuracy in approximating f(::.) by a paI1ial sum of ( I ) by choosing n large enough. This is the practical aspect of formula (9). Singularity, Radius of Convergence. On the circle of convergence of 0) there is at least one singular point of fez), that is, a point 2 = c at which fez) is not analytic (but such that every disk with center c contains points at which fez) is analytic). We also say that f(::.) is singular at c or has a singUlarity at c. Hence the radius of convergence R of (1) is usually equal to the distance from z.() to the nearest singular point of f(::.). (Sometimes R can be greater than that distance: Ln.: is singular on the negative real axis, whose distance from Zo = - 1 + i is ], but the Taylor series of Ln ::. with center Zo = -] + i ha<; radius of convergence V2.) Power Series as Taylor Series Taylor series are power series-Df course! Conversely, we have THEOREM 2 Relation to the Last Section A pml'er series with a non::,ero radills of convergence is the Taylor series of its SUI1I. PROOF Given the power series Then f(zo) = ao. By Theorem 5 in Sec. 15.3 we obtain f' (::.) f"(z) + 2a2(Z - ':0) + 3a3(;:' = 2a 2 + 3' 2(::. - ':0) + ... , = al ZO)2 + ... , thus f'(.::o) = (/1 thus f"(::.o) = 2! a2 SEC. 15.4 687 Taylor and Maclaurin Series and in generalln)(zo) = n! an' With these coefficients the given series becomes the Taylor • series of fez) with center zoo Comparison with Real Functions. One surpnsmg property of complex analytic functions is that they have derivatives of all orders, and now we have discovered the other surprising property that they can always be represented by power series of the form (I). This is not true in general for real/unctions; there are real functions that have derivatives of all orders but cannot be represented by a power series. (Example: f(x) = exp ( - l/x 2 ) if x*"O and f(O) = 0; this function cannot be represented by a Maclaurin series in an open disk with center 0 because all its derivatives at 0 are zero.) Important Special Taylor Series These are as in calculus, with x replaced by complex z. Can you see why? (Answer. The coefficient formulas are the same.) X AMP L E 1 Geometric Series LeI I(z) = 11(1 - z). Then we have In)(::;) 11(1 - ::;) is the geometric series = n!/(1 - ::;)n+l, In)(O) = II!. Hence the Maclaurin expansion of 00 (11) I(::;) is singular at . LE 1- z = 2: zn = I + z + z2 + ... (Izl < I). n=O z = I: this point lies on the circle of convergence. • Exponential Function We know that the exponential function eZ (Sec. 13.5) is analytic for all z, and (e z )' = eZ • Hence frum (I) with Zo = 0 we obtain the Maclaurin series (12) This series is also obtained If we replace x In the familiar Maclaurin series of eX by z. Funhermore. by setting z = iy in (12) and separating the series into the real and imaginary pans (see Theorem 2. Sec. 15.1) we obtain Since the series on the right are the familiar Maclaurin series of the real functions cos y and sin .1', this shows that we have rediscovered the Euler formula (13) e iy = cos y + i sin y. Indeed, one may use (12) for definillg eZ and derive from (12) the basic propenies of eZ • For instance, the Z • differentiation formula (e ) ' = eZ follows readily from (12) by termwise differentiation. R CHAP. 15 E X AMP L E:I Trigonometric and Hyperbolic Functions Power Series, Taylor Series By substituting (12) into (1) of Sec. 13.6 we obtain x COSz = :L .,2n ~2 ~ (_1)n + '" 2! 4! (_11). n=O A '" - 1 + ... - (14) z2n+l 00 sin z = :L (_l)n + 1)! (2n n=O ==z- Z3 Z5 + 3! 5! -+ When ~ = \. these are the familiar Maclaurin series of the real functions cos x and sin x. Similarly, by substituting (12) into (II), Sec. 13.6. we obtain cosh z = 2.: n=O ~2 Z2n 1+ (2n)! 2! _4 + 4! + ... (15) :x; sinh Z = 2.: n=O )( AMP L E 4 z2n+l + (211 Z3 I)! =z+ 3! _5 + • 5! Logarithm From (\) it follows that _2 Ln (1 (16) Replacing;;; by + z) = z- _3 + "'2 3 - + ... Clzi < 1). -z and multiplying both sides by -1, we get 1 (17) -Ln(l -;;;) = Ln ~ .2_3 = z + '2 + '3 + .. (kl < 1). By adding both series we obtain 1 + ;;; Ln - - = 2 (18) 1- z ( z + -~3 + -<;5 + ... ) 3 5 (1;;;1 < 1). • Practical Methods The following examples show ways of obtaining Taylor series more quickly than by the use of the coefficient formulas. Regardless of the method used. the result will be the same. This follows from the uniqueness (see Theorem 1t ., L E r Substitution Find the Maclaurin series of f(;;;) = 1I( I Solution. (19) By substituting -Z2 + ;;;2). for;: in (11) we obtain SEC. 15.4 689 Taylor and Maclaurin Series E X AMP L E 6 Integration Find the Maclaurin series of fez) Solution. We have f' (z) arctan z. = = 11(1 + oc arctan Z (_I)n Z3 2n+l ~ ~I z = flO) = 0 we get Z2). Integrating (19) term by term and using = Z5 (izi < z - -3 + -5 - + ... I); n=O _II u + iv = arctan z defined as that value for which this series represents the principal value of w lui < E X AMP L E 7 -rr12. Development by Using the Geometric Series Develop lI(e - z) in powers of z - zo, where e - 20 *' O. Solutioll. This was done in the proof of Theorem I, where e = z*. The beginning was simple algebra and then the use of (II) with z replaced by (z - zo)/(e - zo): ~ (Z-Zo)n e-z e - Zo - (z - zo) (e-ZQ) ( zI-~ 7 c - Zo ) e - Zo n~O c - Zo e-:: Z--)2 ( e - Zo ) + .... This series converges for z -- Zo e - Zo I E X AMP L E 8 I <I, that is, Iz - 201 < • Ie - zol· Binomial Series, Reduction by Partial Fractions Find the Taylor series of the following function with center Zo f(z) = Solution. We develop f(~) (20) 1with 111 f(z) = 3 Z + 2 Z - L = 8z - 12 in partial fractions and the first fraction in a binomial series ___ = (I + Zr (1 + 1) mz + m(m + Z)-m (-m) ~ = 2! Z2 - m(m Zn n n=O + l)(m + 2) 3! Z3 + ... 2 and the second fraction in a geometric series, and then add the two series term by term. This gives = __1 _ + (z + 2l _2_ i ~ (-2) (z ~ n~O 8 9 = z - 3 I [3 I + n~O 23 108 (z - n~O 2 I) - = 2 2 - (z - I) )n _~ (z; I )n = ~ Il 31 54 (z - 1) - _ (z - 1)]2 2. ( 9 [(-1)~:2+ I [I + !(z I) - ) _ - 1)]2 2~ ] (Z _ I I - ~(z - 1) I)n 3 275 3 1944 (z - I) We see that the first series converges for Iz - II < 3 and the second for Iz - II < 2. This had to be expected because I/(z + 2)2 is singular at -2 and 2/(z - 3) at 3. and these points have distance 3 and 2. respectively, • from the center Zo = L Hence the whole series converges for Iz - II < 2. CHAP. 15 690 Power Series, Taylor Series •••.......... w .... · . _ ... lA--.. " _ ...... . - . ........ • ..-•_ .· ..... ___ I~~ TAYLOR AND MACLAURIN SERIES Find the Taylor or Maclaurin series of the given function with the given point as center and detennine the radius of convergence. 1. e -2z , 3. e Z , 0 2. I/(I - (3), -2i 4. cos2 Z, 0 5. sin z, 7. 1/(1 7r12 6. 1/z. z), 8. Ln (I - z). - -z2(2 , 9. e 10. e Z 2 0 f Z6 - + Z4 I, Z2 - 12. sinh (z - 2i), 0 2i Find the Maclaurin series by tennwise integrating the integrand. (The integrals cannot be evaluated by the usual methods of calculus. They define the error function erf z, sine integral Si(z). and Fresnel integrals4 S(z) and C(z). which occur in statistics, heat conduction. optics, and other applications. These are special so-called higher transcendental functions.) 13. erfz = • 2f V7r L 15. S(z) = f e- t 2 dt 14. Si(z) = f o 0 B2 = I B4 = - 30 B5 - I B3 6 = 0, = O. 1 B6 = 42 t 4i 2i tan z = e 2iz - i _ n=l 18. (Inverse sine) Developing show that arcsin z = z+ • SID t -- uV I - Z2 and integrating, (±) ~ + (~:!) ~ + (~) dt 2'4' 6 _7 +. 7 (izl < 1). Z sin t 2 dt Show that this series represents the principal value of arcsin z (defined in Team Project 30. Sec. 13.7). o 17. CAS PROJECT. sec, tan, arcsin. (a) Euler numbers. The Maclaurin series E22 (21) sec z = Eo - z 2! + -E44 z 4! - + ... defines the Elller numbers E 2n- Show that Eo = 1, E2 = -I, E4 = 5, E6 = -61. Write a program that computes the E 2n from the coefficient formula in (1) or extracts them as a list from the series. (For tables see Ref. [GRI]. p. 810. listed in App. 1.) (b) Bernoulli numbers. The Maclaurin series (22) , (c) Tangent. Using (1), (2), Sec. 13.6, and (22), show that tan z has the following Maclaurin series and calculate from it a table of Bo, ... , B2O: (24) HIGHER TRANSCENDENTAL FUNCTIONS Z 2 Write a program for computing Bn. e- t 2 dt, Z Bl = (23) 0 0 11. defines the Bernoulli numbers Bn. Using undetermined coefficients, show that z e' - 1 19. (Undetennined coefficients) Using the relation sin z = tan Z cos Z and the Maclaurin series of sin z and cos z, find the first four nonzero terms of the Maclaurin series of tan z. (Show the details.) 20. TEAM PROJECT. Properties from Maclaurin Series. Clearly, from series we can compute function values. In this project we show that properties of functions can often be discovered from their Taylor or Maclaurin series. Using suitable series, prove the following. (a) The fonnulas for the derivatives of e2 , cos z, sin z, cosh Z, sinh z, and Ln (1 + z) (b) 4(i + e-iz ) Z (c) sin z =1= = cos Z 0 for all pure imaginalY Z = iy *" 0 4AUGUSTIN FRESNEL (1788-1827), French physicist and engineer, known for his work in optics SEC. 15.5 15.5 Uniform Convergence. 691 Optional Optional Uniform Convergence. We know that power series are absolutely convergent (Sec. 15.2, Theorem 1) and, as another basic property, we now show that they are ul1ifOlmly convergent. Since uniform convergence is of general importance, for instance, in connection with termwise integration of series, we shall discuss it quite thoroughly. To define uniform convergence, we consider a series whose terms are any complex functions f o(z), f 1 (z) . ... : oc L (1) fm(z) = fo(z) + fl(z) + f2(z) + .... m~O (This includes power series as a special case in which f m(Z) = am (Z - Z-O)"m.) We assume that the series (1) converges for all z in some region G. We call its sum s(z) and its nth partial sum sn(z); thus Convergence in G means the following. If we pick a z = ZI in G, then, by the definition of convergence at 210 for given E > 0 we can find an N 1(E) such that If we pick a 22 in G, keeping E as before, we can find an N2 ( E) such that and so on. Hence, given an E > 0, to each Z in G there corresponds a number Nzt E). This number tells us how many terms we need (what Sn we need) at a Z to make Is(:) - sn(z)1 smaller than E. Thus this number NiE) measures the speed of convergence. Small Ni E) means rapid convergence. large NzC E) means slow convergence at the point z considered. Now, if we can find an N(E) larger than all these NzCE) for all z in G, we say that the convergence of the series (1) in G is uniform. Hence this basic concept is defined as follows. DEFINITION Uniform Convergence A series (1) with sum s(z) is called uniformly convergent in a region G if for every E > 0 we can find an N = N( E), not depelldillg Oil Z, such that for all n > N( E) alld all z ill G. UniformilY of convergence is thus a property that always refers to an it~tillite set in the z-plane, that is, a set consisting of infinitely many points. CHAP. 15 692 E X AMP L E 1 Power Series, Taylor Series Geometric Series Show that the geometric series 1 + Z + Z2 + ... is (a) uniformly convergent in any closed disk (b) not uniformly convergent in its whole disk of convergence Izl < 1. Solution. 1111 - zl ~ (a) For z in that closed disk we have 11 - zl ~ I - r (sketch it). Izl ~ r < I. This implies that l/(1 - r). Hence (remember (8) in Sec. 15.4 with q = z) Is(z) - sn(z)1 = I I I I~ L 00 n+l zm = ; _ z n+l ; _ r . m=n+l Since r < I, we can make the right side as small as we want by choosing n large enough, and since the right side does not depend on z (in the closed disk considered), this means that the convergence is uniform. (b) For given real K (no maner how large) and zn+l I1 - z 11 I we can always find a z in the disk = Izln+l zl 11 - Izl < 1 such thaI > K ' simply by taking z close enough tu 1. Hence no single N( E) will suffice to make Is(z) - sn(z)1 smaller than a given E > 0 throllghollt the whole disk. By definition, this shows that the convergence of the geometric series • in Izl < I is not uniform. This example suggests thatfor a power series, the unifomlity of convergence may at most be disturbed near the circle of convergence. This is true: 1 'UOREM 1 Uniform Convergence of Power Series A power series (2) m=O with a nonzero radius of convergence R is uniformly convergent in every circular disk Iz - Zol 2 I' of radius I' < R. PROOF For Iz - zol 2 I' and any positive integers nand p we have Now (2) converges absolutely if Iz - zol = I' < R (by Theorem 1 in Sec. 15.2). Hence it follows from the Cauchy convergence principle (Sec. 15.1) that. an E> 0 being given. we can find an N( E) such that for n > N( E) and p = I, 2, .... From this and (3) we obtain for all z in the disk Iz - zol 2 r. every n > N(E), and every p = 1,2, .... Since N(E) is independent of z, this shows uniform convergence, and the theorem is proved. • Theorem 1 meets with our immediate need and concern, which is power series. The remainder of this section should provide a deeper understanding of the concept of uniform convergence in connection with arbitrary series of variable terms. SEC. 15.5 Uniform Convergence. 693 Optional Properties of Uniformly Convergent Series Unifonu convergence derives its main importance from two facts: 1. [f a series of continuous tenus is unifonuly convergent, its sum is also continuous (Theorem 2, below). 2. Under the same assumptions, tenuwise integration is permissible (Theorem 3). This raises two questions: 1. How can a converging series of continuous tenus manage to have a discontinuous sum? (Example 2) 2. How can something go wrong in termwise integration? (Example 3) Another natural question is: 3. What is the relation between absolute convergence and unifonu convergence? The surprising answer: none. (Example 5) These are the ideas we shall discuss. If we add finitely many continuous functions, we get a continuous function as their sum. Example 2 will show that this is no longer true for an infinite series, even if it converges absolutely. However, if it converges uniformly, this cannot happen, as follows. THEOREM 2 Continuity of the Sum Let the series 2: f",(z) = fo(z) + fl(Z) + ... m=O be ull(fonnly convergent in a region C. Let F(z) be its sum. Then if each term f m(Z) is continuous at a point ZI in C, the junction F(z) is continuous at ZI' PROOF Let sn(:::) be the nth partial sum of the series and Rn(::,J the corresponding remainder: sn = f0 + f1 + ... + f n' Rn = fn+l + fn+2 + ... Since the series converges uniformly, for a given E> 0 we can find an N = N(E) such that for all Z in C. Since SN(Z) is a sum of finitely many functions that are continuous at continuous at ZI' Therefore. we can find a [) > 0 such that Using F = SN + RN and the triangle inequality (Sec. This implies that F(z) is continuous at Zl. 13.2), for these and the theorem is proved. 21' this sum is z we thus obtain CHAP. 15 4 2 Power Series, Taylor Series Series of Continuous Terms with a Discontinuous Sum Consider the series (x real) This is a geometric series with q = 1/(1 sn(x) = x 2[ I + + x 2 ) times a factor x 2 . Its /lth partial sum is I + ---2 I +x (I + I 2 2 + ... + x ) I 2] ' (I + x )n We now use the trick by which one finds the sum of a geometric series, namely, we multiply sn(x) by -q = -I/(I - + 2 x ), __ 1-2 ST/(x) = -x2 [ _ _ 1-2 I + I x + + ... + x (I 2 + x )n + (1 + 1 ] 12 X )n+ . Adding thIs to the previous formula, simplifying on the left, and canceling most terms on the right, we obtain X2 2 [ - - - 2 Sn(x) = x I+x I thus *' 0, the sum is The exciting Fig. 365 "explains" what is going on. We see that if x Sex) = lim sn(x) = I °1l_X + 2 x , but for x = 0 we have s,,(O) = I - I = 0 for all /1, hence s(O) = O. So we have the surprising fact that the sum is discontinuous (at x = 0), although all the tenns are continuous and the series converges even absolutely (its terms are nonnegative. thus equal to their ab~olute value!). Theorem 2 now tells us that the convergence cannot be uniform in an interval containing x = O. We can also verify this directly. Indeed. for x 0/= 0 the remainder ha, the absolute value I IRn(x)1 = Is(x) - sn(x)1 = (I + 2 x) n and we see that for a given E « I) we cannot find an N depending only on and all x, say, in the interval 0 ~ x ~ I. E y 2 8 8 8 4 1.5 8 64 ) I 8 16, ~ 8 1 ,1 1/ -1 "'iR.365. 0 Partial sums in Example 2 x such that IRnl < E for all n > Nf E) SEC. 15.5 I 695 Optional Uniform Convergence. ermwise Integration This is our second topic in connection with unifonn convergence, and we begin with an example to become aware of the danger of just blindly integrating tenn-by-tenn. X AMP L E 3 Series for which Termwise Integration is Not Permissible Let IImtX) = IIIxe -.".,?- and consider the series where in the interval 0 ~ x ~ 1. The nth partial sum is Hence the series has the sum F(x) = lim sn(x) = lim II n (x) = 0 n-+oo n--+oo I (0 ~ x ~ 1). From this we obtain I F(x) dx = O. o On the other hand. by integrating term by term and using I oc :L -rn=1 Now sn = £I" n 1 :L n--+oom=l f mIx) dx = lim 0 I 0 fI + f2 + ... + Ii 1 f mIx) ~x = lim fn = sn' we have n--+x Sn(X) dx. 0 and the expression on the right becomes lim n--+x I 1 unCx) 0 dx = lim n--+oo I 1 0 Ilxe -n.i' dx 1 = lim -2 (1 - e -n) n--+Xi = 21 . but not O. This shows that the serie, under consideration cannot be integrated term by term from x = 0 to = I. • X The series in Example 3 is not unifonnly convergent in the interval of integration, and we shall now prove that in the case of a unifonnly convergent series of continuous functions we may integrate term by tenn. "HOREM Termwise Integration Let 00 Hz) = 2: I m(z) = Io(z) + fl(Z) + m=O be a uniformly convergent series of continuous functions in a region G. Let C be any path in G. Then the series (4) is convergent and has the sum Ic F(z) dz.. CHAP.15 696 PROOF Power Series, Taylor Series From Theorem 2 it follows that F(z) is continuous. Let s,.(::) be the 11th partial sum of the given series and R,,(::') the corresponding remainder. Then F = Sn + Rn and by integration, J c F(z) elz = J c + s,.(z) liz J c Rn (::.) elz. Let L be the length of C. Since the given series converges uniformly, for every given E > 0 we can find a number N such that IRn(z)1 < ElL for all Il > N and all ::. in G. By applying the ML-inequality (Sec. 14.1) we thus obtain E < - L L Since Rn = F - Sn, = for all n > N. E this means that for all fl > N. • Hence, the series (4) converges and has the sum indicated in the theorem. Theorems 2 and 3 characterize the two most important properties of uniformly convergent selies. Also, since differentiation and integration are inverse processes, Theorem 3 implies THEOREM 4 Termwise Differentiation Let the series f o(z) + f 1 (z) + f 2(Z) + ... be convergent in a region G and let F(z) be its sum. Suppose that the series f~(z) + f~(::.) + f~(z) + ... converges ulliformly in G anel its terms are cOluinuollS in G. Then F'(z) = f~(::.) + f~(::.) + f~(z) + for all ::. in G. ~------------------------------------------------------------------~ Test for Uniform Convergence Uniform convergence is usually proved by the following comparison test. THEOREM 5 Weierstrass' M-Test for Uniform Convergence COil sider a series oftlze f01711 (1) ill a region G of the ::.-plane. Suppose that one can find a convergent series qf cOllstallf terms, (5) such that If...(z)I ~ M", for all z in G and every unifo17nZy cOllvergent ill G. 111 = 0, 1, .. , Then (1) IS 5 KARL WEIERSTRASS (1815-1897), great German mathematician. who~e lifework was the development of complex analysis based on the concept of power selies (see the footnote in Sec. 13.4). He aho made basic contributions to the calculus. the calculus of variations. approximation theory. and differential geometry. He obtained the concept of uniform convergence in 1841 (published 1894. sid); the first publication on the concept was by G. G. STOKES (see Sec 10.9) in 1847. SEC. 15.5 Uniform Convergence. 697 Optional The simple proof is left to the student (Team Project 18). E X AMP L E 4 Weierstrass M-Test Does the following series converge uniformly in the disk Izl : ":' I? Uniform convergence follows by the Weierstrass M-test and the convergence of LIIm 2 (see Sec. 15.1. in the proof of Theorem 8) because Solution. _'m I /11 2 + I + cosh mlzl • 2 < 2 III No Relation Between Absolute and Uniform Convergence We finally show the surprising fact that there are series that converge absolutely but not uniformly, and others that converge uniformly but not absolutely, so that there is no relation between the two concepts. E X AMP L E 5 No Relation Between Absolute and Uniform Convergence The series in Example 2 converges absolutely but not uniformly, as we have shown. On the other hand, the series (_l)m-l m=l .\"2 + + m (x real) converge, uniformly on the whole real line but not absolutely. Proof By the familiar Leibniz test of calculus ~see App. A3.3) the remainder Rn does not exceed its first term in absolute value, since we have a series of alternating terms whose absolute values form a monotone decreasing sequence with limit zero. Hence given E > 0, for all x we have ifn > I N~E) ~ - E This proves uniform convergence, since N~ E) does not depend on x. The convergence is not ab,olute because for any fixed x we have k >m • where k is a suitable consrant. and kL 11m diverges. /1-8/ UNIFORM CONVERGENCE Prove that the given series converges uniformly indicated region. 1. .L n=O (z - 2i)2n, /z - 2i/ ::":' 0.999 00 In the 2.2: n=Q ...,2n+l (2n + I)! ' CHAP. 15 698 Power Series, Taylor Series (d) Example 2. Find the precise region of convergence of the series in Example 2 with x replaced by a complex variable z. (e) Figure 366. Show that x 2 ~;;'~1 (1 + x 2 )-m = 1 if x =F 0 and 0 if x = O. Verify hy computation that the partial sums .1'10 S2' S3, look as shown in Fig. 366. _n 5. L ;12 ' Izl;§: I n~l 6. L n=l 7. L n=O 00 8. L n=l [9-161 zn n 2 cosh Ilkl tanhn /1 2 + Izl ;§: I cos nlzl , y 1 Izl Izl ;§: ~ I Izl ;§: s 1010 10 20 -1 o x Fig. 366. Sum 5 and partial sums III Team Project 18(e) POWER SERIES Find the region of uniform convergence. (Give reason.) x (~+ I - 2;)n x (Z _ ;)2n 9. 4" 10. (2/l)! L n=O L n=O 12. 14. L (n) n-2 2 <Xl L (3n (2z - i)n tanh 11);;:2n 119-201 HEAT EQUATION Show that (9) in Sec. 12.5 with coefficients (10) is a solution of the heat equation for t > 0, assuming that f(x) is continuous on the interval 0 ;§: x ;§: L and has one-sided derivatives at all interior points of that interval. Proceed as follows. 19. Show that IB"I is bounded, say Conclude that n=l 15. L n-l _2n ~ 16. 2 Y'n L n=O (_l)n?n (2n)! 17. CAS PROJECT. Graphs of Partial Sums. (a) Figure 365. Produce this exciting figure using your software and adding fm1her curves. say, those of S256' SI024' etc. (b) Power series. Study the nonuniformity of convergence experimentally by plotting partial sums near the endpoints of the convergence interval for real z = x. 18. TEAM PROJECT. Uniform Convergence. (a) Weierstrass M-test. Give a proof. (b) Termwise differentiation. Derive Theorem 4 from Theorem 3. (c) Subregions. Prove that uniform convergence of a series in a region C implies unifonn convergence in any portion of C. Is the converse true? . .--. ..-. to. _ _ _ _ _ .. 1. What are power series? Why are these series very important in complex analysis? 2. State from memory the ratio test, the root test, and the Cauchy-Hadamard fomlLlla for the radius of convergence. 3. What is absolute convergence? Conditional convergence? Uniform convergence? if IB"I < t ~ K for all n. to > 0 and. by the Weierstrass test. the series (9) converges uniformly with respect to x and t for f ~ fo, 0 ;§: x ;§: L. Using Theorem 2. show that II(X, t) is continuous for 1 ~ 10 and thus satisfies the boundary conditions (2) for f ~ fo. 20. Show that Iillln/iltl < An2 Ke-An2to if 1 ~ to and the series of the expressions on the right converges. by the ratio test. Conclude from this. the Weierstrass test, and Theorem 4 that the series (9) can be differentiated term by term with respect to t and the resulting series has the sum duliN. Show that (9) can be differentiated twice with respect to x and the resulting series has the sum a2 u/ilx 2 . Conclude from this and the result to Prob. 19 that (9) is a solution of the heat equation for all t ~ to. (The proof that (9) satisfies the given initial condition can be found in Ref. [CIO] listed in App. 1.) STIONS AND PROBLEMS 4. What do you know about the convergence of power series? 5. What is a Taylor series? What was the idea of obtaining it from Cauchy's integral formula? 6. Give examples of practical methods for obtaining Taylor series. 7. What have power series to do with analytic functions? 699 Summary of Chapter 15 8. Can propel1ies of functions be discovered from their Maclaurin series? If so, give examples. 9. Make a list of Maclaurin series of c. cos z. sin z, cosh z, sinh z, Ln (1 - z) from memory. 10. What do you know about adding and multiplying power series? 111-201 RADIUS OF CONVERGENCE TAYLOR AND MACLAURIN SERIES Find the Taylor or Maclaurin series with the given point as center and determine the radius of convergence. (Show details.) 21. e". 22. Ln z. 7ri 23. 1/(1 - z), 25. 11(1 - Find the radius of convergence. Can you identify the sum as a familiar function in some of the problems? (Show the details of your work.) ~ ~1-301 27. liz, 29. cos Z, d, 2 24. 11(4 - 3z), -1 28. I"t- 1(e t -i o 30. sin2 ~7r +i i 26. l1z2, 0 1 - 1) dt, 0 z, 0 (3z)n 11. L . . . - n! n=O n=l Z2n+l 13'L n~O 2n + 14. 1 16. L (-I)n zn n~O (2n)! L n~O (-I)n (z - + (2n 2y2n+l I)! (2z)2n 18. n=O 20. L -- n~O (217)! L n~O (z - (3 i)" + 4i)n 31. Does every function fez) have a Taylor series? 32. Does there exist a Taylor series in powers of z - 1 - i that diverges at 5 + 5i but converges at 4 + 6i? 33. Do we obtain an analytic function if we replace x by z in the Maclaurin series of a real function f(x)? 34. Using Maclaurin series. show that if fez) is even. its integral (with a suitable constant of integration) is odd. 35. Obtain the first few terms of the Maclaurin series of tan z by using the Cauchy product and sin z = cos z tan z. Power Series, Taylor Series Sequences. series, and convergence tests are discussed in Sec. 15.1. A power series is of the form (Sec. 15.2) ...., (I) n~O Zo is its center. The series (1) converges for Iz - zol < R and diverges for Iz - ;:';01 > R, where R is the radius of convergence. Some power series converge for all z (then we write R = (0). In exceptional cases a power series may converge only at the center; such a series is practically useless. Also, R = lim la"lan + 11 if this limit exists. The series (I) converges absolutely (Sec. 15.2) and uniformly (Sec. 15.5) in every closed disk Iz - zol ~ r < R (R > 0). It represents an analytic function fez) for Iz - Zol < R. The derivatives t(z). f"(;::.), ... are obtained by termwise differentiation of (1 ). and these series have the same radius of convergence R as (1). See Sec. 15.3. 700 CHAP. 15 Power Series, Taylor Series Conversely, every analytic function .f(::.) can be represented by power series. These Taylor series of .f(z) are of the form (Sec. 15.4) x (2) .f(z) = L 1 I"" tnl(zo)(z - z.o)n (Iz - zol < R), n=O 11. as in calculus. They converge for all z in the open disk with center Zo and radius generally equal to the distance from :::0 to the nearest singularity of .f(:::) (point at which .f(z) ceases to be analytic as defined in Sec. 15.4). If .f(z) is entire (analytic for all :::; see Sec. 13.5). then (2) converges for all ;:. The functions eZ , cos z, sin:::, etc. have Maclaurin series, that is, Taylor series with center 0, similar to those in calculus (Sec. 15.4). CHAPTER / 16 Laurent Series. Residue Integration Laurent series generalize Taylor series. Indeed, whereas a Taylor series has positive integer powers (and a constant term) and converges in a disk, a Laurent series (Sec. 16.1) is a series of positive and negative integer powers of z - '::0 and converges in an annulus (a circular ring) with center Zoo Hence by a Laurent series we can represent a given function f(z) that is analytic in an annulus and may have singularities outside the ring as well as in the "hole" of the annulus. We know that for a given function the Taylor series with a given center '::0 is unique. We shall see that, in contrast, a function f(z) can have several Laurent series with the same center '::0 and valid in several concentric annuli. The most important of these series is that which converges for 0 < Iz - zol < R. that is, everywhere near the center ::'0 except at Zo itself. where Zo is a singular point of f(z). The series (or finite sum) of the negative powers of this Laurent series is called the principal part of the singularity of f(z) at Zo, and is used to classify this singularity (Sec. 16.2). The coefficient of the power 1/(;: - zo) of this series is called the residue of f(z) at zoo Residues are used in an elegant and powerful integration method, called residue integration, for complex contour integrals (Sec. 16.3) as well as for certain complicated real integrals (Sec. 16.4). Prerequisite: Chaps. 13, 14, Sec. 15.2. Sections that may be omitted in a shorter course: 16.2, 16.4. References and Answers to Problems: App. 1. Part 0, App. 2. 16.1 Laurent Series Laurent series generalize Taylor series. If in an application we want to develop a function f(z) in powers of Z - Zo when f(z) is singular at Zo (as defined in Sec. 15.4). we cannot use a Taylor series. Instead we may use a new kind of series, called Laurent series, 1 consisting of positive integer powers of::. - Zo (and a constant) as well as negative integer powers of z - ':::0; this is the new feature. Laurent series are also used for classifying singularities (Sec. 16.2) and in a powerful integration method ("residue integration", Sec. 16.3). A Laurent series of f(::.) converges in an annulus (in the "hole" of which f(.:::) may have singularities), as follows. IPIERRE ALPHONSE LAURENT (1813-1854). French military engineer and mathematician, published the theorem in 1843. 701 702 THEOREM 1 CHAP. 16 Laurent Series. Residue Integration Laurent's Theorem Let fez) be analytic in a domain c()ntaining two concentric circles C1 and C2 with center Zo and the annulus betrveen them (blue in Fig. 367). Then fez) can be represented by the Laurent series (1) ... + z - Zo consisting of nonnegative lind negative powers. The coefficients of this Laurent series are given by the integrals (2) taken coullterclockwise around allY simple closed path C that lies in the annulus and encircles the inner circle, as in Fig. 367. [The variable of integration is denoted by z* since z is used in (1).] This series converges and represents fez) in the enlarged open allnulus obtained from the given annulus by continuously increasing the outer circle C1 and decreasing C2 until each of the fiFO circles reaches a point where fez) is singular. III the important special case that :.':0 is the ollly singular point of fez) inside C2 , this circle can be shrunk to the point zo, giving convergence in a disk except at the center. In this case the series (or finite sum) of the negative powers of (1) is called the principal part of the singularity of fez) at zoo .-- - I \ \ \ Fig. 367. Laurent's theorem COMMENT. Obviollsly, instead of (1). (2) we may write (denoting bn by a_ n ) (1') n=-:JO SEC. 16.1 703 Laurent Series where all the coefficients are now given by a single integral formula, namely, a (2') PROOF n = -1- T f(z*) d-* c (z* - Zo)n+1 ~ 2wi (n = 0, ±l, ±2, .. '). We prove Laurent's theorem. (a) The nonnegative powers are those of a Taylor series. To see this, we use Cauchy's integral fOlmula (3) in Sec. 14.3 with z* (instead of z) as the variable of integration and z instead of ~o. Let g(z) and h(::.) denote the functions represented by the two terms in (3), Sec. 14.3. Then fez) (3) = g(z) + hU;) = -1. 2Wl T -f(z*)- d::* c, z* - ::. I -. 2m T -f(z*)- dz*. C z* - 2 Z Here::. is any point in the given annulus and we integrate counterclockwise over both C1 and C2 , so that the minus sign appears since in (3) of Sec. 14.3 the integration over C2 is taken clockwise. We transform each ofthese two integrals as in Sec. 15.4. The first integral is precisely as in Sec. 15.4. Hence we get precisely the same result, namely, the Taylor series of g(z), I g(z) = --. (4) 2Wl T -f(:;::*) - - dz* c, z* - Z = = an(z :L zo)n n~O with coefficients [see (2), Sec. 15.4, counterclockwise integration] an (5) = I -? _WI T c, ( f(z*) d-* z· - Zo )n+1 ~. ok Here we can replace C 1 by C (see Fig. 367), by the principle of deformation of path, since Zo, the point where the integrand in (5) is not analytic, is not a point of the annulus. This proves the formula for the an in (2). (b) The negative powers in (1) and the formula for bn in (2) are obtained if we consider h(z) (the second integral times -J/(2wi) in (3). Since z lies in the annulus, it lies in the exterior of the path C2 . Hence the situation differs from that for the first integral. The essential point is that instead of [see (7*) in Sec. 15.4] (6) (a) I z-::'°I<1 z* - Zo we now have z* - Zo (b) 1 1 < 1. z - Zo Consequently, we must develop the expression I/(z* - z) in the integrand of the second integral in (3) in powers of (::.* - Zo)/(z - Zo) (instead of the reciprocal of this) to get a convergent series. We find 1 z* - z -1 ::.* - Zo - (z - ::'0) (z - Zo) (1 _ z*z -- Zozo) . 704 CHAP. 16 Laurent Series. Residue Integration Compare this for a moment with (7) in Sec. 15.4. to really understand the difference. Then go on and apply formula (8), Sec. 15.4. for a finite geometric sum. obtaining 1-=--1-Z* - Z ;: - Zo { 1+ z* - Zo + (z* - Zo )2 + ... + z - ':0 ::. - Zo __ 1 (Z* - zO)n+l z-z* Z-2o Multiplication by -f(.:*)/27Ti and integration over C2 on both sides now yield 1 l1(z) = - --. 27Tl _1_. 27TI f -f(z*) - - dz* e2 z* - z {_1_ r1 Z - Zo f(z*) dz* + e2 + 1 (z - ZO)2 1 (z - zo)n+l + ... 1 (z* - 2o)f(z*) dz* 1 r (z* - zo)nf(z*) dZ""'-} re2 + Rn*(z) e2 with the last term on the right given by R*( (7) n z) = 1 ? ( )n+l _7rIZ-2o 1 re 2 (z* - ::o)n+l f(z*) dz*. z-z'ok As before. we can integrate over C instead of C2 in the integrals on the right. We see that on the right, the power 1/(z - zo)n is multiplied by b n as given in (2). This establishes Laurent's theorem, provided lim R~(z) (8) ~x = O. (c) COllvergellce proofof (8). Very often (1) will have only finitely many negative powers. Then there is nothing to be proved. Otherwise, we begin by noting that f(z*)/(z - z*) in (7) is bounded in absolute value, say. I f(z*) z - z* I< M for all z* on C2 because f(z*) is analytic in the annulus and on C2 , and z* lies on C2 and z outside, so that z - z* =/= O. From this and the ML-inequality (Sec. 14.1) applied to (7) we get the inequality (L = length of C2 , Iz* - zol = radius of C2 = const) ~ oki lR;"(z) 1 ~ 2 1 7T Z - Zo ~ In+l Iz* - zoln+l ML ML = - 27T I z* - 20 In+l Z - Zo SEC. 16.1 705 Laurent Series From (6b) we see that the expression on the right approaches zero as n approaches infinity. This proves (8). The representation (1) with coefficients (2) is now established in the given annulus. (d) C01lverge1lce of (1) i1l the e1llarged a1l1lulus. The first series in (1) is a Taylor series [representing g(z)]; hence it converges in the disk D with center Zo whose radius equals the distance of the singularity (or singularities) closest to zoo Also, g(z) must be singular at all points outside C I where fez) is singular. The second series in (I), representing h(z), is a power series in Z = 1/(z - Zo). Let the given annulus be 1"2 < Iz - zol < r l , where 1"1 and r2 are the radii of C I and C2, respectively (Fig. 367). This corresponds to 1/r2 > Izi > lirl' Hence this power series in Z must converge at least in the disk Izi < 1/r2' This corresponds to the exterior Iz - Zol > r2 of C2. so that h(z) is analytic for all z outside C2. Also, h(z) must be singular inside C2 where fez) is singular, and the series of the negative powers of (I) converges for all z in the exterior E of the circle with center Zo and radius equal to the maximum distance from <'0 to the singularities of fez) inside C2. The domain common to D and E is the enlarged open annulus characterized near the end of Laurent's theorem, whose proof is now complete. • Uniqueness. The Laurent series of a given analytic function fez) in its annulus of cOllvergence is ullique (see Team Project 24). However. fez) may have different Laurent selies ill two anlluli with the same center; see the examples below. The uniqueness is essential. As for a Taylor series, to obtain the coefficients of Laurent series, we do not generally use the integral formulas (2); instead, we use various other methods, some of which we shall illustrate in our examples. If a Laurent series has been found by any such process, the uniqueness guarantees that it must be the Laurent series of the given function in the given annulus. E X AMP L E 1 Use of Maclaurin Series Find the Laurent series of z-5 sin:: with center O. Solutio1l. By (14). Sec. 15.4. we obtain -5. :: ~ (-I)n SIn Z =:::0 (2n + I 2n-4 I)! Z = I 6;? ;:4 - I + I 120 - 2 5040 z + - ... (Izl > 0). Here the "annulus" of convergence is the whole complex plane without the origin and the pl'incipal part of the series at 0 is Z-4 - ~Z-2. • E X AMP L E 2 Substitution Find the Laurent series of z2 e1/z with center O. Solution. From l12) in Sec. 15.4 with.:: replaced by liz we obtain a Laurent senes whose principal part is an infinite series, (Izl > E X AMP L E 3 0). • Development of 1/(1 - z) Develop 1/(1 - z) (a) in nonnegative powers of~, (b) in negative powers of z. Solutio1l. I x = L -- n 1- z (a) (valid if Izl < I). n=O (b) I - z -] z(l - Z-l) = - = n~o I zn+l I = - ~ I - :;2 - . . . (valid if Izl > I). • 706 E X AMP L E 4 CHAP. 16 Laurent Series. Residue Integration Laurent Expansions in Different Concentric Annuli Find all Laurent series of 1I(~3 - ~ 4) with center O. Solution. Multiplying by IIz 3 we get from Example 3 , I a:J L: -3--4 = (I) Z -z ;;:n-3 = n=O E X AMP L E 5 z - L: - n+4 n-O Z Z Use of Partial Fractions < Izl < 1). ;; _ ... 3 (Izl > 1). • +3 -2z Solution. (0 Z Z x I -3--4 = (II) I +-+I+z+'" 2 + Z3 In terms of partial fractions, I f(::.) = -~ ::.-2 (a) and (b) in Example 3 take care of the first fraction. For the second fraction, x (c) ~ z) z-2 2 (I - (d) z - 2 z (I) From (a) and (c), valid for fez) Izl < = 1(z) = L: (1 + < Izl < 1 a:J 2n+l ;;:n - n=O L: n=O (Ill) From ~d) and (b). valid for Izl > 2nl+1) zn = - L: (1::.1 < 2), n=O = - L: 2n (izi > ';"n+l 2). n=O ..... %+ %z + ~ = I 7n+1 + ... = 2 ~ 2, 1 (2n + I) n=O Z2 2, 00 fez) I 2n+l zn L: I (see Fig. 368), = n~o (II) From (c) and (b), valid for 1 (I -~) = n+l 2 = - - Z 9 3 Z • y ...... III II I Fig. 368. " x Regions of convergence in Example 5 If fez) in Laurent's theorem is analytic inside C2 , the coefficients b n in (2) are zero by Cauchy's integral theorem, so that the Laurent series reduces to a Taylor series. Examples 3(a) and 5(1) illustrate this. SEC. 16.2 ~1=6J 707 Singularities and Zeros. Infinity 115-231 LAURENT SERIES NEAR A SINGULARITY ATO Expand the given function in a Laurent series that converges for 0 < Id < R and determine the precise region of convergence. (Show the details of your work.) 1. 4. _3 Z-3 e llz 17-141 17. 2 6. Z2 - Z 21. Z3 LAURENT SERIES NEAR A SINGULARITY AT Zo Expand the given function in a Laurent series that converges for 0 < I:: - ::01 < Rand detennine the precise region of convergence. (Show details.) eZ 7. - - , z- 9. ll. Z2 (z I + + I 8. = i 10. Zo = ':::0 I 2 i) - (z + i) ' ::0 (Z + cos.::: i) 2. 1 14. ::: smh - , Z 16.2 Zo = -i ::0 0 = 7T)4 , Zo = 13. ~7T 7T ~4 4 I 0 16. , :::0 = 0 18. :::0 = 1 - Z2 = ;:0 and , :::0 = - (;: - i)2 4z - I I sin Z 23. Z , Zo = I - sinh;:; 2;::.2 Z4 - Zo = ~ + ~7T , ::0 = :::0 = 0 i 20. 22. (:: _2 1)4 ::0 = ::0 = ; ~. ':::0 = -~7T 24. TEAM PROJECT. Laurent Series. (a) Uniqueness. Prove that the Laurent expansion of a given analytic function in a given annulus is unique. (b) Accumulation of singularities. Does tan (II:) have a Laurent series that converges in a region o < Izl < R? (Give a reason.) (c) Integrals. Expand the following functions in a Laurent series that converges for Izl > 0: ::: Z - , " I 2 Z2 2 Zo = = -; Z3 12. sin ::: (z - ~7T)3 (::: - Z3 I (. 19. Z2 e I - _3 cosh 2;: -<. 5. 15. Z e- z 3. Find all Taylor and Laurent series with center:: determine the precise regions of convergence. Z2 l 2. ;: cos- Z5 Z4 - TAYLOR AND LAURENT SERIES (et-I L--dr, 0 1 -I Z3 I Z 0 sin t --dt. t 25. CAS PROJECT. Partial Fractions. Write a program for obtaining Laurent series by the use of partial fractions. Using the program, verify the calculations in Example 5 of the text. Apply the program to two other functions of your choice. Singularities and Zeros. Infinity Roughly, a singular point of an analytic function fez) is a ::0 at which f(::) ceases to be analytic, and a ::ero is a z at which fez) = O. Precise definitions follow below. In this section we show that Laurent series can be used for classifying singularities and Taylor series for discussing zeros. Singularities were defined in Sec. 15.4, as we shall now recall and extend. We also remember that. by definition, a function is a single-valued relation, as was emphasized in Sec. 13.3. We say that a function fez) is singular or has a singularity at a point;:: = Zo if fez) is not analytic (perhaps not even defined) at z = zo, but every neighborhood of z = Zo contains points at which fez) is analytic. We also say that z = Zo is a singular point of fez). We call z = Zo an isolated singularity of fez) if z = Zo has a neighborhood without further singularities of fez). Example: tan z has isolated singularities at ± 7T12, ±37T/2, etc.; tan (lIz) has a nonisolated singularity at o. (Explain!) 708 CHAP. 16 Laurent Series. Residue Integration Isolated singularities of fez) at z = Zo can be classified by the Laurent series (1) (Sec. 16.1) valid in the immediate neighborhood of the singular point z is, in a region of the form o < Iz - = zo, except at Zo itself, that zol < R. The sum of the first series is analytic at z = zo, as we know from the last section. The second series, containing the negative powers, is called the principal part of (1), as we remember from the last section. If it has only finitely many terms, it is of the form + ... + (2) Then the singularity of fez) at z = Zo is called a pole, and m is called its order. Poles of the first order are also known as simple poles. If the principal part of (I) hac; infinitely many terms, we say that fez) has at z = Zo an isolated essential singularity. We leave aside nonisolated singularities. E X AMP L E 1 Poles. Essential Singularities The function fez) = z(z - 3 + (z - 2)2 2)5 has a simple pole at z = 0 and a pole of fifth order at z singularity at z = 0 are = 2. Examples of functions having an isolated essential and sin - - z - L n-O (211 ~-l)'" I + 1)!in + 1 z 3!Z3 + 5' .Z 5 - + .... Section 16.1 provides further examples. For instance, Example I shows that z-5 sin;: has a fourth-order pole at O. Example 4 shows that l/(z3 - Z4) has a third-order pole at 0 and a Laurent series with infinitely many negative powers. This is no contradiction, since this series is valid for Izl > 1; it merely tells us that in classifying singularities it is quite important to consider the Laurent series valid ill tile immediate Ileigllborllood of a singular • point. In Example 4 this is the series (I), which has three negative powers. The classification of singularities into poles and essential singularities is not merely a formal matter, because the behavior of an analytic function in a neighborhood of an essential singularity is entirely different from that in the neighborhood of a pole. E X AMP L E 2 Behavior Near a Pole fez) = I/z2 has a pole at z = 0, and If(z)1 ~ theorem. x as ;;; ....... 0 in any manner. This illustrates the foIlowin" • eo SEC. 16.2 709 Singularities and Zeros. Infinity THEOREM 1 Poles If f(z) is analytic and has a pole at z = Zo, then If(z)1 ~ (Xl as Z ~ Zo i17 anv manner. The proof is left to the student (see Prob. 12). E X AMP L E 3 Behavior Near an Essential Singularity The function fez) = el/z has an essential singularity at z = O. It has no limit for approach along the imaginary axis; it becomes infinite if z ..... 0 through positive real values, but it approaches zero if <: --+ 0 through negative real values. It takes on any given value c = coia 0 in an arbitrarily small E-neighborhood of;:: = O. To see the letter. we set z = reill, and then obtain the following complex equation for rand 8. which we must ~olve: '* ellz = e<'cos 0 - i sin tJ)/r = cOeia Equating the absolute values and the arguments, we have e'cos mh· = co' that is cos8= rlnco, respectively. From these two equations and cos2 8 and -sin 8 = ar + sin2 8 = r2(ln cO)2 + and 2 a r2 = I we obtain the formulas a tan8= - - - . Inca Hence r can be made arbitrarily small by adding multiples of 27T to a, leaving c unaltered. This illustrates the very famous Picard's theorem (with z = 0 as the exceptional value). For the rather complicated proof. see Ref. • [D4J. voL 2. p. 258. For Picard. see Sec. 1.7. THEOREM 2 Picard's Theorem If f(z) is analytic alld has all isolated essential singularity at a point zo, it takes Oil eve I}' value, with at most olle exceptional value, in an arbitrarily small E-neighborhood oJzo· Removable Singularities. We say that a function f(::) has a removable sillgulartty at z = Zo if f(z) is not analytic at z = Zo, but can be made analytic there by assigning a suitable value f(zo). Such singularities are of no interest since they can be removed as just indicated. Example: fez) = (sin z)/z becomes analytic at z = 0 if we define f(O) = I. Zeros of Analytic Functions A zero of an analytic function fez) in a domain D is a :: = :::0 in D such that f(zo) = O. A zero has order n if not only f but also the derivatives f', fIt, ... , f n - ll are all 0 at Z = Zo but fn)(Zo) *- O. A fIrst-order zero is also called a simple zero. For a second-order zero, f(Zo) = f' (zo) = 0 but f"(zo) *- O. And so on. E X AMP L E 4 Zeros The function L + ;::2 has simple zeros at :!:i. The function (1 - -;;4)2 has second-order zeros at:!: I and :!:i. The function (::: - a)3 has a third-order zero at Z = a. The function eZ has no zeros (see Sec. 13.5). The function sin z has simple zeros at 0, :!:7T, :!:27T, ... , and sin2 z has second-order zeros at these points. The function I - cos Z has second-order zeros at 0, :!:27T. :!:47T, ... , and the function (I - cos Z)2 has fourth-order zeros at these points. • CHAP. 16 Laurent Series. Residue Integration Taylor Series at a Zero. At an nth-order zero ::: = :::0 of f(:::), the derivatives f' (Zo), ..• , ['n-1)(:::o) are zero, by definition. Hence the first few coefficients (/o, . . . , a n -l of the Taylor series (1), Sec. 15.4, are zero, too, whereas lin =1= 0, so that this series takes the form f(:::) (3) = lIn(:: = (z - zo)n + {/n+l(:: - ::o)n [an + - ':0)'1+1 (/n+l(Z - <::0) + + ... (/n+2(::: - :::0)2 + ... ] This is characteristic of such a zero, because if f(::) has such a Taylor series, it has an nth-order zero at ::: = :::0' as follows by differentiation. Whereas nonisolated singularities may occur, for zeros we have - THEOREM 3 ---------------------------------------------------------------, Zeros The zeros of an analytic filllction f(;::) (¥= 0) are isolated; that is, each of them has a neighborhood that c01l1aills no further :::eros of fez). I ROO F The factor (::: - :::0)" in (3) is zero only at ::: = :::0' The power series in the brackets [ ... ] represents an analytic function (by Theorem 5 in Sec. 15.3), call it g(z). Now g(Zo) = an =1= 0, since an analytic function is continuous, and because of this continuity, also g(:::) =1= 0 in some neighborhood of z = :::0' Hence the same holds of f(:::). • This theorem is illustrated by the functions in Example 4. Poles are often caused by zeros in the denominator. (Example: tan z has poles where cos::: is zero.) This is a major reason for the imp0l1ance of zeros. The key to the connection is the following theorem, whose proof follows from (3) (see Team Project 24). --R": • 4 Poles and Zeros Let fez) be analytic at z = Zo and have a zero of nth order at z = :::0' Then lIf(z) has a pole of 1I1h order at .: = :::0; and so does h(:::)lf(:::), provided he:::) is allalytic at Z = 20 and 17(:::0) =1= 0. Riemann Sphere. Point at Infinity When we want to study complex functions for large Izl, the complex plane will generally become rather inconvenient. Then it may be better to use a representation of complex numbers on the so-called Riemann sphere. This is a sphere S of diameter 1 touching the complex z-plane at z = (Fig. 369), and we let the image of a point P (a number z in the plane) be the intersection P* of the segment PN with S, where N is the "North Pole" diametrically opposite to the origin in the plane. Then to each z there corresponds a point on S. Conversely, each point on S represents a complex number z, except for N, which does not con'espond to any point in the complex plane. This suggests that we introduce an additional point, called the point at infinity and denoted CG ("infinity") and let its image be N The complex plane together with :JO is called the extended complex plane. The complex plane is often called the finite complex plane, for distinction, or simply the ° SEC. 16.2 711 Singularities and Zeros. Infinity N Fig. 369. Riemann sphere complex plane as before. The sphere S is called the Riemann sphere. The mapping of the extended complex plane onto the sphere is known as a stereographic projection. (What is the image of the Northern Hemisphere? Of the Western Hemisphere? Of a straight line through the origin?) Analytic or Singular at Infinity If we want to investigate a function .fez) for large 1z1, we may now set.;: = 1111" and investigate .f(z) = .fO/w) == g(w) in a neighborhood of w = O. We define .f(z) to be analytic or singular at infinity if g(w) is analytic or singular. respectively, at w = O. We also define (4) g(O) = lim zo->o g(w) if this limit exists. Furthermore, we say that f(z.) has an nth-order zero at infinity if f(l/w) has such a zero at w = O. Similarly for poles and essential singularities. E X AMP L E 5 Functions Analytic or Singular at Infinity. Entire and Meromorphic Functions The function f(z.) = 11z2 is analytic at x since g(w) = f(l/w) = .r 2 is analytic at w = 0, and fez) has a secondorder zero at x. The function .t(;:) = 2 3 is singular at x and has a third-order pole there since the function 3 Z g(w) = .to/w) = 1Iw has such a pole at w = O. The function e has an essential singularity at Cf) since eV ", has such a singularity at II' = O. Similarly, cos z and sin z have an essential singularity at x. Recall that an entire function is one that is analytic everywhere in the (finite) complex plane. Liouville's theorem (Sec. l4...l) tells us that the only boullded entire functions are the constants, hence any nonconstant entire function must be unbounded. Hence it has a singularity at x, a pole if it is a polynomial or an essential singularity if it is not. The functions just considered are typical in this respect. An analytic function whose only singularities in the finite plane are poles is called a meromorphic function. Examples are rational function, with nonconstant denominator, tan ;:, cot z, sec z, and eSc z. • In this section we used Laurent series for investigating singularities. In the next section we shall use these series for an elegant integration method . .... [1-101 SINGULARITIES Determine the location and kind of the singularities of the following functions in the finite plane and at infinity. In the case of poles also state the order. 1. tan 2 7TZ 2. z + 2 3 3. cot Z2 4. 5. cos z - sin z 6. lI(cos z - sin z) Z3 e l/(Z-1l CHAP. 16 712 Laurent Series. Residue Integration 21. (1 - cos Z)2 sin 3z 7. (Z4 _ 1)4 4 8. - - + 1 Z - 9. cosh [lie 8 2 (z - 1) 2 (;: - l)3 10. e ll(Z-l)/(e Z - + 1)] 1) 11. (Essential singularity) Discuss e llz2 in a similar way as e llz is discussed in Example 3. 12. (Poles) Verify Theorem I for f(:) Theorem 1. 113-221 = :::-3 - Prove Z-I. ZEROS Determine the location and order of the zeros. 14. (Z4 - 16)4 13. (z + 16i)4 15. :::-3 17. (3z 2 sin 3 + l)e- 19. (,2 + 4)(eZ 16.3 16. cosh 2 ::: 7fZ 18. (Z2 - 1)2(eZ2 Z - - L) 20. (sin z - 1)3 l)2 23. (Zeros) If f(:) is analytic and has a zero of order 11 at z = :0' show that f2(Z) has a zero of order 211. 24. TEAM PROJECT. Zeros. la) Derivative. Show that if f(:) has a zero of order 11 > I at: = :0' then I' (:) has a zero of order 11 - 1 at ::'0. (b) Poles and zeros. Prove Theorem 4. (e) Isolated k-points. Show that the points at which a nonconstant analytic function fez) has a given value k are isolated. (d) Identical functions. If ftC;:) are analytic in a domain D and equal at a sequence of points Zn in D that converges in D, show that fl(:) == .f2(::') in D. 25. (Riemann sphere) Assuming that we let the image of the x-axis be meridians 0° and 180°, describe and sketch (or graph) the images of the following regions on the Riemann sphere: (a) Izl > LOO. (b) the lower half-plane, (c) ! ~ 1::.1 ~ 2. Residue Integration Method The purpose of Cauchy's residue integration method is the evaluation of integrals T.c fez) dz taken around a simple close path C. The idea is as follows. If fez) is analytic everywhere on C and inside C, such an integral is zero by Cauchy's integral theorem (Sec. 14.2), and we are done. If fez) has a singularity at a point z = Zo inside C, but is otherwise analytic on C and inside C, then fez) has a Laurent series fez) = 2:: an(z - zo)n + n~O z- Zo that converges for all points near z = Zo (except at z = Zo itself), in some domain of the form 0 < zol < R (sometimes called a deleted neighborhood, an old-fashioned term that we shall not use). Now comes the key idea. The coefficient hI of the first negative power lI(z - zo) of this Laurent series is given by the integral formula (2) in Sec. 16.1 with 11 = 1, namely, 1 hI = - 2. fez) dz. Iz - 7ft T. C Now, since we can obtain Laurent series by various methods, without using the integral formulas for the coefficients (see the examples in Sec. 16.1), we can find hI by one of those methods and then use the formula for hI for evaluating the integral, that is, (1) SEC 16.3 713 Residue Integration Method Here we integrate conunterclockwise around a simple closed path C that contains z in its interior (but no other singular points of fez) on or inside C!). The coefficient hi is called the residue of fez) at z = Zo and we denote it by = Res hI (2) = Zo fez). Z=Zo E X AMP L E 1 Evaluation of an Integral by Means of a Residue Integrate the function f(z) = Z-4 sin z counterclockwise around the unit circle C. Solution. From (14) in Sec. 15.4 we obtain the Laurent series sin z 1 I z z z3 Z -3'z + . f(z.) = - 4 - = "3 - -5' - . 71. + - ... which converges for Izl > 0 (that is, for all z 1= 0). This series shows that J(z) has a pole of third order at z = 0 and the residue b i = -113!. From (1) we thus obtain the answer J. rc E X AMP L E 2 CAUTION! z • TTi = 27fib 1 = - ""3 dz Use the Right Laurent Series! Integrate f(z) = I/(z3 Solution. sin z -4- Izl Z4) clockwise around the circle C: = 112. = .:3(1 - z) shows that J(z) is singular at z = 0 and z = l. Now z = 1 lies outside C. Hence it is of no interest here. So we need the residue of ftz) at O. We find it from the Laurent series that converges for 0 < Izl < l. This is series (I) in Example 4, Sec. 16.1, z3 - z4 1 1 I 1 ---=-+-+ +I+z+'" Z3 - l Z4 Z2 (0 Z < Izl < I). (Izl > 1), We see from it that this residue is 1. Clockwise integration thus yields J. r dz -3--4 = -27fi Res f(z) = -27fi. z-o cZ-z C4UTlON! Had we used the wrong series (II) in Example 4, Sec. 16.1, • we would have obtained the wrong answer, 0, because this series has no power liz. Formulas for Residues To calculate a residue at a pole, we need not produce a whole Laurent series, but, more economically, we can derive formulas for residues once and for all. Simple Poles. (3) Two formulas for the residue of f(:::;) at a simple pole at Res fez) = hI = lim Zo are (z - zo)f(z) Z----7Zo Z=Zo and, assuming that f(;:,) = p(z)lq(z), p(zo) =1= 0, and q(z) has a simple zero at Zo (so that fez) has at;:,o a simple pole, by Theorem 4 in Sec. 16.2), (4) Res fez) Z=20 p(z) = Res 2=20 q(z) 714 CHAP. 16 PRO 0 F Laurent Series. Residue Integration For a simple pole at z= Zo the Laurent series (1), Sec. 16.1, is (0 '* Here b l O. (Why?) Multiplying both sides by z the formula (3): lim (z - ;;;o)f(z) = bi + lim (z - Zo)[ao Z-+Zo Z---i>Zo ':0 < Iz - zol < and then letting z ~ ':0' R). we obtain + al(Z - zo) + ... ] = b i where the last equality follows from continuity (Theorem L. Sec. 15.3). We prove (4). The Taylor series of q(::.) at a simple zero ':0 is , 2! f = plq and then f into Substituting this into . q"(zo) (z - ':o)p(,:) hm ::'0) - - = q(z) Z-+Zo + (3) gives p(z). Res fez) = lim (z Z~Zo (.: - zol + q(z) = (z - zo)q (zo) Z-+Zo (.: - zo}[q'(;::o) + (z - ::.o)q"(zo)!2 + ... ] z - Zo cancels. By continuity, the limit of the denominator is q' (zo) and (4) follows . • E X AMP L E 3 Residue at a Simple Pole f(:) = (9: + 0/(:3 + ;::) has a simple pole at i because :2 + I Res 9;:: + i z~i ;::{;::2 By (4) with p{i) = + = lim (: - 9; + i and z~; (5) 9: + i ;::(;:: l/ (;::) ~ 3;::2 + Res Poles of Any Order. i) z~i I) 9;;:2 + i :(;:: + I) . I){:: - + . I) = = (: [ + i )(z 9: + i ] ---. ;::{;:: + I) - i). and (3) gives the residue !Oi z~i = - -2 = -5;. I we confirm the result. = [9;:: +i ] -2-3~ + 1 !Oi Fi = - - 2 = -5i. The residue of fez) at an mth-order pole at Zo is ~~~ fez) = 1 (m _ I)! !~~o In particular, for a second-order pole (m {d"'-I [ ]} dzm-I (z - zoynf(z) . = 2), (5*) PROOF The Laurent series of f(z) converging near Zo (except at where b1n '* O. The residue wanted is b l. Zo itself) is (Sec. 16.2) Multiplying both sides by (z - zoyn gives • SEC. 16.3 715 Residue Integration Method We see that hI is now the coefficient of the power (z - ;::0)',,-1 of the power series of = (z - ::'o),"f(;::). Hence Taylor's theorem (Sec. 15.4) gives (5): g(;::) hI = (m - l)! g'm-ll (:0) d"'-I dzm-I (m - l)! 1 E X AMP L E 4 - - [(7 - 7 )'''f(7)] "" ~o '.' • Residue at a Pole of Higher Order f(::;) = (;: + + 2::;2 - 7::; + 4) has a pole of second order at ;: = 1 because the denominator equals 1)2 (verify!). From (5*) we obtain the residue 50::;/(::;3 4)(;: - = lim -d .~1 d;: (50;: -) z+4 • 200 = - 2 = 8. 5 Several Singularities Inside the Contour. Residue Theorem Residue integration can be extended from the case of a single singularity to the case of several singularities within the contour C. This is the purpose of the residue theorem. The extension is surprisingly simple. THEOREM 1 Residue Theorem Let f(;::) he analytic imide a simple closed path C alld 011 C. except forfillite!y many singular points ::'1, Z2, •.. , z" inside C. Then the integral of f(z.) taken cO/lIlterclockll"ise around C equals 27Ti times the sum of the residues of f(;::) {{f Z.I, ••• , Zk: k (6) fefc::,) d::. = 27Ti 2:: ~~s fez). j~I J c Fig. 370. Residue theorem 716 CHAP. 16 PROOF Laurent Series. Residue Integration We enclose each of the singular points Zj in a circle Cj with radius small enough that those k circles and C are all separated (Fig. 370). Then fez) is analytic in the multiply connected domain D bounded by C and C h . . . , Ck and on the entire boundary of D. From Cauchy's integral theorem we thus have f fez) dz + f fez) dz + f fez) dz + ... + f fez) dz = 0, (7) ~ C ~ ~ the integral along C being taken counterclockwise and the other integrals clockwise (as in Figs. 351 and 352, Sec. 14.2). We take the integrals over C1 , . . . , Ck to the right and compensate the resulting minus sign by reversing the sense of integration. Thus, f fez) dz = f fez) dz + f fez) dz + ... + f fez) dz (8) C C, C2 Ck where all the integrals are now taken counterclockwise. By (1) and (2), f fez) dz = 27Ti j Res fez), ~ = 1, ... , k, z=~ • so that (8) gives (6) and the residue theorem is proved. This important theorem has various applications in connection with complex and real integrals. Let us first consider some complex integrals. (Real integrals follow in the next section.) E X AMP L E 5 Integration by the Residue Theorem. Several Contours Evaluate the following integral counterclockwise around any simple closed path such that (a) 0 and 1 are inside C, (b) 0 is inside, I outside, (c) I is inside, 0 outside, (d) 0 and I are outside. r1,C Solution. d;: - Z ;: The integrand has simple poles at 0 and I, with residues [by (3)] 4-3;: = [4-3;:J --- Res - - Z~O z(z - 1) z - I [Confirm this by (4).] Ans. (a) 21Ti(-4 E X AMP L E 6 4 - 3;: -2-- Z~O + = 4-3z Res -- = z~l z(z - I) -4. [4-3zJ --Z 1) = -61Ti, (b) -81Ti, (c) 21Ti, (d) O. z~l = I. • Another Application of the Residue Theorem Integrate (tan Z)/(Z2 - I) counterclockwise around the circle C: Izl = 312. Solution. tan;:; is not analytic at ±1T/2, ±31T12, ... , but all these points lie outside the contour C. Because of the denominator Z2 - I = (z - 1)(z + I) the given function has simple poles at ± I. We thus obtain from (4) and the residue theorem f tanz -2-C Z - I dz = = (tanz 21Ti Res z~l -2-- Z - tanzl 21Ti ( 2;: z~l 1 + Res tan;: + - = 21Ti tan 1 = 9.7855i. ~an;: z~-l 2;:: ) z - I I ) z~-l • SEC. 16.3 717 Residue Integration Method E X AMP L E 7 Poles and Essential Singularities Evaluate the following integral, where C is the ellipse 9x 2 + 1 (4::. Jc eTrZ z - i = 9 (counterclockwise, sketch it). + ze'n'/Z) 16 d::. 4 Solution. Since::. - 16 = 0 at ±2i and ±2. the first tenn of the integrand has simple poles at ±2i inside C, with residues [by (4); note that e27Ti = 1] Res ze= 1 16 z=-2i and simple poles at ±2, which lie outside C, so that they are of no interest here. The second term of the integrand has an essential singularity at 0, with residue 71'2/2 as obtained from (Izl > 16 + AilS. 271';(-16 - 2 !71' ) = 7r! 71"2 - !)i • = 30.22 Ii by the residue theorem. . : c ; 1. Verify the calculations in Example 3 and find the other residues. 2. Verify the calculations in Example 4 and find the other residue. !3:@ 1 5. cos 4+z z 16. sin z Z2 6. + Z2 - I z 10. z 12. Z4 - l 13. CAS PROJECT. Residue at a Pole. Write a program for calculating the residue at a pole of any order. Use it for solving Probs. 3-8. ~ f c sin 7rZ --4- z dz, C: /z - i/ = 2 dz. c: Izl = tan 7rZ dz. C: Izl = 2 eZ --dz. c: cos "- tc tc e' --dz. fc tan 7rZ -_-3- f 25. f c cos coshz 3iz I = 4.5 C: Iz - il = 1.5 7rZ Z2 - Izl C: Izl = ] dz, dz, C: Iz <. + ~il = I I - 4::. + 6z 2 (Z2 1.4 C: Izl = I coth z dz. 24. RESIDUE INTEGRATION Evaluate (counterclockwise). (Show the details.) 14. 23. 11 = Iz - TTZ fc 22. c: tan 20. Z2 C: Izl = 1 dz, d7 tc 1 lfz sinh !7rz 19. 21. 11. tan f f c 1/3 ;:4 - e tc c 8. sec z 7. cot Z 17. 18. 4.~ 2 f c RESIDUES Find all the singular points and the corresponding residues. (Show the details of your work.) 3. 15. + !){2 0). - z) dz. - 23z + 5 (2z ])2(3z - 1) dz, c c: /z/ = I 30::. 2 C: /z/ = I CHAP. 16 718 16.4 Laurent Series. Residue Integration Residue Integration of Real Integrals It is quite surprising that certain classes of complicated real integrals can be integrated by the residue theorem, as we shall see. Integrals of Rational Functions of cos () and sin () We first consider integrals of the type (1) J = f 2 ... e, sin e) de F(cos o where F(cos e, sin 6) is a real rational function of cos e and sin e [for example, (sin2 e)/(5 - 4 cos e)] and is finite (doe5. not become infinite) on the interval of integration. Setting eill = z, we obtain (z + +) (2) (z - +) Since F is rational in cos e and sin e, Eq. (2) shows that F is now a rational function of ;;;, say, f(.:). Since d;;;lde = ieill, we have de = cl.:/i;:. and the given integral takes the form = J. f(::) ~.: Jc IZ J (3) and, as e ranges from 0 to 27T in (I), the vaIiable z = eil! ranges counterclockwise once around the unit circle Izi = 1. (Review Sec. 13.5 if necessary.) E X AMP L ElAn Integral of the Type (1) Show b} the pre,ent method that Solution. J,-271" o We use cos fJ = ~(: de V2 - cos + 1/:) and de = e = 17T. = d:/i::.. f Then the integral becomes d::. i c __ (_2- 7V2 -+ 2 .. I) ~ = - 2 J. i Jc d- (::. - V2 - 1)(;: - V2 + 1) . We see that the integrand has a simple pole at ::'1 = V2 + I outside the unit circle C. so that it is of no interest here. and another simple pole at::2 = '\ '2 - I (where::. - V2 + I = 0) inside C with residue [by (3), Sec. 16.3] z~~~ (~- V2 - I):Z - V2 + I) = [ Z - ~ - I 1~V'2-1 2 Answer: 27Ti( -2/i)( -1/2) = 27T. (Here -21i is the factor in front of the last integral.) • SEC. 16.4 719 Residue Integration of Real Integrals As anOlher large class, let us consider real integrals of the form IX f(x) dx:. (4) -x Such an integral, whose interval of integration is not finite is called an improper integral, and i( has the meaning I (5') X -x f(x) dx = lim a_-:c I 0 f(x) dx: a + b-----')ox lim f b f(x) dt. 0 If both limits exist, we may couple (he (wo independent passages I (5) DO f(x) dx = lim R-----')occ -00 I and (0 - 0 0 x. and write R f(x) dx. -R The limit in (5) is called the Cauchy principal value of the integral. It is written I= pr. v. f(x) dx. -x It may exist even if the limits in (5') do not. EXllmple: lim R~x I R x dol = lim R~x -R (R2 2 R2) - - = 0, 2 but lim b_x f b 0 xdx = x. We assume that the function f(t) in (4) is a real rational function whose denominator is different from zero for all real x and is of degree at least (Wo units higher than the degree of (he numerator. Then the limits in (5') exist. and we may start from (5). We consider the corresponding contour integral f (5*) fez) d::. c around a path C in Fig. 371. Since .f(x) is rational, fez) has finitely many poles in the upper half-plane, and if we choose R large enough, then C encloses ali these poles. By the residue theorem we then obtain fc f(:o d::. f = f(::.) d::. + s I R f(x) dx = 27Ti 2: Res f(::.) -R Yj _L,T\ -R Fig. 371. I R x] Path C of the contour integral in (5*) 720 CHAP. 16 Laurent Series. Residue Integration where the sum consists of all the residues of f(z) at the points in the upper half-plane at which f(z) has a pole. From this we have I (6) R f(x) dx -R = 27Ti ~ Res f(z) - I f(z) dz. S We prove that, jf R --') x, the value of the integral over the semicircle S approaches zero. If we set.: = Rei/!, then S is represented by R = const, and as z ranges along S. the variable ranges from 0 to 7T. Since. by assumption, the degree of the denominator of f(z) is at least two units higher than the degree of the numerator, we have e k (Izl = R > If(z)1 < Izl2 Ro) for sufficiently large constants k and Ro. By the ML-inequality in Sec. 14.1, II I s f(z) dz k < '2 R TTR k7T =- R Hence, as R approaches infinity. the value of the integral over S approaches zero. and (5) and (6) yield the result {>O f(x) dx = 27Ti ~ Res f(z) (7) -00 where we sum over all the residues of f(::.) at the poles of f(z) in the upper half-plane. E X AMP L E 2 An Improper Integral from 0 to 00 Using (7), show that d-.: oe J o 1+ x4 7T = 2\12 . y x Fig. 372. Solutioll. Indeed. fez) = 11(1 + Zl = e ....iJ4~ has four simple poles at the poims Imake a sketch) Z4) "7 Example 2 _ "'2 - e 3wiJ4 , The first two of these poles lie in the upper half-plane (Fig. 372). From (4) in the last section we find the residues SEC. 16.4 721 Residue Integration of Real Integrals . [ Res fez) = Z~Zl Res fez) = [ Z~Z2 1 (1 + (1 + 4 , ] Z~Zl z) 1 4' ] z) Z~Z2 [ = 1 ] _ 1 -3"';/4 -_ - -1 e rri/4. - - e 4z: Z~Zl 4 4 ~ [~J = 4z Z~Z2 = e- 97Ti/ 4 = ..!.. 4 ..!.. 4 e- wi/ 4 • (Here we used e"'; = -I and e -2"'; = 1.) By (1) in Sec. 13.6 and (7) in this section, f cc dr: 27Ti '/4 '/4 27Ti 7T 7T -ro 1 + x4 = - 4 (e= - e--m ) = - 4 '2i'sin '4 = 7Tsin '4 = 7T V2 . Since 1/(1 + x 4) is an even function, we thus obtain, as asserted, • 7T 2V2 . Fourier Integrals The method of evaluating (4) by creating a closed contour (Fig. 371) and "blowing it up" extends to integrals (8) fro f(x) cos sx dx fro f(x) sin sx and tb (s real) -cc -00 as they occur in connection with the Fourier integral (Sec. 11.7). If f(x) is a rational function satisfying the assumption on the degree as for (4), we may consider the corresponding integral f c fez) e isz (s real and positive) dz over the contour C in Fig. 371 on p. 719. Instead of (7) we now get f=-rof(x)e (9) isx dx = 27Ti ~ Res [f(z)e isZ ] (s > 0) where we sum the residues of f(z)e isz at its poles in the upper half-plane. Equating the real and the imaginary parts on both sides of (9), we have foo f(x) cos sx dx = -27T ~ 1m Res [f(z)eisZ ], -00 (10) (s> 0) fro f(x) sin sx d>o: -co = 27T ~ Re Res [f(z)e isZ ]. To establish (9), we must show [as for (4)] that the value of the integral over the semicircle S in Fig. 371 approaches 0 as R -7 00. Now s > 0 and S lies in the upper half-plane y ~ O. Hence (s> 0, y ~ 0). From this we obtain the inequality /f(z)e isz / = /f(z)//e isz / ~ /f(z)/ (s > 0, y ~ 0). This reduces our present problem to that for (4). Continuing as before gives (9) and (10). • 722 E X AMP L E 3 CHAP.16 Laurent Series. Residue Integration An Application of (10) I"" Show that -x cos SX -2 - 2 k +x d~= 7T e - I -ks k cc SIn.1X - 2 - - 2 d~=O -0:; k +x (S > O. k In fact, e isz/(k 2 + ;:2) has only one pole in the upper half·plane, namely. a simple pole at:: and from (4) in Sec. 16.3 we obtain Solution. e isz Res - - z~ik k 2 + Z2 [ = e isz ] - z~ik 2:: > 0). = ik. e -ks = -.- . 2fk Thus I isx :x: -cc Since eisx = -ks e . e J... 2 + x 2 d-.: = 27Tf 2ik -ks 7T = k e . cos sx + ; sin nc. this yield_ the above results lsee also (5) in Sec. 11.7.] • Another Kind of Improper Integral We consider an improper integral I dx B (11) f(x) A whose mtegrand becomes infinite at a point a in the interval of integration. = lim If(x)1 x __ a 00. By definition. this integral (11) means B (12) I f(x) d"l: B Q-E = A lim E_O I f(x) dx + lim I f(x) dx ~-O A a+~ where both E and TJ approach zero independently and through positive values. It may happen that neither of these two limits exists if E and TJ go to 0 independently, but the limit lim (13) E_O [IO-;(x) dx + IB f(x) dX] A a+E exists. This is called the Cauchy principal value of the integral. It is written pro V. I B f(x) dL A For example, pI. V. II {~~ -1 .t = E lim [I- E_O -1 dr + x3 II xdx ] E 3 = 0: the principal value exists, although the integral itself has no meaning. In the case of simple poles on the real axis we shall obtain a formula for the principal value of an integral from -00 to 00. This formula will result from the following theorem. SEC. 16.4 723 Residue Integration of Real Integrals THEOREM 1 Simple Poles on the Real Axis If f(:::;) has a simple poLe at z = a on the real axis, then (Fig. 373) lim 7---+0 I C = 7ri Res fez). fez) dz 2=a 2 0, a-r PROOF a+r a Fig. 373. x Theorem 1 By the definition of a simple pole (Sec. 16.2) the integrand fez) has for 0 < the Laurent series f(:::;) = :::;-a + gC:), bi Iz - al < R = Res fez). 2=a Here g(z) is analytic on the semicircle of integration (Fig. 373) and for all z between C 2 and the x-axis, and thus bounded on C2 , say, Ig(z) I ~ M. By integration. I C2 f(:::;) d:::; = f7i" b!e ireiB dB ore + I g(;;::) d:::; = b I 7ri C2 + I g(:::;) dz. C2 The second integral on the right cannot exceed M7rr in absolute value. by the ML-inequality (Sec. 14.1). and ML = M7rr~ 0 as r~ o. • Figure 374 shows the idea of applying Theorem l to obtain the principal value of the integral of a rational function f(x) from -:lJ to:xl. For sufficiently large R the integral over the entire contour in Fig. 374 has the value J given by 27ri times the sum of the residues of f(:::;) at the singularities in the upper half-plane. We assume that f(x) satisfies the degree Fig. 374. Application of Theorem 1 724 CHAP. 16 laurent Series. Residue Integration condition imposed in connection with (4). Then the value of the integral over the large semicircle S approaches 0 as R ~ x. For r ~ 0 the integral over C2 (clockwise!) approaches the value K = 7Ti Res f(z) - 2=a by Theorem I. Together this shows that the principal value P of the integral from -00 to 00 plus K equals J; hence P = J K = J + 7Ti Resz~a f(z). [f f(z) has several simple poles on the real axis, then K will be -7Ti times the sum of the corresponding residues. Hence the desired formula is pro (14) V. I oc f(x) dx = 27Ti -oc L + Res f(z) 7Ti L Res f(z) where the first sum extends over all poles in the upper half-plane and the second over all poles on the real axis, the latter being simple by assumption. E X AMP L E 4 Poles on the Real Axis Find the principal value d, x pr. Solutioll. V. I_= + (x2 _ 30t 2)(x2 + I) Since x 2 - 3x the integrand f(x), considered for complex z= ~, + 2 = (x - has simple poles at Res f(:d = I, I )(x - 2), z~l 1 [ 2)(~2 + I) (z - ] z~l 2 ' z= 2. Res f(::) = [ z~2 12 1)(:: (:: - ] + I) z~2 I 5 ' :: = Resj(::) = i, z~i [2 (:: - 3.;: 1 + 2)(;: + i) ] z~i 3 - i =6+2;= 20' and ar .: = -; in the lower half-plane, which is of no interest here. From (14) we get the answer pro V. I d, x -x (x2 _ 3x + 2)(x 2 ( + I) = 2wi 3- i) + wi (1 w-"2 + I) 5" W = 10 • More integrals of the kind considered in this section are included in the problem set. Try also your CAS, which may sometimes give you false results on complex integrals. SEC. 16.4 .... -.-.. 1 ...1.1 ....... __ ... 11-81 725 Residue Integration of Real Integrals _-=_ . _....... INTEGRALS INVOLVING COSINE AND SINE Evaluate the following integrals. (Show the details of your work.) 1. 3. r~ o t~ o r dO dO 0 7. f t" o dO o 2 + cos 0 t 23. I oo dO 8 - 2 sin 0 25. I 6. (! t~ o sin 2 0 5 - 4 cos (! x+2 dx -3-- -cc w 0 dO 5 - 4 sin oo x + x 24. I 2 :>0 _:>0 x+5 -!--dx x - I dx -3-- x - x dO 26. cos fJ -----dO. 13 - 12 cos 20 Him. cos 20 = 8. IW IMPROPER INTEGRALS: POLES ON THE REAL AXIS Find the Cauchy principal value (showing details): -00 27< o -'. 37 - 12 cos 0 w 5. 2. 7 + 6 cos 0 123-271 27. CC dx -00 -x-;;4-+-3-x"""2---4 I L: 2 +4cosO dO 17 - 8 cos (! 28. TEAM PROJECT. Comments on Real Integrals. (a) Formula nO) follows from (9). Give the details. 2 (b) Use of auxiliary results. Integrating e- z around IMPROPER INTEGRALS: INFINITE INTERVAL OF INTEGRATION the boundary C of the rectangle with vertices -a, a, + ib, -a + ib, letting a --> co, and using a Evaluate (showing the details): 9. tl. L: I: x 13. 15. I "" I -00 dx x6 18. I I I dx -c""C O--+-)""""2 x2 4 co -x 14. X3 ---8 x2 x +x + + dx I -4-- x + 16. I + x (x 2 I dx 2x - + 5)2 L: -x-'-4-~-x-I-6 L: -(X-;2;-+-1-~-;X"""2-+-9-) dx I cosx = -00 22. I -4-- -x 20. + 12. L: -(X-,.2;:----;-~-+-2-,)2;:x t' ~dx L: -x ell: -"" 1 17. 10. x2 + dx 19. I I "" _::>0 dx cos 4x -x"O;"4-+-5-'x2::--+-4 dx 21. . smx -4-- co -00 X + 1 sin 3x -4-- x + 1 dx dx show that "" L 2 -v:,; e- X cos 2bx dx = - - e o 2 _b 2 . (This integral is needed in heat conduction in Sec. 12.6.) (c) Inspection. Solve Probs. 15 and 21 without calculation. 29. CAS EXPERIMENT. Check your CAS. Find out to what extent your CAS can evaluate integrals of the form (1), (4). and (8) correctly. Do this by comparing the results of direct integration (which may come out false) with those of using residues. 30. CAS EXPERIMENT. Simple Poles on the Real Axis. Experiment with integrals f~co f(x) dx. f(x) = [(x - al)(x - a2) ... (x - ak)r 1. aj real and all different, k > L Conjecture that the principal value of these integrals is O. Try to prove this for a special k, say, k = 3. For general k. 726 CHAP. 16 Laurent Series. Residue Integration = .1 1. Laurent series generalize Taylor senes. Explain the details. S T ION SAN 0 21. c~sh 5z z + k- , C: il = 2 4 2. Can a function have several Laurent series with the same center? Explain. If your answer is yes, give examples. 22. 3. What is the principal part of a Laurent series? Its significance? 23. cot 8;:., C: 1:.::1 = 0.2 4. What is a pole? An essential singularity,? Give examples. 5. What is Picard's theorem? Why did it occur in this chapter? 4z 3 + 7z cos Z z'n ,11 k+ , C: cos ;:. ,:2 sin z 24. 4_2 _ 1 ,C: 25. PRO B L EMS Iz - 11 = 1 11 = 2 I .,., 2 ... ., C-..... H = 6. What is the Riemann sphere? The extended complex plane? Tts significance? 26. 7. Is e lk2 analytic or singular at infinity? cosh;:.? (;:. - 4)3? Explain. 27. 15z +9 , C: Z3 - 9z Iz - 31 28. 15;:. +9 , C: _3 _ 9z Izi Z2 8. What is the residue? Why is it important? 9. State formulas for residues from memory. 10. State some further methods for calculating residues. + ;;.2 _ 1 1 2 + v2 = C: -x 2 . 2;:. = 2 =4 11. What is residue integration? To what kind of complex integrals does it apply? 129- 35 1 12. By what idea can we apply residue integration to real integrals from -x to x,? Give simple examples. Evaluate by the methods of this chapter (showing the details): 13. What is a zero of an analytic function? How are zeros classified? 29. 15. Can the residue at a singular point be O? At a simple pole'? 16. What is a meromorphic function? An entire function? Give examples. COMPLEX INTEGRALS 117-281 30. 17. 31. 32. z c: Izl 18. ~, + i ' C: iz+ Z2 _ i;:. 34. c: 1::1 IOz 20. [: L k- +2 ' C: 2il = 3 /z - 1/ = 3 35. f~ e de k + cos e , k > 1 de 1 - ~ sin sin e e 3 + cos e (l + x 2 )2 x de dx dx ::>0 sin 2;:. 19. 2;:. 25 - 24 cos {7T 0 = 1 de {7T 0 33. -4 ' f7T 0 Integrate counterclockwise around C. (Show the details.) tan ;:. [7T 0 14. What are improper integrals? Cauchy principal values? Give examples. REAL INTEGRALS (1 + 2 X )2 + 2X2 + 4X4 dr 36. Obtain the answer to Prob. 18 in Sec. 16.4 from the present Prob. 35. 727 Summary of Chapter 16 Laurent Series. Residue Integration A Laurent series is a series of the form (Sec. 16.1) (I) or, more briefly written [but this means the same as (1)!] I f(:;;*) on = - - ,( n+l dz* 27Ti Jc (z* - Zo) (1 *) n=-oo where n = 0, ± I, ±2, .... This series converges in an open annulus (ring) A with center Zoo In A the function fez) is analytic. At points not in A it may have singularities. The first series in (1) is a power series. In a given annulus, a Laurent ~eries of fez) is unique. but fez) may have different Laurent series in different annuli with the same center. Of particular importance is the Laurent series (I) that converges in a neighborhood of Zo except at:;;o itself, say, for 0 < Iz - ':01 < R (R > 0, suitable). The series (or finite sum) of the negative powers in this Laurent series is called the principal part of fez) at Zo. The coefficient hI of 1/(:;; - zo) in this series is called the residue of f(::) at Zo and is given by [see (1) and (1 *)] (2) hI = Res fez) Z~Zo = f _1_. f(z*) dz*. 27T1 C Thus ,( f(z"') d.:* = 27Ti Res fez). ~ Z=~ hI can be used for integration as shown in (2) because it can be found from ~~~ fez) = (3) (m ~ I)! !~~o (~:~11 [(z - zo)'''f(:;;)]), (Sec. 16.3), provided f(z.) has at Z.O a pole of order m; by definition this means that that principal part has 1/(z - zo)'n as its highest negative power. Thus for a simple pole (111 = 1), Res fez) = lim (z - ':o)f(z); Z=Zo Z~Zo also, p(.:) Res - - 2~2U q(.:) p(zo) = -,--. q (zo) If the principal part is an infinite series, the singularity of fez) at Zo is called an essential singularity (Sec. 16.2). Section 16.2 also discusses the extended complex plane, that is, the complex plane with an improper point x (,'infinity") attached. Residue integration may also be used to evaluate certain classes of complicated real integrals (Sec. 16.4). CHAPTER 17 Conformal Mapping If a complex function w = f(~) is defined in a domain D of the ~-plane, then to each point in D there corresponds a point in the lI'-plane. In this way we obtain a mapping of D onto the range of values of .f(::;) in the w-plane. We shall see that if f(::;) is an analytic function, then the mapping given by tv = fez) is conformal (angle-preserving), except at points where the derivative (z) is zero. t' Conformality appeared early in history in connection with constructing maps of the globe, which can be conformal (can give directions correctly) or "equiareal" (give areas correctly, except for a scale factor). but cannot have both properties, as can be proved (see [GR8] in App. 1). Conformality is the most important geometric property of analytic functions and gives the possibility of a geometric approach to complex analysis. Indeed, just as in calculus we use curves of real functions y = f(x) for studying "geometric" propelties of functions, in complex analysis we can use conformal mappings for obtaining a deeper understanding of properties of functions, notably of those discussed in Chap. 13. Indeed. we shall first define the concepts of conformal mapping and then consider mappings by those elementary analytic functions in Chap. 13. This is one purpose of this chapter. A second purpose, more important to the engineer and physicist, is the use of conformal mapping in connection with potential problems. In fact, in this chapter and in the next one we shall see that conformal mapping yields a standard method for solving boundary value problems in (two-dimensional) potential theory by transforming a complicated region into a simpler one. Corresponding applications will concern problems from electrostatics, heat flow. and fluid flow. In the last section (17.5) we explain the concept of a Riemann surface, which fits well into the present discussion of "geometric" ideas. Prerequisite: Chap. 13. Sections that may be omitted ill a shorter course: 17.3 and 17.5 References and Answers to Problems: App. I Part D, App. 2. 728 729 SEC. 17.1 Geometry of Analytic Functions: Conformal Mapping 17.1 Geometry of Analytic Functions: Conformal Mapping A complex function (I) w = fez) = u(x. y) + iv(x. y) (z = x + iy) of a complex variable z gives a mapping of its domain of definition D in the complex ~-plane illto the complex w-plane or Ol1to its range of values in that plane. l For any point Zo in D the point Wo = f(zo) is called the image of z{) with respect to f. More generally, for the points of a curve C in D the image points form the image of C; similarly for other point sets in D. Also, instead of the mapping by a fUllction w = f(z) we shall say more briefly the mappillg w = fez). EXAMPLE 1 Mapping w = I(z) = Z2 Using polar forms z ~ ,.eiH and w ~ Rei</>, we have w ~ ;:2 = ,.2e2ifl. Comparing moduli and arguments gives R = ,.2 and cf> = 20. Hence circles r = "0 are mapped onto circles R = "02 and rays 0 = 00 onto rays cf> = 200 , Figure 375 shows this for the region I ~ Izl ~ 3/2. rr/6 ~ 0 ~ TT/3. which is mapped onto the region I ~ Iwl ~ 9/4. TT/3 ~ 0 ~ 2TT13. In Cartesian coordinates we have ::: Hence vertical lines x = obtain y2 = c 2 - u and ~ x + iy and C = COilS! are mapped onto 1I = 2 2 = 4c y2. Together, c2 - y2, V = 2cy. From this we can eliminate y. We v (Fig. 376). ~ These parabolas open to the left. Similarly. hmizuntallines y to the right. k = COliS! are mapped onto parabolas opening (Fig. 376). • v I \ \ I I \ / \ y I \ \ / 2 / / / / x u (z-plane) Fig, 375. Mapping w = (w-plane) Z2. Lines Izi = const, arg z = const and their images in the w-plane IThe general terminology is as follows. A mapping of a set A into a set B is called surjective or a mapping of A onto B if every element of B is the image of at least one element of A. It is called injective or one-to-one ifdi~erent elements ot A have different images in B. Finally, it is called bijective if it is both sUljective and mJeclIve. CHAP. 17 730 Conformal Mapping v y=2 y=l \ -5 / / x=2 Fig. 376. Images of x = const, Y = const under w = Z2 Conformal Mapping A mapping w = fez) is called conformal if it pre~erves angles between oriented curves in magnitude as well as in sense. Figure 377 shows what this means. The angle a (0 ~ a ~ 7T) between two intersecting curves C 1 and C2 is defined to be the angle between their oriented tangents at the intersection point z{). And conformalit), means that the images C 1 * and C2 * of C1 and C2 make the same angle as the curves themselves in both magnitude and direction. THEOREM 1 Conformality of Mapping by Analytic Functions The mapping w = f(:;:') by an a1lalytic jimctioll f is confonnal, except at critical points, that is, poiTlts at which the derivative f I is zero. PROOF n· = :2 has a critical point at z = O. where (see Fig. 375), so that conformality fails. The idea of proof is to consider a curve (2) f' (z) C: z(1) = x(t) = 2z = 0 and the angles are doubled + i)'(t) in the domain of fez) and to show that w = .Hz) rotates all tangents at a point Zo (where 0) through the same angle. Now z(1) = dzldt = .i(1) + i .\i(t) is tangent to C in (2) becau'ie this is the limit of (::1 - zo)/!.lt (which has the direction of the secant 21 - ::0 f' (zo) "* -(z-plane) (w-plane) Fig. 377. Curves C1 and C2 and their respective images ct and C2* under a conformal mapping w = [(z) SEC. 17.1 731 Geometry of Analytic Functions: Conformal Mapping in Fig. 378) as ZI approaches Zo along C. The image C* of C is w = f(z(1)). By the chain rule, Ii· = t' (z(t»z(t). Hence the tangent direction of C* is given by the argument (use (9) in Sec. 13.2) (3) w= arg t' arg + arg z z where arg gives the tangent direction of C. This shows that the mapping rotates all directions at a point Zo in the domain of analyticity of f through the same angle arg f' (:::0), which exists as long as f' (zo) O. But this means conformality, as Fig. 377 illustrates for an angle a between two curves. whose images C1 * and C2 * make the same angle • (because of the rotation). "* I. CurveC Tangent / Fig. 378. Secant and tangent of the curve C In the remainder of this section and in the next ones we shall consider various conformal mappings that are of practical interest, for instance, in modeling potential problems. E X AMP L E 2 Conformality of w = zn . The mapping w = zn, n = 2,3, ... , is conformal, except at Z = 0, where ,/ = llZn-l = O. For n = 2 this is shown in Fig. 375: we see that at 0 the angles are doubled. For general n the angles at 0 are multiplied by a factor II under the mapping. Hence the sector 0 :;" e :;" 'Trln is mapped by ;::n onto the upper half·plane u ~ 0 ~~. x Fig. 379. E X AMP L E 3 Mapping w u Mapping by w = zn = z + 1/z. Joukowski Airfoil In terms of polar coordinates this mapping is tv = tl 1 -t iu = r(cos fI -t i sin fI) -t - (cos fI - i sin fI). r By separating the real and imaginary parts we thus obtain l/ = a cos e, u=bsinfl * where a=r+ Hence circles Izl = r = const I are mapped onto ellipses x 1a onto the segment -2 :;" u :;" 2 of the u-axis. See Fig. 380. 2 2 r r + ilb = 2 l. The circle r = 1 is mapped 732 CHAP. 17 Conformal Mapping v y u Fig. 380. Example 3 Now the derivative of w is (::: + 1)(;:: - 1) _2 which is 0 at Z = ± I. These are the points at which the mapping is not conformal. The two circles in Fig. 381 pas, through z = -I. The larger is mapped onto a Jo"kowski ui/.foi/. The dashed circle passes through both -I and I and is mapped onto a curved segment. Another interesting application of w = Z + liz lthe flow around a cylinder) will be considered in Sec. 18.4. • y , J ' "\ C / II I --0-- x 2 Fig. 381. E X AMP L E 4 Conformality of w = e u Joukowski airfoil Z From (10) in Sec. 13.5 xwe have lezi = eX and Arg::: = y. Hence e Z maps a vertical straight line x = Xo = COllst onto the circle Iwl = e 0 and a horizontal straight line)" = )"0 = CO/1St onto the ray arg ". = )"0. The rectangle in Fig. 382 i, mapped onto a region bounded by circles and rays as shown. The fundamental region -71 < Arg;:: ~ 71 of eZ in the :::-plane is mapped bijectively and conformally onto the entire w-plane without the origin w = 0 (because e Z = 0 for no :::). Figure 383 shows that the upper half o < y ~ 71 of the fundamental region is mapped onto the upper half-plane 0 < arg w ~ 71. the left half being mapped inside the unit disk Iwl ~ 1 and the right half outside (why"!). • J Y~~~~l_ 1 D_____ C 05 _____ . A B oo 1 X Fig. 382. -3 Mapping by w = e Z y 1t o -1 (z-planeJ 0 (w-plane) Fig. 383. Mapping by w = e Z u SEC. 17.1 733 Geometry of Analytic Functions: Conformal Mapping E X AMP L E 5 Principle of Inverse Mapping. Mapping w = Ln z = f-\w) of w = f(::) is obtained by illterchlillging the roles of the z-p/ane and the w-p/ane in the mapping by II' = 1(:;;). Now the principal value w = f(::) = Ln :;; of the natural logarithm has the inverse z = [-1(11") = eW • From Example -=I (with the notations::: aud lI" interchanged!) we know that [-\w) = e W maps the fundamental region of the exponential function onto the :;;-plane without:;; = 0 (becau~e eW 0 for every w). Hence II" = fl:;;) = Ln :;; maps the ::-plane without the origin and cut along the negative real axis (where (j = 1m Ln::: jumps by 21T) Principle. The mapping by the inverse z * confonnally onto the horizontal strip -1T < V :§ 1T of the II"-plane. where w = II + iv. Since the mapping If = Ln::: + 21Ti differs from w = Ln z by the translation 21Ti (veI1icalIy upward). this function maps the z-plane (cut as before and 0 omitted) onto the strip 1T < V :§ 31T. Similarly for each of the infinitely many mappings II' = In:;; = Ln:: :':: 21l1Ti (11 = O. I. 2 .... ). The corresponding horizontal strips of width 21T (images of the :;;-plane under these mappings) together cover the whole w-plane without overlapping. • Magnification Ratio. By the definition of the derivative we have lim If(Z) - .f(zo) (4) Z - Z~Zu 1 = If' (20)1· Zo Therefore, the mapping w = f(:::.) magnifies (or shortens) the lengths of short lines by approximately the factor (zo)l. The image of a small figure conforms to the original figure in the sense that it has approximately the same shape. However, since (z) varies from point to point, a large figure may have an image whose shape is quite different from that of the original figure. More on the Condition f'(z) -=I=- O. From (4) in Sec. 13.4 and the Cauchy-Riemann equations we obtain If' (5') f' au + i au - 12 If ,(z) 12 = 1-. ax ax ( ~1l)2 a.l + (~U)2 all au ax iJy au au ax ay ax all au au au a.l ay ay ax that is, (5) If' (z)1 2 a(lI, u) --- = a(x, y) This determinant is the so-called Jacobian (Sec. 10.3) of the transformation w = f(z) written in real form u = u(x, y), u = u(x, y). Hence (zo) =1= 0 implies that the Jacobian is not 0 at ::0' This condition is sufficient that the mapping w = f(z) in a sufficiently small neighborhood of:::.o is one-to-one or injective (different points have different images). See Ref. [GR4] in App. 1. f' ==== -... = ~. =".-.--==- 1. Verify all calculations in Example I. 14-6/ 2. Why do the images of the curves /;;:1 = COIlsl and arg :: = COllst under a mapping by an analytic function f(:) intersect at right angles, except at points at which f'(:) = O? Find and sketch or graph the image of the given curves under the given mapping. 4. x = I. 2. 3, 4, y = I, 2, 3, 4; w = :2 3. Doe, the mapping w = Z = x - iy preserve angles in size as well as in sense? MAPPING OF CURVES 5. Curves as in Prob. 4, w = iz (Rotation) 6. Izl = 1/3.112. 1,2,3; Arg z = 0, ::'::17/4, ::'::1712, ::'::31712, ::'::17; w = liz CHAP. 17 734 17-151 Conformal Mapping 11?,-23I FAILURE OF CONFORMALITY Find all points at which the following mappings are not conformal. 17. ;:(Z4 - 5) 18. ~2 + 1/;:2 MAPPING OF REGIONS Find and sketch or graph the image of the given region under the given mapping. 7. -7T/4 < Arg z < 7T/4. 8. x ~ 1, \I' = liz Izl < 1/2. w = Z3 19. cos 9. Izl > I, W = 3~ 10. 1m ~ > O. II" = I 11. x ~ 0, y ~ 0, 1;:1 ~ 4; w = Z2 12. -1 ~x~ 1,-7T<Y<7T:H'=ez 13. In 3 < x < In 5, lI" = e' 14. -7T < Y ~ 37T. II" = e Z 15. 2 ~ 1:::1 ~ 3, 7T/4 ~ e ~ 7T/2; w = Ln ~ 21. ;:2 23. = Z4, (b) .Hz) = a)3. (.;::3 - 22. exp (Z5 - 80;:) a)2 MAGNIFICATION RATIO, JACOBIAN Find the magnification ratio M. Describe what it tell, you about the mapping. Where is M equal to I? Find the Jacobian J. 24. w = !Z2 25. w= e Z 26. = 11;::, (c) f(z) Il" 27. w= Lnz = Z3 l/~ 28. 11'= 29. Magnification of Angles. Let f(:::) be analytic at ;:0' Suppose that .f' (zo) = O•... , tlc-IJ(~o) = O. Then the mapping lI" = f(z) magnifies angles with vertex at ~o by a factor k. Illustrate this with examples for k = 2, 3, 4. 11:::2, (d) f(~) = (~ + i)/(l + d. Why do these curves generally intersect at right angles? In your work. experiment to get the best possible graphs. Also do the same for other functions of your own choice. Observe and record shortcomings of your CAS and means to overcome such deficiencies. 17.2 (z - 124-281 16. CAS EXPERI:\IENT. Orthogonal Nets. Graph the orthogonal net of the two families of level curves Re f(z) = COIISt and 1m f(~) = comt, where (a) f(z) 20. cosh 2::: 7T::: + a;: + b 30. Prove the statement in Prob. 29 for general k = I. 2, .... Hint. Use the Taylor series. Linear Fractional Transformations Conformal mappings can help in modeling and solving boundary value problems by first mapping regions confommlly onto another. We shall explain this for standard regions (disks. half-planes, strips) in the next section. For this it is useful to know properties of special basic mappings. Accordingly, let us begin with the following very important class. Linear fractional transformations (or Mobius transformations) are mappings w= (1) a::. cz +b +d (ad - be =1= 0) where a, b, c, d are complex or real numbers. Differentiation gives a(c::. I (2) + d) H" (c::. - c(az + + b) ad - be (c;:: d)2 + d)2 This motivates our requirement ad - be =1= O. It implies conformality for all z and excludes the totally uninteresting case w' == 0 once and for all. Special cases of (I) are tr=::.+b w (3) = a::: w = a::: w = liz with +b (Tmns/atio11S ) lal = 1 (Rotations) (Linear tmmjorl1latiolls) (Inversion in the IInit circle). SEC 17.2 735 Linear Fractional Transformations E X AMP L E 1 = l/z (Fig. 384) Properties of the Inversion w In polar forms z = rew and w = Rei" the inversion .~ Rel~ I I. e-'w = -.- = retE} 11' = liz. is R= and gives r ,. <1>= -e. Hence the unit circle k:1 = r = I is mapped onto the unit circle Iwl = R = I; II" = i<b = e -ill. For a general ~. the image w = 1/;: can be found geometrically by marking Iwl = R = IIr on the segment from 0 to ;: and then reflecting the mark in the real axis. (MaJ-.e a sketch.) Figure 384 shows that w = 1/z maps horizontal and vertical straight lines onto circles or straight lines. Even the following is true. n' = 1/;: maps el'en' stmight lille or circle Dllto {/ circle Dr straight lille. v y I I I L ---L...l .--T I I I 1 +--+--1---.1-----1 -2 L -1 I I 1 I I x 2 ---+-+ +-I-+_---"I ~-l I I I I Fig. 384. Mapping (Inversion) w = l/z Proof. Every straight line or circle in the ;:-plane can be written (A. B C, [) real). A = 0 gives a straight line and A '* 0 a circle. In terms of z and:': this equation becomes -+" A-::+B~+ C~+D=O. -- Now lI' = 2 2i 11;:. Substitution of:: = IIII' and multiplication by II'W gives the equation w+w w-w A + B - - - + C - - - + [)ww 2 2i = 0 or. in terms of II and v. A + This represents a circle (if D BII - Cv + DC/(2 '* 0) or a straight line (if D = + v2 ) = O. 0) in the w-plane. • The proof in this example suggests the use of z and Z instead of x and y. agelleral prillciple that is often quite useful in practice. Surprisingly, evn)' linear fractional transformation has the properly just proved: THEOREM 1 Circles and Straight Lines Every linear fractional trani/ormation (I) maps the totality of circles and STraighT lines in the z-plane onto the totalitv of circles lind straight lines in the w-plane, 736 CHAP. 17 PROOF Conformal Mapping This is trivial for a translation or rotation. fairly obvious for a uniform expansion or contraction. and true for w = 1/z, as just proved. Hence it also holds for composites of these special mappings. Now comes the key idea of the proof: represent (1) in terms of these special mappings. When e = 0, this is easy. When e =1= 0, the representation is 1 w=K--cz + d + a ad - be K= - - - e where e This can be verified by substituting K. taking the common denominator and simplifying: this yields (1). We can now set WI = ez, and see from the previous formula that then 11' = lV4 + ale. This tells us that (1) is indeed a composite of those special mappings and completes the proof. • Extended Complex Plane The extended complex plane (the complex plane together with the point IX in Sec. 16.2) can now be motivated even more naturally by linear fractional transformations as follows. To each z for which ez + d =1= 0 there conesponds a unique w in (I). Now let e =1= O. Then for z = -dIe we have ez + d = 0, so that no w conesponds to this ;:. This suggests that we let IV = ex.; be the image of z = -dIe. Also, the inverse mapping of (1) is obtained by solving (I) for z; this gives again a linear fractional transformation dw - b z= (4) -ew +a When e =1= 0, then en' - a = 0 for w = ale, and we let ale be the image of;: = 00. With these settings, the linear fractional transformation (1) is now a one-to-one mapping of the extended z-plane onto the extended w-plane. We also ~ay that every linear fractional transformation maps "the extended complex plane in a one-to-one manner onto itself." Our discussion suggests the following. General Remark. If z = 00. then the right side of (1) becomes the meaningless expression (a· oo + b)/(c· 00 + d). We assign to it the value w = ale if e =1= 0 and w = 00 if e = O. Fixed Points Fixed points of a mapping w = .f(z) are points that are mapped onto themselves, are "kept fixed" under the mapping. Thus they are obtained from w = .f(z) = z. The identity mapping w = z has every point as a fixed point. The mapping w = Z has infinitely many fixed points, w = liz has two, a rotation has one. and a translation none in the finite plane. (Find them in each case.) For (1), the fixed-point condition w = z is (5) z= az + b ez +d ' thus ez 2 - (a - d)z - b = O. SEC. 17.3 Special Linear Fractional Transformations 737 This is a quadratic equation in z whose coefficients all vanish if and only if the mapping is the identity mapping w = z (in this case, a = d oF 0, b = c = 0). Hence we have Fixed Points THEOREM 2 fr A linear fractional transformation, not the identity, has at most two fixed points. a linear fi'actional transformation is known to have three or more fixed points, it must be the identity mapping w = z. To make our present general discussion of linear fractional transfonnations even more useful from a practical point of view, we extend it by further facts and typical examples, in the problem set as well as in the next section. -..-.............. . ...... - ........ , 1. Verify the calculations in the proof of Theorem 1. 18-141 2. (Composition ofLFTs) Show that substituting a linear fractional transformation (LFf) into a LFf gives a LFT. Find the fixed points. 3. (Matrices) If you are familiar with 2 X 2 matrices, prove that the coefficient matrices of (1) and (4) are inverses of each other, provided ad - be = I, and that the composition of LFfs corresponds to the multiplication of the coefficient matrices. Find the inverse;: = ;:(w). Check the result by solving ;:(w) for II". 4;: 1\' = H' - +i + -3;;: 7 6. = 81z 10. w = 5 9. w = (4 z + 4i z - I 12. w = - z+ 1 + i)z 11. w = (z - i)2 13. w= 2iz - 1 z+ 2i +2 14. w = - - z- 1 3z INVERSE 14-71 4. 8. w FIXED POINTS + Z - 17.3 i ; 15. Find a LFT whose (only) fixed points are -2 and 2. 16. Find a LFT (not w = z) with fixed points 0 and l. 17. Find all LFfs with fixed points -I and 1. 3;: 5. IV = 7. II' = --- 18. Find all LFfs whose only fixed point is O. 2- - 2;: + 5; 19. Find all LFfs with fixed points 0 and 4z 20. Find all LFfs without fixed points in the finite plane. i 00. Special Linear Fractional Transformations In this section we shall see how to determine linear fractional transformations (I) w= az + eZ b +d (ad - be *- 0) for mapping certain standard domains onto others and how to discuss properties of (I). A mapping (1) is determined by a. b, c, d. actually by the ratios of three of these constants to the fourth because we can drop or introduce a common factor. This makes it plausible that three conditions determine a unique mapping (I): 738 CHAP. 17 THE 0 REM 1 Conformal Mapping Three Points and Their Images Given Three given distinct poil1ts :1' Z2, :::3 can always be mapped onto three prescribed distinct poillts WI> W2, W3 by one, and only Olle, linear fractional transforlllation W = .f(z). This mapping is given implicitly by the equation (2) tV - WI H'2 - H'3 W - W3 W2 - WI (rt" one of these poillts is the point this point must be replaced by l.) PROOF x, the qllotiellt of the two differences cOl/taining Equation (2) is of the form F(w) = G(:) with linear fractional F and G. Hence w = F-\G(z» = .f(z) , where F- 1 is the inverse of F and is linear fractional (see (4) in Sec. 17.2) and so is the composite F- 1(G(z» (by Prob. 21). that is. w = fez) is linear fractional. Now if in (2) we set w = Il·l. W2, W3 on the left and :: = Zl. Z2, :3 on the right. we see that F(Wl) = 0. F(W2) = G(:l) = 0, G(:2) = 1. I, From the first column, F(Wl) = G(':l), thus WI = F-l(G(Zl» = f(Zl)' Similarly, W2 = f(Z2), W3 = f(::3)' This proves the existence of the desired linear fractional transfOimation. To prove uniqueness, let w = g(z) be a linear fractional transformation, which also maps Zj onto Wj' j = 1,2,3. Thus Wj = g(Zj)' Hence g-llw) = Zj. where Wj = f(zj)' Together, g-lUlzj » = Zj' a mapping with the three fixed points Zl, :2, :3' By Theorem 2 in Sec. 17.2, this is the identity mapping, g -l<f(:Z» = z for all z. Thus fl:) = g(z) for all z, the uniqueness. The last statement of Theorem I follows from the General Remark in Sec. 17.2. • Mapping of Standard Domains by Theorem 1 Using Theorem 1. we can now find linear fractional transformations according to the following Principle. Prescribe three boundary points ZI> Z2, Z3 of the domain D in the z-plane. Choose their images WI> W2, tl'3 on the boundary of the image D* of D in the w-plane. Obtain the mapping from (2). Make sure that D is mapped onto D*, not onto its complement. In the latter case, interchange two w-points. (Why does this help?) E X AMP L E 1 Mapping of a Half-Plane onto a Disk (Fig. 385) Find the linear fractional transformation (1) that maps 11'3 = I. respectively. Solutioll. Z1 = -1, Z2 = 0':3 = I onto From (2) we obtain w-(-I) -i-I ~-(-l) 11"-1 -;-(-1) z-I thus z -; w=---iz + 1 . 0-1 0-(-1)' W1 = -I. H'2 = -i, SEC. 17.3 Special Linear Fractional Transformations 739 u ,,-y=o ~ /X=~ x=o I Fig. 385. Linear fractional transformation in Example 1 Let us show that we can determine the specific properties of such a mapping without much calculation. For x we have \I" = (x - il/(-ir + I). thus Iwl = I. so that the x-axis maps onto the unit circle. Since;: = i gives IV = O. the upper half-plane maps onto the interior of that circle and the lower half-plane onto the exterior. ;: = O. i. x go onto \I" = -i. O. i. so that the positive imaginmy axis maps onto the segment S: II = 0, -1 ~ v ~ 1. The vertical lines r = COllst map onto circles (by Theorem 1, Sec. 17.2) through w = i (the image of;: = x) and perpendicular to 111'1 = I (by conformality; see Fig. 385). Similarly. the horizontal lines y = cOllstmap onto circles through lI' = i and perpendicular to S (by conforrnality). Figure 385 gives these circles for y ~ O. and for y < 0 they lie outside the unit disk shown. • : = E X AMP L E 2 Occurrence of 00 Determine the linear fractional tram'[ormation that maps 11'3 = 1. respectively. Solution. ZI = 0, Z2 = 1, <=3 = JO onto 11'1 = -1, 11'2 = -i, From (2) we obtain the desired mapping U" = ;: - i z+i This is sometimes called the Cayley trallsformatioll. 2 In this case. (2) gave at first the quotient (I - x)/(:: - x), which we had to replace by 1. • E X AMP L E 3 Mapping of a Disk onto a Half-Plane Find the linear tractional transformation that maps ::1 = -1, ::2 = i. Z3 = I onto "'I = O. '1'2 = i, "'3 respectively. such that the unit disk is mapped onto the right half-plane. (Sketch disk and half-plane.) Solution. = x, From (2) we obtain. after replacing (i - x)/(\\' - x) by I ;: + 1 \1"=--- ;: - 1 • Mapping half-planes onto half-planes is another task of practical interest. For instance, we may wish to map the upper half-plane y ~ 0 onto the upper half-plane v ~ O. Then the x-axis must be mapped onto the u-axis. 2 ARTHUR CAYLEY (1821-1895). English mathematician and professor at Cambridge. is known for his important work in algebra. matrix theory. and ditferential equations. ~ . 740 E X AMP L E 4 CHAP. 17 Conformal Mapping Mapping of a Half-Plane onto a Half-Plane Find the linear fractional transfonnation that maps = 3/8, respectively. ZI = -2, :2 = 0, ;:3 = 2 onto WI :x:, '1'2 = 1/4, w3 Solution. You may verify that (2) gives the mapping function Z + 1 w=-2z + 4 • What is the image of the x-axis? Of the y-axis? Mappings of disks onto disks is a third class of practical problems. We may readily verify that the unit disk in the z-plane is mapped onto the unit disk in the w-plane by the following function, which maps Zo onto the center w = O. z - Zo w= - - ez - 1 ' (3) = To see this, take Izl I, obtaining, with e e = zo, IZol < I. = Zo as in (3), = Iz - el Iz - zol = Izllz - el = Izz - c zl = 11 - e zl = Ie z - II. Hence Iwl = Iz - Zol/lcz - II = I from (3), so that Izl = 1 maps onto Iwl = 1, as claimed, with Zo going onto O. as the numerator in (3) shows. Formula (3) is illustrated by the following example. Another interesting case will be given in Prob. 10 of Sec. 18.2. E X AMP L E 5 Mapping of the Unit Disk onto the Unit Disk Taking Zo = ~ in (3), we obtain (verify!) 2z - 1 w=--- (Fig. 386). z-2 /: ~ I , I J1o----r-I I : I I ~ v \ \ o~~ i / u 1 x ~\-,~:--..--+!-;;;~ Fig. 386. Mapping in Example 5 • SEC. 17.3 741 Special Linear Fractional Transformations E X AMP L E 6 Mapping of an Angular Region onto the Unit Disk Certain mapping problems can be solved by combining linear fractional transformations with others. For instance, to map the angular region D: -71/6 ~ arg z ~ 71/6 (Fig. 387) onto the unit disk Iwl ~ I, we may map D by Z = Z3 onto the right Z-half-plane and then the latter onto the disk 111'1 ~ I by ~3 _ Z- I w=i Z + I ' I • w=i--~3 + 1 combined ,/ \ 11:/6 (z-plane) (Z-plane) Fig. 387. (w-plane) Mapping in Example 6 This is the end of our discussion of linear fractional transformations. In the next section we tum to conformal mappings by other analytic functions (sine, cosine, etc.). 1. Derive the mapping in Example 2 from (2). 2. (Inverse) Find the inverse of the mapping in Example 1. Show that under that inverse the lines x = COlist are the images of circles in the w-plane with centers on the line v = 1. 3. Verify the formula (3) for disks. 4. Derive the mapping in Example 4 from (2). Find its inverse and prove by calculation that it has the same fixed points as the mapping itself. Is this surprising? 5. (Inverse) If w = f(z) is any transformation that has an inverse, prove the (trivial!) fact that f and its inverse have the same fixed points. 6. CAS EXPERIMENT. Linear Fractional Transformations (LFTs). (a) Graph typical regions (squares, disks, etc.) and their images under the LFTs in Examples 1-5. (b) Make an experimental study of the continuous dependence of LFTs on their coefficients. For instance, change the LFT in Example 4 continuously and graph the changing image of a fixed region (applying animation if available). 7. -1. 0, I onto -0.6 - 0.8i, -1, -0.6. + 0.8i 8. 0, 1,2 onto I,!, ! + 9. 2i. -2;.4 onto -4 2i, -4 - 2i, 0 10. i, -I, I onto -1, -i. i 11. O. I, <Xl onto 00, 1,0 12. 0, -i, i onto -1. 0, 13. 2i, i, 0 onto ~i, 2i, 00 00 14. O. 2i. -2i onto - I, O. 00 15. -1,0, I onto 0, I, -I 16. Find all LFTs w(:.':) that map the x-axis onto the £I-axis. 17. Find a LFT that maps 1:.0:1 ~ I onto Iwl ~ 1 so that z = i/2 is mapped onto w = O. Sketch the images of the lines x = COlist and y = COllst. 18. Find an analytic function that maps the second quadrant of the z-plane onto the interior of the unit circle in the w-plane. LFTs FROM THREE POINTS AND THEIR IMAGES 19. Find an analytic function w = i(z) that maps the region o ~ arg z ~ 71/4 onto the unit disk Iwl ~ 1. Find the LFT that maps the given three points onto the three given points in the respective order. 20. (Composite) Show that the composite of two LFrs is a LFT. /7-15/ 742 17.4 CHAP. 17 Conformal Mapping Conformal Mapping by Other Functions So far we have discussed the mapping by zn, eZ (Sec. 17.1) and linear fractional transformations (Secs. 17.2, 17.3), and we shall now tum to the mapping by trigonometric and hyperbolic analytic functions. v y r rl -2" f--~ f-f-- Rl u " x 2 . I I I (z-plane) (w-plane) Fig. 388. Sine Function. (I) Mapping w = u + iv = sin z Figure 388 shows the mapping by w = u + iv = sin z = sin x cosh y + i cos x sinh y (Sec. 13.6). Hence (2) II = sin x cosh y, v = cos x sinh y. Since sin ~ is periodic with period 21T, the mapping is certainly not one-to-one if we consider it in the full z-plane. We restrict:: to the vertical strip S: -~1T ~ X ~ ~1T in Fig. 388. Since f' (z) = cos z = 0 at z = ±!1T, the mapping is not conformal at these two critical points. We claim that the rectangular net of straight lines x = const and y = const in Fig. 388 is mapped onto a net in the w-plane consisting of hyperbolas (the images of the vertical lines x = const) and ellipses (the images of the horizontal lines y = const) intersecting the hyperbolas at right angles (confOimality!). Corresponding calculations are simple. From (2) and the relations sin 2 x + cos 2 X = I and cosh2 y - sinh2 y = I we obtain (Hyperbolas) (Ellipses). Exceptions are the vertical lines x = ±!1T, which are "folded" onto u ~ - 1 and = 0), respectively. Figure 389 illustrates this further. The upper and lower sides of the rectangle are mapped onto semi-ellipses and the vertical sides onto -cosh I ~ If ~ -I and I ~ u ~ cosh I (v = 0), respectively. An application to a potential problem will be given in Prob. 5 of Sec. 18.2. u ~ I (v SEC. 17.4 743 Conformal Mapping by Other Functions v y 1 C B A D n; n; -2 2 -1 E x B" C* E* F* U F Fig. 389. Mapping by w = sin z The mapping w = cos.: could be discussed independently, but since Cosine Function. (3) = cos Z = sin (z + !7T), w we see at once that this is the same mapping as sin z preceded by a translation to the right through !7T units. Hyperbolic Sine. Since (4) \I" = sinh z = -i sin (i.:), the mapping is a counterclockwise rotation Z = i.: through!7T (i.e .. 90 0 ). followed by the sine mapping Z* = sin Z. followed by a clockwise 90 0 -rotation w = -iZ*. Hyperbolic Cosine. This function (5) w = cosh z = cos (d defines a mapping that is a rotation Z = i.: followed by the mapping tv = cos Z. Figure 390 shows the mapping of a semi-infinite strip onto a half-plane by w = cosh z. Since cosh 0 = I, the point z = 0 is mapped onto w = I. For real.: = x ~ 0, cosh.: is real and increases with increasing x in a monotone fashion. starting from I. Hence the positive x-axis is mapped onto the portion 1I ~ I of the lI-axis. For pure imaginary z = iy we have cosh iy = cos y. Hence the left boundary of the strip is mapped onto the segment I ~ u ~ - I of the lI-axis, the point z = 7Ti conesponding to w = co.,h i7T = cos 7T = -I. On the upper boundary of the strip. y = 7T, and since sin 7T = 0 and cos 7T = -I, it follows that this part of the boundary is mapped onto the portion 1I ~ -1 of the u-axis. Hence the boundary of the strip is mapped onto the lI-axis. It is not difficult to see that the interior of the strip is mapped onto the upper half of the w-plane. and the mapping is one-to-one. This mapping in Fig. 390 has applications in potential theory, as we shall see in Prob. 12 of Sec. 18.3. ~b ~,h; -x Fig. 390. -1 Mapping by w 0 = 1 cosh z U 744 CHAP. 17 Conformal Mapping Tangent Function. Figure 391 shows the mapping of a vertical infinite strip onto the unit circle by w = tan z, accomplished in three steps as suggested by the representation (Sec. 13.6) (e iz - e-iz)/i (e 2iZ - I )/i smz w=tanz= e iz + e- iz cos Z e 2iz + I Hence if we set Z = e 2iz and use lIi = -i, we have (6) w = tan z= - Z-1 W= Z+ iW. I . We now see that w = tan z is a linear fractional transformation preceded by an exponential mapping (see Sec. 17.1) and followed by a clockwise rotation through an angle!7T (90 0 ) . The strip is S: -~7T < X < ~7T, and we show that it is mapped onto the unit disk in the w-plane. Since Z = e 2iz = e-2y+2ix, we see from (10) in Sec. 13.5 that Izi = e- 2y , Arg Z = 2x. Hence the vertical lines x = - 7T/4, 0, 7T/4 are mapped onto the rays Arg Z = -7T!2, 0, 7T/2, respectively. Hence S is mapped onto the right Z-half-plane. Also IZl = e- 2y < 1 if y > 0 and Izi > 1 if y < O. Hence the upper half of S is mapped inside the unit circle Izi = 1 and the lower half of S outside Izi = L as shown in Fig. 391. Now comes the linear fractional transformation in (6), which we denote by g(Z): (7) W = g(Z) Z-l = Z+l For real Z this is real. Hence the real Z-axis is mapped onto the real W-axis. FUl1hermore, the imaginary Z-axis is mapped onto the unit circle IwJ = I because for pure imaginary Z = iY we get from (7) Iwi = Ig(iY)1 = I iY - I iY + 1 I= 1. The right Z-half-plane is mapped inside this unit circle Iwi = 1, not outside, because Z = I has its image g(l) = 0 inside that circle. Finally, the unit circle IZI = 1 is mapped y ,, . I - -,,, I I I \ (z-plane) (Z-plane) Fig. 391. ,, . I \ I x v ,, I '- .' ~ I (W-plane) Mapping by w = tan z - -,,, \ I I I \ \ ,, I '- .' (w-plane) ~ I u SEC. 17.4 745 Conformal Mapping by Other Functions onto the imaginary W-axis. because this circle is Z expression. namely, eie!>/2 _ eicb/2 = eie!>. so that (7) gives a pure imaginary e- icbl2 i sin (¢/2) + e- i e!>/2 cos (¢12) From the W-plane we get to the w-plane simply by a clockwise rotation through nIl; see (6). Together we have shown that w = tan.: maps S: - nl4 < Re z < nl4 onto the unit disk = 1, with the four quarters of S mapped as indicated in Fig. 391. This mapping is conformal and one-to-one. Iwl = 11-71 CONFORMAL MAPPING w e' Find and sketch the image of the given region under w = e 2 • 1. 0;;; x;;; 2, -71';;; y ;;; 71' 2. - I ;;; x ;;; 0, 0 ;;; y ;;; 71'12 3. -0.5 < x < 0.5, 371/4 < Y < 571'/4 4. -3 < x < 3, 71'14 < Y < 371'/4 5. 0 < x < 1. 0 < Y < 71' 6. x < 0, - 71'12 < Y < 71'12 7. x arbitrary, 0 ~ y ~ 2IT 8. CAS EXPERIMENT. Conformal Mapping. If your CAS can do conformal mapping, use it to solve Prob. 5. Then increase y beyond IT, say, to 5071' or 10071. State what you expected. See what you get as the image. Explain. 19-121 CONFORMAL MAPPING w = sinz Find and sketch or graph the image of the given region under w = sin z. 9. 0 ;;; x ;;; 71', 0 ;;; y ~ J 10. 0 < x < 71'/6, y arbitrary 11. 0 < x < 271', J < Y < 5 12. - 71'/4 < x < 71'/4, 0 < Y < 3 13. Determine all points at which H" = sin Z IS not conformal. 14. Find and sketch or graph the images of the lines x = O. ±71'/6, ±71'/3. ±71'/2 under the mapping H" = sin z. 15. Find an analytic function that maps the region R bounded by the positive x- and .v-axes and the hyperbola X)' = 71'12 in the first quadrant onto the upper half-plane. Hint. First map the region onto a horizontal strip. 16. Describe the mapping H" = cosh z in terms of the mapping w = sin z and rotations and translations. 17. Find all points at which the mapping w = cosh 71;;: is not conformal. 118-221 CONFORMAL MAPPING w = cos z Find and sketch or graph the image of the given region under w = cos ;;:. 18. 0 < x < 71'12. 0 < Y < 2 19. 0 < x < 71'. 0 < \. < I 20. - I ~ x ;;; 1. 0 ~ y ~ 21. 71' < x < 271'. y < 0 22. 0 < x < 271. l/2 < Y < 23. Find the images of the lines \' mapping w = cos .<:. zz+ = C = COllst under the I 24. Show that w = Ln - - maps the upper half-plane I onto the horizontal strip 0 the figure. A ~ 1m w ~ 71' BCD ! (=) -1 ) E I 0 as shown in (=) (z-plane) 1[i 6---- c" D*(=J E" = A * B*(=J 6---- o (w-planeJ Problem 24 25. Find and sketch the image of R: 2 ;;; Izl ~ 3, 71'/4 ~ () ~ 71'12 under the mapping w = Ln z. 746 17.5 CHAP. 17 Conformal Mapping Optional Riemann Surfaces. Riemann sUifaces are sUifaces on willch multivalued relations, such as w = v':: or w = In z, become single-valued, that is, functions in the usual sense. We explain the idea. which is simple-but ingenious, one of the greatest in complex analysis. The mapping given by w (I) = u + iv = .:2 (Sec. 17.1) is conformal. except at ;: = 0, where w' = 2.:: = O. At Z = 0, angles are doubled under the mapping. Thus the right .:-half-plane (including the positive y-axis) is mapped onto the full w-plane, cut along the negative half of the £I-axis; this mapping is one-to-one. Similarly for the left z-half-plane (including the negative y-axis). Hence the image of the full .:-plane under w = Z2 "covers the w-plane twice" in the sense that every w *- 0 is the image of two z-points: if ZI is one. the other is -Zl. For example. z = i and -i are both mapped onto w = - 1. Now comes the crucial idea. We place those two copies of the cut w-plane upon each other so that the upper sheet is the image of the right half .::-plane R and the lower sheet is the image of the left half .::-plane L. We join the two sheets crosswise along the cuts (along the negative u-axis) so that if;: moves from R to L. its image can move from the upper to the lower sheet. The two origins are fastened together because w = 0 is the image of just one .::-point, z = o. The surface obtained is called a Riemann surface (Fig. 392a). w = 0 is called a "winding point" or branch point. w = Z2 maps the full z-plane onto tills surface in a one-to-one manner. By interchanging the roles of the variables relation z and w it follows that the double-valued w=~ (2) (Sec. 13.2) becomes single-valued on the Riemann surface in Fig. 392a, that is, a function in the usual Its image is sense. We can let the upper sheet conespond to the principal value of the right w-half-plane. The other sheet is then mapped onto the left w-half-plane. -vz. (a) Riemann surface of Vi Fig. 392. (b) Riemann surface of "Vz Riemann surfaces \YZ Similarly, the triple-valued relation w = becomes single-valued on the three-sheeted Riemann surface in Fig. 392b, which also has a branch point at z = o. Chapter 17 Review Questions and Problems 747 The infinitely many-valued natural logarithm (Sec. 13.7) w = = In ~ + Ln ~ (11 21l'lTi = 0, ±l. ±2.... ) becomes single-valued on a Riemann surface consisting of infinitely many sheets. w = Ln::. corresponds to one of them. This sheet is cut along the negative x-axis and the upper edge of the slit is joined to the lower edge of the next sheet, which con·esponds to the argument 'IT' < () ~ 317. that is, to w = Ln::. + 2ITi. The principal value Ln ~ maps its sheet onto the horizontal strip -17 < v ~ 17. The function l1' = Ln ~ + 2 m maps its sheet onto the neighboring strip 17 < V ~ 317, and so on. The mapping of the points z *- 0 of the Riemann surface onto the points of the w-plane is one-to-one. See also Example 5 in Sec. 17.1. H· = ~. Find the path of the image point U' of a point::. thal moves twice around the unit circle. starting from the initial position::. = 1. two sheets that may be cut along the line segment from I to 2 and joined crosswise. HillT. Introduce polar coordinates:: - I = rleiH1. : - 2 = r 2 e i62 . 1. Consider 2. Show that the Riemann surface of w = ~ consists of 11 sheets and has a branch point at ::. = o. 3. Make a sketch. similar to Fig. 392, of the Riemann surface of ~. 15-101 7. 5 4. Shov. that the Riemann surtace of II· = Y(:: - 1)(: - 2) has branch points at ::. = I and::. = 2 and consi~ts of £ .-- RIEMANN SURFACES Find the branch points and the number of sheets of the Riemann surface. 6. \/(1 - ?)(4 - ::2) 5. \/3: + 5 + V'2: + 8. In (3: - 4i) i 9. e V2 10. W =..IfES T ION SAN 0 PRO B L EMS 1. How did we define the angle of imersection of two oriented curves, and what does it mean to say that a mapping is conformal? 2. At what points is a mapping w = f(::) by an analytic function not confonnal? Gi ve examples. 3. What happens to angles at::o under a mapping w = f(:) if J' (Zo) = 0, f"(::o} = o. f"'(::.o) *- O? 4. What do "surjective." "injective." and "'bijective" mean? 5. What mapping gave the 10ukowski airfoil? 6. What are linear fractional transformations (LFTs)? Why are they important in connection with the extended complex plane? 7. Why did we require that ad - be *- 0 for a LFT? 8. What are fixed points of a mapping? Give examples. 9. Can you remember mapping properties of II· = sin::.? cos:? e Z ? 10. What is a Riemann surface? Why was it imroduced? Explain the simplest example. 111-161 MAPPING w = Z2 Find and sketch the image of the given curve or region under 1V = Z2. 13. Izl = 4.5, larg zl < nl4 12. xy = -4 14.0 < Y < 2 15. ! < x < 16. 117-22/ MAPPING w = l/z 11. Y = -1, y = I 1 1m:: > 0 Find and sketch the image of the gi ven curve or region under w = II::. 17. x = -1 18. Y = 1 CHAP. 17 748 19. Iz - ~I = ~ 21. larg zl < 7T/4 20. 22. Conformal Mapping Izl Izl < ~, )' < 0 < I, x < 0, Y > 0 135-40 1 Fixed Points. Find all fixed points of z+2 35. w = - - 36. Where is the mapping by the given function not conformal? (Give reason.) 23. 5;:7 + 7;:5 24. cosh 2~ 3z + 2 37. u · = - - - i~ + 5 38.11'= - - 25. sin 2~ + cos 2~ 27. exp (;:4 + Z2) 39. "',. = 123-281 129-341 26. cos 71'Z 2 28. z + l/~ (z =/= 0) LINEAR FRACTIONAL TRANSFORMATIONS (LFTs) 29. 0, 1, 2 onto 0, i, 2i, respectively 30. -1. 1, 2 onto O. 2, 312, respectively 31. 1, -1, -i onto I, -I, i, respectively 32. -1, -I, i onto 1 - i, 2, 0, respectively GO. 1 z-I 141-451 (2 + i) z + : + 2i 5~ I z- ; 40. lI' = ~4 +i + ~ - 81 GIVEN REGIONS Find an analytic function II' = .f(z) that maps: 41. The infinite strip 0 < Y < 71'/3 onto the upper half-plane v> O. Find the LFT that maps 33. 0, + ~ FAILURE OF CON FORMALITY - 2i~ 1 \1'= - - - -2 onto O. 1. "". respectively 34. O. i, 2i onto 0, x, 2i 42. The intelior of the unit circle Izl = I onto the exterior of the circle Iw + 11 = 5. 43. The region x > 0, Y > 0, xy < k omo the strip O<v<1. 44. The semi-disk Izl < 1. x > 0 onto the exterior of the unit circle 111'1 = 1. 45. The sector 0 < arg Z < 71'/3 onto the region u < 1. Conformal Mapping A complex function w = f(::.) gives a mapping of its domain of definition in the complex z-plane onto its range of values in the complex no-plane. If fez) is analytic, this mapping is conformal, that is, angle-preserving: the images of any two intersecting curves make the same angle of intersection, in both magnitude and sense, as the curves themselves (Sec. 17.1). Exceptions are the point" at which!' (z) = 0 ("critical points," e.g. z = 0 for w = Z2). For mapping properties of eZ , cos z, sin z, etc. see Secs. 17.1 and I7.4. Linear fractional transformations, also called Mobius tran~formations (1) w= az + b cz + d (Secs. 17.2, 17.3) (ad - bc *- 0) map the extended complex plane (Sec. 17.2) onto itself. They solve the problems of mapping half-planes onto half-planes or disks, and disks onto disks or half-planes. Prescribing the images of three points determines (I) uniquely. Riemann surfaces (Sec. 17.5) consist of several sheets connected at certain points called brallch poil1ls. On them, multi valued relations become single-valued, that is, functions in the usual sense. Examples. For w = Vz we need two sheets (with branch point 0) since this relation is doubly-valued. For no = In::. we need infinitely many sheets since this relation is infinitely many-valued (see Sec. 13.7). 18 / Complex Analysis and Potential Theory Laplaces's equation V2~ = 0 is one of the most important PDEs in engineering mathematics, because it occurs in gravitation (Secs. 9.7, 12.10), electrostatics (Sec. 9.7), steady-state heat conduction (Sec. 12.5), incompressible fluid flow, etc. The theory of solutions of this equation is called potential theory (although "potential" is also used in a more general sense in connection with gradients; see Sec. 9.7). In the "two-dimensional case" when ~ depend~ only on two Cartesian coordinates x and y, Laplace's equation becomes From Sec. 13.4 we know that then its solutions ~ are closely related to complex analytic functions ~ + i '\{t. This relation is the main reason for the importance of complex analysis in physics and engineering. (We use the notation <l> + i'\{t since u + iv will be needed in conformal mapping.) In this chapter we shall consider this connection and its consequences in detail and illustrate it by modeling typical examples from electrostatics (Secs. 18.1. 18.2), heat conduction (Sec. 18.3). and hydrodynamics (Sec. 18.4). This will lead to boundary value problems, some of which involving functions whose mapping properties we have studied in Chap. 17. Further relating to that chapter. in Sec. 18.2 we explain conformal mapping as a method in potential theory. In Sec. 18.5 we derive the important Poisson formula for potentials in a circular disk. Finally, in Sec. 18.6 we show that results on analytic functions can be used to characterize general properties of harmonic functions (solutions of Laplace's equation whose second partial derivatives are continuous). Prerequisite: Chaps. 13, 14. 17. References and Answers to Problems: App. 1 Part D, App. 2. 749 CHAP. 18 750 18.1 Complex Analysis and Potential Theory Electrostatic Fields The electrical force of attraction or repulsion between charged particles is governed by Coulomb's law. This force is the gradient of a function ¢, called the electrostatic potential. At any points free of charges, ¢ is a solution of Laplace's equation The surfaces ¢ = const are called equipotential surfaces. At each point P at which the gradient of ¢ is not the zero vector, it is perpendicular to the surface ¢ = const through P; that is. the electrical force has the direction perpendicular to the equipotential surface. (See also Secs. 9.7 and 12.10.) The problems we shall discuss in this entire chapter are two-dimensional (for the reason just given in the chapter opening), that is, they model physical systems that lie in three-dimensional space (of course!). but are such that the potential <P is independent of one of the space coordinates. so that ¢ depends only on two coordinates. which we call x and y. Then Laplace's equation becomes (1) Equipotential surfaces now appear as equipotential lines (curves) in the xy-plane. Let us illustrate these ideas by a few simple basic examples. E X AMP L E 1 Potential Between Parallel Plates Find the potential If> of the field between two parallel conducting plates extending to infinity (Fig. 393). which are kept at potentials If>l and <1>2. respectively. Soluti01l. From the shape of the plates it follows that <I> depends only on x. and Laplace', equation becomes <1>" = O. By integrating twice we obtain <I> = 0\' + b. where the constants 1I and b are determined by the given boundary values of (I> on the plates. For example. if the plates correspond to x = -I and x = I, the solution is • The equipotential surfaces are parallel planes. x Fig. 393. Potential in Example 1 x Fig. 394. Potential in Example 2 SEC. 18.1 751 Electrostatic Fields E X AMP L E 2 Potential Between Coaxial Cylinders Find the potential <I> between two coaxial conducting cylinders extending to infinity on both ends (Fig. 394) and kept at potentials <1>1 and <1>2' respectively. \1.,2 l, Solution. Here <I> depends only on r = + for reasons of symmetry, and Laplace's equation lIee = 0 [(5). Sec. 12.9] with tlee = 0 and II = <I> becomes r'I)" + <1>' = O. By separating variable~ and integrating we obtain r2lf.,"'r + ru,. + <1>" <1>' r In <1>' = -In r + a. <I> , a =-. r <I> = a In,. + b and a and b are determined by the given values of <l> on the cylinders. Although no infinitely extended conductors exist, the field in our idealiLed conductor will approximate the field in a long finite conductor in that part which • is far away from the two ends of the cylinders. E X AMP L E 3 Potential in an Angular Region Find the potential <I> between the conducting plates in Fig. 395. which are kept at potentials <1>1 <the lower plate) and <1>2' and make an angle a. where 0 < a::;;: 7T. lin the figure we have a = 120 0 = 27T/3.) Y * Arg ~ (: = 1: + iy 0) b constant on rays e = conST. It is harmonic since it is the imaginary part of an analytic function, Ln ~ (Sec. 13.7). Hence the solution is Solution. e = <I>(x. y) = 0+ b Arg<: x with a and b determined from the two boundary conditions (given values on the plates) Fig. 395. Potential in Example 3 e= )' arctan _. x • Complex Potential Let CP(x, y) be hannonic in some domain D and qr(x, y) a hannonic conjugate of cP in D. (See Sec. 13.4. where we wrote 1I and v, now needed in conformal mapping from the next section on: hence the change to cP and qr.) Then (2) F(z) = CP(x, y) + iqr(x, y) is an analytic function of z = x + iy. This function F is called the complex potential corresponding to the real potential CPo Recall from Sec. 13.4 that for given CP, a conjugate qr is uniquely determined except for an additive real constant. Hence we may say the complex potential, without causing misunderstandings. The use of F has two advantages. a technical one and a physical one. Technically. F is easier to handle than real or imaginary parts, in connection with methods of complex analysis. Physically, -qr has a meaning. By conformality, the curves qr = const intersect the equipotential lines cP = const in the xy-plane at right angles [except where F' (z) = 0]. Hence they have the direction of the electrical force and. therefore, are called lines of force. They are the paths of moving charged particles (electrons in an electron microscope, etc.). E X AMP L E 4 Complex Potential In Example I, a conjugate is 'I' = 01'. It follows that the complex potential is F(;:) = a: + b = ax + b + iay, 752 CHAP. 18 Complex Analysis and Potential Theory • and the lines of force are horizontal straight lines y = comt parallel to the t-axis. E X AMP L E 5 Complex Potential In Example 2 we have <1> potential is = a In r + b = a In Izl + h. A conjugate is q, F(z) = a Ln z + = a Arg z. Hence the complex b and the lines of force are straight lines through the origin. F(;:) may also be interpreted as the complex potential of a source line (a wire perpendicular to the xy-plane) whose trace in the xr-plane is the origin. • E X AMP L E 6 Complex Potential In Example 3 we get F(:::) by noting that i Ln ::: = i In kl - F(:;;) = a - ib Ln z = a + Arg:::, multiplying this by -b. and adding a: b Arg::: - ib In We see from this that the lines of force are concentric circles Izi = COllst. Id. Can you sketch them? • Superposition More complicated potentials can often be obtained by superposition. E X AMP L E 7 Potential of a Pair of Source Lines (a Pair of Charged Wires) Determine the potential of a pair of oppositely charged source lines of the same 5trength at the -c on the real axis. point~ - = c and z= Soluti01l. From Examples 2 and 5 it follows that the pOlential of each of the source lines is <1>1 = Kin Iz - cl and <1>2 = -K In Iz + cl, re5pectively. Here the real constant K measures the strength (amount of charge). These are the real parts of the complex potentials and Hence the complex pOlential of the combination of the two source lines is (3) F(z) = F 1 (:::) + F 2 (:::) = K fLn (~ - c) - Ln (::: + c»). The equipotential lines are the curves <1> = Re F(z) = K In I Z Z c + c I= consc. I~7- -+ I thus ... (c.· = ("01lSt. These are circles, as you may show by direct calculation. The lines of force are "IjJ = [m F(z) = K[ Arg (z - c) - Arg (z + c)l = const. We write this briefly (Fig. 396) Now 81 - 82 is the angle between the line segments from:: to c and -c (Fig. 396). Hence the lines of force are the curves along each of which the line segment S: -c ~ t ~ C appears under a consram angle. These curves are the totality of circular arcs over S. as is (or should be) known from elementary geometry. Hence the lines of force are circles. Figure 397 shows some of them together with some equipotential lines. In addition to the interpretation as the potential of two source lines, this potential could also be thought of a~ the potential between two circular cylinders whose axes are parallel but do not coincide. or a5 the potential between two equal cylinders that lie outside each other. or as the potential hetween a cylinder and a plane waH. Explain this, using Fig. 397. • The idea of the complex potential as just explained is the key to a close relation of potential theory to complex analysis and will recur in heat flow and fluid flow. SEC. 18.1 753 Electrostatic Fields z -c c Arguments in Example 7 Fig. 396. ........ =..·'''''''-'·W· Fig. 397. Equipotential lines and lines of force (dashed) in Example 7 =:1- .. - - Z- - •. - •• 11-41 POTENTIAL Find and sketch the potential. Find the complex potential: 1. Between parallel plates at x = - 3 and 3. potentials 140 V and 260 V. respectively 2. Between parallel plates at t = -4 and 10. potentials 4.4 kV and 10 kV. respectively 3. Between the axes (potential 110 V) and the hyperbola xy = I (potential 60 V) 113-151 POTENTIALS FOR OTHER CONFIGURATIONS 13. Show that F(z) = arccos z (defined in Problem Set 13.7) gives the potential in Figs. 398 and 399. 4. Between parallel plates at y = x and x + k. potentials o and 15-81 100 V. respectively COAXIAL CYLINDERS Find the potential between two infinite coaxial cylinders of radii r 1 and rz having potentials VI and V z , respectively. Find the complex potential. 5. rl = 0.5, rz = 2.0, VI = -IIOV. V 2 = llOV 6. 7. 8. 1'1 1'1 1'1 = 1, r2 = 10. VI = 100 V. V 2 = I kV = I, 1"z = 4. VI = 200 V. V2 = 0 = 0.1. 1'2 = 10. VI = 150 V. V 2 = 50 V Fig. 398. Slit 9. Show that <1> = el'Tr = (l/'Tr) arctan (ylx) is harmonic in the upper half-plane and satisfies the boundary condition <1>(x, 0) = I if x < 0 and 0 if x > O. and the cOlTesponding complex potential is F(z) = -(i/'Tr) Ln z. 10. Map the upper half z-plane onto the unit disk Iwl ~ I so that 0, x. - I are mapped onto I, i, -i, respectively. What are the boundary conditions on Iwl = I resulting from the potential in Prob. 9'! What is the potential at w = O? n. Verify by calculation that the equipotential lines in Example 7 are circles. 12. CAS EXPERIMENT. Complex Potentials. Graph the equipotential lines and lines offorce in (a)-(d) (four graphs, Re F(z) and 1m F(z) on the same axes). Then explore further complex potentials of your choice with the purpose of discovering configurations that might be of practical interest. (a) F(z) = :;:2 (b) F(z) = iz 2 (e) F(z) = liz (d) F(:;:) = ifz Fig. 399. Other apertures 14. Find the real and complex potentials in the sector -'Tr16 ~ e ~ 'Tr16 between the boundary e = ± 'Tr16 (kept at 0) and the curve X3 - 3xy2 = I, kept at 110 V. 15. Find the potential in the first quadrant of the x)'-plane between the axes (having potential 220 V) and the hyperbola xy = I (having potential 110 V). CHAP. 18 754 18.2 Complex Analysis and Potential Theory Use of Conformal Mapping. Modeling Complex potentials relate potential theory closely to complex analysis, as we have just seen. Another close relation results from the use of conformal mapping in modeling and solving boundary value problems for the Laplace equation, that is, in finding a solution of the equation in some domain assuming given values on the boundary ("Dirichlet problem"; see also Sec. 12.5). Then conformal mapping is used to map a given domain onto one for which the solution is known or can be found more easily. This solution is then mapped back to the given domain. This is the idea. That it works is due to the fact that harmonic functions remain harmonic under conformal mapping: Harmonic Functions Under Conformal Mapping THEOREM 1 Let <1>* be harmonic in a domain D* il1 the tv-plane. Suppose that H' = u + iv = fez) is analytic in a domain D in the z-plane and maps D cOflformally onto D*. Then the jUllction <1>(x, y) (1) = <1>*(II(X, y), vex, y» is harmollic in D. PROOF E X AMP L E 1 The composite of analytic functions is analytic, as follows from the chain IUle. Hence, taking a harmonic conjugate 1]r*(u, v) of <1>*, as defined in Sec. 13.4, and forming the analytic function F*(w) = <1>*(u, v) + i1]r*(u, v), we conclude that F(z) = F*(J(z» is analytic in D. Hence its real part <1>(x, y) = Re F(z) is hannonic in D. This completes the proof. We mention without proof that if D* is simply connected (Sec. 14.2), then a harmonic conjugate of <1>* exists. Another proof of Theorem 1 without the use of a harmonic • conjugate is given in App. 4. Potential Between Noncoaxial Cylinders Model the electrostatic potential between the cylinders C I : IzI = I and C2 : J;: - 2/51 = 2/5 in Fig. 400. Then give the solution for the case that C I is grounded, VI = 0 V, and C2 has the potential V 2 = 110 v. We map the unit disk Izl = I onto the unit disk Iwl = I in such a way that C2 is mapped onto some cylinder C2 *: Iwl = '-0' By (3), Sec. 17.3, a linear fractional transfonTlation mapping the unit disk onto the unit disk is Solution. z- b u·= - - b::: - 1 (2) x (a) z-plane u (bl w·plane Fig. 400. Example 1 SEC. 18.2 Use of Conformal Mapping. Modeling 755 where we have chosen b = ~o real without restriction. ~o is of no immediate help here because centers of circles do not map onto centers of the images. in general. However. we now have two free constants band 1'0 and shall succeed by imposing two reasonable conditions, namely, that 0 and 4/5 (Fig. 400) should be mapped onto 1'0 and -ro, respectively. This gives by (2) O-b '"0 = 0 _ I = h. 4/5 - b and with this, 415 - ro -ro = 4bl5 - I 4"0/5 - I a quadratic equation in "0 with solutions "0 = 2 (no good because "0 < I) and "0 = 112. Hence our mapping function (2) with b = 112 becomes that in Example 5 of Sec. 17.3, (3) From Example 5 in Sec. 18.1. writing lI" for z we have as the complex potential in the lI"-plane the function a Ln w + k and from this the real potential F*{II') = <1>''(u. v) = Re F*{II') = a In [wi + k. This is our model. We now determine a and k from the boundary conditions. If [wi = I, then <1>* = {/ In I + k = O. hence k = O. If [wi = ro = 112. then <1>" = {/ In (I/2) = 110. hence (/ = 110/ln (l12) = -158.7. Substitution of (3) now gives the desired ~olution in the given domain in the :::-plane F(:::) = F*(f(~» = 1I ")- - I Ln ~- _ 2 . The real potential is <1>{x. y) = Re F(:::) = {/ In I ?, - I ~"_ 2 I . (/ = -158.7. Can we "see" this re,ult? Well. <1>(x. Y) = COIlst if and only if [(2::: - 1)/(::: - 2)[ = COllst. that is. [wi = CUllst by (2) with b = 112. These circles are images of circles in the :::-plane because the inverse of a linear fractional transformation is linear fractional (see (4). Sec. 17.2). and any such mapping maps circles onto circles (or straight lines). by Theorem I in Sec. 17.2. Similarly for the rays arg II' = COllst. Hence the equipotential lines <1>{x, y) = const are circles. and the lines of force are circular arcs (dashed in Fig. 400). These two familie< of curves intersect orthogonally. that is, at right angles. as shown in Fig. 400. • E X AMP L E 1 Potential Between Two Semicircular Plates Model the potential between two semicircular plates PI and P 2 in Fig. 40Ia having potentials -3000 V and 3000 V. respectively. Use Example 3 in Sec. 18.1 and conformal mapping. Solution. Step 1. We map the unit disk in Fig. 40la onto the right half of the II'-plane (Fig. 401b) by using the linear fractional transformation in Example 3, Sec. 17.3: II' = I(:::) = I+z. 1-::: v 2 kV 3 kV - 1 kV o u , -3 kV -_..-'\ -2 kV (a) z-plane (b) w-plane Fig. 401. Example 2 756 CHAP. 18 Complex Analysis and Potential Theory The boundary Izl = I is mapped onto the boundary II = 0 lthe v-axis). with z = -I. i. I going onto w = O. i, "", respectively, and: = -i onto w = - i. Hence the upper semicircle of Izl = I is mapped onto the upper half, and the lower semicircle onto the lower half of the v-axis. so that the boundary conditions in the w-plane are as indicated in Fig. 401b. Step 2. We determine the potential <1>*(1/, v) in the right half-plane of the w-plane. Example 3 in Sec. 18.1 with a = 7T, VI = -3000. and V 2 = 3000 [with <1>*(11. v) instead of <1>(x. y)] yields 6000 v cp = arctan - . <1>*(u, v) = - - cpo 7T 11 On the positive half of the imaginary axis (cp = 7T12), this equals 3000 and on the negative half -3000. as it should be. <1>' is the real part of the complex potential " F'(lI') = - 6000 i --- 7T Ln w. Step 3. We substitllle the mapping function into F* to get the complex potential F(::.) in Fig. 40la in the form 6000i 1+;: F(;:) = F*(f(;:» = - - - Ln - - . 'if 1-:;: The real part of this is the potential we wanted to determine: 6000 I+z <1>(x, y) = Re F(;:) = - - 1m Ln - 7T I - ;: 6000 - - A rn 71 0 I+z 1 -:: -- As in Example I we conclude that the equipotential lines <1>(x. y) = const are circular arcs because they correspond to Arg [(I + ::.)/(1 - ;:)] = COIISt. hence to Arg w = COliS/' Also, Arg w = COIISt are rays trom 0 to x, the images of Z = -I and;: = I. respectively. Hence the equipotential lines all have -I and I (the points where the boundary potential jumps) as their endpoints lFig. 401a). The lines of fonce are circular aln, too, and since they must be orthogonal to the equipotential lines, their centers can be obtained as intersections of tangents to the unit circle with the ,,-axis, (Explain!) • Further examples can easily be constructed. Just take any mapping w = J(:) in Chap. 17. a domain D in the z-plane, its image D* in the w-plane. and a potential <1>* in D~'. Then (1) gives a potential in D. Make up some examples of your own. involving, for instance. linear fractional transformations. Basic Comment on Modeling We formulated the examples in this section as models on the electrostatic potential. It is quite important to realize that this is accidental. We could equally well have phrased everything in terms of (time-independent) heat flow; then instead of voltages we would have had temperatures, the equipotential lines would have become isotherms (= lines of constant temperature), and the lines of the electrical force would have become lines along which heat flows from higher to lower temperatures (more on this in the next section). Or we could have talked about fluid flow; then the electrostatic lines of force would have become streamlines (more on this in Sec. 18.4). What we again see here is the unifyillg power of mathematics: different phenomena and systems from different areas in physics having the same types of model can be treated by the same mathematical methods. What differs from area to area is just the kinds of problems that are of practical interest. SEC 18.3 757 Heat Problems =:.:I:_=.: ..... 'A~.====: .... 1. Verify Theorem 1 for <1>':'(11. U) = 112 - U 2 • Z W = fez) = e and any domain D. 2. Verify Theorem I for <1>*(lI. u) = lIU, W = .Hz) = e Z • and D: x ::::2 0,0::::2 Y ::::2 7T, Sketch D and D*. 3. Carry out all steps of the second proof of Theorem I (given in App. 4) in detaiL 4. Derive (3) from (2). 5. Let D'~ be the image of the rectangle D: o ~ x ::::2 ~ 7T, 0 ::::2 Y ::::2 1 under IV = sin z, and <1>*(11, U) = 112 - u 2 • Find the corresponding potential <1> in D and its boundary values. 6. What happens in Prob. 5 if you replace the potential by the conjugate <1>* = 2ltu? Sketch or graph some of the equipotential lines <1> = canst. 7. CAS PROJECT. Graphing Potential Fields. (a) Graph equipotential lines in Probs. 1 and 2. (b) Graph equipOtential lines if the complex potential is F(z) = i;:.2. F(z) = e Z • F(;:.) = ie z , F(z) = eiz . (c) Graph equipotential surfaces corresponding to F(z) = In z as cylinders in space. 8. TEAM PROJECT. Noncoaxial Cylinders. Find the potential between the cylinders C 1 : Izl = I (potential VI = 0) and C2 : Iz - cl = c (V2 = 110 V), where o < c < ~. Sketch or graph the equipotential curves and their orthogonal trajectories for c = 0.1, 0.2, 0.3, OA. Try to think of the further extension C 1 : Izi = 1, C2 : I::: - cI = p *- c. 9. Find the potential <1> in the region R in the first quadrant of the z-plane bounded by the axes (having potential VI) and the hyperbola y = l/x (having potential 0) in two ways. (i) directly, (ii) by mapping R onto a suitable infinite strip, 18.3 10. (Extension of Example 2) Find the linear fractional transfonnation z = g(Z) that maps IZI ::::2 I onto Izi ~ 1 with Z = il2 being mapped onto Z = O. Show that ZI = 0.6 + 0.8i is mapped onto z = -I and Z2 = -0.6 + 0.8i onto::: = I, so that the equipotential lines of Example 2 look in Izi ::::2 1 as shown in Fig. 402. x Fig. 402. Problem 10 11. The equipotential lines in Prob. 10 are circles. Why? 12. Show that in Example 2 the .v-axis is mapped onto the unit circle in the w-plane. 13. Find the complex and real potentials in the upper half-plane with boundary values 0 if x < 4 and 10 kV if x > 4 on the x-axis. 14. (Angular region) Applying a suitable conformal mapping. obtain from Fig. 401 b the potential <1> in the angular region -!7T < Arg z < !7T such that <l> = - 3 kV if Arg z = -!7T and <1> = 3 kV if Arg z = !7T. 15. At z = ± 1 in Fig. 401 a the tangents to the equipotential lines shown make equal angles (7T/6). Why? Heat Problems Laplace's equation also governs heat flow problems that are steady, that is, time-independent. Indeed, heat conduction in a body of homogeneous material is modeled by the heat equation where the function T is temperature, Tt = aT/at, t is time. and c 2 is a positive constant (depending on the material of the body; see Sec. 12.5). Hence if a problem is steady, so that Tt = O. and two-dimensional, then the heat equation reduces to the two-dimensional Laplace equation (1) so that the problem can be treated by our present methods. 758 CHAP. 18 Complex Analysis and Potential Theory T(x, y) is called the heat potential. It is the real part of the complex heat potential F(::.) = T(x. y) + i'llF(x. y). The curves T(x, y) = cOllsl are called isotherms (= lines of constant temperature) and the curves 'IlF(x, y) = const heat flow lines, because along them, heat flows from higher to lower temperatures. It follows that all the examples considered so far (Secs. 18.1. 18.2) can now be reinterpreted as problems on heat now. The electrostatic equipotential lines <I>(x, Y) = COIlSt now become isotherms T(x, y) = C01lst, and the lines of electrical force become lines of heat flow, as in the following two problems. E X AMP L E 1 Temperature Between Parallel Plates Find the temperature between two parallel plates x = 0 and x = d in Fig. 403 having temperatures 0 and 100°e. respectively. Solution. As in Example I of Sec. 18.1 we conclude that T(x, .1') = ax + b. From the boundary conditions, h = 0 and a = 1001d. The answer is T(x.yl IOU = d \" [0C]. The corresponding complex potential is F(:::) = (1 001d) z. Heat tlows hori70ntally in the negative x-direction along the lines y = cOllsr. • E X AMP L E 1 Temperature Distribution Between a Wire and a Cylinder Find the temperature field around a long thin wire of radius /'1 = I mm that is electrically heated to Tl = 500°F and is sUlTounded by a circular cylinder of radius 1'2 = 100 mm. which is kept at temperature T2 = 60°F by cooling it with air. See Fig. 404. (The wire is at the origin of the coordinme system.) Solution. T depends only on r. for reason~ of symmetry. Hence. as in Sec. 18.1 (Example 2). T(x, .1') = a In r + h. The boundary conditions are Tl = 500 = a In 1 Hence b = 500 (since In 1 = 0) and a = + b, T2 = 60 = a In 100 + b. (60 - b)/ln JOO = -95.54. The answer is T(x, y) = 500 - 95.54 In I' [OF]. The isotherms are concentric circles. Heal flows from the wire radially outward to the cylinder. Sketch T as a • function of r. Does it look physically reasonable? YI Insulated I I --f®J-, , x u~-- ~\ ~-I,X hl~ I Fig. 403. Example 1 Fig. 404. Example 2 o T=50°C Fig. 405. 1 Example 3 x SEC. 18.3 759 Heat Problems Mathematically the calculations remain the same in the transition to another field of application. Physically, new problems may arise, with boundary conditions that would make no sense physically or would be of no practical interest. This is illustrated by the next two examples. E X AMP L E 3 A Mixed Boundary Value Problem Find the temperature distribution in the region in Fig. 405 (cross section of a solid quaI1er-cylinder), whose vertical pOI1ion of the boundary is at 20De, the horizontal pOI1ion at 50 De, and the circular portion is insulated. Solution. The insulated portion of the boundary must be a heat flow line, since by the insulation, heat is prevented from crossing such a curve, hence heat must flow along the curve. Thus the isotherms must meet such a curve at right angles. Since T is constant along an isotherm, this means that aT (2) all =0 along an insulated poI1ion of the boundary. Here aT/iin is the nonnal derivative of T, that is, the directional derivative (Sec. 9.7) in the direction normal (perpendicular) to the insulated boundary. Such a problem in which Tis prescribed on one portion of the boundary and aT/an on the other portion is called a mixed boundary value problem. In our case, the normal direction to the insulated circular boundary curve is the radial direction toward the origin. Hence (2) becomes aT/ilr = 0, meaning that along this curve the ~olution must not depend on r. Now Arg;: = 0 satisfies (1). as well as this condition, and is constant (0 and r./2) on the straight portions of the boundary. Hence the solution is of the form T(x, y) The boundary conditions yield a . r./2 +h aO + b. = = 20 and a.0 + b T(x, y) 60 50 - 0, = = 50. This gives 0= arctan r. y x The isotherms are portions of rays 0 = COllSt. Heat flows from the x-axis along circles r = const (daShed in Fig. 405) Lo the y-axis. • y! ;;~\t~ Zf t"lt =~ . u 0 '"II "Eo, T = O°C -1 "---Insulated 1 T = 20°C x " 2 (a) z-plane 2" u (b) w-plane Fig. 406. E X AMP L E 4 "---Insulated Example 4 Another Mixed Boundary Value Problem in Heat Conduction Find the temperature field in the upper half-plane when the x-axis is kept at T for -I < x < I, and is kept at T = 20 e for x> I (Fig.406a). = oDe for x < -I, is insulated 0 Solution. We map the half-plane in Fig. 406a onto the veI1ical strip in Fig. 406b. find the temperature T*(u. v) there, and map it back to get the temperature T(x. y) in the half-plane. The idea of using that strip is suggested by Fig. 388 in Sec. 17.4 with the roles of;: = x + iy and w = u + iv interchanged. The figure shows that z = sin w maps our present strip onto our half-plane in Fig. 406a. Hence the inverse function w = I(:::) = arcsin::: 760 CHAP. 18 Complex Analysis and Potential Theory maps that half-plane onto the strip in the w-plane. This is the mapping function that we need according to Theorem I in Sec. 18.2. The insulated segment -] < x < ] on the x-axis maps onto the segment -wl2 < II < wl2 on the lI-axi~. The rest of the x-axis maps onto the two vertical boundary portions 1I = - wl2 and w/2, u > 0, of the strip. This gives the transformed boundary conditions in Fig. 406b for T*(u. u). where on the insulated hori70ntal boundary, iJT*/iJ/I = iJT*/iJu = 0 because u is a coordinate normal to that ~egment. Similarly to Example I we obtain 20 T*(u. u) = 10 + - 1/ which satisfies all the boundary conditions. This is the real part of the complex potential F"(I\') Hence the complex potential in the ~-plane is F(~) = F*<.f(~» = 10 + 20 w 10 + (20/w)w. = arcsin ~ and T[x, y) = Re F(z) is the solution. The isotherms are 1/ = const in the strip and the hyperbolas in the :.-plane. perpendicular to which heat flows along the dashed ellipses from the 20°-portion to the cooler 0°-portion of the boundary. a physically very reasonable result. • This section and the last one show the usefulness of conformal mappings and complex potentials. The latter will also playa role in the next section on fluid flow. 1. CAS PROJECT. Isotherms. Graph isothenns and lines of heat flow in Examples 2-4. Can you see from the graphs where the heat flow is very rapid? 2. Find the temperature and the complex potential in an infinite plate with edges y = x - 2 and y = x + 2 kept at - 10°C and 20 De, respectively. 3. Find the temperature between two parallel plates \" = 0 and )" = d kept at temperatures ODC and 100De, respectively. (i) Proceed directly. (ii) Use Example I and a suitable mapping. 4. Find the temperature T in the sector 0 ~ Arg z ~ w/3, Izl ~ 1 if T = 20°C on the x-axis, T = 50°C on y = V3 x, and the curved portion is insulated. 5. Find the temperature in Fig. 405 if T = -20 DC on the y-axis, T = 100DC on the x-axis, and the circular portion of the boundary is insulated as before. 6. Interpret Prob. 10 in Sec. 18.2 as a heat flow problem (with boundary temperatures, say, 20 DC and 300°C). Along what curves does the heat flow? 7. Find the temperature and the complex potential in the first quadrant of the ;:-plane if the y-axis is kept at 100°C, the segment 0 < x < I of the x-axis is insulated and the portion x > I of the x-axis is kept at 200 De. Hint. Use Example 4. 8. TEAM PROJECT. Piecewise Constant Boundary Temperatures. (a) A basic building block is shown in Fig. 407. Find the corresponding temperature and complex potential in the upper half-plane. (b) Conformal mapping. What temperature in the first quadrant of the ;:-plane is obtained from la) by the mapping w = a + Z2 and what are the transfonned boundary conditions? (c) Superposition. Find the temperature T'" and the complex potential F* in the upper half-plane satisfying the boundary condition in Fig. 408. (d) Semi-infinite strip. Applying H' = cosh;: to (c), obtain the solution of the boundary value problem in Fig. 409. v o a Fig. 407. ----~ T*=T 1 U Team Project 8(a) v -1 - - o· T*=O Fig. 408. :1 1 0--- T' =To T*=O u Team Project 8(c) T=O T=Tol 00 Fig. 409. T=O x Team Project 8(d) SEC. 18.4 19-141 761 Fluid Flow 11. TEMPERATURE DISTRIBUTIONS IN PLATES 12.Y~ ~COO~~ Find the temperature T(x, y) in the given thin metal plate whose faces are insulated and whose edges are kept at the indicated temperatures or are insulated as shown. 10. 9. o y 13. oc., VlnSulated ~ ~45D T= lOO°C x 14. :\. ';t T = 50 C""""''''--''''''x D 18.4 Fluid Flow Laplace's equation also plays a basic role in hydrodynamics, in steady nonviscous fluid flow under physical conditions discussed later in this section. In order that methods of complex analysis can be applied, our problems will be two-dimensional, so that the velocity vector V by which the motion of the fluid can be given depends only on two space variables x and y and the motion is the same in all planes parallel to the xy-plane. Then we can use for the velocity vector V a complex function (1) giving the magnitude IVI and direction Arg V of the velocity at each point z = x + iy. Here VI and V2 are the components of the velocity in the x and y directions. V is tangential to the path of the moving particles, called a streamline of the motion (Fig. 410). We show that under suitable assumptions (explained in detail following the examples), for a given flow there exists an analytic function (2) P(z) = <I>(x, y) + i"'l'(x, y), called the complex potential of the flow, such that the streamlines are gIVen by = const, and the velocity vector or, briefly, the velocity is given by "'l'(x, y) (3) V = VI + iV2 Fig. 410. = p' (z) Velocity 762 CHAP. 18 Complex Analysis and Potential Theory where the bar denotes the complex conjugate. \)f is called the stream function. The function ¢ is called the velocity potential. The curves ¢(x. y) = const are called equipotential lines. The velocity vector V is the gradient of ¢; by definition, this means thaI (4) Indeed, for F = ¢ + i\)f, Eq. (4) in Sec. 13.4 is F' = ¢x + second Cauchy-Riemann equation. Together we obtain (3): Furthermore. since F(z) is analytic, ¢ and \)f i\)fx with \)fx = -¢y by the satisfy Laplace's equation (5) Whereas in electrostatics the boundaries (conducting plates) are equipotential lines, in fluid flow the boundaries across which fluid cannot flow must be streamlines. Hence in fluid flow the stream function is of particular importance. Before discussing the conditions for the validity of the statements involving (2)-(5), let us consider two flows of practical interest, so that we first see what is going on from a practical point of view. Further flows are included in the problem set. E X AMP L E 1 Flow Around a Corner The complex potential F(:::) = :::2 = x 2 - y2 + 2ixy models a flow with Equipotential lines <I> = x Streamlines 'It = 2xv = 2 (Hyperbolas) - y2 = COllst (Hyperbolas). const From (3) we obtain the velocity vector v= 2~ = 2(x - iy), that is, The speed (magnitude of the velocity) is The flow may be interpreted as the flow in a channel bounded by the positive coordinates axes and a hyperbola. say, xy = I (Fig. 411). We note that the speed along a streamline S has a minimum at the point P where the • cross section of the channel is large. o Fig. 411. -- x Flow around a corner (Example 1) SEC. 18.4 761 Fluid Flow E X AMP L E 2 Flow Around a Cylinder Consider the complex potential F(::.) = <D(x, y) + 1 jqr(x. y) = ::. + -=- Using the polar form:: = reiO. we obtain F(::.) = re iO + -; e -ill (,. + ~) cos () +i (,. - +) sin (). Hence the 'treamlines are qr(x, y) = (r - ~ ) sin () = COIISt. In particular. '1'(.1. y) = 0 gives r - 11,. = 0 or sin () = O. Hence this streamline consists of the unit circle (r = IIr gives r = 1) and the x-axis «(I = 0 and () = 1T). For large 1::1 the term II: in F(:) is small in absolute value, so that for these ~ the flow is nearly uniform and parallel to the x-axis. Hence we can interpret this as a t10w around a long circulm cylinder of unit radius that is perpendicular to the ::-plane and intersects it in the unit circle Izl = I and whose axis cOlTesponds to ::. = O. The flow has two stagnation points (that is, points at which the velocity V is zero). at;: = :!:: 1. This follows from (3) and , I F (::.) = I - "2' (See Fig. 412,) hence • -- x Fig. 412. Flow around a cylinder (Example 2) Assumptions and Theory Underlying (2)-(5) Complex Potential of a Flow THEOREM 1 If the dOlllaill of flow is simply connected and the flow is irrotatiollal lind i1lcompressible, tben the statements inl'Olving (2)--(5) bold. In particular, then the flow has a complex potential F(;:). which is an analytic function. (Explanation of tenns below.) PROOF We prove this theorem, along with a discussion of basic concepts related to fluid flow. (a) First Assumption: Irrotational. Let C be any smooth curve in the z-plane given by .::(s) = xes) + iy(s), where s is the arc length of C. Let the real variable Vt be the component of the velocity V tangent to C (Fig. 413). Then the value of the real line integral (6) 764 CHAP. 18 Complex Analysis and Potential Theory y x Fig. 413. Tangential component of the velocity with respect to a curve C taken along C in the sense of increasing s is called the circulation of the fluid along C, a name that will be motivated as we proceed in this proof. Dividing the circulation by the length of C, we obtain the mean velocit:/ of the flow along the curve C. Now Vt Ivi cos a = (Fig. 413). Hence Vt is the dot product (Sec. 9.2) of V and the tangent vector dzlds of C (Sec. 17.1); thus in (6), Vt ds = (VI -dxds dY) ds = VI dx + V + V2 - 2 ds dy. The circulation (6) along C now becomes JV ds J (7) c = t c (VI dx + V2 dy). As the next idea, let C be a closed curve satisfying the assumption as in Green's theorem (Sec. 10.4), and let C be the boundary of a simply connected domain D. Suppose further that V has continuous partial derivatives in a domain containing D and C. Then we can use Green's theorem to represent the circulation around C by a double integral. fc (8) (VI d"r + V 2 dy) = lj (da: 2 iJ : - l a ) dx dy. The integrand of this double integral is called the vorticity of the flow. The vorticity di vided by 2 is called the rotation w(x, y) = -1 (9) 2 1 DefillitiollS: b J ±J ~a (aV2 ax - -aVI) . ay b f(x) dr = mean value of f on the interval a ~ x ~ b, a f(s) ds = mean value of f on C (L = length of C). f(x, y) dr dy = mean value of f on D (A = area of D). c ±JJ D SEC. 18.4 765 Fluid Flow We assume the flow to be irrotational, that is, w(x, y) aVI (10) ay = == 0 throughout the flow; thus, o. To understand the physical meaning of vorticity and rotation, take for C in (8) a circle. Let r be the radius of C. Then the circulation divided by the length 27Tr of C is the mean velocity of the fluid along C. Hence by dividing this by r we obtain the mean angular velocity Wo of the fluid about the center of the circle: Wo = -127Tr2 Jf (- - D aV2 ax aVI) dx dy = -1-.- 7Tr2 rly Jf w(x. y) dx dy. D If we now let r - 0, the limit of Wo is the value of w at the center of C. Hence w(x, y) is the limiting angular velocity of a circular element of the fluid as the circle shrinks to the point (x, y). Roughly speaking, if (l spherical element of the fluid were suddenly solidified and the surrou1lding fluid simultaneously annihilated, the elemellt would rotate WiTh the angular velocity w. (b) Secolld Assumption: Incompressible. Our second assumption is that the fluid is incompressible. (Fluids include liquids, which are incompressible, and gases, such as air, which are compressible.) Then (11) in every region that is free of sources or sinks, that is, points at which fluid is produced or disappears. respectively. The expression in (11) is called the divergence of V and is denoted by div V. (See also (7) in Sec. 9.8.) (c) Complex Velocity Potelltial. If the domain D of the flow is simply connected (Sec. 14.2) and the flow is irrotational, then (10) implies that the line integral (7) is independent of path in D (by Theorem 3 in Sec. lO.2, where FI = VI, F2 = V2, F3 = 0, and.: is the third coordinate in space and has nothing to do with our present z). Hence if we integrate from a fixed point (a, b) in D to a variable point (x. y) in D, the integral becomes a function of the point (x, y), say, el>(x. y): lx, yJ (12) (Nx. y) = J (VI dx + V2 dy). (a, b) We claim that the flow has a velocity potential el>, which is given by (12). To prove this. all we have to do is to show that (4) holds. Now since the integral (7) is independent of path. VI dt + V2 dy is exact (Sec. lO.2). namely, the differential of el>. that is. VI dx + V2 dy ael> = - ax dx ael> + - ay dy. From this we see that VI = ael>/ax and V2 = ael>/ay, which gives (4). That el> is harmonic follows at once by substituting (4) into (11), which gives the first Laplace equation in (5). 766 CHAP. 18 Complex Analysis and Potential Theory We finally take a harmonic conjugate 'l! of <D. Then the other equation in (5) holds. Also, since the second partial derivatives of ¢ and 'l! are continuous, we see that the complex function F(:;) = ¢(x. y) + i'l!(x. y) is analytic in D. Since the curves 'l!(x, y) = const are perpendicular to the equipotential curves ¢(x, y) = const (except where F' (z) = 0), we conclude that 'l!(x, y) = const are the streamlines. Hence 'l! is the stream function and F(::) is the complex potential of the flow. This completes the proof of Theorem I as well as our discussion of the important • role of complex analysi<; in compressible fluid flow. L ..... _-. _ . . . . -ill . . ..._ ....... lA -: FLOW PATTERNS: STREAMLINES, COMPLEX POTENTIAL 7. What F(z) would be suitable in Example I if the angle of the comer were 7f/3? problems should encourage you to experiment with various functions FC:), many of which model interesting flow patterns. 8. Sketch or graph the streamlines and equipotential lines of F(:;.) = ;:3. Find V. Find all points at which V l~ parallel to the x-axis. 11-151 The~e 1. (Parallel flow J Show that F(~) = -; K~ (K positive real) describes a uniform flow upward. which can be interpreted as a uniform flow between two parallel lines (parallel planes in three-dimensional space). See Fig. 414. Find the velocity vector, the streamlines. and the equipotential lines. 9. Find and graph the streamlines of F(:) = :2 + 2:. Interpret the flow. 10. Show that F(:) = i: 2 models a flow around a comer. Sketch the streamlines and equipotential lines. Find V. = 11. (Potential F(z) lIz) Show that the streamlines of F(~) = l/~ are circles through the origin. 12. (Cylinder) What happens in Example 2 if you replace ::: by Z2? Sketch and interpret the resulting flow in the first quadrant. 13. Change F(~) in Example 2 slightly to obtain a flow around a cylinder of radius 1"0 that gives the flow in Example 2 if 1"0 ~ I. 14. (Aperture) Show that F(:) = arccosh : gives confocal hyperbolas as streamlines, with foci at ::: = ::':: I. and the flow may be interpreted as a flow through an aperture (Fig. 415). x Fig. 414. Parallel flow in Problem 1 2. (Conformal mapping) Obtain the now in Example I from that in Prob. 1 by a suitable conformal mapping. 15. (Elliptical cylinder) Show that F(~) = arccos: gives confocal ellipses as streamlines. with foci at : = ::':: 1. and that the now circulates around an elliptic cylinder or a plate (the segment from -1 to I in Fig. 416). 3. Find the complex potential of a uniform flow parallel to the x-axis in the positive x-direction. 4. What happens to the flow in Prob. I if you replace by ~e-ia with constant 0'. e.g., 0' = 7f/4? z 5. What is the complex potential of an upward parallel flow in the direction of y = 2x? 6. (Extension of Example l) Sketch or graph the flow in Example I on the whole upper half-plane. Show that you can interpret it as as flow against a horizontal wall (the x-axis). Fig. 415. Flow through an aperture in Problem 14 SEC. 18.4 767 Fluid Flow (d) Source and sink combined. Find the complex potentials of a now with a source of strength 1 at ::: = - 0 and of a flow with a sink of strength I at ::: = 1I. Add both and sketch or graph the streamlines. Show that for small 101 these lines look similar to those in Prob. II. Fig. 416. Flow around a plate in Problem 15 (e) Flow with circulation around a cylinder. Add the potential in (b) to that in Example 2. Show that this gives a flow for which the cylinder wall 1::.:1 = I is a streamline. Find the speed and show that the stagnation points are iK 47T x Fig. 417. Point source Fig. 418. Vortex flow 16. TEAM PROJECT. Role of the Natural Logarithm in Modeling Flows. (a) Basic flo\\s: Source and sink. Show that F(:;;) = (cl27T) In:;; with constant positive real c gives a flow directed radially outward (Fig. 417), so that F models a point source at ~. = 0 (that is, a source line x = 0, y = 0 in space) at which t1uid is produced. c is called the strength or discharge of the source. If c is negati ve real. show that the flo,,' is directed radially inward. so that F models a sink at :;; = 0, a point at which fluid disappears. Note that :;; = 0 is the singular point of F(:;;). (b) Basic flows: Vortex. Show that n:;;) = -(Ki/27T) In::: with positive real K gives a t10w circulating counterclockwise around::: = 0 (Fig. 418), :;; = 0 is called a vortex. Note that each time we travel around the vonex. the potential increases by K. (c) Addition of flOl\s. Show that addition of the velocity vectors of two flows gives a flow whose complex potential is obtained by adding the complex potentials of those flows. if K = 0 they are at ± 1; as K increases they move up on the unit circle until they unite at :;; = i (K = 47T, see Fig. 4(9), and if K > 47T they lie on the imaginary axis (one lies in the field of flow and the other one lies inside the cylinder and has no physical meaning). Fig. 419. Flow around a cylinder without circulation (K = 0) and with circulation 768 18.5 CHAP. 18 Complex Analysis and Potential Theory Poisson's Integral Formula for Potentials So far in this chapter we have seen that complex analysis offers powerful methods for modeling and solving two-dimensional potential problems based on conformal mappings and complex potentials. A further method results from complex integration. As a most important result it yields Poisson's integral formula (5) for potentials in a standard domain (a circular disk) and from (5) a useful series (7) for these potentials. Hence we can solve problems for disks and then map solutions conforma\ly onto other domains. Poisson's formula will follow from Cauchy's integral formula (Sec. 14.3) 1 F(::J = - . (I) 27Tt f -F(z*) C z* - ;: d~*. Here C is the circle z* = Reia lcounterclockwise. 0 ~ a ~ 27T), and we assume that F(z*) is analytic in a domain containing C and its full interior. Since dz* = iReia da = iz* da, we obtain from (l) I 2~ F(z) = - ] F(z*) (2) 27T 0 * _z_ z* - Z da Now comes a little trick. If instead of z inside C we take a Z outside C. the integrals (1) and (2) are zero by Cauchy's integral theorem (Sec. 14.2). We choose Z = z* E* /~ = R2/~, which ic; outside C because Izi = R2/lzl = R2/r > R. From (2) we thus have o= -1 27T ]2,,-F(z*) 0 -* da <. Z* - Z I ]2~ 27T 0 = - z* F(z*) - - - - da z*E* z* - - Z and by straightforward simplification of the last expression on the right, o= 1 2~ - - ] F(::*) 27T 0 ~ Z - z* da. We subtract this from (2) and use the following formula that you can verify by direct calculation (zz* cancels): (3) z* z z* - Z .;: - z* .:*z* - z.;: (z* - z)(z* - f) We then have (4) F(.::) 1 ]2~ = -2 7T 0 F(z*) .::*2"* - z.;: (z* - z)(~* - ~) da. From the polar representations of z and z* we see that the quotient in the integrand is real and equal to R2 - 2Rr cos «() - a) + r2 SEC. 18.5 Poisson's Integral Formula for Potentials 769 We now write F(z) = cI>(r, 8) + i'l'(r, 8) and take the real part on both sides of (4). Then we obtain Poisson's integral formula 2 I (5) cI>(,., 8) = 27T R2 - r2 cI>(R, a) R2 _ 2Rr cos (8 _ a) L 27T + ,.2 da. This formula represents the harmonic function cI> in the disk Izl ~ R in terms of its values cI>(R, a) on the boundary (the circle) Izl = R. Formula (5) is still valid ifthe boundary function cI>(R. a) is merely piecewise continuous (as is practically often the case: see Fig. 401 in Sec. 18.2 for an example). Then (5) gives a function harmonic in the open disk. and on the circle Izl = R equal to the given boundary function. except at points where the latter is discontinuous. A proof can be found in Ref. [D1] in App. 1. Series for Potentials in Disks From (5) we may obtain an important series development of cI> in terms of simple harmonic functions. We remember that the quotient in the integrand of (5) was derived from (3). We claim that the right side of (3) is the real pat10f z* + (z* Z + z)(':* z*z* - - z) zz - z*z + ~z* (z* - z)(z* - Z) z* - z Indeed, the last denominator is real and so is z*z* zz m the numerator, whereas -z*z + zz* = 2i 1m (zz*) in the numerator is pure imaginary. This verifies our claim. Now by the use of the geometric series we obtain (develop the denominator) (6) Since z 1 + (z/z*) 1 - (z/z*) z* + z -* <. = ~ ,. rei(J = and z* Re [( ReiD<. z~ TJ = Re [ ; : On the right, cos (118 - na) ,.* + Re ----,. ,.-* we have - - (. 1 einfie-mD< = cos n8 cos na + 2 ~1 Re (z~ + 2 ~1 (6*) (; r + J (; r cos (n8 - na). sin n8 sin na. Hence from (6) we obtain r (cos n8 cos lla + sin n8 sin na). 2SLMEON DENIS POISSON (1781-1840), French mathematician and physicist, professor In Paris from 1809. His work includes potential theory, partial differential equations (Pois~on equation, Sec. 12.1), and probability (Sec. 24.7). 770 CHAP. 18 Complex Analysis and Potential Theory This expression is equal to the quotient in (5), as we have mentioned before, and by inserting it into (5) and integrating term by term with respect to a from 0 to 27T we obtain (7) cI>(r, 8) = + ~1 ao (~ r (lln cos nfl + bn sin nfl) where the coefficients are [the 2 in (6*) cancels the 2 in 1/(27T) in (5)] ao = 27T f 1 2.". an cI>(R, a) da, = 0 7T (8) 1 f = - bn 7T f 27T cI>(R, a) cos lIa da, 0 11 27T = 1,2..... cI>(R. a) sin na da. 0 the Fourier coefficients of cI>(R, a); see Sec. 11.1. Now for r = R the series (7) becomes the Fourier series of cI>(R, a). Hence the representation (7) will be valid whenever the given cI>(R. a) on the boundary can be represented by a Fourier series. E X AMP L E 1 Dirichlet Problem for the Unit Disk Find the electrostatic potential <P(r, 0) in the unit disk r < I having the boundary values <P(l, Solution. { -Cd'7T if al7T if -7T < a < 0 (Fig. 420). 0 < a < = = [- 2 2 -4/(1l 7T ) I~.". -; cos da + L""-; cos da] I 2 - - = l1a (cos 117T 112:2 - 1). = () if 11 = 2.4, ... , and the potential is if 11 is odd. an <P(r 0) = , /la To =! and Since <PO, a) is even. bn = O. and from (8) we obtain ao an ~ Hence. an a) = 4 [ 7T 2 5 ,. cos 0 + -r32 3 cos 38 + -r 52 cos 58 + ... ] Figure 421 shows the unit disk and some of the equipotential lines (curves <P = const). • x -71 Fig. 420. o 71 a Boundary values in Example 1 Fig. 421. Potential in Example 1 _ SEC. 18.6 771 General Properties of Harmonic Functions ..... ---.-!7T 1. Verify (3). 14. TEAM PROJECT. Potential in a Disk. (a) Mean 3. Give the details of the derivation of the series (7) from value property. Show that the value of a harmonic function <P at the center of a circle C equals the mean of the value of <P on C (see Sec. 18.4, footnote I, for definitions of mean values). the Poisson formula (5). 14-131 HARMONIC FUNCTIONS IN A DISK Using (7), find the potential <P(r, 8) in the unit disk r < I having the given boundary values <P(l, 8). Using the sum of the fIrst few terms of the series. compute some values of <P and sketch a figure of the equipotential lines. 4. <P(l, 8) = (b) Separation of variables. Show that the terms of (7) appear as solutions in separating (he Laplace equation in polar coordinates. sin 28 (c) Harmonic conjugate. Find a series for a harmonic conjugate 'It of <P from (7). S. <P(1, e) = 2 sin 2 fJ 6. <P(l, 8) = cos 2 58 7. <P(\, 8) = 8 if (d) Power series. Find a series for F(::.) = <P -7T < 8< 8. <P(I, 8) = 8 if 0 < 8 < 9. <P(I. 8) = sin 3 28 10. <P(l, 8) 11. <P(I, 8) = cos 4 12. <P(L 8) <P(l, 8) = 18.6 = = 8 if -7T < 8< -!7T + i'lt. 15. CAS EXPERIMENT. Series (7). Write a program for series developments (7). Experiment on accuracy by computing values from partial sums and comparing them with values that you obtain from your CAS graph. Do (his (a) for Example I and Fig. 421, (b) for <P in Prob. 8 (which is discontinuous on the boundary!), (c) for a <P of your choice with continuous boundary values. (d) for <P with discontinuous boundary values. 7T 27T 8 2 !7T, 13. <P(L 8) = 8 if < 8< <P(1. 8) = 7T - 8 it!7T < 8 < ~7T 2. Show that every term in (7) is a harmonic function in the disk r < R. 7T !7T. 1 if < 8< 0 it!7T < 8 < ~7T General Properties of Harmonic Functions General properties of hamlOnic functions can often be obtained from properties of analytic functions in a simple fashion. Specifically, important mean value properties of harmonic functions follow readily from those of analytic functions. The details are as follows. THEOREM 1 Mean Value Property of Analytic Functions Let fez) be analytic in a simply connected domain D. Then the value of F(z) at a point Zo in D is equal to the mean value of Hz) on any circle in D with center at zoo PROOF In Cauchy's integral formula (Sec. 14.3) (1) I F(Zo) = - - . 27Tt we choose for C the circle (1) becomes z= (2) F(zo) ::'0 + F(z.) r1 ----=-=dz Z C ";0 re ia in D. Then::. - 1 = -2 7T f 0 z.o = 27T F(::,o + re ia ) dO'. reia, dz = ire ia dO', and 772 CHAP. 18 Complex Analysis and Potential Theory The right side is the mean value of F on the circle (= value of the integral divided by the length 27T of the interval of integration). This proves the theorem. • For harmonic functions, Theorem 1 implies THEOREM 2 Two Mean Value Properties of Harmonic Functions Let <P(x, y) be harmonic in a simply connected d()main D. Then the value of <P(x, y) at a point (xo, Yo) in D is equal to the mean value of <P(x, Y) Oil any circle ill D with center at (xo, Yo). This value is also equal to the mean value of <P(x, y) 0/1 any circular disk in D with center (xo, Yo). [See footnote I in Sec. 18.4.] PROOF The first part of the theorem follows from (2) by taking the real parts on both sides, I <P(xo, Yo) = Re F(xo + (\'0) = -2 7T i 27T <P(xo + r cos 0', Yo + r sin 0') dO'. 0 The second part of the theorem follows by integrating this formula over r from 0 to ro (the radius of the disk) and dividing by r0212. 1 (3) <P(xo, Yo) = --2 7Tro i Lr7T "0 0 0 <P(xo + ,. cos 0', Yo + r sin 0')" dO' dr. The right side is the indicated mean value (integral divided by the area of the region of integration). • Returning to analytic functions, we state and prove another famous consequence of Cauchy's integral formula. The proof is indirect and shows quite a nice idea of applying the ML-inequality. (A bounded regio/1 is a region that lies entirely in some circle about the origin.) THEOREM 3 Maximum Modulus Theorem for Analytic Functions Let F(z) be analytic and nonCOllstant i/1 a dOlllain containing (/ bounded region R and its boundary. Then the absolute value IF(z)1 cannot have a maximum at a/1 il1terior point of R. Consequently, the maximulIl of IF(z)1 is taken on the boundnry of R. If F(z) -=1= 0 in R, the same is true with respect to the minimum l!lIF(z)l. PROOF We assume that IF(z)1 has a maximum at an interior point:::o of R and show that this leads to a contradiction. Let IF(zo)1 = M be this maximum. Since F(z) is not constant, IF(:::)I is not constant, as follows from Example 3 in Sec. 13.4. Consequently. we can find a circle C of radius r with center at zo such that the interior of C is in R and IF(z)1 is smaller than M at some point P of C. Since iF(z) I is continuous, it will be smaller than M on an arc C 1 of C that contains P (see Fig. 422), say, IF(z) I ~ M - k (k > 0) for all z on C1 . SEC. 18.6 General Properties of Harmonic Functions 773 Fig. 422. Proof of Theorem 3 Let C 1 have the length L 1 . Then the complementary arc C2 of C has the length 27TT - L 1 . We now apply the ML-inequality (Sec. 14.1) to (I) and note that Iz - zol = r. We then obtain (using straightforward calculation in the second line of the formula) M = IF(zo) I ;;:: -1 27T IJ -=-=-F(::.) dz I + - 1 IJ -_F(z) d::. I C1 " Zo 27T C2 Z Zo that is, M < M, which is impossible. Hence our a<;sllmption is false and the first statement is proved. Next we prove the second statement. If F(z) *- 0 in R, then lIF(z) is analytic in R. From the statement already proved it follows that the maximum of 11 IF(z) I lies on the boundary of R. But this maximum corresponds to the minimum of IF(z)l. This completes the proof. • This theorem has several fundamental consequences for harmonic functions, as follows. THEOREM 4 Harmonic Functions Let <1>(x, y) be lza17110nic in a domain containing a simply connected bounded region R and its bOllndary curve C. Then: (I) (Maximum principle) If <1>(x, y) is not COli stant, it has neither a maximum nor a minimulIl in R. Consequently, the I1ULyimU111 and the minimulll are taken on the boundary of R. (II) ~f <1>(.1', y) is constant on C, thell <1>(.1", y) is a constant. (III) If h(x, y) is harmonic in R alld on C and h(x, y) = <1>(x, y) everywhere in R. PROOF !f h(x, y) = <1>(x, y) on C, then (I) Let 'IJf(x, y) be a conjugate harmonic function of <1>(x, y) in R. Then the complex function F(::;) = <1>(x, y) + i'lJf(x, y) is analytic in R, and so is G(z) = eF(Z). Its absolute value is IG(;:) I = eRe F(z) = e'Nx. y). From Theorem 3 it follows that IG(z)1 cannot have a maximum at an interior point of R. Since e<P is a monotone increasing function of the real variable <1>, the statement about the CHAP. 18 774 Complex Analysis and Potential Theory maximum of <D follows. From this, the statement about the minimum follows by replacing <1> by -<D. (II) By (I) the function <D(.\, y) takes its maximum and its minimum on C. Thus, if y) is constant on C, its minimum must equal its maximum, so that <D(.\", y) must be a constant. <D(.\", (III) If II and <D are harmonic in R and on C. then h - <D is also harmonic in Rand on C, and by assumption, II - <D = 0 everywhere on C. By (II) we thus have /J - <D = 0 everywhere in R. and (Ill) is proved. • The last statement of Theorem 4 is very important. It means that a IW17110nic jilllction is uniquely determi1led in R by its values Oil the boundary (?f" R. Usually, <1>(.\, y) is required to be harmonic in R and continuous on the boundary of R. that is, lim <D(x, y) = <D(xo, Yo), where (xo, Yo) is on the boundary and (x, y) is in R. X-Xo y~yo Under these assumptions the maximum principle (I) is still applicable. The problem of determining <D(x, y) when the boundary values are given is called the Dirichlet problem for the Laplace equation in two variables, as we know. From (Ill) we thus have, as a highlight of our discussion. THEOREM 5 Uniqueness Theorem for the Dirichlet Problem Iflor a given region and given boulldary mlues the Dirichlet problem for the Laplace equatio/l in two l'ariables has a solutioll. the solution is ullique. ---.. .... -~ ...-.----- kl 2 around the unit circle. Does your result contradict Theorem I? 1. Integrate 12-" 1 VERIFY THEOREM 1 for the given F(::), ::0' and circle of radius l. 2. (::: + 1)3':::0 = 2 3. (::: - 2)2. ::0 = ~ 4. 10::: 15-71 4 '':0 = 0 VERIFY THEOREM 2 for the given cI>(x. y). (xo. Yo) and Circle of radius I. 5. (x - 2)(y - 2), (4. -4) 6. x 2 - .'"2. (3, 8) 7. x 3 - 3xy2, (I, I) 8. Derive Theorem 2 from Poisson's integral formula. 9. CAS EXPERIMENT. Graphing Potentials. Graph the potentials in Probs. 5 and 7 and for three other functions of your choice as smfaces over a rectangle or a disk in the xy-plane. Find the locations of maxima and minima by inspecting these graphs. 10. TEAM PROJECT. Maximum Modulus of Analytic Functions. (a) Verify Theorem 3 for (i) F(:::) = ::2 and the square 4 ~ x ~ 6. 2 ~ Y ~ 4, (ii) F(:::) = e1z and any bounded domain. (iii) F(:::) = sin:: and the unit disk. (b) F(x) = cos x (x real) has a maximum I at O. How does it follow that this cannot be a maximum of IFez.) I = Icos:::1 in a domain containing Z = O? (c) F(::) = 1 + 1::12 is not Lero in the disk Izi ~ 4 and has a minimum at an interior point. Does this contradict Theorem 3? (d) If F(:::) is analytic and not constant in the closed unit disk D: 1<:1 ~ 1 and IF(.;;)I = C = COIlSt on the unit circle. show that F(.:) must have a zero in D. Can you extend this to an arbitrary simple closed curve? 775 Chapter 18 Review Questions and Problems Ill-13 I MAXIMUM MODULUS Find the location and si7e of the maximum of IF(;:)I in the unit disk 1:<:1 ;::; I. 11. F(;:) = ;:2 - 12. F(;:) = a;: 1 + b la, b complex) 13. F(;:) = cos 2;: 14. Verify the maximum principle for <1>(x. y) = eX cos y and the rectangle a ;::; x ;::; b. 0 ;::; y ;::; 2'iT. . . = .-.... •.-. ... : .-====== -- .~ 15. (Conjugate) Do ¢ and a harmonic conjugate \)! of (I) in a region R have their maximum at the same point of R? 16. (Conformal mapping) Find the location (u l . VI) of the maximum of <1>* = ell cos V in R*: Iwl ;::; 1. V ~ 0, where w = 1I + iv. Find the region R that is mapped onto R* by w = f(;:) = ;:2. Find the potential in R resulting from <1>* and the location (xl' .\'1) of the maximum. Is (lib VI) the image of (Xl' YI)? If so, is this ju~t by chance? •• , -:..ll£STIONS AND PROBLEMS 1. Why can potential problems be modeled and solved by complex analysis? For what dimensions? 2. What is a harmonic function? A harmonic conjugate? 3. Give a few example~ of potential problems considered in this chapter. 4. What is a complex potential? What does it give physically'? 5. How can conformal mapping be used in connection with the Dirichlet problem? 6. What heat problems reduce to potential problems? Give a few examples. 7. Write a short essay on potential theory in fluid flow from memory. 8. What is a mixed boundary value problem? Where did it occur? 9. State Poisson's formula and its derivation from Cauchy's formula. 10. State the maximum modulus theorem and mean value theorems for harmonic functions. 11. Find the potential and complex potential between the plates y = x and y = x + 10 kept at 10 V and 110 V, respecti vel y. 12. Find the potential between the cylinders 1:1 = I cm having potential 0 and Id = 10 em having potential 20 kV. 13. Find the complex potential in Prob. 12. 14. Find the equipotential line U = 0 V between the cylinders Id = 0.2S cm and 1:1 = 4 cm kept at -220 V and 220 V. respectively. (Guess first.) 15. Find the potential between the cylinders Izl = 10 cm and 1:::1 = 100 cm kept at the potentials 10 kV and 0, respectively. 16. Find the potential in the angular region between the plates Arg ::: = 'iT16, kept at 8 k V, and Arg ;: = 'iT13, kept dt 6kV. 17. Find the equipotential lines of F(;:) = i Ln z.. 18. Find and sketch the equipotential lines of F(:::) = (l + i)/:::. 19. What is the complex potential in the upper half-plane if the negative half of the x-axis has potential I kV and the positive half is grounded? 20. Find the potential on the ray)' = x, x > 0, and on the positive half of the x-axis if the positive half of the .v-axis is at 1200 V and the negative half IS grounded. 21. Interpret Prob. 20 as a problem in heat conduction. 22. Find the temperature in the upper half-plane if the portion x > 2 of the x-axis is kept at SO"C and the other portion at O°C. 23. Show that the isotherm~ of Fr;:) = hyperbolas. _;:::2 + ;: are 24. If the region between two concentric cylinders of radii 2 cm and 10 cm contains water and the outer cylinder is kept at 20°C, to what temperature must we heat the inner cylinder in order [0 have 30°C at distance 5 cm from the axis? 25. What are the streamlines of Fr:::) = i/;:? 26. What is the complex potential of a flow around a cylinder of radius 4 without circulation? 27. Find the complex potential of a source at ::: = 5. What are the streamlines? 28. Find the temperature in the unit disk 1:1 ;::; I in the form of an infinite series if the left semicircle of Izi = 1 has the temperature of SO°C and the right semicircle has the temperature DoC. 29. Same task as in Prob. 2~ if the upper semicircle is at 40°C and the lower at DoC. 30. Find a series for the potential in the unit disk with boundary values <1>( I, 0) = 0 2 (-'iT < 0 < 'iT). 776 CHAP. 18 Complex Analysis and Potential Theory ...... • ~ ,,~-- Complex Analysis and Potential Theory Potential theory is the theory of solutions of Laplace's equation (1) Solutions who~e second pattial derivatives are continuous are called harmonic functions. Equation (I) is the most important PDE in physics, where it is of interest in two and three dimensions. It appears in electrostatics (Sec. 18.1), steady-state heat problems (Sec. 18.3), fluid flow (Sec. 18.4), gravity, etc. Whereas the three-dimensional case requires other methods (see Chap. 12), two-dimensional potential theory can be handled by complex analysis. since the real and imaginary parts of an analytic function are harmonic (Sec. 13.4). They remain harmonic under conformal mapping (Sec. 18.2), so that conformal mapping becomes a powerful tool in solving boundary value problems for (1), as is illustrated in this chapter. With a real potential <P in (1) we can associate a complex potential (2) F(z) = <P + i\If (Sec. 18.1). Then both families of curves <P = const and \If = COllst have a physical meaning. In electrostatics, they are equipotential lines and lines of electrical force (Sec. 18.1). In heat problems. they are isotherms (curves of constant temperature) and lines of heat flow (Sec. 18.3). In fluid flow, they are equipotential lines of the velocity potential and stream lines (Sec. 18.4). For the disk. the solution of the Dirichlet problem is given by the Poisson formula (Sec. 18.5) or by a series that on the boundary circle becomes the Fourier series of the given boundary values (Sec. 18.5). Hatmonic functions. like analytic functions, have a number of general properties: particularly important are the mean value property and the maximum modulus property (Sec. 18.6), which implies the uniqueness of the solution of the Dirichlet problem (Theorem 5 in Sec. 18.6). , PA RT •• E • •1 , t - .( ., .."" ~ ( Numeric Analysis Software (p. 778-779) CHAPTER 19 Numerics in General CHAPTER 20 Numeric Linear Algebra CHAPTER 21 Numerics for ODEs and PDEs Numeric analysis, more briefly also called numerics, concerns numeric methods. that is, methods for solving problems in terms of numbers or corresponding graphical representations. It also includes the investigation of the range of applicability and of the accuracy and stability of these methods. Typical tasks for numerics are the evaluation of definite integrals. the solution of equations and linear systems, the solution of differential or integral equations for which there are no solution formulas, and the evaluation of experimental data for which we want to obtain. for example. an approximating polynomial. Numeric methods then provide the transition from the mathematical model to an algorithm, which is a detailed stepwise recipe for solving a problem of the indicated kind to be programmed on your computer, using your CAS (computer algebra system) or other software, or on your programmable calculator. In this and the next two chapters we explain and illustrate the most frequently used basic numeric methods in algorithmic form. Chapter 19 concerns numerics in general; Chap. 20 numeric linear algebra. in particular, methods for linear systems and matrix eigenvalue problems; and Chap. 21 numerics for ODEs and PDEs. The algorithms are given in a form that seems best for showing how a method works. We suggest that you also make use of programs from public-domain or commercial software listed on pp. 778-779 or obtainable on the lnternet. 778 PART E Numeric Analysis Numerics has increased in importance to the engineer more than any other field of mathematics owing to the ongoing development of powerful software resulting from great research activity in numerics: new methods are invented, existing methods are improved and adapted. and old methods-impractical in precomputer times-are rediscovered. A main goal in these activities is the development of well-stmctured software. And in largescale work-millions of equations or steps of iteration--even small algorithmic improvements may have a large effect on computing time, storage demand, accuracy, and stability. On average this makes the algorithms used in practice more and more complicated. However, the more sophisticated modem software will become, the more important it will be to understand concepts and algorithms in a basic form that shows original motivating ideas of recent developments. To avoid misunderstandings: Various simple classical methods are still very useful in many routine situations and produce satisfactory results. In other words, not everything has become more sophisticated. Software See also http://www.wiley.com/college/kreyszig/ The following list will help you if you wish to find software. You may also obtain information on known and new software from magazines, such as Byte Magazine or PC Maga::,ille, from articles published by the Americall Mathematical Society (see also their website at www.ams.org).theSocietyforlndustrialandAppliedMathematics(SIAM.at www.siam.org). the Association for COl1lpllting Machillel:r (ACM. at www.acm.org), or the Institute of Electrical and Electronics Engineers (IEEE, at www.ieee.org). Consult also your library, Computer Science Department, or Mathematics Department. Derive. Texas Instntments, Inc .. Dallas, TX. Phone 1-800-842-2737 or (972) 917-8324, website at www.derive.com or www.education.ti.com. EISPACK. See LAPACK. GAMS (Guide to Available Mathematical Software). Website at http://gams.nist.gov. On-line cross-index of software development by NIST. with links to IMSL. NAG, and NETUB. IMSL (International Mathematical and Statistical Library). Visual Numerics, Inc., Housron. TX. Phone \-800-222-4675 or (713) 784-3131. website at www.vni.com. Mathematical and statistical Fortran routines with graphics. LAPACK. Fortran 77 routines for linear algebra. This software package supersedes UNPACK and EISPACK. You can download the routines (see http://cm.bell-labs.comlnetliblbib/minors.html) or order them directly from NAG. The LAPACK User's Guide is available at www.netlib.org. PART E Numeric Analysis 779 LINPACK see LAPACK Maple. Waterloo Maple, Inc., Waterloo, ON, Canada. Phone 1-800-267-6583 or (519) 747-2373, website at www.maplesoft.com. Maple Computer Guide. For Advanced Engineering Mathematics, 9th edition. By E. Kreyszig and E. J. Norminton. J. Wiley and Sons. [nc .. Hoboken. NJ. Phone 1-800-225-5945 or (201) 748-6000. Mathcad. MathSoft, Inc., Cambridge, MA. Phone 1-800-628-4223 or (617) 444-8000. website at www.mathcad.com or www.mathsofLcom. Mathematica. Wolfram Research, Inc., Champaign. IL. Phone 1-800-965-3726 or (217) 3lJ8-0700, website at www.wolframresearch.com. Mathematica Computer Guide. For Advanced Engineering Mathematics. lJth edition. By E. Kreyszig and E. J. Norminton. J. Wiley and Sons. Inc .. Hoboken. NJ. Phone 1-800-225-5945 or (201) 748-6000. Matlab. The MathWorks, Inc., Natick, MA. Phone (508) 647-7000, website at www.mathworks.com. NAG. Numerical Algorithms Group, Inc., Downders Grove. IL. Phone (630) 971-2337, website at www.nag.com. Numeric routines in Fortran 77, Fortran 90, and C. NETLIB. Extensive library of public-domain software. See at www.netIib.org and http://cm.bell-Iabs.comlnetlib/. NIST. National Institute of Standards and Technology, Gaithersburg, MD. Phone (301) 975-2000, website at www.nist.gov. For Mathematical and Computational Science Division phone (301) 975-3800. See also http://math.nist.gov. Numerical Recipes. Cambridge University Press, New York. NY. Phone (212) 924-3900. website at www.us.cambridge.org. Books (also source codes on CD ROM and discettes) containing numeric routines in C, C++. Fortran 77, and F0l1ran 90. To order. call office at West Nyack. NY, at 1-800-872-7423 or (845) 353-7500 or online at www.numerical-recipes.com. FURTHER SOFTWARE IN STATISTICS. See Part G. ~ I :·~f ~.._I , .... ~ "1 . ... . /, -. "«:- CHAPTER 19 i ....J Numerics in General This first chapter on numerics begins with an explanation of some general concepts. such as floating point, roundoff errors, and general numeric errors and their propagation. In Sec. 19.2 we discuss methods for solving equations. Interpolation methods, including splines, follow in Secs. 19.3 and 19.4. The last section (19.5) concerns numeric integration and differentiation. The purpose of this chapter is twofold. First, for all these tasks the student should become familiar with the most basic (but not too complicated) numeric solution methods. These are indispensable for the engineer, because for many problems there is no solution formula (think of a complicated integral or a polynomial of high degree or the interpolation of values obtained by measurements). In other cases a complicated solution formula may exist but may be practically useless. Second. the student should learn to understand some ba<;ic ideas and concepts that are important throughout numerics, such as the practical form of algorithms. the estimation of errors, and the order of convergence. Prerequisite: Elementary calculus References and Answers to Problems: App. J Part E, App. 2 19.1 Introduction Numeric methods are used to solve problems on computers or calculators by numeric calculations, resulting in a table of numbers andlor graphical representations (figures). The steps from a given situation (in engineering, economics. etc.) to the final answer are usually as follows. 1. Modeling. We set up a mathematical model of our problem. such as an integral, a system of equations. or a differential equation. 2. Choosing a numeric method and parameters (e.g., step size), perhaps with a preliminary error estimation. 3. Programming. We u~e the algorithm to write a corresponding program in a CAS, such as Maple, Mathematica, Matlab, or Mathcad, or, say, in Fortran, C, or C++, selecting suitable routines from a software system as needed. 4. Doing the computation. S. Interpreting the results in physical or other terms, also deciding to rerun if further results are needed. 780 SEC. 19.1 Introduction 781 Steps and 2 are related. A slight change of the model may often admit of a more efficient method. To choose methods, we must first get to know them. Chapters 19- 21 contain efficient algorithms for the most important classes of problems OcculTing frequently in practice. In Step 3 the program consists of the given data and a sequence of instructions to be executed by the computer in a certain order for producing the answer in numeric or graphic form. To create a good understanding of the nature of numeric work, we continue in this section with some simple general remarks. Floating-Point Form of Numbers We know that in decimal notation, every real number is represented by a finite or an infinite sequence of decimal digits. Now most computers have two ways of representing numbers, called fixed poillf and floating point. In a fixed-point system all numbers are given with a fixed number of decimals after the decimal point; for example, numbers given with 3 decimals are 62.358, 0.014, 1.000. In a text we would write, say, 3 decimals as 3D. Fixed-point representations are impractical in most scientific computations because of their limited range (explain!) and will not concern us. In a floating-point system we write. for instance. 0.1735' 10- 13 , -0.2000' 10- 1 1.735' 10- 14, - 2.000' 10- 2 . or sometimes also We see that in this system the number of significant digits is kept fixed, whereas the decimal point is "t1oating." Here, a significant digit of a number c is any given digit of c, except possibly for zeros to the left of the first nonzero digit; these zeros serve only to fix the position of the decimal point. (Thus any other zero is a significant digit of c.) For instance, each of the numbers 1360, 1.360, 0.001360 has 4 significant digits. In a text we indicate, say, 4 significant digits, by 4S. The use of exponents permits us to represent very large and very small numbers. Indeed. theoretically any nonzero number a can be written as 0.1 ~ (I) Iml < 1, 11 integer. On the computer, 111 is limited to k digits (e.g., k = 8) and n is limited. giving representations (for finitely many numbers only!) (2) a = :!:.m· 10", These numbers a are often called k-digit decimal macbine numbers. Their fractional part m (or Iii) is called the mamis.m. This has nothing to do with "mantissa" as used for logarithms. 11 is called the exponent of a. 782 CHAP. 19 Numerics in General Underflow and Overflow. The range of exponents that a typical computer can handle is very large. The IEEE (Institute of Electrical and Electronics Engineers) floating-point standard for single precision (the usual number of digits in calculations) is about - 38 < 11 < 38 (about - 125 < n* < 125 for the exponent in binary representations, i.e., representations in base 2). [For so-called double precision it is about - 308 < n < 308 (about -1020 < n* < 1020 for binary).] If in a computation a number outside that range occurs, this is called underflow when the number is smaller and overflow when it is larger. In the case of underflow the result is usually set to zero and computation continues. Overflow causes the computer to halt. Standard codes (by IMSL, NAG, etc.) are written to avoid overflow. Error messages on overflow may then indicate progran1ming errors (incorrect input data, etc.). Roundoff An error is caused by chopping (= discarding all decimals from some decimal on) Or rounding. This error is called roundoff error, regardless of whether we chop or round. The rule for rounding off a number to k decimals is as follows. (The rule for rounding off to k significant digits is the same, with "decimal" replaced by "significant digit.") Roundoff Rule. Discard the (k + 1)th and all subsequent decimals. (a) If the number thus discarded is less than half a unit in the Ath place, leave the kth decimal unchanged (" rounding down"). (b) If it is greater than half a unit in the A1h place, add one to the kth decimal ("rounding lip"). (c) If it is exactly half a unit, round off to the nearest el'en decimal. (Example: Rounding off 3.45 and 3.55 to I decimal gives 3.4 and 3.6, respectively.) The last part of the rule is supposed to ensure that in discarding exactly half a decimal, rounding up and rounding down happens about equally often, on the average. If we round off 1.2535 to 3, 2, I decimals, we get 1.254. 1.25. 1.3, but if 1.25 is rounded off to one decimal, without fmther information, we get 1.2. Chopping is not recommended because the corresponding error can be larger than that in rounding, and is systematic. (Nevertheless, some computers use it because it is simpler and faster. On the other hand, some compurers and calculators improve accuracy of results by doing intermediate calculations using one or more extra digits, called guarding digits.) Error in Rounding. Let a = fl(a) in (2) be the floating-point computer approximation of a in (I) obtained by rounding. where fl suggests floating. Then the roundoff rule gives (by dropping exponents) 1111 - ilil ~ ~. lO- k . Since Iml ~ 0.\, this implies (when a *- 0) (3) II - a a I I 1111 - iii 11/ I ~ I l-k _.\0 2 The right side u = ~. 10 1 - k is called the rounding unit. If we write a = ll( I + 8), we have by algebra (li - a)/a = 8, hence 181 ~ 1I by (3), This sholl"s that the rounding l/17it u is all error bOl/nd in rounding. Rounding errors may ruin a computation completely, even a small computation. In general, these errors become the more dangerous the more arithmetic operations (perhaps several millions!) we have to perform, It is therefore important to analyze computational programs for expected rounding eITors and to find an arrangement of the computations such that the effect of rounding errors is as small as possible. The arithmetic in a computer is not exact either and causes further errors; however, these will not be relevant to our discussion. SEC. 19.1 783 Introduction Accuracy in Tables. Although available software has rendered various tables of function values superfluous, some tables (of higher functions, of coefficients of integration formulas, etc.) will still remain in occasional use. [f a table shows k significant digits, it is conventionally assumed that any value a in the table deviates from the exact value a by at most ±! unit of the kth digit. Algorithm. Stability Numeric methods can be formulated as algorithms. An algorithm is a step-by-step procedure that states a numeric method in a form (a "pseudocode") understandable to humans. (Turn pages to see what algorithms look like.) The algorithm is then used to write a program in a programming language that the computer can understand so that it can execute the numeric method. Important algorithms follow in the next sections. For routine tasks your CAS or some other software system may contain programs that you can use or include as pat1s of larger programs of your own. Stability. To be useful, an algorithm should be stable; that is, small changes in the initial data should cause only small changes in the final results. However, if small changes in the initial data can produce large changes in the final results, we call the algorithm unstable. This "numeric instability," which in most cases can be avoided by choosing a better algorithm, must be distinguished from "mathematical instability" of a problem, which is called "ill-conditiolling, " a concept we discuss in the next section. Some algorithms are stable only for certain initial data, so that one must be careful in such a case. Errors of Numeric Results Final results of computations of unknown quantities generally are approximations; that is, they are not exact but involve errors. Such an error may result from a combination of the following effects. Roundoff errors result from rounding, as discussed on p. 782. Experimental errors are en-ors of given data (probably arising from measurements). Truncating errors result from truncating (prematurely breaking off), for instance, if we replace a Taylor series with the sum of its first few terms. These errors depend on the computational method used and must be dealt with individually for each method. ["Truncating" is sometimes used as a term for chopping off (see before), a terminology that is not recommended.] Formulas for Errors. If a is at1 approximate value of a quantity whose exact value is a, we call the difference (4) E=a-a the error of a. Hence (4*) a = a + E, True value = Approximation + Error. For instance, if a = 10.5 is an approximation of a = 10.], its error is error of an approximation a = 1.60 of a = 1.82 is E = 0.22. E = -0.3. The CHAP. 19 784 Numerics in General CAUTION! In the literature la - al ("ab~olute error") or used as definitions of error. The relative error E,. of is defined by a- a are sometimes also a (5) Er = E a-a Error a a True value (a =f:. 0). This looks useless because a is unknown. But if Itl is much less than a instead of a and get lal. then we can use (5') This still looks problematic hecause E is unknown-if it were known, we could get a = a + E from (4) and we would be done. But what one often can ohtain in practice is an error bound for a, that is, a number f3 such that Itl ~ f3, la - al hence ~ f3. This tells us how far away from our computed a the unknown a can at most lie. Similarly. for the relative error, an en-or bound is a number f3r such that a - a --a- hence I I ~ f3r· Error Propagation This is an imp0l1ant matter. It refers to how en-ors at the beginning and in later steps (roundoff, for example) propagate into the computation and affect accuracy, sometimes very drastically. We state here what happens to en-or bounds. Namely, bounds for the error add under addition and subtraction, whereas bounds for the relative error add under multiplication and division. You do well to keep this in mind. THEOREM 1 Error Propagation (a) In addition and sllbtraction, an error bound for the results is given by the sum of the error bounds for the terms. (b) In multiplication and division, all error bound for the relative error of the results is given (approximately) by the sum of the bounds for the relative errors of the gil'en llumbers. PROOF (a) We use the notations x = E of the difference we obtain Itl = = x + EI' Y = Y+ E2, IEII ~ f31.IE21 ~ f32' Then for the en-or Ix Ix - y - (x - y)1 x - (y - y)1 = h - E21 ~ hi + k21 ~ f31 + f32' SEC. 19.1 785 Introduction The proof for the sum is similar and is left to the student. (b) For the relative error and bounds f3rI, f31'2 IErI = Er of xy we get from the relative errors Ixy - xy I Ixy = xy (x - EI)(Y - E2 ) xY Erl I IElY + E 2X - = and Er2 of X, y EI E21 x)' This proof shows what "approximately" means: we neglected EI E2 as small in absolute value compared to lEI I and IE21. The proof for the quotient is similar but slightly more tricky (see Prob. 15). • Basic Error Principle Every numeric method should be accompanied by an error estimate. [f such a formula is lacking, is extremely complicated, or is impractical because it involves information (for instance, on derivatives) that is not available, the following may help. Error Estimation by Comparison. Do a calculation twice with different accuracy. Regard the difference a2 - al of the results aI, a2 as a (perhaps en/de) estimate of the error EI of the inferior result al' Indeed, al + EI = a2 + E2 by formula (4*). This implies a2 - al = EI - E2 = EI because a2 is generally more accurate than aI, so that IE21 is small compared to IEII. Loss of Significant Digits This means that a result of a calculation has fewer correct digits than the numbers from which it was obtained. This happens if we subtract two numbers of about the same size, for exan1ple. 0.1439 - 0.1426 ("subtractive cancellation"). It may occur in simple problems, but it can be avoided in most cases by simple changes of the algorithm-if one is aware of it! Let us illustrate this with the following basic problem. E X AMP L E 1 Quadratic Equation. Loss of Significant Digits Find the roots of the equation X2 - 40x +2 = 0, using 4 significant digits (abbreviated 4S) in the computation. Solution. A formula for the roots Xl' -"2 of a quadratic equation ax2 + bx +c = 0 is (6) Furthermore, since xlx2 (7) = cIa, another formula for those roots is Xl a~ before. Fro~ (6) we obtain X = 20 :+: V398 = ~O.OO :+: 19.95. This gives Xl = 20.00 + 19.95 = 39.95, involving no dIffIculty, whereas X2 = 20.00 - 19.95 = 0.05 is poor because it involves loss of significant digits. In contrast, .(7) gives Xl = 39.?5, X2 = 2.000/39.95 = 0.05006, in error by less than one unit of the last digit, as a computatIon Wlth more digIts shows. (The lOS-value is 0.05006265674.) 786 CHAP. 19 Numerics in General Comment. To avoid misunderstandings: 4S was used for convenience; (7) is beller than (6) regardless of the number of digits used. For instance. the 8S-computation by (6) is xl = 39.949937. X2 = 0.050063. which i~ poor. and by (7) it is Xl as before. X2 = 2/.\"1 = 0.050062657. In a quadratic equation with real root,. if X2 is absolutely largest (because b > OJ. use (6) for X2 dnd then xl = -_ _-........... c!(m'2)' ... • .... - •••.. - -.... - .. y-~ . . .-. •• - . _". 1. (Floating point) Write 98.17, -100.987, 0.0057869, - 13600 in floating-point form, rounded to 4S (4 significant digits). 2. Write -0.0286403. 11.25845. - 3168\.55 in f1oatingpoint form rounded to 6S. 3. Small differences of large numbers may be particularly strongly affected by rounding errors. l\Iu~trate this by computing 0.36443/(17.862 - 17.798) as given with 5S. then rounding stepwise to 4S. 3S. and 2S. where "stepwise" means: round the rounded numbers. not the given ones. 4. Do the work in Prob. 3 with numbers of your choice that give even more drastically different results. How can you avoid such difficulties? 5. The quotient in Prob. 3 is of the form a/(b - c). Write it as alb + c:)/(b 2 - c 2 ). Compute it first with 5S, then rounding numerator 12.996 and denominator 2.28 stepwise as in Prob. 3. Compare and comment. 6. (Quadratic equation) Solve.\"2 - 20x + I = 0 by (6) and by (7). using 6S in the computation. Compare and comment. 7. Do the computations in Prob. 6 with 4S and 2S. 8. Solve.\"2 + 100x + 2 = 0 by (6) and (7) with 5S and compare. 9. Calculate lIe = 0.367879 (6S) from the partial sums of 5 to 10 terms of the Maclaurin series (a) of e-" with x = 1, (b) of eX with x = 1 and then taking the reciprocal. Which is more accurate? 10. Addition with a fixed number of significant digits depends on the order in which you add the numbers. Illustrate this with an example. Find an empirical rule for the best order. 11. Approximations of 7T = 3.141 592 653 589 79 ... are 22/7 and 355/1 13. Determine the corresponding errors and relative errors to 3 significant digits. 12. Compute 7T by Machin's approximation 16 arctan ( 115) - 4 arctan (1/239) to lOS (which are correct). (In 19X6, D. H. Bailey computed almost 30 million decimals of 7T on a CRA Y-2 in less than 30 hours. The race for more and more decimals is continuing. ) 13. (Rounding and adding) Let al' . . . , an be numbers with aj correctly rounded to Dj decimals. In calculating the sum (II + ... + (In' retmmng D = min DJ decimals, is it essential that we first add and then round the result or that we first round each number to D decimals and then add? 14. (Theorems on errors) Prove Theorem I(a) for addition. 15. Prove Theorem I(b) for division. 16. Show that in Example 1 the absolute value of the error of X2 = 2.000/39.95 = 0.05006 is less than 0.00001. 17. Overflow and underflow can sometimes be avoided by simple changes in a formula. Explain this in terms of V.~ + y2 = I + (ylx)2 with x 2 ~ y2 and x so large that x 2 would cause ovelt10w. Invent examples of your own. 18. (Nested form) Evaluate I(x) = x 3 - 7.5x 2 + 1l.2x + 2.8 xV = «x - 7.5)x + 11.2)x + 2.8 at x = 3.94 using 3S arithmetic and rounding. in both of the given forms. The latter, called the nestedfamz, is usually preferable since it minimizes the number of operations and thus the effect of rounding. 19. CAS EXPERIMENT. Chopping and Rounding. (a) Let x = 4/7 andy = 113. Find the error~ Echop, Eround and the relative errors Er.ch, Er.rd of x + y. x - y. xy. x I)" in chopping and rounding to 5S. Experiment with other fractions of your choice. (b) Graph Echop and Eround (for 5S) of k121 as a function of k = I, 2, . , . , 21 on common axes. What average value can you read from the graph for Echop? For Eround? Experiment with other integers that give similar graphs. Different types of graphs. Can you characteri7e the ditTerent types in terms of prime factors? (c) How does the situation in (b) change if you take 4S instead of 5S? (d) Write programs for the work in (a)-(c). 20. WRITI.'JG PROJECT. Numerics. In your own words write about the overall role of numeric methods in applied mathematics, wby they are impol1ant, where and when they must be used or can be used, and how they are influenced by the use of the computer in engineering and other work. SEC. 19.2 19.2 787 Solution of Equations by Iteration by Iteration Solution of Equations From here on. each ~ection will be devoted to some basic kind of problem and corresponding solution methods. We begin with methods of finding solutions of a single equation (1) f(x) = 0 where f is a given function. For this task there are practically no formulas (except in a few :,imple cases). so that one depends almost entirely on numeric algorithms. A solution of (1) is a number x = s such that f(s) = O. Here. s suggests "solution." but we shall also use other letters. Examples are x 3 + x = I, sin x = O.5x. tan x = x, cosh x = sec x, cosh x cos x = -1, which can all be written in the form (1). The first concerns an algebraic equation because the corresponding f is a polynomial, and in this case the solutions are also called roots of the equation. The other equations are transcendental equations because they involve transcendental functions. Solving equations (1) is a task of prime importance because engineering applications abound: some occur in Chaps. 2.4. 8 (characteristic equations), 6 (partial fractions), 12 (eigenvalues. zeros of Bessel functions). and 16 (integration). but there are many, many others. To solve (1) when there is no formula for the exact solution. we can use an approximation method. in plli1icular an iteration method, that is, a method in which we start from an initial guess .1'0 (which may be poor) and compute step by step (in general better and better) approximations Xl' X2' ... of an unknown solution of (1). We discuss three such methods that are of particular practical importance and mention two others in the problem set. These methods and the underlying principles are basic for understanding the diverse methods in software packages. In general, iteration methods are easy to program because the computational operations are the smne in each step-just the data change from step to step--and. more important. if in a concrete case a method converges. it is stable (see Sec. 19.1) in generaL Fixed-Point Iteration for Solving Equations f{x) = 0 Our present use of the word "fixed point"" has absolutely nothing to do with that in the last section. In one way or another we transform (1) algebraically into the form (2) Then we choose an Xo and compute (3) Xl x = = g(xo), X2 g(x). = g(Xl)' and in general (11 = 0, 1, .. '). A solution of (2) is called a fixed point of g, motivating the name of the method. This is a solution of 0). since from X = g(x) we can return to the original form f(x) = O. From (I) we may get several different forms of (2). The behavior of corresponding iterative sequences Xo. x l ' . . . may differ, in particular. with respect to their speed of convergence. Indeed, some of them may not converge at all. Let us illustrate these facts with a simple example. 788 CHAP. 19 Numerics in General E X AMP L ElAn Iteration Process (Fixed-Point Iteration) Set up an iteration process for the equation f(x) = x 2 x = 1.5 ± vT.25. thus - 3x +I O. Since we know the solutions = and 2.618034 0.381966. we can watch the behavior of the error as the iteration proceeds. Solution. The equation may be written thu~ (4a) Xn+l = ~(X71 2 + I). If we choose Xo = I, we obtain the sequence (Fig. -l23a: computed with 6S and then rounded) Xo = 1.000, Xl = 0.667, X2 = 0.481, X3 = which seems to approach the smaller solution. If we choose Xo ~o = 3, we obtain the sequence (Fig. 423a, upper part) Xo = 3.000. Xl = 3.333. X2 = 4.037. X3 = 0.411. X4 0.390," = . 2, the situation is similar. If we choose = 5.766. = X4 11.415 •... which diverges. Our equation may also be written (divide by x) thu~ (4b) xn+l = 3- and if we choose Xo = I, we obtain the sequence (Fig. 423b) Xo = 1.000, Xl = 2.000, X2 = 2.500, X3 = 2.600, X4 2.615,· .. = which seems to approach the larger solution. Similarly, if we choose Xo = 3, we obtain the sequence (Fig. 423b) Xo = 3.000, Xl = 2.667, X2 = 2.625, X3 = 2.619, X4 = 2.618,· ". Our figures show the following. In the lower part of Fig. 423a the slope of gl(x) is less than the slope of < I, and we seem to have convergence. In the upper part, gl(x) is steeper (g~(x) > I) and we have divergence. In Fig. 423b the slope of g2lx) is less near the intersection point Ix = 2.618, fixed point of g2, solution of f(x) = 0), and both sequences seem to converge. From all this we conclude that convergence seems to depend on the fact that in a neighborhood of a solution the curve of g(x) is less steep than the straight line y = x. and we shall now see that thi~ condition Ig' (x)1 < I (= slope of y = x) is sufficient for convergence. • y = x, which is I, thus Ig~(x)1 5 5 ~/ / c.l / x (a) Fig. 423. 5 0 0 x (h) Example 1, iterations (4a) and (4b) g2(x) 5 SEC. 19.2 789 Solution of Equations by Iteration An iteration process defined by (3) is called convergent for an Xo if the corresponding sequence xo, Xb ••• is convergent. A sufficient condition for convergence is given in the following theorem, which has various practical applications. THEOREM 1 Convergence of Fixed-Point Iteration Let x = s be a solution of x = g(x) and suppose that g has a continuous derivative in some interval J containing s. Then !f Ig' (x) I ~ K < 1 in J, the iteration process defined by (3) converges jor any Xo in J. and the limit of the sequence {xn} is s. PROOF By the mean value theorem of differential calculus there is a t between x and s such that g(x) - g(s) Since g(s) = s and Xl = g(xo), Ig' (x) I in the theorem X2 = g'(t) (x - s) (x in J). = g(Xl)' ... , we obtain from this and the condition on Applying this inequality n times, for n, n - 1, ... , 1 gives Since K < I, we have K n -----? 0; hence IXn sl-----? 0 as n -----? - • 00. We mention that a function g satisfying the condition in Theorem 1 is called a contraction because Ig(x) - g(v)1 ~ Klx - vi, where K < 1. Furthermore, K gives information on the speed of convergence. For instance, if K = 0.5, then the accuracy increases by at least 2 digits in only 7 steps because 0.5 7 < 0.01. E X AMP L E 2 An Iteration Process. Illustration of Theorem 1 + x-I ~ a by iteration. Find a solution of I(x) = ..3 Solution. A sketch shows that a solution lies near x = 1. We may write the equation as (x 2 1 x = gl(x) = - - - 2 . 1 + SO x that xn+ 1 = ---2 . 1 +xn + l)x = lor Also for any x because 4x2 /(1 + x 2 )4 = 4x 2 /(1 + 4x 2 + ... ) < 1, so that by Theorem 1 we have convergence for any Xo. Choosing Xo = 1. we obtaio (Fig. 424 on p. 790) Xl ~ 0.500, X2 = 0.800, X3 = 0.610, X4 = 0.729, X5 = 0.653, X6 = 0.701, .... The solution exact to 6D is s = 0.682 328. The given equation may also be written Then and this is greater than 1 near the solution, so that we cannot apply Theorem I and as~ert convergence. Try = 0.5. Xo = 2 and see what happens. ,The example shows that the transformation of a given i(x) = a into the form x = g(x) with g satisfying Ig (x)1 3 K < 1 may need some experimentation. • Xo = 1. Xo 790 CHAP. 19 Numerics in General 1.0 0.5 x Fig. 424. Iteration in Example 2 Newton's Method for Solving Equations t(x) = 0 Newton's method, also known as Newton-Raphson's method,! is another iteration method for solving equations f(x) = 0, where f is assumed to have a continuous derivative f'. The method is commonly used because of its simplicity and great speed. The underlying idea is that we approximate the graph of f by suitable tangents. Using an approximate value Xo obtained from the graph of f, we let Xl be the point of intersection of the x-axis and the tangent to the curve of f at Xo (see Fig. 425). Then hence In the second step we compute X2 = Xl - f(xIV!' (Xl), in the third step X3 from X2 again by the same formula, and so on. We thus have the algorithm shown in Table 19.1. Formula (5) in this algorithm can also be obtained if we algebraically solve Taylor's formula (5*) y x Fig. 425. Newton's method IJOSEPH RAPHSON (]648-l715), English mathematician who published a method similar to Newton's method. For historical details. see Ref. [GR2], p. 203. listed in App. I. SEC. 19.2 791 Solution of Equations by Iteration Table 19.1 Newton's Method for Solving Equations I(x) ALGORITHM NEWTON (j, t' , XO, E, =0 N) This algorithm computes a solution of f(x) = 0 given an initial approximation Xo (starting value of the iteration). Here the function f(x) is continuous and has a continuous derivative f' tx). f, f', initial approximation xo, tolerance INPUT: € > O. maximum number of iterations N. OUTPUT: For n Approximate solution Xn (n ~ N) or message of failure. 0, I, 2, ... , N - 1 do: = Compute t' (x,,). If f' (xn) = 0 then OUTPUT "Failure". Stop. 2 [Procedure completed unsuccessfully] Else compute 3 (5) 4 If IXn+l - xnl ~ Elx,J then OUTPUT xn + 1. Stop. [Procedure completed sllccessfullv] End 5 OUTPUT "Failure". Stop. [Procedure completed unsuccessfully after N iterations] End NEWTON If it happens that t' (x n ) = 0 for some n (see line 2 of the algorithm), then try another statting value Xo. Line 3 is the heart of Newton's method. The inequality in line 4 is a termination criterion. [f the sequence of the Xn converges and the criterion holds, we have reached the desired accuracy and stop. In this line the factor IXnl is needed in the case of zeros of very small (or very large) absolute value because of the high density (or of the scarcity) of machine numbers for those x. WARNING! The criterion by itself does not imply convergence. Example. The harmonic series diverges, although its partial sums Xn = Lk~l 11k satisfy the criterion because lim (Xn+l - xn) = lim (1/(n + 1) = O. Line 5 gives another termination criterion and is needed because Newton's method may diverge or, due to a POOT choice of xo, may not reach the desired accuracy by a reasonable number of iterations. Then we may try another Xo. If I(x) = 0 has more than one solution, different choices of Xo may produce different solutions. Also, an iterative sequence may sometimes converge to a solution different from the expected one. 792 E X AMP L E 3 CHAP. 19 Numerics in General Square Root Set up a Newton iteration for computing the square root x of a given positive number c and apply it to c = 2. Solutioll. We have x = Vc. hence f(x) = r2 - c = O. f' (x) = X3 = 1.414216, 2r. and (5) takes the form For c = 2, choosing '\0 = I. we obtain Xl X4 E X AMP L E 4 = 1.500000, = X2 1.416667, X4 = 1.414214.... • is exact to 6D. Iteration for a Transcendental Equation Find the positive solution of 2 sin x = x. Solution. = Setting f(x) xn+l r - 2 sin x. we have = Xn - Xn - f' (x) = 2 sinxn I - 2 cos x. and (5) gives 2( sin xn - xn cos xn) 1- 2cosxn 1- 2cosxn From the graph of f we conclude that the solution is near Xo = 2. We compute: I: 1 2 3 X4 = E X AMP L E 5 2.00000 1.90100 1.89552 1.89550 1.83229 1.64847 1.63809 1.63806 3.48318 3.12470 3.10500 3.10493 1.90100 1.89552 1.89550 1.89549 1.89549 is exact to 5D since the solution to 6D is 1.~95 • 494. Newton's Method Applied to an Algebraic Equation Apply Newton's method to the equation f(x) = x3 Solution. +X- I = o. From (5) we have Starting from Xo = 1. we obtain Xl = 0.750000, X2 = 0.686047, X3 = 0.682 340, X4 = 0.682 328, ... where .'4 ha~ the error -I . 10- . A comparison with Example 2 shows that the present convergence is much • more rapid. This may motivate the concept of the order of all iteration process, to be discussed next. 6 Order of an Iteration Method. Speed of Convergence The quality of an iteration method may be characterized by the speed of convergence, as follows. Let xn+ 1 = = g(x'Y!) g(x). Then define an iteration method, and let Xn approximate a solution s of = S - En' where En is the error of X n . Suppose that g is differentiable a number of times, so that the Taylor formula gives x Xn Xn+l (6) = g(xn ) = g(s) = g(s) + g'(s)(xn - g'(S)En - s) + h"(s)(xn + !g"(S)En 2 + - S)2 + SEC. 19.2 793 Solution of Equations by Iteration The exponent of En in the first oonvanishing term after g(s) is called the order of the iteration process defined by g. The order measures the speed of convergence. To see this, subtract g(s) = s on both sides of (6). Then on the left you get Xn+l - S = -E,,+l, where E,,+1 is the error of Xn+l' And on the right the remaining expression equals approximately its first nonzero term because IEnl is small in the case of convergence. Thus (a) in the case of first order, E,,+l=+g'(S)En (7) in the case of second order, etc. Thus if En = lO-k in some step, then for second order. En+l = C· (l0-k)2 = c· 1O-2k, so that the number of significant digits is about doubled in each step. Convergence of Newton's Method In Newton's method, g(x) = x - f(x)/f' (x). By differentiation, g'(x) = 1 - f' (X)2 - f (8) f(.T)f"(X) , 2 (x) f(x)f"(x} f' (X)2 Since f(s) = 0, this shows that also g'(s) = O. Hence Newton's method is at least of second order. If we differentiate again and set x = s, we find that (8*) g"(S) = f"(S) f'(s) which will oot be zero in general. This proves THEOREM 2 Second-Order Convergence of Newton's Method if f(x) is three times differentiable and f' and f" are not z.ero at a solution s of f(x) = 0, then for Xo sufficiently close to s, Newton's method is of second order. Comments. For Newton's method, (7b) becomes, by (8*), For the rapid convergence of the method indicated in Theorem 2 it is important that s be a simple zero of f(x) (thUS (s) =I=- 0) and that Xo be close to s, because in Taylor's formula we took only the linear term [see (5*)], assuming the quadratic term to be negligibly small. (With a bad Xo the method may even diverge!) f' 794 E X AMP L E 6 CHAP. 19 Numerics in General Prior Error Estimate of the Number of Newton Iteration Steps Use .1"0 = 2 and xl = I.YUl 111 Example 4 for estimating how many Iteration steps we need to produce the solution to 5D accuracy. This is an a priori estimate or prior estimate because we can compute it after only one iteration, prior to further iterations. Solution. We have I(x} = x- 2 sin X = O. Differentiation gives {'(s) {'(Xl) 2 sin x] = - - - = ---.....:.--- = , 2J (s) 2I' (.~I) 0.57. Hence (Y) gives 1"n+ll = 0.57"n2 = 0.57(0.57"~_IJ2 = where M = 2n + 2 n condition becomes I + ... + 2 + I = 2n+I - 0.573"~t_l = ... = 0.57M,,~+I ;;: 5' 10-6 I. We show below that Eo = -0.11. Consequently. our Hence II = 2 is the smallest possible n. according to this cnlde estimate. in good agrecment with Example 4. "0 = -0.11 is obtained from "1 - Eo = ("1 - s) - (Eo - s) = -Xl + Xo = 0.10. hence 2 2 "1 = "0 + 0.10 = -0.57"0 or 0.57"0 + "0 + 0.10 = O. which gives "0 = -0.11. • Difficulties may arise if It' (x) I is very small near a solution s of f(x) = o. for instance, if s is a zero of f(x) of second (or higher) order (so that Newton's method converges only linearly, as an application of I'H6pital's rule to (8) shows). Geometrically, small It' (x) I means that the tangent of f(x) near s almost coincides with the x-axis (so that double precision may be needed to get f(x) and f' (x) accurately enough). Then for values x = far away from s we can still have small function values Difficulties in Newton's Method. s R(s) = fCS). In this case we call the equation f(x) = 0 ill-conditioned. R(s) is called the residual of f(x) = 0 at S. Thus a small residual guarantees a small error of only if the equation is not iII-conditioned. s E X AMP L E 7 An Ill-Conditioned Equation J(x) = X5 + 1O- 4.t = 0 is ill·conditioned. x = 0 is a solution . .t' (0) = 10-4 is small. At s = 0.1 the residual J(O.I) = 2· 10- 5 is small, but the en-or -0.1 is laIger in absolute value by a factor 5000. Invent a more drastic example of your own. • Secant Method for Solving {(x) = 0 Newton's method is very powerful but has the disadvantage that the derivative f' may sometimes be a far more difficult expression than f itself and its evaluation therefore computationally expensive. This situation suggests the idea of replacing the derivative with the difference quotient t' (xn ) = f(x n ) - Xn - f(x n - 1) Xn - 1 Then instead of (5) we have the formula of the popular secant method SEC. 19.2 Solution of Equations by Iteration 795 y x Fig. 426. Secant method (10) Geometrically, we intersect the x-axis at -',,+1 with the secant of f(x) passing through Pn - 1 and Pn in Fig. 426. We need two starting values Xo and xl' Evaluation of derivatives is now avoided. [t can be shown that convergence is superlinear (that is, more rapid than linear, IEn+11 = const 'IEnIL62; see [E5] in App. 1), almost quadratic like Newton's method. The algorithm is similar to that of Newton's method, as the student may show. CAUTION! It is not good to write (0) as xn-d(xn) - -'llf(Xn -1) f(xn) - f(X n -1) because this may lead to loss of significant digits if Xn and Xn-l are about equal. (Can you see this from the formula?) EXA M P L E 8 Secant Method Find the positive solution of f(x) = x - 2 sin x = 0 by the secant method, starting from .Io = 2, Solution. Xl = 1.9. Here, (10) is Numerical values are: n X n -1 xn Nn D" 2 2.000000 1.900000 1.895747 1.900000 1.895747 1.895494 -0.000740 -0.000002 0 -0.174005 -0.006986 3 X3 = 1.895 494 is exact to 6D. See Example 4. -0.004253 -0.000252 o • Summary of Methods. The methods for computing solutions s of f(x) = 0 with given continuous (or differentiable) f(x) start with an initial approximation Xo of s and generate a sequence Xl, X2, . . . by iteration. Fixed point methods solve f(x) = 0 written as x = g(x), so that s is a fixed point of g, that is, s = g(s). For g(x) = X - f(x)li' (X) this is Newton's method, which for good Xo and simple zeros converges quadratically (and for multiple zeros linearly). From Newton's method the secant method follows by replacing f' (x) by a difference quotient. The bisection method and the method of false position in Problem Set 19.2 always converge, but often slowly. CHAP. 19 796 Numerics in General ...... - . --11-71 FIXED-POINT ITERATION Apply fixed-point iteration and answer related questions where indicated. Show details of your work. 1. x = 1.4 sin x, Xo = 1.4 2. 00 the iterations indicated at the end of Example 2. Sketch a figure similar to Fig. 424. 3. Why do we obtain a monotone sequence in Example I, but not in Example 2? f = X4 - X + 0.2 = 0, the root near I, Xo 5. f as in Prob. 4, the root near 0, Xo = 0 = 6. Find the smallest positive solution of sin x e- 0 . 5X , 4. Xo = = I. 7. (Bessel functions, drumhead) A partial sum of the Maclaurin series of JoCr) (Sec. 5.5) is f(x) = I - !x 2 + i'4x4 - 23~4X6. Conclude from a sketch that f(x) = 0 near x = 2. Write f(x) = 0 as x = g(x) (by dividing f(x) by !x and taking the resulting x-term to the other side). Find the zero. (See Sec. 12.9 for the importance of these zeros.) 8. CAS PROJECT. Fixed-Point Iteration. (a) Existence. Prove that if g is continuous in a closed interval f and its range lies in f, then the equation x = g(x) has at least one solution in f. Illustrate that it may have more than one solution in f. (b) Convergence. Let f(x) = x 3 + 2r2 - 3x - 4 = o. Write this asx = g(x), for g choosing (1) (x 3 - /)113, (2) (r2 - !n1l2, (3) x + !f, (4) x(l + !f), (5) (x 3 - f)lx 2 , (6) (2x 2 - f)/2x, (7) x - flf' and in each case Xo = 1.5. Find out about convergence and divergence and the number of steps to reach exact 6S-values of a root. 19-181 NEWTON'S METHOD Apply Newton's method (60 accuracy). First sketch the function(s) to see what is going on. ~ ~nx = cot x, Xo = 12. x - 5x + In x + 3 = 2, 16. (Legendre polynomials) Find the largest root of the Legendre polynomial P 5 (x) given by 5 P5(X) = k(63x - 70x 3 + 15x) (Sec 5.3) (to be needed in Gauss integration in Sec. 19.5) (a) by Newton's method, (b) from a quadratic equation. 17. Design a Newton iteration for cube roots and compute ~ (60, Xo = 2). 18. Design a Newton iteration for -\7c" (c > 0). Use it to 3-.4(::;- ,5(::;compute \12, V'2, v 2, v 2 (60, Xo = I). 19. TEAM PROJECT. Bisection Method. This simple but slowly convergent method for finding a solution of f(x) = 0 with continuous f is based on the intermediate value theorem, which states that if a continuous function f has opposite signs at some x = a and x = b (> a). that is, either f(a) < 0, feb) > 0 or f(a) > 0, feb) < 0, then f must be 0 somewhere on [a, b]. The solution is found by repeated bisection of the interval and in each iteration picking that half which also satisfies that sign condition. (a) Algorithm. Write an algorithm for the method. (b) Comparison. Solve x = cosx by Newton's method and by bisection. Compare. (e) Solve e- x = In x and eX + x4 + X 2 by bisection. = 20. TEAM PROJECT. Method of False Position (Regula falsi). Figure 427 shows the idea. We assume that f is continuous. We compute the x-intercept Co of the line through (ao, f(ao), (b o , f(bo». If f(co) = 0, we are done. If f(ao)f(c o ) < 0 (as in Fig. 427), we set a 1 = ao, b 1 = Co and repeat to get Cl, etc. If f(ao)ffco) > 0, then f(co)f(b o ) < 0 and we set al = Co, b 1 = b o, etc. (a) Algorithm. Show that 10. x = cos x, xo 11. x 3 15. (Associated Legendre functions) Find the smallest positive zero of 2 P4 = (I - X2)p~ = ¥(-7x 4 + 8x 2 - 1) (Sec. 5.3) (a) by Newton's method, (b) exactly, by solving a quadratic equation. 0, Xo = 2 Xo = 2 13. (Vibrating beam) Find the solution of cos x cosh x = I near x = ~ 7T. (This determines a frequency of a vibrating beam: see Problem Set 12.3.) 14. (Heating. cooling) At what time x (4S-accuracy only) will the processes governed by fl(X) = 100(1 - e- O.2x ) and f2(X) = 4Oe- O.Olx reach the same temperature? Also find the latter. Co = aof(bo ) - bof(ao) f(bo) - f(ao) and write an algorithm for the method. (b) Comparison. Solve x 3 = 5x + 6 by Newton's method, the secant method, and the method of false position. Compare. (C) Solve X4 = 2, cos x = V~, and by [he method of false position. x + In x 2 SEC. 19.3 797 Interpolation y 121-241 ~/ SECANT METHOD Solve. using ~ y=f(x) x Xo and 23. Prob. 9, = 0.5 Xl = J Fig. 427. 19.3 indicated. = 2.0 = I. Xl I. Xo = 24. Prob. 10, Xo /( Xl i1~ 21. Prob. ll, Xo = 0.5, Xl 22. e- x - tan X = o. Xo = Xl 0.5, 0.7 I 25. WRITING PROJECT. Solution of Equations. Compare the methods in this section and problem set, discussing advantages and disadvantages using examples of your own. Method of false position Interpolation Interpolation means finding (approximate) values of a function f(x) for an X between Xl' • • . , xn at which the values of f(x) are given. These values may come from a "mathematical" function, such as a logarithm or a Bessel function, or, perhaps more frequently, they may be measured or automatically recorded values of an "empirical" function, such as the air resistance of a car or an airplane at different speeds, or the yield of a chemical process at different temperatures, or the size of the U.S. population as it appears from censuses taken at IO-year intervals. We write these given values of a function f in the form different x-values xo, fo = f(xo), fn = f(x,J or as ordered pairs A standard idea in interpolation now is to find a polynomial Pn(x) of degree n (or less) that assumes the given values; thus (1) We call this Pn an interpolation polynomial and xo, ... , xn the nodes. And if f(x) is a mathematical function, we call Pn an approximation of f (or a polynomial approximation, because there are other kinds of approximations, as we shall see later). We use Pn to get (approximate) values of f for x's between Xo and xn ("interpolation") or sometimes outside this interval Xo ~ x ~ Xn ("extrapolation"). Motivation. Polynomials are convenient to work with because we can readily differentiate and integrate them. again obtaining polynomials. Moreover, they approximate continuous functions with any desired accuracy. That is. for any continuous f(x) on an interval 1: a ~ x ~ b and error bound f3 > 0, there is a polynomial Pn(x) (of sufficiently high degree n) such that If(x) - Pn(X) I < f3 for all x on 1. This is the famous Weierstrass approximation theorem (for a proof see Ref. [GR7], p. 280; see App. O. 798 CHAP. 19 Numerics in General Existence and Uniqueness. Pn satisfying (I) for given data exists-we give formulas for it below. Pn is unique. Indeed, if another polynomial qn also satisfies lJn(.ro) = fo, ... , lJn(xn) = f n • then Pn(x) - lJn(x) = 0 at xo . ...• X n , but a polynomial Pn - qn of degree 11 (or less) with 11 + I roots must be identically zero, as we know from algebra; thus Pn(x) = q,,{x) for all x, which means uniqueness. • How to Find Pn? This is the important practical question. We answer it by explaining several standard methods. For given data. these methods give the same polynomial. by the uniqueness just proved (which is thus of practical interest!), but expressed in several forms suitable for different purposes. Lagrange Interpolation Given (xo, f 0), (Xl' f 1). ' , , , (Xn , f n) with arbitrarily spaced Xj' Lagrange had the idea of multiplying each f j by a polynomial that is I at Xj and 0 at the other 1l nodes and then taking the sum of these 11 + 1 polynomials. Clearly, this gives the unique interpolation polynomial of degree 11 or less. Beginning with the simplest case, let us see how this works. Linear interpolation is interpolation by the straight line through (xo, fo), (Xl' II): see Fig. 428. Thus the linear Lagrange polynomial PI is a sum PI = Lof0 + Ld 1 with Lo the linear polynomial that is I at Xo and 0 at Xl; similarly, Ll is 0 at Xo and I at Xl' Obviously, Lo(x) x - = Xl - Xo Xo This gives the linear Lagrange polynomial (2) . fo + Error~ ~ /1" / f .o p(x) 1 1--'" }'=f(x) f) x x Fig. 428. E X AMP L E 1 Linear interpolation Linear Lagrange Interpolation Compute a 4D-value of In \1.2 from In 9.0 = 2.1\172. In 9.5 = 2.2513 by linear Lagrange interpolation and determine the error, llsing In \1.2 = 2.2192 (4Dl. Solutioll. Xo = 9.0. Xl = 9.5.10 = In 9.0. it x - 9.5 Lo(x) = --=0:5 = - 2.0(x L 1 (x) = ----0:5 = 2.0(x - x - 9.0 = In 9.5. In (2) we need - 1).5). 9.0). Lo«).2) = - 2.0( -0.3) = 0.6 L 1 (9.2) = 2· 0.2 = 0.4 (see Fig. 429) and obtain the answer In 9.2 ~ Pl(9.2) = L o(9.2)1 0 + L 1 (9.2)1 1 = 0.6' 2.1972 + 0.4' 2.2513 = 2.2188. SEC. 19.3 799 Interpolation The error is E = a - a = 2.2192 - 2.2188 = OJlO04. Hence linear interpolation is not sufficient here to get 4D-accuracy; it would suffice for 3D-accuracy. • ~~- --_-"'oX'L, o ... --~I~I~'~I--~--~~ 9 9.2 9.5 Fig. 429. 11 x 10 Lo and L, in Example 1 Quadratic interpolation is interpolation of given (xo, fo), second-degree polynomial P2(X), which by Lagrange's idea is (Xl, f1), (X2, f2) by a (3a) Lo(x) = [ (x) (x - Xl)(X - X2) _0_ [o(xo) (3b) LI(x) (xo - [1 (x) (x - Xo)(x - [1(X1) (Xl - XO)(XI - = -- ~(x) = Xl)(XO - (x - Xo)(X - = [2(X) [2(X2) X2) X2) X2) Xl) (X2 - XO)(X2 - Xl) How did we get this? Well, the numerator makes Lk(xj) = 0 ifj makes Lk(xk) = 1 because it equals the numerator at X = Xk' E X AMP L E 2 =1= k. And the denominatOi Quadratic Lagrange Interpolation Compute In 9.2 by (3) from the data in Example I and the additional third value In 11.0 Solution. = 2.3979. In (3), Lo(x) = (x - 9.5)(x - 11.0) (9.0 - 9.5)(9.0 - 11.0) = x 2 - (x - 9.0)(x - 1 1.0) 20.5x I = (9.5 _ 9.0)(9.5 _ 11.0) = - 0.75 ~(x) = (11.0 _ 9.0)(11.0 _ 9.5) = 1 "3 Lo(9.2) = 0.5400, 2 L 1(x) (x - 9.0)(x - 9.5) + 104.5, (x + 99), L1(9.2) = 0.4800, + 85.5), ~(9.2) = -0.0200, - 20x 2 (x - 18.5x (see Fig. 430), so that (3a) gives, exact to 4D, In 9.2 = P2(9.2) = 0.5400' 2.1972 + 0.4800' 2.2513 - 0.0200' 2.3979 x Fig. 430. Lo, L" L2 in Example 2 = 2.2192. • 800 CHAP. 19 Numerics in General General Lagrange Interpolation Polynomial. For general n we obtain (4a) where Lk(Xk) = 1 and Lk is 0 at the other nodes. and the Lk are independent of the function f to be interpolated. We get (4a) if we take (4b) lo(x) = (x - X1)(X - X2) ... (x - xn), Ik(x) = (x - xo) ... (x - Xk-1 )(x - In(x) = (x - xo)(x - xl) ... (x - X,,-l)' Xk+1) ... (x - 0< k < n, x n ), We can easily see that P,,(Xk) = h. Indeed, inspection of (4b) shows that Ik(xj) = 0 if =1= k, so that for x = Xk' the sum in (4a) reduces to the single tenn (lk(Xk)llk (Xk»fk = h. j Error Estimate. If f is itself a polynomial of degree n (or less), it must coincide with p" because the n + 1 data (xo. fo), ... , (x11' fIt) determine a polynomial uniquely. so the error is zero. Now the special f has its (n + 1)st derivative identically zero. This makes it plausible that for a ~ene/"{t! fits (11 + l)st derivative tn+ll should measure the error En(X) = f(x) - p,,(x). t It can be shown that this is true if n + ll exists and is continuous. Then, with a suitable t between Xo and x" (or between Xo. x n ' and x if we extrapolate), f'n+l>(t) (5) E.,(.l) = f(x) - Pn(x) = (x - .lo)(x - Xl) .•• (X - ern) -=-------'-'-- (11 + I)! Thus !En(x)i is 0 at the nodes and small near them, because of continuity. The product (x - Xo) ... (x - xu) is large for \" away from the nodes. This makes extrapolation risky. And interpolation at an x will be best if we choose nodes on both sides of that x. Also, we get error bounds by taking the smallest and the largest value of f'n+ll(t) in (5) on the interval Xo ~ t ~ Xn (or on the interval also containing x if we extrapolate). Most importantly, since Pn is unique, as we have shown, we have THEOREM 1 Error of Interpolation Formula (5) gil'es the error for allY po/molllia/ interpolation lIIet/lOd (f f(x) has a continuous (11 + I )st deri\'{/tive. Practical error estimate. If the derivative in (5) is difficult or impossible to obtain, apply the Error Principle (Sec. 19.1), that is, take another node and the Lagrange polynomial Pn+1(X) and regard PII+1(X) - Pn(x) as a (crude) error estimate for Pn(x), SEC. 19.3 Interpolation E X AMP L E 3 801 Error Estimate (5) of Linear Interpolation. Damage by Roundoff. Error Principle Estimate the error in Example I first by (5) directly and then by the Error Principle (Sec. 19.1). Soluti01l. (A) Estimatioll by (5), We have 11 = 1. fit) = In 1. {(t) = lIt. {'(t) = -2- 2t Hence 0.03 (-1) "l(x) = (x - 9.0){x - 9.5) -1112. thus "1(9.2) = -2- . t 2 = 9.0 gives the maximum 0.03/9 = 0.00037 and f = 9.5 gives the minimum 0.03/9.5 2 = 0.00033. SO that we get 0.00033 ~ "1(9.2) ~ 0.00037, or better, 0.00038 because 0.3/81 = 0.003 703 .... But the error 0.0004 in Example I disagrees, and we can learn something! Repetition of the computation there with 5D instead of 4D gives 1 In 9.2 = 1'1(9.2) = 0.6' 2.19722 + 0.4 . 2.25129 = 2.21885 with an actual en'or " = 2.21920 - 2.21885 = 0.00035. which lies nicely near the middle between our two error bounds. This shows that the discrepancy 10.0004 vs. 0.00035) was caused by rounding, which is not taken into account in (5). (B) EstinUltioll by the Error Pri1lciple. We calculate 1'1(9.2) = 2.21885 as before and then 1'2(9.2) as in Example 2 but with 5D, obtaining 1'2(9.2) = 0.54' 2.19722 + 0.48· 2.25129 - 0.02' 2.39790 = 2.21916. The difference P2(9.2) - 1'1(9.2) = 0.00031 is the approximate error of 1'1(9.2) that we wanted to obtain: this is an approximation of the actual en'or 0.00035 given above. • Newton's Divided Difference Interpolation For given data (xo, f 0)' ... , (xn , f n) the interpolation polynomial Pn(x) satisfying (l) is unique, as we have shown. But for different purposes we may use Pn(x) in different forms. Lagrange's form just discussed is useful for deriving formulas in numeric differentiation (approximation formulas for derivatives) and integration (Sec. 19.5). Practically more imp0l1ant are Newton's forms of Pn(x), which we shall also use for solving ODEs (in Sec. 21.2). They involve fewer arithmetic operations than Lagrange's fonn. Moreover, it often happens that we have to increase the degree 11 to reach a required accuracy. Then in Newton's forms we can use all the previous work and just add another term. a possibility without counterpart for Lagrange's form. This also simplifies the application of the Error Principle (used in Example 3 for Lagrange). The details of these ideas are as follows. Let Pn-l(X) be the (n - l)st Newton polynomial (whose form we shall detelmine); thus Pn-l(XO) = fO,Pn-1(X1) = f1,' .. ,Pn-1(Xn -l) = fn-l' FUl1hermore, let us write the nth Newton polynomial as (6) hence (6') Here gn(x) is to be determined so that Pn(xO) = fo, Pn(Xl) = flo ... , Pn(xn ) = f n' Since p" and Pn-l agree at xo, ... , X n -l' we see that gn is zero there. Also, gn will generally be a polynomial of nth degree because so is Pn' whereas Pn-l can be of degree n - I at most. Hence gn must be of the form (6") CHAP. 19 802 Numerics in General We determine the constant an- For tIns we set x = Xn and solve (6") algebraically for Replacing gn(xn ) according to (6') and using Pn(xn ) = In' we see that this gives (In- (7) We write {lk instead of {In and show that (lk equals the kth divided difference, recursively denoted and defined as follows: and in general (8) PROOF If II = 1, then Pn-l (xn ) = Po(xl ) of f(x) at Xo. Hence (7) gives = Io because Po(x) is constant and equal to Io, the value and (6) and (6") give the Newton interpolation polynomial of the first degree If II = 2, then this PI and (7) give where the last equality follows by straightforward calculation and comparison with the definition of the right side. (Verify it: be patient.) From (6) and (6") we thus obtain the second Newton polynomial For II = k, formula Withpo(x) = (6) gives fo by repeated application with k = I,· .. , II this finally gives Newton's divided difference interpolation formula (10) f(x) = fo + (x - xo)f[xo, XI] + (x - xo)(x - xI)flxo, XI' X2] + ... + (x - xo)(x - XI) ... (x - xn-I)f[xo, ... , xn]. SEC. 19.3 803 Interpolation An algorithm is shown in Table 19.2. The first do-loop computes the divided differences and the second the desired value Pn(x). Example 4 shows how to arrange differences near the values from which they are obtained: the latter always stand a half-line above and a half-line below in the preceding • column. Such an arrangement is called a (divided) difference table. Table 19.2 Newton's Divided Difference Interpolation ALGORITHM INTERPOL txo, ... , Xn; f 0, ••• , f n; x) This algorithm computes an approximation p,,(.f) of f(.O at .f. INPUT: Data (xo- f 0), (x b .fI) •...• (xno f n); Approximation Pn(.x) of fCO OUTPUT: Set f[-':i1 = For I.. . .. 111 = x For j U = 0, ... , n). fj 11 - I do: 0, .... = 11 - Tn do: End End Set Po(x) For k = = f o· 1, ... , 11 do: End OUTPLT Pn(i) End INTERPOL E X AMP L E 4 Newton's Divided Difference Interpolation Formula Compute f(9.2) from the values shown in the fust two columns of the following table. Xj fj = ft9 !to (2.079442" .... 9.0 2.197225 f[Xj. Xj+l] ". (0.117 78~ ,----- - \-U.006 433 0.lO8134 9.5 2.251 292 -0.005200 0.097735 1l.0 2.397895 804 CHAP. 19 Numerics in General Soluti01l. We compute the divided differences as shown. Sample computation: (0.097735 - O.lO8 134)/(11 - 9) = -0.005200. The values we need in (lO) are circled. We have f(x) = P3(x) = 2.079442 + 0.117783(,' - KO) - 0.006433(x - 8.0)(x - 9.0) + 0.000 4Il(x - 8.0)(x - 9.0)(x - 9.5). Alx=9.2, J(9.2) = 2.079442 + 0.141 340 - 0.001544 - 0.000030 2.219208. = The value exact to 6D is J(9.2) = In 9.2 = 2.219203. Note that we can nicely see how the accuracy increases from term to term: Pl(9.2) = 2.220782. P2(9.2) = 2.219 238, • P3(9.2) = 2.219208. Equal Spacing: Newton's Forward Difference Formula Newton's formula (10) is valid for arbitrarily spaced nodes as they may occur in practice in expe1iments or observations. However, in many applications the x/s are regularly spaced-for instance, in measurements taken at regular intervals of time. Then, denoting the distance by 11, we can write (11) Xn = Xo We show how (8) and (10) now simplify considerably! To get started, let us define the first forward difference of I at the second forward difference of I at Xj Xj + I1h. by by and, continuing in tins way, the kth forward difference of I at Xj by (12) (k = 1,2, .. '). Examples and an explanation of the name "forward" follow on p. 806. What is the point of this? We show that if we have regular spacing (11), then (13) We prove (13) by induction. It is true for k = 1 I7 I because Xl = .\"0 + /z, 1 (II - Io) = -111 I::.Io· • 7 SO that SEC. 19.3 805 Interpolation Assuming (13) to be true for all forward differences of order k. we show that (13) holds for k + 1. We use (8) with k + I instead of k; then we use (k + I)h = Xk+1 - .\"0' resulting from (II). and finally (12) with) = 0, that is. j,k+lfo = !lokfl - j.kfo. This gives (k + which is (13) with" + ___ 1 j.k [ k!hk· fl I)h ____ I ~k fo k!hk ] • 1 instead of ". Formula (13) is proved. In (10) we finally set x = .\"0 + rh. Then x - Xo = rh, x - XI = (r - 1)h since Xo = h, and so on. With this and (13), formula (10) becomes Newton's (or Gregory2-Newtoll's) forward difference interpolation formula Xl f(x) = Pn(x) = ~ (:) ~sfo (x = Xo + r = (x - xo)/17) rh, s=o (14) = fo + rj.fo + 1) r(r - 2! 2 j. fo r(r - + ... + 1) ... (r - 11 + Il! 1) j.nfo where the binomial coefficients in the first line are defined by (15) (~) = I. (:) = _r(_r_-_1)_(r_-_2)_s;_'_'_(r_-_s_+_I_) (s > O. integer) and s! = 1 . 2 ... s. Error. (16) From (5) we get. with.\" - En(X) = f(x) - Pn(x) = .\"0 = rh, X - hn + I (n + 1)! r(r - Xl = (r - 1)h. etc .. 1) ... (r - n)tn + ll (t) with t as characterized in (5). Formula (16) is an exact formula for the error, but it involves the unknown t. In Example 5 (below) we show how to use (16) for obtaining an enor estimate and an interval in which the true value of f(x) must lie. Comments on Accuracy. (A) The order of magnitude of the enor to that of the next difference not used in Pn(x), En(X) is about equal (B) One should choose xo, .... x 71 such that the x at which one interpolates is as well centered between xo, . . . , Xn as possible. 2JA~ES GREGORY (1638-1675). SC?ts mathematician. professor a[ SI. Andrews and Edinburgh. ~ in (14) and'\ (on p. 807) have nothmg [0 do WIth the Laplacian. 806 CHAP. 19 Numerics in General The reason for (A) is that in (16), II"(r __ - J)__________ ... (r - 11)1 -cc ~~ ~ 1 ·2 ... (11 + ~ J) if Irl ~ ] (and actually for any I" as long as we do not extrapolate). The reason for (B) is that 11"(1" - 1) ... (r - n)1 becomes smallest for that choice. E X AMP L E 5 Newton's Forward Difference Formula. Error Estimation Compute cosh 0.56 from (14) and the four values in the following table and estimate the error. cosh Xj j Xj ° 0.5 )~IP 626; 0.6 1.185465 fj = 2 t:. fj t:.fj 1 (O.U5783')~ p.Ol] 865, 0.069704 2 t:,.3fj 0.7 ,0 ..900697/ 1.255169 0.012562 0.082266 0.8 I 1.337435 Solution. We compute the forward differences as shown in the table. The values we need are circled. In (14) we have r = (0.56 - 0.50)/0.1 = 0.6, so that (14) gives cosh 0.56 = 1.127626 = + 0.6' 0.057839 + 0.6(~OA) . 0.011 ROS + 0.6(-0.:)(-1.4) . 0.000 697 1.127 626 + 0.034 703 - 0.001 424 + 0.000039 = 1.16U 944. Error estimate. From (16), since the founh derivative is cosh(4l t 0.1 1"3(0.56) = where A = -0.000003 36 and 0.5 ~ t and smallest cosh t in that interval: = cosh t, 4 4! ·0.6(-OA)(-1.4)(-2A) cosh I = A cosh t, ~ 0.8. We do not know t, but we get an inequality by taking the largest A cosh 0.8 ~ 1"3(0.62) ~ A cosh 0.5. Since f(xl = P3(x) + ~(x), this gives P3(0.56) + A cosh 0.8 ~ cosh 0.56 ~ P3(0.56) + A cosh 0.5. Numeric values are 1.I60 939 ~ cosh 0.56 ~ 1.160 941. The exact 6D-value is cosh 0.56 = 1.160941. It lies within these bounds. Such bounds are not always so tight. • Also, we did not consider roundoff errors, which will depend all the number of operations. This example also explains the name ':forward difference formula": we see that the differences in the formula slope forward in the difference table. SEC. 19.3 807 Interpolation Equal Spacing: Newton's Backward Difference Formula Instead of forward-sloping differences we may also employ backward-sloping differences. The difference table remains the same as before (same numbers, in the same positions), except for a very harmless change of the running subscript j (which we explain in Example 6, below). Nevertheless, purely for reasons of convenience it is standard to introduce a second name and notation for differences as follows. We define the first backward difference of fat Xj by the second bac/..:ward difference of f at Xj by and, continuing in this way, the kth backward difference of f at Xj by (17) (k = 1,2, .. '). A formula similar to (14) but involving backward differences is Newton's (or Gregory-Newton's) backward difference interpolation fonnula f(x) = Pn(x) = ~ (r + : - (18) = E X AMP L E 6 fo + rVfo + r(r + ]) 1) 2! VSfo (x = Xo 2 r(1" V fo + ... + + + rh, r = (x - xo)lh) 1) ... (r n! +n- 1) Vnfo· Newton's Forward and Backward Interpolations Compute a 7D-value of the Bessel function Jo(x) for x = 1.72 from the four values in the following table, using (a) Newton's forward formula (14). (b) Newton's backward formula (18). jfor jbaclr .Ij Jo(.\j) 0 -3 1.7 0.3979849 1 -2 l.8 0.3399864 ] st Diff. 2nd Diff. 3rd Diff. -0.0579985 -0.0001693 -0.0581678 -] 2 l.9 0.2818186 0.0004093 0.0002400 -0.0579278 3 0 Solution. I 2.0 0.2238908 The computation of the differences is the same in both cases. Only their notation differs. (a) Forward. In (14) we have r = (1.72 - 1.70)/0.1 = 0.2, andj goes from each column we need the first given number, and (14) thus gives Jo(1.7 2) = 0.3979849 = + 0.2( -0.0579985) + 0.3979849 - 0.011 5997 a to 3 (see first column). In 0.2( -0.8) 0.2( -0.8)( -1.8) 2 (-0.000 1693) + 6 . 0.000 4093 + 0.000 0135 + 0.000 0196 which is exact to 6D, the exact 7D-va1ue being 0.3864185. = 0.3864183, 808 CHAP. 19 r Numerics in General (b) 8ack\\ard. For (18) we usej shown in the second column. and in each column the last number. Since = (1.72 - 2.00)/0.1 = -2.8. we thus gel from lI8) 1 0 (1.72) = 0.223 89U8 - 2.8( -0.0579278) + -2.8(- 1.8) 2 . 0.000 2400 + -2.8( = 0.223 8908 + 0.162 1978 + 0.000 6048 - 0.000 2750 = 0.386 4184. \.8)(- 6 0.8) . 0.000 4093 • Central Difference Notation This is a third notation for differences. The first central difference of f(x) at Xj is defined by and the kth central difference of f(x) at <:okf - (19) v j - Xj by <:ok-If v j+1/2 - <:ok-If v (j j-1/2 = 2,3, .. '). Thus in this notation a difference table, for example, for f -1' fo, fI, f2, looks as follows: X-I f Xo fo -I [)f -1/2 [)2fo [)3f1/2 [)fI/2 Xl fl X2 f2 [)2fl [)f3/2 Central differences are used in numeric differentiation (Sec. 19.5), differential equations (Chap. 21), and centered interpolation formulas (e.g., Everett's formula in Team Project 22). These are formulas that use function values "symmetrically" located on both sides of the interpolation point x. Such values are available near the middle of a given table, where centered interpolation formulas tend to give better results than those of Newton's formulas, which do not have that "symmetry" property. 1. (Linear interpolation) Calculate PI (x) in Example I. Compute from it In 9.4 = PI(9.4). 2. Estimate the enor in Prob. 1 by (5). 3. (Quadratic interpolation) Calculate the Lagrange polynomial 1'2(X) for the 4D-values of the Gamma function [(24), App. 3.1J r(l.oO) = 1.0000. r(l.02) = 0.9888, r(l.04) = 0.9784, and from it approximations of r{x) for x = 1.005, 1.010. 1.015. 1.025. 1.030. 1.035. 4. (Error bounds) Derive enor bounds for P2(9.2) in Example 2 from (5). S. (Error function) Calculate the Lagrange polynomial P2(X) for the 50-values of the enor function j(x) = erf x = (2/\'";) J~ e- dw, namely. 1(0.25) = 0.27633 ..«0.5) = 0.52050. f( I) = 0.84270. and fromp2 an approximation of j(O.75) (= 0.71116. 50). w2 6. Derive an error bound in Prob. 5 from (5). 7. (Sine integral) Calculate the Lagrange polynomial P2(X) for the 40-values of the sine integral Si(x) [(40) in App. 3.1], namely, Si(O) = O. Si(l) = 0.9461. Si(2) = 1.6054, and from P2 approximations of Si(O.5) (= 0.4931. 40) and Si(1.5) (= 1.3247,401. 8. (Linear and quadratic interpolation) Find e- O.25 and O 75 e- . by linear interpolation with xo = 0, Xl = 0.5 and Xo = 0.5, Xl = I. respectively. Then find pix) SEC. 19.3 809 Interpolation interpolating e- x with Xo = O. Xl = 0.5. X2 = 1 and from it e- O.25 and e- O.75 • Compare the errors of these linear and quadratic interpolations. Use 4D-values of e- x . 9. (Cubic Lagrange interpolation) Calculate and sketch or graph L o, L 1 . L 2 , L3 for x = O. 1.2.3 on common axes. Find P3(X) fOT the data (0. I) (1,0.765198) (2, 0.223891 ) (3. -0.260052) [values of the Bessel function 10(x)]. Find = 0.5, 1.5. 2.5 by interpolation. P3 for X 10. (Interpolation and extrapolation) Calculate P2(X) in Example 2. Compute from it approximations of In 9.4, In 10, In 10.5. In 11.5. In 12, compute the errors by using exact 4D-values. and comment. 11. (Extrapolation) Does a sketch or graph of the product of the (x - Xj) in (5) for the data in Prob. 10 indicate that extrapolation is likely to involve larger errors than interpolation does? 12. (Lower degree) Find the degree of the interpolation polynomial for the data (-2.33) (0.5) (2.9) (4,45) (6, 113). 13. (Newton's forward difference formula) Set up (14) for the data in Prob. 7 and derive P2(x) from (14). 14. Set up Newton's forward difference formula for the data in Prob. 3 and compute [(1.01), [(1.03), [(1.05). 15. (Newton's divided difference formula) Compute f(0.8) and f(0.9) from f(0.5) = 0.479 f(I.O) = 0.841 17. (Central differences) Write the difference in the table in Example 5 in central difference notation. 18. (Subtabulation) Compute the Bessel function 11(X) for X = 0.1. 0.3, .... 0.9 from 1 1(0) = 0,11(0.2) = 0.09950. h(O.4) = 0.19603. 1 1 (0.6} = 0.28670.11(0.8) = 0.36884, 11 (1.0) = 0.44005. Use (14) with II = 5. 19. (Notations) Compute a difference table of f(x) = x3 for X = O. 1, 2. 3.4. 5. Choose Xo = 2 and write all occurring numbers in tenTIS of the notations (a) for central differences, (b) for forward differences, (c) for backward differences. 20. CAS EXPERIMENT. Adding Terms in Newton Formulas. Write a program for the forward formula ( 14). Experiment on the increase of accuracy by successively adding terms. As data use values of some function of your choice for which your CAS gives the values needed in determining errors. 21. WRITING PROJECT. Interpolation: Comparison of Methods. Make a list of 5-6 ideas that you feel are most basic in this section. Arrange them in the best logical order. Discuss them in a 2-3 page report. 22. TEAM PROJECT. Interpolation and Extrapolation. (a) Lagrange practical error estimate (after Theorem I). Apply this to PI (9.2) and P2(9.2) for the data Xo = 9.0. Xl = 9.5. X2 = 11.0. fo = In Xo. fl = In XI> f2 = In X2 (6S-values). (b) Extrapolation. Given (.~i' f(x) = (0.2.0.9980). (0.4. 0.9686). (0.6. 0.8443), (0.8, 0.5358), (1.0. 0). Find f(0.7) from the quadratic interpolation polynomials based on (a) 0.6, 0.8. 1.0, ({3) 0.4. 0.6. 0.8. (y) 0.2, 0.4, 0.6. Compare the errors and comment. [Exact f(x) = cos (!7TX 2 ), f(O.7) = 0.7181 (45).1 (c) Graph the product of factors (x - xi> in the error formula (5) for 11 = 2.' . '. 10 separately. What do these graphs show regarding accuracy of interpolation and extrapolation? (d) Central differences. Show that 8 2 fm = fm+l - 2fm + fm-I> and. furthermore 8 3 fm+I12 = fm+2 - 3fYII f 1 + 3fm - f m - 1 • n 8 fm = t!.."fm-n/2 = vnfm+n/2' (e) Everett's interpolation formula f(2.0) = 0.909 f(x) = (I - r)fo by quadratic interpolation. (20) 16. Compute f(6.5) from + (2 - + rf1 r)O - 3! r)( -r) 2 8 fo f(6.0) = O. J 506 f(7.0) = 0.3001 f(7.5) = 0.2663 fO.7) = 0.2346 by cubic interpolation, using (10). is an example of a formula involving only even-order differences. Use it (0 compute the Bessel function Jo(x) for x = 1.72 from 10(l.60} = 0.4554022 and 10 (1.7). 10(1.8),10(1.9) III Example 6. 810 19.4 CHAP. 19 Numerics in General Spline Interpolation Given data (function values, points in the .\y-plane) (.\"0' f 0), (XI- f 1)' . . . , (Xn , f n) can be interpolated by a polynomial P,/x) of degree 11 or less so that the curve of P,ix) passes through these 11 + 1 points (Xj, fj); here fo = f(xo), ...• fn = f(x,.). See Sec. 19.3. Now if 11 is large, there may be trouble: P n(x) may tend to oscillate for x between the nodes xo, ... , x",. Hence we must be prepared for numeric instability (Sec. 19.1). Figure 431 shows a famous example by C. Runge 3 for which the maximum error even approaches x as 11 ~ x (with the nodes kept equidistant and their number increased). Figure 432 illustrates the increase of the oscillation with 11 for some other function that is piecewise linear. Those undesirable oscillations are avoided by the method of splines initiated by I. J. Schoenberg in 1946 (Quarterly of Applied Mathematics 4, pp. 45-99, 112-141). This method is widely used in practice. It also laid the foundation for much of modem CAD (computer-aided design). its name is borrowed from a draftman's spline, which is an elastic rod bent to pass through given points and held in place by weights. The mathematical idea of the method is as follows: Instead of using a single high-degree polynomial Pn over the entire interval a ~ x ~ b in which the nodes lie, that is, a (1) we use 11 = Xo < Xl < ... < Xn = b, low-degree, e.g., cubic, polynomials one over each subinterval between adjacent nodes. hence qo from Xo to Xl' then q1 from YI -5 "'--/ Fig. 431. Fig. 432. 3 -~. o Runge's example {(x) = 1/(1 + x 2 ) and "'-../ 0 5 x interpolating polynomial PlQ(x) Piecewise linear function {(x) and interpolation polynomials of increasing degrees CARL RUNGE (1856-1927). German mathematician, also known for his work on ODEs (Sec. 21.1). SEC. 19.4 811 Spline Interpolation to X2' and so on. From this we compose an interpolation function g(x), called a spline. by fitting these polynomials together into a single continuous curve passing through the data points, that is. Xl Note that g(x) = qo(.\") when Xo ~ x ~ Xl' then g(x) = q1(X) when Xl ~ X ~ X2, and so on, according to our construction of g. Thus spline interpolation is piecewise polY1lomial intel]Jolatioll. The simplest q/ s would be linear polynomials. However. the curve of a piecewise linear continuous function has corners and would be of little interest in general-think of designing the body of a car or a ship. We shall consider cubic splines because these are the most important ones in applications. By definition, a cubic spline g(x) interpolating given data (xo, f 0), ... , (xn , fn) is a continuous function on the interval a = Xo ~ X ~ X." = b that has continuous first and second derivatives and satisfies the interpolation condition (2); furthermore, between adjacent nodes, g(x) is given by a polynomial %(x) of degree 3 or less. We claim that there is such a cubic spline. And if in addition to (2) we also require that (3) (given tangent directions of g(x) at the two endpoints of the interval a ~ X ~ b), then we have a uniquely detennined cubic spline. This is the content of the following existence and uniqueness theorem. whose proof will also suggest the actual determination of splines. (Condition (3) will be discussed after the proof.) THEOREM 1 Existence and Uniqueness of Cubic Splines Let (xo, f 0), (xt> f1)' .•• , (xn , f.,,) with arbitrarily spaced givell Xj [see (I)] and given fj = f(x),.i = 0, I, ... , Il. Let ko alld k n be allY given numbers. Then there is olle and only one cubic spline g(x) corresponding to (1) alld satisfying (2) and (3). PROOF By definition, on every subintervallj given by Xj ~ x ~ Xj+ 1 the spline g(x) must agree with a polynomial %lX) of degree not exceeding 3 such that (4) (j = 0, (j = 0, 1, ... , n - I) I, ... , 17 - I). For the derivatives we write (5) with ko and kn given and k1, ••• , kn - 1 to be detemllned later. Equations (4) and (5) are four conditions for each qj(x), By direct calculation, using the notation (6*) 1 (j = 0, 1, ... , Il - 1) we can verify that the unique cubic polynomial %(x) (j = 0, 1, ... , n - I) satisfying 812 CHAP. 19 Numerics in General (4) and (5) is %(X) = f(xj)c/(x - Xj+l)2[1 (6) + 2clx - x)] + f(Xj+l)C/(X - X)2[1 - 2cj(x - Xj+l)] + kjc/(x - Xj)(x - Xj+l)2 + kj + 1C/(X - X)2(X - Xj+l)' Differentiating twice, we obtain (8) By definition, g(x) has continuous second derivatives. This gives the conditions (j = I,' .. , 11 - I). If we use (8) withj replaced by j - I, and (7), these n - I equations become where ViJ = f(xj) - f(Xj-l) and Vfj+l = f(Xj+l) - f(xj) and j = 1,' .. , n - I, as before. This linear system of 11 - I equations has a unique solution k1 , ••• , kn - 1 since the coefficient matrix is strictly diagonally dominant (that is. in each row the (positive) diagonal entry is greater than the sum ofthe other (positive) entries). Hence the determinant of the matrix cannot be zero (as follows from Theorem 3 in Sec. 20.7), so that we may determine unique values k] • ... , kn - 1 of the first derivative of g(x) at the nodes. This proves the theorem. • Storage and Time Demands in solving (9) are modest. since the matrix of (9) is sparse (has few non7ero entries) and tridiagonal (may have nonzero entries only on the diagonal and on the two adjacent "parallels" above and below it). Pivoting (Sec. 7.3) is not necessary because of that dominance. This makes splines efficient in solving large problems with thousands of nodes or more. For some literature and some critical comments, see American Mathematical Monthly 105 (1998), 929-941. Condition (3) includes the clamped conditions (10) in which the tangent directions t' (xo) and t' (x,,) at the ends are given. Other conditions of practical interest are the free or natural conditions (11) (geometJically: zero curvature at the ends, as for the draftman's spline), giving a natural spline. These names are motivated by Fig. 290 in Problem Set 12.3. SEC. 19.4 813 Spline Interpolation Determination of Splines. Let ko and kn be given. Obtain kl' .... k n - l by solving the linear system (9). Recall that the spline g(x) to be found consists of n cubic polynomials qo, ... , qn-l' We write these polynomials in the form (12) where j = O•...• 11 - I. Using Taylor's formula. we obtain ajO = q/x) = Ij by (2), q; (Xj) = kj by (5), ajl = by (7), (13) with aj3 obtained by calculating q;'(Xj+l) from (12) and equating the result to (8), that is, and now subtracting from this 2aJ 2 as given in (13) and simplifying. Note that for equidistant nodes of distance hj = h we can write cJ = and have from (9) simply = 1111 in (6*) (j = 1, ... , (14) E X AMP L E 1 C 11 - I). Spline Interpolation. Equidistant Nodes Interpolate f(x) = .,.4 on the interval -I :::;:: x :::;:: I by the cubic spline g( r) corresponding to the nodes Xo = -1. Xl = O. X2 = 1 and satisfying the clamped condition, g' (-1) = f' (- I), g '(1) = f' (I). Solution. We have h In our ,tandard notation the given data are fo = f( - \) = 1. II = frO) = O. = 1 and 11 = 2, so that our spline consists of 11 = 2 polynomials qo(x) = 1I00 ql(X) = 1I1O + lIOl(" + + 1Il1-~ + 1) + 1I12X2 lI02(x + 1)2 + lI03(x + 1)3 + {/13..3 f2 = f( 1) (-\ :::;:: x:::;:: 0). (0 :::;:: x:::;:: I). We determine the kj from (14) (equidistance!) and then the coefficients of the spline from (13). Since the sp,tem (14) is a ,ingle equation (with j = 1 and h = I) ko Here fo = f2 = \ (the value of x4 at the ends) and ko - I and 1. Hence + 4kl 11 = 2, + 4kl + k2 = 3(f2 - fo)· = -4. k2 = 4. the value~ of the derivative 4\·3 at the end~ -4 = 1. + 4 = 3(1 - I) = 0, From (13) we can now obtain the coefficients of qo, namely, {loo = fo = 1, {lOI = ko = -4. and 814 CHAP. 19 Numerics in General 3 a02 = ]2 (h - Io) a03 = 2 3(Io I 1 -1 (k1 + h) + 2ko) I "2(kl + ko) I = 3(0 - 1) - (0 - 8) = 2(1 - Similarly, for the coefficients of qi we obtain from (13) the values a12 = 3(I2 - h) - a13 = 2(h - I2) (k2 0) + (0 - 0lO = II 4) = 0, = 5 = -2. all = kl = 0, and + 2k1 ) = 3(1 - 0) - (4 + 0) = -1 = 2(0 + (k2 + k 1) - I) + (4 + 0) = 2. This gives the polynomials of which the spline g(x) consists. namely, if if -I ~x~O O~x~l. Figure 433 shows .f(x) and this spline. Do you see that we could have saved over half of our work by using symmetry? • ((x) x Fig. 433. E X AMP L E 2 Function fIx) = X4 and cubic spline g(x) in Example 1 Natural Spline. Arbitrarily Spaced Nodes Find a spline approximation and a polynomial approximation for the curve of the cross section of the circularshaped Shrine of the Book in Jerusalem shown in Fig. 434. '" ~ • • ;; I T: Jl ~ -3 Fig. 434. -2 0 Shrine of the Book in Jerusalem (Architects F. Kissler and A. M. Bartus) SEC. 19.4 Spline Interpolation 815 Solution. Thirteen points. about equally distributed along the contour (nut along the x-axis!), give these data: Xj 5.8 Ij 0 -5.0 -4.0 -2.5 -1.5 -0.8 1.5 1.8 2.2 2.7 3.5 0 0.8 1.5 2.5 4.0 5.0 5.8 3.9 3.5 2.7 2.2 1.8 1.5 0 The figure shows the conesponding interpolation polynomial of 12th degree, which is useless because of its oscillation. (Because of roundoff your software will also give you small enor terms involving odd powers of x.) The polynomial is P12(X) = 3.9000 - 0.65083..2 + 0.033858x4 + 0.01l04lx 6 + 0.000055595x 10 - - 0.00I40I0x 8 0.00000071 867x 12 . The spline follows practically the contour of the roof, with a small error near the nodes -O.li and 0.8. The spline is symmetric. Its six polynomials corresponding to positive x have the fullowing coefficieuts of their represeutations (12). (Note well that (12) is in terms of powers of x - Xj, uot x!) I I j x-interval ajO ajl aj2 aj3 0 0.0... 0.8 0.8. .. 1.5 [.5 .. .2.5 2.5 .. .4.0 4.0 ... 5.0 5.0 ... 5.8 3.9 3.5 2.7 2.2 1.8 1.5 0.00 -1.01 -0.95 -0.32 -0.027 -1.13 -0.61 -0.65 0.73 -0.09[ 0.29 -1.39 -0.015 0.66 -0.27 0.084 -0.56 0.58 1 2 3 4 5 - -- ....._...._-. ..... _........... -. ~ 1. WRITING PROJECT. Splines. In your own words, and using as few formulas as possible, write a short report on spline interpolation, its motivation, a comparison with polynomial interpolation, and its applications. 2. (Individual polynomial qj) Show that qj(x) in (6) satisfies the interpolation condition (4) as well as the derivative condition (5). 3. Verify the differentiations that give (7) and (8) from (6). 4. (System for derivatives) Derive the basic linear system (9) for k1 , . . . , k n - 1 as indicated in the text. 5. (Equidistant nodes) Derive (14) from (9). 6. (Coefficients) Give the details of the derivation of aj2 and aj3 in (13). 7. Verify the computations in Example I. 8. (Comparison) Compare the spline g in Example I with the quadratic interpolation polynomial over the whole interval. Find the maximum deviations of g and P2 from f. Comment. 9. (Natural spline condition) Using the given coefficients, verify that the spline in Example 2 satisfies g"(x) = 0 at the ends. 110-161 DETERMINATION OF SPLINES Find the cubic spline g(x) for the given data with ko and k n as given. 10. f( -2) = .f( -1) ko = k4 = 0 = f(1) = f(2) = O. f(O) = I. 11. If we started from the piecewise linear function in Fig. 435. we would obtain g(x) in Prob. 10 as the spline satisfying g' (-2) = f' (-2) = 0, g' (2) = f' (2) = O. Find and sketch or graph the corresponding interpolation polynomial of 4th degree and compare it with the spline. Comment. I ~2_-&-----=--2 -----/ 0 ---- x Fig. 435. Spline and interpolation polynomial in Problems 10 and 11 12. fo = f(O) h = f(6) 1, fl = f(2) = 9, f2 = f(4) 41, ko = 0, k3 = -12 = 41, 816 CHAP. 19 Numerics in General 13. fo = f(-I) = O. fl = f(O) = 4. f2 = f(1) = O. ko = 0, k2 = O. Is g(x) even? (Give reason.) 14. fo = f(O) = o . .ft = fO) = L f2 = f(Z) = 6. f3 = f(3) = 10. ko = 0, k3 = 0 15. fo = f(O) = \. fl = fO) = O. f2 = feZ) = - \ . f3 = f(3) = 0.1..0 = O. k3 = -6 16. It can happen that a spline is given by the same polynomial in two adjacent subintervals. To illustrate this, find the cubic spline g(x) for f(x) = sin x corresponding to the partition Xo = -7TI1. Xl = 0, X2 = 7T/2 of the interval -7T/Z ~ x ~ 7T/Z and satisfying g'(-7T/Z) = f'(-7T/2) and in components. \"(t) = Xo + y(t) = Yo + + X~f + (2(xo - + Y~f (2(yo - (3(Xl - Xl) + + X~ (3(Yl YI) xo) - (2x~ + 3 X;)t l'O) - + y~ + -t- X;»f2 (2y~ + y;»t 2 y;)£3. Note that this is a cubic Hennite interpolation polynomiaL and 11 = I because we have two nodes (the endpoints of C). (This has nothing to do with the Hermite polynomials in Sec. 5.S.) The two points g'(7T/Z) = f'(7T/2). 17. (Natural conditions) Explain the remark after (II). 18. CAS EXPERIMENT. Spline versus Polynomial. [f your CAS gives natural splines, find the natural splines when x is integer from -111 to Ill, and yeO) = I and all other y equal to O. Graph each such spline along with the interpolation polynomial P2m' Do this for 111 = Z to 10 (or more). What happens with increasing 111? 19. If a cubic spline is three times continuously differentiable (that is, it has continuous first, second. and third derivatives). show that it must be a single polynomial. 20. TEAM PROJECT. Hermite Interpolation and Bezier Curves. In Hermite interpolation we are looking for a polynomial p(x) (of degree ZI1 + I or less) such that p(x) and its derivative p' (x) have given values at 11 + I nodes. (More generally, p(x). p' (x), p"(xJ, ... may be required to have given values at the nodes.) (a) Curves with given endpoints and tangents. Let C be a curve in the x),-plane parametrically represented by ret) = [x(t), y(t)], 0 ~ t ~ I (see Sec. 9.5). Show that for given initial and terminal points of a curve and given initial and terminal tangents. say, A: ro = [x(O). yeo)] [xo, Yo]. [Xl' Vo = x~. Yo + Y~] = [Xl - x;, YI - Y;] are called guidepoints because the segments AGA and BGB specify the tangents graphically. A, 8, GA , GB determine C. and C can be changed quickly by moving the points. A curve consisting of such Hemlite interpolation polynomials is called a Bezier curve. after the French engineer P. Bezier of the Renault Automobile Company. who introduced them in the early 1960s in designing car bodies. Bezier curves (and surfaces) are used in computer-aided design (CAD) and computer-aided manufacturing (CAM). (For more detaib, ~t!e Ref. [E21] in App. \.) (b) Find and graph the Bezier curve and its guidepoints if A: [0, 0]. 8: [I, 0], Vo = [i, VI = [-i, -!vi3]. n (c) Changing guidepoints changes C. Moving guide points farther away makes C "staying near the tangents for a longer time." Confirm this by changing Vo and VI in (b) to Zvo and 2Vl (see Fig. 436). (d) Make experiments of your own. What happens if you change VI in (b) to -VI' If you rotate the tangents? If you multiply Vo and VI by positive factors less than I? [x(l). y(J)] r1 8: = [xo + and yd [x'(O), ),'(0)] [x~, y~]. VI [x'(I), [x~. y'(I)] y;] we can find a curve C, namely, reT) (15) = ro + vot + + (3(r 1 ro) - (ZVo (Z(r o - r 1 ) + Vo + + vtl)t 2 V 1 )t 3 ; B Fig. 436. x Team Project 20(b) and (c): Bezier curves SEC. 19.5 19.5 817 Numeric Integration and Differentiation Numeric Integration and Differentiation Numeric integration mean<; the numeric evaluation of integrals J = I b f(x) dx a where a and b are given and f is a function given analytically by a formula or empirically by a table of values. Geometrically, J is the area under the curve of f between a and b (Fig. 437). We know that if f is such that we can find a differentiable function F whose derivative is f. then we can evaluate J by applying the familiar formula J = I b f(x) dx = [F' (x) = F(h) - F(a) f(x)j. a Tables of integrals or a CAS (Mathematica. Maple, etc.) may be helpful for this purpose. However, applications often lead to integrals whose analytic evaluation would be very difficult or even impossible, or whose integrand is an empirical function given by recorded numeric values. Then we may obtain approximate numeric values of the integral by a numeric integration method. Rectangular Rule. Trapezoidal Rule Numeric integration methods are obtained by approximating the integrand f by functions that can easily be integrated. The simplest formula. the rectangular rule. is obtained if we subdivide the interval of integration a;:::; x;:::; b into /l subintervals of equal length II = (b - a)//l and in each subinterval approximate f by the constant f(x/), the value of f at the midpoint x/ of the jth subinterval (Fig. 438). Then f is approximated by a step function (piecewise constant function). the 11 rectangles in Fig. 438 have the areas f('\·I*)I1 • ... , f(xn*)I1, and the rectangular rule is J = (1) I b f(x)dx = h[f(Xl*) + f(X2*) + ... + f(xn*)] a ( b-ll) /7=--n . The trapezoidal rule is generally more accurate. We obtain it if we take the same subdivision as before and approximate f by a broken line of segments (chords) with endpoints [a, f(a)], [Xl> f(Xl)], ... , [b, feb)] on the curve of f (Fig. 439). Then the area under the curve of f between a and b is approximated by n trapezoids of areas Mf(a) + f(Xl)]h, ![f(Xl) + f(X2)]h, ![f(Xn-l) y y y={(x) ~ R a b X Fig. 437. Geometric interpretation of a definite integral + f(b)]h. r/oK 11 '" ) I I a xt X· 2 Fig. 438. X I * b n Rectangular rule x CHAP. 19 818 Numerics in General y !~~ ) 'I o( 00 x Trapezoidal rule Fig. 439. By taking their slim we obtain the trapezoidal rule (2) ] I = b lex) dx = h[!l(a) + f(X1) + f(X2) + ... + f(Xn-l) + !f(b)] a = where h E X AMP L E 1 (b - a)/n, as in (1). The x/s and a and b are called nodes. Trapezoidal Rule = Evaluate J f 1 e -:? dx by means of (2) with 11 = 10. o Solution. Table 19.3 j 0 2 3 4 5 6 7 8 9 10 J = 0.1(0.5·1.367879 + 6.778167) = 0.746211 from Table 19.3. • Computations in Example 1 Xj X/ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 0.01 0.04 0.09 0.16 0.25 0.36 0.49 0.64 0.81 1.00 Sums 1.000000 0.367879 1.367879 Error Bounds and Estimate for the Trapezoidal Rule An error estimate for the trapezoidal rule can be derived from (5) in Sec. 19.3 with n = 1 by integration as follows. For a single subinterval we have f(x) - P1(X) = f'(t) (x - Xo)(X - Xl) - - 2 SEC 19.5 819 Numeric Integration and Differentiation with a suitable t depending on x, between Xo and Xl. Integration over X from a Xl = Xo + h gives J Iz xo+h ~ f(x) dx - - 2 J + f(XI)] = (y - xo)(x - Xo - II) ~ Xo to f"(t(x» xo+h [fCyo) = 2 dx. Setting X - Xo = v and applying the mean value theorem of integral calculus, which we can use because (x - xo)(x - Xo - h) does not change sign, we find that the right side equals {'Ci) v(v - III dv - o 2 f (3*) h = 3 ( h 3 - 3 3 h ) f"Ci) h -- = - 2 2 12 - _ f " (t) where t is a (suitable, unknown) value between Xo and Xl. This is the error for the trapezoidal rule with n = 1, often called the local error. Hence the error E of (2) with any n is the sum of such contributions from the n subintervals; since /z = (b - a)/n. nh 3 = neb - a)3/1l 3• and (b - a)2 = n 2/z 2• we obtain (3) E=- (b - a)3 II ~ (b - a) 2 II ~ 12n 2 fU)=12 hf(t) with (suitable, unknown) i between a and b. Because of (3) the trapezoidal rule (2) is also written (2*) 1 = b [1 1 J f(x) dx = /z "2f(a) + f(XI) + ... + f(Xn-I) + "2f(b) a ]- b - a 2" ~ - h f (t). 12 Error Bounds are now obtained by taking the largest value for f", say, M 2 , and the smallest value, M 2 *. in the interval of integration. Then (3) gives (note that K is negative) (4) where (b - a)3 K = - ---=-2 12n Error Estimation by Halving h is advisable if h" is very complicated or unknown, for instance, in the case of experimental data. Then we may apply the Error Principle of Sec. 19.1. That is, we calculate by (2), first with h. obtaining, say, 1 = 1h + Eh, and then with ~/z, obtaining 1 = h/2 + Eh/2. Now if we replace /z2 in (3) with (~h)2, the error is multiplied by 1/4. Hence Eh/2 = iEh (not exactly because i may differ). Together, 1h/2 + Eh/2 = 1h + Eh = 1h + 4Eh/2. Thus 1h/2 - 1h = (4 - l)Eh/2. Division by 3 gives the error formula for 1h/2 (5) E X AMP L E 2 Error Estimation for the Trapezoidal Rule by (4) and (5) Estimate the error of the approximate value in Example I by (4) and (5). Solution. o (A) Error boullds by (4). By differentiation, f"(x) = 2(2x 2 - l)e- x2 . Also, {"(x) > 0 if < x < 1, so that the minimum and maximum occur at the ends of the interval. We compute 820 CHAP. 19 M2 = Numerics in General f"(I) = 0.735759 and M2* = ('(0) = -2. Furthenl1ore. -0.000 614 ~ E ~ K = -1/1200. and (4) gives 0.001 (,67. Hence the exact value of 1 must lie between 0.74(,211 - 0.000 fil4 = 0.745597 0.74(, 21 1 and + 0.001 6(,7 = 0.747 87R. Actually, 1 = 0.746 824, exact 10 60, (8) Error estimate by (5). lh = 0,746211 in Example I. Also. lh/2 = 0,05 [~ e-cil2ol' + + + 0.367R79)] (1 0,74(,671. = J~l Hence Eh/2 = l(Jh/2 - 1,,) = 0,000153 and lh/2 + E"/2 • = 0,746824. exact (0 60, Simpson's Rule of Integration Piecewise constant approximation of f led to the rectangular rule (I), piecewise linear approximation to the trapezoidal rule (2), and piecewise quadratic approximation will lead to Simpson's rule. which is of great practical importance because it is sufficiently accurate for most problems, but still sufficiently simple. To derive Simpson's rule, we divide the interval of integration a 2 x 2 b into an even Ilumber of equal subintervals. say, into 11 = 2m subintervals of length h = (b - a)/(2m), with endpoints Xo (= a), Xl, ... , X2m-1' X2m (= b); see Fig. 440. We now take the first two subintervals and approximate f(x) in the interval Xo 2 X 2 X2 = Xo + 211 by the Lagrange polynomial P2(X) through (xo, fo). (XIo f1), (X2, f2), where f j = f(xj). From (3) in Sec. 19.3 we obtain The denominators in (6) are 2h2, -112, and 2172, respectively. Setting s have X - Xl = sl1, Xo = x - (Xl - h) = (s X - X - X2 = X - (Xl + 11) = + l}h 1)17 = (s - and we obtain P2(X) = !s(s - l)fo - (s y + 1)(S - l)f1 + !(S + l)sf2' r; -...:.: -J First parabola _ ,./'' Second parabola rl~ rI ~ Lo'T~" .. I~Y, I I I x Fig. 440. Simpson's rule (x - x1)111, we SEC. 19.5 821 Numeric Integration and Differentiation We now integrate with respect to x from Xo to X2' This corresponds to integrating with respect to s from -1 to 1. Since dx = h ds, the result is J X2 (7*) = f(x) dx Xv J""2P2(X) dx = h ( -31fo + -34f1 + "3I) f2 . Xv A similar formula holds for the next two subintervals from X2 to -'"4, and so on. By summing all these 111 formulas we obtain Simpson's rule4 (7) I h b f(x) dx = - 3 a + 4f1 + (fo where h = (b - a)/(211l) and fj rule. Table 19.4 2f2 + 4f3 + ... + 2f2m-2 + 4f2m-1 + f2m), f(xj)' Table 19.4 shows an algorithm for Simpson's = Simpson's Rule of Integration ALGORITHM SIMPSON (a, b, 111, fo, f1, ... , f2m) This algorithm computes the integral j = Jgf(x) dx from given values fj = f(xj) at equidistant Xo = {/, Xl = Xo + h . .... X2m = Xo + 2mh = b by Simpson's rule (7). where h = (b - a)/(2m). INPUT: a, b, fo • ... , f2m Approximate value J of j OUTPUT: Compute 111. fo + f2m = f2 + f4 So = S2 + ... + f2m-2 h = (b - a)/2111 _ j h = - 3 (so + 4s1 + 2s2 ) OUTPUT J Stop. End SIMPSON Error of Simpson's Rule (7). If the fourth derivative a ::::; x ::::; b, the error of (7), call it ES' is (8) ES =- (b - a)5 180(2171)4 (4) At _ f ()- t<4) exists and is continuous on _(b_-_G_J h 4f(4)(i)' 180 ' ~HOMAS SIMPSON (1710-1761), self-taught English mathematician. author of several popular textbooks. Simpson's rule was used much earlier by Torricelli, Gregory (in 1668), and Newton (in 1676). 822 CHAP. 19 Numerics in General here t is a suitable unknown value between a and b. This is obtained similarly to (3). With this we may also write Simpson's rule (7) as (7**) Error Bounds. By taking for t<4) ill (8) the maximum M4 and minimum M4 * on the interval of integration we obtain from (8) the error bounds (note that C is negative) C where (9) (b - a)5 = - ----:180(2m)4 Degree of Precision (DP) of an integration f0I111ula. This is the maximum degree of arbitrary polynomials for which the formula gives exact values of integrals over any intervals. Hence for the trapezoidal rule, DP = I because we approximate the curve of f by portions of straight lines (linear polynomials). For Simpson's rule we might expect DP = 2 (why?). Actually, DP = 3 by (9) because /4) is identically zero for a cubic polynomiaL This makes Simpson's nile sufficiently accurate for most practical problems and accounts for its popUlarity. Numeric Stability with respect to rounding is another impOltant property of Simpson'~ nile. Indeed, for the sum of the roundoff errors Ej of the 2m + I values f j in (7) we obtain. since Iz = (b - a)12111, h -3 lEo + 4El + ... + E21111 ~ (b - a) 3·2m 6111u = (b - a)u where u is the rounding unit (u = ~. 10- 6 if we round off to 6D; see Sec. 19.1). Also 6 = I + 4 + I is the sum of the coefficients for a pair of intervals in (7); take 111 = I in (7) to see this. The bound (b - (I) u is independent of Ill, so that it cannot increase with increasing 11l, that is, with decreasing h. This proves stability. • Newton-Cotes Formulas. We mention that the trapezoidal and Simpson rules are special closed Newton-Cafes formulas, that is, integration formulas in which f(x) is interpolated at equalIy spaced nodes by a polynomial of degree n (n = I for trapezoidal, Il = 2 for Simpson), and closed means that a and b are nodes (a = .ro, b = xn). Il = 3 (the three-eighths nile; Review Prob. 33) and a higher n are used occasionally. From n = 8 on, some of the coefficients become negative, so that a positive f· could make a ~egative contribution to an integral, which is absurd. For more on this topi~ see Ref. [E25] m App. I. SEC 19.5 823 Numeric Integration and Differentiation E X AMP L E 3 Simpson's Rule. Error Estimate Evalume J = f Solution. Since h = 0.1, Table 19.5 give, 1 e -:i'- dx by Simpson's rule with 2m = 10 and estimate the error. o 0.1 3 J = (1.367879 + 4· 3.740 266 + 2· 3.037 901) = 0.746825. Estimate of error. Differentiation gives ( 4 )(x) = 4(4..4 - 121.2 + 3)e-:i'-. By considering the derivative /5) of /4) we find that the largest value of /4) in the interval of integration occurs at 0 and the smallest value at r'" = (2.5 - 0.5"\' loi J2 . Computation gives the values M4 = /4)(0) = 12 and 1114* = /4)(x*) = -7.419. Since 2m = 10 and b - a = I, we obtain C = - 111 800 000 = -0.000 000 56. Therefore. from (9). -0.000 007 ~ ES ~ 0.000 005. Hence J must lie between 0.746825 - 0.000 007 = 0.746818 and 0.746825 + 0.000005 = 0.746830, so that at least four digits of our approximate value are exact. Actually. the value 0.746825 is exact to 5D because J = 0.746824 (exact to 6D). Thus our result is much better than that in Example 1 obtained by the trapeLOidal rule. whereas the number of operations is nearly the same in both cases. • Table 19.5 j 0 Computations in Example 3 Xj xl 0 0 0.1 0.01 2 0.2 0.04 3 4 5 6 7 8 9 [0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.09 0.16 0.25 0.36 0.49 0.64 0.81 1.00 e- Xj 1.000000 0.990050 0.960789 0.913931 0.852144 0.778801 0.697676 0.612626 0.527 292 0.444 858 0.367879 1.367879 Sums 2 3.74U 266 3.037 YUl Instead of picking an n = 2m and then estimating the error by (9), as in Example 3, it is better to require an accuracy (e.g., 6D) and then determine 11 = 2111 from (9). EX AMP L E 4 Determination of n What II = 2m in Simpson's Rule from the Required Accuracy should we choose in Example 3 to get 6D-accuracy? Solution. Using M4 = 12 (which i, bigger in absolute value than M 4 *). we get from (9), with b and the required accuracy. thus 2'106'12 m = [ 180·2 4 J1I4 a= 1 = 9.55. Hence we should choose 11 = 2111 = 20. Do the computation, which paralleb that in Example 3. Note that the error bounds in (4) or (9) may sometimes be loose, so that in such a case a smaller 11 = 2m may already suffice. • 824 CHAP. 19 Numerics in General Error Estimation for Simpson's Rule by Halving h. and gives The idea is the same as in (5) (10) lit is obtained by using hand lh/2 by using ~h. and Eh/2 is the error of l h12 . Derivati()n. In (5) we had ~ as the reciprocal of 3 = 4 - I and ~ = (~)2 resulted from 1z2 in (3) by replacing II with ~h. In (0) we have l~ as the reciprocal of 15 = 16 - 1 and {6 = (~)4 results from /1 4 in (8) by replacing h with ~II. E X AMP L E 5 Error Estimation for Simpson's Rule by Halving Integrate flX) = l'1TX4 coslm- from 0 to 2 with" = I and apply (10). Solution. The exact 5D-value of the integral is 1 = 1.25953. Simp,on's rule gives h = Hf(O} + 4f(l) + f(2)] = i [.f(0) 11</2 = I = "6 LO + + l(O + 4· 0.555360 + 0) = 0.740480. 4.f( +) + 2.f(l) + 4J (%) + .f(2)] 4' 0.045351 + 2 ·0.555361 + 4· 1.521579 + 0] = 1.21974. Hence (10) gives e1>/2 = if;(1.21974 - 0.74048) = 0.032617 and thus 1 = 1h/2 + E1</2 = 1.26236. with an error -0.00283. which is less in absolute value than of the error 0.02979 of 1,,/2' Hence the use of (10) was well worthwhile. • to Adaptive Integration The idea is to adapt step h to the variability of f(x). That is, where f varies but little, we can proceed in large steps without causing a substantial error in the integraL but where .f varies rapidly, we have to take small steps in order to stay everywhere close enough to the curve of f. Changing h is done systematically, usually by halving 11, and automatically (not "by hand"') depending on the size of the (estimated) error over a subinterval. The subinterval is halved if the cOlTesponding error is still too large, that is, larger than a given tolerance TOL (maximum admissible absolute en-or), or is not halved if the error is less than or equal to TOL. Adapting is one of the techniques typical of modern software. In connection with integration it can be applied to various methods. We explain it here for Simpson's rule. In Table 19.6 a star means that for that subinterval, TOL has been reached. E X AMP L E 6 Adaptive Integration with Simpson's Rule 4 Integrate J(x) = l'1Tv cos l'1TX from x = 0 to 2 by adaptive integration and with Simpson's rule and TOLfO. 2J = 0.0002. Solutioll. Table 19.6 shows the calculations. Figure 441 show, the integrand ffx) and the adapted intervals used. The first two intervals ([0. 0.5], [0.5, 1.0j) have length 0.5. hence h = 0.25 [because we use 2111 = 2 subintervals in Simpson's rule (7**)1· The next two interval~ ([1.00. 1.25J. [1.25. 1.50]) have length 0.25 (hence h = 0.125) and the la~t four intervals have length 0.125. Sample computatio"s. For 0.7-10480 ~ee Example 5. Formula (10) gives (0.123716 - 0.122794)115 = 0.000061. Note that 0.123716 refers to [0. 0.5] and [0.5. 11. so that we must subtract the value conesponding to [0, I] in the line before. Etc. TOL[O, 2] = 0.0002 gives SEC. 19.5 825 Numeric Integration and Differentiation 0.0001 for subintervals oflength 1,0.00005 for length 0.5. etc. The value of the integral obtained is the sum of the values marked by an asterisk (for which the error estimate has become less than TaL). This gives J = 0.123716 + 0.52~8l)5 + 0.388263 + 0.218483 = 1.25936. The exact 5D-value is J = 1.25953. Hence the error is 0.00017. This is about 11200 of the absolute value of that in Example 5. Our more extensive computation has produced a much better result. • Table 19.6 Computations in Example 6 Interval Integral Error (10) TOL Comment [0,2] 0.740480 [0. 11 [1, 2] 0.122794 1.10695 Sum = 1.22974 0.032617 0.0002 Divide fut1her 0.UU4782 0.118934 Sum = 0.123716* 0.000061 0.0001 TOL reached 0.528176 0.605821 Sum = 1.13300 0.001803 0.0001 Divide further 0.200544 0.328351 Sum = 0.528895* 0.000041< 0.00005 TOL reached 0.388235 0.218457 Sum = 0.606692 0.000058 0.00005 Divide further 0.1%244 0.192019 Sum = 0.388263* 0.000002 0.000025 TOL reached 0.153405 0.065078 Sum = 0.218483* 0.000002 0.000025 TOL reached [U.O, 0.5] [0.5, 1.0] [1.0. 1.5] [1.5, 2.0] [1.UU, 1.25] l1.25. 1.50) [1.50. 1.75] [1.75, 2.00] [1.500, 1.625] [1.625, 1.750] 0.0002 - [1.750. 1.875] [1.875, 2.000] {(xl 1.5 1.0 0.5 o o Fig. 441. 0.5 x Adaptive integration in Example 6 826 CHAP. 19 Numerics in General Gauss Integration Formulas Maximum Degree of Precision Our integration fonnulas discussed so far use function values at predetermined (equidistant) x-values (nodes) and give exact results for polynomials not exceeding a certain degree [called the degree of precision; see after (9)]. But we can get much more accurate integration formulas as follows. We set I (11) n 1 f(t) dt = ~ AJfj j~l -1 with fixed 11, and t = ± I obtained from x = a, b by setting x = HaCt - I) + bet + 1)]. Then we detennine the 11 coefficients AI' ... , AT! and 11 nodes t l •••• , tn so that (II) gives exact results for polynomials of degree k as high as possible. Since 11 + 11 = 211 is the number of coefficients of a polynomial of degree 211 - 1, it follows that k ~ 2n - I. Gauss has shown that exactne!>s for polynomials of degree not exceeding 2n - 1 (instead of n - I for predetermined nodes) can be attained, and he has given the location of the tj (= the jth zero of the Legendre polynomial P n in Sec. 5.3) and the coefficients Aj which depend on 11 but not on f(t), and are obtained by using Lagrange' s interpolation polynomial, as shown in Ref. [E5] listed in App. I. With these tj and Aj , formula (11) is called a Gauss integration formula or Gauss quadrature formllla. Its degree of precision is 211 - I, as just explained. Table 19.7 gives the values needed for n = 2 .... , 5. (For larger 11. see pp. 916-919 of Ref. [GRI] in App. 1.) Table 19.7 n 2 3 4 5 E X AMP L E 7 Gauss Integration: Nodes tj and Coefficients Aj Nodes -0.57735 02692 1 0.57735 02692 1 -0.7745966692 0 0.7745966692 0.55555 55556 0.88888 88889 0.55555 55556 -0.8611303116 -0.33998 10436 0.3478548451 0.6521451549 0.33998 10436 0.8611363116 0.6521-l 51549 0.34785 4845 I -0.90617 98459 -0.5384693101 0 0.5384693101 0.9061798459 0.23692 68851 0.47862 86705 0.56888 88889 0.47862 86705 0.23692 6885 I Gauss Integration Formula with n Degree of Precision Coefficients Aj Ij = 3 5 7 9 3 Evaluate the mtegral in Example 3 by the Gauss integration formula (I I) with /I = 3. Solutio'!.." 1 w~ have to c~nvert our integral from 0 to I into an integral from -\ to I. We set r = Then dx - 2 dr, and (I I) WIth /I ~ 3 and the above values of the nodes and the coefficients yields !U + I). SEC. 19.5 827 Numeric Integration and Differentiation f~Xp(_X2)dX = ~ { o = ~[%exp(-±(I - exp (- ± (I + 1)2) dl -1 /fr) + %exp(-~) + %exP(-±(1 + Hf)] =0.746815 (exact to liD: 0.746 825), which is almost as accurate as the Simpson result obtained in Example 3 with a much larger number of arithmetic operations. With 3 function values (as in this example) and Simpson's rule we would • get ~(l + 4e- O.25 + e -1) = 0.747 180. with an error over 30 times that of the Gauss integration. E X AMP L E 8 Gauss Integration Formula with n = 4 and 5 4 Integrate f(x) = !7TX cos !7TX from x = 0 to 2 by Gau". Compare with the adaptive integration in Example 6 and comment. Solutioll. x = 1 + 1 gives f(t) = J = Ad1 + ... + A4f4 !7T(t + 1)4 cos (!7T(t + 1». as needed in (11). For 11 = 4 we calculate (6S) = A 1{f1 + f4) + A 2 (f2 + i3) = 0.347855(0.000290309 + 1.02570) + 0.652145(0.129464 + 1.25459) = 1.25950. The error is 0.00003 because J = 1.25953 (6S). Calculating with lOS and 11 = 4 gives the same result; so the error is due to the formula. not rounding. For Il = 5 and lOS we get J = 1.25952 6185, too large by the amount 0.000000250 because J = 1.259525935 (lOS). The accuracy is impressive. particularly if we compare the • dmount of work with that in Example 6. Gauss integration is of considerable practical importance. Whenever the integrand f is given by a formula (not just by a table of numbers) or when experimental measurements can be set at times tj (or whatever t represents) shown in Table 19.7 or in Ref. [GRl], then the great accuracy of Gauss integration outweighs the disadvantage of the complicated tj and Aj (which may have to be stored). Also, Gauss coefficients Aj are positive for all n, in contrast with some of the Newton-Cotes coefficients for larger 11. Of course, there are frequent applications with equally spaced nodes, so that Gauss integration does not apply (or has no great advantage if one first has to get the tj in (11) by interpolation). Since the endpoints -1 and 1 of the interval of integration in ell) are not zeros of Pn' they do not occur among to, ... , tn. and the Gauss fonnula (11) is called. therefore, an open formula, in contrast with a closed formula, in which the endpoints of the interval of integration are to and tn- [For example. (2) and (7) are closed formulas.] Numeric Differentiation Numeric differentiation is the computation of values of the derivative of a function f from given values of f. Numeric differentiation should be avoided whenever possible, because, whereas integration is a smoothing process and is not affected much by small inaccuracies in function values, differentiation tends to make matters rough and generally gives values of f' much le~s accurate than those of f-remember that the derivative is the limit of the difference quotient. and in the latter you u!>ually have a small difference oflarge quantities that you then divide by a small quantity. However, the formulas to be obtained will be basic in the numeric solution of differential equations. We use the notations fi = t' (x), f;' = f"exj), etc., and may obtain rough approximation formulas for derivatives by remembering that t' (x) = lim f(x h~O This suggests + 11) - f(x) h CHAP. 19 828 Numerics in General t' - (12) fl - fo 8f1/2 h 1/2 - h Similarly, for the second derivative we obtain etc. (13) More accurate approximations are obtained b) differentiating suitable Lagrange polynomials. Differentiating (6) and remembering that the denominators in (6) are 2h2, -h 2 , 2h 2 , we have Evaluating this at (14) Xo, Xl> X2, we obtain the "three-point formula,," I (a) f~ = 211 (-3fo (b) f~ = 2h (-fo (c) f2 , I 1 = -21z (fo - + 4f1 + - f2), f2)' 4fI + 3f2)' Applying the same idea to the Lagrange polynomial P4(X), we obtain similar formulas, in particular. (15) Some examples and further formulas are included in the problem set as well as in Ref. [E5] listed in App. I. 1. (Rectangular rule) Evaluate the integral in Example I by the rectangular rule (1) with a subinterval of length 0.1. 2. Derive a formula for lower and upper bounds for the rectangular rule and apply it to Prob. I. !3-8! TRAPEZOIDAL AND SIMPSON'S RULES Evaluate the integrals numelically as indicated and determine the error by using an integration formula known from calculus. F(x) = J x 1 dx* - x* H(x) = . G(x) IX 0 1~\:*e-~* dx* o dx* 2 cos x* ' 3. F(2l by (2). Il = 10 4. F(2) by (7), n = [0 5. Gn) by (2).11 = 10 6. G(I) by (7),11 = 10 7. H(4) by (2), Il 10 8. H(4) by (7), Il = 10 !9-121 HALVING Estimate the error by halving. 9. In Prob. 5 10. In Prob. 6 11. In Prob. 7 12. In Prob. 8 Chapter 19 Review Questions and Problems 829 NON ELEMENTARY INTEGRALS If IE3d ~ TOL, stop. The result is is3 = 132 + E32 . (Why does 24 = 16 come in?) Show that we obtain 1"32 = -0.000266, so that we can stop. Arrange your 1- and E-values in a kind of "difference table." 113-191 The following integrals cannot be evaluated by the usual methods of calculus. Evaluate them as indicated. Si(x) = Sex) I sinx* x o dx~'. - - .x~ = {'Sin (X*2) dx*. C(x) = {"cos (X*2) dx* o 0 Si(x) is the sine integral. S(x) and C(x) are the Fresnel integrals. (See App. 3.1.) en. 13. SiO) by II = 5. II = 10 14. Using the values in Prob. 13. obtain a better value for Si(l). Hint. Use (5). 15. Si(l) by (7), 2111 = 2. 2m = 4 16. Obtain a better value in Prob. 15. Hint. Use (10). 17. Si(l) by (7). 2m = 10 18. S(1.25) by (7). 2111 = 10 19. C( 1.25) by (7). 2111 = 10 If IE3d were greater than TOL, you would have [0 go on and calculate in the next step 141 from (1) with h = ~: then 143 = 142 + 1"42 with q2 = 20. (Stability) Prove that the trapezoidal rule is stable with respect to rounding. 144 = 143 + E43 with E43 = 63 142 = 141 + I with E41 1"41 = "3 (141 - 1 31 ) I 15 (142 - 1 32 ) I (J43 - is3) where 63 = 26 - 1. (How does this come in?) Apply the Romberg method to the integral of f( 1:) = ~7TX4 cos !7TX from x = 0 to 2 with TOL = 10-4 . 121-241 GAUSS INTEGRATION Integrate by (11) with II = 5: 21. IIx from I to 3 22. co~ x from 0 to!7T 23. e-x" from 0 to I DIFFERENTIATION 24. sin <x 2 ) from 0 to 1.25 25. (Given TOL) Find the smallest 11 in computing the integral of 1Ix from I to 2 for which 50-accuracy is guaranteed (a) by (4) in the use of (2). (b) by (9) in the use of (7). Compare and comment. 26. TEAM PROJECT. Romberg Integration (W. Romberg. Norske Videllskab. Trolldheil11, FfJrh. 28, Nr. 7, 1955). This method uses the trapezoidal rule and gains precision stepwise by halving h and adding an error estimate. Do this for the integral of f(x) = e- X ti'om x = 0 to x = 2 with TOL = 10-3 , as follows. Step 1. Apply the trapezoidal rule (2) with h = 2 (hence n = I) to get an approximation In. Halve II and use (2) to gel 121 and an error estimate 27. Consider f(x) = X4 for Xo = 0, Xl = 0.2, X2 = 0.4, X3 = 0.6, X4 = 0.8. Calculate f~ from (14a), (14b), (14c), (15). Determine the errors. Compare and comment. 28. A "four-point formula" for the derivative is Apply it to f(x) = X4 with Xl' ... , X4 as in Prob. 27, determine the error. and compare it with that in the case of (15). 29. The derivative f' (x) can also be approximated in terms of first-order and higher order differences (see Sec. 19.3): 1 E21 = 22 _ I (121 - 1u)· If IE211 ~ TOL, stop. The result is 122 = 121 + E21' Step 2. Show that E21 = -0.066596, hence 1"211 > TOL and go on. Use (2) with hl4 to get 131 and add to it the error estimate 1"31 = i(131 - 1 21 ) to get the better 132 = 131 + 1"31- Calculate E32 = I 24 _ I I (132 - 1 22) = 15 (132 - 1 22), + "3I .1.3 fo - 4"1 ...\ 4 fo + - . .. ) . Compute t' (0.4) in Prob. 27 from this fOlmula usino differences up to and including first order, ~econd order, third order, fourth order. 30. Derive the formula in Prob. 29 from (14) in Sec. 19.3. 830 CHAP. 19 Numerics in General ==::== : ::&W :::.... 1. What is a numeric method? How has the computer influenced numeric methods? 2. What is floating-point representation of nwnbers? Overflow and underflow? 3. How do error and relative enor behave under addition? Under multiplication? 4. Why are roundoff errors important? State the rounding rules. 5. What is an algorithm"! Which of its properties are important in software implementation? 6. Why is the selection of a good method at least as important on a large computer as it is on a small one? 7. Explain methods for solving equations, in particular fixed-point iteration and its convergence. 8. Can the Newton (-Raphson) method diverge? Is it fast? Same questions for the bisection method. S T ION SAN D PRO B L EMS 23. What is the relative error of l1a in terms of that of a? 24. Show that the relative error of 25. Compute the solution of x 5 = x + 0.2 near J. = 0 by transforming the equation into the f01Tl1 x = g(x) and starting from Xo = O. (Use 6S.) 26. Solve cos x = x by iteration (6S, Xo = I), writing it as x = (O.74x + cosx)/1.74. obtainingx4 = 0.739085 (exact to (is!). Why does this converge so rapidly? 27. Solve X4 - x 3 - 2x - 34 = 0 by Newton's method with Xo = 3 and 6S accuracy. 28. Solve cos x - x 22. Answer the question in Prob. 21 for the difference 4.81 - 11.752. 0 by the method of false position. .fO.Q) = 3.00000 = 1.98007 f(l.2) f(1.4) = 2.92106 f( 1.6) = 1.111534 10. What do you remember about errors in polynomial imerpolation? 13. In what sense is Gau~s integration optimal? Explain details. 14. What does adaptive imegration mean? Why is it useful? 15. Why is numeric differentiation generally more delicate than numeric integration? 16. Write -0.35287. 1274.799, -0.00614. 14.9482. 113, 8517 in floating-point form with 5S (5 significant digits, properly rounded). 17. Compute (5.346 - 3.644)/(3.454 - 3.055) as given and then rounded stepwise to 3S. 2S. I S. ("Stepwise" means rounding the four rounded numbers. not the given ones.) Comment on your results. 18. Compute 0.29731/(4.1132 - 4.0872) with the numbers as given and then rounded stepwise (that is. rounding the rounded numbers) to 4S. 3S, 2S. Comment. 19. Solve x 2 - 50x + 1 = 0 by (6) and by (7) in Sec. 19.1, using 5S in the computation. Compare and comment. 20. Solve x 2 - 100x + 4 = 0 by (6) and by (7) in Sec. 19.1, using 5S in the computation. Compare and comment. 21. Let 4.81 and 12.752 be correctly rounded to the number of digits shown. Determine the smallest interval in which the sum (using the true instead of the rounded values) must lie. = 29. Compute f(1.28) from 9. What is the advamage of Newton's interpolation formulas over Lagrange's? 11. What is spline interpolation? Tts advantage over polynomial interpolation? 12. List and compare numeric integration methods. When would you appl} them'! a2 is about twice that of a. f( 1.8) = 2.69671 f(l.O) = 2.54030 by linear interpolation. By quadratic interpolation. using f( 1.2), fOA-), IO.6). 30. Find the cubic spline for the data f(-I) = 3 f(1) = I f(3) = 23 f(5) = -1-5 ko = k3 = 3. 31. Compute the integral of X3 from 0 to I by the trapezoidal rule with 11 = 5. What error bounds are obtained from (4) in Sec. 19.5? What is the actual error of the result? Why is this result larger than the exact value? 32. Compute the integral of cos (X2) from 0 to I by Simpson's rule with 2111 = 2 and 2171 = 4 and estimate the error by (10) in Sec. 19.5. (This is the Fresnel integral (38) in App. 3.1 with x = 1.) 33. Compute the integral of cos x from 0 to three-eights rule f the 3 b a -!rr by f(x) dx = "8 hUo + 3.fl + 3f2 + f3) - - 1 80 . (b - a)h 4 !'IV)(t) and give error bounds; here a ~ f ~ band Xj = a + (b - a)jI3, j = 0, ... , 3. 831 Summary of Chapter 19 , .... ....._...... - .. II • Numerics in General [n this chapter we discussed concepts that are relevant throughout numeric work as a whole and methods of a general nature, as opposed to methods for linear algebra (Chap. 20) or differential equations (Chap. 21). In scientific computations we use the floating-point representation of numbers (Sec. 19.1); fixed-point representation is less suitable in most cases. Numeric methods give approximate values a of quantities. The error E of a is (Sec. 19.1) (1) where a is the exact value. The relatil'e error of Ii is E/a. Errors arise from rounding, inaccuracy of measured values, truncation (that is. replacement of integrals by sums, series by partial sums), and so on. An algorithm is called numerically stable if small changes in the initial data give only cOiTespondingly small changes in the final results. Unstable algorithms are generally useless because errors may become so large that results will be very inaccurate. Numeric instability of algorithms must not be confused with mathematical instability of problems ("ill-conditioned problems," Sec. 19.2). Fixed-point iteration is a method for solving equations f(x) = 0 in which the equation is first transformed algebraically to x = g(x), an initial guess Xo for the solution is made, and then approximations X10 X2 • • • • , are successively computed by iteration from (see Sec. 19.2) (2) Xn+l = g(xn ) Newton's method for solving equations f(x) (3) X,,+l = Xn - (n = O. 1. ... ). = 0 is an iteration (Sec. 19.2). Here Xn+1 is the x-intercept of the tangent of the curve y = f(x) at the point X n • This method is of second order (Theorem 2, Sec. 19.2). If we replace f' in (3) by a difference quotient (geometrically: we replace the tangent by a secant). we obtain the secant method; see (10) in Sec. 19.2. For the bisection method (which converges slowly) and the method offalse position, see Problem Set 19.2. Polynomial interpolation means the detelTnination of a polynomial P>l(x) such that Pn(Xj) = Ii, where.i = 0, ... , 11 and (xo, f 0), ••. , (xn , fn) are measured or observed values, values of a function, etc. Pn(x) is called an intelpo/afioll poZvl1omial. For given data. Pn(X) of degree 11 (or less) is unique. However. it can be written in different forms, notably in Lagrange's form (4). Sec. 19.3, or in Newton's divided difference form (10). Sec. 19.3, which requires fewer operations. For regularly spaced xo, Xl = Xo + h, .... xn = Xo + l1h the latter becomes Newton's forward difference formula (formula (14) in Sec. 19.3) 832 CHAP. 19 Numerics in General f(x) = Pn(X) = fo (4) where r = + rt:.fo + ... + r(r - 1) ... (r - n + I) t:.nfo n! (x - xo)lh and the forward differences are t:.fj = f j +1 - fj and (k = 2,3, .. '). A similar formula is Newton's backward difference illterpolationfo1711ula (formula (18) in Sec. 19.3). Interpolation polynomials may become numerically unstable as 11 increases, and instead of interpolating and approximating by a single high-degree polynomial it is preferable to use a cubic spline g(x). that is. a twice continuously differentiable interpolation function [thus. g(Xj) = fjJ, which in each subinterval Xj ~ x ~ Xj+1 consists of a cubic polynomial qj(x): see Sec. 19.4. Simpson's rule of numeric integration is [see (7), Sec. 19.5] h b (5) fa f(x) dx = 3" (fo + 4fL + 2f2 + 4j3 + ... + 2f2711-2 + ~f2m-l + f2711) with equally spaced nodes Xj = Xo + jh. j = 1.... ,2m, h = (b - a)/(2m). and f j = f(xj)' It is simple but accurate enough for many applications. Its degree of precision is DP = 3 because the error (8), Sec. 19.5. involves h4. A more practical error estimate is (10), Sec. 19.5. obtained by first computing with step h, then with step /1/2, and then taking III 5 of the difference of the results. Simpson's rule is the most important of the Newton-Cotes formulas, which are obtained by integrating Lagrange interpolation polynomials, linear ones for the trapezoidal rule (2), Sec. 19.5, quadratic for Simpson's mle. cubic for the three-eights rule (see the Chap. 19 Review Problems). etc. Adaptive integration (Sec. 19.5, Example 6) is integration that adjusts ("adapts") the step (automatically) to the variability of f(x). Romberg integration (Team Project 26, Problem Set 19.5) starts from the trapezoidal rule (2). Sec. 19.5. with h. h12, h/4, etc. and improves results by systematically adding error estimates. Gauss integration (II), Sec. 19.5, is important because of its great accuracy (DP = 217 - 1, compared to Newton-Cotes's DP = 11 - 1 or 11). This is achieved by an optimal I;hoice of the nodes, which are not equally spaced; see Table 19.7, Sec. 19.5. Numeric differentiation is discussed at the end of Sec. 19.5. (Its main application (to differential equations) follows in Chap. 21.) , CHAPTER J 20 Numeric Linear Algebra In this chapter we explain "orne of the most important numeric methods for solving linear systems of equations (Secs. 20.1-20.4), for fitting straight lines or parabolas (Sec. 20.5), and for matrix eigenvalue problems (Secs. 20.6-20.9). These methods are of considerable practical importance because many problems in engineering, statistics, and elsewhere lead to mathematical models whose solution requires methods of numeric linear algebra. COM MEN T. This chapter is independent of Chap. 19 and can be studied immediately after Chap. 7 or 8. Prerequisite: Secs. 7.1. 7.2, 8.1. Sections that may be omitted in a shorter course: 20.4. 20.5. 20.9 References and Answers to Problems: App. I Part E. App. 2 20.1 Linear Systems: Gauss Elimination A linear system of n equations in E I , . . . , En of the form 11 unknowns Xl ••• , Xn is a set of equations (1) where the coefficients ajk and the bj are given numbers. The system is called homogeneous if all the bj are zero; otherwise it is called nonhomogeneous. Usmg matrix multiplication (Sec. 7.2), we can write (1) as a single vector equation (2) Ax where the coefficient matrix A A= = =b [ajk] is the 11 X 11 matrix au al2 al n a21 a22 a2n and anI an2 ann bl Xl x= and Xn and b = bn 833 834 CHAP. 20 Numeric Linear Algebra are column vectors. The following matrix system (I): A= [A b] A is called the augmented matrix of the = A solution of (I) is a set of numbers Xl' • . • , Xn that satisfy all the II equations, and a solution vector of (1) is a vector x whose components constitute a solution of (1). The method of solving such a system by determinants (Cramer's rule in Sec. 7.7) is not practical, even with efficient methods for evaluating the determinants. A practical method for the solution of a linear system is the so-called Gal/ss eliminatioll, which we shall now discuss (proceeding illdependently of Sec. 7.3). Gauss Elimination This standard method for solving linear systems (I) is a systematic process of elimination that reduces (1) to "triangular form" because the system can then be easily solved by "back substitution." For instance, a triangular system is and back substitution gives X3 = 3/6 = 112 from the third equation, then from the second equation, and finally from the first equation How do we reduce a given system (I) to triangular form? In the first step we elimil1ate from equation E2 to En in (I). We do this by adding (or subtracting) suitable multiples of E1 from equations E2 , ••• , En and taking the resulting equations, call them E~, ... , E~ as the new equations. The first equation, E 1 , is called the pivot equation in this step, and llU is called the pivot. This equation is left unaltered. In the second step we take the new second equation E~ (which no longer contains Xl) as the pivot equation and use it to eliminate X2 from E; to E~. And so on. After Il - I steps this gives a triangular system that can be solved by back substitution as just shown. In this way we obtain precisely all solutions of the given system (as proved in Sec. 7.3). The pivot llkk (in step k) must be different from zero and should be large in absolute value, to avoid roundoff magnification by the multiplication in the elimination. For this we choose as our pivot equation one that has the absolutely largest ajk in column k on or below the main diagonal (actually, the uppermost if there are several such equations). This popular method is called partial pivoting. It is used in CASs (e.g., in Maple). Xl SEC 20.1 835 Linear Systems: Gauss Elimination Partial pivoting distinguishes it from total pivoting, which involves both row and column interchanges but is hardly used in practice. Let us illustrate this method with a simple example. EXAMPLE 1 Gauss Elimination. Partial Pivoting Solve the system Solutioll. We must pivot since El ha, no xl-term. In Column 1. equation E3 ha, the large,t coefficient. Hence we interchange El and E3, 6Xl + 2x2 + 8x3 = 3X1 26 + 5X2 + 2X3 = 8.1:2 + 2X3 ~ 8 -7. Step 1. Elimillatioll of Xl It would suftlce to show the augmented matrix and operate on it. We show both the equations and the augmented matrix. In the first step. the first equation is the pivot equation. Thus Pivot 6----+(§i)+ 2X2 + 8x3 = 26 Eliminate - - ' > ~ + 5x2 + 2r3 = 8X2 + 2x3 8 = -7 [: 2 8 5 2 8 2 ':] -7 To eliminate Xl from the other equations (here. from the second equation). do: Subtract 3/6 ~ 112 times the pivot equation from the second equation. The result is ntl + + = 26 2 8 4X2 - 2'-3 = -5 4 -2 -5 8x2 + 2'3 = -7 8 2 -7 2x2 8x3 26] Step 1. Elimillatioll of X2 The largest coefficient in Column 2 is 8. Hence we take the nell' third equation as the pivot equation, interchanging equations 2 and 3, 26 2 8 Pivot 8----+ ~+ 2x3 =-7 8 2 -7 Eliminate I§l- 2X3 = 4 -2 -5 6Xl + 2x2 + 8x3 = --'> -5 [: 26] To eliminate .1:2 from the third equation. do: Subtract 112 times the pivot equation from the third equation. The resulting triangular system i, shown below. Thi, is the end of the forward elimination. Now comes the back substitution. Back substitutioll. Determillatioll of X3, X2, Xl The trIangular system obtained in Step 2 is 2 8 8 2 -7 o -3 _.2 26] 2 836 CHAP. 20 Numeric Linear Algebra From this system. taking the last equation, then the second equation. and finally the fust equation. we compute the solution x 3 -- 12 X2 = ~(-7 - 2X3) = -I Xl = ~(26 - 2~2 - 8x3) = 4. This agrees with the values given above. before the beginning of the example. • The general algorithm for the Gauss elimination is shown in Table 20.1. To help explain the algorithm, we have numbered some of its lines. bj is denoted by aj,n+l' for uniformity. In lines I and 2 we look for a possible pivot. [For k = I we can always find one; otherwise Xl would not occur in (l ).] In line 2 we do pivoting if necessary, picking an ajk of greatest absolute value (the one with the smallest j if there are several) and interchange the corresponding rows. If lakkl is greatest, we do no pivoting. 11ljk in line 3 suggests multiplier, since these are the factors by which we have [0 multiply the pivor equation E~ in Step k before subtracting it from an equation Ef below E~' from which we want to eliminate Xk' Here we have written E~ and E/ to indicate that after Step I these are no longer the given equations in (1), but these underwent a change in each step, as indicated in line 4. Accordingly, ajk etc. in lines 1-4 refer to the most recent equations, andj ~ k in line I indicates that we leave un[Ouched all the equations that have served as pivot equations in previous steps. For p = k in line 4 we get 0 on the right. as it should be in the elimination. In line 5, if the last equation in the triangular system is 0 = b~ *- 0, we have no = 0, we have no unique solution because we then have fewer solution. If it is 0 = equations than unknowns. b: E X AMP L E 2 Gauss Elimination in Table 20.1, Sample Computation In Example 1 we had all = O. so that pivoting wa, necessary. The greatest coefficient in Column I was a3l' Thus J = 3 in line 2. and we interchanged El and E 3 . Then in lines 3 and 4 we computed 11121 = 3/6 = ~ and G22 = 5 - ~ . 2 = 4. G23 = 2 - ~. 8 = -2. G24 = 8 - ~ . 26 = -5. and then 11131 = 0/6 = O. so that the third equation 8X2 + 2X3 = -7 did not change in Step I. In Step 2 = 2) we had 8 as the greatest coefficient in Column 2. hence J = 3. We interchanged equations 2 and 3. computed 11132 = -4/8 = - ! in line 4. and the a33 = -2 - !·2 = -3. a34 = -5 - ~(-7) = -~. This produced the triangular fOim used in the back substitution. • (k If akk = 0 in Step k, we must pivot. If lakkl is small. we should pivot because of roundoff error magnification that may seriously affect accuracy or even produce nonsensical re~ults. E X AMP L E 3 Difficulty with Small Pivots The solution of the 'ystem 0.0004Xl + 1.402x2 = 1.406 0.4003xl - 1.502x2 = 2.501 is Xl = 10'"'2 = 1. We solve this system by the Gauss elimination. using four-digit floating-point arithmetic. (4D is for simplicity. Make an 8D-arithmetic example that shows the same.) In (a) Picking the first of the given equations as the pivot equation. we have to mUltiply this equation by 0.4003/0.0004 = 1001 and subtract the result from the second equation. obtaining = SEC. 20.1 837 Linear Systems: Gauss Elimination Table 20.1 Gauss Elimination ALGORITHM GAUSS (A = [ajd = LA b]) This algorithm computes a unique solution x = [Xj] of the system (1) or indicates that (1) has no unique solution. INPUT: Augmented n X (n = matrix A= [ajk]' where aj,n+l = bj 1. . . . , n - 1. do: 0 for all j exists."' Stop If ajk = 1 I) Solution x = lXj] of (I) or message that the system (1) has no unique solution OCTPUT: For k + ~ k then OUTPUT "No unique solution [Procedure completed unsuccessfully; A is singular] 2 Else exchange the contents of rows J and k of A with J the smallest j ~ k such that lajkl is maximum in column k. 3 For j k = + I. .... n. do: . _ ajk 1njk- - - akk For p = k 4 I + ajp: I. . . . , n + I. do: = ajp - 1njk a kp End End End If ann = 0 then OUTPUT "No unique solution exists." Stop Else 5 6 [Start back substitution] For i 7 = n - 1. . . . . 1, do: Xi = ....!... (ai,n+l - aii i aijXi) j=i+l End OUTPUT x = [Xj]. Stop End GAUSS ~ 1405x2 = Hence x2 = ~ 1404/( ~ 1405) Xl This failure occurs because large error in Xl' = ~ 1404. = 0.9993, and from the first equation. 1 00004 (l.406 ~ 1.402· 0.9993) . instead of Xl U.oo5 =-= 0.0004 = 10. we get 12.5. iani is small compared with ia12i. so that a small roundoff error in X2 leads to a 838 CHAP. 20 Numeric Linear Algebra (b) Picking the second of the given equations as the pivot equation. we have 10 multiply this equation by 0.0004/0.4003 = 0.0009993 and subtract the result from the first equation. obtaining 1.404x2 = 1.404. Hence .\"2 = I, and from the pivot equation xl = 10. This success occur~ because la2l1 is not very small compared to la221. so that a small roundoff enOf in .\"2 would not lead to a large error in Xl' Indeed. for instance. if we had the value x2 = 1.002. we would still have from the pivot equation the good value Xl = (2.501 + 1.505)10.4003 = 10.01. • Error estimates for the Gauss elimination are discussed in Ref. [E5] listed in App. I. Row scaling means the multiplication of each Row j by a suitable scaling factor ~J- It is done in connection with partial pivoting to get more accurate solutions. Despite much research (see Refs. [E9]. [E24] in App. I) and the proposition of several principles, scaling is still not well understood. As a possibility, one can !>cale for pivot choice only (not in the calculation, to avoid additional roundoff) and take as first pivot the entry aj1 for which IlljlltlAjl is largest: here Aj is an entry of largest absolute value in Row j. Similarly in the further steps of the Gauss elimination. For instance, for the system 4.0000X1 + 14020.\"2 = 14060 0.4003.\"1 - 1.502.\"2 = 2.501 we might pick 4 as pivot, but dividing the first equation by 104 gives the system in Example 3, for which the second equation is a better pivot equation. Operation Count Quite generally, important factors in judging the quality of a numeric method are Amount of storage Amount of time (== number of operations) Effect of roundoff elTOL For the Gauss elimination, the operation count for a full matrix (a matrix with relatively many nonzero entries) is as follows. In Step k we eliminate Xk from n - k equations. This needs n - k divisions in computing the IIljk (line 3) and (n - k)(11 - k + 1) multiplications and as many subtractions (both in line 4). Since we do 11 - 1 steps. k goes from I to n - I and thus the total number of operations in this forward elimination is n-1 fen) = L n-1 (n - k) +2L k~l n-1 =L s~l (11 - k)(n - k + (write Il - k = s) 1) k~l n-1 S + 2 L s(s + 1) = !(Il - 1)11 + 1(11 2 - 1)n = 111 3 S=l where 2n 3/3 is obtained by dropping lower powers of 11. We see that f(ll) grows about proportional to /1 3 . We say that f(n) is of order n 3 and write SEC. 20.1 839 Linear Systems: Gauss Elimination where 0 suggests order. The general definition of 0 is as follows. We write f(ll) = O(h(n)) if the quotient If(n)lh(Iz)1 remains bounded (does not trail off to infinity) as 11 ~ x. In our present case, hen) = 11 3 and, indeed. f(n)ln 3 ~ 2/3 because the omitted telms divided by 11 3 go to zero as n _ h . In the back substitution of Xi we make 11 - i multiplications and as many subtractions, as well as 1 division. Hence the number of operations in the back substitution is n b(ll) = 2 2: (11 - i) + 11 = 2 2: s + n = n(n + I) + n = /1 2 + 211 = 2 0(11 ). 5=1 We see that it grows more slowly than the number of operations in the forward elimination of the Gauss algorithm, so that it is negligible for large systems because it is smaller by a factor 11, approximately. For instance, if an operation takes 10- 9 sec, then the times needed are: Algorithm n = 1000 Elimination Back substitution n=10000 0.7 sec 11 min 0.001 sec 0.1 sec .P RO 8 E=E M- S E.E10:;-L For applicatiolls of linear systems see Secs. 7.1 and 8.2. 11-31 5. 2X1 GEOMETRIC INTERPRETATION 6.\1 - 8x2 = -4 + 2x2 = 14 Solve graphically and explain geometrically. 1. 4.\1 + X2 = 3.\1 - 5.\2 2. = -4.3 6. 25.38x1 -33.7 -7. 05x 1 1.820.\1 - 1.183x2 = 0 -12.74.\1 3. + 7.2.\1 -21.6.\1 14-141 + 8.281.\2 = 0 3.5x2 = 10.5.\2 = -4R.5 6X2 X2 = -3 = 30.60 = 4.30.\2 + 8. 5x 1 + 3X2 - 4x2 IOx1 - 6.\2 = 137.86 8X3 = -85.88 178.54 + X3 15x2 = 2 + 8X3 = -3 + 26x3 9. 4.\1 + IOx2 - 2X3 -Xl - -8.50 13x3 13x1 - 8X2 GAUSS ELIMINATION + + 6X1 16.0 Solve the following linear systems by Gauss elimination. with partial pivoting if necessary (but without scaling). Show the intermediate steps. Check the result by substitution. If no solution or more than one solution exists, give a reason. 4. 6.\"1 7. 15.48x2 + = = 3X3 = 0 -20 30 25x2 - 5X3 = -50 CHAP. 20 840 10. Xl + 2X2 lOx 1 + X2 IOx2 11. 3.4x 1 -Xl 3X3 -II + X3 8 i- 2X3 = (b) Gauss elimination and nonexistence. Apply the Gauss elimination to the following two systems and compare the calculations step by step. Explain why the elimination fails if no solution exists. 2 6.l2x2 - 2. 72x3 =0 + 1.80X2 2.7x1 - 4.86x2 12. Numeric Linear Algebra 3X2 + + 0.80X3 - 2. 16x3=0 5X3 = 1.20736 + 6X3 = 1.6x2 -1.6 + 4. 8x3 4.8x2 - 9.6x3 7.2x3 14. 4.4x2 + 32.0 + 7.2x4 = -78.0 + 4.8x4 20.4 = 3.0x3 - 6.6x4 = -4.65 + 8.4X4 = 4.62 -4.35 - 7.6x3 + 3.0x4 = 5.97 15. CAS EXPERIMENT. Gauss Elimination. Write a program for the Gauss eliminarion with pivoting. Apply it to Probs. 11-14. Experiment with systems whose coefficient determinant is small in absolute value. Also investigate the perfonnance of your program for larger systems of your choice. including sparse systems. 2X2 - x3 5 9X1 + 5X2 - X3 13 Xl + + 3 X2 X3 = (c) Zero determinant. Why maya computer program give you the result that a homogeneous linear system has only the trivial solution although you know its coefficient determinant to be zero? (d) Pivoting. Solve System lA) (below) by the Gauss elimination first without pivoting. Show that for any fixed machine word length and sufficiently small E > 0 the computer gives X2 = I and then Xl = O. What is the exact solution? Its limit as E ~ O? Then solve the system by the Gauss elimination with pivoting. Compare and comment. (e) Pivoting. Solve System (H) by the Gauss elimination and three-digit rounding arithmetic. choosing (i) the first equation, (ii) the second equation as pivot equation. (Remember to round to 3S after each operation before doing the next, just as would be done on a computer!) Then use four-digit rounding arithmetic in those two calculations. Compare and comment. 2 16. TEAM PROJECT. Linear S~'stems and Gauss Elimination. (a) Existence and uniqueness. Find a and b such that aX1 + X2 = b, Xl + X2 = 3 has (i) a unique solution. (ii) infinitely many solutions, (iii) no solutions. 20.2 + 4X1 X2 -0.329193 13. 6.4x 1 + 3.2x2 3.2x1 - 3 + -2.34066 3X1 - 4X2 5x1 0 + X3 Xl (E) -4.61 6.21x1 + 3.35x2 = -7.19 Linear Systems: LU-Factorization, Matrix Inversion We continue our discussion of numeric methods for solving linear systems of n equations in 11 unknowns Xl' ••• , X n , (1) Ax = b where A = [ajk] is the 11 X 11 coefficient matrix and x T = [Xl' .. xn] and b T = [hI' .. bnJ. We present three related methods that are modifications of the Gauss elimination, which SEC. 20.2 841 Linear Systems: LU-Factorization, Matrix Inversion require fewer arithmetic operations. They are named after Doolittle, Crout, and Cholesky and use the idea of the LV-factorization of A, which we explain first. An LV-factorization of a given square matrix A is of the form (2) A LV = where L is lower triangular and U i, upper triallgular. For example, A = [2 3J = 8 [1 OJ [2 3J LU = 5 4 I 0-7 It can be proved that for any nonsingular matrix (see Sec. 7.8) the rows can be reordered so that the resulting matrix A has an LU-factorization (2) in which L turns out to be the matrix of the multipliers Injk of the Gauss elimination, with main diagonal 1, ... , 1, and V is the matrix of the triangular system at the end of the Gauss elimination. (See Ref. [E5], pp. 155-156, listed in App. 1.) The crucial idea now is that Land U in (2) can be computed directly, without solving simultaneous equations (thus, without using the Gauss elimination). As a count shows. this needs about n 3 /3 operations. about half as many as the Gauss elimination. which needs about 21l 3 /3 (see Sec. 20.1). And once we have (2), we can use it for solving Ax = b in two steps. involving only about 11 2 operations. simply by noting that Ax = LUx = b may be written (3) (a) Ly =b where Ux (b) = y and solving first (3a) for y and then (3b) for x. Here we can require that L have main diagonal 1, ... ,las stated before; then this is called Doolittle's method. Both systems (3a) and (3b) are triangular, so we can solve them as in the back substitution for the Gauss elimination. A similar method, Crout's method, is obtained from (2) if V (instead of L) is required to have main diagonal 1. ...• 1. In either case the factorization (2) is unique. EXAMPLE 1 Doolittle's Method Solve the system in Example I of Sec. 20.1 by Doolittle's method. S Diu ti~Il. The decomposition (2) is obtained from 5 a12 A = [ajk] = [ "n a21 a22 {[23 = [ 30 ""] 8 a31 a32 a33 6 2 '] [' 2 = 8 ~]F 0 1Il21 nJ31 11132 U12 ""] 1123 "22 0 li33 by determining the 111jk and liJ1<' using matrix multiplication. By going through A row by rov. we get successively all = 3 = I . lIll = 11121 = lill a12 = 5 = I • lI12 0 = 1112 a13 =2 = I . 1/13 = a23 = 2 = 111211113 1123 -I li23 =2 =2.2 11132 = + li13 I •2 + li33 842 CHAP. 20 Numeric Linear Algebra Thus the factorization (2) is : : :] = 6 2 8 [ LU [:1 0 = 2 -I ~] [: : :] . I 0 0 6 We first solve Ly = b, determining '"I = 8. then.\"2 = -7, then Y3 from 2Y1 - )"2 + thu~ (note the interchange in b because of the interchange in A!) [~ 2 0 :] [::::] = [ _ : ] . -I 26 1)3 Then we solve Ux = y. determining Solution x3 = )"3 = 16 + 7 + Y3 = 26; ,=[:] 3/6. then x2, then xl' that is, Solution x= [J • This agrees with the solution in Example I of Sec. 20.1. Our formulas in Example I suggest that for general 11 the entries of the matrices L = rmjk] (with main diagonal l. ...• 1 and mjk suggesting "multiplier") and U = [Ujk] in the Doolittle method are computed from k=l,"',n j = 2,···, n j-1 (4) Ujk = lljk - L k 11ljs U s k = j .... , n; j ~ 2 s=l j = k Row Interchanges. + 1, ... , n; k ~ 2. Matrices, such as [~ :J or [~ ~J have no LU-factorization (try!). This indicates that for obtaining an LU-factorization. row interchanges of A (and corresponding interchanges in b) may be necessary. Cholesky's Method '* For a symmetric, positive definite matrix A (thUS A = AT, X TAx > 0 for all x 0) we can in (2) even choose U = L T, thus Ujk = n1kj (but cannot impose conditions on the main diagonal entries). For example, SEC. 20.2 843 Linear Systems: LU-Factorization. Matrix Inversion The popular method of solving Ax = b based on this factorization A = LLT is called Cholesky's method. In terms of the entries of L = [ljkl the formulas for the factorization are = 2,"',11 j j-l (6) L Gjj - Ij8 2 j = 2.···.11 8=1 = j + I, . . . , 11; p j ~ 2. If A is symmetric but not positive definite, this method could still be applied, but then leads to a complex matrix L, so that the method becomes impractical. E X AMP L E 1 Cholesky's Method Solve by Cholesky"s method: 4.\"1 + 2X2 + 14.\"3 2"1 + 17.\"2 14xl - Solution. 5x2 = 14 5.\"3 = -!o1 + 83.\"3 = 155. From (6) or from the form of the factorization I~ ~:l : [ -5 14 o : 1[~1 = [::: 83 133 131 o 0 we compute. in the given order. 111 = ~ = 2 121 = a21 - 2 = - 2 111 a31 = I 131 = - 14 = - 111 = 2 7 This agrees with (5). We now have to solve Ly = b, that is. 0 [; 4 -3 0] ["'] [ 14] ~ ~~ As the second step. we have to solve Ux [: 4 0 = = -~:~ LT X Solution . y~ [-':J = y, that is. -;] [:} [-,;] Solution x~ [-:J • CHAP. 20 844 Numeric Linear Algebra Stability of the Cholesky Factorization THEOREM 1 The Clwlesky LLT -!actorizatio/1 is numerically stable (as defined in Sec. 19.1). PRO 0 F We have ajj = lj12 + Ij22 + ... + ljf by squaring the third formula in (6) and solving it for ajj. Hence for allljk (note that ljk = 0 for k > j) we obtain (the inequality being trivial) That is, ljk 2 is bounded by an entry of A, which means stability against rounding. • Gauss-Jordan Elimination. Matrix Inversion Another variant of the Gauss elimination is the Gauss-Jordan elimination, introduced by W. Jordan in 1920, in which back substitution is avoided by additional computations that reduce the matrix to diagonal form. instead of the triangular form in the Gauss elimination. But this reduction from the Gauss triangular to the diagonal form requires more operations than back substitution does, so that the method is disadvantageous for solving systems Ax = h. But it may be used for matrix inversion, where the situation is as follows. The inverse of a nonsingular square matrix A may be determined in principle by solving the 11 systems (j (7) = 1, ... , 11) where bj is the jth column of the 11 X 11 unit matrix. However. it is preferable to produce A -1 by operating on the unit matrix I in the same way as the Gauss-Jordan algorithm, reducing A to I. A typical illustrative example of this method is given in Sec. 7.8. ..........- . ........ !'!I!.~.!I!.!IIII~III!.':!I!_~_lJ!!I!!y~~::P--- 11-71 •• _ - ...,. DOOLITTLE'S METHOD Show the factorilation and solve by Doolittle's method. 1. 3Xl + 15xl + 2x2 = 15.2 Ilx2 = 77.3 6. -9.88 3.3x3 = -16.54 0.5Xl - 3.0X2 + -1.5Xl - 3.5x2 - 7. 3Xl 18xl + 9X2 + + 48x2 + 9Xl - 27x2 1O.4x3 = 6X3 = 21.02 2.3 39x3 = 13.6 + 42x3 = 4.5 8. TEAM PROJECT. Crout's method factorizes A = LV, where L is lower ttiangular and V is upper SEC. 20.3 845 Linear Systems: Solution by Iteration triangular with diagonal entnes lljj = l,j = 1, ... ,11, (a) Formulas. Obtain formulas for Crout's method similar to (4). 20 (b) Examples. Solve Probs. I and 7 by Crout's method. (e) Factor the following matrix by the Doolittle. Crout. and Cholesky methods. -4 l-: 14. CAS PROJECT. Cholesky's Method. (a) Write a program for solving linear systems by Cho)esky's method and apply it to Example 2 in the text, to Probs. 9-11, and to systems of your choice. 25 4 (b) Splines. Apply the factorization part of the program to the following matrices (as they occur in (9), Sec. 19.4 (with cJ = 1), in connection with splines). (d) Give the formulas for factoring a tridiagonal matrix by Crout's method. (e) When can you obtain Crout's factorization from Doolittle's by transposition? 19-131 4 CHOLESKY'S METHOD [ Show the factorization and solve. 9. 9X1 + 6X2 + 12x3 = 6Xl + 13x2 + IIx3 = 118 12x1 + IIx2 + 26x3 = 154 0.64X2 + 0.32x3 = 1.6 0.12x1 + 0.32x2 + 0.56x3 = 5.4 + 6x 1 + 8X1 + 12. Xl -Xl - + 6X2 + 8X3 = 0 34x2 + 52x3 = -80 52x2 + 129x3 = -226 X2 + 3X3 5X2 - + 3X1 - 5X2 + 19x3 + 2X1 - 2X2 + 3X3 + 20.3 2X4 = 4 o 2 INVERSE Find the inverse by the Gauss-Jordan method. showing the details. 16. In Prob 4. 17. In Prob. 5. 18. In Prob. 6. 19. 30 3X4 = 188 = 2 21x4 o o 116-191 2x4 = -70 5X3 ]. o o 15. (Definiteness) Let A and B be positive definite 11 X /1 matrices. Are - A, A f. A + B, A - B positive definite? 0. 12x3 = 1.4 11. 4xl 4 87 + 10. 0. 04X 1 o 2 In Prob. 7. 20. (Rounding) For the following matrix A find det A. What happens if you round off the given entries to (a) 5S, (b) 4S, (c) 3S, (d) 2S, (e) IS? What is the practical implication of your work? 114 -3/28 Linear Systems: Solution by Iteration The Gauss elimination and its variants in the last two sections belong to the direct methods for solving linear systems of equations; these are methods that give solutions after an amount of computation that can be specified in advance. In contrast. in an indirect or iterative method we start from an approximation to the true solution and. if successful, obtain better and better approximations from a computational cycle repeated as often as may be necessary for achieving a required accuracy, so that the amount of arithmetic depends upon the accuracy required and varies from case to case. 846 CHAP. 20 Numeric Linear Algebra We apply iterative methods if the convergence is rapid (if matrices have large main diagonal entries, as we shall see), so that we save operations compared to a direct method. We also use iterative methods if a large system is sparse, that is, has very many zero coefficients. so that one would waste space in storing zeros, for instance, 9995 zeros per equation in a potential problem of 104 equations in 104 unknowns with typically only 5 nonzero terms per equation (more on this in Sec. 21.4). Gauss-Seidel Iteration Method l This is an iterative method of great practical importance. which we can simply explain in terms of an example. E X AMP L E 1 Gauss-Seidel Iteration We consider the linear system = -0.25x I + 50 - 0.25x4 = 50 (I) + -0.25.1'1 - 0.25.1'2 - 0.25.1'3 (Equations of this form arise in the numeric in the fonn ~olution + of PDEs and in spline interpolation.) We write the system 0.25x2 + 0.25x3 Xl = + 50 X2 = 0.25.1'1 + 0.25x4 + 50 X3 = 0.25xl + 0.25x4 + 25 (2) 0.25 t2 + 0.25x3 X4 = + 25. These equations are now used for iteration: that is. we start from a (possibly poor) approximatIOn to the solution. say xiO) = 100, x~O> = 100. x~O) = 100. x~O) = 100. and compute from (2) a perhaps better approximation Use "old" values ("New" values here not yet available) t 0.25x~o) (3) X 2(1)- 0.25x~O) + 50.00 = 100.00 0. 25x ll i 0.25x~0l + 50.00 = 100.00 ill 0. 25xiO) + 25.00 = 75.00 0. 25x lll X4 - + 0.25x~1l + 0.25x~1l + 25.00 = 68.75 t Use "new" values These equations (3) are obtained from (2) by substituting on the right the //lost recellt approximation for each unknown. In fact, correspondmg value~ replace previous ones as soon as they have been computed. so that in IPHlLIPP LUDWIG VON SEIDEL (1821-1896), German mathematician. For Gauss see foomore 5 in Sec. 5.4. SEC. 20.3 847 Linear Systems: Solution by Iteration the second and third equations we use xill (not xiOJ), and in the last equation of (3) we use x~l) and x~l) (not x~Ol and x~OJ). Using the same principle. we obtain in the next step 0.25x~ll + 0.25x~ll + 50.00 X~2) = 0.25xi2 ) + 0.25x~1l + 50.00 x~2) = 0.25xi + 2 ) 0.25x~1) (2) _ X4 - = 93.750 = 90.625 + 25.00 = 65.625 + 25.00 = M.062 Further steps give the values Xl X2 X3 X4 89.062 87.891 87.598 87.524 87.506 88.281 87.695 87.549 87.512 87.503 63.281 62.695 62.549 62.512 62.503 62.891 62.598 62.524 62.506 62.502 Hence convergence to the exact solution Xl = X2 = 87.5, x3 = X4 = 62.5 (verify!) seems rather fast. • An algorithm for the Gauss-Seidel iteration is shown on the next page. To obtain the algorithm, let us derive the general formulas for this iteration. We assume that ajj = I for j = 1, ... , n. (Note that this can be achieved if we can realTange the equations so that no diagonal coefficient is zero; then we may divide each equation by the corresponding diagonal coefficient.) We now write (4) A=I+L+U (ajj = 1) where I is the n x n unit matrix and Land U are respectively lower and upper triangular matrices with zero main diagonals. If we substitute (4) into Ax = b. we have Ax = (I + L + U) x = b. Taking Lx and Ux to the right, we obtain, since Ix = x, (5) x = b - Lx - Ux. Remembering from (3) in Example I that below the main diagonal we took "new" approximations and above the main diagonal "old" ones, we obtain from (5) the desired iteration formulas (6) where x(m) [_l}m)] is the mth approximation and x(m+ 1) = [X}m+ 1)] is the (m + I )st approximation. In components this gives the formula in line I in Table 20.2. The matrix A must satisfy Gjj =1= 0 for allj. In Table 20.2 our assumption ajj = I is no longer required, but is automatically taken care of by the factor l/ajj in line 1. 848 CHAP. 20 Table 20.2 Numeric Linear Algebra Gauss-Seidel Iteration ALGORITHM GAUSS-SEIDEL (A. b, XlV), E, N) This algorithm computes a solution x of the system Ax = b given an initial approximation x W), where A = [ajd is an 11 X 11 matrix with lljj =1= 0, j = I, ... , 11. A, b, initial approximation of iterations N INPUT: OUTPUT: x(Q), tolerance E > 0, maximum number Approximate solution x(m) = [x7")] or failure message thal X(N) does not satisfy the tolerance condition For 111 = 0..... N - 1. do: For j = L ... , /I, do: r~m+ll = •.1 - I (j-l b. - ~ £.J J (l .. 2 € £.J Jk k k-] JJ End If m':IX IX)7n+ll - xj",JI < It ~ (/. x(m+lJ - (/. x(mJ Jk k ) k~j+l then OUTPUT xcm+ll. Stop J [Procedure completed successtittly 1 End OUTPUT: "No solution satisfying the tolerance condition obtained after N iteration steps." Stop [Procedure completed U11.\"Ucces.~f~ttly] End GAUSS-SEIDEL Convergence and Matrix Norms An iteration method for solving Ax = b is :-'did to converge for an initial x(V) if the corresponding iterative sequence xW>, x(1), X(2 ), . . . converges to a solution of the given system. Convergence depends on the relation between x(m) and x(m+lJ. To get this relation for the Gauss-Seidel method, we use (6). We first have (I and by multiplying by (I (7) x{m+lJ = Cx{m) + + + L)x(m+lJ = b - Ux(m) L)-1 from the left, (I + L) -1 b where C = -(I + L) -1 U. The Gauss-Seidel iteration converges for every x(O) if and only if all the eigenvalues (Sec. 8.1) of the "iteration matrix" C = [CjkJ have absolute value less than l. (Proof in Ref. [E5]. p. 191, listed in App. 1.) CAUTION! If you want to get C, first divide the rows of A by au to have main diagonal 1, ... , 1. If the spectral radius of C (= maximum of those absolute values) is smalL then the convergence is rapid. Sufficient Convergence Condition. A sufficient condition for convergence is (8) IICII < 1 SEC. 20.3 849 Linear Systems: Solution by Iteration Here IIcll is some matrix norm, such as (9) (Frobenius norm) or the greatest of the sums of the !cjkl in a colulIlll of C (10) IICIi = n max k or the greatest of the sums of the 2: !cjkl (Column "sum" norm) j=l !cjkl in a row of C n (Row "sum" norm). (11) These are the most frequently used matrix norms in numerics. In most cases the choice of one of these norms is a matter of computational convenience. However, the following example shows that sometimes one of these norms is preferable to the others. E X AMP L E 1 Test of Convergence of the Gauss-Seidel Iteration Test whether the Gauss~"eidel iteration converges for the system 2x+ y+ ;::=4 x + 2y + :: = \" = 2 - ~x -~;:: written 4 :: = 2 - ~x x+ y+2::=4 Solution. - h. The decomposition (multiply the matrix by 112 - why?) is In] [l~ 112 112 1/2 1 -U + L)-'U ~ 1/2 =I+L+U=I [0 ° + 112 112 1/2 112 ~][: 1/2 In] 112 0 0 :] + [: 0 0 It shows that c ~ 0 - [ )12 -1/4 -112 We compute the Frobenius norm of e lIell = ( -4I + -41 I 16 +- I 1 + -9 + -16+ -64 64 0 0 In] -1/2 112 [: 0 114 -In] -114 118 318 r Y'2 = (-5064 2= 0884 < I . amI conclude from (8) that this Gauss-Seidel iteration converges. It is interesting that the other two norms would permit no conclusion. as you ~hould verify. Of course. this points to the fact that (8) tS sufficient for convergence rather than necessary. • Residual. Given a system Ax defined by (12) b, the residual r of x with respect to this system is r = b - Ax. CHAP. 20 850 Numeric Linear Algebra Clearly, r = 0 if and only if x is a solution. Hence r *- 0 for an approximate solution. In the Gauss-Seidel iteration, at each stage we modify or relax a component of an approximate solution in order to reduce a component of r to zero. Hence the Gauss-Seidel iteration belongs to a class of methods often called relaxation methods. More about the residual follows in the next section. Jacobi Iteration The Gauss-Seidel iteration is a method of successive corrections because for each component we successively replace an approximation of a component by a corresponding new approximation as soon as the latter has been computed. An iteration method is called a method of simultaneous corrections if no component of an approximation x(m) is used until all the components of x(m) have been computed. A method of this type is the Jacobi iteration, which is similar to the Gauss-Seidel iteration but involves not using improved values until a step has been completed and then replacing x cm) by x(m+D at once, directly before the beginning of the next step. Hence if we write Ax = b (with ajj = I as before!) in the form x = b + (I - A)x, the Jacobi iteration in matrix notation is x(7n-t-l) = b + (I - A)x(m) (13) (ajj = I). This method converges for every choice of x(O) if and only if the spectral radius of 1 - A is less than 1. It has recently gained greater practical interest since on parallel processors alln equations can be solved simultaneously at each iteration step. For Jacobi, see Sec. 10.3. For exercises. see the problem set. - ....-----..... . , _~ _........ ..-. ..---.. • . . . . . . . A _ _ _ . . . ~ ... • . . . 1. Verify the claim at the end of Example 2. 2. Show that for the system in Example 2 the Jacobi iteration diverges. Him. Use eigenvalues. 13-81 -Xl + 4X2 X2 Xl + X2 + 6X3 = 7. -61.3 8. 185.8 4. X2 5Xl + .1.'2 Xl + 6X2 + 7X3 = 25.5 0 + Xl + 4X2 - 2x 1 + 3x2 + X3 = -10.5 2X3 = -2 8X3 = 39 21 X2 GAUSS-SEIDEL ITERATION Do 5 steps, starring [rom Xo = [I l]T and using 6S in the computation. Hint. Make sure that you solve each equation for the variable that has the largest coefficient (why?). Show the details. 3. 6. 4Xl - X3 = -45 + 4x3 = 33 lOx] + X2 + .\"3 6 Xl + lOx2 + X3 6 x] + x2 + IOx3 6 4xl + 5x3 = 12.5 Xl + 6.\"2 + 2X3 = 18.5 8xl + 2X2 + X3 = -1l.5 9. Apply the Gauss-Seidel iteration (3 steps) to the system in Prob. 7, starting from (a) 0, 0, 0, (b) 10. 10, 10. Compare and comment. 10. In Prob. 7, compute C (a) if you solve the rust equation for Xl. the second for .\"2' the third for X 3 , proving convergence; (b) if you nonsensically solve the third equation for Xlo the first for X 2 , the second for X3, proving divergence. SEC 20.4 851 Linear Systems: Ill-Conditioning, Norms improvement of convergence. (Spectacular gains are made with larger systems.) 11. CAS PROJECT. Gauss-Seidel Iteration. ta) Write a program for Gauss-Seidel iteration. (b) Apply the program to A(t)x = b, starting from [0 0 O]T. where A(t) = r: t J b = rJ For t = 0.2.0.5.0.8.0.9 determine the number of steps to obtain the exact solution to 6S and the corresponding spectral radius of C. Graph the number of steps and the spectral radius as functions of t and comment. (c) Successive overrelaxation (SOR). Show that by adding and subtracting x Cm) on the right, formula (6) can be written XCm+l) = X Cm ) + b - LxCm + 1 ) - (U + l)x Cm ) Anticipation of further corrections motivates the introduction of an overrelaxation factor w > I to get the SOR formula for Gauss-Seidel X Cm + ll = X Cm ) + web - 112-151 Do 5 steps. starting from Xo = [I 1 IJT. Compare with the Gauss-Seidel iteration. Which of the two seems to converge faster? (Show the details of your work.) 12. The system in Prob. 6 13. The system in Prob. 5 14. The system in Prob. 8 15. Show convergence in Prob. 14 by verifying that I - A, where A is the matrix in Prob. 14 with the rows divided by the corresponding main diagonal entries. has the eigenvalues -0.519589 and 0.259795 ::'::: 0.246603i. I 116-20 NORMS Compute the norms (9). (10), (II) for the following (square) matrices. Comment on the reasons for greater or smaller differences among the three numbers. 16. The matrix in Prob. 3 17. The matrix in Prob. 7 18. The matrix in Prob. 8 19. Lx Cm + ll (14) 20.4 r~ -k (ajj = 1) intended to give more rapid convergence. A recommended value is w = 2/(1 + ,h - p). where p is the spectral radius of C in (7). Apply SOR to the matrix in (b) for t = 0.5 and 0.8 and notice the JACOBI ITERATION 20. r-: 17 -k -2k -:1 -k 2k -3 -12 -:1 Linear Systems: Ill-Conditioning, Norms One does not need much experience to observe that some systems Ax = b are good, giving accurate solutions even under roundoff or coefficient inaccuracies. whereas others are bad. so that these inaccuracies affect the solution strongly. We want to see what is going on and whether or not we can "trust" a linear system. Let us first formulate the two relevant concepts (ill- and well-conditioned) for general numeric work and then tum to linear systems and matrices. A computational problem is called ill-conditioned (or ill-posed) if "small" changes in the data (the input) cause "large" changes in the solution (the output). On the other hand, a problem is called well-conditioned (or well-posed) if "small" changes in the data cause only "small" changes in the solution. These concepts are qualitative. We would certainly regard a magnification of inaccuracies by a factor 100 as "large," but could debate where to draw the line between "large" and "small." depending on the kind of problem and on uur viewpoint. Double precision may sometimes help, but if data are measured inaccurately, one should attempt changing the mathematical setting of the problem to a well-conditioned one. 852 CHAP. 20 Numeric Linear Algebra Let us now tum to linear systems. Figure 442 explains that ill-conditioning occurs if and only if the two equations give two nearly parallel lines. so that their intersection point (the solution of the system) moves substantially if we raise or lower a line just a little. For larger systems the situation is similar in principle, although geometry no longer helps. We shall see that we may regard ill-conditioning as an approach to singularity of the matrix. y y x x (a) (b) Fir 442. (a) Well-conditioned and (b) ill-conditioned linear system of two equations in two unknowns c X AMP L ElAn III-Conditioned System You may verify that the system 0.9999x - 1.000ly = 1 xhas the solution x = 0.5. Y = -0.5, wherea~ y=l the syMem 0.9999x - 1.000ly x- = 1 )"=I+E has the solution x = 0.5 + 5000.5 E. Y = -0.5 + 4999.5 E. Thi~ shows that the system is ill-conditioned because a change on the right of magnitude E produces a change in the solution of magnitude 5000E. approximately. We see that the lines given by the equations have nearly the same slope. • Well-conditioning can be asserted if the main diagonal entries of A have large absolute values compared to those of the other entries. Similarly if A-I and A have maximum entries of about the same absolute value. Ill-conditioning is indicated if A-I has entries of large absolute value compared to those of the solution (about 5000 in Example I) and if poor approximate solutions may still produce small residuals. Residual. The residual r of an approximate solution (2) b is defined as r = b - Ax. (1) Now b xof Ax = = Ax, so that r = A(x - x). Hence r is small if x has high accuracy, but the converse may be false: SEC. 20.4 85} Linear Systems: Ill-Conditioning, Norms E X AMP L E 1 Inaccurate Approximate Solution with a Small Residual The system 1.0001.\"1 + Xl + I.OOOlx2 = 2.0001 has the exact solution xl = 1, -'"2 = I. Can you see this by inspection? The very inaccurate approximation Xl = 2.0000, X2 = 0.0001 has the very small residual (to 4D) 2.0001 J [ 1.0001 I.OOOOJ [2.0000J [2.000IJ _ [2.0003J = [-O.0002J . [2.0001 - 1.0000 1.0001 0.0001 2.0001 2.0001 0.0000 = r = From this. a naive per,on might draw the false conc\u,ion that the appfoximation should be accumte to 3 Of 4 decimals. Our result is probably unexpected. but we shall see that it has to do with the fact that the system is ill-conditioned. • Our goal is to show that ill-conditioning of a linear system and of its coefficient matrix A can be measured by a number. the "condition number" K(A). Other measures for ill-conditioning have also been proposed, but K(A) is probably the most widely used one. K(A) is defined in terms of norm. a concept of great general interest throughout numerics (and in modem mathematics in general!). We shall reach our goal in three steps. discussing 1. Vector norms 2. Matrix norms 3. Condition number K of a square matrix. Vector Norms A vector norm for column vectors x = [Xj] with 11 components (11 fixed) is a generalized length or distance. It is denoted by II xII and is defined by four properties of the usual length of vectors in three-dimensional space. namely. (3) (a) IIxll is a nonnegative real number. (b) IIxll =0 (c) IIkxll = Iklllxll Cd) Ilx + if and only if yll ~ Ilxll x = O. for all k. + Ilyll (Triangle inequality). If we use several norms, we label them by a subscript. Most important in connection with computations is the p-Ilorm defined by (4) where p is a fixed number and p ~ 1. In practice, one usually takes p third norm, Ilxlix (the latter as defined below), that is, + ... + (5) IIxlll = IX11 (6) IIxl12 = YX12 + ... + xn2 (7) IIxlix = ~x J Ixjl = 1 or 2 and. as a IXnl ("Euclidean" or "12-norm") ("lx-norm"). 854 CHAP. 20 Numeric Linear Algebra For n = 3 the 12-norm is the usual length of a vector in three-dimensional space. The lrnorm and lx-norm are generally more convenient in computation. But all three norms are in common use. E X AMP L E 3 Vector Norms IfxT = [2 -3 0 L -4], then IIxll! = IIxll2 = V30, Ilxll oo = 10, • 4. In three-dimensional space, two points with position vectors x and xhave distance Ix - xl from each other. For a linear system Ax = b, this suggests that we take IIx - xii as a measure of inaccuraty and call it the distance between an exact and an approximate solution. or the error of x. Matrix Norm If A is an n X n matrix and x any vector with n components, then Ax is a vector with 11 components. We now take a vector norm and consider Ilxll and IIAxll. One can prove (see Ref. [E17]. p. 77, 92-93, listed in App. 1) that there is a number c (depending on A) such that IIAxll (8) ~ cllxll for all x. Let x*- O. Then IIxll > 0 by (3b) and division gives IIAx11/1lx11 ~ c. We obtain the smallest possible c valid for all x (=1= 0) by taking the maximum on the left. This smallest c is called the matrix norm of A corresponding to the vector norm we picked and is denoted by IIAII. Thus (9) IIAII the maximum being taken over all x (10) =1= =max IIAxl1 (x =1= 0), W O. Alternatively [see (c) in Team Project 24], IIAII = max IIxll= 1 IIAxll· The maximum in (10) and thus also in (9) exists. And the name "matrix Ilorm" is justified because IIAII satisfies (3) with x and y replaced by A and B. (Proofs in Ref. [EI7] pp. 77,92-93.) Note carefully that IIAII depends on the vector norm that we selected. In particular, one can show that for the lrnorm (5) one gets the column "sum" norm (10), Sec. 20.3, for the lx-norm (7) one gets the row "sum" norm (11), Sec. 20.3. By taking our best possible (our smallest) (11) IIAxl1 ~ c = IIAII we have from (8) IIAllllxll· This is the formula we shall need. Formula (9) also implies for two Ref. [E17], p. 98) (12) thus 11 X n matrices (see SEC. 20.4 855 Linear Systems: III-Conditioning, Norms See Refs. [E9] and [E 17] for other useful formulas on norms. Before we go on. let us do a simple illustrative computation. E X AMP L E 4 Matrix Norms Compute the matrix norms of the coefficient matrix A in Example I and of its inver,e A-I. assuming that we use (a) the lrvector norm, (b) the loo-vector norm. Solutioll. We use (4*). Sec. 7.!l, for the inverse and then (10) and (II) in Sec. 20.3. Thus 0.9999 -1.0001 ] A-I = A= [ 1.0000 - 1.0000 -5000.0 [ -5000.0 5000.5J. 4999.5 (a) The Irvector norm gives the column "sum" norm (10). Sec. 20.3: from Column 2 we thus obtain = 1-1.00011 + 1-1.00001 = 2.0001. Similarly. IIA -111 = 10000. IIAII (b) The lx-vector norm gives the row "sum" norm (11), Sec. 20.3: thus IIAII = 2, IIA -111 = 10000.5 from Row 1. We notice that IIA -111 is surprisingly large. which makes the product IIAII IIA -111 large (20001). We shall see below that this is typical of an ill-conditioned system. • Condition Number of a Matrix We are now ready to introduce the key concept in our discussion of ill-conditioning, the condition number K(A) of a (nonsingular) square matrix A, defined by (13) K(A) = IIAII IIA -III· The role of the condition number is seen from the following theorem. THEOREM 1 Condition Number A linear system of equations Ax = b and its matrix A whose condition number (13) is small are well-cmzditioned. A large condition number indicates ill-conditioning. PROOF b = Ax and (11) give IIbll ::; IIAII IIxll. Let b =1= 0 and x =1= O. Then division by Ilbll IIxll gives 1 IIxll "All lib II -<-- (14) = Multiplying (2) r = Atx - x) by A-I from the left and interchanging sides, we have x - x = A -Ir . Now (11) with A -1 and r instead of A and x yields IIx - xII = IIA -Irll ~ IIA -III IIrll. Division by IIxll [note that IIxll =1= 0 by (3b)] and use of (14) finally gives (15) M IIx - xII < _1_ -1 < IIAII -1 _ IIxll = IIxll IIA IIlIrll = IIbll IIA II IIrll - K(A) IIbil . Hence if K(A) is small, a small IIrll/llbll implies a small relative error IIx - xll/llxll, so that the system is well-conditioned. However, this does not hold if K(A) is large; then a small IIrli/llbil does not necessarily imply a small relative error IIx - xll/llxll • 856 E X AMP L E 5 CHAP. 20 Numeric Linear Algebra Condition Numbers. Gauss-Seidel Iteration 5 [ A =: I] has the : : Since A is symmetric. (10) anll inver~e (II) in Sec. 20.3 give the same condition number We see that a linear system Ax = b "ith this A is well-conditioned. 9]T (confirm this). For instance. if b = [14 0 28]T. the Gauss algorithm gives the solution x = [2 -5 Since the main diagonal entries of A are relatively large. we can expect reasonably good convergence of the Gauss-Seidel iteration. Inlleed, qaning from. say, Xo = [I I qT. we obtain the first 8 steps (3D values) EXAMPLE 6 Xl X2 X3 1.000 2.400 1.630 1.870 1.967 1.993 1.998 2.000 2.000 1.000 -1.100 -3.882 -4.734 -4.942 -4.988 -4.997 -5.000 -5.000 1.000 6.950 8.534 8.900 8.979 8.996 8.999 9.000 9.000 • III-Conditioned Linear System Example 4 gives by (10) or (II). Sec. 20.3, for the matrix in Exmnple I the very large condition number K(A) = 2.0001 . 10 000 ~ 2· 10 000.5 ~ 200001. This confirms that the sy,tem is very ill-conditioned. Similarl) in Example 2. where by (4""). Sec. 7.8 and 6D-computation. A-I = _1_ [ 0.0001 -1.oo00J = [ 1.0001 -1.0000 l.UOOI 5000.5 -5000.0 -5.000.0J 5000.5 so that (10), Sec. 20.3, gives a very large K(A), explaining the surprising result in Example 2. K(A) = (1.0001 + 1.0(00)(5000.5 + 5000.0) = 20002. • In practice, A-I will not be known, so that in computing the condition number K(A), one must estimate IIA -111. A method for this (proposed in 1979) is explained in Ref. [E9] listed in App. I. Inaccurate Matrix Entries. KlA) can be used for estimating the effect ox of an inaccuracy oA of A (errors of measurements of the Qjb for instance). Instead of Ax = b we then have (A + oA)(x + ox) Multiplying out and subtracting Ax Aox = = b. b on both sides, we obtain + oA(x + ox) = O. Multiplication by A-I from the left and taking the second term to the right gives SEC. 20.4 857 Linear Systems: III-Conditioning, Norms + Applying (II) with A-I and vector oA(x Applying (11) on the right, with oA and x ox) instead of A and x, we get - ox instead of A and x, we obtain lIoxll ~ IIA -111 IISAII IIx + Sxll . Now IIA -111 = K(A)/IIAil by the definition of K(A), so that division by IIx + oxll shows that the relative inaccuracy of x is related to that of A via the condition number by the inequality II ox II W (16) = IIx II ox II + oxll <: IIA-IIIII = ~AII b _ - A K( IIoAil ) IIAII . Conclusion, If the system is well-conditioned, small inaccuracies IISAII/IIAil can have only a small effect on the :>olution. However. in the case of ill-conditioning. if II oA 11/11 A II is small. IISxll/llxll11lay be large. Inaccurate Right Side. You may show that. similarly, when A is accurate, an inaccuracy Sb of b causes an inaccuracy ox satisfying (17) Hence lIoxll/llxll must remain relatively small whenever K(A) is small. E X AMP L E 7 Inaccuracies. Bounds (16) and (17) If each of the nine entries of A in Example 5 i_ measured with an inaccuracy of 0.1. then 115A:: (16) gives 115xll 3 '0.1 lJ;f ~ 7.5' - 7- = 0.321 115xll ~ 0.321 IIxll thus = 0.321 . 16 = 9' 0.1 and = 5.14. By experimentation you will find that the actual inaccuracy Iloxll is only about 30% of the bound 5.14. This is typicaL Similarly. if 5b = [0.1 0.1 O.I]T. then 115bll = 0.3 and Ilbll = 42 in Example 5. so that (17) gives 115xll TxlI ~ 7.5' 0.3 42 = 0.0536. hence 115xll ~ 0.0536' 16 = 0.857 but this bound is again much greater than the actual inaccuracy. which is about 0.15. Further Comments on Condition Numbers. may be helpful. • The following additional explainarions 1. There is no sharp dividing line hetween "well-conditioned" and "ill-conditioned," but generally the situation will get worse as we go from systems with small K(A) to systems with larger K(A). Now always K(A) ~ I, so that values of 10 or 20 or so give no reason for concern, whereas K(A) = 100, say, calls for caution. and systems such as those in Examples I and 2 are extremely ill-conditioned. CHAP. 20 858 Numeric Linear Algebra 2. If K(A) is large (or small) in one norm, it will be large (or small, respectively) in any other norm. See Example 5. 3. The literatme on ill-conditioning is extensive. For an introduction to it, see [E9]. This is the end of our discussion of numerics for solving linear systems. In the next section we consider curve fitting, an important area in which solutions are obtained from linear systems. ........ -- -.... .. ......... ==:=:=::: =........_-,._==.-_----,- - ~~ VECTOR NORMS Compute (5), (6), (7). Compute a corresponding unit vector (vector of norm 1) with respect to the too-norm. 1. [1 -6 5] 2. [0.4 3. [-4 4. 5. 6. 7. 8. - 1.2 4 3 0 8.01 -3] 18. Verify the calculations in Examples 5 and 6 of the text. 119-2°1 ILL-CONDITIONED SYSTEMS Solve Ax = bb Ax = b 2 • compare the soLutions. and comment. Compute the condition number of A. [0 0 0 0] -0.1 [0.3 0.5 LO] [L6 21 54 -119] [1 1 1 1 1 1] [3 0 0 -3 0] 9. Show that 16. Verify (II) for x = [4 -5 2]T taken with the loe-norm and the matrix in Prob. L5. 17. Verify (12) for the matrices in Probs. to and 1 L 19. A = Ilxll oo ~ IIxl12 ~ Ilxlli. [ LAJ . b i -- [1.4J 2 1.4 1 1 • b2 -- [1.44J 1 20. A= [ 5 -7J. b= [-2J .b = [-2J -7 110-151 10 i 3 2 3.1 MATRIX NORMS. CONDITION NUMBERS Compute the matrix norm and the condition number corresponding to the II-vector norm. 21. (Residual) For Ax = b] in Prob. 19 guess what the residuaL of = [113 -160]T might be (the solution II. 22. Show that K(A) ~ 1 for the matrix norms (10), (11), Sec. 20.3, and K(A) ~ V;; for the Frobenius nOim (9), Sec. 20.3. 13. [ 57 ro~, 21 10.5 7 5.25 10.5 7 5.25 4.2 7 5.25 4.2 3.5 5.25 4.2 3.5 3 14. 15. [0~1 01 0.1 0] o 100 o '~l x being x = [0 I]T). Then calculate and comment. 23. CAS EXPERIMENT. Hilbert Matrices. The 3 X 3 Hilbert matrix is The n X n HiLbert matrix is Hn = fhjk], where hjk = l/(j + k - 1). (Similar matrices occur in curve fitting by Least squares.) Compute the condition number K(Hn) for the matrix norm corresponding to the loc- (or ld vector norm, for fl = 2, 3, ... ,6 (or further if you wish). Try to find a formula that gives reasonable approximate values of these rapidly growing numbers. Solve a few linear systems involving an Hn of your choice. SEC. 20.5 859 Least Squares Method 24. TEAM PROJECT. Norms. (a) Vector norms in our text are equivalent, that is, they are related by double inequalities; for instance, (a) I\xl\x ~ I\XI\I ~ IIl\xIL,,: (b) -lIxllI ~ IIxlix ~ IIxll l (18) 1 · 11 Hence if for some x, one norm is large (or small), the other norm must also be large (or small). Thus in many investigations the particular choice of a noml is not essential. Prove (18). (b) The Cauchy-Schwarz inequality is (19b) (c) Formula (10) is often more practical than (9). Derive (10) from (9). (d) Matrix normS. lllustrate (11) with examples. Give examples of (12) with equality as well as with strict inequality. Prove that the matrix norms (10), (II) in Sec. 20.3 satisfy the axioms of a nonn IIL\II ~ IIAI\ = 0 if and only if A = o. IIkAil IIA It is very important. (Proof in Ref. [GR7] listed in App. 1.) Use it to prove (l9a) 20.5 o. + BII = IkIIiAII, ~ IIAII + IIBII· 25. WRITING PROJECT. Norms and Their Use in This Section. Make a list of the most important of the many ideas covered in this section and write a two-page report on them. Least Squares Method Having discussed numerics for linear systems, we now turn to an important application. curve fitting. in which the solutions are obtained from linear systems. In curve fitting we are given 11 points (pairs of numbers) (Xb "1)' ... , (Xn , Yn) and we want to determine a function f(x) such that approximately. The type of function (for example. polynomials. exponential functions, sine and cosine functions) may be suggested by the nature of the problem (the underlying physical law, for instance), and in many cases a polynomial of a certain degree will be appropriate. Let us begin with a motivation. If we require strict equality [(Xl) = )'1, . . . , f(xn) = Yn and use polynomials of sufficiently high degree, we may apply one of the methods discussed in Sec. 19.3 in connection with interpolation. However, in certain situations this would not be the appropriate solution of the acmal problem. For instance. to the four points (1) (-1.3.0.103), (-0.1. 1.099), (0.2, 0.808), there corresponds the interpolation polynomial [(x) :~ Fig. 443. = X3 - X (l.3, 1.897) + I (Fig. 443), but if we / t=ilx ~ ApprOXimate fitting of a straight line 860 CHAP. 20 Numeric Linear Algebra graph the points, we see that they lie nearly on a straight line. Hence if these values are obtained in an experiment and thus involve an experimental error, and if the nature of the experiment suggests a linear relation, we better fit a straight line through the points (Fig. 443). Such a line may be useful for predicting values to be expected for other values of x. A widely used principle for fitting straight lines is the method of least squares by Gauss and Legendre. In the present situation it may be formulated as follows. Method of Least Squares. The straight line (2) y =a+ bx should be fitted through the givell points (Xl, Yl)' ...• (Xn . Yn) so that the sum of the squares of the distances of those points from the straight line is minimum. where the distance is measured in the vertical direction (the .v-direction). The point on the line with abscissa Xj has the ordinate a + bXj. Hence its distance from (Xj, .\) is h1 - a - bxjl (Fig. 444) and that sum of squares is n q = L ()J - (/ - bXjl. j~l q depends on a and b. A necessary condition for q to be minimum is (3) oq ob = - 2L Xj (\J - a - b;f) = 0 (where we sum over j from 1 to n). Dividing by 2, writing each sum as three sums, and taking one of them to the right, we obtain the result (4) an +b a~ L.J x·J + ~ L.J X·J b~ L.J = ~ y. X·2 = J L.J .J ~ L.J X·V· J-J' These equations are called the normal equations of our problem. Vertical distance of a point (xi' Yj) from a straight line y = a + bx Fig. 444. SEC. 20.5 861 Least Squares Method E X AMP L E 1 Straight Line Using the method of lea~t squares. fit a straight line to the four points given in formula (I). Solutioll. 11 We obtain = 4, 2::>'J = 2: x/ = 0.1, 2: Yj = 3.43, 2: XjYj = 3.907, 2.3839. Hence the normal equations are 4ll + O.JOb = 3.9070 O.lll + 3.43b = 2.3839. The solution <rounded to 4D) is a = 0.9601, b = 0.6670. and we obtain the straight line (Fig. 443) y = 0.9601 • + 0.6670T. Curve Fitting by Polynomials of Degree m Our method of curve fitting can be generalized from a polynomial y polynomial of degree III (5) where p(x) III ~ 11 - a + bx to a = b o + bix + ... + b",x'Tn l. Then q takes the form " q = 2: (Yj - p(Xj»2 j~I and depends on conditions III + I parameters bo, ... , bm . Instead of (3) we then have aq (6) abo aq abm = 0, = 0 which give a system of 111 + I normal equations. In the case of a quadratic polynomial (7) the normal equations are (summation from I to n) + b 2: Xj + b2 2: x/ = 2: )j ol1 boL + b 2: xl + b 2: x/ = 2: bo 2: xl + b] 2: x/ + b 2: x/ = 2: xl,vj· b (8) i Xj i 2 2 The derivation of (8) is left to the reader. XjYj III + I 862 CHAP. 20 E X AMP L E 2 Numenc Linear Algebra Quadratic Parabola by Least Squares Fit a parabola through the data (0. 5), (2. 4), (4, 1), (6, 6). (8, 7). Solul;on. For the nonnal equations we need 11 = 5, L.\) = 20, L)"j = 23, LXj)] = 104. LX/)'j = 696. Hence these equations are 5bo + 20b! + 120b2 20bo = LX/ = 120, LX/ = 800. 2.x/ = 5664. 23 + 120b! + 800b2 = 104 120bo + 800b! + 5664b 2 = 696. Solving them we obtain the quadratic least squares parabola (Fig. 445) y = 5.11429 - 1.4l429x • + O.21429x2. y 8 • 2 • o Fig. 445. 2 4 6 8 x Least squares parabola in Example 2 For a general polynomial (5) the normal equations form a linear system of equations in the unknowns b o, ... , b"n" When its matrix M is nonsingular, we can solve the system by Cholesky's method (Sec. 20.2) because then M is positive definite (and symmetric). When the equations are nearly linearly dependent, the normal equations may become illconditioned and should be replaced by other methods; see [E5], Sec. 5.7, listed in App. 1. The least squares method also plays a role in statistics (see Sec. 25.9). 11-61 FITTING A STRAIGHT LINE Fit a straight line to the given points (x, y) by least squares. Show the details. Check your result by sketching the points and the line. Judge the goodness of fit. 1. (2,0), (3, 4), (4. lO), (5. 16) 2. How does the line in Prob. I change if you add a point far above it, say. (3. 20)? 3. (2.5, 8.0), (5.0, 6.9), (7.5. 6.2), (10.0, 5.0) 4. (Ohm's law U = Ri) Estimate the resistance R from the least squares line that fits (i, U) = (2.0, 104). (4.0,206), (6.0, 314), (10.0, 530). 5. (Average speed) Estimate the average speed vav of a car traveling according to s = v • t [km] (s = distance traveled. t [h] = time) from (t, s) (11, 3lO). (\2. 4lO). = (9, 140), (10, 220), 6. (Hooke's law F = ks) Estimate the spring modulus k from the force F [lb] and extension s [cm], where (F, s) = (1, 0.50), (2, 1.02), (4, 1.99), (6, 3.01), (10.4.98), (20. 10.03). 7. Derive the normal equations (8). 18-10 1 FITTING A QUADRATIC PARABOLA Fit a parabola (7) to the given points Cx, y) by least squares. Check by Sketching. 8. (-1,3), (0, 0), (I. 2). (2. 8) 9. (0,4), (2, 2), (4, -1), (6, -5) SEC 20.6 863 Matrix Eigenvalue Problems: Introduction 10. Worker's time on duty x [h] 2 Worker's reaction lime [sec] 1.50 1.28 lAO 1.85 2.20 16. TEAM PROJECT. The least squares approximation of a function f(x) on an interval a ;;a x ;;a h by a function 11. Fit (2) and (7) by lea<;t squares to (-1.0,5.4), (-0.5,4.1), (0,3.9), (0.5,4.8), (1.0, 6.3), (l.5, 9.3). Graph the data and the curves on common axes and comment. where Yo(x), ... ,y",(x) are given functions, requires the determination of the coefficients ao, .•• , am such that 3 4 5 12. (Cubic parabola) Derive the formula for the normal equations of a cubic least squares parabola. 13. Fit curves (2) and (7) and a cubic parabola by least squares to (-2, -35), (-1. -9), (0, -I), (I, -I), (2, 17). (3. 63). Graph the three curves and the points on common axes. Comment on the goodness of fit. 14. CAS PROJECT. Least Squares. Write programs for calculating and solving the normal equations (4) and (8). Apply the programs to Probs. 3, 5, 9, 11. If your CAS has a command for fitting (Maple and Mathematica do), compare your results with those by your CAS commands. 15. CAS EXPERIMENT. Least Squares versus Interpolation. For the given data and for data of your choice find the interpolation polynomial and the least squares approximations (linear. quadratic, etc.). Compare and comment. (a) (-2,0), (-I, 0), (0, I), (I, 0), (2, 0) (b) (-4. 0), (-3. 0), (-2. 0). (-1. 0), (0, I). (I, 0). (2. 0), (3. 0). (4. 0) I (9) b [f(x) - Fm(x)]2 dx a becomes mmlmum. This integral is denoted by Ilf - Fm1l 2 , and IIf - Fmll is called the L 2 -norm of f - F m (L suggesting Lebesgue2 ). A necessary condition for that minimum is given by allf - F ml12/aaj = 0, j = 0, ... , m [the analog of (6)]. (a) Show that this leads to m + 1 normal equations (j = 0, ... , m) m 2: hjkak = b j where k=O (10) hjk = I I b )'j(X)Yk(X) dx, a bj = b !(X)'\".i(x) dx. Q (C) Choose five points on a straight line, e.g., (0, 0), (l, 1), ... , (4, 4). Move one point 1 unit upward and (b) Polynomial. What form does (lO) take if Fm(x) = a o + lllX + ... + amx"'? What is the coefficient matrix of (10) in this case when the interval is 0 ;;a x;;a I? find the quadratic least squares polynomial. Do this for each point. Graph the five polynomials on Cornmon axes. Which of the five motions has the greatest effect? (c) Orthogonal functions. What are the solutions of (10) if .\'o(x), ... , Ym(x) are orthogonal on the interval a ;;a x ;;a b? (For the definition, see Sec. 5.7. See also Sec. 5.8.) 20.6 Matrix Eigenvalue Problems: Introduction In the remaining sections of this chapter we discuss some of the most important ideas and numeric methods for matrix eigenvalue problems. This very extensive part of numeric linear algebra is of great practical importance, with much research going on, and hundreds, if not thousands of papers published in various mathematical journals (see the references in [ES], [E9], [Ell], [E29]). We begin with the concepts and general results we shall need in explaining and applying numeric methods for eigenvalue problems. (For typical models of eigenvalue problems see Chap. 8.) 2HENRI LEBESGUE (1875-1941). great French mathematician. creator of a modern theory of measure and integration in his famous doctoral thesis of 1902. 864 CHAP. 20 Numeric Linear Algebra An eigenvalue or characteristic value lor latent root) of a given Il X n matrix A = [ajk] is a real or complex number A such that the vector equation Ax (1) = Ax has a nontrivial solution, that is, a solution x "* 0, which is then called an eigenvector or characteristic vector of A corresponding to that eigenvalue A. The set of all eigenvalues of A is called the spectrum of A. Equation (I) can be written lA - AI)x = 0 (2) where I is the n X n unit matrix. This homogeneous system has a nontrivial solution if and only if the characteristic determinant det (A - AI) is 0 (see Theorem 2 in Sec. 7.5). This gives (see Sec. 8.1) Eigenvalues THEOREM 1 The eigenvalues of A are the solutions A of the characteristic equation all - A a21 (3) a 1n a12 a22 - A a2n det (A - AI) = = O. ann - an2 anI A Developing the characteristic determinant, we obtain the characteristic polynomial of A, which is of degree n in A. Hence A has at least one and at most n numerically different eigenvalues. If A is real, so are the coefficients of the characteristic polynomial. By familiar algebra it follows that then the roots (the eigenvalues of A) are real or complex conjugates in pairs. We shall usually denote the eigenvalues of A by with the understanding that some (or all) of them may be equal. The sum of these n eigenvalues equals the sum of the entries on the main diagonal of A, called the trace of A; thus n (4) trace A = L j~1 n ajj = L Ak . k~1 Also, the product of the eigenvalues equals the determinant of A, (5) Both formulas follow from the product representation of the characteristic polynomial, which we denote by f(A), SEC. 20.6 865 Matrix Eigenvalue Problems: Introduction If we take equal factors together and denote the numerically distinct eigenvalues of A by Ab ... , AI" (r ~ 11), then the product become~ (6) The exponent 111j is called the algebraic multiplicity of Aj • The maximum number of linearly independent eigenvectors corresponding to Aj is called the geometric multiplicity of Aj . It is equal to or smaller than 111j. A subspace S of R n or en (if A is complex) is called an invariant subspace of A if for every v in S the vector A,- is also in S. Eigenspaces of A (spaces of eigenvectors; Sec. 8.1) are important invaJiant subspaces of A. An 11 X 11 matrix B is called similar to A if there is a nonsingular n X 11 matrix T such that (7) Similarity is important for the following reason. THEOREM 2 Similar Matrices Similar matrices have the same eigellmlues. If x is an eigenvector of A, then y = T-Ix is all eigel1l'ector of B in (7) wrrespollding to the sallie eigellmille. (Proof in Sec. 8.4.) Another theorem that has vaJious applications in numerics is as follows. THEOREM 1 Spectral Shift IfA has the eigenvalues A1> •.• , An' then A - kI with arbitrary k has the eigenvalues Al - k . .. '. An - k. This theorem is a special case of the following spectral mapping theorem. THEOREM 4 Polynomial Matrices If A is all eigenvalue of A, then is all eigenvalue of the polynomial matrix q(A)x + O's_lAs - 1 + ... ) x = asAsx + as_lAs-Ix + .. . = asAsx + as_lAs-Ix + ... = q(A) x. = (asAS • CHAP. 20 866 Numeric Linear Algebra The eigenvalues of important special matrices can be characterized as follows. Special Matrices THEOREM 5 The eigenvalues of Hermitian matrices (i.e., -AT = A), hence of real symmetric matrices (i.e., AT = A), lire real. The eigenvalues of skew-Hermitian matrices (i.e., AT = -A), hence of real skew-symmetric matrices (i.e., AT = -A) are pure imaginary or O. The eigenvalues of unifllry matrices (i.e., AT = A-I). hence of orthogonalmlltrices (i.e .• AT = A-I), have absolute value l. (Proofs in Secs. 8.3 and 8.5.) The choice of a numeric method for matrix eigenvalue problems depends essentially on two circumstances, on the kind of matrix (real symmetric, real general, complex, sparse, or full) and on the kind of information to be obtained, that is, whether one wants to know all eigenvalues or merely specific ones, for instance, the largest eigenvalue, whether eigenvalues lind eigenvectors are wanted. and so on. It is clear that we cannot enter into a systematic discussion of all these and further possibilities that arise in practice, but we shall concentrate on some basic aspects and methods that will give us a general understanding of this fascinating field. 20.7 Inclusion of Matrix Eigenvalues The whole numerics for matrix eigenvalues is motivated by the fact that except for a few trivial cases we cannot determine eigenvalues exactly by a finite process because these values are the roots of a polynomial of Ilth degree. Hence we must mainly use iteration. In this section we state a few general theorems that give approximations and error bounds for eigenvalues. Our matrices will continue to be real (except in formula (5) below). but since (nonsymmetric) matrices may have complex eigenvalues. complex numbers will playa (very modest) role in this section. The important theorem by Gerschgorin gives a region conSisting of closed circular disks in the complex plane and including all the eigenvalues of a given matrix. Indeed. for each j = 1. .... n the inequality (1) in the theorem detennines a closed circular disk in the complex A-plane with center {ljj and radius given by the right side of (1); and Theorem 1 states that each of the eigenvalues of A lies in one of these n disks. Gerschgorin's Theorem THEOREM 1 Let A be all eigenvalue qf all arbitrary integer j (l ~ j ~ 11) we have PROOF Il X Il matrix A [ajk]. Then for some Let x be an eigenvector corresponding to an eigenvalue A of A. Then (2) Ax = Ax or (A - AI)x = O. Let Xj be a component of x that is largest in absolute value. Then we have /x11 /Xj/ ~ I for = 1, ... , n. The vector equation (2) is equivalent to a system of 11 equations for the I7l SEC. 20.7 867 Inclusion of Matrix Eigenvalues n components of the vectors on both sides. The jth of these n equations with j as just indicated is Division by Xj (which cannot be zero; why?) and reshuffling terms gives By taking absolute values on both sides of this equation, applying the triangle inequality la + bl ~ lal + Ibl (where a and b are any complex numbers), and observing that because of the choice of j (which is crucial !), IX1lxjl ~ 1, ... , IXn1xjl ~ I, we obtain (l), and the theorem is proved. • E X AMP L E 1 Gerschgorin's Theorem For the eigenvalues of the matrix 112 5 we get the Gerschgorin disks (Fig. 446) ° 1: Center 0, radiu~ I, 02: Center 5. radius 1.5. 03: Center 1. radius 1.5. The centers are the main diagonal entries of A. These would be the eigenvalues of A if A were diagonal. We can take these values as crude approximation~ of the unknown eigenvalues (3D values) Al = -0.209, A2 = 5.305. A3 = 0.904 (verify this): then the radii of the disks are corresponding error bounds. Since A is symmetric, it follow, from Theorem 5, Sec. 20.6, that the spectrum of A must actually lie in the intervals [-1. 2.5] and [3.5, 6.5]. It is interesting that here the Gerschgorin disks form two disjoint sets, namely, 01 U 3 , which contains two • eigenvalues, and 02. which contains one eigenvalue. This is typical. as the following theorem shows. ° y x Fig. 446. THEOREM 2 Gerschgorin disks in Example 1 Extension of Gerschgorin's Theorem If p Gerschgorin disks fom! a set S that is disjoint from the n - p other disks of a given matrix A. t!zen S contains precisely p eigenl'Q/ues of A (each cOllnted with its algebraic multiplicity. as defined in Sec. 20.6). Idea of Proof. Set A apply Theorem I to At = B + C, where B is the diagonal matrix with entries = B + tC with real t growing from 0 [0 1. ajj. and • 868 E X AMP L E 2 CHAP. 20 Numeric Linear Algebra Another Application of Gerschgorin's Theorem. Similarity Suppose that we have diagonalized a matrix by some numeric method that left us with some off-diagonal entries of size 10- 5 , say, 5 10- ] 10-5 2 . 4 What can we conclude about deviation~ of the eigenvalue~ from the main diagonal entrie~? By Theorem 2. one eigenvalue must lie in the disk of radius 2· 10- 5 centered at 4 and two 5 eigenvalues (or an eigenvalue of algebraic mulliplicity 2) in the disk of radius 2· centered at 2. Actually, since the matrix is symmetric. these eigenvalues must lie in the intersections of these disks and the real axis. by Theorem 5 in Sec. 20.6. We sho" how an isolated disk can always be reduced in size by a similarity transformation. The matrix Solution. 10- o 5 10- [I0 10- l~ o 5] o 4 2 0 o ] ;] is similar to A. Hence by Theorem 2. Sec. 20.6. it has the same eigenvalues as A. From Row 3 we get the smaller disk of radius 2· 10- 10. Note that the other disks got bigger. approximately by a factor of 105 . And in choosing T we have to watch lhat the new disks do not overlap with the disk whose size we want to decrease. For funher interesting facts, see the new book [E28j. • By definition, a diagonally dominant matrix A = [ajk] is an n X n matrix such that lajjl (3) ~ L lajkl j = I,"', n k*j where we sum over all off-diagonal entries in Row j. The matrix is said to be strictly diagonally dominant if > in (3) for all j. Use Theorem I to prove the following basic property. THEOREM 3 Strict Diagonal Dominance Strictly diagonally dominant matrices are nOllSingular. Further Inclusion Theorems An inclusion theorem is a theorem that specifies a set which contains at least one eigenvalue of a given matlix. Thus. Theorems I and 2 are inclusion theorems; they even include the whole spectrum. We now discuss some famous theorems that yield further inclusions of eigenvalues. We state the first two of them without proofs (which would exceed the level of this book). SEC. 20.7 869 Inclusion of Matrix Eigenvalues THEOREM 4 Schur's Theorem 3 Let A = [ajk] be an 11 X n matrix. Then for each of its eigenvalues A1> •.• , An' n (4) n n \AmJ 2 ~ L \Ai\2 ~ L L i=l In (4) the second equality sign holds \ajk\2 (Schur's inequality). j=l k=l if alld ollly if A is such that (5) Matrices that satisfy (5) are called normal matrices. It is not difficult to see that Hermitian, skew-Hermitian, and unitary matrices are normal, and so are real symmetric, skew-symmetric, and orthogonal matlices. E X AMP L E 1 Bounds for Eigenvalues Obtained from Schur's Inequality For the matrix we obtain from Schur's inequality IAI ;;: v'I949 = 44.1475. You may verify that the eigenvalues are 30. 25, • and 20. Thus 302 + 25 2 + 202 = 1925 < 1949; in fact, A is not normal. The preceding theorems are valid for every real or complex square matrix. Other theorems hold for special classes of matrices only. Famous is the following. THEOREM 5 Perron's Theorem4 Let A be a real n X n matrix whose e1l1ries are all positive. Then A has a positive real eigenvalue A = p of multiplicity I. The corresponding eigenvector can be chosen with all components positive. (The other eigenvalues are less than p in absolute value.) For a proof see Ref. [B3], vol. II, pp. 53-62. The theorem also holds for matrices with nonnegative real entries ("Perron-Frobenius Theorem,,4) provided A is irreducible, that is, it cannot be brought to the following form by interchanging rows and columns: here Band F are square and 0 is a zero matrix. :ISSAI SCHUR (1875-1941), German mathematician, also known by his important work in group theory. OSKAR PERRON (1880-1975), GEORG FROBENIUS (1849-1917), LOTHAR COLLATZ (1910-1990), German mathematicians, known for their work in potential theory, ODEs (Sec. 5.4) and group theory, and numerics. respectively. 870 CHAP. 20 Numeric Linear Algebra Perron's theorem has various applications, for instance, in economics. It is interesting that one can obtain from it a theorem that gives a numeric algorithm: Collatz Inclusion Theorem4 THEOREM 6 Let A = [ajk] be a real n X 11 matrLl: whose elements are all positive. Let x be a1l\' real vector whose components Xl' • • . • Xn are positive, alld let )'1> •••• Yn be the components of the vector y = Ax. Then the closed imen'al 011 the real axis bounded by the smallest and the largest of the n quotients % = )'/Xj contains at least one eigenvalue of A. PROOF We have Ax = yor (6) y - Ax = O. The transpose AT satisfies the conditions of Theorem 5. Hence AT has a positive eigenvalue A and, corresponding to this eigenvalue. an eigenvector u whose components Uj are all positive. Thus ATu = Au, and by taking the transpose we obtain uTA = AUT. From this and (6) we have or written out n L u/Yj - AXj) = O. j~l Since all the components Uj are positive, it follows that Yj - A.r:j ~ O. that is. for at least one j. Yj - ,\xj ~ 0, that is, for at least one j. (7) and Since A and AT have the same eigenvalues. A is an eigenvalue of A. and from (7) the statement of the theorem follows. • E X AMP L E 4 Bounds for Eigenvalues from Collatz's Theorem. Iteration For a given matrix A with positive entries we choose an x = "0 and iterate, that is, we compute Xl = Axo· X2 = AXI • . . . • X20 = AX19' In each step, taking x = ":i and y = AXj = Xj+l we compute an inclusion interval by Collatz's theorem. This gives (6S) 0.02 A = [ 049 0.02 0.28 0.22] 0.20 0.22 0.20 0.40 •• ' . xl9 = ,xo ~ [I] 1 0.002 16309J 0.00108155 [ 0.00216309 ,Xl = I ,x20 [0.73] 0.50 ,X2 ~ 0.82 = [0.00155743 J 0.000778713 0.00155743 [0.5481] 0.3186 , 0.5886 SEC. 20.7 871 Inclusion of Matrix Eigenvalues and the intervals 0.5 have length ~ j Length 0.32 A ~ 0.82, 0.3186/0.50 = 0.6372 ~ A ~ 0.548l/0.73 = 0.750822, etc. These intervals 2 3 10 15 20 0.113622 0.0539835 0.0004217 0.0000132 0.0000004 Using the characteristic polynomial. you may verify that the eigenvalues of A are 0.72. 0.36. 0.09. so that those intervals include the largest eigenvalue. 0.72. Their lengths decreased withj. so that the iteration was worthwhile. • The reason will appear in the next ~ection, where we discuss an iteration method for eigenvalues. PROBLEM SET 2<h.7 GERSCHGORIN DISKS 11-61 Find and sketch disks or interval~ that contain the eigenvalues. If you have a CAS, find the spectrum and compare. 1.[-~ 2. -: ] [1~_2 10- 2 10- 2 10- 2 9. If a symmetric 11 X 11 matrix A = [ajkl has been diagonalized except for small off-diagonal entries of size 10- 6 , what can you say about the eigenvalues? 10. (Extended Gerschgorin theorem) Prove Theorem 2. 11. Prove Theorem 3. 2 10- ] 10-2 8 8. By what integer factor can you at most reduce the Gerschgorin circle with center 3 in Prob. 6? 12. (Normal matrices) Show that Hermitian, skewHermitian, and unitary matrices (hence real symmetric. skew-symmetric, and orthogonal matrices) are normal. Why is this of practical interest? peA»~ Show that p(A) cannot be greater than the row sum norm of A. 13. (Spectral radius 9 14. (Eigenvalues on the circle) Illustrate with a 2 X 2 matrix that an eigenvalue may very well lie on a Gerschgorin circle (so that Gerschgorin disks can generally not be replaced with smaller disks without losing the inclusion property). 1 4. + 0.3 i 0.3i -3 + 2i 115--171 0.5i] 0.1 [ 5. [ O.li 0.2 4i 0.1i O.li o o -1 + 10 6. [ i 4-i 0.1 0.1 6 -0.2 o SCHUR'S INEQUALITY Use (4) to obtain an upper bound for the spectral radius: 15. In Prob. I 16. In Prob. 6 17. In Prob. 3 118-191 COLLATZ'S THEOREM Apply Theorem 6, choosing the given vectors as vectors x. 10 I 18. 9 [ 3 2 7. (Similarity) Find T-TAT such that in Prob. 2 the radius of the Gerschgorin circle with center 5 is reduced by a factor III 00. '9. [: 4 2 J[J[J[J CHAP. 20 872 Numeric Linear Algebra (b) Apply the program to symmetric matrices of your choice. Explore how convergence depends on the choice of initial vectors. Can you construct cases in which the lengths of the inclusion intervals are not monotone decreasing? Can you explain the reason? Can you experiment on the effect of rounding? 20. CAS EXPERIMENT. Collatz Iteration. (a) Write a program for the iteration in Example 4 (with any A and xo) that at each step prints the midpoint (why?). the endpoints. and the length of the inclusion interval. 20.8 Power Method for Eigenvalues A simple standard procedure for computing approximate values of the eigenvalues of an n X n matrix A = [ajd is the power method. In this method we start from any vector xo (* 0) with n components and compute successively Xs = AX.<_l· For simplifying notation, we denote Xs-l by x and Xs by y. so that y = Ax. The method applies to any n X n matrix A that has a dominant eigenvalue (a A such that is greater than the absolute values of the other eigenvalues). If A is symmetric. it also gives the error bound (2), in addition to the approximation (1). IAI THEOREM 1 Power Method, Error Bounds Let A be an n X n real symmetric matrix. Let x (* 0) be any real vector with n components. Furthermore. let y = Ax, mo = 1111 XTX, = xTy, Then the quotient (I) (Rayleigh 5 quotient) q= is lin lIpproximation for an eigenvalue A of A (usually that which is greatest in absolute value, but no general statements are possible). FlIrthe17llore. if we set q = A-E. so that E is the error of q. then (2) 5 LORD RAYLEIGH (JOHN WILLIAM STRUTT) (1842-1919), great English physicist and mathematician. professor at Cambridge and London, known for his important contributions to various branches of applied mathematics and theoretical physics. in particular. the theory of waves. elasticity. and hydrodynamics. In 1904 he received a Nobel Prize in physics. SEC. 20.8 873 Power Method for Eigenvalues PROOF fJ2 denotes the radicand in (2), Since 1111 = qmo by ( I). we have Since A is real symmetric, it has an orthogonal set of n real unit eigenvectors ZI' . . . , zn corresponding to the eigenvalues AI, . . . , An' respectively (some of which may be equal). (Proof in Ref. [B3], vol. I, pp. 270-272, listed in App. I.) Then x has a representation of the fmm Now AZI = AIZI, and. since the Zj etc., and we obtain are orthogonal unit vectors. (4) It follows that in (3), Since the Zj are orthogonal unit vectors, we thus obtain from (3) Now let Ae be an eigenvalue of A to which q is closest, where c suggests ·'closest"". Then (Ae - q)2 ~ (AJ - q)2 for j = 1, ... , n. From this and (5) we obtain the inequality Dividing by 1110, taking square roots. and recalling the meaning of This shows that fJ is a bound for the error A and completes the proof. E fJ2 gives of the approximation q of an eigenvalue of • The main advantage of the method is its simplicity. And it can handle sparse matrices too large to store as a full square array. Its disadvantage is its possibly slow convergence. From the proof of Theorem ] we see that the speed of convergence depends on the ratio of the dominant eigenvalue to the next in absolute value (2:] in Example I, below). If we want a convergent sequence of eigenvectors, then at the beginning of each step we scale the vector, say, by dividing its components by an absolutely largest one, as in Example 1, as follows. 874 E X AMP L E 1 CHAP. 20 Numeric Linear Algebra Application of Theorem 1. Scaling For the symmetric matrix A in Example 4, Sec. 20.7. and Xo indicated scaling A = O.4Y 0.02 om 0.28 0.22 0.20 [I] 0.22] [ ~::~ , Xo = XlO = ~ ~.504682 [0.890244] , Xl [0.931193] ~.609756 , X2 = 0,999707] 0,990663] X5 = I] T we obtain from (I) and (2) and the = [I . ~.500146 0,999991] . ~.500005 Xl5 = . [ [ [ ~.541284 = Here AXV = [0.73 0.5 0.S2IT, scaled to Xl = [0.73/0.82 0.5/0.82 lIT. etc. The dominam eigenvalue is 0.72, an eigenvector [I 0.5 lIT. The corresponding q and 8 are computed each time before the next scaling. Thus in the first step. XOTAxO 2.05 - - - T - = -3- = 0.683333 Xo Xo ,, __ (1112 _ q2)1/2 __ v 1110 (AXo)TAxo 2)112 T - q = (1.4553 - - - q 2)112 = 0.134743. Xo Xo 3 This gives the foIlowing values of q, 8, and the error j € = 0.72 - q (calculations with 10D, rounded to 6D): 2 5 10 q 0.683333 0.716048 0.719944 0.720000 8 0.134743 0.036667 0.038887 0.003952 0.004499 0.000056 0.000l41 5' 10-8 € The error bounds are much larger than the actual errors. This is typicaL although the bounds cannot be improved: that is, for special symmetric matrices [hey agree with the errors. Our present results are somewhat better than those of CoIlatz'~ method in Example 4 of Sec. 20.7, at the expense of more operations. • Spectral shift, the transition from A to A - kI, shifts every eigenvalue by -k. Although finding a good k can hardly be made automatic, it may be helped by some other method or small preliminary computational experiments. In Example I. Gerschgorin' s theorem gives -0.02 ~ A ~ 0.82 for the whole spectrum (verify!). Shifting by -0.4 might be too much (then -0.42 ~ A ~ 0.42), so let us try -0.2. E X AMP L E 2 Power Method with Spectral Shift For A - 0.21 with A as in Example I we obtain the following substantial improvements (where the index 1 refers to Example I and the index 2 to the present example). j 81 82 €l €2 0.134743 0.134743 0.036667 0.036667 2 5 10 0.038887 0.034474 0.003952 0.002477 0.004499 0.000693 0.000056 1.3. 10-6 0.000141 1.8· 10-6 5 '10- 8 9' 10- 12 • SEC. 20.9 875 Tridiagonalization and QR-Factorization PRO B. L EMS E1::::..2 D-;-8_ 11-71 ° POWER METHOD WITH SCALING Apply the power method (3 steps) with scaling. using = [I I]T or [I I I]T or [I I I I]T. as applicable. Give Rayleigh quotients and error bounds. Show the details of your work. Xo 3.5 1. [ 3. 0.6 2.0J 2. [ U.5 2.0 11. (Spectral shift, smallest eigenvalue) In Prob. 5 set B = A - 31 (as perhaps suggested by the diagonal entries) and try whether you may get a sequence of q's converging to an eigenvalue of A that is smallesT (not largest) in absolute value. Use Xv = [I I nT. Do 8 step". Verify that A has the spectrum (0.3.51. 12. CAS EXPERIMENT. Power Method with Scaling. Shifting. (a) Write a program for 11 X 11 matrices that prints every step. Apply it to the (nonsymmetric!) matrix (20 steps), starting from [1 I If. 0.8J -0.6 0.8 [~ ~J .{; :J 5'[-~ -~ ~] I 2 9. Prove that if x is an eigenvector, then 8 = in (2). Give two examples. 10. (Rayleigh quotient) Why does q generally approximate the eigenvalue of greatest absolute value? When will q be a good approximation? A 3 = ° 4 4 -I ° 2 5 8 3 7. 6. 0 2 3 2 o 8 2 -2 ° ° ° ° 3 o 5 8. (Optimality of 8) In Prob. 2 choose Xo = [3 _l]T and show that q = 0 and 8 = I for all steps and that the eigenvalues are ::'::: 1. so that the interval [q - 8, q + 8] cannot be shortened in general! Experiment with other xo. 20.9 [ :: -19 ~: I:] . - 36 -7 (b) Experiment in (a) with shifting. Which shift do you find optimal? (c) Write a program as in (a) but for symmetric matrices that prints vectors. scaled vectors, q, and 8. Apply it to the matrix in Prob. 6. (d) Find a (nonsymmetric) matrix for which 8 in (2) is no longer an error bound. (e) Experiment systematically with speed of convergence by choosing matrices with the second greatest eigenvalue (i) almost equal to the greatest, (ii) somewhat different, (iii) much different. Tridiagonalization and QR-Factorization We consider the problem of computing all the eigenvalues of a real symmetric matrix A = [ajk]' discussing a method widely used in practice. In the first stage we reduce [he given matrix stepwise to a tridiagonal matrix, that is, a matrix having all its nonzero entries on the main diagonal and in the positions immediately adjacent to the main diagonal (such as A3 in Fig. 447, Third Step). This reduction was invented by A. S. Householder (1. Assn. Comput. Machinery 5 (1958), 335-342). See also Ref. [E29] in App. I. This Householder tridiagonalization will simplify the matrix without changing its eigenvalues. The latter will then be determined (approximately) by factoring the tridiagonalized matrix, as discussed later in this section. 876 CHAP. 20 Numeric Linear Algebra Householder's Tridiagonalization Method An 11 X 11 real symmetric matrix A = [ajk] being given, we reduce it by Il - 2 successive similarity transformations (see Sec. 20.6) involving matrices PI' ... , P n-2 to tridiagonal form. These matrices are orthogonal and symmetric. Thus Pi l = PIT = PI and similarly for the others. These transformations produce from the given Ao = A = [lIjk] the matrices Al = raJ};]. A2 = [aj~)], .... A n - 2 = [a3k- 2 )] in the form Al = PIAoPl A2 (1) = P 2A I P 2 The transformations (1) create the necessary zeros, in the first step in Row I and Column 1, in the second step in Row 2 and Column 2, etc., as Fig. 447 illustrates for a 5 X 5 matrix. B is tridiagonal. How do we determine PI' P 2 , . . . Pn - 2 ? Now, all these P r are of the form where I is the 0; thus (3) = I •... , 11 (r (2) 11 X 11 - 2) unit matrix and Vr = [Vjr] is a unit vector with its first r components o o o * o o * * V n -2 = * * * * where the asterisks denote the other components (which will be nonzero in general). Step 1. VI has the components Vn = 0 (a) (4) j = 3. 4 .... , (b) 11 where (c) Sl = \/a212 + a3I 2 + ... + a nI 2 where SI > 0, and sgn a21 = + 1 if a21 ~ 0 and sgn lI21 = -1 if a21 compute PI by (2) and then Al by (I). This wa<; the first step. < O. With this we SEC 20.9 Tridiagonalization and QR-Factorization 877 [.~~; .. *"'J ,. o.!_ [:~:::** _..... [::::* * :::*.:=;::J ..~.!'*: * *. 1 r: :;: - First Step Second Step Third Step At =PtAP t A" = P,At P 2 ~ =P3 A"P j Fig. 447. Householder's method for a 5 X 5 matrix. Positions left blank are zeros created by the method. Step 2. We compute v 2 by (4) with all subscripts increased by 1 and the aj~), the entries of Al just computed. Thus [see also (3)] ajk replaced by -I ( 1 +la~~1 - -) 2 S2 (4*) j = 4,5, ... , n where + a~~ 2 With this we compute P2 by (2) and then A2 by (1). Step 3. We compute V3 by (4*) with all subscripts increased by 1 and the ajil replaced by the entries aj'fl of A2 , and so on. E X AMP L E 1 Householder Tridiagonalization Tridiagonalize the real symmetric matrix 4 6 5 2 Solution. Step sgn G21 = 2 1. We compute Sl2 = 4 + ]2 + ]2 = l!l from (4c). Since (4) by straightforward computation Q21 = + 1 in (4bl and get tram 0 1 [00.119573 161. [ 0.985 598 56 V21 VI = V31 V41 = 0.119 573 16 From this and (2). o o -~.235 -0.942 809 04 -0.235 702 27 -0.235 702 27 0.97140452 -0.G28 595 48 -0.235 702 27 -0.028 595 48 0.97140452 70227J 4 > 0, we have 878 CHAP. 20 Numeric Linear Algebra From the first line in (I) we now get -Vt8 Step 2. From (4*) we compute S22 o 7 -I -( -I 9/2 312 -I 3/2 912 01 = 2 and [:~] V2 = = [:.923879 ,,]. 0.38268343 V42 From this and (2). P, ~ [~ 0 -l~ 0 0 0 0 -Itv'2 0 -1V2 ] Itv'2 The ",cond line in (I) now gives [, -Vt8 0 7 ,/2 Y2 6 0 0 -v'18 B = A2 = P2 A I P2 = 0 0 ~l This matrix B is tridiagonal. Since our given matrix has order 11 = 4, we needed 11 - 2 = 2 steps to accomplish this reduction. as claimed. (Do yOll see that we got more zeros than we can expect in general?) B is similar to A, as we now show in general. This is essential because B thus has the same spectrum as A, by Theorem 2 in Sec. 20.b. • We assert that Bin (1) is similar to A = Ao. The matrix PT is symmetric; B Similar to A. indeed, T TT T TT T Pr = (I - 2v,.vr) = 1 - 2(v rvr ) = [ - 2v vr = Pro T Also, PT is orthogonal because T P."Pr = Pr 2 = Vr is a unit vector. so that T2 (I - 2vrvr ) T V,T V T = 1 and thus T T = 1 - 4v rvr + 4v.rv r V."Vr T T T = 1 - 4v,.vr + 4vAvr vr)Vr = I. Hence p;l = pr T = Pr and from (1) we now obtain B = P11-2A11-3Pn-2 = ... . . . = Pn - 2 Pn - 3 -1 ••• -} = Pn - 2 P n - 3 P I AP1 ... -1 •.• PI API" Pn - 3 Pn - 2 . Pn - 3 Pn - 2 1 = P- AP where P = P IP 2 ... P n-2' This proves our assertion. • SEC. 20.9 879 Tridiagonalization and QR-Factorization QR-Factorization Method In 1958 H. Rutishauser of Switzerland proposed the idea of using the LU-factorization (Sec. 20.2; he called it LR-factorization) in solving eigenvalue problems. An improved version of Rutishauser's method (avoiding breakdown if certain submatrices become singular, etc.; see Ref. [E291) is the QR-method, independently proposed by the American I. G. F. Francis (Computer 1. 4 (1961-62), 265-271, 332-345) and the Russian V. N. Kublanovskaya (Zhllmal v.ych. Mal. i Mat. Fi::.. 1 (1961),555-570). The QR-method uses the factorization QR with orthogonal Q and upper triangular R. We discuss the QR-method for a real symmetric matrix. (For extensions to general matrices see Ref. rE29] in App. 1.) In this method we first transform a given real symmetric f1 X 11 matrix A into a tridiagonal matrix Bo = B by Householder's method. This creates many Leros and thus reduces the amount of further work. Then we compute B 1 , B 2 , . . . stepwise according to the following iteration method. Step 1. Factor Bo = QoRo with orthogonal Ro and upper triangular Bl = RoQo· Ro. Then compute Step 2. Factor Bl = QIRI' Then compute B2 = R 1 Ql' Ge1leral Step s + 1. (5) Here Qs is orthogonal and Rs upper triangular. The factorization (Sa) will be explained below. B 8 + 1 Similar to B. Convergence to a Diagonal Matrix. From (5a) we have Rs = Q;lBs' Substitution into (5b) gives (6) Thus Bs+l is similar to Bs' Hence Bs+l is similar to Bo = B for all s. By Theorem 2, Sec. 20.6, this implies that Bs+l has the same eigenvalues as B. Also, Bs+l is symmetric. This follows by induction. Indeed. Bo = B is symmetric. Assuming Bs to be symmetric, that is, BsT = Bs, and using Q;l = Qs T (since Qs is orthogonal), we get from (6) the symmetry, If the eigenvalues of B are different In absolute value, say, then lim s_oc Bs = IAll > IA21 > ... > IAnl, D where D is diagonal, with main diagonal entries listed in App. I.) "-]0 "-2, ... , An. (Proof in Ref. [E29J 880 CHAP. 20 Numeric Linear Algebra How to Get the QR-Factorization, say, B = Bo = [bjk ] = QoRo. The tridiagonal matrix B has 11 - I generally nonzero entries below the main diagonal. These are b2l • b32 , . . . • bn,n-l' We multiply B from the left by a matrix C 2 such that C 2 B = [b)~] has bW = O. We multiply this by a matrix C3 such that C3 C 2 B = [bj'r] has b~~ = O. etc. After 11 - I such multiplications we are left with an upper triangular matrix Ro. namely. (7) These 11 X f1 matrices Cj are very simple. Cj has the 2 X 2 submatrix (OJ suitable) in Rows j - 1 and j and Columns j - 1 and j; everywhere else on the main diagonal the matrix Cj has entries 1; and all its other entries are O. (This submatrix is the matrix of a plane rotation through the angle OJ; see Team Project 28. Sec. 7.2.) For instance. if fl = 4. writing Cj = cos OJ. sJ = sin OJ. we have C2 = ~ - 'C, 2 S2 0 C2 0 0 0 0 0 0 ~l4 ~ ~ 0 0 Cg S3 -.1'3 C3 0 0 o 0 o These Cj are orthogonal. Hence their product in (7) is orthogonal, and so is the inverse of this product. We call this inverse Qo. Then from (7), (8) where, with Ci 1 = C/o (9) This is our QR-factorization of Bo. From it we have by (5b) with s = 0 (10) We do not need Qo explicitly, but to get Bl from (10). we first compute R OC 2 T. then (ROC 2 T)C3 T, etc. Similarly in the further steps that produce B2 • B 3 , • • • • Determination of cos OJ and sin OJ. We finally show how to find the angles of rotation. cos f)2 and sin ~ in C 2 must be such that b~) = 0 in the product o o SEC 20.9 881 Tridiagonalization and QR-Factorization Now b~21 is obtained by multiplying the second row of C 2 by the first column of B, cos 82 = v'l (II) E X AMP L E 2 VI] 82 tan 82 sin 82 = Similarly for 83 , 84 , + tan 2 V] + + (b 21 /b n )2 b21lb n V] tan 2 82 + (b 21 /b n )2 The next example illustrates all this. .. '. QR-Factorization Method Compute all the eigenvalues of the matrix 4 6 5 Solution. We fust reduce A to tridiagonal foml. Applying Householder's method. we obtain (see Example I) [ A2 = -~!J8 -~T8 o ,\;'2 o v2 6 o 0 o From the characteristic determinant we see that A 2 • hence A, has the eigenvalue 3. (Can you see this directly from A2?1 Hence it suffices to apply the QR-method to the tridiagonal 3 X 3 matrix '. ~ ~ -~Is • -,\;18 7 [ v2 Step I. We multiply B from the lefl by C2 = [COO ~ sin f)2 -s~ cos f)2 f)2 ] 0 Here (-sin O2 )' 6 + (cos 02)(-~) With these values we compute and lhen C 2 B by = 0 gives 7,34846<) 23 C2 B = [ C, ~ 0 [: cos 1.154 700 54 o 1.414213 56 6.00000000 (cos 7.348469 23 = C g C2 B = [ f)3) • f)3 = -0.57735027. . 1.414 213 56 = 0 the values cos -7.50555350 f)3]' -0.816 496 58] 3.26598632 .!. cos f)3 (II) cos O2 = 0.81649658 and sin f)2 -7.50555350 Si: f)3 - ,in 0 In C 3 we get from (- sin f)3) • 3.265 <)86 32 and sin f)3 = 0.39735971. This gives Ro 0] f)3 -0.8\649658J 0 3.55902608 3.443 784 13 o o 5.04714615 . = 0.917 662 94 CHAP. 20 882 Numeric Linear Algebra From this we compute Bl = RoC2 T C3 T = 10.333 333 33 - 2.054 804 67 -:.05480467 4.03508772 :.005 532 51 ] 2.00553251 4.63157895 [ which is symmetnc and tridiagonal. The off-diagonal entries in BI are still large in absolute value. Hence we have to go on. Step 2. We do the same computations as in the first step. with Bo = B replaced by Bl and C z and C 3 changed accordingly, the new angles being 62 = 0.196291 533 and 63 = 0.513415589. We obtain -2.802322 -II [ 1053565375 Rl ~ -0.391 145 88] 0 4.08329584 3.98824028 0 0 3.06832668 1 and from this B2 = [ 10.8" 879 88 -0.79637918 -:.796379 18 5.44738664 ~50702500 1.50702500 2.672 733 48 We see that the off-diagonal entries are somewhat smaller in absolute value than those of B I . but ,till much too large for the diagonal entries to be good approximations of the eigenvalues of B. Further Steps. Ib~~)1 = Ib~i)1 in all We list the main diagonal entries and the absolutely largest off-diagonal entry, which is steps. You may show that the given matrix A has the spectrum I 1.6,3,2. (J) Stepj b\·i) b<.i) HJ) maXj*k Jbjkl 3 5 7 9 10.966892 9 10.9970872 10.999742 I 10.999977 2 5.94589856 6.00] 81541 6.00024439 6.00002267 2.08720851 2.00109738 2.00001355 2.00000017 0.58523582 0.12065334 0.035911 07 0.01068477 11 22 33 • Looking back at our discussion, we recognize that the purpose of applying Householder's tridiagonalization before the QR-factorization method is a substantial reduction of cost in each QR-factorization. in particular if A is large. Convergence acceleration and thus further reduction of cost can be achieved by a spectral shift, that is, by taking Bs - ksI instead of Bs with a suitable ks . Possible choices of ks are discussed in Ref. [E291, p. 510. ---. .•. "1-41 -...... HOUSEHOLDER TRIDIAGONALIZATION Tridiagonalize. showing the details: 1. [ 3.5 1.0 1.0 5.0 3.0 1.5 3.0 3.5 0.98 3. 0.04 1.5J ~[ 0.04 0.56 OAO 0.44 0.40 0.80 [ 8 2 2 8 8 2 2 2 2 6 4 2 2 4 6 4. 0 ] 15-91 0.44] 8 QR-FACTORIZATION Do three QR-steps to find approximations of the eigenvalues of: 5. The matrix in the answer to Prob 1 883 Chapter 20 Review Questions and Problems 6. The matrix in the answer to Prob. 3 9. 0,:] 0.2 4.1 ,, 1. What are the main problem algebra? 0.1 0.1 4.0 ro O.~] 1.0 0.1 10. CAS EXPERIMENT. QR-Method. Try to find out experimentally on what properties of a matrix the speed of decrease of off-diagonal entries in the QR-method depends. For this purpose write a program that first tridiagonalizes and then does QR-steps. Try the program out on the matrices in Probs. 1. 3. and 4. Summarize your findings in a short report. -U.I -4.3 7.0 area~ =1 : in numeric linear S T ION SAN D PRO B L EMS Xl + X2 + X3 S 2. What is pivoting? When and how would you apply it? Xl + 2X2 + 2x3 6 3. What happens if you apply Gauss elimination to a system that has no solutions? Xl + 2x2 t- 3X3 8 4. What is Doolittle's method? Its connection to Gauss elimination? 17. 18. 5X1 5. What is Cholesk)"s method? When would you apply it? 19. 10. Why are similarity transformations of matrices important in designing numeric methods? Give examples. 11. What is the power method for eigenvalues? What are its advantages and disadvantages? 13. State Schur's inequality and give some applications of it. GAUSS ELIMINATION 116-]91 Solve: 16. 4x2 5Xl + 6Xl - 3X2 + 7X2 + = 11.8 X3 = 34.2 3X3 2X3 = -3.1 -10 3X2 + + 21:1 X2 9X3 0 = 3X3 = IS X3 = -13 + SX3 = 26 20. Solve Prob. 17 by Doolittle's method. 21. Solve Prob. 17 by Cholesky's method. 122-24 1 INVERSE MATRIX Compute the inverse of: 22. 14. What is tridiagonalization? When would you apply it? 15. What is the idea of the QR-method? When would you apply the method? - SX 2 + ISx 3 3x 1 - 12. State Gerschgorin's theorem from memory. Can you remember its proof? 17 X2 4X2 - 8. What is least squares approximation? What are the normal equations? 9. What is an eigenvalue of a matrix? Why are eigenvalue problems important? Give typical examples. 3x3 2Xl - 6. What do you know about the convergence of the Gauss-Seidel method? 7. What is ill-conditioning? What is the condition number and its significance? + 23. r" 2.0 05] O.S 0.5 1.0 1.5 2.0 1.0 2.0 10] r' 2.0 3.S 1.0 1.S I.S 9.0 CHAP. 20 884 Numeric Linear Algebra 125-261 GAUSS-SEIDEL METHOD Do 3 steps without scaling, starting from [1 25. + 15x2 - Xl X3 + 3x2 lOx I 11 = -17 5 34. In Prob. 18 35. In Prob. 19 136-381 CONDITION NUMBER Compute the condition number (corresponding to the Coo-vector norm) of the coefficient matrix: 36. In Prob. 22 37. In Prob. 23 38. In Prob. 24 @9-40 1 FITTING BY LEAST SQUARES Fit: VECTOR NORMS 127-321 Compme the f\-, C2 -, and Ceo-norms of the vectors 27. [0 4 -8 3]T 28. [3 29. 30. 31. 32. 8 [-4 [0 -ll]T 1 0 0 2]T O]T [-5 -2 7 [0.3 1.4 0.2 0 O]T -0.6]T 133-351 MATRIX NORM Compute the matrix norm corresponding to the Ccc-vector norm for the coetlicient matrix: 33. In Prob. 17 39. A straight line to (- 2. O. \). (0. 1.9). (2. 3.8), (4, 6.\), (6,7.8) 40. A quadmtic pambola to O. 9). (2, 5), (3. 4), (4. 5). (5. 7) 141-431 EIGENVALUES Find three circular disks that must contain all the eigen values of the matrix: 41. In Prob. 22 42. In Prob. 23 43. Tn Prob. 24 44. (Power method) Do 4 sleps of the power method for the matrix in Prob. 24. starting from [I I I]T and computing the Rayleigh quotients and error bounds. 45. (Householder and QR) Tridiagonalize the matrix in Prob. 23. Then apply 3 QR steps. (Spectrum (6S): 9.65971, 4.07684. 0.263451) Numeric Linear Algebra Main tasks are the numeric solution of linear systems (Secs. 20.1-20.4), curve fitting (Sec. 20.5). and eigenvalue problems (Secs. 20.6-20.9). Linear systems Ax = b with A = [ajk]' written out (1) can be solved by a direct method (one in which the number of numeric operations can be specified in advance. e.g., Gauss's elimination) or by an indirect or iterative method (in which an initial approximation is improved stepwise). 885 Summary of Chapter 20 The Gauss elimination (Sec. 20.1) is direct, namely, a systematic elimination process that reduces (1) stepwise to triangular fonn. In Step I we eliminate Xl from equations E2 to En by subtracting (a2I/an) EI from E 2, then (a3I/an) EI from E 3. etc. Equation EI is called the pivot equation in this step and an the pivot. In Step 2 we take the new second equation as pivot equation and eliminate X2, etc. If the triangular fonn is reached, we get Xn from the last equation, then Xn-I from the second last. etc. Partial pivoting (= interchange of equations) is necessary if candidates for pivots are zero, and advisable if they are small in absolute value. Doolittle's, Crout's, and Cholesky's methods in Sec. 20.2 are variants of the Gauss elimination. They factor A = LV (L lower triangular, U upper triangular) and solve Ax = LUx = b by solving Ly = b for y and then Ux = y for x. In the Gauss-Seidel iteration (Sec. 20.3) we make an = a22 = ... = ann = I (by division) and write Ax = (I + L + U)x = b; thus x = b - (L + U)x, which suggests the iteration formula (2) XCrn + I ) =b - Lx(m+l) - UX Crn) in which we always take the most recent approximate x./s on the right. If Ilell < l. where C = -(I + L)-IU, then this process converges. Here. IICII denotes any matrix norm (Sec. 20.3). If the condition number K(A) = IIAII IIA -111 of A is large. then the system Ax = b is ill-conditioned (Sec. 20.4), and a small residual r = b - Ax does 1Iot imply that x is close to the exact solution. The fitting of a polynomial p(x) = b o + blx ., ... + bmx'" through given data (points in the J\"}·-plane) (Xl' YI), ... , (Xm Yn) by the method of least squares is discussed in Sec. 20.5 (and in statistics in Sec. 25.9). Eigenvalues A (values A for which Ax = Ax has a solution x =1= 0, called an eigenvector) can be characterized by inequalities (Sec. 20.7), e.g. in Gerschgorin's theorem, which gives 11 circular disks which contain the whole spectrum (all eigenvalues) of A, of centers ajj and radii Llajkl (sum over k from I to 11, k =1= j). Approximations of eigenvalues can be obtained by iteration. starting from an Xo =1= 0 and computing Xl = Axo, x 2 = Ax!> ... , xn = AXn-i. In this power method (Sec. 20.8) the Rayleigh quotient (3) q= gives an approximation of an eigenvalue (usually that of the greatest absolute value) and, if A is symmetric, an error bound is (4) Convergence may be slow but can be improved by a speCTral shift. For determining all the eigenvalues of a symmetric matrix A it is best to first rridiagonalize A and then to apply the QR-method (Sec. 20.9), which is based on a factorization A = QR with 0I1hogonai Q and upper triangular R and uses similarity transformations. CHAPTER 21 >~- Numerics for ODEs and PDEs Numeric methods for differential equation" are of great practical importance to the engineer and physicist because practical problems often lead to differential equations that cannot be solved by one of the methods in Chaps. 1-6 or 12 or by similar methods. Also, sometimes an ODE does have a solution fonnula (as the ODEs in Secs. 1.3-1.5 do), which, however, in some specific cases may become so complicated that one prefers to apply a numeric method instead. This chapter explains and applies basic methods for the numeric solution of ODEs (Secs. 21.1-21.3) and POEs (Secs. 21.4-21.7). Sections 21.1 alld 21.2 may be studied immediately after Chap. 1 alld Sec. 21.3 immediately after Chap. 2, because these sections are independent of Chaps. 19 and 20. Sections 21.4-21.7 Oil PDEs may be studied immediately after Chap. 12 if students have some knowledge of linear systems of algebraic equations. Prerequisite: Secs. 1.1-1.5 for ODEs, Secs. 12.1-12.3, 12.5, 12.10 for POEs. References and Answers to Problems App. 1 Part E (see also Parts A and C), App. 2. 21.1 Methods for First-Order ODEs From Chap. 1 we know that an ODE of the first order is of the form F(x, y. y') = 0 and can often be written in the explicit form y' = f(x. y). An initial value problem for this equation is of the fonn y' = f(x. yt (1) )'(xo) = Yo where .1.'0 and Yo are given and we assume that the problem has a unique solution on some open interval (/ < x < b containing .1.'0' In thi~ section we shall discu~s methods of computing approximate numeric values of the solution y(x) of (1) at the equidistant points on the x-axis Xl = .1.'0 + h. where the step size h is a fixed number, for instance, 0.2 or O. I or 0.01. whose choice we discuss later in this section. Those methods are step-by-step methods, using the same formula in each step. Such formulas are suggested by the Taylor series 2 (2) 886 y(x + h) = y(x) + hy' (x) + h ""2 y"(X) + .... SEC 21.1 887 Methods for First-Order ODEs For a small h the higher powers h 2 , h 3 , approximation y(x + h) .•. = y(x) = y(x) are very small. This suggests the crude + hy' (x) + hf(.\", y) (with the second line obtained from the given ODE) and the following iteration process. In the first step we compute + Yl = Y(.\"o + h). In the second step we compute which approximates )'(X2) = 1'(xo + 2h), etc., and in general which approximates Y(XI) = Yo hf(xo, Yo) (3) (n = 0, 1, .. '). This is called the Euler method or the Euler-Cauchy method. Geometrically it is an approximation of the curve of y(x) by a polygon whose first side is tangent to this curve at xo (see Fig. 448). y Fig. 448. Euler method This crude method is hardly ever u~ed in practice, but since it is simple, it nicely explains the principle of methods based on the Taylor series. Taylor's formula with remainder has the foml y(X + h) = y(x) + hy' (x) + ~h2y"W (where x ~ l; ~ x + h). It shows that in the Euler method the tru/lcation error il1 each step or local truncation error is propOitional to h 2 , written 0(11 2 ), where 0 suggests order (see also Sec. 20.1). Now over a fixed x-interval in which we want to solve an ODE the number of steps is proportional to 1111. Hence the total error or global error is propOitional 2 to h 0/h) = hI. For this reason. the Euler method is called a first-order method. In addition, there are roundoff errors in this and other methods, which may affect the accuracy of the values Yb .\'2, ..• more and more as 11 increases, as we shall see. 888 CHAP. 21 Table 21.1 Numerics for ODEs and PDEs Euler Method Applied to (4) in Example 1 and Error + xn )'n 0 I 0.0 0.000 0.000 0.2 0.4 0.6 0.8 1.0 0.000 0.040 0.128 0.274 0.489 0.040 2 3 4 5 E X AMP L E 1 0.2(xn 11 Exact Values )'n) 0.088 0.146 0.215 Error En 0.000 0.021 0.000 0.092 0.222 0.426 n.7U< 0.052 0.094 0.152 0.229 O.02J Euler Method Apply the Euler method to the following initial value problem. choosing II = 0.2 and computing )'1 • . . . y' = (4) Solution. Here f(x. y) = x + y; hence X + y, yeO) = o. f(x n , Yn) = xn Yn+1 = )"n , \'5: + Yn' and we see that (3) + 0.2lxn -j- become~ Yn)' Table 21.1 shows the computations. the values of the exact solution y(x) = eX - x-I obtained from (4) in Sec. 1.5. and the error. Tn practice the exact solution is unknown, but an indication of the accuracy of the values can be obtained by applying the Euler method once more with step 211 = 0.4. letting )'n* denote the approximation now obtained. and comparing corresponding approximations. This computation is: xn vn * 0.0 0.4 0.8 0.000 0.000 0.160 O.4(Xn + )'n) 0.000 0.160 Yn in Table 21.1 Difference Yn - )'n* 0.000 0.040 0.274 0.000 0.040 0.1l4 Let En and En" be the errors of the computations with II and 211. respectively. Since the error is of order 112. in a switch from II to 217 it is multiplied by 22 = 4, but since we need only half as many steps as before, it will be multiplied only by 412 = 2. Hence En" = 2En SO that the difference is En * - En = 2En - En = En' Now Y = )'n + En = Yn * + En * by the definition of error; hence En" - En = )'n - )'n* indicates En qualitatively. Tn our computations, Y2 - )'2* = 0.04 - 0 = 0.04 (actual error 0.052. see Table 21.1) and )"4 - )'4* = 0.274 - 0.160 = 0.114 (actually 0.152). • E X AMP L E 2 Euler Method for a Nonlinear ODE Figure 449 concerns the initial value problem (5) yeO) = OA and shows the curve of the solution y = 1/[2.5 - Sex)] + 0.01x 2 where Sex) is the Fresnel integral (38) in App. 3.1. It also shows 80 approximate values for 0 ~ x ~ 4 obtained by the Euler method from (3). Although Ii = 0.05 is smaller than Ii in Example 1, the accuracy is still not good. It is interesting that the error is not monotone increasing. obviously since the solution is not monOlone. We shall return to this ODE in the problem set. • SEC. 21.1 889 Methods for First-Order ODEs y 0.70 0.60 0.50 0.40 o Fig. 449. 2 3 4 x Solution curve and Euler approximation in Example 2 Automatic Variable Step Size Selection in Modern Numeric Software The idea of adaptive integration as motivated and explained in Sec. 19.5 applies equally well to the numeric solution of ODEs. It now concerns automatically changing the step size h depending on the variability of y' = f determined by (6*) Accordingly, modern software automatically selects variable step sizes h n so that the error of the solution will not exceed a given maximum size TOL (suggesting tolerance). Now for the Euler method. when the step size is h = h m the local error at Xn is about ~hn21y"(xn)l. We require that this be equal to a given tolerance TOL, (6) (a) 21"(Xn) 1= I 2hn )' 2TOL thus TOL, Iy''lxn) 1 y"(X) must not be zero on the interval J: Xo ~ x = xN on which the solution is wanted. Let K be the minimum of 1y" (X) 1on J and assume that K > O. Minimum 1,/'(x) 1corresponds to maximum h = H = Y2 TOLIK by (6). Thus. Y2 TOL = HVK. We can insert this into (6b). obtaining by straightforward algebra K (7) where For other methods, automatic step size selection is based on the same principle. Improved Euler Method By taking more terms in (2) into account we obtain numeric methods of higher order and precision. But there is a practical problem. If we substitute y' = f(x, y(x)) into (2), we have (2*) 890 CHAP. 21 Numerics for ODEs and PDEs Now y in .f depends on x, so that we have f' as shown in (6*) and .f", .fill even much more cumbersome. The general strategy now is to avoid the computation of these derivatives and to replace it by computing .f for one or several suitably chosen auxiliary values of (x. y). "Suitably" means that these values are chosen to make the order of the method as high as possible (to have high accuracy). Let us discuss two such methods that are of practical imp0l1ance. namely. the improved Euler method and the (classical) Runge-Kutta method. In the improved Euler method or improved Euler-Cauchy method (sometimes also called Heun method), in each step we compute first the auxiliary value (8a) and then the new value (8b) Thi~ method ha" a simple geometric interpretation. In fact. we may say that in the we approximate the solution y by the straight line through interval from x" to Xn + (xn , Yn) with slope f(x n , Yn), and then we continue along the straight line with slope f(X n +l, Y:;+I) until x reaches X,,+I' The improved Euler-Cauchy method is a predictor-corrector method, because in each step we first predict a value by (8a) and then correct it by (Sb). In algorithmic form. using the notations kl = hf(xn, Yn) in (Sa) and k2 = hf(.'n+b Y~+I) in (8b), we can write this method as shown in Table 21.2. !h Table 21.2 Improved Euler Method (Heun's Method) ALGORITHM EULER (f, xo. Yo, h. N) This algorithm compute<; the solution of the initial value problem y' = f(x. y) . .\"(xo) = Yo at equidistant points xl = Xo + 17, X2 = Xo + 211, ... , XN = Xo + Nfl; here f is such that this problem has a unique solution on the mterval [xo. xNl (see Sec. 1.7). INPUT: Initial values xo, Yo, step size II, number of steps N OUTPUT: Approximation Yn+l to the solution Y(Xn +l) at where 11 = 0, . . . , N - 1 For 11 = o. I. .... N - 1 do: Xn+l = Xn + 11 kl = hf(xn, y,,) k2 = hJ(Xn+b Y,,+l = y" Yn + k1 ) + 2(k1 + k2 ) OUTPUT Xn+l, ."n+l End Stop End EULER xn+l = Xo + (n + l)h. SEC 21.1 891 Methods for First-Order ODEs E X AMP L E 3 Improved Euler Method Apply the improved Euler method to the initial vallie problem (4). choosmg h Solution. = 0.2. as before. For the present problem we have in Table 21.2 k2 .1',,+1 = Yn + = 0.2 ""2 0.2(x" + 0.2 + "n + 0.2(xn + .1',,» (2.2xn + 2.2Yn + 0.2) = Yn + 0.22(xn + .1',,) + 0.02. Table 2 1.3 show~ thai ollr present results are more accurate than tho~e in Example I: see also Table 21.6. • Table 21.3 Improved Euler Method Applied to (4) and Error + .1'n) + 0.02 U.22(xn n xn •vn 0 0.0 0.2 0.4 0.6 0.8 1.0 0.0000 0.0200 0.0884 0.2158 0.4153 0.7027 2 3 4 5 Exact Values (4D) Error 0.0000 0.0214 0.0918 0.2221 0.4255 0.7183 0.0000 0.0014 0.0034 0.0063 0.0102 0.0156 0.0200 0.0684 0.1274 0.1995 0.2874 Error of the Improved Euler Method. The local error is of order h 3 and the globed error of order h 2 , so that the method is a second-order method. PROOF Setting In = Ierm .1'(xn » and using (r), we have (9a) Y(Xn + h) - .r(xn ) = - hI,. 1 2-' 1 3-" + :z./I In + 6 h In + .... Approximating the expression in the brackets in (8b) by Taylor expansion, we obtain from (8b) 1 [)'n+l - )'n ="2h In _ (where' 1 [- -"2 h In (9b) = dldx." + ] In+l + (fn - + -, 1 1" + 1"+1 2 -If hIn +"2h In and again using the + ... )] etc.). Subtraction of (9b) from (9a) gives the local error h 3 _If h 3 _If 11 3 -If -6 .f n - -4 I n + ... -- - -12 .f" + .... Since the number of steps over a fixed x-interval is proportional to I11z, the global error 3 2 • is of order h /11 = h , so that the method is of second order. 892 CHAP. 21 Numerics for ODEs and PDEs Runge-Kutta Methods (RK Methods) A method of great practical importance and much greater accuracy than that of the improved Euler method is the classical Runge-Kutta method of fourth order, which we call briefly the Runge-Kutta method. 1 It is shown in Table 21.4. We see that in each step we first compute four auxiliary quantities k 1, k2 , k 3 , k4 and then the new value Yn+1' The method is well suited to the computer because it needs no special starting procedure, makes light demand on storage, and repeatedly uses the same straightforward computational procedure. It is numerically stable. Note that if f depends only on x, this method reduces to Simpson's rule of integration (Sec. 19.5). Note further that kJ, ... , k4 depend on 11 and generally change from step to step. Table 21.4 Classical Runge-Kutta Method of Fourth Order ALGORITHM RUNGE-KUTTA (f, x o, )'0' h, N). This algorithm computes the solution of the initial value problem y' at equidistant points Xl = Xo + h. X2 = Xo + 2h . . . . , XN = Xo = f(x, y), y(Xo) = Yo + Nil: here f is such that this problem has a unique solution on the interval [xo, XN] (see Sec. 1.7). INPUT: Function f, initial values xo, Yo, step size h, number of steps N OUTPUT: Approximation Yn+1 to the solution 'y(Xn +1) at X,,+1 where n = 0, I, ... , N - 1 For n = = Xo + (n + l)h. 0, 1, ... , N - 1 do: k1 = hf(xn, y,,) + ~h, Yn + ~kl) hf(xn + ~h, Yn + ~k2) hf(xn + h, Yn + k3) = Xn + h = Yn + ~(k1 + 2k2 + 2k3 + k2 = hf(xn k3 = k4 = Xn+1 Yn+1 k 4) OUTPUT Xn +], )'1/+1 End Stop End RUNGE-KUTTA INamed after the German mathematicians KARL RUNGE (Sec. 19.4) and WILHELM KUTTA (1867-1944). Runge [Math. Annalen 46 (1895),167-178], KARL HEUN [Zeitschr. Math. Phys. 45 (1900), 23-38], and Kutta [Zeitschr. Math. Phys. 46 1901l. 435-453] developed various such methods. Theoretically, there are infinitely many fourth-order methods using four function values per step. The method in Table 21.4 is most popular [rom a practical viewpoint because of its "symmetrical" form and its simple coefficients. It was given by Kutta. SEC 21.1 893 Methods for First-Order ODEs E X AMP L E 4 Classical Runge-Kutta Method Apply the Runge-Kutta method to the initial "alue problem (4) in Example 1, choosing II computing five steps. Solutioll. For the present problem we have f(x. v} x = + y. 0.2, as before, and = Hence kl = 0.2(xn + y,,), k2 = 0.2(x" + 0.1 + y" + O.Sk l ), k3 = 0.2(x" + 0.1 + Yn + 0.Sk2 }. k4 = 0.2(x" + 0.2 + y" + k3 ). Table 21.5 shows the results and their errors, which are smaller by factors 103 and 104 than those for the two Euler methods. See also Table 21.6. We mention in passing that since the present k 1• . . . . k4 ate simple, operations were saved by substituting kl into k2. then k2 into k3 . etc.: the resulting formula is shown in Column • 4 of Table 21.5. Table 21.5 Runge-Kutta Method Applied to (4) n x" v ." 0 0.0 0.2 0.4 0.6 0.8 1.0 0 0.021400 0.091 818 0.222107 0.425521 0.718251 2 3 4 5 0.2214(x" + Yn) + 0.0214 bxact Values (6D) y = eX - x-I 106 X Error 0.021400 0.070418 0.130289 0.203 414 0.292730 0.000000 0.021403 0.091825 0.222 119 0.425541 0.718282 0 3 7 12 :W 31 ofy" Table 21.6 Comparison of the Accuracy of the Three Methods Under Consideration in the Case of the Initial Value Problem (4), with h = 0.2 Error ! x r=ex-x-I Euler (Table 21.1) Improved Euler (Table 21.3) Runge-Kutta (Table 21.5) 0.2 0.4 0.6 0.8 1.0 0.021403 0.091825 0.222119 0.425541 0.718282 0.021 0.052 0.094 0.152 0.229 0.U014 0.0034 0.0063 0.0102 0.0156 0.000003 0.000007 0.000011 0.000020 0.000031 I Error and Step Size Control. RKF (Runge-Kutta-Fehlberg) The idea of adaptive integration (Sec. 19.5) has analog<; for Runge-Kutta (and other) methods. In Table 21.4 for RK (Runge-Kutta), if we compute in each step approximations y and .17 with step sizes hand 2h. respectively, the latter has error per step equal to 5 2 = 32 times that of the former; however, since we have only half as many steps for 2h, the actual factor is 2 5 /2 = 16. so that, say, and thus 894 CHAP. 21 Numerics for ODEs and PDEs Hence the error E = d- h ) for step size h is abour (10) where y problem y = lh) - y(2hl, as said before. Table 21.7 illustrates (10) for the initial value y' = (1L) (y - x-l)2 + 2, yeo) = I. the step size h = 0.1 and 0 ~ x ~ 0.4. We see that the estimate is close to the actual error. This method of error estimation is simple but may be unstable. Table 21.7 Runge-Kutta Method Applied to the Initial Value Problem (11) and Error Estimate (10). Exact Solution y = tan x + x + 1 \: 0.0 0.1 0.2 0.3 0.4 v y (Step size /7) (Step si7e 211) Error Estimate (10) Actual Error Exact Solution (9D) 1.000 000 000 1.200 334 589 1.402 709 878 1.609 336 039 1.822 792 993 l.000 000 000 0.000000000 1.402 707 408 0.000 000 165 1.822 788 993 0.000000 267 0.000000000 0.000 000 083 0.000 000 157 0.000 000 210 0.000 000 226 1.000000000 1.200334672 1.402 710 036 1.609 336 250 l.822 793 219 RKF. E. Fehlberg [Computing 6 (1970), 61-71] proposed and developed error control by using two RK methods of different orders to go from (xn , Yn) to (xn +1> Yn+l)' The difference of the computed y-values at Xn+l gives an error estimate to be used for step size controL Fehlberg discovered two RK formulas that together need only 6 function evaluations per step. We present these formulas here because RKF has become quite popular. For instance. Maple uses it (also for systems of ODEs). Fehlberg's fifth-order RK method is (12a) with coefficient vector y = [Yl ... Y6]. (12b) y= U;5 o 28561 56430 6656 12825 i5]' His fourth-order RK method is (13a) with coefficient vector (13b) Y * -_ [25 216 o 1408 2565 2197 4104 _1] 5' SEC 21.1 895 Methods for First-Order ODEs In both formulas we use only 6 different function evaluations altogether. namely. (14) k1 = hf(x." k2 = hf(x., + k3 = Izf(x n + ~h. y., + k4 = hf(xn + ~~h, Yn + ~~~~kl k5 = hf(x., + h. J'n + i~~k1 k6 = hf(x., + ~h. .\" n - J'n) !Iz, Yn + !k1) ~k1 + i2k2) ~~g~k2 + ~~~~k3) 8k2 + 3:~~k3 - 481t~k4) 2k2 - ~~~k3 + ~~g:k4 - ~~k5)' + 287 k1 The difference of (12) and (13) gives the error estimate (15) E X AMP L E 5 En+l -- .\'n+1 - * --...Lk Yn+1 360 1 128 k 4275 3 - 2197 k 75240 4 +..!.k +..2..k 50 5 55 6' Runge-Kutta-Fehlberg For the initial value problem (II) we obtain from (12)-(14) with 11= 0.1 in the first step the 0.200062 500000 k1 = 0.200000 000000 k2 k3 = 0.200140756867 k4 = 0.2008.'i6926154 k5 = 0.201006676700 k6 = 0.200250418651 .\'1* = = 12S-value~ 1.20033466949 .\'1 = 1.20033467253 and the error e~timate E1 =."1 - .\'~ = 0.000000 00304. The exact 12S-value is .1'(0.1) = 1.20033467209. Hence the actual error of."1 i~ -4.4· 10- 10, smaller than that in Table 21.7 by a factor 200. • Table 21.8 summarizes essential features of the methods in this section. It can be shown that these methods are Ilumerically stahle (definition in Sec. 19.1). They are one-step methods because in each step we use the data of just one preceding step. in contrast to multistep methods where in each step we use data from several preceding steps. as we shall see in the next section. Table 21.8 Methods Considered and Their Order (= Their Global Error) Method Euler Improved Euler RK (fourth order) RKF Function Evaluation per Step 2 4 6 Global Error Local Error 0(11) 0(17 2 ) 2 0(h ) 0(114) 0(h 5 ) 0(h 3 ) 0(11 5 ) 0(h 6 ) 896 CHAP. 21 Numerics for ODEs and PDEs Backward Euler Method. Stiff ODEs The backward Euler formula for numerically solving (I) is (n = 0, 1, .. '). (16) This formula is obtained by evaluating the right side at the new location (Xn +1, )'n+1); this is called the backward Euler scheme. For known )'n it gives )'n+l implicitly, so it defines an implicit method, in contrast to the Euler method (3), which gives Yn+l explicitly. Hence (16) must be solved for )'n+l' How difficult this is depends on f in (1). For a linear ODE this provides no problem, as Example 6 (below) illustrates. The method is particularly useful for "stiff' ODEs. as they occur quite frequently in the study of vibrations, electric circuits. chemical reactions. etc. The situation of stiffness is roughly as follows; for details, see, for example, [E5]. [E25], [E26] in App. I. Error terms of the methods considered so far involve a higher derivative. And we ask what happens if we let h illcrease. Now if the error (the derivative) grows fast but the desired solution also grows fast. nothing will happen. However. if that solution does not grow fast, then with growing h the error term can take over to an extent that the numeric result becomes completely nonsensical. as in Fig. 450. Such an ODE for which h must thus be restricted to small values, and the physical system the ODE models. are called stiff. This term is suggested by a mass-:-.pring system with a stiff ~pring (spring with a large k; see Sec. 2.4). Example 6 illustrates that implicit methods remove the difficulty of increasing II in the case of stiffness: it can be shown that in the application of an implicit method the solution remains stable under any increase of h, although the accuracy decreases with increasing h. E X AMP L E 6 Backward Euler Method. Stiff ODE The initial value prublem y' = f(x. y) = -20.,' + 20x 2 + 2x. y(O) = I has the solution (verify!) y = e- 20x + x 2. The backward Euler formula (16) is ),,,+1 = Noting that xn+1 y" + hJ(xn+b Yn+1) = Yn = X" + h. tak:ing the term Yn (l6*) + h( -20Yn+1 + 20X~+1 + 2Xn +1)' -20hY,,+1 to the left. and dividing. we obtain + h[20(xn Yn+l = + h}2 + 2(x" + h)j 1 + 20h The numeric re,ulr- in Table 21.9 show the following. Stability of the backward Euler method for h = 0.05 and also for h = 0.2 with an error increase by about a factOT 4 for h = 0.2. Stability of the Euler method fOf h = 0.05 but instability for h = 0.1 (Fig. 450). Stability of RK for h = 0.1 but instability for h = 0.2. 1l1b illustrates that the ODE is stiff. Note thm even in the case of stdbility the appruximation of the ~olution near x = 0 is poor. • Stiffness will be considered further in Sec. 21.3 in connection with systems of ODEs. SEC. 21.1 897 Methods for First-Order ODEs y • 2.0 I I I I I I 1.0 , , , I I I I ,I o " lO.2~ :0.4', I.. 0.6 I , \ I \: -1.0 'I 1, ~: 0.8 4' LO x ' " • Fig. 450. Euler method with h = 0.1 for the stiff ODE in Example 6 and exact solution Table 21.9 Backward Euler Method (BEM) for Example 6. Comparison with Euler and RK BEM x h 0.05 = 1.00000 0.26188 0.10484 0.10809 0.16640 0.25347 0.36274 0.49256 0.64252 0.81250 1.00250 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 BEM h = 0.2 Euler h = 0.05 Euler h = 0.1 1.00000 1.00000 0.00750 0.03750 0.08750 0.15750 0.24750 0.35750 0.48750 0.63750 0.80750 0.99750 1.00000 -1.00000 1.04000 -0.92000 1.16000 -0.76000 1.36000 -0.52000 1.64000 -0.20000 2.00000 0.24800 0.20960 0.37792 0.65158 1.01032 RK h = RK 0.1 1.00000 0.34500 0.15333 0.12944 0.17482 0.25660 0.36387 0.49296 0.64265 0.81255 1.00252 h = 0.2 1.000 5.093 25.48 127.0 634.0 316H Exact 1.00000 0.14534 0.05832 0.09248 0.16034 0.25004 0.36001 0.49001 0.64000 0.81000 1.00000 ===== .... -........ ,,_ . . . . ..-..-. = - - ......- ............... -- .... ---... .. 11-41 EULER METHOD 7. Do 10 steps. Solve the problem exactly. Compute the error. (Show the details.) 1. y' 2. )"' = y, yeO) = = y, yeO) 3. y' = (y 4. y' = (y + IS-lO I = .d. X)2. 1. h = 0.1 1. h = 0.01 yCO) = 0, h yCO) = o. h 1. h 6. (Logistic population»)." = = 1, h = O. I 8. y' + )' tan x = sin 2x, yeO) = I, h = 0.1 9. Do Prob. 7 using the Euler method with h = 0.1 and = 0.1 = 0.1 111-171 0.1. Compare with Prob I y - )'2, yeO) = 0.2, h = 0.1 h 2 ,y(0)=0,h=0.1 CLASSICAL RUNGE-KUTTA METHOD OF FOURTH ORDER Do 10 steps. Compare as indicated. Comment. (Show the details. ) 11. y' - xy2 = O• .1'(0) = 1, h = 0.1. Compare with Prob. 7. Apply (10) to )'10' 12. y' = yeo) compare the accuracy. Do 10 steps. Solve exactly. Compute the error. (Show the details.) = xy2 = O. 10./ = I IMPROVED EULER METHOD S. y' = y. -,,(0) and comment. y' - = y - y2 • .1'(0) = 0.2, h = 0.1. Compare with Prob. 6. Apply (10) to )'10' 898 13. 14. 15. CHAP. 21 Numerics for ODEs and PDEs (b) Graph solution curves of the ODE in (5) for various positive and negative initial values. (c) Do a similar experimem as in (a) for an initial value problem that has a monotone increasing or monotone decreasing solution. Compare the behavior of the error with that in (a). Comment. y' = (l + X-I»),. yO) = e. h = 0.2 y' = !U'lx - xl),). ),(2) = 2. h = 0.2 y' + -" tan x = sin 2x, yeo) = 1. II = 0.1 16. In Prob. 15 use h = 0.2 (5 steps) and compare the error. 17. y' + 5x 4 y2 = 0, yeO) = 1. h = 0.2 20. CAS EXPERIMENT. RKF. (a) W]ite a program for RKF that gives .tn' y". the estimate (10). and if the 18. Kulla's third-order method is defined by Yn+l = y" + irk] + 4k2 + k3*) with kl and k2 as in RK (Table 21.4) and k3* = 1If(xn +].),,, - k] + 2k2 ). Apply this method to (4) in Example I. Choose h = 0.2 and do 5 steps. Compare with Table 21.6. 19. CAS EXPERIMENT. Euler-Cauchy vs. RK. (a) Solve (5) in Example 2 by Euler. Improved Euler. and RK for 0 ~ x ~ 5 with step h = 0.2. Compare the errors for x = 1,3,5 and comment. solution is known. the actual error En' (b) Apply the program to Example 5 in the text (10 steps, 11 = 0.1). (c) E" in (b) gives a relatively good idea of the size of the actual error. Is this typical or accidental? Find out by experimentation with other problems on what properties of the ODE or solution this might depend. 21.2 Multistep Methods In a one-step method we compute )'n+1 using only a single step, namely, the previous value y.,. One-step methods (Ire "self-starting," they need no help to get going because they obtain ."1 from the initial value ."0' etc. All methods in Sec. 21.1 are one-step. In contrast, a multistep method uses in each step values from two or more previous steps. These methods are motivated by the expectation that the additional information will increase accuracy and stability. But to get staJ1ed, one needs values. say. YO,."b ."2'."3 in a 4-step method, obtained by Runge-Kutta or another accurate method. Thus, multistep methods are not self-starting. Such methods are obtained as follows. Adams-Bashforth Methods We consider an initial value problem (1) y' = f(x, y), as before. with f such that the problem has a unique solution on some open interval containing .1'0' We integrate y' = f(x, y) from x" to xn+1 = Xn + h. This gives Now comes the main idea. We replace f(x. y(x» by an interpolation polynomial p(x) (see Sec. 19.3), so that we can later integrate. This gives approximations )'n+1 of y(xn +1) and Yn of y(xn ), x n+ 1 (2) )'n+1 = )'n + J x" p(x) dr. SEC. 21.2 899 Multistep Methods Different choices of p(x) will now produce different methods. We explain the principle by taking a cubic polynomial, namely, the polynomial P3(x) that at (equidistant) has the respective values In = I(xn , Yn) (3) This will lead to a practically useful formula. We can obtainp3(x) from Newton's backward difference formula (18), Sec. 19.3: where x - Xn h r= We integrate p3(X) over x from Xn to Xn+l x = Xn The integral oqdr (4) I x". Xn + + hr. = Xn + 11. thus over r from 0 to we have 1) is 5/12 and that ofir(r 1 = 11 dr. + l)(r + 2) is 3/8. We thus obtain (1~VIn+ 5 P3dX=lzlp3dr=h In+ 0 dx 1. Since ~) - V2In+ ':"V- 3fl1 . 2 12 8 It is practical to replace these differences by their expressions in terms of I: \In V2In = In - In-l = In - 2In-l + 1",-2 We substitute this into (4) and collect terms. This gives the multistep formula of the Adams-Bashforth method of fourth order (5) )'n+l = Yn + h 24 (55In - 59In-l + 37In-2 - 9In-3)' It expresses the new value Yn+ I [approximation of the solution y of (I) at Xn+ 1] in terms of 4 values of.f computed from the y-values ohtained in the preceding 4 steps. The local truncation error is of order 11 5 , as can be shown, so that the global error is of order 114; hence (5) does define a fourth-order method. 900 CHAP. 21 Numerics for ODEs and PDEs Adams-Moulton Methods Adams-Moulton methods are obtained if for p(x) in (2) we choose a polynomial that interpolates f(x, .v(x» at x n +l' x n , Xn_1> ... (as opposed to Xn , X"'-I' ... used before; this is the main point). We explain the principle for the cubic polynomial P3(X) that interpolates at X n +l' X'" Xn -l' X.,-2' (Before we had x n ' Xn -l, X,,-2' X.,-3') Again using (18) in Sec. 19.3 but now setting r = (x - x n +1)lh, we have We now integrate over x from Xn to Xn+l as before. This corresponds to integrating over r from -I to O. We obtain Replacing the differences as before gives (6) Yn-ll = Yn + f lz Xn+l P3(X) dx = Yn + 24 (9f"+1 + 19fn - 5f,,-1 + fn-2)' Xu This is usually called an Adams-Moulton formula. It is an implicit formula because fn+l = f(X,,+b )'n+l) appears on the right, so that it defines Yn+l only implicitly, in contrast to (5), which is an explicit formula, not involving Yn+l on the right. To use (6) we must predict a value y ~;+ I> for instance, by using (5), that is, (7a) Y~+1 = Yn + h 24 (55fn - 59fn-l + 37fn-2 - 9fn-3)' The corrected new value Yn+ 1 is then obtained from (6) with f n+ 1 replaced by f~+1 = J(x n +1, Y~+1) and the other fs as in (6); thus, (7b) h Yn+1 = Yn + 24 (9f~+1 + 19fn - 5fn-l + f n-2)' This predictor-corrector method (7a). (7b) is usually called the Adams-Moulton method offourth order. It has the advantage mer RK that (7) gives the error estimate as can be shown. This is the analog of (10) in Sec. 21.1. Sometimes the name 'Adams-Moulton method' is reserved for the method with several corrections per step by (7b) until a specific accuracy is reached. Popular codes exist for both versions of the method. Getting Started. In (5) we need fo, fl, f2, f3' Hence from (3) we see that we must first compute Yr. -"2, .1'3 by some other method of comparable accuracy, for instance, by RK or by RKF. For other choices see Ref. [E26] listed in App. I. SEC. 21.2 901 Multistep Methods E X AMP L E 1 Adams-Bashforth Prediction (7a), Adams-Moulton Correction (7b) Solve the initial value problem y' (8) = T + y, by (7a), (7b) on the interval 0 ::;; x ::;; 2. choosing h = y(O) = 0 0.2. Solutio". The problem is the same as in Examples 1-3. Sec. 21.1. so that we can compare the results. We compute starting values Yl . .'"2• .'"3 by the classical Runge-Kutta method. Then in each step we predict by (7a) and make one correction by (7b) before we execute the next step. The result~ are shown and compared with the exact values in Table 21.10. We see that the corrections improve the accuracy considerably. This is typicaL • Table 21.10 Adams-Moulton Method Applied to the Initial Value Problem (8); Predicted Values Computed by (7a) and Corrected values by (7b) Il Xn 0 1 2 3 4 5 6 7 8 9 10 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 106 • Error Yn Exact Values 0.425529 0.718270 1.120106 1.655191 2.353026 3.249646 4.389062 0.000 000 0.021403 0.091825 0.222119 0.425541 0.718282 1.120117 1.655200 2.353032 3.249647 4.389056 0 3 7 12 12 12 11 9 6 1 -6 Starting Predicted Corrected Yn Yn* 0.000000 0.021 400 0.091 818 0.222 107 0.425361 0.718066 1.119855 1.654885 2.352653 3.249 190 4.388505 ofYn Comments on Comparison of Methods. An Adams-Moulton formula is generally much more accurate than an Adams-Bashforth formula of the same order. This justifies the greater complication and expense in using the former. The method (7a), (7b) is Ilumerically stable, whereas the exclusive use of (7a) might cause instability. Step size control is relatively simple. If ICorrector - Predictorl > TOL, use interpolation to generate "old" results at half the current "tep size and then try /1/2 as the new step. Wherea~ the Adams-Moulton fOIillula (7a), (7b) needs only 2 evaluations per step, Runge-Kutta needs 4; however, with Runge-Kutta one may be able to take a step size more than twice as large, so that a comparison of this kind (widespread in the literature) is meaningless. For more details, see Refs. [E25], LE26] listed in App. 1. ........ -.. -.. ,,_ ...... ..-...... -.... ......-............. --... 1. Carry out and show the details of the calculations leading to (4 H7) in the text. 12-111 ADAMS-MOULTON METHOD (7a), (7b) Solve the initial value problems by Adams-Moulton. 10 steps with I correction ~r step. Solve exactly and compute the enOL (Use RK where no starting values are given.) 2.),' = J, yeO) = I, II = 0.1 (1.105171, 1.221403, 1.349859) = -0.2xy. yeO) = I, h = 0.2 4. y' = 2xy, yeo) = I. h = 0.1 5. y' = I + y2, )"(0) = O. h = 0.1 3. y' 6. Do Prob. 4 by RK, 5 steps, h = 0.2. Compare the errors. CHAP. 21 902 Numerics for ODEs and PDEs solution and comment. 7. Do Prob. 5 by RK. 5 steps. h = 0.2. Compare the errors. 8. 9. 14. How much can you reduce the error in Prob. 13 by halving II (20 steps. h = 0.05)? First guess, then compute. y' = xly. y(l) = 3, h = 0.2 y' = (x + Y - 4)2, y(O) = 4, h = 0.2. only 7 steps (why?) 15. CAS PROJECT. Adams-Moulton. (a) Accurate starting is important in (7a), (7b). Illustrate this in Example I of the text by using starting values from the improved Euler-Cauchy method and compare the results with those in Table 21.9. (b) How much does the error in Prob. 11 decrease if you use exact starting values (instead of RK-values)? (c) Experiment to find out for what ODEs poor starting is very damaging and for what ODEs it is not. (d) The classical RK method often gives the same accuracy with step 211 as Adams-Moulton with step h. so that the total number of function evaluations is the same in both cases. Illustrate this with Prob. 8. (Hence corresponding comparisons in the literature in favor of Adams-Moulton are not valid. See also Probs. 6 and 7.) y' = I - 4.\'2, y(O) = O. II = 0.1 11. y' = x + y . .1'(0) = O. h = 0.1 (0.00517083. 10. 0.0214026.0.04(8585) 12. Show that by applying the method in the text to a polynomial of second degree we obtain the predictor and corrector formulas II Y~+1 = Yn + 12 Yn+l = Yn + 12 (23fn - h (5fn+l 16f,,_1 + + 5f,,-2) 8fn - fn-l)' 13. Use Prob. 12 to solve y' = 2x.\". y(O) = I (10 steps, It = 0.1, RK starting values). Compare with the exact 21.3 Methods for Systems and Higher Order ODEs Initial value problems for first-order (1) y' = system~ f(x. of ODEs are of the form y). in components f is assumed to be such that the problem has a unique solution y(x) on some open x-interval containing Xo. Our discussion will be independent of Chap. 4 on system~. Before explaining solution methods it is important to note that (l) includes initial value problem:-- for single mth-order ODEs, (2) y'>n) = f(x, y, y', and initial conditions )'(xo) = Kb y' (xo) .v"..... y(Tn-D) = K 2 , ••• , y<m-l)(xo) = Km as special cases. SEC. 21.3 Methods for Systems and Higher Order ODEs 903 Indeed, the connection is achieved by setting (3) )' = _ 1: ", I _\. 1 = _\' , _3 )'2 =)' Then we obtain the system , )'1 =)'2 (4) , Y'111-1 ~ Y·uz. Y~t = I(x. Yl ....• Ym) Euler Method for Systems Methods for single first-order ODEs can be extended to systems (1) simply by writing vector functions Y and f instead of scalar functions y and I, whereas x remains a scalar variable. We begin with the Euler method. Just as for a single ODE this method will not be accurate enough for practical purposes, but it nicely illustrates the extension principle. E X AMP L E 1 Euler Method for a Second-Order ODE. Mass-Spring System Solve the initial value problem for a damped mass-spring system y" + 2/ + 0.75.1' = 0, 1'(0) = 3. /(0) = -2.5 by the Euler method for systems with step h = 0.2 for x from 0 to I (where x is time). Solutioll. The Euler method (3). Sec. 21.1. generalizes to systems in the form (5) in Yn+l = Yn + hf(xn' Yn), componelll~ and similarly for systems of more than two equations. By (4) the given ODE converts to the system y~ = y~ = .f2(X• ."1' .1'2) = -2Y2 - 0.75.\"1' h(x. Yl' .1'2) = Y2 Hence (5) becomes Yl.n+l = )'1,11. + O.2Y2.n )'2.n+l = )'2,n + 0.2(-2)'2,11. - 0.75Yl,n)' The initial conditions are y(O) = )'1(0) = 3, -'" (0) = Y2(O) = -2.5. The calculation~ are shown in Table 21.11 on the next page. As for single ODEs. the results would not be accurate enough for practical purposes. The example merely serves to illustrate the method because the problem can be readily solved exactly. thu~ • 904 CHAP. 21 Table 21.11 Numerics for ODEs and PDEs Euler Method for Systems in Example 1 (Mass-Spring System) ."1 Exact .1" 1.71 0 2 3 4 0.0 0.2 0.4 0.6 0.8 5 1.0 (5D) 3.00000 2.50000 2.11000 1.80100 1.55230 1.34905 buor .\'2 Exact )"2.11 = )"1 - )"I.n EI 3.00000 2.55049 2.18627 1.88821 1.64183 1.43619 -2.50000 - 1.95000 -1.54500 -1.24350 -1.01625 -0.84260 0.00000 0.05049 0.76270 0.08721 0.08953 0.08714 Euor (5D) E2 -2.50000 -2.01606 -1.64195 -1.35067 -1.12211 -0.94123 = )"2 - )"2." 0.00000 -0.06606 -0.09695 -0.10717 -0.10586 -0.09863 Runge-Kutta Methods for Systems As for Euler methods, we obtain RK methods for an initial value problem (1) simply by writing vector formulas for vectors with III components, which for 111 = 1 reduce to the previous scalar formulas. Thus for the classical RK method of fourth order in Table 21.4 we obtain (6a) (Initial values) and for each step 11 = 0, 1, ... , N - 1 we obtain the 4 auxiliary quantities ki = II f(x n , k2 = hf(xn + !h, Yn + !kI ) k3 = hf(xn + !h, Yn + !k2) Yn) (6b) and the new value [approximation of the solution y(x) at x n + 1 = Xo + (n + I)h1 (6c) E X AMP L E 2 RK Method for Systems. Airy's Equation. Airy Function Ai(x) Solve the initial value problem y" = X-,"~ .1"(0) ~ 1/(32/3. rO/3J) = 0.35502805. /(0) = -1I(3113·r(lf3)) = -0.25881940 by the Runge-Kutta method for systems with h = 0.2: do 5 steps. This is Airy's equation,2 which arose in optics (see Ref. [Al3]. p. 188. listed in App. I). r is the gamma function (see App. A3.11. The initial conditions are such that we obtain a standard solution, the Airy function Ai(x), a special function that has been thoroughly investigated: for numeric values. see Ref. [GRI], pp. 446. 475. 2Named after Sir GEORGE BIDELL AIRY (1801-1892), English mathematician. who is known for his work in elasticity and in PDEs. SEC. 21.3 905 Methods for Systems and Higher Order ODEs Solution. For y" = xy. setting YI = Y,)"2 = yi = v' we obtain the system (4) , )"1 =)"2 Hence f = [h f2]T in (1) has the components hex. y) = .\'2' f2(X. Y) = XYI' We now write (6) in components. The initial conditions (6a) are YI.O = 0.35502805. )"2.0 = -0.25881 940. In 16b) we have fewer subscripts by simply writing kl = a, k2 = b. k3 = C, ~ = d. so that a = [01 02]T. etc. Then (6b) takes the form a=h Y2,n [ ] X·n Yl.1l (6b*) For example. the second component of b Now in b (= k 2 ) the first argument is i~ obtained as follows. fIx. y) has the second component f2(x, y) = x = x" + lh. Y = Yn + !a. X\'I' The second argument in b is and the first component of this is Together. Similarly for the other components in (6b*). Finally. (6c*) Yn+! = Yn + !(a + 2b + 2c + d). Table 21.12 shows the values y(x) = YI(x) of the Airy function Ai(x) and of its derivative as of the (rather small!) error of y(x). y' (x) = )'2(x) a~ well • Table 21.12 RK Method for Systems: Values Yl,n(X n) of the Airy Function Ai(x) in Example 2 II Xn Yl,n(X n ) Yl(X n ) Exact (8D) 108 • Error of .V I Y2,n(Xn ) 0 0.0 0.2 0.4 0.6 0.8 1.0 0.35502805 0.30370303 0.25474211 0.20979973 0.16984596 0.13529207 0.35502805 0.30370315 0.25474235 0.20980006 0.16984632 0.13529242 0 12 24 33 36 35 -0.25881 940 -0.25240464 -0.23583 073 -0.21279 185 -0.18641 171 -0.15914687 2 3 4 5 906 CHAP. 21 Numerics for ODEs and PDEs Runge-Kutta-Nystrom Methods (RKN Methods) RKN methods are direct extensions of RK methods (Runge-Kutta methods) to secondorder ODEs y" = f(x, y. y'), as given by the Finnish mathematician E. J. Nystrom [Acta Soc. Sci. fenn., 1925, L, No. 131. The best known of these uses the following formulas, where 11 = 0, I, ... , N - I (N the number of steps): kl ~h.f(xn' -".", y~) = k2 = ~hf(xn (7a) + ~h. )'n + K. y~ + k3 = ~Izf(xn + ~h . ."n + k4 = ~hf(xn + h, ."n K. k1) where K = ~h (y~ where L = h (y~ + ~kl) Y:L + k 2) + L, .":1 + 2k3) From this we compute the approximation )'n+l of Y(Xn+lJ at Xn+l = + Xo k3)' + (n + l)h, (7b) and the approximation y~+ 1 of the derivative y' Ltn+ 1) needed in the next step. (7c) = RKN for ODEs y" f(x, y) Not Containing y'. Then k2 the method particularly advantageous and reduces (7) to (7*) = kl = ~Izf(xn' Yn) k2 = ~hf(xn + ~h. Yn + ~h(y~ + ~kl» = k4 = Yn+l ~hf(xn + h, Yn + h(Y:1 = Yn + h(y~ + !(k1 + + k3 in (7), which makes k3 k 2) 2k2 » Y~+1 = Y:1 + -!(k1 + 4k2 + k 4 )· E X AMP L E 3 RKN Method. Airy's Equation. Airy Function Ai(x) For the problem k2 = k3 In = Example 2 and" O.l(xn + 0.))(.\"" + = 0.2 as betore we obtain from (Y) silllply O.IY;l + 0.05k1 !. k4 = O.I(xn + "1 0.2)(y" = O.I.~nYn and + O.2y~ + 0.2k2)· Table 21.13 ,how, the result,. The accuracy is the 'ame as in Example 2. but the work wa' much less. • Table 21.13 Runge-Kutta-Nystrom Method Applied to Airy's Equation, Computation of the Airy Function y = Ai(x) , Xn Yn \' ,n 0.0 0.2 0.4 0.6 0.8 0.35502805 0.303 703 04 0.254742 II 0.20979974 0.16984599 0.135292 18 -0.258 81940 -0.252404 64 -0.235 830 70 -0.21279172 -0.18641134 -0.15914609 LO y(x) Exact (8D) 0.35502805 0.303 703 15 0.25474235 0.209800 06 0.16984632 0.13529242 lOB. Error of -"n 0 II 24 32 33 24 I SEC. 21.3 907 Methods for Systems and Higher Order ODEs Our work in Examples 2 and 3 also illustrates that usefulness of methods for ODEs in the computation of values of "higher transcendental functions." Backward Euler Method for Systems. Stiff Systems The backward Euler formula (16) in Sec. 21.1 generalizes to systems in the form (8) + Yn+l = Yn h f(Xn+l' (11 Yn+l) = O. I ... '). This is again an implicit method. giving Yn+l implicitly for given Yn- Hence (8) must be solved for Yn+l' For a linear system this is shown in the next example. This example also illustrates that. similar to the case of a single ODE in Sec. 21.1, the method is very useful for stiff systems. These are systems of ODEs whose matrix has eigenvalues A of very different magnitudes, having the effect that, just as in Sec. 21.1, the step in direct methods, RK for example, cannot be increased beyond a certain threshold without losing stability. (A = -I and -10 in Example 4, but larger differences do occur in applications.) E X AMP L E 4 Backward Euler Method for Systems of ODEs. Stiff Systems Compare the backward Euler method (8) with the Euler and the RK methods for numerically solving the initial value problem y" + 11/ + lOy = lOx + II. )"'(0) = "(0) = 2. -10 converted to a system of first-order ODEs, Solution. The given problem "an easily be '<JIved, obtaining y = e- x + e- lOx + x so that we can compute errors, Conversion to a system by setting Y = Yl.)"' = Y2 [see (4)] gives , Yl )"1(0) = =)"2 y~ = -IOYI - 11.'"2 + lOx + II 2 )"2(0) = -10, The coefficient matrix -A has the characteristic determinant whose value is A2 + IIA + lO = (A The backward Euler formula is Yn+l = .'"I,n+lJ [ = ."2.n+l + I)(A I-10 + 10), Hence the eigenvalues are -I and -lO as claimed above. _"I'''J [ "2,n+l J [ Y2.11 + Ii -10)"I,n+1 - IlY2,n+l + lOXn+l + II - Reordering terms gives the linear system in the unknowns )"I,n+l and )"2.n 11 11)'2,n+l )"1.,,+1 - IOh)'1.,,+1 + (1 + 1 Ih)Y2_11+ 1 = )"I,n = )'2;n + 1011!-,,, + h) -t 1111. The coefficient determinant is D = I + IIIz + IOh 2 , and Cramer's rule (in Sec 7_6) gives the solution 1I1i)Yl.n + hY2,n + 10112xn + 11112 -lOhY1,,, + Y2,n + lOlnn + 1111 0 + Yn+l = D [ + 3 1011 ] . + 1011 2 CHAP. 21 908 Numerics for ODEs and PDEs Table 21.14 Backward Euler Method (BEM) for Example 4. Comparison with Euler and RK BEM h = 0.2 x h BEM = 0.4 0.0 0.2 2.00000 1.36667 2.00000 0.4 0.6 0.8 1.0 1.2 1.4 1.20556 1.21574 1.29460 1.40599 1.53627 1.67954 L.83272 1.99386 2.16152 1.31429 l" 1.8 2.0 Euler = 0.1 Euler h = 0.2 RK RK h = 0.2 Iz = 0.3 2.00000 1.01000 1.56100 1.13144 1.23047 1.34868 1.48243 1.62877 1.78530 1.95009 2.12158 2.00000 0.00000 2.04000 0.11200 2.20960 0.32768 2.46214 0.60972 2.76777 0.93422 3.10737 2.00000 1.35207 1.18144 1.18585 1.26168 1.37200 1.50257 1.64706 1.80205 1.96535 2.13536 2.00000 h 1.35020 1.57243 1.86191 2.18625 Exact 2.00000 L.l5407 L.08864 1.15129 1.24966 1.36792 1.50120 1.64660 1.80190 1.96530 2.13534 3.03947 5.07561} 8.72329 Table 21.14 shows the following. Stability of the backward Euler method for h accuracy for increasing Ii. ~ Stability of the Euler method for Ii Stability of RK for h = ~ 0.2 and 0.4 (and in fact for any Ii; try Ii = 5.0) with decreasing 0.1 but instability for h ~ 0.2. 0.2 but instability for II = 0.3. Figure 451 show. the Euler method for Ii ~ 0.18, an interesting case with initial jumping (for about x < 3) but later monotone following the solution curve of y = Y1' See also CAS Experiment 21. • y .. 4.0 -..- 3.0 ~ .- 2.0 1.0 0 Fig. 451. -===== -.•. -.. -. -.... • _ - ~ 4. Y~ Y2, Y1(0) = x 0.18 in Example 4 = I, )'2(0) = Yb Y; = -J'2, .VI(O) = 2, Y2(0) = 2, h = 0.1, 10 steps 5. y" 2. y~ = -3)"1 + )"2' y~ = Y1 - 3)"2' )"1(0) = 2, )"2(0) = 0, h = 0.1,5 steps = 4 3 5 steps EULER FOR SYSTEMS AND SECOND-ORDER ODES Solve by the Euler method: Y; 2 ........ L~ )"1, • Euler method with h 1. Verify the calculations in Example I. 3. y~ = I \I \I . . .-..-- .. _ - . · . . . . . . . A -~ __ ..... _ " "" "I = -1, h = 0.2, + 4y = 0, yeO) = 1, y'(O) = 0, h 0.2, 1, y' (0) 0.1, 5 steps 6. y" - )' 5 steps x, nO) -2, h SEC. 21.4 909 Methods for Elliptic PDEs 7. y~ = -.\"1 + Y2' Y; = -.\"1 - Y2, ."1(0) = .\"2(0) = 4, h = 0.1, 10 steps o. 17. ,," - i, ~ -141 RK FOR SYSTEMS Solve by the classical RK: 9. The system in Prob. 7. How much smaller is the error? 10. The ODE in Prob. 6. By what factor did the error decrease? 11. Undamped Pendulum. Y" + siny = 0, yt77) = O. y' (77) = 1. h = 0.2, 5 steps. How doe~ your result fit into Fig. 92 in Sec. 4.5? 12. Bessel Function 10 , xy" + y' + xy = 0, y(l) = 0.765198,/ (I) = -0.-1-40051, h = 0.5,5 steps. (This gives the standard solution 10(x) in Fig. 107 in Sec. 5.5.) 13• .r ~ = -4Y1 + )'2' y~ = .\'1 - 4)'2' YI(O) = 0, Y2(0) = 2, h = 0.1, 5 steps 14. The system in Prob. 2. Ho'W much smaller is the error? 15. Verify the calculations for the Airy equation in Example 3. fi.-I91 RUNGE-KUTTA-NYSTROM METHOD = - 3, ,,' (0) 6x 2 -+ 3.) (x 2 I - = ! 0, In 2. 20. CAS EXPERIMENT. Comparison of Methods. (a) Write program~ for RKN and RK for systems. (b) Try them out for second-order ODEs of your choice to find out empirically which is better in specific cases. (c) In using RKN, would it pay to first eliminate y' (see Prob. 29 in Problem Set 5.5)? Find out experimentally. 21. CAS EXPERIMENT. Backward Euler Stiffness. Extend Example 4 as follows. and (a) Verify the values in Table 21.14 and show them graphically as in Fig. 451. (b) Compute and graph Euler values for h near the "critical" h = 0.18 to determine more exactly when instability starts. (c) Compute and graph RK values for values of h between 0.2 and 0.3 to find h for which the RK approximation begins to increase away from the exact solution. (d) Compute and graph backward Euler values for large h: confirm stability and investigate the error increase for growing h. Do by RKN: 16. Prob. 12 (Bessel function Jo). Compare the results. 21.4 n" + 4,' = 0, ,'(0) 0.2: 5 steps '(Exact: y '= x4 - x)y" - xy' + y = 0, y(!) = y'(!) = I - In 2, h = 0.1. 4 steps 19. Prob. II. Compare the results. 18. 8. Verify the formulas and calculations for the Airy equation in Example 2. = Methods for Elliptic PDEs The remaining sections of this chapter are devoted to numerics for PDEs (partial differential equations), particularly for the Laplace, Poisson, heat, and wave equations. These POEs are basic in applications and, at the same time, are model cases of elliptic, parabolic, and hyperbolic PDEs, respectively. The definitions are as follows. (recall also Sec. 12.4). A POE is called quasilinear if it is linear in the highest derivatives. Hence a secondorder quasilinear PDE in two independent variables x, y is of the form (I) 1I is an unknown function of x and y (a solution sought). F is a given function of the indicated variables. Depending on the discriminant ac - b 2 , the PDE (I) is said to be of elliptic type if ac - b 2 > 0 (example: Lap/ace e£juation) parabolic type if ae - b 2 = 0 (example: beat equation) hyperbolic type if lie - b 2 < 0 (example: wal'e equation). 910 CHAP. 21 Numerics for ODEs and PDEs Here. in the heat and wave equations, y is time t. The coefficients G, b, c may be functions of x, y. so that the type of (I) may be different in different regions of the xy-plane. This classification is not merely a formal matter but is of great practical importance because the general behavior of solutions differs from type to type and so do the additional conditions (boundary and initial conditions) that must be taken into account. Applications involving elliptic equatio1ls usually lead to boundary value problems in a region R. called a first boundary value problem or Dirichlet problem if u is prescribed on the boundary curve C of R, a second bOllndary I'llllle problem or Neumann problem if lln = au/all (normal derivative of lI) is prescribed on C. and a third or mixed problem if II is prescribed on a part of C and Un on the remaining part. C usually is a closed curve (or sometimes consists of two or more such curves). Difference Equations for the Laplace and Poisson Equations In this section we consider the Laplace equation (2) and the Poisson equation (3) These are the most important elliptic PDEs in applications. To obtain methods of numeric solution. we replace the pmtial derivatives by conesponding difference quotients, as follows. By the Taylor formula, (4) (b) We subtract (4b) from (4a), neglect terms in h3 , 114, ..• , and solve for (5a) ux(x, y) = 1 21z [u(x + ", y) - llx. Then h. y)]. lI(x - Similarly, u(x. y + k) = u(x. y) + /...uy(x, y) + ~k2Uyy(X, y) + .. - and u(x, y - k) = lI(X. y) - kuy{.r. y) + ~k2Uyy(x, y) + - ... By subtracting, neglecting tenns in k 3 , k4, ... , and solving for u y we obtain ~ (5b) I lIy(x, y) = 2k [u(x, Y + k) - u(x, Y - k)]. SEC. 21.4 911 Methods for Elliptic PDEs We now turn to second derivatives. Adding (4a) and (4b) and neglecting terms in h 5 , . • . , we obtain u(x + h, y) + u(x - h, y) = 2u(x, y) + h 2 ux .",(x, y). Solving for u xx ' we have h4, (6a) I l/xx(x, Y) = fl uyy(x. y) = k2 + [u(x h, y) - 2l/(x, y) + + k) - + u(x. y It(x - /z, y)]. Similarly, (6b) 1 [u(x. y 2u(x. y) - k)]. We shall not need (see Prob. 1) 1 + h. \' '4hk' + k) UXy(x. v) = - - [u(x (6c) . - u(x - h. v . - u(x + + k) h, y - k) + u(x - h, Y - k)]. Figure 452a shows the points (x + h, y), (x - h, y), ... in (5) and (6). We now substitute (6a) and (6b) into the Poisson equation (3), choosing k a simple formula: (7) u(x + h, y) + It(x, Y + h) + u(x - h, y) + u(x, Y - = = h) - 4lt(x, y) h to obtain h 2 f(x, y). This is a difference equation corresponding to (3). Hence for the Laplace equation (2) the corresponding difference equation is (8) u(x + h, y) + u(x, y + h) + u(x - h, y) + u(x, y - h) - 4u(x, y) = O. h is called the mesh size. Equation (8) relates u at (x. y) to u at the four neighboring points shown in Fig. 452b. It has a remarkable interpretation: u at (x, y) equals the mean of the values of u at the four neighboring points. This is an analog of the mean value property of harmonic functions (Sec. 18.6). Those neighbors are often called E (East), N (North), W (West), S (South). Then Fig. 452b becomes Fig. 452c and (7) is (7*) u(E) + u(N) + N X x X kl hi h (x-h,y) x~x (x+h,y) (x,y) k u(S) - 4u(x, y) = h 2 f(.'(, y). (x,y+hJ (x,y+k) h + u(W) I h (x-h,y) x-- r" . hi h --x (x,y) (x+h,y) h x (x,y-k) (aJ Points in (5) and (6) Fig. 452. W X __ h_ -6 _h_X (x,Y) h X x (x,y-h) s (b) POints In (7) and (8) (e) Notation in (7*) Points and notation in (5)-(8) and (7*) E 912 CHAP. 21 Numerics for ODEs and PDEs nn Our approximation of h 2 ',pu in (7) and is a 5-point approximation with the coefficient scheme or stencil (also called pattern. lIloleeule. or star) (9) r -4 I}' We may now write (7) a, {I -4 I} u ~ h'!,<, y). Dirichlet Problem In numerics for the Dirichlet problem in a region R we choose an h and introduce a sljuare grid of horizontal and vertical straight lines of distance h. Their intersections are called mesh points (or lattice poillts or nodes). See Fig. 453. Then we approximate the given PDE by a difference equation [(8) for the Laplace equation], which relates the unknown values of u at the mesh points in R to each other and to the given boundary values (details on p. 913). This gives a linear system of algebraic equations. By solving it we get approximations of the unknown values of u at the mesh points in R. We shall see that the number of equations equals the number of unknowns. Now come" an important point. If the number of internal mesh points. call it p, is small, say, p < 100, then a direct solution method may be applied to that linear system of p < 100 equations in p unknowns. However, if p is large, a storage problem will arise. Now since each unknown u is related to only 4 of its neighbors, the coefficient matrix of the system is a sparse matrix, that is. a matrix with relatively few nonzero entries (for instance, 500 of 10000 when p = 100). Hence for large p we may avoid storage difficulties by using an iteration method. notably the Gauss-Seidel method (Sec. 20.3), which in PDEs is also called Liebmann's method. Remember that in this method we have the storage convenience that we can overwrite any solution component (value of ll) as soon as a "new" value is available. Both cases, large p and small p, are of interest to the engineer, large p if a fine grid is used to achieve high accuracy, and small p if the boundary values are known only rather inaccurately, so that a coarse grid will do it because in this case it would be meaningless to try for great accuracy in the interior of the region R. We illustrate this approach with an example. keeping the number of equations small, for simplicity. As convenient llOllitiOIlS for mesh poillts lllld correspondillg Vlt/ues of the solution (and of approximate solutions) we use (see also Fig. 453) Pij (10) = (ih. jh), llij = uUh, jll). y x Fig. 453. Region in the xy-plane covered by a grid of mesh h, also showing mesh points Pll = (h, h), ... , Pij = (ih, jh), ... SEC. 21.4 Methods for Elliptic PDEs 913 With this notation we can write (8) for any mesh point (11) EXAMPLE 1 Ui+l,j + Ui,j+l + Ui-l,j + Pi) in the form 4uij Ui,j-l - O. = Laplace Equation. Liebmann's Method The four sides of a square plate of side 12 em made of homogeneous material are kept at constant temperature ooe and IDOoe as shown in Fig. 454a. Using a (very wide) grid of mesh.j. cm and applying Liebmann's method (that is, Gauss-Seidel iteration), find the (steady-state) temperature at the mesh points. Solution. In the case of independence of time. the heat equation (see Sec. 10.8) reduces to the Laplace equation. Hence our problem i, a Dirichlet problem for the latter. We ehoo,e the grid shown in Fig. 454b and consider the mesh points in the order Pn, P 21 . P 12 • P 22 . We use (11) and, in each equation. take 10 the right all the terms resulting from the given boundary values. Then we obtain the system = + u22 = -200 - 4"12 + U22 = -100 (12) Un U21 -200 + U12 - 4U22 = -100. In practice, one would solve such a small system by the Gauss elimination, finding "n = "21 = 87.5. "12 = U22 = 62.5. More exact values (exact to 3S) of the solution of the actual problem [as opposed to its model (12)] are 88.1 and 61.9. respectively. (These were obtained by using Fourier series.) Hence the error is about 1%. which is surprisingly accurate for a grid of such a large mesh size h. If the system of equations were large, one would solve it by an indirect method, such as Liebmann's method. For (12) this is as follows. We write (12) in the form (divide by -4 and take terms to the right) [/n = 0.25"21 + 1112 = 0.25"n 0.25u21 + + 50 0.25u12 + 0.25u22 + 50 + 0.25u22 + 25 + 25. 0.25u12 These equations are now used for the Gauss-Seidel iteration. They are identical with (2) in Sec. 20.3, where un = Xl' U21 = X2, "12 = X3, U22 = X4. and the iteration is explained there, with 100, 100, 100, 100 chosen as starting values. Some work can be saved by better starting values, usually by taking the average of the bound:.u)' values that enter into the linear system. The exact solution of the system is Ull = "21 = 87.5. u12 = U22 = 62.5, as you may verify. u=o y u=o 121------'"'1 P 22 P12 *-4 u = 100 --, R POl Ipll u = 100 I p21 -e - . IPlO 'P I 20 U (a) Given problem = 100 (b) Grid and mesh points Fig. 454. Example 1 914 CHAP. 21 Numerics for ODEs and PDEs Remark. It is interesting to note that if we choose mesh h = LlII (L = side of R) and consider the (II - 1)2 internal me~h points (i.e .. mesh points not on the boundary) row by row in the order then the system of equations has the (11 - X (II - 1)2 1)2 coefficient matrix -4 B -4 B (13) Here A = B= -4 B -4 B is an (II - I) X (11 - 1) matrix. (In (12) we have 11 = 3. (11 - 1)2 = 4 internal mesh points. two submatrices B. and two submatrices I.) The matrix A is nonsingular. This follows by noting that the off-diagonal entries in each row of A have the sum 3 (or 2). wherea~ each diagonal entry of A equals -4. so that non~ingularity is implied by Gerschgorin's theorem in Sec. 20.7 because no Gerschgorin disk can include O. • A matrix is called a band matrix if it has all its nonzero entries on the main diagonal and on sloping lines parallel to it (separated by sloping lines of zeros or not). For example. A in (13) is a band matrix. Although the Gauss elimination does not pre~erve zeros between bands. it does not introduce nonzero entries outside the limits defined by the original bands. Hence a band structure is advantageous. In (13) it has been achieved by carefully ordering the mesh points. ADI Method A matrix is called a tridiagonal matrix if it has all its nonzero entries on the main diagonal and on the two sloping parallels immediately above or below the diagonal. (See also Sec. 20.9.) In this case the Gauss elimination is particularly simple. This raises the question of whether in the solution of the Dirichlet problem for the Laplace or Poisson equations one could obtain a system of equations whose coefficient matrix is tridiagonal. The answer is yes, and a popular method of that kind, called the ADI method (alternating direction implicit method) was developed by Peaceman and Rachford. The idea is as follows. The stencil in (9) shows that we could obtain a tridiagonal matrix if there were only the three points in a row (or only the three points in a column). This suggests that we write (II) in the form (l4a) Ui-l,j - 4Uij + ui+I,j = -Ui,j-l - lIi,j+I so that the left side belongs to y-Row j only and the right side to x-Column i. Of course, we can also write (11) in the form (l4b) lIi.j-1 - 4Uij + Ui.j+l = -Ui-I,j - ui+l,j so that the left side belongs to Column i and the right side to Row j. In the AD! method we proceed by iteration. At every mesh point we choose an arbitrary starting value u~T. In each step we compute new values at all mesh points. In one step we use an iteration SEC 21.4 91S Methods for Elliptic PDEs formula resulting from (14a) and in the next step an iteration formula resulting from (14b), and so on in alternating order. In detail: suppose approximations lI~j) have been computed. Then, to obtain the next approximations lI~'JHl), we substitute the 1I~'J') on the right side of (l4a) and solve for the l/~j'H1) on the left side; that is, we use (lSa) We use (lSa) for a fixed j, that is, for a fixed row j, and for all internal mesh points in this row. This gives a linear system of N algebraic equations (N = number of internal mesh points per row) in N unknowns, the new approximations of l/ at these mesh points. Note that (ISa) involves not only approximations computed in the previous step but also given boundary values. We solve the system (15a) (j fixed!) by Gauss elimination. Then we go to the next row, obtain another system of N equations and solve it by Gauss. and so on, until all rows are done. In the next step we alternate direction, that is, we compute the next approximations u~jn+2) column by column from the ugn + ll and the given boundary values, using a fonnula obtained from (l4b) by substituting the u~jn+]) on the right: (lSb) For each fixed i, that is. for each colulIlll. this is a system of M equations (M = number of internal mesh points per column) in M unknowns. which we solve by Gauss elimination. Then we go to the next column. and so on, until all columns are Jone. Let us consider an example that merely serves to explain the entire method. E X AMP L E 2 Dirichlet Problem. ADI Method Explain the procedure and formulas of the AD! method in terms of the problem in Example I. using the same grid and starting values 100. 100. 100. 100. Solution. While working. we keep an eye on Fig. 454b on p. 913 and the given boundary values. We obtain 11\]1.. Iti~. lI\]d from (15a) with 111 = O. We w,ite boundary values contained in (15a) first approximations without an upper index. for better identification and to indicate that these given values remain the same during the iteration. From (15a) with 111 = a we have for j = I (first row) the system "ttl.. (i~1) (i The solution is ilill = = 2) u~li ~ 100. Fori = 2 (second row) we obtain fium (15a) the system (i = I) (i = 2) The solution is /li~ ~ /I\]d ~ 66.667. Secolld approximatiolls 1tW,. iI~1. Iti~. /I~i are now obtained from (l5b) with 111 = I by using the first approximations just computed and the boundary values. For i ~ I (first column) we obtain from (15b) the system (J = 1) = -1101 - II'll (J = 2) The solution is /lfl. = 91.11. ,,cli = (J = I) (J 2) ~ 64.44. For i = 2 (second column) we obtain from (15b) the system The solutIOn is /I~l = 91.11, /I~i = 64,44. = -1t'iV. - It:ll Numerics for ODEs and PDEs CHAP. 21 916 In this example. which merely serves to explain the practical procedure in the AD! method. the accuracy of the second approximations is about the same as that of two Gauss-Seidel steps in Sec. 20.3 (where IIU = xl' lt2l = .\"2' lt12 = X3' 1122 = X4)' as the following table shows. -Method lin lI21 AD!. 2nd approximations Gauss-Seidel, 2nd approximations Exact solution of (12) 91.11 93.7S 87.S0 91.11 90.62 87.50 - 64.44 6S.62 62.50 64.44 64.06 62.50 • Improving Convergence. Additional improvement of the convergence of the ADI method results from the following interesting idea. Introducing a parameter p, we can also write (11) in the form (a) Ui-l,j - (2 + P)Uij + ui+l,j = -Ui,j-l + (2 - (b) lli,j-l - (2 + p)uij + ui.j+l = -Ui-l,j + (2 - P) lI ij P)Uij - Ui,j+l (16) - lIi+l,j' This gives the more general ADI iteration formulas (a) (17) (b) u\m+2) _ 1,,)-1 (') .... + p)1l\1n+2) + 1,) 1l\,,!-+2) 1.,)+1 = _u\m+D z-I,) + ("") L. p)u(m+D 'lJ u(m+~) 1,+1,) . For P = 2. this is (IS). The parameter p may be used for improving convergence. Indeed, one can show that the ADI method converges for positive p. and that the optimum value for maximum rate of convergence is Po (18) = 'iT 2 sin K where K is the larger of M + I and N + 1 (see above). Even better results can be achieved by letting P vary from step to step. More details of the ADI method and variants are discussed in Ref. rE2S] listed in App. 1. ---.•.--.......-........ .. -.-~ -- .. .......... ,,-~--.-. 1. Derive (Sb). (6b). and (6c). 12-71 GAUSS ELIMINATION, GAUSS-SEIDEL ITERATION For the grid in Fig. 455 compute the potential at the four internal points by Gauss and by 5 Gauss-Seidel steps with starting value~ 100. 100, 100, 100 (showing the details of your work) if the boundary values on the edges are: 2. 3 II = 0 on the left. x on the lower edge, 27 - 9.\'2 on the righl. x 3 - 27x on the upper edge. ue. lIe I, 0) = 60. 0) = 300. II = lOO on the other three edges. 4. 1/ = X4 on the lower edge. 81 - 54.\'2 + )"4 on the right. x 4 - 54x 2 + 81 on the upper edge, y4 on the left. Verity the exact solution X4 - 6x 2y2 + y4 and determine Ihe elTor. 3. 5. II 6. U = no on the upper and lower edges, lID on the left and right. = sin !7TX on the upper edge, 0 on the other edges. 10 steps. SEC 21.5 Neumann and Mixed Problems. Irregular Boundary 7. Vo on the upper and lower edges, - Vo on the left and right. Sketch the equipotentIal lines. 917 use symmetry; take II = 0 as the boundary value at the two points at which the potential has a jump. u = 110 V "~110V~"~11", 2 y x Fig. 455. 2 "~-110Vt.fj "~-110V 3 Problems 2-7 u=-110V 8. Verify the calculations in Example 1. Find out experimentally how many steps are needed to obtain the solution of the linear system with an accuracy of 3S. 9. tUse of symmetry) Conclude from the boundary values in Example I that U21 = Un and U22 = U12' Show that this leads to a system of two equations and solve it. 10. (3 X 3 grid) Solve Example I, choosing h = 3 and starting values 100, 100, .... 11. For the square 0 ~ x ~ 4, 0 ~ y ~ 4 let the boundary temperatures be O°C on the horizontal and 50°C on the vertical edges. Find the temperatures at the interior points of a square grid with h = I. 12. Using the answer to Prob. II, try to sketch some isotherms. 13. Find the isotherms for the square and grid in Prob. II if U = sin ~'iTx on the horizontal and -sin ~'iTY on the vertical edges. Try to sketch some isotherms. 14. (Intluence of starting values) Do Prob. 5 by Gauss-Seidel, starting from O. Compare and comment. 15. Find the potential in Fig. 456 using (a) the coarse grid, (b) the fine grid, and Gauss elimination. Hint. In (b), 21.5 Fig. 456. Region and grids in Problem 15 16. (ADI) Apply the ADI method to the Dirichlet problem in Prob. 5, using the grid in Fig. 455, as before and starting values zero. 17. What Po in (18) should we choose for Prob. 16? Apply the ADI formulas (17) with Po = l. 7 to Prob. 16, performing I step. lllustrate the improved convergence by comparing with the corresponding values 0.077, 0.308 after the first step in Prob. 16. (Use the starting values zero.) 18. CAS PROJECT. Laplace Equation. la) Write a program for Gauss-Seidel with 16 equations in 16 unknowns, composing the matrix (13) from the indicated 4 X 4 submatrices and including a transfOimation ofthe vector ofthe boundary values into the vector b of Ax = b. (b) Apply the progranl to the square grid in 0 ~ x ~ 5. o ~ y ~ 5 with h = I and u = 220 on the upper and lower edges, U = 110 on the left edge and u = -10 on the right edge. Solve the linear system also by Gauss elimination. What accuracy is reached in the 20th Gauss-Seidel step? Neumann and Mixed Problems. Irregular Boundary We continue our discussion of boundary value problems for elliptic PDEs in a region R in the xy-plane. The Dirichlet problem was studied in the last section. In solving Neumann and mixed problems (defined in the last section) we are confronted with a new situation, because there are boundary points at which the (outer) normal derivative Un = au/an of the solution is given, but u itself is unknown since it is not given. To handle such points we need a new idea. This idea is the same for Neumann and mixed problems. Hence we may explain it in connection with one of these two types of problems. We shall do so and consider a typical example as follows. 918 E X AMP L E 1 CHAP. 21 Numerics for ODEs and PDEs Mixed Boundary Value Problem for a Poisson Equation Solve the mixed boundary value problem for the Poisson equation shown in Fig. ~57a. 1.5 x u=o (a) Regio~ (b) Grid (h = 0.5) R and boundary values Fig. 457. Mixed boundary value problem in Example 1 Solution. We u~e the grid shown in Fig. 457b. where" = 0.5. We recall that (7) in Sec. 21A has the right side f (x. y) = 0.5 2 • 12xy = 3xy. From the formulas II = 3,·3 and I/ n = 6x given on the boundary we compute the bounda.ry data ,,2 (1) PH 1/31 = 0.375. and P21 ilU12 U32 = 3. ely = 6·0.5 = 3, = 6·1 = 6. are internal mesh points and can be handled as in the last section. Indeed. from (7), Sec. 21.4. with {,2 = 0.25 and ,,2f (X. y) = 3xy and from the given boundary values we obtain two equations corresponding to PH and P 21 • as follows (with -0 resulting from the left boundary). 12(O.5·0.5)·! - 0 = 0.75 (2a) + 1122 = 12(1 • 0.5)·! - 0.375 = 1.125 The only difficulty with these equations seems to be that they involve the unknown values U12 and 1/22 of 1/ at P 12 and P22 on the boundary. where the normal derivative I/ n = ol//iln = flu/o)" is given. instead of 1/: but we shall overcome this difficulty as follows. We consider P12 and P 22 . The idea that will help us here is thb. We imagine the region R to be extended above to the first row of external mesh points (corresponding to y = 1.5), and we assume that the Poisson equation also holds in the extended region. Then we can write down two more equations as before (Fig. 457b) = 1.5 - 0 = 1.5 (2b) + "23 =3 - 3 = O. On the right. 1.5 is 12xy,,2 at (0.5. I) and 3 is 12x.'"{,2 at (I. 1) and 0 (at P02 ) and 3 (at P32 ) are given boundary values. We remember that we have not yet used the boundary condition on the upper part of the boundary of R, and we also notice that in (2b) we have introduced two more unknowns 1/13' 1/23' But we can now use that comlilion and get rid of "13' 1/23 by applying the central difference formula for dl/ Idy. From (I) we then obtain (see Fig. 457b) 3= 6= flll12 oy 01/22 oy 1/13 - 1/11 = 1113 - lIll. 21z 1/23 2h hence 1/13 hence 1123 = 1111 + 3 u21 = 1I23 - lI21' = "21 + 6. SEC. 21.5 Neumann and Mixed Problems. Irregular Boundary 919 Substituting these results into (2b) and simplifying, we have 2"11 2"21 41112 + + 1112 - 1122 4U22 = 1.5 - 3 = -1.5 = 3 - 3 - 6 = -6. Together with (2a) this yields, written in matrix form, (3) r -~ -4 0 ~] r::::] r~:::5] r ~:::5]. = 2 0 o 2 1 1112 -4 1122 -4 1.5 - 3 -1.5 ~0-6 -6 (The entnes 2 come from U13 and l/23, and so do -3 and -6 on the right). The solution of (3) (obtained by Gauss elimination) is as follows; the exact values of the problem are given in parentheses. U12 = 0.R66 un = 0.077 = I.RI2 (exact I) U22 lexact 0.125) "21 = 0.191 (exact 2) (exact 0.25). • Irregular Boundary We continue our discussion of boundary value problems for elliptic PDEs in a region R in the xy-plane. If R has a simple geometric shape, we can usually arrange for certain mesh points to lie on the boundary C of R, and then we can approximate partial derivatives as explained in the last section. However, if C intersects the grid at points that are not mesh points, then at points close to the boundary we must proceed differently, as follows. The mesh point 0 in Fig. 458 is of that kind. For 0 and its neighbors A and P we obtain from Taylor'S theorem (a) UA = + uh lIo - h Uo (4) (b) lip = auo ax auo ax + + 1 - 2 - I 2 a2uo (uhf -'-2- ax h2 2 a uo 2 ax + ... + We disregard the terms marked by dots and eliminate alia lax. Equation (4b) times a plus equation (4a) gives p Q Fig. 458. c Curved boundary C of a region R, a mesh point 0 near C, and neighbors A, B, P, Q 920 CHAP. 21 Numerics for ODEs and PDEs We solve this last equation algebraically for the derivative, obtaining ;PU o ax 2 2 [I = + a) a(l h2 I +~ UA I Up - -;; U o lIQ - I -b ] Similarly, by considering the points O. B. and Q. a2Uo 2 iJy 2 h = -2 [1+ b(\ b) + - -I I +b LIB Uo ] . By addition, (5) y 2 U o = -22 [UA h For example, if a a (l = + a) + + liB b (1 + Up 1+a b) + ~ _ ta + b)uo ] I +b ab . !, b = !, instead of the stencil (see Sec. 21.4) we now have because 1/[a(1 + a)l = ~, etc. The sum of all five terms still being zero (which is useful for checking). Using the same ideas, you may show that in the case of Fig. 459. 2 (6) V U o 2 1z2 = - [UA + a(a + p) UB b(b + q) + Up pep + a) + ap + bq abpq uQ --~- q(q + b) ] Uo , a formula that takes care of all conceivable cases. B bh 0 ph ah P~--""---~A qh Q Neighboring points A. B. P, Q of a mesh point 0 and notations in formula (6) Fig. 459. E X AMP L E 2 Dirichlet Problem for the Laplace Equation. Curved Boundary Find the potential II in the region in Fig. 460 that has the boundary values given in that figure: here the curved portion of the boundary is an arc of the circle of radius 10 about (0, 0). Use the grid in the figure. 3 II is a solution of the Laplace equation. From the given formulas for the boundary values II = x , 512 - 24y2 . ... we compute the values at [he points where we need them: the result is shown in [he figure. For Pll and P 12 we have the usual regular' stencil. and for P21 and P 22 we use (6), obtaining Solution. 1/ = 0.5 (7) -2.5 0.5 0+ 0.9 -3 0.6 0+ SEC. 21.5 Neumann and Mixed Problems. Irregular Boundary 921 y 6 u=512-24/ u=o u=o 3 u = 296 u= 216 u = 27 °O~----~3~~--~6--~8~~x 3 U =X Region, boundary values of the potential, and grid in Example 2 Fig. 460. We use this and the boundary obtain the system value~ and take the mesh points in the o- - 4u n + 0.6ll n + - 2. 5u21 0.6U21 0.5U22 = + 0.61112 - u~ual order P lI • P 21 , P 12 • P 22 . Then we 27 = -27 -0.9' 296 - 0.5 . 216 = -374.4 +0 U22 = 702 31122 = 0.9 . 352 702 + 0.9 . 936 = 1159.2. Tn matrix form, -2.5 (8) o 0 -4 0.6 0.6 :.5] I r:: :] r-~~:.4]. -3 1112 702 1122 1159.2 Gauss elimination yields the (ruundedl value, Uu = -55.6, U21 = 49.2, ll12 = -298.5, ll22 = -436.3. Clearly, from a grid with so few mesh points we cannot expect great accuracy. The exact solution of the PDE (not of the difference equation) having the given boundary values is u = x 3 - 3xy2 and yields the values Ull = -54, !l21 = 54, U12 = -297, U22 = -432. Tn praclice one would use a much finer grid and solve the resulting large system by an indirect method. • 1. Verify the calculation for the Poisson equation in Example 1. Check the values for (3) at the end. 2. Delive (5) in particular when a = h = !. 3. Deri ve the general stencil formula (6) in all detail. 4. Verify the calculation for the boundary value problem in Example 2. 5. Do Example I in the text for "\7 2 u = 0 with grid and boundary data as before. MIXED BOUNDARY VALUE PROBLEMS 6. Solve the mixed boundary value problem for the Laplace equation "\7 2 u = 0 in the rectangle in Fig. 457a (using the grid in Fig. 457b) and the boundary conditions u~ = 0 on the left edge, U x = 3 on the right edge, u = x 2 on the lower edge, and u = x 2 - 1 on the upper edge. 7. Solve Prob. 6 when un = 1 on the upper edge and u = 1 on the other edges. 922 CHAP. 21 Numerics for ODEs and PDEs 8. Solve the mixed boundary value problem for the Poisson equation ,2u = 2(x 2 + y2) in the region and for the boundary conditions shown in Fig. 461, using the indicated grid. y /u=o 3 u 2 0 Problems 8 and 10 9. CAS EXPERIMENT. Mixed Problem. Do Example 1 in the text with finer and finer grids of your choice and study the accuracy of the approximate values by comparing with the exact solution u = 2.\"y3. Verify the latter. 10. Solve , 211 = -1T2y sin! 1T.\' for the grid in Fig. 461 and lIy(l, 3) = lI y (2, 3) = !vW, II = 0 on the other three sides of the square. 1/ 0 Fig. 462. 11--<>--=--<r""--~ u=o P21 3 x Problem 11 12. If in Prob. II the axes are grounded (II = 0), what constant potential must the other portion of the boundary have in order to produce 100 volts at Pu? 13. What potential do we have in Prob. II if L/ = 190 volts on the axes and L/ = 0 on the other portion of the boundary? 14. Solve the Poisson equation y 2L/ = 2 in the region and for the boundary values shown in Fig. 463, using the grid also shown in the figure. y 3 u=/-3y----....... 1.5 n-<>----I~ u = / IRREGULAR BOUNDARIES 11. Solve the Laplace equation in the region and for the boundary values shown in Fig. 462. using the indicated grid. (The sloping portion of the boundary is y = 4.5 - x.) 21.6 2 u =3x u=o---" Fig. 461. - 1.5x ~u=9-3y P ll 2 I----{>--='::.....,:>---='--~ 2 P12 u=o--- y =X - 1.5y u=o Fig. 463. Problem 14 Methods for Parabolic PDEs The last two sections concerned elliptic POEs. and we now tum to parabolic POEs. Recall that the definitions of elliptic, parabolic, and hyperbolic POEs were given in Sec. 21.4. There it was also mentioned that the general behavior of solutions differs from type to type, and so do the problems of practical interest. This reflects on numerics as follows. For all three types, one replaces the POE by a corresponding difference equation, but for parabolic and hyperbolic POEs this does not automatically guarantee the convergence of the approximate solution to the exact solution as the mesh h ~ 0; in fact, it does not even guarantee convergence at all. For these two types of POEs one needs additional conditions (inequalities) to assure convergence and stability, the latter meaning that ~mall perturbations in the initial data (or small errors at any time) cause only small changes at later times. In this section we explain the numeric solution of the prototype of parabolic PDEs, the one-dimensional heat equation (c constant). SEC. 21.6 923 Methods for Parabolic POEs This POE is usually considered for x in some fixed interval, say, 0 :;;; x :;;; L, and time t ~ 0, and one prescribes the initial temperature u(x, 0) = f(x) (f given) and boundary conditions at x = 0 and x = L for all t ~ 0, for instance 1/(0. t) = 0, I/(L, t) = O. We may assume c = I and L = I; this can always be accomplished by a linear transformation of x and t (Prob. I). Then the heat equation and those conditions are O:;;;x:;;;Lt~O (1) (2) II(X, (3) u(O. t) = 0) = (Initial condition) f(x) u(], t) =0 (Boundary conditions). A simple finite difference approximation of (1) is [see (6a) in Sec. 21.4;j is the number of the time step] (4) ] k 1 (Ui,j+l - Uij) = /7 2 (Ui-l,j - 2Uij + Ui-l,j)' Figure 464 shows a corresponding grid and mesh points. The mesh size is h in the x-direction and k in the t-direction. Formula (4) involves the four points shown in Fig. 465. On the left in (4) we have used ajonmrd difference quotient since we have no information for negative t at the start. From (4) we calculate ui.j+l' which corresponds to time row j + 1, in terms of the three other II that correspond to time row j. Solving (4) for lIi,j+I' we have (5) lIi,j+1 = (1 - + 2r)uij r(ui+l,j + Ui-l,j), u = 0 --....,..,I---o-----<J---o----l (j = 2) ~u=O I---o-----<J---o----l(j=l) k h u Fig. 464. ! x = {(x) Grid and mesh points corresponding to (4), (5) (i,j + 1) X /k ( i - l . j ) X - - - h - - X - - - h - - X (i + l,j) (i,j) Fig. 465. The four points in (4) and (5) r= 924 CHAP. 21 Numerics for ODEs and PDEs Computations by this explicit method based on (S) are simple. However, it can be shown that crucial to the convergence of this method is the condition (6) r= That is. Uij should have a positive coefficient in (S) or (for r =~) be absent from (S). Intuitively, (6) means that we should not move too fast in the (-direction. An example is given below. Crank-Nicolson Method Condition (6) is a h<mdicap in practice. Indeed, to attain sufficient accuracy, we have to choose h small, which makes k very small by (6). For example, if h = 0.1, then k ::::; O.OOS. Accordingly, we should look for a more satisfactory discretization of the heat equation. A method that imposes no restriction on r = klh 2 is the Crank-Nicolson method, which uses values of u at the six points in Fig. 466. The idea of the method is the replacement of the difference quotient on the right side of (4) by ~ times the sum of two such difference quotients at two time rows (see Fig. 466). Instead of (4) we then have I k 1 (Ui,j+l - Uij) = 2h2 (Ui+l,j (7) + - 2uij I 2h2 (Ui+l,j+l - 2Ui,j+1 + Ui-l,j+l)' Multiplying by 2k and wliting r = klh 2 as before, we collect the terms corresponding to time row j + I on the left and the terms corresponding to time row j on the right: (8) (2 + 2r)Ui,j+1 - rtUi+I,j+l + lIi-l,j+l) = (2 - 2r)uij + r(ui+l,j + Ui-l,j)' How do we use (8)? In general, the three values on the left are unknown, whereas the three values on the right are known. If we divide the x-interval 0 ::::; x ::::; I in (I) into II equal intervals, we have II 1 internal mesh points per time row (see Fig. 464, where II = 4). Then for j = 0 and i = I,···. II I. formula (8) gives a linear system of II - I equations for the II - I unknown values Ull, 11 2 1, ••• , U n -I,1 in the first time row in terms of the initial values lIoo, UlO, . • • , UnO and the boundatl' values Um (= 0). lInl (= 0). Similarly for j = l,.i = 2, and so on; that is, for each time row we have to solve such a linear system of II - I equations resulting from (8). Although r = klh 2 is no longer restricted, smaller r will still give better results. In practice, one chooses a k by which one can save a considerable amount of work, without making r too large. For instance, often a good choice is r = 1 (which would be impossible in the previous method). Then (8) becomes simply (9) 411i ,j+l - ui+l,j+l - Ui-l,j+l = Time rowj + 1 X X Time rowj X X ui+l,j + Ui-l,j' X Ik h h Fig. 466. The six points in the CrankNicolson formulas (7) and (8) X SEC. 21.6 925 Methods for Parabolic PDEs j=5 0.20 0.16 . .. . .. 0.12 P12 j=3 P 22 r- j=2 ----- ----I Ip d ----- - - ----- e - - j = 1 I IplO Ip40 P20 P30 j=O 0.2 0.4 0.6 0.8 1.0 ;=1 ;=3 i = 2 i =4 i= 5 0.08 --- IP 0.04 t= 0 x=O u Grid in Example 1 Fig. 467. E X AMP L E 1 j=4 Temperature in a Metal Bar. Crank-Nicolson Method, Explicit Method Consider a laterally insulated metal bar of length I and such that c 2 = I in the heat equation. Suppose that the ends of the bar are kept at temperature" = O°C and the temperature in the bar at some instant -call it I = 0i, .f(x) = sin 1T.\' Applying the Crank-Nicolson method with h = 0.2 and r = I, find the temperature u(x. f) in the bar for 0 ~ t ~ 0.2. Compare the results with the exact solution. Also apply (5) with an r satisfying (6), say, r = 0.25. and with values nO! satisfying (6). say. ,. = I and,. = 2.5. Solution by Crank-Nicolson. Since,. = I. formula (8) takes the form (9). Since h = 0.2 and klh 2 = I. we have k = h 2 = 0.04. Hence we have to do 5 steps. Figure 467 shows the grid. We shall need the initial values ,. = 1110 = sin 0.277 = 0.587785. "20 = sin 0.41T = 0.951 057. Also, "30 = "20 and "40 = lI1O' (Recall that lI10 means" at P10 in Fig. 467. etc.) In each time row in Fig. 467 there are 4 internal mesh points. Hence in each time step we would have to solve 4 equations in 4 unknowns. But since the initial temperature distribution is symmetric with respect to x = 0.5, and II = 0 al both ends for all t. we have lI31 = lI21, "41 = lIll in the first time row and similarly for the other rows. This reduces each system to 2 equations in 2 unknowns. By (9). since 1/31 = 1/21 and 1101 = O. for j = 0 these equations are (i = I) = 1100 + "20 = 0.951 057 (i = 2) The solution is lin = 0.399274. lI21 = (i = I) (i = 2) The solution is (Fig. 468): I U12 t x=O 0.00 0.04 0.08 0.12 0.16 0.20 0 0 0 0 0 0 0.646039. Similarly, for time row j = I we have the system 4"12 -£l12 + 1122 = "01 + £l21 = 0.646039 + U21 = 1.045313. 31/22 = "11 0.271 221. 1/22 = 0.438844, and so on. This gives the temperature distribution X = u.2 0.588 0.399 0.271 0.184 0.125 0.085 x = U.4 x = U.6 x = u.8 0.951 0.646 0.439 0.298 0.202 0.138 0.951 0.646 0.439 0.298 0.202 0.138 0.588 0.399 0.271 0.184 0.125 0.085 x= 0 0 0 0 0 0 926 CHAP. 21 Numerics for ODEs and PDEs u(x. t) 1 Fig. 468. Temperature distribution in the bar in Example 1 Comparison witll the exact solutio1l. The present problem can be solved exactly by separating variables (Sec. 12.5): the result is (10) = 0.25. For h = 0.2 and r = klh 2 = 0.25 we have k = rh = 0.25 . 0.04 = 0.0 I. Hence we have to perform 4 times as many step' as with the Crank-Nicolson method! Formula (5) with,. = 0.25 is Solution by the explicit method (5) with r 2 (11) We can again make use of the symmetry. For j "20 = "30 = 0.951 057 and compute = 0 we need "00 = O. 1110 = 0.5877!lS (see p. 925). lin = 0.25(1100 + 21110 + "20) = 0.531 657 U21 = 0.25 (1110 + 2"20 + Of cou[,e we can omit the boundary terms compute 1130) O. 1101 = 1112 = 0.25(21111 1122 = 0.25 (1110 + + "02 = 31120) = O.R(iO 239. 0.··· from the formulas. For j I we 1121) = 0.480888 = 0.25(1111 + 31121) = 0.778094 and so on. We have to perform 20 steps in,tead of the S CN steps. but the numeric values ,how that the accuracy is only about the same as that of the Crank-Nicolson values CN. ll1e exact 3D-values follow from (10). x = 0.2 x = 0.4 t 0.04 0.08 0.12 0.16 0.20 CN By (11) Exac[ CN By (11) Exac[ 0.399 0.271 0.184 0.125 0.085 0.393 0.263 0.176 0.118 0.079 0.396 0.267 0.180 0.121 0.646 0.439 0.298 0.202 0.637 0.426 0.285 0.191 0.641 0.432 0.291 0.196 0.082 0.138 0.128 0.132 Failure of (5) with r violati1lg (6). Formula (5) with h = 0.2 and r = I-which violates (6)-is SEC. 21.6 Methods for Parabolic PDEs and give~ 927 very poor values: ,orne of the,e are T = 0.04 0.12 0.20 FOlmula (5) with an even larger r these are 0.1 0.3 0.363 0.139 0.053 = (I - 2r)uoj + Exact T 0.396 0.180 0.082 = 0.4 0.588 0.225 0.086 Exact 0.641 0.291 0.132 = 2.5 (and h = 0.2 as before) gives completely nonsensical results: some of x = 0.2 Exact 0.1)265 0.0001 0.2191 0.0304 1. (Nondimensional form) Show that the heat equation ut = c 2 uj'j', 0 ~ X ~ L, can be transformed to the "nondimensionar' standard fonn lit = l(ox. 0 ~ x ~ I. by setting x = .WL. t = c 271L2. II = uluo, where lto is any constant temperature. 2. Derive the difference approximation (4) of the heat equation. 3. Derive (5) from (4). 4. Using the explicit method [(5) with II = I and k = 0.5]. find the temperature at t = 2 in a laterally insulated bar of length 10 with ends kept at temperature 0 and initial temperature fIx) = x - 0.lx 2 • 5. Solve the heat problem (I )-(3) by Crank-Nicolson for 0 ~ t ~ 0.20 with II = 0.2 and k = 0.04 when fIx) = x if 0 ~ x < !. lex) = I - x if ! ~ x ~ I. Compare with the exact values for t = 0.20 obtained from the series (2 terms) in Sec. 12.5. 6. Solve Prob. 5 by the explicit method with h = 0.2 and k = 0.01. Do II steps. Compare the last values with the Crank-Nicolson 3S-values 0.107. 0.175 and the exact 3S-values 0.108, 0.175. 7. The accuracy of the explicit method depends on r (~!). Illustrate this for Prob. 6. choosing r = ~ (and h = 0.2 as before). Do 4 steps. Compare the values for t = 0.04 and 0.08 with the 3S-values in Prob. 6, which are 0.156. 0.254 (t = 0.04).0.105,0.170 (t = 0.08). 8. If the left end of a laterally insulated bar extending from x = 0 to x = I is insulated. the boundary condition at x = 0 is 11,,(0. t) = u.,.(O. t) = O. Show that in the application of the explicit method given by (5), we can compute IIO,j+ I by the fomlllla 1I0. j + I 0.2 2ruij' Apply this with II = 0.2 and r = 0.25 to determine the temperature II(X. t) in a laterally insulated bar extending fi'om x = 0 to I if lI(X, 0) = O. the left end is insulated 9. 10. 11. 12. x = 0.4 0.0429 0.0001 Exact 0.3545 0.0492. • and the right end is kept at temperature get) = sin ¥'lTt. Hint. Use 0 = Buo/ilx = (IIIj - IL1.) I2h. In a laterally insulated bar of length I let the initial temperature be fIx) = x if 0 ~ x ~ 0.2, fIx) = O.lS( I - x) if 0.2 ~ x ~ I. Let u(O, t) = O. 11(1. t) = 0 for all t. Apply the explicit method with II = 0.2, k = 0.01. Do 5 steps. Solve Prob. 9 for f(x) = x if 0 ~ x ~ 0.5, I(x) = I - x if 0.5 ~ x ~ l, all the other data being as before. Can you expect the solution to satisfy lI(x. t) = 11(1 - x, t) for all t? Solve Prob. 9 by (9) with II = 0.2, 2 steps. Compare with exact values obtained from the series in Sec. 12.5 (2 tenns) with suitable coefficients. CAS EXPERIMENT. Comparison of Methods. (a) Write programs for the explicit and the Crank-Nicolson methods. (b) Apply the programs to the heat problem of a laterally insulated bar of length I with lI(x, 0) = sin 'lTX and u(O, t) = 11(1. t) = 0 for all t, using h = 0.2, k = 0.01 for the explicit method (20 steps), II = 0.2 and (9) for the Crank-Nicolson method (5 steps). Obtain exact 6D-values from a suitable series and compare. (e) Graph temperature curves in (b) in two figures similar to Fig. 296 in Sec. 12.6. (d) Expeliment with smaller h (0.1,0.05. etc.) for both methods to find out to what extent accuracy increases under systematic changes of II and k. 113-151 CRANK-NICOLSON Solve (I )-(3) b) Crank-Nicolson with r = 1 (5 steps). where: 13. fIx) = X( I - x), lz = 0.2 14. f(x) = x( I - x), h = O. I (Compare with Prob. 13.) 15. f(x) = 5x if 0 ~ x < 0.2. f(x) = 1.25(1 - x) if 0.2 ~ x ~ 1, h = 0.2 928 21.7 CHAP. 21 Numerics for ODEs and PDEs Method for Hyperbolic PDEs In thi~ section we consider the numeric solution of problems involving hyperbolic POEs. We explain a standard method in terms of a typical setting for the prototype of a hyperbolic POE. the wave equation: = o~ x ~ 1, 1 ~ 0 (1) U tt (2) u(x. 0) = j(x) (Given initial displacement) (3) ut(x. 0) = g(x) (Given initial velocity) (4) u(O, t) = U xx u(l, t) =0 (Boundary conditions). Note that an equation U tt = c 2 u-r;x and another x-interval can be reduced to the form (I) by a linear transformation of x and 1. This is similar to Sec. 21.6, Prob. I. For instance, (1)-(4) is the model of a vibrating elastic string with fixed ~nds at x = 0 and x = I (see Sec. 12.2). Although an analytic solution of the problem is given in (13). Sec. 12.4, we lise the problem for explaining basic ideas of the numeric approach that are also relevant for more complicated hyperbolic PDEs. Replacing the derivatives by difference quotients as before, we obtain from (1) [see (6) in Sec. 21.4 with y = t] (5) I k2 (Ui,j+l - 2Uij + Ui,j-I) = I h2 (Ui+l,j - 2Uij + Ui-I,j) where h is the mesh size in x, and k is the mesh size in t. This difference equation relates 5 points as shown in Fig. 469a. It suggests a rectangular grid similar to the grids for parabolic equations in the preceding section. We choose r* = k 2flz2 = L Then Uij drops out and we have (6) (Fig. 469b). It can be shown that for 0 < r* ~ 1 the present explicit method is stable, so that from (6) we may expect reasonable results for initial data that have no discontinuities. (For a hyperbolic POE the latter would propagate into the solution domain-a phenomenon that would be difficult to deal with on our present grid. For unconditionally stable implicit methods see [El] in App. L) Equation (6) still involves 3 time steps j - I,j,j + J, whereas the formulas in the parabolic case involved only 2 time steps. Furthermore, we now have 2 initial conditions. x Time rowj + 1 • Time rowj x-,-x Ik x--x--x h Ik h x (a) Formula (5) Fig. 469. Time rowj-l I X (bl Formula (6) Mesh points used in (5) and (6) SEC. 21.7 Method for Hyperbolic PDEs 929 So we ask how we get stal1ed and how we can use the initial condition (3) This can be done as follows. From lItex. 0) = g(x) we derive the difference formula I 2k (7) = where & (uil - g(ih). For t Into this we substitute and by simplification Ui,-l) hence = gi. = O. that is, j = 0, equation (6) is Ui.-l as given in (7). We obtain Ua = Ui-l,O + Ui+l,O - + Uil 2kgi (8) This expresses E X AMP L E 1 lta in terms of the initial data. It is for the beginning only. Then use (6). Vibrating String, Wave Equation Apply the present method "ith" = k = 0.1 to the problem (1)-(4). where i(x) = sin g(x) = O. 77X. Solutio1l. The grid is the same as in Fig. 467, Sec. 21.6, except for the values of t, which now are 0.2. 004, ... (instead of 0.04. 0.08, ... ). The initial values From (8) and gtx) = 0 we have liDO, lIlO' . . • From this we compute. using 0.587785.1120 = lI30 = and ltOI = 0.951 057, 1111 = ~(1I00 + 1120) = ~. 0.951 057 = 0.475528 2) "21 = ~(1I1O + "30) = ~. 1.538841 = 0.769421 "31 = 1121,1141 = 1111 using sin 0.277 = (i = I) (i = and lI10 = 1140 = are the same as in Example 1, Sec. 21.6. 1102 = . . . = by symmetry as in Sec. 21.6, Example I. From (6) withj = I we now compute, 0, (i = I) U12 = 1101 + lI21 - £lID = 0.76') 421 - 0.587785 (i = 2) U22 = "11 + £131 - 1I20 = 0.475528 + 0.769421 - 0.951057 = U.UH 636 = 0.293 !l92, by symmetry; and so on. We thus obtain the following values of the dIsplacement over the first half-cycle: "32 = 1I22, "42 = lI12 II(X. t) of the ~tring x=O 0.0 0.2 0.4 0.6 0.8 1.0 0 0 0 0 0 0 x = 0.2 x = 0.4 X 0.588 0.476 0.182 -0.182 -0.476 -0.588 0.951 0.769 0.294 -0.294 -0.769 -0.951 0.951 0.588 0 0.769 0.294 -0.294 -0.769 -0.951 0.476 0.182 -0.182 -0.476 -0.588 0 0 0 0 0 X = 0.6 = 0.8 x= CHAP. 21 930 The~e Numerics for ODEs and POEs values are exact to 3D (3 decimals), the exact solution of the problem being (see Sec. 12.3) II(X, I) = sin TlX cos TIl. The reason for the exactness follows from d'Alemberfs solution (4), Sec. 12.4. (See Prob. -I, below.) • This is the end of Chap. 21 on numerics for ODEs and POEs, a rapidly developing field of basic applications and interesting research. in which large-scale and complicated practical problems can now be attacked and solved by the computer. ,@ VIBRATING STRING Solve (I )-(4) by the present method with h = k = 0.2 for the given initial deflection fix) and initial velocity 0 on the given (-interval. ~ ( ~ 2 I - x). 0 ~ ( ~ I O.OlxO - x), 0 1. f(x) = 2. f(x) 3. f(x) = x2( if 0 0.2 < x ~ I = x ~ x ~ 0.2. f(x) 0.25(1 - x) if 4. Show that from d' Alemberfs solution (13) in Sec. 11.4 with c = I it follows that (6) in the present section gives the exact value 1I;,i+l = lI(ih, (j + l)ft). 5. If the string governed by the wave equation (I) starts from its equilibrium position with initial velocity g(x) = sin 'In. what is its displacement at time ( = 0.4 and x = 0.2. 0.4. 0.6. 0.8? (Use the present method with h = 0.2. k = 0.2. Use (8). Compare with the exact values obtained from (I2) in Sec. 12.4.) ................. ... ....... _. __ .... -.. _ .... ......... __"a.-.-.~ 1. Explain the Euler and Improved Euler methods in geometrical terms. 2. What are the local and global orders of a method? Give examples. 3. What do you know about error estimates? Why are they important? 4. How did we obtain numeric mcthods by using the Taylor series'? 5. In each Runge-Kutta step we computed auxiliary values. How many? Why? 6. What are one-step and multistep methods? Give examples. 7. What is the idea of a predictor--corrector method? Mention some of these methods. 6. Compute approximate values in Prob. 5. using a finer grid (ft = 0.1. k = 0.1), and notice the increase in accuracy. 7. Illustrate the starting procedure when both f and g are not identically zero. say, f(x) = I - cos 2'lTx, g(X) = x - ,\'2. Choose ft = k = 0.1 and do 2 time steps. 8. Show that (I2) in Sec. 12.4 gives as another starting formula lin I = ~ (Ui+l.O + I lIi-1.0) + ? - - I k Xi + g(s) ds Xi- k (where one can evaluate the integral numerically if necessary). In what case is this identical with (8)? 0.1, 9. Compute u in Prob. 7 for r = 0.1 and x 0.2, .. " 0.9, using the formula in Prob. 8. and compare the values. 10. Solve (I)-(3) (h = k = 0.2, 5 time steps) subject to f(x) = .\.2. g(x) = 2x. 11,.(0. t) = 2t. lIe I. t) = (l + 1)2. STIONS AND PROBLEMS 10. What is automatic step size control? How is it done in practice? 11. Why and how did we use finite differences in this chapter,? 12. Make a list of types of PDEs. corresponding problems, and methods for their numeric solution. 13. How did we approximate the Laplace equation? The Poisson equation? 14. Will a difference equation give exact solutions of a PDE? 15. How did we handle (a) irregularly shaped domains. (b) given normal derivatives at the boundary? 16. Solve y' = 2x.\', yeO) = I, by the Euler method with ft = 0.1. 10 steps. Compute the en·or. B- What is the idea of the Rungc-Kutta-Fehlberg method? 17. Solve y' = 1 + y2. yCO) = O. by the improved Euler method with h = 0.1,5 steps. Compute the error. 9. How can Runge-Kutta be generalized to systems of ODEs? 18. Solve y' = (x + y h = 0.2, 7 steps. 4)2, yeO) = 4. by RK with 931 Chapter 21 Review Questions and Problems 19. Solve Prob. 17 by RK with II = 0.1, 5 steps. Compute the error. Compare with Prob. 17. 20. (Fair comparison) Solve y' = 2x- i Vy - In x + X-I, y(l) = 0 for I ~ x ~ 1.8 (a) by the Euler method with h = 0.1, (b) by the improved Euler method with h = 0.2. (c) by RK with h = 0.4. Verify that the exact solution is y = (In X)2 + In x. Compute and compare the errors. Why is the comparison fair? 21. Compute eX for x = 0, 0.1, .... 1.0 by applying RK to y' = r. yeO) = I. h = 0.1. Show that the result is 5D-exact. I~O-321 POTENTIAL Find the potential in Fig. 471, using the given grid and the boundary values: 30. u = 70 on the upper and left sides, U = 0 on the lower and right sides 31. u{PlO) = u(P 30 ) = 960. u(P20 ) = -480, U = 0 elsewhere on the boundary 32. u(P Ol ) u(P oa ) u{p 41) = u(P 43) = 200, u(P lO ) = u(P 30 ) 400, u(P 20 ) = 1600, u(P 02 ) = U(P 42 ) U(P 34 ) = 0 U(P I4 ) = lI(P 24 ) = 22. Solve y' = (x + y)2, yeO) = 0 by RK with h = 0.2. 5 steps. 23. Show thaI by applying the method in Sec. 21.2 to a polynomial of first degree we obtain the multistep predictor and corrector formulas )'~~+l = Yn + Yn.1 = )'n + h (3fn - 2 fn-I) Fig. 471. where.f~ 11 = f(Xn+l~ Y:+1)' 24. Apply the multistep method in Prob. 23 to the initial value problem y' = x + y. yeO) = 0, choosing h = 0.2 and doing 5 steps. Compare with the exact values. 25. Solve y' = (y - x - 1)2 + 2. y(O) = 1 for 0 ~ x ~ I by Adam~-Moulton with h = 0.1 <md starting values I. 1.200334589. 1.402709878. 1.609336039. 26. Solve y" + Y = O. y(O) = O. y' (0) = I by RKN with h = 0.2, 5 steps. Find the elTOr. 27. Solve y~ = -4."1 + 3.\'2' y~ = 5)'1 - 6Y2' )'1(0) = 3. ."2(0) = -5. by RK for systems. h = O.L 5 steps. 28. Solve y~ = -5.\"1 + 3)'2')'~ = -3.\"1 - 5.\"2' .\"1(0) = 2. Y2(0) = 2 by RK for systems. h = 0.1. 5 steps. 29. Find rough approximate values of the electrostatic potential at Pu. P 12 , P 13 in Fig. 470 that lie in a field between conducting plates (in Fig. 470 appearing as sides of a rectangle) kept at potentials 0 and I IO volts as shown. (Use the indicated grid.) YI 4 u = 110 V ~_"""I"",1_. 'P13 2 u=o u=o- u=o Fig. 470. Problem 29 Problems 30-32 33. Verify (13) in Sec. 21.4 for the sy~tem (12) and show that A in (\ 2) is nonsingular. 34. Derive the difference approximation of the heat equation. 35. Solve the heat equation (l). Sec. 21.6. for the initial condition J(x) = x if 0 ~ x ~ 0.2, J(x) = 0.25(1 - x) if 0.2 < x ~ I and boundary condition (3). Sec. 21.6. by the explicit method [formula (5) in Sec. 21.6] with h = 0.2 and k = 0.0 I so that you get values of the temperature at time t = 0.05 as the answer. 36. A laterally insulated homogeneous bar with ends at x = 0 and x = I has irlitial temperature O. Its left end is kept at O. whereas the temperature at the right end varies sinusoidally according to u(t, I) = get) = sin ~1T1. Find the temperature u(x. t) in the bar [solution of (1) in Sec. 21.6] by the explicit method with h = 0.2 and r = 0.5 (one period, that is, 0 ~ t ~ 0.24). 37. Find lI(X, 0.12) and lI(X, 0.24) in Prob. 36 if the left end of the bar is kept at -g(t) (instead of 0). all the other data being as before. 38. Find out how the results of Prob. 36 can be used for obtaining the results in Prob. 37. Use the values 0.054, 0.172, 0.325, 0.406 (r = 0.12, x = 0.2,0.4.0.6, 0.8) and -0.009, -0.086. -0.252. -0.353 (r = 0.24) from the answer to Prob. 36 to check your answer to Prob. 37. 39. Solve lit = lIxx (0 ~ X ~ I. r ~ 0), 2 lI(X. 0) = x ( I - x), u(O. t) = lI( I, r) = 0 by Crank-Nicolson with h = 0.2. k = 0.04. 5 time steps. 40. Find the solution of the vibrating string problem lit{ = lIxx• U(x. 0) = x(l - x), lit = 0, u(O, t) = lI( I. t) = 0 by the method in Sec. 21.7 with h = 0.1 and k = 0.1 for t = 0.3. 932 CHAP. 21 Numerics for ODEs and PDEs Numerics for ODEs and PDEs fn this chapter we discussed numerics for ODEs (Secs. 21.1-21.3) and PDEs (Secs. 21.4-21.7). Methods for initial value problems v' = f(x, y), (I) Y(Xo) = Yo involving a first-order ODE are obtained by truncating the Taylor series y(x + 11) = y(x) + hr , (x) /7 2 +2 /I y (x) + ... where, by (I), y' = f, y" = f' = aflilx + (ilflay»)"'. etc. Truncating after the term Izy', we get the Euler method, in which we compute step by step (2) J'n+ 1 = Yn + hf(x", Yn) (11 = 0, I, ... ). Taking one more term into account, we obtain the improved Euler method. Both methods show the basic idea but are too inaccurate in most cases. Truncating after the term in /7 4 , we get the important classical Runge-Kutta (RK) method of fourth order. The crucial idea in this method is the replacement of the cumbersome evaluation of derivatives by the evaluation of f(x. y) at suitable points (x, y); thus in each step we first compute four auxiliary quantities (Sec. 21.1) kl = hf(xno Yn) k2 = hf(xn + 41z. )'n + 4k1 ) k3 = hf(xn + 4h. Yn + 4k2) (3a) and then the new value (3b) Error and step size control are possible by step halving or by RKF (Runge-Kutta-Fehlberg). The methods in Sec. 21.1 are one-step methods since they get )"n+l from the result Yn of a single step. A multistep method (Sec. 21.2) uses the values of Ym Yn-l' ... of severa) steps for computing Yn+l. Integrating cubic interpolation polynomials gives the Adams-Bashforth predictor (Sec. 21.2) (4a) = Yn + I 24 h(55fn - 59fn-l + 37fn-2 - 9fn-3) 933 Summary of Chapter 21 where (4b) h f(Xj, }J)' and an Adams-Moulton corrector (the actual new value) Yn+I = Yn + 1 24 h(9f~+1 + 19fn - 5fn-I + fn-2)' where f~+1 = f(Xn+h ."~+I)· Here, to get started, .1'1, ."2' Y3 must be computed by the Runge-Kutta method or by some other accurate method. Section 19.3 concerned the extension of Euler and RK methods to systems y' = f(x, y), j thus = 1, ... , 111. This includes single mth order ODEs, which are reduced to systems. Second-order equations can also be solved by RKN (Runge-Kutta-Nystrom) methods. These are particularly advantageous for .v" = f(x • .1') with f not containing y'. Numeric methods for PDEs are obtained by replacing pattial derivatives by difference quotients. This leads to approximating difference equations, for the Laplace equation to (Sec. 21.4) (5) for the heat equation to (6) 1 k I (Ui,j+I - Uij) = h2 (Ui+I,j - 2Uij + Ui-I,j) (Sec. 21.6) and for the wave equation to (7) (Sec. 21.7); here hand k are the mesh sizes of a grid in the x- and y-directions, respectively. where in (6) and (7) the variable .1' is time t. These PDEs are elliptic, parabolic, and h}perbolic, respectively. Con'esponding numeric methods differ. for the following reason. For elliptic PDEs we have boundary value problems, and we discussed for them the Gauss-Seidel method (also known as Liebmann's method) and the ADI method (Sees. 21.4, 21.5). For parabolic PDEs we are given one initial condition and boundary conditions, and we discussed an nplicit method and the Crank-Nicolson method (Sec. 21.6). For hyperbolic PDEs, the problems are similar but we are given a second initial condition (Sec. 21.7). .. ~\'f . J PA RT .& , ... F ,, - ", '1 - -- I - IIii • ) - -' • ~ Optimization, Graphs C HAP T E R 22 Unconstrained Optimization. Linear Programming C HAP T E R 23 Graphs. Combinatorial Optimization Ideas of optimization and application of graphs play an increasing role in engineering. computer science, systems theory. economics. and Olher areas. In the first chapter of this part we explain some basic concepts. methods. and results in unconstrained and constrained optimization. The second chapter is devoted to graphs and the corresponding so-called combinatorial optimization, a relatively new interesting area of ongoing applied and theoretical research. 935 ~\-:' ...: CHAPTER , . Ii. 22 -.IIt:!!! , .:i!fi3!!iI .... - _. . . Jill"' 1 Unconstrained Optimization. Linear Programming Optimization principles are of basic importance in modern engineering design and systems operation in various areas. The recent development has been influenced by computers capable of solving large-scale problems and by the creation of conesponding new optimization techniques. so that the entire field has become a large area of its own. In the present chapter we give an introduction to the more impOltant concepts, methods, and results on unconstrained optimization (the so-called gradient method) and constrained optimization (linear programming). Prerequisite: a modest working knowledge of linear systems of equations References and Answers to Problems: App. 1 Part F, App. 2. 22.1 Basic Concepts. Unconstrained Optimization In an optimization problem the objective is to opti11li~e (l11axil1li~e or 1I1ini111i~e) some function f. This function f is called the objective function. For example, an objective function f to be maximized may be the revenue in a production of TV sets, the yield per minute in a chemical process, the mileage per gallon of a certain type of car, the hourly number of customers served in a bank, the hardness of steel, or the tensile strength of a rope. Similarly, we may want to minimize f if f is the cost per unit of producing certain cameras, the operating cost of some power plant, the daily loss of heat in a heating system. the idling time of some lathe. or the time needed to produce a fender. In most optimization problems the objective function f depends on several variables These are called control variables because we can "control" them, that is. choose their values. For example, the yield of a chemical process may depend on pressure Xl and temperature x 2 • The efficiency of a cel1ain air-conditioning system may depend on temperature XI' air pressure X2' moisture content X3, cross-sectional area of outlet X4. and so on. Optimization theory develops methods for optimal choices of XI • ••• , x'" which maximize (or minimize) (he objective function f, that is, methods for finding optimal values of Xl • • • • , X n . 936 SEC. 22.1 937 Basic Concepts. Unconstrained Optimization In many problems the choice of values of Xl, . . . , Xn is not entirely free but is subject to some constraints, that is. additional restrictions arising from the nature of the problem and the variables. For example, if Xl is production cost, then Xl ~ 0, and there are many other variables (time. weight, distance traveled by a salesman, etc.) that can take nonnegative values only. Constraints can also have the form of equations (instead of inequalities). We first consider unconstrained optimization in the case of a function I(x}> ... , X,J. We also wlite x = (Xl, . . . , xn) and I<x), for convenience. By definition, I has a minimum at a point x = Xo in a region R (where I is defined) if I(x) for all x in R. Similarly, ~ I(X o) I has a maximum at Xo in R if for all x in R. Minima and maxima together are called extrema. Furthermore, I is said to have a local minimum at Xo if for all x in a neighborhood of Xo, say, for all x satisfying where Xo = (Xl, ... , Xn) and r > 0 is sufficiently small. Similarly, I has a local maximum at Xo if I(x) ~ I(X o ) for all x satisfying Ix - xol < r. If I is differentiable and has an extremum at a point Xo in the interior of a region R (that is, not on the boundary), then the partial derivatives ai/axI' ... , aflax., must be zero at Xo. These are the components of a vector that is called the gradient of I and denoted by grad lor VI. (For 11 = 3 this agrees with Sec. 9.7.) Thus (1) A point Xo at which (1) holds is called a stationary point of I. Condition (1) is necessary for an extremum of I at Xo in the interior of R, but is not sufficient. Indeed, if 11 = 1. then for -" = Ilx), condition ll) is y' = !' (Xo) = 0; and, for instance, )' = x 3 satisfies y' = 3x2 = 0 at X = Xo = 0 where I has no extremum but a point of inflection. Similarly, for I(x) = XIX2 we have VI(O) = 0, and I does not have an extremum but has a saddle point at O. Hence after solving (I), one must still find out whether one has obtained an extremum. In the case 11 = I the conditions y' (Xo) = 0, -,,"(Xo) > 0 guarantee a local minimum at Xo and the conditions y' (xo) = 0, -,,"(Xo) < 0 a local maximum, as is known from calculus. For 11 > 1 there exist similar criteria. However. in practice even solving (1) will often be difficult. For this reason, one generally prefers solution by iteration, that is, by a search process that starts at some point and moves stepwise to points at which I is smaller (if a minimum of I is wanted) or larger (in the case of a maximum). 938 CHAP. 22 Unconstrained Optimization. Linear Programming The method of steepest descent or gradient method is of this type. We present it here in its standard form. (For refinements see Ref. [E25] listed in App. 1.) The idea of this method is to find a minimum of f(x) by repeatedly computing minima of a function get) of a single vruiable t, as follows. Suppose that .f has a minimum at Xo and we strut at a point x. Then we look for a minimum of f closest to x along the straight line in the direction of - vf(x). which is the direction of steepest descent (= direction of maximum decrease) of fat x. That is, we determine the value of t and the corresponding point (2) z(t) = x - tV f(x) at which the function g(t) (3) = f(z(t» has a minimum. We take this z(t) as our next approximation E X AMP L E 1 to Xo' Method of Steepest Descent Determinc a minimum of (4) starting ti'01TI Xo = (6. 3) = 6i + 3j and applying the method of steepest descent. Solutioll. Clearly. inspection shows that j(x) has a minimum at tI. Knowing the solution gives u~ a better feel of how the method worh. We obtain,fIx) = 2Xl i + 6x2j and from this z(t) = x - t'f(x) = (\ - 2tlx1 i g(t) = f(zv» + (I - 6rlx2 j 2 2 2 = (1 - 2t) x]2 + 3(1 - 6t) X2 . We now calculate the derivative set g' (t) = O. and solve for t. finding Starting from Xo = 6i + 3j, we compute the value~ in Table 22.1. which are shown in Fig. 472. Figure 472 suggests that in the case of slimmer ellipses C'a long narrow valley"). convergence would be poor. You may confirm this by replacing the coefficient 3 in (4) with a large coefficieIlt. For more sophisticated descent and other methods. some of them also applicable to vector functions of vector variables. we refer to the references lIsted in Part F of App. 1: see also [E25j. • x2 Fig. 472. Method of steepest descent in Example 1 Linear Programming SEC. 22.2 939 Table 22.1 Method of Steepest Descent, Computations in Example 1 X II 0 6.000 3.484 1.327 0.771 0.294 0.170 0.065 I 2 3 4 5 6 3.000 -0.774 0.664 -0.171 0.147 -0.038 0.032 0.210 0.310 0.210 0.310 0.210 0.310 STEEPEST DESCENT Do 3 steepest descent steps when: 3. f(x) = 3X I 2 + 2X22 - 12xl S. f(X) = Xo = 6. + 16x2, Xo = [I l]T + 2x/ - Xl - 6x2 , Xo = [0 OJT 0.5X12 + 0.7X22 - Xl + 4.2x2 + I, X1 [-I f(x) = .\"0 = [2 2 I]T x/ + 0.lx22 + 8Xl + X2 + 22.5. -I]T 7. f(x) = 0.2x/ + x/ - 0.08Xl' Xo = [4 4]T 8. fIx) = X1 2 - X22. Xo = [2 I ]T. 5 steps. First guess. Then compute. Sketch your path. 22.2 0.581 0.381 0.581 0.38] 0.581 0.381 -0.258 -0.S57 -0.258 -0.857 -0.258 -0.857 10. fIx) = Xl 2 - X2' Xo = [I l]T. Sketch your path. Predict the outcome of further steps. 11. .f(x) = aXI + bX2' any "0. First guess. then compute. ,3-111 = I - 6t 9. f(x) = X1 2 + cx/. Xo = [c I ]T. Show that 2 steps give [c llT times a factor. -4c2 /(c 2 - 1)2. What can you conclude from this about the speed of convergence? 1. What happens if you apply the method of steepesl descent to fIx) = X1 2 + X2 2 ? 2. Verify that in Example I. successive gradients are orthogonal. What is the reason? 4. f(x) I - 2t 12. CAS EXPERIMENT. Steepest Descent. (a) Write a program for the method. (b) Apply your program to f(x) = X1 2 + 4.\"22. experimenting with respect to speed of convergence depending on the choice of Xo . (c) Apply your program to fIx) = .\"12 + X24 and to fIx) = X14 + X24, Xo = [2 I]T. Graph level curves and your path of descent. <Try to include graphing directly in your program.) Linear Programming Linear programming or linear optimization consists of methods for solving optimization problems with constrai1Zts, that is, methods for finding a maximum (or a minimum) x = [Xl' ••• , xnl of a li1Zear objective function satisfying the constraints. The latter are linear inequalities, such as 3x1 + 4X2 :;; 36, or 0, etc. (examples below). Problems of this kind arise frequently, almost daily, for instance, in production, inventory management, bond trading, operation of power plants. routing delivery vehicles, airplane scheduling, and so on. Progress in computer technology has made it possible to solve programming problems involving hundreds or thousands or more variables. Let us explain the setting of a linear programming problem and the idea of a "geometric" solution, so that we shall see what is going on. Xl ~ 940 EXAMPLE 1 CHAP. 22 Unconstrained Optimization. Linear Programming Production Plan Energy Savers. Inc .. produces heaters of types Sand L. The wholesale price is $40 per heater for Sand $88 for L. Two time constraints result from the use of two machines M1 and M 2. On M1 one needs 2 min for an S heater and g min for an L heater. On M2 one needs S min for an S heater and 2 min for an L heater. Determine production figures Xl and X2 for Sand L. respectively (number of heaters produced per hour) so that the hourly revenue ~ = fer) = 40X1 + 88x2 is maximum. Solution. Production figures Xl and X2 must be nonnegative. Hence the objective function (to be maximized) and the four constraints are (0) (1) 2\~1 + 8x2 ~ 60 min time on machine (2) SX1 + 2x2 ~ 60 min time on machine M2 ~ (3) Ml 0 (4) Figure 473 shows (0)--(4) as follows. Con~tancy lines Z = COllst are marked (0). These are lines of constant reHnue. Their slope is -40/88 = - SIll. To increase ~ we must move the line upward (parallel to itself), as the arrow shows. Equation (I) with the equality sign is marked (1) It intersects the coordinate axes at Xl = 6012 = 30 (set x2 = 0) and x2 = 60/8 = 7.S (set Xl = 0). The arrow marks the side on which the points (Xl. x2) lie that satisfy the inequality in (I). Similarly for Eqs. (2)-(4). The blue quadrangle thus obtained is called the feasibilit) region. It is the set of all feasible solutions. meaning solutions that satisfy all four constraints. The figure also lists the revenue at O. A. B. C. The optimal solution is obtained by moving the line of constant revenue up as much as possible without leaving the feasibility region completely. Obviously. this optimum is reached when that line pas~es through B. the intersection (10. S) of (I) and (2). We see that the optimal revenue ';:max = 40· 10 is obtained by producing twice as many S heater~ + 88· S = $840 • as L heaters. x2 0: z=O A: z = 40·12 = 480 B: z = 40 . 10 + 88 . 5 = 840 C: z=88·7.5=660 20 (3) (4) " o " 10 A 20 """"(0) ~""O "",~O ~ ""'(O)~ "'"" ""8</0 Fig. 473. Linear programming in Example 1 SEC 22.2 941 Linear Programming Note well that the problem in Example I or similar optimization problems call1lot be solved by setting certain partial derivatives equal to zero. because crucial to such problems is the region in which the control variables are allowed to vary. Furthermore, our "geometric" or graphic method illustrated in Example I is confined to two variables Xl, x2' However, most practical problems involve much more than two variables, so that we need other methods of solution. Normal Form of a Linear Programming Problem To prepare for general solution methods. we show that constraints can be written more uniformly. Let us explain the idea in terms of (1), This inequality implies 60 - 2XI - 8x2 ;;:; 0 (and conversely), that is, the quantity is nonnegative. Hence, our original inequality can now be written as an equation where is a nonnegative auxiliary variable introduced for converting inequalities to equations. Such a variable is called a slack variable, because it "takes up the slack" or difference between the two sides of the inequality. X3 E X AMP L E 2 Conversion of Inequalities by the Use of Slack Variables With the help of two slack variables following form. Maximize \-3' .\"4 we can write the linear programming problem in Example I in the subject to the constraints 1, ... , 4). (i = We now have /I = 4 variables and III = 2 (linearly independent) equations. so that two of the four variables. for example. Xl' x2. determine the others. Also note that each of the four sides of the quadrangle in Fig. 473 now has an equation of the form Xi = 0: OA: x2 = O. AB: x4 = BC: X3 0, = 0, CO: Xl = O. A vertex of the quadrangle is the intersection of two sides. Hence at a vertex. n - 1/1 = 4 - 2 variables are /ero and the others are nonnegallve. Thus at A we have x2 = 0, \"4 = 0, and so on. = 2 of the • 942 CHAP. 22 Unconstrained Optimization. Linear Programming Our example suggests that a general linear optimization problem can be brought to the following normal form. Maxill1i~e (5) subject to tlte cOllStrail1ts (6) Xi ~ 0 (i = I, ... , 11) with all bj nonnegative. (If a bj < O. multiply the equation by -1.) Here Xl' .... Xn include the slack variables (for which the c/s in f are zero). We assume that the equations in (6) are linearly independent. Then, if we choose values for 11 - m of the variables. the system uniquely detelmines the others. Of course. since we must have Xl ~ 0, ... , Xu :;; 0, this choice is not entirely free. Our problem also includes the minimization of an objective function f since this corresponds to maximizing - f and thus needs no separate consideration. An n-tuple (Xl' ••• ,xn ) that satisfies all the constraints in (6) is called afeasible poillt or feasible solution. A feasible solution is called an optimal solution iffor it the objective function .f becomes maximum. compared with the values of f at all feasible solutions. Finally, by a basic feasible solution we mean a feasible solution for which at least 11 - III of the variables Xl' ••• , X" are zero. For instance, in Example 2 we have 11 = 4, III = 2. and the basic feasible solutions are the four vertices 0, A. B. C in Fig. 473. Here B is an optimal solution (the only one in this example). The following theorem is fundamental. THEOREM 1 ,--------------------------------------Optimal Solution Some optimal solutioll of a lil1ear progra/1lmil1g problem (5). (6) is also a basic feasible .I'nlutimz of (5), (6). J For a proof, see Ref. [F5], Chap. 3 (listed in App. I). A problem can have many optimal solutions and not all of them may be basic feasible solutions: but the theorem guarantees that we can find an optimal solution by searching through the basic feasible solutions only. This is a great simplification; but since there are ( Il) (Il) 1/ - I l l III different ways of equating 11 - III of the 11 variables to zero, considering all these possibilities. dropping those which are not feasible and then searching through the rest would still involve very much work. even when 11 and /1l are relatively small. Hence a systematic search is needed. We shall explain an important method of this type in the next section. SEC 1l.2 943 Linear Programming 1. What is the meaning of the slack variables X3' X4 in Example 2 in terms of the problem in Example I? 2. Can we always expect a unique solution (as is the case in Example L)? 3. Could we find a profit f(XI' X2) = QIXI + Q2X2 whose maximum is at an interior point of the quadrangle in Fig. 473? (Give a reason for your answer.) 4. Why are slack variables always nonnegative? How many of them do we need? REGIONS AND CONSTRAINTS 15-101 Describe and graph the region in the first quadrant of the x l x2-plane determined by the inequalities: 5. Xl + 2X2 ~ IO X2 ~ 0 X2 ::;; 2 Xl - + 7. 2.0XI 5.0XI + 8. 2XI 4XI Xl 9. 6.0X2 ~ 18.0 2.5x2 ~ 20.0 X2 ::;; 6 + 5X2 ~ 40 - 2X2 ::;; -3 - Xl + X2 ::;; 3 Xl + X2 ~ 9 10. + X2 ~ X2 ::;; 2 + 5X2 ::;; 15 X2 ::;; -2 2X2 ~ LO 3x I 2x) - -Xl + 111-151 MAXIMIZATION AND MINIMIZATION Maximize the given objective function given constraints. 11. f = -lOx) + 2X2, 16. (Maximum output) Giant Ladders, Inc., wants to maximize its daily total output of large step ladders by producing Xl of them by a process PI and X2 by a process P 2 , where PI requires 2 hours of labor and 4 machine hours per ladder, and P2 requires 3 hours of labor and 2 machine hours. For this kind of work, 1200 hours of labor and 1600 hours on the machines are at most available per day. Find the optimal Xl and X2' 18. (Minimum cost) Hardbrick, Inc., has two kilns. Kiln I can produce 3000 grey bricks, 2000 red bricks, and 300 glazed bricks daily. For Kiln II the corresponding figures are 2000, 5000, and 1500. Daily operating costs of Kilns I and II are $400 and $600. respectively. Find the number of days of operation of each kiln so that the operation cost in filling an order of 18000 grey, 34000 red, and 9000 glazed blicks is minimized. 3 Xl + 15. Minimize f in Prob. 11. 17. (Maximum profit) Universal Electric, Inc .. manufactures and sells two models of lamps, LI and L 2 , the profit being $150 and $100, respectively. The process involves two workers WI and W2 who are available for this kind of work 100 and 80 hours per month, respectiveLy. WI assembles LI in 20 min and L2 in 30 min. W 2 paints LI in 20 min and L2 in 10 min. Assuming that all lamps made can be sold without difficulty. determine production figures that maximize the profit. -Xl +'\2 ::;; -3 -Xl 14. Minimize fin Prob. 13. Xl::;; 0, f subject to the X2::;; 0, 19. (Maximum profit) United MetaL Inc .. produces alloys B) (special brass) and B2 (yellow tombac). BI contains 50% copper and 50% zinc. (Ordinary brass contains about 65% copper and 35% zinc.) B2 contains 75% copper and 25% zinc. Net profits are $120 per ton of Bl and $100 per ton of B 2 . The daily copper supply is 45 tons. The daily zinc supply is 30 tons. Maximize the net profit of the daily production. 20. (Nutrition) Foods A and B have 600 and 500 calories, contain L5 g and 30 g of protein, and cost $1.80 and $2.10 per unit, respectively. Find the minimum cost diet of at least 3900 calories containing at least 150 g of protein. 944 22.3 CHAP. 22 Unconstrained Optimization. Linear Programming Simplex Method From the last section we recall the following. A linear optimization problem (linear proglamming problem) can be written in normal form; that is: Maximize (1) subject to the cOllstraillts (2) Xi ~ 0 (i = I. .... 11). For finding an optimal solution of this problem. we need to consider only the basic feasible solutions (defined in Sec. 22.2), but there are still so many that we have to follow a systematic search procedure. In 1948 G. B. Dantzig published an iterative method, called the simplex method, for that purpose. In this method, one proceeds stepwise from one basic feasible solution to another in such a way that the objective function f always increases its value. Let us explain this method in terms of the example in the last section. In its original form the problem concerned the maximization of the objective function z subject to = 40x 1 + 88x2 2'\'1 + &-2 ~ 60 X2 ~ O. Converting the first two inequalities to equations by introducing two slack variables X3, we obtained the normal form ofthe problem in Example 2. Together with the objective function (written as an equation;:: - 40X1 - 8!h2 = 0) this normal fonn is X4' = (3) 0 = 60 where Xl ~ 0, ... , .1.'4 ~ O. This is a linear system of equations. To find an optimal solution of it, we may consider its augmented matrix (see Sec. 7.3) b (4) SEC. 22.3 945 Simplex Method This matrix is called a simplex tableau or simplex table (the illitial simplex table). These are standard names. The dashed lines and the letters - b ~, are for ease in further manipulation. Every simplex table contains two kinds of variables .r} By basic variables we mean those whose columns have only one nonzero entry. Thus X3' X4 in (4) are basic variables and Xl' -'"2 are nonbasic variables. Every simplex table gives a basic feasible solution. It is obtained by setting the nonbasic variables to zero. Thus (4) gives the basic feasible solution .\"3 = 60/1 = 60, X4 = z=o 60/1 = 60, with .\"3 obtained from the second row and .\"4 from the third. The optimal solution (its location and value) is now obtained stepwise by pivoting, designed to take us to basic feasible solutions with higher and higher values of::: until the maximulll of z is reached. Here, the choice of the pivot equation and pivot are quite different from that in the Gauss elimination. The reason is that Xl> X2, .\"3, X4 are restricted to nonnegative values. Step 1. Operatioll 0 1 : Selection of the Column of the Pivot Select as the colunrn of the pivot the first column with a negative entry in Row 1. In (4) this is Column 2 (because of the -40). Operatioll O 2 : Selection of the Row of the Pivot. Divide the right sides [60 and 60 in (4)] by the corresponding entries of the column just selected (6012 = 30. 60/5 = 12). Take as the pivot equation the equation that gives the smallest quotient. Thus the pivot is 5 because 60/5 is smallest. Operatioll 0 3 : Elimination by Row Operations. pivot (as in Gauss-Jordan, Sec. 7.8). This gives zeros above and below the With the notation for row operations as introduced in Sec. 7.3, the calculations in Step give from the simplex table To in (4) the following simplex table (augmented matrix). with the blue letters referring to the previous table. z (5) [ b -~--t--~--~z~.;-t-~---~~:-t-j!~-l I o : I 5 Row I = 60/5 = 12, 8 Row 3 Row 2 - 0.4 Row 3 I 2: 0 : 60 We ,ee that basic variables are now Xh X3 and nonbasic variables are latter to zero, we obtain the basic feasible solution given by T 1, Xl + X3 = 36/1 = 36, -1."2, .\"4. Setting the ::: = 480. This is A in Fig. 473 (Sec. 22.2). We thus have moved from 0: (0, 0) wIth::: = 0 to A: (12, 0) with the greater::: = 480. The reason for this increase is our elimination of a term (-40.1."1) with a negative coefficient. Hence elimillation is applied only to Ilegative e1ltries in Row I but to no others. This motivates the selection of the COIUIIlIl of the pivot 946 CHAP. 22 Unconstrained Optimization. Linear Programming We now motivate the selection of the row of the pivot. Had we taken the second row of To instead (thus 2 as the pivot), we would have obtained z = 1200 (verify!), but this line of constant revenue ~ = 1200 lies entirely outside the feasibility region in Fig. 473. This motivates our cautious choice of the entry 5 as our pivot because it gave the smallest quotient (60/5 = 12). Step 2. The basic feasible solution given by (5) is not yet optimal because of the negative entry -72 in Row 1. Accordingly, we perform the operations 0 1 to 0 3 again, choosing a pivot in the column of -72. Operatioll 0 1 , Select Column 3 of T 1 in (5) as the column of the pivot (because -72 < 0). Operation 02' We have 3617.2 = 5 and 60/2 = 30. Select 7.2 as the pivot (because 5 < 30). Operation 0 3 , Elimination by row operations gives z T2 (6) = b ~-~-i--~--~~-f--J~----~~.~-i-~:~-] o :5 I I 0 : I I 1: 3.6 We see that now Xl' X2 are basic and x 3 , from T 2 the basic feasible solution Xl = 50/5 = 10, X2 + 10 Row 2 1 50 I I 0.9 Row I Row 3 ""'::"-Row 2 7.'2 nonbasic. Setting the latter to zero, we obtain X4 = 36/7.2 = 5, .: = 840. This is B in Fig. 473 (Sec. 22.2). In this step . .: has increased from 480 to 840. due to the elimination of -72 in T l . Since T2 contains no more negative entries in Row I, we conclude that z = f(10, 5) = 40 . 10 + 88 . 5 = 840 is the maximum po:"sible revenue. It is obtained if we produce twice as many S heaters as L heaters. This is the solution of our problem by the simplex method of linear programming. • Minimization. If we want to millimize <: = f(x) (instead of maximize), we take as the columns of the pivots those whose entry in Row I is positive (instead of negative). In such a Column k we consider only positive entries Tjk and take as pivot a ~ik for which b/tjk is smallest (as before). For examples, see the problem set . •... _-_ .. _ _.......-- --- ........... - ......... _....... .... --. ~ -=9l L _ SIMPLEX METHOD Write in nonnal form and solve by the simplex method, assuming all xJ to be nonnegative. 1. Maximize.f = 3Xl + 2X2 subject to 3x1 + 4.l2 ~ 60. 4Xl + 3X2 ~ 60. IOxl + 2X2 ~ 120. 2. Prob. 16 in Problem Set 22.2. 3. Maximize the profit in the daily production of XI metal frames Fl ($90 profit/frame) and X2 frames F2 ($50 profit/frame) under the restrictions XI + 3X2 ;§; 1800 (material). Xl + .\"2 ;§; 1000 (machine hours), 3Xl + X2 ;§; 2400 (labor). 4. Maximize f = X\ + X2 + 2Xl X3 ;§; + 3X2 + 4.8, lOxI + \"3 subject to 9.9, -"2 - X3 ~ X3 ;§; 0.2. 5. The problem in the text with the order of the constmims interchanged. 6. Minimize f = 4Xl - 3.\"1 + 4X2 + 5.\"3 2Xl + 3X3 ~ 30. ~ IOx2 - 20.\"3 subject to 60, 2.\"1 + .\"2 ~ 20, 7. MinimiLe f = 5Xl - 20X2 subject to 2Xl + 5x2 ~ 10. 8. Prob. 20 in Problem Set 22.2. 2Xl + 1Ox2 ;§; 5, SEC. 22.4 Simplex Method: 9. Maximize f 34xI = + 2X2 + Xl + X2 + 5X3 X3 ~ 8Xl ~ + 947 Difficulties + 29x2 54. 3xI + 32x3 8X2 + (b) Write a program for maximizing::: in R. subject to 59. 2X3 ~ 39. = {{IXI + 1I2X2 (el Write a program for maximi7ing ::: = {{lXI + ... + lInX" subject to linear constraints. 10. CAS PROJECT. Simplex Method. (a) Write a program for graphing a region R in the first quadrant (d) Apply your programs to problems in this problem of the xlx2-plane determined by linear constraints. set and the previous one. 22.4 Simplex Method: Difficulties We recall from the last section that in the simplex method we proceed stepwise from one basic feasible solution to another, thereby increasing the value of the objective function f until we reach an optimal solution. Occasionally (but rather infi'equently in practice). two kinds of difficulties may occur. The first of these is degeneracy. A degenerate feasible solution is a feasible solution at which more than the usual number 11 - 111 of variables are zero. Here 11 is the number of variables (slack and others) and 111 the number of constraints (not counting the '~i ~ 0 conditions). Tn the last section, 11 = 4 and 111 = 2, and the occurring basic feasible solutions were nondegenerate; n - 111 = 2 variables were zero in each such solution. In the case of a degenerate feasible solution we do an extra elimination step in which a basic variable that is zero for that solution becomes nonbasic (and a nonbasic variable becomes basic instead). We explain this in a typical case. For more complicated cases and techniques (rarely needed in practice) see Ref. [F5J in App. I. E X AMP L E 1 Simplex Method, Degenerate Feasible Solution AB Steel. Inc .. produces two kinds of iron 11' '2 by using three kinds ot raw matenal R I . R2• R3 (scrap iron and two kind, of ore) as shown. Maximi7e the daily profit. Raw Material Needed per Ton Raw Material [ron 11 iron 12 2 I Rl R2 R3 Raw Material Available per Day (tons) 16 I 8 3.5 0 Net profit per ton $150 $300 Let Xl and X2 denote the amount (in tons) of iron '} and 12 , respectively. produced per day. Then our problem is as follows. Maximize SoIZlti01l. ;: = I(x) = subject to the constraints Xl ~ O. -'"2 ~ 150Xl + 300X2 0 and (raw material R I ) (raw material R 2 ) (raw material R3 ). 948 CHAP. 22 Unconstrained Optimization. Linear Programming By introducing slack variable~ X3' X4. X5 we obtain the normal form of the con~tr.lints 16 8 (1) Xi;;;; I. .... 5). (i = 0 As in the last section we obtain from (\ ) and (2) the initial simplex tahle b [ -~l-~~Q-~J~O-l-~---~---~-r~~-l To = (3) I I o: o II : 0 I I 0 0 I . 0 : Thi~ = 1611 = 16. is 0: (0.0) in Fig. 474. We have 11 X5 = 5 variables Xj' m 3.5 = x2 = 0 we have frolll (3) the basic We see that xl. x2 are nonbasic variables and x3' x4' x!) are basic. With Xl feasible solution X3 H I I 0 = 3.511 = 3.5. = 3 constraints, and 11 - 111 ;: = O. = 2 variables equal to zero in our ",Iution. which thus is nondegenerate. Step 1 of Pivoting Operatioll 0 1: Column Selection of Pivot. Column 2 hince -150 < 0). Operatioll O 2: Row Selection of Pivot. 1612 = H, 811 = 8; 3.510 is not possible. Hence we could choose Row 2 or Row 3. We choose Row 2. The pivot is L Operation 0 3: Elimination by Row Operations. This gives the simplex table Xl X3 -"2 X4 b X5 ------------------------------75 0 I 0 -225 0 (4) Tl = [~ 0 2 0 0 1 2 1 0 -2 0 0 0 I I I I I I I I We see that the basic variables are X], X4.'\5 and the nonbasic are x2' we obtain from T] the basic feasible solution I~ 1 Rm, I '5 Rm\ Rm 4 Ro' 16 0 3.5 t3' Row 4 Setting the nonoosic variables to rero, SEC. 22.4 Simplex Method: 949 Difficulties Xl = 1611 = 8. X4 = 0/1 = O. X5 = 3.5/1 = 3.5. ~ = 1200 This is A: (8. 0) in Fig. 474. This solution in degenerate because X4 = 0 (in addition to X2 = O. x3 = 0): geometrically: the straight line x4 = 0 also pa"es through A. This requires the next step. in which x4 will become nonbasic. Step 2 of Pivoti1lg Operatioll 0 1 : Column Selection of Pivot. Column 3 (since -225 < 0). Operatioll O 2: Row Selection of Pivot. 16/1 = 16.0/! = O. Hence! must serve as the piVOL Operatioll 0 3 : Elimination by Row Operations. This gives the following simplex table. h ~ _:_ Q- - - .9_:_-::1JQ. - - ~-?.O- - - - Q. _~!~Q_] Ro", o : 0 : Row (5) 2 I 2 I [ 0: 0 !: o : 0 : 0 -2 0: I() I .! Rm . I -! I 0: - 2 : 0 3.5 Row -I- - 2 Row .• We see that the basic variables are Xl. x2. x5 and the nonbasic are x3' x4' Hence x4 has become nonbasic. as intended. By equating the nonbasic variables to /ero we obtain from T 2 the basic feasible solution Xl = 16/2 = 8. -'2 = O/! = o. -'5 = 3.511 = 3.5. ;: = 1200. This is still A: (8. 0) in Fig. 474 and;: has not increa,ed. But this opens the way to the maximum. which we reach in the next step. Step J of Pivoting Operatioll 0 1 : Column Selection of Pivot. Column 4 (since -150 < 0). Operatioll O 2: Row Selection of Pivot. 16/2 = 8. O/(-!) = O. 3.511 = 3.5. We crultakc I as the pivot. (With -! as the pivot we would nm leave A. Try iL) Operatioll 0 3 : Elimination by Row Operations. This gives the simplex table b _1_ ~_ Q(6) [ o : I o II o : 2 0: 0 0 ! I I I 0 0: We see that basic variables are basic feasible solution Xl = - - Q- ~ _0__ J~O___ !?Q - ~ J 2?J__ ] 2 Xl' X2' X3 0 2 -2: 0 1 -2 -2 and non basic X4' 912 = 4.5. X3 = X5' 9 I I I 1.75 : 3.5 Row 150 Row-l- Row 2 Row-l- Equating the latter to zero we obtain from T3 the 3.5/1 = 3.5. ;: = 1725. This is B: (-1-.5.3.5) in Fig. 474. Since Row of T3 has no negative entries. we have reached the maximum daily profit ::max = f(4.5. 3.5) = 150' 4.5 + 300' 3.5 = SI725. This is obtained by u~ing 4.5 tons of iron h ~U~cl~~ • Difficulties in Starting As a second kind of difficulty, it may sometimes be hard to find a basic feasible solution to start from. In such a case the idea of an artificial variable (or several such variables) is helpful. We explain this method in terms of a typical example. 950 E X AMP L E 2 CHAP. 22 Unconstrained Optimization. Linear Programming Simplex Method: Difficult Start, Artificial Variable Maximize (7) subject to the constraints Xl Solutioll. ~ O. X2 ~ 0 and (Fig. 475) By means of ~Iack variables we achieve the normal foml of the constrrnnts =0 = 1 =2 (8) Xi ~ 0 (i = I, .... 51. Note that the first slack variable is negative (OT zero). which makes \3 nonnegative within the (and negative outside). From (7) and (8) we obtain the simplex table fea~ibilit) region b r--L- -~:-l- -!o I I o II ro : I -2 I -I _Q ___<L __ ~_ T-~-1 I I I I -1 0 0 0 I 0 : 0 0 I 1 I . I :2 I : 4 are nonbasic. and we would like to take x3. '\4. X5 as basic variables. By our usual process of equating the nonbasic variables to zero we obtain from this table Xl. X2 Xl = O. .1."3 = 1I(-l) = -1, X4 = 211 = 2. "5 = 4/1 = 4 . ~ = O. "3 < 0 indicates that (0, 0) lies outside the feasibility region. Since X3 < O. we cannot proceed immediately. Now. instead of searching for other basic variables. we use the following idea. Solving the second equation in (8) for .1."3. we have To this we now add a variable X6 on the right. B 2 f=7 \ \ \ \ \ ~ C o ,------A, o Fig. 475. 1 ~ 2 3 Xl Feasibility region in Example 2 SEC. 22.4 Simplex Method: Difficulties 951 (9) is called an artificial variable and is subject to the constraint X6 ~ O. We must take care that x6 (which is not part of the given problem!) will disappear eventually. We shall see that we can accomplish this by adding a term -MX6 with very large M to the objective function. Because of (7) and (9) (solved for x6) this gives the modified objective function for this "extended problem" X6 (10) We see that the simplex table corresponding to (10) and (8) is b \"1 I L I -2 - M -I +!M __ ______________ MOO 0 _________________ -! : -I -I I I I 0 0: : 0 I I o: I oI To = I ~ I -! oI I 0 0 0: 0 0 0 0 - I I ~ -M_ ___ I I I 2 0: 4 I 0 I The last row of this table results from (9) written as Xl - !X2 - x3 + x6 = 1. We see that we can now start. taking x4. x5. X6 as the basic variables and Xl' X2. X3 a, the nonbasic variables. Column 2 has a negative first entry. We can take the second entry (1 in Row 2) as the pivot. This gives b I I 0 -2 o : I -!: - I -2 0 0 0 I 2 0 0 0: I I ---,-------,------------------T--I T1 = I -!: I 0: 0 0 0: 0 O:O~: 0:3 I 10 I I 01001000 This corresponds to Xl = 1,.1"2 = 0 (point A in Fig. 475). x3 = 0, -"4 = I, Row 5 and Column 7. In this way we get rid of \"6' as wanted. and obtain X5 = 3. X6 = O. We can now drop b -~ -t -~- -~! -t- ~~ ---~ ---~-l- ~-j r I I I 0 _1 I 2 I I I 3 I 01021 o 1 I 0 0 I I I I I 13 In Column 3 we choose 312 as the next pivot. We obtain b This corresponds to Xl = 2. -'"2 = 2 (this is B in Fig. 475). X3 as the pivot, by the usual principle. This gives = O. X4 = 2, X5 = O. In Column 4 we choose 4/3 CHAP. 22 952 Unconstrained Optimization. Linear Programming b [ This cone5ponds to Xl = 3. -~-1-~----~-t-~---!---!-1-~-1 I I 0:0 O:~ o X2 = I I 0 3 I "I I . ~:2 1 3 I 3 "I" 3 () 1 (point C in Fig. 475). -" -"3 = ~. x4 = O. x5 = O. This is the maximum fmax = I(3. 1) = 7. -............... _..... -.•........ • 2E4 ~ If in a step you have a choice between pivots. take the one that comes first in the column considered. 1. Maximize:;; = fleX) = 6xl + 12x2 subject to o ~ Xl ::0; 4. 0::0; X2 ~ 4. 6x1 + 12x2 ::0; 72. 2. Do Prob. I with the last two constraints interchanged. 3. Maximi7e the daily output in producing Xl glass plates by a process PI and X2 glass plates by a process P2 subject to the constraints (labor hours. machine hours, raw material supply) to input constraints /limitation of machine time) 4Xl 8X1 + 2X2 + 4X2 X2 ;;; 4. Maximize:;; = 300.\"1 + 500.\"2 subject to 2Xl + 8X2 ~ 60. 2X1 + .\"2 ~ 30. 4XI + 4.\"2 ~ 60. 5. Do Prob. 4 with the last two constraints interchanged. Comment on the resulting simplification. 6. Maximize the total output f = Xl + X2 + X3 (production figures of three different production processes) subject 5X2 + + 8X3 4X3 ::0; 12, ::0; 12. ::0; 40. 9. Maximize f ::0; 140. 5X2 7. Maximize f = 6.\"1 + 6.\"2 + 9.\"3 subject to .\"j ;;; 0 (j = I. .... 5), and Xl + .\"3 + X4 = 1. X2 + .\"3 + \"5 = I. 8. Using an artificial variable. minimize f = 2x 1 - X2 subject to Xl ;;; O. X2 ;;; 0, X1 + X2 ;;; 5, -Xl + X2 ~ I, 5XI 4Xl + + O. X3 = ::0; 0, + X2 + + X2 + x3 4Xl 2X3 Xl ::0; 1. subject to Xl ;;; Xl + X2 .\"3 o. ~ O. 10. If one uses the method of artificial variables in a problem without solution. this nonexistence will become apparent by the fact that one cannot get rid of the artificial variable. TIIustrate this by trying to maximize f = 2Xl + X2 subject to Xl ;;; 0, X2 ;;; 0, 2Xl + X2 ::0; 2, Xl + 2X2 ;;; 6, Xl + X2 ~ 4. ==::,.:.,::.===:=:--:e-".;::&':==.:U S T ION SAN D PRO B L EMS 1. What is the difference between constrained and unconstrained optimization? 2. State the idea and the basic formulas of the method of steepest descent. 3. Write down an algorithm for the method of steepest descent. 4. Design a "method of steepest ascent" for determining maxima. 5. What i~ linear programming? objective function? lt~ ba~ic idea? An 6. Why can we not use methods of calculus for extrema in linear programming? f(x) = X1 2 + 1.5X22. starting from (6. 3). Do 3 steps. Why is the convergence faster than in Example 1. Sec. 22.1? 9. What does the method of steepe~t de~cent amount to in the case of a single variable? 10. In Prob. 8 start from Xo = [1.5 I ]T. Show that the next even-numbered approximations are X2 = kXo. X4 = k 2xo. etc .. where k = OJl4. 11. What happens in Example I of Sec. 22.1 if you replace the function fex) = X 12 + 3X22 by fex) = x 12 + 5X22? Do 5 steps, starting from Xo = [6 3]T. Is the convergence faster or slower? 7. Whar are slack variables? Artificial variables? Why did we use them'? 12. Apply the method of steepest descent to f(x) = 9X12 + X2 2 + 18.\'1 - 4X2, 5 steps. starting from Xo = [2 4]T. 8. Apply the method of steepest descent to 13. In Prob. 12, could you start from [0 O]T and do 5 steps? Summary of Chapter 22 953 121-25] 14. Show that the gradients in Prob. 13 are orthogonal. Give a reason. Xl 115-20 1 Graph or sketch the region in the first quadrant of the X1x2-plane detelwined by the following inequalities. 15. 17. Xl + 2X1 + 3X2 ~ 6 X2 ~ 4 -Xl + X2 ~ 0 XI + X2 ~ 4 16. 0.8X1 18. 2X2 ~ Xl - Xl - + X2 2X1 -4 2xI + X2 ~ 12 XI + X2 ~ 8 Xl + X2 ~ 2 3X2 ~ -12 ~ 15 f = ~ + X2 f ~ + x2 -XI + 20. X2 ~ 5 X2 ~ 3 2X1 - 2 Xl + X2 ~ Xl X2 20X2 subject to Xl ~ 5. + ~ X2 subject to Xl + 2X2 ~ + + 40X2 ~ 1800 (Machine hours), 20X2 ~ 6300 (Labor). 25. Maximize the daily output in producing XI chairs by a process PI and X2 chairs by a process P 2 subject to 3x 1 + 4X2 ~ 550 (machine hours), 5xI + 4X2 ~ 650 (labor). --:!:.,: .'fI.=.:.,=:~"'=="==:._,=:':==:=' Unconstrained Optimization. Linear Programming In optimization problems we maximize or minimize an objective function.: = f(x) depending on control variables Xl' ••• , X'" whose domain is either unrestricted ("unconstrained optimization," Sec. 22. I) or restricted by constraints in the fonn of inequalities or equations or both ("constrained optimization," Sec. 22.2). If the objective function is linear and the constraints are linear inequalities in X10 ••• , -'""" then by introducing slack variables X m +l, . • . , Xn we can write the optimization problem in normal form with the objective function given by (I) (where c m + 1 = ell 10. 4. ~ 200X1 XI = LO. + 4. = 2X1 IOx2 subject to Xl - X2 ~ 4. 14. Xl + X2 ~ 9. -Xl + 3x2 ~ 15. 24. A factory produces two kinds of gaskets, G I • G 2 • with net profit of $60 and $30. respectively. Maximize the total daily profit subject to the constraints (Xj = number of gaskets Gj produced per day) 2xI 40X1 19. lOx. X2 ~ 6. 23. Minimize f 6 2x2 ~ + X2 22. Maximize -2 ~ Maximize or minimize as indicated. 21. Maximize = 0) and the constraints given by (2) Xl ~ o... '. x., ~ O. [n this case we can then apply the widely used simplex method (Sec. 22.3), a systematic stepwise search through a very much reduced subset of all feasible solutions. Section 22 4 shows how to overcome difficulties with this method t --\ ... ~ 23 ---, C HAP T E R Graphs. Combinatorial Optimization Graphs and digraphs (= directed graphs) have developed into powerful tools in areas, such as electrical and civil engineering, communication networks, operations research, computer science, economics, industrial management, and marketing. An essential factor of this growth is the use of computers in large-scale optimization problems that can be modeled by graphs and solved by algorithms provided by graph theory. This approach yields models of general applicability and economic imporlance. II lies in the center of combinatorial optimization, a term denoting optimization problems that are of pronounced discrete or combinatorial structure. This chapter gives an introduction to this wide area, which constitutes a shift of emphasis away from differential equation~. eigenvalues, and so on, and is full of new ideas as well as open problems-in connection, for instance, with efficient computer algorithms. The classes of problems we shall consider include transportation of minimum cost or time, best assignment of workers to jobs, most efficient use of communication networks, and many others. Problems for these classes often form the core of larger and more involved practical problems. Prerequisite: none. References and Answers 10 Problems: App. I Parl F, App. 2. 23.1 Graphs and Digraphs Roughly, a graph consists of points, called vertices, and lines connecting them, called edges. For example, these may be four cities and five highways connecting them, as in Fig. 476. Or the points may represent some people, and we connect by an edge those who do business with each other. Or the vertices may represent computers in a network and the edges connections between them. Let us now give a fonnal definition . ./JLOOP ~ flsolated \ ......------... " vertex '-------"" Double edge Fig. 476. Graph consisting of 4 vertices and 5 edges 954 Fig. 477. Isolated vertex, loop, double edge. (Excluded by definition.) SEC. 23.1 955 Graphs and Digraphs DEFINITION Graph A graph G consists of two finite sets (sets having finitely many elements), a set V of points, called vertices, and a set E of connecting lines, called edges, such that each edge connects two vertices, called the endpoints of the edge. We write G = (V, E). Excluded are isolated vertices (vertices that are not endpoints of any edge), loops (edges whose endpoints coincide), and lIlultiple edges (edges that have both endpoints in common. See Fig. 477. CAUTION! Our three exclusions are practical and widely accepted, but not uniformly. For instance, some authors permit multiple edges and call graphs without them simple graphs. • We denote vertices by letters, Lt, v, ... or VI' V 2 , .•• or simply by numbers 1,2, ... (as in Fig. 476). We denote edges by el, e2, ... or by their two endpoints; for instance, el = (1, 4), e2 = (1, 2) in Fig. 476. An edge (Vi' V) is called incident with the vertex Vi (and conversely); similarly, (Vi' Vj) is incident with Vj. The number of edges incident with a vertex V is called the degree of v. Two vertices are called adjacent in G if they are connected by an edge in G (that is, if they are the two endpoints of some edge in G). We meet graphs in ditlerent fields under different names: as "networks" in electrical engineering, "structures" in civil engineering, "molecular structures" in chemistry, "organizational structures" in economics, "sociograms," "road maps," "telecommunication networks," and so on. Digraphs (Directed Graphs) Nets of one-way streets, pipeline networks, sequences of jobs in construction work, flows of computation in a computer, producer-consumer relations, and many other applications suggest the idea of a "digraph" (= directed graph), in which each edge has a direction (indicated by an arrow, as in Fig. 478). Fig. 478. DEFINITION Digraph Digraph (Directed Graph) A digraph G = (V, E) is a graph in which each edge e its "initial point" i to its "terminal point" j. = (i,j) has a direction from Two edges connecting the same two points i, j are now permitted, provided they have opposite directions, that is, they are (i,j) and (j, i). Example. (1,4) and (4, 1) in Fig. 478. 956 CHAP. 23 Graphs. Combinatorial Optimization A subgraph or subdigraph of a given graph or digraph G = (V. E), respectively, is a graph or digraph obtained by deleting some of the edges and vertices of G. letaining the other edges of G (together with their pairs of endpoints). For instance, el' e3 (together with the vertices I, 2. 4) form a subgraph in Fig. 476. and e3, e4, e5 (together with the vertices l. 3. 4) fonn a subdigraph in Fig. 478. Computer Representation of Graphs and Digraphs Drawings of graphs are useful to people in explaining or illustrating specific situations. Here one should be aware that a graph may be <;ketched in various ways; see Fig. 479. For handling graphs and digraphs in computers. one uses matrices or lists as appropriate data stmctures. as follows. (a) (b) Fig. 479. Different sketches of the same graph Adjacency Matrix of a Graph G: {/ij = (c) Matrix A = {Io [{lU] with entries if G has an edge (i, j), else. Thus £Iij = 1 if and only if two veItices i and j are adjacent in G. Here, by definition, no vertex is considered to be adjacent to itself; thus, {/i; = O. A is symmetric, {/ij = {lji- (Why?) The adjacency matrix of a graph is generally much smaller than the so-called illcidellce matrix (see Probs. 21, 22) and is preferred over the latter if one decides to store a graph in a computer in matrix form. EXAMPLE 1 Adjacency Matrix of a Graph Vertex Vertex I 2 3 4 Adjacency Matrix of a Digraph G: {/ij = {Io 2 [~ 3 4 0 0 0 J Matrix A = [aij] with entries if G has a directed edge (i, j), else. This matrix A is not symmetric. (Why?) • SEC. 23.1 957 Graphs and Digraphs E X AMP L E 2 Adjacency Matrix of a Digraph To vertex 2 From vertex I 2 3 4 3 4 0 [i 0 0 0 0 U ;] • Lists. The vertex incidence list of a graph shows for each vertex the incident edges. The edge incidence list shows for each edge its two endpuints. Similarly for a digraph; in the vertex list, outgoing edges then get a minus sign, and in the edge list we now have ordered pairs of vertices. E X AMP L E 3 Vertex Incidence List and Edge Incidence List of a Graph This graph is the same as in Example I. except for notation. Vertex Incident Edges Edge Endpoints VI el' e5 el VI, V 2 V2 el' C2, e3 C2 V 2 , V3 V3 e2' C4 C3 V 2 , V4 V4 C3 • C4 , e5 e4 V 3• V4 e5 VI. V 4 • "Sparse graphs" are graphs with few edges (far fewer than the maximum possible number n(n - I )/2. where n is the number of vertices). For these graphs. matrices are not efficient. Lists then have the advantage of requiring much less storage and being easier to handle; they can be ordered, sorted, or manipulated in various other ways directly within the computer. For instance, in tracing a "walk" (a connected sequence of edges with pairwise common endpoints), one can easily go back and furth between the two lists just discussed, instead of scanning a large column of a matrix for a single 1. Computer science has deVeloped more refined lists, which, in addition to the actual content, contain "pointers" indicating the preceding item or the next item to be scanned or both items (in the case of a "walk": the preceding edge or the subsequent one). For details, see Refs. [E 16] and LF7]. This section was devoted to basic concepts and notations needed throughuut this chapter, in which we shall discuss some of the most impurtant classes of combinatorial optimization problems. This will at the same time help us to become more and more familiar with graphs and digraphs. 958 CHAP. 23 -.•.--..-- ..... -.... --~-- Graphs. Combinatorial Optimization .. ~.-. ........ j--. .... - -.....-. -- 1. Sketch the graph consisting of the vertices and edges of a square. Of a tetrahedron. 2. Worker WI can do jobs 11 and 13 , worker W2 job 14 , worker W3 jobs 12 and J 3 . Represent this by a graph. 3. Explain how the following may be regarded as graphs or digraphs: flight connections between given cities: memberships of some persons in some committees; relations between chapters of a book: a tennis tournament; a family tree. 4. How would you represent a net of one-way and two-way streets by a digraph? 5. Give further examples of situations that could be represented by a graph or digraph. 6. Find the adjacency matrix of the graph in Fig. 476. Sketch the graph whose adjacency matrix is: 0 0 14. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 16. ADJACENCY MATRIX Find the adjacency matrix of the graph or digraph. 8. 0 15. 7. When will the adjacency matrix of a graph be symmetric? Of a digraph? 18-131 0 0 Sketch the digraph whose adjacency matrix is: 17. The matrix in Prob. 14. 10. 18. The matrix in Prob. 16. 19. (Complete graph) Show that a graph G with 11 vertices can have at most 11(11 - 1)/2 edges, and G has exactly n(1l - I )12 edges if G is complete, that is, if every pair of vertices of G is joined by an edge. (Recall that loops and multiple edges are excluded.) 20. In what case are all the off-diagonal entries of the adjacency matrix of a graph G equal to I? 11. Incidence Matrix of a Graph: Matrix B = [bjk ] with entries if vertex j is an endpoint of edge ek otherwise. Find the incidence matrix of: 21. The graph in Prob. 9. 12. 22. The graph in Prob. 8. Incidence Matrix of a Digraph: Matrix B = entries a;jk] with if edge ek leaves vertex j if edge ek enters vertex j 13. othelwise. Find the incidence matrix of: 23. The digraph in Prob. II. 24. The digraph in Prob. 13. 25. Make a vertex incidence list ofthe digraph in Prob. 13. SEC. 23.2 Shortest Path Problems. 23.2 959 Complexity Shortest Path Problems. Complexity Beginning in this section, we shall discuss some of the most important classes of optimization problems that concern graphs and digraphs as they arise in applications. Basic ideas and algorithms will be explained and illustrated by small graphs, but you should keep in mind that real-life problems may often involve many thousands or even millions of vertices and edges (think of telephone networks, worldwide air travel, companies that have offices and stores in all larger cities). Then reliable and efficient systematic methods are an absolute necessity-solution by inspection or by trial and error would no longer work, even if "'nearly optimal" solutions are acceptable. We begin with shortest path problems, as they arise, for instance, in designing shortest (or least expensive, or fastest) routes for a traveling salesman. for a cargo ship. etc. Let us first explain what we mean by a path. In a graph G = (V, E) we can walk from a vertex VI along some edges to some other vertex vk- Here we can (A) make no restrictions. or (B) require that each edge of G be traversed at most once, or (C) require that each vertex be visited at most once. In case (A) we call this a walk. Thus a walk from VI to Vk is of the form (1) where some of these edges or vertices may be the same. In case (B), where each edge may occur at most once, we call the walk a trail. Finally, in case (C), where each vertex may occur at most once (and thus each edge automatically occurs at most once), we call the trail a path. We admit that a walk. trail. or path may end at the vertex it started from. in which case we call it closed; then Vk = VI in (1). A closed path is called a cycle. A cycle has at least three edges (because we do not have double edges; see Sec. 23.1). Figure 480 illustrates all these concepts. Walk, trail, path, cycle Fig. 480. 14 11- 21 22- 3233- 2 is 34 4- a walk (not a trail). 4 - 5 is a trail (not a path). 5 is a path (not a cycle). 1 is a cycle. Shortest Path To define the concept of a shortest path, we assume that G = (V, E) is a weighted graph, that is, each edge (Vi, V) in G has a given weight or length lij > O. Then a shortest path VI ----;. Vk (with fixed VI and Vk) is a path (1) such that the sum of the lengths of its edges 112 (/12 = length of VI to Vk)' + 123 + 134 + ... + Ik-I,K (Vb V 2 ), etc.) is minimum (as small as possible among all paths from Similarly, a longest path VI ----;. Vk is one for which that sum is maximum. 960 CHAP. 23 Graphs. Combinatorial Optimization Shortest (and longest) path problems are among the most important optimization problems. Here, "length" lij (often also called "cost" or "weight") can be an actual length measured in miles or travel time or gasoline expenses, but it may also be something entirely different. For instance, the "traveling salesman problem" requires the determination of a shortest Hamiltonian i cycle in a graph, that is, a cycle that contains all the vertices of the graph. As another example, by choosing the "most profitable" route VI --4 Vk, a salesman may want to maximize "Llij , where lij is his expected commission minus his travel expenses for going from town i to townj. In an investment problem, i may be the day an investment is made,j the day it matures, and lij the resulting profit, and one gets a graph by considering the various possibilities of investing and reinvesting over a given period of time. Shortest Path if All Edges Have Length I = 1 Obviously, if all edges have length I, then a shortest path VI - Vk is one that has the smallest number of edges among all paths VI - Vk in a given graph G. For this problem we discuss a BFS algorithm. BFS stands for Breadth First Search. This means that in each step the algorithm visits all neighhnring (all adjacent) vertices of a vertex reached. as opposed to a DFS algorithm (Depth First Search algorithm), which makes a long trail (as in a maze). This widely used BFS algorithm is shown in Table 23.1. We want to find a shortest path in G from a vertex s (start) to a vertex t (terminal). To guarantee that there is a path from s to t, we make sure that G does not consist of separate portions. Thus we assume that G is connected, that is, for any two vertices V and w there is a path V _ w in G. (Recall that a vertex V is called adjacent to a vertex 11 if there is an edge (u, v) in G.) Table 23.1 Moore's BFS for Shortest Path (All Lengths One) Proceedings of the Illfemationai Symposillm for Switching TIleory. Part II. pp. 285-292. Cambridge: Harvard University Press. 1959. ALGORITHM MOORb [G = (V, E), s, t] This algorithm determines a shortest path in a connected graph G s to a vertex t. INPUT: OUTPUT: 1. 2. 3. 4. 5. = (V, E) from a vertex Connected graph G = (V, E), in which one vertex is denoted by sand one by t, and each edge (i, j) has length lij = 1. Initially all vertices are unlabeled. A shortest path s ---+ t in G = (V. £) Label s with O. Set i = O. Find all unlabeled vertices adjacent to a vertex labeled i. Label the vertices just found with i + l. If vertex t is labeled. then "backtracking" gives the shortest path k (= label of t), k - 1, k - 2, ... , 0 OUTPUT k, k - 1, k - 2, ... , O. Stop Else increase i by I. Go to Step 3. End MOORE lWILLIAM ROWAN HAMILTON (1805-1865), Irish mathematician, known for his work in dynamics SEC. 23.2 Shortest Path Problems. E X AMP L E 1 961 Complexity Application of Moore's BFS Algorithm Find a shortest path s ---+ t in the graph G shown in Fig. 481. Solution. Figure 481 shows the labels. The blne edges form a shortest path (length 4). There is another shortest path s ---+ t. (Can yon find it?) Hence in the program we must introduce a rule that makes backtracking unique because otherwise the computer would not know what to do next if at some step there is a choice (for instance, in Fig. 481 when it got back to the vertex labeled 2) The following rule seems to be natural. Backtracking rule. Using the numbeIing of the vertices from I to II (not the labeling'). at each step, if a vertex labeled i i, reached, take as the next ve11ex that with the smallest number (not label!) among all the • vertices labeled i - I . 2 Fig. 481. Example 1, given graph and result of labeling Complexity of an Algorithm Complexity of Moore's algorithm. To find the vertices to be labeled 1, we have to scan all edges incident with s. Next, when i = 1, we have to scan all edges incident with vertices labeled I, etc. Hence each edge is scanned twice. These are 2m operations (Ill = number of edges of G). This is a function c(m). Whether it is 2171 or 5111 + 3 or 12m is not so essential; it is essential that c(m) is proportional to 111 (not nz 2 , for example); it is of the "order" 1II. We write for any function alll + b simply Oem), for any function am 2 + bm + d simply 0(11/ 2 ), and so on; here, 0 suggests order. The underlying idea and practical aspect are as follows. lnjudging an algorithm, we are mostly interested in its behavior for very large problems (large m in the present case), since these are going to determine the limits of the applicability of the algorithm. Thus, the essential item is the fastest growing term (am 2 in alll 2 + bill + d, etc.) since it will overwhelm the others when III is large enough. Also, a constant factor in this term is not very essential; for instance, the difference between two algorithms of orders. say, 5111 2 and 8m 2 is generally not very essential and can be made irrelevant by a modest increase in the speed of computers. However, it does make a great practical difference whether an algorithm is of order 11/ or m 2 or of a still higher power lII P • And the biggest difference occurs between these "polynomial orders" and "exponential orders," such as 2'11. For instance, on a computer that does I 09 operations per second. a problem of size m = 50 will take 0.3 second with an algorithm that requires /715 operation~, but 13 days with an algorithm that requires 2m operations. But this is not our only reason for regarding polynomial orders as good and exponential orders as bad. Another reason is the gain in using afaster computer. For example let two algorithms be Oem) and 0(111 2 ). Then, since 1000 = 31.62 , an increase in speed by a factor 1000 has the effect that per hour we can do problems 1000 and 31.6 times as big, respectively. But since 1000 = 2 9 .97 • with an algorithm that is 0(2"'), all we gain is a relatively modest increase of 10 in problem size because 2 9 . 97 • 2 m = 2>n+9.97. 962 CHAP. 23 Graphs. Combinatorial Optimization The symbol 0 is quite practical and commonly used whenever the order of growth is essential. but not the specific form of a function. Thus if a function g(m) is of the form gem) = kh(m) + more slowly growing terms (k *" 0, constant), we say that g(m) is of the order h(111) and write g(m) = O(lZ(111)). For instance, am + b = 5 . 2m 0(111). + 3m 2 = 0(21n). We want an algorithm .iI to be "efficient." that is. "good" with respect to (i) Time (number C.,l(111) of computer operations), or (ii) Space (storage needed in the internal memory) or both. Here c. J suggests "complexity" of .iI. Two popular choices for cjm) = longest time 9'1 takes for a problem of size (Worst case) (Average case) C~'I are /11, c.cim) = average time sl takes for a problem of size 111. In problems on graphs, the "size" will often be 11l (number of edges) or 11 (number of vertices). For our present simple algorithm, cJm) = 2m in both cases. For a "good" algorithm.iI, we want that c.jm) does not grow too fast. Accordingly, we call .cJ1 efficient if C,,'1(111) = O(mk) for some integer k ~ 0; that is, C". I may contain only powers of m (or functions that grow even more slowly, such as In m), but no exponential functions. Furthermore, we call sJ polynomially bounded if .9l is efficient when we choose the "worst case" C,,«111). These conventional concepts have intuitive appeal. as our discussion shows. Complexity should be investigated for every algorithm, so that one can also compare different algorithms for the same task. This may often exceed the level in this chapter; accordingly, we shall confine ourselves to a few occasional comments in this direction. [1=6] SHORTEST PATH 5. Find a shortest path P: s --> t and its length by Moore's BFS algorithm; si--erch the graph with the labels and indicate P by heavier lines (as in Fig. 481). 1. 2't~\ \ ·V~ . . . . . /"-J s 4. / "':::) ' " 0 ~~-''b t/ 6. S /\_1-1- . . . . V\/_. /" __ t /\/~( ! sQ\ • • 7. (Nonuniqueness) A shonest path s --> t for given sand t need not be unique. Illustrate this by finding another shortest path s --> t in Example I in the text. 8. (Maximum length) If P is a shortest path between any two vertices in a graph with n vertices, how many edges can P at most have'? In a complete graph (with all edges of length 1)7 Give a reason. 9. (Moore's algorithm) Show that if a vertex v has label .A(v) = k, then there is a path s --> v of length k. SEC. 23.3 Bellman's Principle. 10. Call the length of a shortest path s ~ v the distance of v from s. Show that if v has distance /, it has label A(v) = 963 Dijkstra's Algorithm 16. Find 4 different closed Euler trails in Fig. 484. 2 I. /\/\ 11. (Hamiltonian cycle) Find and sketch a Hamiltonian cycle in the graph of Prob. 3. 12. Find and sketch a Hamiltonian cycle in the graph of a dodecahedron. which has 12 pentagonal faces and 20 vertices (Fig. 482). This is a problem Hamilton himself considered. Fig. 482. Problem 12 13. Find and sketch a Hamiltonian cycle Sec. 23.1. In Fig. 479. 14. (Euler graph) An EllIeI' graph G is a graph that has a clo:-.ed Euler trail. An Euler trail is a trail that contains every edge of G exactly once. Which subgraph with four edges of the graph in Example I, Sec. 23.1. is an Euler graph? 15. Is the graph in Fig. 483 an Euler graph? (Give a reason.) 4 3 Fig. 484. 5 Problem 16 17. The postman problem is the problem of finding a closed walk W: s ~ s (s the post office) in a graph G with edges (i,j) of length lij > 0 such that every edge of G is traversed at least once and the length of W is minimum. Find a solution for the graph in Fig. 483 by inspection. (The problem is also called the Chinese postman problem since it was published in the journal Chinese MathenlOtic.I' 1 (1962),273-277.) 18. Show that the length of a shortest postman trail is the same for every starting verteX. 19. (Order) Show that 0(111 3 ) kO(111 P ) = + 0(111 3 ) = 0(111 3 ) and O(m P )' 20. Show that ~ = 0(111), O.02e m + 100m2 = O(e m ). 21. If we switch from one computer to another that is 100 times as fast. what is our gain in problem size per hour in the use of an algorithm that is 0(111), 0(111 2 ). 0(111 5 ). O(e"")? 4 3}-------{ Fig. 483. 23.3 Problems 15, 17 Bellman's Principle. 22. CAS PROBLEM. Moore's Algorithm. Write a computer program for the algorithm in Table 23.1. Test the program with the graph in Example I. Apply it to Probs. 1-3 and to some graphs of your own choice. Dijkstra's Algorithm We continue our discussion of the shorrest path problem in a graph G. The last section concerned the special case that all edges had length 1. But in most applications the edges (i, j) will have any lengths lij > 0, and we now turn to this general case, which is of greater practical importance. We write lij = :x; for any edge (i,j) that does not exist in G (setting 'Xl + a = :x; for any number a, as usual). We consider the problem of finding shortest paths from a given veltex. denoted by I and called the origin, to all other vel1ices 2. 3 ..... Il of C. We let Lj denote the length of a shortest path p/ I ~ j in G. THEOREM 1 Bellman's Minimality Principle or Optimality Principle2 {l Pj : I ~ j is a ShOrTeST path from 1 TO j ill G lllld (i. j) is tlze llist edge of Pj (Fig. 485), Then Pi: 1 ~ i [obtained by droppi1lg (i, j) from Pj ] is a sh017est path I~i. 2RICHARD BELLMAN (1920--1984). American mathematician, known for his WOIX in dynamic programming. 964 CHAP. 23 Graphs. Combinatorial Optimization P.I ~ __________~A~__________~\ ~ "'/-v~ Fig. 485. PROOF j Paths P and Pi in Bellman's minimality principle Suppose that the conclusion is false. Then there is a path P;"': I _ i that is shorter than Pi' Hence if we now add U. j) to Pi*, we get a path I _ j that is shorter than Pj' This • contradicts our assumption that Pj is "hortest. From Bellman's principle we can derive basic equations as follows. For fixed j we may obtain various paths 1 _ j by taking shortest paths Pi for vmious i for which there is in G an edge (i,j), and add U,.i) to the conesponding Pi' These paths obviously have lengths Li + lij (Li = length of Pi)' We can now take the minimum over i. that is. pick an i for which Li + lij is smallest. By the Bellman principle. this gives a shortest path I ~ j. It has the length Ll = 0 (1) L·:J = min (L· i*j 1, + I) 1,J' j = 2, ... , fl. These are the Bellman equations. Since Iii = 0 by definition, instead of mini'1'j we can simply write mini' These equations suggest the idea of one of the best-known algorithms for the shortest path problem, as follows. Dijkstra's Algorithm for Shortest Paths Dijkstra's3 algorithm is shown in Table 23.2, where a connected graph G is a graph in which for any two vertices v and 1I' in G there is a path v _ w. The algorithm is a labeling procedure. At each stage of the computation. each vertex v gers a label, either (PU a permallent label = length Lv of a shortest path 1 ----7 V or (TL) a temporary label = upper bound Lv for the length of a shortest path 1 ~ v. We denote by '!P;£ and 2J;£ the sets of vertices with a permanent label and with a temporary label, respectively. The algorithm has an initial step in which vertex I gets the permanent label Ll = 0 and the other vertices get temporary labels, and then the algorithm alternates between Steps 2 and 3. In Step 2 the idea is to pick k "minimally." In Step 3 the idea is that the upper bounds will in general improve (decrease) and must be updated accordingly. Namely, the new temporary label I j of vertex j will be the old one if there is no improvement or it will be Lk + Ikj if there is. 3 EDSGER WYBE DIJKSTRA (1930-2002), Dutch computer scientist. 1972 recipient of the ACM TurinG" Award. His algorithm appeared in Numerische Mathematik J (1959),269-271. b SEC. 23.3 Bellman's Principle. 965 Dijkstra's Algorithm Table 23.2 Dijkstra's Algorithm for Shortest Paths ALGORITHM DIJKSTRA [G = (V, E), V = {L ... , 17}, lij for all (i, j) in E] Given a connected graph G = (V, i',") with vertices 1, ... , n and edges (i, j) having lengths liJ > 0, this algorithm determines the lengths of shortest paths fi'om vertex I to the vertices 2, ... , n. INPUT: Number of vertices 17, edges (i, j), and lengths lij OUTPUT: Lengths Lj of sh011est paths I -? j, j = 2, ... , n 1. lnitiul step Vertex I gets PL: LI = O. Vertexj (= 2, .. ',1/) gets TL: ~ = Ilj (= Set qp5£ = {I}, 'j5£ = {2, 3, ... , n}. C/J if there is no edge (1,j) in G). 2. Fixing a permanent label Find a k in 'j5£ for which Lk is miminum, set Lk = L k . Take the smallest k if there are several. Delete k from 'j5£ and include it in f!P5£. If 'j5£ = 0 (that is, 'j5£ is empty) then OUTPUT L 2 , Ln. Stop ••. , Else continue (that is, go to Step 3). 3. Updating tempnral}" labels For all j in 'j5£, set L j ~ mink {L j , Lk Lk + I kj as your new L j ). + lk.i} (that is, take the smaller of L j and Go to Step 2. End DIJKSTRA E X AMP L E 1 Application of Dijkstra's Algorithm Applying Dijkstra's algorithm to the graph in Fig. 4b6a, find shortest paths from vertex I to vertices 2, 3, 4. Solution. = 2. L3 = 3. L2 = L4 = 2. L2 = 1. Ll We list the steps and computations. 0, L2 = 8, La = 5, L4 = 7, [L 2 .L3 , L4 } = 5. k = 3. min min {8,L3 + /32J min [7, L3 min 3. L4 = 2. 4 = 7. {L2' L4 ) = min {7,L2 = + 134} = + min [6,7) /24) = = 6 = 2. = 7 min {8, 5 + IJ min [7, x} = = 6, k min [7,6 + '!J5f. = [2,3, 4} '!J5f. = {2, 4} 'If':£ '!J5f. = [4} '!J:£ = 0. 7 2} = {I, 2, 3], 'd''£= [1,2,3,4}, k = 4 Figure 486b shows the resulting shortest paths. of lengths (a) Given graph G Lz = 6. L3 = 5.4 = 7. (b) Shortest paths in G Fig. 486. Complexity. qp5f. = {I}, '!P5f.={1.3}, Dijkstra's algorithm is 0(n2). Example 1 • CHAP. 23 966 PROOF Graphs. Combinatorial Optimization Step 2 requires comparison of elements, first II - 2, the next time 11 - 3, etc., a total of (11 - 2)(11 - 1)/2. Step 3 requires the same number of comparisons, a total of (11 - 2)(11 - I )/2, as well m; additions, fIrst Il - 2, the next time 11 - 3, etc., dgain a [otal of 2 (ll - 2)(11 - I )/2. Hence the total number of operations is 3(11 - 2)(11 - 1)/2 = O(1l ) . • 1. The net of roads in Fig. 487 connecting four villages is to be reduced to minimum length. but so that one can still reach every village from every other village. Which of the roads should be retained? Find the solution (a) by inspection. (b) by Dijkstra's algorithm. ·t 5. 6. 18 Fig. 487. Problem 1 2-=7] DIJKSTRA'S ALGORITHM Find shortest paths for the following graphs. 2. 3. 8. Show that in Dijkstra's algorithm, for L" there is a path P: I ~ k of length L k . 9. Show that in Dijkstra's algorithm. at each instant the demand on storage is light (data for less than II edges) 10. CAS PROBLEM. Dijkstra's Algorithm. Write a program and apply it to Probs. 2--4. 23.4 Shortest Spanning Trees: Greedy Algorithm So far we have discussed shortest path problems. We now turn to a particularly important kind of graph. called a tree. along with related optimization problems that arise quite often in practice. By definition. a tree T is a graph that is connected and has no cycles. "Connected" was defined in Sec. 23.3; it means that there is a path fmm any vertex in T to any other veltex in T. A cycle is a path s ~ t of at least three edges that is closed (t = s); see also Sec. 23.2. Figure 488a shows an example. CAUTION! The terminology varies; cycles are sometimes also called circuits. SEC. 23.4 967 Shortest Spanning Trees: Greedy Algorithm A spanning tree T in a given connected graph G = (V, E) is a tree containing all the vertices of G. See Fig. 488b. Such a tree has 17 - 1 edges. (Proof?) A shortest spanning tree T in a connected graph G (whose edges (i, j) have lengths lij> 0) is a spanning tree for which 'Llij (sum over all edges of 7) is minimum compared to 'Llij for any other spanning tree in G. 17 Trees are among the most important types of graphs, and they occur in various applications. Familiat examples ate family trees and organization charts. Trees can be used to exhibit, organize, or analyze electrical networks, producer-consumer and other business relations, infonnation in database systems, syntactic structure of computer programs, etc. We mention a few specific applications that need no lengthy additional explanations. The set of shortest paths from vertex I to the vertices 2..... 17 in the last section forms a spanning tree. Railway lines connecting a number of cities (the vertices) can be set up in the form of a spanning tree, the "length" of a line (edge) being the construction cost, and one wants to minimize the total construction cost. Similarly for bus lines, where "length" may be the average annual operating cost. Or for steamship lines (freight lines), where "length" may be profit and the goal is the maximization of total profit. Or in a network of telephone lines between some cities, a shortest spanning tree may simply represent a selection of lines that connect all the cities at minimal cost. In addition to these examples we could mention others from distribution networks. and so on. We shall now discuss a simple algorithm for the problem of finding a shortest spanning tree. This algorithm (Table 23.3) is patticularly suitable for spatse graphs (graphs with very few edges: see Sec. 23.1). Table 23.3 Kruskal's Greedy Algorithm for Shortest Spanning Trees Proceedings of the American Mathematical Society 7 (1956), 48-50. ALGORITHM KRUSKAL [G = (V, E), lij for all (i,j) in EJ Given a connected graph G = (V, E) with edges (i.j) having length lij > O. the algorithm detelmines a shortest spanning tree Tin G. INPUT: Edges (i, j) of G and their lengths lij OUTPUT: ShOltest spanning tree T in G 1. Order the edges of G in ascending order of length. 2. Choose them in this order as edges of T, rejecting an edge only if it forms a cycle with edges already chosen. If 17 - I edges have been chosen. then OUTPUT T (= the set of edges chosen). Stop EndKRUSKAL CaJ A cycle Fig. 488. (b) A spanning tree Example of (a) a cycle, (b) a spanning tree in a graph 968 E X AMP L E 1 CHAP. 23 Graphs. Combinatorial Optimization Application of Kruskal's Algorithm Using Kmskars algorithm. we shall determine a shortest spanning tree in the gmph in Fig. 489. Fig. 489. Graph in Example 1 Solution. See Table 23.4. In some of the intennediate stages the edges chosen form a disconnected gmph (see Fig. 490); this is typical. We stop after n - I = 5 choices since a spanning tree has II - I edges. In our problem the edges chosen are in the upper part of the list. This is typical of problems of any ~ize: in general, • edges farther down in the list have a smaller chance of being chosen. Table 23.4 Edge Solution in Example 1 Length (3,6) (I. 2) (1,3) (4.5) (2.3) (3,4) (5.6) (2,4) 2 4 6 7 8 Choice 1st 2nd 3rd 4th Reject 5th 9 11 The efficiency of Kruskars method is greatly increased by Double Labeling of Vertices. Each vertex i carries 1I ri = Root of the subtree to which Pi = Predecessor of i = 0 for roots. Pi double label (ri, Pi), where i belongs, ill its subtree, This simplifies Rejecting. If (i. j) is next in the list to be considered, reject (i. j) if ri = ') (that is. i and j are in the same subtree. so that they are already joined by edges and (i. j) would thus create a cycle). If ri i= I} include (i, j) ill T. If there are several choices for ri, choose the smallest. If subtrees merge (become a single tree), retain the smallest root as the root of the new subtree. For Example I the double-label list is shown in Table 23.5. In storing it, at each instant one may retain only the latest double label. We show all double labels in order to exhibit the proces~ in all its stages. Labels that remain unchanged are nO[ listed again. Underscored are the two I' s that are the common root of vertices 2 and 3, the reason for rejecting the edge (2, 3). By reading for each vertex the latest label we can read from this list that I is the vertex we have chosen as a root and the tree is as shown in the last part of Fig. 490. SEC. 23.4 969 Shortest Spanning Trees: Greedy Algorithm 1 2 --"'I 3~ f 6 Second First Third Fig. 490. 4 3/ Fourth Choice process in Example 1 " Fifth This is made possible by the predecessor label that each vertex carries. Also, for accepting or rejecting an edge we have to make only one comparison (the roots of the two endpoints of the edge). Ordering is the more expensive part of the algorithm. It is a standard process in data processing for which various methods have been suggested (see Sorting in Ref. [E25] listed in App. 1). For a complete list of 111 edges, an algorithm would be Oem log211/), but since the 17 - 1 edges of the tree are most likely to be found earlier, by inspecting the q « 1/1) topmost edges, for such a list of q edges one would have O(q log2 /11). Table 23.5 List of Double Labels in Example 1 Choice 1 (3,6) Vertex Choice 2 n,2) Choice 3 (1,3) Choice 4 (4,5) Choice 5 (3,4) l4,0) (4,4) n, 3) (1,0) 2 3 4 5 6 11-61 (1, 1) (3,0) (3, 3) KRUSKAL'S ALGORITHM Find a shortest spanning tree by Kruskal' s algorithm. 1. (1, 1) 2. (1,3) 3. (1,4) CHAP. 23 970 Graphs. Combinatorial Optimization 8. Design an algorithm for obtaining longest spanning trees. 9. Apply the algorithm in Prob. 8 to the graph in Example I. Compare with the result in Example I. 10. To get a minimum spanning tree, instead of adding shortest edges, one could think of deleting longest edges. For what graph5 would this be feasible? Describe an algorithm for this. 11. Apply the method suggested in Prob. IO to the graph in Example 1. Do you get the same tree? 7. CAS PROBLEM. Kruskal's Algorithm. Write a corresponding program. (Sorting is discussed in Ref. [E25] listed in App. I.) Chicago Dallas Denver Los Angeles New York Dallas Denver Los Angele<; New York Washington, DC 800 900 650 1800 700 1350 1650 2500 650 1200 1500 2350 200 13. (Forest) A (not necessarily connected) graph without cycles is called a forest. Give typical examples of applications in which graphs occur that are forests or trees. I]4-20 I GENERAL PROPERTIES OF TREES Prove: 14. (l:niqueness) The path connecting any two vertice~ 1I and u in a tree is unique. 15. If in a graph any two vertices are connected by a unique path, the graph is a tree. 23.5 12. Find a shortest spanning tree in the complete graph of all possible 15 connections between the six cities given (distances by airplane. in miles. rounded). Can you think of a practical application of the result? 1300 850 I 16. If a graph has no cycles, it must have at least 2 vertices of degree 1 (definition in Sec. 23.\). 17. A tree with exactly two vertices of degree 1 must be a path. 18. A tree with induction.) 11 vertices has 11 - I edges. (Proof by 19. If two vertices in a tree are joined by a new edge. a cycle is formed. 20. A graph with 11 vertices is a tree if and only if it has 11 1 edges and has no cycles. Shortest Spanning Trees: Prim's Algorithm Prim's algorithm shown in Table 23.6 is another popular algorithm for the shortest spanning tree problem (see Sec. 23.4). This algorithm avoids ordering edges and gives a tree T at each stage. a property that Kruskal's algorithm in the last section did not have (look back at Fig. 490 if you did not notice it). In Plim's algorithm, starting from any ~ingle vertex, which we call I, we "grow" the tree T by adding edges to it, one at a time, according to some rule (in Table 23.6) until T finally becomes a spanning tree, which is shortest. We denote by U the set of vertices of the growing tree T and by S the set of its edges. Thus, initially U = {I} and S = 0; at the end, U = V. the vertex set of the given graph G = (V. E), whose edges (i, j) have length lij > 0, as before. SEC. 23.5 Shortest Spanning Trees: 971 Prim's Algorithm Thus at the beginning (Step 1) the labels 2.... , of the vertices are the lengths of the edges connecting them to vertex I (or 00 11 if there is no such edge in G). And we pick (Step 2) the shortest of these as the first edge of the growing tree T and include its other endj in U (choosing the smallestj if there are several, to make the process unique). Updating labels in Step 3 (at this stage and at any later stage) concerns each vertex k not yet in U. Vertex k has label Ak = li(k),k from before. If Ijk < Ak , this means that k is closer to the new member j just included in U than k is to its old "closest neighbor" i(k) in U. Then we update the label of k, replacing Ak = li(k).k by Ak = Ijk and setting i(k) = j. If. however, Ijk :;;:: Ak (the old label of k), we don't touch the old label. Thus the label Ak always identifies the closest neighbor of k in U, and this is updated in Step 3 as U and the tree T grow. From the final labels we can backtrack the final tree, and from their numeric values we compute the total length (sum of the lengths of the edges) of this tree. Table 23.6 Prim's Algorithm for Shortest Spanning Trees Bell System Technical Joltl1l1l136 (1957). 1389-I·WL For an improved version of the algorithm. see Cheriton and Tmjan. SIAM Jolt,."al on COmplltlition 5 (1976).724-7-12. ALGORITHM PRIM [G = (V, E), V = {I, ... , 11}, lij for all (i, j) in E] Given a connected graph G = (V, E) with vertices 1,2, ... ,11 and edges (i,j) having length lij > O. this algorithm dete1l11ine~ a shortest spanning tree Tin G and its length L(n INPUT: n. edges (i. j) of G and their lengths lij OUTPUT: Edge set 5 of a shortest spanning tree T in G: L(T) [Initially, alll'ertices are lInlabeled.] 1. Initial step Set i(k) = I, U = {I}, 5 Label vertex k (= 2, ... = 0. ,11) with Ak = Ii/e 1= :G if G has no edge n. k)j. 2. Addition of an edge to the cree T Let Aj be the smallest Ale for vertex k not in U. Include veltex j in U and edge (i(j), j) in 5. If U = V then compute L(T) = "'2:/ij (sum over all edges in 5) OUTPUT 5. UT). Stop [5 is the edge set of a shortest spanning tree T ill G.] Else continue (that is. go to Step 3). 3. Label updating For every k not in U. if Ijle < Ak • then set Ale = lile and i(k) = j. Go to Step 2. End PRIM CHAP. 23 972 Graphs. Combinatorial Optimization Fig. 491. E X AMP L E 1 Graph in Example 1 Application of Prim's Algorithm Find a shortest spanmng tree in the graph can <:ompare). In Fig. 4Yl (which is the ~ame as in Example I. Sec. 23.4, so that we Solutio1/. The steps are a~ follows. 1. irk) = I, U = IlJ, S = 0, initial labels see Table 23.7. 2. A2 = 112 = 2 is smallest, U = 11. 2J. S = {(I, 2)/ 3. Update labels as shown in Table 23.7. column (I). 2. A3 = 113 = 4 is smallest. U = II. 2. 3}. S = {(I, 2), (I. 3») 3. Update labels a~ shown in Table 23.7. column (Il). 2. A6 = 136 = I is smallest. U = II. 2. 3. 6J. S = 1(1. 21. (I. 3). (3. 6)J 3. Update labels as shown in Table 23.7, column (III). 2. A4 = 134 = 8 is smallest, U = II, 2, 3, 4, 6J, S = 10,2), (I. 3), (3. 4), (3, 6») 3. Update labels a~ shown in Table 23.7, column (IV). 2. A5 = 145 = 6 is smallest. U = V. S = (I, 2). (I. 3). (3. 4). (3. 6). (4, 5). Stop. The tree is the same as in Example I. Sec. 23.4. Its length is 21. You will find it interesting to compare the • growth process of the present tree with that in Sec. 23.4. Table 23.7 Vertex 2 3 4 11-71 Labeling of Vertices in Example 1 Relabeling Initial Label 112 113 = 2 =4 (ll) (I) 1 13 = 4 :x; 124 = 5 :x; x 6 :x; x 11 PRIM'S ALGORITHM Find a sh0l1est spanning tree by Prim's algorithm. Sketch it. 1. For the graph in Prob. I, Sec. 23.4 2. For the graph in Prob. 2. Sec. 23.4 3. For the graph in Prob. 4, Sec. 23.4 4. 6. (Ill) (IV) SEC. 23.6 973 Flows in Networks 7. 8. (Complexity) Show that Prim's algorithm has 9. 10. 11. 12. complexity 0(n2). How does Prim's algorithm prevent the formation of cycles as one grows T? For a complete graph (or one that is almost complete), if our dara is an 11 X 11 distance table (as in Prob. 12, Sec. 23.4). sho'" that the present algorithm [which is 2 0(11 )] cannot easily be replaced by an algorithm of order less than 0(/1 2 ). In what case will Prim's algorithm give S = E as the final result? TEAM PROJECT. Center of a Graph and Related Concepts. (a) Distance, eccentricity. Call the length of a shortest path u ~ v in a graph C = (V. E) the distance d(u, v) from II to v. For fixed u, call the greatest £1(11. u) as u ranges over V the ecce1ltricity E(II) of u. Find the eccentricity of vertices I, 2, 3 in the graph in Prob. 7. 23.6 (b) Diameter, radius, center. The diameter d(C) of a graph C = (V, E) is the maximum of li(li. u) as u and u vary over V. and the radius r(C) is the smallest eccentricity E(V) of the vertices v. A vertex v with E(V) = r(C) is called a ce1ltral rertex. The set of all central vertices is called the center of C. Find d(C), r(C) and the center of the graph in Prob. 7. (c) What are the diameter, radius, and center of the spanning tree in Example I? (d) Explain how the idea of a center can be used in setting up an emergency service facility on a transportation network. In setting up a fire station. a shopping center. How would you generalize the concepts in the case of two or more such facilities? (e) Show that a tree T whose edges all have length I has center consisting of either one vertex or two adjacent vel1ices. <0 Set up an algorithm of complexity 0(11) for finding the center of a tree T. 13. What would the result be if you applied Prim's algorithm to a graph that is not connected? 14. CAS PROBLEM. Prim's Algorithm. Write a program and apply it to Probs. 4--6. Flows in Networks After shortest path problems and problems for trees. as a third large area in combinatorial optimization we discuss flow problems in networks (electrical, water, communication, traffic, business connections, etc.), turning from graphs to digraphs (directed graphs; see Sec. 23.1). By definition, a network is a digraph G = (V, E) in which each edge (i,j) has assigned > 0 = maximum possible flow along (i, j)], and at one vertex, s, called the source, a flow is produced that flows along the edges of the digraph G to another vertex, t, called the target or sink, where the flow disappears. to it a capacity Cij r In applications, this may be the flow of electricity in wires, of water in pipes. of cars on roads, of people in a public transportation system. of goods from a producer to consumers, of e-mail from senders to recipients over the Internet, and so on. We denote the flow along a (directed!) edge (i, j) by fij and impose two conditions: 1. For each edge (i, j) in G the flow does not exceed the capacity (1) Cij, ('"Edge condition"). 2. For each vertex i, not s or t, Inflow = Outflow ("Vertex condition," "Kirchhoff's law"); 974 CHAP. 23 Graphs. Combinatorial Optimization in a formula, 0 if vertex i k { j lnnuw s. i =1= t. - f at the source s, = (2) =1= f at the target (sink) t, where f is the total flow (and at s the inflow is zero. whereas at t the outflow is zero). Figure 492 illustrates the notation (for some hypothetical figures). Fig. 492. Notation in (2): inflow and outflow for a vertex i (not 5 or t) Paths By a path of edges VI ~ Vk from a ve11ex VI to a ve11ex Vk in a digraph G we mean a sequence regardless of their directiolls ill G, that forms a path as in a graph (see Sec. 23.2). Hence when we travel along this path from VI to Vk we may traverse some edge ill its given direction-then we call it a forward edge of our path-or opposite to its given directionthen we call it a backward edge of our path. In other words. our path consists of oneway streets. and forward edges (backward edges) are those that we travel in the right direction (in the wrong direction). Figure 493 shows a forward edge (u. v) and a backward edge (w. v) of a path VI ~ Vk' CAUTION! Each edge in a network has a given direction, which we COllnot change. Accordingly, if (u, v) is a forward edge in a path VI ~ Vk, then (u, v) can become a backward edge only in another path Xl ~ Xj in which it is an edge and is traversed in the opposite direction as one goes from Xl to.\); see Fig. 494. Keep this in mind. to avoid misunderstandings. Fig. 493. Forward edge (u. v) and backward edge (w. v) of a path v, ~ Vk Fig. 494. Edge (u, v) as forward edge in the path and as backward edge in the path X, ~ Xj V, ~ Vk Flow Augmenting Paths Our goal will be to maximize thejlow from the sourCe s to the target t of a given network. We shall do this by developing mcthods for increasing an existing flow (including the special case in which the latter is zero). The idea then is to find a path P: s ~ t all of whose edges are not fully used, so that we can push additional flow through P. This suggests the following concept. SEC. 23.6 975 Flows in Networks DEFINITION Flow Augmenting Path A flow augmenting path in a network with a given flow path P: s ~ t such that (i) no forward edge is used to capacity; thus (ii) no backward edge has flow 0; thus EXAMPLE 1 Iij Iij < Iij Cij on each edge (i, j) is a for these; > 0 for these. Flow Augmenting Paths Find flow augmenting paths in the network in Fig. 495, where the first number is the capacity and the second number a given flow. Fig. 495. Network in Example 1 First number = Capacity, Second number = Given flow Solution. In practical problems. networks are large and one needs a sy.•tematic method for augmenting flows, wllich we discllss ill tile next sectioll. In our small network, which should help to illustrate and clarify the concepts and ideas, we can find flow augmenting paths by inspection and augment the existing flow f = 9 in Fig. 495. (The outtlow from s is 5 + 4 = 9, which equals the inflow 6 + 3 into t.) We use the notation for forward edges llij = hj for backward II = min ti ij edge~ taken over all edges of a path. From Fig. 495 we see that a flow augmenting path PI: s --> t is Pt= 1 - 2 - 3 - 6 (Fig. 496). with j.I2 = 20 - 5 = 15. etc .. and j. = 3. Hence we can use PI to increase the given flow 9 to f = 9 + 3 = 12. All three edges of PI are forward edges. We augment the flow by 3. Then the flow in each of the edges of PI is increased by 3. so that we now have .fI2 = 8 (instead of 5), f23 = 11 (instead of 8), and h6 = 9 (instead of 6). Edge (2. 3) is now used to capacity. The flow in the other edges remains as before. We shall now try to increase the flow in thi~ network in Fig. 495 beyond f = 12. There is another flow augmenting path P 2 : s --> t. namely. P 2 : J - 4 - 5 - 3 - 6 (Fig. 496). [t shows how a backward edge comes in and how it is handled. Edge (3. 5) is a backward edge. It has now 2, so that tl.35 = 2. We compute tl.14 = 10 - 4 = 6. etc. (Fig. 496) and .i = 2. Hence we can use P 2 for another augmentation to get f = 12 + 2 = 14. The new flow is shown in Fig. 497. No further augmentation is possible. We shall confirm later that f = 14 is maximum. • "'23 = 3 )-------i!~3 r ~6"'4 s~ "'35=2 @t ~4~5 ""45 = Fig. 496. 3 Flow augmenting paths in Example 1 CHAP. 23 976 Graphs. Combinatorial Optimization Cut Sets A "cut set" is a set of edges in a network. The underlying idea is simple and natural. If we want to find out what i~ flowing from s to t in a network, we may cut the network somewhere between sand t (Fig. 497 shows an example) and see what is t10wing in the edges hit by the cut. because any flow from s to t must sometimes pass through some of these edges. These form what is called a cut set. [In Fig. 497, the cut set consists of the edges (2, 3), (5, 2), (4, 5).] We denote this cut set by (S, T). Here S is the set of vertices on that side of the cut on which s lies (S = {s, 2, 4} for the cur in Fig. 497) and T is the set of the other vertices (T = {3, 5, t} in Fig. 497). We say that a cut "partitiolls" the vertex set V into two parts Sand T. Obviously, the corresponding cut set (S, T) consists of all the edges in the network with one end in S and the other end in T. Fig. 497. Maximum flow in Example 1 By definition, the capacity cap (S, T) of a cut set (S, T) is the sum of the capacities of all forward edges in (S, T) (forward edges only!), that is, the edges that are directed from S to T, (3) cap (S, T) = [sum over the forward edges of (S, T)]. LCij Thus, cap (S, T) = II + 7 = 18 in Fig. 497. The other edges (directedji-ol11 T to S) are called backward edges of the cut set (S, T), and by the net flow through a cut set we mean the sum of the t10ws in the forward edges minus the sum of the flows in the backward edges of the cut set. CAUTION! Distinguish well between forward and backward edges in a cut set and in a path: (5, 2) in Fig. 497 is a backward edge for the cut shown but a forward edge in the path 1 - 4 - 5 - 2 - 3 - 6. For the cut in Fig. 497 the net flow is II + 6 - 3 = 14. For the same cut in Fig. 495 (not indicated there), the net flow is 8 + 4 - 3 = 9. In both cases it equals the flow f. We claim that this is not just by chance, but cuts do serve the purpose for which we have introduced them: THEOREM 1 Net Flow in Cut Sets AllY gil'e1l flow ill a network G is the net flow through a1lY cut set (S, T) of G. PROOF By Kirchhoff's law (2), multiplied by - L at a vertex i we have (4) L fij - j L fh 1 ~~ Oulilow Inflow = [0f if i =I- 05, if i = s. t, SEC. 23.6 977 Flows in Networks Here we can sum over j and I from 1 to 11 (= number of vertices) by putting fij = 0 for j = i and also for edges without flow or nonexisting edges; hence we can write the two sums as one, if i =I- s, t, if i = s. We now sum over all i in S. Since s is in S, this sum equals f: :L :L (fij - (5) fji) = f· iES jEV We claim that in this sum, only the edges belonging to the cut set contlibute. Indeed, edges with both ends in T cannot contribute, since we sum only over i in S; but edges (i,j) with both ends in S contribute + fij at one end and - fij at the other, a total contribution of O. Hence the left side of (5) equals the net flow through the cut set. By (5), this is equal • to the flow f and proves the theorem. This theorem has the following consequence. which we shall also need later in this section. THEOREM 2 Upper Bound for Flows A pow PROOF f ill a network G cannot exceed the capacity of any cut set (S, 1) in G. By Theorem I the flow f equals the net flow through the cut set. f = f 1 - f 2' where f 1 is the sum of the flows through the forward edges and f2 (~ 0) is the sum of the flows through the backward edges of the cut set. Thus f ~ fl' Now f 1 cannot exceed the sum of the capacities of the forward edges; but this sum equals the capacity of the cut set, hy definition. Together, f ~ cap (S, 1), as asserted. • Cut sets will now bring out the full importance of augmenting paths: Main Theorem. Augmenting Path Theorem for Flows THEOREM 3 A pow from s to t in a network G is maximum (f and only (f there does not exist a flow augmenting path s ~ t ill G. PROOF (a) If there is a flow augmenting path P: s ~ t, we can use it to push through it an additional flow. Hence the given t10w cannot be maximum. (b) On the other hand, suppose that there is no flow augmenting path s ~ t in G. Let So be the set of all vertices i (induding .1') such that there is a flow augmenting path s ~ i, and let To be the set of the other vertices in G. Consider any edge (i, j) with i in So and j in To. Then we have a t10w augmenting path s ~ i since i is in So, but s ~ i ~ j is not t10w augmenting because j is not in So. Hence we must have forward (6) if (i, j) is a [ edge of the path s backward ~ i ~ j. 978 CHAP. 23 Graphs. Combinatorial Optimization Otherwise we could use (i, j) to get a flow augmenting path s ---+ ; ---+ j. Now (So, To) defines a cut set (since I is in To: why?). Since by (6), forward edges are used to capacity and backward edges carry no flow, the net flow through the cut set (So, To) equals the sum of the capacities of the forward edges. which is cap (So. To) by definition. This net t10w equals the given flow f by Theorem 1. Thus f = cap (So, To). We also have f ~ cap (So, To) by Theorem 2. Hence f must be maximum since we have reached ~w~. • The end of this proof yields another basic result (by Ford and Fulkerson, Canadian JOlln1al of Mathematics 8 (1956), 399-404), namely. the so-called THEOREM 4 Max-Flow Min-Cut Theorem The maximum flow ill any network G equals the capacity of a "minimum cut set"' (= a cut set of minimum capacity) in G. PROOF We have just seen that f = cap (So, To) for a maximum flow f and a suitable cut set (So, To). Now by Theorem :2 we also have f ~ cap (S. T) for this f and any cut set (S, T) in G. Together, cap (So, To) ~ cap (S, n. Hence (So, To) is a minimum cut set. The existence of a maximum flow in this theorem follows for rational capacities from the algorithm in the next section and for arbitrary capacities from the Edmonds-Karp BFS also in that section. • The two basic tools in connection with networks are flow augmeming paths and cut sets. In the nexl section we show how flow augmenting paths can be used in an algorithm for maximum flows. - _ •... _.... _....lA__.........._____ .-....... ..-- = _ _ .....- . · . . . . . 11-41 FLOW AUGMENTING PATHS 3. Find flow augmenting paths: 1. 2. 4. SEC. 23.7 Maximum Flow: Ford-Fulkerson Algorithm ~! MAXIMUM FLOW Find the maximum flow by inspection: S. In Prob. 1. 6. In Prob. 2. 7. In Prob. 3. 8. In Prob. 4. !9-11! CAPACITY In Fig. 495 find T and cap (5. T) if 5 equals 9. [1,2.31 10. [I. 2.4.51 11. [1, 3, 51 12. Find a minimum cut set in Fig. 495 and verify that its capacity equals the maximum flow I = 14. 13. Find examples of flow augmenting paths and the maximum flow in the network in Fig. 498. lB~ CAPACITY In Fig. 498 find T and cap (5. T) if 5 equals 14. [1,2,41 23.7 Maximum Flow: 979 15. [L 2. 4. 6 I 16. [1,2.3.4,51 17. In Fig. 498 find a minimum cut set and its capacity. Fig. 498. 18. Why are backward edge~ not considered III the definition of the capacity of a cut set? 19. In which case can an edge U, j) be used as a forward as well as a backward edge of a path in a network with a given flow? 20. (Incremental network) Sketch the network in Fig. 498, and on each edge (i,j) write Cij - Iij and Iij' Do you recognize that from this "incremental network" one can more easily see flow augmenting paths? Ford-Fulkerson Algorithm Flow augmenting paths, as discussed in the last section. are used as the basic tool in the Ford-Fulkerson4 algorithm in Table 23.8 on the next page in which a given flow (for instance, zero flow in all edges) is increased until it is maximum. The algOlithm accomplishes the increa-;e by a stepwise construction of flow augmenting paths. one at a time. until no further such paths can be constructed, which happens precisely when the tlow is maximum. In Step I, an initial t10w may be given. In Step 3, a vertex j can be labeled if there is an edge (i. j) with i labeled and ("forward edge") or if there is an edge (j, i) with i labeled and f ·· > 0 • JZ ("backward edge"). To scan a labeled vertex i means to label every unlabeled vertex j adjacent to i that can be labeled. Before scanning a labeled vertex i, scan all the vertices that got labeled before i. This BFS (Breadth First Search) strategy was suggested by Edmonds and Karp in 1972 (Journal ofrhe Associariollfor Compllring Machinery 19, 248-64). It has the effect that one gets shortest possible augmenting paths. 4LESTER RANDOLPH FORD (horn 1927) and DELBERT RAY FULKERSON (1924-1976), American mathematicians known for their pioneering work on flow algorithms. 980 CHAP. 23 Graphs. Combinatorial Optimization Table 23.8 Ford-Fulkerson Algorithm for Maximum Flow Cllllllcllll1l JOl/mlll of Mathematics 9 (1957),210--218 I ALGORITHM FORD-FULKERSON [G = (V, E), vertices 1 (= s) . .... 11 (= t). edges (i.j), Cij] This algorithm computes the maximum flow in a network G with source s. sink t. and capacities Cij > 0 of the edges (i, j). INPUT: 11, s = I, t = 11. edges (i, j) of G. OUTPUT: Maximum flow f in G 1. Assign an initial flow fij (for instance, Cij fij = 0 for all edges), compute f. 2. Label s by 0. Mark the other veltices "lIlllabeled." 3. Find a labeled vertex i that has not yet been scanned. Scan i as follows. For every unlabeled adjacent veltexj, if Cij > fij' compute if i = 1 L11j and j,j = [ . mm (Ll·l' Ll 1J.) and labelj with a ':forward label" (i+, L1j ); or if fji if i > I > 0, compute and labelj by a "backward label"' (C, Ll). If no such j exists then OUTPUT f. Stop [f is the maximum flmr.] Else continue (that is, go to Step -l). 4. Repeat Step 3 until t is reached. [This gives a flow allgmellting path P: s ~ t.] [f it is impossible to reach t then OUTPUT f. Stop [f is the maximum flow.J Else continue (that is. go to Step 5). 5. Backtrack the path P, using the labels. 6. Using P, augment the existing flow by !::"t. Set f 7. Remove all labels from veltices 2, ... , 11. = .f + Ll,. Go to Step 3. End FORD-FULKERSON E X AMP L ElFord-Fulkerson Algorithm Applying the Ford-Fulkerson algorithm, determine the maximum flow for the network in Fig. 499 (which is the same as that in Example I. Sec. 23.6. ~o that we can compare). Solu lion. The algorithm proceeds as follows. 1. An initial flow.f = 9 is given. 2. Label of (= I) by 0. Mark 2, 3, 4. 5, 6 "unlabeled." SEC. 23.7 Maximum Flow: 981 Ford-Fulkerson Algorithm Fig. 499. Network in Example 1 with capacities (first numbers) and given flow 3. Scan I. Compute J. 12 = 20 - 5 = IS = J.2. Label 2 by (1 +, IS). Compute J. 14 = 10 - 4 = 6 = ,).4' Label 4 by (1 +. 6). 4. Scan 2. Compute .1.23 = II - 8 = 3. ~3 = min (~2' 3) = 3. Label 3 by (2+. 3). Compute ~5 = min (tl. 2• 3) = 3. LabelS by (T, 3). Scan 3. Compute .1.36 = 13 - 6 = 7, J. 6 = ~t = min (~3' 7) = 3. Label 6 by (3+, 3). 5. P: I - 2 - 3 - 6 6. J.t 1 (= t) is a flow augmenting path. = 3. Augmentation gives = 9 + 3 = 12. h2 = 8. 123 = II. 136 9. other /;j unchanged. Augmented flow 7. Remove labels on vertices 2..... 6. Go 10 Step 3. 3. Scan I. Compute .112 = 20 - 8 = 12 = .:).2' Label 2 by (I +, 12). Compute ~14 = 10 - 4 = Ii = tl. 4. Label 4 by 11 +, 6). 4. Scan 2. Compute J. 5 = min (.1. 2, 3) = 3. LabelS by (2-, 3). Scan 4. [No I'ertex left forlabelillg.] Scan 5. Compute,).3 = min (.15 , 2) = 2. Label 3 by (5-. 1). Scan 3. Compute J.36 = 13 - 9 = 4 . .16 = min (.1. 3 , 4) = 2. Label Ii by (3+, 2). 5. P: I - 2 - 5 - 3 - 6 (= t) 6. .1.r. = 2. Augmentalion gives flow 1 = 12 + 2 = 14. 7. Remove labels on vertice~ is a flow augmenting path. h2 = 10./52 = I. i35 = 0, 136 = II, other Ji] unchanged. Augmented 2, ... , Ii. Go to Step 3. One can now scan I and then scan 2, as before, but in scanning 4 and then 5 one finds that no vertex is left for labeling. Thus one can no longer reach 1. Hence the flo" obtained (Fig. 500) is maximum, in agreement with our result in the last section. • Fig. 500. Maximum flow in Example 1 982 CHAP. 23 Graphs. Combinatorial Optimization 1. Do the computations indicated near the end of Example 1 in detail. 2. Solve Example 1 by Ford-Fulkerson with initial now O. Is it more work than in Example I? 3. Which are the "bottleneck" edges by which the flow in Example 1 is actually limited? Hence which capacities could be decreased without decreasing the maximum How? 14. If the Ford-Fulkerson algorithm stops without reaching t. sho~ that the edges with one end labeled and the other end unlabeled form a cut set (S. T) whose capacity equals the maximum flow. 15. (Several sources and sinks) If a network has several sources Sl' . . . , Sk' sho\\ that it can be reduced to the case of a single-source network by introducing a new vertex S and connecting S to Slo • . • • Sk by k edges of capacity Similarly if there are several sinks. lllustrate tlus idea by a network with two sources and two sinks. 16. Find the maximum flow in the network in Fig. 50 I with two sources (factories) and two sinks (consumers). 17. Find a minimum cut set in Fig. 499 and its capacity. 18. Show that in a network G with all Cij = I, the maximum flow equals the number of edge-disjoint paths s ~ t. 19. In Prob. 17, the cut set contains precisely all forward edges used to capacity by the maximum How (Fig. 500). Is this just by chance? 20. Show that in a network G with capacities all equal to I, the capacity of a minimum cut set (S, T) equals the minimum number q of edges whose deletion destroys all directed paths S ~ t. (A directed path v ~ w is a path in which each edge has the direction in which it is traversed in going from v to w.) 0/:). 14-71 MAXIMUM FLOW Find the maximum How by Ford-Fulkerson: 4. In Prob. 2, Sec. 23.6. 5. In Prob. I, Sec. 23.6. 6. In Prob. 4, Sec. 23.6. 7. In Prob. 3, Sec. 23.6. 8. What is the (simple) reason that Kirchhoffs law is preserved in augmenting a flow by the use of a flow augmenting path? 9. How does Ford-Fulkerson prevent the fOimation of cycles? 10. How can you see that Ford-Fulkerson follows a BFS technique? 11. Are the consecutive How augmenting paths produced by Ford-Fulkerson unique"! 12. (Integer flow theorem) Prove that if the capacities in a network G are integers. then a maximum How exists and is an integer. 13. CAS PROBLEM. Ford-Fulkerson. Write a program and apply it to Probs. 4-7. 23.8 Bipartite Graphs. Fig. 501. Problem 16 Assignment Problems From digraphs we return to graphs and discuss another impOitant class of combinatOlial optimization problems that arises in assignment problems of workers to jobs, jobs to machines, goods to storage, ships to piers. classes to classrooms, exams to time periods, and so on. To explain the problem, we need the following concepts. A bipartite graph G = (V, E) is a graph in which the veltex set V is partitioned into two sets 5 and T (without common elements, by the definition of a partition) such that every edge of G has one end in 5 and the other in T. Hence there are no edges in G that have both ends in 5 or both ends in T. Such a graph G = (V, E) is also written G = (5, T; E). Figure 502 shows an illustration. V consists of seven elements, three workers a, b, c, making up the set 5, and four jobs I, 2. 3, 4, making up the set T. The edges indicate that worker {/ can do the jobs 1 and 2, worker b the jobs I, 2, 3, and worker c the job 4. The problem is to assign one job to each worker so that every worker gets one job to do. This suggests the next concept, as follows. SEC. 23.8 Bipartite Graphs. DEFINITION 983 Assignment Problems Maximum Cardinality Matching A matching in G = (S. T; E) is a set M of edges of G such that no two of them have a vertex in common. If M consists of the greatest possible number of edges. we call it a maximum cardinality matching in G. For in<;tance, a matching in Fig. 502 is Ml = {(a. 2), (b. l)}. Another is M2 = {(a, 1), (b, 3), (c, 4)}; obviously, this is of maximum cardinality. s T :~: c ____ 4 Fig. 502. Bipartite graph in the assignment of a set 5 = {a, b, c} of workers to a set T = {l, 2. 3. 4} of jobs A vertex v is exposed (or /lot covered) by a matching M if v is not an endpoint of an edge of M. This concept. which always refers to some matching, will be of interest when we begin to augment given matchings (below). If a matching leaves no vertex exposed. we call it a complete matching. Obviously, a complete matching can exist only if Sand T consist of the same number of vertices. We now want to show how one can stepwise increase the cardinality of a matching M until it becomes maximmn. Central in this task is the concept of an augmenting path. An alternating path is a path that consists alternately of edges in M and not in M (Fig. 503A). An augmenting path is an alternating path both of whose endpoints (a and b in Fig. 503B) are exposed. By dropping from the matching M the edges that are on an augmenting path P (two edges in Fig. 503B) and adding to M the other edges of P (three in the figure), we get a new matching, with one more edge than M. This is how we use an augmenting path in augmenting a given matching by one edge. We assert that this will always lead, after a number of steps, to a maximum cardinality matching. Indeed, the basic role of augmenting paths is expressed in the following theorem. (Al Alternating path (8) Augmenting path P Fig. 503. Alternating and augmenting paths. Heavy edges are those belonging to a matching M. CHAP. 23 984 THEOREM 1 Graphs. Combinatorial Optimization Augmenting Path Theorem for Bipartite Matching A matching M ill a bipartite graph G = (S, T; E) is of mllximum cardillality only if there does not exist an augmenting path P witlz respect to M. PROOF if alld (a) We show that if such a path P exists. then M is not of maximum cardinality. Let P have q edges belonging to M. Then P has q + I edges not belonging to M. (In Fig. 503B we have q = 2.) The endpoints a and b of P are exposed. and all the other vertices on P are endpoints of edges in M. by the definition of an alternating path. Hence if an edge of M is not an edge of P. it cannot have an endpoint on P since then M would not be a matching. Consequently. the edges of M not on P. together with the q + 1 edges of P not belonging to M form a matching of cardinality one more than the cardinality of M because we omitted q edges from M and added q + I instead. Hence M cannot be of maximum cardinality. (b) We now show that if there is no augmenting path for M, then M is of maximum cardinality. Let M';' be a maximum cardinality matching and consider the graph H consisting of all edges that belong either to M or to M*. but not to both. Then it is possible that two edges of H have a vertex in common. but three edges cannot have a vertex in common since then two of the three would have to belong to M (or to M*), violating that M and M-;' are matchings. So every v in V can be in common with two edges of H or with one or none. Hence we can characterize each "component" (= maximal cOllllected subset) of H as follows. (A) A component of H can be a closed path with an even number of edges (in the case of an odd number, two edges from M or two from M* would meet. violating the matChing property). See (A) in Fig. 504. (B) A component of H can be an open path P with the same number of edges from M and edges from M*, for the following reason. P must be alternating. that is, an edge of M is followed by an edge of M*. etc. (since M and M':' are matchings). Now if P had an edge more from M*, then P would be augmenting for M [see (B2) in Fig. 504]. contradicting our assumption that there is no augmenting path for M. If P had an edge more from M, it would be augmenting for M'~ [see (B3) in Fig. 504]. violating the maximum cardinality of M*. by part (a) of this proof. Hence in each component of H. the two matchings have the same number of edges. Adding to this the number of edges that belong to both M and M* (which we left aside when we made up H), we conclude that M and M* must have the same number of edges. Since M* is of maximum cardinality, this shows that the same holds for M, as we wanted to prove. • , (Al . ...... '"._---_. -EdgefromM --. - - - - Edge from M'" (81) . - - - -_ _ . - - - . _ - '---"-'_--.. (82) . - - - ...__- (83) _ _ _ . - - - . ~'---''''''''''--4. Fig. 504. __ - - - . - .--- .. (Possible) (Augmenting for M) (Augmenting for M") Proof of the augmenting path theorem for bipartite matching SEC. 23.8 Bipartite Graphs. Assignment Problems 985 This theorem suggests the algorithm in Table 23.9 for obtaining augmenting paths. in which vertices are labeled for the purpose of backtracking paths. Such a label is ill additiol1 to the number of the vertex, which is also retained. Clearly, to get an augmenting path. one must start from an exposed vertex. and then trace an alternating path until one arrives at another exposed vertex. After Step 3 all vertices in S are labeled. In Step 4. the set T contains at least one exposed vertex. since otherwise we would have stopped at Step I. Table 23.9 Bipartite Maximum Cardinality Matching ALGORITHM MATCHING [G = (S, T; E), M, n] This algorithm determines a maximum cardinality matching M in a bipartite graph G by augmenting a given matching in G. INPUT: Bipartite graph G = (S, T; E) with ve11ices I, ... instance, M = 0) OUTPUT: Maximum cardinality matching M in G ,11, marching Min G (for 1. If there is no exposed vertex in S then OUTPUT M. Stop [M is of maxilllulIl cardillality ill G.] Else label all exposed vertices ill S with 0. 2. For each i in S and edge U, j) l10t in M, label j with i, unless already labeled. 3. For each 11011exposed j in T. label i with j, where i is the other end of the unique edge U. j) in M. -I. Backtrack the alternating paths P ending on an exposed vertex in T by using the labels on the ve11ices. 5. If no P in Step 4 is augmenting then OUTPUT M. Stop [M is of maximum cardinality ill G.] Else augment M by using an augmenting path P. Remove all labels. Go to Step I. End MATCHING E X AMP L E 1 Maximum Cardinality Matching Is the matching Ml in Fig. SOSa of maximum cardinality? If not. augment it until maximum cardinality is reached. S T 3 3 (aJ Given graph and matChing M J Fig. 505. 5 3 7 2 o B 3 4 (b) Matching M2 and new labels Example 1 986 CHAP. 23 Graphs. Combinatorial Optimization Solution. We apply the algorithm. 1. Label I and.J. with 0. 2. Label 7 with I. Label 5. 6. !l with 3. 3. Label 2 with 6, and 3 with 7. [All I'afices are now labeled (IS shown in Fig. 474a.] 4. PI: 1 - 7 - 3 - 5. [By backtracking, PI is augmenting.] P 2: I - 7 - 3 - 8. [P2 is lIugmellfillg.j 5. Augment MI by using Pl' dropping (3,7) from MI and including (I, 7) and (3. 5). Remove all label~. Go to Step I. Figure 474b shows the resulting matching M2 = {(I. 7). (2, 6). (3. 5)j. 1. Label.J. with 0. 2. Label 7 with 2. Label 6 and !l with 3. 3. Label I with 7. and 2 with 6. and 3 with 5. 4. P 3 : 5 - 3 - 8. [P3 is aitematillg but /lot aug11lellfing.] • 5. Stop. M2 is otl1lllxi11l11111 cardillality (namely, 3). --.. 11-6/ IS-lO I BIPARTITE OR NOT? Are the following graphs bipartite? If you answer is yes, find S and T. 1.~ 2. AUGMENTING PATHS hnd an augmenting path: cp------<f s.:\: dr---4 ~ 9. ~ ~ 3.~ ~ 10. 7 111-131 MAXIMUM CARDINALITY MATCHING Augmenting the given matching, find a maximum cardinality matching: 11. In Prob. 9. 12. In Prob. 8. 7. Can you obtain the answer to Prob. 3 from that to Prob. I? 13. In Prob. 10. 987 Chapter 23 Review Questions and Problems 14. (Scheduling and matching) Three teachers Xl, X2' -'3 teach four classes )'1, Y2, Y3, Y4 for these numbers of periods: Y1 Y2 y .3 1 0 0 I 1 1 1 1 Xl X2 -'"3 " .4 1 Show that this arrangement can be represented by a bipartite graph G and that a teaching schedule for one period corresponds to a matching in G. Set up a teaching schedule with the smallest possible number of periods. 15. (Vertex coloring and exam scheduling) What is the smallest number of exam periods for six subjects a, b, c, d, e, t if some of the students simultaneously take a, b, t, some c, d, e, some a, c, e, and some c, e? Solve this as follows. Sketch a graph with six vertices a, ... , t and join vertices if they represent subjects simultaneously taken by some students. Color the vertices so that adjacent vertices receive different colors. (Use numbers L 2, ... instead of actual colors if you want) What is the minimum number of colors you need? For any graph G, this minimum number is called the (vertex) chromatic number Xv(G). Why is this the answer to the problem? Write down a possible schedule. 16. How many colors do you need in vertex coloring the graph in Prob. 5? 17. Show that all trees can be vertex colored with two colors. 18. (Harbor management) How many piers does a harbor master need for accommodating six cruise ships 51, " ' , 56 with expected dates of arrival A and departure D in July, (A, D) = (10, 13), (13, 15), (14, 17), (\2, 15), (16, 18), (\4, 17), respectively, if each pier can accommodate only one ship, arrival being at 6 a:m and departures at 11 p:m? Hint. Join 5i and 5j by an edge if their intervals overlap. Then color vertices. ship~ 51> ... , 55 had to be accommodated? 20. (Complete bipartite graphs) A bipartite graph G = (5, T: E) is called complete if every vertex in 5 is joined to every vertex in Tby an edge, and is denoted by K n1 ,,%, where n1 and n2 are the numbers of vertices in 5 and T, respectively. How many edges does this graph have? 21. (Planar graph) A planar graph is a graph that can be drawn on a sheet of paper so that no two edges cross. Show that the complete graph K4 with four vertices is planar. The complete graph K5 with five vertices is not planar. Make this plausible by attempting to draw K5 so that no edges cross. Interpret the result in terms of a net of roads between five cities. 22. (Bipartite graph K 3 ,3 not planar) Three factories 1, 2, 3 are each supplied underground by water, gas, and electricity, from points A, E, C, respectively. Show that this can be represented by K 3 •3 (the complete bipartite graph G = (5. T; £) with 5 and T consisting of three vertices each) and that eight of the nine supply lines (edges) can be laid out without crossing. Make it plausible that K 3 .3 is not planar by attempting to draw the ninth line without crossing the others. 23. (Four- (vertex) color theorem) The famous Jour-color theorem states that one can color the vertices of any planar graph (so that adjacent vertices get different colors) with at most four colors. It had been conjectured for a long time and was eventually proved in 1976 by Appel and Haken [Illinois J. Math 21 (1977), 429-5671. Can you color the complete graph K5 with four colors? Does the result contradict the four-color theorem? (For more details, see Ref. [F8] in App. I.) 24. (Edge coloring) The edge chromatic number xeCG) of a graph G is the minimLUll number of colors needed for coloring the edges of G so that incident edges get different colors. Clearly, Xe(G) ;;; max d(u), where d(lI) is the degree of vertex u. If G = (5, T; E) is bipartite, the equality sign holds. Prove this for K n .n . 25. Vizing's theorem states that for any graph G (without multiple edges!). max d(u) ~ Xe(G) ~ max d(u) + I. Give an example of a graph for which Xe( G) does exceed max d(u). 19. What would be the answer to Prob. 18 if only the five .. - :«;::.,w. :::''''==::== S T ION SAN D PRO B L EMS 1. What is a graph? A digraph? A tree? A cycle? A path? 2. State from memory how you can handle graphs and digraphs on computers. 3. Describe situations and problems that can be modeled using graphs or digraphs. 4. What is a shortest path problem? Give applications. 5. What is BFS? DFS? In what connection did these concepts occur? 6. Give some applications in which spanning trees playa role. 7. What are bipartite graphs? What applications motivate this concept? CHAP. 23 988 Graphs. Combinatorial Optimization 8. What is the traveling salesman problem? 9. What is a network? What optimization problems are connected with it? 10. Can a forward edge in one path be a backward edge in another path? [n a cut set? Explain. 11. There is a famous theorem on cut sets. Can you remember and explain it? 26. MATRICES FOR GRAPHS OR DIGRAPHS 112-171 }gt482 017 3 Find shortest paths by Dijkstra's algorithm: 2 810 27. 10 6 2 28 Find the adjacency matrix of: 12. 4 3 4 2 3 4 28. 15. 14. fA \!dJ 16. 17.~ ~ B ~ 29. (Shortest spanning tree) Find a shortest spanning tree for the graph in Prob. 26. 30. Find a shortest ~panning tree in Prob. 27. 31. Cayley's theorem states that the number of spanning trees in a complete graph with II vertices is nn-2. Verify this for f1 = 2. 3. 4. 32. Show that 0(1Il 3) + 0(111 2) = 0(11/3). 133-341 MAXIMUM FLOW. Find the maximum flow. where the given numbers are capacities: 33. 34. 21. Make a vertex incidence list of the digraph in Prob. 13. 22. Make a vertex incidence list of the digraph in Prob. 14. /23-28/ SHORTEST PATHS Find a shortest path and its length by Moore's BFS algorithm. assuming that all the edges have length I: 23. / ......... /.:~~>---:\ sVI~/ t 24. 35. Company A has offices in Chicago. Los Angeles. and New York. Company B in Boston and New York. Company C in Chicago, Dallas, and Los Angeles. Represent this by a bipartite graph. 36. (Maximum cal'dinality matching). Augmenting the given matching. find a maximum cardinality matching: 989 Summary of Chapter 23 . --.- ..... . - _ -. .-II .. -.. ...... . . . . . . . . . _ _ .-.. Graphs and Combinatorial Optimization Combinatorial optimization concerns optimIzation problems of a discrete or combinatorial stmcture. It uses graphs and digraphs (Sec. 23.1) as basic tools. A graph G = (V. E) consists of a set V of vertices VI, V2, .•.• V n • (often simply denoted by I. 2 .... , /I) and a set E of edges el' e2' .... em. each of which connects two ve11ices. We also write (i. j) for an edge with vertices i and j as endpoints. A digraph (= directed graph) is a graph in which each edge has a direction (indicated by an arrow). For handling graphs and digraphs in computers. one can use matrices or lists (Sec. 23.1). This chapter is devoted to important classes of optimization problems for graphs and digraphs that all arise from practical applications. and corresponding algorithms, as follows. In a shortest path problem (Sec. 23.2) we determine a path of minimum length (consisting of edges) from a vertex s to a ve11ex t in a graph whose edges (i.j) have a "'length" lij > O. which may be an actual length or a travel time or cost or an electrical resistance [if (i, j) is a wire in a net], and so on. Dijkstra's algorithm (Sec. 23.3) or, when all lij = I, Moore's algorithm (Sec. 23.2) are suitable for these problems. A tree is a graph that is connected and has no cycles (no closed paths). Trees are very important in practice. A ~P(/l111illg tree in a graph G is a tree containing all the vertices of G. If the edges of G have lengths, we can detenrune a shortest spanning tree, for which the sum of the lengths of all its edges is minimum, by Kruskal's algorithm or Prim's algorithm (Sees. 23.4, 23.5). A network (Sec. 23.6) is a digraph in whieh each edge (i. j) has a capacity Cij > 0 [= maximum possible flow along (i. j)] and at one ve11ex, the source s. a flow is produced that flows along the edges to a vertex t, the sink or target, where the flow disappears. The problem is to maximize the tlow, for instance. by applying the Ford-Fulkerson algorithm (Sec. 23.7), which uses flow augmenting paths (Sec. 23.6). Another related concept is that of a cut set, as defined in Sec. 23.6. A bipartite graph G = (V, E) (Sec. n.8) is a graph whose vertex set V consists of two parts Sand T such that every edge of G has one end in S and the other in T, so that there are no edges connecting vertices in S or ve11ices in T. A matching in G is a set of edges. no two of which have an endpoint in common. The problem then is to find a maximum cardinality matching in G. that is. a matching M that has a maximum number of edges. For an algorithm. see Sec. 23.8. r PA RT G Probability, Statistics ." CHAPTER 24 Data Analysis. Probability Theory CHAPTER 25 Mathematical Statistics Probability theory (Chap. 24) provides models of probability distributions (theoretical models of the observable reality involving chance effects) to be tested by statistical methods, and it will also supply the mathematical foundation of these methods in Chap. 25. Modern mathematical statistics (Chap. 25) has various engineering applications, for instance, in testing materials, control of production processes, quality control of production outputs, perfOlmance tests of systems, robotics. and automatization in general, production planning, marketing analysis, and so on. To this we could add a long list of fields of applications, for instance, in agriculture, biology, computer science, demography, economics, geography, management of natural resources, medicine, meteorology, politics, psychology, sociology, traffic control, urban planning, etc. Although these applications are very heterogeneous, we shall see that most statistical methods are universal in the sense that each of them can be applied in various fields. Additional Software for Probability and Statistics See also the list of software at the beginning of Part E on Numerical Analysis. DATA DESK. Data Description, Inc., Ithaca, NY. Phone 1-800-573-5121 or (607) 257-1000, website at www.datadescription.com. MINITAB. Minitab, Inc., College Park, PA. Phone 1-800-448-3555 or (814) 238-3280, website at www.minitab.com. SAS. SAS Institute, Inc., Cary, NC. Phone 1-800-727-0025 or (919) 677-8000, website at www.sas.com. 991 992 PART G Probability, Statistics S-PLUS. Insightful Corporation, Inc., Seattle, W A. Phone 1-800-569-0123 or (206) 283-8802, website at www.insightful.com. SPSS. SPSS, Inc., Chicago, [L. Phone 1-800-543-2185 or (312) 651-3000, website at www.spss.com. STATISTIC.I\. StatSoft, Inc., Tulsa, OK. Phone (918) 749-1119, website at www.statsoft.com. 24 CHAPTER ". Data Analysis. Probability Theory We first show how to handle data numelically or in terms of graphs, and how to extract information (average size. spread of data, etc.) from them. If these data are influenced by "chance," by factors whose effect we cannot predict exactly (e.g., weather Jata, stock prices, lifespans of tires, etc.), we have to rely on probability theory. This theory originated in games of chance, such as flipping coins, rolling dice, or playing cards. Nowadays it gives mathematical models of chance processes called random experiments or, briefly, experiments. In such an experiment we observe a random variable X, that is, a function whose values in a trial (a pelfOimance of an experiment) occur "by chance" (Sec. 24.3) according to a probability distribution that gives the individual probabilities with which possible values of X may occur in tlle long run. (Example: Each of the six faces of a die should occur witll the same probability. l/6.) Or we may simultaneously observe more than one random variable, for instance. height alld weight of persons or hardness alld tensile strength of steel. This is discussed in Sec. 24.9, which will abo give the basis for tlle mathematical justification of the statistical methods in Chap. 25. Prereqllisite: Calculus. References and Answers to Problems: App. L Part G, App. 2. 24.1 Data Representation. Average. Spread Data can be represented numelically or graphically in various ways. For instance, your daily newspaper may contain tables of stock plices and money exchange rates, curves or bar charts illustrating economical or political developments, or pie charts showing how your tax dollar is spent. And there are numerous other representations of data for special purposes. In this section we discuss the use of standard representations of data in statistics. (For these, software packages, such as DATA DESK and MINITAB, are available, and Maple or Mathematica may also be helpful; see pp. 778 and 991) We explain corresponding concepts and methods in terms of typical examples, beginning with (1) 89 84 87 81 89 86 91 90 78 89 87 99 83 89. These are 11 = 14 measurements of the tensile strength of sheet steel in kg/mm2, recorded in the order obtained and rounded to integer values. To see what is going on, we sort these data, that is, we order them by size, (2) 78 81 83 84 86 87 87 89 89 89 89 90 91 99. S0l1ing is a standard process on the computer; see Ref. [E25], listed in App. 1. 993 994 CHAP. 24 Data Analysis. Probability Theory Graphic Representation of Data We shall now discuss standard graphic representations used in statistics for obtaining information on properties of data. Stem-and-Leaf Plot This is one of the simplesl but most useful representations of data. For (I) it is shown in Fig. 506. The numbers in (1) range from 78 to 99; see (2). We divide these numbers into 5 groups, 75-79, 80-84, 85-89, 90-94, 95-99. The integers in the tens position of the groups are 7,8,8,9,9. These form the stem in Fig. 506. The first lealis 8 (representing 78). The second leaf is 134 (representing 81, 83, 84), and so on. The number of times a value occurs is called its absolute frequency. Thus 78 has absolute frequency 1, the value 89 has absolute frequency 4, etc. The column to the extreme left in Fig. 506 shows the cumulative absolute frequencies, that is, the sum of the absolute frequencies of the values up to the line of the leaf. Thus, the number 4 in the second line on the left shows that (1) has 4 values up to and including 84. The number 11 in the next line shows that there are II values not exceeding 89, etc. Dividing the cumulative absolute frequencies by 11 (= 14 in Fig. 506) gives the cumulative relative frequencies. Histogram For large sets of data, histograms are better in displaying the distribution of data than stem-and-leaf plots. The principle is explained in Fig. 507. (An application to a larger data set is shown in Sec. 25.7). The bases of the rectangles in Fig. 507 are the x-intervals (known as class intervals) 74.5-79.5, 79.5-84.5, 84.5-89.5, 89.5-94.5, 94.5-99.5, whose midpoints (known as class marks) are x = 77, 82, 87. 92, 97, respectively. The height of a rectangle with class mark x is the relative class frequency frel(X), defined as the number of data values in that class interval, divided by 1l (= 14 in our case). Hence the areas of the rectangles are proportional to these relative frequencies, so that histograms give a good impression of the distribution of data. Center and Spread of Data: Median, Quartiles As a center of the location of data values we can simply take the median, the data value that falls in the middle when the values are ordered. In (2) we have 14 values. The seventh of them is 87, the eighth is 89, and we split the difference, obtaining the median 88. (In general, we would get a fraction.) The spread (vmiabil ity) of the data values can be measured by the range R = Xmax - xmin' the largesl minus the smallest data values, R = 99 - 78 = 21 in (2). 0.5 Leaf unit = 1.0 1 4 11 13 14 7 8 8 9 9 8 134 6779999 01 9 Fig. 506. Stem-and-Ieaf plot of the data in (I) and (2) 0.4 0.3 0.2 0.1 o x Fig. 507. Histogram of the data in (I) and (2) (grouped as in Fig. S06) SEC. 24.1 Data Representation. Average. 995 Spread Better information gives the interquartile range IQR = qu - qv Here the upper quartile qu is the middle value among the data values above the median. The lower quartile qL is the middle value among the data values below the median. Thus in (2) we have qu = 89 (the fourth value from the end), qL = 84 (the fOlllth value from the beginning). and IQR = 89 - 84 = 5. The median is also called the middle quartile and is denoted by qM. The rule of "splitting the difference" (just applied to the middle quartile) is equally well used for the other quartiles if necessary. Boxplot The boxplot of (I) in Fig. 508 is obtained from the five numbers xmin, qv qM, qu, Xmax just determined. The box extends from qL to quo Hence it has the height IQR. The position of the median in the box shows that the data distribution is not symmetric. The two lines extend from the box to Xmin below and to Xmax above. Hence they mark the range R. Boxplots are particularly suitable for making comparisons. For example, Fig. 508 shows boxplots of the data sets (I) and (3) 91 89 93 91 87 94 92 85 91 90 96 93 89 91 92 93 93 94 96 (consisting of 11 = 13 values). Ordeling gives (4) 85 87 89 89 90 91 91 (tensile strength, as before). From the plot we immediately see that the box of (3) is shorter than the box of (I) (indicating the higher quality of the steel sheets!) and that qM is located in the middle of the box (showing the more symmetric form of the distribution). Finally, '"max is closer to qu for (3) than it is for (l), a fact that we shall discuss later. For plotting the box of (3) we took from (4) the values xmin = 85, qL = 89, qM = 91, qu = 93, Xmax = 96. Outliers An outlier is a value that appears to be uniquely different from the rest of the data set. It might indicate that something went wrong with the data collection process. In connection with qumtiles an outlier is conventionally defined as a value more than a distance of 1.5 IQR from either end of the box. 100 95 gqU qM 90 qu qM 85 I qL qL 80 75 Data set 0) Fig. 508. Data set (3) Boxplots of data sets (1) and (3) CHAP. 24 996 Data Analysis. Probability Theory For the data in (1) we have lQR = 5, qL = 84, qu = 89. Hence outliers are smaller than 84 - 7.5 or larger than 89 + 7.5, so that 99 is an outlier [see (2)]. The data (3) have no outliers, as you can readily verify. Mean. Standard Deviation. Variance Medians and quartiles are easily obtained by ordering and counting. practically without calculation. But they do not give full information on data: you can change data values to some extent without changing the median. Similarly for the qUaItiles. The average size of the data values can be measured in a more refined way by tlle mean 1 .t = - (5) n 2: Xj 11 j=l I = - (xl + X2 + ... + X,.). 11 This is the aritllmetic mean of the data values, obtained by taking their sum and dividing by the data si::e 11. Thus in (I). x= l~ (89 + 84 + ... + 89) = 6~1 = 87.3. Every data value contributes, and changing one of them will change the mean. Similarly, the spread (variability) of tlle data values can be measured in a more refined way by the standard deviation s or by its square, the variance (6) Thus, to obtain the variance of the data, take the difference .\1 - .r of each data value fi·om the mean, square it, take the sum of these 1l squares, and divide it by /I - 1 (not 11, as we motivate in Sec. 25.2). To get the standard deviation s, take the square root of S2. For example, using.t = 61117, we get for the data (I) tlle variance S2 = 113 [(89 - 6~1)2 + (84 - 6i l )2 + ... + (89 - 6i l )21 = 1~6 = 25.14. Hence the standard deviation is s = V176/7 = 5.014. Note that the standard deviation has the same dimension as the data values (kg/mm2, see at the beginning), which is an advantage. On the other hand, the variance is preferable to the standard deviation in developing statistical metl1Ods. as we shall see in Chap. 25. CAUTION! Your CAS (Maple, for instance) may use lin instead of 1/(n - 1) in (6), but the latter is better when 11 is small (see Sec. 25.2). --.... -_...........--- --....--................ .•. '1_1 ~ °1 DATA REPRESENTATIONS Represent the data by a stem-and-leaf plot, a histogram, and a boxplot: 1. 20 2. 7 21 6 20 4 0 19 7 20 1 2 19 21 4 6 19 6 3.56 58 54 33 38 38 49 39 4. 12.1 10 12.4 14.7 9.9 5. 70.6 70.9 69.1 71.1 68.9 70.3 41 10.5 30 44 9.2 37 17.2 51 46 56 11.4 11.8 71.3 70.5 69.7 7l.5 69.2 71.2 70.4 72.8 69.8 SEC. 24.2 997 Experiments, Outcomes, Events 6. -0.52 0.11 -0.48 0.94 0.24 -0.19 -0.55 7. Reaction time [sec] of an automatic switch 2.3 2.2 2.4 1.5 2.3 2.3 2.4 2.1 2.5 2.4 2.6 1.3 2.5 2.1 2.4 2.2 2.3 2.5 2.4 2.4 8. Carbon content ['k J of coal 89 87 90 86 89 82 84 85 !i8 89 80 76 90 87 89 86 88 86 90 !i5 9. Weight offilled bottles [g] in an automatic tilling process 403 399 398 401 400 401 401 10. Gasoline consumption [gallons per mile] of six cars of the same model 14.0 111-161 14.5 13.5 14.0 14.5 14.0 AVERAGE AND SPREAD Find the mean and compare it with the median. Find the standard deviation and compare it with the interquartile range. 24.2 11. The data in Prob. I. 12. 13. 14. 15. The data in The data in The data in The data in 16. 5 22 Prob. Prob. Prob. Prob. 7 23 2. 5. 6. 9. 6. Why is Ix - qMI so large? 17. Construct the simplest possible data with.t = 100 but qM = O. 18. (Mean) Prove that t must always lie between the smallest and the largest data values. 19. (Outlier, reduced data) Calculate s for the data 4 I 3 10 2. Then reduce the data by deleting the outlier and calculate s. Comment. 20. WRITING PROJECT. Average and Spread. Compare QM' IQR and .t, s, illustrating the advantages and disadvantages with examples of your own. Experiments, Outcomes, Events We now turn to probability theory. This theory has the purpose of providing mathematical models of situations affected or even governed by "chance effects," for instance, in weather foreca~ting, life insurance, quality of technical products (computers. batteries, steel sheets, etc.). traffic problems, and, of course, games of chance with cards or dice. And the accuracy of these models can be tested by suitable observations or experiments-this is a main purpose of statistics to be explained in Chap. 25. We begin by defining some standard terms. An experiment is a process of measurement or observation, in a laboratory, in a factDlY, on the street, in nature, or wherever; so "experiment" is used in a rather general sense. Our interest is in expeliments that involve randomness, chance effects, so that we cannot predict a result exactly. A trial is a single performance of an experiment. Its result is called an outcome or a sample point. n trial" then give a sample of size 11 consisting of n sample points. The sample space S of an experiment is the set of all pussible outcumes. E X AMP L E S 1 - 6 Random Experiments. Sample Spaces (1) In'peeting a IightbuIb. S = {Defective. Nondefeetive}. (2) RoIling a die. S = {I. 2. 3.4,5. Ii}. (3) Measuring tensile strength of wire. S the numbers in some interval. (4) Measuring coppel' content of brass. S: 50% to 909<, say. (5) Counting daily traffic acciden(, in New York. S the integers in some interval. (6) Asking for opinion about a new car model. S = (Like. Dislike. Undecided). • The subsets of S are called events and the outcomes simple events. E X AMP L E 7 Events In (2). evcms are A = {l. 3, 5} events are {I}. {2} .... , {6}. ("Odd /lllmbe,.··), B = {2. 4. 6} ("El'ell Illlmbe,."). C = {5. 6}, etc. Simple 998 CHAP. 24 Data Analysis. Probability Theory If in a trial an outcome a happens and a E A (a is an element of A), we say that A happens. For instance, if a die turns up a 3, the event A: Odd number happens. Similarly, if C in Example I happens (meaning 5 or 6 turns up), then D = {4, 5, 6} happens. Also note that S happens in each triaL meaning that some event of S always happens. All this is quite natural. Unions, Intersections, Complements of Events In connection with basic probability laws we shall need the following concepts and facts about events (subsets) A, B, C, ... of a given sample space S. The union A U B of A and B consists of all points in A or B or both. The intersection A n B of A and B consists of all points that are in both A and B. If A and B have no points in common. we write AnB=0 where 0 is the empty set (set with no elements) and we call A and B mutually exclusive (or disjoint) because in a trial the occurrence of A excludes thal of B (and conversely)if your die turns up an odd number, it cannot turn up an even number in the same trial. Similarly, a coin cannot turn up Head and Tail at the same time. Complement A C of A. This is the set of all the points of S Ilot in A. Thus, An A C = 0, AU N = S. In Example 7 we have A C = B, hence A U A C = {l, 2, 3, 4, 5, 6} = S. Another notation for the complement of A is A (instead of A C ), but we shall not use this because in set theory A is used to denote the closure of A (not needed in our work). Unions and intersections of more events are defined similarly. The union of events AI' ... , Am consists of all points that are in at least one Aj . Similarly for the union A I U A2 U ... of infinitely many subsets A b A 2, ... of an ififinite sample space S (that is, S consists of infinitely many points). The intersection of AI> ... , Am consists of the points of S thar are in each of these events. Similarly for the intersection Al n A2 n ... of infinitely many subsets of S. Working with events can be illustrated and facilitated by Venn diagrams I for showing unions, intersections. and complements, as in Figs. 509 and 510, which are typical examples thal give the idea. E X AMP L E 8 Unions and Intersections of 3 Events In rolling a die, consider the events A: Number greater thall 3, B: Number less thall 6. c: Evell /lumber. Then A n B = (4, 5), B n c = (2. 4). C n A = (4, 6), An B n c = {4). Can you sketch a Venn diagram of this? Furthermore. A U B = S. hence A U B U C = S (why?). • I JOHN VENN (1834-1923), English mathematician. SEC. 24.2 999 Experiments, Outcomes, Events s s Intersection A n B UmonAuB Fig. 509. Venn diagrams showing two events A and B in a sample space 5 and their union A U B (colored) and intersection A n B (colored) Fig. 510. A -- ---- ........ -- .... .. ........... 11-91 _....... = Venn diagram for the experiment of rolling a die, showing 5, {1, 3, 5}, C = {5, 6}, A U C = {l, 3, 5, 6}, A n C = {5} ...... ---- .-. ~.----- SAMPLE SPACES, EVENTS Graph a sample space for the expeliment: 1. Tossing 2 coins 2. Drawing 4 screws from a lot of right-handed and left-handed screws 3. Rolling 2 dice VENN DIAGRAMS 115-201 15. In connection with a trip to Europe by some students, consider the events P that they see Paris, G that they have a good time. and M that they run out of money. and describe in words the events 1, .. " 7 in the diagram. G 4. Tossing a coin until the fIrst Head appears 5. Rolling a die until the first "Si,," appears 6. Drawing bolts from a lot of 20, containing one defective D, until D is drawn. one at a time tmd assuming sampling without replacement. that is, bolts drawn are not returned to the lot 7. Recording the lifetime of each of 3 lightbulbs 8. Choosing a committee of 3 from a group of 5 people Problem 15 9. Recording the daily maximum tempemture X and the maximum air pressure Y at some point in a city 16. Using Venn diagrams, graph and check the mles 10. In Prob. 3, circle and mark the events A: Equal faces, B: Sum exceeds 9, C: SUIll equals 7. 11. In rolling 2 dice, are the events A: SIIII1 divisible by 3 and B: Sum dil'isible by 5 mutually exclusive? 12. Answer the question in Prob. 11 for rolling 3 dice. 13. In Prob. 5 list the outcomes that make up the event E: First "Six" in rolling at most 3 times. Describe E C • 14. List all 8 subsets of the sample space S = (a, b, c}. A U (B A n n C) = (A U B) (B U C) = (A n n (A U C) B) U (A n C) 17. (De Morgan's laws) Using Venn diagrams. graph and check De Morgan's laws (A U B)c = A C (A n n BC B)c = A C U B C • CHAP. 24 1000 Data Analysis. Probability Theory (A C)C 18. Using a Venn diagram. show that A <;;; B if and only if An B = A. SC A UN = S, 19. Show that, by the definition of complement, for any subset A of a sample space S, 24.3 = A, = 0, 0 c = S, AnN = 0. 20. Using a Venn diagram, show that A <;;; B if and only if AU B = B. Probability The "probability" of an event A in an experiment is supposed to measure how frequently A is about to occur if we make many trials. If we flip a coin, then heads H and tails T will appear about equally often-we say that Hand T are "equally likely." Similarly, for a regularly shaped die of homogeneous material ("fair die") each of the six outcomes 1, ... , 6 will be equally likely. These are examples of experiments in which the sample space S consists of finitely many outcomes (points) that for reasons of some symmetry can be regarded as equally likely. This suggests the following definition. First Definition of Probability DEFINITION 1 If the sample space S of an experiment consists of finitely many outcomes (points) that are equally likely, then the probability peA) of an event A is (1) peA) = Number of points in A Number of points in S From this definition it follows immediately that. in particular, (2) E X AMP L E 1 peS) = 1. Fair Die In rolling a fair die once. what is the probability peA) of A of obtaining a 5 or a 6? The probabilIty of B: "El'en 11l1llzber"? Solutioll. The six outcomes are equally likely. so that each has probability 1/6. Thus peA) = 2/6 = 1/3 • because A = (5, 6J has 2 points, and PCB) = 3/6 = 112. Definition 1 takes care of many games as well as some practical applications, as we shall see, but celtainly not of all experiments, simply because in many problems we do not have finitely many equally likely outcomes. To arrive at a more general definition of probability, we regard probability as the coullterpart of relative frequellcy. Recall from Sec. 24.1 that the absolute frequency f(A) of an event A in n trials is the number of times A occurs, and the relative frequency of A in these trials is f(A)ln; thus (3) Number of times A occurs Number of trials SEC 24.3 1001 Probability Now if A did not occur, then f(A) = O. If A always occurred. then f(A) = the extreme cases. Division by 11 gives 11. These are (4*) In particular, for A = S we have f(S) = 11 because S always occurs (meaning that some event always occurs; if necessary, see Sec. 24.2, after Example 7). Division by 11 gives (5*) fre1(S) = I. Finally. if A and B are mutually exclusive, they cannot occur together. Hence the absolute frequency of their union A U B must equal the sum of the absolute frequencies of A and B. Division by 11 gives the same relation for the relative frequencies, (6*) f re1(A U B) = f rei (A) + n (A fre1(B) B = 0). We are now ready to extend the definition of probability to experiments in which equally likely outcomes are not available. Of course, the extended definition should include Definition 1. Since probabilities are supposed to be the theoretical cOllnterpmt of relative frequencies, we choose the properties in (4*), (5*), (6"") as axioms. (Historically, such a choice is the result of a long process of gaining experience on what might be best and most practical.) DEFINITION 2 General Definition of Probability Given a sample space S, with each event A of S (subset of S) there is associated a number P(A), called the probability of A. such that the following axioms of probability are satisfied. 1. For every A in S, (4) 0:0;: peA) :0;: I. 2. The entire sample space S has the probability (5) peS) = 1. 3. For mutually exclusive events A and B (A (6) peA U B) n = peA) + B = 0; see Sec. 24.2), PCB) (A n B = 0). If S is infinite (has infinitely many points). Axiom 3 has to be replaced by 3'. For mutually exclusive events AI> A 2 • • . • , (6') In the infinite case the subsets of S on which peA) is defined are restricted to form a so-called u-algebra. as explained in Ref. (GR6j (not (G6]!) in App. 1. This is of no practical consequence to us. CHAP. 24 1002 Data Analysis. Probability Theory Basic Theorems of Probability We shall see that the axioms of probability will enable us to build up probability theory and its application to statistics. We begin with three basic theorems. The first of them is useful if we can get the probability of the complement A C more easily than peA) itself. THEOREM 1 Complementation Rule For an event A and its complemellt A C in a sample space S, (7) PROOF peN) I - peA). By the definition of complement (Sec. 24.2). we have S Hence by Axioms 2 and 3, I EXAMPLE 2 = = peS) = peA) + P(A C ). = A U A C and A n AC 0. • thus Coin Tossing Five coin~ are tossed simultaneously. Find the probability of the event A: At least one head tums up. Assume that the coins are fair. Solution. Since each coin can [Urn up heads or mils, the sample space consists of 25 = 32 Olilcomes. Since the coins are fair. we may assign the same probability (1/32) to each outcome. Then the event A C (No heads c c tum LIp) consists of only 1 outcome. Hence P(A ) = 1/32, and the answer is peA) = 1 - P(A ) = 31/32. • The next theorem is a simple extension of Axiom 3, which you can readily prove by induction. THEOREM 2 Addition Rule for Mutually Exclusive Events For mutually exclusive events AI, ... , Am in a sample space S, E X AMP L E 3 Mutually Exclusive Events If the probability that on any workday a garage will get 10-20,21-30,31-40, over 40 cars to service is 0.20, 0.35, 0.25, 0.12, respectively, what is the probability that on a given workday the garage gets at least 21 cars to service? Solution. Since these are mutually exclusive events, Theorem 2 gives the answer 0.35 + 0.25 + 0.12 = 0.72. Check this by the complementation rule. • In many cases, events will not be mutually exclusive. Then we have THEOREM 3 Addition Rule for Arbitrary Events For events A and B in (9) 1I sample space, peA U B) = peA) + PCB) - peA n B). SEC. 24.3 1003 Probability PROOF C, D. E in Fig. 511 make up A U B and are mutually exclusive (disjoint). Hence by Theorem 2. peA U B) = P( C) + P(D) + This gives (9) because on the right P(C) and peE) = PCB) - PW) = PCB) - peA + P(D) = peA) by Axiom 3 and disjointness; n B). also by Axiom 3 and disjointness. • B A Fig. 511. peE). Proof of Theorem 3 Note that for mutually exclusive events A and B we have A by comparing (9) and (6), (10) P(0) (Can you also prove this by (5) and E X AMP L E 4 n B = 0 by definition and. = O. om Union of Arbitrary Events In tossing a fair die. what is the probability of getting an odd number or a number less than 4? Solutioll. Let A be the event "Odd number" and B the event "Numberless than 4." Then Theorem 3 gives the answer P(AUB)=~+~-~=~ because A n • B = .. Odd lIl//uber less thall 4" = {I, 3}. Conditional Probability. Independent Events Often it is required to find the probability of an event B under the condition that an event A occurs. This probability is called the conditional probability of B given A and is denoted by p(BIA). In this case A serves as a new (reduced) sample space, and that probability is the fraction of peA) which corresponds to A n B. Thus (11) p(BIA) = PeA n B) peA) [peA) *- 01. [PCB) *- Similarly, the cunditiO/wl probability of A gil'en B is p(AIB) (12) PCB) Solving (II) and (12) for peA THEOREM 4 = peA n B) n B), we obtain Multiplication Rule If A and B lire events in a sample space Sand peA) *- 0, PCB) *- O. then (13) peA n B) = P(A)P(BIA) = P(B)P(AIB). 0]. 1004 E X AMP L E 5 CHAP. 24 Data Analysis. Probability Theory Multiplication Rule In producing screw~.letA mean "screw too slim" and B "screw too short." Let peA) = 0.1 and let the conditional probability that a slim sere", is also too short be p(BIA) = n.2. What is the probability that a scre'" that we pick randoml) from the lot produced will be both too slim and too short? Solution. PIA n B) = p(A)p(BIA) = 0.1 • 0.2 = 0.02 = 2'7c. by Theorem 4. Independent Events. • If events A and B are such that (14) P(A n B) = P(A)P(B), they are called independent events. Assuming P(A) that in this case p(AIB) = peA), * O. P(B) * 0, we see from (II )-( 13) P(BIA) = PCB). This means that the probability of A does not depend on the occurrence or nonoccurrence of B, and conversely. This justifies the term "independent." Independence of III Events. Similarly, 111 events A I, ••• , Am are called independent if (ISa) as well as for every k different events A jl , Ah , ... , Aj ". (ISb) where k = 2. 3, ... , 111 - l. Accordingly. three events A. B. C are independent if and only if peA n B) = P(A)P(B), n 0 = P(B)P(0, P(C n A) = P(C)P(A). PCB (16) peA n B n0 = P(A)P(B)P(O. Sampling. Our next example has to do with randomly drawing objects, one at a time, from a given set of objects. This is called sampling from a popUlation, and there are two ways of sampling, as follows. 1. In sampling with replacement, the object that was drawn at random is placed back to the given set and the set is mixed thoroughly. Then we draw the next object at random. 2. In sampling without replacement the object that was drawn is put aside. E X AMP L E 6 Sampling With and Without Replacement A box contains 10 screws. three of which are defective. Two screws are drawn at random. Find the probability that none of the lWO screws is defective. Solution. We consider the event~ A: First drawn screll' 1100ldefectil'e. B: Second drawl1 screw /101ulefectil'e. SEC. 24.3 1005 Probability Clearly. PIA) = -k becau,e 7 of the 10 screw, are nomlefective and we sample at ramI om. so thm each screw has the same probability (to) of being picked. If we sample with replacement. the situation before the second drawing is the same as at the beginning. and PCB) = -k. The events are independent. and the answer is PeA n B) = P(A)P(B) = 0.7' 0.7 = 0.49 = 49<)}. If we sample without replacement. then PIA) = in the box, 3 of which are defective. Thus P(B/A) PIA n -k, a~ before. If A has occurred. then there are 9 ~crews left = ~ = ~. and 1l1eorem 4 yields the answer B) = -k . ~ = 47<)}. • Is it intuitively clear that this value mu,t be smaller than the preceding one'! --.•.. .... - ••• "'A 1. Three screws are drawn at random from a lot of 100 screws. 10 of which are defective. Find the probability that the screws drawn will be nondefective in drawing (a) with replacement. (b) without replacement. 11. In roIling two fair dice. what is the probability of 2. In Prob. I find the probability of E: At least I defective (i) directly. (ii) by using complements; in both cases (a) and (b). 13. A motor drives an eiectIic generator. During a 30-day period. the motor needs repair with probability 8%- and the generator needs repair with probability 49'r. What is the probability that during a given period. the entire apparatus (consisting of a motor and a generator) will need repair? 3. If we inspect paper by drawing 5 sheets without replacement from every batch of 500. what is the probability of getting 5 clean sheets although 2% of the sheets contain spots') First guess. 4. Under what conditions will it make practically no difference whether we sample with or witholll replacement? Give numeric examples. 5. If you need a right -handed screw from a box containing 20 right-handed and 5 left-handed screws. what is the probability that you get at least one right-handed screw in drawing 2 screws with replacement? 6. If in Prob. 5 you draw without replacement. does the probability decrease or increase? First think, then calculate. 7. What gives the greater probability of hitting some target at least once: (a) hitting in a shot with probability 112 and firing I shot. or (b) hitting in a shot with probability 1I4 and firing 2 shots? First guess. Then calculate. 8. Suppose that we draw cards repeatedly and with replacement from a file of 100 cards. 50 of which refer to male and 50 to female persons. What is the probability of obtaining the second "female" card before the third "male" card? 9. What is the complementary event of the event considered in Prob. 8'1 Calculate its probability and use it to check your result in Prob. 8. 10. In rolling two fair dice. what is the probability of obtaining a sum greater than 4 but not exceeding 7'1 obtaining equal numbers or numbers with an even product? 12. Solve Prob. II by considering complements. 14. If a circuit contains 3 automatic switches and we want that. with a probability of 959'c. during a given time interval they are all working. what probability of failure per time interval can we admit for a single switch? 15. If a certain kind of tire has a life exceeding 25 000 miles with probability 0.95. what is the probability that a set of 4 of these tires on a car will last longer than 25000 miles? 16. In Prob. 15. what is the probability that at least one of the tires will not last for 25 000 miles? 17. A pressure control apparatus contains 4 valves. The apparatus will not work unless all valves are operative. If the probability of failure of each valve during some interval of time is 0.03. what is the corresponding probability of failure of the apparatus? 18. Show that if B is a subset of A, then P(B) 19. Extending Theorem 4. show that PeA n B n C) = P(A)P(BIA)P(CiA n ~ peA). B). 20. You may wonder whether in (16) the last relation follows from the others. but the answer is no. To see this. imagine that a chip is drawn from a box containing 4 chips numbered 000,01 I. 101, 110. and let A. B. C be the events that the first. second. and third digit. respectively. on the drawn chip is 1. Show that then the first three formulas in (16) hold but the last one does not hold. CHAP. 24 1006 Data Analysis. Probability Theory 24.4 Permutations and Combinations Permutations and combinations help in finding probabilities peA) = alk by systematically counting the number a of pointf> of which an event A consists; here, k is the number of points of the sample space S. The practical difficulty is that a may often be surprisingly large, so that actual counting becomes hopeless. For example, if in assembling some instrument you need 10 different screws in a certain order and you want to draw them randomly from a box (which contains nothing else) the probability of obtaining them in the required order is only 1/3 628800 because there are 10! = I . 2' 3 ·4·5·6·7' 8 . 9 . 10 = 3628800 orders in which they can be drawn. Similarly, in many other situations the numbers of orders, anangements, etc. are often incredibly large. (If you are unimpressed. take 20 screws-how much bigger will the number be?) Permutations A permutation of given things (elements or objects) is an arrangement of these things in a row in some order. For example, for three letters a, b, c there are 3! = I . 2 . 3 = 6 permutations: abc, acb, bac, bca, cab, cba. This illustrates (a) in the following theorem. THEOREM 1 Permutations (a) Different t/zings. The lUt1llber of permutations of n different things taken all at a time is (1) n! = I . 2 . 3 ... l (read "n.factorial"). Il (b) Classes of equal things. If n given things can be divided into c classes of alike things differing from class to class, then the number of permutations of these things taken all at a time is (2) n! (Ill + n2 + ... + llc = 11) where nj is the 1lumber of things ill the jth class. PROOF (a) There are 11 choices for filling the first place in the row. Then n - I things are still available for filling the second place, etc. (b) 171 alike things in class 1 make n 1 ! permutations collapse into a single permutation (those in which class I things occupy the same 111 positions), etc., so that (2) follows • from (I). SEC. 24.4 1007 Permutations and Combinations E X AMP L E 1 Illustration of Theorem l(b) If a box contains 6 red and 4 blue balls. the probability of drawing first the red and then the blue balls is p ~ • 6!4!110! = 11210 = 0.5%. A permutation of n things taken k at a time is a permutation containing only k of the n given things. Two such permutations consisting of the ~ame k elements. in a different order, are different, by definition. For example, there are 6 different permutations of the three letters a, b, e, taken two letters at a time, ab, ae, be, ba, ea, eb. A permutation of Il things taken k at a time with repetitions is an arrangement obtained by putting any given thing in the first position, any given thing, including a repetition of the one just used, in the second, and continuing until k positions are filled. For example, there are 32 = 9 different such permutations of a, b, e taken 2 letters at a time, namely, the preceding 6 permutations and aa, bb, ec. You may prove (see Team Project 18): THEOREM 2 Permutations TIle number of different pennutations of n different things taken k at a time without repetitions is (3a) n(n - 1)(n - 2) ... (n - k + 1) = n! (n - k)! alld with repetitions is (3b) E X AMP L E 2 Illustration of Theorem 2 In a coded telegram the letters are arranged in groups of five letters, called words. From (3b) we see that the number of different such words is 265 = II 8!B 376. From (3a) it follows that the number of different such words containing each letter no more tiMn once is 26!/(26 - 5)! = 26' 25 . 24 . 23' 22 = 7893600. • Combinations In a permutation, the order of the selected things is essential. In contrast, a combination of given things means any selection of one or more things without regard to order. There are two kinds of combinations, as follows. The number of combinations of Il different things, taken k at a time, without repetitions is the number of sets that can be made up from the n given things, each set containing k different things and no two sets containing exactly the same k things. The number of combinations of Il different things, taken k at a time, with repetitions is the number of sets that can be made up of k things chosen from the given n things, each being used as often as desired. CHAP. 24 1008 Data Analysis. Probability Theory For example, there are three combinations of the three letters a, b, c, taken two letters at a time, without repetitions. namely, lib, lIC, bc, and six such combinations with repetitions, namely, lib, lIC, be, aa, bb, cc. Combinations THEOREM 3 The Ilumber of d~fferent combinations of 11 d!fferent things taken. k at a time. withollt repetitiollS. is (4a) C) = 1) ... (n - k n(n - n! + I) I ' 2· .. k k!(n - k)! lind the number of those combinlltions with repetitions is (4b) PROOF E X AMP L E 3 The statement involving (4a) follows from the first part of Theorem 2 by noting that there are k! permutations of k things from the given n things that differ by the order of the elements (see Theorem I). but there is only a single combination of those k things of the type characterized in the first statement of Theorem 3. The last statement of Theorem 3 can be proved by induction (see Team Project 18). • Illustration of Theorem 3 The number of samples of five lightbulbs that can be selected from a 1m of 500 bulbs is rsee (4a 11 500 . 499 . 498 . -1-97 . 496 500) 500! = 255 244 687 600. ( 5 = 5!495! = 1·2·3'4·5 • Factorial Function In (I )-(4) the factorial function is basic. By definition. (5) O! = L Values may be computed recursively from given values by (6) (n + l)! = (II + l)n!. For large 11 the function is very large (see Table A3 in App. 5). A convenient approximation for large 11 is the Stirling formula 2 (7) 2 jAMES (e STIRUI\G 0692-1770). Scots mathematician. = 2.718 ... ) SEC. 24.4 1009 Permutations and Combinations where - is read "asymptotically equal" and means that the ratio of the two sides of (7) approaches 1 as 11 approaches infinity. EXAMPLE 4 Stirling Formula [ n! By (7) Exact Value Relative Error 4! I IO! 23.5 3598696 2.422 79 . 1018 24 3628800 2 432 902 008 176 640 000 2.1% 0.8% 0.4% l 20! • Binomial Coefficients The binomial coefficients are defined by the formula 0) = o(a (k (8) 1)(0 - 2) ... (0 - k + 1) ~ (k k! 0, integer). The numerator has k factors. FUlthermore. we define (~) (9) For integer II = 11 (~) in particular, = I, = 1. we obtain from (8) (n ~ 0, 0 ~ k ~ n). (10) Binomial coefficients may be computed recursively, because 0 (11) ( k + + 1) 1 ~ (k 0, integer). Formula (8) also yields (k (12) ~ 0, integer) (m > 0). There are numerous further relations; we mention two important ones, (13) ~ s=o (k+S) = (l1+k) k k + (k ~ O. n ~ 1, both integer) 1 and (14) (r ~ 0, integer). 1010 CHAP. 24 Data Analysis. Probability Theory 1. List all pennutations of four digits I, 2, 3, 4, taken all at a time. 2. List (a) all permutations, (b) all combinations without repetitions, (c) all combinations with repetitions, of 5 letters G, e, i, 0, 1I taken 2 at a time. 3. In how many ways can we assign 8 workers to 8 jobs (one worker to each job and conversdy)? 4. How many samples of 4 objects can be drawn from a lot of 80 objects? 5. In how many different ways can we choose a committee of 3 from 20 persons? First guess. 6. In how many different wa) s can we select a committee consisting of 3 engineers. 2 biologists. and 2 chemists from 10 engineers, 5 biologist~, and 6 chernists? First guess. 7. Of a lot of 10 items, 2 are defective. (a) Find the number of different samples of 4. Find the number of samples of 4 containing (b) no defectives, (c) I defective, (d) 2 defectives. 8. If a cage contains 100 mice, two of which are male, what is the probability that the two male mice will be included if 12 mice are randomly selected? 9. An urn contains 2 blue, 3 green, and 4 red balls. We draw I ball at random and put it aside. Then we draw the next ball, and so on. Find the probability of drawing at first the 2 blue balls, then the 3 green ones, and finally the red ones. 10. By what factor is the probability in Prob. 9 decreased if the number of balls is doubled (4 blue, etc.)? 11. Detennine the number of different bridge hands. (A bridge hand consists of 13 cards selected from a full deck of 52 cards.) 12. In how many different ways can 5 people be seated at a round table? 13. If 3 suspects who committed a burglary and 6 innocent persons are lined up, what is the probability that a witne~s who is not sure and has to pick three persons will pick the three suspects by chance? That the witness picks 3 innocent persons by chance? 24.5 14. (Birthday problem) What is the probability that in a group of 20 people (that includes no twins) at least two have the same birthday. if we assume that the probability of having birthday on a given day is 11365 for every day. First guess. 15. How many different license plates showing 5 symbols, nanlely, 2 letters followed by 3 digits. could be made? 16. How many automobile registrations may the police have to check in a hit-and-run accident if a witness reports KDP5 and cannot remember the last two digits on the license plate but is certain that all three digits were different? 17. CAS PROJECT. Stirling formula. (a) Using (7), compute approximate values of n! for 11 = I, ... ,20. (b) Detennine the relative error in (a). Find an empirical formula for that relative error. (el An upper bound for that relative error is e 1/ 12n Try to relate your empirical formula to this. - L. (d) Search through the literature for further infonnation on Stirling's fonnula. Write a short report about your findings, arranged in logical order and illustrated with numeric examples. 18. TEAM PROJECT. Permutations, Combinations. (a) Prove Theorem 2. (b) Prove the last statement of Theorem 3. (e) Derive (11) from (8). (d) By the binomial theorem, so that llkbn - k has the coefficient (~). Can you conclude this from Theorem 3 or is this a mere coincidence? (e) Prove (14) by using the binomial theorem. (f) Collect further formulas for binomial coefficients from the literature and illustrate them numerically. Random Variables. Probability Distributions In Sec. 24.1 we considered frequency distributions of data. These distributions show the absolute or relative frequency of the data values. Similarly, a probability distribution or, briefly, a distribution, shows the probabilities of events in an experiment. The quantity that we observe in an experiment will be denoted by X and called a random variable (or SEC 24.5 1011 Random Variables. Probability Distributions stochastic variable) because the value it will assume in the next trial depends on chance, on randomness-if you roll a dice, you get one of the numbers from I to 6, but you don't know which one will show up next. Thus X = Number a die tUnIS up is a random variable. So is X = Elasticity of rubber (elongation at break). ("Stochastic" means related to chance.) If we COUllt (cars on a road, defective screws in a production, tosses until a die shows the first Six), we have a discrete random variable and distribution. If we measure (electric voltage, rainfall, hardness of steel), we have a continuous random variable and distribution. Precise definitions follow. In both cases the distribution of X is determined by the distribution function F(x) (1) = P(X ~ x); this is the probability that in a trial, X will assume any value not exceeding x. CAUTION! The terminology is not uniform. F(x) is sometimes also called the cumulative distribution function. For (I) to make sense in both the discrete and the continuous case we formulate conditions as follows. DEFINITION r Random Variable A random variable X is a function defined on the sample space S of an experiment. Its values are real numbers. For every number 11 the probability P(X = a) with which X assumes a is defined. Similarly, for any interval 1 the probability P(X E I) with which X assumes any value in 1 is defined. Although this definition is very general, practically only a very small number of distributions wil1 occur over and over again in applications. From (I) we obtain the fundamental formula for the probability corresponding to an interval a < x ~ b, Pea < X (2) ~ b) = F(b) - F(a). This follows because X ~ a ("X assumes any value not exceeding a") and a < X ~ b ("X assumes any value in the illterval a < x ~ b") are mutually exclusive events, so that by (1) and Axiom 3 of Definition 2 in Sec. 24.3 F(b) = P(X ~ b) = P(X ~ = F(a) + and subtraction of F(a) on both sides gives (2). a) + Pea < X Pea < X ~ b) ~ b) 1012 CHAP. 24 Data Analysis. Probability Theory Discrete Random Variables and Distributions By definition, a random variable X and its distribution are discrete if X assumes only finitely many or at most countably many values Xl' -'"2' X3, .... called the possible values of X, with positive probabilities PI = P(X = Xl), P2 = P(X = X2), P3 = P(X = X3), ... , whereas the probability P(X E 1) is zero for any interval 1 containing no possible value. Clearly, the discrete distribution of X is also determined by the probability function f(x) of X, defined by p (3) f(x) = { ~ (j = 1,2. ... ), otherwise From this we get the values of the distribution function F(x) by taking sums, (4) where for any given x we sum all the probabilities Pj for which Xj is smaller than or equal to that of x. This is a step function with upward jumps of size Pj at the possible values Xj of X and constant in between. E X AMP L E 1 Probability Function and Distribution Function Figure 512 shows the probability function f(x) and the distribution function F(x) of the discrete random variable x= Number a fair die tums up. X has the possible values x = 1, 2, 3, 4, 5, 6 with probability 116 each. At these x the distribution function has upward jumps of magnitude 1/6. Hence from the graph of flx) we can construct the graph of F(x), and conversely. In Figure 512 (and the next one) at each jump the fat dot indicates theftmctioll value at the jump! • t(xl Y6[ I I I I I I o 5 x o 5 10 12 x 12 x F(x) F(x) 1 30 36 20 36 1 2" 10 36 o , 5 x Fi-. 512. Probability function {(x) and distribution function F(x) of the random variable X = Number obtained in tossing a fair die once ! [ 10 Fig. 513. Probability function {(x) and distribution function F(x) of the random variable X = Sum of the two numbers obtained in tossing two fair dice once SEC. 24.5 1013 Random Variables. Probability Distributions E X AMP L E 2 Probability Function and Distribution Function The random variable X = Slim of the two Illlll/bers t ....o fair dice tum "I' is discrete and has the possible values 2 (= 1 + II. 3.4..... 12 (= 6 + 6). There are 6' 6 = 36 equally likely outcomes (I. 1) (I. 2)....• (6. 6). where the first number is lhat shown on the first die and the second number that on the other die. Each such outcome has probability 1/36. Now X = 2 occurs in the case of the outcome O. I): X = 3 in the case of the lwo outcomes (I. 2) and (2. I \: X = 4 in the case of the three outcomes (1. 3), (2. 2). (3. I): and so on. Hence fix) = PIX = x) and F(x) = PIX ~ x) have the values I i x 2 3 4 5 f(x) 1136 1136 2/36 3/36 3/36 6/36 I F(x) 6 7 8 9 10 II 12 4/36 5/36 15/36 6/36 21/36 5/36 26/36 4/36 10136 30136 3/36 33136 2136 35/36 1136 36/36 Figure 513 shows a bar chan of this function and the graph of the distribution funcllon. which is again a srep function. with jumps (of different height!) at the possible values of X. Two useful formulas for discrete distributions are readily obtained as follows. For the probability cOlTesponding to intervals we have from (2) and (4) (5) P(a < X~ = b) = F(b) - F(a) 2: Pj (X discrete). a<xJ~b This is the sum of all probabilities Pj for which Xj satisfies a < Xj ~ b. (Be careful about < and ~!) From this and peS) = I (Sec. 24.3) we obtain the following formula. 2: Pj = (6) (sum of all probabilities). I j E X AMP L E 3 illustration of Formula (5) In Example 2. compme [he probability of a sum of at least 4 and at Solution. E X AMP L E 4 P(3 < X ~ 8) = F(8) - F(3) = ~ - mo~t 8. • :k = ~. Waiting Time Problem. Countably Infinite Sample Space In to~sing a fair coin. let X = Number of trials limit the first head appears. Then. by independence of events (Sec. 24.3). PiX = 1) = P(H) _ 1 = ~.~ -2 PIX = 2) = P(TH) PIX = 3) = P(TTH) = and in general PiX series. = II) = (!)n. II ! .~ .! = ~. (H = Head) (T= Tail) etc. = 1. 2•.... Also. (6) can be confirmed by the slim form lila for the geometric 1 2 + 1 4 1 + 8 + ... = -I + 1- ~ =-1+2=1. • 1014 CHAP. 24 Data Analysis. Probability Theory Continuous Random Variables and Distributions Discrete random variables appear in experiments in which we count (defectives in a production, days of sunshine in Chicago. customers standing in a line, etc.). Continuous random variables appear in experiments in which we measure (lengths of screws, voltage in a power line, Brinell hardness of steel, etc.). By definition. a random variable X and its distribution are of continuous type or, briefly. continuous, if its distribution function F(x) [defined in (1)] can be given by an integral (7) = F(x) r f(v) dv -00 (we write v because x is needed as the upper limit of the integral) whose integrand f(x), called the density of the distribution, is nonnegative, and is continuous, perhaps except for finitely many x-values. Differentiation gives the relation of f to F as (8) f(x) = F'(x) for every x at which f(x) is continuous. From (2) and (7) we obtain the very important formula for the probability corresponding to an interval: (9) Pta <X~ b) = F(b) - F(a) = I b f(v) dv. a This is the analog of (5). From (7) and peS) = I (Sec. 24.3) we also have the analog of (6): I (10) oo f(v) dv = 1. _00 Continuous random variables are simpler than discrete ones with respect to intervals. Indeed, in the continuous case the four probabilities corresponding to a < X ~ b, a < X < b, a ~ X < b, and a ~ X ~ b with any fixed a and b (> a) are all the same. Can you see why? (Answer. This probability is the area under the density curve, as in Fig. 514, and does not change by adding or subtracting a single point in the interval of integration.) This is different from the discrete case! (Explain.) The next example illustrates notations and typical applications of our present formulas. Curve of density f(X)~ / 1~<bJ a Fig. 514. b x Example illustrating formula (9) SEC. 24.5 1015 Random Variables. Probability Distributions E X AMP L E 5 Continuous Distribution Let X have the density function I(x) = O.75( I - x 2 ) if -I ~ x ~ I and 7ero otherwi~e. Find the distribution function. Find the probabilities P( -~ ~ X ;;; ~) and PC! ~ X ~ 2). Find x such that P(X ~ x) = 0.95. Solutioll. From (7) F(x) we obtain F(x) = 0 if x ~ -I, = 0.75 IX (I - v 2 ) dv = 0.5 + 0.75x - 0.25x3 if-I <x~ I, -1 and F( t) = I if x > l. From this and (9) we get P{-~ ~ X ~~) = F@ - F{-~) = 0.75 I 1/2 (I - v 2 ) dv = 68.75% -1/2 (because P( -~ ~ X ~~) = P{ -~ < X ;;; ~) for a continuous distribution) and P{~ ~ X ~ 2) = F(2) - F(!) = 0.75 f 1 (1 - v 2 ) dv = 31.64%. 1/4 (Note that the upper limit of integration is I, not 2. Why?) Finally, P{X ~ x) = F(x) = 0.5 + 0.75x - 0.25,3 = 0.95. 3 Algebraic simplification gives 3x - x = 1.8. A solution is x = 0.73, approximately. Sketch fet) and mark x = -~,~, !. and 0.73. so that you can see the results (the probabilities) as areas under the curve. Sketch also Fex). • Further examples of continuous distributions are included in the next problem set and in later sections. -_·.·_.w· ........-...____ - ..... ,--.-----....,;. ..... y _ ....... . - . - - , _ h 2 (x = 1, 2, 3,4, 5; k suitable) and the distribution function. Graph the density function fIx) = b: 2 (Q ~ X ~ 5: k suitable) and the distribution function. (Uniform distribution) Graph f and F when the density is f(x) = k = const if -4 ~ x ~ 4 and 0 elsewhere. In Prob. 3 find P(O ~ x ~ 4) and c such that P( -c < X < c) = 95%. Graph f and F when f{ -2) = f(2) = 1/8, f( -1) = f(1) = 3/8. Can f have further positive values? Graph the distribution function F(x) = I - e- 3x if x > 0, F(x) = 0 if x ~ 0, and the density f(x). Find x such that F(x) = 0.9. Let X be the number of years before a particular type of machine will need replacement. Assume that X has the probability function f(l) = 0.1, f(2) = 0.2, f(3) = 0.2, 1(4) = 0.2. 1(5) = 0.3. Graph I and F. Find the probability that the machine needs no 1. Graph the probability function f(x) 2. 3. 4. 5. 6. 7. = replacement during the first 3 years. 8. If X has the probability function f(x) = k/2x (x = O. l. 2 ... '). what are k and P(X ~ 4)? 9. Find the probability that none of the three bulbs in a traffic signal must be replaced during the first 1200 hours of operation if the probability that a bulb must be replaced is a random variable X with density f(x) = 6[0.25 - (x - 1.5)2] when 1 ~ x ~ 2 and fIx) = 0 otherwise. where x is time mea~ured in mUltiples of 1000 hours. 10. Suppose that certain bolts have length L = 200 + X mm, where X is a random variable with density f(x) = ~(l - x 2 ) if - 1 ~ x ~ I and 0 otherwise. Determine c so that with a probability of 95% a bolt will have any length between 200 - c and 200 + c. Hint: See also Example 5. 11. Let X [millimeters] be the thickne~~ of washers a machine turns out. Assume that X has the density f(x) = h if 1.9 < x < 2.1 and 0 otherwise. Find k. What is the probability that a washer will have thickness between 1.95 mm and 2.05 mm? 1016 CHAP. 24 Data Analysis. Probability Theory that X is between ::!.5 (40% profit) and 5 (20% profit)? 12. Suppose that in an automatic process of filling oil into cans, the content of a can (in gallons) is Y = 50 + X. where X is a random variable with density fIx) = I - Ixl when Ixl :=; 1 and 0 when Ixl > I. Graph fIx) and F(x). In a lot of 100 cans, about how many will contain 50 gallons or more'? What is the probability that a can will contain less than 49.5 gallons? Less than 49 gallons? 13. Let the random variable X with density fIx) = ke-;c if o ~ x :=; 2 and 0 otherwise (x = time measured in years) be the time after which cel1ain ball bearings are worn out. Find k and the probability that a bearing will last at least I year. 14. Let X be the ratio of sales to profits of some fiml. Assume that X has the distribution function F(x) = 0 if x < 2, F(x) = (x 2 - 4)/5 if 2 :=; x < 3. F(x) = I if x ~ 3. Find and graph the density. What is the probability 15. Show that b < c implies P(X:=; b) :=; PIX :=; c). 16. If the diameter X of axles has the density fIx} = k if 119.9 :=; x ~ 120.1 and 0 otherwise, how many defectives will a lot of 500 axles approximately contain if defectives are axles slimmer than 119.92 or thicker than 120.08? 17. Let X be a random variable that can as~ume evelY real value. What are the complements of the events X ~ b. X<~X~~X>~b:=;X~~b<X~~ 18. A box contains 4 right-handed and 6 left-handed screws. Two screws are drawn at random without replacement. Let X be the number of left-handed screws drawn. Find the probabilities PIX = m. PIX = I). PIX = 2). P( I < X < 2), P(X :=; I). PIX ~ I). PIX > I), and P(0.5 < X < 10). 24.6 Mean and Variance of a Distribution The mean J-L and variance 0'2 of a random variable X and of its distribution are the theoretical counterpalts of the mean x and variance S2 of a frequency distribution in Sec. 24.1 and serve a similar purpose. Indeed, the mean characterizes the central location and the variance the spread (the variability) of the distribution. The mean J-L (mu) is defined by (a) /1- = 2: xjf(xj) (Discrete distribution) j (1) (b) J-L = {" xf(x) dx (Continuous distribution) -x and the variance 0'2 (sigma square) by (a) 0'2 = 2: L~j - J-L)2f(xj) (Discrete distribution) j (2) (b) 0'2 = f"" (x - J-L)2f{x) dx (Continuous distribution). -= a (the positive square root of (T2) is called the standard deviation of X and its distribution. f is the probability function or the density, respectively, in (a) and (b). The mean J-L is also denoted by E(X) and is called the expectation of X because it gives the average value of X to be expected in many trials. Quantities such as J-L and 0'2 that measure certain properties of a distribution are called parameters. J-L and 0'2 are the two most important ones. From (2) we see that (3) (except for a discrete "distribution" with only one possible value, so that 0'2 = 0). We assume that J-L and 0'2 exist (are finite), a<; is the ca<;e for practically all distributions that are useful in applications. SEC. 24.6 1017 Mean and Variance of a Distribution E X AMP L E 1 Mean and Variance The random variable X = Number o.{heads ina single toss o{ a lair coin has the possible values X = 0 and X = I + 1 = ~. and with probabiliues PIX = 0) = ~ and PIX = 1) = From (1 a) we thus obtain the mean f.L = (2a) yields the variance o·l i. ·l • E X AMP L E 1 Uniform Distribution. Variance Measures Spread The distribution with the den~ity fIx) = b - a<x<b if 1I and.f = 0 otherwise is called the uniform distribution on [he interval a < x < h. From (I b) (or from Theorem I. below) we find that f.L = (0 + b)l2. and (2b) yields the variance I (. _(/ b 2 _ (T - a .~ + b)2 2 2 _1_ b-l/ ._ d.\- (b - l/) 12 • Figure 515 illustrates that the spread is large if and only if (T2 is large. ((x) {(x) If-----, o 1 (0 2 = 1/12) x 1 o -1 F(x) 2 x 2 x F(x) 1 -1 Fig. 515. Uniform distributions having the same mean (0.5) but different variances u' Symmetry. We can obtain the mean JL without calculation if a distribution is symmetric. Indeed, you may prove THEOREM 1 Mean of a Symmetric Distribution If a distribution is symmetric ~rith re!oJpect to x thell JL = c. (Examples I and 2 illustrate this.) = c, that is, f(c - x) = f(c + x), Transformation of Mean and Variance Given a random variable X with mean JL and variance (]"2. we want to calculate the mean and variance of X* = al + a 2X. where al and a2 are given constants. This problem is important in statistics, where it appears often. CHAP. 24 1018 THE 0 REM 2 Data Analysis. Probability Theory Transformation of Mean and Variance (a) If a random variable X has mean J.L and variance (]"2, then the random variable (4) has the mean J.L* and variance (]"*2. where (5) and (b) In particular, the standardized random variable Z corresponding to X, given by X-J.L Z=-- (6) (]" has the mean 0 and the variance 1. PROOF We prove (5) for a continuous distribution. To a small interval I of length .it" on the x-axis there corresponds the probability f(X)flx [approximately; the area of a rectangle of base .i,. and height f(x)]. Then the probability f(x).::lx must equal that for the corresponding interval on the x*-axis, that is, f*(x*)Llx*, where f* is the density of X* and Llx* is the length of the interval on the x*-axis corresponding to l. Hence for differentials we have f*(x*) dx* = f(x) dx. Also. x* = al + a2x by (4). so that (lb) applied to X* gives J.L* = I"" x*f*(x*) dx* -co = I x (al + a2x )f(x) dx -x = al {Xl f(x) dx + a2 I= xf(x) dx. -x -CD On the right the first integral equals 1, by (10) in Sec. 24.5. The second integral is J.L. This proves (5) for J.L *. It implies From this and (2) applied to X*, again using f*(x*) dx* = f(x) dx. we obtain the second formula in (5), (]"*2 = IX (x* -~ J.L*)21*(x*) dx* = a 22 !"C (x - J.L)2f(x) dx = a2 2(]"2. -x For a discrete distribution the proof of (5) is similar. Choosing a l = - J.LI(]" and a2 = I/(]" we obtain (6) from (4), writing X* {II, a2 formula (5) gives J.L* = 0 and (]"*2 = 1, as claimed in (b). = Z. For these • SEC. 24.6 1019 Mean and Variance of a Distribution Moments Expectation, Recall that (1) defines the expectation (the mean) of X, the value of X to be expected on the average, written IL = E(X). More generally, if g(x) is non constant and continuous for all x, then g(X) is a random variable. Hence its mathematical expectation or, briefly, its expectation E(g(X» is the value of g(X) to be expected on the average. defined [similarly to (I)] by (7) E(g(X» = 2: g(.l)f(xj) E(g(X» or = fC g(x)f(x) dx. -x j In the first fonnula, f is the probability function of the discrete random variable X. In the second formula, f is the density of the continuous random variable X. Important special cases are the kth moment of X (where k = L 2.... ) (8) E(Xk) = 2: x/f(xj) or j and the kth central moment of X (k (9) E([X - IL]k) = 2: (Xj = L, 2, ... ) or - ILlf(X.i) {Xl (x _ ILlf(x) dx. -x j This includes the first moment. the mean of X (10) IL = E(X) [(8) with k = I]. It also includes the second central moment, the variance of X [(9) with k = 2]. (11) For later use you may prove (12) 11-:.6J E(l) MEAN, VARIANCE Find the mean and the variance of the random variable X with probability function or density f(x). 1. f(x) = 2x (0;:;; x ;:;; I) 2. f(O) = 0.512, f(3) = 0.008 f(l) = 0.384, f(2) = 0.096, 3. X = Number a fair die turns up 4. Y = -4X + 5 with X as in Prob. 1 5. Uniform distribution on [0, 8] 6. f(x) = 2e- 2x (x ~ 0) = 1. 7. What is the expected daily profit if a store sells X air conditioners per day with probability f(lO) = 0.1, fell) = 0.3, f(12) = 0.4, f(13) = 0.2 and the profit per conditioner is $55? 8. What is the mean life of a light bulb whose life X [hours] has the density f(x) = O.OOle- O.OOlx (x ~ O)? 9. If the mileage (in multiples of 1000 mi) after which a tire must be replaced is given by the random variable X with density f(x) = ()e- flx (x > 0). what mileage can you expect to get on one of these tires? Let = 0.04 and find the probability that a tire will last at least 40000 mi. e 1010 CHAP. 24 Data Analysis. Probability Theory 10. What sum can you expect in rolling a fair die 10 times? Do it. Repeat this experiment 20 times and record how the sum varies. 11. A small filling station is supplied with gasoline every Saturday afternoon. Assume that its volume X of sales in ten thousands of gallons has the probability density f(x) = 6x( I - x) if 0 ~ x ~ 1 and 0 otherwise. Determine the mean. the vaIiance. and the standardized variable. 12. What capacity must the tank in Prob. II have in order that the probability that the tank will be emptied in a given week be 5%'1 13. Let X [cm] be the diameter of bolts in a production. Assume that X has the density f(x) = k(x - 0.9)( 1.1 - x) if 0.9 < x < 1.1 and 0 otherwise. Detern1ine k. sketch fIx). and find JL and u 2 . 14. Suppose that in Prob. 13. a bolt is regarded as being defective if its diameter deviates from 1.00 cm by more than 0.09 cm. What percentage of defective bolts should we then expect? 15. For what choice of the maximum possible deviation c from 1.00 cm shall we obtain 3o/c defectives in Probs. 13 and 14? 24.7 16. TEAM PROJECT. Means, Variances, Expectations. (a) Show that E(X - IL) = 0, (]"2 = E(X 2 ) - IL2. (b) Prove (10)-(12). (c) Find all the moments of the uniform distribution on an interval a ~ x ~ b. (d) The skewness 'Y of a random variable X is defined by (13) 'Y = 3 1 3 E([X - ILl ). (]" Show that for a symmetric distribution (whose third central moment exists) the skewness is 7ero. (e) Find the skewness of the distribution with density fIx) = xe- x when x > 0 and fIx) = 0 otherwise. Sketch f(x). (I) Calculate the skewness of a few simple discrete distributions of your own choice. (g) Find a non symmetric discrete distribution with 3 possible values. mean O. and skewness O. Binomial, Poisson, and Hypergeometric D istri butions These are the three most important discrete distributions. with numerous applications. Binomial Distribution The binomial distribution occurs in games of chance (rolling a die, see below, etc.), quality inspection (e.g., counting of the number of defectives), opinion polls (counting number of employees favoring certain schedule changes, etc.), medicine (e.g., recording the number of patients recovered by a new medication), and so on. The conditions of its OCCUlTence are as follows. We are interested in the number of times an event A occurs in n independent trials. In each trial the event A has the same probability P(A) = p. Then in a trial, A will not occur with probability q = I - p. rn 11 trials the random variable that interests us is x = Number oj times the event A occurs in 11 trials. X can assume the values 0, I. .... 11. and we want to detennine the corresponding probabilities. Now X = x means that A occurs in l" trials and in n - x trials it does not occur. This may look as follows. SEC. 24.7 1021 Binomial, Poisson. and Hypergeometric Distributions A A··· A B '---v----' (1) B··· B. ~ x times 11 - x times Here B = A C is the complement of A, meaning that A does not Occur (Sec. 24.2). We now use the assumption that the trials are independent, that is. they do not influence each other. Hence (I) has the probability (see Sec. 24.3 on independent events) pp ... p . qq'" '-------v-----' (1 *) x times q = p"·qn-x. '-------v-----' x times 11 - Now (l) is just one order of ananging x A's and 11 - x B's. We now use Theorem l(b) in Sec. 24.4, which gives the number of permutations of 11 things (the 11 outcomes of the II trials) consisting of 2 classes, class I containing the III = x A's and class 2 containing the 11 - III = 11 - x B's. This number is n! x!(n - x)! = (:) . Accordingly, (1 *) multiplied by this binomial coefficient gives the probability P(X = x) of X = x, that is, of obtaining A precisely x times in 11 trials. Hence X has the probability function (\' = 0, I, ... , /l) (2) and I(x) = 0 otherwise. The distribution of X with probability function (2) is called the binomial distribution or Be1'1loll11i distriblltion. The occunence of A is called sllccess (regardless of what it actually is: it may mean that you miss your plane or lose your watch) and the nonoccunence of A is calledfailllre. Figure 516 shows typical examples. Numeric values can be obtained from Table A5 in App. 5 or from your CAS. The mean of the binomial distribution is (see Team Project 16) (3) J.L = np and the variance is (see Team Project 16) (4) (]'2 = npq. 0.5 Fig. 516. °O!:---'--'-~----:C5 O!--,--'-...L~5 p=O.l p=O.2 jll O:--,--'-...L.1....:!5 0~~-'-.......,.5 p=O.5 p=O.8 p =0.9 0 5 Probability function (2) of the binomial distribution for n = 5 and various values of p 1022 CHAP. 24 Data Analysis. Probability Theory For the symmetric case of equal chance of success and failure (p the mean nl2, the variance nJ4, and the probability function (2*) E X AMP L E 1 f(x) = (:) = q (~r = 112) this gives (x = 0, 1, ... , n). Binomial Distribution Compute the probability of obtaining at least two "Six" in rolling a fair die 4 times. Solution. p = peA) = P("Six") = 1/6, q 2 or 3 or 4 "Six." Hence the answer is P = J(2) + J(3) + J(4) = (~) = 5/6. n = 4. The event "At leasr two r(r 'Six'" occurs if we obtain r( (i % + (~) (i %) + (:) (i I = - 4 (6·25 + 4·5 + 6 I) r • 171 = - - = 13.2%. 1296 Poisson Distribution The discrete distribution with infinitely many possible values and probability function (5) (x = 0, 1, . , .) f(x) = is called the Poisson distribution, named after S. D. Poisson (Sec, 18.5). Figure 517 shows (5) for some values of f.L. It can be proved that this distribution is obtained as a limiting case of the binomial distribution, if we let p ~ and n ~ 00 so that the mean f.L = np approaches a finite value. (For instance, f.L = np may be kept constant.) The Poisson distribution has the mean f.L and the variance (see Team Project 16) ° (6) Figure 517 gives the impression that with increasing mean the spread of the distribution increases, thereby illustrating formula (6), and that the distribution becomes more and more (approximately) symmetric. 0.5 o 11 = 0.5 Fig. 517. 11=1 11=2 5 10 11=5 Probability function (S) of the Poisson distribution for various values of J1, SEC. 24.7 1023 Binomial, Poisson. and Hypergeometric Distributions E X AMP L E 2 Poisson Distribution If the probability of producing a defective screw is I' = 0.01. what is the probability that a lot of 100 will contain more than 2 defectives'! screw~ Solutioll. The complementary event is AC : Not more thaI! 2 defeethoes. For its probability we get from the binomial distribution with mean M = 111' = I the value [see (2)] P(A c ) = ( 2 98 100) 0 (J.YY lOO + (100) I (J.OI • 0.yy99 + ( 100) 2 0.01' 0.Y9 . Since p is very small. we can approximale this by the much more conveniem Poisson distriblllion with mean J1. = 111' = 100· 0.01 = I. obtaining [see (5)] = 91.97%. Thu, PIA) = 8.03%. Show that the binomial distribution gives PIA) = 7.94%, so that the is quite good. E X AMP L E 3 Poi~son approximation • Parking Problems. Poisson Distribution If on the average. 2 cars enter a certain parking lot per minute. what is the probability that during any given minute 4 or more car, will enter the lot? Solutioll. To understand that the Poisson distribution is a model of the situation. we imagine the minute to be divided into very many short time intervals, let p be the (constant) probability that a car will enter the lot during any such short interval. and assume independence of the events that happen during those imervals. Then we are dealing with a binomial distribution with very large 11 and very small 1', which we can approximate by the Pois~on distriblllion with M = 1117 = 2, because 2 cars enter on the average. The complementary event of the event ·'4 cars or more during a given minute" is "3 cars orfewer elller the lot" and ha~ the prohability 0 f(O) + f(l) + f(2) + f(3) = e- = Answer: l·t3"k. 2 2 ( -0 1 21 + - I! 22 + - 2! 3 2 ) + - 3! 0.857. • (Why did we consider that complement"!) Sampling with Replacement This mean, that we draw things from a given set one by one, and after each trial we replace the thing drawn (put it back to the given set and mix) before we draw the next thing. This guarantees independence of trials and leads to the binomial distribution. Indeed, if a box contains N things, for example. screws. M of which are defective, the probability of drawing a defective screw in a trial is p = MIN. Hence the probability of drawing a nondefective screw is q = 1 - p = 1 - MIN, and (2) gives the probability of drawing x defectives in n trials in the form (7) _ (11) (M)X (I - M)n-X - f(x) - x N N (x = 0, I, ... , 11). 1024 CHAP. 24 Data Analysis. Probability Theory Sampling without Replacement. Hypergeometric Distribution Sampling without replacement means that we return no screw to the box. Then we no longer have independence of trials (why?). and instead of (7) the probability of drawing x defectives in n trials is (x I(x) = (8) = 0, I, ... , 11). The distribution with this probability function is called the hypergeometric distribution (because its moment generating function (see Team Project 16) can be expres~ed by the hypergeometric function defined in Sec. 5.4, a fact that we shall not use). Derivation of (8). By (4a) in Sec. 24.4 there are (a) (:) different ways of picking 11 things from N. (b) (~) (c) (Nn -- M\ different ways of picking n xJ different ways of picking x defectives from M, x nondefectives from N- M, and each way in (b) combined with each way in (c) gives the total number of mutually exclusive ways of obtaining x defectives in 11 drawings without replacement. Since (a) is the total number of outcomes and we draw at random. each such way has the probability l/(Ij . From this, (8) follows. • The hypergeometric distribution has the mean (Team Project 16) M (9) J.L=/l- N and the variance nM(N - MeN - 11) (10) E X AMP L E 4 N 2 (N - I) Sampling with and without Replacement We want to draw random samples of two gaskets from a box containing 10 gaskets. three of which are defective. Find the probability function of the random variable X = Number of defectives ill the sample. Solution. We have N = 10. M = 3. N - M = 7. f(x) = (~) I~ ( r( r7 10 Il = 2. For sampling with replacement. (7) yields x J(O) = 0.49, f(l) = 0.42, J(2) = 0.09. For sampling without replacement we have to use (8), finding 21 fCO) = f(l) = 45 "" 0.47. 3 f(2) = 45 = 0.07. • SEC. 24.7 1025 Binomial, Poisson, and Hypergeometric Distributions If N, M, and N - M are large compared with n, then it does not matter too much whether we sample with or without replacemellt, and in this case the hypergeometric distribution may be approximated by the binomial distribution (with p = MIN), which is somewhat simpler. Hence in slimp ling from an indefinitely large population ("infinite population") we 1Il11\' use the binomial distribution, regardless of whether we sample with or withollI replacement. ===== -... -...-.. -.. : .. ..".-~ ... -- 1. Four fair coins are tossed simultaneously. Find the probability function of the random variable X = Number of heads and compute the probabilities of obtaining no heads, precisely I head, at least I head, not more than 3 heads. 2. If the probability of hitting a target in a single shot is 10% and 10 shots are fired independently. what is the probability that the target will be hit at least once? 3. In Prob. 2, if the probability of hitting would be 5% and we fired 20 shots. would the probability of hitting at least once be less than. equal to, or greater than in Prob. 2? Guess first, then compute. 4. Suppose that 3% of bolts made by a machine are defective, the defectives occurring at random during production. If the bolts are packaged 50 per box, what is the Poisson approximatIOn of the probabilit) that a given box will contain x = 0, 1, ... , 5 defeClives? 5. Let X be the number of cars per minute passing a certain point of some road between 8 A.M. and 10 A.M. on a Sunday. Assume that X has a Poisson disuibution with mean 5. Find the probability of observing 3 or fewer cars during any given minute. 6. Suppose thar a telephone switchboard of some company on the average handles 300 calls per hour, and that the board can make at most 10 connections per minute. Using the Poisson disU'ibution, estimate the probability that the board will be overtaxed during a given minute. (Use Table A6 in App. 5 or your CAS.) 7. (Rutherford-Geiger experiments) In 1910. E. Rutherford and H. Geiger showed experimenrally that the number of alpha particles emitted per second in a radioactive process is a random variable X having a Poisson distribution. If X has mean 0.5, whar is the probability of observing two or more particles during any given second? 8. A process of manufacturing screws is checked every hour by inspecting 11 screws selected at random from that hour's production. If one or more screws are defective, the process is halted and carefully examined. How large should n be if the manufacturer wants the probability to be about 95% that the process will be haIted when 10% of the screws being produced are defective? (Assume independence of the quality of any screw of that of the other screws.) 9. Suppose that in the production of 50-n resistors, nondefective items are those that have a resistance between 45 nand 55 n and the probability of a resistor's being defective is 0.2%. The resistors are sold in lots of 100, with the guarantee that all resistors are nondefective. What is the probability that a given lot will violate this guarantee? (Use the Poisson distribution.) 10. Let P = lo/c be the probability that a certain type of lightbulb will fail in a 24-hr test. Find the probability that a sign consisting of 10 such bulbs will bum 24 hours with no bulb failures. 11. Guess how much less the probability in Prob. 10 would be if the sign consisted of 100 bulbs. Then calculate. 12. Suppose that a certain type of magnetic tape contains. on the averdge, 2 defects per 100 meters. What is the probability that a roll of tape 300 meters long will contain (a) x defects, (b) no defects? 13. Suppose thar a test for extrasensory perception consists of naming (in any order) 3 cards randomly drawn from a deck of 13 cards. Find the probability that by chance alone, the person will correctly name (a) no cards, (b) I card, (c) 2 cards, (d) 3 cards. 14. A carton contains 20 fuses, 5 of which are defective. Find the probability that. if a sample of 3 fuses is chosen from the carton by random drawing without replacement, x fuses in the sample will be defective. 15. (Multinomial distribution) Suppose a trial can result in precisely one of k mutually exclusive events Ab ... , Ak with probabilities PI' .•• , Pk' respectively, where PI + ... + Pk = 1. Suppose that /I independent trials are performed. Show that the probability of getting Xl AI's, . . • , Xk Ak's is where Xl 0 ~ Xj ~ II, + ... + Xk = 11. j = I, ... , k. and The distribution having this CHAP. 24 1026 Data Analysis. Probability Theory (b) Shov. that the binomial distribution has the moment generating function probability function is called the lIlultinomial distribution. 16. TEAM PROJECT. Moment Generating Function. The moment generating function G(t) is defined by ~ tX tx· G(t) = E(e J) = .L.J e 'f(xj) = (pet + q)". or G(t) = E(etx ) = (c) Using (b), prove (3). fX et·t:f(x) dx (d) -x where X is a discrete or continuous random variable, respectively. (a) Assuming that termwise differentiation and differentiation under the integral sign are permissible, show that E(Xk) = dkl(O), where d k ) = dkG/dtk. in particular, /L = G' (0). 24.8 Prove (4). (e) Show that the Poisson distribution has the moment generating function G{t) = e-lLe lLe ' and prove (6). (f) Prove x (~) Using this. prove = M (~ =!) . (9). Normal Distribution Turning from discrete to continuous distributions, in this section we discuss the normal distribution. This is the most important continuous distribution because in applications many random variables are normal random variables (that is, they have a normal distribution) or they are approximately normal or can be transformed into normal random variables in a relatively simple fashion. Furthermore, the normal distribution is a useful approximation of more complicated distributions. and it also occurs in the proofs of various statistical tests. The normal distribution or Gauss distribution is defined as the distribution with the density (1) f(x) = -I- exp [_ 21 d\!2; (x -cr )2J J.L (cr> 0) where exp is the exponential function with base e = 2.718 .... This is simpler than it may at first look. fex) has these features (see also Fig. 518). 1. J.L is the mean and cr the standard deviation. 2. l/(dV2;) is a constant factor that makes the area under the curve of f(x) from to x equal to L as it must be by (0), Sec. 24.5. -x 3. The curve of f(x) is symmetric with respect to x = J.L because the exponent is quadratic. Hence for J.L = 0 it is symmetric with respect to the y-axis x = 0 (Fig. 518, "bell-shaped curves"). 4. The exponential function in (1) goes to zero very fast-the faster the smaller the standard deviation cr is, as it should be (Fig. 518). SEC. 24.8 1027 Normal Distribution ((x) cr = 1.0 2 Fig. 518. x Density (1) of the normal distribution with /L = 0 for various values of u Distribution Function F(x) From (7) in Sec. 24.5 and (I) we see that the normal di<;tribution has the distribution function (2) .~ IX 27T Fex) = (TV exp [- 21 (u - J.L (T -00 )2J du. Here we needed x as the upper limit of integration and wrote u (instead of x) in the integrand. For the conesponding standardized normal distribution with mean 0 and standard deviation I we denote F(x) by <P(.:). Then we simply have from (2) I Vh I (3) Z e- u2/2 duo cI>(.:) = - - -x This integral cannot be integrated by one of the methods of calculus. But this is no serious handicap because its values can be obtained from Table A 7 in App. 5 or from your CAS. These values are needed in working with the normal distribution. The curve of cI>(z) is S-shaped. It increases monotone (why?) from 0 to 1 and intersects the vertical axis at 112 (why?), as shown in Fig. 519. Relation Between F(x) and «I>(z). Although your CAS will give you values of F(x) in (2) with any J.L and (T directly, it is important to comprehend that and why any such an F(x) can be expressed in terms of the tabulated standard cI>(z), as follows. y ,",(xl ~ 1.0 0.8 / 0.6 o. /0.2 -3 Fig. 519. -2 -1 0 2 3 x Distribution function <I>(z) of the normal distribution with mean 0 and variance 1 1028 CHAP. 24 THE 0 REM 1 Data Analysis. Probability Theory Use of the Normal Table A7 in App. 5 r The distributioll fimctioll FCr) of the nonnal distriblltioll with allY J.L and 0" see (2)] is related to the stalldardi-;.ed distribution filllCtiOIl <1>(.:) ill (3) by the formula (4) PROOF F(x) = <I> ( -x-J.L) 0"- . Comparing (2) and (3) we see that we should set u= Then v = x gives u= 0" 0" as the new upper limit of integration. Also v - J.L 0" drops out. I I(X-P)/u u2 2 F(x) = --O"~ e- / = 0" O"U, du thus dv = 0" du. Together, since X ) = <I> ( _ _J.L_ 0" -ex: • Probabilities corresponding to intervals will be needed quite frequently in statistics in Chap. 25. These are obtained as follows. THEOREM 2 Normal Probabilities for Intervals The probability that a normal random variable X with mean J.L and standard del'iatioll 0" assume allY vallie ill all inten'al a < x ~ b is (5) PROOF Pea < X ~ b) = F(b) - F(a) = <I> ( -b-J.L) 0 " - - <J:> (a-J.L) -0"- . . Formula (2) in Sec. 24.5 gives the first equality in (5). and (4) in this section gives the ~~~~~. Numeric Values In practical work with the normal disnibution it is good to remember that about 2/3 of all values of X to be observed will lie between J.L ± 0", about 95% between J.L ± 20", and practically all between the three-sigma limits J.L ± 30". More precisely, by Table A7 in App. 5, (6) <X (a) P(J.L - 0" (b) P(J.L - 20" < (c) P(J.L - 30" <X ~ J.L X ~ J.L ~ J.L + 0") = 68% + 20") = + 95.5% 30") = 99.7%. Formulas (6a) and (6b) are illustrated in Fig. 520. The formulas in (6) show that a value deviating from JL by more than 0", 20", or 30" will occur in one of about 3, 20, and 300 trials, respectively. SEC. 24.8 1029 Normal Distribution 2.25% -\ r .u-20 (a) 95.5% '/ 2.25% 'tl ! .u .u+20 (b) Fig. 520. Illustration of formula (6) In tests (Chap. 25) we shall ask conversely for the intervals that corre<;pond to certain given probabilities; practically most important are the probabilities of 95%, 99%, and 99.9%. For these, Table AS in App. 5 gives the answers J.L ± 2u, J.L ± 2.5u, and J.L ± 3.3u, respectively. More precisely, (7) (a) P(J.L - 1.96u <X ~ J.L + 1.96u) = 95% (b) P(J.L - 2.5Su <X ~ J.L + 2.5Su) = 99% (e) P(J.L - 3.29u <X ~ J.L + 3.29u) = 99.9%. Working With the Normal Tables A7 and AS in App. 5 There are two normal tables in App. 5, Tables A7 and A8. If you want probabilities, use Table A7. If probabilities are given and corresponding intervals or x-values are wanted, use Table AS. The following examples are typical. Do them with care. verifying all values, and don't just regard them as dull exercises for your software. Make sketches of the density to see whether the results look reasonable. E X AMP L E 1 Reading Entries from Table A7 If X is standardi7ed nonnal (so that /L = O. a = 1), then P(X ~ 2.'14) = 0.9927 = 99~% P(X~ P(X ~ PO.O E X AMP L E 2 -1.16) = 1 - <1>(1.16) = 1 - 0.8770 = 0.1230 = 12.3'J1: 1) = 1 - P(X ~ X ~ ~ I) = 1 - 0.H413 = 0.1587 by (7), Sec. 24.3 1.8) = <1>(1.8) - <1>(1.0) = 0.9641 - 0.8413 = 0.1228. • Probabilities for Given Intervals, Table A7 Let X be normal with mean 0.8 and variance 4 (so that a = 2). Then by (4) and (5) P(X ~ 2.44) = F(2.44) = q) ( 2.44 - 0.80 ) 2 = <1>(0.82) = 0.7939 = 80% or if you like it better (similarly in the other cases) P(X ~ 2.44) PIX ~ 1) = = P( 2.44 - 0.80 ) X - 0.80 2 ~ 2 = 1-08) I - P(X ~ 1) = I - <I> ( --2-'- P(1.0 ~ X ~ 1.8) = <1>(0.5) - <1>(0.1) = P(Z ~ 0.82) = = 0.7939 I - 0.5398 = 0.4602 0.6915 - 0.5398 = 0.1517. • 1030 E X AMP L E 3 CHAP. 24 Data Analysis. Probability Theory Unknown Values c for Given Probabilities, Table AS Let X be nonnal with mean 5 and variance 0.04 (hence standard deviation 0.1). Find c or k corresponding to the given probability PIX ~ c) = P(5 - k P(X E X AMP L E 4 ~ 950/(. X ~ c) = ~ 19C. 5) c - <P ( 0:2 = 5 + k) = 90%, thus PIX c - 5 0.1 = 1.645. 95%. 5 + k = 5.319 ~ c) = (as before: why?) c-5 0:2 99%. c = 5.319 = c = 5.46S. 2.326. • Defectives In a production of iron rods let the diameter X be nunnally distributed with mean 2 in. ilnd standard deviation 0.008 in. (a) What percentage of defectives can we expect if we ,et the tolerance limit, at 2 ::':: 0.01 in.? (b) How should we set the tolerance limits to allow for 4'lt defectives? Solutioll. (a) I!'k because from (S) and Table A7 we obtain for the complementary event the probability P( 1.98 ~ X ~ 2.02) = <I) ( ( 1.9H - 2.00 ) 2.02 - 2.00) 0.008 - <I) 0.008 = <1)(2.5) - <P( -2.S) = 0.9938 - (I - 0.9938) = 0.9876 = 98~%. (b) 2 ::':: 0.0164 because for the complementary event we have 0.96 = P(2 - c ~ X ~ 2 + c) or 0.98 = P(X ~ 2 + c) so that Table A8 gives 0.98 = <P ( 1+C-2) 0 . 0.08 2+c-2 0.008 = 2.0S4. • c = 0.0164. Normal Approximation of the Binomial Distribution The probability function of the binomial distribution is (Sec. 24.7) (8) (x = 0, 1, ... , n). If 11 is large, the binomial coefficients and powers become very inconvenient. It is of great practical (and theoretical) importance that in this case the normal distribution provides a good approximation of the binomial distribution, according to the following theorem, one of the most important theorems in all probability theory. SEC. 24.8 1031 Normal Distribution THEOREM 3 Limit Theorem of De Moivre and Laplace For large n. f(x) ~ f*(x) (9) Here f (x = 0, 1, ... , n). is given by (8). The function (10) f* (x) = x - np ---c=---== V2;v,;pq v;pq is the density of the n01711al distribution with mean J.L = np and variance (j2 = npq (the mean and variance of the billomial distribution). The symbol ~ (read asymptotically equal) means that the ratio of both sides approaches 1 as n approaches so. Flirthe17nore. fbr any llOllIlegative integers a and b (> a). Pea ~ X ~ b) = ~a (:) p~:qn-x ~ cI>(f3) - cI>(ex), (11 ) ex= a - IIp - 0.5 v,;pq 13= b - I1p + 0.5 v,;pq A proof of this theorem can be found in [03] listed in App. 1. The proof shows that the term 0.5 in ex and 13 is a correction caused by the change from a discrete to a continuous distribution. 11-131 NORMAL DISTRIBUTION 1. Let X be normal with mean 80 and variance 9. Find P(X > 83). P(X < 81), P(X < 80), and P(78 < X < 82). 2. Let X be normal with mean 120 and variance 16. Find P(X;;;; 126), P(X > 116), P(l25 < X < 130). 3. Let X be normal with mean 14 and variance 4. Determine c such that P(X ;;;; c) = 95%, P(X ;;;; c) = 5%, P(X;;;; c) = 99.5%. under the guarantee? 6. If the standard deviation in Prob. 5 were smaller. would that percentage be smaller or larger? 7. A manufacturer knows from experience that the resistance of resistors he produces is normal with mean /L = 150 n and standard deviation (T = 5 n. What percentage of the resistors will have resistance between 148 nand 152 n? Between 140 nand 160 n? 4. Let X be normal with mean 4.2 and variance 0.04. Find c such that P(X ;;;; c) = 50%, P(X > C) = 10%, P(-c < X - 4.2;;;; c) = 99%. 8. The breaking strength X [kg] of a certain type of plastic block is normally distributed with a mean of 1250 kg and a standard deviation of 55 kg. What is the maximum load such that we can expect no more than 5% of the blocks to break? 5. If the lifetime X of a certain kind of automobile battery is normally distributed with a mean of 4 yr and a standard deviation of I yr, and the manufacturer wishes to guarantee the battery for 3 yr, what percentage of the batteries will he have to replace 9. A manufacturer produces airmail envelopes whose weight is normal with mean /L = 1.950 grams and standard deviation if = 0.025 grams. The envelopes are sold in lots of 1000. How many envelopes in a lot will be heavier than 2 grams? 1032 CHAP. 24 Data Analysis. Probability Theory 10. If the resistance X of ce11ain wires in an electrical network is normal with mean 0.01 D and standard deviation 0.001 D, how many of 1000 wires will meet the specification that they have resistance between 0.009 and 0.011 Q? (d) Considering cI>2(ao) and introducing polar coordinates in the double integral (a standard trick worth remembering), prove (12) 11. If the mathematics scores of the SAT college entrance exams are normal with mean 480 and standard deviation 100 (these are about the actual values over the past years) and if some college sets 500 as the minimum score for new students, what percent of students will not reach that score? 1 ~ \. 2'11" IX e- ll 2 12 dll = 1. -x (I) Bernoulli's law oflarge nwnbers.ln an experiment let an event A have probability p (0 < P < I), and let X be the number of time~ A happens in /I independent trials. Show that for any given E > 0, as /1-+ x. (g) Transformation. If X is normal with mean J.L and variance u 2 , show that X* = clX + C2 (cI > 0) is normal with mean J.L* = CIJ.L + C2 and variance U*2 = C1 2 U 2 . 15. WRITING PROJECT. Use of Tables. Give a systematic discussion of the use of Tables A7 and A8 for obtaining P(X < b), P(X > a), P(a < X < fA PIX < c) = k. P(X > c) = k. as well as P(J.L - C < X < J.L + c) = k: include simple examples. If you have a CAS. describe to what extent it makes the use of those tables superfluous; give examples. 14. TEAM PROJECT. Normal Distribution. (a) Derive the formula~ in (6) and (7) from the appropriate normal table. (b) Show that cI>(-:) = I - cI>(:). Give an example. (e) Find the points of inflection of the curve of (1). 24.9 = (e) Show that u in (1) is indeed the standard deviation of the normal distribution. [Use (12).] 12. If the monthly machine repair and maintenance cost X in a ce11ain factory is known to be normal with mean $12000 and standard deviation $2000, what is the probability that the repair cost for the next month will exceed the hudgeted amount of $150007 13. [f sick-leave time X used by employees of a company in one month is (very roughly) normal with mean 1000 hours and standard deviation 100 hours. how much time t should be budgeted for sick leave during the next month if t is to be exceeded with probability of only 20o/c? cI>(x) Distributions of Several Random Variables Distributions of two or more random variables are of interest for two reasons: 1. They occur in experiments in which we observe several random variables, for example, carbon content X and hardness Y of steel, amount of fertilizer X and yield of corn Y, height Xl' weight X 2 , and blood pressure X3 of persons, and so on. 2. They will be needed in the mathematical justification of the methods of statistics in Chap. 25. In this section we consider two random variables X and Yor, as we also say, a twodimensional random variable (X, Y). For (X, Y) the outcome of a trial is a pair of numbers X = x, Y = y, briefly (X, Y) = (x, y), which we can plot as a point in the XY-plane. The two-dimensional probability distribution of the random variable (X, Y) is given by the distribution function (1) F(:.:, y) = P(X ~ x, Y ~ y). This is the probability that in a trial, X will assume any value not greater than x and in the same trial, Y will assume any value not greater than y. This corresponds to the blue region in Fig. 521, which extends to -00 to the left and below. F(x, y) determines the SEC. 24.9 1033 Distributions of Several Random Variables Fig. 521. Formula (1) probability distribution uniquely, because in analogy to formula (2) in Sec. 24.5, that is, < X ~ b) = F(b) - F(a), we now have for a rectangle (see Prob. 14) Pea As before, in the two-dimensional case we shall also have discrete and continuous random variables and distributions. Discrete Two-Dimensional Distributions In analogy to the case of a single random variable (Sec. 24.5), we call (X, Y) and its distribution discrete if (X, Y) can assume only finitely many or at most countably infinitely many pairs of values (XI' YI), (X2' )'2), '" with positive probabilities, whereas the probability for any domain containing none of those values of (X, Y) is zero. Let (x;, -':) be any of those pairs and let P(X = Xi' Y = Yj) = Pij (where we admit that Pij may be 0 for certain pairs of subscripts i, j). Then we define the probability function f(x. y) of ex, Y) by (3) f(x, y) = Pij if x = Xi, Y = Yj f(x, y) = 0 and otherwise; here, i = 1,2, ... andj = 1,2, ... independently. In analogy to (4), Sec. 24.5, we now have for the distribution function the formula (4) Instead of (6) in Sec. 24.5 we now have the condition 2: 2: f(Xi' Yj) (5) = 1. j E X AMP LEI Two-Dimensional Discrete Distribution If we ~imilltaneously toss a dime and a nickel and consider x= Number of heads the dime turns up, Y = Number of heads the nickel turns up, then X and Y can have the values 0 or 1. and the probability function is fCO, 0) = fO, 0) = f(O, 1) = f(1, 1) =~, f(x, y) = ° otherwise. • 1034 CHAP. 24 Data Analysis. Probability Theory y Fig. 522. Notion of a two-dimensional distribution Continuous Two-Dimensional Distributions In analogy to the case of a single random variable (Sec. 24.5) we caIl (X, Y) and its distribution continuous if the corresponding distribution function F(x, \') can be given by a double integral (6) F(x, y) = I ,j IX -00 f(x*, y*) dx* dy* -':X:l whose integrand f, called the density of (X, Y), is nonnegative everywhere, and is continuous, possibly except on finitely many curves. From (6) we obtain the probability that (X, Y) assume any value in a rectangle (Fig. 522) given by the formula (7) E X AMP L E 2 Two-Dimensional Uniform Distribution in a Rectangle Let R be the rectangle "1 < x (8) ~ f31' "2 f(x. y) = Ilk < Y ~ f32' The density (see Fig. 523) if (x. y) i~ in R. f(x, y) = 0 otherwise defines the so-called uniform distribution ill the rectallgle R: here k = (f31 - "1)({32 - "2) is the area of R. The distribution function is shown in Fig. 524. • y x x o Fig. 523. Density function (8) of the uniform distribution o Fig. 524. Distribution function of the uniform distribution defined by (8) Marginal Distributions of a Discrete Distribution This is a rather natural idea, without counterpart for a single random variable. It amounts to being interested only in one of the two variables in (X, Y), say, X, and asking for its SEC. 24.9 1035 Distributions of Several Random Variables distribution, called the marginal distribution of X in (X, V). So we ask for the probability = x, Yarbitrary). Since (X, Y) is discrete, so is X. We get its probability function, call it f1(x), from the probability function f(x, y) of (X, Y) by summing over y: P(X (9) f1(x) = P(X = = x, Yarbitrary) 2: f(x, y) y where we sum all the values of f(x, y) that are not 0 for that x. From (9) we see that the distribution function of the marginal distribution of X is (10) Similarly, the probability function (11) f2(Y) = P(X arbitrary. Y = = y) 2: f(x. y) :r determines the marginal distribution of Y in (X, V). Here we sum all the values of f(x, y) that are not zero for the conesponding y. The distribution function of this marginal distribution is (12) y*~y E X AMP L E 3 Marginal Distributions of a Discrete Two-Dimensional Random Variable In drawing 3 cards with replacement from a bridge deck let us consider (X. Yl. x= NlIl11ber of queens. Y = Number of kings or lIces. The deck has 52 cards. These include 4 queens. 4 kings. and 4 aces. Hence in a single trial a queen has probability 4/52 = 1/13 and a king or ace 8/52 = 2/13. This gives the probability function of (X, Y), 3! (-I /<x,\") = . x! y! (3 - x - y)! 13 )X ( - 2 )Y ( - 10 )3-T-Y 13 (x 13 + y ~ 3) and fIx. y) = 0 otherwise. Table 24.1 shows in the center the values of fIx, y) and on the right and lower margins the values of the probability functions hex) am] hey) of the marginal distributions of X and Y, respectively . • Table 24.1 Values of the Probability Functions f{x, y), fl{X), f 2{Y) in Drawing Three Cards with Replacement from a Bridge Deck, where X is the Number of Queens Drawn and Y is the Number of Kings or Aces Drawn x 0 y 0 I 2 3 f1(x) 1000 2197 600 2197 120 2197 8 2197 1728 2197 ,IOU 1 2197 120 2197 12 2197 0 432 2197 2 30 2197 6 2197 0 0 36 2197 3 1 2197 0 0 0 2197 f2(Y) 1331 2197 726 2197 132 2197 8 2197 1 1036 CHAP. 24 Data Analysis. Probability Theory Marginal Distributions of a Continuous Distribution This is conceptually the same as for discrete distributions. with probability functions and sums replaced by densities and integrals. For a continuous random variable (X, Y) with density f(x, y) we now have the marginal distribution of X in (X. Yl. defined by the distribution function (13) Fl(x) = P(X ;:;; x, -QO < Y < ce) = IX fl(X*) dx* -co with the density f 1 of X obtained from f(x. y) by integration over y, (14) fleX) = I= f(x, y) dy. -x Interchanging the roles of X and Y, we obtain the marginal distribution of Y in (X, Y) with the distribution function (15) F 2(y) = P( - x < X< ce, y;:;; y) = I y f2(V*) dy* -x and density (16) f2(Y) = fX f(x, y) dx. -co Independence of Random Variables X and Y in a (discrete or continuous) random variable (X, Y) are said to be independent if (17) holds for all (x, y). Otherwise these random variables are said to be dependent. These definitions are suggested by the corresponding definitions for events in Sec. 24.3. Necessary and sufficient for independence is (IS) for all x and y. Here the f's are the above probability functions if (X, Y) is discrete or those densities if (X, Y) is continuous. (See Prob. 20.) E X AMP L E 4 Independence and Dependence In tossing a dime and a nickel. X = Number of heads all the dillie, Y = Number of headf all the nickel may • assume the values 0 or I and are independent. The random variables in Table 24.1 are dependent. Extension of Independence to II-Dimensional Random Variables. This will be needed throughout Chap. 25. The distribution of such a random variable X determined by a distribution function of the form = (Xl> ... , Xn) is SEC. 24.9 1037 Distributions of Several Random Variables The random variables Xl> .... Xn are said to be independent if (19) for all (Xl> ••• , xn). Here of Xj in X, that is, FiXj) is the distribution function of the marginal distlibution Otherwise these random variahles are said to be dependent. Functions of Random Variables When 11 = 2, we write Xl = X, X 2 = Y, Xl = X, X2 = y. Taking a nonconstant continuous function g(x, y) defined for all x, y, we obtain a random variable Z = g(X, Y). For example, if we roll two dice and X and Yare the numbers the dice turn up in a trial, then Z = X + Y is the sum of those two numbers (see Fig. 513 in Sec. 24.5), In the case of a discrete random variable (X, Y) we may obtain the probability function f(:::.) of Z = g(X. Y) by summing all f(x, y) for which g(x, y) equals the value of :::. considered; thus (20) = f(:::.) P(Z LL f(x, y). = :::.) = g(x.y)~z Hence the distribution function of Z is (21) = F(::.) P(Z ~ :::.) = LL f(x, y) g(x,Y)""Z where we sum all values of f(x, y) for which g(x, y) ~ z. In the case of a continuous random variable eX, Y) we similarly have (22) F(z) = P(Z ~ z) = ff f(x, y) dx dy g(x,Y)""z where for each z we integrate the density f(x, y) of (X, Y) over the region g(x, y) the xy-plane. the boundary curve of this region being g(x, v) = z. ~ z in Addition of Means The number LL (23) x E(g(X, Y» = { x g(x, y)f(x. y) [(X, Y) discrete] Y x ix ixg(X, y)f(x, y) dx dy [(X, Y) continuous] CHAP. 24 1038 Data Analysis. Probability Theory is called the mathematical expectatio/1 or, briefly, the expectation of g(X, Y). Here it is assumed that the double series converges absolutely and the integral of Ig(x, y)I.f(x, y) over the xy-plane exists (is finite). Since summation and integration are linear processes, we have from (23) (24) E(ag(X, Y) + bh(X, Y» = aE(g(X, Y» + bE(h(X, Y». An imponam special case is E(X + Y) = E(X) + E( Y), and by induction we have the following result. THEOREM 1 Addition of Means The mean (expectation) of a sum of random variables equals the (expectations), that is, Sll1l1 of the means Furthermore, we readily obtain THEROEM 2 Multiplication of Means The mean (e.\pectation) of the product (~lilldependellt random variables equals the product qf the meam (expectatiolls), that is, (26) PROOF If X and Yare independent random variahles (both discrete or hoth continuous), then E(XY) = E(X)E(Y). In fact, in the di~crete case we have E(XY) = 2: 2: xyf(x, y) x y = 2: xfl(x) 2: yf2(Y) = E(X)E(y), y and in the continuous case the proof of the relation is similar. Extension to random variables gives (26), and Theorem 2 is proved. /1 independent • Addition of Variances This is another matter of practical impOltance that we shall need. As before, let Z = X + Y and denote the mean and Valiance of Z by I-t and u 2 • Then we first have (see Team Project 16(a) in Problem Set 24.6) SEC. 24.9 1039 Distributions of Several Random Variables From (24) we see that the first term on the right equals For the second term on the right we obtain from Theorem I [E(z)f = [E(X) + E(y)]2 = lE(X)f + 2E(X)E(Y) + [E(y)]2. By substituting these expressions into the formula for u 2 we have u2 = E(X2) - [E(X)]2 + + E(y2) - [E(Y)]2 2[E(XY) - E(X)E(Y)]. From Team Project 16, Sec. 24.6, we see that the expression in the first line on the right is the sum of the variances of X and Y, which we denote by U1 2 and U2 2, respectively. The quantity in the second line (except for the factor 2) is (27) UXY = E(XY) - E(X)E(Y) and is called the covariance of X and Y. Consequently, our result is (28) If X and Y are independent, then E(XY) = E(X)E(Y): hence UXY = 0, and (29) Extension to more than two variables gives the basic THEOREM 3 Addition of Variances The variance of the sum of independent random variables equals the sum of the variances of these variables. CAUTION! In the numerous applications of Theorems 1 and 3 we must always remember that Theorem 3 holds only for independent variables. This is the end of Chap. 24 on probability theory. Most of the concepts, methods, and special distributions discussed in this chapter will play a fundamental role in the next chapter, which deals with methods of statistical inference, that is, conclusions from samples to populations. whose unknown properties we want to know and try to discover by looking at suitable properties of samples that we have obtained. CHAP. 24 1040 Data Analysis. Probability Theory 1. Let f(x, y) = k when 8 ~ x ~ 12 and 0 ~ y ~ 2 and zero elsewhere. Find k. Find P(X ~ 11, 1 ~ Y ~ 1.5) and P(9 ~ X ~ 13, Y ~ I). 2. Find P(X > 2, Y> 2) and P(X ~ 1. Y ~ 1) if (X. Y) has the density f{x, y) = 1/8 if x ~ O. Y ~ 0, X -t- Y ~ 4. = k if x > O. y f(x, y) = 2500 if 0.99 < x < 1.01. 1.00 < Y < 1.02 > 0, x + y < 3 and 0 otherwise. Find k. Sketch f(x, y). Find P(X + Y ~ I), P(Y> X). 3. Let f(x, y) 12. Let X [cm1 and Y [cm 1 be the diameter of a pin and hole. respectively. Suppose that (X, Y) has the density 4. Find the density of the marginal distribution of X in Prob 2. 5. Find the density of the marginal distribution of Y in Fig. 523. and 0 otherwise. (a) Find the marginal distributions. (b) What is the probability that a pin chosen at random will fit a hole whose diameter is l.00? 13. An electronic device consists of two components. Let X and Y [months] be the length of time until failure of the first and second component, respectively. Assume that (X, Y) has the probability density 6. If certain sheets of wrapping paper have a mean weight of 10 g each. with a standard deviation of 0.05 g. what are the mean weight and standard deviation of a pack of IO 000 sheets? 7. What are the mean thickness and the standard deviation of transformer cores each consisting of 50 layers of sheet metal and 49 insulating paper layers if the metal sheets have mean thickness 0.5 mm each with a ~tandard deviation of 0.05 mm and the paper layers have mean 0.05 mm each with a standard deviation of O.02mm? S. If the weight of certain (empty) containers has mean 2 Ib and standard deviation 0.1 lb. and if the filling of the containers has mean weight 751b and standard deviation 0.8 lb. what are the mean weight and standard deviation of filled containers'! 9. A 5-gear assembly is put together with spacers between the gears. The mean thickness of the gear~ is 5.020 em with a standard deviation of 0.003 cm. The mean thickness of the spacers is 0.040 cm with a standard deviation of 0.002 cm. Find the mean and standard deviation of the a~sembled units consisting of 5 randomly selected gears and 4 randomly selected spacers. 10. Give an example of two different discrete distributions that have the same marginal distributions. f(x, y) = 0.01 e -O.l(x+y) if x > 0 and y > 0 and 0 otherwise. (a) Are X and Y dependent or independent? (b) Find the densities of the marginal distributions. (c) What is the probability that the first component has a lifetime of 10 months or longer? 14. Prove (2). 15. Find P(X > Y) when (X. Y) has the density ftx, Y) = 0.25e- O.5 (x+y) if x ~ 0, y ~ 0 and 0 otherwise. 16. Let (X. Y) have the density f(x, y) = k if x 2 + y2 < 1 and 0 otherwise. Determine k. Find the densities of the marginal distributions. Find the probability P(X 2 + y2 < 1/4). 17. Let (X, Y) have the probability function f(O.o) = f(1. I) = lIS. f(O, 1) = f(l. 0) = 3/8. 11. Show that the random vatiables with the densities Are X and Y independent? f(l:, y) = X +Y and g(x, y) = (x if 0 ~ x ~ g(x. y) = distribution. + ~)(y + ~) I, 0 ~ Y ~ I and f(x. y) = 0 and 0 elsewhere. have the same marginal IS. Using Theorem 1, obtain the formula for the mean of the hypergeometrie distribution. Can you use Theorem 3 to obtain the variance of that distribution? 19. Using Theorems I and 3, obtain the formulas for the mean and the variance of the binomial distribution. 20. Prove the statement involving (18). 1041 Chapter 24 Review Questions and Problems . ::a.::iI ...':==IU STIONS AND PROBLEMS 1. Why did we begin the chapter with a section on handling data? 23. Find the mean, standard deviation, and variance in Prob.21. 2. What are stem-and-Ieaf plots? Boxplots? Histograms? Compare their advantages. 24. Find the mean, standard deviation, and variance Prob.22. 3. What quantities measure the average size of data? The spread? 25. What are the outcomes of the sample space of 4. Why did we consider probability theory? What is its 26. What are the outcomes in the sample space of the experiment of simultaneously tossing three coins? role in statistics? In X: Tossing a coin until the first Head appears? 5. What do we mean by an experiment? By a random variable related with it? What are outcomes? Events? 27. A box contains 50 screws, five of which are defective. Find the probability function of the random variable 6. Give examples of experiments in which you have equally likely cases and others in which you don't. X = Number of defective screws in drawing tl1'O screws without replacement and compute its values. 7. State the definition of probability from memory. 28. Find the values of the distribution function in Prob. 27. 8. What is the difference between the concepts of a permutation and a combination? 29. Using a Venn diagram, show that A <::;; B if and only if AUB=B. 9. State the main theorems on probability. IIIustmte them by simple examples. 30. Using a Venn diagram, show that A <::;; B if and only if 10. What is the distribution of a random variable? The distribution function? The probability function'! The density? 31. If X has the density f(x) = 0.5x (0 :;::: x :;::: 2) and o otherwise, what are the mean and the Vallance of X* = -2X + 5? 11. State the definitions of mean and variance of a random variable from memory. 32. If 6 different inks are available, in how many ways can we select two colors for a printing job? Four colors? 12. If peA) = PCB) and A <::;; B, can A =1= B? 13. If E =1= S (= the sample space), can P(E) = I? 14. What distributions correspond to sampling with replacement and without replacement? 15. When will an experiment involve a binomial distribution? A hypergeometric distribution? 16. When will the Poisson distribution be a good approximation of the binomial distribution? 17. What do you know about the approximation of the binomial distribution by the normal distribution? An B =A. 33. Compute 5! by the Stirling formula and find the absolute and relative errors. 34. Two screws are randomly drawn without replacement from a box containing 7 right-handed and 3 lefthanded screws. Let X be the number of left-handed screws drawn. Find P(X = 0), P(X = I), P(X = 2), P(I < X < 2), P(O < X < 5). 35. Find the mean and the variance of the distribution having the density f(x) = ~e-Ixl. 36. Find the skewness of the distribution with density f(x) = 2(1 - x) if 0 < x < I, f(x) = 0 otherwise. 18. Explain the use of the tables of the normal distribution. If you have a CAS, how would you proceed without the tables? 37. Sketch the probability function f(x) = x 2 /30 (x = 1, 2, 3,4) and the distribution function. Find 19. Can the probability function of a discrete random variable have infinitely many positive values? 38. Sketch F(x) = 0 if x :;::: 0, F(x) = 0.2x if 0 F(x) = 1 if x > 5, and its density f(x). 20. State the most important facts about distributions of two random variables and their marginal distributions. 39. If the life of tires is normal with mean 25 000 km and variance 25 000 000 km 2 , what is the probability that a given one of those tires will last at least 30 000 km? At least 35000 km? 21. Make a stem-and-Ieaf plot, histogram, and boxplot of the data 22.5. 23.2, 22.1, 23.6, 23.3, 23.4, 24.0, 20.6, 23.3. 22. Do the same task as in Prob. 21, for the data 210, 213, 209,218,210,215,204,211,216,213. < /-L. x :;::: 5, 40. If the weight of bags of cement is normal with mean 50 kg and standard deviation 1 kg, what is the probability that 100 bags will be heavier than 5030 kg? 1042 CHAP. 24 Data Analysis. Probability Theory : ' ·-11 : Data Analysis. Probability Theory A random experiment, briefly called experiment, is a process in which the result ("outcome") depends on "chance" (effects of factors unknown to us). Examples are games of chance with dice or cards, measuring the hardness of steel. observing weather conditions, or recording the number of accidents in a city. (Thus the word "experiment" is used here in a much wider sense than in common language.) The outcomes are regarded as points (elements) of a set S, called the sample space, whose subsets are called events. For events E we define a probability PtE) by the axioms (Sec. 24.3) o~ (1) peE) u £2 U ... ) = I = I peS) peEl ~ peEl) + P(£2) + These aXiOms are motivated by properties of frequency distributions of data (Sec. 24.1). The complement ~ of E has the probability (2) = P(EC ) 1 - peE). The conditional probability of an event B under the condition that an event A happens is (Sec. 24.3) (3) p(BIA) = peA n B) peA) [peA) > 0]. Two events A and B are called independent if the probability of their simultaneous appearance in a trial equals the product of their probabilities, that is, if (4) PtA n B) = P(A)P(B). With an experiment we associate a random variable X. This is a function defined on S whose values are real numbers; furthermore, X is such that the probability p(X = a) with which X assumes any value a, and the probability pea < X ~ b) with which X assumes any value in an interval a < X ~ b are defined (Sec. 24.5). The probability distribution of X is determined by the distribution function (5) F(x) = P(X ~ x). In applications there are two important kinds of random variables: those of the discrete type, which appear if we count (defective items, customers in a bank, etc.) and those of the continuous type, which appear if we measure (length, speed, temperature, weight, etc.). 1043 Summary of Chapter 24 A discrete random variable has a probability function f(x) = P(X = x}. (6) Its mean 11- and variance a 2 are (Sec. 24.6) (7) 11- = L u2 and xjf(xj) L = j (Xj - 11-}2f(xj) j where the Xj are the values for which X has a positive probability. Important discrete random variables and distributions are the binomial. Poisson. and hypergeometric distributions discussed in Sec. 24.7. A continuous random variable has a density (8) [see (5)j. f(x} = F'(x} Its mean and variance are (Sec. 24.6) (9) 11- = fO xf(x) dx u2 = and -= fC (x - 11-)2f(x) dt:. -oc Very important is the normal distribution (Sec. 24.8), whose density is I f(1;) = - - exp (l0) u\,f2; [1 (x - 11_._u - 2 )2J and whose distribution function is (Sec. 24.8: Tables A 7. A8 in App. 5) (II) A two-dimensional random variable (X, Y) occurs if we simultaneously observe two quantities (for example, heightX and weight Yof adults}. Its distribution function is (Sec. 24.9) ( 12) F(x, y} = P(X ~ x, Y ~ y}. X and Y have the distribution functions (Sec. 24.9) (13) FI(x} = P(X ~ x, Yarbitrary) and F 2 (y) = P(x arbitrary, Y ~ y) respectively; their distribution-; are called marginal distributions. If both X and Y are discrete. then (X. Y) has a probability function f(x, y} = P(X = x, Y = y}. If both X and Yare continuous. then (X. Y) has a density f(x, y). CHAPTER Of 2 5 Mathematical Statistics In probability theory we set up mathematical models of processes that are affected by "chance". In mathematical statistics or, briefly. statistics, we check these models against the observable reality. This is called statistical inference. It is done by sampling, that is, by drawing random samples, briefly called samples. These are sets of values from a much larger set of values that could be studied, called the popUlation. An example is 10 diameters of screws drawn from a large lot of screws. Sampling is done in order to see whether a model of the population is accurate enough for practical purposes. If this is the case, the model can be used for predictions, decisions, and actions, for instance, in planning productions, buying equipment, investing in business projects, and so on. Most important methods of statistical inference are estimation of parameters (Secs. 25.2). determination of confidence intervals (Sec. 25.3), and hypothesis testing (Secs. 25.4. 25.7. 25.8). with application to quality control (Sec. 25.5) and acceptance sampling (Sec. 25.6). In the last section (25.9) we give an introduction to regression and correlation analysis, which concern experiments involving two variables. Prerequisite: Chap. 24. Sections that may be omitted in a shorter course: 25.5, 25.6. 25.8. References, Answers to Problems. a1ld Statistical Tables: App. I Part G, App. 2, App.5. 25.1 Introduction. Random Sampling Mathematical statistics consists of methods for designing and evaluating random experiments to obtain information about practical problems. such as exploring the relation between iron content and density of iron ore, the quality of raw material or manufactured products, the efficiency of air-conditioning systems, the performance of certain cars, the effect of advertising, the reactions of consumers to a new product, etc. Random variables occur more frequently in engineering (and elsewhere) than one would think. For example, properties of mass-produced articles (screws, lightbulbs. etc.) always show random variation, due to small (uncontrollable!) differences in raw material or manufacturing processes. Thus the diameter of screws is a random variable X and we have nOlldefecfive screws. with diameter between given tolerance limits, and defective screws, with diameter outside those limits. We can ask for the distribution of X, for the percentage of defective screws to be expected, and for necessary improvements of the production process. 1044 SEC. 25.1 Introduction. 1045 Random Sampling Samples are selected from populations-20 screws from a lot of 1000, 100 of 5000 voters, 8 beavers in a wildlife conservation project-because inspecting the entire population would be too expensive, time-consuming, impossible or even senseless (think of destructive testing of lightbulbs or dynamite). To obtain meaningful conclusions, samples must be random selections. Each of the 1000 screws must have the same chance of being sampled (of being drawn when we sample), at least approximately. Only then will the sample mean x = (Xl + ... + X20)120 (Sec. 24.1) of a sample of size 11 = 20 (or any other 11) be a good approximation of the population mean JL (Sec. 24.6); and the accuracy of the approximation will generally improve with increasing II, as we shall see. Similarly for other parameters (standard deviation, variance, etc.). Independent sample values will be obtained in experiments with an infinite sample space S (Sec. 24.2), certainly for the normal distribution. This is also true in sampling with replacement. It is approximately true in drawing small samples from a large finite population (for instance. 5 or 10 of 1000 items). However. if we sample without replacement from a small population, the effect of dependence of sample values may be considerable. Random numbers help in obtaining samples that are in fact random selections. This is sometimes not easy to accomplish because there are many subtle factors that can bias sampling (by personal interviews, by poorly working machines, by the choice of nontypical observation conditions, etc.). Random numbers can be obtained from a random number generator in Maple, Mathematica, or other systems listed on p. 991. (The numbers are not truly random, as they would be produced in flipping coins or rolling dice, but are calculated by a tricky formula that produces numbers that do have practically all the essential features of true randomness.) E X AMP L E 1 Random Numbers from a Random Number Generator To select a sample of size n = 10 from 80 given ball bearings, we number the bearings from I to 80. We then let the generator randomly produce 10 of the integers from I to 80 and include the bearings with the numbers obtained in our sample. for example. 44 55 53 03 52 61 67 78 39 54 or whatever. Random numbers are also contained in (older) statistical tables. • Representing and processing data were considered in Sec. 24.1 in connection with frequency distributions. These are the empirical counterparts of probability distributions and helped motivating axioms and properties in probability theory. The new aspect in this chapter is randomness: the data are samples selected randomly from a population. Accordingly, we can immediately make the connection to Sec. 24.1. using stem-and-leaf plots, box plots. and histograms for representing samples graphically. Also, we now call the mean x in (5), Sec. 24.1, the sample mean (1) We call (2) 11 the sample size, the variance S2 in (6), Sec. 24.1, the sample variance 1046 CHAP. 25 Mathematical Statistics and its positive square root s the sample standard deviation..\', 52, and parameters ~l a sample; they will be needed throughout this chapter. 25.2 5 are called Point Estimation of Parameters Beginning in this section, we shall discuss the most basic practical tasks in statistics and corresponding statistical methods to accomplish them. The first of them is point estimation of parameters, that is, of quantities appearing in distributions, such as p in the binomial distribution and JL and u in the normal distribution. A point estimate of a parameter is a number (point on the real line). which is computed from a given sample and serves as an approximation of rhe unknown exact value of the parameter of the population. An interval estimate is an interval ("confidence il1terval"') obtained from a sample; such estimates will be considered in the next section. Estimation of parameters is of great practical importance in many applications. As an approximation of the mean JL of a popUlation we may take the mean .X' of a corresponding sample. This gives the estimate /L = .\' for JL, that is, (1) JL=X= tXl + ... + x.,,) 11 where n is the sample size. Similarly. an estimate &2 for the variance of a popUlation is the variance S2 of a corresponding sample, that is, (2) Clearly, (1) and (2) are estimates of parameters for distributions in which JL or u 2 appear explicity as parameters. such as the normal and Poisson distributions. For the binomial distribution, p = JLln lsee (3) in Sec. 24.71. From (1) we thus obtain for p the estimate (3) jJ= x 1l We mention that (1) is a special case of the so-called method of moments. In this method the parameters to be estimated are expressed in terms of the moments of the distribution (see Sec. 24.6). In the resulting formulas those moments of the distribution are replaced by the corresponding moments of the sample. This gives the estimates. Here the kth moment of a sample X10 •••• Xn is SEC. 25.2 1047 Point Estimation of Parameters Maximum Likelihood Method Another method for obtaining estimates is the so-called maximum likelihood method of R. A. Fisher [Messellger Math. 41 (1912). 155-160]. To explain it we consider a discrete (or continuous) random variable X whose probability function (or density) f(x) depends on a single parameter e. We take a corresponding sample of 11 illdepelldent values Xl' • • • • X1]" Then in the discrete ca~e the probability that a sample of size 1l consists precisely of those 11 values is (4) In the continuous case the probability that the sample consists of values in the small intervals}.; ~ x ~ Xj + tu (j = 1,2, .. ',11) is (5) Since f(xj) depends on e, the function I in (5) given by (4) depends on Xl • . • . , Xn and e. We imagine Xl, • • . , Xn to be given and fixed. Then I is a function of e. which is called the likelihood function. The basic idea of the maximum likelihood method is quite simple, as follows. We choose that approximation for the unknown value of e for which I is as large as possible. If I is a differentiable function of e, a necessary condition for I to have a maximum in an interval (not at the boundary) is ae = o. (6) (We write a partial derivative. because I depends also on Xlo • • • • xn") A solution of (6) depending onx1' ... , Xn is called a maximum likelihood estimate for e. We may replace (6) by a In I (7) ae = 0, because f(xj) > 0, a maximum of I is in general positive, and In I is a monotone increasing function of I. This often simplifies calculations. Several Parameters. If the distribution of X involves r parameters e1 , . . • , en then instead of (6) we have the r conditions al/ae1 = 0, ... , rJllae,. = 0, and instead of (7) we have (8) E X AMP L E 1 a In I ae,. = o. = O. Normal Distribution Find maximum likelihood estimates tor Solutioll. (JI = /L and (J2 = u in the case of the normal distribution. From (1). Sec. 24.8. and (4) we obtain the likelihood function I = (~)n (~)n "271" u e- h where 1048 CHAP. 25 Mathematical Statistics Taking logarithms. we have In 1= -11 In v T;;- - /I In u - h. The first equation in (8) is vOn I)lvf.L = O. written out hence L" '\J - /If.L = O. j~l The ~olution is the desired estimate [L for f.L: we find LXj=.r. Il j=l The ~econd equation in (8) i~ a(ln l)trlu = O. written out v In 1 11 ill! u au Replacing f.L by [Land solving for u 2 • we obtain the estimate -2 U 1 ~ _2 = - L.. (Xj - xl n j=l which we shall use in Sec. 25.7. Note that this differs from (2). We cannot discuss criteria for the goodness of estimates but want to mention that for 'mall 11. formula (2) is preferable. • --•...---_............ ......... _...... ... ... ... ..-. ..-. ~--.-~ 1. Find the maximum likelihood estimate for the parameter f.L of a nOlmal distribution with known variance u 2 = uo2 . 2. Apply the maximum likelihood method to the normal distribution with f.L = O. 3. (Binomial distribution) Derive a maximum likelihood estimate for p. 4. Extend Prob. 3 as follows. Suppose that 111 times 11 trials were made and in the first 11 trials A happened kl times, in the second n trials A happened k2 times, ... , in the mth 11 triab A happened k m times. Find a maximum likelihood estimate of p based on this information. 5. Suppose that in Prob. 4 we made 4 times 5 trials and A happened 2, I, ..k 4 times, respectively. Estimate p. 6. Consider X = Number of independent trials IImil all el'ent A occurs. Show that X has the probability function f(x) = pqX-l. X = l. 2 ..... where p is the probability of A in a single trial and q = I - p. Find the maximum likelihood estimate of p corresponding to a sample .\'1, ••• , Xn of observed values of X. 7. In Prob. 6 find the maximum likelihood estimate of p corresponding to a single observation x of X. 8. In rolling a die. suppose that we get the first Six in the 7th trial and in doing it again we get it in the 6th trial. Estimate the probability p of getting a Six in rolling that die once. 9. (Poisson distribution) Apply the maximum likelihood method to the Poisson distribution. 10. (Uniform distribution) Show that in the case of the parameters a and b of the uniform distribution (see Sec. 24.6), the maximum likelihood estimate cannot be obtained by equating the first derivative to zero. How can we obtain maximum likelihood estimates in this case? 11. Find the maximum likelihood estimate of e in the density f(x) = ee- HX if x ~ 0 and f(x) = 0 if x < O. 12. In Prob. I I. find the mean f.L. substitute it in fex). find the maximum likelihood estimate of IL. and show that It is identical with the estimate for f.L which can be obtained from that for e in Prob. I I. 13. Compute in Prob. 11 from the sanlple 1.8, 0.4. 0.8. 0.6. 1.4. Graph the sample distribution function Fcx) and the distribution function F(x) of the random variable, with e = on the same axes. Do they agree reasonably well? (We consider goodness of fit systematically in Sec. 25.7.) e e. SEC 25.3 Confidence Intervals 14. Do the same task as in Frob. 13 if the given sample is 0.5.0.7.0.1. 1.1. 0.1. 15. CAS EXPERIMENT. Maximum Likelihood Estimates. (MLEs). Find experimentally how much 25.3 1049 MLEs can differ depending on the ~ample ~ize. Hillf. Generate many samples of the same size II. e.g.. of the standardized normal distribution. and record i and S2. Then increase /I. Confidence Intervals Confidence intervals 1 for an unknown parameter 8 of some distribution (e.g .. 8 = J-L) are intervals 81 :2: 8 :2: 82 that contain 8, not with certainty but with a high probability 'Y. which we can choose (95% and 99% are popular). Such an interval is calculated from a sample. 'Y = 95% means probability I - 'Y = 5% = 1/20 of being wrong--Dne of about 20 such intervals will not contain 8. Instead of writing 81 :2: e ~ 82 , we denote this more distinctly by writing (1) Such a special symbol, CONE seems worthwhile in order to avoid the misunderstanding that 8 mllst lie between 81 and 82 , 'Y is called the confidence level, and 81 and 82 are called the lower and upper confidence limits. They depend on 'Y. The larger we choose 'Y. the smaller is the error probability 1 - 'Y, but the longer is the confidence interval. If 'Y ---7 I, then its length goes to infinity. The choice of 'Y depends 011 the kind of application. In taking no umbrella, a 5% chance of getting wet is not tragic. In a medical decision of life or death, a 5% chance of being wrong may be too large and a I % chance of being wrong ('Y = 99%) may be more desirable. Confidence intervals are more valuable than point estimates (Sec. 25.2). Indeed. we can take the midpoint of (1) as an approximation of 8 and half the length of ( I) as an "error bound" (not in the strict sense of numerics. but except for an error whose probability we know). 81 and 82 in (1) are calculated from a sample Xl, . . . • X n . These are 11 observations of a random variable X. Now comes a standard trick. We regard Xl, ••• , X 11 as single observations of n random variables Xl' ... , Xn (with the same distribution, namely, that ofX)· Then 81 = 81 (X1, ••• , xn) and 82 = 82 (x 1 , . • • , xn) in (I) are observed values of two random variables 8 1 = 8 1 (X1 , . . • , Xn) and 8 2 = 8 2 (X1 , ••• , X,,). The condition (I) involving 'Y can now be written (2) Let us see what all this means in concrete practical cases. In each case in this section we shall first state the steps of obtaining a confidence interval in the form of a table, then consider a typical example, and finally justify those steps theoretically. 1 JERZY NEYMAN (1894-1981 l. American statistician, developed the theory of confidence intervals (Alll/als of Mathematical Statistics 6 (1935). 111-116). CHAP. 25 1050 Mathematical Statistics Confidence Interval for JL of the Normal Distribution with Known (J"2 Table 25.1 Determination of a Confidence Interval for the Mean p. of a Normal Distribution with Known Variance u 2 Step 1. Choose a confidence level y (95%, 99%, or the like). Step 2. Determine the conesponding c: 'Y c I 0.90 1.645 0.95 0.99 0.999 1.1)60 2.576 3.21) 1 Step 3. Compute the mean .f of the sample Xl> •••• Xu- Step -I. Compute k = cuIyr-;;. The confidence interval for jL is (3) E X AMP L E 1 CONFy Ix - k ~ p. ~ x+ k). Confidence Interval for jL of the Normal Distribution with Known u 2 Deterimine a 95'ff confidence interval for the mean of a normal distribution with variance sample of 11 = 100 values with mean x = 5. 0- 2 = 9. using a Solution. Step 1. l' = 0.95 is required. Step 2. The corresponding c equals 1.960; see Table 25.1. Step 3..r = 5 is ghen. Step 4. We need k = 1.960' 3/v'lOo = 0.588. Hence r - k = 4.412..1' + k = 5.588 and the confidence interval is CONFo.95 [4.412 ::'" /L::'" 5.588}. This is sometimes written /L = 5 ::':: 0.588, but we shall not lise this notation, which can be misleading. With your CAS YOll can determine this interval more directly. Similarly for the other examples in this ,eetion... Theory for Table 25.1. THEOREM 1 The method in Table 25.1 follows from the basic Sum of Independent Normal Random Variables Let Xl' ... , X" be illdependent nonnal random variables each jL and I'ariallce u 2 . Theil the followillg holds. la) The Slllll Xl + ... + has mean Xn is Ilonlla! with meall IljL alld variallce IlU 2 . (b) The following random variable (4) (~fwhich X I = - n X is normal with mean (Xl jL and variallce u 2 /n. + ... + Xn) le) The followillg ralldom variable Z is Ilonllal with melill 0 alld variallce l. (5) PROOF The statements about the mean and variance in (a) follow from Theorems 1 and 3 in Sec. 24.9. From this and Theorem 2 in Sec. 24.6 we see that X has the mean (I/1l)lljL = jL and the variance (l11ly2nu 2 = u 2 /n. This implies that Z has the mean 0 and variance I. by Theorem 2(b) in Sec. 24.6. The nonnality of Xl + ... + Xn is proved in Ref. [031 listed in App. 1. This implies the normality of (4) and (5). • SEC. 25.3 Confidence Intervals 1051 Derivation of (3) in Table 25.1. Sampling from a normal distribution gives independent sample values (see Sec. 25.1). so that Theorem I applies. Hence we can choose 'Yand then determine c such that (6) P( -c ~ Z ~ c) = P ( -c X-11- ~ ~ ~ a/v 11 c ) = <D(c) - (1:>( -c) = 'Y. For the value 'Y = 0.95 we obtain ::.(D) = 1.960 from Table AS in App. 5. as used in Example 1. For 'Y = 0.9. 0.99. 0.999 we get the other values of c listed in Table 25.1. Finally. all we have to do is to convert the inequality in (6) into one for 11- and insert observed values obtained from the sample. We multiply -c ~ Z ~ c by -I and then by a/V;;. writing car\!;; = k (as in Table 25.1), P( - c ~ Z ~ c) = P( c ~ - Z ~ - c) = P (c = PCk Adding X gives P(X (7) + k ~ JL ~ X - k) P(X - k = 'Y ~ JL ~ ~ JL - X :> - a/V;; = ~ JL - X ~ c) -k) = 'Y. or X + k) = y. Inserting the observed value.X' of X gives (3). Here we have regarded Xl> • • • • Xn as single observations of Xl' ...• X" (the standard trick!). so tha!..xI + ... + Xn is an observed value of Xl + ... + Xn and .X' is an observed value of X. Note further that (7) is of the • form (2) with 8 1 = X - k and 8 2 = X + k. E X AMP L E 2 Sample Size Needed for a Confidence Interval of Prescribed Length How large must Solution. 11 be in Example I if we want to obtain a 95% confidence interval of length L = OA? ll1e interval (3) has the length L = 2k = 'leu'v;,. Solving for 11. we obtain In the present case the answer is 11 = (2 . 1.960' 310.4)2 = 870. Figure 525 shows how L decreases as 11 increases and that for l' = 99% the confidence interval is substantially longer than for l' = 95% (and the same sample size Ill. • 0.6 r - , , - - - - - - - - - , 0.4 Llu 0.2 -+00 n L-- _ 500 Fig. 525. Length of the confidence interval (3) (measured in multiples of tT) as a function of the sample size n for 'Y = 95% and y = 99% 1052 CHAP. 25 Mathematical Statistics Confidence Interval for J.L of the Normal Distribution With Unknown 0-2 In practice a 2 is frequently unknown. Then the method in Table 25.1 does not help and the whole theory changes, although the steps of determining a confidence interval for /-L remain quite similar. They are shown in Table 25.2. We see that k differs from that in Table 25.1, namely, the sample standard deviation s has taken the place of the unknown standard deviation a of the popUlation. And c now depends on the sample size 11 and must be determined from Table A9 in App. 5 or from your CAS. That table lists values z for given values of the distribution function (Fig. 526) F(~) = (8) I_= Z K", u2 ( I + - )-(1n+ 1)/2 du III of the t-distribution. Here, I1l (= I, 2, ... ) is a parameter, called the number of degrees of freedom of the distribution (abbreviated d.f.). [n the present case. I7l = 1l I; see Table 25.2. The constant Km is such that F(x) = 1. By integration it turns out that Km = r(~111 + ~)/[V;;;;: r(~1Il)]. where r is the gamma function (see (24) in App. A3.1). T 'lIe 25.2 Determination of a Confidence Interval for the Mean I.t of a Normal Distribution with Unknown Variance 0"2 Step 1. Choose a confidence level y (95%.99%. or the like). Step 2. Determine the solution c of the equation (9) F(c) = ~(l + y) from the table of the t-distribution with 11 - I degrees of freedom (Table A9 in App. 5; or use a CAS; 11 = sample size). Step 3. Compute the mean x and the variance S2 of the sample X b · · · 'Xno Step 4. Compute k (10) = cst\;;;. The confidence interval is CONFl' {x - k :2i /-L :2i x+ k}. y 3 d.f. r.~~f~ l.°l~ r y 0.8 0.6 o d' -3 -2 Fir 526. -1 0 2 3 x Distribution functions of the tdistnbution with 1 and 3 dJ. and of the standardized normal distribution (steepest curve) Fig. 527. Densities of the t-distribution with 1 and 3 dJ. and of the standardized normal distribution SEC. 25.3 Confidence Intervals 1053 Figure 527 compares the curve ofthe density ofthe t-distribution with that of the normal distribution. The latter is steeper. This illustrates that Table 25.1 (which uses more information, namely, the known value of ( 2 ) yields shorter confidence intervals than Table 25.2. This is confirmed in Fig. 528, which also gives an idea of the gain by increasing the sample size. 2,--,-rr-----,----,----, \\ L'lL l.5 lL-____~____~_____ L_ _ _ _~ o 10 20 11 Fig. 528. Ratio of the lengths L' and L of the confidence intervals (10) and (3) with y = 95% and y = 99% as a function of the sample size n for equals and (T E X AMP L E 3 Confidence Interval for p. of the Normal Distribution with Unknown u 2 Five independent measurements of the point of int1ammation (flash point) of Diesel oil (D-2) gave the values (in OF) 144 147 146 142 144. Assuming normality. determine a 99% confidence interval for the mean. Solutioll. Step 1. y = Step 2. FCc) = ~(1 Step 3.."i' = 144.6, + O.l)l) is required. y) = 0.995. and Table A9 in App. 5 with s2 = 11 - I = 4 d.f. gives c = 4.60. 3.8. Step 4. k = V3.8. 4.60/Vs = 4.01. The confidence interval is CONFo.99 {140.5 ;;;; fL ;;;; 148.7J. If the variance (T2 were known and equal 10 the sample variance s2. thus 0"2 = 3.8. then Table 25.1 would give k = culV-;' = 2.576V3.8/V5 = 2.25 and CONFR99 {142.35;;;; fL;;;; 146.85J. We see that the present interval is almost twice as long as that obtained from Table 25.1 (with 0"2 = 3.8). Hence for small sample, the ditlerence is considerable! Sec also Fig. 528. • Theory for Table 25.2. THEOREM 2 For deriving (10) in Table 25.2 we need from Ref. lG3] Student's t-Distribution Let Xl ..... Xn be independent normal random variables with the same mean p. and the slime raricmce a 2 • Then the rllndom variable X-p. (11) has a t-distribution [see (8)] with by (4) alld (12) T=-- S/V;; 11 I degrees oj ji·eedo/Jl (d.f.): here X is given 1054 CHAP. 25 Mathematical Statistics Derivation of (10). This is similar to the derivation of (3). We choose a number 'Y between 0 and I and determine a number (' from Table A9 in App. 5 with 11 - I dJ. (or from a CAS) such that (13) P(-c ~ T~ = c) F(c) - F(-c) = ')'. Since the t-distribution is symmetric, we have F( -c) = I - F(c), and (13) assumes the form (9). Substituting (11) into (13) and transforming the result as before, we obtain (14) ~ P(X - K /-L ~ X + K) = 'Y where K = cS/Y;;. By inserting the observed values .'t of X and S2 of S2 into (14) we finally obtain (10) . • Confidence Interval for the Variance of the Normal Distribution 0- 2 Table 25.3 shows the steps. which are similar to those in Tables 25.1 and 25.2. Table 25.3 Determination of a Confidence Interval for the Variance 0"2 of a Normal Distribution, Whose Mean Need Not Be Known Step 1. Choose a confidence level 'Y (95%, 99%. or the like). Step 2. Determine solutions Cl and C2 of the equations (15) from the table of the chi-square distribution with 11 - I degrees of freedom (Table AlO in App. 5; or use a CAS: 11 = sample size). Step 3. Compute (n - 1 )S2, where Step 4. Compute kl = (11 confidence interval is S2 l)s2/c ] is the variance of the sample and k2 = (/1 - I)S2/C2 . The (16) E X AMP L E 4 Confidence Interval for the Variance of the Normal Distribution Detennine a 95'i!' confidence interval (16) for the variance. using Table 25.3 and a sample (tensile strell"th of . 2 . e sheet steel m kg/mm . rounded to mteger values) 89 84 R7 81 89 R6 91 90 78 89 R7 99 83 R9. SEC. 25.3 1055 Confidence Intervals Solution. Step 1. Step 2. For 11 - 'Y = 0.95 is required. I = 13 we find ("I = Step 3. 13s2 = Step -I. l3s 2 1c1 5.01 and c2 = 24.7~. 326.9. = 65.25. l3s 21c2 = 13.21. The confidence interval is CONFo.95 {l3.2l ~ This IS fT2 ~ 65.25}. rather large, and for obtaining a more precise result, one would need a much larger sample. • Theory for Table 25.3. In Table 25.1 we used the normal distribution. in Table 25.2 the t-distribution. and now we shall use the X2-distribution (chi-square distribution), whose distlibution function is F(:::) = 0 if.: < 0 and F(.:) = em f z e-u/211cm-2)/2 du if.: ~ 0 (Fig. 529). o y 0.8 0.6 0.4 0.2 o Fig. 529. 4 6 10 8 x Distribution function of the chi-square distribution with 2, 3, 5 dJ. The parameter III (= 1. 2, ... ) is called the number of degrees of freedom (d.L), and Note that the distribution is not symmetric (see also Fig. 530). For deriving (16) in Table 25.3 we need the following theorem. THEOREM 3 Chi-Square Distribution Under the assllmptions il1 Theorem 2 the random variable (17) with S2 giren by (12) has a chi-square distribution with Proof in Ref. [03]. listed in App. I. 11 - I degrees offreedom 1056 CHAP. 25 Mathematical Statistics y 0.5 \ 0.4 2 d.f. 0.3 0.2 I I ,~ 3 d.f. 0.1 : o Fig. 530. 2 6 4 8 10 x Density of the chi-square distribution with 2, 3, 5 dJ. Derivation of (16). This is similar to the derivation of (3) and (10). We choose a number l' between 0 and 1 and determine c] and C2 from Table AIO, App. 5, such that [see (15)] Subtraction yields Transforming CI ~ Y ~ C2 with Y given by (17) into an inequality for u 2 , we obtain n - 1 2 ---s C2 By inserting the observed value S2 ~u 2 n - I 2 ~---s. CI of S2 we obtain (16). • Confidence Intervals for Parameters of Other Distributions The methods in Tables 25.1-25.3 for confidence intervals for J.1- and u 2 are designed for the normal distribution. We now show that they can also be applied to other distrihutions if we use large samples. We know that if Xl' ... , Xn are independent random variables with the same mean JL and the same valiance u 2 , then their sum Yn = Xl + ... + X" has the following properties. (A) Yn has the mean nJL and the variance nu 2 (by Theorems I and 3 in Sec. 24.9). (B) If those variables are normal, then Yn is normal (by Theorem 1). If those random variables are not normal, then (B) is not applicable. However, for large n the random variable Yn is still approximately normal. This follows from the central limit theorem, which is one of the most fundamental results in probability theory. SEC. 25.3 1057 Confidence Intervals Central Limit Theorem THEOREM 4 Let Xl> ... , X", ... be independent random variables that have the same distribution function and therefore the same mean /-L and the same variance a 2 . Let Yn = Xl + ... + Xn . Then the random variable (18) is asymptotically normal with mean 0 and variance 1; that is. the distribution function Fn(X) of Zn satisfies lim Fn(x) n~oo = I -yI2; I <I>(x) = - - x e- u2/2 duo -00 A proof can be found in Ref. [03] listed in App. 1. Hence when applying Tables 25.1-25.3 to a non normal distribution, we must use sllfficiently large samples. As a rule of thumb, if the sample indicates that the skewness of the distribution (the asymmetry; see Team Project 16(d), Problem Set 24.6) is small, use at least Il = 20 for the mean and at least n = 50 for the variance. _...-.. ..... _ ••• w ..... • _ _ .... _ 11-71 ... --- ....... ..-. • ..-. --. ..., MEAN (VARIANCE KNOWN) 1. Find a 95% confidence interval for the mean JL of a normal population with standard deviation 4.00 from the sample 30. 42, 40, 34, 48, 50. 2. Does the interval in Prob. I gel longer or shorter if we take 'Y = 0.99 instead of 0.95? By what factor? 3. By what factor does the length of the interval in Prob. 1 change if we double the sample size? 4. Find a 90% confidence interval for the mean JL of a nonnal population with variance 0.25, using a sample of 100 values with mean 212.3. 5. What sample size would be needed for obtaining a 95% confidence interval (3) of length 2u? Of length u"? 18-121 MEAN (VARIANCE UNKNOWN) Fwd a 99% confidence interval for the mean of a nonnal popUlation from the sample: 8. 425, 420. 425, 435 9. Length of 20 bolts with sample mean 20.2 cm and sample variance 0.04 cm2 10. Knoop hardness of diamond 9500, 9800, 9750, 9200, 9400, 9550 11. Copper content (%) of brass 66. 66. 65. 64. 66. 67, 64. 65,63,64 12. Melting point eC) of aluminum 660, 667. 654, 663, 662 6. (Use of Fig. 525) Find a 95% confidence interval for a sample of 200 values with mean 120 from a normal distribution with variance 4, lIsing Fig. 525. 13. Find a 95% confidence interval for the percentage of cars on a certain highway that have poorly adjusted brakes. using a random sample of 500 cars stopped at a roadblock on that highway. 87 of which had poorly adjusted brakes. 7. What sample size i~ needed to obtain a 99% confidence interval of length 2.0 for the mean of a normal population with variance 25? Use Fig. 525. Check by calculation. 14. Find a 99% confidence interval for p in the binomial distribution from a classical result by K. Pearson, who in 24000 trials oftossing a coin obtained 12012 Heads. Do you think that the coin was fair? CHAP. 25 1058 Mathematical Statistics [ii-20 I VARIANCE Find a Y5% confidence interval for the variance of a normal population from the sample: 15. A sample of 30 values with variance 0.0007 16. The sample in Prob. 9 17. The sample in Prob. II 18. Carbon monoxide emission (grams per mile) of a certain type of passenger car (cruising at 55 mph): 17.3,17.8,18.0,17.7,18.2,17.4. 17.6. 18.1 19. Mean energy (keV) of delayed neutron group (Group 3, half-life 6.2 sec.) for uranium U235 fission: 435, 451, 430,444,438 20. Ultimate tensile strength (k psi) of alloy steel (Maraging H) at room temperature: 251, 255, 258, 253, 253,252,250,252,255,256 21. If X is normal with mean 27 and variance 16, what distributions do -X, 3X, and 5X - 2 have? 22. If Xl and X2 are independent normal random variables 25.4 with mean 23 and 4 and variance 3 and I, respectively, what distribution does 4X1 - X2 have? Hint. Use Team Project 14(g) in Sec. 24.8. 23. A machine fills boxes weighing Y lb with X lb of salt, where X and Yare normal with mean 100lb and 51b and standard deviation 1 lb and 0.5 lb, respectively. What percent of filled boxes weighing between 104 Ib and 1061b are to be expected? 24. If the weight X of bags of cement is normally distributed with a mean of 40 kg and a standard deviation of 2 kg, how many bags can a delivery truck carry so that the probability of the total load exceeding 2000 kg will be 5%? 25. CAS EXPERIMENT. Confidence lntervals. Obtain 100 samples of size 10 of the standardized normal distribution. Calculate from them and graph the corresponding 95o/{' confidence intervals for the mean and count how many of them do not contain O. Does the result suppon the theory? Repeat the whole experiment. compare and comment. Testing of Hypotheses. Decisions The ideas of confidence intervals and of tests 2 are the two most important ideas in modern statistics. In a statistical test we make inference from sample to population through testing a hypothesis, resulting from experience or observations, from a theory or a quality requirement, and so on. In many cases the result of a test is used as a basis for a decision, for instance, to buy (or not to buy) a certain model of car, depending on a test of the fuel efficiency (miles/gal) (and other tests, of course), to apply some medication, depending on a test of its effect; to proceed with a marketing strategy, depending on a test of consumer reactions, etc. Let us explain such a test in terms of a typical example and introduce the corresponding standard notions of statistical testing. E X AMP L E 1 Test of a Hypothesis. Alternative. Significance Level a We want to buy 100 coils of a certain kind of wire, provided we can verify the manufacturer's claim that the wire has a breaking limit /-t = /-to = 200 Ih (or more). This is a test ofthe hypothesis [also called /lull hypothesis} /-t = /-to = 200. We shall not buy the wire if the (statistical) test shows that actually /-t = /-t1 < /-to, the wire is weaker, the claim does not hold. /-tl is called the alternative (or alternative iz)lJOtizesis) of the test. We shall accept the hypothesis if the test suggests that it is true, except for a small error probability a, called the significance level of the test. Otherwise we reject the hypothesis. Hence a is the probability of rejecting a hypothesis although it is hue. The choice of a is up to us. SCk and I % are popular values. For the test we need a sample. We randomly select 25 coils of the wire, cut a piece from each coil, and determine the breaking limit experimentally. Suppose that this sample of n ~ 25 values of the breaking limit has the mean x = 1971b (somewhat less than the claim!) and the standard dcviation s = 6 lb. 2Beginning around 1930, a systematic theory of tests was developed by NEYMAN (see Sec. 25.3) and EGON SHARPE PEARSON (1895-1980), English statistician, the son of Karl Pearson (see the footnote on p. 1066). SEC. 25.4 Testing of Hypotheses. 1059 Decisions At this point we could only speculate whether this difference 197 - 200 = - 3 is due to randomness, is a chance effect. or whether it is significant, due to the actually inferior quality of the wire To continue beyond ,peculation requires probability theory. as follows. We assume thaI the breaking limit is normall} distributed. (This as,umption could be tested by the method in Sec. 25.7. Or we could remember the central limit theorem (Sec. 25.3) and take a still larger sample.) Then T= x- /Lo SI\ ';; in (II). Sec. 25.3, with JL = /Lo has a t-distribution with 1/ - 1 degrees of freedom (1/ - I = 24 for our sample). Also X = 197 and s = 6 are observed values of X and S to be used later. We can now choose a significance level, say, a = 5%. From Table A9 in App. 5 or from a CAS we then obtain a critical value c such that peT ~ c) = a = 5%. For PIT ~ c) = I - a = 95'it the table gives c = 1.71. so that c = -c = -1.71 because of the symmetry of the distribution (Fig. 531). We now reason as follows-this is the crucial idea of the test. If the hypothesis is true, we have a chance of only a (= 5'it) that we observe a value t of T (calculated from a sample) that will fall between -:x; and -1.71. Hence if we nevertheless do observe such a t, we assert that the hypothesis cannot be true and we reject it. Then wc accept the alternative. If. however. t ~ c. wc accept the hypothesis. A simple calculation finally give~ t = (197 - 200)/(6rV25) = -2.5 as an observed value of T. Since -2.5 < -1.71. we reject the hypothesis (the manufacturer's claim) and accept the alternative /L = /Ll < 200, • the wire ~eems to be weaker than claimed. I Reject hypotheSiS~; Do not reject hypothesis : 95% a~~~ c =-1.71 0 Fig. 531. t-distribution in Example 1 This example illustrates the steps of a test: 1. Formulate the hypothesis {} = 80 to be tested. (80 = /-Lo in the example.) 2. Formulate an alternative 8 = 8}. (81 = /-Ll in the example.) 3. Choose a significance level a (5%,1%,0.1%). e 4. Use a random variable = g(Xl , . . . , Kn) whose distribution depends on the hypothesis and on the alternative, and this distribution is known in both cases. Determine assuming the hypothesis to be true. (In the a critical ':.alue c from the distribution of example. e = T. and c is. obtained from peT ~ c) = a.) e, e of e. Accept or reject the hypothesis. depending on the size of erelative to c. (t < c in 5. Use a sample Xl • . . . , Xn to determine an observed value = g(x}, ... , xn) (t in the example.) 6. the example, rejection of the hypothesis.) Two important facts require further discussion and careful attention. The first is the choice of an alternative. In the example, /-L] < /-Lo, but other applications may require /-Ll > /-Lo or /-Ll /-Lo· The second fact has to do with errors. We know that a (the significance level of the test) is the probability of rejecting a true hypothesis. And we shall discuss the probability {3 of accepting afalse hypothesis. '* 1060 CHAP. 25 Mathematical Statistics One-Sided and Two-Sided Alternatives (Fig. 532) Let 8 be an unknown parameter in a distribution, and suppose tbat we want to test the hypothesis 8 = 80 , Then there are three main kinds of alternatives. namely, (1 ) (2) (3) (1) and (2) are one-sided alternatives, and (3) is a two-sided alternative. We call rejection region (or critical region) the region such that we reject the hypothesis if the observed value in the test falls in this region. In CD the critical c lies to the right of 80 because so does the alternative. Hence the rejection region extends to the right. This is called a right-sided test. In @ the critical c lies to the left of 80 (as in Example 1), the rejection region extends to the left, and we have a left-sided test (Fig. 532, middle part). These are one-sided tests. In @ we have two rejection regions. This is called a two-sided test (Fig. 532, lower part). All three kinds of alternatives occur in practical problems. For example, (1) may arise if 80 is the maximum tolerable inaccuracy of a voltmeter or some other instrument. Alternative (2) may occur in testing strength of material, as in Example 1. Finally, 80 in (3) may be the diameter of axle-shafts, and shafts that are too thin or too thick are equally undesirable, so that we have to watch for deviations in both directions. Acceptance Region Do not reject hypothesis (Accept hypothesis) Rejection Region (Critical Region) Reject hypothesis c Acceptance Region Do not reject hypothesis (Accept hypothesis) Rejection Region (Critical Region) Reject hypothesis ~ __________________ ~----~I-----------------------80 c Rejection Region (Critical Region) Reject hypothesis Acceptance Region Do not reject hypothesis (Accept hypothesis) 0------ ----- Re jectlon Region (Critical Region) Reject hypothesIs 80 Fig. 532. Test in the case of alternative (1) (upper part of the figure), alternative (2) (middle part), and alternative (3) Errors in Tests Tests always involve risks of making false decisions: (I) Rejecting a true hypothesis (Type 1 error). ll' = Probability of making a Type I error. (II) Accepting a false hypothesis (Type II error). f3 = Probability of making a Type II error. SEC 25.4 Testing of Hypotheses. 1061 Decisions Clearly, we Cannot avoid these errors because no absolutely certain conclusions about populations can be drawn from samples. But we show that there are ways and means of choosing suitable levels of risks, that is, of values a and {3. The choice of a depends on the nature of the problem (e.g., a small risk a = 1% is used if it is a matter of life or death). Let us discuss this systematically for a test of a hypothesis ti = tio against an alternative that is a single number el , for simplicity. We let el > eo, so that we have a right-sided test. For a left-sided or a two-sided test the discussion is quite similar. We choose a critical c > eo (as in the upper part of Fig. 532, by methods discussed below). From a given sample Xl> ••• , Xn we then compute a value with a suitable g (whose choice will be a main point of our further disc~ssion; for instance, take g = (Xl + ... + Xn)l11 in the case in which e is the mean). If e > c, we reject the hypothesis. If ~ c, we accept it. Here, the value can be regarded as an observed value of the random variable e e (4) becam,e Xj may be regarded as an observed value of Xj' j = L ... , 11. In this test there are two possibilities of making an error, as follows. Type I Error (see Table 25.4). The hypothesis is true but is rejected (hence the alternative is accepted) because e assumes a value > c. Obviously, the probability of making such an error equals e (5) a is called the significance level of the test, as mentioned before. Type II Error (see Table 25.4). The hypothesis is false but is accepted because assumes a value ~ c. The probability of making such an error is denoted by {3; thus e e (6) 7] = I - {3 is called the power of the test. Obviously, the power TJ is the probability of avoiding a Type I1 error. Table 25.4 Type I and Type II Errors in Testing a Hypothesis ()o Against an Alternative () ()J () = = Unknown Truth '0 Cl) E. '-' '-' e = eo '-' <C e = el e = eo e = el True decision Type II error P=l-a P=f3 Type 1 error True decision P=I-{3 P=a 1062 CHAP. 25 Mathematical Statistics Formulas (5) and (6) show that both (]' and f3 depend on c, and we would like to choose c so that these probabilities of making errors are as small as possible. But the important Figure 533 shows that these are conflicting requirements because to let (]' decrease we must shift c to the right, but then f3 increases. In practice we first choose (]' (5%, sometimes 1%), then determine c, and finally compute f3. If f3 is large so that the power 1] = 1 - f3 is small, we should repeat the test, choosing a larger sample, for reasons that will appear shortly. " Density of e if the hypothesis is true : & Density of if the alternative / l;f ,,. . . . -T-.. . . . . . . is true // " // I I " ... I I I I " ... " a-::::----...! eo ........ _ " el c Acceptance region ~ Rejection region (Critical regIOn) Fig. 533. Illustration of Type I and II errors in testing a hypothesis e = eo against an alternative e = e, (> eo, right-sided test) If the alternative is not a single number but is of the form (1}-(3), then f3 becomes a function of e. This function f3( e) is called the operating characteristic (OC) of the test and its curve the OC curve. Clearly, in this case 1] = 1 - f3 also depends on e. This function 1](e) is called the power function of the test. (Examples will follow.) Of course, from a test that leads to the acceptance of a certain hypothesis Bo, it does not follow that this is the only possible hypothesis or the best possible hypothesis. Hence the terms "not reject" or "fail to reject" are perhaps better than the term "accept." Test for IL of the Normal Distribution with Known u 2 The following example explains the three kinds of hypotheses. E X AMP L E 2 Test for the Mean of the Normal Distribution with Known Variance Let X be a normal random variable with variance a 2 = 9. Using a sample of size hypothesis /L = /Lo = 24 against the three kinds of alternatives. namely. (a) Solution. /L> /Lo (b) /L < /Lo /L (c) '* 11 = 10 with mcan.Y, test the /Lo' We choose the significance level a = 0.05. An estimate of the mean will be obtained from I X = - (Xl + ... + X17)' 11 If the hypothesis is true. X is normal with mean /L = 24 and variance Hence we may obtain the critical value (' from Table A8 in App. 5. Case (a). (T 2 Right-Sided Test. We determine e from PIX > e)fL~24 = a - P(X ;" C)fL~24 = <P (e-24) vo:9 /n = 0.9. see Theorem I, Sec. 25.3. = 0.05, that is. = I - a = 0.95. Table A~ in App. 5 _gives (e - 24)/VO.9 = 1.645, and e = 25.56. which is grcater than /Lo, as in the upper part of FIg. 532. If x ;" 25.56, the hypothesis is accepted. If.ct' > 25.56, it is rejected. The power function of the test is (Fig. 534) SEC. 25.4 Testing of Hypotheses. Decisions 1063 0.8 0.6 O.Ll 0.2 20 Fig. 534. > 25.56)" = I - (7) = I - <P ( P(X ~ 25.56)" 25.56 - fL) vo.9 = I - <P(26.94 - 1.05fL} 0.9 Left-Sided Test. The critical value c is obtained from the equation - PiX ~ C),,-24 = <P (cYo.9 - 24) Table A8 in App. 5 yields c = 24 - 1.56 = 22.44. If x reject it. Ihe power function of the test is (8) Case (c). fL Power function 1)(fL) in Example 2, case (a) (dashed) and case (c) 7}(fL) = P(X Case (b). 28 fLO 7}(fL) - = P(X ~ 22.44)" = <1> ~ ~ = a = 0.05. 22.44. we accept the hypothesis. If x < 22.44. we ( 22.44 - fL ) ~ = <1>(23.65 VO.9 Two-Sided Test. Since the normal distribution is symmetric. we choose + k. and detennine k from cl and c2 equidistant from fL = 24. say. cl = 24 - k and c2 = 24 P(24 - k ~ X ~ 24 + k), ,~24 = <1>( k VO.9 ) - <p(- k VO.9 ) = I - a = 0.95. Table A8 in App. 5 gives k/Y0.9 = 1.960. hence k = 1.86. This gives the values cl = 24 - 1.86 = 22.14 and c2 = 24 + 1.86 = 25.86. If x is not smaller than cl and not greater than c2. we accept the hypothesis. Otherwise we reject it. The power function of the test is (Fig. 534) 7}(fL} (9) = P(X < 22.14)" + P(X =1+<1> ( = I + > 25.86)" = 22.14 - fL) Yo.9 P(X -<P < 22.14)" + I - P(X ~ 25.86)" (25.86 - fL) VO.9 <1>(23.34 - 1.05fL) - <P(27.26 - 1.05fL). Consequently. the operating characteristic f3(.fL) = I - 7}(fL) (see before i is (Fig. 535) If we take a larger sample. ~ay. of size 11 = 100 (instead of 10). then a 2 /1l = 0.09 (instead of 0.9) and the critical values are Cl = 23.41 and c2 = 24.59, as can be readily verified. Ihen the operating characteristic of the test is f3(fL) = <p( 24.59 VO.09 = fL) _ <1>( 23.41 . fL) V'0.09 <P(81.97 - 3.33fL) - <P(78.03 - 3.33fL}. 1064 CHAP. 25 Mathematical Statistics Figure 535 shows that the corresponding OC curve is steeper than that for 11 = 10. l11is means that the increase of 11 has led to an improvement of the test. In any practical case, 11 is chosen as small as possible but SO large that the test brings out deviations bet",een J-L and J.Lo that are of practical interest. For instance. if deviations of ±2 units are of interest. we see from Fig. 535 that 11 = 10 is much too small because when J-L = 24 - 2 = 22 or J-L = 24 + 2 = 26 f3 is almost 50%. On the other hand, we see that" = I 00 is sufticient for that purpose. • (3(p.) 1.0 \ \ 0.8 0.6 0.4 n = 10 \ 0.2 n= 100 , ....... 6- 28 p. 26 P.o Curves of the operating characteristic (OC curves) in Example 2, case (e), for two different sample sizes n Fig. 535. Test for J.L When u 2 is Unknown, and for u 2 E X AMP L E 3 Test for the Mean of the Normal Distribution with Unknown Variance The tensile strength of a sample of /I = 16 manila ropes (diameter 3 in.) was measured. The sample mean was 4482 kg, and the sample standard deviation was s = 115 kg (N. C. Wiley, 41st Annual Meeting of the American Society for Testing Materials). Assuming that the tensile strength is a normal random variable, test the hypothesis J.Lo = 4500)..g against the alternative J-LI = 4400 kg. Here J.Lo may be a value given by the manufacturer. while J-Ll may result from previous experience. x= We choose the significance level a = S%. If the hypothesis i~ true, it follows from Theorem 2 in Sec. 25.3, that the random variable Solution. T= x- x- J-Lo Sry';; 4500 S/4 has at-distribution ",ith 11 - I = IS d.f. The test is left-sided. TIle critical value c is obtained from peT < c)I'O = a = 0.05. Table A9 in App. 5 gives c = -1.7S. As an observed value of T we obtain from the sample t = (4482 - 4S00)/( 11S/4) = -0.626. We see that t > c and accept the hypothesis. For obtaining numeric values of the power of the test. we would need tables called noncentral Student t-tables: we shall not discuss this question here. • E X AMP L E 4 Test for the Variance of the Normal Distribution Using a sanlple of size n = 15 and '<lmple varilmce s2 = 13 from a normal population, test the hypothe,is 2 2 2 2 u = Uo = 10 against the alternative u = Ul = 20. Solution. We choose the significance level a = 5%. If the hypothesis is true, then S2 y= (/I - S2 1)(J',2 1410 = lAS 2 o has a chi-square distribution with n - I = 14 d.f. by Theorem 3, Sec. 2S.3. From P(Y> c) = a = O.OS. that is. PlY ~ c) = 0.9S, and Table AIO in App. 5 with 14 degrees of freedom we obtain c = 23.68. This is the critical value of Y. Hence SEC. 25.4 Testing of Hypotheses. Decisions 1065 to S2 = Uo 2Y1(11 - 1) = 0.7I4Y there corresponds the critical value c* = 0.714' 23.68 = 16.91. Since < c*. we accept the hypothesis. If the alternative is true, the random variable Y1 = l4S2/U12 = 0.7S2 has a chi-square distribution with 14 dJ. Hence our test has the power .1'2 From a more extensive table of the chi-square distribution (e.g. in Ref. [G3] or [G8]) or from your CAS, you see that 7J = 62%. Hence the Type II risk is very large. namely. 38%. To make this risk smaller. we would have to increase the sample size. • Comparison of Means and Variances EXAMPLE 5 Comparison of the Means of Two Normal Distributions Using a sample Xl> ••. , x n , from a normal distribution with unknown mean /L,- and a sample Yl, ... , Yn from 2 another normal distribution with unknown mean /Ly, we want to test the hypothesis that the means are equal, 3 /Lx = /Ly, against an altemative, say, /Lx > /Ly. The variances need not be known but are assumed to be equal. Two cases of comparing means are of practical importance: The samples have the same size. Furthermore, each value of the first sample corre~ponds to precisely one value of the otlzer. because conesponding values result from the same person or thing (paired comparison)for example, two measurements of the same thing by two different methods or two measurements from the two eyes of the same person. More generally, they may result from pairs of similar individuals or things, for example, identical twins, pairs of used front tires from the same car, etc. Then we should form the differences of conesponding values and test the hypothesis that the population conesponding to the differences has mean 0, using the method in Example 3. If we have a choice. this method is better than the following. Case A. Case B. The tl\'O samples are indepel1dent and not necessarily of the same size. Then we may proceed as follows. Suppose that tbe altemative is /Lx > /Ly. We choose a significance level a. Then we compute the sample means X and y as well as (nl - I)s,< 2 and (n2 - l)sy 2, where '\·x2 and Sy 2 are the sample variances. Using Table A9 in App. 5 with 111 + n2 - 2 degrees of freedom. we now determine (" from peT ~ c) = I-a. (10) We finally compute (11) to = nln2(111 + n2 "1 + 112 x-y - 2) V(nl It can be shown that this is an observed value of a random variable that has a t-distribution with nl + 112 - 2 degrees of freedom, provided the hypothesis is true. If to ~ c, the hypothesis is accepted. If to > G, it is rejected. If the alternative is /Lx /Ly' then (] 0) must be replaced by * (10*) peT ~ Gl) = 0.5a, peT ~ c2) = I - O.Sa. Note that for sanlples of equal size "1 = n2 = n, formula (11) reduces to (12) 3 This assumption of equality of variances can be tested, as shown in the next example. If the test shows that they differ significantly, choose two samples of tbe same size nl = n2 = n (not too small, > 30, say), use tbe test in Example 2 together with the fact that (12) is an observed value of an approximately stillldardized normal random variable. 1066 CHAP. 25 Mathematical Statistics To illustrate the computations. let us consider the two samples 105 lOR 89 92 (xl' . . . • .lnl ) and (.'"1 ....• -""2) given by 103 103 107 124 105 97 103 107 III 97 and 84 showing the relative output of tin plate worker~ under two different working conditions p. J. B. Worth. JUllrnol ofJndll.<triol Engineering 9. 2--19-253). Assuming that the conesponding populations are normal and have the smne variance. let us test the hypothesis iJ-:c = iJ-y against the alternative iJ-x iJ-y' (Equality of varimlCes will be tested in the next example.) *' Solution. We find y .1' = 105.125. ~ sx2 = 106.125. 97.500. Sy 2 = 84.000. We choose the significance level a = 5'k. From (10") with 0.5a = 2.5'if.. 1 - O.5a = 97.5'k and Table A9 in App.5 with 14 degrees of freedom we obtain ("1 = -2.1--1 and ("2 = 2.14. Fonnula (12) with n = 8 giYes the value to = v'8. 7.625/\ '"i9o.i25 = 1.56. Since ("1 ~ to ~ ("2' we accept the hypothesis iJ-x = iJ-y that under both conditions the mean output is the same. Case A applies to the example becau~e the two first sample values conespond to a certain type of work. the next two were obtained in another kind of work. etc. So we may use the differences 16 16 6 o o 13 8 of corresponding sample values and the method in Example 3 to test the hypothesis,.,. = O. where,.,. is the memt of the population cOITcsponding to the differences. As a logical alternativc we take,.,. O. The sample mean is d = 7.625. and the sample variance is -,2 = --15.696. Hence *' t = v'8 (7.625 - O)/V --15.696 = 3.19. From P( T ~ ("1) = 2.5'k. P( T ~ ("2) = 97.5'7c and Table A9 in App. 5 with 11 - I = 7 degrees of freedom we obtain ("1 = - 2.36. ("2 = 2.36 and reject the hypothesis because t = 3.19 does not lie between ("1 and ("2' Hence our pre~ent test. in which we used more information (but the ,ame samples). shows that the difference in output is ~ignificant. • E X AMP L E 6 Comparison of the Variance of Two Normal Distributions Using the fWO ~amples in the last example. test the hypothesis u.t .2 = U y 2: a"ume thm the corresponding populations are normal and the nature of the experiment suggests the altemative u x 2 > u y 2 . Sollttioll. We find -'x2 = 106.125, 8--1.000. We choose the signiticance level a = 5'k. Using I. n2 - 1) = (7. 7) degrees of freedom. we = s:r2/Sy2 = 1.26. Since Vo ~ (". we accept the hypothesis. If Sy2 = p( V ~ (") = I - a = 95'if. and Table All in App. 5. with (n1 - determine (" = 3.79. We finally compute Vo > (". we would reject it. This test is justified by the fact that Vo i~ an ob,erved value of a random variable that ha, a ,o-called F-distribution with (n1 - 1.112 - I) degrees of freedom. provided the hypothesis is true. (Proof in Ref. IG3] li,ted in App. I.J The F-distribution with (111. n) degree, of freedom was introduced by R. A. Fisher4 and has the distribution function F(:::) = 0 if : < 0 and Vo (13) F(:::) = Kmn f z ,<'n-2)/2(l1It + n)-<11t+n)/2 dt (::: ~ 0). o • 4 After the pioncering work of the English statistician and biOlogist. KARL PEARSON (1857-1936). the founder of the English school of statistics. and WILLIAM SEALY GOSSET (]876-1937). who discovered the t-di,tribution (mtd published under the name ·Student"). the English statistician Sir RONALD AYLMER FISHER (1890-1962). professor of eugenic~ in London (1933-1943) and professor of genetics in Cambrid"e Enuland (19~3:-1957) and Adelaide, Australia (1957-1962), had great influence on the ~fllfther developmen~;f m~dern statIstics. SEC 25.4 Testing of Hypotheses. 1067 Decisions This long section contained the basic ideas and concepts of testing, along with typical applications and you may perhaps want to review it quickly before going on, because the next sections concern an adaption of these ideas to tasks of great practical importance and resulting tests in connection with quality control, acceptance (or rejection) of goods produced. and so on. ... -.. _.. 1. Test JL = a against JL > 0, assuming normality and using the sample 1. -\. I. 3. -8. 6. a (deviations of the azimuth [multiples of 0.01 radian] in some revolution of a satellite). Choose a = 5'k. 2. In one of his classical experiments Buffon obtained 2048 heads in tossing a coin 4040 times. Was the coin fair'? 3. Do the same test as in Prob. 2. using a result by K. Pearson. who obtained 6 019 heads in 12000 trials. 4. Assuming normality and known variance u 2 = 4. test the hypothesis fL = 30.0 again~t the alternative (a) JL = 28.5. (b) JL = 30.7. using a sample of size 10 with mean x = 28.5 and choosing a = 5%. 5. How does the result in Prob. 4(a) change if we use a smaller sample. say. of size 4. the other data (.ct = 28.5, a = 5%. etc.) remaining as before? 6. Detemine the power of the test in Prob. 4(a). 7. What is the rejection region in Prob. 4 in the case of a two-sided test with a = 5'k? 8. Using the sample 0.80. 0.81, 0.81. 0.82, 0.81. 0.82, 0.80,0.82. 0.81. 0.81 (length of nails in inches), test the hypothesis JL = 0.80 in. (the length indicated on the box) against the alternative JL 0.80 in. (A"sume normality. choose a = 5%.) *' 9. A firm sells oil in cans containing 1000 g oil per can and is interested to know whether the mean weight differs significantly from 1000 g at the 5% level. in which case the filling machine has to be adjusted. Set up a hypothesis and an alternative and perform the test. assuming normality and using a sample of 20 fillings with mean 996 g and standard deviation 5 g. 10. If a sample of 50 tires of a certain kind has a mean life of 32 000 mi and a standard deviation of 4000 mi. can the manufacturer claim that the true mean life of such tires is greater than 30000 mi? Set up and test a corresponding hypothesis at a 5% level, assuming normality. 11. If simultaneous measurements of electric voltage by two different types of voltmeter yield the differences (in volts) 0.8. 0.2, -0.3.0.1. 0.0. 0.5, 0.7. 0.2, can we assert at the 5% level that there is no Significant difference in the calibration of the two types of instruments'! (Assume normality.) 12. If a standard medication cures about 70% of patients with a certain disease and a new medication cured 148 of the first 200 patients on whom it was hied, can we conclude that the new medication is better? (Choose a = 5%.) 13. Suppose that in the past the standard deviation of weights of certain 25.0-oz packages tilled by a machine was 0.4 oz. Test the hypothesiS Ho: u = 0.4 against the alternative HI: U> 0.4 (an undesirable increase). using a sample of 10 packages with ~tandard deviation 0.507 and assuming normality. (Choose a = 5%.) 14. Suppose that in operating battery-powered electrical equipment, it is less expensive to replace all batteries at fixed intervals than to replace each battery individually when it breaks down, provided the standard deviation of the lifetime is less than a certain limit. say. less than 5 hours. Set up and apply a suitable test. using a sample of 28 values of lifetime~ with standard deviation s = 3.5 hours and assuming normality: choose a = 5%. 15. Brand A gasoline was used in 9 automobiles of the same model under identical conditions. The corresponding sample of 9 values (miles per gallon) had mean 20.2 and standard deviation 0.5. Under the same conditions. high-power brand B gasoline gave a sample of 10 values with mean 21.8 and standard deviation 0.6. Is the mileage of B significantly better than that of A'? (Test at the 5% level; assume normality.) 16. The two samples 70. 80. 30. 70. 60. 80 and 140. 120, 130. 120. 120. 130. 120 are values of the differences of temperatures (OC) of iron at two stages of casting. taken from two different crucibles. Is the variance of the first population larger than that of the second? (Assume normality. Choose a = 5%.) 17. Using samples of sizes 10 and 16 with variances 2 2 Sx = 50 and Sy = 30 and assuming normality of the corresponding popUlations, test the hypothesis Ho: a./ = u y2 against the alternative u/ > u y2 . Choose a = 5%. 18. Assuming normality and equal varIance and usmg independent samples with 1/1 = 9, .r = 12 . .\'x = 2, 1/2 = 9, Y = 15, Sy = 2. test Ho: JLx = JL y against JLx fLy; choose a = 5%. *' 1068 CHAP. 25 Mathematical Statistics 19. Show that for a nonnal distribution the two types of en'ors in a test of a hypothesis Ho: IL = JLo against an alternative HI: IL = ILl can be made as small as one pleases (not zero) by taking the sample sufficiently large. 20. CAS EXPERIMENT. Tests of Means and Variances. (a) Obtain 100 samples of size 10 each from 25.5 the normal distribution with mean 100 and variance 25. For each sample test the hypothesis ILo = 100 against the alternative ILl > 100 at the level of a = 10% Record the number of rejections of the hypothesis. Do the whole experiment once more and compare. (b) Set up a similar experiment for the variance of a normal distribution and perform it 100 times. Quality Control The ideas on testing can be adapted and extended in various ways to serve basic practical needs in engineering and other fields. We show this in the remaining sections for some of the most important tash; solvahle by statistical methods. As a first such area of problems, we discuss industrial quality control, a highly successful method used in various industries. No production process is so perfect that all the products are completely alike. There is always a small variation that is caused by a great number of small. uncontrollable factors and must therefore be regarded as a chance variation. It is important to make sure that the products have required values (for example, length. strength, or whatever property may be essential in a particular case). For this purpose one makes a test of the hypothesis that the products have the required property. say. fL = fLo, where fLo is a required value. If this is done after an entire lot has been produced (for example, a lot of 100 000 screws), the test will tell us how good or how bad the products are, but it it obviously too late to alter undesirable results. It is much better to test during the production run. This is done at regular intervals of time (for example, every hour or half-hour) and is called quality control. Each time a sample of the same size is taken, in practice 3 to 10 times. If the hypothesis is rejected. we stop the production and look for the cause of the trouble. If we stop the production process even though it is progressing properly, we make a Type I error. If we do not stop the process even though something is not in order. we make a Type II error (see Sec. 25.4). The result of each test is marked in graphical form on what is called a control chart. This was proposed by W. A. Shew hart in 1924 and makes quality cuntrol particularly effective. Control Chart for the Mean An illustration and example of a control chart is given in the upper part of Fig. 536. This control chart for the mean shows the lower control limit LCL, the center control line CL, and the upper control limit UCL. The two control limits correspond to the critical values Cl and C 2 in case (c) of Example 2 in Sec. 25.4. As soon as a silmple mean falls outside the range between the control limits, we reject the hypothesis and assert that the production process is "out of control"; that is, we assert that there has been a shift in process level. Action is called for whenever a point exceeds the limits. If we choose control limits that are too loose, we shall not detect process shifts. On the other hand, if we choose control limits that are too tight, we shall be unable to run the process because of frequent searches for nonexistent trouble. The usual significance level SEC 25.5 Quality Control 1069 is Q' = 1%. From Theorem I in Sec. 25.3 and Table A8 in App. 5 we see that in the case of the normal distribution the corresponding control limits for the mean are (1) 0" LCL = lLo - 2.58 ~! UCL ' = lLo Vll + 0" 2.58 ---;= . \'11 Here 0" is assumed to be known. If 0" is unknown, we may compute the standard deviations of the first 20 or 30 samples and take their arithmetic mean as an approximation of 0". The broken line connecting the means in Fig. 536 is merely to display the results. Additional, more subtle controls are often used in industry. For instance, one observes the motions of the sample means above and below the centerline, which should happen frequently. Accordingly, long runs (conventionally of length 7 or more) of means all above (or all below) the centerline could indicate trouble. 4.20 ~ I I I 0.5% ue' 4.15 I c ~ :::;;: 4.10 / ~ ,Y II 1\ CL f" 'f I I l 99% lel_{ 4.05 \ 0.5% / [ ) 4.00 Sample no. 10 5 0.04 0.0365 I I I .! I 1% II UCL- / 0.03 I c a :;::; '" .;; 1 W "0 "Cl ro "Cl 0.02 0.01 Sample no. 't.\ 1\ I 1\II c '" ill \ I \ \ 99 % \ 1/ o 5 10 Control charts for the mean (upper part of figure) and the standard deviation in the case of the samples on p. 1070 Fig. 536. 1070 CHAP. 25 Mathematical Statistics Table 25.5 Twelve Samples of Five Values Each (Diameter of Small Cylinders, Measured in Millimeters) Sample Number - Sample Values I 2 3 4 5 6 7 8 9 10 II 12 x s R 4.06 4.10 4.06 4.06 4.08 4.08 4.10 4.06 4.08 4.10 4.08 4.12 4.08 4.08 4.12 4.08 4.12 4.10 4.10 4.12 4.10 4.12 4.12 4.12 4.12 4.080 4.112 4.084 4.088 4.108 0.014 0.011 0.026 0.023 0.018 0.04 0.02 0.06 0.06 0.04 -1-.08 4.06 4.08 4.06 4.06 4.10 4.08 4.08 4.08 4.08 4.10 4.08 4.10 4.10 4.10 4.10 4.10 4.10 4.12 4.12 4.12 4.12 4.12 4.14 4.16 4.100 4.088 4.096 4.100 4.104 0.014 0.023 0.017 0.032 0.038 0.04 0.06 0.04 0.08 0.10 4.12 4.14 4.14 4.14 4.14 4.16 4.14 4.16 4.16 4.16 4.140 4.152 0.014 0.011 0.04 0.02 Control Chart for the Variance In addition to the mean, one often controls the variance. the standard deviation, or the range. To set up a control chart for the variance in the case of a normal distribution, we may employ the method in Example 4 of Sec. 25.4 for detennining control limits. It is customary to use only one control limit. namely. an upper control limit. Now from Example 4 of Sec. 25.4 we have S2 = uo2 Y/(n - I), where because of our normality assumption the random variable Y hao.; a chi-square distribution with 11 - 1 degrees of freedom. Hence the desired control limit is (2) UCL = n-I where (' is obtained from the equation P(Y> c) = a, that is, P(Y ~ c) = I - a and the table of the chi-square distribution (Table AIO in App. 5) with 11 - 1 degrees of freedom (or from your CAS); here a (51k or I st. say) is the probability that in a properly running process an observed value S2 of S2 is greater than the upper control limit. If we wanted a control chart for the variance with both an upper control limit UCL and a lower control limit LCL. these Iimi[s would be u 2c1 LCL= _ _ (3) where (4) 1/-1 Cl and C2 and UCL= are obtained from Table A I 0 with and 11 - P(Y I d.f. and the equations ~ C2) = a I - -. 2 SEC. 25.5 Quality Control 1071 Control Chart for the Standard Deviation To set up a control chart for the standard deviation, we need an upper control limit uVc UCL= (5) Vn"=l obtained from (2). For example, in Table 25.5 we have 11 = 5. Assuming that the corresponding population is normal with standard deviation u 0.02 and choosing a = I %, we obtain from the equation P(Y ~ c) = I - a = 99% and Table A lOin App. 5 with 4 degrees of freedom the critical value c (5) the corresponding value UCL= 0.02vrns V4 = 13.28 and from = 0.0365, which is shown in the lower part of Fig. 536. A control chart for the standard deviation with both an upper and a lower control limit is obtained from (3). Control Chart for the Range Instead of the variance or standard deviation, one often controls the range R (= largest sample value minus smallest sample value). It can be shown that in the case of the normal distribution, the standard deviation u is proportional to the expectation of the random variable R* for which R is an observed value, say, u = AnE(R*), where the factor of proportionality An depends on the sample size 11 and has the values II An = uIE(R*) 11 2 3 4 5 6 7 8 9 10 0.89 0.59 0.49 0.43 0.40 0.37 0.35 0.34 0.32 12 14 16 18 20 30 40 50 0.31 0.29 0.28 0.28 0.27 0.25 0.23 0.22 Since R depends on two sample values only, it gives less information about a sample than s does. Clearly, the larger the sample size 11 is, the more information we lose in using R instead of s. A practical rule is to use s when 11 is larger than 10. I. Suppose a machine for filling cans with lubricating oil is set so that it will generate fillings which form a normal population with mean I gal and standard deviation 0.03 gal. Set up a conrrol chart of the type shown in Fig. 536 for controlling the mean (that is. find LCL and VCL). a~suming that the sample size is 6. 2. (Three-sigma control chart) Show that in Prob. I, the requirement of the significance level a = 0.3% leads to LCL = J.L - 3ulY;; and VCL = J.L + 3ulY;;, and find the corresponding numeric values. 3. What sample size should we choose in Prob. 1 if we want LCL and VCL somewhat closer together. say. VCL - LCL = 0.05. without changing the significance level'! CHAP. 25 1072 Mathematical Statistics 4. How does the meaning of the control limits ( I ) change if we apply a control chart with these limits in the case of a population that is not normal? 5. How should we change the sample size in controlling the mean of a normal population if we want the difference UCL - LCL to decrease to half its original value? 6. What LCL and UCL should we use instead of (1) if instead of x we use the sum Xl + ... + xn of the sample values? Detennine these limits in the case of Fig. 536. 7. Ten samples of size 2 were taken from a production lot of bolts. The values (length in mm) are as shown. Assuming that the population is normal with mean 27.5 and variance 0.024 and using (I ), set up a control chart for the mean and graph the sample means on the chart. Sample Length 4 3 2 No. 5 7 6 8 9 10 27.4 27.4 '17.5 27.3 27.9 27.6 27.6 27.8 27.5 27.3 14. How would progressive tool wear in an automatic lathe operation be indicated by a control chart of the mean? Answer the same question for a sudden change in the position of the tool In that operation. 15. (Number of defects per unit) A so-called c-chart or defects-per-unit chart is used for the control of the number X of defects per unit (for instance, the number of defects per 10 meters of paper. the number of missing rivets in an airplane wing, etc.) (a) Set up formulas for CL and LCL, UCL corresponding to J.L ::':: 3u. assuming that X has a Poisson distribution. (b) Compute CL, LCL, and UCL in a control process ofthe number of imperfections in sheet glass; assume that this number is 2.5 per sheet on the average when the process is under control. 16. (Attribute control charts). Twenty samples of size 100 were taken from a production of containers. The numbers of defectives (leaking containers) in those samples (in the order observed) were 27.6 27.4 '17.7 27.4 27.5 27.5 27.4 27.3 27.4 27.7 8. Graph the means of the following 10 samples (thickness of washers. coded values) on a control chart for means, assuming that the population is normal with mean 5 and standard deviation 1.55. Time 8:008:309:00 9:30 10:00 10:30 11:00 11:3012:00 12:30 13 Sample 4 Values 18 .4 3 6 6 5 2 5 8 6 7 5 4 4 7 3 6 5 4 4 3 6 5 6 4 6 6 4 6 4 5 5 6 4 5 2 5 3 9. Graph the ranges of the samples in Prob. 8 on a control chart for ranges. 10. What effect on UCL - LCL does it have if we double the sample size? Ifwe switch from £l' = 1% to £l' = 5%? 11. Since the presence of a point outside control limits for the mean indicates trouble ("the process is out of contro]"), how often would we be making the mistake oflooking for nonexistent trouble if we used (a) I-sigma limits. (b) 2-sigma limits? (Assume normality.) 12. Graph An = uIE(R*) as a function of 11. Why is monotone decreasing function of 11? An a 13. (Number of defectives) Find formulas for the UCL, CL, and LCL (corresponding to 3u-limits) in the case of a control charI for the number of defectives, assuming that in a state of statistical control the fraction of defectives is p. 376 45497056134 902 12 8. From previous experience it was known that the average fraction defective is p = 5% provided that the process of production is mnning properly. Using the binomial distribution. ~et up afmction defectil'e chart (also called a p-chart). that is. choose the LCL = 0 and determine the UCL for the fraction defective (in percent) by the use of 3-sigma limits. where u 2 is the variance of the random variable X = Fractioll defectil'e ill a sample of size 100. Is the process under control? 17. CAS PROJECT. Control Charts. (a) Obtain 100 samples of 4 values each from the normal distribution with mean 8.0 and variance 0.16 and their means. variances, and ranges. (b) Use these samples for making up a control chart for the mean. (c) Use them on a control chart for the standard deviation. (d) Make up a control chart for the range. (e) Describe quantitative properties of the samples that you can see from those charts (e.g., whether the corresponding process is under control, whether the quantities observed vary randomly, etc.). SEC. 25.6 1073 Acceptance Sampling 25.6 Acceptance Sampling Acceptance sampling is usually done when products leave the factory (or in some cases even within the factory). The standard situation in acceptance sampling is that a producer supplies to a consumer (a buyer or wholesaler) a lot of N items (a carton of screws. for instance). The decision to accept or reject the lot is made by determining the number x of defectives (= defective items) in a sample of size 11 from the lot. The lot is accepted if x ~ c, where c is called the acceptance number, giving the allowable number of defectives. If x > c, the consumer rejects the lot. Clearly, producer and consumer must agree on a ce11ain sampling plan giving 11 and c. From the hypergeometric distribution we see that the event A: "Accept the lot" has probability (see Sec. 24.7) c (1) P(A) = P(X ~ c) = 2: x=o where M is the number of defectives in a lot of N items. In terms of the fraction defective e = MIN we can write (1) as (2) P(A; e) can assume n + 1 values conesponding to e = 0, liN, 21N, ... , NIN; here, n and c are fixed. A monotone smooth curve through these points is called the operating characteristic curve (OC curve) of the sampling plan considered. E X AMP L E 1 Sampling plan Suppose that certain tool bits are packaged 20 to a box. and the following sampling plan is u~ed. A sample of two tool bits is drawn, and the corresponding box is accepted if and only if both bits in the sample are good. In Ihis case, N = 20, 11 = 2, C = O. and (2) t<lies the form la factor 2 drops out) (20 - 208)(19 - 200) 380 The values of peA. 0) for 8 = O. 1120.2120. ...• 20/20 and the re,ulting OC curve are shown in Fig. 537 on p. 1074. (Verify!) • In most practical cases e will be small (less than 10%). Then if we take small samples compared to N, we can approximate (2) by the Poisson distribution (Sec. 24.7); thus (3) (J.-t = ne). 1074 CHAP. 25 p(A;e) Mathematical Statistics 0.5 p(Ae) 0.5 0.2 Ii Fig. 537. E X AMP L E 2 Ii OC curve of the sampling plan with n = 2 and c = 0 for lots of size N = 20 Fig. 538. OC curve in Example 2 Sampling Plan. Poisson Distribution Suppose that for large lots the following sampling plan is used. A sample of size II = 20 is taken. If it contains not more than one defective. the lot is accepted. If the sample contains two or more defectives. the lot is rejected. In this plan, we obtain from (3) e- 2o o(1 + 20/J). peA: /J) - • The corresponding OC curve is shown in Fig. 538. Errors in Acceptance Sampling We -;how how acceptance sampling fits into general test theory (Sec. 25.4) and what this means from a practical point of view. The producer wants the probability a of rejecting an acceptable lot (a lot for which 6 does not exceed a certain number 60 on which the two pm1ies agree) to be small. 60 is called the acceptable quality level (AQL). Similarly, P(A:Ii) 95% S]~ \ Producer's risk a = 5° 50% \ 15% :;ollsumer's risk {3= 15°1-. -1"--------I I o 60 iiI = 1% = 5% Good : Indifference ' Poor material, zone : material Fig. 539. OC curve, producer's and consumer's risks SEC. 25.6 1075 Acceptance Sampling the consumer (the buyer) wants the probability f3 of accepting an unacceptable lot (a lot for which e is greater than or equal to some e1 ) to be small. e1 is called the lot tolerance percent defective (LTPD) or the rejectable quality level (RQL). a is called producer's risk. It corresponds to a Type [ error in Sec. 25.4. f3 is called consumer's risk and corresponds to a Type II error. Figure 539 shows an example. We see that the points (eo, I - a) and (e1 • f3) lie on the OC curve. It can be shown that for large lots we can choose eo, el (> eo), a, f3 and then determine 11 and c such that the OC curve runs very close to those prescribed points. Table 25.6 shows the analogy between acceptance sampling and hypothesis testing in Sec. 25.4. Table 25.6 Acceptance Sampling and Hypothesis Testing Acceptance Sampling Hypothesis Testing ---- . - - - - ----+-------=-------------1 Acceptable quality level (AQL) e = 80 Hypothesis tI = tlo Lot tolerance percent defectives (LTPD) Alternative 8 = til 8 = 81 Allowable number of defectives c Critical value c Producer·s risk a of rejecting a lot Probability a of making a Type Terror (significance level) with 8 ~ 80 Consumer's risk {3 of accepting a lot Probability {3 of making a Type II error with 8 ~ 81 Rectification Rectification of a rejected lot means that the lot is inspected item by item and all defectives are removed and replaced by nondefective items. (This may be too expensive if the lot is cheap; in this case the lot may be sold at a cut-rate price or scrapped.) If a production turns out 100e% defectives, then in K lots of size N each, KN8 of the KN items are defectives. Now KP(A; 8) of these lots are accepted. These contain KPNe defectives, whereas the rejected and rectified lots contain no defectives, because of the rectification. Hence after the rectification the fraction defective in all K lots equals KPNeIKN. This is called the average outgoing quality (AOQ); thus (4) AOQ(e) = ep(A; e). \ OC curve \ 0.5 \ AOQL.-~ 0/" o Fig. 540. ..f e* .. 0.5 ~ e OC curve and AOQ curve for the sampling plan in Fig. 537 CHAP. 25 1076 Mathematical Statistics Figure 540 on p. 1075 shows an example. Since AOQ(O) = 0 and peA; 1) = 0, the AOQ curve has a maximum at some = 8*, giving the average outgoing quality limit (AOQL). This is the worst average quality that may be expected to be accepted under rectification. e --_....---........ -........... .. -~ ...... .-.. 1. Lots of knives are inspected by a sampling plan that uses a sample of size 20 and the acceptance number c = I. What are probabilitIes of accepting a lot with 1%, 2%, 10% defectives (dull blades)? Use Table A6 in App. 5. Graph the OC curve. 2. What happens in Prob. I if the sample size is increased to 50? First guess. Then calculate. Graph the OC curve and compare. 3. How will the probabilities in Prob. I with 11 = 20 change (up or down) if we decrease c to zero? First guess. compare with Example I. 11. Samples of 5 screws are drawn from a lot with fraction defective O. The lot is accepted if the sample contains (a) no defective screws, (b) at most 1 defective screw. Using the binomial distribution, find, graph, and compare the OC curves. 12. Find the risks in the single sampling plan with 11 = 5 and c = 0, assuming that the AQL is 00 = I % and the RQL is 01 = 15%. 4. What are the producer's and consumer's risks in Prob. I if the AQL is 1.5% and the RQL is 7.5'70? 13. Why is it impossible for an OC curve to have a vertical portion separating good from poor quality? 5. Large lots of batterie~ are inspected according to the following plan. 11 = 30 batteries are randomly drawn from a lot and tested. If this sample contains at most c = I defective battery, the lot is accepted. Otherwise it is rejected. Graph the OC curve of the plan, using the Poisson approximation. 6. Graph the AOQ curve in Prob. 5. Determine the AOQL, assuming that rectification is applied. 14. If in a single sampling plan for large lots of spark plugs, the sample size is 100 and we want the AQL to be 5% and the producer's risk 2%, what acceptance number c should we choose? (Use the normal approximation.) 7. Do the work required in Prob. 5 if 11 = 50 and c = O. 8. Find the binomial approximation of the hypergeometric distribution in Example 1 and compare the approximate and the accurate values. 15. What is the consumer's risk in Prob. 14 if we want the RQL to be 12%? 16. Graph and compare sampling plans with c = I and increasing values of II, say, 11 = 2. 3, 4. (Use the binomial distribution.) 9. In Example 1, what are the producer's and consumer's risks if the AQL is 0.1 and the RQL is 0.6? 17. Samples of 3 fuses are drawn from lots and a lot is accepted if in the corresponding sample we find no more than I defective fuse. Criticize this sampling plan. In particular, find the probability of accepting a lot that is 509r defective. (Use the binomial distribution.) 10. Calculate peA; 0) in Example 1 if the sample size is increased from 11 = 2 to 11 = 3, the other data remaining as before. Compute peA; 0.10) and peA; 0.20) and 18. Graph the OC curve and [he AOQ curve for the single sampling plan for large lots with 11 = 5 and c = 0, and find the AOQL. 25.7 Goodness of Fit. To test for goodness of fit means that we wish to test that a certain function F(x) is the distribution function of a distribution from which we have a sample Xl' . • . ,xn . Then we test whether the sample distribution function F(x) defined by F(x) = SUIIl of the relative frequencies of all sample l'lllues xJ not exceedbza x b fits F(x) "sufficiently well." If this is so, we shall accept the hypothesis that F(x) is the distribution function of the population; if not, we shall reject the hypothesis. SEC. 25.7 Goodness of Fit. x 2 -Test 1077 This test is of considerable practical importance, and it differs in character from the tests for parameters (IL. a 2 • etc.) considered so far. To test in that fashion, we have to know how much F(x) can differ from F(x) if the hypothesis is true. Hence we must first introduce a quantity that measures the deviation of F(x) from F(x), and we must know the probability distribution of this quantity under the assumption that the hypothesis is true. Then we proceed as follows. We determine a number c such that if the hypothesis is true, a deviation greater than c has a small preassigned probability. If, nevertheless, a deviation greater than c occurs, we have reason to doubt that the hypothesis is true and we reject it. On the other hand. if the deviation does not exceed c, so that F(x) approximates F(x) sufficiently well. we accept the hypothesis. Of course. if we accept the hypothesis, this means that we have insufficient evidence to reject it, and this does not exclude the possibility that there are other functions that would not be rejected in the test. In this respect the situation is quite similar to that in Sec. 25.4. Table 25.7 shows a test of that type, which was introduced by R. A. Fisher. This test is justified by the fact that if the hypothesis is true, then Xo 2 is an observed value of a random variable whose distribution function approaches that of the chi-square distribution with K - I degrees of freedom (or K - r - 1 degrees of freedom if r parameters are estimated) as 11 approaches infinity. The requirement that at least five sample values lie in each interval in Table 25.7 results from the fact that for finite 11 that random variable has only approximately a chi-square distribution. A proof can be found in Ref. [G3] listed in App. 1. If the sample is so small that the requirement cannot be satisfied. one may continue with the test, but then use the result with caution. Table 25.7 Chi-square Test for the Hypothesis That F(x) is the Distribution Fundion of a Population from Which a Sample XlJ ••• , Xn is Taken Step 1. Subdivide the x-axis into K intecvab 110 12 , . . . , IK such thm each interval contains at least 5 values of the given smuple Xl, . . . , x n . Determine the number bj of smupJe values in the interval I j , where j = I .... , K. If a sample value lies at a common boundary point of two intervals, add 0.5 to each of the two corresponding b;r Step 2. U:-ing F(x), compute the probability Pj that the random variable X under consideration assumes any value in the interval I j , where j = 1, ... , K. Compute ej = IIPj. (This is the number of sample values theoretically expected in I j if the hypothesis is true.) Step 3. Compute the deviation (1) Step 4. Choose a significance level (5%. I %. or the like). Step 5. Determine the solution c of the equation P(X 2 ~ c) = I - Q' from the table of the chi-sqare distribution with K - I degrees of fi'eedom (Table AIO in App. 5). If rparameters of F(x) are unknown and their maximum likelihood estimates (Sec. 25.2) are used. then use K - r - I degrees of freedom (instead 2 of K - I). If Xo ~ c, accept the hypothesis. If Xo 2 > c, reject the hypothesis. 1078 CHAP. 25 Mathematical Statistics Table 25.8 Sample of 100 Values of the Splitting Tensile Strength (Ib/in?) of Concrete Cylinders 320 350 370 320 400 420 390 360 370 340 380 340 390 350 360 400 330 390 400 360 340 350 390 360 350 350 360 350 360 390 410 360 440 340 390 370 380 370 350 400 340 350 390 350 350 320 330 350 380 410 380 370 330 340 400 330 350 370 380 370 350 370 360 390 340 380 300 370 340 400 360 380 330 350 360 390 360 390 360 360 320 300 400 380 370 400 360 370 330 340 370 420 370 340 420 370 360 340 370 360 D. L. IVEY. Splitting tensile tests on structural lightweight aggregate concrete. Texas Transp0l1ation Institute. College Station. Texas. EXAMPLE 1 Test of Normality Test whether the population from which the sample in Table 25.8 wa~ taken is normal. SolutiOIl. Table 25.8 show~ the values (column by column) in the order obtained in the experiment. Table 25.9 gives the frequency distribution and Fig. 541 the histogram. It is hard to guess the outcome of the test-does the histogram resemble a normal density curve sufficiently well or not? The maximum likelihood estimates for IL and cr2 are jL = X = 364.7 and ;;2 = 712.9. The computation in Table 25.10 yields Xo2 = 2.942. It is very intereMing that the interval 375 ... 385 contributes over 501ft of X02. From the histogram we see that the corresponding frequency looks much too small. The second largest contribution comes from 395 ... 405. and the histogram shows that the frequency seems somewhat too large. which is perhaps not obvious from inspection. Table 25.9 Tensile Strength x [lb/in. 2 ] 300 310 320 330 340 Frequency Table of the Sample in Table 25.8 2 Absolute Frequency 3 4 5 Relative Frequency Cumulative Absolute Frequency Cumulative Relative Freguency FlX) lex) 2 o 4 6 II 350 360 370 380 390 14 16 15 8 400 410 420 430 440 8 2 3 10 o 1 0.02 0.00 0.04 0.06 0.11 2 6 12 23 0.02 0.02 0.06 0.12 0.23 0.14 0.16 0.15 0.08 0.10 37 53 68 76 86 0.37 0.53 0.68 0.76 0.86 0.08 0.02 0.03 0.00 0.01 94 96 99 99 100 0.94 0.96 0.99 0.99 2 LOO SEC. 25.7 1079 Goodness of Fit. x2-Test 0.20,--------,---,----...,-----, 0.15 ((x) 0.10 0.05 o~ 250 __~~~~~--~~~~ 350 2 [lb.lin. ] Fig. 541. Frequency histogram of the sample in Table 25.8 We choose a = 5%. Since K = 10 and we e5timated r = 2 parameters we have to usc Table AlO in App. 5 with K - ,. - I = 7 degrees of freedom. We find c = 14.07 as the solution of P(X 2 ~ c) = 95%. Since 2 Xo < c, we accept the hypothesis that the population is normal. • Table 25.10 Computations in Example 1 Xj - Xj 26.7 ej hj Ternl in (1) 0.0000 ... 0.0681 6.81 6 0.0% 0.0681 ... 0.1335 6.54 6 0.045 -1.11 ... -0.74 0.1335 ... 0.2296 0.2296 ... 0.3594 0.3594 ... 0.4960 9.61 12.98 13.66 11 14 16 0.201 0.080 0.401 0.4960 ... 0.6517 0.6517 ... 0.7764 15.57 12.47 15 8 0.021 1.602 0.7764 ... 0.8708 9.44 10 0.033 0.8708 ... 0.9345 0.9345 . . . 1.0000 6.37 8 0.417 6.55 6 0.046 335 ... 345 345 ... 355 355 ... 365 385 ... 395 364.7) 26.7 -1.49 ... - 1.11 -x 395· .. 405 405 ... co Xj - ... -1.49 -:x;···325 325 ... 335 365 ... 375 375 ... 385 ¢( 364.7 -0.74' .. -0.36 -0.36· . 0.01 0.01 ... 0.39 0.39' . 0.76 0.76" . 1.13 1.13 ... 1.51 1.51 .. co X02 = 1. If 100 flips of a coin result in 30 heads and 70 tails. can we assel1 on the 5% level that the coin is fair? 2. If in 10 flips of a coin we get the same ratio as in Prob. I (3 heads and 7 tails), is the conclusion the same as in Prob. I? First conjecture. then compute. 3. What would be the smallest number of heads in Prob. I under which the hypothesis "Fair coin" is still accepted (with ex = 5%)? 4. If in rolling a die 180 times we get 39. 22. 41. 26. 20, 32. can we claim on the 5% level that the die is fair? 2.942 5. Solve Prob. 4 if the sample is 25, 31. 33, 27, 29. 35. 6. A manufacturer claims that in a process of producing kitchen knives, only 2.5% of the knives are dull. Test the claim against the alternative that more than 2.5% of the knives are dull, using a sample of 400 knives containing 17 dull ones. (Use ex = 5%.) 7. Between 1 P.M. and 2 P.M. on five consecutive days (Monday through Friday) a certain service station has 92,60. 66. 62. and 90 customers, respectively. Test the hypothesis that the expected number of customers during that hour is the same on lhose days. (Use ex = 591:.) CHAP. 25 1080 Mathematical Statistics 8. Test for normality at the I % level using a sample of /I = 79 (rounded) values x (tensile strength [kg/mm2] of steel sheets of 0.3 nun thickness). a = a(x) = absolute frequency. (Take the first two values together, also the last three, to get K = 5.) 58 59 60 61 62 63 10 17 27 8 9 3 64 17. 9. In a sample of 100 patients having a certain disease 45 are men and 55 women. Does this support the claim that the disease is equally common among men and women? Choose a = 5%. 10. In Prob. 9 find the smallest number (>50) of women that leads to the rejection ofthe hypothesis on the levels 5%, 1%, 0.5%. 11. Verify the calculations in Example 1 of the text. 12. Does the random variable X = Number of accide/lTs per week in 1I certain foundry have a Poisson distribution if within 50 weeks. 33 were accident-free. I accident occurred in II of the 50 weeks. 2 in 6 of the weeks and more than 2 accidents in no week? (Choose a = 5%.) 13. Using the given sample, test that the corresponding population has a Poisson distribution. x is the number of alpha particles per 7.5-sec intervals observed by E. Rutherford and H. Geiger in one of their classical experiments in 1910, and a(x) is the absolute frequency (= number of time periods during which exactly x particles were observed). (Use a = 5%.) x 0 a I 57 t a I 16. 2 3 4 5 6 203 383 525 532 408 273 7 8 9 10 II 12 ~B 139 45 27 10 4 2 0 18. 19. load of 5000 lb, can we claim that a new process yields the same breakage rate if we find that in a sample of 80 rods produced by the new process, 27 rods broke when subjected to that load? (Use a = 5%.) Three samples of 200 livets each were taken from a large production of each of three machines. The numbers of defective rivets in the samples were 7, 8, and 12. Is this difference significant? (Use a = 5%.) In a table of properly rounded function values, even and odd last decimals should appear about equally often. Test this for the 90 values of lI(:':) in Table Al in App. 5. Are the 5 tellers in a ceI1ain bank equally time-efficient if during the same time interval on a certain day they serve 120.95, 110, 108, 102 customers? (Use a = 5%.) CAS EXPERIMENT. Random Number Generator. Check your generator expelimentally by imitating results of 11 trials of rolling a fair die, with a convenient 11 (e.g .. 60 or 300 or the like). Do this many times and see whether you can notice any "nonrandomness" features, for example. too few Sixes, too many even numbers. etc .. or whether your generator ~eems to work properly. Design and perform other kinds of checks. 20. TEAM PROJECT. Difficulty with Random Selection. 77 students were asked to choose 3 of the imegers II, 12, 13, ... ,30 completely arbitrarily. The amazing result was as follows. 14. Can we assert that the traffic on the three lanes of an expressway (in one direction) is about the same on each lane if a count gives 910. 850. 720 cars on the right. middle. and left lanes, respectively. during a particular time interval? (Use a = 5%.) 15. If it i5 known that 25% of certain steel rod~ produced by a standard process will break when subjected to a '\lumber Il Frequ. 11 Number 21 rrequ. 12 12 13 14 15 16 17 18 19 20 10 20 8 13 9 21 9 16 8 22 23 24 25 26 27 28 8 15 10 10 9 12 8 29 30 13 9 If the selection were completely random, the following hypotheses should be true. (a) The 20 numbers are equally likely. (b) The 10 even numbers together are as likely as the 10 odd numbers together. (c) The 6 prime numbers together have probability 0.3 and the 14 other numbers together have probability 0.7. Te~t these hypotheses. using a = 5%. Design further experiments that illustrate the difficulties of random selectiOn. 25.8 Nonparametric Tests Nonparametric tests, also called distribution-free tests, are valid for any distribution. Hence they are used in cases when the kind of distribution is unknown, or is known but such that no tests specifically designed for it are available. In this section we shall explain the basic idea of these tests, which are based on "order statistics" and are rather simple. SEC. 25.8 1081 Nonparametric Tests If there is a choice, then tests designed for a specific distribution generally give better results than do nonparametric tests. For instance. this applies to the tests in Sec. 25.4 for the normal distribution. We shall discuss two tests in terms of typical examples. In deriving the distributions used in the test, it is essential that the distributions from which we sample are continuous. (Nonparametric tests can also be derived for discrete distributions, but this is slightly more complicated. ) E X AMP L E 1 Sign Test for the Median A median of the population is a solution x ~ Ii of the equation F(x) ~ 0.5. where F is the distribution function of the population. Suppose that eight radio operators were tested, first in rooms without air-conditioning and then m air-conditioned rooms over the same period of time. and the difference of errors (unconditioned minus conditioned) were 9 o 4 o 4 11. 7 Test the hypothesis Ii ~ 0 (that is, air-conditioning has no effect) against the alternative ji > 0 (that is. inferior performance in unconditioned rooms). Solution. We choose the significance level a ~ 5%. If the hypothesis is true. the probability p of a positive difference is the same as that of a negative difference. Hence in this case. l' ~ 0.5. and the random variable x = NLlmber of positil'e \'a/LIes omollg 11 I'a/ues has a binomial distribution with p = 0.5. Our sample has eight values. We omit the values O. which do not contribute to the decision. Then six values are left. all of which are positive. Since P(X ~ Ii) = (~) (0.5)6(0.5)0 = 0.0156 = 1.56% we do have observed an event whose probability is very small if the hypothesis is true: in fact 1.56% < a = 5%. Hence we assert that the alternative Ii > 0 is true. Thal is. the number of errors made in unconditioned rooms is significantly higher, so that installation of air conditioning should be considered. • E X AMP L E 2 Test for Arbitrary Trend A certain machine is used for cutting lengths of wire. Five successive piece, had the lengths 29 31 28 30 32. Using this sample. test the hypothesis that there is no trend, that is. the machine does not have the tendency to produce longer and longer pieces or shorter and sh0l1er pieces. Assume that the type of machine suggests the alternative that there is positil'e trend, that is. there is the tendency of successive pieces to get longer. Solution. We count the number of transpositions in the sample. that is. the number of times a larger value precedes a smaller value: 29 precedes 28 (1 transposition), 31 precedes 28 and 30 (2 transpositions). The remaining three sample values follow in ascendmg order. Hence in the sample there are I transpositions. We now consider the random valiable .L 2 3 T = Number of transpositiolls. If the hypothesis is true (no trend). then each of the 5! = 120 permutations of five elements I 2 3 4 5 has the same probability (11120). We alTange these permutations according to their number of transpositions: 1082 CHAP. 25 Mathematical Statistics 2 3 T=2 T=I T=O 4 5 2 2 3 2 3 4 2 3 5 4 3 5 4 5 4 5 2 5 3 5 4 2 2 3 5 3 2 4 2 3 3 4 5 2 1 3 2 2 I 4 3 1 2 3 T=3 4 4 2 3 3 4 4 5 3 4 4 5 5 4 5 5 5 2 2 2 3 2 3 2 3 3 4 3 2 4 5 4 3 5 2 5 2 4 2 5 3 3 2 5 2 3 4 4 5 3 5 3 4 etc. I 5 4 4 5 I 3 5 2 5 4 4 2 5 4 5 2 3 5 4 From this we obtain P(T:o:; 3) = l~O + lio + l~O + )1;0 = )2io = 24%. We accept the hypothesis because we have ob~erved an event that ha, a relatively large probability (certainly much more than 5'k) if the hypothesis i~ true. Values of the distribution function of T in the case of no trend are shoINn in Table A12. App. 5. For insrance. if /I = 3. then FlO) = 0.167. F(I) = 0.500, F(2) = I - 0.167. If /I = 4. then F(O) = 0.042. F(l) = 0.167, F(2) = 0.375. F(3) = I - 0.375, F(4) = I - 0.167, and SO on. Our method and those values refer to contilluous distributions. Theoretically. we may then expect that all the \'alues of a sample arc different. Practically. some sample values may still be equal. because of rounding: If 111 "alues are equal. add m(m - 1)14 (= mean value of thc transpo,itions in the case of the perIl1Ulalion~ of 111 elements). that is.l for each pair of equal values. ~ for each triple. etc. • ., - 1. What would change in Example 1. had we observed only 5 positive values? Only 4? 2. Does a process of producing plastic pipes of length f.L = 2 meters need adjustment if in a sample. 4 pipes have the exact length and 15 are shorter and 3 longer than 2 meters? (Use the nonnal approximation of the binomial distribution.) 3. Do the computations in Prob. 2 without the use of the DeMoivre-Laplace limit theorem (in Sec. 24.8). 4. Test whether a thermostatic switch is properly set to 200 e against the alternative that its setting is too low. Use a sample of9 values, 8 of which are less than 20 0 e and I is greater than 20°e. 5. Are air filters of type A better than type B filters if in IO trials. A gave cleaner air than B in 7 cases, B gave cleaner air than A in I case. whereas in 2 of the trials the results for A and B were practically the same? 6. In a clinical experiment. each of 10 patients were given two different sedatives A and B. The following table shows the effect (increase of sleeping time. measured in hours). Using the sign test. find out whether the difference is significant. A B 1.9 0.8 1.I 0.1 -0.1 4.4 5.5 1.6 4.6 3.4 0.7 -1.6 -0.2 -1.2 -0.1 3.4 3.7 0.8 0.0 2.0 Difference 1.2 2.4 1.3 1.3 0.0 1.0 1.8 0.8 4.6 1.4 7. Assuming that the populations corresponding to the samples in Prob. 6 are nOlmal. apply a suitable test for the normal distribution. 8. Thirty new employees were grouped into 15 pairs of similar intelligence and expeIience and were then instructed in data processing by an old method (A) applied to one (randomly selected) person of each pair. and by a new presumably better method (B) applied to SEC. 25.9 Regression. 1083 Fitting Straight Lines. Correlation the other person of each pair. Test for equality of methods against the alternative that (B) is better than (A), using the following scores obtained after the end of the training period. A 60 70 80 85 75 40 70 45 95 80 90 60 80 75 65 I B 65 85 85 80 95 65 100 60 90 85 100 75 90 60 80 9. Assuming normality. solve Prob. 8 by a suitable test from Sec. 25.4. 10. Set up a sign test for the lower quartile q25 (defined by the condition F(q25) = 0.25). 11. How would you proceed in the sign test if the hypothesis is fi = fio (any number) instead of fi = O? Temperature T rCJ Reading V [volts] 10 30 20 40 50 99.5 101.1 100.4 100.8 101.6 15. In a swine-feeding experiment. the following gains in weight [kg] of 10 animals (ordered according to increasing amounts of food given per day) were recorded: 20 17 19 18 23 16 25 28 24 22. Test for no trend against positive trend. 16. Apply the test explained in Example 2 to the following data (x = diastolic blood pressure [mm Hgl. y = weight of hemt [in grams] of 10 patients who died of cerebral hemorrhage). 12. Check the table in Example 2 of the text. A'121 120 95 13. Apply the test in Example 2 to the following data (x = disulfide content of a certain type of wool, measured in percent of the content in umeduced fibers; y = saturation water content of the wool. measured in percent). Test for no trend against negative trend. ,) 521 465 352 455 490 388 301 395 375 418 y 50 17. Does an increase in temperature cause an increase of the yield of a chemical reaction from which the following sample was taken"! 15 30 40 50 55 80 100 Temperature tOe] 10 20 30 40 60 80 46 43 42 36 39 37 33 Yield [kg/min] O.b 1.1 0.9 1.6 1.2 2.0 14. Test the hypothesis that for a certain type of voltmeter. readings are independent of temperature T [0C] against the alternative that they tend to increase with T. Use a sample of values obtained by applying a constant voltage: 25.9 123 140 112 92 100 102 91 Regression. Correlation 18. Does the amount of feltilizer increase the yield of wheat X [kg/plot]? Use a sample of values ordered according to increasing amounts of fertilizer: 41.4 43.3 39.6 43.0 44.1 45.6 -l4.5 46.7. Fitting Straight Lines. So far we were concerned with random experiments in which we observed a single quantity (random variable) and got samples whose values were single numbers. In this section we discuss experiments in which we observe or measure two quantities simultaneously, so that we get samples of pairs of values (Xl, .\'1), (X2' )'2), ... , (x"' JII). Most applications involve one of two kinds of experiments, as follows. 1. In regression analysis one of the two variables. call it x. can be regarded as an ordinary variable because we can measure it without substantial en'or or we can even give it values we want. x is called the independent variable. or sometimes the controlled variable because we can control it (set it at values we choose). The other variable, Y. is a random variable, and we are interested in the dependence of Yon x. Typical examples are the dependence of the blood pressure Y on the age x of a person or, as we shall now say, the regression of Yon x. the regression of the gain of weight Y of certain animals on the daily ration of food x. the regression of the heat conductivity Y of cork on the specific weight x of the cork. etc. 1084 CHAP. 25 Mathematical Statistics 2. In correlation analysis both quantities are random variables and we are interested in relations between them. Examples are the relation (one says "correlation") between wear X and wear Y of the front tires of cars, between grades X and Y of students in mathematics and in physics, respectively. between the hardness X of steel plates in the center and the hardness Y near the edges of the plates, etc. Regression Analysis In regression analysis the dependence of Y on x is a dependence of the mean /.L of Yon x, so that /.L = /.L(x) is a function in the ordinary sense. The curve of /.L(x) is called the regression curve of Y on x. In this section we discuss the simplest case, namely, that of a straight regression line (1) Then we may want to graph the sample values as n points in the xY-plane, fit a straight line through them, and use it for estimating /.L(x) at values of x that interest us, so that we know what values of Y we can expect for those x. Fitting that line by eye would not be good because it would be sUbjective; that is, different persons' results would come out differently, particularly if the point<; are scattered. So we need a mathematical method that gives a unique result depending only on the Il points. A widely used procedure is the method of least squares by Gauss and Legendre. For our task we may fOlmulate it as follows. Least Squares Principle The straight line should be fitted through the given points so that the slim of the squares of the distances of those points from the straight line is minimum, where tile distance is measured in the vertical direction (the y-direction). (Formulas below.) To get uniqueness of the straight line, we need some extra condition. To see this, take the sample (0, I), (0, -I). Then all the lines y = k1x with any kl satisfy the principle. (Can you see it?) The following assumption will imply uniqueness, as we shall find out. General Assumption (Al) The x-values Xl, . . . , Xn in OLlr sample (Xl' Yl), . . . , (Xn, Yn) are not all equal. From a given sample (Xl. Yl)' ••.. (X". Yn) we shall now determine a straight line by least ~quares. We write the line as (2) and call it the sample regression line because it will be the counterpart of the population regression line (1). Now a sample point (Xj, )J) has the vertical distance (distance measured in the y-direction) from (2) given by (see Fig. 542). SEC. 25.9 Regression. 1085 Fitting Straight Lines. Correlation y x x. J Fig. 542. Vertical distance of a point (Xj' Yj) from a straight line Y = ko + k,x Hence the sum of the squares of these distances is n (3) q = ~ (Yj - ko - klXj)2. j~I In the method ofleast squares we now have to determine ko and kl such that q is minimum. From calculus we know that a necessary condition for this is aq = 0 (4) ilq and ilko ilkl = o. We shall see that from this condition we obtain for the sample regression line the formula (5) Here i and.v are the means of the x- and the y-values in our sample, that is, (a) j: = - I (Xl + ... + xn) (."1 + . . . + Yn)· 11 (6) I Y= (b) 11 The slope kl in (5) is called the regression coefficient of the sample and is given by SXY (7) S 2 x Here the "sample covariance" 1 (8) and Sxy Sx 2 n = -11 - I ~ (x· J is x)(,·· -J y) - = 11 - J~l is given by 1 (9a) Sxy S 2 x = -f)-I n ~ j~I (Xo J x)2 = f)-I I 1086 CHAP. 25 Mathematical Statistics :n. From (5) we see that the sample regression line passes through the point Ct', by which it is detennined, together with the regression coefficient (7). We may call Sx 2 the variance of the x-values, but we should keep in mind that x is an ordinary variable. not a random variable. We shall soon also need S 2 Y (9b) 71 = -1-I ~ ~ 11 - j=l Derivation of (5) and (7). (y. _ 1')2 -J. = -I1 Il - [n ~ 1'.2 ~ -J j~l -I (71 ~ n j=l )2] 1'. ~-J . Differentiating (3) and using (4), we first obtain iJq ako aq ak = -2 ~ )'i)j - ko - k1xj) =0 l where we sum over j from 1 to 11. We now divide by 2, write each of the two sums as three sums, and take the sums containing )j and XjYj over to the right. Then we get the "normal equations" = ~ , .. ~.J (10) This is a linear system of two equations in the two unknowns ko and k1 . Its coefficient determinant is [see (9)] 11 and is not zero because of Assumption (A I). Hence the system has a unique solmion. Dividing the first equation of (10) by 11 and using (6), we get ko = Y - k1x. Together with y = ko + k1x in (2) this gives (5). To get (7), we solve the system (10) by Cramer's rule (Sec. 7.6) or elimination, finding (II) This gives (7)-(9) and completes the derivation. [The equality of the two expressions in (8) and in (9) may be shown by the student: see Prob. 14]. • E X AMP L E 1 Regression Line The decrease of volume y ['i!-] of leather for certain fixed values of high pres~ure x [atmospheres I was measured. The resulh are shown in the first mo columns of Table 25.1 L Find the regression line of .'. on x. Solutioll. and (X) We see that 11 = 4 and obtain the values.r = 2800014 = 7000. -" = II}.O/4 = 4.75. and from (9) SEC. 25.9 Regression. 1087 Fitting Straight Lines. Correlation Table 25.11 Regression of the Decrease of Volume y [%] of Leather on the Pressure x [Atmospheres] I Given Values Auxiliary Values x} Xj .v· J 4000 6000 8000 10000 2.3 4.1 5.7 6.9 16000000 36000000 64000000 100000000 9200 24600 45600 69000 28000 19.0 216000000 148400 2 Sx . ="3I ~XIJ ( 2 28 000 ) 216000000 - - - 4 - . _ ~ ( 3 148400 - xjYj = 20 000 000 3 28000'19) _ 4 - 15400 3 . Hence k] = 15 ·:100120000000 = 0.00077 from (7). and the regression line is y - 4.75 = 0.000 77(x - 7000) or y = 0.000 77r: - 0.64. Note that y(O) = -0.64. which is physically meaningless. but typically indicates that a linear relation is merely an approximation valid 011 some restricted interval. • Confidence Intervals in Regression Analysis If we want to get confidence intervals, we have to make assumptions about the distribution of Y (which we have not made so far; least squares is a "geometric principle," nowhere involving probabilities!). We assume normality and independence in sampling: Assumption (A2) For each fixed x the random )'Qriable Y is /lonnalwith mean (I), that is, (12) and varial/ce (]"2 independent of x. Assumption (A3) The 11 pel.1lJ/7llaIlCeS of the experi1llellt by which we obtain a sample a re independent. K1 in (12) is called the regression coefficient of the population because it can be shown that under Assumptions (AI)-(A3) the maximum likelihood estimate of KI is the sample regression coefficient kl given by (11). Under Assumptions (A1)-(A3) we may now obtain a confidence interval for Kb as shown in Table 25.12. 1088 CHAP. 25 Table 25.12 Mathematical Statistics Determination of a Confidence Interval for Kl in (1) under Assumptions (Al)-(A3) Step 1. Choose a confidence level ')'(95%,99%, or the like). Step 2. Determine the solution c of the equation (13) F(c) ~(1 = + ')') from the table of the t-distribution with n - 2 degrees of freedom (Table A9 in App. 5; 11 = sample size). Step 3. Using a sample (Xlo Y1), .•• , (xn , Yn), compute (n - l)sx 2 from (9a), (n - l)S.TY from (8), kl from (7), n (14) I )Sy 2 = ~ Yj 2 (n - - n [as in (9b)], and (15) Step 4. Compute K=c (11 - 2)(n - I )s,.2 . The confidence interval is (16) E X AMP L E 2 Confidence Interval for the Regression Coefficient Using the sample in Table 25.1 L determine a confidence interval for Solution. Step 1. We choose l' = /(1 by the method in Table 25.12. 0.95. Step 2. Equation (13) takes the form HC) = 0.975, and Table A9 in App. 5 with 11 gives c = 4.30. - 2 = 2 degrees offreedom Step 3. From Example I we have 3s",2 = 20000000 and k1 = 0.00077. From Table 25.11 we compute 3s y2 = 102.2 = 192 4 11.95, qo = I L')5 - 20 (X)O om . OJ)(J0772 = 0.092. = 4.30v'0.092/(2 . 20 000 (00) Step 4. We thus obtain K = 0.000206 and CONFo.95 (0.00056 ~ K1 ~ 0.000981. • SEC 25.9 Regression. 1089 Fitting Straight Lines. Correlation Correlation Analysis We shall now give an introduction to the basic facts in correlation analysis: for proofs see Ref. [G2J or [G8] in App. I. Correlation analysis is concerned with the relation between X and Y in a two-dimensional random variable (X, Y) (Sec. 24.9). A sample consists of 11 ordered pairs of values (Xl' .\"1)' . . . , (xn , y,,), as before. The interrelation between the \" and y values in the sample is measured by the sample covariance Sxy in (8) or by the sample correlation coefficient (17) with Sx and Sy given in (IJ). Here r has the advantage that it does not change under a multiplication of the X and y values by a factor (in going from feet to inches, etc.). THEOREM 1 Sample Correlation Coefficient The sample correlation coefficient r sati.~fies - I ~ r ~ 1. In particular. r and onl.r if the sample values lie on a straight line. (See Fig. 543.) = :::'::: 1 if The theoretical counterpart of r is the correlation coefficient p of X and Y, p= (18) where JLx = E(X) , JLy = E(Y), ux2 = E([X - JLxf), Uy2 = E([Y - JLy]2) (the means and variances of the marginal distributions of X and Y; see Sec. 24.9), and UXy is the r=l 10 00 •• • •• •• • 10 00 .~. 543. • • • r = 0.98 •• r=O 10 00 10 00 •• • • 10 ••• •• • 20 r = 0.6 • • •• • • • • • 10 20 • • • • • • • 20 10 r = -0.3 10 00 • • • • • • • • • • 10 20 r = -0.9 10 00 • • • • 10 •• • • 20 •• Samples with various values of the correlation coefficient r 1090 CHAP. 25 Mathematical Statistics covariance of X and Y given by (see Sec. 24.9) (19) UXY = E([X - J.Lx][Y - J.Ly]) = E(XY) - E(X)E(Y). The analog of Theorem 1 is THEOREM 2 Correlation Coefficient The correlation coefficient p satisfies - 1 ~ P ~ 1. In particular. p = :::': 1 ollly if X alld Yare linearly related, that is. Y = yX + 8. X = y* Y + 8*. if alld X and Yare called uncorrelated if p = O. THEOREM 3 Independence. Normal Distribution (a) Indepelldent X and Y (see Sec. 24.9) are uncorrelated. (b) If (X, Y) is nOl1llal (see below), then uncorrelated X alld Yare independent. Here the two-dimensional normal distribution can be introduced by taking two independent standardized normal random variables X*. Y*, whose joint distribution thus has the density (20) f*(x*. y*) = _1_ e-<x*2+y*2)/2 27T (representing a surface of revolution over the x*y*-plane with a bell-shaped curve as cross section) and setting X = J.Lx Y = J.Ly + + uxX * + ~ uyY*. pUyX* This gives the general two-dimensional normal distribution with the density f(x, y) = (21a) 1 2 e- h (x,y)/2 27TUXUy~ where (21b) hex. y) = In Theorem 3(b), normality is important, as we can see from the following example. E X AMP L E 3 Uncorrelated but Dependent Random Variables If X assumes -1, 0, I with probability 113 and Y = X2. then EO() = 0 and in (3) __ 3 3 1 CTXY - E(XY) - E(X ) ~ (-1) . - 3 + 1 3 0 . 3 1 = 0 3' +] 3 . - . so that p = 0 and X and Yare uncorrelated. But they are cenainly not independent since they are even functlonally ~~ SEC. 25.9 Regression. Fitting Straight Lines. Correlation 1091 Test for the Correlation Coefficient p Table 25.13 shows a test for p in the case of the two-dimensional normal distribution. t is an observed value of a random variable that has a t-distribution with n - 2 degrees of freedom. This was shown by R. A. Fisher (Biometrika 10 (1915), 507-521). Table 25.13 Test of the Hypothesis p = 0 Against the Alternative p of the Two-Dimensional Normal Distribution > 0 in the Case Step 1. Choose a significance level a (5%, 1%, or the like). Step 2. Determine the solution c of the equation P(T:;;::: c) I - a = from the t-distribution (Table A9 in App. 5) with n - 2 degrees of freedom. Step 3. Compute r from (17), using a sample (XIo Yl), ... , (x", Yn)' Step 4. Compute t=r(~). ~~ If t E X AMP L E 4 ~ c, accept the hypothesis. If t > c, reject the hypothesis. Test for the Correlation Coefficient p Test the hypothesis p = 0 (independence of X and Y, because of Theorem 3) against the alternative p > 0, using the data in the lower left corner of Fig. 543. where r = 0.6 (manual soldering errors on 10 two-sided circuit boards done by 10 workers; x = front, y = back of the boards). Solution. We choose a = 5%; thus 1 - a = 95%. Since n = 10, n - 2 = 8, the table gives c = 1.86. Also. t = 0.6VS/0.64 = 2.12 > c. We reject the hypothesis and assert that there is a positive correlation. A worker making few (many) errors on the front side also tends to make few (many) errors on the reverse side of the board. • 11-101 SAMPLE REGRESSION LINE Find and sketch or graph the sample regression line of Y and x and the given data as points on the same axes. 1. (-1, 1), (0, 1.7), (1, 3) 2. (3, 3.5), (5, 2), (7, 4.5), (9, 3) 3. (2, 12), (5, 24), (9. 33), (14, 50) 4. (11, 22), (15, 18), e17, 16), (20, 9), (22, 10) x 6 9 II 13 22 7. x = Revolutions per minute. y engine [hpJ 5. Speed x [mph] of a car 30 40 50 60 Stopping distance y [tt] 150 195 240 295 Also find the stopping disrance ar 35 mph. 6. x = Deformation of a certain steel [mm], y hardness [kg/mm 2 ] 26 28 = Brinell 33 35 = Power of a Diesel x 400 500 600 700 750 y 580 1030 1420 18!m 2100 1092 CHAP. 2S 8. Humidity of air x [%] Expansion of gelatin y [%] Mathematical Statistics 10 20 30 40 111-131 0.8 1.6 2.3 2.8 Find a 95% confidence interval for the regression coefficient Kl, assuming that (A2) and (A3) hold and using the sample: 9. Voltage x [V] 40 40 80 80 110 110 Current)' [A] 5.1 4.8 10.0 LO.3 13.0 12.7 CONFIDENCE INTERVALS 11. In Prob. 6 Also find the resistance R [il] by Ohms' law (Sec. 2.9]. 10. Force x [Ib] Extension y [in] of a spring 2 4 4.1 7.8 6 8 12.3 15.8 Also find the spring modulus by Hooke's law (Sec. 2.4). ~ .. ··1_'.'=01..· 1. What is a sample? Why do we take samples? 2. What is the role of probability theory in statistics? 3. Will you get better results by taking larger samples? Explain. 4. Do several samples from a certain popUlation have the same mean? The same variance? 5. What is a parameter? How can we estimate it? Give an example. 6. What is a statisticaL test? What errors occur in testing? 7. How do we test in quality control? 8. What is the x2-test? Give a simple example hom memory. 9. What are nonparametric tests? When would you apply them? 10. In what tests did we use the I-distribution? The X 2 -distribution? 11. What are one-sided and two-sided tests? Give typical examples. 12. List some areas of application of statistical tests. 13. What do we mean by "goodness of fiC? 14. Acceptance sampling uses principles of testing. Explain. 15. What is the power of a test? What can you do if the power is low? 12. In Prob. 7 13. In Prob. 8 14. Derive the second expression for first one. 2 in (9a) from the Sx 15. CAS EXPERIMENT. Moving Data. Take a sample, for instance, that in Prob. 6, and investigate and graph the effect of changing y-values (a) for small.\', (b) for large x, (c) in the middle of the sampLe. S T ION SAN D PRO B L EMS 18. Couldn't we make the error in interval estimation zero simply by choosing the confidence level I? 19. What is the Least squares principle? Give applications. 20. What is the difference between regression and cOIl'elation analysis? 21. Find the maximum likelihood estimates of mean and variance of a normal distribution using the sample 5, 4, 6,5,3,5,7,4,6,5,8,6. 22. Determine a 95% confidence interval for the mean fL of a normal population with variance 0"2 = 16, using a sample of size 400 with mean 53. 23. What will happen to the length of the interval in Prob. 22 if we reduce the sample size to 100? 24. Determine a 99% confidence interval for the mean of a normal population with standard deviation 2.2, using the sample 28,24,31,27,22. 25. What confidence interval do we obtain in Prob. 24 if we assume the variance to be unknown? 26. Assuming normality, find a 95% confidence interval for the variance hom the sample 145.3, 145.1, 145.4, 146.2. 127-291 Find a 95% confidence interval for the mean fL, assuming normality and using the sample: 27. Nitrogen content [%] of steel 0.74. 0.75. 0.73, 0.75, 0.74.0.72 16. Explain the idea of a maximum likelihood estimate from memory. 28. Diameters of 10 gaskets with mean 4.37 em and standard deviation 0.157 cm 17. How does the length of a confidence interval depend on the sample size? On the confidence level? 29. Density [g/cm 3 J of coke 1.40, 1.45, 1.39, 1.44, 1.38 1093 Summary of Chapter 25 30. What sample size should we use in Prob. 28 if we want to obtain a confidence interval of length 0.1. assuming that the standard deviation of the samples is (about) the same? 131-321 Find a 99'1t confidence interval for the variance 2 0- • assuning normality and using the sample: 31. Rockwell hardness of tool bits 64.9. 64.1, 63.8. 64.0 32. A sample of size II = 128 with variance s2 = 1.921 33. Using a sample of IO values with mean 14.5 from a normal population with variance 0-2 = 0.25. test the hypothesis flo = 15.0 against the alternative fLl = 14.4 on the 5% level. 34. In Prob. 33. change the alternative to fL =1= 15.0 and test as before. 35. Find the power in Prob. 33. 36. Using a sample of 15 values with mean 36.2 and variance 0.9. te~t the hypothesis fLo = 35.0 against the alternative fLl = 37.0. assuming normality and taking a = 1%. 37. Using a sample of 20 values with variance 8.25 from a normal population. test the hyothesis 0-02 = 5.0 against the alternative 0-1 2 = 8.1. choosing a = 5%. 38. A firm sells paint in cans containing I kg of paint per can and is interested to know whether the mean weight differs significantly from I kg, in which case the filling machine must be adjusted. Set up a hypothesis and an alternative and perform the test. assuming normality and using a sample of 20 fillings having a mean of 991 g and a standard deviation of 8 g. (Choose a = 5%.) 39. Using samples of sizes to and 5 with variances s/ = 50 and Sy2 = 20 and assuming normality of the conesponding populations. test the hypothesis Ho: 0".,,2 = o-y2 against the alternative 0".",2 > o-y2. Choose a = 5%. 40. Assume the thickness X of washers to be normal with mean 2.75 mm and variance 0.00024mm2. Set up a control chaIt for fL. choosing a = I %, and graph the means of the five samples (2.74. 2.76). <2.74. 2.74). (2.79.2.81), (2.78, 2.76), (2.71. 2.75) on the chart. 41. What effect on UCL - LCL in a control chart for the mean does it have if we double the sample size? If we switch from a = I % to a = 5'70? 42. The following sample~ of screws (length in inches) were taken from an ongoing production. Assuming that the population is normal with mean 3.500 and variance 0.0004, set up a control chart for the mean, choosing a = I %, and graph the sample means on the chart. Sample No. Length 2 3 4 5 6 7 3.49 3.48 3.52 3.50 3.51 3.49 3.52 3.53 3.50 3.47 3.49 3.51 3.48 3.50 3.50 3.49 43. A purchaser checks gaskets by a single sampling plan that uses a sample size of 40 and an acceptance number of I. Use Table A6 in App. 5 to compute the probability of acceptance of lots containing the following percentages of defective gaskets !'It. !%. I 'It, 2%. 5%. 10%. Graph the OC curve. (Use the Poisson approximation.) 44. Does an automatic cutter have the tendency of cutting longer and longer pieces of wire if the lengths of subsequent pieces [in.] were to.l. 9.8. 9.9, 10.2, 10.6, to.5? 45. Find the least squares regression line to the data (-2. I). n. (6. 5). (0. 1). (2, 3), (4 ... Mathematical Statistics We recall from Chap. 24 that with an experiment in which we observe some quantity (number of defectives. height of persons, etc.) there is associated a random variable X whose probability distribution is given by a distribution function (I) F(x) = P(X ~ x) (Sec. 24.5) which for each x gives the prObability that X assumes any value not exceeding x. 1094 CHAP. 2S Mathematical Statistics In statIstIcs we take random samples Xl, . . . , Xn of size n by performing that experiment n times (Sec. 25.1) and draw conclusions from properties of samples about properties of the distribution of the con-esponding X. We do this by calculating point estimates or confidence intervals or by peIiorming a test for parameters (/-L and (T2 in the normal distribution. p in the binomial distribution. etc.) or by a test for distribution functions. A point estimate (Sec. 25.2) is an approximate value for a parameter in the distribution of X obtained from a sample. Notably, the sample mean (Sec. 25.1) 1 (2) X = - n n 2: -'J = j~l 1 - n (Xl + ... + XII) is an estimate of the mean /-L of X, and the sample variance (Sec. 25.1) (3) is an estimate of the variance (T2 of X. Point estimation can be done by the basic maximum likelihood method (Sec. 25.2). Confidence intervals (Sec. 25.3) are intervals 81 ~ 8 ~ 82 with endpoints calculated from a sample such that with a high probability 'Y we obtain an interval that contains the unknown true value of the parameter 8 in the distribution of X. Here, 'Y is chosen at the beginning, usually 95% or 99%. We denote such an interval by CONF y {8l ~ 8 ~ 82 }. In a test for a parameter we test a h)pothesis 8 = 80 against an alte171ative 8 = 81 and then, on the basis of a sample, accept the hypothesis. or we reject it in favor of the alternative (Sec. 25.4). Like any conclusion about X from samples, this may involve en-ors leading to a false decision. There is a small probability a (which we can choose, 5% or 1%, for instance) that we reject a true hypothesis, and there is a probability f3 (which we can compute and decrease by taking Larger samples) that we accept a false hypothesis. a is called the significance level and I - f3 the power of the test. Among many other engineeIing applications, testing is used in quality control (Sec. 25.5) and acceptance sampling (Sec. 25.6). If not merely a parameter but the kind of distribution of X is unknown, we can use the chi-square test (Sec. 25.7) for testing the hypothesis that some function F(x) is the unknown distribution function of X. This is done by determining the discrepancy between F(x) and the distribution function F(x) of a given sample. "Distribution-free" or nonparametric tests are tests that apply to any distribution, since they are based on combinatorial ideas. These tests are usually very simple. Two of them are discussed in Sec. 25.8. The last section deals with samples of pairs of values, which arise in an experiment when we simultaneously observe two quantities. In regression analysis, one of the quantities, x, is an ordinary variable and the other, Y, is a random variable whose mean /-L depends on x, say, /-L(x) = Ko + KIX, In correlation analysis the relation between X and Yin a two-dimensional random variable (X, Y) is investigated. notably in terms of the correlation coefficient p. I APPENDIX 1 References Software see at the beginning of Chaps. 19 and 24. General References [GR I] Abramowitz, M. and 1. A. Stegun (eds.), Handbook of Mathematical Functiol1s. 10th printing, with corrections. Washington, DC: National Bureau of Standards. 1972 talso New York: Dover, 1965). [GR2] Cajori, F., History of Mathematics. 5th ed. Reprinted. Providence. RI: American Mathematical Society. 2002. [GR3] Courant, Rand D. Hilbert, Methods of Mathematical Physics. 2 vols. Hoboken, NJ: Wiley, 2003. [GR4] Courant, R, Differelltial and Integral Calculus. 2 vols. Hoboken, NJ: Wiley, 2003. IGR5] Graham. R L. et aI., Concrete Mathematics. 2nd ed. Reading. MA: Addison-Wesley. 1994. IGR6] Ito, K. (ed.), Encyclopedic Dictionmy of Mathematics. 4 vols. 2nd ed. Cambridge, MA: MIT Press, 1993. [GR7] Kreyszig, E., Introductory Functional Analysis with Applications. New York: Wiley, 1989. [GR8] Krey~zig, E.. Differemial Geometry. Mineola. NY: Dover, 1991. IGR9] Kreyszig. E. Introduction to Diflerentilll Ceometr.V alld Riemannian CeO/netr". Toronto: University of Toronto Press, 1975. IGRlOl Szeg6, G., Orthogonal Polyno/lliaL~. 4th ed. Replinted. New York: American Mathematical Society, 2003. [GRII] Thomas, G. et aI., Thomas' Calculus, Early TranscendelltaL~ Update. 10th ed. Reading, MA: Addison-Wesley, 2003. Part A. Ordinary Differential Equations (ODEs) (Chaps. 1-6) See also Part E: Numeric Analysis [AI] Arnold, V. L Ordinm) D(fferential Equations. 3rd ed. New York: Springer. 1997. [A2] Bhatia, N. P. and G. P. Szego, Stability Theory of" Dynamical Systems. New York: Splinger, 2002. [A3] Birkhofl", G. and G.-c. Rota, Ordinary D(fferential Equations. 4th ed. New York: Wiley, 1989. [A4] Brauer. F. and J. A. Nohel, Qualitative Theory of Ordinary D!f/erelltial Equations. Mineola, NY: Dover, 1994. [A5] Churchill, R V .. Operatiollal Mathematics. 3rd ed. New York: McGraw-Hill, 1972. [A6] Coddington, E. A. and R Carlson, Linear Ordinary D(fferelltial Equations. Philadelphia: SIAM, 1997. [A7] Coddington, E. A. and N. Levinson, Theory of Ordinary Differential Equations. Malabar, FL: Krieger, 1984. [A8] Dong, T.-R et al., Qualitative Theory ofD(fferential Equations. Providence, RI: American Mathematical Society, 1992. [A9] Erdelyi, A. et aI., Tables of Integral Transfonns. 2 vols. New York: McGraw-Hill, 1954. IAlO] Hartman. P .. Ordinary Differential Equations. 2nd ed. Philadelphia: SIAM, 2002. IAII] lnce, E. L., Ordinary Differential Equations. New York: Dover, 1956. [A12] Schiff, J. L., The Laplace Tran.~fonll: Theory and Applications. New York: Springer, 1999. [A13] Watson. G. N., A Treatise on the Theory of Bessel Functions. 2nd ed. Reprinted. New York: Cambridge University Press, 1995. [AI4] Widder, D. V., The Laplace Transfol1ll. Princeton, NJ: Princeton University Press, 1941. [A 15] Zwillinger, D., Handhook of Differential Equations. 3rd ed. New York: Academic Press, 1997. Part B. Linear Algebra, Vector Calculus (Chaps. 7-10) For books on numeric linear algebra, see also Part E: Numeric Analysis. IBI] Bellman, R., Introduction to Matrix Analysis. 2nd ed. Philadelphia: SIAM, 1997. IB2] Chatelin, F., Eigenvalues of Matrices. New York: Wiley-Interscience, 1993. [B3] Gantmacher. F. R, The Theory of Matrices. 2 vols. Providence, RI: American Mathematical Society. 2000. [B4] GOhberg, I. P. et a!., Invariant Subspaces ofMatrices with Applications. New York: Wiley, 1986. [B5] Greub, W. H., Linear Algebra. 4th ed. New York: Springer, 1996. [B6] Herstein. I. N., Ahstract Algebra. 3rd ed. New York: Wiley. 1996. [B7] John, A. W .. Matrices and Tensors in Physics. 3rd ed. New York: Wiley, 1995. Al A2 References [B8J Lang. S .. Linear Algebra. 3rd ed. New York: Springer. 1996. [B9] Nef. W .. Linear Algebra. 2nd ed. New York: Dover. 1988. [B 10] Parlett. B.. The Symmetric Eigenmlue Problem. Philadelphia: SIAM. 1997. Part C. Fourier Analysis and POEs (Chaps. 10-11) For books on numerics for PDEs see also Part [D7] Krantz. S. G .. Complex Analysis: The Geometric ViCllpoi11t. Washington. DC: The Mathematical Association of America. 1990. [D8] Lang, S .. Complex Analysis. 3rd ed. New York: Springer, 1993. [D9] Narasimhan, R. Compact Riell/ann SlIIfaces. New York: Springer, 1996. [D 10] Nehari. Z.. C01!f017lwl Mapping. Mineola. NY: Dover. 1975. [DIl] Springer. G., Introduction to Riemann SlIIfaces. Providence, RI: American Mathematical Society, 1002. E: Numeric Analysis. [CII Antimirov. M. Ya .. Applied Imegral Tramfol1ns. Providence. RI: American Mathematical Society, 1993. [Cl] Bracewell, R, Tile Fourier TralJSforlll and Its Applications. 3rd ed. New York: McGraw-Hili. 2000. [C3] Carslaw. H. S. and J. C Jaeger, COllduction (~f Heat in Solids. 2nd ed. Reprinted. Oxford: Clarendon, 1986. [C4J Churchill. R V. and J. W. Brown, Fourier Series and BOllndary Value Problellls. 6th ed. New York: McGraw-HilL 2000. IC5] DuChateau. P. and D. Zachmann. Applied Partial Differential Equations. Mineola. NY: Dover. 2001. [C6] Hanna, J. Rand J. H. Rowland. Fourier Series, Tran.~fo1711s. and Boundary Value Problems. 2nd ed. New York: Wiley. 1990. [C7] leni. A. L The Gibbs Phenomenon ill Fourier Analysis. Splines. and WCII'e!et Approximations. Boston: Kluwer. 1998. [C8] John. E. Partial Differelltial Equatiolls. New York: Springer. 1995. [C9] Tolstov. G. P .. Fourier Series. New York: Dover, 1976. [CIO] Widder. D. V .. The Heat Equation. New York: Academic Press. 1975. [CII] Zauderer, E.. Partial Dif(eremial Equatio/lS of Applied Mathematics. 2nd ed. New York: Wiley, 1989. [CI2] Zygmund, A. and R Fefferman. Trigonometric Series. 3rd ed. New York: Cambridge University Press, 2003. Part o. Complex Analysis (Chaps. 13-18) [D I] Ahlfors. L. V.. Complex Analysis. 3rd ed. New York: McGraw-Hill. 1979. [D2] Bieberbach. L.. C01!formal Mapping. Providence, RI: American Mathematical Society. 1000. [D3] Hemici. P .. Applied and Computational Complex Analysis. 3 vols. New York: Wiley. 1993. [D4] Hille, E., Analytic Function Theory. 1 vols. 2nd ed. Providence. Rl: American Mathematical Society. 1997. [D5] Knopp. K.. Eleme11ts at tbe Theory of Functions. Ne\\' York: Dover. 1951. [D6] Knopp, K .. TheOlY of Functions. 2 parts. New York: Dover, 1945, 1947. Part E. Numeric Analysis (Chaps. 19-21) [EI] Ames, W. E, humerical Methods for Partial Ditferelltial Equations. 3rd ed. New York: Academic Press, 1992. [E2] Anderson, E., et aI., LAPACK User's Guide. 3rd ed. Philadelphia: SIAM, 1999. [E3J Bank, R E., PLTMG. A Software Package for Solring Elliptic Partial Differential Equaticms: Users' Guide 7.0. Philadelphia: SIAM. 1994. [E4] Constanda. C, Solution Techniques for Elementary Partial Differelltial Eqllations. Boca Raton, R: CRe Press. 2001. [E5] Dahlquist, G. and A. Bjorck. NUII/erical Methods. Mineola. NY: Dover. 2003. [E6] DeBoor. C .. A Practical Guide to Splilles. Reprinted. New York: Springer. 1991. [E7] Dongarra. J. J. et aL LlNPACK Users Guide. Philadelphia: SIAM. 1978. (See also at the beginning of Chap. 19.) [E8] Garbow, B. S. et al., Matrix Eige11system Routi11es: EISPACK Gllide Extensioll. Reprinted. Kew York: Springer, 1990. rE9] Golub, G. H. and C F. Van Loan, Matrix Computatio11s. 3rd ed. Baltimore, MD: Johns Hopkins University Press, 1996. [E 10] Higham, N. J., Accuracy a11d Stability o.f Numerical Algorithms. 2nd ed. Philadelphia: SIAM, 2002. [E 11] IMSL (International Mathematical and Statistical Libraries). FORTRAN Numerical Library. Houston, TX: Visual Numerics, 2001. (See also at the beginning of Chap. 19.) [E 12] IMSL IMSL for Jam. Houston, TX: Visual Numerics, 2001. [El3] IMSL C Library. Houston. TX: Visual Numelics. 1002. [E14] Kelley, C T., Iteratil'e Methods for Li11ear alld NOlllillear Equatio11s. Philadelphia: SIAM. 1995. [EI5J Knabner. P. and L. Angerman, Numerical Methods .for Partial Differelltial Equatiolls. New York: Springer, 1003. [EI6] Knuth. D. E., The Art of Computer Programmillg. 3 voIs. 3rd ed. Reading, MA: Addison-Wesley, 1005. AJ App.l [E 17] Kreyszig. E., Imroductory Funcrimwl Analysis with Applications. New York: Wiley, 1989. [E181 Kreyszig. E.. On methods of Fourier analysis in muItigrid theory. Lecture Notes in Pure and Applied Mathematics IS7. New York: Dekker, 1994, pp. 22S-242. [EI9] Kreyszig, E., Basic ideas in modem numerical analysis and their origins. Proceedings of the Annnal COIlference (dthe Canadian Society for the History (llId Philosophy of Mathematics. 1997. pp. 34-4S. [E20J Kreyszig. E.. and J. Todd. QR in two dimensions. Elelllellte da Mathematik 31 (1976). pp. 109-114. [E21] Mortensen. M. E., Geometric Modelin!!,. 2nd ed. New York: Wiley. 1997. [E22] Morton, K. W., and D. F. Mayers, Numerical Solution of Partial Differential Equations: An Imroduction. New York: Cambridge University Press, 1994. [E23] Ot1ega. J. M .. Introduction to Parallel and Vector Solution of Linear Systems. New York: Plenum Press, 19HK [E24J Overton, M. L., Numerical Computing I\'ith IEEE Floating Poillt Arithmetic. Philadelphia: SIAM. 2001. [E2S] Pre~s, W. H. et aL Numerical Recipes in C: The Art of Scientific Computing. 2nd ed. New York: Cambridge University Press, 1992. [E26] Shampine, L. F., Numerical Solutions of Ordinary Differential Equations. New York: Chapman and Hall, 1994. [E27] Varga, R. S., Matrix Iterative Analysis. 2nd ed. New York: Springer, 2000. [E28] Varga, R. S .. GerJgorin and His Circles. New York: Springer. 2004. [E29] Wilkinson. J. H.. The Algebraic Eigenl'{lilte Problem. Oxford: Oxford University Press, 1988. Part F. O"ltimization, Gra"lhs (Chaps. 22-23) [f-J] Bondy, J. A., Graph Them:,' with Applications. Hoboken, NJ: Wiley-Interscience, 2003. [F2] Cook. W. J. et aL Combinatorial Optimi;::ation. New York: Wiley. 1993. [F3] Diestel, R.. Graph Theory. 2nd ed. New York: Springer, 2000. [F4] Diwekar. U. M., Introduction to Applied Optimi;::atimz. Boston: Kluwer. 2003. [FS] Gass. S. L.. Linear Programming. Method and Applications. 3rd ed. New York: McGraw-Hill. 1969. [F6] Gross. J. T., Handbook of Graph Them:" and Applications. Boca Raton. FL: CRC Press, 1999. [F7) Goodrich. M. T., and R. Tamassia, Algorithm Design: Foundations, Analysis. and Imemet Examples. Hoboken, NJ: Wiley, 2002. [FH] Harm)" F., Graph TheOl}". Reprinted. Reading, MA: Addison-Wesley, 2000. [F91 Merris, R.. Graph Theory. Hoboken. NJ: WileyInterscience, 2000. [FIOJ Ralston, A .. and P. Rabinowitz. A First Coune in Numerical Analysis. 2nd ed. Mineola, NY: Dover. 200 I. [FIIJ Thulasiraman. K., and M. N. S. Swamy, Graph Theory and Algorithms. New York: Wiley-Interscience, 1992. [FI2J Tucker, A.. Applied Combinarorics. 4th ed. Hoboken. NJ: Wiley, 2001. Part G. Probability and Statistics (Chaps. 24-25) [G I] Amelican Society for Testing Materials, Manual on Presentation of Data alld Control Chart Analysis. 7th ed. Philadelphia: ASTM. 2002. [G2] Anderson. T. W., An Imroduction to Multivariate Statistical Analysis. 3rd ed. Hoboken. NJ: Wiley, 2003. [G3] Cramer. H .. Mathematical Methods of Statistics. Reprinted. Princeton, NJ: Princeton University Press, 1999. [G41 Dodge, Y., The Oxford Dictionary of Statistical Terms. 6th ed. Oxford: Oxford University Press. 2003. [GS] Gibbons. J. D., NOl/parametric Statistical Irlference. 4th ed. New York: Dekker. 2003. [G6] Grant. E. L. and R. S. Leavenworth. Statistical Quality Control. 7th ed. New York: McGraw-HilI. 1996. [G7] IMSL, Fortran Numerical Librm:r. Houston. TX: Visual Numelics, 2002. [G8] Kreyszig, E .. Illtroductory Mathematical Statistics. Principles {UJd Methods. New York: Wiley, 1970. [G9] O'Hagan, T. et aI., Kendal/'s Advanced TheOlY of Statistics 3-Volllllle Set. Kent, U.K.: Hodder Arnold, 2004. [GIO] Rohatgi, V. K. and A. K. MD. E. Saleh, All Imrodllction to Probability and Statistics. 2nd ed. Hoboken, NJ: Wiley-Interscience. 200\. 'r: A P PEN D I X ,,1, 2 'I Answers to Odd-Numbered Problems Problem Set 1.1, page 8 1. (cos 7T.X)/7T + 7. Second order 11. y = ~ tan ('2x 13. y = e-:L:l C + 5. First order 9. Third order n7T), n = 0, ±I, ±2, ... 15. (A) No. (B) No. Only y = O. 17. )''' = g, y' = gt, Y = gt 2 /2 19. )''' = k, y' = kt + 6, y = ~kt2 + 6t, y(60) = 1800k + 360 y' (60) = 1.47·60 + 6 = 94 [rnlsec] = 210 [mph] 21. ekH = ~,H = (1n ~)/k = (1011 In 2)/1.4 = 1570 [yearsl = 3000, k = 1.47, Problem Set 1.2, page 11 15. y = x(l - In x) + c 2 17. Verify the general solution y2 + t = c. Circle of radius 3Vz 19. 111V ' = 111g - bv 2 , v' = 9.8 - v 2 , v(O) = 10. v' = 0 gives the limit V9.8 = 3.1 [meter/sec]. 11. y = -(2/7T) cos ~7TX I- c Problem Set 1.3, page 18 + 3. cos 2y dy = '2 dx, y = ~ arcsin (4x + c) 7. dy/y = cot 77X dx, )' = c(sin 7TX)lhT t2 11. r = roe- 36x 2 = c, ellipses 9. y = tan (c - e- 7rx/7T) 13. I = Ioe- RtiL 15.y=ex/~ 17. y = 4ln x 5. )'2 = Vln (X - 2x + e) y' = (y - b)/(x - a), y - b = c(x - a) yoe k = 2yo, e k = 2 (l week), e2k = 22 (2 weeks), e4k = 24 y = yoe kt = yoe- 0,OO01213t = )'oe-O.OO01213.4000 = 0.62yo; 62%; cf. Example 2. 27. y' = -ky, y = Yoe-1<t, e- 5k = 0.5, k = -(1n 0.5)/5 = 0.139, f = -(1n 0.05)/0.139 = 22 [min] 29. T(O) = 10. T = 23 - 13ek t, T(2) = 23 - 13e2k = 18. k = -0.478, T = 22.8 gives t = [In (-0.2/-13)]/(-0.478) = 8.73 [min]. 19. 21. 23. 25. y 2 V2ih 31. h = gt 2 /2, t = Y2h/g, v = gt = gY2h/g = 33. y' = 0 - (2/800)y, y = 200e- 0,0025t, f = 300 [min], y(300) = 94.5 [lb] 35. (A) is related to the enor function and (C) concerns the Fresnel integral C(x); see App.3.1. (D) y' = '2.\y + I, yeO) = 0 A4 App. 2 AS Answers to Odd-Numbered Problems Problem Set 1.4, page 25 1. Exact. x4 + y4 = e 3. Exact. u = cos TTX sinh Y + key), u y = cos TTX cosh y + k', k' = O. Ans. cos TTX sinh y = e 5. Exact. 9x2 + 4y2 = e 2f1 2f1 7. Exact, Mil = NT = -2e- , u = re- 2(J + k«(}), U e = -2re- + k', k' = O. 2fJ Ans. re- 21i = e, r = ee 9. Exact. u = ylx + sin 2x + key). lIy = lIx ...!.. k' = IIx - 2 sin 2y. AlIS. ylx + sin 2x + cos 2y = e 11. Not exact. F = 1/x2 by Theorem I. -ylx 2 dx + IIx dy = d(ylx) = O. Ails. Y = ex 13. - 3y 2/x 4 dx + 2ylx3 dy = d(\,2/x 3) = O. Y = e).2./2 (semicubical parabolas) 15. Exact, U = e 2x cos Y + key), lly = _e2x sin y + k', k' = O. AlIS. e2x cos y = e, e = 1 17. Not exact. Try R. F = e- x , e-X(cos wx + w sin wx) dx + dy = 0, U = Y + [(x), x U x = [' = e-X(cos wx + w sin wx), U = Y + [ = Y - e- cos wx = c, c = 0 Y Y 19. U = eX + key), u y = k' = -1 + e , k = -y + e • Ans. eX - y + e Y = e 21. B = C, !Ax2 + Cxy + !Dy2 = e Problem Set 1.5, page 32 3. y 7. 9. 13. 17. = Cf;,-3.5x + 0.8 ee- kx 5. y = 2.6e-1.25x + 4 2 e /(x13k if k =1= 0 11. y = 2xe cos 2x 15. Y = e llX (x2 + c), c = 4.1 + c (if k = 0). y = + Separate. y - 2.5 = c cosh4 1.5x y = sin 2x + c/sin2 2x, e = 1 y = (c + cosh lOx)/x3 • Note (X 3 y)' = 5 sinh lOx. y = x 19. )' = l/u ! , 1I = ce- 57 .x - 6.5 5.7 - 21. u = y-2 = ecc\ l + ce2x ), c = 3, u(O) = 4 23. Separate variables. y2 = 1 - ce cos x, e = - 1/e 25 • .v' = Ry + k. y = ce Rt - klR. c = Yo + /.JR. Yo = 1000, R = 0.06. t = 65 - 25 = 40, k = 1000. Y = $178076.12. StaJt at 45 gives Yo[(I + 1I0.06)eo.o6.20 - 110.06] = 41.988732yo = 178076.12, Yo = k = $4241.05. 27. y' = 175(0.0001 - y/450), yeO) = 450· 0.0004 = 0.18, y = 0.135e-O.3889t + 0.045 = 0.1812, e-O.3889t = (0.09 - 0.045)/0.135 = 1/3. t = (In 3)/0.3889 = 2.82. AilS. About 3 years 29. y' = A - ky, yeO) = o. y = A(l - e-kt)/k 31. y' = By2 - Ay = By(y - AlB), A> 0, B > O. Constant solutions y = 0, y = AlB. y' > 0 if y > AlB (unlimited growth),.v' < 0 if 0 < y < AlB (extinction). y = AI(ceAt + B), yeO) > AlB if c < 0, yeo) < AlB if e > O. 33. y' = y - y2 - 0.2y, Y = 11(1.25 - 0.75e- O . 8t ), limit 0.8, limit 1 35. y' = y - 0.25y2 - O.ly = 0.25y(3.6 - y). Equilibrium harvest 3.6, y = 18/(5 + ce-O. 9t ) 37. (YI + )'2)' + P(YI + Y2) = c.v/ 39. (YI + .\"2)' + P(YI + Y2) = (YI' 41. Solution of eyt' + PQ'l = e(y/ + PYI) + (1'2' + Ph) = 0 + 0 = 0 + PYI) + (Y2' + PY2) = r + 0 = r + PYl) = cr A6 App. 2 Answers to Odd-Numbered Problems 43. CAS Experiment (a) y = x sin (lIx) + c\. c = 0 if y(2/rr) = 2/rr. y is undefined at x = 0, the point at which the "waves" of sin (11x) accumulate; the factor x makes them smaller and smaller. Experiment with various x-intervals. (b) )' = x"Lsin (llx) + c]. y(2/rr) = (2/rr)n. n need not be an integer. Try n = ~. Try n = - 1 and see how the "waves" near 0 become larger and larger. 45. y = uy*, y' + py = u'y* + uy*' + puy* = u'y* + u(y*' + py*) = u'y* + U' 0 = r, u' = r/y* = re Jp da', U = I eJp d.e r dx + c. Thus, y = UYh gives (4). We shall see that this method extends to higher-order ODEs (Secs. 2.10 and 3.3). Problem Set 1.6, page 36 1. y' = 4, .v' = -1/4, Y = -x/4 + c* 3. y/x = c, y'lx = Y/X2, y' = ylx,)" = -X/y,)'2 + x 2 = c*, circles 5.2xy + x\" = 0, y' = -2y/x, y' = x/(2y), y2 - x2/2 = c*. hyperbolas 7. ye-.l:2/2 = c, y' = xy, y' = -1I(xy), yy' = -l/x, y2/2 = -In Ixl + c**, x = c*e- y2/2 , bell-shaped curves (with x and y interchanged) 9. y' = -4x/y. y' = )il4x, 4 In l}il = In Ixl + c':'*. x = C*J4. parabolas 11. xe- yl4 = c, y' = 4/x,}i' = -x/4, y = -x 2 /8 + c* 13. Use dy/d\ = I/(dx/dy). (y - 2x)e'" = c, tv' - 2 + Y - 2x)e" = 0, y' = 2 - Y + 2x, dxld"y = -2 + v - 2:r is linear, dx/dy + 2x = Y - 2, x = c*e- 2y + y/2 - 5/4 15. II = c, uxdx + uydy = 0, y' = -u,Ju y. TrajectOlies y' = uy/ux- Now v = c*. v,rdx + vydy = 0, y' = -v:r/Vy. This agrees with the trajectory ODE in u if U.l , = uy (equal denominators) and u y = -v.~ (equal numerators). But these are just the Cauchy-Riemann equations. 17.2r + 2y.v' = o. y' = -x/yo Trajectories y' = 5h. In IJI = In Ixl + c**, y = c*x. 19. y' = -4.\19)'. Trajectories}i' = 9}'14x. y = c*x9/4 (c* > 0). Sketch or graph these curves. Problem Set 1.7, page 41 1. In Ix - xol < a; just take b in ex = b/K large. namely, b = aK. 3. No. At a common point (Xl> .vI) they would both satisfy the "initial condition"" .v(Xl) = Yl, violating uniqueness. 5.)" = f(x. y) = rex) - pCr))': hence af/ay = -p(x) is continuous and i~ thus bounded in the closed interval Ix - xol ~ iI. 7. R has sides 2a and 2b and center (1, 1) since y(l) = 1. In R, f = 2y2 ~ 2(b + 1)2 = K, a = b/K = b/(2(b + 1)2). da/db = 0 gives b = 1, and a opt = b/K = 1/8. Solution by dy/)'2 = 2 dx, etc., y = 11(3 - 2x). 9. 11 + .v 2 1 ~ K = I + b 2 , a = b/K. da/db = O. b = 1. a = 1/2. Chapter 1 Review Questions and Problems, page 42 + ~) = 4 dx. 2 arctan 2y = 4x + c*. y = ! tan (21' + c) = l/u, y' = -u' /u 2 = 4/u - 1/112, 1/ = c*e-4.r + ~ 15. dy/(y2 + 1) = x 2 dx, arctan y = x 3/3 + c, y = tan (x 3/3 + c) 17. Bernoulli.),' + xy = x/y, u = )'2, II' = 2)y' = 2x - 2ru linear, 2 J X2") U = e' (e' ~X dr + c) = 1 + ce- x , l' = V I u. Or write , (2 1) and separate. . )y = -x y 11. dy/(y2 13. Logistic ODE. y -:t: Z --\. App. 2 A7 Answers to Odd-Numbered Problems 19. Linear, y = e Cos xU e- cos x sin x dx + ce cos x + 1. Or by separation. 21. Not exact. Use Theorem 1, Sec. 1.4: R = 2/x, F = x 2 : the resulting exact ODE is 3x 2 sin 2y dx + 2x 3 cos 2y dy = d(x 3 sin 2y), x 3 sin 2y = c. Or by separation, cot 2y dy = - 3/(2x) d.r. etc., sin 2y = n·- 3 . 23. Exact. /I = I M dx = sin xy - x 2 + k, lIy = X cos xy + k' = N, k = y2, sin .l)" - x 2 + y2 = c. 25. Not exact. R* = 1 in Theorem 2, Sec. 1.4, F* = e Y • Exact is e Y sin (y - x) dx + eY[cos (y - x) - sin (y - x)J dy = O. II = I M dx = eY cos (y - x) + k, lIy = eY(cos (y - x) - sin (y - x» + k' = N, eY cos (y - x) = c. 27. Separation. )'2 + x 2 = 25 29. Separation. )' = tan (x + c), c = 31. Exact. u = X\2 + cos X + 2-" = c, c = 1I(0, 1) = 3 33. y' = x/yo Trajectories y' = -Yfx. y = c*/x by separation. Hyperbolas. 35. Y = Yoekt, e 4k = 0.9, k = ! In 0.9, e kt = 0.5, f = (In 0.5)/k = (In 0.5)/[(ln 0.9)/4] = 26.3 [daysl 37. ekt = 0.01, t = On O.OI)/k = 175 [days] 39. y' = -4x/)'. Trajectories y = CIX1l4 or X = C2y4 41. Logistic ODE y' = Ay - By2, Y = 1/1/, U' + All = +B, 1I = ce- At + B/A 43. A = amount of incident light. A thin layer of thicknes!> ,lx absorbs M = -kALh (-k = constant of proportionality). Thus ,lA/b.x = -kA. Let b.x ~ O. Then A' = -kA. A = Aoe-kl: = amount of light in a thick layer at depth x from the surface of incidence. c) = -!7T Problem Set 2.1, page 52 1. \" = 2.5e4x + 0.5e- 4x 3. y = e- x cos x 9. Yes if a 1= 0 15. F(x, z, z') = 0 7. Yes 13. No 19. .'" dddy = 4:::, y = (CIX + C2)-1I3 21. (dddy)~ = _.::3 sin y, -11.:: = -dx/dy 23. y"y' = 2. y = ~(t + 1)3/2 - i. ),(3) = 25. y" = ky',.::' = k.:: . .:: = cle kx = y', Cl 5. Y = 4x2 11. No = cos y + CI. X = -sin y + 3l, /(3) = 4 = 1, Y = (e kx - l)/k + 7/x 2 ClY + C2 Problem Set 2.2, page 59 1. Y = cle7X + C2 e - x 5. y = cleO.9X + C2e-L1X 9. y = cle3.5X + C2e-1.5X 13. y = Cl e12,- + C2e-12X 3. Y = (ci + c2x )e2.5x 7. y = eO. 5X(A cos I.5x + B sin 1.5.\) 11. y = A cos 3rr.r + B sin 3rr.r 15. y" - 3y' + 2y = 0 19. y" - 16y = 0 23. Y = e- 2x(2 cos x - sin x) 17. -" " - 2'[;;3' v.J y + 3Y = 0 3x 21. Y = 4e - 2e- X 27. y = (2 - 4x)e-O. 25x 25. \" = 2 + e- 7TX 29. y = e-o.1:r(3.2 cos 0.2x + 1.6 sin 0.2x) 31. \" = 4e5X - 4e- 5J; x x x 33')"1 = e- '.'"2 = O.OOle + e· E • , 1 , 35. W nte = e -a:"C/2 , c = cos wx, s = S1l1 wx. Note that £ = -"2a£, C = -ws, 2 s' = we. Substitute, drop £, collect c-terms, then s-terms, and use w = b - !a 2 , to get c(b - !a 2 + !02 - w 2) + s( -ow + !aw + ~aw) = 0 + 0 = O. A8 App. 2 Answers to Odd-Numbered Problems Problem Set 2.3, page 61 1. O. 0, -2 cos x 3. -0.8 X 3 S. -12x 3 + 9x2 + 8x - 2. -28 sin 4x - 4 cos 4x. 0 7. Y 11. y = = 2X (Cl + c2x )eCle-3.1X + C2e-x + 6x 2 + 0.4, O. eO. 4x = e- 3X(A cos 2x + B sin 2x) 13. y = A cos 4.2wx + B sin 4.2wx 9. y Problem Set 2.4, page 68 1. Y = Yo cos wof + (uo/wo) sin wof. At integer f (if Wo = 7T), because of periodicity. 3. lIlLe" = -lIlg sin e = -mge (tangential component of W = lIlg). e" + w02e = O. WO/(27T) = \/i/i/(27T). 5. No. because the frequency depends only on kIm. 7. (i) Greater by a factor vi (ii) Lower 9. w* = [w0 2 - c 2/(41112)1112 = wo[1 - c 2/(411lk)]112 = wo(l - c 2/8111k) = 2.9583 11. 27T/W* since Eq. (10) and y' = 0 give tan (w';'f - 8) = -a/w*; tan is periodic with period 7T/W"'. 13. Case (II) of (5) with c = "\, '417lk = V 4· 500' 4500 = 3000 [kg/secJ. where 500 kg is the mass per wheel. 15. y = [Yo + (uo + aYo)f1e- at , Y = [1 + (uo + l)t]e- t ; (ii) u o = -2. -3/2, -4/3, -5/4, -6/5 17. Y = 0 gives Cl = -C2e-2{Jt, which has one or no positive zero, depending on the initial conditions. Problem Set 2.5, page 72 1. CIX3 + C2X-2 5. xlA cos (In Ixl) 9. CIXO.1 + C2XO.9 + 13. x-o. 5 [2 cos (10 In B sin (In Ixl) - Ixl)] sin (10 In Ixl)] 3. (Cl + C2 In Ixl )x4 7. CIX1.4 + C2X1.6 11.3x2 - 2x 3 15.2.\"-3 + 10 Problem Set 2.6, page 77 1. y" - 0.25.'" = 0, W = -1 3. y" - 21..;/ + k 2 y = 0, W = e2kx 5. x 2/ ' + 0.5x/ + 0.0625), = 0, W = x-O. 5 7. x2y" + xy' + 4y = 0, W = 2/x 9. x2y" 0.75.'" = O. W = -2 11. y" - 6.25." = O. W = 2.5 13. y" + 2/ + 1.64." = O. W = 0.8e- 2x 15. y" + 5/ + 6.34y = 0, W = 0.3e- 5X 7 6rr 17. y" + 7.67Ty' + 14.44~\" = 0, W = e- . :,. Problem Set 2.7, page 83 1. cIe- x + c 2e- 2x + 2.5e2x 3. cIe4X + C2 e - 4X + 2.4xe4.'t 3 5. Cle2X + C2e-3X - x - 3x - 0.5 7. e- 3X(A cos 8x + B sin 8x) + eX(cos 4x + ! sin 4x) 9. c 1e-O. 4x + C2eo.4.'t + 20xeo.4.'t - 2Q,e-O. 4x 11. Cl cos 1.2x + C2 sin 1.2x + lOx sin 1.2x 2T 13. e- (A cos x + B sin x) + 5x 2 - 8x + 4.4 - 1.6 cos 2x + 0.2 sin 2x 15. 4x sin 2x - 4e x App. 2 A9 Answers to Odd-Numbered Problems 17. e-O. IX ( 1.5 cos 0.5.t - sin 0.5.\:) + 2eO. 5x 19. 2e- 3X + 3e4x - 12.\"3 + 3x 2 - 6.5x Problem Set 2.S, page 90 1. -0.4 cos 3t + 7.2 sin 3t 3. -12.8 cos 4.5t + 3.6 sin 4.5t 5.0.16 cos 2t + 0.12 sin 2t 7. 475 cos 3t - 415 sin 3t 9 • c 1 e- t/2 + C 2 e- 3tJ2 - 32 5 cos t - 1 5 sin t 11. (ci + c2t)e-3t/2 - ~ cos 31 - sin 3t 13. e-1. 5t (A cos t + B sin t) + 4 + 0.8 cos 2t - 6.4 sin 2t 15. 0.32e- t cos 5t + 0.68 cos 3t + 0.24 sin 3t 17. 5e- 41 - 4e- 2t - 0.3 cos 2t + 0.1 sin 2t 19. e-1. 5t (O.2 cos t - 1.1 sin t) + 0.8 cos t + 0.4 sin t Problem Set 2.9, page 97 1. LI' 3. Rl' + Rl = E,1 = (E/R) + ce- RtJL = 2.4 + + lIC = O. I = ce-tJ(RC) ce- 50t 5. I = 5(cus t - cos 1Ot)/99 7.10 is maximum when S = 0; thus C = 1/(w2 L). 9. R > Rcrit = 2VLIC is Case I. etc. 11.0 13. c l e- 20t + C2e-lOt + 16.5 sin lOt + 5.5 cos lOt 15. £' = -e- 4t (7.605 cos ~t + 1.95 sin ~t), I = e-o.lt(A cos ~t + B sin ~t) - e- 4t cos ~t 17. £(0) = 600. I' (0) = 600, 1 = e- 3t ( - 100 cos 4t + 75 sin 4t) + 100 co'> t 19. (b) R = 2 fl, L = 1 H, C = 1112 F, £ = 4.4 sin lOt V Problem Set 2.10, page 101 + B sin x - x cos x + (sin x) In Isin xl 3. CIX + C2X2 - X cos x 5. (cos X)(ci + sin .\" - In Isec x + tan xl) + (sin X)(C2 - cos x) = (cl - In Isec x + tan xl) cos x + C2 sin x 7. (CI + ~x) sin x + (C2 + In Icos xl) cos x 9. (Cl + c2x)ex + x 2 + 4x + 6 - eXOn Ixl + I) 11. c] cos 2x + C2 sin 2x + ~.\: cosh 2x 13. CIX + C2X2 - x sin x 1. A cos x 15. A cos x + B sin r + Ypi + Y p 2. Ypi as .\"p in Example 1, .\"p2 = ~~ sin 5x 17. lI" + 1I = 0 by substitution of Y = lIX- I/2 . YI = x- 1I2 cos x . .\"2 = x- 1I2 1I2 sin from (2) with the ODE in standard sin x. Yp = _~x1l2 cos X + form. ix- x Chapter 2 Review Questions and Problems, page 102 9. 4 cle ." 11. e- 4X + C2e-2X - (A cos 3r + 1.1 cos 6x - 0.3 sin 6x B sin 3x) - ~ cos 3x + ~ sin 3x A10 App. 2 Answers to Odd-Numbered Problems 13. Y1 = x 3 , Y2 = x- 4 , r = x- 5 , W = -7x- 2 , Yp = - 412X-3 - ~X-3 = -*x- 3 15. )'1 = eX, Y2 = xe x , W = e2.T, Yp = e X I(2x) 17. -'"1 = eX cos X')'2 = eX sin x, W = e 2x , yp = -xe x cos x + eX(sin x) In Isin xl 19. y = 4e 2x + 2e- 7x 21. Y = 9x- 4 + 6x 6 2 3x 23. Y = e-2.1' - 2e- + 18x - 30t" + 19 25. Y = ~X3 + 4x 2 - 5x- 2 27. Y = -16 cos 2t + 12 sin 2t + 16(cos 0.5t - sin 1.5t). Resonance for wl(27T) = 2/(27T) = 1/7T 29. w = 3.1 is close to Wo = -vkj;, = 3, Y = 25(cos 3t - cos 3.1t). 31. R = 9,0, L = 0.5 H. C = 0.025 F, E = 17 sin 6t V, hence 0.5/" + 91' + 401 = 102 cos 6t, 1= -8.16e- 8t + 7.5e- lOt + 0.66 cos 6t + 1.62 sin 61 33. E' = 220·314 cos 314t, I = e- 50t(A cos 150t + B sin 150r) + 0.847001 sin 3141 - 1.985219 cos 314t Problem Set 3.1, page 111 7. Linearly independent 11. xlxl = x2 if x > 0, linearly dependent 13. Linearly independent 17. Linearly independent 9. Linearly dependent 15. Linearly independent 19. Linearly dependent Problem Set 3.2, page 115 + 11y' - = 3. yiv - Y = 0 7. C1 + C2 cos x -I:=. C3 sin x 9. C1ex + (c2 + c3x)e-·" 11. C1 ex + c2il+v7lX + c 3e(1-\ 7)x 13. eO. 25x + 4.3e-O. 7X + 12.1 cos O. Lt - 0.6 sin O.Ix 15. 2.4 + e-1.6x(cos 1.5x - :2 sin 1.5x) 17. y = cosh 5x - cos 4x 19. y = c 1x- 2 + C2X + C3X2. W = 121x2 1. Y'" - 6y" 5. /v + 4-,"" = 6-," 0 0 Problem Set 3.3, page 122 1. (Cl + c2x)e2x + C3e-2.T - 0.04e- 3x + x 2 + X + 3. c] cos ~x + C2 sin ~x + X(C3 cos ~x + C4 sin ~x) - ~e-x sin !x 5. C1XO.5 + C2X + C3X1.5 + 0.1.\.5·5 7. C] cos X + C2 sin x + C3 cos 3x + C4 sin 3x 9. Y = (4 - x 2)e 3X - 0.5 cos 3x + 0.5 sin 3x 11. x- 2 - x 2 + 5x 4 + x(In x + I) 13. 3 + ge- 2x cos 9x - (1.6 - 1.5x)e X + 0.2 cosh 2x Chapter 3 Review Questions and Problems, page 122 7. Cl + C 2x 1/2 + C3X-1/2 9. cle-O. 5x + C2 e O. 5X + C3e-L5x 2 11. Clx (i In x - ~) + C2X2 + c3.,· + C4 + tx 7 13. C1e-x + e X / 2(c2 cos (~V'3x) + C3 sin (~v'3x» + 8ext2 15. (c1 + c2x)eX + C3e-x + 0.25x 2e x 17. -0.5x- 1 + 1.5x- 5 19. cos 7x + e 3x - 0.02 cosh x App. 2 An Answers to Odd-Numbered Problems Problem Set 4.1, page 135 1. Yes 5. y~ = 0.02(-)'1 + )'2), y~ = 0.02(\'1 - 2)'2 + )'3)')'~ = 0.02(.1'2 - )'3) 7. Cl = 1, C2 = -5 9.3 and 0 2t 11. )'~ = )'2, y~ = 4Yb .1'1 = c 1e- + C2 e2t = y, Y2 = v~ 13. y~ = Y2' y~ = )'2, eigenvalues 0, 1')'1 = Cl + C2 e t, Y2 = y~ = y' IS.)'~ = )'2, )'~ = 0.109375.1'1 + 0.75)'2 (divide by 64). Yl = cle-O.125t + c2eO.875t Problem Set 4.3, page 146 + C2e 6t , )'2 = -2Cle-6t + 2c2e6t 3')'1 Cle2t + C2, )'2 = Cl e2t - C2 5. Yl = Cle4it + C2e-4it = (Cl + C2) cos 41 + i(CI - C2) sin 41 = A cos 4t + B sin 4t, Y2 = iC1e4it - iC2e-4it = (iCI - iC2) cos 4t + i(iCI + iC2) sin 4t = B cos 4t - A sin 4t, A 1. )"1 = = c 1e- 6t = Cl + C2. B = i(CI - C2) 2e 1 + C2e-6t • .1'2 = -Cl + C3e-6t, )'3 = -Cl + 2(C2 + c 3 )e-6t 9f + 2c e-1. 8t , )'2 = 2Cle1.8t + c e- O.9t - 2C3e-1.8t, )"1 = c1e1.8t + 2c2e-O. 2 3 )'3 = 2Cle1.8t - 2C2e-O.9t + C3 e -1. 8t 11')'1 = 10 + 6e 2t , )'2 = -5 + 3e 2t 13. )"1 = 2.4e- t - 2e2.5t , .1'2 = 1.8e- t + 2e2.5t 15')'1 = 2e 14.5t + 10, Y2 = 5e 14.5t - 4 7. 9. )'1 = 17. )'2 = .1'~ + Yb Y~ = )'~ + )'~ = -)'1 - Y2 = -Yl - ()'~ + .1'1), y~ + 2y~ + 2Y1 = 0, Y1 = e-t(A cos 1 + B sin t), )'2 = y~ + )'1 = e-t(B cos t - A sin t). Note that r2 = Y12 + .1'22 = e- 2t (A 2 + B2). 19.11 = 4c1e-200t + C2 e - 50t , 12 = -Cle-200t - 4C2e-50t Problem Set 4.4, page 150 1. Saddle point, unstable, .1'1 = Cle-4t + c 2e 4t, )'2 = -2c 1e- 4t + 2c2 e 4t 3. Unstable node . .1'1 = Clet + C2e3t, )'2 = -Clet + C2e3t 5. Stable and attractive node, .'11 = Cle-3t + C2e-5t, .'12 = c 1e- 3t - c 2 e- 5t 7. Center, stable'.\'1 = A cos 41 + B sin 4t, .1'2 = -2B cos 4t + 2A sin 4t 9. Saddle point, unstable, .1'1 = Cle3t + C2e-t, .1'2 = C1e3t - C2e-t 11.)\ = Y = c 1e kt + c 2 e- kt ')'2 = y', hyperbolas k 2.'112 - Y2 2 = const 13. )' = e- 2t (A cos t + B sin t), stable and attractive spirals 17. For instance, (a) -2, (b) -1, (c) -~, (d) 1, (e) 4. Problem Set 4.5, page 158 (9, 0), y~ = )'2, Y~ = 3.1'1, saddle poim; (0, -1)')'1 = 5\')'2 = -1 + Y2, y~ = -Y2, )1 = 35\. center 3. (0, 0), Y~ = 4.1'2. Y~ = 2Yl, saddle point; (2, 0), Yl = 2 + )lb )'2 = )12' Y~ = 4Y2, y~ = -2Yl' center 5. (0, 0), Y~ = -Yl + )'2, y~ = - Yl - .1'2' stable and attractive spiral point; (-2, 2), )'1 = -2 + )11> Y2 = 2 + Y2, Y~ = -.h - 3.V2, y~ = -j\ - )12, saddle point 1. App. 2 7 • .\'~ Answers to Odd-Numbered Problems = .\"2' Y~ = -.\"10 - 4)'1), (0, O),)'~ = .\"2, .,.~ = -."1, center; + Yb )'2 = 5'2, Y~ = 5'2, Y~ = (-! - 5'1)( -4Yl)' y~ = (!, 0)'.\'1 = ! Yl' saddle 9. (~7T ::!: 2117T, 0) saddle points; (-~7T ::!: 2117T, 0) centers. Use -cos (::!:~7T + :VI) = sin (::!:5'1) = ::!:Yl' 11 . .\"~ = .\'2, y~ = -)'1(2 + )'1)(2 - .\"1)' (0, 0), y; = -4.\"1' center; (-2, 0), y~ = 85'1' saddle point; (2, 0), y~ = 85'1 saddle point 13. )'''/y' + 2y' /)' = 0, In y' + 2 In Y = c, y' y2 = Y2Y] 2 = const 15.), = A cos t + B sin t, radius YA 2 + B2 Problem Set 4.f page 162 3. Yt = A cos 4t + B sin 41 + ~~, )'2 = B cos 4t - A sin 4t - ~t 5. )'1 = Cle4t + C2e-3t + 4, Y2 = c 1e 4t - 2.5c2e-3t - 10 7. Y1 = 2cle-9t + C2e-4t - 90t + 28'.\"2 = Cle-9t + C2e-4t - J26t + 14 9')'1 = Clet + 4c2e 2t - 3t - 4 - 2e- t ')'2 = -Clet - 5C2e2! + 5t + 7.5 + e- t 11')'1 = 3 cos 2t - sin 2t + t + 1,)'2 = cos 2t + 3 sin 2t + 2t - ~ 13'."1 = 4e- t - 4et + e 2t'."2 = -4e- t + t 15'."1 = 7 - 2e 2t + e 3t - 4e- 3t , )'2 = _e 2t + 3e- 3t 17. I~ + 2.5(ft - 12) = 845 sin t, 2.5(1~ - I~) + 2512 = 0, 11 = (95 + 162.5t)C 5t - 95 cos t + 312.5 sin t, 12 = (-30 - 162.5t)e-5t + 30 cos t + 12.5 sin t 19. I~ + 2(11 - 12) = 200, 2(12 - It) + 812 + 2 I 12 dt = O. II = 2cleA,t + 2C2e"~2t + 100, 12 = (1.1 + Vo:4T)cleA,t + 0.1 - \"0.41)C2eAzt, Al = -0.9 + \,'0.41, A2 = -0.9 - v'6AT Chapter 4 Review Questions and Problems, page 163 11')'1 = Cle8t + C2e-8t, )'2 = 2c 1e 8t - 2c2e-8t. Saddle point 13. ."1 = Clet + C2e-6t, ."2 = Clet - 6c2e-6t. Saddle point 15. )'1 = Cle7.5t + C2e-3t, Y2 = -Cle7.5t + 0.75c2e-3t. Saddle point 17. ."1 = Cle5t + C2et, .\'2 = Cle5t - c2e t . Unstable node 19. Yl = e-t(A cos 2t + B sin 2t), )'2 = e-t(B cos 2t - A sin 2t). Stable and attractive spiral point 21. )'1 = Clet + C2e-t + e 2t + e- 2t , )'2 = -C2e-t - 1.5e- 2t 23')'1 = Clet + C2e-2t - 6e- t - 5. Y2 = -Clet - 2c2e-21- + 10e- t + 6 25• .\"1 = Cle3t + C2e-t + t 2 - 2t + 2, )'2 = c 1e 3t - C2e-t - t 2 + 2t - 2 27. A saddle point at (0, 0) 29. I] = 4e- 40t - e- lOt , 12 = _e- 40t + 4e- lOt 31. (117T, 0) center for even 11 and saddle point for odd n 33. Saddle points at (0, 0) and (~, ~), centers at (0, ~) and (~, 0) (oblem Set 5. 1 page 170 1. aoO + x + ~X2 + ... ) = aoe x 3. aoO - 2X2 + ~X4 - + ... ) + al(x - ~x3 = ao cos 2,r + ~al sin 2x + I~X5 - + ... ) App. 2 AU Answers to Odd-Numbered Problems 5. ao(1 + ~x) 7. ao + aox + (~ao + ~)x2 + ... = aoe x + eX - x - I = ce x 9. ao + a1x + ~alx2 + ... = ao - a l + alex 11. s = ~ - 4x + 8x 2 - 3ix 3 + 3ix 4 - I;:X5, s(O.2) = 0.69900 13. s = ~ + ~x - 1s x 3 + ~OX5, sCI) = 0.73125 15. s = I + x - x 2 - ~X3 + ~X4 + ~!X5, s@ = ~~~ x-I, c = ao + 1 Problem Set 5.2, page 176 1. lei 5.0 3. 2 (as function of t = lx - 3)2). Ans. 9. 1 13. V2 7. 2 11. 7T 15. L (_1)S-l L xS'R 5(s - 2) s~3 = 1 , (s - 4)2 s~5 (s-3)! xS ' R = Cf) ' + tx 3 + 2~X4 - :14x5 - ... ) 19. ao + aI(x - ~X3 + ~X5 - 2i x 7 + 227X9 - I~5Xll + - ... ) 21. lIo(l - ~X2 - :14x4 + 7I~ox6 + ... ) + aI(x - tx 3 - 214X5 + 1O~SX7 + ... ) 23. ao(1 + x 2 + x 3 + X4 + x 5 + x 6 + ... ) + alx 17. ao(l - l2X4 _lo-\:5 - ... ) + a1(x + ~x2 Problem Set 5.3, page 180 3. P 6lx) = I~l231x6 - 315x4 + 105x2 - 5), P 7lx) = I~(429x7 - 693x 5 + 315x 3 - 35x) 7. Set x = az.. y = cIPn(x/a) + C2Qn(x/a) =~. P 21 = 3x~, P22 2 P 4 = (l - x 2)(105x2 - 15)/2 15. P l 1 = 3(1 - x 2), Problem Set 5.4, page 187 1. VI = 1+ 3•."1 = 1 - x2 ~ 3! ~ 12 5. r(r - 1) + 4r + X4 ~ 5! I ~-.r4 384 + + 1 2 + ... = - sinh x x . .. Y2 v = 9". 1 In '.2 = x x - 2 = 0, r l = -1, r2 = -2; Y1 = + 24 720 + x 2! 1M 3 + x + .,. ~ 4! 36 x x 6 + x2 2 rl 25x 4 1024 + ... , 120 + - ... 7. Euler-Cauchy equation with t = x + 3, h = (x + 3)5, Y2 = ."1 In (x 9. ho = 1, Co = 0, r2 = 0, Yl = e- x , ."2 = e- x In x 11'."1 = 1I(x + 1), .1'2 = 1Ix 13. ho = ~, Co = 0, cosh x x -2+ X4 = + 3) = ~, r2 = 0'."1 = x I/2 (1 + 2x + 2X2 + ~X3 + ... ), Y2=I +2x+2x + ... 15'."1 = (x - 4)7, .1'2 = (x - 4)-5 (Euler-Cauchy with t = x - 4) 3 17'."1 = X + x - I~X4 + I~X5 - 2~X6 + .. " Y2 = 1 + 3x 2 - tx 3 2 + ~~x6 -+ ... + ~X4 - ~x5 + ... 14 App. 2 Answers to Odd-Numbered Problems = c1F(~, ~, ~; x) + C2 -yt;:F(l, 1,~; x) 21. y = A(l - 4x + ~X2) + B-yt;:F( -~, ~, ~: x) 23. y = c1F(2, -2, -~; t - 2) + C2(t - 2)3/2F(~, 19. Y -~,~; t - 2) Problem Set 5.5, page 197 1. Use (7b) in Sec. 5.2. 3.0.77958 (exact 0.76520), 0.19674 (0.22389). -0.27651 (-0.26005). -0.39788 (-0.39715), -0.17038 (-0.17760), 0.15680 (0.15065), 0.30086 (0.30008),0.16833 (0.17165) 5. Y = C1I,,(Ax) + c2L,,(Ax:), v*" 0, ± I, .. . 7. Y = C1I,,(-yt;:) + C2LvC-yt;:), v O. ± L .. . 9. Y = C1xI1(2x), II, I_I linearly dependent 11. y = X-"lC1I,,(X) + C2I_,,(X)], 0, ±1, .. . 13. Y = C1I,,(x3) + C2L,,(x3), v =1= 0, ± 1. .. . 15. Y = c]-yt;:1 1(2-yt;:), I]. I_I linearly dependent 17. Y = Xl/4ft(~Xl/4), II' L1 linearly dependent *' v*' 19. y = .~/\C1I8/5(4xl/4) + c2I_s/5(4x l/4» = 0, (24a) with v = I, (24d) with v = 2, respectively. 23. I n (X1) = I,,(x2) = 0 implies x 1- n 1,,(X1) = X2 -71 I,,(x2) = 0 and [x- n 1 n (x)]' = 0 somewhere between Xl and X2 by Rolle's theorem. Now use (24b) to get 1 n + l (x) = 0 there. Conversely, 1,,+ 1(X3) = 1,,+I(X4) = 0, thus X3n-lIn+I(X3) = X4n+11n+1(X4) = 0 implies 1n(x) = 0 in between by Rolle's theorem and (24a) with v = 11 + L 25. Integrate the formulas in (24). 27. Use (24a) with JJ = 1, partial integration, (24b) with v = 0, partial integration. 33. CAS Experiment. (b) Xo = I. Xl = 2.5, X2 = 20, approximately. It increases with (c) (14) is exact. (d) It oscillates. (e) Formula (24b) with v = 0 21. Use (24b) with v Problem Set 5.6, page 202 1. )' = C]15(X) + C2 Y5(X) 3. Y = C1I O(-yt;:) + C2 Yo(-yt;:) 2 5. Y = C112(X2) + C2 Y2(X ) 7. y = X-\CI15(X) + C2Y5(X» 11. Set H(l) = kH c21, use (10). 9. Y = X3(c1I3(X3) + C2 Y3(X3» 13. Set x = is in (l), Sec. 5.5, to get the present ODE (12) in terms of s. Use (20), Sec. 5.5. Problem Set 5.7, page 209 3. Set x = ct + k. 5 ..\ = cos e. dx = -sin e de, etc. 7. Am = (11lr./5)2, 111 = 1,2, ... ; Ym = sin tll1r.x/5) 9. Am = [(2m + 1)r.12L]2, m = 0, 1. ... : Ym(x) = sin [(2m + l)m/2L] 2 11. Am = 111 , m = 0, 1, ... ; Yo = 1, Ym = cos II1X, sin IIlX, III = 1,2, ... 13. k = k m from tan k = -k. Am = k m 2 , III = 1,2, ... ; Ym = sin kmx 15. Am = 1112, 111 = 1, 2, ... ; Ym = x sin (m In /x/) 17• j'J -- e Sx, q -- 0, r -- e Sx, Am , . IIIX, III = I , 2,... = III 2 ; Ym = e -4x SIn 19. Am = (1Ilr.)2, Ym = X cos lIlr.x, X sin 111r.X, m = 0, 1, ... n. App. 2 A15 Answers to Odd-Numbered Problems Problem Set 5.8, page 216 3. ~P3(X) - ~P2(X) 1. 1.6P4 (x) - 0.6Po(x) 7. -0.4775P l (x) - 0.6908P3(x) + ~Pl(X) - ~Po(.x) + 0.1544Pg (x) + ... , + 1.844P5 (x) - 0.8234P7 (x) = 9. Rounding seems to have considerable influence in Probs. 6-15. 9. 0.3799P2(x) + 1.673P4 (x) - 1.397P6 (x) + 0.3968Ps(x) + ... . 1110 = 8 11. 1.175Po(x) + 1.l04Pl (x) + 0.3575P2(x) + 0.0700P3(x) - .... 1Il0 = 3 or 4 1110 13. 0.7855Po(x) - 0.3550P2(x) + 0.0900P4 (x) - ... , = 4 1110 15. 0.1212Po(x) - 0.7955P2(x) + 0.9600P4 (x) - 0.3360P6 (x) 2 17. (c) am = (2IJ1 (aO;rn»(J1(aO. m )/ao,m) = 2/(ao. m ll(aO. m » + .... 1Il0 = 8 Chapter 5 Review Questions and Problems, page 217 11. e 3x , e- 3x , or cosh 3x, sinh 3x 13. eX, 1 + x 15. e-X'-, xe- x2 17. e- x , e- x In x 2 2 19. I/(l - x ), x/(l - x ) or 1/(1 - x), 1/(1 + x) 21. Y = ("11V2(6x) + c21 _ v2(6x) 23. Y = Clit (x2) + 25. y = ~[clll/4(~kx2) + ['21 _l/4(!kx 2)J 27. Am = (21117Tl, Yo = I, Ym = cos nl17TX, sin 2m7Tx, I, 2, ... 111 = 2 C2 Yl (x ) o. 29. Y = c l l l (kx) + C2Yl(kx). C2 = O. y(l) = c l l l (k) = k = k m = al,m (the positive zeros of 1 1 ), Ym = l l (a1.",x) 31. 1.813Po(x) + 2.923P l (x) + 1.759P2(x) + 0.663P3(x) + 0.185P4 (x) + ... 33. 0.693Po(x) - 0.285P2(x) + 0.144P4 (x) - 0.09IP6 (x) + ... 35.0.25Po(x) + O.5P1 (x) + 0.3 125P2(x) - 0.0938P4 (x) + O.0508P6 (x) + ... Problem Set 6.1, page 226 2 2 s s s cos () - w sin () 7. ---;;2:------:2;:--- + S k 13. s (1 - s-2 s 1."""3-"""2 w e- bs ) 5. 15. 1 1 9.--s + 2b 1 - (1 2 (s - 2) - 11.-S2 + 4 1 - e- bs be- bs 17. ---;;:--- - - - + 2s)e- 2S ----;;2--- 2s s S2 (I - e- s )2 19. - - - s 23. Set ct = p. Then :£(f(ct» = {'oC e-stf(ct) dt L=e-(s'C)Pf(p) dp/c = = F(s/c)/c. 0 29. 4 cos 7Tt - 3 sin 7Tt 35.2 - 2e- 4t 39. 45. 1 Vs sin ,,1st - 37. (ev'3t - e-V5t)/(V3 e- 5t 41. (s - 3.8 2 2.4) + k) + b 2 + k) + 1 a(s (s 51. 3e- 2t sin 5t 53. e- 5 "1Tt sinh 7Tt + Vs) 5w 43. (s 2 + a) + w 2 A16 App. 2 Answers to Odd-Numbered Problems Problem Set 6.2, page 232 1 1. (s - k) 2 7. (S2 + ~17"2)2 9. Use shifting. Use cos 2 0:' = ~ + ~ cos 20:'; use cos 2 0:' + sin2 0:' = 1. Ans. (2S2 + 1) 1[2s(S2 + 1)] 11. (s + ~)Y = -1 + 17· 21(s2 + 4), y = 7e-t{2 + 2 sin 2r - 8 cos 2t 13. (S2 - ~)Y = 4s, y = 4 cosh ~t 15. (S2 + 2s + 2)Y = s - 3 + 2 . 1. Y = (s + 1 - 2)/[(s + 1)2 + 11, y = e-t(cos t - 2 sin t) 17. (S2 + 7s + 12)Y = 3.58 - 10 + 24.5 + 211(s - 3), Y = ~e3t + ~e-4t 19. (s + 1.5fY = s + 31.5 + 3 + 541s 4 + 64ls, Y = lI(s + 1.5) + lI(s + 1.5)2 y = (1 + t)e-1. 5t + 4t3 - 16t2 + ~e-3t 241s4 - 321s 3 + 321s2, 32t 21. t = t + 2, l' = 4/(s - 6), ji = 4e 6t, y = 4e 6Ct - 2) 23. t = t + 1, (s - l)(s + 4)1' = 4s + 17 + 6/(s - 2), Y = 3et - 1 + e 2Ct - ll 25. (b) In the proof, integrate from 0 to 1I and then from II to 0: and see what happens. (c) Find 3:;(f) and 3:;(f') by integration and substitute them into (1 *). 27.2 - 2e- t/2 29. + + 1 ""k.2 (e kt ~ r;:: 31. cosh v5t - 1 t k - 1) - 33. t sinh 2t - ~t Problem Set 6.3, page 240 3. (l - e 2 - 2S (~ S3 ~ S2 5. 7. + s S2 + 17"2 ) I(s - 1) + ~) e- s - (~ S S3 + ~ S2 e (-S21 + -10) s ( -e -s - e -4s) -20s (e- 3s S2 + 17"2 1 13. - - (e- 2s + 27T _ e- 4s + 47T ) s - 17" 15. 0 if t < 4, t - 4 if t > 4 17. sin t if 217" < r < 817", 0 elsewhere 19. 0 if t < 2, (t - 2)4/24 if t > 2 21. lI(t - 3) cosh (2r - 6) t 23. e- sin t 25. e- 2t cos 3t + 9 cos 2t + 8 sin 2t 27. sin 3t + sin r if 0 < t < 17" and ~ sin 3t if t > 17" 29. t - sin t if 0 < t < 1, cos (t - 1) + sin (t - 1) - sin t if t > I 31. et - sin t + u(t - 217")(sin t - ~ sin 2t) 33. t = 1 + t, 'f" + 4ji = 8(1 + t)2(1 - lI({ - 4», cos 2t + 2r2 - I if t < 5, cos 2r + 49 cos (2r - 10) + 10 sin (2t - 10) if t > S 35. Rq' + qlC = 0, Q = 3:;(q), q(O) = CVo, i = q' (r). R(sQ - CVo) + QIC = 0, q = CVoe- tlCRC ) 11. 37. IO[ 100 + -[ + e- 6s ) 100 = - - e- 2s , [ s S2 1 - e- lOCt - 2 ) if t > 2 1= e- 2s ( S s ] + 10 ), i = 0 if t < 2 and -lOs App. 2 Answers to Odd-Numbered Problems = 39. i e- 20t + + lI(t - 2)[ - 20t + 1 + 3ge-20Ct-2J] 5t 490e- [l - lI(t - 1)], i = 20(e- 5t - e- 250t ) 20t - 1 41. O.Ii' + 25i = + e-250t+245] 43. i (10 sin lOt = 45. i' + + 2i 2 A17 f + 20£1(1 - 1)[-e- 5t + 100 sin t)(lI(t - 7T) - u(t - 37T)) t i(T) dT = 1 - 1I(t - 2), I = (1 - e- 2S )/(s2 + + 2s 2), ° i = e- t sin t - lI(t - 2) e-t+2 sin (t - 2) = 27 cos t + 6 sin t + lI(t - 27T) [- 27 cos t 47. i e- t (27 cos 3t + 11 sin 3t) - 6 sin t + e-(t-2TI"J(27 cos 3t + 11 sin 3t)1 Problem Set 6.4, page 247 1. Y = 10 cos t if 0 < t < 27T and 10 cos t + sin t if t > 27T 3. Y = 5.5e t + 4.5e- t + 5(e t - 1I2 - e- t +1/2 )u(t - ~) - 50(e t - 1 - e-t+l)lI(t - 1) 5. Y = O.lfe t + e- 2 t(-cos t + 7 sin t)] + O.llI(t - lO)[-e t + e- 2t+ 3 0(cos (t - LO) - 7 sin (t - 10))] 7. Y = 1 + le- t sin 3t + lI(t - 4)[-1 + e-t+ 4 (cos (3t - 12) + l sin (3t - 12))] l~ lI(t - 5)e -t+5 sin (3t - 15) 9. Y = 5t - 2 - 501l(t - 7T)e- t +7r sin 2t. Straight line. sharply deformed between 7T and about 8 11. Y = (O.4t + 1.52)et + 0.48e- 4t + 1.611(t - 2)[ -et + e- 4t +1O ] Problem Set 6.5, page 253 1. t 1 5. - sin wt I kt 7. 2k (e w 9. ~(e3t - e- 5t ) ll. 13. t - sin t 15. 19. Y = 3/((S2 + 4)(S2 + 9», y = 0.3 sin 2t 21. (S2 + 9)Y = 4 + 8(1 + e-=)/(s2 + 1), y = 23. 0 if 0 < t < 1, ~ f t sin (2( T - 1)) dT = ~(~ - e- - kt ) ! cos 2t) = L k = ~ sinh kl sin 2 t t(cosh 3t - 1) 0.2 sin 3t sin t + sin 3t if t < 7T, ~ sin 3t if t -~ cos (2t - 2) + ~ if t > > 7T 1 1 25. Y = 2e- 2t - e- 4t + (e- 2t + 2 - e- 4t + 4)u(t t 27. Y - 1 * y = 1, y = e 31. Y(1 + l/s2) = lis, y = cos t L) + (e- 2t + 4 - e- 4t +S )1I(t - 2) 29. y - y * sin t = cos t, Y = lis, y = 1 33. Y(1 + 2/(s - 1) = (s - 1)-2, Y = sinh t Problem Set 6.6, page 257 4 1. 2 (s - 1) +4 4s + 5)2 2w(3s 2 - w 2) 9. (S2 + W2)3 2s 5. (S2 + 3. (S2 2ws + w2)2 24s2 + 128 7. (S2 - 16y3 2s cos k 11. + (s 2 (S2 - I) sin k + 2 1) A18 App. 2 Answers to Odd-Numbered Problems 15. te- 2t sin t 19. In s - In (5 - 1); (e t 13.6te- t 2 kt 17. t e - l)/t Problem Set 6.7, page 262 1. .'"1 = -e- t sin t, )"2 = e- t cos t 3 ..'·1 = 2e- 4t - 4e 2t , )"2 = e- 4t - 8e 2t 5'.'"1 = 2e- t + 4e- 2t +!t -~, )"2 = -3e- t - 4e- 2t - ! t + ~ 7. )"1 = e- t (2 cos 2t + 6 sin 2t) + t 2• .'"2 = lOe- t sin '2t - t 2 9• .'"1 = 4 cos 5t + 6 sin 5t - 2 cos t - 25 sin t. .'"2 = 2 cos 5t - 10 sin 5t + 20 sin t 11'.'"1 = -cos t + sin t + I + 1I(t - 1)1-1 + cos (t - I) - sin (t - 1)1 )"2 = cos t + sin t -I + 1I(t - l)ll - cos (t - 1) - sin (t - 1)] 13. Y1 = 2u(t - 2)(e 4t - e t + 6 ), Y2 = e 2t + ll(t - 2)(e 4t - 3e2t + 4 + 2e t + 6) 15. Yl = _e- 2t + et + ~llU - I)(-e- 2t + 3 + et ), Y2 = _e- 2t + 4e t + ~lllt - l)l-e- 2t + 3 + et ) 17')"1 = 3 sin 2t + 8e- 3t , )'2 = -3 sin 21 + 5e- 3t 19')'1 = e t - e- t , Y2 = et , Y3 = e- t 25.4i 1 + 8(il - i2) + 2i~ = 390 cos t, 8i2 + 8(i2 - i 1 ) + 4i~ = 0, il = - 26e- 2t - 16e- 8t + 42 cos t + 15 sin t, i2 = -26e- 2t + 8e- 8t + 18 cos t + 12 sin t Chapter 6 Review Questions and Problems, page 267 11. 17. 13. (s - 3)2 S (s - 1)(5 2 23. 10 cos + 4) tV2 29. te- 2t sin t 19. 2 S(S2 + 4) 15. e-m ( 7T 5 25. 3e- 2t sin 4t 31. (t2 - l)u(t - 1) 21. s12 ) a-b 2S2 -4-5 - I + (5 - (1)(5 - b) 27. lI(t - 2)(5 + 4(t - 2» 7T 33. ---:3 (wt - sin wt) w 35.20 sin t + li(1 - 1)[1 - cos (t - I)] 37. 10 cos 21 - ~ sin 2t + 4u(t - 5) sin (2t - 10) 39. e- t (7 cos 3t + 2 sin 3t) t 41. e- + u(t - 7T)[1.2 cos t - 3.6 sin t + 2e-t+7T - 0.8e2t - 2"] 43. u(t - l)(t - l)e 2t - 2 + 4u(t - 2)(2 - t)e 2t - 4 45. Y1 = e t + ~e-t - ~ cos t - ~ sin t, )'2 = -et + ~e-t + ~ cos t + ~ sin t 47. Y1 = ~e-t sin 2t, )"2 = e-t(cos 2t - ~ sin 2t) 49. )'1 = e 2t , )'2 = e 2t + et 51. I = (l - e- 2S )/[s(s + 10)], i = 0.1 (I - e- lOt ) + O.lu(t - 2)[-1 + e- lOt + 20 ] 53. I = e- 2t(76 cos 4t - 42 sin 4t) - 76 cos 20t + 16 sin 201 55. i~ + IOUI - i2) = 100 t 2• 30i~ + 1O(i~ - i~) + 100i2 = 0, ;1 = (~ + 4t)e- 5t + 10t2 -~, i2 = (~ + 2t)e- 5t + 2t - ~ Problem Set 7.1, page 277 App. 2 A19 Answers to Odd-Numbered Problems -:l[-: -1] [-0 3. Undef., ~ 6 o 9 5. [-48 -2] [ 38 -44 67 -15 . -33 9. 24 72 60 smne, . same. -5.0 -4.9 1.8 -3.6 3.5 -48 -34] 3.8 0.4 [_~::] -2.2 2X2 = 4 = 0 4X2 24 [-03 = -3 -5X2 -3Xl 48] 0 3(, -4.2 + + -5Xl -9 -12 7. [ 6~], [ :.2]' , undef. Problem Set 7.2, page 286 64 [ 230 -92 -92 38 -12 6 5. [20 -3 -114 -114 126 -1:] -7], [-62 34 2], [50S] 525 ,same 790 5 7. r : 0 ~] [-310 8 ,31, -62 0 16 0 170 34 -124 337 8 9. [ 252 49 -100] -308 52 233 -68 , same, 10]; ,same 68 257 68 232 97 -248 -16 [ -188J -96 265 -72] , 110 A20 App. 2 11. [ 13. 19. 21. 23. 27. Answers to Odd-Numbered Problems 324 32 2M 38 -244 -10 216 -104 280 -132 -280 140 -320] [ -322. 366 4324 1520 [ 3636 -4816] 1242 -451!S -3700 -1046 5002 7060 960 -5120] 7548 1246 -5434 -8140 -1090 6150 1M] [ -68, 76 83, 166, 593, 0 (d) AB = (AB)T = BTAT = BA; etc. (e) AilS. If AB = -BA. Triangular are U 1 + U 2 , U 1 U2 , U 12 , L1 + L 2 , L 1 L 2 , L12. [0.8 1.2]T, [0.76 1.24]T, [0.752 1.248]T P = [110 45 801 T, v = [92000 863001 T Problem Set 7.3, page 295 = 2.5, Y = -4.2 x = 0, y = - 2, z = 9 1. x 3. x = 0.2, y = 1.6 7. x = 4, Y = 0, z = -2 x = 3y + 2, y arb.,.: = -y + 6 11. Y = 2x + 3.: + I, x, .: arb. w = I, Y = 2.: - x, x, .: arb. 15. w = 3, x = 0, y = -2, ;;: = 8 h = (R 1 + R2)Eo/(R 1R 2), 12 = Eo/R1' 13 = Eo/R2 [Amps] 11 - 12 - 13 = O. (3 + 2 + 5)1t + 10/2 = 95 + 35, 10/2 - 5/3 = 35, 11 = 8, 12 = 5. 13 = 3 Amps 21. Xl + X4 = 500, Xl + X2 = 800. X2 + X3 = 1100, X3 + X4 = 800, Xl = 500 - X4, X2 = 300 + X4, X3 = 800 - X4, X4 arbitrary 5. 9. 13. 17. 19. Problem Set 7.4, page 301 1. L [I 3.3, [I [0 0 5. 2, [3 7.2, r8 9. 3, II 11. 4, [I 13. No 19. Yes 29. No 35. I, [5 -2]; [1 4 0 -3]T 7], [0 -2 0 I 3], rO 0 5 105]; r -2 -I- 5]T, [0 I 5]T, I]T 0 0 0 0 5], 4], 3 0 [0 3 [0 2 0], [0 0], [0 41; [3 0 5]T, [0 3 4]T OJ; [8 0 4 O]T, [0 2 0 41T 5 8 -371, [0 0 -74 296]; same transposed o 0], [0 0 1 0], [0 0 0 I]; same transposed 15. No 17. Yes 21. (c) I 27.2, [I -I OJ, lO 0 1] 31. I, [-i l 1] 33. No Problem Set 7.7, page 314 5. 107 9. -66.88 13. 113 + v 3 + w 3 - 311VW 19. x = -1.2, Y = 0.8, .: = 3.1 23.3 7. cos (a + f3) 11.0 15.4 21. 1 App. 2 Answers to Odd-Numbered Problems A2l Problem Set 7.8, page 322 1. [ 1.80 3. - 2.32J -0.25 0.60 [COS '28 -sin 28J sin 28 9. A-I 15. (A'j-L (A-'j' ~ [I: -~ cos 28 =A 11. No inver~e =] 19. AA -1 = I, (AA -1)-1 = (A -l)-lA -1 = I. Multiply by A from the right. 21. det A = - I. C 12 = C21 = C33 = - I, the other Cjk are zero. 23. uet A = 1. Cl l = I, C 12 = -2, C22 = 1, C 13 = 3. C23 = -4, C33 = I Problem Set 7.9, page 329 1. Yes, 2, [3 5. Yes, 2, [0 5 0 -51 T 1 OIT, [0 0 OIT, [2 0 11. Yes, 2, xe-:r , e- x 13. [I O]T, [0 I]T; [I I]T, [-1 15. Xl = -0.6Yl + 0A.Y2 X2 = -0.8y! + 0.2.'"2 19. Xl = 5-"1 + 3.'"2 - 3)'3 X2 = o 0 [_~ ~J 7. Yes, 1, X3 3. No 0 3.'"1 + = 2.'"1 - 2.1'2 - )"2 + l]T 9. No I]T; [I O]T, [0 17. Xl = x2 = -l]T 2.1"1 5\" . 1 + Y2 + 3\".2 2)"3 2.'"3 21. Vs6 25. 2 23. 16Vs 29. 4V1 - 3V2 = 0, V = ± [53 15]T Chapter 7 Review Questions and Problems, page 330 11. X = 4, Y = 7 15. x = ~. Y = -~. z = ~ 19. x = 22. Y = 4, 2 arbitrary 23.638.0,0 13. x = v + 6, z = y, )" arbitrary 17.x=7.)"= -3 21. 0 25. 8.0 -3.6 1.2 -3.6 2.6 2.4 1.2 2.4 9.0 [ 12 27. 14, 14. 2; [ o o o r J 20 29. [-20 9 -3], _: ] A22 App. 2 Answers to Odd-Numbered Problems 31.2,2 33.2.2 35.2,2 37. 39. 4~ [~~ 1~ 23 _ 5] 42 41. -10 5\ [ 4 -5 ,~ r :] 72 -72 31 -3:2 -19 20 132] 59 -35 43. It = 33 A, 12 = 11 A, 13 = 22 A 45. II = 12 A, 12 = 18 A, 13 = 6 A Problem Set 8.1, page 338 1. -2, [1 O]T; 0.4, [0 I]T 3.4,2\"1 + (-4 - 4)X2 = 0, say, Xl = 4. X2 = 1; -4, [0 I]T 5. -4, [2 9]T; 3, [1 I]T 7.0.8 + 0.6i, [1 _i]T; 0.8 - 0.6i, [l i]T 9.5, [1 2]T; 0, [-2 I]T 11.4, [I 0 O]T; 0, [0 1 OlT; -I, [0 0 I]T 13. -(A3 - I8A2 + 99A - I62)/(A - 3) = -(A2 - I5A + 54); 3. [2 -2 I]T; 6, [l 2 2]T; 9, [2 I _2]T 15. I, [-3 2 lO]T; 4, [0 1 2]T; 2, [0 0 I]T 17. -(A3 - 7A2 - 5A + 75)/(A + 3) = -(A2 - lOA + 25); -3, [I 2 _I]T; 5, [3 0 l]T, [-2 I O]T 19. -(A - 9)3; 9, [2 -2 I]T; defect 2 21. MA 3 - 8A2 - 16A + I28)/(A - 4) = A(A 2 - 4A - 32); 4. [-1 3 1 l]T; -4,[1 1 -I -l]T;O.[I I L 1]T;8, [1 -3 1 -3]T 23.2, [8 8 -16 l]T; 1, [0 7 0 4]T; 3, [0 0 9 2]T, -6, [0 0 0 l]T 25. (A + IfcA2 + 2A - 15); -I, [I 0 0 O]T, [0 1 0 O]T; -5, [- 3 -3 1 l]T, 3, [3 -3 I -I]T 29. Use that real entries imply real coefficients of the characteristic polynomial. Problem Set 8.2, page 343 1. [ - ~ ~] ; -1, [~] ; I, [~] ; any point (x, 0) on the x-axis is mapped onto (-x, 0), ~u that [l 3. (x, y) O]T is an eigenvector corresponding to A = -1. maps onto (x, 0). [~ ~] ; 1, [~] ; 0, [~] . A point on the x-axis maps onto itself, a point on the y-axis maps onto the OI;gin. 5. (x. y) maps onto (5x, 5)'). 2 X 2 diagonal matrix with entries 5. 7. -2, [I -l]T, -45°; 8, [1 1]T,45° 9.2, [3 -l]T, -18.4°; 7, [I 3]T.71.6° 11. I. [-l/V6 1],112.2°; 8, [1 1IV6]. 22.2° 13. 1. [l I]T, 45°; -5, [I _l]T, -45° 15. c[l5 24 50l T , c > 0 17. x = (I - A)-l y = [0.73 0.59 1.04]T (rounded) 19. [I I 1JT 21. 1.8 23. 2.1 App. 2 A23 Answers to Odd-Numbered Problems Problem Set 8.3, page 348 5. A-I = (_AT)-1 = _(A- 1 )T 3. No 7. No since det A = det (AT) = det (-A) = (-1)3 det A = -det A = O. 11. Neither, 2, 2, defect 1 9. Orthogonal, 0.96 ± 0.28i Symmetric, 18, 15.0rthugunal, 1, i, -i Symmetric, 1I + 2b, a - b, 1I - b 13. 9, 1R 17. Problem Set 8.4, page 355 2]T. [2 1.[1 - I]T, [1 3. [1 [~ -1]T:X= 2J -1 D= [7 0 :J I]T, D = [: :J 5. [2 7. [1 9. [0 _l]T, [2 1]T, diag (-2, 4) 0 O]T, [1 -2 1]T, [0 1 O]T, diag (I, 2, 3) 3 2]T, [5 3 O]T, [1 0 2]T, diag (45. 9, -27) 13. [~: -~:J: -5. [~J :2. [~J : = [-~J . [-~J -72J. [-3J x -30 15. [ 4~ 17'[-:: ~: 66 19. C = [ 1 12 3 21. C = [ -4 100 = 27. C = [ -3 12J ' IOY1 2 -6 -4J , 5Y1 2 [ 16 ~]; -2 15yl = 5, x = [0.8 0.6 - 5.\"22 - = 0, X = - 3 + 5"_22 = 10, \' 2 '.1 1 -6J , 7y . 12 -6 1 12 -4 6 1:] 1~] 23.C~ [~ V3] 2 25. C . [-3J 3 -1::];4,[_:]:_2'[_:]:1'[ -12 [=:] F [9J. .2 . . O. [12J.. x = -4 -5 32 5y .22 - 16J ' 28Y1 2 12 - x 0.6J y, hyperbola -0.8 [2tV5 -ltv's [ = = 35 , x = [ 112 - v 3/2 ~r:; 1IV2 -l/V2 4yl = 112, x = ltv's] 2tv's y, straight lines V312] y, ellipse 112 lIV2J lIV2 N2J [lIV2 1 ItV2 - lIV2 y, hyperbola y, hyperbola A24 App. 2 Answers to Odd-Numbered Problems Problem Set 8.5, page 361 3. (ABC)T = CTBTj\T = C- 1(-B)A 5. Hermitian, 3 + v'2. [-i I - v'2]T; 3 - 0. [-; I + 7. Hermitian. unitary, 1, [1 ; - iV2]T; -1, [1 i + ;"\,'2]T 9. Skew-Hermitian. 5;. [I 0 O]T, [0 I I]T; -5i, [0 I -1]T 11. Skew-Hermitian, unitary, ;, [I 0 IJ T. [0 1 01 T: -i. [l 0 _lIT 13. Skew-Hennitian. -66i 15. Hermitian, 10 v2T Chapter 8 Review Questions and Problems, page 362 [ 3 2J [3 2J = [5 OJ 11. [-1/3 2/3J A [1 9. -4 _3 2/3 oJ!lo ~ -9 A -1/3 [ 4.8 19. I.lY1 2 + - 3 0 = A -9 -7 ~J [~ -~J 2 ;l rl-~ -:l -1.0 15. -4 -2 2 4 = - I = 0 0 17r~ OJ,s. -I 5.0 yl rlo~ -: ~l I. ellipse Problem Set 9.1, page 370 1. 5. 9. 13. 17. 23. 27. 31. 2, -4,0; V20; [ltv's, -2tv's, 0] -8. -6.0; 10; [-0.8, -0.6,0] ~, £); V37/8 [4, -2,0], [-2, 1,0], [-I,~, 0] [2!S, -14, -14] (5.5, 5.5, 0), (t, ~, IS'» l-8, -2,4]; V84 [-9. 0, 0]. [0, -2, 0]. [0, 0, -11]. Yes. ci, 35. [ ~ , ~] - [- ~ , ~] 3. 7. 11. 15. 19. 25. 29. 33. = [ -1,0,5; V26: [-I/V26, 0, 5/V26] (7. 5, 0); ViO (0, 1, ~); V37/2 [10, -5, -15J [-2, 1,8], [6, -3, -24] [0,0,91; 9 v = [0. O. -9] Ip + q + ul ~ 6. Nothing ~ , ~] 37. Iwl/(2 sin a) Problem Set 9.2, page 376 1. 4 3. -v24T 5. [12. -8.4], [-18. -9. -36] 9. -4.4 11. -24 17·la 19.0 2 7. 17 IS. Use (I) and Icos + bl + la - bl 2 = aoa + 2aob + bob + (aoa - 2a°b + bob) = 21. 15 yl ~ 1. 2/a/2 + 2/b/ 2 23. Orthogonality. Yes App. 2 A25 Answers to Odd-Numbered Problems 27. 79.11 ° 25.2,2,0, -2 33.54.79°,79.11°,46.10° 31. 54.74° 39. 1.4 37.3 41. If lal = Ibl or if a and bare OIthogonal 29. 82.45° 35.63.43°.116.57° Problem Set 9.3, page 383 1. [0,0, -I OJ, [0, o. 10J 3. [-4, -8, 26J 5. [0. 0, - 60J 7. -20, -20 9.240 11. [19, -21. 24], V1378 13. [10, -5, -I] 15.2 17.30, -30 19. -20. -20 25. [-2. 2. 0] x [4.4. OJ = -16k. 16 27. [I, -1.2] x [1,2,3] = [-7, -1, 3J, v'59 29. [0, 10, 0] x [4, 3, 0] = [0, 0, -40], speed 40 31.1[7. O. OJ x [I. 1.0]1 = 7 33. ~V3 35. [18,14,26]; 9x + 7.v + 13z = c, 9·4 + 7·8 + 13·0 = 92 = c 37. 16 39. c = 2.5 Problem Set 9.4, page 389 5. Circles 1. Hyperholas 3. Hyperbolas 7. Ellipses; 288, 100, 409; elliptic ring between the ellipses and 9. Ellipsoids 11. Cones 23. [8x. 0, .\''::], [0,0, x.::], LO, 18;:, xy]; [0, ;:, yl, [z, O. x], 13. Planes b', x, 0] Problem Set 9.5, page 398 1. 5. 9. 13. 17. 23. 25. [4 + 3 cos t, 6 + 3 sin t] 3. L2 - t. 0, 4 + t] [3, -2 + 3 cos t. 3 sin f] 7. [a + 3t, b - 21, c + 5t1 11. Helix on (x - 2)2 + (y - 6)2 = r2 [v'2 cos t, sin t. sin t] Circle (x - 2)2 + (y + 2)2 = 1, z = 5 IS. X4 + .l = 1 Hyperbola xy = 1 r' = [-5 sin 1. 5 cos t. 0], U = [-sin t. cos t. 0]. q = [4 - 3w. 3 + 4w. 0] r' = [sinh t. cosh f], U = (cosh 21)-112 [sinh t, cosh t]. q = Li + 4H'. ~ + 5wJ v;:r:;r = cosh t. l = sinh I = 1.175 Stan from ret) = [t, f(t)]. v = r' = ll. 2t. O].lvl = VI + 4t 2 • a = [0. 2. 0] v(O) = 2wRi, a(O) = -w2 Rj I year = 365' 86400 sec, R = 30' 365' 86400121T = 151· 106 [km]. lal = w 2R = Iv/ 2 JR = 5.98· 10-6 [kmlsec2 ] 39. R = 3960 + 80 mi = 2.133' 107 ft. g = lal = w 2 R = IvI 2 /R, Ivl = = V6.61· 108 = 25700 [ft/sec] = 17500 [mph] 43. ret) = [t, yet), 0], r' = [1, y', 0], r' • r' = I + y'2, r" = [0, y", 0], etc. 47. 3/(1 + 9t2 + 9t4 ) 27. 29. 33. 35. 37. ViR A26 App. 2 Answers to Odd-Numbered Problems Problem Set 9.6, page 403 1. w' 3. w' = 2V2(sinh 4t)/(cosh 4t)1I2 = (cosh t)sinh t-\(cosh2 t) In (cosh t) + sinh 2 t) 7. e 4u sin 2 2v, ~e4u sin 4v 5. w' = 3(2t4 + t 8 f(8t 3 + 8t7 ) 9. -2(u 2 + V 2)-3U, -2(u 2 + V 2 )-3V Problem Set 9.7, page 409 1. [2x, 2y1 7. [6. 4. 4] 13. [-4, 2] 19. [6. 4] 23. [-0.0015. O. 29. [8, 6. 0] 35.7/3 3. [1Iy, - X/y2] 9. [-1.25. 01 15. [-18, 24] 21. [-6, -12] -0.0020] 31. [108, 108, 108] 37.2e 2 /V13 41. X4 + y3 5. Lv + 2. x - 2] 11. [0. -e] 17. [48, -36] 27. [a, h. 33.V2t3 c] 3~2 - Problem Set 9.8, page 413 1. 3(x + y)2 3. 2(x + x~ + 5. (y z) +x + 1) cos xy 7.9x2y2.::2 9. [Vt, V2, V31 = r' = [x', y', ~'] = b', o. 0].::.' = 0, z = C3, y' = 0, y = C2, x' = Y = C2, X = C2t + Cl' Hence as t increases from 0 to 1, this "shear flow" transforms the cube into a parallelepiped of volume l. 11. div (w x r) = 0 because Vb V2, V3 do not depend on x, y, z, respectively. 13. (b) (fVl)x + (fV2)y + (fv 3 )z = ![lVI)x (c) Use (b) with v = Vg. 15. 4(x + )')/(y - x)3 17.0 + (V2)y + (v 3 )z] + fxv] + fyV2 19. e XY\v 2z2 + !zV3, etc. + X 2.: 2 + x2y2) Problem Set 9.9, page 416 1. [0. 0, 4x - I] 3. [0, O. 2e x sin y1 5. [0, 0, -4y/(x 2 + )'2)] 9. curl v = [- 2~. 0, 01, incompressible, v = r' = [x'. y'. z'] = [0. .:2, 0]. x = CI, Z = C3, y' = Z2 = C32, Y = c32t + C2 11. curl v = [0, 0, -2], incompressible. x' = y, y' = -x, z' = 0, Z = C3, )' dy + x dx = 0, x 2 + )'2 = C 13. Irrotational, div v = 1, compressible, r = ['let, C2e-t. C3etJ 17.0,0, [xy - .:x, yz - xy, ,:x' - yz] 19. 0, 0, 0, - 2.\'Z2 - 2.:x 2 - 2.\"y2 Chapter 9 Review Questions and Problems, page 416 11. [-1, 9, 24] 15. [0, 0, -740], [0, 0, 19. -495, -495 23. If u x v = 0. Always 27.3.4 -740] 13. 0, [-43, 54, 3], [43, -54, 17. [-24, 3, -398], [114, 95, 21.90°, 95.4° 25. [VI, V2, -3] 29. If 'Y > ~7T, ~7T -3] -76] App. 2 A27 Answers to Odd-Numbered Problems 33.45/6 37. 0, 2y2 41. 0, 2X2 + (z + xl + 4y2 + 2z 2 + 35. No 39. l-l, 1, - l], [-2z, 43.4881"\13323 45.0 4xz -2x, -2y] Problem Set 10.1, page 425 1. 5. 7. 9. 11. 15. F(r(t» F(r(t» F(r(t» F(r(t» F(r(t» 17/3 = 1125t6 , t 3 , 0], 1644817 = 2350 3.0 + 160 = [cosh t sinh 2 t. cosh 2 t sinh tl, 93.09 cos t, sin t], 67T lcosh ~t, sinh !t, ett8 ], 0.6857 = let, et2 , e t2 ], e 2 + 2e 4 - 3 17. [367T. = It, = ~(87Tl, 367T] Problem Set 10.2, page 432 3• _12 e -cx2+y2) , 0 11. sinh ae 1. sin xy, 1 7. x 2 )' + cosh z, 392 15. cea - aeb 5. e Xz 13. No 19. No 17.~a2bc2 + y, -2 Problem Set 10.3, page 438 f Ix 7. f ~(e3x I + i) 1 x3 3. - (x 2 - x 5 )] dx o 4 - e- X ) dx 5. ~ cosh 6 - cosh 3 = l2 = !e 12 + !e-4 - o 1 (2x 2 11. dx -1 15. = i 19. Ix = (a ~ + b)h 124, Iy = ~ (e Sin y cos y - cos y) dy 9. 0 1 0 x = ib, Y = ~h 3 f 13. f f +~ x 0 = e- 2 ~ d)' dx = ~ 17. Ix = bh 3 /l2. Iy = b 3 h/4 h(a 4 - 4 b )/(48(a - b» Problem Set 10.4, page 444 1. 2x 3 ), - 2.\}"3, 81 - 36 = 45 5. e'C- Y - e X + Y, -~e3 + !e 2 + e- 1 - ! 9. 0 (why?) 13. Y from 0 to ~x. x from 0 to 2. Ans. cosh 15. Y from 1 to 5 - x 2 • Ans. 56 3. 3x 2 + 3y2, 18757T/2 = 2945 7. 2x - 2v, -56/15 11. Integrand 4. Ans. 407T 2 - ! sinh 2 19. 4e 4 - 4 Problem Set 10.5, page 448 1. Straight lines, k 3. x 2/a 2 + y2/b 2 = 1, ellipses. straight lines, [-b CllS v, a sin v, 0] 5. z = (cla)Y + y2, circles. straight lines, [-aeu cos v, -aeu sm v, a 2u] 7. x 2/9 + ."2116 = z, ellipses, parabolas. [-811 2 cos V, -6112 sin v, 12u] 2 9. x /4 + )'2/9 + z2/]6 = I, ellipses, [12 cos 2 v cos u, 8 cos 2 v sin u, 6 cos v sin v] 13. [lOu, IOv, 1.6 - 4u + 2v], [40, -20, LOO] 15. [-2 + cos v cos u, cos v sin £I, 2 + sin v], 2 [cos V cos u, cos 2 v sin u, cus v sin v] r A2B App. 2 Answers to Odd-Numbered Problems 17. [Lt, v, 3v 2 ], [0, -6v, II 19. [u cos V, 311 sin v, 3111, [-911 cos v, -3u sin v. 311] 21. Because r1< and rl" are tangent to the coordinate curves v = respectively. 23. [iT. v, li 2 + v 2 ]. N = [-2u, -2v, 1] COllsT and 1I = COIlST, Problem Set 10.6, page 456 1.-64 7.27r 15. 140V6/3 3. -18 9. ~a3 17. 1287TV213 = 189.6 19. 6 7T 2 (373/2 - 53/2 ) = 22.00 27. 7Th4/V2 29. 7Th + 27Th 3 /3 i 5. -1287T 11. 17h14 25. 27Th Problem Set 10.7, page 463 1. 8a 3b 3c 3127 7. 2347T 13.7Th5 /lO 21. 0 3.6 9.2a 5 /3 17. 1087T 23.8 5. 42~7T 11. ha 4 rr12 19. 2167T 25. 3847T Problem Set 10.8, page 468 1. 3. 5. 7. Integrab ..j.. I . 1 (x = 1). ..j.· 1 . 1 (y = 1), -8' 1 . I (z = 1).0 (x = v = 2 (volume integral of 6y2), 2 (surface integral over x = 1). Others 0 Volume integral of 6)'2 - 6x2 is O. 2 (x = 1), -2 (v = 1), others O. F = [x. y. z], div F = 3. In (2). Sec. 10.7. Fon = IFllnl cos cp = z = 0) Vx 2 + y2 + Z2 cos cp = r cos cp. 9. F = [x, 0, 0], div F = 1, use (2*), Sec. 10.7, etc. Problem Set 10.9, page 473 1. 3. 5. 7. 11. 15. 19. [0, 16]0[0, -1, 1], :::':12 -ex. eY ] [-1. -1. 1], ±(e2 - 1) S: [u, v, v 2 1, (curl F)oN = _4ve 2v2 , ±(4 - 4e 2 ) (curl F)on = 312, :::':3a 2 /2 9. The sides contribute a, 3a 212, -a, O. curl F = [0, o. 6],247T 13. (curl F)on = 2x - 2y, 1/3 -7T/4 17. (curl F)oN = 7T(COS TTX + sin -n:v), 2 For = [-sin e, cos eJo[ -sin e, cos e) = I. 27T, 0 8z, r-ez , 0 Chapter 10 Review Questions and Problems, page 473 11. Exact. -542/3 17. By Stokes, ± 18rr 23. 0, 4a137T 29. By Gauss, 1007T 35. Direct, 5(e 2 - 1) 13. Not exact, e 4 - 7 19. By Stokes, :::': 127T 25. 817. 118/49 31. By Gauss, 40abc 15. By Green, 1152rr 21. 4/5, 8/15 27. Direct. 5 33. Direct, rrh App. 2 A29 Answers to Odd-Numbered Problems Problem Set 11.1, page 485 3. 13. 15 k, k, 2mn, 27T11l. ±+ ! (COS + x - 7T '2 9 7T I sin - sin 2x 2 y - 19. - -4 ( cos x 7T 2 (sin x I 21. "3 7T I 23. 6 7T 2 - t' + ..!.. 2 (cos x 4 + cos 5x - 9 7T 7T + + + cos 3x + ... ) + 4 (cos x + ..!.. cos 3x + _1_ cos 5x + ... ) 17. - 29. kin kill, 2 + 4 + - sin 3x - + ... 3 + + sin 3x + -I cos 5x 25 + + ... ) + ... ) sin 5x "4I cos LX + 9"1 cos 3x - + ... ) I cos x - - cos 2x 2 - 7T = 25 I cos 3x 9 + _1_ cos 5x + ... ) cos 3x + -1 4 ( cos x - - 25 2x, f" = 2,jl = 4 1 277T 8 + - - cos 3x + - cos 4x - ... O,j~ = -47T,j~ = 0, an = _1_ (n7T ..!..) cos (-47T) 117T, etc. 11 Problem Set 11.2, page 490 1. : 3. ..!.. 3 ~x (sin ~ 2 7T + (cos i sin 3~X ..!.. TTX - cos 4 + 2+4 (cos 9. "3 7T2 3 11. 4 4 7T ( 2 ( co:> TTX 2 + ..!.. cos 9 27TX + -1 9 3 1 + - 2 cus 2r 15. Translate by!. + 37TX - + ... ) + -I- cos 47TX + -1- cos 3·5 cos 1 2 + - cos TTX 37TX 5·7 + -1 25 cos 1 9 1 -8 cos 4x 17. Setx 37TX 2 57TX = O. 67TX + ... ) + ... ) 47Tt" + - ... ) I 57TX cos - 25 2 + - cos - - + - + .. -) 13. 8 + ... ) I cos "41 cos 27TX + 9"1 cos 37TX - 16 TTX - cos - TTX 5~X sin 27TX 5. Rectifier, -2-4 - (1 - - cos 7T 7T 1·3 7. Rectifier, -1 - 24 2 7T + 1 cos 37lX 18 + - A30 App. 2 Answers to Odd-Numbered Problems Problem Set 11.3, page 496 1. 3. 5. 7. 9. Even, odd, neither, even, neither, odd Odd Neither Odd Odd . 2 13. ~7r + 11 7r ~7r ~9 cos 3x + + (cos x ~9 sin 3x + (Sin x - 4 ( 71X 15. 1 - --:;; sin ""2 ~ + ... ) _1_ sin 5x 25 37rx "31 sin -2- + + ... ) _1_ cos 5x + 51 + ... + 8 ( 7T.X 19. (a) 1 + 7r2 cos ""2 + 9 cos -2- + 25 cos -2- + ... 4 ( 21. (a) n"( sin 2 "23 - 23. (a) "2L - 2 7rX L 2 371X 1 1 9 7r + 9 ~ ) 4 I 57rx 1 37rx cos 3m:- L 1 + 25 cos 77rx - --:;1 cos -2- + - ... I 57rx sin -2- - "9 sin 371X 571X ) L + ... - "21 sin L27rx + "31 sin L37rx - + ... ) + -4 ( cos x + -I cos (b) 2 ( sin x 3 1 + ) 571X - "3 cos -2- + 5 cos 2 4L ( 7rX 7r2 cos + ... 37rx I ) + -1 sin - + - sin 27rx + ... 1 7H 1 37rx ""2 - "3 sin 71X + "3 sin -2- + 5 2L ( 71X (b) --;; sin L 25. (a) -7r 2 + -1 sin 7rX 2 ( 7r cos ""2 6 ( (b) 7r sin 1 + 5nr 51 sin -2- ) 4 ( 71X 17. (a) 1. (b) 7r sin ""2 (b) 7r 37rx "31 sin -2- 57T.X sin -2- sin 2x + 3x + + - I cos 5x + ... ) sin 3x 25 + . . .) Problem Set 11.4, page 499 3. Use (5). (-1)n 00 L... 9 .1· " n=-oo - - einx ' n 2i 7 . - - ~ _ _ _ e(2n+ 1)ix 7r n=-oo 2n + 1 2 1)n 11. ~ + 2 ~ ---- einx 3 n2 n=-(X: 00 n*O co 13. 7r + i ~ n=-x ( ) + ... ) App. 2 A3l Answers to Odd-Numbered Problems Problem Set 11.5, page 501 3. (0.0511)2 in Dn changes to (0.02n)2. which gives C5 = 0.5100. leaving the other coefficients almost unaffected. 5. Y = Cl cos wt + C2 sin wt + A{w) cos t, A(w) = 1/(w2 - L) < 0 if w2 < I (phase shift!) and> 0 if w 2 > I N 7. y = Cl cos wt + C2 sin wt + an L n=l 9. )' = c cos wt .+ .. ) 11. Y = Cl cos wt + c sin , + C2 1 - 2 3· 5(w 16) - wt sin wl cos4t- w 2 - n + -7T- + -4 26" ~ 2 cos nt ( 1 -----0;--- cos t + 'u'-I oo - 1· 3(w2 cos 2t 4) - cos 3t 6i'-9 1 + 2w2 1/9 • 13. The situation is the same as in Fig. 53 in Sec. 2.8. 3c 15. y = - 64 + 9c 2 cos 3t - 8 64 + 9c 2 sin 3t Problem Set 11.6, page 505 ± 1. F = 2 ( sin x - sin Ir + ... + 4 ( 1 3. F = -7T 2 - -7T cos x + -9 cos 3x {_l)N+l N ) sin Nx , E* = 8.1,5.0,3.6,2.8,2.3 1 cos 5x + . .. ) E* = 00748 00748 + -25 ...., 0.01 19, 0.01 19. 0.0037 5. F = -2-4 - (1 - - cos 2x + -1- cos 4x 7T IT 1·3 3·5 E* = 0.5951, 0.0292. 0.0292, 0.0066, 0.0066 7. F = - 4 ( sin x + -1 7T 0.6243. 0.4206 9. -8 ( sin x 7T + -1 27 0.0015, 0.00023 sin 3x + -1 sin 5.1' 3 5 (0.1272 when N = 20) sin 3x + 1~5 sin 5x + -1- cos 6x + . .. ) , 5·7 + . .. ) . E* + .. J = 1.1902, 1.1902, 0.6243, E* = 0.0295, 0.0295, 0.0015, A32 App. 2 Answers to Odd-Numbered Problems Problem Set 11.7, page 512 LX 1. f(x) = 7Te- x (x > 0) gives A = e- V cos wv dv = , B = __t_v----=-2 + w2 1 + tr (see Example 3), etc. 0 2 X 3.I(x) = !7Te- gives A = 1/(1 + w ). 5. Use f = (7TI2) cos v and (11) in App. 3.1 to get A = (cos (ml'!2»/(1 - w 2 ). 2 7. 7T 2 11. - Icc sin 7T cos XW --dw 2 9 - • 7T W 0 LX cos TTW + I l-w 7To 2 17. - (ltv LX tv w2 0 LX 7To 7TH' - sin 7TW sin 2 0 2 15. - cos xw dw 2 I"" -cos- -w-+- , ,sin- -w--- - cos XIV 2 19. rr dw IV HI' sin mv 2 sin xw dw l-w L wa cc 0 sin wa 2 sin xw dw IV Problem Set 11.8, page 517 1. 7. {2 ( sin 2w - V--;; 2 sin w ) 5. w v:;;n cos w if 0 < 11. V(2/7T) w/(w 2 + 7T tv 2 < 7T12 and 0 if w > 7T12 v:;;n e-x (x> 0) 9. Yes, no ) 19. In (5) for f(ax) set a.>.: = v. Problem Set 11.9, page 528 3. ik(e- ibw 7. [(I - l)/(v'2;w) iw)e- iw - 1]/(v'2;w2 ) + 11 ·21 e -w 13. (e ibw 5. V(2/7T)k (sin w)/w 9. V(21'7T)i(cos w - 1)/w 2 /2 - e- ibw )/(i}vv'2;) = v'2/;-(sin bw)/w Chapter 11 Review Questions and Problems, page 532 11. 4k (sin TTX + 7T 13. 4 ( sin 15. 8 ( 7T2 "2x- I"2 sin !3 sin 31TX sin x + 2nt - '9I sin + !5 sin 57TX + ... ) '31 sin 23x - 4"1 sin 4x + "51 sin 25x 3nt -2- + I 25 5nr sin -2- - 17. -2-4 - (1 - - cos 16rrx + -1- cos 327TX 7T 7T 1·3 3·5 + ... ) ) + -1- cos 5·7 + ... 481TX + ...) dw App. 2 A33 Answers to Odd-Numbered Problems 19. 7r 2 12 - I l L cos 6x + 16 cos 8 \" - "4 cos 4x - 9 + cos 2x 23. 7r 2 /8 by Prob. 15 21. rr/4 by Prob. 11 25. 1 "2 [f(x) + + ... 1 "2 Lt(x) - f(-x)1, fe-x)] 27. rr - -8 ( cos -x + -I cos -3x 7r 2 9 2 + -1 25 cos -5x + ... ) 2 29.8.105,4.963,3.567,2.781,2.279, 1.929. 1.673, 1.477 31. y C] cos wI + C2 sin wt = L"" (cos 1 7r 3w + w sin w - 1) cos W LX sin tV - 0 7r (cos I 2 w - 1 1 4 - - • cos 2t _ 4 w2 + 1 9 cos 3t w2 - cos H' 2 W + (sin w - w cos w) sin 2 tVX dH' . •• • Sin tU d~~ W LX sin 2w - 4 37. - tVx w 0 2 35. 7r 4 --2 - + _ ... ) __1_. cos 41 16 w 2 - 16 33. - 7? + 2w cos 2w w 0, 3 39. cos wx dl\' ~.----::--+4 ,,-; I\,2 t' , Problem Set 12.1, page 537 1. 1I 5. 1I = Cl(X) = c(x)e- Y 9. 1I = Cl(X»)' cos 4)' + + + C2(X) eXY/(x + 3. 7. sin 4)' 1) C2(.1'»),-2 15. C = 1/4 19. 7r/4 27. 1I = 110 - (lIO/in 100) In (X2 + )'2) + 1I = Cl(X) 1I = c(x) exp C2(X)y (h 2 cosh x) 11. u = c(x)eY + h(y) 17. Any C 21. Any C and w 29. 1I = ClX + C2(Y) Problem Set 12.3, page 546 1. k cos 2m sin 2m; 3. '8k 3 ( cos 7r1 sin 7rX + - 5. 7r 27 cos 37r1 sin 37rx + 4 57r ( COS 7r1 sin 7rX - -1 cos 37rf sin 37rx 9 + --2 2 7. 7r 2 + ( cv'2 - ~ Cv'2 + 1) cos m sin 7rX + 1~5 cos 57rISin57r,+ ... ) _1_ cos 57rf sin 57rx 25 "21 cos 2m sin 27rx 1) cos 3m sin 31TX - .• -) + ...) 9 A34 App. 2 Answers to Odd-Numbered Problems 9. 22 ~ ( (2 - V2) cos m sin = - -I (2 _1_ (2 25 it = 3= + + V2) cos 5m sin 5= + ... ) 2 17. + v'2) cos 3m sin 9 8L ~3 ( cos [(~)2J C L t L7TX sin 1 cos [(3~)2J 33 C L t + sin 371:\ + ...) L 19. (a) u(O, t) = 0, (b) lI(L, t) = 0, (c) 1l:t"(0. t) = 0, (d) uxCL, t) = O. C = -A, D = -B from (a), (b). Insert this. The coefficient determinant resulting b'om (c), (d) must be zero to have a nontrivial solution. This gives (22). Problem Set 12.4, page 552 3. 11. 13. 15. 17. 19. c 2 = 300/[0.9/(2·9.80)] = 80.83 2 [m2 /sec2 ] Hyperbolic. u = fl(X) + f2(X + y) Elliptic, U = fl(Y + 3ix) + f2(Y - 3ix) Parabolic, 1I = Xfl(X - Y) + f2(x - y) Parabolic. 1I = Xfl(2x + y) + f2(2x + y) Hyperbolic, u = (l/y)fl(XY) + f2(Y) Problem Set 12.5, page 560 5. It = sin OAnt e-1.752.167T2t/lOO 7. II = 9. u = 11. II = ! (~ sin O.lnt 2:~ (sin 0.17T.t e-O.017527T2t til + Un. where Un = II - UI ~ + e-O.017527T2t + sin 0.27TX e-O.01752(2m2t - . . . ) ~ sin 0.3n\" e-O.01752(37T)2t - - ... ) satisfies the boundary conditions of the text. so fL oc 117T.t 2 2 that u II = L... ~ B sin - - e-lCn7TIL) t B = n' L . II L n~1 0 13. F = A cospx + B sinpx, F'(O) p = n~/L, etc. 15. u = 1 17. II = 2;2 + ~2 19. u = 12 23 . - -K~ L l1~l [f(x) (x)] sin -L - dx . - II I' ~ 4 (cos x e- t - = Bp = O. B = 0, F'(L) = -Ap sinpL = 0, cos 2x e- 4t + i I I 4 9 cos 3x e- 9t - + ... ) + cos 2,' e- 4t + - cos 4x e- 16t + - cos 6x e- 36t + ... . L OC nB II e- A " 2 t 25. w = e-{3t 1I~1 27. v t - c 2v xx = 0, w" = -Ne- ux/c 2, w = C~2 [_e- ux - ~ (J - e-uL)x so that w(O) = n'(L) = O. 29. 1I = (sin i~x sinh ~~y)/sinh 1T 80 ~ 31. II = - L... ~ n~l (2n - 1)= (211 - 1) sinh (211 - l)~ Sll1 24 (211 - 1)7TV sinh - - - - 24 + 1] ' App. 2 A35 Answers to Odd-Numbered Problems 33. = Aox + II = Ao I -2 24 L sinh (I171:T/24) 117TY An - - - - - cos -24 ' sinh 117T f fey) dy, 24 117TX oc 35. ~ An sin - - sinh f24 0 117T(b - y) a n~l I 12 = - An 0 , An a 117TV f( \") cos - - ' dv 24- = 2 fa a sinh (l17Tb/a) 0 117TX f(x) sin - - dx a Problem Set 12.6, page 568 2 sin ap 1. A = , B = 0, fX -sin-ap- cos px e- c 2 2 = - II rr ~ L cos 2 p t dp p 0 oo = e- P , B = 0, u = 3. A px dp e-p-cZp2t o 5. Set = s. A = I if 0 < TTV p/7T < I, B = 0, u {T = cos px e-c2p2t dp o ~ 7. A 2[cos P + P sin p 1)/(~2)]. B - O. = f" u = A cos px e-c2p2t dp o Problem Set 12.8, page 578 1. 3. 5. 7. 11. (a), (b) It is multiplied by v'2. (c) Half = 16/(ml17T2) if 111. 11 odd. 0 otherwise Bmn = (-l)n+18/(mn7T2) if 111 odd, 0 if m even Bmn = Bmn (_l)m+n4/(mn7T2) k cos v'29 1 sin 2x sin 5y 6.4 "" "" I 13. - 2 ~ ~ -----:33 cos (1 m~l n~l 117 II 111, n odd 7T 17. V 117 2 + 2 /1 ) sin l/IX sin 11)" (colTesponding eigenfunctions F 4,16 and F 1614 ), etc. 0 (m or 11 even). Bmn = 16k/(l1l1l7T2 ) (m, 11 odd) C7TY'260 19. Bmn = 21. Bmn = (-l)m+nI44a 3 b 3 /(m 3 11 3 1fl) 9 23. cos ( 7Tt 2 a + 216) 47TV 37TX sin - - sin - - ' a b b Problem Set 12.9, page 585 7. 30r cos 8 + 10,-3 cos 38 9.55 220 + - ( r cos 8 - -1 ,.3 cos 38 3 7T 11. 7T _ 2 ~ 7T (r cos 8 + ~9 ,.3 cos 38 15. Solve the problem in the disk,. < and -lio on the lower semicircle. u = 4uo 7T (!.. a sin 8 17. Increase by a factor + ~ 3a v'2 r3 a + -1 5 + _1_ 25 r 5 cos 58 ,.5 cos 58 :-.ubject to sin 38 + Uo ~ r 5a 19. T 5 + ... ) + ... ) (given) on the upper semicircle sin 58 + ...) = 6.826pR2 f1 2 A36 App. 2 Answers to Odd-Numbered Problems 21. No 23. Differentiation brings in a factor lIAm = RI(eCi m ). Problem Set 12.10, page 593 11. v = F(r)C(t), F" + k 2F = 0, G + e 2k 2C 2 R C n = Bn exp (-e 21l 27T 2tIR2), Bn = - = 0, fR Fn = sin (I1'mfR), WITT rf(r) sin - - dr R 0 13. u = 100 15. u = ~r3P3(CoS 1;) - ~rPl(cos 1;) 17. 64r4P4(COS 1;) 21. Analog of Example 1 in the text with 55 replaced by 50 23. v = r(cos ())lr2 = r/(x 2 + )"2), V = xyl(x 2 + )"2)2 Problem Set 12.11, page 596 e(s) = - X S + x , W(O, s) = 0, e(s) = 0, w(x, t) = x(t - 1 + e- t ) + 1) 7. w = f(x)g(t), xI' g + fi = xt, take f(x) = x to get g = ce- t + t - 1 and e = 1 from w(x, 0) = x(e - L) = o. 9. Set x 2/(4c 27) = Z2. Use z as a new variable of integration. Use erf(x) = I. 5. W S2(S Chapter 12 Review Questions and Problems, page 597 19. u = + el(y)e X c2(y)e- 2X 21. u = g(x)(1 - e- Y ) + f(x) 25. u = ~ cos 1 sin x - -! cos 3t sin 3x 23. u = cos t sin x - ~ cos 21 sin 2x 27. l/ = sin (0.02'i1:\") e-0.0045721 200 29 u = 7T 2 • 31. u = (7TX sin 50 · e-0.004572t - 37T.\" -1 sin - e-0.04115t 9 50 100 cos 4x e- 16t 7T 16 33. u = - - 2 7T (1- 4 cos 2x e- 4t + - 1 36 cos 6.1: e- 36t + ... ) + -I- cos lOx e- lOOt 100 + ... ) 37. u = fl(Y) + f2(X + y) 39. II = fl(Y - 2ix) + f2(Y + 2ix) 41. l/ = Xfl(Y - x) + f2(Y - x) 49. II = (111 - 1l0l(1n r)lIn (rl/ro) + (110 In rl - Ul In ro)lIn (rl/rO) Problem Set 13.1, page 606 5. x - iy = -(x + iy), x = 0 9. -5/169 11. -7/13 -(22/l3)i 15. -7/17 - (l1117)i 17.xl(x2 + y2) 7.484 13. - 273 + 136i 19. (x 2 - y2)/(x 2 + y2)2 Problem Set 13.2, page 611 1. 3V2(cos (--!7T) + i sin (-!7T)) 3. 5 (cos 7T + i sin 7T) = 5 cos 7T 1 5• cos "27T . 1 + I.sm "27T App. 2 Answers to Odd-Numbered Problems A37 9. -37r/4 7. ~v'6I (cos arctan ~ + i sin arctan ~) 15. 37r/4 11. arctan (±3/4) 13. ±7r/4 17.2.94020 + 0.59601i 19.0.54030 - 0.84147i 21. cos (-~7r) + i sin (-~7r), cos ~7r + i sin ~7r 23. ±O ±i)rV2 25. -I, cos!7r ::!: i sin !7r, cos ~7r ± i sin ~7r 27. 4 + 3i, 4 - 8i 29. ~ - i, 2 + ~i 35. 1<:1 + z21 2 = (:1 + Z2)(Zl + Z2) = (:1 + z'2)(':1 + ':2)' Multiply out and use Re <:1':2 ~ IZ1z21 (Prob. 32): 2 :1':1 + :1':2 + :2':1 + ::2':2 = hl + 2 Re :1':2 + IZ212 ~ 1:112 + 2blk:21 + IZ212 = (1:11 + IZ21)2. Take the square root to get (6). Problem Set 13.3, page 617 1. 3. 5. 9. 13. 15. 19. 23. Circle of radius ~, center 3 + 2i Set obtained from an open disk of radius I by omitting its center z = 1 Hyperbola xy = 1 7 . .v-axis The region above y = x f = I - 11(::; + 1) = 1 - (x + 1 - iy)/l(x + 1)2 + y21; 0.9 - O.li (x 2 - y2 - 2.ixr)/(x2 + y2)2, -i/2 17. Yes since r2(sin 2e1/r ~ 0 Yes 21. 6;::2(;::3 + i) 2i (1 - Z)-3 Problem Set 13.4, page 623 5. Yes 1. Yes 3. No 9.Yesforz*0 7. No 11. rx = x/r = cos e"y = sin e, ex = -(sin e)/r, ey = (cos e)/r, (a) 0 = Ux - Vy = It,. cos e + tllI(-sin e)/r - Vr sin e - vo(cos e)/r. (b) 0 = lty + Vx = 1I,. sin e + lle(COS e)/r + Vr cos e + v e ( -sin e)/r. Multiply (a) by cos e, (b) by sin e, and add. Etc. 13. z2/2 15. In Izl + i Arg : 17. Z3 19. No 21. No 23. c = I, cos x sinh y 27. Use (4), (5), and (1). Problem Set 13.5, page 626 3. -1.13120 + 2.47173i, e = 2.71828 5. -i, 1 9. e- 2x cos 2.y, _e- 2T sin 2.y 7. eO.s(cos 5 - i sin 5), 2.22554 2 2 11. exp (x - )"2) cos 2xy. exp (x - y2) sin 2.\)" 13. e i7r/4 , e 5m/4 15. Vr exp [ice + 2k7r)/Il], k = 0," ',11 - 1 17. ge m 19. z = In 2. + tri + 21l7ri (11 = 0, ±l,"') 21.: = In 5 - arctan ~i ± 21l7ri (11 = 0,1,"') Problem Set 13.6, page 629 3. Use (II), then (5) for e iy , and simplify. 5. Use (II) and simplify. 7. cos 1 cosh 1 - i sin 1 sinh 1 = 0.83373 - 0.98890i A38 App. 2 Answers to Odd-Numbered Problems 11. -3.7245 - 0.51182i 15. cosh 4 = 27.308 19. z = ~(21l + 1)7T - (-l)n1.4436i 9.74.203, 74.210 13. -1 17. z = :::':::(211 + 1)7Ti/2 21. ::: = :::':::lI7Ti 25. Insert the definitions on the left, multiply out, simplify. Problem Set 13.7, page 633 1. 5. 7. 11. 15. 17. 19. 21. 25. In 10 + 7Ti 3. ~ In 8 - ~7Ti In 5 + (arctan ~ - 7T)i = 1.609 - 2.214i 0.9273i 9. ~ In 2 - ~7Ti :::':::(211 + I )7Ti. 11 = 0, I. . . . 13. In 6 :::'::: (21Z + 117Ti, 11 = O. 1. ... (7T - I :::'::: 2117T)i, Il = 0, 1, ... In (;2) = (:::':::211 + i)7Ti. 21n i = :::':::(411 + l)'ITi. n = O. 1. ... 3 eO. (cos 0.7 + i sin 0.7) = 1.032 + 0.870i 2 e (l + i)/V2 23. 64(cos (In 4) + i sin (In 4» 2.8079 + 1.3179i 27. (I + ;)/,\0. Chapter 13 Review Questions and Problems, page 634 19. -~ - ~i 25. 12e- 77i/2 31. fez) = liz 37. (-x 2 + )"2)/2 43.0.6435i 17. -32 - 24i 23. 6V2e37ri/4 29. (:::':::1 :::'::: i)/V2 z2 35. f(:::) = e 41.0 21.5 - 3i 27. :::':::(2 + 2i) 33. fez) = (l + i):::2 39. No 45. -1.5431 Problem Set 14.1, page 645 1. 3. 7. 9. Straight segment from I + 3i to 4 + 12i Circle of radius 3, center 4 + i 5. Semicircle. radius 1. center 0 Ellipse, half-axes 6 and 5 from - 1 - ~i to 2 + 4i Parabola)' = ¥3 11. e- it (0 ~ t ~ 27T) 13. t + ilt (1 ~ t ~ 4) 17. -Q - ib + re- it (0 ~ t ~ 27T) 21. 0 25. il2 29.2 sinh ~ 15. t + (4 - 4t2 )i (-1 ~ t ~ I) 19.~ + ~i 23. 7Ti + ~i sinh 27T 27. -I + i tanh ~7T = -I + 0.6558i Problem Set 14.2, page 653 1. 7Ti. no 3.0, yes 7. O. yes 9.0, no 15. Yes, by the deformation principle 21. 7Ti 23.27Ti 27. (a) 0, (b) 7T 29.0 5.0, yes 11. O. yes 19. 7Ti by path deformation 25. 0 App. 2 A39 Answers to Odd-Numbered Problems Problem Set 14.3, page 657 1.-4~ 7.0 13. ~ 17. ~i cosh2(l 5.8m 3.4~ + 11.m 9. -~i 15. 2m Ln 4 = 8.710i i) = ~(-0.2828 + 1.6489i) Problem Set 14.4, page 661 1. 27~i/4 7. 2~i if lal < 2.0 if lal 11. 2~2i 5. ma 3 /3 >2 9. m(cos i-sin ~) 13. ~iea/2124 if la - 2 - il < 3 0 if la - 2 - il > 3 3. -2~ierr/2 Chapter 14 Review Questions and Problems, page 662 17. -6~i 23. ~i sin 8 21. 27. 19.0 25.0 -~i ~ 29.0 Problem Set 15.1, page 672 Bounded, divergent, ± J 3. Bounded, convergent, 0 Unbounded Bounded. divergent, ±llV'2" ± i, O. I, -2 Convergent, 0 13. IZn - II < ~E, Iz~ - [*1 < ~E (n > N(E)), hence IZn + z~ - (l + [*)1 < ~E + ~E 17. Convergent 19. Divergent 21. Conditionally convergent 23. Divergent by Theorem 3 27.11= 1100 + 75il = 125 (why?); 1100 + 75iI 125/125! = 125125/[\h50~ (l251el 25J = e125tV250~ = 6.91 . 1052 1. 5. 7. 9. Problem Set 15.2, page 677 1. ~ a n z2n = ~ a n(z2)n, Iz21 < R = lim lanlan+ll, hence Izl <\'R. 3. -i, I 5. -1, e by (6) and (1 + I1n)n ~ e. 7.0, Iblal 9. O. 1 11.0, 1 15. i, 1IV2 17. 0, V2 13. 3 - 2i. 1 Problem Set 15.3, page 682 3. V2 9. ] 1. 3 7. 1/\/7 5. V5i3 Problem Set 15.4, page 690 1. 1 - 2z + 2z 2 - ~Z3 + ~Z4 - + .. ', R = 3. e- 2i (1 + (z + 2i) + hz + 2i)2 + t(z + 2i)3 5. I - ~(z - ~~)2 7. ~ + ~i + !i(z + - i) l4(Z - !~)4 - 7~O(Z + (-~ + ~O(Z - - ;)2 - 00 + + 2i)4 + ... , R = 00 !~)6 + - ... , R = oc ~(z - ;)3 + - ... , R = V2 l4(Z A40 App. 2 Answers to Odd-Numbered Problems + t:: 4 - .i8::.6 + 3~4<.8 - + .. " R = x 11.4(:: - 1) + \O(z - 1)2 + 16(:: - 1)3 + 14(:: - \)4 + 6(:: 13. (2/v;.)(:: - z3/3 + ::5(2!5) - z7/(3!7) + ... ), R = GC 15. ;:3/(l!3) - ::7/(3!7) + ;:1l/(5!1l) - + . ". R = x 19.:: + ~Z3 + 1~::5 + i{5:: 7 + . ", R = !7T 9. 1 - !::2 - 1)5 + (z - 1)6 Problem Set 15.5, page 697 3. R = \[\1'; > 0.56 7. Itanhn Izll ~ I, 1/(/1 2 + 11. Izl ~ 2 - 8 (8 > 0) 15.lzl ~ \,'5 - 8 (8) 0) 1. Use Theorem 1. 5. Iz nl ~ 1 and ~ 1//1 2 converges. 9. Iz + <R= L - 2il ~ r 4 13. Nowhere \) < \//1 2 Chapter 15 Review Questions and Problems, page 698 13. 1. ! Ln[(1 + z)/(1 - .:)] 17. 1/v;., [l - 7T(Z - 2i)2r 1 11. x, e 3 - 15. GC 19. 1/3 21. -1 - (::. - 7Ti) - (:: - 7Ti)2/2! - .. " 23.! + ~(:: + I) + t(z + R = x 1 1 6(Z 1)2 + + 1)3 + . ", R = 2 + 3;: + 6;:2 + 10;:3 + .. '. R = I. Differentiate the geometric + (:: + i) - i(z + i)2 - (z + i)3 + .. " R = 1 25. \ 27. i 1 + 29. -(:: - 2 7T) 3!\ (:: - 13 27T) - 1 5! (;: - 15 27T) + - .. '. R = series. ex; Problem Set 16.1, page 707 \ 1. .:4 1 1 1 + -::3 + -Z2 + -z + 1 + z + 1111 + - - Z2 2;:: 6 3. -3 - 5. -_3 I 24 .(. + - Z \ 1 + - ::5 1 2::7 ez-1e 7. - - = e Z - 9. - \ (i 2:"2 x n~O 1 6;:9 + oc + - (::: - 2: n~O I (.: - 12 120 ~ I)n-l = e i)n-l = - R = \ + - ... R = , [1 -Z- \ + 1 z- \ + - - +"'],R=X 2! il2 \; --=-=-=+ "4 + -8 (:: <. i) - I R=2 x 11. - 2: (.: + ;)n-l = - n~O 3 13. - - z- 1 1 ~ Z + 2 + (z - \) I x = ox ' /1! + .. '. - - + ... R n -l-l Z2 - \ - (.: + ;) - .. " R = 1 1 16 (:: - ;)2 - .. " App. 2 A41 Answers to Odd-Numbered Problems 15. L -L < I. :x; 1:::1 ;::4n+2, < I, > 1 ~ l x -L 4n+2' Izl > 1 '11=0 Z '>1=0 19. Izl _3n+3' n~O 17. L 1 Xl ~3n, 1::1 i I + - - + i + (;:: ~ - (::: - i)2 i L ~4n, Izl < x 21. (I - 4:::) i) (4 I) L I 1, 3 n=O x - 'I Z Z 4n' n~O Z Izl > I Problem Set 16.2, page 711 1. :::,:::~. :::':::~, ... (poles of 2nd order), :x (essential singularity) 3.0, :::,:::v:;.;.. :::,:::\12;. ... (simple poles), :x (essential singularity) 5. :x (essential singularity) 7. :::'::: I, :::':::i (fourth-order poles), :x (essential singularity) 9. :::':::i (essential singularities) 13. -16i (fourth order) 15. :::'::: 1, :::':::2, ... (third order) 17. :::':::irV3 (simple) 19. :::':::2i (simple), 0, ::':.2 m, :!::4'ITi, ... (second order) 21. O. :::':::271. :::':::471• ... (fourth order) 23. f(~) = (~ - ~o)ng(:::), g(zo) "* 0, hence p(:~) = (::: - ;::0) 2n l(z). Problem Set 16.3, page 717 1. i. 4i 5. 1/5! (at::: = 0) (at ~ = I), ~ (at z = -1) 9. 15. el/z = 1 + 1/;:: + ... , AilS. 2m 19. -4'ITi sinh ~7I -! 23.0 3. -~i (at ~ = 2i), ~i (at - 2i) 7. I (at :::':::ll'1T) 11. -1 (at z = :::':::~71", :::':::~71", ... ) 17. Simple poles at :!::~. Am. -4i 21. -4i sinh ~ 25. ~ (at ~ = ~), 2 (at ~ = ~). Ans. 5m Problem Set 16.4, page 725 1. 271/\;TI 7.0 13.71"/16 19.0 25.0 5. 271/3 3.271/35 9. 1f 15.0 21. 0 27. -71"/2 11. 271/3 17.71"12 23.71" Chapter 16 Review Questions and Problems, page 726 17. 271"i/3 23. m/4 27.6m 33.0 19. 571" 25.0 (11 even), 29. 271/7 35. 71/2 (_I)<n-1)/2 21. ~'IT cos 10 271"i/(/1 - I)! (/1 odd) 31. 4'IT/V3 A42 App. 2 Answers to Odd-Numbered Problems Problem Set 17.1, page 733 3. Only the size 5. x = e, W = -y + ie, y = k. w = -k 7. -371"/4 < Arg w < 31714, Iwl < 1/8 11.lwl ~ 16, v ~ 0 15. In 2 ~ II ~ In 3, 71"/4 ~ v ~ 17/2 19.0, ±1, ±2, . . . + ix 9.lwl > 3 13. Annulus 3 < 17. ±1, ±i 21. -a/2 Iwl < 5 23. (/ and 0, ~ 25. M = eX = 1 when x = 27. M = 1IIzi = 1 on the unit circle, J = 111<:12 o. J = e 2x Problem Set 17.2, page 737 _ __ - ~.~. -nv -2w + 3 9. z = 0 13. z = ±i 17. (/ - d = 0, ble = 1 by (5) 5i 7.;::= - - 4w - 2 n. <: = ~ + i ± Vi + i 15. w = 4/z, etc. 19. w = add (a *- 0, d *- 0) Problem Set 17.3, page 741 5. Apply the inverse g of f on both sides of Zl = f(zl) to get g(Zl) 7. w = (<: + 2i)/(z - 2i) 9. w = z - 4 11. w = 15. w = 19. w = liz (z + 1)/(-3z + I) 4 (Z4 - i)/(-i: + 1) = 13. w = (3iz + 1)lz 17. w = (2z - i)/(-iz - 2) Problem Set 17.4, page 745 1. Annulus 1 ~ Iwl ~ e 2 3. liVe < Iwl < Ye, 371"/4 < arg w < 571"/4 5. 1 < Iwl < e. u > 0 7. w-plane without 0 9. u 2 /cosh 2 1 + v 2 /sinh 2 1 ~ 1, u ~ 0 11. Elliptic annulus bounded by u2/cosh2 I + v 2 /sinh 2 1 = I and u 2/cosh 2 5 + v 2 /sinh 2 5 = \ 13. ::!:(211 + 1)17/2, II = 0, \, ... 15.0 < [m t < 71" is the image of R under t = Z2. AilS. e t = e z2 17. O. ±i, ±2i, ... 19. 112/cosh 2 1 + v 2 /sinh 2 1 ~ \, v < 0 21. v < 0 23. -I ~ l/ ~ I. v = 0 (c = 0). u 2 /cosh2 e + u2 /sinh 2 e = I (e *- 0) 25. In 2 ~ l/ ~ In 3, 71"/4 ~ v ~ 7[/2 Problem Set 17.5, page 747 1. w moves once around the unit circle. 7. - i/2, 3 sheets g(f(zl» 5. - 5/3, 2 sheets 9. 0, 2 sheets = Zl· App. 2 Answers to Odd-Numbered Problems A43 Chapter 17 Review Questions and Problems, page 747 11. 15. 17. 23. 27. 33. 39. 45. 1I = ~V2 I, ~V2 - I The domain between II Iw + ~I = ~ 0, (:::'::: 1 :::'::: i)/V2 0, :::':::i/V2 tV = zi(z + 2) I + i : :': : v'1+2i iz 3 + 1 = 13.lwl = 20.25, larg wi < n12 ~ - v 2 and It = 1 - ~V2 19. 1I = 1 21.larg wi < 7T/4 25. n/8 :::'::: 117T/2. 11 = O. I, ... 29. tV = iz 31. w = liz 35. ::!:V2 37. 2 : :': : v'6 41. w = e3z 43. z2 12k Problem Set 18.1, page 753 3. 110 - SOx)" 110 + 25iz 2 1. 20x + 200, 20z + 200 7. F = 200 - (lOOlln 2) Ln z 5. F = (1lO/ln 2)Ln z 13. Use Fig. 388 in Sec. 17.4 with the z- and w-planes interchanged, and cos z = sin (z + !7T). 15. <1> = 220 - 110xy Problem Set 18.2, page 757 2 )') c]:> 2 2 2 1• 1I 2 - v 2 = e2 X(cos 2 ] \' - sin, xx = 4e X(cos )' - sin .v) = -c]:> YY' V2c]:> = 0 3. Straightforward calculation, involving the chain rule and the Cauchy-Riemann equations 5. See Fig. 389 in Sec. 17.4. c]:> = sin 2 x cosh2 J - cos 2 x sinh 2 y. 9. (i) c]:> = U1(l - "\y). (ii) w = iz 2 maps R onto -2 ~ 1I ~ O. thus <1>* = U1(l + !1I) = U1(l + -2x),». 11. By Theorem 1 in Sec. 17.2 13. <1> = 10[1 - (lin) Arg (z - 4)J, F = lOll + (i/n) Ln (z - 4)1 15. Corresponding rays in the w-plane make equal angles, and the mapping is conformal. i( Problem Set 18.3, page 760 3. (lOO/d»'. Rotate through 7T12. 5. 100 - 2408/7T 7. Re F(z) = 100 + (200/n) Re (arcsin z) 9. (240/n) Arg z 11. To + (2/7T)(TI - To) Arg z 13.50 + (400/7T) Arg z Problem Set 18.4, page 766 1. V = iV2 = iK, 'It = - Kx = canst, c]:> 3. F(z) = Kz (K positive real) 5. V = (I + 2i)K. F = (l - 2i)K::. 7. F(z) = Z3 = Ky = canST 9. Hyperbolas (x + I)y = COllst. Flow around a corner formed by x x-axis. 11. Y/(X2 + .1'2) = C or X2 + (y - k)2 = k 2 13. F(z) = ::.Iro + rolz = -} and the A44 App. 2 Answers to Odd-Numbered Problems z with the roles of the z- and w-planes 15. Use that w = arccos Z is w = cos interchanged. Problem Set 18.5, page 771 5. I - ,.2 cos 28 7. 2(r sin 8 - !r2 sin 28 + ~,.3 sin 38 - + ... ) 9. ~r2 sin 28 - ~r6 sin 68 11. ~7T2 - 4(,. cos 8 - ~r2 cos 28 + ,.3 sin 38 + -1 13. -4 ( r sin 8 - -I 9 7T + ... ) ~r3 cos 38 - ~ r5 + ... ) sin 58 - Problem Set 18.6, page 774 1. 7. 13. 15. No; Izl2 is not analytic. 3. Use (2). F(~) = 2i 5. cJ:>(4, -4) = -12 Use (3). cJ:>(L 1) = -2 11. IF(ei 1)1 2 = 2 - 2 cos 'lA, e = 7T12, Max = 2 IF(z) I = [cos 2 2x + sinh2 2y]1I2, z = ±i, Max = [1 + sinh 2 2]112 = cosh 'l = 3.7622 No Chapter 18 Review Questions and Problems, page 775 11. 13. 17. 23. cJ:> = 10(1 - x + y), F = 10 - 10(1 (201ln 10) Ln ;:: Arg z = const T(x. y) = x(2y + I) = const 27. F(;::) = - c Ln (;:: - 5), Arg (;:: - 5) + i)z. 15. (101In 1O)(ln 100 - In r) 19. (-i/7T) Ln z 25. Circles (x - C)2 + )'2 = c 2 = C 27T 29.20 + -80 ( r sin 7T e + -I ,.3 sin 3fJ + -I ,.5 sin 5fJ + ... ) 3 5 Problem Set 19.1, page 786 1. 0.9817' 102, -0.1010' 103 ,0.5787' 10- 2, -0.1360' 105 3.0.36443/(17.862 - 17.798) = 0.36443/0.064 = 5.6942, 0.3644/(17.86 - 17.80) = 0.3644/0.06 = 6.073, 0.364/(17.9 - 17.8) = 3.64, impossible 5. 0.36443(17.862 + 17.798) 17.8622 _ 17.7982 = 0.36443' 35.660 12.996 319.05 _ 316.77 = ~ = 5.7000, 13.00 13.0 13 10 -22 = 5.702, ~2 = 5.70. = 5.7. . 8 _. 8 2.3 2 = 5 7. 19.95,0.049,0.05013: 20, 0, 0.05 9. In the present calculation, (b) is more accurate than (a). 11. -0.126' 10-2, -0.402' 10-3 ; -0.267' 10-6 , -0.847' 10- 7 13. Add first, then round. 15. ~ Q2 = ~I + EI Q2 + 102 = a l : a2 EI (1 _ ~2 + ~2: a2 a2 _ + ...) = ~I + ~l a2 a2 _ ~2 . ~1 a2 Q2 , App. 2 A45 Answers to Odd-Numbered Problems hence l al)/I -all = lEII( -a - -:::::a2 a2 al {/2 - E21 ~ IErll + IEr21 ~ {3rl + {3r2 a2 19. (a) 19121 = 0.904761905, Echop = Eround = 0.1905.10- 5 , 5 E,·.chop = Er.round = 0.2106.10- , etc. Problem Set 19_2, page 796 1. g = 1.4 sin x, 1.37263 (= X5) 5. g = X4 + 0.2, 0.20165 (= x 3 ) 7. 2.403 (= X5' exact to 3S) 9.0.904557 (= x 3 ) 11. 1.834243 (= X4) 13. Xo = 4.5, X4 = 4.73004 (6S exact) 15. (a) 0.5, 0.375, 0.377968, 0.377964; (b) IIV7 = 0.377964 473 17. Xn+l = (2xn + 7/xn2 )/3, 1.912931 (= X3) 19. (a) Algorithm Bisect (f. ao_ bo, N) Bisection Method This algorithm computes an interval [an, b1 J containing a solution of f(x) = 0 (f continuous) or it computes a solution Cn' given an initial interval lao, bol such that f(ao)f(b o) < O. Here N is determined by (b - a)I2 N ~ {3, {3 the required accuracy. INPUT: Initial interval lao, boL maximum number of iterations N. OUTPUT: Interval LaN, bNl containing a solution, or a solution Cn. For 11 = 0, I, .. " N - I do: Compute Cn = ~(an + b,J. If f(c n ) = 0 then OUTPUT Cn. Stop. [Procedure completed] Else continue. If f(an)f(c n ) < 0 then an+l = an and bn +l = cn. Else set {/n+l = Cn and b n + 1 = bn . End OUTPUT LUN' bN ]. Stop. [Procedure completed] End BISECT Note that [aN' bNl gives (aN + bN )!2 as an approximation of the zero and (b N - aN)!2 as a cOlTesponding elTor bound. (b) 0.739085; (c) 1.30980, 0.429494 21. 1.834243 23.0.904557 Problem Set 19.3, page 808 1. Lo(x) = -2.r + 19, L 1 (x) = 2x - 18, Pl(X) = 0.1082x + 1.2234, Pl(9.4) = 2.2405 3.0.9971,0.9943.0.9915 (0.9916 4D), 0.9861 (0.9862 4D), 0.9835, 0.9809 5. P2(X) = -0.44304x 2 + 1.30906x - 0.02322, P2(0.75) = 0.70929 7. P2(X) = -0.1434x2 + 1.0895x, P2(0.5) = 0.5089, P2(1.5) = 1.3116 9. La = -t(x - I)(x - 2)(x - 3), Ll = !x(x - 2)(x - 3), ~ = -~x(x - 1)(x - 3), ~ = tx(x - I)(x - 2); P3(X) = 1 + 0.039740x - 0.335187x2 + 0.060645x 3; P3(O·5) = 0.943654 (6S-exact 0.938470), P3(1.5) = 0.510116 (0.511828), P3(2.5) = -0.047993 (-0.048384) A46 App. 2 Answers to Odd-Numbered Problems 13. P2(X) = 0.9461x - 0.2868x(x - 1)/2 = -0.1434x2 + 1.0895x 15.0.722,0.786 17. 8f1/2 = 0.057839, 8f3/2 = 0.069704. etc. Problem Set 19.4, page 815 9. [-1.39(x - 5)2 + 0.58(x - 5)3]" = 0.004 at x = 5.8 (due to roundoff; should be 0). 11. 1 - ~X2 + ~X4 13.4 - 12x2 - 8x 3 ,4 - 12x2 + 8x 3 . Yes 15. I - x 2 , -2(x - l) - (x - 1)2 + 2(x - 1)3, -1 + 2(x - 2) + 5(x - 2)2 - 6(x - 2)3 17. Curvature f"/(I + j'2)3/2 = f" if If'l is small 19. Use that the third derivative of a cubic polynomial is constant, so that gill is piecewise constant, hence constant throughout under the present assumption. Now integrate three times. Problem Set 19.5, page 828 1. 0.747131 3. 0.69377 (5S-exact 0.69315) 7.0.894 (3S-exact 0.908) 5. 1.566 (4S-exact 1.557) 9.1h /2 + Eh/2 = 1.55963 - 0.00221 = 1.55742 (6S-exact 1.55741) 11. J h /2 + Eh/2 = 0.90491 + 0.00349 = 0.90840 (5S-exact 0.90842) 13. 0.94508, 0.94583 (5S-exact 0.94608) 15.0.94614588, 0.94608693 (8S-exact 0.94608307) 17. 0.946083 (6S-exact) 19. 0.9774586 (7S-exact 0.9774377) 21. x - 2 = t, 1.098609 (7S-exact 1.098612) 23. x = !U + 1),0.7468241268 (lOS-exact 0.7468241330) 25. (a) M2 = 2. M2* = ~, IKM21 = 2/(J2n 2). 11 = 183. (b) iv) = 24/x 5 , 2m = 14 27. 0.08, 0.32, 0.176, 0.256 (exact) 29.5(0.1040 - !. 0.1760 + i· 0.1344 - ~. 0.0384) = 0.256 r Chapter 19 Review Questions and Problems, page 830 17.4.266,4.38, 6.0, impossible 19.49.980,0.020; 49.980, 0.020008 21. 17.5565 ~ s ~ 17.5675 23. The same as that of a. 25. -0.2, -0.20032, -0.200323 27.3,2.822785,2.801665,2.801386, VlOI386 29.2.95647.2.96087 31. 0.26, M2 = 6, M2* = 0, -0.02 ~ E ~ 0, 0.24 ~ a ~ 0.26 33. 1.001005, -0.001476 ~ E ~ 0 Problem Set 20.1, page 839 1. Xl = -2.4, X2 = 5.3 5. Xl = 2, X2 = 1 3. No solutlOn 7. Xl = 6.78, x2 = -11.3, X3 = 15.82 App. 2 A47 Answers to Odd-Numbered Problems 9. Xl = 0, x 2 = T1 arbitrary, X3 = 5f1 + 10 11. Xl = fl , x 2 = t2, both arbitrary, X3 = 1.25fl 13. Xl = 1.5, X2 = -3.5, X3 = 4.5, X4 = -2.5 - 2.25t2 Problem Set 20.3, page 850 Exact 21.5, 0, -13.8 5. Exact 2, I, 4 7. Exact 0.5, 0.5, 0.5 (3 )T = [0.49982 0.50001 0.50002], (b) X (3)T = [0.50333 0.49985 0.49968] 6, 15, 46, 96 steps; spectral radius 0.09, 0.35, 0.72, 0.85, approximately [1.99934 1.00043 3.99684]T (Jacobi, step 5): [2.00004 0.998059 4.00072jT (Gauss-Seidel) 17. v'306 = 17.49, 12, 12 19. V 18k2 = 4.24Ikl, 41kl, 41kl 3. 9. 11. 13. (a) X A48 App. 2 Answers to Odd-Numbered Problems Problem Set 20.4, page 858 1. 5. 7. 13. 17. 21. [t -] ~] -0.1 1] 12, v'62 = 7.87,6, 1.9, V1.35 = 1.16, 1. lO.3 6, \''6, 1. [1 1 K = 100· 100 46 ~ 6· 17 or 7 . 17 [-0.6 2.8]T 0.5 3.14, V56 = 7.07,4, [-1 I ~ -~] 1.0] 11. II AliI = 17, II A-I 111 = 17, K = 289 15. K = 1.2' 1~~ = 1.469 19. [0 11T, [1 -OAIT,289 23.27,748,28375,943656,29070279 Problem Set 20.5, page 862 1. 5. 11. 13. 3. 8.95 - 0.388x -11.4 + 5Ax s = -675 + 901, Vav = 90 kmlh 9.4 - 0.75x - 0.125x2 5.248 + 1.543x, 3.900 + 0.5321x + 2.021x2 -2.448 + 16.23x, -9.114 + 13.73x + 2.500x 2, -2.270 + 1.466x -1.778x 2 + 2.852x 3 Problem Set 20.7, page 871 1. 5 ~ A ~ 9 3.5,0, 7; radii 4, 6, 6 5.IA - 4il ~ v'2 + 0.1. IAI ~ 0.1, IA - 9il ~ v'2 7.111 = 100,122 = 133 = I 9. They lie in the intervals with endpoints ajj :::t::: (11 - 1)10- 6 . (Why?) 11. 0 lies in no Gerschgorin disk, by (3) with >; hence det A = Al ... An 13. peA) ~ Row sum norm II A IIx = max L J 15. Vi53 = 12.37 17. lajkl = max (Iaiil k Vl22 = 11.05 =1= O. + GerschgOlin radius) J 19. 6 ~ A ~ 10. 8 ~ A ~ 8 Problem Set 20.S, page 875 1. q = 4,4.493.4.4999: lEI ~ 1.5.0.1849,0.0206 3. q = 8,8.1846,8.2252; lEI ~ L 0.4769, 0.2200 5. q = 4,4.786,4.917; lEI ~ 1.63,0.619,0.399 7. q = 5.5,5.5738.5.6018: lEI ~ 0.5. 0.3115, 0.1899: eigenvallle~ (4S) 1.697,3.382, 5.303.5.618 9. )' = Ax = Ax, yTx = AxTx, yTy = A2xTx, E2 ~ '~/Y/XTX - (yTX/XTl\.)2 = A2 - A2 = 0 11. q = 1. ... , - 2.8993 approximates - 3 (0 of the given matrix), lEI ~ 1.633, .. ',0.7024 (Step 8) Problem Set 20.9, page 882 [ 1. 3~ - ~.!lO2776 -UlO1776 6.730769 l.H-l6154 L~61~1 ~[ 1.769230 0.9g0000 -0.-1--1-1814 -O.-1-·HS1-1- 0.H70l64 U37iOO:] 0 0.371803 0.4H9H36 App.2 A49 Answers to Odd-Numbered Problems 5. Eigenvalues 8, 3, 1 -2.50867 r 5.M516 01 r 7.45139 0.374953 , -1.56325 5.307219 -2.5086~ 0.374953 r 7.91494 1.04762 0.0983071 1.00446 03124691 :1.000482 3.08458 0.0312469 r18.3171 IB3786 o0.360275 1 , r 0.396511 0.881767 ~.881767 8.29042 0.360275 r18.3910 1.39250 0.396511 8.24727 0 0.0600924 :O6~241· 1.37414 0.177669 ~.177669 9. 1 0098307 3.544142 -0.646602 -~.646602 7. 0 -1.56325 8.23540 :01lm141 0.0100214 1.37363 7an24 0.0571287 ~.0571287 4.00088 r 7OO322 1 r7~n98 0.0249333 , 0.0326363 0.0326363 o 0.0249333 0.996875 4.U0034 0 0.00621221 :=21221 0.996681 0.0186419 ~.0186419 r 4.00011 :.001547821 U.00154782 0.Y96669 Chapter 20 Review Questions and Problems, page 883 17. r4 -I 2{ 19. L6 -3 I{ 21. All nonzero entries of the factors are 1. 23. 2.8193 -1.5904 -1.5904 1.2048 -0.0482 -0.0241 [ 27. 15, ,189, 8 =~:::~l ,/21, 4 31. 14, ,178, 7 37. 11.5 . 4.4578 = 51.2651 35.9 39. Y = 1.98 1 0.1205 29.7, 33. 6 25. Exact [-2 (4D-values) + 0.98x 41. Centers 1. 1, 1. radii 2.5. 1. 2.5 (A = 2.944.0.028 ::':: 0.290i, 3D) 43. Centers 5, 6, 8: radii 2,1, I, (A = 4.1864. 6.4707, 8.3429. 5S) -2.23607 [ 45. 15 -~.23607 o 5.8 -3.1 -3.1 6.7 J ,Step3: [ 9.M913 -1.06216 0 -1.06216 4.28682 -0.00308 -:00308J 0.26345 A50 App. 2 Answers to Odd-Numbered Problems Problem Set 21.1, page 897 1. Y = eX, 0.0382, 0.1245 (elTor of X5, .\"10) 3. Y = x - tanh x (set y - x = til, 0.009292, 0.0188465 (elTor of .\"5, XIO) 5. y = eX, 0.001275, 0.004200 (elTor of X5' XIO) 7. Y = 11(1 - x 2 /2), 0.00029, 0.01187 (elTor of X5' xw) 9. y = 11(1 - x 212), 0.03547. 0.28715 (elTor of X5' XIO) 11. Y = 11(1 - x 2 /2); error -10- 8 , -4' 10-8 , . . " -6' 10-7 , + 10- 5; about 1.3' 10-5 by (10) 13.y = xe x ; eITor'l~ (for x = L"" 3) 19,46,85,139,213,315,454,640,889,1219 15. Y = 3 cos x - 2 cos 2 x; error' 107 : 0.18, 0.74, 1.73,3.28,5.59,9.04, 14.33,22.77, 36.80, 61.42 17. Y = 1I(x5 + 1), 0.000307, -0.000259 (error of X5' XIO) 19. The elTors are for E.-c. 0.02000, 0.06287. 0.05076. for Improved E.-C. -0.000455, 0.012086,0.009601, for RK 0.0000012,0.000016,0.000536. Problem Set 21.2, page 901 3. y = e- O. 1x2 ; elTors 10-6 to 6· 10-6 5. y = tan x; .\'4' .. " )'10 (error' 105): 0.422798 (-0.48),0.546315 (-1.2),0.684161 (-2.4),0.842332 (-4.4),1.029714 (-7.5),1.260288 (-13),1.557626 (-22) 7. RK-elTor smaller. elTor' 105 = 0.4, 0.3, 0.2. 5.6 (for x = 0.-1-, 0.6, 0.8, 1.0) 9• .\'4 = 4.229690, Y5 = 4.556 859, )'6 = 5.360657. Y7 = 8.082 563 11. ElTors between -6 . 10-7 and + 3 . 10- 7 . Solution eX - x - I 13. Errors' 105 from x = 0.3 to 0.7: -5, -II, -19, -31, -47 15. (a) 0, 0.02, 0.0884, 0.215 848,)'4 = 0.417818'.\'5 = 0.708887 (poor). (b) By 30-50% Problem Set 21.3, page 908 3. )'1 = eX, )'2 = -ex. elTors range from ±0.02 to ±0.23, monotone. 5. )'~ = Y2, )'~ = -4)'1' )' = Yl = I, 0.84, 0.52, 0.0656, -0.4720; y = cos 2x x x 7')'1 = 4e- sin x')'2 = 4e- cos x; elTors from 0 to about ±O.I 9. ElTors smaller by about a factor 104 11. Y = 0.198669,0.389494,0.565220,0.719632,0.847790; y' = 0.980132,0.922062,0.830020,0.709991,0.568572 13. Y1 = e- 3 .1: - e- 5 :r')'2 = e- 3 .1: + e- 5 .1:;)'1 = 0.1341. 0.1807, 0.1832, 0.1657, 0.1409;)'2 = 1.348,0.9170.0.6300,0.4368.0.3054 17. You gel the exact solution, except for a roundoff elTor [e.g., Yl = 2.761 608, y(0.2) = 2.7616 (exact), etc.]. Why? 19. Y = 0.198669,0.389494,0.565220,0.719631. 0.847789; y' = 0.980132,0.922061. 0.830019, 0.709988, 0.568568 Problem Set 21.4, page 916 3. 105, 155, 105, 115; Step 5: 104.94, 154.97, 104.97, 114.98 5.0.108253, 0.108253,0.324760,0.324760; Step 10: 0.108538, 0.108396, 0.324902, 0.324831 App. 2 A51 Answers to Odd-Numbered Problems 7. 0, O. O. O. All equipotentia11ines meet at the comers (why?). Step 5: 0.29298. 0.14649,0.14649.0.073245 9. - 3un + U12 = -200, Un - 3U12 = -100 11. U12 = U32 = 31.25, U21 = U23 = 18.75, ujk = 25 at the others 13. U21 = U23 = 0.25, U12 = U32 = -0.25, Ujk = 0 else 15. (a) Un = -U12 = -66. (b) Reduce to 4 equations by symmetry. Un = U31 = -U15 = - 1I35 = -92.92, U21 = -U25 = -87.45, U12 = U 32 = -U 14 = -U 34 = -64.22, U 2 2 = -U24 = -53.98, U13 17. = U23 = U33 = 0 \13, Un = U21 = 0.0849, U12 = U22 = 0.3170. (0.1083, 0.3248 are 4S-values of the solution of the linear system of the problem.) Problem Set 21.5, page 921 5. Un = 0.766. U21 = 1.109. U12 = 1.957. U22 = 3.293 7. A as in Example I, right sides -2, -2, -2, -2. Solution Un = U21 = 1.14286, U 1 2 = U22 = 1.42857 11. -4un + U21 + U12 = - 3. Un - 4U21 + U22 = -12, Un - 4U12 + U22 = 0, 2U21 + 2U12 - 12u22 = - 14. Un = U22 = 2, U21 = 4. U12 = 1. Here -14/3 = -~(1 + 2.5) with 4/3 from the stencil. 13. b = [-380 -190, -190, O]T; Un = 140, U21 = U 1 2 = 90, U22 = 30 Problem Set 21.6, page 927 5.0.1636.0.2545 (t = 0.04. x = 0.2,0.4).0.1074.0.1752 (t = 0.08),0.0735.0.1187 (t = 0.12),0.0498,0.0807 (t = 0.16),0.0339,0.0548 (t = 0.2; exact 0.0331,0.0535) 7. Substantially less accurate, 0.15, 0.25 (1 = 0.04),0.100,0.163 (t = 0.08) 9. Step 5 gives 0,0.06279,0.09336,0.08364,0.04707, O. 11. Step 2: 0 (exact 0),0.0453 (0.0422),0.0672 (0.0658), 0.0671 (0.0628),0.0394 (0.0373), 0 (0) 13.0.1018,0.1673,0.1673,0.1018 (t = 0.04),0.0219,0.0355, ... (t = 0.20) 15.0.3301,0.5706.0.4522.0.2380 (t = 0.04).0.06538.0.10604,0.10565.0.6543 (t = 0.20) Problem Set 21.7, page 930 1. For x = 0.2, 0.4 we obtain 0.012, 0.02 (t = 0.2), 0.004, 0.008 (t = 0.4), -0.004, -0.008 (t = 0.6). etc. 3. u(x, 1) = 0, -0.05, -0.10, -0.15, -0.075,0 5.0.190,0.308,0.308,0.190 (0.178, 0.288, 0.288, 0.178 exact to 3D) 7.0,0.354.0.766, 1.271, 1.679. 1.834.... (t = 0.1); 0.0.575.0.935, 1.135, 1.296. 1.357, ... (t = 0.2) Chapter 21 Review Questions and Problems, page 930 17. y = tan x; 0 (0),0.10050 (-0.00017). 0.20304 (-0.00033), 0.30981 (-0.00047), 0.42341 (-0.00062), 0.54702 (-0.00072) ASl App. 2 Answers to Odd-Numbered Problems 19. 0.1 003349 (0.8 . 10-7 ) 0.2027099 (I.6 . 10- 7 ), 0.3093360 (2.1 . 10-7 ). 0.4227930 (2.3· 10-7 ),0.5463023 (1.8' 10-7 ) 25.y(0.4) = 1.822798,.\'(0.5) = 2.046315,)'(0.6) = 2.284161,)'(0.7) = 2.542332, y(0.8) = 2.829714, y(0.9) = 3.160288, .v(1.0) = 3.557626 27'.\"1 = 3e- 9x,.I"2 = -5e- 9:r, [1.23251 -2.05419J, [0.506362 -0.843937],···, [0.035113 -0.058522] 29. 1.96, 7.86, 29.46 31. II(P l l ) = 1I(P31) = 270. U(P21 ) = U(P13 ) = U(P23 ) = U(P33 ) = 30, U(P 12 ) = U(P32 ) = 90, 1I(P22 ) = 60 35.0.06279,0.09336,0.08364,0.04707 37.0, -0.352, -0.153,0.153,0.352,0 if t = 0.12 and 0,0.344,0.166, -0.166, -0.344, 0 if t = 0.24 39.0.010956.0.017720.0.017747,0.010964 if t = 0.2 Problem Set 22.1, page 939 3. f = 3(-"1 - 2)2 + 2(X2 + 4)2 - 44. Step 3: [2.0055 - 3.9Y75]T 5. f = 0.5(x1 - 1)2 + 0.7(X2 + 3)2 - 5.8, Step 3: rO.99406 -3.0015]T 7. f = 0.2(X1 - 0.2)2 + X2 2 - 0.008. Step 3: [0.493 -O.Oll]T, Step 6: [0.203 0.004]T Problem Set 22.2, page 943 1. X3, X4 unused time on MI' M 2. respectively 3. No 11. fmax = f(O. 5) = 10 13. f max = f(9, 6) = 36 15. fmin = f(3.5, 2.5) = - 30 17. X1/3 + x2/2 ~ 100, x1/3 + -"2/6 ~ 80, f = 150X1 + fmax = f(210, 60) = 37500 19. 0.5X1 + 0.75x2 ~ 45 (copper), 0.5X1 + 0.25x2 ~ 30, f = 120x1 + 100x2, fmax = f(45, 30) = 8400 Problem Set 22.3, page 946 1. f(12011 I, flOIl 1) = 48011 I 2100 200) 3. f ( - 3 - ' 2/3 = 78000 5. Matrices with Rows 2 and 3 and Columns 4 and 5 interchanged 7. f(O, [0) = -10 9. f(5, 4, 6) = 478 Problem Set 22.4, page 952 1. f(-l-. -1-) = 72 7. f(l, 1, 0) = 12 3. f(10, 30) = 50 f(!, 0, ~) = 3 5. f(lO, 5) = 5500 9. Chapter 22 Review Questions and Problems, page 952 n. Step 5: [0.353 -0.028]T. Slower 13. Of course! Step 5: [-1.003 1.897]T 21. f(2, -1-) = 100 23. f(3, 6) = -54 25. /(50, 100) = 150 App. 2 Answers to Odd-Numbered Problems A53 Problem Set 23.1, page 958 o o 9. 0 o o 1 15. 0 o o o o 0 CD-------® 11. 1 f: 0 15'm 3 ] 0 0 0 0 0 0 0 0 0 0 13. 0 4 Edge o 21. >< 2 > 3 ~<ll 0 25. 0 E 23. 0 4 1 >< <ll > 0 2 3 0 Vertex 1 2 3 4 Incident Edges -eh -e2, e3, -e4 el e2, -e3 e4 Problem Set 23.2, page 962 1.4 5.4 3.5 9. The idea is to go backward. There is a VIc-I adjacent to Vk and labeled k - 1, etc. Now the only vertex labeled 0 is s. Hence A(vo) = 0 implies Vo = s, so that Vo - VI - ... - Vk-l - Vic is a path s ~ Vic that has length k. 15. No; there is no way of traveling along (3, 4) only once. 21. From Tn to 100m, 10m, 2.5m, 111 + 4.6 Problem Set 23.3, page 966 1. 3. 5. 7. (1. 2). (1, 2), (1,4), (1, 5), (2,4). (I, 4), (2, 4), (2. 3), (4, (2, (3, (2. 3); 3); 4), 6), L2 = 6. L3 = 18. L4 = 14 L2 = 2, L3 = 5, L4 = 5 (3, 5); L2 = 4, L3 = 3, L4 = 2, L5 = 8 (3. 4), (3, 5); ~ = 9, ~ = 7, L4 = 8, L5 = 4, L6 = 14 Problem Set 23.4, page 969 2 1. 1 :, / 3 L = 12 I / 3. 4 ,\""2 3- 5 4"\ 5 L = 10 8 / 5. 1 - 2"\ 5 3~ 6- 4 "\ 7 L = 28 A54 App. 2 Answers to Odd-Numbered Problems 2 / L = 38 11. Yes 5-6, 15. G is connected. If G were not a tree, it would have a cycle, but this cycle would provide two paths between any pair of its vertices, contradicting the uniqueness. 19. If we add an edge (u, u) to T, then since T is connected, there is a path U ~ u in T which. together with (II, u), forms a cycle. 9. I - 3 - 4 '\. Problem Set 23.5, page 972 1. (I, 2), (1.4), (3, 4), (4,5). L = 12 3. (I. 2). (2, 8), (8, 7), (8. 6), (6, 5), (2, 4), (4, 3), L 5. (1,4), (3,4). (2,4). (3,5), L = 20 = 40 7. (I, 2), (I, 3), (I, 4), (2, 6), (3, 5), L = 32 11. If G is a tree 13. A shortest spanning tree of the largest connected graph that contains vertex 1 Problem Set 23.6, page 978 1. 3. 5. 7. I - 2 - 5, Ilf = 2; 1 - 4 - 2 - 5, Ilf = 2, etc. I - 2 - 4 - 6 . .1f = 2; I - 2 - 3 - 5 - 6. Ilf = I, etc. f12 = 4, f13 = 1. f14 = 4. f42 = 4. f43 = 0,125 = 8, f35 = 1, f = 9 f12 = 4. f13 = 3, f24 = 4, f35 = 3, f54 = 2, f4fj = 6, f56 = 1, f = 7 ~ {~5,6},28 11. {2,~ 6},50 13. I - 2 - 3 - 7, lJ.f = 2; I - 4 - 5 - 6 - 7, Ilf = 1; 1 - 2 - 3 - 6 - 7, Ilf = 1; fmax = 14 15. {3, 5, 7}. 22 17. S = {I, 41. cap (S. 19. If fii < Cij as well as fii > 0 n= 6 + 8 = 14 Problem Set 23.7, page 982 3. 5. 7. 9. (2, 3) and (5. 6) 1 - 2 - 5, .:It = 2; 1 - 4 - 2 - 5, ~t = I; f = 6 + 2 + 1 = 9 1 - 2 - 4 - 6, .:It = 2: 1 - 3 - 5 - 6. Il t = 1; f = 4 + 2 + I = 7 By considering only edges with one labeled end and one unlabeled end 17. S = {I, 2,4, 51. T = {3, 6}, cap (S, n = 14 Problem Set 23.8, page 986 1. No 3. No 5. Yes, S = { I, 4, 5, 8} 7. Yes; a graph is not bipartite if it has a nonbipartite subgraph. 9.1 - 2 - 3 - 5 11. (1, 5), (2, 3) by inspection. The augmenting path I - 2 - 3 - 5 gives I - 2 - 3 - 5, that is, (\, 2), (3, 5). 13. (1,4), (2. 3). (5, 7) by inspection. Or (1, 2), (3, 4), (5, 7) by the use of the path 1 - 2 - 3 - 4. 15. 3 25. K3 19. 3 23. No; K5 is not planar. App. 2 ASS Answers to Odd-Numbered Problems Chapter 23 Review Questions and Problems, page 987 0 0 13. r~ 1 ] 0 0 0 0 15. 0 0 0 0 o o o CD---0 o o o o 17. 21. o 1 19.\ o o CD o 23.4 Incident Edges Ve11ex / e2, -e3 2 3 -eb e3 'eb -e2 25.4 29. I - 4 - 3 - 2, L = 27. L2 = 10. L3 33. f = 7 16 = 15, L4 = 13 Problem Set 24.1, page 996 1. qL 19, qM = 20, qu = 20.5 5. qL = 69.7, qM = 70.5, qu = 71.2 9. qL = 399, qM = 401, qu = 401 13..r = 70.49, s = 1.047,IQR = 1.5 17. 0 0 300 = 3.qL = 38,QM = 44,Qu = 54 7. qL = 2.3, qM = 2.4, qu = 2.45 11. x = 19.875, s = 0.835, IQR = 1.5 15. x = 400.4, s = 1.618, IQR = 2 19. 3.54, 1.29 Problem Set 24.2, page 999 1. 4 outcomes: HH, HT, TH, TT (H = Head, T = Tail) 3.62 = 36 outcomes (1, 1), (1, 2), .. " (6, 6) 5. Infinitely many outcomes S, SCS, ScScS, ... (S = "Six") 7. The space of ordered triples of nonnegative numbers 9. The space of ordered pairs of numbers 11. Yes 13. E = IS, scs, SCSCS}, E" = {SCSCSCS, ScScScScS, ... } (S = "Six") Problem Set 24.3, page 1005 1. (a) 0.9 3 = 72.9%, (b) 190~ • ~~. ~~ = 72.65% 3 490. 489 • 488 • 487 • 486 - 90 3501. • 500 499 498 497 496 - 5. 1 - 2~ = 0.96 9. P(MMM) + P(MMFM) . /0 7. I - 0.75 2 = 0.4375 < 0.5 + P(MFMM) + P(FMMM) = ~ + 3 . 1~ = 1~ A56 App. 2 Answers to Odd-Numbered Problems 11. 3~ + ~~ - 3~ = ~~ by Theorem 3, or by counting outcomes 13. 0.08 + 0.04 - 0.08 . 0.04 = 11.68% 17. 1 - 0.97 4 = 11.5% 15.0.95 4 = 81.5% Problem Set 24.4, page 1010 5. (2~) 3. In 40320 ways 7.210,70. 112.28 11. (~~) = 635013559600 15.676000 = 1140 = 1260. AIlS. 111260 9. 9!/(2!3!4!) 13. 1184, 5121 Problem Set 24.5, page 1015 1. k = 1/55 by (6) 3. k = II8 by (10) 7. 1 - P(X ~ 3) = 0.5 5. No because of (6) 9. P(X > 1200) f = 2 6[0.25 - (x - 1.5)2] dx = 0.896. AilS. 0.8963 = 72% 1.2 11. k 17. X = 2.5; 50% > b, X ~ b. 13. k = 1.1565; 26.9% X < c. X ~ c. etc. Problem Set 24.6, page 1019 1. 2/3, 1118 3.3.5.2.917 7. $643.50 5.4, 16/3 9. JL = lie = 25; P 13. 750, 1, 0.002 = 20.2% 11.~, 2~' (X - ~)V20 15. 15c - 500c 3 = 0.97. c = O.ms55 Problem Set 24.7, page 1025 1. 0.0625, 0.25, 0.9375, 0.9375 3.64% 5.0.265 7. f(x) = OS"e- o.5 /x!, f(O) + f(l) = e- O.5 ( 1.0 + 0.5) = 0.91. Am. 9% 9. I - e- O.2 = 18% 11. 0.99100 = 36.6% 13. ~~~, ~~~, 2~~' :di6 Problem Set 24.8, page 1031 1. 0.1587, 0.6306, 0.5, 0.4950 5.16% 9. About 23 13. t = 1084 hours 3. 17.29, 10.71, 19.152 7. 31.1 %, 95.5% 11. About 58st Problem Set 24.9, page 1040 1. 1/8, 3/16, 3/8 3. 2/9, 2/9, 1/2 5. f2(Y) = 11(/32 - ll'2) if ll'2 < Y < /32 and 0 elsewhere 7. 27.45 mm, 0.38 mm 9. 25.26 cm, 0.0078 cm App. 2 Answers to Odd-Numbered Problems 13. lndependent, .f1(X) 15. 50lJf A57 = O.le- O.lx if X> 0, f2(Y) = O.le- Oly if Y > 0, 36.8% 17. No Chapter 24 Review Questions and Problems, page 1041 23. x = 22.89, s = 1.028. 21. QL = 22.3, QM = 23.3, Qu = 23.5 25. H, TH, TTH, etc. 27. f(O) = 0.80816. f(l) = 0.18367. f(2) = 0.00816 29. Always B !: A U B. If also A !: B, then B = A U B, etc. 31. 7/3, 8/9 33. 118.019, 1.98, 1.65% 35.0, 2 37. JL = 100/30 39. 16%, 2.3% (see Fig. 520 in Sec. 24.8) S2 = 1.056 Problem Set 25.2, page 1048 3. 1 = pk(1 - p)n-k, 5. 11120 7. 1 = f(x), aOn l)/ap 9. it = x 13. = 1 p = kin, k = number of Sllccesses in n trials = lip - e p = 0, = 1/x 11. = nl'i. Xj = l!x 15. Variability larger than perhaps expected (x - 1)10 - p) e Problem Set 25.3, page 1057 1. CONFo.95 [37.47 ~ JL ~ 43.87} 3. Shorter by a factor v'2 5.4, 16 7. Cf. Example 2. n = 166 11. CONFo.99 [63.71 ::::; JL ~ 66.29} 9. CONFo.99 [20.07 ~ JL ~ 20.33} 13. c = 1.96, x = 87, S2 = 87' 413/500 = 71.86, k = cslVn = 0.743, CONFo.95 {86 ~ JL ~ 88}, CONF095 (0.17 ~ p ~ 0.18} 15. CONFo.95 (0.00045 ~ a 2 ~ 0.00l31} 17. CONFO.95 [0.73 ::::; a 2 ~ 5.19} 19. CONFO.95 [23 ~ a 2 ~ 548}. Hence a larger sample would be desirable. 21. Normal distributions, means -27,81, 133, variances 16, 144,400 23. Z = X + Y is normal with mean 105 and variance 1.25. Ans. P(l04 ~ Z ~ 106) = 63% Problem Set 25.4, page 1067 t = V7(0.286 - 0)/4.31 = 0.18 < c = 1.94; do not reject the hypothesis. c = 6090 > 6019; do nol reject the hypothesis. ifln = I, c = 28.36: do not reject the hypothesis. JL < 28.76 or JL > 31.24 Alternative JL =1= 1000, t = v'2O (996 - 1000)/5 = -3.58 < c = - 2.09 (Table A9, 19 degrees of freedom). Reject the hypothesis JL = 1000 g. 11. Test JL = 0 against JL =1= O. t = 2.11 < c = 2.36 (7 degrees of freedom). Do not reject the hypothesis. 13. ll' = 5%, c = 16.92 > 9.0.5 2 /0.4 2 = 14.06; do not reject hypothesis. 15. to = \1'10' 9·17119 (21.8 - 20.2)1\1'9.0.62 + 8.0.5 2 = 6.3 > c = 1.74 (17 degrees of freedom). Reject the hypothesis and assert that B is better. 1. 3. 5. 7. 9. ASS App. 2 17. Answers to Odd-Numbered Problems = 50/30 = 1.67 < c = 2.59 [(9, 15) degrees of freedom]: do not reject the hypothesis. Vo Problem Set 25.5, page 1071 1. LCL = I - 2.58 . 0.03/v6 = 0.968, VCL = 1.032 3.11 = 10 5. Choose 4 times the original sample size (why?). 7. 2.58VO.024/\12 = 0.283. VCL = 27.783, LCL = 27.217 11. In 30% (5%) of the cases, approximately 13. VCL = Ill' + 3Vllp(\ - p), CL = Ill', LCL = "l' - 3Vl1p(1 - p) 15. CL = JL = 2.5, VCL = JL + 3~ = 7.2, LCL = JL - 3~ is negative in (b) and we set LCL = O. Problem Set 25.6, page 1076 1. 0.9825, 0.9384, 0.4060 5. peA; e) = e- 30H(1 + 308) 7. P(A; 8) = e- 50 (J 11. (l - 8)5, (l - 8)5 + 5e(l - 8)4 15. <1>«9 - 12 + ~)/V12(1 - 0.12)) = 17. (l - ~)3 + 3 . ~(l - ~)2 = ~ 3.0.8187,0.6703.0.1353 9. 19.5%, 14.7% 13. Because /I is finite 0.22 (if c = 9) Problem Set 25.7, page 1079 1. X02 = (30 - 50)2/50 + (70 - 50)2150 = 16 > c = 3.84: no 3.41 5. X02 = 2.33 < c = 11.07. Yes 7. ej = I1Pj = 370/5 = 74. X02 = 984174 = 13.3. c = 9.49. Reject the hypothesis. 9. X02 = I < 3.84; yes 13. Combining the results for x = lo, II, 12, we have K - r - 1 = 9 (r = I since we estimated the mean. 1~ci:f = 3.87). Xo2= 12.98 < c = 16.92. Do not reject. 15. X02 = 49/20 + 49/60 = 3.27 < c = 3.84 (1 degree of freedom, a = 5%), which supports the claim. 17. 42 even digits, accept. Problem Set 25.8, page 1082 3. (~l8(l + 18 + 153 + 816) = 0.0038 5. Hypothesis: A and B are equally good. Then the probability of at least 7 trials favorable to A is ~8 + 8 . ~8 = 3.5%. Reject the hypothesis. 7. Hypothesis JL = O. Alternative JL > 0, .r = 1.58, t = \liD. 1.58/1.23 = 4.06 > c = 1.83 (a = 5%). Hypothesis rejected. 9. x = 9.67, s = 11.87, to = 9.67/(11.871'\115) = 3.15 > c = 1.76 (a = 5%). Hypothesis rejected. 11. Consider .'J = Xj - Po. 13. peT ~ 2) = 0.1% from Table A12. Reject. App. 2 A59 Answers to Odd-Numbered Problems 15. P(T ~ 15) = 10.8%. Do not reject. 17. P{T ~ 2) = 2.8%. Reject. Problem Set 25.9, page 1091 1. Y = 1.9 + x 3. y = 6.7407 + 3.068x 5. y = 4 + 4.8x. 172 ft 7. y = -1146 + 4.32x 9. y = 0.5932 + 0.1l38x, R = 1/0.1138 11. qo = 76, K = 2.36V76/(7· 944) = 0.253, CONFo.95 { -1.58 ~ Kl ~ -1.06} 13. 3sx 2 = 500, 3sxy = 33.5, kl = 0.067, 3s y2 = 2.268. qo = 0.023. K = 0.02] CONFo.95 {0.046 ~ Kl ~ 0.088} Chapter 25 Review Questions and Problems, page 1092 i/ 21. fl = 5.33. = 1.722 25. CONFo.99 { 19.1 ~ J.L ~ 33.7} 29. CONFo.95 { 1.373 ~ J.L ~ 1.4511 33. c = 14.74 > 14.5; reject J.Lo. 23. It will double. 27. CONFo.95 {0.726 ~ J.L ~ 0.75]} 31. CONFo.99 {0.05 ~ u 2 ~ 10} 14.74 - 14.40) 35. cD ( •~ = 0.9842 v 0.025 37.30.14/3.8 = 7.93 < 8.25. Reject. 39. Vo = 2.5 < 6.0 [(9.4) degrees of freedom]; accept the hypothesis. 41. Decrease by a factor v'2. By a factor 2.58/1.96 = 1.32. 43.0.9953,0.9825,0.9384, etc. 45. y = 1.70 + 0.55x ;;~".·APPENDIX II p ',.. j ---- .... 3 n" Auxiliary Material A3.1 Formulas for Special Functions For tables of IlUllleric values. see Appelldix 5. Exponential function e eX (Fig. 544) = 2.71828 1828459045235360287471353 (1) Natural logarithm (Fig. 545) In (xy) = In x (2) + In y. In (xly) = Inx - 1ny. In x is the inverse of eX, and e ln x = x, e- ln x = e1n nIx) = IIx. Logarithm of base ten 10glOx or simply log x (3) log x = M In x, (4) In x = I X M loa b" M I M = log e = 0.434294481903251 82765 11289 18917 = In 10 = 2.30258509299404568401 7991454684 log x is the inverse of lOT, and I Olog x = x, I O-log X = IIx. Sine and cosine functions (Figs. 546.547). In calculus, angles are measured in radians, so that sin x and cos x have period 27['. sin .1' is odd. sin (-x) = - sin x, and cos x is even. cos (-x) = cos x. y I y 5 2 o x Fig. 544. A60 Exponential function eX I x -2, Fig. 545. Natural logarithm In x SEC. A3.1 Formulas for Special Functions A61 y y / x / ~l \. X sin x Fig. 546. Fig. 547. 1° = 0.01745 32925 19943 radian radian = 57° 17' 44.80625" cos x = 57.29577 95131 ° sin 2 x (5) sin (x + y) = sin (x - y) (6) cos (x { sin (7T (9) cos 2 (10) X = { cos x cos y - sin x sin y = cos x cos y + sin x sin y cos 2x x) = sinx. - ~(1 cos = cos2 X - sin2 x = ~[-cos (x cos x cos y = ~[cos (x sin x cos y = Msin (x cos u + sin v x) = -cosx (7T - sin 2 x + cos 2x), = + y) + + y) + + y) + ~(l - cos 2x) cos (x - y)] em (x - y)] sin (x - y)] u+v u-v 2 2 u+v u-v u+v u-v 2 2 = 2 sin - - - cos - - - + cos v = 2 cos --2- cos --2- cos v - cos u (14) - cos x sin y x= cos (x - ;) = cos ( ; - x) cos x= sin (x + ;) = sin ( ; - x) sin u (12) sin y sin sin x sin y (1 L) + cus x = sin x cos y sin 2x = 2 sin x cos x, (8) (13) cos 2 X = 1 sin x cos y + y) = cos (x - y) (7) + A cos x +B sin x = VA 2 A cos x +B sin x = VA 2 = 2 sin - - - sin - - - + B2 cos (x ± + B2 sin (x 0), ± 8), tan 8 = tan 8 = sin 8 B +- cos 8 sin 8 eas 8 A = A -T- B A62 APP. 3 Auxiliary Material y y 5 5 ) ) -Tr / Tr -Tr x ( ( \ -5 \ -5 tan x Fig. 548. cot x Fig. 549. Tangent, cotangent, secant, cosecant (Figs. 548, 549) (15) tanx (16) = sinx cosx tan (x + y) = cot x = tan x + cosx sec x sinx tany = tan (x - y) = 1 - tanxtany cscx cos x = sin x tan x - tan y + 1 tan x tan v Hyperbolic functions (hyperbolic sine sinh x, etc.; Figs. 550, 551) (17) tanh x (18) cosh x (19) + = sinh x coshx ' sinh x sinh2 x (21) = i(cosh 2x - 1), = sinh x cosh x - sinh x = eX, COSh2 X (20) cosh x coth x - sinh2 x = I COSh2 X = i(cosh 2x y y 4 4 2 -2 g. 550. / 2 x ;-2 -4 -4 Fig. 551. + l) \ 2 x -2 -2 sinh x (dashed) and cosh x e- x = tanh x (dashed) and coth x SEC A3.1 A63 Formulas for Special Functions { (22) sinh (x ± y) sinh x cosh y ± cosh x sinh y = cosh (x ± y) = cosh x cosh y ± sinh x sinh y tanh (x ± y) (23) tanh x ± tanh y = ------I ± tanh x tanh y Gamma function (Fig. 552 and Table A2 in App. 5). The gamma function f(a) is defined by the integral (24) (a> 0) which is meaningful only if a> 0 (or, if we consider complex a, for those a whose real part is positive). Integration by parts gives the importantfullctional relatio1l of the gamma function, (25) f(a + 1) = af(a). From (24) we readily have r(l) = I: hence if a is a positive integer. say k. then by repeated application of (25) we obtain + r(k (26) 1) = (k = 0, 1, .. '). k! This shows that the ga11l111afunction can be regarded as a generalization of the elementary factorial jilllction. [Sometimes the notation (a - L)! is used for rea), even for noninteger values of a, and the gamma function is also known as the factorial function.] By repeated application of (25) we obtain f(a)= rea + 1) f(a a(a + + 2) f(a + k + 1) + I )(a + 2) ... (a + ----'-- a(a 1) nw 4 I I I I I I I I -2 in Fig. 552. -4 Gamma function ex k) A64 APP. 3 Auxiliary Material and we may use this relation rea) (27) rea + k + I) = -------a(a + I) ... (a + k) (a oF 0, -1, -2,· .. ) for defining the gamma function for negative a (oF -1, -2, " .), choosing for k the smallest integer such that a + k + I > O. Together with (24), this then gives a definition of rca) for all a not equal to zero or a negative integer (Fig. 552). It can be shown that the gamma function may also be represented as the limit of a product namely. by the formula (28) rea) n! nl> = lim n~!XJ ( a a + I )(a + 2) ... (a (a oF 0, - 1, .. '). + 11) From (27) or (28) we see that, for complex a, the gamma function r( a) is a meromurphic function with simple poles at a = 0, - 1, - 2, .... An approximation of the gamma function for large positive a is given by the Stirling formula (29) where e is the base of the natural logarithm. We finally mention the special value (30) Incomplete gamma functions Q(a. x) (31) = f=e-tt U - 1 dt (a> 0) x (32) rca) = pea, x) + Q(a. x) Beta function (33) B(x. y) = I 1 1 (] - tX- t)y-l (x> 0, y > 0) dt o Representation in terms of gan1ma functions: (34) B(x. y) = f(x)f()') rex. + .y) Error function (Fig. 553 and Table A4 in App. 5) (35) erf x = -2- v:;;: IXe- t2 dt 0 7 (36) x 3!7 + _ ... ) SEC. A3.1 A65 Formulas for Special Functions erfx 1 0.5 -2 x / ~ /0.5 -1 Error function Fig. 553. erf (x) 1, C017lple171CllTal}, error jill1ction = erfc x = I - erf x (37) = ~ 2I V'Tr I""e - t 2 dt x Fresnel integrals! (Fig. 554) x C(x) = {cos (t o (38) C(x) = -v:;;;s, S(X) = vi'Tr/S, 2 ) Set) dt, = Jo sin (t2) dt co171plemelllary fimctiollS !¥ - c(x) = s(x) = C(x) (39) LXcos (t = r; - to S(x) = \ S 2 ) dt sin (12) dt x Sine integral (Fig. 555 and Table A4 in App. 5) Sitx) (40) = x J o sin t -~ t dt y 1 C(x) / -' '- _/_' ~----4'~ ~ '-_/ / o - ~.//-- x Fig. 554, Fresnel integrals lAUGUSTIN FRESNEL (1788-1827), French physicist and mathematician. For tables ~ee Ref. [GRI]. A66 APP. 3 Auxiliary Material Si(X~l ,,/2 - 1 O~'--~I---L--~--L-~5~-L--~--~~L-~1~~~~x Fig. 555. Si(:x:) = Sine integral 7T/2. ("(}17lplemento ry jimction (41) si(x) = 7T - f Si(x) = - 2 x . SIll t - - dt t x Cosine integral (Table A4 in App. 5) (42) citx} f = cc cos t -- dt (x> 0) dt (x> 0) t x Exponential integral (43) f Ei(x) = x -t ~ t x Logarithmic integral (44) A3.2 lie\") = x Io - dt In t Partial Derivatives For differentiation formulas, see inside of front cover. Let ~ = f(x, y) be a real function of two independent real variables, x and y. If we keep y constant. say, y = )'1' and think of x as a variable, then f(x, )'1) depends on x alone. If the derivative of f(x, Yl) with respect to x for a value x = Xl exists. then the value of this derivative is called the partial derivative of f(x. \") u'ith respect to x at tbe point (Xl' .'"1) and is denoted by or by Other notations are and these may be used when subscripts are not used for another purpose and there is no danger of confusion. SEC. A3.2 A67 Partial Derivatives We thus have, by the definition of the derivative, (1) The partial delivative of.: x constant, say, equal to Xl, = f(x. y) with respect to y is defined similarly; we now keep and differentiate f(XI, y) with respect to y. Thus (2) Other notations are fy(x}. YI) and ~y(XI' )'1)' It is clear that the values of those two partial derivatives will in general depend on the point (Xl, YI)' Hence the partial delivatives a~/ax and iJz/iJ.v at a variable point (x, y) are functions of x and y. The function az/iJx is obtained as in ordinary calculus by differentiating z = f(x, y) with respect to x. treating y as a constant, and Bz/By is obtained by differentiating z with respect to y, treating x as a constant. E X AMP L E 1 Let::: 2 = I(x. y) = x y + x sin y. Then iiI ~ ilx = 2n + ~in'·. . ilj 2 = x ily ~ + \. COS Y. . • The partial derivatives iJ:;:/i)x and a~/ay of a function;: = f(x, y) have a very simple geometric interpretation. The function ;:: = f(x, y) can be represented by a surface in space. The equation y = Yl then represents a vertical plane intersecting the surface in a curve. and the partial derivative a::Ji)x at a point (Xl' Yl) is the slope of the tangent (that is, tan a where a is the angle shown in Fig. 556) to the curve. Similarly. the partial derivative iJ;:/ay at (Xl, Yl) is the slope of the tangent to the curve X = Xl on the surface z = f(x, y) at (Xl' YI)' Fig. 556. Geometrical interpretation of first partial derivatives A68 APP. 3 Auxiliary Material The partial derivatives azlax and a-::.My are called first partial derivatives or partial derivatives of first order. By differentiating these derivatives once more, we obtain the four second partial derivatives (or partial derivatives of second order)2 a2 f a ax 2 ax a2f a ax ay ax a2f a (3) ayax a2 f ay2 (:~ ) = fxx (::.) = fyx f (a ) = fxy ay ax a aI' f -ay ) =fyy' (a It can be shown that if all the derivatives concerned are continuous, then the two mixed partial derivatives are equal, so that the order of differentiation does not matter (see Ref. [GR4] in App. 1), that is, (4) E X AMP L E 2 For the function in Example I. fxx = 2y, f xy = 2x + cos Y = f yx, fyy = -x siny. • By differentiating the second partial derivatives again with respect to x and y, respectively, we obtain the third partial derivatives or partial derivatives of the third order of f, etc. If we consider a function f(x, y, z) of three independent varutbles, then we have the three first partial derivatives fAx, y, z), fy{x, y, z), and fz(x, y, z). Here Ix is obtained by differentiating f with respect to x, treating both y and z as constallts. Thus, analogous to (I), we now have etc. By differentiating derivatives of f, etc. E X AMP L E 3 f.p f y, fz again in this fashion we obtain the second partial Let f(x, y, z) = x 2 + y2 + Z2 + xy eZ • Then f" = 2x + )' eZ , fy = 2y + x eZ , fxx = 2, fxy = f y ,' = e fyy = 2. fyz = fzy = fz = 2z + xy eZ , Z • z xe , • 2 LAUTIO]'l In the subscript notation the subscripts are written in the order in which we differentiate whereas in the "iY' notation the order is opposite. ' SEC. A3.3 A69 Sequences and Series A3.3 Sequences and Series See also Chap. 15. Monotone Real Sequences We call a real sequence xl, X2, increasing, that is, . • • ,Xn , .•• a monotone sequence if it is either monotone or monotone decreasing, that is, We call Xl, for all n. THE 0 REM 1 PROOF X2, ••• a bounded sequence if there is a positive constant K such that IXnl <K (f a real sequence is bounded and monOTOne, it converges. Let Xl> X2' . • • be a bounded monotone increasing sequence. Then its terms are smaller than some number B and, since Xl ~ Xn for all n. they lie in the interval Xl ~ X." ~ B. which will be denoted by 1o, We bisect 1o; that is, we subdivide it into two parts of equal length. If the right half (together with its endpoints) contains terms of the sequence, we denote it by 11' If it does not contain terms of the sequence, then the left half of 10 (together with its endpoints) is called 11 , This is the first step. In the second step we bisect h, select one half by the same rule, and call it 12 , and so on (see Fig. 557 on p. A 70). In this way we obtain shorter and shorter intervals 1o, 11 , 12 , • • . with the following prope11ies. Each 1m contains all In for n > 111. No term of the sequence lies to the right of 1m , and, since the sequence is monotone increasing. all Xn with Il greater than some number N lie in 1m; of course, N will depend on 111. in general. The lengths of the 1m approach zero as 111 approaches infinity. Hence there is precisely one number, call it L, that lies in all those intervals,3 and we may now easily prove that the sequence is convergent with the limit L. In fact, given an E > 0, we choose an I1l such that the length of 1m is less than E. Then L and all the Xn with n > N(m) lie in 1m' and. therefore, IXn - LI < E for all those n. This completes the proof for an increasing sequence. For a decreasing sequence the proof is the same, except for a suitable interchange of "left" and "right" in the construction of those intervals. • 3 This statement seems to be obvious, but actually it is not; it may be regarded as an axiom of the real number system in the following form. Let h. 12 , .•• be closed intervals such that each 1m contains all 1n with 11 > m. and the lengths of the 1m approach zero as III approaches intinity. Then there is precisely one real number that is contained in all those intervals. This is the so-called Cantor-Dedekind axiom, named after the German mathematicians GEORG CANTOR (1845-1918). the creamr of set theory, and RICHARD DEDEKIND (1831-1916), known for his fundamental work in nwnber theory. For further details see Ref. [GR2] in App. I. (An interval 1 is said to be closed if its two endpoints are regarded as points belonging to J. It is said to be open if the endpoints are not regarded as points of I.) APP. 3 A70 Auxiliary Material 10 ---------1 B I I~<~--I---:: I I'~"III. :1 Fig. 557. Proof of Theorem 1 Real Series THEOREM 2 Leibniz Test for Real Series Let Xl' X2, ••• be real and monotone decreasing to zero, that is, (1) lim (b) Xm = O. 1U----+X Then the series witll terms of altematillg signs converges, and for the remainder Rn after the nth term we have the estimate (2) PROOF Let Sn so that be the 11th partial sum of the series. Then, because of (1 a), S2 ~ S3 ~ Sl' Proceeding in this fashion, we conclude that (Fig. 558) (3) which shows that the odd partial sums form a bounded monotone sequence, and so do the even partial sums. Hence. by Theorem L both sequences converge, say, lim n_x S2n+l = S, I 2 Fig. 558. S2n = s*. -X 2 IE 8 lim n_x 8 C-" =:j 4 8 3 8 1 Proof of the Leibniz test SEC. A3.4 Grad, Div, Curl, V2 in Curvilinear Coordinates Now. since S271+1 - s - s* S2n = lim = X2n+1' A71 we readily see that Ob) implies lim S271+1 - n~x = n_x lim (S271+1 S2n n~x - = n_x lim '2n+1 = S2") Hence s* = s. and the series converges with the sum s. We prove the estimate (2) for the remainder. Since s" - o. s, it follows from (3) that and also By subtracting S2n and respectively, we obtain S2n-1' In these inequalities, the first expression is equal to X2,,+1' the last is equal to -X2m and the expressions between the inequality signs are the remainders R2n and R 2n - 1 . Thus the inequalities may be written • and we see that they imply (2). This completes the proof. A3.4 Grad, Div, Curl, V 2 in Curvilinear Coordinates To simplify formulas we write Cartesian coordinates -' = Xl' Y = -'2' Z = -'3' We denote curvilinear coordinates by qb q2, q3' Through each point P there pass three coordinate surfaces q1 = const, q2 = COllSt. q3 = COllSt. They intersect along coordinate curves. We assume the three coordinate curves through P to be orthogonal (perpendicular to each other). We write coordinate transformations as (I) Corresponding transformations of grad, div, curl, and V2 can all be written by using (2) Next to Cartesian coordinates. most important are cylindrical coordinates = ;: (Fig. 559a on p. A 72) defined by q1 = r, q2 = e. lJ3 (3) Xl = q1 cos q2 = r cos e, and spherical coordinates tiI (4) Xl = q1 cos q2 sin q3 = r, = r cos -'3 = = = e, q3 = 4> (Fig. 559b) defined by4 q2 q1 sin q2 = e, X2 e sin 4>, q1 cos -'2 q3 = r = cos r sin q1 sin q2 sin q3 = r sin e sin 4> 4>. 4This is the notation used in calculus and in many other books. It is logical since in it. 8 play, the ,arne role as in polar coordinates. C-\L'TIOM Some books interchange the roles of IJ and <p. A72 APP. 3 Auxiliary Material z z h~ e. z) -- -z e x (a) Y Y r Cylindrical coordinates Fig. 559. (b) Spherical coordinates Special curvilinear coordinates In addition to the general formulas for any orthogonal coordinates qh additional formulas for these important special cases. Linear Element ds. q2, Q3, we shall give In Cartesian coordinates, (Sec. 9.5). For the q-coordinates, (5) (5') (Cylindrical coordinates). For polar coordinates set d-;,2 = o. (5") (Spherical coordinates). Gradient. grad f = vf = [f Xl' f X2' f X) (partial derivatives; Sec. 9.7). In the q-system, with u, v, w denoting unit vectors in the positive directions of the Ql, Q2, Q3 coordinate curves, respectively, (6) (6') (6") divF = V.F 1 ar r = Vf = - u + - !!fad! =""f 1 a -:-{rF1 ) r ilr = - af = -u <0 (7') ilf gr ad! a,. af aR --'--v (Cylindrical coordinates) iJz 1 af 1 a! + - - -y + --w rsin <I> i/O I iJF aF iJO az + _ ~ + __3 r iiI +-w ,. 0<1> (Spherical coordinates). (Cylindrical coordinates) SEC. A3.4 Grad, Div, Curl, (7") ,2 A73 in Curvilinear Coordinates div F = v· F = I iJ 2 -;- (,.2 F1 ) r "r I iJF2 rsm </J UV I iJ + - . - --:;-;- + - . - -.- (sin </J F3 ) (Spherical coordinates). rsm </J rJ</J (8') (Cylindrical coordinates) (8") (Spherical coordinates). Curl (Sec. 9.9): ] (9) curlF=,xF=--lz]h2 h 3 hlu h2 v h3W a a a aq1 aq2 aq3 hlFI h2F2 h3 F 3 For cylindrical coordinates we have in (9) (as in the previous formulas) and for spherical coordinates we have hI = h,. = I, 1z2 = h" = qI sin q3 = r sin </J, ( .. I ,, oS. " 1 4 APPENDIX Additional Proofs ~ .. Section 2.6, page 73 Uniqueness 1 Assuming that the problem consisting of the ODE PROOF OF THEOREM 1 )''' + p(x)y' + q(x)y = 0 (1) and the two initial conditions (3) has two solutions Yllx) and difference on the interval 1 in the theorem, we show that their Y2(X) is identically zero on 1: then Yl == )'2 on 1, which implies uniqueness. Since (I) is homogeneous and linear. y is a solution of that ODE on 1. and since )"2 satisfy the !>ame initial conditions, y satisfies the conditions (10) = 0, z(x) = ylxo) y' (xo) VI and = O. We consider the function y(x)2 + y' (X)2 and its derivative z' = 2yy' + 2)"')"". From the ODE we have y" = , -py - q)'. By substituting this in the expression for z.' we obtain (11) Now, since y and y' are real. IThis proof was suggested by my colleague. Prof. A. D. Ziebur. In this proof we use formula numbers that have not yet been used in Sec. 2.6. A74 APP. 4 A75 Additional Proofs From this and the definition of :: we obtain the two inequalities (12) (a) 2yy' ~ )'2 From (l2b) we have 2)')" ~ obtain + = ;::, y'2 -z. Together, 12)')"] ~ -2qyy' -2y)"' (b) 1-2qyy'l ~ ~ y2 + y'2 = z. z. For the last term in (II) we now Iq112y/1 ~ Iqlz. = Using this result a" well as -p ~ Ipl and applying (I2a) to the term 2yy' in (11), we find Since / 2 ~ )'2 + )"2 = .<;, from this we obtain z' ~ (l + 21pl + Iql)z or, denoting the function in parentheses by 11, z' (l3a) ~ 11:: for all x on 1 Similarly. from (11) and (12) it follows that (l3b) ~ Z + 2Ip!;:: + Iq\::: = 11;::. The inequalities (I3a) and (13b) are equivalent to the inequalities (14) .<;' - h:: ~ 0, z' + h.<; ~ o. Integrating factors for the two expressions on the left are and The integrals in the exponents exist because Iz is continuous. Since Fl and F2 are positive. we thus have from (14) and This means that F1z is non increasing and F 2 z is nondecreasing on I. Since ::(xo) = 0 by (10). when x ~ Xo we thus obtain and similarly, when x ~ Xo, Dividing by Fl and F2 and noting that these functions are positive. we altogether have z~ · 1·IeS that;:: = y 2+' l'2 ThisImp "'" o. z~o 0 on I. Hence y "'" 0 or YI "'" Y2 on I. for all x on I. • A76 APP. 4 Additional Proofs Section 5.4, pages 184 PROOF 0 F THE 0 REM 2 Frobenius Method. Basis of Solutions. Three Cases The formula numbers in this proof are the same as in the text of Sec. 5.4. An additional formula not appearing in Sec. 5.4 will be called (A) (see below). The ODE in Theorem 2 is b(x), where c(x) )''' + -x -·\' + -x )' = 0• 2 (1) b(,) and c(x) are analytic functions. We can write it x 2 y" (1') + xb(x)),' + dx)), + Co = O. The indicial equation of (l) is (4) r(r - 1) bor + = O. The roots 1"1' 1"2 of this quadratic equation determine the general form of a basis of solutions of 0). and there are three possible cases as follows. Case 1. Distinct Roots not Differing by an Integer. form A first solution of 0) is of the (5) and can be determined as in the power series method. For a proof that in this case, the ODE (1) has a second independent solution of the form (6) see Ref. [All] listed in App. 1. Case 2. Double Root. The indicial equation (4) has a double root r if and only if (b o - 1)2 - 4co = 0, and then r = !( I - b o). A first solution (7) r = !(l - boY, can be determined as in Case I. We show that a second independent solution is of the form (x> 0). (8) We use the method of reduction of order (see Sec. 2.1), that is, we determine ll(X) such that )'2(X) = U(X)Yl(X) is a solution of (I). By inserting this and the derivatives = "" )'2 II )'1 + I , 2u Yl + UYIII into the ODE (I') we obtain x 2(" U)'1 +')" _U .'"1 + " + xb(ll ,)'1 + UYI) , + UYI) cUYI = O. APP. 4 A77 Additional Proofs Since )'1 is a solution of (I r), the sum of the terms involving u is zero, and this equation reduces to By dividing by 2 X )'1 and inserting the power series for b we obtain u " ( + 2 -)'1, + -b o + . .. x Yt ) u , o. = Here and in the following the dots designate terms that are constant or involve positive powers of x. Now from (7) if follows that y~ + l)alx + ... ] x [aO+alX +···] xT - 1 ll"ao + (I" T Yt I" x + Hence the previous equation can be written (A) U rr + (21": bo + ... ) o. Ur = Since r = (l - b o)/2, the term (21" + bo)/x equals I/x, and by dividing by u' we thus have u" u , x + By integration we obtain In u' = -In x + ... , hence u' = (Ilx)e<·· .J. Expanding the exponential function in powers of x and integrating once more, we see that u is of the form Inserting this into)'2 = UY1, we obtain for )'2 a representation of the form (8). Case 3. Roots Differing by an Integer. positive integer. A first solution We write 1"1 = I" and 1"2 = I" - P where p is a (9) can be determined as in Cases 1 and 2. We show that a second independent solution is of the form (10) where we may have k -=I=- 0 or k = O. As in Case 2 we set)'2 literally as in Case 2 and give Eq. (A), u" + (21": b o + .. -) u' = O. = 1t)'1. The first steps are APP.4 A7B Additional Proofs Now by elementary algebra. the coefficient b o - I of r in (4) equals minus the sum of the roots, = bo - I Hence 2r + r2) = -(r + r - p) = -2r + p. -(r1 + bo = p + L and division by u' gives The further ')teps are as in Case 2. Integrating, we find In u' = -(p + 1) In x + ... , thus I u = x -(p+ll ( ... J e where dots stand for some series of nonnegative integer powers of x. By expanding the exponential function as before we obtain a series of the form I U We integrate once more. Writing the resulting logarithmic term first, we get u = k Inx p + (- _1_P - ... - kp- 1 + kp+IX + ... ) Hence. by (9) we get for Y2 = px UYI x the formula But this is of the form (10) with k = kp since rl - P = r2 and the product of the two • series involves nonnegative integer powers of x only. Section 5.7, page 205 THEOREM Reality of Eigenvalues If p, q, r, alld p' ill the Sturm-Liouville equation (I) of Sec. 5.7 are real-valued and continuous on the interval a ~ x ~ band rex) > 0 throughout that interval (or rex) < 0 throughout that interval). then all the eigenvalues of the Stunl1-Liouville problem (1). (2). Sec. 5.7. are real. PROOF Let A = 0' + i{3 be an eigenvalue of the problem and let y(x) = u(x) be a corresponding eigenfunction: here Sec. 5.7, we have ( pu I 0', + iu(x) {3. u. and u are real. Substituting this into (1), +.lpU ')' + (q + O'r + i{3r)(u + iu) = o. APP. 4 Additional Proofs A79 This complex equation is equivalent to the following pair of equations for the real and the imaginary parts: (pu')' + (q + ar)u - {3rv = 0 (pU')' + (q + ar)u + 13m = O. Multiplying the first equation by u, the second by -(3(1I 2 + u 2 )r -ll and adding, we get = u(pu')' - u(pu')' = [(pu')lI - (pu')uJ'. The expression in brackets is continuous on a ~ x ~ b. for reasons similar to those in the proof of Theorem I, Sec. 5.7. Integrating over x from a to b. we thus obtain Because of the boundary conditions the right side is zero; this is as in that proof. Since y is an eigenfunction, u 2 + u2 O. Since y and r are continuous and r > 0 (or r < 0) on the interval a ~ x ~ b, the integral on the left is not zero. Hence, (3 = 0, which means that A = a is reaL This completes the proof. • *' Section 7.7, page 308 THEOREM Determinants The definition of a detennillant (7) D = detA = as given in Sec. 7.7 is unambiguous, that is, it yields the same value of D no matter which rows or columns we choose in developings. PROOF In this proof we shall use fonnula numbers not yet used in Sec. 7.7. We shall prove first that the same value is obtained no matter which row is chosen. The proof is by induction. The statement is true for a second-order determinant. for which the developments by the first row aU{/22 + a 1 2( -(21) and by the second row a21(-a12) + {/22 a ll give the same value alla 22 - a12a21' Assuming the statement to be true for an (n - l)st-order determinant, we prove that it is true for an nth-order determinant. ABO APP. 4 Additional Proofs For this purpose we expand D in terms of each of two arbitrary rows, say, the ith and the jth, and compare the results. Without loss of generality let us assume i < j. First Expansioll. We expand D by the ith row. A typical term in this expansion i~ The minor Mik of aik in D is an (11 - 1)st-order determinant. By the induction hypothesis we may expand it by any row. We expand it by the row corresponding to the jth row of D. This row contains the entries ajl (/ =1= k). It is the (j - I )st row of M ik • because Mik does not contain entries of the ith row of D. and i < j. We have to distinguish between two ca<;es as follows. Case I. If I < k, then the entry ajl belongs to the lth column of Mik (see Fig. 560). Hence the term involving ajl in this expansion is (20) where M ikj / is the minor of (ljl in M ik . Since this minor is obtained from Mik by deleting the row and column of ajl, it is obtained from D by deleting the ith and jth rows and the kth and lth columns of D. We insert the expansions of the Mik into that of D. Then it follows from (19) and (20) that the terms of the resulting representation of D are of the form (2Ia) (l < k) where b=i+k+j+l-l. 11. If I > k, the only difference is that then ajl belongs to the (l - I )st column of because Mik does not contain entries of the kth column of D, and k < I. This causes an additional minus sign in (20). and. instead of (21 a). we therefore obtain Case M ik , (2Ib) (l where b is the same as before. lth kth kth lth col. col. col. col. I I I I --~------@---I ith row I --6}-----~---I I I I I I Case I Fig. 560. - -&-----~--I jth row I ---+------@--I I I I I I Case II Cases I and II of the two expansions of D > k) APP. 4 A81 Additional Proofs Second Expansion. expansion is We now expand D at first by the jth row. A typical term in this (22) By the induction hypothesis we may expand the minor Mjl of ajl in D by its ith row, which corresponds to the ith row of D. since j > i. > I, the entry 0ik in that row belongs to the (k - I )st column of Mjl' because does not contain entIies of the Ith column of D. and I < k (see Fig. 560). Hence the term involving aik in this expansion is Case T. If k Mjl (23) aik· · Mjl) (cof actor 0 f aik In = llik· (- l)i+(k- D M ikjZ, where the minor M ikjl of aik in Mjl is obtained by deleting the ith and jth rows and the kth and Ith columns of D [and is, therefore, identical with M ikj /, in (20), so that our notation is consistentJ. We insert the expansions of the Mjl into that of D. It follows from (22) and (23) that this yields a representation whose terms are identical with those given by (21 a) when I < k. < I, then 0ik belongs to the kth column of Mjl' we obtain an additional minus sign, and the result agrees with that characterized by (21 b). Case 1I. If k We have shown that the two expansions of D consist of the same terms, and this proves our statement concerning rows. The proof of the statement concerning colu11l1ls is quite similar; if we expand D in terms of two arbitrary columns, say, the kth and the !th. we find that the general term involving 0jlaik is exactly the same as before. This proves that not only all column expansions of D yield the same value, but also that their common value is equal to the common value of the row expansion), of D. This completes the proof and shows that ollr definitioll of all mil-order detel7ninalll is unambiguolls. • Section 9.3, page 377 PROOF OF FORMULA (2) We prove that in right-handed Carte~ian coordinates. the vector product has the components (2) We need only consider the case v =1= O. Since v is perpendicular to both a and b, Theorem 1 in Sec. 9.2 gives a • v = 0 and b • v = 0; in components [see (2), Sec. 9.2], (3) A82 APP. 4 Additional Proofs Multiplying the first equation by b3 , the last by a3. and subtracting, we obtain Multiplying the first equation by b l , the last by ill' and subtracting, we obtain We can ea~ily verify that these two equations are ~atisfied by (4) where c is a constant. The reader may verify by inserting that (4) also satisfies (3). Now each of the equations in (3) represents a plane thruugh the origin in VIV2v3-space. The vectors a and b are normal vectors of these planes (see Example 6 in Sec. 9.2). Since v =1= 0, these vectors are not parallel and the two planes do not coincide. Hence their intersection is a straight line L through the origin. Since (4) is a solution of (3) and, for varying c, represents a straight line, we conclude that (4) represents L, and every solution of (3) must be of the form (4). Tn particular, the components of v must be of this form, where c is to be determined. From (4) we obtain This can be written as can be verified by performing the indicated multiplications in both formulas and comparing. Using (2) in Sec. 9.2, we thus have By comparing this with formula (12) in Team Project 24 of Problem Set 9.3 we conclude that c = ±1. We show that c = + 1. This can be done a" follows. If we change the lengths and directions of a and b continuously and so that at the end a = i and b = j (Fig. l86a in Sec. 9.3), then v will change its length and direction continuously, and at the end, v = i X j = k. Obviously we may effect the change so that both a and b remain different from the zero vector and are not parallel at any instant. Then v is never equal to the zero vector, and since the change is continuous and c can only assume the values + 1 or -I, it follows that at the end c must have the same value as before. Now at the end a = i, b = j. v = k and, therefore, al = 1. b2 = I, V3 = L and the other components in (4) are zero. Hence from (4) we see that V3 = c = + 1. This proves Theorem I. For a left -handed coordinate system, i X j = -k (see Fig. 186b in Sec. 9.3), resulting in c = -1. This proves the statement right after formula (2). • APP. 4 A83 Additional Proofs Section 9.9, page 416 PROOF OF THE INVARIANCE OF THE CURL This proof will follow from two theorems (A and B), which we prove first. THEOREM A Transformation Law for Vector Components For any vector v the componenTs VI, V2 , V3 and Vl*' V2*' V3* in allY two systems of Cartesian coordinates Xl> x3, X3 and Xl*' X2*' x3*' respectively, are related by 0) and conversely (2) with coefficients C 13 = i*·k (3) C31 = k*·i C32 = c 33 = k*·k k*· j satisfying 3 (4) L CkjCmj = i5km (k,1I/ = 1, 2, 3), j~1 where the Kronecker delta2 is given by (k * m) (k = 11/) and i, j, k and i*, j*, k* denote the unit vectors ill the positive X2*-' x3*-directions, respectively. Xl-, X2-, X3- and Xl*-' 2LEOPOLD KRONECKER (l823-18YI), German mathematician at Berlin, who made important contributions to algebra. group theory. and number theory. We shall keep our discussion completely independent of Chap. 7, but readers familiar with matrices should rccognize that we are dealing with orthogonal transformations and matrices and that our present theorem follows from Theorem 2 in Sec. 8.3. APP. 4 A84 PROOF Additional Proofs The representation of v in the two systems are (5) Since i* • i* = L i* • j* from this and (Sa) = 0, i* • k* = 0, we get from (5b) simply i* • v = Vl* and Because of (3), this is the first formula in (1), and the other two fommlas are obtained similarly. by considering j* • v and then k* • v. Formula (2) follows by the same idea. taking i • v = VI from (Sa) and then from (5b) and (3) and similarly for the other two components. We prove (4). We can write (1) and (2) briefly as 3 (6) (a) Vj 3 = ~ cl1lJ v m *, (b) Vk *= Vj into Vk *, CkjVj. j=1 m=1 Substituting ~ we get 3 3 3 Vk* =~ Ckj j=1 ~ CmjV n .* = ~ m=1 Vm * m=l where k = 1, 2, 3. Taking k = 1, we have For this to hold for eve1), vector v, the first sum must be I and the other two sums O. This proves (4) with k = 1 for m = 1, 2, 3. Taking k = 2 and then k = 3. we obtain (4) with • k = 2 and 3, form = 1,2,3. THEOREM B Transformation Law for Cartesian Coordinates The trclllc~f01111{{tioll of allY Cartesian XIX2x3-coordinate system into any other Cartesian XI *X2 *X3 *-coordillate system is of the fonn 3 (7) Xm* = ~ CnljXj + b m , 111 = 1.2.3, j=1 with coefficients (3) and COllstants bI> b 2, b3; cOllversely, 3 (8) Xk = :L 1£=1 CnkXn * + bk , k = 1.2,3. APP. 4 A8S Additional Proofs Theorem B follows from Theorem A by noting that the most general transformation of a Cartesian coordinate system into another such system may be decomposed into a tranSf0I111ation of the type just considered and a translation; and under a translation, cOlTesponding coordinates differ merely by a constant. PROOF OF THE INVARIANCE OF THE CURL We write again Xl, X2, X3 instead of x, y,~, and similarly Xl*' X2*' X3i< for other Cartesian coordinates. assuming that both systems are right-handed. Let al. a 2 • 03 denote the components of curl v in the xIx2x3-coordinates. as given by (l), Sec. 9.9. with v=X2, . Similarly, let al*' a2*' a 3* denote the components of curl v in the xl*x2*x3*-coordinate system. We prove that the length and direction of curl v are independent of the particular choice of Cartesian coordinates. as asserted. We do this by showing that the components of curl v satisfy the transformation law (2), which is characteristic of vector components. We consider {[I' We use (6a), and then the chain rule for functions of several variables (Sec. 9.6). This gives 3 = 3 L L ( aV.,/ c m3 (Jx/' m=I j=l From this and (7) we obtain Note what we did. The double sum had 3 X 3 = 9 terms, 3 of which were zero (when = j). and the remaining 6 terms we combined in pairs as we needed them in getting a I *, a2*' {/3* 111 We now use (3), Lagrange's identity (see Team Project 24 in Problem Set 93) and k* x j* = -i* and k X j = -i. Then = (k* x j*) • (k x j) = i* • i = Cn. etc. A86 APP. 4 Additional Proofs Hence a] = clla]* + c2]a2* + c3]a3*' This is of the form of the first formula in (2) in Theorem A, and the other two formulas of the form (2) are obtained similarly. This proves the theorem for right-handed systems. If the .\"lx2'\·3-coordinates are left-handed, then k X j = +i, but then there is a minus sign in front of the determinant in (1), Sec. 9.9. • Section 10.2, pages 426-427 PROOF 0 F THE 0 REM 1, PA R T (b) f (1) We prove that if f F(r) • dr = (Fl dx c C + F2 dy + F3 dz) with continuous F I , F 2, F3 in a domain D is independent of path in D, then F = grad f in D for some f; in components (2' ) We choose any fixed A: (xo, Yo, zo) in D and any B: (x, )" z) in D and define f by (3) f(x, y, z) = fo + I B (F] dx* + F2 dy* + F3 dz*) A with any constant fo and any path from A to BinD. Since A is fixed and we have independence of path. the integral depends only on the coordinates x. y. z. so that (3) defines a function f(x, y. z) in D. We show that F = grad f with this f, beginning with the first of the three relations (2'). Because of independence of path we may integrate from A to B]: (x], y, z) and then parallel to the x-axis along the segment B]B in Fig. 561 with B] chosen so that the whole segment lies in D. Then f(x, y, z) = fo + I Bl (FI d.x:* + F2 dy* A + F3 dz*) + f B (FI dx* + F2 dy* + F3 dz*). ~ We now take the partial derivative with respect to x on both sides. On the left we get iJf/iJx. We show that on the right we get F]. The derivative of the first integral is zero because A: (xo, Yo, zo) and B 1 : (x], y, z) do not depend on x. We consider the second integral. Since on the segment B]B, both y and z are constant, the terms F2 dy* and z y x Fig. 561. Proof of Theorem 1 APP.4 A87 Additional Proofs F3 d::.* do not contribute to the detivative of the integral. The remaining part can be written as a definite integral. Hence its partial derivative with respect to x is Fl(X, y, ::.), and the first of the relations (2') is proved. The other two formulas in (2') follow by the same argument. • Section 13.4, page 620 Cauch~-Riemann Equations We prove that Cauchy-Riemann equations PROOF 0 F THE 0 REM 1 (1) are sufficient for a complex function fez) = u(x, y) + iv(x, y) to be analytic; precisely. (f the real parI u and the inzaginaJ~v part v of f(z.) satisfy (I) ill a domain D ill the complex plane and if the ponied derivatives i11 (I) are COlltillllOUS in D, then fez) is analytic in D. In this proof we write D.::. = .lx is as follows. + iD.y and D.f = fez. + D.z.) - f(z.). The idea of proof (a) We express .If in terms of first partial derivatives of II and v. by applying the mean value theorem of Sec. 9.6. (b) We get rid of partial derivatives with respect to y by applying the Cauchy-Riemann equations. (c) We let .l.:: approach zero and show that then D.fl.l::. as obtained approaches a limit which is equal to U x + iv x , the right side of (4) in Sec. 13.4. regardless of the way of approach to zero. (a) Let P: (x, y) be any fixed point in D. Since D is a domain, it contains a neighborhood of P. We can choose a point Q: (x + D.x, y + D.)') in this neighborhood such that the straight-line segment PQ is in D. Because of our continuity a~sumptions we may apply the mean value theorem in Sec. 9.6. This yields u(x + v(x + D.x, y + .ly) - vex, y) D.x, y + .ly) - lI(X, y) = (.lx)ux(M 1 ) + (D.y)uyCM 1 ) = (D.x)v x (M 2) + (D.y)vyCM2) where Ml and M2 (01= Ml in general!) are suitable points on that segment. The first line is Re D.f and the second is 1m .If, so that (b) uy = -vx and Vy = IIx by the Cauchy-Riemann equations, so that A88 APP. 4 Additional proofs Also 11::. = 11.\' = 0::. - ~x + il1.\', so that we can write I1x = ~::. - il1."'" in = -i(.1.:: - .1x) in the second term. This gives the first term and ~x)li By performing the multiplications and reordering we obtain 111 = (.1::')lI x (M l ) - iI1Y{lIx (Ml) - lI x (M2 )} + i[(~::.Wr(Ml) - ..lx{ux(Ml ) - u x(M2 )}]· Division by 11::. now yields i:lx i.1y .1~ . {lI x (Ml) - ~- tl x(M2 )} - - {ux(M l ) - u x(M2 )}· (e) We finally let 11::. approach zero and note that 111-,,111::.1 ~ I and Il1xll1zl ~ I in (A). Then Q: (x + ~x, y + ~y) approaches P: (x, y), so that Ml and M2 must approach P. Also, since the partial derivatives in (A) are assumed to be continuous, they approach their value at P. In particular, the differences in the braces {... } in (A) approach zero. Hence the limit of the right side of (A) exists and is independent of the path along which 11::. ---7 O. We see that this limit equals the right side of (4) in Sec. 13.4. This means that 1(::.) is analytic at every point.:: in D, and the proof is complete. • Section 14.2, pages 647-648 Goursat proved Cauchy's integml theorem without assuming that f' (.::) is continuous, as follows. We start with the case when C is the boundary of a triangle. We orient C counterclockwise. By joining the midpoints of the sides we subdivide the triangle into four congruent triangles (Fig. 562). Let CI . C n . C m . CIV denote their boundaries. We claim that (see Fig. 562). GOURSAT'S PROOF OF CAUCHY'S INTEGRAL THEOREM (I) fI G d.:: = f G, I d.:: + f Gn I dz + f GIll I d.:: + f I d::.. G,v Indeed, on the right we integrate along each of the three segments of subdivision in both possible directions (Fig. 562), so that the corresponding integrals cancel out in pairs, and the sum of the integrals on the right equals the integral on the left. We now pick an integral on the right that is biggest in absolute value and call its path Cl' Then. by the triangle inequality (Sec. 13.2), Fig. 562. Proof of Cauchy's integral theorem APP. 4 A89 Additional proofs We now subdivide the triangle bounded by C 1 as before and select a triangle of subdivil>ion with boundary C2 for which Then Continuing in this fashion, we obtain a sequence of triangles T 1 , T2 , ••• with boundaries Cl> C2 , . • • that are similar and such that Tn lies in Tm when 11 > Ill, and n = 1,2, .. '. (2) Let ':::0 be the point that belongs to all these triangles. Since the derivative J' (':::0) exists. Let (3) h(.:::) = fez) - f(zo) z -:'::0 f is differentiable at :.:: = :'::0, J' (zo)· - Solving this algebraically for f(:.::) we have fez) = f(:.::o) + (z - zo)J' (:'::0) + 11(:::)(:.:: - ':::0)· Integrating this over the boundary Cn of the triangle Tn gives Since f(.:::o) and J' (zo) are constants and Cn is a dosed path. the first two integrals on the right are zero, as follows from Cauchy's proof, which is applicable because the integrands do have continuous derivatives (0 and const, respectively). We thus have f fez) d::: en = f h(z)(z - :'::0) d:.::. en Since J' (:'::0) is the limit of the difference quotient in (3), for given 8 > 0 such that (4) Ih(z)1 < e when Iz - E > 0 we can find a :'::01 < 8. We may now take 11 so large that the tliangle Tn lies in the disk Iz - :::01 < 8. Let Ln be the length of Cn' Then I::: - zol < Ln for all :: on Cn and::oin Tn. From this and (4) we have 111(.:::)(z - :(0)1 < eLn . The ML-inequality in Sec. 14.1 now gives (5) Now denote the length of C by L. Then the path C1 has the length Ll = Ll2, the path C2 has the length ~ = Ll/'2 = Ll4, etc., and Cn has the length L" = Ll2n. Hence Ln 2 = L2/4n. From (2) and (5) we thus obtain A90 APP. 4 Additional Proofs By choosing E (> 0) sufficiently small we can make the expression on the right as small as we please, while the expression on the left is the definite value of an integral. Consequently. this value must be zero, and the proof is complete. The proof for the case ill which C is the boundary of a polygon follows from the previous proof by subdividing the polygon into triangles (Fig. 563). The integral corresponding to each such triangle is zero. The sum of these integrals is equal to the integral over C, because we integrate along each segment of subdivision in both directions, the corresponding integrals cancel out in pairs, and we are left with the integral over C. The case of a general simple closed path C can be reduced to the preceding one by inscribing in C a closed polygon P of chords, which approximates C "sufficiently accurately," and it can be shown that there is a polygon P such that the integral over P differs from that over C by less than any preassigned positive real number E, no matter how small. The details of this proof are somewhat involved and can be found in Ref. [D6] listed in App. 1. • Fig. 563. Proof of Cauchy's integral theorem for a polygon Section 15.1, page 667 PROOF 0 F THE 0 REM 4 Cauchy's Convergence Principle for Series (a) [n this proof we need two concepts and a theorem, which we list first. 1. A bounded sequence SI' S2, • • . is a sequence whose terms all lie in a disk of (sufficiently large, finite) radius K with center at the origin; thus ISnl < K for alln. 2. A limit point a of a sequence Sb S2, . • • is a point such that, given an E> 0, there are infinitely many terms satisfying ISn - al < E. (Note that this does not imply convergence, since there may still be infinitely many tenns that do not lie within that circle of radius E and center a.) Example: ~. ~, ~. ~, 1~' ~~ • . . . has the limit points 0 and 1 and diverges. 3. A bounded sequence in the complex plane has at least one limit point. (Bolzano-Weierstrass theorem: proof below. Recall that "sequence" always mean infinite sequence.) (b) We now turn to the actual proof that every E > 0 we can find an N such that (1) Izn+l ZI + ... + zn+pl < E + ~2 + ... converges if and only if for for every n > Nandp Here, by the definition of partial sums, Sn+p - Sn = 2n+l + ... + zn+p' = 1,2, . ". APP. 4 A91 Additional proofs Writing II +P= r, we see from this that (1) is equivalent to for all r > Nand n (1*) > N. Suppose that SI' S2, ... converges. Denote its limit by s. Then for a given E> 0 we can find an N such that for every n Hence, if r > Nand n > > N. N, then by the triangle inequality (Sec. 13.2), that is, (I *) holds. (c) Conversely, assume that SI, S2, • . . satisfies (1 *). We first prove that then the sequence must be bounded. Indeed, choose a fixed E and a fixed n = no > N in (1 *). Then (1 *) implies that all s,. with r > N lie in the disk of radius E and center s"o and only fillitely many tel7l1S SI, ... , SN may not lie in this disk. Clearly, we can now find a circle so large that this disk and these finitely many terms all lie within this new circle. Hence the sequence is bounded. By the Bolzano-Weierstrass theorem, it has at least one limit point, call it s. We now show that the sequence is convergent with the limit s. Let E > 0 be given. Then there is an N* such that 1ST - snl < E/2 for all r > N* and 11 > N*, by (1 *). Also, by the definition of a limit point, ISn - sl < E/2 for infinitely lI1allY n, so that we can find and fix an Il > N* such that ISn - sl < El2. Together, for e\'el)' r > N*, ST - 1 sl = I(ST - S.,} + (Sn - s)1 ~ Is,. - snl + ISn - sl E E < 2 + 2 = E; that is. the sequence SI, S2' ... is convergent with the limit s. THEOREM • Bolzano-Weierstrass Theorem 3 A bounded infillite sequellce Z1> Z2, 23, . . • in the complex plane has at least one limit point. PROOF It is obvious that we need both conditions: a finite sequence cannot have a limit point. and the sequence I, 2, 3.... , which is infinite but not bounded. has no limit point. To prove the theorem, consider a bounded infinite sequence ZI. Z2 • ... and let K be such that < K for all n. If only finitely many values of the Zn are different, then. since the sequence is infinite, some number z must occur infinitely many times in the sequence, and, by definition, this number is a limit point of the sequence. We may now tum to the case when the sequence contains infinitely many differem terms. We draw a large square Qo that contains all Zw We subdivide Qo into four congruent squares, which we number 1, 2. 3, 4. Clearly, at least one of these squares (each taken Iznl 3BERNARD BOLZANO (1781-1848). Austrian mathematician and professor of religious studies, was a pioneer in the study of point sets, the foundation of analysis, and mathematical logic. For Weierstrass. see Sec. 15.5. A92 APP. 4 Additional Proofs with its complete boundary) must contain infinitely many terms of the sequence. The square of this type with the lowest number (1. 2, 3, or 4) will be denoted by Q1' This is the first step. In the next step we subdivide Q1 into four congruent squares and select a square Q2 by the same rule, and so on. This yields an infinite sequence of squares Qo. Q1, Q2, ... , Qn, ... with the property that the side of Qn approaches zero as 11 approaches infinity, and Qm contains all Qn with 11 > m. It is not difficult to see that the number which belongs to all these squares,4 call it:::: = a, is a limit point of the sequence. In fact, given an E > O. we can choose an N so large that the side of the square QN is less than € and, since QN contains infinitely many Zn. we have Izn - aJ < E for infinitely many 11. This completes the proof. • Section 15.3, pages 681-682 T (b) OF THE PROOF OF THEOREM 5 We have to show that = L an LlZ[(Z + + LlZ)",-2 2z(z + uz)n-3 + ... + (11 - 1) zn-2], 11.=2 thus, (z + LlZ)n - zn Llz = LlZ[(Z + Llz)n-2 + If we set Z + .1z = band z 2z(z + LlZ)n-3 = a, thuf, ,lz = + ... + (n - l)z11.-2]. h - a, this becomes simply (7a) (11 = 2,3, ... ), where An is the expression in the brackets on the right, (7b) thus, A2 = 1, A3 = b since then + 2a. etc. We prove (7) by induction. When n - 2a = (b + a)(b - a) b-a - 2a = 2. then (7) holds. = b - a = (b - a)A 2 . Assuming that (7) holds for 11 = k, we show that it holds for n = k + 1. By adding and subtracting a term in the numerator and then dividing we first obtain bk + 1 b-a - ba k + bak - ak + 1 b-a 4The fact that such a unique number;:; = a exists seems [0 be obvious, but it actually follows from an axiom of the real number system, the so-called CantOl~Dedekind axiom: see footnote 3 in App. A3.3. APP. 4 Additional Proofs A93 By the induction hypothesis, the right side equals b[(b - a)Ak calculation shows that this is equal to From (7b) with Il + ka k - 1 ] + a k . Direct = k we see that the expression in the braces {... } equals bk - 1 + 2ab k - 2 + ... + bk + 1 ak + 1 (k - l)ba k - 2 + ka k - 1 = A k + 1 • + (k + Hence our result is _ b-a = (b - a)A k + 1 Taking the last term to the left, we obtain (7) with n integer n ~ 2 and completes the proof = k + l)a k . I. This proves (7) for any • Section 18.2, page 754 without the use of a harmonic conjugate A NOT HER PROOF 0 F THE 0 REM 1 We show that if w = u + iu = i(z) is analytic and maps a domain D conformally onto a domain D* and <I>*(u, u) is harmonic in D*, then (1) <I>(x, y) = <I>*(u(x, y), u(x, y)) is harmonic in D. that is, y2<1> = 0 in D. We make no use of a hmmonic conjugate of <1>*, but use straightforward differentiation. By the chain rule, We apply the chain rule again. underscoring the terms that will drop out when we form ,2<1>: <l>yy is the same with each x replaced by y. We form the sum y2<1>. In it, <I>~u = <I>~v is multiplied by which is 0 by the Cauchy-Riemann equations. Also y 2 U = 0 and y 2 U = O. There remains By the Cauchy-Riemann equations this becomes and is 0 since <1>* is harmonic. • APPENDIX 5 Tables For Tables of Laplace transforms see Secs. 6.8 and 6.9. For Tables of Fourier transforms see Sec. 11.10. If you have a Computer Algebra System (CAS), you may not need the present tables, but you may still find them convenient from time to time. Table Al Bessel Functions For more extensive tables see Ref. [GRll in App. I. x 1 0(x) 1 1 (x) x 1 0(:0:) 0.0 0.1 0.2 0.3 0.4 1.0000 0.9975 0.9900 0.9776 0.9604 0.0000 0.0499 0.0995 0.1483 0.1960 3.0 3.1 3.2 3.3 3.4 -0.2601 -0.2921 -0.3202 -0.3443 -0.3643 0.3391 0.3009 0.2613 0.2207 0.1792 6.0 6.1 6.2 6.3 6.4 0.1506 0.1773 0.2017 0.2238 0.2433 -0.2767 -0.2559 -0.2329 -0.2081 -0.1816 0.5 0.6 0.7 0.8 0.9 0.9385 0.9120 0.8812 0.8-1-63 0.8075 0.2423 0.2867 0.3290 0.3688 0.4059 3.5 3.6 3.7 3.8 3.9 -0.3801 -0.3918 -0.3992 -0.4026 -0.4018 0.1374 0.0955 0.0538 0.0118 -0.0272 6.5 6.6 6.7 6.8 6.9 0.2601 0.2740 0.2851 0.2931 0.2981 -0.1538 -0.1250 -0.0953 -0.0652 -0.0349 1.0 1.1 1.2 1.4 0.7652 0.7196 0.6711 0.6201 0.5669 0...\401 0.4709 OA983 0.5220 0.5419 -l.0 4.1 4.2 4.3 4.4 -0.3971 -0.3887 -0.3766 -0.3610 -0.3423 -0.0660 -0.1033 -0.1386 -0.1719 -0.2028 7.0 7.1 7.2 7.3 7.4 0.3001 0.2991 0.2951 0.2882 0.2786 -0.0047 0.0152 0.0543 0.0826 0.1096 1.5 1.6 1.7 1.8 1.9 0.5118 0.4554 0.3980 0.3400 0.2818 0.5579 0.5699 0.5778 0.5815 0.5812 4.5 -l.6 4.7 4.8 4.9 -0.3205 -0.1961 -0.1693 -0.2404 -0.2097 -0.2311 -0.2566 -0.2791 -0.2985 -0.3147 7.5 7.6 7.7 7.8 7.9 0.2663 0.2516 0.2346 0.2154 0.1944 0.1352 0.1592 0.1813 0.2014 0.2192 2.0 2.1 2.2 2.3 2.4 0.2239 0.1666 0.1104 0.0555 0.0025 0.5767 0.5683 0.5560 0.5399 0.5202 5.0 5.1 5.2 5.3 5.4 -0.1776 -0.1443 -0.1103 -0.0758 -0.0412 -0.3276 -0.3371 -0.3431 -0.3460 -0.3453 8.0 8.1 8.2 8.3 8.4 0.1717 0.1475 0.1222 0.0960 0.0692 0.2346 0.2476 0.2580 0.2657 0.2708 2.5 2.6 2.7 2.8 2.9 -0.0484 -0.0968 -0.1424 -0.1850 -0.2143 0.4971 0.-1-708 0.4416 0.4097 0.3754 5.5 5.6 5.7 5.8 5.9 -0.0068 0.0270 0.0599 0.0917 0.1220 -0.3414 -0.3343 -0.3241 -0.3110 -0.2951 8.5 8.6 8.7 8.8 8.9 0.0419 0.0146 -0.0125 -0.0392 -0.0653 0.2731 0.2728 0.2697 0.2641 0.2559 1.3 - hex) I x 1 1 (x) 10(x) I fo(x) ~ 0 for x = 2.40483. 5.52008, 8.65373, 11.7915, 14.9309, 1~.0711, 21.2116, 24.3525. 27.4935, 30.6346 11(x) = 0 for x = 3.83171, 7.01559,10.1735, 13.3237. 16.4706. 19.6159,22.7601,25.9037,29.0468.32.1897 A94 A95 APP. 5 Tables (continued) Table A1 - x :I - I I 0.0 0.5 Y1 (xj YolX} I :~ I 2.0 (-x) (-x) -0.445 0.088 0.382 0.510 -1.471 -0.781 -0.412 -0.107 Table Al x Yo\.l:) Yl~\') x YoIX) 2.5 3.0 3.5 4.0 4.5 0.498 0.377 0.189 -0.017 -0.195 0.146 0.325 0.410 0.398 0.301 5.0 5.5 6.0 6.5 7.0 -0.309 -0.339 -0.288 -0.173 -0.026 Yl~\") I 0.148 -0.024 -0.175 -0.274 -0.303 I I Gamma Function [see (24) in App. A3.1] a f(a) a na) a r(a) a na) Q' f(a) 1.00 1.000000 1.20 0.911l169 1.40 0.1l1l7264 1.60 0.1l93515 1.1l0 0.93131l4 1.02 1.04 1.06 LOll 0.988844 0.978438 0.968744 0.959725 1.22 1.24 1.26 1.28 0.913106 0.908521 0.904397 0.900718 1.42 1.44 1.46 1048 0.886356 0.885805 0.885604 0.885747 1.62 1.64 1.66 1.68 0.895924 0.898642 0.901668 0.905001 1.82 1.84 1.86 1.88 0.936845 0.942612 0.948687 0.955071 1.10 0.951351 1.30 0.897471 1.50 0.886227 1.70 0.908639 1.90 0.961766 1.12 1.14 1.16 1.18 0.943590 0.936416 0.929803 0.923728 1.32 1.34 1.36 1.38 0.894640 0.892216 0.890185 0.888537 1.52 1.54 1.56 1.58 0.887039 0.888178 0.889639 0.891420 1.72 1.74 1.76 1.78 0.912581 0.916826 0.921375 0.926227 1.92 1.94 1.96 1.98 I 0.976099 0.983743 1.20 0.918 169 1.40 0.887264 1.60 0.893515 1.80 0.931 384 2.00 11.000 000 : I I I '- Table Al 0.968774 0.991 708 Factorial Function and Its Logarithm with Base 10 I I II 11! log (II!) 11 11! log (II!) 11 11' log VI!) 1 2 3 4 5 1 2 6 24 120 u.uuuOOO 0.301 030 0.778 151 1.380211 2.079 181 6 7 8 9 10 72u 5040 4032u 362880 3628800 2.857332 3.702431 4.605521 5.559763 6.559763 11 12 13 14 15 39916800 479001 600 6227 u2u IlUO 87178291200 1 307674368000 7.6Ul 156 8.680337 9.794280 10.940408 12.116500 I Table A4 I erfx Silxj CIIXI x erfx Si(xi ci(x) U.U u.uuuu u.uuuu :JO 2.u 0.9953 1.6054 -0.4230 0.2 0.4 0.6 0.8 1.0 0.2227 0.4284 0.6039 0.7421 0.8427 0.1996 0.3965 0.5881 0.7721 0.9461 1.0422 0.3788 0.0223 -0.1983 -0.3374 2.2 2.4 2.6 2.8 3.0 0.9981 0.9993 0.9998 0.9999 1.0000 1.6876 1.7525 1.8004 1.8321 1.8487 -0.3751 -0.3173 -0.2533 -0.1865 -0.1196 1.2 0.9103 0.9523 0.9763 0.9891 0.9953 1.1080 1.2562 1.3892 1.5058 1.6054 -0.4205 -0.4620 -0.4717 -0.4568 -0.4230 3.2 3.4 3.6 3.8 4.0 1.0000 1.0000 1.0000 1.0000 1.0000 1.8514 1.8419 1.8219 1.7934 1.7582 -0.0553 0.0045 0.0580 0.1038 0.1410 I I 1.8 2.0 I Error Function, Sine and Cosine Integrals [see (35), (40), (42) in App. A3.1] l: 104 1.6 I II A96 APP. 5 Tables Table AS Binomial Distribution Probability function f(x) [see (2), Sec. 24.7] and distribution function F(x) p - " x = 0.1 I(~F(X)_ o. p - J(x) - o. = 0.2 F(x) r-- P - = 0.3 F(x) I(x) - - -I - - o. p I(x) = 0.4 P = 0.5 F(x) I(x) o. I F(x) o. 0 I 9000 1000 0.9000 1.0000 8000 2000 0.8000 1.0000 7000 3000 0.7000 1.0000 6000 4000 0.6000 1.0000 5000 5000 0.5000 1.0000 '1 0 I 2 !HOO 1800 0100 OJHOO 0.9900 1.0000 6400 3200 0400 0.6400 0.9600 1.0000 4'J00 4200 0900 OA'JOO 0.9100 1.0000 3600 4800 1600 0.3600 0.8400 1.0000 2500 5000 2500 0.2500 0.7500 1.0000 7290 2430 0270 0010 0.7290 0.9720 0.9990 1.0000 5120 3840 0960 0080 0.5120 3 0 1 2 3 0.9920 1.0000 3430 4410 I!NO 0270 0.3430 0.7840 0.9730 1.0000 2160 4320 2880 0640 0.2160 0.6480 0.9360 1.0000 1250 3750 3750 1250 0.1250 0.5000 0.8750 1.0000 6561 2916 0486 0036 0001 0.6561 0.9477 0.9963 0.9999 1.0000 40Y6 0.4096 0.8192 0.9728 0.9984 1.0000 2401 4116 2646 0756 0081 0.2401 0.6517 0.9163 0.9919 1.0000 12'J6 0.12'J6 4 0 I 2 3 4 3456 3456 1536 0256 0.4752 0.8208 0.9744 1.0000 0625 2500 3750 2500 0625 0.0625 0.3125 0.6875 0.9375 1.0000 0 5'}u5 3281 2 3 4 5 0081 0005 0000 u.5905 0.9185 0.9'J14 0.9995 1.0000 1.0000 2048 0512 0064 0003 u.3277 0.7373 0.9421 0.9933 0.9997 1.0000 Ib81 3602 3087 1323 0284 0024 u.lb81 0.5282 0.8369 0.9692 0.9976 1.0000 lJ778 I 3456 2304 0768 0102 (J.l)778 0.3370 0.6826 0.9130 0.9898 1.0000 lJ313 1563 3125 3125 1563 0313 0.u313 0.1875 0.5000 0.8125 0.9688 1.0000 0 I 2 3 4 5 6 5314 3543 0984 0146 0012 0001 0000 0.2621 0.6554 0.9011 0.'J830 0.9984 0.9999 1.0000 1176 3025 3241 1852 0595 0102 0007 0.1176 0.4202 0.7443 0.92'J5 1.0000 1.0000 2621 3932 2458 0819 0154 0015 0001 0467 1866 3110 2765 1382 0369 0041 0.0467 0.2333 0.5443 0.8208 0.9590 0.9959 1.0000 0156 0938 2344 3125 2344 0938 0156 0.0156 0.1094 0.3438 0.6563 0.8906 0.9844 1.0000 0 I 2 3 4 5 6 7 4783 3720 1240 0230 0026 0002 0000 0000 0.4783 0.8503 0.9743 0.9973 0.9998 1.0000 1.0000 1.0000 2097 3670 2753 1147 0287 0043 0004 0000 0.2097 0.5767 0.8520 0.'J667 0.9'J53 0824 2471 3177 2269 0972 0250 0036 0002 0.0824 0.3294 0.6471 0.8740 0.9712 0.9962 0.9998 1.0000 0280 1306 2613 2903 0.0280 0.1586 0.4199 0.7102 0.9037 0.9812 0.9984 1.0000 0078 0547 1641 2734 2734 1641 0547 0078 0.0078 0.0625 0.2266 0.5000 0.7734 0.9375 0.9922 1.0000 U I 2 3 4 :; 6 7 8 4305 3826 1488 0331 0046 0004 0000 0000 0000 U.4305 0.8131 0.9619 1678 3355 0.1678 0.5033 0576 1977 2965 2541 1361 0467 0100 0012 0001 U.0576 0.2553 0.5518 0.8059 0.9420 0.9887 0.9987 0.9999 1.0000 0168 0896 2090 2787 2322 1239 0413 0079 0007 U.0168 0.1064 0.3154 0.5941 0.8263 0.'J502 0.9915 om'J 0313 1094 2188 2734 2188 1O'J4 0313 0039 o.um'J 0.0352 0.1445 0.3633 0.6367 0.8555 0.9648 0.9961 1.0000 I' - 4096 1536 0256 0016 0.8Y60 - I 5 072'J 3277 40Y6 25Y2 - 6 I I 7 8 0.5314 0.8857 0.9841 0.9987 0.99'J'J 0.'J'J50 0.9996 1.0000 1.0000 1.0000 1.0000 0.9996 I.oooo 1.0000 2'J36 0.7Y69 1468 0459 0092 0011 0001 0000 0.9437 0.98'J6 0.9988 0.9999 1.0OOO 1.0000 0.9891 0.9993 1.0000 1'J35 0774 0172 0016 0.9'J93 1.0000 I APP.5 A97 Tables Table A6 Poisson Distribution Probability function f(x) [see (5), Sec. 24.7] and distribution function F(x) ix f.L I(x) = 0.1 F(x) o. f.L I(x) = 0.2 F(x) f.L I(x) = 0.3 F(x) o. 0 9048 0.9048 O. 8187 O.ll 1117 7408 0.74011 1 2 3 4 5 0905 0045 0002 0000 0.9953 0.9998 1.0000 1.0000 1637 0164 0011 0001 0.9825 0.9989 0.9999 1.0000 2222 0333 0033 0003 0.9631 0.9964 0.9997 1.0000 x I(~) 0 I f.L = 0.6 f.L = 0.8 F(x) I(x) F(x) 5488 0.5488 O. 4966 0.4966 O. 4493 0.4493 3293 0988 0198 0030 0004 0.8781 0.9769 0.9966 0.9996 1.0000 3476 1217 0284 0050 0007 0.1l442 0.9659 0.9942 0.9992 0.9999 3595 1438 0383 0077 0012 0001 1.0000 0002 6 7 x = 0.7 J(x) o. 2 3 4 5 f.L F(x) f.L = 1.5 F(x) I(r) II. I(x) f.L=2 F(x) II. I(x) f.L I(x) O. 6703 2681 0536 oon 0007 0001 f.L I(x) = 0.4 F(x) f.L I(x) 0.6703 O. 6065 0.9384 0.9921 0.9992 0.9999 1.0000 3033 0758 0126 0016 0002 = 0.9 = 0.5 F(x) 0.6065 0.9098 0.9856 0.9982 0.9998 1.0000 I(x} 4066 0.4066 3679 0.3679 0.8088 0.9526 0.9909 0.9986 0.9998 3659 1647 0494 0111 0020 0.7725 0.9371 0.9865 0.9977 0.9997 3679 1839 0613 0153 0031 0.7358 0.9197 0.9810 0.9963 0.9994 1.0000 0003 1.0000 0005 0001 0.9999 1.0000 f.L=3 F(x) I(x} o. f.L=4 F(x) o. II. I(x) f.L=5 F(x) II. 0 2231 0.2231 1353 0.1353 0498 0.0498 0183 0.0183 0067 0.0067 I 3347 2510 1255 0471 0141 0.5578 0.8088 0.9344 0.9814 0.9955 2707 2707 1804 0902 0361 0.4060 0.6767 0.8571 0.9473 0.9834 1494 2240 2240 1680 1O01l 0.1991 0.4232 0.6472 0.8153 0.9161 0733 1465 1954 1954 1563 0.0916 0.2381 0.4335 0.6288 0.7851 0337 0842 1404 1755 1755 0.0404 0.1247 0.2650 0.4405 0.6160 0035 0008 0001 0.9991 0.9998 1.0000 0120 0034 0009 0002 0.9955 0.9989 0.9998 1.0000 0504 0216 0081 0027 0008 0.9665 0.9881 0.9962 0.9989 0.9997 1042 0595 0298 0132 0053 0.8893 0.9489 0.9786 0.9919 0.9972 1462 1044 0653 0363 0181 0.7622 0.8666 0.9319 0.9682 0.9863 0002 0001 0.9999 1.0000 0019 0006 0.9991 0.9997 0082 0034 0.9945 0.9980 13 0002 0.9999 0013 0.9993 14 15 0001 1.0000 0005 0002 0.9998 0.9999 0000 1.0000 2 3 4 5 6 7 8 9 10 II 12 16 I f.L=1 F(x) F(x) o. I I A98 APP. 5 Tables Table A7 Normal Distribution Values of the distribution function <1>(.::) [see (3). Sec. 24.81. <1>( -~) = 1 - <1>(z) z <1>(;:) 5040 5080 5120 5160 5199 0.51 0.52 0.53 0.54 0.55 0.06 0.07 0.08 0.09 0.10 5239 5279 5319 5359 5398 0.11 0.12 0.13 0.14 0.15 z <1>(~) 0.01 0.02 0.03 0.04 0.05 z <1>(~) z <1>(~) 6950 6985 7019 7054 7088 loOI lo02 1.03 1.04 1.05 9778 9783 9788 9793 9798 2.51 2.52 2.53 2.54 2.55 9940 9941 9943 9945 9946 0.56 0.57 0.58 0.59 0.60 7123 7157 7190 7224 7257 2.06 2.07 2.08 2.09 2.10 9R03 9808 9812 9817 9821 2.5ti 2.57 2.58 2.59 2.60 9948 9949 9951 9952 9953 5438 5478 5517 5557 5596 0.61 0.62 0.63 0.64 0.65 94ti3 9474 9484 9495 9505 HI 2.12 2.13 2.14 2.15 982ti 9830 9834 9838 9842 2.61 2.62 2.63 2.64 2.65 9955 9956 9957 9959 9960 0.16 0.17 0.18 0.19 0.20 5636 5675 5714 5753 5793 1.66 1.67 1.68 1.69 1.70 9515 9525 9535 9545 9554 2.16 2.17 2.18 2.19 2.20 9846 9850 9854 9857 9861 2.66 2.67 2.68 2.69 2.70 9961 9962 9963 9964 9965 0.21 0.22 0.23 0.24 0.25 8869 8888 8907 8925 8944 1.71 1.72 1.73 1.74 1.75 9564 9573 9582 9591 9599 2.21 2.22 2.23 2.24 2.25 9864 9868 9871 9875 9878 2.71 2.72 2.73 2.74 2.75 9966 9967 9968 9969 9<)70 1.26 1.27 1.28 1.29 1.30 8962 8980 8997 9015 9032 1.76 1.77 1.78 1.79 1.80 9608 9616 9625 9633 9641 2.26 2.27 2.28 2.29 2.30 9RRI 9884 9887 98<)0 9893 2.76 2.77 2.78 2.79 2.80 9971 9972 9973 9974 9974 7910 7939 7967 7995 8023 1.31 1.32 1.33 1.34 1.35 9049 9066 9082 9099 9115 1.81 1.82 lo83 1.84 1.85 9649 9656 9664 9671 9678 2.31 2.32 2.33 2.34 2.35 9896 9898 9901 9904 9906 2.81 2.82 2.83 2.84 2.85 9975 9976 9977 9977 9<)78 0.86 0.87 0.88 0.89 0.90 8051 8078 8106 8133 8159 1.36 1.37 1.38 1.39 lAO 9131 <)147 9162 9177 9192 1.86 1.87 1.88 1.89 1.90 9686 9693 9699 9706 9713 2.36 2.37 2.38 2.39 2AO 9909 9911 9913 9916 9918 2.86 2.87 2.88 2.89 2.90 9979 9979 9980 9981 9981 6591 6628 6664 6700 6736 0.91 0.92 0.93 0.94 0.95 8186 8212 8238 8264 8289 1.41 1.42 1.43 1.44 lA5 9207 9222 9236 9251 9265 1.91 1.92 1.93 1.94 1.95 9719 9726 9732 9738 9744 2.41 2A2 2.43 2.-14 2A5 9920 9922 9925 9927 <)929 2.91 2.92 2.93 2.94 2.95 9982 9982 9983 9984 9984 6772 6808 6844 6879 6915 0.96 0.97 0.98 0.99 1.00 8315 8340 8365 8389 8413 1.46 1.47 1.48 1.49 1.50 9279 <)292 9306 9319 9332 1.96 1.97 1.98 1.99 2.00 9750 9756 9761 9767 9772 2A6 2A7 2A8 2.49 2.50 9931 9932 9934 9936 9938 2.96 2.97 2.98 2.99 3.00 9985 9<)85 9986 9986 9987 ~ <P(~) <. <1>(~) 8438 8461 8485 8508 8531 1.51 1.52 1.53 1.54 1.55 9345 9357 9370 9382 9394 2.01 2.02 2.03 2.04 2.05 1.06 1.07 1.08 1.09 1.10 8554 8577 8599 8621 8643 1.56 1.57 1.58 1.59 1.60 9406 9418 9429 9441 9452 7291 7324 7357 7389 7422 1.11 1.12 1.13 1.14 1.15 8665 8686 8708 8729 8749 1.61 1.62 1.63 1.64 1.65 0.66 0.67 0.68 0.69 0.70 7454 7486 7517 7549 7580 1.16 1.17 1.18 1.19 1.20 8770 8790 8810 8830 8849 5832 5871 5910 5948 5987 0.71 0.72 0.73 0.74 0.75 7611 7642 7ti73 7704 7734 1.21 1.22 1.23 1.24 1.25 0.26 0.27 0.28 0.29 0.30 6026 6064 6103 6141 6179 0.76 0.77 0.78 0.79 0.80 7764 7794 7823 7852 7881 0.31 0.32 0.33 0.34 0.35 6217 6255 6293 6331 ti3tiR 0.81 0.82 0.83 0.84 0.85 0.36 0.37 0.38 0.39 0.40 6406 6443 6480 6517 6554 OAI OA2 OA3 0.44 OA5 OA6 OA7 0.48 OA9 I 0.50 u. u. u. u. I - u. u. A99 APP. 5 Tables Table AS Normal Distribution Values of z for given values of cI>(z) [see (3). Sec. 24.8] and D(-;.) Example: -;. = 0.279 if cI>(-;.) = 61 %; z = 0.860 if D(-;.) = 61 %. % ~(<1» z(D) % z(<1» cI>(-;.) - cI>( - z) ~(D) % :(<1» :(D) 1 2 3 4 5 -2.326 -2.054 -1.881 -1.751 -1.645 0.013 0.025 0.038 0.050 0.063 41 42 43 44 45 -0.22S -0.202 -0.176 -0.151 -0.126 0.539 0.553 0.568 0.583 0.598 SI 82 83 84 85 0.S7S 0.915 0.954 0.994 1.036 1.311 1.341 1.372 1.405 1.440 6 7 8 9 10 -1.555 -1.476 -1.405 -1.341 -1.282 0.075 0.088 0.100 0.113 0.126 46 47 48 49 50 -0.100 -0.075 -0.050 -0.025 0.000 0.613 0.628 0.643 0.659 0.674 86 S7 88 89 90 1.080 1.126 1.175 1.227 1.282 1.476 1.514 1.555 1.598 1.645 11 12 13 14 15 -1.227 -1.175 -1.126 -1.080 -1.036 0.138 0.151 0.164 0.176 0.IS9 51 52 53 54 55 0.025 0.050 0.075 0.100 0.126 0.690 0.706 0.722 0.739 0.755 91 92 93 94 95 1.341 1.405 1.476 1.555 1.645 1.695 1.751 1.812 1.881 1.960 16 17 18 19 20 -0.994 -0.954 -0.915 -0.878 -0.842 0.202 0.215 0.228 0.240 0.253 56 57 58 59 60 0.151 0.176 0.202 0.228 0.253 0.772 0.789 0.806 0.824 0.842 96 97 97.5 98 99 1.751 1.881 1.9{)() 2.054 2.326 2.054 2.170 2.241 2.326 2.576 21 22 23 24 25 -0.806 -0.772 -0.739 -0.706 -0.674 0.266 0.279 0.292 0.305 0.319 61 62 63 64 65 0.279 0.305 0.332 0.358 0.385 0.860 0.878 0.896 0.915 0.935 99.1 99.2 99.3 99.4 99.5 2.366 2.409 2.457 2.512 2.576 2.612 2.652 2.697 2.748 2.807 26 27 2R 29 30 - 0.643 - 0.613 -0.5S3 -0.553 -0.524 0.332 0.345 0.358 0.372 0.385 66 67 68 69 70 0.412 0.-140 0.468 0.496 0.524 0.954 0.974 0.994 1.015 1.036 99.6 99.7 99.8 99.9 2.652 2.748 2.878 3.090 2.878 2.968 3.090 3.291 31 32 33 34 35 -0.496 -0.468 -0.-140 -0.412 -0.385 0.399 0.412 0.-1-26 0.440 0.454 71 72 73 74 75 0.553 0.583 0.613 0.643 0.674 1.058 1.080 1.103 1.126 1.150 99.91 99.92 99.93 99.94 99.95 3.121 3.156 3.195 3.239 3.291 3.320 3.353 3.390 3.432 3.481 36 37 38 39 40 -0.358 -0.332 -0.305 -0.279 -0.253 0.468 0.482 0.496 0.510 0.524 76 77 78 79 80 0.706 0.739 0.772 O.S06 0.842 1.175 1.200 1.227 1.254 1.282 99.96 99.97 99.98 99.99 3.353 3.432 3.540 3.719 3.540 3.615 3.719 3.891 - -- , A100 APP. 5 Tables Table A9 t-Distribution Values of z for given values of the distribution function F(z) (see (8) in Sec. 25.3). Example: For 9 degrees of freedom, z = 1.83 when F(z) = 0.95. Number of Degrees of Freedom F(:) 2 3 4 5 6 7 8 9 10 0.00 0.32 0.73 1.38 3.08 0.00 0.29 0.62 1.06 1.89 0.00 0.28 0.58 0.98 1.64 0.00 0.27 0.57 0.94 1.53 0.00 0.27 0.56 0.92 1.48 0.00 0.26 0.55 0.91 1.44 0.00 0.26 0.55 0.90 1.41 0.00 0.26 0.55 0.89 1.40 0.00 0.26 0.54 0.88 1.38 0.00 0.26 0.54 0.88 1.37 6.31 12.7 31.8 63.7 318.3 2.92 4.30 6.96 9.92 22.3 2.35 3.18 4.54 5.84 10.2 2.13 2.78 3.75 4.60 7.17 2.02 2.57 3.36 4.03 5.89 1.94 2.45 3.14 3.71 5.21 1.89 2.36 3.00 3.50 4.79 1.86 2.31 2.90 3.36 4.50 1.83 2.26 2.82 3.25 4.30 1.81 2.23 2.76 3.17 4.14 19 20 L 0.5 0.6 0.7 0.8 0.9 0.95 0.975 I 0.99 0.995 0.999 l Number of Degrees of Freedom F(~) II 12 0.00 0.26 0.54 0.88 1.36 U.OO 0.6 0.7 0.8 0.9 0.26 0.54 0.87 1.36 0.95 0.975 0.99 0.995 0.999 1.80 2.20 2.72 3.11 4.02 1.78 2.18 2.68 3.05 3.93 1.77 2.16 2.65 3.01 3.85 U.S I 13 14 15 17 18 U.OO U.OO !l.OO 0.26 0.54 0.87 1.35 0.26 0.54 0.87 1.35 0.26 0.54 0.87 1.34 O.UU U.OU 0.26 0.54 0.86 1.34 0.26 0.53 0.86 1.33 0.00 0.26 0.53 0.86 1.33 1.76 2.14 2.62 2.98 3.79 1.75 2.13 2.60 2.95 3.73 1.75 2.12 2.58 2.92 3.69 1.74 2.11 2.57 2.90 3.65 16 II U.OO U.UU 0.26 0.53 0.86 1.33 0.26 0.53 0.86 1.33 1.73 2.10 2.55 2.88 3.61 1.73 2.09 2.54 2.86 3.58 1.72 2.09 2.53 2.85 3.55 I Number of Degrees of Freedom F(:) 22 24 26 28 30 40 50 100 200 ex: 0.5 0.6 0.7 0.8 0.9 0.00 0.26 0.53 0.86 1.32 0.00 0.26 0.53 0.86 1.32 0.00 0.26 0.53 0.86 1.31 0.00 0.26 0.53 0.85 1.31 0.00 0.26 0.53 0.85 1.31 0.00 0.26 0.53 0.85 1.30 0.00 0.25 0.53 0.85 1.30 0.00 0.25 0.53 0.85 1.29 0.00 0.25 0.53 0.84 1.29 0.00 0.25 0.52 0.84 1.28 0.95 0.975 0.99 1.72 2.07 2.51 2.82 3.50 1.71 2.06 2.49 2.80 3.47 1.71 2.06 2.48 2.78 3.43 1.70 2.05 2.47 2.76 3.41 1.70 2.04 2.46 2.75 3.39 1.68 2.02 2.42 2.70 3.31 1.68 2.01 2.40 2.68 3.26 1.66 1.98 2.36 2.63 3.17 1.65 1.97 2.35 2.60 3.13 1.65 1.96 2.33 2.58 3.09 I 0.995 0.999 I APP. 5 A10l Tables Table A10 Chi-square Distribution Values of x for given values of the distribution function F(z) (see Sec. 25.3 before (17)). Example: For 3 degrees of freedom, z = 1l.34 when F(z.) = 0.99. Number of Degrees of Freedom F(z) 1 2 3 0.005 0.01 0.025 0.05 0.00 0.00 0.00 0.00 0.01 0.02 0.05 0.10 0.07 0.11 0.22 0.35 0.95 0.975 0.99 0.995 3.84 5.02 6.63 7.88 5.99 7.38 9.21 10.60 7.81 9.35 11.34 12.84 4 5 6 0.21 0.30 0.48 0.71 0.41 0.55 0.83 US 0.68 0.87 1.24 1.64 9.49 11.14 13.28 14.86 11.07 12.83 15.09 16.75 12.59 14.45 16.81 18.55 7 8 - 0.99 1.34 1.24 1.65 2.18 1.69 2.17 "2.73 14.07 16.01 15.51 17.53 20.09 21.95 18A8 20.28 9 10 1.73 2.09 2.70 3.33 2.16 2.56 3.25 3.94 16.92 19.02 21.67 23.59 18.31 20.48 23.21 25.19 - Number of Degrees of Freedom F(z) 11 12 13 14 IS 16 17 18 19 20 0.005 0.01 0.025 0.05 2.60 3.05 3.82 4.57 3.07 3.57 4.40 5.23 3.57 4.11 5.01 5.89 4.07 4.66 5.63 6.57 4.60 5.23 6.26 7.26 5.14 5.81 6.91 7.96 5.70 6.41 7.56 8.67 6.26 7.01 8.23 9.39 6.84 7.63 8.91 10.12 7.43 8.26 9.59 10.85 0.95 0.975 0.99 0.995 19.68 21.92 24.72 26.76 21.03 23.34 26.22 28.30 22.36 24.74 27.69 29.82 23.68 26.12 29.14 31.32 25.00 27.49 30.58 32.80 26.30 28.85 32.00 34.27 27.59 30.19 33.41 35.72 28.87 31.53 34.81 37.16 30.14 32.85 36.19 38.58 31.41 34.17 37.57 40.00 -- -'-- ~ -- -~ -~ Number of Degrees of Freedom F(::J 21 22 23 24 25 26 27 28 29 30 0.005 0.01 0.0"25 0.05 8.0 8.9 10.3 11.6 8.6 9.5 11.0 12.3 9.3 10.2 11.7 13.1 9.9 10.9 12.4 13.8 10.5 11.5 13.1 14.6 11.2 12.2 13.8 15.4 11.8 12.9 14.6 16.2 12.5 13.6 15.3 16.9 13.1 14.3 16.0 17.7 13.8 15.0 16.8 18.5 0.95 0.975 0.99 0.995 32.7 35.5 38.9 41..1- 33.9 36.8 40.3 42.8 35.2 38.1 41.6 44.2 36.4 39.4 43.0 45.6 37.7 40.6 44.3 46.9 38.9 41.9 45.6 48.3 40.1 43.2 47.0 49.6 41.3 44.5 48.3 51.0 42.6 45.7 49.6 52.3 43.8 47.0 50.9 53.7 Number of Degrees of Freedom F(z) > 100 (Approximation) 40 50 60 70 80 90 100 0.005 0.01 0.025 0.05 20.7 22.2 24.4 26.5 28.0 29.7 32.4 34.8 35.5 37.5 40.5 43.2 43.3 45.4 48.8 51.7 51."2 53.5 57.2 60.4 59."2 61.8 65.6 69.1 67.3 70.1 74.2 77.9 !(h !(h !(h !(h - 0.95 0.975 0.99 L 0.995 55.8 59.3 63.7 66.8 67.5 71.4 76.2 79.5 79.1 83.3 88.4 92.0 90.5 95.0 100.4 104.2 101.9 106.6 112.3 116.3 113.1 118.1 124.1 1"28.3 124.3 129.6 135.8 140.2 !(h !(h + 1.(4)2 + 1.96)2 + "2.33)2 + 2.58)2 -- In the last column, h = ~, where III IS the nUlllbel of degree~ of freedom. ~(h WI 2.58)2 2.33)2 1.96)2 1.64)2 Al02 APP.5 Tables Table All F-Distribution with (m, n) Degrees of Freedom Values of z for which the distribution function F(z) [see (13), Sec. 25.4] has the value Example: For (7, 4) d.f., .::: = 6.09 if F(.:::) = 0.95. 111=1 II 111=2 111=3 111=4 - = 5 111=6 111 = 7 111=8 111=9 230 19.3 9.01 6.26 5.05 234 19.3 S.94 6.16 4.95 237 19.4 8.89 6.09 4.88 239 19.4 8.85 6.04 4.82 241 19.4 8.81 6.00 4.77 III I 2 3 4 5 161 18.5 10.1 7.71 6.61 200 19.0 9.55 6.94 5.79 216 19.2 9.2S 6.59 5.41 225 19.2 9.12 6.39 5.19 6 7 8 9 10 5.99 5.59 5.32 5.12 4.% 5.14 4.74 4.46 4.26 4.10 4.76 4.35 4.07 3.86 3.71 4.53 4.12 3.84 3.63 3.4R 4.39 3.97 3.69 3A8 3.33 4.28 3.87 3.5S 3.37 3.22 4.21 3.79 3.50 3.29 3.14 4.15 3.73 3.44 3.23 3.07 4.10 3.68 3.39 3.18 3.02 II 12 13 14 15 4.84 4.75 4.67 4.60 4.54 3.98 3.89 3.SI 3.74 3.68 3.59 3.49 3.41 3.34 3.29 3.36 3.26 3.18 3.11 3.06 3.20 3.11 3.03 2.96 2.90 3.09 3.00 2.92 2.85 2.79 3.01 2.91 2.83 2.76 2.71 2.95 2.85 2.77 2.70 2.64 2.90 2.80 2.71 2.65 2.59 16 17 18 19 20 4.49 4.45 4.41 4.38 4.35 3.63 3.59 3.55 3.52 3.49 3.24 3.20 3.16 3.13 3.10 3.01 2.96 2.93 2.90 2.87 2.85 2.81 2.77 2.74 2.71 2.74 2.70 2.66 2.63 2.60 2.66 2.61 2.5S 2.54 2.51 2.59 2.55 2.51 2.48 2.45 2.54 2.49 2.46 2.42 2.39 22 24 26 28 30 4.30 4.26 4.23 4.20 4.17 3.44 3.40 3.37 3.34 3.32 3.05 3.01 2.98 2.95 2.92 2.82 2.78 2.74 2.71 2.69 2.66 2.62 2.59 2.56 2.53 2.55 2.51 2.47 2.45 2A2 2.46 2.42 2.39 2.36 2.33 2.40 2.36 2.32 2.29 2.27 2.34 2.30 2.27 2.24 2.21 32 34 36 38 40 4.15 4.13 4.11 4.10 4.08 3.29 3.28 3.26 3.24 3.23 2.90 2.88 2.87 2.85 2.84 2.67 2.65 2.63 2.62 2.61 2.51 2.49 2.48 2.46 2.45 2.40 2.38 2.36 2.35 2.34 2.31 2.29 2.28 2.26 2.25 2.24 2.23 2.21 2.19 2.18 2.19 2.17 2.15 2.14 2.12 50 60 70 80 90 4.03 4.00 3.98 3.96 3.95 3.18 3.15 3.13 3.11 3.10 2.79 2.76 2.74 2.72 2.71 2.56 2.53 2.50 2.49 2.47 2.40 2.37 2.35 2.33 2.32 2.29 2.25 2.23 2.21 2.20 2.20 2.17 2.14 2.13 2.11 2.13 2.10 2.07 2.06 2.04 2.07 2.04 2.02 2.00 1.99 100 150 200 IOUO 3.94 3.90 3.89 3.85 3.84 3.09 3.06 3.04 3.00 3.00 2.70 2.66 2.65 2.61 2.60 2.46 2.43 2.42 2.38 2.37 2.31 2.27 2.26 2.22 2.21 2.19 2.16 2.14 2.11 2.10 2.10 2.07 2.06 2.02 2.01 2.03 2.00 1.98 1.95 1.94 1.97 1.94 1.93 1.89 1.88 I :x: 0.95 APP. 5 A103 Tables Table All F-Distribution with (m, n) Degrees of Freedom (continued) Values of z for which the distribution function F(z) [see (13), Sec. 25.4] has the value 111=10 n 111 = 15 III = 20 111 = 30 111 = 40 III = 50 III = 100 0.95 00 2 3 4 5 242 19.4 8.79 5.96 4.74 246 19.4 8.70 5.86 4.62 248 19.4 8.66 5.80 4.56 250 19.5 8.62 5.75 4.50 251 19.5 8.59 5.72 4.46 252 19.5 8.58 5.70 4.44 253 19.5 8.55 5.66 4.41 254 19.5 8.53 5.63 4.37 6 7 8 9 10 4.06 3.64 3.35 3.14 2.98 3.94 3.51 3.22 3.01 2.85 3.87 3A4 3.15 2.94 2.77 3.81 3.38 3.08 2.86 2.70 3.77 3.34 3.04 2.83 2.66 3.75 3.32 3.02 2.80 2.64 3.71 3.27 2.97 2.76 2.59 3.67 3.23 2.93 2.71 2.54 11 12 13 14 15 2.85 2.75 2.67 2.60 2.54 2.72 2.62 2.53 2.46 2.40 2.65 2.54 2.46 2.39 2.33 2.57 2.47 2.38 2.31 2.25 2.53 2.43 2.34 2.27 2.20 2.51 2.40 2.31 2.24 2.18 2.46 2.35 2.26 2.19 2.12 2.40 2.30 2.21 2.13 2.07 16 17 18 19 20 2.49 2.45 2.41 2.38 2.35 2.35 2.31 2.27 2.23 2.20 2.28 2.23 2.19 2.16 2.12 2.19 2.15 2.11 2.07 2.04 2.15 2.10 2.06 2.03 1.99 2.12 2.08 2.04 2.00 1.97 2.07 2.02 1.98 1.94 1.91 2.01 1.96 1.92 1.88 1.84 22 24 26 28 30 2.30 2.25 2.22 2.19 2.16 2.15 2.11 2.07 2.04 2.01 2.07 2.03 1.99 1.96 1.93 1.98 1.94 1.90 1.87 1.84 1.94 1.89 1.85 1.82 1.79 1.91 1.86 1.82 1.79 1.76 1.85 1.80 1.76 1.73 1.70 1.78 1.73 1.69 1.65 1.62 32 34 36 38 40 2.14 2.12 2.11 2.09 2.08 1.99 1.97 1.95 1.94 1.92 1.91 1.89 1.87 1.85 1.84 1.82 1.80 1.78 1.76 1.74 1.77 1.75 1.73 1.71 1.69 1.74 1.71 1.69 1.68 1.66 1.67 1.65 1.62 1.61 1.59 1.59 1.57 1.55 1.53 1.51 50 60 70 80 90 2.03 1.99 1.97 1.95 1.94 un 1.84 1.81 1.79 1.78 1.78 1.75 1.72 1.70 1.69 1.69 1.65 1.62 1.60 1.59 1.63 1.59 1.57 1.54 1.53 1.60 1.56 1.53 1.51 1.49 1.52 1.48 1.45 1.43 1.41 1.44 1.39 1.35 1.32 1.30 100 150 200 1000 1.93 1.89 1.88 1.84 1.83 1.77 1.73 1.72 1.68 1.67 1.68 1.64 1.62 1.58 1.57 1.57 1.54 1.52 1.47 1.46 1.52 1.48 1.46 1.41 1.39 1.48 1.44 1.41 1.36 1.35 1.39 1.34 1.32 1.26 1.24 1.28 1.22 1.19 1.08 1.00 I cc I Al04 APP. 5 Tables Table All F-Distribution with (m, n) Degrees of Freedom (continued) Values of:z for which the distribution function F(::.) [see (13), Sec. 25.4] has the value 0.99 m=1 111=2 m=3 111=4 111=5 111=6 111=7 111=8 111=9 1 2 3 4 5 4052 98.5 34.1 21.2 16.3 4999 99.0 30.8 18.0 13.3 5403 99.2 29.5 16.7 12.1 5625 99.2 28.7 16.0 11.4 5764 99.3 28.2 15.5 11.0 5859 99.3 27.9 15.2 10.7 5928 99.4 27.7 15.0 10.5 5981 99.4 27.5 14.8 10.3 6022 99.4 27.3 14.7 10.2 6 7 8 9 10 13.7 12.2 11.3 10.6 10.0 n 10.9 9.55 8.65 8.02 7.56 9.78 8.45 7.59 6.99 6.55 9.15 7.85 7.01 6.42 5.99 8.75 7.46 6.63 6.06 5.64 8.47 7.19 6.37 5.80 5.39 8.26 6.99 6.18 5.61 5.20 8.10 6.84 6.03 5.47 5.06 7.98 6.72 5.91 5.35 4.94 12 13 14 15 9.65 9.33 9.07 8.86 8.68 7.21 6.93 6.70 6.51 6.36 6.22 5.95 5.74 5.56 5.42 5.67 5.41 5.21 5.04 4.89 5.32 5.06 4.86 4.69 4.56 5.07 4.82 4.62 4.46 4.32 4.89 4.64 4.44 4.28 4.14 4.74 4.50 4.30 4.14 4.00 4.63 4.39 4.19 4.03 3.89 16 17 18 19 20 8.53 8.40 8.29 8.18 8.10 6.23 6.11 6.01 5.93 5.85 5.29 5.18 5.09 5.01 4.94 4.77 4.67 4.58 4.50 4.43 4.44 4.34 4.25 4.17 4.10 4.20 4.10 4.01 3.94 3.87 4.03 3.93 3.84 3.77 3.70 3.89 3.79 3.71 3.63 3.511 3.78 3.68 3.60 3.52 3.46 22 24 26 28 30 7.95 7.82 7.72 7.64 7.56 5.72 5.61 5.53 5.45 5.39 4.82 4.72 4.64 4.57 4.51 4.31 4.22 4.14 4.07 4.02 3.99 3.90 3.82 3.75 3.70 3.76 3.67 3.59 3.53 3.47 3.59 3.50 3.42 3.36 3.30 3.45 3.36 3.29 3.23 3.17 3.35 3.26 3.18 3.12 3.07 32 34 36 38 40 7.50 7.44 7.40 7.35 7.31 5.34 5.29 5.25 5.21 5.18 4.46 4.42 4.38 4.34 4.31 3.97 3.93 3.89 3.86 3.83 3.65 3.61 3.57 3.54 3.51 3.43 3.39 3.35 3.32 3.29 3.26 3.22 3.18 3.15 3.12 3.13 3.09 3.05 3.02 2.99 3.02 2.98 2.95 2.92 2.89 50 60 70 80 90 7.17 7.08 7.01 6.96 6.93 5.06 4.98 4.92 4.88 4.85 4.20 4.13 4.07 4.04 4.01 3.72 3.65 3.60 3.56 3.54 3.41 3.34 3.29 3.26 3.23 3.19 3.12 3.07 3.04 3.01 3.02 2.95 2.91 2.87 2.84 2.89 2.82 2.78 2.74 2.72 2.78 2.72 2.67 2.64 2.61 100 150 200 1000 6.90 6.81 6.76 6.66 6.63 4.82 4.75 4.71 4.63 4.61 3.98 3.91 3.88 3.80 3.78 3.51 3.45 3.41 3.34 3.32 3.21 3.14 3.11 3.04 3.m 2.99 2.92 2.89 2.82 2.80 2.82 2.76 2.73 2.66 2.64 2.69 2.63 2.60 2.53 2.51 2.59 2.53 2.50 2.43 2.41 11 '" A10S APP. 5 Tables F-Distribution with (m, n) Degrees of Freedom (continued) Values of ~ for which the distribution function F(;:) [see (13), Sec. 25.4] has the value Table All 11 III III 6056 99.4 27.2 14.5 10.1 I 2 3 4 5 I = 10 = 15 III = 20 111 = 30 111 = 40 III = 50 III = 100 0.99 co 6157 99.4 26.9 14.2 9.72 6209 99.4 26.7 14.0 9.55 6261 99.5 26.5 13.8 9.38 6287 99.5 26.4 13.7 9.29 6303 99.5 26.4 13.7 9.24 6334 99.5 26.2 13.6 9.13 6366 99.5 26.1 13.5 9.G:! 6 7 8 9 10 7.87 6.62 5.81 5.26 4.85 7.56 6.31 5.52 4.96 4.56 7.40 6.16 5.36 4.81 4.41 7.23 5.99 5.20 4.65 4.25 7.14 5.91 5.12 4.57 4.17 7.09 5.86 5.07 4.52 4.12 6.99 5.75 4.96 4.42 4.01 5. 65 4.86 4.31 3.91 11 12 13 14 15 4.54 4.30 4.10 3.94 3.80 4.25 4.01 3.66 3.52 4.10 3.86 3.66 3.51 3.37 3.94 3.70 3.51 3.35 3.21 3.86 3.62 3.43 3.27 3.13 3.81 3.57 3.38 3.22 3.08 3.71 3.47 3.27 3.11 2.98 3.60 3.36 3.17 3.00 2.87 16 3.69 3.59 3.51 3.43 3.37 3.31 3.23 3.15 3.09 3.26 3.16 3.08 3.00 2.94 3.10 3.00 2.92 2.84 2.78 3.02 2.92 2.84 2.76 2.69 2.97 2.87 2.78 2.71 2.64 2.86 2.76 2.68 2.60 2.54 2.75 2.65 2.57 2.49 2.42 3.26 3.17 3.09 3.03 2.98 2.98 2.89 2.81 2.75 2.70 2.83 2.74 2.66 2.60 2.55 2.67 2.58 2.50 2.44 2.39 2.58 2.49 2.42 2.35 2.30 2.53 2.44 2.36 2.30 2.25 2.42 2.33 2.25 2.19 2.13 2.31 2.21 2.13 2.06 2.01 34 36 38 40 2.93 2.89 2.86 2.83 2.80 2.65 2.61 2.58 2.55 2.52 2.50 2.46 2.43 2.40 2.37 2.34 2.30 2.26 2.23 2.20 2.25 2.21 2.18 2.14 2.11 2.20 2.16 2.12 2.09 2.06 2.08 2.04 2.00 1.97 1.94 1.96 1.91 1.87 1.84 1.80 50 60 70 80 90 2.70 2.63 2.59 2.55 2.52 2.42 2.35 2.31 2.27 2.24 2.27 2.20 2.15 2.12 2.09 2.10 2.03 1.98 1.94 1.92 2.01 1.94 1.89 1.85 1.82 1.95 1.88 1.83 1.79 1.76 1.82 1.75 1.70 1.65 1.62 1.68 1.60 1.54 1.49 1.46 100 150 200 1000 2.50 2.44 2.22 2.16 2.13 2.06 2.04 2.07 2.00 1.97 1.89 1.83 1.79 1.72 1.70 1.80 1.73 1.69 1.61 1.59 1.74 1.66 1.63 1.54 1.52 1.60 1.52 1.48 1.38 1.36 1.43 1.33 1.28 1.11 1.00 171 18 19 20 HZ 3.-l1 6.88 I 22 24 26 28 :~ I I I 2.-l1 2.34 2.32 JC I I I 1.90 1.88 I I 1 A106 APP. 5 Tables Table All Distribution Function F(x) in Section 25.8 ~ I x =3 o 167 I £ I 1=20 o. 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 I 67 68 69 70 71 72 73 74 75 1 76 '77 '78 179 '80 '81 82 83 84 85 86 87 88 89 90 91 92 93 94 1 001 002 002 003 004 005 006 007 008 010 012 014 017 020 023 027 1 032 037 043 049 056 064 073 082 093 104 117 130 144 159 176 193 211 I 230 250 271 293 315 339 1 362 387 411 1 436 462 487 =6 0 1 2 3 4 5 6 7 001 008 028 068 136 235 360 500 o. 008 1 1042 2 117 0421 1 167 2 , 375 x o. o o ~ x x 134 12421 408 ~ x =19 , , I , 11 x 1 2 3 4 5 6 001 0051 015 035 068 119 71 191 8 281 9 386 10,500 O. , 73 411 74 441 , 75 1 -1-70 : 76 500 1 6 7 8 9 10 11 II =8 O. I001 0031 007 016 031 054 089 138 199 274 360 452 IZ 001 002 003 003 004 005 007 009 1 011 I 013 016 020 024 029 0341 041 048 056 066 076 088 100 115 130 147 165 64 184 65 1 205 66 227 67 250 68 275 69 300 70 327 71 354 ,72 383 , 2 3 4 5 12 13 =18 38 39 40 41 '42 143 '144 45 46 '47 148 '49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 I X , o. I O. 43 001 44 002 45 002 46 003' 47 003 48 004, 49 005 50 006 51 008 52 010 53 012 54 014 1 155 017 56 021 57 025 58 029 I 59 034 I 60 040' 61 0471 62 054 163 062: 64 072 I 65 082 66 I 093 : 67 1105 68 119 69 133 1 70 149' 71 166 72 184' 73 203 ~ 74 223 75 245, 76 267 1 77 290 I 78 314 79 339 1 80 3651 181 391 82 4181 83 445 84 473 85 500 I =71 II 11 /I =4 1 O. o. ~ 11 = P(T ~ x) of the Random Variable T x =17 I I ~ /I x ~ 7 8 9 10 11 12 13 14 15 16 ,17 IZ x =10 6 001 7 002 8 005 4 001 5 003 6' 006 O. 32 001 33 002 34 002 1 27 35 003 28 002 36 1004 29 002 37 005 30 003 38 007 31 004 39 009 32 006 40 011 33 008 41 014 34 010 42 017 35 013 43 021 44 026 361016 37 021 45 032 381026 46 1038 39 032 47 046 40 039 48 054 41 048 49 064 42,058 50 076 51 088 431070 44 083 52 102 45 097 53 118 46 114 54 135 1 47 133 55 154 1 48 153 56 174 49 175 57 196 50 199 58 220 51 1 225 59 245 52 253 60 271 53 282 61 299 54' 313 62 328 55 345 63 358 64 388' 56 378 65 420 57 412/ 66 452/ 58 1 447 59 482 , / 67 484 I rE 8 001 9 002 10 003 11 0051 12 008 13 013 1 9 0081 0121 10 014 022 038 11 023 060' 12 0361 14 020 13 054 090 15 030 16 043 130, 1 14 078 17 060 179 151108 238 I 16 146 18 082 "106' 1 171190 19 109 18 242 20 141 381 460 119 300 121 179 1 1 20 364 22 2231 '23 271 431 500 '24 324 , I I~~ l x 27 =15 o. ~! ~~I 25 003 26 004 27006 28 008 29 010 30 014 31 018 32 023 £ =14 18 19 20 21 22 23 24 25, 001 002 O. ~ =';3 x om' O. 003 005' 007 I 010 013 , 14 15 16 17 18 001 001 002 003 005 11 x I =12 O. 11 001 12 002 1 28 29 30 31 132 33 34 35 36 37 38 39 40 141 42 43 44 '45 I ~ ,~! ~~~ ~~ ,~~! I ~~ ~~ I!!~! 35 046 136 057 137 070 1 38 0841 , 391101 : 401120 411141 42 164 43 190 44 218 45 248 46 279 47 313 I 48 3491 : 49 385 ,50 423 51 461 '52 500 I =11 031 21 015 15 007 116 010 040 122 , 021 23 029 17 016 051 063 124 038 18 022 25 050 19 031 079 096 26 064 20 043 117, 27 082 21 058 140 28 102 22 076 165 29 126/ 23 098 194: 30 153 24 125 225 31 1 184 25 155 259 32 218/ 26 190 295 33 255 I 27 230 334 I 34 295 28 273 374 35 338 129 319 415 36 383 30 369 457 137 4291 131 420 500 I 38 476' 32 473 PHOTO CREDITS Part A Opener: John SohmJChromoshohmJPhoto Researchers. Part B Opener: Lester Lefkowitz/Corbis Images. Part C Opener: Science Photo Library/Photo Researchers. Part D Opener: Lester Lefkowitz/Corbis Images. Part E Opener: Walter Hodges/Stone/Getty Images. Chapter 19, Figure 434: From Dennis Sharp, ARCHITECKTUR. ©1973 der deutschsprachigen Ausgabe Edition Praeger GmbH, Munchen. Part F Opener: Charles O'Rear/Corbis Images. Part G Opener: Greg Pease/Stone/Getty Images. Appendix 1 Opener: Patrick Bennett/Corbis Images. Appendix 2 Opener: Richard T. Nowitz/Corbis Images. Appendix 3 Opener: Chris KapolkaiStone/Getty Images. Appendix 4 Opener: John OlsoniCorbis Images. Pl IN 0 E X Page numbers AI, A2, A3, ... refer to App. I to App. 5 at the end of the book. A Abel 78 Absolute convergence 667,697 frequency 994. 1000 value 607 Absolutely integrable 508 Acceleration 395, 995 Acceptance sampling 1073 Adams-Bashforth methods 899 Adams-Moulton methods 900 Adaptive 824 Addition of complex numbers 603 matrices 275 means 1038 normal random variables 1050 power series 174. 680 variances 1039 vectors 276, 324, 367 Addition rule 1002 Adiabatic 561 ADI method 915 Adjacency matrix 956 Adjacent vertices 955 Airfoil 732 Airy's equation 552. 904 Algebraic multiplicity 337, 865 Algorithm 777, 783 Dijkstra 964 efficient 962 Ford-Fulkerson 979 Gauss 837 Gauss-Seidel 848 Greedy 967 Kruskal967 Moore 960 polynomially bounded 962 Prim 971 Runge-Kutta 892,904 stable, unstable 783 Aliasing 526 Allowable number of defectives 1073 Alternating direction implicit method 915 path 983 Alternative hypothesis 1058 Ampere 92 Amplification 89 Amplitude spectrum 506 Analytic function 175,617,681 Analytic at infinity 711 Angle between curves 35 between vectors 372 Angular speed 38 L 765 Annulus 613 Anticommutative 379 AOQ, AOQL 1075-1076 Approximate solution of differential equations 9, 886-934 eigenvalue problems 863-882 equations 787-796 systems of equations 833-858 Approximation least squares 860 polynomial 797 trigonometric 502 A priori estimate 794 AQL 1074 Arc of a curve 391 Archimedian principle 68 Arc length 393 Arctan 634 Area 435, 442, 454 Argand diagram 605 Argument 607 Artificial variable 949 Assignment problem 982 Associated Legendre functions 182 Asymptotically equal 191, 1009 normal 1057 stable 148 n 12 Index Attractive 148 Augmented matrix 288, 833 Augmenting path 975, 983 theorem 977, 984 Autonomous 31, 151 Average (see Mean value) Average outgoing quality 1075 Axioms of probability 100 I B Back substitution 289, 834 Backward differences 807 edge 974, 976 Euler method 896, 907 Band matrix 914 Bashforth method 899 Basic feasible solution 942. 944 variable~ 945 Basis 49, 106. 113, 138, 300, 325, 360 Beam 120,547 Beats 87 Bellman optimality principle 963 Bell-shaped curve 1026 Bernoulli 30 distribution 1020 equation 30 law of large numbers 1032 numbers 690 Bessel 189 equation 189, 204 functions 191, 198, 202, 207, A94 functions, tables A94 inequality 215, 504 Beta function A64 Bezier curves 816 BFS 960 Bijective mapping 729 Binary 782 Binomial coefficients 1009 distribution 1020, A96 series 689 theorem 1010 Binormal (Fig. 210) 397 Bipartite graph 982, 985 Birthday problem 1010 Bisection method 796 Bolzano-Weierstrass theorem A91 Bonnet 181 Boundary 433 conditions 203, 540, 558, 571, 587 point 433, 613 value problem 203, 558 Bounded domain 646 function 38 region 433 sequence A69 Boxplot 995 Branch cut 632 point 746 Breadth first search 960 Buoyance force 68 C Cable 52, 198, 593 CAD (Computer aided design) 810 Cancellation law 321 Cantor-Dedekind axiom A69 Capacitance 92 Capacitor 92 Capacity of a cut set 976 of an edge 973 Cardano 602 Cardioid 443 Cm1esian coordinates 366, 604 CAS (Computer algebra system) vii, 777 Catenary 399 Cauchy 69 convergence principle 667, A90 determinant 112 -Goursat theorem 647 -Hadamard formula 676 inequality 660 integral formula 654 integral theorem 647, 652 method of steepest descent 938 principal value 719, 722 product 680 -Riemann equations 37, 618, 621 -Schwarz inequality 326, 859 Cayley transformation 739 13 Index Center 143 of a graph 973 of gravity 436, 457 of a power series 171 Central differences 808 limit theorem 1057 moments 1019 Centrifugal, centripetal 396 Cgs system: Front cover Chain rule 401 Characteristic determinant 336. 864 equation 59, Ill, 336, 551, 864 function 542, 574 of a partial differential equation 551 polynomial 336, 864 value 324, 864 vector 324, 864 Chebyshev polynomials 209 Chinese postman problem 963 Chi-square 1055, 1077, AlOl Cholesky's method 843 Chopping 782 Chromatic number 987 Circle 391 of convergence 675 Circuit 95 Circular disk 613 helix 391, 394 membrane 580 Circulation 764 Cissoid 399 Clairaut equation 34 Class intervals 994 Closed disk 613 integration formula 822, 827 interval A69 path 959 point set 613 region 433 Coefficient matrix 288. 833 Coefficients of a differential equation 46 power series 171 system of equations 287, 833 Cofactor 309 Collatz's theorem 870 Column 275 space 300 sum norm 849 vector 275 Combination 1007 Combinatorial optimization 954-986 Comparison test 668 Complement 613, 988 Complementary enor function A64 Fresnel integrals A65 sine integral A66 Complementation rule 1002 Complete bipartite graph 987 graph 958 matching 983 orthonormal set 214 Complex conjugate numbers 605 exponential function 57, 623 Fourier integral 519 Fourier series 497 function 614 hyperbolic functions 628, 743 impedance 98 indefinite integral 637 integration 637~663, 7l2~725 line integral 633 logarithm 630, 688 matrices 356 number 602 plane 605 plane. extended 710 potential 761 sequence 664 series 666 sphere 710 trigonometric functions 626. 688 trigonometric polynomial 524 variable 614 vector space 324. 359 Complexity 961 Component 366,374 Composite transformation 281 Compound interest 9 Compressible fluid 412 Computer aided design (CAD) 810 algebra system (CAS) vii, 777 14 Index Computer (Com.) graphics 287 software (see Software) Conchoid 399 Condition number 855 Conditionally convergent 667 Conditional probability 1003 Conduction of heat 465, 552, 757 Cone 406, 448 CONF 1049 Confidence intervals 1049-1058 level L049 limits 1049 Conformal mapping 730. 754 Conic sections 355 Conjugate complex numbers 605 harmonic function 622 Connected graph 960 set 613 Conservative 415. 428 Consistent equations 292. 303 Constraints 937 Consumer's risk 1075 Continuity of a complex function 615 equation 413 of a vector function 387 Continuous distribution 1011 random variable 101 L. 1034 Contour integral 647 Contraction 789 Control chart 1068 limit 1068 variables 936 Convergence absolute 667 circle of 675 conditional 667 interval 172. 676 of an iterative process 793, 848 mean 214 mean-square 214 in norm 214 principle 667 radius 172 Convergence (Cont.) of a sequence 386, 664 of a series ] 71, 666 superlinear 795 tests 667-672 uniform 691 Conversion of an ODE to a system 134 Convolution 248. 523 Cooling 14 Coordinate transformations A 71. A84 Coordinates Cartesian 366, 604 curvilinear A71 cylindrical 587, A71 polar 137. 443, 580, 607 spherical 588, A71 Coriolis acceleration 396 COlTector 890, 900 COlTelation analysis 1089 Cosecant 627, A62 Cosine of a complex variable 627, 688, 743 hyperbolic 688 integral A66. A95 of a real variable A60 Cotangent 627. A62 Coulomb 92 law 409 Covariance 1039, 1085 Cramer's rule 306-307, 312 Crank-Nicolson method 924 Critical damping 65 point 31, 142,730 region 1060 Cross product 377 Crout's method 841 Cubic spline 811 Cumulative distribution function 1011 frequency 994 Curl 414.430,472, A71 Curvature 397 Curve 389 arc length of 393 fitting 859 orientation of 390 piecewise smooth 421 rectifiable 393 simple 391 15 Index Curve (Cont.) smooth 421, 638 twisted 391 Curvilinear coordinates A71 Cut set 976 Cycle 959 Cycloid 399 Cylinder 446 flow around 763, 767 Cylindrical coordinates 587, A71 o D' Alembert's solution 549, 551 Damping 64, 88 Dantzig 944 DATA DESK 991 Decay 5 Decreasing sequence A69 Decrement 69 Dedekind A69 Defect 337 Defective item 1073 Definite complex integral 639 Definiteness 356 Deformation of path 649 Degenerate feasible solution 947 Degree of precision 822 a vertex 955 Degrees of freedom 1052, 1055, 1066 Deleted neighborhood 712 Delta Dirac 242 Kronecker 210, A83 De Moi vre 610 formula 610 limit theorem 1031 De Morgan's laws 999 Density 1014 Dependent linearly 49, 74, 106,21.)7, 300, 325 random variables 1036 Depth first search 960 Derivative of a complex function 616, 658 directional 404 left-hand 484 right-hand 484 of a vector function 387 DERIVE 778 Descartes 366 Determinant 306, 308 Cauchy 112 characteristic 336. 864 of a matrix 305 of a matrix product 321 Vandermonde 112 DFS 960 Diagonalization 351 Diagonal matrix 284 Diagonally dominant matrix 868 Diameter of a graph 973 Differences 802, 804. 807-808 Difference table 803 Differentiable complex function 616 Differential 19, 429 form 20, 429 geometry 389 operator 59 Differential equation (ODE and PDE) Airy 552, 904 Bernoulli 30 Bessel 189. 204 Cauchy-Riemann 37, 618. 621 with constant coefficients 53, III elliptic 551, 909 Euler-Cauchy 69, 185 exact 20 homogeneous 27, 46, 105, 535 hyperbolic 551,928 hypergeometric 188 Laguerre 257 Laplace 407. 465. 536. 579. 587. 910 Legendre 177, 204. 590 linear 26, 45, 105, 535 nonhomogeneous 27, 46, 105, 535 nonlinear 45, 151,535 numeric methods for 886-934 ordinary 4 parabolic 551, 909, 922 partial 535 Poisson 910. 918 separable 12 Sturm-Liouville 203 of vibrating beam 547, 552 of vibrating mass 61,86, 135, 150,243,261, 342,499 of vibrating membrane 569-586 of vibrating string 538 16 Index Differentiation analytic functions 691 complex functions 616 Laplace transforms 254 numeric 827 power series 174, 680 series 696 vector functions 387 Diffusion equation 464. 552 Diffusivity 552 Digraph 955 Dijkstra's algorithm 964 Dimension of vector space 300, 325. 369 Diodes 399 Dirac's delta 242 Directed graph 955 line segment 364 path 982 Directional derivative 404 Direction field 10 Direct method 845 Dirichlet 467 discontinuous factor 509 problem 467, 558, 587,915 Discharge of a source 767 Discrete Fourier transform 525 random variable 1011. 1033 spectrum 507, 524 Disjoint events 998 Disk 613 Dissipative 429 Distribution 1010 Bernoulli 1020 binomial 1020, A96 chi-square 1055. AIOI continuous 1011 discrete 1011.1033 Fisher's F- 1066, AI02 -free test 1080 function 1011. 1032 Gauss 1026 hypergeometric 1024 marginal 1035 multinomial 1025 normal 1026. 1047-1057, 1062-1067, A98 Poisson 1022, 1073. A97 Student's t- 1053. AIOO two-dimensional 1032 Distribution (COli f.) uniform 1015, 1017,1034 Divergence theorem of Gauss 459 of vector fields 410, A 72 Divergent sequence 665 series 171, 667 Di vided differences 802 Division of complex numbers 604, 609 Domain 401,613,646 Doolittle's method 841 Dot product 325, 371 Double Fourier series 576 integral 433 labeling 968 precision 782 Driving force 84 Drumhead 569 Duffing equation 159 Duhamel's formula 597 E Eccentricity of a vertex 973 Echelon form 294 Edge 955 coloring 987 incidence list 957 Efficient algorithm 962 Eigenbasis 349 Eigenfunction 204, 542, 559, 574 expansion 210 Eigenspace 336, 865 Eigenvalue problems for matrices (Chap. 8) 333-363 matrices, numerics 863-882 ODEs (Sturm-Liouville problems) 203-216 PDEs 540-593 systems of ODEs 130-165 Eigenvector 334, 864 EISPACK 778 Elastic membrane 340 Electrical network (see Networks) Electric circuit (see Circuit) Electromechanical analogies 96 Electromoti ve force 91 Electrostatic field 750 17 Index Electrostatic (COllt.) potential 588-592, 750 Element of matrix 273 Elementary matrix 296 operations 292 Elimination of first derivative 197 Ellipse 390 Ellipsoid 449 Elliptic cylinder 448 paraboloid 448 partial differential equation 551, 909 Empty set 998 Engineering system: Front cover Entire function 624,661, 711 Entry 273, 309 Equality of complex numbers 602 matrices 275 vectors 365 Equally likely 1000 Equipotential lines 750, 762 surfaces 750 Equivalence relation 296 Equivalent linear systems 292 Erf 568, 690. A64, A95 Error 783 bound 784 estimation 785 function 568, 690. A64. A95 propagation 784 Type I, Type II 1060 Essential singularity 708 Estimation of parameters 1046-1057 Euclidean norm 327 space 327 Euler 69 backward methods for ODEs 896, 907 beta function A64 -Cauchy equation 69, 108. 116. 185. 589 -Cauchy method 887, 890, 903 constant 200 formula 58, 496, 624, 627, 687 formulas for Fourier coefficients 480, 487 graph 963 numbers 690 method for systems 903 Euler (Cont.) trail 963 Evaporation L8 Even function 490 Event 997 Everett interpolation formula 809 Exact differential equation 20 differential form 20, 429 Existence theorem differential equations 37, 73, 107, 109, 137, 175 Fourier integral 508 Fourier series 484 Laplace transforms 226 Linear equations 302 Expectation 10 16 Experiment 997 Explicit solution 4 Exponential decay 5 function, complex 57, 623 fUnction, real A60 growth 5, 31 integral A66 Exposed vertex 983 Extended complex plane 710, 736 Extension. periodic 494 Extrapolation 797 Extremum 937 F Factorial function 192, 1008, A95 Failure 1021 Fair die 1000 Falling body 8 False position 796 Family of curves 5, 35 Faraday 92 Fast Fourier transform 526 F-distribution 1066. A 102 Feasible solution 942 Fehlberg 894 Fibonacci numbers 683 Field conservative 415. 428 of force 385 gravitational 385,407,411,587 18 Index Field (Cant.) irrotational 415, 765 scalar 384 vector 384 velocity 385 Finite complex plane 710 First fundamental form 457 Green's formula 466 shifting theorem 224 Fisher, R. A. 1047, 1066, 1077 F -distribution I 066, A 102 Fixed decimal point 781 point 736, 781, 787 Flat spring 68 Floating point 781 Flow augmenting path 975 Flows in networks 973-981 Fluid flow 412,463, 761 Flux 412, 450 integral 450 Folium of Descarte" 399 Forced oscillations 84, 499 Ford-Fulkerson algorithm 979 Forest 970 Form Hermitian 361 quadratic 353 skew-Hermitian 361 Forward differences 804 edge 974. 976 Four-color theorem 987 Fourier 477 -Bessel series 213, 583 coefficients 480, 487 coefficients, complex 497 constants 210 cosine integral 511 cosine series 491 cosine transform 514, 529 double series 576 half-range expansions 494 integral 508, 563 integral, complex 519 -Legendre series 212, 590 matrix 525 series 211, 480. 487 series, complex 497 Fourier (Cant.) series, generalized 210 sine integral 511 sine series 491, 543 sine transform 514.530 transform 519. 531, 565 transform, discrete 525 transform, fast 526 Fractional linear transformation 734 Fraction defective J073 Fredholm 20 I Free fall 18 oscillations 61 Frene! formulas 400 Frequency 63 of values in samples 994 Fresnel integrals 690, A65 Friction 18-19 Frobenius 182 method 182 norm 849 theorem 869 Fulkerson 979 Full-wave rectifier 248 Function analytic 175, 617 Bessel 191, 198,202,207, A94 beta A64 bounded 38 characteristic 542, 574 complex 614 conjugate harmonic 622 entire 624, 661. 71 I error 568, 690, AM, A95 even 490 exponential 57, 623, A60 factorial 192, 1008, A95 gamma 192, A95 Hankel 202 harmonic 465, 622, 772 holomorphic 617 hyperbolic 628, 743. A62 hypergeometric 188 inverse hyperbolic 634 inverse trigonometric 634 Legendre 177 logarithmic 630. A60 meromorphic 711 Neumann 201 Index Function (Collt.) odd 490 orthogonal 205, 482 orthonormal 205, 210 periodic 478 probability 1012. 1033 rational 617 scalar 384 space 382 staircase 248 step 234 trigonometric 626, 688, A60 unit step 234 vector 384 Function space 327 Fundamental form 457 matrix 139 mode 542 period 485 system -1-9, 106, 113, 138 theorem of algebra 662 G Galilei 15 Gamma function 192, A95 GAMS 778 Gau~s 188 distribution 1026 divergence theorem 459 elimination method 289. 834 hypergeometric equation 188 integration formula 826 -Jordan elimination 317, 844 least squares 860, 1084 quadrature 826 -Seidel iteration 846, 913 General powers 632 solution 64. 106. 138, 159 Generalized Fourier series 210 function 242 solution 545 triangle inequality 608 Generating function 181,216,258 Geometric multiplicity 337 series 167,668,673,687,692 19 Gerschgorin's theorem 866 Gibbs phenomenon 490, 510 Global error 887 Golden Rule 15,23 Goodness of fit 1076 Gosset 1066 Goursat 648, A88 Gradient 403, 415. 426, A72 method 938 Graph 955 bipartite 982 complete 958 Euler 963 planar 987 sparse 957 weighted 959 Gravitation 385, 407, 411, 587 Greedy algorithm 967 Greek alphabet: Back cover Green 439 formulas 466 theorem 439, ..J.66 Gregory-Newton formulas 805 Growth restriction 225 Guldin's theorem 458 H Hadamard's formula 676 Half-life time 9 Half-plane 613 Half-range Fourier series 494 Half-wave rectifier 248, 489 Halving 819, 824 Hamiltonian cycle 960 Hanging cable 198 Hankel functions 202 Hard spring 159 Harmonic conjugate 622 function 465, 622. 772 oscillation 63 series 670 Heat equatiun 464, 536, 553, 757, 923 potential 758 Heaviside 221 expansions 245 formulas 247 110 Index Heaviside (Cont.) function 234 Helicoid 449 Helix 391, 394, 399 Helmholtz equation 572 Henry 92 Hermite interpolation 816 polynomials 216 Hermitian 357, 361 Hertz 63 Hesse's normal form 375 Heun's method 890 High-frequency line equations 594 Hilbert 201, 326 matrix 858 space 326 Histogram 994 Holomorphic 617 Homogeneous differential equation 27, 46, 105. 535 system of equations 288, 304, 833 Hooke's law 62 Householder's tridiagonalization 875 Hyperbolic differential equation 551,909,928 functions, complex 628, 743 functions, real A62 paraboloid 448 partial differential equations 551, 928 spiral 399 Hypergeometric differential equation 188 distribution 1024 functions 188 series 188 Hypocycloid 399 Hypothesis 1058 Idempotent matrix 286 Identity of Lagrange 383 matrix (see Unit matrix) theorem for power ~eries 679 transformation 736 trick 351 Ill-conditioned 851 Image 327, 729 Imaginary axis 604 part 602 unit 602 Impedance 94, 98 Implicit solution 20 Improper integral 222. 719. 722 Impulse 242 IMSL 778 Incidence list 957 matrix 958 Inclusion theorem ~68 Incomplete gamma function A64 Incompressible 765 Inconsistent equations 292, 303 Increasing sequence A69 Indefinite integral 637, 650 integration 640 Independence of path 426, 648 Independent events 1004 random variables 1036 Indicial equation 184 Indirect method 845 Inductance 92 Inductor 92 Inequality Bessel 215, 504 Cauchy 660 ML- 644 Schur 869 triangle 326. 372. 608 Infinite dimensional 325 population 1025, 1045 sequence 664 series (see Series) Infinity 710, 736 Initial condition 6, 48, 137, 540 value problem 6. 38.48. 107,886,902 Injective mapping 729 Inner product 325. 359, 371 space 326 lnput 26, 84. 230 Instability (see Stability) Integral contour 647 111 Index Integral (Cant.) definite 639 double 433 equation 252 Fourier 508, 563 improper 719. 722 indefinite 637. 650 line 421, 633 surface 449 theorems, complex 647, 654 theorems, real 439, 453, 469 transform 221, 513 triple 458 Integrating factor 23 Integration complex functions 637-663, 701-727 Laplace transforms255 numeric 8l7-827 power series 680 series 695 Integra-differential equation 92 Interest 9, 33 Interlacing of zeros 197 Intermediate value theorem 796 Interpolation 797-815 Hermite 816 Lagrange 798 Newton 802, 805, 807 spline 811 Interquartile range 995 Intersection of events 998 Interval closed A69 of convergence 172, 676 estimate 1046 open 4, A69 Invariant subspace 865 Inverse hyperbolic functions 634 mapping plinciple 733 of a matrix 315, 844 trigonometric functions 634 Inversion 735 Investment 9, 33 Irreducible 869 Irregular boundary 919 Irrotational415,765 Isocline 10 Isolated singularity 707 Isotherms 758 Iteration for eigenvalues 872 for equations 787-794 Gauss-Seidel 846. 913 Jacobi 850 Picard 41 J Jacobian 436, 733 Jacobi iteration 850 Jerusalem, Shrine of the Book 814 Jordan 316 Joukowski airfoil 732 K Kirchhoff's laws 92, 973 Kronecker delta 210, A83 Kruskal's algorithm 967 Kutta 892 L l10 l2, lex; 853 L 2 863 Labeling 968 Lagrange 50 identity of 383 interpolation 798 Laguerre polynomials 209, 257 Lambert's law 43 LAPACK 778 Laplace 221 equation 407, 465, 536, 579, 587, 910 integrals 512 limit theorem 1031 operator 408 transform 221, 594 Laplacian 443, A73 Latent root 324 Laurent series 701, 712 Law of absorption 43 cooling 14 gravitation 385 large numbers 1032 mass action 43 the mean (see Mean value theorem) LC-circuit 97 III Index LCL 1068 Least squares 860, 1084 Lebesgue 863 Left-hand derivative 484 limit 484 Left-handed 378 Legendre 177 differential equation 177, 204, 590 functions 177 polynomials 179, 207, 590, 826 Leibniz 14 convergence test A 70 Length of a curve 393 of a vector 365 Leonardo of Pisa 638 Leontief 344 Leslie model 341 Libby 13 Liebmann's method 913 Likelihood function 1047 Limit of a complex function 615 cycle 157 left-hand 484 point A90 right-hand 484 of a sequence 664 vector 386 of a vector function 387 Lineal element 9 Linear algebra 271-363 combination 106, 325 dependence 49, 74, 106, 108, 297, 325 differential equation 26, 45, 105, 535 element 394, A 72 fractional transformation 734 independence 49, 74, 106, 108, 297, 325 interpolation 798 operator 60 optimization 939 programming 939 space (see Vector space) system of equations 287, 833 transformation 281, 327 Linearization of systems of ODEs 151 Line integral 421, 633 Lines of force 751 UNPACK 779 Liouville 203 theorem 661 Lipschitz condition 40 List 957 Ljapunov 148 Local error 887 minimum 937 Logarithm 630, 688, A60 Logarithmic decrement 69 integral A66 spiral 399 Logistic population law 30 Longest path 959 Loss of significant digits 785 Lotka-Volterra population model 154 Lot tolerance per cent defective 1075 Lower control limit 1068 triangular matlix 283 LTPD 1075 LU-factorization 841 M Maclaurin 683 series 683 trisectrix 399 Magnitude of a vector (see Length) Main diagonal 274, 309 Malthus's law 5, 31 MAPLE 779 Mapping 327, 729 Marconi 63 Marginal distributions 1035 Markov process 285, 341 Mass-sPling system 61, 86, 135, 150, 243, 252, 261,342,499 Matching 985 MATHCAD 779 MATHEMATICA 779 Mathematical expectation 1019, 1038 MATLAB 779 Matlix addition 275 augmented 288, 833 band 914 Index Matrix (COllt.) diagonal 284 eigenvalue problem 333-363. 863-882 Hermitian 357 identity (see Unit matrix) inverse 315 inversion 315, 844 multiplication 278, 321 nonsingular 315 norm 849. 854 normal 362. 869 null (see Zero matrix) orthogonal 345 polynomial 865 scalar 284 singular 315 skew-Hermitian 357 skew-symmetric 283, 345 sparse 812, 912 square 274 stochastic 285 symmetric 283, 345 transpose 282 triangular 283 tridiagonal 812. 875, 914 unit 284 unitary 357 zero 276 Max-flow min-cut theorem 978 Maximum 937 flow 979 likelihood method 1047 matching 983 modulus theorem 772 principle 773 Mean convergence 214 Mean-square convergence 214 Mean value of a (an) analytic function 771 distribution IO 16 function 764 harmonic function 772 sample 996 Mean value theorem 402, 434, 454 Median 994, 1081 Membrane 569-586 Meromorphic function 711 Mesh incidence matrix 278 Method of false position 796 113 Method of (Com.) least squares 860. 1084 moments 1046 steepest descent 938 undetermined coefficients 78, 117, 160 variation of parameters 98, 118, 160 Middle quartile 994 Minimum 937, 942, 946 MINITAB 991 Minor 309 Mixed boundary value problem 558. 587. 759. 917 triple product 381 Mixing problem 13, 130, ]46, 163,259 Mks system: Front cover ML-inequality 644 Mobius 453 strip 453, 456 transformation 734 Mode 542. 582 Modeling 2. 6. 13. 61. 84. 130. 159. 222, 340, 499, 538,569.750-767 Modified Bessel functions 203 Modulus 607 Molecule 912 Moivre's formula 610 Moment central 1019 of a distribution 1019 of a force 380 generating function 1026 of inertia 436. 455. 457 of a sample 1046 vector 380 Monotone sequence A69 Moore's shortest path algorithm 960 Morera's theorem 661 Moulton 900 Moving trihedron (see Trihedron) M-test for convergence 969 Multinomial distribution 1025 Multiple point 391 Multiplication of complex numbers 603, 609 determinants 322 matrices 278, 321 means ]038 power series 174, 680 vectors 2]9, 371, 377 Multiplication rule for events 1003 114 Index Multiplicity 337, 865 Multiply connected domain 646 Multistep method 898 "Multivalued function" 615 Mutually exclusive events 998 N Nabla 403 NAG 779 Natural frequency 63 logarithm 630, A60 spline 812 Neighborhood 387, 613 Nested form 786 NETLIB 779 Networks 132, 146, 162,244,260.263.277,331 in graph theory 973 Neumann, C. 201 functions 20 I problem 558, 587, 917 Newton 14 -Cotes formulas 822 interpolation formulas 802, 805, 807 law of cooling 14 law of gravitation 385 method 800 -Raphson method 800 second law 62 Neyman 1049, 1058 Nicolson 924 Nicomedes 399 Nilpotent matrix 286 NIST 779 Nodal incidence matrix 277 line 574 Node 142, 797 Nonbasic variables 945 Nonconservative 428 Nonhomogeneous differential equation 27, 78, 116, 159, 305, 535 system of equations 288. 304 Nonlinear differential equations 45. 151 Nonorientable smface 453 Nonparametric test 1080 Nonsingular matrix 315 Nonn 205, 326, 346, 359, 365, 849 Normal acceleration 395 asymptotically 1057 to a curve 398 derivative 444, 465 distribution 1026, 1047-1057, 1062-1067, A98 two-dimensional 1090 equations 860. 1086 form of a PDE 551 matrix 362, 869 mode 542, 582 plane 398 to a plane 375 random variable 1026 to a surface 447 vector 375, 447 Null hypothesis 1058 matrix (see Zero matlix) space 301 vector (see Zero vector) Nullity 301 Numeric methods 777-934 differentiation 827 eigenvalues 863-882 equations 787-796 integration 817-827 interpolation 797-816 linear equations 833-858 matrix inversion 315, 844 optimization 936-953 ordinary differential equations (ODEs) 886-908 partial differential equations (PDEs) 909-930 Nystrom method 906 o 0962 Objective function 936 OC curve 1062 Odd function 490 ODE 4 (see also Differential equations) Ohm's law 92 One-dimensional heat equation 553 wave equation 539 One-sided derivative 484 test 1060 115 Index One-step method 898 One-to-one mapping 729 Open disk 613 integration formula 827 interval A69 point set 613 Operating characteristic 1062 Operational calculus 59, 220 Operation count 838 Operator 59. 327 Optimality principle. Bellman's 963 Optimal solution 942 Optimization 936-953, 959-990 Orbit 141 Order 887, 962 of a determinant 308 of a differential equation 4, 535 of an iteration process 793 Ordering 969 Ordinary differential equations 2-269. 886-908 (see also Differential equation) Orientable surface 452 Orientation of a curve 638 surface 452 Orthogonal coordinates A 71 curves 35 eigenvectors 350 expansion 210 functions 205, 482 matrix 345 polynomials 209 series 210 trajectories 35 transformation 346 vectors 326, 371 Orthonormal functions 205, 210 Oscillations of a beam 547,552 of a cable 198 in circuits 9 I damped 64, 88 forced 84 free 61, 500, 547 harmonic 63 of a mass on a spring 61, 86, 135, 150, 243, 252, 261, 342.499 of a membrane 569-586 Oscillations (Cont.) self-sustained 157 of a string 204, 538. 929 undamped 62, Osculating plane 398 Outcome 997 Outer product 377 Outlier 995 Output 26, 230 Overdamping 65 Overdetermined system 292 Overflow 782 Overrelaxation 851 Overtone 542 p Paired comparison 1065 Pappus's theorem 458 Parabolic differential equation 551, 922 Paraboloid 448 Parachutist 12 Parallelepiped 382 Parallel flow 766 Parallelogram equality 326, 372, 612 law 367 Parameter of a distribution 1016 Parametric representation 389, 446 Parking problem 1023 Parseval's equality 215. 504 Partial derivative 388, A66 differential equation 535, 909 fractions 231, 245 pivoting 291, 834 sum 171,480, 666 Particular solution 6, 48. 106, 159 Pascal 399 Path in a digraph 974 in a graph 959 of integration 421, 637 PDE 535, 909 (see a/so Differential equation) Peaceman-Rachford method 915 Pearson. E. S. 1058 Pearson, K. 1066 Pendulum 68, 152, 156 Period 478 116 Index Periodic extension 494 function 478 Permutation 1006 Perron-Frobenius theorem 344, 869 Pfaffian form 429 Phase angle 88 of complex number (see Argument) lag 88 plane 141, 147 pOltrait 14 I, 147 Picard iteration method 41 theorem 709 Piecewise continuous 226 smooth 421, 448. 639 Pivoting 291, 834 Planar graph 987 Plane 315, 375 Plane curve 391 Poincare 216 Point estimate 1046 at infinity 710. 736 set 613 source 765 spectrum 507,524 Poisson 769 distribution 1022, 1073, A97 equation 910, 918 integral formula 769 Polar coordinates 437, 443, 580, 607-608 form of complex numbers 607 moment on inertia 436 Pole 708 Polynomial approximation 797 matrix 865 Polynomially bounded 962 Polynomials 617 Chebyshev 209 Hermite 216 Laguerre 207, 257 Legendre 179. 207. 590. 826 trigonometric 502 Population in statistics 1044 Population models 31, 154. 341 Position vector 366 Positive definite 326. 372 Possible values 1012 Postman problem 963 Potential 407, 427,590, 750, 762 complex 763 theory 465, 749 Power method for eigenvalues 872 of a test 1061 series 167, 673 series method 167 Precision 782 Predator-prey 154 Predictor-corrector 890, 900 Pre-Hilbert space 326 Prim's algorithm 971 Principal axes theorem 354 branch 632 diagonal (see main diagonal) directions 340 normal (Fig. 210) 397 part 708 value 607. 630. 632. 719. 722 Prior estimate 794 Probability 1000, 10m conditional 1003 density 1014. 1034 distribution 10 10, 1032 function 1012, 1033 Producer's risk 1075 Product (see Multiplication) Projection of a vector 374 Pseudocode 783 Pure imaginary number 603 Q QR-factorization method 879 Quadratic equation 785 form 353 interpolation 799 Qualitative methods 124. 139-165 Quality control 1068 Quartile 995 Quasilinear 551, 909 Quotient of complex numbers 604 117 Index R Rachford method 915 Radiation 7, 561 Radiocarbon dating 13 Radius of convergence 172, 675, 686 of a graph 973 Random experiment 997 numbers 1045 variable 1010. 1032 Range of a function 614 sample 994 Rank of a matrix 297, 299. 31 I Raphson 790 Rational function 617 Ratio test 669 Rayleigh 159 equation 159 quotient 872 RC-circuit 97, 237, 240 Reactance 93 Real axis 604 part 602 sequence 664 vector space 324, 369 Rectangular membrane 571 pulse 238. 243 rule 817 wave 21 1, 480, 488, 492 Rectifiable curve 393 Rectification of a lot 1075 Rectifier 248, 489, 492 Rectifying plane 398 Reduction of order 50, 116 Region 433,614 Regression 1083 coefficient Im;5, L088 line 1084 Regula falsi 796 Regular point of an ODE 183 Sturm-Liouville problem 206 Rejectable quality level 1075 Rejection region 1060 Relative class frequency 994 Relati ve (Cont.) error 784 frequency 1000 Relaxation 850 Remainder 171, 666 Removable singularity 709 Representation 328 Residual 849, 852 Residue 713 Residue theorem 715 Resistance 91 Resonance 86 Response 28, 84 Restoring force 62 Resultant of forces 367 Riccati equation 34 Riemann 618 sphere 710 surface 746 Right-hand derivative 484 limit 484 Right-handed 377 Risk 1095 RL-circuit 97, 240 RLC-circuit 95, 241, 244, 499 Robin problem 558, 587 Rodrigues's formula 181 Romberg integration 829 Root 610 Root test 671 Rotation 381, 3R5, 414, 734, 764 Rounding 782 Row echelon form 294 -equivalent 292, 298 operations 292 scaling 838 space 300 sum norm 849 vector 274 Runge-Kutta-Fehlberg 893 Runge-Kutta methods 892, 904 Runge-Kutta-Nystrom 906 s Saddle point 143 Sample 997, 1045 covariance 1085 118 Index Sample (Cont.) distribution function 1076 mean 996, 1045 moments 1046 point 997 range 994 size 997, 1045 space 997 standard deviation 996, 1046 variance 996, 1045 Sampling 1004, 1023, 1073 SAS 991 Sawtooth wave 248, 493, 505 Scalar 276, 364 field 384 function 384 matIix 284 multiplication 276, 368 triple product 381 Scaling 838 Scheduling 987 Schoenberg 810 Scrnodinger 242 Schur's inequality 869 Schwartz 242 Schwarz inequality 326 Secant 627, A62 method 794 Second Green's formula 466 shifting theorem 235 Sectionally continuous (see Piecewise continuous) SeIdel 846 Self-starting 898 Self-sustained oscillations 157 Separable differential equation 12 Separation of variables 12, 540 Sequence 664, A69 Series 666, A69 addition of 680 of Bessel functions 213, 583 binomial 689 convergence of 171, 666 differentiation of 174. 680 double Fourier 576 of eigenfunctions 210 Fourier 211,480.487 geometIic 167.668,673,687.692 harmonic 670 Series (Cont.) hypergeometric 188 infinite 666, A70 integration of 680 Laurent 701, 712 MaclauIin 683 multiplication of 174. 680 of orthogonal functions 210 partial sums of 17 L 666 power 167, 673 real A69 remainder of 171, 666, 684 sum of 171,666 Taylor 683 tligonometric 479 value of 171, 666 Serret-Frenet formulas (see Frenet formulas) Set of points 613 Shifted data problem 232 Shifting theorems 224, 235, 528 Shortest path 959 spanning tree 967 Shrine of the Book 814 Sifting 242 Significance level 1059 Significant digit 781 in statistics 1059 Sign test 1081 Similar matrices 350 Simple curve 391 event 997 graph 955 pole 708 zero 709 Simplex method 944 table 945 Simply connected 640, 646 Simpson's rule 821 Simultaneous corrections 850 differential equations 124 linear equations (see Linear systems) Sine of a complex variable 627, 688, 742 hyperbolic 688 119 Index Sine (Cont.) integral 509, 690, A65, A95 of a real variable A60 Single precision 782 Single-valued relation 615 Singular at infinity 711 matrix 315 point 183, 686, 707 solution 8, 50 Sturm-Liouville problem 206 Singularity 686, 707 Sink 464, 765, 973 SI system: Front cover Size of a sample 997. 1045 Skew-Hermitian 357, 361 Skewness L020 Skew-symmetric matrix 283, 345 Skydiver 12 Slack variable 941 Slope field 9 Smooth curve 42 I, 638 piecewise 421, 448, 639 surface 448 Sobolev 242 Soft spring 159 Software 778, 991 Solution of a differential equation 4, 46, 105, 536 general 6, 48, 106, 138 particular 6, 48, 106 singular 8, 50 space 304 steady-state 88 of a system of differential equations 136 of a system of equation~ 288 vector 288 SOR 851 Sorting 969, 993 Source 464. 765, 973 Span 300 Spanning tree 967 Sparse graph 957 matrix 812, 912 system of equations 846 Spectral density 520 mapping theorem 344,865 Spectral (Cont.) radius 848 representation 520 shift 344, 865, 874 Spectrum 324, 542, 864 Speed 394 Sphere 446 Spherical coordinates 588, A 71 Spiral 399 point 144 Spline 81 I S-PLUS, SPSS 992 Spring 62 Square error 503 matrix 274 membrane 575 root 792 wave 211, 480, 488, 492 Stability 3 I, 148, 783, 822, 922 chart 148 Stagnation point 763 Staircase function 248 Standard basis 328, 369 deviation 10 I 6 form of a linear ODE 26, 45, 105 Standardized random variable 1018 Stationary point 937 Statistical inference 1044 tables A96-A I 06 Steady 413, 463 state 88 Steepest descent 938 Steiner 399, 457 Stem-and-leaf plot 994 Stencil 912 Step-by-step method 886 Step function 234 Step size control 889 Stereographic projection 7 I I Stiff ODE 896 system of ODEs 907 Stirling formula 1008, A64 Stochastic matrix 285 variable 10 11 Stokes's theorem 469 120 Index Straight line 375, 391 Stream function 762 Streamline 761 Strength of a source 767 Stlictly diagonally dominant 868 String 204, 538, 594 Student's (-distribution 1053, A I 00 Sturm-Liouville problem 203 Subgraph 956 Submarine cable equations 594 Submatrix 302 Subsidiary equation 220, 230 Subspace 300 invariant 865 Success [021 Successive conections 850 overrelaxation 851 Sum (see Addition) Sum of a selies 171,666 Superlinear convergence 795 Superposition principle 106, 138 Surface 445 area 435, 442, 454 integral 449 normal 406 Surjective mapping 729 Symmetric matrix 283, 345 System of differential equations 124-165,258-263,902 linear equations (see Linear system) units: Front cover T Tables on differentiation: Front cover of Fourier transforms 529-531 of functions A94-A 106 of integrals: Front cover of Laplace transforms 265-267 statistical A96-A106 Tangent 627, A62 to a curve 392, 397 hyperbolic 629, A62 plane 406, 447 vector 392 Tangential acceleration 395 Target 973 Tarjan 971 Taylor 683 formula 684 series 683 Tchebichef (see Chebyshev) (-distribution 1053, A100 Telegraph equations 594 Termination critelion 791 Termwise differentiation 696 integration 695 multiplication 680 Test chi-square 1077 for convergence 667-672 of hypothesis 1058-1 06R nonparametric 1080 Tetrahedron 382 Thermal conductivity 465, 552 diffusivity 465, 552 Three -eights rule 830 -sigma limits 1028 Time tabling 987 Torricelli's law 15 Torsion of a curve 397 Torsional vibrations 68 Torus 454 Total differential 19 Trace of a matrix 344, 355, !:I64 Trail 959 TrajectOlies 35, 133, 141 Transfer function 230 Transformation of Cartesian courdinates A84 by a complex function 729 of integrals 437, 439, 459, 469 linear 281 linear fractional 734 orthogonal 346 similarity 350 unitary 359 of vector components A83 Transient state 88 Translation 365, 734 Transmission line equations 593 Transpose of a matrix 282 Transpositions 1081. AI06 Trapezoidal rule 817 Traveling salesman problem 960 Index Tree 966 Trend 1081 Trial 997 Triangle inequality 326, 372. 608 Triangular matlix 283 Tricomi equation 551. 55~ Tridiagonalization 875 Tridiagonal matrix 812, 875. 914 Trigonometric approximation 502 form of complex numbers 607, 624 functions, complex 626, 688 functions, real A60 polynomial 502 series 479 system 479. 482 Trihedron 398 Triple integral 458 product 381 Tlivial solution 27, 304 Truncation error 783 Tuning 543 Twisted curve 391 Two-dimensional distribution 1032 normal distribution 1090 random variable 1032 wave equation 571 Two-sided test 1060 Type of a differential equation 551 Type I and II errors 1060 U UeL 1068 Unconstrained optimization 937 Uncorrelated I 090 U mlamped system 62 Underdamping 66 Underdetemlined system 292 Underflow 782 Undetermined coefficients 79, 117, 160 Uniform convergence 691 distribution 1015, 1017. 1034 Union of events 998 Uniqueness differential equations 37, 73, 107, 137 Dirichlet problem 774 121 Uniqueness (Cont.) Laurent series 705 linear equations 303 power series 678 Unit binormal vector 398 circle 611 impulse 242 matrix 284 normal vector 447 principal normal vector 391:\ step function 234 tangent vector 392. 398 vector 326 Unitary matrix 357 system of vectors 359 transformation 359 Unstable (see Stability) Upper control limit 1068 v Value of a series 171. 666 Vandermonde determinant I 12 Van der Pol equation 157 Variable complex 614 random 10 I0, 1032 standardized random 1018 stochastic I 0 11 Variance of a distribution 1016 sample 996, 1045 Variation of parameters 98. 118, 160 Vector 274, 364 addition 276. 327, 367 field 384 function 384 moment 380 norm 853 product 377 space 300, 323, 369 subspace 300 Velocity 394 field 385 potential 762 vector 394 Venn diagram 998 Verhulst 31 122 Index Vertex 955 coloring 987 exposed 983 incidence list 957 Vibrations (see Oscillations) Violin string 538 Vizing's theorem 987 Volta 92 Voltage drop 92 Volterra 154,201, 253 Volume 435 Vortex 767 Vorticity 764 w Waiting time problem 10 13 Walk 959 Wave equation 536, 539, 569, 929 Weber 217 equation 217 . Weber (Cont.) functions 201 Website see Preface Weierstrass 618. 696 approximation theorem 797 M-test 696 Weighted graph 959 Weight function 205 Well-conditioned 852 Wessel 605 Wheatstone bridge 296 Work 373, 423 integral 422 Wronskian 75, 108 z Zero of analytic function 709 matrix 276 vector 367 Systems of Units. Some Important Conversion Factors The most important systems of units are shown in the table below. The mks system is also known as the bztemational System of Units (abbreviated S/), and the abbreviations s (instead of sec), g (instead of gm), and N (instead of nt) are also used. Length System of units Mass Time Force cgs system centimeter (cm) gram (gm) second (sec) dyne mks system meter (m) kilogram (kg) second (sec) newton (nt) Engineering system foot (ft) slug second (sec) pound (lb) I inch (in.) = 2.540000 cm I yard (yd) I foot (ft) = 3 ft = 91.440000 cm = 12 in. = 30.480000 cm I statute mile (mi) = 5280 ft = 1.609344 km I mi 2 = 640 acres = 2.5899881 I nautical mile = 6080 ft = 1.853184 km I acre = 4840 yd2 = 4046.8564 m 2 = IIl28 U.S. gallon = 2311128 in. 3 = 29.573730 cm 3 I fluid ounce I U.S. gallon = 4 quarts (liq) = 8 pints (liq) I British Imperial and Canadian gallon I slug = = 128 fl oz = 3785.4118 cm3 1.200949 U.S. gallons = 4546.087 cm 3 = 14.59390 kg I pound (lb) = 4.448444 nt I British thermal unit (Btu) I calorie (cal) I newton (nt) = 1054.35 joules = 105 dynes I joule = 107 ergs = 4.1840 joules I kilowatt-hour (kWh) = 3414.4 Btu = 3.6' 106 joules I horsepower (hp) = 2542.48 Btu/h I kilowatt (kW) OF km 2 = 178.298 cal/sec = 0.74570 kW = 1000 watts = 3414.43 Btu/h = 238.662 cal/sec = °C . 1.8 + 32 1° = 60' = 3600" = 0.01 7453293 radian For further details see, for example, D. Halliday, R. Resnick, and 1. Walker, FUl1damel1tals of Physics. 7th ed., New York: Wiley. 2005. See also AN American National Standard. ASTMIlEEE Standard Metric Practice. Institute of Electrical and Electronics Engineers, Inc., 445 Hoes L~ne. Piscataway, N. 1. 08854. Integration Differentiation (eu)' = eu' (e constant) f uv' dx = f u'v dx ltV - xn+l (u + v)' = u' + v' (uv)' = u'v + v'u , , (: )' 2 du dy du -=-.dx dy dx (Chain rule) f sin x dx = -cos x + f cos dx = sin + x = f sec x dx (tan x)' = sec2 x (cotx), = -csc 2 x (sinh x)' = cosh x (cosh x)' = sinh x , I (lnx) = x +e -In Icos xl In Isec x = e e x f cotxdx = In Isinxl (cosx)' = -sinx (n oF -I) feaxdx=~eax+e f tan x dx (sin x)' = cos x e e uv-vu v + f f ~ dx = In Ixl + xndx = - - n + 1 + e + tan xl + f csc x dx = In Icsc x - cot \:1 + dx x 2 2 = - arctan - + f + f Ya dx x = arcsin'::'a + e f Yx dx+ = arcsinh .::. + f Yx dx a = arccosh .::.a + I x a 2 a e e 2 - 2 2 a e a e 2 a e 2 - f sin2 x dx = 12 x - 14 sin 2x + e ! x + ~ sin 2x + e f cos 2 X dx = f tan 2 x dx = tan x - x + e x (arcsin x)' (arccos x)' f cot2 = -cotx - + fIn x dx = x In x - x + e xdx x f eax sin bx dx = 2 eax a (arctan x)' e f + b 2 (a sin bx - b cos bx) eax cos bx dx eax (arccotx)' +e = a 2 + b 2 (a cos bx + b sin bx) + e Polar Coordinates Some Constants x = rcos 0 e = 2.71828 182845904523536 Ve = 1.64872 127070012814685 e 2 = 7.38905 60989 30650 22723 y tanO=x lOglO = 7T Series = 0.49714987269413385435 = I 1.14472 98858 4940017414 IOglO e = 0.434294481903251 82765 In 10 = 2.30258 50929 94045 68402 In 7T v'2 = = rdrdO dxdy 3.14159265358979323846 ~ = 9.86960 440lO 89358 61883 y:;;:- = 1.77245385090551602730 7T y=rsinO 00 (Ixl - - = ~ xm 1- x < 1) m~O 1.41421 356237309504880 -{Y2 = 1.25992 1049894873 16477 v'3 = V'3 = • 1.73205 08075 68877 29353 1.44224 95703 07408 38232 In 2 = 0.69314718055994530942 In 3 = 1.09861 228866810969140 cos x = Alpha IJ Nu p Beta g Xi Gamma 0 Omicron Delta 7T Pi Epsilon p Rho e €, ~ Zeta T} Eta 0, tt, e Theta L Iota K Kappa A,A fL cr, L T v, Y l)! (_I)mx 2m ~ (2m)! m~O xm In (1 - x) = - ~ m 00 (Ixl < 1) m~l (_l)m 2m+1 x 00 arctan x = ~ + 2m m=O (Ixl < I 1) Vectors 0' 0, !J. + (2m 00 Greek Alphabet r _l)mx 2m+ I ( =~ m~O 'Y = 0.57721566490153286061 In 'Y = -0.549539312981644 82234 (see Sec. 5.6) 0 1 = 0.01745 32925 19943 29577 rad 1 rad = 57.29577 95130 82320 87680 0 = 57° 17' 44.806" 'Y, 00 Sill X Sigma a-b = alb l + a2b2 + a3b3 j k axb= al a2 a3 bl b2 b3 i af af at ax ay az gradf = Vf = - i + - j + - k Tau Upsilon . cP. cp,<I> Phi X Chi Lambda I/J, 'It Psi Mu w,n Omega aVl diV v = V-v = - curl v = V x v = ax aV2 + - ay i j a - a - ax ay V2 VI aV 3 + - az k a az V3