a a Fourier Senes and Boundary Value Problems FIFTH EDITION — — -. I I .- —. I- — -. - • -- - F - -- - - . "I - -— -. . A JAMES WARD BROWN RUELY. CHURCHILL FOURIER SERIES AND BOUNDARY VALUE PROBLEMS International Series in Pure and Applied Mathematics Ahifors: Complex Analysis Bender and Orszag: Advanced Mathematical Methods for Scientists and Engineers Boas: Invitation to Complex Analysis Buchanan and Turner Numerical Methods and Analysis Buck: Advanced Cakulus Chartrand and Oellermann: Applied and Algorithmic Graph Theory Colton: Partial Differential Equations Conte and de Boon Elementary Numerical Analysis: An Algorithmic Approach Edeistein-Keshet: Mathematical Models in Biology Goldberg: Matrix Theory with Applications Gulicic Encounters with Chaos Hill: Experiments in Computational Matrix Algebra Kurtz: Foundations ofAbstract Mathematics Lewin and Lewin: An Introduction to Mathematical Analysis Morash: Bridge to Abstract Mathematics: Mathematical Proof and Structures Parzynski and Zipse: Introduction to Mathematical Analysis Pinsky: Partial Differential Equations and Boundary-Value Problems with Applications Pinter: A Book ofAbstract Algebra Ralston and Rabinowitz: A First Course in Numerical Analysis Ritger and Rose: Differential Equations with Applications Rudin: Functional Analysis Rudin: Principles of Mathematical Analysis Rudin: Real and Complex Analysis Simmons: Differential Equations with Applications and Historical Notes Small and Hosack: Cakulus: An Integrated Approach Small and Hosacle Explorations in Cakulus with a Computer Algebra System Vanden Eynden: Elementary Number Theory Walker: Introduction to Abstract Algebra Churchill/ Brown Series Complex Variables and Applications Fourier Series and Boundary Value Problems Operational Mathematics Also Available from McGraw-Hill Schaum's Outline Series in Mathematics & Statistics Most outlines include basic theory, definitions, and hundreds of solved problems and supplementary problems with answers. 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A complete list of Schaum titles may be obtained by writing to: Schaum Division McGraw-Hill, Inc. Princeton Road S-i Hightstown, NJ 08520 , * . # FOURIER SERIES AND BOUNDARY VALUE PROBLEMS Fifth Edition James Ward Brown Professor of Matheina tics The University of Michigan —Dearborn Ruel V. Churchill Late Professor of Mathematics The University of Michigan McGraw-Hill, Inc. New York St. Louis San Francisco Auckland Bogota Caracas Lisbon London Madrid Mexico Milan Montreal New Delhi Paris San Juan Singapore Sydney Tokyo Toronto This book was set in Times Roman by Science Typographers, Inc. The editors were Richard H. Wallis, Maggie Lanzillo, and James W. Bradley; the production supervisor was Denise L. Puryear. The cover was designed by John Hite. R. R. Donnelley & Sons Company was printer and binder. FOURIER SERIES AND BOUNDARY VALUE PROBLEMS Copyright © 1993, 1987, 1978, 1963, 1941 by McGraw-Hill, Inc. All rights reserved. Copyright renewed 1959 by Rue! V. Churchill. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 1234567890D0C 9098765432 ISBN 0-07-008202-2 Library of Congress Cataloging-in-Publication Data Brown, James Ward. Fourier series and boundary value problems / James Ward Brown, Ruel V. Churchill.—Sth ed. p. cm. Churchill's name appears first on the earlier editions. Includes bibliographical references and index. ISBN 0-07-008202-2 2. Functions, Orthogonal. 3. Boundary value 1. Fourier series. II. Title. I. Churchill, Ruel Vance, (date). problems. 1993 0A404.B76 515'.2433—dc2O 92-12194 ABOUTTHE AUTHORS JAMES WARD BROWN is Professor of Mathematics at The University of Michigan—--Dearborn. He earned his A.B. in physics from Harvard University and his AM. and Ph.D. in mathematics from The University of Michigan in Ann Arbor, where he was an Institute of Science and Technology Predoctoral Fellow. He was coauthor with Dr. Churchill of Complex Variables and Applica.. tions, now in its fifth edition. He has received a research grant from the National Science Foundation as well as a Distinguished Faculty Award from the Michigan Association of Governing Boards of Colleges and Universities. Dr. Brown is listed in Who's Who in America. RUEL V. CHURCHILL was, at the time of his death in 1987, Professor Emeritus of Mathematics at The University of Michigan, where he began teaching in 1922. He received his B.S. in physics from the University of Chicago and his M.S. in physics and Ph.D. in mathematics from The University of Michigan. He was coauthor with Dr. Brown of Complex Variables and Applications, a classic text that he first wrote over 45 years ago. He was also the author of Operational Mathematics, now in its third edition. Dr. Churchill held various offices in the Mathematical Association of America and in other mathematical societies and councils. To THE MEMORY OF MY FATHER, George H. Brown, AND OF MY LONG-TIME FRIEND AND COAUTHOR, Rue! V. Churchi!!. THESE DISTINGUISHED MEN OF SCIENCE FOR YEARS INFLUENCED THE CAREERS OF MANY PEOPLE, INCLUDING MYSELF. J.w.Ii JOSEPH FOURIER JEAN BAPTISTE JOSEPH FOURIER was born in Auxerre, about 100 miles south of Paris, on March 21, 1768. His fame is based on his mathematical theory of heat conduction, a theory involving expansions of arbitrary functions in certain types of trigonometric series. Although such expansions had been investigated earlier, they bear his name because of his major contributions. Fourier series are now fundamental tools in science, and this book is an introduction to their theory and applications. Fourier's life was varied and difficult at times. Orphaned by the age of 9, he became interested in mathematics at a military school run by the dictines in Auxerre. He was an active supporter of the Revolution and narrowly escaped imprisonment and execution on more than one occasion. After the Revolution, Fourier accompanied Napoleon to Egypt in order to set up an educational institution in the newly conquered territory. Shortly after the French withdrew in 1801, Napoleon appointed Fourier prefect of a department in southern France with headquarters in Grenoble. It was in Grenoble that Fourier did his most important scientific work. Since his professional life was almost equally divided between politics and science and since it was intimately geared to the Revolution and Napoleon, his advancement of the frontiers of mathematical science is quite remarkable. The final years of Fourier's life were spent in Paris, where he was Secretary of the Académie des Sciences and succeeded Laplace as President of the Council of the Ecole Polytechnique. He died at the age of 62 on May 16, 1830. CONTENTS Preface xv Partial Differential Equations of Physics Linear Boundary Value Problems. Conduction of Heat. Higher Dimensions and Boundary Conditions. The Laplacian in Cylindrical and Spherical Coordinates. A Vibrating String. Vibrations of Bars and Membranes. Types of Equations and Boundary Conditions. Methods of Solution. On the Superposition of Separated Solutions. 2 Fourier Series 39 Piecewise Continuous Functions. Inner Products and Orthonormal Sets. Generalized Fourier Series. Fourier Cosine Series. Fourier Sine Series. Fourier Series. Best Approximation in the Mean. One-Sided Derivatives. Two Lemmas. A Fourier Theorem. Discussion of the Theorem and Its Corollary. Fourier Series on Other Intervals. Uniform Convergence of Fourier Series. Differentiation and Integration of Fourier Series. Convergence in the Mean. 3 The Fourier Method 105 Linear Operators. Principle of Superposition. A Temperature Problem. Verification of Solution. A Vibrating String Problem. Verification of Solution. Historical Development. 4 Boundary Value Problems 129 A Slab with Various Boundary Conditions. The Slab with Internally Generated Heat. Dirichlet Problems. Other Types of Boundary Conditions. A String with Prescribed Initial Velocity. An Elastic Bar. Resonance. Fourier Series in Two Variables. Periodic Boundary Conditions. XIII xiv S CONTENTS Sturm-Liouville Problems and Applications 168 Modifications. Orthogonality of Eigenfunctions. Uniqueness of Eigenfunctions. Methods of Solution. Examples of Eigenfunction Expansions. Surface Heat Transfer. Polar Coordinates. Modifications of the Method. A Vertically Hung Elastic Bar. Regular Sturm-Liouville Problems. 6 Fourier Integrals and Applications 217 The Fourier Integral Formula. An Integration Formula. Two Lemmas. A Fourier Integral Theorem. The Cosine and Sine Integrals. More on Superposition of Solutions. Temperatures in a Semi-Infinite Solid. Temperatures in an Unlimited Medium. 7 Bessel Functions and Applications 242 Bessel Functions J,,. General Solutions of Bessel's Equation. Recurrence Relations. Bessel's Integral Form of Consequences of the Integral Representations. The Zeros of J0(x). Zeros of Related Functions. Orthogonal Sets of Bessel Functions. The Orthonormal Functions. Fourier-Bessel Series. Temperatures in a Long Cylinder. Heat Transfer at the Surface of the Cylinder. Vibration of a Circular Membrane. 8 Legendre Polynomials and Applications 293 Solutions of Legendre's Equation. Legendre Polynomials. Orthogonality of Legendre Polynomials. Rodrigues' Formula and Norms. Legendre Series. Dirichlet Problems in Spherical Regions. Steady Temperatures in a Hemisphere. 9 Uniqueness of Solutions 321 Abel's Test for Uniform Convergence. Uniqueness of Solutions of the Heat Equation. Solutions of Laplace's or Poisson's Equation. Solutions of a Wave Equation. Bibliography Index 336 341 PREFACE This is an introductory treatment of Fourier series and their applications to boundary value problems in partial differential equations of engineering and physics. It is designed for students who have completed a first course in ordinary differential equations and the equivalent of a term of advanced calculus. In order that the book be accessible to as great a variety of students as possible, there are footnotes referring to texts which give proofs of the more delicate results in advanced calculus that are occasionally needed. The physical applications, explained in some detail, are kept on a fairly elementary level. The first objective of the book is to introduce the concept of orthogonal sets of functions and representations of arbitrary functions in series of functions from such sets. Representations of functions by Fourier series, involving sine and cosine functions, are given special attention. Fourier integral representations and expansions in series of Bessel functions and Legendre polynomials are also treated. The second objective is a clear presentation of the classical method of separation of variables used in solving boundary value problems with the aid of those representations. Some attention is given to the verification of solutions and to uniqueness of solutions; for the method cannot be presented properly without such considerations. Other methods are treated in the authors' book Complex Variables and Applications and in Professor Churchill's book Operational Mathematics. This book is a revision of the 1987 edition. The first two editions, published in 1941 and 1963, were written by Professor Churchill alone. While improvements appearing in earlier editions have been retained with this one, there are a number of major changes in this edition that should be mentioned. The introduction of orthonormal sets of functions is now blended in with the treatment of Fourier series. Orthonormal sets are thus instilled earlier and are reinforced immediately with available examples. Also, much more attention is now paid to solving boundary value problems involving nonhomogeneous partial differential equations, as well as problems whose nonhomogeneous boundary conditions prevent direct application of the method of separation of variables. To be specific, considerable use is made, both in examples and in problem sets, of the method of variation of parameters, where the coefficients in xv xvi PREFACE certain eigenfunction expansions are found by solving ordinary differential equations. Other improvements include a simpler derivation of the heat equation that does not involve vector calculus, a new section devoted exclusively to examples of eigenfunction expansions, and many more figures and problems to be worked out by the reader. There has been some rearrangement of the early material on separation of variables, and the exposition has been improved throughout. The chapters on Bessel functions and Legendre polynomials, Chapters 7 and 8, are essentially independent of each other and can be taken up in either order. The last three sections of Chapter 2, on further properties of Fourier series, and Chapter 9, on uniqueness of solutions, can be omitted to shorten the course; this also applies to some sections of other chapters. The preparation of this edition has benefited from the continued interest of various people, many of whom are colleagues and students. They include Jacqueline R. Brown, Michael A. Lachance, Ronald P. Morash, Joyce A. Moss, Frank J. Papp, Richard L. Patterson, Mark A. Pinsky, and Sandra M. Razook. Ralph P. Boas, Jr., and George H. Brown furnished some of the references that are cited in the footnotes; and the derivation of the laplacian in spherical coordinates that is given was suggested by a note of R. P. Agnew's in the American Mathematical Monthly, vol. 60 (1953). Finally, it should be emphasized that this edition could not have been possible without the enthusiastic editorial support of people at McGraw-Hill, most especially Richard H. Wallis and Maggie Lanzillo. They, in turn, obtained the following reviewers of both the last edition and the present one in manuscript form: Joseph M. Egar, Cleveland State University; K. Bruce Erickson, University of Washington; William W. Farr, Worcester Polytechnic Institute; Thomas L. Jackson, Old Dominion University; Charles R. MacCluer, Michigan State University; Robert Piziak, Baylor University; and Donald E. Ryan, Northwestern State University of Louisiana. James Ward Brown FOURIER SERIES AND BOUNDARY VALUE PROBLEMS CHAPTER 1 PARTIAL D IFFERENTIAL EQUATIONS OF PHYSICS This book is concerned with two general topics: (a) One is the representation of a given function by an infinite series involving a prescribed set of functions. (b) The other is a method of solving boundary value problems in partial differential equations, with emphasis on equations that are prominent in physics and engineering. Representations by series are encountered in solving such boundary value problems. The theories of those representations can be presented independently. They have such attractive features as the extension of concepts of geometry, vector analysis, and algebra into the field of mathematical analysis. Their mathematical precision is also pleasing. But they gain in unity and interest when presented in connection with boundary value problems. The set of functions that make up the terms in the series representation is determined by the boundary value problem. Representations by Fourier series, which are certain types of series of sine and cosine functions, are associated with a large and important class of boundary value problems. We shall give special attention to the theory and application of Fourier series. But we shall also consider extensions and generalizations of such series, concentrating on Fourier integrals and series of Bessel functions and Legendre polynomials. 1 2 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP. I A boundary value problem is correctly set if it has one and only one solution within a given class of functions. Physical interpretations often suggest boundary conditions under which a problem may be correctly set. In fact, it is sometimes helpful to interpret a problem physically in order to judge whether the boundary conditions may be adequate. This is a prominent reason for associating such problems with their physical applications, aside from the opportunity to illustrate connections between mathematical analysis and the physical sciences. The theory of partial differential equations gives results on the existence and uniqueness of solutions of boundary value problems. But such results are necessarily limited and complicated by the great variety of types of differential equations and domains on which they are defined, as well as types of boundary conditions. Instead of appealing to general theory in treating a specific problem, our approach will be to actually find a solution, which can often be verified and shown to be the only one possible. 1. LINEAR BOUNDARY VALUE PROBLEMS In the theory and application of ordinary or partial differential equations, the dependent variable, denoted here by u, is usually required to satisfy some conditions on the boundary of the domain on which the differential equation is defined. The equations that represent those boundary conditions may involve values of derivatives of u, as well as values of u itself, at points on the boundary. In addition, some conditions on the continuity of u and its derivatives within the domain and on the boundary may be required. Such a set of requirements constitutes a boundary value problem in the function u. We use that terminology whenever the differential equation is accompanied by some boundary conditions, even though the conditions may not be adequate to ensure the existence of a unique solution of the problem. EXAMPLE 1. The three equations (1) u(x) = —1 u'(O) = 0, u(1) = 0 u"(x) (0 <x < — 1), make up a boundary value problem in ordinary differential equations. The differential equation is defined on the domain 0 < x < 1, whose boundary points are x = 0 and x = 1. A solution of this problem which, together with its derivative, is continuous on the closed interval 0 c x (2) u(x)=1- cosh x cosh 1 Solution (2) is easily verified by direct substitution. c 1 is LINEAR BOUNDARY VALUE PROBLEMS SEC. I 3 Frequently, it is convenient to indicate partial differentiation by writing independent variables as subscripts. If, for instance, u is a function of x and y, we may write ôu Ux oru(x,y)for—, x for u dx ax2 u , for ayax etc. We shall always assume that the partial derivatives of u satisfy conditions y), allowing us to write = ufl,. Also, we shall be free to use the symbol for example, to denote values of the function au/ax on the line x = c. EXAMPLE 2. The problem consisting of the partial differential equation (x>O,y>O) (3) and the two boundary conditions (4) (y > y) u(O, y) = 0), u(x,0)=sinx+cosx is a boundary value problem in partial differential equations. The differential equation is defined in the first quadrant of the xy plane. As the reader can readily verify, the function (5) x+ u(x, y) = cos x) is a solution of this problem. The function (5) and its partial derivatives of the first and second order are continuous in the region x 0, y 0. A differential equation in a function u, or a boundary condition on u, is linear if it is an equation of the first degree in u and derivatives of u. Thus the terms of the equation are either prescribed functions of the independent variables alone, including constants, or such functions multiplied by u or a derivative of u. Note that the general linear partial differential equation of the second order in u = u(x, y) has the form (6) +Fu = + G, where the letters A through G denote either constants or functions of the independent variables x and y only. The differential equations and boundary conditions in Examples I and 2 are all linear. The differential equation f'7\ 'ii is w ii C linear in u = u(x, y, z), but the equation u= is ff + the first degree as an algebraic expression in the two variables u and uy [compare equation (6)]. A boundary value problem is linear if its differential equation and all its boundary conditions are linear. The boundary value problems in Examples 1 4 CHAP I PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS and 2 are, therefore, linear. The method of solution presented in this book does not apply to nonlinear problems. A linear differential equation or boundary condition in u is homogeneous if each of its terms, other than zero itself, is of the first degree in the function u and its derivatives. Homogeneity will play a central role in our treatment of linear boundary value problems. Observe that equation (3) and the first of conditions (4) are homogeneous but that the second of those conditions is not. Equation (6) is homogeneous in a domain of the xy plane only when the function G is identically zero (G 0) throughout that domain; and equation (7) is nonhomogeneous unless f(y, z) 0 for all values of y and z being considered. CONDUCTION OF HEAT 2. Thermal energy is transferred from warmer to cooler regions interior to a solid body by means of conduction. It is convenient to refer to that transfer as a flow of heat, as if heat were a fluid or gas that diffused through the body from regions of high concentration into regions of low concentration. Let P0 denote a point (x0, y0, z0) interior to the body and S a plane or smooth curved surface through P0. Also, let n be a unit vector that is normal to S at the point P0 (Fig. 1). At time t, the flux ofheat t(x0, y0, z0, t) across S at in the direction of n is the quantity of heat per unit area per unit time that is being conducted across S at P0 in that direction. Flux is, therefore, measured in such units as calories per square centimeter per second. FIGURE 1 If u(x, y, z, t) denotes temperatures at points of the body at time t and if n is a coordinate that represents distance in the direction of n, the flux 4(x0, y0, z0, it) is positive when the directional derivative du/dn is negative at P0 and negative when du/dn is positive there. A fundamental postulate, known as Fourier 's law, in the mathematical theory of heat conduction states that the magnitude of the flux t(x0, y0, z0, t) is proportional to the magnitude of the directional derivative du/dn at P0 at time t. That is, there is a coefficient K, known as the thermal conductivity of the material, such that (1) at P0 and time t. du t=—K— dn (K>O) SEC. 2 CONDUCTION OF HEAT 5 Another thermal coefficient of the material is its specific heat a-. This is the quantity of heat required to raise the temperature of a unit mass of the material one unit on the temperature scale. Unless otherwise stated, we shall always assume that the coefficients K and a are constants and that the same is true of 6, the mass per unit volume of the material. With these assumptions, a second postulate in the mathematical theory is that conduction leads to a temperature function u which, together with its derivative and those of the first and second order with respect to x, y, and z, is continuous throughout each domain interior to a solid body in which no heat is generated or lost. Suppose now that heat flows only parallel to the x axis in the body, so that and temperatures U depend on only x and t. Thus 1 = F(x, t) and flux U u(x, t). We assume at prçsent that heat is neither generated nor lost within the body and hence that heat enters or leaves only through the surface. We then construct a small rectangular parallelepiped, lying in the interior of the body, with one vertex at a point (x, y, z) and with faces parallel to the coordinate planes. The lengths of the edges are Ax, Ay, and Az, as shown in Fig. 2. Observe that, since the parallelepiped is small, the continuous function Ut little in that region and has approximately the value it) throughout it. This approximation improves, of course, as A x tends to zero. varies z C B D F E H y FIGURE 2 The mass of the element of material occupying the parallelepiped is 8 Ax Ay Az. So, in view of the definition of specific heat a stated above, we know that one measure of the quantity of heat entering that element per unit time at time t is approximately (2) Ax Ltsy t). Another way to measure that quantity is to observe that, since the flow of heat is parallel to the x axis, heat crosses only the surfaces ABCD and EFGH of the element, which are parallel to the yz plane. If the direction of the flux CP(x, it) 6 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP I is in the positive direction of the x axis, it follows that the quantity of heat per unit time crossing the surface ABCD into the element at time it is t(x, t) Ay Az. Because of the heat leaving the element through the face EFGH, the net quantity of heat entering the element per unit time is, then, t(x,t) AyAz — t(x + Ax,t) AyAz. In view of Fourier's law (1), this expression can be written (3) Equating expressions (2) and (3) for the quantity of heat entering the element per unit time and then dividing by o-5 Ax Ay Az, we find that K = — go lim = Ax K aS The temperatures in the solid body, when heat flows only parallel to the x axis, thus satisfy the one-dimensional heat equation K (4) k = k here is called the thermal d4ffusivity of the material. In the derivation of equation (4), we assumed that there is no source (or sink) of heat within the solid body, but only heat transfer by conduction. If there is a uniform source throughout the body that generates heat at a constant rate Q per unit volume, where Q denotes the quantity of heat generated per unit volume per unit time, it is easy to modify the derivation to obtain the nonhomogeneous heat equation (5) = +q Q q= — as This is accomplished by simply adding the term Q A x A y A z to expression (3) and proceeding in the same way as before. The rate Q per unit volume at which heat is generated may, in fact, be any continuous function of x and t, in which case the term q in equation (5) also has that property. The heat equation describing flow in two and three dimensions is cussed in Sec. 3. HIGHER DIMENSIONS AND BOUNDARY CONDITIONS 3. When the direction of heat flow in a solid body is not restricted to be simply parallel to the x axis, temperatures u in the body depend, in general, on all the space variables, as well as it. By considering the rate of heat passing through each of the six faces of the element in Fig. 2 (Sec. 2), one can derive (see HIGHER DIMENSIONS AND BOUNDARY CONDITIONS SEC. 3 7 Problem 6, Sec. 4) the three-dimensional heat equation, satisfied by u = u(x, y, z, t): ( 1) Ut = k is the thermal diffusivity of the material, appearing in equation (4), Sec. 2. When the laplacian ( 2) is used, equation (1) takes the compact form (3) kV2U. Note that when there is no flow of heat parallel to the z axis, so that = 0 and U U(x, y, tO, equation (1) reduces to the heat equation for two-dimensional flow parallel to the xy plane: + (4) The one-dimensional heat equation Ut = in Sec. 2 for temperatures U(X, it) follows, of course, from this when there is, in addition, no flow U parallel to the y axis. If temperatures are in a steady state, in which case U does not vary with time, equation (1) becomes Laplace's equation ( 5) Equation (5) is often written as V2u = 0. The derivation of equation (1) in Problem 6, Sec. 4, takes into account the possibility that heat may be generated in the solid body at a constant rate Q per unit volume, and the generalization (6) of equation (5), Sec. 2, is obtained. If the rate Q is a continuous function of the space variables x, y, and z and temperatures are in a steady state, equation (6) becomes Poisson's equation (7) V2u=f(x,y,z), where f(x, y, z) = —q(x, y, z)/k. that describe thermal conditions on the surfaces of the solid body and initial temperatures throughout the body must accompany the heat equation if we are to determine the temperature function u. The conditions on the surfaces may be other than just prescribed temperatures. Suppose, for example, that the flux F into the solid at points on a surface S is some constant That is, at each point P on 5, units of heat per unit area per unit time Equations flow across S in the opposite direction of an outward unit normal vector n at P. From Fourier's law (1) in Sec. 2, we know that if du/dn is the directional derivative of u at P in the direction of n, the flux into the solid across S at P is 8 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP the value of Kdu/dn there. Hence du (8) on the surface S. Observe that if S is perfectly insulated, and condition (8) becomes = 0 at points on 5; du (9) &=o. On the other hand, there may be surface heat transfer between a bound- ary surface and a medium whose temperature is a constant T. The inward flux which can be negative, may then vary from point to point on 5; and we assume that, at each point P, the flux is proportional to the difference between the temperature of the medium and the temperature at P. Under this assumption, which is sometimes called Newton's law of cooling, there is a positive constant H, known as the surface conductance of the material, such that = H(T — u) at points on S. Condition (8) is then replaced by the condition du K—=H(T—u), (10) dn or H du (11) EXAMPLE. Consider a semi-infinite slab occupying the region 0 c x c c, 0 of three-dimensional space. Figure 3 shows the cross section of the slab in y into the slab at points on the xy plane. Suppose that there is a constant flux the face in the plane x = and that there is surface heat transfer (possibly inward) between the face in the plane x = c and a medium at temperature zero. Also, the surface in the plane y = 0 is insulated. Since du/dn = — au/ax and 0 du/dn = au/ax on the faces in the planes x = yI 00 0 .J x=c x FIGURE3 0 and x = c, respectively, a THE LAPLACIAN IN CYLINDRICAL AND SPHERICAL COORDINATES SEC. 4 9 temperature function u(x, y, z, t) evidently satisfies the boundary conditions = —hu(c,y,z,t). The insulated surface gives rise to the boundary condition 0, z, 0 = 0. It should be emphasized that the various partial differential equations in this section are important in other areas of applied mathematics. In simple diffusion problems, for example, Fourier's law t = —Kdu/dn applies to the flux ci' of a substance that is diffusing within a porous solid. In that case, ct' represents the mass of the substance that is diffused per unit area per unit time through a surface, u denotes concentration (the mass of the diffusing substance per unit volume of the solid), and K is the coefficient of diffusion . Since the mass of the substance entering the element of volume in Fig. 2 (Sec. 2) per unit time is A x A y A z , one can replace the product aö in the derivation of the heat equation by unity to see that the concentration satisfies the diffusion equation (12) u = u(x, y, z) that is continuous, together with its partial derivatives of the first and second order, and satisfies Laplace's equation (5) is called a harmonic function. We have seen in this section that the steady-state temperatures at points interior to a solid body in which no heat is generated are represented by a harmonic function. The steady-state concentration of a diffusing substance is also represented by such a function. Among the many physical examples of harmonic functions, the velocity potential for the steady-state irrotational motion of an incompressible fluid is prominent in hydrodynamics and aerodynamics. An important harmonic function in electrical field theory is the electrostatic potential V(x, y, z) in a region of space that is free of electric charges. The potential may be caused by a static distribution of electric charges outside that region. The fact that V is harmonic is a consequence of the inverse.square law of attraction or repulsion between charges. Likewise, gravitational potential is a harmonic function in regions of space not occupied by matter. In this book, the physical problems involving the laplacian, and Laplace's equation in particular, are limited mostly to those for which the differential equations are derived in this chapter. Derivations of such differential equations in other areas of applied mathematics can be found in books on hydrodynamics, elasticity, vibrations and sound, electrical field theory, potential theory, and other branches of continuum mechanics. A number of such books are listed in the Bibliography at the back of this book. THE LAPLACIAN IN CYLINDRICAL AND SPHERICAL COORDINATES 4. We recall that the heat equation, derived in Sec. 2, and its modifications (Sec. 3), including Laplace's equation, can be written in terms of the laplacian ( 1) V2u = + uyy + 10 CHAP. I PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS Often, because of the geometric configuration of the physical problem, it is more convenient to use the laplacian in other than rectangular coordinates. In this section, we show how the laplacian can be expressed in terms of the variables of two important coordinate systems already encountered in calculus. The cylindrical coordinates p, 4', and z determine a point P(p, 4, z) whose rectangular coordinates are (Fig. 4)t x=pcos4, ( 2) y=psin4, z=z. Thus p and 4, are the polar coordinates in the xy plane of the point Q, where Q is the projection of P onto that plane. Relations (2) can be written z=z, ( 3) where the quadrant to which the angle 4 belongs is determined by the signs of x and y, not by the ratio y/x alone. z P(p,4,z) y p Q x FIGURE 4 Let u denote a function of x, y, and z. Then, in view of relations (2), it is also a function of the three independent variables p, 4, and z. If u is continuous and possesses continuous partial derivatives of the first and second orders, the following method, based on the chain rule for differentiating composite functions, can be used to express the laplacian (1) in terms of p, cb, and z. Relations (3) enable us to write du dudp du x du y xdu ydu Hence, by relations (2), (4) Bu du dx ap —=cos4i—— sin4ôu p t In calculus, the symbols r and 0 are often used instead of p and tfr, but the notation used here is common in physics and engineering. The notation for spherical coordinates, treated later in this section, may also differ somewhat from that learned in calculus. SEC 4 THE LAPLACIAN IN CYLINDRICAL AND SPHERICAL COORDINATES 11 Replacing the function u in equation (4) by au/ax, we see that a2u (5) ax2 sin4 au a =cos4— — ax ap a — au ax We may now use expression (4) to substitute for the derivative au/ax appearing on the right-hand side of equation (5): au a a2u cosØ—— ap ax sin4 sinçt au ap a — p au cos4—--ap sin4 au p By applying rules for differentiating differences and products of functions and using the relation = which is ensured by the continuity of the partial derivatives, we find that (6) sin24 a2u 2sin4cos4i a2u a2u = p + sin2Ø au —+ ap p p2 2sin4cos4 au p2 In the same way, one can show that (7) xau yau pap au ay p2a4 or cos4au au au —=sin4—+ ap ay (8) p and also that a2u a2u (9) ay2 + 2sin4cos4 cos24 a2u a2u p p2 p By adding corresponding sides of equations (6) and (9), we arrive at the identity (10) a2u ax2 + a2u ay2 = a2u ap2 I a2u I au +——-+-— p ap p2 Since rectangular and cylindrical coordinates share the coordinate z, it follows 12 CHAP. I PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS that the laplacian of u in cylindrical coordinates is V2u= (11) 32u 3p2 1 32u 1 Bu +——+— p dp p2 + We can group the first two terms and use the subscript notation for partial derivatives to write this in the form (12) V2u = + + (10) gives us the two-dimensional laplacian in polar coordinates. Note that Laplace's equation V2u = 0 in that coordinate system in the xy plane can be written Expression (13) p2upp + = 0. + Note, too, how it follows from expression (1 1) that when temperatures u in a = kV2U solid body vary only with p, and not with 4 and z, the heat equation becomes I ( 14) p Equations (13) and (14) will be of particular interest in the applications. The spherical coordinates r, 4, and 0 of a point P(r, 4, 0) (Fig. 5) are related to x, y, and z as follows: y = rsinOsin4, x = rsinOcosØ, ( 15) z = rcosO. The coordinate 4 is common to cylindrical and spherical coordinates, and the coordinates in those two systems are related by the equations z=rcosO, (16) p=rsinO, 4=4. Expression (1 1) for the laplacian can be transformed into spherical coordi- nates quite readily by means of the proper interchange of letters, without any z P(r,4,O) r 0 y x FIGURE 5 SEC. 4 PROBLEMS 13 further application of the chain rule. This is accomplished in three steps, described below. First, we observe that, except for the names of the variables involved, transformation (16) is the same as transformation (2). Since transformation (2) gave us equation (10), we know that 32u (17) 3z2 + 82u 32u = 3p2 I du 1 32u r dr r2 ao2 +——+— Br2 we note that when transformation (16) is used, the counterpart of equation (7) is Second, With pdis zdu rdr r230 this and relations (16), we are able to write ldu I 32u ldu cotOdu =——+—+ ——+—- (18) p dp r p2 Br r2 1 r2sin2O ao Third, by grouping the first and last terms in expression (11), and also the second and third terms there, we see that, according to equations (17) and (18), the laplacian of u in spherical coordinates is 32u " I V2 — 23u 1 r dr r2sin2O +——+ Br2 32u 1 32u +— r2 + cotOBu r2 — ao Other forms of this expression are (20) V2u=—(ru)rr+ r 1 1 1 r 2 sin r 2sinO 0 (sinou9)0, 1 1 1 V2u=—(r2u) r 2 (sinOu0) 0 r2sinO Many of our applications later on will involve Laplace's equation V2u = 0 in spherical coordinates when u is independent of 4. According to expression (21) r2 r + r2sin2O u + (20), that equation can then be written r—(ru)+ ar2 (22) 1 . 3 au — sinO— =0. sinO ao ao PROBLEMS u(x) denote the steady-state temperatures in a slab bounded by the planes x = 0 and x = c when those faces are kept at fixed temperatures u = 0 and u = u0, 1. Let respectively. Set up the boundary value problem for u(x) and solve it to show that and where is the flux of heat to the left across each plane x = x0 (0 x0 c c). 14 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP. 1 2. A slab occupies the region 0 c x c c. There is a constant flux of heat into the slab through the face x = 0. The face x = c is kept at temperature u = 0. Set up and solve the boundary value problem for the steady-state temperatures u(x) in the slab. Answer u(x) = — x). 3. Let a slab 0 c x c c be subjected to surface heat transfer, according to Newton's law of cooling, at its faces x = 0 and x = c, the surface conductance H being the same on each face. Show that if the medium x < 0 has temperature zero and the medium x > C has the constant temperature T, then the boundary value problem for steady-state temperatures u(x) in the slab is (O<x<c), u"(x)=O Ku'(c) = H[T Ku'(O) = Hu(O), — u(c)], where K is the thermal conductivity of the material in the slab. Write h = H/K and derive the expression u(x)= T ch + 2 (hx+1) for those temperatures. 4. Let u(r) denote the steady-state temperatures in a solid bounded by two concentric spheres r = a and r = b (a c b) when the inner surface r = a is kept at temperature zero and the outer surface r = b is maintained at a constant temperature u0. Show why Laplace's equation for u = u(r) reduces to —(ru) = dr2 0, and then derive the expression u(r)= bu0 ( '\1__) Sketch the graph of u(r) versus r. 5. In Problem 4, replace the condition on the outer surface r = b with the condition that there is surface heat transfer into a medium at constant temperature T according to Newton's law of cooling. Then obtain the expression hb2T the steady-state temperatures, where h is the ratio of the surface conductance H to the thermal conductivity K of the material. 6. Let u = u(x, y, z, t) denote temperatures in a solid body throughout which there is a uniform heat source. Derive the heat equation for for those temperatures, where the constants k and q are the same ones as in equation (5), Sec. 2. SEC 4 PROBLEMS 15 Suggestion: Modify the derivation of equation (5), Sec. 2, by also considering the net rate of heat entering the element in Fig. 2 (Sec. 2) through the faces parallel to the xz and xy planes. Since the faces are small, one may consider the needed flux at points on a given face to be constant over that face. Thus, for instance, the net rate of heat entering the element through the faces parallel to the xy plane is to be taken as y, z + y, z, it)] Ax Ay. zXz, it) — 7. A slender wire lies along the x axis, and surface heat transfer takes place along the wire into the surrounding medium at a fixed temperature T. Modify the procedure in Sec. 2 to show that if u = u(x, it) denotes temperatures in the wire, then + b(T = — where b is a positive constant. Suggestion: Let r denote the radius of the wire, and apply Newton's law of cooling to see that the quantity of heat entering the element in Fig. 6 through its cylindrical surface per unit time is approximately H[T — u(x, t)]2rr Ax. T" t 4 I V 1' x+iXx T° FIGURE 6 8. Suppose that the thermal coefficients K and a are functions of x, y, and z. Modify the derivation in Problem 7 to show that the heat equation takes the form u3u = + + in a domain where all functions and derivatives involved are continuous. 9. Show that the substitution = kit can be used to write the where k = 1. equation u1 = + in the form + = 10. Derive heat expressions (8) and (9) in Sec. 4 for du/By and 32u/3y2 in cylindrical coordinates. 11. In Sec. 4, show how expressions (20) and (21) for V2u in spherical coordinates follow from expression (19). 12. (a) Show that if u is a function of the polar coordinates p and 4', where x = p cos 4' and y= p sin 4', then 0u du 84, Bx ay (b) Let u(p, 4') denote temperatures, independent of the cylindrical coordinate z, in a long rod, parallel to the z axis, whose cross section in the xy plane is the 16 PARTIAL DiFFERENTIAL EQUATIONS OF PHYSICS CHAP I y p=1 x FIGURE 7 sector 0 c p C 1, 0 i,-/2 of a disk (Fig. 7). Use the result in part (a) to 4' show that if the rod is insulated on its planar surfaces, where 4 = = ir/2, then u must satisfy the boundary conditions u4p,O) = 0, ii) = 0 0 and (O<p<l). 13. Suppose that temperatures u in a solid hemisphere r 1, 0 0 ( ir/2 are independent of the spherical coordinate 4, so that u = u(r, 0), and that the base of the hemisphere is insulated (Fig. 8). Use transformation (16), Sec. 4, relating cylindrical to spherical coordinates, to show that du Bu 30 dz Thus show that u must satisfy the boundary condition u0(r, ir/2) = 0. 1 FiGURE 8 14. Show that the physical dimensions of thermal conductivity k (Sec. 2) are L2T where L denotes length and T time. Suggestion: Observe first that the dimensions of thermal conductivity K and specific heat a are AL T 1B and AM 'B , respectively, where M denotes mass, A quantity of heat, and B temperature. Then recall that k = K/(aS), where 5 is density (ML3). 5. A VIBRATING STRING A tightly stretched string, whose position of equilibrium is some interval on the is vibrating in the xy plane. Each point of the string, with coordinates x (x, 0) in the equilibrium position, has a transverse displacement y = y(x, it) at time t. We assume that the displacements y are small relative to the length of AVIBRATINGSTRING SEC 5 17 the string, that slopes are small, and that other conditions are such that the movement of each point is parallel to the y axis. Then, at time t, a point on the string has coordinates (x, y), where y = y(x, t). Let the tension of the string be great enough that the string behaves as if it were perfectly flexible. That is, at a point (x, y) on the string, the part of the string to the left of that point exerts a force T, in the tangential direction, on the part to the right; and any resistance to bending at the point is to be neglected. The magnitude of the x component of the tensile force T is denoted by H. See Fig. 9, where that x component has the same positive sense as the x axis. Our final assumption here is that H is constant. That is, the variation of H with respect to x and t can be neglected. y V(x + -H t) (x,y) H V(x, it) 0 x x + Ax FIGURE 9 These idealizing assumptions are severe, but they are justified in many applications. They are adequately satisfied, for instance, by strings of musical instruments under ordinary conditions of operation. Mathematically, the assumptions will lead us to a partial differential equation in y(x, it) that is linear. Now let V(x, it) denote the y component of the tensile force T exerted by the left-hand portion of the string on the right-hand portion at the point (x, y). We take the positive sense of V to be that of the y axis. If cx is the angle of inclination of the string at the point (x, y) at time t, then —V(x,t) H (1) it) > 0. If V(x, it) > This is indicated in Fig. 9, where V(x, it) < 0 and then 'w/2 < a < 'w and t) < 0; and a similar sketch shows that V(x, it) H 0, =tan(r—a) = —tana= Hence relations (1) still hold. Note, too, that it) = 0 when V(x, it) = 0, since a = 0 then. It follows from relations (1) that the y component V(x, it) of the force exerted at time it by the part of the string to the left of a point (x, y) on 18 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP. I the part to the right is given by the equation (2) (H> 0), V(x,t) = is basic for deriving the equation of motion of the string. Equation (2) is also used in setting up certain types of boundary conditions. which Suppose that all external forces such as the weight of the string and resistance forces, other than forces at the end points, can be neglected. Consider a segment of the string not containing an end point and whose projection onto the x axis has length Ax. Since x components of displacements are negligible, the mass of the segment is 8 A x , where the constant 6 is the mass per unit length of the string. At time t, the y component of the force end (x, y) is V(x, it), given exerted by the string on the segment at the by equation (2). The tangential force S exerted on the other end of the segment by the part of the string to the right is also indicated in Fig. 9. Its y component V(x + Ax, t) evidently satisfies the relation V(x + Ax,t) H where /3 = tanfl is the angle of inclination of the string at that other end of the segment. That is, (3) (H>0). Note that, except for a minus sign, this is equation (2) when the argument x there is replaced by x + Ax. it). Now the acceleration of the end (x, y) in the y direction is Consequently, by Newton's second law of motion (mass times acceleration equals force), it follows from equations (2) and (3) that (4) = approximately, when Ax is small. Hence H = — lim y(x+Ax,t)—y(x,t) X X H Ax 15 ôAx—*O whenever these partial derivatives exist. Thus the function y(x, t), which represents the transverse displacements in a stretched string under the conditions stated above, satisfies the one-dimensional wave equation (5) t) = H a2 a has the physical dimensions of velocity. One can choose units for the time variable so that a = 1 in the wave equation. More precisely, if we make the substitution 'r = at, the chain rule 19 A VIBRATING STRING SEC. 5 shows that dy By at ar —=a-— and 82y at2 a ay =a— a— =a ar ar 2a2y ar2 Equation (5) then becomes (A similar observation was made in = Problem 9, Sec. 4, with regard to the heat equation.) When external forces parallel to the y axis act along the string, we let F denote the force per unit length of string, the positive sense of F being that of the y axis. Then a term FAx must be added on the right-hand side of equation (4), and the equation of motion is F • (6) axis vertical and its positive sense upward, suppose that the external force consists of the weight of the string. Then F A x = — 3 A x g, where the positive constant g is the acceleration due to gravity; and equation (6) becomes the linear nonhomogeneous equation y (7) —g. equation (6), F may be a function of x, z', y, or derivatives of y. If the external force per unit length is a damping force proportional to the velocity in the y direction, for example, F is replaced by —Bye, where the positive constant B is a damping coefficient. Then the equation of motion is linear and homogeIn neous: B (8) t) = If an end x = t) — b = of the string is kept fixed at the origin at all times t the boundary condition there is clearly 0 (9) 0, y(0,t)=0 But if the end is permitted to slide along the y axis and is moved along that axis with a displacement fQ), the boundary condition is the linear nonhomogeneous one y(0,t) =f(t) (10) (t 0). that the left-hand end is attached to a ring which can slide along the y axis. When a force F(t) (t > 0) in the y direction is applied to that end, FQ) is the limit, as x tends to zero through positive values, of the force V(x, t) Suppose described earlier in this section. According to equation (2), the boundary condition at x = 0 is then t) = F(t) (t > 0). The minus sign disappears, however, if x = 0 is the right-hand end, in view of equation (3). 30 SJISAHd OZ WLfllVd 9 SNOLLVthIIA 40 SUVIE UNV SNOLLVflO3 °M oquosop dVHD SOdA3 JO SUOJ3I3JqJA JOJ OflUflUOD 1!UJ!j JflO AJSflI3S OAPM SuOflI3nbo qo!qM t S3U0W03131dS!p o:j AjJrnJ ojdwis tuomouoqd jo sigh WU!PflJ!8110'I lO'J Oqi O1PU!pJ003 X adrqs rn 'UJS!Jd 03 JO E JOpU!IAD S1UOUJaOEIdS!U JO JO popoJ!p ur JO SO3UP2S!p UJOJJ OUO suonoos IP!1!u! sluowoorJds!p io U! oqi SSOJD S! oqi POUWJISUfl SOfl!DOjOA U! pUP UJJOJ!Ufl JOAO qzwo SSOJD UO930S 'pOAjOAU! OSflRD 0q1 SUOLJDOS 03 jojjrnid 02 0q3 X 'SJXI3 2V OUJfl 'I uoqoos oq2 3uOwODPjdS!p JO 2u!od x S! Aq 'x){ JO 0q1 3UOWODPjdS!p JO oqi UO!330S 313 X S! 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SMOjjOJ 0143 UO Aq oqi oJsPq UOflI3flbO (s'x)j (1) 'Ajnpw!s oqi ODJOJ UO = 0q1 pUO S! x)j + (i'xv (z) oqi q aip ssrm + (i'xv 2!Ufl OUJfljOA MPj 03 0q1 UO!lOm JO UP Jo 1P!JOIEW Jo oqi Jo SNOIIVNUIA JO 9 OM Arm 03!JM (c) °M 'pug = (I'x)'a21V— 10JJ13 V9 Aq 2uuioj + xv puoi 0q2 'x)( U s3uomocrnjds!p ur (I = oqj Anpunoq uo!ppuoo 'ji 'S! '0)1 (1 (ç) nq O9SEjO . AJS!P3S 0q3 Joq3o Ofl? jji?uJs 31343 pooEjds!p uo!prnba (ii) puo 0 21 = uo oqi SO3JOJ spuo OM pOuJflSSE 3Eq3 AWj soqddR puii piq3 suonoos urnmoi nuiijd OUSPjO Arm oq pozwjdoJ Aq r uwnjoo jo 'J!E rn suo!p33!IddP £iooqi Jo punos pjoq 'oJoz wqi 03 Zn u! ou OARM (;'xv (i (i (v) S! U: UNV = 0 Jo ou OOJOJ X puo x = o S! oatj I < '0 = 0 'pun ui MO!A Jo uOfll?flbo '(T) = 0 JEq S201? 31? Jo sauiuqwaj&j 1O'] uo 'x)z '{ (i oiouop jjnus ojqixojj 3Rq3 S! Jdd qiXUdi OM uoqooloJd jo ospi qUM OM MOu ou!uJI3xo 0q2 ji3uia3u! iuomojo slu!od Jo 0q3 oAino pur jrunou uv qons wv 'I jo s3u!od •OUIEJJ ui oqi fl? OTIHI sou?u!pJ003 'x) u! eCx oi oqi zx ouEjd sposJo3u! 0q2 s3u!od pojoqrj V u! uuoj UJWV Jo pur 'suouonJ3suoD jjrms 'uogoanp JOAO U! P OAJflD Aq S! Z Apq8ij 1rnod uo 0q2 irqi 1u!od wnuqijinbo 'uo!psod ourjd °qi poorfds!p ouriqmaw Cr ourjd S! S3uOmOzn3jds!p u! Jo q32uoj xv aip •'cv uo 0q2 SODJOJ P341 SODJOJ r 'oaioj OM ioj Jo 0q3 ii oiouop 043 t(x z 'C 3M1Tl9Id H oi.p oqi 'ourjd °M 22 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP. I that H is constant, regardless of what point or curve on the membrane is being discussed. In view of expressions (2) and (3) in Sec. 5 for the forces V on the ends of a segment of a vibrating string, we know that the force in the z direction exerted over the curve AB is approximately y, t) Ax and that the corresponding force over the curve DC is approximately y + Ay, t) Ax. Similar expressions are found for the vertical forces exerted over AD and BC when the tensile forces on those curves are considered. It then follows that the sum of the vertical forces exerted over the entire boundary of the element is approximately (6) y, t) Ax + y + Ay, t) Ax y, t) Ay + + Ax, y, t) Ay. If Newton's second law is applied to the motion of the element in the z direction and if 5 denotes the mass per unit area of the membrane, it follows from expression (6) for the total force on the element that z(x, y, it) satisfies the two-dimensional wave equation ( 7) (a2 + ztt = Details of these final steps are left to the problems, where it is also shown that if an external transverse force F(x, y, t) per unit area acts over the membrane, the equation of motion takes the form ( 8) = + + F Equation (8) arises, for example, when the z axis is directed vertically upward and the weight of the membrane is taken into account in the derivation of equation (7). Then F/6 = — g , where g is the acceleration due to gravity. From equation (7), one can see that the static transverse displacements z(x, y) of a stretched membrane satisfy Laplace's equation (Sec. 3) in two dimensions. Here the displacements are the result of displacements, perpendicular to the xy plane, of parts of the frame that support the membrane when no external forces are exerted except at the boundary. PROBLEMS 1. A stretched string, with its ends fixed at the points 0 and 2c on the x axis, hangs at rest under its own weight. The y axis is directed vertically upward. Point out how it follows from the nonhomogeneous wave equation (7), Sec. 5, that the static displacements y(x) of points on the string must satisfy the differential equation a2y"(x)=g SEC 6 PROBLEMS 23 on the interval 0 < x < 2c, in addition to the boundary conditions y(O) = y(2c) = 0, 0. By solving this boundary value problem, show that the string hangs in the parabolic arc 2a2 (x—c) 2 g gc2 Y+—j• 2a and that the depth of the vertex of the arc varies directly with c2 and S and inversely with H. 2. Use expression (2), Sec. 5, for the vertical force V and the equation of the arc in which the string in Problem 1 lies to show that the vertical force exerted on that string by each support is ôcg, half the weight of the string. 3. Let z(p) represent static transverse displacements in a membrane, stretched between the two circles p = 1 and p = Po (Po > 1) in the plane z = 0, after the outer displaced by a distance z = z0. State why the boundary value support p = Po problem in z(p) can be written d z(1) = and dz (l<p<pO), =0 z(p0) = 0, obtain the solution z(p)=z0 ln p ln Po 4. Show that the steady-state temperatures u(p) in a hollow cylinder 1 — 00 < Z < 0O also p Po' satisfy the boundary value problem written in Problem 3 if u = 0 on the inner cylindrical surface and u = z0 on the outer one. Thus show that Problem 3 is a membrane analogy for this temperature problem. Soap films have been used to display such analogies. 5. Give needed details in the derivation of equation (6), Sec. 5, for the forced vibrations of a stretched string. 6. The physical dimensions of H, the magnitude of the x component of the tensile force in a string, are those of mass times acceleration: MLT2, where M denotes mass, L length, and T time. Show that, since a2 = H/3, the constant a has the 7. dimensions of velocity: LT A strand of wire 1 ft long, stretched between the origin and the point 1 on the x axis, weighs 0.032 lb (Sg = 0.032, g = 32 ft/s2) and H = 10 lb. At the instant it = 0, the strand lies along the x axis but has a velocity of 1 ft/s in the direction of the y axis, perhaps because the supports were in motion and were brought to rest at that instant. Assuming that no external forces act along the wire, state why the displacements y(x, t) should satisfy this boundary value problem: t) = y(0,t)=y(1,t)=0, t) y(x,0)=0, (0 < x < 1, t > 0), 24 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 8. The end x = CHAP. 1 of a cylindrical elastic bar is kept fixed, and a constant compressive force of magnitude F0 units per unit area is exerted at all times it > 0 over the end x = C. The bar is initially unstrained and at rest, with no external forces acting along it. State why the function y(x, it) representing the longitudinal displacements of cross sections should satisfy this boundary value problem, where a2 = 0 it) = y(0,t) = 0, = (0 c x c c, it > 0), it) y(x,0) =y(x,0) =0. —F0, 9. The left-hand end x = 0 of a horizontal elastic bar is elastically supported in such a way that the longitudinal force per unit area exerted on the bar at that end is proportional to the displacement of the end, but opposite in sign. State why the end condition there has the form it) = by(0, (b > 0). it) Use expression (6), Sec. 6, to derive the nonhomogeneous wave equation (8), Sec. 6, for a membrane when there is an external transverse force F(x, y, it) per unit area acting on it. [Note that if this force is zero (F 0), the equation reduces to equation (7), Sec. 6.] 11. Let z(x, y) denote the static transverse displacements in a membrane over which an external transverse force F(x, y) per unit area acts. Show how it follows from the nonhomogeneous wave equation (8), Sec. 6, that z(x, y) satisfies Poisson's equation: 10. [Compare equation (7), Sec. 3.] 12. A uniform transverse force of F0 units per unit area acts over a membrane, stretched between the two circles p = I and p = Po (Po > 1) in the plane z = 0. From Problem 11, show that the static transverse displacements z(p) satisfy the equation (pz')'+f0p=0 and derive the expression 10 2 z(p)=—(p0--1) 4 7. lnp lnp0 p2—1 — (lcpcpo). 2 Po' TYPES OF EQUATIONS AND BOUNDARY CONDITIONS The second-order linear partial differential equation (Sec. 1) (1) + in u = u(x, y), where A, B, . + Cur, + . . , G + + Fu = G are constants or functions of x and y, is classified in any given region of the xy plane according to whether B2 — 4AC is SEC 7 TYPES OF EQUATIONS AND BOUNDARY CONDITIONS 25 positive, negative, or zero throughout that region. Specifically, equation (1) is (a) Hyperbolic if B2 — 4AC > 0; (b) Elliptic if B2 — 4AC 0; (c) Parabolic if B2 — 4AC = 0. For each of these categories, equation (1) and its solutions have distinct features. Some indication of this is given in Problems 15 and 16, Sec. 9. The terminology used here is suggested by the fact (Problem 6, Sec. 9) that when A, B, . , F are constants and G 0, equation (1) always has solutions of the form U = exp (Ax + ,ay), where the constants A and ,a satisfy the algebraic . . equation (2) From analytic geometry, we know that such an equation represents a conic plane and that the different types of conic sections arising are section in the similarly determined by B2 — 4AC. EXAMPLES. Laplace's equation uxx+uyy=0 is a special case of equation (1) in which A = C = 1 and B = elliptic throughout the xy plane. Poisson's equation (Sec. 3) uxx + uyy 0. Hence it is =f(x,y) in two dimensions is elliptic in any region of the xy plane where f(x, y) is defined. The one-dimensional heat equation + = 0 in U = u(x, t) is parabolic in the xt plane, and the one-dimensional wave equation + in y = y(x, it) = 0 hyperbolic there. Another special case of equation (1) is the telegraph eqUationt is vxx = + (KR + + RSv. Here v(x, t) represents either the electrostatic potential or current at time t at a point x units from one end of a transmission line or cable that has t A derivation of this equation is outlined in the book by Churchill (1972, pp. 272—273) that is listed in the Bibliography. 26 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP. I electrostatic capacity K, self-inductance L, resistance R, and leakage conductance S, all per unit length. The equation is hyperbolic if KL > 0. It is parabolic if either K or L is zero. As indicated below, the three types of second-order linear equations just described require, in general, different types of boundary conditions in order to determine a solution. Let u denote the dependent variable in a boundary value problem. A condition that prescribes the values of u itself along a portion of the boundary is known as a Dirichiet condition. The problem of determining a harmonic function on a domain such that the function assumes prescribed values over the entire boundary of that domain is called a Dirichiet problem. In that case, the values of the function can be interpreted as steady-state temperatures. Such a physical interpretation leads us to expect that a Dirichlet problem may have a unique solution if the functions considered satisfy certain requirements as to their regularity. A Neumann condition prescribes the values of normal derivatives du/dn on a part of the boundary. Another type of boundary condition is a Robin condition. It prescribes values of hu + du/dn at boundary points, where h is either a constant or a function of the independent variables. If a partial differential equation in y is of the second order with respect to are one of the independent variables it and if the values of both y and prescribed when t = 0, the boundary condition is one of Cauchy type with such a condition respect to t. In the case of the wave equation = corresponds physically to that of prescribing the initial values of the transverse in a stretched string. Initial values for both y displacements y and velocities and appear to be needed if the displacements y(x, t) are to be determined. + = 0 or the heat equation however, conditions of Cauchy type on u with respect to x = cannot be imposed without severe restrictions. This is suggested by interpreting When the equation is Laplace's equation u physically as a temperature function. When the temperatures u in a slab to the 0 c x C c are prescribed on the face ; = 0, for example, the flux left through that face is ordinarily determined by the values of u there and by is prescribed at other conditions in the problem. Conversely, if the flux x = 0, the temperatures there are affected. 8. METHODS OF SOLUTION Some boundary value problems in partial differential equations can be solved by a method corresponding to the one usually used to solve such problems in ordinary differential equations, namely the method of first finding the general solution of the differential equation. SEC. 8 METHODS OF SOLUTION 27 EXAMPLE 1. Let us solve the boundary value problem y) = (1) u(1, y) = I u(O, y) = y2, 0, on the domain 0 < x < 1, —ac < y < Successive integrations of the equation = kept fixed, lead to the equations = 4&v) and 0 with respect to x, with y u=xØ(y)+i,4(y), (2) where 0 and are arbitrary functions of y. The boundary conditions in problem (I) require that =y2, Thus = 1 — y2, and the solution of the problem is u(x,y) =x(l —y2) ±y2. (3) EXAMPLE 2. We next solve the wave equation t) = (4) it) ( —oc < x < oc, t > 0), subject to the boundary conditions (5) (—ao<x<oo), y(x,0)=f(x), in terms of the constant a and the function f. The differential equation (4) can be simplified as follows by introducing the new independent variables (6) u=x+at, v=x—at. According to the chain rule for differentiating composite functions, dy dydu dydu — = —— + ——. auat at avat That is, ay ay ay —=a——a——. at av au (7) Replacing the function y by ay/at in equation (7) yields the expression a2y a ay a ay —=a—— —a——; au at av at at2 and using equation (7) again, this time to substitute for ay/at on the right here, 28 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP 1 we see that 32y at 2 a a au ay ay a ay ay au av av au av a——a— —a— a——a— or a2y (8) at2 =a2 a2y a2y au2 —2 avau + a2y av2 We have, of course, assumed that _ a2y a2y avau auav In like manner, one can show that (9) = a2y au2 +2 a2y avau + a2y av2 In view of expressions (8) and (9), then, equation (4) becomes (10) yuv=o with the change of variables (6). Equation (10) can be solved by successive integrations to give y = 4i(u) + where the arbitrary functions 4, and are twice-differentiable. The general solution of the wave equation (4) is, therefore, y=4i(x+at)+i/,(x—at). (11) In this example, the boundary conditions are simple enough that we can actually determine the functions 4 and ift. Observe that the function (11) satisfies conditions (5) when and a41(x) — ai/i'(x) = 0. 4(x) + ifi(x) =f(x) Thus 4(x) — i/4x) = c, where c is a constant; and it follows that 24(x)=f(x)+c and 2if,(x)=f(x)—c. Consequently, y(x,t) = (12) +f(x-at)]. solution (12) of the boundary value problem consisting of equations (4) and (5) is known as d'Alembert's solution. It is easily verified under the The assumption that f'(x) and f"(x) exist for all x. The method for solving boundary value problems illustrated in the two examples here has severe limitations. The general solutions (2) and (11), SEC 9 ON THE SUPERPOSITION OF SEPARATED SOLUTIONS 29 involving arbitrary functions, were obtained by successive integrations, a procedure that applies to relatively few types of partial differential equations. But, even in the exceptional cases in which such general solutions can be found, the determination of the arbitrary functions directly from the boundary conditions is often difficult. Among a variety of other methods, the one to be developed in this book will be suggested by the example in Sec. 9. That method, which is sometimes called the Fourier method, is a classical and powerful one. Before turning to it, however, we mention some other important ones. Methods based on Laplace, Fourier, and other integral transforms, all included in the subject of operational mathematics, are especially effective.t The classical method of conformal map- ping in the theory of functions of a complex variable applies to a prominent class of problems involving Laplace's equation in two dimensions.t There are still other ways of reducing or solving such problems, including applications of so-called Green's functions and numerical, or computational, methods. Even when a problem yields to more than one method, however, different methods sometimes produce different forms of the solution; and each form may have its own desirable features. On the other hand, some problems require successive applications of two or more methods. Others, including some fairly simple ones, have defied all known exact methods. The development of new methods is an activity in present-day mathematical research. ON THE SUPERPOSITION OF SEPARATED SOLUTIONS 9. The purpose of this section is to motivate the two main topics of the book, indicated at the beginning of the chapter. Namely, in seeking a solution of the boundary value problem in the example below, we shall find it necessary to expand an arbitrary function in a series of trigonometric functions (Chap. 2), as well as to formalize the Fourier method for solving boundary value problems in partial differential equations (Chap. 3). EXAMPLE. The Dirichlet problem (see Sec. 7) (O<x<1,y>O), (1) (2) (3) u(1,y)=O u(x,O)=f(x) u(O,y)=O, (y>O), (O<x<1) is satisfied by steady-state temperatures u(x, y), subject to the indicated boundary conditions, in a semi-infinite slab occupying the region 0 x 1, y 0 of t See the book by Churchill (1972), listed in the Bibliography. 4: See the authors' book (1990), also listed in the Bibliography. 30 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS u=O 0 V2u=O CHAP 1 u=O u=f(x) x=1 x FIGURE12 three-dimensional space (Fig. 12). We shall not be concerned here with precise conditions on the function f. We assume only that f is bounded and observe that it is then physically reasonable to seek solutions of equation (1) that tend to zero as y tends to infinity. Since equation (1) has constant coefficients and is of the type treated below in Problem 6, we know from that problem that the function u(x, y) = ( 4) a solution, where A and equation is eAX+ILY eAXe/.LY are any constants (real or complex) related by the (5) Such a solution is separated in the sense that it is the product of two functions one of which depends only on x and the other of which depends only on y) In anticipation of real-valued solutions of equation (1) that tend to zero as y tends must be a negative_real number and write to infinity, we stipulate that I , according to relation —U, where u > 0. Then A = ±i"i, where i = /L (5). So, in view of Euler's formulat 0 + i sin 0, we have these two families of solutions: &° U1(x, y) = U2(x,y) = cos = jix + i sin ux), — isinvx). t This terminology is borrowed from the book by Pinsky (1991) that is listed in the Bibliography. While the method used here is well known, that book has an especially good variety of problems to which it is applied. t Complete justification of basic facts from complex analysis that are used in this section can be found in the authors' book (1990), listed in the Bibliography. SEC 9 ON THE SUPERPOSITION OF SEPARATED SOLUTIONS 31 Now the partial differential equation (1) is linear and homogeneous; and, as is the case with linear homogeneous ordinary differential equations, any linear combination of solutions of equation (1) is also a solution (see Problem 7). Accepting the fact that such a superposition principle remains valid when complex-valued functions and complex constants are involved, we arrive at the real-valued solutions U3(x,y) = U1(x,y) + U2(x,y) 2 =e 2i of equation (1), where u has any positive value. It is, of course, a simple matter to verify directly that U3(x, y) and U4(x, y) actually satisfy equation (1). Turning now to the boundary conditions (2), we see that the solutions U3(x, y) cannot satisfy the first of those conditions since cos 0 = 1. But the solutions U4(x, y) satisfy both conditions, provided that sin u = 0. Since the only (positive) values of v having that property are i' = (n = 1, 2, . ), it follows that the functions . . (n = 1,2,...) ( 6) all satisfy conditions (1) and (2). Referring once again to Problem 7, we see that any linear combination N u(x, y) = E ( 7) sin n=1 of the first N of the functions (6) satisfies equation (1). It is, moreover, obvious that this sum also satisfies conditions (2). must be determined so that As for condition (3), the constants N f(x)= (O<x<1). n=1 If the function f(x) is itself a linear combination of the sine functions sin2rx, sin'n-x, the needed values of b1, b2, . , bN sinN'irx, are evident. If, for instance, f(x) = 2sinrx + sin3n, (8) we may take N = . . ..., 3 and write b1 = u(x,y) = 2, b2 = 0, b3 = 1. The function + thus satisfies all the conditions (1)—(3) when f(x) is the particular function (8). 32 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP I But suppose that f(x) is an arbitrary function. The following generalization of the above method is suggested. We might replace the sum (7) by a generalized linear combination, or infinite series, u(x,y) = ( 9) We must, of course, assume that this series converges and that the superposition principle verified in Problem 7 can be extended so as to apply. Then condition such that (3) requires us to find values of the constants (O<x<1). (10) The problem of finding such coefficients, as well as ones in series involving cosines, is the problem, noted just prior to this example, that is to be treated in Chap. 2. Separated solutions and a superposition principle, extended to infinite series, are the foundation of the Fourier method for solving boundary value problems that is described in Chap. 3. PROBLEMS 1. Solve each of these boundary value problems by first finding the general solution of the partial differential equation involved. (O<x<1,—oo<y<oo), u(O, y) = y, y) = 0; (x>0,y>0), u(x,0) =x2. Answers: (a) u(x, y) = (x3 u(0, y) = 0, — 3x + l)y; (b) u(x, y) = x2(1 + y). 2. Whether a second-order linear partial differential equation in u = u(x, y) is hyperbolic, elliptic, or parabolic (Sec. 7) can vary from region to region in the xy plane when at least one of the coefficients is a nonconstant function of x and y. Classify each of the following differential equations in various regions, and sketch those regions. (a) (c) + + = 0; + = 0; (b) — = 0. = 2; (d) + +(i — — Answers: (a) Parabolic on the x axis, elliptic above it, and hyperbolic below it; (b) parabolic on the curve y = x4, elliptic above it, and hyperbolic below it; (d) parabolic on the circle x2 + y2 = 1, elliptic inside it, and hyperbolic outside it. 3. In Example 2, Sec. 8, d'Alembert's solution y(x,t) = 1 + at) +f(x - at)] represents transverse displacements in a stretched string of infinite length, initially released at rest from a position y = f(x) ( — oc < x < oc). Use that solution to show how the instantaneous position of the string at time t can be displayed graphically by adding ordinates of two curves, one obtained by translating the curve y = ff(x) to SEC 9 PROBLEMS 33 the right through the distance at, the other by translating it to the left through the same distance. As t varies, the curve y = moves in each direction as a wave with velocity a. Sketch some instantaneous positions when f(x) is zero except on a small interval about the origin. 4. Use the general solution (11) in Example 2, Sec. 8, to solve the boundary value problem (—oc<x<oo,t>O), y,(x,O)=g(x) y(x,O)=O, (—oc<x<oc). Suggestion: Note that one can write I g(x) 5. fXg(5) ds + C. x+at 1 Answer: = y(x,t) = —J g(s)ds. 2a x—at Let Y(x, t) denote d'Alembert's solution (12) in Example 2, Sec. 8, of the boundary value problem solved there, and let Z(x, it) denote the solution found in Problem 4 for a related boundary value problem. Verify directly that the sum y(x, t) = Y(x, t) + Z(x, t) is a solution of the boundary value problem (—oc<x<oc,t>O), y(x,O)=f(x), (—oc<x<oc). Thus show that y(x,t) is ;[f(x + at) +f(x — at)] L x+at g(s)ds La x—at 1 1 = + a solution of the problem here. Interpret the problem physically (see Problem 3). 6. Let the coefficients A, B, . . . , F in the linear homogeneous partial differential equation + + + + + Fu = 0 be constants, rather than more general functions of x and y. By substituting the exponential function u = exp (Ax + p'y), where A and p. are constants, into that differential equation, show that it is always a solution when A and satisfy the algebraic equation values of p, arise from the that result. The values of ji are, of course, not necessarily A quadratic equations in real even when A is real.t Similar remarks apply when values of p. are first selected. t The usual rules for differentiating the exponential function in calculus also hold when complex numbers are involved. See the footnote in Sec. 9. 34 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP. I (The possibility of such exponential solutions is suggested by experience with ordinary differential equations that have constant coefficients.) 7. Suppose that the functions = y) (n = 1, 2, . ) are all solutions of Laplace's equation . . 32u 32u —+—=o. fly2 3x2 (n = Show that, for any constants 1, 2, . . . , N), the linear combination N U= n=1 is also a solution. Do this by substituting the sum here into the left-hand side of the differential equation and grouping terms appropriately. [This result is a special case of the principle of superposition of solutions, to be developed more fully in Chap. 3 (Sec. 26).] 8. Show that if each of the functions = it) (n = 1, 2, . . . ) satisfies the heat equation 32u flu =k—, fix2 — fit then the same is true of any linear combination N L u= n=1 (Compare Problem 7.) 9. Let umn umn(x, y, z) (m = equation 1, 2, 32u fix2 n= . . . ; + 32u fly2 1, 2, 32u + 3z2 . . . ) denote solutions of Laplace's =0. Verify that any linear combination NM u= L m=1 E n==1 is also a solution. (Compare Problem 7.) 10. Verify that each of the functions (n= 1,2,...) u0(x,y)=y, satisfies Laplace's equation y) + y) = 0 (0 < x < ir, 0 < y < 2) and the three boundary conditions y) = y) = 0, u(x,0) = 0. Then, with the aid of the superposition principle in Problem 7, note that any linear SEC 9 PROBLEMS 35 combination N u(x,y)=A0y+ n=1 the same differential equation and boundary conditions. Take N = 2 and find values of the coefficients A0, A1, A2 such that a fourth boundary condition satisfies u(x, 2) = 4 + 3 cos x — cos 2x is satisfied. Interpret the result physically. 3 Answer:A0=2,A1= . sinh2 —1 ,A2= . sinh4 11. When the unit of time is chosen so that the diffusivity k in the heat equation is unity (see Problem 9, Sec. 4), the boundary value problem u(0,t) = (0 c x < it) it) 1, it > 0), u(x,0) =f(x) = 0, describes temperatures in a slab 0 c x c 1 whose face x = 0 is kept at temperature zero, whose face x = 1 is insulated, and whose initial temperatures depend only onx. (a) Assuming that the function f is bounded, modify the treatment of the Dirichlet problem in the example in Sec. 9 to discover the following solutions of the above heat equation and the first two boundary conditions: (2n — (2n — t 1)irx — (n=1,2,...). sin 2 4 Verify these solutions directly. Then, with the aid of the result in Problem 8, point out how it follows that any linear combination u(x,t)= N (2n — 1)2r2 (2n —t — 4 sin — 1)irx 2 is also a solution. (b) Use the final result in part (a) to obtain the solution u(x,t)=2exp ——t sin——exp 4 25ir2 — 5irx it 4 2 sin 2 of the stated boundary value problem when ,;Tx f(x)=2sin-j——sin 5irx 2 12. Verify that each of the products umn(x,y,z)exp(_zVm2+n2)cosmysinnx satisfies (m=0,1,2,...;n= 1,2,...) Laplace's equation y, z) + y, z) + y, z) = 0 (0 < x c r, 0 < y c ir, z > 0) 36 PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS CHAP. 1 and the boundary conditions u(O, y, z) = u(ir, y, z) = 0, z) = z) = 0. Then, with the aid of the result in Problem 9, obtain a function u(x, y, z) that satisfies not only Laplace's equation and the stated boundary conditions, but also the condition = (—6 + 5cos4y)sin3x Interpret that function physically. Answer: u(x, y,z) = (2e_3Z 13. Let y(x, it) represent transverse displacements in a long stretched string, one end of which is attached to a ring that can slide along the y axis. The other end is so far out on the positive x axis that it may be considered to be infinitely far from the origin. The ring is initially at the origin and is then moved along the y axis (Fig. 13) so that x y = f(t) when x = 0 and t 0, where f is a prescribed continuous function and f(0) = 0. We assume that the string is initially at rest on the x axis; thus y(x, t) as x -÷ The boundary value problem for y(x, it) is (x t) t) = y(x,0)=0, y(0,t) =f(t) (x 0), (tO). (a) Apply the first two of these boundary conditions to the general solution (Sec. 8) y = Ø(x + at) + i/i(x — at) of the wave equation to show that there is a constant C qS(x)=C and çls(x) = such —C that when x Then apply the third boundary condition to show that =f(-) - C where C is the same constant. when 0. PROBLEMS SEC. 9 37 (b) With the aid of the results in part (a), derive the solution whenxnt, 0 y(x,t)= I fit——I " Note that the part of the string to the right of the point x = at on the x axis is unaffected by the movement of the ring prior to time it, as shown in Fig. 13. 14. Use the solution obtained in Problem 13 to show that if the ring at the left-hand end of the string in that problem is moved according to the function sinirt when 0 when the ring is lifted up 1 unit and then returned to the origin, where it remains after time t = 1. The expression for y(x, 0 here shows that when it > 1, the string coincides with the x axis except on an interval of length a, where it forms one arch of a sine curve (Fig. 14). Furthermore, as right with speed a. y (it it increases, the arch moves to the > 1) 0 aU—i) at x FiGURE 14 15. Consider the partial differential equation where A, B, and C are constants, and assume that B2 — 4AC > 0 it is hyperbolic, so that (Sec. 7). (a) Use the transformation U=X+at, v=x+f3t to obtain the new differential equation (A + Ba + + [2A + B(a + /3) + +(A + Bf3 + = 0. (a*fJ) 'lVIDIVd WIIN3N3JJIG SNOIIVIIOH JO SDLSAHd v pur d pEqJ U- dVHD I oqj + =od =01, jEflUOJaJJ!p UOflEflbO 4i OlE MOq = •0 UOflfljOS Jo o:i jEUi8iJO S! = pun ' uotwnba Ut pEd (V) UI liEd (q) :I1ll.j:J 'AjOA!podsaJ (3) apnpuoj i.uoJj 4 :3v17-zg/k-g- 3V17 + (i°xi + x)4i + 'aJqrn1uOJOJJtp-031M1 £IE.fl!qJE SUOflOUflJ 'UT) oos 'g jo UOflfljOS + OAEM "itT uaqj Moqs uoqnnba = 0 SMOjjOJ SE E IE!OOdS 9J .iopun u0FIEWJ0JSuEJ:1 ,d+rv=a 'x=n uoA!8 jnquoJajj!p uouBnbo + uaqj Moqs fluji (v) + S! oqdqjo = 0 JVfr > (0 = uoflEnba — GVt' = (0 flg + = •0 o:j U! OiWj SUO!SUOIfl!p uoqM oqj jBU!8L10 UOflEflbO pUl? ' I (q) + uoip?nba saonpoi uo!3Bnbo — cj somoooq U! vvz) MOU '(o*d) U! ouo =d vz I UO!SUOUHp PUB vz=d oqi jnur2uo UOflnflba S! CHAPTER 2 FOURIER SERIES In this chapter, we shall present the basic theory of Fourier series, which are expansions of arbitrary functions in series of sine and cosine functions. In so doing, we shall introduce the concept of orthonormal sets of functions. That will not only clarify underlying concepts behind the various types of Fourier series treated here but also lay the foundation for finding other types of series expansions that are needed in later chapters. 10. PIECEWISE CONTINUOUS FUNCTIONS If u1 and u2 are functions and c1 and c2 are constants, the function c1u1 + c2u2 is called a linear combination of u1 and u2. Note that u1 + u2 and c1u1, as well as the constant function 0, are special cases. A linear space of functions, or f unction space, is a class of functions, all with a common domain of definition, such that each linear combination of any two functions in that class remains in it; that is, if u1 and u2 are in the class, then so is c1u1 + c2u2. Before developing the theory of Fourier, or trigonometric, series, we need to specify function spaces containing the functions to be represented. Let a function f be continuous at all points of a bounded open interval a<x < b except possibly for a finite set of points x1, x2, . . . , x,,_1, where a<x1<x2< ... If we write x0 = a and = b, then f is continuous on each of the n open subintervals xo<x<x1, x1<x<x2, ..., xn_1<x<xn. 39 40 FOURIER SERIES CHAP. 2 It is not necessarily continuous, or even defined, at their end points. But if, in each of those subintervals, f has finite limits as x approaches the end points from the interior, f is said to be piecewise continuous on the interval a < x < b. More precisely, the one-sided limits (1) and f(xk—)= limf(x) f(xk_l+)== lim f(x) X<Xk X>Xk_1 (k=l,2,...,n) are required to exist. Note that if the limiting values from the interior of a subinterval are assigned to f at the end points, then f is continuous on the closed subinterval. Since any function that is continuous on a closed bounded interval is bounded, it follows that f is bounded on the entire interval a x b. That is, there exists a nonnegative number M such that If(x)I M for all points x (a c x c b) at which f is defined. EXAMPLE 1. Consider the function f that has the values x f(x)= —1 1 whenO <x <1, <2, whenl when2 <x <3. (See Fig. 15.) Although f is discontinuous at the points x = 1 and x = 2 in the interval 0 < x < 3, it is nevertheless piecewise continuous on that interval. This is because the one-sided limits from the interior exist at the end points of each of the three open subintervals on which f is continuous. Note, for instance, that the right-hand limit at x = 0 is f(O + ) = 0 and that the left-hand limit at x = I isf(1—)= 1. f( x) 1 I 1 0 Ii ' I —1 12 I 1 3 x I —---h FIGURE 15 A function is piecewise continuous on an interval a < x < b if it is continuous on the closed interval a c x b. Continuity on the open interval a < x < b does not, however, imply piecewise continuity there, as the following example illustrates. SEC 10 PIECEWISE CONTINUOUS FUNCTIONS 41 EXAMPLE 2. The function f(x) = 1/x is continuous on the interval 0 c x < 1, but it is not piecewise continuous there since f(O + ) fails to exist. When a function f is piecewise continuous on an interval a < x < b, the integral of f(x) from x = a to x = b always exists. It is the sum of the integrals of f(x) over the open subintervals on which f is continuous: (2) + + = +7 f(x)dx. The first integral on the right exists since it is defined as the integral over the interval a C X C Xi of the continuous function whose values are f(x) when a <x <x1 and whose values at the end points x =a and x =x1 are f(a+) and — ), respectively. The remaining integrals on the right in equation (2) are similarly defined and therefore exist. EXAMPLE 3. If f is the function in Example I and Fig. 15, then I f(x)dx=fxdx+f 3 0 1 0 1 2 3 2 1 1 2 2 Observe that the value of the integral of f(x) over each subinterval is unaffected by the values of f at the end points. The function is, in fact, not even defined at x = 0, 2, and 3. If two functions f1 and f2 are each piecewise continuous on an interval a < x < b, then there is a finite subdivision of the interval such that both functions are continuous on each closed subinterval when the functions are given their limiting values from the interior at the end points. Hence a linear combination c1f1 + c2f2, or the product f1f2, has that continuity on each subinterval and is itself piecewise continuous on the interval a < x c b. Consequently, the integrals of the functions c1f1 + c2f2, f1f2, and [f1(x)]2 all exist on that interval. Since any linear combination of functions that are piecewise continuous also has that property, we may use the terminology at the beginning of this section and refer to the class of all piecewise continuous functions defined on an interval a c x < b as a function space; we denote it by b). It is analogous to three-dimensional space, where linear combinations of vectors are welldefined vectors in that space. In Sec. 11, we shall extend the analogy by developing the concept of inner products of functions in b). Other function spaces occur in the theory of Fourier series. An especially important subspace of b) will be introduced in Sec. 17. More advanced texts treat the space of all integrable functions f on an interval a < x < b whose products, including squares [f(x)]2, are integrable. Then a more general type of integral, known as the Lebesgue integral, is often used. 42 CHAP. 2 FOURIER SERIES Our treatment of Fourier series involves more elementary concepts in mathematical analysis. Except when otherwise noted, in this book we shall restrict our attention to functions that are piecewise continuous on all bounded intervals under consideration. When it is stated that a function is piecewise continuous on an interval, it is to be understood that the interval is bounded; and the notion of piecewise continuity clearly applies regardless of whether the interval is open or closed. INNER PRODUCTS AND ORTHONORMAL SETS 11. Let f and g denote any two functions that are continuous on a closed bounded interval a x b. Dividing that interval into N closed subintervals of equal length Ax = (b — a)/N and letting Xk denote any point in the kth subinterval, we recall from calculus that when N is large, fbf(x)g(x) the symbol Ax, here denoting approximate equality. That is, & (1) kl where ak=f(xk){&i and The left-hand side of expression (1) is, then, approximately equal to the inner product of two vectors in N-dimensional space when N is large. The approximation becomes exact in the limit as N tends to infinity.t This suggests defining an inner product of the functions f and g: (2) (f, g) = ff(x)g(x) The integral here is, of course, also well-defined when f and g are allowed to be piecewise continuous on the fundamental interval a < x < b. Equation (2) can, therefore, be used to define an inner product of any two functions f and g b), introduced in Sec. 10. in the function space b), with inner product (2), is analogous to The function space ordinary three-dimensional space. Indeed, the following counterparts of familiar properties of vectors in three-dimensional space hold for any functions f, g, t See the book by Lanczos (1966, pp. 210ff), listed in the Bibliography, for an elaboration of this idea. INNER PRODUCTS AND ORTHONORMAL SETS SEC. I and h in 43 b): (f,g)=(g,f), (f,g+h)=(f,g)+(f,h), (3) (4) (cf,g) =c(f,g), (5) where c is any constant, and (f,f);O. (6) The analogy is continued with the introduction of the norm (7) IfU = (f,f)l/2 b). It is evident from equation (2) that the norm of f of a function f in can be written I (8) IIflI = b [f(x)] 2dx) The norm of the difference of two functions f and g, If - (9) = {fb[f() - g(x)}2 dx) 1/2 is a measure of the area of the region between the graphs of y = f(x) and y = 4x) (Fig. 16). To be specific, the quotient hf — g112/(b — a) is the mean, or average, value of the squares of the vertical distances If(x) — g(x)I between points on those graphs over the interval a < x < b. The quantity hf — gM2 is called the mean square deviation of one of the functions f and g from the other. y 0 b a Two functions f and g in x FIGURE16 b) are orthogonal when (f,g) =0, or (10) ff(x)g(x)dx=0. 44 FOURIER SERIES CHAP 2 the function f is said to be normalized. We have carried our analogy too far to preserve the original meaning of the geometric terminology. The orthogonality of two functions f and g signifies nothing about perpendicularity, but instead that the product fg assumes both positive and negative values on the fundamental interval in such a manner that equation (10) holds. Also, if IIfM = 1, A set of functions a<x < (n = 1, 2, ) is orthogonal on an interval 0 when m # n. Assuming that none of the functions i/i,, b if 0km' . . . has zero norm (see Problem 7), one can normalize each of them by dividing it by the positive constant The new set so formed, where (11) = ifi(x) (f/n is orthonormal on the fundamental interval; that is, (m=1,2,...;n=1,2,...), (12) where Kronecker's 5. Written in full, the characterization (12) of an orthonormal set becomes whenm#n, whenm=n. çb\ (0 dx = ç 4)mLt)&(.1) Ja (13) EXAMPLE 1. From the trigonometric identity 2sinAsinB = cos(A —B) — +B), cos(A we know that 1 sin mx sin nx = cos (m — n)x 1 cos — (m + n)x, where m and n are positive integers. It is thus easy to verify that (14) (sinmxsinnxdx= .10 0 whenm*n, — whenm=n. 2 Evidently, then, the set of sine functions (n = 1,2,...) = sinnx (15) is orthogonal on the interval of 0 < x < 'w ; and the norm V'n U of each of these functions is . Hence the corresponding orthonormal set {Øn(x)} consists of the functions (16) = (n = 1,2,...). iIYr — It is sometimes more convenient to index an infinite orthogonal or orthonormal set by starting with n = 0, rather than n = 1 This is the case in the following example. Verification that the given set is orthonormal is left to the . problems. SEC 1 PROBLEMS 1 45 EXAMPLE 2. The functions 1 = constitute a set r (n = = , 0, 1, 2, . . . ) Ii cosnx (n = V 1,2,...) that is orthonormal on the interval PROBLEMS 1. (a) Use the trigonometric identity 2cos to A cos B = cos(A — B) + cos(A + B) show that if m and n are positive integers, fcosmxcosnxdx= 0 0 whenm*n, — whenm=n. 2 (b) With the aid of the integration formula obtained in part (a), verify that the set {q5,7(x)} (n = 0, 1, 2, . . . ) in Example 2, Sec. 11, is orthonormal on the interval Suggestion: Note that, in order to establish orthogonality in part (b), it is necessary to show that = 0 and = 0 (m * n) for positite integers m and n. 2. (a) Use the fact that the functions (16) in Example 1, Sec. 11, constitute an orthonormal set on the interval 0 < x < to show that the functions 'n- (n=1,2,...) a set that is orthonormal on the interval — ii- < x < in (b) Use the fact that the set in Example 2, Sec. 11, and Problem 1 is orthonormal on the interval 0 <C x < ir to show that the functions form 1 ØQ(X) fl_, y2ir 1 (n=1,2,...) \f1T form an orthonormal set on the interval — ir < x < ii-. (See the suggestion with Problem 1.) Suggestion: Observe that if f is an even integrable function, one where f(—x) =f(x), then rJ(X) iv = 2ff(x) dx since the graph of y = f(x) is symmetric with respect to the y axis. 9J, SZIRHS jo at.p PW = '0 U) 'z 'T (,)0p . . = IVHD ' pournqo ( 8UflS!SU03 Jo T UO S! °qi IEMOIU! — it > T itA .LtA > X T = 'wqt OWN U! MO!A JO OUO imp paoU AIU0 or qsijqinso ApjrUo8oqpo U! {(x)"4} SUOflOUflJ = (x)J_UQp '__J 'z Moqs U! z wq, :uousdXXnS f J! Ur ppo ojqrJ8aw! 'UOUOUflJ S! oJoqM OUO U0qJ qdiu8 jo a = (x)f S! podsai o:j ot.p U!8!JO wqi oqi SUOflOUflJ (x)tm = '[ (x)Z4i = x an jEUO8OqpO UO aqj > X > 'J OU!mJOpp SJUEISUOO V g qons aqr UOflOUflJ OOU!S — J ('YIP = S! jEUo8oqpo c osoddnS pEqp o:j V = '0 U = + Zn! 'Eqi IwapU! zib UO •E— SflOflUflUOO SUOflOUflJ OAVJ rv+ 11 I UO oqj IrMOIU! quM OAfl!SOd '(x)f (x)'m AjJEOUq IUOPUOdOPU! UO UE IEAJOIU! V JEOU!j UOUEU!qWOD Sq 8U!URUJOpp jEUo8OqpO WMOW! V f > 'S! OUO S! JOU E JUEISUO3 + Tf/iv Jo SUOUOUflJ X > 'q air 'SULIOU UE sawn S! J!Ed 'Zif? (x)Zm (x)J= _ (x)'m sap Uo!ssaJdxa oords 'ç osoddns •9 UI aqj xusoo=(x)'m u S! U! '14j jEMOJU! S! pur wqi it— > :r > 'ruuis+rusoo=(x)f pun uonounj (x)Z4? oioqj sujn:j poxg oAqLsod o:j aq urs •ru piOM? ps U! L 4TPSflr (v) S! Aur jnuo8oqpo uo oqi IrMOlU! — IL > n (flf = o jdooxo Ajqissod uaqi lift = •0 (q) 'AjosJaAuoJ B Ilfil = '0 V > x > SJU!Od U! aqj L1 > SUOfl3UflJ 8U!MOjjOJ 'IL, f jo S1U!Od (flJ = •q 0 0q1 :iorj :irq:i oqi Aq U! :(q oords U! oqr IEAJaIU! n V > x > 'q opug iaqmnu Jo SEC 12 GENERALIZED FOURIER SERIES 47 Suggestion: In part (b), use the fact that a definite integral of a nonnegative continuous function over a closed bounded interval has positive value if the function has a positive value somewhere in that interval. 8. Verify that, for any two functions f and g in the space b), 1 bb [f(x)g(y) - g(x)f(y)]2 dxdy = ': 11f11211g112 (f, g)2. Thus establish the Schwarz inequality I(f,g)I c IIfIHIgII, which is also valid when f and g denote vectors in three-dimensional space. In that case, it is known as Cauchy's inequality. 9. Let f and g denote any two functions in the space b). Use the Schwarz inequality (Problem 8) to show that if either of these functions has zero norm, then (f,g) = 10. Prove 0. that if f and g are functions in the space C IIfII + ugh. If f and g denote, instead, vectors in three-dimensional space, this is the familiar triangle inequality, which states that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides. Suggestion: Start the proof by showing that hf + ghh2 = hlfhh2 + 2(f, g) + !hglh2, and then use the Schwarz inequality (Problem 8). 12. GENERALIZED FOURIER SERIES Let f be any given function in b), the space of piecewise continuous functions defined on an interval a c x < b. When an orthonormal set of functions b) is specified, it may be possible to (n = 1, 2, ) in represent f(x) by a linear combination of those functions, generalized to an . . . infinite series that converges to f(x) at all but possibly a finite number of points in the fundamental interval a < x < b: (1) (a<x<b). f(x)= n=1 This is analogous to the expression for any vector in three-dimensional space in terms of three mutually orthogonal vectors of unit length, such as i, j, and k. In order to discover an expression for the coefficients c,, in representation (1), if such a representation actually exists, we use the index of summation m, rather than n, to write (2) f(x)= (a<x<b). m=1 We also assume that after each of the terms here is multiplied by a specific 48 FOURIER SERIES CHAP. 2 the resulting series is integrable term by term over the interval a < x < b. This enables us to write = dX, m=1 a a or (3) = m=1 But (4am' 4'n) = 0 for all values of m here except when m = n, in which case and is = 1. Hence equation (3) becomes (f, = (4'm' = evidently the inner product of f and 4,,. As indicated above, we cannot be certain that representation (1), with coefficients = (f, 4,,), is actually valid for a specific f and a given orthonorHence we write mal set f(x) (4) where the tilde symbol (a < x < ' merely denotes correspondence when (5) (n=1,2,...). To strengthen the analogy with vectors, we recall that if a vector A in three- dimensional space is to be written in terms of the orthonormal set {i,j, k} as A=a1i+a2j+a3k, the components can be obtained by taking the inner product of A with each of the vectors of that set. That is, the inner product of A with i is a1, etc. The series in correspondence (4) is the generalized Fourier series, with respect to the orthonormal set (4j, for the function f on the interval a < x < b. The coefficients c,, are known as Fourier constants. EXAMPLE. Let f denote any function in the space from Example 2, Sec. 11, that the set (n = 0, 1, 2, . . 'w). We know . ) consisting of the functions I (6) is (n=1,2,...) , orthonormal on the interval 0 < x < summation starting from n = f(x) 0, and correspondence (4), with the becomes c0 i_ VP;,. °° +E n=1 c ir), SEC. 12 GENERALIZED FOURIER SERIES 49 where co= 1-i-IT 1 \f,n_ vii- 0 (n=1,2,...). 0 By writing 1-i- 2 (n=1,2,...), v,;T V 'IT we thus arrive at the Fourier cosine series correspondence (7) where (n=O,1,2,...). (8) Fourier cosine series will be developed further in the next section. The generalized Fourier series that we shall encounter will always involve orthonormal sets and functions f in a space of the type b) or subspaces of it, and we say that representation (1) is valid for functions f in a given space if equality holds at all but possibly a finite number of points x in the fundamental interval a < x < b. Representation (1) will not, however, always be valid even in very restricted function spaces. We may anticipate this limitation by considering vectors in three-dimensional space. For if only the two vectors i and j make up the orthonormal set, any vector A that is not parallel to the xy plane fails to have a representation of the form A = a1i + a2j. In particular, the nonzero vector k is orthogonal to both i and j, in which case the components a1 = k j and a2 = k j would both be zero. Similarly, an orthonormal set may not be large enough to write a generalized Fourier series. To be specific, if the function f(x) in correspondence (4) is orthogonal to each function in the orthonormal set then the Fourier constants = (f, are all zero. This means, of course, that the sum of the series is the zero function. Consequently, if f has a positive norm, the series is not equal to f(x) at all but possibly a finite number of points in the fundamental interval [see Problem 7(a), Sec. 11]. An orthonormal set is closed in b), or a subspace of it, if there is no function in the space, with positive norm, that is orthogonal to each of the Thus, according to the preceding paragraph, if an orthonormal set is not closed, then representation (1) cannot be valid for each function fin the space. In Sec. 20, we shall identify a certain subspace of 'iv-) such that series functions (7) is valid when f is in that subspace. Note that if the function 40(x) is not included with the other functions (n = 1, 2, . ) in the set (6), the . . 50 FOURIER SERIES CHAP. 2 resulting set is not closed in the subspace since 40(x) is orthogonal to each of the functions in that smaller set. Hence the term a0/2 is nSded, in general, for Fourier cosine series representations to be valid in the subspace. 13. FOURIER COSINE SERIES In the example in Sec. 12, we introduced the concept of a Fourier cosine series corresponding to a function f(x) in 'w): a0 f(x)-'.'—+ 2 (1) where n=1 2r (2) (n=O,1,2,...). The fact that f is piecewise continuous on the interval 0 < x < ensures the As already existence of the integrals in expression (2) for the coefficients noted at the end of the preceding section, we shall, in Sec. 20, establish further conditions on f under which the cosine series actually converges to f(x) when 0 < x < 'IT, in which case correspondence (1) becomes an equality. Observe that correspondence (1), with coefficients (2), can be written more compactly as (3) f(x) 1 —ff(s) ds 2°° + — E cos nxff(s) cos nsds, 0 where s is used for the variable of integration in order to distinguish it from the free variable x. If f is defined on the interval 0 x c and series (1) converges to f(x) for all x in that interval, the series also converges to the even periodic extension, of f on the entire x axis. That is, it converges to a function with period F(x) having the properties F(x)=f(x) (4) and (5) F(—x)=F(x), forallx. The reason for this is that each term in series (1) is itself even and periodic with period 2in The graph of the extension y = F(x) is obtained by reflecting the graph of y = f(x) in the y axis, to give a graph for the interval — ir c and then repeating that graph on the intervals r c x c 3'n-, 3r c x c 5'n-, etc., —3w, etc. It follows as well as on the intervals —'w, —5r c x cx from these observations that if one is given a function f that is both even and periodic with period 2'w, then the cosine series corresponding to f(x) on the interval 0 < x < represents f(x) for all x when that series converges to it on FOURIER COSINE SERIES SEC. 13 51 the interval 0 c x c 'in Clearly, a cosine series cannot represent a function f(x) for all x if f(x) is not both even and periodic with period 2r. EXAMPLE. Let us find the Fourier cosine series for the function f(x) = sin x on the interval 0 < x < 'ir. The trigonometric identity 2sinAcosB = sin(A +B) + sin(A —B) enables us to write an=—f sinxcosnxdr lIT =—f[sin(l+n)x+sin(l—n)x]dr (n=O,l,2,...). Hence, when n * I, n l+n 'IT and when n = 1, I—n 1—n2 the coefficient is ai=—f sin2xcfr=O. Correspondence (1) then becomes 2 2 °° sinx"—+—E 2 'IT cosnx (O<x<'w). = 0 when n is odd and that this series can be + (— written more efficiently by summing only the terms that occur when n is even. I This is accomplished by replacing n by 2n wherever n appears after the summation symbol and starting the summation from n = I . The result is 2 4 °° sinx"———E (6) 'IT cos2nx 2 —1 (O<x<'w). The function sin x will, in fact, satisfy conditions in Sec. 20 ensuring that the correspondence here is an equality for each value of x in the interval 0 c x C 'iT. Thus, at each point on the x axis, the series converges to the even periodic extension, with period 2'ir, of sin x (0 in Fig. 17, is the function y = I sin x I. x ii-). That extension, shown 3, % a eS y = sin xl. FIGURE 17 52 FOURIER SERIES 14. CHAP. 2 FOURIER SINE SERIES We saw in Example 1, Sec. 11, that the sine functions = constitute — Y';r (n = 1,2,...) sinnx an orthonormal set on the interval 0 < x < 'ir. The generalized Fourier series (Sec. 12) corresponding to a function f(x) in f(x) sin ir) is (0 < x < where (n=1,2,...). Upon writing IT yew we have the Fourier sine series correspondence (1) where (n=1,2,...). (2) The correspondence can, of course, also be written (3) fGr) 2°° — i: sinnxff(s)sinnsds. 0 Suppose that f is defined on the open interval 0 c x < ew and that series (1) converges to f(x) there. Since series (1) clearly converges to zero when x = 0 and x = 'w, it converges to f(x) for all x in the closed interval 0 c x c if f is assigned the values f(O) = 0 and f('n-) = 0. Remarks similar to ones in Sec. 13, regarding cosine series, show that series (1) then converges to the odd periodic extension, with period 2'w, of f for all values of x. This time, the extension is the function F(x) defined by the equations (4) F(x)=f(x) and (5) F(—x) = —F(x), F(x+2'w) =F(x) forallx. The extension F is odd and periodic with period 2ew since the terms sin at in series (1) have those properties. The graph of y = F(x) is symmetric with SEC 14 FOURIER SINE SERIES 53 respect to the origin and can be obtained by first reflecting the graph of y = f(x) in the y axis, then reflecting the result in the x axis, and finally r repeating the graph found for the interval — 'w x c every 2'w units along the entire x axis. Evidently, a Fourier sine series on the interval 0 < x < 'w can also be used to represent a given function that is defined for all x and is both odd and periodic with period 2'w, provided the representation is valid when EXAMPLE 1. To find the sine series corresponding to the function f(x) = x on the interval 0 < x c we refer to expression (2) for the coeffiand use integration by parts to write cients 2 21 xcosnx ITO 'w[ ii (1)fl+1 sinnxlr + n Thus x-'.'2L (6) (0 <x <'ir). sinnx n=1 Our theory will show that the series actually converges to f(x) when 0 c x < 'iT. Hence it converges to the odd periodic function y = F(x) that is graphed in Fig. 18. The fact that the series converges to zero at the points x = 0, ± ± 3r, ± 5w, . is in agreement with our theory, which will tell us . . that it must converge to the mean value of the one-sided limits of F(x) at each of those discontinuities. y ,p p ,t4# -in. / // 'p ITT x FIGURE 18 In the evaluation of integrals representing Fourier coefficients, it is sometimes necessary to apply integration by parts more than once. We now give an example where this can be accomplished by means of a single formula due to 54 FOURIER SERIES CHAP. 2 L. Kronecker (1823—1891). We preface the example with a statement of that formula.t Let p(x) be a polynomial of degree m, and suppose that f(x) is continuous. Then, except for an arbitrary additive constant, (7) =pF1 —p'F2+p"F3 — p is successively differentiated until it becomes zero, where F1 denotes an indefinite integral of f, F2 an indefinite integral of F1, etc., and where alternating signs are affixed to the terms Note that the differentiation of p begins with the second term, whereas the integration of f begins with the first where term. The formula, which is readily verified by differentiating its right-hand side to obtain p(x)f(x), could even have been used to evaluate the integral in Example 1, where only one integration by parts was needed. EXAMPLE 2. To illustrate the advantage of formula (7) when successive integration by parts is required, let us find the Fourier sine series for the function f(x) = x3 on the interval 0 < x < 'n-. With the aid of that formula, we may write 2 =; (x3)(- =2(_1)n+1 cos ía )_(3X2)(_ n (nir)2 — sin nx n2 )+(6x)( cos ía sin nx )_(6)( ) 6 (n=1,2,...). n3 Hence 00 (8) x 3 n=1 n+i\ \26 I ii 3 sinnx (O<x<w). As was the case in Example 1, the series converges to the given function on the interval 0 < x < 'ii-. Since is an odd function whose value is zero when 3 x = 0, this series represents on the interval — ii- < x < 'ii- also. We conclude this section by pointing out a computational aid that is useful in finding the coefficients (n = 1, 2, . . . ) in the Fourier sine series for a linear combination c1f1(x) + c2f2(x) of two functions f1(x) and f2(x) whose sine t Kronecker actually treated the problem more extensively in papers that originally appeared in the Berlin Sitzungsberichte (1885, 1889). PROBLEMS SEC. 14 55 series are already known. Namely, since the expression [c1f1(x) + c2f2(x)] = can be written as = it sin nxdr ci—f f1(x)sin nxdx + c2—f f2(x)sin nxdx, is simply the same linear combination of the nth is clear that each coefficients in the sine series for the individual functions f1(x) and f2(x). Such an observation applies as well in finding coefficients in cosine and other types of series encountered in the present and later chapters. EXAMPLE 3. In view of the sine series for x and x3 found in Examples I in the sine series corresponding to the and 2, respectively, the coefficients function (O<x<r) are I 2(1)fl+1 —=12 n (n=l,2,...). Thus 00 x('n-2—x2)".'12E (9) sinnx 3 n=1 PROBLEMS Find (a) the Fourier cosine series and (b) the Fourier sine series on the interval 0 c x < iT that corresponds to each of the functions in Problems 1 through 4. Lf(x)=1 (O<x<,r). Answers: (a) 4 1; (b) — 00 sin(2n—1)xt 2n—1 2.f(x)=r—x (O<x<w). 'iT Answers: 4 (a)— + — 2 cos(2n—1)x (2n—1)2 °° ; sinnx (b)2 n=1 are zero when n is even. The index n in the series can, therefore, be the coefficients wherever it appears after the summation symbol. (Compare the example in replaced by 2n — I Sec. 13.) part(b) 56 FOURIER SERIES CHAP. 2 IT 1 2' 0 when—<x<IT. 3.f(x)= IT I cos(2n—1)x 2n—1 °° 2 (_1)n4.1 Answers: (a) — + — 2 2 (b) — fl'TT\SiflflX I °° L ii 4.f(x)=x2 (O<x<r) cr IT2 Answers: (a) — + 4 3 2i — (—if 2 n=1 n COS ía; n+1 00 (b)2IT —2-- fliT n=1 _ —1 SlflflX. (nr) 5. By referring to the sine series for x in Example I, Sec. 14, and the one found for x2 in Problem 4(b), show that 8 °° sin(2n—i)x n=1 (2n — 6. What is the Fourier sine series corresponding to the function f(x) = interval 0 < x < IT? (O<x<IT). IT sin x on the Suggestion: To find the coefficients in the series, refer to the integration formula (iO), Sec. ii. Answer sin x. 7. Find the Fourier cosine series for x on the interval 0 < x < IT. Then, given that the correspondence obtained is actually an equality when 0 c x c IT, point out how it follows that 4 °° cos(2n—1)x IT IxI=———E 2 8. Show that (nIT)2 IT4 5 (—IT!cxcIT). (2n—i) 2 — (O<x<IT). n n=1 Given that this correspondence is actually an equality when 0 c x c IT, sketch the function represented by the series for all x. 9. Verify Kronecker's formula (7), Sec. 14.' 10. Let (n = i, 2, . . ) denote an orthogonal, but not necessarily orthonormal, set on a fundamental interval a c x 'C b. Show that the correspondence between a piecewise continuous function f(x) and its generalized Fourier series with respect to the orthonormal set of functions = (n = 1, 2, . ) can be written . . . f(x) ii. i: n=1 where = ' 114//nil In the space of continuous functions on the interval a c x c b, prove that if two functions I and g have the same Fourier constants with respect to a closed FOURIER SERIES SEC. 15 57 then f and g must be identical. Thus show that f is orthonormal set uniquely determined by its Fourier constants. Suggestion: Show that the norm of the difference f(x) — g(x) is zero. Then point out how it follows that f(x) — g(x) 0 (see the suggestion with Problem 7, Sec. 11). and write f(x) = g(x) + h(x), where g and h 12. Let f be a function in are defined by the equations f(x) +f(-x) g(x)=— and 2 f(x) -f(--x) h(x)=— 2 The functions g and h are evidently even and odd, respectively, on the interval —iT<X<7r. (a) Explain why it is reasonable to expect that f(x) = a0 -;;-- (—ir<x <ir), + n=1 (n = 0, 1, 2, ) are the coefficients in the Fourier cosine series for g(x) on the interval 0 < x < ir and (ii = 1, 2, . ) are the coefficients in the where . . . . . Fourier sine series for h(x) on that same interval. (b) Show that the coefficients and in part (a) can be written in the form ir ir bn;ff(x)sinnxdX (n=O,1,2,...), f(x)cosnxdx (n=1,2,...). of the type introduced here are discussed in the next section.) Suggestion: In part (b), write (Series 1 r ff(x)cosnxdx+ff(—s)cosnsds IT 0 0 then make the substitution x = are to be treated similarly. and 15. —s in the second integral here. The coefficients FOURIER SERIES In Problem 3, Sec. 1 1, we found that the functions (1) 40(x)= 1 1 1 rsinnx (n=1,2,...) form an orthonormal set on the fundamental interval — alized Fourier series corresponding to a function f in + E[c2fl_102fl_l(x) < x < r. The gener— ir, 'r) is, therefore, SHflHS dVHD z '(L>x>L-_) (z) J IL LZA It- = T_UZ3 IL rprusoo(x)J f—'ILLA LU 17?XUU!S(X)J Lit 05 J! LU- OM 02!JM 1_Uza '°d—ft=°V LA (c) _I ' UZd —a -Ltft Uq .LtA ooUopUodsauoo (Z) souioooq oJOqM IUJ (sO J_="v xprusoo(x)J J_Uq 1731UU!S(X)f ,, IL— pUt? (9) ,, sougs IEAJO2U! w > x > .Lt = (flg UO!20UflJ — qdn2 '(9) S! 0q3 RLIflOJ SdlJdS7 — it > X > SO!JOS osOddng U! MO!A jo 0q2 J0 S2! 'SWJ02 3! s2u0!oqJ003 (x)J UO 0q2 E L"z'i=u) -IL— eC 8! SOp!OU!OO ippvi £JOAO AT = (x)J s2!un ywn poydcl 'itt Jo 'oJojoJoq2 0q2 d!poudd so!Jos UOA!2 znpouod 'uouounj qpii& PO!JOd 2! SO2JOAUOO oi (x)J uo 0q2 jThUO3U! — It ) L1 (x)J SO&IOAUOO 03 UO jRAJO3U! pUt? X SjXE oqj UOflDUflJ j 'jj uo oqi (x)f ) •.Lt oSoqM 'S! 'punj J S! OJOqMAJOAO SEC 15 f(x) FOURIER SERIES 59 EXAMPLE 1. Let us find the Fourier series corresponding to the function which is defined on the fundamental interval —r C x c 'w as follows: when —iT < X 10 f(x) = (7) 0, when 0<x<w. The graph of y = f(x) is indicated by bold line segments in Fig. 19. y /'p IT aa p 0 -2IT / • aaa 2ir IT 3ir 4i,' /, x FIGURE 19 According to expression (5), + = = 0lTx I cos at tir) lIT (n=0,l,2,..). —I xcosnxdr By applying integration by parts, or Kronecker's method (Sec. 14), one can show that -1 'lTfl when n = 1, 2 To avoid division by zero, we must evaluate the integral for 2 a0 separately: IT a0=—J xdr=—. 2 Expression (6) tells us that = i(f0osinnntr 1 çrr =—I xsinnxdx= for all positive integers n = IT (8) •1' 1, 2 n Hence, on the interval — 'w < x < IT, — E + I (_l)n+1 cosnx + sinnx n=1 [ This series will be shown to converge to f(x) on the fundamental interval, as well as to the periodic extension F(x) that is indicated in Fig. 19, where the 60 FOURIER SERIES CHAP. 2 graph of y = F(x) is sketched. As in Example 1, Sec. 14, the series must converge to the mean value of the one-sided limits of the periodic extension at each of the discontinuities x = ± 'ii-, ± Here the mean values are ± 5r all 'ir/2. • —IT It may be that the given function f in — 'ii-, r) is even on the interval < X < 'ir; that is, f(—x) = f(x) for all such values of x. Then f(—x)cos(—nx)=f(x)cosnx (n=O,1,2,...) and when —'7T <x < f(—x)sin(—nx) = —f(x)sinnx (n = 1,2,...) and we see that f(x)cosnx and f(x)sinnx are even and odd, respectively. Hence expressions (5) and (6) reduce to (n=O,1,2,...) = 0 (n = 1, 2, . ) (see the suggestions with Problems 2 and 3, Sec. 11). Series (4) thus becomes a Fourier cosine series (Sec. 13) for f(x) (0 < x < r). Similarly, if f is odd on the interval — < x < it follows from expressions (5) and (6) that = 0 (n = 0, 1, 2, . ) and and . . . . (n=1,2,...). In this case, series (4) becomes a Fourier sine series (Sec. 14) for f(x) (O<x<'i,). EXAMPLE 2. The function f(x) = Isin xl (—ir < x < r) is even. Hence the Fourier series corresponding to f(x) on the interval — ir < x < r is actually the Fourier cosine series for the function f(x)=IsinxH=sinx (O<x<'w). That series has already been found in the example in Sec. 13; and, rewriting correspondence (6) there, we see that 2 4 °° cos2nx 2 'IT —1 EXAMPLE 3. Since the function f(x) = x (—'ii- < x < 'w) is odd, the Fourier series for f on — c x < is simply the Fourier sine series for that function on 0 < x < 'in Hence correspondence (6) in Example 1, Sec. 14, is also a correspondence on the larger interval — < x < 00 n=1 n sinnx (—r<x<'i,). SEC 16 BEST APPROXIMATION IN THE MEAN 61 Similarly, correspondence (8) in Example 2, Sec. 14, can be written x n+1 3 I Ii n=1 3 sinnx (—'ir<x<r). Correspondence (4), when combined with expressions (5) and (6) for the and lip, becomes constants I f(s)ds +— E —IT —IT The trigonometric identity cos(A —B) =cosAcosB + sinAsinB then enables us to write the correspondence in the form 1 Lff(s)cosn(s-x)ds. (9) IT IT IT Note that the term 1 IT f(s)ds here, which is the same as the term a0/2 in series (4), is the mean, or average, value of f(x) over the interval —'w < x < r. Form (9) of correspondence (4) will be the starting point of the proof in Sec. 19 of our theorem ensuring the convergence of the Fourier series to f(x) on the interval —r < x < 'ir. 16. BEST APPROXIMATION IN THE MEAN Let SN(x) (N = function f in 1, 2, . . . ) denote partial sums of the Fourier series for a 'ir): a0 N (—'n-<x<n-). (1) n=1 We consider here the matter of approximating the function f by these partial sums. While the main result of this section is of interest in itself, it yields a and property of the coefficients that we shall need in our treatment of 62 CHAP 2 FOURIER SERIES convergence of Fourier series in the next several sections. Namely, (2) and n—)oo n—)oo As is often the case, the discussion is simplified by first treating any orthonormal set (n = 1, 2, . . ) on a fundamental interval a < x < b. of that set, and we We consider the first N functions 41(x), 42(x), . , . . . denote any linear combination of them: let = y141(x) + y242(x) + . (3) . . The norm {fb[f() - If- (4) 1/2 a measure of the deviation of the sum 1N from a given function f in b) in expression (3) that (see Sec. 1 1). Let us determine values of the constants or the quantity make If — is fb[f() E = If - tNII (5) tN(x)]2 as small as possible. The nonnegative number E represents the mean square error in the approximation by the function 'FN to the function f; and we seek the best approximation in the mean.t We start with the observation that 2 N (f— = (f— n=l) N =f2 2 N — N N N E EE n=1 m=1 t The approximation sought here is also called a least squares approximation. 2 SEC. 16 BEST APPROXIMATION IN THE MEAN 63 and this enables us to write (f-tN)2=f2+ n=1 1( m=1 Integrating each side here over the interval a < x < b and then using the relations 8mn and (f, çb,,) = c,,, where i5 Kronecker's 5 (Sec. 11) and the are Fourier constants (Sec. 12), we arrive at the following expression for the error E, defined above: E= MfM2 + - n=1 That is, E= (6) IlfM2 + In view of the squares in the first summation appearing in equation (6), the smallest possible value of E is, then, obtained when = (n = 1, 2, . , N), . . that value being N (7) E=11f112— n=1 We state the result as a theorem. (n = be the Fourier constants for a function f in b) with respect to an orthonormal set (n = 1, 2, . ) in that space. Then, ofalipossible linear combinations of the functions cfr1(x), 42(x), . . , Theorem. Let 1, 2, . . . ) . . . the combination c141(x) + c242(x) + . . +cN4N(x) . is the best approximation in the mean to f(x) on the fundamental interval a < x < b. In that case, the mean square error E is given by equation (7). This theorem is analogous to, and even suggested by, a corresponding result in three-dimensional space. Namely, suppose that we wish to approximate a vector A = a1i + a2j + a3k by a linear combination of just the two basis vectors i and j. If we interpret A and any linear combination a1i + a2j as radius vectors, it is geometrically evident that the shortest distance d between their tips occurs when a1i + a2j is the vector projection of A onto the plane of i and j. That projection is, of course, the vector a1i + a2j (see Fig. 20), the components a1 and a2 being the inner products of A with i and j, respectively. 64 FOURIER SERIES CHAP. 2 A k d 0 I a1i + a2j a1i + a2j FIGURE 20 Corollary. If (n = 1, 2, . . . ) are the Fourier constants for a function f in b) with respect to an orthonormal set (n = 1, 2, . ) in that space, . . then (8) 0o n The proof of this corollary is based on Bessel's inequality N (9) n=1 which is an immediate consequence of expression (7) for the mean square error E and the fact that E 0. We observe that the right-hand side of Bessel's inequality is independent of the positive integer N; and as N increases on the left-hand side, the sums there form a sequence that is bounded and nondecreasing. Since such a sequence must converge and since this particular sequence is the sequence of partial sums of the series whose terms are (n = 1, 2, . ..), that series must converge. Limit (8) now follows from the fact that the nth term of a convergent series always tends to zero as n tends to infinity. We recall from Sec. 15 that when the orthonormal set of functions 1 40(x) = 1 = v&' 1/; 1 = cosnx, sinnx (n=1,2,...) in — 'ir, 'ir) is used, the generalized Fourier series 00 (10) L 00 = c040(x) + + n=1 n=O (—'ir<x<w) corresponding to a function f in (11) a0 —;;- + L n==1 — r, 'ir) is the ordinary Fourier series (—jr < x < SEC. 16 PROBLEMS 65 where C- I-i- 12) ( 2n = — C0, a0 = r 1 = C 2n (n %c = 1,2,...). The above theorem now tells us that, of all possible linear combinations of the functions 4)2(x), . . . , the partial sum N 2N E n=0 of — + = C0çb0(X) + n=1 series (10) is the best approximation in the mean to f on the interval < X < r. That is, the partial sum (1) is the best approximation of all linear combinations of the functions 1 cosnx, sinnx (n=1,2,...,N). The corollary, together with relations (12), also yields limits (2). Those limits are, in fact, valid when and are the coefficients in the Fourier cosine and sine series, respectively, for a function f in 'rr). To see that the coefficients in the cosine series tend to zero as n tends to infinity, we need only observe that the cosine series is the same as the Fourier series on — < x < for the even extension of f onto the interval — c x < 0 (see Sec. 15). Similarly, the sine series can be thought of as the Fourier series for the odd extension of f onto — 'ii- < x < 0. Hence the coefficients in the sine series also tend to zero as n tends to infinity. PROBLEMS Find the Fourier series on the interval — ir < x c ir that corresponds to each of the functions in Problems 1 through 6. when 2 00 Answer: 2 n=1 sin(2n O<X<'ir. — 1)x 2n—1 t See the footnote with Problem 1(b), Sec. 14. 66 FOURIER SERIES CHAP. 2 2. f(x) is the function such that the graph of y = f(x) consists of the two line segments shown in Fig. 21. Answer: + 2 — 2 (_1)n+1 _ 00 3 costa + (n'ir) 2 n=1 nir sinnx y (ir,2) (0,2) x 0 (—ir,O) 1 FIGURE21 (—ir<x<,r). Suggestion: Use the series for x in Example 3, Sec. 15, and the one for x2 in Problem 4(a), Sec. 14. 12 2sinnx /cosrzx Answer: — + 2 n=1 n 4.f(x)=e" (—ir<x<w),wherea*O. Suggestion: Use Euler's formula e'° = + cos U + i sin 0, where i = lIT = 1 , to write (n = 1,2,...). —f Then, after evaluating this single integral, equate real and imaginary parts.t Answer 2sinhair 1 — 2a + (—if (acosnx—nslnnx) 5. f(x)= sinhax (—'ir<x <r). Suggestion: Use the series found in Problem 4. Answer 2sinhair IT (_1)n+1 n=1 n a+n 2 2 sinnx. 6.f(x)=cosax Suggestion: With the aid of Euler's formula, stated in Problem 4, write cosax= + 2 t For a justification of Euler's formula and background on complex-variable methods, see the authors' book (1990), listed in the Bibliography. ONE-SIDED DERIVATIVES SEC. 17 67 Then use the series already obtained in that earlier problem. Answer °° 2a sin air 1 IT 2a2 (_1)n+1 n=1 7. Find the Fourier series on the interval — ir < x < ir for the function f defined by the equations f(x) = {O sln x when —'ir c x c 0, when 0<xc'ir. Then, given that the series converges to f(x) when —ir x c ir, describe graphically the function that is represented by the series for all x ( — oc < x < cc). Suggestion: To find the series, write the function in the form f(x) = sinx+ Isinxl 2 (—ir Cr) and then use the results in Problem 6, Sec. 14, and Example 2, Sec. 15. cos2nx 2 1 1 Answer: — + — 2 2 —1 and denote the coefficients in the Fourier cosine and sine series, respec8. Let tively, corresponding to a function f(x) in ir). (a) By referring to the example in Sec. 12, obtain from Bessel's inequality (9), Sec. 16, the inequality N 2 2 (N=1,2,...). 2 o n=1 (b) By referring to Sec. 14, show how it follows from Bessel's inequality (9), Sec. 16, that N n=1 9. Show that when function f(x) in 2r[f(x)]2dx (N=1,2,...). 0 are the coefficients in the Fourier series corresponding to a and — ir, r) (Sec. 15), the inequality N 2 + ;f[f(x)]2dx 1 C (N=1,2,...) follows from Bessel's inequality (9), Sec. 16, for Fourier constants. 17. ONE-SIDED DERIVATIVES In developing sufficient conditions on a function f such that its Fourier series on the interval — ir < x < ir converges to f(x) there, we need to generalize the 68 FOURIER SERIES CHAP. 2 concept of the derivative f(x0)= tim x—)xo , (1) offatapoint . f(x) —f(x0) X—Xo x=x0. Suppose that the right-hand timit f(x0 + ) exists at x0 (see Sec. 10). The right-hand derivative, or derivative from the right, of f at x0 is defined as fottows: , (2) fR(xo) = . tim f(x) -f(x0+) X—Xo x—)xo x>xo provided the timit here exists. Note that, atthough f(x0) need not exist, f(x0 +) does. When the ordinary, or two-sided, derivative f'(x0) must exist if exists, f is continuous at x0; and it is obvious that fk(x0) = f'(x0). Simitarty, if f(x0 — ) exists, the left-hand derivative of f at x0 is given by the equation , (3) . fL(xo) = tim f(x) -f(x0-) X—Xo x—)xo x < x0 when this timit exists; and if f'(x0) exists, f equations denote the continuous function defined by the I whenxcO, whenx>O. x2 With the aid of t'Hospitat's rute, we see that = tim x—*O x>O sinx X = 1; = 0. Since these one-sided derivatives have different vatues, the ordinary derivative f'(O) cannot exist. furthermore, The ordinary derivative f'(x0) can fait to exist even when f(x0) is defined and and fL(x0) have a common value. EXAMPLE 2. If f is the step function (0 whenx<O, U = then fk(O) = continuous at x = 0. 0. But the derivative f'(O) does not exist since f is not SEC. 17 ONE-SIDED DERIVATIVES 69 As is the case with ordinary derivatives, the mere continuity of f at a point xo does not ensure the existence of either one-sided derivative there. 6 EXAMPLE 3. The function f(x) = (x 0) has no right-hand derivative at the point x = 0, although it is continuous there. A number of properties of ordinary derivatives remain valid for one-sided derivatives. If, for example, each of two functions f and g has a right-hand derivative at a point x0, then so does their product. A direct proof is left to the problems. But a proof can be based on the corresponding property of ordinary derivatives in the following way. We use f(x0 + ) and g(x0 + ) as the values of f and g at x0, and we define those functions when x c x0 as the linear functions represented by the tangent lines at the points (x0, f(x0 + )) and (x0, g(x0 +)) with slopes fk(x0) and respectively (Fig. 22). Those modifications of f and g are differentiable at x0, with derivatives that are also right-hand derivatives. Thus the derivative of their product exists there, and its value is also the right-hand derivative of f(x)g(x) at x0. f( x) g(x) / / -e (x0, g(x0 + )) (x0,f(x0 + 0 )) x 0 x FIGURE 22 exist, the left-hand derivative of the and Likewise, if product f(x)g(x) exists at x0. Finally, we turn to a property of one-sided derivatives that is particularly important in the theory of convergence of Fourier series. It concerns the b) consisting of all piecewise continuous functions f b) of on an interval a < x < b whose derivatives f' are also piecewise continuous on subspace that interval. Such a function is said to be piecewise smooth because, over the subintervals on which both f and f' are continuous, any tangents to the graph of y = f(x) that turn do so continuously. Theorem. If a function f is piecewise smooth on an interval a < x < b, then at each point x0 in the closed interval a c x b the one-sided derivatives of f, f rom the interior at the end points, exist and are the same as the corresponding one-sided limits off': (4) fk(x0) =f'(x0+), =f'(x0—). 70 CHAP. 2 FOURIER SERIES To prove this, we assume for the moment that f and f' are actually continuous on the interval a c x < b and that the one-sided limits of f and f' from the interior exist at the end points x = a and x = b. If x0 is a point in exist, and both are and that open interval, f'(x0) exists. Hence equal to f'(x0). Because f' is continuous at x0, then, equations (4) hold. The exists and is equal to following argument shows that it is also true that f'(a + ). If we let denote any number in the interval a < x < b and define f (a) as f(a + ), then f is continuous on the closed interval a c x c f. Since f' exists in the open interval a c x < , the mean value theorem for derivatives such that applies. To be specific, there exists a number c, where a < c < f(x*) -f(a+) =f'(c). x*_a (5) and therefore c, tend to a in equation (5), we see that since f'(a +) Letting exists, the limit of f'(c) exists and has that value. Consequently, the limit of the difference quotient on the left in equation (5) exists, its value being fk(a). Thus = f'(b —). fk(a) = f'(a + ). Similarly, any piecewise smooth function f is continuous, along with its derivative f', on a finite number of subintervals at whose end points the one-sided limits of f and f' from the interior exist. If the results of the Now preceding paragraph are applied to each of those subintervals, the theorem is established. The following example illustrates the distinction between one-sided derivatives and one-sided limits of derivatives. EXAMPLE 4. Consider the function f whose values are 1 f(x) = x whenx=0. 0 Since 0 x2sin(1/x)I f (0 — ) exist and have value x*0, when x # both one-sided limits f(0 + ) and zero. Moreover, since 0 c Ix sin(I/x)I c lxi when 0, 1 1 limxsin—=0 x—*O x>O X limxsin—=O. and x-'O x<O X But, from the expression f'(x) = 2x sin 1 — x I — cos — x (x*O), we see that the one-sided limits f'(O + ) and f'(O — ) do not exist. Note that although its one-sided derivatives exist everywhere, the function f is not piecewise smooth on any bounded interval containing the origin. Hence the theorem is not applicable to this function on such an interval. SEC. 18 TWO LEMMAS 71 18. TWO LEMMAS We begin our discussion of the convergence of Fourier series with two lemmas, or preliminary theorems. The first is a special case of what is known as the Riemann-Lebesgue lemma. That lemma appears later on in Chap. 6 (Sec. 53), where it is needed in full generality. Lemma 1. If a function G(u) is piecewise continuous on the interval then limfG(u)sin (1) (2N+1)u 2 N—ooo where N denotes positive integers. To prove this, we recall the trigonometric identity sin(A +B) = sinAcosB+ cosAsinB and write 'IT fG(u)sin (2N+1)u u ,Tr 2 IT =10 u u G(u)sin—cosNudu+f G(u)cos—sinNudu. 2 0 2 Now, except for a factor of 2/r, the first of these last two integrals is the coefficient aN in the Fourier cosine series for the piecewise continuous function G(u) sin (u/2) on the interval 0 < u < 'r. The other integral is, except for a factor of 2/r, the coefficient bN in the Fourier sine series for G(u) cos (u/2) on the same interval. Thus, according to the last paragraph in Sec. 16, the numbers aN and bN tend to zero as N tends to infinity; and Lemma I is established. Our second lemma involves the Dirichlet kernel 1 (2) where DN(u) = — + 2 N cosnu, N is any positive integer. Note that DN(u) is continuous, even, and periodic with period 2r. The Dirichlet kernel plays a central role in our theory, and two other properties will also be useful: (3) (4) DN(u)= sin [(2N + 1)u/2] 2sin(u/2) 72 FOURIER SERIES CHAP. 2 Property (3) is obvious upon integrating each side of equation (2). Expression (4) can be derived with the aid of a certain trigonometric identity (Problem 14, Sec. 20). Lemma 2. Suppose that a function g(u) is piecewise continuous on the interval 0 < u < 'ir and that the right-hand derivative exists. Then 'IT (5) du = —g(0 +), 2 lim f N-.ooo where DN(u) is defined by equation (2). To start the proof', we write jg(u)DN(u)du = 'N (6) where fg(0+)DN(u)du. and 'N= f[g(u) In view of expression (4), the first of these two integrals can be put in the form 1ng(u)—g(0+) 'N= Jo 2sin(u/2) (7) . (2N+1)u . sin G(u)= g(u) —g(0+) 2sm(u/2) I 2 du. Observe that the function is a quotient of two functions that are piecewise continuous on the interval 0 < u < 'in Although the denominator vanishes at the point u = 0, the existence of ensures the existence of G(0 +): limG(u) = lim u—*O u>O g(u) —g(0+) u-'O u>O lim (u/2) u>O G(u) is itself piecewise continuous on the interval 0 c u < r. Applying Lemma 1 to integral (7), we therefore conclude that Hence (8) = 0. lim With property (3) of the Dirichlet kernel, we know that = +), or IT (9) N—'oo 2 The desired result (5) now follows from equation (6) and limits (8) and (9). A FOURIER THEOREM SEC. 19 19. 73 A FOURIER THEOREM A theorem that gives conditions under which a Fourier series converges to its function is called a Fourier theorem One such theorem will now be established. Although it is stated for periodic functions of period 2'w, it applies as well to functions defined only on the fundamental interval — < x < 'rr; for we need only consider the periodic extensions of those functions, with period 2'w. This will be done in the corollary to follow. . Theorem. Let f denote a function that is piecewise continuous on the interval — 'iT < X < and periodic with period 2 ir . Its Fourier series a0 (1) n=1 L# where (2) f(x)cosnxdx (n=O,1,2,...) f(x)sinnxdx (n=l,2,...), and lIT (3) converges to the mean value f(x+) ±f(x—) (4) of the one-sided limits of f at each point x ( — oc < x < one-sided derivatives ) and oc) where both of the f is actually continuous at x, the quotient (4) becomes f(x). Hence a0 he n=1 and exist. at that point, provided The integrals in expressions (2) and (3) for the coefficients and always exist since f is piecewise continuous; and we begin our proof of the theorem by writing series (I) as (see Sec. 15) 1 —ff(s)ds+ 2r — E fIT IT with those coefficients incorporated into it. Then, if SN(x) denotes the partial 74 FOURIER SERIES CHAP. 2 sum consisting of the sum of the first N + 1 (N 1N I (5) f(s)cosn(s-x)ds. f(s)ds+— 2r 1) terms of the series, Using the Dirichiet kernel (Sec. 18) N I DN(u) = — + 2 cosnu n=1 we can put equation (5) in the form lIT SN(x) = —I f(s)DN(s - x) ds. periodicity of the integrand here allows us to change the interval of integration to any interval of length 2 'iv- without altering the value of the The integral (see Problem 13, Sec. 20). Thus 1 SN(x)=—f (6) X+'TT 1TX-IT f(s)DN(s-x)ds, where the point x is at the center of the interval we have chosen. It now follows from equation (6) that 1 (7) SN(x) = ;[IN(x) +JN(x)], where X+IT f(s)DN(s-x)ds (8) and (9) = If we replace the variable of integration s in integral (8) by the new variable U = (10) ff(s)DN(s - x) ds. s — x, that integral becomes IN(x) = fIT + u)DN(u) du. f is piecewise continuous on the fundamental interval — ir < x < 'ir and also periodic, it is piecewise continuous on any bounded interval of the x axis. So, for a fixed value of x, the function g(u) = f(x + u) in expression (10) is piecewise continuous on any bounded interval of the u axis and, in particular, on the interval 0 < u < 'r. Let us assume that the right-hand derivative fk(x) exists. After observing that Since g(O+)= u—O limg(u)=u—O limf(x+u)= limf(v)=f(x+), u>O u>O v>x SEC. 19 A FOURIER THEOREM one can show that the right-hand derivative of g at u = 0 g(u)-g(O+) . = lim u>O exists: f(x+u)-f(x+) u-O u—O u—O = tim 75 U u>O f(v) -f(x+) = tim V—X v-)x v>x According to Lemma 2 in Sec. 18, then, tim IN(x) = —g(O+) = —f(x+). (11) 2 N—oo 2 If, on the other hand, we make the substitution u = x — s in integrat (9) and recatt from our discussion in Sec. 18 that DN(U) is an even function of U, we find that JN(x) = (12) ff(x U)DN(U) dii. — exists; and we note that This time, we assume that the teft-hand derivative the function g(u) = f(x — u) in expression (12) is piecewise continuous on the intervat 0 < is < r. Furthermore, g(O+)= timg(x)= timf(x—U)= timf(v)=f(x—) u—O u—M V<X u>O u>O and . = tim u-O g(U)-g(O+) ii0 u>O = — tim Li So . = tim f(x-ii)-f(x-) U u>O f(v) -f(x-) = <X once again by Lemma 2 in Sec. 18, (13) tim JN(x) = —g(O+) = —f(x—). 2 2 Finatty, we may conctude from equation (7) and timits (1 1) and (13) that tim SN(x) = N—oo f(x+) +f(x—) 2 and the theorem is proved. This theorem is especiatty suited to functions f that are piecewise smooth on the fundamentat intervat — 'ir < x < 'ir, ones such that both f and f' are piecewise continuous there. For, when f is piecewise smooth on — 'ir < x < we know from the theorem in Sec. 17 that its one-sided derivatives, from the interior at the end points x = ± r, exist everywhere in the ctosed intervat — '7T X C w. Hence if F denotes the periodic extension of f, with period 2 76 CHAP. 2 FOURIER SERIES the one-sided derivatives of F exist at each point x ( — oc < x < oc). According to the theorem just proved, then, the Fourier series for f on — 'n- < x < everywhere to the mean value of the one-sided limits of F. We state this useful result as follows. converges Corollary. Suppose that a function f is piecewise smooth on the interval — 'IT < X < r and let F denote its periodic extension , with period 2 ir. Then , for , each x ( — oc < x < oc), the Fourier series (1), with coefficients (2) and (3), converges to the mean value (14) F(x+) +F(x—) of the one-sided limits of F at x. DISCUSSION OF THE THEOREM AND ITS COROLLARY 20. It should be emphasized that the conditions in the theorem in Sec. 19, as well as the corollary there, are only sufficient, and there is no claim that they are necessary conditions. More general conditions are given in a number of the references listed in the Bibliography. Indeed, there are functions that even become unbounded at certain points but nevertheless have valid Fourier series representations.t The corollary in Sec. 19 will be adequate for most of the applications in this book, where the functions are generally piecewise smooth. We note that if f and F denote the functions in the corollary, then F(x + ) = f(x + ) and the F(x—)=f(x—) for —'w <x < 'ir. Consequently, when —'n- <x corollary tells us that the Fourier series (1) with coefficients (2) lIT rITf(x)cosnxdr (n=O,l,2,...) and (3) I See, for instance, the book by Tolstoy (1976, pp. 91—94) that is listed in the Bibliography. DISCUSSION OF THE THEOREM AND ITS COROLLARY SEC. 20 77 converges to the number f(x+) +f(x—) (4) 2 which becomes f(x) if x is a point of continuity of f. At the end points x =± however, the series converges to f(-ir+) +f(ir—) (5) 2 To see that this is so, consider first the point x = F(-r+) =f(-ir+) — 'yr. Since F(-r-) and as is evident from Fig. 23, the quotient F(x+) +F(x—) 2 in the corollary becomes the quotient (5) when x = — r. Because of the periodicity of the series, it also converges to the quotient (5) when x = 'ir. Observe how it follows that the series converges to f( — 'w + ) at x = — 'w and to f&—)at x=r ifandonlyiff(—'r+)=f&—). F( x) f( x) S 3ir x FIGURE 23 EXAMPLE 1. In Example 1, Sec. 15, we obtained the Fourier series -1 (6) on the interval — 'w < x < n 'w sln nx I for the function f defined by the equations f(x) = 10 when when f'(x) = 10 when —r < x < <x 0, 0<x<w. Since \.1 0, when 0<x<r, 78 FOURIER SERIES CHAP. 2 r f is clearly piecewise smooth on the fundamental interval — < x < ii-. In view of the continuity of f when — ir < x < r, the series converges to fUr) at each point x in that open interval. Since f( — + ) = 0 and f(r — ) = r, it converges to r/2 at the end points x = ± 'w. The series, in fact, converges to at each of the points x = ±'w, ±3'n-, ±5'w, . . , as indicated in Fig. 19 (Sec. 15), where the sum of the series for all x is described graphically. In particular, since series (6) converges to when x = 'ii-, we have the identity r . r °° (—1) 2 which can be written as 1)2 n=1 (2n ç. This illustrates how Fourier series can sometimes be used to find the sums of convergent series encountered in calculus. Note that setting x = also yields this particular summation. 0 in series (6) The corollary in Sec. 19 tells us that a function f in the space — of piecewise smooth functions on the interval — ii- < x < has a valid Fourier series representation on that interval, or one that is equal to f(x) at all but possibly a finite number of points there. It also ensures that a function f in the space 'iv-) has valid Fourier cosine and sine series representations on the interval 0 c x < r. For, according to Sec. 15, the cosine series 00 (7) -f + where (8) the same as the Fourier series corresponding to the even extension of f on the interval — ii- < x c r; and the sine series is (9) where (10) is the Fourier series for the odd extension of f on that same interval. SEC. 20 PROBLEMS 79 In view of the even periodic function represented by the series (7), that series converges to f(O + ) at the point x = 0 and to f('r — ) at x = 'r. The sum of the series (9) is, of course, zero when x = 0 and x = 'in EXAMPLE 2. In the example in Sec. 13, we found the Fourier cosine series corresponding to the function f(x) = sin x on the interval 0 < x < 'ir: 2 (11) sin x 0 4 °° cos2nx on 'ir, correspondence (1 1) is evidently an equality when 0 x xc x 'in Our final example illustrates how the theorem in Sec. 19 can be useful when the corollary there fails to apply. is continuous on the closed interval EXAMPLE 3. The odd function — and, therefore, piecewise continuous on the interval — 'ir < x < 'in 'iT C X C If we let f denote that function, we see that f is not piecewise smooth on —r < X < not apply. since f'(O + ) and f'(O — ) do not exist. Hence the corollary does If, however, f denotes the periodic extension, with period 2ir, of x 1/3 < x < err), the theorem can be used. To be precise, since the one-sided (— derivatives of f exist everywhere in the interval — 'ir < x < 'ir except at x = 0, we find that the Fourier series for x 1/3 on — 'w < x c 'w converges to x 1/3 when —'7. < x < 0 or 0 < x < 'in That series representation is valid even at x = 0 since x1"3 is odd and the series is actually a Fourier sine series on 0 < x < 'ir, which converges to zero when x = 0. We may conclude, then, that the Fourier < x < 'ir) is valid throughout the entire series representation for x 1/3 ( < x < 'in interval PROBLEMS 1. For each of the following functions, point out why its Fourier series on the interval x ir, and state the sum of the series when —IT < X < IT i5 convergent when —ir x= (a) The function when 2 O<X<IT. whose series was found in Problem 1, Sec. 16. 0), whose series was found in Problem 4, Sec. 16. (b) The function f(x) = (a Answers: (a) Sum = 0; (b) Sum = cosh aIT. 08 S3fl135 r= 'z Ag 2uq!JM IVHD r o u! oqj uOflEluOsOJdoJ 500 ruz it pOqsijqElsO aqi JOUflOd (r)J oJaqMLIaAa U! 'r 0) > Ajinj r> uoqM ) r ) 'it OjdWRx3 'T °°S 'frT JOJ r p!jEA UOfllflUOSaJdOJ joj Oqi JOJ fiR )— U! ST •QiJ (it r oqp jEMOJU! it— SO!JOS U! AqM uoqounj '9j Joj 50!105 ut uts SO2JaAUOO 02 it T_ — i: J=u AqM ) r '(it :suoqtaumns 8U!MOIjOJ T = o) — i: T=uit 'oz uitnqo oqi oidmrnca 'z T 't: Z S! r UOiiDUflj >r > UO aqT jI?AJZflU! >r > (r)f = r snqj 41!JOA 0q2 (oo 02 S! i!d 'S (N it z S! J=UJL uo oqi IThuolu! AjiEnpE — r(j—uz)soo Z(J_uz) r it i: andmoj) ojduiExj 'T r> > PN.fl .w z.Lt = — uz) it—) Z(T •°°s Coz 9 (v) -) z.Lt — Zr 1=" 'DOS U! 'jij o:i Moqs = Zu T=u (q) Aq 8Ufl!JM I= U! ruso3 u zit zit ZT ' u oqj oouapuodsonoo j7 — (itu) vu puooos uOflflUWflS U! l7it u (N '8 T==u 8U!JJOJaJ 9 06 . 9 soo pnd '(ii) Moqs o) x> (iij> PROBLEMS SEC. 20 81 7. With the aid of the correspondence (Problem 6, Sec. 16) cos ax where a 2a sin air + 'ir E n=1 n2—a2 cosnx (—ir<x<ir), 0, ±1, ±2,...,show that air sin air 00 =1+2a2E (_1)n+1 n=1 n2 a2 from the orthonormal set in the 8. If we exclude the constant function 40(x) = example in Sec. 12, we still have an orthonormal set, consisting of the functions = cos - (n=1,2,...). V7T State why this set is closed (Sec. 12) in the space of all functions f that are piecewise smooth on the interval 0 < x < ir and satisfy the condition IT f(x)dx=O. 0 Suggestion: Refer to the statement in italics near the end of Sec. 12. 9. Without actually finding the Fourier series for f(x) = x < 'ir, point out how the theorem in Sec. 19 ensures the convergence of that series to f(x) when x=O. 10. With the aid of l'Hospital's rule, find f(O + ) and f(x)= for the function ex x (x#0). Answer f(O + ) = 1, fk(O) = 11. Show that the function f(x)= I x sin — x 0 whenx=0 is continuous at x = 0 but that neither fk(0) nor exists. This provides another illustration (see Example 3, Sec. 17) of the fact that the continuity of a function f at a point x0 is not a sufficient condition for the existence of the one-sided derivatives offat x0. 12. Given that the right-hand derivatives of two functions f and g exist at a point x0, prove that the product f(x)g(x) of those functions has a right-hand derivative there by inserting the term f(x)g(x0 + ) and its negative in the numerator of the difference quotient f(x)g(x) —f(x0+)g(x0+) x — x0 82 FOURIER SERIES CHAP. 2 13. Let f denote a function that is piecewise continuous on an interval — c c x < c and periodic with period 2c. Show that, for any number a, L!(X) dx = ff(x) dx. Suggestion: Write L!(X) dx = L:f(x) dx + ff(s) ds then make the substitution x = side of this equation. 14. Derive the expression and DN(u)= s — 2c in the second integral on the right-hand sin [(2N + 1)u/2] 2sin(u/2) — — for the Dirichlet kernel (Sec. 18) 1 h_ N cosnu n=1 by writing A = u/2 and B = nu in the trigonometric identity 2 sin A cos B = sin (A + B) + sin (A — B) and then summing each side of the resulting equation from n = Suggestion: Note that N L n=O n=1 21. to n = N. N—i \ L 1 FOURIER SERIES ON OTHER INTERVALS Suppose that a function f is piecewise smooth on an interval — c < x < c and periodic with period 2c, where c is any positive number. For convenience in the discussion below, we assume that f(x) at each point of discontinuity of f is the mean value of the one-sided limits f(x + ) and f(x — ), as is the case at a point of continuity. Let us define the function g(s)=f(—) (2) g(s)=f(x) where x=— (—oo<s<oc), is periodic with period 2'w. The equation x = cs/r, or s = n/c, establishes a one to one correspondence between points on the x axis and which points on the s axis; and it is evident from relations (2) that if a specific point x FOURIER SERIES ON OTHER INTERVALS SEC. 21 corresponds 83 to a point s, then g(s+) =f(x+), g(s—) =f(x—). Since f(x) is always the mean value of f(x + ) and f(x — ), it follows from these relations between one-sided limits that the number g(s) = f(x) is always the mean value of g(s + ) and g(s — ). In particular, g is continuous at s when f is continuous at x. Since f is piecewise continuous on the interval — c < x < c, then, g is piecewise continuous on the interval — C s < in The derivative f' is also piecewise continuous, and a similar argument shows that g' is piecewise continuous. So g is piecewise smooth on the interval — 'ii- c s < ii-; and it is its own periodic extension, with period 2'w, on the entire s axis. An application of the corollary in Sec. 19 now shows that function (1) is represented by its Fourier series everywhere on the s axis. That is, for each s, /CS\ a0 + L (3) r = he n=1 where /CS\ I (4) (n=O,1,2,...) fI—Icosnsds 1T-7r'1T/ and CS\ 1 (5) (n=1,2,...). I —Jsinnsds 'WI With the substitution x = cs/'ir, representation (3) becomes (6) f(x) a0 = —i- I + fl'ITX\ fl'WX C C n=1 and this is valid for all x. Expressions (4) and (5) can be written as follows, where the new variable of integration x = Cs/u- is used: 1 c an=—ff(x)cos (7) 1 c C —c (8) f(x)sin fl'ITX (n=O,1,2,...), fl'JTX dx (n=1,2,...). C We state the result as a theorem that is sufficient for our applications. It is possible, by appealing directly to the theorem in Sec. 19, to obtain a more general result for periodic functions f that are only piecewise continuous on the x ) and fundamental interval — c < x < c but have one-sided derivatives f a funCtion that is piecewise smooth on an interval < X < C and periodic with period 2C. Iff(x) at each point of discontinuity off is defined as the mean value of the one..sided limits f(x + ) and f(x — ), then the —C 84 FOURIER SERIES CHAP. 2 Fourier series representation (6), with coefficients (7) and (8), (—oc <x < oc). is valid for all x If the function f is even , then so is the function g that is defined by equation (1); and we know from Sec. 15 that expression (4) for the coefficients may be written ics\ 2 (n=O,1,2,...). 0 Fir Furthermore, the coefficients are all zero. Hence series (6) reduces to a cosine series: (9) a0 °° L1 n=1 n'irx c where 2 (10) n-in c Co dx (n=O,1,2,...). C Similarly, if f is odd, we have a sine series: 00 n'irx n=1 C f(x)= (11) where 2 (12) c CO n'in C (n=1,2,...). It is easy to adapt the above theorem to any piecewise smooth function f defined only on the interval — c < x < c. To do this, we introduce the periodiC extension, with period 2c, of f and denote it by F. The graph of y = F(x) is the graph of y = f(x) repeated every 2c units along the x axis. After defining F(x) at each point of discontinuity of F as the mean value of the one-sided limits F(x + ) and F(x — ), one can apply the theorem to that extension. The same procedure may be used to verify representations (9) and (1 1) for piecewise smooth functions defined only on the interval 0 < x < C, once the even and odd extensions, respectively, are made on the interval — c < x < c. The following corollary, which summarizes these results, applies to any function f that has the following properties: (a) f is piecewise smooth on the stated interval; (b) f(x) at each point of discontinuity of f in that interval is the mean value of the one-sided limits f(x + ) and f(x —). Corollary. If a function f has properties (a) and (b) on an interval —C (8), < X < C, then the Fourier series representation (6), with coefficients (7) and is valid for each x (—c < x < c). 1ff has those properties on an interval PROBLEMS SEC. 21 85 0 c x < C, then the Fourier cosine series representation (9), with coefficients (10), is valid for each x (0 c x < c); and the same is true of the Fourier sine series representation (1 1), with coefficients (12). EXAMPLE The function f(x ) = 2 is piecewise smooth on any interval 0 < x < c, and the sine series representation (_1)n+1 ( 13) x2 2c2E I- n'irx n'r (0 <x < c) c can be obtained by evaluating the integrals in expression (12) when f(x) = x2. But, since we already know from Problem 4(b), Sec. 14, and the corollary here that (1\fl+1 00 I k (14) —2 n'n- n=1 1— (_1\fl sinnx (nr) (0<x<ir), it is simpler to start with that special case. To be specific, 0<—<r c when 0<x<c; and so we can replace x by 'irx/c on each side of representation (14) to obtain an equation that is valid when 0 < x < c. Then, by multiplying through that equation by c2/'rr2, we arrive at representation (13), which is actually valid on theintervalO cx <c. PROBLEMS In Problems 1 through 3, use formulas in Sec. 21 to find the coefficients in the Fourier series involved. 1. Show that if when —3 <x <0, f(x) = and f(0) = when 0<x<3 then 1 2 °° f(x)=—+--L---2 1 sin (2n—1)irx 3 (—3 <x < 3). Describe graphically the function that is represented by this series for all x (—oc <x < oc). 2. Let f denote the function whose values are f(x)= when —2 <x < when 1 1<x<2 86 FOURIER SERIES CHAP. 2 and f( — 2) = f(1) = f(2) = —2 c x for each x Show that, the closed interval in 2, f(x) = 1 1 °° PZfl 1 — — — E — sin 4 ,lTn=ln 2 cos fliT flITX 2 2 + (cos nir — cos—) sin 2 3. Obtain the Fourier sine series representation n cosv-x=— E 2 (O<x<1). sin2nirx —1 Suggestion: To evaluate the integrals that arise, recall the trigonometric identity 2sinAcosB = sin(A +B) + sin(A —B). 4. Let f denote the periodic function, of period 2, where (cosirx f(x)= 4; whenO<x<1 whenl<x<2 " and where f(O) = and f(1) = in Problem 3, show that With the aid of the series representation found — 1 f(x)=—cosn+—L mn=1 n ii 2 sin2mrx (—cc<x<oo). 5. (a) Use the Fourier sine series found in Example 1, Sec. 14, for f(x) = x (0 < x < 'ii-) to show that 2 °° x=—E mn=1 n (—1<x<1). —sinnirx (b) By referring to the Fourier cosine series in Problem 4(a), Sec. 14, for f(x) = (0 < x < ir), derive the expansion 4c2 c2 °° x2=—+---i-L 3 IT n=1 (_1)fl n2 n'n-x cos C 6. Show how it follows from the expansions obtained in Problem 5 that x+x2= + _L(—lf(2cosnrx— isinnn) (—1 <x< 1). 7. (a) Use the Fourier sine series in Example 3, Sec. 14, for the function f(x)=x(r2—x2) (O<x<n-) to establish the representation 12 n3 sinnirx (Ocxcl). PROBLEMS SEC. 21 87 (b) Replace x by 1 — x on each side of the representation found in part (a) to show that °° 12 sin (Ocxcl). 3 n=1 17• nirx 8. Show how it follows from the Fourier sine series for the function (O<x<ir), f(x)=x('n-—x) found in Problem 5, Sec. 14, that x(2c—x)= p77.3 (2n — 1)'irx oo 32c2 E sin (Ocxc2c). 2c n=1 (2n — 9. Let M(c, it) denote the square wave (Fig. 24) defined by the equations M(c,t)=1 and M(c, it + 2c) = M(c, t) (it 4 M(c,t) = — E when O<t<c, 1 whenc<t<2c > 0). Show that (2n—1)irt 1 sin c (it c,2c,3c,...). Suggestion: The sine series found in Problem 1(b), Sec. 14, for the function f(x) = 1 (0 c x 'C 'ir) can be used here. M(c,t) 9 I I I I I I 13c IC -1- 12c & -r;4c I I t 15C 6— FIGURE 24 10. Let F denote the periodic function, of period c, where C -4:—x 3C x—— whenOcx C (a) Describe the function F(x) graphically, and show that it is, in fact, the even periodic extension, with period C, of the function 88 FOURIER SERIES CHAP. 2 (b) Use the result in part (a) and the Fourier cosine series found in Problem 2(a), Sec. 14, for f(x) = — x (0 < x < ir) to show that r 2c n=1 (2n — 1)2 11. (4n—2)irx 1 F(x)=—L ¶2 cos (—oc<x<oc). C Suppose that a function f is piecewise smooth on the interval 0 < x < C, and let F denote this extension of f to the interval 0 c x < 2C: (f(x) F(X)=\f(2) when O<x<C, whenC<x<2C. [The graph of y = F(x) is evidently symmetric with respect to the line x = C.] Show that the coefficients in the Fourier sine series for F on the interval 0 < x < 2C can be written = Thus [i I c — (n = 1,2,...). 2 show that (2n f(x)= — 1)irx n=1 where 2 (2n—1)'rrx (n=1,2,...), dx C 2C 0 for each point x (0 < x < C) at which f is continuous. Suggestion: Write I c bn_[if(x)sin nirx 2c f(2C—s)sin 2C mrs 2C and make the substitution x = 2C — s in the second of these integrals. 12. Use the result in Problem 1 1 to establish the representation 8C (2n — 1)irx . x=—L sln 2C IT 13. Show that in Sec. 21 the Fourier series (6), with coefficients (7) and (8), can be written in the compact form -x)] ds. + 14. (See Sec. 15, where this form was obtained when C = (a) Verify that the set of functions 40(x) = I mrx 1 = ' cos C nrx 1 4i2,1(x) = sin C (n==1,2,...), UNIFORM CONVERGENCE OF FOURIER SERIES SEC. 22 89 which becomes the set (1), Sec. 15, when c = ir, is orthonormal on the interval —c<x<c. (b) Show that the generalized Fourier series corresponding to a function f(x) in — c, c) with respect to the orthonormal set in part (a) can be written as the ordinary Fourier series, with coefficients (n = 0, 1, 2, . . ) and (n = 1, 2, . ), for f on the interval —c < x < c (Sec. 21). . . . (c) Derive Bessel's inequality N a2 1 c (N=1,2,..,) and b1,, in part (b) from the general form (9), Sec. 16, of the coefficients that inequality for Fourier constants. (Compare Problem 9, Sec. 16.) for Suggestion: In part (a), the integrals involved may be transformed by means of the substitution s = in/c into integrals whose values are known, since the set is already known to be orthonormal on —ir < x < ir when c = ir. 15. After writing the Fourier series representation (6), Sec. 21, as N a0 f(x) = — + 2 use C C / the exponential formst cosO= of nirx\ nirx lim e'° + 2 sinO= , e'° — 2i the cosine and sine functions to put that representation in exponential form: fn'irx\ N f(x)= lim N-OOn_N \ C I where a +ib " a —ib a0 Then use expressions (7) and (8), Sec. 21, for the coefficients the single formula 1 c nirx C (n=1,2,...). and to obtain (n=O,±1,±2,...). UNIFORM CONVERGENCE OF FOURIER SERIES 22. The reader may at this time pass directly to Chap. 3 without serious disruption. This section, regarding the uniformity of the convergence of Fourier series, and the remaining two sections of the present chapter, dealing with other aspects of t For background on these forms and an introduction to series and integrals involving complexvalued functions, see the authors' book (1990), listed in the Bibliography. 90 FOURIER SERIES CHAP. 2 the theory of convergence of such series, will be used only occasionally later on. The reader may refer to these sections for results that will be specifically cited as needed. For convenience, we treat only Fourier series for which the fundamental interval is — 'w < x < in Applications of the theorems to series on any fundamental interval — c < x < c can be made by the method used in Sec. 21. We preface our theorem on the uniform convergence of Fourier series with an important lemma. Lemma. Suppose that f is continuous on the interval — 'ir x c r, where f( — 'IT) = fe;r), and that its derivative f' is piecewise continuous on the interval —IT and < X < 'IT. Then are the Fourier coefficients hr = ;ff(x)cosnxdr, (1) = tin —ff(x)sinnxdx, the series (2) converges. The class of functions satisfying the conditions in this theorem is, of course, a subspace of the space of piecewise smooth functions on the interval —IT<X<IT. We begin the proof of the lemma with the observation that the Fourier coefficients (3) lIT In f'(x)sinnxdx f'(x)cosnxdr, for f' exist because of the piecewise continuity of f'. Note that ln 1 f is continuous and f( — whenn=1,2,..., 'ir) = f(ir), integration by parts reveals that 1 = + = IT nn —ff(x)sinnxc/x [f(ir) + = and 1 ir —[f(x)slnnx]_r IT f(x)cosnxcfr = J IT-n —nan. SEC. 22 That UNIFORM CONVERGENCE OF FOURIER SERIES 91 is, = (4) (n = Ii 1,2,...). IL In view of relations (4), the sum SN of the first N terms of the infinite series (2) becomes Ni N (5) Cauchy 's inequality N 2 N N n=1 n=1 C n=1 which applies to any two sets of real numbers (n = 1, 2, . . , N) and (n = 1, 2, . , N) (see Problem 6, Sec. 24, for a derivation), can now be used to . . . write NI (6) N E—1 n n=1 (N=1,2,...). n=1 The sequence of sums Ni E —i (N= 1,2,...) here is clearly bounded since each sum is a partial sum of the convergent series whose terms are i/n2 [see Problem 6(a), Sec. 20]. The sequence (N1,2,...) (n = 0, 1, 2, . . ) and (n = 1, 2, . . . ) are the Fourier coefficients for f' on the interval — 'w < x c and must, therefore, satisfy is also bounded since . Bessel's inequality: N lIT (N1,2,...). (See Problem 9, Sec. 16.) It now follows from inequality (6) that the sequence (N = 1, 2, . ) is both bounded and nondecreasing. Hence it converges; and this means that the sequence 5N (N = 1, 2, . . . ) converges. Thus series (2) . . converges. 92 FOURIER SERIES CHAP. 2 We turn now to the uniformity of convergence of Fourier series. We begin by recalling some facts about uniformly convergent series of functions.t where the Let s(x) denote the sum of an infinite series of functions series is convergent for all x in some interval a x b. Thus (7) s(x)= limsN(x) N—*oo n=1 where sN(x) is the partial sum consisting of the sum of the first N terms of the series. The series converges uniformly with respect to x if the absolute value of its remainder rN(x) = s(x) — sN(x) can be made arbitrarily small for all x in the interval by taking N sufficiently large; that is, for each positive number e, there exists a positive integer independent ofx, such that (8) Is(x)—sN(x)I <F whenever (acxcb). A sufficient condition for uniform convergence is given by the Weierstrass M-test. Namely, if there is a convergent series 00 (9) n=1 of positive constants such that (acxcb) (10) for each n, then series (7) is uniformly convergent on the stated interval. We include here a few properties of uniformly convergent series that are are continuous and if series (7) is uniformly often useful. If the functions convergent, then the sum s(x) of that series is a continuous function. Also, the series can be integrated term by term over the interval a x b to give the and their derivatives integral of s(x) from x = a to x = b. If the functions fh are continuous, if series (7) converges, and if the series whose terms are fh(x) is uniformly convergent, then s'(x) is found by differentiating series (7) term by term. Theorem. When the conditions stated in the above lemma are satisfied, the Fourier series (11) a0 L1 n=1 with coefficients (1), converges absolutely and uniformly to f(x) on the interval -17W in t See, for instance, the book by Kaplan (1991, chap. 6) or the one by Taylor and Mann (1983, chap. 20), both listed in the Bibliography. UNIFORM CONVERGENCE OF FOURIER SERIES SEC. 22 93 To prove this, we first note that the conditions on f ensure the continuity of the periodic extension of f for all x. Hence it follows from the corollary in r Sec. 19 that series (1 1) converges to f(x) everywhere in the interval — Observe that since both are less than or equal to + x ii-. I I c + (n=1,2,...). series (2) converges, the comparison test and the Weierstrass M-test thus apply to show that the convergence of series (1 1) is absolute and uniform on the interval —r C x ( 'ir, as stated. In like manner, one can establish the absolute and uniform convergence of the series of cosine or sine terms only. Series (11) is, in fact, the sum of those Since series: a0 °° z' n=1 f(x)=—--+ (12) (—rcxc'w). n=1 Modifications of the statements in both the lemma and the theorem are apparent. For instance, it follows from the theorem that the Fourier cosine series on 0 < x < ii- for a function f that is continuous on the interval converges absolutely and uniformly to f(x) if f' is piecewise 0cxC continuous on the interval 0 < x < 'n-. For the sine series, however, the tional conditions f(O) = f€n-) = 0 are needed. Since a uniformly convergent series of continuous functions always con- verges to a continuous function, a Fourier series for a function f cannot converge uniformly on an interval that contains a point at which f is discontinuous. Hence the continuity of f, assumed in the theorem, is necessary in order for the series there to converge uniformly. Suppose that x0 is a point at which a piecewise smooth function f is discontinuous. The nature of the deviation near x0 of the values of the partial sums of a Fourier series for f from the values of f is commonly referred to as the Gibbs phenomenon and is illustrated below.t EXAMPLE. Consider the (odd) function f defined by the equations f(x) = 7 ii and f(O) = 0. when O<x<r According to Problem 1, Sec. 16, and the corollary in Sec. 21, the t For a detailed analysis of this phenomenon, see the book by Carsiaw (1952, chap. 9) that is listed in the Bibliography. 94 CHAP. 2 FOURIER SERIES Fourier series 00 2L sin(2n n=1 — 1)x (—r<x<r) 2n—1 for f converges to f(x) everywhere in the interval — r < x < in Let SN(X) denote the sum of the first N terms of this series. The sequence SN(X) (N = 1, 2, . . . ) thus converges to f(x) when —'w < x < in In particular, it converges to the number 'w/2 = 1.57 . . . when 0 < x < in But, as shown in Problem 7, Sec. 24, there is a fixed number a = 1.85 . . such that sN('w/(2N)) . tends to a. See Fig. 25, which indicates how "spikes" in the graphs of the partial sums y = SN(X), moving to the left as N increases, are formed, their tips tending to the point a on the y axis. The behavior of the partial sums is similar on the interval —r C x < 0. This illustrates that special care must be taken when a function is approximated by a partial sum of its Fourier series near a point of discontinuity. y y = SN(X) a ir/2 0 IT x IT 2N FIGURE 25 DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES 23. According to the corollary in Sec. 21, the Fourier series in Example 3, Sec. 15, for the function f(x) = x (—iv- < x < 'iv-) converges to f(x) at each point in the interval —'w c x < ii-: 00 (1) x=2E n=1 n —sinnx (—iT < x < But the series here is not differentiable. The differentiated series 00 2 n=1 DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES SEC. 23 95 does not even converge since its nth term fails to approach zero as n tends to infinity. The periodic function represented by series (1) for all x has discontinw. ities at the points x = ±'w, ±1w, ±5ir, . . . , as shown in Fig. 18 (Sec. 14). Continuity of the periodic extension of the function represented is an important condition for differentiability of a Fourier series on the fundamental interval. Sufficient conditions can be stated as follows. Theorem 1. Suppose that the conditions stated in the lemma in Sec. 22 are satisfied. Namely, f is continuous on the interval — c x 'ii-, where f( — = and f' is piecewise continuous on the interval — < x < 'in Then the Fourier series in the representation r f a0 f(x)=--+ (2) h1 n=1 where = lIT —f f(x)cosnxdx, = is differentiable at each point x in the interval — —f f(x)sinnxdx, rCxC at which f"(x) exists: f'(x) = ( 3) (— lIT Our proof of this theorem is especially brief. To start with, we let x < x < r) be a point at which f" exists; and we note that f' is, therefore, continuous at x. Hence an application of the Fourier theorem in Sec. 19 to the function f' shows that f'(x) = (4) are the coefficients (3) in Sec. 22. But since f and f' satisfy all the conditions stated in the lemma in Sec. 22, we know from that section that where and f3,,= —nan ( 5) (n=l,2,...). When these substitutions are made, equation (4) takes the form (3); and the proof is complete. At a point x where f"(x) does not exist, but where f' has one-sided derivatives, differentiation is still valid in the sense that the series in equation (3) converges to the mean of the values f'(x + ) and f'(x — ). This is also true for the periodic extension of f. Theorem 1 applies, with obvious changes, to other Fourier series. For instance, if f is continuous when 0 c x c 'ir and f' is piecewise continuous on the interval 0 < x < r, then the Fourier cosine series for f on 0 c x < is differentiable at each point x (0 < x c 'r) where f"(x) exists. r 96 FOURIER SERIES CHAP. 2 Integration of a Fourier series is possible under much more general conditions than those for differentiation. This is to be expected because an integration of the series in the correspondence a0 (6) (—'w<x<rr), n==1 where (7) = lIT —ff(x) cos nxdx, = lIT —ff(x) sin nxdx, introduces a factor n in the denominator of the general term. In the following theorem, it is not even essential that the original series converge in order that the integrated series converge to the integral of the function. The integrated series is not, however, a Fourier series if a0 * 0; for it contains a term (a0/2)x. Theorem 2. Let f be a function that is piecewise continuous on the interval — < x < iT. Regardless of whether series (6) converges, the following equation is validwhen '17W LIT2fllfl{k]} 00 (8) (8) is obtained by replacing x by s in series (6) and then integrating term by Series termfroms= Our —'irtos=x. proof starts with the fact that since f is piecewise continuous, the function (9) is continuous; moreover, a0 F'(x)=f(x)——2 (—ir<x<r), except at points where f is discontinuous. Hence F' is piecewise continuous on the interval — c x < in Since F is piecewise smooth, then, it follows from the corollary in Sec. 21 that r A (10) where (11) = lIT F(x)cosnxdx, —f 'TIT = —f F(x)sinnxdx. SEC. 24 CONVERGENCE IN THE MEAN 97 We note from expression (9) and the first of expressions (7) that F(r) = ff(s) ds - = a0ir - -fir = F( — r) = (a0/2)'n-; thus F( — 'ri) = F(w). This shows that representation (10) is also valid at the end points of the interval — < x < 'ir (see Sec. 20) and, therefore, at each point x in the closed interval — 'ir x c Let us now write the coefficients and in terms of and b,. When and n r 1, we may integrate integrals (11) by parts, using the fact that F is continuous and F' is piecewise continuous. Thus 1 . IT An=—[F(x)srnnx]_r— —JF'(x)sinnxdx 'IT I a0 f(x)—-— =——I 2 nIT-IT a0 1 =—-----f mr—ITf(x)sinnxdr+ b f n = ar/n. As for A0, we know from the preceding paragraph that F('n-) = (a0/2)'ir and that representation (10) is also valid when x = ii-. So by Similarly, writing x = ii in that representation and then solving for A0, we see that 00 00 (1\fl+1 I n=1 n=1 these expressions for takes the form F(x) including A0, representation (10) and With a = + — + Finally, if we use expression (9) to substitute for F(x) here, we arrive at the desired result (8). The theorem can be written for the integral from x0 to x, where —'n-cx0cirand ff(s) ds = fJ(s) ds - Cf(s) ds. 24. CONVERGENCE IN THE MEAN A sequence sN(x) (N = 1, 2, . . . ) of piecewise continuous functions, defined on an interval a c x < b, is said to converge in the mean, or in the norm, to a b) if the mean square error (Sec. 16) in the approximation function f in 98 FOURIER SERIES by SN(X) to CHAP. 2 f(x) tends to zero as N tends to infinity: urn f [f(x) (1) N—oo a - SN(X)] That is, (2) tim If—sNII=O. N-oo Sometimes condition (2) is atso written t.i.m.sN(x) =f(x), (3) N—oo where the abbreviation "t.i.m." stands for limit in the mean. Suppose now that the function SN are the partiat sums of a generatized Fourier series (Sec. 12) corresponding to f on the fundamentat intervat a N (4) SN(x)= n=1 This is the linear combination 4?N(x) in Sec. 16 when = c,, there. If condition (2) is satisfied by each function f in our function space b), or we say that possibly a subspace containing the orthonormal set is complete in that space or subspace. Thus each function f can be approximated arbitrarily closely in the mean by some linear combination of functions of a complete set, namely the linear combination (4) when N is large enough.t According to equation (7), Sec. 16, the mean square error in the approxi- mation by N (5) Hence, when (6) IIf-SNM2 = 11f112 - n=1 L is complete, it is always true that = If 112. Equation (6) is known as ParSeval'S equation. It identifies the sum of the squares of the components of f, with respect to the generalized reference set as the square of the norm of f. Conversely, if each function f in the space satisfies Parseval's equation, the set is complete in the sense of mean convergence. This is because, in view of equation (5), the limit (2) is merely a restatement of equation (6). t In the mathematical literature, including some earlier editions of this text, the terms complete and closed are sometimes applied to sets that we have called closed (Sec. 12) and complete, respectively. CONVERGENCE IN THE MEAN SEC. 24 99 = (f, in equation (6), we now have a theorem that provides an alternative characterization of complete sets. Writing Theorem 1. A necessary and sufficient condition for an orthonorinal set (n = 1, 2, . ) to be complete is that, for each function f in the space . . considered, Parseval's equation E(ffl2 (7) be If 12 satisfied. For an application of Theorem 1 to Fourier series on the interval — <x< we recall from Sec. 15 that the functions 1 1 1 (8) (n=1,2,...) form an orthonormal set on the fundamental interval — < x < w. The Fourier constants (n = 0, 1, 2, . ) in the generalized Fourier series for a function f in — ir, ir) with respect to this set were then used to define the constants . . c2 ( 9) c2 (n=1,2,...). IT v— IT This gave rise to the Fourier series correspondence (10) a n=1 where (11) lIT f(x)cosnxdx (n=O,l,2,...) and (12) lIT (n=l,2,...). following theorem states that the set of functions (8) is complete in a certain subspace of the space — 'w, r) of piecewise smooth functions (Sec. 17) on the interval —ir < x < r. The Theorem 2. The orthonormal set (8) is complete in the space of functions f that are continuous on the interval — 'w x c r, where f( — 'r) = f('w), and have piecewise continuous derivatives f' on the interval — ir < x < r. 100 FOURIER SERIES CHAP. 2 Note that, in view of relations (9), Parseval's equation (6) for the orthonormal set in question can be written (13) a2 °° 2 n=1 1 'IT Hence, once we show that the coefficients (1 1) and (12) actually satisfy equation (13), Theorem 2 is proved. The fact that Parseval's equation (13) is satisfied is an easy consequence of the theorem in Sec. 22, which tells us that, for functions f in the space considered here, the series in correspondence (10) converges uniformly to f(x) on the interval —ii- x c ii-: (14) a0 LI (—'n-cxcr). n==1 Now a uniformly convergent series of continuous functions can be integrated term by term (Sec. 22). Hence we may multiply each term in equation (14) by f(x) itself, thus leaving the series still uniformly convergent, and then integrate over the fundamental interval: L [anff(x) cos nxdx + sin In view of expressions (1 1) and (12), the integrals on the right here can be and written in terms of and we find that = i44 + + Since this is the same as Parseval's equation (13), the proof is finished. Theorem 2 is readily modified so as to apply to the orthonormal sets leading to Fourier cosine and sine series on the interval 0 < x < ii-. More specifically, the set of normalized cosine functions in Example 2, Sec. 1 1 , is complete in the space consisting of continuous functions f, on the interval x :( 'IT, whose derivatives f' are piecewise continuous. When the normal.0 ized sine functions in Example 1 , Sec. 1 1 , are used to obtain a sine series, the conditions f(O) = f('ir) = 0 are also needed for the set to be complete. The function space in the theorem we have just proved is quite restricted. It can be shown that Parseval's equation (13) holds for any function f whose square is integrable over the interval — 'ii- < x < ir) t See, for instance, the book by Tolstoy (1976, pp. 54—57 and 117—120) that is listed in the Bibliography. SEC. 24 PROBLEMS 101 This bare introduction to the theory of convergence in the mean will not be developed further here. It should be emphasized, however, that statement (3) is not the same as the statement limsN(x)=f(x) (15) N-oo (a<x<b), even if a finite number of points in the interval are excepted.t In fact, neither of statements (3) and (15) implies the other (see Problems 10 and 1 1). PROBLEMS 1. Show that the function when—irzcxcO, f(x)=<10 when satisfies all the conditions in the lemma and the theorem in Sec. 22. Then, with the aid of the Weierstrass M-test (Sec. 22), verify directly that the Fourier series 1 2 1 °° —+—sinx——L IT 2 cos2nx (—ir<x<ir) for f, found in Problem 7, Sec. 16, converges uniformly on the interval —ir x as the theorem in Sec. 22 tells us. Also, state why this series is differentiable in the interval — ir < x < r, except at the point x = 0, and describe graphically the function that is represented by the differentiated series for all x. 2. We know from Problem 7, Sec. 14, that the series 4 n Z E cos(2n—1)x (2n—1) 2 is the Fourier cosine series for the function f(x) = x on the interval 0 < x c 'ir. Differentiate this series term by term to obtain a representation for the function f'(x) = on that interval. State why the procedure is reliable here. 3. State Theorem 1 in Sec. 23 as it applies to Fourier sine series. Point out, in particular, why the conditions f(O) = fGr) 0 are present in this case. and 4. Let denote the Fourier coefficients in the lemma in Sec. 22. Using the fact that the coefficients in the Fourier series for a function in — ir, 'ir) always tend to zero as n tends to infinity (Sec. 16), show why 1 and n—)cc 5. Integrate from s = 0 to s = x n—,cc ( — ir 2L x ir) the Fourier series — n+1 sinns, n=1 An example of a sequence of functions that converges in the mean to zero but diverges at each point of the interval is given in the book by Franklin (1964, p. 408), listed in the Bibliography. 1: 102 FOURIER SERIES CHAP. 2 mentioned at the beginning of Sec. 23, and the one sin(2n — 1)s 2n—1 n=1 2L appearing in the example in Sec. 22. In each case, describe graphically the function that is represented by the new series. 6. Let (n = 1, 2, . . , N) and (n = 1, 2, . . . , N) denote real numbers, where at least one of the numbers say Pm' 15 nonzero. By writing the quadratic equation . N N n=1 n=1 N in the form N =0, n=1 show that the number x0 = —qm/pm i5 the only possible real root. Conclude that, since there cannot be two distinct real roots, the discriminant of this quadratic equation is negative or zero. Thus derive Cauchy's inequality (Sec. 22) 2 N LPnqn n=1 N N n=1 n=1 which is clearly valid even if all the numbers are zero. 7. As in the example in Sec. 22, let sN(x) denote the partial sum consisting of the sum of the first N terms of the Fourier series sin(2n 00 2E n=1 for the function f there. (a) By writing A = x and B = (2n — ,, 1)x — (—'ir<x<w) Ii 1)x in the trigonometric identity 2sinAcosB = sin(A +B) + sin(A —B) and then summing each side of the resulting equation from n = 1 to n = N, derive the summation formula N sin2Nx 2E cos(2n—1)x= —;--sinx n=1 (x#0,±w,±2'rr,...). Use this formula to write the derivative of sN(x) on the interval 0 c x < ir as a simple quotient: sin 2Nx . Mn x (0<x<w). (b) With the aid of the expression for the derivative 44x) in part (a), show that the first extremum of sN(x) in the interval 0 < x C 'ir is a relative maximum occurring when x = w/(2N). 103 PROBLEMS SEC. 24 (c) By integrating each side of the summation formula in part (a) from x = x = ir/(2N), show that 0 to '2' = '1 where jlTi'(2N) X — .sin x 'r'(2N) sin 2Nx 1* '1= xsinx sin2Nxdx and I Jo " x dx. Verify that the integrands of these two integrals are piecewise continuous on the interval 0 < x < ir/(2N) and hence that the integrals actually exist. in part (c) is bounded (see (d) Using the fact that the integrand of the integral tends to zero as N tends to infinity. Then Sec. 10), show that the value of conclude that I ,,rsint \ limsN(----I=I \2V1 jo t N—'co dt. The value of this last integral is the nu' iber a in the example in Sec. 22) 8. Use Theorem 1, Sec. 24, to show that a:i orthonormal set is closed (Sec. 12) in a given function space if it is complete 'z that space. be an orthonormal set in the space of continuous functions on the 9. Let interval a c x c b, and suppose that the generalized Fourier series for a function f in that space converges uniformly to a sum s(x) on that interval. (a) Show that s and f have the same Fourier constants with respect to is closed (b) Use results in part (a) and Problem 11, Sec. 14, to show that if (Sec. 12), then s(x) = f(x) on the interval a x b. Suggestion: Recall from Sec. 22 that the sum of a uniformly convergent series of continuous functions is continuous and that such a series can be integrated term by term. 10. Consider the sequence of functions sN(x) (N = 0 x 1, 2, . . . ) defined on the interval 1 by means of the equations 1 0 when 1 2 5N@) when Show that this sequence converges pointwise to the function f(x) = 0 (0 c x c 1) 1) or any subspace but that it does not converge in the mean to f in the space of 1). t The integral occurs as a particular value of the sine integral function Si(x), which is tabulated in, for instance, the handbook edited by Abramowitz and Stegun (1972, p. 244) that is listed in the Bibliography. Approximation methods for evaluating definite integrals can also be used to find a. dVHD z S3flIEES jj ( oq (xyVs N) = 'T Aq SUEOIU Jo aqi SuOflEnbo wMalu! x T i: i: 0 T — 2i?qi oouonbos inq 'iluji ioj qoRa (x)f = SO8JOAUOO U! OAfl!SOd J080JU! 'd — d :UO!18a133n57 oAiosqo (d/ftVs = •0= N ( 'ci J U! I ... .— N, (1: CHAPTER 3 THE FOURIER METHOD We turn now to a careful presentation of the Fourier method for solving boundary value problems involving partial differential equations, which was touched on in Sec. 9 (Chap. 1). While the example there served to motivate the method, at that time we were seriously limited by our inability to expand functions in Fourier series. Chapter 2 has, of course, addressed that problem. Once the basics of the Fourier method have been presented, we shall, in Chap. 4, use it to solve a variety of boundary value problems whose solutions entail Fourier series representations. Then, in subsequent chapters, we shall apply the method to problems with solutions involving other, but closely related, types of representations. 25. LINEAR OPERATORS We recall (Sec. 10) that any two functions u1 and u2 in a given function space have the same domain of definition and that each linear combination c1u1 + c2u2 is also in the space. A linear operator on the space is an operator L that transforms each function u of that space into a function Lu, which need not be in the space, and has the property that, for each pair of functions u1 and u2, ( 1) L(c1u1 + c2u2) = c1Lu1 + c2Lu2 whenever c1 and c2 are constants. In particular, (2) L(u1 + u2) = Lu1 + Lu2, L(c1u1) = c1Lu1. 105 106 THE FOURIER METHOD CHAP. 3 The function Lu may be a constant function; we note that L(O) =L(00) =OL(O) =0. If u3 is a third function in the space, then L(c1u1 + c2u2 + c3u3) = L(c1u1 + c2u2) + L(c3u3) = c1Lu1 + c2Lu2 + c3Lu3. Proceeding by induction, we find that L transforms linear combinations of N functions in this manner: N (3) L i: n=1 N i: n=1 EXAMPLE 1. Suppose that both u1 and u2 are functions of the dent variables x and y. According to elementary properties of derivatives, a derivative of any linear combination of the two functions can be written as the same linear combination of the individual derivatives. Thus ( 4) a au1 ax ax —(c1u1 + c2u2) = + au2 ax provided au1/ax and au2/ax exist. In view of property (4), the class of functions of x and y that have partial derivatives of the first order with respect to x in the xy plane is a function space. The operator a/ax is a linear operator on that space. It is naturally classified as a linear differential operator. EXAMPLE 2. Consider a space of functions u(x, y) defined on the xy plane. If f(x, y) is a fixed function, also defined on the xy plane, then the operator L that multiplies each function u(x, y) by f(x, y) is a linear operator, where Lu =fu. If linear operators L and M, distinct or not, are such that M transforms each function u of some function space into a function Mu to which L applies and if u1 and u2 are functions in that space, it follows from equation (1) that ( 5) LM(c1u1 + c2u2) = L(c1Mu1 + c2Mu2) = c1LMu1 + c2LMu2. That is, the product LM of linear operators is itself a linear operator. The sum of two linear operators is defined by the equation (L+M)u=Lu+Mu. (6) If we replace u here by c1u1 + c2u2, we can see that the sum L + M is a linear operator and hence that the sum of any finite number of linear operators is linear. EXAMPLE 3. Let L denote the linear operator a2/ax2 defined on the space of functions u(x, y) whose derivatives of the first and second order with SEC 26 PRINCIPLE OF SUPERPOSITION 107 f respect to x exist in a given domain of the xy plane. The product M = a/ax of the linear operators in Examples 1 and 2 is linear on the same space, and the sum L+M= is, 26. a2 a ax2 ax therefore, linear. PRINCIPLE OF SUPERPOSITION Each term of a linear homogeneous differential equation in u (Sec. 1) consists of a product of a function of the independent variables by one of the derivatives of u or by u itself Hence every linear homogeneous differential equation has the form Lu=O, (1) where L is a linear differential operator. For example, if (2) L=A a2 ax2 +B a2 a2 ayax +C ay2 a a ax ay where the letters A through F denote functions of x and y only, equation (1) is the linear homogeneous partial differential equation =0 (3) in u = u(x, y). Linear homogeneous boundary conditions also have the form (1). Then the variables appearing as arguments of u and as arguments of functions that serve as coefficients in the linear operator L are restricted so that they represent points on the boundary of a domain. Now let (n = 1, 2, . , N) denote functions that satisfy equation (1); that is, = 0 for each n. It follows from property (3), Sec. 25, of linear operators that each linear combination of those functions also satisfies equation (1). We state that principle of superposition of solutions, which is fundamental to the Fourier method for solving linear boundary value problems, as follows. . . Theorem 1. If each of N functions u1, u2, . . , uN satisfies a linear homogeneous differential equation Lu = 0, then every linear combination . ( 4) u= c1u1 + c2u2 + +cNuN, where the c's are arbitrary constants, satisfies that differential equation. If each of the N functions satisfies a linear homogeneous boundary condition Lu = 0, ihen every linear combination (4) also satisfies that boundary condition. The principle of superposition is useful in ordinary differential equations. For example, from the two solutions y = ex and y = e_X of the linear homoge- 108 THE FOURIER METHOD CHAP. 3 neous equation y" — y = 0, the general solution y ciex + c2e_x can be written. In this book, we shall be concerned mainly with applying the principle of superposition to solutions of partial differential equations. EXAMPLE. Consider the linear homogeneous heat equation (Sec. 2) (5) t) = (0 c x it) c c, z' > 0), together with the linear homogeneous boundary conditions (6) = 0, It is easy to verify that if L = k82/8x2 I — 2 (n = 0 C 0, 1, 2, . . . (t n'rrx (n=1,2,...), > 0). a/at and n2'ir2k uo=—, then Lu,, = — = 0 cos 2 C ). Hence it follows from Theorem 1 that Lu = 0 for every linear combination u = a0u0 + a1u1 + a2u2 + That is, when N 1, the function a0 (7) +aNuN. N — n=1 mrx n2ir2k C cos 2 C satisfies the heat equation (5). Although it seems more natural to write u0 = 1, with a0 instead of a0/2 in expression (7), our choice of u0 = f is simply for convenience in notation later on (Sec. 27). As for boundary conditions (6), we write L = a/ax and observe that (n = 0, 1, 2, . . ) has value zero when x = 0 and x = C. So, again by Theorem 1, Lu has value zero when x = 0 and x = C. This shows that the linear combination (7) also satisfies boundary conditions (6). . In order to state a general principle of superposition, similar to Theorem 1, that applies to an infinite set of functions u1, u2, . . , one must deal with the convergence and differentiability of infinite series involving those functions. This is indicated below. Suppose that the functions and the constants c,, are such that the infinite series whose terms are converges throughout some domain of the independent variables. The sum of that series is a function . ( 8) u = Let x represent one of the independent variables. The series is differentiable, or termwise differentiable, with respect to x if the derivatives au/ax PRINCIPLE OF SUPERPOSITION SEC. 26 exist converges to ôu/ôx: and if the series of functions c, ôu ôu (9) 109 ox Ox n=1 Note that a series must be convergent if it is to be differentiable. Sufficient conditions for differentiability were noted in Sec. 22. If, in addition, series (9) is differentiable with respect to x, then series (8) is differentiable twice with respect to x. Let L be a linear operator where Lu is a product of a function of the independent variables by u or by a derivative of u, or where Lu is a sum of a finite number of such terms. We now show that if series (8) is differentiable for in series (8) all the derivatives involved in L and if each of the functions = 0, then so does u; satisfies the linear homogeneous differential equation that is, Lu = 0. To accomplish this, we first note that, according to the definition of the sum of an infinite series, N Ou f— =f lim aX N-oo Ou Ox n=1 when series (8) is differentiable with respect to x. Thus N Ou f— = lim f— N-oo OX (10) c,,u,. Here the operator 0/Ox can be replaced by other derivatives if the series is so differentiable. Then, by adding corresponding sides of equations similar to equation (10), including one that may not have any derivative at all, we find that N Lu = lim L (11) N—oo n=1 The sum on the right-hand side of equation (11) is a linear combination of the functions u1, u2, . , uN; and if = 0 (n = 1, 2,...), Theorem 1 allows us to . . write N =0 L n=1 for every N. Hence, from equation (11), we have the desired result Lu = 0. A linear homogeneous boundary condition is also represented by an equation Lu = 0. In that case, we may require the function Lu to satisfy a condition of continuity at points on the boundary so that its values there will 110 THE FOURIER METHOD CHAP 3 represent limiting values as those points are approached from the interior of the domain. The following generalization of Theorem 1 is now established. Theorem 2. Suppose that each function of an infinite set u1, u2,... satisfies a linear homogeneous differential equation or boundary condition Lu = 0. Then the infinite series u= (12) n=1 where the is are constants, also satisfies Lu = 0, provided the series converges and differentiable for all derivatives involved in L and provided any required continuity condition at the boundary is satisfied by Lu when Lu = condition. 0 is a boundary With Theorem 2, we are now ready to begin our illustration of the Fourier method for solving boundary value problems. 27. A TEMPERATURE PROBLEM The linear boundary value problem t) = =0, u(x,0) =f(x) (0 <x < c, > 0), t) t =0 (t>0), (O<x<c) is a problem for the temperatures u(x, t) in an infinite slab of material, bounded by the planes x = 0 and x = c, if its faces are insulated and the initial temperature distribution is a prescribed function f(x) of the distance from the face x = 0. (See Fig. 26). It is also a problem of determining temperatures in a 11 II FIGURE 26 A TEMPERATURE PROBLEM SEC. 27 u(x,O) =f(x) = 111 = x x=c x=O FIGURE27 bar of uniform cross section, such as one in the shape of a right circular cylinder (Fig. 27), when its bases in the planes x = 0 and x = c and its lateral surface, parallel to the x axis, are insulated and its initial temperatures are f(x) <x <c). We assume that the thermal diffusivity k of the material is constant throughout the slab, or bar, and that no heat is generated within it. We saw in the example in Sec. 26 that the functions (0 u0 = nri-x n2ri-2k 1 (4) = exp t cos (n = 1,2,...) all satisfy the homogeneous conditions (1) and (2) in the stated temperature problem. In this section, we describe the method to be used in subsequent chapters for finding such functions. We shall also take into account the nonhomogeneous condition (3) and complete the solution of the boundary value problem. A number of the steps to be taken are formal, or manipulative. The validity of the solution obtained will be established in Sec. 28. To determine nontrivial (u 0) solutions of the homogeneous conditions (1) and (2), we seek separated functions (Sec. 9), or functions of the form u=X(x)T(t), (5) that satisfy those conditions. Note that X is a function of x alone and T a function of t alone. Note, too, that X and T must be nontrivial (X If u = XT satisfies equation (1), then 0, T 0). X(x)T'(t) = kX"(x)T(t); and, for values of x and t such that X(x)T(t) is nonzero, we can divide by kX(x)T(t) to separate the variables: T'(t) kT(t) — — X"(x) X(x) the left-hand side here is a function of t alone, it does not vary with x. However, it is equal to a function of x alone, and so it cannot vary with t. Since Hence the two sides must have some constant value — A in common. That is, T'(t) kT(t) — A, — X"(x) — X(x) — Our choice of —A, rather than A, for the separation constant is, of course, a minor matter of notation. It is only for convenience later on (Chap. 5) that we have written — A. 112 THE FOURIER METHOD CHAP. 3 If u = XT is to satisfy the first of conditions (2), then X'(O)T(t) must vanish for all t (t > 0). With our requirement that T * 0, it follows that X'(O) = 0. Likewise, the second of conditions (2) is satisfied by u = XT if = X'(c) 0. Thus u = XT satisfies conditions (1) and (2) when X and T satisfy these two homogeneous problems: X"(x) + AX(x) = 0, X'(O) = 0, T'(t) + AkT(t) = 0, (8) (9) where X'(c) = 0, the parameter A has the same value in both problems. To find nontrivial solutions of this pair of problems, we first note that problem (9) has no boundary conditions. Hence it has nontrivial solutions for all values of A. Since problem (8) has two boundary conditions, it may have nontrivial solutions for only particular values of A. Problem (8) is called a Sturm-Liouville problem. The general theory of such problems is developed in Chap. 5, where it is shown that A must be real-valued in order for there to be nontrivial solutions. If A = 0, the differential equation in problem (8) becomes X"(x) = 0. Its general solution is X(x) = Ax + B, where A and B are constants. Since X'(x) = A, the boundary conditions X'(O) = 0 and X'(c) = 0 are both satisfied when A = 0. So X(x) = B; and, except for a constant factor, problem (8) has the solution X(x) = if A = 0. Note that any nonzero value of B might have been selected here. If A > 0, we may write A = a2 (a > 0). The differential equation in problem (8) then takes the form X"(x) + a2X(x) = 0, its general solution being X(x) = C1 cos ax + C2 sin ax. Writing X'(x) = —C1a sin ax + C2acosax keeping in mind that a is positive and, in particular, nonzero, we see that the condition X'(O) = 0 implies that C2 = 0. Also, from the condition X'(c) = 0, and it follows that C1a sin ac = 0. Now if X(x) is to be a nontrivial solution of problem (8), C1 * 0. Hence a must be a positive root of the equation sin ac = That is, 0. (n=1,2,...). So, except for the constant factor C1, nlrx X(x)=cos— If A <0, we write A = —a2 (a > 0). This time, the differential equation in problem (8) has the general solution X(x) = (n=1,2,...). + A TEMPERATURE PROBLEM SEC. 27 113 Since X'(x) = — the condition X'(O) = 0 implies that C2 = C1. Hence X(x) = + or X(x) = 2C1coshax. But the condition X'(c) = 0 requires that C1 sinh ac = 0; and, since sinh ac * 0, it follows that C1 = 0. So problem (8) has only the trivial solution X(x) 0 if A <0. A for which problem The values A0 = 0, (8) has nontrivial solutions are called eigenvalues of that problem, and the solutions X0(x) = f, Xn(x) = cos (nirx/c) (n = 1, 2,...) are the corresponding eigenfunctions. We turn now to the differential equation (9) and determine its solution when A is an eigenvalue. The solution when A = factor, T0(t) = 1. When A = A,1 = (nir/c)2 (n = A0 = 0 is, except for a constant 2,...), any solution of equation (9) is evidently a constant multiple of Tn(t) = exp Hence each of the products 1, 1 u0 = X0(x)T0(t) = and (13) =Xn(x)Tn(t) = t exp — nirx cos— (n = 1,2,...) satisfies the homogeneous conditions (1) and (2). These are the solutions (4). The procedure just used to obtain them is called the method of separation of variables. Assuming that the conditions in Theorem 2 of Sec. 26 are satisfied, we may now use that theorem to see that the generalized linear combination a0 nirx °° — n=1 C 2 cos— C of the functions (12) and (13) also satisfies the homogeneous conditions (1) and (2). The remaining (nonhomogeneous) condition U(x, 0) = f(x) requires that a0 00 n=1 nirx (0<x<c), 114 THE FOURIER METHOD or that the constants CHAP. 3 be, in fact, the coefficients 2 (15) nirx c (n=O,1,2,...) C in the Fourier cosine series for f on the interval 0 <x <c (Sec. 21). Our formal solution of the temperature problem (1)—(3) is now complete. It consists of series (14) together with expression (15) for the coefficients The method used, involving separation of variables, superposition, and Fourier series, is the Fourier method. Note that the steady-state temperatures, occurring when t tends to infinity, are a0/2. That constant temperature is evidently the mean, or average, value of the initial temperatures f(x) over the interval 0 <x <c. EXAMPLE. Suppose that the thickness c of the slab is unity and that the initial temperatures are f(x) = x (0 x 1). Here a0 = 2f1xdr = 1. Using integration by parts and observing that sin nir = when n is an integer, we find that xsinni,-x Jo + cosnirx fliT 2 fliT 0 and cos nir = 1 =2 2 (_i)nl 2 (— —1 2 o (n=1,2,...). We have evaluated a0 separately in order to avoid dividing by zero. When c = 1 and these values for (n = 0, 1, 2,...) are used, expression (14) becomes u(x,t) 2 1 = —+ 2 or [see the footnote with Problem 1(b), 1 (16) 28. 2 4 °° exp(—n2ir2kt)cosnirx, 2 n=1 Sec. exp[—(2n — 14] 1)2lr2kt] cos(2n—1)irx. (2n—1) 2 n=1 VERIFICATION OF SOLUTION We turn now to the full verification of the solution of the boundary value problem t) = =0, u(x,0) =f(x) (0 t) <x < c, t > 0), =0 (0<x<c) VERIFICATION OF SOLUTION SEC. 28 115 that was obtained in Sec. 27. We recall that the continuous functions n2ir2k 1 (4) = = exp t cos nirx (n = 1,2,...) — were found to satisfy the homogeneous conditions (1) and (2). As already pointed out in the example in Sec. 26, Theorem 1 in that section ensures that any linear combination N u= n=O of those functions also satisfies conditions (1) and (2). The generalization u= (5) n=O of that linear combination to an infinite series is, of course, the solution nirx a0 (6) — n==1 cos— C are assigned the values in Sec. 27 when the coefficients 2 (7) C 2 nirx c (n=O,1,2,...). CO C We assume that f is piecewise smooth (Sec. 17) on the interval 0 <x <c. Also, at a point of discontinuity of f in that interval, we define f(x) as the mean value of the one-sided limits f(x +) and f(x —). Note how it follows from expression (7) that (n=0,1,2,...) and hence that there is a positive constant M, independent of n, such that (n=O,1,2,...). We begin our verification by showing that series (5), with coefficients (7), actually converges in the region 0 x C, t > 0 of the xt plane and that it satisfies the homogeneous conditions (1) and (2). To accomplish this, we first note from expressions (4) and inequalities (8) that, if t0 is a fixed positive number, — C2 (n = 0,1,2,...) 116 THE FOURIER METHOD x 0 c (t CHAP. 3 t0) to o C whenever 0 x that the series FIGURE28 X c and t t0 (Fig. 28). An application of the ratio test shows 00 n2i,-2k (10) — n=O C 2 to of constants converges when i is any nonnegative integer and, in particular, when i = 0. So we know from the comparison and absolute convergence tests that the series (5) converges when 0 x c, t to. One can use series (10) and the Weierstrass M-test (Sec. 22) to show that the series (11) n=O n=O of derivatives converge uniformly on the interval 0 (t t0). Likewise, the series c for any fixed t x (12) n=O converges uniformly on the semi-infinite interval t (0 x t0 for any fixed x c). The uniformity of the convergence of these series ensures that the series (5) is differentiable twice with respect to x and once with respect to t when x c, t 0 to. Consequently, if we write L = ka2/ax2 — ô/c9t and recall from the example in Sec. 26 that = 0 (n = 0, 1,2,...), we know from Theorem 2 there that Lu = 0 when 0 x c, t to. Thus series (5) converges and satisfies the heat equation (1) in the domain 0 <x <c, t > 0 since the positive number to can be chosen arbitrarily small. Writing L = a/ax and again using Theorem 2 in Sec. 26, we see that series (5) also satisfies boundary conditions (2). Observe that since the first of series (11) is uniformly convergent on the interval 0 x c for any fixed t t) of series (5) is continuous in x on that interval. (t t0), the derivative VERIFICATION OF SOLUTION SEC. 28 (See Fig. 28.) Hence the one-sided limits ,t) = lim x—*c ,t) = lim X<C x>O at 117 the end points of the interval 0 x c (t t0) exist and have the values t) and t), respectively. Since conditions (2) are satisfied and since t0 can be chosen arbitrarily small, then, ,t) (13) = 0, ,t) =0 (t >0). In seeking solutions of boundary value problems, we shall tacitly require that those solutions satisfy such continuity conditions at boundary points. Thus, when conditions (2) are part of a boundary value problem, it is understood that conditions (13) must also be satisfied. As we have just seen, series (5) has that property. The nonhomogeneous condition (3) is clearly satisfied by our solution since series (6) reduces to the Fourier cosine series 00 nirx a0 (0<x<c) (14) n==1 for f when t = 0; and the corollary in Sec. 21 ensures that series (14) converges to f(x) when 0 <x <c. It remains to show that (15) u(x,0+)=f(x) (0<x<c). This is a continuity requirement that must be satisfied when t = 0, just as conditions (13) must hold when x = 0 and x = c. One can show that solution (6) has this property by appealing to a convergence theorem, due to Abel,t that is to be proved in Chap. 9 (Sec. 79). According to that theorem, the series formed by multiplying the terms of a convergent series of constants, such as series (14) with x fixed, by corresponding terms of a bounded sequence of functions of t whose values never increase with n, such as exp (n = 0, 1, 2,...), is uniformly convergent with respect to t. So, for any fixed x (0 <x <c), the series in expression (6) converges uniformly with respect to t 0 and thus represents a function that is continuous in t 0 0). This when t shows that our solution u(x, t) is continuous in t when t 0, in particular when t = ft That is, limu(x,t) ==u(x,0), t>O or u(x, 0 +) = u(x, 0), for each fixed x (0 <x <c). Property (15) now follows from the fact that u(x, 0) = f(x) (0 <x <c). This completes the verification that the function (6) is a solution of the boundary value problem (1)—(3). Henrik Abel, Norwegian, 1802—1829. 118 THE FOURIER METHOD CHAP. 3 PROBLEMS 1. Show that the solution of the temperature problem in Sec. 27 reduces to 2 when c = 'r and f(x) = x2 (—ir <x Suggestion: Refer to the Fourier cosine series for x2 that was found in Problem 4(a), Sec. 14. 2. In Problem 10, Sec. 9, the functions (n=1,2,...) u0=y, were shown to satisfy Laplace's equation and the homogeneous boundary conditions u(x,0) = 0. y) = 0, y) = After writing u = X(x)Y(y) and separating variables, use the solutions of the Sturm-Liouville problem in Sec. 27 to show how these functions can be discovered. Then, by proceeding formally, derive the following solution of the boundary value problem resulting when the nonhomogeneous condition u(x, 2) = f(x) is included: u(x, y) = A0y + sinh ny cos nx, n=1 where A0= (n= 1,2,...). = (The final result in Problem 10, Sec. 9, is a special case of this.) 3. For each of the following partial differential equations in u = u(x, t), determine if it is possible to write u = X(x)T(t) and separate variables to obtain two ordinary differential equations in X and T. If it can be done, find those ordinary differential equations. = 0; (b) (x + = 0; (a) — — + + = 0; (d) = 0. (c) — — 4. Suppose that equation (6), Sec. 27, had been written in the form T'(t) T(t) - X(x) Set each side here equal to —A and show how the functions (12) and (13) in Sec. 27 still follow. (This illustrates how it is generally simpler to keep the physical constant in the heat equation out of the Sturm-Liouville problem, as we did in Sec. 27.) 5. Show that if an operator L has the two properties L(u1 + u2) = Lu1 + Lu2, L(c1u1) = c1Lu1 for all functions u1, u2 in some space and for every constant c1, then L is linear; that is, show that it has property (1), Sec. 25. A VIBRATING STRING PROBLEM SEC. 29 119 6. Use special cases of linear operators, such as L = x and M = 8/8x, to illustrate that products LM and ML are not always the same. 7. Let u and v denote functions of x and t that satisfy the one-dimensional heat equation: and u1 = v1 = Multiply each side of these two equations by constants c1 and c2, respectively, and add to show that the linear combination c1u + c2v also satisfies the heat equation. This illustrates a variation in the proof of Theorem 1 in Sec. 26. 8. Show that each of the functions y1 = 1/x and y2 = 1/(1 + x) satisfies the nonlinear differential equation y' + y2 = 0. Then show that the sum y1 + y2 fails to satisfy that equation. Also show that if c is a constant, where c 0 and c :/: 1, neither cy1 nor cy2 satisfies the equation. 9. Let u1 and u2 satisfy a linear nonhomogeneous differential equation Lu = f, where f is a function of the independent variables only. Prove that the linear combination c1u1 + c2u2 fails to satisfy that equation when c1 + c2 :/: 1. 10. Let L denote a linear differential operator, and suppose that f is a function of the independent variables. Show that the solutions u of the equation Lu = are of the form u = u1 + u2, where the u1 are the solutions of the equation Lu1 = 0 and u2 is any particular solution of Lu2 = f. (This is a principle of superposition of solutions for nonhomogeneous differential equations.) is a bounded sequence of constants, prove that the series 11. Assuming that f u(x, y) = sin nx n=1 converges and is twice-differentiable with respect to x and y when y y0, where y0 + uyy = 0 is any positive constant. Then show that u satisfies Laplace's equation in the half plane y > 0. 12. Prove that if M (n = 1,2,...), where M is a positive constant, then the series y(x,t)= n=1 converges and satisfies the wave equation y11 = 29. for all x and t. A VIBRATING STRING PROBLEM To illustrate further the Fourier method, we now consider a boundary value problem for displacements in a vibrating string. This time, the nonhomogeneous condition will require us to expand a function f(x) into a Fourier sine series, rather than a cosine series. Let us find an expression for the transverse displacements y(x, t) in a string, stretched between the points x = 0 and x = c on the x axis and with no external forces acting along it, if the string is initially displaced into a position y = f(x) and released at rest from that position. The function y(x, t) must 120 THE FOURIER METHOD CHAP. 3 satisfy the wave equation (Sec. 5) (1) t) = (0 <x < c, > 0). t) t It must also satisfy the boundary conditions (2) y(O,t) = 0, y(c,t) = 0, = 0, y(x,0)=f(x) (3) where the prescribed displacement function f is continuous on the interval o c and f(O) =f(c) = 0. We assume a product solution y=X(x)T(t) (4) of the homogeneous conditions (1) and (2) and substitute it into those conditions. This leads to the two homogeneous problems (5) X"(x) + AX(x) = 0, X(0) = 0, X(c) = 0, lta2T(t) = T'(O) = 0. Problem (5) is another instance of a Sturm-Liouville problem. The method of solution that was used to solve the one in Sec. 27 can be applied here. It turns out (Problem 5, Sec. 30) that the eigenvalues are = (nir/c)2 (n = 1, 2,...) and that the corresponding eigenfunctions are = sin (n17-x/c). When A = 1k,1, problem (6) becomes (6) T"(t) + T"(t) nira 0, 2 — T(t) = 0, T'(O) = 0; and it follows that, except for a constant factor, the solution is cos (fir at/c). Consequently, each of the products nirx = sin—cos = (n = 1,2,...) satisfies the homogeneous conditions (1) and (2). According to Theorem 2 in Sec. 26, the generalized linear combination 00 nirx ni,-at y(x,t)= C n=1 also satisfies the homogeneous conditions (1) and (2), provided the constants can be restricted so that the infinite series is suitably convergent and differentiable. That series will satisfy the nonhomogeneous condition (3) if the are such that nirx f(x)= (0<x<c). n=1 Note that this series converges to zero at the end points x = 0 and x = c. Hence if representation (9) is valid, it also holds on the closed interval 0 x c. SEC. 29 A VIBRATING STRING PROBLEM 121 The constants in expression (9) are evidently the coefficients in the Fourier sine series for f on the interval 0 <x < c (Sec. 21): = (10) 2 c (n=1,2,...). -ff(x)sin—-—dx The formal solution of our boundary value problem for the displacements in a vibrating string is, therefore, series (8) with coefficients (10). EXAMPLE. Suppose that the string has length c = 2 and that its mid- point is initially raised to a height h above the horizontal axis. The rest position from which the string is released thus consists of two line segments (Fig. 29). (1,h) 0 FIGURE 29 x (2,0) The function f, which describes the initial position of this plucked string, is given by the equations when and the coefficients in the Fourier sine series for that function on the interval 0 <x <2 can be written = f2f(x) sin —i-— dx = hf1x sin —i-— dx — hf2(x — 2) sin dx. After integrating by parts and simplifying, we find that 8h = fliT fl'Tr sin (n=1,2,...). — 2 Series (8) then becomes y( x, t) Since sin (nir/2) = sin (2n—1)ir 2 0 8h 1 sin 2 = nir sin —i-- cos nirat 2 when n is even and since • = sin 17-\ I = —cosn'Tr = (—1) n+1 (n = 1,2,...), — expression (12) for the displacements of points on the string in question can also 122 THE FOURIER METHOD CHAP. 3 be written (13) y(x,t) = 8h 00 2 sin (2n — 1)irx (2n — 1)irat cos 2 2 Before verifying our solution of the boundary value problem (1)—(3), we comment briefly on its physical interpretation. From expression (8), we can see that, for each fixed x, the displacement y(x, t) is a period function of time t, with period 2c T0=—. (14) a The period is independent of the initial displacement f(x). Since a2 = H is the magnitude of the x component of the tensile force and is the mass per unit length of the string (Sec. 5), the period varies directly with c and and inversely with i/it. It is also evident from expression (8) that, for a given length c and initial displacement f(x), the displacement y depends on only the value of x and the value of the product at. That is, y = at) where the function 4) is the same function regardless of the value of the constant a. Let a1 and a2 denote different values of that constant, and let y1(x, t) and y2(x, t) be the corresponding displacements. Then a1t1 = a2t2 (15) if (0 x c). y1(x, t1) = y2(x, t2) In particular, suppose that only the constant H has different values, H1 and 112. The same set of instantaneous positions is taken by the string when H = H1 and when H = H2. But the times t1 and t2 required to reach any one position have the ratio (16) Except for the nonhomogeneous condition (3), our boundary value problem is satisfied by any partial sum N (17) YN(X, t) = sin nirx cos nirat C n=1 of the series solution (8). Instead of meeting requirement (3), however, it satisfies the condition N y(x,O)== fl7TX n=1 The sum on the right-hand side of equation (18) is, of course, the partial sum consisting of the sum of the first N terms of the Fourier sine series for f on the interval 0 <x <c. Since the odd periodic extension of f is clearly continuous VERIFICATION OF SOLUTION SEC. 30 123 and f' is piecewise continuous, that series converges uniformly to f(x) on the interval 0 x c (Sec. 22). Hence, if N is taken sufficiently large, the sum YN(X, 0) can be made to approximate f(x) arbitrarily closely for all values of x in that interval. The function YN(X, t), which is everywhere continuous together with all its partial derivatives, is therefore established as a solution of the approximating problem obtained by replacing condition (3) in the original problem by condition (18). Corresponding approximations can be made to other problems. But a remarkable feature in the present case is that YN(X, t) never deviates from the actual displacement y(x, t) by more than the greatest deviation of YN(X, 0) from f(x). To see this, we need only recall the trigonometric identity 2 sin A cos B = sin (A + B) + sin (A — B) and write (19) 2 sin nirx cos nirat = sin nir(x + at) C C + sin C — at) C Expression (17) then becomes 1 N N yN(x,t) = + n=1 n=1 and the two sums here are those of the first N terms of the sine series for the odd periodic extension F of the function f, with arguments x + at and x — at. But the greatest deviation of the first sum from F(x + at), or of the second from F(x — at), is the same as the greatest deviation of YN(X, 0) from f(x). VERIFICATION OF SOLUTION 30. In this section, we shall verify the formal solution that we found in Sec. 29 for the boundary value problem t) = y(O, t) = 0, y(c, t) = y(x,0) =f(x), o x (0 <x < c, t > 0), t) 0, = 0. The given function f was assumed to be continuous on the interval c; also, f(0) = f(c) = 0. Assuming further that f' is at least piecewise continuous, we know (Sec. 21) that f(x) is represented by its Fourier sine series when 0 x c. The coefficients 2 c co nirx c (n=1,2,...) 124 THE FOURIER METhOD CHAP. 3 in that series are the ones in the series solution nirx nirat (5) y(x, t) = sin cos C n=1 that we obtained. Hence, when t = 0, the series in expression (5) converges to f(x); that is, y(x, 0) = f(x) when 0 x c. The nature of the problem calls for a solution y(x, t) that is continuous in x and t when 0 x c and t 0 and is such that t) is continuous in t at t = 0. Hence the prescribed boundary values in conditions (2) and (3) are also limiting values on the boundary of the domain 0 <x <c, t > 0: y(O+,t)=O, y(c—,t)=O =0 y(x,O+) =f(x), To verify that expression (5) represents a solution, we must prove that the series there converges to a continuous function y(x, t) which satisfies the wave equation (1) and all the boundary conditions. But series (5), with coefficients (4), can fail to be twice-differentiable with respect to x and t even when it has a sum that satisfies the wave equation. This was, in fact, the case with the solution in the example in Sec. 29, where the coefficients were 8h nir (n=1,2,...). 2 nir series (5) is differentiated twice with respect to x or t when those values are used, it is apparent that the resulting series cannot converge since its nth term does not tend to zero. It is possible, however, to write series (5) in a closed form, which does not involve infinite series. That will enable us to verify our solution. To do this, we first refer to identity (19) in Sec. 29 and write series (5) as After of (6) 1 00 nir(x+at) n7r(x—at) 00 + n=1 n=1 Now the odd periodic extension F of f, with the properties (7) and (8) F(x) = f(x) when 0 F(x+2c) =F(x) F(—x) = —F(x), is represented for all x by the sine series for f: 00 nirx F(x) = sin (9) Consequently, expression (6) can be written y(x,t) = + at) + F(x c forallx, (—oo <x <oo). n=1 (10) x — at)]. SEC. 30 VERIFICATION OF SOLUTION 125 Note that the convergence of series (5) and (6) is ensured for all x and t by the convergence of series (9) for all x. y y =f(x) 2c —C y x FIGURE 30 = F(x) We turn now to the verification of our solution in the form (10). From our assumption that f is continuous when 0 x c and that f(0) = f(c) = 0, we see that the odd periodic extension F is continuous for all x (Fig. 30). Let us also assume that f' and f" are continuous when 0 x c and that f"(O) = f"(c) = 0. It is then easy to show that the derivatives F' and F" are continuous for all x. For, by recalling that F(x) = —F(—x) and then applying the chain rule, we can write d F'(x) = =F'(-x), where F'( —x) denotes the derivative of F evaluated at —x. Thus F' is an even periodic function; likewise, F" is an odd periodic function. Consequently, F' and F" are continuous, as indicated in Fig. 31. F'(x) \c I —c/p F"(x) 2c x / — 0 — x FIGURE 31 To show that the function (10) satisfies the wave equation, we write it as 1 where u = x + at and v = x — 1 at. The chain rule for differentiating composite functions reveals that 3y 3y3u 9y a 0y3v —=——+—-— ôt ôu 3t 3v 3t or — = a — 126 THE FOURIER METHOD CHAP. 3 and, by letting (9y/(9t play the role of y in this last expression, we find that (92y (11) at2 = (9 (9y a2 ôt at 2 — — = —F"(u) a2 + —F"(v). 2 Similarly, (92y 1 = (12) 1 + In view of expressions (11) and (12), the function (10) satisfies the wave equation (1). Furthermore, because F is continuous for all x, the function (10) is continuous for all x and t, in particular when 0 x c and t 0. While it is evident from series (5) that our solution y(x, t) satisfies the conditions y(0, t) = y(c, t) = 0 and y(x, 0) = f(x), expression (10) can also be used to verify this. For example, when x = c in expression (10), one can write F(c — at) = —F(—c + at) = —F(—c + at + 2c) = —F(c + at). Therefore, y(c,t) = As for the final boundary condition —F(c+at)] =0. 0) 0, we observe that —F'(x—at)J. Hence 0) = 0, and the continuity of F' ensures that t) is continuous. The function (10) is now fully verified as a solution of the boundary value problem (1)—(3). In Chap. 9 (Sec. 82), we shall show why it is the only possible solution which, together with its derivatives of the first and second order, is continuous throughout the region 0 x c, t 0 of the xt plane. If the conditions on f' and f" are relaxed by merely requiring those two functions to be piecewise continuous, we find that at each instant t there may be a finite number of points x (0 x c) where the partial derivatives of y fail to exist. Except at those points, our function satisfies the wave equation and the condition 0) = 0. The other boundary conditions are satisfied as before, but we have a solution of our boundary value problem in a broader sense. PROBLEMS 1. A string is stretched between the fixed points 0 and 1 on the x axis and released at rest from the position y = A sin 'rx, where A is a constant. Obtain from expression (10), Sec. 30, the subsequent displacements y(x, t), and verify the result fully. Sketch the position of the string at several instants of time. Answer: y(x, t) = A sin cos 'rat. 2. Solve Problem 1 when the initial displacement there is changed to y = B sin 2'rx, where B is a constant. Answer: y(x, t) = B sin 2'rx cos SEC 31 HISTORICAL DEVELOPMENT 127 3. Show why the sum of the two functions y(x, t) found in Problems 1 and 2 represents the displacements after the string is released at rest from the position y =Asin'rx +Bsin2'rx. 4. By assuming a product solution y = X(x)T(t), obtain conditions (5) and (6) on X and T in Sec. 29 from the homogeneous conditions (1) and (2) of the string problem there. 5. Derive the eigenvalues and eigenfunctions, stated in Sec. 29, of the Sturm-Liouville problem X"(x) + AX(x) = 0, X(0) = 0, X(c) = 0. 6. For the initially displaced string of length c considered in Secs. 29 and 30, show why the frequency v of the vibration, in cycles per unit time, has the value a 1) = —= 2c 2cy — t5 that if H = 200 lb, the weight per foot is 0.01 lb = 0.01, g = 32), and the length is 2 ft, then u = 200 cycles/s. 7. In Secs. 29 and 30, the position of the string at each instant can be shown graphically by moving the graph of the periodic function to the right with velocity a and an identical curve to the left at the same rate and then adding ordinates, on the interval c, of the two curves so obtained at the instant t. Show how this follows from 0 x Show expression (10), Sec. 30. 8. Plot some positions of the plucked string considered in the example in Sec. 29 by the method described in Problem 7 to verify that the string assumes such positions as those indicated by the bold line segments in Fig. 32. y (1,h) 0 (2,0) x FIGURE 32 9. Write the boundary value problem (1)—(3), Sec. 29, in terms of the two independent variables x and r = at to show that the problem in y as a function of x and T does not involve the constant a (see Sec. 5). Thus, without solving the problem, deduce that the solution has the form y = 4(x, T) = 4(x, at) and hence that relation (15), Sec. 29, is true. 31. HISTORICAL DEVELOPMENT Mathematical sciences experienced a burst of activity following the invention of calculus by Newton (1642—1727) and Leibnitz (1646—1716). Among topics in mathematical physics that attracted the attention of great scientists during that 128 THE FOURIER METHOD CHAP. 3 period were boundary value problems in vibrations of strings, elastic bars, and columns of air, all associated with mathematical theories of musical vibrations. Early contributors to the theory of vibrating strings included the English mathematician Brook Taylor (1685—1731), the Swiss mathematicians Daniel Bernoulli (1700—1782) and Leonhard Euler (1707—1783), and Jean d'Alembert (1717—1783) in France. By the 1750s d'Alembert, Bernoulli, and Euler had advanced the theory of = vibrating strings to the stage where the partial differential equation was known and a solution of a boundary value problem for strings had been found from the general solution of that equation. Also, the concept of funda- mental modes of vibration led those men to the notion of superposition of solutions, to a solution of the form (8), Sec. 29, where a series of trigonometric functions appears, and thus to the matter of representing arbitrary functions by trigonometric series. Euler later found expressions for the coefficients in those series. But the general concept of a function had not been clarified, and a lengthy controversy took place over the question of representing arbitrary functions on a bounded interval by such series. The question of representation was finally settled by the German mathematician Peter Gustav Lejeune Dirichlet (1805—1859) about 70 years later. The French mathematical physicist Jean Baptiste Joseph Fourier (1768—1830) presented many instructive examples of expansions in trigonometric series in connection with boundary value problems in the conduction of heat. His book Théorie analytique de la chaleur, published in 1822, is a classic in the theory of heat conduction. It was actually the third version of a monograph that He he originally submitted to the Institut de France on December 21, effectively illustrated the basic procedures of separation of variables and superposition, and his work did much toward arousing interest in trigonometric series representations. But Fourier's contributions to the representation problem did not include conditions of validity; he was interested in applications and methods. As noted above, Dirichlet was the first to give such conditions. In 1829 he firmly established general conditions on a function sufficient to ensure that it can be represented by a series of sine and cosine functions.t Representation theory has been refined and greatly extended since Dirichlet's time. It is still growing. Freeman's early translation of Fourier's book into English was reprinted by Dover, New York, in 1955. The original 1807 monograph itself remained unpublished until 1972, when the critical edition by Grattan-Guinness that is listed in the Bibliography appeared. supplementary reading on the history of these series, see the articles by Langer (1947) and Van Vleck (1914) that are listed in the Bibliography. CHAPTER 4 BOUNDARY VALUE PROBLEMS This chapter is devoted to the application of Fourier series in solving various types of boundary value problems that are mathematical formulations of problems in physics. The basic method has already been described in Chap. 3. Except for the final section of this chapter (Sec. 40), we shall limit our attention to problems whose solutions follow from the solutions of the two SturmLiouville problems encountered in Secs. 27 and 29 of Chap. 3. To be specific, we saw there that the Sturm-Liouville problem (1) X"(x) + AX(x) = on the interval 0 x 0, X'(O) = X'(c) = 0, 0, c, has nontrivial solutions only when A is one of the eigenvalues 2 A0 = 0, = (n = 1,2,...) (—) and that the corresponding solutions, or eigenfunctions, are X0(x) = nlrx 1 = cos (n = 1,2,...). For the Sturm-Liouville problem X"(x) + AX(x) = 0, X(0) = 0, X(c) = 0, 129 130 CHAP 4 BOUNDARY VALUE PROBLEMS on the same interval 0 x c, 2 (n = 1,2,...) An and = fl7TX sin— (n = 1,2,...). As illustrated in Chap. 3, the solutions of problems (1) and (2) lead to Fourier cosine and sine series representations, respectively. A third SturmLiouville problem, to be solved in Sec. 40, leads to Fourier series with both cosines and sines. Boundary value problems whose solutions involve terms other than cos (nirx/c) and sin are taken up in Chap. 5, where the general theory of Sturm-Liouville problems is developed, and in subsequent chapters. In Chap. 3, we indicated ways of proving that a solution found for a given boundary value problem truly satisfies the partial differential equation and all the boundary conditions and continuity requirements. When that is done, the solution is rigorously established. But, even for many of the simpler problems, the verification of solutions may be lengthy or difficult. The boundary value problems in this chapter will be solved only formally in the sense that we shall not always explicitly mention needed conditions on functions whose Fourier series are used and we shall not verify the solutions. We shall also ignore questions of uniqueness, but the physics of a given boundary value problem that is well posed generally suggests that there should be only one solution of that problem. In Chap. 9 we shall give some attention to uniqueness of solutions. 32. A SLAB WITH VARIOUS BOUNDARY CONDITIONS We consider here the problem of finding temperatures in the same slab (or bar) as in Sec. 27 when its boundary surfaces are subjected to other simple thermal conditions. For convenience, however, we take the thickness of the slab as Eigenvalues A,1 = (n = 1, 2,...) then become simply A,1 = n2. As illustrated in Problem 4 of this section, temperature formu- units, so that c = las for a slab of arbitrary thickness c follow readily once they are found when c = ir. In each of the three examples below, the temperature function u = u(x, t) is to satisfy the one-dimensional heat equation (0 <x <ir, t> = 0). EXAMPLE 1. If both faces of the slab are kept at temperature zero and the initial temperatures are f(x) (Fig. 33), then u(O,t) = 0, = 0, and u(x,0) =f(x). Conditions (1) and (2) make up the boundary value problem; and, by separation SEC. 32 u = 0 A SLAB WITH VARIOUS BOUNDARY CONDITIONS u(x, 0) = f(x) 131 u=0 0 x x FIGURE 33 of variables, we find that a function u = X(x)T(t) satisfies the homogeneous conditions if X"(x) + AX(x) = (3) and (4) X(0) = 0, T'(t) + AkT(t) = 0. 0, = 0 According to Sec. 29, the Sturm-Liouville problem (3) has eigenvalues and eigenfunctions = sin nx (n = 1,2,...). The corresponding functions of t arising from equation (4) are, except for constant factors, A= = exp ( —n2kt). Formally, then, the function (5) u(x,t) satisfies all the conditions in the boundary value problem, including the nonhomogeneous condition u(x, 0) = f(x), if f(x)= (6) n=1 Let us assume that f is piecewise smooth on the interval 0 <x is represented by its Fourier sine series (6), where Then f(x) (n=1,2,...). ITO The function (5), with coefficients (7), is our formal solution of the boundary value problem (1)—(2). It can be expressed more compactly in the form u(x, t) = sin nxf f(s)sin nsds, where the variable of integration s is used to avoid confusion with the free variable x. 132 BOUNDARY VALUE PROBLEMS CHAP. 4 EXAMPLE 2. If the slab is initially at temperature zero throughout and is kept at a the face x = 0 is kept at that temperature while the face x = constant temperature u0 when t > 0, then u(0,t) =0, (8) u(x,O) =0. =u0, The boundary value problem consisting of equations (1) and (8) is not in proper form for the method of separation of variables because one of the two-point boundary conditions is nonhomogeneous. If we write u(x,t) = U(x,t) (9) + however, those equations become t) + t) = and U(0,t) + = 0, U(x,O) + = u0, + = 0. Suppose now that (10) = and 0 = 0, = u0. Then U satisfies the conditions u = U(0,t) = U(i,-,t) = 0, U(x,0) = Conditions (10) tell us that '1'(x) = (u0/Tr)x. Hence problem (11) is a special case of the one in Example 1, where f(x) = (— When f(x) is this particular function, the coefficients in solution (5) can be found by evaluating the integrals in expression (7). But since we already know from (11) Example 1, Sec. 14, that 00 x= (12) (0<x<ii-) sinnx n=1 and since the numbers function f(x) = (— are the coefficients in the Fourier sine series for the on the interval 0 <x <n-, we can see at once that n+1 Uo = ——2 IT fl = —2 (—1) fl fl IT (n = 1,2,...). Consequently, U0 00 (—1) e_nktsinnx. 2 n=1 By letting t tend to infinity in solution (13), we see that the function represents the steady-state temperatures in the slab. In fact, conditions (10) consist of Laplace's equation in one dimension together with the 1(x) = conditions that the temperature be 0 and u0 at x = 0 and x = IT, respectively. SEC 32 A SLAB WITH VARIOUS BOUNDARY CONDITIONS 133 Expression (9), in the form U(x,t) = u(x,t) — reveals that U(x, t) is merely the desired solution minus the steady-state temperatures. Finally, note that one can replace the term x in solution (13) by its representation (12) and write that solution as (14) u(x, t) = (1)fl+1 2u0 (1 2 — kt) sin nx. This alternative form can be more useful in approximating u(x, t) by a few terms of the series when t is small. For the factors 1 — exp (—n2kt) are then small compared to the factors exp (—n2kt) in expression (13). Hence the terms that are discarded are smaller. The terms in series (13) are, of course, smaller when t is large. EXAMPLE 3. Suppose that the face x = and that the face x = is insulated. Then (15) u(O,t) = 0 and 0 is kept at temperature zero t) = 0 (t> 0). Also, let the initial temperatures be u(x,0) =f(x) (16) (0 <x <n-), where f is piecewise smooth. By writing u = X(x)T(t) and separating variables, we find that X"(x) + X(0) = 0, X'('n-) = 0. Although this problem in X can be treated by methods to be developed in Chap. 5, we are not fully prepared to handle it at this time. The stated AX(x) = 0, temperature problem can, however, be solved here by considering a related problem in a larger slab 0 x 2ir (Fig. 34). U =0 u(x,O)=f(21T—x) iu=O u(x,0) =f(x) x x=ir FIGURE34 134 BOUNDARY VALUE PROBLEMS CHAP. 4 Let the two faces x = 0 and x = of that larger slab be kept at temperature zero; and let the initial temperatures be u(x,O) =F(x) (17) where F(x) = If(x) (18) — when when x) 0 <x <x The function F is a piecewise smooth extension of the function f on the interval 0 <x <2i,-, and the graph of y = F(x) is symmetric with respect to the line x = ir. This procedure is suggested by the fact that, with the initial condition (17), no heat will flow across the midsection x = of the larger slab. So, when the variable x is restricted to the interval 0 <x <ir, the temperature function for the larger slab will be the desired one for the original slab. According to Problem 4(b), which gives the solution of the boundary value problem in Example 1 for a slab of arbitrary thickness, the temperature function for the larger slab is 00 n2k u(x,t)= (19) nx ——-,—t n=1 where the are the coefficients in the Fourier sine series for the function F on the interval 0 <x <2ir: 1 2 This expression can be written in terms of the original function f(x) by simply referring to Problem 11, Sec. 21, which tells us that = that is, = 0 [i flX (n = 1,2,...); — 2 and 2 (20) iTO f(x)sin (2n—1)x 2 (n=1,2,...). dx Solution (19) then becomes (21) (2n — 1)2k u(x,t) = — with coefficients (20). t sin (2n — 1)x 2 SEC. 32 PROBLEMS 135 PROBLEMSt 1. Let the initial temperature distribution be uniform over the slab in Example 1, Sec. 32, so that f(x) = U0. Find u(x, t) and the flux — t) across a plane x = 17-) when t > 0. Show that no heat flows across the center plane x = 'r/2. (0 x0 2. Suppose that f(x) = sin x in Example 1, Sec. 32. Find u(x, t) and verify the result fully. Suggestion: Use the integration formula (10), Sec. 11. sin x. Answer: u(x, t) = 3. Let v(x, t) and w(x, t) denote the solutions found in Examples 1 and 2 in Sec. 32. Assuming that those solutions are valid, show that the sum u = v + w gives a temperature formula for a slab 0 x 'r whose faces x = 0 and x = 'r are kept at temperatures 0 and u0, respectively, and whose initial temperature distribution is f(x). 4. The faces x = 0 and x = c of a slab 0 x c, which is initially at temperatures f(x), are kept at temperature zero. Use the following method to derive an expression for the temperatures u = u(x, t) throughout the slab when t > 0. (a) After writing the boundary value problem for the temperatures, make the substitution s = to show that u1 = u = 0 when s = 0 and s = 'r, and u = f(cs/'r) when t = 0. (b) By referring to the solution (5), with coefficients (7), of the problem in Example 1, Sec. 32, write an expression for u in terms of s and t. Then, with the aid of the relation s = 'rx/c that was used in part (a), show that n2'r2k u(x,t)= — 2 sin— n= 1 where 2 mrx c Co 5. (n=1,2,...). C (a) Show that if A is a constant and C f(x)= A when—<x<c 0 2 the temperature formula in Problem 4(b) becomes u(x, t) = 4A sin2 (mr/4) exp — C 2 sin n'rx C (b) Two slabs of iron (k = 0.15 cgs unit), each 20 cm thick, are such that one is at 100°C and the other at 0°C throughout. They are placed face to face in perfect contact, and their outer faces are kept at 0°C. Use the result in part (a) here to tOnly formal solutions of the boundary value problems here and in the sets of problems to follow are expected, unless the problem specifically states that the solution is to be fully verified. Partial verification is often easy and helpful. 136 BOUNDARY VALUE PROBLEMS CHAP 4 show that the temperature at the common face 10 mm after contact has been made is approximately 36°C. Then show that if the slabs are made of concrete (k = 0.005 cgs unit), it takes approximately 5 h for the common face to reach that temperature of 36°C. [Note that u(x, t) depends on the product kt.] 6. Let u(r, t) denote temperatures in a solid sphere r a, where r is the spherical coordinate (Sec. 4), when that solid is initially at temperatures f(r) and its surface r = a is kept at temperature zero (Fig. 35). The function u u(r, t) satisfies the conditions k82 8u u(a,t)=0, u(r,0)=f(r). u =0 r=a FIGURE 35 Introduce the new function v(r, t) = ru(r, t), and note that v(0, t) = 0 because u is continuous at the center r = 0. Set up a new boundary value problem in v; and, with the aid of the solution in Problem 4(b), derive the expression u(r,t) = n2'r2k 00 2 — a2 n'rr fllT5 a sin—_—fsf(s)sin__—ds. 7. A solid spherical body 40 cm in diameter, initially at 100°C throughout, is cooled by keeping its surface at 0°C. Use the temperature formula in Problem 6, and also the fact that (sin O)/O tends to unity as 0 tends to zero, to show formally that u(0 + ,t) = 200 ( Thus find the approximate temperature at the center of the sphere 10 mm after cooling begins when the material is (a) iron, for which k = 0.15 cgs unit; (b) concrete, for which k = 0.005 cgs unit. Answers: (a) 22°C; (b) 100°C. 8. The initial temperature of a slab 0 x 'r is zero throughout, and the face x = 0 is kept at that temperature. Heat is supplied through the face x = at a constant rate t) = A (see Sec. 3). Use the solution of the A (A > 0) per unit area, so that problem in Example 3, Sec. 32, to derive the expression A 8 u(x,t)=— x+—E K for the temperatures in this slab. 2exp — (2n — 1)2k 4 (2n — 1)x t sin 2 ____ SEC 33 9. THE SLAB WITH INTERNALLY GENERATED HEAT 137 Let v(x, t) denote temperatures in a slender wire lying along the x axis. Variations of the temperature over each cross section are to be neglected. At the lateral suriace, the linear law of surface heat transfer between the wire and its surroundings is assumed to apply (see Problem 7, Sec. 4). Let the surroundings be at temperature zero; then t), t) — bv(x, v1(x, t) = where b is a positive constant. The ends x = 0 and x = c of the wire are insulated (Fig. 36), and the initial temperature distribution is f(x). Solve the boundary value problem for v by separation of variables. Then show that v(x, t) = u(x, where u is the temperature function found in Sec. 27. t°°t t x=c 0 00 FIGURE 36 10. Use the substitution v(x, t) = u(x, t)exp ( —bt) to reduce the boundary value problem in Problem 9 to the one in Sec. 27. 11. Assuming that the ends of the wire in Problem 9 are not insulated, but kept at temperature zero instead, find the temperature function. 12. Solve the boundary value problem consisting of the differential equation where b is a positive constant, and the boundary conditions u(O,t) = u('r,t) = 0, 1, u(x,0) = 0. Also, give a physical interpretation of this problem (see Problem 9). Suggestion: The Fourier series for sinh ax in Problem 5, Sec. 16, is useful here. Answer: u(x, t) = • ,_ + 2 n (—1) n 2 e" 'sin nx. 2 33. THE SLAB WITH INTERNALLY GENERATED HEAT We consider here the same infinite slab 0 x as in Sec. 32, but we assume that there is a source that generates heat at a rate per unit volume which depends on time. The slab is initially at temperatures f(x), and both faces are maintained at temperature zero. According to Sec. 2, the temperatures u(x, t) in the slab must satisfy the modified form (1) t) = t) + q(t) (0 <x <n-, t> 0) 138 BOUNDARY VALUE PROBLEMS CHAP. 4 of the one-dimensional heat equation, where q(t) is assumed to be a continuous function of t. The conditions =0, u(O,t) =0, u(x,0) =f(x) and complete the statement of this boundary value problem. Since the differential equation (1) is nonhomogeneous, the method of separation of variables cannot be applied directly. We shall use here, instead, a method known as the method of variation of parameters. Also called the method of eigenfunction expansions, it is often useful when the differential equation is nonhomogeneous, especially when the term making it so is time-dependent. To be specific, we seek a solution of the boundary value problem in the form u(x,t)= n=1 of a Fourier sine series whose coefficients are differentiable functions of t. The form (3) is suggested by Example 1, Sec. 32, where the problem is the same as this one when q(t) 0 in equation (1). We anticipate that the function q(t) in solution (5), Sec. 32, of the in equation (1) will cause the coefficients homogeneous part of that earlier problem to depend on t. Instead of writing exp (—n2kt), we combine with the exponential function and denote the product by So our approach here is, in fact, to start with a generalized linear combination, with coefficients depending on t, of the eigenfunctions sin nx (n = 1,2,...) of the Sturm-Liouville problem arising in Example 1, Sec. 32. The reader will note that the method of finding a solution of the form (3) is similar in spirit to the method of variation of parameters which is used in solving linear ordinary differential equations that are nonhomogeneous. We assume that series (3) can be differentiated term by term. Then, by substituting it into equation (1) and recalling [Problem 1(b), Sec. 141 that i sinnx (0<x<ir), F n=1 we may write 00 i F sinnx + q(t) = k n=1 n=1 sinnx, — n=1 or 00 i F sinnx = + q(t)sinnx. — n=1 n=1 By identifying the coefficients in the sine series on each side of this last equation, we now see that + = 2[1 - q(t) (n = 1,2,...). THE SLAB WITH INTERNALLY GENERATED HEAT SEC 33 139 Moreover, according to the third of conditions (2), sin nx=f(x) n=1 and this means that (5) where (n = Bn(O) = 1,2,...), are the coefficients (6) = —f iTo f(x)sinnxdx (n = 1,2,...) in the Fourier sine series for f(x) on the interval 0 <x < For each value of n, equations (4) and (5) make up an initial value problem in ordinary differential equations. To solve the linear differential equation (4), we observe that an integrating factor ist exp fn2k dt = exp n2kt. Multiplication through equation (4) by this integrating factor puts it in the form = 2[1 (1)] en2 ktq(t) where the left-hand side is an exact derivative. If we replace the variable t here by T and integrate each side from r = 0 to T = t, we find that [etl 2 kTB(T)] t 2 f en kTq(r) dr. flTr o In view of condition (5), then, 2[1 — = + (_1y] dT. Finally, by substituting this expression for Bn(t) into series (3), we arrive at the formal solution of our boundary value problem: (8) u(x,t) = E1)ne_nl2ktsi11nx + 4 — sin(2n—1)x 2n—1 f e_(2n_1)2k(t_T)q(T) dT. o tThe reader will recall that any linear first-order equation y' + p(t)y = g(t) has an integrating factor of the form expfp(t) dt. See, for instance, the book by Rainville and Bedient (1989, chap. 2) that is listed in the Bibliography. 140 BOUNDARY VALUE PROBLEMS CHAP. 4 Observe that the first of these series represents the solution of the boundary value problem in Example 1, Sec. 32, where q(t) 0. To illustrate how interesting special cases of solution (8) are readily obtained, suppose now that f(x) 0 in the third of conditions (2) and that q(t) is the constant function q(t) = q0. Since = 0 (n = 1,2,...) and f e_(2n_1)2/c(t_T)q0 dT = q 1— exp[_(2n — 1)2kt] k (2n—1)2 solution (8) reduces t0t u(x,t) = (9) 00 4q0 exp[—(2n 1 — — 1)2kt] sin(2n (2n — 1) n=1 — 1)x. In view of the Fourier sine series representation (Problem 5, Sec. 14) 8 sin(2n—1)x °° x(i,-—x)=— (0<x<n-), (2n—1) 3 solution (9) can also be written (10) u(x,t) = 4q0 q0 — 2k exp[—(2n 00 irk n=1 1)2kt} — sin(2n (2n — 1) — 1)x. (See remarks at the end of Example 2, Sec. 32.) PROBLEMS 1. The boundary value problem t) + xp(t) u1(x, t) = u(1,t)=0, u(O,t)=0, (0 <x < 1, t > 0), u(x,O)=0 describes temperatures in an internally heated slab, where the units for t are chosen so that the thermal conductivity k of the material can be taken as unity (compare Problem 10, Sec. 4). Solve this problem by recalling [Problem 5(a), Sec. 21] the expansion 2 00 (0<x<1) sinmrx and using the method of variation of parameters. Answer: u(x, t) = 2 — (1)fl+1 sin n7rxf 2 dr. 0 tThis result occurs, for example, in the theory of gluing wood with the aid of radio-frequency heating. See G. H. Brown, Proc. Inst. Radio Engrs., vol. 31, no. 10, pp. 537—548, 1943, where operational methods are used. SEC. 33 2. 141 PROBLEMS Show that when the function p(t) in Problem 1 is the constant function p(t) = solution obtained there reduces to 2a (—1) n+1 a, the 22 1 3. Let u(x, t) denote temperatures in a slab 0 x 1 that is initially at temperature zero throughout and whose faces are at temperatures u(O,t)=0 u(1,t)=F(t), and where F(t) and F'(t) are continuous when t 0 and where F(0) = 0. The unit of time is chosen so that the one-dimensional heat equation has the form u1(x, t) = t). Write u(x,t) = U(x,t) +xF(t), and observe how it follows from the stated conditions on the faces of the slab that U(1,t)=0. and Transform the remaining conditions on u(x, t) into conditions on U(x, t), and then refer to the solution found in Problem 1 to show that (—1Y u(x, t) = xF(t) + 4. dr. sin Show that when F(t) = At, where A is a constant, the expression for u(x, t) derived in Problem 3 becomes u(x,t)=A xt+32 1— 5. By writing u(x,t)= and recalling [Problem 5(b), Sec. c2 A0(t) 2 mrx + 21] that 4c 2 (—1) 2 n=i solve the following temperature problem for a slab 0 t) u1(x, t) = = 0, = 0, where x (0<x<c), c with insulated faces: (0 <x <c, > 0), + ax2 u(x,0) = t 0, a is a constant. Thus show that u(x, 6. inrx cos— c 4c 2 t) = ac2 — 22 (—1) + —a-— 3 n 1 — exp — c 2 cos n'rx C bar, with its lateral surface insulated, is initially at temperature zero, and its ends x = 0 and x = c are kept at that temperature. Because of internally generated heat, the temperatures in the bar satisfy the differential equation A u1(x, t) = t) + q(x, t) (0 <x <c, t> 0). 142 CHAP 4 BOUNDARY VALUE PROBLEMS Use the method of variation of parameters to derive the temperature formula 2 mrx u(x,t) = — where denotes the iterated integrals = f c exp — C 0 (t —T) fq(x,T)sin_—dxdr 2 (n = 1,2,...). C 0 Suggestion: Write n'rx q(x,t) = 2 where c = —f CO n=1 q(x,t)sin —dx. C 7. Use the method of variation of parameters to solve the temperature problem t) = b(t)u(x, t) + q0 t) — u('r,t)=O, u(O,t)=0, (0 <x <jr, t> 0), u(x,0)=0, where q0 is a constant.t (See Problem 7, Sec. 4.) Answer:u(x,t)= sin(2n — 1)x 4q0 'ra(t) 2n — n=1 1 o where a(t) = exp ftb(u) do-. 8. When the term making the heat equation nonhomogeneous is a constant or a function of x only, the substitution u(x, t) U(x, t) + used in Example 2, Sec. 32, is often a convenient alternative to the method of variation of parameters. Use that substitution and the solution of the problem in Example 1, Sec. 32, to derive the following solution of the boundary value problem (1)—(2) in Sec. 33 when q(t) = u(x, t) = q0 there: — x) + sin nx, where 2 9. Show that when f(x) q0 sinnxdx (n=1,2,...). 2k 0, the solution obtained in Problem 8 can be put in the form (9), Sec. 33. 10. A solid sphere r 1 is initially at temperature zero, and its surface is kept at that temperature. Heat is generated at a constant uniform rate per unit volume throughout the interior of the sphere, so that the temperature function u = u(r, t) satisfies tIn finding an integrating factor for the ordinary differential equation that arises, it is useful to note that b(cr) do is an antiderivative of b(t). DIRICHLET PROBLEMS SEC 34 143 the nonhomogeneous heat equation 3u = k82 (0 < r < 1, t > 0), +q0 where q0 is a positive constant. Make the substitution u(r,t) = U(r,t) + in the temperature problem for this sphere, where U and 1 are to be continuous tends to zero as when r = 0. [Note that this continuity condition implies that r tends to zero.] Then refer to the solution derived in Problem 6, Sec. 32, to write the solution of a new boundary value problem for U(r, t) and thus show that u(r, t) = kr —r(1 — r2) + 2 6 n=1 Suggestion: It is useful to note that, in view of the formula for the coefficients in a Fourier sine series, the values of certain integrals that arise are, except for a constant factor, the coefficients in the series [Problem 7(a), Sec. 211 12 sinn'rx (0<x<1). DIRICHLET PROBLEMS 34. As already noted in Sec. 7, a boundary value problem in u is said to be a Dirichlet problem when it consists of Laplace's equation V2u = 0, which states that u is harmonic in a given domain, together with prescribed values of u on the boundary of that domain. We now illustrate the use of the Fourier method in solving such problems for certain domains in the plane. EXAMPLE 1. Let u be harmonic in the interior of a rectangular region so that 0 0 y) + y) = (0 <x <a, 0 <y <b). 0 These values are prescribed on the boundary (Fig. 37): u(0, y) = 0, u(x,0)=f(x), y U =0 U =0 0 a FIGURE 37 0 (0 <y <b), u(x,b)=0 (0<x<a). u(a, y) 144 CHAP 4 BOUNDARY VALUE PROBLEMS Separation of variables, with u = X(x)Y(y), leads to the Sturm-Liouville problem X"(x) + AX(x) = (4) X(a) = 0, X(O) = 0, 0, whose eigenvalues and eigenfunctions are (Sec. 29) fliT2 (n = 1,2,...), = , = and to the conditions Y(b) = Y"(y) — AY(y) = 0, (5) 0. When A is a particular eigenvalue of the Sturm-Liouville problem (4), the function satisfying conditions (5) is found to be y) = C1 exp niry — exp nir(2b — y) a where C1 denotes an arbitrary nonzero constant. Instead of setting C1 = we have always done in such cases, let us write 1, as ci= Then takes the compact form — y) a Thus the function 00 u(x, y) = sinh n7T(b—y) a n=1 sin niTx a formally satisfies all the conditions (1) through (3), provided that f(x) = n'n-x sinh a n=1 sin (0 <x <a). a We assume that f is piecewise smooth. Then series (7) is the Fourier sine series representation of f(x) on the interval 0 <x <a if sinh (nt-b/a) = where 2 a (n=1,2,...). The function defined by equation (6), with coefficients (8) 2 is, therefore, our formal solution. a ff(x)sin—dr a (n=1,2,...), SEC. 34 DIRICHLET PROBLEMS 145 If y is replaced by the new variable b — y in the problem above, as well as its solution, and if f(x) is replaced by g(x), the nonhomogeneous condition satisfied by u becomes u(x, b) = g(x). An interchange of x and y then places nonhomogeneous conditions on the edge x = 0 or x = a. Superposition of the four solutions thus gives the harmonic function whose values are prescribed as functions of position along the entire boundary of the rectangular domain, except for the corners. From equations (1) through (3), we note that u(x, y) represents the steady-state temperatures in a rectangular plate, with insulated faces, when u = f(x) on the edge y = 0 and u = 0 on the other three edges. The function u also represents the electrostatic potential in a space formed by the planes x = 0, x = a, y = 0, and y = b when the space is free of charges and the planar surfaces are kept at potentials given by conditions (2) and (3). EXAMPLE 2. Let u(p, 4)) denote a function of the cylindrical, or polar, coordinates p and 4) that is harmonic in the domain 1 <p <b, 0 <4) <7r of the plane z = 0 (Fig. 38). Thus (Sec. 4) (9) + + = (1 <p <b, 0 <4) 0 <17-). Suppose further that u(p,0) = 0, u(1,4)) = 0, u(p,i,-) = u(b,4)) = (1 <p <b), (0<4) <i,-), 0 u0 where u0 is a constant. U = U0 FIGURE 38 Substituting u = ing variables, we find that into the homogeneous conditions and separat- p2R"(p) + pR'(p) — AR(p) = 0, R(1) 0 and + = 0, 1(O) = 0, 1(ir) = 0. Except for notation, the problem in 1 is the Sturm-Liouville problem in whose eigenvalues and eigenfunctions are Sec. 29, with c = = n2, = sin n4) (n = 1,2,...). 146 BOUNDARY VALUE PROBLEMS CHAP. 4 The corresponding functions equation are determined by solving the differential p2R"(p) + pR'(p) — n2R(p) = (1 <p <b), 0 where R(1) = 0. This is a Cauchy-Euler equation (see Problem 3, Sec. 35), and the substitution p = exp s transforms it into the differential equation d2R ds Hence R= + = C1p'2 + Because R(1) = 0, it follows that, except for constant factors, the desired functions of p are =pfl (n = 1,2,...). Thus, formally, u(p,4)) = n=1 where, according to the second of conditions (11), the constants that u0 = — are such (0 <4 <n-). sin n4 n=1 Since this is in the form of a Fourier sine series representation for the constant function u0 on the interval 0 <4) <ir, 2 2u0 = = —f IT fl ITO The complete solution of our Dirichlet problem is, therefore, — u(p,4) = (n = 1,2,...). pfl_p_fl 2u0 ITn=1 sinn4, — or u(p,4) = 4u0 00 p2T1_l — 2n—1 —b sin (2n — 1)çb —(2n—1) 2n—1 35. OTHER TYPES OF BOUNDARY CONDITIONS Boundary value problems consisting of Laplace's equation V2u = 0 and boundary conditions not all of which are of the Dirichlet type are also important in applications. In the following example, values of a derivative of the function u, OTHER TYPES OF BOUNDARY CONDITIONS SEC. 35 147 rather than values of u itself, are prescribed along a portion of the boundary of the dotnain in which u is harmonic. EXAMPLE. Using cylindrical coordinates, let us derive an expression for the steady temperatures u = u(p, 4)) in a long rod, with uniform semicircular cross section and occupying the region 0 p a, 0 4) IT, which is insu- lated on its planar surface and maintained at temperatures f(4)) on the semicircular part (Fig. 39). u =f(4)) 0 a FIGURE 39 As in Example 2, Sec. 34, u(p, 4)) satisfies Laplace's equation =0, (1) but now in the domain 0 <p <a, 0 <4) It also satisfies the homogeneous conditions [see Problem 12(b), Sec. 4] (2) = 0, = (0 <p <a), 0 as well as the nonhomogeneous one (O<çb<ir). u(a,4)) The function f is understood to be piecewise smooth and, therefore, bounded. We assume further that Iu(p, 4))I M, where M denotes some positive constant. The need for such a boundedness condition is physically evident and has been only tacitly assumed in earlier problems. Here it serves as a condition at the origin, which may be thought of as the limiting case of a smaller semicircle (compare Fig. 38) as its radius tends to zero. The substitution u = R(p)t(4)) in the homogeneous conditions (1) and (2) leads to the condition p2R"(p) + pR'(p) — AR(p) = (0 <p <a) 0 on R(p) and to the Sturm-Liouville problem V'(4)) + = 0, V(O) = 0, = 0, whose eigenvalues and eigenfunctions are A0 = 0, A= (n = 1,2,...) dVHD T = 005 = 0q2 + ! =V oq oip duj OOU!S v OAU!SOd \' = = j mo oAwsod = U = U) = 'J jo + (j7) ioj pOxLj S! uou!puoo sOJ!flbOJ i 11d uj d + 'g oioqM oq oo— spuoj d oq OJOZ S! d> 'v> 0 omoz = — S! OOU!S U) = '(d)°j UOUOuflJ LZ j7 U! oq OOUOH OM suOUipuOO '( (j) (z) Aq (4'd)n oioqM UV U) = V = '0 = 0p + 1=u ( uOij!puOo 'T = 'Apuonbosuoj 00 Up o) U) .LtZ = — J IL O, = '0 soo jo mo o2ojdwoo '(s) + = Up oq 'j 'z (IL> '( 4u 4p wojqoid 's! 'uoip 00 oq JO U o 4 .j y sp =o X = 0) x> (it> = 0 idoj x = 0) (it> '(flJ SEC 35 PROBLEMS respectively. 149 Let u(x, y) denote its steady temperatures. Derive the expression sinhny a0 u(x,y)= —y+ 2'r sinhmr n==1 cosnx, where (n = 0,1,2,...). = —f 770 Find u(x, y) when f(x) = u0, where u0 is a constant. 2. One edge of a square plate with insulated faces is kept at a uniform temperature u0, and the other three edges are kept at temperature zero. Without solving a boundary value problem, but by superposition of solutions of like problems to obtain the trivial case in which all four edges are at temperature u0, show why the steady temperature at the center of the given plate must be u0/4. 3. If A, B, and C are constants, the differential equation Ax2y" + Bxy' + Cy = 0 is called a Cauchy -Euler equation. Show that, with the substitution x = exp s, it can be transformed into the constant-coefficient differential equation d2y dy ds ds A—T + (B -A)— + Cy = 0. 4. Let p, 4), z be cylindrical coordinates. Find the harmonic function u(p, 4)) in the of the plane z = 0 when u = 0 and u = f(4)) on the domain 1 <p <b, 0 <4) arcs p = 1 and p = b (0 <4) respectively, and 'r/2) = 0 0) = (1 <p <b). Give a physical interpretation of this problem. = a0 1 np —i-- p 2n —p —2n + afl cos2n4), where 4 = ir/2 —f (n = 0,1,2,...). f(4))cos2n4)d4) 5. Let the faces of a plate in the shape of a wedge 0 p a, 0 4) a be insulated. Find the steady temperatures u(p, 4)) in the plate when u = 0 on the two rays 4) = 0, 4) = a (0 <p <a) and u = f(4)) on the arc p = a (0 <4) <a). Assume that f is piecewise smooth and that u is bounded. Answer: u(p,4)) = 2 — fp\PI7T/a n'r4) a a a of a rectangular plate 0 x 'r, 0 y y0 6. The faces and edge y = 0 (0 <x are insulated. The other three edges are maintained at the temperatures indicated in 150 CHAP. 4 BOUNDARY VALUE PROBLEMS y u u=O[ =0 V2u=O ju=i 0 ////////////////////// x FIGURE 40 Fig. 40. By making the substitution u(x, y) = U(x, y) + t(x) in the boundary value problem for the steady temperatures u(x, y) in the plate and using the method described in Example 2, Sec. 32, derive the temperature formula coshny sinnx coshny0 1 u(x,y)=— x+2E n=i n Suggestion: The series representation (Example 1, Sec. 14) ( 1\fl+1 sinnx (0<x<17-) n=1 is useful in finding U(x, y). 7. Let u(x, y) denote the bounded steady temperatures in the semi-infinite plate x 0, 'r, whose faces are insulated, when the edges are kept at the temperatures o y shown in Fig. 41. (The boundedness condition serves as a condition at the missing right-hand end of the plate.) Assuming that the function f is piecewise smooth, derive the temperature formula u(x,y) = n=1 where (n=1,2,...). 'TO U =0 0 U =0 x FIGURE 41 SEC 36 8. A STRING WITH PRESCRIBED INITIAL VELOCITY 151 Suppose that in the plate described in Problem 7 there is a heat source depending on the Variable y and that the entire boundary is kept at temperature zero. According to Sec. 3, the steady temperatures u(x, y) in the plate must now satisfy Poisson's equation (x>O,O<y <r). (a) By assuming a (bounded) solution of the form u(x,y) = n=1 of this temperature problem and using the method of variation of parameters (Sec. 33), show formally that are the coefficients in the Fourier sine series for q(y) on the interval where o (n = 1,2,...), — e_nx) = <)' <iT. (b) Show that when q(y) is the constant function q(y) = q0, the solution in part (a) becomes u(x,y) = 4q0 IT 1—exp[—(2n—1)x] sin(2n (2n—1) — l)y. Suggestion: In part (a), recall that the general solution of a linear second-order is any particuequation y" + p(x)y = g(x) is of the form y = + yr,, where lar solution and is the general solution of the complementary equation y" +p(x)y = O.t 9. Derive an expression for the bounded steady temperatures u(x, y) in a semi- x c, y 0 whose faces in the planes x = 0 and x = c are insulated and where u(x, 0) = f(x). Assume that f is piecewise smooth on the interval infinite slab 0 0 <x <c. 36. A STRING WITH PRESCRIBED INITIAL VELOCITY When, initially, the string in Sec. 29 has some prescribed distribution of velocities g(x) parallel to the y axis in its position of equilibrium y = 0, the boundary value problem for the displacements y(x, t) becomes t) = y(0,t) = y(x,0) = t) 0, y(c,t) = 0, = (0 <x < c, t > 0), 0, g(x). tSee, for instance, the book by Boyce and DiPrima (1992, sec. 3.6), listed in the Bibliography. 152 CHAP 4 BOUNDARY VALUE PROBLEMS If the xy plane, with the string lying on the x axis, is moving parallel to the y axis and is brought to rest at the instant t = 0, the function g(x) is a constant. The hammer action in a piano may produce approximately a uniform initial velocity over a short span of a piano wire, in which case g(x) may be considered to be a step function. As in Sec. 29, we seek functions of the type y = X(x)T(t) that satisfy all the homogeneous conditions in the boundary value problem. The SturmLiouville problem that arises is the same as the one in Sec. 29: X"(x) + AX(x) = X(0) = 0, X(c) = 0, 0. We recall that the eigenvalues are A,1 = (nir/c)2 (n = 1, 2,...), with eigenfunctions = sin Since the conditions on T(t) are T"(t) + Aa2T(t) = T(O) = 0, 0, the corresponding functions of t are = sin (nirat/c). The homogeneous conditions in the boundary value problem are, then, formally satisfied by the function 00 n'n-x y(x, t) = sin sin ni,-at C n=1 where the constants are to be determined from the second of conditions (3): (4) nira nirx C C n=1 (0<x<c). Under the assumption that g is piecewise smooth, the series in equation (4) is the Fourier sine series representing g(x) on the interval 0 <x <c if where = = Thus = 2 nirx c _fg(x)sin_—dr. and y( x, t) = c n'n-at 00 — sin c sin c We can sum the series here by first writing nirx = n=1 nirat = 1 +G(x—at)], C where G is the odd periodic extension, with period 2c, of the given function g SEC. 37 AN ELASTIC BAR 153 (compare Sec. 30). Then, since y(x, 0) = y(x,t) = it = ds t + JG(x + aT) dT — atG( — aT) dTI ) dsj; and, in terms of the periodic function I(x)=fG(s)ds (7) y(x,t) (8) If = i (—oo<x<oo), -I(x-at)]. points on the string are given both nonzero initial displacements and nonzero initial velocities, so that y(x,0) =f(x) and =g(x), the displacements y(x, t) can be written as a superposition of solution (iO), Sec. 30, and solution (8) above: (10) y(x,t) = +F(x—at)] + —I(x—at)]. Note that both terms satisfy the homogeneous conditions (i) and (2), while their sum clearly satisfies the nonhomogeneous conditions (9). (Compare Problem 5, Sec. 9.) In general, the solution of a linear problem containing more than one nonhomogeneous condition can be written as a sum of solutions of problems each of which contains only one nonhomogeneous condition. The resolution of the original problem in this way, though not an essential step, often simplifies the process of solving it. 37. AN ELASTIC BAR A cylindrical bar of natural length c is initially stretched by an amount bc (Fig. 42) and is at rest. The initial longitudinal displacements of its sections are then proportional to the distance from the fixed end x = 0. At the instant t = 0, both ends are released and left free. The longitudinal displacements tSee also the footnote with Problem 7, Sec. 33, regarding antiderivatives. _______ 154 CHAP. 4 BOUNDARY VALUE PROBLEMS i x=c FIGURE42 x y(x, t) satisfy the following boundary value problem, where a2 = E/6 (Sec. (1) t) (0 <x < c, t > 0), t) = 6): t) = 0, t) = 0, = 0. (3) y(x,O) = bx, The homogeneous two-point boundary conditions (2) state that the force per (2) unit area on the end sections is zero. The function y(x, t) can also be interpreted as representing transverse displacements in a stretched string, released at rest from the position y(x, 0) = bx, when the ends are looped around perfectly smooth rods lying along the lines x = 0 and x = c. In that case, a2 = H/6; and the boundary conditions (2) state that no forces act in the y direction at the ends of the string (see Sec. 5). Functions y = X(x)T(t) satisfy all the homogeneous conditions above when X(x) is an eigenfunction of the problem X'(c) = 0 (4) X'(O) = 0, X"(x) + AX(x) = 0, and when, for the same eigenvalue A, (5) T"(t) + Aa2T(t) = T'(O) = 0, 0. = (n7,-/c)2 (n = 1, 2,...), with The eigenvalues are (Sec. 27) A0 = 0 and The corresponding func= cos eigenfunctions X0(x) = and tions of t are T0(t) = 1 and = cos Formally, then, the generalized linear combination y( x, t) = a0 00 + n'n-x cos cos nirat satisfies conditions (1) through (3), provided that (0<x<c). The function bx is such that it is represented by the Fourier cosine series (6) on the interval 0 x c, where 2b (n=0,2,...). CO C Consequently, a0=bc, 2bc fl 2 SEC. 38 and 155 RESONANCE we arrive at the solution bc (9) 4bc (2n — 1 (2n — cos 2cos 2 By C C a method already used in Sees. 30 and 36, we can put this series solution in closed form, involving the even periodic extension P(x), with period 2c, of the function bx (0 x c). To be specific, we know from the trigonometric identity 2 cos A cos cos (A + B) + B cos (A B) that 2 cos nTrx nTrat = cos C + cos C cos C C Hence expression (6) can be written as y(x,t) (10) 1 a0 = nir(x+at) + n=1 a0 n=1 But series (7) represents P(x) for all values of x when the values (8) of the coefficients (n = 0, 1, 2,...) are used. Hence expression (10), with those values of (11) reduces to +P(x—at)]. y(x,t) This is the desired closed form of solution (9). 38. RESONANCE A stretched string, of length unity and with fixed ends, is initially at rest in its position of equilibrium. A simple periodic transverse force acts uniformly on all elements of the string, so that the transverse displacements y(x, t) satisfy this modified form (see Sec. 5) of the wave equation: +Asinwt (0 <x < 1, t >0), where A is a constant. Equation (1), together with the boundary conditions y(0,t) = y(x,0) = 0, y(1,t) = 0, 0, = 0, just described, make up a boundary value problem to which the method of variation of parameters (Sec. 33) can be applied. We note that if the constant A were actually zero, the Sturm-Liouville problem arising would have eigenfunctions sin nirx (n = 1,2,...). Hence we 156 CHAP. 4 BOUNDARY VALUE PROBLEMS seek a solution of our boundary value problem having the form y(x,t) = n=1 where the coefficients are to be determined. Substituting series (4) into equation (1) and using the Fourier sine series representation 2A[1 00 A= — (0 <x <1), sin n=1 we write sin sin 1Z7TX = 00 2A[1 — sinwt (n = 1,2,...), nIT and conditions (3) yield the initial conditions = and 0 = 0 (n = 1,2,...) on 0 when n is even; that is, It is easy to see that (n = 1, 2,...). It remains to find 0 when n is odd. If we write (n=1,2,...), the initial value problem in ordinary differential equations for (n = 1,2,...) (7) (8) - 1(t) is = + 4A sin = 0, wt, = 0. We may now refer to Problem 14 below, where methods learned in an introductory course in ordinary differential equations are used to solve the initial value problem consisting of the second-order equation y"(t) + a2y(t) = b sin wt, where a and b are constants, and the conditions y(O) = 0, y'(O) = 0. SEC 38 PROBLEMS 157 To be specific, if w * a, b (U) y(t) = sin at — w —a2 I\a— (11) 2 Sin Thus we see that if w * w,1 for any value of n (n = wt 1, 2,...), the solution of problem (7)—(8) is 4A 1(t) = and (12) 2 sin w,1t — sin cut 2 — cu,1 it follows from equation (4) that y( x, t) = sinwx 4A 2 n=1 cu sin 2 t — sin cut — It is also shown in Problem 14 that if w = a, the solution of differential equation (9), with conditions (10), is y(t) = hi — sin at — t 2a a cos at Hence, when there is a value N of n (n = i, 2,...) such that cu = B2N_ t) = 2A i — sin WNt — t cos WNt Because of the factor t with the cosine function here, this means that series (4) contains an unstable component. Such an unstable oscillation of sections of the string is called resonance. The periodic external force is evidently in resonance with the string when the frequency w of that force coincides with any one of the resonant frequencies cu,1 = (2n — (n = i, 2,...). Those frequencies de- pend, in general, on the physical properties of the string and the manner in which it is supported. PROBLEMS 1. Show that, for each fixed x, the displacements y given by equation (10), Sec. 36, are periodic functions of t, with period 2c/a. 2. 3. Show that the motion of each cross section of the elastic bar treated in Sec. 37 is periodic in t, with period 2c/a. A string, stretched between the points 0 and on the x axis, is initially straight with velocity y1(x, 0) = b sin x, where b is a constant. Write the boundary value problem in y(x, t), solve it, and verify the solution. Answer: y(x, t) = b — sin x sin at. 4. Display graphically the periodic functions G(x) and 1(x) in Sec. 36 when all points of the string there have the same initial velocity g(x) = v0. Then, using expression 158 BOUNDARY VALUE PROBLEMS CHAP. 4 (8), Sec. 36, indicate some instantaneous positions of the string by means of line segments similar to those in Fig. 32 (Problem 8, Sec. 30). 5. In Sec. 36, we used the fact that G(s)= (—oo<s<oo). n=1 Integrate that series (Sec. 23) to show that c I(x)=— fl'TX\ —11--cos-—--—1 C (—oo<x<oo) / and hence that the function (8) in Sec. 36 is represented by the series (6) there. 6. From expression (11), Sec. 37, show that y(O, t) = P(at) and hence that the end x = 0 of the bar moves with the constant velocity ab during the half period 0 <t <c/a and with velocity —ab during the next half period. 7. A string, stretched between the points 0 and 'r on the x axis and initially at rest, is released from the position y = f(x). Its motion is opposed by air resistance, which is proportional to the velocity at each point (Sec. 5). Let the unit of time be chosen so that the equation of motion becomes t) = (0 <x <v-, t> 0), t) — 2/3y1(x, t) where /3 is a positive constant. Assuming that 0 <13 < 1, derive the expression y(x, t) = e_Pt cos + sin sin n=1 where (n=1,2,...), for the transverse displacements. 8. Suppose that the string in Problem 7 is initially straight with a uniform velocity in the direction of the y axis, as if a moving frame supporting the end points is brought to rest at the instant t = 0. The transverse displacements y(x, t) thus satisfy the same differential equation, where 0 </3 < 1, and the boundary conditions y(0,t) =y(ir,t) = 0, y(x,0) = 0, y1(x,0) = v0. Derive this expression for those displacements: y(x,t) = sin(2n — 1)x 4v0 n=i (2n — sin art, — 1)2 _/32 where = on the 9. The ends of a stretched string are fixed at the origin and at the point x = horizontal x axis. The string is initially at rest along the x axis and then drops under its own weight. The vertical displacements y(x, t) thus satisfy the differential equation (Sec. 5) where g is the acceleration due to gravity. PROBLEMS SEC. 38 159 (a) Use the method of variation of parameters to derive the expression y(x, t) = 4g 2 n=1 sin(2n—1)x cos (2n — 1)at (2n—1) — — 8 x) for these displacements. (b) With the aid of the trigonometric identity 2 sin A cos B = sin (A + B) + sin (A — B), show that the expression found in part (a) can be put in the closed form g y(x, t) = P(x + at) + P(x — at) — 2 x('r where P(x) is the odd periodic extension, with period x('r — x) (0 x — x) of the function jr). Suggestion: In both parts (a) and (b), the Fourier sine series representation (Problem 5, Sec. 14) sin(2n—1)x (2n—1) 3 8 10. F I is needed. Also, for part (a), see the suggestion with Problem 8, Sec. 35. The end x = 0 of an elastic bar is free, and a constant longitudinal force F0 per unit area is applied at the end x = c (Fig. 43). The bar is initially unstrained and at rest. Set up the boundary value problem for the longitudinal displacements y(x, t), the conditions at the ends of the bar being t) = F0/E (Sec. 6). t) = 0 and After noting that the method of separation of variables cannot be applied directly, follow the steps below to find y(x, t). i- F0 x=c x FIGURE43 t) differ by a linear t) and (a) By writing y(x, t) = Y(x, t) + Ax2, so that function of x (compare Problem 3, Sec. 33), determine a value of A that leads to the new boundary value problem F0 2 }ç(0,t) =0, F0 Y(x,0)= ————x2 , 2cE =0, }ç(x,0)=0. (0<x<c,t>0), 160 BOUNDARY VALUE PROBLEMS CHAP. 4 (b) Point out why it is reasonable to expect that the boundary value problem in part (a) has a solution of the form n'rx A0(t) Y(x,t)= + 2 Then use the method of variation of parameters to find Y(x, t) and thereby derive the solution y(x, t) = F0 3(x2 + a2t2) — 12c2 c2 —r — cos mrat cos of the original problem. (c) Use the trigonometric identity 2 cos A cos B = cos (A + B) + cos (A — B) and the series representation [Problem 5(b), Sec. 21] (_1)fl 4c2 c2 n c to write the expression for y(x, t) in part (b) as F0 x +at22 P(x+at)+P(x—at) 2 2 — where P(x) is the periodic extension, with period 2c, of the function x2 (—c 11. Show x= 0 x c). how it follows from the expression for y(x, t) in Problem 10(c) that the end of the bar remains at rest until time t = c/a and then moves with velocity = 2aF0/E when c/a <t < 3c/a, with velocity 2v0 when 3c/a <t <5c/a, etc. 12. The value problem t) = y(0,t) =y(c,t) = t) + Ax sin wt 0, y(x,0) =y1(x,0) = (0 <x <c, t > 0), 0 describes transverse displacements in a vibrating string (compare Sec. 38). Show that resonance occurs when w has one of the values = (n'ra)/c (n = 1,2,...). 13. Let a, b, and w denote nonzero constants. The general solution of the ordinary differential equation y"(t) + a2y(t) = b sin wt is of the form y = + yr,, where is the general solution of the complementary equation y"(t) + a2y(t) = 0 and is any particular solution of the original nonhomogeneous equation.t tFor the method of solution to be used here, which is known as the method of undetermined coefficients, see, for instance, the book by Boyce and DiPrima (1992) or the one by Rainville and Bedient (1989). Both books are listed in the Bibliography. FOURIER SERIES IN TWO VARIABLES SEC. 39 161 = A cos wt + B sin wt, where A and B are constants, into the given differential equation, determine values of A and B such that y,, is a solution. Thus derive the general solution (a) Suppose that w # a. After substituting y(t) = C1 cos at + C2 sin at + b a 2 —w 2 sin wt of that equation. = At cos wt + (b) Suppose that w = a, and find constants A and B such that Bt sin wt is a particular solution of the given differential equation. Thus obtain the general solution y(t) = C1 b cos at + C2 sin at — cos at. 14. Use the general solutions derived in Problem 13 to obtain the following solutions of the initial value problem y"(t) + a2y(t) = bsinwt, b w2a2 y(t)= y(O) = y'(O) = 0, 1w — sin at — sin a 0: wt) when w # a; — sin at — t cos at —I 2a \a 39. when w = a. FOURIER SERIES IN TWO VARIABLES Let z(x, y, t) denote the transverse displacement at each point (x, y) at time t in a membrane that is stretched across a rigid square frame in the xy plane. To simplify the notation, we select the origin and the point (i,-, ir) as ends of a diagonal of the frame. If the membrane is released at rest with a given initial displacement f(x, y) that is continuous and vanishes on the boundary of the square, then (Sec. 6) = + in the three-dimensional domain 0 <x <ir, 0 <y <ir, t > 0; and z(0, y, t) = z('ii-, y, t) = z(x, 0, t) = z(x, ir, t) = z(x,y,0) =f(x,y), =0, ir. We assume that the partial derivatives x 0 y y) are also continuous. Functions of the type z = X(x)Y(y)T(t) satisfy equation (1) if where 0 and 0, T"(t) — a2T(t) - X"(x) X(x) Y"(y) — + Y(y) - - y) 162 BOUNDARY VALUE PROBLEMS CHAP. 4 Separating variables again, in the second of equations (4), we find that Y"(y) Y(y) where — — X"(x) — — X(x) — A — is another separation constant. So we are led to two Sturm-Liouville problems, X"(x) + (A X(0) = = 0, — X(n-) = 0, 0 and Y"(y) + Y(O) = 0, = 0, = 0, and to the conditions T"(t) + )ta2T(t) = 0, T'(O) 0 on T. We turn to the Sturm-Liouville problem in V first since it involves only = one of the separation constants. According to Sec. 29, the values sin my; and, when (m = 1,2,...) of ,a give rise to the eigenfunctions A— = n2 (n = sin nx of the problem in 1,2,...), the eigenfunctions X are obtained. The conditions on T thus become T"(t) + a2(m2 + n2)T(t) = 0, T'(O) = 0, where m = 1,2,... and n = 1,2 For any fixed positive integers m and n, the solution of this problem in T is, except for a constant factor, Tmn(t) = cos(atVm2 + n2). The formal solution of our boundary value problem is, therefore, z(x,y,t) Bmn need to be determined so that f(x,y) = sin nx sin my n=1 m=1 when 0 x ir and 0 y ir. By grouping terms in this double sine series so as to display the total coefficient of sin nx for each n, one can write, formally, f(x,y) = sinnx. n=1 m=1 For each fixed y (0 y ir), equation (7) is a Fourier sine series representation of the function f(x, y), with variable x (0 x v-), provided that Bmn sin my m=1 = —f0 f(x,y)sinnxdx The right-hand side here is a sequence of functions (n = 1,2,...). (n = 1, 2,...), each PERIODIC BOUNDARY CONDITIONS SEC. 40 represented by its Fourier sine series (8) on the interval 0 BmnfFn(Y)sinmydy 163 ir when y (m=1,2,...). ITO Hence the coefficients Bmn have the values (9) 'iTO sinmyff(x,y)sinnxdxdy. 0 A forma of our membrane problem is now given by equation (5) with the coefficients defined by equation (9). Since the numbers Vm2 + n2 do not change by integral multiples of some fixed number as m and n vary through integral values, the cosine functions in equation (5) have no common period in the variable t; so the displacement z is not generally a periodic function of t. Consequently, the vibrating membrane, in contrast to the vibrating string, generally does not produce a musical note. It can be made to do so, however, by giving it the proper initial displacement. If, for example, z(x,y,0) =Asinxsiny, where A is a constant, the displacements (5) are given by a single term: z(x,y,t) Then z is periodic in t, with period 40. PERIODIC BOUNDARY CONDITIONS The solutions of the boundary value problems in this chapter have been based on the solutions of just two Sturm-Liouville problems, which lead to Fourier cosine and sine series representations of prescribed functions. Although Chap. 5 is devoted to the theory and application of many other Sturm-Liouville problems, as well as to the precise definition of such a problem, we conclude this chapter by considering a third problem that arises in certain boundary value problems for regions with circular boundaries: (1) X"(x) + AX(x) = 0, X( —ir) = X'( = We include it here since its solutions also lead to Fourier series representations, but now involving both cosines and sines on the interval — IT <x and since the general theory of Sturm-Liouville problems is not actually required. We need accept only the fact, to be verified in Chap. 5 (Sec. 43), that each eigenvalue, or value of A for which problem (1) has a nontrivial solution, is a real number. In anticipation of Chap. 5, we continue to refer to such values of A as eigenvalues and to the nontrivial solutions as eigenfunctions. To solve problem (1), we consider first the case in which A = 0. Then X(x) = Ax + B, where A and B are constants; and the boundary conditions are 164 CHAP 4 BOUNDARY VALUE PROBLEMS satisfied if A = 0. Since the conditions in problem (1) are all homogeneous, we thus find that, except for a constant factor, X(x) = When A > 0, we write A = a2 (a > 0) and note that the general solution of the differential equation in problem (1) is X(x) = C1 cos cxx + C2 sin ax. It is straightforward to show that, in order for the boundary conditions to be satisfied, C2 sin air = and 0 C1 sin = 0. Since C1 and C2 cannot both vanish if X(x) is to be nontrivial, it follows that the positive number a must, in fact, be a positive integer n. Thus A = n2 (n = 1, 2,...), and the corresponding general solution of problem (1) is an arbitrary linear combination of two linearly independent eigenfunctions, cos nx and sin nx. It is left to the reader (Problem 4) to show that there are no negative eigenvalues. We now illustrate the use of this Sturm-Liouville problem, involving periodic boundary conditions. EXAMPLE. Let u(p, 4) denote the steady temperatures in a thin disk 1, with insulated surfaces, when its edge p = 1 is kept at temperatures f(4). The variables p and 4) are, of course, polar coordinates, and u satisfies p Laplace's equation V2u = 0. That is, (2) = + where (3) 0 (0 <p <1, u(1,4))=f(4)) <4) <ir), (—ir<4i<rr). Also, u and its partial derivatives of the first and second order are continuous and bounded in the interior of the disk. In particular, u and its first-order partial derivatives are continuous on the ray 4) = IT (Fig. 44). u =f(4) p= 1 FIGURE 44 SEC 40 PROBLEMS If functions of the type u = continuity requirements, then are to satisfy condition (2) and the p2R"(p) + pR'(p) — AR(p) = (4) 165 (0 <p < 1) 0 and + A'1'(d) (5) where A is = 0, a separation constant. We now recognize that conditions (5) constitute a Sturm-Liouville problem in t, with eigenvalues A0 = 0 and = n2 (n = 1, 2,...). The corresponding eigenfunctions are and linear combinations of cos n4 and sin n4. Equation (4) is a Cauchy-Euler equation, and we know from Sec. 35 that its bounded solutions are R0(p) = 1 when A = 0 and = pfl when A = n2 (n = 1, 2,...). Hence the generalized linear combination of our continuous functions can be written (6) + = including a0, are the and This satisfies the boundary condition (3) if coefficients in the Fourier series for f on the interval <x < lIT (7) = —f f is piecewise smooth. PROBLEMS 1. All four faces of an infinitely long rectangular prism, formed by the planes x = 0, x = a, y = 0, and y = b, are kept at temperature zero. Let the initial temperature distribution be f(x, y), and derive this expression for the temperatures u(x, y, t) in the prism: u(x, y, t) = m2 exp — ( a n=lm=1 + n2 sin mrx a sin m'ry where 4 Bmn = b a sin f(x, y) sin ThTX dxdy. 2. Write f(x, y) = g(x)h(y) in Problem 1 and show that the double series obtained there for u reduces to the product u(x, y, t) = v(x, t)w(y, t) x a and b with faces at temperature zero and with initial temperatures g(x) and of two series, where v and w represent temperatures in the slabs 0 o y h(y), respectively. 166 BOUNDARY VALUE PROBLEMS 3. Let the functions v(x, t) and CHAP. 4 w(y, t) satisfy the heat equation for one-dimensional flow: = = Show by differentiation that their product u = equation vw satisfies the two-dimensional heat + = Use this result to arrive at the expression for U(x, y, t) in Problem 2. 4. Write A = —a2 (a > 0), and show that the Sturm-Liouville problem X"(x) +AX(x) =0, X(—n-) =X('r), =X'('r) in Sec. 40 has no negative eigenvalues. 5. Using the cylindrical coordinates p, 4), and z, let U(p, 4)) denote steady temperatures in a long hollow cylinder a p b, —00 <z <oo when the temperatures on the inner surface p = a are and the temperature of the outer surface p = b is zero. (a) Derive the temperature formula U(p, ln(b/p) = ln (b/a) 00 a0 2+ where the coefficients and (b/p)'1 — (p/b)" cos n4) + (b/a)" — (a/b)'1 sin n4), including a0, are given by equations (7), Sec. 40. (b) Use the result in part (a) to show that if f(4)) = A + B sin 4), where A and B are constants, then =A ln(b/p) ln(b/a) + Bab b2 — b a2 p sin4). — 6. Solve the boundary value problem —ir, t) = U(n-, t), (—'r <x < 'r, t > 0), t) t) = t), —ir, t) = U(x,O) = f(x). The solution u(x, t) represents, for example, temperatures in an insulated wire of length 2'r that is bent into a unit circle and has a given temperature distribution along it. For convenience, the wire is thought of as being cut at one point and laid on the x and x = The variable x then measures the distance along axis between x = the wire, starting at the point x = — 'r, and the points x = — and x = 'r denote the same point on the circle. The first two boundary conditions in the problem state that the temperatures and the flux must be the same for each of those values of x. This problem was of considerable interest to Fourier himself, and the wire has come to be known as Fourier's ring. Answer: U(x, t) = + sin nx)e_n2kt, cos nx + where lIT = —f = —f f(x)sinnxdx. PROBLEMS SEC. 40 7. (a) By writing A = nO and B = 0 167 in the trigonometric identity 2 cos A cos B = cos (A + B) + cos (A — B), multiplying through the resulting equation by aPZ (— 1 <a < 1), and then summing each side from n = 1 to n = oo, derive the summation formula acos0—a2 (—1<a<1). 1—2acos0+a2 n=1 [One can readily see that this series is absolutely convergent by comparing it with the geometric series whose terms are aPZ (n = 1, 2,. ).] (b) Write expression (6), with coefficients (7), in Sec. 40 for steady temperatures in a disk as • - + = Then, with the aid of the summation formula in part (a), derive Poisson's integral formula for those temperatures: 1 u(p,cb)=—f 2'r ir 1—p2 (p<l). CHAPTER 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS We turn now to a careful presentation of the rudiments of the theory of Sturm-Liouville problems and their solutions. Once that is done, we shall illustrate the Fourier method in solving physical problems involving eigenfunctions not encountered in earlier chapters. 41. REGULAR STURM-LIOUVILLE PROBLEMS In Chap. 4, we found solutions of various boundary value problems by the Fourier method. Except in Sec. 40, the method always led to the need for a Fourier cosine or sine series representation of a given function. The cosine and sine functions in the series were the eigenfunctions of one of the following two Sturm-Liouville problems on an interval 0 x C: X"(x) + AX(x) = 0, X"(x) + AX(x) = 0, X'(O) = 0, X(0) = 0, X'(c) = 0, X(c) = 0. When applied to many other boundary value problems in partial differen- tial equations, the Fourier method continues to involve a Sturm-Liouville problem consisting of a homogeneous ordinary differential equation of the type X"(x) +R(x)X'(x) + [Q(x) +AP(x)]X(x) =0 168 REGULAR STURM-LIOUVILLE PROBLEMS SEC 41 169 on a finite interval a <x <b, together with a pair of homogeneous boundary conditions at the end points of the interval. The functions P, Q, and R and the boundary conditions are prescribed by the original boundary value problem involving a partial differential equation. Values of the parameter A, which appears in equation (3) only as indicated, and corresponding nontrivial solutions X(x) are to be determined. We now give a precise definition of such a Sturm-Liouville problem, which includes problems (1) and (2) as special cases. Note that a function r(x) =expfR(x)dx is an integrating factor for the sum of the first two terms in equation (3); that is, r(x)[X"(x) + R(x)X'(x)] [r(x)X'(x)]'. Consequently, when each of its terms is multiplied by r(x), equation (3) takes the standard form (4) [r(x)X'(x)]' + [q(x) + Ap(x)]X(x) =0 (a <x <b), where the functions p, q, and r are independent of A. We assume here that p, q, r, and r' are real-valued functions of the real variable x which are continuous on the closed bounded interval a x b and that p(x)> 0 and r(x)> 0 when x b. Also, X(x) is required to satisfy the homogeneous separated bounda ary conditions (5) a1X(a) + a2X'(a) = b1X(b) + b2X'(b) = 0, 0. The constants a1, a2, b1, b2 are all real numbers, independent of A. Moreover, a1 and a2 are not both zero; and the same is true of the constants b1 and b2. The differential equation (4) and boundary conditions (5) make up a regular Sturm-Liouville problem.t Sturm-Liouville problems other than regular ones will be noted in Sec. 42. EXAMPLES. Problems (1) and (2) are both regular Sturm-Liouville problems. Two other examples, to be solved later in this chapter, are X"(x) +AX(x) —0 hX(c) + X'(c) = X'(O) = 0, (0 <x <c), 0, where h denotes a positive constant, and X(1) = 0, + AX(x) = 0 X(b) = 0. (1 <x <b), tPapers by J. C. F. Sturm and J. Liouville giving the first extensive development of the theory of thi problem appeared in vols. 1-3 of the Journal de mathématique (1836—1838). 170 CHAP. 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS As was the case with problems (1) and (2), a value of A for which problem (4)—(5) has a nontrivial solution is called an eigenvalue; and the nontrivial solution is called an eigenfunction. Note that if X(x) is an eigenfunction, then so is CX(x), where C is any nonzero constant. It is understood that in order for X(x) to be an eigenfunction, X(x) and X'(x) must be continuous on the closed interval a x b. Such continuity conditions are usually required of solutions of boundary value problems in ordinary differential equations. The set of eigenvalues of problem (4)—(5) is called the spectrum of the problem. It can be shown that the spectrum of a regular Sturm-Liouville We state this problem consists of an infinite number of eigenvalues A1, A2 fact without proof, which is quite involved.t In special cases, the eigenvalues will be found; and so their existence will not be in doubt. When eigenvalues are sought, however, it is useful to know that they are all real and hence that there is no possibility of discovering others in the complex plane. The proof that the eigenvalues must be real is given in Sec. 43; and we agree that they are to be (n = 1, 2,...). It arranged in ascending order of magnitude, so that can be shown that A,1 oc as n —b If we define the differential operator 2' by means of the equation = [r(x)X'(x)]' + q(x)X(x), (6) then the Sturm-Liouville equation (4) takes the form 2'[X(x)] + Ap(x)X(x) = (7) 0. is a special case of the general second-order linear differential This operator operator L defined by the equation (8) L[X(x)] =A(x)X"(x) +B(x)X'(x) +C(x)X(x). In discussing such operators, we tacitly agree that they are to be defined on function spaces (Sec. 10) in which the functions X are suitably differentiable. The adjoint of L is the operator L* such that (9) L*[X(x)] = [A(x)X(x)]" — [B(x)X(x)}' + C(x)X(x), and L is called self-adjoint when L* = L. It can be shown (Problem 2, Sec. 43) that a necessary and sufficient condition for L to be seif-adjoint is that B(x) = A'(x), in which case equation (8) becomes L[X(x)] = {A(x)X'(x)]' + C(x)X(x). Except for notation, this is equation (6); and so we use 2' to denote the general self-adjoint linear differential operator of the second order. The Sturm-Liouville differential equation (4) is said to be in seif-adjoint form since it is the same as equation (7), which involves the self-adjoint operator verification of statements in this section that we do not prove, see the book by Churchill (1972, chap. 9), which contains proofs when a2 = b2 = 0 in conditions (5), and the one by Birkhoff and Rota (1989). Also, extensive treatments of Sturm-Liouville theory appear in the books by Ince (1956) and Titchmarsh (1962). These references are all listed in the Bibliography. SEC 42 MODIFICATIONS This operator 42. 171 form turns out to be especially useful because of properties of the Some of those properties are noted explicitly in the problems. MODIFICATIONS Although we are mainly concerned in this chapter with the theory and application of regular Sturm-Liouville problems, described in Sec. 41, certain important modifications are also of interest in practice. We mention them here since some of their theory is conveniently included in the discussion of regular Sturm-Liouville problems in Sec. 43. A Sturm-Liouville problem (1) (2) (a <x <b), [r(x)X'(x)]' + [q(x) + Ap(x)JX(x) =0 a1X(a) + a2X'(a) = 0, b1X(b) + b2X'(b) = 0 is singular when at least one of the regularity conditions stated in Sec. 41 fails to be satisfied. The function q, for example, may have an infinite discontinuity at an end point of the interval a x b. The problem is also singular if p(x) or r(x) vanishes at an end point. When r(x) does this, we drop the boundary condition at the end point in question. Note that the dropping of the boundary condition at x a is the same as letting both of the coefficients a1 and a2 in that condition be zero; a similar remark can be made when the condition at b is to be dropped. x EXAMPLE 1. One singular Sturm-Liouville problem to be studied in Chap. 7 consists of the differential equation [xX'(x)]' + —— +Ax X(x) = 0 (0 <x <c), where n = 0, 1,2,..., and the single boundary condition X(c) = 0. Observe that the functions p(x) = x and r(x) = x both vanish at x = 0 and that the function q(x) = —n2/x has an infinite discontinuity there when n is positive. EXAMPLE 2. The differential equation [(1 — x2)X'(x)J' + AX(x) = 0 (—1 <x < 1), with no boundary conditions, constitutes a singular Sturm-Liouville problem. Here the function r(x) = 1 — x2 vanishes at both ends x = ± 1 of the interval —1 x 1. This problem is the main one that is solved and used in Chap. 8. Although it will turn out that the problems in Examples 1 and 2 have discrete spectra, where the eigenvalues may be indexed with the positive or nonnegative integers, this is not always the case with singular problems. Such problems may, in fact, have no eigenvalues at all. Moreover, other types of singular problems, defined on infinite or semi-infinite intervals and to be 172 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS CHAP. 5 encountered in Chap. 6, have continuous spectra containing all nonnegative values of A. As indicated in Sec. 41, the nature of the spectrum of any particular problem will be determined by actually finding the eigenvalues. Finally, in addition to singular problems, another modification of problem (1)—(2) occurs when r(a) = r(b) and conditions (2) are replaced by the periodic boundary conditions (3) X'(a) =X'(b). X(a) —X(b), EXAMPLE 3. The problem X"(x) +AX(x) =0, X(—i,-) =X(ir), X'(—ir) =X'(ir), already solved in Sec. 40, has periodic boundary conditions. ORTHOGONALITY OF EIGENFUNCTIONS 43. As pointed out in Sec. 41, a regular Sturm-Liouville problem (a <x <b), [r(x)X'(x)]' + [q(x) + Ap(x)]X(x) = 0 a1X(a) + a2X'(a) = 0, b1X(b) + b2X'(b) = 0 always has an infinite number of eigenvalues A1, A2 In this section, we shall establish the orthogonality of eigenfunctions corresponding to distinct eigenval- ues. The concept of orthogonality to be used here is, however, a slight generalization of the one originally introduced in Sec. 11. To be specific, a set (n = 1,2,...) is orthogonal on an interval a <x <b with respect to a weight function p(x), which is piecewise continuous and positive on that interval, if dx = 0 The integral here represents an inner product weight function. The set is normalized by dividing each = () when m n. with respect to the by where = 0. This type of orthogonality can, of course, be reduced to that in Sec. 11 by using the products x) as functions of the set. Orthogonal sets with respect to weight functions that are not piecewise continuous, or ones where the fundamental interval is unbounded, also occur in applied mathematics. and where it is assumed that IIç(,,jI The theorem below states that eigenfunctions associated with distinct eigenvalues are orthogonal on the interval a <x <b with respect to the weight function p(x), where p(x) is the same function as in equation (1). If such ORTHOGONALITY OF EIGENFUNCTIONS SEC. 43 eigenfunctions are denoted by (n = 1, 173 2,...), the normalized eigenfunc- tions are X(x) where , = = f b 2 and a generalized Fourier series corresponding to a given function f(x) in is (compare Sec. 12) (a<x<b), n==1 where (n=1,2,...). Examples of such series will be given in Sec. 46. In presenting the theorem, we relax the conditions of regularity on the coefficients in the differential equation (1) so that the result can also be applied to eigenfunctions that are found for the modifications of regular Sturm-Liouville problems mentioned in Sec. 42. We retain all the conditions for a regular problem, stated in Sec. 41, except that now q may be discontinuous at an end point of the interval a x b, and p(x) and r(x) may vanish at an end point. That is, p, r, and r' are continuous on the closed interval a x b, q is continuous on the open interval a <x <b, and p(x)> 0 and r(x)> 0 when a <x <b. Theorem. If Am and A,1 are distinct eigenvalues of the Sturm-Liouville problem (1)—(2), then corresponding eigenfunctions Xm(X) and are orthog- onal with respect to the weight function p(x) on the interval a <x <b. The orthogonalily also holds in each of the following cases: (a) When r(a) = 0 and the first of boundary conditions (2) is dropped from the problem; (b) When r(b) = 0 and the second of conditions (2) is dropped; (c) When r(a) = r(b) and conditions (2) are replaced by the conditions X(a) =X(b), X'(a) =X'(b). Note that both cases (a) and (b) here may apply to a given Sturm-Liouville problem (see Example 2, Sec. 42). To prove the theorem, we first observe that + = — Am PXm, (rX,ç)' + = — since each eigenfunction satisfies equation (1) when A is the eigenvalue to which it corresponds. We then multiply each side of these two equations by and 174 Xm, CHAP. 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS respectively, and subtract: (3) (Am — 1tn)PXmXn = — d = While the final reduction here to an exact derivative is elementary, it is made possible by the special nature of the self-adjoint operator 22 defined in equation (6), Sec. 41. Details regarding this point are left to the problems. The function q has been eliminated, and the continuity conditions on the remaining functions allow us to write (Am — An)ibpXmXn dx (4) is the determinant where Xm(X) (5) = That is, (6) (Am — An) JbXX dx = — The first of boundary conditions (2) requires that aiXm(a) + = 0, aiXn(a) + = 0; and for this pair of linear homogeneous equations in a1 and a2 to be satisfied by be numbers a1 and a2, not both zero, it is necessary that the determinant zero. Similarly, from the second boundary condition, where b1 and b2 are not both zero, we see that = 0. Then, according to equation (6), (7) and, since Am (Am — An)ibpXmXn dx = 0; An, the desired orthogonality property follows: dx =0. (8) 0, or If r(a) = 0, property (8) follows from equation (6) even when when a1 = a2 = 0, in which case the first of boundary conditions (2) disappears. Similarly, if r(b) = 0, the second of those conditions is not used. When r(a) = r(b) and the periodic boundary conditions X(a)=X(b), X'(a) =X'(b) SEC 43 ORTHOGONALITY OF EIGENFUNCTIONS 175 are used in place of conditions (2), then = and, again, property (8) follows. This completes the proof of the theorem. EXAMPLE 1. The eigenfunctions of the regular Sturm-Liouville problem X"(x) + AX(x) X(O) = = 0, (0 <x <ir), 0 0 = sin nx (n = 1, 2,...), and they correspond to the distinct eigenvalare are ues it,1 = n2 (Sec. 29). The theorem tells us that the functions orthogonal on the interval 0 <x <rn- with weight function p(x) = f sin mx sin nx dx = (9) 1: (m * n). 0 We recall that this orthogonality was established earlier in Example 1, Sec. 11, where integral (9) was evaluated directly. EXAMPLE 2. Eigenfunctions corresponding to distinct eigenvalues of the regular Sturm-Liouville problem [xX'(x)]' + —X(x) = 0 X(b) = 0 X(1) = 0, (1 <x <b), are, according to the theorem, orthogonal on the interval 1 <x <b with weight function p(x) = 1/x. In Problem 1 the eigenfunctions are actually found, and the orthogonality is verified. The following corollary is an immediate consequence of the theorem. Corolkuy. If A is an eigenvalue of the Sturm-Liouville problem (1)—(2), then it must be a real number; and the same is true in cases (a), (b), and (c), treated in the theorem. We begin the proof by writing the eigenvalue as it = a + if3, where a and 13 are real numbers. If X denotes a corresponding eigenfunction, which is nontrivial and may be complex-valued, conditions (1) and (2) are satisfied. Now the complex conjugate of A is the number it a — i/3; and X = u — iv and X' = u' + iv' if X = u + iv. Also, the conjugate of a sum or product of two complex numbers is the sum or product, respectively, of the conjugates of those numbers. Hence, by taking the conjugates of both sides of the equations in conditions (1) and (2) and keeping in mind that the functions p, q, and r are real-valued and that the coefficients in conditions (2) are real numbers, we see 176 CHAP 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS that (rX')' + (q a1X(a) + a2X'(a) = 0, 0, b1X(b) + b2X'(b) = 0. Thus the nontrivial function X is an_eigenfunction corresponding to A. — If we assume that /3 0, then A A; and the theorem tells us that X and X are orthogonal on the interval a <x <b with respect to the weight function even in cases (a), (b), and (c): (10) But p(x)> 0 when a <x <b. Moreover, X=u2+v2= x b; and 1X12is not identically zero since X is an eigenfunction. So integral (10) has positive value, and our assumption that f3 * 0 has led us to a contradiction. Hence we must conclude that /3 = 0, or that A is real. when a PROBLEMS 1. (a) After writing the differential equation in the regular Sturm-Liouville problem A [xX'(x)]' + —X(x) = 0 X(1) = 0, X(b) = (1 <x <b), 0 in Cauchy-Euler form (see Problem 3, Sec. 35), use the substitution x = transform the problem into one consisting of the differential equation exp s to d2X —+AX=0 ds2 and the boundary conditions X=0 when s=Oands=lnb. Then, by simply referring to the solutions of the Sturm-Liouville problem in Sec. 29, show that the eigenvalues and eigenfunctions of the original problem here are = where = (n = 1,2,...), b. (b) By making the substitution s = ('r/ln b)ln x in the integral involved and then referring to the integration formula (9) in Example 1, Sec. 43, give a direct verification that the eigenfunctions obtained in part (a) are orthogonal on the interval 1 <x <b with weight function p(x) = 1/x, as ensured by the theorem in Sec. 43. 2. Let L be the general second-order linear differential operator defined by the equation L[X] =AX"+BX' + CX, where A, B, and C are functions of x. In Sec. 41, the adjoint of L was defined as the SEC. 43 PROBLEMS 177 operator L* such that L*[X] = (AX)" — (BX)' + CX. Show that a necessary and sufficient condition for L to be self-adjoint (L* = L) is that B =A'. Suggestion: Note that if L*[X] = L[X], then L*[1] = L[1] and L*[x] = L[x], in particular. The condition B = A' follows from these last two equations. 3. (a) Verify the identity X(rY')' — Y(rX')' = — X'Y)], where r, X, and V denote functions of x. (b) Show that if 2' is the self-adjoint differential operator defined by the equation (Sec. 41) 2'[X] = (rX')' + qX, then the identity in part (a) can be written X2'[Y] - Ysr[X] = d - X'Y)]. This is called Lagrange's identity for the operator 4. (a) Suppose that the seif-adjoint operator 2' in Problem 3(b) is defined on a space of functions satisfying the conditions a1X(a) + a2X'(a) = b1X(b) + b2X'(b) = 0, 0, where a1 and a2 are not both zero and where the same is true of b1 and b2. Use Lagrange's identity, obtained in that problem, to show that = where these inner products are on the interval a <x <b and with weight function unity. denote distinct eigenvalues of a regular Sturm-Liouville problem, (b) Let Am and whose differential equation is (Sec. 41) 2'[X] +ApX—O. Use the result in part (a) to prove that if Xm and corresponding to Am and are eigenfunctions then (pXm,Xn) = 0. Thus show that Xm and are orthogonal on the interval a <x <b with weight function p, as already demonstrated in Sec. 43. 5. Show that if L is the general second-order linear differential operator, where L[X] =AX" +BX' + CX, and if L* is its adjoint, defined by the equation (Sec. 41) L*[X] = (AX)" — (BX)' + CX, then the adjoint of L* is L. That is, show that L** = L. 178 6. STURM-LIOUVILLE PROBLEMS AND APPLICATIONS Show that if is the operator CHAP. 5 then Xcir[Y] — = d — x'Y" + Y'X"). — Thus show that if X1 and X2 are eigenfunctions of the fourth-order eigenvalue problem + AX = X(0) = X"(O) = 0, X(c) = X"(c) = 0, 0, corresponding to distinct eigenvalues A1 and A2, then X1 is orthogonal to X2 on the interval 0 <x <c with weight function unity. UNIQUENESS OF EIGENFUNCTIONS 44. The theory of ordinary differential equations ensures the existence and uniqueness of solutions of certain types of initial value problems, problems in which all boundary data are given at one point. We state here as a lemma, without proof, a fundamental result from the theory of ordinary second-order linear equations that we shall use in discussing the uniqueness of eigenfunctions.t Lemma. Let A, B, and C denote continuous functions of x on an interval b. If x0 is a point in that interval and y0 and y'0 are prescribed constants, then there is one and only one function y, continuous together with its derivative y' when a x b, that satisfies the differential equation a x (a <x <b) y"(x) +A(x)y'(x) +B(x)y(x) = C(x) and the two initial conditions y(x0) =y0, Note that since y" = C — Ày' y'(x0) =y'0. — By, y" is continuous when a <x <b. the general Also, since any values can be assigned to the constants y0 and solution of the differential equation has two arbitrary constants. Suppose now that X and V are two eigenfunctions corresponding to the same eigenvalue A of the regular Sturm-Liouville problem (rX')' + (q + Ap)X — a1X(a) + a2X'(a) = 0, 0 b1X(b) + b2X'(b) = (a <x <b), 0. As stated in Sec. 41, the functions p, q, r, and r' are continuous on the interval a x b; also, p(x)> 0 and r(x)> 0 when a x b. The above lemma tA proof of the lemma can be found in, for instance, the book by Coddington (1989, chap. 6) that is listed in the Bibliography. UNIQUENESS OF EIGENFUNCTIONS SEC 44 enables 179 us to prove that X and Y can differ by at most a constant factor; that is, Y(x) = CX(x), (3) where C is a nonzero constant. We start the proof by observing that, in view of the principle of superposition, the linear combination (4) Z(x) = Y'(a)X(x) — X'(a)Y(x) satisfies the linear homogeneous differential equation (rZ')'+(q+Ap)Z=O in addition, Z'(a) = 0. (a<x<b); Since X and V satisfy the conditions a1X(a) + a2X'(a) = 0, a1Y(a) + a2Y'(a) = 0, where a1 and a2 are not both zero, and since Z(a) is the determinant of that pair of linear homogeneous equations in a1 and a2, we also know that Z(a) = 0. According to the lemma, then, Z(z) = 0 when a x b. That is, (a x b). Y'(a)X(x) — X'(a)Y(x) = 0 Since eigenfunctions cannot be identically zero, it is clear from relation (6) that if either of the values X'(a) or Y'(a) is zero, then so is the other. Relation (3) now follows from equation (6), provided that X'(a) and Y'(a) are nonzero. Suppose, on the other hand, that X'(a) = Y'(a) = 0. Then X(a) and Y(a) are nonzero since, otherwise, X and V would be identically zero, according to the lemma; and zero is not an eigenfunction. The procedure applied to Z(x) may now be used to show that the linear combination W(x) = Y(a)X(x) —X(a)Y(x) is zero when a x b and hence that relation (3) still holds. It follows immediately from relation (3) that, except possibly for a nonzero constant factor, any eigenfunction X of problem (1)—(2) is real-valued. To show this, we first recall from the corollary in Sec. 43 that the eigenvalue A to which X corresponds must be real. So if we make the substitution X = U + iV, where U and V are real-valued functions, in problem (1)—(2) and separate real and imaginary parts, we find that U and V are themselves eigenfunctions corresponding to A. Hence there is a nonzero constant /3 such that V = /3U. Here /3 is real since U and V are real-valued, and we may conclude that X= U+if3U= (1 +i/3)U. That is, X can be expressed as a nonzero constant times a real-valued function. Note, too, how one can write U= 1 1+ We collect our results as follows. if3 X. 180 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS CHAP. 5 Theorem. If X and Y are eigenfunctions corresponding to the same eigenvalue of a regular Sturm-Liouville problem, then Y = CX, where C is a nonzero constant. Also, each eigenfunction can be made real-valued by multiplying it by an appropriate nonzero constant. According to the theorem, a regular Sturm-Liouville problem cannot have two linearly independent eigenfunctions corresponding to the same eigenvalue. For certain modifications of regular Sturm-Liouville problems, however, it is possible to have an eigenvalue with linearly independent eigenfunctions (see Sec. 40). The following corollary, which uses the fact that there is always a realvalued eigenfunction corresponding to a given eigenvalue of problem (1)—(2), is an additional aid in determining eigenvalues since it often eliminates the possibility that there are negative ones. We already know from the corollary in Sec. 43 that each eigenvalue must be real. Corollary. If A is an eigenvalue of the regular Sturm-Liouville problem (1)—(2) and if the conditions q(x) 0 (a b) and a1a2 0, b1b2 0 are satisfied, then A 0. To prove this, we let X denote a real-valued eigenfunction corresponding to the eigenvalue A. Equation (1) is thus satisfied, and we multiply each term of that equation by X and integrate each of the resulting terms from x = x= (8) jb, + fbqx2dx a to +Af"pX2 dr =0. After applying integration by parts to the first of these integrals, we can write equation (8) in the form (9) AfbpX2dr = fb(_qx2)dx + fbr(XF)2d): + r(a)X(a)X'(a) — r(b)X(b)X'(b). Let us now assume that the conditions stated in the corollary are satisfied. Since —q(x) 0 and r(x)> 0 when a x b, the values of the two integrals on the right in equation (9) are clearly nonnegative. As for the third term on the right, we note that if a1 = 0 or a2 = 0 in the first of conditions (2), then X'(a) = 0 or X(a) = 0, respectively. In either case, the third term is zero. If, on the other hand, neither a1 nor a2 is zero, then r(a)X(a)X'(a) = Similarly, — r(b)X(b)X'(b) r( a) [a1X( a)] —a1a2 2 0. 0; and it follows that all the terms on the right- SEC 45 METHODS OF SOLUTION 181 hand side of equation (9) are nonnegative. Consequently, o. But this integral has positive value, and so A 0. 45. METHODS OF SOLUTION We turn now to two examples that illustrate methods to be used in finding the eigenvalues and eigenfunctions of the problems that follow this section. The basic method has already been touched on in Secs. 27 and 40, where simpler Sturm-Liouville problems were solved. It consists of first finding the general solution of a differential equation and then applying the boundary conditions in order to determine the eigenvalues. EXAMPLE 1. Let us solve the regular Sturm-Liouville problem (0<x<c), X"+AX=O (1) X'(O) = (2) hX(c) + X'(c) = 0, 0, where h is a positive constant. From the corollary in Sec. 44, we know that there are no negative eigenvalues. If A = 0, the general solution of equation (1) is X(x) = Ax + B, where A and B are constants; and it follows from boundary conditions (2) that A = 0 and B = 0. But eigenfunctions cannot be identically zero. Consequently, the number A = 0 is not an eigenvalue. This leaves only the possibility that A >0. If A > this time is 0, we write A = a2 X(x) = (a > C1 0). The general solution of equation (1) cos ax + C2 sin ax. It reduces to X(x) = (3) C1 cos ax when the first of boundary conditions (2) is applied. The second boundary condition then requires that C1(h cos ac — a sin ac) = 0. If the function (3) is to be nontrivial, the constant C1 must be nonzero. Hence the factor in parentheses in equation (4) must be equal to zero. That is, if we are to have an eigenvalue A = a2 (a > 0), the number a must be a positive root of the equation tan ac = h — a 182 CHAP 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS FIGURE 45 Figure 45, where the graphs of y = tan ac and y h/a are plotted, shows that equation (5) has an infinite number of positive roots a1, a2,..., where (n = 1,2,...); they are the positive values of a for which those graphs intersect. The eigenvalues are, then, the numbers = (n = 1,2,...). We identify them by simply writing = where tan h = — > 0). Note that the dashed vertical lines in Fig. 45 are equally spaced ir/c units tend to be the positive roots apart. Also, as n tends to infinity, the numbers of the equation tan ac = 0. More precisely, we see from Fig. 45 that when n is For various values of the constant large, is approximately (n — a = hc, the first few positive roots x1, x2,... of the equation tan x = a/x have been tabulated, and it follows from equation (5) that a1 = x1/c, a2 = x2/c are approximately = In view of the above remarks, the eigenvalues when n is large. This is in agreement with the statement, made [(n — earlier in Sec. 41, that if (n = 1, 2,...) are the eigenvalues of a regular Sturm-Liouville problem, arranged in ascending order of magnitude, then it is 00• oo as n always true that Roots of this and the related equation tan x = ax, arising in some of the problems of this section, are tabulated in, for example, the handbook edited by Abramowitz and Stegun (1972, pp. 224—225) that is listed in the Bibliography. SEC 45 METHODS OF SOLUTION Expression 183 (3) now tells us that, except for constant factors, the corre- sponding eigenfunctions are = cos (n = 1, 2,...). Let us put these eigenfunctions in normalized form, the form that we shall need in the applications. To accomplish this, we note that the functions are orthogonal on the interval 0 <x <c with weight function unity, according to the theorem in Sec. 43. Thus = c 1 f 1 —1(1 + 2 1,2 in write this expression for 2 and + = h/tan sin2ac enable us to the form hc + (7) c = sin2 ac 2h which is obviously positive since h and c are positive. Dividing each we then arrive at the normalized eigenfunctions 2h I by (n = 1,2,...). = V hc + sin2 Sometimes the solutions of a given Sturm-Liouville problem are most easily obtained by transforming the problem into one whose solutions are known. This has already been indicated in Problem 1(a), Sec. 43, and the next example illustrates the method more fully. EXAMPLE 2. We consider here the problem A (1<x<b), (xX')'+—X=O X'(l) hX(b) + X'(b) = 0, 0, where h is a positive constant. Since equation (9) can be put in Cauchy-Euler form (see Problem 3, Sec. 35), x2X" + xX' + AX = 0, the substitution x = exp s transforms it into the equation d2X (0<s<lnb). Also, since dX = dX 184 CHAP 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS the boundary conditions (10) become (12) dX —a— = 0 when s= (hb)X 0, dX + = 0 when s = in b. Hence, by referring to Example 1, we see immediateiy that the eigenvaiues of problem (11)—(12), and therefore of probiem (9)—(1O), are the numbers (13) hb = — where A,1 = >0). The corresponding eigenfunctions are evidently (n = 1,2,...). = cosa,1s = From equations (9) and (11), we know that the weight functions for the eigenfunctions are 1/x and 1, respec= cos (a,1 ln x) and = cos tively. The value of the norm is, however, the same regardiess of whether we think of as a function of x or s. For the substitution x = exp s (s = in x) shows that rlnb rb J 1X So, 0 in view of expression (7), the normalized eigenfunctions of probiem (9)—(10) are 2hb I (14) (n = 1,2,...). = V hblnb + sin2 PROBLEMS In Problems 1 through 5, solve directly (without referring to any other problems) for the eigenvalues and normalized eigenfunctions. 1. X" + AX = 0, Answer: 2. X" + AX = X(0) = = = X(0) = 0, X'(l) = 0, Answer: sin (n = 1,2,...); hX(1) + X'(l) = 0, = 3. X"+AX=O, 0. — cos = = 2 (h > 0). + X(c)=0. IT X'(O)=O, 0 (2n — = (n = 1,2,...); = 2c SEC. 45 4. PROBLEMS 185 + AX = 0, X(O) = 0, X(1) — X'(l) = 0. Suggestion: The trigonometric identity cos2 A = 1/(1 + tan2 A) is useful in putting in a form that leads to the expression for in the answer below. = Answer: A0 = 0, (n = 1,2,...); tan 5. X" + AX = = = X(1) = 0 (h > 0). (n = 1,2,...); —x) + = tan sin > 0). hX(0) — X'(O) = 0, 0, 1) — = > 0). — -h-- 6. In Problem 1(a), Sec. 43, the eigenvalues and eigenfunctions of the Sturm-Liouville problem A (xX')' + —X = 0, X(1) = 0, X(b) = 0 were found to be (n = 1,2,...), = where = mr/in b. Show that the normalized eigenfunctions are &(x) (n = 1,2,...). = Suggestion: The integral that arises can be evaluated by making the substitution s = (17-/in b)ln x and referring to the integration formula (10) in Example 1, Sec. 11. 7. Find the eigenvalues and normalized eigenfunctions of the Sturm-Liouville problem X(0)=0, X"+AX=O, X'(c)=O by making the substitution s = x/c and referring to the solutions of Problem 1. Answer: 8. IT = = (n = 1,2,...); — sin 2c (a) Show that the solutions obtained in Problem 2 can be written I where cos = —h sin > 0). + h2) (n=1,2,...), 186 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS CHAP. 5 (b) By referring to the solutions of Problem 1, point out why the Solutions in part (a) here are actually valid solutions of Problem 2 when h 0, not just when h > Suggestion: In part (a), use the trigonometric identity cos2 A = 1 0. + tan2 A 9. Use the solutions obtained in Problem 3 to find the eigenvalues and normalized eigenfunctions of the Sturm-Liouville problem A (xX')' + —X = X'(l) = 0, 0, X(b) = 0. Answer: (2n-1)'r 2lnb 10. By making an appropriate substitution and referring to the known solutions of the same problem on a different interval in the section indicated, find the eigenfunctions of the Sturm-Liouville problem (a) X" + AX = 0, = 0, X'(n-) = 0 (Sec. 27); (b) X" + AX = 0, X(—c) = X(c), X'(—c) = X'(c) (Sec. 40). 1 Answers: (a) -i-, cos n'rx 1 (b) —, cos 2 c 1, 2,...); (n 2 sin n'rx —(n = 1,2,...). c 11. (a) By making the substitutions Y 1 X=—7= and 4 'six transform the regular Sturm-Liouville problem X(1) = (x2X')' + AX = 0, 0, X(b) = 0, where b> 1, into the problem Y(1)=O, Y(b)=0. (b) Write the eigenvalues and normalized eigenfunctions of the new problem in part (a) by referring to Problem 6. Then substitute back to show that, for the original problem in part (a), the eigenvalues and normalized eigenfunctions are 1 = where [2 = V = inr/ln b. xlnb (n = 1,2,...), SEC 46 EXAMPLES OF EIGENFUNDTION EXPANSIONS 187 12. Find the eigenfunctions of each of the following Sturm-Liouville problems: (h < —1); (a)X"+AX=O, X(O)=0, hX(1)+X'(l)=O (b) (x3X')' + )txX = 0, X(1) = 0, X(e) = 0. Answers: (a) X0(x) = sinh a0x, where tanh a0 = — (a0 >0), (n = 1,2,.. = sin 1 (b) X (x) = — sin x . ), where tan > 0); — —j-- (n'r ln x) (n = 1,2,...). 13. Give details showing that the function W(x) defined by equation (7), Sec. 44, is identically zero on the interval a x b. EXAMPLES OF EIGENFUNCTION EXPANSIONS 46. We now illustrate how generalized Fourier series representations (a <x <b) f(x) = (1) n=1 are obtained when the functions (n = 1, 2,...) are the normalized eigenfunctions of specific Sturm-Liouville problems. We have, of course, already illustrated the method when the eigenfunctions are ones leading to Fourier cosine and sine series on the interval 0 <x <ir, as well as Fourier series on — <x <ir (see Secs. 12, 14, and 15). Except for a few cases in which it is easy to establish the validity of an expansion by transforming it into a known Fourier series representation, in this book we do not treat the convergence of series (1) for other eigenfunctions. We merely accept the fact that results analogous to the Fourier theorem and its corollary in Sec. 19 exist when specific eigenfunctions are used. Such results are often obtained with the aid of the theory of residues of functions of a complex variable.t Proofs are complicated by the fact that explicit solutions of the Sturm-Liouville differential equation with arbitrary coefficients cannot be written. EXAMPLE 1. According to Problem 6, Sec. 45, the Sturm-Liouville problem A (xX')' + —X = 0, X(1) = 0, X(b) = 0 has eigenvalues and normalized eigenfunctions = (n = 1,2,...), theory of eigenfunction expansions is extensively developed in the volumes by Titchmarsh (1962, 1958) that are listed in the Bibliography. 188 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS CHAP. 5 = n17-/ln b. Since the orthogonality of the set (n = 1,2,...) is with respect to the weight function p(x) = 1/x, the coefficients in the expan- where sion (1<x<b) 1= (2) n=1 are = jb (f, = in x) dx. sin Making the substitution s = in x here and noting that cos in b) = cos nir = we readily see that f b'— mb 1—(—1) sin = Thus 1— (n=1,2,...), mb and expansion (2) becomes 00 4 (3) (1<x<b). The validity of this representation is evident if we make the substitution (2n — 1)s = a2n_1 in x and note that —1 = (2n — 1)w/ln b. For the resuit, 4 00 sin(2n—1)s 2n—1 (O<s<ir), is a known [Problem 1(b), Sec. 14] Fourier sine series representation on the indicated interval. EXAMPLE 2. The eigenvalues and normaiized eigenfunctions of the Sturm-Liouville problem X"+AX=O, X(O)=O, X'(c)=O are (Problem 7, Sec. 45) = (n = 1,2,...), 189 EXAMPLES OF EIGENFUNCTION EXPANSIONS SEC 46 where (2n — 1)ir 2c The weight function is p(x) = 1, and we may find the coefficients in the expansion (O<x<c) x= n=1 by writing = Since cos = = 0 — xcosax f and sin ( = V [— flfl+ 1, this expression for cnV_ + sin cr,1x C 2 0 reduces to (n=1,2,...). 2 Hence 1\fl+1 2 (0<x<c). After putting expansion (4) in the form 8c (—1) n+1 (2n — 1)2 (0 <x <c), 2c we see from Problem 12, Sec. 21, that it is actually valid on the closed interval —c + 2c) = — sin x c. Furthermore, since sin (n = 1, 2,...), series (4) converges for all x; and if H(x) denotes the sum of that series for each value of x, it is clear that H(x) represents the triangular wave function defined by means of the equations (see Fig. 46) (5) — 3c FIGURE 46 H(x) =x H(x + 2c) = —H(x) (—c x c), (_oc <x <oc). x 190 CHAP 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS Thus H(x) is an antiperiodic function, with period 2c. It is also periodic, with period 4c, as is seen by writing H(x + 4c) = H(x + 2c + 2c) = —H(x + 2c) = H(x). Note, too, that H(2c — x) = —H(x — 2c) = —H(x + 2c) H(x). We conclude with an example in which the series obtained is a sine series that cannot be transformed into an ordinary Fourier sine series. We must, therefore, accept the representation without verification. EXAMPLE 3. We consider here the eigenvalues and normalized eigenfunctions of the Sturm-Liouville problem X(0) = X" + AX = 0, X(1) — 0, X'(l) = 0. According to Problem 4, Sec. 45, they are (n=1,2,...) A0=O, and + 1) 40(x) = where tan a,1 = the representation = sin a,2x (n = 1,2,...), > 0), the weight function being unity. The coefficients in f(x) (0 <X <1) + n=1 of a piecewise smooth function f(x) are = = and = i/2(a2 + 1) = f(x) = B0x + f 1 sin n=1 SEC. 46 PROBLEMS 191 where B0 = 1 3f xf(x) dx 2(a2+1) = and 2 0 f f(x) 1 sin dx 0 (n=1,2,...). PROBLEMS 1. Use the normalized eigenfunctions in Problem 3, Sec. 45, to derive the representa- tion 2 1=— (O<x<c), where (2n — 1)nn 2. Lc Derive the expansion 2 (O<x<c), a where = (2n 1)n- — 2c using the normalized eigenfunctions in Problem 7, Sec. 45. 3. Use the normalized eigenfunctions in Problem 2, Sec. 45, to derive the expansion 1—cosa, 1 = 2h where tan = —ar/h (0 <x < 1), \ sin + cos 2 n=1 > 0). 4. Using the normalized eigenfunctions in Problem 3, Sec. 45, when c = 'r, show that 4 (0<x<n-), where = (2n — 1)/2. 5. (a) Use the normalized eigenfunctions in Problem 7, Sec. 45, to obtain the expan- sion 4 x(2c—x)=—-E sina where (2n — 2c 1)'r 3 x (O<x<c), 192 CHAP 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS (b) Show how it follows from the result in Problem 8, Sec. 21, that the series in part (a) converges for all x and that its sum is the antiperiodic function Q(x) (see Example 2, Sec. 46), with period 2c, that can be described by means of the equations Q(x)=x(2c—x) Q(x+2c)= —Q(x) (—oo<x<oo). 6. Using the normalized eigenfunctions in Problem 2, Sec. 45, derive the representation f2+h \1+h xI \ (O<x<1), —xI =4h / where tan = —ar/h > 0). Suggestion: In the simplifications, it is useful to note that = —h sin cos 7. Use the normalized eigenfunctions in Problem 1, Sec. 45, to show that \fl / sin wx = 2w cos w where 2 sin (0 <x < 1), for any value of n. w# Suggestion: The trigonometric identity 2 sin A sin B = cos (A — B) — cos (A + B) is useful in evaluating the integrals that arise. 8. Find the Fourier constants for the function f(x) = x (1 <x <b) with respect to the normalized eigenfunctions in Problem 6, Sec. 45, and reduce them to the form n'r[l + 2 (lnb) + (mT) Suggestion: The 2 (n=1,2,...). integration formula ex(sin I sin ax dx = ax — a cos ax) 1+a2 J derived in calculus, is useful here. 9. Let f be a piecewise smooth function defined on the interval 1 <x <b. (a) Use the normalized eigenfunctions in Problem 6, Sec. 45, to show formally that if = inr/ln b, then f(x)= (1<x<b), n=1 where 2 —f lnb (b) i b1 x (n= 1,2,...). By making the substitution x = exp s in the series and integral in part (a) and then referring to the corollary in Sec. 21, verify that the series representation in SURFACE HEAT TRANSFER SEC. 47 193 part (a) is valid at all points in the interval 1 <x < b at which f is continuous. (Compare Example 1, Sec. 46.) 10. Suppose that a function f, defined on the interval 0 <x <c, is piecewise smooth there. (a) Use the normalized eigenfunctions (Problem 7, Sec. 45) IT (n = 1,2,...), = where (2n — 1)'ir 2c to show formally that f(x)= (0<x<c), n=1 where 11. 2c (n=1,2,...). (b) Note that, according to Problem 11, Sec. 21, the series in part (a) is actually a Fourier sine series for an extension of f on the interval 0 <x < 2c. Then, with the aid of the corollary in Sec. 21, state why the representation in part (a) is valid for each point x (0 <x <c) at which f is continuous. (a) Use the normalized eigenfunctions (n = 1,2,...), where (2n — - 2 in Problem 1, Sec. 45, to show formally that / ( 1 x(1__x2 (0<x<1). " \ n=i an (b) Note that, according to Problem 10(a) above and Problem 11, Sec. 21, the series in part (a) here is a Fourier sine series on the interval 0 <x <2. Then, with the aid of the corollary in Sec. 21, show that the series in part (a) converges for all x and that its sum is the antiperiodic function Q(x), with period 2, that is described by the equations Q(x) =x(1 47. - (-1 1), Q(x + 2) = -Q(x) (_oo <x <oo). SURFACE HEAT TRANSFER The following two examples illustrate the Fourier method in solving temperature problems in rectangular coordinates when Sturm-Liouville problems other 194 CHAP 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS than those used in Chap. 4 arise. Here, and in the rest of the chapter, we seek only formal solutions of the boundary value problems. EXAMPLE 1. Let u(x, t) denote temperatures in a slab 0 x 1, initially at temperatures f(x), when the face x = 0 is insulated and surface heat transfer takes place at the face x = 1 into a medium at temperature zero (Fig. 47). According to Newton's law of cooling (Sec. 3), the condition on u at the face x = 1 is t) = —hu(1, t), where h is a positive constant. The boundary value problem to be solved is, then, (1) = (2) (0 = —hu(1,t), = 0, <x < 1, t > 0), u(x,O) =f(x). I x=O x=1 FIGURE47 Writing u = X(x)T(t) and separating variables, we arrive at the SturmLiouville problem X"(x) + AX(x) = X'(O) = 0, along with the condition T'(t) + itkT(t) = 0. (3) 0, hX(1) + X'(l) = 0, The eigenvalues and normalized and = eigenfunctions of problem (3) are, according to Example 1, Sec. 45, I = = h/ar where tan constant multiples of 2h V h + (n = 1,2,...), > 0). The corresponding functions of t are evidently = Hence the formal solution of our temperature problem is u(x,t) = n=1 (n = 1,2,...). SURFACE HEAT TRANSFER SEC. 47 where, in order that u(x, 0) = f(x) (6) (0 <x < 1), = = /;+sin2an = Observe (n fl = nLi h = 1,2,...). is substituted, is that series (5), when expression (4) for u(x,t) 195 2h cn) + Hence the solution just obtained can be written u(x,t) = (7) where 2h (8) 1 (n = 1,2,...). = h + It is easy to show that solution (7), with coefficients (8), also satisfies the boundary value problem (9) (—1 <x < 1, t>0), = (10) = hu(—1,t), (11) u(x,O) = —hu(1,t) =f(x) (t >0), (—1 <x < 1) when f is an even function, or when f( —x) = f(x) (—1 <x < 1). For we already know that u satisfies the heat equation and the second of boundary conditions (10). Since the cosine function is even, it is clear from expression (7) is odd in x. Hence the first of that u is even in x; and its partial derivative boundary conditions (10) is also satisfied: = Finally, when — =hu(1,t) =hu(—1,t). we already know that u(x, 0) = f(x) when 0 <x < 1; furthermore, 1 <x < 0, the fact that u and f are even in x enables us to write u(x,0) =u(—x,0) =f(—x) =f(x). The boundary value problem (9)—(11) is, of course, a temperature problem 1 with initial temperatures (11) and with surface heat 1 x for a slab — transfer at both faces into a medium at temperature zero (Fig. 48). The solution when f is not necessarily even is obtained in the problems. 196 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS u(x,O) =f(x) 00 CHAP. 5 00 x x= 1 x = —1 FIGURE 48 EXAMPLE 2. Let u(x, y) denote the bounded steady temperatures in a semi-infinite slab bounded by the planes x = 0, x = ii-, and y = 0 (Fig. 49), whose faces are subject to the following conditions. The face in the plane x = 0 is insulated, the face in the plane x = is kept at temperature zero, and the flux inward through the face in the plane y = 0 (see Sec. 3) is a prescribed function f(x). The boundary value problem for this slab is y) = (y > 0), 0 =f(x) (0 <x <IT), where K is a positive constant. U =0 x f(x) FIGURE 49 By assuming a product solution u = X(x)Y(y) of conditions (12) and (13) and separating variables, we find that X"(x) + AX(x) = 0, X'(O) = 0, =0 SEC 47 and SURFACE HEAT TRANSFER 197 that Y(y) is to be a bounded solution of the differential equation Y"(y) — AY(y) = (16) 0. According to Problem 3, Sec. 45, the eigenvalues and normalized eigenfunctions of the Sturm-Liouville problem (15) are (n = 1,2,...), — = vir (2n — 1)/2. The corresponding bounded solutions of equation (16) are constant multiples of the functions where (n = 1,2,...). Consequently, u(x,y) = (17) n=1 Applying the nonhomogeneous condition (14) to this expression, we see that the constants must be such that f(x)= (O<x<ir). n==1 That is, (18) = (n = 1,2,...). = Finally, it follows from expressions (17) and (18) that u(x, y) = 1 cos — where (20) Since (n=1,2,...). ITO (2n — 1)/2, equations (19) and (20) can, of course, be written in the form u(x, y) = exp[—(2n — 1)y/2] 2 — (2n — 1)x cos 2n—1 2 where 2 ITO (2n—1)x 2 dx (n=1,2,...). 198 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS 48. POLAR COORDINATES CHAP 5 We consider here a Dirichlet problem (Sec. 7) for a function u(p, 4)), involving polar coordinates, that satisfies Laplace's equation =0 (1) (1 <p <b, 0 and the boundary conditions (Fig. 50) u(p,O) = 0, u(1,4) = 0, (1 <p <b), (0 <g& <rr), =u0 u(b,4) = 0 where u0 is a constant. U=0 FIGURE 50 Substitution of the product u = tions here yields these conditions on R and 1: A [pR'(p)]' + —R(p) = into the homogeneous condi- R(1) = 0, 0, R(b) = 0, p = 0, — c1(O)=0. Conditions (4) on R make up a Sturm-Liouville problem whose eigenvalues are = = nir/ln b, and whose normalized eigenfunc(n = 1,2,...), where tions are = V lnb (n=1,2,...). (See Problem 6, Sec. 45.) Note that the weight function for these eigenfunctions is i/p. Except for constant factors, the corresponding functions of 4, arising from conditions (5), are = Hence u(p,4)) = n=1 (n=1,2,...). SEC. 48 PROBLEMS 199 Turning to the nonhomogeneous condition u(p, ir) = u0, we set 4) = ir in expression (6) and write u0 = (1 <p <b). sinh n=1 Evidently, then, = (u0, sinh ln p) dp. = uoV lnb f 1— p This integral is readily evaluated by making the substitution p = exp s; and, by recalling that = nir/ln b, one can simplify the result to show that sinh = u0V2lnb fl IT So, in view of expressions (6) and (7), u(p,4) = 2u0 00 IT n=1 1— (_i)nl n That is, 00 4u0 2n — 1 n=i It is interesting to contrast this solution with the one obtained in Example 2, Sec. 34, for a Dirichlet problem involving the same region but with the nonhomogeneous condition u = u0 occurring when p = b instead of when = IT. PROBLEMSt 1. Show that when f(x) = 1 (0 <x < 1) in the boundary value problem in Example 1, Sec. 47, the solution (7)—(8) there reduces to 00 u(x,t) = 2h E = h/ar where tan \ + sin 2 • n=1 > 0). 2. Show that if the condition u(p, 'r) = u0 (1 <p <b) in Sec. 48 is replaced by the condition u(p, 'r) = p (1 <p <b), then I n11 + (—1) n+11 bj (mb) 2 where = +(n'r) 2 • • b. tThe eigenvalues and (normalized) eigenfunctions of any Sturm-Liouville problem that arises have already been found in Sec. 45 or in one of the problems of that section. 200 CHAP 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS Suggestion: The Fourier constants found in Problem 8, Sec. 46, can be used here. 3. Use the normalized eigenfunctions of the Sturm-Liouville problem X"+AX=O, X'('r)=O X(O)=O, to solve the boundary value problem (0 <x < t) u1(x, t) = u(0,t)=0, t > 0), u(x,0)=f(x). Show that the solution can be written (2n — 1)x (2n — 1)2k u(x,t)= t sin 2 — where 2 = —f f(x)sin (2n—1)x (n = 1,2,...). 2 (The solution in this form was obtained in another way in Example 3, Sec. 32.) 4. Solve the boundary value problem (0<x<a,O<y<b), y) = y) = 0, u(x,O) = 0, —hu(a, y) u(x,b) =f(x) (0 <y <b), (0 <x <a), where h is a positive constant, and interpret u(x, y) physically. 00 Answer: u(x,y) = 2h 2 n=1 ha + sin • sinh • f a o where tan > 0). = h/ar 5. A bounded harmonic function u(x, y) in the semi-infinite strip x> 0, 0 <y < 1 is to satisfy the boundary conditions u(x,0)=0, —hu(x,1), u(0,y)=u0, where h (h > 0) and u0 are constants. Derive the expression n=1 where tan = —ar/h + cos 2 > 0). Interpret u(x, y) physically. 6. Find the bounded harmonic function u(x, y) in the semi-infinite strip 0 <x < 1, y> 0 that satisfies the boundary conditions = —hu(1,y), = 0, u(x,0) =f(x), where h is a positive constant, and interpret u(x, y) physically. Answer: u(x, y) = exp (—any) cos n=1 201 PROBLEMS SEC. 48 where tan = h/ar > 0) and An=h+ 2h (n=1,2,...). 1 in2 7. Find the bounded solution of this boundary value problem, where b and h are positive constants: u(O,y) = (O<x< l,y>O), u(x,0) =f(x). = —hu(1,y), 0, sinax 00 Answer: u(x, y) = ,, n=1 where tan > 0) and = —ar/h 2h (n=1,2,...). 1 h+cos2a 8. Let p, 4', z denote cylindrical coordinates, and solve the following boundary value problem in the region 1 p b, 0 4' 'r of the plane z = 0: (1 <p <b, 0 <4) 4)) = 0 4)) + 4)) + (0 <4) <jr), = —hu(b,4)) = 0, <17-), (1<p<b), u(p,'r)=u0 where h (h > 0) and u0 are constants. Interpret the function u(p, 4)) physically. Answer: u(p, 4)) = 2hbu0 cosh ln b) cos sin n=1 cosha,1n- ln p) + ln b) = where tan > 0). 9. Give a full physical interpretation of the following temperature problem, involving a time-dependent diffusivity, and derive its solution: t) (t + 1)u1(x, t) = u(0,t) = Answer: u(x, t) = u(x,0) = = 0, 0, 2sinax (t + i)' 2 (0 < x < 1, t > 0), , = where 1. (2n—1)'r - 2 n=1 10. (a) Give a physical interpretation of the boundary value problem u(0, t) = 0, t) = —hu(1, (0 <x < 1, t > 0), t) u1(x, t) = t), u(x,0) = f(x), where h is a positive constant. Then derive the solution u(x,t) = 202 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS where tan = —ar/h CHAP. 5 > 0) and 2h 1 (n= 1,2,...). (b) Use an argument similar to the one at the end of Example 1 in Sec. 47 to show that the solution found in part (a) formally satisfies the boundary value problem (9)—(11) in that example when the function f there is odd, or when f(—x) = -f(x) (-1<x<1). 11. Use the following method to solve the temperature problem (see Fig. 48 in Sec. 47) (—1 <x < 1, t >0), = = hu(—1,t), = —hu(1,t) (t >0), (-1 <x <1) u(x,0) =f(x) when the function f is not necessarily even or odd, as it was in Example 1, Sec. 47, and Problem 10(b). (a) Show that if v(x, t) is the solution of the problem when f(x) is replaced by the function G(x) f(x) +f(—x) and if w(x, t) is the solution when f(x) is replaced by H(x) = f(x) —f(—x) then the sum u(x, t) = v(x, t) + w(x, t) satisfies the above boundary value problem. (b) After noting that the functions G and H in part (a) are even and odd, respectively, apply the result there, together with results in Example 1, Sec. 47, and Problem 10, to show that u(x,t) where tan = h/ar > 0), tan h h + sin 2 • h = h + cos2 > 0) and = f 1 —1 ff(x)sin 1 Suggestion: A procedure similar to the one in the suggestion with Problem 12, and in part (b). Sec. 14, can be used in obtaining the expressions for 49. MODIFICATIONS OF THE METHOD In this section, we illustrate two modifications of the Fourier method involving normalized eigenfunctions. Both such modifications were used in Chap. 4 when only ordinary Fourier cosine and sine series arose. MODIFICATIONS OF THE METHOD SEC 49 203 EXAMPLE 1. If heat is introduced through the face x = 1 of a slab 1 at a uniform rate A (A > 0) per unit area (Sec. 3) while the face x = 0 is kept at the initial temperature zero of the slab, then the temperature 0 x function u(x, t) satisfies the conditions (1) (0 <x <1, t >0), = u(0,t) = (2) =A 0, u(x,O) = (3) (0 <x < 0 1). Because the second of conditions (2) is nonhomogeneous, we do not have two-point boundary conditions leading to a Sturm-Liouville problem. But, by writing u(x,t) = U(x,t) (4) + (compare Example 2, Sec. 32), we find that conditions (1)—(3) become t) + t) = u(O, t) + = A, K[Lç(1, t) + = 0, and U(x,O) + 0. Hence, if we require that K'V(1) =A, we have a boundary value problem for U(x, t) that does have two-point = 0 and = 0, boundary conditions leading to a Sturm-Liouville problem: =0, U(0,t) =0, (6) U(x,0) = It follows readily from conditions (5) that A = —x. K Also, by assuming a product solution U = X(x)T(t) of the homogeneous conditions in problem (6), we see that X"(x) + AX(x) = X(0) = 0, X'(l) = 0 0, and T'(t) + AkT(t) = 0. According to Problem 1, Sec. 45, the Sturm-Liouville problem (8) has the eigenvalues and normalized eigenfunctions A,2 = where exp = (2n — (n = 1, = sin (n = 1,2,...), and the corresponding functions of t 2,...). Hence U(x,t) = are = 204 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS CHAP. 5 where, in view of the last of conditions (6), (O<x<1). n=1 Now the Fourier constants for x (0 <x < 1) with respect to the normalized eigenfunctions here are already known to us (see Example 2, Sec. 46, when c = 1), and that earlier result tells us that After substituting these values of into expression (9) and simplifying and combining the result with expression (7), as indicated in equation (4), we arrive at the desired temperature function: 00 (10) u(x,t) = x +2 2 n=1 where a,1 = (2n — 1)ir/2. EXAMPLE 2. Let u(x, t) denote temperatures in a slab 0 x ir (Fig. 51) that is initially at temperature zero and whose face x = 0 is insulated, while the face x = ir has temperatures t) = t (t 0). If the unit of time is chosen so that the thermal diffusivity k in the heat equation is unity, the boundary value problem for u(x, t) is (11) t) = (12) II = 0, (0 <x < ii-, t > 0), t) u(T7-,t) = t, u(x,O) 0. I 111=t x=O FIGURES! Observe that if u(x, t) satisfies the first two of conditions (12), then the related function U(x, t) = u(x, t) — t satisfies the conditions and U(ir,t)=0, both of which are homogeneous. In fact, by writing u(x, t) = U(x, t) + t, MODIFICATIONS OF THE METHOD SEC. 49 205 we have the related boundary value problem consisting of the differential equation (14) —1 = and conditions (13), along with the condition U(x,O) (15) 0. The nonhomogeneity in the second of conditions (12) is now transferred to the differential equation in the new boundary value problem, consisting of equations (13)—(15), and this suggests applying the method of variation of parameters, first used in Sec. 33. We begin by noting that when the method of separation of variables is t), which is applied to the homogeneous differential equation L1(x, t) = equation (14) with the — term deleted, and conditions (13), the Sturm- 1 Liouville problem X"(x) + AX(x) X'(O) = 0, X(ir) 0, = 0 arises. Furthermore, from Problem 3, Sec. 45, we know that the eigenfunctions of this problem are the cosine functions cos = (n = 1, 2,...), where (2n — 1)/2. We thus seek a solution of the boundary value problem (13)—(15) having the form U(x,t) (16) = n=1 By substituting series (16) into equation (14) and referring to Problem 1, Sec. 46, for the expansion 1= (_1)n+1 2 (0 <x <ir), cos — n + cos = — cos n=1 n=1 Then, by identifying the coefficients in the eigenfunction expansions on each side here, we have the differential equation + 2 (n = = Also, condition (15) tells us that cos n=1 or = 0 (n = 1,2,...). = 0, 1,2,...). 206 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS CHAP. 5 Now an integrating factor for the linear first-order differential equation (17) is exp dt t. = exp Hence, if we multiply through the differential equation by this integrating factor, we have d = By replacing the variable t here by T, integrating the result from T = and keeping in mind the requirement that = 0, we see that 2 2 = 0 to T = — 1), or = [1- (n Finally, by substituting this expression for then into = 1,2,...). equation (16) and recalling that u(x, t) = U(x, t) + t, we obtain the solution of the original boundary value problem: u(x,t) = + (18) [1 — where = (2n — 1)/2. Note that, in view of the representation found in Problem 4, Sec. 46, this solution can also be written as (_.1)fl 00 2 (19) u(x,t) = 2 — — 3 — PROBLEMSt 1. With the aid of representation (4) in Example 2, Sec. 46, Show that the temperature function (10) in Example 1, Sec. 49, can be written in the form n+1 u(x,t) where = (2n — = [i — 1)'r/2. footnote with the problem set for Sec. 48 also applies here. sin 207 PROBLEMS SEC. 49 2. Heat transfer takes place at the surface x = 0 of a slab 0 x 1 into a medium at temperature zero according to the linear law of surface heat transfer, so that (Sec. 3) t) = hu(0, t) (h > 0). The other boundary conditions are as indicated in Fig. 52, and the unit of time is chosen so that k = 1 in the heat equation. By proceeding as in Example 1, Sec. 49, derive the temperature formula hx+1 u(x,t) h+ 2hE 1 + > 0). = —ar/h where tan Suggestion: In simplifying the expression for the Fourier constants that arise, it is useful to note that — — 00 u(x,O)=O u=1 x x=0 x=1 FIGURE52 3. Use the method of variation of parameters to solve the boundary value problem t) u1(x, t) = u(1,t) = = 0, for temperatures in an internally heated slab. Suggestion: The representation, with c = 46, is needed here. (0 <x < + q(t) u(x,0) = 0, t 1, > 0), 0 that was found in Problem 1, Sec. 1, 1n+1 Answer: u(x, t) = cos aflxft exp 2 n=1 fl — r)] q(r) dr, 0 where = (2n — 1)'r/2. 4. Solve the temperature problem u1(x, t) = = 0, u(1,t) = F(t), t) (0 u(x,0) = 0, <x < 1, t > 0), 208 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS where F is continuous and F(O) = 0. CHAP. 5 (Compare Example 2, Sec. 49.) Express the answer in the form u(x,t) = F(t) + 2 r)] F'(r) dr, — o n==1 where = (2n — 1>n-/2. 5. Use the method in Example 1, Sec. 49, to solve the boundary value problem (t + 1)u1(x, t) = u(1,t) = = —1, 0, t) (0 u(x,0) = 0. <x < 1, t > 0), Interpret this problem physically (compare Problem 9, Sec. 48). 00 Answer: u(x, t) = 1 — x — (t + 1) 2 2cosax 2 n=1 where = (2n — 1)ir/2. 6. By using the method of variation of parameters, derive the bounded solution of this problem: +q0=0 u(0,y) = (h >0), = —hu(1,y) 0, (0<x< l,y>O), u(x,0) = 0, where q0 and h are constants. Interpret the problem physically. Suggestion: The representation found in Problem 3, Sec. 46, is needed here. Also, for general comments on solving a nonhomogeneous linear second-order differential equation that arises, see the suggestion with Problem 8, Sec. 35. 00 Answer:u(x,y) = + cos2 an) [1 — sin > 0). where tan = —ar/h 7. With the aid of the representation found in Problem 6, Sec. 46, write the solution in Problem 6 above as 2+h q0 1 where tan 8. + h = —an/h 00 — + > 0). Then observe how it follows that The boundary r = 1 of a solid sphere is kept insulated, and that solid is initially at temperatures f(r). If u(r, t) denotes subsequent temperatures, then = k82 ur(1,t) = 0, u(r,0) =f(r). By writing v(r, t) = ru(r, t) and noting that u is continuous when r = 0 (compare Problem 6, Sec. 32), set up a boundary value problem in v, involving the boundary SEC. 50 A VERTICALLY HUNG ELASTIC BAR 209 conditions v(O, t) = v(1, t) 0, Vr(1, t), v(r,0) = rf(r). Then derive the temperature formula sinar 00 u(r,t)=B0+ n=1 where tan B0 = = > 0) and 3f'r2f(r)dr, (n = 1,2,...). = 2(an + (An eigenfunction expansion similar to the one required here was found in Example 3, Sec. 46.) 50. A VERTICALLY HUNG ELASTIC BAR An unstrained elastic bar, or heavy coiled spring, is clamped along its length c so as to prevent longitudinal displacements and then hung from its end x = 0 (Fig. 53). At the instant t = 0, the clamp is released and the bar vibrates longitudinally because of its own weight. If y(x, t) denotes longitudinal displacements in the bar once it is released, then y(x, t) satisfies the modified form t) = t) +g (0 <x <c, t> 0) of the wave equation, where g is the acceleration due to gravity. The stated conditions at the ends of the bar tell us that y(0,t) =0, =0, the initial conditions being y(x,0) = 0, = 0. The fact that equation (1) is nonhomogeneous suggests that we use the method of variation of parameters. More precisely, we seek a solution of our I I FIGURE 53 210 CHAP. 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS boundary value problem having the form (4) y(x, t) = sin cr,1x, n=1 where (2n — 1)ir afl 2c We have chosen the sine functions sin (n = 1, 2,...) here since they are the eigenfunctions (Problem 7, Sec. 45) of the Sturm-Liouville problem X"(x) + AX(x) = X(O) 0, X'(c) = 0, 0, which arises when the method of separation of variables is applied to the homogeneous wave equation t) = t) and conditions (2). Substituting series (4) into equation (1) and recalling the representation (Problem 2, Sec. 46) 2 00 sin sin That is, 2g (n = 1,2,...). = + (5) It follows, moreover, from conditions (3) that (6) = 0, = (n = 1,2,...). 0 Now the general solution of the complementary equation + = 0 = C1 a particular and C2 solution of equation (5) is 2g = Hence the general solution of equation (5) is = C1 cos + C2 sin + 2g A VERTICALLY HUNG ELASTIC BAR SEC. 50 The constants C1 and C2 are 211 readily determined by imposing conditions (6) on expression (7). The result is 2g 3(1 = a2 — and, in view of equation (4), this means that y(x,t) = (8) sina x 2g (1 — This solution can actually be written in closed form in the following way. We first recall from Problem 5, Sec. 46, that 4 -E (9) sifla X =Q(x) (—oc<x<oo), where Q(x) is the antiperiodic function, with period 2c, described by means of the equations Q(x) =x(2c —x) (0 (—oc<x<oo). Q(x+2c)=—Q(x) Thus we can put expression (8) in the form (11) g y(x,t) = x(2c —x) — 4 °° — 3 As for the remaining series here, the trigonometric identity 2 sin A cos B = sin (A + B) + sin (A — B) enables us to write sin sin cos + at) 00 = n=1 or Q(x + at) + Q(x 4 — at) 2 Finally, then, (12) y(x,t) = g —x) — Q(x 2c), +at) + Q(x —at)1 2 212 CHAP. 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS PROBLEMSt 1. A horizontal elastic bar, with its end x = 0 kept fixed, is initially stretched so that its longitudinal displacements are y(x, 0) = bx (0 x c). It is released from rest in that position at the instant t = 0; and its end x = c is kept free, so that t) = 0. Derive this expression for the displacements: y(x, t) = b + at) + H(x — at)], where H(x) is the triangular wave function (5) in Example 2, Sec. 46. (Except for the condition at x = 0, the boundary value problem here is the same as the one solved in Sec. 37.) 2. Suppose that the end x = 0 of a horizontal elastic bar of length c is kept fixed and that a constant force F0 per unit area acts parallel to the bar at the end x = c. Let all parts of the bar be initially unstrained and at rest. The displacements y(x, t) then satisfy the boundary value problem t) = t) y(O, t) = 0, y(x,0) = t) <x < c, t > 0), = F0, y1(x,0) = 0, (0 0, E is Young's modulus of elasticity, and ô is the mass per unit where a2 = volume of the material (see Sec. 6). (a) Write y(x, t) = Y(x, t) + c1'(x) (compare Example 1, Sec. 49) and determine such that Y(x, t) satisfies a boundary value problem whose solution is obtained by simply referring to the solution in Problem 1. Thus show that H(x+at)+H(x—at) F0 2 where H(x) is the same triangular wave function as in Problem 1. (b) Use the expression for y(x, t) in part (a) to show that those displacements are periodic in t, with period 4c T0= — 3. That is, show that y(x, t + T0) = y(x, t). Show that the displacements at the end x = c of the bar in Problem 2 are y(c, t) = F0 + H(at — c)] and that the graph of this function is as shown in Fig. 54. The footnote with the problem set for Sec. 48 applies here as well. SEC. 50 PROBLEMS 213 y(c,t) 2cF() E 0 4c 8c a a t FIGURE 54 4. Show that the force per unit area exerted by the bar in Problem 2 on the support at the end x = 0 is the function (see Sec. 6) t) = F0{1 — H'(at)] and that the graph of this function is as shown in Fig. 55. (Note that this force becomes twice the applied force during regularly spaced intervals of time.) t) 2 F0 9 I 0 I 1— —9 I ( I i I I I I I I I I I I I -[ I C 3c 5c 7c 9c a a a a a I i FIGURE 55 5. Let the constant F0 in Problem 2 be replaced by a finite impulse of duration 4c/a: F0 when 4c 0 < t < —, a F(t) = when 0 4c t>—. a (a) State why the displacements y(x, t) are the same as in Problem 2 during the time interval 0 < t <4c/a. Then, after showing that y(x,4c/a) = 0 and y1(x,4c/a) = 0 when y(x, t) is the solution in Problem 2, state why there is no motion in the bar here after the time t = 4c/a. 214 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS CHAP. 5 (b) Use results in part (a) and Problem 3 to show that the end x = c of the bar moves with constant velocity v0 = aF0/E during the time interval 0 <t <2c/a and then with velocity — v0 when 2c/a <t <4c/a and that it remains station- ary after time t = 4c/a. 1 of a stretched string is elastically supported (Fig. 56) so that the transverse displacements y(x, t) satisfy the condition t) = —hy(1, t), where h is a positive constant. Also, 6. The end x = y(0,t)=O, y(x,0)=bx, y1(x,0)=O, where b is a positive constant; and the wave equation y11 = the following expression for the displacements: y(x,t) = is satisfied. Derive sinci sina x 2bh(h + 1) cos 2 2 where tan = —ar/h (ar> 0). Suggestion: In simplifying the solution to the form given here, note that = (1,b) x FIGURE 56 7. An unstrained elastic bar of length c, whose cross sections have area A and whose modulus of elasticity (Sec. 6) is E, is moving lengthwise with velocity v0 when at the instant t = 0 its right-hand end x = c meets and adheres to a rigid support (Fig. 57). and the end = The displacements y(x, t) thus satisfy the wave equation conditions t) = y(c, t) = 0, as well as the initial conditions y(x,0) = 0, y1(x,0) = V0 x x=c FIGURE 57 v0. PROBLEMS SEC. 50 215 (a) Derive this expression for the displacements: y(x,t) = (—1) 2v0 n+1 2 ac n=1 where = (2n — 1)ir/2c (n = 1,2,...). (b) Use the expression for y(x, t) in part (a) to show that =0 and = —v0 (O<x<c). According to these two equations, if the end x = c of the bar is suddenly freed from the support at time t = 2c/a, the bar will move after that time as a rigid unstrained body with velocity — v0. (c) Show how it follows from the expression in part (a) that, as long as the end of the bar continues to adhere to the support, the force on the support can be written = AEv0 I 2c where M(c, t) (t> 0) is the square wave represented by the series (see Problem 9, Sec. 21) M(c,t) = 4 1 sin — (2n—1)n-t c 8. Let y(x, t) denote longitudinal displacements in an elastic bar of length unity whose end x = 0 is fixed and at whose end x = 1 a force proportional to t2 acts longitudinally (Fig. 58), so that y(0,t)=O and The bar is initially unstrained and at rest, and the unit of time is such that a = the wave equation. I y(x,t)I 1 in I_øAt2 FIGURE 58 (a) Write out the complete boundary value problem for y(x, t); and observe that if Y(x, t) = y(x, t) — At2x, then Y(O,t)=0 and Set up the complete boundary value problem for Y(x, t), the differential equation being (O<x< 1,t>0). 216 CHAP 5 STURM-LIOUVILLE PROBLEMS AND APPLICATIONS Then, with the aid of representation (4) in Example 2, Sec. 46, apply the method of variation of parameters to solve the boundary value problem for Y(x, t) and thus derive this solution of the original problem: 00 y(x, t) = A 1\fl+1 ( xt2 — 4 (1 — cos art) sin n=1 where = (2n — 1)'r/2. (b) Use the result in Problem 11(b), Sec. 46, to write the solution in part (a) here i the form 1 y(x,t) =A x(t2 — 1) + + Q(x+t)+Q(x—t) 2 where Q(x) is the antiperiodic function, with period 2, described by the equations - Q(x) =x(1 (-1 1), (-oo<x<oo). Q(x+2)= -Q(x) 9. Consider the same boundary value problem as in Problem 8 except that the condition at the end x = 1 of the bar is now replaced by the condition = sinwt. (a) By proceeding in the same manner as in Problem 8, show that if (2n — (n=1,2,...) 2 and w # for any value of n, then fw 00 y(x, t) = x sin wt + 2w n=1 2 2 I — sin wt — sin sin wax. / — for any (b) Modify part (a) to show that resonance (Sec. 38) occurs when w = value of N. Suggestion: In each part of this problem, it is helpful to refer to the general solution of a certain ordinary differential equation in Problem 13, Sec. 38. 10. By referring to expansion (4) in Example 2, Sec. 46, and the expansion found in Problem 7, Sec. 46, write the solution in Problem 9(a) here in the form y(x, t) = where = (2n — sinwxsinwt wcosw 1)'r/2. + 2w (—1) 2 n+1 2 —wa) sin sin CHAPTER 6 FOURIER INTEGRALS AND APPLICATIONS In Chap. 2 (Sec. 21), we saw that a periodic function, with period 2c, has a Fourier series representation that is valid for all x when it satisfies certain conditions on the fundamental interval — c <x <c. In this chapter, we develop the theory of trigonometric representations for functions, defined for all x, that are not periodic. Such representations, which are analogous to Fourier series representations, involve improper integrals, rather than infinite series. 51. THE FOURIER INTEGRAL FORMULA From Problem 13, Sec. 21, we know that the Fourier series corresponding to a function f(x) on an interval —c <x <c can be written 1JC ds + JC cos - x)] ds; and, from the corollary in Sec. 21, we know conditions under which this series converges to f(x) everywhere in the interval — c <x <c. Namely, it is sufficient that f be piecewise smooth on the interval and that the value of f at each of its points of discontinuity be the mean value of the one-sided limits f(x +) and f(x —). Suppose now that f satisfies such conditions on eveiy bounded interval — c <x <c. Here c may be given any fixed positive value, arbitrarily large but 217 218 FOURIER INTEGRALS AND APPLICATIONS CHAP. 6 finite, and series (1) will represent f(x) over the large segment —c <x <c of the x axis. But that series representation cannot apply over the rest of the x axis unless f is periodic, with period 2c, because the sum of the series has that periodicity. In seeking a representation that is valid for all real x when f is not periodic, it is natural to try to modify series (1) by letting c tend to infinity. The first term in the series will then vanish, provided that the improper integral f f(s)ds = ir/c, the remaining terms take the form exists. If we write 100 C — f(s) cos [n — x)] ds, —c which is the same as 100 / — CJ where x) = ff(s) cos a(s - x) ds. Let the value of x be fixed and c be large, so that a small positive number. The points n (n = 1,2,...) are equally spaced along the entire positive a axis; and, because of the resemblance of the series in expression (3) to a sum of the type used in defining definite integrals, one might expect that the sum of that series tends to I x) da, or possibly x) da, as tends to zero. As integral (6) indicates, however, the function x) changes with because c = Also, the limit of the series in expression tends to zero is not, in fact, the definition of the improper integral (5) (3) as even if c could be kept fixed. The above manipulations merely suggest that, under appropriate conditions on f, the function may have the representation 10000 f(x)=_f f f(s)cosa(s—x)dsda IT 0 This (—oo<x<oo). —00 is the Fourier integral formula for the function f, to be established in Sec. 54. AN INTEGRATION FORMULA 219 SEC. 52 The formula can be written in terms of separate cosine and sine functions as follows: f(x)=f [A(cr)coscrx+B(a)sinax]da (—oo<x<oo), where (9) loo B(a)=-_f f(x)sinaxdr. loo A(a)=_f f(x)coscrxdx, Expressions (8) and (9) bear a resemblance to Fourier series representations on <x and formulas for the coefficients and derived in Sec. 15. 52. AN INTEGRATION FORMULA Just as we prefaced our Fourier theorem in Sec. 19 with some preliminary theory, we include here and in Sec. 53 background that is essential to our proof of a Fourier integral theorem, which gives conditions under which representation (7) in Sec. 51 is valid. This section is devoted to the evaluation of an improper integral that is prominent in applied mathematics. We show here thatt oosinx (1) Our method of evaluation requires us to first show that the integral actually converges. We note that the integrand (sin x)/x is piecewise continuous on any bounded interval; and since sin x (2) X 0 ,i sin x 0 ,c sin x X X 1 where c is any positive number, it suffices to show that the integral toosinx I —dx= lim X c-÷oo-'1 (3) I X )i converges. To accomplish this, we use the method of integration by parts and write (4) csinX 1 Since I(cos c)/c I X cosc C ccOsX 1 X 2 1/c, the second term on the right-hand side here tends to tFor another approach that is fairly standard, see, for instance, the book by Buck (1978, pp. 293—295). A method of evaluation involving complex variables is indicated in the authors' book (1990, pp. 197—198). Both books are listed in the Bibliography. 220 FOURIER INTEGRALS AND APPLICATIONS CHAP 6 zero as c tends to infinity. Also, since I(cos x)/x21 X 1 is (absolutely) convergent. The limit of the left-hand side of equation (4) as c tends to infinity therefore exists; that is, integral (3) converges. Now that we have established that integral (1) converges to some number L, or that sin x limJ —=L, c-+oo 0 X we note that, in particular, rf(2N+l)1T/2 dx = L as N passes through positive integers. That is, 1)u/2] hmJ N-÷000 where U du=L, the substitution x = (2N + 1)u/2 has been made for the variable of integration. Observe that equation (7) can be written du = L, lim f0 where g(u) sin (u/2) = (u/2) and where DN(u) is the Dirichlet kernel (Sec. 18) DN(u) sin [(2N + 1)u/2] = 2sin(u/2) The function g(u), moreover, satisfies the conditions in Lemma 2, Sec. 18 (see Problem 1, Sec. 54), and g(O + ) = 1. So, by that lemma, limit (8) has the value ir/2; and, by uniqueness of limits, L = ir/2. Integration formula (1) is now established. 53. TWO LEMMAS The two lemmas in this section are analogues of the ones in Sec. 18, leading up to a convergence theorem for Fourier series. The statement and proof of the corresponding theorem for Fourier integrals follow in Sec. 54. SEC. 53 TWO LEMMAS 221 Lemma 1. If a function G(u) is piecewise continuous on the interval 0 <x <c, then sin rudu = (1) 0. This is the general statement of the Riemann-Lebesgue lemma involving a sine function. Lemma 1 in Sec. 18 is a special case of it, where c = IT and r tends to infinity through the half integers r = (2N + 1)/2 (N = 1,2,...), rather than continuously as it does here. This lemma also holds when sin ru is replaced by cos ru; and the proof is similar to the one below involving sin ru. To verify limit (1), it is sufficient to show that if G(u) is continuous at each point of an interval a u b, then jb (2) = 0. For, in view of the discussion of integrals of piecewise continuous functions in Sec. 10, the integral in limit (1) can be expressed as the sum of a finite number of integrals of the type appearing in limit (2). Assuming, then, that G(u) is continuous on the closed bounded interval a u b, we note that it must also be uniformly continuous there. That is, for each positive number e, there exists a positive number 6 such that IG(u) — G(v)I <e whenever u and v lie in the interval and satisfy the inequality lu — v I Writing Eo E = 2(b — a)' where is an arbitrary positive number, we are thus assured that there is a positive number 6 such that (3) IG(u) — < 2(b— a) whenever lu — vI <6. To obtain the limit (2), divide the interval a u b into N subintervals of equal length (b — a)/N by means of the points a = u0, u1, u2,. . ., UN = where u0 <u1 <u2 < <uN, and let N be so large that the length of each tSee, for instance, the book by Taylor and Mann (1983, pp. 529—531), listed in the Bibliography. 222 FOURIER INTEGRALS AND APPLICATIONS CHAP 6 subinterval is less than the number 6 in condition (3). Then write jn fbG() sin rudu a G(u) sin rudu n=1 = N n=1 N sin rudu, + from which it follows that (4) fbG() sin >ffl n=1 n=1 In view of condition (3) and the fact that sin I - Isinruldu ru I 1, it is easy to see that b-a < 2(b-a) (n=1,2,...,N). Also, since G(u) is continuous on the closed interval a u b, it is bounded there; that is, there is a positive number M such that IG(u)I M for all u in that interval. Furthermore, J + sinrudu 2 r r where it is understood that r> 0. With these observations, we find that inequality (4) yields the statement 2MN b r Now write R = 4MN/e0 and observe that if r > R, then 2MN/r <e0/2. Consequently, b f G(u) sin rudu < Eo + = whenever r > R; and limit (2) is established. Our second lemma makes direct use of the first one. Lemma 2. Suppose that a function g(u) is piecewise continuous on every bounded interval of the positive u axis and that the right-hand derivative TWO LEMMAS SEC 53 exists. 223 If the improper integral f g(u)Idu (5) converges, then iimfg(u) (6) sinru Observe that the integrand appearing in equation (6) is piecewise continuous on the same intervals as g(u) and that when u 1, sin ru g(u) U the convergence of integral (5) ensures the existence of the integral in Thus equation (6). We begin the proof of the lemma by demonstrating its validity when the range of integration is replaced by any bounded interval 0 <x <c. That is, we first show that if a function g(u) is piecewise continuous on a bounded interval exists, then o <x <c and sinru lim f g(u) du = —g(O +). (7) r—000 2 U To prove this, we write fg(u) 0 sinru du = 1(r) U +J(r), where 1(r) = f cg(u) —g(O+) J(r) and sin ru du = f c g(0 +) sinru du. Since the function G(u) = [g(u) — g(O + )]/u is piecewise continuous on the interval 0 <x <c, where G(0 + ) = we need only refer to Lemma 1 to see that liml(r) = (8) 0. On the other hand, if we substitute x = ru in the integral representing J(r), the integration formula in Sec. 52 tells us that (9) limJ(r) =g(O +) lim f o crsinx x —g(0 +). 2 Limit (7) is evidently now a consequence of limits (8) and (9). To actually obtain limit (6), we note that f g(u) sinru du f Ig(u)Idu, 224 FOURIER INTEGRALS AND APPLICATIONS where we assume that c p00 J g(u) CHAP. 6 1. We then write sinru IT du——g(O+) u 2 sinru to choosing c the be so large 00 that the value of the last integral on the right, which is is less than e/2, where e is an arbitrary positive number independent of the value of r. In view of limit (7), there exists a positive remainder of integral (5), number R such that whenever r > R, the first absolute value on the right-hand side of inequality (10) is also less than e/2. It then follows that + fg(u) whenever r du — < = > R, and this is the same as statement (6). 54. A FOURIER INTEGRAL THEOREM The following theorem gives conditions under which the Fourier integral representation (7), Sec. 51, is valid.t Theorem. Let f denote a function which is piecewise continuous on every bounded interval of the x axis, and suppose that it is absolutely integrable over that axis; that is, the improper integral f converges. I Then the Fourier integral 10000 —ff f(s)cosa(s—x)dsda IT 0 converges to the mean value f(x +) +f(x -) 2 of the one-sided limits of f at each point x (—oc <x <oo) one-sided derivatives where both of the and fL(x) exist. For other conditions, see the books by Carslaw (1952, pp. 315ff) and Titchmarsh (1986, pp. 13ff), both listed in the Bibliography. AFOURIERINTEGRALTHEOREM SEC. 54 225 We begin our proof with the observation that integral (1) represents the limit as r tends to infinity of the integral 1 r°° _fff(s)coscr(s —x) (3) 1 dsda = —[I(r,x) +J(r,x)], where I(r, x) = J(r, x) = ff f(s) cos a(s — x) dsda and fff(s) cos a(s — x) dsdcr. We now show that the individual integrals I(r, x) and J(r, x) exist; and, assuming that and as r tends to infinity. exist, we examine the behavior of these integrals Turning to I(r, x) first, we introduce the new variable of integration u= s — x and write that integral in the form I(r, x) = (4) fff(X + u)cos crududa. Since If(x +u)cosaul and because f If(x+u)Idu=f the Weierstrass M-test for improper integrals applies to show that the integral f(x + u) cos audu converges uniformly with respect to the variable a. Consequently, not only does the iterated integral (4) exist, but also the order of integration there can be reversed:t I(r,x)=fff(x+u)cosaudadu=ff(x+u) 00 0 Now sinru du. U the function g(u) = f(x + u) satisfies all the conditions in Lemma 2, Sec. 53 (compare Sec. 19). Hence, if we apply that lemma to this last integral, we find tTheorems on improper integrals used here are developed in the book by Buck (1978, sec. 6.4), listed in the Bibliography, as well as in most other texts on advanced calculus. The theorems are usually given for integrals with continuous integrands, but they are also valid when the integrands are piecewise continuous. 226 CHAP 6 FOURIER INTEGRALS AND APPLICATIONS that liml(r,x) = +). The limit of J(r, x) as r tends to infinity is treated similarly. Here we make the substitution u = x — s and write J(r,x)=fff(x—u)cosaududcr=ff(x—u) When g(u) = f(x — du. u), the limit limJ(r,x) = also sinru —) follows from Lemma 2 in Sec. 53. Finally, in view of limits (5) and (6), we see that the limit of the left-hand side of equation (3) as r tends to infinity has the value (2), which is, then, the value of integral (1) at any point where the one-sided derivatives of f exist. Note that since the integrals in expressions (9), Sec. 51, for the coefficients A(a) and B(a) exist when f satisfies the conditions stated in the theorem, the form (8), Sec. 51, of the Fourier integral formula is also justified. PROBLEMS 1. Show that the function g(u) sin (u/2) = (u/2) used in equation (8), Sec. 52, satisfies the conditions in Lemma 2, Sec. 18. To be precise, show that g is piecewise continuous on the interval 0 <x and that exists. Suggestion: To obtain show that = urn 2sin(u/2) — u—*O U u>O u 2 Then apply l'Hospital's rule twice. 2. Verify that all the conditions in the theorem in Sec. 54 are satisfied by the function f that is defined by the equations f(x) = and f( ±1) = when lxi <1, when lxi >1, (1 Thus show that, for every x (— oo <x <oo), f(x)=_j 'ro 1 sin a(1 + x) + a sin a(1 — x) 2 da=_J sin a cos ax a da. SEC. 54 PROBLEMS 227 3. Show that the function defined by the equations f(x) and f(O) = (0 when x <0, when x> 0, = satisfies the conditions in the theorem in Sec. 54 and hence that 1 f(x)=_f 'ro oocosax+asinax 1+a 2 (_oo <x < da oo). Verify this representation directly at the point x = 0. 4. Show how it follows from the result in Problem 3 that 2 'n (_oo <x <oo). o 1+a 5. Use the theorem in Sec. 54 to show that if f(x) = {0 when sin x x <0 or x > 'r, when then (_oo <x <oo). 1—a 0 In particular, write x = 'r/2 to show that I 1—az Jo da=—. 2 6. Show why the Fourier integral formula fails to represent the function f(x) = 1 (—oo <x <oo). Also, point out which condition in the theorem in Sec. 54 is not satisfied by that function. 7. Give details showing that the integral J(r, x) in Sec. 54 actually exists and that limit (6) in that section holds. 8. Let f be a nonzero function that is periodic, with period 2c. Point out why the integrals ff(x)dx and flf(x)I dx fail to exist. 9. Prove Lemma 1 in Sec. 53 when sin ru is replaced by cos ru in integral (1) there. 10. Assume that a function f(x) has the Fourier integral representation (8), Sec. 51, which can be written +B(a)sinax]da. f(x) = Use the exponential forms (compare Problem 15, Sec. 21) cosO= e'° + 2 e'° sinO= — 2i 228 CHAP 6 FOURIER INTEGRALS AND APPLICATIONS of the cosine and sine functions to show formally that f(x) = where a(a) A(a) — iB(a) = ' 2 a(—a) A(a) + iB(a) = 2 (a >0). Then use expressions (9), Sec. 51, for A(a) and B(a) to obtain the single formulat a(a) = (—oo <a 55. THE COSINE AND SINE INTEGRALS Let f denote a function satisfying the conditions stated in the theorem in Sec. 54. As noted at the end of the proof of that theorem, the Fourier integral representation of f(x) remains valid when written f(x) = f [A(a) cos ax + B(a) sin ax] da, where loo loo (2) A(a)=_f f(x)cosaxdr, IT-oo B(cr)=_-f IT—oo Also, in view of the theorem in Sec. 17, representation (1) is valid for any function f that is absolutely integrable over the entire x axis and piecewise smooth on every bounded interval of it. Observe that if the function f is even, then f(x) sin ax is odd in the variable x. The graph of y = f(x) sin ax is, therefore, symmetric with respect to the origin. Hence B(a) = 0, and representation (1) reduces to f(x) = f A(a) cos axda. The function f(x) cos ax is even in x, and so the graph of y = f(x) cos ax is symmetric with respect to the y axis. Consequently, 2oo A(a) = —f 17•0 tThe function a(a) is known as the exponential Fourier transform of f(x) and is of particular importance in electrical engineering. For a development of this and other types of Fourier transforms, see, for example, the book by Churchill (1972) that is listed in the Bibliography. THECOSINEANDSINEINTEGRALS SEC. 55 229 The Fourier cosine integral formula (3) can, of course, be written f(x) (5) 2oo = cosaxf —f 17-0 0 00 f(s)cosasdsda. If, on the other hand, f is an odd function, then A(a) = tation (1) becomes f(x) (6) f B(a) = sin 0; and represen- axda, where B(a) (7) 200 = —f Integral (6) is known as the Fourier sine integral formula. The compact form is f(x) (8) Suppose 200 —f sin axf f(s)sincrsdsda. 00 17-0 0 now that f is defined only when x> 0 and that it has the following properties: (a) is absolutely integrable over the positive x axis and piecewise smooth on every bounded interval of it; f (b) f(x) at each point of discontinuity of f is the mean value of the one-sided limits f(x + ) and f(x —). When the even extension of f is made, integral (3) represents that extension for every nonzero x and equals f(O + when x = 0. Likewise, ) integral (6) represents the odd extension of f for every nonzero x and has value zero when x = 0. The Fourier integral theorem in Sec. 54 thus provides us with a theorem that will be especially useful in the applications. Theorem. Let f denote a function that is defined on the positive x axis and satisfies conditions (a) and (b). Then the Fourier cosine integral representation (3), where the coefficient A(a) is defined by equation (4), is valid for each x (x > 0); and the same is true of the Fourier sine integral representation (6), where the coefficient B(a) is given by equation (7). Representation (3) is needed in the applications because of the solutions of the eigenvalue problem X'(O) = 0, IX(x)I <M (x>0), where M is a positive constant. This problem is singular (Sec. 42) because its fundamental interval x> 0 is unbounded. If A = 0, X(x) is any constant multiple of unity. If A is a real number such that A > 0 and we write A = a2 X"(x) + (a > 0), AX(x) = 0, we readily find that, except for constant factors, the eigenfunctions are 230 CHAP. 6 FOURIER INTEGRALS AND APPLICATIONS X(x) = cos ax, where a takes on all positive values. The eigenvalues A = a2 are continuous rather than discrete. If A <0, or A = —a2 (a > 0), the solution of the differential equation and the boundary condition at x = 0 is X(x) = 2C1 cosh ax. This is, however, unbounded on the half line x > 0 unless C1 = 0. So the case A <0 yields no eigenfunctions. Cases other than those in which A is real need not be considered since they yield unbounded solutions of the differential equation (see Problem 7). Although the eigenfunctions X(x) = cos ax (a 0) have no orthogonality property, the Fourier cosine integral formula (3) gives representations of functions f(x) on the interval x> 0 which are generalized linear combinations of those eigenfunctions. Likewise, A = a2 and X(x) = sin ax (a> 0) are the eigenvalues and eigenfunctions of the singular problem X"(x)+AX(x)=O, X(O)=0, IX(x)I<M (x>O); and formula (6) represents functions f(x) in terms of sin ax. PROBLEMS 1. By applying the Fourier sine integral formula and the theorem in Sec. 55 to the function defined by the equations f(x) = and f(b) = whenO<x<b, (1 when x>b, obtain the representation 2 ,ool—cosba f(x)=_J a iro sinaxda (x>O). 2. Verify that the function exp( — bx), where b is a positive constant, satisfies the conditions in the theorem in Sec. 55, and show that the coefficient B(a) in the Fourier sine integral representation of that function is a 2 B(a) = — I sin 'rJo axdx = — a2+b2 Thus prove that ooasinax da 'r-'o a2+b2 2 (b>O x>O). 3. Verify the Fourier sine integral representation X x2+b2 2 çoossinas coo sinaxlJo =—I v-Jo dsda by first observing that, according to the final result in Problem 2, the value of the inner integral here is ('r/2) exp (— ba). Then, by referring to the expression for B(a) in Problem 2, complete the verification. Show that the function x/(x2 + b2) is not, however, absolutely integrable over the positive x axis. ______ SEC. 55 4. PROBLEMS 231 As already verified in Problem 2, the function exp (— bx), where b is a positive constant, satisfies the conditions in the theorem in Sec. 55. Show that the coefficient A(a) in the Fourier cosine integral representation of that function is A(a) = 200 I — 2 b — a2+b2 Thus prove that 00cosax 2b 'r Jo a2 + b2 da (b>0 5. By regarding the positive constant b in the final equation obtained in Problem 4 as a variable and then differentiating each side of that equation with respect to b, show formally that 4 cosax coo " 0 (a2 + 1) 6. 2da Verify that the function e cos x satisfies the conditions in the theorem in Sec. 55, and show that the coefficient A(a) in the Fourier cosine integral representation of that function can be written 100 A(a) = —f e_Xcos(a + 770 + —f 'TO e_xcos(a — Then use the expression for the corresponding coefficients in Problem 4 to prove that e_xcosx = 2 — ( ooa2+2 cosaxda (x 0). a complex number that is not real, so that the square roots of —A are of i13), where a and 13 are real numbers and a # 0. Use the identitiest the form ±(a + cosh (x + iy)12 = sinh2 x + cos2 y, lsinh (x + iy)12 = sinh2 x + sin2 y, where x and y are real numbers, to show that such a A cannot be an eigenvalue of (a) the singular eigenvalue problem (9), Sec. 55; (b) the singular eigenvalue problem (10), Sec. 55. 8. Show that the eigenvalues of the singular eigenvalue problem X"(x)+AX(x)=O, IX(x)I<M (—oo<x<oo), where M is a positive constant, are A = a2 (a 0) and that the corresponding eigenfunctions are constant multiples of unity when a = 0 and arbitrary linear combinations of cos ax and sin ax when a > 0. (Use the method in Problem 7 to show that the eigenvalues must be real.) tFor results from the theory of functions of a complex variable used here, see the authors' book (1990, secs. 7 and 25), listed in the Bibliography. 232 CHAP 6 FOURIER INTEGRALS AND APPLICATIONS 9. Let A(a) and B(a) denote the coefficients (9), Sec. 51, in the Fourier integral representation (8) in that section for a function f(x) (—oo <x <oo) that satisfies the conditions in the theorem in Sec. 54. (a) By considering even and odd functions of a, point out why f [A(a)cos ax + B(a)sin ax] da = 2f(x) and f (b) [B(a)cosax +A(a)sinax]da = 0. By adding corresponding sides of the equations in part (a), obtain the following symmetric form of the Fourier integral formula:t f(x)= 1 00 1 00 —i=-f g(a)(cosax+sinax)da (—oo<x<oo), where g(a) = —,==f f(x)(cos ax + sin ax) dx. V2ir —00 56. MORE ON SUPERPOSITION OF SOLUTIONS In Sec. 26 we showed that linear combinations of solutions of linear homogeneous differential equations and boundary conditions are also solutions. In the same section we also extended that result to include infinite series of solutions, thus providing the basis of the technique for solving boundary value problems that was used in Chaps. 4 and 5. Another useful extension is illustrated by the following example, where superposition consists of integration with respect to a parameter a instead of summation with respect to an index n. It will enable us to solve certain boundary value problems in which Fourier integrals, rather than Fourier series, are required. EXAMPLE. Consider the set of functions exp (—ay) sin ax, where each function corresponds to a value of the parameter a (a > 0) and where a is independent of x and y. Each function satisfies Laplace's equation y) + and y) = 0 (x > 0, y > 0) the boundary condition u(O,y)=O (y>0). These functions are bounded in the domain x> 0, y > 0 (Fig. 59) and are tThis form is useful in certain types of transmission problems. See R. V. L. Hartley, Proc. Inst. Radio Engrs., vol. 30, no. 3, pp. 144—150, 1942. MORE ON SUPERPOSITION OF SOLUTIONS SEC. 56 233 y U =0 0 u=f(x) x FIGURE 59 obtained from conditions (1) and (2) by the method of separation of variables when that boundedness condition is included (Problem 1, Sec. 58). We now show that their combination of the type u(x,y) (3) = f (x>O, y >0) also represents a solution of the homogeneous conditions (1) and (2) which is bounded in the domain x> 0, y > 0 for each function B(a) that is bounded and continuous on the half line a > 0 and absolutely integrable over it. To accomplish this, we use tests for improper integrals that are analogous to those for infinite series.t The integral in equation (3) converges absolutely and uniformly with respect to x and y because (x 0, y 0) sin axi and B(a) is independent of x and y and absolutely integrable from zero to infinity with respect to a. Moreover, since (4) u(x,y)kf (5) IB(a)Ida, is bounded. It is also a continuous function of x and y (x 0, y 0) because of the uniform convergence of the integral in equation (3) and the continuity of the integrand. Clearly, u = 0 when x = 0. u When y> 0, (6) and y y0, where y0 is some small positive number, then the absolute value of the integrand of the integral on the far right does not for if IB(a)I exceed B0a exp (—ay0), which is independent of x and y and integrable from the book by Kaplan (1991, pp. 471ff) or Taylor and Mann (1983, pp. 682ff), listed in the Bibliography. 234 a= FOURIER INTEGRALS AND APPLICATIONS to a = CHAP 6 Hence that integral is uniformly convergent. Integral (3) is then differentiable with respect to x, and similarly for the other derivatives involved in the laplacian operator V2 = 32/3x2 + 92/9y2. Therefore, 0 00• (7) V2u = f B(a)V2(e_aYsinax)da = 0 (x >0, y >0). Suppose now that the function (3) is also required to satisfy the nonhomogeneous boundary condition u(x,0)=f(x) (8) (x>0), where f is a given function satisfying the conditions stated in the theorem in Sec. 55. We need to determine the function B(a) in equation (3) so that f(x) (9) = f B(a)sinaxda (x >0). This is easily done since representation (9) is the Fourier sine integral formula (6), Sec. 55, when 2oo (10) (a>O). We have shown here that the function (3), with B(a) given by equation (10), is a solution of the boundary value problem consisting of equations (1), (2), and (8), together with the requirement that u be bounded. 57. TEMPERATURES IN A SEMI-INFINITE SOLID The face x = 0 of a semi-infinite solid x 0 is kept at temperature zero (Fig. 60). Let us find the temperatures u(x, t) in the solid when the initial temperature distribution is f(x), assuming at present that f is piecewise smooth on each bounded interval of the positive x axis and that f is bounded and absolutely integrable from x = 0 to x = 00• u=O 0 u(x,O)=f(x) x FIGURE 60 If the solid is considered as a limiting case of a slab 0 x c as c increases, some condition corresponding to a thermal condition on the face x = c seems to be needed. Otherwise, the temperatures on that face may be increased in any manner as c increases. We require that our function u be bounded; that condition also implies that there is no instantaneous source of TEMPERATURES IN A SEMI-INFINITE SOLID SEC. 57 heat on the face x = 0 at the instant t = 0. Then (x > 0, t >0), (1) = (2) u(O,t) = (t>0), (x>0), 0 u(x,O)=f(x) (3) 235 and Iu(x, t)I <M, where M is some positive constant. Linear combinations of functions u = X(x)T(t) will not ordinarily be bounded unless X and T are themselves bounded. Upon separating variables, we thus have the conditions (4) X"(x) + AX(x) = 0, X(0) = 0, IX(x)I <M1 (x >0) < M2 (t > 0), and (5) T'(t) + AkT(t) = 0, where M1 and M2 are positive constants. As pointed out at the end of Sec. 55, the singular eigenvalue problem (4) has continuous eigenvalues A = a2 where a represents all positive real numbers; X(x) = sin ax are the eigenfunctions. In this case, the corresponding functions T(t) = exp (—a2kt) are bounded. The generalized linear combination of the functions X(x)T(t) for all positive a, u(x,t) = f B(a)e_a2/ctsinaxda, will formally satisfy all the conditions in the boundary value problem if the function B(a) can be determined so that f(x) = f B(a)sinaxda (x>0). As in Sec. 56, we note that representation (7) is the Fourier sine integral formula (6), Sec. 55, for f(x) if 2oo B(a) = —f f(x) sin axdx ITO (a> 0). Our formal solution (6), with B(a) defined by equation (8), can also be written u(x,t) = —f e_a2/ctsinaxf f(s)sinasdsda. We can simplify this result by formally reversing the order of integration, replacing 2 sin as sin ax by cos a(s — x) — cos a(s + x), and then applying the integration formula (Problem 19, Sec. 58) = (a >0). 236 FOURIER INTEGRALS AND APPLICATIONS CHAP 6 Equation (9) then becomes (11) u(x,t) 1 (s—x) 2 oo exp = (s+x) —exp 4kt — — 4kt 2 ds t> 0. An alternative form of equation (11), obtained by introducing new variables of integration, is when (12) u(x,t) f(x = + 2 dor. Our use of the Fourier sine integral formula in obtaining solution (9) suggests that we apply the theorem in Sec. 55 in verifying that solution. The forms (11) and (12) suggest, however, that the condition in the theorem that f(x)I be integrable from zero to infinity can be relaxed in the verification. More precisely, when s is kept fixed and t> 0, the functions (s±x)2 1 7=-exp 4kt — satisfy the heat equation (1). Then, under the assumption that f(x) is continuous and bounded when x 0, it is possible to show that the function (11) is bounded and satisfies the heat equation when x0 <x <x1 and t0 <t <t1, where x0, x1, t0, and t1 are any positive numbers. Conditions (2) and (3) can be verified by using expression (12). By adding step functions to f (see Problem 4, Sec. 58), we can permit f to have a finite number of jumps on the half line x > 0. Except for special cases, details in the verification of formal solutions of this problem are, however, tedious. When f(x) = 1, it follows from equation (12) that u(x, t) = 1 f 00 2 do- — f 00 2 dff In terms of the error function erf(x) = where 2 \I'7T x 2 do-, 0 erf(x) tends to unity as x tends to infinity (see Problem 18, Sec. 58), expression (13) can be written u(x,t) The full verification of that result is not difficult. TEMPERATURES IN AN UNLIMITED MEDIUM 237 SEC 58 58. TEMPERATURES IN AN UNLIMITED MEDIUM For an application of the general Fourier integral formula, we now derive expressions for the temperatures u(x, t) in a medium that occupies all space, where the initial temperature distribution is f(x). We assume that f(x) is bounded and, for the present, that it satisfies conditions under which it is represented by its Fourier integral formula. The boundary value problem consists of a boundedness condition Iu(x, t)I (1) t) <M and the conditions t) = (-oo<x<oo). u(x,O)=f(x) (2) t > 0), (—oo <x < Separation of variables leads to the singular eigenvalue problem X"(x) +AX(x) = IX(x)I <M1 0, (—oo <x <00), eigenvalues are A = a2, where a 0, and to the two linearly independent eigenfunctions cos ax and sin ax corresponding to each nonzero value of a (Problem 8, Sec. 55). Our generalized linear combination of functions X(x)T(t) is whose u(x,t) =f exp(—a2kt)[A(a)cosax +B(a)sinax]da. (3) The coefficients A(a) and B(a) are to be determined so that the integral here represents f(x) (— oo <x < oo) when t 0. According to equations (8) and (9) in Sec. 51 and our Fourier integral theorem (Sec. 54), the representation is valid lao B(a)=_f f(x)sinaxdr. IT—oo loo A(a)=_f f(x)cosaxdx, IT-oo Thus u(x, t) If 1 = 2 —f ktf f(s) cos a(s — x) dsda. we formally reverse the order of integration here, the integration formula (10) in Sec. 57 can be used to write equation (4) in the form u(x,t) 1 oo = (s—x) 2 f(s)exp — 4kt ds (t>0). An alternative form of this is 10° u(x,t) = 7=-f f(x + 2 do-. Forms (5) and (6) can be verified by assuming only that f is piecewise continuous over some bounded interval Ix I <c and continuous and bounded 238 ('HAP 6 FOURIER INTEGRALS AND APPLICATIONS over the rest of the x axis, or when lxi c. If f is an odd function, u(x, t) becomes the function found in Sec. 57 for positive values of x. PROBLEMS details showing how the functions exp(—ay) sin ax (a > 0) arise by means of separation of variables from conditions (1) and (2), Sec. 56, and the condition that the function u(x, y) there be bounded when x > 0, y > 0. 2. (a) Substitute expression (10), Sec. 56, for the function B(a) into equation (3) of that section. Then, by formally reversing the order of integration, show that the solution of the boundary value problem treated in Sec. 56 can be written 1. Give u(x,y) (b) 1 = —f0 f(s) (s—x) 2 +y2 1 - (s+x) 2 ds. +y2 Show that when f(x) = 1, the form of the solution obtained in part (a) can be written in terms of the inverse tangent function as 2 u(x, y) = — tan' x — 31 3. Verify that the function u = in Sec. 57 satisfies the heat equation when x > 0, t> 0 as well as the conditions = u(O+,t)=0 u(x,0+)=1 lu(x, t)I < 1 (t>0), (x>0), (x > 0, t> 0). 4. Show that if (0 when0<x<c, when expression x>c, (12), Sec. 57, reduces to u(x,t) - = Verify this solution of the boundary value problem in Sec. 57 when f is this function. 5. The face x = 0 of a semi-infinite solid x 0 is kept at a constant temperature U0 after its interior x > 0 is initially at temperature zero throughout. Obtain an expression for the temperatures u(x, t) Answer: u(x, t) = u0 [i in the body. — erf ) 6. (a) The face x 0 of a semi-infinite solid x 0 is insulated, and the initial temperature distribution is f(x). Derive the temperature formula u(x,t) = 1f f(x + +—f f(—x + PROBLEMS SEC 5S 239 (b) Show that if the function f in part (a) is defined by the equations when0<x<c, (1 when x>c, then u(x,t) = + 7. Let the initial temperature distribution f(x) in the unlimited medium in Sec. 58 be defined by the equations whenx<0, when x> 0. f(x) = (0 Show that u(x t) 1 1 = — + — erf 2 fx I 2 Verify this solution of the boundary value problem in Sec. 58 when f is this function. (—00 <x < oc, 8. Derive this solution of the wave equation y11 = > 0), which satisfies the conditions y(x, 0) = f(x), y,(x, 0) = 0 when —00 <x < oo: y(x,t)=_f cosaatf f(s)cosa(s—x)dsda. 0 Also, reduce the solution to the form obtained in Example 2, Sec. 8: y(x,t) = + at) +f(x - at)]. 9. A semi-infinite string, with one end fixed at the origin, is stretched along the positive half of the x axis and released at rest from a position y = f(x) (x 0). Derive the expression y(x,t) —f0 cosaatsinaxf0 f(s)sinasdsda for the transverse displacements. Let F(x) (—oo <x < oo) denote the odd extension = of f(x), and show how this result reduces to the form y(x,t) = + at) + F(x — at)]. [Compare solution (10), Sec. 30, of the string problem treated in that section.] 10. Find the bounded harmonic function u(x, y) in the horizontal semi-infinite strip x> 0, 0 <y < 1 that satisfies the boundary conditions y) = 2 coo cos c_V. ax cosh ay da. (1+a2)cosha Find u(x, y) when the boundary conditions in Problem 10 are replaced by the Answer: u(x y) = — I 11. u(x, 1) = 0, conditions = 0, = —u(x,1), u(x,0) =f(x), 240 FOURIER INTEGRALS AND APPLICATIONS CHAP. 6 where when0<x<1, f(x) = (1 x>1. when Interpret this problem physically. 2 Answer: u(x, y) = — I 'r Jo acosha(1 —y) + sinha(1 —y) 2 cosh a + a sinh a sin . a cos ax da. 12. Find the bounded harmonic function u(x, y) in the semi-infinite strip 0 <x < 1, y> 0 that satisfies the conditions = 0, Answer: 13. Find u(x, y) u(0, sinh 2 = —J 'r o 0 0, y) =f(y). ax cos ay acosha the bounded harmonic function such that u(x,0) = y) = u(x, y) Jo f(s)cos as ds da. in the strip —00 and u(x,b) =f(x) (—oo <x <oo), represented by its Fourier integral. 1 ,oosinhay ,oo Answer: u(x, y) = — I I f(s) cos a(s 'rJo . — <x where <oo, 0 <y <b f is bounded and x) ds da. 14. Let a semi-infinite solid x 0, which is initially at a uniform temperature, be cooled or heated by keeping its boundary at a uniform constant temperature (Sec. 57). Show that the times required for two interior points to reach the same temperature are proportional to the squares of the distances of those points from the boundary plane. 15. Solve the following boundary value problem for steady temperatures u(x, y) in a thin plate in the shape of a semi-infinite strip when heat transfer to the surroundings at temperature zero takes place at the faces of the plate: y) — bu(x, y) = y) + 0 (x> 0,0 <y < 1), (0 <y < 1), y) = 0 u(x,0)=0, (x>0), u(x,1)=f(x) where b is a positive constant and when0<x<c, f(x) = (1 Answer: 16. u(x, y) = 2 f when x>c. oosinaccosaxsinh(yVa2+b) — 0 asinhVa2+b da. Verify that, for any constant C, the function v(x, t) = Cxt3"2 exp —x2 when x> 0 and t> 0. Also verify that satisfies the heat equation = v(0 + , t) = 0 when t> 0 and that v(x, 0 + ) = 0 when x> 0. Thus v can be added to the function u found in Sec. 57 to form other solutions of the problem there if the temperature function is not required to be bounded. But note that v is unbounded as x and t tend to zero, as can be seen by letting x vanish while t = SEC. 58 241 PROBLEMS y u =0 x FIGURE 61 = u(x, y, t) denote the bounded solution of the two-dimensional temperature 17. Let u problem indicated in Fig. 61, where (x>0,0<y<1,t>O), and v = v(x, t) and w = w(y, t) denote let the bounded solutions of the following one-dimensional temperature problems: v(0,t) = = 0, = v(x,0) w(0,t)=w(1,t)=0, (x > 0, 1 of > 0), w(y,0)=1(0<y<1,t>0). (a) With the aid of the result obtained in Problem 3, Sec. 40, (b) t show that u = vw. By referring to the solution (15), Sec. 57, to the temperature function found in Problem 4(b), Sec. 32, write explicit the temperature problem there and expressions for v and w. Then use the result in part (a) to show that u(x,y,t) 18. Let I denote 4 x 00 sin(2n—1)'ry = exp[—(2n —1) 2n —1 2 the integral of exp (—x2) from zero to infinity, and write j2 = f°°e_x2dxfO°e_2ciy = Evaluate this iterated integral by using polar coordinates, and show that I = Thus verify that erf(x), defined in equation (14), Sec. 57, tends to unity as x tends to infinity. 19. Derive the integration formula (10), Sec. 57, by first writing y(x) = f e_a2(lcosaxda (a > 0) and differentiating the integral to find y'(x). Then integrate the new integral by parts to show that 2ay'(x) = —xy(x), point out why = (see Problem 18), and solve for y(x). The desired result is the value of y when x = b.t tAnother derivation is indicated in the authors' book (1990, p. 199), listed in the Bibliography. CHAPTER 7 BESSEL FUNCTIONS AND APPLICATIONS In boundary value problems that involve the laplacian V2u expressed in cylindri- cal or polar coordinates, the process of separating variables often produces a differential equation of the form (1) d2y + dy + — = 0, where y is a function of the coordinate p. In such a problem, — A is a separation constant; and the values of A are the eigenvalues of a SturmLiouville problem involving equation (1). The parameter is a nonnegative number determined by other aspects of the boundary value problem. Usually, v is either zero or a positive integer. In our applications, it turns out that A 0; and, when A > 0, the substitution x = can be used to transform equation (1) into a form that is free of (2) x2y"(x) + xy'(x) + (x2 — ii2)y(x) = 0. This differential equation is known as Bessel's equation. Its solutions are called Bessel functions, or sometimes cylindrical functions. Equation (2) is an ordinary differential equation of the second order that is linear and homogeneous; and, upon comparing it with the standard form y"(x) +A(x)y'(x) +B(x)y(x) =0 242 SEC. 59 BESSEL FUNCTIONS 243 of such equations, we see that A(x) = 1/x and B(x) = 1 — (ii/x)2. These coefficients are continuous except at the origin, which is a singular point of Bessel's equation. The lemma in Sec. 44, regarding the existence and uniqueness of solutions, applies to that equation on any closed bounded interval that does not include the origin. But for boundary value problems in regions p c, bounded by cylinders or circles, the origin x = 0 corresponds to the axis or center p = 0, which is interior to the region. The interval for the variable x then has zero as an end point. We limit our attention primarily to the cases ji = n, where n = 0, 1, 2 For such a case, we shall discover a solution of Bessel's equation that is represented by a power series which, together with all its derivatives, converges and its for every value of x, including x = 0. That solution, denoted by derivatives of all orders are, therefore, everywhere continuous functions. In referring to any power series, we shall always mean a Maclaurin series, or a Taylor series about the origin. 59. BESSEL FUNCTIONS J,, We let n denote any fixed nonnegative integer and seek a solution of Bessel's equation (1) x2y"(x) +xy'(x) + (x2 —n2)y(x) = (n = 0,1,2,...) 0 in the form of a power series multiplied by XC, where the first term in that series is nonzero and c is some constant. That is, we propose to determine c and the coefficients so that the function j=O j=O satisfies equation (1)•t Assume for the present that the series is differentiable. Then, upon substituting the function (2) and its derivatives into equation (1), we obtain the equation [(c +j)(c +1 — 1) + (c +j) — j=O But (c + jXc + j — 1) n2]aJXc+i + = 0. j=O + (c + I) = (c + j)2, and the second summation here can be written j=2 tThe series method used here to solve equation (1) is often referred to as the method of Frobenius and is treated in introductory texts on ordinary differential equations, such as the one by Boyce and DiPrima (1992) or the one by Rainville and Bedient (1989). Both are listed in the Bibliography. 244 BESSEL FUNCTIONS AND APPLICATIONS CHAP. 7 Hence [(c +j)2 — n2jaxc+i + = 0. j=2 1=0 Multiplying through this equation by XC and writing out the j = 0 and I = 1 terms of the first series separately, we have the equation (3) (c — n)(c + n)a0 + (c + +j — n)(c {(c + 1 — n)(c + 1 +1 + n)a1 + + n)a1x a1_2}x-' = 0. j=2 Equation (3) is an identity in x if the coefficient of each power of x vanishes. Thus c = n or c = case, a1 = 0. Furthermore, —n if the constant term is to vanish; and, in either (j=2,3,...). We make the choice c = n. Then the recurrence relation (4) —1 a1 = j(2n ±J)03_2 (I = 2,3,...) obtained, giving each coefficient (j = 2, 3,...) in terms of the second coefficient preceding it in the series. Note that when n is positive, the choice c = — n does not lead to a well-defined relation of the type (4), where the denominator on the right is never zero. Since a1 = 0, relation (4) requires that a3 = 0; then a5 = 0, etc. That is, is (5) (k=0,1,2,...). a2k+1=O To obtain the remaining coefficients, we let k denote any positive integer and use relation (4) to write the following k equations: —1 a2 = 1(n + a4= —1 —1 a2k = k(n + Upon equating the product of the left-hand sides of these equations to the product of their right-hand sides, and then canceling the common factors a2, a4,. . ., a2k_2 on each side of the resulting equation, we arrive at the SEC 59 245 BESSEL FUNCTIONS expression (k=1,2,...). a2k= k!(fl+1)(fl+2)...(fl+k)22ka0 (6) In view of identity (5) and since c = n, series (2) now takes the form a2kx, y = a0x'2 + (7) k= 1 where the coefficients a2k (k = 1, 2,...) are those in expression (6). This series is absolutely convergent for all x, according to the ratio test: lim a2(k+l)x n+2(k+1) = lim a2kx k-4oo k-÷oo 2 x 1 — (k + 1)(n + k + 1) = 0. 2 Hence it represents a continuous function and is differentiable with respect to x any number of times. Since it is differentiable and its coefficients satisfy the recurrence relation needed to make its sum satisfy Bessel's equation (1), series (7) is, indeed, a solution of that equation. The coefficient a0 in series (7) may have any nonzero value. If we substitute expression (6) into that series and write y—a0x — (—1) + k (x\2k we see that the choice 1 simplifies our solution of Bessel's equation to y = J0(x), where 1 (8) °o x (1)k x n+2k is known as the Bessel function of the first kind of order n 1, 2,...). With the convention that 0! = 1, it is written more compactly This function (n = 0, as (1)k (9) = k=0 x n+2k k!(n + k)! From expression (9), we note that that is, it is an even function if n = 0, 2, 4, is clear from expression (8) that J0(0) = (n = 0,1,2,...); = (10) 1. ... = but odd if n = 0 when n = 1, 1, 3,5 Also, 2,... but that 246 BESSEL FUNCTIONS AND APPLICATIONS CHAP. 7 The case in which n = 0 will be of special interest to us in the applications. Bessel's equation (1) then becomes xy"(x) + y'(x) + xy(x) = 0; (11) and expression (9) reduces to 1 J0(x) = (12) pk x 2k k=O (k.) (2) Since (k!)222k when k (13) = 224262 [(1)(2)(3) 1, another form is x2 x4 224262 Expressions (12) and (13) bear some resemblance to the power series for cos x. There is also a similarity between the power series representations of the odd functions J1(x) and sin x. Similarities between the properties of those functions include, as we shall see, the differentiation formula = —J1(x), corresponding to the formula for the derivative of cos x. Graphs of y = J0(x) and v = J1(x) are shown in Fig. 62. More details regarding these graphs, especially the nature of the zeros of J0(x) and J1(x), will be developed later in the chapter. y 1.0 0.5 —0.5 FIGURE 62 GENERAL SOLUTIONS OF BESSEL'S EQUATION 60. A function linearly independent of that satisfies Bessel's equation x2y"(x) +xy'(x) +(x2—n2)y(x) =0 (n = 0,1,2,...) can be obtained by various methods of a fairly elementary nature. GENERAL SOLUTIONS OF BESSEL'S EQUATION SEC. 60 247 The singular point x = 0 of equation (1) is of a special type and is known as a regular singular point. The power series procedure, extended so as to give general solutions near regular singular points, applies to Bessel's equation. We do not give further details here but only state the results. When n 0, the general solution is found to be (2) y=AJ0(x)+B J0(x)lnx x2 x4 +_____1+_ + 22 2242 2 1 1+_+_ 2 3 1 224262 1 where A and B are arbitrary constants and x > 0. Observe that, as long as 0, any choice of A and B yields a solution which is unbounded as x tends to zero through positive values. Such a solution cannot, therefore, be expressed as a constant times J0(x), which tends to unity as x tends to zero. So J0(x) and the solution (2) are linearly independent when B * 0. It is most common to use Euler's constant y = 0.5772..., which is defined as the limit of the sequence B 1 1 1 (3) and to write 2 A = —(y — 1n2) IT When and 2 B = —. IT A and B are assigned those values, the second solution that arises is Weber's Bessel function of the second kind of order zero:t (4) Y0(x) = x2 x4 +____1+_ + 2242 2 1 22 1 224262 1 1+_+_ 2 3 More generally, when n has any one of the values n = 0, 1,2,..., equation (1) has a solution that is valid when x> 0 and is unbounded as x tends to zero. Since is continuous at x = 0, then, and are linearly independent; and when x > 0, the general solution of equation (1) can be written + C2Y,1(x) y= (n = 0,1,2,...), tThere are other Bessel functions, and the notation varies widely throughout the literature. The treatise by Watson (1952) that is listed in the Bibliography is, however, usually regarded as the standard reference. 248 CHAP 7 BESSEL FUNCTIONS AND APPLICATIONS where C1 and C2 are arbitrary constants. The theory of the second solution and we shall limit our is considerably more involved than that of is applications to problems in which it is only necessary to know that discontinuous at x = 0. To write the general solution of Bessel's equation (6) x2y"(x) +xy'(x) + (x2 — p2)y(x) = 0 >0; * 1,2,...), where ii is any positive number other than 1,2,..., we mention here some elementary properties of the gamma function, defined when ij'> 0 by means of the equationt I'(p) = (7) An integration by parts shows that + 1) = (8) 0. This property is assigned to the function when <0, SO that <0, —2 < < —1, etc. Thus equations (7) F(u) = I'(p + 1)/u when —1 and(8), together, define I'(p) for all ii except ii = 0, — 1, —2,... (Fig. 63). We find from equation (7) that I'(l) = 1; also, it can be shown that F(u) is continuous and positive when ii > 0. It then follows from the identity = I'(u + 1)/u that F(O + ) = oc and, furthermore, that IF(v)I becomes infinite as v —n (n = 0, 1, 2,...). This means that 1/FU') tends to zero as J1 tends to — n (n = 0, 1,2,...); and, for brevity, we write 1/F( — n) = 0 when n = Note that the reciprocal 1/I'(u) is then continuous for all ii. 0, 1, 2 when 1'(v) 1234 p FIGURE 63 tThorough developments of the gamma function appear in the books by Lebedev (1972, chap. 1) and Rainville (1971, chap. 2) that are listed in the Bibliography. RECURRENCE RELATIONS SEC. 61 249 When ji = 1, 2, 3,. . ., FU') reduces to a factorial: (n = 0,1,2,...). F(n + 1) = n! (9) The verification of property (9) and the further property that (10) is left to the problems. The Bessel function of the first kind of order (1)k (11) = k=O 0) is defined as (ij' x z'+2k k!F(u + k + 1) note that this becomes expression (9), Sec. 59, for when ii = n = The Bessel function 1,2 (i'> 0) is also well defined when ii is replaced by — ii in equation (11). If ii = n = 1, 2,..., however, the summation in the resulting series starts from k = n since 1/F( — n + k + 1) is zero when and J_1, o k n — 1. It is not difficult to verify by direct substitution that We 0, are solutions of equation (6). Those solutions are arrived at by a modification, involving property (8) of the gamma function, of the procedure used in Sec. 59. When ii > 0 and ii * 1,2,..., the Bessel function is the product of 1/f and a power series in x whose initial term (k = 0) is nonzero; hence is unbounded as x 0. Since tends to zero as x — 0, it is evident and that are linearly independent functions. The general solution of Bessel's equation (6) is, therefore, y= (12) (ii >0; ii * 1,2,...), + where C1 and C2 are arbitrary constants. [Contrast this with solution (5) of equation (1).] It can be shown that are linearly dependent because and (n = 0,1,2,...) = (13) (see Problem 1, Sec. 61). So if ii = n = general solution of equation (6). 0, 1,2,..., solution (12) cannot be the 61. RECURRENCE RELATIONS Starting with the equation 00 1 (1)k x 2k + k)! = (n = 0,1,2,...), we write d 1 = If we replace k by x 2k-i °° 2k1 k(k - 1)!(n + k)! (2) k + 1 here, so that k runs from zero to infinity again, it 250 CHAP. 7 BESSEL FUNCTIONS AND APPLICATIONS follows that d (1)k+1 x = k=O —x (x\n+1+2k —n k=O 2k+1 k!(n + k + 1)! (1)k = x k!(n + 1 + k)! or d (n=O,1,2,...). = (1) The special case = —J1(x) (2) was mentioned at the end of Sec. 59. Similarly, from the power series representation of that d (3) Relations one can show (n = 1,2,...). (1) and (3), which are called recurrence relations, can be written =nJn(x) +xJn_i(x). = Eliminating from these equations, we find that (4) = —xJn_1(x) This recurrence relation expresses + 1 in terms of the functions lower orders. From equation (3), we have the integration formula f ds (n = 1,2,...). and —1 of (n = 1,2,...). An important special case is (6) IX Relations (1), (3), and (4) are valid when n is replaced by the unrestricted parameter ii. Modifications of the derivations simply consist of writing or in place of (n + k)!. SEC 61 251 PROBLEMS PROBLEMS 1. Using series (11), Sec. 60, and recalling that certain terms are to be dropped, show that (n = 1,2,...) = and are linearly dependent. and hence that the functions 2. Derive the recurrence relation (3), Sec. 61. 3. Establish the differentiation formula (n = 0,1,2,...). = (n2—n 4. (a) Derive the reduction formula fx ds + (n — = — (n — 1)2f ds (n=2,3,...) by applying integration by parts twice and using the relations (Sec. 61) d d = sJ0(s), = —J1(s) in the first and second of those integrations, respectively. (b) Note that, in view of equation (6), Sec. 61, the identity obtained in part (a) can be applied successively to evaluate the integral on the left-hand side of that identity when the integer n is odd.t Illustrate this by showing that fX5 ds = x(x2 + (x2 — 8)Ji(x)]. — 5. Let y be any solution of Bessel's equation of order zero, and let denote the seif-adjoint (Sec. 41) differential operator defined by the equation 2'[X(x)] = [xX'(x)]' + xX(x). (a) By writing X = J0 and V = y in Lagrange's identity [Problem 3(b), Sec. 43] YSf[X] = XSf[Y] d - X'Y)} for that operator, show that there is a constant B such that d y(x) B x[J0(x)] tNote too, that when n is even, the reduction formula can be used to transform the problem of ds into that of evaluating evaluating ds, which is tabulated for various values of x in, for example, the book edited by Abramowitz and Stegun (1972, pp. 492—493) that is listed in the Bibliography. Further references are given on pp. 490—491 of that book. 252 BESSEL FUNCTIONS AND APPLICATIONS CHAP. 7 (b) Assuming that the function 1/[J0(x)]2 has a Maclaurin series expansion of the formt 1 [J0(x)J 2= 1+ k=1 and that the expansion obtained by multiplying each side of this by 1/x can be integrated term by term, use the result in part (a) to show formally that y can be written in the form y=AJ0(x)+B J0(x)lnx+ Edkx2k k=1 where A, B, and dk (k = 1,2,...) are constants. [Compare equation (2), Sec. 60.] 2,...) be the sequence defined in equation (3), Sec. 60. Show that 5n > 0 and 5n — 5n +1 > 0 for each n. Thus show that the sequence is bounded and 6. Let Sn (n = 1, decreasing and that it therefore converges to some number y. Also, point out how it follows that 0 y < 1. Suggestion: Observe from the graph of the function y = 1/x that n—il iX k=lk and 1 n+1 <1 —=ln(n+1)—lnn x 7. (a) Derive the property F(v + 1) = vF(v) of the gamma function, as stated in Sec. 60. (b) Show that 12(1) = when n = 0, 1,2 1 and, using mathematical induction, verify that fln + 1) = (,i 0), defined by equation (11), Sec. 60, satisfies 8. Verify that the function Bessel's equation (6) in that section. Point out how it follows that J_,, is also a solution. 9. Derive the differentiation formula d = where ii 0, and point out why it is also valid when ii is replaced by — [Compare relation (1), Sec. 61.] 10. Refer to the result obtained in Problem 18, Sec. 58, and show that = 2f (i.' > 0). e_x2dT = tThis valid assumption is easily justified by methods from the theory of functions of a complex variable. See the author's book (1990, chap. 5), listed in the Bibliography. BESSEL'S INTEGRAL FORM OF SEC. 62 11. With the aid of mathematical induction, verify that (2k)! 1 F(k (k = = k!22k + 12. J1/2(x) I-Y— sin x; I-Y- (b) J_l/2(x) = = 13. Use results in Problems 9 and 12 to show that V 0,1,2,...). and the identity in Problem 11 Use the series representation (11), Sec. 60, for to show that (a) 253 — cos x. 'TX \ 14. Show that if y is a differentiable function of x and if s = ax, where a is a nonzero constant, then dy dy — 2d2y d2y —=a-—--— and ds2 Thus show that the substitution s = ax transforms the differential equation d2y + dy + (a2x2 — n2)y = (n=0,1,2,...) 0 into Bessel's equation 2d2y s ds2 dy + s— + (s2 — ds n2)y = 0 which is free of a. Conclude that the general solution of the first differential equation here is y= + 15. From the series representation (9), Sec. = k=O show that for 1 x n+2k (n = 0,1,2,...). k!(n + k)! = is the modified Bessel function of the first kind of order The function n. Show that the series here converges for all x, that 0 when x> 0, and that Also, by referring to the result in Problem 14, point out why —x) = (— is a solution of the modified Bessel equation x2y"(x) + xy'(x) 62. — (x2 + n2)y(x) = 0. BESSEL'S INTEGRAL FORM OF now derive a useful integral representation for note that the series in the expansions We To do this, we first kk (1) exp( = JO — = k=O k t_k 254 BESSEL FUNCTIONS AND APPLICATIONS CHAP. 7 are absolutely convergent when x is any number and t 0. Hence the product of these exponential functions is itself represented by a series formed by multiplying each term in one series by every term in the other and then summing the resulting terms in any order.t Clearly, the variable t occurs in each of those resulting terms as a factor t'1 (n = 0, 1, 2,. . .) or t-'1 (n = 1, 2,. ..); and the terms involving any particular power of t may be collected as a sum. In the case of t'1 (n = 0, 1, 2,...), that sum is obtained by multiplying the kth term in the second series by the term in the first series whose index is = n + k and then summing from k = 0 to k = oo• The result is j x n+2k k=O k!(n + k)! Similarly, the sum of the terms involving 1,2,...) is found by multiply- (n = ing the jth term in the first series by the term in the second series with index k = n + j and summing from j = 0 to j = oo. That sum may be written x n+2j 00 = +j)! A series representation for the product of the exponential functions (1) is, therefore, (2) Let us write t = =J0(x) + + in equation (2). In view of Euler's formula = cos 4 + i sin we know that — = 2i sin 4) and = cos n4 + i sin n4), = cos n4) — i sin n4. It thus follows from equation (2) that (3) =J0(x) + + [i — n=1 For a justification of this procedure, see, for example, the book by Taylor and Mann (1983, pp. 601—602) that is listed in the Bibliography. BESSEL'S INTEGRAL FORM OF SEC. 62 255 Now, again by Euler's formula, exp (ix sin 4)) = cos (x sin 4)) + i sin (x sin 4)); and if we equate the real parts on each side of equation (3), we find that cos(xsin4)) ==J0(x) + + Holding x fixed and regarding this equation as a Fourier cosine series representation of the function cos (x sin 4)) on the interval 0 <4) <n-, we need only recall the formula for the coefficients in such a series to write (4) [i + (n = 0,1,2,...). = —f 17•0 If, on the other hand, we equate the imaginary parts on each side of equation (3), we obtain the Fourier sine series representation sin(xsinçb) = — (_1)nh]Jn(x)sinncb for sin (x sin 4) on the same interval. Consequently, (5) [i (n = 1,2,...). = — According to expressions (4) and (5), then, (6) (n = 0,1,2,...) = —f iTO and (7) = —f 17•0 sin(xsin4))sin(2n — (n = 1,2,...). A single expression for can be obtained by adding corresponding sides of equations (4) and (5) and writing = —f ITO [cos n4) cos (x sin 4)) + sin sin (x sin 4)] d4). That is, (8) = —f This is known as Bessel's integral form of special cases of it. (n = 0,1,2,...). and expressions (6) and (7) are 256 BESSEL FUNCTIONS AND APPLICATIONS CHAP. 7 CONSEQUENCES OF THE INTEGRAL REPRESENTATIONS 63. From the integral representation (1) = —f 17•0 (n=O,1,2,...) cos(nçb—xsinçb)dçb just obtained, it follows that = ITO —f sin (n4) — x sin 4)) sin 4 d4. Continued differentiation yields integral representations for J,'1'(x), etc. In each case, the absolute value of the integrand that arises does not exceed unity. The following boundedness properties are, then, consequences of Bessel's integral form (1): (2) 1, (k = 1,2,...). 1 The first of these inequalities, for example, is obtained by writing ITO ITO d4)=1. Sometimes it is useful to write the integral representations (n=O,1,2,...) 17•0 and 17•0 sin(xsin4))sin(2n—1)4d4 obtained in Sec. 62, as 2 = —f cos(xsin4))cos2n4)d4) (n = 0,1,2,...) and (4) 2 = —f sin(xsin4)sin(2n — 1)4)d4) (n = 1,2,...). Expressions (3) and (4) follow from the fact that, when x is fixed, the graphs of the integrands y = g( 4) = cos (x sin 4)) cos 2n4, y = h(4)) = sin(xsin4))sin(2n — 1)4 CONSEQUENCES OF THE INTEGRAL REPRESENTATIONS SEC. 63 257 are symmetric with respect to the line 4 = g(ir — 4)) = g(4)), 4)) = h(çb). — We note the special case 2 J0(x) = —f 17•0 (5) cos(xsin4))d4) of representation (3), which can also be written J0( x) = (6) by 2 f — 17•0 cos (x cos 0) dO means of the substitution 0 = — 4). Representations (3) and (4) may be used to verify that, for each fixed n (n = 0,1,2,...), = (7) To give the details when n = 0, we substitute u = sin —J0(x)=I cosxu 2 du+I 1 4) in equation (5) to write cosxu V1_u2 du where 0 <c < 1. The second integral here is improper but uniformly convergent with respect to x. Corresponding to any positive number e, the absolute value of that integral can be made less than E/2, uniformly for all x, by selecting c so that 1 — c is sufficiently small and positive. The RiemannLebesgue lemma involving a cosine function (Sec. 53) then applies to the first integral with that value of c. That is, there is a number XE such that the absolute value of the first integral is less than E/2 whenever X > XE. Therefore, <e whenever x > XE; this establishes property (7) when n = 0. Verification when n is a positive integer is left to the problems. It is interesting to contrast limit (7) with the limit and = 0, which is valid for each fixed X (— oc <X < oo). This limit follows from the fact that the coefficients = and = in the Fourier cosine and sine series for certain functions of 4) in Sec. 62 must tend to zero as n tends to infinity, according to Sec. 16. Limit (8) can also be obtained by applying the Riemann-Lebesgue lemma to the integral representations for and 258 CHAP. 7 BESSEL FUNCTIONS AND APPLICATIONS PROBLEMS 1. Use integral representations for to verify that = 0 (n = 1,2,...); (c) (a) J0(O) = 1; (b) = —J1(x). 2. Derive representation (3), Sec. 63, for by writing the Fourier cosine series for cos (x sin 4)) in Sec. 62 as cos (x sin 4)) = .J0(x) + cos 2n4) 2 n=1 and then interpreting it as a Fourier cosine series on the interval 0 <4) 3. Deduce from expression (3), Sec. 63, that = (n = 0,1,2,...). 4. Deduce from expression (4), Sec. 63, that = — 1)OdO (n = 1,2,...). 5. Complete the verification of property (7), Sec. 63, that = 0 for each fixed n (n = 0, 1, 2,...). 6. Apply integration by parts to representations (3) and (4) in Sec. 63 and then use the Riemann-Lebesgue lemma (Sec. 53) to show that = 0 for each fixed x. 7. Verify directly from the representation (Sec. 63) çir/2 J0(x) = —J 'TO cos(xsin4))d4) that .J0(x) satisfies Bessel's equation xy"(x) + y'(x) + xy(x) = 0. 8. According to Sec. 24, if a function f and its derivative f' are continuous on the interval — x 'r and if f( — 'r) = f('r), then Parseval's equation dx = + + holds, where the numbers (n = 0, 1,2,...) and (n = 1,2,...) are the Fourier coefficients lIT lIT = —f f(x)cosnxdx, = —f f(x)sinnxdx. SEC 64 259 THE ZEROS OF J0(x) (a) By applying that result to f(4)) = referring cos (x sin 4), an even function of 4), and in Sec. 62, show that to the Fourier (cosine) series for (—oo<x<oo). (b) Similarly, by writing f(4) = sin(x sin series expansion in Sec. 62, show that 4) and referring to the Fourier (sine) (—oo<x<oo). (c) Combine the results in parts (a) and (b) to = [J0(x)]2 + and show that (_oo 1 <x <oo), point out how it follows from this identity that (n=1,2,...) and IJ0(x)kl for all x. 9. By writing t = i in the series representation (2), Sec. 62, derive the expansions cosx =J0(x) + 2 (—1) n=1 10. which are valid for all x. Show that series representation (2), Sec. 62, can be written in the form N xi exp —(t——) = lim (t#O). tj This exponential function is, then, a generating function for the Bessel functions 64. THE ZEROS OF J0(x) A modified form of Bessel's equation (1) x2y"(x) + xy'(x) + (x2 — ii2)y(x) = 0 in which the term containing the first derivative is absent is sometimes useful. That form is easily found (Problem 1, Sec. 66) by making the substitution y(x) = xcu(x) in equation (1) and observing that the coefficient of u'(x) in the resulting differential equation x2u"(x) + (1 + 2c)xu'(x) + (x2 — p2 + c2)u(x) = 0 260 is BESSEL FUNCTIONS AND APPLICATIONS zero if c = — CHAP. 7 The desired modified form of equation (1) is, then, 1 x2u"(x) + u(x) = + x2 — 0; and the function u = is evidently a solution of it. In particular, when ii = 0, we see that the function u = satisfies the differential equation 1 u"(x) + 1 + (4) u(x) = 0. We shall now use equation (4) to show that the positive zeros of J0(x), or 0, form an increasing sequence of numbers roots of the equation J0(x) = (1= We start with the observation that the differential operator 2' = d2/dv2 is self-adjoint (Sec. 41) and that Lagrange's identity [Problem 3(b), Sec. 431 for this operator is (5) U(x)V"(x) - V(x)U"(x) = d [U(x)V'(x) - U'(x)V(x)], where U(x) and V(x) are any functions that are twice-differentiable. We write U(x) = (6) V(x) = sinx. and From equation (4), we know that U"(x) + U(x) U(x) = 4x2 furthermore, V"(x) + V(x) = 0. The left-hand side of identity (5) then becomes U(x)V(x)/(4x2), and it follows that (7) bU(X)V(X) dX = [U(x)V'(x) b — where 0 <a <b. It is now easy to show that our function U(x) = ViJ0(x), and hence J0(x), has at least one zero in each interval (k=1,2,...). method is a modification of the one used by A. Czarnecki, Amer. Math. Monthly, vol. 71, no. v 4, pp. 403—404, 1964, who considers Bessel functions 4(x), where — THEZEROSOFJ0(x) SEC. 64 261 0 anywhere in an interval We do this by assuming that U(x) x 2kir + ir and obtaining a contradiction. According to that assumption, either U(x)> for all x in the interval or U(x) < 0 for all such x, since U(x) is continuous and thus cannot change sign without having a zero value at some point in the interval. Suppose that U(x)> 0 when 2kv- x 2kv- + ir. In identity (7), write Since V(a) = V(b) = 0, V'(a) = 1, and V'(b) = a = 2kv- and b = 2kir + — 1, that identity becomes (8) '2kTr The integrand here is positive when 2kv- <x < 2kir + Hence the left-hand side of this equation has a positive value while the right-hand side is negative, giving a contradiction. If, on the other hand, U(x) < 0 when 2ki,- x 2ki,- + ii-, those two sides of equation (8) are negative and positive, respectively. This is again a contradiction. Thus J0(x) has at least one zero in each interval (k=1,2,...). Actually, J0(x) can have at most a finite number of zeros in any closed bounded interval a x b. To see that this is so, we assume that the interval a x b does contain an infinite number of zeros. From advanced calculus, we know that if a given infinite set of points lies in a closed bounded interval, there is always a sequence of distinct points in that set which converges to a point in the interval.t In particular, then, our assumption that the interval a x b contains an infinite number of zeros of J0(x) implies that there exists a sequence Xm (m = 1, 2,...) of distinct zeros such that Xm c as m —4 00, where c is a point which also lies in the interval. Since the function J0(x) is continuous, J0(c) = 0; and, by the definition of the limit of a sequence, every interval centered about c contains other zeros of J0(x). But the fact that J0(x) is not identically zero and has a Maclaurin series representation which is valid for all x means that there exists some interval centered at c which contains no other Since this is contrary to what has just been shown, the number of zeros in the interval a x b cannot, then, be infinite. It is now evident that the positive zeros of J0(x) can, in fact, be arranged as an increasing sequence of numbers tending to infinity. The table below gives the values, correct to four significant figures, of the first five zeros of J0(x) and the corresponding values of J1(x) [see Fig. 62 (Sec. 59)1. Extensive tables of tSee, for example, the book by Taylor and Mann (1983, pp. 515—519), listed in the Bibliography. That is, the zeros of such a function are isolated. An argument for this is given in a somewhat more general setting in the authors' book (1990, p. 181), listed in the Bibliography. 262 BESSEL FUNCTIONS AND APPLICATIONS CHAP. 7 numerical values of Bessel and related functions, together with their zeros, will be found in books listed in the Bibliography.t J0(x3) = 65. j 1 x3 2.405 .I1(x3) 0.5191 2 — 0 4 3 5.520 8.654 0.3403 0.2715 5 11.79 — 0.2325 14.93 0.2065 ZEROS OF RELATED FUNCTIONS = 0 and = 0 for two distinct positive numbers a and b, then also vanishes when x = a and when x = b. It thus follows from Rolle's If vanishes for at least one value of x theorem that the derivative of between a and b. But (Sec. 61) d = (n=O,1,2,...); and so, when n = 0, 1, 2,..., there is at least one zero of + 1(x) between any two positive zeros of Also, just as in the case of J0(x) (Sec. 64), the function + 1(x) can have at most a finite number of zeros in each bounded interval. We have already shown that the positive zeros of J0(x) form an unbounded increasing sequence of numbers. It now follows that the zeros of J1(x) must form such a set. The same is then true for J2(x), etc. That is, for each fixed =0 nonnegative integer n, the set of all positive roots of the equation 00• oo as j forms an increasing sequence x = (j = 1,2,...), where satisfies Bessel's equation, which is a linear homoThe function y = geneous differential equation of the second order with the origin as a singular point. According to the lemma in Sec. 44 on the uniqueness of solutions of second-order linear differential equations, there is just one solution that satisfies the conditions y(c) = y'(c) = 0, where c > 0; that solution is identically zero. = 0. That Consequently, there is no positive number c such that = must change its thus is, cannot vanish at a positive zero of sign at that point. 0, If Let a and b (0 <a <b) be two consecutive zeros of is decreasing at its zero b; that is, then > 0 when a <x < b and alternate 0. So the values of <0. Similarly, if <0, then in sign at consecutive positive zeros of h is a nonnegative We now consider the function constant. The zeros of this function will also arise in certain boundary value it follows that problems. If a and b are consecutive positive zeros of + must have the values a and the and Stegun (1972) and the ones by Jahnke, Emde, and Lösch (1960), Gray and Mathews (1966), and Watson (1952). ORTHOGONAL SETS OF BESSEL FUNCTIONS SEC. 66 263 x= b, respectively. Since one of those values is positive and the other negative, the fupction vanishes at some point, or at some finite number of points, between a and b. It therefore has an increasing sequence of positive zeros tending to infinity.t We collect our principal results as follows. Theorem. For each fixed n (n = 0, 1,2,...), the positive roots of the equa- tion (1) form an increasing sequence x = (j = 1,2,...) such that also, the positive roots of the equation 0o as j 00; (2) h is a constant, always form a sequence of that type. Observe that x = 0 is a root of both equations (1) and (2) if n is a positive integer, since = 0 (n = 1, 2,...). It is also a root of equation (2) when h= n = 0. If x = c is a root of equation (1), then x = —c is also a root since That statement is true of equation (2) as well; for, in view of the recurrence relation (Sec. 61) — c) = (— equation (2) can be written (h + (3) — = 0, and we note that (h + — = + — Finally, although our discussion leading up to the theorem need not have excluded the possibility that h be negative, those values of h will not arise in our applications. 66. ORTHOGONAL SETS OF BESSEL FUNCTIONS As indicated at the beginning of the chapter, where somewhat different notation was used, the physical applications in this chapter will involve solutions of the In the important special case n = 0, the first few zeros are tabulated for various positive values of h in, for example, the book on heat conduction by Carslaw and Jaeger (1959, p. 493) that is listed in the Bibliography. 264 CHAP. 7 BESSEL FUNCTIONS AND APPLICATIONS differential equation d2X (1) dX +(Ax2—n2)X=O (n=O,1,2,...), whose self-adjoint form (Sec. 41) is (2) (__ +Ax)X=O (n=O,1,2,...). + More specifically, we shall need to solve a singular (Sec. 42) Sturm-Liouville problem, on an interval 0 x c, consisting of the differential equation (1) and a boundary condition of the type b1X(c) + b2X'(c) = 0. The constants b1 and b2 are real and not both zero, and X and X' are to be (3) continuous on the entire interval 0 x c. In the important special case when b2 = (5) the boundary condition (3) is X(c)=O. (4) When b2 * 0, 0, we may multiply through condition (3) by c/b2 and write it as hX(c) + cX'(c) = 0, where h = cb1/b2. In solving our Sturm-Liouville problem, we shall find it convenient to use the boundary condition (3) in its separate forms (4) and (5); and, when using condition (5), we shall always assume that h 0. The corollary in Sec. 43, applied to case (a) of the theorem in that section, ensures that any eigenvalue of our singular Sturm-Liouville problem must be a real number. We now consider the three possibilities of A being zero, positive, or negative. When A = 0, equation (1) is a Cauchy-Euler equation (see Problem 3, Sec. 35): (6) d2X + dX — n2X = 0. To solve it, we write x = exp s and put it in the form (7) d2X B are A If n is positive, it follows that X = Since our solution must be continuous, constants; that is, X(x) = Ax'1 + and therefore bounded, on the interval 0 x c, we require that B = 0. Hence X(x) = Ax'1. It is now easy to see that A = 0 if either condition (4) or (5) is to be satisfied, and we arrive at only the trivial solution X(x) zero is not an eigenvalue if n is positive. 0. Thus If, on the other hand, n = 0, equation (7) has the general solution X = As + B; and the general solution of equation (6) when n = 0 is, therefore, ORTHOGONAL SETS OF BESSEL FUNCTIONS SEC. 66 265 B. According to the continuity requirements, then, X(x) = B. When condition (4) is imposed, B = 0; the same is true of condition (5) when h > 0. But when h = 0, condition (5) becomes simpiy X'(c) = 0, and B can remain arbitrary. So if n = 0 and condition (5) is used when h = 0, we have the X(x) = A in x + eigenfunction corresponding X( x) = 1, (8) to A = 0. any eigenfunction corresponding to that eigenvalue is a constant muitiple of the function (8). We consider next the case in which A > 0 and write A = a2 (a > 0). Equation (1) is then This is the only case in which A = 0 is an eigenvalue, and d2X (9) + dX + — n2)X = 0, and we know from Probiem 14, Sec. 61, that its generai soiution is X(x) = (10) + C2 = 0, since lç(ax) is discontinuous at (see Sec. 60). Hence any nontriviai soiution of equation (9) that meets x= 0 those requirements must be a constant multipie of the function X(x) = In appiying the boundary condition at x = c, we emphasize that the symbol s = stands for the derivative of Jr(s) with respect to s, evaluated at ax. Then d/dx and conditions (4) and (5) require that = = (11) and (12) 0 = 0, + respectiveiy. Note that since equation (2), Sec. 65, can be written in the form (3) in that section, equation (12) can aiso be written as (h + — = 0. According to the theorem in Sec. 65, each of equations (11) and (12) has an infinite number of positive roots X;i (j = 1,2,...), (13) = where (I = 1, 2,...) is the unbounded increasing sequence in the statement here depend, of course, on the vaiue of n and of that theorem. The numbers aiso on the vaiue of h in the case of equation (12). Our Sturm-Liouvilie probiem thus has eigenvaiues = (I = 1, 2,...), and the corresponding eigenfunctions are (14) (I = 1,2,...). are the positive roots of equation (12) We note that if the numbers when n = h = 0, which is the oniy case where A = 0 was found to be an 266 BESSEL FUNCTIONS AND APPLICATIONS CHAP. 7 eigenvalue, that equation can be written as (15) J1(crc) = 0. The numbers are then given more directly as the positive roots of equation (15). Also, making a minor exception in our notation, we let the subscript j range over the values j = 2, 3,..., instead of starting from unity. The subscript j = 1 is reserved for writing a1 = 0 and A1 = = 0. This allows us to include the eigenvalue A1 = 0 and the eigenfunction X1(x) = J0(a1x) = 1, obtained earlier for the case n = h = 0. Note that it is also possible to describe the numbers a (j = 1, 2,...) here as the nonnegative roots of equation (15). Finally, we consider the case in which A <0, or A = — a2 (a > 0), and write equation (1) as d2X (16) + dX (a2x2 + n2)X = 0. — The substitution s = ax can be used here to put equation (16) in the form (compare Problem 14, Sec. 61) d2X (17) ds dX + s— — ds (s2 + n2)X = 0. From Problem 15, Sec. 61, we know that the modified Bessel function X = satisfies equation (17); and, since it has a power series repreIa(s) = sentation that converges for all s, X = satisfies the continuity requirements in our problem. As was the case with equation (9), equation (16) has a second solution which is discontinuous at x = 0, that solution being analogous to Thus we know that, except for an arbitrary constant factor, X(x) = We now show that, for each positive value of a, the function X(x) = fails to satisfy either of the boundary conditions (4) and (5). In each case, our proof rests on the fact that > 0 when x > 0, as demonstrated in Problem 15, Sec. 61. Since 0 when a > 0, it is obvious that condition (4), which requires that = 0, fails to be satisfied for any positive number a. Also, in view of the alternative form (3), Sec. 65, of equation (2) in that section, condition (5), when applied to our function X(x) = = becomes (h + + = 0, or (h + + = 0. tFor a detailed discussion of this, see, for example, the book by Tranter (1969, pp. 16ff) that is listed in the Bibliography. SEC 66 ORTHOGONAL SETS OF BESSEL FUNCTIONS 267 Since a > 0, the left-hand side of this last equation is positive; and, once again, no positive values of a can occur as roots. We conclude, then, that there are no negative eigenvalues. We have now completely solved our singular Sturm-Liouville problem consisting of equations (1) and (3), where, since we have assumed that the constant h = cb1/b2 is nonnegative, it is understood that the constants b1 and b2 in equation (3) have the same sign when neither is zero. The eigenvalues are all represented by the numbers A3 = where the are given by equation (13) and where A3 > 0, except that A1 = 0 in the case used in equation (13) form an unbounded n = h = 0. Since the numbers increasing sequence, it is clear that the same is true of the eigenvalues = That is, A3 and A3 oc as j —p oo• We summarize our results in the following theorem. Theorem. Let n have one of the values n = 0, 1,2 Sturm-Liouville problem consisting of the differential equation d2X (19) + dX + (Ax2 — n2)X = For the singular (0 <x <c), 0 which reduces to d2X when n = 0, dX (0<x<c) and one of the boundary conditions (21) X(c) = 0, hX(c) + cX'(c) = (22) X'(c) = (20) 0, h + (h 0 n > 0), (n = 0), 0 the eigenvalues and corresponding eigenfunctions are (j= 1,2,...), where the numbers a3 are defined as follows: (a) When condition (20) is used, the equation = 1, 2,...) are the positive roots of (j = 0. (b) When condition (21) is used, (I = 1, 2,...) are the positive roots of the equation + (ac)J,ç(crc) = 0, which can also be written as (h + — + 1(ac) = 0. 268 BESSEL FUNCTIONS AND APPLICATIONS (c) When condition (22) positive roots of the equation is CHAP. 7 used, a1 = and a3 (j = 2,3,...) are the 0 = 0, which can also be written as J1(ac) = 0. Note that when n is positive (n = 1, 2,...), the constant h in condition (21) can be zero. For that value of h, condition (21) is simply X'(c) = 0, and the condition in case (b) that is used to define the becomes = 0. Note, too, that condition (22) is condition (21) when h = 0 and n = 0. Since the first eigenvalue is then a1 = 0, as stated in case (c), the first eigenfunction is actually X1 =J0(a1x) =J0(0) = 1. For each of the cases in this theorem, the orthogonality property JC (23) follows from case (a) of the theorem in Sec. 43. Observe that this orthogonality of the eigenfunctions with respect to the weight function x, on the interval on o <x <c, is the same as ordinary orthogonality of the functions that same interval. Also, many orthogonal sets are represented here, depending on the values of c, n, and h. In the next two sections, we shall normalize these eigenfunctions and find formulas for the coefficients in generalized Fourier series expansions involving the normalized eigenfunctions. PROBLEMS 1. By means of the substitution y(x) = xcu(x), transform Bessel's equation x2y"(x) + xy'(x) + (x2 = — 0 into the differential equation x2u"(x) + 2. (1 + 2c)xu'(x) + (x2 — + c2)u(x) = 0, which becomes equation (3), Sec. 64, when c = — Use equation (3), Sec. 64, to obtain a general solution of Bessel's equation when ii = Then, using the expressions (Problem 12, Sec. 61) Jl/2(x) = sin x, J_1/2(x) = cos x, point out how Jl/2(x) and J_ 1/2(x) are special cases of that solution. 3. By referring to the theorem in Sec. 66, show that the eigenvalues of the singular Sturm-Liouville problem xX"+X'+AxX=O, X(2)=0, SEC 66 PROBLEMS 269 the interval 0 x 2, are the numbers = (j = 1, 2,...), where are the positive roots of the equation J0(2a) = 0, and that the corresponding eigenfunctions on are = (j = 1, 2, ..). With the aid of the table in Sec. 64, obtain the numerical values a1 = 1.2, a2 = 2.8, a3 = 4.3, valid to one decimal place. 4. Write U(x) where n has any one of the values n = 0, 1,2,... and a is a positive constant. (a) Use equation (3), Sec. 64, to show that U"(x) + (a2 + 1 4n2 2 )U(x) = 0. (b) Let c denote any fixed positive number and write (J(x) = 1, 2,...), where are the positive roots of the equation (j = = 0. Use the result in part (a) to show that (aJ — = — UkUJ". (c) Use the result in part (b) and Lagrange's identity [Problem 3(b), Sec. 431 for the self-adjoint operator 2' = d2/dx2 to show that the set (j 1, 2,...) in part (b) is orthogonal on the interval 0 <x <c with weight function unity. Thus give another proof that the set (j = 1, 2,...) in case (a) of the theorem in Sec. 66 is orthogonal on that interval with weight function x. 5. Let n have any one of the fixed values n = 0, 1, 2 (a) Suppose that J,(ib) = 0 (b * 0) and use results in Problem 15, Sec. 61, to reach a contradiction. Thus show that the function J,(z) has no pure imaginary zeros z = ib (b * 0). (b) Since our series representation of J,7(x) (Sec. 59) converges when x is replaced by any complex number z and since the coefficients of the powers of z in that representation are all real, it follows that J,7(2) = , where 2 denotes the complex conjugate x — iy of the number z = x + jy•t Also, the proof of orthogonality in Problem 4 above remains valid when a is a nonzero complex number and when the set of roots there is allowed to include any nonzero complex roots of the equation J,,(ac) = 0 that may occur. Use these facts to show that if the complex number a + ib (a * 0, b * 0) is a zero of then a — ib is also a zero and that + ib)x) dx = 0. Point out why the value of the first integral here is actually positive and, with this contradiction, deduce that J,,(z) has no zeros of the form a + ib (a * 0, b 0). Conclude that if z = (j = 1, 2,...) are the positive zeros of then the only other zeros, real or complex, are the numbers z = also z = 0 when n is positive. (j = 1, 2,...), and For a discussion of power series representations in the complex plane, see the authors' book (1990, chap. 5), listed in the Bibliography. 270 BESSEL FUNCTIONS AND APPLICATIONS 67. THE ORTHONORMAL FUNCTIONS CHAP. 7 From Problem 14, Sec. 61, we know that if a is a positive constant, the function X(x) cxx) satisfies the equation (n=O,1,2,...). (xX')'+ a2x—— X=O (1) We multiply each side by 2xX' and write 2)d + (a2x2 — (X2) = 0. After integrating both terms here and using integration by parts in the second term, we find that [(xX')2 + (a2x2 — n2)x21C where c is any positive number. When n = clearly vanishes at x = X(0) = (2) = 0 0; — 0, 2a2f = 0, the quantity inside the brackets and the same is true when n = 1, 2,..., since then. We thus arrive at the expression 2a2fx[Jn(ax)}2 = + (a2c2 — n2)[j(ac)]2 which we now use to find the norms of all the eigenfunctions in the theorem in Sec. 66, except for the one corresponding to the zero eigenvalue in case (c) there. That norm is treated separately. (j = 1,2,...) are the positive roots of the equation (a) Suppose that = 0. (3) Expression (2) tells us that 2fx[Jn(ajx)]2 dx = on the interval 0 <x <c, The integral here is the square of the norm of with weight function x. Also (Sec. 61) and therefore = Hence + 2 (b) Suppose that C = 2 (j = 1,2,...). (j = 1,2,...) are the positive roots of the equation SEC. 68 FOURIER-BESSEL SERIES 271 We find from equation (2) that 2 (6) = a2c2—n2+h2 2 = (j (c) Suppose that a1 = 0 and that a3 (j = 2,3,...) are 1,2,...). the positive roots of the equation = 0. (7) Since J0(a1x) = J0(0) = (8) 1, MJo(aix)II Expressions for 11J0(a3c)112 2 cC j xdx = —. 0 (j = 2, 3,...) are obtained by writing n h= 0 in equation (6): (I = = 2,3,...). For equation (7) is simply equation (5) when n = h = 0, and the restriction h + n > 0 is not actually needed in deriving expression (6). of our singular SturmThe orthogonal eigenfunctions X3(x) = Liouville problem can now be written in normalized form as (j = 1,2,...). = The norms here are given for the eigenfunctions in cases (a) and (b) of the theorem in Sec. 66 by equations (4) and (6), respectively. In case (c), they are given by equations (8) and (9). The fact that the set (10) is orthonormal on the interval 0 <x <c with weight function x is, of course, expressed by the equations (0 c 68. whenj*k whenj=k. FOURIER-BESSEL SERIES Let f be any piecewise continuous function defined on an interval 0 <x <c. We consider here the generalized Fourier series j=1 272 CHAP. 7 BESSEL FUNCTIONS AND APPLICATIONS for f with respect to the normalized eigenfunctions (10) in Sec. 67. According to Sec. 12, the coefficients c3 in that series are the numbers fc Writing we thus have the correspondence = (O<x<c), (1) j=1 where 1 (2) (j=1,2,...). The norms found in Sec. 67 can now be used to adapt expression (2) to each of the three cases (a), (b), and (c) treated in that section and in the theorem in Sec. 66. be the coefficients in correspondence (1). Theorem 1. Let (a) If (j = 1, 2,...) are the positive roots of the equation = 0, then 2 (3) c2 (b) If a3 (j = (1= 1,2,...). 0 1, 2,...) are the positive roots of the equation h + n > 0), which can also be written as (h + + — 1(ac) = 0, then 2a2 (4) — n2 + 0 (j=1,2,...). (c) If n = 0 in series (1) and if a1 = positive roots of the equation = 0, which can also be written as J1(ac) = (5) 0, then 0 and (I = 2,3,...) are the FOURIER-BESSEL SERIES SEC. 68 273 and 2 (6) (j=2,3,...). 0 1, expression (6) becomes expression (5) when It is, however, more convenient in the applications to treat A1 separately j= and to write correspondence (1) as Note that since J0(O) = 1. (O<x<c) j=2 when case (c) is to be used. Proofs that correspondence (1) is actually an equality, under conditions similar to those used to ensure the representation of a function by its Fourier cosine or sine series, usually involve the theory of functions of a complex variable. We state, without proof, one form of such a representation theorem and refer the reader to the Bibliography.t Theorem 2. Let f denote a function that is piecewise smooth on an interval 0 <x <c, and suppose that f(x) at each point of discontinuity of f in that interval is defined as the mean value of the one-sided limits f(x + ) and f(x — Then (7) (0 <x <c), f(x) = j= 1 are defined by equation (3) or (4) or the pair of equations (5) and (6), depending on the particular equation that determines the numbers where the coefficients a Fourier-Bessel series representation of f(x). EXAMPLE 1. Let us expand the function f(x) = series of the type 1 (0 <x <c) into a j= 1 where a3 (I = 1, 2,...) are the positive roots of the equation J0(ac) = 0. Case (a) in Theorem 1 is evidently applicable here, and expression (3) tells us that 2 (8) c 2 0 tThis theorem is proved in the book by Watson (1952). Also see the work by Titchmarsh (1962), as well as the books by Gray and Mathews (1966) and Bowman (1958). These are all listed in the Bibliography. 274 BESSEL FUNCTIONS AND APPLICATIONS CHAP. 7 This integral is readily evaluated by substituting s = tion formula (Sec. 61) jx ds and using the integra- = xJ1( x). To be specific, 1 dx fo = C = a3 aj Consequently, where J0(a.x) 00 2 (10) (O<x<c), c 3=i a1J1(a3c) = 0 (a1> 0). EXAMPLE 2. To represent the function f(x) = x (0 <x < of the form 1) in a series A1+ j=2 = 2,3,...) are the positive roots of the equation J1(a) = fer to case (c) in Theorem 1. According to expression (5), where a1 (j A1 = 2f 1 0, we re- 2 x2dx = —; 3 0 and we find from expression (6) that (j=2,3,...). This last integral can be evaluated by referring to the reduction formula (see Problem 4, Sec. 61, and the footnote with that problem) jX2 and then recalling that s) ds = x 2J1( x) + xJ0( x) — = JJ( s) ds 0: f1x 2J0( aix) dx = = f s) ds [aiJo(ai) — dsj. PROBLEMS SEC. 68 275 Thus 00 2 x= —+ (11) 3 where J1(a1) = [aiJo(ai) 2 j=2 0 — f J0(s) ds] J0(a.x) 2 0 (0 <x < 1), (a'> 0). The theorem in Sec. 65 and the two in this section are also valid when n although we have not is replaced by an arbitrary real number ji (ii > — far enough to establish any such developed properties of the function generalizations. For functions f on the unbounded interval x > 0, there is an integral analogous to the Fourier cosine and sine integral representation in terms of is formulas. The representation, for a fixed ji (i'> — f(x) = f (x> 0) dsdcr and is known as Hankel's integral formula.t It is valid if f is piecewise smooth on each bounded interval, if Vif(x) is absolutely integrable from zero to infinity, and if f(x) is defined as its mean value at each point of discontinuity. If the interval 0 <x <c is replaced by some interval a <x <b, where a> 0, the Sturm-Liouville problem treated in Sec. 66 is no longer singular when the same differential equation is used and boundary conditions of type (3) in that section are applied at each end point. In general, the resulting eigenfuncand tions then involve both of the Bessel functions PROBLEMS 1. Show that J0(a.x) a3 1(a3) a 1 1— 2 (O<x<1), where a3 (j = 1, 2,...) are the positive roots of the equation .J0(a) = 0. 2. Derive the representation c2 x2=—+4E 2 j=2 where J0(a.x) 2 (0<x<c), (j = 2,3,...) are the positive roots of the equation J1(ac) = 0. tSee the book by Sneddon (1951, chap. 2) that is listed in the Bibliography. For a summary of representations in terms of Bessel functions, see the work edited by Erdélyi (1981, vol. 2, chap. 7) that is also listed. 276 3. BESSEL FUNCflONS AND APPLICATIONS CHAP. 7 Show that if whenO<x<1, when 1<x<2, (1 and f(1) = then °° 1 — f(x) = 2 J1(a.) (0 <x <2), j=i a1[Ji(2a1)} where a (j = 1,2,...) are the positive roots of the equation J0(2a) = 0. (j = 1,2,...) denote the positive roots of the equation J0(ac) = 0, where c is 4. Let a fixed positive number. (a) Derive the expansion 2 X2 = (a1c)2 —4 — c (0<x<c). J0(a1x) (b) Combine expansion (10) in Example 1, Sec. 68, with the one in part (a) to show that 8 °° J0(a1x) c j=1 a.J1(a1c) 5. Find the coefficients (j = 1,2,...) in the expansion (0 <x <c) A.J0(a1x) 1= 1=1 when a1 = 0 and a (j = 2,3,...) are the positive roots of the equation 1, A. = 0 (j = 2,3,...). = 0. Answer: A1 = 6. (a) Obtain the representation 00 (0<x<c) + h2)[Jo(ajc)]2 j=i where a3 (j = 1, 2,...) are the positive roots of the equation hJ0(ac) + = (h > 0). 0 [Contrast this representation with representation (10) in Example 1, Sec. 68, and the one in Problem 5.] (b) Show how the result in part (a) can be written in the form 2 1 C a3 [Jo(a1c)]2 + [Ji(a3c)J 2 (0<x<c). 7. Show that if (x when0<x<1, when 1<x<2 SEC. 68 PROBLEMS and f(1) = 277 then a.J2(a.) j=1 (O<x<2), — = 0. where a3 (j = 1,2,...) are the positive roots of the equation Suggestion: Note that when h = 0 in case (b) of Theorem 1 in Sec. 68, the equation defining the a3 there becomes = 0. Show that 8. Let n have any one of the positive values n = 1,2 a1 n+1 a3 2 j=i (0<x<1), —n 2 (j = 1,2,...) are the positive roots of the equation where = 0. (See the suggestion with Problem 7.) 9. Point out why the eigenvalues of the singular Sturm-Liouville problem X(1) = x2X" + xX' + (Ax2 — 1)X = 0, 0, on the interval 0 <x < 1, are the numbers A. = (j = 1,2,...), where a3 are the positive roots of the equation J1(a) = 0, and why the corresponding eigenfunctions are X3 = J1(a3x) (j = 1,2,...). Then obtain the representation Ji(ax) x=2E (0<x<1) j=1 terms of those eigenfunctions. indicated in Sec. 68, there exist conditions on f under which representation (7) there is valid when n is replaced by ii where ii is not necessarily an >— integer. In particular, suppose that in 10. As f(x) = (0 <x <c), E AJJ1/2(afx) j=1 where Jl/2(ac) = 0, is piecewise smooth, where a3 are the positive roots of the equation and where [compare expression (3), Sec. 68] 2 (j= 2 c [J3/2(afc)] 1,2,...). 0 Using the expressions [Problems 12(a) and 13, Sec. 61] Jl/2(x) = sin x and J3/2(x) = — cos x) to substitute for the Bessel functions involved, show that this Fourier-Bessel representation is actually the Fourier sine series representation of on the interval 0 <x <c. 278 BESSEL FUNCTIONS AND APPLICATIONS 69. TEMPERATURES IN A LONG CYLINDER CHAP. 7 In both of the following examples, we shall use Bessel functions to find temperatures in an infinitely long circular cylinder p c whose lateral surface p = c is kept at temperature zero (Fig. 64). Other thermal conditions will be such that the temperatures will depend only on the space variable p, which is the distance from the axis of the cylinder, and time t. We assume that the material of the solid is homogeneous. z I" I I I NI r V FIGURE 64 EXAMPLE 1. When the cylinder is as shown in Fig. 64 and the initial temperatures vary only with p, the temperatures u = u(p, t) in the cylinder satisfy the special case (Sec. 4) 1 (O<p<c,t>O) of the heat equation in cylindrical coordinates and the boundary conditions u(c,t)=O (t>O), u(p,O) =f(p) (0 <p <c). when t > 0, the function u is to be continuous throughout the cylinder and, in particular, on the axis p = 0. We assume that f is piecewise smooth on the interval 0 <p <c and, for convenience, that f is defined as the mean value of its one-sided limits ateach point in that interval where it is discontinuous. Any solutions of the homogeneous equations (1) and (2) that are of the type u = R(p)T(t) must satisfy the conditions Also, 1 RT' = kT R" + —R' p , R(c)T(t) = 0. TEMPERATURES IN A LONG CYLINDER SEC. 69 279 Separating variables in the first equation here, we have T' 1 1 R p —=—R"+—R' =—A kT where — (4) A is the separation constant. Thus pR"(p) + R'(p) + ApR(p) = R(c) = 0, 0 (0 <p <c), and T'(t) + AkT(t) (5) = 0 (t > 0). The differential equation in R is Bessel's equation, with the parameter A in which n = 0. Problem (4), together with continuity conditions on R and R' on the interval 0 p c, is a special case of the singular Sturm-Liouville problem in the theorem in Sec. 66 when n = 0 and the boundary condition (20) there is taken. According to that theorem, the eigenvalues A3 of problem (4) are the numbers A3 = (j = 1, 2,...), where a3 are the positive roots of the equation J0(ac) = (6) 0; are the corresponding eigenfunctions. = equation (5) is satisfied by = exp (— When A = products are and So the desired (1= 1,2,...); and the generalized linear combination of these functions, u(p,t) = j=1 formally satisfies the homogeneous conditions (1) and (2) in our boundary value problem. It also satisfies the nonhomogeneous initial condition (3) when the coefficients are such that f(p)= (0<p<c). j=1 This is a valid Fourier-Bessel series representation (Theorem 2, Sec. 68) if the coefficients have the values 2 0 (j= 1,2,...), obtained by writing n = 0 in equation (3), Sec. 68. The formal solution of the boundary value problem is, therefore, given by equation (7) with the coefficients (8), where a3 are the positive roots of equation 280 (6). CHAP 7 BESSEL FUNCTIONS AND APPLICATIONS Thus our temperature formula can be written (9) u(p,t) = EXAMPLE 2. Suppose now that heat is generated in the cylinder in Example 1 at a constant rate per unit volume and that the surface and initial temperatures are both zero. The temperatures u = u(p, t) must satisfy the conditions 1 (10) (0<p<c,t>O), +q0 p where q0 is a positive constant (Sec. 2), and u(c,t) (11) u(p,0) = 0, = 0. The function u is, of course, required to be continuous in the cylinder, as was the solution in Example 1. The differential equation (10) is nonhomogeneous, because of the constant term q0, and this suggests that we apply the method of variation of parameters, first used in Sec. 33. To be specific, we know from Example 1 above that, without the term q0, the eigenfunctions = where J0(a3c) = 0 (as> 0), arise. Hence we seek a solution of the present boundary value problem having the form u(p,t) = j= 1 where the a3 are as just stated. Substituting this series into equation (10) and noting how the representation (0 <p <c) q0 = c 3=i immediately from the one obtained in Example 1, Sec. 68, we find that if series (12) is to satisfy equation (10), then follows + + f j=1 But, according to the theorem in Sec. 66, d2 + ld = SEC. 69 TEMPERATURES IN A LONG CYLINDER 281 Thus 00 2q 00 = + caf j=1 and, 1 by equating coefficients on each side of this equation, we arrive at the differential equation 2q (13) + (j = = 1,2,...). Furthermore, in view of the second of conditions (11), (0 <p <c). 0 j=1 Consequently, (14) = 0 (1 = 1,2,...). To solve the linear differential equation (13), we multiply each side by the integrating factor dt = exp exp This enables us to write the differential equation as = dt Ca3 After replacing t by T here, we then integrate each side from T = and recall condition (14). The result is 2q0 2 — = 0 to T = t 1), ck Finally, by substituting this expression for the coefficients (12), we arrive at the desired temperature formula: u( p, t) where J0(a1c) = 0 2q0 = ck (a'> 0). °° 1 — exp (— alJi( J0( a1p), into series 282 CHAP. 7 BESSEL FUNCTIONS AND APPLICATIONS HEAT TRANSFER AT THE SURFACE OF THE CYLINDER 70. Let us replace the condition that the surface of the infinite cylinder in Example 1, Sec. 69, be at temperature zero by the condition that heat transfer take place there into surroundings at temperature zero. As in Sec. 3, where Newton's law for surface heat transfer was discussed, the flux through the surface is assumed to be proportional to the difference between the temperature of the surface and that of its surroundings. That is, K is the thermal conductivity of the material of the cylinder and H is its surface conductance. The boundary value problem for the temperature function u(p, t) is now 1 (1) (O<p<c,t>O), p = —hu(c,t) (2) (t >0), u(p,O)=f(p) (O<p<c). We have written h = cH/K; and, for convenience, we allow the possibility that the otherwise positive constant h be zero. In that case, condition (2) simply states that the surface p = c is insulated. When u = R(p)T(t), separation of variables produces the eigenvalue (3) problem hR(c) + cR'(c) = 0 (0 <p <c). pR"(p) + R'(p) + ApR(p) = 0, If h > 0, the eigenvalues are, according to the theorem in Sec. 66, A3 = (j = 1, 2,...), where are the positive roots of the equation (5) hJ0(ac) + = 0. The corresponding eigenfunctions are = J0(a3p) (I = 1, 2,...); and, since T'(t) + itkT(t) = 0, we arrive at the products (j= 1,2,...). (4) The formal solution of our problem is, then, u(p,t) = (6) j=1 where, in view of expression (4), Sec. 68, 2a2 (7) = If h = R'(c) = 0. 2 0, 2 +h 2 210 dp (j = 1,2,...). the boundary condition in our eigenvalue problem becomes (I = 1, 2,...), In that case, the theorem in Sec. 66 tells us that A3 = SEC 70 PROBLEMS where a1 = = 0, 0 (j and 2, 283 3,...) are the positive roots of the equation or J1(ac) = (8) 0. The corresponding eigenfunctions are, moreover, R1 = 1 and R3 = J0(a3p) (j= 2, 3, .. ). Thus . u(p,t) (9) =A1 + j=2 referring to expressions (5) and (6) in Sec. 68, we see that the coefficients here are By 2c (10) A1 = (11) A1= (j=2,3,...). c[J0(a3c)]o A number of steady-state temperature problems in cylindrical coordinates, giving rise to Bessel functions, appear in the problems to follow. In those problems, the temperatures will continue to be independent of 4). The function u(p, z) will, then, be harmonic and satisfy Laplace's equation V2u = u 0, where (see Sec. 4) V2u = (12) + 1 + p PROBLEMS 1. Let u(p, t) denote the solution found in Example 1, Sec. 69, when c = 1 and f(p) = U0, where U0 is a constant. With the aid of the table at the end of Sec. 64, show that the first three terms in the series for U(p, t) are, approximately, as follows: u(p, t) = 2u0[O.80J0(2.4p)e58k1 — O.53J0(5.5p)e_30k1 +O.43J0(8.7p)e_76k1 — 2. ... ] Show that the solution of the temperature problem in Example 2, Sec. 69, can be written q0 4k where c 2 —p 2 8 c > 0). J0(a1c) = 0 Suggestion: Note that, according to Problem 4(b), c Sec. =—(c —p) 2 2 68, (0<p<c). 284 CHAP. 7 BESSEL FUNCFIONS AND APPLICATIONS 3. In Example 2, Sec. 69, suppose that the rate per unit volume at which heat is internally generated is q(t), rather than simply q0. Derive the following generalization of the solution found in that example: u(p,t) = where J0(a1c) = 0 2 2 fq(r)exp[_afk(t_T)]dr, — c j1 o (a1> 0). 4. Derive an expression for the steady temperatures u(p, z) in the solid cylinder formed by the three surfaces p = 1, z = 0, and z = 1 when u = 0 on the side, the bottom is insulated, and u = 1 on the top. J0(a1p) cosha3z = 0 (a3> 0). , where Answer: u(p, z) = 2 cosh a 5. Find the bounded steady temperatures u(p, z) in the semi-infinite cylinder p 1, z 0 when u = 1 on the base and there is heat transfer into surroundings at temperature zero, according to Newton's law (see Sec. 70), at the surface p = z>O. Answer: u(p, z) = 2h J0(a1p)exp(—a3z) L1 + h2) j=1 where — a3J1(a3) = 1, 0 (a3> 0). 1, z = 0, and z = b (b> 0). The side is insulated, the bottom kept at temperature zero, and the top at 6. (a) A solid cylinder is formed by the three surfaces p = temperatures f(p). Derive this expression for the steady temperatures u(p, z) in the cylinder: sinha.z 2z 1 1 u(p,z) = _1—f sf(s)ds + where a2, a3,... are the positive roots of the equation J1(a) = 0. (b) Show that when f(p) = 1 (0 <p < 1) in part (a), the solution there reduces to u(p, z) = z/b. 7. A function u(p, z) is harmonic interior to the cylinder formed by the three surfaces p = c, z = 0, and z = b (b > 0). Assuming that u = 0 on the first two of those surfaces and that u(p, b) = f(p) (0 <p <c), derive the expression sinha.z u(p, z) = b' sin h j=1 where a are the positive roots of the equation J0(ac) = 0 and the coefficients A3 are given by equation (8), Sec. 69. 8. Solve this Dirichlet problem (Sec. 7) for u(p, z): V2u=0 u(p,0) = u(1,z) = 0, and u is to be bounded in the domain p < 1, z> 0. Answer: u(p, z) = 2 j=1 a.J1(a1) exp (0<p<1,z>0), 1, where = 0 (as> 0). SEC. 70 PROBLEMS 285 9. Solve the following problem for temperatures u(p, t) in a thin circular plate with heat transfer from its faces into surroundings at temperature zero: (0 <p < 1, u(1,t) = 0, u(p,0) = 0), 1, where b is a positive constant. exp(—aJt), Answer: u(p, t) = 2 exp (—bt) j=1 where J0(a3) = 0 (a3> 0). 10. Solve Problem 9 after replacing the condition u(1, t) = 0 by this heat transfer condition at the edge: t) = —hu(1, t) (h > 0). 11. Give a physical interpretation of the following boundary value problem for a function u(p, t) (see Example 2, Sec. 69): (0 <p < 1, t) = 0, u(p, 0) = 0), ap2, where q0 and a are positive constants. Then, after pointing out why it is reasonable to seek a solution of the form u(p, t) = A1(t) + j=2 where (j = 2,3,...) are the positive roots of the equation J1(a3) = method of variation of parameters to actually find that solution. a J0(a3p) exp (— ait) Answer: u(p, t) = — + q0t + 4a 2 2 j=2 0, use the a are as stated above. 12. Interpret this boundary value problem as a temperature problem in a cylinder (see Sec. 3): Ut = = B, + (0 <p < 1, t > 0), up u(p,0) = 0, where B is a positive constant. Then, after making the substitution u(p,t) = U(p,t) + B to obtain a boundary value problem for U(p, t), refer to the solution in Problem 11 286 (HAP 7 BESSEL FUNCTIONS AND APPLICATIONS to derive the temperature formula J0(a.p)exp(—cr2t) B u(p,t)=— 2 crJJ()(crf) j=2 where a1 (j = 2, 3,...) are the positive roots of the equation J1(cr) = 0. [Note that the substitution for u(p, t) made here is suggested by the fact that t) = 0.] 13. Over a long solid cylinder p 1, at uniform temperature A, there is tightly fitted a long hollow cylinder 1 p 2 of the same material at temperature B. The outer surface p = 2 is then kept at temperature B. Let u(p, t) denote the temperatures in the cylinder of radius 2 SO formed, and set up the boundary value problem for those temperatures. Then, after making the substitution u(p, t) = U(p, t) + B to obtain a boundary value problem for U(p, t), refer to the solution in Example 1, Sec. 69, to derive the temperature formula u(p,t) =B + J1(a) A—B 2 j=1 where a1 are the positive roots of the equation J0(2a) = 0. (This is a temperature problem in shrunken fittings.) 14. Solve this boundary value problem for u(x, t): xu1 = — (0 <x <c, t > 0), —u (t>0), u(c,t)=O u(x,0) =f(x) where u is continuous for 0 c, t> x Answer: u(x, t) = 0 (0 <x <c), and where n is a nonnegative integer. exp ( —ait), where and are the con- j=1 stants in case (a) of Theorem 1 in Sec. 68. 15. Let u(p, z) denote a function which is harmonic interior to the cylinder formed by the three surfaces p = c, z = 0, and z = b (b > 0). Given that u = 0 on both the top and bottom of the cylinder and that u(c, z) = f(z) (0 <z < b), derive the expression u(p,z) 10(mrp/b) = niTz sin—s----, where = 2 b ThTZ [See Problem 15, Sec. 61, as well as the comments immediately following equation (17), Sec. 66, regarding the solutions of that modified form of Bessel's equation.] VIBRATION OF A CIRCULAR MEMBRANE SEC 71 the steady temperatures u(p, z) in a semi-infinite cylinder p base is insulated, be such that u(1, z) = f(z), where 1, z 16. Let f(z) 0, whose when0<z<1, (1 = 287 z>1. when With the aid of the Fourier cosine integral formula (Sec. 55), derive the expression 2 u(P,z)=_J IT a10(a) o cosazsinada for those temperatures. (See the remarks at the end of Problem 15.) 17. Given a function f(z) that is represented by its Fourier integral formula (Sec. 51) for all real z, derive the following expression for the harmonic function u(p, z) inside the infinite cylinder p c, —00 <z < 00 such that u(c, z) = f(z) (—00 <z < oo): u(p,z) (See 1 fQdI0(ap) = —J IT 0 10(ac) f(s)cosa(s —z)dsda. j the remarks at the end of Problem 15.) 71. VIBRATION OF A CIRCULAR MEMBRANE A membrane, stretched over a fixed circular frame p = c in the plane z 0, is given an initial displacement z = f(p, 4)) and released at rest from that position. The transverse displacements z(p, 4), t), where p, 4), and z are cylindrical coordinates, are described by the continuous function that satisfies this boundary value problem: 1 1 z11 = a2 + p p z(c,4),t)=0 (2) (3) + z(p,4),0) =f(p,4)), z1(p, 4), 0) = 0 (0 c, p 4) where the function z(p, 4), t) is periodic, with period 2ir; in the variable 4). A function z = R(p)1(4))T(t) satisfies equation (1) if T" a2T = 1 1 R p — R"+ —R' + 1 t" = —A where —A is any constant. We separate variables again in the second of = —jt. Then we find that the function equations (4) and write satisfies the homogeneous conditions and has the necessary periodicity with respect to 4) if R and 1 are eigenfunctions of the Sturm-Liouville problems p2R"(p) + pR'(p) + (Ap2 — (6) V'(4)) + = 0, = 0, = R(c) = 0, 288 and CHAP 7 BESSEL FUNCTIONS AND APPLICATIONS T is such that T"(t) + lta2T(t) = T'(O) = 0, 0. If p. has one of the values (n = 0,1,2,...), p. = the theorem in Sec. 66 can be applied to problem (5); and if we consider problem (6) first, we see that the constant p. must, in fact, have one of those values. For, according to Sec. 40, they are the eigenvalues of problem (6). To be precise, = when n = 0; and when n = 1,2,. . ., can be any linear combination of cos n4 and sin n4. From the theorem in Sec. 66, we now see that the eigenvalues of problem (5) are the numbers = (I = 1, 2,...), where are the positive roots of the equation (7) = (n = 0,1,2,...), 0 the corresponding eigenfunctions being R(p) = Then T(t) = cos The generalized linear combination of our functions (8) = + cos n4) + sin n4) cos n=1 j=1 formally satisfies all the homogeneous conditions. It also satisfies the condition and are such that z(p, 4), 0) = f(p, 4)) if the coefficients (9) f(p,4)) = cos n4)2 + + n=1 sin n4) j=1 j=1 For each fixed value of p, series (9) is the Fourier series for f(p, 4)) on the interval —i,- ii- if 4) = _f (n = 0,1,2,...), = —f (n = 1,2,...). j=1 00 For every fixed n, the series on the left-hand side of each equation here furnishes the Fourier-Bessel series representation, on the interval 0 <p <c, of SEC. 71 PROBLEMS 289 the corresponding function of p on the right-hand side, provided that (Sec. 68) 2 (10) = (11) = 2 10 f(p, 4)) cos n4 d4 dp, 2 f f(p, 4)) sin n4) d4 dp. 2 0 The displacements z(p, 4, t) are, then, given by equation (8) when the coefficients have the values (10) and (11). We assume, of course, that the function f is such that the series in expression (8) has adequate properties of convergence and differentiability. PROBLEMS 1. Suppose that in Sec. 71 the initial displacement function f(p, 4) is a linear combination of a finite number of the functions and cos n4, sin (n = 1,2,...). Point out why the iterated series in expression (8) of that section then contains only a finite number of terms and represents a rigorous solution of the boundary value problem. 2. Let the initial displacement of the membrane in Sec. 71 be f(p), a function of p only, and derive the expression z(p, t) = 2 J0(a1p)cosa1at cC [Ji(a3c)] C 2 J0 sf(s)J0(a3s) ds, where a3 are the positive roots of the equation J0(ac) = 0, for the displacements when t> 0. 3. Show that if the initial displacement of the membrane in Sec. 71 is AJO(akp), where A is a constant and ak is some positive root of the equation J0(ac) = 0, then the subsequent displacements are z(p,t) that these displacements are all periodic in t with a common period; thus the membrane gives a musical note. 4. Replace the initial conditions (3), Sec. 71, by the conditions that z = 0 and z1 = 1 Observe when t = 0. This is the case if the membrane and its frame are moving with unit velocity in the z direction and the frame is brought to rest at the instant t = 0. Derive the expression 2 z(p,t) = — ac sin a1at 2 where a3 are the positive roots of the equation J0(ac) = when t> 0. 0, for the displacements 290 CHAP. 7 BESSEL FUNCTIONS AND APPLICATIONS 5. Suppose that the damped transverse displacements z(p, t) in a membrane, stretched over a circular frame, satisfy the conditions = z(1,t) = The + z(p,0) 0, (0 <p < — 2bz1 z1(p,0) = = 0, 1, t > 0), v0. constant coefficient of damping 2b is such that 0 <b <a1, where a1 is the smallest of the positive zeros of J0(a). Derive the solution sin z(p,t)=2v0e —bt tIa2_b2 y 3 L1 — b2 where .J0(a3) = 0 (a3> 0), of this boundary value problem. 6. Derive the following expression foi the temperatures u(p, 4), t) in an infinite cylinder p c when u = 0 on the surface p = c and u = f(p, 4)) at time t = 0: = + cos n4) + sin n4)) exp n=1 j=1 are the numbers defined in Sec. 71. and where c, in a solid cylinder p ir whose entire surface is kept at temperature zero and whose initial temperature is a constant A. Show that it can be written as the product 7. Derive an expression for the temperatures u(p, z, t) o z u(p, z, t) = Av(z, t)w(p, t) of A and the functions v(z,t) — sin(2n—1)z exp[—(2n 2n—1 2 J0(a3p) 4 = 00 — 1) 2 kt] and w(p,t) where a3 are — c j=1 a3 1(a3c) roots of the equation J0(ac) = 0. Also, show that v(z, t) 'r and w(p, t) temperatures in an infinite z c, both with zero boundary temperature and unit initial temperature the positive represents temperatures in a slab 0 cylinder p (see Example 1, Sec. 32, and Example 1, Sec. 69). 8. Derive the following expression for temperatures u(p, 4), t) in the long right-angled cylindrical wedge formed by the surface p = 1 and the planes 4) = 0 and 4) = when u = 0 on its entire surface and u = f(p, 4)) at time t = 0: u(p,4),t) = E n=1 j=1 PROBLEMS SEC. 71 are the positive roots of the equation where = = 0 291 and sin 2n4)f'pf(p, dp 9. Show that if the plane 4) = 'r/2 in Problem 8 is replaced by a plane 4) = the expression for the temperatures in the wedge will, in general, involve Bessel functions of nonintegral orders. 10. Solve Problem 8 when the entire surface of the wedge is insulated, instead of being kept at temperature zero. 11. Solve the boundary value problem 1 n2 p p u(1, z) = (0<p<1,z>0), u(p,0) = 0, where u(p, z) is bounded and continuous for 0 p < 1, z > 0 and where n is a positive integer. (When n = 0, this problem becomes the Dirichlet problem that was solved in Problem 8, Sec. 70.) Answer: u(p, z) = where exp 2 = 0 (ai> 0). j=1 12. Let the function u(p, 4), z) satisfy Poisson's equation (Sec. 3) V2u + ay = 0, where a is a constant, inside a semi-infinite half cylinder 0 p 1, 0 4) 'r, z 0, and suppose that u = 0 on the entire surface. The function u, which is assumed to be bounded and continuous for 0 p < 1, 0 <4 <jr, z > 0, thus satisfies the boundary value problem 1 + 1 + p + apsin4) = + (0 <p <1,0< 4 <jr, z >0), 0 p u(1,4),z) = 0, u(p,4),O) = 0, u(p,0,z) = = 0. Use the following method to solve it. (a) By writing u(p, 4), z) = a sin 4) v(p, z), reduce the stated problem to the one 1 1 p p v(1, z) = (0<p<1,z>0), v(p,0) = 0, 0 in v(p, z), where v is bounded and continuous for 0 p < 1, z > 0. (b) Note how, when n = 1, the solution in Problem 11 suggests that the method of variation of parameters (see Example 2, Sec. 69) be used to seek a solution of the form v(p,z) = j=1 where = 0 (a3 > 0), for the problem in part (a). Apply that method to obtain the initial value problem — = 2 — , A3(0) = 0 292 BESSEL FUNCTIONS AND APPLICATIONS CHAP. 7 in ordinary differential equations. Then, by adding a particular solution of this differential equation, which is a constant that is readily found by inspection, to the general solution of the complementary equation = 0 (com— pare Problem 13, Sec. 38), find v(p, z). Thus arrive at the solution 1—exp(—a.z) j=1 0 (as> 0), of the original problem. Suggestion: In obtaining the ordinary differential equation for A.(z) in part (b), one can write the needed Fourier-Bessel expansion for p by simply referring to the expansion already found in Problem 9, Sec. 68. Also, it is necessary to observe how the identity where J1(cr1) = d2 + id — follows immediately from that problem. 1 — = CHAPTER 8 LEGEND RE POLYNOMIALS AND APPLICATIONS As we shall see later in this chapter (Secs. 77 and 78), an application of the method of separation of variables to Laplace's equation in the spherical coordi- nates r and 0 leads, after the substitution x = cos 0 is made, to Legendre's equation + Ay(x) = 0, [(1 — (1) where A is the separation constant. The points x = 1 and x = —1 correspond respectively, and we begin the chapter by using series to to 0 0 and 0 = discover solutions of equation (1) that can be used when —1 x 1. SOLUTIONS OF LEGENDRE'S EQUATION 72. To solve Legendre's equation, we write it as (1 — x2)y"(x) — 2xy'(x) + Ay(x) = and observe that, while x = ±1 are singular points, x = 0 0 is an ordinary point. 293 294 CHAP. 8 LEGENDRE POLYNOMIALS AND APPLICATIONS We thus seek a solution of the formt y= (2) j=o Substitution of series (2) into equation (1) yields the identity Since [j(j -1) + 2j - - = 0. j=O j=O the first two terms in the first series here are actually zero and since j(j - + 2j 1) =j(j + 1) in the second series, we may write - 1) -A]a1x3 = 0. j=O j=2 Finally, by putting the first of these series in the form (I + 2)(j + j=O we arrive at the equation (3) j=O {(j + 2)(j + l)aJ+2 - [j(j + 1) - = 0, involving a single series. Equation (3) is an identity in x if the coefficients a3 satisfy the recurrence relation j(j+ 1) -A (4) (j=O,1,2,...). The power series (2) thus represents a solution of Legendre's equation within its interval of convergence if its coefficients satisfy relation (4). This leaves a0 and a1 as arbitrary constants. If = 0, it follows from relation (4) nontrivial (5) = 0. Thus one that a3 = a5 = solution of Legendre's equation, containing only even powers of x, is y1 = a0 + (a0 * 0), k=1 where a0 is an arbitrary nonzero constant and where the remaining coefficients a2, a4,... are expressed in terms of a0 by successive applications of relation (4). tFor a discussion of ordinary points and a justification for this substitution, see, for example, the books referred to earlier in the footnote in Sec. 59. SEC 73 LEGENDRE POLYNOMIALS 295 Problem 8, Sec. 75.) Another solution, containing only odd powers of x, is obtaining by writing a0 = 0 and letting a1 be arbitrary. More precisely, the (See series (a * 0) y2 = a1x + (6) k= 1 satisfies Legendre's equation for any nonzero value of a1 when a3, a5,... are written in terms of a1 in accordance with relation (4). These two solutions are, of course, linearly independent since they are not constant multiples of each other. From relation (4), it is clear that the value of A affects the values of all but the first coefficients in series (5) and (6). As we shall see in Sec. 73, there are certain values of A that cause series (5) and (6) to terminate and become polynomials. Assuming for the moment that series (5) does not terminate, we note from relation (4), with I = 2k, that lim k-4oo a2(k+l)x 2(k+1) —1- 1\ — = lim a2kx A (2k + 2)(2k + 1) x2 =x2. So, according to the ratio and absolute convergence tests, series (5) converges when x2 < 1 and diverges when x2> 1. Although it is somewhat more difficult to show, series (5) diverges when x = ± Similar arguments apply to series (6). In summary, then, if A is such that either of the series (5) or (6) does not terminate and become a polynomial, that series converges only when — 73. 1 <x < 1. LEGENDRE POLYNOMIALS When Legendre's equation (1 — x2)y"(x) 2xy'(x) + Ay(x) = — 0 arises in the applications, it will be necessary to have a solution which, along with its derivative, is continuous on the closed interval — 1 x 1. But we know from Sec. 72 that, unless it terminates, neither of the series solutions (1) y1 = a0 (a0 * 0), + k=1 y2 = a1x + a2k+lx k=1 obtained there satisfies those continuity conditions. tSee for instance, the book by Bell (1968, pp. 230—231), listed in the Bibliography. (a1 * 0) 296 CHAP 8 LEGENDRE POLYNOMIALS AND APPLICATIONS Suppose now that the parameter A in Legendre's equation has one of the integral values (n = 0,1,2,...), A = n(n + 1) (3) in which case the recurrence relation (4), Sec. 72, becomes j(j+ 1) -n(n+ 1) (j+2)(j+1) (4) (j=0,1,2,...). Since = 0 and, consequently, = = = 0, it follows that one of the solutions y1, y2 is actually a polynomial. Note that if n = 0, then a2 = a4 = a6 = = 0; and series (1) becomes simply y1 = a0. If, moreover, ii is any one of the even integers 2,4,..., so that 0. Series n = 2m (m = 1,2,...), then a2m 0 and a2(m+1) = a2(m+2) (1) thus reduces to a polynomial whose degree is 2m, or n. On the other hand, if n = 1, we see that y2 = a1x; and if n is any one of the odd integers n = 2m + 1, then a2m+1 * 0 and a2(m+1)+1 = series (2) becomes a polynomial of degree n a2(m+2)+1 = 0. Hence if n is odd. Thus, if A has any one of the values (3), solution (1) reduces to the polynomial (5) y1 = a0 + a2x2 + * 0) when n is even; and solution (2) becomes y2 = a1x + a3x3 + (6) * 0) when n is odd. The coefficients a0 and a1 are arbitrary nonzero constants, and the others are determined by successive applications of relation (4). Observe that when n is even, solution (2) remains an infinite series and that when n is odd, the same is true of solution (1). If n is even, it is customary to assign a value to a0 such that when the in expression (5) are determined by means of relation (4), coefficients a2,.. . , the final coefficient a = The reason for this requirement is that the polynomial (5) will then have the value unity when x = 1, as will be shown in Sec. 75. The precise value of a0 that is needed is not important to us here. Using the convention that 0! = 1, we note that a0 = 1 if n = 0. In that case, y1 = 1. If n is odd, we choose a1 so that the final coefficient in expression (6) is also given by equation (7). The reason for this choice is similar to the one above regarding the value assigned to a0. Note that y2 = x if n = 1, since a1 = 1 for that value of n. When n = 2, 3, . , relation (4) can be used to write all the coefficients that precede To accomplish this, we in expressions (5) and (6) in terms of . . LEGENDRE POLYNOMIALS SEC. 73 297 first observe that the numerator on the right-hand side of relation (4) can be written j(j+1)-n(n+1)=_[(n2_j2)+(n_j)]=_(n_j)(n+j+1). the result is We then solve for (8) - a3 (j+2)(j+1) — (n —j)(n +1 (n)(n 1) (n—2)(n—3) = (4)(2n — 3) — afl_2k — 1) (2)(2n — = 1)aJ+2. we now use relation (8) to write the following in terms of To express k equations: + (n—2k+2)(n—2k+1) = (2k)(2n — — 2k + afl_2k+2. 1) Equating the product of the left-hand sides of these equations to the product of their right-hand sides and then canceling the common factors • . on , each side of the resulting equation, we find that (—1) afl_2k = 2kk! k (2n — 1)(2n — 3) (2n — 2k + 1) into equation (9) and combining Then, upon substituting expression (7) for various terms into the appropriate factorials (see Problem 5, Sec. 75), we arrive at the desired expression: 1 afl_2k (10) As usual, 0! = (—1)" = k! (2n—2k)! (n — 2k)!(n — 1. In view of equation (10), the polynomials (5) and (6), when the nonzero has values (7), can be written constants a0 and a1 are such that 1 (11) = where m = for m (1)k k! (n (2n—2k)! — 2k)!(n — n/2 if n is even and m = (n — 1)/2 if n is odd. will be given in Sec. 75. Note that since = 0,1,2,...), Another expression is a polynomial containing only even powers of x if n is even and only odd powers if n is odd, it 298 CHAP 8 LEGENDRE POLYNOMIALS AND APPLICATIONS is an even or an odd function, depending on whether n is even or odd; that is, (12) (n = 0,1,2,...). = The polynomial is called the Legendre polynomial of the first several values of n, expression (11) becomes P0(x) = P3(x) = P5(x) = Observe 1, P1(x) =x, — 1), P4(x) = — 3x), — P2(x) = degree n. For — 30x2 + 3), 70x3 + 15x). that the value of each of these six polynomials is unity when x = 1, as anticipated. See Fig. 65, where the first four are displayed graphically on the interval — 1 x 1. x FIGURE 65 We have just seen that Legendre's equation (1 —x2)y"(x) — 2xy'(x) + n(n + 1)y(x) = (13) always has the polynomial solution y = 0 (n = 0,1,2,...) which is solution (5) (n even) or solution (6) (n odd) when appropriate values are assigned to the arbitrary constants a0 and a1 in those solutions. Details regarding the standard form of and is called a the accompanying series solution, which is denoted by Legendre function of the second kind, are left to the problems. We, of course, know from the statement in italics at the end of Sec. 72 that the series is convergent only when —1 <x < 1. It will, however, be representing fail to be a pair of continuous sufficient for us to know that and functions on the closed interval — 1 x 1 (Problem 9, Sec. 76). Since are linearly independent, the general solution of equation (13) is and C1 C2 y= are arbitrary constants. ORTHOGONALITY OF LEGENDRE POLYNOMIALS SEC 74 299 ORTHOGONALITY OF LEGENDRE POLYNOMIALS 74. Let X(x) denote the dependent variable in Legendre's equation, with arbitrary A: (1) (1 — x2)X"(x) — 2xX'(x) + AX(x) = 0. Writing this equation in its self-adjoint form (Sec. 41) [(1 — (2) + AX(x) = 0, we see that we have a special case of the Sturm-Liouville differential equation [r(x)X'(x)]' + [q(x) + Ap(x)]X(x) =0, where p(x) = 1, q(x) = 0, and r(x) = 1 — x2. The function r(x) vanishes at x = ±1; thus, as already pointed out in Example 2, Sec. 42, equation (2) here, without boundary conditions, is a singular Sturm-Liouville problem on the interval —1 x 1, where X and X' are required to be continuous on that closed interval. The following theorem provides us with all the solutions of this problem. Theorem. The eigenvalues and corresponding eigenfunctions of the singular 1, are Sturm-Liouville problem (2), on the interval — 1 x where (n = 0,1,2,...), A,1 =n(n + 1), (3) the are the Legendre polynomials. and are We start the proof by recalling from Sec. 73 that linearly independent solutions of equation (2) when A has any one of the values and its derivative are A = n(n + 1) (n = 0, 1, 2,...). Since the polynomial continuous on the entire interval — 1 x 1 and since this is not true of the it is clear that the continuity requirements on X and Legendre function Hence the and X' are met only when X is a constant multiple of in the statement of the theorem are, in fact, eigenvalues and eigenfunctions. It remains to show that there are no other eigenvalues. To accomplish this, we digress for a moment and observe that, since the eigenfunctions just noted all correspond to different eigenvalues, the set (n = 0, 1,2,...) is orthogonal on the interval —1 p(x) = 1. (See the theorem in Sec. 43.) That is, <x < 1, with weight function (4) In the notation used for inner products, property (4) reads (em' = 0 (m n). Later on (Sec. 76), there will be a theorem telling us that if a function f is piecewise smooth on the interval — 1 <x < 1, then the generalized Fourier 300 CHAP. 8 LEGENDRE POLYNOMIALS AND APPLICATIONS series for f with respect to the orthonormal set of functions = (5) P(x) (n = 0,1,2,...) n converges to f(x) at all but possibly a finite number of points in the interval — 1 <x < 1. The set is, therefore, closed (Sec. 12) in the function space C,( — 1, 1) (see Sec. 17). Suppose now that A is another eigenvalue, different from those listed in the statement of the theorem, and let X denote an eigenfunction corresponding to A. Because of the orthogonality of eigenfunctions corresponding to distinct are those in = 0 (n = 0, 1,2,...), where the functions eigenvalues, (X, equation (5). But the fact that is closed requires that X, which is continuous on the entire interval — 1 x 1, have value zero for each x in that interval. Consequently, since an eigenfunction cannot be identically zero, X is not an eigenfunction. In view of this contradiction, there are no other eigenvalues; and the proof of the theorem is finished. If the interval 0 x 1, rather than — 1 x 1, is used, the differential equation (2) along with either one of the boundary conditions X'(O) = X(0) = 0 0, is also a singular Sturm-Liouville problem (Sec. 42). Corollary. The eigenvalues and corresponding eigenfunctions of the singular Sturm-Liouville problem consisting of the differential equation (2), on the interval o <x < 1, and the boundary condition X'(O) = 0 are (n=0,1,2,...). (6) If the condition X(0) = (7) 0 is used instead, the eigenvalues and eigenfunctions are = (2n + 1)(2n + 2), (n = 0,1,2,...). To see how these solutions follow, we consider first the solutions in the =0 theorem when the condition X'(O) = 0 is imposed on them. Since only when n is an even integer (Problem 7, Sec. 75), the polynomials + 1(x) (n = 0, 1, 2,...) must be eliminated. This leaves the eigenvalues and eigenfunctions (6). If, on the other hand, the condition X(0) = 0 is imposed, the fact that = 0 only when n is an odd integer leads us to the eigenvalues and eigenfunctions (7). The theorem in Sec. 43, regarding the orthogonality of eigenfunctions, tells us that each of the sets (n = 0, 1,2,...) and (n = 0, 1,2,...) is orthogonal on the interval 0 <x < 1 with weight function unity. That is, (8) f1P2m(X)P2n(X)d)C0 RODRIGUES' FORMULA AND NORMS SEC. 75 301 and (9) f1P2m+i(X)P2n+i(X)dr=O where m = 0, 1, 2,... and n = 0, 1, 2 Valid representations of piecewise smooth functions on the interval 0 <x < 1 will follow (Sec. 76) from representations on the interval — 1 <x < 1 in terms of the set (n = 0, 1, 2,...), just as Fourier cosine and sine series follow from Fourier series involving both cosines and sines. Hence the same argument, involving closed sets, that was used in the proof of the theorem above can be used to show that there are no other eigenvalues of the Sturm-Liouville problems in the corollary. 75. RODRIGUES' FORMULA AND NORMS According to expression (11), Sec. 73, 1 (1) m — = k!(n — k)! k=O where m = n/2 if n is even and m = (n d'2 (2n—2k)! n! 1)k 2n—2k — — — (n — 2k)! xn2k, 1)/2 if n is odd. Since (2n—2k)! (n—2k)! n—2k '0 < k < and because of the linearity of the differential operator d'1/dx'1, expression (1) can be written '\ ) " The — dnm 1 ( 2'1n! dx's k=O n! k!(n 2n—2k — powers of x in the sum here decrease in steps of 2 as the index k increases; and the lowest power is 2n — 2m, which is n if n is even and n + 1 if n is odd. Evidently, then, the sum can be extended so that k ranges from 0 to n. For the additional polynomial that is introduced is of degree less than n, and its nth derivative is, expansion of (x2 — therefore, zero. Since the resulting sum is the binomial 1)nl, it follows from equation (2) that 1 = (3) — (n = 0,1,2,...). formula for the Legendre polynomials. Various useful properties of Legendre polynomials are readily obtained from Rodrigues' formula with the aid of Leibnitz' rule for the nth derivative Dnl[f(x)g(x)] of the product of two functions: This is Rodrigues' n = k=O k'.(n— k ). 302 CHAP. 8 LEGENDRE POLYNOMIALS AND APPLICATIONS where it is understood that all the required derivatives exist and that the zero-order derivative of a function is the function itself. We note, for example, that if we write u = x2 — 1, so that it follows from Leibnitz' rule that = k=O + 1)n]Dn_k[(x k!(n— — Now the first term in this sum is D0[(x + = (x + — and the remaining terms all contain the factor (x — 1) to some positive power. Hence the value of the sum when x = 1 is 2'1n!, and it follows from Rodrigues' formula (3) that (n = 0,1,2,...). (5) = 1 Observe how it follows from this and the relation (n = 0, 1, 2,...), obtained in Sec. 73, that —x) = (— (6) (n = 0,1,2,...). = For another application of Rodrigues' formula (3), we use it to write where u = x2 — 1. But = 2(n + 1)xu'2, and so = 2(n + 1)(u'2 + = 2(n + + 2n(x2 + — = 2(n + 1)[(2n + 1)u'2 + Consequently, = (2n + Substituting — 1(n + here, we find that 1 — 2n + 1 (7) = 2'2n! On the other hand, Leibnitz' rule (4) enables us to write Pfl+1(x)= = + 1)! D'2(xu'2) 2'2n! = xD'2u'2 + 2'2n! SEC 75 and, RODRIGUES' FORMULA AND NORMS 303 since D"u" = 2n!DU* (8) between this and equation (7) gives the recurrence Elimination of relation (n + (9) + = (2n + (n = 1,2,...). = (2n + (n Note, too, that the relation (10) 1,2,...) is an immediate consequence of equation (7). We now show how relation (9) and its form (11) + (n (n = 2,3,...), can be used to find the norms = = (2n — obtained by replacing n by n — 1, — of the orthogonal polynomials Keeping in mind that = 0, we find from equations (9) and (11), = 0 and respectively, that (Pa, = (2n + (12) and = (2n — and are identical, and we The integrals representing need only eliminate those quantities from equations (12) and (13) to see that (2n + = (2n — or = (2n (2n + - (n = 2,3,...). It is easy to verify directly that equation (14) is also valid when n 1. Next, we let n be any fixed positive integer and use equation (14) to write the following n equations: (2n + (2n 2 — = (2n — (2n — 2 (5)11P2112 = (3)11P1112, (3)11P1112 = (1)11P0112. Setting the product of the left-hand sides of these equations equal to the product of their right-hand sides and then canceling appropriately, we arrive at the result (2n + = 11P0112 (n = 1,2,...). 304 LEGENDRE POLYNOMIALS AND APPLICATIONS Since 11P0112 = 2, this means that (15) IIP,jI CHAP. 8 /2 = V 2n + (n = 0,1,2,...). 1 The set of polynomials 12n + =V 2 (16) 1 (n = 0,1,2,...) is, therefore, orthonormal on the interval — 1 <x < 1. PROBLEMS 1. From the orthogonality of the set state why = O(n = 1,2,...); (a) (b) f1(Ax + dx = 0 (n = 2,3,...), where A and B are constants. 2. Verify directly that the Legendre polynomials P0(x) = 1, P2(x) = P1(x) = x, — form an orthogonal set on the interval —1 <x < 1), 1. P3(x) = — 3x) Show that their graphs are as indicated in Fig. 65 (Sec. 73). 3. Use the fact that the set defined by equation (16), Sec. 75, is orthonormal on the interval — 1 <x < 1 to show that the following sets are orthonormal on the interval 0 <x < 1: (a) {V4n + (b) {V4n + (n = 0,1,2,...); (n = 0,1,2,...). Suggestion: Use the suggestion with Problem 2, Sec. 11, modified so as to apply when the even function f there is defined on the interval — 1 <x < 1. 4. From recurrence relation (10), Sec. 75, obtain the integration formula f1Pn(x)dx = (n = 1,2,...). 2 1 5. Give details showing how expression (10) in Sec. 73 for the coefficients afl_2k in the Legendre polynomials are obtained from equations (7) and (9) there. Suggestion: Observe that the factorials in equation (7), Sec. 73, can be written (2n)!= (2n)(2n — 1)(2n — 2) (2n — 2k + 1)(2n — 2k)!, n!=n(n— 1)"(n—2k+ 1)(n—2k)!, n!= n(n — 1) ... (n — k + 1)(n — k)!. the aid of expression (11), Sec. 73, for show that when n = 2,3,. . ,the constants a0 and a1 in equations (5) and (6) in that section must have the following 6. With . PROBLEMS SEC. 75 values 305 to have the value specified in equation (7) in order for the final constant there: (-1) ( 7. 1 n/2 (1)(3)(5) (n • — 1) (n = 2,4,...), (2)(4)(6) "(n) (n-1)/2 (1)(3)(5) (n) (n —- (n-i) (2)(4)(6) Establish these properties of Legendre polynomials, where n = (2n)! (a) 2 2n (n!) 2' 0; (b) 0, = (c) , ,...). 1,2,... 0; = (2n + (d) Suggestion: For parts (a) and (d), refer to Problem 6. 8. Legendre's equation (1), Sec. 72, is often written (1 — x2)y"(x) + — + 1)y(x) = 0, where i' is an unrestricted complex number. Show that when A = rence relation (4), Sec. 72, can be put in the form — + 1), recur- (j = 2,3,...). (f — 1) a3 = i.'(i.' Then, by proceeding as we did in solving Bessel's equation (Sec. 59), use this relation to obtain the following linearly independent solutions of Legendre's equation: y1 = (1)k + k=1 — 2) ... — 2k + + 3) ... + + 2k — 1)]2k (2k)! (—1) k=1 — — 3) ... + 2)(v + 4) ... (v + — 2k + 2k)]2kl (2k+1)! 9. where a0 and a1 are arbitrary nonzero constants. (These two series converge when — 1 <x < 1, according to Sec. 72.) Show that if i.' is the complex number i (a>0), Legendre's equation in Problem 8 becomes (1 — x2)y"(x) — + 2xy'(x) a2)y(x) = 0. — Then show how it follows from the solutions obtained in Problem 8 that the 306 LEGENDRE POLYNOMIALS AND APPLICATIONS CHAP. 8 functions 2 1 a2 pa(X) = + 2 + k1 a2 + + 2 =x + k=1 ( a2 a2 + + 2 + ( 2 x2k (2k)!' ) 4k—i 2 a2 2 4k—3 a2 2 ) (2k + i)! are linearly independent solutions of this differential equation, valid on the interval — i <x < 1. These particular Legendre functions arise in certain boundary value problems in regions bounded by cones. 10. Note that the solutions y1 and y2 obtained in Problem 8 are solutions (5) and (6) in + i). They remain infinite series when ii = n = i, 3,5,... and Sec. 72 when A = = n = 0,2,4,. , respectively. When ii = n = 2m (m = 0, 1,2,...), the Legendre of the second kind is defined as y2, where function . . = (_i)rn22m(m!)2 a1 and when i.' = n = 2m + 1 (2m)! (m = 0, i, 2,...), is defined as y1, where !)2 a0 (2m+1)! Using the fact that i+x x2k 00 i—x (—i<x<i) k=O2k+i show that Q0(x) = i i+x Q1(x) = and 11. Use mathematical induction on the integer n 12. Write F(x, t) = (i where lxi — i+x x to —1 =xQ0(x) — i. verify Leibnitz' rule (4), Sec. 75. 2xt + i and t is as yet unrestricted. cos 0 for some uniquely determined value of 0 (a) Note that x = (0 0 'r), and show that F(x,t) = (i — e'°t) — Then, using the fact that (1 — zY 1/2 has a valid Maclaurin series expansion when Izi < i, point out why the functions (i — e ± '°t) 1/2, considered as functions of t, can be represented by Maclaurin series which are valid when ti < i. It follows that the product of those two functions also has such a representation LEGENDRE SERIES SEC 76 when < (n = That is, there are functions 0, 1, 2, .) F(x, t) = such 307 that < 1). n=O (b) Show that the function F(x, t) satisfies the identity aF (1 — 2xt + = (x and use this result to show that the functions — t)F, in part (a) satisfy the recurrence relation (n + (n = 1,2,...). = (2n + + (c) Show that the first two functions f0(x) and f1(x) in part (a) are 1 and x, respectively, and notice that the recurrence relation obtained in part (b) can Compare that relation with then be used to determine f,,(x) when n = 2, 3 are, in fact, the relation (9), Sec. 75, and conclude that the functions Legendre polynomials P,7(x); that is, show that (1 — 2xt + < 1). = The function F is thus a generating function for the Legendre polynomials. = 1 (n = 0, 1, 2,...), using 13. Give an alternative proof of the property (Sec. 75) (a) recurrence relation (9), Sec. 75, and mathematical induction; (b) the generating function obtained in Problem 12(c). 76. LEGENDRE SERIES In Sec. 75, we saw that the set of polynomials /2n + 1 = V (n = 0,1,2,...) 2 is orthonormal on the interval — 1 <x < 1. The Fourier constants (Sec. 12) with respect to that set, for a function f defined on the interval — 1 <x < 1, are /2n+1 1 2 and the generalized Fourier series corresponding to f(x) is = £ 2n+ ds. For a discussion of this point, see, for example, the authors' book (1990, pp. 161—162), listed in the Bibliography. 308 LEGENDRE POLYNOMIALS AND APPLICATIONS CHAP. 8 That is, (2) (-1<x<1), n=O where 2n+1 (3) 1 (n=O,1,2,...). 2 Series(2), with coefficients (3), is a Legendre series. We state here, without proof, a representation theorem that is applicable to piecewise smooth functions.t Theorem. Let f denote a function that is piecewise smooth on the interval —1 <x < 1, and suppose that f(x) at each point of discontinuity of f in that interval is defined as the mean value of the one-sided limits f(x +) and f(x —). Then (-1<x<1), f(x)= (4) n=O where the coefficients are given by equation to Sec. 73, (n = 0, 1,2,...) is odd. That is, (n = According and 0, (3). 1,2,...) is even and = Evidently, then, if the function f in the statement of theorem is even, the product + 1(x) is odd and the graph of y = + 1(x) is symmetric with respect to the origin. On the other hand, is even, and the graph of y = is symmetric with respect to the y axis. Consequently, dx = 0 and dx = dx. Hence it follows from expression (3) that the coefficients in representation (4) become = 0 (n = 0,1,2,...) and (n=0,1,2,...). Thus if we apply the theorem to the even extension of a function f that is piecewise smooth on the interval 0 <x < 1 and whose value f(x) at each point proof, which is rather lengthy, can be found in, for example, the book by Kreider, Kuller, Ostberg, and Perkins (1966, pp. 425—432), listed in the Bibliography. A simplified proof of a special case of the theorem appears in the book by Rainville (1971, pp. 177—179), also listed there. LEGENDRE SERIES SEC. 76 309 of discontinuity is the mean value of f(x +) and f(x —), we find that f(x) = (6) (0 <x < 1), n=O have the values (5). the coefficients where Similarly, if + 1(x) f an odd is function, the products and are odd and even, respectively. It is then easy to show that when the value f(x) of a piecewise smooth function f on 0 <x < 1 is defined as the mean value of f(x +) and f(x —) at each point of discontinuity, f(x) has the representation f(x) = (7) (0 <x < 1), where (n=0,1,2,...). EXAMPLE. Let us expand the function f(x) = 1 (0 <x < 1) in a series of type (7), involving Legendre polynomials of odd degree. According to expression (8), (n=0,1,2,...). The integral here is readily evaluated with the aid of the integration formula (Problem 4, Sec. 75) dx = (n = 1,2,...), — 2 1 which tells us that (9) + 1)(0). — +1= Thus (10) 1= (0 — <x < 1). n=O Since [Problem 7(a), Sec. 75] (2n)! 2 2 (n!) = the 2n coefficients (9) can also be written as 4n+3 = + (2n)! 2) 22n(n!)2 (n = 0,1,2,...), 310 LEGENDRE POLYNOMIALS AND APPLICATIONS CHAP. 8 This alternative form of representation (10) is then obtained: 4n+3 00 (12) 1= (2n)! <x <1). (0 + 2)22flfl!)2P2n+1x) PROBLEMS 1. Let F denote the odd extension of the function f(x) = 1 (0 <x < 1) to the interval —1 <x < 1, where F(0) = 0. Also, let g be the function defined by means of the equations g(x) and g(0) = = when —1 <x <0, (0 O<x<1, when Then, by observing that g(x)= 1 + 1 (-1<x<1) and referring to expansion (10), Sec. 76, show that g(x) = (—1 <x <1). + 2. Let f denote the function defined by the equations f(x) = (0 when 0<x<1. when (a) State why f(x) is represented by its Legendre series (4), Sec. 76, at each point of the interval —1 <x < 1. (b) Show that + = 0 (n = 1, 2,...) in the series in part (a). (c) Find the first four nonvanishing terms of the series in part (a) to show that 1 f(x) = 1 1 + + 5 3 + — (—1 <x< 1). 3. Verify that, for all x, (a) x2 = 1 + 2 (b) = 2 3 + 4. Obtain the first three nonzero terms in the series of Legendre polynomials of even degree representing the function f(x) = x (0 <x < 1) to show that x= 1 3 5 + —P2(x) + — Point out why this expansion remains valid when x = series represents on the interval — 1 <x < 1. 0, (0 <x <1). and state what function the 5. 311 PROBLEMS SEC. 76 By applying the corollary in Sec. 16 to the Fourier constants + f 1 in Sec. 76, state why f is piecewise continuous on the interval —1 <x < 1. 6. Let f denote a function that is piecewise smooth on the interval 0 <x < 1, and suppose that f(x) at each point of discontinuity there is the mean value of the one-sided limits f(x +) and f(x —). (a) By finding the Fourier constants for f with respect to the orthonormal set +1 (n = 0, 1,2,...) [Problem 3(a), Sec. 75], derive the coefficients (5), Sec. 76, appearing in expansion (6) in that section. (b) Apply the corollary in Sec. 16 to the Fourier constants in part (a) to show that (compare Problem 5) dx = lim%14n + 1 0. 7. (a) By recalling that Pm(X) is a polynomial of degree m containing only alternate powers of x (Sec. 73), state why X m_ DI \ — C Cm_2X m—2 Cm_4X m—4 where the coefficients are constants. Apply the same argument to etc., to conclude that Xm is a finite linear combination of the polynomials Pm(X), Pm_2(X), Pm_4(X) (b) With the aid of the result in part (a), point out why dx = 0, where is a Legendre polynomial of degree n polynomial whose degree is less than n. (n = 1, 2,...) and p(x) is any 8. Let n have any one of the values n = 1,2 (a) By recalling the result in Problem 1(a), Sec. 75, state why must change sign at least once in the open interval —1 <x < 1. Then let x1, x2,. . ., Xk denote the changes sign. Since any totality of distinct points in that interval at which polynomial of degree n has at most n distinct zeros, we know that 1 k n. in part (a) is such that k <n, (b) Assume that the number of points x1, x2,. . , and consider the polynomial . p(x) = (x -xi)(x Use the result in Problem 7(b) to (x -Xk). X2) show that the integral dx has value zero; and, after noting that and p(x) change sign at precisely the same points in the interval — 1 <x < 1, state why the value of the integral cannot be zero. Having reached this contradiction, conclude that k = n and hence that the zeros of a interval — 1 Legendre polynomial <x < 1. are all real and distinct and lie in the open 312 CHAP 8 LEGENDRE POLYNOMIALS AND APPLICATIONS 9. Show in the following way that, for each value of n (n = 0, 1,2,...), the Legendre function of the second kind (Sec. 73) and its derivative Q',1(x) fail to be a pair of continuous functions on the closed interval —1 x 1. Suppose that there is an integer N such that QN(x) and Q'N(x) are continuous on that interval. The functions (n * N) are, then, eigenfunctions corresponding to different eigenQN(x) and values of the Sturm-Liouville problem (2), Sec. 74. Point out how it follows that =0 (n * N), and then use the theorem in Sec. 76 to show that QN(x) = ANPN(x), where AN some constant. This is, however, impossible since PN(x) and QN(x) are linearly independent. DIRICHLET PROBLEMS IN SPHERICAL REGIONS 77. For our first application of Legendre series, we shall determine the harmonic function u in the region r < c such that u assumes prescribed values F(O) on the spherical surface r = c (Fig. 66). Here r, 4, and 0 are spherical coordinates, and u is independent of 4). Thus u satisfies Laplace's equation (Sec. 4) (1) a2 1 a au r—(ru)+--—---— sin0— =0 ar2 sin o ao ao (r<c O<0<ii-) and the condition u(c,0) =F(0) (2) (O<0<ir). The function u and its partial derivatives of the first and second order are to be continuous throughout the interior (0 r <c, 0 0 ir) of the sphere. r=c FIGURE66 Physically, the function u may denote steady temperatures in a solid sphere r c whose surface temperatures depend only on 0; that is, the surface temperatures are uniform over each circle r = c, 0 = Also, u represents electrostatic potential in the space r < c, which is free of charges, when u = F(0) on the boundary r = c. DIRICHLET PROBLEMS IN SPHERICAL REGIONS SEC 77 313 Consider now a solution of equation (1) of the form u = R(r)®(O) that satisfies the stated continuity requirements. Separation of variables show that, for some constant A, rd2 dO d 1 = —A. Consequently, R must satisfy the ordinary differential equation d2 (3) r <c. Also, for the same constant A, and be continuous when 0 d 1 (4) (r<c) —---—— dO sinO— +AO=O sinOdO dO 0 where 0 and 0' are to be continuous on the closed interval 0 If, in equation (4), we make the substitution x = cos 0, sø that dO dO sin0— = (1 — cos20)—------—-- = —(1 —x2)—--, sin0 dO dx dO it follows readily that dO d (—1<x<1), +AO=0 (5) id® where 0 and its derivative with respect to x are continuous on the entire closed interval — 1 x 1. Equation (5) is Legendre's equation in self-adjoint form; and we know from the theorem in Sec. 74 that A must be one of the eigenvalues A,1 = n(n + 1) (n = 0, 1,2,...) and that the corresponding eigenfunction is = 0) (n = 0, 1, 2,...) thus satisfy equa- = The functions tion (4) when A = n(n + 1): 1 (6) [sin 0)] + n(n + 0) = (n=0,1,2,...). Writing equation (3) in the form r2R"+2rR'—AR=0, we see that it is a Cauchy-Euler equation, which reduces to a differential equation with constant coefficients after the substitution r = exp s is made (see Problem 3, Sec. 35). When A = n(n + 1), its general solution is R = C1r'1 + as is easily verified. The continuity of R at r = 0 requires that C2 = 0, and so = r'1 (n = 0, 1, 2,...). the desired functions of r are = The functions 0) (n = 0, 1, 2, . .), therefore, satisfy . Laplace's equation (1) and the continuity conditions accompanying it. Formally, 314 LEGENDRE POLYNOMIALS AND APPLICATIONS CHAP. 8 their generalized linear combination u(r,O) = (7) n=O is a solution of our boundary value problem if the constants u(c, 0) = F(O), or are such that (8) (0 < F(O) = E 0) 0 <i,-). n=O To find these constants, we introduce the new function f(x) =F(cos1x) (9) (—1 <x <1), where principal values of the inverse cosine are taken. Then if we make the substitutions = and 0 = cos1 x in equation (8), that equation becomes f(x) = (10) (—1 <x < 1); n=O and, according to the theorem in Sec. 76, 2n+1 (11) We 1 (n=O,1,2,...). 2 assume that f is piecewise smooth on the interval — 1 <x < 1, so that expansion (10) is valid for each point x at which f is continuous. In view of definition (9) of the function f(x), the substitution x = cos 0 in integral (11) enables us to write in terms of the original function F(0): 2n+1 = 2 f (n = 0,1,2,...). = the required values of the constants in expression (7) are thus obtained; and the formal solution of our Dirichlet problem can be written in terms of the coefficients (12) as Since u(r,O) = We note that the harmonic function v in the unbounded region r> c, exterior to the spherical surface r = c, which assumes the values F(0) on that surface and which is bounded as r oc can be found in like manner. Here C1 = 0 in our solution R = C1r'1 + C2 of equation (3) if R is to remain bounded as r oo; and the solutions of equation (1) are = 0) — (n = 0, 1,2,...). Thus v(r,O) = n=O T STEADY TEMPERATURES IN A HEMISPHERE SEC. 78 are this time related to the coefficients (12) by means of the where the equation 315 = That is, c n+1 v(r,O) = (15) STEADY TEMPERATURES IN A HEMISPHERE 78. The base r < 1, 0 = of a solid hemisphere r 1, 0 0 ir/2, part of which is shown in Fig. 67, is insulated. The flux of heat inward through the hemispherical surface is kept at prescribed values F(O). In order that temperatures be steady, those values are such that the resultant rate of flow through the hemispherical surface is zero. That is, F satisfies the condition sin 0 dO = 0, which, in terms of the function f(x) = F(cos1 x) (1) can (2) (0 <x <1), also be written f1f(x)dr=0. FIGURE 67 If u denotes temperatures as a function of r and 0, then the condition that the base be insulated is, according to Problem 13, Sec. 4, j) = (0 <r < 1). 316 LEGENDRE POLYNOMIALS AND APPLICATIONS CHAP 8 The boundary value problem in u(r, 0) consists of Laplace's equation 10 a2 (4) sinOaO sin0— ao =0 / fr<1,0<0<—2 condition (3), and the flux condition (see Sec. 3) KUr(l,0) = F(0) (5) (o <o< where K is thermal conductivity. We assume that the function f, defined by equation (1), is piecewise smooth on the interval 0 <x < 1. Also, u must satisfy the usual continuity conditions when 0 r < 1 and 0 0 ir/2. Writing u = R(r)O(0) and separating variables in equations (3) and (4), we obtain the conditions (r<1), r(rR)"—AR=O where R must be continuous when 0 r < 1, and (6) 1 dO d —---—— sin0— +AO=0, sin0d0 dO (7) 0,1— =0 10<0<— 2 where 0 and 0' are to be continuous when 0 0 The substitution x = cos 0 transforms equations (7) into the singular Sturm-Liouville problem consisting of Legendre's equation (see Sec. 77) dO d (0<x<1) and the condition that dO when x = 0, where 0 and dO/dv are to be continuous when 0 1. According to x the corollary in Sec. 74, this problem has eigenvalues A,, = 2n(2n + 1) (n = 0, 1, 2, .) and eigenfunctions = hence = P2,,(cos 0). The . . corresponding bounded solution of the Cauchy-Euler equation (6) is Formally, then, = r2". u(r,O) = n=O if the constants B,, are such that condition (5) is satisfied. That condition requires that ==f(x) (0 <x <1), n=1 where x = cos 0. This is the representation for f(x) on the interval 0 <x < 1 in a series of Legendre polynomials of even degree (Sec. 76) if 2KnB,, = PROBLEMS SEC. 78 317 where = (4n + (9) (n = 1,2,...), dx and if f is such that the condition A0 0, which is precisely condition (2), is satisfied. Thus B0 is left arbitrary; and (10) u(r,O) = B0 (r + 1,0 have the values (9). where the coefficients The constant B0 is the temperature at the origin r = 0. Solutions of such problems with just Neumann conditions (Sec. 7) are determined only up to such an arbitrary additive constant because all the boundary conditions prescribe only values of derivatives of the harmonic functions. PROBLEMS 1. Suppose that u is harmonic throughout the regions r <c and r> c, that u 0 as oo, and that u = 1 on the spherical surface r = c. Show from results found in r Sec. 77 that u = 1 when r c and u = c/r when r c. 2. Suppose that, for all 4), the steady temperatures u(r, 0) in a solid sphere r 1 are such that u(1, 0) = F(O), where when0<O<—, 1 2 F(O)= 0 Derive the expression u(r,O) = + for those temperatures. 3. The base r < 1, 0 = temperature u = 0, while u = 1 r 1, 0 U ir/2 is kept at on the hemispherical surface r = 1, 0 < U < 'r/2. Derive the expression u(r,U) = 4n+3 E(_1Y(2 (2n)! + for the steady temperatures in that solid. is insulated. The 4. The base r < c, U = 'r/2 of a solid hemisphere r c, 0 U temperature distribution on the hemispherical surface is u = F(U). Derive the expression u(r,U) = (4n + n==0 ds, C 0 318 LEGENDRE POLYNOMIALS AND APPLICATIONS CHAP. S where f(x) = F(cos1 x) (0 <x < 1), for the steady temperatures in the solid. Also, show that u(r, 0) = 1 when F(O) = 1. 5. A function u is harmonic and bounded in the unbounded region r> c, 0 4 2i7-, 0 0 Also, u = 0 everywhere on the flat boundary surface r> c, 0= and u = F(0) on the hemispherical boundary surface r = c, 0 < 0 Derive the expression u(r,0) = (4n + 3)(_) r n=O o where f(x) = F(cos1 x) (0 <x < 1). 6. The flux of heat KUr(1, 0) into a solid sphere at its surface r = 1 is a prescribed function F(0), where F is such that the net time rate of flow of heat into the solid is zero. Thus (see Sec. 78) =0, where f(x) = F(cos1 x) (— 1 <x < 1). Assuming that u = 0 at the center r = 0, derive the expression u(r,0) = 2n+1 00 1 1 for the steady temperatures throughout the entire sphere 0 r u(r, 0) denote steady temperatures in a hollow sphere a r 7. Let u(a,0) =F(0) and u(b,0) = 0 1. b when (0<0 Derive the expression 00 ia \fl+1 — u(r,0) = — n=O a 2n+1 r where 2n+1 2 f IT (n=0,1,2,...). o 8. Let u(x, t) represent the temperatures in a nonhomogeneous insulated bar — 1 x 1 along the x axis, and suppose that the thermal conductivity is proportional to 1 — x2. The heat equation takes the form 0u 8t 8u 8 — = b— 8x (1 — (b > 0). 8x Here b is constant since we assume that the product of the physical constants o- and 5 used in Sec. 2 and in Problem 8, Sec. 4, is constant. Note that the ends x = ± 1 are insulated because the conductivity vanishes there. Assuming that u(x, 0) = f(x) (—1 <x < 1), derive the expression °° u(x, t) = 2n+1 2 exp[—n(n + 1 ds. SEC. 78 9. PROBLEMS Show that if f(x) = x2 (— 319 <x < 1) in Problem 8, then 1 u(x,t) + (x2 = — 10. Heat is generated at a steady and uniform rate throughout a solid hemisphere r 1, 0 0 and the entire surface is kept at temperature zero. Thus 0 the steady temperatures u = u(r, 0) satisfy the nonhomogeneous differential equation 132 ——(ru)+ r ar2 and 8/ 1 8u\ I0<r<10<O<—2 I —lsino—I+q0=O r2 sin 0 00 ', 80 J the boundary conditions u(1, 0) = 0, u(r, 'r/2) = 0. Also, u(r, 0) is continuous 0. Point out how Problem 3 suggests seeking a solution of the form at r = u(r,0) = n=O and applying the method of variation of parameters, which was first used in Sec. 33. Follow the steps below to find the solution by that method. (a) Observe how it follows immediately from equation (6), Sec. 77, that d 1 d = —(2n + 1)(2n + sin (n == 0, 1, 2, ... ). Then, with the aid of this identity and expansion (10), Sec. 76, obtain the initial value problem + — (2n + 1)(2n + = = 0 (n=0,1,2,...), where interval 0 + 1 r = — + 2(0) and where is to be continuous on the 1. (b) Solve the differential equation in part (a) by adding a particular solution of it to the general solution of the complementary equation (compare Problem 13, Sec. 38). Then apply the required conditions on stated in part (a), to complete the solution of the initial value problem in ordinary differential equations there. Thus arrive at the desired temperature function: u(r,0) = Suggestion: Observe that the differential equation in part (a) has a particular = ar2, where a is a constant. Also, note that the solution of the form complementary equation in part (b) is of Cauchy-Euler type, and solve it by the method described in Problem 3, Sec. 35. 320 LEGENDRE POLYNOMIALS AND APPLICATIONS CHAP. S a physical interpretation of the following boundary value problem in spherical coordinates for a harmonic function u(r, 0): 11. Give 82 1 81 8u\ r—(ru) + ——--—IsinO—I =0 80/ sinO3O\ u(b,0) = 0, u(1,0) = 0, u(r,01) =f(r), u(r,02) = 0, (1 <r<b,01 <0<02), where 0 < Then, using the normalized eigenfunctions found in Prob<02 lem 11, Sec. 45, and the functions Pa and in Problem 9, Sec. 75, derive the expression u(r,0) = 1 °° yr n=1 where n'r a= lnb B = 2 I lnbJ1 sin (a ln r) dr and = (cos 02) — (cos 0). CHAPTER 9 UNI QUENESS OF SOLUTIONS In this chapter, we examine in greater detail the question of verifying solutions of boundary value problems of certain types and, more particularly, the question of establishing that a solution of a given problem is the only possible solution. A multiplicity of solutions may actually arise when the statement of the problem does not demand adequate continuity or boundedness of a solution and its derivatives. This was illustrated in Problem 16, Sec. 58. The theorem in Sec. 79 below enables us to establish uniform convergence of solutions obtained in the form of series and is useful in both verifying a solution and proving that it is unique. The remaining theorems in the chapter give conditions under which a solution is unique. They apply only to specific types of problems, and their applications are further limited because they require a rather high degree of regularity of the functions involved. ABEL'S TEST FOR UNIFORM CONVERGENCE 79. We begin with some needed background on uniform convergence. Let denote the sum of the first n terms of a series of functions X,(x) which converges to the sum s(x): s,1(x) = s(x) = Suppose that the series converges uniformly with respect to x for all x in some interval. Then (see Sec. 22), for each positive number E, there exists a 321 322 CHAP. 9 UNIQUENESS OF SOLUTIONS positive integer n6, independent of x, such that whenever n > x) I < Is( x) — for every x in the interval. Let j denote any positive integer. Then s — s— Is — + Is — < provided n > n6. Thus a necessary condition for uniform convergence of the series is that, for all positive integers j, (2) Is,, x) I <.e x) — whenever n > Condition (2) is also a sufficient condition, known as the Cauchy criterion, for is not independent of x. convergence of the series for each fixed x even if Hence it implies that the sum s(x) exists. Then, for any fixed n and x and for exists such that the given number a positive integer (3) <e Is(x) — wheneverj >jE(x). To show that condition (2) is sufficient for uniform convergence, let n corresponds to the given where denote any fixed integer greater than number E in the sense that condition (2) is satisfied for all x. Then, for each and, since fixed x, condition (3) is satisfied when j > Is — = Is — + Is — — + — sRI, it follows from conditions (3) and (2) that ls(x) — <28 which is independent of I, is and n > Thus Is(x) — is independent of x, uniform Since arbitrarily small for each x when n > convergence is now established. of Note that x here may equally well denote elements (x1, x2,. . ., some set in N-dimensional space. The uniform convergence is then with respect to all N variables x1, x2,.. ., XN together. We now derive a test for the uniform convergence of infinite series whose terms are products of certain types of functions. Its application in verifying formal solutions of boundary value problems was illustrated in Sec. 28. The test, known as Abel's test, involves functions in a sequence (i = 1, 2, .) which when j > . . is uniformly bounded for all points t in an interval. That is, there exists a constant M, independent of i, such that 1(t)I <1W (i= 1,2,...) for all t in the interval. The sequence is, moreover, monotonic with respect to i. Thus, for every t in the interval, either 1(t) (i = 1,2,...) ABEL'S TEST FOR UNIFORM CONVERGENCE SEC. 79 323 or 1(t) (7) We (i = 1,2,...). state the test as a theorem which shows that when the terms of a uniformly convergent series are multiplied by functions T1(t) of the type just described, the new series is also uniformly convergent. Theorem. The series (8) i=1 converges uniformly with respect to the two variables x and t together in a region R of the xt plane if the series i=1 converges uniformly with respect to x for all x such that (x, t) is in R and if the are uniformly bounded and monotonic with respect to i (i = functions 1, 2,...) for all t such that (x, t) is in R. To start the proof, we let denote partial sums of series (8): = As indicated above, the uniform convergence of that series will be established if there corresponds a positive integer we prove that to each positive number independent of x and t, such that x, t) — x, t) I <e whenever n > nE, for all integers m = n + 1, n + 2,... and for all points (x, t) in R. We write the partial sum =X1(x) +X2(x) + ... Then, for each pair of integers m and n (m > n), Sm — can be written +XmTm = — + — = — + — + +(5m — 5n)Tm + — = — + — — + 5m_i)Tm — — (5m-i — By pairing alternate terms here, we find that (9) +(5m 5n)Tm — 5n)(Tn+2 — +(5m_15n)(Tm_iTm) +(5m5n)Tm. 324 CHAP. 9 UNIQUENESS OF SOLUTIONS Suppose now that the functions T1 are nonincreasing with respect to i, so that they satisfy condition (6), and that they also satisfy the uniform boundedness condition (5). Then the factors + 1 — +2' +2 — etc., in equation <M. Since the series with terms X1(x) con(9) are nonnegative, and verges uniformly, an integer exists such that I I < whenever n> for all positive integers j, where e is any given positive number and n6 is independent of x. Then if n > nE and m > n, it follows from equation (9) that + - Tn+2) + - SnI ISm = + + - Tm + Therefore, ISm( and x, t) — x, t) I <E whenever m > n > is established. the functions T, are nondecreasing the uniform convergence of series (8) The proof is similar when with respect to i. When x is kept fixed, the series with terms X, is a series of constants; and only requirement placed on that series is that it be convergent. Then the theorem shows that when T, are bounded and monotonic, the series of terms the uniformly convergent with respect to t. of the theorem to cases in which X, are functions of x and t, or both X1 and T, are functions of several variables, become evident when it is observed that our proof rests on the uniform convergence of the series of terms X, and the bounded monotonic nature of the functions X,T,(t) is Extensions UNIQUENESS OF SOLUTIONS OF THE HEAT EQUATION 80. Let D denote the domain consisting of all points interior to a closed surface S; and let D be the closure of that domain, consisting of all points in D and all points on S. We assume always that the closed surface S is piecewise smooth. That is, it is a continuous surface consisting of a finite number of parts over each of which the outward unit normal vector exists and varies continuously from point to point. Then if U is a function of x, y, and z which is continuous in D, together with its partial derivatives of the first and second order, a special case of Green's identity that we shall need here states that dA = fff(u V2U + U2 + U2 + dv. UNIQUENESS OF SOLUTIONS OF THE HEAT EQUATION SEC. 80 325 Here d4 is the area element on S, dV represents dx dy dz, and dU/dn is the derivative in the direction of the outward unit normal to S.t Consider a homogeneous solid whose interior is the domain D and whose temperatures at time t are denoted by u(x, y, z, t). A fairly general problem in heat conduction is the following: [(x,y,z)inD,t>O], u( x, y, z, 0) = u [(x, y, z) in f( x, y, z) = g(x,y,z, t) [(x,y,z) on S, t 0]. This is the problem of determining temperatures in a body, with prescribed initial temperatures f(x, y, z) and surface temperatures g(x, y, z, t), interior to which heat may be generated continuously at a rate per unit volume proportional to q(x, y, z, t). Suppose that the problem has two solutions u = u2(x, y, z, t), u = u1(x, y, z, t), where both u1 and u2 are continuous functions in the closed region D when 0, while their derivatives of the first order with respect to t and of the first t and second orders with respect to x, y, and z are continuous in D when t> 0. Since u1 and u2 satisfy the linear equations (2), (3), and (4), their difference U(x, y, z, t) = u1(x, y, z, t) — u2(x, y, z, t) satisfies the homogeneous problem (5) (6) U(x,y,z,O) =0 (7) U=0 [(x,y,z)inD,t>O], [(x,y,z)inD], Moreover, U and its derivatives have the continuity properties of u1 and u2 assumed above. We shall now show that U = 0 in D when t> 0, so that the two solutions u1 and u2 are identical. That is, not more than one solution of the boundary value problem in u can exist if the solution is required to satisfy the stated continuity conditions. The continuity of U with respect to x, y, z, and t together in the closed region D when t 0 implies that the integral 1 is 1(t) = —fff [U(x,y,z,t)J2dV 2 D a continuous function of t when t 0; and, according to equation (6), 1(0) = 0. Also, in view of the continuity of (4 when t> 0, we may use equation Identity (1) is found by applying Gauss's divergence theorem to the vector field U grad U. See the book by Taylor and Mann (1983, pp. 492—493), listed in the Bibliography. 326 UNIQUENESS OF SOLUTIONS CHAP. 9 (5) to write (t>0). I'(t) =fff Identity (1) applies to the last integral here because of the continuity of the derivatives of U when t > 0. Thus (9) fffu V when t> 0. But U = = 0 ffu dA on S, and k > I'(t) = 0; - fff + + U2) dV consequently, + + dV 0. The mean value theorem for derivatives applies to 1(t). That is, for each positive t, a number t1 (0 < t1 < t) exists such that 1(t) — 1(0) = tI'(t1); and, since 1(0) = 0 and I'(t1) 0, it follows that 1(t) (8) of the integral shows that 1(t) 0. Therefore, 0. However, definition I(t)=0 and so the nonnegative integrand U2 cannot have a positive value at any point in D. For if it did, the continuity of U2 would require that U2 be positive throughout some neighborhood of the point, and that would mean 1(t)> 0. Consequently, U(x,y,z,t) =0 and we arrive at the following theorem on uniqueness. Theorem 1. Let u satisfy these conditions of regularity: (a) It is a continuous function of the variables x, y, z, and t together when the point (x, y, z) is in the closed region D and t 0; (b) the derivatives of u appearing in the heat equation (2) are continuous in the same sense when t> 0. Then if u is a solution of the boundary value problem (2)—(4), it is the only possible solution satisfying conditions (a) and (b). When conditions (a) and (b) in Theorem 1 are added to the requirement that u is to satisfy the heat equation and the boundary conditions, our boundary value problem is completely stated, provided it has a solution; for that will be the only possible solution. — The condition that u be continuous in D when t = 0 restricts the usefulness of our theorem. It is clearly not satisfied if the initial temperature function f in condition (3) fails to be continuous throughout D, or if at some point on S the initial value g(x, y, z, 0) of the prescribed surface temperature SEC. 80 UNIQUENESS OF SOLUTIONS OF THE HEAT EQUATION differs from the value f(x, y, z). The continuity requirement at t = 0 327 can be relaxed in some cases.t The proof of Theorem 1 required that the integral dU JJU—dA dn s in equation (9) either vanish or have a negative value. It vanished because U = 0 on S. But it is never positive if condition (4) is replaced by the boundary condition du (10) +hu=g(x,y,z,t) {(x,y,z)onS,t>0], where h 0. For, in that case, dU/dn our theorem can be modified as follows. —hU on S; and UdU/dn 0. Thus Theorem 2. The conclusion in Theorem 1 is true if boundary condition (4) is replaced by condition (10), or if condition (4) is satisfied on part of the surface S and condition (10) is satisfied on the rest. EXAMPLE. In the problem of temperature distribution in a slab with insulated faces x = 0 and x = c and initial temperatures f(x) (Secs. 27 and 28), write c = and assume that f is continuous and f' is piecewise continuous on the interval 0 x Then the Fourier cosine series for f converges uni- formly to f(x) on that interval (Sec. 22). Let u(x, t) denote the sum of the series a0 obtained as a formal solution of the problem, a0 and (n = 1, 2,. .) being the coefficients in the Fourier cosine series for f. As noted in Sec. 27, this temperature problem for a slab is the same as the problem for a bar of uniform cross section whose bases, in the planes x = 0 and . x = and whose lateral surface, parallel to the x axis, are all insulated (du/dn 0). Let the domain D consist of the interior points of the bar. We can see from Abel's test (Sec. 79) that series (11) converges uniformly with respect to x and t together in the region 0 x t 0 of the xt plane; thus u is continuous there. When t t0, where t0 is any positive number, the series obtained by differentiating series (11) term by term any number of times with respect to x or t is uniformly convergent, according to the Weierstrass M-test (Sec. 22). Consequently, we now know that u satisfies all the equations in Integral transforms can sometimes be used to prove uniqueness of solutions of certain types of boundary value problems. This is illustrated in the book by Churchill (1972, sec. 79), listed in the Bibliography. 328 UNIQUENESS OF SOLUTIONS CHAP 9 the boundary value problem (compare Sec. 28) and also that continuous functions in the region 0 t> x 0. Thus U are and satisfies the regularity conditions (a) and (b) stated in Theorem 1, and Theorem 2 applies to show that the sum U(x, t) of series (11) is the only solution that satisfies those conditions. SOLUTIONS OF LAPLACE'S OR POISSON'S EQUATION 81. Let U be a harmonic function in a domain D of three-dimensional space bounded by a continuous closed surface S that is piecewise smooth. Assume also that U and its partial derivatives of the first order are continuous in the closure D of the domain. Then, since V2U(x, y, z) = (1) [(x, y, z) 0 in D], Green's identity (1), Sec. 80, becomes ffu (2) dA = fff + U2 + dv. This equation is valid for our function U even though we have required the derivatives of !he second order to be continuous only in D, and not in the closed region D. It is not difficult to prove that, because V2U 0, modification of the usual conditions in the divergence theorem from which Green's identity follows is possible.t Suppose that U = 0 at all points on the surface S. Then the first integral in equation (2), and therefore the second, vanishes. But the integrand of the second is nonnegative and continuous in D. Hence it must vanish throughout D; that is, Consequently, U(x, y, z) is constant; but it is zero on S and continuous in D, and so U = 0 throughout D. Suppose that dU/dn, instead of U, vanishes on S; or, to make the condition more general, suppose that dU -a-— where h +hU=0 [(x,y,z)onS], 0 and h is either a constant or a function of x, y, and z. Then dU U— = dn —hU2 0 on 5, so that the first integral in equation (2) is less than or equal to zero. But tSee the book by Kellogg (1953, p. 119), listed in the Bibliography. SOLUTIONS OF LAPLACE'S OR POISSON'S EQUATION SEC. 81 329 the second integral is greater than or equal to zero. Hence it must vanish; and, again, conditions (3) follow. So U is constant in D. If U vanishes over part of S and satisfies condition (4) on the rest of that surface, our argument still shows that U is constant in D. In such a case, the constant must be zero. — Now let u denote a function which is continuous in D, together with its partial derivatives of the first order. Suppose that it also has continuous derivatives of the second order in D and satisfies these conditions: (5) [(x,y,z)inD], V2u(x,y,z) =f(x,y,z) du [(x,y,z)onS]. (6) f, p, h, and g denote prescribed constants or functions of x, y, and z. assume that p 0 and h 0. Equation (5) is known as Poisson's equation and is a generalization of Laplace's equation, which occurs when f(x, y, z) is identically zero throughout Here We D. It was encountered in Chaps. 1 and 4. Boundary condition (6) includes important special cases. When p = 0 on S, or on part of S, the value of u is assigned there. When h = 0, the value of du/dn is assigned. Of course, p and h must not vanish simultaneously. If u = u1(x, y, z) and u = u2(x, y, z) are two solutions of this problem, their difference U(x,y,z) =u1(x,y,z)—u2(x,y,z) satisfies Laplace's equation in D and the condition dU p— + hU = dn 0 S. Moreover, U satisfies the conditions of regularity required of u1 and u2. Thus it is harmonic in D, and U and its derivatives of the first order are continuous in D. It follows from the results established above for harmonic functions that U must be constant throughout D. Thus dU/dn = 0 onS. If h 0 at some point on 5, then U vanishes there; and U = 0 throughout D. For the harmonic function U vanishes over the closed surface S and hence cannot have values other than zero interior to S. We have now established the following uniqueness theorem for problems in electrostatic or gravitational potential, steady temperatures, and other boundon ary value problems involving Laplace's or Poisson's equation. Theorem. Let u(x, y, z) satisfy these conditions of regularity in a domain D by a closed surface S: (a) It is continuous, together with its derivatives of the first order, in D; (b) its derivatives of the second order are continuous in D. Then if u is a solution of the boundary value problem (5)—(6), it is the only solution bounded satisfying conditions (a) and (b), except possibly for u + C, where C is an 330 UNIQUENESS OF SOLUTIONS arbitrary constant. Unless h = CHAP. 9 0 at every point on S, C = 0 and the solution is unique. It is possible to show that this theorem also applies when D is the unbounded domain exterior to the closed surface S, provided u satisfies the additional requirement that the absolute values of ru, r r is the distance from (x, y, z) to the origin.t Then, since u vanishes as r oo, the constant C is zero; and the solution is unique. But note that S is a closed surface, so that this extension of the theorem does not apply, for instance, to unbounded domains between two planes or inside a cylinder. Condition (a) in the theorem is severe because it requires u and its derivatives of the first order to be continuous on the surface S. For problems in which p = 0 on S, so that u is prescribed on the entire boundary, the condition can be relaxed so as to require only the continuity of u itself in D if derivatives are continuous in D. This follows directly from a fundamental result in potential theory: If a function other than a constant is harmonic in D and continuous in D, then its maximum and minimum values are assumed at points on S, and never in EXAMPLE. To illustrate the use of the theorem, consider the problem in Example 1, Sec. 34, of determining steady temperatures u(x, y) in a rectangular plate with three edges at temperature zero and an assigned temperature distribution on the fourth edge. The faces of the plate are insulated. For convenience, we shall consider the plate to be square, with edge length IT. As long as du/dn = 0 on the faces, the thickness of the plate does not affect the problem. The domain D is the interior of the finite region bounded by the planes z = z1, and z = z2, where z1 and z2 are constants. x = 0, x = y = 0, y = Then S is the boundary of that domain. The required function u is harmonic in D. It vanishes on the three parts x = 0, x = ii-, and y = ir of S; and u = f(x) on the part y = 0. Also, = 0 on the parts z = z1 and z = z2. Thus the theorem applies if u and its derivatives of the first order are continuous in D. First, suppose that u is independent of z. Then y) = y) + u(0, y) = u(ir, y) = 0 u(x,0) =f(x), 0 (0 <x <IT, 0 <y <ir), (0 y u(x,rr) =0 tSee the book by Kellogg (1953), referred to earlier in this section. Physically, the result seems evident since it states that steady temperatures cannot have maximum or minimum values interior to a solid in which no heat is generated. For a proof in three dimensions, see the book by Kellogg (1953), cited earlier in this section; and, for two dimensions, see the authors' book (1990, sec. 42), which is also listed in the Bibliography. SOLUTIONS OF LAPLACE'S OR POISSON'S EQUATION SEC. 81 331 The formal solution found in Sec. 34 becomes u(x,y)= (10) sinnx, n=1 are the coefficients in the Fourier sine series for f on the interval 0 <x <IT. where To show that the function (10) satisfies the regularity conditions, let us require that f and f' be continuous and f" piecewise continuous and that = 0. Results found in Secs. 22 and 23 then show that f(0) = f'(x) = f(x) = (11) n=1 n=1 and that both of these series converge uniformly on the interval 0 x ir. The second series, obtained by differentiating the first, is the Fourier cosine series for f' on 0 <x and, since f' is continuous and f" is piecewise continu- ous, not only is that series uniformly convergent, but also the series of absolute values of its coefficients converges. Hence it follows from the Weierstrass M-test (Sec. 22) that the series I n=1 also converges uniformly with respect to x. Let us show that, for each fixed y, the sequence of functions sinh appearing in series (10) is monotonic and nonincreasing as n increases. This is evident when y = 0 and when y = ir. It is also true when 0 <y <ir, provided that the function sinh/3t T(t)= sinh at (t>0,a>f3>0) decreases as t increases. To see that this is so, we write . T'( t) sinh2 at = /3 sinh at cosh f3t — a sinh f3t cosh at 1 1 = a+f3 a —f3 2 a—/3 (a+f3) (2n + 1)! 332 UNIQUENESS OF SOLUTIONS CHAP. 9 Since the terms of this series are positive, T'(t) < 0; thus T(t) decreases as t increases. Likewise, the positive-valued functions cosh (14) — y) sinh nir never increase in value as n grows because their arising in the series for squares can be written 1 sinh and each term here is nonincreasing. The values of the functions (13) clearly vary only from zero to unity for all n and y involved. The functions (14) are also uniformly bounded. Hence those functions can be used in Abel's test for uniform convergence (Sec. 79). From the uniform convergence of the series in equations (11) and of series (12), on the interval 0 x ir, we conclude not only that series (10) converges uniformly with respect to x and y together in the region 0 x ii-, 0 y IT of the xy plane, but also that the uniform convergence holds true for the series obtained by differentiating series (10) once termwise with respect to either x or y. Consequently, series (10) is differentiable with respect to x and y; also, its sum u(x, y) and and are continuous in the closed region 0 x 0 y Clearly, u(x, y) satisfies boundary conditions (8) and (9). The derivatives of the second order, with respect to either x or y, of the terms in series (10) have absolute values not greater than sinh (16) y0) sinh when 0 I — IT and y0 x y IT, where y0> 0. Let M be chosen such that < M for all n. Then, from the inequalities 2sinh n(IT — y0) — y0) and 2sinhnir — it follows that the numbers (16) are less than 1— exp( — 2IT) n2 exp ( — ny0). The series with these terms converges, according to the ratio test, since y0 > 0; and so the Weierstrass M-test ensures the uniform convergence of the series of second-order derivatives of terms in series (10) when y0 y ir. Thus series are continuous in the region and (10) is twice-differentiable; also, 0 0 X <y IT. the terms in series (10) satisfy Laplace's equation (7), the same is true of the sum u(x, y) of that series. Thus u is established as a solution of our boundary value problem. Moreover, u satisfies our regularity conditions, even Since SEC 82 SOLUTIONS OF A WAVE EQUATION 333 with respect to z since it is independent of z and = 0 everywhere and on the parts z = z1 and z = z2 of S, in particular. According to the above theorem, the function defined by series (10) is, then, the only possible solution that satisfies the regularity conditions. 82. SOLUTIONS OF A WAVE EQUATION Consider the following generalization of the problem solved in Sec. 29 for the transverse displacements in a stretched string: (1) = t) y(0,t) =p(t), y(x,0) ==f(x), (2) (3) t) + çb(x, t) y(c,t) = (0 < x < c, t > 0), q(t) (t 0), =g(x) But we now require y to be of class C2 in the region R: 0 x c, t 0, by which we shall mean that y and its derivatives of the first and second order, including and are to be continuous functions in R. As indicated in Sec. 30, the prescribed functions 4), p, q, f, and g must be restricted if the problem is to have a solution of class C2. Suppose that there are two solutions y1(x, t) and y2(x, t) in that class. Then the difference Y(x,t) =y1(x,t) —y2(x,t) is of class C2 in R and satisfies the homogeneous problem (0 <x <c, t >0), (4) Y(0,t) =0, Y(c,t) =0 Y(x,0)=0, We shall prove that V = thus y1 = 0 throughout R; y2, as stated in the following theorem. Theorem. The boundary value problem (1)—(3) cannot of class C2 in R. have more than one solution To start the proof, we note that the integrand of the integral lc 1 satisfies conditions such that I'(t) = f 1 C + 334 CHAP 9 UNIQUENESS OF SOLUTIONS Since = the integrand here can be written as a ax in view of equations (5), from which it follows that So = 0, = 0, we can write (9) I'(t) = t)Y(c, t) — YX(O, t)Y(O, t) = 0. is a constant. But equation (7) shows that 1(0) = 0 because 0) = 0; also, Y(x, 0) = 0, and so 0) = 0. Thus 1(t) = 0. The nonnegative continuous integrand of that integral must, therefore, vanish; that is, Hence 1(t) =0 So V is constant. In fact, Y(x, t) = 0 since Y(x, 0) = 0; and the proof of the theorem is complete. If instead of y, is prescribed at the end point in either or both of conditions (2), the proof of uniqueness is still valid because condition (9) is again satisfied. The requirement of continuity on derivatives of y is severe. Solutions of many simple problems in a wave equation have discontinuities in their derivatives. PROBLEMS 1. In Problem 8, Sec. 33, on temperatures u(x, t) in a slab, initially at temperatures f(x) and throughout which heat is generated at a constant rate per unit volume, assume that f is continuous and f' is piecewise continuous (0 x and that f(0) = f('r) = 0. Prove that the function u(x, t) obtained there is the only solution of the problem which satisfies the regularity conditions (a) and (b) stated in Theorem 1, Sec. 80. 2. Verify the solution of Problem 8, Sec. 32, and prove that it is the only possible solution satisfying the regularity conditions (a) and (b) stated in Theorem 1, Sec. 80. Note that, in this case, the Weierstrass M-test suffices for all proofs of uniform convergence. 3. In the Dirichlet problem for a rectangle in Example 1, Sec. 34, let f be piecewise smooth on the interval 0 <x <a. Verify that the formal solution found there satisfies the condition u(x, 0 + ) = f(x) when 0 <x <a if f(x) is defined as the mean of f(x + ) and f(x — ) at its points of discontinuity. 4. Formulate a complete statement of the boundary value problem for steady tempera- tures in a square plate with insulated faces when the edges x = 0, x = 'r, and y = 0 are insulated and the edge y = 'r is kept at temperatures u = f(x). Show that if f, f', and f" are continuous on the interval 0 x 'r and f'(O) = f'('r) = 0, the problem SEC. 82 has PROBLEMS 335 the unique solution coshny a0 u(x,y)=—+ 2 n=1 coshn'r cosnx, where 'TO (n=0,1,2,...). 5. Replace the thin disk in the example in Sec. 40 by a cylinder bounded by the surfaces = 0 on the last two parts. Also, let p = 1, z = z1, and z = z2, where be a periodic function of period 2'r with a continuous derivative of the second order everywhere. Then show that the function u given by equation (6), Sec. 40, is the unique solution, satisfying our conditions of regularity, of the problem in steady temperatures. 6. Use the uniqueness that was established in Sec. 82 to show that the solution of Problem 1, Sec. 30, is the only solution of class C2 in the region 0 x 1, t 0 of the xt plane. 7. Show that the solution of Problem 3, Sec. 38, is unique in the class C2. 8. In Sec. 30, let f(x) be such that its odd periodic extension F(x) has a continuous second derivative F"(x) for all x (— oo <x < oo). Then show that the solution (10) there is unique in the class C2. BIBLIOGRAPHY The following list of books and articles for supplementary study of the various topics that have been introduced is far from exhaustive. Further references are given in the ones mentioned here. Abramowitz, M., and I. A. Stegun (Eds.): Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Dover Publications, Inc., New York, 1972. Apostol, T. M.: Mathematical Analysis (2d ed.), Addison-Wesley Publishing Company, Inc., Reading, MA, 1974. Bell, W. W.: Special Functions for Scientists and Engineers, D. Van Nostrand Company, Ltd., London, 1968. Berg, P. W., and J. L. McGregor: Elementary Partial Dzffereiztial Equations (Holden-Day, Inc., San Francisco, 1966), McGraw-Hill, Inc., New York. Birkhoff, G., and G.-C. Rota: Introduction to Ordinary Differential Equations (4th ed), John Wiley & Sons, Inc., New York, 1989. Bowman, F.: Introduction to Bessel Functions, Dover Publications, Inc., New York, 1958. Boyce, W. E., and R. C. DiPrima: Elementary Differential Equations (5th ed.), John Wiley & Sons, Inc., New York, 1992. Broman, A.: Introduction to Partial Differential Equations, Dover Publications, Inc., New York, 1989. Brown, J. W., and F. Farris: On the Laplacian, Math. Mag., vol. 59, no. 4, pp. 227—229, 1986. Buck, R. C.: Advanced Calculus (3d ed.), McGraw-Hill, Inc., New York, 1978. Budak, B. M., A. A. Samarskii, and A. N. Tikhonov: A Collection of Problems on Mathematical Physics, Dover Publications, Inc., New York, 1988. 336 BIBLIOGRAPHY 337 Byerly, W. E.: Fourier's Series and Spherical Harmonics, Dover Publications, Inc., New York, 1959. Cannon, J. T., and S. Dostrovsky: The Evolution of Dynamics, Vibration Theory from 1687 to 1742, Springer-Verlag, New York, 1981. Carslaw, H. S.: Introduction to the Theory of Fourier's Series and Integrals (3d ed.), Dover Publications, Inc., New York, 1952. and J. C. Jaeger: Conduction of Heat in Solids (2d ed.), Oxford University Press, London, 1959. Churchill, R. V.: Operational Mathematics (3d ed.), McGraw-Hill, Inc., New York, 1972. and J. W. Brown: Complex Variables and Applications (5th ed.), McGraw-Hill, Inc., New York, 1990. Coddington, E. A.: An Introduction to Ordinary Differential Equations, Dover Publications, Inc., New York, 1989. and N. Levinson: Theory of Ordinary Differential Equations, Krieger Publishing Co., Inc., Melbourne, FL, 1984. Courant, R., and D. Hilbert: Methods of Mathematical Physics, vols. 1 and 2, John Wiley & Sons, Inc., New York, 1989. Davis, H. F.: Fourier Series and Orthogonal Functions, Dover Publications, Inc., New York, 1989. Erdélyi, A. (Ed.): Higher Transcendental Functions, vols. 1—3, Krieger Publishing Co., Inc., Melbourne, FL, 1981. Farrell, 0. J., and B. Ross: Solved Problems in Analysis as Applied to Gamma, Beta, Legendre, and Bessel Functions, Dover Publications, Inc., New York, 1971. Fourier, J.: The Analytical Theory of Heat, translated by A. Freeman, Dover Publications, Inc., New York, 1955. Franklin, P.: A Treatise on Advanced Calculus, Dover Publications, Inc., New York, 1964. Grattan-Guinness, I.: Joseph Fourier, 1 768—1 830, The MIT Press, Cambridge, MA, and London, 1972. Gray, Alfred, and M. A. Pinsky: Gibbs' Phenomenon for Fourier-Bessel Series, Expo. Math. (in press). Gray, Andrew, and G. B. Mathews: A Treatise on Bessel Functions and Their Applications to Physics (2d ed.), with T. M. MacRobert, Dover Publications, Inc., New York, 1966. Hanna, J. R., and J. H. Rowland: Fourier Series, Transforms, and Boundary Value Problems (2d ed.), John Wiley & Sons, Inc., New York, 1990. Hayt, W. H., Jr.: Engineering Electromagnetics (5th ed.), McGraw-Hill, Inc., New York, 1989. Herivel, J.: Joseph Fourier: The Man and the Physicist, Oxford University Press, London, 1975. Hewitt, E., and R. E. Hewitt: The Gibbs-Wilbraham Phenomenon: An Episode in Fourier Analysis, Arch. Hist. Exact Sci., vol. 21, pp. 129—160, 1979. Hobson, E. W.: The Theory of Spherical and Ellipsoidal Harmonics, Chelsea Publishing Company, Inc., New York, 1955. Hochstadt, H.: The Functions of Mathematical Physics, Dover Publications, Inc., New York, 1986. Ince, E. L.: Ordinary Differential Equations, Dover Publications, Inc., New York, 1956. Jackson, D.: Fourier Series and Orthogonal Polynomials, Carus Mathematical Monographs, no. 6, Mathematical Association of America, 1941. 338 BIBLIOGRAPHY Jahnke, E., F. Emde, and F. Lösch: Tables of Higher Functions (6th ed.), McGraw-Hill, Inc., New York, 1960. Kaplan, W.: Advanced Calculus (4th ed.), Addison-Wesley Publishing Company, Inc., Reading, MA, 1991. Advanced Mathematics for Engineers, Addison-Wesley Publishing Company, Inc., Reading, MA, 1981. Kellogg, 0. D.: Foundations of Potential Theory, Dover Publications, Inc., New York, 1953. Kevorkian, J.: Partial Differential Equations, Wadsworth, Inc., Belmont, CA, 1990. Körner, T. W.: Fourier Analysis, Cambridge University Press, Cambridge, England, 1988. Kreider, D. L., R. G. Kuller, D. R. Ostberg, and F. W. Perkins: An Introduction to Linear Analysis, Addison-Wesley Publishing Company, Inc., Reading, MA, 1966. Lanczos, C.: Discourse on Fourier Series, Hafner Publishing Company, New York, 1966. Langer, R. E.: Fourier's Series: The Genesis and Evolution of a Theory, Slaught Memorial Papers, no. 1, Am. Math. Monthly, vol. 54, no. 7, part 2, pp. 1—86, 1947. Lebedev, N. N.: Special Functions and Their Applications, Dover Publications, Inc., New York, 1972. I. P. Skalskaya, and Y. S. Uflyand: Worked Problems in Applied Mathematics, Dover Publications, Inc., New York, 1979. McLachlan, N. W.: Bessel Functions for Engineers (2d ed), Oxford University Press, London, 1955. MacRobert, T. M.: Spherical Harmonics (3d ed.), with I. N. Sneddon, Pergamon Press, Ltd., Oxford, England, 1967. Magnus, W., F. Oberhettinger, and R. P. Soni: Formulas and Theorems for the Special Functions of Mathematical Physics (3d ed.), Springer-Verlag, New York, 1966. Oberhettinger, F.: Fourier Expansions: A Collection of Formulas, Academic Press, New York and London, 1973. M. N.: Boundary Value Problems of Heat Conduction, Dover Publications, Inc., New York, 1989. Papp, F. J.: Two Equivalent Properties for Orthonormal Sets of Functions in Complete Spaces, mt. J. Math. Educ. Sci. Technol., vol. 22, pp. 147—149, 1991. Pinsky, M. A.: Partial Differential Equations and Boundary-Value Problems with Applications (2d ed.), McGraw-Hill, Inc., New York, 1991. Powers, D. L.: Boundary Value Problems (3d ed.), Academic Press, Orlando, FL, 1987. Rainville, E. D.: Special Functions, Chelsea Publishing Company, Inc., New York, 1971. and P. E. Bedient: Elementary Differential Equations (7th ed), Macmillan Publishing Co., New York, 1989. Rogosinski, W.: Fourier Series (2d ed.), Chelsea Publishing Company, Inc., New York, 1959. Sagan, H.: Boundary and Eigenvalue Problems in Mathematical Physics, Dover Publications, Inc., New York, 1989. Sansone, G., and A. H. Diamond: Orthogonal Functions (rev. ed), Dover Publications, Inc., New York, 1991. Seeley, R. T.: An Introduction to Fourier Series and Integrals, W. A. Benjamin, Inc., New York, 1966. Smirnov, V. I.: Complex Variables—Special Functions, Addison-Wesley Publishing Company, Inc., Reading, MA, 1964. Sneddon, I. N.: Elements of Partial Differential Equations, McGraw-Hill, Inc., New York, 1957. BIBLIOGRAPHY 339 Fourier Transforms, McGraw-Hill, Inc., New York, 1951. Special Functions of Mathematical Physics and Chemistry (3d ed.), Longman, New York, 1980. The Use of Integral Transforms, McGraw-Hill, Inc., New York, 1972. Streeter, V. L., and E. B. Wylie: Fluid Mechanics (8th ed.), McGraw-Hill, Inc., New York, 1985. Taylor, A. E., and W. R. Mann: Advanced Calculus (3d ed.), John Wiley & Sons, Inc., New York, 1983. Timoshenko, S. P., and J. N. Goodier: Theory of Elasticity (3d ed.), McGraw-Hill, Inc., New York, 1970. Titchmarsh, E. C.: Eigenfunction Expansions Associated with Second-Order Differential Equations, Oxford University Press, London, part I (2d ed), 1962; part II, 1958. Introduction to the Theory of Fourier Integrals (3d ed), Chelsea Publishing Company, Inc., New York, 1986. Tolstoy, G. P.: Fourier Series, Dover Publications, Inc., New York, 1976. Tranter, C. J.: Bessel Functions with Some Physical Applications, Hart Publishing Company, Inc., New York, 1969. Trim, D. W.: Applied Partial Differential Equations, PWS-KENT Publishing Company, Boston, 1990. Van Vleck, E. B.: The Influence of Fourier's Series upon the Development of Mathematics, Science, vol. 39, pp. 113—124, 1914. Watson, G. N.: A Treatise on the Theory of Bessel Functions (2d ed.), Cambridge University Press, Cambridge, England, 1952. Weinberger, H. F.: A First Course in Partial Differential Equations, John Wiley & Sons, Inc., New York, 1965. Whittaker, E. T., and G. N. Watson: A Course of Modern Analysis (4th ed.), Cambridge University Press, Cambridge, England, 1963. Zygmund, A.: Trigonometric Series (2d ed.), 2 vols. in one, Cambridge University Press, Cambridge, England, 1977. INDEX Abel, N. H., 117 Abel's test for uniform convergence, 117 proof of, 322 Adjoint of operator, 170 Antiperiodic functions, 189, 192, 193, 211, 216 Approximation: best in mean, 62 least squares, 62n. of solutions, 123, 133 Average value of function, 61 norms of, 270—271 orthogonal sets of, 263, 269 orthonormal sets of, 270 recurrence relations for, 249 series of, 273 zeros of, 259—263, 269 of second kind, Weber's, 247 Bessel's equation, 242 modified, 253 self-adjoint form of, 264 Bessel's inequality, 64, 67 Bessel's integral form, 255 Best approximation in mean, 62 Bibliography, 336—339 Bernoulli, D., 128 Bessel functions, 242 of first kind, 245, 249 boundedness of, 256, 259 derivatives of, 250, 256 generating function for, 259 graphs of, 246 integral forms of, 255, 256 integrals of, 250, 251 modified, 253, 266, 286, 287 Boundary conditions, 2 Cauchy type, 26 Dirichlet type, 26 linear, 3 linear homogeneous, 4 Neumann type, 26, 317 periodic, 163, 172 Robin, 26 separated, 169 types of, 26 341 342 INDEX Boundary value problems, 2, 129 approximate solutions of, 123, 133 formal solutions of, 111, 130, 135 methods of solving, 26—32 solutions verified, 114, 123 Boundedness: of Bessel functions, 256, 259 of piecewise continuous functions, 40 Brown, G. H., 140n. Cauchy criterion for uniform convergence, 322 Cauchy-Euler equation, 149 Cauchy's inequality, 47, 91, 102 Closed orthonormal sets, 49, 98n., 103 of Legendre polynomials, 300 Complete orthonormal sets, 98 of trigonometric functions, 99 Component, unstable, 157 Conductance, surface, 8 Conductivity, thermal, 4 Conformal mapping, 29 Continuous eigenvalues, 172, 230 Continuous functions: piecewise, 40 uniformly, 221 Continuous spectrum, 172 Convergence: of Fourier-Bessel series, 273 of Fourier cosine series, 78, 84—85 of Fourier series, 73, 76, 83—84 of Fourier sine series, 78, 84—85 of Legendre series, 308 in mean, 97, 103, 104 pointwise, 103, 104 (See also Uniform convergence) Cylindrical coordinates, 10 laplacian in, 12 Czarnecki, A., 260n. d'Alembert, J., 128 d'Alembert's solution, 28 Derivative: left-hand, 68 right-hand, 68 Differential equations: Cauchy-Euler, 149 linear, 3 linear homogeneous, 4 nonhomogeneous, 4, 119, 138 ordinary, 178 (See also Partial differential equations) Differential operator, 106 Differentiation of series, 92, 95, 108, 116 Diffusion: coefficient of, 9 equation of, 9 Diffusivity, thermal, 6 Dirichlet, P. G. L., 128 Dirichlet condition, 26 Dirichlet kernel, 71, 220 Dirichlet problems, 26 in rectangle, 143, 334 in regions of plane, 143—146 in spherical regions, 312 Discrete spectrum, 171 Eigenfunctions, 113, 170 linear independence of, 164, 180 uniqueness of, 178 Eigenvalue problems (see Sturm-Liouville problems) Eigenvalues, 113, 170 continuous, 172, 230 nonnegative, 180 Elastic (vibrating) bar, 20, 24, 153, 159, 209—216 Elasticity, modulus of, 20 Elliptic type, partial differential equations of, 25, 38 Error function, 236 343 INDEX Euler, L., 128 Euler's constant, 247 Euler's formula, 30, 66, 254 Even function, 45, 57, 60, 195, 202 Even periodic extension, 50 Exponential form of Fourier series, 89 Extension, periodic, 58, 76, 84, 160 even, 50, 155 odd, 52, 159 Flux of heat, 4 Formal solutions, 111, 130, 135 Fourier, J. B. J., 128 Fourier-Bessel series, 273 in two variables, 288, 290 Fourier constants, 48 Fourier cosine integral formula, 229 Fourier cosine series, 49, 50, 84 convergence of, 78, 84—85 differentiation of, 95 Fourier integral formula, 218 applications of, 232—241 exponential form of, 228 symmetric form of, 232 Fourier integral theorem, 224 Fourier method, 29, 105 Fourier series, 39 convergence of, 73, 76, 83—84 differentiation of, 95 exponential form of, 89 generalized, 48 integration of, 96 in two variables, 161 uniform convergence of, 92 Fourier sine integral formula, 229 Fourier sine series, 52, 84 convergence of, 78, 84—85 Fourier theorem, 73 Fourier transform, 29 exponential, 228n. Fourier's law, 4 Fourier's ring, 166 Frobenius, method of, 243n. Function space, 39, 41, 69, 105 Functions: antiperiodic, 189, 192, 193, 211, 216 average value of, 61 Bessel (see Bessel functions) error, 236 even, 45, 57, 60, 195, 202 gamma, 248 generating: for Bessel functions, 259 for Legendre polynomials, 307 harmonic (see Harmonic functions) inner product of, 42, 172 Legendre, of second kind, 298, 306, 312 mean value of, 61 normalized, 44, 172 norms of, 43 odd, 46, 57, 60, 202 periodic, 73, 82 piecewise continuous, 40 boundedness of, 40 piecewise smooth, 69, 75, 273, 308 potential, 9 sine integral, 103n. trigonometric, orthonormal sets of, 44—46, 48, 52, 57, 99 closed, 49 complete, 98 uniformly continuous, 221 weight, 172, 173 Fundamental interval, 42 Gamma function, 248 Generalized Fourier series, 48 Generating function: for Bessel functions, 259 for Legendre polynomials, 307 Gibbs phenomenon, 93 344 INDEX Hankel's integral formula, 275 Harmonic functions, 9 in circular regions, 145, 149, 286 in rectangular region, 143 in spherical regions, 312—315, 317, 318 in strip, 239, 240 Hartley, R. V. L., 232n. Heat conduction, postulates of, 4, 5 Heat equation, 6 solutions of: product of, 166, 241 uniqueness of, 324 Heat transfer at surface, 7—8, 14, 137, 193, 282 Homogeneous equations, 4, 107, 110 Hooke's law, 20 Hyperbolic type, partial differential equations of, 25, 37 Improper integral, uniform convergence of, 225 Inequality: Bessel's, 64, 67 Cauchy's, 47, 91, 102 Schwarz, 47 Initial value problem, 178 Inner product, 42, 172 Insulated surface, 8 Integral forms of Bessel functions, 255, 256 Integral formula: Hankel's, 275 (See also Fourier integral formula) Integral theorem, Fourier, 224 Integration of series, 92, 96 Kronecker, L., 54 Kronecker's 6, 44 Lagrange's identity, 177, 269 Laplace transforms, 29 Laplace's equation, 7, 9 in polar coordinates, 12 in spherical coordinates, 13 Laplacian, 7 in cylindrical coordinates, 12 in polar coordinates, 12 in spherical coordinates, 13 Least squares approximation, 62n. Left-hand derivative, 68 Legendre functions of second kind, 298, 306, 312 Legendre polynomials, 298 closed orthonormal sets of, 300 derivatives of, 303, 305 generating function for, 307 graphs of, 298 integrals of, 304 norms of, 303—304 orthogonal sets of, 300—301 orthonormal sets of, 304 recurrence relations for, 303 Rodrigues' formula for, 301 series of, 308 zeros of, 311 Legendre series, 308 Legendre's equation, 293 self-adjoint form of, 299 Leibnitz, G. W., 127 Leibnitz' rule, 301 Limits: in mean, 98 one-sided, 40 Linear boundary condition, 3 homogeneous, 4 Linear combinations, 39, 106, 107 extensions of: by integrals, 232 by series, 110 Linear differential equation, 3 homogeneous, 4 Linear independence of eigenfunctions, 164, 180 345 INDEX Linear operators, 105 adjoints of, 170 differential, 106 product of, 106 self-adjoint, 170, 177 sum of, 106 Liouville, J., 169n. M-test, Weierstrass, for uniform convergence, 92, 225 Mapping, conformal, 29 Mean: best approximation in, 62 limit in, 98 Mean convergence, 97 Mean square error, 62 Mean value of function, 61 Membrane: analogy, 23 partial differential equations for, 22 static, 22, 24 (See also Vibrating membrane) Method: Fourier, 29, 105 of Frobenius, 243n. of separation of variables, 113 of undetermined coefficients, 160n. of variation of parameters, 138, 140, 142, 151, 155, 160, 207—209, 216, 280, 285, 291, 319 Modified Bessel functions, 253, 266, 286, 287 Modulus of elasticity, 20 Neumann condition, 26, 317 Newton, I., 127 Newton's law of cooling, 8, 14, 194, 282 Newton's second law of motion, 18, 20 Nonhomogeneous partial differential equations, 4, 119, 138, 142, 280 Normalized functions, 44, 172 Norms: of Bessel functions, 270—271 of functions, 43 of Legendre polynomials, 303—304 Odd function, 46, 57, 60, 202 Odd periodic extension, 52 One-sided derivatives, 67 One-sided limits, 40 Operators: linear, 105 product of, 106 sum of, 106 linear differential, 106 adjoints of, 170 self-adjoint, 170, 177 Ordinary point, 293 Orthogonal sets, 44, 172 of Bessel functions, 263, 269 of Legendre polynomials, 300—301 Orthogonality, 43 of eigenfunctions, 173 with respect to weight function, 172 Orthonormal sets, 44 of Bessel functions, 270 closed, 49, 98n, 103, 300 complete, 98, 99 of Legendre polynomials, 304 of trigonometric functions, 44—46, 48, 49, 52, 57, 98, 99 Parabolic type, partial differential equations of, 25, 38 Parseval's equation, 98, 100, 258 346 INDEX Partial differential equations: of diffusion, 9 for elastic bar, 21 general linear, of second order, 3, 24 general solutions of, 26 for membrane, 22 for stretched string, 18, 19 types of, 25, 38 Periodic boundary conditions, 163, 172 Periodic extension, 58, 76, 84, 160 even, 50, 155 odd, 52, 159 Periodic functions, 73, 82 Piecewise continuous functions, 40 boundedness of, 40 Piecewise smooth functions, 69, 75, 273, 308 Piecewise smooth surface, 324, 328 Poisson's equation, 7, 24, 151 uniqueness of solution of, 328 Poisson's integral formula, 24, 167 Polynomials, Legendre (see Legendre polynomials) Postulates of heat conduction, 4, 5 Potential, 9 in space bounded by planes, 145 in spherical region, 312 Principle of superposition of solutions, 107 Recurrence relations: for Bessel functions, 249 for Legendre polynomials, 303 Regular Sturm-Liouville problems, 169 Resonance, 157, 216 Riemann-Lebesgue lemma, 71, 221, 257 Right-hand derivative, 68 Robin condition, 26 Rodrigues' formula, 301 Schwarz inequality, 47 Self-adjoint operator, 170, 177 Semi-infinite solid, temperatures in, 234, 238, 240 Semi-infinite vibrating string, 36—37, 239 Separated boundary conditions, 169 Separated solutions, 30 Separation constant, 111 Separation of variables, method of, 113 Series: differentiation of, 92, 95, 108, 116 Fourier-Bessel, 273 integration of, 92, 96 Legendre, 308 Sturm-Liouville, 173, 187—193 superposition of solutions by, 32, 110 (See also Fourier cosine series; Fourier series; Fourier sine series; Uniform convergence) Sine integral function, 103n. Singular point, 243 regular, 247 Singular Sturm-Liouville problems, 171, 229, 235, 237, 264, 299, 300 Solution: approximation of, 123, 133 d'Alembert's, 28 separated, 30 Space of functions, 39, 41, 69, 105 Specific heat, 5 Spectrum, 170 continuous, 172 discrete, 171 Spherical coordinates, 12 laplacian in. 13 Spherical regions: Dirichlet problems in, 312 harmonic functions in, 312—315, 317, 318 347 INDEX Steady temperatures, 7 in cylinder, 284, 287 hollow, 166 in disk, 164, 335 in hemisphere, 315, 317, 319 in plate: rectangular, 145, 149, 150, 330, 334 semi-infinite, 150 in rod, 147 in slab, 13 in sphere, 312, 317, 318 hollow, 14 in wedge-shaped plate, 149 String (see Vibrating string) Sturm, J. C. F., 169n. Sturm-Liouville equation, 169 Sturm-Liouville problems, 112, 120, 129, 163, 168—187 regular, 169 singular, 171, 229, 235, 237, 264, 299, 300 Sturm-Liouville series, 173, 187—193 Superposition of solutions, 34, 107, 110, 119 by integrals, 232 by series, 32, 110 Surface: insulated, 8 piecewise smooth, 324, 328 Surface conductance, 8 Taylor, B., 128 Telegraph equation, 25 Temperatures: in bar, 111, 141, 318 in cylinder, 278—287 in prism, 165 in semi-infinite solid, 234, 238, 240 in slab, 8, 110, 130—141, 194, 203—207, 327 in sphere, 136, 208 in unlimited medium, 237 in wedge, 290, 291 in wire, 15, 137 (See also Steady temperatures) Tension: in membrane, 21 in string, 17 Thermal conductivity, 4 Thermal diffusivity, 6 , 48 Tilde symbol, Transform: Fourier, 29, 228n. Laplace, 29 Trigonometric functions, orthonormal sets of, 44—46, 48, 49, 52, 57, 98, 99 Undetermined coefficients, method of, 160n. Uniform convergence: Abel's test for, 117, 322 Cauchy criterion for, 322 of Fourier series, 92 of improper integrals, 225 of series, 92 Uniformly continuous functions, 221 Uniqueness: of eigenfunctions, 178 of solutions, 321 of heat equation, 324 of Laplace's equation, 328 of ordinary differential equations, 178 of Poisson's equation, 328 of wave equation, 333 Unstable component, 157 Variation of parameters, method of, 138, 140, 142, 151, 155, 160, 207—209, 216, 280, 285, 291, 319 348 INDEX Vectors, 42, 48—49, 63—64 Vibrating (elastic) bar, 20, 24, 153, 159, 209—216 Vibrating membrane, 21 circular, 287—290 Vibrating string, 16—19, 36—37, 119—127, 151, 155—158, 214 approximating problem for, 123 end conditions for, 19 equation of motion of, 18, 19 initially displaced, 32, 119—127, 153, 158, 214 with air resistance, 158 semi-infinite, 36—37, 239 Wave equation, 18—22, 239 solution of: general, 28 uniqueness of, 333 Weber's Bessel function of second kind, 247 Weierstrass M-test for uniform convergence, 92, 225 Weight function, 172, 173 Zeros: of Bessel functions, 259—263, 269 of Legendre polynomials, 311