OSCILLATION OF HIGHER ORDER
NEUTRAL TYPE
DIFFERENCE EQUATIONS
A. ZAFER
Department of Mathematics
Middle East Technical University
06531 Ankara, TURKEY
Abstract In this work we are concerned with oscillation
of solutions of the neutral difference equation of the form
∆p [xn − pn xn−l ] + δqn f (xn−k ) = hn ,
n ≥ n0 , δ = ±1,
where ∆ is the forward difference operator ∆xn = xn+1 −xn ,
{pn }, {qn } and {hn } are real sequences, and l and k are
integers. A necessary and sufficient condition is obtained
under which every bounded solution {xn } is oscillatory
when (−1)p δ = −1, and is either oscillatory or such that
limn→∞ xn = 0 when (−1)p δ = 1.
1
INTRODUCTION
We consider the neutral difference equation
∆p [xn − pn xn−l ] + δqn f (xn−k ) = hn
n ≥ n0
(1.1)
where δ = ±1 and n ∈ N(n0 ) = {n0 , n0 + 1, . . .}, n0 a fixed positive
integer. We assume that l and k are integers with l > 0, x, p, q, and f
are real valued functions and f is continuous.
By a solution of Eq. (1.1) we mean a real sequence {xn } satisfying
Eq. (1.1) so that supn≥m |xn | > 0 for any m ∈ N(n0 ). We always
assume that such solutions of Eq. (1.1) exist. A solution of Eq. (1.1)
is called oscillatory if there is no end of n1 and n2 (n1 < n2 ) in N(n0 )
such that xn1 xn2 ≤ 0; otherwise it is called nonoscillatory. Clearly, a
nonoscillatory solution must be eventually of fixed sign.
A. ZAFER
The oscillatory behavior of solutions of first and second order difference equations has been extensively studied by many authors (see
[1–8] and the references cited therein). However, much less has been
done for higher order equations. For some results regarding the oscillation and as well as the asymptotic behavior of higher order difference
equations, we refer in particular to [9–13]. In [11] the present author
established sufficient conditions for the existence of positive solutions
and oscillation of solutions of Eq. (1.1) when pn = c, c %= ±1. Here, an
attempt is made to extend to Eq. (1.1) some of the results obtained
previously for pn = c > 1.
2
MAIN RESULTS
In what follows n(s) denote the usual factorial function; that is, n(0) = 1
and n(s) = n(n − 1) . . . (n − s + 1).
Theorem 2.1 Let there exist two real numbers c and C such that
1 < c ≤ pn ≤ C, and assume that for some positive real numbers
a and b such that a/b < (c − 1)/C the function f satisfies a Lipschitz
condition with Lipschitz constant L on [a, b]. If
∞
!
s(p−1) |qs | < ∞,
(2.1)
s(p−1) |hs | < ∞,
(2.2)
s=n
∞
!
s=n
then Eq. (1.1) has a nonoscillatory bounded solution {xn }.
Proof Let K = max{|f (x)|/|x| : a ≤ x ≤ b} and M = max{K, L}.
Due to condition (2.1) and (2.2) we can find a sufficiently large n1 ≥ n0
such that
∞
!
(p − 1)!βc
s(p−1) |qs | <
,
(2.3)
2Mb
s=n+l
∞
!
s(p−1) |hs | <
s=n+l
(p − 1)!βc
,
2
(2.4)
and n ≥ max{n0 − l, n0 − l + k} for n ≥ n1 , where β is defined as
β=
(c − 1)b − Ca
.
c+C
NEUTRAL TYPE DIFFERENCE EQUATIONS
We introduce the Banach space Y of all functions x : N(n0 ) → R
such that
||x|| = sup |xn |.
n∈N (n0 )
Define
X = {x ∈ Y : a ≤ xn ≤ b, n ∈ N(n0 )}.
Clearly X is bounded, convex, and closed subset of Y . Let T be an
operator defined on X by
T xn =
1
"
α + xn+l
pn+l
∞
(−1)p !
+
(s + p − 1 − n − l)(p−1) δqs f (xs−k )
(p − 1)! s=n+l
∞
(−1)p−1 !
(s + p − 1 − n − l)(p−1) hs ,
+
(p − 1)! s=n+l
= T xn1 ,
for n ≥ n1 .
for n0 ≤ n ≤ n1 ,
(2.5)
where α = (a + β)C.
We shall show that T is a contraction mapping on X. It is easy to
see that T maps X into itself. In fact, if x ∈ X, then because of (2.3)
and (2.4) it follows that
T xn ≤
α 1
β β
+ ·b+ + =b
c
c
2
2
and
T xn ≥
α β β
− − = a.
C
2
2
Therefore T X ⊆ X.
To show that T is a contraction, let x, y ∈ X. Using (2.3), we easily
see from (2.5) that
1
|xn+l − yn+l |
c
∞
!
M
+
s(p−1) |qs ||xs−k − ys−k |
c(p − 1)! s=n+l
|T xn − T yn | ≤
≤
M
(p − 1)!βc
1
||x − y|| +
||x − y||,
c
c(p − 1)! 2Mb
A. ZAFER
and hence
|T xn − T yn | ≤
%
&
1
β
||x − y||.
+
c 2b
(2.6)
It is easy to verify that
1
β
+
< 1.
(2.7)
c 2b
In view of (2.6) and (2.7) it follows that T is a contraction on X, and
therefore there exists a fixed point x ∈ X such that T x = x. It can
easily be seen that x is a bounded positive solution of Eq. (1.1). This
completes the proof.
Theorem 2.2 Let 1 < c ≤ pn < C, {qn } be eventually positive and
xf (x) > 0 for x %= 0. Suppose that there exists an oscillatory function
ρ on N such that ∆p ρn = hn and limn→∞ ρn = 0. If
∞
!
s(p−1) qs = ∞,
(2.8)
then every bounded solution {xn } of Eq. (1.1) is oscillatory when
(−1)p δ = −1, and is either oscillatory or else limn→∞ xn = 0 when
(−1)p δ = 1.
Proof Suppose that Eq. (1.1) has a nonoscillatory solution {xn }. We
may assume that {xn } is eventually positive. Set
zn = xn − pn xn−l
and
yn = zn − ρn
We claim that zn is eventually negative; otherwise for sufficiently large
values of n
xn > pn xn−l
and by induction
xn+ml > cm xn .
(2.9)
But inequality (2.9) as m → ∞ contradicts the fact that xn is bounded,
and so zn must be eventually negative. Since ρn is oscillatory, it follows
that yn should also be eventually negative. Furthermore from Eq. (1.1),
δ∆p yn = −qn f (xn−k ) < 0.
NEUTRAL TYPE DIFFERENCE EQUATIONS
Since yn < 0 and δ∆p yn < 0, applying a lemma of Agarwal [9] it follows
that there is an integer l ∈ {0, 1} with (−1)p−l δ = 1 so that
∆j yn < 0 for j = 0, 1, . . . , l
(−1)j−l ∆j yn < 0 for j = l + 1, . . . , p − 1.
(2.10)
Multiplying (1.1) by n(p−1) and summing from n1 to n − 1 we obtain
n−1
!
s(p−1) δ∆p ys +
s=n1
n−1
!
s(p−1) qs f (xs−k ) = 0.
(2.11)
s=n1
Applying the summation by parts formula [9] to the first term on the
left hand side of (2.11) we have
n−1
!
s
(p−1)
p
δ∆ ys =
s=n1
p−1
!
(−1)k+1 ∆k−1 n(p−1) δ∆p−k yn+k−1 − K
k=1
+ (−1)p+1δ(p − 1)![yn+p−1 − yn1 +p−1 ]
(2.12)
where, in view of (2.10),
K=
p−1
!
(p−1)
(−1)k+1 δ∆k−1 n1
∆p−k yn1 +k−1 > 0.
k=1
Using (2.10) and (2.12) in (2.11) leads to
n−1
!
s(p−1) qs f (xs−k ) ≤ K + (−1)p δ(p − 1)![yn+p−1 − yn1 +p−1 ]. (2.13)
s=n1
Since {yn } is bounded and (2.8) holds it follows from (2.13) that
lim
inf f (xn−k ) = 0,
n→∞
and hence
lim inf xn = 0.
n→∞
On the other hand, the bounded sequence {yn } being negative and
monotone has a finite limit L ≤ 0. Moreover, L = 0 is possible only
when l = 0, and L < 0 may hold if l = 1.
Suppose L < 0. Clerly limn→∞ zn = L and therefore we can choose
n2 ∈ N(n0 ) such that for n ∈ N(n2 ),
2L < xn − pn xn−k <
L
.
2
A. ZAFER
But then
−L
2C
for n ∈ N(n2 ), a contradiction with lim inf n→∞ xn = 0. Thus, we must
have L = 0 and l = 0. Using the arguments developed in [12] one can
easily show that
lim xn = 0.
xn−k >
n→∞
This completes the proof.
In view of Theorem 2.1 and Theorem 2.2 we obtain the following
necessary and sufficient condition for the oscillation of bounded solutions of Eq. (1.1).
Theorem 2.3 Let 1 < c ≤ pn < C, {qn } be eventually positive and
xf (x) > 0 for x %= 0. Assume that
(i) there exists an oscillatory function ρ on N such that
∆p ρn = hn
and
limn→∞ ρn = 0
(ii) 1 < c ≤ pn ≤ C
(iii) (2.2) is satisfied
(iv) for some positive real numbers a and b, a/b < (c−1)/C, f satisfies
a Lipschitz condition on the interval [a, b].
Then a necessary and sufficient condition for every bounded solution
{xn } of Eq. (1.1) to be oscillatory when (−1)p δ = −1, and be either
oscillatory or limn→∞ xn = 0 when (−1)p δ = 1 is that
∞
!
s(p−1) qs = ∞.
Example. Consider the difference equation
∆4 [xn + (3 + (−1)n )xn−3 ] − xn−4 = 33(−1)n−1 ,
(2.14)
so that p = 4, pn = 3 + (−1)n , qn = 1, δ = −1, and hn = 33(−1)n−1 .
It is easy to see that c = 2, C = 4, and the conditions of Theorem 2.3
are satisfied. Therefore, all bounded solutions of (2.14) are oscillatory.
Indeed, xn = (−1)n is such a solution of (2.14).
NEUTRAL TYPE DIFFERENCE EQUATIONS
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