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Some suﬃcient conditions for the existence of bounded nonoscillatory solutions of this equation are established by using Schauder fixed point theorem and Krasnoselskii fixed point theore

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Volume 2010, Article ID 767620, 14 pages

doi:10.1155/2010/767620

Research Article

Solvability of a Higher-Order Nonlinear Neutral Delay Difference Equation

Min Liu and Zhenyu Guo

School of Sciences, Liaoning Shihua University, Fushun, Liaoning 113001, China

Correspondence should be addressed to Zhenyu Guo,guozy@163.com

Received 19 March 2010; Revised 10 July 2010; Accepted 5 September 2010

Academic Editor: S Grace

Copyrightq 2010 M Liu and Z Guo This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

The existence of bounded nonoscillatory solutions of a higher-order nonlinear neutral delay diﬀerence equation Δakn · · · Δa 2n Δa 1n Δx n b n x n−d fn, x n−r 1n , x n−r 2n , , x n−r sn 0, n ≥ n0,

where n0 ≥ 0, d > 0, k > 0, and s > 0 are integers, {a in}n≥n0 i 1, 2, , k and {b n}n≥n0are real sequences,s

j1 {r jn}n≥n0 ⊆ Z, and f : {n : n ≥ n0} ×Rs → R is a mapping, is studied Some

suﬃcient conditions for the existence of bounded nonoscillatory solutions of this equation are established by using Schauder fixed point theorem and Krasnoselskii fixed point theorem and expatiated through seven theorems according to the range of value of the sequence{b n}n≥n0 Moreover, these suﬃcient conditions guarantee that this equation has not only one bounded nonoscillatory solution but also uncountably many bounded nonoscillatory solutions

1 Introduction and Preliminaries

Recently, the interest in the study of the solvability of diﬀerence equations has been increasing

see 1 17

diﬀerence equations For example,

ΔanΔxn pn x g n 0, n ≥ 0 1.1

see 14

ΔanΔxn qn x n1, ΔanΔxn qn f xn1, n ≥ 0 1.2

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see 11

Δ2

x n pxn −m pn x n −k − qn x n −l 0, n ≥ n0 1.3

see 6

Δ2

x n pxn −k fn, xn 0, n ≥ 1 1.4

see 10

Δ2

x n − pxn −τm

i1

q i f ixn −σ i , n ≥ n0 1.5

see 9

ΔanΔxn bxn −τ fn, xn −d 1n , x n −d 2n , , x n −d kn cn , n ≥ n0 1.6

see 8

Δm xn cxn −k pn x n −r 0, n ≥ n0 1.7

see 15

Δm xn cn x n −k pn f xn −r 0, n ≥ n0 1.8

see 3,4,12,13

Δm xn cxn −k u

s1

p n s f sxn −r s qn , n ≥ n0 1.9

see 16

Δm xn cxn −k pn x n −r − qn x n −l 0, n ≥ n0 1.10

see 17

Motivated and inspired by the papers mentioned above, in this paper, we investigate the following higher-order nonlinear neutral delay diﬀerence equation:

Δakn · · · Δa 2n Δa 1n Δxn bn x n −d fn, xn −r 1n , x n −r 2n , , x n −r sn 0, n ≥ n0, 1.11

where n0≥ 0, d > 0, k > 0, and s > 0 are integers, {ain} n ≥n0i 1, 2, , k and {bn} n ≥n0are real sequences,s

j1{rjn} n ≥n ⊆ Z, and f : {n : n ≥ n0} × Rs → R is a mapping Clearly, diﬀerence

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equations1.1–1.10 are special cases of 1.11 By using Schauder fixed point theorem and Krasnoselskii fixed point theorem, the existence of bounded nonoscillatory solutions of1.11

is established

Lemma 1.1 Schauder fixed point theorem Let Ω be a nonempty closed convex subset of a Banach

space X Let T : Ω → Ω be a continuous mapping such that TΩ is a relatively compact subset of X.

Then T has at least one fixed point in Ω.

Lemma 1.2 Krasnoselskii fixed point theorem Let Ω be a bounded closed convex subset of a

Banach space X, and let T1, T2 : Ω → X satisfy T1x T2y ∈ Ω for each x, y ∈ Ω If T1 is a contraction mapping and T2is a completely continuous mapping, then the equation T1x T2x x has

at least one solution in Ω.

The forward diﬀerence Δ is defined as usual, that is, Δx n xn1− xn The higher-order diﬀerence for a positive integer m is defined as Δ m x n ΔΔm−1x n, Δ0x n xn Throughout this paper, assume that R −∞, ∞, N and Z stand for the sets of all positive integers and integers,

respectively, α inf{n − rjn : 1 ≤ j ≤ s, n ≥ n0}, β min{n0− d, α}, limn→ ∞n − rjn ∞,

1≤ j ≤ s, and l∞

β denotes the set of real sequences defined on the set of positive integers lager than β where any individual sequence is bounded with respect to the usual supremum norm x sup n ≥β |xn|

for x {xn} n ≥β ∈ l∞

β It is well known that l∞β is a Banach space under the supremum norm A subset

Ω of a Banach space X is relatively compact if every sequence in Ω has a subsequence converging to

an element of X.

for every ε > 0, there exists an integer N0such that

whenever i, j > N0for any x {xk} k ≥βinΩ

is relatively compact.

Let

A M, N x {xn} n ≥β ∈ l∞

β : M ≤ xn ≤ N, ∀n ≥ β for N > M > 0. 1.13

Obviously, A M, N is a bounded closed and convex subset of l∞

β Put

b lim sup

n→ ∞ b n , b lim inf

n→ ∞ b n 1.14

By a solution of 1.11, we mean a sequence {xn} n ≥β with a positive integer N0 ≥

n0 d |α| such that 1.11 is satisfied for all n ≥ N0 As is customary, a solution of1.11 is

said to be oscillatory about zero, or simply oscillatory, if the terms x nof the sequence{xn} n ≥β

are neither eventually all positive nor eventually all negative Otherwise, the solution is called nonoscillatory

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2 Existence of Nonoscillatory Solutions

In this section, a few suﬃcient conditions of the existence of bounded nonoscillatory solutions

of1.11 are given

Theorem 2.1 Assume that there exist constants M and N with N > M > 0 and sequences

{ain} n ≥n0 1 ≤ i ≤ k, {bn} n ≥n0, {hn} n ≥n0, and {qn} n ≥n0such that, for n ≥ n0,

b n ≡ −1, eventually, 2.1

f n, u1, u2, , u s − fn, v1, v2, , v s ≤ hnmax{|ui− vi| : ui , v i

2.2

f n, u1, u2, , u s ≤ qn , u i 2.3

∞

t n0

max

1

|ait| , h t , q t: 1≤ i ≤ k < ∞. 2.4

Then1.11 has a bounded nonoscillatory solution in AM, N.

Proof Choose L ∈ M, N By 2.1, 2.4, and the definition of convergence of series, an

integer N0> n0 d |α| can be chosen such that

b n ≡ −1, ∀n ≥ N0, 2.5

∞

j1

∞

t1N0jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i

≤ min{L − M, N − L}. 2.6

Define a mapping T L : AM, N → X by

TL xn

⎧

⎪

L− −1k∞

j1

∞

t1njd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

f t, xt −r 1t , x t −r 2t , , x t −r st

k

i1a it i

, n ≥ N0,

TL xN0, β ≤ n < N0

2.7

for all x ∈ AM, N.

i It is claimed that TL x ∈ AM, N, for all x ∈ AM, N.

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In fact, for every x ∈ AM, N and n ≥ N0, it follows from2.3 and 2.6 that

TL xn ≥ L −∞

j1

∞

t1njd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

f t, xt −r

1t , x t −r 2t , , x t −r st

k

i1a it i

≥ L −∞

j1

∞

t1N0jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i

≥ M,

TL xn ≤ L ∞

j1

∞

t1N0jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i

≤ N.

2.8

That is,TL x AM, N ⊆ AM, N.

ii It is declared that TLis continuous

Let x {xn} ∈ AM, N and x u {x n u } ∈ AM, N be any sequence such that

x u n → xn as u → ∞ For n ≥ N0,2.2 guarantees that

TL x u n − TL x n

≤∞

j1

∞

t1njd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

ft, x u t −r

1t , x u t −r 2t , , x u t −r

st

− ft, xt −r 1t , x t −r 2t , , x t −r st

k

i1a it i

≤∞

j1

∞

t1njd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

h tmaxx u

t −r jt − xt −r jt

 : 1 ≤ j ≤ s

k

i1a it i

≤x u − x∞

j1

∞

t1N0jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

h t

k

i1a it i.

2.9

This inequality and2.4 imply that TLis continuous

iii It can be asserted that TL A M, N is relatively compact.

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By2.4, for any ε > 0, take N3≥ N0large enough so that

∞

j1

∞

t1N3jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i

<

ε

2. 2.10

Then, for any x {xn} ∈ AM, N and n1, n2≥ N3,2.10 ensures that

|TL x n1− TL x n2| ≤∞

j1

∞

t1n1jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

f t, xt −r

1t , x t −r 2t , , x t −r st

k

i1a it i

∞

j1

∞

t1n2jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i

≤∞

j1

∞

t1N3jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i

∞

j1

∞

t1N3jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i

< ε

2 ε

2  ε,

2.11

which means that TL A M, N is uniformly Cauchy Therefore, byLemma 1.4, TL A M, N is

relatively compact

ByLemma 1.1, there exists x {xn} ∈ AM, N such that TL x x, which is a bounded

nonoscillatory solution of1.11 In fact, for n ≥ N0 d,

x n L − −1 k∞

j1

∞

t1njd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i ,

x n −d L − −1 k∞

j1

∞

t nj−1d

∞

t t · · · ∞

t t−1

∞

t t

k

i1a it ,

2.12

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which derives that

x n − xn −d −1k∞

j1

n jd−1

t1nj−1d

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i ,

Δxn − xn −d −1k∞

j1

njd

t1n1j−1d

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i

− −1k∞

j1

n jd−1

t1nj−1d

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i

−−1k∞

j1

∞

t2nj−1d

∞

t3t2

· · · ∞

t k t k−1

∞

t t k

a1nj−1dk

i2a it i

−1k∞

j1

∞

t2njd

∞

t3t2

· · · ∞

t k t k−1

∞

t t k

a1njdk

i2a it i

−1k−1∞

t2n

∞

t3t2

· · · ∞

t k t k−1

∞

t t k

a 1n k

i2a it i .

2.13

That is,

a 1n Δxn − xn −d −1k−1∞

t2n

∞

t3t2

· · · ∞

t k t k−1

∞

t t k

k

i2a it i , 2.14

by which it follows that

Δa 1n Δxn − xn −d −1k−1 ∞

t2n1

∞

t3t2

· · · ∞

t k t k−1

∞

t t k

k

i2a it i

−−1k−1∞

t2n

∞

t3t2

· · · ∞

t k t k−1

∞

t t k

k

i2a it i

−1k−2∞

t3n

∞

t4t3

· · · ∞

t k t k−1

∞

t t k

a 2n k

i3a it i

,

Δakn · · · Δa 2n Δa 1n Δxn bn x n −d −1k −k1 f n, xn −r 1n , x n −r 2n , , x n −r sn

 −fn, xn −r 1n , x n −r 2n , , x n −r sn .

2.15

Therefore, x is a bounded nonoscillatory solution of1.11 This completes the proof

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Remark 2.2 The conditions of Theorem 2.1 ensure the 1.11 has not only one bounded nonoscillatory solution but also uncountably many bounded nonoscillatory solutions

In fact, let L1, L2 ∈ M, N with L1/ L2 For L1 and L2, as the preceding proof

in Theorem 2.1, there exist integers N1, N2 ≥ n0  d |α| and mappings TL1, T L2

satisfying 2.5–2.7, where L, N0 are replaced by L1, N1 and L2, N2, respectively, and ∞

j1∞

t1N4jd∞

t2t1· · ·∞t k t k−1∞t t k ht /|k

i1a it i | < |L1 − L2|/2N for some N4 ≥ max{N1, N2} Then the mappings TL1and TL2have fixed points x, y ∈ AM, N, respectively,

which are bounded nonoscillatory solutions of1.11 in AM, N For the sake of proving

that1.11 possesses uncountably many bounded nonoscillatory solutions in AM, N, it is only needed to show that x / y In fact, by 2.7, we know that, for n ≥ N4,

x n L1− −1k∞

j1

∞

t1njd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i

,

y n L2− −1k∞

j1

∞

t1njd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

f

t, y t −r 1t , y t −r 2t , , y t −r st k

i1a it i

.

2.16

Then,

x n − yn ≥ |L1− L2|

−∞

j1

∞

t1njd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

f t, xt −r 1t , x t −r 2t , , x t −r st − f

t, y t −r 1t , y t −r 2t , , y t −r st

k

i1a it i

≥ |L1− L2| −x − y∞

j1

∞

t1N4jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

h t

k

i1a it i

≥ |L1− L2| − 2N∞

j1

∞

t1N4jd

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

h t

k

i1a it i

> 0, n ≥ N4,

2.17

that is, x / y.

Theorem 2.3 Assume that there exist constants M and N with N > M > 0 and sequences

{ain} n ≥n0 1 ≤ i ≤ k, {bn} n ≥n0, {hn} n ≥n0, {qn} n ≥n0, satisfying2.2–2.4 and

b n ≡ 1, eventually. 2.18

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Proof Choose L ∈ M, N By 2.18 and 2.4, an integer N0> n0 d |α| can be chosen such

that

b n ≡ 1, ∀n ≥ N0,

∞

j1

N02jd−1

t1N02j−1d

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i ≤ min{L − M, N − L}. 2.19 Define a mapping T L : AM, N → X by

TL xn

⎧

⎪

L −1k∞

j1

n 2jd−1

t1n2j−1d

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i

, n ≥ N0,

TL xN0, β ≤ n < N0

2.20

The proof that TL has a fixed point x {xn} ∈ AM, N is analogous to that in

Theorem 2.1 It is claimed that the fixed point x is a bounded nonoscillatory solution of1.11

In fact, for n ≥ N0 d,

x n L −1 k∞

j1

n 2jd−1

t1n2j−1d

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i ,

x n −d L −1 k∞

j1

n 2j−1d−1

t1n2j−1d

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i

,

2.21

by which it follows that

x n xn −d 2L −1 k∞

j1

n jd−1

t1nj−1d

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i . 2.22 The rest of the proof is similar to that inTheorem 2.1 This completes the proof

Theorem 2.4 Assume that there exist constants b, M, and N with N > M > 0 and sequences

{ain} n ≥n01 ≤ i ≤ k, {bn} n ≥n0, {hn} n ≥n0, {qn} n ≥n0, satisfying2.2–2.4 and

|bn| ≤ b < N − M

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Proof Choose L ∈ M bN, N − bN By 2.23 and 2.4, an integer N0> n0 d |α| can be

chosen such that

|bn| ≤ b < N − M

2N , ∀n ≥ N0,

∞

t1N0

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i

≤ min{L − bN − M, N − bN − L}. 2.24 Define two mappings T 1L , T 2L : AM, N → X by

T 1L xn

⎧

⎨

⎩

L − bn x n −d , n ≥ N0,

T 1L xN0, β ≤ n < N0,

T 2L xn

⎧

⎪

−1k∞

t1n

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

k

i1a it i

, n ≥ N0,

T 2L xN0, β ≤ n < N0

2.25

i It is claimed that T 1L x T 2L y ∈ AM, N, for all x, y ∈ AM, N.

In fact, for every x, y ∈ AM, N and n ≥ N0, it follows from2.3, 2.24 that

T 1L x T 2L y

n ≥ L − bN − ∞

t1N0

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i

≥ M,

T 1L x T 2L y

n ≤ L bN  ∞

t1N0

∞

t2t1

· · · ∞

t k t k−1

∞

t t k

q t

k

i1a it i

≤ N.

2.26

That is,T 1L x T 2L y AM, N ⊆ AM, N.

ii It is declared that T 1L is a contraction mapping on AM, N.

In reality, for any x, y ∈ AM, N and n ≥ N0, it is easy to derive that

T 1L xn−T 1L y

n ≤ |bn|x n −d − yn −d ≤ bx − y, 2.27 which implies that

T 1L x − T 1L y ≤ bx − y. 2.28

Then, b < N − M/2N < 1 ensures that T 1L is a contraction mapping on AM, N.

iii Similar to ii and iii in the proof ofTheorem 2.1, it can be showed that T 2Lis completely continuous

ByLemma 1.2, there exists x {xn} ∈ AM, N such that T 1L x T 2L x x, which is a

bounded nonoscillatory solution of1.11 This completes the proof

Tiêu đề	Solvability of a Higher-Order Nonlinear Neutral Delay Difference Equation
Tác giả	Min Liu, Zhenyu Guo
Trường học	Liaoning Shihua University
Chuyên ngành	Mathematics
Thể loại	bài báo
Năm xuất bản	2010
Thành phố	Fushun

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