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SOLUTIONS OF A HIGHER ORDER NEUTRALDIFFERENCE EQUATION ZHI-QIANG ZHU, GEN-QIANG WANG, AND SUI SUN CHENG Received 25 May 2005; Revised 21 September 2005; Accepted 28 September 2005 Nonosc

SOLUTIONS OF A HIGHER ORDER NEUTRAL

DIFFERENCE EQUATION

ZHI-QIANG ZHU, GEN-QIANG WANG, AND SUI SUN CHENG

Received 25 May 2005; Revised 21 September 2005; Accepted 28 September 2005

Nonoscillatory solutions of a nonlinear neutral type higher order difference equations are classified by means of their asymptotic behaviors By means of the Kranoselskii’s fixed point theorem, existence criteria are then provided for justification of such classification Copyright © 2006 Zhi-qiang Zhu et al 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

1 Introduction

Classification schemes for nonoscillatory solutions of nonlinear difference equations are important since further investigations of some of the qualitative behaviors of nonoscilla-tory solutions can then be reduced to only a number of cases There are several studies which provide such classification schemes for difference equations, see, for example, [4–

11] In particular, in [7], a class of nonlinear neutral difference equations of the form

Δmx n+c n x n − k

+ fn,x n − l

wherem, k and l are integers such that m ≥2,k > 0 and l ≥0 is studied and classification schemes are given when{ c n }is a nonnegative constant sequence{ c0}, and in [10], the same equation is studied with odd integerm ≥1, positive integerk, integer l and { c n } = {−1}

In this paper, we continue our investigation on the possible types of nonoscillatory solutions when{ c n } ⊆(−1, 0] and limn →∞ c n = c0(while the case{ c n } ⊆(−∞,−1] will be discussed elsewhere) Besides the assumption that{ c n } ⊆(−1, 0], we will assume further that f is a continuous function defined on {0, 1, } × R such that f = f (n,x) is

nonde-creasing in the second variablex and satisfies x f (n,x) > 0 for x =0 andn ≥0

We will accomplish two things in this paper: to provide a classification scheme for the nonoscillatory solutions of (1.1) inSection 2and establish inSection 3sufficient and/or necessary criteria for the existence of solutions in each class There are no overlapping

Hindawi Publishing Corporation

Advances in Di fference Equations

Volume 2006, Article ID 47654, Pages 1 19

DOI 10.1155/ADE/2006/47654

results between our paper and [4–11], although some proofs are similar However, the existence proofs are different in that Cheng-Patula existence theorem is applied in [7], monotone method is used in [10] while we use Krasnoselskii fixed point theorem here

We remark further that classification scheme is also provided for neutral differential equa-tions in [2]

Before we go into details, we will need some preparatory terminologies and results First of all, given initialx ifor−max{ k,l } ≤ i ≤0, we may calculate from (1.1)x1,x2,x3,

in a recursive manner Such a sequence{ x n }is said to be a solution of (1.1) Among the solutions of (1.1), one is said to be nonoscillatory if it is eventually positive or eventually negative

Given an integera, it is convenient to set

Given an integerα ≥0, the generalized factorial functiong(x) = x(α)is defined as fol-lows

x(α) =

⎧

⎨

⎩

x(x −1)(x −2)···(x − α + 1) α > 0

It is well known thatΔn(α) = αn(α −1)forα > 0 (see, e.g., [3])

Let

l ∞

N0=



x =x n

n ≥ 0: sup

n ≥ 0

x n

r n < ∞

where N0> 0 is an integer and { r n } n ≥ 0 is a positive sequence with a uniform posi-tive lower bound When endowed with the usual linear structure and the norm x  =

supn ≥ 0(| x n | /r n), (l ∞

N0, · ) is a Banach space A setB ⊆ l ∞

N0is said to be uniformly Cauchy

if for anyε > 0 there exists an integer M ≥ N0such that

x i

r i − x j

for allx = { x n } ∈ B.

Lemma 1.1 A bounded and uniformly Cauchy subset B ⊆ l ∞

N0is relatively compact Proof By assumption, we know that for any such ε > 0, there exists an integer M ≥ N0> 0

such that for anyx ∈ B, we have

x i

r i − x j

r j < ε

LetΓ > 0 be a bound for B That is  x  ≤ Γ for all x ∈ B Choose integers M n,n = N0,N0+

1, ,M, and numbers y(1)

n < y(2)

n < ··· < y(M n)

n such thaty(1)

n = − r n Γ, y(M n)

n = r nΓ and

y(j+1) n

r n − y(j)

n

r n < ε

Now define a sequence{ v k } kN0as follows Letv N0be one of the values{ y(1)

N0, , y(M N0)

N0 },

v N0 +1be one of the values{ y(1)

N0 +1, , y(M N0+1)

N0 +1 } In general, forN0≤ k ≤ M, let v kequal one

of the values{ y(1)

k , , y(M k)

k } Fork > M, let v k =(r k /r M)v M It is clear that the sequence

{ v k } kN0 belongs tol ∞

N0 LetL be the set of all possible sequences { v k } k ≥ 0 defined as above Note thatL has M N0M N0 +1··· M M elements

We assert thatL is a finite ε-net for B It is sufficient to show that for any x = { x k } kN0∈

B, L contains a sequence v = { v k } kN0such that

 x − v  =sup

n ≥ 0

x n − v n

Indeed, by definition ofL, we can choose a sequence { v k } kN0inL such that

x k

r k − v k

r k < ε

Furthermore, by (1.6), (1.9), and the definition ofv = { v k } kN0, fork > M, we have

x k

r k − v k

r k

x k

r k − v M

r M

x k

r k − x M

r M + x M

r M − v M

r M

ε

3+

ε

3=2ε

From (1.9) and (1.10), we see that (1.8) holds The proof is complete 

Lemma 1.2 Suppose that lim n →∞ c n = c0with c0∈(− 1, 0] and the sequence { x n /n(i) } is eventually positive or eventually negative, where i is a nonnegative integer Suppose further that z n = x n+c n x n − k and lim n →∞(z n /n(i))= b Then lim n →∞(x n /n(i))= b/(1 + c0).

Proof Without loss of generality, we assume that x n /n(i) > 0 for any positive integer n.

In caseb is finite, we assert that { x n /n(i) }is bounded Otherwise, there would exist a sequence{ n λ }of integers withn λ → ∞forλ → ∞such that

lim

λ →∞

x n λ

n(i) λ

= ∞, x n ≤ x n λ, 0< n ≤ n λ (1.11)

On the other hand, we have

z n λ

n(i) λ

= x n λ

n(i) λ

+c n λ

x n λ − k

n(i) λ

≥1 +c n λ

x n λ

n(i) λ

asλ → ∞ This is contrary to the fact thatb is finite.

Let lim supn →∞(x n /n(i))= Q and liminf n →∞(x n /n(i))= q Then 0 ≤ q ≤ Q < ∞ More-over, there exist{ n λ }and{ n λ }such that limλ →∞ n λ =∞, limλ →∞ n λ =∞, limλ →∞(x n λ /n λ(i))=

Q and lim λ →∞(x n λ /n λ(i))= q Note that

b =lim

λ →∞

z n λ

n λ(i) =lim

λ →∞

x n λ

n λ(i)+c n λ

x n λ − k

n λ(i)

λ →∞

x n λ

n λ(i)+ lim

λ →∞infc n λ

x n λ − k



n λ − k(i)

n λ − k(i)

n λ(i) ≥ Q + c0Q,

b =lim

λ →∞

z n λ

n λ(i) =lim

λ →∞

x n λ

n λ(i)+c n λ

x n λ − k

n λ(i)

λ →∞

x n λ

n λ(i)+ lim

λ →∞supc n λ

x n λ − k



n λ − k(i)



n λ − k(i)

n λ(i) ≤ q + c0q,

(1.13)

we have (1 +c0)q ≥(1 +c0)Q It follows that q ≥ Q Hence q = Q and it implies that

limn →∞(x n /n(i)) exists In view ofz n = x n+c n x n − kand limn →∞(z n /n(i))= b, we have

lim

n →∞

x n

n(i) = b

In caseb is infinite, then b = ∞ orb = −∞ We assert thatb = −∞cannot hold In fact, for givenc1 with− c0< c1< 1, there exists a large integer N0such that− c n ≤ c1for

n ≥ N0 Hence, ifb = −∞, thenz n = x n+c n x n − k < 0 for n ≥ N and

x n < − c n x n − k ≤ c1x n − k, n ≥ N, (1.15) whereN ≥ N0is some positive integer It implies that

0< x N+λk < c1x N+(λ −1)k < ··· < c λ

So that limλ →∞ x N+λk =0 Thus

lim

which implies thatb = −∞is impossible

Now, for arbitraryM > 0, there exists a sufficiently large integer N such that

z n

n(i) = x n

n(i)+c n x n − k

It follows that

x n

That is limn →∞(x n /n(i))= ∞ The proof is complete  The following two propositions are respectively in [1, Theorems 1.7.9 and 1.7.11]

Lemma 1.3 Suppose that the sequence { x n } and { y n } satisfy the following conditions,

(i) y n > 0 and Δy n > 0 for all large integers n and lim n →∞ y n = ∞ , and

(ii) limn →∞(Δxn /Δy n)= b.

Then lim n →∞(x n /y n)=limn →∞(Δx n /Δy n)= b, where b can be finite or infinite.

Lemma 1.4 Let u = u(n) be a sequence defined for n ∈ N(a), u(n) > 0 with Δ m u(n) of constant sign on N(a) and not identically zero Then, there exists an integer m ∗ , 0 ≤ m ∗ ≤ m with m + m ∗ odd forΔm u(n) ≤ 0 or, m + m ∗ even forΔm u(n) ≥ 0 and such that

m ∗ ≤ m − 1 implies ( −1)m ∗+iΔi u(n) > 0 ∀ n ∈ N(a), m ∗+ 1≤ i ≤ m,

m ∗ ≥ 1 impliesΔi u(n) > 0 ∀ large n ∈ N(a), 1 ≤ i ≤ m ∗ (1.20) Remark 1.5 If u(n) < 0 inLemma 1.4, then there existsm ∗, 0≤ m ∗ ≤ m with m + m ∗

odd forΔm u(n) ≥0 or,m + m ∗even forΔm u(n) ≤0 and such that

m ∗ ≤ m −1 implies (−1)m ∗+iΔi u(n) < 0 ∀ n ∈ N(a), m ∗+ 1≤ i ≤ m,

m ∗ ≥1 impliesΔi u(n) < 0 ∀largen ∈ N(a), 1 ≤ i ≤ m ∗ (1.21) Lemma 1.6 (Kranoselskii’s fixed point theorem) Suppose B is a Banach space and Ω is a bounded, convex and closed subset of B Let U,S : Ω → B satisfy the following conditions.

(i)Ux + Sy ∈ Ω for any x, y ∈ Ω,

(ii)U is a contraction mapping, and

(iii)S is completely continuous.

Then U + S has a fixed point in Ω.

2 Classifications of nonoscillatory solutions

In the following discussions, we assume throughout that

lim

We set

whenever it is defined Equation (1.1) can now be written as

Δm z n = − fn,x n − l

We will propose a classification scheme for the nonoscillatory solutions of (1.1) For this purpose, we first note that ifx = { x n }is an eventually negative solution of (1.1), then

y = { y n }defined byy n = − x nwill satisfy

Δm

y n+cy n − k

+fn, y n − l

where



f (n,u) = − f (n, − u), n ∈ N(0), u ∈ R (2.5)

has the same properties satisfied by f , that is, f is a continuous function defined on

{0, 1, } × R such that f=  f (n,u) is nondecreasing in the second variable u and satisfies

u f (n,u) > 0 for u =0 andn ≥0 Therefore, we may restrict our attention to the setS+of all eventually positive solutions of (1.1) Motivated by the classification scheme in [2], we make use of the following notations for classifying our eventually positive solutions:

A k(α,β) =

x n

∈ S+: limn

→∞

x n

n(k −1) = α, lim n

→∞

x n

n(k) = β, k ≥1,

A0(α) =

x n

∈ S+: limn

Theorem 2.1 (a) Suppose that m is even If x = { x n } is an eventually positive solution

of (1.1), then either x ∈ A0(0) or there are some j ∈ {1, 2, ,m/2 } and a > 0 such that x belongs to A2j −1(∞,a), A2j −1(∞ , 0) or A2j −1(a,0) (b) Suppose that m is odd If x = { x n }

is an eventually positive solution of (1.1), then either x belongs to A0(α) for some α ≥ 0, or there are j ∈ {1, 2, ,(m −1)/2 } and a > 0 such that x belongs to A2j(∞,a), A2j(∞ , 0) or

A2j(a,0).

Proof Let m is even and x = { x n }be an eventually positive solution of (1.1) Then, in view of (2.3), there exists some integerN > 0 such that Δ m z n < 0 for n ≥ N Therefore, z n

is eventually of fixed sign For the sake of simplicity, we may assume that{ z n }is of fixed sign forn ≥ N.

First of all, supposez n < 0 for n ≥ N By the same reasoning as in the proof ofLemma 1.2, we may show that

lim

On the other hand, in view ofLemma 1.4, there exists some evenm ∗with 0≤ m ∗ ≤ m

such that eventuallyΔi z n < 0 for 0 ≤ i ≤ m ∗ and (−1)m ∗+iΔi z n < 0 for m ∗+ 1≤ i ≤ m.

There are now two cases to consider

Case 1 (m ∗ =0) Then we have eventually

By (2.8), we can set

lim

In view of (2.7), we find that limn →∞ z n =0 ByLemma 1.2, we have limn →∞ x n =0 Hence

x belongs to A0(0)

Case 2 (m ∗ ≥2) Then we have eventually

It implies limn →+∞ z n < 0 which is contrary to (2.7) Hencem ∗ ≥2 does not hold

Now we supposez n > 0 for n ≥ N Similar to the proof in [7, Theorem 1], we may see thatx belongs to A2j −1(∞,a), A2j −1(∞, 0) orA2j −1(a,0) for some j ∈ {1, 2, ,m/2 }and

a > 0.

Whenm is odd, the proof is similar to those above and hence is skipped The proof is

3 Existence criteria

Eventually positive (and by analog eventually negative) solutions of (1.1) have been clas-sified according toTheorem 2.1 We now justify our classification schemes by finding existence criteria for each type of solutions

Theorem 3.1 Suppose that m is even If (1.1) has a solution in A2j −1(∞,a) for some j ∈ {1, 2, ,m/2 } and a > 0, then there exists some K > 0 such that

∞



i =0

(i + m −2j)(m 2j)

(m −2j)! f



i,K(i − l)(2j −1) 

The converse is also true.

Proof First of all, we remark that

∞



i λ = n

∞



i λ −1= i λ

···

∞



i2= i3

∞



i1= i2

fi1,x i1− l

=

∞



i = n

(i − n + λ −1)(λ −1)

(λ −1)! fi,x i − l

Let x = { x n } be an eventually positive solution of (1.1) inA2j −1(∞,a) Then we may

suppose that there exists an integerN0> 0 such that x n > 0 and x n − l > 0 for n > N0 In view of (2.3), we haveΔm z n < 0 for n > N0 Thereby{Δi z n }is eventually monotonic for

i =0, 1, 2, ,m −1 Since limn →∞(x n /n(2j −1))= a > 0, there exists some integer N1> N0

such that

1

2an(2j −1)≤ x n ≤3

2an(2j −1), n ≥ N1. (3.3) Note that limn →∞(z n /n(2j −1))=(1 +c0)a implies

lim

n →∞Δ2j −1z n =1 +c0



a2j −1

By (3.4) and the monotonicity ofΔi z n, we have

lim

n →∞Δi z n =0, i =2j,2j + 1, ,m −1. (3.5) Summing (2.3)m −2j times from n to N1and invoking (3.5) in each time, we obtain

Δ2j z n = −

∞



i m −2j = n ···

∞



i2= i3

∞



i1= i2

fi1,x i1− l

= −∞

i = n

(i − n + m −2j −1)(m 2j −1)

(m −2j −1)! fi,x i − l

, n ≥ N1.

(3.6)

Summing the above equation again fromN1ton, we obtain

Δ2j −1z n+1 =Δ2j −1z N1−

n



i2= 1

∞



i1= i2



i1− i2+m −2j −1 (m 2j −1)

(m −2j −1)! fi1,x i1− l

By (3.4), the above equation implies that

∞



i2= n

∞



i1= i2



i1− i2+m −2j −1 (m 2j −1)

(m −2j −1)! fi1,x i1− l< ∞, n ≥ N1. (3.8) That is,

∞



i = n

(i − n + m −2j)(m 2j)

(m −2j)! f



i,x i − l

< ∞, n ≥ N1. (3.9)

LetK = a/2 In view of (3.3), (3.9) and the monotonicity of f (n,x) in x, we see that (3.1) holds

Conversely, suppose (3.1) holds for someK > 0 Set R n = n(2j −1) In view of (3.2), we have

∞



i m −2j+1 = n

∞



i m −2j = i m −2j+1

··· ∞

i2= i3

∞



i1= i2

fi1,Ki1− l(2j −1) 

=

∞



i = n

(i − n + m −2j)(m 2j)

(m −2j)! f



i,K(i − l)(2j −1) 

.

(3.10)

Note that (2.1), there are two cases to consider

In case−1< c0< 0, take c1so that− c0< c1< (1 −4c0)/5 < 1 Then (1 −5c1)/(4c0)<

1 Note that limn →∞(| c n | R n /R n − k − l)= | c0|, limn →∞(R n − k /R n)=1 and (3.1) holds Thus there exists an integerN > k + l such that when n ≥ N, we have

c n R n

R n − k

R n ≥1−5c1

∞



i m −2j+1 =

∞



i m −2j = i m −2j+1

···

∞



i1= i2

fi1,Ki1− l(2j −1) 

<



1− c1



K

TakeN0= N − k − l, r n = R2

nand define the Banach spacel ∞

N0as in (1.4) Let

Ω=



x ∈ l ∞

N0:1

2KR n ≤ x n ≤ KR n



Then it is obvious thatΩ is a bounded, convex and closed subset of l ∞

N0, and for anyx ∈

Ω and n ≥ N0+l, we have

fn,x n − l

≤ fn,K(n − l)(2j −1) 

Define operatorsU and S on Ω as follows:

(Ux) n =

⎧

⎪

⎨

⎪

⎩

−3

4c1KR n − c N x N k

R N R n N0≤ n < N

−3

4c1KR n − c n x n − k n ≥ N,

(Sx) n =

⎧

⎪

⎨

⎪

⎩

3

4KR n N0≤ n < N

3

4KR n+F(n) n ≥ N,

(3.17)

where

F(n) =

n−1

i m =

···

i m −2 j+4 −1

i m −2j+3 =

i m −2 j+3 −1

i m −2j+2 =

∞



i m −2j+1 = i m −2j+2

∞



i m −2j = i m −2j+1

···∞

i1= i2

fi1,x i1− l

. (3.18)

In view of (3.16) and (3.14),we have

F(n) ≤

n−1

i m =

···

i m − 2j+4 −1

i m −2j+3 =

i m − 2j+3 −1

i m −2j+2 =



1− c1



K



1− c1



K(n − N)(2j −1)

8(2j −1)! ≤



1− c1



K

(3.19) forn ≥ N.

Next, we will show that the operatorsU and S satisfy the conditions of Kranoselskii’s

fixed point theorem

First, we claim thatUx + Sy ∈ Ω for any x, y ∈ Ω Indeed, for N0≤ n < N, in view of

(3.13) and (3.12), we have

(Ux) n+ (Sy) n =

 3 4



1− c1



K − c N x N k

R N R n ≥

 3 4



1− c1



K − c N K R N k

2R N R n ≥1

2KR n,

(Ux) n+ (Sy) n ≤

 3 4



1− c1



K − c N K R R N k

 3 4



1− c1

 +c1 KR n ≤ KR n

(3.20) Whenn ≥ N, invoking (3.13) again, we have

(Ux) n+ (Sy) n ≥3

4



1− c1 KR n − c n x n − k ≥3

4



1− c1 KR n − c n1

2K R n − k

R n R n ≥1

2KR n

(3.21)

and, in view of (3.19) and (3.12), we have

(Ux) n+ (Sy) n ≤3

4



1− c1



KR n − c n x n − k+



1− c1



K

4



1− c1



KR n − c n KR n − k+



1− c1



K

8 R n ≤ KR n

(3.22)

That is,Ux + Sy ∈ Ω for any x, y ∈Ω

Letx, y ∈Ω In view of (3.11), we have

1

R2

n (Ux) n −(U y) n c N x N k − y N k

R N R n

= x N k − y N k

R2

N k

c N R2

N k

R N R n ≤ c1sup

n ≥ 0

x n − y n

R2

n

(3.23)

forN0≤ n < N And, for n ≥ N, we have

1

R2

n (Ux) n −(U y) n c n sup

n ≥ 0

x n − y n

R2

Therefore, we have

for anyx, y ∈ Ω Hence, U is a contraction mapping.

Next, we will prove thatS is a completely continuous mapping Indeed, it is obvious

that (Sx) n ≥(K/2)R nforn ≥ N0and (Sx) n ≤ KR nforN0≤ n < N When n ≥ N, by means

of (3.19), we have

(Sx) n ≤3

4KR n+



1− c1



K

That is, the operatorS maps Ω into Ω.

Now we consider the continuity ofS Let x(λ) ∈Ω and x(λ) − x  →0 whenλ → ∞, we assert thatSx(λ) converges toSx by  ·  Indeed, x(λ) − x  →0 implies thatx ∈Ω and

| x(λ)

n − x n | →0 whenλ → ∞for any integern ≥ N0 Thereby, we have

fn,x(λ)

n − l



for any integern ≥ N0+l By definition of S, we have

Sx(λ)

forN0≤ n < N and

Sx(λ)