# Solvability of a Class of Generalized Neumann Boundary Value Problems for Second-Order Nonlinear Difference Equations

Authors: Jianye Xia; Yuji Liu
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

This paper is motivated by Rachnkovab and Tisdell (2006) and Anderson et al. (2007). New sufficient conditions for the existence of at least one solution of the generalized Neumann boundary value problems for second order nonlinear difference equations∇Δx(k)=f(k,x(k),x(k+1)), k∈[1,n−1], x(0)=ax(1), x(n)=bx(n−1), are established.


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## Body

## 1. Introduction

Recently, there have been many papers discussed the solvability of two-point or multipoint boundary value problems for second-order or higher-order difference equations, we refer the readers to the text books [1, 2] and papers [3–8] and the references therein.In a recent paper [3], Anderson et al. studied the following problem:
(1.1)∇Δy(k)=f(k,y(k),Δy(k)),k=1,…,n-1,Δy(0)=Δy(n)=0,
where
(1.2)Δy(k)={y(k+1)-y(k),fork=0,…,n-1,0,fork=n,∇Δy(k)={y(k+1)-2y(k)+y(k-1),fork=1,…,n-1,0,fork=0orn.
The following result was proved.Theorem ART
Suppose thatf is continuous and there exist constants α≤0, K≥0 such that
(*)|f(t,p,q)-p|≤α[2pf(t,p,q)+q2]+K,(t,p,q)∈{1,…,n-1}×R2.
Then BVP(1.1) has at least one solution.The methods in [3] involved new inequalities on the right-hand side of the difference equation and Schaefer's Theorem in the finite-dimensional space setting.In [7], the following discrete boundary value problem (BVP) involving second order difference equations and two-point boundary conditions
(1.3)∇Δykh2=f(tk,yk,Δykh),k=1,…,n-1,y0=0,yn=0,
was studied, where n≥2 an integer, f is continuous, scalar-valued function, the step size is h=N/n with N a positive constant, the grid points are tk=kh for k=0,…,n. The differences are given by
(1.4)Δyk={yk+1-yk,k=0,…,n-1,0,k=n,∇Δyk={yk+1-2yk+yk-1,k=1,…,n-1,0,k=0ork=n.The following two results were proved in [7].Theorem RT
Letf be continuous on [0,N]×R2 and α,β, and, K be nonnegative constants. If there exist c,d∈[0,1) such that
(1.5)|f(t,u,v)|≤α|u|c+β|v|d+K,(t,u,v)∈[0,N]×R2,
then the discrete BVP(1.3) has at least one solution.Theorem RT
Letf be continuous on [0,N]×R2 and α,β, and K nonnegative constants. If
(1.6)|f(t,u,v)|≤α|u|+β|v|+K,(t,u,v)∈[0,N]×R2,(1.7)αN28+βN2<1,
then the discrete BVP(1.3) has at least one solution.In paper [8], Cabada and Otero-Espinar studied the existence of solutions of a class of nonlinear second-order difference problem with Neumann boundary conditions by using upper and lower solution methods. Assuming the existence of a pair of ordered lower and upper solutions γ and β, they obtained optimal existence results for the case γ≤β and even for γ≥β.In this paper, we study the following boundary value problem for second-order nonlinear difference equation(1.8)∇Δx(k)=f(k,x(k),x(k+1)),k∈[1,n-1],x(0)=ax(1),x(n)=bx(n-1),
where a,b∈R, n≥2is an integer, and f is continuous, scalar-valued function. We note that when a=b=1, BVP(1.8) becomes the following BVP:
(1.9)∇Δx(k)=f(k,x(k),x(k+1)),t∈[1,T-1],Δx(0)=0=Δx(n-1),
which is called Neumann boundary value problem of difference equation and is a special case of BVP(1.1). When a=b=0, BVP(1.8) is changed to
(1.10)∇Δx(k)=f(k,x(k),x(k+1)),t∈[1,T-1],x(0)=0=x(n),
which is the so-called Dirichlet problem for discrete difference equations and is a special case of BVP(1.3).The purpose of this paper is to improve the assumptions (*), (1.5), and (1.6) in the results in paper [3, 5, 7–9], by using Mawhin′s continuation theorem of coincidence degree, to establish sufficient conditions for the existence of at least one solution of BVP(1.8). It is interesting that we allow f to be sublinear, at most linear or superlinear.This paper is organized as follows. In Section2, we make the main results, and in Section 3, we give some examples, which cannot be solved by theorems in [5, 7, 9], to illustrate the main results presented in Section 3.


## 2. Main Results

To get the existence results for solutions of BVP(1.8), we need the following fixed point theorems.LetX and Y be Banach spaces, L:D(L)⊂X→Y a Fredholm operator of index zero, and P:X→X, Q:Y→Y projectors such that ImP=KerL,KerQ=ImL,X=KerL⊕KerP,Y=ImL⊕ImQ. It follows that L|D(L)∩KerP:D(L)∩KerP→ImL is invertible; we denote the inverse of that map by Kp.IfΩ is an open bounded subset of X, D(L)∩Ω¯≠∅, the map N:X→Y will be called L-compact on Ω¯ if QN(Ω¯) is bounded and Kp(I-Q)N:Ω¯→X is compact.Lemma 2.1 . (see [9]).
LetL be a Fredholm operator of index zero, and let N be L-compact on Ω. Assume that the following conditions are satisfied:
(i)Lx≠λNx for every (x,λ)∈[(D(L)∖KerL)∩∂Ω]×(0,1);
(ii)Nx∉ImL for every x∈KerL∩∂Ω;
(iii)deg(∧QN|KerL,Ω∩KerL,0)≠0, where ∧:KerL→Y/ImL is the isomorphism.
Then the equationLx=Nx has at least one solution in D(L)∩Ω¯.Lemma 2.2 . (see [9]).
LetX and Y be Banach spaces. Suppose L:D(L)⊂X→Y is a Fredholm operator of index zero with KerL={0}, N:X→Y is L-compact on any open bounded subset of X. If 0∈Ω⊂X is an open bounded subset and Lx≠λNxforallx∈D(L)∩∂Ωandλ∈[0,1], then there is at least one x∈Ω so that Lx=Nx.
LetX=Rn+1,Y=Rn-1 be endowed with the norms
(2.1)‖x‖=max⁡n∈[0,n]|x(n)|,‖y‖=max⁡k∈[1,n-1]|y(k)|
for x∈X and y∈Y, respectively. It is easy to see that X and Y are Banach spaces. Choose D(L)={x∈X:x(0)=ax(1),x(n)=bx(n-1)}. Let L:X→Y,Lx(k)=∇Δx(k),x∈D(L), and N:X→Y by Nx(k)=f(k,x(k),x(k+1)).Consider the following problem:(2.2)∇Δx(k)=0,x(0)=ax(1),x(n)=bx(n-1).
It is easy to see that problem (2.2) has a unique solution x(k)=0 if and only if
(2.3)(1-a)[(n-1)b-n]≠a(1-b).
If (2.3) holds, we call BVP(1.8) at nonresonance case. If (1-a)[(n-1)b-n]=a(1-b), then problem (2.2) has infinite nontrivial solutions. At this case, we call BVP(1.8) at resonance case. In this paper, we establish sufficient conditions for the existence of solutions of BVP(1.8) at nonresonance case, that is, (1-a)[(n-1)b-n]≠a(1-b), and at resonance case, a=b=1. It is similar to get existence results for the existence of solutions at resonance case when (1-a)[(n-1)b-n]=a(1-b) and a≠1,b≠1.Lemma 2.3.
Supposea=b=1. Then the following results are valid.
(i)KerL={x=(c,…,c)∈X:c∈R}.
(ii)ImL={y∈Y:∑i=1n-1y(i)=0}.
(iii)L is a Fredholm operator of index zero.
(iv) There are projectorsP:X→X and Q:Y→Y such that KerL=ImP,KerQ=ImL. Furthermore, let Ω⊂X be an open bounded subset with Ω¯∩D(L)≠∅; then N is L-compact on Ω¯.
(v)x∈D(L) is a solution of L(x)=N(x) which implies that x is a solution of BVP(1.8).The projectorsP:X→X and Q:Y→Y, the isomorphism ∧:KerL→Y/ImL, and the generalized inverse Kp:ImL→D(L)∩ImP are as follows:
(2.4)Px(n)=x(1),Qy(n)=1n-1∑i=1n-1y(i),∧(c)=c,Kpy(n)=∑s=1k∑i=1sy(i).Lemma 2.4.
Suppose(1-a)[(n-1)b-n]≠a(1-b). Then the following results are valid.
(i)x∈D(L) is a solution of L(x)=N(x) which implies that x is a solution of BVP(1.8).
(ii)KerL={0}.
(iii)L is a Fredholm operator of index zero, N is L-compact on each open bounded subset of X.Suppose(A)
there exist numbersβ>0, θ≥1, nonnegative sequences p(n),q(n),r(n), functions g(n,x,y), h(n,x,y) such that f(n,x,y)=g(n,x,y)+h(n,x,y) and
(2.5)g(n,x,y)x≥β|x|θ+1,|h(n,x,y)|≤p(n)|x|θ+q(n)|y|θ+r(n),
for all n∈{1,…,n-1},(x,y)∈R2;(B)
there exists a constantM>0 so that
(2.6)c[∑i=1n-1f(n,c,c)]>0
for all |c|>M or
(2.7)c[∑i=1n-1f(n,c,c)]<0
for all |c|>M.Theorem L
Supposea2≤1,b2≤1, and that (A) and (B) hold. Then BVP(1.8) has at least one solution if
(2.8)‖p‖+‖q‖max⁡{|b|θ+1,1}<β.Proof.
To apply Lemma2.1, we consider Lx=λNx for λ∈[0,1].
Step 1.
LetΩ1={x∈X:Lx=λNx,λ∈[0,1]}. For x∈Ω1, we have
(2.9)x(k+1)-2x(k)+x(k-1)=λf(k,x(k),x(k+1)),k∈[1,n-1],x(0)=ax(1),x(n)=bx(n-1).
So
(2.10)[x(k+1)-2x(k)+x(k-1)]x(k)=λf(k,x(k),x(k+1))x(k),k∈[1,n-1].
It is easy to see that
(2.11)2∑n=1n-1[x(k+1)-2x(k)+x(k-1)]x(k)=∑n=1n-1(-[x(k+1)]2+2x(k)x(k+1)-[x(k)]2-[x(k-1)]2+2x(k-1)x(k)-[x(k)]2+[x(k+1)]2-2[x(k)]2+[x(k-1)]2)=∑n=1n-1(-[x(k+1)-x(k)]2-[x(k-1)-x(k)]2+[x(k+1)]2-2[x(k)]2+[x(k-1)]2)=∑n=1n-1(-[x(k+1)-x(k)]2-[x(k-1)-x(k)]2)+([x(n)]2-[x(n-1)]2-[x(1)]2+[x(0)]2)=∑n=1n-1(-[x(k+1)-x(k)]2-[x(k-1)-x(k)]2)+((b2-1)[x(n-1)]2+(a2-1)[x(1)]2).
Since a2≤1,b2≤1, we get
(2.12)∑n=1n-1[x(k+1)-2x(k)+x(k-1)]x(k)≤0.
So, we get
(2.13)λ∑n=1n-1f(k,x(k),x(k+1))x(k)≤0.
Then
(2.14)∑n=1n-1[g(k,x(k),x(k+1))+h(k,x(k),x(k+1))]x(k)≤0.
It follows that
(2.15)β∑k=1n-1|x(k)|θ+1≤-∑n=1n-1h(k,x(k),x(k+1))x(k)≤∑n=1n-1[p(k)|x(k)|θ+1+q(k)|x(k+1)|θ|x(k)|+r(k)|x(k)|]≤‖p‖∑n=1n-1|x(k)|θ+1+‖q‖∑k=1n-1|x(k+1)|θ|x(k)|+∑k=1n-1r(k)|x(k)|.
For xi≥0,yi≥0, Holder′s inequality implies
(2.16)∑i=1sxiyi≤(∑i=1sxip)1/p(∑i=1syiq)1/q,1p+1q=1,q>0,p>0.
It follows that
(2.17)β∑k=1n-1|x(k)|θ+1≤‖p‖∑n=1n-1|x(k)|θ+1+‖q‖(∑k=1n-1|x(k+1)|θ+1)θ/(θ+1)(∑k=1n-1|x(k)|θ+1)1/(θ+1)+‖r‖∑k=1n-1|x(k)|=‖r‖(n-1)θ/(θ+1)(∑k=1n-1|x(k)|θ+1)1/(θ+1)+‖p‖∑n=1n-1|x(k)|θ+1+‖q‖(|b|θ+1|x(1)|θ+1+∑k=1n-2|x(k+1)|θ+1)θ/(θ+1)(∑k=1n-1|x(k)|θ+1)1/(θ+1)≤‖r‖(n-1)θ/(θ+1)(∑k=1n-1|x(k)|θ+1)1/(θ+1)+‖p‖∑n=1n-1|x(k)|θ+1+‖q‖max⁡{|b|θ+1,1}(∑k=1n-1|x(k)|θ+1)θ/(θ+1)(∑k=1n-1|x(k)|θ+1)1/(θ+1)=‖r‖(n-1)θ/(θ+1)(∑k=1n-1|x(k)|θ+1)1/(θ+1)+‖p‖∑n=1n-1|x(k)|θ+1+‖q‖max⁡{|b|θ+1,1}∑k=1n-1|x(k)|θ+1.
It follows from (2.8) that there exists a constant M1>0 such that
(2.18)∑k=1n-1|x(k)|θ+1≤M1.
Hence |x(k)|≤(M1/(n-1))1/(θ+1) for all k∈{1,…,n-1}. Hence ||x||≤(M1/(n-1))1/(θ+1). So Ω1 is bounded.
Step 2.
Prove that the setΩ2={x∈KerL:N(x)∈ImL} is bounded.
Forx∈KerL, we have x(k)=c for k∈[0,n]. Thus we have Nx(k)=f(k,c,c). N(x,y)∈ImL implies that
(2.19)∑k=1n-1f(n,c,c)=0.
It follows from condition (B) that |c|≤M. Thus Ω2 is bounded.
Step 3.
Prove the setΩ3={x∈KerL:±λ∧(x)+(1-λ)QN(x)=0,∃λ∈[0,1]} is bounded.
If the first inequality of(B) holds, let
(2.20)Ω3={x∈Ker⁡L:λ∧(x)+(1-λ)QN(x)=0,∃λ∈[0,1]}.
We will prove thatΩ3 is bounded. For x(k)=c for k∈[0,n] such that x∈Ω3, and λ∈[0,1], we have
(2.21)-(1-λ)∑k=1n-1f(n,c,c)=λc(n-1).
Ifλ=1, then c=0. If λ≠1, then
(2.22)0>-(1-λ)c∑k=1n-1f(n,c,c)=λc2T≥0,
a contradiction.
If the second inequality of(B) holds, let
(2.23)Ω3={x∈Ker⁡L:-λ∧(x)+(1-λ)QN(x)=0,∃λ∈[0,1]}.
Similarly, we can get a contradiction. So Ω3 is bounded.
Step 4.
Obtain open bounded setΩ such (i), (ii), and (iii) of Lemma 2.1.
In the following, we will show that all conditions of Lemma2.1 are satisfied. Set Ω an open bounded subset of X such that Ω⊃⋃i=13Ωi¯. We know that L is a Fredholm operator of index zero and N is L-compact on Ω¯. By the definition of Ω, we have Ω⊃Ω1¯ and Ω⊃Ω2¯, thus L(x)≠λN(x) for x∈(D(L)∖KerL)∩∂Ω and λ∈(0,1); N(x)∉ImL for x∈KerL∩∂Ω.
In fact, letH(x,λ)=±λ∧(x)+(1-λ)QN(x). According the definition of Ω, we know Ω⊃Ω3¯, thus H(x,λ)≠0 for x∈∂Ω∩KerL, thus by homotopy property of degree,
(2.24)deg⁡(QN|Ker⁡L,Ω∩Ker⁡L,0)=deg⁡(H(⋅,0),Ω∩Ker⁡L,0)=deg⁡(H(⋅,1),Ω∩Ker⁡L,0)=deg⁡(±∧,Ω∩Ker⁡L,0)≠0.
Thus by Lemma 2.1, L(x)=N(x) has at least one solution in D(L)∩Ω¯, which is a solution of BVP(1.8). The proof is completed.Theorem L
Supposea2≤1,b2≤1, (1-a)[(n-1)b-n]≠a(1-b), and that (A) holds. Then BVP(1.8) has at least one solution if (2.8) holds.Proof.
To apply Lemma2.2, we consider Lx=λNx for λ∈[0,1]. Let Ω1={x∈X:Lx=λNx,λ∈[0,1]}. For x∈Ω1, we get (2.9) and (2.10). using the methods in the proof of Theorem LX1, we get that Ω1 is bounded. Let Ω be a nonempty open bounded subset of X such that Ω⊃Ω1¯ centered at zero. It is easy to see that L is a Fredholm operator of index zero and N is L-compact on Ω¯. One can see that Lx≠λNxforallx∈D(L)∩∂Ωandλ∈[0,1]. Thus, from Lemma 2.2, Lx=Nx has at least one solution x∈D(L)∩Ω¯, so x is a solution of BVP(1.8). The proof is complete.


## 3. An Example

In this section, we present an example to illustrate the main results in Section2.Example 3.1.
Consider the following problem:(3.1)x(k+1)-2x(k)+x(k-1)=β[x(k)]2m+1+p(k)[x(k)]2m+1+q(k)[x(k+1)]2m+1+r(k),k∈[1,n-1],x(0)=ax(1),x(n)=bx(n-1),
where n≥2,m≥1 are integers and β>0, p(n), q(n), r(n) are sequences. Corresponding to the assumptions of Theorem L1, we set
(3.2)f(k,x,y)=βx2m+1+p(k)x2m+1+q(k)y2m+1+r(k),g(k,x,y)=βx2m+1,h(k,x,y)=p(k)x2m+1+q(k)y2m+1+r(k),
and θ=2m+1. It is easy to see that (A) holds, and
(3.3)f(n,c,c)=c2m+1β+p(k)c2m+1+q(k)c2m+1+r(k)
implies that there is M>0 such that c∑i=1n-1[c2m+1β+p(k)c2m+1+q(k)c2m+1+r(k)]>0 for all n∈[1,n-1] and |c|>M.It follows from Theorem L2 that (3.1) has at least one solution if a2≤1,b2≤1,(1-a)[(n-1)b-n]≠a(1-b) and ||p||+||q||max{|b|θ+1,1}<β. BVP(3.1) has at least one solution if a=b=1 and ||p||+||q||max{|b|θ+1,1}<β.Remark 3.2.
It is easy to see that BVP(3.1) when a=b=0 cannot be solved by using theorems obtained in paper [7]. BVP(3.1) when a=b=1 cannot be solved by the results obtained in paper [3].



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