# On Nonlinear Boundary Value Problems for Functional Difference Equations withp-Laplacian

Authors: Yong Wan; Yuji Liu
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

Sufficient conditions for the existence of solutions of nonlinear boundary value
problems for higher-order functional difference equations withp-Laplacian are established by making of continuation theorems.
We allow f to be at most linear, superlinear, or sublinear in obtained results.


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## Body

## 1. Introduction

The existence of solutions of boundary value problems for finite difference equations were studied by many authors, one may see the text books [1, 2], the papers [3–5] and the references therein. We present some representative ones, which are the motivations of this paper.In papers [3, 4], using Krasnoselskii fixed point theorem and Leggett-Williams fixed point theorem, respectively, Karakostas studied the existence of three positive solutions of the problems consisting of the functional differential equation(1.1)[Φ(x′)]′+c(t)f(t,x(g1(t)),x(g2(t)),…,x(gn(t)))=0,a.a.t∈[0,1]
and one of the following three pairs of conditions:(1.2)x(0)-B0(x′(0))=0,x(1)+B1(x′(1))=0,x(0)-B0(x′(0))=0,x′(1)=0,x′(0)=0,x(1)+B1(x′(1))=0.
Here Φ is a Sup-Multiplicative-Like function, see [3, 4]. In [4] to get the main existence theorems the author assumes the validity of(H1)
for eachi=0,1 the function Bi is continuous nondecreasing and such that αBi(α)≥0 and at least one of the following:(H2)
lim supα→0+(B0(α)/α)<+∞,(H3)
lim supα→0+(-B1(-α)/α)<+∞,(H4)
supα>0(B0(α)/α)<+∞,(H5)
supα>0(-B1(-α)/α)<+∞.The discrete simulation of BVPs studied in [3, 4] is as follows:(1.3)Δ[Φ(Δx(k))]+c(k)f(k,x(g1(k)),…,x(gn(k)))=0,k∈[0,T]
subject to one of the following boundary conditions:(1.4)x(0)-B0(Δx(0))=0,x(T+2)+B1(Δx(T+1))=0,x(0)-B0(Δx(0))=0,Δx(T+1)=0,Δx(0)=0,x(T+2)+B1(Δx(T+1))=0.
The question follows: under what conditions above BVP (1.3) has solutions if (H1)–(H5) are not satisfied?Particular significance lies in the fact that when a BVP is discretized, strange and interesting changes can occur in the solutions. For example, properties such as existence, uniqueness and multiplicity of solutions may not be shared between the continuous differential equation and its related discrete difference equation. Moreover, when investigating difference equations, as opposed to differential equations, basic ideas from calculus are not necessarily available to use, such as the intermediate value theorem, the mean value theorem and Rolle's theorem. Thus, new challenges are faced and innovation is required [5].In recent paper [5], Liu studied the solvability of the following problem consisting of the higher-order functional difference equation and boundary conditions(1.5)Δ2x(n)=f(n,x(n+1),x(n-τ1(n)),…,x(n-τm(n))),n∈[0,T-1],ax(0)-bΔx(0)=0,cx(T+1)+dΔx(T)=0,x(i)=ϕ(i),i∈[-τ,-1],x(i)=ψ(i),i∈[T+2,T+δ],
where T≥1, a,b,c,d∈R with a2+b2≠0 and c2+d2≠0, τi(n), i=1,…,m, are sequences,(1.6)τ=max{0,maxn∈[0,T-1]{τi(n)}:i=1,…,m},δ=-min{0,minn∈[0,T-1]{τi(n)}:i=1,…,m},f(n,u) is continuous in u=(x0,…,xm,xm+1) for each n. Two cases, that is, bc+ad+ac(T+1)=0 or bc+ad+ac(T+1)≠0 are considered in [6].Motivated by [3–5], we study the nonlinear boundary value problems for higher-order functional difference equation with p-Laplacian, that is, the equation(1.7)Δ[ϕ(Δx(n))]=f(n,x(n+1),x(n-τ1(n)),…,x(n-τm(n))),n∈[0,T-1],
subject to the following boundary conditions(1.8)x(0)-B0(Δx(0))=0,B1(x(T+1))+Δx(T)=0,x(i)=γ(i),i∈[-τ,-1],x(i)=ψ(i),i∈[T+2,T+δ],
where [0,T-1]={0,1,…,T-1}, ϕ(x)=|x|p-2x for x≠0 and ϕ(0)=0 with p>1, the inverse function is denoted by ϕ-1, T≥1 an integer, B0,B1 are continuous and satisfy xBi(x)≥0 for all x∈R, i=0,1, {τi(n)}, i=1,…,m, are integer vectors,(1.9)τ=max{0,maxn∈[0,T-1]{τi(n)}:i=1,…,m},δ=-min{0,minn∈[0,T-1]{τi(n)}:i=1,…,m},f(n,u) is continuous about u=(x0,…,xm) for each n. Boundary condition (1.8) is called nonlinear Sturm-Liouville type conditions.The purposes of this paper are to establish sufficient conditions for the existence of at least one solutions of BVP (1.7)-(1.8). It is interesting that we allow that f to be sublinear, at most linear or superlinear. We do not need the assumptions (H2)–(H5) imposed on B0,B1.This paper is organized as follows. In Section2, we give the main results of this paper, and in Section 3, examples to illustrate the main results will be presented.


## 2. Main Results

To get existence results for solutions of BVP (1.7)-(1.8), we need the following fixed point theorem, which was used to solve multi-point boundary value problems for differential equations in many papers but not used to solve boundary value problems for difference equations.LetX and Y be real Banach spaces, let L:DomL⊂X→Y be a Fredholm operator of index zero. If Ω is an open bounded subset of X, DomL∩Ω¯≠∅, the map N:X→Y will be called L-compact on Ω¯ if N(Ω¯) is bounded and compact.Lemma 2.1 (see [6]).
LetX and Y be Banach spaces. Suppose L:DomL⊂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 Ω⊂X is an open bounded set with 0∈Ω and
(2.1)Lx≠λNx,∀x∈DomL∩∂Ω,λ∈[0,1],
then there is at least one x∈Ω so that Lx=Nx.
LetX=RT+τ+δ+1×RT+1 be endowed with the norm
(2.2)∥(x,y)∥=max{maxn∈[1,T+τ+δ+1]|x(n)|,maxn∈[0,T]|y(n)|}for(x,y)∈X.
LetY=RT+1×RT×R2×Rτ×Rδ-1 be endowed with the norm
(2.3)∥(u,v,a,b,s,t)∥=max{maxn∈[0,T]|u(n)|,maxn∈[0,T-1]|v(n)|,|a|,|b|,maxn∈[-τ,-1]|s(n)|,maxn∈[T+2,T+δ]|t(n)|}
for (u,v,a,b,s,t)∈Y.
It is easy to see thatX and Y are real Banach spaces.
LetL:X→Y, and
(2.4)L(x(n)y(n))={Δx(n),n∈[0,T]Δy(n),n∈[0,T-1]x(1)y(T)x(i),i∈[-τ,-1]x(j),j∈[T+2,T+δ],
and N:X→Y by
(2.5)N(x(n)y(n))={ϕ-1(y(n)),n∈[0,T]f(n,x(n+1),x(n-τ1(n)),…,x(n-τm(n))),n∈[0,T-1]B0(ϕ-1(y(0)))-ϕ(B1(x(T+1)))γ(i),i∈[-τ,-1]ψ(j),j∈[T+2,T+δ]
for all (x,y)∈X.
Sincef, B0,B1 are continuous, it is easy to show that(i)
(x,y)∈DomL is a solution of L(x,y)=N(x,y) implies that x is a solution of BVP (1.7)-(1.8),(ii)
KerL={0},(iii)
L is a Fredholm operator of index zero and N is L-compact on each Ω¯ with Ω being an open bounded subset of X.Lemma 2.2.
(∑i=1mai)σ≤lσm-1(∑i=1maiσ) for all ai≥0 and σ>0, where lσ is defined by lσ=2σ-1 for σ≥1 and lσ=1 for σ∈(0,1).Proof.
We have the following cases.Case 1 (m=2).
Without loss of generality, supposea1≥a2. Let
(2.6)g(x)=(1+x)σ-lσ(1+xσ),x∈[1,+∞),
then
(2.7)g(1)=2σ-2lσ={2σ-2σ=0,σ≥1,2σ+1-2≤0,σ∈(0,1)
and for x∈[1,∞), we get
(2.8)g'(x)=σxσ-1[(1+1x)σ-1-lσ]={σxσ-1[(1+1x)σ-1-2σ-1]≤0,σ≥1,σxσ-1[(1+1x)σ-1-1]≤0,σ∈(0,1).
We get that g(x)≤g(1) for all x≥1 and so (1+x)σ≤lσ(1+xσ) for all x∈[1,+∞). Hence (a1+a2)σ=a2σ(1+(a1/a2))σ≤lσa2σ[1+(a1/a2)σ)=lσ(a1σ+a2σ).Case 2 (m>2).
It is easy to see that(2.9)(∑i=1mai)σ=(a1+∑i=2mai)σ≤lσ(a1σ+(∑i=2mai)σ)≤lσ[a1σ+lσ(a2σ+(∑i=3mai)σ)]≤lσ2(a1σ+a2σ+(∑i=3mai)σ)≤⋯≤lσm-1(∑i=1maiσ).
The proof is complete.Theorem 2.3.
Suppose that(A)
f, B0,B1 are continuous, and xBi(x)≥0 for all x∈R and i=0,1;(B)
there exist numbersβ>0, θ>1, nonnegative sequences pi(n),r(n)(i=0,…,m), functions g(n,x0,…,xm), h(n,x0,…,xm) such that(2.10)f(n,x0,…,xm)=g(n,x0,…,xm)+h(n,x0,…,xm),g(n,x0,x1,…,xm)x0≥β|x0|θ+1,|h(n,x0,…,xm)|≤∑i=0mpi(n)|xi|θ+r(n),
for all n∈{1,…,T}, (x0,x1,…,xm)∈Rm+1.
Then BVP (1.7)-(1.8) has at least one solution if
(2.11)∥p0∥+Tθ/(θ+1)∑i=1m∥pi∥<β.Proof.
To apply Lemma2.1, we divide the proof into two steps.
LetΩ1={(x,y):L(x,y)=λN(x,y),((x,y),λ)∈[DomL]×[0,1]}. For (x,y)∈Ω1, we have L(x,y)=λN(x,y), λ∈[0,1], then
(2.12)Δx(n)=λϕ-1(y(n)),n∈[0,T],Δy(n)=λf(n,x(n+1),x(n-τ1(n)),…,x(n-τm(n))),n∈[0,T-1],x(0)=λB0(ϕ-1(y(0))),y(T)=-λϕ(B1(x(T+1))),x(i)=λγ(i),i∈[-τ,-1],x(j)=λψ(j),j∈[T+2,T+δ].
Step 1.
We will show that ifL(x,y)=λN(x,y), for some λ∈[0,1], then x is bounded. Indeed, we see that
(2.13)Δ[ϕ(Δx(n))]x(n+1)=λϕ(λ)f(n,x(n+1),x(n-τ1(n)),…,x(n-τm(n)))x(n+1).
Since
(2.14)x(0)=λB0(ϕ-1(y(0))),y(T)=-λϕ(B1(x(T+1))),
we get
(2.15)x(0)=λB0(Δx(0)λ),Δx(T)λ=-ϕ-1(λ)B1(x(T+1)).
It is easy to see from (2.12) and the definition of B0,B1 and ϕ that
(2.16)∑n=0T-1Δ[ϕ(Δx(n))]x(n+1)=∑n=0T-1[ϕ(Δx(n+1))-ϕ(Δx(n))]‍[x(n+2)-Δx(n+1)]=∑n=0T-1[ϕ(Δx(n+1))x(n+2)-ϕ(Δx(n))x(n+1)]-∑n=0T-1ϕ(Δx(n+1))Δx(n+1)=ϕ(Δx(T))x(T+1)-ϕ(Δx(0))x(1)-∑n=0T-1ϕ(Δx(n+1))Δx(n+1)=-ϕ(λϕ-1(λ)B1(x(T+1)))x(T+1)-ϕ(Δx(0))Δx(0)-ϕ(Δx(0))x(0)-∑n=0T-1ϕ(Δx(n+1))Δx(n+1)=-ϕ(λϕ-1(λ)B1(x(T+1)))x(T+1)-ϕ(Δx(0))λB0(Δx(0)λ)-∑n=0T-1ϕ(Δx(n+1))Δx(n+1)-ϕ(Δx(0))Δx(0)≤0.
So we get
(2.17)∑n=0T-1f(n,x(n+1),x(n-τ1(n)),…,x(n-τm(n)))x(n+1)≤0.
It follows from the assumptions that
(2.18)β∑n=0T-1|x(n+1)|θ+1≤∑n=0T-1g(n,x(n+1),x(n-τ1(n)),…,x(n-τm(n))x(n+1)≤-∑n=0T-1h(n,x(n+1),x(n-τ1(n)),…,x(n-τm(n))x(n+1)≤∑n=0T-1|h(n,x(n+1),x(n-τ1(n)),…,x(n-τm(n))||x(n+1)|‍≤∑n=0T-1p0(n)|x(n+1)|θ+1+∑i=1m∑n=0T-1pi(n)|x(n-τi(n))|θ|x(n+1)|+∑n=0T-1r(n)|x(n+1)|≤∥p0∥∑n=0T-1|x(n+1)|θ+1+∑i=1m∥pi∥∑n=0T-1|x(n-τi(n))|θ|x(n+1)|+∥r∥∑n=0T-1|x(n+1)|.
For xi≥0,yi≥0, we have Holder’s inequality
(2.19)∑i=1sxiyi≤(∑i=1sxip)1/p(∑i=1syiq)1/q,1p+1q=1,q>0,p>0.
It follows from Lemma 2.2 that
(2.20)β∑n=0T-1|x(n+1)|θ+1≤∥p0∥∑n=0T-1|x(n+1)|θ+1+∥r∥Tθ/(θ+1)(∑n=0T-1|x(n+1)|θ+1)1/(θ+1)+∑i=1m∥pi∥(∑n=0T-1|x(n-τi(n))|θ+1)θ/(θ+1)(∑n=0T-1|x(n+1)|θ+1)1/(θ+1)=∥p0∥∑n=0T-1|x(n+1)|θ+1+∥r∥Tθ/(θ+1)(∑n=0T-1|x(n+1)|θ+1)1/(θ+1)+∑i=1m∥pi∥(∑u∈{n-τi(n)-1:n=0,…,T-1}|x(u+1)|θ+1)θ/(θ+1)(∑n=0T-1|x(n+1)|θ+1)1/(θ+1)≤∥p0∥∑n=0T-1|x(n+1)|θ+1+∥r∥Tθ/(θ+1)(∑n=0T-1|x(n+1)|θ+1)1/(θ+1)+[Tθ/(θ+1)∑i=1m∥pi∥(∑n=0T-1|x(n+1)|θ+1)θ/(θ+1)+Tθ/(θ+1)∑i=1m∥pi∥(∑n=TT+δ|λψ(n+1)|θ+1)θ/(θ+1)+Tθ/(θ+1)∑i=1m∥pi∥(∑n=-τ-1|λγ(n+1)|θ+1)θ/(θ+1)](∑n=0T-1|x(n+1)|θ+1)1/(θ+1)≤∥p0∥∑n=0T-1|x(n+1)|θ+1+∥r∥Tθ/(θ+1)(∑n=0T-1|x(n+1)|θ+1)1/(θ+1)+Tθ/(θ+1)∑i=1m∥pi∥∑n=0T-1|x(n+1)|θ+1+Tθ/(θ+1)∑i=1m∥pi∥(∑n=TT+δ|ψ(n+1)|θ+1)θ/(θ+1)(∑n=0T-1|x(n+1)|θ+1)1/(θ+1)+Tθ/(θ+1)∑i=1m∥pi∥(∑n=-τ-1|γ(n+1)|θ+1)θ/(θ+1)(∑n=0T-1|x(n+1)|θ+1)1/(θ+1).
It follows from (2.11) that there is M1>0 such that ∑u=0T-1|x(u+1)|θ+1≤M1.
Hence|x(n+1)|≤M11/(θ+1) for all n∈{0,…,T-1}, which proves Step 1.Step 2.
We will show that the setΩ1 is bounded. We first prove that there exists k∈[0,T-1] such that y(k)y(k+1)≤0. In fact, if y(i)>0 for all i∈[0,T], then
(2.21)Δx(n)=λϕ-1(y(n)),n∈[0,T]
implies that x(i) is increasing on [0,T+1], so x(0)<x(T+1). Then assumption (A) and
(2.22)x(0)=λB0(ϕ-1(y(0))),y(T)=-λϕ(B1(x(T+1)))
imply that x(0)>0 and x(T+1)<0. This contradicts x(0)<x(T+1). If y(i)<0 for all i∈[0,T], the similar contradiction can be deduced. Hence there exists k∈[0,T-1] such that y(k)y(k+1)≤0.
It follows that there is a constantξ∈[k,k+1] such that
(2.23)y(k-1)-y(k)k+1-k=0-y(k)ξ-k.
Then |y(k)|≤|Δy(k)|. Hence
(2.24)|y(k)|≤|Δy(k)|=λ|f(k,x(k+1),x(k-τ1(k)),…,x(k-τm(k)))|≤maxn∈[0,T-1],|xi|≤M11/(θ+1),i=0,…,m|f(n,x0,…,xm)|=:M2.
Then for n∈[k+1,T] one sees that
(2.25)|y(n)|=|y(k)+∑s=kn-1Δy(s)|≤M2+∑s=kn-1|Δy(s)|≤M2+Tmaxn∈[0,T-1],|xi|≤M11/(θ+1),i=0,…,m|f(n,x0,…,xm)|=(T+1)M2.
For n∈[0,k-1], we get
(2.26)|y(n)|=|y(k)-∑s=nk-1Δy(s)|≤M2+∑s=nk-1|Δy(s)|≤M2+Tmaxn∈[0,T-1],|xi|≤M11/(θ+1),i=0,…,m|f(n,x0,…,xm)|=(T+1)M2.
It follows that
(2.27)∥y∥≤(T+1)M2.
Then
(2.28)|x(0)|=|λB0(ϕ-1(y(0)))|≤|B0(ϕ-1(y(0)))|≤max|x|≤(T+1)M2|B0(ϕ-1(x))|.
On the other hand, since Δx(T)=λϕ-1(y(T)) implies that
(2.29)|Δx(T)|≤ϕ-1(|y(T)|)≤ϕ-1((T+1)M2),
then
(2.30)|x(T+1)|=|Δx(T)+x(T)|≤|Δx(T)|+|x(T)|≤ϕ-1((T+1)M2)+M11/(θ+1).
It follows that
(2.31)∥x∥≤max{maxi∈[T+2,T+δ]|ψ(i)|,maxi∈[-τ,-1]|γ(i)|,ϕ-1((T+1)M2)+M11/(θ+1)}.
Hence
(2.32)∥(x,y)∥≤max{(T+1)M2,maxi∈[T+2,T+δ]|ψ(i)|,maxi∈[-τ,-1]|γ(i)|,ϕ-1((T+1)M2)+M11/(θ+1)}.Step 2 is proved, namely Ω1 is bounded. Let Ω⊃Ω1¯ be an open bounded subset of X centered at zero, it is easy to see that L(x,y)≠λN(x,y) for all (x,y)∈∂Ω and λ∈(0,1). It follows from Lemma 2.1 that equation L(x,y)=N(x,y) has at least one solution (x,y), then x is a solution of BVP (1.7)-(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 BVP:(3.1)Δ(ϕ(Δx(n)))=β[x(n+1)]2k+1+∑i=1mpi(n)[x(n-i)]2k+1+r(n),n∈[0,T-1],x(0)-5[Δx(0)]5=0,6[x(T+1)]3+Δx(T)=0,x(n)=γ(n),n∈[-m,-1],
where ϕ(x)=|x|p-2x for x≠0 and ϕ(0)=0 with p>1 a constant, T≥1 is an integer, β>0, pi(n),r(n) are sequences. Corresponding to the assumptions of Theorem 2.3, we set
(3.2)g(n,x0,…,xm)=β[x0]2k+1,h(n,x0,…,xm)=∑i=1mpi(n)xi2k+1+r(n),θ=2k+1,B0(x)=5x5,B1(x)=6x3.
It is easy to see that assumptions (A) and (B) in Theorem 2.3 hold. It follows from Theorem 2.3 that (3.1) has at least one solution if ∥p0∥+Tθ/(θ+1)∑i=1m∥pi∥<β.Remark 3.2.
The BVP in Example3.1 cannot be solved by the theorems in [3–5] since the difference equation in (3.1) is a p-Laplacian equation and the boundary conditions superlinear.



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