# Positive Solutions to a Second-Order Discrete Boundary Value Problem

Authors: Xiaojie Lin; Wenbin Liu
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

We are concerned with  second-order discrete boundary value problems and obtain some sufficient conditions for the existence of at least one positive solution by using the fixed point theorem due to Krasnosel'skii on a cone.


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## Body

## 1. Introduction

Boundary value problems for difference equations have been studied extensively by many authors, for example, [1–10] to name a few.  Many techniques arose in the studies of this kind of problem. For example, Agarwal et al. [1] employed the critical point theory to establish the existence of multiple solutions of some regular as well as singular discrete boundary value problems. Cai and Yu [2] applied the Linking Theorem and the Mountain Pass Lemma in the critical point theory to study second-order discrete boundary value problems and obtained some new results for the existence of solutions. Li and Sun [3, 4] were concerned with discrete system boundary value problems and gave some sufficient conditions for the existence of one or two positive solutions by using a nonlinear alternative of Leray-Schauder type and Krasnosel'skii's fixed point theorem in a cone. Pang et al. [5] provided sufficient conditions for the existence of at least three positive solutions for quasilinear boundary value problems for finite difference equations by using a generalization of the Leggett-Williams fixed point theorem due to Avery and Peterson. Du [6], Lin and Liu [7] discussed triple positive solutions of some second-order discrete boundary value problems by making use of the Leggett-Williams fixed-point theorem, respectively.This paper deals with the following three-point boundary value problem for second-order difference equation of the form(1.1)Δ2y(k-1)+h(k)f(y(k))=0,k∈{1,…,T},y(0)-αΔy(0)=0,y(T+1)=βy(n),
where Δy(k-1)=y(k)-y(k-1), Δ2y(k-1)=y(k+1)-2y(k)+y(k-1),k∈{1,…,T}.Throughout this paper, we will assume that the following conditions are satisfied:(A1)
T≥3 is a fixed positive integer, n∈{2,…,T-1}, constant α,β>0 such that H:=T+1-βn+α(1-β)>0 and T+1-βn>0;(A2)
f∈C([0,+∞),[0,+∞)), f is either superlinear or sublinear, that is, either f0=0, f∞=∞ or f0=∞, f∞=0, where(1.2)f0=lim⁡u→0+f(u)u,f∞=lim⁡u→∞f(u)u;(A3)
h(k) is nonnegative on {1,…,T} and h(k)≡0 does not hold on {n,…,T}.In the paper, we show the existence of positive solutions of (1.1) under some assumptions. We also establish the associate Green's function. Readers may find that it is useful to define a cone on which a positive operator was defined, and a fixed point theorem due to Krasnosel'skii [11] will be applied to yield the existence of at least one positive solution.


## 2. Preliminary and Green's Function

LetN be the nonnegative integers; we let Ni,j={k∈N:i≤k≤j} and Np=N0,p.By apositive solutiony of problem (1.1), we mean y:NT+1→R, y satisfies the first equation of (1.1) on N1,T, y fulfills y(0)-αΔy(0)=0, y(T+1)=βy(n), and y is nonnegative on NT+1 and positive on N1,T.We shall need the following fixed point theorem due to Krasnosel'skii [8, 11].Theorem A.
LetE be a Banach space, and let K⊂E be a cone in E. Assume that Ω1 and Ω2 are open subsets of E with 0∈Ω1, Ω1¯⊂Ω2, and let A:K⋂(Ω2¯∖Ω1)→K be a completely continuous operator such that either(1)
∥Au∥≤∥u∥, u∈K⋂∂Ω1 and ∥Au∥≥∥u∥, u∈K⋂∂Ω2 or(2)
∥Au∥≥∥u∥, u∈K⋂∂Ω1 and ∥Au∥≤∥u∥, u∈K⋂∂Ω2.Then A has a fixed point in K⋂(Ω2¯∖Ω1).Lemma 2.1 (see [7]).
The function(2.1)G(k,l)=1H{(l+α)[T+1-k-β(n-k)],l∈N1,k-1∩N1,n-1,(l+α)(T+1-k)+β(n+α)(k-l),l∈Nn,k-1,(k+α)[T+1-l-β(n-l)],l∈Nk,n-1,(k+α)(T+1-l),l∈Nk,T∩Nn,T,
is the Green's function of the problem
(2.2)-Δ2y(k-1)=0,k∈N1,T,y(0)-αΔy(0)=0,y(T+1)=βy(n).Remark 2.2.
We observe that the conditionH>0 and T+1-βn>0 implies G(k,l) is positive on NT+1×N1,T, which means that the finite set
(2.3){G(k,l)G(k,k):k∈NT+1,l∈N1,T}
takes positive values. Then we let
(2.4)M1=min⁡{G(k,l)G(k,k):k∈NT+1,l∈N1,T},(2.5)M2=max⁡{G(k,l)G(k,k):k∈NT+1,l∈N1,T}.


## 3. Main Results

Theorem 3.1.
Assume that(A1)–(A3) hold, then problem (1.1) has at least one positive solution.Proof.
In the following, we denote(3.1)m=min⁡k∈Nn,TG(k,k),M=max⁡k∈NT+1G(k,k).
Then 0<m<M.
LetE be the Banach space defined by E={y:NT+1→R}. Define(3.2)K={y∈E:y(k)≥0,fork∈NT+1andmin⁡k∈Nn,Ty(k)≥σ‖y‖},
where σ=M1m/M2M∈(0,1), ∥y∥=maxk∈NT+1|y(k)|. It is clear that K is a cone in E.
We define the operatorS:K→E by
(3.3)(Sy)(k)=∑l=1TG(k,l)h(l)f(y(l)),k∈NT+1.
It is clear that problem (1.1) has a solution y if and only if y∈E is a solution of the operator equation y(k)=(Sy)(k). We shall now show that the operator S maps K into itself. For this, let y∈K; from (A2),(A3), we find
(3.4)(Sy)(k)=∑l=1TG(k,l)h(l)f(y(l))≥0,fork∈NT+1.
From (2.5), we obtain
(3.5)(Sy)(k)=∑l=1TG(k,l)h(l)f(y(l))≤M2∑l=1TG(k,k)h(l)f(y(l))≤M2M∑l=1Th(l)f(y(l)),fork∈NT+1.
Therefore
(3.6)‖Sy‖≤M2M∑l=1Th(l)f(y(l)).
Now from (A2), (A3), (2.4), and (3.6), for k∈Nn,T, we have
(3.7)(Sy)(k)≥M1∑l=1TG(k,k)h(l)f(y(l))≥M1m∑l=1Th(l)f(y(l))≥M1mM2M‖Sy‖=σ‖y‖.
Then
(3.8)min⁡k∈Nn,T(Sy)(k)≥σ‖Sy‖.
From (3.4) and (3.6), we obtain Sy∈K. Hence S(K)⊆K. Also standard arguments yield that S:K→K is completely continuous.Case 1.
Supposef is superlinear. Now since f0=0, we may choose C1>0 such that f(u)≤δ1u, for 0<u≤C1, where δ1 satisfies
(3.9)δ1M2M∑l=1Th(l)≤1.
Let y∈K be such that ∥y∥=C1; by using (2.5) and (3.9), we have
(3.10)(Sy)(k)≤M2∑l=1TG(k,k)h(l)f(y(l))≤δ1M2M∑l=1Th(l)y(l)≤δ1M2M∑l=1Th(l)‖y‖≤‖y‖.
Now if we let
(3.11)Ω1={y∈E:‖y‖<C1},
then
(3.12)‖Sy‖≤‖y‖,fory∈K∩∂Ω1.
Next since f∞=∞, there exists C2¯>0, such that f(u)≥δ2u, for u≥C2¯, where δ2>0 satisfying
(3.13)δ2M1σ∑l=nTG(n,n)h(l)≥1.
Let C2=max{2C1,C2¯/σ} and Ω2={y∈E:∥y∥<C2}, and let y∈K and ∥y∥=C2, then
(3.14)min⁡k∈Nn,Ty(k)≥σ‖y‖≥C2¯.
Applying (2.4) and (3.13), one has
(3.15)(Sy)(n)=M1∑l=1TG(n,l)h(l)f(y(l))≥M1∑l=nTG(n,n)h(l)f(y(l))≥δ2M1∑l=nTG(n,n)h(l)y(l)≥δ2M1σ∑l=nTG(n,n)h(l)‖y‖≥‖y‖.
Thus
(3.16)‖Sy‖≥‖y‖,y∈K∩∂Ω2.
In view of (3.12) and (3.16), it follows from Theorem A that S has a fixed point y∈K⋂(Ω2¯∖Ω1) such that C1≤∥y∥≤C2.Case 2.
Supposef is sublinear case. Since f0=∞, we may choose C3>0 such that f(u)≥δ3u for 0<u≤C3, where δ3>0 satisfying
(3.17)δ3M1σ∑l=nTG(n,n)h(l)≥1,Ω3={y∈E:∥y∥<C3}; let y∈K and ∥y∥=C3. Using (2.4) and (3.17), one has
(3.18)(Sy)(n)≥M1∑l=nTG(n,n)h(l)f(y(l))≥δ3M1∑l=nTG(n,n)h(l)y(l)≥δ3M1σ∑l=nTG(n,n)h(l)‖y‖≥‖y‖.
Then ∥Sy∥≥∥y∥, y∈K⋂∂Ω3.
In view off∞=0, there exists C4¯>0 such that f(u)≤δ4u for u≥C4¯, where δ4>0 satisfying
(3.19)δ4M2M∑l=nTh(l)≤1.
There are two subcases to consider, that is, f is bounded and f is unbounded.
Subcase 2.1.
Supposef is bounded, that is, f(y)≤L for all y∈[0,∞) for some L>0. Let
(3.20)C4=max⁡{2C3,LM2M∑l=1Th(l)}.
Then, for y∈K and ∥y∥=C4, one has
(3.21)(Sy)(k)≤M2∑l=1TG(k,k)h(l)f(y(l))≤LM2M∑l=1Th(l)≤C4=‖y‖.
Hence ∥Sy∥≤∥y∥.
Subcase 2.2.
Supposef is unbounded, that is, there exists C4>max{2C3,C4¯/σ} such that f(u)≤f(C4) for all 0<u≤C4. Then, for y∈K with ∥y∥=C4, from (2.5) and (3.19), we have
(3.22)(Sy)(k)≤M2∑l=1TG(k,k)h(l)f(y(l))≤M2M∑l=1Th(l)f(C4)≤δ4M2M∑l=1Th(l)C4≤C4=‖y‖.
Thus in both Subcases 2.1 and 2.2, we may put Ω4={y∈E:∥y∥<C4}. Then
(3.23)‖Sy‖≤‖y‖,y∈K∩∂Ω4.
By using the fixed point Theorem A, it follows that problem (1.1) has at least one positive solution, such that C3≤∥y∥≤C4. The proof is finished.
Finally, we give an example to demonstrate our main result.Example 3.2.
Consider the following three-point boundary value problem:(3.24)Δ2y(k-1)+2(-k2+10k+33)1.5(y+20)1.5=0,k∈N1,8,y(0)-139Δy(0)=0,y(9)=2237y(4),
where T=8, n=4, α=13/9, β=22/37, T+1-βn+α(1-β)=800/111>0, T+1-βn=245/37>0, h(k)=2/(-k2+10k+33)1.5, k∈N1,8, f(y) = (y+20)1.5, then f is superlinear. Conditions of Theorem 3.1 are all satisfied. Then problem (3.24) has at least one positive solution y. Indeed y=-k2+10k+13 is one such positive solution.



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*Source: 596437-2011-09-29.xml*