Discrete Dynamics in Nature and Society

Volume 2011 (2011), Article ID 596437, 8 pages

http://dx.doi.org/10.1155/2011/596437

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

^{1}School of Mathematical Sciences, Xuzhou Normal University, Xuzhou, Jiangsu 221116, China^{2}Department of Mathematics, China University of Mining and Technology, Xuzhou, Jiangsu 221008, China

Received 28 June 2011; Accepted 4 August 2011

Academic Editor: Zengji Du

Copyright © 2011 Xiaojie Lin and Wenbin Liu. 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.

#### 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.

#### 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 where , .

Throughout this paper, we will assume that the following conditions are satisfied:(A1) is a fixed positive integer, , constant such that and ;(A2), is either superlinear or sublinear, that is, either , or , , where (A3) is nonnegative on and does not hold on .

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

Let be the nonnegative integers; we let and .

By a *positive solution * of problem (1.1), we mean , satisfies the first equation of (1.1) on , fulfills , , and is nonnegative on and positive on .

We shall need the following fixed point theorem due to Krasnosel'skii [8, 11].

Theorem A. *Let be a Banach space, and let be a cone in . Assume that and are open subsets of with , , and let be a completely continuous operator such that either*(1)*, and , or*(2)*, and , . **Then has a fixed point in .*

Lemma 2.1 (see [7]). *The function
**
is the Green's function of the problem
*

*Remark 2.2. *We observe that the condition and implies is positive on , which means that the finite set
takes positive values. Then we let

#### 3. Main Results

Theorem 3.1. *Assume that – hold, then problem (1.1) has at least one positive solution.*

*Proof. *In the following, we denote
Then .

Let be the Banach space defined by . Define
where , . It is clear that is a cone in .

We define the operator by
It is clear that problem (1.1) has a solution if and only if is a solution of the operator equation . We shall now show that the operator maps into itself. For this, let ; from , we find
From (2.5), we obtain
Therefore
Now from (A2), (A3), (2.4), and (3.6), for , we have
Then
From (3.4) and (3.6), we obtain . Hence . Also standard arguments yield that is completely continuous.

*Case 1. *Suppose is superlinear. Now since , we may choose such that , for , where satisfies
Let be such that ; by using (2.5) and (3.9), we have
Now if we let
then
Next since , there exists , such that , for , where satisfying
Let and , and let and , then
Applying (2.4) and (3.13), one has
Thus
In view of (3.12) and (3.16), it follows from Theorem A that has a fixed point such that .

*Case 2. *Suppose is sublinear case. Since , we may choose such that for , where satisfying
; let and . Using (2.4) and (3.17), one has
Then , .

In view of , there exists such that for , where satisfying
There are two subcases to consider, that is, is bounded and is unbounded.*Subcase 2.1. *Suppose is bounded, that is, for all for some . Let
Then, for and , one has
Hence .*Subcase 2.2. *Suppose is unbounded, that is, there exists such that for all . Then, for with , from (2.5) and (3.19), we have
Thus in both Subcases 2.1 and 2.2, we may put . Then
By using the fixed point Theorem A, it follows that problem (1.1) has at least one positive solution, such that . 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:
where , , , , , , , , = , then is superlinear. Conditions of Theorem 3.1 are all satisfied. Then problem (3.24) has at least one positive solution . Indeed is one such positive solution.

#### Acknowledgments

This project is supported by the National Natural Science Foundation of China (11101349, 11071205), the NSF of Jiangsu Province (BK2008119gundong), and the NSF of the Education Department of Jiangsu Province (11KJB110013).

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