#### Abstract

We consider the fourth-order difference equation: , subject to the boundary conditions: , , , where and for , is continuous. is nonnegative ; is nonnegative for . Using fixed point theorem of cone expansion and compression of norm type and Hölder’s inequality, various existence, multiplicity, and nonexistence results of positive solutions for above problem are derived, which extends and improves some known recent results.

#### 1. Introduction

Boundary value problems (BVPs) for ordinary differential equations arise in different areas of applied mathematics and so on. The existence of solutions for second order and higher order nonlocal boundary value problems has been studied by several authors; for example, see [1–11] and the references therein. Many authors have also discussed the existence of positive solutions for higher order difference equation BVPs [12, 13], by using fixed point theorem of cone expansion and compression of norm type, sufficient conditions for the existence of positive solutions for fourth-order and third-order difference equation BVPs are established, respectively. Recently, there has been much attention on the existence of positive solutions for the fourth-order differential equations with integral boundary conditions [14–17]. In [17], Zhang and Ge considered the differential equation BVP: where, is symmetric on , for some , and it is symmetric on the interval,is continuous, andfor all, andare nonnegative, symmetric on. The authors made use of fixed point theorem of cone expansion and compression of norm type and Hölder inequality to prove the existence of positive solutions for the above problem.

Motivated by the above works, we intend to study the existence and nonexistence of positive solutions of the following fourth-order difference BVP with sum form boundary conditions:

Throughout this paper, we make the following assumptions: is symmetric onand; there is asuch thatfor;is continuous;is nonnegative onand,is nonnegative onand, where,;,.

In order to establish the existence of positive solutions of the problem (2), we need the following definitions, theorem, and lemma.

*Definition 1. *A functionis said to be a solution of problem (2) ifsatisfying BVP (2).

*Definition 2 (see [18]). *Letbe a real Banach space over . A nonempty closed setis said to be a cone provided that(i)for alland alland(ii)implies.

Every coneinduces an ordering ingiven byif and only if.

Theorem 3 (see [18]). *Letbe a cone in a real Banach space. Assume thatare bounded open sets inwith. If
**
is completely continuous such that either*(i)*,and,,* *or*(ii)*,and,,* *thenhas at least one fixed point in.*

Lemma 4 (Hölder). *Suppose thatis a real-valued column; let
**, satisfy the conditionwhich are called conjugate exponent, andfor. If, then
**
which can be recorded as
*

#### 2. Preliminaries

Let; is a real Banach space with the normdefined by

Letbe a cone of, and where.

In our main results, we will use the following lemmas and properties.

Lemma 5. *Suppose thatandhold and; then, for all, the BVP
**
has unique solutiongiven by
**
where
**
where,.*

Proposition 6. *Assume that; then
*

Proposition 7. *Assume that; then
**
where
*

Proposition 8. *Suppose that; then
**
where
*

*Proof. *From Lemma 5 and Proposition 7, we have

On the other hand, using, we get

Lemma 9. *Suppose that, for all; the BVP
**
has unique solutiongiven by
**
where
*

*Proof. *From the properties of difference operator, we can get
then we have
It can imply that

Let; we have
That is
thus,

We can get

From the boundary conditions, we have
thus,
Because
we have
Thus, we get
where

Multiplying the above equation with, summing them fromto, we have
So, we can get

Proposition 10. *Assume that; then
*

Proposition 11. *For, one has
**
where.*

Proposition 12. *If, then, for all, one has
**
where, .*

We construct a cone onby where Obviously,is a closed convex cone of.

Define an operatoras Let Then we can obtain the following properties.

Proposition 13. *Ifandhold, then
*

Proposition 14. *Ifandhold, then, for all , one has
*

Lemma 15. *Suppose thathold; ifis a solution of the equation
**
thenis a solution of the BVP (2).*

Lemma 16. *Assume thathold; thenandis completely continuous.*

*Proof. *From above works, for all, we have
Because
That is to sayfor.

On the other hand, for, we have
Similarly, we can get
So,and. It is easy to see thatis completely continuous.

#### 3. The Existence of One Positive Solution

In this part, we apply Theorem 3 and Lemma 4 to prove the existence of one positive solution for BVP (2). We need consider the following cases for:,, and.

Let wheredenotesor, and

*Remark 17. *If we only consider the case, then we can take

Firstly, the following theorem deals with the case .

Theorem 18. *Suppose thathold. If there exist two constants, withsuch that**for allandfor all;* *or**for allandfor all, then BVP (2) has at least one positive solution.*

*Proof. *We only consider condition. For, from the definition ofwe obtain that
So, for, we have, which implies that. Thus, for, fromwe get
that is,implies that

On the other hand, for, we have, which implies that; therefore for, fromwe have
that is,implies that

From the above works, we apply (i) of Theorem 3 to yield thathas a fixed point, and. Thus, it follows that BVP (2) has at least one positive solution.

The following theorem deals with the case.

Theorem 19. *Suppose thathold and or holds. Then BVP (2) has at least one positive solution.*

*Proof. *Letreplace and repeat the argument of Theorem 18.

Finally we consider the case.

Theorem 20. *Suppose thathold and or holds. Then BVP (2) has at least one positive solution.*

*Proof. *Similar to the proof of Theorem 18, for, we have
So, for, we have. And from the proof of Theorem 18,for. Thus Theorem 20 is proved.

Theorem 21. *Assume thathold. If one of the following conditions is satisfied**and( particularly,and);*

*and (*

*particularly*,and),*then BVP (2) has at least one positive solution.*

*Proof. *We only consider the case. The proof of caseis similar to case. Considering, there existssuch thatfor,, wheresatisfies. Then, for,, fromwe have
that is,implies that.

Next, considering, then there existssuch that
wheresatisfies; assume that
then
Choosing.

Thus, for, we obtain