# Existence and Multiple Solutions for Nonlinear Second-Order Discrete Problems with Minimum and Maximum

- Ruyun Ma
^{1}Email author and - Chenghua Gao
^{1}

**2008**:586020

**DOI: **10.1155/2008/586020

© R. Ma and C. Gao. 2008

**Received: **15 March 2008

**Accepted: **19 July 2008

**Published: **29 July 2008

## Abstract

## 1. Introduction

It is clear that the above are norms on and , respectively, and that the finite dimensionality of these spaces makes them Banach spaces.

where is a continuous function, are fixed numbers satisfying and satisfying .

Functional boundary value problem has been studied by several authors [1–7]. But most of the papers studied the differential equations functional boundary value problem [1–6]. As we know, the study of difference equations represents a very important field in mathematical research [8–12], so it is necessary to investigate the corresponding difference equations with nonlinear boundary conditions.

where is a bounded function, that is, there exists a constant , such that . The proofs in [1] are based on the technique of monotone boundary conditions developed in [2]. From [1, 2], it is clear that the results of [1] are valid for functional differential equations in general form and for some cases of unbounded right-hand side of the equation (see [1, Remark 3 and (5)], [2, Remark 2 and (8)]).

*ě*k [3] worked on the existence of two different solutions to the nonlinear differential equation with nonlinear boundary conditions

It is not difficult to see that when we take , (1.8) is to be (1.7), and may not be bounded.

But as far as we know, there have been no discussions about the discrete problems with minimum and maximum in literature. So, we use the Borsuk theorem [13] to discuss the existence of two different solutions to the second-order difference equation boundary value problem (1.5), (1.6) when satisfies

## 2. Preliminaries

Definition 2.1.

Remark 2.2.

So, in the rest part of this paper, we only deal with BVP (1.5), (2.4).

Lemma 2.3.

Proof.

Without loss of generality, we suppose .

- (i)
- (ii)
- (iii)

Similarly, we can obtain the following lemma.

Lemma 2.4.

Lemma 2.5.

then there exist , such that .

Proof.

We only prove that there exists , such that , and the other can be proved similarly.

Suppose for . Then . Furthermore, , which contradicts with .

Lemma 2.6.

Proof.

and be the number of elements in the number of elements in .

If , then ; if , then . Equation (2.24) is obvious.

Since , by Lemma 2.5, there exist , such that . Without loss of generality, we suppose .

For any , there exits satisfying one of the following cases:

Case 1.

Case 2.

We only prove that (2.27) holds when Case 1 occurs, (if Case 2 occurs, it can be similarly proved).

If Case 1 holds, we divide the proof into two cases.

Subcase 1.1.

Without loss of generality, we suppose . Then will be discussed in different situations.

From (2.26), (2.55), and (2.56), the assertion is proved.

Remark 2.7.

Lemma 2.8.

where denotes Brouwer degree, and the identity operator on .

Proof.

Obviously, is a bounded open and symmetric with respect to subset of Banach space .

By Borsuk theorem, to prove , we only need to prove that the following hypothesis holds.

- (a)
- (b)
- (c)

Since and are continuous, is a continuous operator. Then is a completely continuous operator.

At last, we prove (c).

Case 1.

So, , which contradicts with .

Case 2.

Similarly, , then we get and , which contradicts with .

Case 3.

If , then . Furthermore, , which contradicts with .

If , then . Furthermore, , which contradicts with .

Then (c) is proved.

## 3. The Main Results

Theorem 3.1.

Proof.

So, is a solution of (1.5) and (2.4), that is, is a solution of (1.5) and (1.6).

So, is a solution of (1.5) and (2.4).

Next, we need to prove BVPs (1.5), (3.2), and (1.5) and (3.3) have solutions, respectively.

Now, we prove (3.10) has a solution, when .

By Lemma 2.8, . Now we prove the following hypothesis.

Since is finite dimensional, is a completely continuous operator.

From (3.13), is a solution of second-order difference equation . By Remark 2.7, . And from (3.14), there exist , such that . Now, we can prove it in two cases.

Case 1.

If there exists , such that , then

Case 2.

- (i)
- (ii)

we can obtain a solution of BVP (1.5) and (3.3).

Theorem 3.2.

Proof.

are two different solutions of (1.5) and (2.4), then are the two different solutions of (1.5) and (1.6).

## Declarations

### Acknowledgments

This work was supported by the NSFC (Grant no. 10671158), the NSF of Gansu Province (Grant no. 3ZS051-A25-016), NWNU-KJCXGC, the Spring-sun Program (no. Z2004-1-62033), SRFDP (Grant no. 20060736001), and the SRF for ROCS, SEM (2006[311]).

## Authors’ Affiliations

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