# A Global Description of the Positive Solutions of Sublinear Second-Order Discrete Boundary Value Problems

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

**2009**:671625

**DOI: **10.1155/2009/671625

© Ruyun Ma et al. 2009

**Received: **12 February 2009

**Accepted: **20 August 2009

**Published: **7 October 2009

## Abstract

## 1. Introduction

Here is a positive parameter, and are continuous. Denote and .

There are many literature dealing with similar difference equations subject to various boundary value conditions. We refer to Agarwal and Henderson [1], Agarwal and O'Regan [2], Agarwal and Wong [3], Rachunkova and Tisdell [4], Rodriguez [5], Cheng and Yen [6], Zhang and Feng [7], R. Ma and H. Ma [8], Ma [9], and the references therein. These results were usually obtained by analytic techniques, various fixed point theorems, and global bifurcation techniques. For example, in [8], the authors investigated the global structure of sign-changing solutions of some discrete boundary value problems in the case that . However, relatively little result is known about the global structure of solutions in the case that , and no global results were found in the available literature when . The likely reason is that the Rabinowitz's global bifurcation theorem [10] cannot be used directly in this case.

- (A1)
- (A2)
- (A3)
- (A4)

Let denote the Banach space defined by

Let denote the closure of set of positive solutions of (1.1) in .

Let be a subset of . A component of is meant a maximal connected subset of , that is, a connected subset of which is not contained in any other connected subset of .

The main results of this paper are the following theorem.

Theorem 1.2.

for some . Moreover, there exists such that (1.1) has at least two positive solutions for .

## 2. Some Preliminaries

In this section, we give some notations and preliminary results which will be used in the proof of our main results.

Definition 2.1 (see [12]).

Definition 2.2 (see [12]).

A *component* of a set
is meant a maximal connected subset of
.

Lemma 2.3 ([12, Whyburn]).

Suppose that is a compact metric space, and are nonintersecting closed subsets of , and no component of interests both and . Then there exist two disjoint compact subsets and , such that , , .

Using the above Whyburn's lemma, Ma and An [13] proved the following lemma.

Lemma 2.4 ([13, Lemma ]).

Then there exists an unbounded component in and .

Using the standard arguments, we may prove the following lemma.

Lemma 2.5.

Assume that (A1)–(A2) hold. Then and is completely continuous.

Lemma 2.6.

Proof.

## 3. Proof of the Main Results

Similarly we may extend to be an odd function for each .

as a bifurcation problem from the trivial solution .

Equation (3.8) can be converted to the equivalent equation

Further we note that for near in .

The results of Rabinowitz [10] for (3.8) can be stated as follows. For each integer , , there exists a continuum of solutions of (3.8) joining to infinity in . Moreover,

Lemma 3.1.

Let (A1)–(A4) hold. Then, for each fixed , joins to in .

Proof.

We divide the proof into two steps.

Step 1.

Step 2.

(where ), which yields that is bounded. However, this contradicts (3.19).

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.

Proof of Theorem 1.2.

Notice that satisfies all conditions in Lemma 2.4, and consequently, contains a component which is unbounded. However, we do not know whether joins with or not. To answer this question, we have to use the following truncation method.

We claim that satisfies all of the conditions of Lemma 2.4.

that is, condition (ii) in Lemma 2.4 holds. Condition (iii) in Lemma 2.4 can be deduced directly from the Arzelà -Ascoli theorem and the definition of . Therefore, the superior limit of contains a component joining with infinity in .

and the superior limit of contains a component joining with infinity in .

From Lemma 2.4, it follows that is closed in , and furthermore, is compact in .

Let

If for some , , then Theorem 1.2 holds.

Assume on the contrary that for all .

where and are the boundary and closure of in , respectively.

where and are the boundary and closure of in , respectively.

However, this contradicts (3.50).

Therefore, there exists such that which is unbounded in both and .

Finally, we show that joins with . This will be done by the following three steps.

Step 1.

This is impossible by (A3) and the assumption .

Step 2.

where . By (A2), it follows that . Obviously, (3.65) implies that is bounded. This is a contradiction.

Step 3.

where . By (A2), it follows that . Obviously, (3.68) implies that is bounded. This is a contradiction.

## Declarations

## Authors’ Affiliations

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