## Boundary Value Problems

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# Structure of positive solution sets of differential boundary value problems

Boundary Value Problems20132013:100

DOI: 10.1186/1687-2770-2013-100

Accepted: 8 March 2013

Published: 22 April 2013

## Abstract

In this paper, we first obtain some results on the structure of positive solution sets of differential boundary value problems. Then by using the results, we obtain an existence result for differential boundary value problems. The method used to show the main result is the global bifurcation theory.

### Keywords

structure of positive solution sets differential boundary value problems bifurcation theory

## 1 Introduction

This paper considers the differential boundary value problem
(1.1)

where f is ϕ-superlinear at ∞ and maybe negative and p is a positive continuous function, is a parameter.

Equations of form (1.1) occur in the study for the p-Laplacian equation, non-Newtonial fluid theory and the turbulent flow of a gas in a porous medium. The case where

i.e., perturbations of the p-Laplacian, has received much attention in the recent literature. Also, problem (1.1) with has been studied by several authors in recent years (see [1] and the references therein). Here, we are interested in the case when may be negative (the so-called semipositone case) (see [2] and its references for a review). As pointed out by Lions in [3], semi-positone problems are mathematically very challenging. During the last ten years, finding positive solutions to semi-positone problems has been actively pursued and significant progress on semi-positone problems has taken place; see [48] and the references therein. For instance, Hai et al. [9] considered the existence positive solution of (1.1). Under some super-linear conditions on the non-linear term f, they proved that there exists such that (1.1) has one positive solution for . The main method in [9] used to show the main result are the fixed-point theorems.

The main purpose of this paper is going to study the structure of the positive set of (1.1). Rabinowitz [10] gave the first important results on the structure of the solution sets of non-linear equations and obtained by the degree theoretic method. Amamn [11] studied the structure of the positive solution set of non-linear equations; the reader is referred to [12, 13] for other results concerning the structure of solution sets of non-linear equations. In our paper, we will study the existence results for an unbounded connected component of a positive solution set for the differential boundary value problem of (1.1). This paper generalizes some results from the literature [9]. The paper is arranged as follows. In Section 2, we will give some preliminary lemmas. The main results will be given in Section 3.

## 2 Some lemmas

For convenience, we make the following assumptions:

(A1) ϕ is an odd, increasing homeomorphism on R with concave on .

(A2) For each , there exists such that , and (note that (A2) implies the existence of such that , and ).

(A3) is continuous.

(A4) is continuous and

uniformly for .

Let , the usual real Banach space of continuous functions with the maximum norm . Let for and . Then P is a cone of E. Define
Then for (here is a constant). For , , define
here C is a constant such that
We know that C exists and is unique for every (see [14]). Then if and only if u is a solution of

From [9], we have the following Lemmas 2.1 and 2.2.

Lemma 2.1 Let ϕ satisfy (A1) and (A2). Then for each , there exist constants and such that

Further, as and as .

Lemma 2.2 Let ϕ be as in Lemma  2.1. Then there exist and such that for , .

Lemma 2.3 Let such that for . Let and ω be a solution of
(2.1)
here is continuous function with (here is a constant) for , if , then

where .

Proof By integrating, it follows that (3.1) has the unique solution given by
where C is such that . Let for some . Then
where . By Lemma 2.2, we get
Now
And so
(2.2)
here
Note that satisfies
In fact, for . We next prove that for , here ν satisfies
Suppose it is not true, then has a negative absolute minimum at . Since
there exist such that
and
Then
Let , , then
On the other hand, using the inequality
and the fact that there exists such that , we have
which is a contradiction. So, for . Obviously, , , since for each . From (2.2), we have
Similarly,
If , then

The proof is complete. □

Let for each , where and are defined as that in Lemma 2.2 and Lemma 2.3, respectively. Then is also a cone of E. From Lemma 2.3, we know that is completely continuous.

Let

From [[15], Lemma 29.1], we have Lemma 2.4.

Lemma 2.4 Let X be a compact metric space. Assume that A and B are two disjoint closed subsets of X. Then either there exist a connected component of X meeting both A and B or where , are disjoint compact subsets of X containing A and B, respectively.

Let U be an open and bounded subset of the metric space . We set , whose boundary is denoted by . Consider a map , such that is compact and . Such a map h will also be called an admissible homotopy on U. If h is an admissible homotopy, for every and every , one has that and it makes sense to evaluate .

Lemma 2.5 If h is an admissible homotopy on , the is constant for all .

Lemma 2.6 Let such that . Then for arbitrary , there exists such that for each , and ,

where .

Proof From (A4), for , such that
there exists such that for . Let
(2.3)
for some , , and . Let
From Lemma 2.3, we know that . Namely,
(2.4)
From (2.3) and (2.4), we have
so . For , we have
for . Therefore, let , , assume that , then
where
and
Thus,

so , which is a contradiction. Then (2.3) holds. The proof is complete. □

## 3 Main results

For convenience, let us introduce the following symbols. For any r,

Now we give our main results of this paper.

Theorem 3.1 Suppose (A1) to (A4) hold. Then possesses an unbounded connected component such that for some and

where denotes the projection of onto the λ-axis.

Proof We divide our proof into four steps.

Step 1. Let
(3.1)

where . Obviously, is a completely continuous operator.

Note that for all and with . From Lemma 2.5, there exists large enough such that and
(3.2)
Obviously,
(3.3)
Therefore,
(3.4)

So, and .

Step 2. Let
From Lemma 2.6, we have
This implies that T has no bifurcation point on . From step 1,we have , then for each , denote by the connected component of the metric space emitting from . Now we will show that, there must exist such that is unbounded. Assume on the contrary that is bounded for each . Take a bounded open neighborhood in for each such that
(3.5)
and , where denotes the closure of in the metric space . Let denote the boundary of in the metric space . Obviously, is a compact subset. Assume that . From the maximal connectedness of , there is no connected subset of meeting both and . From Lemma 2.4, there exist compact disjoint subsets and of such that and , and . Let and be the -neighborhood of in the metric space . Set
(3.6)
Then we have and
(3.7)
Obviously, the collection of the subsets
is an open cover of . Since is compact, then there exist finite points, namely
such that
Let . Then U is a bounded open subset of . From (3.7), we have
(3.8)
Thus,
From (3.5) and (3.8), we have
Then from Lemma 2.5, we have
(3.9)
Since , then
(3.10)
Since , then from (3.4) we have
(3.11)

which contradicts to (3.10) and (3.11). Therefore, there must exist such that is bounded.

Step 3. Obviously, the projection of is a interval, denote it by , then . Then is a bounded connected component of . Take , let
Obviously, . For each , denote by the connected component of the metric space , which passes the point p. Now we shall prove that there must exist a such that is an unbounded connected component of the metric space . On the contrary, assume that is bounded for each . Then, for each , in the same way as in the construction of in (3.6) we can show that there exists a neighborhood of in such that
(3.12)
Obviously, the set of is an open cover of the set and is a compact set. Thus, there exist finite subsets of , say
which is also an open cover of , that is,
(3.13)
Let , then is a bounded open subset of . Since
then by (3.12) we have
(3.14)
Let
and . It is easy to see that

From (3.12) and (3.13), we see that . Obviously, and . Note the unboundedness of , then . Now we have , which is a contradiction of the connectedness of . Therefore, there must exist such that is an unbounded connected component of .

Step 4. Since for each , we have
So, for each . This implies is an unbounded subset. Let be the connected component of containing . Obviously, there exists such that . As is unbounded, we easily see that

The proof is complete. □

Corollary 3.1 Let (A1) to (A4) hold. Then there exists such that problem (1.1) has a positive solution for with as .

## Declarations

### Acknowledgements

This paper is supported by Innovation Project of Jiangsu Province postgraduate training project (CXLX12_0979).

## Authors’ Affiliations

(1)
Department of Mathematics, Jiangsu Normal University

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