Research

Structure of positive solution sets of differential boundary value problems

Xu Xian and Zhu Xunxia*

Author Affiliations

Department of Mathematics, Jiangsu Normal University, Xuzhou, Jiangsu, 221116, P.R. China

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Boundary Value Problems 2013, 2013:100  doi:10.1186/1687-2770-2013-100

 Received: 26 September 2012 Accepted: 8 March 2013 Published: 22 April 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 [4-8] 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 2.1Letϕsatisfy (A1) and (A2). Then for each, there exist constantsandsuch that

Further, asandas.

Lemma 2.2Letϕbe as in Lemma 2.1. Then there existandsuch thatfor, .

Lemma 2.3Letsuch thatfor. Letandωbe a solution of

(2.1)

hereis continuous function with (hereis 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 2.4LetXbe a compact metric space. Assume thatAandBare two disjoint closed subsets ofX. Then either there exist a connected component ofXmeeting bothAandBorwhere, are disjoint compact subsets ofXcontainingAandB, respectively.

LetUbe an open and bounded subset of the metric space. We set, whose boundary is denoted by. Consider a map, such thatis compact and. Such a maphwill also be called an admissible homotopy onU. Ifhis an admissible homotopy, for everyand every, one has thatand it makes sense to evaluate.

Lemma 2.5Ifhis an admissible homotopy on, theis constant for all.

Lemma 2.6Letsuch that. Then for arbitrary, there existssuch 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 3.1Suppose (A1) to (A4) hold. Thenpossesses an unbounded connected componentsuch thatfor someand

wheredenotes the projection ofonto theλ-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 3.1Let (A1) to (A4) hold. Then there existssuch that problem (1.1) has a positive solutionforwithas.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

The authors declare that the study was realized in collaboration with the same responsibility. All authors read, checked and approved the final manuscript.

Acknowledgements

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

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