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Discrete Dynamics in Nature and Society
Volume 2014 (2014), Article ID 614376, 7 pages
http://dx.doi.org/10.1155/2014/614376
Research Article

Global Structure of Positive Solutions for a Singular Fourth-Order Integral Boundary Value Problem

Department of Basic Courses, Lanzhou Institute of Technology, Lanzhou 730050, China

Received 11 November 2013; Revised 14 December 2013; Accepted 16 December 2013; Published 8 January 2014

Academic Editor: Gabriele Bonanno

Copyright © 2014 Wenguo Shen and Tao He. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We consider fourth-order boundary value problems , where is a Stieltjes integral with being nondecreasing and being not a constant on may be singular at and , with on any subinterval of ; and for all , and We investigate the global structure of positive solutions by using global bifurcation techniques.

1. Introduction

Recently, fourth-order boundary value problem has been investigated by the fixed point theory in cones, see [14] (). By applying bifurcation techniques, see Rynne [5] (), Korman [6] (), Xu and Han [7] (), Shen [8, 9] (), and references therein. However, these papers only studied the nonsingular boundary value problems.

In 2008, Webb et al. [10] studied the existence of multiple positive solutions of nonlinear nonlocal boundary value problems (BVPs) for equations of the form where are continuous and nonnegative functions and is a function of bounded variation. They treat many boundary conditions appearing in the literature in a unified way. The main tool they used is the fixed point index theory in cones. In 2009, Ma and An [11] studied the global structure for second-order nonlocal boundary value problem involving Stieltjes integral conditions by applying bifurcation techniques.

Motivated by above papers, in this paper, we will use global bifurcation techniques to study the global structure of positive solutions of the singular problem where may be singular at and , and is a parameter.

In order to prove our main result, let us make the assumptions as follows:(A1) is nondecreasing and is not a constant on , for , and with ;(A2) with on any subinterval of , and ;(A3) satisfies for all ;(A4);(A5).

Remark 1. For other results on the existence and multiplicity of positive solutions and nodal solutions for the boundary value problems of fourth-order ordinary differential equations based on bifurcation techniques, see Ma et al. [1215] and Bai and Wang [16] and their references.
The rest of the paper is arranged as follows: In Section 2, we state some properties of superior limit of certain infinity collection of connected sets. In Section 3, we will give some preliminary results. In Section 4, we state and prove our main results.

2. Superior Limit and Component

In order to treat the case , we will need the following definition and lemmas.

Definition 2 (see [17]). Let be a Banach space and let be a family of subsets of . Then the superior limit of is defined by

Lemma 3 (see [17]). Each connected subset of metric space is contained in a component, and each connected component of is closed.

Lemma 4 (see [11]). Let be a Banach space and let be a family of closed connected subsets of . Assume that(i)there exist , and , such that ;(ii); (iii)for all , is a relative compact set of , where Then there exists an unbounded connected component in and .

3. Preliminaries

We consider the problem as follows:

Lemma 5. For any , the problem (6) has a unique solution where .

Proof. By [10], the problem (6) can be equivalently written as Applying to both sides of (10), we obtain Thus, we have Furthermore, it follows that So, we obtain

Lemma 6 (see [24]). Green's function defined by (9) satisfies the following:(i) is continuous for all ;(ii), , and for any and , such that where

Lemma 7. Green's function defined by (8) satisfies the following:(i) is continuous for all ;(ii), , and for any , there exists a constant , for any , such that where is defined by (9), , .

Proof. (i) From Lemma 6 (i), we get the proof of Lemma 7 (i) immediately.
(ii) By Lemma 6 (ii), we get By Lemma 6 (ii), for any and , we obtain

Lemma 8. For and , the unique solution of the problem (6) satisfies the following:(i),  ;(ii),where is defined by Lemma 7 (ii), .

Proof. (i) From Lemma 7 (i), we get the proof of Lemma 8 (i) immediately.
(ii) From (7) and Lemma 7, we have Therefore, the proof of Lemma 8 is complete.

Let be the Banach space with the norm .

Let with the norm

Let and for , let .

In order to use bifurcation technique to study the problem (3), we consider the linear eigenvalue problem

Let

By [18], it is easy to show the following lemma.

Lemma 9. Assume that (A1)–(A3) hold the following.
is a completely continuous linear operator and , and the fixed points of the operator in are the positive solutions of the BVP (23).
is a completely continuous operator and , and the fixed points of the operator in are the positive solutions of the BVP (3).
By virtue of Krein-Rutman theorem (Theorem 2.5 in [19]), one has (see [18] or [20]) the following lemma.

Lemma 10. Suppose that is a completely continuous linear operator and . If there exist and a constant such that , then the spectral radius and has a positive eigenfunction corresponding to its first eigenvalue , that is, .

Lemma 11. Suppose (A1) and (A2) are satisfied, then for the operator defined by (24), the spectral radius and has a positive eigenfunction corresponding to its first eigenvalue .

Proof. It is easy to see that there is such that . Thus there exists such that and . Take such that , and . Then for , So there exists a constant such that . From Lemma 10, we know that the spectral radius and has a positive eigenfunction corresponding to its first eigenvalue .

Lemma 12. Let (A1)–(A3) hold. The solution of the problem (3) satisfies

Proof. From , there exists , such that . Using a similar proof of (10) in [21, page 212], it is easy to show that Furthermore, we obtain

Lemma 13. Let (A1)–(A3) hold. Assume that is a sequence of positive solutions of (3). Assume that for some constant , and Then

Proof. Assume on the contrary that for some constant .
Since is a solution of the problem (3), we have Thus, where , (see [3]).
Furthermore, it follows that where , together with , which implies that is bounded whenever is bounded.
Together with Lemma 12, we obtain for some constant . This is a contradiction.

4. Main Results

Let be the closure of the set of positive solutions of (3) in . The main results of this paper are the following.

Theorem 14. Let (A1)–(A5) hold, then (3) has at least one solution for any .
Let be an operator defined by with Then is a closed operator and is completely continuous.
For each , define by Then with By , it follows that To apply the global bifurcation theorem, one extends to an odd function by Similarly one may extend to an odd function for each .
Now let one consider the auxiliary family of the equations Let be such that Then Let one consider as a bifurcation problem from the trivial solution .
From Lemma 5, (46) can be converted to the equivalent equation Further one has that
Indeed, (8) implies that, for all , where , , (see [3]).
So, the compactness of together with (45) and (A2) imply that and consequently
Let , , then . Let .
By Lemma 11 and the fact , the global bifurcation result (see Rabinowitz [22]) for (46) can be stated as follows: there exists a continuum of positive solutions of (46) joining to infinity in . Moreover, is the only positive bifurcation point of (46) lying on trivial solutions line . Moreover, .
Since condition (i) in Lemma 4 is satisfied with . Obviously and accordingly (ii) in Lemma 4 holds. (iii) in Lemma 4 can be deduced directly from the Arzela-Ascoli theorem and the definition of . Therefore, the superior limit of , , contained an unbounded connected component with . Since , one concludes . Moreover, by (3).

Proof of Theorem 14. We firstly prove
Assume on the contrary that Then there exists a sequence such that for some positive constant with doing not depend on . From Lemma 13, we have
This together with the fact imply that for arbitrary Since , we have that
Set . Then
Now, choosing a subsequence and relabeling if necessary, it follows that there exists with such that
By (A3), let Then is nondecreasing and Further it follows from (66) that
Thus, Notice that (62) is equivalent to
Furthermore, by (59), (68), and (69), together with the Lebesgue dominated convergence theorem, it follows that It follows that
This contradicts (63). Therefore
Noticing that is the only solution of the problem (3), thus
Furthermore, it follows the proof of Theorem 14.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

Thanks are given to the anonymous referee for his/her valuable suggestions. The authors were supported by the NSFC (no. 11261052) and the Scientific Research Foundation of the Education department of Gansu Province (no. 1114-04).

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