- Open Access
Positive solutions for singular boundary value problems involving integral conditions
© Hu; licensee Springer 2012
- Received: 20 December 2011
- Accepted: 1 June 2012
- Published: 5 July 2012
We are interested in the following singular boundary value problem:
where is a parameter and is the Stieltjes integral. The function and w may be singular at and/or , and . Some a priori estimates and the existence, multiplicity and nonexistence of positive solutions are obtained. Our proofs are based on the method of global continuous theorem, the lower-upper solutions methods and fixed point index theory. Furthermore, we also discuss the interval of parameter μ such that the problem has a positive solution.
- global continuous theorem
- solution of boundedness
- fixed point index
- positive solution
where is a parameter and is a Stieltjes integral. The function and w may be singular at and/or , and .
Integral boundary conditions and multi-point boundary conditions for differential equations come from many areas of applied mathematics and physics [1–7]. Recently, singular boundary value problems have been extensively considered in a lot of literature [1, 2, 5, 8], since they model many physical phenomena including gas diffusion through porous media, nonlinear diffusion generated by nonlinear sources, chemically reacting systems as well as concentration in chemical or biological problems. In all these problems, positive solutions are very meaningful.
involving a Stieltjes integral with a signed measure, that is, A has bounded variation. They dealt with many boundary conditions given in the literature in a unified way by utilizing the fixed point index theory in cones.
Recently, many researchers were interested in the global structure of positive solutions for the nonlinear boundary value problem (see, e.g., [3, 6, 7]). In 2009, Ma and An  considered the problem (1.1). Assume that
(A0) is nondecreasing and is not a constant on , with , and for (for the definition of , see (2.1) below).
(A1) is continuous and on any subinterval of , and , where , .
(A2) and for .
(A3) , where and .
They obtained the following main result:
Theorem 1.1 (, Theorem 4.1])
Here, ∑ is the closure of the set of positive solutions of (1.1) on , , and the component of a set M is a maximal connected subset of M.
A natural problem arises: How can we consider the global structure of positive solutions for the case ?
In this paper, we first obtain the global structure of positive solutions by the use of global continuous theorem, and some a priori estimates. Applying the analysis technique, we construct the lower and upper solutions. Those combined with the fixed point index theory, the existence, multiplicity and nonexistence of positive solutions to (1.1) in the case are investigated. Finally, we discuss the interval of parameter μ such that the problem (1.1) has positive solutions. The proof of the method which is based on the construction of some bounds of the solution together with global continuous theorem and fixed point index is of independent interest, and is different from the other papers.
This paper is arranged as follows. We will give some hypotheses and lemmas in Section 2. In Section 3, new criteria of the existence, multiplicity and nonexistence of a positive solution are obtained. Moreover, an example is given to illustrate our result.
Throughout this paper, we suppose that the following conditions hold:
(H0) is nondecreasing, on and with .
(H2) ( obviously holds).
Remark 2.1 It is easy to see from (H0) that . Therefore, if we assume that , then (2.3) obviously holds.
Assume that the conditions (H0)-(H2) hold, then it is easy to verify that is well defined and completely continuous.
Lemma 2.1 ( Global continuation theorem)
whereand. Suppose thatis completely continuous andfor all, then, the component of the solution set of (2.4) containingis unbounded.
We note that u is a positive solution of the problem (1.1) if and only if on K.
If for and for all , then we get from Lemma 2.1 that there exists an unbounded continuum emanating from in the closure of the set of positive solutions (1.1) in .
Lemma 2.2 ()
Let X be a Banach space, K an order cone in X andan open bounded set in X with. Suppose thatis a completely continuous operator. If, for alland all, then.
Lemma 3.1 Let (H 0)-(H 3) hold and letwith. Then there exists a constantsuch that for alland all possible positive solutionsof (1.1), the inequalityholds.
This is a contradiction. □
Lemma 3.2 Assume that the hypotheses (H 0)-(H 3) hold. Then there existssuch that if the problem (1.1) has a positive solution for parameter μ, then.
This completes the proof. □
Thus, it implies that . This is contradiction. □
Theorem 3.1 Assume that the conditions (H 0)-(H 3) hold and. Then there exists a constantsuch that the problem (1.1) has at least two positive solutions for, and at least one positive solution for, and no positive solution for.
From Lemma 2.1 and Remark 2.2, we can find that there exists an unbounded continuum emanating from in the closure of the set of positive solutions in and for all . Meanwhile, Lemma 3.1 and Lemma 3.3 respectively imply that is bounded (, ) and unbounded (, and ). Therefore, we conclude that the set of (3.5) is nonempty. Those combined with Lemma 3.2 follows that is well defined and . From the definition of , it is easy to see that the problem (1.1) has at least two positive solutions for . Again, since the continuum is a compact connected set and T is a completely continuous operator, the problem (1.1) has at least one positive solution at .
Next, we only show that the problem (1.1) has no positive solution for any . Suppose on the contrary that there exists some () such that the problem (1.1) has a positive solution corresponding to . Then we will prove that the problem (1.1) has at least two positive solutions for any which contradicts the definition of (3.5).
For the sake of obtaining the contradiction, we divide the proof into four steps.
Step 1. Constructing a modified boundary value problem.
Step 2. We will show that if u is a positive solution of (3.9), then .
This is a contradiction.
This contradicts (3.13).
Case III. Since , for , we have that , for . Again since , we know from the maximum principle that , for . This contradicts the assumption of Case III.
Therefore, we conclude that the claim (3.10) holds.
Step 3. (the definition of see below).
Step 4. We conclude that the problem (1.1) has at least two positive solutions corresponding to μ.
Then we conclude that the problem (1.1) has at least two positive solutions corresponding to μ. □
- (i)From the hypotheses (H2) and (H3), it implies that there exists such that(3.17)
- (ii)If u is a positive solution of the equation (1.1) corresponding to μ, then we have
Therefore, we get that .
where μ is a positive parameter, , , and. Then there exists a constantsuch that the problem (3.18) has at least two positive solutions for, and at least one positive solution for, and no positive solution for.
Then the boundary condition of (1.1) reduces to the m-point boundary condition of (3.18). Applying the method of Theorem 3.1, we get the conclusion. □
Thus, it implies that (2.3) holds. It is easy to verify that the conditions (H2) and (H3) hold. Therefore, by Theorem 3.1, we obtain that there exists a constant such that the problem (3.19) has at least two positive solutions for , and at least one positive solution for , and no positive solution for .
The author typed, read and approved the final manuscript.
The author would like to thank the anonymous referees very much for helpful comments and suggestions which led to the improvement of presentation and quality of the work. The work was supported partly by NSCF of Tianyuan Youth Foundation (No. 11126125), K. C. Wong Magna Fund of Ningbo University, Subject Foundation of Ningbo University (No. xkl11044) and Hulan’s Excellent Doctor Foundation of Ningbo University.
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