Positive solutions for singular boundary value problems involving integral conditions

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.

We are concerned with the second order nonlocal boundary value problem:

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.

In [1,2], Webb and Infante considered the existence of positive solutions of nonlinear boundary value problem:

where hg are nonnegative functions, subjected to the nonlocal boundary conditions

Here is a linear functional on given by

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 [3] 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 ([3], Theorem 4.1])

Assume that (A0)-(A3) hold. Then there exists a componentin ∑ which joinsto, and

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

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.

Let X denote the Banach space with the maximum norm

Define , then K is a cone. Let

Denote

Throughout this paper, we suppose that the following conditions hold:

(H0) is nondecreasing, on and with .

(H1) and w may be singular at and/or 1, satisfying

(H2) ( obviously holds).

(H3) .

Remark 2.1 It is easy to see from (H0) that . Therefore, if we assume that , then (2.3) obviously holds.

Define an operator as follows:

Assume that the conditions (H0)-(H2) hold, then it is easy to verify that is well defined and completely continuous.

Lemma 2.1 ([9] Global continuation theorem)

LetXbe a Banach space and letKbe an order cone inX. Consider the equation

whereand. Suppose thatis completely continuous andfor all, then, the component of the solution set of (2.4) containingis unbounded.

Remark 2.2

(1) We note that u is a positive solution of the problem (1.1) if and only if on K.

(2) 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 ([10])

LetXbe a Banach space, Kan order cone inXandan open bounded set inXwith. Suppose thatis a completely continuous operator. If, for alland all, then.

Lemma 3.1Let (H0)-(H3) hold and letwith. Then there exists a constantsuch that for alland all possible positive solutionsof (1.1), the inequalityholds.

Proof Suppose on the contrary that there exist a sequence and a sequence of the positive solutions of (1.1) corresponding to such that

Denote . From the concavity of , it follows that

Choose , where . Then we find from (H3) that there exists a constant such that

Since , we get

This, together with (3.1) and (3.2), implies

Put , for . Hence, we know from (3.3) that

On the other hand, multiplying (1.1) by ψ and integrating by parts, we obtain that

leads to

i.e.,

This is a contradiction. □

Lemma 3.2Assume that the hypotheses (H0)-(H3) hold. Then there existssuch that if the problem (1.1) has a positive solution for parameterμ, then.

Proof Let u be a positive solution of (1.1) corresponding to μ. From the hypotheses (H2) and (H3), it follows that there exists such that for all . Consequently, we have

Let be the first eigenvalue of

and let be the positive eigenfunction corresponding to (see [8]). It is easy to see that . Multiplying (3.4) by and integrating by parts, we get that

implies

This completes the proof. □

Lemma 3.3Assume that (H0)-(H3) hold. Then we have

Proof Suppose this fails, that is, there exists such that

where is a positive constant. Since , we get that for all

where

Hence, we find that

Thus, it implies that . This is contradiction. □

Theorem 3.1Assume that the conditions (H0)-(H3) 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.

Proof Define

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.

Choose arbitrarily a constant . Since f is uniformly continuous on , there exists a constant such that

where

Denote , for . Then we claim that

and

Indeed, using (3.6), we have

Define a set . Then the set is bounded and open in . Now, we construct the modified second-order boundary value problem:

where is defined by

Step 2. We will show that if u is a positive solution of (3.9), then .

Let u be a positive solution of (3.9), then we claim that

Suppose this fails, that is, . Clearly, we only show that , for . Comparing the boundary conditions (3.8) and (3.9), the only following three cases need to be considered. Case I. There exists such that and , for and some ; Case II. There exists such that and for . Case III. There exists such that , , and . See the three Figures 1, 2 and 3.

Case I. From (3.7), it implies that there exists a constant such that

Since f is uniformly continuous on , there exists a such that if and , then we get

where . From the assumption of Case I, it follows that there exists a subinterval with such that

and

This together with (3.11) and (3.12) leads to

This is a contradiction.

Case II. Let , for . Then we have that for

Obviously, we have . From (3.13), it implies that , for any . Hence, the function is strictly increasing in . From (3.8) and the boundary condition (3.9), we find

We obtain a contradiction. In particular, if , then we obtain

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).

Using (2.2), we define an operator by

Then is completely continuous and u is a positive solution of (3.9) if and only if on K. From the definition of , it implies that there exists such that , for all . Consequently, we get from Lemma 2.2 that

where . Applying the conclusion of Step 2 and the excision property of fixed point index, we find that

Step 4. We conclude that the problem (1.1) has at least two positive solutions corresponding to μ.

Since the problem (1.1) is equivalent to the problem (3.9) on , we get that the problem (1.1) has a positive solution in . Without loss of generality, we may suppose that T has no fixed point on (otherwise the proof is completed). Then is well defined and (3.15) implies

On the other hand, from Lemma 3.2, we choose such that the problem (1.1) has no positive solution in K. By apriori estimate in , there exists such that for all possible positive solutions of (1.1) with , we know that . Define by

Then it is easy to verify that is completely continuous on and for all . From the property of homotopy invariance, it follows that

Hence, by the additivity property and (3.16), we have

Then we conclude that the problem (1.1) has at least two positive solutions corresponding to μ. □

Remark 3.1

(i) From the hypotheses (H2) and (H3), it implies that there exists such that

Let f attain its maximum at the point of . If , then adopting the similar method as in [11], Theorem 1], we get that for , the problem (1.1) has at least two positive solutions and such that by the use of compression of conical shells in [12], Corollary 20.1]. Consequently, we know that .

(ii) If u is a positive solution of the equation (1.1) corresponding to μ, then we have

i.e., from (3.17),

Therefore, we get that .

Corollary 3.1Assume that (H1)-(H3) hold. Consider the followingm-point boundary value problem

whereμis 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.

Proof In the boundary condition of (1.1), if we let

where is the characteristic function on , i.e.,

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. □

Example 3.1 We consider the following singular boundary value problem:

where .

Computing yields

We find that for any ,

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 declares that they have no competing interests.

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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