Solvability of Conjugate Boundary Value Problems with Integral Boundary Conditions at Resonance

Abstract

We investigate a conjugate boundary value problem with integral boundary conditions. By using Mawhin continuation theorem, we study the solvability of this boundary value problem at resonance. It is shown that the boundary value problem , , , , , has at least one solution under some suitable conditions.

1. Introduction

In the past years, many authors have investigated the existence of solutions for conjugate boundary value problems at nonresonance (see [1–8]). They got the existence of solutions by using varying methods such as upper and lower solution method and fixed point theorem. For example, by using fixed point index theory, Zhang and Sun [6] investigated the existence of positive solutions for the following problem: with the boundary conditionsThe research on the solvability of boundary value problems at resonance has also been done by many people (see [9–15]), but only a few people study the solvability of conjugate boundary value problems at resonance. For instance, in [9], Jiang and Qiu studied the existence of solutions for the following conjugate boundary value problem at resonance:where ,

Inspired by [6, 9], we shall discuss the solvability of the following conjugate boundary value problem at resonance:where ,, is right continuous on and left continuous at ; denotes the Riemann-Stieltjes integrals of with respect to .

Different from the above results, the boundary condition we study is . As far as we are concerned, it is innovative to study the solvability of conjugate boundary value problem at resonance in the case .

The organization of this paper is as follows. In Section 2, we provide Mawhin continuation theorem which will be used to prove the main results. In Section 3, we will give some lemmas and prove the solvability of problem (4).

2. Preliminaries

Firstly, for the convenience of the reader, we recall some definitions and notations.

Definition 1. Assume that and be real Banach spaces and be a Fredholm operator of index zero, if the following conditions hold: (1) is a closed subspace of ; (2) .

Let , be real Banach spaces and let be a Fredholm operator of index zero. , are continuous projectors such that It follows that is invertible. We denote the inverse of the mapping by (generalized inverse operator of ). If is an open bounded subset of such that , the mapping will be called -compact on if is bounded and is compact.

Theorem 2 (see [16] (Mawhin continuation theorem)). Let be a Fredholm operator of index zero and let be L-compact on . The equation has at least one solution in if the following conditions are satisfied:(1) for every ;(2) for every ;(3), where is a projection such that .

Let with norm in which . Let with norm . Operator is defined as with

Define operator as follows: So problem (4) becomes .

3. Main Results

Assume that the following conditions hold in this paper:(H1), , where (H2) satisfies Caratháodory conditions.(H3)There exist functions with such that (H4)There exists a constant such that if , then (H5)There is a constant such that either or holds if

Then we can present the following theorem.

Theorem 3. Suppose that (H1)–(H5) are satisfied; then there must be at least one solution of problem (4) in .

To prove the theorem, we need the following lemmas.

Lemma 4. Assume that (H1) holds; then is a Fredholm operator with index zero. And a linear continuous projector can be defined by Furthermore, define a linear operator as follows:such that .

Proof. It follows from (10) that Thus we have Moreover, we can obtain that On one hand, suppose ; then there exists such that Then we have Furthermore, for , By this together with (H1) we can get which means So we obtain that On the other hand, if satisfies , we let Then we conclude that Besides, therefore That is, ; then . In conclusion, We define a linear operator as It is obvious that and For any , together with , we have It is easy to obtain that which implies Next operator is defined as follows: Noting that it means is a projection operator. And obviously, . For any , because , we have . Moreover, by simple calculation, we can get . Above all, .
To sum up we can get that is a closed subspace of ; ; that is, is a Fredholm operator of index zero.
We now define operator as follows: For any , we have . Consequently, So In addition, it is easy to know that then Therefore Next we will prove that is the inverse of . It is clear that For each , we have and This implies that So Thus the lemma holds.

Lemma 5. is L-compact on if , where is a bounded open subset of .

Proof. We can get easily that is bounded. By Lebesgue dominated convergence theorem and condition (H2), we have that is bounded. In addition, for is equicontinuous, by Ascoli-Arzela theorem, we get is compact. Thus, is -compact. The proof is completed.

Lemma 6. The set , is bounded if (H1)–(H4) are satisfied.

Proof. Take ; then . Thus we haveand , Hence there exists at least a point , such that Thus, we get So,It follows from (42) and condition (H4) that there exists one point such that From it follows that Therefore, we can obtain that By (43) we know Then is bounded. The proof of the lemma is completed.

Lemma 7. The set is bounded if (H1), (H2), and (H5) hold.

Proof. Let ; then and , so we can get According to (H5) we have ; that is to say, is bounded. We complete the proof.

Lemma 8. The set is bounded if conditions (H1), (H2), and (H5) are satisfied, where is a linear isomorphism, and

Proof. Suppose that ; we have , and If , by condition (H5), we have . If , then If , we suppose , and then which contradicts . So the lemma holds.

Then Theorem 3 can be proved now.

Proof of Theorem 3. Suppose that is a bounded open subset of ; from Lemma 5 we know that is -compact on . In view of Lemmas 6 and 7 we can get(1), for every ;(2), for every Set . It follows from Lemma 8 that we have for any . So by the homotopy of degree, we have All the conditions of Theorem 2 are satisfied. So there must be at least one solution of problem (4) in . The proof of Theorem 3 is completed.

4. Example

To illustrate our main theorem, we present the following example. Consider the boundary value problemObviously, ,,. Let ; then ; thus it is at resonance. Let then where , , , Taking , then we have , and we can obtain . This together with the fact that for implies That is, (H4) holds. Finally, taking , when , we obtain then condition (H5) is satisfied. It follows from Theorem 3 that the boundary value problem (54) has at least one solution in .

Competing Interests

The authors declare that they have no competing interests.

Acknowledgments

The project was supported by NNSF of China (11371221, 11571207), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001), the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province, and the Tai’shan Scholar Engineering Construction Fund of Shandong Province of China.