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Abstract and Applied Analysis

Volume 2014 (2014), Article ID 393187, 9 pages

http://dx.doi.org/10.1155/2014/393187
Research Article

Green’s Function and Positive Solutions for a Second-Order Singular Boundary Value Problem with Integral Boundary Conditions and a Delayed Argument

1Department of Mathematics and Physics, North China Electric Power University, Beijing 102206, China

2School of Applied Science, Beijing Information Science & Technology University, Beijing 100192, China

Received 16 May 2014; Revised 15 July 2014; Accepted 16 July 2014; Published 6 August 2014

Academic Editor: Dumitru Baleanu

Copyright © 2014 Xuemei Zhang and Meiqiang Feng. 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

This paper investigates the expression and properties of Green’s function for a second-order singular boundary value problem with integral boundary conditions and delayed argument , where and, may be singular at or/and at . Furthermore, several new and more general results are obtained for the existence of positive solutions for the above problem by using Krasnosel’skii’s fixed point theorem. We discuss our problems with a delayed argument, which may concern optimization issues of some technical problems. Moreover, the approach to express the integral equation of the above problem is different from earlier approaches. Our results cover a second-order boundary value problem without deviating arguments and are compared with some recent results.

1. Introduction

Boundary value problems with integral boundary conditions arise naturally in thermal conduction problems [1], semiconductor problems [2], hydrodynamic problems [3], and so on. It is interesting to point out that such problems include two-, three-, and multipoint and nonlocal boundary value problems as special cases and have been extensively studied in the last ten years; see, for example, [418]. Recently, Feng et al. [19] applied the fixed point theory in a cone for strict set contraction operators to study the existence and multiplicity of positive solutions for the problem given by where is the zero element of a real Banach space .

At the same time, a class of boundary value problems with deviating arguments are receiving much attention. For example, in [20], Yang et al. studied the existence and multiplicity of positive solutions to a three-point boundary value problem with an advanced argument: where , , and . The main tool is the fixed point index theory. It is clear that the solution of [20] is concave when on and on . However, few papers have reported the same problems where the solution is without concavity; for example, see some recent excellent results and applications of the case of ordinary differential equations with deviating arguments to a variety of problems from Jankowski [2123], Jiang and Wei [24], Wang [25], Wang et al. [26], and Hu et al. [27]. This paper will resolve this problem.

Consider the second-order singular boundary value problem with integral boundary conditions and a delayed argument: where denotes the linear operator and where , , , and may be singular at or/and at .

Throughout this paper, we assume that on . In addition, , , , and satisfy the following:) with and does not vanish on any subinterval of ;(), with on ;() is nonnegative with , where where satisfies

Remark 1. Generally, when on , the solution is not concave for the linear equation This means that the method depending on concavity is no longer valid, and we need to introduce a new method to study this kind of problems.

Remark 2. For simplicity we only consider Neumann boundary conditions since all the results obtained in this paper can also be adapted with minor changes to the other boundary conditions.

Some special cases of (3), such as boundary value problems with delay, have been investigated [2832]. It is not difficult to see that the corresponding function appearing on the right-hand side depends on , , where initial function is given on the initial set, for example, . T. Jankowski and R. Jankowski [33, 34] pointed out that, in such cases , there are some problems with a constant delay . If we consider the differential problem on intervals , where , then it means that we have no delays; we have such a situation in [32]. If , then it is easy to solve the differential equation on , since we have the solution on the initial set . Continuing this process, we can find a solution on the whole interval , by using the method of steps. In our paper, for example, the deviating argument can have a form with a fixed number , so the delay is a function of . In this case, the initial set reduces to one point , and we cannot apply the step method. To our knowledge, it is the first paper in which positive solution has been investigated for a second-order singular differential equation with a delayed argument under the case that .

Being directly inspired by [5, 12, 20, 21], the authors will prove several new and more general results for the existence of positive solutions for problem (3) by using fixed point theories in a cone. Another contribution of this paper is to study the expression and properties of Green’s function associated with problem (3). The expression of the integral equation is simpler than that of [5, 12].

The organization of this paper is as follows. In Section 2, we present the expression and properties of Green’s function associated with the problem (3). In Section 3, we present some definitions and lemmas which are needed throughout this paper. In Section 4, we use fixed point theorem to obtain the existence of positive solutions for problem (3) with a delayed argument . In particular, our results in these sections are new when on . Finally, in Section 5, three examples are also included to illustrate the main results.

2. Expression and Properties of Green’s Function

Theorem 3. Assume that . Then for any , the boundary value problem has a unique solution where Here and satisfy (6) and respectively.

Proof. The proof is similar to that of Lemma 2.3 in [5] and Lemma 2.1 in [35].

Remark 4. It is not difficult from [5, 12] to show that and (i) is nondecreasing on and on and (ii) is strictly decreasing on .

Remark 5. The expression of the integral equation (9) is different from that of (2.10) in [5] and that of (2.9) in [12].

Remark 6. Noticing , it follows from the definition of that where .

Lemma 7. Let , , , and be given as in Theorem 3. Then one has the following results: where

Proof. Noticing Remark 4, it follows from the definition of , , and that (13) holds. Now, we show that (14) also holds.

In fact, for and , we have

Similarly, we can prove that for and . This and (15) imply that This gives the proof of Lemma 7.

Remark 8. It follows from (13) and (14) that

3. Preliminaries

In this section, we first present some definitions and lemmas which are needed throughout this paper.

Definition 9 (see [36]). Let be a real Banach space over . A nonempty closed set is said to be a cone provided that(i) for all and all , ;(ii) , implies .

Every cone induces an ordering in given by if and only if .

Let . Then is a real Banach space with the norm defined by

Definition 10. A function is called a solution of (3) if it satisfies (3). If and on , then is called a positive solution of (3).

Define a cone in by Also, define, for a positive number , by Note that .

Define by

Lemma 11. Assume that ()–() hold. Then, and are completely continuous.

Proof. For , it follows from (13) and (22) that It follows from (14), (22), and (23) that Thus, .

Next, by standard methods and Ascoli-Arzela theorem, one can prove that is completely continuous. So it is omitted, and the lemma is proved.

Remark 12. From (22), we know that is a solution of problem (3) if and only if is a fixed point of operator .

In the rest of this section, we state a well known fixed point theorem which we need later.

Lemma 13 (see [36]). Let be a cone in a real Banach space . Assume and are bounded open sets in with , . If is completely continuous such that either (i), for all , and , for all , or(ii), for all , and , for all ,

then has at least one fixed point in .

4. Existence of Single or Twin Positive Solutions

For convenience, we introduce the following notations:

We also define as [37] number of zeros in the set and number of infinities in the set . Sun and Li [38] pointed out that , , or 2, and there are six possible cases: (i) and ; (ii) and ; (iii) and ; (iv) and ; (v) and ; and (vi) and . By using Krasnoseliis fixed point theorem in a cone, some results are obtained for the existence of at least one or two positive solutions of problem (3) for on under the above six possible cases.

4.1. For the Case and

In this subsection, we discuss the existence of single positive solution for problem (3) under and .

For convenience, we introduce the following notations:

Theorem 14. Assume that ()–() hold. If and , then problem (3) has at least one positive solution.

Proof. First, we consider the case and . Since , then there exists such that

Since on , it follows from on that

Consequently, for any and , (13) and (22) imply which implies

Next, turning to , there exists satisfying such that

Since on , it follows from on that Hence, for , it follows from Remark 8 and (22) that which implies

Thus, by (i) of Lemma 13, it follows that has a fixed point in with Remark 12 shows that problem (3) has at least one positive solution with

Next, we consider the case and . Since , we can choose such that

Since on , it follows from on that

Consequently, for , it follows from Remark 8 and (22) that which implies

If , we can choose and such that

Letting , then

Since on , it follows from or on that

Let . Then, for and , (13) and (22) imply which implies

Thus, by (ii) of Lemma 13, it follows from (41) and (46) that has a fixed point in with . Remark 12 shows that problem (3) has at least one positive solution with . This gives the proof of Theorem 14.

4.2. For the Case and

In this subsection, we discuss the existence for the positive solutions of problem (3) under and . For convenience, we introduce the following notation: Now, we will state and prove the following main result.

Theorem 15. Suppose ()–() hold. In addition, let the following two conditions hold: ()there exists such that ;()there exist and such that for ; furthermore, .

Then problem (3) has at least one positive solution.

Proof. Without loss of generality, we may assume that . Considering , we have for , .

Since on , it follows from on that

Consequently, for any and , (13) and (22) imply which implies

On the other hand, from , when a is fixed, then there exists a such that for and . Since on , it follows from on that

Hence, for , it follows from Remark 8 and (22) that which implies

Thus, by (i) of Lemma 13, it follows that has a fixed point in with Thus, it follows from Remark 12 that problem (3) has at least one positive solution with . This finishes the proof of Theorem 15.

We remark that condition in Theorem 15 can be replaced by the following condition:()′,

which is a special case of .

Corollary 16. Suppose ()–(), , and hold. Then problem (3) has at least one positive solution.

Proof. We show that implies . Suppose that holds. Then, there exists a positive number such that Hence, we obtain

Therefore, holds. Hence, by Theorem 15, problem (3) has at least one positive solution.

Theorem 17. Suppose ()–() hold. In addition, let the following condition hold: ().

Then problem (3) has at least one positive solution.

Proof. The proof is similar to that of (50) and (35), respectively.

Corollary 18. Suppose ()–(), , and hold. Then problem (3) has at least one positive solution.

4.3. For the Case and or and

In this subsection, we discuss the existence for the positive solutions of problem (3) for the case and or and . For convenience, we introduce the following notation:

Theorem 19. Suppose ()–() hold, , and . Then problem (3) has at least one positive solution.

Proof. The proof is similar to that of Theorem 15.

Theorem 20. Suppose ()–() hold, , and . Then problem (3) has at least one positive solution.

Proof. Considering , then there exists such that for , .

Since on , it follows from on that

Consequently, for , it follows from Remark 8 and (22) that which shows

Next, we turn to . In fact, we can show that implies .

Let . Then, there exists such that for . Let

Then, we have This implies that . Hence, implies that .

Similar to the proof of (46), we have

Thus, by (ii) of Lemma 13, it follows that has a fixed point in with This finishes the proof of Theorem 20.

From Theorems 19 and 20, we have the following result.

Corollary 21. Assume that ()–() hold. Furthermore, suppose that and condition in Theorem 15 hold. Then problem (3) has at least one positive solution.

Theorem 22. Suppose ()–() hold, , and . Then problem (3) has at least one positive solution.

Proof. The proof is similar to that of Theorem 15.

Theorem 23. Suppose ()–() hold, , and . Then problem (3) has at least one positive solution.

Proof. The proof is similar to that of Theorem 15.

From Theorems 22 and 23, the following corollaries are easily obtained.

Corollary 24. Assume that ()–() hold. Furthermore, suppose that and condition in Theorem 15 hold. Then problem (3) has at least one positive solution.

Corollary 25. Assume that ()–() hold. Furthermore, suppose that and condition in Theorem 15 hold. Then problem (3) has at least one positive solution.

4.4. For the Case and or and

In this subsection, we study the existence of multiple positive solutions for the problem (3) for the case and or and .

Combining the proof of Theorems 14 and 15, the following theorem is easily proven.

Theorem 26. Suppose that ()–(), , and , and the condition of Theorem 15 hold. Then problem (3) has at least two positive solutions.

Corollary 27. Suppose that ()–(), , and , and the condition of Corollary 16 hold. Then problem (3) has at least two positive solutions.

Theorem 28. Suppose that ()–(), , and , and the condition of Theorem 15 hold. Then problem (3) has at least two positive solutions.

Corollary 29. Suppose that ()–(), , and , and the condition of Theorem 17 hold. Then problem (3) has at least two positive solutions.

5. Three Examples

To illustrate how our main results can be used in practice, we present three examples.

Example 1. Consider the following boundary value problem: where , on , and here is a positive integral number.

This means that problem (66) involves the advanced argument . For example, we can take . It is clear that may be singular at and/or and is both nonnegative and continuous.

Problem (66) can be regarded as a problem of the form (3), where , , and .

Let and satisfy where and

It follows from the definition of , , and that ()–() hold, and

Hence, by Theorem 14, problem (66) has at least one positive solution.

Example 2. If we replace in Example 1 by in this case, by Theorem 20, we obtain that if and , then problem (66) admits at least one positive solution.

In fact, we can prove that which shows that and .

Example 3. If we replace in Example 1 by in this case, by Theorem 26, we obtain that problem (66) admits at least two positive solutions.

In fact, we can prove that which shows that and .

Conflict of Interests

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

Acknowledgments

The authors are indebted to the referee’s suggestions. These have greatly improved this paper. This work is sponsored by the Project NSFC (11301178, 11171032), the Fundamental Research Funds for the Central Universities (2014MS58), and the Improving Project of Graduate Education of Beijing Information Science and Technology University (YJT201416).

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