Research Article | Open Access

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

**Academic Editor:**Dumitru Baleanu

#### 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, [4–18]. 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 [21–23], 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 [28–32]. 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