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`Abstract and Applied AnalysisVolume 2013 (2013), Article ID 129640, 7 pageshttp://dx.doi.org/10.1155/2013/129640`
Research Article

## Positive Solutions for Boundary Value Problems of Singular Fractional Differential Equations

1College of Information Science and Engineering, Shandong University of Science and Technology, Qingdao 266590, China
2School of Information and Communication Engineering, Beijing University of Posts and Telecommunications, Beijing 100876, China

Received 25 March 2013; Accepted 20 May 2013

Copyright © 2013 Zhanbing Bai et al. 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

In this paper, by using a fixed point theorem, we investigate the existence of a positive solution to the singular fractional boundary value problem , , where , , , is Caputo fractional derivative, is singular at the value 0 of its arguments , and satisfies the Lipschitz condition.

#### 1. Introduction

In recent years, as an extended concept of integral differential equations, fractional differential equations are widely concerned in various fields of science. For examples, see [114]. Many results, such as [1, 2, 6, 15, 16], discuss singular fractional boundary value problems.

In [1], the authors discuss positive solutions to the singular Dirichlet problem where , and is a Carathéodory function on . Here, is the standard Riemann-Liouville fractional derivative. The existence of positive solutions is obtained by the combination of regularization and sequential techniques with the Guo-Krasnosel’skii fixed point theorem on cone.

The singular problem was discussed in [16], where and () are positive. The existence results of positive solutions are acquired by the use of regularization and sequential techniques with a fixed point theorem for mixed monotone operators on normal cones.

Paper [6] investigates positive solutions of singular fractional boundary value problem where , , , is the standard Riemann-Liouville fractional derivative, and is a Carathédory function. The existence and multiplicity of positive solutions are obtained by means of Guo-Krasnosel'skii fixed point theorem on cones.

In this paper, we are concerned with the following singular fractional boundary value problem: where , , and are real numbers. is the Caputo fractional derivative of order . satisfies the Carathéodory condition on , , may be singular at the value 0 of all its space variables , , , and satisfies the Lipschitz condition.

A function is called a positive solution of problems (4), (5) if on , , and satisfies boundary condition (5) and equality (4) for a.e. .

Throughout the paper, denote which is the norm of , and is the norm of space , while is the norm of . and are sets of absolutely continuous functions and functions having absolutely continuous th derivatives on , respectively.

The following conditions on and in (4) will be used.() is a Carathéodory function on , where , and there exists a positive constant such that, for a.e. and all , () satisfies the following inequality, for a.e. and all , : with ()For a.e. and all , where , , and are positive, and are nonincreasing and nondecreasing in all their arguments, respectively,

We will use regularization and sequential techniques to prove the existence of a positive solution of problems (4), (5). Define and by the following formulas: for a.e. and all ,

Then, condition () gives that is a Carathéodory function on ,

Condition () gives

In Section 3, We will firstly investigate the regular fractional differential equation

#### 2. Preliminaries

Definition 1. The Caputo fractional derivative of order of a function is defined by provided that the right-hand side is pointwise defined on , where and means the integer part of the number . is the Euler function.

Definition 2. The fractional integral of order of a function is defined by provided the right-hand side is pointwise defined on .

Lemma 3 (see [10]). One has where means the integral part of and .

Lemma 4 (see [10]). Suppose that , . If and , then where and , .

Lemma 5. Given , then for , is the unique solution in of the equation satisfying the boundary condition (5), where and

Proof. By Lemma 4, are all solutions of (22) in , where . Lemma 3 guarantees that , for ; therefore, are all solutions of (22) in , where . Considering that the solutions should satisfy , we get that . Consequently, is the unique solution of problems (22), (5).

Lemma 6. Let be as defined in (2.3). Then, (1) and on , (2) for , (3) for , (4) and on , (5) for , (6) for , (7) and on , (8) for , (9) for .

Proof. (1), (4), and (7) are as follows
Because , therefore and .
It is obvious that (2), (3), (5), (6), (8), and (9) hold.

#### 3. Auxiliary Regular Problems (16), (5)

Let , and let

For , we can obtain that

We define the operators and on as

Lemma 7. is a completely continuous operator.

The proof is similar to Lemma 3.1 of [6], so we omit it.

Lemma 8 (see [17]). Let be a closed convex and nonempty subset of a Banach space . Let A, B be the operators such that (i) wherever , (ii) A is compact and continuous, and (iii) B is a contraction mapping. Then, there exists such that .

Theorem 9. Let () and () hold. Then, problems (16), (5) have a solution such that

Proof. By Lemma 7, is a completely continuous operator. Now, for , we obtain that
Therefore, is a contraction mapping, and it is obvious that , for . Thus, all the assumptions of Lemma 8 are satisfied, and the conclusion of Lemma 8 implies that the boundary value problems (16), (5) have at least one solution.

Lemma 10. Suppose that (), (), and () hold and be a solution of problems (16), (5). Then, the sequence is relatively compact in X.

Proof. Note that, for and , And fulfills (31).
Lemma 6 and (13) imply that
So,
Since then,
Let
It follows from (31) and (37) that, for , ,
Therefore,
By Lemma 6, (15), and (39), there hold that where , , and .
It follows from () and the assumption that . Hence, where . Since , there exists such that, for ,
Consequently, for , so that is bounded in . We are now in a position to prove that is equicontinuous on . Let
Then, and, for a.e. , all , holds. Suppose that , then
The proof is similar to that of Lemma 7. We choose . Then, there exists such that , for any , , and . Suppose that . Then, for , , , we have
Thus, is equicontinuous on .

#### 4. Main Result

Theorem 11. Suppose that (), (), and () hold. Then, problems (4), (5) has a positive solution and, for ,

Proof. Theorem 9 shows that problems (16), (5) have a solution . In addition, Lemma 10 gives that is relatively compact in and satisfies inequality (39) for . Assume that itself is convergent in and . Then, satisfies the boundary condition (5), and and in . Consequently, satisfies (49). Furthermore,
Let . Then, it follows from and
Hence, for a.e. and all , we have where is defined by (45). Putting , by the Lebesgue dominated convergence theorem, we have, for ,
Consequently, is a positive solution of problems (4), (5) and satisfies inequality (49). The proof is complete.

#### Acknowledgments

This work is supported by SDNSF (ZR2010AM035), NNSF of China (61174078), a project of Shandong province Higher Educational Science and Technology Program (J11LA07), the Research Fund for the Taishan Scholar Project of Shandong Province of China, Research Award Fund for Outstanding Young Scientists of Shandong Province (BS2010SF023, BS2012SF022) and SDUST Research Fund (2011KYTD105).

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