Abstract and Applied Analysis

Volume 2009, Article ID 314656, 18 pages

http://dx.doi.org/10.1155/2009/314656

## The Existence of Positive Solution to Three-Point Singular Boundary Value Problem of Fractional Differential Equation

^{1}Department of Mathematics, Xiangnan University, Chenzhou 423000, China^{2}School of Mathematics and Computational Science, Xiangtan University, Xiangtan 411105, China

Received 13 May 2009; Accepted 23 June 2009

Academic Editor: Yong Zhou

Copyright © 2009 Yuansheng Tian and Anping Chen. 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

We investigate the existence of positive solution to nonlinear fractional differential equation three-point singular boundary value problem: , , , , where is a real number, , and is the standard Riemann-Liouville fractional derivative, and (i.e., is singular at ). By using the fixed-point index theory, the existence result of positive solutions is obtained.

#### 1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and the applications of such constructions in various sciences such as physics, mechanics, chemistry, and engineering. For details, see [1–10] and the reference therein. However, up to our knowledge, most of those papers have studied the existence and multiplicity of solution (or positive solution) to the initial value problem of nonlinear fractional differential equations; see [1, 4, 10, 11].

Recently, there are a few paper considering the Dirichlet-type boundary value problem for nonlinear ordinary differential equations of fractional order; see [12–14]. Delbosco [13] has investigated the nonlinear Dirichlet-type problem

He has proved that if is Lipschitizan function, then the problem has at least one solution in a certain subspace of in which fractional derivative has a Holder property. When is continuous on ), by the use of some fixed-point theorem on cones, Bai and Lü [12] and Zhang [14] have given the existence of positive solutions to the equation

with boundary condition respectively.

This paper is to study the existence of positive solution for the three-point singular boundary value problem of nonlinear fractional differential equation By using the fixed-point index theory, where is a real number, , satisfy that , is the standard Riemann-Liouville fractional derivative, and the function satisfies the following condition:

, , there exists a constant such that is continuous function on .The organization of this paper is as follows. In Section 2, we present some necessary definitions and Preliminary results that will be used to prove our main results. The proof of our main result is given in Section 3. In Section 4, we will give an example to ensure our main result.

#### 2. Preliminaries

The material in this section is basic in some sense. For the reader’s convenience, we present some necessary definitions from fractional calculus theory and preliminary results.

*Definition 2.1. *The fractional integral of order of a function is given by
provided that the right side is pointwise defined on .

*Definition 2.2. *The fractional derivative of order of a continuous function is given by
where , provided that the right side is pointwise defined on .

Lemma 2.3 (see [7]). * If , , then .*(2)*If , , then *

Lemma 2.4 (see [12]). *Assume that with a fractional derivative of order that belongs to . Then
**, , where is the smallest integer greater than or equal to .*

Lemma 2.5. *If and , , satisfy that , then the problems
**
have the unique solution
*

*Proof. *By applying Lemma 2.4, we may reduce (2.4) to an equivalent integral equation
for some . Consequently the general solution of (2.4) is
Note that , we have and
On the other hand, by (2.6) and Lemma 2.3, we have
Therefore
By , combine with (2.8) and (2.10), we obtain
So, the unique solution of problem (2.4) is
The proof is completed.

Lemma 2.6. *If and , , satisfy that , then the unique solution of the problem (2.4)
**
is nonnegative on , where
*

*Proof. *By Lemma 2.5, the unique solution of problem (2.4) is

Observing the expression of , it is clear that for . On the other hand, by , we have
Hence
It implies that
Therefore is nonnegative on .

Lemma 2.7. *Let , , satisfy that and is continuous, and . Suppose that there exists a constant such that is continuous function on . Then the unique solution of (2.4)
**
is continuous on .*

*Proof. *Since is continuous in , thus there exists a constant such that for all and
For any , we will prove (, ). For the convenience, the proof is divided into three cases.*Case 1 (). *It is easy to know that . For all , we have
where denotes beta function.*Case 2 (, for all ). *We have
*Case 3 (, for all ). *The proof is similar to the step 2. Here, we omit it.

The main tool of this paper is the following well-known fixed-point index theorem (see [15]).

Lemma 2.8. *Let be a Banach space, is a cone. For , define . Assume that is a completely continuous such that for .*(1)*If for , then *(2)*If for , then *

#### 3. Main Results

For the convenience we introduce the following notations:

*Remark 3.1. *If , , , , and , then

Let be a Banach spaces with the maximum norm . Define the cone by

The positive solution which we consider in this paper is the form , , , .

Define an operator by

Lemma 3.2. *Assume that condition holds. Then the operator is completely continuous.*

*Proof. *If condition holds, by Lemmas 2.6 and 2.7, we have . Let and ; if and , then . By the continuous of , we know that is uniformly continuous on . Thus, for all , there exists a () such that
for all and with . Obviously, if , then and for each . Hence, we have
for all with . It follows from (3.6) that
By the arbitrariness of , we have that is continuous.

Let be bounded; that is, there exists a positive constant such that . Since is continuous on , we let
For all , we have
Hence is bounded.

On the other hand, given , set
For each , we will prove that if and , then
In fact, similar to the proof of Lemma 2.7, we have
In the following, the proof is divided into three cases. *Case 1 (, ). **Case 2 (, ). **Case 3 (, ). *

Therefore, is equicontinuous. The Arzela-Ascoli Theorem implies that is compact. Thus, the operator is completely continuous.

We obtain the following existence results of the positive solution for problem (1.4).

Theorem 3.3. *If condition holds and assume further that there exist two positive constants such that**, for ;**, for ,**then problem (1.4) has at least one positive solution such that .*

*Proof. *Problem (1.4) has a solution if and only if is a solution of the operator equation . In order to apply Lemma 2.8, we separate the proof into the following two steps.*Step 1. *Let . For any , we have and for all . Observing the expression of (see (2.14), it is clear that . By assumption , we have
So
By Lemma 2.8, we have
*Step 2. *Let . For any , we have and for all . By assumption , for , we get
Therefore
By Lemma 2.8, we have
Combine with (3.18) and (3.21), we have
Therefore, has a fixed point . Then problem (1.4) has at least one positive solution such that .

#### 4. Example

Let , . We consider the following boundary value problem: where Let . By simple computation, we have Choosing , , we have