Research Article | Open Access
The Existence of Positive Solution to Three-Point Singular Boundary Value Problem of Fractional Differential Equation
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.
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  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  and Zhang  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.
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 ). If , , then .(2)If , , then
Lemma 2.4 (see ). 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
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 ).
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 (,