#### Abstract

We establish the existence and uniqueness of a positive solution for the fractional boundary value problem , with the condition , where , and is a nonnegative continuous function on that may be singular at or .

#### 1. Introduction

Fractional differential equations arise in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, and electromagnetism. They also serve as an excellent tool for the description of hereditary properties of various materials and processes (see). In consequence, the subject of fractional differential equations is gaining much importance and attention. Motivated by the surge in the development of this subject, we consider the following problem: where , , is a nonnegative continuous function on that may be singular at or and is the Riemann-Liouville fractional derivative. Then we study the existence and exact asymptotic behavior of positive solutions for this problem.

We recall that for a measurable function , the Riemann-Liouville fractional integral and the Riemann-Liouville derivative of order are, respectively, defined by provided that the right-hand sides are pointwise defined on . Here and means the integral part of the number and is the Euler Gamma function.

Moreover, we have the following well-known properties (see [2, 4]):(i) for , , ,(ii) for a.e. , where , ,(iii) if and only if , where and .

Several results are obtained for fractional differential equation with different boundary conditions (see  and the references therein), but none of them deals with the existence of a positive solution for problem (1).

Our aim in this paper is to establish the existence and uniqueness of a positive solution for problem (1) with a precise asymptotic behavior, where is the set of all functions such that is continuous on . Note that for , the solution for problem (1) blows up at .

To state our result, we need some notations. We will use to denote the set of Karamata functions defined on by for some , where and such that . It is clear that a function is in if and only if is a positive function in such that For two nonnegative functions and defined on a set , the notation , , means that there exists such that , for all . We denote also for .

Throughout this paper we assume that is nonnegative on and satisfies the following condition.

such that where , , satisfying In the sequel, we introduce the function defined on by Our main result is the following.

Theorem 1. Let and assume that satisfies . Then problem (1) has a unique positive solution satisfying for

This paper is organized as follows. Some preliminary lemmas are stated and proved in the next section, involving some already known results on Karamata functions. In Section 3, we give the proof of Theorem 1.

#### 2. Technical Lemmas

To keep the paper self-contained, we begin this section by recapitulating some properties of Karamata regular variation theory. The following is due to [16, 17].

Lemma 2. The following hold.(i)Let and , then one has (ii)Let , and . Then one has , , and .

Example 3. Let . Let , and let be a sufficiently large positive real number such that the function is defined and positive on , for some , where ( times. Then .

Applying Karamata’s theorem (see [16, 17]), we get the following.

Lemma 4. Let and let be a function in defined on . One has the following.(i)If , then diverges and . (ii)If , then converges and .

Lemma 5. Let be defined on . Then one has If further converges, then one has

Proof. We distinguish two cases.
Case  1. We suppose that converges. Since the function is nonincreasing in , for some , it follows that, for each , we have It follows that . So we deduce (11).
Now put Using that , we obtain This implies that So (12) holds.
Case  2. We suppose that diverges. We have, for some , This implies that This proves (11) and completes the proof.

Remark 6. Let defined on ; then, using (4) and (11), we deduce that If further converges, we have by (11) that

Lemma 7. Given and is such that the function is continuous and integrable on , then the boundary value problem has a unique solution given by where is the Green function for the boundary value problem (21).

Proof. Since is a solution of the equation , then . Consequently there exist two constants , such that . Using the fact that and , we obtain and . So In the following, we give some estimates on the function . So, we need the following lemma.

Lemma 8. For , and , one has

Proposition 9. On , one has

Proof. For , we have Since for , then, by applying Lemma 8 with and , we obtain which completes the proof.

In the sequel we put where and we aim to give some estimates on .

Proposition 10. Assume that with Then for ,

Proof. Using Proposition 9, we have For , we use Lemma 4 and hypotheses (30) to deduce that Hence, it follows from Lemma 2 and hypothesis (30) that, for , we have Now, for , we use again Lemma 4 and hypothesis (30) to deduce that Hence, it follows from Lemmas 2 and 5 that, for , we have
This together with (34) implies that (36) holds on .

#### 3. Proof of Theorem 1

We begin this section by giving a preliminary result that will play a crucial role in the proof of Theorem 1.

Proposition 11. Assume that the function satisfies and put for . Then one has, for ,

Proof. For , we have So, we can see that where , and, according to Lemma 2, the functions and are in . Moreover, using Lemma 4, we have and . So the result follows from Proposition 10.

Proof of Theorem 1. From Proposition 11, there exists such that for each where .
Put and let In order to use a fixed point theorem, we denote and we define the operator on by For this choice of , we can easily prove that for , we have and .
Now, we have Since the function is continuous on and the function is integrable on , we deduce that the operator is compact from to itself. It follows by the Schauder fixed point theorem that there exists such that . Put . Then and satisfies the equation Since the function is continuous and integrable on , then by Lemma 7 the function is a positive continuous solution of problem (1).
Finally, let us prove that is the unique positive continuous solution satisfying (8). To this aim, we assume that (1) has two positive solutions satisfying (8) and consider the nonempty set and put . Then and we have . It follows that and consequently Which implies by Lemma 7 that . By symmetry, we obtain also that . Hence and . Since , then and consequently .

Example 12. Let and be a positive continuous function on such that where and or and . Then, using Theorem 1, problem (1) has a unique positive continuous solution satisfying the following estimates.(i)If or and , then for , (ii)If , then for , (iii)If , then for , (iv)If , then for , (v)If , then for ,

#### Acknowledgment

The authors thank the anonymous referees for a careful reading of the paper and for their helpful suggestions.