Abstract and Applied Analysis

Volume 2014 (2014), Article ID 120781, 7 pages

http://dx.doi.org/10.1155/2014/120781

Research Article

Existence and Estimates of Positive Solutions for Some Singular Fractional Boundary Value Problems

Habib Mâagli, Noureddine Mhadhebi, and Noureddine Zeddini

Department of Mathematics, College of Sciences and Arts, King Abdulaziz University, Rabigh Campus, P.O. Box 344, Rabigh 21911, Saudi Arabia

Received 25 December 2013; Accepted 14 February 2014; Published 1 April 2014

Academic Editor: Samir Saker

Copyright © 2014 Habib Mâagli 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.

We establish the existence and uniqueness of a positive solution for the following fractional boundary value problem, with the conditions , , where , , and is a nonnegative continuous function on that may be singular at or . We also give the global behavior of such a solution.

1. Introduction

Recently, the theory of fractional differential equations has been developed very quickly and the investigation for the existence of solutions of these differential equations has attracted considerable attention of researchers in the last few years (see [1–11] and the references therein).

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 [12–14]). In consequence, the subject of fractional differential equations is gaining much importance. Motivated by the surge in the development of this subject, we consider the following singular Dirichlet problem:

where , , and is a nonnegative continuous function on that may be singular at or . 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 integer part of the number and is the Euler Gamma function.

Moreover, we have the following well-known properties (see [3, 13, 15]):

(i) for , , ;(ii) for a.e. , where , ;(iii)if and , then , where and is the smallest integer greater than or equal to .

Several results are obtained for fractional differential equations with different boundary conditions, but none of them deal with the existence of a positive solution to 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 .

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 by for and by the set of all nonnegative measurable functions on .

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

such that for

where , , satisfying

In the sequel, we introduce the function defined on by

where

Our main result is the following.

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

Remark 2. Note that, for , we have

This implies in particular that, for and , the solution blows up at and 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 let the paper be self-contained, we begin this section by recapitulating some properties of Karamata regular variation theory. The following is due to [16, 17].

Lemma 3. The following hold.

(i)Letting and , then one has

(ii)Let and . Then one has , , and .

Example 4. Let be a positive integer. Let , , and 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 5. Let and be a function in defined on . One has the following:

(i)if , then diverges and ;(ii)if , then converges and .

Lemma 6. 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 (13).

Now put

Using that , we obtain

This implies that

So (14) holds.

Case 2. We suppose that diverges. We have, for some ,

This implies that

This proves (13) and completes the proof.

Remark 7. Let be defined on ; then using (4) and (13), we deduce that

If further converges, we have by (13) that

Lemma 8. Given and , then the unique continuous solution of

is given by

where

is Green's function for the boundary value problem (23).

Proof. Since , then is a solution of the equation . Hence . 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 Green function . So, we need the following lemma.

Lemma 9. For and one has

Proposition 10. On , one has

(i); (ii) there exist two constants

Proof. (i) For we have

Since for , then by applying Lemma 9 with and , we obtain

(ii) Using the following inequalities for ,

we deduce the result from (i).

As a consequence of Proposition 10, we obtain the following.

Corollary 11. Let and put for . Then

Proposition 12. Given and such that the function is continuous and integrable on , then is the unique solution in of the following boundary value problem:

Proof. From Corollary 11, the function is defined on and by Proposition 10, we have

which implies that is bounded on . Now, using Fubini's theorem, we have

On the other hand, we have

Now, suppose that ; then we have

By considering the substitution , we obtain

Moreover if and , we have .

So, it follows that

This implies that

Moreover, using part of Proposition 10 and the dominated convergence theorem, we conclude that and .

Finally, we prove the uniqueness. Let be two solutions of (33) and put . Then and . Hence, it follows that . Using the fact that , we conclude that and so .

In the sequel, we assume that and and we put

where satisfy

So, we aim to give some estimates on the potential function .

We define the Karamata functions , by

Then, we have the following.

Proposition 13. For ,

Proof. Using Proposition 10, we have

For , we have . So, using Lemma 5 and hypothesis (42), we deduce that

Now, we have

which implies by Lemma 5 that

Hence, it follows by Lemma 6 and hypothesis (42) that, for , we get

That is

Now, for , we use again Lemma 5 and hypothesis (42) to deduce that

Hence, it follows from Lemma 3 that, for , we get

That is

This together with (51) implies that, for , we have

3. Proof of Theorem 1

In order to prove Theorem 1, we need the following Lemma.

Lemma 14. Assume that the function satisfies and put for . Then one has, for ,

Proof. Put and . Then for , we have

Let , , , and . Then, using Proposition 13, we obtain by a simple computation that

Proof of Theorem 1. From Lemma 14, 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 by Proposition 10, Corollary 11, and Lemma 14, 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 Proposition 12, the function is a positive continuous solution of problem (1).

Finally, let us prove that is the unique positive continuous solution satisfying (9). To this aim, we assume that (1) has two positive solutions satisfying (9) and consider the nonempty set and put . Then and we have . It follows that and consequently

which implies by Proposition 12 that . By symmetry, we also obtain that . Hence, and . Since , then and consequently .

Example 15. Let and be a positive continuous function on such that

where and . Then, using Theorem 1, problem (1) has a unique positive continuous solution satisfying the following estimates:

where

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This paper was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The authors, therefore, acknowledge with thanks DSR technical and financial support.

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