Abstract and Applied Analysis

Volume 2013, Article ID 368659, 15 pages

http://dx.doi.org/10.1155/2013/368659

## The Existence of Positive Solutions for a New Coupled System of Multiterm Singular Fractional Integrodifferential Boundary Value Problems

^{1}Department of Chemical and Materials Engineering, Faculty of Engineering, King Abdulaziz University, P.O. Box 80204, Jeddah 21589, Saudi Arabia^{2}Department of Mathematics, Cankaya University, Ogretmenler Caddesi 14, Balgat, 06530 Ankara, Turkey^{3}Institute of Space Sciences, RO 76900, Magurele-Bucharest, Romania^{4}Department of Mathematics, Azarbaijan Shahid Madani University, Azarshahr, Tabriz 9177948974, Iran

Received 24 June 2013; Accepted 15 August 2013

Academic Editor: Juan J. Trujillo

Copyright © 2013 Dumitru Baleanu 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 discuss the existence of positive solutions for the coupled system of multiterm singular fractional integrodifferential boundary value problems , and for all , for and , for and , where , , , , are continuous functions and Here is the standard Riemann-Liouville fractional derivative, is a Caratheodory function, and is singular at the value 0 of its variables.

#### 1. Introduction

During the last decade, there were a lot of manuscripts on fractional differential equations (see, e.g., [1–19] and the references therein). Fractional equations have been discussed extensively as valuable tools in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in fluid mechanics, viscoelasticity, edge detection, porous media, and electromagnetism, as well as in various other areas. For more examples and details, see [3, 13–15] and references therein. On the other hand, there are many works about the existence of positive solutions of fractional differential equations (see, e.g., [1, 2, 6–8, 10, 17, 19] and the references therein).

#### 2. The Problem

In this paper, we investigate the following coupled system of multiterm singular fractional integrodifferential boundary value problem: where , , , , , (), and are positive-valued continuous functions on (), and . Also, and satisfy the local Caratheodory condition on , where is a subset of and may be singular at the value 0 of all its variables. Here, is the standard Riemann-Liouville fractional derivative. In fact, a function satisfies the local Caratheodory condition on whenever(i) is measurable for all ,(ii) is continuous for almost all ,(iii)for each compact subset , there is a function such that for almost all and all . The functions are called positive solutions of the problem (1), (2), and (3) whenever and on , , , , satisfy boundary conditions (2), (3), and (1) holds for almost all . In 2010, Agarwal et al. reviewed positive solutions of the singular Dirichlet problem , where , and satisfies the local Caratheodory condition on [1]. In 2011, Stanek reviewed the singular problem , , where , and satisfies the local Caratheodory condition on [19]. Here, is a subset of . In 2012, Bai and Sun reviewed the singular problem , where , , and satisfies the local Caratheodory condition on [8]. Again, is a subset of . In this year, Agarwal et al. reviewed the singular problem , , where , and is a -Caratheodory function on [2]. Here, , and is the Caputo fractional derivative. Many researchers have established the existence and uniqueness of solutions for some systems of nonlinear fractional differential equations, but there are few works about coupled system of multiterm nonlinear fractional differential equations (see, for example, [4, 5, 20, 21]). Also, there are a few papers discussing singular system of fractional differential equations (see, e.g., [6, 16]).

Let be the norm of , the norm of , and the norm of . As you know, and are Banach spaces, where . Suppose that and are the spaces of absolutely continuous functions and functions having absolutely continuous derivatives on , respectively. In this paper, we suppose that the functions and in (1) satisfy the following conditions:(*H*_{1}), where . Also, there exists a positive constant such that for almost all and ;(*H*_{2}) there exist positive mappings , , and such that and are nonincreasing and and are nondecreasing in all their variables,
for almost all and . Since we suppose that the problem (1) is singular, we use regularization and sequential techniques for the existence of positive solutions of the problem. In this way, for each natural number define the function () by
for almost all and , where
Then condition implies that and for almost all and , . Also, the condition implies
for almost all and and
for almost all and . For obtaining solutions of the system (1), (2), and (3), for each natural number , we will obtain solutions of the system
via the boundary conditions (2) and (3) and by using solutions of this system, we will obtain solution of the system (1), (2), and (3). It has been proved that the fractional integral maps into whenever and maps into whenever [19]. Here, means the integral part of and . Suppose that and is not a natural number. If and , then for all , where and for [7]. Suppose that , , , and . Then, , , , and [2].

#### 3. Main Results

Now, we are ready to state and prove our main results. One can find main idea of next result in [17].

Lemma 1. *Let and . For each , is the unique solution of the equation in which satisfies the boundary condition (2), where
*

*Proof. *It is easy to see that the functions are solutions of in for all . Since , and so are solutions of in , where . The boundary condition (2) implies that and
Thus, is the unique solution of the problem in .

Note that the Green function in Lemma 1 has some properties. For example, , and are continuous functions on , on , for all , for all , on , for all , for all , on , for all , and for all . Let . Define the cone on by It is easy to see that for each , and we have and , , , . Now for each natural number , define the operator on by

Lemma 2. *For each natural number , is a completely continuous operator on .*

*Proof. *Let be a natural number, ,
and =, , , , , , , , . Then, and there exist positive constants and such that and for almost all . Since , , , , , and are nonnegative and continuous functions on , , and for all , we get , , , , , , , and on . Thus, maps into . Suppose that is a convergent sequence in and . Then, and uniformly on for . Since for , for all ,
for and , we have , , , for , and , uniformly on for . Also, we have
and also and for and . Now, put
and,,, , for all . Then, for almost all and . Since and , are bounded sequences in and also by using the above inequalities, , , , for and , for are bounded sequences in , there exist such that for almost all and . Thus, by using the Lebesgue dominated convergence theorem and the following relations:
we get and uniformly on for . This implies that and are continuous operators and so is a continuous operator. Let be a bounded sequence in . We show that the sequence is relatively compact in . By using the Arzela-Ascoli theorem, it is sufficient to prove that is bounded in and is equicontinuous on . Choose a positive number such that , , , , , for all . By using the above inequalities, one can see that , , , and for and and for . Since
. Similarly, we can verify that . Thus,
for all . Hence, is bounded in . Also, for we have
Similarly, we have
Let be given. Since the functions and are uniformly continuous on and the functions and are uniformly continuous on , there exists such that , , , and for all , and . If and , then
and . Therefore, is equicontinuous on and so is a completely continuous operator on .

We need the following result (see [10]).

Lemma 3. *Let be a Banach space, a cone in , and and bounded open balls in centered at the origin with . Suppose that is a completely continuous operator such that for all and for all . Then has a fixed point in .*

Theorem 4. *For each natural number , the system (15), (2), and (3) has a solution such that and for all .*

*Proof. *Let and be given. Then, and for all . Hence, and and so and . Thus, . Now, put
Then, for all . Let for . Then, we have
for all and . Hence,
Similarly, we have
for all . Since for , there exists such that
Put . Then, the above inequalities imply that for all . Now by using Lemma 3, we conclude that the operator has a fixed point in