Abstract and Applied Analysis

Volume 2014, Article ID 896871, 13 pages

http://dx.doi.org/10.1155/2014/896871

## The Existence of Solution for a -Dimensional System of Multiterm Fractional Integrodifferential Equations with Antiperiodic 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, Magurele, Bucharest, Romania^{4}Department of Mathematics, Azarbaijan Shahid Madani University, Azarshahr, Tabriz, Iran

Received 8 January 2014; Revised 18 March 2014; Accepted 20 March 2014; Published 24 April 2014

Academic Editor: Hossein Jafari

Copyright © 2014 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

There are many published papers about fractional integrodifferential equations and system of fractional differential equations. The goal of this paper is to show that we can investigate more complicated ones by using an appropriate basic theory. In this way, we prove the existence and uniqueness of solution for a -dimensional system of multiterm fractional integrodifferential equations with antiperiodic boundary conditions by applying some standard fixed point results. An illustrative example is also presented.

#### 1. Introduction

Fractional differential equations have recently been studied by many researchers for a variety of problems (see, e.g., [1–33] and the references therein). Antiperiodic boundary value problems occur in the mathematical modeling of a variety of physical processes (see, e.g., [3–6, 29, 30] and the references therein). On the other hand, the study of a coupled system of fractional order is also very significant because this kind of system can often occur in applications (see, e.g., [7, 15, 24, 28, 29] and the references therein). We are going to investigate a complicated case in this work. Let and . In this paper, we study the existence and uniqueness of solution for the -dimensional system of multiterm fractional integrodifferential equations with antiperiodic boundary conditions , , and for , where denotes the Caputo fractional derivative, , , for , , and , are continuous functions for all . Hereafter, we will use vector notations. Define the space endowed with the norm . In fact, and the product space endowed with the norm are Banach spaces. The Riemann-Liouville fractional integral of order is defined by ( and ), provided the integral exists. The Caputo derivative of order for a function is defined by for and [25]. Recently, Wang et al. proved the following result [30].

Lemma 1. *For each , the unique solution of the boundary value problem
**
is given by , where is Green's function defined as
*

*One can find the next result in [34].*

*Theorem 2. Let be a Banach space and a completely continuous operator. Suppose that the set is bounded. Then has a fixed point in .*

*We will use the last two results for solving the problem (1).*

*2. Main Results*

*Now, we are ready to state and prove our main results. For each , put
and , where for all . Define the operator by
where and
for , where
Thus, for each , we have
*

*Theorem 3. The operator is completely continuous.*

*Proof. *First, we show that the operator is continuous. Let and for and let be a sequence in such that . Then, we have
for . Since for , the sequences , , and converge uniformly on and also , , and converge uniformly on for . Since
by using the above inequalities and the continuity of (), we get
Thus, is continuous in . Let be a bounded subset of . Choose positive constants such that for all and . Thus, for each we have
for all . Hence,
for all and so . This implies that the operator is uniformly bounded. Now, we show that is an equicontinuous set. Let . Then, we have
for all . As , the right-hand side of the above inequalities tends to zero. Thus, by using the Arzela-Ascoli theorem one can conclude that the operator is completely continuous. This completes the proof.

*Theorem 4. Assume that there exist positive constants , , , , and () such that
and for all , , and . Then problem (1) has at least one solution.*

*Proof. *First, we show that for some is bounded. Let . Then, for each we have
Hence,
for all . Thus, we get
Hence, for . This implies that
and so . Therefore, the set is bounded. Now by using Theorem 2, the operator has at least one fixed point. This implies that the problem (1) has at least one solution.

*Theorem 5. Suppose that there exist nonnegative constants , , , and for such that
for all , , and . Then the problem (1) has a unique solution.*

*Proof. *Let for , , and
We show that . Let . Then
for . Hence,