#### Abstract

In 2010, Agarwal et al. studied the existence of a one-dimensional fractional neutral functional differential equation. In this paper, we study an initial value problem for a class of -dimensional systems of fractional neutral functional differential equations by using Krasnoselskii’s fixed point theorem. In fact, our main result generalizes their main result in a sense.

#### 1. Introduction

As you know, many researchers are interested in developing the theoretical analysis and numerical methods of fractional equations, because different applications of this area have been foundedin various fields of sciences and engineering (see, e.g., [137]). In this paper, we investigate the initial value problem of a -dimensional system of fractional neutral functional differential equations with bounded delay: where , , and are constants, , , for , is the standard Caputo’s fractional derivative, are given functions satisfying some assumptions that will be specified later,, and for . If , then for each define by for all . One-dimensional version of the problem has been studied by Agarwal et al. (see [4]). We show that the problem (1) is equivalent to an integral equation and by using Krasnoselskii's fixed point theorem, we conclude that the equivalent operator has (at least) a fixed point. This implies that the problem (1) has at least one solution. One can find the following lemma in [38].

Lemma 1 (Krasnoselskii’s fixed point theorem). Let be a Banach space and a closed convex subset of . Suppose that and are two maps of into such that for all . If is a contraction and is completely continuous, then the equation has a solution on .

Let be an interval in and with the norm , where denotes a suitable complete norm on . Consider the product Banach space with the norm . The fractional integral of order with the lower limit for a function is defined by for and , provided the right-hand side is pointwise defined on . Here, is the gamma function. Also, Caputo’s derivative of order with the lower limit for a function is defined by for and ([34]).

#### 2. Main Results

Consider the problem (1). Let and be positive constants, , and where . For obtaining our results, we need the following conditions:(H1) is measurable with respect to on for all ,(H2) is continuous with respect to on for all ,(H3)there exist and a real-valued function such that for all , , and ,(H4) for all ,(H5) is continuous and for all , and , where is a constant, for all ,(H6) is completely continuous and the family is equicontinuous on for all bounded set in and .

Lemma 2. Suppose that there exist and such that hold. Then the problem (1) for is equivalent to the equation with conditions for and .

Proof. It is easy to see that is Lebesgue measurable on by using conditions (H1) and (H2) for all . Also, a direct calculation shows that for . By using Holder's inequality and condition (H3), we get that is Lebesgue integrable with respect to for all , , and , and It is easy to see that if is a solution of the problem (1), then is a solution of . Now, suppose that is a solution of the equation and . Then and for all and . Thus, is a solution of the problem (1). This completes the proof.

Theorem 3. Suppose that there exist and such that hold. Then the problem (1) has at least one solution on for some positive number .

Proof. Since condition (H4) holds, the equation is equivalent to the equation and for all and . Let be defined by and for all and . If is a solution of problem (1) and for and , then for and . Thus, for and . Since , are continuous and is continuous in for all , there exists such that and for and . Put , where and for all . Define In fact, is a closed, bounded, and convex subset of . Define the operators and on by where for . It is easy to check that the operator equation has a solution if and only if is a solution for for all . In this case, will be a solution of the problem (1) on . Thus, the existence of a solution of the problem (1) is equivalent to the existence of a fixed point for the operator on . Hence, it is sufficient that we show that has a fixed point in . We prove it in three steps.
Step  I. for all .
Let be given. Then, for all . It is easy to check that for all . Also, we have for all and . Thus, for all . Hence, for all .
Step  II. is a contraction on .
Let . Then, and so for all . This implies that , where . Since , is a contraction on .
Step  III. is a completely continuous operator.
Suppose that for . It is clear that
Since is completely continuous for all , is continuous and also is uniformly bounded. By using condition (H6), it is easy to check that is equicontinuous. On the other hand, for all and . This implies that is uniformly bounded. Now, we prove that is equicontinuous. Let and be given. Then, we have for all . Thus, is equicontinuous. Moreover, it is clear that is continuous for all . This implies that is a completely continuous operator. Now, by using Krasnoselskii's fixed point theorem we get that has a fixed point on and so the problem (1) has a solution , where for all and .

If we put for all , then we obtain next result.

Corollary 4. Suppose that there exist and such that conditions hold, is continuous for all , and for all , , and , where is a constant for all . Then the problem (1) has at least one solution on for some positive number .

If we put for all , then we obtain next result.

Corollary 5. Suppose that there exist and such that conditions hold, is completely continuous for all , and the family is equicontinuous on for all bounded set in . Then the problem (1) has at least one solution on for some positive number .

#### 3. Conclusions

In this work, we study an initial value problem for a class of -dimensional systems of fractional neutral functional differential equations by using Krasnoselskii’s fixed point theorem. Our result generalizes some old related results in a sense.

#### Conflict of Interests

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

#### Acknowledgment

The research of the second and third authors was supported by Azarbaidjan Shahid Madani University.