#### Abstract

In this article we present some results concerning the existence of solutions for a system of Hadamard integral equations. Our investigation is conducted with an application of an extension of the fixed point theorem of Burton-Kirk in Fréchet spaces.

#### 1. Introduction

In recent years, fractional differential and integral equations are emerging as a useful tool in modeling of the dynamics of many physical systems and electrical phenomena, which has been demonstrated by many researchers in the fields of mathematics, science, and engineering; see monographs [1–7] and a series of papers [8–18] and the references cited therein. Some properties of the Hadamard fractional calculus were investigated by Butzer et al. [19, 20]. Additional papers can be found in [7, 21] and the references therein. Recently, some interesting results on the existence of the solutions of some classes of integral equations have been obtained by Abbas et al. [22–24], Banaś et al. [25, 26], and Pachpatte [27, 28] and the references therein.

This paper deals with the existence of solutions of the following system of Hadamard fractional integral equations of the form That is,where , , , , is the Hadamard integral of order , are given continuous functions, , and is the (Euler’s) gamma function defined by

#### 2. Preliminaries

Let ; for , be the space of Lebesgue-integrable functions with the standard norm

*Definition 1 (see [4]). *The Hadamard fractional integral of order for a function is defined as

*Example 2. *The Hadamard fractional integral of order for the function , defined by with , is

*Definition 3. *Let , , , and For , define the Hadamard partial fractional integral of order by the expression

Let be a Fréchet space with a family of seminorms such that A subset of is bounded if, for every , there exists such that We associate a sequence of Banach spaces to as follows: For every , we consider the equivalence relation defined by if and only if for We denote by the quotient space, the completion of with respect to . To every , we associate a sequence of subsets as follows: For every , we denote by the equivalence class of of subset and we defined We denote by , , and , respectively, the closure, the interior, and the boundary of with respect to in For more details, we refer the reader to [29].

*Definition 4. *Let be a Fréchet space. A function is said to be a contraction if for each there exists such that

We need the following extension of the Burton-Kirk fixed point theorem in the case of a Fréchet space.

Theorem 5 (see [30]). *Let be a Fréchet space and let be two operators such that *(a)* is a compact operator*(b)* is a contraction operator with respect to a family of seminorms *(c)*the set is bounded** Then the operator equation has a solution in *

#### 3. Existence Results

Set

*Definition 6. *A function is said to be a solution of (2) if satisfies (2) on

For each we consider the set , and we define in the seminorms by Then is a Fréchet space with the family of seminorms

Also, the product space is a Fréchet space with the family of seminorms

Now, we are concerned with the existence of solutions for system (2). Let us introduce the following hypotheses:There exist continuous functions , with such that for each and each There exist continuous functions ; , , and nondecreasing functions such that for , ; , and Moreover, assume that for each with

For any , set and

Theorem 7. *Assume that hypotheses and hold. Ifthen system (2) has at least one solution in the space *

*Proof. *Define the operators ; , by and consider the continuous operators defined byWe shall show that operators and satisfied all the conditions of Theorem 5.

The proof will be given in several steps.*Step 1* ( is compact). To this aim, we must prove that is continuous and it transforms every bounded set into a relatively compact set. Recall that is bounded if and only if and is relatively compact if and only if, , the family is equicontinuous and uniformly bounded on The proof will be given in several claims.*Claim 1* ( is continuous). Let be a sequence in such that in Then, for each and , we haveSince and is continuous, then for each ; , and , (27) gives *Claim 2* ( maps bounded sets into bounded sets in ). Let be the bounded set in as in Claim 1. Then, for each , there exists , such that for all we have

For arbitrarily fixed and each , we have whereHence*Claim 3* ( maps bounded sets into equicontinuous sets in* C*). Let , , , and let ; thus for each , we have From the continuity of functions and as and , the right-hand side of the above inequality tends to zero. As a consequence of Claims 1–3 and from the Arzelá-Ascoli theorem, we can conclude that is continuous and compact.*Step 2* ( is a contraction). Consider Then, by , for any and each and , we have Thus, Hence, Inequality (21) implies that is a contraction operator.*Step 3* (the set is bounded). Let , such that , for some Then, for any and each and , we have where Thus, Hence, the set is bounded. As a consequence of Steps 1–3 and from Theorem 5, we deduce that has a fixed point which is a solution of system (2).

#### 4. An Example

Consider the following coupled system of Hadamard fractional integral equations of the formwhere ; , , and and Set The functions ; are continuous and satisfy assumption , with Also, the functions are continuous and satisfy assumption , with Then, Finally, we shall show that condition (21) holds with Indeed, for each , we get Hence by Theorem 7, system (39) has a solution defined on .

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

#### Acknowledgments

The work was supported by the National Natural Science Foundation of China (No. 11671339).