Research Article | Open Access
Existence Results for a System of Coupled Hybrid Fractional Differential Equations
This paper studies the existence of solutions for a system of coupled hybrid fractional differential equations with Dirichlet boundary conditions. We make use of the standard tools of the fixed point theory to establish the main results. The existence and uniqueness result is elaborated with the aid of an example.
Fractional calculus is the study of theory and applications of integrals and derivatives of an arbitrary (noninteger) order. This branch of mathematical analysis, extensively investigated in the recent years, has emerged as an effective and powerful tool for the mathematical modeling of several engineering and scientific phenomena. One of the key factors for the popularity of the subject is the nonlocal nature of fractional-order operators. Due to this reason, fractional-order operators are used for describing the hereditary properties of many materials and processes. It clearly reflects from the related literature that the focus of investigation has shifted from classical integer-order models to fractional-order models. For applications in applied and biomedical sciences and engineering, we refer the reader to the books [1–4]. For some recent work on the topic, see [5–25] and the references therein. The study of coupled systems of fractional-order differential equations is quite important as such systems appear in a variety of problems of applied nature, especially in biosciences. For details and examples, the reader is referred to the papers [26–33] and the references cited therein.
Hybrid fractional differential equations have also been studied by several researchers. This class of equations involves the fractional derivative of an unknown function hybrid with the nonlinearity depending on it. Some recent results on hybrid differential equations can be found in a series of papers (see [34–37]).
Motivated by some recent studies on hybrid fractional differential equations, we consider the following Dirichlet boundary value problem of coupled hybrid fractional differential equations: where denote the Caputo fractional derivative of orders , respectively, and , .
The aim of this paper is to obtain some existence results for the given problem. Our first theorem describes the uniqueness of solutions for the problem (1) by means of Banach’s fixed point theorem. In the second theorem, we apply Leray-Schauder’s alternative criterion to show the existence of solutions for the given problem. The paper is organized as follows. Section 2 contains some basic concepts and an auxiliary lemma, an important result for establishing our main results. In Section 3, we present the main results.
Definition 1. The Riemann-Liouville fractional integral of order for a continuous function is defined as provided that the integral exists.
Definition 2. For at least -times continuously differentiable function , the Caputo derivative of fractional-order is defined as where denotes the integer part of the real number .
Lemma 3 (auxiliary lemma). Given , the integral solution of the problem is
Proof. It is well known that the general solution of the fractional differential equation in (4) can be written as where are arbitrary constants. Alternatively, we have Using the given boundary conditions in (7), we find that Substituting the values of in (7) yields the solution This completes the proof.
3. Main Results
Let denote a Banach space equipped with the norm , where . Notice that the product space with the norm , is also a Banach space.
In view of Lemma 3, we define an operator by where In the sequel, we need the following assumptions.)The functions are continuous and bounded; that is, there exist positive numbers such that , .()There exist real constants and such that and , , .For brevity, let us set
Now we are in a position to present our first result that deals with the existence and uniqueness of solutions for the problem (1). This result is based on Banach’s contraction mapping principle.
Theorem 4. Suppose that condition holds and that are continuous functions. In addition, there exist positive constants , such that If and are given by (12), then the problem (1) has a unique solution.
Proof. Let us set and define a closed ball: , where
Then we show that . For , we obtain
Working in a similar manner, one can find that
From (17) and (18), it follows that .
Next, for and for any , we have which yields Similarly, one can get From (20) and (21), we deduce that In view of condition , it follows that is a contraction. So Banach’s fixed point theorem applies and hence the operator has a unique fixed point. This, in turn, implies that the problem (1) has a unique solution on . This completes the proof.
Example 5. Consider the following coupled system of hybrid fractional differential equations: Here , , , , and . Note that Thus all the conditions of Theorem 4 are satisfied and, consequently, there exists a unique solution for the problem (23) on .
In our second result, we discuss the existence of solutions for the problem (1) by means of Leray-Schauder alternative.
Lemma 6 (Leray-Schauder alternative [38, page 4]). Let be a completely continuous operator (i.e., a map that is restricted to any bounded set in is compact). Let . Then either the set is unbounded or has at least one fixed point.
Proof. We will show that the operator satisfies all the assumptions of Lemma 6. In the first step, we prove that the operator is completely continuous. Clearly, it follows by the continuity of functions , and that the operator is continuous.
Let be bounded. Then we can find positive constants and such that Thus for any , we can get which yields In a similar manner, one can show that From the inequalities (27) and (28), we deduce that the operator is uniformly bounded.
Now we show that the operator is equicontinuous. For that, we take with and obtain which tend to independently of This implies that the operator is equicontinuous. Thus, by the above findings, the operator is completely continuous.
In the next step, it will be established that the set is bounded. Let ; then we have . Thus, for any , we can write Then which imply that In consequence, we have which, in view of (13), can be expressed as This shows that the set is bounded. Hence all the conditions of Lemma 6 are satisfied and consequently the operator has at least one fixed point, which corresponds to a solution of the problem (1). This completes the proof.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
This paper was supported by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia. The authors, therefore, acknowledge technical and financial support of KAU.
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