Analysis of Multiterm Initial Value Problems with Caputo–Fabrizio Derivative

Abstract

In this paper, we discuss the solvability of a class of multiterm initial value problems involving the Caputo–Fabrizio fractional derivative via the Laplace transform. We derive necessary and sufficient conditions to guarantee the existence of solutions to the problem. We also obtain the solutions in closed forms. We present two examples to illustrate the validity of the obtained results.

1. Introduction

Recently, there is great interest to develop new types of fractional derivatives of nonsingular kernel. Motivated by applications, Caputo and Fabrizio were the first to introduce such types of fractional derivatives with nonlocal and nonsingular kernel [1]. The Caputo–Fabrizio derivative is connected with a variety of applications (see [2–6]). Stability analysis of fractional differential equations without inputs was studied in [7], where exponential stability is obtained for the Caputo–Fabrizio derivative. Since their kernels are nonlocal, fractional derivatives preserve memories, and therefore, they have been used to model several (SIR) epidemic models (see [8–14]). Several analytical techniques have been implemented to study various fractional equations with fractional derivatives without singular kernels, such as the Laplace transform, reduction to initial value problems with integer derivatives, maximum principles, and fixed point theorems (see [15–21]), just to mention a few out of many in the literature. The following definitions are required to state our problem.

Definition 1. A function is said to be absolutely continuous on , if there exists a function such thatIn the following, we will use the notation to denote the space of absolutely continuous functions on .

Definition 2. A function is said to be of exponential order, if there exists three constants such that .
Then, we define the following space.

Definition 3. The space is defined byIn this paper, we consider the multiterm fractional initial value problem:where , and is the Caputo–Fabrizio fractional derivative of Caputo sense. Here, we assume that , so their Laplace transforms are well defined.

Definition 4. (see [1]). For , and , the Caputo–Fabrizio fractional derivative of Caputo sense is defined bywhere is a normalization function satisfying .
For the corresponding fractional integral and more properties of the derivative, we refer the readers to [1, 5, 22–24]. In [15, 17], the fractional initial value problems were transformed to equivalent initial value problems with integer derivatives. However, the technique is not valid for the multiterm initial value problems. The single term of the problem with was discussed in [23], and the solution of the problem was obtained in a closed form. In this paper, we apply the Laplace transform to analyze the solutions of the fractional initial value problems (3) and (4). This paper is organized as follows: in Section 2, we present some preliminary results about the Caputo–Fabrizio fractional derivative and derive necessary and sufficient conditions to guarantee the existence of solutions to problems (3) and (4). We also obtain the exact solutions in closed forms using the Laplace transform. In Section 3, we present two examples to illustrate the validity of our results. Finally, we close up with some concluding remarks in Section 4.

2. Main Results

We start with the definition and main results concerning the Caputo–Fabrizio derivative. Then, we present necessary and sufficient conditions for the solution of problems (3) and (4). Since , equation (3) can be written aswhereand denotes to the convolution of two functions.

Remark 1. Let where and are polynomials with degrees and , respectively, and are with real coefficients. If , then the inverse Laplace transform of exists and can be evaluated using partial fractions.

Lemma 1. For , the following hold:(1)(2)

Proof. (1)For any function , is the convolution product of a continuous function and function, that is continuous(2)Since for any , is continuous; then,

Definition 5. Let the space be defined by

Lemma 2. The following holds for multiterm fractional initial value problems (3) and (4).(1)If a solution exists such that for any , then (2)If , and , then a unique solution exists(3)If a solution exists for and , then

Proof. (1)Let be a solution to multiterm fractional initial value problems (3) and (4) with for any . By Lemma 1, for any Then, from equation (3), one gets(2)Let , , and . Then, applying the Laplace transform to equation (6) and using the convolution result, we have where . The above equation yields We have where and are polynomials of degrees and , respectively. Substituting in equation (11), we have or Now, the leading coefficient of the polynomial is , and the leading coefficient of the polynomial is 1. If , then is a polynomial of degree . Let so that are well defined. Then, Using the uniqueness result of the inverse Laplace transform, we have which completes the proof.(3)Let a solution exists for and . Starting from equation (14), let us defineDefine admits an inverse Laplace transform since is of degree . Moreover, since , also admits an inverse Laplace transform, thus also (by the convolution rule). One can rewrite equation (14) asHowever, since is of degree and , the right-hand side does not admit an inverse Laplace transform while the left-hand side does, which is a contradiction.

Lemma 3. Let , be a solution to the multiterm fractional initial value problem (3) with , then .

Proof. From Lemma 1, since , then .

Remark 2. If , and , then a solution exists by Lemma 2. If , a solution exists even if , and it can be evaluated using the Laplace transform in the following manner.

Lemma 4. Let be a possible solution to multiterm fractional initial value problems (3) and (4), with , , and . If has an inverse Laplace transform, then , where and is defined in equation (22).

Proof. Substituting , in equation (14), we haveSince , we have is a polynomial of degree . Letso that is well defined. Then, equation (14) yieldsIf , then in equation (23) can be written asApplying the inverse Laplace transform, we haveWe summarize the obtained results in the following main theorem.

Theorem 1. Consider multiterm fractional initial value problems (3) and (4).(1)For , if a solution , exists, then (2)If , and , then a solution exists(3)If and , then a solution exists if and only if (4)If , and , then a solution exists

3. Illustrative Examples

We discuss two main examples. The first one is a two-term problem where it holds that . We show the existence of a solution for a specific case by imposing extra conditions. The second example is a three-term problem where it holds that . So, the existence of a unique solution is guaranteed. We present the solution in a closed form and discuss several special cases.

Example 1. Consider the two-term fractional initial value problem:Since , then the problem has no solution for . For and has Laplace inverse, the problem admits a solution. To verify, let us find the solution given by equation (25). We havewhereThus,and the solution is obtained, provided that .
As a special case, let us consider . Then, , , and have Laplace inverse. Thus,For and , we have , and

Example 2. As a second example, we consider the three-term initial value problem:Since and , the problem has a unique solution given by equation (17) provided that . We haveLetthenwithand the solution is given byFor , we have , and thusFor , we have , and thusAs a specific case, if we consider , we havewhere , , and .
For , we have .

4. Concluding Remarks

We obtained the solutions of a class of multiterm fractional initial value problems in closed forms using the Laplace transform. We have also discussed several necessary and sufficient conditions to guarantee the existence of solutions to the problem. Whether the results are extendable to wider classes of multiterm initial value problems or systems of fractional equations is left for future work.

Data Availability

The data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that there are no conflicts of interest.

Authors’ Contributions

The first author has initiated the idea and obtained the analytical results. The second author did the numerical examples, shared the discussion, and approved the final draft of the manuscript.