Recent Trends in Special Functions and Analysis of Differential EquationsView this Special Issue
A Study of Fractional Differential Equation with a Positive Constant Coefficient via Hilfer Fractional Derivative
The aim of the paper is to consider the existence and uniqueness of solution of the fractional differential equation with a positive constant coefficient under Hilfer fractional derivative by using the fixed-point theorem. We also prove the bounded and continuous dependence on the initial conditions of solution. Besides, Hyers–Ulam stability and Hyers–Ulam–Rassias stability are discussed. Finally, we provide an example to demonstrate our main results.
In recent years, the study of the fractional differential and integral equation (FDE and IDE for short) has become the topic of the applied mathematics. FDE and IDE have been used as a tool mathematical to the modeling of many phenomena in various fields for example, in theory of signal processing, physics, economics, and chaotic dynamics. The reader can refer to the books (see Varsha , Kilbas et al. , Miller et al. , Abbas et al. , and Zhou et al. ) or the papers (see Zin et al. , Ma et al. , P. Agarwal et al. [8, 9], Rameshet al. , Vivek et al. , O’Regan et al. , and Duc et al. ).
Tate and Dinde  proved the existence of solution of the problem:where the symbol is Caputo fractional derivative and . Besides, the authors also considered the properties of solutions of this problem such as the boundedness of solution and the continuous dependence of solutions on the initial conditions. In , Tate et al. also performed the same study as  for the class of fractional integro-differential equations with positive constant coefficient.
In 2018, Sousa and Oliveira  have introduced a new fractional derivative with respect to another function the so-called -Hilfer fractional derivative, and the properties of this concept were also presented. Besides, the authors considered the relationship between the Hilfer fractional derivative and the other fractional derivative such as Riemann–Liouville fractional derivative, Caputo fractional derivative, Hadamard fractional derivative, Katugampola fractional derivative, and Chen fractional derivative. By using the Hilfer fractional derivative, Sousa and Oliveira  studied the existence and uniqueness of solution of the initial valued problem for FDEs. The continuous dependence of solution on the initial condition was also considered. The stability theory for FDIs and IDEs via Hilfer fractional derivative have also been discussed (see [18–22]). In , by using Gronwall inequality and Picard operator theory, Kharade and Kucche proved the existence and uniqueness of solutions for impulsive implicit delay Hilfer fractional differential equations. The Ulam–Hyers–Mittag–Leffler stability was also considered.
Motivated by Tate el al. [14, 15], Sousa et al. , and Kharade et al. , in this paper, we investigate the existence and uniqueness of solutions and some properties of solutions of the following fractional differential equation with the constant coefficient :with the initial conditionwhere the symbol is the Hilfer fractional derivative of with and is a continuous function, is a continuous function with respect to and on , is –Riemann–Liouville fractional integral with , and is a given constant.
Let be the space of all continuous functions and be the space of all times continuously differentiable functions on . We will introduce the weighted spaces of all continuous functionswith the normwith the norm
Let function be integrable on and function be increasing on with for all . Then, the Riemann–Liouville fractional integral of with respect to is defined bywhere for all .
Let , , and . Then, the Hilfer fractional derivative of the function of order and is given bywhere .
Lemma 1. (see ). Let be integrable functions and . Let such that for any . Assume that are nonnegative and is nonnegative and nondecreasing. Ifthenwhere is Mittag–Leffler function is defined by
Definition 1. (see [21, 23]). Problem (2) is called Hyers–Ulam stable if there exists a positive constant such that, for any and for each satisfying the inequality,there exists a solution of problem (2) satisfying
Definition 2. (see [21, 23]). Problem (2) is called Hyers–Ulam–Rassias stable, with respect to , if there exists a positive constant such that, for any and for each satisfying the inequality,there exists a solution of problem (2) satisfying
3. Main Results
Firstly, we note that applying the fractional integral operator to both sides of equation (2), we obtain
By Theorem 1, we have
Taking in equation (21), we have
3.1. Existence and Uniqueness Solution for Problem (2)
In this section, we will prove the existence and uniqueness solution of equation (2) with the initial condition (3). Firstly, we assume that the function satisfies the following assumption: there exists a constant such that
Proof. We set for any . Let us define the setwithIt is easy to see that is a nonempty, closed, bounded, and convex subset of Banach space .
Consider the operator given byfor any .
Firstly, we prove that the fixed point of the operator is a solution of equations (2) and (3). For any and for each , we have the following estimate:for any .
Combining the estimation above with the definition of , we infer thatHence, we conclude the operator maps into itself.
Secondly, is uniformly bounded since .
Thirdly, we will prove that the operator is continuous. Let the sequence and such thatFor any , we have the following estimate:Hence, for any , we haveBased on the continuity of function , we obtainSo, for any , we obtainwhich leads to operator which is continuous.
Next, we will show that is equicontinuous. Let such that and , and we haveAs , we imply the right-hand side of the estimation above tend to 0, that is, is equicontinuous.
Finally, we see that all conditions of Schauder fixed point theorem are satisfied. So, we can conclude that problem (2) and (3) has at least one solution. The proof is complete.
Proof. To prove this theorem, divide the proof into two steps. Now, we define the following set:whereLet us define the operator as follows: Step 1: similarl to the proof of Theorem 2, we can infer that the functions of are uniformly bounded in . Step 2: we will show that is a contraction on . For any and , we haveCombining the estimation above with assumption (36), we obtainthat is, is a contraction on .
Here, we see that all conditions in the Banach fixed point theorem are satisfied. Therefore, there exists a unique solution of problems (2) and (3). This proof is completed.
3.2. Continuous Dependence and Boundedness of Solution of Problem (2)
In this section, we will study the continuous dependence of solutions on initial conditions and the boundedness of solution of equations (2) and (3). Now, we consider the following problems:where the functions satisfy assumption (H1), for any .
Theorem 4. Assume that functions satisfy assumption (H1). Let be the solutions of problems (42), respectively. Then, we have the following estimate:for any .
Proof. Since are the solutions of problems (42), respectively. For any , we haveUsing assumption (H1) and for any , we have the following estimate:If we putfor any , then inequality (45) becomesApplying Gronwall Lemma 1 to (47), we obtainThis gives inequality (43).
Proof. Let be any solution of problems (2) and (3). Then, we haveUsing assumption (H1) and for any , we haveWe putfor any , which leads to the following estimate:Applying Gronwall Lemma 1, we obtainThe proof is completed.
3.3. Hyers–Ulam Stability and Hyers–Ulam–Rassias Stability for Problem (2)
Proof. Let be a solution of (12) and let be a unique solution of (2). Then, for any, we haveFor any , we haveMoreover, due to the function which satisfies inequality (12), there exists a function such that , for any andApplying the fractional integral to both sides of equation (57) and by using Theorem 1, we obtainThus, we haveCombining inequality (57) with inequality (59), we obtainApplying Gronwall Lemma 1 to (60), we obtainwhere
Based on the inequality above and Definition 2, we infer that problem (2) is Hyers–Ulam stable. The proof is completed.
Proof. Let be a solution of (12) and let