# A Study of Fractional Differential Equation with a Positive Constant Coefficient via Hilfer Fractional Derivative

Authors: Hoa Ngo Van; Vu Ho
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

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.


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## Body

## 1. Introduction

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 [1], Kilbas et al. [2], Miller et al. [3], Abbas et al. [4], and Zhou et al. [5]) or the papers (see Zin et al. [6], Ma et al. [7], P. Agarwal et al. [8, 9], Rameshet al. [10], Vivek et al. [11], O’Regan et al. [12], and Duc et al. [13]).Tate and Dinde [14] proved the existence of solution of the problem:(1)Dc0+αξt=λξt+ζt,ξt,ξ0=x0∈R,λ>0,where the symbol DC0+α is Caputo fractional derivative and ζ∈C0,a,R. 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 [15], Tate et al. also performed the same study as [14] for the class of fractional integro-differential equations with positive constant coefficient.In 2018, Sousa and Oliveira [16] have introduced a new fractional derivative with respect to another function the so-called ψ-Hilfer fractional derivative, and the properties of this concept were also presented. Besides, the authors considered the relationship between the ψ−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 ψ−Hilfer fractional derivative, Sousa and Oliveira [17] 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 ψ−Hilfer fractional derivative have also been discussed (see [18–22]). In [23], by using Gronwall inequality and Picard operator theory, Kharade and Kucche proved the existence and uniqueness of solutions for impulsive implicit delay ψ−Hilfer fractional differential equations. The Ulam–Hyers–Mittag–Leffler stability was also considered.Motivated by Tate el al. [14, 15], Sousa et al. [16], and Kharade et al. [23], 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 λ>0:(2)D0+ℍh,υ;gξt=λξt+ζt,ξt,with the initial condition(3)I0+1−γ,gξ0=A,where the symbol D0+ℍh,υ;gξt is the g−Hilfer fractional derivative of ξ with 0<h<1,0≤υ≤1 and ξ:0,a⟶R is a continuous function, ζ is a continuous function with respect to t and ξ on 0,a×R, I0+1−γ,g⋅ is g–Riemann–Liouville fractional integral with 0≤γ=h+υ1−h<1, and A is a given constant.


## 2. Preliminaries

In this section, we introduce some notations and some concepts which are used throughout this paper. This result can be found in the books [3, 8] and the papers [16, 17].LetC0,a,R be the space of all continuous functions ζ:0,a⟶R and Cn0,a,R be the space of all n−times continuously differentiable functions on 0,a. We will introduce the weighted spaces of all continuous functions(4)Cγ,g0,a≔ζ:0,a⟶R:gt−g0γζt∈C0,a,with the norm(5)ζCγ,g0,a=gt−g0γζtC0,a=maxt∈0,agt−g0γζt,C1−γ,g0,a≔ζ:0,a⟶R:gt−g01−γζt∈C0,a,with the norm(6)ζC1−γ,g0,a=gt−g01−γζtC0,a=maxt∈0,agt−g01−γζt.Let functionf be integrable on 0,a and function g:0,a⟶R+ be increasing on 0,a with g′t≠0 for all t∈0,a. Then, the g−Riemann–Liouville fractional integral of f with respect to g is defined by(7)I0+h;gξt=1Γh∫0tQght,sξsds,h>0,where Qght,s=g′sgt−gsh−1 for all t,s∈0,a.Letn−1<h<n, n∈N, and f∈Cn0,a,R. Then, the Hilfer fractional derivative DH0+h,υ;gξ⋅ of the function ξ of order h and υ∈0,1 is given by(8)Dℍ0+h,υ;gξt=I0+n−hυ;g1g′tddtnI0+n−h1−υ;gξt,where n=h+1.Theorem 1.
(see [16, 17]). If ζ∈C10,a,R, h>0, and υ∈0,1, then(i)
Dℍ0+h,υ;gI0+h;gξt=ξt and ∀t∈0,a(ii)
I0+h;gℍD0+h,υ;gξt=ξt−∑k=0ngt−g0γ−k/Γγ−k+1I0+1−h1−υ;gξ0 and γ=h+υ1−hLemma 1.
(see [17]). Let atandbt be integrable functions and c∈C0,a,R. Let g∈C10,a,R such that g′t>0 for any t∈0,a. Assume that atandbt are nonnegative and g is nonnegative and nondecreasing. If(9)at≤bt+ct∫0tQght,sasds,then(10)at≤bthctΓhgt−g0h,∀t∈0,a,where hξ is Mittag–Leffler function is defined by(11)hξ=∑k=0∞ξkΓ1+kh,h>0,k∈N.Definition 1.
(see [21, 23]). Problem (2) is called Hyers–Ulam stable if there exists a positive constant Cζ such that, for any ɛ>0 and for each ξ∈C1−γ,g0,a satisfying the inequality,(12)D0+ℍh,υ;gξt−λξt−ζt,ξt≤ɛ,∀t∈0,a,there exists a solution ξ^∈C1−γ,g0,a of problem (2) satisfying(13)ξt−ξ^t≤Cζɛ,∀t∈0,a.Definition 2.
(see [21, 23]). Problem (2) is called Hyers–Ulam–Rassias stable, with respect to ϕ∈C0,a,R, if there exists a positive constant Cζ,ϕ such that, for any ɛ>0 and for each ξ∈C1−γ,g0,a satisfying the inequality,(14)D0+ℍh,υ;gξt−λξt−ζt,ξt≤ϵϕt,∀t∈0,a,there exists a solution ξ^∈C1−γ,g0,a of problem (2) satisfying(15)ξt−ξ^t≤Cζ,ϕϵϕt,∀t∈0,a.


## 3. Main Results

Firstly, we note that applying the fractional integral operatorI0+h;g· to both sides of equation (2), we obtain(16)I0+h;gD0+ℍh,υ;gξt=λI0+h;gξt+I0+h;gζt,ξt,∀t∈0,a.Using Theorem1 and the initial condition (3), we have the following integral equation:(17)ξt=gt−g0γ−1ΓγI0+h;gξ0+λI0+h;gξt+I0+1−h1−υ;gζt,ξt=gt−g0γ−1ΓγA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,for any t∈0,a.On the contrary, ifξ satisfies equation (17), then ξ satisfies equations (2 and 3. Moreover, operating the fractional derivative operator D0+ℍh,υ;g⋅ on both sides of equation (17), we obtain(18)D0+ℍh,υ;gξt=D0+ℍh,υ;ggt−g0γ−1ΓγI0+h;gξ0+D0+ℍh,υ;gλI0+h;gξt+D0+ℍh,υ;gI0+1−h1−υ;gζt,ξt,∀t∈0,a.By Theorem1, we have(19)D0+ℍh,υ;ggt−g0h−1ΓγI0+h;gξ0=0,0<h<1.Combining (18) and (19), we imply(20)D0+ℍh,υ;gξt=λξt+ζt,ξt,∀t∈0,a.Next, we verify that the initial condition3 holds. Indeed, applying the Riemann–Liouville fractional integral I0+1−υ⋅ on both sides of equation (17), we have(21)I0+1−υξt=I0+1−υgt−g0γ−1ΓγI0+h;gξ0+I0+1−υλI0+h;gξt+I0+1−υI0+1−h1−υ;gζt,ξt,t∈0,a.Takingt⟶0 in equation (21), we have(22)I0+1−υξt=A.In summary, we can conclude thatξ satisfies equations (2 and 3 if and only if ξ satisfies the following integral equation:(23)ξt=gt−g0γ−1ΓγA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,for any t∈0,a.


### 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 ζ satisfies the following assumption: there exists a constant L>0 such that(24)H1ζt,ξ1−ζt,ξ2≤Lξ1−ξ2,t∈0,a.Theorem 2.
Assume that assumption (H1) is satisfied. Then, problems (2) and (3) have at least one solutionProof.
We setζ˜≔ζt,0 for any t∈0,a. Let us define the set(25)Ur≔ξ∈C1−γ;g0,a:ξC1−γ;g≤r,with(26)r≔AΓγ+Bh,γga−g0hΓhλ+LξC1−γ;g+Bh,γga−g0hΓhζ˜C1−γ;g,∀t∈0,a.
It is easy to see thatUr is a nonempty, closed, bounded, and convex subset of Banach space C1−γ,g0,a.
Consider the operatorK:Ur⟶Ur given by(27)Kξt=gt−g0γ−1ΓhA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,for any t∈0,a.
Firstly, we prove that the fixed point of the operatorK is a solution of equations (2) and (3). For any ξ∈C1−γ;g0,a and for each t∈0,a, we have the following estimate:(28)Kξtgt−g01−γ=AΓγ+λgt−g01−γΓh∫0tQght,sξsds+gt−g01−γΓh∫0tQght,sζs,ξsds≤AΓγ+λgt−g01−γΓh∫0tQght,sξsds+gt−g01−γΓh∫0tQght,sζs,ξsds≤AΓγ+λgt−g01−γΓh∫0tQght,sξsds+gt−g01−γΓh∫0tQght,sζs,ξs−ζs,0+ζs,0ds≤AΓγ+λgt−g01−γΓhBh,γgt−g0h+γ−1ξC1−γ;g+gt−g01−γΓhBh,γgt−g0h+γ−1LξC1−γ;g+gt−g01−γΓhBh,γgt−g0h+γ−1ζ˜C1−γ;g,for any t∈0,a.
Combining the estimation above with the definition ofUr, we infer that(29)Kξtgt−g01−γ≤AΓγ+Bh,γga−g0hΓhλ+LξC1−γ;g+Bh,γga−g0hΓhξ˜C1−γ;g=r.
Hence, we conclude the operatorK maps into itself.
Secondly,KUr is uniformly bounded since KUr⊂Ur.
Thirdly, we will prove that the operatorK is continuous. Let the sequence ξn∈Ur and ξ∈Ur such that(30)ξn−ξC1−γ;g⟶0,asn⟶∞.
For anyt∈0,a, we have the following estimate:(31)Kξnt−Kξtgt−g01−γ≤λgt−g01−γΓh∫0tQght,sξns−ξsds+gt−g01−γΓh∫0tQght,sζs,ξns−ζs,ξsds≤Bh,γΓhgt−g0hλξn⋅−ξ⋅C1−γ;g+Bh,γΓhgt−g0hζ⋅,ξn⋅−ζ⋅,ξ⋅C1−γ;g,∀t∈0,a.
Hence, for anyt∈0,a, we have(32)Kξn−KξC1−γ;g≤Bh,γΓhga−g0hλξn⋅−ξ⋅C1−γ;g+Bh,γΓhga−g0hζ⋅,ξn⋅−ζ⋅,ξ⋅C1−γ;g.
Based on the continuity of functionζ, we obtain(33)ζ⋅,ξn⋅−ζ⋅,ξ⋅C1−γ;g⟶0,asn⟶∞.
So, for anyt∈0,a, we obtain(34)Kξn−KξC1−γ;g⟶0,asn⟶∞,which leads to operator K which is continuous.
Next, we will show thatKUr is equicontinuous. Let t1,t2∈0,a such that 0<t1<t2<a and ξ∈Ur, and we have(35)Kξt2gt2−g01−γ−Kξt1gt1−g01−γ=λgt2−g01−γΓh∫0t2Qght2,sξsds+gt2−g01−γΓh∫0t2Qght2,sζs,ξsds−λgt1−g01−γΓh∫0t1Qght1,sξsds−gt1−g01−γΓh∫0t1Qght1,sζs,ξsds≤1ΓhBh,γλξC1−γ;g+ζC1−γ;ggt2−g0h−gt1−g0h.
Ast2⟶t1, we imply the right-hand side of the estimation above tend to 0, that is, KUr 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.Theorem 3.
Leta˜∈0,a. Assume that assumption (H1) is satisfied. If(36)Bh,γga˜−g0hλ+LΓh<1,then problems (2) and (3) have a unique solution on 0,a˜.Proof.
To prove this theorem, divide the proof into two steps. Now, we define the following set:(37)Uρ≔ξ∈C1−γ,g0,a˜:ξC1−γ,g0,a˜≤ρ,where(38)ρ≔AΓhΓγΓh−1+λ+Lga˜−g0h.
Let us define the operatorB:C1−γ,g0,a˜⟶C1−γ,g0,a˜ as follows:(39)Bξt=gt−g0γ−1ΓγA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,∀t∈0,a˜.Step 1: similarl to the proof of Theorem2, we can infer that the functions of BUρ are uniformly bounded in C1−γ,g0,a˜.Step 2: we will show thatB is a contraction on C1−γ,g0,a˜. For any t∈0,a˜ and ξ1,ξ2∈C1−γ,g0,a˜, we have(40)Kξ1t−Kξ2tgt−g01−γ≤λgt−g01−γΓh∫0tQght,sξ1s−ξ2sds+gt−g01−γΓh∫0tQght,sζs,ξ1s−ζs,ξ2sds≤Bh,γΓhgt−g0hλ+Lξ1−ξ2C1−γ,g0,a˜.
Combining the estimation above with assumption (36), we obtain(41)Kξ1−Kξ2C1−γ,g0,a˜≤ξ1−ξ2C1−γ,g0,a˜,that is, B is a contraction on C1−γ,g0,a˜.
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:(42)D0+ℍh,υ;gξt=λξt+ζt,ξt,I0+1−γξ0=A,D0+ℍh,υ;gξ∗t=λξ∗t+ζt,ξ∗t,I0+1−γξ∗0=A∗,where the functions ζ⋅,ξ⋅,ζ⋅,ξ∗⋅∈C1−γ,g0,a satisfy assumption (H1), for any ξ,ξ∗∈C1−γ,g0,a.Theorem 4.
Assume that functionsζandζ∗ satisfy assumption (H1). Let ξtandξ∗t be the solutions of problems (42), respectively. Then, we have the following estimate:(43)ξt−ξ∗t≤A−A∗gt−g01−γΓγhλ+Lgt−g0h,for any t∈0,a.Proof.
Sinceξtandξ∗t are the solutions of problems (42), respectively. For any t∈0,a, we have(44)ξt=gt−g0γ−1ΓhA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,ξ∗t=gt−g0γ−1ΓhA∗+λΓh∫0tQght,sξ∗sds+1Γh∫0tQght,sζs,ξ∗sds.
Using assumption (H1) and for anyt∈0,a, we have the following estimate:(45)ξt−ξ∗t≤A−A∗gt−g01−γΓγ+λΓh∫0tQght,sξs−ξ∗sds+1Γh∫0tQght,sζs,ξs−ζs,ξ∗sds≤A−A∗gt−g01−γΓγ+λ+LΓh∫0tQght,sξs−ξ∗sds.
If we put(46)v˜t≔ξt−ξ∗t,u˜t≔A−A∗gt−g01−γΓγ,g˜t≔λ+LΓh,for any t∈0,a, then inequality (45) becomes(47)v˜t≤u˜t+g˜t∫0tQght,sv˜sds.
Applying Gronwall Lemma1 to (47), we obtain(48)v˜t≤u˜thλ+Lgt−g0h.
This gives inequality (43).Theorem 5.
Assume that assumption (H1) is satisfied. Ifξ is any solution of problems (2) and (3), then(49)ξt≤Agt−g01−γΓγ+ζ^gt−g0hΓ1+h×hλ+Lgt−g0h,∀t∈0,a,where ζ^=maxs∈0,aζs,0.Proof.
Letξ be any solution of problems (2) and (3). Then, we have(50)ξt=gt−g0γ−1ΓγA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,∀t∈0,a.
Using assumption (H1) and for anyt∈0,a, we have(51)ξt≤Agt−g01−γΓγ+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds≤Agt−g01−γΓγ+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξs−ζs,0+ζs,0ds≤Agt−g01−γΓγ+λ+LΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,0ds≤Agt−g01−γΓγ+ζ^gt−g0hΓ1+h+λ+LΓh∫0tQght,sξsds.
We put(52)vt≔Agt−g01−γΓγ+ζ^gt−g0hΓ1+h,ut≔λ+LΓh,for any t∈0,a, which leads to the following estimate:(53)ξt≤vt+ut∫0tQght,sξsds,∀t∈0,a.
Applying Gronwall Lemma1, we obtain(54)ξt≤vthutΓhgt−g0h,∀t∈0,a.
The proof is completed.


### 3.3. Hyers–Ulam Stability and Hyers–Ulam–Rassias Stability for Problem (2)

Theorem 6.
Assume that assumption (H1) and (36) are satisfied. Then, problem (2) is Hyers–Ulam stable.Proof.
Letξt∈C1−γ,g0,a be a solution of (12) and let ξ^t∈C1−γ,g0,a be a unique solution of (2). Then, for any, we have(55)ξ^t=gt−g0γ−1ΓγA+λΓh∫0tQght,sξ^sds+1Γh∫0tQght,sζs,ξ^sds,∀t∈0,a.
For anyt∈0,a, we have(56)ξt−ξ^t=ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξ^sds−1Γh∫0tQght,sζs,ξ^sds≤ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξ^sds−1Γh∫0tQght,sζs,ξ^sds+λΓh∫0tQght,sξ^sds+λΓh∫0tQght,sζs,ξsds−λΓh∫0tQght,sξ^sds−λΓh∫0tQght,sζs,ξsds≤ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξ^sds−1Γh∫0tQght,sζs,ξ^sds+1Γh∫0tQght,sξs−ξ^tds+1Γh∫0tQght,sζs,ξs−ζs,ξ^sds=ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξ^sds−1Γh∫0tQght,sζs,ξ^sds+λ+LΓh∫0tQght,sξs−ξ^tds.
Moreover, due to the functionξt which satisfies inequality (12), there exists a function h:0,a⟶R such that ht≤ɛ, for any t∈0,a and(57)D0+ℍh,υ;gξt−λξt−ζt,ξt=ht,∀t∈0,a.
Applying the fractional integralI0+h;g⋅ to both sides of equation (57) and by using Theorem 1, we obtain(58)I0+h;gD0+ℍh,υ;gξt−λξt−ζt,ξt=I0+h;ght,∀t∈0,a.
Thus, we have(59)ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξsds−1Γh∫0tQght,sζs,ξsds≤I0+h;ght≤I0+h;gϵ=gt−g0hΓhϵ,∀t∈0,a.
Combining inequality (57) with inequality (59), we obtain(60)ξt−ξ^t≤gt−g0hΓhϵ+λ+LΓh∫0tQght,sξs−ξ^tds,∀t∈0,a.
Applying Gronwall Lemma1 to (60), we obtain(61)ξt−ξ^t≤gt−g0hΓhhλ+Lgt−g0hϵ≤ga−g0hΓhhλ+Lgt−g0hϵ≤Cζϵ,∀t∈0,a,where Cζ≔ga−g0h/Γhhλ+Lgt−g0h.
Based on the inequality above and Definition2, we infer that problem (2) is Hyers–Ulam stable. The proof is completed.Theorem 7.
Assume that assumption (H1) and (36) are satisfied. If there exists function ϕ∈C0,a,R and the positive constant Cϕ such that(62)1Γh∫0tQght,sϕsds≤Cϕϕt,∀t∈0,a,then problem (2) is Hyers–Ulam–Rassias stable.Proof.
Letξt∈C1−γ,g0,a be a solution of (12) and let ξ^t∈C1−γ,g0,a be a unique solution of (2). Performing the same calculations as in Theorem 3, we have the estimation as follows:(63)ξt−ξ^t≤ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξsds−1Γh∫0tQght,sζs,ξsds+λ+LΓh∫0tQght,sξs−ξ^tds.
On the contrary, applying the fractional integralI0+h;g⋅ to both sides of inequality (14) and by Theorem 1 and assumptions (61), we obtain(64)ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξsds−1Γh∫0tQght,sζs,ξsds≤ϵI0+h;gϕt≤Cϕϕtϵ,∀t∈0,a.
From estimations (63) and (64), we imply(65)ξt−ξ^t≤Cϕϕtϵ+λ+LΓh∫0tQght,sξs−ξ^tds,∀t∈0,a.
Using Gronwall Lemma1 and (64), we obtain(66)ξt−ξ^t≤Cϕhλ+Lgt−g0hϕtϵ≤Cζ,ϕϵ,∀t∈0,a,where Cζ,ϕ≔Cϕhλ+Lgt−g0h.
Similarly, based on the inequality above and Definition2, we infer that problem (2) is Hyers–Ulam–Rassias stable. The proof is completed.


### 3.4. Example

Let us consider the following problem:(67)D0+ℍh,υ;gξt=13ξt+1+ξt9+et,∀t∈0,1/3,with the initial condition I0+1−γ,gξ0=0.For this example, we only consider the two situation as below. The other one is considered similarly.Situation 1.
Letgt=t, taking the limit υ⟶1 on both sides of (8). Then, by Sousa et al. [16], we have(68)D0+ℍh,υ;gξt=0+CDhξt,∀t∈0,1.
Combining (67) with (68), we infer that(69)D0+Chξt=13ξt+1+ξt9+et,∀t∈0,1/3,with the initial condition ξ0=0.
We considerh=1/2, λ=1/3, and(70)ζt,ξt≔1+ξt9+et,∀t∈0,1/3.
It is easy to see thatζ is a continuous function and it satisfies assumption (H1) with Lipschitz constant L=1/10. Indeed, for any ξ1,ξ∈C0,a,R, we have the following estimate:(71)ζt,ξ1t−ζt,ξ2t=1+ξ1t9+et−1+ξ2t9+et≤19+etξ1t−ξ2t≤110ξ1t−ξ2t,∀t∈0,1/3.
Moreover, we have(72)Bh,γga˜−g0hλ+LΓh=B1/2,1/21/3−01/21/3+1/10Γ1/2≈0.45<1.
We see that all the assumptions of Theorem2 is satisfied. So, we infer that problem (69) has a unique solution on 0,1/3.Situation 2.
Leth=2/3, υ=1/2, and gt=tρ, ρ>0, for any t∈0,1/3. Then, by Sousa et al. [16], we have(73)D0+ρ2/3,1/2ξt=13ξt+1+ξt9+et,∀t∈0,1/3,with the initial condition Iρ0+5/6ξ0=0.
Performing the same calculations as in Case 1, then it is also easy to check that problem (73) has a unique solution on 0,1/3.
We putϕt=t3ρ/2−1 for any t∈0,1/3. Now, we will prove that problem (73) is Hyers–Ulam–Rassias stable. For any t∈0,1/3, we have(74)1Γh∫0tQght,sϕsds=ρ1/3Γ2/3∫0ts−ρ/2tρ−sρ−1/3ϕsds=ρ1/3Γ2/3∫0ts−ρ/2tρ−sρ−1/3s3ρ/2−1ds=ρ1/3Γ2/3∫0ttρ−sρ−1/3sρ−1ds≤3ρ−2/32Γ2/3131−5ρ/6ϕt,∀t∈0,1/3.
Hence, assumption (73) of Theorem 3 is satisfied by Cϕ≔3ρ−2/3/2Γ2/31/31−5ρ/6>0 with ρ>0. All the assumptions of Theorem 3 are satisfied. So, we imply problem (73) is Hyers–Ulam–Rassias stable.


## 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 ζ satisfies the following assumption: there exists a constant L>0 such that(24)H1ζt,ξ1−ζt,ξ2≤Lξ1−ξ2,t∈0,a.Theorem 2.
Assume that assumption (H1) is satisfied. Then, problems (2) and (3) have at least one solutionProof.
We setζ˜≔ζt,0 for any t∈0,a. Let us define the set(25)Ur≔ξ∈C1−γ;g0,a:ξC1−γ;g≤r,with(26)r≔AΓγ+Bh,γga−g0hΓhλ+LξC1−γ;g+Bh,γga−g0hΓhζ˜C1−γ;g,∀t∈0,a.
It is easy to see thatUr is a nonempty, closed, bounded, and convex subset of Banach space C1−γ,g0,a.
Consider the operatorK:Ur⟶Ur given by(27)Kξt=gt−g0γ−1ΓhA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,for any t∈0,a.
Firstly, we prove that the fixed point of the operatorK is a solution of equations (2) and (3). For any ξ∈C1−γ;g0,a and for each t∈0,a, we have the following estimate:(28)Kξtgt−g01−γ=AΓγ+λgt−g01−γΓh∫0tQght,sξsds+gt−g01−γΓh∫0tQght,sζs,ξsds≤AΓγ+λgt−g01−γΓh∫0tQght,sξsds+gt−g01−γΓh∫0tQght,sζs,ξsds≤AΓγ+λgt−g01−γΓh∫0tQght,sξsds+gt−g01−γΓh∫0tQght,sζs,ξs−ζs,0+ζs,0ds≤AΓγ+λgt−g01−γΓhBh,γgt−g0h+γ−1ξC1−γ;g+gt−g01−γΓhBh,γgt−g0h+γ−1LξC1−γ;g+gt−g01−γΓhBh,γgt−g0h+γ−1ζ˜C1−γ;g,for any t∈0,a.
Combining the estimation above with the definition ofUr, we infer that(29)Kξtgt−g01−γ≤AΓγ+Bh,γga−g0hΓhλ+LξC1−γ;g+Bh,γga−g0hΓhξ˜C1−γ;g=r.
Hence, we conclude the operatorK maps into itself.
Secondly,KUr is uniformly bounded since KUr⊂Ur.
Thirdly, we will prove that the operatorK is continuous. Let the sequence ξn∈Ur and ξ∈Ur such that(30)ξn−ξC1−γ;g⟶0,asn⟶∞.
For anyt∈0,a, we have the following estimate:(31)Kξnt−Kξtgt−g01−γ≤λgt−g01−γΓh∫0tQght,sξns−ξsds+gt−g01−γΓh∫0tQght,sζs,ξns−ζs,ξsds≤Bh,γΓhgt−g0hλξn⋅−ξ⋅C1−γ;g+Bh,γΓhgt−g0hζ⋅,ξn⋅−ζ⋅,ξ⋅C1−γ;g,∀t∈0,a.
Hence, for anyt∈0,a, we have(32)Kξn−KξC1−γ;g≤Bh,γΓhga−g0hλξn⋅−ξ⋅C1−γ;g+Bh,γΓhga−g0hζ⋅,ξn⋅−ζ⋅,ξ⋅C1−γ;g.
Based on the continuity of functionζ, we obtain(33)ζ⋅,ξn⋅−ζ⋅,ξ⋅C1−γ;g⟶0,asn⟶∞.
So, for anyt∈0,a, we obtain(34)Kξn−KξC1−γ;g⟶0,asn⟶∞,which leads to operator K which is continuous.
Next, we will show thatKUr is equicontinuous. Let t1,t2∈0,a such that 0<t1<t2<a and ξ∈Ur, and we have(35)Kξt2gt2−g01−γ−Kξt1gt1−g01−γ=λgt2−g01−γΓh∫0t2Qght2,sξsds+gt2−g01−γΓh∫0t2Qght2,sζs,ξsds−λgt1−g01−γΓh∫0t1Qght1,sξsds−gt1−g01−γΓh∫0t1Qght1,sζs,ξsds≤1ΓhBh,γλξC1−γ;g+ζC1−γ;ggt2−g0h−gt1−g0h.
Ast2⟶t1, we imply the right-hand side of the estimation above tend to 0, that is, KUr 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.Theorem 3.
Leta˜∈0,a. Assume that assumption (H1) is satisfied. If(36)Bh,γga˜−g0hλ+LΓh<1,then problems (2) and (3) have a unique solution on 0,a˜.Proof.
To prove this theorem, divide the proof into two steps. Now, we define the following set:(37)Uρ≔ξ∈C1−γ,g0,a˜:ξC1−γ,g0,a˜≤ρ,where(38)ρ≔AΓhΓγΓh−1+λ+Lga˜−g0h.
Let us define the operatorB:C1−γ,g0,a˜⟶C1−γ,g0,a˜ as follows:(39)Bξt=gt−g0γ−1ΓγA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,∀t∈0,a˜.Step 1: similarl to the proof of Theorem2, we can infer that the functions of BUρ are uniformly bounded in C1−γ,g0,a˜.Step 2: we will show thatB is a contraction on C1−γ,g0,a˜. For any t∈0,a˜ and ξ1,ξ2∈C1−γ,g0,a˜, we have(40)Kξ1t−Kξ2tgt−g01−γ≤λgt−g01−γΓh∫0tQght,sξ1s−ξ2sds+gt−g01−γΓh∫0tQght,sζs,ξ1s−ζs,ξ2sds≤Bh,γΓhgt−g0hλ+Lξ1−ξ2C1−γ,g0,a˜.
Combining the estimation above with assumption (36), we obtain(41)Kξ1−Kξ2C1−γ,g0,a˜≤ξ1−ξ2C1−γ,g0,a˜,that is, B is a contraction on C1−γ,g0,a˜.
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:(42)D0+ℍh,υ;gξt=λξt+ζt,ξt,I0+1−γξ0=A,D0+ℍh,υ;gξ∗t=λξ∗t+ζt,ξ∗t,I0+1−γξ∗0=A∗,where the functions ζ⋅,ξ⋅,ζ⋅,ξ∗⋅∈C1−γ,g0,a satisfy assumption (H1), for any ξ,ξ∗∈C1−γ,g0,a.Theorem 4.
Assume that functionsζandζ∗ satisfy assumption (H1). Let ξtandξ∗t be the solutions of problems (42), respectively. Then, we have the following estimate:(43)ξt−ξ∗t≤A−A∗gt−g01−γΓγhλ+Lgt−g0h,for any t∈0,a.Proof.
Sinceξtandξ∗t are the solutions of problems (42), respectively. For any t∈0,a, we have(44)ξt=gt−g0γ−1ΓhA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,ξ∗t=gt−g0γ−1ΓhA∗+λΓh∫0tQght,sξ∗sds+1Γh∫0tQght,sζs,ξ∗sds.
Using assumption (H1) and for anyt∈0,a, we have the following estimate:(45)ξt−ξ∗t≤A−A∗gt−g01−γΓγ+λΓh∫0tQght,sξs−ξ∗sds+1Γh∫0tQght,sζs,ξs−ζs,ξ∗sds≤A−A∗gt−g01−γΓγ+λ+LΓh∫0tQght,sξs−ξ∗sds.
If we put(46)v˜t≔ξt−ξ∗t,u˜t≔A−A∗gt−g01−γΓγ,g˜t≔λ+LΓh,for any t∈0,a, then inequality (45) becomes(47)v˜t≤u˜t+g˜t∫0tQght,sv˜sds.
Applying Gronwall Lemma1 to (47), we obtain(48)v˜t≤u˜thλ+Lgt−g0h.
This gives inequality (43).Theorem 5.
Assume that assumption (H1) is satisfied. Ifξ is any solution of problems (2) and (3), then(49)ξt≤Agt−g01−γΓγ+ζ^gt−g0hΓ1+h×hλ+Lgt−g0h,∀t∈0,a,where ζ^=maxs∈0,aζs,0.Proof.
Letξ be any solution of problems (2) and (3). Then, we have(50)ξt=gt−g0γ−1ΓγA+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds,∀t∈0,a.
Using assumption (H1) and for anyt∈0,a, we have(51)ξt≤Agt−g01−γΓγ+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξsds≤Agt−g01−γΓγ+λΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,ξs−ζs,0+ζs,0ds≤Agt−g01−γΓγ+λ+LΓh∫0tQght,sξsds+1Γh∫0tQght,sζs,0ds≤Agt−g01−γΓγ+ζ^gt−g0hΓ1+h+λ+LΓh∫0tQght,sξsds.
We put(52)vt≔Agt−g01−γΓγ+ζ^gt−g0hΓ1+h,ut≔λ+LΓh,for any t∈0,a, which leads to the following estimate:(53)ξt≤vt+ut∫0tQght,sξsds,∀t∈0,a.
Applying Gronwall Lemma1, we obtain(54)ξt≤vthutΓhgt−g0h,∀t∈0,a.
The proof is completed.


## 3.3. Hyers–Ulam Stability and Hyers–Ulam–Rassias Stability for Problem (2)

Theorem 6.
Assume that assumption (H1) and (36) are satisfied. Then, problem (2) is Hyers–Ulam stable.Proof.
Letξt∈C1−γ,g0,a be a solution of (12) and let ξ^t∈C1−γ,g0,a be a unique solution of (2). Then, for any, we have(55)ξ^t=gt−g0γ−1ΓγA+λΓh∫0tQght,sξ^sds+1Γh∫0tQght,sζs,ξ^sds,∀t∈0,a.
For anyt∈0,a, we have(56)ξt−ξ^t=ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξ^sds−1Γh∫0tQght,sζs,ξ^sds≤ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξ^sds−1Γh∫0tQght,sζs,ξ^sds+λΓh∫0tQght,sξ^sds+λΓh∫0tQght,sζs,ξsds−λΓh∫0tQght,sξ^sds−λΓh∫0tQght,sζs,ξsds≤ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξ^sds−1Γh∫0tQght,sζs,ξ^sds+1Γh∫0tQght,sξs−ξ^tds+1Γh∫0tQght,sζs,ξs−ζs,ξ^sds=ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξ^sds−1Γh∫0tQght,sζs,ξ^sds+λ+LΓh∫0tQght,sξs−ξ^tds.
Moreover, due to the functionξt which satisfies inequality (12), there exists a function h:0,a⟶R such that ht≤ɛ, for any t∈0,a and(57)D0+ℍh,υ;gξt−λξt−ζt,ξt=ht,∀t∈0,a.
Applying the fractional integralI0+h;g⋅ to both sides of equation (57) and by using Theorem 1, we obtain(58)I0+h;gD0+ℍh,υ;gξt−λξt−ζt,ξt=I0+h;ght,∀t∈0,a.
Thus, we have(59)ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξsds−1Γh∫0tQght,sζs,ξsds≤I0+h;ght≤I0+h;gϵ=gt−g0hΓhϵ,∀t∈0,a.
Combining inequality (57) with inequality (59), we obtain(60)ξt−ξ^t≤gt−g0hΓhϵ+λ+LΓh∫0tQght,sξs−ξ^tds,∀t∈0,a.
Applying Gronwall Lemma1 to (60), we obtain(61)ξt−ξ^t≤gt−g0hΓhhλ+Lgt−g0hϵ≤ga−g0hΓhhλ+Lgt−g0hϵ≤Cζϵ,∀t∈0,a,where Cζ≔ga−g0h/Γhhλ+Lgt−g0h.
Based on the inequality above and Definition2, we infer that problem (2) is Hyers–Ulam stable. The proof is completed.Theorem 7.
Assume that assumption (H1) and (36) are satisfied. If there exists function ϕ∈C0,a,R and the positive constant Cϕ such that(62)1Γh∫0tQght,sϕsds≤Cϕϕt,∀t∈0,a,then problem (2) is Hyers–Ulam–Rassias stable.Proof.
Letξt∈C1−γ,g0,a be a solution of (12) and let ξ^t∈C1−γ,g0,a be a unique solution of (2). Performing the same calculations as in Theorem 3, we have the estimation as follows:(63)ξt−ξ^t≤ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξsds−1Γh∫0tQght,sζs,ξsds+λ+LΓh∫0tQght,sξs−ξ^tds.
On the contrary, applying the fractional integralI0+h;g⋅ to both sides of inequality (14) and by Theorem 1 and assumptions (61), we obtain(64)ξt−gt−g0γ−1ΓγA−λΓh∫0tQght,sξsds−1Γh∫0tQght,sζs,ξsds≤ϵI0+h;gϕt≤Cϕϕtϵ,∀t∈0,a.
From estimations (63) and (64), we imply(65)ξt−ξ^t≤Cϕϕtϵ+λ+LΓh∫0tQght,sξs−ξ^tds,∀t∈0,a.
Using Gronwall Lemma1 and (64), we obtain(66)ξt−ξ^t≤Cϕhλ+Lgt−g0hϕtϵ≤Cζ,ϕϵ,∀t∈0,a,where Cζ,ϕ≔Cϕhλ+Lgt−g0h.
Similarly, based on the inequality above and Definition2, we infer that problem (2) is Hyers–Ulam–Rassias stable. The proof is completed.


## 3.4. Example

Let us consider the following problem:(67)D0+ℍh,υ;gξt=13ξt+1+ξt9+et,∀t∈0,1/3,with the initial condition I0+1−γ,gξ0=0.For this example, we only consider the two situation as below. The other one is considered similarly.Situation 1.
Letgt=t, taking the limit υ⟶1 on both sides of (8). Then, by Sousa et al. [16], we have(68)D0+ℍh,υ;gξt=0+CDhξt,∀t∈0,1.
Combining (67) with (68), we infer that(69)D0+Chξt=13ξt+1+ξt9+et,∀t∈0,1/3,with the initial condition ξ0=0.
We considerh=1/2, λ=1/3, and(70)ζt,ξt≔1+ξt9+et,∀t∈0,1/3.
It is easy to see thatζ is a continuous function and it satisfies assumption (H1) with Lipschitz constant L=1/10. Indeed, for any ξ1,ξ∈C0,a,R, we have the following estimate:(71)ζt,ξ1t−ζt,ξ2t=1+ξ1t9+et−1+ξ2t9+et≤19+etξ1t−ξ2t≤110ξ1t−ξ2t,∀t∈0,1/3.
Moreover, we have(72)Bh,γga˜−g0hλ+LΓh=B1/2,1/21/3−01/21/3+1/10Γ1/2≈0.45<1.
We see that all the assumptions of Theorem2 is satisfied. So, we infer that problem (69) has a unique solution on 0,1/3.Situation 2.
Leth=2/3, υ=1/2, and gt=tρ, ρ>0, for any t∈0,1/3. Then, by Sousa et al. [16], we have(73)D0+ρ2/3,1/2ξt=13ξt+1+ξt9+et,∀t∈0,1/3,with the initial condition Iρ0+5/6ξ0=0.
Performing the same calculations as in Case 1, then it is also easy to check that problem (73) has a unique solution on 0,1/3.
We putϕt=t3ρ/2−1 for any t∈0,1/3. Now, we will prove that problem (73) is Hyers–Ulam–Rassias stable. For any t∈0,1/3, we have(74)1Γh∫0tQght,sϕsds=ρ1/3Γ2/3∫0ts−ρ/2tρ−sρ−1/3ϕsds=ρ1/3Γ2/3∫0ts−ρ/2tρ−sρ−1/3s3ρ/2−1ds=ρ1/3Γ2/3∫0ttρ−sρ−1/3sρ−1ds≤3ρ−2/32Γ2/3131−5ρ/6ϕt,∀t∈0,1/3.
Hence, assumption (73) of Theorem 3 is satisfied by Cϕ≔3ρ−2/3/2Γ2/31/31−5ρ/6>0 with ρ>0. All the assumptions of Theorem 3 are satisfied. So, we imply problem (73) is Hyers–Ulam–Rassias stable.



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