# Existence Results for Impulsive Fractional Integrodifferential Equations Involving Integral Boundary Conditions

Authors: Kulandhaivel Karthikeyan; Jiraporn Reunsumrit; Panjaiyan Karthikeyan; Sadhasivam Poornima; Dhatchinamoorthy Tamizharasan; Thanin Sitthiwirattham
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

This research paper is devoted to investigating the existence results for impulsive fractional integrodifferential equations in the form of Atangana - Baleanu - Caputo (ABC) fractional derivative, by using Gronwall–Bellman inequality and Krasnoselskii’s fixed point theorem to study the existence and uniqueness of the problem with integral boundary conditions. At the end, the examples are illustrated to verify results.


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## Body

## 1. Introduction

Fractional differential equations play a key role to describe various problems in different areas of science. Fractional models are more useful than the classical models. Fractional differential equations are used in economics, image processing, physics, and so on. For detailed information on fractional differential equations and their applications, see [1–9].A. Baleanu and D. Baleanu [10] introduces a nonsingular Caputo and Riemann–Liouville version of the fractional differential operator using the Mittag–Leffler function as its kernel. Bonyah et al. [11] developed a mathematical model for cancer and hepatitis coinfection diseases using the AB-fractional derivative. The AB-fractional derivative of the fractional-order tumor-immune-vitamin model was provided in [12] and discussed the existence, uniqueness, and Hyres–Ulam stability results also. Through the fractional tumor-immune dynamical model with AB-fractional derivative, researchers [13] created a chaotic and comparative work of tumour and effector cells. The fractional AB derivative was used to explore the numerical solution of the fractional immunogenetic tumour model in [14]. Like this, several applications were mentioned in reference [15–19].In [20], the authors deal with the transmission dynamics of the COVID-19 mathematical model under ABC-fractional-order derivative. With the AB-fractional derivative, Logeswari et al. [21] investigated the mathematical model for COVID-19 infection propagation over the world. They also developed a framework for generating numerical outputs in order to forecast the effect of the illness spreading across India. [22–24] are a few other key publications that sought to address the issue of various diseases modeled as FDEs involving AB-fractional derivative.In [25], Tidke introduced the following linear evolution equation:(1)dqxdtq=Atxt+ft,xt,t∈J=0,b,x0=x0,where 0<q<1, the unknown x. takes values in the Banach space X; f∈CJ×X,X, and At is a bounded linear operator on a Banach space X, and x0 is a given element of X. The operator Dq denotes the Caputo fractional derivative of order q and he has investigated about the existence, uniqueness, and other properties of solutions of fractional semilinear evolution equations in Banach spaces.In [26], the authors Kucche et al. dealt with the following nonlinear implicit fractional differential equations and investigated the existence and interval of existence of solutions, uniqueness, and properties of continuous dependance.(2)cDαxt=ft,tc,Dαt,0=x0∈ℛ,t∈0,T,where cDα0<α<1 denotes the Caputo fractional derivative and f∈0,T×ℛ×ℛ is a given continuous function.In [27], Guo et al. discussed the impulsive fractional differential equations with boundary value problems of the form:(3)DtαCxt=ft,t,t∈J′:Jt1,t2,…,tm,J=0,T,Δxtk=utk+−tk−=Iktk−,k=0,1,2,…,m,ax0+bxT=c,where DtαC is the Caputo fractional derivative of order α∈0,1 with the lower limit zero, f:J×ℝ⟶ℝ is jointly continuous and tk satisfy 0=t0<t1<…<tm<tm+1=T,xtk+=limϵ⟶0+xtk+ϵ, and xtk−=limϵ⟶0−xtk+ϵ represent the right and left limits of xt at t=tk.Ik∈Cℝ,ℝ, and a,b,c are real constants with a+b≠0.In [28], Yukunthorn et al. studied the impulsive Hadamard fractional differential equations with boundary value problems of the form:(4)DtkpkCxt=ft,xt,t∈Jk⊂t0,T,t=tk,Δxtk=φkxtk,k=1,2,…,m,αxt0+βxT=∑i=0mγiJtiqixti+1,where DtkpkC is the Hadamard fractional derivative of order 0<pk≤1 on intervals Jk=tk,tk+1,k=1,2,…,m, with J0=t0,t1,0<t1<t2<t3<…<tk<…<tm<tm+1=T which are the impulse points, J:=t0,T,f:J×ℝ⟶ℝ is a continuous function, and φk∈Cℝ,ℝ,Jtiqi is the Hadamard fractional integral of order qi>0,i=1,2,…,m. The jump conditions are defined by Δxtk=xtk+−xtk,xtk+=limϵ⟶0+xtk+ϵ,k=1,2,3,…,m.Motivated by the above works, we study multiderivative nonlinear impulsive FDEs including AB-fractional derivative (AB derivative) of the following form:(5)Dτα0∗τ=fτ,ωτ,Bωτ,τ∈J,(6)ωtk+=tk−+yk,yk∈ℝ,with integral boundary condition of the form:(7)ω0=∫0TT−σα−1Γαℋσ,ωσdσ,where J=0,T,T>0,0<α<1,Dτα indicates the ABR-fractional differential operator of order α and f∈CJ×ℝ×ℝ⟶ℝ,ℋ∈CJ×ℝ⟶ℝ are nonlinear functions. Where Bωτ=∫0τkτ,s,ωsds and k:Δ×0,T⟶ℝ,Δ=τ,s:0≤s≤τ≤T. 0=τ0<τ1<τ2<…<τm=1,Δω|τ=τk=ωτk+−ωτk−, and ωτk+=limh⟶0+ωτk+h and ωτk−=limh⟶0−ωτk+h represent the right and left hand limits of τt at τ=τk. The fixed point theorem of Krasnoselskii is used to prove the existence of a solution. The Gronwall–Bellman inequality and the properties of the fractional integral operator are used to prove that the solution is unique.The following is the outline of the paper: the required background for the development of the study is presented in Section2. Section 3 discusses the existence and uniqueness of impulsive fractional integrodifferential equations. Section 4 describes the examples.


## 2. Preliminaries

The ABC-fractional derivative and the generalized Mittag–Leffler function are defined and discussed in this section.Definition 1.
(see [29]). For p∈1,∞ and ω form an open subset of ℝ , then the Sobolev space Hpω can be defined as follows:(8)Hpω=f∈L2ω:Dβf∈L2ω,forallβ≤ω.Definition 2.
(see [10]). We assume x∈H10,1 and 0<α<1. Then, in Riemann–Liouville perspective, the left Atangana–Baleanu fractional derivative of ω of order α (ABR derivative) is defined by(9)Dτα=Bα1−αddσ∫0τEα−α1−ατ−σαωσdσ,where Bα>0 is a normalization function satisfying B0=B1=1 and E is one parameter Mittag–Leffler function.Definition 3.
(see [10]). Let ω∈H10,1 and 0<α<1. Then, the left Atangana–Baleanu fractional derivative of x of order α in Caputo sense is defined by(10)Dtα=Bα1−α∫0tEα−α1−ατ−σαω′σdσ.
A normalization function which conforms toB0=B1=1 is Bα>0, and the Mittag–Leffler function is E.Definition 4 (see [30, 31]).
A Mittag–Leffler generalized functionγα,βz for the complex α,β with Reα>0 can be defined as follows:(11)Eα,βγz=∑k=0∞γkΓαk+βzkk!,where γk is the Pochhammer symbol given by(12)γ0=1,γk=γγ+1…γ+k−1,k∈ℕ.
We note that(13)Eα,β1z=Eα,βz,Eα,11z=Eαz.
The following Laplace transformation results are required.Lemma 1 (see [10]).
SupposeLfτ;p=F¯p, then Dtαfτ;p =  Bα/1−αpαF¯p/pα+α/1−α.Lemma 2 (see [32]).
Ltkα+β−1Eα,βk±atα;p=k!pα−β/pα±ak+1, Ekt=dk/dtkt.Definition 5 (see [31, 33]).
Letρ,μ,ω,γ∈ℂReρ,Reμ>0,b>a. On a class La,b, the fractional integral operator Eρ,μ,ω;a+γ is defined by(14)Eρ,μ,ω;a+γϕτ=∫atτ−σγ−1Eρ,μγωτ−σρϕσdσ,τ∈a,b.Lemma 3 (see [31, 33]).
Letρ,μ,ω,γ∈ℂReρ,Reμ>0,b>a, then the operator Eρ,μ,ω;a+γ is bounded on Ca,b such that(15)Eρ,μ,ω;a+γϕτ≤Qϕ,where(16)Q=b−aReu∑k=0∞γkΓρk+μReρk+ρμωb−aReρkk!.Lemma 4 (see [31, 33]).
Letρ,μ,ω,γ∈ℂReρ,Reμ>0, then the operator Eρ,μ,ω;a+γ is invertible in the space ℒa,b and for f∈ℒa,b, its left inversion is given by the relation(17)Eρ,μ,ω;a+γ−1fτ=Da+μ+νEρ,μ,ω;a+−γfτ,a<τ≤b,where ν∈C,Reν>0 and Da+μ+ν denotes the Riemann–Liouville fractional differential operator of order μ+ν with a.Lemma 5 (see [31, 33]).
We assumeρ,μ,ω,γ∈ℂReρ,Reμ>0. Suppose(18)∫att−σγ−1Eρ,μγxt−σρϕσdσ=ft,a<t≤b,is solvable in the space La,b, then(19)ϕτ=Da+μ+νEρ,μ,ω;a+−γfτ,a<τ≤b,where ν∈C,Reν>0 and Da+μ+ν denotes the Riemann–Liouville fractional differential operator of order μ+ν with a , which is a unique solution.Lemma 6 (Krasnoselskii’s Fixed Point Theorem [6]).
Letω be a Banach space. Let S be a bounded, closed, convex subset of ω, and let ℱ1,ℱ2 be maps of S into ω such that ℱ1ω+ℱ2υ∈S for every pair ω,υ∈S. If ℱ1 is contraction and ℱ2 is completely continuous, then the equation(20)ℱ1ω+ℱ2υ=ω,has a solution on S.Lemma 7 (Gronwall–Bellman inequality [34]).
Letu and f be continuous and non-negative functions defined on J=0,T and let c be a non negative constant. Then, the inequality(21)uτ≤c+∫0τfσuσdσ,τ∈J,implies that(22)uτ≤cexp∫0τfσdσ,τ∈J.Lemma 8.
The functionω∈CJ is a solution of ABR-FDEs, for any function h∈CJ(23)Dτα0∗ωτ=hτ,τ∈J,(24)ωτk+=ωτk−+yk,yk∈ℝ,with integral boundary condition of the form(25)ω0=∫0TT−σα−1Γαℋσ,ωσdσ,if and only if x is a solution of fractional integral equation(26)ωτ=∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫0τ1Eα−α1−ατ−σαω′σdσ+∫0τhσdσ,forτ∈0,τ1,y1+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τ1τ2Eα−α1−ατ−σαω′σdσ,+∫0τhσdσ,forτ∈τ1,τ2,y1+y2+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τ2τ3Eα−α1−ατ−σαω′σdσ,+∫0τhσdσ,forτ∈τ2,τ3,⋮∑i=1myi+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τmTEα−α1−ατ−σαω′σdσ,+∫0τhσdσ,forτ∈τm,T.Proof.
We assume thatω satisfies (23) and (25). If τ∈0,τ1, then(27)Dτα0∗ωτ=hτ,(28)ω0=∫0TT−σα−1Γαℋσ,ωσdσ,ωτ=∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫0τ1Eα−α1−ατ−σαω′σdσ+∫0τhσdσ.
Ifτ∈τ1,τ2, then we have(29)Dτα0∗ωτ=hτ,ωτk+=ωτk−+yk,yk∈ℝ,and so(30)ωτ=ωτ1+−∫0τ1hσdσ+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τ1τ2Eα−α1−ατ−σαω′σdσ+∫0τhσdσ=ωτ1++y1−∫0τ1hσdσ+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τ1τ2Eα−α1−ατ−σαω′σdσ+∫0τhσdσ=y1+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τ1τ2Eα−α1−ατ−σαω′σdσ+∫0τhσdσ.
Ifτ∈τ2,τ3, then we find(31)ωτ=ωτ2+−∫0τ2hσdσ+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τ2τ3Eα−α1−ατ−σαω′σdσ+∫0τhσdσ=ωτ2++y2−∫0τ1hσdσ+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τ2τ3Eα−α1−ατ−σαω′σdσ+∫0τhσdσ=y1+y2+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τ2τ3Eα−α1−ατ−σαω′σdσ+∫0τhσdσ.
Let us consider the caseτ∈τm,T. Then, we conclude(32)ωτ=∑i=1myi+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫τmTEα−α1−ατ−σαω′σdσ+∫0τhσdσ.
Conversely, we assume thatω satisfies the impulsive equation (26). The fractional integral equation (26) to the ABR-FDEs (23) and (25) is Eρ,μ,ω;a+γ as given below:(33)ωτ=∑i=1myi+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−αEα,1,−α/1−α;0+1ωτ+∫0τhσdσ,τ∈J.Theorem 1.
The functionω∈CJ is a solution of ABR-FDEs (5) and (7), for any f∈CJ×ℛ,ℛ if and only if ω is a solution of fractional integral equations(34)ωτ=∑i=1myi+∫0TT−σα−1Γαℋσ,ωσdσ+Bα1−α∫0τEα−α1−ατ−σαωσdσ+∫0tfσ,ωσdσ,t∈J.Proof.
Proof follows by taking from Lemma8, hτ=fτ,ωτ,τ∈J.
The features of fractional integral operatorEρ,μ,ω;a+γ are used to prove the following theorem.Theorem 2.
We assume0<α<1. A function ℱ on CJ is defined by(35)ℱωτ=Bα1−αEα,1,−α/1−α;0+1ωτ,ω∈CJ,τ∈J.Then,(1)
ℱ is a CJ bounded linear operator(2)
the Lipschitz condition is satisfied byℱ(3)
IfS is any bounded subset of CJ, then ℱS is equicontinuous(4)
ℱ is invertible, and the operator equation ℱω=f has unique solution in CJ, for any f∈CJProof.
(i)
Since the integral operatorEα,1,−α/1−α;0+1 is a bounded and linear operator on CJ by the definition and Lemma 3, the equation becomes(36)Eα,1,−α/1−α;0+ω1≤Qω,τ∈J,where we find(37)Q=∑k=0∞1kΓαk+1αk+1−α/1−αTαkk!=∑k=0∞α/1−αTαkΓαk+2=Eα,2α1−αTα+1.We have(38)ℱω=πBα1−αEα,1,−α/1−α;0+ω1≤QBα1−αω,for allω∈CJ.As a result, the linear operatorℱ on CJ is bounded.(ii)
For someω,σ∈CJ . Then, for some τ∈J, we find utilising the linearity of ℱ and the boundedness of operator Eα,1,−α/1−α;0+1, we have(39)ℱωτ−ℱστ=ℱω−ℱστ=Bα1−αEα,1,−α/1−α;0+1ω−στ≤Bα1−αEα,1,−α/1−α;0+1ω−σ≤QBα1−αω−σ.This gives(40)ℱω−ℱσ≤QBα1−αω−σ,ω,σ∈CJ.With Lipschitz constantTBα/1−αEα,2α/1−αTα, the operator ℱ satisfies Lipschitz condition.(iii)
Any closed, bounded subset ofCJ is S=ω∈CJ:ω≤R. Then, for every ω∈S and any τ1,τ2∈J with τ1<τ2, we get(41)ℱωτ1−ℱστ2=Bα1−αEα,1,−α/1−α;0+1ωτ1−Bα1−αEα,1,−α/1−α;0+1ωτ2=Bα1−α∫0τEα−α1−ατ1−σαωσdσ−∫0τEα−α1−ατ2−σαωσdσ≤Bα1−α∫0τEα−α1−ατ1−σα−∫0τEα−α1−ατ2−σαωσdσ+Bα1−α∫τ1τ2Eα−α1−ατ2−σαωσdσ≤Bα1−α∑k=0∞−α1−αk1Γαk+1∫0τ1τ1−αkα−τ2−αkαωσdσ+Bα1−α∑k=0∞−α1−αk1Γαk+1∫τ1τ2τ2−αkαωσdσ≤RBα1−α∑k=0∞α1−αk1Γαk+1∫0τ1τ2−αkα−τ1−αkαωσdσ+RBα1−α∑k=0∞α1−αk1Γαk+1∫τ1τ2τ2−αkαωσdσ≤RBα1−α∑k=0∞α1−αk1Γαk+2−τ2−τ1kα+1+τ2kα+1−τ1kα+1+τ2−τ1kα+1≤RBα1−α∑k=0∞α1−αk1Γαk+2τ2kα+1−τ1kα+1.From above inequality, it follows that, ifτ1−τ2⟶0, then ℱωτ1−ℱστ2⟶0. This establishes the equicontinuous nature of ℱS on J.(iv)
For anyf∈CJ, we obtain the below by using Lemmas 4 and 5.(42)Eα,1,−α/1−α;0+1f−1τ=D0+1+βEα,1,−α/1−α;0+1f−1τ,τ∈a,b,where β∈ℂ, with Reβ>0.
Then, using definition of operatorℱ and equation (64), we have(43)ℱ−1fτ=Bα1−αEα,1,−α/1−α;0+1f−1τ=1−αBαD0+1+βEα,1,−α/1−α;0+1f−1τ,τ∈a,b.
This demonstrates that the operator equationℱ is invertible on CJ,(44)ℱωτ=fτ,τ∈J,has the unique solution(45)ωτ=1−αBαD0+1+βEα,1,−α/1−α;0+1fτ,τ∈a,b.
This completes the proof.Theorem 3.
Supposef∈CJ×ℛ,ℛ, then the ABR-FDEs Dτα=fτ,ωτ, τ∈J can be solvable on CJ and has a solution given by(46)ωτ=1−αBαD0+1+βEα,1,−α/1−α;0+1f¯τ,τ∈J,where β∈ℂ with Reβ>0 and ∫0τfσ,ωσdσ,τ∈J.


## 3. Main Results

Theorem 4 (Existence theorem).
Supposef∈CJ×ℛ×ℛ,ℛ and the Lipschitz condition ℋ∈CJ×ℛ is satisfied.(1)
fτ,ω,κ1−fτ,υ,κ2≤pτω−υ+κ1−κ2,ω,υ,κ1,κ2∈CJ,(2)
ℋτ,ω−ℋτ,υ≤Lℋω−υ,ω,υ∈CJ,(3)
ℋτ,ω≤Cℋω+Mℋ,ω∈CJ,where p:J⟶ℝ+, with L=suppτ and for the real constants Lℋ,Mℋ,Cℋ>0. If 0<L<min1,1/2T, then the ABR-FDEs (5) and (7) has a solution in CJ provided(47)TαΓα+1Cℋ+BαEα,2α/1−αTα+11−α<1.Proof.
Assume that(48)R=Tα/Γα+1Mℋ+MfT+M∗1−LT−Tα/Γα+1Cℋ−BαEα,2α/1−αTα+1/1−α,where Mf=supfτ,0,0 and M∗>0 is a constant such that ∑i=1myi≤M∗. We have R>0 because of the choice of L and condition (47).
Let us consider the set(49)S=ω∈CJ:ω≤R.
It is possible to prove thatS is a closed, convex, and bounded subset of ω. We consider the operators ℱ1:S⟶ω and ℱ2:S⟶ω, both of which are defined by(50)ℱ1ωτ=∑i=1myi+∫0TT−σα−1Γαℋσ,ωσdσ+∫0τfσ,ωσ,Bωσdσ,τ∈J,ℱ2ωτ=Bα1−αEα,1,−α/1−α;0+1ωτ=ℱωτ,τ∈J.
The operator equation for the equivalent fractional integral equation (55) to the ABR–FDEs (5) and (7) is as follows:(51)ω=ℱ1ω+ℱ2ω,ω∈PCJ.
We show that the operatorsℱ1 and ℱ2 satisfy Lemma 6 requirements. The following steps have been used to demonstrate the same.Step 1.
ℱ1 is contraction.
For everyω,υ∈PCJ and τ∈J, we obtain using the Lipschitz condition on f,(52)ℱτ,ωτ,Bωσ−ℱτ,υτ,Bυτ≤pτω−υ.
This gives(53)ℱ1ω−ℱ1υ≤Lℋ+LTω−υ,ω,υ∈PCJ.Step 2.
Next, we show thatℱ2 is completely continuous.
One can readily demonstrate that the operatorℱ2 is completely continuous using the Ascoli–Arzela theorem and Theorem 2.Step 3.
ℱ1ω+ℱ2υ∈S, for any ω,υ∈S.
For anyω,υ∈S using Theorem 2, we obtain(54)ℱ1ω+ℱ2υτ≤ℱ1ωτ+ℱ2υτ≤∫0TT−σα−1Γvℋσ,ωσdσ+∑i=1myi+∫0τfσ,ωσ,Bωσdσ+Bα1−αEα,2α1−αTα+1υ≤∫0TT−σα−1Γαℋσ,ωσdσ+M∗+∫0τfσ,ωσ,Bωσ−fσ,0,0|dσ+∫0τfσ,0,0dσ+Bα1−αEα,2α1−αTα+1R≤TαΓα+1Cℋω+Mℋ+M∗+L∫0τfωσ,Bωσ+Mf∫0τdσ+Bα1−αEα,2α1−αTα+1R≤TαΓα+1CℋR+Mℋ+M∗+LRτ+Mfτ+Bα1−αEα,2α1−αTα+1R≤TαΓα+1CℋR+Mℋ+M∗+LRT+MfT+Bα1−αEα,2α1−αTα+1R.i.e., by condition (47) with (48), we get(55)TαΓα+1Mℋ+MfT+M∗=R1−LT−TαΓα+1Cℋ−BαEα,2α/1−αTα+11−α.From inequalities (54) and (55), we have(56)ℱ1ω+ℱ2υτ≤R,τ∈J.
This gives(57)ℱ1ω+ℱ2υ≤R,forallω,υ∈S.
This shows thatℱ1ω+ℱ2υ∈S, for ω,υ∈S.
Then, the operator equation,(58)ω=ℱ1ω+ℱ2ω,has a fixed point in S, which is the solution of ABR–FDEs (5) and (7). This completes the proof.
The uniqueness of solutions to ABR-FDEs (5) and (7) is demonstrated in two methods in the following theorem. We prove the result first using the properties of the fractional integral operator Eα,1,−α/1−α;0+1, and then using the Gronwall–Bellman inequality.Theorem 5 (Uniqueness result).
The ABR-FDEs (5) and (7) has a unique solution in CJ under the conditions of Theorem 4.Proof.
In operator equation form, the analogous fractional integral equation to ABR-FDEs (5) and (7) is(59)Eα,1,−α/1−α;0+1ωτ=f˜τ,τ∈J,where(60)f˜τ=1−αBα∫0TT−σα−1Γαℋσ,ωσdσ−ωτ+∫0τfσ,ωσ,Bωσdσ+∑i=1myi,τ∈J.
The operator (59) is solvable in CJ according to Theorem 4. As a result of applying Lemma 5, the operator equation equation (59) has a unique solution in CJ , which is the ABR-FDEs (5) and (7).Proof.
Letω and υ be two ABR-FDEs (5) solutions (7). We derive for any τ∈J using the linearity of the fractional integral operator,(61)ωτ−υτ=∫0TT−σα−1Γαℋσ,ωσdσ−Bα1−αEα,1,−α/1−α;0+1ωτ+∫0τfσ,ωσ,Bωσdσ−∫0TT−σα−1Γαℋσ,υσdσ−Bα1−αEα,1,−α/1−α;0+1υτ+∫0τfσ,υσ,Bσdσ≤TαΓα+1Lℋωσ−υσ+Bα1−α∫0τEα−α1−αT−σαωσ−υσdσ+∫0τpl1σωσ−υσσ≤TαΓα+1Lℋωσ−υσ+Bα1−α∫0τEαα1−αTαωσ−υσdσ+∫0τpl1σωσ−υσdσ≤∫0τT−σα−1ΓαLℋ+Bα1−αEαα1−αTα+pl1σωσ−υσdσ.
Applying Lemma7, we get(62)ωτ−υτ≤0,τ∈J,which shows that ωτ=υτ, for every τ∈J. This demonstrates that the ABR-FDEs (5) solution is unique (7).


## 4. Examples

(1)
We consider the following fractional boundary value problem:(63)Dτ1/20∗ωτ=t3+sinωt45+110∫0Tt2sinsωsds,t∈0,1,(64)Δωt=ω1/2−3+ω1/2−,with integral boundary condition of the form:(65)ω0=∫011−σα−1Γα125exp−ωσdσ,where(66)fτ,ωτ,Bωτ=t3+sinωt45+110∫0Tt2sinsωsds,ℋσ,ωσ=125exp−ωσ.AsT=1,α=1/2 , let ω,υ∈PCJ(67)fτ,ωτ,Bωτ−fτ,υτ,Bυτ≤|t3+sinωt45+110∫0Tt2sinsωsds−t3+sinυt45−110∫0Tt2sinsυsds≤145+1−cosT10ωt−υtℋσ,ωσ−ℋσ,υσ≤125exp−ωσ−125exp−υσ≤125ωt−υtℋσ,ωσ≤125exp−ωσ≤125ωt+125.Thus, we haveL=1/4.5,Cℋ=1/25,Mℋ=1/25,Mf=1/45,M∗=1/3 and consider k=1,(68)TαΓα+1Cℋ+BαEα,2α/1−αTα+11−α=121.5+11.44=0.740<1,(69)R=Tα/Γα+1Mℋ+MfT+M∗1−LT−Tα/Γα+1Cℋ−BαEα,2α/1−αTα+1/1−α=0.4010.036=11.13.Thus, the condition of Theorem4 is satisfied.(2)
We consider the following example for fractional impulsive integrodifferential equations of the form:(70)Dτ1/20∗ωτ=cosωt35+150∫0tt2+s2ωsds,t∈0,1,Δωt=ω1/2−3+ω1/2−,with integral boundary condition(71)ω0=∫01130σ2ωσdσ,where(72)fτ,ωτ,Bωτ=cosωt35+150∫0tt2+s2ωsds,ℋσ,ωσ=130σ2ωσ.AsT=1,α=1/2, let ω,υ∈PCJ(73)fτ,ωτ,Bωτ−fτ,υτ,Bυτ≤|cosωt35+150∫0tt2+s2ωsds−cosυt35−150∫0tt2+s2υsds|≤135+275ωt−υtℋσ,ωσ−ℋσ,υσ≤|130σ2ωσ−130σ2υσ|≤130ωt−υtℋσ,ωσ≤130σ2ωσ≤130ωt+130.Thus, we haveL=0.054,Cℋ=1/30,Mℋ=1/30,Mf=1/35,M∗=1/3 and consider k=1, (74)TαΓα+1Cℋ+BαEα,2α/1−αTα+11−α=0.728<1,R=Tα/Γα+1Mℋ+MfT+M∗1−LT−Tα/Γα+1Cℋ−Bαα,2α/1−αTα+1/1−α=1.835.Thus, the condition of Theorem4 is satisfied.


## 5. Conclusion

In this paper, we examined the impulsive fractional integrodifferential equations involving ABC derivative with integral boundary conditions. Recently ABC-derivative gained much attention due to the nonsingular property of the kernels. The existence of solution is investigated for the proposed equations by using Krasnoselskii fixed point theorem. The uniqueness of the result is derived with the help of Gronwall–Bellman inequality as well as the properties of fractional integral operator. In future, we extend this work with the delay properties involving Mittag–Leffler function.



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*Source: 6599849-2022-09-05.xml*