# Boundary Value Problem for Nonlinear Implicit Generalized Hilfer-Type Fractional Differential Equations with Impulses

Authors: Abdelkrim Salim; Mouffak Benchohra; Jamal Eddine Lazreg; Gaston N’Guérékata
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

This article deals with some existence, uniqueness, and Ulam-Hyers-Rassias stability results for a class of boundary value problem for nonlinear implicit fractional differential equations with impulses and generalized Hilfer Fractional derivative. The results are obtained using the Banach contraction principle and Krasnoselskii’s and Schaefer’s fixed-point theorems.


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## Body

## 1. Introduction

Differential equations of fractional order have been recently proved to be a powerful tool to study many phenomena in various fields of science and engineering such as electrochemistry, finance, hydrology, electromagnetics, and viscoelasticity. There are numerous books and articles focused on linear and nonlinear initial and boundary value problems for fractional differential equations involving different kinds of fractional derivatives, see, for example, [1–6]. Impulsive fractional differential equations have been considered by many authors (see, for instance, [7–12]). Recent results involving different fractional derivatives can be found in [13–21] and the references therein.Ulam was the first who raise the concept of stability of functional equations [22]. In 1941, Hyers [23] provided the first answer to Ulam’s question. Thereafter, this type of stability is called the Ulam-Hyers stability. In 1978, Rassias [24] was able to make a remarkable generalization of Ulam-Hyers stability of mappings by considering variables.Considerable attention has been given to the study of the Ulam-Hyers and Ulam-Hyers-Rassias stability of all kinds of functional equations; one can see the monograph of Abbas et al. [3] and the paper by Rus [25] who discussed the Ulam-Hyers stability for operational equations (see also [26–29]).Recently, in [30], Harikrishnan et al. investigated existence theory and different kinds of stability in the sense of Ulam, for the following initial value problem with nonlinear generalized Hilfer-type fractional differential equation and impulses:
(1)ρDα,βut=ft,ut,t∈I¯≔I\t1,⋯,tm,I≔0,b,ΔρJ1−γutt=tk=Lkutk−,k=1,⋯,m,ρJ1−γu0=u0,where ρDα,βand ρJ1−γ are a generalized Hilfer fractional derivative of order α∈0,1 and type β∈0,1 and generalized fractional integral of order 1−γγ=α+β−αβ, respectively; 0=t0<t1<⋯<tm<tm+1=b<∞, utk+=limε→0+utk+ε, and utk−=limε→0−utk+ε represent the right and left hand limits of ut at t=tk; ΔρJ1−γutt=tk=ρJ1−γutk+−ρJ1−γutk−; f:I×ℝ→ℝ is a given function; and Lk:ℝ→ℝ;k=1,⋯,m are given continuous functions.Motivated by the works mentioned above, in this paper, we establish existence and uniqueness results to the boundary value problem with nonlinear implicit generalized Hilfer-type fractional differential equation and impulses:(2)ρDtk+α,βut=ft,ut,ρDtk+α,βut,t∈Jk,k=0,⋯,m,(3)ρJtk+1−γutk+=ρJtk−1+1−γutk−+Lkutk−,k=1,⋯,m,(4)c1ρJa+1−γua++c2ρJtm+1−γub=c3,where ρDtk+α,β,ρJtk+1−γ are a generalized Hilfer fractional derivative of order α∈0,1 and type β∈0,1 and generalized fractional integral of order 1−γγ=α+β−αβ, respectively; c1,c2,c3 are real with c1+c2≠0; Jk≔tk,tk+1,k=0,⋯,m; a=t0<t1<⋯<tm<tm+1=b<∞; utk+=limϵ→0+utk+ϵ and limϵ→0−utk+ϵ represent the right and left hand limits of ut at t=tk; f:a,b×ℝ×ℝ→ℝ is a given function; and Lk:ℝ→ℝ,k=1,⋯,m, are given continuous functions.The present paper is organized as follows. In Section2, some notations are introduced and we recall some preliminaries about generalized Hilfer fractional derivative and auxiliary results. In Section 3, three results for problems (2)–(4) are presented which are based on the Banach contraction principle and Krasnoselskii’s and Schaefer’s fixed-point theorems. In Section 4, we discuss the Ulam-Hyers-Rassias stability for problems (2)–(4). Finally, we give examples to illustrate the applicability of our main results.


## 2. Preliminaries

In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper. Let0<a<b,J=a,b. By C, we denote the Banach space of all continuous functions from J into ℝ with the norm
(5)u∞=suput:t∈J.We consider the weighted spaces of continuous functions(6)Cγ,ρJ=u:a,b→ℝ:tρ−aρρ1−γut∈CJ,ℝ,0≤γ<1,Cγ,ρnJ=u∈Cn−1J:un∈Cγ,ρJ,n∈ℕ,Cγ,ρ0J=Cγ,ρJ,with the norms
(7)uCγ,ρ=supt∈Jtρ−aρρ1−γut,uCγ,ρn=∑k=0n−1uk∞+unCγ,ρ.Consider the Banach spacePCJ=u:a,b→ℝ:ut∈CJk,k=0,⋯,m, and there exist utk− and ρJtk+1−γutk+,k=0,⋯,m, with utk−=utk}, 0≤γ<1.Also, we consider the weighted space(8)PCγ,ρJ=ut;tρ−tkρρ1−γut∈PCJ,0≥γ<1,PCγ,ρnJ=u∈PCn−1:un∈PCγ,ρJ,n∈ℕ,PCγ,ρ0J=PCγ,ρJ,with the norm
(9)uPCγ,ρ=supt∈Jktρ−tkρρ1−γut,k=0,⋯,m.Consider the spaceXcpa,bc∈ℝ,1≤p≤∞ of those complex-valued Lebesgue measurable functions f on a,b for which fXcp<∞, where the norm is defined by
(10)fXcp=∫abtcftpdtt1/p1≤p<∞,c∈ℝ.In particular, whenc=1/p, the space Xcpa,b coincides with the Lpa,b space: X1/ppa,b=Lpa,b.Definition 1 [31].
Generalized Hilfer fractional integral.
Letα∈ℝ+,c∈ℝ, and g∈Xcpa,b. A generalized Hilfer fractional integral of order α is defined by
(11)ρJa+α+gt=∫atsρ−1tρ−sρρα−1gsΓαds,t>a,ρ>0,where Γ· is the Euler gamma function defined by Γα=∫0∞tα−1e−tdt,α>0.Definition 2 [31].
Generalized Hilfer fractional derivative.
Letα∈ℝ+\ℕ and ρ>0. A generalized Hilfer fractional derivative ρDa+α of order α is defined by
(12)ρDa+α+gt=δρnρJa+n−αgt=t1−ρddtn∫atsρ−1tρ−sρρn−α−1gsΓn−αds,t>a,ρ>0,where n=α+1 and δρn=t1−ρd/dtn.Theorem 3 [31].
Letα>0,β>0, 1≤p≤∞, and 0<a<b<∞. Then, for g∈Xcpa,b, the semigroup property is valid, i.e.,
(13)ρJa+α+ρJa+βgt=ρJa+α+βgt.Lemma 4 [31, 32].
Letα>0, and 0≤γ<1. Then, ρJa+α is bounded from PCγ,ρJ into PCγ,ρJLemma 5 [32].
Let0<a<b<∞, α>0, 0≤γ<1, and u∈PCγ,ρJ. If α>1−γ, then ρJa+α is continuous on J and
(14)ρJa+αua=limt→a+ρJa+αut=0.Lemma 6 [33].
Lett>a. Then, for α≥0 and β>0, we have
(15)ρJa+αsρ−aρρβ−1t=ΓβΓα+βtρ−aρρα+β−1,ρDa+αsρ−aρρα−1t=0,0<α<1.Lemma 7 [32].
Letα>0, 0≤γ<1, and g∈PCγa,b. Then,
(16)ρDa+α+ρJa+αgt=gt,forallt∈a,b.Lemma 8 [32].
Let0<α<1, 0≤γ<1. If g∈PCγ,ρa,b and ρJa+1−αg∈PCγ,ρ1a,b, then
(17)ρJa+αρDa+αgt=gt−ρJa+1−αgaΓαtρ−aρρα−1,forallt∈a,b.Definition 9 [32].
Let orderα and type β satisfy n−1<α<n and 0≤β≤1, with n∈ℕ. A generalized Hilfer fractional derivative to t, with ρ>0 of a function g∈PCγ,ρa,b, is defined by
(18)ρDa+α,βgt=ρJa+βn−αtρ−1ddtnρJa+1−βn−αgt=ρJa+βn−αδρnρJa+1−βn−αgt.In this paper, we consider the casen=1 only, because 0<α<1.Property 10 [32].
The operatorρDa+α,β can be written as
(19)ρDa+α,β=ρJa+β1−αδρρJa+1−γ=ρJa+β1−αρDa+γ,γ=α+β−αβ.Property 11 [32].
The fractional derivativeρDa+α,β is an interpolator of the following fractional derivatives: Hilfer ρ→1, Hilfer–Hadamard ρ→0+, generalized β=0, Caputo–type β=1, Riemann–Liouville β=0,ρ→1, Hadamard β=0,ρ→0+, Caputo β=1,ρ→1, Caputo–Hadamard β=1,ρ→0+, Liouville β=0,ρ→1,a=0, and Weyl β=0,ρ→1,a=−∞.
Consider the following parametersα,β,γ satisfying
(20)γ=α+β−αβ,0<α,β,γ<1.We define the spaces(21)PCγ,ρα,βJ=u∈PCγ,ρJ,ρDtk+α,βu∈PCγ,ρJ,PCγ,ργJ=u∈PCγ,ρJ,ρDtk+γu∈PCγ,ρJ,where k=0,⋯,m.SinceρDtk+α,βu=ρJtk+γ1−αρDtk+γu, it follows from Lemma 4 that
(22)PCγ,ργJ⊂PCγ,ρα,βJ⊂PCγ,ρJ.Lemma 12 [32].
Let0<α<1,0≤β≤1, and γ=α+β−αβ. If u∈PCγ,ργJ, then
(23)ρJa+γρDa+γu=ρJa+αρDa+α,βu,ρDa+γρJa+αu=ρDa+β1−αu.Lemma 13 (Theorem25 [32]).
Letf:a,b×ℝ→ℝ be a function such that f⋅,u⋅,ρDtk+α,βu⋅∈Cγ,ρJ, for any u∈PCγ,ρJ. Then, u∈PCγ,ργJ is a solution of the differential equation:
(24)ρDtk+α,βut=ft,ut,ρDtk+α,βut,foreacht∈Jk,k=0,⋯,m,0<α<1,0≤β≤1,if and only if u satisfies the following Volterra integral equation:
(25)ut=ρJtk+1−γutk+Γγtρ−tkρργ−1+1Γα∫tkttρ−sρρα−1sρ−1fs,us,ρDtk+α,βusds,where γ=α+β−αβ.Theorem 14 ([34]) (Banach’s fixed-point theorem).
LetD be a nonempty closed subset of a Banach space E; then, any contraction mapping N of D into itself has a unique fixed point.Theorem 15 ([34]) (Schaefer’s fixed-point theorem).
LetE be a Banach space and N:E→E be a completely continuous operator. If the set
(26)D=u∈E:u=λNu,forsomeλ∈0,1,is bounded, then N has a fixed point.Theorem 16 ([34]) (Krasnoselskii’s fixed-point theorem).
LetD be a closed, convex, and nonempty subset of a Banach space E and the operators A,B such that
(1)
Ax+By∈D, for all x,y∈D(2)
A is compact and continuous(3)
B is a contraction mappingThen, there existsz∈D such that z=Az+Bz.Now, we consider the Ulam stability for problems ((2))–((4)) that will be used in Section 4. Let u∈PCγ,ρJ, ϵ>0, τ>0, and ϑ:a,b⟶0,∞ be a continuous function.We consider the following inequality:(27)ρDtk+α,βut−ft,ut,ρDtk+α,βut≤ϵϑt,t∈Jk,k=0,⋯,m,ρJtk+1−γutk+−ρJtk−1+1−γutk−−Lkutk−≤ϵτ,k=1,⋯,m.Definition 17 ([35]).
Problems ((2))–((4)) are Ulam-Hyers-Rassias (U-H-R) stable with respect to ϑ,τ if there exists a real number af,m,ϑ>0 such that for each ϵ>0 and for each solution u∈PCγ,ρJ of inequality (27), there exists a solution w∈PCγ,ρJ of (1)–(3) with
(28)ut−wt≤ϵaf,m,ϑϑt+τ,t∈a,b.Remark 18 ([35]).
A functionu∈PCγ,ρJ is a solution of inequality (27) if and only if there exist σ∈PCγ,ρJ and a sequence σk,k=0,⋯,m, such that
(1)
σt≤ϵϑt and σkϵτ,t∈Jk,k=1,⋯,m(2)
ρDtk+α,βut=ft,ut,ρDtk+α,βut+σt,t∈Jk,k=0,⋯,m(3)
ρJtk+1−γutk+=ρJtk−1+1−γutk−+Lkutk−+σk,k=1,⋯,m


## 3. Existence of Solutions

We consider the following linear fractional differential equation:(29)ρDtk+α,βut=ψt,t∈Jk,k=0,⋯,m,where 0<α<1, 0≤β≤1, and ρ>0, with the conditions
(30)ρJtk+1−γutk+=ρJtk−1+1−γutk−+Lkutk−,k=1,⋯,m,and
(31)c1ρJa+1−γua++c2ρJtm+1−γub=c3,where γ=α+β−αβ and c1,c2,c3∈ℝ with
(32)c1+c2≠0,ξ1=c2c1+c2,ξ2=c3c1+c2,and
(33)p∗=suptkρ−tk−1ρργ−1:k=1,⋯,m.The following theorem shows that problems ((29))–((31)) have a unique solution given by
(34)ut=1Γγtρ−aρργ−1ξ2−ξ1∑i=1mLiuti−−ξ1∑i=1mρJti−11−γ+αti−ξ1ρJtm+1−γ+αψb+1Γα∫attρ−sρρα−1sρ−1ψsdsift∈J0,1Γγtρ−tkρργ−1ξ2−ξ1∑i=1mLiuti−−ξ1∑i=1mρJti−11−γ+αψti−ξ1ρJtm+1−γ+αψb+∑i=1kLiuti−+∑i=1kLiρJti−11−γ+αψti+ρJtk+αψtift∈Jk,k=1,⋯,m.Theorem 19.
Letγ=α+β−αβ, where 0<α<1 and 0≤β≤1. If ψ:a,b→ℝ is a function such that ψ·∈Cγ,ρJ, then u∈PCγ,ργJ satisfies problems ((29))–((31)) if and only if it satisfies (34).Proof.
Assumeu satisfies (29)–(31). If t∈J0, then
(35)ρDa+α,βut=ψt.Lemma 20.
implies that(36)ut=ρJa+1−γua+Γγtρ−aρργ−1+1Γα∫attρ−sρρα−1sρ−1ψsds.Ift∈J1, then Lemma 13 implies
(37)ut=ρJt1+1−γut1+Γγtρ−t1ρργ−1+1Γα∫t1tsρ−sρρα−1sρ−1ψsds=ρJa+1−γut1−+L1ut1−Γγtρ−t1ρργ−1+ρJt1+αψt=tρ−t1ργ−1Γγργ−1ρJa+1−γua++L1ut1−+ρJa+1−γ+αψti+ρJt1+αψt.Ift∈J2, then Lemma 13 implies
(38)ut=ρJt2+1−γut2+Γγtρ−t2ρργ−1+1Γα∫t2ttρ−sρρα−1sρ−1ψsds=ρJt1+1−γut2−L2ut2−Γγtρ−t2ρργ−1+ρJt2+αψt=1Γγtρ−t2ρργ−1ρJa+1−γua++L1ut1−+L2ut2−+ρJa+1−γ+αψt1+ρJt1+1−γ+αψt2+ρJt2+αψt.Repeating the process in this way, the solutionut for t∈Jk, k=1,⋯,m, can be written as
(39)ut=1Γγtρ−tkρργ−1ρJa+1−γua++∑i=1kLiuti−+∑i=1kρJti−1+1−γ+αψti+ρJtk+αψt.ApplyingρJtm+1−γ on both sides of (39), using Lemma 6 and taking t=b, we obtain
(40)ρJtm+1−γub=ρJa+1−γua++∑i=1mLiuti−+∑i=1mρJti−1+1−γ+αψti+ρJtm+1−γ+αψb.Multiplying both sides of (40) by c2 and using condition (31), we obtain
(41)c3−c1ρJa+1−γua+=c2ρJa+1−γua++c2∑i=1mLiuti−+c2∑i=1mρJti−1+1−γ+αψti+c2ρJtm+1−γ+αψb,which implies that
(42)ρJa+1−γua+=ξ2−ξ1∑i=1mLiuti−−ξ1∑i=1mρJti−1+1−γ+αψti−ξ1ρJtm+1−γ+αψb.Substituting (42) into (39) and (36), we obtain (34).Reciprocally, applyingρJtk+1−γ on both sides of (34) and using Lemma 6 and Theorem 3, we get
(43)ρJtk+1−γut=ξ2−ξ1∑i=1mLiuti−−ξ1∑i=1mρJti−1+1−γ+αψti−ξ1ρJtm+1−γ+αψb+ρJa+1−γ+αψtift∈J0,ξ2−ξ1∑i=1mLiuti−−ξ1∑i=1mρJti−1+1−γ+αψti−ξ1ρJtm+1−γ+αψb+∑i=1kLiuti−+∑i=1kρJti−1+1−γ+αψti+ρJtk+1−γ+αψtift∈Jk,k=1,⋯,m.Next, taking the limitt→a+ of (43) and using Lemma 5, with 1−γ<1−γ+α, we obtain
(44)ρJa+1−γua+=ξ2−ξ1∑i=1mLiuti−−ξ1∑i=1mρJti−1+1−γ+αψti−ξ1ρJtm+1−γ+αψb.Now, takingt=b in (43), we get
(45)ρJtm+1−γub=ξ2+1−ξ1∑i=1mLiuti−+∑i=1mρJti−1+1−γ+αψti+ρJtm+1−γ+αψb.From (44) and (45), we find that
(46)c1ρJa+1−γua++c2ρJtm+1−γub=c3,which shows that the boundary condition c1ρJa+1−γua++c2ρJtm+1−γub=c3 is satisfied. Next, apply operator ρDtk+γ on both sides of (34), where k=0,⋯,m. Then, from Lemma 6 and Lemma 12, we obtain
(47)ρDtk+γut=ρDtk+β1−αψt.Sinceu∈PCγ,ργJ and by definition of PCγ,ργJ, we have ρDtk+γu∈PCγ,ρJ, then (47) implies that
(48)ρDtk+γut=δρρJtk+1−β1−αψt=ρDtk+β1−αψt∈PCγ,ρJ.Asψ·∈Cγ,ρJ and from Lemma 4, it follows that
(49)ρJtk+1−β1−αψ∈PCγ,ρJ.From (48) and (49) and by the definition of the space PCγ,ρnJ., we obtain
(50)ρJtk+1−β1−αψ∈PCγ,ρ1J.Applying operatorρJtk+β1−α on both sides of (47) and using Lemma 8, Lemma 5, and Property 10, we have
(51)ρDtk+α,βut=ρJtk+β1−αρDtk+γut=ψt−ρJtk+1−β1−αψtkΓβ1−αtρ−tkρρβ1−α−1=ψt,that is, (29) holds. Also, we can easily show that
(52)ρJtk+1−γutk+=ρJtk−1+1−γutk−+Lkutk−,k=1,⋯,m.This completes the proof.As a consequence of Theorem19, we have the following result.Lemma 21.
Letγ=α+β−αβ, where 0<α<1 and 0≤β≤1; let f:a,b×ℝ×ℝ→ℝ be a function such that f·,u·,w·∈Cγ,ρJ for any u,w∈PCγ,ρJ. If u∈PCγ,ργJ, then u satisfies problems ((2))–((4)) if and only if u is the fixed point of the operator Ψ:PCγ,ρJ→PCγ,ρJ defined by
(53)Ψut=1Γγtρ−tkρργ−1ξ2−ξ1∑i=1mLiuti−−ξ1∑i=1mρJti−1+1−γ+αhti−ξ1ρJtm+1−γ+αhb+∑a<tk<tLkuti−+∑a<tk<tρJtk−1+1−γ+αhtk+ρJtk+αht,t∈Jk,k=0,⋯,m,where h:a,b→ℝ is a function satisfying the functional equation
(54)ht=ft,ut,ht.Assume that the functionf:a,b×ℝ×ℝ→ℝ is continuous and satisfies the following conditions:(H1). The functionf:a,b×ℝ×ℝ→ℝ be such that
(55)f⋅,u⋅,w⋅∈PCγ,ρβ1−α,foranyu,w∈PCγ,ρJ.(H2). There exist constantsK>0 and 0<M<1 such that
(56)ft,u,w−ft,u¯,w¯≤Ku−u¯+Mw−w¯,for any u,w,u¯,w¯∈ℝ and t∈a,b.(H3). There exists a constantl∗>0 such that
(57)Lku−Lku¯≤l∗u−u¯,for any u,u¯∈ℝ and k=1,⋯,m.(H4). There exist functionsp1,p2,p3∈CJ,ℝ+ with
(58)p1∗=supt∈Jp1t,p2∗=supt∈Jp2t,p3∗=supt∈Jp3t<1,such that
(59)ft,u,w≤p1t+p2tu+p3tw,fort∈a,b,u,w∈ℝ.(H5). The functionsLk:ℝ→ℝ are continuous, and there exist constants Φ1,Φ2>0 such that
(60)Lku≤Φ1uΦ2,foreachu∈ℝ,k=1,⋯,m.We are now in a position to state and prove our existence result for problems ((2))–((4)) based on Banach’s fixed point.Theorem 22.
Assume (H1)–(H3) hold. If(61)L≔ξ1+1ml∗p∗Γγ+mK1−MΓ1−αbρ−aρρα+K1−Mξ1Γ1−α+ΓγΓγ+αbρ−aρρα<1,then problems ((2))–((4)) have a unique solution in PCγ,ργJ⊂PCγ,ρα,βJ.Proof.
The proof will be given in two steps.Step 1.
We show that the operatorΨ defined in (53) has a unique fixed point u∗ in PCγ,ρJ. Let u,w∈PCγ,ρJ and t∈a,b; then, we have
(62)Ψut−Ψwt≤tρ−tkργ−1Γγργ−1ξ1∑i=1mLiuti−−Liwti−+ξ1ρJtm+1−γ+αhs−gsb+ξ1∑i=1mρJti−1+1−γ+αhs−gsti+∑a<tk<tLkuti−−Lkwtk−+∑a<tk<tρJti−1+1−γ+αhs−gstk+ρJtk+αhs−gst,where h,g∈Cγ,ρJ such that
(63)ht=ft,ut,ht,gt=ft,wt,gt.By (H2), we have(64)∣ht−gt∣=∣ft,ut,ht−ft,wt,gt∣≤K∣ut−wt∣+Mht−gt.Then,(65)ht−gt≤K1−Mut−wt.Therefore, for eacht∈a,b,
(66)Ψut−Ψwt≤tρ−tkργ−1Γγργ−1ξ1∑i=1ml∗uti−wti+ξ1K1−MρJtm+1−γ+αus−wsb+ξ1K1−M∑i=1mρJti−1+1−γ+αus−wsti+∑i=1ml∗uti−wti+K1−M∑i=1mρJti−1+1−γ+αus−wsti+K1−M∑i=1mρJtk+αus−wst.Thus,(67)Ψut−Ψwt≤tρ−tkρΓγργ−1ξ1ml∗p∗+ξ1K1−MρJtm+1−γ+αtρ−tmρργ−1b+mKξ11−MρJtk−11−γ+αsρ−tk−1ρργ−1tk+ml∗p∗+mK1−MρJtk−11−γ+αsρ−tk−1ρργ−1tk+K1−Mu−wPCγ,ρρJtkαsρ−tkρργ−1t.By Lemma6, we have
(68)Ψut−Ψwt≤1γtρ−tkρργ−1u−wPCγ,ρξ1ml∗p∗+ξ1KΓγ1−MΓ1−αbρ−bmρρα+mKξ1Γγ1−MΓ1−αtkρ−tk−1ρρα+ml∗p∗+mKΓγ1−MΓ1−αtkρ−tk−1ρρα+KΓγ1−MΓγ−αu−wPCγ,ρtρ−tkρρα+γ−1.Hence,(69)tρ−tkρρ1−γΨut−Ψwt≤ξ1+1ml∗p∗Γγ+mK1−MΓ1+αbρ−aρρα+K1−Mξ1Γ1+α+ΓγΓγ+αbρ−aρραu−wPCγ,ρ,which implies that
(70)Ψu−ΨwPCγ,ρ≤ξ1+1ml∗p∗Γγ+mK1−MΓ1+αbρ−aρρα+K1−Mξ1Γ1+α+ΓγΓγ+αbρ−aρραu−wPCγ,ρ.By (61), the operator Ψ is a contraction. Hence, by Theorem 14, Ψ has a unique fixed point u∗∈PCγ,ρJ.Step 2.
We show that such a fixed pointu∗∈PCγ,ρJ is actually in PCγ,ργJ.
Sinceu∗ is the unique fixed point of operator Ψ in PCγ,ρJ, then for each t∈Jk, with k=0,⋯,m, we have
(71)u∗t=1Γγtρ−tkρργ−1ξ2−ξ1∑i=1mLiuti−−ξ1∑i=1mρJti−1+1−γ+αhti−ξ1ρJtm+1−γ+αhb+∑a<tk<tuti−+∑a<tk<tρJti−1+1−γ+αhtk+ρJtk+αht,where h∈Cγ,ρJ such that
(72)ht=ft,u∗t,ht.ApplyingρDtk+γ to both sides and by Lemma 6 and Lemma 12, we get
(73)ρDtk+γu∗t=ρDtk+γρJtk+αfs,u∗s,hst=ρDtk+β1−αfs,u∗s,hst.Sinceγ≥α, by (H1), the right-hand side is in PCγ,ρJ and thus, ρDtk+γu∗∈PCγ,ρJ which implies that u∗∈PCγ,ργJ. As a consequence of Steps 1 and 2 together with Theorem 22, we can conclude that problems ((2))–((4)) have a unique solution in PCγ,ργJ.Our second result is based on Schaefer’s fixed-point theorem.Theorem 23.
Assume (H1), (H4), and (H5) hold. If(74)ξ1+1mΦ1p∗Γγ+mp2∗bρ−aρα1−p3∗Γ1+αρα+ξ1Γ1+α+ΓγΓγ+αp2∗bρ−aρα1−p3∗ρα<1,then problems ((2))–((4)) have at least one solution in PCγ,ργJ⊂PCγ,ρα,βJ.Proof.
We shall use Schaefer’s fixed-point theorem to prove in several steps that the operatorΨ defined in (53) has a fixed point.Step 1.
Ψ is continuous. Let un be a sequence such that un→u in PCγ,ρJ. Then, for each t∈a,b, we have
(75)Ψunt−Ψuttρ−tkρρ1−γ≤1Γγξ1∑i=1mLiunti−−Liuti−+ξ1ρJtm+1−γ+αhns−hsb+ξ1∑i=1mρJti−1+1−γ+αhns−hsti+∑a<tk<tLkuntk−−Lkutk−+∑a<tk<tρJti−1+1−γ+αhns−hstk+tρ−tkρρ1−γρJtk+αhns−hst,where hn,h∈Cγ,ρJ such that
(76)hnt=ft,unt,hnt,ht=ft,ut,ht.Sinceun→u, then we get hnt→ht as n→∞, for each t∈a,b, and since f and Lk are continuous, then we have
(77)Ψun−ΨuPCγ,ρ→0asn→∞.Step 2.
We show thatΨ maps bounded set into bounded set of PCγ,ρJ. For η>0, there exists a positive constant r such that for u∈Bη=u∈PCγ,ρJ:uPCγ,ρ≤η, we have ΨuPCγ,ρ≤r.By (H4) and from (53), we have for each t∈Jk,k=0,⋯,m,
(78)tρ−tkρρ1−γht=tρ−tkρρ1−γft,ut,ht≤tρ−tkρρ1−γp1t+p2tut+p3tht,which implies that
(79)hPCγ,ρ≤p1∗bρ−aρρ1−γ+p2∗η+p3∗hPCγ,ρ.Then,(80)hPCγ,ρ≤p1∗bρ−aρ/ρ1−γ+p2∗η1−p3∗≔Λ.Thus, (53) implies
(81)tρ−tkρρ1−γΨut≤1Γγξ2+ξ1∑i=1mLiuti−+ξ1∑i=1mρJti−1+1−γ+αhsti+ξ1ρJtm∣+1−γ+αhsb+∑a<tk<tLkuti−+∑a<tk<tρJti−1+1−γ+αhstk+tρ−tkρρ1−γρJtk+αhst,≤1Γγξ2+ξ1mΦ1p∗η+Φ2+ξ1mΛρJti−1+1−γ+αsρ−tk−1ρργ−1tk+ξ1ΛρJtm+1−γ+αsρ−tmρργ−1b+mΦ1p∗η+Φ2+mΛρJti−1+1−γ+αsρ−tk−1ρργ−1tk+Λtρ−tkρρ1−γρJtk+αsρ−tkρργ−1t.By Lemma6, we have
(82)ΨuPCγ,ρ≤ξ1+1mΦ1p∗η+Φ2Γγ+mΛΓ1+αbρ−aρρα+Λξ1Γ1+α+ΓγΓγ+αbρ−aρρα+ξ2Γγ.Step 3.
Ψ maps bounded sets into equicontinuous sets of PCγ,ρ.Letϵ1,ϵ2∈J,ϵ1<ϵ2Bη, be a bounded set of PCγ,ρ as in Step 2, and let u∈Bη.Then,(83)ϵ1ρ−tkρρ1−γΨuϵ1−ϵ2ρ−tkρρ1−γΨuϵ2≤1Γγ∑ϵ1<tk<ϵ2Lkuti−+∑ϵ1<tk<ϵ2ρJti−1+1−γ+αhstk+ΛΓγΓγ+αϵ1ρ−tkρρα−ϵ2ρ−tkρρα.Asϵ1→ϵ2, the right-hand side of the above inequality tends to zero. From Steps 1 to 3 with Arzela-Ascoli theorem, we conclude that Ψ:PCγ,ρ→PCγ,ρ is continuous and completely continuous.Step 4.
A priori bound. Now it remains to show that the set(84)G=u∈PCγ,ρ:u=λ∗Ψu,forsome0<λ∗<1,is bounded. Let u∈G, then u=λ∗Ψu for some 0<λ∗<1.By (H4), we have for eacht∈J(85)tρ−tkρρ1−γht=tρ−tkρρ1−γft,ut,ht≤tρ−tkρρ1−γp1t+p2tut+p3tht,which implies that
(86)hPCγ,ρ≤p1∗bρ−aρρ1−γ+p2∗uPCγ,ρ+p3∗hPCγ,ρ,then
(87)hPCγ,ρ≤p1∗bρ−aρ/ρ1−γ+p2∗uPCγ,ρ1−p3∗.This implies, by (53) and (H5) and by letting the estimation of Step 2, that for each t∈J, we have
(88)uPCγ,ρ≤ξ1+1mΦ1p∗uPCγ,ρ+Φ2Γγ+mp1∗bρ−aρ/ρ1+α−γ+mp2∗uPCγ,ρbρ−aρ/ρα1−p3∗Γ1−α+p1∗bρ−aρ/ρ1+α−γ+p2∗uPCγ,ρbρ−aρ/ρα1−p3∗ξ1Γ1−α+ΓγΓ1−α+ξ2Γγ,≤ξ1+1mΦ1p∗Γγ+mp2∗bρ−aρα1−p3∗Γ1+αρα+ξ1Γ1+α+ΓγΓγ+αp2∗bρ−aρα1−p3∗ραuPCγ,ρ+ξ2Γγξ1+1mΦ2Γγ+mp1∗bρ−aρ1+α−γ1−p3∗Γ1+αρ1+α−γ+ξ1Γ1+αΓγΓγ+αp1∗bρ−aρ1+α−γ1−p3∗ρ1+α−γ.By (74), we have
(89)uPCγ,ρ≤ξ2+1mΦ2/Γγmp1∗bρ−aρ1+α−γ/1−p3∗Γ1+αρ1+α−γ+ξ1/Γ1+α+Γγ/Γλ+αp1∗bρ−aρ1+α−γ/1−p3∗ρ1+α−γ1−ξ1+1mΦ1p∗/Γγ+mp2∗bρ−aρα/1−p3∗Γ1+αρα+ξ1/Γ1+α+Γγ/Γλ+αp2∗bρ−aρα/1−p3∗ρα≔R.As a consequence of Theorem15 and using Step 2, we deduce that Ψ has a fixed point which is a solution of problems (2)–(4).Our third result is based on Krasnoselskii’s fixed-point theorem.Theorem 24.
Assume (H1), (H4), and (H5) hold. If(90)ξ1∣+1mΦ1p∗Γγ+mp2∗bρ−aρα1−p3∗Γ1+αρα+p2∗ξ1bρ−aρα1−p3∗Γ1+αρα<1,then problems ((2))–((4)) have at least one solution in PCγ,ργJ⊂PCγ,ρα,βJ.Proof.
Consider the set(91)Bη=u∈PCγ,ρJ:uPCγ,ρ≤η,where
(92)η≥ξ1+1mΦ2/Γγ+mΦ2/Γ1+αbρ−aρ/ρα+Λξ1/Γ1+α+Γγ/Γγ+αbρ−aρ/ραξ2/Γγ1−ξ1+1mΦ1p∗/Γγ.We define the operatorsQ1 and Q2 on Bη by
(93)Q1ut=1Γγtρ−tkρργ−1ξ2−ξ1∑i=1mLiuti−−ξ1∑i=1mρJti−1+1−γ+αhti−ξ1ρJtm+1−γ+αhb+∑a<tk<tLkutk−+∑a<tk<tρJtk−1+1−γ+αhtk,(94)Q2ut=ρJtk+αht,where k=0,⋯,m, and h:a,b→ℝ be a function satisfying the functional equation
(95)ht=ft,ut,ht.Then, the fractional integral equation (53) can be written as operator equation
(96)Ψut=Q1ut+Q2ut,u∈PCγ,ρJ.The proof will be given in several steps.Step 1.
We prove thatQ1u+Q2w∈Bη, for any u,z∈Bη.
By (H4), (H5), and Lemma6, for each t∈a,b, we have
(97)Q1u+Q2wPCγ,ρ≤Q1uPCγ,ρ+Q2uPCγ,ρ≤ξ1+1mΦ1p∗η+Φ2Γγ+mΛΓ1+αbρ−aρρα+Λξ1Γ1+α+ΓγΓγ+αbρ−aρραξ2Γγ.Since(98)η≥ξ1+1mΦ2/Γγ+mΛ/Γ1+αbρ−aρ/ρα+Λξ1/Γ1+α+Γγ/Γγ+αbρ−aρ/ρα+ξ2/Γγ1−ξ1+1mΦ1p∗/Γγ,we have
(99)Q1y+Q2zPCγ,ρ≤η,which infers that Q1u+Q2w∈Bη.Step 2.
Q1 is a contraction.
Letu,w∈PCγ,ρJ and t∈a,b.
By (H4), we have(100)ht−gt=ft,ut,ht−ft,wt,gt≤p2tut−wt+p3tht−gt.Then,(101)ht−gt≤p2t1−p3tut−wt≤p2∗t1−p3∗tut−wt,where p1∗=supt∈Jp1t, p2∗=supt∈Jp2t, and h,g∈Cγ,ρJ such that
(102)ht=ft,ut,ht,gt=ft,wt,gt.Then, by (H5), we have(103)Q1yt−Q1zt≤1Γγtρ−tkρργ−1u−wPCγ,ρξ1mΦ1p∗+p2∗ξ1Γγ1−p3∗Γ1−αbρ−tmρρα+mp2∗ξ1Γγ1−p3∗Γ1−αtkρ−tk−1ρρα+mΦ1p∗+mp2∗Γγ1−p3∗Γ1−αtkρ−tk−1ρρα.Hence,(104)Q1u−Q1wPCγ,ρ≤ξ1+1mΦ1p∗Γγ+mp2∗1−p3∗Γ1+αbρ−aρρα+p2∗ξ11−p3∗Γ1+αbρ−aρραu−wPCγ,ρ.By (90), the operator Q1 is a contraction.Step 3.
Q2 is continuous and compact.
The continuity ofQ2 follows from the continuity of f. Next, we prove that Q2 is uniformly bounded on Bη. Let any w∈Bη. (94) imply
(105)tρ−tkρρ1−γQ2zt≤tρ−tkρρ1−γρJtk+αgst,≤Λtρ−tkρρ1−γρJtk+αsρ−tkρργ−1t,where k=0,⋯,m, and h:a,b→ℝ is a function satisfying the functional equation
(106)gt=ft,wt,gt.By Lemma6, we have
(107)Q2zPCγ,ρ≤ΛΓγΓγ+αbρ−aρρα.This means thatQ2 is uniformly bounded on Bη. Next, we show that Q2Bη is equicontinuous. Let any w∈Bη and a<ϵ1<ϵ2≤b. Then,
(108)ϵ1ρ−tkρρ1−γQ2zϵ1−ϵ2ρ−tkρρ1−γQ2zϵ2≤ΛΓγΓγ+αϵ1ρ−tkρρα−ϵ2ρ−tkρρα.Note that(109)ϵ1ρ−tkρρ1−γQ2zϵ1−ϵ2ρ−tkρρ1−γQ2zϵ2→0asϵ1→ϵ2.This shows thatQ2Bη is equicontinuous. Therefore, Q2Bη is relatively compact. By PCγ-type Arzela-Ascoli theorem, Q2 is compact. As a consequence of Theorem 16, we deduce that Ψ has at least a fixed point u∗∈PCγ,ρJ and by the same way of the proof of Theorem 22, we can easily show that u∗∈PCγ,ρJ. Using Lemma 21, we conclude that problems ((2))–((4)) have at least one solution in the space PCγ,ργJ.


## 4. Ulam-Hyers-Rassias Stability

Now, we are concerned with the Ulam-Hyers-Rassias stability of our problems (2)–(4).Theorem 25.
Assume that in addition to (H1)–(H3) and (61), the following hypothesis holds:(H6). There exist a nondecreasing functionϑ∈PCγ,ρJ and λϑ,λ~ϑ>0 such that for each t∈a,b, we have
(110)ρJa+αϑt≤λϑϑt.and
(111)ρJa+1−γϑt≤λ~ϑϑt.Then, Equation (2) is U-H-R stable with respect to ϑ,τ.Proof.
Consider the operatorΨ defined in (53). Let u∈PCγ,ρJ be a solution of inequality (27), and let us assume that w is the unique solution of the problem
(112)ρDtk+α,βwt=ft,wt,ρDtk+α,βwt;t∈Jk,k=0,⋯,m,ρJtk+1−γwtk+=ρJtk−1+1−γwtk−+Lkwti−,k=1,⋯,m,c1ρJa+1−γwa++c2ρJtm+1−γwb=c3,ρJa+1−γwa+=ρJa+1−γua+.By Lemma21, we obtain for each t∈a,b(113)wt=1Γγtρ−tkρργ−1ρJa+1−γwa++∑a<tk<tLkwti−+∑a<tk<tρJti−1+1−γ+αhtk+ρJtk+αht,t∈Jk,k=0,⋯,m,where h:a,b→ℝ is a function satisfying the functional equation
(114)ht=ft,wt,ht.Sinceu is a solution of inequality (27), by Remark 2.18, we have
(115)ρDtk+α,βut=ft,ut,ρDtk+α,βut+σt,t∈Jk,k=0,⋯,m,ρJtk+1−γtk+=ρJtk−1+1−γutk−+Lkuti−+σk,k=1,⋯,m.Clearly, the solution of (115) is given by
(116)ut=1Γγtρ−tkρργ−1ρJa+1−γua++∑a<tk<tLkuti−+∑a<tk<tσk+∑a<tk<tρJti−1+1−γ+αgtk+∑a<tk<tρJti−1+1−γ+ασtk+ρJtk+αgt+ρJtk+ασt,t∈Jk,k=0,⋯,m,where g:a,b→ℝ is a function satisfying the functional equation
(117)gt=ft,ut,gt.Hence, for eacht∈a,b, we have
(118)ut−wt≤1Γγtρ−tkρργ−1∑k=1mLkutk−−Lkwtk−+∑k=1mσk+∑k=1mρJti−1+1−γ+αgs−hstk+∑k=1mρJti−1+1−γ+ασstk+ρJtk+αgs−hst+ρJtk+ασst.Thus,(119)u−wPCγ,ρ≤1Γγmϵτ+mλ~ϑ+1ϵλϑϑt+∑k=1ml∗uti−−wti−+∑k=1mρJtk−1+1−γ+αgs−hstk+tρ−tkρρ1−γρJtk+αgs−hst.By (H2) and Lemma6, for t∈a,b, we have
(120)u−wPCγ,ρ≤1Γγmϵτ+mλ~ϑ+1ϵλϑϑt+ml∗p∗u−wPCγ,ρ+mK1−MΓ1+αtkρ−tk−1ρρα+KΓγ1−MΓγ+αtρ−tkρραu−wPCγ,ρ.Thus,(121)u−wPCγ,ρ≤1Γγmϵτ+mλ~ϑ+1ϵλϑϑt+ml∗p∗Γγ+K1−MmΓ1+α+ΓγΓλ+αbρ−aρραu−wPCγ,ρ.Then, by (61), we have
(122)u−wPCγ,ρ≤aϑϵτ+ϑt,where
(123)aϑ=1Γγm+mλ~ϑ+1λ01−ml∗p∗Γγ+K1−MmΓ1+α+ΓγΓγ+αbρ−aρρα−1.Hence, Equation (2) is U-H-R stable with respect to ϑ,τ.


## 5. Examples

Example 1.
Consider the following impulsive BVP of generalized Hilfer Fractional differential equation(124)1/2Dtk+1/2,0ut=197et+21+ut+1/2Dtk+1/2,0ut+lne+te2t−1,foreacht∈J0∪J1,(125)1/2Je+1/2ue+−1/2J1+1/2ue−=ue−3+ue−,(126)31/2J1+1/2u1+−21/2Je+1/2u3=0,where J0=1,e,J1=e,3,t0=1, and t1=e.Set(127)ft,u,w=197et+21+u+w+lne+te2t−1,t∈1,3,u,w∈ℝ.We have(128)PCγ,ρβ1−α1,3=PC1/2,1/201,3=g:1,3→ℝ:2t−tk1/2g∈PC1,3,with γ=α=1/2, ρ=1/2, β=0, and k∈0,1. Clearly, the continuous function f∈PC1/2,1/201,3.Hence, the condition (H1) is satisfied.For eachu,u¯,w,w¯,∈ℝ and t∈1,3,
(129)ft,u,w−ft,u¯,w¯≤197et+2u−u¯+w−w¯≤197e3u−u¯+w−w¯.Hence, condition (H2) is satisfied withK=M=1/97e3.And let(130)L1u=u3+u,u∈0,∞.Letu,w∈0,∞. Then, we have
(131)L1u−L1w=u3+u−w3+w=3u−w3+u3+w≤13u−w,and so the condition (H3) is satisfied and l∗=1/3.A simple computation shows that condition (61) of Theorem 22 is satisfied, for
(132)L=12πe−1+323−11/297e3−1Γ3/2+23−11/297e3−12Γ3/2+π≈0.52720987569<1.Then, problems ((124))–((126)) have a unique solution in PC1/2,1/21/21,3⊂PC1/2,1/21/2,01,3.Also, hypothesis (H6) is satisfied withϑt=e5, τ=1, and λϑ=λ~ϑ=2/Γ3/2. Indeed, for each t∈J0∪J1, we get
(133)ρJ1+1/2ϑt≤2e5Γ3/2=λϑϑt=λ~ϑϑt.Consequently, Theorem25 implies that Equation (124) is U-H-R stable.Example 2.
Consider the following impulsive initial value problem of generalized Hilfer fractional differential equation:(134)1Dtk+1/2,0ut=3+ut+1Dtk+1/2,0ut53e−t+41+ut+1Dtk+1/2,0ut,foreacht∈J0∪J1,(135)1Je+1/2ue+−1J1+1/2ue−=ue−2+ue−,(136)1J1+1/2u1+=0,where J0=1,e,J1=e,3,t0=1, and t1=e.Set(137)ft,u,w=3+u+w53e−t+41+u+w,t∈1,3,u,w∈ℝ.We have(138)PCγ,ρβ1−α1,3=PC1/2,101,3=g:1,3→ℝ:t−tkg∈PC1,3,with γ=α=1/2, ρ=1, β=0, and k∈0,1. Clearly, the continuous function f∈PC1/2,101,3 .Hence, the condition (H1) is satisfied. For eachu,w∈ℝ and ∈1,3,
(139)ft,u,w≤153e−t+43+u+w.Hence, condition (H4) is satisfied withp1t=3/53e−t+4, p2t=p3t=3/53e−t+4, p1∗=3/53e, and p2∗=p3∗=1/53e.And let(140)L1u=u2+u,u∈0,∞.Letu∈0,∞. Then, we have
(141)L1u≤12u+2,and so the condition (H5) is satisfied with Φ1=1/2 and Φ2=2.The condition (74) of Theorem 23 is satisfied, for
(142)ξ1+1mΦ1p∗Γγ+mp2∗bρ−aρα1−p3∗Γ1+αρα+ξ1Γ1+α+ΓγΓγ+αp2∗bρ−aρα1−p3∗ρα=122π+253e−1Γ3/2+2π53e−1≈0.22814541069≤1.Then, problems ((134))–((136)) have at least one solution in PC1/2,11/21,3⊂PC1/2,11/2,01,3. Also, hypothesis (H6) is satisfied with ϑt=t−1,τ=1, and λϑ=λ~ϑ=2Γ2/Γ5/2. Indeed, for each t∈J0∪J1, we get
(143)ρJ1+1/2ϑt≤2Γ2Γ5/2t−1=λϑϑt=λ~ϑϑt.Consequently, by a simple change of the constantsl∗, K, and M from (H1) and (H2) to Φ1,p2∗, and p3∗ from (H4) and (H5), Theorem 25 implies that Equation (134) is G.U-H-R stable.Example 3.
Consider the following impulsive antiperiodic boundary value problem of generalized Hilfer Fractional differential equation:(144)ρDtk+1/2,0ut=e2+ut+1Dtk+1/2,0ut77e−t+21+ut+1Dtk+1/2,0ut,foreacht∈Jk,k=0,⋯,4,(145)1Jtk+1/2utk+−ρJtk−1+1/2uti−=utk−10k+utk−,k=1,⋯,4,(146)1J1+1/2u1+=−1J9/5+1/2u2,where Jk=tk,tk+1, tk=1+k/5, for k=0,⋯,4,m=4, a=t0=1, and b=t5=2.Set(147)ft,u,w=e2+u+w77e−t+21+u+w,t∈1,2,u,w∈ℝ.We have(148)PCγ,ρβ1−α1,2=PC1/2,101,2=g:1,2→ℝ:t−tkg∈C1,2,with γ=α=1/2,ρ=1,β=0, and k=0,⋯,4. Clearly, the continuous function f∈PC1/2,101,2 .Hence, the condition (H1) is satisfied. For eachu,w∈ℝ and t ∈ (1,2],
(149)ft,u,w≤177e−t+2e2+u+w.Hence, condition (H4) is satisfied withp1t=e2/77e−t+2,p2t=p3t=1/77e−t+2, p1∗=e2/77, and p2∗=p3∗=1/77.And let(150)Lku=u10k+u,k=1,⋯,4,u∈0,∞.Letu∈0,∞. Then, we have
(151)Lku=110u+1,k=1,⋯,4,and so the condition (H5) is satisfied with Φ1=1/10 and Φ2=1.The condition (90) of Theorem 24 is satisfied, for
(152)ξ1+1mΦ1p∗Γγ+mp2∗bρ−aρα1−p3∗Γ1+αρα+p2∗ξ1bρ−aρα1−p3∗Γ1+αρα=355π+1251463Γ3/2<1.Then, problems ((144))–((146)) have at least one solution in PC1/2,11/21,2⊂PC1/2,11/2,01,2. Also, hypothesis (H6) is satisfied with ϑt=1−t2,τ=1, and λϑ=λ~ϑ=Γ3/Γ7/2. Indeed, for each t∈Jk,k=0,⋯,4, we get
(153)ρJ1+1/2ϑt≤Γ3Γ7/2t−12=λϑϑt=λ~ϑϑt.The same as Example2, Theorem 25 implies that Equation (144) is U-H-R stable.


## 6. Conclusion

In this paper, we are concerned with existence, uniqueness, and Ulam-Hyers-Rassias stability results for a class of boundary value problem for nonlinear implicit fractional differential equations with impulses and generalized Hilfer fractional derivative. We gave three results using the Banach contraction principle and Krasnoselskii’s and Schaefer’s fixed-point theorems. Two illustrative examples about the uniqueness and the stability were presented. An interesting extension of our research would be to address Ulam-Hyers-Mittag-Leffler stability for a class and coupled system of implicit fractional differential equations. This subject will be discussed in the forthcoming papers.



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