# Existence of Mild Solutions for a Class of Impulsive Hilfer Fractional Coupled Systems

Authors: Karim Guida; Khalid Hilal; Lahcen Ibnelazyz
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

The aim of this paper is to give existence results for a class of coupled systems of fractional integrodifferential equations with Hilfer fractional derivative in Banach spaces. We first give some definitions, namely the Hilfer fractional derivative and the Hausdorff’s measure of noncompactness and the Sadovskii’s fixed point theorem.


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## Body

## 1. Introduction

Fractional differential equations have been a good tool in many research areas in the last decade, such as engineering, mathematics, physics, and many other sciences [1, 2]. For some basic results on this theory, we refer the readers to the papers [3, 4] and the references therein.There are many different definitions of fractional derivatives, each one with its importance and application, which helped justify the importance of fractional calculus. We mention here a few of the most notable definitions of fractional derivatives: Hadamard, Caputo–Hadamard, Hilfer,ψ-Hilfer, Caputo–Riesz, Grünwald–Letnikov, for more details we refer the readers to [5–9].Recently, a lot of attention has been devoted to the existence of fractional differential problems with Hilfer fractional derivative, see [10–12]. The Hilfer fractional derivative, which is a generalization of the Riemann-Liouville fractional derivative, was introduced by nonother than Hilfer [1, 13].The first results on the existence of general value problems involving Hilfer fractional derivative were investigated in [14] and after that in [12]. Following these results, Gu and Trujilo [15] gave the existence of solutions for fractional differential equations with Hilfer fractional derivative using the notion of measure of noncompactness. These equations are widely employed in the biomedical field.On the other hand, the concept of noninstantaneous impulses was first introduced in [16] by Hernandez; these conditions appeared in the mathematical description of problems that experience abrupt changes during their evolution in time. In the established works, fractional differential equations (FDEs) involving Caputo’s fractional derivative are commonly considered with impulsive conditions for obtaining mild solutions [17–20]. However, in [21], Sousa obtained for the first time the mild solutions for Hilfer fractional differential equations with noninstantaneous impulses. To the best of our knowledge, there are few papers dealing with coupled systems, and on top of that, even fewer existence results for neutral Hilfer fractional differential equations. That is why, to make a little contribution to the already existing results, we consider in this paper a class of coupled systems of Hilfer fractional differential equations with not instantaneous impulses in a Banach space as follows:
(1)HD0+β1,γ1x1t+f1t,x1t,x2t=A1x1t+g1t,x1t,x2t,∫0tu1t,sφ1t,s,x1s,x2sds,t∈si,ti+1,i=0,1,2,⋯,m,HD0+β2,γ2x2t+f2t,x1t,x2t=A2x2t+g2t,x1t,x2t,∫0tu2t,sφ2t,s,x1s,x2sds,t∈si,ti+1,i=0,1,2,⋯,m,x1t=mit,x1t,t∈ti,si,i=1,2,⋯,m,x2t=nit,x2t,t∈ti,si,i=1,2,⋯,m,I0+1−α1x10+f10,x10,x20=x01,I0+1−α2x20+f20,x10,x20=x02,where for j=1,2,I0+1−αj are the RL fractional integrals, HDβj,γj are the Hilfer fractional derivatives of order (βj,γj) with 0≤βj<1,0≤γj≤1, and 0≤αj=βj+γj−βjγj≤1, the linear operators Aj:DAj⊂X−→X are the infinitesimal generators of strongly continuous semigroups Tjtt≥0 in a Banach space X,0=t0=s0<t1<s1<t2<⋯<ti<si<ti+1=T is a partition of 0,T,T>0,i=1,2,⋯,m.The functionsfj:0,T×X×X−→X and gj:0,T×X×X×X−→X are satisfying some assumptions that will be given later, and the functions mi: ti,si×X−→X and ni:ti,si×X−→X characterize the impulsive conditions and x0j∈X. The conditions on uj:0,T×0,T→ℝ and φj:0,T×0,T×X×X−→X are given in a later part.This paper is organized as follows: we first give some preliminaries and notions that will be used throughout the work; after that, we will establish the existence results by means of the fixed point theory; last but not least, we will give an example that illustrates the results.


## 2. Preliminaries and Notations

LetCJ,X be the complete normed linear space of all continuous functions xt defined on the interval J=0,T with xt=supt∈jxt. We define the Banach space C1−αJ,X introduced in [14] by C1−αJ,X=x:J→Xsuchthatt1−αxt∈CJ,X with norm xC1−α=sup0≤t≤Tt1−αxt.We also define the Banach space(2)PC1−αJ,X=xt:t−ti1−αxt∈Cti,ti+1,ℝandlimt→tit−ti1−αxtexists,for i=1,2,⋯,m with the norm xPC1−α=maxsupt∈Jt1−αxt+,supt∈Jt1−αxt−.The spaceX=PC1−α1×PC1−α2 equipped with the norm x1,x2=maxx1PC1−α1,x2PC1−α2 is also a Banach space.ByLX, we denote the family of bounded linear operators defined on X, and for j=1,2,Rβj,γjtt≥0 are the (βj−γj)-resolvent operators generated by Aj.Definition 1 (see [1]).
The Hilfer fractional derivative of ordern−1≤β<n, n∈ℕ; 0≤γ≤1, with lower limit a is defined as follows:
(3)Da+β,γft=Ia+βn−γddtIa+1−βn−γft=Ia+βn−γDa+γ+βn−γβft,where Ia+βn−γ is the RL integral, and Da+γ+βn−γβ is the RL derivative.Lemma 2. (see [14]).
Let0<β<1, 0≤γ≤1, and α=β+γ−βγ. If f∈C1−αa,b is such that Da+αf∈C1−αa,bD, then
(4)Ia+αDa+αf=Ia+βDa+β,γfandDa+αIa+αf=Da+γ1−βf.Lemma 3. (see [3]).
Let0<β<1 and 0≤α≤1. If f∈C1−αa,b and Ia+1−βf∈Cα1a,b, then
(5)Ia+βDa+βfx=fx−Ia+1−βfaΓβx−aβ−1,∀x∈a,b.Definition 4. (see [22]).
The Hausdorff measure of noncompactness on a bounded subsetΩX of Banach space X is the mapping μ:B⊂ΩX→0,∞ defined by
(6)μB=infε>0:B=∪BiwithradiusofBi≤εfori=1,2,..,m.
We are going to look back on some properties of the measure of noncompactness.Lemma 5 (see [22, 23]).
The measure of noncompactnessμ defined on bounded subsets AandB of a Banach space X has the properties:
(1)
μA=0 if and only if A is a relatively compact set(2)
A⊂B implies that μA≤μB(7)μA¯=μA(3)
μA∪B=maxμA,μB(4)
μλA=λμA,∀λ∈ℝLemma 6 (see [22, 23]).
For a bounded setD⊂X, there is a countable set D0⊂D such that μD0≤μD.Lemma 7 (see [24]).
For a bounded and equicontinuous functionG∈CJ,X, the Hausdorff measure of noncompactness μGt is continuous on J, and μG=maxt∈JμGt.Lemma 8 (see [23]).
LetD=xn⊂CJ,X be a bounded countable subset of X. Then, μDtis Lebesque integrable on X, and
(8)μ∫xntdt:n∈ℕ≤∫μxntdt.Lemma 9. (see [21]).
We apply lemmas2 and 3; then, we obtain an equivalent system of equations to the system 1 as follows:
(9)x1t=tα1−1x01Γα1−f1t,x1t,x2t+1Γβ1∫0tt−sβ1−1A1x1s+g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdtds,t∈0,t1,mit,x1t,t∈ti,si,i=1,2,⋯,mmit,x1t−f1t,x1t,x2t+1Γβ1∫0tt−sβ1−1A1x1s+g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdtds,t∈si,ti+1,i=1,2,⋯,m,x2t=tα2−1x02Γα1−f2t,x1t,x2t+1Γβ2∫0tt−sβ2−1A2x2s+g2s,x1s,x2s,∫0τu2s,τφ2s,τ,x1τ,x2τdtds,t∈0,t1,nit,x2t,t∈ti,si,i=1,2,⋯,mnit,x2t−f2t,x1t,x2t+1Γβ2∫0tt−sβ2−1A2x2s+g2s,x1s,x2s,∫0τu2s,τφ2s,τ,x1τ,x2τdtds,t∈si,ti+1,i=1,2,⋯,m,Remark 10.
The Laplace transform of the Hilfer fractional derivative of a functionft of order 0<β<1 and 0<γ<1 is given in [13] by
(10)LDa+β,γfts=sβLfts−sγβ−1I0+1−β1−γf0+,where I0+1−β1−γf0+ is the Riemann-Liouville derivative of order 1−β1−γ.
Now, we give a definition of a pair of mild solution to the problem 1, which is obtained by applying the Laplace transform of the Hilfer fractional derivative.Definition 11 (see [15, 21]).
A pairx,y∈PC1−α1J,X×PC1−α2J,X=X is said to be a pair of mild solutions of the system (1) if the couple (x,y) satisfies the following coupled system:
(11)x1t=Rβ1,γ1tx01−f1t,x1t,x2t−∫0tSβ1t−sA1f1s,x1s,x2sds+∫0tSβ1t−sg1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds,t∈0,t1,mit,x1t,t∈ti,si,i=1,2,⋯,mRβ1,γ1tmisi,x1si−f1t,x1t,x2t−∫SitSβ1t−sA1f1s,x1s,x2sds+∫sitSβ1t−sg1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds,t∈si,ti+1,i=1,2,⋯,m,(12)x2t=Rβ2,γ2tx02−f2t,x1t,x2t−∫0tSβ2t−sA2f2s,x1s,x2sds+∫0tSβ2t−sg2s,x1s,x2s,∫0τu2s,τφ2s,τ,x1τ,x2τdτds,t∈0,t1,nit,x2t,t∈ti,si,i=1,2,⋯,mRβ2,γ2tnisi,x2si−f2t,x1t,x2t−∫sitSβ2t−sA2f2s,x1s,x2sds+∫sitSβ2t−sg2s,x1s,x2s,∫0τu2s,τφ2s,τ,x1τ,x2τdτds,t∈si,ti+1,i=1,2,⋯,m,for j=1,2, we have
(13)Rβj,γjt=Ivγj1−βjSβjt,Sβjt=tαj−1Zβjt,Zβjt=∫0∞βjvjWβjvjTjtβjvjdvj,where Wβjvj are the Wright functions defined as follows:
(14)Wβjvj=∑n=1∞−vjn−1n−1Γ1−μjn,0<μj<1,vj∈ℂ,and satisfying
(15)∫0∞vjσjWβjvjdvj=Γ1+σjΓ1+βjσj,vj,σj≥0.
From [15, 25], we can assume that for j=1,2(i)
The linear operatorsWβj,γjtt>0 and Sβtt>0 are strongly continuous and verify:(16)Wβj,γjt≤Mjtαj−1ΓαjandSβjt≤Mjtβj−1Γβjfort>0,(ii)
The norm continuity of the familyTjt for t>0


## 3. Existence Results

In this section, we make some assumptions that are necessary to obtain our results:(H1) For j=1,2, the functions fj:J×X×X→X are bounded and Lipschitz continuous, that is, there exist Lfj>0 and Mfj∈0,1 such that
(17)fjt,u,v≤Lfjfjt,u1,v1−fjt,u2,v2≤Mfju1−u2+Nfjv1−v2.(H2) For j=1,2, the functions gj:J×X×X×X→DB⊂X are Caratheodory, that is, gj·,x1,x2,x3:J→DB⊂X is measurable for all x1,x2,x3∈X×X×X,gjt,·,·,·:X×X×X→DB⊂X is continuous a.e for t∈J, and there exist ψ1,ψ2∈L1/δJ,ℝ+,1/δ>1, and a continuous function ψ3 such that
(18)gj⋅,x1,x2,x3≤ψj1tx1+ψj2tx2+ψj3tx3,for almost all t∈J.(H3) For j=1,2, there exist functions ψ^j1,ψ^j2∈L1/δJ,ℝ+ and constants Mgj1,Mgj2>0 such that µgjt,D1,D2≤Mgj1ψ^j1µD1+Mgj2ψ^j2µD2,t∈J, for any bounded, equicontinuous, and countable sets Di⊂X,i=1,2.(H4) The impulsive functions mi,ni:ti,si×X→X are Lipschitz continuous, that is, there exist Kmi,Kni>0,i=1,2,⋯,m, such that for all x,y∈X, we have:
(19)mit,x−mit,y≤Kmix−y,andnit,x−nit,y≤Knix−y.(H5) For j=1,2,ϕjt,s,·,·:X→X are caratheodory functions, and there exist ξj:0,T×0,T→ℝ with
(20)ξj∗=supt∈J∫0tξjt,s<∞suchthatφt,s,x1,x2≤ξjt,sx1,x2.(H6) For j=1,2, and for any bounded set D1⊂X and 0≤s≤t≤T, there exist functions ν^j:0,T×0,T→ℝ such that
(21)μφjt,s,D1≤v^jt,sμD1wherev^j∗=sup∫0tv^jt,sds<∞.(H7) For j=1,2,uj∗=supujt,s,0≤s≤t is bounded and measurable on J along with the continuity of ujt:J→L∞J,ℝ defined by ujt=ujt,s.Theorem 12.
The system (1) has a pair of solutions x1t,x2t in the space X=PC1−α1×PC1−α2 if the assumptions (H1)−(H7) hold and the following conditions are verified, for j=1,2:
(22)MjKjΓαj+T1−αj+βj+δΓβj1−δβj−δ1−δψj1+ψj2+ψj3Tβj+1−αjuj∗ξj∗βjΓβj<1,andmaxj=1,2MjKj+Mfj+Nfj+maxj=1,2MjTβj−δΓβj1−δβj−δ1−δMg1ψ^j1L1δ0,T+Mgj2uj∗v^j∗ψj2L1δ0,T+MjLfjTδΓβj+1<1
Note thatK1=Kmi and K2=Kni.Proof.
To prove the existence of solutions for system (1), we only have to prove the existence of solutions for the system (11) and (12) because they are equivalent.
Let us defineΩr=x1,x2∈PC1−α1J,X×PC1−α2J,X=X:x1,x2≤r with fix radius r,Ωr is a nonempty closed convex bounded subset of X.
Define the operatorS:X→X such that Sx1,x2t=Px1,x2t,Qx1,x2tT,∀x1,x2∈X,t∈0,T, where
(23)Px1,x2t=Rβ1,γ1tx01−f1t,x1t,x2t−∫0tSβ1t−sA1f1s,x1s,x2sds+∫0tSβ1t−sg1s,x1s,x2s,∫0τu1s−τφ1s,τ,x1τ,x2τdτds,t∈0,t1,mit,x1t,t∈ti,si,i=1,2,⋯,mRβ1,γ1tm1si,x1si−f1t,x1t,x2t−∫sitSβ1t−sA1f1s,x1s,x2sds+∫sitSβ1t−sg1s,x1s,x2s,∫0τu1s−τφ1s,τ,x1τ,x2τdτds,t∈si,ti+1,i=1,2,⋯,m,(24)Qx1,x2t=Rβ2,γ2tx02−f2t,x1t,x2t−∫0tSβ2t−sA2f2s,x1s,x2sds+∫0tSβ2t−sg2s,x1s,x2s,∫0τu2s−τφ2s,τ,x1τ,x2τdτds,t∈0,t1,nit,x2t,t∈ti,si,i=1,2,⋯,mRβ2,γ2tn1si,x2si−f2t,x1t,x2t−∫sitSβ2t−sA2f2s,x1s,x2sds+∫sitSβ2t−sg2s,x1s,x2s,∫0τu2s−τφ2s,τ,x1τ,x2τdτds,t∈si,ti+1,i=1,2,⋯,m,by splitting both (23) and (24), we have:
(25)P1x1,x2t=Rβ1,γ1tx01−f1t,x1t,x2t,t∈0,t1,mit,x1t,t∈ti,si,i=1,2,⋯,mRβ1,γ1tm1si,x1si−f1t,x1t,x2t,t∈si,ti+1,i=1,2,⋯,m,P2x1,x2t=∫0tSβ1t−sg1s,x1s,x2s,∫0τu1s−τφ1s,τ,x1τ,x2τdτds−∫0tSβ1t−sA1f1s,x1s,x2sds,t∈0,t1,∫sitSβ1t−sg1s,x1s,x2s,∫0τu1s−τφ1s,τ,x1τ,x2τdτds−∫sitSβ1t−sA1f1s,x1s,x2sds,t∈si,ti+1,i=1,2,⋯m,Q1x1,x2t=Rβ2,γ2tx02−f2t,x1t,x2t,t∈0,t1,nit,x2t,t∈ti,si,i=1,2,⋯,mRβ2,γ2tn1si,x2si−f2t,x1t,x2t,t∈si,ti+1,i=1,2,⋯,m,Q2x1,x2t=∫0tSβ2t−sg2s,x1s,x2s,∫0τu2s−τφ2s,τ,x1τ,x2τdτds−∫0tSβ2t−sA2f2s,x1s,x2sds,t∈0,t1,∫sitSβ2t−sg2s,x1s,x2s,∫0τu2s−τφ2s,τ,x1τ,x2τdτds−∫sitSβ2t−sA2f2s,x1s,x2sds,t∈si,ti+1,i=1,2,⋯m,
The upcoming part of the proof needs us to rewrite the operatorS as follows:
(26)Sx1,x2t=S1x1,x2t+S2x1,x2t,where
(27)S1x1,x2t=P1x1,x2,Q1x1,x2t,S2x1,x2t=P2x1,x2,Q2x1,x2t.Step 1.
We first show thatSx1,x2∈X for x1,x2∈X, that is, t1−α1Px1,x2t and t1−α2Qx1,x2t are continuous functions for t∈titi+1,i=0,1,2,⋯,m.
For0≤t2≤t≤t1, we have:
(28)t1−α1Px1,x2t−Px1,x2t2≤Rβ1,γ1tx01−Rβ1,γ1t2x01+f1t,x1t,x2t−f1t2,x1t2,x2t2+∫0t2t1−α1Sβ1t−s−Sβ1t2−s×g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds+∫t2tt1−α1Sβ1t−s×g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds+∫0t2t1−α1Sβ1t−s−Sβ1t2−sA1f1s,x1s,x2sds+∫t2tt1−α1Sβ1t−sA1f1s,x1s,x2sds.
We make the substitutiont2−s=s1 in the third and fifth terms, we get
≤Rβ1,γ1tx01−Rβ1,γ1t2x01+Lt−t2+∫0t2T1−α1Sβ1t−t2+s1−Sβ1s1×g1t2−s1,x1t2−s1,x2t2−s1,∫0τu1t2−s1,τφ1t2−s1,τ,x1τ,x2τdτds1+∫t2tT1−α1Sβ1t−s×g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds+∫0t2T1−α1Sβ1t−t2+s1−Sβ1s1A1f1t2−s1,x1t2−s1,x2t2−s1ds1+∫t2tT1−α1Sβ1t−sA1f1s−x1s,x2sds.→0 as t→t2.
This proves the continuity oft1−α1Px1,x2tfort∈0,t1, and similarly, we prove the continuity of t1−α2Qx1,x2t.
To show that the operatorS is continuous on the intervals tisi and si,ti+1 for i=1,2,⋯,m, we use the continuity of noninstantaneous impulsive functions mit,x1t and nit,x2t. Thus, we conclude that Sx1,x2∈X.Step 2.
We show thatS:Ωr→Ωr, that is, Sx1,x2∈Ωr, for x1,x2∈Ωr.
We first show that the operatorP is bounded, which means, Px1,x2∈Ωr, for x1,x2∈Ωr. Suppose the opposite, so there exist x1,x2∈Ωr and t∈J such that Px1,x2C1−α1>r.
Fort∈0,t1, we have
(29)Px1,x2tC1−α1=t1−α1Px1,x2t≤t1−α1Rβ1,γ1tx01+t1−α1f1t,x1t,x2t+∫0tSβ1t−s×t1−α1g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds+∫0tSβ1t−sA1t1−α1f1s,x1s,x2sds≤Lf1+M1x01Γα1+M1Lf1A1Tβ1Γβ1+1+M1t1−α1Γβ1∫0tt−sβ1−1g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds≤Lf1+M1x01Γα1+M1Lf1A1Tβ1Γβ1+1+M1t1−α1Γβ1∫0tt−sβ1−1ψ11sr+ψ12sr+ψ13su1∗r∫0sξ1s−τds≤Lf1+M1x01Γα1+M1Lf1A1Tβ1Γβ1+1+M1ψ13u1∗rTβ+1−α1ξ1∗β1Γβ1+rM1T1−α1Γβ1∫0tt−sβ1−11−δds1−δ∫0tψ11s1δdsδ+rM1T1−α1Γβ1∫0tt−sβ1−11−δds1−δ∫0tψ12s1δdsδ=Lf1+M1x01Γα1+M1Lf1A1Tβ1Γβ1+1+rM1T1−α1Γβ11−δβ1−δ1−δTβ1−δψ11L1δ0,T1+1−δβ1−δ1−δTβ1−δψ12L1δ0,T1+ψ13Tβ1u1∗ξ1∗β1.which implies that
(30)t1−α1Px1,x2t≤Lf1+M1x01Γα1+M1Lf1A1Tβ1Γβ1+1+rM1T1−α1Γβ11−δβ1−δ1−δTβ1−δψ11L1δ0,T1+1−δβ1−δ1−δTβ1−δψ12L1δ0,T1+ψ13Tβ1u1∗ξ1∗β1.
Fort∈ti,si,i=1,2,⋯,m, we have
(31)t1−α1Px1,x2t=mit,x1tC1−α1=mit,x1t−mit,0+mit,0C1−α1≤Kmix1tC1−α1+mit,0C1−α1≤Kmir+V,whereV=supmit,0C1−α1.which implies that
(32)t1−α1Px1,x2t≤Kmir+V.
Fort∈si,ti+1,
(33)t1−α1Px1,x2t=t1−α1Rβ1,γ1tmisi,x1si−f1t,x1t,x2t−∫sitSβ1t−sA1f1s,x1s,x2sds+∫sitSβ1t−sg1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds≤M1Kmir+VΓα1+Lf1+M1Lf1A1t−siβ1Γβ1+1+rM1T1−α1Γβ11−δβ1−δ1−δTβ1−δψ11L1δ0,T1+1−δβ1−δ1−δTβ1−δψ12L1δ0,T1+ψ13Tβ1u1∗ξ1∗β1which implies that
(34)t1−α1Px1,x2t≤Lf1+M1Kmir+VΓα1+M1Lf1A1t−siβ1Γβ1+1+rM1T1−α1Γβ11−δβ1−δ1−δTβ1−δψ11L11δ0,T+1−δβ1−δ1−δTβ1−δψ12L11δ0,T+ψ13Tβ1u1∗ξ1∗β1.
Combining the expressions (30), (32) and (34), we obtain
(35)Px1,x2tC1−α1≤Lf1+M1x01Γα1+M1Kmir+VΓα1+M1Lf1A1Tβ1Γβ1+1+rM1Γβ11−δβ1−δ1−δT1−α1+β1−δψ11L11δ0,T+1−δβ1−δ1−δTβ1−δ+1−α1ψ12L11δ0,T+ψ13Tβ1−δ+1−α1u1∗ξ1∗β1.
By our assumptions, we have
Px1,x2tC1−α1>r which implies that
(36)Lf1+M1x01Γα1+M1Kmir+VΓα1+M1Lf1A1Tβ1Γβ1+1+rM1Γβ11−δβ1−δ1−δT1−α1+β1−δψ11L11δ0,T+1−δβ1−δ1−δTβ1−δ+1−α1ψ12L11δ0,T+ψ13Tβ1+1−α1u1∗ξ1∗β1≥Px1,x2tC1−α1>r.
Dividing both sides byr and taking δ→∞, we obtain
(37)M1KmiΓα1+T1−α1+β1+δΓβ11−δβ1−δ1−δψ11+ψ12+ψ13Tβ1+1−α1u1∗ξ1∗β1Γβ1>1which is a contradiction. Hence, Px1,x2tC1−α1<r.
Similarly, we show thatQx1,x2tC1−α2<r.
Finally,Sx1,x2t=maxPx1,x2tC1−α1,Qx1,x2tC1−α2<r.
That shows that the operatorS maps bounded sets to bounded sets.Step 3.
We prove that the operatorS1 is Lipschitz continuous.
Fort∈0,t1, we have
(38)P1x1,x2−P1y1,y2C1−α1≤Mf1+Nf1x1−y1,x2−y2.
Fort∈ti,si,i=1,2,⋯,m, we have
(39)P1x1,x2−P1y1,y2C1−α1=mit,x1t−mit,y1tC1−α1≤Kmix1−y1C1−α1,so we get
(40)P1x1,x2−P1y1,y2C1−α1≤Kmix1−y1,x2−y2,and for t∈ti,si, we have
(41)P1x1,x2−P1y1,y2C1−α1≤M1Kmi+Mf1+Nf1x1−y1,x2−y2.
From (38), (40) and (41), we can say that the operator P1 is Lipschitz continuous with constant M1Kmi+Mf1+Nf1, and similarly, we show that the operator Q1 is Lipschitz continuous with constant (M2Kni+Mf2+Nf2).
Finally, the operatorS1 is Lipschitz continuous with the constant maxM1Kmi+Mf1+Nf1;M2Kni+Mf2+Nf2.Step 4.
We show thatS2 is a continuous operator.
Letx1n,x2n be a sequence in Ωr such that x1n,x2n→x1,x2 as n→∞.
The functionsf1,g1 are continuous with respect to the second, third, and fourth variables, it follows that
(42)limt→∞f1t,x1nt,x2nt=f1t,x1t,x2tandlimt→∞g1t,x1nt,x2nt,∫0τu1s,τϕ1t,τ,x1nτ,x2nτdτ=g1t,x1t,x2t,∫0τu1s,τϕ1t,τ,x1τ,x2τdτby (H1) and (H2), we have
(43)f1t,x1nt,x2nt−f1t,x1t,x2t≤2Lf1g1t,x1nt,x2nt∫0τu1s,τϕ1t,τ,x1nτ,x2nτdτ−g1,t,x1t,x2t,∫0τu1s,τϕ1t,τ,x1τ,x2τdτ≤2rψ11t+ψ12t+ψ13tu1∗ξ∗.
Sinceψ11,ψ12∈L1/δ0,T and ψ13t is continuous, the functions on the right hand side are integrable.
For allt∈J,x1n,x2n,x1,x2∈Ωr, we have
(44)t1−α1P2x1n,x2nt−P2x1,x2t2≤∫Sitt1−α1Sβ1t−s×g1s,x1ns,x2ns,∫0τu1s,τϕ1s,τ,x1nτ,x2nτdτ−g1s,x1s,x2s,∫0τu1s,τϕ1s,τ,x1nτ,x2nτdτds+∫Sitt1−α1Sβ1t−sA1f1s,x1ns,x2ns−f1s,x1s,x2sds≤M1T1−α1Γβ1∫sitt−sβ1−1×g1s,x1ns,x2ns,∫0τu1s,τϕ1s,τ,x1nτ,x2nτdτ−g1s,x1s,x2s,∫0τu1s,τϕ1s,τ,x1nτ,x2nτdτds+M1A1T1−α1Γβ1×∫sitt−sβ1−1f1s,x1ns,x2ns−f1s,x1s,x2sds
By the Lebesgue dominated convergence theorem, we have(45)t1−α1P2x1n,x2nt−P2x1,x2t2→0asn→∞,and by the same method, we show that
(46)t1−α2Q2x1n,x2nt−Q2x1,x2t2→0asn→∞,consequently, the operator S2 is continuous.Step 5.
We show that the operatorS2 is equicontinuous.
For anyx1,x2∈Ωr and si<t1<t2<ti+1,i=0,1,⋯,m, we have
(47)t1−α1P2x1,x2t2−P2x1,x2t1=∫sit2t1−α1Sβ1t−s×g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds−∫sit1t1−α1Sβ1t−s×g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds−∫sit2t1−α1Sβ1t−sA1f1s,x1s,x2sds+∫sisit1−α1Sβ1t−sA1f1s,x1s,x2sds≤∫sit1Sβ1t2−s−Sβ1t1−st1−α1×g1s,x1s,x2s,∫0τu1s,τφ1s,τ,x1τ,x2τdτds+∫sit1Sβ1t2−s−Sβ1t1−st1−α1A1f1s,x1s,x2sds+∫t1t2Sβ1t2−st1−α1A1f1s,x1s,x2sds=I1+I2+I3+I4.
We substitutet1−s=s1 in I1 and I3, we obtain
(48)I1≤∫0t1T1−α1Sβ1t2−t1+s1−Sβ1s1×g1t1−s1,x1t1−s1,∫0τu1t1−s1,τφ1t1−s1,τ,x1τ,x2τdτds1,I3≤∫0t1T1−α1T1−α1Sβ1t2−t1+s1−Sβ1s1A1f1t1−s1,x1t1−s1,x2t1−s1ds1,
by the equicontinuity ofβ1−γ1-resolvent operator nad Lebesgue dominated convergence theorem, the integrals I1,I3→0ast2→t1.
And we haveI2,I4→0ast2→t1; it follows that
(49)t1−α1P2x1,x2t2−P2x1,x2t1→0ast2→t1,we show with a similar method that
(50)t1−α2Q2x1,x2t2−Q2x1,x2t1→0ast2→t1.
This proves the equicontinuity of the operatorS2.Step 6.
We prove that the operatorS is condensing. Hence, we have to show that for any bounded subset D⊂Ωr,μSD<μD.
SinceS2 is continuous, for any bounded set D⊂Ωr, there exists a countable set D0=x1n,x2n⊂D such that μS2D=μS2D0.
We know thatS2 is bounded and equicontinuous; it follows that
(51)μs2D0=maxt∈si,ti+1μS2D0t,i=0,1,⋯,m,we recall that S2x1,x2t=P2x1,x2,Q2x1,x2twe have
(52)μP2D0t=μ∫sitSβ1t−sg1s,x1ns,x2ns,∫0τu1s,τφ1s,τ,x1nτ,x2nτdτ−∫sitSβ1t−sA1f1s,x1ns,x2nsds≤M1Γβ1∫sitt−sβ1−1μg1s,x1ns,x2ns,∫0τu1s,τφ1s,τ,x1nτ,x2nτdτ+M1Γβ1∫sitt−sβ1−1μA1,f1s,x1ns,x2nsds≤M1Γβ1μD0s∫sitt−sβ1−1Mg11Ψ^11s+Mg12su1∗v^1∗ds+MΓβ1μD0s∫sitt−sβ1−1A1Lf1ds≤M1Tβ1−δΓβ11−δβ1−δ1−δMg11ψ^11L1δ0,T+Mg12u1∗v^1∗ψ^12L1δ0,TμD+M1Lf1Tβ1−δΓβ1+1μD,which implies that
(53)μP2D≤M1Tβ1−δΓβ11−δβ1−δ1−δMg11ψ^11L1δ0,T+Mg12u1∗v^1∗ψ^12L1δ0,TμD+M1Lf1Tβ1−δΓβ1+1μD.
Similarly, we show that(54)μQ2D≤M2Tβ2−δΓβ21−δβ2−δ1−δMg21ψ^21L1δ0,T+Mg22u2∗v^2∗ψ^22L1δ0,TμD+M2Lf2Tβ2−δΓβ2+1μD.
Hence, we get(55)μS2D≤maxμP2D,μQ2D≤maxj=1,2MjTβj−δΓβj1−δβj−δ1−δMgj1ψ^j1L1δ0,T+Mgj2uj∗v^j∗ψ^j2L1δ0,T+MjLfjTβj−δΓβj+1μD.
Since the operatorS1 is a Lipschitz operator with constant maxM1Kmi+Mfi+Nfi;M2Kni+Mf2+Nf2 for any bounded set D⊂Ωr, we have
(56)μS1D0≤maxM1Kmi+Mfi+Nfi;M2Kni+Mf2+Nf2μD.
As the operatorS=S1+S2, we obtain μSD≤μS1D+μS2D≤maxM1Kmi+Mf1+Nf1;M2Kni+Mf2+Nf2+maxj=1,2MjTβj−δ/Γβj1−δ/βj−δ1−δMgj1ψ^j1L1δ/1δ0,T+Mgj2ψ^j2L1δ/1δ0,T+MjLfjTβj−δ/Γβj+1μD<μD
Thus,S:Ωr→Ωr is a condensing operator. Hence, by Sadovskii’s fixed point theorem [26], the operator S has at least a pair of solutions x1t,x2t. Therefore, the problem (1) has a pair of solutions x1t,x2t. This completes the proof.


## 4. Example

We consider in this example the following problem onJ=0,1(57)DH012,18x1t,w+sinx1t,w+sinx2t,w40=∂x1t,w∂w2+x1t,w101+et1/4+x2t,w101+et1/4,+e−t/25∫0te−tt2sinx1s,w+sinx2t,wds,∈0,13∪23,1DH012,18x2t,w+sinx1t,w+sinx2t,w40=∂x2t,w∂w2+x1t,w101+et1/4+x2t,w101+et1/4,+e−t/25∫0te−tt2sinx1s,w+sinx2t,wds,t∈0,13∪23,1x1t,w=costx1t,w25+x1t,w,t∈13,2,3,x2t,w=costx2t,w25+x2t,w,t∈13,2,3,x1t,0=x1t,1=0,t∈0,1,x2t,0=x2t,1=0,t∈0,1,I0+1−α1x10,w+f10,x10,w,x20,w=x01,I0+1−α2x20,w+f20,x10,w,x20,w=x02,LetX=L20,1 and for j=1,2,Ajxj=xj′′ withDAj=xj∈X:xj,xj′ are absolutely continuous, and xj0=xj1=0.The operatorsAj generates equicontinuous C0-semigroups Tjt,t≥0 on X with Tjt≤1fort≥0.We have the following forj=1,2(58)βj=12,γj=18,αj=916,T=1gjt,x1t,x2t,∫0tujt,sφjt,s,x1s,wx2s,wds=x1t,w101+et1/4+x2t,w101+et1/4+e−t/25∫0te−tt2sinx1s,w+sinx2t,wds,φjt,s,x1s,w,x2s,w=e−tt2sinx1s,w+sinx2t,w,ujt,s=1fjt,x1t,w,x2t,w=sinx1t,w+sinx2t,w40mit,x1t,w=costx1t,w25+x1t,w,nit,x2t,w=costx2t,w25+x2t,w.Takingxjt,w=xjt, we can see that
(59)gjt,x1t,x2t,∫0tujt,sφjt,s,x1s,w,x2s,wds≤ψj1tx1t+ψj2tx2t+ψj3t∫0tφjt,s,x1s,x2sdsψj1t=ψj2t=1101+et1/4,ψj3t=e−t/25.Similarly, we have(60)μgjt,x1t,x2t,∫0tujt,sφjt,s,x1s,w,x2s,wds≤ψ^j1tμx1t,x2t+ψ^j2tμ∫0tφjt,s,x1s,x2sds,ψ^j1t=1101+et1/4,ψ^j2t=e−t/25.φjt,s,x1s,w,x2s,w≤ξjt,sx1s,w,x2s,wwith ξjt,s=e−s/t2 and ξj∗=supt∈0,1∫0tξjt,sds=0.63212,We takeδ=1/4(61)ψj1L1δ0,1=ψj2L1δ0,1=∫011101+et1/4dt414=0.063andψj3tL1δ0,1=∫01e−t/25dt414=0.102.Forj=1,2,i=1,2,⋯,m, the functions fjt,x1t,x2t,mit,x1t and nit,x2t are Lipschitz functions with constants:Lfj=1/20,Mfj=Nfj=1/40 and Kmi=Kni=1/25. Thus, we have
(62)βj=12,γj=18,αj=916,δ=14,Mj=1,uj∗=1,Mgj1=Mgj2=1,Lfj=140,Lmi=Lni=125,ψj1=ψj2=ψ^j1=0.063,ψj3=ψ^j2=0.102,v^j∗=ξj∗=0.63212.For these values, the condition (1) of theorem 12 is satisfied:we have forj=1,2:
(63)MjKjΓαj+T1−αj+βj+δΓβj1−δβj−δ1−δψj1+ψj2+ψj3Tβj+1−αjuj∗ξj∗βjΓβj≃0.260<1.The second condition is also verified:(64)maxj=1,2MjKj+Mfj+Nfj+maxj=1,2MjTβj−δΓβj1−δβj−δ1−δMgj1ψ^j1L1δ0,T+Mgj2uj∗v^j∗ψ^j2L1δ0,T+MjLfjTβj−δΓβj+1≃0.3103<1Consequently, both conditions are satisfied, which means that the problem (1) has a couple of solutions x1t,x2t in the spaceX=PC1−α1×PC1−α2.


## 5. Conclusion

In this paper, we achieved the existence of solutions for a class of impulsive Hilfer fractional coupled systems by converting the problem to an integral form and then using the Sadovskii’s fixed point theorem. For future works, we can consider other fractional operators for example theψ-Hilfer fractional operator for its new results and applications.



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