# Existence and Uniqueness of Solutions for Coupled Impulsive Fractional Pantograph Differential Equations with Antiperiodic Boundary Conditions

Authors: Karim Guida; Lahcen Ibnelazyz; Khalid Hilal; Said Melliani
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this paper, we investigate the solutions of coupled fractional pantograph differential equations with instantaneous impulses. The work improves some existing results and contributes toward the development of the fractional differential equation theory. We first provide some definitions that will be used throughout the paper; after that, we give the existence and uniqueness results that are based on Banach’s contraction principle and Krasnoselskii’s fixed point theorem. Two examples are given in the last part to support our study.


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## Body

## 1. Introduction

Fractional differential equations (FDEs) involve fractional derivatives of the formdα/dxα, which are defined for α>0, where α is not necessarily an integer. They are generalizations of the ordinary differential equations to a random (noninteger) order. These FDEs have attracted considerable interest due to their ability to model complex phenomena. The fractional differential operators are global, and they are used to model several physical phenomena because they give accurate results. For the new readers that are interested in the fractional calculus theory in a more general concept, please see [1–6] and the references therein. Our work is concerned with impulsive coupled systems of pantograph FDEs. Impulsive FDEs have found applications in many areas such as business mathematics, management sciences, and population dynamics. Some physical problems have sudden changes and discontinuous jumps. To model these problems, we impose impulsive conditions on the differential equations at discontinuity points; for more details about impulsive fractional differential equations, we give the following references [7–13].Many papers have studied impulsive fractional differential equations with antiperiodic boundary conditions, and results on the existence and uniqueness have been given (see [14–17]). For example, recently, Zuo et al. [18] investigated the existence results for an equation with impulsive and antiperiodic boundary conditions given by
(1)Dcαxt+γxt=ft,xt,Axt,Bxt,t∈J=0,1,t≠ti,i=1,2,⋯,m,Δxt=ti0=Iixti,i=1,⋯,m,x0=−x1,where Dcα is the Caputo fractional derivative of order α∈0,1, γ>0, Ii∈ℝ, 0=t0<t1<⋯<ti+1=1, the function f is in CJ×ℝ3,ℝ, Δx∣t=ti denotes the jump of xt at t=ti, and A and B are linear operators. The authors established the existence and uniqueness under some conditions using Banach’s and Krasnoselskii’s fixed point theorems.On the other hand, in the deterministic situation, there is a very special case of delay differential equations known as the pantograph equations given by(2)g′t=kgt+lgλt,t∈0,b,b>0,0<λ<1,g0=g0.These equations are also called equations with proportional delays. Pantographs are special devices mounted on electric trains to collect current from one or several contact wires. They consist of a pantograph head, frame, base, and drive system, and their geometrical shape is variable. But it is recently being used in electric trains. Many researchers have investigated the pantograph differential equations and their properties; see [19–21].Motivated by all the previous works, we consider in this paper coupled impulsive fractional pantograph differential equations with antiperiodic boundary conditions as follows:(3)Dcα1xt+γ1xt=f1t,xt,xλ1t,yt,t∈J,t≠ti,i=1,2,⋯,m,0<α1<1,0<λ1<1,Dcα2yt+γ2yt=f2t,xt,yt,yλ2t,t∈J,t≠tj,j=1,2,⋯,n,0<α2<1,0<λ2<1,Δxt=ti0=Iixti,i=1,⋯,m,Δyt=tj0=Ijxtj,j=1,⋯,n,a1x0+b1x1=0,a1≥b1>0,a2y0+b2y1=0,a2≥b2>0,where Dcα1 and Dcα2 are the Caputo fractional derivatives of orders α1 and α2, respectively, f1,f2:J×ℝ3→ℝ are two continuous functions, J=0,1, Δx∣t=ti=xti+−xti−, with xti+ and xti− representing the right and left limits of xt at t=ti, i=1,⋯,m, and also Δy∣t=tj=ytj+−ytj−, with ytj+ and ytj− representing the right and left limits of yt at t=tj, j=1,⋯,n.The objective of this paper is to establish the existence and uniqueness results of the solutions of problem (3) by means of Banach’s contraction principle and Krasnoselskii’s fixed point theorem.The main contributions of this paper are as follows:(i)
We consider a new system of impulsive pantograph fractional differential equations(ii)
We consider antiperiodic boundary value conditions with a more general formThis paper contributes toward the development of qualitative analysis of impulsive fractional differential equations.This paper is organized as follows: in Section2, we give some definitions and useful lemmas that will be used throughout the work; after that, in Section 3, we will establish the existence and uniqueness results by means of the fixed point theorems; last but not least, in Section 4, we give two illustrative examples.


## 2. Preliminaries and Lemmas

LetJ0=0,t1,J1=t1,t2,⋯,Jm=tm,1, and PCJ,X=x:J→ℝ:x∈CJi,ℝ, where i=0,1,2,⋯,m, xti+ and xti− exist, i=1,⋯,m, is a space of continuous real-valued functions on the interval J, and xti−=xti.Similarly, we definePCJ,Y=y:J→ℝ:y∈CJj,ℝ, where j=0,1,2,⋯,n, ytj+ and ytj− exist, j=1,⋯,n, is a space of continuous real-valued functions on the interval J, and ytj−=ytj.Then, clearly,PCJ,X and PCJ,Y are two Banach spaces with the norms x=supt∈0,1xt and y=supt∈0,1yt, respectively.Consequently, the spacePCJ,X×PCJ,Y is a Banach space with the norm x,y=x+y.We note that the spaceLpJ,ℝ is a Banach space of Lebesque measurable functions with .LpJ<∞.Definition 1 (see [1]).
The fractional integral of orderα with the lower limit zero for a function f is defined as
(4)I0+αft=1Γα∫0tt−sα−1fsds,α>0,n∈ℕ,provided the right-hand side is pointwise defined on 0,∞, where Γ. denotes the Gamma function.Definition 2 (see [1]).
The Riemann-Liouville derivative of orderα with the lower limit zero for a function f is defined as
(5)DL0+αft=1Γn−αdndtn∫0tt−sn−α−1fsds,α>0,n−1<α<n,provided the function f is absolutely continuous up to order n−1 derivatives, where Γ. denotes the Gamma function.Definition 3 (see [1]).
The Caputo derivative of orderα>0 with the lower limit zero for a function f is defined as
(6)Dc0+αft=1Γn−αdndtn∫0tt−sn−α−1fs−∑k=0n−1tkk!fk0ds,n=α+1,n∈ℕ,provided the function f:0,∞→R, where Γ. denotes the Gamma function.Definition 4.
A couplex,y is a solution of problem (3) if it satisfies the equations
(7)Dcα1xt+γ1xt=f1t,xt,xλ1t,yt,t≠ti,i=1,2,⋯,m,Dcα2yt+γ2yt=f2t,xt,yt,yλ2t,t≠tj,j=1,2,⋯,n,a.e. on J, and the conditions Δx∣t=ti0=Iixti,Δy∣t=tj0=Ijytj,i=1,⋯,m,j=1,⋯,n and a1x0+b1x1=0,a2y0+b2y1=0,a1≥b1>0,a2≥b2>0.Lemma 5 (see [22]).
The nonnegative functionsEα and Eα,α given by
(8)Eαz=∑k=0∞zkΓαk+1,Eα,αz=∑k=0∞zkΓαk+α,have the following properties:
(1)
For anyγ>0 and t∈J,(9)Eα−tαγ≤1,Eα,α−tαγ≤1Γα.
In addition, we haveEα0=1 and Eα,α0=1/Γα.
(2)
For anyγ>0 and t1,t2∈J,(10)Eα−t2αγ→Eα−t1αγ,ast2→t1,Eα,α−t2αγ→Eα,α−t1αγ,ast2→t1.(3)
For anyγ>0 and t1,t2∈J such that t1≤t2,(11)Eα−t2αγ≤Eα−t1αγ,Eα,α−t2αγ≤Eα,α−t1αγ.Lemma 6 (see [23]).
LetM be a closed, convex, and nonempty subset of a Banach space X, and let F1 and F2 be operators such that
(1)
F1x+F2y∈M whenever x,y∈M(2)
F1 is compact and continuous(3)
F2 is a contraction mapping
Then, there existsz∈M such that z=F1z+F2z.Lemma 7 (see [24]).
LetX be a Banach space, and let J=0,T. Suppose that W⊂PCJ,X satisfies the following conditions:
(1)
W is a uniformly bounded subset of PCJ,X(2)
W is equicontinuous in ti,ti+1, i=0,1,⋯,m, where t0=0 and tm+1=T(3)
Itst-sections Wt=xt:x∈W,t/t1,⋯,tm, Wti+=xti+:x∈W, and Wti−=xti−:x∈W are relatively compact subsets of X
Then,W is a relatively compact subset of PCJ,X.Lemma 8 (see [25]).
Letf1,f2→ℝ be two continuous functions. The couple x,y given by
(12)xt=−Eα1−γ1Eα1−tα1γ11+Eα1−γ1∑k=1mukEα1−tkα1γ1+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1sds−Eα1−tα1γ11+Eα1−γ1∫011−sα1−1Eα1,α1−1−sα1γ1f1sds,t∈J0,Eα1−tα1γ11+Eα1−γ1∑k=1mukEα1−tkα1γ1−∫011−sα1−1Eα1,α1−1−sα1γ1f1sds−Eα1−tα1γ1∑p=i+1mupEα1−tpα1γ1+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1sds,t∈Ji,i=1,2,⋯,m−1,Eα1−tα1γ11+Eα1−γ1∑k=1mukEα1−tkα1γ1−∫011−sα1−1Eα1,α1−1−sα1γ1f1sds+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1sds,t∈Jm,(13)yt=−Eα2−γ2Eα2−tα2γ21+Eα2−γ2∑k=1nvkEα2−tkα2γ2+∫0tt−sα2−1Eα2,α2−t−sα2γ2f2sds−Eα2−tα2γ21+Eα2−γ2∫011−sα2−1Eα2,α2−1−sα2γ2f2sds,t∈J0,Eα2−tα2γ21+Eα2−γ2∑k=1nvkEα2−tkα2γ2−∫011−sα2−1Eα2,α2−1−sα2γ2f2sds−Eα2−tα2γ2∑p=j+1mvpEα2−tpα2γ2+∫0tt−sα2−1Eα2,α2−t−sα2γ2f2sds,t∈Jj,j=1,2,⋯,n−1,Eα2−tα2γ21+Eα2−γ2∑k=1nvkEα2−tkα2γ2−∫011−sα2−1Eα2,α2−1−sα2γ2f2sds+∫0tt−sα2−1Eα2,α2−t−sα2γ2fssds,t∈Jn,is a solution of the impulsive problem
(14)Dcα1xt+γ1xt=f1t,t∈J,t≠ti,i=1,2,⋯,m,Dcα2yt+γ2yt=f2t,t∈J,t≠tj,j=1,2,⋯,n,Δxt=ti0=ui,i=1,⋯,m,Δyt=tj0=vj,j=1,⋯,n,x0+x1=0,y0+y1=0.It follows from Lemma8, and by using the boundary conditions a1x0+b1x1=0 and a2y0+b2y1=0, that the solution of (3) can be expressed as follows:
(15)xt=−Eα1−γ1Eα1−tα1γ11+Eα1−γ1∑k=1mIkxtkEα1−tkα1γ1+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1s,xs,xλ1s,ysds−Eα1−tα1γ11+σ1Eα1−γ1σ1∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ysds,t∈J0,Eα1−tα1γ11+σ1Eα1−γ1∑k=1mIkxtkEα1−tkα1γ1−σ1∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ysds−Ea1−tα1γ1∑p=i+1mIpxtpEα1−tpα1γ1+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1s,xs,xλ1s,ysds,t∈Ji,i=1,2,⋯,m−1,Eα1−tα1γ11+σ1Eα1−γ1∑k=1mIkxtkEα1−tkα1γ1−σ1∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ysds+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1s,xs,xλ1s,ysds,t∈Jm,(16)yt=−Eα2−γ2Eα2−tα2γ21+Eα2−γ2∑k=1nIkytkEα2−tkα2γ2+∫0tt−sα2−1Eα2,α2−t−sα2γ2f2s,xs,ys,yλ2sds−Eα2−tα2γ21+σ2Eα2−γ2σ2∫011−sα2−1Eα2,α2−1−sα2γ2f2s,xs,ys,yλ2sds,t∈J0,Eα2−tα2γ21+σ2Eα2−γ2∑k=1nIkytkEα2−tkα2γ2−σ2∫011−sα1−1Eα2,α2−1−sα2γ2f2s,xs,ys,yλ2sds−Eα2−tα2γ2∑p=j+1nIpytpEα2−tpα2γ2+∫0tt−sα2−1Eα2,α2−t−sα2γ2f2s,xs,ys,yλ2sds,t∈Jj,j=1,2,⋯,n−1,Eα2−tα2γ21+σ2Eα2−γ2∑k=1nIkytkEα2−tkα2γ2−σ2∫011−sα2−1Eα2,α2−1−sα2γ2f2s,xs,ys,yλ2sds+∫0tt−sα2−1Eα2,α2−t−sα2γ2f2s,xs,ys,yλ2sds,t∈Jn,where σ1=b1/a1 and σ2=b2/a2.


## 3. Main Results

Theorem 9.
We consider the following hypotheses:
H1. The functions f1,f2:J×ℝ3→ℝ are continuous, and there exist two constants L1,L2>0 such that
(17)f1t,x,y,z−f1t,u,v,w≤L1x−u+y−v+z−w,f2t,x,y,z−f2t,u,v,w≤L2x−u+y−v+z−w,for all t∈J,x,y,z,u,v,w∈ℝ.
H2. Iix−Iiy≤C1x−y and
(18)Ijx−Ijy≤C2x−y,for all x,y∈ℝ, i=1,2,⋯,m, and j=1,2,⋯,n.
H3. We suppose that μ1+μ2<1.
Then, problem (3) has a unique solution x∗,y∗.Remark 10.
The expressions ofμ1 and μ2 are given in the proof.Proof.
We define the operatorT:PCJ,X×PCJ,Y→PCJ,X×PCJ,Y by
(19)Tx,yt=Ux,yt,Vx,yt, where
(20)Ux,yt=Eα1−tα1γ11+σ1Eα1−γ1∑k=1mIkxtkEα1−tkα1γ1−σ1∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ysds−Eα1−tα1γ1∑p=i+1mIpxtpEα1−tpα1γ1+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1s,xs,xλ1s,ysds,t∈ti,ti+1,i=0,1,2,⋯,m,Vx,yt=Eα2−tα2γ21+σ2Eα2−γ2∑k=1nIkytkEα2−tkα2γ2−σ2∫011−sα2−1Eα2,α2−1−sα2γ2f2s,xs,ys,yλ2sds−Eα2−tα2γ2∑p=j+1nIpytpEα2−tpα2γ2+∫0tt−sα2−1Eα2,α2−t−sα2γ2f2s,xs,ys,yλ2sds,t∈tj,tj+1,j=0,1,2,⋯,n.
We show now that the operatorT has a fixed point, which is a solution of problem (3).
LetK1=supt∈Jf1t,0,0,0, K2=supt∈Jf2t,0,0,0, K1∗=maxIi0:i=1,2,⋯,m, and K2∗=maxIj0:j=1,2,⋯,n.
We choose(21)r≥∑k=1mK1∗/Eα1−tkα1γ1+K1/Γα1+11+σ1Eα1−γ1/4+2σ1−∑k=1mC1/Eα1−tkα1γ1+6L1σ1+1/2+σ1Γα1+1+∑k=1nK2∗/Eα2−tkα2γ2+K2/Γα2+11+σ2Eα2−γ2/4+2σ2−∑k=1nC2/Eα2−tkα2γ2+6L2σ2+1/2+σ2Γα2+1.
Firstly, we show thatTBr⊂Br, where Br=x,y∈PCJ,X×PCJ,Y:x,y≤r. It follows from the hypotheses above and Lemma 5 that for any x,y∈Br, we have
(22)Ux,yt≤Eα1−tα1γ111+σ1Eα1−γ1∑k=1mIkxtkEα1−tkα1γ1−σ1∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ysds−∑p=i+1mIpxtpEα1−tpα1γ1+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1s,xs,xλ1s,ysds≤11+σ1Eα1−γ1∑k=1mIkxtkEα1−tkα1γ1+σ1Γα1∫011−sα1−1f1s,xs,xλ1s,ysds+∑k=1mIkxtkEα1−tkα1γ1+1Γα1∫0tt−sα1−1f1s,xs,xλ1s,ysds≤1+1+σ1Eα1−γ11+σ1Eα1−γ1∑k=1mIkxtk−Ik0+K1∗Eα1−tkα1γ1+σ1Γα11+σ1Eα1−γ1×∫011−sα1−1f1s,xs,xλ1s,ys−fs,0,0,0+fs,0,0,0ds+1Γα1∫0tt−sα1−1f1s,xs,xλ1s,ys−fs,0,0,0+fs,0,0,0ds≤2+σ11+σ1Eα1−γ1∑k=1mC1r+K1∗Eα1−tkα1γ1+K1Γα1+11+σ1Eα1−γ1+K1Γα1+1+σ1Γα11+σ1Eα1−γ1∫011−sα1−1L1xs+xλ1s+ysds+1Γα1∫0tt−sα1−1L1xs+xλ1s+ysds≤2+σ11+σ1Eα1−γ1∑k=1mC1r+K1∗Eα1−tkα1γ1+K1Γα1+1+3rL1σ1Γα1+11+σ1Eα1−γ1+3rL1Γα1+1≤2+σ11+σ1Eα1−γ1∑k=1mK1∗Eα1−tkα1γ1+K1Γα1+1+∑k=1mC1Eα1−tkα1γ1+6L1σ1+12+σ1Γα1+1r≤r2.
Similarly, we show that(23)Vx,yt≤2+σ21+σ2Eα2−γ2∑k=1nK2∗Eα2−tkα2γ2+K2Γα1+1+∑k=1nC2Eα2−tkα2γ2+6L2σ2+12+σ2Γα2+1r≤r2.
Finally,(24)Tx,yt≤Ux,yt+Vx,yt≤r,which implies that TBr⊂Br.
Next, we show that the operatorT is a contraction; we let x,y,x¯,y¯∈X×Y; then, for t∈J, we have
(25)Ux,yt−Ux¯,y¯t=Eα1−tα1γ11+σ1Eα1−γ1∑k=1mIkxtk−Ikx¯tkEα1−tkα1γ1−σ1∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ys−f1s,x¯s,x¯λ1s,y¯sds−Eα1−tα1γ1∑p=i+1mIpxtp−Ipx¯tpEα1−tpα1γ1+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1s,xs,xλ1s,ys−f1s,x¯s,x¯λ1s,y¯sds≤11+σ1Eα1−γ1+1∑k=1mC1xtk−x¯tkEα1−tkα1γ1+σ1Γα11+σ1Eα1−γ1∫011−sα1−1L1xs−x¯s+xλ1s−x¯λ1s+ys−y¯sds+1Γα1∫0tt−sα1−1L1xs−x¯s+xλ1s−x¯λ1s+ys−y¯sds≤2+σ11+σ1Eα1−γ1∑k=1mC1Eα1−tkα1γ1x−x¯+3L1σ1Γα1+11+σ1Eα1−γ1x−x¯+y−y¯+3L1Γα1+1x−x¯+y−y¯≤2+σ11+σ1Eα1−γ1∑k=1mC1Eα1−tkα1γ1+6L1σ1+12+σ1Γα1+1x−x¯+y−y¯=μ1x−x¯+y−y¯.
With a similar method, we also get(26)Vx,yt−Vx¯,y¯t≤2+σ21+σ2Eα2−γ2∑k=1nC2Eα2−tkα2γ2+6L2σ2+12+σ2Γα2+1x−x¯+y−y¯=μ2x−x¯+y−y¯.
Finally, we can obtain(27)Tx,yt−Tx¯,y¯t≤μ1+μ2x−x¯+y−y¯.
And sinceμ1+μ2<1, then the operator T is a contraction.
Therefore, we conclude by Banach’s contraction mapping principle thatT has a fixed point which is the unique solution x∗,y∗ of problem (3). The proof is now completed.Next, we present a result based on Krasnoselskii’s fixed point theorem.Theorem 11.
Assume that the conditionH2 and the following additional conditions are satisfied:
H4. Two functions φ1,ψ1∈L1/ρ10,+∞0<ρ1<α1<1, φ2,ψ2∈L1/ρ20,+∞0<ρ2<α2<1, and ω1,ω2,η1,η2∈C0,+∞ are nondecreasing functions satisfying the following inequalities:
(28)f1t,xs,xλ1s,ys≤φ1tω1x+ψ1tη1y,f2t,xs,ys,yλ2s≤φ2tω2x+ψ2tη2y,for all x,y∈PCJ,X×PCJ,Y,t∈J.
H5. We suppose that ε1+ε2<1.
Then, problem (3) has at least one solution.Remark 12.
The expressions ofε1 and ε2 are given in the proof.Proof.
The setBr=x,y∈PCJ,X×PCJ,Y:x,y≤r is a closed, bounded, and convex set in PCJ,X×PCJ,Y for all r>0.We define the operatorT by Tx,yt=Ux,yt,Vx,yt for any x,y∈Br and t∈a,b, where
(29)Ux,yt=Eα1−tα1γ11+σ1Eα1−γ1∑k=1mIkxtkEα1−tkα1γ1−σ1∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ysds−Eα1−tα1γ1∑p=i+1mIpxtpEα1−tpα1γ1+∫0tt−sα1−1Eα1,α1−t−sα1γ1f1s,xs,xλ1s,ysds,t∈ti,ti+1,i=0,1,2,⋯,m,Vx,yt=Eα2−tα2γ21+σ2Eα2−γ2∑k=1nIkytkEα2−tkα2γ2−σ2∫011−sα2−1Eα2,α2−1−sα2γ2f2s,xs,ys,yλ2sds−Eα2−tα2γ2∑p=j+1nIpytpEα2−tpα2γ2+∫0tt−sα2−1Eα2,α2−t−sα2γ2f2s,xs,ys,yλ2sds,t∈tj,tj+1,j=0,1,2,⋯,n.By splitting the two operators above, we have(30)U1x,yt=∫0tt−sα1−1Eα1,α1−t−sα1γ1f1s,xs,xλ1s,ysds−σ1Eα1−tα1γ11+σ1Eα1−γ1∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ysds,U2x,yt=Eα1−tα1γ11+σ1Eα1−γ1∑k=1mIkxtkEα1−tkα1γ1−Eα1−tα1γ1∑p=i+1mIpxtpEα1−tpα1γ1,V1x,yt=∫0tt−sα2−1Eα2,α2−t−sα2γ2f2s,xs,ys,yλ2sds−σ2Eα2−tα2γ21+σ2Eα2−γ2∫011−sα2−1Eα2,α2−1−sα2γ2f2s,xs,ys,yλ2sds,V2x,yt=Eα2−tα2γ21+σ2Eα2−γ2∑k=1nIkytkEα2−tkα2γ2−Eα2−tα2γ2∑p=j+1nIpytpEα2−tpα2γ2.This upcoming part of the proof requires us to rewrite the operatorT as
(31)Tx,yt=T1x,yt+T2x,yt,where
(32)T1x,yt=U1x,yt,V1x,yt,(33)T2x,yt=U2x,yt,V2x,yt.It follows fromH4 and Holder’s inequality that for any x,y∈Br and each t∈J, we have
(34)∫0tt−sα1−1Eα1,α1−t−sα1γ1f1s,xs,xλ1s,ysds≤1Γα1∫0tt−sα1−1φ1sω1r+ψ1sη1rds≤1Γα1∫0tt−sα1−1φ1sω1r+∫0tt−sα1−1ψ1sη1rds≤1Γα1∫0tt−sα1−1/1−ρ1ds1−ρ1∫0tω1rφ1s1/ρ1dsρ1+1Γα1∫0tt−sα1−1/1−ρ1ds1−ρ1∫0tη1rψ1s1/ρ1dsρ1≤1Γα1φ1L1/ρ1Jα1−ρ1/1−ρ11−ρ1ω1r+1Γα1ψ1L1/ρ1Jα1−ρ1/1−ρ11−ρ1η1r.By the same method used above, we get(35)∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ysds≤1Γα1φ1L1/ρ1Jα1−ρ1/1−ρ11−ρ1ω1r+1Γα1ψ1L1/ρ1Jα1−ρ1/1−ρ11−ρ1η1r.Next, we show thatTx,y=Ux,y+Vx,y is a bounded operator, which means that Tx,y∈Br, for x,y∈Br.Suppose the opposite; so there existx,y∈Br and t∈J such that Ux,y>r/2 and Vx,y>r/2. Assumption H2 implies that Iixti≤Iixti−Ii0+Ii0≤C1r+K1∗ and Ijytj≤Ijytj−Ij0+Ij0≤C2r+K2∗.Thus,(36)r2<Ux,y≤φ1L1/ρ1JΓα1α1−ρ1/1−ρ11−ρ1ω1r+ψ1L1/ρ1JΓα1α1−ρ1/1−ρ11−ρ1η1r+φ1L1/ρ1JΓα11+σ1Eα1−γ1α1−ρ1/1−ρ11−ρ1ω1r+ψ1L1/ρ1JΓα11+σ1Eα1−γ1α1−ρ1/1−ρ11−ρ1η1r+11+σ1Eα1−γ1∑k=1mC1r+K1∗Eα1−tkα1γ1+∑k=1mC1r+K1∗Eα1−tkα1γ1≤2+σ11+σ1Eα1−γ11+2σ1φ1L1/ρ1J2+σ1Γα1α1−ρ1/1−ρ11−ρ1ω1r+1+2σ1ψ1L1/ρ1J2+σ1Γα1α1−ρ1/1−ρ11−ρ1η1r+∑k=1mC1r+K1∗Eα1−tkα1γ1.Dividing both sides byr and taking the lower limit as r→+∞, we get
(37)12≤2+σ11+σ1Eα1−γ11+2σ1φ1L1/ρ1J2+σ1Γα1α1−ρ1/1−ρ11−ρ1limr→+∞infω1r+1+2σ1ψ1L1/ρ1J2+σ1Γα1α1−ρ1/1−ρ11−ρ1limr→+∞infη1r+∑k=1mC1Eα1−tkα1γ1=ε1.Similarly, we have(38)12≤2+σ21+σ2Eα2−γ21+2σ2φ2L1/ρ2J2+σ2Γα2α2−ρ2/1−ρ21−ρ2limr→+∞infω2r+1+2σ2ψ2L1/ρ2J2+σ2Γα2α2−ρ2/1−ρ21−ρ2limr→+∞infη2r+∑k=1nC2Eα2−tkα2γ2=ε2.So we getε1+ε2≥1, which contradicts H5. Hence, Tx,y=Ux,y+Vx,y is a bounded operator, for all x,y∈Br.Now, we show thatT2 is a contraction mapping.Thus, for allt∈J and x,y,x¯,y¯∈PCJ,X×PCJ,Y, we get
(39)U2x,yt−U2x¯,y¯t≤Eα1−tα1γ11+σ1Eα1−γ1∑k=1mIkxtk−Ikx¯tkEα1−tkα1γ1+Eα1−tα1γ1∑k=1mIkxtk−Ikx¯tkEα1−tkα1γ1≤∑k=1mIkxtk−Ikx¯tkEα1−tkα1γ11+11+σ1Eα1−γ1≤2+σ11+σ1Eα1−γ1∑k=1mC1Eα1−tkα1γ1x−x¯+y−y¯=ε1∗x−x¯+y−y¯.Similarly, we show that(40)V2x,yt−V2x¯,y¯t≤2+σ21+σ2Eα2−γ2∑k=1nC2Eα2−tkα2γ2x−x¯+y−y¯=ε2∗x−x¯+y−y¯.It follows fromH5 that 0<ε1∗+ε2∗<1, and
(41)T2x,yt−T2x¯,y¯t≤ε1∗+ε2∗x−x¯+y−y¯.Thus,T2 is a contraction mapping.Since the functionsf1 and f2 are continuous, this implies that the operator T1 is also continuous. Now, we show that T1=U1,V1 is compact; we apply the same method as in Theorem 9. One can verify easily that T1Br is uniformly bounded on PCJ,X×PCJ,Y. We now show that T1Br is equicontinuous on J. Let f1∗=supt∈Jf1t,xt,xλ1t,yt and f2∗=supt∈Jf2t,xt,yt,yλ2t; then, for any ti<ξ2<ξ1<ti+1, we have
(42)U1x,yξ2−U1x,yξ1≤∫0ξ2ξ2−sα1−1Eα1,α1−ξ2−sα1γ1f1s,xs,xλ1s,ysds−∫0ξ1ξ1−sα1−1Eα1,α1−ξ1−sα1γ1f1s,xs,xλ1s,ysds+σ1Eα1−ξ2α1γ1−Eα1−ξ1α1γ11+σ1Eα1−γ1×∫011−sα1−1Eα1,α1−1−sα1γ1f1s,xs,xλ1s,ysds≤∫0ξ2ξ2−sα1−1Eα1,α1−ξ2−sα1γ1f1s,xs,xλ1s,ysds−∫0ξ2ξ1−sα1−1Eα1,α1−ξ2−sα1γ1f1s,xs,xλ1s,ysds+∫0ξ2ξ1−sα1−1Eα1,α1−ξ2−sα1γ1f1s,xs,xλ1s,ysds−∫0ξ2ξ1−sα1−1Eα1,α1−ξ1−sα1γ1f1s,xs,xλ1s,ysds−∫ξ2ξ1ξ1−sα1−1Eα1,α1−ξ1−sα1γ1f1s,xs,xλ1s,ysds+σ1Eα1−ξ2α1γ1−Eα1−ξ1α1γ1Γα11+σ1Eα1−γ1∫011−sα1−1f1s,xs,xλ1s,ysds≤∫0ξ2ξ2−sα1−1−ξ1−sα1−1Eα1,α1−ξ2−sα1γ1f1∗ds+∫0ξ2ξ1−sα1−1Eα1,α1−ξ2−sα1γ1−Eα1,α1−ξ1−sα1γ1f1∗ds+f1∗Γα1∫ξ2ξ1ξ1−sα1−1+σ1Eα1−ξ2α1γ1−Eα1−ξ1α1γ1Γα1+11+σ1Eα1−γ1f1∗≤f1∗Γα1∫0ξ2ξ2−sα1−1−ξ1−sα1−1ds+ξ1−ξ2α1f1∗Γα1+1+σ1Eα1−ξ2α1γ1−Eα1−ξ1α1γ1Γα1+11+σ1Eα1−γ1f1∗+f1∗∫0ξ2ξ1−sα1−1Eα1,α1−ξ2−sα1γ1−Eα1,α1−ξ1−sα1γ1ds≤ξ1−ξ2α1+ξ1α1−ξ2α1Γα1+1f1∗+ξ1−ξ2α1f1∗Γα1+1+σ1Eα1−ξ2α1γ1−Eα1−ξ1α1γ1Γα1+11+σ1Eα1−γ1f1∗+f1∗∫0ξ2ξ1−sα1−1Eα1,α1−ξ2−sα1γ1−Eα1,α1−ξ1−sα1γ1ds.By (3) of Lemma 5, it follows that Eα1,α1−tα1γ1 is continuous on t∈J, and thus, Eα1,α1−tα1γ1 is uniformly continuous on t∈J; hence, for every real number ε>0, there exists a small κ>0 such that for t1,t2∈J with t1−t2≤κ, we have
(43)Eα1,α1−t1α1γ1−Eα1,α1−t2α1γ1<εξ2α1/2−α1.Letp=2−α1/21−α1 and q=2−α1/α1. Then, p>1, q>1, and 1/p+1/q=1. Applying Holder’s inequality yields
(44)∫0ξ2ξ1−sα1−1Eα1,α1−ξ2−sα1γ1−Eα1,α1−ξ1−sα1γ1ds≤∫0ξ2ξ1−sα1−12−α1/21−α1ds21−α1/2−α1×∫0ξ2Eα1,α1−ξ2−sα1γ1−Eα1,α1−ξ1−sα1γ12−α1/α1dsα1/2−α1≤2ξ1α1/2−2ξ1−ξ2α1/2α121−α1/2−α1ε.Hence,(45)U1x,yξ2−U1x,yξ1≤ξ1−ξ2α1+ξ1α1−ξ2α1Γα1+1f1∗+ξ1−ξ2α1f1∗Γα1+1+σ1Eα1−ξ2α1γ1−Eα1−ξ1α1γ1Γα1+11+σ1Eα1−γ1f1∗+f1∗2ξ1α1/2−2ξ1−ξ2α1/2α121−α1/2−α1ε→0,as ξ2→ξ1, which implies that U1 is equicontinuous on J. Similarly, the operator V1 is also equicontinuous on J. Hence, T1Br is relatively compact on J. It follows by Arzela-Ascoli’s theorem that T1 is compact. Therefore, we conclude from Lemma 6 that problem (3) has at least one solution.


## 4. Examples

Example 1.
Consider the following coupled fractional pantograph differential equations with instantaneous impulses given by(46)Dc1/2xt+xt=costt+82x2t1+x2t+sintet+72x13t+costt+82y2t1+y2t,t∈J\12,Dc1/2yt+yt=sintt+82x2t1+x2t+costt+72y2t1+y2t+sintet+8y13t,t∈J\12,Δxt=1/20=x1/240+x1/2,Δyt=1/20=y1/240+y1/2,x0=−x1,y0=−y1.
By comparing (46) with problem (3), we get
(47)f1t,xt,xλ1t,yt=costt+82x2t1+x2t+sintet+72x13t+costt+82y2t1+y2t,f2t,xt,yt,yλ2t=sintt+82x2t1+x2t+costt+72y2t1+y2t+sintet+82y13t.
We havea1=a2=b1=b2=σ1=σ2=1, α1=α2=1, and γ1=γ2=1.
(48)Iix=xt40+xt,Iiy=yt40+yt.
Then, for anyx,x¯,y,y¯∈ℝ and t∈J, we have
(49)f1t,xt,xλ1t,yt−f1t,x¯t,x¯λ1t,yt≤132x−x¯+y−y¯,f2t,xt,yt,yλ2t−f2t,x¯t,y¯t,y¯λ2t≤132x−x¯+y−y¯,Iix−Iix¯≤140x−x¯+y−y¯,Ijy−Ijy¯≤140x−x¯+y−y¯.
Then, by simple calculations, we can easily see that(50)L1=L2=132,C1=C2=140,E1/2−1≈0.42,E1/2−121/2≈0.52,Γ32≈0.89,μ1+μ2=2×31+E1/2−1∑k=1nC1E1/2−1/21/2+12L13Γ3/2≈0.796<1.
Therefore, all the hypotheses of Theorem9 are satisfied. Hence, problem (46) has a unique solution x,y on 0,1.Example 2.
Consider the following impulsive fractional coupled equations with antiperiodic conditions given by(51)Dc1/2xt+2xt=t+13sin2t14xt1+xt+t+137et2x32t+t+13cos2t14yt1+yt,t∈j\12,Dc1/2yt+2yt=t+13cos2t14xt1+xt+t+13sin2t14yt1+yt+t+137et2y32t,Δxt=1/20=x1/290+x1/2,Δyt=1/20=y1/290+y1/2,x0+x1=0,y0+y1=0.
By comparing (51) with problem (3), we have
(52)f1t,xt,xλ1t,yt=t+13sin2t14xt1+xt+t+137et2x32t+t+13cos2t14yt1+yt,f2t,xt,yt,yλ2t=t+13cos2t14xt1+xt+t+13sin2t14yt1+yt+t+137et2y32t.
This implies that(53)f1t,xt,xλ1t,yt≤t+13sin2t14x+1+t+13sin2t14y,f2t,xt,yt,yλ2t≤t+13sin2t14x+t+13sin2t14y+1,for any x,x¯,y,y¯∈ℝ and t∈J.
Thus, applying the same procedure as in Example1, we get
(54)C1=C2=190,E1/2−2≈0.25,E1/2−2121/2≈0.40,Γ12≈3.14,φ1t=φ2t=ψ1t=ψ2t=t+1314.
This implies that(55)limr→+∞infω1r=limr→+∞infη1r=limr→+∞infω2r=limr→+∞infη2r=1,(56)ε1+ε2≈0.84<1.
Thus, all the hypotheses of Theorem11 are satisfied. Hence, problem (51) has at least a solution x,y on 0,1.



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