# Existence of Solutions for a Periodic Boundary ValueProblem with Impulse and Fractional Derivative Dependence

Authors: Yaohong Li; Yongqing Wang; Donal O’Regan; Jiafa Xu
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, we present some theorems on impulsive periodic boundary value problems with fractional derivative dependence. In particular, we discuss the existence of solutions of a class of fractional-order impulsive periodic boundary values with nonlinear terms and impulsive terms satisfying certain growth conditions. Three examples are provided to illustrate our results.


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## Body

## 1. Introduction

This paper considers the existence of solutions of the following fractional-order impulsive periodic boundary value problem:(1)cDtqut=ft,ut,cDtγut,t∈J′,Δutk=Ikutk,cDtγutk,ΔcDtγutk=Jkutk,cDctγutk,k=1,2,…,m,au0−bu1=0,aDtγu0−bDtγu1=0,where Dctq and Dctγ represent the common Caputo derivatives of orders q and γ, and 1<q<2,0<γ<1, and J=0,1,0=t0<t1<t2<⋯<tm<tm+1=1,J′=J\t1,t2,…,tm. Here, f:J×ℝ×ℝ⟶ℝ and Ik,Jk:ℝ×ℝ⟶ℝ are continuous functions. Now, Δutk=utk+−utk−, where utk+andutk− denote the right limit and the left limit of ut at the impulsive point tk. Also, ΔDctγutk=Dctγutk+−Dctγutk−, where Dctγutk+andDctγutk− denote the right limit and the left limit of Dctγut at the impulsive point tk. If utk−andDctγutk− exist, we let utk=utk−andDctγutk=Dctγutk−,wherek=1,2,…,m. Also, aandb are two real constants with b>a>0.The theory of fractional differential equation has received a lot of attention because of its wide application in mathematical models (see [1–27] and the references therein). Fractional-order impulsive differential equations are a natural generalization of the case of nonimpulses and are used to describe sudden changes in their states, such as in optimal control, population dynamics, biological systems, financial systems, and mechanical systems with impact. We refer the reader to [28–36] and the references therein. In particular, Bai et al. [37] investigated a mixed boundary value problem of nonlinear impulsive fractional differential equation:(2)Dc0+qut=ft,ut,t∈J′,Δutk=Ikutk,Δu′tk=Jkutk,k=1,2,…,m,u0+u1=0,u′0+u′1=0,and some sufficient conditions on the existence and uniqueness of solutions for problem (2) are obtained under Lipschitz conditions. In [38], Zhang and Xu studied the following impulsive periodic boundary value problem with the Caputo fractional derivative:(3)Dctqut=ft,ut,t∈J′,Δutk=Ikutk,Δu′tk=Jkutk,k=1,2,…,m,au0−bu1=0,au′0−bu′1=0,using Green’s function in [36], and via the symmetry property of Green’s function and topological degree theory, the authors obtained the existence of positive solutions for (3) when the growth of f is superlinear and sublinear.Inspired by the above research studies, in this paper, we consider fractional-order impulsive differential equations with generalized periodic boundary value conditions (1), where the nonlinear term, impulse terms, and periodic boundary conditions all depend on unknown functions and the lower-order fractional derivative of unknown functions. This is obviously more general and more widely applied, but it is also more complex and difficult to solve. Compared with (1), the nonlinear term, pulse term, and periodic boundary conditions of (3) are all independent of fractional derivatives, so it is a special form of (1). In this paper, we first give an equivalent integral form of solutions for problem (1) using some new Green’s functions. Next, we present some sufficient conditions for the existence of solutions for problem (1), where the nonlinear and impulse terms satisfy some nonlinear and linear growth conditions, which are different from the conditions in [36–38]. Finally, we present three examples to illustrate our main results.


## 2. Preliminaries and Lemmas

In this section, we only present some necessary definitions and lemmas about fractional calculus.Definition 1.
(see [39, 40]). The Riemann–Liouville fractional integral of order α>0 for a function f:0,∞⟶ℝ is defined as(4)I0+αft=1Γα∫0tt−sα−1fsds,where Γ⋅ is the Euler gamma function.Definition 2.
(see [39, 40]). The Caputo fractional derivative of order α>0 for a continuous and n-order differentiable function f:0,∞⟶ℝ is defined as(5)Dctαft=1Γn−α∫0tt−sn−α−1fnsds,where Γ⋅ is the Euler gamma function and n=α+1,whereα is the smallest integer greater than or equal to α.Lemma 1 (see [39, 40]).
Letα>0. The differential equation Dctαut=0 has a unique solution:(6)ut=c0+c1t+⋯+cn−1tn−1,for some ci∈ℝi=0,1,2,…,n−1, where n=α+1.Lemma 2.
Lety∈CJ and 1<q<2. The unique solution of the following periodic boundary value problem(7)Dctqut=yt,t∈J′,Δutk=Ik,ΔcDtγutk=Jk,k=1,2,…,m,au0−bu1=0,aDtγu0−bDtγu1=0,0<a<b,is expressed by(8)ut=∫01K1t,sysds+∑i=1mK2t,tiJi+∑i=1mK3t,tiIi,t∈J,where(9)K1t,s=t−sq−1Γq−b1−sq−1b−aΓq−Γ2−γ1−sq−γ−1Γq−γt−bb−a,0≤s≤t≤1,−b1−sq−1b−aΓq−Γ2−γ1−sq−γ−1Γq−γt−bb−a,0≤t≤s≤1,K2t,ti=aΓ2−γtib−ati1−γ,0<ti<t≤1,i=1,2,…,m,Γ2−γti1−γbtib−a−t,0≤t≤ti<1,i=1,2,…,m,K3t,ti=−aa−b,0<ti<t≤1,i=1,2,…,m,−ba−b,0≤t≤ti<1,i=1,2,…,m.Furthermore,(10)Dctγut=∫0tH1t,sysds+∑i=1mH2t,tiJi,where(11)H1t,s=t−sq−γ−1Γq−γ−1−sq−γ−1t1−γΓq−γ,0≤s≤t≤1,−1−sq−γ−1t1−γΓq−γ,0≤t≤s≤1,H2t,ti=0,0<ti<t≤1,i=1,2,…,m,tti1−γ,0≤t≤ti<1,i=1,2,…,m.Proof.
Supposeu is a general solution of (7) on each interval tk,tk+1k=0,1,2,…,m. Then, using Lemma 1, (7) can be transformed into the following equivalent integral equation:(12)ut=I0+qyt−ck−dkt,t∈tk,tk+1,where t0=0andtm+1=1. Also, we have(13)Dctγut=I0+q−γyt−dkt1−γΓ2−γ,t∈tk,tk+1.
From (12) and (13), according to (7), we obtain(14)ac0−bcm=bdm−bI0+qy1,(15)dm=Γ2−γI0+q−γy1.
Applying the right fractional-order impulsive condition of (7), we obtain(16)dk−1−dk=Γ2−γtk1−γJk,(17)ck−1−ck=Ik−Γ2−γtk1−γJk.
From (15) and (16), after a recursive calculation, we have(18)d0=dm+Γ2−γ∑i=1mJiti1−γ=Γ2−γI0+q−γy1+Γ2−γ∑i=1mJiti1−γ.
Similar to (18), we see that(19)dk=d0−Γ2−γ∑i=1kJiti1−γ=Γ2−γI0+q−γy1+Γ2−γ∑i=k+1mJiti1−γ.
From (13), (14), and (16), we have(20)c0=bb−aI0+qy1−Γ2−γI0+q−γy1+∑i=1mIi−Γ2−γtiγJi.
From (17) and (20), after a recursive calculation, we have(21)ck=c0−∑i=1kIi−Γ2−γtiγJi=bb−aI0+qy1−Γ2−γI0+q−γy1+ab−a∑i=1kIi−Γ2−γtiγJi+bb−a∑i=1mIi−Γ2−γtiγJi.
Fort∈J0=t0,t1, substituting (18) and (20) into (12) and (13), we obtain(22)ut=I0+qyt−bb−aI0+qy1−Γ2−γt−bb−aI0+q−γy1−∑i=1mΓ2−γti1−γt−btib−aJi−bb−a∑i=1mIi=∫0tt−sq−1Γqysds−bb−a∫0t+∫t11−sq−1Γqysds−Γ2−γt−bb−a∫0t+∫t11−sq−γ−1Γq−γysds+∑i=1mΓ2−γti1−γbtib−a−tJi−bb−a∑i=1mIi=∫0tK1t,sysds+∑i=1mK2t,tiJi+∑i=1mK3t,tiIi,Dctγut=I0+q−γyt−I0+q−γy1+∑i=1mJiti1−γt1−γ=∫0tt−sq−γ−1Γq−γysds−∫0t+∫t11−sq−γ−1t1−γΓq−γysds−∑i=1mtti1−γJi=∫0tH1t,sysds+∑i=1mH2t,tiJi,where K1t,s,K2t,ti,K3t,ti,H1t,s,andH2t,ti are defined by (7) and (9).
ForJk=tk,tk+1,k=1,2,…,m, substituting (20) and (18) into (11) and (12), we have(23)ut=I0+qyt−bb−aI0+qy1−Γ2−γt−bb−aI0+q−γy1+ab−a∑i=1kΓ2−γtkγJi−∑i=1mΓ2−γti1−γt−btib−aJi−ab−a∑i=1kIi−bb−a∑i=1mIi=∫0tt−sq−1Γqysds−bb−a∫0t+∫t11−sq−1Γqysds−Γ2−γt−bb−a∫0t+∫t11−sq−γ−1Γq−γysds+ab−a∑i=1kΓ2−γtiti1−γJi+∑i=1mΓ2−γti1−γbtib−a−tJi−ab−a∑i=1kIi−bb−a∑i=1mIi=∫0tK1t,sysds+∑i=1mK2t,tiJi+∑i=1mK3t,tiIi,Dctγut=I0+q−γyt−I0+q−γy1+∑i=1mJiti1−γt1−γ=∫0tt−sq−γ−1Γq−γysds−∫0t+∫t11−sq−γ−1t1−γΓq−γysds−∑i=1mtti1−γJi=∫01H1t,sysds+∑i=1mH2t,tiJi,where K1t,s,K2t,ti,K3t,ti,H1t,s,​andH2t,ti are defined by (9) and (11). The proof is completed.Lemma 3.
Let0<a<b<+∞. Then, K1t,s+K2t,tiandK3t,ti and H1t,sandH2t,ti defined as in (9) and (11) are continuous, and the following inequalities hold:(i)
K1t,s≤2b−a1−sq−1/b−aΓq+2b−aΓ2−γ1−sq−γ−1/b−aΓq−γ,H1t,s≤21−sq−γ−1/Γq−γ,t,s∈J(ii)
K2t,ti≤bΓ2−γ/b−a,H2t,ti≤1,K3t,ti≤b/b−a,t,ti∈J.Proof.
Directly observe that(24)K1t,s≤1−sq−1Γq+b1−sq−1b−aΓq+Γ2−γ1−sq−γ−1Γq−γ1+bb−a≤2b−a1−sq−1b−aΓq+2b−aΓ2−γ1−sq−γ−1b−aΓq−γ,t,s∈J,H1t,s≤1−sq−γ−1Γq−γ+1−sq−γ−1t1−γΓq−γ≤1+t1−γ1−sq−γ−1Γq−γ,≤21−sq−γ−1Γq−γ,t,s∈J,K2t,ti≤Γ2−γti1−γbtib−a≤bΓ2−γb−atiγ≤bΓ2−γb−a,t,ti∈J,K3t,ti≤ba−b,t,ti∈J,H2t,ti≤tti1−γ≤1,t,ti∈J.
LetE=u:J⟶ℝ|u∈CJ′,Dctγut∈CJ′,andutk−,utk+,Dctγutk−, and Dctγutk+ exist, where k=1,2,…,m. Note [35] that E is a Banach space equipped with the norm(25)u=supt∈Jut+supt∈JDctγut.Lemma 4.
If the functionft,u,Dctγut is continuous, then u∈E is a solution of (1) if and only if u∈E is a solution of the following integral equation:(26)ut=∫01K1t,sfs,us,Dctγusds+∑i=1mK2t,tiJiuti,Dctγuti+∑i=1mK3t,tiIiuti,Dctγuti.Proof.
Assume thatu satisfies (1). From Lemma 2, we see that u satisfies integral equation (26).
Conversely, assume thatu satisfies integral equation (26). Applying Definition 2, by a direct fractional derivative computation, it follows that the solution given by (26) and (2) satisfies (1).
Define an operatorT:E⟶E as(27)Tut=∫01K1t,sfs,us,Dctγusds+∑i=1mK2t,tiJiuti,Dctγuti+∑i=1mK3t,tiIiuti,Dctγuti,(28)DctγTut=∫0tH1t,sfs,us,Dctγusds+∑i=1mH2t,tiJiuti,Dctγuti.
It is easy to prove that the functionu is a solution of (1) if and only if u is a fixed point of the operator T.
For convenience, we list some hypotheses:(B1)0<a<b<+∞,1<q<2,0<γ<1 with q−γ>1(B2)f:J×ℝ×ℝ⟶ℝ and Ik,Jk:ℝ×ℝ⟶ℝ are continuous functionsLemma 5.
Assume that (B1) and (B2) hold. Then, the operatorT:E⟶E defined as in (27) is completely continuous.Proof.
We divide the proof into three steps. SetΩr=u∈E,u≤r for some r>0. The steps are as follows:(i)
Step 1. T is continuous from the continuity of the functions K1,K2,K3,H1,H2,f,Ik,Jk.(ii)
Step 2. T is uniformly bounded. Now, for u∈Ωr we have ft,u,Dctγu≤m1,Jk≤m2,Ik≤m3, where mi>0,i=1,2,3.In fact, for eacht∈Jk=tk,tk+1,u∈Ωr,k=0,1,2,…,m, from Lemma 3, we have(29)Tut≤∫01K1t,sfs,us,Dctγusds+∑i=1mK2t,tiJiuti,Dctγuti+∑i=1mK3t,tiIiuti,Dctγuti≤m1∫01K1t,sds+m2∑i=1mK2t,ti+m3∑i=1mK3t,ti,DctγTut≤∫01H1t,sfs,us,Dctγusds+∑i=1mH2t,tiJiuti,Dctγuti≤m1∫01H1t,sds+m2∑i=1mH2t,ti,which and Lemma4 imply that(30)Tu=supt∈JTut+supt∈JDctγTut≤m1∫01K1t,sds+m2∑i=1mK2t,ti+m3∑i=1mK3t,ti+m1∫01H1t,sds+m2∑i=1mH2t,ti≤m1∫01K1t,s+H1t,sds+m2∑i=1mK2t,ti+H2t,ti+m3∑i=1mK3t,ti≤m1∫012b−a1−sq−1b−aΓq+2b−aΓ2−γ1−sq−γ−1b−aΓq−γ+21−sq−γ−1Γq−γds+bΓ2−γ+b−amm2b−a+bmm3b−a=m12b−ab−aΓq+1+2b−aΓ2−γb−aΓq−γ+1+2Γq−γ+1+bΓ2−γ+b−amm2b−a+bmm3b−a=M.(iii)
Step 3. T is equicontinuous. For any t1,t2∈Jk,k=0,1,…,m, fixed s∈J and for any ϵ>0, there exists a constant δ>0 such that for t1−t2<δ, we have(31)K1t1,s−K1t2,s<ϵ6m1,K2t1,ti−K2t2,ti<ϵ6m2m,K3t1,ti−K3t2,ti<ϵ6m3m,H1t1,s−H1t2,s<ϵ4m1,H2t1,ti−H2t2,ti<ϵ4m2m.Then,(32)Tut1−Tut2=∫01K1t1,s−K1t2,sfs,us,Dctγusds+∑i=1mK2t1,ti−K2t2,tiJiuti,Dctγuti+∑i=1mK3t1,ti−K3t2,tiIiuti,Dctγuti≤m1∫01K1t1,s−K1t2,sds+m2mK2t1,ti−K2t2,ti+m3mK3t1,ti−K3t2,ti<ϵ6+ϵ6+ϵ6=ϵ2,DctγTut1−DctγTut2=∫01H1t1,s−H1t2,sfs,us,Dctγusds+∑i=1mH2t1,ti−H2t2,tiJiuti,Dctγuti≤m1∫01H1t1,s−H1t2,sds+m2mH2t1,ti−H2t2,ti<ϵ4+ϵ4=ϵ2.Thus,(33)Tut1−Tut2<ε.which implies thatTΩr is equicontinuous on any subinterval Jk,k=0,1,…,m.From the Arzela–Ascoli theorem, we deduce thatT:E⟶E is completely continuous.Lemma 6 (Schauder fixed-point theorem, see [41, 42]).
LetX be a real Banach space, C⊂X be a nonempty closed bounded and convex subset, and F:C⟶C be compact. Then, T has at least one fixed point in C.Lemma 7 (Krasnoselskii fixed point theorem, see [41, 42]).
LetΩ be a closed convex and nonempty subset of a Banach space X. Let ΦandΨ be the operators such that (i) Φx+Ψy∈Ω whenever x,y∈Ω; (ii) Φ:Ω⟶X is compact and continuous; and (iii) Ψ is a contraction mapping. Then, there exists an z∈Ω such that z=Φz+Ψz.Lemma 8 (Banach’s fixed point theorem, see [43]).
LetE be a Banach space, Ω⊂E be closed, and F:Ω⟶Ω be a strict contraction, i.e., Fx−Fy≤kx−y for some k∈0,1 and all x,y∈Ω. Then, F has a unique fixed point in Ω.


## 3. Existence of the Solutions

For convenience, we give the following symbols:(34)Ai=∫012b−a1−sq−1b−aΓq+2b−aΓ2−γ1−sq−γ−1b−aΓq−γ+21−sq−γ−1Γq−γaisds,Bi=bΓ2−γ+b−ambib−a,Ci=mbcib−a,i=0,1,2.Now, we present our main theorems.Theorem 1.
Assume that (B1) and (B2) hold, and the following hypotheses are satisfied:(C1) There exist three nonnegative functionsa0,a1,a2∈LJ and two constants λ1,λ2∈0,1 such that(35)ft,u,v≤a0t+a1tuλ1+a2tvλ2,∀t∈J,u,v∈ℝ,(C2) There exist eight positive constantsb1,b2,c1,c2≥0 and μ1,μ2,ν1,ν2∈0,1 such that(36)Iiu,v≤b0+b1uμ1+b2vμ2,Jiu,v≤c0+c1uν1+c2vν2,i=1,2,…,m,∀u,v∈ℝ.Then, (1) has at least one solution in E.Proof.
Let(37)R1≥max7A0+B0+C0,7A11/1−λ1,7A21/1−λ2,7B11/1−μ1,7B21/1−μ2,7C11/1−ν1,7C11/1−ν2,ΩR1=u∈E:u≤R1.
Now,ΩR1 is a closed bounded convex subset of E.
For eachu∈ΩR1, from (C1) and (C2), we have(38)Tut≤∫01K1t,sfs,us,Dctγusds+∑i=1mK2t,tiJiuti,Dctγuti+∑i=1mK3t,tiIiuti,Dctγuti≤∫01K1t,sa0s+a1susλ1+a2sDctγusλ2ds+∑i=1mK2t,tib0+b1usμ1+b2Dctγusμ2+∑i=1mK3t,tic0+c1usν1+c2Dctγusν2,DctγTut≤∫01H1t,sfs,us,Dctγusds+∑i=1mH2t,tiJiuti,Dctγuti≤∫01H1t,sa0s+a1susλ1+a2sDctγusλ2ds+∑i=1mH2t,tib0+b1usμ1+b2Dctγusμ2.
From Lemma3, we obtain that(39)Tu=supt∈Jut+supt∈JDctγTut≤∫01K1t,s+H1t,sa0s+a1suλ1+a2suλ2ds+∑i=1mK2t,ti+H2t,tib0+b1uμ1+b2uμ2+∑i=1mK3t,tic0+c1uν1+c2uν2≤A0+A1R1λ1+A2R1λ2+B0+B1R1μ1+B2R1μ2+C0+C1R1ν1+C2R1ν2≤R1,which implies that TΩR1⊂ΩR1.
From Lemmas5 and 6, T has at least one fixed point in ΩR1, so (1) has at least one solution in E.Theorem 2.
Assume that (B1) and (B2) hold, and the following hypotheses are satisfied:(C3) There exists a nonnegative functiona0∈LJ, such that(40)ft,u,v≤a0t,∀t∈J,u,v∈ℝ.(C4) There exist four positive constantsb1,b2,c1,c2≥0 such that(41)Iiu1,v1−Iiu2,v2≤b1u1−u2+b2v1−v2,∀u1,u2,v1,v2∈ℝ,i=1,2,…,m,Jiu1,v1−Jiu2,v2≤c1u1−u2+c2v1−v2,∀u1,u2,v1,v2∈ℝ,i=1,2,…,m.IfΛ=∑i=12Bi+Ci<1/2, then (1) has at least one solution in E.Proof.
We first define the operators. Foru∈E, let(42)Φut=∫01K1t,sfs,us,Dctγusds,Φ1ut=∫01H1t,sfs,us,DctγusdsΨut=∑i=1mK2t,tiJiuti,Dctγuti+∑i=1mK3t,tiIiuti,Dctγuti,Ψ1ut=∑i=1mH2t,tiJiuti,Dctγuti.
Now,(43)Tut=Φut+Ψut,DctγTut=DctγΦut+DctγΨut=Φ1ut+Ψ1ut.
Let(44)M1=max1≤i≤mIi0,0,M2=max1≤i≤mJi0,0.
Let(45)R2≥max2A0,2Θ1−2Λ,ΩR2=u∈E:u≤R2.
Note thatΩR2 is a nonempty bounded closed convex subset of E.
From Lemma5, Φ is completely continuous (i.e., condition (ii) of Lemma 7 is satisfied).
For anyu,v∈ΩR2, from hypothesis (C4), we have(46)Ψut−Ψvt≤∑i=1mK2t,tiJiuti,Dctγuti−Jivti,Dctγvti+∑i=1mK3t,tiIiuti,Dctγuti−Iivti,Dctγvti≤∑i=1mK2t,tic1us−vs+c2Dctγus−Dctγvs∑i=1mK3t,tib1us−vs+b2Dctγus−Dctγvs≤∑i=1mK2t,tic1+c2u−v+∑i=1mK3t,tib1+b2u−v,Ψ1ut−Ψ1vt≤∑i=1mH2t,tiJiuti,Dctγuti−Jivti,Dctγvti≤∑i=1mH2t,tic1us−vs+c2Dctγus−Dctγvs≤∑i=1mH2t,tic1+c2u−v.
Therefore,(47)Ψu−Ψv=maxt∈JΨu−Ψv+maxt∈JΨ1u−Ψ1v≤∑i=1mK2t,ti+H2t,tic1+c2u−v+∑i=1mK3t,tib1+b2u−v≤C1+C2+B1+B2u−v=Λu−v,and since Λ<1/2, Ψ is a contraction (so condition (iii) of Lemma 7 is satisfied).
For eachu∈ΩR2, from hypothesis (C3), we have(48)Φut≤∫01K1t,sfs,us,Dctγusds≤∫01K1t,sa0sds,Φ1ut≤∫01H1t,sfs,us,Dctγusds≤∫01H1t,sa0sds.
Consequently,(49)Φu≤∫01|H1t,s|+H1t,sa0sds≤A0≤R22.
For eachv∈ΩR2, we have(50)Ψv≤Ψ0−Ψv+Ψ0≤Λ0−v+∑i=1m|K2t,ti|+H2t,timaxt∈JJi0,0+∑i=1mK3t,timaxt∈JIi0,0≤Λv+Θ,where(51)Θ=mbΓ2−γ+b−ab−aM2+bb−aM1.
Thus, for anyu,v∈ΩR2, we obtain(52)Φu+Ψv≤Φu+Ψv≤A0+Λv+Θ≤R2,which implies that Φu+Ψv∈ΩR2 (so condition (i) of Lemma 7 is satisfied).
In view of Lemma7, there exists a u∈ΩR2 such that Φu+Ψu=u, so (1) has at least one solution in E.Theorem 3.
Assume that (B1), (B2), and (C4) hold and the following hypothesis is satisfied:(C5)
There exist two nonnegative functionsa1,a2∈LJ such that(53)ft,u1,v1−ft,u2,v2≤a1tu1−u2+a2tv1−v2,∀t∈J,u1,u2,v1,v2∈ℝ.IfΠ=A1+A2+bΓ2−γ+b−ac1+c2+bb1+b2/b−a<1, then (1) has a unique solution in E.Proof.
Choose(54)R3≥11−ΠA′M0+B′,where(55)M0=maxt∈Jft,0,0,M1=max1≤i≤mIi0,0,M2=max1≤i≤mJi0,0,A′=2b−ab−aΓq+1+2b−aΓ2−γb−aΓq−γ+1+2Γq−γ+1,B′=bΓ2−γM2+bM1+b−ab−a.
First, we show thatTΩR3⊂ΩR3, where ΩR3=u∈E,u≤R3. For u∈ΩR3, from hypotheses (C4) and (C5), we obtain(56)Tut≤∫01K1t,sfs,us,Dctγus−ft,0,0+ft,0,0ds+∑i=1mK2t,tiJiuti,Dctγuti−Ji0,0+Ji0,0+∑i=1mK3t,tiIiuti,Dctγuti−Ii0,0+Ii0,0≤∫01K1t,sa1t+a2tu+M0ds+∑i=1mK2t,tic1+c2u+M2+∑i=1mK3t,tib1+b2u+M1,DctγTut≤∫01H1t,sfs,us,Dctγus−ft,0,0+ft,0,0ds+∑i=1mH2t,tiJiuti,Dctγuti−Ji0,0+Ji0,0≤∫01H1t,sa1t+a2tu+M0ds+∑i=1mH2t,tic1+c2u+M2.
Then,(57)Tu≤∫01K1t,s+H1t,sa1t+a2tu+M0ds+∑i=1mK2t,ti+H2t,tic1+c2u+M2+∑i=1mK3t,tib1+b2u+M1≤Πu+A′M0+B′≤R3,so TΩR3⊂ΩR3.
Furthermore, from hypotheses (C4) and (C5), for allu,v∈ΩR3, we have(58)Tut−Tvt≤∫01K1t,sfs,us,Dctγus−fs,vs,Dctγvsds+∑i=1mK2t,tiJiuti,Dctγuti−Jivti,Dctγvti+∑i=1mK3t,tiIiuti,Dctγuti−Iivti,Dctγvti≤∫01K1t,sa1t+a2tu−vds+∑i=1mK2t,tic1+c2u−v+∑i=1mK3t,tib1+b2u−v,DctγTut−DctγTvt≤∫01H1t,sfs,us,Dctγus−fs,vs,Dctγvsds+∑i=1mH2t,tiJiuti,Dctγuti−Jivti,Dctγvti≤∫0tH1t,sa1t+a2tu−vds+∑i=1mH2t,tic1+c2u−v.
Thus,(59)Tu−Tv≤∫01K1t,s+H1t,sa1t+a2tu−vds+∑i=1mK2t,ti+H2t,tic1+c2u+∑i=1mK3t,tib1+b2u≤Πu−v,where Π<1, so T is a contraction. Lemma 8 guarantees that T has a unique fixed point in ΩR3, which is the unique solution of (1) in E. This completes the proof.


## 4. Examples

In (1), let q=1.25,γ=0.15,a=1,b=2,t1=0.5,andk=1 and then, we obtain the following fractional-order impulsive differential equation:(60)Dct1.25ut=ft,ut,Dct0.15ut,t∈0,1,t≠0.5,Δu0.5=I1u0.5,Dct0.15u0.5,ΔDct0.15u0.5=J1u0.5,Dct0.15u0.5,u0−2u1=0,Dct0.15u0−2cDt0.15u1=0.By a direct observation, note that0<a<b<+∞,1<q<2,0<γ<1 with q−γ>1, so hypothesis (B1) is satisfied.Example 1.
In (60), let(61)ft,ut,Dct0.15ut=et50+1−t2ut0.2100+e2tDct0.15ut0.3200,Δu0.5=sin1+2u0.50.5+3Dct0.15u0.50.4150,ΔDct0.25u0.5=sin1+3u0.50.2+2Dct0.15u0.50.1120,so hypothesis (B2) is satisfied. Set Dct0.15ut=vt, and then, we obtain(62)ft,u,v≤et50+1−t2100u0.2+e2t200v0.3=a0t+a1tu0.2+a2tv0.3,I1u,v≤1150+175u0.5+150v0.4=b0+b1u0.5+b2v0.4,J1u,v≤1120+140u0.2+160v0.1=c0+c1u0.2+c2v0.1,which implies that (C1) and (C2) are satisfied. Thus, all the hypotheses in Theorem 1 are satisfied, so (60) has at least one solution in E.Example 2.
In (60), let(63)ft,ut,Dct0.15ut=1−s250×ut+Dct0.15ut1+ut+cDt0.15ut,Δu0.5=1+2u0.5+3Dct0.15u0.5150,ΔDct0.15u0.5=1+3u0.5+2Dct0.15u0.5120,so hypothesis (B2) is satisfied. Set Dct0.15ut=vt, and then, we obtain(64)ft,u,v≤1−s250=a0t,I1u1,v1−I1u2,v2≤175u1−u2+150v1−v2=b1u1−u2+b2v1−v2,J1u1,v1−J1u2,v2≤140u1−u2+160v1−v2=c1u1−u2+c2v1−v2,which implies that (C3) and (C4) are satisfied. Also, note that Λ=0.179707<0.5. Then, all the hypotheses in Theorem 2 are satisfied, so (60) has at least one solution in E.Example 3.
In (60), let(65)ft,ut,Dct0.15ut=et50+1−s2ut100+1−sDct0.15ut200,Δu0.5=1+2u0.5+3Dct0.15u0.5150,ΔDct0.15u0.5=1+3u0.5+2Dct0.15u0.5120,so hypothesis (B2) is satisfied. Set Dct0.15ut=vt, and then, we obtain(66)ft,u1,v1−ft,u2,v2≤1−s2100u1−u2+1−s200v1−v2=a1tu1−u2+a2tv1−v2,I1u1,v1−I1u2,v2≤175u1−u2+150v1−v2=b1u1−u2+b2v1−v2,J1u1,v1−J1u2,v2≤140u1−u2+160v1−v2=c1u1−u2+c2v1−v2,which implies that (C4) and (C5) are satisfied. Note that Π=0.787135<1. Then, all the hypotheses in Theorem 3 are satisfied, so (60) has a unique solution in E.


## 5. Conclusion

In this paper, we use fixed-point theorems to study fractional-order impulsive differential equation (1) with generalized periodic boundary value conditions. Very little is known on fractional-order impulsive differential equations with generalized periodic boundary value conditions where nonlinear terms and impulse terms depend on the unknown function and the lower-order fractional derivative of the unknown function. Our main results are obtained under some nonlinear and linear growth conditions corresponding to the relevant linear operators where the symmetry property of a Green’s function is not required, so our results generalize and improve works in the literature.



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