# Boundary Value Problems of Fractional  Order Differential Equation with Integral Boundary  Conditions and Not Instantaneous Impulses

Authors: Peiluan Li; Changjin Xu
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We investigate the existence of mild solutions for fractional order differential equations with integral boundary conditions and not instantaneous impulses. By some fixed-point theorems, we establish sufficient conditions for the existence and uniqueness of solutions. Finally, two interesting examples are given to illustrate our theory results.


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## Body

## 1. Introduction

Impulsive differential equations are used to describe many practical dynamical systems including evolutionary processes characterized by abrupt changes of the state at certain instants. Such processes are naturally seen in biology, physics, engineering, and so forth. Due to their significance, many authors have established the solvability of impulsive differential equations. Nowadays, the theory of impulsive differential equations has received great attention. Differential equations with instantaneous impulses have been treated in several works (see, e.g., the monographs [1–3], the works on time variable impulses problem [4–7], and the references therein).However, in almost all the papers concerning impulsive differential equations, the impulses are all instantaneous impulses, and the classical models with instantaneous impulses cannot characterize many practical problems, for example, the dynamics of evolution processes in pharmacotherapy. Let us consider the hemodynamic equilibrium of a person. The introduction of the drugs in the bloodstream and the consequent absorption for the body are gradual and continuous processes. In fact, this situation should be characterized by a new case of impulsive action, which starts at an arbitrary fixed pointti and stays active on a finite time interval [ti,si]. To this end, Hernández and O’Regan [8] initially offered to study a new class of abstract semilinear impulsive differential equations with not instantaneous impulses in a PC-normed Banach space. In [8], the authors discussed the following problem: (1)u′t=Aut+ft,ut,t∈si,ti+1,i=0,…,N,u(t)=gi(t,u(t)),t∈(ti,si],i=1,…,N,u(0)=x0,where A:D(A)⊂X→X is the generator of a C0-semigroup of bounded linear operators (T(t))t>0 defined on a Banach space (X,·), 0=t0=s0<t1≤s1≤⋯≤tN≤sN≤tN+1=a are prefixed numbers, x0∈X, gi∈C((ti,si]×X,X) for all i=1,…,N, and f:[0,a]×X→X is a suitable function. Meanwhile, Pierri et al. [9] continued the work in [8] in a PCα-normed Banach space.On the one hand, the absorption of drugs has a memory effect; thus, the new class of impulsive conditions introduced by [8] may not explain this phenomenon very well. On the other hand, fractional calculus provides a powerful tool for the description of hereditary properties of various materials and memory processes [10, 11]. Fractional differential equations have recently proved to be strong tools in the modeling of medical, physics, economics, and technical sciences. For more details on fractional calculus theory, one can see the monographs of Diethelm [12], Kilbas et al. [13], Lakshmikantham et al. [14], Miller and Ross [15], Podlubny [16], and Tarasov [17]. Fractional differential equations involving the Riemann-Liouville fractional derivative or the Caputo fractional derivative have been paid more and more attention (see [10, 11, 18–22]).The theory of boundary value problems (BVPs) with integral boundary conditions for differential equations arises in different areas of applied mathematics and physics. For example, heat conduction, chemical engineering, underground water flow, thermo-elasticity, and plasma physics can be reduced to the nonlocal problems with integral boundary conditions. For BVPs with integral boundary conditions and comments on their importance, we refer the readers to the papers by Gallardo [23], Karakostas and Tsamatos [24], Lomtatidze and Malaguti [25], and the references therein. For more information about the general theory of integral equations and their relation with BVPs, we refer to the books of Corduneanu [26] and Agarwal and O’Regan [27]. Moreover, BVPs with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multipoint, and nonlocal BVPs as special cases. The existence and multiplicity of positive solutions for such problems have received a great deal of attention. To identify a few, we refer the readers to [28–31] and references therein.In [32], the authors consider the following problem:(2)Dtquct=ft,ut,∫0tkt,s,usds,t∈J′∶=J∖t1,t2,…,tm,J∶=0,T,utk+=utk-+uk,uk∈R,k=0,1,…,m,u0=∫01gsusds,where Dtqc is the Caputo fractional derivative of order q∈(0,1) with the lower limit zero, f:J×X×X→X is a given function, k:Δ×X→X, g∈L1([0,1],R+), gt∈0,1,  Δ={(t,s):0≤s≤t≤1}, and tk satisfy 0=t0<t1<⋯<tm+1=T, u(tk+)=limh→0+⁡utk+h and u(tk-)=limh→0-⁡u(tk+h) represent the right and left limits of u(t) at t=tk. Obviously, the impulses in (2) are instantaneous. Motivated by the work in [8, 9, 32], in this paper, we consider the following impulsive fractional differential equations with integral boundary conditions and not instantaneous impulses:(3)Dtqcut=ft,ut,∫0tkt,s,usds,t∈si,ti+1,i=0,…,N,ut=git,ut,t∈ti,si,i=1,…,N,u0=∫0Twsusds,where Dtqc is the Caputo fractional derivative of order q∈(0,1) with the lower limit zero, 0=t0=s0<t1≤s1≤⋯≤tN≤sN≤tN+1=T are prefixed numbers, gi∈C((ti,si]×R,R), for i=1,…,N, J=[0,T], f:J×R×R→R, k:Δ×R→R, w∈L1([0,T],R+), and w∈[0,1/T), Δ={(t,s):0≤s≤t≤T}.The rest of this paper is organized as follows. In Section2, some lemmas which are essential to prove our main results are stated. In Section 3, we give the main results. In Section 4, two interesting examples are given to illustrate our theory results.


## 2. Preliminaries

At first, we present the necessary definitions for the fractional calculus theory.Definition 1 (see [13]).
The Riemann-Liouville fractional integral of orderα>0 of a suitable function y:(0,∞)→R is given by(4)I0+αyt=1Γα∫0tt-sα-1ysds,where the right side is pointwise defined on (0,+∞).Definition 2 (see [13]).
The Caputo fractional derivative of orderα>0 of a suitable function y:(0,∞)→R is given by(5)Dαcyt=1Γn-α∫0tt-sn-α-1ynsds,where n=[α]+1, [α] denotes the integer part of number α and the right side is pointwise defined on (0,+∞).Lemma 3 (see [13]).
Letα>0; then the fractional differential equation Dαcu(t)=0 has solutions(6)ut=c0+c1t+c2t2+⋯+cn-1tn-1,where ci∈R, i=0,1,…,n-1, n=[α]+1.Lemma 4 (see [13]).
Letα>0, then one has (7)I0+αDαcut=ut+c0+c1t+c2t2+⋯+cn-1tn-1,where ci∈R, i=0,1,…,n-1, n=[α]+1.Lemma 5 (Krasnoselskii’s fixed point theorem [33]).
LetM be a closed convex and nonempty subset of a Banach space X. Let A and B be two operators such that(1)
Ax+By∈M whenever x,y∈M;(2)
A is compact and continuous;(3)
B is a contraction mapping.Then there exists z∈M such that z=Az+Bz.
In order to study problem (3), we define X=PC(J,R)={x:J→R;x∈C((tk,tk+1],R), k=0,1,…,N, and x(tk+), x(tk-) exist with x(tk-)=x(tk), k=1,…,N}.
It is easy to check thatX is a Banach space with the norm xPC=sup⁡t∈J⁡xt.LetBu(t)=∫0tk(t,s,u(s))ds, σ=∫0t1w(t)dt; then 0≤σ≤∫0Tw(t)dt<1.Ifu∈PC(J,R) satisfies problem (3), then for t∈(si,ti+1], i=1,…,N, integrating the first equation of (3) from si to t by virtue of Definition 1, one can obtain(8)ut=usi+1Γq∫sitt-sq-1fs,us,Busds.From the second equation in (3), we know u(si)=gi(si,u(si)). Then, for t∈(si,ti+1], we have(9)ut=gisi,usi+1Γq∫sitt-sq-1fs,us,Busds.Fort∈[0,t1], integrating the first equation in (3) from 0 to t by virtue of Definition 1, one can obtain(10)ut=u0+1Γq∫0tt-sq-1fs,us,Busds.By the boundary conditions, we have(11)ut=∫0Twtutdt+1Γq∫0tt-sq-1fs,us,Busds.Multiplying (11) with w(t) and integrating from 0 to t1, we have(12)∫0t1wtutdt=∫0t1wtdt∫0Twsusds+1Γq∫0t1wt∫0tt-sq-1fs,us,Busdsdt,Multiplying (9) with w(t) and integrating from si to ti+1, i=1,…,N, we have(13)∫siti+1wtutdt=gisi,usi∫siti+1wtdt+1Γq·∫siti+1wt1Γq∫sitt-sq-1fs,us,Busdsdt.Multiplying the second equation of (3) with w(t) and integrating from ti to si, i=1,…,N, we can obtain(14)∫tisiwtutdt=∫tisiwtgit,utdt.Adding (12), (13), and (14), one has(15)∫0Twtutdt=∫0t1wtdt∫0Twsusds+∑i=1Ngisi,usi∫siti+1wtdt+1Γq∑i=0N∫siti+1wt∫sitt-sq-1fs,us,Busdsdt+∑i=1N∫tisiwtgit,utdt.Hence(16)∫0Twtutdt=11-σ·∑i=1Ngisi,usi∫siti+1wtdt+1Γq∑i=0N∫siti+1wt∫sitt-sq-1fs,us,Busdsdt+∑i=1N∫tisiwtgit,utdt.So, for t∈[0,t1], we have(17)ut=1Γq∫0tt-sq-1fs,us,Busds+11-σ·∑i=1Ngisi,usi∫siti+1wtdt+1Γq·∑i=0N∫siti+1wt∫sitt-sq-1fs,us,Busdsdt+∑i=1N∫tisiwtgit,utdt.Then similar to Definition  2.1 in [9], we can define the mild solution for (3).Definition 6.
A functionu∈PC(J,R) is a mild solution of problems (3) if(18)u0=∫0Twsusds,ut=git,ut,∀t∈ti,si,i=1,…,N,ut=1Γq∫0tt-sq-1fs,us,Busds+11-σ·∑i=1Ngisi,usi∫siti+1wtdt+1Γq·∑i=0N∫siti+1wt∫sitt-sq-1fs,us,Busdsdt+∑i=1N∫tisiwtgit,utdtfor all t∈[0,t1] and(19)ut=gisi,usi+1Γq∫sitt-sq-1fs,us,Busdsfor all t∈(si,ti+1], i=1,…,N.


## 3. Main Results

This section deals with the existence of mild solutions for (3). Before stating and proving the main results, we make the following hypotheses.(H1)
f:J×R×R→R is jointly continuous. There exists a function e∈C(J,R) such that(20)ft,u1t,v1t-ft,u2t,v2t≤etu1t-u2t+v1t-v2t,for allu1,v1,u2,v2∈PC(J,R), for all t∈J.(H2)
k:Δ×R→R is continuous and there exists a function d∈C[J,R+] such that(21)kt,s,us-kt,s,vs≤dtut-vt,∀u,v∈PCJ,R,t,s∈Δ.(H3)
gi∈C((ti,si]×R,R) and there exist li∈C[J,R], i=1,…,N such that(22)git,xt-git,yt≤litxt-yt,∀x,y∈PCJ,R,∀t∈J.Let(23)L=max⁡1≤i≤Nsup⁡t∈Jlit,M=sup⁡t∈Jet,D=sup⁡t∈Jdt.Now we are in the position to establish the main results. Our first theorem is based on contraction mapping principle.Theorem 7.
Let(H1)–(H3) hold and n<1; then the problem (3) has a unique mild solution, where (24)n=TqM1+TD2-σΓq+11-σ+2L1-σ.Proof.
LetA:PC(J,R)→PC(J,R) be the map defined by(25)Aut=git,ut,fort∈ti,si,i=1,…,N,Aut=1Γq∫0tt-sq-1fs,us,Busds+11-σ·∑i=1Ngisi,usi∫siti+1wtdt+1Γq·∑i=0N∫siti+1wt∫sitt-sq-1fs,us,Busdsdt+∑i=1N∫tisiwtgit,utdtfor t∈[0,t1] and(26)Aut=gisi,usi+1Γq∫sitt-sq-1fs,us,Busdsfor t∈(si,ti+1], i=1,…,N. Clearly, A is well defined.
Next we show thatA is contraction on PC(J,R).
Fixx,y∈PC(J,R); we consider three cases.
Case 1. If t∈(si,ti+1], i=1,…,N, by the assumptions (H1)–(H3) and the property Γ(q+1)=qΓ(q), we have(27)Axt-Ayt≤gisi,xsi-gisi,ysi+1Γq∫0tt-sq-1·fs,xs,Bxsds-fs,ys,Bysds≤Lx-yPC+MΓq-(t-s)qq0t·x-yPC+Bx-ByPC=Lx-yPC+MΓqtqqx-yPC+Bx-ByPC≤Lx-yPC+TqMΓq+1x-yPC+Bx-ByPC≤L+TqM1+TDΓq+1x-yPC.
Case 2. If t∈[0,t1], by (H1), (H2), one can obtain(28)Axt-Ayt≤1Γq∫0tt-sq-1·fs,xs,Bxs-fs,ys,Bysds+11-σ∑i=1Ngisi,xsi-gisi,ysi∫siti+1wtdt+1Γq∑i=0N∫siti+1wt∫sitt-sq-1·fs,xs,Bxs-fs,ys,Bysdt+∑i=1N∫tisiwtgit,xt-git,ytdt.Then, by a similar argument, we can get(29)Axt-Ayt≤TqM1+TD2-σΓq+11-σ+2L1-σx-yPC.
Case 3. If t∈ti,si, i=1,…,N, from the assumption (H3), we get(30)Axt-Ayt≤Lx-yPC.
Therefore,Ax-AyPC≤nx-yPC, for all x,y∈PC(J,R), which implies that A is a contraction mapping. Then, there exists a unique mild solution of (3).
In order to get the second main result, we give assumption(H4).(H4)
The functionf:J×R×R→R is jointly continuous and strongly measurable on J. There exist mf∈C(J,R+) and a nondecreasing function hf∈C(R+,R+) such that(31)ft,x,y≤mfthfx+y∀t∈J,x,y∈R.
Our second result is based on Krasnoselskii’s fixed point theorem.Theorem 8.
Assume that(H2)–(H4) hold; if 2L/(1-σ)<1 and there exists a constant r>0 such that(32)r1-σ-2L≥Tq2-σΓq+1mfPChfr1+TD+TK+2a,where a=max⁡i=1,…,N⁡gi(t,0), then the problem (3) has at least a mild solution.Proof.
LetAu(t) be the map introduced in the proof of Theorem 7. We consider the decomposition Au(t)=A1u(t)+A2u(t), where(33)A1ut=∑i=1NAi1ut,A2ut=∑i=1NAi2ut,Ai1ut=1Γq∫sitt-sq-1fs,us,Busds,ift∈si,ti+1,i≥1,1Γq∫0tt-sq-1fs,us,Busds+1(1-σ)Γ(q)·∑i=0N∫siti+1wt∫sitt-sq-1·fs,us,Busdsdt,ift∈[0,t1],0,else,Ai2ut=git,ut,ift∈ti,si,i≥1,gisi,usi,ift∈si,ti+1,i≥1,11-σ∑i=1Ngisi,usi∫siti+1wtdt+∑i=1N∫tisiwtgit,utdtift∈0,t1,0,else.
LetBr={x∈PC(J,R):xPC≤r}. We divide our proof into three steps.
Step 1. First we show that A1x+A2y∈Br whenever x,y∈Br.
From(H2), we know that k:Δ×X→X is continuous, and then k(t,s,0) is bounded, for (t,s)∈Δ={(t,s):0≤s≤t≤T}. Let K:=max⁡{k(t,s,0):(t,s)∈Δ}.
Letx∈Br; if t∈(si,ti+1], we have(34)Ai1xt≤1Γq∫sitt-sq-1fs,xs,Bxsds≤1Γq∫sitt-sq-1mfsdshfx+BxPC≤1Γ(q)∫sitt-sq-1dsmfPChf(x+BxPC)≤1Γq∫0tt-sq-1dsmfPChfx+BxPC=1Γq-t-sqq0tmfPChfx+BxPC=1ΓqtqqmfPChfx+BxPC≤TqqΓqmfPChfxPC+BxPC.By the definition of B, the integral mean value theorem, (H2), and the property Γ(q+1)=qΓ(q), we have(35)Ai1xt≤TqΓq+1mfPC·hfxPC+Tkt,ξ,x-kt,ξ,0+kt,ξ,0≤TqΓq+1mfPC·hfxPC+DTxPC+Tkt,ξ,0≤TqΓq+1mfPChfr1+TD+TK,where 0≤ξ≤t≤T.
By a similar argument, letx∈Br, if t∈[0,t1]; we have(36)Ai1xt≤1Γq∫0tt-sq-1fs,xs,Bxsds+11-σΓq·∑i=0N∫siti+1wt∫sitt-sq-1fs,xs,Bxsdsdt≤TqΓq+1mfPChfr1+TD+TK+11-σΓq·∫0TwtTqqmfPChfr1+TD+TKdt=TqΓq+11+∫0Twtdt1-σmfPC·hfr1+TD+TK.From the condition w∈L1([0,T],R+), w∈[0,1/T), we can get(37)Ai1xt≤Tq2-σΓq+11-σmfPChfr1+TD+TK.
For the other cases, from the definition ofAi1u, one can get Ai1x(t)=0.
From the proof above, letx∈Br; then for all t∈[0,T], we have(38)Ai1xt≤Tq2-σΓq+11-σmfPChfr1+TD+TK.
Lety∈Br; if t∈(ti,si], i≥1, we can obtain(39)Ai2yt≤git,yt-git,0+git,0≤Lr+a.
Lety∈Br; if t∈[0,t1], we can get(40)Ai2yt≤11-σ∑i=1N∫siti+1wtgisi,ysi-gisi,0+gisi,0∫siti+1wtdt+∑i=1N∫tisiwtgit,yt-git,0∑i=1Nwtgit,yt-git,0+git,0dt≤2Lr+a1-σ.
Proceeding as above, we obtain thatAi2y(t)≤Lr+a, ∀y∈Br, for t∈(si,ti+1], i≥1.
Then, for allx,y∈Br, we have that(41)Ax+ByPC≤Tq2-σΓq+11-σmfPChfr1+TD+TK+2Lr+a1-σ≤r.
Step 2. We show that A2=∑i=1NAi2 is a contraction mapping.
From the definition ofA2u(t), Ai2u(t), and (H3), we can easily get(42)Ai2x(t)-Ai2yt≤2L1-σx-yPC,∀x,y∈Br,∀t∈J,which implies that A2 is a contraction mapping.
Step 3. Next we will prove that A1 is compact and continuous.
We also divide the proof into 3 steps.
(I) We show thatA1 is continuous.
Let{xn} be a sequence such that xn→x in PC(J,R). Then for all t∈J, by the definition of A1u(t), Ai1u(t), we have(43)A1xnt-A1xt≤Tq2-σΓq+11-σ·ft,xnt,Bxnt-ft,xt,Bxt≤Tq2-σΓq+11-σ·ft,xnt,Bxnt-ft,xt,Bxnt+ft,xt,Bxnt-ft,xt,Bxt.
From(H2), (H4), we can get the continuity of f and B. Then one has(44)A1xn-A1xPC⟶0,asxn⟶xn⟶∞,which shows that the operator A1 is continuous.
(II) We show thatA1 maps bounded sets into bounded sets in PC(J,R).
Indeed, it is enough to show that, for anyR>0, there exists a R′>0 such that, for each x∈BR={u∈PC(J,R):uPC≤R}, we have A1xPC≤R′.
For allt∈J, from the definition of A1u(t), Ai1u(t), and (H2), (H4), one can obtain(45)A1xt≤Tq2-σΓq+11-σmfPChfR1+TD+TK∶=R′.
Then we conclude thatA1 maps bounded sets into bounded sets in PC(J,R).
(III) At last, we prove thatA1 maps bounded sets into equicontinuous sets in PC(J,R).
For intervalt∈si,ti+1, si≤l1<l2≤ti+1, i=1,…,N, ∀x(t)∈Br, by definition of A1u(t) and (H3), we have(46)A1xl2-A1xl1=1Γq∫sil2l2-sq-1fs,xs,Bxsds-1Γq∫sil1l1-sq-1fs,xs,Bxsds≤1Γq∫l1l2l2-sq-1fs,xs,Bxsds+1Γq∫sil1fs,xs,Bxs·l1-sq-1-l2-sq-1ds≤mfPChfr1+TD+TKΓq·∫l1l2l2-sq-1ds+∫sil1l1-sq-1-l2-sq-1ds≤mfPChfr1+TD+TKΓq+1·l2-l1q+l1-siq-l2-siq,which is independent of x. As l1→l2, the right-hand side of the above inequality tends to zero. Therefore A1 is equicontinuous on interval (si,ti+1], i≥1.
Proceeding as above, we can also prove thatA1 is equicontinuous for the time interval [0,t1]. From the definition of Ai1, it is easy to see that A1 is equicontinuous for the other cases.
By Arzela-Ascoli Theorem,A1 is continuous and compact.As a consequence of Lemma5, we deduce that the operator A has at least a fixed point on Br which means that problem (3) has at least a mild solution.


## 4. Examples

Consider the following impulsive system of fractional differential equations.Example 1.
Consider(47)Dt1/2uct=116utsint21+et2+∫0tsin⁡t2+s2usds,t∈si,ti+1,i=1,…,N,ut=ut18et1+ut,t∈ti,si,i=1,…,N,u0=∫01ut3dt,where 0=t0=s0<t1=1/2≤s1≤⋯≤tN≤sN≤tN+1=1 are pre-fixed numbers, J=[0,1], q=1/2, and w(t)=1/3<1.
We prove that Example1 satisfies all the assumptions of Theorem 7.
In Example1, set(48)ft,u,v=usint2161+et2+v16,But=∫0tsint2+s2usds,kt,s,u=sin⁡t2+s2u.
It is easy to see thatf is jointly continuous. We can also check that (H1) and (H2) are satisfied with e(t)=1/16 and d(t)=1/2.
Fort∈[0,1], u∈R, and gi(t,u)=u/18et(1+u), then we know (49)git,u-git,v≤118u-v,with li(t)=1/36, so (H3) is also satisfied.
From (47), we can get L=1/18, T=1, M=1/16, D=1/2, q=1/2, Γ(3/2)=π/2, σ=1/6, and then(50)n=TqM1+TD2-σΓq+11-σ+2L1-σ=3380π+215<1.
So all the conditions of Theorem7 are satisfied. As a consequence of Theorem 7, Example 1 has a unique mild solution.Example 2.
Consider(51)Dt1/2uct=1151+esintut+∫0tsin⁡t2+s2usds,t∈si,ti+1,i=1,…,N,ut=ut181+ut,t∈ti,si,i=1,…,N,u0=∫01ut2dt,where 0=t0=s0<t1=1/3≤s1≤t2≤⋯≤tN≤sN≤tN+1=1 are prefixed numbers, J=[0,1], q=1/2, and w(t)=1/2<1.
It is easy to see thatgi(t,0)=0 is bounded. Set(52)ft,u,v=u+v151+esint,But=∫0tsin⁡t2+s2usds,kt,s,u=sin⁡t2+s2u.
Then we have(53)ft,u,Bu≤115u+Bu,with mf=1/15∈C(J,R+), hf(u+Bu)=u+Bu∈C(R+,R+) being nondecreasing. So (H4) is satisfied. Similarly to the proof of Example 1, we know that (H2) and (H3) are satisfied.
From (51), we can obtain T=1, σ=1/6, D=1/2, L=1/18<1, a=0, K=0, Γ(3/2)=π/2, mf=1/15, q=1/2, and 2L/(1-σ)=2/15<1, and then the inequality (32) becomes 13r/18≥(11/30π)r. Hence, inequality (32) holds for all r>0.
Thus, all the assumptions in Theorem8 are satisfied, and our results can be applied to Example 2. So Example 2 has at least one mild solution.



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