# Existence Results for Fractional Differential Equations with Separated Boundary Conditions and Fractional Impulsive Conditions

Authors: Xi Fu; Xiaoyou Liu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper is concerned with the fractional separated boundary value problem of fractional differential equations with fractional impulsive conditions. By means of the Schaefer fixed point theorem, Banach fixed point theorem, and nonlinear alternative of Leray-Schauder type, some existence results are obtained. Examples are given to illustrate the results.


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## Body

## 1. Introduction

Recently, much attention has been paid to study fractional differential equations due to the fact that they have been proven to be valuable tools in the mathematical modeling of many phenomena in physics, biology, mechanics, and so forth, (see [1–3]).The theory of impulsive differential equations of integer order has found its extensive applications in realistic mathematical modeling of a wide variety of practical situations and has emerged as an important area of investigation in recent years. For the general theory and applications of impulsive differential equations, see [4–10] and so forth. However, impulsive fractional differential equations have not been much studied, and many aspects of these equations are yet to be explored. For some recent work on impulsive fractional differential equations, we can refer to [11–26] and the references therein.In this paper, we consider the existence and uniqueness of solutions for the following fractional separated boundary value problem with fractional impulsive conditions:(1)cDαx(t)=f(t,x(t)),t∈J:=[0,T],t≠tk,k=1,2,…,m,Δx(tk)=Ik(x(tk-)),Δ(cDγx(tk))=Ik*(x(tk-)),k=1,2,…,m,a1x(0)+b1(cDγx(0))=c1,a2x(T)+b2(cDγx(T))=c2,
where cDα is the Caputo fractional derivative of order α∈(1,2) with the lower limit zero, 0<γ<1, f∈C(J×ℝ,ℝ), Ik,Ik*∈C(ℝ,ℝ), 0=t0<t1<⋯<tm<tm+1=T, Δx(tk)=x(tk+)-x(tk-) with x(tk+)=limϵ→0+x(tk+ϵ), x(tk-)=limϵ→0-x(tk+ϵ) representing the right and left limits of x(t) at t=tk, Δ(cDγx(tk)) has a similar meaning for cDγx(tk), and ai,bi,ci, i=1,2, are real constants with a1≠0 and a2TγΓ(2-γ)≠-b2.We note that the papers on this topic cited above except [24] all deal with the Caputo derivative and the impulsive conditions only involve integer order derivatives. Here we study the fractional differential equations with fractional impulsive conditions and fractional separated boundary conditions.In [24], the author considered the following two impulsive problems:
(2)cDδx(t)=f(t,x(t)),t∈(0,1]∖{t1,t2,…,tm},cDγx(tk+)-Dcγx(tk-)=Jk(x(tk)),k=1,2,…,m,x(0)=x0,x′(0)=x1,
where cDδ is the Caputo fractional derivative of order δ∈(1,2) with the lower limit zero, 0<γ<1, and
(3)LDδx(t)=f(t,x(t)),t∈(0,1]∖{t1,t2,…,tm},LDγx(tk+)-LDγx(tk-)=Jk(x(tk)),k=1,2,…,m,I1-αx(0)=x0,
where LDδ is the Riemann-Liouville fractional derivative of order δ∈(0,1) with the lower limit zero and 0<γ<δ.In [25], Fečkan et al. studied the impulsive problem of the following form:
(4)cDδx(t)=f(t,x(t)),t∈(0,T]∖{t1,t2,…,tm},δ∈(0,1),Δx(tk)=Ik(x(tk-)),k=1,2,…,m,x(0)=x0,
where f:[0,T]×ℝ→ℝ is jointly continuous, Ik:ℝ→ℝ and tk satisfy 0=t0<t1<⋯<tm+1=T, x(tk+)=limϵ→0+x(tk+ϵ), and x(tk-)=limϵ→0-x(tk+ϵ).Furthermore, Wang et al. [26] considered the impulsive fractional differential equations with boundary conditions as follows:
(5)cDδu(t)=h(t),t∈J′,δ∈(1,2),Δu(tk)=yk,Δu′(tk)=y-k,k=1,2,…,m,u(0)=0,u′(1)=0,
where yk,y-k∈ℝ.To the best of our knowledge, there are few papers concerning fractional differential equations with separated boundary conditions [27, 28].The rest of the paper is organized as follows. In Section2 we introduce some preliminary results needed in the sequel. In Section 3 we present the existence results for the problem (1). Two examples are given in Section 4 to illustrate the results.


## 2. Preliminaries

Let us setJ0=[0,t1], J1=(t1,t2],…, Jm-1=(tm-1,tm], and Jm=(tm,tm+1], J′:=J∖{t1,t2,…,tm} and introduce the space PC(J,ℝ):={u:J→ℝ∣u∈C(Jk,ℝ), k=0,1,2,…,m, and there exist u(tk+) and u(tk-), k=1,2,…,m, with u(tk-)=u(tk)}. It is clear that PC(J,ℝ) is a Banach space with the norm ∥u∥=sup{|u(t)|:t∈J}.Definition 1 (see [3]).
The Riemann-Liouville fractional integral of orderq for a continuous function f:[0,∞)→ℝ is defined as
(6)Iqf(t)=1Γ(q)∫0t‍(t-s)q-1f(s)ds,q>0,
which provided that the integral exists.Definition 2 (see [3]).
Forn-1 times an absolutely continuous function f:[0,∞)→ℝ, the Caputo derivative of order q is defined as
(7)cDqf(t)=1Γ(n-q)∫0t‍(t-s)n-q-1f(n)(s)ds,n-1<q<n,n=[q]+1,
where [q] denotes the integer part of the real number q.Lemma 3 (see [3]).
Letα>0. Then the differential equation
(8)cDαh(t)=0
has solutions h(t)=c0+c1t+c2t2+⋯+cn-1tn-1 and
(9)IαDcαh(t)=h(t)+c0+c1t+c2t2+⋯+cn-1tn-1
which hold for almost all points on the interval [0,∞), here ci∈ℝ, i=0,1,2,…,n-1, n=[α]+1.Definition 4.
A functionx∈PC(J,ℝ) with its α-derivative existing on J′ is said to be a solution of the problem (1) if x satisfies the equation cDαx(t)=f(t,x(t)) on J′ and the conditions
(10)Δx(tk)=Ik(x(tk-)),Δ(cDγx(tk))=Ik*(x(tk-)),k=1,2,…,m,a1x(0)+b1(cDγx(0))=c1,a2x(T)+b2(cDγx(T))=c2
are satisfied.By using a similar discussion of [25], we have the following lemma.Lemma 5.
Lety∈PC(J,ℝ). A function x is a solution of the fractional integral equation:
(11)x(t)={∫0t‍(t-s)α-1Γ(α)y(s)ds+c1a1-Λtv-a2Πtv-a2c1tva1+c2tv-Γ(2-γ)t∑i=1m‍Ii*(x(ti-))ti1-γ,t∈J0;∫0t‍(t-s)α-1Γ(α)y(s)ds+c1a1+I1(x(t1-))-Γ(2-γ)t1γI1*(x(t1-))-Λtv-a2Πtv-a2c1tva1+c2tv-Γ(2-γ)t∑i=2m‍Ii*(x(ti-))ti1-γ,t∈J1;∫0t‍(t-s)α-1Γ(α)y(s)ds+c1a1+∑i=1k‍Ii(x(ti-))-Γ(2-γ)∑i=1k‍tiγIi*(x(ti-))-Λtv-a2Πtv-a2c1tva1+c2tv-Γ(2-γ)t∑i=k+1m‍Ii*(x(ti-))ti1-γ,t∈Jk,k=2,…,m,
where
(12)v=a2TΓ(2-γ)+b2T1-γΓ(2-γ),Λ=a2∫0T‍(T-s)α-1Γ(α)y(s)ds+b2∫0T‍(T-s)α-γ-1Γ(α-γ)y(s)ds,(13)Π=∑i=1m‍Ii(x(ti-))-Γ(2-γ)∑i=1m‍tiγIi*(x(ti-)),
if and only if x is a solution of the impulsive fractional BVP:
(14)cDαx(t)=y(t),t∈J′,1<α<2,Δx(tk)=Ik(x(tk-)),Δ(cDγx(tk))=Ik*(x(tk-)),k=1,2,…,m,a1x(0)+b1(cDγx(0))=c1,a2x(T)+b2(cDγx(T))=c2.Proof.
For1<α<2, by Lemma 3, we know that a general solution of the equation cDαx(t)=y(t) on each interval Jk (k=0,1,2,…,m) is given by
(15)x(t)=Iαy(t)+dk+ekt=∫0t‍(t-s)α-1Γ(α)y(s)ds+dk+ekt,t∈Jk,
where dk,ek∈ℝ are arbitrary constants. Since cDγC=0 (C is a constant), cDγt=t1-γ/Γ(2-γ), and cDγIαy(t)=Iα-γy(t) (see [3]), then from (15), we have
(16)cDγx(t)=Iα-γy(t)+ekt1-γΓ(2-γ)=∫0t‍(t-s)α-γ-1Γ(α-γ)y(s)ds+ekt1-γΓ(2-γ),
for t∈Jk. Applying the boundary conditions of (14), we get
(17)a1×d0+b1×0=c1,a2×(∫0T‍(T-s)α-1Γ(α)y(s)ds+dm+emT)+b2×(∫0T‍(T-s)α-γ-1Γ(α-γ)y(s)ds+emT1-γΓ(2-γ))=c2.
Next, using the impulsive conditions in (14), we obtain that for k=1,2,…,m(18)dk-dk-1+(ek-ek-1)tk=Ik(x(tk-)),(ek-ek-1)tk1-γΓ(2-γ)=Ik*(x(tk-)).
Now we can derive the values of dk,ek, k=0,1,2,…,m from formulae (17)-(18). That is,
(19)d0=c1a1,dk=d0+∑i=1k‍Ii(x(ti-))-Γ(2-γ)∑i=1k‍tiγIi*(x(ti-))
for k=1,2,…,m and
(20)em=-1v(a2∫0T‍(T-s)α-1Γ(α)y(s)ds+b2∫0T‍(T-s)α-γ-1Γ(α-γ)y(s)ds)-a2dmv+c2v,ek=em-Γ(2-γ)∑i=k+1m‍Ii*(x(ti-))ti1-γ,fork=0,1,2,…,m-1.
Hence for k=0,1,2,…,m, we have
(21)dk+ekt=c1a1+∑i=1k‍Ii(x(ti-))-Γ(2-γ)∑i=1k‍tiγIi*(x(ti-))-tv×(a2∫0T‍(T-s)α-1Γ(α)y(s)ds+b2∫0T‍(T-s)α-γ-1Γ(α-γ)y(s)ds)-a2tv(∑i=1m‍Ii(x(ti-))-Γ(2-γ)∑i=1m‍tiγIi*(x(ti-)))-a2c1tva1+c2tv-Γ(2-γ)t∑i=k+1m‍Ii*(x(ti-))ti1-γ.
Now it is clear that a solution of the problem (14) has the form of (11).
Conversely, assume thatx satisfies the fractional integral equation (11). That is, for t∈Jk, k=0,1,2,…,m, we have
(22)x(t)=∫0t‍(t-s)α-1Γ(α)y(s)ds+c1a1+∑i=1k‍Ii(x(ti-))-Γ(2-γ)∑i=1k‍tiγIi*(x(ti-))-(Λv+a2Πv+a2c1va1-c2v)t-Γ(2-γ)t∑i=k+1m‍Ii*(x(ti-))ti1-γ.
Since 1<α<2, we have cDαC=0 (C is a constant) and cDαt=0. Using the fact that cDα is the left inverse of Iα, we get
(23)cDαx(t)=y(t),t∈J′,
which means that x satisfies the first equation of the impulsive fractional BVP (14). Next we will verify that x satisfies the impulsive conditions. Taking fractional derivative cDγ of (22), we have, for t∈Jk,
(24)cDγx(t)=∫0t‍(t-s)α-γ-1Γ(α-γ)y(s)ds-(Λv+a2Πv+a2c1va1-c2v)t1-γΓ(2-γ)-t1-γ∑i=k+1m‍Ii*(x(ti-))ti1-γ.
From (22), we obtain
(25)x(tk+)=∫0tk‍(tk-s)α-1Γ(α)y(s)ds+∑i=1k‍Ii(x(ti-))-Γ(2-γ)∑i=1k‍tiγIi*(x(ti-))-(Λv+a2Πv+a2c1va1-c2v)tk-Γ(2-γ)tk∑i=k+1m‍Ii*(x(ti-))ti1-γ+c1a1,(26)x(tk-)=∫0tk‍(tk-s)α-1Γ(α)y(s)ds+∑i=1k-1‍Ii(x(ti-))-Γ(2-γ)∑i=1k-1‍tiγIi*(x(ti-))-(Λv+a2Πv+a2c1va1-c2v)tk-Γ(2-γ)tk∑i=km‍Ii*(x(ti-))ti1-γ+c1a1.
Hence we have, for k=1,2,…,m,
(27)Δx(tk)=Ik(x(tk-))-Γ(2-γ)tkγIk*(x(tk-))+Γ(2-γ)tkIk*(x(tk-))tk1-γ=Ik(x(tk-)).
Similarly, from (24), we can obtain that, for k=1,2,…,m,
(28)Δ(cDγx(tk))=tk1-γIk*(x(tk-))tk1-γ=Ik*(x(tk-)).
Finally, it follows from (22) and (24) that (since 0∈J0, T∈Jm) x(0)=c1/a1, cDγx(0)=0, and
(29)x(T)=∫0T‍(T-s)α-1Γ(α)y(s)ds+c1a1+Π-(Λv+a2Πv+a2c1va1-c2v)T,cDγx(T)=∫0T‍(T-s)α-γ-1Γ(α-γ)y(s)ds-(Λv+a2Πv+a2c1va1-c2v)T1-γΓ(2-γ).
Now we get
(30)a1x(0)+b1(cDγx(0))=c1,a2x(T)+b2(cDγx(T))=Λ+a2c1a1+a2Π-a2T(Λv+a2Πv+a2c1va1-c2v)-b2T1-γΓ(2-γ)(Λv+a2Πv+a2c1va1-c2v)=c2.
Therefore x given by (11) satisfies the impulsive fractional boundary value problem (14). The proof is complete.Remark 6.
We notice that the expression of (11) does not depend on the parameter b1 appearing in the boundary conditions of the problem (14). Thus by Lemma 5, we conclude that the parameter b1 is of arbitrary nature of the problem (14).LetX,Y be Banach spaces and f:X→Y, and we say that f is a compact if the image of each bounded set in X under f is relatively compact. The following are two fixed point theorems which will be used in the sequel.Theorem 7 (nonlinear alternative of Leray-Schauder type [29]).
LetX be a Banach space, C a nonempty convex subset of X, and U a nonempty open subset of C with 0∈U. Suppose that P:U¯→C is a continuous and compact map. Then either (a) P has a fixed point in U¯ or (b) there exist a x∈∂U (the boundary of U) and λ∈(0,1) with x=λP(x).Theorem 8 (Schaefer fixed point theorem [30]).
LetX be a normed space and P a continuous mapping of X into X which is compact on each bounded subset B of X. Then either (I) the equation x=λPx has a solution for λ=1 or (II) the set of all such solutions x, for 0<λ<1, is unbounded.


## 3. Main Results

This section deals with the existence and uniqueness of solutions for the problem (1).In view of Lemma5, we define an operator F:PC(J,ℝ)→PC(J,ℝ) by
(31)(Fx)(t)=∫0t‍(t-s)α-1Γ(α)f(s,x(s))ds+c1a1+∑i=1k‍Ii(x(ti-))-Γ(2-γ)∑i=1k‍tiγIi*(x(ti-))-Λxtv-a2Πxtv-a2c1tva1+c2tv-Γ(2-γ)t∑i=k+1m‍Ii*(x(ti-))ti1-γ,t∈Jk,k=0,1,2,…,m,
with
(32)Λx=a2∫0T‍(T-s)α-1Γ(α)f(s,x(s))ds+b2×∫0T‍(T-s)α-γ-1Γ(α-γ)f(s,x(s))ds,Πx=∑i=1m‍Ii(x(ti-))-Γ(2-γ)∑i=1m‍tiγIi*(x(ti-)).
Here Λx, Πx mean that Λ, Π defined in Lemma 5 are related to x∈PC(J,ℝ). It is obvious that F is well defined because of the continuity of f, Ik, and Ik*. Observe that the problem (1) has solutions if and only if the operator equation Fx=x has fixed points.Lemma 9.
The operatorF:PC(J,ℝ)→PC(J,ℝ) defined by (31) is completely continuous.Proof.
Sincef, Ik, and Ik* are continuous, it is easy to show that F is continuous on PC(J,ℝ).
LetB⊆PC(J,ℝ) be bounded. Then there exist positive constants Ni, i=1,2,3, such that |f(t,x(t))|≤N1, |Ik(x(tk-))|≤N2, and |Ik*(x(tk-))|≤N3 for all t∈J, x∈B, k=1,2,…,m. Thus, for x∈B and t∈J, we have
(33)|(Fx)(t)|≤N1TαΓ(α+1)+|c1||a1|+mN2+Γ(2-γ)N3∑i=1m‍tiγ+|Λx|T|v|+|a2Πx|T|v|+|a2c1|T|va1|+|c2|T|v|+Γ(2-γ)TN3∑i=1m‍tiγ-1,(34)|Λx|≤|a2|N1TαΓ(α+1)+|b2|N1Tα-γΓ(α-γ+1),|Πx|≤mN2+Γ(2-γ)N3∑i=1m‍tiγ.
Now we can obtain that, for all x∈B, and t∈J,
(35)|(Fx)(t)|≤N1TαΓ(α+1)+T|v|(|a2|N1TαΓ(α+1)+|b2|N1Tα-γΓ(α-γ+1))+(1+|a2|T|v|)mN2+Γ(2-γ)N3×((1+|a2|T|v|)∑i=1m‍tiγ+T∑i=1m‍tiγ-1)+|a2c1|T|va1|+|c2|T|v|+|c1||a1|,
which implies that the operator F is uniformly bounded on B.
On the other hand, letx∈B and for any τ1·τ2∈Jk, k=0,1,2,…,m, with τ1<τ2, we have
(36)|(Fx)(τ2)-(Fx)(τ1)|≤|∫0τ2‍(τ2-s)α-1Γ(α)f(s,x(s))ds-∫0τ1‍(τ1-s)α-1Γ(α)f(s,x(s))ds|+(|Λx||v|+|a2Πx||v|+|a2c1||va1|+|c2||v|)(τ2-τ1)+Γ(2-γ)∑i=k+1m‍tiγ-1|Ii*(x(ti-))|(τ2-τ1)≤N1(τ2α-τ1α)Γ(α+1)+(|Λx||v|+|a2Πx||v|+|a2c1||va1|+|c2||v|)×(τ2-τ1)+Γ(2-γ)N3∑i=k+1m‍tiγ-1(τ2-τ1).
By (34) and the above inequality, we deduce that
(37)|(Fx)(τ2)-(Fx)(τ1)|⟶0asτ2⟶τ1.
This implies that F is equicontinuous on the interval Jk. Hence by PC-type Arzela-Ascoli theorem (see Theorem 2.1 [10]), the operator F:PC(J,ℝ)→PC(J,ℝ) is completely continuous.Theorem 10.
Assume that (1) there existh∈L∞(J,ℝ+) and φ:[0,∞)→(0,∞) continuous, nondecreasing such that |f(t,x)|≤h(t)φ(|x|) for (t,x)∈J×ℝ; (2) there exist ψ,ψ*:[0,∞)→(0,∞) continuous, nondecreasing such that |Ik(x)|≤ψ(|x|), |Ik*(x)|≤ψ*(|x|) for all x∈ℝ and k=1,2,…,m; (3) there exists a constant M>0 such that
(38)MPφ(M)∥h∥L∞+Qψ(M)+Rψ*(M)+H>1,
where
(39)P=TαΓ(α+1)+T|v|(|a2|TαΓ(α+1)+|b2|Tα-γΓ(α-γ+1)),Q=m(1+|a2|T|v|),R=Γ(2-γ)[(1+|a2|T|v|)∑i=1m‍tiγ+T∑i=1m‍tiγ-1],H=|a2c1|T|va1|+|c2|T|v|+|c1||a1|.
Then, BVP (1) has at least one solution.Proof.
We will show that the operatorF defined by (31) satisfies the assumptions of the nonlinear alternative of Leray-Schauder type.
From Lemma9, the operator F:PC(J,ℝ)→PC(J,ℝ) is continuous and completely continuous.
Letx∈PC(J,ℝ) such that x(t)=λ(Fx)(t) for some λ∈(0,1). Then using the computations in proving that F maps bounded sets into bounded sets in Lemma 9, we have
(40)|x(t)|≤∥h∥L∞φ(∥x∥)×[TαΓ(α+1)+T|v|(|a2|TαΓ(α+1)+|b2|Tα-γΓ(α-γ+1))]+Γ(2-γ)ψ*(∥x∥)((1+|a2|T|v|)∑i=1m‍tiγ+T∑i=1m‍tiγ-1)+(1+|a2|T|v|)mψ(∥x∥)+|a2c1|T|va1|+|c2|T|v|+|c1||a1|.
Consequently, we have
(41)∥x∥P∥h∥L∞φ(∥x∥)+Qψ(∥x∥)+Rψ*(∥x∥)+H≤1.
Then by condition (38), ∥x∥≠M. Let us set
(42)U={x∈PC(J,ℝ):∥x∥<M}.
The operator F:U¯→PC(J,ℝ) is continuous and compact. From the choice of the set U, there is no x∈∂U such that x=λFx for some λ∈(0,1). Therefore by the nonlinear alternative of Leray-Schauder type (see Theorem 7), we deduce that F has a fixed point x in U¯ which is a solution of the problem (1). The proof is complete.Theorem 11.
Assume that there existh∈L∞(J,ℝ+) and positive constants H1,H2 such that, for t∈J, x∈ℝ, k=1,2,…,m,
(43)|f(t,x)|≤h(t),|Ik(x)|≤H1,|Ik*(x)|≤H2.
Then, BVP (1) has at least one solution on [0,T].Proof.
Lemma9 tells us that the operator F:PC(J,ℝ)→PC(J,ℝ) defined by (31) is continuous and compact on each bounded subset B of PC(J,ℝ).
LetV={u∈PC(J,ℝ):u=λFu,0<λ<1}. Since, for each t∈J,
(44)|x(t)|=|λ(Fx)(t)|≤∥h∥L∞×[TαΓ(α+1)+T|v|(|a2|TαΓ(α+1)+|b2|Tα-γΓ(α-γ+1))]+Γ(2-γ)H2((1+|a2|T|v|)∑i=1m‍tiγ+T∑i=1m‍tiγ-1)+(1+|a2|T|v|)mH1+|a2c1|T|va1|+|c2|T|v|+|c1||a1|,
we know that V is bounded. Thus, by Theorem 8, the operator F has at least one fixed point. Hence the problem (1) has at least one solution. The proof is completed.Theorem 12.
Assume that there existh∈L∞(J,ℝ+) and positive constants L,L* such that, for t∈J, x,y∈ℝ, k=1,2,…,m,
(45)|f(t,x)-f(t,y)|≤h(t)|x-y|,|Ik(x)-Ik(y)|≤L|x-y|,|Ik*(x)-Ik*(y)|≤L*|x-y|.
Moreover
(46)∥h∥L∞TαΓ(α+1)+T∥h∥L∞|v|(|a2|TαΓ(α+1)+|b2|Tα-γΓ(α-γ+1))+(1+|a2|T|v|)mL+Γ(2-γ)L*×[(1+|a2|T|v|)∑i=1m‍tiγ+T∑i=1m‍tiγ-1]<1.
Then, BVP (1) has a unique solution on J.Proof.
Letx,y∈PC(J,ℝ). Then for each t∈J, we have
(47)|(Fx)(t)-(Fy)(t)|≤∫0t‍(t-s)α-1Γ(α)|f(s,x(s))-f(s,y(s))|ds+∑i=1m‍|Ii(x(ti-))-Ii(y(ti-))|+Γ(2-γ)∑i=1m‍tiγ|Ii*(x(ti-))-Ii*(y(ti-))|+T|v||Λx-Λy|+|a2|T|v||Πx-Πy|+Γ(2-γ)T×∑i=1m‍tiγ-1|Ii*(x(ti-))-Ii*(y(ti-))|.
Since
(48)|Λx-Λy|≤|a2|∫0T‍(T-s)α-1Γ(α)|f(s,x(s))-f(s,y(s))|ds+|b2|∫0T‍(T-s)α-γ-1Γ(α-γ)|f(s,x(s))-f(s,y(s))|ds≤∥h∥L∞|a2|TαΓ(α+1)∥x-y∥+∥h∥L∞|b2|Tα-γΓ(α-γ+1)∥x-y∥,|Πx-Πy|≤∑i=1m‍|Ii(x(ti-))-Ii(y(ti-))|+Γ(2-γ)∑i=1m‍tiγ|Ii*(x(ti-))-Ii*(y(ti-))|≤mL∥x-y∥+Γ(2-γ)L*∑i=1m‍tiγ∥x-y∥,
then combining these two estimations with (47), we obtain
(49)∥Fx-Fy∥≤[T∥h∥L∞|v|(|a2|TαΓ(α+1)+|b2|Tα-γΓ(α-γ+1))+Γ(2-γ)L*((1+|a2|T|v|)∑i=1m‍tiγ+T∑i=1m‍tiγ-1)+∥h∥L∞TαΓ(α+1)+(1+|a2|T|v|)mL]∥x-y∥.
Therefore, by (46), the operator F is a contraction mapping on PC(J,ℝ). Then it follows Banach's fixed point theorem that the problem (1) has a unique solution on J. This completes the proof.


## 4. Examples

Finally we give two simple examples to show the applicability of our results.Example 1.
Consider the following impulsive fractional separated BVP:(50)cD7/4x(t)=cost(t+6)2(x(t)+arctanx(t)),t∈[0,1],t≠12,Δx(12)=|x((1/2)-)|(17+|x((1/2)-)|),Δ(cD1/4x(12))=|x((1/2)-)|(20+|x((1/2)-)|),x(0)+2(cD1/4x(0))=12,12x(1)+13(cD1/4x(1))=2.
Hereα=7/4, γ=1/4, T=1, and m=1. Clearly, we can take h(t)=2cost/(t+6)2, L=1/17 and L*=1/20 such that the relations (45) hold. Moreover
(51)∥h∥L∞TαΓ(α+1)+T∥h∥L∞|v|(|a2|TαΓ(α+1)+|b2|Tα-γΓ(α-γ+1))+(1+|a2|T|v|)mL+Γ(2-γ)L*×[(1+|a2|T|v|)∑i=1m‍tiγ+T∑i=1m‍1ti1-γ]≈118×1.2728+117×1.5796+120×2.7666<1.
Thus, all the assumptions of Theorem 12 are satisfied. Hence, by the conclusion of Theorem 12, the impulsive fractional BVP (50) has a unique solution on [0,1].Example 2.
Consider the following impulsive fractional separated BVP:(52)cD3/2x(t)=5t2+e-|x(t)|+sinx(t),t∈[0,1],t≠14,Δx(14)=2|x((1/4)-)|(1+|x((1/4)-)|),Δ(cD1/2x(14))=cosx(14-)+3,3x(0)+12(cD1/2x(0))=1,2x(1)+3(cD1/2x(1))=2.5.
In the context of this problem, we have(53)|f(t,x)|=|5t2+e-|x|+sinx|≤7,t∈[0,1],x∈ℝ,|Ik(x)|≤2,|Ik*(x)|≤4,x∈ℝ.
Put h(t)≡7, H1=2, and H2=4. Then from Theorem 11, the impulsive fractional BVP (52) has at least one solution on [0,1].



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*Source: 785078-2013-09-16.xml*