# An Alternative Method for the Study of Impulsive Differential Equations of Fractional Orders in a Banach Space

Authors: Asma Bouzaroura; Saïd Mazouzi
Journal: International Journal of Differential Equations
 (2013)

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## Abstract

This paper is concerned with the existence, uniqueness, and stability of the solution
of some impulsive fractional problem in a Banach space subjected to a nonlocal condition.
Meanwhile, we give a new concept of a solution to impulsive fractional equations of multiorders.
The derived results are based on Banach's contraction theorem as well as Schaefer's
fixed point theorem.


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## Body

## 1. Introduction

It is well known that the theory of fractional calculus deals with the concepts of differentiation and integration of arbitrary orders, real and complex. Actually, the real importance of fractional derivatives lies in their nonlocal character which gives rise to a long memory effect and thus to a better insight into the modelled processes. On the other hand, since models using classical derivatives are just a special case of those using fractional derivatives, then most of the investigators in different areas such as electronics, viscoelasticity, satellite guidance, medicine, anomalous diffusion, signal processing, and many other branches of science and technology have revisited some classical dynamic systems in the framework of fractional derivatives to get better results; see the references [1–8]. We point out that most of dynamic systems are naturally governed by fractional differential equations; for further applications of fractional derivatives in other areas and useful backgrounds we refer the reader to the works [1–5, 7–12].As far as we are concerned with impulsive fractional differential equations, we intend to improve and correct in this paper some existence results established earlier in [4, 13–18] for impulsive fractional differential equations. There have been in the last couple of years several concepts of solutions satisfying some fractional equations subjected to impulsive conditions, see [13, 14, 18, 19], while the authors of [18] claimed that their new concept is the more realistic than the existing ones. Actually, we believe that nobody holds all the truth about this subject and a lot of dark sides of these approaches are not yet well elucidated.Regarding the concept of a solution for impulsive fractional equations introduced by [18] we point out that Lemma 2.6 which has been used by the authors to obtain the equivalence between an impulsive fractional problem and an integral equation is false as we see in the following counterexample.In the famous book of Nagy and Riesz [20, page 48], there is an example of monotonic continuous function F:[0,1]→ℝ which is not constant in any subinterval of [0,1] and satisfies F′=0, almost everywhere in [0,1]. So, in terms of Caputo’s derivative we would have formally for any α∈(0,1)(1)CD0+αF(t)=1Γ(1-α)∫0t(t-s)-αF′(s)ds=0,t∈[0,1],F(a)=F0,0<a<1,withF0beingthevalueofFata.
However, there is no apparent equivalence between this problem and the fractional integral representation of F defined in Lemma 2.6 [18]; otherwise the function F(t) would be constant and equal to F0 throughout the interval [0,1] which is a contradiction! Moreover, since in the same work Lemma 2.7 is based on the latter lemma then it is not correct and may lead to apparent contradiction.Our main contribution in this paper is the study of new fractional problems of several orders in a Banach space subjected to some impulsive conditions of the form(2)Dtk+αkCu(t)=A(t,u)u(t)+F(∫ath(t,s,u(σ(s)))ds,t,u(t),∫ath(t,s,u(σ(s)))ds,∫atk+1k(t,s,u(τ(s)))ds),t∈Jk,k=0,…,m,u(a)=u0∈E,u(tk+)=u(tk-)+Ik(u(tk-)),k=1,…,m.
Let us first give a concrete example of such a problem in ℝ; namely;
(3)DC0+αu(t)=tp-1,t∈J0=[0,1],DC1+βu(t)=(t-1)q,t∈J1=(1,T],u(0)=1,u(1+)=u(1-)+2,
where 0<α<1, 0<β<1, T>1, and p,q∈ℝ+. So, we look for a piecewise continuous function u:[0,T]→ℝ satisfying (3). Solving the subproblem
(4)DC0+αu(t)=tp-1,t∈J0,u(0)=1,
we obtain
(5)u(t)=1+1Γ(α)∫0t(t-s)α-1spds-1Γ(α)∫0t(t-s)α-1ds=1+Γ(p+1)Γ(α+p+1)tα+p-1Γ(α+1)tα,
from which we get u(1)=1+(Γ(p+1)/Γ(α+p+1))-(1/Γ(α+1)).On the other hand, the solution of the subproblem(6)DC1+βu(t)=(t-1)q,t∈J1,u(1+)=u(1)+2=3+Γ(p+1)Γ(α+p+1)-1Γ(α+1)
is given by
(7)u(t)=u(1+)+1Γ(β)∫1t(t-s)β-1(s-1)qds=3+Γ(p+1)Γ(α+p+1)-1Γ(α+1)+Γ(q+1)Γ(β+q+1)(t-1)β+q.Hence, the piecewise continuous function(8)u(t)={1+Γ(p+1)Γ(α+p+1)tα+p-1Γ(α+1)tα,t∈J0,3+Γ(p+1)Γ(α+p+1)-1Γ(α+1)+Γ(q+1)Γ(β+q+1)(t-1)β+q,t∈J1,
is a solution to the impulsive fractional problem (3).A particular problem of (3) is as follows:
(9)DC0+γu(t)=tp-1,t∈J0,DC1+γu(t)=(t-1)q,t∈J1,u(0)=1,u(1+)=u(1-)+2
corresponding to the case γ=α=β whose solution is
(10)u(t)={1+Γ(p+1)Γ(γ+p+1)tγ+p-1Γ(γ+1)tγ,t∈J0,3+Γ(p+1)Γ(γ+p+1)-1Γ(γ+1)+Γ(q+1)Γ(γ+q+1)(t-1)γ+q,t∈J1.
The paper is organized as follows. We present in Section 2 our problem as we establish some equivalence between the the given problem and a nonlinear integral equation. Next, we state a piecewise-continuous type of the Ascoli-Arzela theorem as well as Schaefer’s fixed point theorem in order to apply them subsequently in our proofs. In Section 3 we use the Banach contraction theorem to establish an existence and uniqueness theorem of a quasilinear impulsive fractional problem in an abstract Banach space. In Section 4 we apply Schaefer’s fixed point theorem to some semilinear impulsive fractional problem in a finite dimensional Banach space to obtain the existence of a piecewise continuous solution; on the other hand we prove the stability of the obtained solution with respect to the initial value. Finally, we conclude the paper by a concrete example illustrating one of our results.


## 2. Preliminaries

The main purpose of this paper is the investigation of the existence and uniqueness of solution corresponding to the following impulsive fractional integrodifferential equation in a Banach space(E,∥·∥)(11)DCtk+αku(t)=A(t,u)u(t)+F(∫ath(t,s,u(σ(s)))dst,u(t),∫ath(t,s,u(σ(s)))ds,∫atk+1k(t,s,u(τ(s)))ds),t∈Jk,k=0,…,m,u(a)=u0∈E,u(tk+)=u(tk-)+Ik(u(tk-)),k=1,…,m,
where(i)
J=[a,T] with 0≤a<T<∞ and J0=[a,t1],  Jk=(tk,tk+1];  k=1,…,m,(ii)
DCtk+αk is Caputo’s fractional derivative of order αk∈(0,1),  k=0,…,m,(iii)
A:J×E→ℬ(E) is a continuous operator, where ℬ(E)  is the Banach space of  bounded linear operators on E  in itself,(iv)
Ik:E→E,  t0=a<t1<⋯<tm<tm+1=T; u(tk+)=limh→0+u(tk+ϵ) and u(tk-)=limh→0-u(tk+ϵ)=u(tk) are, respectively, the right and left limits of u(t)  at the discontinuity point t=tk.We set the following hypotheses:(j)
the functionsσ,τ:J→J are continuous with a≤σ(t)≤t  and a≤τ(t)≤t, for every t∈J,(jj)
the nonlinear functionF:J×E×E×E→E  is continuous, and
(12)h:D×E→E,D={(t,s)∈ℝ2:a≤s≤t≤T},k:D0×E→E,whereD0={(t,s)∈ℝ2:t∈J,a≤s≤T},are two continuous functions over D×E  and D0×E, respectively.We will use in the sequel the following notation:(13)Hu(t)=∫ath(t,s,u(σ(s)))ds,Ktk+1u(t)=∫atk+1k(t,s,u(τ(s)))ds,k=0,1,…,m,Φtk+1(t,u(t))=F(t,u(t),Hu(t),Ktk+1u(t)),k=0,1,…,m.
We recall that 𝒞=𝒞(J;E)  is the Banach space of continuous functions u:J→E  endowed with the norm
(14)∥u∥𝒞=supt∈J∥u(t)∥.
Next, we introduce the definition of the fractional derivative in the sense of Caputo. We have the following.Definition 1.
We define the left-sided fractional Riemann-Liouville integral of orderα∈(0,1) of a function f:[c,d]→E  as follows:
(15)Jc+αf(t)=1Γ(α)∫ct(t-s)α-1f(s)ds,t>c.
We define the left-sided fractional derivative of orderα∈(0,1) of a function f:[c,d]→E  in the sense of Caputo by
(16)DCc+αf(t)=1Γ(1-α)∫ct(t-s)-αf′(s)ds,t>c.Remark 2.
(1) We point out that the previous integrals are understood in the sense of Bochner.
(2) We assume of course that the functionf satisfies the necessary conditions for which those integrals are well defined.Next, we consider the linear functional space(17)𝒫𝒞(J;E)={u:J→E,u∈𝒞((tk,tk+1];E),k=0,…,ms.t.u(tk-)andu(tk+)existwithu(tk-)=u(tk),k=1,…,m}
equipped with the norm
(18)∥u∥𝒫𝒞=supt∈J∥u(t)∥.
We obtain a Banach space (𝒫𝒞(J;E),∥·∥𝒫𝒞).Now, we recall the definition of the solution of the problem (11).Definition 3.
A functionu∈𝒫𝒞(J;E) is said to be a solution of the problem (11) if DCtk+αku(t) exists in Jk, for k=0,…,m,  and satisfies(i)
the equationDCtk+αku(t)=A(t,u)u(t)+Φtk+1(t,u(t)) in Jk, k=0,…,m,(ii)
the initial conditionu(a)=u0,(iii)
the impulsive conditionsu(tk+)=u(tk-)+Ik(u(tk-)), k=1,…,m.Lemma 4.
A functionu∈𝒫𝒞(J;E) satisfies the following nonlinear integral equation
(19)u(t)={u0+1Γ(α0)∫at‍(t-s)α0-1A(s,u)u(s)ds+1Γ(α0)∫at‍(t-s)α0-1Φt1(s,u)ds,t∈[a,t1],u0+∑i=1kIi(u(ti-))+1Γ(αk)∫tkt(t-s)αk-1A(s,u)u(s)ds+∑i=1k1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1A(s,u)u(s)ds+1Γ(αk)∫tkt(t-s)αk-1Φtk+1(s,u)ds+∑i=1k‍1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1Φti(s,u)ds,t∈Jk,k=1,…,m,
if and only if it is a solution to problem (11).Proof.
Since we haveu(tk-)=u(tk), then u(tk+)=u(tk)+Ik(u(tk-)).
Now, fort∈J0=[a,t1], the solution of the problem
(20)Dca+α0u(t)=A(t,u)u(t)+Φt1(t,u),t∈J0,u(a)=u0∈E
is given by
(21)u(t)=u0+1Γ(α0)∫at(t-s)α0-1A(s,u)u(s)ds+1Γ(α0)∫at(t-s)α0-1Φt1(s,u)ds,t∈J0.
We have for t=t1 the following relationu(t1+)=u(t1)+I1(u(t1-)), and so
(22)u(t1+)=1Γ(α0)∫at1(t1-s)α0-1A(s,u)u(s)ds+1Γ(α0)∫at1(t1-s)α0-1Φt1(s,u)ds+u0+I1(u(t1-)).
Next, fort∈J1=(t1,t2], we have
(23)u(t)=u(t1+)+1Γ(α1)∫t1t(t-s)α1-1A(s,u)u(s)ds+1Γ(α1)∫t1t(t-s)α1-1Φt2(s,u)ds=u0+1Γ(α0)∫at1‍(t1-s)α0-1A(s,u)u(s)ds+1Γ(α0)∫at1(t1-s)α0-1Φt1(s,u)ds+1Γ(α1)∫t1t(t-s)α1-1A(s,u)u(s)ds+1Γ(α1)∫t1t(t-s)α1-1Φt2(s,u)ds+I1(u(t1-))
from which we infer that
(24)u(t2+)=u0+1Γ(α0)∫at1(t1-s)α0-1A(s,u)u(s)ds+1Γ(α0)∫at1(t1-s)α0-1Φt1(s,u)ds+1Γ(α1)∫t1t2(t2-s)α1-1A(s,u)u(s)ds+1Γ(α1)∫t1t2(t2-s)α1-1Φt2(s,u)ds+I1(u(t1-))+I2(u(t2-)).
Arguing as before we obtain fort∈J2(25)u(t)=u(t2+)+1Γ(α2)∫t2t(t-s)α2-1A(s,u)u(s)ds+1Γ(α2)∫t2t(t-s)α2-1Φt3(s,u)ds=u0+1Γ(α0)∫at1(t1-s)α0-1A(s,u)u(s)ds+1Γ(α0)∫at1(t1-s)α0-1Φt1(s,u)ds+1Γ(α1)∫t1t2(t2-s)α1-1A(s,u)u(s)ds+1Γ(α1)∫t1t2(t2-s)α1-1Φt2(s,u)ds+1Γ(α2)∫t2t(t-s)α2-1A(s,u)u(s)ds+1Γ(α2)∫t2t(t-s)α2-1Φt3(s,u)ds+I1(u(t1-))+I2(u(t2-)).
Reasoning by induction we get, for any t∈Jk, k=1,…,m, the general expression
(26)u(t)=u0+1Γ(αk)∫tkt(t-s)αk-1A(s,u)u(s)ds+∑i=1k1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1A(s,u)u(s)ds+1Γ(αk)∫tkt(t-s)αk-1Φtk+1(s,u)ds+∑i=1k1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1Φti(s,u)ds+∑i=1kIi(u(ti-)).
Conversely, we assume that u satisfies (19). If t=a, then u(a)=u0.
Now, using the fact that Caputo’s derivative of a constant is zero, then, for everyt∈Jk,  k=0,…,m, we get
(27)DCtk+αku(t)=DCtk+αk[1Γ(αk)∫tkt(t-s)αk-1A(s,u)u(s)ds]+DCtkαk[1Γ(αk)∫tkt‍(t-s)αk-1Φtk+1(s,u)ds]=DCtk+αk(Jtk+αkA(t,u)u(t))+DCtk+αk(Jtk+αkΦtk+1(t,u)).
So(28)DCtk+αku(t)=A(t,u)u(t)+Φtk+1(t,u),
for every t∈Jk, k=0,…,m.
Also we can easily show that(29)u(tk+)=u(tk)+Ik(u(tk-)),k=1,…,m.We conclude this section by introducing some useful theorems which will be used in the sequel.Theorem 5 (𝒫𝒞-type Ascoli-Arzela theorem [21]).
LetE  be a Banach space and 𝒲⊂𝒫𝒞(J,E). If the following conditions are satisfied(i)
𝒲 is a uniformly bounded subset of 𝒫𝒞(J,E);(ii)
𝒲 is equicontinuous in (tk,tk+1), k=0,1,2,…,m;(iii)
𝒲(t)={u(t):u∈𝒲,t∈J∖{tk}}, 𝒲(tk+)={u(tk+):u∈𝒲}, and 𝒲(tk-)≡{u(tk-):u∈𝒲} are relatively compact subsets of E,then𝒲 is a relatively compact subset of 𝒫𝒞(J,E).Theorem 6 (Schaefer’s fixed point theorem).
LetE  be a Banach space and let 𝒯:E→E  be a completely continuous operator. If the set
(30)X={u∈E:u=λ𝒯u,λ∈(0,1)}
is bounded, then 𝒯  has at least a fixed point.


## 3. A Quasilinear Impulsive Fractional Problem

We begin our investigation by the following result which ensures the existence and the uniqueness of the solution of the following impulsive quasilinear problem:(31)DCtk+αku(t)=A(t,u)u(t)+Φtk+1(t,u(t)),t∈Jk,k=0,…,m,u(a)=u0∈E,u(tk+)=u(tk)+Ik(u(tk-)),k=1,…,m.
We assume that A:J×E→ℬ(E) is continuous and there exists a constant M>0 such that
(32)∥A(t,u)-A(t,v)∥≤M∥u-v∥,∀t∈J,∀u,v∈E.
We set M′=maxt∈J∥A(t,0)∥.It is not hard to establish the following estimates.Lemma 7.
Let the functionsh(t,s,u)  and k(t,s,u)  be continuous with respect to the variables s and t, and there are two positive constants C1 and C2 such that
(33)∥h(t,s,u)-h(t,s,v)∥≤C1∥u-v∥,∀t,s∈J,∀u,v∈E,∥k(t,s,u)-k(t,s,v)∥≤C2∥u-v∥,∀t,s∈J,∀u,v∈E.
Then, there exist two positive constants C1′ and  C2′ so that
(34)∥Hu(t)∥≤(T-a)(C1∥u∥𝒫𝒞+C1′),∥Hu(t)-Hv(t)∥≤C1(T-a)∥u-v∥𝒫𝒞,
and, for k=0,…,m, one has
(35)∥Ktk+1u(t)∥≤(T-a)(C2∥u∥𝒫𝒞+C2′),∥Ktk+1u(t)-Ktk+1v(t)∥≤C2(T-a)∥u-v∥𝒫𝒞,
for every u,v∈𝒫𝒞(J,E) and t∈J.We assume the following hypotheses:(H1)
α0,…,αm∈(0,1). We set T′=max0≤i≤m{(T-a)αi} and Γ′=min0≤i≤m{Γ(αi+1)}.(H2)
There is a positive constantL1 such that
(36)∥Φtk+1(t,u)-Φtk+1(t,v)∥≤{L1+(C1+C2)(T-a)}∥u-v∥𝒫𝒞,∀t∈J,∀u,v∈𝒫𝒞,k=0,…,m.We setL=L1+(C1+C2)(T-a) and L2=supt∈J∥F(t,0,0,0)∥.(H3)
There is a positive constantμ>0 such that
(37)∥Ik(u)-Ik(v)∥≤μ∥u-v∥,∀u,v∈E,k=1,…,m.(H4)
The positive real number(38)γ=mμ+(m+1)(L+2rM+M′)T′Γ′satisfies0<γ<1.Next, we state and prove the existence and uniqueness result for the quasilinear integrodifferential problem (31); we have the following.Theorem 8.
If the assumptions (H1)–(H4) are satisfied, then problem (31) has one and only one solution u∈𝒫𝒞(J,E).Proof.
Since we are concerned with the existence and uniqueness of the solution of (31) then, it is wise to use the Banach contraction principle in order to establish such results.
Letℬr={u∈𝒫𝒞(J,E):∥u∥𝒫𝒞≤r} be the closed ball of 𝒫𝒞(J,E)  centered at 0 with radius r satisfying the following inequality:
(39)φ(r)≔∥u0∥+(m+1)T′Γ′L′+[(m+1)T′Γ′(rM+M′+L)+mμ]r≤r,
where
(40)L′=(C1′+C2′)(T-a)+L2.
Endowing ℬr with the metric d(u,v)=∥u-v∥𝒫𝒞, for every u,v∈ℬr, we obtain a complete metric space (ℬr,d).  Next, we define the operator Ψ:ℬr→ℬr by
(41)Ψu(t)=u0+1Γ(αk)×∫tkt(t-s)αk-1A(s,u)u(s)ds+∑a<ti<t‍1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1A(s,u)u(s)ds+1Γ(αk)∫tkt(t-s)αk-1Φtk+1(s,u)ds+∑a<ti<t‍1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1Φti(s,u)ds+∑a<ti<t‍Ii(u(ti-)),t∈Jk,k=0,…,m.
It is understood that the sum ∑a<ti<t is zero if t∈J0.
First, we prove that ifu∈𝒫𝒞(J;E), then Ψu∈𝒫𝒞(J;E).
Indeed, for eacht∈(tk,tk+1), u∈𝒞((tk,tk+1),E), and any sufficiently small δ>0, we have
(42)∥Ψu(t+δ)-Ψu(t)∥≤(M∥u∥𝒫𝒞+M′+L)∥u∥𝒫𝒞+L′Γ(αk)×∫tkt[(t-s)αk-1-(t+δ-s)αk-1]ds+(M∥u∥𝒫𝒞+M′+L)∥u∥𝒫𝒞+L′Γ(αk)×∫tt+δ(t+δ-s)αk-1ds.
Calculating the integrals we find that
(43)∥Ψu(t+δ)-Ψu(t)∥≤3[(Mr+M′+L)r+L′]Γ′δαk.
Thus, the right-hand side tends to zero as δ→0. Likewise one gets limδ→0∥Ψu(t)-Ψu(t-δ)∥=0; this shows that Ψu is continuous at t. Hence Ψu∈𝒞((tk,tk+1),E).
Next, for the right endpointt=tk+1 we get for any sufficiently small δ>0(44)∥Ψu(tk+1)-Ψu(tk+1-δ)∥≤3[(Mr+M′+L)r+L′]Γ′δαk,
which shows that the right-hand side tends to zero as δ→0, and accordingly, Ψu  is continuous at tk+1. Therefore Ψu∈𝒫𝒞(J,E).
To prove thatΨℬr⊂ℬr  we see that, for any u∈ℬr  and t∈Jk,  k=0,…,m, we have
(45)∥Ψu(t)∥≤∥u0∥+1Γ(αk)×∫tkt(t-s)αk-1∥A(s,u)∥·∥u(s)∥ds+∑a<ti<t1Γ(αi-1)×∫ti-1ti(ti-s)αi-1-1∥A(s,u)∥·∥u(s)∥ds+1Γ(αk)∫tkt(t-s)αk-1∥Φtk+1(s,u)∥ds+∑a<ti<t1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1∥Φti(s,u)∥ds+∑a<ti<t∥Ii(u(ti-))∥.
Estimating the right-hand side we find(46)∥Ψu(t)∥≤∥u0∥+(m+1)T′Γ′×(M∥u∥𝒫𝒞+M′)·∥u∥𝒫𝒞+1Γ(αk)×∫tkt(t-s)αk-1[L∥u∥𝒫𝒞+(T-a)(C1′+C2′)+L2]ds+∑a<ti<t1Γ(αi-1)×∫ti-1ti(ti-s)αi-1-1[L∥u∥𝒫𝒞+(T-a)(C1′+C2′)+L2]ds+mμ∥u∥𝒫𝒞.
So
(47)∥Ψu(t)∥≤∥u0∥+(m+1)T′Γ′L′+[(m+1)T′Γ′(rM+M′+L)+mμ]r≤φ(r)≤r.
Therefore, ∥Ψu∥𝒫𝒞≤r, and consequently Ψℬr⊂ℬr.
Next, we prove thatΨ is a contraction mapping; indeed, for any u,v∈ℬr   and t∈Jk,  k=0,…,m, we have
(48)∥Ψu(t)-Ψv(t)∥≤1Γ(αk)∫tkt(t-s)αk-1∥A(s,u)u(s)-A(s,v)v(s)∥ds+∑a<ti<t1Γ(αi-1)×∫ti-1ti(ti-s)αi-1-1∥A(s,u)u(s)-A(s,v)v(s)∥ds+1Γ(αk)∫tkt(t-s)αk-1∥Φtk+1(s,u)-Φtk+1(s,v)∥ds+∑a<ti<t‍1Γ(αi-1)×∫ti-1ti(ti-s)αi-1-1∥Φti(t,u)-Φti(s,v)∥ds+∑a<ti<t‍∥Ii(u(ti-))-Ii(v(ti-))∥.
Taking into account the previous assumptions we get the following estimate:
(49)∥Ψu(t)-Ψv(t)∥≤mμ∥u-v∥𝒫𝒞+∥u-v∥𝒫𝒞Γ(αk)∫tkt(t-s)αk-1[Mr+M′+Mr]ds+∑a<ti<t∥u-v∥𝒫𝒞Γ(αi-1)×∫ti-1ti(ti-s)αi-1-1[Mr+M′+Mr]ds+L∥u-v∥𝒫𝒞Γ(αk)∫tkt(t-s)αk-1ds+∑a<ti<t‍L∥u-v∥𝒫𝒞Γ(αi-1)∫ti-1ti(ti-s)αi-1-1ds.
Thus,(50)∥Ψu(t)-Ψv(t)∥≤[mμ+(m+1)(L+2rM+M′)T′Γ′]∥u-v∥𝒫𝒞≤γ∥u-v∥𝒫𝒞.
Accordingly, the mapping Ψ has a unique fixed point u=Ψu∈ℬr, which completes the proof.


## 4. A Semilinear Impulsive Fractional Problem

In this section we consider a semilinear impulsive fractional integrodifferential problem subjected to a nonlocal condition in a finite dimensional normed space(E,∥·∥). Actually, the finite dimension requirement is due to some technical difficulties in order to prove some compactness properties. The problem is as follows:
(51)DCtk+αku(t)=A(t)u(t)+Φtk+1(t,u(t)),t∈Jk,k=0,…,m,u(a)+g(u)=u0∈E,u(tk+)=u(tk)+Ik(u(tk-)),k=1,…,m.We assume that the mappingA:J→ℬ(E) is continuous and we put
(52)M′′=maxt∈J∥A(t)∥.
We need the following hypothesis:(H5)
there exists a constantG>0  such that the mapping g:𝒫𝒞(J,E)→E satisfies
(53)∥g(u)-g(v)∥≤G∥u-v∥𝒫𝒞,∀u,v∈𝒫𝒞(J,E).Now, we are ready to state and prove the following result.Theorem 9.
If the assumptions (H1)–(H3) and (H5) are satisfied, then problem (51) has at least one solution u∈𝒫𝒞(J,E).Proof.
Let us define the operatorQ:𝒫𝒞(J,E)→𝒫𝒞(J,E) by
(54)Qu(t)=u0-g(u)+1Γ(αk)×∫tkt(t-s)αk-1A(s)u(s)ds+∑a<ti<t1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1A(s)u(s)ds+1Γ(αk)∫tkt(t-s)αk-1Φtk+1(s,u)ds+∑a<ti<t1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1Φti(s,u)ds+∑a<ti<tIi(u(ti-)),t∈Jk,k=0,…,m.
First, we notice that by using the same technique as that in the proof of the Theorem 8 we can establish that if u∈𝒫𝒞(J,E), then Qu∈𝒫𝒞(J,E); that is, the operator Q  maps the space 𝒫𝒞(J,E) into itself.
To prove thatQ  has a fixed point we use Schaefer’s fixed point theorem. We proceed in four steps.
Step  1 (Q  is continuous). Let {un}n≥1⊂𝒫𝒞(J,E) such that un→u in 𝒫𝒞(J,E); then
(55)∥Qun(t)-Qu(t)∥≤∥g(un)-g(u)∥+1Γ(αk)×∫tkt(t-s)αk-1∥A(s)(un(s)-u(s))∥ds+∑a<ti<t1Γ(αi-1)×∫ti-1ti(ti-s)αi-1-1∥A(s)(un(s)-u(s))∥ds+1Γ(αk)∫tkt(t-s)αk-1∥Φtk+1(s,un)-Φtk+1(s,u)∥ds+∑a<ti<t1Γ(αi-1)×∫ti-1ti(ti-s)αi-1-1∥Φti(s,un)-Φti(s,u)∥ds+∑a<ti<t∥Ii(un(ti-))-Ii(u(ti-))∥,t∈Jk,k=0,…,m.
Taking into account the assumptions (H2)-(H3) and (H5) and using Lemma7 we get
(56)∥Qun(t)-Qu(t)∥≤G∥un-u∥𝒫𝒞+M′′Γ(αk)∥un-u∥𝒫𝒞×∫tkt(t-s)αk-1ds+M′′∥un-u∥𝒫𝒞×∑a<ti<t1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1ds+LΓ(αk)∥un-u∥𝒫𝒞∫tkt(t-s)αk-1ds+L∥un-u∥𝒫𝒞∑a<ti<t1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1ds+mμ∥un-u∥𝒫𝒞,t∈Jk,k=0,…,m.
Calculating the integrals in the right-hand side we obtain
(57)∥Qun-Qu∥𝒫𝒞≤[G+mμ+(m+1)T′Γ′(M′′+L)]∥un-u∥𝒫𝒞.
So
(58)∥Qun-Qu∥𝒫𝒞→0,asn→∞,
and accordingly, Q  is continuous.
Step  2. Let ε>0 and Bε={u∈𝒫𝒞(J,E):∥u∥𝒫𝒞≤ε}. Define 𝒲={Qu:u∈Bε}; then for any u∈Bε we have
(59)∥Qu(t)∥≤∥u0-g(u)∥+1Γ(αk)∫tkt(t-s)αk-1∥A(s)∥∥u(s)∥ds+∑a<ti<t‍1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1∥A(s)∥∥u(s)∥ds+1Γ(αk)∫tkt(t-s)αk-1∥Φtk+1(s,u)∥ds+∑a<ti<t1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1∥Φti(s,u)∥ds+∑a<ti<t∥Ii(u(ti-))∥,t∈Jk,k=0,…,m.
Estimating the right-hand side we obtain(60)∥Qu(t)∥≤∥u0∥+G∥u∥𝒫𝒞+∥g(0)∥+(m+1)T′Γ′M′′·∥u∥𝒫𝒞+1Γ(αk)∫tkt(t-s)αk-1[L∥u∥𝒫𝒞+L′]ds+∑a<ti<t‍1Γ(αi-1)∫ti-1ti(ti-s)αk-1[L∥u∥𝒫𝒞+L′]ds+mμ∥u∥𝒫𝒞,
implying that
(61)∥Qu∥𝒫𝒞≤∥u0∥+∥g(0)∥+(m+1)T′Γ′(M′′ε+Lε+L′)+Gε+mμε:=ρ.
Hence 𝒲  is uniformly bounded.
Step  3 (we prove that 𝒲 is equicontinuous). Let u∈Bε; then, for any tk<τ1<τ2≤tk+1, we have
(62)∥Qu(τ2)-Qu(τ1)∥≤1Γ(αk)×∫tkτ1[(τ1-s)αk-1-(τ2-s)αk-1]∥A(s)∥∥u(s)∥ds+1Γ(αk)∫τ1τ2(τ2-s)αk-1∥A(s)∥∥u(s)∥ds+1Γ(αk)∫tkτ1[(τ1-s)αk-1-(τ2-s)αk-1]∥Φtk+1(s,u)∥ds+1Γ(αk)∫τ1τ2(τ2-s)αk-1∥Φtk+1(s,u)∥ds.
So
(63)∥Qu(τ2)-Qu(τ1)∥≤(M′′+L)ε+L′Γ(αk)∫tkτ1[(τ1-s)αk-1-(τ2-s)αk-1]ds+(M′′+L)ε+L′Γ(αk)∫τ1τ2(τ2-s)αk-1ds≤3[(M′′+L)ε+L′]Γ′(τ2-τ1)αk.
As τ1→τ2, the right-hand side of the previous inequality tends to zero, which means that 𝒲 is equicontinuous.
We point out that the closures of the subsets𝒲(t):={Qu(t):u∈Bε,t∈J∖{tk},k=1,…,m}, 𝒲(tk-):={Qu(tk-):u∈Bε}, and 𝒲(tk+):={Qu(tk+):u∈Bε}, k=1,…,m, are bounded in E  (dimE<∞); hence they are compact.
As a consequence of the previous steps and the𝒫𝒞-type Arzela-Ascoli theorem we conclude that Q  is completely continuous.
Step  4. Now, we show that the set
(64)X={u∈𝒫𝒞(J,E):u=λQu,λ∈(0,1)}
is bounded.
Letu∈X; then u=λQu, for some λ∈(0,1). Thus, for each t∈J,
(65)∥u(t)∥≤λ∥u0∥+λ∥g(u)∥+λΓ(αk)∫tkt(t-s)αk-1∥A(s)∥·∥u(s)∥ds+λ∑a<ti<t1Γ(αi-1)×∫ti-1ti(ti-s)αi-1-1∥A(s)∥·∥u(s)∥ds+λΓ(αk)∫tkt(t-s)αk-1∥Φtk+1(s,u)∥ds+λ∑a<ti<t1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1∥Φti(s,u)∥ds+λ∑a<ti<t∥Ii(u(ti-))∥≤λ[∥u0∥+∥g(0)∥+(m+1)T′Γ′×(M′′ε+Lε+L′)+Gε+mμε(m+1)T′Γ′]<∞.
This shows that the set X is bounded.
We conclude by Schaefer’s fixed point theorem that the operatorQ  has a fixed point u∈𝒫𝒞(J,E) such that Qu=u, which means that u  is a solution to problem (51).Next, we establish the continuous dependence of the solution upon the initial value. We have the following.Proposition 10.
Under the hypotheses (H1)–(H3) and (H5) the solution of problem (51) depends continuously upon its initial value if
(66)G+mμ+(m+1)(L+M′′)T′Γ′<1.Proof.
Sinceu is a solution to (51), then it satisfies the integral equation (19). Let v be a solution to problem (51) with initial value v(a)=v0-g(v). Then v(t) satisfies the integral equation
(67)v(t)=v0-g(v)+1Γ(αk)∫tkt(t-s)αk-1A(s)v(s)ds+∑a<ti<t1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1A(s)v(s)ds+1Γ(αk)∫tkt(t-s)αk-1Φtk+1(s,v)ds+∑a<ti<t1Γ(αi-1)∫ti-1ti(ti-s)αi-1-1Φti(s,v)ds+∑a<ti<tIi(v(ti-)),t∈Jk,k=0,…,m.
Estimating the difference between solutions u(t) and v(t) to (19) and (67), respectively, we get
(68)∥u(t)-v(t)∥≤∥u0-v0∥+G∥u-v∥𝒫𝒞+M′′(m+1)T′Γ′∥u-v∥𝒫𝒞+L(m+1)T′Γ′∥u-v∥𝒫𝒞+mμ∥u-v∥𝒫𝒞,t∈Jk,k=0,…,m.
Taking the supremum over the interval J  we find that
(69)∥u-v∥𝒫𝒞≤1ρ∥u0-v0∥,
where
(70)ρ=1-G-mμ-(m+1)(L+M′′)T′Γ′,
which proves that the mapping u0↦u  is continuous from E→𝒫𝒞(J,E).


## 5. Example

Consider the following impulsive fractional integrodifferential problem(71)D0+1/2u(t)=t24u(t)cosu(t)+|u(t)|(|u(t)|+2)(t2+12)+∫0te-|u(s)|/6(t+2)2ds+∫01/2sin(u(s)/4)et+6ds,t∈J0,D(1/2)+3/4u(t)=t24u(t)cosu(t)+|u(t)|(|u(t)|+2)(t2+12)+∫0te-|u(s)|/6(t+2)2ds+∫01sin(u(s)/4)et+6ds,t∈J1,u(0)=1,u(12+)=u(12-)+|u((1/2)-)||u((1/2)-)|+6.
We set J0=[0,1/2],  J1=(1/2,1], and J=[0,1].  We take E=ℝ, a=t0=0, t1=1/2, T=1, α0=1/2, α1=3/4, and ℬr={u∈𝒫𝒞(J,E),∥u∥𝒫𝒞≤r}. Define
(72)A(t,u)=t24(cosu)I,Hu(t)=∫0te-|u(s)|/6(t+2)2ds,Ktk+1u(t)=∫0tk+1sin(u(s)/4)et+6ds,k=0,1,Φtk+1(t,u(t))=|u(t)|(|u(t)|+2)(t2+12)+∫0te-|u(s)|/6(t+2)2ds+∫0tk+1sin(u(s)/4)et+6ds,k=0,1,I1u(12-)=|u((1/2)-)||u((1/2)-)|+6.
For any u,v∈ℬr  and t∈J,  we have
(73)|Hu(t)-Hv(t)|≤124∥u-v∥𝒫𝒞.
Hence C1=1/24. Likewise, one has, for k=0,1,
(74)|Ktk+1u(t)-Ktk+1v(t)|≤124∥u-v∥𝒫𝒞;
then C2=1/24.By (H1), we have(75)T′=max{(T-a)α0,(T-a)α1}=1,Γ′=min{Γ(α0+1),Γ(α1+1)}=min{Γ(32),Γ(74)}=Γ(32)=π2.
On the other hand, using (H2) we obtain
(76)|Φtk+1(t,u)-Φtk+1(t,v)|≤||u(t)|(|u(t)|+2)(t2+12)-|v(t)|(|v(t)|+2)(t2+12)|+|Hu(t)-Hv(t)|+|Ktk+1u(t)-Ktk+1v(t)|≤324∥u-v∥𝒫𝒞.Thus(77)L1=124,L=324.
Assumption (H3) gives
(78)|I1(u)-I1(v)|≤16∥u-v∥𝒫𝒞,
so μ=1/6.Due to the definition ofA(t,u) we have M=M′=1/24. Let us now find a threshold for the value of r  for which condition (H4) is satisfied. We should have
(79)0<γ=16+23π+r3π<1,
so r  is any positive number such that r<(5π-4)/2=2.4311. We conclude by Theorem 8 that problem (71) has a unique solution u∈𝒫𝒞([0,1],ℝ) such that ∥u∥𝒫𝒞≤r.


## 6. Concluding Remarks

In this work we have first noticed that most of the published papers dealing with impulsive differential equations of fractional orders are not mathematically correct, so we have proved through a concrete counterexample that the concept of solution proposed recently by some authors is not realistic. On the other hand, we introduced a new class of impulsive fractional problems with several fractional orders and we established an equivalence with some integral equation. Moreover, we derived two existence results by using two different fixed point theorems as we proved the stability of the solution of the given problem with respect to the initial value. Finally, we illustrated our first theorem of existence and uniqueness by a concrete example inℝ.



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