IS S N 1 3 0 3 –5 9 9 1
LOCAL AND EXTREMAL SOLUTIONS OF SOME FRACTIONAL INTEGRODIFFERENTIAL EQUATION WITH IMPULSES*
ATMANIA RAHIMA
Abstract. The subject of this work is to prove existence, uniqueness, and con-tinuous dependence upon the data of solution to integrodi¤erential hyperbolic equation with integral conditions. The proofs are based on a priori estimates and Laplace transform method. Finally, the solution by using a numerical technique for inverting the Laplace transforms is obtained.
Introduction
The concept of fractional analysis like di¤erentiation and integration can be considered as a generalization of ordinary ones with integer order. However, it re-mains a lot to be done before assuming that this generalization is really established. Fractional di¤erential equations have been extensively applied in many …elds, for example, in probability, viscoelasticity and electrical circuits. Di¤erent theoretical studies about the subject were done by many famous mathematicians over the years like Liouville, Riemann, Fourier, Abel, Leibniz. For more details, we refer to the books [6, 9, 10].
On the other side, the interest in studying impulsive di¤erential equations is related to their utility for modeling phenomena subject to considerable short-term changes during their evolution. The fact that the duration of the perturbations is negligible in comparison with the duration of the phenomena requires us to consider them in the form of impulses. The theory of impulsive di¤erential equations has been well developed during these twenty last years; to know more see [2, 7]
Received by the editors Nov. 02, 2012; Accepted: J¬ne 24, 2013;
2010 Mathematics Subject Classi…cation. Primary 34A08, Secondary, 34A12, 34A37. Key words and phrases. Local existence, extremal solution, integrodi¤erential equation, Ca-puto fractional derivative, impulsive conditions, …xed point theory, impulsive fractional inequality. The main results of this paper were presented in part at the conference Algerian-Turkish International Days on Mathematics 2012 (ATIM’2012) to be held October 9–11, 2012 in Annaba, Algeria at the Badji Mokhtar Annaba University.
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It is therefore interesting that the impulsive e¤ects may be a part of studies of fractional di¤erential problems. This topic has awoken the curiosity of many re-searchers in recent years. Recently, some authors [1, 3, 5, 11] discussed existence results of solutions of impulsive fractional di¤erential equations under di¤erent con-ditions, boundary ones, non local ones etc. The results are obtained by using …xed point principles. We point out that in the papers [1, 3, 5, 11] the authors used an incorrect formula of solutions; for this reason in [4] the authors introduced the right formula for solutions of some given impulsive Cauchy problem with Caputo fractional derivative.
In this paper, we study the existence of local and extremal solutions for some integrodi¤erential fractional equation involving Caputo’s derivative subject to im-pulses in …xed moments by using …xed-point theory and fractional analysis under suitable assumptions. This, taking into account the discontinuous nature of impul-sive di¤erential problems compared with non impulimpul-sive di¤erential ones precisely for the fractional order. A non impulsive fractional problem was treated in [8].
The paper is divided into three sections. In Section 2 we recall some basic notions which will be used in the remainder of the paper. In Section 3 we establish existence results, …rst, of local solution based on Schauder …xed point theorem then of extremal solutions by using impulsive fractional inequalities.
1. Preliminaries
1.1. Fractional calculus. We will introduce notations and de…nitions that are used in this paper and can be found in [6]. Let a ( 1 < a < 1) a constant on the real axis R, the Riemann-Liouville fractional integral operator of order > 0 is de…ned by Ia+f (t) = 1 ( ) Z t a (t s) 1f (s) ds; t > a:
Among the great amount of de…nitions dealing with fractional derivatives of order 0 we recall the Riemann-Liouville one which is de…ned by
Da+f (t) = DnIan+ f (t) = 1 (n ) dn dtn Z t a (t s)n 1f (s) ds; (1.1) provided that the right-hand-side exists; where n = [ ] + 1; ( ) is the classical Gamma function. Remark that Da+K 6= 0, for any constant K.
The following Caputo’s de…nition is also widely used due to its practical formu-lation in real world problems:
cD a+f (t) = Ian+ D nf (t) = 1 (n ) Z t a (t s)n 1f(n)(s) ds: (1.2) It’s clear thatcD
a+K = 0, for any constant K: Thus, the following properties hold cDn
The function f (t) = c0 + c1(t a) + : + cn 1(t a)n 1 is a solution of the
equation cD
a+f (t) = 0 with c0; c1; : : : ; cn 1 arbitrary real constants, then for
f 2 Cn[a; b] we have
Ia+ cDa+f (t) = f (t) + c1(t a) + : + cn 1(t a)n 1; (1.3)
for n 1 < < n. In particular, when 0 < < 1, we have
cD a+f (t) = 1 (1 ) Z t a f0(s) (t s) ds; Ia+cDa+f (t) = f (t) + c0; c02 R:
1.2. Impulsive e¤ects. The most real case of the instants of impulsive e¤ects is as follows: A …nite or in…nite number of …xed moments noted tk given by an
increasing sequence without accumulation points, i.e., t1< t2< < tk< and
lim
k!1tk= +1.
Let us denote the right and left limits of x (t) at t = tk respectively by
x t+k = lim
h!0+x (tk+ h) ; x tk = limh!0 x (tk+ h) :
De…nition 1.1. The impulsive e¤ects said impulsive condition is measured by the di¤erence between the limits of the state function x(t) on the right and left of the moment of impulses tk and is noted
x (tk) = x t+k x tk ; k = 1; 2; 3; : : : :
Remark 1.2. Submitting a system to such conditions deprives the state function of its continuity but improves signi…cantly other properties especially the numerical results.
To study an impulsive initial value problem on the interval [t0; t0+ T ], where
the number of impulses is m; we proceed as follows: [t0; t0+ T ] is subdivided
into m + 1 intervals and we act as if we had a classical Cauchy problem on each interval (tk; tk+1], k = 0; : : : ; m, where tm+1 = t0+ T . To ensure the
ex-istence of a solution we must assume the continuity of x (t) on (tk; tk+1], k =
0; : : : ; m and its right limit exists at tk for k = 0; : : : ; m. Hence, the solutions
should belong to the space of piece continuous functions denoted by PC and de-…ned by PC ([t0; t0+ T ] ; R) = fx : [t0; t0+ T ] ! R : x (t) is continuous for t 6=
tk; left continuous at t = tk and x t+k exists for k = 1; : : : ; mg which is a Banach
space once endowed with the norm kxkPC= max ( sup t2(tk;tk+1] jx (t)j ; k = 0; 1; : : : ; m ) :
2. Main results
2.1. Impulsive fractional integrodi¤erential initial value problem. We are concerned by the following scalar integrodi¤erential equation of fractional order 0 < < 1; cD t+0 x (t) + G (t; x (t)) = Z t t0 K (t; s; x (s)) ds; t 6= tk; k = 1; : : : ; m; (2.1)
with the initial condition
x (t0) = x0; t0 0; (2.2)
and the impulsive conditions
x (tk) = Jk x tk ; k = 1; : : : ; m: (2.3)
We set the following assumptions:
(A1) The instants of impulsive e¤ects tk; k = 1; : : : ; m are such that t0 < t1
< < tk< tk+1< < tm< t0+ T . (A2) G(t; x) 2 C([t0; t0+T ] R; R); K(t; s; x) 2 C([t0; t0+T ] [t0; t0+T ] R; R) and Jk 2 C(R; R), k = 1; : : : ; m. (A3) The integrals Rt t0(t s) 1Rs t0K(s; ; x( ))d ds and t R t0 (t s) 1G(s ; x(s))ds
are pointwise de…ned on (t0; t0+ T ].
De…nition 2.1. A function x 2 PC ([t0; t0+ T ] ; R) with its -derivatives
exist-ing on [t0; t0+ T ] n ftkgk=1;:::;m for 0 < < 1; is said to be a solution of the
problem (2.1)-(2.3) if x satis…es the fractional integrodi¤erential equation (2.1) on [t0; t0+ T ] n ftkgk=1;:::;m , the impulsive conditions (2.2) for t = tk; k = 1; : : : ; m;
and the initial condition (2.3) for t = t0.
2.2. Impulsive fractional integral equation. We begin with the following lemma which allows us to discuss the properties of the impulsive fractional integral equation (2.4) rather than the impulsive fractional integrodi¤erential problem (2.1)-(2.3). Lemma 2.2. A function x 2 PC ([t0; t0+ T ] ; R) is a solution of the problem
(2.1)-(2.3) if and only if x satis…es the integral equation of the form x (t) = x0+ X t0<tk<t Jk(x (tk)) + 1 ( ) (2.4) Z t t0 (t s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds: Proof. Let x (t) be a solution of the problem (2.1)-(2.3). Using appropriate prop-erties of fractional calculus for 0 < 1, after applying I
t+0
obtain It+c 0 Dt + 0x (t) = x(t) + c0 (2.5) = It+ 0 Z t t0 K (t; s; x (s)) ds G (t; x (t)) : From (2.2)we get c0= x0; then
x(t) = x0+ 1 ( ) Z t t0 (t s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds: Doing the same thing on (t1; t2] we obtain from (2.5)
c0= x t+1 + 1 ( ) Z t1 t0 (t1 s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds; where x t+1 = x t1 + J1 x t1 = x0+ 1 ( ) Z t1 t0 (t1 s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds +J1(x(t1)) ;
with x tk = x (tk) ; k = 1; : : : ; m. So, on (t1; t2] we get
x(t) = x0+ J1(x(t1)) + 1 ( ) Z t t0 (t s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds:
Then, we obtain by induction for t 2 (tm; t0+ T ] the form of integral equation
satis…ed by x (t) x (t) = x0+ m X k=1 Jk(x(tk)) + 1 ( ) Zt t0 (t s) 1 Zs t0 K (s; ; x ( )) d G (t; x (t)) ds:
This shows the …rst implication. For the other implication, we applycD t+
0
to (2.4) to get (2.1). Conditions (2.2) and (2.3) are obtained easily from (2.4) respectively for t = t0and t = tk; k = 1; : : : ; m:
2.3. Local existence. From the …xed-point theory, we recall the following theorem which will be used in the sequel.
Schauder’s …xed-point theorem :
If E is a closed, bounded, convex subset of a Banach space and the mapping A : E ! E is completely continuous, then A has a …xed point in E:
Theorem 2.3. We assume that
(i) jK (t; s; x)j h (t; s) ' (jxj) ; where h (t; s) 2 C ([t0; t0+ T ] [t0; t0+ T ] ; R+)
and ' 2 C (R+; R+) is nondecreasing;
(ii) jG (t; x)j a (t) g (jxj) ; where a 2 C ([t0; t0+ T ] ; R+) and g 2 C (R+; R+)
is nondecreasing.
(iii) jJk(x)j 'k(jxj) ; where 'k2 C (R+; R+) is nondecreasing, k = 1; : : : ; m:
Then there exists at least one solution x (t) of the problem (2.1)-(2.3) in PC([t0; t0
+ ]; R) for some positive number . Proof. We introduce the following notation
= fx 2 PC ([t0; t0+ ] ; R) such that kx x0kP C bg ;
for some such that 0 < T and 0 < X
t0<tk<
ak+
( ) M1 + 1+ M2 b: (2.6)
From the continuity of the functions given in (A4) on their domains we can …nd
positive constants M1; M2 and ak, k = 1; : : : ; m such that for x 2 we have
jK (t; s; x (s))j sup t0<s t t0+ h (t; s) ' (kxkPC) := M1; jG (t; x (t))j sup t2[t0;t0+ ] a (t) g (kxkPC) := M2; j Jk(x(tk))j 'k(kxkPC) := ak; k = 1; : : : ; m:
For applying Schauder’s theorem we need to check that is a non empty closed, bounded and convex subset of the Banach PC ([t0; t0+ ] ; R) which is an easy task.
Let us de…ne the operator A on by Ax (t) = x0+ X t0<tk<t Jk(x(tk)) + 1 ( ) t Z t0 (t s) 1 (2.7) Z s t0 K (s; ; x ( )) d G (s; x (s)) ds: It is clear that for each x 2 we have from (2.7)
jAx (t) x0j X t0<tk<t jJk(x(tk))j + 1 ( ) Z t t0 (t s) 1[M1(s t0) + M2] ds X t0<tk< ak+ 1 ( ) M1 Z t t0 (t s) 1(s t0) ds + (t t0) M2 X t0<tk< ak+ 1 ( ) " M1 (t t0) +1 + 1 + M2 # ;
then,
kAx x0kPC b: (2.8)
Hence A maps into itself. To show that A is completely continuous we will show it is continuous and A is relatively compact in PC ([t0; t0+ ] ; R). Let (yn)n 0
be a sequence such that yn! y in when n ! 1. Then, for each t 2 [t0; t0+ ]
jAyn(t) Ay (t)j X t0<tk<t jJk(yn(tk)) Jk(y (tk))j + 1 ( ) Z t t0 (t s) 1 Z s t0 jK (s; ; yn( )) K (s; ; y ( ))j d ds + 1 ( ) Z t t0 (t s) 1jG (s; yn(s)) d G (s; y (s))j ds:
Since the functions K; G, Jk with k = 1; : : : ; m are continuous and by the
dom-inated convergence theorem we have kAyn AykPC ! 0; when n ! 1. Thus,
A is continuous. In view of Arzela-Ascoli theorem it su¢ ces to show that A is uniformly bounded and equicontinuous in PC ([t0; t0+ ] ; R) indeed to show that
A is relatively compact. From (2.8) we get the following kAxkPC jx0j + b:
Thus, the functions of A are uniformly bounded in PC ([t0; t0+ ] ; R). To prove
that the functions of A are equicontinuous, we consider 1; 22 [t0; t0+ ] such
that 1< 2, it follows that
jAx ( 2) Ax ( 1)j X 1 tk< 2 jJk(x(tk))j + 1 ( ) Z 2 t0 ( 2 s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds Z 1 t0 ( 1 s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds ; so jAx ( 2) Ax ( 1)j X 1 tk< 2 ak+ M2 ( ) Z 1 t0 h ( 2 s) 1 ( 1 s) 1 i ds + Z 2 1 ( 2 s) 1ds + M1 ( ) Z 1 t0 h ( 2 s) 1 ( 1 s) 1 i (s t0) ds + Z 2 1 ( 2 s) 1(s t0) ds :
Therefore kAx ( 2) Ax ( 1)kPC X 1 tk< 2 ak+ M2 ( )[( 2 t0) ( 1 t0) ] + M1 ( + 1) ( ) h ( 2 t0) +1 ( 1 t0) +1 i + M1 ( )( 1 t0) [( 2 t0) ( 1 t0) ] ; from which we get kAx ( 2) Ax ( 1)kPC! 0, when 2! 1, that is, fAx (t)g is
an equicontinuous family on [t0; t0+ ] : Hence A is compact and so A is
com-pletely continuous. Finally, we conclude by virtue of Schauder’s theorem that A has at least one …xed-point in which is a solution of the problem ( 2.1)-(2.3). The proof is complete.
2.4. Extremal solutions. We shall prove the existence of extremal solution of the problem (2.1)-(2.3) through the following steps by using comparison principles and the notion of convergence.
First we give results regarding the impulsive fractional inequalities in the follow-ing lemmas.
Lemma 2.4. Further (A1)-(A4), assume that for t0< s t t0+ T ; x 2 R;
(A5) K (t; s; x) is nondecreasing with respect to x and G (t; x) is nonincreasing
with respect to x:
Let x; y 2 PC ([t0; t0+ T ] ; R) satisfying respectively the following inequalities
x (t) < x (t0) + X t0<tk<t Jk(x (tk)) (2.9) + 1 ( ) Z t t0 (t s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds and y (t) y (t0) + X t0<tk<t Jk(y (tk)) (2.10) + 1 ( ) Z t t0 (t s) 1 Z s t0 K (s; ; y ( )) d G (s; y (s)) ds: If x (t0) < y (t0) ; (x (tk)) < (y (tk)) ; k = 1; : : : ; m; then x (t) < y (t) for every t 2 [t0; t0+ T ] : (2.11)
Proof. Suppose that the inequality (2.11) does not hold. Then, there exists some
12 (t0; t0+ T ] such that
Using the fact that x (t0) < y (t0) ; Jk(x (tk)) < Jk(y (tk)) ; k = 1; : : : ; m; and
the inequalities (2.9), ( 2.10) we get y ( 1) y (t0) + X t0<tk< 1 Jk(y (tk)) + 1 ( ) Z 1 t0 ( 1 s) 1 Z s t0 K (s; ; y ( )) d G (s; y (s)) ds > x (t0) + X t0<tk< 1 Jk(x (tk)) + 1 ( ) Z 1 t0 ( 1 s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds > x ( 1) :
This contradicts (2.14), which completes the proof.
Remark 2.5. One of the two inequalities (2.9) or (2.10) strictly holds.
Remark 2.6. The condition on the jumps of the state functions (x (tk)) < (y (tk)),
k = 1; : : : ; m; can be replaced by the nondecrease of Jk(x) ; k = 1; : : : ; m for x 2 R:
Lemma 2.7. Let (A1)-(A4) hold and x; y 2 PC ([t0; t0+ T ] ; R) be solutions of
(2.1)-(2.3) satisfying the following
cD t+0x (t) < cD t+0y (t) ; t 6= tk; k = 1; : : : ; m; t 2 (t0; t0+ T ]: (2.13) If x (t0) < y (t0) ; (x (tk)) < (y (tk)) ; k = 1; : : : ; m; then inequality (2.11) holds.
Proof. Suppose that the inequality (2.11) is not true. Then, there exists some
12 (t0; t0+ T ] such that
x ( 1) = y ( 1) : (2.14)
Using the fact that x (t0) < y (t0) ; Jk(x (tk)) < Jk(y (tk)) ; k = 1; : : : ; m; and
the inequalities (2.9), (2.10) we get y ( 1) y (t0) + X t0<tk< 1 Jk(y (tk)) + 1 ( ) Z 1 t0 ( 1 s) 1 Z s t0 K (s; ; y ( )) d G (s; y (s)) ds > x (t0) + X t0<tk< 1 Jk(x (tk)) + 1 ( ) Z 1 t0 ( 1 s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds > x ( 1) :
This contradicts (2.14) which completes the proof.
Now, we can give the main result of this subsection dealing with existence of minimal and maximal solutions called extremal solutions.
Theorem 2.8. Assume that (A1)-(A5) are satis…ed.
Then the problem (2.1)-(2.3) has extremal solutions on [t0; t0+ ] for 0 < T:
Proof. First, prove the existence of a maximal solution. Consider the impulsive fractional initial value problem
8 > > > > < > > > > : cD t+0 x (t) + G (t; x (t)) =Rtt 0K (t; s; x (s)) ds + ; t 6= tk; k = 1; : : : ; m; 0 < " b 2 (m + 1); x (t0) = x0+ "; t0 0 ; x (tk) = Jk x tk + ; tk2 [t0; t0+ T ] ; k = 1; : : : ; m: (2.15)
De…ne the closed bounded set in
= fx 2 PC ([t0; t0+ ] ; R) such that kx (x0+ )kP C b=2g ;
for some 0 < T chosen such that 0 < X t0<tk< ak+ mb 2 (m + 1)+ ( ) M1 + 1+ M2+ b 2 (m + 1) b 2: It is clear that all assumptions of Theorem 2.3 are satis…ed, then problem (2.15) has a solution x (t; ") 2 PC ([t0; t0+ ] ; R) : From Lemma 2.2 the solution of (2.15)
satis…es the integral equation of the form x (t; ) = x (t0; ") + X t0<tk<t Jk(x (tk; )) + 1 ( ) Z t t0 (t s) 1 (2.16) Z s t0 K (s; ; x ( ; )) d + G (s; x (s; )) ds; with x (t0; ") = x0+ and Jk(x (tk; )) = Jk(x (tk)) + ; k = 1; : : : ; m.
Let fx (t; ")g be a family of functions in PC ([t0; t0+ ] ; R) for > 0 satisfying
(2.16) . So, under the assumptions (A1) (A3) we have
jx (t; )j jx0j + + 1 ( ) Zt t0 (t s) 1f[M1(s t0) + + M2]g ds + X t0<tk< (ak+ ) jx0j + (m + 1) + X t0<tk< ak+ 1 ( ) M1 +1 + 1+ M2+ jx0j + X t0<tk< ak+ b 2+ ( + 1) M1 + 1+ M2+ b 2 (m + 1) jx0j + b:
Then, fx (t; ")g is a uniformly bounded family in PC ([t0; t0+ ] ; R). It is also
equicontinuous on [t0; t0+ ]. Indeed, for 1; 22 [t0; t0+ ] such that 1< 2we
have jx ( 2; ) x ( 1; )j X 1 tk< 2 Jk x(tk) + + 1 ( ) Z 2 t0 ( 2 s) 1 Z s t0 K (s; ; x ( ; )) d + G (s; x (s; )) ds Z 1 t0 ( 1 s) 1 Z s t0 K (s; ; x ( ; )) d + G (s; x (s; )) ds : Then, by the same arguments used in the proof of Theorem 2.3, we obtain
kx ( 2; ) x ( 1; )kPC ! 0 when 1! 2:
On the other hand, we point out that for each "1; "2 such that 0 < "1< "2 " we
have x (t0; "1) = x0+ "1< x0+ "2= x (t0; "2) ; and (x (tk; "1)) = x (tk) + 1< x (tk) + 2= x (tk; 2) ; k = 1; : : : ; m: Let F (t; x (t; )) = t R t0 K (t; s; x (s; )) ds G (t; x (t; )) ; F (t; x) = F (t; x) + and Jk x tk + "1= Jk x tk; "1 = Jk(x (tk; "1)) ; we get x (t; "1) < x (t0; "1) + X t0<tk<t Jk(x (tk; "1)) + 1 ( ) Z t t0 (t s) 1 Z s t0 K (s; ; x ( ; 1)) d + "2 G (s; x (s; 1)) ds < x (t0; "1) + X t0<tk<t Jk(x (tk; "1)) + 1 ( ) Z t t0 (t s) 1F"2(s; x (s; "1)) ds and x (t; "2) x (t0; "2) + 1 ( ) Z t t0 (t s) 1F"2(s; x (s; "2)) ds + X t0<tk<t Jk(x (tk; "2)) :
We conclude by Arzela-Ascoli lemma there is a decreasing sequence f"ngn 1
such that lim
n!1"n = 0 and x (t; "n) satis…es the form
x (t; "n) = x0+ "n+ X t0<tk<t (Jk(x (tk)) + "n) + 1 ( ) Z t t0 (t s) 1 Z s t0 K (s; ; x ( ; "n)) d + "n G (s; x (s; "n)) ds:
Since K and G are uniformly continuous, we get the following integral equation by letting n ! 1 x (t) = x0+ X t0<tk<t Jk(x (tk)) + 1 ( ) Z t t0 (t s) 1 Z s t0 K (s; ; x ( )) d G (s; x (s)) ds: Then lim
n!1x (t; "n) = x (t) uniformly on [t0; t0+ ] with x (t0) = x0. Therefore,
x (t) is a solution of (2.1)-(2.3) on [t0; t0+ ]. Now, we have to prove that x (t)
is the maximal solution of (2.1)-(2.3)on [t0; t0+ ]. Let y (t) be any solution of
(2.1)-(2.3) on [t0; t0+ ] : It is clear that
y (t0) = x0< x0+ " = x (t0; ") ;
and y (tk) = x (tk) < x (tk) + = x (tk; ") ; k = 1; : : : ; m:
The fact that y (t) satis…es (2.1) implies that
cD t0y (t) < cD t0x (t; ") ; t 6= tk; k = 1; : : : ; m; t 2 [t0; t0+ ] ; 0 < " b 2 (m + 1): Then we have by Lemma 2.7 , y (t) < x (t; ") ; t 2 [t0; t0+ ] : Since the maximal
solution is unique, then lim
"!0x (t; ") = x (t) uniformly on [t0; t0+ ] :
Likewise, we can prove by the same arguments the existence of a unique minimal solution; this completes the proof.
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Current address : LMA, Department of Mathematics, University of Badji-Mokhtar, Annaba, P.O. box, Annaba 23000, ALGERIA
E-mail address : atmanira@yahoo.fr
