Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 107, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF MILD SOLUTIONS FOR IMPULSIVE
FRACTIONAL-ORDER SEMILINEAR EVOLUTION EQUATIONS
WITH NONLOCAL CONDITIONS
ARCHANA CHAUHAN, JAYDEV DABAS
Abstract. In this work we consider a class of impulsive fractional-order semilinear evolution equations with a nonlocal initial condition. By means of solution operator and application of fixed point theorems we established the
existence and uniqueness of a mild solution.
1. Introduction
Recently fractional differential equations attracted many authors (see for instance [3, 8, 9, 12, 18, 19, 20, 21, 25, 27] and references in these papers). Many
phenomena in engineering, physics,continuum mechanics, signal processing, electromagnetics, economics and science describes efficiently by fractional order differential
equations. Impulsive differential equations have become important in recent years
as mathematical models of phenomena in both physical and social sciences (see
for instance [2, 7, 15, 16, 19, 26] and references in these papers). There has been
a significant development in impulsive theory especially in the area of impulsive
differential equations with fixed moments.
In this article, we are concerned with the existence and uniqueness of the solution
for the fractional order differential equation in a complex Banach space X,
dα
x(t) + Ax(t) = f (t, x(t), x(a1 (t)), . . . , x(am (t))),
dtα
x(0) + g(x) = x0 ,
∆x(ti ) =
t ∈ J = [0, T ], t 6= ti , (1.1)
Ii (x(t−
i )),
α
(1.2)
(1.3)
d
where dt
α is Caputo’s fractional derivative of order 0 < α < 1, i = 1, 2, . . . , p,
0 = t0 < t1 < t2 < · · · < tp < tp+1 = T . Linear operator A, defined from
the domain D(A) ⊂ X into X, is such that −A generates α−resolvent family
{Sα (t) : t ≥ 0} of bounded linear operators in X, the nonlinear map f is defined
from J × X m+1 into X, for each of i the map ai is defined on [0, T ] into [0, T ]
−
+
−
and ∆x(ti ) = x(t+
i ) − x(ti ), x(ti ), x(ti ) denotes the right and the left limit of x
2000 Mathematics Subject Classification. 34K05, 34A12, 34A37, 26A33.
Key words and phrases. Fractional order differential equation; nonlocal conditions;
contraction mapping; mild solution; impulsive conditions.
c
2011
Texas State University - San Marcos.
Submitted April 29, 2011. Published August 24, 2011.
1
2
A. CHAUHAN, J. DABAS
EJDE-2011/107
at ti , respectively. In general the derivatives x0 (ti ) do not exist, we assume that
x0 (ti ) = x0 (ti − 0) at the point of discontinuity ti of the solution
t → x(t).
Pp
The nonlocal condition g : X → X, defined as g(x) = k=1 ck x(tk ), where ck ,
k = 1, . . . , p, are given constants and 0 < t1 < t2 < · · · < tp < T . Let us recall that
such nonlocal conditions were first used by Deng [13]. In this paper, Deng indicated
that using the nonlocal condition x(0) + g(x) = x0 to describe, for instance, the
diffusion phenomenon of a small amount of gas in a transparent tube can give better
result than using the usual local Cauchy Problem x(0) = x0 .
The study of the impulsive fractional order semilinear functional differential
problem of the type (1.1) is motivated by the paper of Byszewski and Akca [11]
and Sui, Lai and Chen [26]. In [11] the authors have considered the nonlocal Cauchy
problem
u0 (t) + Au(t) = f (t, u(t), u(a1 (t)), . . . , u(am (t))),
t ∈ J = [0, T ],
u(0) + g(u) = u0 ,
(1.4)
where −A is the generator of a compact semigroup in X, g : C(J, X) into X, u0 ∈ X
and for each i = 1, 2, . . . , m, ai : J → J. Further, the results obtained in [11] have
been extended by Bahuguna in [4]. For more results on nonlocal conditions we refer
the papers [4, 5, 6, 10, 11, 13, 14] and references therein.
In [26], the authors have investigated the existence of mild solutions of the following system
Dα x(t) = Ax(t) + f (t, x(t)),
t ∈ [0, T ], t 6= tk ,
x(0) = x0 ∈ X,
∆x|t=tk = Ik (x(t−
k )),
k = 1, . . . , m,
and corrected the errors in Mophu paper [19], and generalized some previous results.
The organization of this paper is as follows. In Section 2, we present some
necessary definitions and preliminary results that will be used to prove our main
results. The proofs of our main results are given in Section 3.
2. Preliminaries
Throughout, in this paper X will be a complex Banach space provided with the
norm k · kX and L(X) is the Banach space of bounded linear operators from X
into X. In addition, Br (x, X) represents the closed ball in X with the center at
x and the radius r. −A is the infinitesimal generator of an analytic α−resolvent
family {Sα (t)}t≥0 of operators on X . For the theory of resolvent operator one
can see the monograph by Pazy [22]. The Mittag-Leffler function is an important
function that finds widespread use in the world of fractional calculus. Just as
the exponential naturally arises out of the solution to integer order differential
equations, the Mittag-Leffler function plays an important role in the solution of
non-integer order differential equations. The standard definition of the MittagLeffler function (see[24]) is given as
Eα (z) =
∞
X
k=0
zk
.
Γ(αk + 1)
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EXISTENCE OF MILD SOLUTIONS
3
It is also common to represent the Mittag-Leffler function in two arguments, α and
β, such that
Z
∞
X
zk
1
µα−β
Eα,β (z) =
=
dµ, α, β > 0, z ∈ C,
eµ α
Γ(αk + β)
2πi Ha µ − z
k=0
where Ha is a Hankel path, that is a contour which starts and ends at −∞ and
1
encircles the disc |µ| ≤ |z| α counter clockwise. It is an entire function which provides a generalization of several usual functions, for example: Exponent function:
E1,1 (z) = ez ; cosine functions: E2,1 (z 2 ) = cosh(z) and E2,1 (−z 2 ) = cos(z); Sine
functions: zE2,2 (z 2 ) = sinh(z) and zE2,2 (−z 2 ) = sin(z). The Laplace transform of
the Mittag-Leffler function is given as:
λα−β
, Re λ > ρ1/α , ρ > 0.
λ α + ρα
To begin with the analysis we need some basic definitions and properties from the
fractional calculus theory (see [24]).
L(tβ−1 Eα,β (−ρα tα )) =
Definition 2.1. Caputo’s derivative of order α for a function f : [0, ∞) → R is
defined as
Z t
dα f (t)
1
=
(t − s)n−α−1 f (n) (s)ds,
dtα
Γ(m − α) 0
for n − 1 < α < n, n ∈ N . If 0 < α ≤ 1, then
Z t
1
dα f (t)
=
(t − s)−α f (1) (s)ds.
dtα
Γ(1 − α) 0
The Laplace transform of the Caputo derivative of order α > 0 is given as
L{Dtα f (t); λ} = λα fb(λ) −
n−1
X
λα−k−1 f (k) (0);
n − 1 < α ≤ n.
k=0
Definition 2.2 ([3, Definition 2.3]). Let A be a closed and linear operator with
domain D(A) defined on a Banach space X and α > 0. Let ρ(A) be the resolvent
set of A. We call A the generator of an α−resolvent family if there exists ω ≥ 0 and
a strongly continuous function Sα : R+ → L(X) such that {λα : Reλ > ω} ⊂ ρ(A)
and
Z ∞
(λα I − A)−1 x =
e−λt Sα (t)xdt, Re λ > ω, x ∈ X.
0
In this case, Sα (t) is called the α−resolvent family generated by A.
Definition 2.3 ([1, Definition 2.1]). Let A be a closed and linear operator with
domain D(A) defined on a Banach space X and α > 0. Let ρ(A) be the resolvent
set of A, then we say that A is the generator of a solution operator if there exists
ω ≥ 0 and a strongly continuous function Sα : R+ → L(X) such that {λα : Reλ >
ω} ⊂ ρ(A) and
Z ∞
λα−1 (λα I − A)−1 x =
e−λt Sα (t)xdt, Re λ > ω, x ∈ X.
0
In this case, Sα (t) is called the solution operator generated by A.
The concept of solution operator is closely related to the concept of a resolvent
family (see [23, Chapter 1]). For more details on α-resolvent family and solution
operators, we refer to [17, 23] and the references therein.
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A. CHAUHAN, J. DABAS
EJDE-2011/107
3. Main results
In [26], if α ∈ (0, 1) and A ∈ Aα (θ0 , ω0 ), then for any x ∈ X and t > 0, we have
kTα (t)kL(X) ≤ M eωt ,
kSα (t)kL(X) ≤ Ceωt (1 + tα−1 ),
t > 0, ω > ω0 .
Let
fT = sup kTα (t)kL(X) ,
M
fS = sup Ceωt (1 + t1−α ),
M
0≤t≤T
0≤t≤T
where L(X) is the Banach space of bounded linear operators from X into X
equipped with its natural topology. So we have
fT ,
kTα (t)kL(X) ≤ M
fS .
kSα (t)kL(X) ≤ tα−1 M
Let us consider the set of functions
P C(J, X) = {x : J → X : x ∈ C((tk , tk+1 ], X), k = 0, 1, . . . p and there exist
+
−
x(t−
k ) and x(tk ), k = 1, . . . , p with x(tk ) = x(tk )}.
Endowed with the norm
kxkP C = sup kx(t)kX ,
t∈J
the space (P C(J, X), k · kP C ) is a Banach space.
Lemma 3.1 ([26]). Consider the Cauchy problem
Dtα x(t) + Ax(t) = f (t, x(t), x(a1 (t)), . . . , x(am (t))),
t > t0 , t0 ≥ 0, 0 < α < 1,
x(t0 ) = x0 ∈ X,
if f satisfies the uniform Holder condition with exponent β ∈ (0, 1] and A is a
sectorial operator, then the unique solution of this Cauchy problem is
Z t
+
x(t) = Tα (t − t0 )x(t0 ) +
Sα (t − θ)f (θ, x(θ), x(a1 (θ)), . . . , x(am (θ)))dθ,
t0
where
λα−1
dλ,
λα + A
br
B
Z
1
1
Sα (t) = tα−1 Eα,α (−Atα ) =
dλ,
eλt α
2πi Bbr
λ +A
1
Tα (t) = Eα,1 (−At ) =
2πi
α
Z
eλt
br denotes the Bromwich path. Sα (t) is called the α−resolvent family and
where B
Tα (t) is the solution operator, generated by −A.
Proof. Let t − t0 = s, then
Dsα x(s + t0 ) + Ax(s + t0 ) = f (s + t0 , x(s + t0 ), x(a1 (s + t0 )), . . . , x(am (s + t0 ))),
for s > 0. Now, applying the Laplace transform, we have
λα L{x(s + t0 )} − λα−1 x(t+
0 ) + AL{x(s + t0 )}
= L{f (s + t0 , x(s + t0 ), x(a1 (s + t0 )), . . . , x(am (s + t0 )))}.
(3.1)
Since (λα I + A)−1 exists, that is λα ∈ ρ(A), from (3.1), we obtain
α
−1
L{x(s + t0 )} = λα−1 (λα I + A)−1 x(t+
0 ) + (λ I + A)
× L{f (s + t0 , x(s + t0 ), x(a1 (s + t0 )), . . . , x(am (s + t0 )))}.
EJDE-2011/107
EXISTENCE OF MILD SOLUTIONS
5
Therefore, by the inverse Laplace transform, we have
Z s
x(s + t0 ) = Eα,1 (−Asα )x(t+
)
+
(s − τ )α−1 Eα,α (−A(s − τ )α )
0
0
× f (τ + t0 , x(τ + t0 ), x(a1 (τ + t0 )), . . . , x(am (τ + t0 )))dτ.
Let s + t0 = t, we obtain
x(t) = Eα,1 (−A(t − t0 )α )x(t+
0)+
Z
t−t0
(t − t0 − τ )α−1 Eα,α (−A(t − t0 − τ )α )
0
× f (τ + t0 , x(τ + t0 ), x(a1 (τ + t0 )), . . . , x(am (τ + t0 )))dτ.
This is the same as
α
x(t) = Eα,1 (−A(t − t0 )
)x(t+
0)
Z
t
+
(t − θ)α−1 Eα,α (−A(t − θ)α )
t0
× f (θ, x(θ), x(a1 (θ)), . . . , x(am (θ)))dθ.
Let Tα (t) = Eα,1 (−Atα ) and Sα (t) = tα−1 Eα,α (−Atα ), then we have
Z t
+
x(t) = Tα (t − t0 )x(t0 ) +
Sα (t − θ)f (θ, x(θ), x(a1 (θ)), . . . , x(am (θ)))dθ.
t0
This completes the proof of the Lemma.
Now, we define the definition of mild solution of (1.1).
Definition 3.2. A function x ∈ P C(J, X) solution of the fractional integral equation

Tα (t)(x0 − g(x))


Rt




+ 0 Sα (t − s)f (s, x(s), x(a1 (s)), . . . , x(am (s)))ds, t ∈ [0, t1 ];


−

Tα (t − t1 )[x(t−

1 ) + I1 (x(t1 ))]
Rt
x(t) = + t Sα (t − s)f (s, x(s), x(a1 (s)), . . . , x(am (s)))ds, t ∈ (t1 , t2 ];
1




...



−

Tα (t − tp )[x(t−

p ) + Ip (x(tp ))]


 Rt
+ tp Sα (t − s)f (s, x(s), x(a1 (s)), . . . , x(am (s)))ds, t ∈ (tp , T ].
will be called a mild solution of problem (1.1). From Lemma 3.1 we can verify this
definition.
Now we introduce the following assumptions:
(H1) There exists a constant Lg > 0 such that kg(x) − g(y)kX ≤ Lg kx − ykX .
(H2) The nonlinear map f : [0, T ] × X m+1 → X is continuous and there exist a
constant Lf such that
kf (t, x1 , x2 , . . . , xm+1 ) − f (s, y1 , y2 , . . . , ym+1 )kX
m+1
X
≤ Lf |t − s| +
kxi − yi kX
i=1
for all (x1 , . . . , xm+1 ) and (y1 , . . . , ym+1 ) in X m+1 and t ∈ [0, T ].
(H3) The function Ik : X → X are continuous and there exists Lk > 0 such
that kIk (x) − Ik (y)kX ≤ Lk kx − ykX , x, y ∈ X, k = 1, 2, . . . , p, L =
max {Lk } > Lg .
6
A. CHAUHAN, J. DABAS
EJDE-2011/107
Theorem 3.3. Assume (H1)–(H3) are satisfied and
α
fT (1 + L) + M
fS Lf (m + 1) T
M
< 1.
α
Then impulsive problem (1.1) has a unique mild solution x ∈ P C(J, X).
Proof. Define a mapping N from P C(J, X) into itself by


TαR(t)(x0 − g(x))

t


+ 0 Sα (t − s)f (s, x(s), x(a1 (s)), . . . , x(am (s)))ds, t ∈ [0, t1 ];




−
−


TαR(t − t1 )[x(t1 ) + I1 (x(t1 ))]
t
(N x)(t) = + t Sα (t − s)f (s, x(s), x(a1 (s)), . . . , x(am (s)))ds, t ∈ (t1 , t2 ];
1




.
.
.



−

Tα (t − tp )[x(t−
p ) + Ip (x(tp ))]


 Rt
+ tp Sα (t − s)f (s, x(s), x(a1 (s)), . . . , x(am (s)))ds, t ∈ (tp , T ].
Now we show that N is a contraction on P C(J, X). We have
kN x(t) − N y(t)kX
Rt


kTα (t)kL(X) (kg(x) − g(y)kX ) + 0 kSα (t − s)kL(X)



×kf (s, x(s), x(a1 (s)), . . . , x(am (s)))



−f (s, y(s), y(a1 (s)), . . . , y(am (s)))kX ds,




−
−
−


kTα (t − t1 )kL(X) (kx(t−
1 ) − y(t1 )kX + kI1 (x(t1 )) − I1 (y(t1 ))kX )


 Rt
+ t1 kSα (t − s)kL(X) kf (s, x(s), x(a1 (s)), . . . , x(am (s)))
≤

−f (s, y(s), y(a1 (s)), . . . , y(am (s)))kX ds,





...




−
−
−

kTα (t − tp )kL(X) (kx(t−

p ) − y(tp )kX + kIp (x(tp )) − Ip (y(tp ))kX )

Rt



+ tp kSα (t − s)kL(X) kf (s, x(s), x(a1 (s)), . . . , x(am (s)))



−f (s, y(s), y(a1 (s)), . . . , y(am (s)))kX ds,
t ∈ [0, t1 ];
t ∈ (t1 , t2 ];
t ∈ (tp , T ];
Applying Assumptions (H1)–(H3), we obtain

fT [Lg + M
fS Lf (m + 1) T α ]kx − ykP C ,
[M
t ∈ [0, t1 ];

α


[M
fT (1 + L1 ) + M
fS Lf (m + 1) T α ]kx − ykP C , t ∈ (t1 , t2 ];
α
kN x(t) − N y(t)kX ≤

.
.
.


 f
fS Lf (m + 1) T α ]kx − ykP C , t ∈ (tp , T ].
[MT (1 + Lp ) + M
α
Which implies that for t ∈ [0, T ],
α
fT (1 + L) + M
fS Lf (m + 1) T ]kx − ykP C .
kN x − N ykP C ≤ [M
α
Tα
f
f
Since [MT (1 + L) + MS Lf (m + 1) α ] < 1, N is a contraction. Therefore, N has
a unique fixed point by Banach contraction principle. This completes the proof of
the theorem.
Our second result is based on the following Krasnoselkii’s fixed point theorem.
Theorem 3.4. Let B be a closed convex and nonempty subset of a Banach space
X. Let P and Q be two operators such that:
EJDE-2011/107
EXISTENCE OF MILD SOLUTIONS
7
(1) P x + Qy ∈ B whenever x, y ∈ B;
(2) P is compact and continuous;
(3) Q is a contraction mapping;
Then there exists z ∈ B such that z = P z + Qz.
Now, we make the following assumptions:
(H4) f ∈ C(J × X m+1 , X), g ∈ C(X, X), and bi ∈ C(J, J) (i = 1, . . . , m).
Moreover, there are Ci > 0 (i = 1, 2) such that kf (s, z0 , z1 , . . . , zm )k ≤ C1
for s ∈ J, zi ∈ Br (i = 0, 1, . . . , m) and kg(w)k ≤ C2 for w ∈ X.
(H5) The function Ik : X → X are continuous and there exists ρ > C2 such that
ρ=
max
1≤k≤m,x∈Br
{kIk (x)kX }.
Theorem 3.5. Assume (H2), (H4), (H5) are satisfied and
α
fS Lf (m + 1) T ] < 1.
[M
α
Then the impulsive problem (1.1) has at least one mild solution on J.
fT (r + ρ) + M
fS C1 T α ] and consider Br = {x ∈ P C(J, X) :
Proof. Choose r ≥ [M
α
kxkP C ≤ r, } then Br is a bounded, closed convex subset in P C(J, X). Define on
Br the operators P and Q by:

Tα (t)(x0 − g(x)),
t ∈ [0, t1 ];



T (t − t )[x(t− ) + I (x(t− ))], t ∈ (t , t ];
α
1
1
1 2
1
1
(P x)(t) =

.
.
.



−
Tα (t − tp )[x(t−
p ) + Ip (x(tp ))], t ∈ (tp , T ],
R t

S (t − s)f (s, x(s), x(a1 (s)), . . . , x(am (s)))ds, t ∈ [0, t1 ];

0 α


R t S (t − s)f (s, x(s), x(a (s)), . . . , x(a (s)))ds, t ∈ (t , t ];
1
m
1 2
t1 α
(Qx)(t) =

.
.
.


R

 t Sα (t − s)f (s, x(s), x(a1 (s)), . . . , x(am (s)))ds, t ∈ (tp , T ].
tp
Now we present the proof in five steps:
Step 1. We show that P x + Qy ∈ Br whenever x, y ∈ Br . Let x, y ∈ Br , then
kP x + QykP C

kTα (t)kL(X) (kx0 kX + kg(x)kX )


Rt




+ 0 kSα (t − s)kL(X) kf (s, y(s), y(a1 (s)), . . . , y(am (s)))kX ds, t ∈ [0, t1 ];


−

kTα (t − t1 )kL(X) [kx(t−

1 )kX + kI1 (x(t1 ))kX ]
Rt
≤ + t kSα (t − s)kL(X) kf (s, y(s), y(a1 (s)), . . . , y(am (s)))kX ds, t ∈ (t1 , t2 ];
1




...



−

kTα (t − tp )kL(X) [kx(t−
p )kX + kIp (x(tp ))kX ]


 Rt
+ tp kSα (t − s)kL(X) kf (s, y(s), y(a1 (s)), . . . , y(am (s)))kX ds, t ∈ (tp , T ].

fT (r + C2 ) + M
fS C1 T α , t ∈ [0, t1 ];
M

α


M
fT (r + ρ) + M
fS C1 T α ,
t ∈ (t1 , t2 ];
α
≤

.
.
.


f
fS C1 T α ,
MT (r + ρ) + M
t ∈ (tp , T ].
α
8
A. CHAUHAN, J. DABAS
EJDE-2011/107
Which implies
Tα
] ≤ r.
α
Step 2. Continuity of P . For this purpose, let {xn }∞
n=0 be a sequence in Br
with lim xn → x in Br . Then for every t ∈ J, we have

kTα (t)kL(X) kg(xn ) − g(x)kX ,
t ∈ [0, t1 ];




−
n −

kTα (t − t1 )kL(X) [kx (t1 ) − x(t1 )kX



+kI (xn (t− )) − I x(t− )k ],
t ∈ (t1 , t2 ];
1
1
X
1
1
k(P xn )(t) − (P x)(t)kX ≤

.
.
.



−

kTα (t − tp )kL(X) [kxn (t−
p ) − x(tp )kX



n −
−
+kIp (x (tp )) − Ip x(tp )kX ],
t ∈ (tp , T ].
fT (r + ρ) + M
fS C1
kP x + QykP C ≤ [M
Since the functions g and Ik , k = 1, . . . , p are continuous, limn→∞ kP xn −P xkP C =
0 in Br . This implies that the mapping P is continuous on Br .
Step 3. P maps bounded sets into bounded sets in P C(J, X). So, let us prove
that for any r > 0 there exists a γ > 0 such that for each x ∈ Br = {x ∈ P C(J, X) :
kxkP C ≤ r}, we have kP xkP C ≤ γ. Indeed, we have for any x ∈ Br ,

kTα (t)kL(X) (kx0 kX + kg(x)kX ),
t ∈ [0, t1 ];



kT (t − t )k
−
−
α
1 L(X) [kx(t1 )kX + kI1 (x(t1 ))kX ], t ∈ (t1 , t2 ];
kP x(t)kX ≤

.
.
.



−
kTα (t − tp )kL(X) [kx(t−
p )kX + kIp (x(tp ))kX ], t ∈ (tp , T ].

fT (r + C2 ), t ∈ [0, t1 ];
M



M
fT (r + ρ),
t ∈ (t1 , t2 ];
≤

...


f
MT (r + ρ),
t ∈ (tp , T ].
fT (r + ρ) = γ.
Which implies that kP xkP C ≤ M
Step 4. We prove that P (Br ) is equicontinuous with Br . For 0 ≤ u < v ≤ T ,
we have
k(P x)(v) − (P x)(u)kX


kTα (v) − Tα (u)kL(X) [kx0 kX + kg(x)kX ], 0 ≤ u < v ≤ t1 ;






kTα (v − t1 ) − Tα (u − t1 )kL(X)



−
×[kx(t−
t1 < u < v ≤ t2 ;
1 )kX + kI1 (x(t1 ))kX ],
≤

.
.
.





kTα (v − tp ) − Tα (u − tp )kL(X)



×[kx(t− )k + kI (x(t− ))k ,
tp < u < v ≤ T.
X
p
X
p
p

(r + C2 )kTα (v) − Tα (u)kL(X) ,
0 ≤ u < v ≤ t1 ;



(r + ρ)kT (v − t ) − T (u − t )k
α
1
α
1 L(X) , t1 < u < v ≤ t2
≤

.
.
.



(r + ρ)kTα (v − tp ) − Tα (u − tp )kL(X) , tp < u < v ≤ T.
Therefore, the continuity of the function t 7→ kT (t)k allows us to conclude that
limu→v kTα (v − ti ) − Tα (u − ti )kL(X) = 0, i = 1, . . . , p and limu→v kTα (v) −
EJDE-2011/107
EXISTENCE OF MILD SOLUTIONS
9
Tα (u)kL(X) = 0. Finally, combining Step 2 to Step 4 with the Ascoli’s Theorem,
we deduce that the operator P is a compact.
Step 5. We show that Q is a contraction mapping. Let x, y ∈ Br and we have
k(Qx)(t) − (Qy(t))(t)kX
R t

kS (t − s)kL(X) kf (s, x(s), x(a1 (s)), . . . , x(am (s)))

 0 α



t ∈ [0, t1 ];
−f (s, y(s), y(a1 (s)), . . . , y(am (s)))kX ds,


Rt



 t1 kSα (t − s)kL(X) kf (s, x(s), x(a1 (s)), . . . , x(am (s)))
t ∈ (t1 , t2 ];
≤ −f (s, y(s), y(a1 (s)), . . . , y(am (s)))kX ds,



. . .


Rt



kSα (t − s)kL(X) kf (s, x(s), x(a1 (s)), . . . , x(am (s)))

tp


−f (s, y(s), y(a (s)), . . . , y(a (s)))k ds,
t ∈ (tp , T ].
1
m
X

α
T
fS Lf (m + 1) kx − ykP C , t ∈ [0, t1 ];
M

α


M
fS Lf (m + 1) T α kx − ykP C , t ∈ (t1 , t2 ];
α
≤

.
.
.


α
f
MS Lf (m + 1) Tα kx − ykP C , t ∈ (tp , T ].
α
fS Lf (m + 1) T ) < 1 then Q is a contraction mapping. Hence, by the
Since (M
α
Krasnoselkii theorem, we can conclude that (1.1) has at least one solution on [0, T ].
This completes the proof of the theorem.
Acknowledgements. The authors wish to express their deep gratitude to the
anonymous referees for their valuable suggestions and comments for improving the
original manuscript.
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Archana Chauhan
Department of Mathematics, Motilal Nehru National Institute of Technology, Allahabad - 211 004, India
E-mail address: archanasingh.chauhan@gmail.com
Jaydev Dabas
Department of Paper Technology, IIT Roorkee, Saharanpur Campus, Saharanpur 247001, India
E-mail address: jay.dabas@gmail.com