Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 273, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
IMPULSIVE NEUTRAL FRACTIONAL
INTEGRO-DIFFERENTIAL EQUATIONS WITH STATE
DEPENDENT DELAYS AND INTEGRAL CONDITION
JAYDEV DABAS, GANGA RAM GAUTAM
Abstract. In this article, we establish the existence of a solution for an impulsive neutral fractional integro-differential state dependent delay equation
subject to an integral boundary condition. The existence results are proved
by applying the classical fixed point theorems. An example is presented to
demonstrate the application of the results established.
1. Introduction
Let X be a Banach space and P Ct := P C([−d, t]; X), d > 0, 0 ≤ t ≤ T < ∞,
be a Banach space of all such functions φ : [−d, t] → X, which are continuous
everywhere except for a finite number of points ti , i = 1, 2, . . . , m, at which φ(t+
i )
−
and φ(t−
)
exists
and
φ(t
)
=
φ(t
),
endowed
with
the
norm
i
i
i
kφkt =
sup kφ(s)kX , φ ∈ P Ct ,
−d≤s≤t
where k · kX is the norm in X.
In this article we study an impulsive neutral fractional integro-differential equation of the form
Z t
h
i
(t − s)α−1
Dtα x(t) +
g(s, xρ(s,xs ) )ds
Γ(α)
(1.1)
0
= f (t, xρ(t,xt ) , B(x)(t)), t ∈ J = [0, T ], T < ∞, t 6= tk ,
∆x(tk ) = Ik (x(t−
k )),
∆x0 (tk ) = Qk (x(t−
k )),
k = 1, 2, . . . , m,
x(t) = φ(t), t ∈ [−d, 0],
Z T
ax0 (0) + bx0 (T ) =
q(x(s))ds, a + b 6= 0, b 6= 0,
(1.2)
(1.3)
(1.4)
0
where x0 denotes the derivative of x with respect to t and Dtα , α ∈ (1, 2) is Caputo’s
derivative. The functions f : J × P C0 × X → X, g : J × P C0 → X, and q :
X → X are given continuous functions where P C0 = P C([−d, 0], X) and for any
x ∈ P CT = P C([−d, T ], X), t ∈ J, we denote by xt the element of P C0 defined by
2000 Mathematics Subject Classification. 26A33, 34K05, 34A12, 34A37, 26A33.
Key words and phrases. Fractional order differential equation; nonlocal condition;
contraction; impulsive condition.
c
2013
Texas State University - San Marcos.
Submitted May 3, 2013. Published December 17, 2013.
1
2
J. DABAS, G. R. GAUTAM
EJDE-2013/273
xt (θ) = x(t + θ), θ ∈ [−d, 0]. In the impulsive conditions for 0 = t0 < t1 < · · · <
tm < tm+1 = T , Qk , Ik ∈ C(X, X), (k = 1, 2, . . . , m), are continuous and bounded
−
0
0 +
0 −
functions. We have ∆x(tk ) = x(t+
k ) − x(tk ) and ∆x (tk ) = x (tk ) − x (tk ). The
term Bx(t) is given by
Z t
Bx(t) =
K(t, s)x(s)ds,
(1.5)
0
where K ∈ C(D, R+ ), is the set of all positive functions which are continuous on
Rt
D = {(t, s) ∈ R2 : 0 ≤ s ≤ t < T } and B ∗ = supt∈[0,t] 0 K(t, s)ds < ∞.
The study of fractional differential equations has been gaining importance in
recent years due to the fact that fractional order derivatives provide a tool for
the description of memory and hereditary properties of various phenomena. Due to
this fact, the fractional order models are capable to describe more realistic situation
than the integer order models. Fractional differential equations have been used in
many field like fractals, chaos, electrical engineering, medical science, etc. In recent
years, we have seen considerable development on the topics of fractional differential
equations, for instance, we refer to the articles [8, 10, 26, 27].
The theory of impulsive differential equations of integer order is well developed
and has applications in mathematical modelling, especially in dynamics of populations subject to abrupt changes as well as other phenomena such as harvesting,
disease, and so forth. For general theory and applications of fractional order differential equations with impulsive conditions, we refer the reader to the references
[1, 7, 11, 17, 21, 22, 28, 29, 30].
Integral boundary conditions have various applications in applied fields such as
blood flow problems, chemical engineering, thermoelasticity, underground water
flow, population dynamics, etc. For a detailed description of the integral boundary
conditions, we refer the reader to some recent papers [4, 5, 6, 13, 16, 17] and the
references therein. On the other hand, we know that the delay arises naturally in
systems due to the transmission of signal or the mechanical transmission. Moreover, the study of fractional order problems involving various types of delay (finite,
infinite and state dependant) considered in Banach spaces has been receiving attention, see [2, 3, 8, 12, 14, 15, 18, 19, 20, 23, 24, 25] and references cited in these
articles.
In [17] authors have established the existence and uniqueness of a solution for
the following system
Dtα x(t) = f (t, xt , Bx(t)),
∆x(tk ) = Qk (x(t−
k )),
∆x0 (tk ) = Ik (x(t−
k )),
t ∈ J = [0, T ], t 6= tk ,
k = 1, 2, . . . , m,
k = 1, 2, . . . , m,
t ∈ (−∞, 0],
Z T
ax0 (0) + bx0 (T ) =
q(x(s))ds,
(1.6)
x(t) = φ(t),
0
the results are proved by using the contraction and Krasnoselkii’s fixed point theorems. This paper is motivated from some recent papers treating the boundary
value problems for impulsive fractional differential equations [4, 5, 13, 17, 30].
To the best of our knowledge, there is no work available in literature on impulsive neutral fractional integro-differential equation with state dependent delay and
with an integral boundary condition. In this article, we first establish a general
EJDE-2013/273
FRACTIONAL INTEGRO-DIFFERENTIAL EQUATIONS
3
framework to find a solution to system (1.1)–(1.4) and then by using classical fixed
point theorems we proved the existence and uniqueness results.
2. Preliminaries
In this section, we shall introduce notations, definitions, preliminary results
which are required to establish our main results. We continue to use the function spaces introduced in the earlier section. For the following definitions we refer
to the reader to see the monograph of Podlunby [27].
Definition 2.1. Caputo’s derivative of order α for a function f : [0, ∞) → R is
defined as
Z t
1
(t − s)n−α−1 f (n) (s)ds = J n−α f (n) (t),
(2.1)
Dtα f (t) =
Γ(n − α) 0
for n − 1 ≤ α < n, n ∈ N . If 0 ≤ α < 1, then
Dtα f (t) =
1
Γ(1 − α)
Z
t
(t − s)−α f (1) (s)ds.
(2.2)
0
Definition 2.2. The Riemann-Liouville fractional integral operator for order α >
0, of a function f : R+ → R and f ∈ L1 (R+ , X) is defined by
Jt0 f (t)
= f (t),
Jtα f (t)
t
Z
1
=
Γ(α)
(t − s)α−1 f (s)ds,
α > 0, t > 0,
(2.3)
0
where Γ(·) is the Euler gamma function.
Lemma 2.3 ( [30]). For α > 0, the general solution of fractional differential equations Dtα x(t) = 0 is given by x(t) = c0 + c1 t + c2 t2 + c3 t3 + · · · + cn−1 tn−1 where
ci ∈ R, i = 0, 1, . . . , n − 1, n = [α] + 1 and [α] represent the integral part of the real
number α.
Lemma 2.4 ([21, Lemma 2.6]). Let α ∈ (1, 2), c ∈ R and h : J → R be continuous
function. A function x ∈ C(J, R) is a solution of the following fractional integral
equation
Z
x(t) =
0
t
(t − s)α−1
h(s)ds −
Γ(α)
Z
0
w
(w − s)α−1
h(s)ds + x0 − c(t − w),
Γ(α)
(2.4)
if and only if x is a solution of the following fractional Cauchy problem
Dtα x(t) = h(t),
t ∈ J, x(w) = x0 , w ≥ 0.
(2.5)
As a consequence of Lemma 2.3 and Lemma 2.4 we have the following result.
Lemma 2.5. Let α ∈ (1, 2) and f : J ×P C0 ×X → X be continuously differentiable
function. A piecewise continuous differential function x(t) : (−d, T ] → X is a
4
J. DABAS, G. R. GAUTAM
EJDE-2013/273
solution of system (1.1)–(1.4) if and only if x satisfied the integral equation


φ(t), t ∈ [−d, 0],

n R

R t (t−s)α−1


bt
1 T

g(s,
x
)ds
+
φ(0)
−

ρ(s,x
)
s
Γ(α)
a+b b 0 q(x(s))ds
0


R T (T −s)α−2
Pk



− i=1 Qi (x(t−

ρ(s,xs ) )ds
i )) + 0
Γ(α−1) g(s, xo


R T (T −s)α−2


− 0 Γ(α−1) f (s, xρ(s,xs ) , B(x)(s))ds




R t (t−s)α−1



+ 0 Γ(α) f (s, xρ(s,xs ) , B(x)(s))ds, t ∈ [0, t1 ],
x(t) = . . .

Pk
Pk

−

φ(0) + i=1 Ii (x(t−

i )) +
i=1 (t −nti )Qi (x(ti ))


R
R
α−1

t
bt
1 T


− 0 (t−s)

Γ(α) g(s, xρ(s,xs ) )ds + a+b b 0 q(x(s))ds


R
α−2
P

T (T −s)
k

− i=1 Qi (x(t−

ρ(s,xs ) )ds
i )) + 0

Γ(α−1) g(s, xo

 R T (T −s)α−2


− 0 Γ(α−1) f (s, xρ(s,xs ) , B(x)(s))ds




+ R t (t−s)α−1 f (s, x
, B(x)(s))ds, t ∈ (t , t
].
0
k
ρ(s,xs )
Γ(α)
(2.6)
k+1
Proof. If t ∈ [0, t1 ], then
Z t
(t − s)α−1
Dtα [x(t) +
g(s, xρ(s,xs ) )ds] = f (t, xρ(t,xt ) , B(x)(t)),
Γ(α)
0
x(t) = φ(t), t ∈ [−d, 0].
(2.7)
Taking the Riemann-Liouville fractional integral of (2.7) and using the Lemma 2.4,
we have
Z t
(t − s)α−1
x(t) +
g(s, xρ(s,xs ) )ds
Γ(α)
0
(2.8)
Z t
(t − s)α−1
= a0 + b0 t +
f (s, xρ(s,xs ) , B(x)(s))ds,
Γ(α)
0
using the initial condition, we get a0 = φ(0), then (2.8) becomes
Z t
(t − s)α−1
g(s, xρ(s,xs ) )ds
x(t) +
Γ(α)
0
(2.9)
Z t
(t − s)α−1
= φ(0) + b0 t +
f (s, xρ(s,xs ) , B(x)(s))ds.
Γ(α)
0
Similarly, if t ∈ (t1 , t2 ], then
Z t
(t − s)α−1
Dtα [x(t) +
g(s, xρ(s,xs ) )ds] = f (t, xρ(t,xt ) , B(x)(t))
Γ(α)
0
−
−
x(t+
1 ) = x(t1 ) + I1 (x(t1 )),
x
0
(t+
1)
=x
0
(t−
1)
+
Q1 (x(t−
1 )).
(2.10)
(2.11)
(2.12)
Again apply the Riemann-Liouville fractional integral operator on (2.10) and using
the lemma 2.4, we obtain
Z t
(t − s)α−1
x(t) +
g(s, xρ(s,xs ) )ds
Γ(α)
0
(2.13)
Z t
(t − s)α−1
f (s, xρ(s,xs ) , B(x)(s))ds,
= a 1 + b1 t +
Γ(α)
0
EJDE-2013/273
FRACTIONAL INTEGRO-DIFFERENTIAL EQUATIONS
5
rewrite (2.13) as
t1
(t1 − s)α−1
g(s, xρ(s,xs ) )ds
Γ(α)
0
Z t1
(t1 − s)α−1
= a1 + b1 t1 +
f (s, xρ(s,xs ) , B(x)(s))ds,
Γ(α)
0
x(t+
1)+
Z
(2.14)
due to impulsive condition (2.11) and the fact that x(t1 ) = x(t−
1 ), we may write
(2.14) as
Z t1
(t1 − s)α−1
−
g(s, xρ(s,xs ) )ds
x(t1 ) + I1 (x(t1 )) +
Γ(α)
0
(2.15)
Z t1
(t1 − s)α−1
= a1 + b1 t1 +
f (s, xρ(s,xs ) , B(x)(s))ds.
Γ(α)
0
Now from (2.9), we have
Z t1
(t1 − s)α−1
g(s, xρ(s,xs ) )ds
x(t1 ) +
Γ(α)
0
(2.16)
Z t1
(t1 − s)α−1
= φ(0) + b0 t1 +
f (s, xρ(s,xs ) , B(x)(s))ds.
Γ(α)
0
From (2.15) and (2.16), we get a1 = φ(0) + b0 t1 − b1 t1 + I1 (x(t−
1 )), hence (2.14) can
be written as
Z t
(t − s)α−1
x(t) +
g(s, xρ(s,xs ) )ds
Γ(α)
0
Z t
(t − s)α−1
= φ(0) + b0 t1 + b1 (t − t1 ) + I1 (x(t−
))
+
f (s, xρ(s,xs ) , B(x)(s))ds.
1
Γ(α)
0
(2.17)
On differentiating (2.13) with respect to t at t = t1 , and incorporate second impulsive condition (2.12), we obtain
Z t1
(t − s)α−2
−
0 −
x (t1 ) + Q1 (x(t1 )) +
g(s, xρ(s,xs ) )ds
Γ(α − 1)
0
(2.18)
Z t1
(t1 − s)α−2
= b1 +
f (s, xρ(s,xs ) , B(x)(s))ds,
Γ(α − 1)
0
Now differentiating (2.9), with respect to t at t = t1 , we get
Z t1
(t1 − s)α−2
0
x (t1 ) +
g(s, xρ(s,xs ) )ds
Γ(α − 1)
0
Z t1
(t1 − s)α−2
= b0 +
f (s, xρ(s,xs ) , B(x)(s))ds.
Γ(α − 1)
0
(2.19)
From (2.18) and (2.19), we obtain b1 = b0 + Q1 (x(t−
1 )). Thus, (2.17) become
Z t
(t − s)α−1
x(t) +
g(s, xρ(s,xs ) )ds
Γ(α)
0
−
= φ(0) + b0 t + I1 (x(t−
1 )) + (t − t1 )Q1 (x(t1 ))
Z t
(t − s)α−1
+
f (s, xρ(s,xs ) , B(x)(s))ds.
Γ(α)
0
(2.20)
6
J. DABAS, G. R. GAUTAM
EJDE-2013/273
Similarly, for t ∈ (t2 , t3 ], we can write the solution of the problem as
Z t
(t − s)α−1
x(t) +
g(s, xρ(s,xs ) )ds
Γ(α)
0
−
−
−
= φ(0) + b0 t + I1 (x(t−
1 )) + I2 (x(t2 )) + (t − t1 )Q1 (x(t1 )) + (t − t2 )Q2 (x(t2 ))
Z t
(t − s)α−1
+
f (s, xρ(s,xs ) , B(x)(s))ds.
Γ(α)
0
In general, if t ∈ (tk , tk+1 ], then we have the result
Z t
(t − s)α−1
g(s, xρ(s,xs ) )ds
x(t) +
Γ(α)
0
= φ(0) + b0 t +
Z
+
0
t
k
X
Ii (x(t−
i )) +
i=1
α−1
(t − s)
Γ(α)
k
X
(t − ti )Qi (x(t−
i ))
(2.21)
i=1
f (s, xρ(s,xs ) , B(x)(s))ds.
RT
Finally, we use the integral boundary condition ax0 (0) + bx0 (T ) = 0 q(x(s))ds,
where x0 (0) calculated from (2.9) and x0 (T ) from (2.20). On simplifying, we get
the following value of the constant b0 ,
Z
Z T
m
X
(T − s)α−2
b n1 T
−
Qi (x(ti )) +
g(s, xρ(s,xs ) )ds
q(x(s))ds −
b0 =
a+b b 0
Γ(α − 1)
0
i=1
Z T
o
(T − s)α−2
f (s, xρ(s,xs ) , B(x)(s))ds .
−
Γ(α − 1)
0
On summarizing, we obtain the desired integral equation (2.6). Conversely, assuming that x satisfies (2.6), by a direct computation, it follows that the solution given
in (2.6) satisfies system (1.1)–(1.4). This completes the proof of the lemma.
3. Existence result
The function ρ : J × P C0 → [−d, T ] is continuous and φ(0) ∈ P C0 . Let the
function t→ ϕt be well defined and continuous from the set <(ρ− ) = {ρ(s, ψ) :
(s, ψ) ∈ [0, T ] × P C0 } into P C0 . Further, we introduce the following assumptions
to establish our results.
(H1) There exist positive constants Lf 1 , Lf 2 , Lq and Lg , such that
kf (t, ψ, x) − f (t, χ, y)kX ≤ Lf 1 kψ − χkP C0 + Lf 2 kx − ykX ,
kg(t, ψ) − g(t, χ)kX ≤ Lg kψ − χkP C0 , t ∈ J, ∀ ψ, χ ∈ P C0 , ∀ x, y ∈ X,
kq(x) − q(y)kX ≤ Lq kx − ykX , ∀x, y ∈ X.
(H2) There exist positive constants LQ , LI , Lq , such that
kQk (x) − Qk (y)kX ≤ LQ kx − ykX ,
kIk (x) − Ik (y)kX ≤ LI kx − ykX .
(H3) The functions Qk , Ik , q are bounded continuous and there exist positive
constants C1 , C2 , C3 , such that
kQk (x)kX ≤ C1 ,
kIk (x)kX ≤ C2 ,
kq(x)kX ≤ C3 ,
Our first result is based on the Banach contraction theorem.
∀x ∈ X.
EJDE-2013/273
FRACTIONAL INTEGRO-DIFFERENTIAL EQUATIONS
7
Theorem 3.1. Let the assumptions (H1)–(H2) are satisfied with
n
T α Lg
bT T Lq
4 = m(LI + T LQ ) +
+
+ mLQ
Γ(α + 1) a + b
b
T α−1 Lg
T α−1 (Lf 1 + Lf 2 B ∗ ) T α (Lf 1 + Lf 2 B ∗ ) o
+
+
< 1.
+
Γ(α)
Γ(α)
Γ(α + 1)
Then (1.1)–(1.4) has a unique solution.
Proof. We transform problem (1.1)–(1.4) into a fixed point problem. Consider the
operator P : P CT → P CT defined by

φ(t), t ∈ [−d, 0],


n R

Rt
α−1

bt
1 T


φ(0) − 0 (t−s)
g(s, xρ(s,xs ) )ds + a+b

Γ(α)
b 0 q(x(s))ds


R
α−2
P

T
(T −s)
−

− m
ρ(s,xs ) )ds

i=1 Qi (x(ti )) + 0
Γ(α−1) g(s, xo


R
α−2
T (T −s)


f (s, xρ(s,xs ) , B(x)(s))ds
−


 R0 Γ(α−1)

+ t (t−s)α−1 f (s, x
t ∈ [0, t1 ]
ρ(s,xs ) , B(x)(s))ds,
Γ(α)
0
P x(t) =

...


Pk
Pk


−

φ(0) + i=1 Ii (x(t−

i )) +
i=1 (t −nti )Qi (x(ti ))

 R t (t−s)α−1
R
Pm
T

−
bt
1

− 0 Γ(α) g(s, xρ(s,xs ) )ds + a+b

i=1 Qi (x(ti ))
b 0 q(x(s))ds −


o

R
R
α−2
α−2
T (T −s)
T (T −s)


g(s,
x
)ds
−
f
(s,
x
,
B(x)(s))ds
+

ρ(s,xs )
ρ(s,xs )
Γ(α−1)
Γ(α−1)
0
0



 R t (t−s)α−1
+ 0 Γ(α) f (s, xρ(s,xs ) , B(x)(s))ds, t ∈ (tk , tk+1 ].
Let x, x∗ ∈ P CT and t ∈ [0, t1 ]. Then
kP (x) − P (x∗ )kX
Z t
(t − s)α−1
≤
kg(s, xρ(s,xs ) ) − g(s, x∗ρ(s,x∗s ) )kX ds
Γ(α)
0
Z
m
X
bt n 1 T
∗
∗ −
+
kq(x(s)) − q(x (s))kX ds +
kQi (x(t−
i )) − Qi (x (ti ))kX
a+b b 0
i=1
Z T
(T − s)α−2
kg(s, xρ(s,xs ) ) − g(s, x∗ρ(s,x∗s ) )kX ds
+
Γ(α − 1)
0
Z T
o
(T − s)α−2
+
kf (s, xρ(s,xs ) , B(x)(s)) − f (s, x∗ρ(s,x∗s ) , B(x∗ )(s))kX ds
Γ(α − 1)
0
Z t
(t − s)α−1
+
kf (s, xρ(s,xs ) , B(x)(s)) − f (s, x∗ρ(s,x∗s ) , B(x∗ )(s))kX ds
Γ(α)
0
n Tα
bT T
T α−1
≤
Lg +
Lq + mLQ +
Lg
Γ(α + 1)
a+b b
Γ(α)
o
T α−1
Tα
+
(Lf 1 + Lf 2 B ∗ ) +
(Lf 1 + Lf 2 B ∗ ) kx − x∗ kP CT .
Γ(α)
Γ(α + 1)
In a similar way for t ∈ (tk , tk+1 ], we have
kP (x) − P (x∗ )kX
8
J. DABAS, G. R. GAUTAM
≤
k
X
∗ −
kIi (x(t−
i )) − Ii (x (ti ))kX +
i=1
EJDE-2013/273
k
X
∗ −
(t − ti )kQi (x(t−
i )) − Qi (x (ti ))kX
i=1
t
(t − s)α−1
kg(s, xρ(s,xs ) ) − g(s, x∗ρ(s,x∗s ) )kX ds
×
Γ(α)
0
Z
m
X
bt n 1 T
∗
∗ −
+
kq(x(s)) − q(x (s))kX ds +
kQi (x(t−
i )) − Qi (x (ti ))kX
a+b b 0
i=1
Z T
α−2
(T − s)
+
kg(s, xρ(s,xs ) ) − g(s, x∗ρ(s,x∗s ) )kX ds
Γ(α − 1)
0
Z T
o
(T − s)α−2
kf (s, xρ(s,xs ) , B(x)(s)) − f (s, x∗ρ(s,x∗s ) , B(x∗ )(s))kX ds
+
Γ(α − 1)
0
Z t
(t − s)α−1
+
kf (s, xρ(s,xs ) , B(x)(s)) − f (s, x∗ρ(s,x∗s ) , B(x∗ )(s))kX ds
Γ(α)
0
n
bT T
T α−1
Tα
Lg +
Lq + mLQ +
Lg
≤ mLI + mT LQ +
Γ(α + 1)
a+b b
Γ(α)
o
Tα
T α−1
(Lf 1 + Lf 2 B ∗ ) +
(Lf 1 + Lf 2 B ∗ ) kx − x∗ kP CT
+
Γ(α)
Γ(α + 1)
≤ ∆kx − x∗ kP CT .
Z
Since ∆ < 1, implies that the map P is a contraction map and therefore has a
unique fixed point x ∈ P CT , hence system (1.1)–(1.4) has a unique solution on the
interval [−d, T ]. This completes the proof of the theorem.
Our second result is based on Krasnoselkii’s fixed point theorem.
Theorem 3.2. Let B be a closed convex and nonempty subset of a Banach space
X. Let P and Q be two operators such that
(i) P x + Qy ∈ B, whenever x, y ∈ B. (ii) P is compact and continuous.
(iii) Q is a contraction mapping. Then there exists z ∈ B such that z = P z+Qz.
Theorem 3.3. Let the function f, g be continuous for every t ∈ [0, T ], and satisfy
the assumptions (H1)–(H3) with
n Tα
bT T
T α−1
T α−1
∆=
Lg +
Lq +
Lg +
(Lf 1 + Lf 2 B ∗ )
Γ(α + 1)
a+b b
Γ(α)
Γ(α)
o
α
T
+
(Lf 1 + Lf 2 B ∗ ) < 1.
Γ(α + 1)
Then system (1.1)–(1.4) has at least one solution on [−d, T ].
Proof. Choose
h
r ≥ kφ(0)k + mLI r + mT LQ r +
Tα
bT T
Lg r +
( Lq r + mLQ r
Γ(α + 1)
a+b b
i
T α−1
T α−1
Tα
+
Lg r +
(Lf 1 r + Lf 2 B ∗ r)) +
(Lf 1 r + Lf 2 B ∗ r) .
Γ(α)
Γ(α)
Γ(α + 1)
Define P CTr = {x ∈ P CT : kxkP CT ≤ r}, then P CTr is a bounded, closed convex
subset in P CT . Consider the operators N : P CTr → P CTr and P : P CTr → P CTr for
EJDE-2013/273
FRACTIONAL INTEGRO-DIFFERENTIAL EQUATIONS
9
t ∈ Jk = (tk , tk+1 ], defined by
N (x) = φ(0) +
k
X
Ii (x(t−
i ))
i=1
+
k
X
m
(t −
ti )Qi (x(t−
i ))
i=1
bt X
Qi (x(t−
−
i )) (3.1)
a + b i=1
Z T
Z
(T − s)α−2
bt n 1 T
q(x(s))ds +
g(s, xρ(s,xs ) )ds
P (x) =
a+b b 0
Γ(α − 1)
0
Z T
o
(T − s)α−2
−
f (s, xρ(s,xs ) , B(x)(s))ds
Γ(α − 1)
0
Z t
Z t
(t − s)α−1
(t − s)α−1
−
g(s, xρ(s,xs ) )ds +
f (s, xρ(s,xs ) , B(x)(s))ds.
Γ(α)
Γ(α)
0
0
(3.2)
We complete the proof in the following steps:
Step 1. Let x, x∗ ∈ P CTr then,
kN (x) + P (x∗ )kX ≤ kφ(0)kX +
k
X
kIi (x(t−
i ))kX +
k
X
(t − ti )kQi (x(t−
i ))kX
i=1
i=1
Z
m
bt n 1 T
bt X
−
kQi (x(ti ))kX +
+
kq(x∗ (s))kX ds
a + b i=1
a+b b 0
Z T
(T − s)α−2
+
kg(s, x∗ρ(s,x∗s ) )kX ds
Γ(α − 1)
0
Z T
o
(T − s)α−2
kf (s, x∗ρ(s,x∗s ) , B(x∗ )(s))kX ds
+
Γ(α − 1)
0
Z t
(t − s)α−1
+
kg(s, x∗ρ(s,x∗s ) )kX ds
Γ(α)
0
Z t
(t − s)α−1
+
kf (s, x∗ρ(s,x∗s ) , B(x∗ )(s))kX ds
Γ(α)
0
h
Tα
bT T
≤ kφ(0)k + mC2 + mT C1 +
Lg r +
( C3
Γ(α + 1)
a+b b
T α−1
T α−1
Lg r +
(Lf 1 r + Lf 2 B ∗ r))
+ mC1 +
Γ(α)
Γ(α)
i
Tα
(Lf 1 r + Lf 2 B ∗ r) ≤ r.
+
Γ(α + 1)
Which shows that P CTr is closed with respect to both the maps.
Step 2. N is continuous. Let xn → x be sequence in P CTr , then for each t ∈ Jk
kN (xn ) − N (x)kX
≤
k
X
i=1
−
kIi (xn (t−
i )) − Ii (x(ti ))kX +
k
X
−
(t − ti )kQi (xn (t−
i )) − Qi (x(ti ))kX
i=1
m
bt X
−
+
kQi (xn (t−
i )) − Qi (x(ti ))kX .
a + b i=1
Since the functions Qk and Ik , k = 1, . . . , m, are continuous, hence kN (xn ) −
N (x)k → 0, as n → ∞. Which implies that the mapping N is continuous on P CTr .
10
J. DABAS, G. R. GAUTAM
EJDE-2013/273
Step 3. The fact that the mapping N is uniformly bounded is a consequence of
the following inequality. For each t ∈ Jk , k = 0, 1, . . . , m and for each x ∈ P CTr , we
have
kN (x)kX ≤ kφ(0)kX +
k
X
kIi (x(t−
i ))kX +
i=1
k
X
(t − ti )kQi (x(t−
i ))kX
i=1
m
bt X
kQi (x(t−
+
i ))kX
a + b i=1
≤ kφ(0)k + mC2 + mT C1 +
bT
mC1 .
a+b
Step 4. Now, to show that N is equi-continuous, let l1 , l2 ∈ Jk , tk ≤ l1 < l2 ≤ tk+1 ,
k = 1, . . . , m, x ∈ P CTr , we have
kN (x)(l2 ) − N (x)(l1 )kX ≤ (l2 − l1 )
k
X
i=1
m
kQi (x(t−
i ))kX +
b(l2 − l1 ) X
kQi (x(t−
i ))kX .
a + b i=1
As l2 → l1 , then kN (x)(l2 ) − N (x)(l1 )k → 0 implies that N is an equi-continuous
map. Combining the Steps 2 to 4, together with the Arzela Ascoli’s theorem, we
conclude that the operator N is compact.
Step 5. Now, we show that P is a contraction mapping. Let x, x∗ ∈ P CTr and
t ∈ Jk , k = 1, . . . , m, we have
kP (x) − P (x∗ )kX
Z t
(t − s)α−1
≤
kg(s, xρ(s,xs ) ) − g(s, x∗ρ(s,x∗s ) )kX ds
Γ(α)
0
Z
bt n 1 T
kq(x(s)) − q(x∗ (s))kX ds
+
a+b b 0
Z T
(T − s)α−2
+
kg(s, xρ(s,xs ) ) − g(s, x∗ρ(s,x∗s ) )kX ds
Γ(α − 1)
0
Z T
o
(T − s)α−2
kf (s, xρ(s,xs ) , B(x)(s)) − f (s, x∗ρ(s,x∗s ) , B(x∗ )(s))kX ds
.+
Γ(α − 1)
0
Z t
(t − s)α−1
kf (s, xρ(s,xs ) , B(x)(s)) − f (s, x∗ρ(s,x∗s ) , B(x∗ )(s))kX ds
+
Γ(α)
0
n Tα
bT T
T α−1
≤
Lg +
Lq +
Lg
Γ(α + 1)
a+b b
Γ(α)
o
T α−1
Tα
+
(Lf 1 + Lf 2 B ∗ ) +
(Lf 1 + Lf 2 B ∗ ) kx − x∗ kP CTr
Γ(α)
Γ(α + 1)
∗
r
≤ ∆kx − x kP CT
As ∆ < 1, it implies that P is a contraction map. Thus all the assumptions of the
Krasnoselkii’s theorem are satisfied. Hence we have that the set P CTr has a fixed
point which is the solution of system (1.1)–(1.4) on (−d, T ]. This completes the
proof of the theorem.
EJDE-2013/273
FRACTIONAL INTEGRO-DIFFERENTIAL EQUATIONS
11
4. Example
Consider the following example to demonstrate the application of the results
established.
Z t
Z t
1
et x(t − σ(x(t)))
xes
Dtα [x(t) +
x(t − σ(x)ds] =
+
cos(t − s)
ds,
2
25 + x (t − σ(x(t)))
4+x
0 47
0
t ∈ [0, T ], t 6= ti ,
Z ti
Z ti
γi (ti − s)x(s)
γi (ti − s)x(s)
∆x(ti ) =
ds, ∆x0 (ti ) =
ds,
25
9
−d
−d
Z T
1
0
0
x(t) = φ(t), t ∈ (−d, 0], x (0) + x (T ) =
sin( x(s))ds,
4
0
where γi ∈ C([0, ∞), X), σ ∈ C(X, [0, ∞)), 0 < t1 < t2 < · · · < tn < T . Set γ > 0,
and choose P C γ as
P C γ = {φ ∈ P C((0, ∞], X) : lim eγt φ(t) exist}
t→−d
γt
with the norm kφkγ = supt∈(0,∞] e |φ(t)|, φ ∈ P C γ . We set
ρ(t, ϕ) = t − σ(ϕ(0)),
(t, ϕ) ∈ J × P C γ ,
et (ϕ)
, (t, ϕ) ∈ J × P C γ ,
25 + (ϕ)2
ϕ
g(t, ϕ) =
ds, ϕ ∈ P C γ ,
47
Z t
xes
ds, (t, x) ∈ I × P C γ ,
B(x)(t) =
cos(t − s)
(4 + x)
0
Z ti
γi (ti − s)x(s)
Qk (x(tk )) =
ds,
25
−d
Z ti
γi (ti − s)x(s)
Ik (x(tk )) =
ds.
9
−d
f (t, ϕ) =
We can see that all the assumptions of Theorem 3.1 are satisfied with
kϕ − χk
∀t ∈ J, ϕ, χ ∈ P C γ ,
25
kx − yk
|B(x) − B(y)| ≤ et
∀t ∈ J, x, y ∈ P C γ ,
4
1
|g(t, ϕ) − g(t, χ)| ≤
kϕ − χk, ∀t ∈ J, ϕ, χ ∈ P C γ ,
47
1
|Qk (x(tk )) − Qk (y(tk ))| ≤ γ ∗ kx − yk, x, y ∈ X,
25
1
|Ik (x(tk )) − Ik (y(tk ))| ≤ γ ∗ kx − yk, x, y ∈ X,
9
1
|q(x) − q(y)| ≤ kx − yk, x, y ∈ X.
4
Further, we observe that
n
Tα
bT T
T α−1
Lg +
Lq + mLQ +
Lg
mLI + mT LQ +
Γ(α + 1)
a+b b
Γ(α)
|f (t, ϕ) − f (t, χ)| ≤ et
12
J. DABAS, G. R. GAUTAM
EJDE-2013/273
o
T α−1
Tα
(Lf 1 + Lf 2 B ∗ ) +
(Lf 1 + Lf 2 B ∗ )
Γ(α)
Γ(α + 1)
≈ 0.513γ ∗ + 0.534 < 1.
Rt
We fix γ ∗ = −d γi (ti − s)ds < 0, 0 < t1 < t2 < t3 < 1, α = 3/2, T = 1. This
implies that there exists a unique solution of the considered problem in this section.
+
Acknowledgements. The research work of J. Dabas has been partially supported
by the Sponsored Research & Industrial Consultancy, Indian Institute of Technology
Roorkee, project No.IITR/SRIC/247/F.I.G(Scheme-A).
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Jaydev Dabas
Department of Applied Science and Engineering, IIT Roorkee, Saharanpur Campus,
Saharanpur-247001, India
E-mail address: jay.dabas@gmail.com
Ganga Ram Gautam
Department of Applied Science and Engineering, IIT Roorkee, Saharanpur Campus,
Saharanpur-247001, India
E-mail address: gangaiitr11@gmail.com