Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 229, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
INTEGRAL BOUNDARY-VALUE PROBLEM FOR IMPULSIVE
FRACTIONAL FUNCTIONAL DIFFERENTIAL EQUATIONS
WITH INFINITE DELAY
ARCHANA CHAUHAN, JAYDEV DABAS, MUKESH KUMAR
Abstract. In this article, we establish a general framework for finding solutions for impulsive fractional integral boundary-value problems. Then, we
prove the existence and uniqueness of solutions by applying well known fixed
point theorems. The obtained results are illustrated with an example for their
feasibility.
1. Introduction
The purpose of this article is to establish the existence and uniqueness of solution
to an integral boundary-value problem for impulsive fractional functional integrodifferential equation with infinite delay of the form:
c
Dtα x(t) = f (t, xt , Bx(t)),
∆x(tk ) = Qk (x(t−
k )),
−
0
∆x (tk ) = Ik (x(tk )),
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,
(1.1)
x(t) = φ(t),
0
where T > 0, α ∈ (1, 2), a, b ∈ R such that a + b 6= 0. c Dtα is the Caputo
fractional derivative. The functions f : J × Bh × X → X and q : X → X
are given functions that satisfy certain assumptions, where Bh is a phase space
defined in details in Section 2. Here 0 = t0 < t1 < · · · < tm < tm+1 = T ,
−
Qk , Ik ∈ C(X, X), (k = 1, 2, . . . , m), are bounded functions, ∆x(tk ) = x(t+
k )−x(tk )
0
0 +
0 −
and ∆x (tk ) = x (tk ) − x (tk ). We assume that xt : (−∞, 0] → X, xt (s) = x(t + s),
s ≤ 0, belong to an abstract phase space Bh . The term Bx(t) is given by
Z t
Bx(t) =
K(t, s)x(s)ds,
0
2000 Mathematics Subject Classification. 26A33, 34K05, 34A12, 34A37.
Key words and phrases. Fractional differential equation; integral boundary condition;
impulsive conditions; infinite delay.
c
2012
Texas State University - San Marcos.
Submitted May 29, 2012. Published December 18, 2012.
1
2
A. CHAUHAN, J. DABAS, M. KUMAR
EJDE-2012/229
where K ∈ C(D, R+ ), 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 < ∞.
Fractional differential equations have attracted considerable interest because of
their ability to model complex phenomena. Due to the extensive applications of
fractional differential equations in engineering and science, research in this area has
grown significantly all around the world. For more details about fractional calculus
and fractional differential equations we refer the interested readers to the books by
Podlubny [14], Hilfer [9] and the papers [1, 3, 4, 10, 13, 15, 16, 17, 20] and references
there in.
The impulsive differential equations arising from the real world problems to
describe the dynamics of processes in which sudden, discontinuous jumps occurs.
Such processes are naturally seen in biology, physics, engineering, etc. Due to
their significance, many authors have been established the solvability of impulsive
differential equations. For the general theory and applications of such equations we
refer the interested reader to see the papers [4, 16, 20] and references therein.
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 [2, 3, 4, 6] and the references
therein. On the other hand, we know that delay arises naturally in practical systems
due to the transmission of signal or the mechanical transmission. Moreover, the
Cauchy problem for various delay equations in Banach spaces has been receiving
more and more attention during the past decades, see [10, 11, 13, 15, 19] and for
boundary value problem with infinite delay one can see these papers[5, 12] and
references therein.
Recently, Michal Feckan et al [7] gave a counter example to show that the formula
of solutions for impulsive fractional differential equations used in previous papers
are incorrect. Since the authors used that the Caputo derivative c Dtα restricted
α
on (a, b], 0 < a < b, is c Da,t
, but unfortunately it does not hold. In [7], the
authors introduced a correct formula of solutions for a impulsive Cauchy problem
with Caputo fractional derivative. In [18], the author discussed some existence
results for boundary value problems for impulsive fractional differential equations.
However, the theory of boundary value problem for impulsive fractional differential
equations is still in the initial stages. Our work is motivated by these papers
[7, 8, 18].
To the best of our knowledge, this is the first paper dealing with integral boundary value problem involving impulsive nonlinear integro-differential equations of
fractional order α ∈ (1, 2) with infinite delay. We organize the rest of this paper 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. Section 4 contains an illustrative example.
2. Preliminaries and assumptions
In this section, we shall introduce some basic definitions, properties and lemmas
which are required for establishing our results. Let (X, k · kX ) be a real Banach
space.
To describe fractional order functional differential equations with infinite delay,
we need to discuss the abstract phase space Bh in a convenient way (see for instance
EJDE-2012/229
IMPULSIVE FRACTIONAL DIFFERENTIAL EQUATIONS
3
in [20]). Assume that h : (−∞, 0] → (0, ∞) is a continuous functions with l =
R0
h(t)dt < ∞. For any a > 0, we define B = {ψ : [−a, 0] → X such that ψ(t) is
−∞
bounded and measurable} and equip the space B with the norm
kψk[−a,0] =
sup kψ(s)kX ,
∀ ψ ∈ B.
s∈[−a,0]
Let us define Bh = {ψ : (−∞, 0] → X, such that for any c > 0, ψ|[−c,0] ∈ B and
h(s)kψk[s,0] ds < ∞}. If Bh is endowed with the norm
−∞
Z 0
kψkBh =
h(s)kψk[s,0] ds, ∀ ψ ∈ Bh ,
R0
−∞
then it is clear that (Bh , k · kBh ) is a Banach space. Now we consider the space
B0h = {x : (−∞, T ] → X such that x|Jk ∈ C(Jk , X) and there exist
−
−
x(t+
k ) and x(tk ) with x(tk ) = x(tk ), x0 = φ ∈ Bh , k = 1, . . . , m},
where x|Jk is the restriction of x to Jk = (tk , tk+1 ], k = 0, 1, 2, . . . , m. Set k · kBh0 to
be a seminorm in B0h defined by
kxkB0h = sup{kx(s)kX : s ∈ [0, T ]} + kφkBh , x ∈ B0h .
Assume that x ∈ B0h , then for t ∈ J, xt ∈ Bh . Moreover,
lkx(t)kX ≤ kxt kBh ≤ l sup kx(s)kX + kx0 kBh ,
0<s<t
where l =
R0
−∞
h(t)dt.
Definition 2.1. The Riemann-Liouville fractional integral operator for order α >
0, of a function f : R+ → R and f ∈ L1 (R+ , X) is defined by
Z t
1
(t − s)α−1 f (s)ds, α > 0, t > 0,
(2.1)
Jt0 f (t) = f (t), Jtα f (t) =
Γ(α) 0
where Γ(·) is the Euler gamma function.
Definition 2.2. 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),
(2.2)
Dtα f (t) =
Γ(n − α) 0
for n − 1 ≤ α < n, n ∈ N . If 0 < α ≤ 1, then
Z t
1
Dtα f (t) =
(t − s)−α f (1) (s)ds.
(2.3)
Γ(1 − α) 0
Obviously, Caputo’s derivative of a constant is equal to zero.
Definition 2.3. A function x ∈ B0h is said to be a solution of the problem-(1.1) if
x satisfies the differential equation c Dtα x(t) = f (t, xt , Bx(t)) a.e. on J \{t1 , . . . , tm }
and the following conditions:
∆x(tk ) = Qk (x(t−
k )),
k = 1, 2, . . . , m,
Ik (x(t−
k )),
k = 1, 2, . . . , m,
0
∆x (tk ) =
t ∈ (−∞, 0],
Z T
0
0
ax (0) + bx (T ) =
q(x(s))ds.
x(t) = φ(t),
0
(2.4)
4
A. CHAUHAN, J. DABAS, M. KUMAR
EJDE-2012/229
Lemma 2.4 ([18, lemma 2.5]). For α > 0, the general solution of fractional differential equation c Dtα x(t) = 0 is given by x(t) = c0 + c1 t + c2 t2 + · · · + cn−1 tn−1 ,
where ci ∈ R, i = 0, 1, 2, . . . , n − 1 (n = [α] + 1) and [α] denotes the integer part of
the real number α.
Note that
Z
ω
x(t) = x0 − ct −
0
(ω − s)α−1
h(s)ds +
Γ(α)
Z
0
t
(t − s)α−1
h(s)ds
Γ(α)
(2.5)
is the solution of the Cauchy problem
c
Dtα x(t) = h(t), t ∈ J, α ∈ (1, 2),
Z ω
(ω − s)α−1
x(0) = x0 −
h(s)ds.
Γ(α)
0
(2.6)
Now, we can obtain the following result.
Lemma 2.5 ([7, 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 fractional integral equation
(2.7)
Z ω
Z t
(ω − s)α−1
(t − s)α−1
h(s)ds −
h(s)ds + x0 − c(t − ω)
(2.7)
x(t) =
Γ(α)
Γ(α)
0
0
if and only if x is a solution of the fractional Cauchy problem
c
Dtα x(t) = h(t),
x(ω) = x0 ,
t ∈ J,
ω ≥ 0.
(2.8)
Lemma 2.6. Let α ∈ (1, 2) and f : J × Bh × X → R be continuously differentiable
function. A piecewise continuously differentiable function x ∈ B0h is a solution of
system (1.1) if and only if x ∈ B0h is a solution of the fractional integral equation

φ(t), if t ∈ (−∞, 0],




R t (t−s)α−1
Pm

−
bt



i=1 Ii (x(ti ))
Γ(α) f (s, xs , Bx(s))ds + φ(0) − a+b
0

R
R
α−2

T
− bt T (T −s)
t


a+b 0
Γ(α−1) f (s, xs , Bx(s))ds + a+b 0 q(x(s))ds,




 if t ∈ [0, t1 ],
(2.9)
x(t) = . . . ,

R t (t−s)α−1

Pk

−


i=1 (t − ti )Ii (x(ti ))
Γ(α) f (s, xs , Bx(s))ds + φ(0) +
0


Pm
 Pk
−
−
bt

Qi (x(t )) −
Ii (x(ti ))
+


 bt i=1
R T (T −s)iα−2 a+b i=1
RT

t

q(x(s))ds,
− a+b 0 Γ(α−1) f (s, xs , Bx(s))ds + a+b


0


if t ∈ (tk , tk+1 ],
where k = 1, . . . , m.
Proof. Assume x satisfies (1.1). If t ∈ [0, t1 ], then
c
Dtα x(t) = f (t, xt , Bx(t)), t ∈ (0, t1 ], x(0) = φ(0).
By using Lemma 2.5, we can write the solution of (2.10) as
Z t
(t − s)α−1
f (s, xs , Bx(s))ds + φ(0) − ct.
x(t) =
Γ(α)
0
(2.10)
(2.11)
EJDE-2012/229
IMPULSIVE FRACTIONAL DIFFERENTIAL EQUATIONS
5
If t ∈ (t1 , t2 ], then
c
Dtα x(t) = f (t, xt , Bx(t)), t ∈ (t1 , t2 ],
−
−
−
0 +
0 −
x(t+
1 ) = x(t1 ) + Q1 (x(t1 )), x (t1 ) = x (t1 ) + I1 (x(t1 )).
(2.12)
Again by lemma 2.5, we have the following form of the solution
Z t1
Z t
(t1 − s)α−1
(t − s)α−1
f (s, xs , Bx(s))ds −
f (s, xs , Bx(s))ds
x(t) =
Γ(α)
Γ(α)
0
0
−
+ x(t−
1 ) + Q1 (x(t1 )) − d(t − t1 )
Z t
(t − s)α−1
=
f (s, xs , Bx(s))ds + φ(0) − ct1 + Q1 (x(t−
1 )) − d(t − t1 ).
Γ(α)
0
−
−
0 −
Since x0 (t+
1 ) = x (t1 ) + I1 (x(t1 )), we obtain d = c − I1 (x(t1 )). Thus
Z t
(t − s)α−1
−
f (s, xs , Bx(s))ds + φ(0) + (t − t1 )I1 (x(t−
x(t) =
1 )) + Q1 (x(t1 )) − ct.
Γ(α)
0
If t ∈ (t2 , t3 ], then by similar way using the lemma 2.5, we have
Z t
Z t2
(t − s)α−1
(t2 − s)α−1
x(t) =
f (s, xs , Bx(s))ds −
f (s, xs , Bx(s))ds
Γ(α)
Γ(α)
0
0
−
+ x(t−
2 ) + Q2 (x(t2 )) − e(t − t2 ).
Z t
(t − s)α−1
f (s, xs , Bx(s))ds + φ(0) − ct2 + (t2 − t1 )I1 (x(t−
=
1 ))
Γ(α)
0
−
+ Q1 (x(t−
1 )) + Q2 (x(t2 )) − e(t − t2 ).
−
−
−
0 −
Since x0 (t+
2 ) = x (t2 ) + I2 (x(t2 )), we obtain e = c − I1 (x(t1 )) − I2 (x(t2 )). Thus
Z t
(t − s)α−1
x(t) =
f (s, xs , Bx(s))ds + φ(0) + (t − t1 )I1 (x(t−
1 ))
Γ(α)
0
−
−
+ (t − t2 )I2 (x(t−
2 )) + Q1 (x(t1 )) + Q2 (x(t2 )) − ct.
Similarly, if t ∈ (tk , tk+1 ], then again from lemma 2.5, we have
Z t
k
X
(t − s)α−1
f (s, xs , Bx(s))ds + φ(0) +
(t − ti )Ii (x(t−
x(t) =
i ))
Γ(α)
0
i=1
+
k
X
Qi (x(t−
i )) − ct.
i=1
RT
By using the integral boundary condition ax0 (0)+bx0 (T ) = 0 q(x(s))ds, we obtain
Z T
b
(T − s)α−1
c=
f (s, xs , Bx(s))ds
a + b 0 Γ(α − 1)
Z T
m
b X
1
−
+
Ii (x(ti )) −
q(x(s))ds.
a + b i=1
a+b 0
Thus for t ∈ [0, t1 ],
Z t
m
bt X
(t − s)α−1
f (s, xs , Bx(s))ds + φ(0) −
Ii (x(t−
x(t) =
i ))
Γ(α)
a + b i=1
0
6
A. CHAUHAN, J. DABAS, M. KUMAR
bt
a+b
−
T
Z
0
EJDE-2012/229
(T − s)α−2
t
f (s, xs , Bx(s))ds +
Γ(α − 1)
a+b
Z
T
q(x(s))ds,
0
and for t ∈ (tk , tk+1 ], k = 1, 2, . . . , m, we have
t
Z
x(t) =
0
+
k
X
(t − s)α−1
f (s, xs , Bx(s))ds + φ(0) +
(t − ti )Ii (x(t−
i ))
Γ(α)
i=1
k
X
m
Qi (x(t−
i )) −
i=1
−
bt
a+b
T
Z
0
bt X
Ii (x(t−
i ))
a + b i=1
(T − s)α−2
t
f (s, xs , Bx(s))ds +
Γ(α − 1)
a+b
Z
T
q(x(s))ds.
0
Conversely, assume that x satisfies (2.9). By a direct computation, it follows that
the solution given in (2.9) satisfies the system (1.1). This completes the proof of
the lemma.
Further we introduce the following assumptions to establish our results.
(H1) There exists constants µ1 , µ2 > 0, such that
kf (t, ϕ, x) − f (t, ψ, y)kX ≤ µ1 kϕ − ψkBh + µ2 kx − ykX ,
t ∈ J, ϕ, ψ ∈ Bh , x, y ∈ X.
(H2) The function q : X → X is continuous and there exists constant Lq > 0,
such that
kq(x(s)) − q(y(s))kX ≤ Lq kx − ykX .
(H3) For each k = 1, . . . , m, there exists L, L > 0, such that
kQk (x) − Qk (y)kX ≤ Lkx − ykX ,
kIk (x) − Ik (y)kX ≤ Lkx − ykX ,
∀x, y ∈ X.
∀x, y ∈ X.
(H4) The function f : J × Bh × X → X is continuous and there exist two continuous functions µ1 , µ2 : J → (0, ∞) such that kf (t, ψ, x)kX ≤ µ1 (t)kψkBh +
µ2 (t)kxkX and µ∗1 = supt∈[0,T ] µ1 (t), µ∗2 = supt∈[0,T ] µ2 (t).
(H5) The functions q : X → X, Ik : X → X and Qk : X → X, k = 1, . . . , m are
continuous and there exist constants C, ρ, Ω such that kq(x)kX ≤ C, x ∈ X,
ρ = max1≤k≤m,x∈Br {kIk (x)kX } and Ω = max1≤k≤m,x∈Br {kQk (x)kX }.
3. Existence and uniqueness results
Theorem 3.1. Suppose that the assumptions (H1)–(H3) hold and
Λ=
i
h (a + (1 + α)b)(µ l + µ B ∗ )T α
(a + 2b)LT m + Lq T 2
1
2
+
+ Lm < 1.
(a + b)Γ(α + 1)
a+b
Then (1.1) has an unique solution.
EJDE-2012/229
IMPULSIVE FRACTIONAL DIFFERENTIAL EQUATIONS
7
Proof. Consider the operator N : Bh0 → Bh0 defined by


φ(t), if t ∈ (−∞, 0],


R t (t−s)α−1

Pm

−
bt


i=1 Ii (x(ti ))
Γ(α) f (s, xs , Bx(s))ds + φ(0) − a+b
0


R
R
α−2

T (T −s)
T
bt
t


− a+b

Γ(α−1) f (s, xs , Bx(s))ds + a+b 0 q(x(s))ds,
0




 if t ∈ [0, t1 ],
N x(t) = . . . ,

R t (t−s)α−1
Pk



f (s, xs , Bx(s))ds + φ(0) + i=1 (t − ti )Ii (x(t−

i ))
Γ(α)
0

Pm
Pk

−
−
bt


+ i=1 Qi (x(ti )) − a+b i=1 Ii (x(ti ))


R T (T −s)α−2
RT


bt
t

− a+b

Γ(α−1) f (s, xs , Bx(s))ds + a+b 0 q(x(s))ds,
0


 if t ∈ (t , t
k k+1 ],
where k = 1, 2, . . . , m. Let y(.) : (−∞, T ] → X be the function defined by
(
φ(t),
y(t) =
0,
t ∈ (−∞, 0];
t ∈ J,
then y0 = φ. For each z ∈ C([0, T ], R) with z(0) = 0, we denote
(
0,
z(t) =
z(t),
t ∈ (−∞, 0];
t ∈ J.
If x(.) satisfies (2.9) then we can decompose x(·) as x(t) = y(t) + z(t), which implies
xt = yt + z t for t ∈ J and the function z(·) satisfies
R t (t−s)α−1


Γ(α) f (s, ys + z s , B(y(s) + z(s)))ds + φ(0)
0

RT
Pm

−
bt
t


− a+b i=1 Ii (z(ti )) + a+b 0 q(y(s) + z(s))ds

R T (T −s)α−2


bt
− a+b

Γ(α−1) f (s, ys + z s , B(y(s) + z(s)))ds,
0


. . . ,
z(t) = R t (t−s)α−1

f (s, ys + z s , B(y(s) + z(s)))ds + φ(0)

0


PkΓ(α)
Pk

−

+
(t
−
ti )Ii (z(t−

i )) +
i=1
i=1 Qi (z(ti ))


R
P
T

m
−
bt
t

− a+b

i=1 Ii (z(ti )) + a+b 0 q(y(s) + z(s))ds


R
α−2
 bt T (T −s)
− a+b 0 Γ(α−1) f (s, ys + z s , B(y(s) + z(s)))ds,
t ∈ [0, t1 ],
t ∈ (tk , tk+1 ],
where k = 1, 2, . . . , m. Set B00h = {z ∈ B0h such that z0 = 0} and let k · kB00h be the
seminorm in B00h defined by
kzkB00h = sup kz(t)kX + kz0 kBh = sup kz(t)kX , z ∈ B00h .
t∈J
t∈J
8
A. CHAUHAN, J. DABAS, M. KUMAR
EJDE-2012/229
Thus (B00h , k · kB00h ) is a Banach space. We define the operator P : B00h → B00h by
R t (t−s)α−1

f (s, ys + z s , B(y(s) + z(s)))ds + φ(0)

 0 Γ(α)
RT
Pm

−
bt
t


− a+b i=1 Ii (z(ti )) + a+b 0 q(y(s) + z(s))ds

R T (T −s)α−2


bt
− a+b
t ∈ [0, t1 ],

Γ(α−1) f (s, ys + z s , B(y(s) + z(s)))ds,
0


. . . ,
P z(t) = R t (t−s)α−1

f (s, ys + z s , B(y(s) + z(s)))ds + φ(0)

0


PkΓ(α)
Pk

−

+
(t
−
ti )Ii (z(t−

i )) +
i=1
i=1 Qi (z(ti ))


R
P
T

m
−
t
bt

− a+b

i=1 Ii (z(ti )) + a+b 0 q(y(s) + z(s))ds


R
α−2
 bt T (T −s)
− a+b 0 Γ(α−1) f (s, ys + z s , B(y(s) + z(s)))ds, t ∈ (tk , tk+1 ],
where k = 1, 2, . . . , m. It is clear that the operator N has a unique fixed point if
and only if P has a unique fixed point. So let us prove that P has a unique fixed
point. Let z, z ∗ ∈ B00h and t ∈ [0, t1 ] we have
k(P z)(t) − (P z ∗ )(t)kX
Z t
(t − s)α−1
kf (s, ys + z s , B(y(s) + z(s))) − f (s, ys + z ∗s , B(y(s)
≤
Γ(α)
0
m
bt X
∗ −
+ z ∗ (s)))kX ds +
kIi (z(t−
i )) − Ii (z (ti ))kX
a + b i=1
Z T
t
kq(y(s) + z(s)) − q(y(s) + z ∗ (s))kX ds
+
a+b 0
Z T
bt
(T − s)α−2
+
kf (s, ys + z s , B(y(s) + z(s))) − f (s, ys + z ∗s , B(y(s)
a + b 0 Γ(α − 1)
+ z ∗ (s))kX ds
h (µ l + µ∗ )(a + (1 + α)b)T α
T (bLm + Lq T ) i
1
B
+
kz − z ∗ kB00h .
≤
(a + b)Γ(α + 1)
a+b
If t ∈ (tk , tk+1 ], k = 1, 2, . . . , m, then
k(P z)(t) − (P z ∗ )(t)kX
Z t
(t − s)α−1
≤
kf (s, ys + z s , B(y(s) + z(s))) − f (s, ys + z ∗s , B(y(s)
Γ(α)
0
+ z ∗ (s)))kX ds +
k
X
∗ −
(t − ti )kIi (z(t−
i )) − Ii (z (ti ))kX
i=1
+
k
X
∗ −
kQi (z(t−
i )) − Qi (z (ti ))kX
i=1
m
bt X
∗ −
kIi (z(t−
i )) − Ii (z (ti ))kX
a + b i=1
Z T
t
kq(y(s) + z(s)) − q(y(s) + z ∗ (s))kX ds
+
a+b 0
Z T
bt
(T − s)α−2
+
kf (s, ys + z s , B(y(s) + z(s))) − f (s, ys + z ∗s , B(y(s)
a + b 0 Γ(α − 1)
+
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IMPULSIVE FRACTIONAL DIFFERENTIAL EQUATIONS
9
+ z ∗ (s)))kX ds
i
h (a + (1 + α)b)(µ l + µ B ∗ )T α
(a + 2b)LT m + Lq T 2
1
2
+
+ Lm kz − z ∗ kB00h .
≤
(a + b)Γ(α + 1)
a+b
Thus for all t ∈ [0, T ], we have the estimate
kP (z) − P (z ∗ )kBh00
i
h (a + (1 + α)b)(µ l + µ B ∗ )T α
(a + 2b)LT m + Lq T 2
1
2
+
+ Lm kz − z ∗ kBh00
≤
(a + b)Γ(α + 1)
a+b
≤ Λkz − z ∗ kB00h .
Since Λ < 1, the map P is a contraction map and has a unique fixed point z ∈ B00h ,
which is obviously a solution of the system (1.1) on (−∞, T ]. This completes the
proof of the theorem.
Our second existence result is based on the following 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 operator 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. Suppose that assumptions (H1), (H4), (H5) are satisfied with
∆=
(µ1 l + µ2 B ∗ )(a + (1 + α)b)T α
< 1.
(a + b)Γ(α + 1)
Then (1.1) has at least one solution on (−∞, T ].
Proof. Choose
h
(bρm + T C)T
r ≥ kφ(0)k + (ρT + Ω)m +
a+b
(µ∗1 (kφk + lr) + µ∗2 B ∗ r)(a + (1 + α)b)T α i
.
+
(a + b)Γ(α + 1)
Define Br = {z ∈ B00h : kzkB00h ≤ r}, then Br is a bounded, closed convex subset in
B00h . Let P1 : Br → Br and P2 : Br → Br be defined as
Z T
k
X
t
q(y(s) + z(s))ds +
(P1 z)(t) = φ(0) +
Qi (z(t−
i ))
a+b 0
i=1
−
t ∈ Jk .
t
(t − s)α−1
f (s, ys + z s , B(y(s) + z(s)))ds
Γ(α)
0
Z T
bt
(T − s)α−2
−
f (s, ys + z s , B(y(s) + z(s)))ds,
a + b 0 Γ(α − 1)
Z
(P2 z)(t) =
m
k
X
bt X
Ii (z(t−
(t − ti )Ii (z(t−
i )) +
i )),
a + b i=1
i=1
t ∈ Jk ,
where J0 = [0, t1 ] and Jk = (tk , tk+1 ], k = 1, . . . , m. Now, we proceed the proof in
following steps:
10
A. CHAUHAN, J. DABAS, M. KUMAR
EJDE-2012/229
Step 1. Let z, z ∗ ∈ Br , then we show that P1 z + P2 z ∗ ∈ Br for t ∈ Jk , k =
0, 1, . . . , m. We have
k(P1 z)(t) + (P2 z ∗ )(t)kX
Z T
t
kq(y(s) + z(s))kX ds
≤ kφ(0)kX +
a+b 0
+
k
X
m
kQi (z(t−
i ))kX +
i=1
t
k
X
bt X
kIi (z(t−
(t − ti )kIi (z(t−
i ))kX +
i ))kX
(3.1)
a + b i=1
i=1
(t − s)α−1
kf (s, ys + z ∗s , B(y(s) + z ∗ (s)))kX ds
Γ(α)
0
Z T
bt
(T − s)α−2
+
kf (s, ys + z ∗s , B(y(s) + z ∗ (s)))kX ds,
a + b 0 Γ(α − 1)
Z
+
we estimate the inequality (3.1), by using (H4) and (H5), as
h
(bρm + T C)T
k(P1 z)(t) + (P2 z ∗ )(t)kX ≤ kφ(0)k + (ρT + Ω)m +
a+b
(µ∗1 (kφk + lr) + µ∗2 B ∗ r)(a + (1 + α)b)T α i
,
+
(a + b)Γ(α + 1)
which implies that kP1 z + P2 z ∗ kBh00 ≤ r.
Step 2. Now, we shall show that the mapping (P1 z)(t) is continuous on Br .
n
For this purpose, let {z n }∞
n=1 be a sequence in Br with lim z → z in Br . Then for
t ∈ Jk , k = 0, 1, . . . , m, we have
k(P1 z n )(t) − (P1 z)(t)kX
Z T
t
kq(y(s) + z n (s)) − q(y(s) + z(s))kX ds
≤
a+b 0
+
k
X
m
−
kQi (z n (t−
i )) − Qi (z(ti ))kX +
i=1
+
k
X
bt X
−
kIi (z n (t−
i )) − Ii (z(ti ))kX
a + b i=1
−
(t − ti )kIi (z n (t−
i )) − Ii (z(ti ))kX .
i=1
Since the functions q, Qk , Ik , k = 0, 1, . . . , m, are continuous, hence limn→∞ P1 z n =
P1 z in Br . Which implies that the mapping P1 is continuous on Br .
Step 3 (P1 z)(t) is uniformly bounded follows by the following inequality. For
t ∈ Jk , k = 0, 1, . . . , m, we have
k(P1 z)(t)kX
t
≤ kφ(0)kX +
a+b
+
bt
a+b
m
X
i=1
Z
T
kq(y(s) + z(s))kX ds +
0
kIi (z(t−
i ))kX +
k
X
kQi (z(t−
i ))kX
i=1
k
X
(t − ti )kIi (z(t−
i ))kX
i=1
T (T C + bρm)
≤ kφ(0)kX +
+ Ωm + ρmT.
a+b
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IMPULSIVE FRACTIONAL DIFFERENTIAL EQUATIONS
11
Step 4 To show that P1 (Br ) is equicontinuous. Let τ1 , τ2 ∈ Jk , tk ≤ τ1 < τ2 ≤
tk+1 , k = 0, 1, . . . , m, z ∈ Br , we have
Z T
1
k(P1 z)(τ2 ) − (P2 z)(τ1 )kX ≤ (τ2 − τ1 )[
kq(y(s) + z(s))kX ds
a+b 0
+
k
X
m
kIi (z(t−
i ))kX +
i=1
b X
kIi (z(t−
i ))kX ],
a + b i=1
which implies that P1 (Br ) is equicontinuous. Finally, combing Step 2 to Step 4
together with the Ascol’s theorem, we conclude that the operator P1 is a compact.
Step 5. Now, we show that P2 is a contraction mapping. Let z, z ∗ ∈ Br and
t ∈ Jk , k = 0, 1, . . . , m, we have
k(P2 z)(t) − (P2 z ∗ )(t)kX
Z t
(t − s)α−1
≤
kf (s, ys + z s , B(y(s) + z(s))) − f (s, ys + z ∗s , B(y(s)
Γ(α)
0
Z T
(T − s)α−2
bt
kf (s, ys + z s , B(y(s)
+ z ∗ (s)))kX ds +
a + b 0 Γ(α − 1)
+ z(s))) − f (s, ys + z ∗s , B(y(s) + z ∗ (s)))kX ds
(µ1 l + µ2 B ∗ )(a + (1 + α)b)T α
kz − z ∗ kBh00
(a + b)Γ(α + 1)
≤ ∆kz − z ∗ kB00h ,
≤
where
∆=
(µ1 l + µ2 B ∗ )(a + (1 + α)b)T α
.
(a + b)Γ(α + 1)
As ∆ < 1, then P2 is a contraction map. Thus all the assumptions of the Theorem
3.2 are satisfied and the conclusion of the Theorem 3.2 implies that the system (1.1)
has at least one solution on (−∞, 0]. This completes the proof of the theorem. 4. Application
We consider the model
Z 0
1
C 3/2
Dt u(t) =
e2θ sin(ku(t + θ)kX )dθ
(t + 9)2 −∞
Z t
1
+
sin k
(t − s)u(s)dskX , t ∈ [0, 1], t 6= ti , i = 1, 2, 3,
(t + 7)2
0
u(t) = φ(t), t ∈ (−∞, 0],
Z 0
ku(ti + θ)kX
∆u(ti ) =
e2θ
dθ,
25 + ku(ti + θ)kX
−∞
Z 0
ku(ti + θ)kX
dθ,
e2θ
∆u0 (ti ) =
27 + ku(ti + θ)kX
−∞
Z 1
1
0
0
u (0) + u (1) =
sin( ku(s)kX )ds,
2
0
(4.1)
12
A. CHAUHAN, J. DABAS, M. KUMAR
EJDE-2012/229
where X is a real Banach space, 0 < t1 < t2 < Rt3 < 1 are prefixed numbers
0
and φ ∈ Bh . Let h(s) = e2s , s < 0 then l = −∞ h(s)ds = 1/2 and define
R0
kφkBh = −∞ h(s) sups<θ<0 kφ(θ)kX ds. Hence for t ∈ [0, 1] and φ ∈ Bh , we set
Z 0
1
1
f (t, φ, Bu(t)) =
h(θ) sin(kφ(θ)kX )dθ +
sin(kBu(t)kX ),
(t + 9)2 −∞
(t + 7)2
Z 0
kφ(θ)kX
Qi (φ) =
dθ,
h(θ)
25 + kφ(θ)kX
−∞
Z 0
kφ(θ)kX
dθ,
Ii (φ) =
h(θ)
27
+ kφ(θ)kX
−∞
Rt
Rt
where Bu(t) = 0 (t − s)u(s)ds, now B ∗ = supt∈[0,1] 0 (t − s)ds = 21 < ∞. Then
the above equations (4.1) can be written in the abstract form as (1.1). Moreover,
1
1
kφ − ψkBh + kBu(t) − Bv(t)kX ,
81
49
1
kQi (φ) − Qi (ψ)kX ≤
kφ − ψkBh ,
25
1
kφ − ψkBh ,
kIi (φ) − Ii (ψ)kX ≤
27
1
kq(u) − q(v)kX ≤ ku − vkX ,
2
therefore, (H1),(H2) and (H3) are satisfied with µ1 = 1/81, µ2 = 1/49, Lq = 1/2,
L = 1/27, L = 1/25. Further,
kf (t, φ, Bu(t)) − f (t, ψ, Bv(t))kX ≤
(a + (1 + α)b)(µ1 l + µ2 B ∗ )T α
(a + 2b)LT m + Lq T 2
+
+ Lm ≈ 0.558 < 1.
(a + b)Γ(α + 1)
a+b
Thus, all the assumptions of Theorem 3.1 are satisfied. Hence, the impulsive fractional boundary-value problem (4.1) has a unique solution.
Acknowledgements. The authors are grateful to the anonymous referees for their
valuable suggestions that led to the improvement of the original manuscript. J.
Dabas has been partially supported by Sponsored Research & Industrial Consultancy, Indian Institute of Technology Roorkee, project IITR/SRIC/247/F.I.G
(Scheme-A).
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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 Applied Science and Engineering, IIT Roorkee, Saharanpur Campus,
Saharanpur-247001, India
E-mail address: jay.dabas@gmail.com
Mukesh Kumar
Department of Mathematics, Motilal Nehru National Institute of technology, Allahabad - 211 004, India
E-mail address: mukesh@mnnit.ac.in