Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 241, pp. 1–15.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF PIECEWISE CONTINUOUS MILD SOLUTIONS
FOR IMPULSIVE FUNCTIONAL DIFFERENTIAL EQUATIONS
WITH ITERATED DEVIATING ARGUMENTS
PRADEEP KUMAR, DWIJENDRA N. PANDEY, DHIRENDRA BAHUGUNA
Abstract. The objective of this article is to prove the existence of piecewise
continuous mild solutions to impulsive functional differential equation with
iterated deviating arguments in a Banach space. The results are obtained by
using the theory of analytic semigroups and fixed point theorems.
1. Introduction
In the theory of differential equations with deviating arguments, we study differential equations involving variables (arguments) as well as unknown functions
and its derivative; generally speaking, under different values of the variables (arguments). It is very important and significant branch of nonlinear analysis with
numerous applications to physics, mechanics, control theory, biology, ecology, economics, theory of nuclear reactors, engineering, natural sciences, and many other
areas of science and technology. A comprehensive coverage can be found in the book
by El’sgol’ts and Norkin [7]. For more details on recent works in this direction, we
refer to [8, 9, 10, 11, 12] and the references cited therein.
Impulsive effects are common phenomena due to short-term perturbations whose
duration is negligible in comparison with the total duration of the original process.
The governing equations of such phenomena may be modelled as impulsive differential equations. In recent years, there has been a growing interest in the study
of impulsive differential equations as these equations approach the simulation processes in the control theory, physics, chemistry, population dynamics, biotechnology,
economics and so on. The investigation of existence and uniqueness of mild solutions for impulsive differential equations have been discussed by many authors (see
[4, 6, 19, 5, 3, 23, 17, 16, 24, 26, 13, 14, 25, 1, 2] and references cited therein).
However, due to theoretical and practical difficulties, the study of impulsive
differential equations with deviating arguments has been developed rather slowly.
Recently, the study of impulsive differential equations with deviating arguments
has studied by some authors. Guobing et al. [26] established the existence solution
of periodic boundary value problems for a class of impulsive neutral differential
2000 Mathematics Subject Classification. 34K45, 34A60, 35R12, 45J05.
Key words and phrases. Impulsive functional differential equation; deviating argument;
analytic semigroup; fixed point theorem.
c
2013
Texas State University - San Marcos.
Submitted May 20, 2013. Published October 31, 2013.
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P. KUMAR, D. N. PANDEY, D. BAHUGUNA
EJDE-2013/241
equations with multi-deviation arguments. Jankowski [14] discussed the existence of
solutions for second-order impulsive differential equations with deviating arguments
(see also [16, 24, 13, 25, 1, 2] and the references cited therein).
Liu [15] studied the existence of mild solutions of the following impulsive evolution equation by using semigroup theory
u0 (t) = Au(t) + f (t, u(t)),
∆u(ti ) = Ii (u(ti )) =
u(t+
i )
−
u(t−
i ),
0 < t < T0 , t 6= ti ,
i = 1, 2, . . . , 0 < t1 < t2 < · · · < T0 , (1.1)
u(0) = u0 ,
−
where A is the infinitesimal generator of a C0 -semigroup, u(t+
i ), u(ti ) represent the
right and left hand limits respectively at t = ti , Ii ’s are some operators and f is a
suitable function.
Chang et al. [4] investigated the existence of mild solutions for the following
impulsive problem
Z t
h
i
d
(u(t) − F (t, u(h1 (t)))) ∈ A u(t) +
f (t − s)u(s)ds
dt
(1.2)
0
+ G(t, u(h2 (t)))), t ∈ [0, a], t 6= tk ,
∆u|t=tk = Ik (u(t−
k )),
k = 1, . . . , m,
(1.3)
u(0) + g(u) = u0 ,
(1.4)
where A is the infinitesimal generator of a compact analytic resolvent operator
R(t), t > 0 in a Banach space H, u0 ∈ H, 0 = t0 < t1 < . . . tm < tm+1 = a,
−
+
−
∆u|t=tk = u(t+
k ) − u(tk ), u(tk ), u(tk ) represent the right and left hand limits
respectively at t = tk . hi : [0, a] → [0, a], i = 1, 2, f (t), t ∈ [0, a] is a bounded linear
operator, and F, G, g are suitable functions.
With strong motivation from the above work, in this article, we consider the following impulsive functional differential equations with iterated deviating arguments
in a Banach space (H, k · k):
d
u(t) + Au(t) = f (t, u(t), u[w1 (t, u(t))]), t ∈ I = [0, T0 ], t 6= tk ,
dt
∆u|t=tk = Ik (u(t−
k )), k = 1, 2, . . . , m,
(1.5)
u(0) = u0 ,
where
w1 (t, u(t)) = h1 t, u(h2 (t, . . . , u(hm (t, u(t))) . . . ))
and −A is the infinitesimal generator of an analytic semigroup of bounded linear
operators, {S(t), t ≥ 0} on H. Functions f and hi are suitably defined and satisfying
certain conditions to be stated later. 0 = t0 < t1 < · · · < tm < tm+1 = T0 ,
−
Ik ∈ C(H, H)(k = 1, 2, . . . , m), are bounded functions and ∆u|t=tk = u(t+
k )−u(tk ),
−
+
u(tk ) and u(tk ) represent the left and right limits of u(t) at t = tk , respectively.
The paper is organized as follows. In Section 2, we provide some basic definitions,
notation, lemmas and proposition which are used throughout the paper. In Sections
3 and 4, we prove the existence and uniqueness results concerning the piecewise
continuous mild solutions. In the last section, we give an example to demonstrate
the application of the main results.
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3
2. Preliminaries and assumptions
In this section, we will introduce some basic definitions, notation, lemmas and
proposition which are used throughout this paper.
It is assumed that −A generates an analytic semigroup of bounded operators,
denoted by S(t), t ≥ 0. It is known that there exist constants M̃ ≥ 1 and ω ≥ 0
such that
kS(t)k ≤ M̃ eωt , t ≥ 0.
If necessary, we may assume without loss of generality that kS(t)k is uniformly
bounded by M ; i.e., kS(t)k ≤ M for t ≥ 0, and that 0 ∈ ρ(−A); i.e., −A is
invertible. In this case, it is possible to define the fractional power Aα for 0 ≤ α ≤ 1
as closed linear operator with domain D(Aα ) ⊆ H. Furthermore, D(Aα ) is dense
in H and the expression
kxkα = kAα xk,
defines a norm on D(Aα ). Henceforth, we denote the space D(Aα ) by Hα endowed
with the norm k · kα . Also, for each α > 0, we define H−α = (Hα )∗ , the dual space
of Hα , is a kxk−α = kA−α xk. For more details, we refer to the book by Pazy [18].
Lemma 2.1 ([18, pp. 72,74,195-196]). Suppose that −A is the infinitesimal generator of an analytic semigroup S(t), t ≥ 0 with kS(t)k ≤ M for t ≥ 0 and 0 ∈ ρ(−A).
Then we have the following:
(i) Hα is a Banach space for 0 ≤ α ≤ 1;
(ii) For any 0 < δ ≤ α implies D(Aα ) ⊂ D(Aδ ), the embedding Hα ,→ Hδ is
continuous;
(iii) The operator Aα S(t) is bounded for every t > 0 and
kAα S(t)k ≤ Cα t−α .
(iv) For α ≤ 0, Aα is bounded.
We define the space
n
X = PC(Hα ) = u : [0, T0 ] → Hα : u ∈ C((tk , tk+1 ], Hα ), k = 0, 1, . . . , m,
o
+
−
u(t−
k ), u(tk ) exist and u(tk ) = u(tk ) ,
is a Banach space endowed with the supremum norm
kukPC := sup ku(t)kα .
t∈I
We shall use the following conditions on f and hi in its arguments:
(H1) Let W ⊂ Dom(f ) be an open subset of R+ × Hα × Hα−1 , where 0 ≤ α < 1.
For each (t, u, v) ∈ W , there is a neighborhood V1 ⊂ W of (t, u, v), such
that the nonlinear map f : R+ × Hα × Hα−1 → H satisfies the condition
kf (t, u, v) − f (s, u1 , v1 )k ≤ Lf {|t − s|θ1 + ku − u1 kα + kv − v1 kα−1 }
for all (t, u, v), (s, u1 , v1 ) ∈ V1 , where Lf = Lf (t, u, v, V1 ) > 0 and 0 < θ1 ≤
1 are constants.
(H2) Let Uhi ⊂ Dom(hi ) be open subsets of R+ × Hα−1 , where 0 ≤ α < 1. For
each (t, u) ∈ Uhi , there is a neighborhood Vhi ⊂ Uhi of (t, u), such that
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P. KUMAR, D. N. PANDEY, D. BAHUGUNA
EJDE-2013/241
hi (0, ·) = 0 for each i = 1, 2, . . . , m, hi : R+ × Hα−1 → R+ satisfies the
condition
|hi (t, u) − hi (s, v)| ≤ Lhi {ku − vkα−1 + |t − s|θ2 }
(2.1)
for all (t, u), (s, v) ∈ Vhi , Lhi = Lhi (t, u, Vhi ) > 0 and 0 < θ2 ≤ 1 are
constants.
(H3) The functions Ik : Hα → Hα are continuous and there exists Υk such that
kIk (u)kα ≤ Υk , k = 1, 2, . . . , m.
(H4) There exist continuous nondecreasing dk : R+ → R+ such that kIk (u) −
Ik (v)kα ≤ dk ku − vkα , k = 1, 2, . . . , m.
New concept of solutions. Here, we prove a new concept of solutions [23, 15]
for the problem
u0 (t) + Au(t) = r(t), t ∈ [0, T0 ], t 6= tk ,
u(0) = u0
u(tk ) =
where r ∈ H. Let
Ik (u(t−
k )),
(2.2)
k = 1, 2 . . . , m,
v 0 (t) + Av(t) = r(t),
t ∈ [0, T0 ],
v(0) = v0 ,
and
w0 (t) + Aw(t) = 0,
t ∈ [0, T0 ], t 6= tk ,
w(0) = 0,
w(tk ) =
Ik (u(t−
k )),
(2.3)
(2.4)
k = 1, 2, . . . , m,
be the decomposition of u(.) = v(.) + w(.), where v is the continuous mild solution
of (2.3) and w is the piecewise continuous mild solution of (2.4).
By a mild solution for (2.3), we mean a continuous function v : [0, T0 ] → H
satisfying the integral equation
Z t
v(t) = S(t)v0 +
S(t − s)r(s)ds, t ∈ [0, T0 ].
(2.5)
0
and by a piecewise continuous mild solution for (2.4), we mean a function w ∈
PC([0, T0 ], D(A)) satisfying the integral equation equivalent to the system (2.4)
([23, see Eq. (3.4)])
 Rt
− 0 Aw(s)ds,
t ∈ [0, t1 ],



I (u(t− )) − R t Aw(s)ds,
t ∈ (t1 , t2 ],
1
0
Rt
(2.6)
w(t) = Pk 1
−


i=1 Ii (u(ti )) − 0 Aw(s)ds, t ∈ (tk , tk+1 ],


k = 1, 2, . . . , m.
The above equation can be expressed as
w(t) =
k
X
i=1
χi (t)Ii (w(t−
i ))
Z
−
t
Aw(s)ds,
(2.7)
0
for t ∈ [0, T0 ], where
(
0, for t ∈ [0, t1 ],
χi (t) =
1, for t ∈ [tk , tk+1 ].
(2.8)
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5
Taking Laplace transform of (2.7), we obtain
w(p) =
k
X
e−ti p
i=1
p
Ii −
Aw(p)
,
p
this gives
w(p) =
We also note that (pI +
solution for (2.4),
k
X
e−ti p (pI + A)−1 Ii .
i=1
R∞
A)−1 = 0 e−pt S(t)dt.
w(t) =
k
X
(2.9)
Thus we can obtain the mild
χi (t)S(t − ti )Ii (w(t−
i )).
(2.10)
i=1
Hence, the mild solution for the problem (2.2) is
u(t) = S(t)u0 +
k
X
χi (t)S(t − ti )Ii (u(t−
i )) +
i=1
Z
t
S(t − s)r(s)ds.
(2.11)
0
We can rewrite equation (2.11) as

Rt
S(t)u0 + 0 S(t − s)r(s)ds,
t ∈ [0, t1 ],



S(t)u + S(t − t )I (u(t− )) + R t S(t − s)r(s)ds,
t ∈ (t1 , t2 ],
0
1 1
1
0
Rt
u(t) =
Pk
−

S(t)u0 + i=1 S(t − ti )Ii (u(ti )) + 0 S(t − s)r(s)ds, t ∈ (tk , tk+1 ],



k = 1, 2, . . . , m.
(2.12)
3. Local existence of mild solutions
In this section, we prove the existence and uniqueness results concerning piecewise continuous-mild solutions for system (1.5). For 0 ≤ α < 1, we define
X1 = {u ∈ X : ku(t) − u(s)kα−1 ≤ L|t − s|, ∀t, s ∈ (tk , tk+1 ], k = 0, 1, . . . , m},
which is needful for proving contraction principle (See [8, 9]). Where L is a suitable
positive constant to be specified later.
Definition 3.1. A function u : [0, T0 ] → H solution of the problem (1.5),

Rt
S(t)u0 + 0 S(t − s)f (s, u(s), u(w1 (s, u(s))))ds, t ∈ [0, t1 ],






S(t)u0 + S(t − t1 )I1 (u(t−
1 ))


Rt


t ∈ (t1 , t2 ],
+ 0 S(t − s)f (s, u(s), u(w1 (s, u(s))))ds,
u(t) = . . .

Pk


−

S(t)u
+
0

i=1 S(t − ti )Ii (u(ti ))

R

t


t ∈ (tk , tk+1 ],
+ 0 S(t − s)f (s, u(s), u(w1 (s, u(s))))ds,


k = 1, 2, . . . , m.
is said to be a piecewise continuous-mild solution.
For a fixed R > 0, we define
W = {u ∈ X ∩ X1 : u(0) = u0 , ku − u0 kPC ≤ R}.
(3.1)
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P. KUMAR, D. N. PANDEY, D. BAHUGUNA
EJDE-2013/241
Clearly, W is a closed and bounded subset of X1 and is a Banach space. We define
a map G : W → W by

Rt
S(t)u0 + 0 S(t − s)f (s, u(s), u(w1 (s, u(s))))ds, t ∈ [0, t1 ],







S(t)u0 + S(t − t1 )I1 (u(t−
1 ))


Rt


+
S(t
−
s)f
(s,
u(s),
u(w
t ∈ (t1 , t2 ],
 0
1 (s, u(s))))ds,
(Gu)(t) = . . .
(3.2)

P

k


S(t)u0 + i=1 S(t − ti )Ii (u(t−

i ))

Rt



+
S(t
−
s)f
(s,
u(s),
u(w
(s,
u(s))))ds,
t ∈ (tk , tk+1 ],
1


 0
k = 1, 2, . . . , m.
Theorem 3.2. Let 0 ≤ α < 1, u0 ∈ Hα and the assumptions (H1)–(H4) hold.
Then problem (1.5) has a mild solution provided that
Cα N
k
X
R
T01−α
+M
Υi ≤ ,
1−α
2
i=1
(3.3)
m
Cα Lf (2 + L Lh )
X
T01−α
+M
di < 1,
1−α
i=1
(3.4)
where N and Lh are positive constants to be specified later.
Proof. Clearly, G : X → X. We begin with showing that Gu ∈ X1 for each u ∈ X1 .
Let u ∈ X1 , then for each τ1 , τ2 ∈ [0, t1 ], τ1 < τ2 and 0 ≤ α < 1, we have
k(Gu)(τ2 ) − (Gu)(τ1 )kα−1
≤ k[S(τ2 ) − S(τ1 )]u0 kα−1
Z τ1
+
kAα−1 [S(τ2 − s) − S(τ1 − s)]k kf (s, u(s), u(w1 (s, u(s))))kds
0
Z τ2
+
kAα−1 S(τ2 − s)k kf (s, u(s), u(w1 (s, u(s))))kds.
(3.5)
τ1
Since f (t, u(t), u(w1 (u(t), t))) is continuous, by assumptions (H1), (H2), there
exists a constant N , such that
kf (t, u(t), u(w1 (t, u(t))))k ≤ N,
u ∈ X, t ∈ [0, T0 ].
For the first term on the right-hand side of (3.5), we have
Z τ2
kAα−1 [S(τ2 ) − S(τ1 )]u0 k ≤
kAα−1 S 0 (s)u0 kds
τ1
Z τ2
=
kAα S(s)u0 kds
(3.6)
(3.7)
τ1
≤ M ku0 kα (τ2 − τ1 ).
For the second term on the right-hand side of (3.5), we have
Z τ2 −τ1
k(S(τ2 − s) − S(τ1 − s))kα−1 ≤
kAα−1 S 0 (l)S(τ1 − s)kdl
0
Z τ2 −τ1
kS(l)Aα S(τ1 − s)kdl
=
0
≤ M Cα (τ2 − τ1 )(τ1 − s)−α .
(3.8)
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7
Then using (3.8), we obtain the following bound for the second term on the righthand side of (3.5),
Z τ1
k(S(τ2 − s) − S(τ1 − s))Aα−1 k kf (s, u(s), u(w1 (s, u(s))))kds
0
(3.9)
T01−α
≤ N M Cα
(τ2 − τ1 ).
1−α
The third term on the right-hand side of (3.5) is estimated as
Z τ2
kS(τ2 − s)Aα−1 kkf (s, u(s), u(w1 (s, u(s))))kds ≤ kAα−1 kM N (τ2 − τ1 ). (3.10)
τ1
Thus from inequalities (3.7), (3.9) and (3.10), we see that
T01−α
+ N kAα−1 k}(τ2 − τ1 ). (3.11)
1−α
For τ1 , τ2 ∈ (t1 , t2 ], τ1 < τ2 and 0 ≤ α < 1, we have
k(Gu)(τ2 ) − (Gu)(τ1 )kα−1 ≤ M {ku0 kα + N Cα
k(Gu)(τ2 ) − (Gu)(τ1 )kα−1
≤ k[S(τ2 − S(τ1 )]u0 kα−1 + kAα−1 [S(τ2 − t1 ) − S(τ1 − t1 )]I1 (u(t−
1 ))k
Z τ1
+
kAα−1 [S(τ2 − s) − S(τ1 − s)]k kf (s, u(s), u(w1 (s, u(s))))kds
0
Z τ2
+
kAα−1 S(τ2 − s)k kf (s, u(s), u(w1 (s, u(s))))kds.
(3.12)
τ1
The second term on the right-hand side of (3.12) is estimated as
kAα−1 [S(τ2 − t1 ) − S(τ1 − t1 )]I1 (u(t−
1 ))k
Z τ2
≤
kAα−1 S 0 (l − t1 )I1 (u(t−
1 ))kdl
τ1
Z τ2
=
kAα S(l − t1 )I1 (u(t−
1 ))kdl
(3.13)
τ1
≤ M kI1 (u(t−
1 ))kα (τ2 − τ1 ).
Thus, from inequalities (3.7), (3.9), (3.10) and (3.13), we obtain
k(Gu)(τ2 ) − (Gu)(τ1 )kα−1
n
o
(3.14)
T 1−α
+ N kAα−1 k (τ2 − τ1 ).
≤ M ku0 kα + Υ1 + N Cα 0
1−α
Similarly, for τ1 , τ2 ∈ (tk , tk+1 ], τ1 < τ2 , k = 1, 2, . . . , m and 0 ≤ α < 1, we have
k(Gu)(τ2 ) − (Gu)(τ1 )kα−1
k
n
o
X
T 1−α
≤ M ku0 kα +
Υi + N C α 0
+ N kAα−1 k (τ2 − τ1 ).
1−α
i=1
(3.15)
Thus, for each τ1 , τ2 ∈ [0, T0 ], τ1 < τ2 and 0 ≤ α < 1, we have
k(Gu)(τ2 ) − (Gu)(τ1 )kα−1 ≤ L(τ2 − τ1 ),
where
L = max{M ku0 kα , M
m
X
i=1
Υi , N M Cα
T01−α
, N M kA1−α k}.
1−α
(3.16)
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P. KUMAR, D. N. PANDEY, D. BAHUGUNA
EJDE-2013/241
Therefore, G is piecewise Lipschitz continuous on [0, T0 ] and so G : X1 → X1 .
Next we will show that G : W → W. Let u ∈ X ∩ X1 , then for each t ∈ [0, t1 ],
k(Gu)(t) − u0 kα ≤ k(S(t) − I)Aα u0 k
Z t
+
kS(t − s)Aα kkf (s, u(s), u(w1 (s, u(s))))kds
0
T01−α
.
1−α
Similarly, let u ∈ X ∩ X1 , then for each t ∈ (tk , tk+1 ], k = 1, . . . , m,
≤ k(S(t) − I)Aα u0 k + Cα N
k(Gu)(t) − u0 kα
α
t
Z
kS(t − s)Aα kkf (s, u(s), u(w1 (s, u(s))))kds
≤ k(S(t) − I)A u0 k +
0
+
k
X
kAα S(t − ti )Ii (u(t−
i ))k
i=1
k
≤ k(S(t) − I)Aα u0 k + Cα N
X
T01−α
+M
Υi .
1−α
i=1
Part (iii) of Lemma 2.1, implies that
k(S(t) − I)Aα (u0 )k ≤
R
.
2
(3.17)
Thus, from (3.3) and (3.17), it is clear that
kGu − u0 kPC ≤ R.
Therefore, G : W → W.
Finally, we will claim that G is a contraction map. If t ∈ [0, t1 ] and u, v ∈ W,
then
k(Gu)(t) − (Gv)(t)kα
Z t
≤
kS(t − s)Aα k kf (s, u(s), u(w1 (s, u(s)))) − f (s, v(s), v(w(s, v(s))))kds.
0
(3.18)
Also, we note that
kf (t, u(t), u(w1 (t, u(t)))) − f (t, v(t), v(w1 (t, v(t))))k
≤ Lf {ku(t) − v(t)kα + ku(w1 (t, u(t))) − v(w1 (t, v(t)))kα−1 }
h
≤ Lf ku(t) − v(t)kα + kA−1 k ku(w1 (t, v(t))) − v(w1 (t, v(t)))kα
i
+ ku(w1 (t, u(t))) − u(w1 (t, v(t)))kα−1
≤ Lf 2ku − vkPC + L|w1 (t, u(t)) − w1 (t, v(t))| .
Now, let
wi (t, u(t)) = hi (t, u(hi+1 (t, . . . u(t, hm (t, u(t))) . . . ))), i = 1, . . . , m,
with wm+1 (t, u(t)) = t. For more details, we refer to [21, p. 2183].
Hence, we have
|w1 (t, u(t)) − w1 (t, v(t))| = |h1 (t, u(w2 (t, u(t)))) − h1 (t, v(w2 (t, v(t))))
(3.19)
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≤ Lh1 ku(w2 (t, u(t))) − u(w2 (t, u(t)))kα−1
n
≤ Lh1 ku(w2 (t, u(t))) − u(w2 (t, v(t)))kα−1
o
+ ku(w2 (t, v(t))) − v(w2 (t, v(t)))kα−1
n
≤ Lh1 L|h2 (t, u(w3 (t, u(t))) − h2 (t, v(w3 (t, v(t)))|
o
+ kAk−1 ku − vkPC
...
≤ (Lm−1 Lh1 . . . Lhm + Lm−2 Lh1 . . . Lhm−1 + . . .
+ LLh1 Lh2 + Lh1 )kAk−1 ku − vkPC .
Thus, we have
kf (t, u(t), u(w1 (t, u(t)))) − f (t, v(t), v(w1 (t, v(t))))k
≤ Lf 2 + LLh kA−1 k ku − vkPC
(3.20)
≤ Lf (2 + LLh )ku − vkPC .
m−1
where Lh = (L
Lh1 . . . Lhm + Lm−1 Lh1 . . . Lhm−1 + · · · + Lh1 ) > 0. We use (3.20)
in (3.18), we obtain
k(Gu)(t) − (Gv)(t)kα ≤ Cα Lf (2 + LLh )
T01−α
ku − vkPC .
1−α
(3.21)
For t ∈ (t1 , t2 ], we have
i
h
T 1−α
+ M d1 ku − vkPC .
k(Gu)(t) − (Gv)(t)kα ≤ Cα Lf (2 + LLh ) 0
1−α
For t ∈ (tk , tk+1 ], k = 1, 2, . . . , m, we have
k
h
i
X
T 1−α
k(Gu)(t) − (Gv)(t)kα ≤ Cα Lf (2 + LLh ) 0
di ku − vkPC .
+M
1−α
i=1
(3.22)
(3.23)
Thus, for each t ∈ [0, T0 ], we have
m
i
h
X
T 1−α
di ku − vkPC .
+M
k(Gu)(t) − (Gv)(t)kα ≤ Cα Lf (2 + LLh ) 0
1−α
i=1
Therefore, the map G is a contraction map, hence G has a unique fixed point u ∈ W.
That is, problem (1.5) has a unique mild solution.
4. Further existence results
Theorem 3.2 can be proved if we omit the hypothesis (H1). In that case the
proof is based on the idea of Wang et al. [23].
Theorem 4.1. Assume the conditions (H3)-(H4) hold, the semigroup {S(t), t ≥ 0}
is compact, and f : I × H × H → H is continuous. For u0 ∈ Hα there exists a
constant r > 0 such that
m
X
T 1−α
M {ku0 kα +
Υi } + Cα Nf 0
≤ r,
(4.1)
1−α
i=1
10
P. KUMAR, D. N. PANDEY, D. BAHUGUNA
EJDE-2013/241
where
Nf =
sup
kf (s, u(s), u(w1 (s, u(s))))k,
s∈[0,T0 ],u∈Ω
and u ∈ Ω = {v ∈ PC(Hα ) : kvkPC ≤ r}. Then there exists a mild solution
u ∈ PC(Hα ) of problem (1.5).
Proof. Let us define a map Υ : PC(Hα ) → PC(Hα ), by

Rt
S(t)u0 + 0 S(t − s)f (s, u(s), u(w1 (s, u(s))))ds,






S(t)u0 + S(t − t1 )I1 (u(t−
1 ))


Rt


+ 0 S(t − s)f (s, u(s), u(w1 (s, u(s))))ds,
(Υu)(t) = . . .

Pk



S(t)u0 + i=1 S(t − ti )Ii (u(t−

i ))

R

t


+ 0 S(t − s)f (s, u(s), u(w1 (s, u(s))))ds,


k = 1, 2, . . . , m.
t ∈ [0, t1 ],
t ∈ (t1 , t2 ],
t ∈ (tk , tk+1 ],
Step 1. First we show that Υ is continuous. It follows from the continuity of f
that
kf (s, un (s), un (w1 (s, un (s)))) − f (s, u(s), u(w1 (s, u(s))))k ≤ ,
as n → ∞,
for s ∈ [0, t], t ∈ [0, T0 ].
Now, for each t ∈ [0, t1 ], we have
k(Υun )(t) − (Υu)(t)kα ≤ Cα
T01−α
→ 0,
1−α
as n → ∞.
(4.2)
For, t ∈ (t1 , t2 ], we have
k(Υun )(t) − (Υu)(t)kα
T01−α
→ 0,
1−α
Similarly, for each t ∈ (tk , tk+1 ], k = 1, 2, . . . , m,
−
≤ M kI1 (un (t−
1 )) − I1 (u(t1 ))kα + Cα
as n → ∞.
(4.3)
k(Υun )(t) − (Υu)(t)kα
≤M
k
X
−
kIi (un (t−
i )) − Ii (u(ti ))kα + Cα
i=1
T01−α
→ 0,
1−α
as n → ∞.
(4.4)
We have that Υ is continuous.
Step 2. Next we show that Υ maps bounded sets into bounded sets in PC(Hα ).
Let u ∈ Ω, then for t ∈ [0, t1 ], we have
k(Υu)(t)kα ≤ M ku0 kα + Nf Cα
T01−α
,
1−α
(4.5)
For, t ∈ (t1 , t2 ], we have
T 1−α
k(Υu)(t)kα ≤ M ku0 kα + Υ1 + Nf Cα 0 .
1−α
Similarly, for each t ∈ (tk , tk+1 ], k = 1, 2, . . . , m, we have
k
n
o
X
T 1−α
k(Υu)(t)kα ≤ M ku0 kα +
Υ i + Nf C α 0 .
1−α
i=1
(4.6)
(4.7)
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11
Thus, from inequality (4.1), we see that Υ : Ω → Ω.
Step 3. In this step, we show that Υ maps bounded sets into equicontinuous sets
in PC(Hα ). Let τ1 , τ2 ∈ [0, t1 ], τ1 < τ2 , we have
k(Υu)(τ2 ) − (Υu)(τ1 )kα
n
o
T 1−α
≤ M ku0 kα + Nf Cα 0
+ kAα−1 kNf (τ2 − τ1 ).
1−α
(4.8)
Similarly, for each τ1 , τ2 ∈ (tk , tk+1 ], τ1 < τ2 , k = 1, 2, . . . , m, we have
k(Υu)(τ2 ) − (Υu)(τ1 )kα
k
o
n
X
T 1−α
+ kAα−1 kNf (τ2 − τ1 ).
≤ M ku0 kα +
Υi + Nf Cα 0
1−α
i=1
(4.9)
As τ2 → τ1 the right-hand side of the above inequality tends to zero. So, Υ(Ω) is
equicontinuous.
Step 4. Υ maps Ω into a compact set in Hα . For this purpose, we decompose Υ
as Υ = Υ1 + Υ2 , where
(Υ1 u)(t) = S(t)u0 + (T1 u)(t),
where (T1 u)(t) =
Rt
0
t ∈ I\{t1 , . . . , tm },
S(t − s)f (s, u(s), u(w1 (s, u(s))))ds. and
(
0,
(Υ2 u)(t) = Pk
i=1
S(t −
ti )Ii (u(t−
i )),
t ∈ [0, t1 ],
t ∈ (tk , tk+1 ], k = 1, 2, . . . , m.
Υ2 is a constant map and hence compact.
Finally, we need to prove that for 0 ≤ t ≤ T0 , (Υ1 u)(t) is relatively compact in
Ω. For each t ∈ [0, T0 ], the set {S(t)u0 } is precompact in Hα since {S(t), t ≥ 0} is
compact.
For t ∈ (0, T0 ], and > 0 sufficiently small, we define
(T1 u)(t) = S()
Z
t−
S(t − − s)f (s, u(s), u(w1 (s, u(s))))ds,
u ∈ Ω.
0
The set {(T1 u)(t) : u ∈ Ω} is precompact in Hα since S() is compact. Moreover,
for any u ∈ Ω, we have
k(T1 u)(t) − (T1 u)(t)kα ≤
Z
t
kAα S(t − s)kkf (s, u(s), u(w1 (s, u(s))))kds
t−
≤ (Cα Nf )1−α .
Therefore, {(T1 u)(t) : u ∈ Ω} is arbitrarily close to the set {(T1 u)(t) : u ∈ Ω}, t > 0.
Hence the set {(Υ1 u)(t) : u ∈ Ω} is precompact in Hα .
Thus, Υ1 is a compact operator by Arzela-Ascoli theorem, and hence Υ is a
compact operator. Then Schauder fixed point theorem ensures that Υ has a fixed
point, which gives rise to a piecewise continuous-mild solution.
12
P. KUMAR, D. N. PANDEY, D. BAHUGUNA
EJDE-2013/241
5. Applications
We consider the following semi-linear heat equation with a deviating argument
(See also [8]).
∂u
∂2u
+ H̃(x, u(x, t)) + G(t, x, u(x, t)),
=
∂t
∂x2
1
1
x ∈ (0, 1), t ∈ (0, ) ∪ ( , 1),
2
2
(5.1)
2u( 12 )−
∆u|t= 12 =
2 + u( 12 )−
u(0, t) = u(1, t) = 0,
u(x, 0) = u0 (x),
x ∈ (0, 1),
where
Z
x
H̃(x, u(x, t)) =
K(x, y)u(y, P (t))dy,
0
P (t) = g1 (t)u x, g2 (t)|u(x, . . . gm (t)|u(x, t)|)| ,
and the function G : R+ × [0, 1] × R → R is measurable in x, locally Hölder
continuous in t, locally Lipschitz continuous in u, uniformly in x. Assume that gi :
R+ → R+ are locally Hölder continuous in t with gi (0) = 0 for each i = 1, 2, . . . , m,
and K ∈ C 1 ([0, 1] × [0, 1]; R). Let
X = L2 ((0, 1); R),
Au =
X1/2 = D((A)1/2 ) = H01 (0, 1),
d2 u
,
dx2
D(A) = H 2 (0, 1) ∩ H01 (0, 1),
X−1/2 = (H01 (0, 1))∗ = H −1 (0, 1) ≡ H 1 (0, 1).
For x ∈ (0, 1), we define the function f : R+ × X1/2 × X−1/2 → X by
f (t, u, ψ)(x) = H̃(x, ψ) + G(t, x, u),
where H̃ : [0, 1] × X−1/2 → X is given by
Z x
H̃(x, ψ) =
K(x, y)ψ(y)dy,
(5.2)
(5.3)
0
and G : R+ × [0, 1] × X1/2 → X satisfies
kG(t, x, u)k ≤ Q(x, t)(1 + kuk1/2 ),
(5.4)
with Q(., t) ∈ X and Q continuous in its second arguments.
For u ∈ D(A) and λ ∈ R with −Au = λu, we have
h−Au, ui = hλu, ui,
ku0 kL2 = λkukL2 ;
so we have λ > 0. The solution u of −Au = λu is
√
√
u(x) = D1 cos( λx) + D2 sin( λx).
Using the boundary condition, we get D1 = 0 and λ = λn = n2 π 2 for n ∈ N.
Thus, for n ∈ N, we have
p
un (x) = D2 sin( λn x).
Also hun , um i = 0, m 6= n and hun , un i = P
1. So, for u ∈ D(A)
P there exists a
sequence αn of real numbers such that u(x) = n∈N αn un (x) with n∈N (αn )2 < ∞
P
and n∈N (αn )2 (λn )2 < ∞.
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13
The semigroup is given by
S(t)u =
X
exp(−n2 t)hu, um ium .
n∈N
Let T > 0 be fixed. Now we will show that assumptions (H1) and (H2) are
satisfied. For (H1), we will show that H̃ : [0, 1] × X−1/2 → X, where
Z x
H̃(x, ψ(x, t)) =
K(x, y)ψ(y, t)dy,
0
with ψ(x, t) = u(x, w1 (t, u(x, t))). For x ∈ [0, 1], we have
Z x
kH̃(x, ψ1 (x, ·)) − H̃(x, ψ2 (x, ·))k ≤
|K(x, y)|k(ψ1 − ψ2 )(y, ·)kdy
0
Z x
≤ kKk∞
|(ψ1 − ψ2 )(y, ·)|dy.
0
Thus for ψ1 , ψ2 ∈ H 1 (0, 1) and by applying the Minkowski’s integral inequality we
obtain
Z 1Z y
kH̃(x, ψ1 (x, ·)) − H̃(x, ψ2 (x, ·))k2 ≤ kKk2∞
|(ψ1 − ψ2 )(y, ·)|2 dxdy
0
≤ kKk2∞
Z
0
1
y|(ψ1 − ψ2 )(y, ·)|2 dy
0
≤ kKk2∞ kψ1 − ψ2 k2−1/2 .
Since,
Z x
∂
∂K
H̃(x, ψ(x, ·)) = K(x, x)ψ(x, ·) +
(x, y)ψ(y, ·),
∂x
0 ∂x
we obtain, in similar way, that
∂
∂
∂K
k H(x, ψ1 (x, ·)) −
H(x, ψ2 (x, ·))k ≤ (kKk∞ + ||
k∞ )kψ1 − ψ2 k−1/2 .
∂x
∂x
∂x
Thus
∂K
kH̃(x, ψ1 (x, ·)) − H̃(x, ψ2 (x, ·))k ≤ (2kKk∞ + k
k∞ )kψ1 − ψ2 k−1/2
∂x
∂K
≤ (2kKk∞ + k
k∞ )kψ1 − ψ2 k−1/2
∂x
= C1 kψ1 − ψ2 k−1/2 ,
where C1 = (2kKk∞ + k ∂K
∂x k∞ ). Again the assumption on G implies that there
exist constants C2 > 0 and 0 < γ ≤ 1 such that
kG(t, x, u) − G(s, x, v)kH01 (0,1) ≤ C2 (|t − s|γ + ku − vk1/2 ),
for t, s ∈ [0, T ], x ∈ (0, 1) and u, v ∈ H01 (0, 1). Thus f : [0, T ]×H01 (0, 1)×H 1 (0, 1) →
L2 (0, 1) defined by f = H̃ + G satisfies the assumption (H1).
Our next aim is to prove that the functions hi : [0, T ] × X−1/2 → [0, T ] defined
by hi (t, φ) = gi (t)|φ(x, ·)| for each i = 1, 2, . . . , m, satisfies (H2). For t ∈ [0, T ], we
obtain
|hi (t, φ)| = |gi (t)| |φ(x, ·)| ≤ kgi k∞ kφkL∞ (0,1) ≤ Ckφk−1/2 ,
where we used the embedding H01 (0, 1) ⊂ C[0, 1] in the last inequity, and C is a
constant depending on the bounds of gi ’s. Since gi ’s are locally Hölder continuous,
14
P. KUMAR, D. N. PANDEY, D. BAHUGUNA
EJDE-2013/241
for some constant Lgi > 0 and 0 < θ ≤ 1 we have |gi (t) − gi (s)| ≤ Lgi |t − s|θ for
t, s ∈ [0, T ]. Moreover t, s ∈ [0, T ] and φ1 , φ2 ∈ X−1/2 , we have
|hi (t, φ1 ) − hi (t, φ2 )| = |gi (t)(|φ1 (x, ·)| − |φ2 (x, ·)|) + (gi (t) − gi (s))|φ2 (x, ·)|
≤ kgk∞ kφ1 − φ2 kL∞ (0,1) + Lgi |t − s|θ kφ2 kL∞ (0,1)
≤ Ckgk∞ kφ1 − φ2 k−1/2 + Lgi |t − s|θ kφ2 k−1/2
≤ max{Ckgk∞ , Lgi kφ2 k∞ }(kφ1 − φ2 k−1/2 + |t − s|θ ).
If u, v ∈ D((−A)1/2 ), then
kIk (u) − Ik (v)k1/2 ≤
2ku − vk1/2
1
≤ ku − vk1/2 .
k(2 + u)(2 + v)k1/2
2
Next for Theorem (4.1) we define
f (t, u(t), ψ(t)))(x) =
e−t (cos(u(x, t)) + sin(ψ(x, t)))
+ e−t ,
(2 + t2 )(et + e−t )
for t ∈ [0, 1/2) ∪ (1/2, 1], u ∈ X, and x ∈ (0, 1). Clearly,
kf (t, u, ψ)k ≤
e−t
(2 +
t2 )(et
+ e−t )
+ e−t = m(t),
with m(t) ∈ L∞ ([0, 1]; R+ ). Thus, we can apply all the results of this section to
obtain the main results.
Acknowledgements. We highly appreciate the valuable comments and suggestions of the referees on our manuscript which helped to considerably improve the
quality of the manuscript. The third author would like to acknowledge the financial
aid from the Department of Science and Technology, New Delhi, under its research
project SR/S4/MS:796.
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Pradeep Kumar
Department of Mathematics and Statistics, Indian Institute of Technology Kanpur,
Pin 208016, India
E-mail address: prdipk@gmail.com
Dwijendra N. Pandey
Department of Mathematics, Indian Institute of Technology Roorkee, Pin 247667, India
E-mail address: dwij.iitk@gmail.com
Dhirendra Bahuguna
Department of Mathematics and Statistics, Indian Institute of Technology Kanpur,
Pin 208016, India
E-mail address: dhiren@iitk.ac.in, Tel. +91-512-2597053, Fax +91-512-2597500