SOME EXISTENCE AND UNIQUENESS RESULTS FOR
FIRST-ORDER BOUNDARY VALUE PROBLEMS FOR
IMPULSIVE FUNCTIONAL DIFFERENTIAL EQUATIONS
WITH INFINITE DELAY IN FRÉCHET SPACES
JOHN R. GRAEF AND ABDELGHANI OUAHAB
Received 31 January 2006; Revised 19 May 2006; Accepted 28 May 2006
A recent nonlinear alternative for contraction maps in Fréchet spaces due to Frigon and
Granas is used to investigate the existence and uniqueness of solutions to first-order
boundary value problems for impulsive functional differential equations with infinite delay. An example to illustrate the results is included.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction
This paper is devoted to the study of the existence and uniqueness of solutions of firstorder boundary value problems for impulsive functional and neutral impulsive functional differential equations with infinite delay. In particular, in Section 3, we will consider the class of first-order boundary value problem for functional differential equations
with impulsive effects,
y (t) = f t, yt
+
−
a.e. t ∈ J := [0, ∞), t = tk , k = 1,2,...,
− y tk − y tk = Ik y tk
Ay0 − x∞ = φ(t),
,
t = tk , k = 1,2,...,
t ∈ (−∞,0],
(1.1)
(1.2)
(1.3)
where f : J × Ꮾ → Rn and Ik : Rn → Rn , k = 1,2,..., are given functions, limt→∞ y(t) = x∞ ,
A = 1, and φ ∈ Ꮾ, which is called the phase space, will be defined later. Section 4 is devoted to the impulsive neutral functional differential equation with boundary conditions,
d
y(t) − g t, yt = f t, yt , t ∈ J, t = tk , k = 1,2,...,
dt
y tk+ − y tk− = Ik y tk− , t = tk , k = 1,2,...,
Ay0 − x∞ = φ(t),
(1.4)
t ∈ (−∞,0],
where f , Ik , A, x∞ , and Ꮾ are as in the problem (1.1)–(1.3) and g : J × Ꮾ → Rn is a given
function.
Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2006, Article ID 31256, Pages 1–16
DOI 10.1155/IJMMS/2006/31256
2
Impulsive functional differential equations
The notion of the phase space Ꮾ plays an important role in the study of both the
qualitative and quantitative theory of functional differential equations. A usual choice is
a seminormed space satisfying suitable axioms as was introduced by Hale and Kato [18]
(also see Kappel and Schappacher [21] and Schumacher [28]). For a detailed discussion
on this topic, we refer the reader to Hino et al. [20]. For the case where the impulses are
absent (i.e., Ik = 0, k = 1,2,...), an extensive theory has been developed for the problem
(1.1), (1.3), and we refer the reader to Corduneanu and Lakshmikantham [8], Hale and
Kato [18], Hino et al. [20], Lakshmikantham et al. [23], and Shin [29].
Impulsive differential equations have become increasingly important in recent years as
mathematical models of real-world processes and phenomena studied in control theory,
physics, chemistry, population dynamics, biotechnology, and economics. There has been
a significant development in impulse theory and this has been especially true in the area
of impulsive differential equations with fixed moments; see, for example, the monographs
of Baı̆nov and Simeonov [3], Lakshmikantham et al. [22], and Samoı̆lenko and Perestyuk
[27], as well as the papers of Agur et al. [2], Ballinger and Liu [4], Benchohra et al. [5, 6],
Franco et al. [9], and the references contained therein.
Boundary value problems on infinite intervals appear in many problems of practical
interest, for example, in linear elasticity, nonlinear fluid flow, and foundation engineering (e.g., see, [1, 13, 16, 24–26]). Recently, fixed point arguments using such approaches
as the Banach contraction principle, fixed point index theory, and monotone iterative
technique have been applied to first- and second-order impulsive differential equations.
We mention here the survey papers by Guo [12, 14, 15], Guo and Liu [17], Yan and Liu
[30], and the references therein. Very recently, a nonlinear alternative due to Frigon and
Granas [10] was applied to impulsive functional differential equation with variable times
[11]; see the paper by Benchohra et al. [7] for first-order equations and Henderson and
Ouahab [19] for second- and higher-order problems. Our goal here is to give existence
and uniqueness results for the problems (1.1)–(1.3) and (1.4) above by using this nonlinear alternative for contraction maps.
2. Preliminaries
In this short section, we introduce notations and definitions that are used throughout the
remainder of this paper. We let C([0,b], Rn ) denote the Banach space of all continuous
functions from [0, b] into Rn with the norm
y ∞ = sup y(t) : 0 ≤ t ≤ b ,
(2.1)
and we let L1 ([0, ∞), Rn ) be the Banach space of measurable functions y : [0, ∞) → Rn
that are Lebesgue integrable with the norm
y L 1 =
∞
0
y(t)dt
∀ y ∈ L1 [0, ∞), Rn .
(2.2)
For more details on the following notions, we refer the reader to Frigon and Granas [10].
Let X be a Fréchet space with a family of seminorms { · n , n ∈ N}. If Y ⊂ X, we say
J. R. Graef and A. Ouahab 3
that Y is bounded if for every n ∈ N there exists Mn > 0 such that
y n ≤ Mn
∀ y ∈ Y.
(2.3)
We associate to X a sequence of Banach spaces {(X n , · n )} as follows: for every n ∈ N,
we consider the equivalence relation ∼n defined by x ∼n y if and only if x − y n = 0 for
all x, y ∈ X. We let X n = (X/ ∼n , · ) denote the quotient space that is the completion
of X n with respect to · n . To every Y ⊂ X, we associate a sequence {Y n } of subsets
Y n ⊂ X n as follows: for every x ∈ X, let [x]n denote the equivalence class of x of subsets
n
X n , and we define Y n = {[x]n : x ∈ Y }. We let Y , intn (Y n ), and ∂n Y n , respectively, denote
the closure, the interior, and the boundary of Y n with respect to · in X n . We will
assume that the family of seminorms { · n } satisfies
x 1 ≤ x 2 ≤ x 3 ≤ · · ·
for every x ∈ X.
(2.4)
Next, we define what we mean by a contraction.
Definition 2.1. A function f : X → X is said to be a contraction if for each n ∈ N there
exists kn ∈ (0,1) such that
f (x) − f (y) ≤ kn x − y n
n
∀x, y ∈ X.
(2.5)
The following nonlinear alternative will be used to prove our main results.
Theorem 2.2 (nonlinear alternative [10]). Let X be a Fréchet space, let Y ⊂ X be a closed
subset in X, and let N : Y → X be a contraction such that N(Y ) is bounded. Then one of the
following statements holds:
(C1) N has a unique fixed point;
(C2) there exists λ ∈ [0,1), n ∈ N, and x ∈ ∂n Y n such that x − λN(x)n = 0.
We will also need the following definition.
Definition 2.3. The map f : [0, ∞) × Ꮾ → Rn is said to be L1 -Carathéodory if
(i) t → f (t,x) is measurable for each x ∈ Ꮾ;
(ii) x → f (t,x) is continuous for almost all t ∈ [0, ∞);
(iii) for each q > 0 there exists hq ∈ L1 ([0, ∞), R+ ) such that
f (t,x) ≤ hq (t)
∀xᏮ ≤ q and for almost all t ∈ [0, ∞).
(2.6)
3. Functional differential equations
In order to define the phase space and a solution of the problem (1.1)–(1.3), we consider
the space
PC = y : (−∞, ∞) −→ Rn | y tk− and y tk+ exist with y tk = y tk− ,
Ay(t) − x∞ = φ(t) for t ≤ 0, yk ∈ C Jk , Rn , k = 1,2,... ,
(3.1)
where yk is the restriction of y to Jk = (tk ,tk+1 ], k = 1,2,.... In the remainder of this paper,
we will assume that Ꮾ satisfies the following properties.
4
Impulsive functional differential equations
(A-1) If y : (−∞, ∞) → Rn and y0 ∈ Ꮾ, then for every t in [0, ∞), we have
(i) yt is in Ꮾ,
(ii) yt Ꮾ ≤ K(t)sup{| y(s)| : 0 ≤ s ≤ t } + M(t) y0 Ꮾ , and
(iii) | y(t)| ≤ H yt Ꮾ ,
where H ≥ 0 is a constant, K : [0, ∞) → [0, ∞) is continuous, M : [0, ∞) → [0, ∞) is locally
bounded, and H, K, and M are independent of y.
(A-2) For the function y in (A-1), yt is a continuous function on [0, ∞) \ {t1 ,t2 ,...}.
(A-3) The space Ꮾ is complete.
Now set
B∗ = y : (−∞, ∞) −→ Rn : y ∈ PC ∩Ꮾ ,
where Jk∗ = − ∞,tk .
Bk = y ∈ B∗ : sup y(t) < ∞ ,
∗
t ∈J k
(3.2)
Let · k be the seminorm in Bk defined by
y k = y0 Ꮾ + sup y(s) : 0 ≤ s ≤ tk ,
y ∈ Bk .
(3.3)
We will next define what we mean by a solution of (1.1)–(1.3).
Definition 3.1. A function y is said to be a solution of the problem (1.1)–(1.3) if y ∈ B∗
and y satisfies (1.1)–(1.3).
To prove our existence results for the problem (1.1)–(1.3), we first establish the following lemma.
Lemma 3.2. Let f : [0, ∞) → Rn be a continuous function with
solution of the impulsive integral equation
⎧
φ(0)
⎪
1
⎪
⎪
+
⎪
⎪
⎪
⎪
⎪ A(A − 1) A − 1
⎪
⎪
⎪
⎨
∞
y(t) = ⎪ φ(0) + 1
⎪
⎪
⎪A−1 A−1
⎪
⎪
⎪
⎪
⎪
⎪
⎪
Ik y tk− ,
⎩+
0
∞
f ys ds +
0
∞
− Ik y tk
k =1
f ys ds +
∞
Ik y tk−
k =1
∞
0
f (s)ds < ∞. Then y is a
+
φ(t)
,
A
t
+
0
t ∈ (−∞,0],
f ys ds
t ∈ [0, ∞),
0<tk <t
(3.4)
if and only if y is a solution of the impulsive boundary value problem
y (t) = f yt ,
+
−
t ∈ [0, ∞), t = tk , k = 1,2,...,
− y tk − y tk = Ik y tk
Ay0 − y∞ = φ(t),
where limt→∞ y(t) = y∞ .
,
k = 1,2,...,
t ∈ (−∞,0],
(3.5)
(3.6)
(3.7)
J. R. Graef and A. Ouahab 5
Proof. Let y be a solution of the impulsive integral equation (3.4). Then for t ∈ [0, ∞)
and t = tk , k = 1,2,..., we have
φ(0)
1
+
y(t) =
A−1 A−1
∞
∞
f ys ds +
0
− Ik y tk
0
k =1
t
+
f ys ds +
Ik y tk− .
0<tk <t
(3.8)
Thus, y (t) = f (yt ) for t ∈ [0, ∞) and t = tk , k = 1,2,.... From the definition of y, we see
that
y tk+ − y tk− = Ik y tk−
for k = 1,2,....
(3.9)
Finally, to see that Ay0 − y∞ = φ(t) for t ∈ (−∞,0], note that
y∞ =
φ(0)
A
+
A−1 A−1
φ(0)
1
y0 =
+
A(A − 1) A − 1
∞
0
∞
Ik y tk−
k =1
f ys ds +
0
∞
f ys ds +
∞
− Ik y tk
k =1
,
φ(t)
+
.
A
(3.10)
Hence,
Ay0 − y∞ =
φ(0)
A
+
A−1 A−1
−
A
A−1
0
∞
0
∞
f ys ds +
∞
Ik y tk−
k =1
f ys ds +
∞
+ φ(t) −
φ(0)
A−1
(3.11)
− = φ(t).
Ik y tk
k =1
Now let y be a solution of the problem (3.5)–(3.7). Then,
y (t) = f yt
for t ∈ 0,t1 ,
(3.12)
and an integration from 0 to t ∈ (0,t1 ] yields
t
y(t) − y(0) =
0
t
y (s)ds =
0
f ys ds
(3.13)
or
y(t) = y(0) +
t
0
f ys ds
(3.14)
for t ∈ [0,t1 ]. If t ∈ (t1 ,t2 ], then
y t1− − y(0) + y(t) − y t1+ =
t1
0
y (s)ds +
t
t1
y (s)ds =
t
0
f ys ds,
(3.15)
6
Impulsive functional differential equations
so
t
y(t) = y(0) +
f ys ds + I1 y t1− .
0
(3.16)
Continuing this procedure, we obtain
y(t) = y(0) +
t
f ys ds +
0
Ik y tk−
(3.17)
0<tk <t
for t ∈ [0, ∞). Since limt→∞ y(t) = y∞ , we have
y∞ = y(0) +
∞
∞
f ys ds +
0
Ik y tk− .
(3.18)
k =1
From (3.7), we have y∞ = Ay(0) − φ(0), and so
y(0) =
φ(0)
1
+
A−1 A−1
∞
0
f ys ds +
∞
Ik y tk−
.
(3.19)
k =1
Substituting (3.19) into (3.17), we obtain
y(t) =
φ(0)
1
+
A−1 A−1
∞
0
f ys ds +
∞
Ik y tk−
t
+
0
k =1
f ys ds +
Ik y tk−
0<tk <t
(3.20)
for t ∈ [0, ∞).
From (3.7), (3.18), and (3.19), we have
φ(t) 1
y(0) +
y(t) =
+
A
A
φ(0)
1
+
=
A(A − 1) A − 1
∞
f (ys )ds +
0
∞
− Ik y tk
k =1
∞
0
f ys ds +
∞
− Ik y tk
k =1
φ(t)
+
A
for t ∈ (−∞,0]. This completes the proof of the lemma.
(3.21)
We are now ready to prove our existence and uniqueness result for the problem (1.1)–
(1.3).
Theorem 3.3. Let f : J × Ꮾ → Rn be an L1 -Carathéodory function and assume that
(H1) there exist constants dk > 0, k = 1,2,..., such that
Ik (x) − Ik (x) ≤ dk x − x
for each k = 1,2,..., ∀x,x ∈ Rn ;
(3.22)
(H2) there exists a function l ∈ L1 ([0, ∞), Rn ) such that
f (t,x) − f (t,x) ≤ l(t)x − xᏮ
for each t ∈ [0, ∞), ∀x,x ∈ Ꮾ;
(3.23)
J. R. Graef and A. Ouahab 7
(H3) there exist a function p ∈ L1 ([0, ∞), R+ ) and positive constants ck , k = 1,2,...,
such that
f (t,u) ≤ p(t) for (t,u) ∈ [0, ∞) × Ꮾ,
Ik (z) ≤ ck ∀z ∈ Rn ,
∞
If
(3.24)
ck < ∞.
k =1
∞
k=1 dk
< 1, then the problem (1.1)–(1.3) has a unique solution.
Proof. We transform the problem (1.1)–(1.3) into a fixed point problem. Consider the
operator N : B∗ → B∗ defined by
⎧
∞
∞ ⎪
− φ(0)
φ(t)
1
⎪
⎪
⎪
f s, ys ds + Ik y tk
+
+
,
⎪
⎪
A
⎪
⎪ A(A − 1) A − 1 0
k =1
⎪
⎪
⎪
⎪
∞
⎨ φ(0)
∞ 1
N(y)(t) = ⎪
f s, ys ds + Ik y tk−
+
A−1 A−1 0
⎪
⎪
k =1
⎪
⎪
⎪
⎪
t ⎪
⎪
⎪
⎪+ f s, ys ds +
Ik y tk− ,
⎪
⎩ 0
t ∈ (−∞,0],
t ∈ [0, ∞).
0<tk <t
(3.25)
It should be clear from Lemma 3.2 that the fixed points of N are solutions of the problem
(1.1)–(1.3). Let x : (−∞, ∞) → Rn be the function defined by
⎧
⎪
φ(0)
1
⎪
⎪
⎪
+
⎪
⎪
⎨A−1 A−1
x(t) = ⎪
∞
0
⎪
φ(0)
1
⎪
⎪
⎪
⎪
⎩ A(A − 1) + A − 1
f s,xs ds +
∞
Ik x tk−
if t ∈ [0, ∞),
k =1
∞
0
f s,xs ds +
∞
− Ik x tk
+
k =1
φ(t)
A
if t ∈ (−∞,0].
(3.26)
Then,
x0 =
φ(0)
1
+
A−1 A−1
∞
0
f s,xs ds +
∞
Ik x tk−
.
(3.27)
k =1
For each z ∈ C([0, ∞), Rn ) with z(0) = 0, denote by z̄ the function given by
⎧
⎨z(t)
z̄(t) = ⎩
0
if t ∈ [0, ∞),
(3.28)
if t ∈ (−∞,0].
If y satisfies the integral equation
φ(0)
1
+
y(t) =
A−1 A−1
∞
0
f s, ys ds +
∞
k =1
− Ik y tk
t
+
0
f s, ys ds +
Ik y tk− ,
0<tk <t
(3.29)
8
Impulsive functional differential equations
we can decompose y into y(t) = z̄(t) + x(t), 0 ≤ t < ∞, which implies that yt = z̄t + xt for
0 ≤ t < ∞, and z satisfies
z(t) =
t
f s, z̄s + xs ds +
0
Ik z̄ tk− + x tk− .
(3.30)
0<tk <t
Let
B∗k = z ∈ Bk : z0 = 0 .
(3.31)
For any z ∈ B∗k , we have
zk = z0 Ꮾ + sup z(s) : 0 ≤ s ≤ tk = sup z(s) : 0 ≤ s ≤ tk .
(3.32)
Thus (B∗k , · k ) is a Banach space. If we set
C0 = z ∈ B∗ : z0 = 0
(3.33)
with the Bielecki-type norm on B∗k defined by
zB∗k = max z(t)e−τ l(t) : t ∈ 0,tk ,
(3.34)
where l(t) = 0 l(s)ds, l(t) = Kk l(t), Kk = sup{|K(t)| : t ∈ [0,tk ]}, k = 1,2,..., and τ > 0 is
a constant, then C0 is a Fréchet space with the family of seminorms · B∗k .
Let the operator P : C0 → C0 be defined by
t
⎧
⎪
⎪
⎨0,
(Pz)(t) = ⎪
⎪
⎩
t
0
t ≤ 0,
f s, z̄s + xs ds +
Ik z̄ tk− + x tk− ,
t ∈ [0, ∞).
(3.35)
0<tk <t
Clearly, the operator N having a fixed point is equivalent to the operator P having one,
and so we turn our attention to proving that P does have a fixed point. We will use the
nonlinear alternative given in Theorem 2.2 to show this.
Let z be a possible solution of the problem z = λP(z) for some 0 < λ < 1. Then, for each
t ∈ [0, ∞),
z(t) = λ
t
0
f s, z̄s + xs ds +
Ik z̄ tk− + x tk−
,
(3.36)
ci := Mk∗ .
(3.37)
0<tk <t
so by (H3), we have
z(t) <
t
0
p(s)ds +
i
=k
Ii z̄ ti + x ti
|≤
i=1
tk
0
p(s)ds +
i
=k
i=1
Hence,
sup z(t) : t ∈ 0,tk
< Mk∗ .
(3.38)
J. R. Graef and A. Ouahab 9
Let
Y = z ∈ C0 | sup zB∗k : 0 ≤ t ≤ tk ≤ Mk∗ ∀k = 1,2,... .
(3.39)
Clearly, Y is a closed subset of C0 . We will show that P : Y → B∗k is a contraction map. To
see this, consider z, z∗ ∈ Y ; then for each t ∈ [0,tk ] and k = 1,2,..., we have
P(z)(t) − P z∗ (t) ≤
t
0
+
f s, z̄s + xs − f s, z̄∗ + xs ds
s
i
=k
−
Ii z̄ t + x t − − Ii z̄∗ t − + x t − i
i
i
i
i =1
≤
≤
≤
t
0
t
0
t
0
+
l(s)z̄s − z̄s∗ Ꮾ ds +
=
≤
0
1
τ
i =1
s∈[0,t]
i
=k
di z ti− − z∗ ti− i =1
l(s)eτ l(s) e−τ l(s) sup z(s) − z∗ (s)ds
s∈[0,t]
i
=k
t
di z ti− − z∗ ti− l(s)Kk sup z(s) − z∗ (s)ds +
di eτ l(t) e−τ l(t) sup z(s) − z∗ (s)
s∈[0,tk ]
i =1
=
i
=k
l(s)eτ l(s) dsz − z∗ t
0
B∗ +
i
=k
k
i=1
eτ l(s) dsz − z∗ B∗k +
di eτ l(t) z − z∗ B∗k
i
=k
i =1
di eτ l(t) z − z∗ B∗k
i
=k
1 τl(t) e dsz − z∗ B∗k + di eτ l(t) z − z∗ B∗k .
τ
i =1
(3.40)
Thus,
i
=k
i =k
1
1 + di z − z∗ B∗k .
e−τ l(t) P(z)(t) − P z∗ (t) ≤ z − z∗ B∗k + di z − z∗ B∗k =
τ
τ i=1
i =1
(3.41)
Therefore,
P(z) − P z∗ i =k
1 + di z − z∗ B∗k ,
B∗k ≤
τ i =1
(3.42)
which shows that if τ is large enough, then P is a contraction. From the choice of Y , there
is no z ∈ ∂Y n such that z = λP(z) for some λ ∈ (0,1). As a consequence of Theorem 2.2,
10
Impulsive functional differential equations
we conclude that P has a unique fixed point. In turn, this implies that the operator N
has a unique fixed point that is a solution to (1.1)–(1.3). This completes the proof of the
theorem.
4. Neutral functional differential equations
This section is concerned with the existence of solutions to the boundary value problem
for first-order neutral functional differential equations with infinite delay and impulses
given in (1.4). Our main result is as follows.
Theorem 4.1. Let f : J × Ꮾ → Rn be an L1 -Carathéodory function. In addition to (H1)–
(H3), assume that M(0) < 1 and
(B) there exist constants c1∗ ,c2∗ ≥ 0, and d¯k > 0, k = 1,2,..., such that
g(t,u) ≤ c∗ uᏮ + c∗ , t ∈ [0, ∞), u ∈ Ꮾ, c∗ Kk < 1,
1
2
1
g(t,u) − g(t,u) ≤ d¯k u − u
Ꮾ for t ∈ 0,tk ,
(4.1)
where
Kk = sup
If
∞
k=1 [dk
K(t) : t ∈ 0,tk ,
k = 1,2,....
(4.2)
+ d¯k Kk ] < 1, then the problem (1.4) has a unique solution.
Proof. We proceed similarly to what we did in the proof of Theorem 3.3 by defining the
operators N1 : B∗ → B∗ and P1 : C0 → C0 by
⎧
φ(0)
φ(t)
⎪
⎪
⎪
g 0,φ(0) − g t,φ(t) +
+
⎪
⎪
A(A
−
1)
A
⎪
⎪
⎪
⎪
⎪
⎪
∞
⎪
∞ ⎪
− 1
⎪
⎪
⎪
f s, ys ds + Ik y tk
,
t ∈ (−∞,0],
+
⎪
⎪
⎨ A−1 0
k =1
N1 (y)(t) = ⎪
∞
∞ ⎪
− 1
⎪
⎪
⎪
f s, ys ds + Ik y tk
g(0,φ) − g t, ys +
⎪
⎪
A−1 0
⎪
⎪
k =1
⎪
⎪
⎪
⎪
t ⎪ φ(0)
⎪
⎪
⎪
Ik y tk− ,
t ∈ [0, ∞),
⎪
⎩+ A − 1 + 0 f s, ys ds +
0<tk <t
⎧
⎪
t ≤ 0,
⎪
⎨0,
t P1 z (t) = ⎪
Ik z̄ tk− + x tk− , t ∈ [0, ∞),
⎪
⎩g(0,φ) − g t,zt +xt + 0 f s,zs + xs ds +
0<tk <t
(4.3)
where zt and xt are defined similar to the way they are in the proof of Theorem 3.3. In
order to apply the nonlinear alternative, Theorem 2.2, we first obtain a priori estimates
for the solutions of the integral equation
z(t) = λ g(0,φ) − g t,zt + xt +
t
0
f s,zs + xs ds +
0<tk <t
−
− Ik z̄ tk + x tk
(4.4)
J. R. Graef and A. Ouahab 11
for any λ ∈ (0,1). We have
z(t) < g 0,φ(0) + g t, z̄t + xt +
≤ g(0,φ) + c1∗ z̄t + xt Ꮾ + c2∗ +
≤ g 0,φ(0) + c1∗ K(t) sup
+ c2∗ +
t
0
p(s)ds +
p(s)ds +
0
k
k
0<tk <t
t
0
p(s)ds +
i
=k
ci
i =1
z(s) + c1∗ K(t) sup x(s) + c1∗ M(t)x0 Ꮾ
s∈[0,t]
k
Ik z̄ t − + x t − t
s∈[0,t]
ci ≤ g 0,φ(0) + c1∗ Kk sup z(s)
s∈[0,t]
i=1
+ Kk c1∗ x∞ + Mk c1∗ x0 Ꮾ + c2∗ +
t
0
p(s)ds +
k
ci ,
i=1
(4.5)
where
Mk = sup M(t) : t ∈ 0,tk .
(4.6)
Hence,
z∞ < g(0,φ) + c1∗ Kk z∞ + Kk c1∗ x∞ + Mk c1∗ x0 Ꮾ + c2∗ +
∞
0
p(s)ds +
k
i=1
ci .
(4.7)
Now,
x0 ≤ K(0) x(0) + M(0)x0 ,
Ꮾ
Ꮾ
(4.8)
so
x0 ≤
Ꮾ
K(0)
x(0) ,
1 − M(0)
(4.9)
and since
x∞ := sup
t ∈[0,tk ]
∞
φ(0)
1
p L 1 +
x(t) ≤
ck := ,
+
A−1 A−1
k =1
(4.10)
we then have
z ∞ <
i
=k
1
g(0,φ) + c∗ + Kk c∗ + Mk c∗ K(0) + pL1 + ci := M k .
2
1
1
1 − c1∗ Kk
1 − M(0)
i =1
(4.11)
12
Impulsive functional differential equations
Now set
Y1 = z ∈ C0 : sup z(t) : 0 ≤ t ≤ tk ≤ M k ∀k = 1,2,... .
(4.12)
Clearly, Y1 is a closed subset of C0 . To show that P1 is a contraction, let z, z∗ ∈ Y1 . Then,
for each t ∈ [0,tk ] and k = 1,2,..., we have
P1 (z)(t) − P1 z∗ (t)
≤ g t, z̄t + xt − g t, z̄t∗ + xt +
+
i
=k
t
0
f s, z̄s + xs − f s, z̄∗ + xs ds
s
−
Ii z̄ t + x t − − Ii z̄∗ t − + x t − ≤ d¯k z̄t − z∗ i
i
i
i
t
i=1
t
i
=k
l(s)z̄s − z̄s∗ Ꮾ ds +
+
0
di z ti− − z∗ ti− ≤ d¯k Kk sup z(t) − z∗ (t)
t ∈[0,tk ]
i =1
l(s)Kk sup z(τ) − z∗ (τ)ds +
t
+
0
τ ∈[0,s]
i
=k
t ∈[0,tk ]
i
=k
i=1
t
+
0
+
1
τ
di z ti− − z∗ ti− i=1
≤ d¯k Kk eτ l(t) e−τ l(t) sup z(t) − z∗ (t) +
+
Ꮾ
t
0
l(s)eτ l(s) e−τ l(s) sup z(τ) − z∗ (τ)ds
τ ∈[0,s]
di eτ l(t) e−τ l(t) sup z(s) − z∗ (s) ≤ d¯k Kk eτ l(t) z − z∗ B∗k
s∈[0,tk ]
l(s)eτ l(s) dsz − z∗ t
B∗ +
i
=k
k
i =1
eτ l(s) dsz − z∗ B∗k +
0
di eτ l(t) z − z∗ B∗k = d¯k Kk eτ l(t) z − z∗ B∗k
i
=k
i =1
di eτ l(t) z − z∗ B∗k ≤ d¯k Kk eτ l(t) z − z∗ B∗k
i
=k
1 + eτ l(t) dsz − z∗ B∗k + di eτ l(t) z − z∗ B∗k .
τ
i=1
(4.13)
Thus,
e−τ l(t) P1 (z)(t) − P1 z∗ (t) ≤
i =k
1 + di + d¯k Kk z − z∗ B∗k .
τ i=1
(4.14)
Therefore,
P1 (z) − P1 z∗ k
B∗
≤
i=k
1 + di + d¯k Kk z − z∗ B∗k ,
τ i =1
(4.15)
J. R. Graef and A. Ouahab 13
and again if τ is large enough, P1 is a contraction. From the choice of Y1 , there is no
z ∈ ∂Y1n such that z = λP1 (z) for some λ ∈ (0,1). As a consequence of the nonlinear alternative (Theorem 2.2), we see that P1 has a unique fixed point which again leads to the
existence of a unique solution to (1.4).
5. Example
In this section we give an example to illustrate the usefulness of our main results. Consider
the problem
y (t) =
e−γt yt
,
(t + 1)(t + 2) 1 + yt
y tk+ − y tk− = bk y tk− ,
y(t) = φ(t),
a.e. t ∈ J := [0, ∞) − t1 ,t2 ,... ,
k = 1,2,...,
(5.1)
t ∈ (−∞,0],
where γ > 0 is a constant and bk > 0 for k = 1,2,.... Let
Ᏸ = ψ : (−∞,0] −→ Rn | ψ is continuous everywhere except for a countable number of
points t̄ at which ψ(t̄ − ), ψ(t̄ + ) ∃,ψ(t̄ − ) = ψ(t̄) .
(5.2)
Let
Bγ = y ∈ Ᏸ : lim eγθ y(θ) exists in Rn
θ −→−∞
(5.3)
with the norm in Bγ given by
y γ = sup eγθ y(θ).
(5.4)
−∞<θ ≤0
Let y : (−∞, ∞) → Rn be such that y0 ∈ Bγ ; then
lim eγθ yt (θ) = lim eγθ y(t + θ) = lim eγ(θ−t) y(θ) = e−γt lim eγθ y0 (θ) < ∞.
θ −→−∞
θ −→−∞
θ −→−∞
θ −→−∞
(5.5)
Hence, yt ∈ Bγ .
Next, we prove that | y(t)| ≤ H yt Bγ and
yt ≤ K(t)sup y(s) : 0 ≤ s ≤ t + M(t) y0 ,
γ
(5.6)
where yt (θ) = y(t + θ), K(t) ≡ M(t) ≡ 1, and H = 1. Now, if θ + t ≤ 0, we have
yt (θ) ≤ sup y(s) : −∞ < s ≤ 0 .
(5.7)
yt (θ) ≤ sup y(s) : 0 < s ≤ t .
(5.8)
For t + θ ≥ 0, we see that
14
Impulsive functional differential equations
Thus, for all t + θ ∈ R, we obtain
yt (θ) ≤ sup y(s) : −∞ < s ≤ 0 + sup y(s) : 0 ≤ s ≤ t .
(5.9)
yt ≤ y0 + sup y(s) : 0 ≤ s ≤ t .
γ
γ
(5.10)
Then,
Now (Bγ , · ) is a Banach space and Bγ will serve as the phase space for our problem.
We see that the function f satisfies
f (t,x) =
e−γt |x|
1
≤
= p(t),
(t + 1)(t + 2)
(t + 1)(t + 2) 1 + |x|
∞
0
p(t)dt =
∞
0
1
dt = ln2.
(t + 1)(t + 2)
(5.11)
Let x, y ∈ Bγ ; then,
f (t,x) − f (t, y) =
≤
e−γt |x| − | y |
| y|
|x|
e−γt
−
=
(t + 1)(t + 2) 1 + |x| 1 + | y |
(t + 1)(t + 2) 1 + |x| 1 + | y |
x − y B γ
e−γt |x − y |
≤
.
(t
+ 1)(t + 2)
(t + 1)(t + 2) 1 + |x| 1 + | y |
(5.12)
Hence, the hypotheses of Theorem 3.3 are satisfied, so if
(5.1) has a unique solution.
∞
k=1 bk
< 1, then the problem
Acknowledgment
The research of J. R. Graef was supported in part by the the Office of Academic and
Research Computing Services of the University of Tennessee at Chattanooga.
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John R. Graef: Department of Mathematics, University of Tennessee at Chattanooga, Chattanooga,
TN 37403, USA
E-mail address: john-graef@utc.edu
Abdelghani Ouahab: Department of Mathematics, University of Sidi Bel Abbes, BP 89,
2000 Sidi Bel Abbes, Algeria
E-mail address: ouahab@univ-sba.dz