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J. Math. Anal. Appl. 301 (2005) 237–248

www.elsevier.com/locate/jmaa

Strict stability of impulsive functional differential

equations✩

Yu Zhang, Jitao Sun∗

Dept. of Appl. Math., Tongji University, 200092, PR China

Received 3 November 2003

Available online 22 September 2004

Submitted by D. O’Regan

Abstract

Strict stability is the kind of stability that can give us some information about the rate of decay

of the solutions. There are some results about strict stability of differential equations. In the present

paper, we shall extend the strict stability to impulsive functional differential equations. By using

Lyapunov functions and Razumikhin technique, we shall get some criteria for the strict stability of

impulsivefunctional differentialequations, and wecan seethatimpulsesdo contributetothesystem’s

strict stability behavior.

2004 Elsevier Inc. All rights reserved.

Keywords: Impulsive functional differential equation; Strict stability; Lyapunov function; Razumikhin technique

1. Introduction

Impulses can make unstable systems stable, so it has been widely used in many fields,

such as physics, chemistry, biology, population dynamics, industrial robotics, and so on.

The impulsive differential equations represents a more natural framework for mathemati-

cal modelling of many real word phenomena than differential equations. In recent years,

✩This work was supported by the National Natural Science Foundation of China (60474008) and the Natural

Science Foundation of Shanghai City, China (03ZR14095).

*Corresponding author. Fax: +86-21-65982341.

E-mail address: sunjt@sh163.net (J. Sun).

0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved.

doi:10.1016/j.jmaa.2004.07.018

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Y. Zhang, J. Sun / J. Math. Anal. Appl. 301 (2005) 237–248

significant progress has been made in the theory of impulsive differential equations [3–13]

and references therein.

Strict stability is analogous to Lyapunov’s uniform asymptotic stability. It can give us

some information about the rate of decay of the solutions. In [1], the authors have further

the definitions of strict stability for differential equations, and have gotten some results.

In [2], the authors have gotten some results about the strict practical stability of differential

equations. But there does not exist impulses in these systems. In book [4], the author has

given us many results about impulsive systems without time delay. But time delay systems

are frequently encountered in engineering, biology, economy, and other disciplines, so it

is necessary to study these systems with time delay. And there are some results about

impulsive functional differential equations and impulsive differential equations with time

delay [3,5–7,9,12]. In the present paper, we shall consider the strict stability of impulsive

functional differential equations. We can see that impulses do contribute to the system’s

strict stability behavior.

This paper is organized as follows. In Section 2, we introduce some basic definitions

and notations. In Section 3, we get some criteria for strict stability of impulsive functional

differential equations. Finally, concluding remarks are given in Section 4.

2. Preliminaries

Consider the following impulsive functional differential equations:

˙ x(t) = f(t,xt),

∆x(t) ? x(τk) −x?τ−

where f ∈ C[R+× D,Rn], D is an open set in PC([−τ,0],Rn), where τ > 0 and

PC([−τ,0],Rn) = {φ:[−τ,0]→ Rn, φ(t) is continuous everywhere except a finite num-

ber of points ˆ t at which φ(ˆ t+) and φ(ˆ t−) exist and φ(ˆ t+) = φ(ˆ t )}. f(t,0) = 0, for all

t ∈ R, Ik(0) = 0, for all k ∈ Z, 0 = τ0< τ1< τ2< ··· < τk< ···, τk→ ∞ for

k → ∞, and x(t+) = lims→t+ x(s), x(t−) = lims→t− x(s). For each t ? t0, xt∈ D is

defined by xt(s) = x(t + s), −τ ? s ? 0. For φ ∈ PC([−τ,o],Rn), |φ|1is defined by

|φ|1= sup−τ?s?0?φ?, |φ|2is defined by |φ|2= inf−τ?s?0?φ?, where ? · ? denotes the

norm of vector in Rn. We can see that x(t) ≡ 0 is a solution of (1), which we call the zero

solution.

For given σ ?t0and ϕ ∈ PC([−τ,0],Rn), the initial value problem of Eq. (1) is

˙ x(t) = f(t,xt),

∆x(t) ? x(τk) −x?τ−

xσ= ϕ,

Throughoutthis paper we let the following hypotheses hold:

t ?t0, t ?= τk,

?= Ik

k

?x?τ−

k

??,k = 1,2,...,

(1)

t ?σ, t ?= τk,

?= Ik

k

?x?τ−

k

??,

k = 1,2,....

(2)

(H1) For each function x(s):[σ − τ,∞) → Rn, σ ? t0, which is continuous everywhere

except the points τkat which x(τ+

k) and x(τ−

k) exist and x(τ+

k) = x(τk), f(t,xt)

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Y. Zhang, J. Sun / J. Math. Anal. Appl. 301 (2005) 237–248

239

is continuous for almost all t ∈ [σ,∞) and at the discontinuous points f is right

continuous.

(H2) f(t,φ) is Lipschitzian in φ in each compact set in PC([−τ,0],Rn).

(H3) The functions Ik:Rn→ Rn, k = 1,2,..., are such that for any H > 0, there exists

a ρ > 0 such that if x ∈ S(ρ) = {x ∈ Rn: ?x? < ρ} implies that ?x +Ik(x)? < H.

Under the hypotheses (H1)–(H3), there is a unique solution of problem (2) throughout

(σ,ϕ).

Let

K =?a ∈ C[R+,R+]: a(t) is monotone strictly increasing and a(0)= 0?,

PC1(ρ) =?φ ∈ PC?[−τ,0],Rn?: |φ|1< ρ?,

K1=?w ∈ C[R+,R+]: w(t) ∈ K and 0 < w(s) < s, s > 0?,

PC2(θ) =?φ ∈ PC?[−τ,0],Rn?: |φ|2> θ?.

We have the following definitions.

Definition 1. The zero solution of (1) is said to be

(A1) strictly stable, if for any σ ? t0and ε1> 0, there exists a δ1= δ1(σ,ε1) > 0 such that

ϕ ∈ PC1(δ1) implies ?x(t;σ,ϕ)? < ε1, t ? σ, and for every 0 < δ2? δ1, there exists

an 0< ε2< δ2such that

ε2<??x(t;σ,ϕ)??,

(A3) strictly attractive, if given σ ? t0and α1> 0,ε1> 0, for every α2? α1, there exists

ε2< ε1and T1= T1(σ,ε1), T2= T2(σ,ε2) such that

ϕ ∈ PC1(α1)∩PC2(α2)

for σ +T1?t ? σ +T2;

(A4) strictly uniformly attractive if T1,T2in (A3) are independent of σ;

(A5) strictly asymptotically stable if (A3) holds and the trivial solution is stable;

(A6) strictly uniformly asymptotically stable if (A4) holds and the trivial solution is uni-

formly stable.

ϕ ∈ PC2(δ2)

implies

t ? σ;

(A2) strictly uniformly stable, if δ1, δ2and ε2is independent of σ;

implies

ε2<??x(t;σ,ϕ)??< ε1,

It is obvious that (A1) and (A3), or (A2) and (A4) cannot hold at the same time.

Definition 2 [3]. The function V :[t0,∞)×S(ρ) → R+belongs to class v0if

(1) the function V is continuous on each of the sets [τk−1,τk) × S(ρ) and for all t ? t0,

V(t,0)≡ 0;

(2) V(t,x) is locally Lipschitzian in x ∈ S(ρ);

(3) for each k = 1,2,..., there exist finite limits

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Y. Zhang, J. Sun / J. Math. Anal. Appl. 301 (2005) 237–248

lim

(t,y)→(τ−

lim

(t,y)→(τ+

k,x) = V(τk,x) satisfied.

Definition 3 [4]. Let V ∈ v0, for (t,x) ∈ [τk−1,τk)×S(ρ), D+V is defined as

D+V?t,x(t)?= lim

k,x)V(t,y)= V?τ−

k,x?,

k,x)V(t,y)= V?τ+

k,x?

with V(τ+

δ→0+sup1

δ

?V?t +δ,x(t +δ)?−V?t,x(t)??.

3. Main results

Now we consider the strict stability of the impulsivefunctionaldifferentialequation(1).

We have the following results.

Theorem 1. Assume that

(i) There exists V1∈ v0such that

b1

??x??? V1(t,x)? a1

that D+V1(t,x(t)) ? 0.

??x??,a1,b1∈ K;

(ii) For any solution x(t) of (1), V1(t +s,x(t +s)) ? V1(t,x(t)) for s ∈ [−τ,0], implies

Also, for all k ∈ Z+and x ∈ S(ρ),

V1

k

?τk,x?τ−

where dk?0 and

?+Ik

?x?τ−

?

k

???? (1+dk)V1

dk< ∞.

?τ−

k,x?τ−

k

??,

∞

k=1

(iii) There exists V2∈ v0such that

b2

??x??? V2(t,x) ?a2

that D+V2(t,x(t)) ? 0.

??x??,a2,b2∈ K;

(iv) For any solution x(t) of (1), V2(t +s,x(t +s)) ? V2(t,x(t)) for s ∈ [−τ,0], implies

Also, for all k ∈ Z+and x ∈ S(ρ),

V2

k

?τk,x?τ−

where 0 ?ck< 1 and

?+Ik

?x?τ−

k

???? (1−ck)V2

?

?τ−

k,x?τ−

k

??,

∞

k=1

ck< ∞.

Then the trivial solution of (1) is strictly uniformly stable.

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241

Proof. Since?∞

Let 0 < ε1< ρ and σ ? t0 be given, and σ ∈ [τk−1,τk) for some k ∈ Z+. Choose

δ1= δ1(ε1) > 0 such that Ma1(δ1) < b1(ε1).

Then we claim that ϕ ∈ PC1(δ1) implies ?x(t)? < ε1, t ? σ.

Obviously for any t ∈ [σ −τ,σ], there exists a θ ∈ [−τ,0] such that

V1

= a1

Then we claim that

?t,x(t)??a1(δ1),

If inequality (3) does not hold, then there is a ˆ t ∈ (σ,τk) such that

V1

which implies that there is a ˇ t ∈ (σ,ˆ t ] such that

D+V1

k=1dk< ∞,?∞

k=1ck< ∞, it follows that?∞

k=1(1 + dk) = M and

?∞

k=1(1−ck) = N, obviously 1 ? M < ∞, 0 < N ? 1.

?t,x(t)?= V1

?σ +θ,x(σ +θ)??a1

???x(σ +θ)???= a1

???xσ(θ)???

???ϕ(θ)???? a1(δ1).

σ ? t < τk.

V1

(3)

?ˆ t,x(ˆ t )?> a1(δ1) ? V1

?ˇ t,x(ˇ t )?> 0

?ˇ t +s,x(ˇ t +s)?? V1

By condition (ii), which implies that D+V1(ˇ t,x(ˇ t )) ? 0. This contradicts inequality (4),

so inequality (3) holds.

From condition (ii), we have

?τk,x(τk)?= V1

Next, we claim that

?t,x(t)??(1+dk)a1(δ1),

If inequality (5) does not hold, then there is an ˆ s ∈ (τk,τk+1) such that

V1

?σ,x(σ)?

(4)

and

V1

?ˇ t,x(ˇ t )?,s ∈ [−τ,0].

V1

?τk,x?τ−

k

?+Ik

?x?τ−

k

???? (1+dk)V1

?τ−

k,x?τ−

k

??

? (1+dk)a1(δ1).

V1

τk? t < τk+1.

(5)

?ˆ s,x(ˆ s)?> (1+dk)a1(δ1) ? V1

which implies that there is an ˇ s ∈ (τk, ˆ s) such that

D+V1

?τk,x(τk)?

?ˇ s,x(ˇ s)?> 0

?ˇ s +s,x(ˇ s +s)?? V1

By condition (ii), which implies that D+V1(ˇ s,x(ˇ s)) ? 0. This contradicts inequality (6),

so inequality (5) holds.

And from condition (ii), we have

?τk+1,x(τk+1)?= V1

? (1+dk+1)V1

(6)

and

V1

?ˇ s,x(ˇ s)?,s ∈ [−τ,0].

V1

?τk+1,x?τ−

k+1

?τ−

?+Ik+1

?x?τ−

k+1

k+1

???

k+1,x?τ−

??? (1+dk+1)(1+dk)a1(δ1).

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By a simple induction, we can easily prove that, in general, for m = 0,1,2,...,

V1

?t,x(t)??(1+dk+m)...(1+dk)a1(δ1),

which together with inequality (3) and condition (i) we have

???x(t)???? V1

??x(t)??< ε1,

Next we claim that ϕ ∈ PC2(δ2) implies ?x? > ε2, t ? σ. If this holds, ϕ ∈ PC1(δ1) ∩

PC2(δ2) implies ε2< ?x? < ε1, t ? σ.

Obviously for any t ∈ [σ −τ,σ], there exists a θ ∈ [−τ,0] such that

V2

= b2

Then we claim that

?t,x(t)??b2(δ2),

If inequality (7) does not hold, then there is a ¯ t ∈ (σ,τk) such that

V2

τk+m? t < τk+m+1,

b1

?t,x(t)?? Ma1(δ1) < b1(ε1),t ? σ.

Thus, we have

t ? σ.

Now, let 0< δ2? δ1and choose 0 < ε2< δ2such that a2(ε2) < Nb2(δ2).

?t,x(t)?= V2

?σ +θ,x(σ +θ)??b2

???x(σ +θ)???= b2

???xσ(θ)???

???ϕ(θ)???? b2(δ2).

σ ? t < τk.

V2

(7)

?¯ t,x(¯ t )?< b2(δ2) ? V2

which implies that there exists a t1∈ (σ,¯ t ) such that

D+V2

?σ,x(σ)?

?t1,x(t1)?< 0

?t1+s,x(t1+s)?? V2

By condition (iv), this implies that D+V2(t1,x(t1)) ? 0, which contradicts inequality (8),

so inequality (7) holds.

From condition (iv), we have

?τk,x(τk)?= V2

Next, we claim that

?t,x(t)??(1−ck)b2(δ2),

If inequality (9) does not hold, then there is an ¯ r ∈ (τk,τk+1) such that

V2

(8)

and

V2

?t1,x(t1)?,s ∈ [−τ,0].

V2

?τk,x?τ−

k

?+Ik

?x?τ−

k

???? (1−ck)V2

?τ−

k,x?τ−

k

??

? (1−ck)b2(δ2).

V2

τk? t < τk+1.

(9)

?¯ r,x(¯ r)?< (1−ck)b2(δ2) ? V2

which implies that there is an ˇ r ∈ (τk, ¯ r) such that

D+V2

?τk,x(τk)?

?ˇ r,x(ˇ r)?< 0(10)

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Y. Zhang, J. Sun / J. Math. Anal. Appl. 301 (2005) 237–248

243

and

V2

?ˇ r +s,x(ˇ r +s)?? V2

By condition (iv), this implies that D+V2(ˇ r,x(ˇ r)) ? 0, which contradicts inequality (10).

So inequality (9) holds.

And from condition (iv), we have

?τk+1,x(τk+1)?= V2

? (1−ck+1)V2

By a simple induction, we can easily prove that, in general, for m = 0,1,2,...,

V2

?ˇ r,x(ˇ r)?,s ∈ [−τ,0].

V2

?τk+1,x?τ−

k+1

?τ−

?+Ik+1

?x?τ−

k+1

k+1

???

k+1,x?τ−

??? (1−ck+1)(1−ck)b2(δ2).

?t,x(t)??(1−ck+m)...(1−ck)b2(δ2),

which together with inequality (7) and condition (iii) we have

???x(t)???? V2

??x(t)??> ε2,

The proof of Theorem 1 is complete.

τk+m? t < τk+m+1,

a2

?t,x(t)?? Nb2(δ2) > a2(ε2), t ? σ.

Thus, we have

t ? σ.

Thus, the zero solution of (1) is strictly uniformly stable.

2

Theorem 2. Assume that

(i) There exists V1∈ v0such that

b1

??x??? V1

?t,x(t)?? a1

??x??,a1,b1∈ K.

(ii) For any solution x(t) of (1), there exists a ψ1∈ K1, such that V1(t + s,x(t + s)) ?

ψ−1

C[R+,R+], locally integrable.

And for all k ∈ Z+, V1(τk,x(τk)) ? ψ1(V1(τ−

1(V1(t,x(t))) for s ∈ [−τ,0],implies that D+V1(t,x(t)) ? g(t)w(V1(t,x)), g,w :

k,x(τ−

k))).

(iii) There exists a constant A > 0 such that

τk

?

τk−1

g(s)ds < A,k ∈ Z+.

Also, for any u > 0,

ψ−1

1(u)

?

u

ds

w(s)?A.

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Y. Zhang, J. Sun / J. Math. Anal. Appl. 301 (2005) 237–248

(iv) There exists V2∈ v0such that

b2

??x???V2

?t,x(t)?? a2

??x??,a2,b2∈ K.

(v) For any solution x(t) of (1), there exists a ψ2∈ K1, such that V2(t + s,x(t + s)) ?

ψ2(V2(t,x(t))) for s ∈ [−τ,0], implies that D+V2(t,x(t)) ? h(t)p(V2(t,x)), h,p :

C[R+,R+], locally integrable.

And for all k ∈ Z+, V2(τk,x(τk)) ? ψ−1

2(V2(τ−

k,x(τ−

k))).

(vi) There exists a constant B > 0 such that

τk

?

τk−1

h(s)ds < B,k ∈ Z+.

Also, for any u > 0,

ψ2(u)

?

u

ds

p(s)? B.

Then the zero solution of (1) is strictly uniformly stable.

Proof. For given 0 < ε1< ρ, σ ? t0. Choose δ1= δ1(ε1) > 0, such that ψ−1

b1(ε1). Let σ ∈ [τk−1,τk) for some k ∈ Z+.

We claim that ϕ ∈ PC1(δ1) implies ?x? < ε1, t ? σ.

First, we claim that

?t,x(t)??ψ−1

Obviously for any t ∈ [σ −τ,σ], there exists a θ ∈ [−τ,0] such that

V1

???xσ(θ)???= a1

V1

1

From the continuity of V1(t,x(t)) at [σ,τk), it follows that there exists an s1∈ (σ, ˆ s) such

that

?s1,x(s1)?= ψ−1

V1

1

σ −τ ? t ? s1,

and also, there exists an s2∈ [σ,s1) such that

V1

V1(t,x(t)) ?a1(δ1),t ∈ [s2,s1].

1(a1(δ1)) <

V1

1

?a1(δ1)?,

?σ +θ,x(σ +θ)??a1

σ ? t < τk.

(11)

?t,x(t)?= V1

???x(σ +θ)???

= a1

???ϕ(θ)???? a1(δ1) < ψ−1

?σ,x(σ)?.

1

?a1(δ1)?.

So if inequality (11) does not hold, then there is an ˆ s ∈ (σ,τk) such that

?ˆ s,x(ˆ s)?> ψ−1

?a1(δ1)?> a1(δ1) ? V1

V1

1

?a1(δ1)?,

?t,x(t)??ψ−1

?s2,x(s2)?= a1(δ1),

?a1(δ1)?,

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Y. Zhang, J. Sun / J. Math. Anal. Appl. 301 (2005) 237–248

245

Therefore, for t ∈ [s2,s1] and −τ ? s ? 0, we have

V1

?t +s,x(t +s)?? ψ−1

In view of condition (ii), we get

D+V1

1

?a1(δ1)?? ψ−1

1

?V1

?t,x(t)??.

?t,x(t)??g(t)w?V1(t,x)?,

And integrate the inequality (12) over (s2,s1), we have by condition (iii),

s2? t ?s1.

(12)

V1(s1,x(s1))

?

V1(s2,x(s2))

On the other hand,

du

w(u)?

s1

?

s2

g(s)ds ?

τk

?

τk−1

g(s)ds < A.

V1(s1,x(s1))

?

V1(s2,x(s2))

a contradiction. So inequality (11) holds.

From condition (ii) and inequality (11), we have

?τk,x(τk)?= V1

Next, we prove that

?t,x(t)??ψ−1

Since a1(δ1) < ψ−1

such that

?ˆ r,x(ˆ r)?> ψ−1

From the continuity of V1(t,x(t)) at [τk,τk+1), it follows that there is an r1∈ (τk, ˆ r) such

that

?r1,x(r1)?= ψ−1

V1

1

σ −τ ? t ? r1,

and also, there exists an r2∈ [τk,r1) such that

V1

?t,x(t)??a1(δ1),

Therefore, for t ∈ [r2,r1] and −τ ? s ? 0, we have

V1

1

In view of condition (ii), we get

D+V1

r2? t ?r1.

Similar as before, we integrate the inequality (14) over (r2,r1), by condition (iii) we can

obtain a contradiction, so inequality (13) holds.

du

w(u)=

ψ−1

1(a1(δ1))

?

a1(δ1)

du

w(u)?A,

V1

?τk,x?τ−

?a1(δ1)?,

k

?+Ik

τk? t < τk+1.

?x?τ−

k

???? ψ1

?V1

?τ−

k,x?τ−

k

???? a1(δ1).

V1

1

(13)

1(a1(δ1)),if inequality(13)doesnot hold,thenthere is an ˆ r ∈ (τk,τk+1)

?a1(δ1)?> a1(δ1) ? V1

V1

1

?τk,x(τk)?.

V1

1

?a1(δ1)?,

?t,x(t)??ψ−1

?r2,x(r2)?= a1(δ1),

?a1(δ1)?,

V1

t ∈ [r2,r1].

?t +s,x(t +s)?? ψ−1

?t,x(t)??g(t)w?V1(t,x)?,

?a1(δ1)?? ψ−1

1

?V1

?t,x(t)??.

(14)

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Y. Zhang, J. Sun / J. Math. Anal. Appl. 301 (2005) 237–248

From condition (ii) and inequality (13),

?τk+1,x(τk+1)?= V1

By simple induction, we can prove, in general, that for i = 0,1,2,...,

V1

1

τk+i? t < τk+i+1,

V1

Since a1(δ1) < ψ−1

??x??? V1

Thus ?x? < ε1, t ? σ.

Now, let 0< δ2? δ1and choose 0 < ε2< δ2such that a2(ε2) < ψ2(b2(δ2)).

Next we claim that ϕ ∈ PC2(δ2) implies ?x? > ε2, t ? σ. If this holds, ϕ ∈ PC1(δ1) ∩

PC2(δ2) implies ε2< ?x? < ε1, t ? σ.

First, we claim that

?t,x(t)??ψ2

Obviously for any t ∈ [σ −τ,σ], there exists a θ ∈ [−τ,0] such that

V2

= b2

So if inequality (16) does not hold, then there is a ¯ t ∈ (σ,τk) such that

V2

From the continuity of V2(t,x(t)) at [σ,τk) there exists a t1∈ (σ,¯ t) such that

V2

?t,x(t)??ψ2

And also, there exists a t2∈ [σ,t1) such that

V2

?t,x(t)??b2(δ2),

Therefore, for t ∈ [t2,t1], and −τ ? s ?0, we have

V2

In view of condition (v), we get

D+V2

t2? t ? t1.

Integrate inequality (17) over (t2,t1), by condition (vi) we have

V1

?τk+1,x?τ−

? ψ1

k+1

?+Ik+1

k+1

?x?τ−

k+1

???

?V1

?τ−

k+1,x?τ−

???? a1(δ1).

?t,x(t)??ψ−1

?a1(δ1)?,

?τk+i+1,x(τk+i+1)?? a1(δ1).

?t,x(t)?? ψ−1

(15)

1(a1(δ1)), by condition (i) and inequalities (11), (15), we have

?a1(δ1)?? b1(ε1),

b1

1

t ?σ.

V2

?b2(δ2)?,

?σ +θ,x(σ +θ)??b2

σ ? t < τk.

(16)

?t,x(t)?= V2

???x(σ +θ)???

???xσ(θ)???= b2

?b2(δ2)?< b2(δ2) ? V2

?b2(δ2)?,

???ϕ(θ)???? b2(δ2) > ψ2

?σ,x(σ)?.

?b2(δ2)?.

?¯ t,x(¯ t )?< ψ2

?t1,x(t1)?= ψ2

V2

?b2(δ2)?,σ −τ ? t ? t1.

?t2,x(t2)?= b2(δ2),

V2

t ∈ [t2,t1].

?t +s,x(t +s)?? ψ2

?t,x(t)??h(t)p?V2(t,x)?,

?b2(δ2)?? ψ2

?V2

?t,x(t)??.

(17)

V2(t1,x(t1))

?

V2(t2,x(t2))

du

p(u)?

t1

?

t2

h(s)ds ?

τk

?

τk−1

h(s)ds < B

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Y. Zhang, J. Sun / J. Math. Anal. Appl. 301 (2005) 237–248

247

and at the same time

V2(t1,x(t1))

?

V2(t2,x(t2))

du

p(u)=

ψ2(b2(δ2))

?

b2(δ2)

du

p(u)? B

a contradiction. So inequality (16) holds.

From condition (v) and inequality (16), we have

?τk,x(τk)?? ψ−1

Next, we prove

?t,x(t)??ψ2

If inequality (18) does not hold, then there is a ˆ q ∈ (τk,τk+1) such that

V2

V2

2

?V2

?τ−

k,x?τ−

k

???? b2(δ2).

V2

?b2(δ2)?,

?b2(δ2)?< b2(δ2) ? V2

?b2(δ2)?,

τk? t < τk+1.

(18)

?ˆ q,x(ˆ q)?< ψ2

?q1,x(q1)?= ψ2

?τk,x(τk)?.

From the continuity of V2(t,x(t)) at [τk,τk+1) there is a q1∈ (τk, ˆ q) such that

V2

?t,x(t)??ψ2

and also, there exists a q2∈ [τk,q1) such that

V2

?t,x(t)??b2(δ2),

Therefore, for t ∈ [q2,q1], and −τ ? s ? 0, we have

V2

V2

?b2(δ2)?,σ −τ ? t ? q1,

?q2,x(q2)?= b2(δ2),

V2

t ∈ [q2,q1].

?t +s,x(t +s)?? ψ2

In view of condition (v), we get

D+V2

?b2(δ2)?? ψ2

?V2

?t,x(t)??.

?t,x(t)??h(t)p?V2(t,x)?,

Similar as before, we integrate the inequality (19) over (q2,q1), by condition (vi) we can

obtain a contradiction, so inequality (18) holds.

From condition (v) and inequality (18), we have

?τk+1,x(τk+1)?= V2

? ψ−1

2

By simple induction, we can prove, in general, that for i = 0,1,2,...,

V2

τk+i? t < τk+i+1,

V2

q2? t ?q1.

(19)

V2

?τk+1,x?τ−

k+1

?+Ik+1

k+1

?x?τ−

k+1

???

?V2

?τ−

k+1,x?τ−

???? b2(δ2).

?t,x(t)??ψ2

?b2(δ2)?,

?τk+i+1,x(τk+i+1)?? b2(δ2).

Since b2(δ2) > ψ2(b2(δ2)), from inequality (16) and (20) we have

?t,x(t)??ψ2

(20)

V2

?b2(δ2)?,t ? σ,

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Y. Zhang, J. Sun / J. Math. Anal. Appl. 301 (2005) 237–248

which together with condition (iv), we have

??x??? V2

So ?x? > ε2, for t ?σ.

Thus the zero solution of (1) is strictly uniformly stable.

The proof of Theorem 2 is complete.

a2

?t,x(t)?? ψ2

?b2(δ2)?> a2(ε2),t ? σ.

2

4. Conclusion

In this paper, we have extended the notion of strict stability to impulsive functional

differential equations. By using Lyapunov functions and Razumikhin technique, we have

gottensomeresultsforthestrict uniformstabilityofthisequation.We cansee thatimpulses

do contribute to system’s strict stability behavior.

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