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

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).