Razumikhin-type theorems of neutral stochastic functional differential equations

Acta Mathematica Scientia 2009,29B(1):181–190 http://actams.wipm.ac.cn

RAZUMIKHIN-TYPE THEOREMS OF NEUTRAL STOCHASTIC FUNCTIONAL DIFFERENTIAL EQUATIONS∗



Zhou Shaobo (

)


)

Hu Shigeng (

School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan 430074, China

Abstract The stability of stochastic functional differential equation with Markovian switching was studied by several authors, but there was almost no work on the stability of the neutral stochastic functional differential equations with Markovian switching. The aim of this article is to close this gap. The authors establish Razumikhin-type theorem of the neutral stochastic functional differential equations with Markovian switching, and those without Markovian switching. Key words Markovian chain, Razumikhin-type theorem, neutral stochastic functional differential equation, exponential stability 2000 MR Subject Classification

1

34F05

Introduction

Stochastic modelling has come to play an important role in many branches of science and industry. An area of particular interest has been the automatic control of stochastic systems, with consequent emphasis placed on the analysis of stability in stochastic models. The stability of stochastic differential equation with Markovian switching was studied by many authors: Li [1–3], Yuan [4, 5], Jiang [6], and Shen [7, 8]. For example, Mao [9] studied the stability of stochastic differential equations with Markovian switching of the form dx(t) = f (x(t), t, r(t)) + g(x(t), t, r(t))dw(t) on t ≥ 0 with initial value x(0) = x0 ∈ Rn , where r(t) is a Markov chain taking values on S = {1, 2, · · · , N }, f : Rn × R+ × S → Rn , g : Rn × R+ × S → Rn×m . This equation may be regarded as the result of the following N equations, dx(t) = f (x(t), t, i) + g(x(t), t, i)dw(t), 1 ≤ i ≤ N, ∗ Received

April 10, 2007; revised January 8, 2008. Sponsored by HUST Foundation (0125011017) and the National NSFC under grant (70671047)

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switching from one to other according to the movement of Markov chain. Mao [9–13] discussed the stability of stochastic delay differential equation with Markovian switching dx(t) = f (x(t), x(t − τ1 (t)), t, r(t)) + g(x(t), x(t − τ2 (t)), t, r(t))dw(t). Kolmanovskii [14] examined the stability of neutral stochastic differential delay equation with Markovian switching d[x(t) − u(x(t − τ ), r(t))] = f (x(t), x(t − τ ), t, r(t)) + g(x(t), x(t − τ ), t, r(t))dw(t). However, there was little work on the stability of neutral stochastic functional differential equation with Markovian switching of the form d[x(t) − u(xt , r(t))] = f (xt , t, r(t)) + g(xt , t, r(t))dw(t). In this article, we establish the Razumikhin-type theorem of the equation and the Razumikhintype theorem of the neutral stochastic functional differential equations without Markovian switching.

2 Neutral Stochastic Functional Differential Equations With Markovian Switchings Throughout this article, unless otherwise specified, let (Ω, F, {Ft }t≥0 , P ) be a complete probability space with a filtration {Ft }t≥0 , satisfying the usual conditions (that is, it is right continuous and F0 contains all P -null sets). Let w(t) = (w1 (t), · · · , wm (t))T be an m-dimensional Brownian motion defined on the probability space. Let τ > 0 and C([−τ, 0]; Rn ) denote the family of continuous functions ϕ from [−τ, 0] to Rn with norm ϕ = sup |ϕ(θ)|, where | · | −τ ≤θ≤0

is the Euclidean norm in Rn . If A is a vector or matrix, its transpose is denoted by AT . If A is  a matrix, its trace norm is denoted by |A| = trace(AT A), while its operator norm is denoted b by A = sup{|Ax| : |x| = 1}(without any confusion with ϕ). Denote by CF ([−τ, 0]; Rn ) the 0 family of all bounded, F0 -measurable, C([−τ, 0]; Rn )-valued random variables. For p > 0 and t ≥ 0, denote by LpFt ([−τ, 0]; Rn ) the family of all Ft -measurable C([−τ, 0]; Rn )-valued random variables φ = {φ(θ) : −τ ≤ θ ≤ 0} such that sup E|φ(θ)|p < ∞. −τ ≤θ≤0

Let r(t)(t ≥ 0) be a right-continuous Markov chain on the probability space taking values in a finite state space S = {1, 2, · · · , N } with generator Γ = (γij )N ×N given by ⎧ ⎨ γ Δ + 0(Δ), if i = j, ij P {r(t + Δ) = j|r(t) = i} = ⎩ 1 + γii Δ + 0(Δ), if i = j, where Δ > 0. Here γij ≥ 0 is the transition rate from i to j if i = j, while γii = −



j=i

γij .

We assume that the Markov chain r(·) is independent of the Brownian motion w(·). It is well known (see Skorohod [15]) that almost every sample path of r(t) is a right-continuous step function with a finite number of simple jumps in any finite subinterval of R+ . In other words, there is a sequence of stopping times 0 = τ0 < τ1 < · · · < τk → ∞ almost surely such that r(t) =

∞  k=0

r(τk )I[τk .τk+1 ) (t),

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where IA denotes the indicator function of set A, that is, r(t) is constant on every interval [τk , τk+1 ), for every k ≥ 0 r(t) = r(τk ), on τk ≤ t < τk+1 . Consider an n-dimensional neutral stochastic functional differential equation with Markovian switching of d(x(t) − u(xt , r(t))) = f (xt , t, r(t))dt + g(xt , t, r(t))dw(t)

(2.1)

b on t ≥ 0 with initial data x0 = ξ. Here ξ ∈ CF ([−τ, 0]; Rn ) and xt = {x(t + θ) : −τ ≤ θ ≤ 0}, 0 n which is regarded as a C([−τ, 0]; R )-valued stochastic process. Moreover,

f : C([−τ, 0]; Rn ) × R+ × S −→ Rn , g : C([−τ, 0]; Rn ) × R+ × S −→ Rn×m , u : C([−τ, 0]; Rn ) × S −→ Rn . By the definition of Ito’s stochastic differential equation (2.1) that for every t > 0  x(t) − u(xt , r(t)) = ξ(0) − u(x0 , r(0)) +

0

t

 f (xs , s, r(s))ds +

0

t

g(xs , s, r(s))dw(s).

(2.2)

We assume that the solution of the equation exists and is unique. Lemma 2.1 [16] (x + y)p ≤ (1 + ε)p−1 (xp + ε1−p y p ), ε > 0. Definition 2.1 [16] The solution of equation (2.1) is said to be exponentially stable in pth moment if 1 lim sup log(E|x(t; ξ)|p ) < 0 t→∞ t b for all ξ ∈ CF ([−τ, 0]; Rn ). The solution of equation (2.1) is said to be almost surely exponen0 tially stable if 1 lim sup log(|x(t; ξ)|) < 0 a.s. t→∞ t b for all ξ ∈ CF ([−τ, 0]; Rn ). 0 = ϕ(θ) − u(ϕ, i), and ϕ = xt , then ϕ(0) = xt (0) = x(t), ϕ(θ) = xt (θ) = x(t + Set ϕ(θ) θ), ϕ(0) = ϕ(0)−u(ϕ, i) = x(t)−u(xt , i). Let C 2,1 (Rn ×[−τ, ∞]×S; R+) denote the family of all

nonnegative functions V (x, t, i) on Rn × [−τ, ∞] × S, which are continuously twice differentiable in x and once differentiable in t. If V (x, t, i) ∈ C 2,1 (Rn × [−τ, ∞] × S; R+ ), define an operator LV [16] from C([−τ, 0]; Rn ) × R+ × S to R by t, i) + Vx (ϕ(0), t, i)f (ϕ, t, i) LV (ϕ, t, i) = Vt (ϕ(0),

n  1 T t, j), γij V (ϕ(0), + trace[g (ϕ, t, i)Vxx (ϕ(0), t, i)g(ϕ, t, i)] + 2 j=1

where Vt =

∂V (x, t, i) ∂V (x, t, i)

∂ 2 V (x, t, i)  ∂V (x, t, i) ∂V (x, t, i)  , Vx = , Vxx = , ,···, . ∂t ∂x1 ∂x2 ∂xn ∂xi ∂xj n×n

b ([−τ, 0]; Rn ). We shall always fix the Markov chain r(t) but let the initial data ξ vary in CF 0 Without loss of generality, we may assume f (0, t, i) = 0, g(0, t, i) = 0, u(0, i) = 0.

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Theorem 2.2 Let c1 , c2 , μ, p be all positive numbers and q > 1. Assume that there exists a function V (x, t, i) ∈ C 2,1 (Rn × [−τ, ∞] × S; R+ ) such that c1 |x|p ≤ V (x, t, i) ≤ c2 |x|p , 1 ≤ i ≤ N,

(2.3)

for all (x, t, i) ∈ Rn × [−τ, ∞) × S and also for all t ≥ 0, E|u(ϕ, i)|p ≤ κp ϕp , 1 ≤ i ≤ N, 0 < κ < 1, ϕ ∈ LpFt ;

(2.4)

t, i)], E[ max LV (ϕ, t, i)] ≤ −μE[ max V (ϕ(0),

(2.5)

for all t ≥ 0,

1≤i≤N

1≤i≤N

provided ϕ = {ϕ(θ) : −τ ≤ θ ≤ 0} ∈ LpFt ([−τ, 0]; Rn ), satisfying t, i)] E[V (ϕ(θ), t + θ, i)] ≤ qE[V (ϕ(0),

(2.6)

b for all −τ ≤ θ ≤ 0. Then, for all ξ ∈ CF ([−τ, 0]; Rn ), t ≥ 0 0

E|x(t; ξ)|p ≤ where γ = γ ∧

1 τ

c2 −γt e Eξp0 , c1

(2.7)

c1 q log c2 [(1−κ) 1−p +qκ] , γ being the root of the following equation

c2 c2 (1 − κ)1−p γ + κeγτ μ = μ. c1 c1

(2.8)

In other words, the trivial solution of equation (2.1) is pth moment exponentially stable and the pth moment Lyapunov exponent is not greater than −γ. b Proof Fix the initial data ξ ∈ CF ([−τ, 0]; Rn ) arbitrarily and write x(t; ξ) = x(t). 0 Extend r(t) to [−τ, 0] by setting r(t) = r(0). Recalling the facts that x(t) is continuous, E( sup |x(s)|p ) < ∞ for all t ≥ 0 and r(t) is right continuous, we see that EV (x(t), t, r(t)) is −τ ≤s≤t

right continuous on t ≥ −τ. Define U (t) =

sup [eγ(t+θ) EV (x(t + θ), t + θ, r(t + θ))] −τ ≤θ≤0

for t ≥ 0.

We claim that D+ U (t) = lim sup

h→0+

U (t + h) − U (t) ≤0 h

for all t ≥ 0.

(2.9)

Note that for each t ≥ 0 (fix for the moment), either U (t) > eγt EV (x(t), t, r(t)) or U (t) = eγt EV (x(t), t, r(t)). (i) If U (t) > eγt EV (x(t), t, r(t)), that is, there exists θ such that θ = max{θ ∈ [−τ, 0] : eγ(t+θ)EV (x(t + θ), t + θ, r(t + θ)) = U (t)}, obviously, θ is well defined, θ < 0 and eγ(t+θ) EV (x(t + θ), t + θ, r(t + θ)) = U (t). Since θ < 0, then eγ(t+θ)EV (x(t + θ), t + θ, r(t + θ)) < eγ(t+θ) EV (x(t + θ), t + θ, r(t + θ))

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for all θ < θ ≤ 0. It is easy to obtain that for all h > 0 sufficiently small, eγ(t+θ+h) EV (x(t + θ + h), t + θ + h, r(t + θ + h)) ≤ eγ(t+θ) EV (x(t + θ), t + θ, r(t + θ)), hence U (t + h) < U (t) and D+ U (t) ≤ 0. (ii) If U (t) = eγt EV (x(t), t, r(t)), we have eγ(t+θ) EV (x(t + θ), t + θ, r(t + θ)) ≤ eγ(t) EV (x(t), t, r(t)) for all − τ ≤ θ ≤ 0, so EV (x(t + θ), t + θ, r(t + θ)) ≤ e−γθ EV (x(t), t, r(t)) ≤ eγτ EV (x(t), t, r(t))

(2.10)

for all −τ ≤ θ ≤ 0. Note that EV (x(t), t, r(t)) = 0 or EV (x(t), t, r(t)) > 0. (iia) If EV (x(t), t, r(t)) = 0, then U (t) = 0 by assumption U (t) = eγt EV (x(t), t, r(t)), EV (x(t + θ), t + θ, r(t + θ)) = 0, by the inequility (2.3), we have x(t + θ) = 0 a.s. for all −τ ≤ θ ≤ 0, that is, xt = 0 a.s. by f (0, t, i) = 0, g(0, t, i) = 0, u(0, i) = 0, and equation (2.1), we have x(t) = 0, that is, x(t) − u(xt , r(t)) = 0, Thus, x(t) = 0. According to the existence and uniqueness theorem [13], x(t + h) = 0 a.s. for all h > 0; hence, U (t + h) = 0 and U (t) = 0, thus D+ U (t) = 0. (iib) If EV (x(t), t, r(t)) > 0, then by the inequility (2.10) and condition (2.3), we have EV (x(t + θ), t + θ, r(t + θ)) ≤ eγτ EV (x(t), t, r(t)) ≤ eγτ c2 |x(t)|p + u(xt , r(t))|p ≤ eγτ c2 |x(t)

by condition(2.3)

1 + ε p−1 p ) |x(t)| + (1 + ε)p−1 |u(xt , r(t))|p ) by Lemma 2.1 ε 1 + ε p−1 p ) |x(t)| + (1 + ε)p−1 c2 eγτ κp |xt |p ≤ c2 eγτ ( ε EV (x(t + θ), t + θ, r(t + θ)) 1 + ε p−1 p ) |x(t)| + (1 + ε)p−1 c2 eγτ κp ≤ c2 eγτ ( ε c1 by condition(2.3) ≤ eγτ c2 ((

that is, EV (x(t + θ), t + θ, r(t + θ)) ≤

1 (1 − κ)1−p c2 eγτ t, r(t)) EV (x(t), c1 1 − cc21 κeγτ

t, r(t)), < qEV (x(t), for all −τ ≤ θ ≤ 0, where ε =

1−κ κ ,

the second inequality comes from

γ≤

1 log τ

c2 c1 [(1

q . − κ)1−p + qκ]

Thus, xt ∈ LpFt ([−τ, 0]; Rn ) satisfies t, i)] E[ max LV (ϕ, t, i)] ≤ −μE[ max V (ϕ(0), 1≤i≤N

1≤i≤N

(2.11)

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by condition (2.6), that is, t, i)], E[ max LV (xt , t, i)] ≤ −μE[ max V (x(t), 1≤i≤N

which implies

1≤i≤N

t, r(t)). ELV (xt , t, r(t)) ≤ −μEV (x(t),

By Itˆo formula, we have eγ(t+h) EV (x(t + h), t + h, r(t + h)) − eγt EV (x(t), t, r(t))  t+h ≤ eγs [γEV (x(s), s, r(s)) + ELV (xs , s, r(s))],

(2.12)

t

by condition(2.3) EV (x(s), s, r(s)) ≤ c2 |x(s)|p p ≤ c2 |x(s) + u(xs , r(s))|

1 + ε p−1 p + c2 (1 + ε)p−1 κp |xs |p ≤ c2 |x(s)| ε

1 + ε p−1 p + c2 (1 + ε)p−1 κp |x(s + θ)|p ≤ c2 |x(s)| ε

1 + ε p−1 p + c2 (1 + ε)p−1 κp EV (x(s + θ), s + θ, r(s + θ)) ≤ c2 |x(s)| ε c1 

1−p γτ p−1 c2 1 + ε s, r(s)) + c2 (1 + ε)p−1 κp (1 − κ) c2 e EV (x(s), s, r(s)) ≤ EV (x(s), γτ c1 ε c1 c1 − c2 κe c2 (1 − κ)1−p s, r(s)) ε = 1 − κ . ≤ EV (x(s), c2 γτ c1 1 − c1 κe κ γEV (x(s), s, r(s)) + ELV (xs , s, r(s)) c2 (1 − κ)1−p s, r(s)) − μEV (x(s), s, r(s)) EV (x(s), ≤γ c1 1 − cc21 κeγτ

c (1 − κ)1−p  2 s, r(s)), ≤ γ − μ EV (x(s), c1 1 − cc21 κeγτ Since γ¯ is the root of equation (2.8) and if 1 −

c2 γτ c1 κe

< 0, then

c2 (1 − κ)1−p EV (x(t), t, r(t)) < 0. c1 1 − cc21 κeγτ If 1 −

c2 γτ c1 κe

> 0, then γ < γ¯ , c2 c2 c2 c2 (1 − κ)1−p γ + κeγτ μ < (1 − κ)1−p γ + κeγτ μ = μ. c1 c1 c1 c1

c (1 − κ)1−p  2 γ − μ EV (x(t), t, r(t)) < 0. c2 γτ c1 1 − c1 κe

Thus, eγ(t+h) EV (x(t + h), t + h, r(t + h)) ≤ eγt EV (x(t), t, r(t)). So U (t + h) = U (t) for all h > 0 sufficiently small holds, and hence D+ U (t) = 0. Inequality (2.9) is proved. It follows from (2.9) that U (t) ≤ U (0) for all t ≥ 0. By the definition of U (t) and (2.3) eγ(t+θ) EV (x(t + θ), t + θ, r(t + θ)) ≤ eγθ EV (x(θ), θ, r(θ)) ≤ c2 eγθ Eξp , for all − τ ≤ θ ≤ 0.

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187

c2 −γt e Eξp , c1 c2 E|x(t)|p ≤ sup E|x(t + θ)|p ≤ e−γt Eξp , c1 −τ ≤θ≤0 E|x(t + θ)|p ≤

that is, E|x(t)|p ≤

c2 −γt e Eξp . c1

The proof is completed. We may easily obtain the following theorem from the above theorem: Theorem 2.3 Assume for p ≥ 2, E|x(t)|p ≤ ce−γt , c > 0, γ is the same as in Theorem 2.2, and for 0 < κ < 1, |u(ϕ, r(t))| < κϕ, |f (ϕ, t, i)|p ∨ |g(ϕ, t, i)|p < Kϕp . Then, 1 γ log |x(t, ξ)|p < a.s. p ≥ 2 t p t→∞ sup

b for any ξ ∈ CF ([−τ, 0]; Rn ). In other words, the solution of equation (2.1) is almost surely exponentially stable.

3

Neutral Stochastic Functional Differential Equations

In this section, we shall establish the Razumikhin-type theorem of the neutral stochastic functional differential equations without Markovian switching d(x(t) − u(xt )) = f (xt , t)dt + g(xt , t)dw(t).

(3.1)

All the assumptions are the same as the above. Theorem 3.1 Let c1 , c2 , μ, p be all positive numbers, and q > 1. Assume that there exists a function V (x, t) ∈ C 2,1 (Rn × [−τ, ∞]; R+ ) such that c1 |x|p ≤ V (x, t) ≤ c2 |x|p ,

(3.2)

for all (x, t) ∈ Rn × [−τ, ∞) and also for all t ≥ 0 E|u(ϕ)|p ≤ κp ϕp and for all t ≥ 0

0 < κ < 1, ϕ ∈ LpFt ;

(3.3)

t)], E[LV (ϕ(θ), t)] ≤ −μE[V (ϕ(0),

(3.4)

ϕ = {ϕ(θ) : −τ ≤ θ ≤ 0} ∈ LpFt ([−τ, 0]; Rn ),

(3.5)

t)] E[V (ϕ(θ), t + θ)] ≤ qE[V (ϕ(0),

(3.6)

provided

satisfying b ([−τ, 0]; Rn ), t ≥ 0 for all −τ ≤ θ ≤ 0. Then for all ξ ∈ CF 0

E|x(t; ξ)|p ≤

c2 (1 + κ)p −γt e Eξp0 , c1 (1 − κ1 )p

(3.7)

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1

γτ

where γ = μ ∧ τ −1 log q1 (1 + κq1p )−p , q1 = cc12 q, q > cc21 (1 − κ)−p , κ1 = κe p . In other words, the trivial solution of equation (3.1) is pth moment exponentially stable and the pth moment Lyapunov exponent is not greater than −γ. b Proof Fix the initial data ξ ∈ CF ([−τ, 0]; Rn ) arbitrarily and write x(t; ξ) = x(t). Re0 calling the facts that x(t), u(xt ) are continuous, E[ sup |x(s)|p ] < +∞ for all t ≥ 0. We see −τ ≤s≤t

easily that EV (x(t), t) and u(xt ) are continuous on t ≥ −τ. Define H(t) ≡ ≡ x(t) − u(xt ). x(t)

sup [eγs E|x(s; ξ)|p ],

−τ ≤s≤t

H(t) ≤ Eξp ∨ sup [eγs E|x(s)|p ] 0≤s≤t

+ u(xs )|p ] = Eξp ∨ sup [eγs E|x(s) 0≤s≤t

p + (1 + ε)p−1 eγs |u(xs )|p ] ≤ Eξ ∨ sup [eγs (1 + ε)p−1 ε1−p E|x(s)| p

0≤s≤t

p] ≤ Eξ ∨ sup [eγs (1 + ε)p−1 ε1−p E|x(s)| p

0≤s≤t

p−1 p γτ

+(1 + ε)

κ e

sup [eγs |x(s)|p ]

−τ ≤s≤t

1 + ε p−1

p + (1 + ε)p−1 κp eγτ H(t). eγs E|x(s)| ε 0≤s≤t

≤ Eξp ∨ sup Let κ1 = κe 1 q1p

1 1+κq1p

, κe

γτ p

γτ p

,ε = 1 κq1p

<

(3.8) changes to

1−κ1 κ1 ,

1 1+κq1p

1

1

(3.8)

since γ < τ −1 log q1 (1 + κq1p )−p , then eγτ < q1 (1 + κq1p )−p , e

γτ p

<

p−1 < 1. Thus, 0 < κ1 < 1, ( 1+ε = (1 − κ1 )1−p , (1 + ε)p−1 κp eγτ = κ1 , ε )

p ], (1 − κ1 )H(t) ≤ (1 − κ1 )1−p sup [eγs E|x(s)| 0≤s≤t

then p ]. H(t) ≤ (1 − κ1 )−p sup [eγs E|x(s)| 0≤s≤t

We claim that p ] ≤ β, sup [eγs E|x(s))|

0≤s≤t

where β =

c2 c1 (1

(3.9)

+ κ)p Eξp0 . By condition (3.2), we should only prove that t) ≤ βc1 . W (t) ≡ eγt EV (x(t),

First, 0) ≤ c2 |x(0) − u(x0 )|p W (0) = EV (x(0), ≤ c2 (1 + ε)p−1 [ε1−p Eξ(0)p + E|u(ξ0 )|p ]

1 + ε p−1 Eξp + c2 (1 + ε)p−1 κp Eξp . ≤ c2 ε Choose ε = κ1 , then

W (0) ≤ c2 (1 + κ)p Eξp = βc1 .

(3.10)

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Therefore, (3.10) satisfies for t = 0. Since EV (x(t), t) is continous, then there must exist the smallest ρ ∈ R+ such that (3.10) is satisfied on [0, ρ], and W (ρ) = βc1 . Given any t ∈ [ρ − τ, ρ]. Thus, according to (3.9) on the interval, we have H(t) ≤

β . (1 − κ1 )p

1

1

1

Since γ < τ −1 log q1 (1 + κq1p )−p , then γτ < log q1 (1 + κq1p )−p , eγτ < q1 (1 + κq1p )−p , 1

eγτ (1 + κq1p )p < q1 , e

γτ p

1

1

(1 + κq1p ) < q1p , e

γτ p

1

< q1p (1 − κe

γτ p

1

) = q1p (1 − κ1 ),

eγτ < q1 (1 − κ1 )p , thus H(t) ≤

β ≤ βq1 e−γτ . (1 − κ1 )p

Given t ∈ [ρ − τ, ρ], that is, ρ − τ ≤ t ≤ ρ, then ρ ≤ t + τ ≤ ρ + τ ; thus, we have EV (x(t), t, r(t)) ≤ c2 E|x(t)|p ≤ c2 e−γt [eγt E|x(t)|p ] ≤ c2 e−γt H(t) < c2 e−γt βq1 e−γτ ρ). ≤ qe−γρ W (ρ) = qV (x(ρ), (3.11) If 0 ≤ t ≤ ρ, (3.11) and (3.10) are satisfied. If ρ − τ ≤ t < 0 then (3.10) is satisfied. By t]. We claim that EV (x(ρ), ρ) > 0. In fact, if condition (3.6), E[LV (ϕ(θ), t)] ≤ −μEV [ϕ(0), EV (x(ρ), ρ) = 0, then c1 E|x(ρ)|p ≤ EV (x(ρ), ρ) = 0, for all ρ − τ ≤ t ≤ ρ, thus xρ = 0 a.s. according to the uniqueness of the solution, x(t) = 0 for all t > ρ, which contradicts the smallest of ρ. Thus EV (ϕ(ρ), ρ) > 0. When t → ρ, ELV (x(t), t) ≤ −μV (x(t), t) ≤ −γV (x(t), t). Thus, when h > 0,  ρ+h W (t)|ρ+h =E L[eγt V (x(t), t)]dt ρ 

ρ

ρ+h

=E ρ

eγt [LV (x(t), t) + γEV (x(t), t)]dt ≤ 0.

Thus, W (ρ + h) ≤ W (ρ), which contradicts the smallest of ρ. Thus, (3.10) must be satisfied for all t ≥ 0. c2 1 By (3.9), H(t) ≤ (1−κ (1 + κ)p Eξp0 , that is p 1 ) c1 E|x(t, ξ)|p <

c2 1 (1 + κ)p Eξp0 e−γt , t ≥ 0. p (1 − κ1 ) c1

We may easily obtain the following theorem from the above theorem: Theorem 3.2 Assume for p ≥ 2, E|x(t)|p ≤ ce−γt , c > 0, γ is the same as in Theorem 3.1, and for 0 < κ < 1, |u(ϕ))| < κϕ, |f (ϕ, t)|p ∨ |g(ϕ, t)|p < Kϕp . Then, 1 γ log |x(t, ξ)|p < a.s. p ≥ 2 t p t→∞ sup

b for any ξ ∈ CF ([−τ, 0]; Rn ). In other words, the solution of equation (3.1) is almost surely exponentially stable.

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Vol.29 Ser.B

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