随机泛函微分方程的稳定性问题一直是许多学者关注的焦点, 已获得很多好的结论^[1-3].关于随机泛函微分方程的稳定性, 许多文献考虑的是方程解的依概率稳定、指数稳定或几乎必然稳定.但事实上, 从任意初始状态开始, 充分长时间后, 方程解可能收敛于一个随机变量, 即依分布稳定.换句话说, 无论系统的初始状态是什么, 长时间后系统将处于平衡状态, 此状态用不变测度来刻画. Hu和Wang在文献[4]中讨论了带Markov切换的中立型随机泛函微分方程的依分布稳定性; Bao等在文献[5]中讨论了带Markov切换的中立型随机延迟微分方程的依分布稳定性.另一方面, 脉冲效应是自然界的普遍现象, 关于脉冲随机泛函微分方程的稳定性已有许多好的成果. Pan和Cao在文献[6]中研究了有限延时的脉冲随机泛函微分方程的$ p $阶矩指数稳定性和几乎必然指数稳定性; Zhu在文献[7]中研究了带Markov切换的脉冲随机泛函微分方程的$ p $阶矩指数稳定性; Kao, Zhu和Qi在文献[8]中研究了带Markov切换的脉冲随机泛函微分方程的$ p $阶矩指数稳定性、几乎必然指数稳定性和$ p $阶矩指数不稳定性, 并指出了该方程可被脉冲指数稳定化的条件; 在文献[9]中, Li进一步讨论了随机泛函微分方程的稳定性问题, 通过利用方程的比较原理等技巧, 放宽了对扩散算子的限制, 引入方程系数的积分平均值和平均脉冲区间, 得到了新的Lyapunov稳定性标准; 在文献[10]和[11]中, Hu和Zhu提出了带有依赖于分布延时的脉冲效应的随机泛函微分方程和脉冲随机延迟微分方程的Razumikhin稳定性定理, 该稳定性标准的特点是Razumikhin函数的导数是不确定的, 且对方程参数的限制进一步弱化.但是, 关于脉冲随机泛函微分方程的分布稳定性的工作很少.本文从一个脉冲随机泛函微分方程出发, 利用弱收敛方法、伊藤公式和随机分析技巧, 给出了该方程依分布稳定的充分条件, 从而恰到好处地填补了这一空白.

设$ (\Omega, \mathcal {F}, \{{\mathcal{F}_t}\}_{t\geq0}, \mathbb{P}) $是一个完备的带流的概率空间, $ \{\mathcal {F}_t\}_{t\geq0} $是满足通常条件的流.本文中所有的随机变量和随机过程都定义在这个概率空间上. $ R^n $为$ n $维欧氏空间, 定义内积和范数分别为: $ \langle\alpha, \beta\rangle = \alpha^T\beta, \alpha, \beta\in R^n $; $ \mid x\mid = \sqrt{x^T x}, x\in R^n $.若$ A $是矩阵, 则其迹范数为$ ||A|| = \sqrt{A^T A} $, 其中$ A^T $表示$ A $的转置. $ B(t) = (B_1 (t), B_2 (t), \cdots, B_m(t))^T $是$ m $维标准布朗运动.设$ \tau > 0 $, $ PC([-\tau, 0];R^n) = \{\varphi:[-\tau, 0]\rightarrow R^n|\varphi(t^+) $和$ \varphi(t^-) $存在, 且$ \varphi(t^-) = \varphi(t)\} $, $ ||\varphi||_\tau: = \sup\limits_{-\tau\leq\theta\leq 0}|\varphi(\theta)| $, 这里$ \varphi(t^+), \varphi(t^-) $分别表示函数$ \varphi(t) $在$ t $处的右极限和左极限.

考虑以下脉冲随机泛函微分方程

$ \begin{equation} \left\{ \begin{array}{llllllll} dx(t) = f(x_t, t)dt+g(x_t, t)dB(t), \ t\geq 0, \ t\neq t_k, \ k = 1, 2, 3, \cdots, \\ \triangle x(t_k) = I_k (t_k, x(t_k)), \ k = 1, 2, 3, \cdots, \\ x_0 = \xi, \end{array}\right. \end{equation} $

(2.1)

其中$ \xi\in M $, $ M $是$ PC([-\tau, 0];R^n) $的有界子集且$ \xi $是$ \mathcal {F}_0 $可测的.方程(2.1)的解$ x(t) $记作$ x^\xi (t) $, $ x^\xi(t) = (x_1^\xi(t), (x_2^\xi(t), \cdots, (x_n^\xi(t))^T $, $ x^\xi _t = \{x^\xi (t+\theta):-\tau\leq\theta\leq 0\} $, $ t_k \geq 0 $是脉冲时刻, $ t_k < t_{k+1}, \lim\limits_{k\rightarrow+\infty}t_k = +\infty $, $ \triangle x^\xi (t_k) = x^\xi(t_k^+)-x^\xi (t_k) $表示$ x^\xi (t) $在$ t_k $处的跳跃幅度. $ f:PC([-\tau, 0];R^n)\times R^+\rightarrow R^n $, $ g:PC([-\tau, 0];R^n)\times R^+\rightarrow R^{n\times m} $, $ I_k:R^+ \times R^n\rightarrow R^n $.

假设$ f $, $ g $和$ I_k $满足一定的条件, 比如Lipschitz条件和线性增长条件, 即存在三个正的常数$ K_1, K_2, K_3 $, 使得

(ⅰ)对任意正数$ t $和$ \phi, \psi\in PC([-\tau, 0];R^n) $, 有$ \mid f(\phi, t)-f(\psi, t)\mid\vee\mid g(\phi, t)-g(\psi, t)\mid\leq K_1 \parallel\phi-\psi\parallel_\tau $.

(ⅱ)对任意正数$ t $和$ \phi \in PC([-\tau, 0];R^n) $, 有$ \mid f(\phi, t)\mid\vee\mid g(\phi, t)\mid\leq K_2(1+\parallel \phi\parallel_\tau $).

(ⅲ)对任意正数$ t $, 自然数$ k $和$ x, y\in R^n $, 有$ \mid I_k (t, x)-I_k (t, y)\mid\leq K_3\mid x-y\mid $.

在这些条件下, 方程(2.1)存在唯一解$ x^\xi (t), t\geq 0, x^\xi (t) $在$ (t_k, t_{k+1}) $上连续, 在$ t_k $处左连续且存在右极限^[6-9], $ f(0, t)\equiv 0, g(0, t)\equiv 0, I_k (t, 0)\equiv 0, k = 1, 2, \cdots $, 这蕴含$ x^\xi (t)\equiv 0 $是(2.1)的一个平凡解. $ f $, $ g $和$ I_k $满足以下条件.

存在常数$ \lambda_1 > \lambda_2 > 0, \lambda_3 > 0 $及$ [-\tau, 0] $上的概率测度$ \mu $, 对任意的$ \varphi, \psi\in PC([-\tau, 0];R^n) $和$ t\geq 0 $, 有

$ \begin{eqnarray} && 2<\varphi(0)-\psi(0), f(\varphi, t)-f(\psi, t)>+\parallel g(\varphi, t)-g(\psi, t)\parallel^2 \\ && \leq -\lambda_1|\varphi(0)-\psi(0)|^2+\lambda_2\int_{-\tau}^0| \varphi(\theta)-\psi(\theta)|^2 \mu(d\theta). \end{eqnarray} $

(2.2)

$ \begin{equation} \|g(\varphi, t)-g(\psi, t)\|^2\leq\lambda_3(|\varphi(0)-\psi(0)|^2+\int_{-\tau}^0|\varphi(\theta)-\psi(\theta)|^2\mu(d\theta)), \end{equation} $

(2.3)

且假设$ \{x_t^\xi\}_{t\geq 0} $满足强Markov性^{[12, 13]}.

设$ B_1 = \inf\limits_{k\geq0}\{t_{k+1}-t_k\} > 0, B_2 = \sup\limits_{k\geq0}\{t_{k+1}-t_k\} < +\infty $, 其中$ t_0 = 0 $, 易见存在非负整数$ N_1, N_2 $, 使得$ t_{N_1}\leq\tau < t_{N_1+1}, B_2\leq N_2B_1 $.记$ B_3 = 2(N_1+1)N_2-1 $, $ P(\mathcal {L}) $是所有概率测度的集合, $ P(\xi, t, \cdot) $是$ x_t^\xi $的概率分布, 对任意的$ P_1, P_2\in P(\mathcal {L}) $, 定义距离

$ d_{\mathcal {L}}(P_1, P_2) = \sup\limits_{\omega\geq 0}\sup\limits_{h\in {\mathcal {L}}}|\int_{PC([-\tau, 0];R^n)}h(\xi, \omega)P_1(d\xi)-\int_{PC([-\tau, 0];R^n)}h(\eta, \omega)P_2(d\eta)|, $

其中$ \mathcal {L} = \{h:PC([-\tau, 0];R^n)\times[0, +\infty)\rightarrow R, |h(\xi, \omega)-h(\eta, \omega)|\leq\|\xi-\eta\|_\tau, |h(\xi, \omega)|\leq1, \xi, \eta\in PC([-\tau, 0];R^n), w\in[0, +\infty)\} $.

下面先给出方程(2.1)依分布稳定的定义.

定义1^{[12, 14]}  随机过程$ \{x_t^\xi\}_{t\geq 0} $是分布稳定的是指存在一个概率测度$ \pi\in P(\mathcal{L}) $, 使得当$ t\rightarrow+\infty $时, $ P(\xi, t, \cdot) $弱收敛于$ \pi $, 即对任意的$ \xi\in PC([-\tau, 0];R^n) $, 有$ \lim\limits_{t\rightarrow +\infty} d_{\mathcal{L}}(P(\xi, t, \cdot), \; \pi(\cdot)) = 0 $, 此时也称方程(2.1)依分布稳定.

为了证明本文的主要结果, 先给出以下引理.

引理1   假设(2.2)和(2.3)式成立, 对任意的$ \xi\in M $, 有

$ \begin{equation} E|x^\xi(t_k^+)|^2 \leq\alpha E|x^\xi(t_k)|^2, \qquad k = 0, 1, 2, \cdots \qquad 0<\alpha<1, \end{equation} $

(3.1)

且

$ \begin{equation} \tau\lambda_2B_3<\frac{1}{e}, \end{equation} $

(3.2)

$ \begin{equation} \lambda_1>\frac{1}{\tau B_3}-\frac{1}{\tau}ln(\tau B_3\lambda_2), \end{equation} $

(3.3)

则存在常数$ C > 0, \delta > 0 $, 使得$ E\|x^\xi_t\|_\tau^2\leq Ce^{-\delta t}E\|\xi\|_\tau^2, t\geq 0 $.

证  对任意的$ \delta_1 > 0, \; t > 0, \; t\neq t_k, \; k = 0, 1, 2, \cdots $, 用伊藤公式, 有

$ \begin{array}{llll} d(e^{\delta_1 t}|x^\xi (t)|^2)& = &\delta_1 e^{\delta_1 t}|x^\xi (t)|^2dt+2e^{\delta_1 t}<x^\xi(t), f(x^\xi_t, t)>dt\\ &&+2e^{\delta_1t}<x^\xi(t), g(x^\xi_t, t)dB(t)> + e^{\delta_1 t}\|g(x^\xi_t, t)\|^2dt. \end{array} $

记$ y(t) = e^{\delta_1 t}E|x^\xi (t)|^2 $, 对$ t\in (t_0, t_1] $

$ \begin{array}{llll} e^{\delta_1 t}|x^\xi (t)|^2& = &|x^\xi(0^+)|^2+\int_0^t \delta_1 e^{\delta_1 s}|x^\xi (s)|^2 ds+\int_0^t 2e^{\delta_1 s}<x^\xi (s), f(x^\xi_s, s)>ds\\ & & +\int_0^t 2e^{\delta_1 s}<x^\xi(s), g(x^\xi_s, s)dB(s)>+\int_0^t e^{\delta_1 s}\|g(x_s^\xi, s)\|^2ds. \end{array} $

两边取期望, 注意到$ \int_0^t 2e^{\delta_1 s} \langle x^\xi(s), g(x^\xi_s, s)dB(s)\rangle $是鞅, 且用式(2.2), 得

$ \begin{equation} \begin{array}{llll} y(t)& = &y(0^+)+E\int_0^t\delta_1 e^{\delta_1 s}|x^\xi(s)|^2ds+E\int_0^t e^{\delta_1s}[\langle2x^\xi(s), f(x^\xi_s, s)\rangle+\|g(x^\xi_s, s)\|^2]ds\\ &\leq &y(0^+)+\delta_1 E\int_0^te^{\delta_1 s}|x^\xi(s)|^2ds-\lambda_1E\int_0^t e^{\delta_1 s}|x^\xi(s)|^2ds\\ & &+\lambda_2 E\int_0^t\int_{-\tau}^0 e^{\delta_1 s}|x^\xi(s+\theta)|^2\mu(d\theta)ds. \end{array} \end{equation} $

(3.4)

而

$ \begin{equation} \begin{array}{llll} &&E\int_0^t\int_{-\tau}^0e^{\delta_1 s}|x^\xi(s+\theta)|^2\mu(d\theta)ds = E\int_{-\tau}^0\mu(d\theta)\int_0^te^{\delta_1 s}|x^\xi(s+\theta)|^2ds\\ & = &E\int_{-\tau}^0 \mu(d\theta)\int_\theta^{t+\theta}e^{\delta_1(r-\theta)}|x^\xi(r)|^2 dr \leq e^{\delta_1 \tau}E\int_{-\tau}^0\mu(d\theta)\int_{-\tau}^t e^{\delta_1 r}|x^\xi (r)|^2dr\\ & = &e^{\delta_1 \tau}E\int_{-\tau}^t e^{\delta_1 r}|x^\xi (r)|^2dr \leq \frac{1}{\delta_1}e^{\delta_1\tau}E\|\xi\|_\tau^2+e^{\delta_1 \tau}E\int_0^t e^{\delta_1r}|x^\xi(r)|^2dr. \end{array} \end{equation} $

(3.5)

把式(3.5)代入式(3.4), 并且用式(3.1), 得

$ \begin{equation} \begin{array}{llll} y(t)\leq E\|\xi\|_\tau^2+\frac{\lambda_2}{\delta_1}e^{\delta_1 \tau}E\|\xi\|_\tau^2+(\delta_1-\lambda_1+\lambda_2e^{\delta_1 \tau})\int_0^t y(s)ds. \end{array} \end{equation} $

(3.6)

当$ t\in (t_1, t_2] $时, 注意到$ x^\xi(t) $在$ (t_{k-1}, t_k) $内连续, 在$ t_k $处左连续有右极限, $ k = 1, 2, 3, \cdots $, 用式(2.2), 得

$ \begin{equation} \begin{array}{llll} y(t)& = &y(t_1^+)+E\int_{t_1}^t \delta_1 e^{\delta_1 s}|x^\xi (s)|^2ds+E\int_{t_1}^t e^{\delta_1 s}[<2x^\xi(s), f(x^\xi_s, s)>+\|g(x^\xi_s, s)\|^2]ds\\ &\leq & y(t_1^+)+\delta_1E\int_{t_1}^t e^{\delta_1 s}|x^\xi (s)|^2ds-\lambda_1E\int_{t_1}^t e^{\delta_1 s}|x^\xi(s)|^2ds\\ &&+\lambda_2 E\int_{t_1}^t\int_{-\tau}^0e^{\delta_1 s}|x^\xi(s+\theta)|^2\mu(d\theta)ds. \end{array} \end{equation} $

(3.7)

用类似得到式(3.5)的方法, 有

$ \begin{equation} E\int_{t_1}^t\int_{-\tau}^0e^{\delta_1 s}|x^\xi(s+\theta)|^2 \mu(d\theta)ds \leq e^{\delta_1 \tau}E\int_{t_1-\tau}^{t_1}e^{\delta_1 r}|x^\xi(r)|^2dr+e^{\delta_1 \tau}E\int_{t_1}^t e^{\delta_1 r}|x^\xi(r)|^2dr. \end{equation} $

(3.8)

由式(3.1)和(3.6), 得

$ \begin{equation} \begin{array}{llll} y(t_1^+)& = &e^{\delta_1 t_1}E|x^\xi(t_1^+)|^2\leq e^{\delta_1 t_1}\alpha E|x^\xi(t_1)|^2 = \alpha y(t_1)\\ &\leq &\alpha(1+\frac{\lambda_2}{\delta_1}e^{\delta_1\tau})E\|\xi\|_\tau^2+\alpha(\delta_1-\lambda_1+\lambda_2e^{\delta_1\tau})\int_0^{t_1}y(s)ds. \end{array} \end{equation} $

(3.9)

把式(3.8)和(3.9)代入式(3.7), 得当$ t\in (t_1, t_2] $时, 有

$ \begin{equation} \begin{array}{llll} y(t) &\leq & \alpha(1+\frac{\lambda_2}{\delta_1}e^{\delta_1 \tau})E\|\xi\|_\tau^2+\alpha(\delta_1-\lambda_1+\lambda_2 e^{\delta_1 \tau})\int_0^{t_1}y(s)ds\\ &+& (\delta_1-\lambda_1+\lambda_2 e^{\delta_1 \tau})\int_{t_1}^t y(s)ds+\lambda_2e^{\delta_1 \tau}\int_{t_1-\tau}^{t_1}y(s)ds. \end{array} \end{equation} $

(3.10)

重复以上步骤, 注意到$ 0 < \alpha < 1 $, 对$ n = 2, 3, 4, \cdots, $当$ t\in (t_n, t_{n+1}] $时, 有

$ \begin{equation} y(t)\leq (1+\frac{\lambda_2}{\delta_1}e^{\delta_1 \tau})E\|\xi\|_\tau^2+(\delta_1-\lambda_1+\lambda_2 e^{\delta_1\tau})\int_0^t y(s)ds+\lambda_2 e^{\delta_1\tau}\sum\limits_{k = 1}^n\int_{t_k-\tau}^{t_k} y(s)ds. \end{equation} $

(3.11)

如果$ \delta_1-\lambda_1+\lambda_2 e^{\delta_1 \tau} < 0 $, 则此时

$ \begin{equation} y(t)\leq (1+\frac{\lambda_2}{\delta_1}e^{\delta_1 \tau})E\|\xi\|_\tau^2+\lambda_2 e^{\delta_1\tau}\sum\limits_{k = 1}^n\int_{t_k-\tau}^{t_k} y(s)ds. \end{equation} $

(3.12)

显然式(3.12)对$ t\in(t_1, t_2] $也成立.下面对任意自然数$ n $, 限制$ t\in(t_n, t_{n+1}] $, 分两种情况讨论.

(1) $ n < (N_1+1)N_2 $.此时

$ \sum\limits_{k = 1}^n\int_{t_k-\tau}^{t_k}y(s)ds\leq\sum\limits_{k = 1}^n\int_{-\tau}^t y(s)ds\leq (N_1+1)N_2\int_{-\tau}^t y(s)ds. $

(2) $ n\geq(N_1+1)N_2 $.对满足$ K\geq(N_1+1)N_2 $的自然数$ k $, 因$ t_{N_1}\leq\tau < t_{N_1+1} $, 故$ t_k-t_{N_1+1} < t_k-\tau\leq t_k-t_{N_1} $, 而$ t_k-t_{N_1+1} > t_k-(N_1+1)B_2\geq t_k-(N_1+1)N_2B_1 > t_{k-(N_1+1)N_2}, $所以

$ \begin{array}{llll} \sum\limits_{k = 1}^n\int_{t_k-\tau}^{t_k}y(s)ds& = &\sum\limits_{k = 1}^{(N_1+1)N_2-1}\int_{t_k-\tau}^{t_k}y(s)ds+\sum\limits_{k = (N_1+1)N_2}^n\int_{t_k-\tau}^{t_k}y(s)ds\\ &\leq& ((N_1+1)N_2-1)\int_{-\tau}^t y(s)ds+\sum\limits_{k = (N_1+1)N_2}^n\sum\limits_{i = 1}^{(N_1+1)N_2}\int_{t_{k-i}}^{t_{k-i+1}}y(s)ds\\ &\leq& ((N_1+1)N_2-1)\int_{-\tau}^ty(s)ds+\sum\limits_{i = 1}^{(N_1+1)N_2}\int_{t_{(N_1+1)N_2-i}}^{t_{n-i+1}}y(s)ds\\ &\leq& (2(N_1+1)N_2-1)\int_{-\tau}^t y(s)ds. \end{array} $

无论哪种情况, 都有

$ \begin{equation} y(t)\leq(1+\frac{\lambda_2}{\delta_1}e^{\delta_1\tau})E\|\xi\|_\tau^2+\lambda_2 e^{\delta_1\tau}(2(N_1+1)N_2-1)\int_{-\tau}^t y(s)ds. \end{equation} $

(3.13)

而$ \int_{-\tau}^t y(s)ds = E\int_{-\tau}^0 e^{\delta_1 s}|x^\xi(s)|^2ds+\int_0^t y(s)ds\leq \frac{1}{\delta_1}E\|\xi\|_\tau^2+\int_0^t y(s)ds $, 故

$ \begin{equation} y(t)\leq(1+2(N_1+1)N_2\frac{\lambda_2}{\delta_1}e^{\delta_1 \tau})E\|\xi\|_\tau^2+\lambda_2 e^{\delta_1\tau}(2(N_1+1)N_2-1)\int_0^t y(s)ds. \end{equation} $

(3.14)

显然, (3.14)式对$ t\in [t_0, t_1] $也成立, 故(3.14)式对$ t\in[0, +\infty) $成立, 利用Gronwall不等式, 得

$ y(t)\leq (1+2(N_1+1)N_2\frac{\lambda_2}{\delta_1}e^{\delta_1\tau})E\|\xi\|_\tau^2 \text {exp}\{\lambda_2 e^{\delta_1\tau}(2(N_1+1)N_2-1)t\}, \quad t\geq 0. $

即

$ \begin{equation} E|x^\xi(t)|^2\leq(1+2(N_1+1)N_2\frac{\lambda_2}{\delta_1}e^{\delta_1\tau})E\|\xi\|_\tau^2 \text {exp} \{(\lambda_2 e^{\delta_1\tau}(2(N_1+1)N_2-1)-\delta_1)t\}, t\geq 0. \end{equation} $

(3.15)

取$ \delta_1 = -\frac{1}{\tau}\text {ln}(\tau B_3 \lambda_2) $, 则由式(3.2)和(3.3)得$ \delta_1-\lambda_1+\lambda_2 e^{\delta_1\tau} < 0 $且$ B_3\lambda_2e^{\delta_1\tau}-\delta_1 < 0 $, 取$ \delta = \delta_1-B_3\lambda_2e^{\delta_1 \tau} $, $ C_1 = 1+2(N_1+1)N_2\frac{\lambda_2}{\delta_1}e^{\delta_1\tau} $, 则有

$ \begin{equation} E|x^\xi (t)|^2\leq C_1 E\|\xi\|_\tau^2 e^{-\delta t}, \qquad t\geq 0. \end{equation} $

(3.16)

若$ t\in[-\tau, 0] $, 则$ E|x^\xi (t)|^2\leq E\|\xi\|_\tau^2 $.而$ C_1 > 1, e^{-\delta t}\geq1 $, 故$ E|x^\xi (t)|^2\leq C_1 E\|\xi\|_\tau^2 e^{-\delta t} $.所以, 使得式(3.16)成立的$ t $的取值范围可以扩充到$ [-\tau, +\infty) $.当$ t_k < s\leq t_{k+1} $时, 利用伊藤公式和式(2.2), 得

$ \begin{eqnarray} \begin{array}{llll} |x^\xi (s)|^2-|x^\xi(t_k^+)|^2& = &\int_{t_k}^s (2<x^\xi(r), f(x^\xi_r, r)>+\|g(x_r^\xi, r)\|^2)dr+M(s)\\ &\leq& \lambda_2\int_{t_k}^s\int_{-\tau}^0|x^\xi(r+\theta)|^2\mu(d\theta)dr+M(s), \end{array} \end{eqnarray} $

(3.17)

其中$ M(s) = 2\int_{t_k}^s x^\xi (r)^T\cdot g(x^\xi_r, r)dB(s) $.利用Birkholder-Davis-Gundy不等式和式(2.3), 得

$ \begin{equation} \begin{array}{llll} E\sup\limits_{t_k<s \leq t_{k+1}}|M(s)| &\leq& C_{2, k}E[\int_{t_k}^{t_{k+1}}|x^\xi (r)^T \cdot g(x^\xi_r, r)|^2dr]^{\frac{1}{2}}\\ &\leq& C_{2, k}E[\sup\limits_{t_k<r\leq t_{k+1}}|x^\xi (r)|^2\cdot\int_{t_k}^{t_{k+1}}\|g(x^\xi_r, r)\|^2 dr]^{\frac{1}{2}}\\ &\leq& \frac{1}{2}E[\sup\limits_{t_k<r\leq t_{k+1}}|x^\xi(r)|^2]+\frac{1}{2}C_{2, k}^2E\int_{t_k}^{t_{k+1}}\|g(x_r^\xi, r)\|^2dr\\ &\leq& \frac{1}{2}E[\sup\limits_{t_k<r\leq t_{k+1}}|x^\xi(r)|^2]+\frac{1}{2}C_{2, k}^2 \lambda_3 E\int_{t_k}^{t_{k+1}}|x^\xi(r)|^2 dr\\ && +\frac{1}{2}C_{2, k}^2 \lambda_3 E\int_{t_k}^{t_{k+1}}\int_{-\tau}^0 |x^\xi(r+\theta)|^2\mu(d\theta)dr\\ &\leq& \frac{1}{2}E[\sup\limits_{t_k<r\leq t_{k+1}}|x^\xi(r)|^2]+\frac{1}{2}C_{2, k}^2 \lambda_3 E\int_{t_k}^{t_{k+1}}|x^\xi(r)|^2 dr\\ &&+\frac{1}{2}C_{2, k}^2 \lambda_3E\int_{t_k-\tau}^{t_{k+1}}|x^\xi (r)|^2 dr, \end{array} \end{equation} $

(3.18)

其中$ C_{2, k} $是正的常数, $ k = 0, 1, 2\cdots $, 由式(3.17)和(3.18)得

$ \begin{equation} \begin{array}{llll} && E[\sup\limits_{t_k<r\leq t_{k+1}}|x^\xi (r)|^2]\\ & \leq &2E|x^\xi(t_k^+)|^2+\lambda_3 C_{2, k}^2E\int_{t_k}^{t_{k+1}}|x^\xi (r)|^2dr+(C_{2, k}^2\lambda_3+2\lambda_2)E\int_{t_k-\tau}^{t_{k+1}}|x^\xi(r)|^2 dr. \end{array} \end{equation} $

(3.19)

再由式(3.1), (3.16)和(3.19), 得

$ \begin{equation} \begin{array}{llll} &&E[\sup\limits_{t_k<r\leq t_{k+1}}|x^\xi (r)|^2]\\ &\leq &2C_1 E\|\xi\|_\tau^2 e^{-\delta t_k}+\lambda_3 C_{2, k}^2C_1 E\|\xi\|_\tau^2\frac{1}{\delta}e^{-\delta t_k } +(\lambda_3 C_{2, k}^2+2\lambda_2)C_1 E\|\xi\|_\tau^2\frac{1}{\delta}e^{\delta \tau}e^{-\delta t_k}. \end{array} \end{equation} $

(3.20)

对任意的$ t > 0 $, 存在非负整数$ n_0 $, 使得$ t_{n_0} < t\leq t_{n_0+1} $, 分三种情况讨论.

(1) $ n_0\geq (N_1+1)N_2. $

$ \begin{equation} \begin{array}{llll} E\|x_t^\xi\|_\tau^2& = &E[\sup\limits_{t-\tau\leq s\leq t}|x^\xi (s)|^2]\leq E|x^\xi (t_{n_0-(N_1+1)N_2})|^2\\ & & +E[\sup\limits_{t_{n_0-(N_1+1)N_2}<s\leq t_{n_0-(N_1+1)N_2+1}}|x^\xi(s)|^2]+\cdots+E[\sup\limits_{t_{n_0}<s\leq t_{n_0+1}}|x^\xi(s)|^2]. \end{array} \end{equation} $

(3.21)

由式(3.16), (3.20)和(3.21)得

$ E\|x_t^\xi\|_\tau^2\leq C_3E\|\xi\|_\tau^2 e^{-\delta t_{n_0-(N_1+1)N_2}}\leq C_3E\|\xi\|_\tau^2e^{\delta((N_1+1)N_2+1)B_2}e^{-\delta t}, $

其中$ C_3 > 0 $是常数.

(2) $ N_1 < n_0 < (N_1+1)N_2. $

$ \begin{equation} \begin{array}{llll} E\|x_t^\xi\|_\tau^2 &\leq& E[\sup\limits_{0< s\leq t_{(N_1+1)N_2}}|x^\xi (s)|^2]\leq E[\sup\limits_{0< s\leq t_1}|x^\xi (s)|^2]\\ & &+E[\sup\limits_{t_1<s\leq t_2}|x^\xi (s)|^2]+\cdots+E[\sup\limits_{t_{(N_1+1)N_2-1}<s\leq t_{(N_1+1)N_2}}|x^\xi (s)|^2]. \end{array} \end{equation} $

(3.22)

与第(1)种情况同理, $ E\|x_t^\xi\|_\tau^2\leq C_4E\|\xi\|_\tau^2\leq C_4E\|\xi\|_\tau^2e^{\delta(N_1+1)N_2B_2}e^{-\delta t}, $其中$ C_4 > 0 $是常数.

(3) $ n_0\leq N_1. $

$ \begin{equation} E\|x_t^\xi\|_\tau^2\leq E[\sup\limits_{-\tau\leq s\leq 0}|x^\xi(s)|^2]+ E[\sup\limits_{0< s\leq t_1}|x^\xi (s)|^2]+\cdots+E[\sup\limits_{t_{N_1}< s\leq t_{N_1+1}}|x^\xi (s)|^2]. \end{equation} $

(3.23)

与第(1)种情况同理, $ E\|x_t^\xi\|_\tau^2\leq C_5 E\|\xi\|_\tau^2\leq C_5E\|\xi\|_\tau^2e^{\delta(N_1+1)B_2}e^{-\delta t}, $其中$ C_5 > 0 $是常数.

以下给出本文的主要结果.

定理1   假设引理1的条件都满足, 则方程(2.1)依分布稳定.

证   设$ x_t^\eta $是方程(2.1)满足初始条件$ x_0 = \eta $的解, 其中$ \eta\in M $.由引理1,

$ \sup\limits_{t\geq 0}\sup\limits_{\xi\in M}\{E\|x_t^\xi\|_\tau^2\}<+\infty, \; \; \lim\limits_{t\rightarrow+\infty}\sup\limits_{\xi\in M}\{E\|x_t^\xi\|_\tau^2\} = 0. $

同理$ \lim\limits_{t\rightarrow+\infty}\sup\limits_{\eta\in M}\{E\|x_t^\eta\|_\tau^2\} = 0. $所以

$ \lim\limits_{t\rightarrow+\infty}\sup\limits_{\xi, \eta\in M}E\|x_t^\xi-x_t^\eta\|_\tau^2 \leq 2\lim\limits_{t\rightarrow+\infty}\sup\limits_{\xi\in M}\{E\|x_t^\xi\|_\tau^2\}+2\lim\limits_{t\rightarrow+\infty}\sup\limits_{\eta\in M}\{E\|x_t^\eta\|_\tau^2\} = 0. $

由文献[12]的引理2.4, $ P(t, \xi, \cdot)_{t\geq 0} $是$ P(\mathcal{L}) $中的Cauchy列, 故存在唯一概率测度$ \pi(\cdot)\in P(\mathcal{L}) $, 使得$ d_{\mathcal{L}}(P(t, 0, \cdot), \pi(\cdot))\rightarrow+\infty, t\rightarrow+\infty. $所以

$ d_{\mathcal {L}}(P(t, \xi, \cdot), \pi(\cdot))\leq d_{\mathcal{L}}(P(t, \xi, \cdot), P(t, 0, \cdot))+d_{\mathcal{L}}(P(t, 0, \cdot), \pi(\cdot))\rightarrow0, \ t\rightarrow+\infty. $

即$ t\rightarrow +\infty $时, 概率测度$ P(t, \xi, \cdot)_{t\geq 0} $弱收敛于$ \pi(\cdot) $, 故方程(2.1)依分布稳定.

下面举例说明第3节中结论的正确性.考虑以下一维脉冲随机泛函微分方程

$ \begin{equation} dx(t) = -ax_t (0)dt+(x_t(0)+b\int_{-\tau}^0x_t(\theta)\mu(d\theta))dB(t), \quad t\geq 0, t\neq t_k, k = 1, 2, 3, \cdots, \end{equation} $

(4.1)

其中$ x_0 = \xi, \xi\in M $, 这里$ a, b $是常数, 且$ a > \frac{1}{2\tau}(2+ln2)+1, 0 < b < \frac{1}{2\sqrt{e\tau}} $.与前两节一样, 假设式(4.1)有唯一解, 设为$ x^\xi(t) $, 且式(3.1)成立.这时

$ f(\varphi, t) = -a\varphi(0), \; \; g(\varphi, t) = \varphi(0)+b\int_{-\tau}^0\varphi(\theta)\mu(d\theta). $

设$ B_1 = B_2 > \tau, N_1 = 0, N_2 = 1, B_3 = 1 $, 取$ \lambda_1\in(\frac{1}{\tau}(2+ln2), 2a-2), $ $ \lambda_2 = \frac{1}{2e\tau} $, $ \lambda_3 > 2+2b^2 $, 则式(3.2)和(3.3)成立, 而

$ \begin{array}{llll} &&2\langle\varphi(0)-\psi(0), f(\varphi, t)-f(\psi, t)\rangle+|g(\varphi, t)-g(\psi, t)|^2\\ &&+\lambda_1|\varphi(0)-\psi(0)|^2-\lambda_2\int_{-\tau}^0|\varphi(\theta)-\psi(\theta)|^2\mu(d\theta)\\ &\leq&(\lambda_1+2-2a)|\varphi(0)-\psi(0)|^2+(2b^2-\lambda_2)\int_{-\tau}^0|\varphi(\theta)-\psi(\theta)|^2\mu(d\theta)\\ &\leq& 0, \\ &&|g(\varphi, t)-g(\psi, t)|^2\\ &\leq& 2|\varphi(0)-\psi(0)|^2+2b^2\int_{-\tau}^0|\varphi(\theta)-\psi(\theta)|^2\mu(d\theta)\\ &\leq& \lambda_3(|\varphi(0)-\psi(0)|^2+\int_{-\tau}^0|\varphi(\theta)-\psi(\theta)|^2\mu(d\theta)). \end{array} $

即式(2.2)和(2.3)成立.由定理1知方程(4.1)依分布稳定.