引言

中立型时滞随机微分方程是描述诸如医学、生态学、经济学、物理学等学科中众多现象的一种重要工具。然而, 由于此类方程的复杂性, 虽然其应用广泛，但是解析解却很难得到, 因此研究其数值解以及数值解的收敛性很有必要。另一方面, 过程的当前状态依赖于其过去的状态, 但是这种依赖未必是常时滞的, 从而使得考虑变时滞情形具有重要的实际意义。Mao等^[1-2]提出截断Euler-Maruyama(EM)方法，并对该方法得到的数值解的收敛性和稳定性进行了研究。Milošević^[3]和Guo等^[4]分别研究了时滞随机微分方程和中立型时滞随机微分方程数值解的收敛性。Lan等^[5-7]提出了修正截断EM方法，并且得到典型随机微分方程和中立型常时滞随机微分方程的数值解的收敛速度和渐进稳定性。由于在变时滞情形下通常的修正截断方法会变为隐式格式，而相比于显式格式隐式格式通常计算量较大，因此本文将研究对应方程的显式格式(改进的修正截断EM方法)的收敛性及收敛速度。

1 基本假设、定义及重要引理

设(Ω, $\mathscr{F}$, {$\mathscr{F}_t$}_t≥0, P)是一个完备的概率空间，考虑中立型变时滞随机微分方程为

$ \begin{array}{l} \ \ \ \ \ \ \ \ \mathrm{d}[x(t)-u(x(t-\delta(t)))]=f(x(t), x(t- \\ \delta(t))) \mathrm{d} t+g(x(t), x(t-\delta(t))) \mathrm{d} B(t) \end{array} $

(1)

初值满足：x₀=ξ={ξ(θ), θ∈[－τ, 0]}∈ $C_{{\mathscr{F}_0}}^b$([－τ, 0];Rⁿ), 其中f, g, u(f: Rⁿ×Rⁿ→Rⁿ, g: Rⁿ×Rⁿ→Rⁿ⊗Rⁿ, u: Rⁿ→Rⁿ) 均为可测函数。为得到本文主要结论，提出以下假设。

H₁  对任意正整数R，存在正常数L_R，使得∀x, x, y, y∈Rⁿ，且|x|∨|x|∨|y|∨|y|≤R，都有

$ \begin{aligned} &\ \ \ \ \ \ \ \ |f(x, y)-f(\bar{x}, \bar{y})| \vee|g(x, y)-g(\bar{x}, \bar{y})| \leqslant \\ &L_{R}(|x-\bar{x}|+|y-\bar{y}|) \end{aligned} $

H₂  存在常数η∈(0, 1)，对∀x, y∈Rⁿ，都有|u(x)-u(y)|≤η|x-y|，当u(0)=0时，|u(x)|≤η|x|。

H₃  函数δ: R₊→R₊连续可微，且满足|δ′(t)| < δ < 1。

H₄(Khasminskii-type条件)  存在常数p≥2和K>0，使得对于∀x, y∈Rⁿ, a∈(0, 1]，有

$ \begin{aligned} &\ \ \ \ \ \ \ \ \left\langle x-a u\left(\frac{y}{a}\right), f(x, y)\right\rangle+\frac{p-1}{2}|g(x, y)|^{2} \leqslant K(1+ \\ &\left.|x|^{2}+|y|^{2}\right) \end{aligned} $

H₅  存在常数C_ξ，对于p≥2有

$ E \sup \limits_{t, s \in[-\tau, 0],|s-t| \leqslant \varDelta}|\xi(s)-\xi(t)|^{p} \leqslant C_{\xi} \varDelta^{\frac{p}{2}} $

需注意的是由文献[3]可知，当H₁~H₄满足时，方程(1)的解x(t)在初值x₀的条件下存在且唯一，并且对T>0有$\mathop {{\rm{sup}}}\limits_{ - \mathit{\tau } \le t \le T} E|x(t){|^p} < \infty $。

定义停时τ_R=inf {t≥0, |x(t)|≥R}, infϕ=∞，则$P\left({{\mathit{\tau }_R} \le T} \right) \le \frac{{C'}}{{{R^p}}}$。

令Δ∈(0, 1)为步长且τ=mΔ，对于充分小的Δ^*>0，令h(Δ)是正的严格递减函数h: (0, Δ^*]→(0, ∞)且满足

$ \lim\limits _{\varDelta \rightarrow 0} h(\varDelta)=\infty, \lim\limits _{\varDelta \rightarrow 0} L_{h(\varDelta)}^{2} \varDelta=0 $

(2)

由文献[6]知若给定L_R，则函数h一定存在。

对任意的Δ>0，定义修正截断函数f_Δ为

$ f_{\varDelta}(x, y)= \begin{cases}f(x, y), & |x| \vee|y| \leqslant h(\varDelta) \\ a f\left(x_{1}, y_{1}\right), & |x| \vee|y|>h(\varDelta)\end{cases} $

式中，$a = \frac{{|x| \vee |y|}}{{h(\mathit{\Delta })}}, {x_1} = \frac{x}{a}, {y_1} = \frac{y}{a}$。

下面定义改进的修正截断EM格式。令t_k=kΔ, N=T/Δ，X_k=ξ(t_k), －m≤k≤0, X_－m－1: =X_－m，若k=0, 1, …, N，令

$ \begin{array}{l} \ \ \ \ \ \ \ \ X_{k+1}=u\left(X_{k-I_{k}}\right)+X_{k}-u\left(X_{k-1-I_{k-1}}\right)+f_{\varDelta}\left(X_{k},\right. \\ \left.X_{k-I_{k}}\right) \varDelta+{g_{\varDelta}}\left(X_{k}, X_{k-I_{k}}\right) \varDelta B_{k} \end{array} $

(3)

式中，g_Δ定义与f_Δ相同，其中${I_k} = \left[{\frac{{\mathit{\delta }\left({{t_k}} \right)}}{\mathit{\Delta }}} \right], k \ge 0$, I_－1: =I₀，[x]为x的整数部分。

为定义连续格式，令

$ {\bar{x}}_{\varDelta}(t)=\sum\limits_{k=0}^{N-1} X_{k} I_{\left[t_{k}, t_{k+1}\right)}(t) $

(4)

$ \bar{y}_{\varDelta}(t)=\sum\limits_{k=0}^{N-1} X_{k-I_{k}} I_{\left[t_{k}, t_{k+1}\right)}(t) $

(5)

$ \begin{aligned} &\ \ \ \ \ \ \ \ Z_{k}(t)=\left(1-\frac{t-t_{k}}{\varDelta}\right) \bar{y}_{\varDelta}\left(t_{k-1}\right)+\frac{t-t_{k}}{\varDelta} \bar{y}_{\varDelta}\left(t_{k}\right), t \in \\ &{\left[t_{k}, t_{k+1}\right)} \end{aligned} $

(6)

$ \bar{z}_{\varDelta}(t)=\sum\limits_{k=0}^{N-1} Z_{k}(t) I_{\left[t_{k}, t_{k+1}\right)}(t) $

(7)

定义x_Δ(t)=ξ(t), t∈[－τ, 0]，对∀t∈[0, T]，令

$ \begin{aligned} &\ \ \ \ \ \ \ \ x_{\varDelta}(t)=\xi(0)+u\left(\bar{z}_{\varDelta}(t)\right)-u\left(x_{\varDelta}\left(-\varDelta-I_{-1} \varDelta\right)\right)+ \\ &\int_{0}^{t} f_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right) \mathrm{d} s+\int_{0}^{t} g_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right) \cdot \\ &\mathrm{d} B(s) \end{aligned} $

(8)

对∀t∈[t_k, t_k+1)，上述格式也可写成如下形式

$ \begin{array}{l} \ \ \ \ \ \ \ \ x_{\varDelta}(t)=x_{\varDelta}\left(t_{k}\right)+u\left(Z_{k}(t)\right)-u\left(x _ { \varDelta } \left(t_{k-1}-I_{k-1} \cdot\right.\right.\\ \varDelta))+\int_{t_{k}}^{t} f_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right) \mathrm{d} s+\int_{t_{k}}^{t} g_{\varDelta}\left(\bar{x}_{\varDelta}(s),\right.\\ \left.\bar{y}_{\varDelta}(s)\right) \mathrm{d} B(s) \end{array} $

(9)

容易验证x_Δ(t_k)=x_Δ(t_k)=X_k, －m≤k≤N。

引理1  假设H₁成立，那么对于任意固定的Δ>0，有

$ \begin{aligned} &\ \ \ \ \ \ \ \ \left|f_{\varDelta}(x, y)-f_{\varDelta}(\bar{x}, \bar{y})\right| \vee\left|g_{\varDelta}(x, y)-g_{\varDelta}(\bar{x}, \bar{y})\right| \leqslant \\ &5 L_{h(\varDelta)}(|x-\bar{x}|+|y-\bar{y}|) \end{aligned} $

证明参见文献[7]中引理3.1。

引理2  假设H₄成立，则有

$ \begin{aligned} &\ \ \ \ \ \ \ \ \left\langle x-u(y), f_{\varDelta}(x, y)\right\rangle+\frac{p-1}{2}\left|g_{\varDelta}(x, y)\right|^{2} \leqslant 2 K(1+ \\ &\left.|x|^{2}+|y|^{2}\right) \end{aligned} $

证明参见文献[7]中引理3.2。

引理3  对于∀t≥0，有

$ E\left|\bar{z}_{\varDelta}(t)\right|^{p} \leqslant \sup\limits _{-\tau \leqslant s \leqslant t} E\left|x_{\varDelta}(s)\right|^{p} $

证明  不妨设t∈[t_k, t_k+1)，由式(6)、(7)可知

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|\bar{z}_{\varDelta}(t)\right|^{p}=E\left|Z_{k}(t)\right|^{p}=E \left| \bar{y}_{\varDelta}\left(t_{k-1}\right)+\frac{t-t_{k}}{\varDelta}\right. \\ &\left.\left(\bar{y}_{\varDelta}\left(t_{k}\right)-\bar{y}_{\varDelta}\left(t_{k-1}\right)\right)\right|^{p} \leqslant \max \left\{E\left|\bar{y}_{\varDelta}\left(t_{k-1}\right)\right|^{p},\right. \\ &\left.E\left|\bar{y}_{\varDelta}\left(t_{k}\right)\right|^{p}\right\} \leqslant \sup \limits_{-\tau \leqslant s \leqslant t} E\left|x_{\varDelta}(s)\right|^{p} \end{aligned} $

引理4  假设H₁、H₂、H₃、H₅成立，且$\mathit{\eta } < \frac{1}{2}$，则存在常数C(p, T)>0使得对任意的Δ∈(0, Δ^*)，有

$ \begin{aligned} &\ \ \ \ \ \ \ \ \sup \limits_{k \leqslant\left[\frac{t}{\varDelta}\right]} E\left|x_{\varDelta}\left(t_{k}\right)-x_{\varDelta}\left(t_{k-1}\right)\right|^{p} \leqslant C(p, T)\left(L_{h(\varDelta)}^{p}\right. \\ &\left.\varDelta^{\frac{p}{2}} \sup \limits_{s \leqslant t} E\left|x_{\varDelta}(s)\right|^{p}+\varDelta^{\frac{p}{2}}\right) \end{aligned} $

证明  由式(9)易知，

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|x_{\varDelta}\left(t_{k}\right)-x_{\varDelta}\left(t_{k-1}\right)\right|^{p}=E \mid u\left(\bar{z}_{\varDelta}\left(t_{k}\right)\right)-u\left(\bar{z}_{\varDelta}\right. \\ &\left.\left(t_{k-1}\right)\right)+\int_{t_{k-1}}^{t_{k}} f_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right) \mathrm{d} s+\int_{t_{k-1}}^{t_{k}} g_{\varDelta}\left(\bar{x}_{\varDelta}(s),\right. \\ &\left.\bar{y}_{\varDelta}(s)\right)\left.\mathrm{d} B(s)\right|^{p} \leqslant(1+c)^{p-1} \eta^{p} E\left|\bar{z}_{\varDelta}\left(t_{k}\right)-\bar{z}_{\varDelta}\left(t_{k-1}\right)\right|^{p}+ \\ &\left(\frac{1+c}{c}\right)^{p-1} E\left[\left| \int_{t_{k-1}}^{t_{k}} f_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right) \mathrm{d} s+\int_{t_{k-1}}^{t_{k}} g_{\varDelta}\right.\right. \\ &\left.\left.\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right) \mathrm{d} B(s)\right|^{p}\right] \end{aligned} $

又由式(6)、(7)可知

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|\bar{z}_{\varDelta}\left(t_{k}\right)-\bar{z}_{\varDelta}\left(t_{k-1}\right)\right|^{p}=E\left|\bar{y}_{\varDelta}\left(t_{k-1}\right)-\bar{y}_{\varDelta}\left(t_{k-2}\right)\right|^{p}= \\ &E\left|X_{k-1-I_{k-1}}-X_{k-2-I_{k-2}}\right|^{p} \end{aligned} $

根据[x]的定义和假设H₃可知

$ \begin{array}{l} \ \ \ \ \ \ \ \ k-1-I_{k-1}-\left(k-2-I_{k-2}\right)=1+I_{k-2}-I_{k-1} \leqslant \\ 1+\left(\left[\frac{\delta\left(t_{k-2}\right)-\delta\left(t_{k-1}\right)}{\varDelta}\right]+1\right) \leqslant 1+\left(\left[\frac{\left|\delta^{\prime}(\theta)\right| \varDelta}{\varDelta}\right]+\right.\\ 1)=2 \end{array} $

故

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|\bar{z}_{\varDelta}\left(t_{k}\right)-\bar{z}_{\varDelta}\left(t_{k-1}\right)\right|^{p} \leqslant 2^{p-1} \sup\limits _{-m \leqslant i \leqslant k} E \mid x_{\varDelta}\left(t_{i}\right)- \\ &\left.x_{\varDelta}\left(t_{i-1}\right)\right|^{p} \end{aligned} $

又由引理1，所以得到

$ \begin{array}{l} \ \ \ \ \ \ \ \ \sup \limits_{k \leqslant\left[\frac{t}{\varDelta}\right]} E\left|x_{\varDelta}\left(t_{k}\right)-x_{\varDelta}\left(t_{k-1}\right)\right|^{p} \leqslant[2(1+c)]^{p-1} \eta^{p} \\ \sup \limits_{i \leqslant\left[\frac{t}{\varDelta}\right]} E\left|x_{\varDelta}\left(t_{i}\right)-x_{\varDelta}\left(t_{i-1}\right)\right|^{p}+\left(\frac{1+c}{c}\right)^{p-1} \sup \limits_{k \leqslant\left[\frac{t}{\varDelta}\right]} E \\ {\left[\left| \int_{t_{k-1}}^{t_{k}} f_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right) \mathrm{d} s+\int_{t_{k-1}}^{t_{k}} g_{\varDelta}\left(\bar{x}_{\varDelta}(s),\right.\right.\right.} \\ \left.\left.\bar{y}_{\varDelta}(s)\right)\left.\mathrm{d} B(s)\right|^{p}\right] \leqslant[2(1+c)]^{p-1} \eta^{p} \sup \limits_{i \leqslant\left[\frac{i}{\varDelta}\right]} E \mid x_{\varDelta}\left(t_{i}\right)-\\ \left.x_{\varDelta}\left(t_{i}\right)\right|^{p}+\left(\frac{3(1+c)}{c}\right)^{p-1}\left(\varDelta^{p}+\varDelta^{\frac{p}{2}}\right) L_{h(\varDelta)}^{p} 2 \sup \limits_{s \leqslant t} E \\ \left|x_{\varDelta}(s)\right|^{p}+\left(\frac{3(1+c)}{c}\right)^{p-1}\left(|f(0,0)|^{p} \varDelta^{p}+|g(0,0)|^{p} \varDelta^{\frac{p}{2}}\right) \end{array} $

由于$\mathit{\eta } < \frac{1}{2}$，故可取c充分小使得[2(1+c)]^p－1η^p < 1，结论得证。

引理5  设引理4中的条件均成立，则

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|x_{\varDelta}(t)-\bar{x}_{\varDelta}(t)\right|^{p} \leqslant C(p, T)\left(L_{h(\varDelta)}^{p} \varDelta^{\frac{p}{2}} \sup \limits_{s \leqslant t} E\right. \\ &\left.\left|x_{\varDelta}(s)\right|^{p}+\varDelta^{\frac{p}{2}}\right) \end{aligned} $

证明  不妨设t∈[t_k, t_k+1)。由式(9)得

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|x_{\varDelta}(t)-\bar{x}_{\varDelta}(t)\right|^{p}=E \left| u\left(Z_{k}(t)\right)-u\left(x _ { \varDelta } \left(t_{k-1}-\right.\right.\right. \\ &\left.\left.I_{k-1} \varDelta\right)\right)+\int_{t_{k}}^{t} f_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right) \mathrm{d} s+\int_{t_{k}}^{t} g_{\varDelta}\left(\bar{x}_{\varDelta}(s),\right. \\ &\left.\bar{y}_{\varDelta}(s)\right)\left.\mathrm{d} B(s)\right|^{p} \leqslant\left(1+c_{0}\right)^{p-1} \eta^{p} E \mid Z_{k}(t)-x_{\varDelta}\left(t_{k-1}-\right. \\ &\left.I_{k-1} \varDelta\right)\left.\right|^{p}+\left[\frac{2\left(1+c_{0}\right)}{c_{0}}\right]^{p-1} \times E\left[\left| \int_{t_{k}}^{t} f_{\varDelta}\left(\bar{x}_{\varDelta}(s),\right.\right.\right. \\ &\left.\left.\bar{y}_{\varDelta}(s)\right)\left.\mathrm{d} s\right|^{p}+\left|\int_{t_{k}}^{t} g_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right) \mathrm{d} B(s)\right|^{p}\right] \end{aligned} $

注意到

$ \begin{aligned} &\ \ \ \ \ \ \ \ \left|Z_{k}(t)-x_{\varDelta}\left(t_{k-1}-I_{k-1} \varDelta\right)\right|=\left|\left(1-\frac{t-t_{k}}{\varDelta}\right)\right. \\ &\left.X_{k-1-I_{k-1}}+\frac{t-t_{k}}{\varDelta} X_{k-I_{k}}-X_{k-1-I_{k-1}}\right|\leqslant| X_{k-I_{k}}- \\ &X_{k-1-I_{k-1}} \mid \end{aligned} $

从而由引理4得

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|x_{\varDelta}(t)-\bar{x}_{\varDelta}(t)\right|^{p} \leqslant C(p, T)\left(L_{h(\varDelta)}^{p} \varDelta^{\frac{p}{2}} \sup \limits_{s \leqslant t} E\right. \\ &\left.\left|x_{\varDelta}(s)\right|^{p}+\varDelta^{\frac{p}{2}}\right) \end{aligned} $

根据引理5不难得到

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|\bar{y}_{\varDelta}(t)-\bar{z}_{\varDelta}(t)\right|^{p} \leqslant C(p, T)\left(L_{h(\varDelta)}^{p} \varDelta^{\frac{p}{2}} \sup \limits_{s \leqslant t} E\right. \\ &\left.\left|x_{\varDelta}(s)\right|^{p}+\varDelta^{\frac{p}{2}}\right) \end{aligned} $

引理6  假设H₁~H₅均成立，则

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|x_{\varDelta}(t)-u\left(\bar{z}_{\varDelta}(t)\right)\right|^{p} \leqslant C(p, T)\left(L_{h(\varDelta)}^{p} \varDelta^{\frac{p}{2}}\cdot\right. \\ &\left.\int_{0}^{t} \sup \limits_{r \leqslant s} E\left|x_{\varDelta}(r)\right|^{p} \mathrm{~d} s+\varDelta^{\frac{p}{2}}\right)+C(p, \eta, T) \end{aligned} $

证明  由伊藤公式得

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|x_{\varDelta}(t)-u\left(\bar{z}_{\varDelta}(t)\right)\right|^{p} \leqslant E \left| \xi(0)-u\left(x _ { \varDelta } \left(-\varDelta-\right.\right.\right. \\ &\left.\left.I_{-1} \varDelta\right)\right)\left.\right|^{p}+p E \int_{0}^{t}\left|x_{\varDelta}(s)-u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p-2}\left[\bar{x}_{\varDelta}(s)-\right. \\ &\left\langle u\left(\bar{y}_{\varDelta}(s)\right), f_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right)\right\rangle+\frac{p-1}{2} \mid g_{\varDelta}\left(\bar{x}_{\varDelta}(s),\right. \\ &\left.\left.\bar{y}_{\varDelta}(s)\right)\left.\right|^{2}\right] \mathrm{d} s+E \int_{0}^{t}\left|x_{\varDelta}(s)-u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p-2}\left\langle x_{\varDelta}(s)-\right. \\ &\left.\bar{x}_{\varDelta}(s), f_{\varDelta}\left(\bar{x}_{\varDelta}(s), \bar{y}_{\varDelta}(s)\right)\right\rangle \mathrm{d} s+E \int_{0}^{t} \mid x_{\varDelta}(s)- \\ &\left.u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p-2}\left\langle u\left(\bar{y}_{\varDelta}(s)\right)-u\left(\bar{z}_{\varDelta}(s)\right), f_{\varDelta}\left(\bar{x}_{\varDelta}(s),\right.\right. \\ &\left.\left.\bar{y}_{\varDelta}(s)\right)\right\rangle \mathrm{d} s=: I_{1}+I_{2}+I_{3}+I_{4} \end{aligned} $

下面分别估计I₁、I₂、I₃、I₄这4项。

$ \begin{aligned} &\ \ \ \ \ \ \ \ I_{1} \leqslant 2^{p-1}\left(E|\xi(0)|^{p}+\eta^{p} E\left|x_{\varDelta}\left(-\varDelta-I_{-1} \varDelta\right)\right|^{p}\right) \leqslant \\ &C(p, \eta) \sup \limits_{-\tau \leqslant s \leqslant 0} E|\xi(s)|^{p} \end{aligned} $

(10)

由引理2和Young不等式得

$ \begin{aligned} &\ \ \ \ \ \ \ \ I_{2} \leqslant p K E \int_{0}^{t}\left|x_{\varDelta}(s)-u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p-2} \times(1+ \\ &\left.\left|\bar{x}_{\varDelta}(s)\right|^{2}+\left|\bar{y}_{\varDelta}(s)\right|^{2}\right) \mathrm{d} s \leqslant C(p, \eta)\left(E \int_{0}^{t} \mid x_{\varDelta}(s)-\right. \\ &\left.\left.u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p} \mathrm{~d} s+\int_{0}^{t} \sup \limits_{r \leqslant s} E\left|x_{\varDelta}(r)\right|^{p} \mathrm{~d} s\right)+C(p, \eta, T) \end{aligned} $

(11)

由引理1和Young不等式得

$ \begin{aligned} &\ \ \ \ \ \ \ \ I_{3} \leqslant p E \int_{0}^{t}\left|x_{\varDelta}(s)-u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p-2}\left(|f(0,0)|^{2}+\right. \\ &\left.L_{h(\varDelta)}^{2}\left|x_{\varDelta}(s)-\bar{x}_{\varDelta}(s)\right|^{2}+\left|\bar{x}_{\varDelta}(s)\right|^{2}+\left|\bar{y}_{\varDelta}(s)\right|^{2}\right) \mathrm{d} s \leqslant \\ &C(p, \eta) E \int_{0}^{t}\left|x_{\varDelta}(s)-u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p} \mathrm{~d} s+C(p, T) \\ &\left(L_{h(\varDelta)}^{p} \varDelta^{\frac{p}{2}} \int_{0}^{t} \sup \limits_{r \leqslant s} E\left|x_{\varDelta}(r)\right|^{p} \mathrm{~d} s+\varDelta^{\frac{p}{2}}\right) \end{aligned} $

(12)

同理，由引理5和Young不等式可得

$ \begin{aligned} &\ \ \ \ \ \ \ \ I_{4} \leqslant p \eta E \int_{0}^{t}\left|x_{\varDelta}(s)-u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p-2}\left(|f(0,0)|^{2}+\right. \\ &\left.L_{h(\varDelta)}^{2}\left|\bar{y}_{\varDelta}(s)-\bar{z}_{\varDelta}(s)\right|^{2}+\left|\bar{x}_{\varDelta}(s)\right|^{2}+\left|\bar{y}_{\varDelta}(s)\right|^{2}\right) \mathrm{d} s \leqslant \\ &C(p, \eta) E \int_{0}^{t}\left|x_{\varDelta}(s)-u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p} \mathrm{~d} s+C(p, T)\\ &\left(L_{h(\varDelta)}^{p} \varDelta^{\frac{p}{2}} \int_{0}^{t} \sup \limits_{r \leqslant s} E\left|x_{\varDelta}(r)\right|^{p} \mathrm{~d} s+\varDelta^{\frac{p}{2}}\right) \end{aligned} $

(13)

由式(10)~(13)得

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|x_{\varDelta}(t)-u\left(\bar{z}_{\varDelta}(t)\right)\right|^{p} \leqslant C(p, \eta) E \int_{0}^{t} \mid x_{\varDelta}(s)- \\ &\left.u\left(\bar{z}_{\varDelta}(s)\right)\right|^{p} \mathrm{~d} s+C(p, T)\left(L_{h(\varDelta)}^{p} \varDelta^{\frac{p}{2}} \int_{0}^{t} \sup \limits_{r \leqslant s} E\left|x_{\varDelta}(r)\right|^{p} \mathrm{~d} s+\right. \\ &\left.\varDelta^{\frac{p}{2}}\right)+C(p, \eta, T) \end{aligned} $

再根据Gronwall引理^[8]即可得证。

引理7  假设H₁~H₅均成立，且$\mathit{\eta } < \frac{1}{2}$, p>2, 则存在0 < Δ₀ < Δ^*和C(p, T)>0，使得对任意Δ∈(0, Δ₀)，有$\mathop {{\rm{sup}}}\limits_{0 < \mathit{\Delta } < {\mathit{\Delta }_0}} \mathop {{\rm{sup}}}\limits_{0 < t < T} E{\left| {{x_\mathit{\Delta }}(t)} \right|^p} \le C < \infty $, ∀T>0。

定义停时ρ_{Δ, R}=inf{t≥0, |x_Δ(t)|≥R}，则$P\left({{\mathit{\rho }_{\mathit{\Delta }, R}} \le T} \right) \le \frac{C}{{{R^p}}}$。

证明

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|x_{\varDelta}(t)\right|^{p}=E\left|x_{\varDelta}(t)-u\left(z_{\varDelta}(t)\right)+u\left(z_{\varDelta}(t)\right)\right|^{p} \leqslant \\ &\left(1+c^{\prime}\right)^{p-1} E\left|x_{\varDelta}(t)-u\left(z_{\varDelta}(t)\right)\right|^{p}+\left(\frac{1+c^{\prime}}{c^{\prime}}\right)^{p-1} \eta^{p} \cdot \\ &E\left|z_{\varDelta}(t)\right|^{p} \end{aligned} $

当c′充分大时，${\left({\frac{{1 + c'}}{{c'}}} \right)^{p - 1}}{\mathit{\eta }^p} < 1$，由引理3、引理6和Gronwall引理得

$ \sup \limits_{0<\varDelta<\varDelta_{0} } \sup \limits_{0<t<T}E\left|x_{\varDelta}(t)\right|^{p} \leqslant C<\infty, \forall T>0 $

(14)

显然将式(14)中的x_Δ(t)换成x_Δ(t∧ρ_{Δ, R})不等式仍成立。

再由R^pP(ρ_{Δ, R}≤T)≤E(|x_Δ(T∧ρ_{Δ, R})|^p)≤C可得$P\left({{\mathit{\rho }_{\mathit{\Delta }, R}} \le T} \right) \le \frac{C}{{{R^p}}}$。

引理8  假设引理7中的假设均成立，则对∀t∈[t_k, t_k+1)和充分小的Δ(< 1)，2 < q < p，有

$ E\left|\bar{y}_{\varDelta}(t-\delta(t))-\bar{z}_{\varDelta}(t)\right|^{q} \leqslant C(q, T) L_{h(\varDelta)}^{q} \varDelta^{\frac{q}{2}} $

证明  显然t－δ(t)∈[t_k－1－I_kΔ, t_k+1－I_k+1·Δ)，由H₃，可得

$ \begin{aligned} &\ \ \ \ \ \ \ \ t_{k+1}-I_{k+1} \varDelta-\left(t_{k-1}-I_{k} \varDelta\right)=2 \varDelta+\left(I_{k} \varDelta-I_{k+1} \varDelta\right) \leqslant \\ &3 \varDelta+\left[\left|\delta^{\prime}(\theta)\right|\right] \varDelta=3 \varDelta \end{aligned} $

故y_Δ(t－δ(t))可能取X_k－1－Ik、X_k－Ik、X_k+1－Ik，由引理4可得如下结果。

1) 当y_Δ(t－δ(t))=X_k－1－Ik时，有

$ \begin{array}{l} \ \ \ \ \ \ \ \ E\left|\bar{y}_{\varDelta}(t-\delta(t))-\bar{z}_{\varDelta}(t)\right|^{q}=E\left|X_{k-1-I_{k}}-\bar{z}_{\varDelta}(t)\right|^{q} \leqslant \\ 2^{q-1}\left(1-\frac{t-t_{k}}{\varDelta}\right)^{q} E\left|X_{k-1-I_{k}}-X_{k-1-I_{k-1}}\right|^{q}+2^{q-1} \cdot\\ \left(\frac{t-t_{k}}{\varDelta}\right)^{q} E\left|X_{k-1-I_{k}}-X_{k-I_{k}}\right|^{q} \leqslant 2^{q-1} \sup \limits_{-m \leqslant j \leqslant k} E \mid X_{j}- \\ \left.X_{j-1}\right|^{q} \leqslant C(q, T) L_{h(\varDelta)}^{q} \varDelta^{\frac{q}{2}}\left(\sup \limits_{r \leqslant t} E\left|x_{\varDelta}(r)\right|^{q}+\varDelta^{\frac{q}{2}}\right) \end{array} $

2) 当y_Δ(t－δ(t))=X_k－Ik时，有

$ \begin{array}{l} \ \ \ \ \ \ \ \ E\left|\bar{y}_{\varDelta}(t-\delta(t))-\bar{z}_{\varDelta}(t)\right|^{q}=E\left|X_{k-I_{k}}-\bar{z}_{\varDelta}(t)\right|^{q} \leqslant\\ E\left|X_{k-1-I_{k}}-X_{k-1-I_{k-1}}\right|^{q} \leqslant \sup \limits_{-m \leqslant j \leqslant k} E\left|X_{j}-X_{j-1}\right|^{q} \leqslant C(q,\\ T) L_{h(\varDelta)}^{q} \varDelta^{\frac{q}{2}}\left(\sup \limits_{r \leqslant t} E\left|x_{\varDelta}(r)\right|^{q}+\varDelta^{\frac{q}{2}}\right)\end{array} $

3) 同理当y_Δ(t－δ(t))=X_k+1－Ik时，有

$ \begin{gathered} \ \ \ \ \ \ \ \ E\left|\bar{y}_{\varDelta}(t-\delta(t))-\bar{z}_{\varDelta}(t)\right|^{q} \leqslant 3^{q-1} \sup \limits_{-m \leqslant j \leqslant k} E \mid X_{j}- \\ \left.X_{j-1}\right|^{q} \leqslant C(q, T) L_{h(\varDelta)}^{q} \varDelta^{\frac{q}{2}}\left(\sup \limits_{r \leqslant t} E\left|x_{\varDelta}(r)\right|^{q}+\varDelta^{\frac{q}{2}}\right) \end{gathered} $

由引理7, 结论得证。

2 主要结果与证明

定理1  设H₁~H₅成立，则由式(8)定义的数值解x_Δ(t)q阶矩强收敛到精确解x(t)，即对任意的p≥2，q∈[2, p)，有$\mathop {\sup }\limits_{0 \le t \le T} E{\left| {{x_\mathit{\Delta }}(t) - x(t)} \right|^q} \le {C_{q, T}}L_{h(\mathit{\Delta })}^q \cdot {\mathit{\Delta }^{\frac{q}{2}}}$。

证明  截断函数定义为F_R(x, y)=f_h^－1(R)(x, y) 和G_R(x, y)=g_h^－1(R)(x, y)，取Δ充分小，易知对任意|x|∨|y|≤R≤h(Δ)，F_R(x, y)=f_h^－1(R)(x, y)=f(x, y)=f_Δ(x, y)，同理G_R(x, y)=g_h^－1(R)(x, y)=g(x, y)=g_Δ(x, y)。

设v(θ)=ξ(θ), θ∈[－τ, 0], 当t≥0时考虑如下中立型变时滞随机微分方程。

$ \begin{array}{l} \ \ \ \ \ \ \ \ \mathrm{d}[v(t)-u(v(t-\delta(t)))]=F_{R}(v(t), v(t- \\ \delta(t))) \mathrm{d} t+G_{R}(v(t), v(t-\delta(t))) \mathrm{d} B(t) \end{array} $

(15)

对任意固定的R, F_R、G_R满足全局Lipschtiz条件，故方程(15)有唯一的解v(t), t≥－τ，所以有

$ P\left(x\left(t \wedge \tau_{R}\right)=v\left(t \wedge \tau_{R}\right), \forall t \in[0, T]\right)=1 $

(16)

类似于方程(1)的数值解，可类似定义v_Δ(t)，${{\bar v}_\mathit{\Delta }}(t) = \sum\limits_{k = 0}^{N - 1} {{V_k}} {I_{\left[{{t_k}, {t_{k + 1}}} \right)}}(t)$，${{\mathit{\bar v'}}_\mathit{\Delta }}(t) = \sum\limits_{k = 0}^{N - 1} {{V_{k - {I_k}}}} $· ${I_{\left[{{t_k}, {t_{k + 1}}} \right)}}(t), {{\bar v'}_k}(t) = \left({1 - \frac{{t - {t_k}}}{\mathit{\Delta }}} \right){{\bar v'}_\mathit{\Delta }}\left({{t_{k - 1}}} \right) + \frac{{t - {t_k}}}{\mathit{\Delta }}{{\bar v'}_\mathit{\Delta }}\left({{t_k}} \right)$，$t \in \left[{{t_k}, {t_{k + 1}}} \right), {{\mathit{\bar v'}}_\mathit{\Delta }}(t) = \sum\limits_{k = 0}^{N - 1} {{{\bar v'}_k}} (t){I_{\left[{{t_k}, {t_{k + 1}}} \right)}}(t)$。

由解的唯一性，方程(1)、(15)对应的数值解满足

$ P\left(x_{\varDelta}\left(t \wedge \rho_{\varDelta, R}\right)=v_{\varDelta}\left(t \wedge \rho_{\varDelta, R}\right), \forall t \in[0, T]\right)=1 $

(17)

令l(t)=v(t)－u(v(t－δ(t)))，l_Δ(t)=v_Δ(t)－u(v′_Δ(t))，则

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|v(t)-v_{\varDelta}(t)\right|^{q} \leqslant\left(1+c_{1}\right)^{q-1} E\left|l(t)-l_{\varDelta}(t)\right|^{q}+ \\ &\left(\frac{1+c_{1}}{c_{1}}\right)^{q-1} \eta^{q} E\left|v(t-\delta(t))-\bar{v}_{\varDelta}^{\prime}(t)\right|^{q} \end{aligned} $

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|l(t)-l_{\varDelta}(t)\right|^{q} \leqslant 2^{q-1} T^{q-1} E \int_{0}^{t} \mid F_{R}(v(s), v(s- \\ &\delta(s)))-\left.F_{R}\left(\bar{v}_{\varDelta}(s), \bar{v}_{\varDelta}^{\prime}(s)\right)\right|^{q} \mathrm{~d} s+2^{q-1} T^{q}-1 E \int_{0}^{t} \\ &\left|G_{R}(v(s), v(s-\delta(s)))-G_{R}\left(\bar{v}_{\varDelta}(s), \bar{v}_{\varDelta}^{\prime}(s)\right)\right|^{q} \mathrm{~d} s \leqslant \\ &2^{q-1}\left(T^{q-1}+T^{q}-1\right) E \int_{0}^{t} L_{R}^{q}\left(\left|v(s)-\bar{v}_{\varDelta}(s)\right|^{q}+\mid v(s-\right. \\ &\left.\delta(s))-\left.\bar{v}_{\varDelta}^{\prime}(s)\right|^{q}\right) \mathrm{d} s \leqslant 2^{q-1}\left(T^{q-1}+T^{q}-1\right) L_{R}^{q}[E \\ &\int_{0}^{t}\left(\left|v(s)-v_{\varDelta}(s)+v_{\varDelta}(s)-\bar{v}_{\varDelta}(s)\right|^{q} \mathrm{~d} s+E \int_{0}^{t} \mid v(s-\right. \\ &\delta(s))-v_{\varDelta}(s-\delta(s))+v_{\varDelta}(s-\delta(s))-\bar{v}_{\varDelta}(s-\delta(s))+ \\ &\left.\bar{v}_{\varDelta}(s-\delta(s))-\left.\bar{v}_{\varDelta}^{\prime}(s)\right|^{q} \mathrm{~d} s\right] \end{aligned} $

根据引理5和引理8得

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|l(t)-l_{\varDelta}(t)\right|^{q} \leqslant C(q, T) L_{R}^{q} E \int_{0}^{t} \mid v(s)- \\ &\left.v_{\varDelta}(s)\right|^{q} \mathrm{~d} s+C(q, T) L_{R}^{q} \varDelta^{\frac{q}{2}}\left(\sup \limits_{r \leqslant t} E\left|x_{\varDelta}(r)\right|^{q}+\varDelta^{\frac{q}{2}}\right) \end{aligned} $

(18)

同理可得

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|v(t-\delta(t))-v_{\varDelta}^{\prime}(t)\right|^{q} \leqslant\left(\frac{1+c_{2}}{c_{2}}\right)^{q-1} E \mid v(t- \\ &\delta(t))-\left.v_{\varDelta}(t-\delta(t))\right|^{q}+C(q, T) L_{h(\varDelta)}^{q} \varDelta^{\frac{q}{2}}\left(\sup \limits_{r \leqslant t} E\right. \\ &\left.\left|x_{\varDelta}(r)\right|^{q}+\varDelta^{\frac{q}{2}}\right) \end{aligned} $

取c₁、c₂充分大，使得${\left({\frac{{1 + {c_1}}}{{{c_1}}}} \right)^{q - 1}}\; \; {\mathit{\eta }^q}$·${\left({\frac{{1 + {c_2}}}{{{c_2}}}} \right)^{q - 1}} < 1$, 则

$ \begin{aligned} &\ \ \ \ \ \ \ \ \sup \limits_{s \leqslant t} E \quad\left|\quad v \quad(s)-v_{\varDelta} \quad(s) \quad\right|^{q} \leqslant \\ &\frac{C(q, T) L_{R}^{q} E \int_{0}^{t}\left|v(s)-v_{\varDelta}(s)\right|^{q} \mathrm{~d} s}{1-\left(\frac{1+c_{1}}{c_{1}}\right)^{q-1} \eta^{q}\left(\frac{1+c_{2}}{c_{2}}\right)^{q-1}}+ \\ &\frac{C(q, T) L_{R}^{q} \varDelta^{\frac{q}{2}}\left(\sup \limits_{r \leqslant t} E\left|x_{\varDelta}(r)\right|^{q}+\varDelta^{\frac{q}{2}}\right)}{1-\left(\frac{1+c_{1}}{c_{1}}\right)^{q-1} \eta^{q}\left(\frac{1+c_{2}}{c_{2}}\right)^{q-1}} \end{aligned} $

(19)

易知将t替换成t∧θ_Δ，R，其中θ_Δ，R=τ_R∧θ_Δ，R, 式(19)仍成立。类似于文献[6]中引理4.1的证明, 由Gronwall引理得

$ \begin{aligned} &\ \ \ \ \ \ \ \ \sup \limits_{s \leqslant t} E\left|v\left(s \wedge \theta_{\varDelta, R}\right)-v_{\varDelta}\left(s \wedge \theta_{\varDelta, R}\right)\right|^{q} \leqslant C_{1} L_{R}^{q} \cdot \\ &\exp \left(C_{2} L_{R}^{q} T\right) \varDelta^{\frac{q}{2}} \end{aligned} $

其中，

$ C_{1}:=\frac{C(q, T)\left(\sup \limits_{r \leqslant t} E\left|x_{\varDelta}(r)\right|^{q}+\varDelta^{\frac{q}{2}}\right)}{1-\left(\frac{1+c_{1}}{c_{1}}\right)^{q-1} \eta^{q}\left(\frac{1+c_{2}}{c_{2}}\right)^{q-1}} $

$ C_{2}=\frac{C(q, T)}{1-\left(\frac{1+c_{1}}{c_{1}}\right)^{q-1} \eta^{q}\left(\frac{1+c_{2}}{c_{2}}\right)^{q-1}} $

令h(Δ^*)=L^－1(L_Rexp(C₂L_R^qT/q))，则∀Δ∈(0, Δ^*)，有

$ \begin{aligned} &\ \ \ \ \ \ \ \ E\left|v\left(t \wedge \theta_{\varDelta, R}\right)-v_{\varDelta}\left(t \wedge \theta_{\varDelta, R}\right)\right|^{q} \leqslant C_{1} L_{R}^{q} \cdot \\ &\exp \left(C_{2} L_{R}^{q} T\right) \varDelta^{\frac{q}{2}} \leqslant C_{1} L_{h(\varDelta)}^{q} \varDelta^{\frac{q}{2}} \end{aligned} $

由式(16)、(17)以及停时的定义可得

$ E\left|x\left(t \wedge \theta_{\varDelta, R}\right)-x_{\varDelta}\left(t \wedge \theta_{\varDelta, R}\right)\right|^{q} \leqslant C(q, T) L_{h(\varDelta)}^{q} \varDelta^{\frac{q}{2}} $

再由标准的截断程序^{[1, 5]}和引理7即得

$ E\left|x(T)-x_{\varDelta}(T)\right|^{q} \leqslant C(q, T) L_{h(\varDelta)}^{q} \varDelta^{\frac{q}{2}} $