引言

随机时滞微分方程可用于刻画有重要应用并与时间相关，且依赖于过去状态的过程。近年来，关于这类方程数值解稳定性的研究取得了很大进展。Higham^[1]研究了线性随机微分方程对应的随机θ方法的渐近稳定性；Zong等^[2]研究了非线性方程对应的随机θ方法的均方指数稳定性；Wu等^[3]得到了中立型随机时滞微分方程相应数值解的指数稳定性；Lan等^[4]得到了带马氏切换的中立型随机时滞微分方程的数值解指数稳定的充分条件。当扩散系数不满足线性增长条件时，Huang^[5]通过研究随机线性θ方法平凡解的渐近稳定性，证明了特定条件下随机线性θ(1/2≤θ≤1)方法的均方渐近稳定性，但是并没有给出收敛速度。

本文研究当扩散系数高度非线性时随机线性θ方法的均方指数稳定性，在给出随机时滞微分方程的随机线性θ方法和两种稳定性定义的基础上，证明了给定条件下随机线性θ方法的均方指数稳定性(从而几乎处处指数稳定)，完善了文献[5]中的结论。

1 基本假设和定义

对于一个抽象的全集Ω，$ \mathscr{F}$是Ω的子集类，P表示概率测度，令(Ω, $\mathscr{F} $, {$\mathscr{F} $_t}_t≥0, P)是一个完全概率空间，σ代数流{$\mathscr{F} $_t}_t≥0满足基本条件(即单增、右连续且$\mathscr{F} $₀包含所有零测集)。定义期望E为Ω上随机变量关于P的积分。W(t)是定义在概率空间上的标准布朗运动。

考虑以下形式的随机时滞微分方程

$ \left\{ \begin{array}{l} {\rm{d}}y(t) = f(t,y(t),y(t - \tau )){\rm{d}}t + \\ \;\;\;\;\;\;g(t,y(t),y(t - \tau )){\rm{d}}W(t),t \ge 0\\ y(t) = \phi (t),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;t \in \left[ { - \tau ,0} \right] \end{array} \right. $

式中，y(t)、f(t, x, y)、g(t, x, y)是定义在相同概率空间上的随机变量，y(t)是未知函数，t∈[0, T]；τ是正常数；ϕ(t)是$\mathscr{F} $₀可测的，且满足$\mathop {\sup }\limits_{ - \tau \le t \le 0} E{\left| {\phi \left( t \right)} \right|^2} < \infty $。

考虑随机线性θ方法(SLT方法)，即划分格子点0=t₀ < t₁ < t₂… < t_n < T，t_n=nΔ，Δ为步长。考虑随机线性θ方法的数值迭代格式如下

$ \left\{ \begin{array}{l} {y_{n + 1}} = {y_n} + \theta \mathit{\Delta} f\left( {{t_{n + 1}},{y_{n + 1}},{{\bar y}_{n + 1}}} \right) + \\ \;\;\;\;\;\;(1 - \theta )\mathit{\Delta} f\left( {{t_n},{y_n},{{\bar y}_n}} \right) + g\left( {{t_n},{y_n},{{\bar y}_n}} \right)\mathit{\Delta} {W_n},\;\;\;n \ge 0\\ {y_n} = \phi \left( {{t_n}} \right),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;n \le 0 \end{array} \right. $

(1)

式中，y_n+1是y(t_n+1)的近似值，y_n是y(t_n-τ)的近似值，Δ>0, θ∈[0, 1], ΔW_n=W(t_n+1)－W(t_n)。

对于τ，存在正整数m和δ(0<δ≤1)，使得τ=(m－δ)Δ，所以式(1)中的y_n可以线性表示为y_n=(1－δ)y_n－m+δy_n－m+1。显然当δ=0时，随机线性θ方法回归为普通随机θ方法，该方法是随机θ方法的一个推广。

注意到当θ>0时式(1)是隐式方法，为使式(1)有定义，通常需要使系数f满足特定条件(如单边Lipschitz条件)。

定义1  对于式(1)中的随机线性θ方法y_n，若存在Δ>0，使得$ \mathop {\lim \;\sup }\limits_{k \to \infty } \frac{{\lg E\left( {{{\left| {{y_n}} \right|}^2}} \right)}}{{k\mathit{\Delta} }} < 0$，则称y_n是均方指数稳定的。

定义2  对于式(1)中的随机线性θ方法y_n，若存在Δ>0，使得$\mathop {\lim \mathit{E}}\limits_{\mathit{k} \to \infty } \left( {{{\left| {{y_n}} \right|}^2}} \right) = 0 $，则称y_n是渐近均方稳定的。

2 主要结果和证明

定理1  假定式(1)有定义，若系数f、g满足

$ \begin{array}{l} 2\langle u,f(t,u,v)\rangle + |g(t,u,v){|^2} \le - {C_1}|u{|^2} + {C_2}\\ |v{|^2},{C_1} > {C_2} > 0 \end{array} $

(2)

则当$\frac{1}{2} $< θ≤1时，随机线性θ方法是均方指数稳定的。

证明：定理1的证明分5步进行。

① 方程(1)可变换为

$ \begin{array}{l} {y_{n + 1}} - \theta \mathit{\Delta} f\left( {{t_{n + 1}},{y_{n + 1}},{y_{n + 1}}} \right) = {y_n} - \theta \mathit{\Delta} f\left( {{t_n},{y_n},{{\bar y}_n}} \right) + \\ f\left( {{t_n},{y_n},{{\bar y}_n}} \right)\mathit{\Delta} + g\left( {{t_n},{y_n},{{\bar y}_n}} \right)\mathit{\Delta} {W_n} \end{array} $

令F_n=y_n-θΔf(t_n, y_n, y_n)，则有

$ \begin{array}{l} \begin{array}{*{20}{r}} {{{\left| {{F_{n + 1}}} \right|}^2} = {{\left| {{F_n}} \right|}^2} + |f\left( {{t_n},{y_n},{{\bar y}_n}} \right)\mathit{\Delta} + g\left( {{t_n},{y_n},{{\bar y}_n}} \right)}\\ {{{\left. {\mathit{\Delta} {W_n}} \right|}^2} + 2\left\langle {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)\mathit{\Delta} ,{F_n}} \right\rangle + 2\left\langle {g\left( {{t_n},{y_n},{{\bar y}_n}} \right)\mathit{\Delta} {W_n},} \right.} \end{array}\\ \begin{array}{*{20}{l}} {\left. {{F_n}} \right\rangle = {{\left| {{F_n}} \right|}^2} + {{\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|}^2}{\mathit{\Delta} ^2} + {{\left| {g\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|}^2}\mathit{\Delta} + }\\ {2\left\langle {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)\mathit{\Delta} ,{y_n}} \right\rangle - 2\theta {{\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|}^2}{\mathit{\Delta} ^2} + {M_n} = } \end{array}\\ \begin{array}{*{20}{l}} {{{\left| {{F_n}} \right|}^2} + (1 - 2\theta ){{\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|}^2}{\mathit{\Delta} ^2} + {{\left| {g\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|}^2}\mathit{\Delta} + }\\ {2\left\langle {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)\mathit{\Delta} ,{y_n}} \right\rangle + {M_n} = {{\left| {{F_n}} \right|}^2} + ((1 - 2\theta )} \end{array}\\ {\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|^2}\mathit{\Delta} + {\left| {g\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|^2} + 2\left\langle {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right.,\\ \left. {\left. {{y_n}} \right\rangle } \right)\mathit{\Delta} + {M_n} \end{array} $

(3)

式(3)中，M_n=|g(t_n, y_n, y_n)|²|ΔW_n|²－|g(t_n, y_n, y_n)|²Δ+ 2〈g(t_n, y_n, y_n)ΔW_n, F_n〉+2〈f(t_n, y_n, y_n)Δ, g(t_n, y_n, y_n)ΔW_n〉为鞅，鞅取期望为0。

② 证明存在0 < C < C₁，使得

$ \begin{array}{l} (1 - 2\theta ){\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|^2}\mathit{\Delta} + {\left| {g\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|^2} + 2\\ \left\langle {f\left( {{t_n},{y_n},{{\bar y}_n}} \right),{y_n}} \right\rangle \le - C{\left| {{F_n}} \right|^2} + {C_2}{\left| {{{\bar y}_n}} \right|^2} \end{array} $

(4)

事实上，有

$ \begin{array}{l} \begin{array}{*{20}{c}} {(2\theta - 1){{\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|}^2}\mathit{\Delta} - C{{\left| {{F_n}} \right|}^2} = (2\theta - 1)}\\ {{{\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|}^2}\mathit{\Delta} - C\left( {{{\left| {{y_n}} \right|}^2} + {{\left| {\theta \mathit{\Delta} f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|}^2} - } \right.} \end{array}\\ \begin{array}{*{20}{l}} {\left. {2\left\langle {\theta \mathit{\Delta} f\left( {{t_n},{y_n},{{\bar y}_n}} \right),{y_n}} \right\rangle } \right) = \left[ {(2\theta - 1)\mathit{\Delta} - C{\theta ^2}{\mathit{\Delta} ^2}} \right]}\\ {{{\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|}^2} + 2C\theta \mathit{\Delta} \left\langle {f\left( {{t_n},{y_n},{{\bar y}_n}} \right),{y_n}} \right\rangle - C{{\left| {{y_n}} \right|}^2} = } \end{array}\\ a{\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right) + b{y_n}} \right|^2} - \left( {a{b^2} + C} \right){\left| {{y_n}} \right|^2} \end{array} $

(5)

式(5)中，a=(2θ－1)Δ－Cθ²Δ², $b = \frac{{C{\theta }\mathit{\Delta} }}{a} $。当 $\frac{1}{2} $ < θ≤1，能取到Δ足够小，使得a足够小且a>0，因此存在a足够小，使得(ab²+C) < C₁，从而式(5)可变为

$ \begin{array}{l} (2\theta - 1){\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|^2}\mathit{\Delta} - C{\left| {{F_n}} \right|^2} \ge - \left( {a{b^2} + } \right.\\ C){\left| {{y_n}} \right|^2} \ge - {C_1}{\left| {{y_n}} \right|^2} \end{array} $

又由式(2)可得

$ \begin{array}{l} (2\theta - 1){\left| {f\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|^2}\mathit{\Delta} - C{\left| {{F_n}} \right|^2} \ge 2\left\langle {{y_n},f} \right.\\ \left. {\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right\rangle + {\left| {g\left( {{t_n},{y_n},{{\bar y}_n}} \right)} \right|^2} - {C_2}{\left| {{{\bar y}_n}} \right|^2} \end{array} $

移项后，式(4)得证。

③ 将式(4)代入式(3)，有

$ \begin{array}{l} {\left| {{F_{n + 1}}} \right|^2} \le {\left| {{F_n}} \right|^2} + \left( { - C{{\left| {{F_n}} \right|}^2} + {C_2}{{\left| {{{\bar y}_n}} \right|}^2}} \right)\mathit{\Delta} + \\ {M_n} \le {\left| {{F_n}} \right|^2} - C{\left| {{F_n}} \right|^2}\mathit{\Delta} + {C_2}{\left| {{{\bar y}_n}} \right|^2}\mathit{\Delta} + {M_n} \le (1 - C\mathit{\Delta} )\\ {\left| {{F_n}} \right|^2} + {C_2}{\left| {{{\bar y}_n}} \right|^2}\mathit{\Delta} + {M_n} \end{array} $

且|y_n|²≤δ|y_n-m+1|²+(1－δ)|y_n-m|²，所以有

$ \begin{array}{l} {\left| {{F_{n + 1}}} \right|^2} \le (1 - C\mathit{\Delta} ){\left| {{F_n}} \right|^2} + {C_2}\mathit{\Delta} \delta {\left| {{y_{n - m + 1}}} \right|^2} + \\ {C_2}\mathit{\Delta} (1 - \delta ){\left| {{y_{n - m}}} \right|^2} + {M_n} \end{array} $

又由于对任意的A>1，都有

$ \begin{array}{l} \begin{array}{*{20}{c}} {{A^{(n + 1)\mathit{\Delta} }}{{\left| {{F_{n + 1}}} \right|}^2} - {A^{n\mathit{\Delta} }}{{\left| {{F_n}} \right|}^2} \le {A^{(n + 1)\mathit{\Delta} }}[(1 - C\mathit{\Delta} )}\\ {\left. {{{\left| {{F_n}} \right|}^2} + {C_2}\mathit{\Delta} \delta {{\left| {{y_{n - m + 1}}} \right|}^2} + {C_2}\mathit{\Delta} (1 - \delta ){{\left| {{y_{n - m}}} \right|}^2} + {M_n}} \right] - } \end{array}\\ \begin{array}{*{20}{l}} {{A^{n\mathit{\Delta} }}{{\left| {{F_n}} \right|}^2} \le {A^{(n + 1)\mathit{\Delta} }}\left( {1 - C\mathit{\Delta} - {A^{ - \mathit{\Delta} }}} \right){{\left| {{F_n}} \right|}^2} + {A^{(n + 1)\mathit{\Delta} }}}\\ {{C_2}\mathit{\Delta} \delta {{\left| {{y_{n - m + 1}}} \right|}^2} + {A^{(n + 1)\mathit{\Delta} }}{C_2}\mathit{\Delta} (1 - \delta ){{\left| {{y_{n - m}}} \right|}^2} + {A^{(n + 1)\mathit{\Delta} }}} \end{array}\\ \begin{array}{*{20}{l}} {{M_n} \le {R_1}{A^{(n + 1)\mathit{\Delta} }}{{\left| {{F_n}} \right|}^2} + {R_2}{A^{(n + 1)\mathit{\Delta} }}{{\left| {{y_{n - m + 1}}} \right|}^2}\mathit{\Delta} + }\\ {{R_3}{A^{(n + 1)\mathit{\Delta} }}{{\left| {{y_{n - m}}} \right|}^2}\mathit{\Delta} + {A^{(n + 1)\mathit{\Delta} }}{M_n}} \end{array} \end{array} $

(6)

式(6)中，R₁=1－CΔ－A^－Δ, R₂=C₂δ, R₃=C₂(1－δ)，R₂+R₃=C₂。

对式(6)从n=0到n=k－1求和，得

$ \begin{array}{l} {A^{k\mathit{\Delta} }}{\left| {{F_k}} \right|^2} \le {\left| {{F_0}} \right|^2} + {R_1}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{F_i}} \right|^2} + \\ {R_2}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_{i - m + 1}}} \right|^2}\mathit{\Delta} + {R_3}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_{i - m}}} \right|^2}\mathit{\Delta} + \\ \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {M_i} \end{array} $

(7)

注意到R₁|_Δ=0=0，且

$ R_1^\prime (\Delta ) = - C + {A^{ - \mathit{\Delta} }}\ln A $

(8)

当1 < A < e^C时，R₁′(Δ) < 0，因此当Δ>0时R₁ < 0。

对F_n=y_n-θΔf(t_n, y_n, y_n)两边平方，并应用式(2)得

$ \begin{array}{l} {\left| {{F_i}} \right|^2} = {\left| {{y_i}} \right|^2} + {\theta ^2}{\mathit{\Delta} ^2}{\left| {f\left( {{t_i},{y_i},{{\bar y}_i}} \right)} \right|^2} - 2\theta \mathit{\Delta} \left\langle {f\left( {{t_i},} \right.} \right.\\ \left. {\left. {{y_i},{{\bar y}_i}} \right),{y_i}} \right\rangle \ge {\left| {{y_i}} \right|^2} - 2\theta \mathit{\Delta} \left\langle {f\left( {{t_i},{y_i},{{\bar y}_i}} \right),{y_i}} \right\rangle \ge {\left| {{y_i}} \right|^2} + \\ {C_1}\theta \mathit{\Delta} {\left| {{y_i}} \right|^2} - {C_2}\theta \mathit{\Delta} {\left| {{{\bar y}_i}} \right|^2} \end{array} $

(9)

所以有

$ {R_1}{\left| {{F_i}} \right|^2} \le {R_1}\left( {{{\left| {{y_i}} \right|}^2} + {C_1}\theta \mathit{\Delta} {{\left| {{y_i}} \right|}^2} - {C_2}\theta \mathit{\Delta} {{\left| {{{\bar y}_i}} \right|}^2}} \right) $

(10)

将式(10)代入式(7)，有

$ \begin{array}{l} \begin{array}{*{20}{c}} {{A^{k\mathit{\Delta} }}{{\left| {{F_k}} \right|}^2} \le {{\left| {{F_0}} \right|}^2} + {R_1}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} \left( {{{\left| {{y_i}} \right|}^2} + {C_1}\theta \mathit{\Delta} } \right.}\\ {\left. {{{\left| {{y_i}} \right|}^2} - {C_2}\theta \mathit{\Delta} {{\left| {{{\bar y}_i}} \right|}^2}} \right) + {R_2}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_{i - m + 1}}} \right|}^2}\mathit{\Delta} + } \end{array}\\ \begin{array}{*{20}{l}} {{R_3}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_{i - m}}} \right|}^2}\mathit{\Delta} + \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {M_i} \le {{\left| {{F_0}} \right|}^2} + }\\ {{R_1}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} \left[ {{{\left| {{y_i}} \right|}^2} + {C_1}\theta \mathit{\Delta} {{\left| {{y_i}} \right|}^2} - {C_2}\theta \mathit{\Delta} (\delta } \right.} \end{array}\\ \begin{array}{*{20}{l}} {\left. {\left. {{{\left| {{y_{i - m + 1}}} \right|}^2} + (1 - \delta ){{\left| {{y_{i - m}}} \right|}^2}} \right)} \right] + {R_2}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} }\\ {{{\left| {{y_{i - m + 1}}} \right|}^2}\mathit{\Delta} + {R_3}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_{i - m}}} \right|}^2}\mathit{\Delta} + \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {M_i} \le } \end{array}\\ \begin{array}{*{20}{l}} {{{\left| {{F_0}} \right|}^2} + {R_1}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} \left( {1 + {C_1}\theta \mathit{\Delta} } \right){{\left| {{y_i}} \right|}^2} + \left( {{R_2} - } \right.}\\ {\left. {{R_1}{C_2}\theta \delta } \right)\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_{i - m + 1}}} \right|}^2}\mathit{\Delta} + \left( {{R_3} - {R_1}{C_2}\theta (1 - } \right.} \end{array}\\ \delta ))\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_{i - m}}} \right|^2}\mathit{\Delta} + \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {M_i} \end{array} $

整理得

$ \begin{array}{l} {A^{k\mathit{\Delta} }}{\left| {{F_k}} \right|^2} \le {\left| {{F_0}} \right|^2} + {k_1}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} + \\ {k_2}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_{i - m + 1}}} \right|^2} + {k_3}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_{i - m}}} \right|^2} + \\ \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {M_i} \end{array} $

(11)

式(11)中

$ \left\{ {\begin{array}{*{20}{l}} {{k_1} = {R_1}\left( {1 + {C_1}\theta \mathit{\Delta} } \right)}\\ {{k_2} = \left( {{R_2} - {R_1}{C_2}\theta \delta } \right)\mathit{\Delta} }\\ {{k_3} = \left( {{R_3} - {R_1}{C_2}\theta (1 - \delta )} \right)\mathit{\Delta} } \end{array}} \right. $

以下考虑式(11)中$ \sum\limits_{i = 0}^{k - 1} {{A^{\left( {i + 1} \right)\mathit{\Delta }}}{{\left| {{y_{i - m + 1}}} \right|}^2}} $和$\sum\limits_{i = 0}^{k - 1} {{A^{\left( {i + 1} \right)\mathit{\Delta }}}{{\left| {{y_{i - m}}} \right|}^2}} $两项。显然有

$ \begin{array}{*{20}{l}} {\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_{i - m + 1}}} \right|}^2} = \sum\limits_{i = - m - 1}^{k - m} {{A^{(i + m)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2} \le }\\ {{A^\tau }\sum\limits_{i = - m - 1}^{k - m} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2} \le {A^\tau }\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2} + }\\ {{A^\tau }\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2} \le {A^{\tau + \mathit{\Delta} }}\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2} + }\\ {{A^{\tau + \mathit{\Delta} }}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2}} \end{array} $

(12)

$ \begin{array}{l} \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_{i - m}}} \right|^2} = \sum\limits_{i = - m}^{k - m - 1} {{A^{(i + m + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} \le \\ {A^{\tau + \mathit{\Delta} }}\sum\limits_{i = - m}^{k - m - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} \le {A^{\tau + \mathit{\Delta} }}\sum\limits_{i = - m}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} + \\ {A^{\tau + \mathit{\Delta} }}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} \le {A^{\tau + \mathit{\Delta} }}\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} + \\ {A^{\tau + \mathit{\Delta} }}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} \end{array} $

(13)

将式(12)、(13)代入式(11)，有

$ \begin{array}{l} {A^{k\mathit{\Delta} }}{\left| {{F_k}} \right|^2} \le {\left| {{F_0}} \right|^2} + {k_1}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} + \\ {k_2}\left( {{A^{\tau + \mathit{\Delta} }}\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2} + {A^{\tau + \mathit{\Delta} }}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2}} \right) + \\ {k_3}\left( {{A^{\tau + \mathit{\Delta} }}\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2} + {A^{\tau + \mathit{\Delta} }}\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {{\left| {{y_i}} \right|}^2}} \right) + \\ \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {M_i} \end{array} $

合并系数得

$ \begin{array}{l} {A^{kA}}{\left| {{F_k}} \right|^2} \le {\left| {{F_0}} \right|^2} + \left( {{k_1} + {k_2}{A^{\tau + \mathit{\Delta} }} + } \right.\\ \left. {{k_3}{A^{\tau + \mathit{\Delta} }}} \right)\sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} + \left( {{k_2}{A^{\tau + \mathit{\Delta} }} + } \right.\\ \left. {{k_3}{A^{\tau + \mathit{\Delta} }}} \right)\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} + \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {M_i} \end{array} $

(14)

式(14)中

$ \begin{array}{l} {k_1} + {k_2}{A^{\tau + \mathit{\Delta} }} + {k_3}{A^{\tau + \mathit{\Delta} }} = {R_1}\left( {1 + {C_1}\theta \mathit{\Delta} } \right) + \left( {{C_2} - } \right.\\ \left. {{R_1}{C_2}\theta \delta } \right)\mathit{\Delta} {A^{\tau + \mathit{\Delta} }} = \mathit{\Delta} \left( {\left( {\frac{{{R_1}}}{\mathit{\Delta} } + {C_2}{A^{\tau + \mathit{\Delta} }}} \right) + {R_1}\left( {{C_1}\theta - } \right.} \right.\\ \left. {\left. {{A^{\tau + \mathit{\Delta} }}{C_2}\theta \delta } \right)} \right) \end{array} $

(15)

④ 证明A^kΔE|F_k|²有界。

考虑式(15)中k₁+k₂A^τ+Δ+k₃A^τ+Δ的正负。

首先考虑R₁(C₁θ－A^τ+ΔC₂θδ)的正负。

令C₁>C₂A^τ+Δ，故C₁>C₂A^τ+Δδ，从而有C₁θ－A^τ+ΔC₂θδ>0，因此有R₁(C₁θ－A^τ+ΔC₂θδ) < 0，进而有k₁+k₂A^τ+Δ+k₃A^τ+Δ≤ $\mathit{\Delta }\left( {\frac{{{R_1}}}{\mathit{\Delta}} + {C_2}{A^{\tau + \mathit{\Delta }}}} \right) $。

接下来考虑$\mathit{\Delta} \left( {\frac{{{R_1}}}{\mathit{\Delta} } + {C_2}{A^{\tau + \mathit{\Delta} }}} \right)$的正负。

由0 < C₂, C < C₁, A>1=e⁰可知，存在足够小的Δ，使得C>C₂A^τ+Δ，因此

$ {{\rm{e}}^{C - {C_2}{A^{\tau + \Delta }}}} > A $

(16)

在式(16)的条件下，有C－C₂A^τ+Δ≥lnA，进而C－C₂A^τ+Δ≥A^－ΔlnA。

结合式(8)有－C₂A^τ+Δ≥R₁′(Δ)，故$ \mathop {\lim }\limits_{\mathit{\Delta } \to 0} \frac{{{R_1}}}{\mathit{\Delta }} + {C_2}{A^{\tau + \mathit{\Delta }}} \le 0$；进一步因式(15)中k₁+k₂A^τ+Δ+k₃A^τ+Δ≤0，则式(14)变为

$ \begin{array}{l} {A^{k\mathit{\Delta} }}{\left| {{F_k}} \right|^2} \le {\left| {{F_0}} \right|^2} + \left( {{k_2}{A^{\tau + \mathit{\Delta} }} + } \right.\\ \left. {{k_3}{A^{\tau + \mathit{\Delta} }}} \right)\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} {\left| {{y_i}} \right|^2} + \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} {M_i} \end{array} $

(17)

对式(17)两边取期望得

$ \begin{array}{l} {A^{k\mathit{\Delta} }}E{\left| {{F_k}} \right|^2} \le E{\left| {{F_0}} \right|^2} + \left( {{k_2}{A^{\tau + \mathit{\Delta} }} + } \right.\\ \left. {{k_3}{A^{\tau + \mathit{\Delta} }}} \right)\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} E{\left| {{y_i}} \right|^2} + \sum\limits_{i = 0}^{k - 1} {{A^{(i + 1)\mathit{\Delta} }}} E{M_i} = \\ E{\left| {{F_0}} \right|^2} + \left( {{k_2}{A^{\tau + \mathit{\Delta} }} + {k_3}{A^{\tau + \mathit{\Delta} }}} \right)\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} E{\left| {{y_i}} \right|^2} \le \\ E{\left| {{y_0} - \theta \mathit{\Delta} f\left( {0,{y_0},{{\bar y}_0}} \right)} \right|^2} + \left( {{k_2}{A^{\tau + \mathit{\Delta} }} + } \right.\\ \left. {{k_3}{A^{\tau + \mathit{\Delta} }}} \right)\sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} E{\left| {\phi \left( {{t_i}} \right)} \right|^2} \le E|\phi (0) - \\ {\left. {\theta \mathit{\Delta} f(0,\phi (0),\overline {\phi (0)} )} \right|^2} + \left( {{k_2}{A^{\tau + \mathit{\Delta} }} + {k_3}{A^{\tau + \mathit{\Delta} }}} \right)\\ \sum\limits_{i = - m - 1}^{ - 1} {{A^{(i + 1)\mathit{\Delta} }}} E{\left| {\phi \left( {{t_i}} \right)} \right|^2}: = K < \infty \end{array} $

由此得到了A^kΔE|F_k|²的有界性。

⑤ 证明$\mathop {\lim }\limits_{\mathit{k} \to \infty } \frac{{\lg E{{\left| {{y_k}} \right|}^2}}}{{k\mathit{\Delta }}} \le - \lg A < 0 $。

整理式(9)得

$ {\left| {{F_k}} \right|^2} \ge \left( {1 + {C_1}\theta \mathit{\Delta} } \right){\left| {{y_k}} \right|^2} - {C_2}\theta \mathit{\Delta} {\left| {{{\bar y}_k}} \right|^2} $

故有

$ {\left| {{y_k}} \right|^2} \le \frac{{{{\left| {{F_k}} \right|}^2} + {C_2}\theta \mathit{\Delta} {{\left| {{{\bar y}_k}} \right|}^2}}}{{1 + {C_1}\theta \mathit{\Delta} }} $

(18)

对式(18)两边乘A^kΔ并取期望，得

$ \begin{array}{l} {A^{k\mathit{\Delta} }}E{\left| {{y_k}} \right|^2} \le \frac{{{A^{k\mathit{\Delta} }}E{{\left| {{F_k}} \right|}^2} + {A^{kA}}{C_2}\theta \mathit{\Delta} E{{\left| {{{\bar y}_k}} \right|}^2}}}{{1 + {C_1}\theta \mathit{\Delta} }} \le \\ \begin{array}{*{20}{l}} {\frac{{K + {A^{k\mathit{\Delta} }}{C_2}\theta \mathit{\Delta} E{{\left| {{{\bar y}_k}} \right|}^2}}}{{1 + {C_1}\theta \mathit{\Delta} }} \le \frac{{K + {A^{k\mathit{\Delta} }}{C_2}\theta \mathit{\Delta} \delta E{{\left| {{y_{k - m + 1}}} \right|}^2}}}{{1 + {C_1}\theta \mathit{\Delta} }} + }\\ {\frac{{{A^{k\mathit{\Delta} }}{C_2}\theta \mathit{\Delta} (1 - \delta )E{{\left| {{y_{k - m}}} \right|}^2}}}{{1 + {C_1}\theta \mathit{\Delta} }}} \end{array} \end{array} $

记

$ \left\{ {\begin{array}{*{20}{l}} {b = \frac{K}{{1 + {C_1}\theta \mathit{\Delta} }}}\\ {{q_1} = \frac{{{C_2}\theta \mathit{\Delta} \delta {A^{(m - 1)\mathit{\Delta} }}}}{{1 + {C_1}\theta \mathit{\Delta} }} \le \frac{{{C_2}\theta \mathit{\Delta} \delta {A^\tau }}}{{1 + {C_1}\theta \mathit{\Delta} }} \le \frac{{{C_2}\theta \mathit{\Delta} \delta {A^{\tau + \mathit{\Delta}}}}}{{1 + {C_1}\theta \mathit{\Delta} }}}\\ {{q_2} = \frac{{{C_2}\theta \mathit{\Delta} (1 - \delta ){A^{m\mathit{\Delta}}}}}{{1 + {C_1}\theta \mathit{\Delta} }} \le \frac{{{C_2}\theta \mathit{\Delta} (1 - \delta ){A^{\tau + \mathit{\Delta}}}}}{{1 + {C_1}\theta \mathit{\Delta} }}} \end{array}} \right. $

则有A^kΔE|y_k|²≤b+q₁A^(k－m+1)ΔE|y_k-m+1|²+q₂A^(k－m)ΔE|y_k-m|²。

令a_k=A^kΔE|y_k|²，有递推式a_k≤b+q₁a_k－m+1+q₂a_k－m，从而得到

$ \begin{array}{l} {a_k} \le b + {q_1}{a_{k - m + 1}} + {q_2}{a_{k - m}} \le b\sum\limits_{i = 0}^1 {{{\left( {{q_1} + {q_2}} \right)}^i}} \\ + C_2^0q_1^2q_2^0{a_{k - 2(m - 1)}} + C_2^1q_1^1q_2^1{a_{k - 2(m - 1) - 1}} + C_2^2q_1^0q_2^2{a_{k - 2m}} \le \\ b\sum\limits_{i = 0}^2 {{{\left( {{q_1} + {q_2}} \right)}^i}} + C_3^0q_1^3q_2^0{a_{k - 3(m - 1)}} + C_3^1q_1^2q_2^1{a_{k - 3(m - 1)}} - 1 + \\ C_3^2q_1^1q_2^2{a_{k - 3(m - 1) - 2}} + C_3^3q_1^0q_2^3{a_{k - 3m}} \le \cdots \le b\sum\limits_{i = 1}^l {\left( {{q_1} + } \right.} \\ {\left. {{q_2}} \right)^i} + \sum\limits_{i = 0}^{l + 1} {C_{l + 1}^i} q_1^{l + 1 - i}q_2^i{a_{k - (l + 1)(m - 1) - i}} \end{array} $

(19)

式(19)中$l = \left\lfloor {\frac{k}{{m - 1}}} \right\rfloor $。显然式(19)最后一个不等式中a的下标均小于等于0。

令C₁>C₂A^τ+Δ，当Δ充分小时，有q₁+q₂≤ $ \frac{{{C_2}\theta \mathit{\Delta }{A^{\tau + \mathit{\Delta }}}}}{{1 + {C_1}\theta \mathit{\Delta }}}$ < 1，从而式(19)变为

$ \begin{array}{*{20}{l}} {{A^{k\mathit{\Delta} }}E{{\left| {{y_k}} \right|}^2} \le \frac{b}{{1 - \left( {{q_1} + {q_2}} \right)}} + \mathop {\sup }\limits_{ - \tau \le t \le 0} E|\phi (t){|^2}}&{}\\ {\sum\limits_{i = 0}^{l + 1} {C_{l + 1}^iq_1^{l + 1 - i}} q_2^i = {K^\prime } < \infty }&{} \end{array} $

即

$ \mathop {\lim }\limits_{k \to \infty } \frac{{\lg E{{\left| {{y_k}} \right|}^2}}}{{k\mathit{\Delta} }} \le - \lg A < 0 $

第⑤步证明完成。至此，得到了随机线性θ方法数值解的均方指数稳定性，定理1得证。证毕。

3 结束语

本文参考文献[4]中θ－EM数值格式的方法，研究了随机时滞微分方程随机线性θ方法的均方指数稳定性。与文献[5]相比，本文处理了更为复杂的滞后项,并在相同条件下得到了随机线性θ方法的均方指数稳定性(1/2<θ≤1)，从而几乎处处指数稳定，比文献[5]中的结论更完善。