中立延迟随机微分方程历年来在生物、工程、金融等各个领域引起了学者的广泛关注.文[1]系统地介绍了不带跳的随机泛函微分方程的基本理论及其在金融、随机游戏、人口问题中的应用; 文[2]给出了带有Lévy跳随机泛函微分方程解的存在唯一性; 文[3]得到了一类带Lévy跳的中立随机泛函微分方程解的存在唯一性; 文[4]研究了带有特殊跳(泊松跳)的中立随机延迟微分方程的数值逼近; 文[5]得到了Lévy噪声扰动的混合随机微分方程的Euler近似解; 文[6]和文[7]研究了带Markov状态转换的跳扩散方程的数值解.

设$ (\Omega, \mathcal F, P) $是完备概率空间, $ (\mathcal F_{t})_{t\geq0} $是其上一个满足通常条件的适应流.设$ \{\bar{p} = \bar{p}(t), t\geq0\} $是一个关于$ (\mathcal F_{t})_{t\geq0} $适应的稳定的$ R^{n} $值泊松点过程.设$ B(R^{n}- \{0\}) $为$ R^{n}-\{0\} $上的波莱尔$ \sigma $-代数, 对$ A\in B(R^{n}- \{0\}) $, 定义与$ \bar{p} $联系的泊松计数测度$ N(t, A) = N((0, t]\times A) $如下

$ N((0, t], A) = \sum\limits_{0<s\leq t}I_{A}(\bar{p}(s)), $

则存在一个$ \sigma $有限测度$ \pi $使得

$ E(N(t, A)) = \pi(A)t, \qquad P(N(t, A) = n) = \frac{\exp(-t\pi(A))(\pi(A)t)^{n}}{n!}, $

这里的测度$ \pi $称为Lévy测度.由Doob-Meyer分解定理, 存在关于$ (\mathcal F_{t})_{t\geq0} $适应的唯一的鞅$ \tilde{N}(t, A) $和唯一的增过程$ \hat{N}(t, A) $, 使得

$ N(t, A) = \tilde{N}(t, A)+\hat{N}(t, A), t\geq0, $

这里的$ \tilde{N}(t, A) $称作补偿Lévy跳且$ \hat{N}(t, A) = \pi(A)t $称作补偿子.

设$ |\cdot| $表示欧式空间$ R^{d} $中的范数, $ \tau $为一个正的固定的延迟, $ C([-\tau, 0], R^{d}) $为$ [-\tau, 0] $到$ R^{d} $上的连续函数类, 其上的范数为$ \|\varphi\| = \sup\limits_{-\tau\leq\theta\leq0}|\varphi(\theta)| $.设$ \xi(t) $为关于$ \mathcal F_{0} $可测的$ C([-\tau, 0], R^{d}) $随机变量且满足$ E\|\xi\|^{p}<\infty $, 其中$ p $为大于等于2的任意正整数.设$ W(t) $是$ (\Omega, \mathcal F, P) $上关于流$ (\mathcal F_{t})_{t\geq0} $适应的标准的$ r $维布朗运动且与Lévy跳$ N $独立.设$ Z\in B(R^{n}-\{0\}) $且$ \pi(Z)<\infty $, 设$ 0<T<\infty, D:[0, T]\times R^{n}\rightarrow R^{n}, $ $ f:[0, T]\times R^{n}\times R^{n}\rightarrow R^{n}, $ $ g:[0, T]\times R^{n}\times R^{n}\rightarrow R^{n\times r}, $ $ h:[0, T]\times R^{n}\times Z\rightarrow R^{n} $.

本文将研究如下带Lévy跳的中立随机微分方程的EM算法

$ \begin{eqnarray} d[x(t)-D(t, x(t-\tau))] & = &f(t, x(t-\tau), x(t))dt+g(t, x(t-\tau), x(t))dW(t) \\ &&+\int_{Z}h(t, x(t-\tau), x(t), \nu)N(dt, d\nu), \ t\in[0, T], \\ x(t)& = &\xi(t), \ t\in[-\tau, 0]. \end{eqnarray} $

(1.1)

在系数满足局部Lipschitz条件和线性增长条件, 中立项$ D(t, x(t-\tau)) $关于第二个分量为压缩映射的条件下, 类似于文[2], 我们可得方程(1.1)存在唯一解.由于解没有显示表达, 因此有必要研究其数值解.如果数值解逼近于真实解, 我们可以用数值解来估计真实解.

本文内容安排如下:第二节介绍了方程(1.1)的Euler的数值算法, 并给出主要结果即定理1;第三节给出定理1的证明.本文推广了文[4]的结果, 考虑的中立项是时间和状态的二元函数, 在逼近的时候对中立项需要加一定的条件才可以放缩.此外, 有中立项时需要把它看成一个整体, 进而用伊藤公式, 再由基本不等式及压缩映射最终得到数值解稳定于真实解.方程(1.1)中$ f, g, h $也依赖于时间, 因此需要三个函数关于时间$ t $是局部Lipschitz的.

在It$ {\rm{\hat o}}$ 意义下方程(1.1)的随机积分形式为

$ \begin{eqnarray} x(t) & = & D(t, x(t-\tau))+\xi(0, x(-\tau))+\int_{0}^{t}f(s, x(s-\tau), x(s))ds \\ &&+ \int_{0}^{t}g(s, x(s-\tau), x(s))dW(s)+\int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)N(ds, d\nu). \end{eqnarray} $

(2.1)

下面给出(1.1)式的EM逼近解.

给定步长$ \Delta\in(0, 1) $且满足$ \Delta = \frac{\tau}{m}, m $为一个大于$ \tau $的正整数.定义$ t_{k} = k\Delta $, 当$ -m\leq k<0 $时, 定义$ y_{k} = \xi(t_{k}) $.当$ k\geq0 $时, 定义$ y_{-1-m} = \xi(t_{m}) $,

$ \begin{eqnarray} y_{k+1} & = & D((k+1)\Delta, y_{k+1-m})+y_{k}-D(k\Delta, y_{k-m})+f(k\Delta, y_{k-m}, y_{k})\Delta \\ &&+g(k\Delta, y_{k-m}, y_{k})\Delta W_{k}+\int_{k\Delta}^{(k+1)\Delta}\int_{Z}h(k\Delta, y_{k-m}, y_{k}, \nu)N(ds, d\nu), \end{eqnarray} $

(2.2)

其中$ \Delta W_{k} = W_{t_{k+1}}-W_{t_{k}} $.假设$ \tilde{y}(t) = y_{k}, \tilde {y}{(t-\tau)} = y_{k-m}, t\in[t_{k}, t_{k+1}) $, 则EM逼近解$ y(t) $的连续形式如下

$ \begin{eqnarray} y(t) = \left\{ \begin{aligned} &D([\frac{t}{\Delta}]\Delta, \tilde{y}(t)) +\xi(0)-D(0, \xi(-\tau))+\int_{0}^{t}f([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))ds \nonumber\\ &+\int_{0}^{t}g([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))ds +\int_{0}^{t}\int_{Z}h([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)N(ds, d\nu), \quad t>0, \nonumber\\ &x(-\tau), \ \quad t\in[-\tau, 0]. \end{aligned} \right. \end{eqnarray} $

本文对系数做如下假设.

(H$ _{1} $)  对任意的正整数$ m $, 存在正整数$ \bar{k}_{m} $, 使得对任意的$ t_{1}, t_{2}\in[0, +\infty) $, 任意的$ x, y, \bar{x}, \bar{y}\in R^{n} $且$ |x|\leq m, |y|\leq m, |\bar{x}|\leq m, \bar{y}\leq m $, 有

$ \begin{eqnarray*} &&|f(t_{1}, x, y)-f(t_{2}, \bar{x}, \bar{y})|^{2}\vee|g(t_{1}, x, y)-f(t_{2}, \bar{x}, \bar{y})|^{2}\vee \int_{Z}|h(t_{1}, x, y, \nu)-h(t_{2}, \bar{x}, \bar{y}, \nu)|^{2}\pi(d\nu)\\ \leq&&\bar{k}_{m}\big(|t_{1}-t_{2}|^{2}+|x-y|^{2}+|\bar{x}-\bar{y}|^{2}\big). \end{eqnarray*} $

(H$ _{2} $)  对任意的$ p\geq2 $, 存在正数$ k_{1} $使得对任意$ t\in[0, +\infty) $, $ x, y\in R^{n} $, 有

$ \begin{equation*} |f(t, x, y)|^{2}\vee|g(t, x, y)|^{2}\vee\big(\int_{Z}|h(t, x, y, \nu)|^{p}\pi(d\nu)\big)^{\frac{2}{p}} \leq k_{1}\big(1+|x|^{2}+|y|^{2}\big). \end{equation*} $

(H$ _{3} $)  存在正数$ k_{2}\in(0, 1) $, 使得对任意$ t_{1}, t_{2}\in[0, +\infty), x, y\in R^{n} $, 有

$ \begin{equation*} |D(t_{1}, x)-D(t_{2}, y)|^{2}\leq k_{2}(|x-y|^{2}). \end{equation*} $

设$ T\in[0, +\infty) $为任一常数, 本文主要结果如下.

定理 1  在(H$ _{1} $)–(H$ _{3} $)条件下, 方程(1.1)的Euler数值解收敛到真实解.即

$ \begin{equation} \lim\limits_{\Delta\rightarrow 0}E(\sup\limits_{0\leq t\leq T}|x(t)-y(t)|^{2}) = 0. \end{equation} $

(2.3)

在证明定理1之前, 需要一些重要的引理.

引理 1  在(H$ _{1} $)–(H$ _{3} $)条件下, 对任意$ p\geq2 $, 存在一个独立于$ \Delta $的常数$ M>0 $, 使得

$ \begin{eqnarray} E(\sup\limits_{-\tau\leq t\leq T}|x(t)|^{p}) \vee E(\sup\limits_{-\tau\leq t\leq T}|y(t)|^{p})<M. \end{eqnarray} $

(3.1)

证  不失一般性, 假定$ x(t) $是有界的, 否则的话, 对每个整数$ n $, 定义停时$ \tau_{n} = \inf \{t\in[0, T]:|x(t)|\geq n\} $, 考虑停止过程$ x(t\vee\tau_{n}) $即可.由基本不等式、假设(H$ _{3} $)及Hölder不等式可得

$ \begin{eqnarray*} &&|x(t)|^{p}\\ & = &|D(t, x(t-\tau))+\xi(0)-D(0, x(-\tau))+\int_{0}^{t}f(s, x(s-\tau), x(s))ds \\ && +\int_{0}^{t}g(s, x(s-\tau), x(s))dW(s)+ \int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)N(ds, d\nu)|^{p} \\ &\leq& 5^{p-1}|D(t, x(t-\tau))-D(0, x(-\tau))|^{p}+5^{p-1}|\xi(0)|^{p}+5^{p-1} |\int_{0}^{t}f(s, x(s-\tau), x(s))ds|^{p}\\ && +5^{p-1}|\int_{0}^{t}g(s, x(s-\tau), x(s))dW(s)|^{p}+5^{p-1} |\int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)N(ds, d\nu)|^{p}\\ &\leq& 5^{p-1}k_{2}^{\frac{p}{2}}|x(t-\tau)-x(-\tau)|^{p}+5^{p-1}|\xi(0)|^{p} +5^{p-1}t^{p-1}\int_{0}^{t}|f(s, x(s-\tau), x(s))|^{p}ds\\ && +5^{p-1}|\int_{0}^{t}g(s, x(s-\tau), x(s))dW(s)|^{p} +5^{p-1}|\int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)N(ds, d\nu)|^{p}\\ &\leq&10^{p-1}k_{2}^{\frac{p}{2}}\left[|x(t-\tau)|^{p}+|x(-\tau)|^{p}\right]+5^{p-1}|\xi(0)|^{p}\\ && +5^{p-1}t^{p-1}\int_{0}^{t}k_{1}^{\frac{p}{2}} \left(1+|x(s-\tau)|^{2}+|x(s)|^{2}\right)^{\frac{p}{2}}ds +5^{p-1}|\int_{0}^{t}g(s, x(s-\tau), x(s))dW(s)|^{p}\\ && +5^{p-1}| \int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)N(ds, d\nu)|^{p}. \end{eqnarray*} $

因此对任意的$ t_{1}\in[0, T] $, 有

$ \begin{eqnarray} E(\sup\limits_{0\leq t\leq t_{1}}|x(t)|^{p}) &\leq&10^{p-1}k_{2}^{\frac{p}{2}} E(\sup\limits_{-\tau\leq t\leq t_{1}}|x(t)|^{p}) +(10^{p-1}k_{2}^{\frac{p}{2}}+5^{p-1})\|\xi\|^{p}\\ &&+5^{p-1}t_{1}^{p-1}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1}E(\sup\limits_{0\leq t\leq t_{1}}\int_{0}^{t}(1+|x(s-\tau)|^{p}+|x(s)|^{p})ds) \\ && +5^{p-1}E(\sup\limits_{0\leq t\leq t_{1}}|\int_{0}^{t}g(s, x(s-\tau), x(s))dW(s)|^{p}) \\ &&+5^{p-1}E(\sup\limits_{0\leq t\leq t_{1}} |\int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)N(ds, d\nu)|^{p}). \end{eqnarray} $

(3.2)

显然有

$ E[\sup\limits_{0\leq t\leq t_{1}}\int_{0}^{t}\big(1+|x(s-\tau)|^{p}+|x(s)|^{p}\big)ds]\leq t_{1}+2\int_{0}^{t_{1}}E(\sup\limits_{-\tau\leq u\leq s}|x(u)|^{p})ds. $

由BDG不等式及假设(H$ _{2} $)可得

$ \begin{eqnarray} &&E(\sup\limits_{0\leq t\leq t_{1}} |\int_{0}^{t}g(s, x(s-\tau), x(s))dW(s)|^{p} ) \leq C_{p}E(\int_{0}^{t_{1}}|g(s, x(s-\tau), x(s))|^{p}ds) \\ &\leq& C_{p}k_{1}^{\frac{p}{2}}E({\int_{0}^{t_{1}}[1+|x(u-\tau)|^{2} +|x(u)|^{2}]^{\frac{p}{2}}ds})\\ &\leq& C_{p}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1} [t_{1}+2\int_{0}^{t_{1}}E(\sup\limits_{-\tau\leq u\leq s}|x(u)|^{p})ds], \end{eqnarray} $

(3.3)

其中$ C_{p} $为与p有关的正的常数.对于跳部分, 由假设(H$ _{2} $)及文[3]引理3.2, 得

$ \begin{eqnarray} &&E(\sup\limits_{0\leq t\leq t_{1}} |\int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)N(ds, d\nu)|^{p} )\\ & = &E(\sup\limits_{0\leq t\leq t_{1}} |\int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)\tilde{N}(ds, d\nu)+\int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)\pi(d\nu)ds|^{p})\\ &\leq& 2^{p-1}E(\sup\limits_{0\leq t\leq t_{1}} |\int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)\tilde{N}(ds, d\nu)|^{p} )\\ &&+2^{p-1}E(\sup\limits_{0\leq t\leq t_{1}}|\int_{0}^{t}\int_{Z}h(s, x(s-\tau), x(s), \nu)\pi(d\nu)ds|^{p} )\\ &\leq&2^{p-1}D_{p} [E(\int_{0}^{t_{1}}\int_{Z}|h(s, x(s-\tau), x(s), \nu)|^{2}\pi(d\nu)ds)^{\frac{p}{2}}\\ &&+E(\int_{0}^{t_{1}}\int_{Z}|h(s, x(s-\tau), x(s), \nu)|^{p}\pi(d\nu)ds)]\\ &&+2^{p-1}\pi(Z)^{p-1}E[\int_{0}^{t_{1}} (\int_{Z}|h(s, x(s-\tau), x(s), \nu)|^{p}\pi(d\nu))^{\frac{1}{p}}ds ]^{p}\\ &\leq&2^{p-1}D_{p} E[ \int_{0}^{t_{1}}(1+\sup\limits_{0\leq u\leq s}|x(u-\tau)|^{2}+\sup\limits_{0\leq u\leq s}|x(u)|^{2})ds ]^{\frac{p}{2}}\\ &&+2^{p-1}D_{p} E( \int_{0}^{t_{1}} (1+\sup\limits_{0\leq u\leq s}|x(u-\tau)|^{2}+\sup\limits_{0\leq u\leq s}|x(u)|^{2})^{\frac{p}{2}}ds )\\ &&+2^{p-1}\pi(Z)^{p-1} E[ \int_{0}^{t_{1}} (1+\sup\limits_{0\leq u\leq s}|x(u-\tau)|^{2}+\sup\limits_{0\leq u\leq s}|x(u)|^{2})^{\frac{1}{2}}ds ]^{p}\\ &\leq& [2^{p-1}D_{p}(t_{1}^{\frac{p-2}{p}}+1)+2^{p-1}\pi(Z)^{p-1}t_{1}^{p-1}] E[\int_{0}^{t_{1}}(1+\sup\limits_{0\leq u\leq s}|x(u-\tau)|^{2} +\sup\limits_{0\leq u\leq s}|x(u)|^{2})^{\frac{p}{2}}ds)]\\ &\leq& 3^{\frac{p}{2}-1} \Big[2^{p-1}D_{p}(t_{1}^{\frac{p-2}{p}}+1)+2^{p-1}\pi(Z)^{p-1}t_{1}^{p-1}\Big] (t_{1}+2\int_{0}^{t_{1}}E(\sup\limits_{-\tau\leq u\leq s}|x(u)|^{p})ds). \end{eqnarray} $

(3.4)

其中$ D_{p} $为正的常数.注意到对任意的$ t_{1}\in[0, T] $, 有

$ \begin{eqnarray} E(\sup\limits_{-\tau\leq t\leq t_{1}}|x(t)|^{p}) \leq E\|\xi\|^{p}+E(\sup\limits_{0\leq t\leq t_{1}}|x(t)|^{p}), \end{eqnarray} $

(3.5)

将(3.2)–(3.4)式代入(3.5)式得

$ \begin{eqnarray*} &&E(\sup\limits_{-\tau \leq t\leq t_{1}}|x(t)|^{p}) \leq E\|\xi\|^{p}+E(\sup\limits_{0\leq t\leq t_{1}}|x(t)|^{p}) \\ &\leq&\big(10^{p-1}k_{2}^{\frac{p}{2}}+5^{p-1}+1\big)E\|\xi\|^{p}+10^{p-1}k_{2}^{\frac{p}{2}} E(\sup\limits_{-\tau\leq t\leq t_{1}}|x(t)|^{p})\\ &&+5^{p-1}t_{1}^{p-1}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1} (t_{1}+2\int_{0}^{t_{1}}E(\sup\limits_{-\tau\leq u\leq s}|x(u)|^{p})ds)\\ &&+5^{p-1}C_{p}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1} [t_{1}+2\int_{0}^{t_{1}}E(\sup\limits_{-\tau\leq u\leq s}|x(u)|^{p})ds]\\&&+5^{p-1}\times 3^{\frac{p}{2}-1} [2^{p-1}D_{p}(t_{1}^{\frac{p-2}{p}}+1)+2^{p-1}\pi(Z)^{p-1}t_{1}^{p-1}] (t_{1}+2\int_{0}^{t_{1}}E(\sup\limits_{-\tau\leq u\leq s}|x(u)|^{p})ds) \end{eqnarray*} $

$ \begin{eqnarray*} &\leq& (10^{p-1}k_{2}^{\frac{p}{2}}+5^{p-1}+1)E\|\xi\|^{p}+10^{p-1}k_{2}^{\frac{p}{2}} E(\sup\limits_{-\tau\leq t\leq t_{1}}|x(t)|^{p}) +5^{p-1}t_{1}^{p}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1} \\ &&+ 5^{p-1}C_{p}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1}t_{1} +20k_{1}5^{p-1}t_{1}+10^{p-1}\times3^{\frac{p}{2}-1} \Big[D_{p}(t_{1}^{\frac{2p-2}{p}}+1)+\pi(Z)^{p-1}t_{1}^{p}\Big]\\ &&+[ 5^{p-1}t_{1}^{p-1}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1}(t_{1}^{p-1}+2C_{p}) +2\times10^{p-1}\times3^{\frac{p}{2}-1} (D_{p}(t_{1}^{\frac{p-2}{p}}+1)+\pi(Z)^{p-1}t_{1}^{p-1}) ]. \end{eqnarray*} $

因此

$ \begin{eqnarray*} &&E(\sup\limits_{-\tau \leq t \leq T}|x(t)|^{p})\\&\leq&\frac{1}{1-10^{p-1}k_{2}^{\frac{p}{2}}} \left\{(10^{p-1}k_{2}^{\frac{p}{2}}+5^{p-1}+1) E\|\xi\|^{p}+5^{p-1}T^{p}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1}\right.\\ &&+ 5^{p-1}C_{p}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1}T+20k_{1}5^{p-1}T+10^{p-1}\times 3^{\frac{p}{2}-1}\Big[D_{p}(T^{\frac{2p-2}{p}}+1)+\pi(Z)^{p-1}T^{p}\Big]\\ &&+ [5^{p-1}T^{p-1}k_{1}^{\frac{p}{2}}3^{\frac{p}{2}-1}(T^{p-1}+2C_{p})+2\times 10^{p-1}3^{\frac{p}{2}-1}(D_{p}(T^{\frac{2p-2}{p}}+1)+\pi(Z)^{p-1}T^{p-1})]\\ &&\times\left. \int_{0}^{T}E\big(\sup\limits_{-\tau\leq u\leq s}|x(u)|^{p}\big)ds \right\}. \end{eqnarray*} $

所以由Gronwall不等式得$ E(\sup\limits_{-\tau\leq t\leq T}|x(t)|^{p})\leq M_{1} $, 用同样的方法可以证明

$ E(\sup\limits_{-\tau\leq t\leq T}|y(t)|^{p})\leq M_{2}. $

从而定理得证.

下面先建立两个停时,

$ \begin{eqnarray} \sigma_{d} = \inf\{t\geq 0, |y(t)|\geq d\}, \ \upsilon_{d} = \inf\{t\geq 0, |x(t)|\geq d\}, \ \mbox{令}\rho_{d} = \sigma_{d}\wedge\upsilon_{d}. \end{eqnarray} $

(3.6)

引理 2  在假设(H$ _{1} $)–(H$ _{3} $)下, 有

$ \begin{eqnarray} E(\sup\limits_{-\tau\leq t\leq T}|y(t\wedge \rho_{d})|^{2})<C_{1}, \end{eqnarray} $

(3.7)

这里$ C_{1} $是独立于$ \Delta $的正的常数.

证  对任意的$ t_{1}\in[0, T] $,

$ \begin{eqnarray} &&E(\sup\limits_{-\tau\leq t\leq T}|y(t\wedge\rho_{d})|^{2}) \leq E\|\xi\|^{2}+E(\sup\limits_{0\leq t\leq T}|y(t\wedge\rho_{d})|^{2}), \\ &&E( \sup\limits_{0\leq t\leq T}\big|y(t\wedge\rho_{d})|^{2} ) = E(\sup\limits_{0\leq t\leq T\wedge\rho_{d}}|D([\frac{t}{\Delta}]\Delta, \tilde{y}(t-\tau)) +\xi(0)-D(0, \xi(-\tau))\\ &&+\int_{0}^{t}f([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))ds+ \int_{0}^{t}g([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))dW(s)\\ &&+\int_{0}^{t}\int_{Z}h([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)N(ds, d\nu)|^{2} ). \end{eqnarray} $

(3.8)

证明方法与引理1的方法相同, 这里其证明省略.

推论 3  在假设(H$ _{1} $)–(H$ _{3} $)下有

$ \begin{eqnarray} E(|y(t)|^{2}I_{[-\tau, T\wedge\rho_{d}]})\leq M_{2}, \end{eqnarray} $

(3.9)

这里的$ M_{2} $是一个正的常数且独立于$ \Delta $.

引理 4  在假设(H$ _{1} $)–(H$ _{3} $)下, 对任意$ t\in [0, T] $, 存在一个正的常数$ M_{3} $, 其中$ M_{3} $独立于$ \Delta $, 使得$ \int_{0}^{t\wedge\rho_{d}}E(|y(s)-\tilde{y}(s)|^{2})ds\leq M_{3}\Delta $.

证  对任意的$ t\in[0, T\wedge \rho_{d}] $, 存在$ k $使得$ t\in[t_{k}, t_{k+1}) $, 注意到

$ \begin{eqnarray*} y_{k} & = & D(k\Delta, y_{k-m})+y_{k-1}-D((k-1)\Delta, y_{k-1-m})\Delta+f((k-1)\Delta, y_{k-1-m}, y_{k-1})\Delta\\ &&+g((k-1)\Delta, y_{k-1-m}, y_{k-1})\Delta W_{k-1}+\int_{(k-1)\Delta}^{k\Delta}\int_{Z}h((k-1)\Delta, y_{k-1-m}, y_{k-1}, \nu)N(ds, d\nu). \end{eqnarray*} $

因此

$ \begin{eqnarray*} y_{k} & = & D(k\Delta, y_{k-m})+y_{0}-D(0, y_{-m})+\sum\limits_{i = 1}^k f((i-1)\Delta, y_{i-1-m}, y_{i-1})\Delta \\ &&+ \sum\limits_{i = 1}^k g((i-1)\Delta, y_{i-1-m}, y_{i-1})\Delta W_{i-1}\\ &&+\sum\limits_{i = 1}^k\int_{(i-1)\Delta}^{i\Delta}\int_{Z}h((i-1)\Delta, y_{i-1-m}, y_{i-1}, \nu)N(ds, d\nu). \end{eqnarray*} $

注意到$ \tilde{y}(t) = y_{k}, \tilde{y}(t-\tau) = y_{k-m}, t\in[t_{k}, t_{k+1}) $, 因此

$ \begin{eqnarray*} y_{k}& = & D(k\Delta, y_{k-m})+\xi(0)-D(0, \xi(-\tau))+\sum\limits_{i = 1}^k\int_{(i-1)\Delta}^{i\Delta}f([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))ds \\ &&+ \sum\limits_{n = 1}^k\int_{(i-1)\Delta}^{i\Delta}g([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))dW_{s}\\ &&+\sum\limits_{n = 1}^k\int_{(i-1)\Delta}^{i\Delta}\int_{Z}h([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)N(ds, d\nu). \\ & = &D([\frac{t}{\Delta}]\Delta, \tilde{y}(s-\tau))+\xi(0)-D(0, \xi(-\tau))+\int_{0}^{k\Delta}f([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))ds\\ &&+\int_{0}^{k\Delta}g([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))dW(s)+\int_{0}^{k\Delta}\int_{Z}h([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)N(ds, d\nu). \end{eqnarray*} $

又

$ \begin{eqnarray*} y(t) & = &D([\frac{t}{\Delta}]\Delta, \tilde{y}(t-\tau))+\xi(0)-D(0, \xi(-\tau))+\int_{0}^{t}f([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))ds\\ &&+ \int_{0}^{t}g([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))dW(s)+\int_{0}^{t}\int_{Z}h([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)N(ds, d\nu), \end{eqnarray*} $

因此

$ \begin{eqnarray*} y(t)-\tilde{y}(t) & = & \int_{k\Delta}^{t}f([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))ds +\int_{k\Delta}^{t}g([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))dW(s)\\ &&+ \int_{k\Delta}^{t}\int_{Z}h([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)N(ds, d\nu)\\ & = &f([\frac{t}{\Delta}]\Delta, \tilde{y}(t-\tau), \tilde{y}(t))(t-t_{k})+g([\frac{t}{\Delta}]\Delta, \tilde{y}(t-\tau), \tilde{y}(t))(W(t)-W(t_{k})\\ &&+ \int_{Z}h([\frac{t}{\Delta}]\Delta, \tilde{y}(t-\tau), \tilde{y}(t), \nu)(N([k\Delta, t], d\nu). \end{eqnarray*} $

由基本不等式及假设(H$ _{2} $)可得

$ \begin{eqnarray} |y(t)-\tilde{y}(t)|^{2} &\leq&3|f([\frac{t}{\Delta}]\Delta, \tilde{y}(t-\tau), \tilde{y}(t))|^{2}\Delta^{2}+3|g([\frac{t}{\Delta}]\Delta, \tilde{y}(t-\tau), \tilde{y}(t))|^{2}|W(t)-W(t_{k})|^{2}\\ &&+3|\int_{Z}h([\frac{t}{\Delta}]\Delta, \tilde{y}(t-\tau), \tilde{y}(t), \nu)N([k\Delta, t], d\nu)|^{2}. \\ &\leq&3k_{1}[1+|\tilde{y}(t-\tau)|^{2}+|\tilde{y}(t)|^{2}](\Delta^{2}+|W(t)-W(t_{k})|^{2})^{2} \\ &&+3|\int_{Z}h([\frac{t}{\Delta}]\Delta, \tilde{y}(t-\tau), \tilde{y}(t), \nu)N([k\Delta, t], d\nu)|^{2}. \end{eqnarray} $

(3.10)

由推论3, 文[2]引理3.2及文[9]中Lyapunov不等式得

$ \begin{eqnarray*} &&E(\int_{0}^{t\wedge\rho_{d}}|y(s)-\tilde{y}(s)|^{2}ds) = E(\int_{0}^{t\wedge\rho_{d}}\sum\limits_{k = -m}^{[\frac{T}{\Delta}]} I_{[t_{k}, t_{k+1})}(s)|y(s)-\tilde{y}(s)|^{2}ds)\\ &\leq& E(\int_{0}^{t\wedge\rho_{d}}3k_{1}\sum\limits_{k = -m}^{[\frac{T}{\Delta}]} I_{[t_{k}, t_{k+1})}(s)[1+|\tilde{y}(s-\tau)|^{2}+|\tilde{y}(s)|^{2}](\Delta^{2}+|W(t)-W(t_{k})|^{2})ds)\\ &&+3E(\int_{0}^{t\wedge\rho_{d}}\sum\limits_{k = -m}^{[\frac{T}{\Delta}]}I_{[t_{k}, t_{k+1})}(s) |\int_{Z}h([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)N([k\Delta, s], d\nu)|^{2} ds)\\ &\leq& E(\int_{0}^{t\wedge\rho_{d}}3k_{1}\sum\limits_{k = -m}^{[\frac{T}{\Delta}]} I_{[t_{k}, t_{k+1})}(s)[1+|\tilde{y}(s-\tau)|^{2}+|\tilde{y}(s)|^{2}](\Delta^{2}+r\Delta)ds )\\ &&+6E(\int_{0}^{t\wedge\rho_{d}}\sum\limits_{k = -m}^{[\frac{T}{\Delta}]} I_{[t_{k}, t_{k+1})}(s)\int_{Z}|h([\frac{s}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)|^{2} \Delta\pi(d\nu)ds)\\ &\leq& E(\int_{0}^{t\wedge\rho_{d}}3k_{1}[1+|\tilde{y}(s-\tau)|^{2} +|\tilde{y}(s)|^{2}](\Delta^{2}+r\Delta+2\Delta\pi(Z))ds)\\ &\leq& 3k_{1}T[1+2M_{2}](\Delta+r+2\pi(Z))\Delta, \end{eqnarray*} $

取$ M_{3} = 3k_{1}T[1+r+2\pi(Z)](1+2M_{2}) $即可, 其中$ r $为布朗运动的维数.

定理 1 的证明  假设$ e(t) = x(t)-y(t) $, 易知

$ \begin{eqnarray} E(\sup\limits_{0\leq t\leq T}|e(t)|^{2}) = E(\sup\limits_{0\leq t\leq T}|e(t)|^{2}I_{\sigma_{d}\leq T}) +E(\sup\limits_{0\leq t\leq T}|e(t)|^{2}I_{\{\rho_{d}>T\}}). \end{eqnarray} $

(3.11)

根据Young不等式, 对于$ \frac{1}{p}+\frac{1}{q} = 1 (p, q>0) $, 有

$ ab = a\delta^{\frac{1}{p}}\frac{b}{\delta^\frac{1}{q}}\leq\frac{(a\delta^{\frac{1}{p}})^{p}}{p}+\frac{b^{q}}{q\delta^{\frac{q}{p}}}, \forall a, b, \delta>0. $

因此对任意的$ \delta>0 $, 有

$ \begin{eqnarray} E(\sup\limits_{0\leq t\leq T}|e(t)|^{2}I_{\{\sigma_{d}\leq T {\hbox{或}}\; \upsilon_{d}\leq T\}}) \leq \frac{2\delta}{p}E(\sup\limits_{0\leq t\leq T}|e(t)|^{p}) +\frac{1-\frac{2}{p}}{\delta^{\frac{2}{p-2}}}P\{\sigma_{d}\leq T {\hbox{或}}\; \upsilon_{d}\leq T\}. \end{eqnarray} $

(3.12)

根据引理1, 有

$ \begin{eqnarray*} P\{\sigma_{d}\leq T\} = E(I_{\{\sigma_{d}\leq T\}}\frac{|y(\sigma_{d})|^{d}}{d^{p}}) \leq \frac{1}{d^{p}}E(\sup\limits_{0\leq t\leq T}|y(t)|^{p}) \leq\frac{M}{d^{p}}. \end{eqnarray*} $

类似可得$ P\{\nu_{d}\leq T\}\leq\frac{2M}{d^{p}}. $因此

$ \begin{eqnarray} P\{\sigma_{d}\leq T {\hbox{或}}\; \nu_{d}\leq T\}\leq P\{\sigma_{d}\leq T\}+P\{\nu_{d}\leq T\}\leq\frac{2M}{d^{p}}. \end{eqnarray} $

(3.13)

由基本不等式得

$ \begin{eqnarray} E(\sup\limits_{0\leq t\leq T}|e(t)|^{p}) \leq2^{p-1} [ E(\sup\limits_{0\leq t\leq T}|x(t)|^{p}) +E(\sup\limits_{0\leq t\leq T}|y(t)|^{p}) ] \leq 2^{p}M. \end{eqnarray} $

(3.14)

将(3.13), (3.14)式代入(3.12)式得

$ \begin{eqnarray} E(\sup\limits_{0\leq t\leq T}|e(t)|^{p}I_{\{\sigma_{d}\leq T {\hbox{或}}\; \upsilon_{d}\leq T\}}) \leq \frac{2^{p+1}\delta M}{p}+\frac{2(p-2)M}{p\delta^{\frac{2}{p-2}}d^{p}}. \end{eqnarray} $

(3.15)

根据$ x(t) $和$ y(t) $的定义, 有

$ \begin{eqnarray*} &&x(t\wedge\rho_{d})- y(t\wedge\rho_{d})\\ & = &D(t\wedge\rho_{d}, x(t\wedge\rho_{d}-\tau))-D([\frac{t\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(t\wedge\rho_{d}-\tau))\\ &&+\int_{0}^{t\wedge\rho_{d}}[f(s, x(s-\tau), x(s)) -f([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))]ds\\ &&+\int_{0}^{t\wedge\rho_{d}}[g(s, x(s-\tau), x(s)) -g([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))]dW(s)\\ &&+\int_{0}^{t\wedge\rho_{d}}\int_{Z} [h(s, x(s-\tau), x(s), \nu) -h([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)]N(ds, d\nu).\\ \end{eqnarray*} $

类似于引理4中的证明可知

$ E(\sup\limits_{0\leq t\leq T}|x(t\wedge\rho_{d})- y(t\wedge\rho_{d})|^{2}) \leq 3k_{1}[1+2M_{2}](\Delta+m+2\pi(Z))\Delta. $

不妨令$ M_{4} = 3k_{1}[1+2M_{2}](\Delta+m+2\pi(Z)) $, 则

$ \begin{eqnarray} E(\sup\limits_{0\leq t\leq T}|x(t\wedge\rho_{d})- y(t\wedge\rho_{d})|^{2}) \leq M_{4}\Delta. \end{eqnarray} $

(3.16)

由假设(H$ _{3} $)及(3.16)式知

$ \begin{eqnarray} &&E(\sup\limits_{0\leq t\leq T} |D(t\wedge\rho_{d}, x(t\wedge\rho_{d}-\tau))- D([\frac{t\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(t\wedge\rho_{d}-\tau))|^{2})\\ &\leq& k_{2}E(\sup\limits_{0\leq t\leq T} \left| x(t\wedge\rho_{d}-\tau)-y(t\wedge\rho_{d}-\tau) +y(t\wedge\rho_{d}-\tau)-\tilde{y}(t\wedge\rho_{d}-\tau) \right|^{2})\\ &\leq&2k_{2}M_{4}\Delta+2k_{2} E(\sup\limits_{0\leq t\leq T}\left|x(t\wedge\rho_{d}-\tau)-y(t\wedge\rho_{d}-\tau)\right|^{2}). \end{eqnarray} $

(3.17)

由假设(H$ _{1} $)及Hölder不等式可得

$ \begin{eqnarray*} &&E(\sup\limits_{0\leq t\leq T}|\int_{0}^{t\wedge\sigma_{d}}f(s\wedge\rho_{d}, x(s-\tau), x(s)) -f([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))ds|^{2})\\ & = &E(\sup\limits_{0\leq t\leq T} |\int_{0}^{t\wedge\sigma_{d}}f(s\wedge\rho_{d}, x(s-\tau), x(s)) -f([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, y(s-\tau), y(s))\\ && +f([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, y(s-\tau), y(s)) -f([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))ds|^{2} )\\ &\leq& 2\bar{k}_{d}T E(\sup\limits_{0\leq t\leq T} [\int_{0}^{t\wedge\sigma_{d}}(|s\wedge\rho_{d}| -[\frac{s\wedge\rho_{d}}{\Delta}]\Delta)^{2} +2|x(s-\tau)-\tilde{y}(s-\tau)|^{2}+2|x(s)-\tilde{y}(s)|^{2}ds ])\\ &\leq& 2\bar{k}_{d}\Delta^{2}T^{2}+16\bar{k}_{d}TM_{4}\Delta+ 32T\bar{k}_{d}\int_{0}^{T}E(\sup\limits_{0\leq u\leq s}|x(u\wedge\rho_{d})-y(u\wedge\rho_{d})|^{2})ds. \end{eqnarray*} $

由假设(H$ _{1} $)及BDG不等式可得

$ \begin{eqnarray*} &&E(\sup\limits_{0\leq t\leq T} |\int_{0}^{t\wedge\sigma_{d}}[g(s\wedge\rho_{d}, x(s-\tau), x(s)) -g([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s))]dW(s) |^{2})\\ &\leq& 4\bar{k}_{d}\Delta^{2}T+32\bar{k}_{d}M_{4}\Delta+32\bar{k}_{d}\int_{0}^{T} E(\sup\limits_{0\leq u\leq s}|x(u\wedge\rho_{d})-y(u\wedge\rho_{d})|^{2})ds. \end{eqnarray*} $

由假设(H$ _{1} $)及文[2]引理3.2可得

$ \begin{eqnarray*} && E(\sup\limits_{0\leq t\leq T} |\int_{0}^{t\wedge\sigma_{d}}\int_{Z}[h(s\wedge\rho_{d}, x(s-\tau), x(s), \nu) -h([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)]N(ds, d\nu)|^{2} )\\ & = &E( \sup\limits_{0\leq t\leq T} |\int_{0}^{t\wedge\sigma_{d}} \int_{Z}[h(s\wedge\rho_{d}, x(s-\tau), x(s), \nu) -h([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)] \\ &&(\tilde{N}(ds, d\nu)-\pi(d\nu)ds)|^{2})\\ &\leq& 2E(\sup\limits_{0\leq t\leq T} |\int_{0}^{t\wedge\sigma_{d}}\int_{Z}[h(s\wedge\rho_{d}, x(s-\tau), x(s), \nu) -h([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)]\tilde{N}(ds, d\nu)|^{2} )\\ &&+2E(\sup\limits_{0\leq t\leq T} |\int_{0}^{t\wedge\sigma_{d}}\int_{Z}[h(s\wedge\rho_{d}, x(s-\tau), x(s), \nu) -h([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s-\tau), \tilde{y}(s), \nu)]\pi(d\nu)ds|^{2} )\\&\leq&4D_{2} E(\int_{0}^{T}\int_{Z}\Big|h(s\wedge\rho_{d}, x(s\wedge\rho_{d}-\tau), x(s\wedge\rho_{d}), \nu)\\ && -h([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s\wedge\rho_{d}-\tau), \tilde{y}(s\wedge\rho_{d}), \nu) |^{2}\pi(d\nu)ds)\\&&+2T\pi(Z) E(\int_{0}^{T}\int_{Z}\Big|h(s\wedge\rho_{d}, x(s\wedge\rho_{d}-\tau), x(s\wedge\rho_{d}), \nu)\\ && -h([\frac{s\wedge\rho_{d}}{\Delta}]\Delta, \tilde{y}(s\wedge\rho_{d}-\tau), \tilde{y}(s\wedge\rho_{d}), \nu) |^{2}\pi(d\nu)ds) \end{eqnarray*} $

$ \begin{eqnarray*} &\leq&[4D_{2}+2T\pi(Z)]\bar{k}_{d} E( \int_{0}^{T}[|s\wedge\rho_{d}-[\frac{s\wedge\rho_{d}}{\Delta}]\Delta|^{2} +|x(s\wedge\rho_{d}-\tau)-\tilde{y}(s\wedge\rho_{d}-\tau)|^{2} \\ &&+|x(s\wedge\rho_{d})-\tilde{y}(s\wedge\rho_{d})|^{2}]ds)\\ &\leq&[4D_{2}+2T\pi(Z)]\bar{k}_{d} [T\Delta+8E(\int_{0}^{T}|x(s\wedge\rho_{d})-\tilde{y}(s\wedge\rho_{d})|^{2}ds) ]\\ &\leq&[4D_{2}+2T\pi(Z)]\bar{k}_{d} [T\Delta+16E(\int_{0}^{T}|x(s\wedge\rho_{d})-y(s\wedge\rho_{d})|^{2}ds)\\ &&+16E(\int_{0}^{T}|y(s\wedge\rho_{d})-\tilde{y}(s\wedge\rho_{d})|^{2}ds) ]\\ &\leq&[4D_{2}+2T\pi(Z)]\bar{k}_{d} [T\Delta+16E(\int_{0}^{T}\sup\limits_{0\leq u\leq s}|x(s\wedge\rho_{d})-y(s\wedge\rho_{d})|^{2}ds) +16TM_{4}\Delta ], \end{eqnarray*} $

因此

$ \begin{eqnarray*} &&E(\sup\limits_{0\leq t\leq T}|x(t\wedge\rho_{d})-y(t\wedge\rho_{d})|^{2})\\ &\leq&2k_{2}M_{4}\Delta+2k_{2} E(\sup\limits_{0\leq t\leq T}|x(t\wedge\rho_{d}-\tau)-y(t\wedge\rho_{d}-\tau)|^{2})\\ &&+2\bar{k}_{d}\Delta^{2}T^{2}+16\bar{k}_{d}TM_{4}\Delta +16T\bar{k}_{d}\int_{0}^{T}E(\sup\limits_{0\leq u\leq s}|x(u\wedge\rho_{d})-y(u\wedge\rho_{d})|^{2})ds\\ &&+4\bar{k}_{d}\Delta^{2}T+32\bar{k}_{d}M_{4}\Delta +32\bar{k}_{d}\int_{0}^{T}E(\sup\limits_{0\leq u\leq s}|x(u\wedge\rho_{d})-y(u\wedge\rho_{d})|^{2})ds\\ &&+[4D_{2}+2T\pi(Z)]\bar{k}_{d} [T\Delta +16E(\int_{0}^{T}\sup\limits_{0\leq u\leq s}|x(s\wedge\rho_{d})-y(s\wedge\rho_{d})|^{2}ds)+16TM_{4}\Delta ]\\ & = &2\Delta[k_{2}M_{4}+\bar{k}_{d}\Delta T^{2}+16\bar{k}_{d}TM_{4} +2T\bar{k}_{d}\Delta+32\bar{k}_{d}M_{4}+[4D_{2} +2T\pi(Z)]\bar{k}_{d}(T+16TM_{4})]\\ &&+2k_{2}E(\sup\limits_{0\leq t\leq T}|x(t\wedge\rho_{d})-y(t\wedge\rho_{d})|^{2})+\bar{k}_{d} [48+64D_{2}+32T\pi(Z)]\\ &&\times\int_{0}^{T}E(\sup\limits_{0\leq u\leq s}|x(u\wedge\rho_{d})-y(u\wedge\rho_{d})|^{2})ds. \end{eqnarray*} $

设

$ L = \frac{2[k_{2}M_{4}+\bar{k}_{d}\Delta T^{2}+16\bar{k}_{d}TM_{4}+2T\bar{k}_{d}\Delta +32\bar{k}_{d}M_{4}+[4D_{2}+2T\pi(Z)]\bar{k}_{d}(T+16TM_{4})]}{1-2k_{2}}, $

则由Gronwall不等式得$ E(\sup\limits_{0\leq t\leq T}|x(t\wedge\rho_{d})-y(t\wedge\rho_{d})|^{2}) \leq L\Delta e^{\bar{k}_{d}[48+64D_{2}+32T\pi(Z)]T} $.即

$ \begin{eqnarray} E(\sup\limits_{0\leq t\leq T}|e(t\wedge\rho_{d}|^{2}) \leq Le^{\bar{k}_{d}[48+64D_{2}+32T\pi(Z)]T}\Delta. \end{eqnarray} $

(3.18)

将(3.15)、(3.18)式代入(3.11)式得

$ \begin{eqnarray*} E(\sup\limits_{0\leq t\leq T}|e(t)|^{2}) &\leq&E(\sup\limits_{0\leq t\leq T}|e(t\wedge\rho_{d})|^{2}) +E(\sup\limits_{0\leq t\leq T}|e(t)|^{2}I_{\{\sigma_{d}\leq T \mbox或 \upsilon_{d}\leq T\}}) \\ &\leq&Le^{\bar{k}_{d}[48+64D_{2}+32T\pi(Z)]T}\Delta +\frac{2^{p+1}\delta M}{p}+\frac{2(p-2)M}{p\delta^{\frac{2}{p-2}}d^{p}}. \end{eqnarray*} $

取充分小的$ \delta $以及充分大的$ d $, 则当$ \Delta\rightarrow 0 $时, $ E(\sup\limits_{0\leq t\leq T}|e(t)|^{2})\rightarrow 0 $.定理得证.