由于在分子动力学、材料科学以及自动化控制等领域的实际应用,随机微分方程(SDE)的平均值原理引起了研究者的广泛兴趣.在文献[8]中, Stoyanov和Bainov首先研究了SDE的平均值原理, Xu和Liu在文献[10]中探讨了一类多值SDE的平均值原理.近来, Guo和Pei在文献[1]中建立了一类由Poisson点过程驱动的多值SDE的平均值原理. Pei及其合作者在系列论文[5-7]中讨论了几种不同类型SDE的平均值原理.

为刻画一些金融市场,在文献[4]中,作者提出了一类时变SDE.特别地, Wu在文献[9]中建立了由时变布朗运动驱动SDE解的随机稳定、随机渐近稳定以及整体随机渐近稳定性. Zhu等在文献[11]中给出了由时变布朗运动驱动SDE解的几乎必然指数稳定性.近来, Nane和Ni在文献[2-3]中推广了文献[9]中的相关结论,讨论了由时变Lévy噪声驱动SDE(LSDE)解的相关稳定性.尚未有文献对LSDE的平均值原理进行探讨.基于此,本文旨在建立LSDE的平均值原理.

全文共分四节.第二节给出一些预备知识,第三节给出LSDE的平均值原理,第四节给出一个具体例子.

本节给出一些预备知识,具体细节参见文献[2-3, 9].假设$(\Omega, {\cal F}, ({\cal F}_{t}), {\Bbb P})$是一个完备概率空间, $N$是定义在$\mathbb R ^{+}\times (\mathbb R -\left\{0 \right\})$上的${\cal F}_{t}$ -适应的Poisson随机测度,其补偿子为$\widetilde{N}$,强度测度为$\nu$,其中$\nu$为一个Lévy测度并且满足

$ \widetilde{N}({\rm d}t, {\rm d}v)= N ({\rm d}t, {\rm d}v)-\nu({\rm d}v){\rm d}t. $

假设$\left\{D(t), t\geq 0 \right\}$是一个右连续且左极限存在的(RCLL)单增Lévy过程.由此, $\left\{D(t), t\geq 0 \right\}$是一个从0点出发的从属过程,其Laplace变换具有如下形式

$ {\Bbb E}(e^{-\lambda D(t)})=e^{-t\phi(\lambda)}, $

其中Laplace指数具有形式$\phi(\lambda)=\int_{0}^{\infty}(1-e^{-\lambda x})\nu ({\rm d}x).$定义其逆过程为

(2.1) $\begin{eqnarray}\label{equation0}{E}_{t}:=\inf\left\{\tau >0:D(\tau)>t\right\}.\end{eqnarray}$

假设$D(t)$为RCLL从属过程, $E_{t}$为由(2.1)式所定义的逆过程.定义流$\left\{{\cal G}_{t}\right\}_{t\geq0}$为${\cal G}_{t}={\cal F}_{E_t}$.本文主要考虑如下LSDE的平均值原理

(2.2) $\begin{eqnarray}\label{equation1}\left\{\begin{array}{ll}{\rm d}X(t)=f(t, {E}_{t}, X(t-)){\rm d}t+\delta(t, {E}_{t}, X(t-)){\rm d}E_{t}+g(t, {E}_{t}, X(t-)){\rm d}B_{E_{t}}\\\displaystyle~~~~~~~~~~~~+\int_{|v|<c}h(t, {E}_{t}, X(t-), v)\widetilde{N}({\rm d}E_{t}, {\rm d}v), ~~~t\in \mathbb R ^{+}, \\X(0)=x_{0}, \end{array}\right.\end{eqnarray}$

其中$f, ~\delta, ~g, ~h$为实值函数.

引理2.1^[9]   假设$E_{t}$为由所从属过程$D(t)$的逆所定义的${\cal F}_{t}$ -可测时间变换, $\upsilon(s)$与$\mu(s)$为两个${\cal F}_{t}$有界可测函数,则对于$t\geq 0$以概率1有下式成立

$ \int_{0}^{t}\upsilon(s){\rm d}E_{s}+\int_{0}^{t}\mu(s){\rm d}B_{E_{s}}=\int_{0}^{E_{t}}\upsilon(D(s)){\rm d}s+\int_{0}^{E_{t}}\mu(D(s)){\rm d}B_{s}. $

条件2.1 (Lipschitz条件)   假定对于$x_{1}, ~x_{2} \in\mathbb R ^{d}, ~t_{1}, ~t_{2}\in \mathbb R ^{+}$,存在常数$K_{1}$使得

$ \begin{eqnarray} &&|f(t_{1}, t_{2}, x_{1})-f(t_{1}, t_{2}, x_{2})|^{2}+|\delta(t_{1}, t_{2}, x_{1})-\delta(t_{1}, t_{2}, x_{2})|^{2}\\&&+|g(t_{1}, t_{2}, x_{1})-g(t_{1}, t_{2}, x_{2})|^{2}+\int_{|v|<c}|h(t_{1}, t_{2}, x_{1}, v)-h(t_{1}, t_{2}, x_{2}, v)|^{2}\nu({\rm d}v)\\&\leq &K_{1}|x_{1}-x_{2}|^{2}. \end{eqnarray} $

条件2.2 (线性增长条件)  假定对于$x\in \mathbb R ^{d}, ~t_{1}, ~t_{2}\in \mathbb R ^{+}$,存在常数$K_{2}$使得

$ \begin{eqnarray} |f(t_{1}, t_{2}, x)|^{2}+|\delta(t_{1}, t_{2}, x)|^{2}+|g(t_{1}, t_{2}, x)|^{2}+\int_{|v|<c}|h(t_{1}, t_{2}, x, v)|^{2}\nu({\rm d}v)\leq K_{2}(1+|x|^{2}). \end{eqnarray} $

条件2.3   假设$X(t)$是一个RCLL, ${\cal G}_{t}$ -适应过程,则

$ f(t, {E}_{t}, X(t)), \delta(t, {E}_{t}, X(t)), g(t, {E}_{t}, X(t)), h(t, {E}_{t}, X(t-), v)\in {\Bbb L}({\cal G}_{t}), $

其中${\Bbb L}({\cal G}_{t})$表示RCLL, ${\cal G}_{t}$ -适应过程的集合.

条件2.4^[11]   假定$E_{t}$是${\cal F}_{t}$ -可测的时间变换并且渐近慢于$t$,即

$ \lim\limits_{t\rightarrow \infty}\frac{E_{t}}{t}=0 ~~{\rm a.s..} $

对于$t\in[0, T]$,考虑$ \mathbb R ^{d}$上LSDE

(3.1) $\begin{eqnarray} \label{equation2}\left\{\begin{array}{ll} \displaystyleX(t)=X_{0}+\int_{0}^{t}f(s, E_{s}, X(s-)){\rm d}s+\int_{0}^{t}\delta(s, E_{s}, X(s-)){\rm d}E_{s} \\[3mm]\displaystyle~~~~~~~~~~~~+\int_{0}^{t}g(s, E_{s}, X(s-)){\rm d}B_{E_{s}}+\int_{0}^{t}\int_{|v|<c}h(s, E_{s}, X(s-), v)\widetilde{N}({\rm d}E_{s}, {\rm d}v), \\X(0)=X_{0}.\end{array}\right.\end{eqnarray}$

如果条件2.1-2.3满足,由文献[2]可知方程(3.2)有唯一解$X(t), t\in [0, T].$

考虑$\mathbb R ^d$上LSDE的平均值原理.为此,定义如下LSDE

(3.2) $\begin{eqnarray}X_{\epsilon}(t)&=X(0)+\epsilon\int_{0}^{t} f(s, E_{s}, X_{\epsilon}(s-)){\rm d}s+\epsilon\int_{0}^{t} \delta(s, E_{s}, X_{\epsilon}(s-)){\rm d}E_{s} \\&+\sqrt{\epsilon}\int_{0}^{t}g(s, E_{s}, X_{\epsilon}(s-)){\rm d}B_{E_{s}} \\&+\sqrt{\epsilon}\int_{0}^{t}\int_{|v|<c}h(s, E_{s}, X_{\epsilon}(s-), v)\widetilde{N}({\rm d}E_{s}, {\rm d}v), t\in[0, T], \end{eqnarray}$

其中上述方程的系数满足条件2.1, 2.2以及2.3, $\varepsilon_{0}$是一个给定的常数, $\epsilon\in(0, \epsilon_{0}]$为小参数.易知,方程(3.2)存在唯一解$X_{\epsilon}(t), ~t\in[0, T].$

考虑如下均值化LSDE

(3.3) $\begin{eqnarray}\label{equation3}Y_{\epsilon}(t)&=&X(0)+\epsilon\int_{0}^{t} \bar{f}(Y_{\epsilon}(s-)){\rm d}s+\epsilon\int_{0}^{t}\bar{\delta}(Y_{\epsilon}(s-)){\rm d}E_{s}+\sqrt{\epsilon}\int_{0}^{t}\bar{g}(Y_{\epsilon}(s-)){\rm d}B_{E_{s}}\nonumber\\&&+\sqrt{\epsilon}\int_{0}^{t}\int_{|v|<c}\bar{h}(Y_{\epsilon}(s-), v)\widetilde{N}({\rm d}E_{s}, {\rm d}v), ~~t\in[0, T].\end{eqnarray}$

假定$\bar{f}, ~\bar{\delta}, ~\bar{g}, ~\bar{h}$满足条件2.1, 2.2以及2.3,则方程(3.3)存在唯一解$Y_{\epsilon}(t), ~0\leq t \leq T.$进一步地,对于$ x\in \mathbb R ^d$, $T_{1}\in[0, T]$,假设

(A5) $\displaystyle\frac {1}{T_{1}}\int_{0}^{T_{1}}|f(s, E_{s}, x)-\bar{f}(x)|^{2}{\rm d}s \leq \varphi_{1}(T_{1})(1+|x|^{2}), $

(A6) $\displaystyle\frac {1}{T_{1}}\int_{0}^{T_{1}}|\delta(s, E_{s}, x)-\bar{\delta}(x)|^{2}{\rm d}s \leq \varphi_{2}(T_{1})(1+|x|^{2}), $

(A7) $\displaystyle\frac {1}{T_{1}}\int_{0}^{T_{1}}|g(s, E_{s}, x)-\bar{g}(x)|^{2}{\rm d}s \leq \varphi_{3}(T_{1})(1+|x|^{2}), $

(A8) $\displaystyle\frac {1}{T_{1}}\int_{0}^{T_{1}}\int_{|v|<c}|h(s, E_{s}, x, v)-\bar{h}(x, v)|^{2}\nu({\rm d}v){\rm d}s \leq \varphi_{4}(T_{1})(1+|x|^{2}), $

其中$\varphi_{i}(T_{1}), (i=1, 2, 3, 4, T_{1}\in[0, T])$为有界函数.

下面主要建立$X_{\epsilon}(t)$与$Y_{\epsilon}(t)$的关系,旨在证明均值化方程(3.3)的解在均方和以概率意义下收敛到方程(3.2)的解.

定理3.1   假定LSDE (3.2)以及均值化LSDE (3.3)均满足条件2.1-2.4以及(A5)-(A8),对于给定的任意常数$\delta_{1}>0, $存在常数$\alpha\in (0, 1), ~L>0, ~\varepsilon_{1} \in(0, \varepsilon_{0}]$使得当$\epsilon \in(0, \epsilon_{1}]$时有

$ {\Bbb E}\Big(\sup\limits_{t\in[0, L\epsilon^{-\alpha}]}|X_{\varepsilon}(t)-Y_{\varepsilon}(t)|^{2}\Big)\leq \delta_{1}. $

证  对于方程(3.2)和(3.3),考虑其差$X_{\epsilon}(t)-Y_{\epsilon}(t)$有

$ \begin{eqnarray} X_{\epsilon}(t)-Y_{\epsilon}(t)&=&\epsilon\int_{0}^{t} \left[f(s, E_{s}, X_{\epsilon}(s))- \bar{f}(Y_{\epsilon}(s))\right]{\rm d}s\\&&+\epsilon\int_{0}^{t} \left[\delta(s, E_{s}, X_{\epsilon}(s))- \bar{\delta}(Y_{\epsilon}(s))\right]{\rm d}E_{s} \\&&+\sqrt{\epsilon}\int_{0}^{t} \left[g(s, E_{s}, X_{\epsilon}(s))-\bar{g}(Y_{\epsilon}(s))\right]{\rm d}B_{E_{s}}\\&&+\sqrt{\epsilon}\int_{0}^{t}\int_{|v|<c}\left[h(s, E_{s}, X_{\epsilon}(s), v)-\bar{h}(Y_{\epsilon}(s), v)\right]\widetilde{N}({\rm d}E_{s}, {\rm d}v). \end{eqnarray} $

由不等式有

$ \begin{eqnarray} \sup\limits_{0\leq t\leq u}|X_{\epsilon}(t)-Y_{\epsilon}(t)|^{2}&\leq&4\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t} |f(s, E_{s}, X_{\epsilon}(s))- \bar{f}(Y_{\epsilon}(s))|{\rm d}s\bigg)^{2}\\&&+4\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}|\delta(s, E_{s}, X_{\epsilon}(s))- \bar{\delta}(Y_{\epsilon}(s))|{\rm d}E_{s}\bigg)^{2}\\&&+4\epsilon\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}|g(s, E_{s}, X_{\epsilon}(s))-\bar{g}(Y_{\epsilon}(s))|{\rm d}B_{E_{s}}\bigg)^{2}\\&&+4\epsilon\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}\int_{|v|<c}\left|h(s, E_{s}, X_{\epsilon}(s), v)-\bar{h}(Y_{\epsilon}(s), v)\right|\widetilde{N}({\rm d}E_{s}, {\rm d}v)\bigg)^{2}\\&:=&I_{1}^{2}+I_{2}^{2}+I_{3}^{2}+I_{4}^{2}. \end{eqnarray} $

进而有

$ \begin{eqnarray} I^{2}_{1}&=&4\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}|f(s, E_{s}, X_{\epsilon}(s))-f(s, E_{s}, Y_{\epsilon}(s))+f(s, E_{s}, Y_{\epsilon}(s))-\bar{f}(Y_{\epsilon}(s))|{\rm d}s\bigg)^{2}\\&\leq&8\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}|f(s, E_{s}, X_{\epsilon}(s))-f(s, E_{s}, Y_{\epsilon}(s))|{\rm d}s\bigg)^{2}\\&&+8\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}|f(s, E_{s}, Y_{\epsilon}(s))-\bar{f}(Y_{\epsilon}(s))|{\rm d}s\bigg)^{2}\\&:=&I^{2}_{11}+I^{2}_{12}. \end{eqnarray} $

对$I^{2}_{11}$取期望,由Cauchy不等式以及条件2.1有

$ \begin{eqnarray} {\Bbb E}|I_{11}|^{2}&=&8\epsilon^{2}{\Bbb E}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}|f(s, E_{s}, X_{\epsilon}(s))-f(s, E_{s}, Y_{\epsilon}(s))|{\rm d}s\bigg)^{2}\\&\leq&8\epsilon^{2}{\Bbb E}\bigg(\sup\limits_{0\leq t\leq u}t\int_{0}^{t}|f(s, E_{s}, X_{\epsilon}(s))-f(s, E_{s}, Y_{\epsilon}(s))|^{2}{\rm d}s\bigg)\\&\leq&8\epsilon^{2}uK_{11}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq s}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}s\\&\leq&8\epsilon^{2}uK_{11}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u, \end{eqnarray} $

其中$K_{11}$为常数.对于$I^{2}_{12}$,取期望并由条件(A5)可得

$ \begin{eqnarray} {\Bbb E}|I_{12}|^{2}&=&8\epsilon^{2}{\Bbb E}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}|f(s, E_{s}, Y_{\epsilon}(s))-\bar{f}(Y_{\epsilon}(s))|{\rm d}s\bigg)^{2}\\&\leq&8\epsilon^{2}{\Bbb E}\sup\limits_{0\leq t\leq u}t^{2}\bigg(\frac{1}{t}\int_{0}^{t}|f(s, E_{s}, Y_{\epsilon}(s))-\bar{f}(Y_{\epsilon}(s))|^{2}{\rm d}s\bigg)\\&\leq&8\epsilon^{2}u^{2}\bigg(\sup\limits_{0\leq t\leq u}\varphi_{1}(t)\bigg)\bigg(1+{\Bbb E}\Big(\sup\limits_{0\leq t\leq u}|Y_{\epsilon}(t)|^{2}\Big)\bigg). \end{eqnarray} $

由文献[3],对于LSDE,如果${\Bbb E}|X(0)|^{2}<\infty$,对于$t\geq0$有${\Bbb E}|X(t)|^{2}<\infty$进而有

$ {\Bbb E}\bigg(\sup\limits_{0\leq t\leq u}|Y_{\epsilon}(t)|^{2}\bigg)<\infty $

以及$\varphi_{1}(T_{1})$是正的有界函数,则有

$ {\Bbb E}|I_{12}|^{2}\leq8\epsilon^{2}u^{2}K_{12}. $

由上可得

(3.4) $\begin{eqnarray}\label{equation4}{\Bbb E}|{I}_{1}|^{2}\leq 8\epsilon^{2}uK_{11}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u+8\epsilon^{2}u^{2}K_{12}.\end{eqnarray}$

由引理2.1以及Cauchy不等式有

$ \begin{eqnarray} I^{2}_{2}&=&4\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}|\delta(s, E_{s}, X_{\epsilon}(s))- \bar{\delta}(Y_{\epsilon}(s))|{\rm d}E_{s}\bigg)^{2}\\&=&4\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{E_{t}}|\delta(D(s), s, X_{\epsilon}(D(s)))- \bar{\delta}(Y_{\epsilon}(D(s)))|{\rm d}s\bigg)^{2}\\&\leq&4\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(E_{t}\int_{0}^{E_{t}}|\delta(D(s), s, X_{\epsilon}(D(s)))- \bar{\delta}(Y_{\epsilon}(D(s)))|^{2}{\rm d}s\bigg)\\&\leq&8\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(E_{t}\int_{0}^{E_{t}}|\delta(D(s), s, X_{\epsilon}(D(s)))- \bar{\delta}(D(s), s, Y_{\epsilon}(D(s)))|^{2}{\rm d}s\bigg)\\&&+8\epsilon^{2}\sup\limits_{0\leq t\leq u}\bigg(E_{t}\int_{0}^{E_{t}}|\delta(D(s), s, Y_{\epsilon}(D(s)))- \bar{\delta}(Y_{\epsilon}(D(s)))|^{2}{\rm d}s\bigg)\\&:=&I^{2}_{21}+I^{2}_{22}. \end{eqnarray} $

对$I^{2}_{21}$取期望并由条件2.1,引理2.1以及$E_{t}$的单调性可得

$ \begin{eqnarray} {\Bbb E}|I_{21}|^{2}&\leq& 8\epsilon^{2}E_{u}\bigg(\int_{0}^{E_{u}}|\delta(D(s), s, X_{\epsilon}(D(s)))- \bar{\delta}(D(s), s, Y_{\epsilon}(D(s))|^{2}{\rm d}s\bigg)\\&\leq& 8\epsilon^{2}E_{u}\bigg(\int_{0}^{u}|\delta(s, E_{s}, X_{\epsilon}(s))- \bar{\delta}(Y_{\epsilon}(s))|^{2}{\rm d}E_{s}\bigg)\\&\leq& 8\epsilon^{2}E_{u}K_{21}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}E_{u}. \end{eqnarray} $

对$I^{2}_{22}$取期望并由条件(A6)可得

$ \begin{eqnarray} {\Bbb E}|I_{22}|^{2}&\leq&8\epsilon^{2}{\Bbb E}\sup\limits_{0\leq t\leq u}E^{2}_{t}\bigg(\frac{1}{E_{t}}\int_{0}^{E_{t}}|\delta(D(s), s, Y_{\epsilon}(D(s)))- \bar{\delta}(Y_{\epsilon}(D(s)))|^{2}{\rm d}s\bigg)\\&\leq&8\epsilon^{2}E^{2}_{u}\varphi_{2}(E_{u})\bigg(1+{\Bbb E}(\sup\limits_{0\leq t\leq u}|Y_{\epsilon}(t)|^{2})\bigg)\\&\leq&8\epsilon^{2}E^{2}_{u}K_{22}. \end{eqnarray} $

由此可得

$ \begin{eqnarray} {\Bbb E}|{I}_{2}|^{2}\leq 8\epsilon^{2}E_{u}K_{21}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}E_{u}+8\epsilon^{2}E^{2}_{u}K_{22}. \end{eqnarray} $

由条件2.4以及过程$E_{t}$渐近慢于$t$,对任意的$\lambda>0$存在常数$T_{0}$使得$t>T_{0}, ~E_{t}\leq \lambda t~~a.s., $由此可得当$u\in(T_{0}, T]$时有

(3.5) $\begin{eqnarray}\label{equation5}{\Bbb E}|{I}_{2}|^{2}\leq 8\epsilon^{2}\lambda^{2}uK_{21}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u+8\epsilon^{2}\lambda^{2}u^{2}K_{22}.\end{eqnarray}$

对于$I^{2}_{3}$,取期望并由Burkholder-Davis-Gundy不等式可得

$ \begin{eqnarray} {\Bbb E}I^{2}_{3}&=&4\epsilon{\Bbb E}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}|g(s, E_{s}, X_{\epsilon}(s))-\bar{g}(Y_{\epsilon}(s))|{\rm d}B_{E_{s}}\bigg)^{2}\\&\leq&16\epsilon{\Bbb E}\int_{0}^{u}|g(s, E_{s}, X_{\epsilon}(s))-\bar{g}(Y_{\epsilon}(s))|^{2}{\rm d}E_{s}\\&\leq&32\epsilon{\Bbb E}\int_{0}^{u}|g(s, E_{s}, X_{\epsilon}(s))-g(s, E_{s}, Y_{\epsilon}(s))|^{2}{\rm d}E_{s}\\&&+32\epsilon{\Bbb E}\int_{0}^{u}|g(s, E_{s}, Y_{\epsilon}(s))-\bar{g}(Y_{\epsilon}(s))|^{2}{\rm d}E_{s}\\&:=&I^{2}_{31}+I^{2}_{32}. \end{eqnarray} $

由条件2.1有

$ I^{2}_{31}\leq32\epsilon K_{31}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}E_{u}. $

由引理2.1,条件(A7)以及$E_{t}$的单调性有

$ \begin{eqnarray} |I_{32}|^{2}&\leq& 32\epsilon{\Bbb E}\int_{0}^{E_{u}}|g(D(s), s, Y_{\epsilon}(D(s)))-\bar{g}(Y_{\epsilon}(D(s)))|^{2}{\rm d}s\\&\leq& 32\epsilon E_{u}\varphi_{3}(E_{u})\bigg(1+{\Bbb E}\sup\limits_{0\leq t\leq u}|Y_{\epsilon}(t)|^{2}\bigg)\\&\leq& 32\epsilon E_{u}K_{32}. \end{eqnarray} $

进而

$ {\Bbb E}I^{2}_{3}\leq 32\epsilon K_{31}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}E_{u}+32\epsilon E_{u}K_{32}. $

由条件2.4, $u\in (T_{0}, T]$有

(3.6) $\begin{eqnarray}\label{equation6}{\Bbb E}I^{2}_{3}\leq 32\epsilon\lambda K_{31}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u+32\epsilon \lambda uK_{32}.\end{eqnarray}$

对$I^{2}_{4}$取期望并由Doob鞅不等式有

$ \begin{eqnarray} {\Bbb E}|I_{4}|^{2}&=&4\epsilon{\Bbb E}\sup\limits_{0\leq t\leq u}\bigg(\int_{0}^{t}\int_{|v|<c}\left|h(s, E_{s}, X_{\epsilon}(s), v)-\bar{h}(Y_{\epsilon}(s), v)\right|\widetilde{N}({\rm d}E_{s}, {\rm d}v)\bigg)^{2}\\&\leq&16\epsilon {\Bbb E}\bigg(\int_{0}^{u}\int_{|v|<c}\left|h(s, E_{s}, X_{\epsilon}(s), v)-\bar{h}(Y_{\epsilon}(s), v)\right|\widetilde{N}({\rm d}E_{s}, {\rm d}v)\bigg)^{2}\\&\leq&32\epsilon {\Bbb E}\int_{0}^{u}\int_{|v|<c}\left|h(s, E_{s}, X_{\epsilon}(s), v)-h(s, E_{s}, Y_{\epsilon}(s), v)\right|^{2}\nu({\rm d}v){\rm d}E_{s}\\&&+32\epsilon {\Bbb E}\int_{0}^{u}\int_{|v|<c}\left|h(s, E_{s}, Y_{\epsilon}(s), v)-\bar{h}(Y_{\epsilon}(s), v)\right|^{2}\nu({\rm d}v){\rm d}E_{s}\\&:=&I^{2}_{41}+I^{2}_{42}. \end{eqnarray} $

由条件2.1有

$ |I_{41}|^{2}\leq32\epsilon k_{41}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u. $

由条件(A8)有

$ \begin{eqnarray} |I_{42}|^{2}&\leq&32\epsilon{\Bbb E}\bigg(u\frac{1}{u}\int_{0}^{u}\int_{|v|<c}\left|h(s, E_{s}, Y_{\epsilon}(s), v)-\bar{h}(Y_{\epsilon}(s), v)\right|^{2}\nu({\rm d}v){\rm d}E_{s}\bigg)\\&\leq&32\epsilon u\varphi_{4}(u)\big(1+{\Bbb E}|Y_{\epsilon}(t)|^{2}\big)\\&\leq&32\epsilon uK_{42}. \end{eqnarray} $

故有

(3.7) $\begin{eqnarray}\label{equation7}{\Bbb E}|I_{4}|^{2}\leq32\epsilon k_{41}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u+32\epsilon uK_{42}.\end{eqnarray}$

由(3.4)-(3.7)式有

$ \begin{eqnarray} &&{\Bbb E}\left(\sup\limits_{0\leq t\leq u}|X_{\epsilon}(t)-Y_{\epsilon}(t)|^{2}\right)\\&\leq& 8\epsilon^{2}uK_{11}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u+8\epsilon^{2}u^{2}K_{12}\\&&+8\epsilon^{2}\lambda^{2}uK_{21}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u+8\epsilon^{2}\lambda^{2}u^{2}K_{22}\\&&+32\epsilon\lambda K_{31}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u+32\epsilon \lambda uK_{32}\\&&+32\epsilon k_{41}\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u+32\epsilon uK_{42}\\&\leq& 8\epsilon(\epsilon uK_{11}+\epsilon\lambda^{2}uK_{21}+4\lambda K_{31}+4K_{41})\int_{0}^{u}{\Bbb E}\bigg(\sup\limits_{0\leq r\leq u}|X_{\epsilon}(r)-Y_{\epsilon}(r)|^{2}\bigg){\rm d}u\\&&+8\epsilon u(\epsilon uK_{12}+\epsilon\lambda^{2}uK_{22}+4\lambda K_{32}+4K_{42}). \end{eqnarray} $

由Gronwall不等式有

$ \begin{eqnarray} {\Bbb E}\left(\sup\limits_{0\leq t\leq u}|X_{\epsilon}(t)-Y_{\epsilon}(t)|^{2}\right)&\leq&8\epsilon u(\epsilon uK_{12}+\epsilon\lambda^{2}uK_{22}+4\lambda K_{32}+4K_{42})\\&&\times\exp\big(8\epsilon u(\epsilon uK_{11}+\epsilon\lambda^{2}uK_{21}+4\lambda K_{31}+4K_{41})\big). \end{eqnarray} $

选取$\alpha\in(0, 1)$以及$L>0$使得当$t\in [0, L\epsilon^{-\alpha}]$时有

$ {\Bbb E}\bigg(\sup\limits_{t\in [0, L\epsilon^{-\alpha}]}|X_{\epsilon}(t)-Y_{\epsilon}(t)|^{2}\bigg) \leq CL\epsilon^{1-\alpha}, $

其中$C=8(L\epsilon^{1-\alpha} K_{12}+L\epsilon^{1-\alpha}\lambda^{2}K_{22}+4\lambda K_{32}+4K_{42}) \exp\big(8L\epsilon^{1-\alpha}(L\epsilon^{1-\alpha} K_{11}+L\epsilon^{1-\alpha} \lambda^{2}K_{21}+4\lambda K_{31}+4K_{41})\big)$为常数, $T_{0}<L\epsilon^{-\alpha}.$因此,对于给定的$\delta_{1}>0$,可以选取$\epsilon_{1} \in(0, \epsilon_{0}]$使得对于任意的$\epsilon \in(0, \epsilon_{1}]$以及$t\in[0, L\epsilon^{-\alpha}]$,如下式子成立

(3.8) $ \label{equation9}{\Bbb E}\bigg(\sup\limits_{t\in [0, L\epsilon^{-\alpha}]}|X_{\epsilon}(t)-Y_{\epsilon}(t)|^{2}\bigg)\leq \delta_{1}. $

定理3.1证毕.

下面进一步给出以概率收敛性.

定理3.2   假定条件2.1-2.4以及(A5)-(A8)满足,对于任意给定的数$\eta_{1}>0, L>0$以及$\alpha\in (0, 1)$,存在常数$\varepsilon_{1} \in(0, \varepsilon_{0}]$使得当$\epsilon \in(0, \epsilon_{1}], ~t\in[0, L\epsilon^{-\alpha}]$时有

$ \lim\limits_{\epsilon\rightarrow 0}{\Bbb P}\bigg(\sup\limits_{t\in[0, L\epsilon^{-\alpha}]}|X_{\epsilon}(t)-Y_{\epsilon}(t)|>\eta_{1} \bigg)=0. $

证  由定理3.1以及Chebyshev不等式有

$ \begin{eqnarray} {\Bbb P}\bigg(\sup\limits_{t\in[0, L\epsilon^{-\alpha}]}|X_{\epsilon}(t)-Y_{\epsilon}(t)|>\eta_{1} \bigg)&\leq& \frac{1}{\eta_{1}^{2}}{\Bbb E}\bigg(\sup\limits_{t\in[0, L\epsilon^{-\alpha}]}|X_{\epsilon}(t)-Y_{\epsilon}(t)|^{2}\bigg)\\&\leq& \frac{ CL\epsilon^{1-\alpha}}{\eta_{1}^{2}}, \end{eqnarray} $

因此,定理结论成立.

注3.1   上述定理表明均值化方程的解$Y_{\epsilon}(t)$在均方和以概率两种意义下收敛到$X_{\epsilon}(t)$.

考虑如下LSDE

(4.1) $\begin{eqnarray}\label{equation10}{\rm d}X_{\epsilon}(t)&=&2\epsilon MX_{\epsilon}(t)\cos^{2}(t){\rm d}t+\epsilon LX_{\epsilon}(t){\rm d}E_{t}-\sqrt{\epsilon}KX_{\epsilon}(t)\sin^{2}(t){\rm d}B_{E_{t}} \nonumber\\&&+\sqrt{\epsilon}\int_{|v|<c}vX_{\epsilon}(t)\widetilde{N}({\rm d}E_{t}, {\rm d}v), \end{eqnarray}$

其中初值$X_{\epsilon}(0)=X_{0}$满足${\Bbb E}|X_{0}|^{2}<\infty, ~t\in[0, T], $系数$f(t, E_{t}, X_{\epsilon}(t))=2MX_{\epsilon}(t)\cos^{2}(t), $$\delta(t, E_{t}, X_{\epsilon}(t))=LX_{\epsilon}(t), $$g(t, E_{t}, X_{\epsilon}(t))=-KX_{\epsilon}(t)\sin^{2}(t), $$h(t, E_{t}, X_{\epsilon}(t), v)=vX_{\epsilon}(t)$.令

$ \begin{eqnarray} &&\displaystyle\bar{f}(X_{\epsilon}(t))=\frac{1}{\pi}\int_{0}^{\pi}f(t, E_{t}, X_{\epsilon}(t)){\rm d}t=MX_{\epsilon}(t), \\&&\displaystyle\bar{\delta}(X_{\epsilon}(t))=\frac{1}{\pi}\int_{0}^{\pi}\delta(t, E_{t}, X_{\epsilon}(t)){\rm d}t=LX_{\epsilon}(t), \\&&\displaystyle \bar{g}(X_{\epsilon}(t))=\frac{1}{\pi}\int_{0}^{\pi}g(t, E_{t}, X_{\epsilon}(t)){\rm d}t=-\frac{K}{2}X_{\epsilon}(t), \\&&\displaystyle \bar{h}(X_{\epsilon}(t))=\frac{1}{\pi}\int_{0}^{\pi}\int_{|v|<c}h(t, E_{t}, X_{\epsilon}(t), v)\nu({\rm d}v){\rm d}s=\int_{|v|<c}vX_{\epsilon}(t)\nu({\rm d}v). \end{eqnarray} $

由上,定义一个均值化LSDE

(4.2) $ \label{equation11}{\rm d}Y_{\epsilon}(t)=\epsilon MY_{\epsilon}(t){\rm d}t+\epsilon LY_{\epsilon}(t){\rm d}E_{t}-\sqrt{\epsilon}\frac{K}{2}Y_{\epsilon}(t){\rm d}B_{E_{t}}+\sqrt{\epsilon}\int_{|v|<c}vY_{\epsilon}(t)\widetilde{N}({\rm d}E_{t}, {\rm d}v). $

由定理3.1以及3.2,方程(4.2)的解$Y_{\epsilon}(t)$在均方和以概率意义下收敛到方程(4.1)的解$X_{\epsilon}(t)$.