数学物理学报, 2019, 39(6): 1532-1544 doi:

论文

Lévy噪声驱动的三维随机LANS-α模型的中偏差原理

黄建华,1, 张再云,2, 陈涌3

The Moderate Deviation Principle for Stochastic 3D LANS-α Model Driven by Multiplicative Lévy Noise

Huang Jianhua,1, Zhang Zaiyun,2, Chen Yong3

通讯作者: 陈涌

收稿日期: 2018-02-19  

基金资助: 国家自然科学基金.  11771449
国家自然科学基金.  11401532
浙江省自然科学基金.  LY18A010027
湖南省自然科学基金.  2016JJ2061
浙江理工大学基本科研业务费专项资金.  2019Q068

Received: 2018-02-19  

Fund supported: the NSFC.  11771449
the NSFC.  11401532
the Zhejiang Provincial NSF.  LY18A010027
the Hunan Provincial NSF.  2016JJ2061
the Fundamental Research Funds of Zhejiang Sci-Tech University.  2019Q068

作者简介 About authors

黄建华,E-mail:jhhuang32@nudt.edu.cn , E-mail:jhhuang32@nudt.edu.cn

张再云,E-mail:zhangzaiyun1226@126.com , E-mail:zhangzaiyun1226@126.com

摘要

该文利用弱收敛方法建立了Lévy噪声驱动的三维随机LANS-α模型的中偏差原理.

关键词: 随机LANS-α模型 ; Lévy噪声 ; 弱收敛方法

Abstract

In this paper, we construct the moderate deviation principle for stochastic 3D LANS-α model driven by multiplicative Lévy noise by the weak convergence method.

Keywords: Moderate deviation principle ; Stochastic Lagrangian averaged Navier-Stokes equations ; Weak convergence method

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本文引用格式

黄建华, 张再云, 陈涌. Lévy噪声驱动的三维随机LANS-α模型的中偏差原理. 数学物理学报[J], 2019, 39(6): 1532-1544 doi:

Huang Jianhua, Zhang Zaiyun, Chen Yong. The Moderate Deviation Principle for Stochastic 3D LANS-α Model Driven by Multiplicative Lévy Noise. Acta Mathematica Scientia[J], 2019, 39(6): 1532-1544 doi:

1 引言

该文考虑如下Lévy噪声驱动的随机LANS-$ \alpha $模型

$ \begin{equation} \left\{\begin{array}{ll} d(u^\epsilon-\alpha\triangle u^\epsilon)-\nu\triangle(u^\epsilon-\alpha\triangle u^\epsilon){\rm d}t +[(u^\epsilon\cdot\nabla)(u^\epsilon-\alpha\triangle u^\epsilon) -\alpha(\nabla u^{\epsilon})^*\cdot\triangle u^\epsilon+\nabla p]{\rm d}t\\ = F(t, u^\epsilon){\rm d}t+\epsilon\int_{{\Bbb Z}}G(u^\epsilon, z)\tilde{N}^{\epsilon^{-1}}({\rm d}t{\rm d}z), \;(t, x)\in (0, T)\times D, \\ \nabla\cdot u^\epsilon = 0, \;(t, x)\in (0, T)\times D, \\ u^\epsilon = 0, \;(t, x)\in (0, T)\times\partial D, \\ u^\epsilon(0, x) = u_{0}(x), \;x\in D, \end{array}\right. \end{equation} $

其中$ D $是带有边界正则性$ C^2 $$ \mathbb{R} ^3 $中有界连通开集, $ T > 0 $是解的最大存在时间, $ u = (u_1, u_2, u_3) $为流体速度, $ p $为压力, $ \nu > 0 $$ \alpha > 0 $为给定的常数, $ (\nabla u^{\epsilon})^* $$ \nabla u^{\epsilon} $的转置, $ {\Bbb Z} $为局部紧的光滑空间, $ G $为可测映射, $ N^{\epsilon^{-1}} $$ [0, T]\times {\Bbb Z} $上的泊松随机测度,其带有$ \sigma $ -有限测度$ \epsilon^{-1}\lambda_T\otimes\vartheta, \lambda_T $$ [0, T] $是勒贝格测度, $ \vartheta $$ {\Bbb Z} $$ \sigma $ -有限测度, $ \tilde{N}^{\epsilon^{-1}} $为补偿泊松随机测度,即对于$ {\cal O}\in {\cal B}({\Bbb Z}) $$ \vartheta({\cal O}) < \infty $,有

LANS-$ \alpha $模型在物理和非线性偏微分方程中具有重要运用[1-3]. LANS-$ \alpha $模型在有界区域中整体弱解在文献[1-2]中建立,整体吸引子在文献[1]中得到.在周期区域得到了类似的结果[2].维纳噪声驱动的三维LANS-$ \alpha $模型解的存在唯一性和强解的渐近行为分别在文献[4-8]中讨论. Lévy噪声驱动的三维LANS-$ \alpha $模型的适定性在文献[9]中建立.

该文考虑当$ \epsilon\to0 $时, $ u^\epsilon $收敛到如下方程的解$ u $

$ \begin{eqnarray} &&d(u-\alpha\triangle u)-\nu\triangle(u-\alpha\triangle u){\rm d}t +[(u\cdot\nabla)(u-\alpha\triangle u) -\alpha(\nabla u)^*\cdot\triangle u+\nabla p]{\rm d}t\\ & = &F(t, u){\rm d}t, \end{eqnarray} $

且建立相应的中偏差原理,即考虑如下随机变量

的大偏差原理,其中$ a(\epsilon) $满足

$ \begin{equation} a(\epsilon)\to0, \epsilon/a^2(\epsilon)\to0\;\;\hbox{as}\;\; \epsilon\to0. \end{equation} $

主要借助文献[10]中的弱收敛方法和文献[11]中解的紧性分析方法.

该文结构如下:第2节给出非线性项的性质和主要定理.第3节给出主要定理的证明.

2 中偏差原理

本节分为两部分.第一部分给出非线性项的性质,第二部分给出本文的主要结果.

2.1 非线性项性质

$ (\cdot, \cdot) $$ |\cdot| $分别为$ (L^2(D))^3 $中内积和范数. $ (H^1_0 (D))^3 $中内积定义为

相应的范数$ \|\cdot\| $等价于通常的梯度范数.令$ H $$ {\cal V} = \{v \in ({\cal D}(D))^3: \nabla v = 0 \; \hbox{in}\; D\} $$ (L^2(D))^3 $中闭集, $ V $$ {\cal V} $$ (H^1_0(D))^3 $闭集. $ H $为带有$ (L^2(D))^3 $内积的希尔伯特空间, $ V $$ (H^1_0(D))^3 $中希尔伯特子空间.令$ A $为Stokes算子,定义域$ D(A) = (H^2(D))^3 \cap V $且定义为$ Aw = -{\cal P}(\bigtriangleup w), w\in D(A) $,其中$ {\cal P} $为从$ (L^2(D))^3 $$ H $的投影算子.因为$ \partial D $$ C^2 $, $ D(A) $相应于内积$ (v, w)_{D(A)} = (Av, Aw) $为希尔伯特空间.对于$ u, v\in D(A) $,定义算子$ \tilde{A} $

显然对于$ v\in D(A) $,有

$ \begin{equation} \langle\tilde{A}v, v\rangle\geq \tilde{\alpha} |Av|^2, \end{equation} $

其中$ \tilde{\alpha} = \nu\alpha $.对于$ u\in D(A) $$ v \in (L^2(D))^3 $,定义$ (u\cdot \nabla)v $$ (H^{-1}(D))^3 $中元素如下

$ \begin{equation} \langle(u\cdot \nabla)v, w\rangle = \sum\limits_{i, j = 1}^3\langle\partial_iv_j, u_iw_j\rangle, \forall w \in (H^1_0 (D))^3. \end{equation} $

如果$ u \in D(A) $,那么$ (\nabla u)^*\in (H^1(D))^{3\times3} \subset (L^6(D))^{3\times3} $,相应的对于$ v\in (L^2(D))^3 $,有$ (\nabla u)^*\cdot v \in(L^{3/2}(D))^3 \subset (H^{-1}(D))^3 $,且

对于$ (u, v, w) \in D(A)\times (L^2(D))^3 \times (H^1_0 (D))^3 $,考虑如下三线性形式

对于$ (u, v, w)\in(D(A))^3 $,令$ \langle\tilde{B}(u, v), w\rangle = b^\sharp(u, v-\alpha\triangle v, w) $.$ \tilde{B}: D(A)\times D(A)\to D(A)' $是双线性映射且对于$ u, v, w\in D(A) $,满足

$ \begin{equation} \langle \tilde{B}(u, v), u\rangle = 0, \end{equation} $

$ \begin{equation} \|\tilde{B}(u, v)\|_{D(A)'}\leq c_1\|u\|\|v\|_{D(A)}, \end{equation} $

$ \begin{equation} \langle \tilde{B}(u, v), w\rangle\leq c_1\|u\|_{D(A)}\|v\|_{D(A)}\|w\|. \end{equation} $

因此(1.1)式中第一个方程等价于如下方程

$ \begin{equation} {\rm d}\hat{u}^\epsilon(t)+\tilde{A}u^\epsilon(t){\rm d}t+\tilde{B}(u^\epsilon(t), u^\epsilon(t)){\rm d}t = F(t, u^\epsilon(t)){\rm d}t+\epsilon\int_{{\Bbb Z}} G (t, u^\epsilon(t-), z)\tilde{N}^{\epsilon^{-1}}({\rm d}t, {\rm d}z), \end{equation} $

其中$ \hat{u}^\epsilon = u^\epsilon+\alpha Au^\epsilon $且初值为$ u^\epsilon(x, 0) = u_{0}(x) $.方程(1.2)等价于

$ \begin{eqnarray} {\rm d}\hat{u}(t)+\tilde{A}u(t){\rm d}t+\tilde{B}(u(t), u(t)){\rm d}t = F(t, u(t)){\rm d}t, \end{eqnarray} $

其中$ \hat{u} = u+\alpha Au $且初值$ u(x, 0) = u_0 $.

假设$ F: [0, T]\times \Omega\times V\to V $$ {\cal B}{\cal F}\otimes {\cal B}(V) $ -可测函数, $ F $对第二个变量可微. $ G: [0, T]\times \Omega\times V\times {\Bbb Z}\to V $$ P\otimes {\cal B}(V)\otimes {\Bbb Z} $ -可测函数.进一步假设$ F $$ G $满足如下假设, P-a.s.

$ \begin{equation} \|F(t, v_1)-F(t, v_2)\|\leq C\|v_1-v_2\|, \end{equation} $

$ \begin{equation} \|F(t, v_1)\|\leq C(1+\|v_1\|), \end{equation} $

$ \begin{equation} \|F'(t, v_1)-F'(t, v_2)\|_{L(V)}\leq C\|v_1-v_2\|, \end{equation} $

$ \begin{equation} \| G (t, v_1, z)- G (t, v_2, z)\|_{V} \leq L_{ G }(z)\|v_1-v_2\|, \end{equation} $

$ \begin{equation} \| G (t, v_1, z)\|_{V} \leq K_0(z)(1+\|v_1\|), \end{equation} $

$ \begin{equation} \| G (t, v_1, z)\|^4_{L^4({\Bbb Z}, \vartheta; V)} \leq C(1+\|v_1\|^4), \end{equation} $

其中$ v_1, v_2 \in V $, $ L_{ G }(z), K_0(z)\in L^2(\vartheta)\cap{\cal H} $$ L(V) $为从$ V $$ V $的有界线性算子.

2.2 中偏差原理

对于局部紧的光滑空间$ S $,令$ {\cal M}(S) $$ (S, {\cal B}(S)) $上所有测度$ \vartheta $构成的空间,且对$ S $的任意紧子集$ K $$ \vartheta(K) < \infty $.赋予$ {\cal M}(S) $最弱的拓扑使得对于$ f \in C_c({\Bbb Z}) $, $ \vartheta\to \langle f, \vartheta\rangle = \int_Sf(x)\vartheta({\rm d}x), \vartheta\in {\cal M}(S) $为连续函数.该拓扑可测使得$ {\cal M}({\Bbb Z}) $为光滑空间.固定$ T > 0 $,令$ S_T = [0, T]\times S $, $ \vartheta_T = \lambda_T\otimes\vartheta $,其中$ \lambda_T $$ [0, T] $上的勒贝格测度, $ \vartheta\in {\cal M}(S) $.

$ {\Bbb Z} $为光滑空间$ {\Bbb M} = {\cal M}({\Bbb Z}_T) $,则$ {\cal B}({\Bbb M}) $代表空间$ {\cal M}({\Bbb Z}) $上的波雷尔$ \sigma $ -场, $ N: {\Bbb M}\to {\Bbb M}, N(m)\doteq m $为带有测度$ \vartheta_T $泊松随机测度.对于$ \theta > 0 $,定义$ P_\theta $$ ({\Bbb M}, {\cal B}({\Bbb M})) $上唯一概率测度使得$ N $是一个带有测度$ \theta\vartheta_T $的泊松随机测度.令$ {\Bbb X} = {\Bbb Z}\times [0, \infty) $, $ {\Bbb X}_T = [0, T]\times {\Bbb X} $, $ \bar{{\Bbb M}} = {\cal M}({\Bbb X}_T) $, $ \bar{N}: \bar{{\Bbb M}}\to \bar{{\Bbb M}}, \bar{N}(\bar{m})\doteq \bar{m} $, $ {\cal F}_t = \sigma(\bar{N}(s, Z): 0\leq s\leq t, Z\in {\cal B}({\Bbb X})) $$ \bar{{\cal F}}_t $$ {\cal F}_t $的完备化.令$ \bar{{\cal P}} $$ [0, T]\times \bar{{\Bbb M}} $上可料$ \sigma $ -场.

$ \bar{{\cal A}}_+ $ ($ \bar{{\cal A}} $)为所有$ (\bar{{\cal P}}\otimes{\cal B}({\Bbb Z}))/{\cal B}[0, \infty) $ ($ (\bar{{\cal P}}\otimes{\cal B}({\Bbb Z}))/{\cal B}(\mathbb{R}) $)可测映射$ \phi: {\Bbb Z}_T\times \bar{{\Bbb M}}\to [0, \infty) $ ($ \phi: {\Bbb Z}_T\times \bar{{\Bbb M}}\to \mathbb{R} $).对于$ \phi\in \bar{{\cal A}}_+ $,定义$ {\Bbb Z}_T $上可数过程$ N^\phi $

这里$ N^\phi $被认为是控制随机测度,其中$ \phi $选择在位置$ z $和时间$ s $处的点的强度.当$ \phi(s, x, \bar{m})\equiv\theta\in(0, \infty) $时,令$ N^\phi = N^\theta. $注意到$ N^\theta $相应于$ \bar{P} $$ N $相应于$ P_\theta $有相同的分布.

定义$ \ell: [0, \infty) \to [0, \infty) $ by $ \ell(r) = r \hbox{log} r - r + 1, r \in [0, \infty) $.对任意$ \phi\in \bar{{\cal A}}_+ $,定义$ L_T(\phi) $

$ \{K_n\subset{\Bbb Z}, n = 1, 2, \cdots\} $为一列递增的紧集使得$ \bigcup\limits_{n = 1}^\infty K_n = {\Bbb Z} $.对任意$ n $,令

$ \bar{{\cal A}}_b = \bigcup\limits_{n = 1}^\infty\bar{{\cal A}}_{b, n}. $

$ \{{\cal G}^\epsilon\}_{\epsilon > 0} $为从$ {\Bbb M} $到光滑空间$ E $的可测映射, $ a(\epsilon) $满足(1.3)式.对于$ \epsilon > 0 $, $ M < \infty, $

空间$ L^2(\vartheta_T) $中的范数表示为$ \|\cdot\|_2 $, $ B_2(R) $代表$ L^2(\vartheta_T) $中半径为$ R $的球且带有弱拓扑$ L^2(\vartheta_T) $.假设$ \phi\in {\cal S}^M_{+, \epsilon}. $由文献[10,引理3.2]知,存在独立于$ \epsilon $$ \kappa_2(1)\in(0, \infty) $使得$ \psi1_{\{|\psi|\leq1/a(\epsilon)\}}\in B_2(\sqrt{M\kappa_2(1)}) $,其中$ \psi = (\phi-1)/a(\epsilon) $.

$ \tilde{Y}^\epsilon = \frac1{a(\epsilon)}(u^\epsilon-u) $.$ \tilde{Y}^\epsilon $满足如下方程

$ \begin{equation} \left\{\begin{array}{ll} {\rm d}\tilde{Y}^\epsilon(t) = -\tilde{A}\tilde{Y}^\epsilon(t){\rm d}t-\tilde{B}(\tilde{Y}^\epsilon(t), u(t)){\rm d}t -\tilde{B}(u^\epsilon(t), \tilde{Y}^\epsilon(t)){\rm d}t\\ +\frac{1}{a(\epsilon)}(F(t, u^\epsilon(t))-F(t, u(t))){\rm d}t+\frac{\epsilon}{a(\epsilon)}\int_{{\Bbb Z}} G (t, u^\epsilon(t-), z) \tilde{N}^{\epsilon^{-1}\phi^\epsilon}({\rm d}t, {\rm d}z) \\ +\frac{1}{a(\epsilon)}\int_{{\Bbb Z}} G (t, u^\epsilon(t), z)(\phi^\epsilon(z, t)-1)\vartheta({\rm d}z){\rm d}t, \\ Y^\epsilon(0) = 0. \end{array}\right. \end{equation} $

方程(2.14)的解给定一个从$ {\Bbb M} $$ {\cal D}([0, T];V) $映射$ {\cal G}^\epsilon $使得$ {\cal G}^\epsilon(\epsilon N^{\epsilon^{-1}}) = \tilde{Y}^\epsilon. $$ {\cal G}_0: L^2(\vartheta_T)\to C([0, T];V)\cap L^2(0, T;D(A)) $,对于$ \psi\in L^2(\vartheta_T) $, $ {\cal G}_0(\psi) = \eta, $其中$ \eta $满足

$ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\eta(t)& = &-\tilde{A}\eta(t)-\tilde{B}(\eta(t), u(t))-\tilde{B}(u(t), \eta(t)) +F'(t, u(t))\eta(t){\rm d}t\\ & & +\int_{{\Bbb Z}}\psi(z, t) G (t, u(t), z)\vartheta({\rm d}z). \end{eqnarray} $

现在给出本文的主要结果.

定理2.1  $ \{\tilde{Y}^\epsilon = {\cal G}^\epsilon(\epsilon N^{\epsilon^{-1}}), \epsilon > 0\} $$ {\cal D}([0, T];V) $上满足大偏差原理,其速度为$ \epsilon/a^2(\epsilon) $且速度函数$ I $

其中如果$ S^0_\eta = \emptyset $,则$ I(\eta) = +\infty $.

3 定理2.1的证明

本节将给出定理2.1的证明.首先,给出如下引理.

引理3.1[10]  令$ h\in L^2(\vartheta)\cap {\cal H} $$ [0, T] $上的可测子集$ I $, $ M > 0 $.那么存在$ \varsigma_h > 0 $, $ \Gamma_h, \rho_h:(0, \infty) \to(0, \infty) $满足$ u\to\infty $$ \Gamma_h(u)\to 0 $,对任意$ \epsilon, \beta\in(0, \infty), $

因为$ {\cal G}^\epsilon(\epsilon N^{\epsilon^{-1}}) = \tilde{Y}^\epsilon = \frac1{a(\epsilon)}(u^\epsilon-u) $,则$ Y^\epsilon = {\cal G}^\epsilon(\epsilon N^{\epsilon^{-1}\phi^\epsilon}) = \frac1{a(\epsilon)}(X^\epsilon-u) $满足如下方程

$ \begin{equation} \left\{\begin{array}{ll} {\rm d}Y^\epsilon(t) = -\tilde{A}Y^\epsilon(t){\rm d}t-\tilde{B}(Y^\epsilon(t), u(t)){\rm d}t -\tilde{B}(X^\epsilon(t), Y^\epsilon(t)){\rm d}t\\ +\frac{1}{a(\epsilon)}(F(t, X^\epsilon(t))-F(t, u(t))){\rm d}t+\frac{\epsilon}{a(\epsilon)}\int_{{\Bbb Z}} G (t, X^\epsilon(t-), z) \tilde{N}^{\epsilon^{-1}\phi^\epsilon}({\rm d}t, {\rm d}z) \\ +\frac{1}{a(\epsilon)}\int_{{\Bbb Z}} G (t, X^\epsilon(t), z)(\phi^\epsilon(z, t)-1)\vartheta({\rm d}z){\rm d}t, \\ Y^\epsilon(0) = 0, \end{array}\right. \end{equation} $

其中$ X^\epsilon\in {\cal D}([0, T], V)\cap L^2(0, T; D(A)) $为如下方程的解

$ \begin{eqnarray} \left\{\begin{array}{ll} {\rm d}x^\epsilon(t) = -\tilde{A}X^\epsilon(t){\rm d}t-\tilde{B}(X^\epsilon(t), X^\epsilon(t)){\rm d}t +F(t, X^\epsilon(t)){\rm d}t\\ +\epsilon\int_{{\Bbb Z}} G (t, X^\epsilon(t-), z) \tilde{N}^{\epsilon^{-1}\phi^\epsilon}({\rm d}t, {\rm d}z) \\ +\int_{{\Bbb Z}} G (t, X^\epsilon(t-), z) (\phi^\epsilon(z, t)-1)\vartheta({\rm d}z){\rm d}t, \\ X^\epsilon(0) = u_0. \end{array}\right. \end{eqnarray} $

下面给出$ u $, $ X^\epsilon $的估计及它们之间的收敛关系.

引理3.2  存在$ \epsilon_0 > 0 $使得

$ \begin{equation} \sup\limits_{0\leq t\leq T}\|u(t)\|^2+\int_0^T\|u(t)\|^2_{D(A)}{\rm d}t\leq C, \end{equation} $

$ \begin{equation} \sup\limits_{\epsilon\in(0, \epsilon_0]}E\bigg[\sup\limits_{0\leq t\leq T}\|X^\epsilon(t)\|^2+\int_0^T\|X^\epsilon(t)\|^2_{D(A)}{\rm d}t\bigg] \leq C, \end{equation} $

$ \begin{equation} \lim\limits_{\epsilon\to0}\bigg(E\sup\limits_{0\leq t\leq T}\|X^\epsilon(t)-u(t)\|^2 +E\int_0^T\|X^\epsilon(t)-u(t)\|_{D(A)}^2{\rm d}t\bigg) = 0. \end{equation} $

   (3.3)式的证明见文献[3]. (3.4)式的证明类似(3.5)式.以下证明(3.5)式.令$ Z^\epsilon = X^\epsilon-u. $

由Itô公式得

由(2.1)式得

由(2.3)-(2.5)式和Young不等式得

由(2.8)式得

$ \psi^\epsilon = (\phi^\epsilon(z, s)-1)/a(\epsilon) $.由(2.11)-(2.12)得

结合以上估计得

由Gronwall引理得

由Burkholder-Davis-Gundy (BDG)不等式得

由引理3.1和(3.3)式得

联合以上估计

由此证毕.

引理3.3   $ \{Y^\epsilon\} $$ {\cal D}([0, T], V) $中是紧的.

  类似(3.4)式的证明有

$ \begin{eqnarray} \sup\limits_{\epsilon\in(0, \epsilon_0]}E\bigg[\sup\limits_{0\leq t\leq T}\|Y^\epsilon(t)\|^2+\int_0^T\|Y^\epsilon(t)\|^2_{D(A)}{\rm d}t\bigg]\leq C. \end{eqnarray} $

$ \{e_i\}_{i\in {\Bbb N}} $$ V $中一组完备正交基.由(3.6)式得

由此对$ \forall\eta > 0, $

$ \begin{eqnarray} \lim\limits_{L\to\infty} \lim\limits_{\epsilon\to0}\sup P(r_L^2(Y^\epsilon)>\eta \; s\in[0, T]) = 0. \end{eqnarray} $

$ {\cal D} = D(A) $.下证$ \{Y^\epsilon\} $$ {\cal D} $ -弱紧.令$ h\in D(A) $,且$ \{\tau_\epsilon, \delta_\epsilon\} $满足(a)对任意$ \epsilon, \tau_\epsilon $相对于$ \sigma $ -场为停时且取有限多的值; (b)常数$ \delta_\epsilon\in [0, T] $$ \epsilon\to0 $时满足$ \delta_\epsilon\to0 $.由(3.6)式可知对$ t\in [0, T] $, $ \{((Y^\epsilon(t), h))\} $是紧的.由(3.1)式得

由(3.3), (3.4)和(3.6)式得$ \lim\limits_{\epsilon\to0} E|((I_1+I_2+I_3, h))|^2 = 0. $由(2.8)及(3.6)式得

注意到$ \psi^\epsilon = (\phi^\epsilon(z, t)-1)/a(\epsilon) $,因此$ \lim\limits_{\epsilon\to0} E|((I_5+I_6, h))|^2 = 0. $联合以上不等式得当$ \epsilon\to0 $,有

$ \begin{eqnarray} Y^\epsilon(\tau_\epsilon+\delta_\epsilon)-Y^\epsilon(\tau_\epsilon)\to 0, \;\;{\Bbb P}\mbox{-a.s.} . \end{eqnarray} $

由(3.7)式, (3.8)式及文献[12]中的紧性判别准则得, $ \{Y^\epsilon\} $$ {\cal D}([0, T], V) $中是紧的.证毕.

将(3.1)式的解分解为$ Z^\epsilon $, $ L^\epsilon $$ U^\epsilon $,且分别满足如下方程

初值为$ Z^\epsilon(0) = 0, $$ L^\epsilon(0) = 0 $$ U^\epsilon(0) = 0. $

引理3.4   $ Z^\epsilon $, $ L^\epsilon $$ U^\epsilon $满足如下估计

$ \begin{equation} \lim\limits_{\epsilon\to0}E\bigg(\sup\limits_{t\in[0, T]}\|Z^\epsilon(t)\|^2+2\tilde{\alpha}\int_0^t\|Z^\epsilon(s)\|_{D(A)}^2{\rm d}s\bigg) = 0. \end{equation} $

$ \begin{equation} \lim\limits_{\epsilon\to0}E\bigg(\sup\limits_{t\in[0, T]}\|L^\epsilon(t)\|^2+2\tilde{\alpha}\int_0^t\|L^\epsilon(s)\|_{D(A)}^2{\rm d}s\bigg) = 0. \end{equation} $

$ \begin{equation} \lim\limits_{\epsilon\to0}E\bigg(\sup\limits_{t\in[0, T]}\|U^\epsilon(t)\|^2+2\tilde{\alpha}\int_0^t\|U^\epsilon(s)\|_{D(A)}^2{\rm d}s\bigg) = 0. \end{equation} $

  由Itô公式得

由BDG不等式, (2.12)及(3.4)式得

结合以上估计可得(3.9)式.类似可得(3.10)式和(3.11)式.证毕.

命题3.1  给定$ M > 0, $$ \{\phi^\epsilon\}_{\epsilon > 0} $使得对$ \epsilon > 0, $$ \phi^\epsilon\in {\cal U}^M_{+, \epsilon} $.$ \psi^\epsilon = (\phi^\epsilon-1)/a(\epsilon) $$ \beta\in(0, 1]. $如果当$ \epsilon\rightarrow 0 $时, $ \psi^\epsilon1_{\{|\psi^\epsilon|\leq\beta/a(\epsilon)\}} $$ B_2(\sqrt{M\kappa_2(1)}) $中依分布收敛到$ \psi $,那么当$ \epsilon\rightarrow 0 $时, $ {\cal G}^\epsilon(\epsilon N^{\epsilon^{-1}\varphi^\epsilon}) $$ {\cal D}([0, T];V) $中依分布收敛到$ {\cal G}_0(\psi) $.

  令$ K^\epsilon = Z^\epsilon+L^\epsilon+U^\epsilon $$ \Upsilon^\epsilon = Y^\epsilon-K^\epsilon. $由(3.1)式得

$ \begin{equation} \left\{\begin{array}{ll} {\rm d}\Upsilon^\epsilon(t) = -\tilde{A}\Upsilon^\epsilon(t){\rm d}t-a(\epsilon)\tilde{B}(\Upsilon^\epsilon(t)+K^\epsilon(t), \Upsilon^\epsilon(t)+K^\epsilon(t)){\rm d}t\\ -\tilde{B}(\Upsilon^\epsilon(t)+K^\epsilon(t), u(t)){\rm d}t -\tilde{B}(u(t), \Upsilon^\epsilon(t)+K^\epsilon(t)){\rm d}t\\ +\frac1{a(\epsilon)}(F(t, a(\epsilon)(\Upsilon^\epsilon(t)+K^\epsilon(t))+u(t))-F(t, u(t))){\rm d}t\\ +\int_{{\Bbb Z}} G (t, u(t), z)\psi^\epsilon(z, t)1_{\{|\psi^\epsilon|\leq\beta/a(\epsilon)\}}\vartheta({\rm d}z){\rm d}t, \\ \Upsilon^\epsilon(0) = 0. \end{array}\right. \end{equation} $

$ \Pi = ({\cal D}([0, T], V); {\cal D}([0, T], V)\cap L^2([0, T], D(A)); B_2(\sqrt{M\kappa_2(1)})). $由引理3.3知$ Y^\epsilon $$ {\cal D}([0, T], V) $紧.由(3.9)-(3.11)式及紧性定义,存在$ {\cal D}([0, T], V)\cap L^2([0, T], D(A)) $中的紧集$ K^\epsilon $.由文献[10,引理3.2],存在$ (\psi^\epsilon(z, t)1_{\{|\psi^\epsilon|\leq\beta/a(\epsilon)\}})_{\epsilon > 0} $且在$ B_2(\sqrt{M\kappa_2(1)}) $中紧.令$ (Y, 0, \psi) $$ (Y^\epsilon, K^\epsilon, \psi^\epsilon(z, t)1_{\{|\psi^\epsilon|\leq\beta/a(\epsilon)\}})_{\epsilon > 0} $$ \Pi $中的任意极限点.下面说明$ Y $$ {\cal G}_0(\psi) $具有同样的分布,且当$ \epsilon\to0 $时, $ Y^\epsilon $$ {\cal D}([0, T];V) $依分布收敛到$ Y $.

由Skorokhod表达定理,存在随机基$ (\Omega^1, {\cal F}^1, \{{\cal F}_t^1\}_{t\in[0, T]}, P^1) $及其上的$ \Pi $ -值随机变量$ (\tilde{Y}^\epsilon, \widetilde{K}^\epsilon, \tilde{\psi}^\epsilon), (\tilde{Y}, 0, \tilde{\psi}), \epsilon\in(0, \epsilon_0) $使得$ (\tilde{Y}^\epsilon, \widetilde{K}^\epsilon, \tilde{\psi}^\epsilon) $$ (Y^\epsilon, K^\epsilon, \psi^\epsilon(z, t)1_{\{|\psi^\epsilon|\leq\beta/a(\epsilon)\}})_{\epsilon > 0} $, $ (\tilde{Y}, 0, \tilde{\psi}) $$ (Y, 0, \psi) $具有相同的分布且在$ \Pi $中, $ P^1 $-a.s. $ (\tilde{Y}^\epsilon, \widetilde{K}^\epsilon, \tilde{\psi}^\epsilon)\to (\tilde{Y}, 0, \tilde{\psi}) $.因此,只需证明如下收敛性

$ \begin{equation} \sup\limits_{t\in[0, T]}\|\tilde{Y}^\epsilon-\tilde{Y}\|^2\to0, P^1\hbox{-a.s., }\; \epsilon\to0. \end{equation} $

$ \begin{equation} \left\{\begin{array}{ll} {\rm d}\tilde{\Gamma}^\epsilon(t) = -\tilde{A}\tilde{\Gamma}^\epsilon(t){\rm d}t +\int_{{\Bbb Z}} G (t, u(t), z)\tilde{\psi}^\epsilon(z, t)\vartheta({\rm d}z){\rm d}t, \\ \tilde{\Gamma}^\epsilon(0) = 0, \end{array}\right. \end{equation} $

$ \tilde{\Gamma} $为(3.14)式中将$ \tilde{\psi}^\epsilon $替换成$ \tilde{\psi} $的解.由类似(3.19)式的估计可得

$ \begin{eqnarray} \lim\limits_{\epsilon\to0}\bigg(\sup\limits_{t\in[0, T]}\|\tilde{\Gamma}^\epsilon(t)-\tilde{\Gamma}(t)\|^2 +2\tilde{\alpha}\int_0^t\|\tilde{\Gamma}^\epsilon(s)-\tilde{\Gamma}(s)\|_{D(A)}^2{\rm d}s\bigg) = 0, \end{eqnarray} $

$ \tilde{M} = \tilde{Y}-\tilde{\Gamma} $$ \tilde{M}^\epsilon = \tilde{Y}^\epsilon-\tilde{K}^\epsilon-\tilde{\Gamma}^\epsilon $.由(3.12)式和(2.15)式得

类似引理3.2的证明可得如下解的估计

$ \begin{eqnarray} \sup\limits_{\epsilon\in[0, \epsilon_0]}\bigg[\sup\limits_{t\in[0, T]}\|\tilde{M}^\epsilon(t)\|^2+\int_0^T\|\tilde{M}^\epsilon(t)\|^2_{D(A)}{\rm d}t\bigg] \leq C, \;\; P^1\hbox{-a.s.} \end{eqnarray} $

$ m^\epsilon = \tilde{M}^\epsilon-\tilde{M} $$ n^\epsilon = \tilde{\Gamma}^\epsilon-\tilde{\Gamma} $.则(3.13)式的证明变为

$ \begin{equation} \sup\limits_{t\in[0, T]}\|m^\epsilon(t)\|^2\to0, P^1\hbox{-a.s., as}\; \epsilon\to0. \end{equation} $

固定$ \omega^1\in \Omega^1. $由(2.4)-(2.5)式, (2.10)和(3.16)式得

因此,由(3.15)式, $ a(\epsilon) \to0 $和当$ \epsilon \to0 $时, $ \sup\limits_{t\in[0, T]} (\|n^\epsilon(t)\|^2+ \|\tilde{K}^\epsilon(t)\|^2)\to0 $,可得(3.17)式.

命题3.2  固定$ \Upsilon > 0 $且当$ g^\epsilon\to g $$ g^\epsilon, g\in B_2(\Upsilon) $.则在$ C([0, T];V)\cap L^2(0, T;D(A)) $$ {\cal G}_0(g^\epsilon)\to {\cal G}_0(g) $.

  令$ f^\epsilon(t) = \int_{{\Bbb Z}}g^\epsilon(z, t) G (t, u(t), z)\vartheta({\rm d}z) $$ f(t) = \int_{{\Bbb Z}}g(z, t) G (t, u(t), z)\vartheta({\rm d}z) $.$ Z^\epsilon $$ Z $分别为以下方程的解

$ \begin{equation} {\rm d}z = -\tilde{A}Z {\rm d}t+f(t){\rm d}t, \;t\in[0, T], \end{equation} $

其初始条件分别为$ Z^\epsilon(0) = 0 $$ Z(0) = 0 $.

第一步  证明如下强收敛性

$ \begin{eqnarray} \lim\limits_{\epsilon\to0}\bigg(\sup\limits_{t\in[0, T]}\|Z^\epsilon(t)-Z(t)\|^2+\tilde{\alpha}\int_0^t\|Z^\epsilon(s)-Z(s)\|_{D(A)}^2{\rm d}s\bigg) = 0. \end{eqnarray} $

因为

所以对于$ v\in V, ((G (t, u(t), z), v))\in L^2(\vartheta_T) $

$ \begin{eqnarray} \lim\limits_{\epsilon\to0}\bigg(\int_0^tf^\epsilon(s) {\rm d}s, v\bigg) = \bigg(\int_0^t\int_{{\Bbb Z}}g(z, t) G (t, u(t), z)\vartheta({\rm d}z), v\bigg). \end{eqnarray} $

$ {\cal O} = \{f^\epsilon, \epsilon > 0\}. $对于$ I\subset[0, T] $,有

$ \begin{eqnarray} \int_I\|f^\epsilon(t)\|{\rm d}t&\leq & \int_I\int_{{\Bbb Z}}|g^\epsilon(z, t)|\| G (t, u(t), z)\|\vartheta({\rm d}z){\rm d}t\\ & \leq&\bigg(\int_0^T\int_{{\Bbb Z}}|g^\epsilon(z, t)|^2\vartheta({\rm d}z){\rm d}t\bigg)^{1/2} \bigg(\int_I\int_{{\Bbb Z}}\| G (t, u(t), z)\|^2\vartheta({\rm d}z){\rm d}t\bigg)^{1/2}\\ &\leq&\Upsilon\sup\limits_{t\in[0, T]}(1+\|u(t)\|)\sqrt{\lambda_T(I)}. \end{eqnarray} $

因此$ {\cal O}\subset L^1([0, T];V) $$ L^1([0, T];V) $中一致可积.由文献[13,命题5.4], $ Z^\epsilon $$ C([0, T];V) $相对紧.由(3.20)式,得

$ \bar{Z^\epsilon} = Z^\epsilon-Z $.注意(3.21)式对$ f(t) $也成立且$ \epsilon\to0 $$ \sup\limits_{t\in[0, T]}\|\bar{Z^\epsilon}\|\to0 $,因此

第二步  令$ L^\epsilon = {\cal G}_0(g^\epsilon)-Z^\epsilon $, $ L = {\cal G}_0(g)-Z $$ \bar{L^\epsilon} = L^\epsilon-L $.

其初值$ \bar{L^\epsilon}(0) = 0 $.由(2.1), (2.3)-(2.5), (2.8), (2.10)式和Hölder不等式得

由Gronwall引理和(3.19)式得

因此有

证毕.

定理2.1的证明  由命题3.1,命题3.2和文献[10,定理2.3],得证定理2.1.

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