该文考虑如下Lévy噪声驱动的随机LANS-$ \alpha $模型

(1.1) $ \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 $的$ \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 $和$ \alpha > 0 $为给定的常数, $ (\nabla u^{\epsilon})^* $为$ \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}} $为$ [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 $为$ [0, T] $是勒贝格测度, $ \vartheta $为$ {\Bbb Z} $上$ \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}) $且$ \vartheta({\cal O}) < \infty $,有

$ \tilde{N}^{\epsilon^{-1}}([0, t]\times {\cal O}) = N^{\epsilon^{-1}}([0, t]\times {\cal O})-\epsilon^{-1}t\vartheta({\cal O}). $

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 $

(1.2) $ \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

$ \tilde{Y}^\epsilon = (u^\epsilon-u)/a(\epsilon) $

的大偏差原理,其中$ a(\epsilon) $满足

(1.3) $ \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.

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.

令$ (\cdot, \cdot) $及$ |\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

$ ((u, v)) = (u, v) + \alpha(\nabla u, \nabla v), \; \hbox{for}\; u, v \in (H^1_0 (D))^3, $

相应的范数$ \|\cdot\| $等价于通常的梯度范数.令$ H $为$ {\cal V} = \{v \in ({\cal D}(D))^3: \nabla v = 0 \; \hbox{in}\; D\} $在$ (L^2(D))^3 $中闭集, $ V $为$ {\cal V} $在$ (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 $为$ (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 $到$ H $的投影算子.因为$ \partial D $是$ 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} $

$ \langle\tilde{A}u, v\rangle = \nu(Au, v)+\nu\alpha(Au, Av). $

显然对于$ v\in D(A) $,有

(2.1) $ \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) $和$ v \in (L^2(D))^3 $,定义$ (u\cdot \nabla)v $为$ (H^{-1}(D))^3 $中元素如下

(2.2) $ \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

$ \langle(\nabla u)^*\cdot v, w\rangle = \sum\limits_{i, j = 1}^3\int_D\partial_ju_iv_iw_j{\rm d}x, \forall w \in (H^1_0 (D))^3. $

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

$ b^\sharp(u, v, w) = \langle(u\cdot \nabla)v, w\rangle+\langle(\nabla u)^*\cdot v, w\rangle. $

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) $.有$ \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

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

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

(2.5) $ \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

(2.6) $ \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

(2.7) $ \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 $.令

$ {\cal H}: = \bigg\{h: {\Bbb Z}\to \mathbb{R} : \exists\delta>0, \;\hbox{s.t.}\; \forall \Gamma\;\; \vartheta(\Gamma)<\infty, \int_\Gamma\exp(\delta |h(z)|)\vartheta({\rm d}z)<\infty\bigg\}. $

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

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

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

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

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

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

(2.13) $ \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} $及$ L(V) $为从$ V $到$ V $的有界线性算子.

对于局部紧的光滑空间$ S $,令$ {\cal M}(S) $为$ (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 $有$ \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 $为$ [0, T] $上的勒贝格测度, $ \vartheta\in {\cal M}(S) $.

令$ {\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 $为$ ({\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})) $且$ \bar{{\cal F}}_t $为$ {\cal F}_t $的完备化.令$ \bar{{\cal P}} $为$ [0, T]\times \bar{{\Bbb M}} $上可料$ \sigma $ -场.

令$ \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 $

$ N^\phi(t, Z) = \int_{(0, t]\times Z}\int_0^\infty 1_{[0, \phi(s, z)]}(r)\bar{N}({\rm d}s, {\rm d}z, {\rm d}r), t\in[0, T], Z\in{\cal B}({\Bbb Z}), $

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} $和$ 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) $

$ L_T(\phi) = \int_{{\Bbb Z}_T}\ell(\phi(t, z, \omega))\vartheta({\rm d}z){\rm d}t. $

令$ \{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

$ \begin{eqnarray*} \bar{{\cal A}}_{b, n} = &\Big\{&\phi\in \bar{{\cal A}}_+:\; (t, \omega)\in [0, T]\times \bar{{\Bbb M}}, n\geq \phi(t, x, \omega)\geq \frac1n\;\\ &&\hbox{若}\; x\in K_n; \phi(t, x, \omega) = 1\;\hbox{若}\; x\in K_n^c\Big\} \end{eqnarray*} $

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

令$ \{{\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, $令

$ {\cal S}^M_{+, \epsilon} = \{\phi: {\Bbb Z}\times [0, T]\to \mathbb{R} _+| L_T(\phi)\leq Ma^2(\epsilon)\}, \nonumber $

$ {\cal S}^M_{\epsilon} = \{\psi: {\Bbb Z}\times [0, T]\to \mathbb{R} | \psi = (\phi-1)/a(\epsilon), \phi\in {\cal S}^M_{+, \epsilon}\}. $

$ {\cal U}^M_{+, \epsilon} = \{\phi\in \bar{{\cal A}}_b: \phi(\cdot, \cdot, \omega)\in{\cal S}^M_{+, \epsilon}, \bar{P}\hbox{-a.s.}\}, $

$ {\cal U}^M_{\epsilon} = \{\psi\in\bar{{\cal A}}: \psi(\cdot, \cdot, \omega)\in{\cal S}^M_{\epsilon}, \bar{P}\hbox{-a.s.}\}. $

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 $的$ \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) $.

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

(2.14) $ \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} $到$ {\cal D}([0, T];V) $映射$ {\cal G}^\epsilon $使得$ {\cal G}^\epsilon(\epsilon N^{\epsilon^{-1}}) = \tilde{Y}^\epsilon. $令$ {\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

(2.15) $ \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\} $在$ {\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 $为

$ I(\eta) = \inf\limits_{\psi\in S^0_\eta: = \{\psi\in L^2(\vartheta_T): \eta = {\cal G}_0(\psi)\}}\bigg[\frac12\|\psi\|_2^2\bigg], $

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

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} $及$ [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 $时$ \Gamma_h(u)\to 0 $,对任意$ \epsilon, \beta\in(0, \infty), $有

$ \sup\limits_{\phi\in {\cal S}^M_{+, \epsilon}}\int_{{\Bbb Z}\times I}h^2(z)\phi(z, s)\vartheta({\rm d}z){\rm d}s \leq\varsigma_h(a^2(\epsilon)+|I|), $

$ \sup\limits_{\phi\in {\cal S}^M_{\epsilon}}\int_{{\Bbb Z}\times I}|h(z)\phi(z, s)|1_{\{|\psi|\geq\beta/a(\epsilon)\}}\vartheta({\rm d}z){\rm d}s \leq\Gamma_h(\beta)(1+\sqrt{|I|}), $

$ \sup\limits_{\phi\in {\cal S}^M_{\epsilon}}\int_{{\Bbb Z}\times I}|h(z)\phi(z, s)|\vartheta({\rm d}z){\rm d}s \leq\rho_h(\beta)\sqrt{|I|}+\Gamma_h(\beta)a(\epsilon), $

$ \lim\limits_{\epsilon\to0}\sup\limits_{\phi\in {\cal S}^M_{\epsilon}}\int_{{\Bbb Z}\times [0, T]}|h(z)\phi(z, s)|1_{\{|\psi|\geq\beta/a(\epsilon)\}}\vartheta({\rm d}z){\rm d}s = 0. $

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

(3.1) $ \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

(3.2) $ \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

(3.3) $ \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} $

(3.4) $ \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} $

(3.5) $ \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} $

证   (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. $则

$ \left\{\begin{array}{ll} {\rm d}z^\epsilon(t) = -\tilde{A}Z^\epsilon(t){\rm d}t -\tilde{B}(X^\epsilon(t), Z^\epsilon(t)){\rm d}t-\tilde{B}(Z^\epsilon(t), u(t)){\rm d}t \\ +F(t, X^\epsilon(t)){\rm d}t-F(t, u(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, \\ Z^\epsilon(0) = 0. \end{array}\right. $

由Itô公式得

$ \begin{eqnarray*} &&\|Z^\epsilon(t)\|^2+2\int_0^t\langle\tilde{A}Z^\epsilon(s), Z^\epsilon(s)\rangle {\rm d}s +2\int_0^t\langle\tilde{B}(X^\epsilon(s), Z^\epsilon(s)), Z^\epsilon(s)\rangle {\rm d}s\nonumber\\ & = &2\int_0^t((F(s, X^\epsilon(s))-F(s, u(s)), Z^\epsilon)) {\rm d}s +2\int_0^t((\int_{{\Bbb Z}} G (s, X^\epsilon(s), z) (\phi^\epsilon(z, s)-1)\vartheta({\rm d}z){\rm d}s, Z^\epsilon))\nonumber\\ &&+2\int_0^t((\int_{{\Bbb Z}}\epsilon G (s, X^\epsilon(s-), z) \tilde{N}^{\epsilon^{-1}\phi^\epsilon}({\rm d}s, {\rm d}z), Z^\epsilon))\nonumber\\ &&+\int_0^t\int_{{\Bbb Z}}\epsilon^2\| G (s, X^\epsilon(s-), z)\|^2N^{\epsilon^{-1}\phi^\epsilon}({\rm d}s, {\rm d}z) \\ & = &:J_0+J_1+J_2+J_3. \end{eqnarray*} $

由(2.1)式得

$ 2\int_0^t\langle\tilde{A}Z^\epsilon(s), Z^\epsilon(s)\rangle {\rm d}s\geq \tilde{\alpha}\int_0^t\|Z^\epsilon(s)\|_{D(A)}^2{\rm d}s. $

由(2.3)-(2.5)式和Young不等式得

$ \begin{eqnarray*} -2\int_0^t\langle\tilde{B}(X^\epsilon(s), Z^\epsilon(s)), Z^\epsilon(s)\rangle {\rm d}s &\leq&2c_1\int_0^t\|Z^\epsilon(s)\|\|Z^\epsilon(s)\|_{D(A)}\|u(s)\|_{D(A)}{\rm d}s\nonumber\\ &\leq&\frac{\tilde{\alpha}}2\int_0^t\|Z^\epsilon(s)\|_{D(A)}^2{\rm d}s+ \frac{2c_1}{\tilde{\alpha}}\int_0^t\|u(s)\|_{D(A)}^2\|Z^\epsilon(s)\|^2{\rm d}s. \end{eqnarray*} $

由(2.8)式得

$ \begin{eqnarray*} J_0 &\leq&2\int_0^t\|F(s, X^\epsilon(s))-F(s, u(s))\|^2+\|Z^\epsilon(s)\|^2{\rm d}s \leq 2(C+1)\int_0^t\|Z^\epsilon(s)\|^2{\rm d}s. \end{eqnarray*} $

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

$ \begin{eqnarray*} J_1 &\leq&2\int_0^t\|Z^\epsilon(s)\|\int_{{\Bbb Z}}\| G (s, X^\epsilon(s), z)- G (s, u(s), z)\| |\phi^\epsilon(z, s)-1|\vartheta({\rm d}z){\rm d}s\nonumber\\ &&+2\int_0^t\|Z^\epsilon(s)\|\int_{{\Bbb Z}}\| G (s, u(s), z)\| |\phi^\epsilon(z, s)-1|\vartheta({\rm d}z){\rm d}s\nonumber\\ &\leq&a(\epsilon)\int_0^t\|Z^\epsilon(s)\|^2\int_{{\Bbb Z}}(2L_{ G }(z) +(1+\sup\limits_{\tau\in[0, T]}\|u(\tau)\|)K_0(z)) |\psi^\epsilon(z, s)|\vartheta({\rm d}z){\rm d}s\nonumber\\ &&+a(\epsilon)(1+\sup\limits_{\tau\in[0, T]}\|u(\tau)\|)\int_0^t\int_{{\Bbb Z}}K_0(z) |\psi^\epsilon(z, s)|\vartheta({\rm d}z){\rm d}s\\ &\leq & \int_0^t\|Z^\epsilon(s)\|^2{\rm d}s+J_4. \end{eqnarray*} $

结合以上估计得

$ \begin{eqnarray*} \|Z^\epsilon(t)\|^2+\frac{\tilde{\alpha}}2\int_0^t\|Z^\epsilon(s)\|_{D(A)}^2{\rm d}s \leq J_2+J_3+J_4+C\int_0^t\|Z^\epsilon(s)\|^2{\rm d}s. \end{eqnarray*} $

由Gronwall引理得

$ \begin{eqnarray*} \|Z^\epsilon(t)\|^2+\frac{\tilde{\alpha}}2\int_0^t\|Z^\epsilon(s)\|_{D(A)}^2{\rm d}s \leq C(J_2+J_3+J_4). \end{eqnarray*} $

由Burkholder-Davis-Gundy (BDG)不等式得

$ \begin{eqnarray*} EJ_2&\leq&E\bigg(\int_0^T\int_{{\Bbb Z}}4\epsilon^2\| G (s, X^\epsilon(s-), z)\|^2\|Z^\epsilon\|^2 N^{\epsilon^{-1}\phi^\epsilon}({\rm d}s, {\rm d}z)\bigg)^{1/2}\nonumber\\ &\leq&\frac12E\sup\limits_{t\in[0, T]}\|Z^\epsilon\|^2+8\epsilon E\int_0^T\int_{{\Bbb Z}}K_0^2(z)(1+\|X^\epsilon(s)\|)^2 \phi^\epsilon(z, s)\vartheta({\rm d}z){\rm d}s\nonumber\\ &\leq&\frac12E\sup\limits_{t\in[0, T]}\|Z^\epsilon(t)\|^2+16C\epsilon \varsigma_h(a^2(\epsilon)+T), \\ EJ_3& = &\epsilon E(\int_0^T\int_{{\Bbb Z}}\| G (s, X^\epsilon(s-), z)\|^2 \phi^\epsilon(z, s)\vartheta({\rm d}z){\rm d}s \leq C\epsilon \varsigma_h(a^2(\epsilon)+T). \end{eqnarray*} $

由引理3.1和(3.3)式得

$ EJ_3\leq Ca(\epsilon) (\rho_{K_0}(\beta)\sqrt{T}+\Gamma_{K_0}(\beta)a(\epsilon)). $

联合以上估计

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

由此证毕.

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

证  类似(3.4)式的证明有

(3.6) $ \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} $

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

$ \lim\limits_{L\to\infty} \lim\limits_{\epsilon\to0}\sup E\sup\limits_{s\in[0, T]}r_L^2(Y^\epsilon) = \lim\limits_{L\to\infty} \lim\limits_{\epsilon\to0}\sup E\sup\limits_{s\in[0, T]}\sum\limits_{i = L+1}^\infty((Y^\epsilon, e_i)) = 0, $

由此对$ \forall\eta > 0, $有

(3.7) $ \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} $

令$ {\cal D} = D(A) $.下证$ \{Y^\epsilon\} $为$ {\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] $当$ \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

$ \begin{eqnarray*} Y^\epsilon(\tau_\epsilon+\delta_\epsilon)-Y^\epsilon(\tau_\epsilon) & = &-\int_{\tau_\epsilon}^{\tau_\epsilon+\delta_\epsilon}\tilde{A}Y^\epsilon(t){\rm d}t -\int_{\tau_\epsilon}^{\tau_\epsilon+\delta_\epsilon}\tilde{B}(Y^\epsilon(t), u(t)){\rm d}t\nonumber\\ & &-\int_{\tau_\epsilon}^{\tau_\epsilon+\delta_\epsilon}\tilde{B}(X^\epsilon(t), Y^\epsilon(t)){\rm d}t +\frac{1}{a(\epsilon)}\int_{\tau_\epsilon}^{\tau_\epsilon+\delta_\epsilon}(F(t, X^\epsilon(t))-F(t, u(t))){\rm d}t\nonumber\\ &&+\frac{\epsilon}{a(\epsilon)}\int_{\tau_\epsilon}^{\tau_\epsilon+\delta_\epsilon}\int_{{\Bbb Z}} G (t, X^\epsilon(t-), z) \tilde{N}^{\epsilon^{-1}\phi^\epsilon}({\rm d}t, {\rm d}z) \nonumber\\ & &+\frac{1}{a(\epsilon)}\int_{\tau_\epsilon}^{\tau_\epsilon+\delta_\epsilon}\int_{{\Bbb Z}} G (t, X^\epsilon(t), z)(\phi^\epsilon(z, t)-1)\vartheta({\rm d}z){\rm d}t\nonumber\\ & = &I_1+I_2+I_3+I_4+I_5+I_6. \end{eqnarray*} $

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

$ \begin{eqnarray*} \lim\limits_{\epsilon\to0} E|((I_4, h))|^2\leq \lim\limits_{\epsilon\to0}E\int_{\tau_\epsilon}^{\tau_\epsilon+\delta_\epsilon}\|Y^\epsilon\|\|h\|{\rm d}t = 0. \end{eqnarray*} $

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

(3.8) $ \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\} $在$ {\cal D}([0, T], V) $中是紧的.证毕.

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

$ \begin{eqnarray*} {\rm d}z^\epsilon& = &-\tilde{A}Z^\epsilon {\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), \\ {\rm d}L^\epsilon& = &-\tilde{A}L^\epsilon {\rm d}t+\int_{{\Bbb Z}} G (t, X^\epsilon(t), z)\psi^\epsilon(z, t)1_{\{|\psi^\epsilon|>\beta/a(\epsilon)\}}\vartheta({\rm d}z){\rm d}t, \\ {\rm d}U^\epsilon& = &-\tilde{A}U^\epsilon {\rm d}t+\int_{{\Bbb Z}}( G (t, X^\epsilon(t), z) - G (t, u(t), z))\psi^\epsilon(z, t)1_{\{|\psi^\epsilon|\leq\beta/a(\epsilon)\}}\vartheta({\rm d}z){\rm d}t, \end{eqnarray*} $

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

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

(3.9) $ \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} $

(3.10) $ \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} $

(3.11) $ \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} $

证  由Itô公式得

$ \begin{eqnarray*} \|Z^\epsilon(t)\|^2+\tilde{\alpha}\int_0^t\|Z^\epsilon(s)\|_{D(A)}^2{\rm d}s &\leq &\frac{2\epsilon}{a(\epsilon)}\int_0^t\int_{{\Bbb Z}}(( G (s, X^\epsilon(s-), z) \tilde{N}^{\epsilon^{-1}\phi^\epsilon}({\rm d}s, {\rm d}z), Z^\epsilon(s)))\nonumber\\ & &+\frac{\epsilon^2}{a^2(\epsilon)}\int_0^t\int_{{\Bbb Z}}\| G (s, X^\epsilon(s-), z)\|^2 N^{\epsilon^{-1}\phi^\epsilon}({\rm d}s, {\rm d}z). \end{eqnarray*} $

由BDG不等式, (2.12)及(3.4)式得

$ \begin{eqnarray*} &&E\sup\limits_{t\in[0, T]}\frac{2\epsilon}{a(\epsilon)}\int_0^t\int_{{\Bbb Z}}(( G (s, X^\epsilon(s-), z) \tilde{N}^{\epsilon^{-1}\phi^\epsilon}({\rm d}s, {\rm d}z), Z^\epsilon(s)))\nonumber\\ & \leq&CE\bigg(\int_0^T\int_{{\Bbb Z}}\frac{\epsilon^2}{a^2(\epsilon)}\| G (s, X^\epsilon(s-), z)\|^2\|Z^\epsilon(s)\|^2 N^{\epsilon^{-1}\phi^\epsilon}({\rm d}s, {\rm d}z)\bigg)^{1/2}\nonumber\\ & \leq&\frac12E\sup\limits_{t\in[0, T]}\|Z^\epsilon(t)\|^2+CE\int_0^T\int_{{\Bbb Z}}\frac{\epsilon^2}{a^2(\epsilon)}\| G (s, X^\epsilon(s-), z)\|^2 N^{\epsilon^{-1}\phi^\epsilon}({\rm d}s, {\rm d}z)\nonumber\\ &\leq&\frac12E\sup\limits_{t\in[0, T]}\|Z^\epsilon(t)\|^2+\frac{C\epsilon}{a^2(\epsilon)}\varsigma_{K_0}(a^2(\epsilon)+T), \\ & &E\frac{\epsilon^2}{a^2(\epsilon)}\int_0^t\int_{{\Bbb Z}}\| G (s, X^\epsilon(s-), z)\|^2 N^{\epsilon^{-1}\phi^\epsilon}({\rm d}s, {\rm d}z)\nonumber\\ & = &\frac{\epsilon}{a^2(\epsilon)}E\int_0^T\int_{{\Bbb Z}}\| G (s, X^\epsilon(s), z)\|^2 \phi^\epsilon(z, s)\vartheta({\rm d}z){\rm d}s\nonumber\\ &\leq&\frac{\epsilon}{a^2(\epsilon)}\varsigma_{K_0}(a^2(\epsilon)+T). \end{eqnarray*} $

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, $令$ \{\phi^\epsilon\}_{\epsilon > 0} $使得对$ \epsilon > 0, $$ \phi^\epsilon\in {\cal U}^M_{+, \epsilon} $.令$ \psi^\epsilon = (\phi^\epsilon-1)/a(\epsilon) $且$ \beta\in(0, 1]. $如果当$ \epsilon\rightarrow 0 $时, $ \psi^\epsilon1_{\{|\psi^\epsilon|\leq\beta/a(\epsilon)\}} $在$ 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}) $在$ {\cal D}([0, T];V) $中依分布收敛到$ {\cal G}_0(\psi) $.

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

(3.12) $ \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} $

令$ \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 $在$ {\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) $为$ (Y^\epsilon, K^\epsilon, \psi^\epsilon(z, t)1_{\{|\psi^\epsilon|\leq\beta/a(\epsilon)\}})_{\epsilon > 0} $在$ \Pi $中的任意极限点.下面说明$ Y $和$ {\cal G}_0(\psi) $具有同样的分布,且当$ \epsilon\to0 $时, $ Y^\epsilon $在$ {\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) $和$ (Y^\epsilon, K^\epsilon, \psi^\epsilon(z, t)1_{\{|\psi^\epsilon|\leq\beta/a(\epsilon)\}})_{\epsilon > 0} $, $ (\tilde{Y}, 0, \tilde{\psi}) $和$ (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

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

令

(3.14) $ \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} $

及$ \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

(3.15) $ \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} $

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

$ \left\{\begin{array}{ll} {\rm d}\tilde{M}(t) = -\tilde{A}\tilde{M}(t){\rm d}t-\tilde{B}(\tilde{M}(t)+\tilde{\Gamma}(t), u(t)){\rm d}t\\ -\tilde{B}(u(t), \tilde{M}(t)+\tilde{\Gamma}(t)){\rm d}t +F'(t, u(t))(\tilde{M}(t)+\tilde{\Gamma}(t)), \\ \tilde{M}(0) = 0, \end{array}\right. $

$ \left\{\begin{array}{ll} {\rm d}\tilde{M}^\epsilon(t) = -\tilde{A}\tilde{M}^\epsilon(t){\rm d}t-\tilde{B}(\tilde{M}^\epsilon(t)+\tilde{\Gamma}^\epsilon(t) +\tilde{K}^\epsilon(t), u(t)){\rm d}t\\ -\tilde{B}(u(t), \tilde{M}^\epsilon(t)+\tilde{\Gamma}^\epsilon(t) +\tilde{K}^\epsilon(t)){\rm d}t\\ -a(\epsilon)\tilde{B}(\tilde{M}^\epsilon(t)+\tilde{\Gamma}^\epsilon(t) +\tilde{K}^\epsilon(t), \tilde{M}^\epsilon(t)+\tilde{\Gamma}^\epsilon(t) +\tilde{K}^\epsilon(t)){\rm d}t\\ +\frac1{a(\epsilon)}(F(t, a(\epsilon)(\tilde{M}^\epsilon(t)+\tilde{\Gamma}^\epsilon(t) +\tilde{K}^\epsilon(t))+u(t)){\rm d}t-F(t, u(t))){\rm d}t, \\ \tilde{M}^\epsilon(0) = 0. \end{array}\right. $

类似引理3.2的证明可得如下解的估计

(3.16) $ \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} $

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

(3.17) $ \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

$ \begin{eqnarray*} &&\|m^\epsilon(t)\|^2+\tilde{\alpha}\int_0^t\|m^\epsilon(s)\|^2_{D(A)}{\rm d}s\nonumber\\ & \leq &-2\int_0^t\langle\tilde{B}(n^\epsilon(s) +\tilde{K}^\epsilon(s), u(s)), m^\epsilon(s)\rangle {\rm d}s\nonumber\\ &&-2\int_0^t\langle\tilde{B}(u(s), m^\epsilon(s)+n^\epsilon(s) +\tilde{K}^\epsilon(s)), m^\epsilon(s)\rangle {\rm d}s\nonumber\\ &&-2a(\epsilon)\int_0^t\langle\tilde{B}(\tilde{M}^\epsilon(s)+\tilde{\Gamma}^\epsilon(s) +\tilde{K}^\epsilon(s), \tilde{M}^\epsilon(s)+\tilde{\Gamma}^\epsilon(s) +\tilde{K}^\epsilon(s)), m^\epsilon(s)\rangle {\rm d}s\nonumber\\ &&+\frac2{a(\epsilon)}\int_0^t((F(s, a(\epsilon)(\tilde{M}^\epsilon(s)+\tilde{\Gamma}^\epsilon(s) +\tilde{K}^\epsilon(s))+u(s))-F(s, u(s)), m^\epsilon(s))) {\rm d}s\nonumber\\ &&-2\int_0^t(F'(s, u(s))(\tilde{M}(s)+\tilde{\Gamma}(s)), m^\epsilon(s))) {\rm d}s\nonumber\\ &\leq& Ca(\epsilon)+C\sup\limits_{t\in[0, T]}(\|n^\epsilon(t)\|^2+ \|\tilde{K}^\epsilon(t)\|^2)+C\int_0^t\|m^\epsilon(s)\|^2 {\rm d}s. \end{eqnarray*} $

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

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

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

(3.18) $ \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 $和$ Z(0) = 0 $.

第一步  证明如下强收敛性

(3.19) $ \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

$ \begin{eqnarray*} \int_0^T\int_{{\Bbb Z}}\| G (t, u(t), z)\|^2\vartheta({\rm d}z){\rm d}t &\leq&\int_{{\Bbb Z}} K_0^2(z)\vartheta({\rm d}z)\int_0^T(1+\|u(t)\|)^2{\rm d}t\\ &\leq&2T\sup\limits_{t\in[0, T]}(1+\|u(t)\|^2)\int_{{\Bbb Z}} K_0^2(z)\vartheta({\rm d}z)<\infty, \end{eqnarray*} $

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

(3.20) $ \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} $

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

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

$ \begin{eqnarray*} ((Z, v)) = -\int_0^t\langle Z(s), \tilde{A}v(s)\rangle {\rm d}s +\bigg(\int_0^t\int_{{\Bbb Z}}g(z, t) G (s, u(s), z)\vartheta({\rm d}z){\rm d}s, v(s)\bigg), \forall v\in D(A). \end{eqnarray*} $

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

$ \begin{eqnarray*} \|\bar{Z^\epsilon}(t)\|^2+\tilde{\alpha}\int_0^t\|\bar{Z^\epsilon}(s)\|_{D(A)}^2{\rm d}s& \leq& 2\int_0^t\bigg(\bar{Z^\epsilon}(s), \int_{{\Bbb Z}}(g^\epsilon(z, s)-g(z, s)) G (s, u(s), z)\vartheta({\rm d}z){\rm d}s\bigg)\nonumber\\ &\leq&4 \Upsilon\sup\limits_{t\in[0, T]}(1+\|u(t)\|)\sqrt{T}\sup\limits_{t\in[0, T]}\|\bar{Z^\epsilon}(t)\| \to0, \; \epsilon\to0. \end{eqnarray*} $

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

$ \begin{eqnarray*} {\rm d}\bar{L^\epsilon}& = &-\tilde{A}\bar{L^\epsilon} {\rm d}t -\tilde{B}(\bar{L^\epsilon}(t)+\bar{Z^\epsilon}(t), u(t)){\rm d}t -\tilde{B}(u(t), \bar{L^\epsilon}(t)+\bar{Z^\epsilon}(t)){\rm d}t\\ &&+F'(t, u(t))(\bar{L^\epsilon}(t)+\bar{Z^\epsilon}(t)), \end{eqnarray*} $

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

$ \begin{eqnarray*} &&\|\bar{L^\epsilon}(t)\|^2+\tilde{\alpha}\int_0^t\|\bar{L^\epsilon}(s)\|_{D(A)}^2{\rm d}s\nonumber\\ &\leq&-2\int_0^t\langle\tilde{B}(\bar{Z^\epsilon}(s), u(s)), \bar{L^\epsilon}(s)\rangle {\rm d}s -2\int_0^t\langle\tilde{B}(u(s), \bar{Z^\epsilon}(s)), \bar{L^\epsilon}(s)\rangle {\rm d}s\nonumber\\ &&+2\int_0^t\langle\tilde{B}(\bar{L^\epsilon}(s), \bar{L^\epsilon}(s)), u(s)\rangle {\rm d}s +2\int_0^t\langle F'(s, u(s))(\bar{L^\epsilon}(s)+\bar{Z^\epsilon}(s)), \bar{L^\epsilon}(s)\rangle {\rm d}s\nonumber\\ &\leq& \frac{6Cc_1^2}{\tilde{\alpha}}\bigg(\sup\limits_{t\in[0, T]}\|\bar{Z^\epsilon}(t)\|^2+\int_0^t\|\bar{Z^\epsilon}(s)\|_{D(A)}^2{\rm d}s\bigg) +\frac{6Cc_1^2}{\tilde{\alpha}}\int_0^t\|u(s)\|_{D(A)}^2\|\bar{L^\epsilon}(s)\|^2{\rm d}s. \end{eqnarray*} $

由Gronwall引理和(3.19)式得

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

因此有

$ \begin{eqnarray*} \lim\limits_{\epsilon\to0}\bigg(\sup\limits_{t\in[0, T]}\|{\cal G}_0(g^\epsilon)-{\cal G}_0(g)\|^2+ \int_0^t\|{\cal G}_0(g^\epsilon)-{\cal G}_0(g)\|_{D(A)}^2{\rm d}s\bigg) \to0. \end{eqnarray*} $

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.