本文研究如下二维带弱阻尼不可压Navier-Stokes方程的初边值问题

(1.1) $ \begin{eqnarray} \left\{\begin{array}{ll} u_t-\mu\Delta u+(u\cdot\nabla)u+\alpha u+\nabla p=f(t, x), \ \ &(t, x)\in {{\Bbb R}} _{\tau}\times\Omega, \\ \nabla\cdot u=0, \ \ &(t, x)\in{{\Bbb R}} _{\tau}\times\Omega, \\ u(t, x)|_{\partial\Omega}=0, \ &t\in {{\Bbb R}} _{\tau}, \\ u(\tau, x)=u(\tau), \ &x\in \Omega, \end{array}\right. \end{eqnarray} $

其中$ \Omega\subset {{\Bbb R}} ^{2} $是具有光滑边界的有界区域, $ u=(u_1, u_2) $为流体的速度向量, $ p $为压力项, $ \tau\in {{\Bbb R}} $为初始时刻, $ {{\Bbb R}} _{\tau}=(\tau, +\infty) $, 系数$ \alpha>0 $, $ f(t, x) $表示外力项.

当$ \alpha=0 $时, 系统(1.1)即为经典Navier-Stokes方程

(1.2) $ \begin{eqnarray} \left\{\begin{array}{ll} u_t-\mu\Delta u+(u\cdot\nabla)u+\nabla p=f(t, x), \ \ &(t, x)\in {{\Bbb R}} _{\tau}\times\Omega, \\ \nabla\cdot u=0, \ \ &(t, x)\in{{\Bbb R}} _{\tau}\times\Omega, \\ u(t, x)|_{\partial\Omega}=0, \ &t\in {{\Bbb R}} _{\tau}, \\ u(\tau, x)=u(\tau), \ &x\in \Omega.\end{array}\right. \end{eqnarray} $

Navier-Stokes方程是代表性的流体力学方程, 刻画了粘性不可压缩流体的运动规律, 在很多领域有着广泛的应用. 有关Navier-Stokes方程解的适定性等相关结果可参考文献[1, 7, 10, 13–14]. 另外, 该系统在吸引子理论方面所取得的成果对研究湍流有着重要意义, 涉及整体吸引子, 一致吸引子及拉回吸引子等的存在性[2–3, 9, 11, 15, 19–20].

近年来, 吸引子的收敛性等相关问题受到了数学工作者的关注. Caraballo等^[4–5]考虑了带小扰动系统拉回(随机)吸引子的上半连续性, 并给出了如下的满足条件

$ \lim\limits_{\varepsilon\rightarrow 0}dist(A_{\varepsilon}(t), K)=0, \; \; \ \forall t\in {{\Bbb R}} , $

其中$ {\cal A}_{\varepsilon}(t)=\{A_{\varepsilon}(t)\}_{t\in {{\Bbb R}} } $表示拉回(随机)吸引子, $ K $为某紧集. 2009年, Wang^[16]给出了验证带随机扰动系统随机吸引子上半连续的方法, 其后, Wang等^[17–18]通过研究板方程及Kirchhoff模型, 设计了一种拉回吸引子上半连续性的证明方法并给出了结论成立的充分条件. 其它有关吸引子收敛性的结果可参考文献[6, 8, 21].

到目前为止, 当$ \alpha\rightarrow 0 $时, 系统(1.1)的拉回吸引子$ {\cal A}_\alpha(t) = \{A_\alpha(t)\}_{t\in {{\Bbb R}} } $的收敛性还没有相关的研究成果, 本文将研究这一问题, 其主要特征如下

(1) 利用Galerkin方法和紧性理论去验证系统解的适定性和正则性;

(2) 基于解的适定性, 通过验证过程$ \{U_{\alpha}(t, \tau)\} $拉回吸收集$ {\cal D}_{\alpha}(t)=\{D_{\alpha}(t)\}_{t\in {{\Bbb R}} } $的存在性及渐近紧性, 利用吸引子理论证明拉回吸引子$ {\cal A}_{\alpha}(t)=\{A_{\alpha}(t)\}_{t\in {{\Bbb R}} } $的存在性;

(3) 借助文献[17–18]中的方法证明系统(1.1)拉回吸引子的上半连续性, 即

$ \lim\limits_{\alpha\rightarrow 0}dist_H(A_\alpha(t), A_0(t))=0, \; \; \ \forall\ t\in {{\Bbb R}} . $

已知$ X $为Banach空间, 范数和距离分别为$ \|\cdot\|_X $和$ d_X(\cdot, \cdot) $, 过程$ \{U(t, \tau)\}_{t\geq\tau} $及其连续性的定义可参阅文献[2, 9].

定义2.1  对任意的$ t\geq\tau $及有界集$ D_b\subset X $, 若紧集族$ {\cal A}(t)=\{A(t)\}_{t\in {{\Bbb R}} } $满足

$ {\rm (i)} \ U(t, \tau)A(\tau)=A(t);\;\;\;{\rm (ii)} \lim\limits_{\tau\rightarrow \infty} dist_{X}(U(t, t-\tau)D_b, A(t))=0, $

则称$ {\cal A}(t)=\{A(t)\}_{t\in {{\Bbb R}} } $为过程$ \{U(t, \tau)\}_{t\geq\tau} $的拉回吸引子, 其中$ dist_{X}(\cdot, \cdot) $表示$ X $中的$ {\rm{Hausdorff}} $半距离.

定义2.2  若对任意的$ t\in {{\Bbb R}} $及有界集$ D_b\subset X $, 存在$ T(t, D_b)>0 $, 使得

$ U(t, t-\tau)D_b\subset D(t), \ \forall\ \tau\geq T(t, D_b), $

则称集合族$ {\cal D}(t)=\{D(t)\}_{t\in {{\Bbb R}} } $为过程$ \{U(t, \tau)\}_{t\geq\tau} $的拉回吸收集.

定义2.3  已知$ {\cal D}(t)=\{D(t)\}_{t\in {{\Bbb R}} } $为$ X $中的子集族. 若对任意$ t\in {{\Bbb R}} $, 序列$ \tau_n\rightarrow \infty\ (n\rightarrow \infty) $及$ x_n\in D(t-\tau_n) $, 序列$ \{U(t, t-\tau_n)x_n\} $在$ X $中预紧, 则称过程$ \{U(t, \tau)\}_{t\geq\tau} $在$ X $中拉回渐近紧.

定理2.1  已知$ {\cal D}(t)=\{D(t)\}_{t\in {{\Bbb R}} } $是连续过程$ \{U(\cdot, \cdot)\} $的拉回吸收集, 且$ \{U(\cdot, \cdot)\} $在$ X $中拉回渐近紧. 则$ {\cal A}(t)=\{A(t)\}_{t\in {{\Bbb R}} } $, 其中

$ A(t)=\mathop\bigcap\limits_{s\geq 0}\overline{\mathop\bigcup\limits_{\tau\geq s}U(t, t-\tau)D(t-\tau)}^X, \ \forall\ t\in {{\Bbb R}} , $

称为过程$ \{U(\cdot, \cdot)\} $在$ X $中的拉回吸引子.

下面, 我们给出一种验证拉回渐近紧的方法, 具体可参考文献[12].

定理2.2  假设过程可分解为$ U(\cdot, \cdot)=U_1(\cdot, \cdot)+U_2(\cdot, \cdot) $, 且对任意的$ t>\tau>0 $及$ u(t-\tau) \in D(t-\tau) $满足

$ ({\rm{i}}) $$ \|U_1(t, t-\tau)u(t-\tau)\|_{X}\leq K(t, \tau) $, 其中$ K(\cdot, \cdot):\ {{\Bbb R}} \times{{\Bbb R}} \rightarrow {{\Bbb R}} ^{+} $满足$ \lim\limits_{\tau\rightarrow +\infty}K(t, \tau)=0 $;

$ ({\rm{ii}}) $$ \{U_2(\cdot, \cdot)\} $在$ X $中紧.

则过程$ \{U(\cdot, \cdot)\} $在$ X $中拉回渐近紧.

为了研究拉回吸引子$ {\cal A}_{\alpha}(t)=\{A_\alpha(t)\}_{t\in {{\Bbb R}} } $之间的收敛关系, 下面给出相关的定义及结论.

定义2.4  已知$ \alpha \in [0, \alpha_0] $, $ \{D_\alpha(t)\}_{t\in {{\Bbb R}} } $是$ X $上的子集族, 对任意的$ t\in {{\Bbb R}} $, $ \Phi_t(\cdot, \cdot, \cdot, \cdot) $是定义在

$ [0, \alpha_0] \times [0, \alpha_0] \times \mathop\bigcup\limits_{\alpha\in[0, \alpha_0]}D_\alpha(t) \times \mathop\bigcup\limits_{\alpha\in[0, \alpha_0]}D_\alpha(t) $

上的函数. 若对任意$ \{\alpha_n\}_{n\in {\mathbb N}} \subset [0, \alpha_0] $及$ \{x_n\}_{n\in {\mathbb N}}\subset D_{\alpha_n}(t) $, 存在子序列$ \{\alpha_{n_k}\}_{k\in {\mathbb N}} \subset \{\alpha_n\} $及$ \{x_{n_k}\}_{k\in {\mathbb N}} \subset \{x_n\}_{n\in {\mathbb N}} $满足$ x_{n_k} \in D_{\alpha_{n_k}} $, 使得

$ \lim\limits_{k\rightarrow \infty}\lim\limits_{l\rightarrow \infty}\Phi_{t}(\alpha_{n_k}, \alpha_{n_l}, x_{n_k}, x_{n_l})=0, $

则称$ \Phi_t $为收缩函数.

定义2.5  若

$ \lim\limits_{\alpha\rightarrow 0}dist_X(A_\alpha(t), A_0(t))=0, \ \forall\ t\in {{\Bbb R}} , $

则称$ {\cal A}_\alpha(t) $和$ {\cal A}_0(t) $具有上半连续性.

下面, 我们给出有关上半连续性的结论, 具体可参见文献[17–18].

定理2.3  已知$ \alpha\in [0, \alpha_0] $, $ {\cal D}(t)=\{D(t)\}_{t\in {{\Bbb R}} } $为连续过程$ \{U_\alpha(t, \tau)\} $在$ X $中的拉回吸收集, $ {\cal A}(t)=\{A(t)\}_{t\in {{\Bbb R}} } $为拉回吸引子. 若

($ C_1 $) 对任意的$ t\in {{\Bbb R}} , \ \tau\in {{\Bbb R}} ^+, \ \alpha_n\rightarrow 0 $, 及$ x_n, x\in X $满足$ x_n\rightarrow x\ (n\rightarrow \infty) $, 成立

$ \lim\limits_{n\rightarrow \infty}U_{\alpha_n}(t, t-\tau)x_n=U_0(t, t-\tau)x; $

($ C_2 $) 有常数$ \theta\in(0, 1) $使得对任意的$ t\in {{\Bbb R}} $, 存在$ T=T(t, \theta) > 0 $, 使得

$ U_0(t-\theta\tau, t-\tau)\bigg(\overline{\mathop\bigcup\limits_{\alpha\in[0, \alpha_0]}A_\alpha(t-\tau)}^X\bigg) \subset D_0(t-\theta\tau), \ \tau > T; $

($ C_3 $) 对任意的$ t\in {{\Bbb R}} $, $ \mathop\bigcup\limits_{\alpha\in[0, \alpha_0]}A_{\alpha}(t) $在$ X $中预紧.

则当$ \alpha\rightarrow 0 $时, $ {\cal A}_\alpha(t) $和$ {\cal A}_0(t) $具有上半连续性.

以下结论是条件($ C_3 $)的替代方案.

引理2.1  已知$ \alpha\in[0, \alpha_0] $, 若对任意的$ t\in {{\Bbb R}} $及$ \delta>0 $, 存在$ T=T(t, \delta, \{A_{\alpha}(t)\}_{t\in {{\Bbb R}} }) $和定义在

$ [0, \alpha_0] \times [0, \alpha_0] \times \mathop\bigcup\limits_{\alpha\in[0, \alpha_0]}A_\alpha(t-T) \times \mathop\bigcup\limits_{\alpha\in[0, \alpha_0]}A_\alpha(t-T) $

上的收缩函数$ \Phi_{t-T}(\cdot, \cdot, \cdot, \cdot) $使得对任意的$ \alpha_1, \alpha_2\in[0, \alpha_0] $, $ x\in A_{\alpha_1}(t-T) $及$ y\in A_{\alpha_2}(t-T) $, 有

$ \|U_{\alpha_1}(t, t-T)x-U_{\alpha_2}(t, t-T)y\|_X\leq \delta+\Phi_{t-T}(\alpha_1, \alpha_2, x, y). $

则对任意的$ t\in {{\Bbb R}} $, $ \mathop\bigcup\limits_{\alpha\in[0, \alpha_0]}A_{\alpha}(t) $在$ X $中预紧.

记$ L^p(\Omega)\ (1\leq p\leq \infty) $为通常的Sobolev空间, 范数为$ \|\cdot\|_p $, 令$ E:=\{u|u\in(C^{\infty}_0(\Omega))^2, $$ {\rm div} u $$ = 0\} $, 则$ H=\overline{E}^{(L^2(\Omega))^2} $和$ V=\overline{E}^{(H^1(\Omega))^2} $均为可分的Hilbert空间, 内积及范数分别为

$ (u, v)=\sum^2\limits_{j=1}\int_{\Omega}u_j(x)v_j(x){\rm d}x, \ |u|=(u, u)^{1/2}, \ \forall\ u, v\in H, $

$ ((u, v))=\sum^2\limits_{i, j=1}\int_{\Omega}\frac{\partial u_j}{\partial x_i}\frac{\partial v_j}{\partial x_i}{\rm d}x, \ \|u\|=((u, u))^{1/2}, \ \forall\ u, v\in V. $

$ V' $表示$ V $的对偶空间, 范数和对偶积分别记为$ \|\cdot\|_{*} $和$ \langle\cdot, \cdot\rangle_{V\times V'} $.

Stokes算子$ A:=-P_L\Delta $, $ P_L:(L^2(\Omega))^2\rightarrow H $是Helmholtz-Leray映照, $ \{\omega_j\}^{\infty}_{j=1} $表示$ A $的对应于特征值$ \{\lambda_j\}^{\infty}_{j=1} $的特征向量. 定义幂运算$ A^{s} $如下

$ A^{s}f=\sum\limits_{j}\lambda_{j}^{s}a_{j}\omega_{j}, \ \ s\in {\mathbb C}, \ \ j\in {{\Bbb R}} , \ \ f=\sum\limits_{j}a_{j}\omega_{j}, $

仍然用$ D(A^{s})=\{f:A^{s}f\in H\} $表示$ \overline{E}^{D(A^{s})} $, 其范数为$ \|u\|_{D(A^s)} $.

对任意的$ u, v, w\in V $, 分别定义双线性算子及三线性算子

$ B(u, v):=P_L((u\cdot\nabla)v), \ b(u, v, w)=\langle B(u, v), w \rangle, $

其中, $ B(u, v) $是从$ V\times V $到$ V' $的映照, 且成立

(3.1) $ \begin{equation} \left\{ \begin{array}{ll} b(u, v, v)= 0, \ &\forall\ u, v, w\in V, \\ b(u, v, w)= -b(u, w, v), \ &\forall\ u, v, w\in V, \\ |b(u, v, w)|\leq C|u|\|v\||w|, \ &\forall\ u, v, w\in V, \end{array}\right. \end{equation} $

其它性质可参见文献[14].

定义3.1  对任意的$ v\in V $及$ t>\tau $, 若$ u\in L^{\infty}(\tau, T;H)\bigcap L^{2}(\tau, T;V)\bigcap C([\tau, T];H) $满足

(3.2) $ \begin{eqnarray} \left\{\begin{array}{ll} { } \frac{\rm d}{{\rm d}t}(u, v)+\mu((u, v))+b(u, u, v)+(\alpha u, v)=(P_Lf, v), \\ u(\tau, x)=u(\tau), \end{array}\right. \end{eqnarray} $

则称$ u $为系统$ (1.1) $在$ [\tau, T]\times\Omega $上的一个整体弱解.

记$ B(u)=B(u, u) $, 则系统(1.1)还可写成如下形式

(3.3) $ \begin{eqnarray} \left\{\begin{array}{ll} u_t+\mu Au+B(u)+\alpha u=P_Lf(t, x), \ \forall\ t>\tau, \\ u(\tau, x)=u(\tau). \end{array}\right. \end{eqnarray} $

假设$ f(t, x) $在$ V' $上一致有界, 且满足

$ \int_{-\infty}^{t}e^{\sigma s}|f(s, x)|^2{\rm d}s<\infty, \ \forall\ t\in {{\Bbb R}} , \ 0<\sigma<\frac{1}{2}\mu\lambda_1. $

定理4.1  已知$ u(\tau)\in H $, 则系统$ (1.1) $存在唯一的弱解$ u(t, x) $满足

$ u\in L^{\infty}(\tau, T;H)\bigcap L^{2}(\tau, T;V)\bigcap C([\tau, T];H). $

证  首先运用Galerkin方法定义$ P_mu = \sum\limits^m_{j=1}(w_j, u)w_j, $ 其中$ \{w_j\}_{j\in {\mathbb N}} $为$ H $中的且在$ V $中稠密的正交基, $ P_m $是从$ H $到$ V_m:= \mbox{span}[w_1, \cdots, w_m] $的投影算子. 令$ u_m = \sum\limits^m_{k=1}\gamma_{mk}(t)w_k $, 其中$ \gamma_{mk}(t) = (u_m(t), w_k)\ (k=1, 2, \cdots, m) $由如下的常微分方程组确定

(4.1) $ \begin{eqnarray} &&(u_m(t), w_k) + \mu\int^t_\tau \langle Au_m, w_k \rangle {\rm d}s + \int^t_\tau \langle B(u_m), w_k \rangle {\rm d}s + \int_\tau^t(\alpha u_m, w_k){\rm d}s{}\\ & =& (u(\tau), w_k) + \int^t_\tau(P_mP_Lf, w_k){\rm d}s, \;\;\;\;\;\;t>\tau, \;\;\forall \;1\leq k\leq m. \end{eqnarray} $

此时

(4.2) $ \begin{equation} \frac{\rm d}{{\rm d}t}(u_m, w_k) + \mu\langle Au_m, w_k\rangle + \langle B(u_m), w_k\rangle + (\alpha u_m, w_k) = (P_mP_Lf, w_k), \end{equation} $

a.e. $ t>\tau $, $ \forall $$ 1\leq k\leq m $, 系统(4.2)在$ [\tau, t_m) $上存在唯一的局部解, 通过先验估计可知解区间可延拓至$ [\tau, +\infty) $.

在(4.2)式两端乘以$ (u_m, w_k) $并对$ k $从$ k=1 $到$ k=m $求和, 得到

(4.3) $ \begin{equation} \frac{\rm d}{{\rm d}t}|u_m|^2 + 2\mu\|u_m\|^2 + 2\alpha|u_m|^2 \leq (\mu\lambda_1+\alpha)|u_m|^2 + \frac{1}{\mu\lambda_1+\alpha}|f|^2, \end{equation} $

i.e.

(4.4) $ \begin{equation} \frac{\rm d}{{\rm d}t}|u_m|^2 + (\mu\lambda_1+\alpha)|u_m|^2 \leq \frac{1}{\mu\lambda_1+\alpha}|f|^2, \end{equation} $

进而

(4.5) $ \begin{equation} \frac{\rm d}{{\rm d}t}(e^{\sigma t}|u_m|^2) + (\mu\lambda_1+\alpha - \sigma)e^{\sigma t}|u_m|^2 \leq \frac{1}{\mu\lambda_1+\alpha}e^{\sigma t}|f|^2. \end{equation} $

将(4.5)式从$ \tau $到$ t $积分, 可得

(4.6) $ \begin{equation} |u_m(t)|^2 \leq e^{-\sigma t}e^{\sigma \tau}|u(\tau)|^2 + \frac{e^{-\sigma t}}{\mu\lambda_1+\alpha}\int_{-\infty}^te^{\sigma s}|f|^2{\rm d}s. \end{equation} $

由(4.3)和(4.6)式可知

$ u_m \in L^\infty(\tau, T;H) \bigcap L^2(\tau, T;V), \;\;\;\;\forall\;\ T>\tau, $

又因$ Au_m, \;B(u_m, u_m)\in L^2(\tau, T;V') $, 由(4.2)式可得

$ \|\frac{\partial}{\partial t}u_m\|_*\leq \nu\|Au_m\|_* + \|B(u_m, u_m)\|_* + \alpha\|u_m\|_* + \|f\|_*, $

i.e., $ \frac{\partial}{\partial t}u_m \in L^2(\tau, T;V'). $

根据Aubin-Lions定理, 存在子列$ \{u_{m'}(t)\} $使得

$ \begin{eqnarray*} &&u_{m'}\rightharpoonup^*\ u\ \mbox{在} L^\infty(\tau, T;H) \mbox{中弱}^*\mbox{收敛};\nonumber\\ &&u_{m'}\rightharpoonup\ u\ \mbox{在} L^2(\tau, T;V) \mbox{中弱收敛};\nonumber\\ &&\frac{\partial}{\partial t}u_{m'} \rightharpoonup\ \eta\ \mbox{在} L^2(\tau, T;V') \mbox{中弱收敛}.\nonumber \end{eqnarray*} $

可证$ \eta=\frac{\partial}{\partial t}u $, 且由Lions-Magenes定理可得$ u\in C([\tau, T];H) $.

下面讨论解的唯一性及关于初值的连续依赖性. 不妨设$ u, v $分别是系统(3.3)关于初值$ u(\tau), v(\tau) $的弱解, 则

(4.7) $ \begin{eqnarray} &&\frac{\partial u}{\partial t} + \nu Au + B(u, u) + \partial u = P_Lf, \end{eqnarray} $

(4.8) $ \begin{eqnarray} &&\frac{\partial v}{\partial t} + \nu Av + B(v, v) + \partial v= P_Lf. \end{eqnarray} $

(4.7)式减(4.8)式, 令$ w=u-v $, 得

(4.9) $ \begin{equation} \frac{\partial}{\partial t}w + \nu Aw + B(w, u) + B(v, w) + \partial w = 0. \end{equation} $

将(4.9)式两端乘以$ w $, 有

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}|w|^2 + 2\nu\|w\|^2 + 2\alpha|w|^2 &\leq &2|b(w, u, w)|\leq C\|u\|\cdot|w|^2\\ &\leq &2(2\lambda_1 + \alpha)|w|^2 + C\|u\|^2\cdot|w|^2, \end{eqnarray*} $

i.e., $ \frac{\rm d}{{\rm d}t}|w|^2 \leq C\|u\|^2\cdot|w|^2 $, 由Gronwall不等式可得

$ |u(t) - v(t)|^2 \leq |u(\tau)-v(\tau)|e^{C\int_\tau^t\|u\|^2{\rm d}s}, $

结论得证.

基于解的整体适定性, 可定义系统(1.1)在$ H $中的过程$ \{U_\alpha(t, \tau)\} $. 下面讨论解$ u(t, x) $的正则性.

定理4.2  已知$ u(\tau)\in H $, $ u(t, x) $是系统$ (1.1) $的解, 则$ u(t, x)\in L^{\infty}(\tau, T;D(A^\frac{1}{4})) $.

证  由(4.2)式可得

(4.10) $ \begin{equation} \langle \frac{\partial}{\partial t}u_m, A^\frac{1}{2}u_m \rangle + \mu\langle Au_m, A^\frac{1}{2}u_m \rangle+ b(u_m, u_m, A^\frac{1}{2}u_m) + \alpha\langle u_m, A^\frac{1}{2}u_m \rangle = \langle P_mP_Lf, A^\frac{1}{2}u_m \rangle, \end{equation} $

利用Sobolev不等式, Hölder不等式及Young不等式, 得到

$ \begin{eqnarray*} &&\frac{\rm d}{{\rm d}t}|A^\frac{1}{4}u_m|^2 + 2\mu|A^\frac{3}{4}u_m|^2 + 2\alpha|A^\frac{1}{4}u_m|^2 \\ &\leq& 2|b(u_m, u_m, A^\frac{1}{2}u_m)| + 2|\langle f, A^\frac{1}{2}u_m \rangle|\\ &\leq& C\|u_m\|_4\cdot|\nabla u_m|\cdot\|A^\frac{1}{2}u_m\|_4 + 2|f|\cdot|A^\frac{1}{2}u_m|\\ &\leq& C|A^\frac{1}{4}u_m|\cdot|A^\frac{1}{2}u_m|\cdot|A^\frac{3}{4}u_m| + 2|f|\cdot|A^\frac{3}{4}u_m|\\ &\leq& \mu|A^\frac{3}{4}u_m|^2 + \frac{C}{\mu}|A^\frac{1}{4}u_m|^2|A^\frac{1}{2}u_m|^2 + \frac{C}{\mu}|f|^2, \end{eqnarray*} $

i.e.

(4.11) $ \begin{equation} \frac{\rm d}{{\rm d}t}|A^\frac{1}{4}u_m|^2 + \mu|A^\frac{3}{4}u_m|^2 \leq \frac{C}{\mu}|A^\frac{1}{2}u_m|^2|A^\frac{1}{4}u_m|^2 + \frac{C}{\mu}|f|^2, \end{equation} $

即

(4.12) $ \begin{equation} \frac{\rm d}{{\rm d}t}(e^{\sigma t}|A^\frac{1}{4}u_m|^2) \leq \frac{C}{\mu}|A^\frac{1}{2}u_m|^2e^{\sigma t}|A^\frac{1}{4}u_m|^2 + \frac{C}{\mu}e^{\sigma t}|f|^2. \end{equation} $

(4.12)式两端在$ [s, t] $积分, 得

(4.13) $ \begin{equation} e^{\sigma t}|A^\frac{1}{4}u_m(t)|^2 \leq \exp \left(\frac{C}{\mu}\int_s^t|A^\frac{1}{2}u_m|^2{{\rm d}p}\right) \times \left(e^{\sigma s}|A^\frac{1}{4}u_m(s)|^2 + \frac{C}{\mu}\int_s^te^{\sigma p}|f|^2{{\rm d}p}\right). \end{equation} $

与(4.3)式相似, 还可推出

(4.14) $ \begin{equation} \frac{\rm d}{{\rm d}t}|u_m|^2 + \mu\|u_m\|^2 \leq \frac{1}{\mu\lambda_1 + \alpha}|f|^2, \end{equation} $

进而

(4.15) $ \begin{equation} \frac{\rm d}{{\rm d}t}(e^{\sigma t}|u_m|^2) + \frac{\mu}{2} e^{\sigma t}\|u_m\|^2 \leq \frac{1}{\mu\lambda_1 + \alpha}e^{\sigma t}|f|^2. \end{equation} $

将(4.15)式两端在$ [s, t] $上积分并利用(4.6)式, 得出

(4.16) $ \begin{eqnarray} \frac{\mu}{2}\int_s^te^{\sigma p}\|u_m(p)\|^2{{\rm d}p} &\leq& e^{\sigma s}|u_m(s)|^2 + \frac{1}{\mu\lambda_1 + \alpha}\int_s^te^{\sigma p}|f|^2{{\rm d}p}\\ &\leq& e^{\sigma \tau}|u(\tau)|^2 + \frac{1}{\mu\lambda_1 + \alpha}\int_{-\infty}^{t}e^{\sigma p}|f|^2{{\rm d}p}, \end{eqnarray} $

进而

(4.17) $ \begin{equation} \int_s^te^{\sigma p}\|u_m\|^2{{\rm d}p} \leq \frac{2}{\mu}e^{\sigma\tau}|u(\tau)|^2 + \frac{2}{\mu(\mu\lambda_1+\alpha)}\int_{-\infty}^{t}e^{\sigma p}|f|^2{{\rm d}p}. \end{equation} $

选取

$ {\cal P}^2(t) = \frac{4}{\mu}e^{\sigma \tau}|u(\tau)|^2 + \frac{4}{\mu(\mu\lambda_1+\alpha)}\int_{-\infty}^{t}e^{\sigma p}|f|^2{{\rm d}p}, $

则

$ |\{p\in[t-1, t];e^{\sigma p}\|u_m(p)\|^2 > {\cal P}^2(t)\}| < 1, $

可知在任意闭区间$ [t-k-1, t-k]\ (k\in {\mathbb N}) $中, 存在$ p $使得

$ e^{\sigma p}\|u_m(p)\|^2 < {\cal P}^2(t), \;\;\;\forall t\in {{\Bbb R}} , $

进而由(4.13)和(4.15)式得到

(4.18) $ \begin{equation} |A^\frac{1}{4}u_m(t)|^2 \leq C_{T, \tau}. \end{equation} $

因为$ u_m\rightarrow u $在$ L^{2}(\tau, T;V) $中弱收敛, $ V $紧嵌入到$ D(A^\frac{1}{4}) $中, 则存在子列$ \{u_m\} $使得

$ u_m\rightarrow u\ \mbox{在} L^{2}(\tau, T;D(A^\frac{1}{4})) \mbox{中强收敛, } $

进而存在另一子列$ \{u_m\} $使得在$ D(A^\frac{1}{4}) $中

$ u_m\rightarrow u, \ a.e.\ t, $

结论得证.

定理5.1  已知$ u(\tau)\in H $, 对任意$ \alpha >0 $, 系统$ (1.1) $在$ H $中存在拉回吸引子$ {\cal A}_\alpha(t) = \{A_\alpha(t)\}_{t\in {{\Bbb R}} } $满足

$ A_\alpha(t) = \mathop\bigcap\limits_{s\geq 0}\overline{\mathop\bigcup\limits_{\tau\geq s} U_\alpha(t, t-\tau)D_\alpha(t-\tau)}^H, \;\;\;\;\forall t\in {{\Bbb R}} , $

其中$ {\cal D}_\alpha(t) = \{D_\alpha(t)\}_{t\in {{\Bbb R}} } $为系统$ (1.1) $的拉回吸收集.

证  第一步  $ H $中拉回吸收集$ {\mathcal D}_\alpha(t) = \{D_\alpha(t)\}_{t\in {{\Bbb R}} } $的存在性.

(3.3)式两端乘以$ u $, 得

$ \frac{\rm d}{{\rm d}t}|u|^2 + (\mu\lambda_1 + \alpha)|u|^2 \leq \frac{1}{\mu\lambda_1 + \alpha}|f|^2, $

进而

$ \frac{\rm d}{{\rm d}t}(e^{\sigma t}|u|^2) + (\mu\lambda_1 + \alpha - \sigma)e^{\sigma t}|u|^2 \leq \frac{1}{\mu\lambda_1 + \alpha}e^{\sigma t}|f|^2. $

上式两端在$ [\tau, t] $上积分, 推出

(5.1) $ \begin{equation} |u(t)|^2 \leq e^{-\sigma t}e^{\sigma \tau}|u(\tau)|^2 + \frac{1}{\mu\lambda_1 + \alpha}e^{-\sigma t}\int_\tau^t e^{\sigma s}|f|^2{\rm d}s, \end{equation} $

及

(5.2) $ \begin{equation} |u(t)|^2 \leq e^{-\sigma \tau}|u(t-\tau)|^2 + \frac{1}{\mu\lambda_1 + \alpha}e^{-\sigma t}\int_{t-\tau}^t e^{\sigma s}|f|^2{\rm d}s, \end{equation} $

即拉回吸收集$ {\cal D}_\alpha(t) = \{D_\alpha(t)\}_{t\in {{\Bbb R}} } $存在, 其中

$ D_\alpha(t) = \left\{u\in H: |u|^2 \leq \frac{2}{\nu\lambda_1 + \alpha}e^{-\sigma t} \int_{-\infty}^te^{\sigma s}|f|^2{\rm d}s \equiv R_\alpha^2(t)\right\}. $

若$ u_{t-\tau} \in D_\alpha(t-\tau) $, 则

$ \begin{eqnarray*} |U_\alpha(t, t-\tau)u_{t-\tau}|^2& \leq& e^{-\sigma \tau}R_\alpha^2(t-\tau) + \frac{1}{\mu\lambda_1 + \alpha}e^{-\sigma t}\int_{-\infty}^te^{\sigma s}|f|^2{\rm d}s\\ &\leq& \frac{2e^{-\sigma t}}{\mu\lambda_1 + \alpha}\int_{-\infty}^{t-\tau}e^{\sigma s}|f|^2{\rm d}s + \frac{e^{-\sigma t}}{\mu\lambda_1 + \alpha}\int_{-\infty}^{t}e^{\sigma s}|f|^2{\rm d}s, \end{eqnarray*} $

存在$ T>0 $使得对任意的$ t\in {{\Bbb R}} $, 有

$ U_\alpha(t, t-\tau)D_\alpha(t-\tau) \subset D_\alpha(t), \;\;\;\forall\ \tau>T. $

第二步  过程$ \{U_\alpha(t, \tau)\} $的渐近紧性.

令

$ U_\alpha(t, \tau)u(\tau) = L_\alpha(t-\tau)u(\tau) +Q_\alpha(t, \tau)u(\tau)= u_1(t) + u_2(t), $

分别满足

(5.3) $ \begin{equation} \left\{ \begin{array}{ll} { } \frac{\partial}{\partial t}u_{1} + \nu Au_1 + \alpha u_1 = 0, \;\;&(t, x)\in {{\Bbb R}} _\tau\times\Omega, \\ {\rm div} u_1 = 0, &(t, x)\in {{\Bbb R}} _\tau\times\Omega, \\ u_1(t, x)|_{\partial\Omega} = 0, &t\in {{\Bbb R}} _\tau, \\ u_1(\tau) = u(\tau), &x\in \Omega \end{array} \right. \end{equation} $

和

(5.4) $ \begin{equation} \left\{ \begin{array}{ll} { } \frac{\partial}{\partial t}u_{2} + \nu Au_2 + B(u, u) + \alpha u_2 = P_Lf, \;\;&(t, x)\in {{\Bbb R}} _\tau\times\Omega, \\ {\rm div} u_2 = 0, &(t, x)\in {{\Bbb R}} _\tau\times\Omega, \\ u_2(t, x)|_{\partial\Omega} = 0, &t\in {{\Bbb R}} _\tau, \\ u_2(\tau) = 0, &x\in \Omega. \end{array} \right. \end{equation} $

(5.3)式乘以$ u_1 $, 得

$ \frac{\rm d}{{\rm d}t}|u_1|^2 + 2(\nu\lambda_1 + \alpha)|u_1|^2 \leq 0, $

由Gronwall不等式可知

$ |u_1(t)|^2 \leq e^{-2(\nu\lambda_1 + \alpha)(t-\tau)}|u(\tau)|^2, $

从而对任意的$ t\in {{\Bbb R}} $, $ \lim\limits_{\tau\rightarrow \infty}|L_\alpha(t-\tau)u(\tau)|^2 = 0 $.

(5.4)式两端乘以$ A^\frac{1}{2}u_2 $, 有

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}|A^\frac{1}{4}u_2|^2 + 2\mu|A^\frac{3}{4}u_2|^2 + 2\alpha|A^\frac{1}{4}u_2|^2 &\leq& 2|b(u, u, A^\frac{1}{2}u_2)| + 2|f|\cdot|A^\frac{1}{2}u_2|\\ &\leq& C\|u\|_4\cdot|\nabla u|\cdot\|A^\frac{1}{2}u_2\|_4 + C|f|\cdot|A^\frac{3}{4}u_2|\\ &\leq& C|A^\frac{1}{4}u|\cdot|A^\frac{1}{2}u|\cdot|A^\frac{3}{4}u_2| + C|f|\cdot|A^\frac{3}{4}u_2|\\ &\leq& 2\mu|A^\frac{3}{4}u_2|^2 + \frac{C}{\mu}|A^\frac{1}{4}u|^2|A^\frac{1}{2}u|^2 + \frac{C}{\mu}|f|^2, \end{eqnarray*} $

i.e.

$ \frac{\rm d}{{\rm d}t}|A^\frac{1}{4}u_2|^2 + \alpha|A^\frac{1}{4}u_2|^2 \leq \frac{C}{\mu}|A^\frac{1}{4}u|^2|A^\frac{1}{2}u|^2 + \frac{C}{\mu}|f|^2, $

利用Gronwall不等式及定理4.2可得

$ \begin{eqnarray*} |A^\frac{1}{4}u_2(t)| &\leq& e^{-\alpha(t-\tau)}|A^\frac{1}{4}u_2(\tau)|^2 + \frac{C}{\mu}\int_\tau^t(|A^\frac{1}{4}u|^2|A^\frac{1}{2}u|^2 + |f|^2){\rm d}s\\ &\leq& \frac{C_{t, \tau}}{\mu}\int_\tau^t|A^\frac{1}{2}u|^2{\rm d}s + \frac{C}{\mu}\int_\tau^t|f|^2{\rm d}s< +\infty. \end{eqnarray*} $

因为$ D(A^{1/4}) $紧嵌入到$ H $中, $ \{Q_{\alpha}(t, \tau)\} $在$ H $是紧的, 由定理2.2可知过程$ \{U_{\alpha}(t, \tau)\} $拉回渐近紧. 根据定理2.1, 可得证系统(1.1)具有拉回吸引子$ {\cal A}_\alpha(t) = \{A_\alpha(t)\}_{t\in {{\Bbb R}} } $. 证毕.

相应地, 也可证明由系统(1.2)生成的过程$ \{U_{0}(t, \tau)\} $具有拉回吸引子$ {\cal A}_0(t) = \{A_0(t)\}_{t\in {{\Bbb R}} } $及拉回吸收集$ {\cal D}_0(t) = \{D_0(t)\}_{t\in {{\Bbb R}} } $, 其中

$ D_\alpha(t) \subset D_0(t), \ \forall\ t\in {{\Bbb R}} . $

下面, 我们讨论拉回吸引子$ {\cal A}_\alpha(t) $的收敛性.

引理6.1  已知$ t\in {{\Bbb R}} $, $ \tau \in {{\Bbb R}} ^+ $, $ 0\leq\alpha_n < \frac{1}{2}\mu\lambda_1 $, $ \alpha_n\rightarrow 0\ (n\rightarrow \infty) $, 且对任意的$ \mathring{u_{n, t-\tau}} $, $ \mathring{u_{t-\tau}} $满足$ \mathring{u_{n, t-\tau}}\rightarrow \mathring{u_{t-\tau}} $, 则

$ \lim\limits_{n\rightarrow \infty}U_{\alpha_n}(t, t-\tau)\mathring{u_{n, t-\tau}} = U_0(t, t-\tau)\mathring{u_{t-\tau}}. $

证  不妨假设$ U_{\alpha_n}(t, t-\tau)\mathring{u_{n, t-\tau}} = u^{\alpha_n}(t) $和$ U_0(t, t-\tau)\mathring{u_{t-\tau}} = u(t) $分别满足

(6.1) $ \begin{equation} \frac{\partial}{\partial t}u^{\alpha_n}(t) + \nu Au^{\alpha_n}(t) + B(u^{\alpha_n}) + \alpha_nu^{\alpha_n} = P_Lf \end{equation} $

和

(6.2) $ \begin{equation} \frac{\partial}{\partial t}u + \nu Au + B(u) = P_Lf. \end{equation} $

令$ z^n(t) = u^{\alpha_n}(t) - u(t) $, 则有

(6.3) $ \begin{equation} \frac{\partial}{\partial t}z^n + \nu Az^n + B(u^{\alpha_n}) - B(u) + \alpha_nu^{\alpha_n} = 0. \end{equation} $

(6.3)式两端乘以$ z^n $, 有

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}|z^n|^2 + 2\nu\|z^n\|^2 &\leq& 2|b(z^n, u, z^n)| + 2\alpha_n|(u^{\alpha_n}, z^n)|\\ &\leq& C\|u\|\cdot|z^n|^2 + 2\alpha_n|u^{\alpha_n}|\cdot|z^n|\\ &\leq& 2\nu\lambda_1|z^n|^2 + C\|u\|^2|z^n|^2 + C\alpha_n^2|u^{\alpha_n}|^2, \end{eqnarray*} $

进而

$ \frac{\rm d}{{\rm d}t}|z^n|^2 \leq C\|u\|^2|z^n|^2 + C\alpha_n^2|u^{\alpha_n}|^2, $

由Gronwall不等式可知

$ |z^n(t)|^2 \leq |\mathring{u_{n, t-\tau}}-\mathring{u_{t-\tau}}|^2e^{C\int_{t-\tau}^t\|u\|^2{\rm d}s}+ C\alpha_n^2\int_{t-\tau}^t|u^{\alpha_n}|^2 e^{C\int_s^t\|u\|^2{\rm d}k}{\rm d}s, $

即

$ \lim\limits_{n\rightarrow \infty}U_{\alpha_n}(t, t-\tau)\mathring{u_{n, t-\tau}} = U_0(t, t-\tau)\mathring{u_{t-\tau}}. $

引理6.1得证.

引理6.2  对任意的$ t\in {{\Bbb R}} $及$ \tau\in {{\Bbb R}} ^+ $, 有常数$ \theta\in(0, 1) $, 使得存在$ T=T(t) $满足

$ U_0(t-\theta t, t-\tau)\bigg(\overline{\mathop\bigcup\limits_{0\leq\alpha\leq\frac{1}{2}\mu\lambda_1}A_\alpha(t-\tau)}^H\bigg) \subset D_0(t-\theta\tau), \;\;\forall\;\;\tau>T. $

证  根据拉回吸引子$ \mathcal {A}_\alpha(t) = \{A_{\alpha}(t)\}_{t\in {{\Bbb R}} } $的结构可得

$ A_\alpha(t) = \mathop\bigcap\limits_{s\geq0}\overline{\mathop\bigcup\limits_{\tau\geq s}U_\alpha(t, t-\tau)D_\alpha(t-\tau)}^H, $

且由定理5.1可知

$ \overline{\mathop\bigcup\limits_{0\leq\alpha\leq\frac{1}{2}\mu\lambda_1}A_\alpha(t)}^H \subset \overline{\mathop\bigcup\limits_{0\leq\alpha\leq\frac{1}{2}\mu\lambda_1}D_\alpha(t)}^H \subset D_0(t). $

在定理5.1中已证得对任意的$ t\in {{\Bbb R}} $, 总存在常数$ T>0 $使得

$ U_\alpha(t, t-\tau)D_\alpha(t-\tau) \subset D_\alpha(t), \;\;\;\;\forall \;\;\tau>T, $

进而

$ U_0(t-\theta\tau, t-\tau)D_0(t-\tau) \subset D_0(t-\theta\tau), \;\;\;\;\forall \theta\in(0, 1), $

即, 当$ \tau>T $时, 对任意的$ \theta\in(0, 1) $, 成立

$ U_0(t-\theta\tau, t-\tau)\overline{\mathop\bigcup\limits_{0\leq\alpha\leq\frac{1}{2}\mu\lambda_1}A_\alpha(t-\tau)}^H \subset D_0(t-\theta\tau), $

引理6.2结论得证.

引理6.3  对任意的$ t\in {{\Bbb R}} $, $ \mathop\bigcup\limits_{0\leq\alpha\leq\frac{1}{2}\mu\lambda_1}A_\alpha(t) $在$ H $中预紧.

证  对任意的$ t\in {{\Bbb R}} $, 固定$ \tau\in {{\Bbb R}} ^+ $, $ \{\alpha_n\}_{n\in {\mathbb N}}\subset[0, \frac{1}{2}\mu\lambda_1] $. 不妨设$ u_n(t) $是对应于初值的解, 且

$ u_n(t-\tau) \subset D_{\alpha_n}(t-\tau), \;\;\;(n=1, 2, \cdots), $

则$ u_m $和$ u_n $分别满足

(6.4) $ \begin{equation} \frac{\partial}{\partial t}u_m + \mu Au_m + B(u_m) + \alpha_mu_m = P_Lf \end{equation} $

和

(6.5) $ \begin{equation} \frac{\partial}{\partial t}u_n + \mu Au_n + B(u_n) + \alpha_nu_n = P_Lf, \end{equation} $

则有

$ \frac{\partial}{\partial t}(u_m-u_n) + \mu A(u_m-u_n) + B(u_m)-B(u_n) + \alpha_mu_m - \alpha_nu_n = 0. $

令$ w=u_m-u_n $, 可得

(6.6) $ \begin{equation} \frac{\partial}{\partial t}w + \mu Aw + B(u_m)-B(u_n) + \alpha_mu_m - \alpha_nu_n = 0, \end{equation} $

(6.6)式两端乘以$ w $, 有

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}|w|^2 + 2\mu\|w\|^2 + 2\alpha_m|w|^2 &\leq& 2|b(w, u_n, w)| + 2|\alpha_m-\alpha_n|\cdot|(u_n, w)|\\ &\leq& C\int_\Omega|w|\cdot|\nabla w|\cdot|u_n|{\rm d}x + 2|\alpha_m-\alpha_n|\cdot|(u_n, w)|\\ &\leq& C\|w\|_4\cdot|\nabla w|\cdot\|u_n\|_4 + 2|\alpha_m-\alpha_n|\cdot|u_n|\cdot|w|\\ &\leq& C|A^\frac{1}{4}w|\cdot|\nabla w|\cdot|A^\frac{1}{4}u_n| + 2|\alpha_m-\alpha_n|\cdot|u_n|\cdot|w|\\ &\leq& \mu\|w\|^2 + \frac{C}{\mu}|A^\frac{1}{4}u_n|^2|A^\frac{1}{4}w|^2 + \frac{(\alpha_m-\alpha_n)^2}{\mu\lambda_1}|u_n|^2, \end{eqnarray*} $

i.e.

$ \frac{\rm d}{{\rm d}t}|w|^2 + \mu\|w\|^2 \leq \frac{C}{\mu}|A^\frac{1}{4}u_n|^2|A^\frac{1}{4}w|^2 + \frac{(\alpha_m-\alpha_n)^2}{\mu\lambda_1}|u_n|^2. $

另外

$ \frac{\rm d}{{\rm d}t}(e^{\sigma t}|w|^2) \leq \frac{C}{\mu}e^{\sigma t}|A^\frac{1}{4}u_n|^2|A^\frac{1}{4}w|^2 + \frac{(\alpha_m-\alpha_n)^2}{\mu\lambda_1}e^{\sigma t}|u_n|^2, $

上式两端在$ [\tau, t] $上积分, 利用定理4.2, 成立

$ e^{\sigma t}|w(t)|^2 \leq e^{\sigma \tau}|w(\tau)|^2 + \frac{C_{t, \tau}}{\mu}e^{\sigma t}\int_\tau^t|A^\frac{1}{4}w|^2{\rm d}s + \frac{(\alpha_m-\alpha_n)^2}{\mu\lambda_1}e^{\sigma t}\int_\tau^t|u_n(s)|^2{\rm d}s, $

i.e.

$ \begin{eqnarray*} |u_m(t)-u_n(t)|^2 &\leq& e^{-\sigma t}e^{\sigma \tau}|u_m(\tau)-u_n(\tau)|^2\\ &&+ \frac{1}{\mu}C_{t, \tau}\int_\tau^t\|w\|^2{\rm d}s + \frac{(\alpha_m-\alpha_n)^2}{\mu\lambda_1}\int_\tau^t|u_n(s)|^2{\rm d}s. \end{eqnarray*} $

由定理4.1可知

$ \begin{eqnarray*} |u_m(t)-u_n(t)|^2 &\leq& e^{-\sigma t}e^{\sigma \tau}|u_m(\tau)-u_n(\tau)|^2\\ && + C_{t, \tau}\int_\tau^t\|u_m(s)-u_n(s)\|^2 {\rm d}s+ C_{t, \tau}(\alpha_m-\alpha_n)^2. \end{eqnarray*} $

根据定理4.1结论, 有

$ u_m(\cdot)\rightharpoonup u(\cdot)\ \mbox{在} L^2(\tau, t;V) \mbox{中弱收敛, } $

及

$ \lim\limits_{m\rightarrow \infty}\lim\limits_{n\rightarrow \infty}\int_\tau^t\|u_m(s)-u_n(s)\|^2{\rm d}s = 0. $

因为序列$ \{\alpha_m\}_{m\in {\mathbb N}}\subset[0, \frac{1}{2}\mu\lambda_1] $有界, 则存在Cauchy序列满足

$ \lim\limits_{m\rightarrow \infty}\lim\limits_{n\rightarrow \infty}C_{t, \tau}(\alpha_m-\alpha_n)^2 = 0. $

记

$ \Phi_\tau(\alpha_m, \alpha_n, u_m(\tau), u_n(\tau)) = C_{t, \tau}\int_\tau^t\|u_m(s)-u_n(s)\|^2{\rm d}s + C_{t, \tau}(\alpha_m-\alpha_n)^2, $

则对任意的$ t\in {{\Bbb R}} $和$ \tau>0 $, $ \Phi_\tau(\cdot, \cdot, \cdot, \cdot) $为定义在

$ [0, \frac{1}{2}\mu\lambda_1]\times[0, \frac{1}{2}\mu\lambda_1]\times\mathop\bigcup\limits_{\alpha\in[0, \frac{1}{2}\mu\lambda_1]}D_\alpha(\tau)\times\mathop\bigcup\limits_{\alpha\in[0, \frac{1}{2}\mu\lambda_1]}D_\alpha(\tau) $

上的收缩函数, 引理6.3结论得证.

基于引理6.1, 引理6.2和引理6.3, 上半连续的充分条件($ C_1 $), ($ C_2 $)和($ C_3 $)得以验证, 以下结论自然成立.

定理6.1  已知$ u(\tau)\in H $, 对任意的$ t\in {{\Bbb R}} $, 拉回吸引子$ {\cal A}_\alpha(t) = \{A_\alpha(t)\}_{t\in {{\Bbb R}} } $具有上半连续性, 即

$ \lim\limits_{\alpha\rightarrow 0} \mbox{dist}(A_\alpha(t), A_0(t)) = 0. $