数学物理学报  2016, Vol. 36 Issue (1): 36-48   PDF (384 KB)    
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马巧珍
徐玲
张彦军
记忆型非经典反应扩散方程在R3中解的渐近性态
马巧珍 , 徐玲, 张彦军    
西北师范大学数学与统计学院 兰州 730070
摘要: 该文证明了非经典反应扩散方程$$u_{t}-\Delta{u_t}-\Delta{u}-\int^{\infty}_{0}k(s)\triangle u(t-s){\rm d}s+f(x,u)=g(x)$$在${\Bbb R}^{3}$ 上的全局吸引子的存在性,其中非线性项f(x,u)满足临界条件.
关键词: 非经典反应扩散方程     衰退记忆     临界非线性项     无界区域     全局吸引子    
Asymptotic Behavior of the Solution for the Nonclassical Diffusion Equations with Fading Memory on the Whole Space R3
Ma Qiaozhen , Xu Ling, Zhang Yanjun    
College of Mathematics and Statistics, Northwest Normal University, Lanzhou 730070
Abstract: We prove the existence of global attractors for the nonclassical diffusion equations with fading memory $$u_{t}-\Delta{u_t}-\Delta{u}-\int^{\infty}_{0}k(s)\triangle u(t-s){\rm d}s+f(x,u)=g(x)$$ defined on the entire space ${\Bbb R}^{3}$ , where nonlinearity f(x, u) is critical.
Key words: Nonclassical diffusion equation     Fading memory     Critical nonlinearity     Unbounded domain     Global attractor    
1 引言

我们讨论下面的非经典反应扩散方程 \begin{equation} \left\{\begin{array}{ll} u_{t}-\Delta{u_t}-\Delta{u}-\int^{\infty}_{0}k(s)\triangle u(t-s){\rm d}s+f(x,u)=g(x),\quad t\geq0,\\[2mm] u(0)=u_{0},x\in{\Bbb R}^{3} \end{array}\right.\tag{1.1} \end{equation} 在 ${\Bbb R}^{3}$ 上解的长时间行为,其中$g\in H^{-1}({\Bbb R}^{3})$.非线性函数 $f\in C^2({\Bbb R}^{4},{\Bbb R})$ 是局部有界可测函数,并且存在 $r_{0}>0$ 和正常数 $c_{i} (i=1,2,3)$ 使得 \begin{equation} f(\cdot,0)\in L^{2}({\Bbb R}^{3}); \tag{1.2} \end{equation} \begin{equation} \left|f'(x,0)\right|\leq c_{1},~~ \forall x\in {\Bbb R}^{3}; \tag{1.3} \end{equation} \begin{equation} \left|f"(x,s)\right|\leq c_{2}(1+|s|^{3}),~~ \forall s\in{\Bbb R},x\in {\Bbb R}^{3}; \tag{1.4} \end{equation} \begin{equation} \liminf\limits_{|s|\rightarrow \infty}\frac{f(x,s)}{s}\geq 0,~~ \forall |x|\leq r_{0}; \tag{1.5} \end{equation} \begin{equation} (f(x,s)-f(x,0))s\geq c_{3}s^{2},~~ \forall s\in{\Bbb R},|x|>r_{0} \tag{1.6} \end{equation} 成立, 其中 $f'$ 表示 $f(\cdot,\cdot)$ 关于第二个变量的偏导数.

方程 (1.1) 中衰退记忆的影响是通过线性时间卷积函数 $\triangle u(\cdot)$ 和记忆核 $k(\cdot)$ 来体现的. 这里 $k(\cdot)\in {\cal C}^{2}({\Bbb R}^{+}),k(s)\geq0,k'(s)\leq0, \forall s\in{\Bbb R}^{+}$. 设 $\mu(s)=-k'(s)$ 满足下面的条件 \begin{equation} \mu\in{\cal C}^{1}({\Bbb R}^{+})\bigcap L^{1}({\Bbb R}^{+}),\mu(s)\geq0, \mu'(s)\leq0,~~ \forall s\in{\Bbb R}^{+}; \tag{1.7} \end{equation} \begin{equation} \mu'(s)+\delta\mu(s)\leq0,~~ \forall s\in{\Bbb R}^{+},\delta>0.\tag{1.8} \end{equation} 问题 (1.1) 在文献 [15] 中首次被提出并研究. 文献 [16] 中,作者在外力项满足平移有界而非平移紧,非线性项满足临界指数增长时讨论了非自治非经典反应扩散方程一致吸引子的存在性. 孙春友等人在文献 [7] 中研究了类似问题在有界域上解的渐近行为.

如果不考虑记忆核的影响,问题 (1.1)$_1$ 即为非经典反应扩散方程, 该方程在流体力学和热传导领域有着广泛的应用,见文献 []. 作为无穷维动力系统中最基本的模型之一,在有界域[4, 8, 9, 10, 11, 12, 13]和无界域[10, 14, 20] 情形下这类问题解的长时间行为都已被广泛研究. 然而,到目前为止还未曾见针对带衰退记忆的非经典反应扩散方程 (1.1) 在无界域上全局吸引子的存在性的任何结论. 本文将文献 [] 在有界域上的结果推广到无界域的情形.

众所周知,无界区域上 Sobolev 嵌入不再是紧的; Poincaré 不等式也不再适用, 这些问题都使得方程 (1.1) 在无界区域上解的长时间行为的研究显得更为有趣和复杂. 为了克服上述困难,我们使用了一些典型的方法,参见文献 [10, ]. 例如, 作者在文献 [] 中使用了截断函数和 Temam[19]的思想研究了无界域上波方程和带衰退记忆的波方程的解的长时间行为. 王碧祥在文献 [18] 中提出尾部估计的方法, 并将其理论应用到了反应扩散方程上.

本文主要证明问题 (1.1) 对应的解半群在 $H^{1}({\Bbb R}^{3})\times L^{2}_{\mu} ({\Bbb R}^{+},H^{1}({\Bbb R}^{3}))$ 中全局吸引子的存在性. 为此,我们把问题分解到有界域和无界域上分别讨论: 对无界域,通过对解的球外估计,证明在充分大的球外解任意小; 在有界域,利用算子半群分解技巧证明解具有渐近光滑性质. 另外,利用平移紧定理得到了记忆项的紧性,从而获得了全局吸引子的存在性.

本文的主要结果是

定理 1.1 假设 $f$ 满足(1.2)--(1.8)式,$g\in L^{2}({\Bbb R}^{3})$. 则问题 (1.1) 对应的解半群$S(t)$在 $H^{1}({\Bbb R}^{3})\times L^{2}_{\mu} ({\Bbb R}^{+},H^{1}({\Bbb R}^{3}))$ 中存在全局吸引子${\cal A}$.

全文安排如下: 第二节是文章的预备工作,回顾了一些将要用到的概念和结论; 在第三节, 我们证明了问题 (1.1) 相应解半群的渐近紧性和本文的主要结论.

2 预备工作

类似文献[5, ],引入变量 $$ \eta^{t}(x,s)=\int^{s}_{0}u(x,t-r){\rm d}r,~~ s\geq0, $$ $$ \partial_{t}\eta^{t}(x,s)=u(x,t)-\partial_{s}\eta^{t}(x,s),~~ s\geq0.$$ 令 $\mu(s)=-k'(s)$ 且 $k(\infty)=0$,则 (1.1)式转化为 \begin{equation} \left\{\begin{array}{ll} u_{t}-\Delta{u_t}-\Delta{u}-\int^{\infty}_{0}\mu(s)\triangle \eta^{t}(s){\rm d}s+f(x,u)=g(x),\\[2mm] \eta^{t}_{t}=-\eta^{t}_{s}+u.\end{array}\right.\tag{2.1} \end{equation} 相应的初始条件为 \begin{equation} \left\{\begin{array}{ll} u(x,0)=u_{0}(x),& x\in{\Bbb R}^{3},\\\ \eta^{0}(x,s)=\eta_{0}(x,s)=\int_{0}^{s}u(x,-r){\rm d}r, \quad &(x,s)\in{\Bbb R}^{3}\times{\Bbb R}^{+}, \end{array} \right.\tag{2.2} \end{equation} 其中 $u(\cdot)$ 满足以下条件: 存在两个正数 ${\mathfrak R}$ 和 $\varrho\leq\delta$,使得 $$\int_{0}^{\infty}{\rm e}^{-\varrho s}\|\nabla u(-s)\|^{2}{\rm d}s\leq{\mathfrak R}.$$

不失一般性,记$H=L^{2}({\Bbb R}^{3}),V=H^{1}({\Bbb R}^{3})$ 和 $V^{-1}= H^{-1}({\Bbb R}^{3})$. 用 $\langle\cdot,\cdot\rangle$ 和 $\|\cdot\|$ 分别表示 $H$ 中的内积和范数. 对空间 $V$,定义范数为: $\|u\|^2_{V}=\|u\|^2+\|\nabla u\|^2$.除此以外,用 $C$ 表示任意的正常数,每行甚至同一行中的 $C$ 都不同.

根据记忆核 $\mu(\cdot)$ 满足的条件,设 $L^{2}_{\mu}({\Bbb R}^{+},V)$ 为定义于 ${\Bbb R}^{+}$ 上取值于 $V$ 的一族 Hilbert 空间,并且赋予相应的内积和范数为 $$\langle\varphi_{1},\varphi_{2}\rangle_{\mu,V}=\int_{0}^{\infty}\mu(s)\langle\nabla\varphi_{1},\nabla\varphi_{2}\rangle{\rm d}s, ~~ \|\varphi\|_{\mu,V}=\int_{0}^{\infty}\mu(s)\|\nabla \varphi(s)\|^{2}{\rm d}s.$$ 记${\cal M}=V\times L^{2}_{\mu}({\Bbb R}^{+},V),$ 并且赋予范数 $\|z\|_{{\cal M}}=\|(u,\eta^{t})\|_{{\cal M}}= [\frac{1}{2}(\|u\|_{V}+\|\eta^{t}\|_{\mu,V})]^{2}.$

引理 2.1[15, 16]  记$I=[0,T],\forall\;T>0$. 设记忆核$\mu(s)$满足(1.7)和(1.8)式,则对于任意的 $\eta^{t}\in {\cal C}(I;L^{2}_{\mu}({\Bbb R}^{+},V))$,存在常数 $\delta>0$ 使得 \begin{equation} \langle\eta^{t},\eta_{s}^{t}\rangle_{\mu,V} \geq\frac{\delta}{2}\|\eta^{t}\|_{\mu,V}.\tag{2.3} \end{equation}

引理 2.2[5, 15, 16]\quad 假设 (1.2)--(1.6) 式成立. 则对每个 $u>0$,存在 $\rho(u)\geq0$ 使得 \begin{equation} {\mathfrak F}(u)\geq -u\|u\|^{2}-\rho(u),~~ \forall\;u\in V, \tag{2.4} \end{equation} 其中 ${\mathfrak F}(u)=\int_{{\Bbb R}^{3}}\int_{0}^{u(x)}f(x,y){\rm d}y{\rm d}x$.且存在 $\alpha>0$ 和 $\beta\geq0$ 使得 \begin{equation} \langle f(x,u),u\rangle\geq \alpha\|u\|^{2}-\frac{1}{2}\|\nabla u\|^{2}-\beta,~~ \forall \;u\in V. \tag{2.5} \end{equation}

引理 2.3[5, 6]  设$\mu\in{\cal C}^{1}({\Bbb R}^{+})\bigcap L^{1}({\Bbb R}^{+})$ 为非负函数且满足: 若存在 $s_{0}\in{\Bbb R}^{+}$, 使得 $\mu(s_{0})=0$,则 $\mu(s)=0$ 对所有的 $s\geq s_{0}$ 均成立. 此外,设 ${\bf X}_{0},{\bf X}_{1},{\bf X}_{2}$ 均为 Banach 空间,其中 ${\bf X}_{0},{\bf X}_{1}$ 自反并满足: ${\bf X}_{0}\hookrightarrow{\bf X}_{1}\hookrightarrow{\bf X}_{2}$, 其中嵌入 ${\bf X}_{0}\hookrightarrow{\bf X}_{1}$ 是紧的. 设 ${\bf C}\subset L^{2}_{\mu}({\Bbb R}^{+},{\bf X}_{1})$ 满足

(1) ${\bf C}$ 在 $L^{2}_{\mu}({\Bbb R}^{+},{\bf X}_{0})\bigcap H^{1}_{\mu} ({\Bbb R}^{+},{\bf X}_{2})$ 中有界;

(2) $\sup\limits_{\eta\in{\bf C}}\|\eta(s)\|^{2}_{{\bf X}_{1}}\leq h(s), \forall\; s\in{\Bbb R}^{+},h(s)\in L_{\mu}^{1}({\Bbb R}^{+}).$\\ 则 ${\bf C}$ 在 $L^{2}_{\mu}({\Bbb R}^{+},{\bf X}_{1})$ 中相对紧.

3 全局吸引子的存在性

引理 3.1[15, 16]  假设 (1.2)--(1.8) 式成立, 并且 $g\in H^{-1}({\Bbb R}^{3})$. 则对任意的 $T>0$ 和 $z_0=(u_{0},\eta_{0})\in {\cal M}$,问题 (2.1)--(2.2) 存在唯一的弱解 $z(t)=(u(t),\eta^{t}(t))$ 满足: $z(t)\in {\cal C}^1(0,T;{\cal M})\cap L^\infty(0,T;{\cal M}),$ 并且 $u(t)$ 连续依赖于 ${\cal M}$ 中的初值 $z_{0}$.

根据引理3.2,我们可定义问题 (2.1)--(2.2) 在 ${\cal M}$ 中的连续解半群 $\{S(t)\}_{t\geq0}$,即$S(t):{\cal M}\rightarrow{\cal M},$ $$z_0=(u_0,\eta_0)\rightarrow(u(t),\eta^t)=S(t)z_0.$$

引理 3.2  假设 (1.2)--(1.8) 式成立,并且 $g\in H^{-1}({\Bbb R}^{3})$. 则存在仅依赖于 $g$ 和 $\delta$ 的正常数 $R$,以及 ${\cal M}$ 中的任意有界子集 $B$,使得 \begin{equation} \|z(t)\|^{2}_{{\cal M}}=\frac{1}{2}(\|u\|^{2}_{V}+\|\eta^{t}\|^{2}_{\mu,V})\leq R^{2},\forall\; t\geq t_{0}=t_0(B).\tag{3.1} \end{equation}

  用$u$ 与 $(2.1)_1$式在$H$中做内积,用$\eta^{t}$与 $(2.1)_2$式在$L^{2}_{\mu}({\Bbb R}^{+},V)$中做内积,然后作和计算得 \begin{equation} \frac{1}{2}\frac{\rm d}{{\rm d}t}(\|u\|^{2}+\|\nabla u\|^{2}+\|\eta^{t}\|_{\mu,V}^{2})+\|\nabla u\|^{2}+\langle\eta^{t},\eta_{s}^{t}\rangle_{\mu,V}^{2}+\langle f(x,u),u\rangle=\langle g(x),u\rangle, \tag{3.2} \end{equation} 由(2.3) 和 (2.5)式,可得 \begin{equation} \frac{1}{2}\frac{\rm d}{{\rm d}t}(\|u\|^{2}+\|\nabla u\|^{2}+\|\eta^{t}\|_{\mu,V}^{2})+\frac{\alpha_{1}}{2}(\|u\|^{2}+\|\nabla u\|^{2}+\|\eta^{t}\|_{\mu,V}^{2})\leq C(1+\|g\|^{2}), \tag{3.3} \end{equation} 其中 $\alpha_{1}=\frac{1}{2}\min\{4\alpha,1,2\delta\}$.利用 Gronwall 引理,我们有 \begin{equation} \|z(t)\|_{{\cal M}}^{2}\leq\|z(0)\|^{2}_{{\cal M}}{\rm e}^{-\alpha_{1} t}+\frac{C}{\alpha_{1}}(1+\|g\|^{2}_{V^{-1}}).\tag{3.4} \end{equation} 因此,设 $\|z_0\|_{{\cal M}}\leq R_0$,取 $t_0=\frac{1}{\alpha_1}\ln\frac{\alpha_1R_0^2}{C(1+\|g\|^{2}_{V^{-1}})}$,则有 $$\|z(t)\|_{{\cal M}}^{2}\leq \frac{2C}{\alpha_1}(1+\|g\|^{2}_{V^{-1}}),~~ \forall t\geq t_0.$$ 令 $R^{2}=\frac{2C}{\alpha_1}(1+\|g\|^{2}_{V^{-1}})$,从而 $\|z(t)\|_{{\cal M}}^{2}\leq R^{2},\forall t\geq t_0.$

由引理 3.2 可知,存在吸收集 $B_R\subset{\cal M}$,使得对任意的有界子集 $B\subset{\cal M}$,有 $S(t)B\subset B_R,\forall t\geq t_0.$ 令 ${\Bbb B}_{0}=\bigcup\limits_{t\geq0}S(t)B_{R_{0}}$, 其中 $B_{R_{0}}$ 表示 $V$ 中原点为圆心,半径为 $R_{0}$ 的球, 则 ${\Bbb B}_{0}$ 是 $V$ 中正不变,有界的吸收集,即 $S(t){\Bbb B}_{0}\subset{\Bbb B}_{0}$. 对 $\forall\; t\geq0$,有界集 $B\subset V$,存在 $t_{0}=t_{0}(B)\geq0$,使得 $S(t)B\subset{\Bbb B}_{0}$,$\forall\;t\geq t_{0}$.

使用文献 [, 10, 18] 中的技巧,对任意的 $r\geq r_{0}$, 引入两个光滑函数 $\varphi^{r}_{i}(\cdot): {\Bbb R}^{3}\rightarrow {\Bbb R}^{+},$ $i=1,2$,使得 $$\varphi^{r}_{1}(x)+\varphi^{r}_{2}(x)=1,\quad \forall x\in{\Bbb R}^{3},$$ $$\varphi^{r}_{1}(x)=0,~~ \forall\;|x|\leq r, \varphi^{r}_{2}(x)=0,~~\forall\;|x|\geq r+1.$$

因为 $f(x,s)$ 是连续的,所以存在 $u_{r}>0$ 使得当 $|s|\leq1,|z|\leq u_{r}$,$|x|\leq r_{0}+1$ 时,有 \begin{equation} |f(x,s)||z|\leq \left(\frac{3}{8\pi(r_0+1)^3}\right)\frac{1}{r} \tag{3.5} \end{equation} 成立,这里不妨设 $u_{r}\leq1$.此外,由 (1.5) 式可得,存在 $M_{r}>0$ 使得当 $|s|\geq u_{r}$ 时, 对每一 $|x|\leq r_{0}$,都有 \begin{equation} \frac{f(x,s)}{s}\geq-M_{r}.\tag{3.6} \end{equation} 因此,我们将 $-f(x,s)+g(x)$ 分解为 $$-f(x,s)+g(x)=-f_{1}(x,s)-f_{2}(x,s)+g_{1}(x)+g_{2}(x), $$ 其中 $$f_{1}(x,s)=(f(x,s)-f(x,0))\varphi^{r}_{1}(x)+(c_{3}s+f(x,s)-f(x,0)+M_{r}s)\varphi^{r}_{2}(x),$$ $$f_{2}(x,s)=-(c_{3}s+M_{r}s)\varphi^{r}_{2}(x), g_{i}(x)=(-f(x,0)+g(x))\varphi^{r}_{i}(x),i=1,2.$$ 显然,$f_{1}(\cdot,0)\equiv0$ 且 $f$ 也满足 (1.3)--(1.5) 式 (常数可以被重新定义),并且对任意的 $s\in{\Bbb R}$ 和几乎处处的 $x\in{\Bbb R}^{3}$,有 \begin{equation}f_{1}(x,s)s\geq c_{4}s^{2},\tag{3.7} \end{equation} 其中 $c_{4}>0$. 此外, $$ f_{2}(x,s)=0,~~ g_{2}(x)=0,~~ \forall s\in {\Bbb R}^{3},~~ |x|\geq r+1,$$ $$\|g_1\|_{V^{-1}},~~ \|g_{1}\|\rightarrow 0,~~ r\rightarrow\infty.$$ 所以,我们分解问题 (2.1)--(2.2) 的解 $S(t)z_{0}=z(t)=(u(t),\eta^{t})$ 为 $$z^{r}(t)=z^{r}_{1}(t)+z^{r}_{2}(t),u=v^{r}+w^{r},\eta^{t}=\zeta^{tr}+\xi^{tr},$$ 其中 $z_{1}^{r}(t)=(v^{r}(t),\zeta^{tr})$ 和 $z^{r}_{2}=(w^{r}(t),\xi^{tr})$ 满足下面方程 \begin{equation} \left\{\begin{array}{ll} v^{r}_{t}-\Delta{v^{r}_t}-\Delta{v^{r}}-\int^{\infty}_{0}\mu(s)\triangle\zeta^{tr}(s){\rm d}s+f_{1}(x,v^{r})=g_{1}(x),\\ [2mm] \zeta_{t}^{tr}=-\zeta_{s}^{tr}+v^{r},\\ v^{r}(x,t)\mid_{t=0}=u_{0}(x),\\\ \zeta^{0}=\zeta^{0r}(x,s)=\eta_{0}(x,s)=\int^{s}_{0}u(x,-r){\rm d}r \end{array} \right.\tag{3.8} \end{equation} 和 \begin{equation} \left\{\begin{array}{ll} w^{r}_{t}-\Delta{w^{r}_t}-\Delta{w^{r}}-\int^{\infty}_{0}\mu(s)\triangle\xi^{tr}(s){\rm d}s+f_{1}(x,u)-f_{1}(x,v^{r})+f_{2}(x,u)=g_{2}(x),\\ [2mm] \xi_{t}^{tr}=-\xi_{s}^{tr}+w^{r},\\ w^{r}(x,t)\mid_{t=0}=0,\\ \xi^{0}=\xi^{0r}(x,s)=0.\end{array} \right.\tag{3.9} \end{equation} 利用与引理 3.1 类似的方法可证明方程 (3.8) 和 (3.9) 解的存在唯一性. 为了方便,我们记方程(3.8) 和 (3.9) 的解算子分别为 $\{S_{1}(t)\}_{t\geq0}$ 和 $\{S_{2}(t)\}_{t\geq0}$,则对于初值 $z_{0}\in{\cal M}$,有 $$z(t)=S(t)z_{0}=S_{1}(t)z_{0}+S_{2}(t)z_{0},\forall\;t\geq0.$$

引理 3.3  假设 (1.2)--(1.8)式成立,且 $g\in H^{-1}({\Bbb R}^{3})$. 对于任意的 $\varepsilon>0$,存在 $t_{\varepsilon}\geq1$ (依赖于 $r$) 和 $r_{\varepsilon}\geq r_{0}$,使得问题 (3.8) 的解 $z_1^{r}$ 满足: 对任意的 $z_{0}\in{\Bbb B}_{0}$ 和 $r\geq r_{\varepsilon}$,成立 $$\|z^{r}_{1}(t_{\varepsilon})\|_{{\cal M}}\leq \frac{\varepsilon}{2}.$$

  用$v^{r}$与$(3.8)_1$式在$H$中做内积,用$\zeta^{tr}$与$(3.8)_2$式的在$L^{2}_{\mu}({\Bbb R}^{+},V)$中做内积,计算后得 \begin{equation} ~~\frac{1}{2}\frac{\rm d}{{\rm d}t}(\|v^{r}\|^{2}+\|\nabla v^{r}\|^{2}+\|\zeta^{tr}\|_{\mu,V}^{2})+\|\nabla v^{r}\|^{2}+\langle\zeta^{tr},\zeta^{tr}_{s}\rangle_{\mu,V}^{2}+\langle f_{1}(x,v^{r}),v^{r}\rangle=\langle g_{1}(x),v^{r}\rangle, \tag{3.10} \end{equation} 因为 $f_{1}(\cdot,0)\equiv0$ 且 $f_{1}$ 也满足 (1.3)--(1.5)式,则由 (3.7) 式,可知 \begin{eqnarray} \langle f_{1}(x,v^{r}),v^{r}\rangle\geq c_{4}\int_{{\Bbb R}^{3}}|v^{r}|^{2}{\rm d}x\geq c_{4} \int_{|x|>r_{0}}|v^{r}|^{2}{\rm d}x\geq\alpha\|v^{r}\|^{2}-\frac{1}{4}\|\nabla v^{r}\|^{2}.\tag{3.11}\end{eqnarray} 应用引理 2.1,将 (3.11) 式代入 (3.10)式,有 \begin{equation} \frac{1}{2}\frac{\rm d}{{\rm d}t}(\|v^{r}\|^{2}+\|\nabla v^{r}\|^{2}+\|\zeta^{tr}\|_{\mu,V}^{2})+\frac{\alpha_{2}}{2}(\|v^{r}\|^{2}+\|\nabla v^{r}\|^{2}+\|\zeta^{tr}\|_{\mu,V}^{2})\leq\|g_1\|^{2}_{V^{-1}}, \tag{3.12} \end{equation} 其中 $\alpha_{2}=\min\{2\alpha,1,\delta\}$.应用 Gronwall 引理,可得 \begin{equation} \|z^{r}_{1}(t)\|_{{\cal M}}^{2}\leq C_{0}\|z^{r}_{1}(0)\|^{2}_{{\cal M}}{\rm e}^{-\alpha_{2} t}+\frac{C}{\alpha_{2}}\|g_{1}\|^{2}.\tag{3.13} \end{equation}

设 $\|z^{r}_{1}(0)\|^{2}_{{\cal M}}=\|z_{0}\|^{2}_{{\cal M}}\leq R_{0}$, 并且注意到当 $r\rightarrow\infty$ 时,$\|g_1\|^{2}_{V^{-1}}\rightarrow 0$, 所以对任意的 $\varepsilon>0$,存在 $t_{\varepsilon}\geq1$ (依赖于 $r$) 和 $r_{\varepsilon}\geq r_{0}$,当 $r\geq r_{\varepsilon}$ 时,有 $\|z^{r}_{1}(t_{\varepsilon})\|_{{\cal M}}\leq \frac{\varepsilon}{2}$.

为了得到动力系统 $S(t)$ 必要的紧性,我们作如下定义: 设 $B_{r}=\{x\in{\Bbb R}^{3}:|x|\leq r\}\subset{\Bbb R}^{3}$ 是具有光滑边界的有界区域. 定义 $L^{2}(B_{r})$ 上的算子 $A$ 为: $A=-\Delta,$ 定义域为 $D(A)=H^{2}(B_{r})\cap H_{0}^{1}(B_{r}).$ 考虑一族 Hilbert 空间 $D(A^{s/2}),s\in{\Bbb R}$,它们的内积和范数分别为 $$\langle\cdot,\cdot\rangle_{D(A^{s/2})}=\langle A^{s/2}\cdot,A^{s/2}\cdot\rangle, \|\cdot\|_{D(A^{s/2})}=\|A^{s/2}\cdot\|.$$ 特别地,$\langle\cdot,\cdot\rangle_{L^{2}(B_{r})}$ 和 $\|\cdot\|_{L^{2}(B_{r})}$ 分别表示 $L^{2}(B_{r})$ 空间的内积和范数. 并且有 $$D(A^{s/2})\hookrightarrow D(A^{r/2}),~~ \forall s>r.$$

为了方便起见,记 ${\cal H}^{s}(B_{r})=D(A^{(1+s)/2}),0\leq s\leq1.$ 特别地, $${\cal H}^{-1}(B_{r})=L^{2}(B_{r}),~~{\cal H}^{0}(B_{r})=V(B_{r}),~~ {\cal H}^{1}(B_{r})=H^{2}(B_{r})\cap H_{0}^{1}(B_{r}).$$ 同理,定义 $L^{2}_{\mu}({\Bbb R}^{+},{\cal H}^{s}(B_{r}))$ 为定义于 ${\Bbb R}^{+}$ 上取值于 ${\cal H}^{s}(B_{r})$ 的一族 Hilbert 空间,并且赋予相应的内积和范数为 $$\langle\varphi_{1},\varphi_{2}\rangle_{\mu,{\cal H}^{s}(B_{r})}= \int_{0}^{\infty}\mu(s)\langle\varphi_{1},\varphi_{2}\rangle_{{\cal H}^{s}(B_{r})}{\rm d}s, \|\varphi\|_{\mu,{\cal H}^{s}(B_{r})}=\int_{0}^{\infty}\mu(s) \|\varphi(s)\|_{{\cal H}^{s}(B_{r})}^{2}{\rm d}s.$$

同时,设 $$H_\mu^1({\Bbb R}^{+},{\cal H}^{s}(B_{r}))=\{\varphi:\varphi(s), \partial_s\varphi(s)\in L_\mu^2({\Bbb R}^{+},{\cal H}^{s}(B_{r}))\}, $$ 其上相应的内积和范数分别为 \begin{eqnarray*} \langle\varphi_{1},\varphi_{2}\rangle_{H_\mu^1({\Bbb R}^{+},{\cal H}^{s}(B_{r}))} &=&\int_{0}^{\infty}\mu(s)\langle\varphi_{1}(s),\varphi_{2}(s)\rangle_{{\cal H}^{s}(B_{r})}{\rm d}s \\ &&+\int_{0}^{\infty}\mu(s)\langle\partial_s\varphi_{1}(s),\partial_s\varphi_{2}(s)\rangle_{{\cal H}^{s}(B_{r})}{\rm d}s, \end{eqnarray*} $$\|\varphi\|_{H_\mu^1({\Bbb R}^{+},{\cal H}^{s}(B_{r}))}^2=\|\varphi\|_{\mu, {\cal H}^{s}(B_{r})}^2+\|\partial_s\varphi\|_{\mu,{\cal H}^{s}(B_{r})}^2.$$

定义一族 Hilbert 空间 $${\cal M}_{s}(B_{r})={\cal H}^{s}(B_{r})\times L^{2}_{\mu}({\Bbb R}^{+}, {\cal H}^{s}(B_{r})),$$ 并且赋予范数 $$\|z\|_{{\cal M}_{s}(B_{r})}=\|(u,\eta^{t})\|_{{\cal M}_{s}(B_{r})} =\Big[\frac{1}{2}(\|u\|_{{\cal H}^{s}(B_{r})}^2+\|\eta^{t}\|_{\mu, {\cal H}^{s}(B_{r})}^2)\Big]^{\frac{1}{2}}.$$ 显然,${\cal M}_{0}(B_{r})={\cal M}(B_{r})$.

引理 3.4  假设 (1.2)--(1.8) 式成立,且 $g\in H^{-1}({\Bbb R}^{3})$. 对任意的 $\varepsilon>0$,存在 $r_{1}$ 和 $r_{2}$,且 $r_{1}\geq r_{\varepsilon}$, 取 $r_{2}=2r_{1}+2$,使得问题 (3.9) 在时刻 $t_{\varepsilon}$ 和相应 $r_{1}$ 的解 $z_{2}(t)=(w^{r_{1}}(t_{\varepsilon}),\xi^{t_{\varepsilon}r_{1}})$ 满足: $\|z^{r_{1}}_{2}(t_{\varepsilon})\|_{{\cal M}(B^{c}_{r_{2}})}\leq \frac{\varepsilon}{2}$, 其中 $z_{0}\in{\Bbb B}_{0}$,$t_{\varepsilon}$ 在引理 3.3 中已经给出, $B^{c}_{r_{2}}={\Bbb R}^{3}\backslash B_{r_{2}}$.

在证明引理 3.4 之前,我们需要以下两个辅助引理.

引理 3.5  在引理 3.4 的条件下,对于引理 3.3 中已给出的 $t_{\varepsilon}\geq1$,存在 $R_{1}=R_{1}(R_{0},t_{\varepsilon})$,使得当 $\|z_{0}\|_{{\cal M}}\leq R_{0}$ 时,问题 (2.1)--(2.2) 的解 $z(t)=S(t)z_{0}$ 满足 \begin{equation} \int_{0}^{t_{\varepsilon}}(\|\nabla u(s)\|^{2}+\|u(s)\|^{2}+\|\eta^{t}(s)\|_{\mu,V}){\rm d}s\leq R_{1}.\tag{3.14} \end{equation}

  对 (3.3)式从 $[0,t_{\varepsilon}]$ 积分即可得证.\rule{0.8mm}{3.5mm} 这样,固定 $\varepsilon>0$,在以上引理中选取合适的 $r\geq r_{\varepsilon}>r_{0}$ 和 $t_{\varepsilon}>0$.

推论 3.6 \begin{equation} \sup\limits_{t\in[0,t_{\varepsilon}]}\sup\limits_{z_{0}\in{\Bbb B}_{0}} \{\|u\|_{V},\|v^{r}\|_{V},\|w^{r}\|_{V},\|\eta^{t}\|_{\mu,V}, \|\zeta^{t}\|_{\mu,V},\|\xi^{t}\|_{\mu,V}\} < \infty.\tag{3.15} \end{equation}

引理 3.7  在引理 3.4 的条件下,给定任意 $R\geq0$,则存在 $\Lambda_{0}=\Lambda_{0}(R_{0},R_{1},t_{\varepsilon})$,使得当 $\|z_{0}\|_{{\cal M}}\leq R_{0}$时,问题 (2.1)--(2.2) 的解 $z(t)=S(t)z_{0}$ 满足 \begin{equation} \int_{0}^{t_{\varepsilon}}\|u_{t}(y)\|_{V}^{2}{\rm d}y=\int_{0}^{t_{\varepsilon}}(\|\nabla u_{t}(y)\|^{2}+\|u_{t}(y)\|^{2}){\rm d}y\leq \Lambda_{0}.\tag{3.16} \end{equation}

  用$u_{t}$与$(2.1)_1$式的在$H$中做内积,用$\eta^{t}$与$(2.1)_2$式在 $L^{2}_{\mu}({\Bbb R}^{+},V)$中做内积,计算后得 \begin{eqnarray} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}(\|\nabla u\|^{2}+\|\eta^{t}\|_{\mu,V}^{2}+2{\mathfrak F}(u)-2\langle g,u\rangle)+\|u_{t}\|^{2}+\|\nabla u_{t}\|^{2} onumber\\ &\leq& \langle\triangle\eta^{t},u_{t}\rangle-\langle\eta^{t},\eta_{s}^{t}\rangle_{\mu,V}^{2}+\langle u,\eta^{t}\rangle_{\mu,V}^{2}.\tag{3.17}\end{eqnarray} 而 \begin{eqnarray} |\langle\triangle\eta^{t},u_{t}\rangle|=\bigg|-\int^{\infty}_{0}\mu(s)\langle\nabla\eta^{t}(s), \nabla u_{t}\rangle {\rm d}s\bigg| \leq\frac{k_{0}}{2}\|\eta^{t}\|^{2}_{\mu,V}+\frac{1}{2}\|\nabla u_{t}\|^{2},\tag{3.18} \end{eqnarray} 其中 $k_0=\int^{\infty}_{0}\mu(s){\rm d}s$. 同理,可得 \begin{eqnarray} \langle u,\eta^{t}\rangle_{\mu,V}^{2}\leq\frac{k_{0}}{2}\|\eta^{t}\|^{2}_{\mu,V}+\frac{1}{2}\|\nabla u\|^{2}.\tag{3.19}\end{eqnarray} 结合 (3.17)--(3.19)式及引理 2.1 得 \begin{equation} \frac{\rm d}{{\rm d}t}(\|\nabla u\|^{2}+\|\eta^{t}\|_{\mu,V}^{2}+2 {\mathfrak F}(u)-2\langle g,u\rangle)+2\|u_{t}\|^{2}+\|\nabla u_{t}\|^{2}\leq \|\nabla u\|^{2}+(2k_{0}-\delta)\|\eta^{t}\|_{\mu,V}^{2}, \tag{3.20} \end{equation} 其中 $\delta < 2k_{0}$.在 (2.4) 式中取 $u=\frac{1}{4}$,从而上式可化为 \begin{eqnarray} \|\nabla u\|^{2}+\|\eta^{t}\|_{\mu,V}^{2}+2{\mathfrak F}(u)-2\langle g,u\rangle \geq\|\nabla u\|^{2}+\|\eta^{t}\|_{\mu,V}^{2}-\|u\|^{2}-2\|g\|^{2}-\rho(\frac{1}{4}).\tag{3.21}\end{eqnarray} 利用 (1.4) 式和广义 Gagliardo-Nirenberg-Sobolev 不等式 $\|u\|_{L^{3}({\Bbb R}^{3})}\leq\|u\|^{\frac{1}{2}}\|\nabla u\|^{\frac{1}{2}}$,有 \begin{eqnarray} &&\|\nabla u\|^{2}+\|\eta^{t}\|_{\mu,V}^{2}+2{\mathfrak F}(u)-2\langle g,u\rangle onumber\\ &\leq&\|\nabla u\|^{2}+\|\eta^{t}\|_{\mu,V}^{2}+C(\|u\|^{\frac{3}{2}}\|\nabla u\|^{\frac{3}{2}}+\|\nabla u\|^{6})+|2\langle g,u\rangle| onumber\\ &\leq& C(\|u\|_{V}^{2}+\|u\|_{V}^{6})+\|\eta^{t}\|_{\mu,V}^{2}+\|g\|^2_{V^{-1}}.\tag{3.22}\end{eqnarray} 结合 (3.21) 和 (3.22)式,利用引理 3.5,对 (3.20) 式从 $[0,t_{\varepsilon}]$ 积分可得证该引理.

引理 3.4 的证明  对于给定的 $r_{1}\geq r_{\varepsilon}$, 定义 $\phi(x):{\Bbb R}^{3}\rightarrow [0,1]$ 为 \begin{equation} \phi(x)= \left\{\begin{array}{ll} 0, |x| < r_{1}+1,\\\ \frac{1}{r_{1}+1}(|x|-r_{1}-1), r_{1}+1\leq|x|\leq2r+2,\\\ 1, |x|>2r_{1}+2.\end{array} \right.\tag{3.23} \end{equation} 则 $|\nabla \phi(x)|\leq \frac{1}{r_{1}+1},|\nabla \phi^{2}(x)|\leq \frac{2}{r_{1}+1}\phi(x),\Delta\phi(x)\equiv0.$

用$2\phi^{2}(x)w^{r_{1}}$ 在$L^{2}({\Bbb R}^{3})$中与$(3.9)_1$ 式作内积, 用$2\phi^{2}(x)\xi^{tr_{1}}$ 在$L^{2}({\Bbb R}^{+},V)$中与$(3.9)_2$式作内积,得 \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\int_{{\Bbb R}^{3}}\phi^{2}(x)(|w^{r_{1}}|^{2}+ |\nabla w^{r_{1}}|^{2}){\rm d}x+\frac{\rm d}{{\rm d}t}\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla \xi^{tr_{1}}|^{2}{\rm d}x{\rm d}s onumber\\ &&+2\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla w^{r_{1}}|^{2} +\frac{4}{r_{1}+1}\int_{{\Bbb R}^{3}}\phi(x)w^{r_{1}}(\nabla w_{t}^{r_{1}} +\nabla w^{r_{1}}){\rm d}x onumber\\ &&+\frac{4}{r_{1}+1}\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi(x)w^{r_{1}}\nabla \xi^{tr_{1}}{\rm d}x{\rm d}s +\frac{4}{r_{1}+1}\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi(x)\xi^{tr_{1}}\nabla \xi_{t}^{tr_{1}}{\rm d}x{\rm d}s onumber\\ && +\langle f_{1}(x,u)-f_{1}(x,v^{r_{1}}),2\phi^{2}(x)w^{r_{1}}\rangle onumber\\ &=&\langle -f_{2}(x,u)+g_{2},2\phi^{2}(x)w^{r_{1}}\rangle-\frac{4}{r_{1}+1}\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi(x)\xi^{tr_{1}}\nabla \xi_{s}^{tr_{1}}{\rm d}x{\rm d}s onumber\\ &&-2\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi^{2}(x)\nabla\xi^{tr_{1}}\nabla \xi_{s}^{tr_{1}}{\rm d}x{\rm d}s +\frac{4}{r_{1}+1}\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi(x)\xi^{tr_{1}}\nabla w^{r_{1}} {\rm d}x{\rm d}s. \tag{3.24} \end{eqnarray}

由$(3.9)_2$式可知: $\nabla\xi_{t}^{tr_{1}}=-\nabla\xi_{s}^{tr_{1}}+\nabla w^{r_{1}}$, 所以对 (3.24) 式整理可得 \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\int_{{\Bbb R}^{3}}\phi^{2}(x)(|w^{r_{1}}|^{2} +|\nabla w^{r_{1}}|^{2}){\rm d}x+\frac{\rm d}{{\rm d}t} \int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla \xi^{tr_{1}}|^{2}{\rm d}x{\rm d}s onumber\\ &&+2\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla w^{r_{1}}|^{2} onumber\\ &\leq&-\frac{4}{r_{1}+1}\int_{{\Bbb R}^{3}}\phi(x)w^{r_{1}}(\nabla w_{t}^{r_{1}}+\nabla w^{r_{1}}){\rm d}x-\frac{4}{r_{1}+1}\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi(x)w^{r_{1}}\nabla \xi^{tr_{1}}{\rm d}x{\rm d}s onumber\\ &&-\langle f_{1}(x,u)-f_{1}(x,v^{r_{1}}),2\phi^{2}(x)w^{r_{1}}\rangle+\langle -f_{2}(x,u)+g_{2},2\phi^{2}(x)w^{r_{1}}\rangle onumber\\ &&-\delta\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla\xi^{tr_{1}}|^{2}{\rm d}x{\rm d}s.\tag{3.25}\end{eqnarray}

由引理 3.2,并利用 H\"{o}lder 和 Young 不等式,对 $\forall t\geq \max\{t_{0},t_{\varepsilon}\}$,有 \begin{eqnarray} \left|\frac{4}{r_{1}+1}\int_{{\Bbb R}^{3}}\phi(x)w^{r_{1}}\nabla w_{t}^{r_{1}}{\rm d}x\right| &\leq&\frac{4}{r_{1}+1}\int_{|x|\geq2r_{1}+2}|w^{r_{1}}||\nabla w_{t}^{r_{1}}|{\rm d}x onumber\\ &\leq&\frac{C}{r_{1}+1}+\frac{2}{r_{1}+1}\|\nabla w_{t}^{r_{1}}\|^{2} \tag{3.26} \end{eqnarray} 且 \begin{equation} \left|\frac{4}{r_{1}+1}\int_{{\Bbb R}^{3}}\phi(x)w^{r_{1}}\nabla w^{r_{1}}{\rm d}x\right|\leq\frac{C}{r_{1}+1}; \tag{3.27} \end{equation} \begin{eqnarray} &&\left|-\frac{4}{r_{1}+1}\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi(x)w^{r_{1}}\nabla \xi^{tr_{1}}{\rm d}x{\rm d}s\right| onumber\\ &\leq&\frac{C}{(r_{1}+1)^{2}}+2k_{0}\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla\xi^{tr_{1}}|^{2}{\rm d}x{\rm d}s.\tag{3.28}\end{eqnarray}

另一方面,由广义 Gagliardo-Nirenberg-Sobolev 不等式和引理 3.4,可得 \begin{eqnarray} &&\left|\langle f_{1}(x,u)-f_{1}(x,v^{r_{1}}),2\phi^{2}(x)w^{r_{1}}\rangle\right| onumber\\ &\leq& 2c_{2}\int_{{\Bbb R}^{3}}(1+|u|^{4}+|v^{r_{1}}|^{4})\phi^{2}(x)|w^{r_{1}}|^{2}{\rm d}x onumber\\ &\leq &2c_{2}\left(\int_{{\Bbb R}^{3}}(1+|u|^{4}+|v^{r_{1}}|^{4})^{\frac{3}{2}}{\rm d}x\right)^{\frac{2}{3}}\left(\int_{{\Bbb R}^{3}}\phi^{6}(x)|w^{r_{1}}|^{6}{\rm d}x\right)^{\frac{1}{3}} onumber\\ &\leq &2c_{2}(1+||\nabla u||^{4}+||\nabla v^{r_{1}}||^{4})\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla w^{r_{1}}|^{2}{\rm d}x onumber\\ &\leq &C(1+\frac{\varepsilon}{2})\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla w^{r_{1}}|^{2}{\rm d}x,\forall t\geq \max\{t_{0},t_{\varepsilon}\}.\tag{3.29}\end{eqnarray} 再利用 $f_{2}$ 和 $g_{2}$ 的性质,可知 \begin{equation} \langle -f_{2}(x,u)+g_{2},2\phi^{2}(x)w^{r_{1}}\rangle=0.\tag{3.30} \end{equation} 结合以上不等式,我们有 \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\left(\int_{{\Bbb R}^{3}}\phi^{2}(x)(|w^{r_{1}}|^{2}+|\nabla w^{r_{1}}|^{2}){\rm d}x+\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla \xi^{tr_{1}}|^{2}{\rm d}x{\rm d}s\right) onumber\\ &\leq&\alpha_{3}\left(\int_{{\Bbb R}^{3}}\phi^{2}(x)(|w^{r_{1}}|^{2}+|\nabla w^{r_{1}}|^{2}){\rm d}x+\int^{\infty}_{0}\mu(s)\int_{{\Bbb R}^{3}}\phi^{2}(x)|\nabla \xi^{tr_{1}}|^{2}{\rm d}x{\rm d}s\right) onumber\\ &&+\frac{C}{(r_{1}+1)^{2}}+\frac{2}{r_{1}+1}\|\nabla w_{t}^{r_{1}}\|^{2},\tag{3.31} \end{eqnarray} 其中 $\alpha_{3}=\max\{C(1+\frac{\varepsilon}{2}),2k_{0}-\delta\}$.

最后对上式从 $[0,t_{\varepsilon}]$ 上积分,根据引理 3.5,引理 3.7 和 $w^{r_{1}}(0)=0$,可得 \begin{eqnarray*} &&\int_{V(B^{c}_{r_{2}})}(|w^{r_{1}}(t_{\varepsilon})|^{2}+|\nabla w^{r_{1}}(t_{\varepsilon})|^{2}){\rm d}x+\int^{\infty}_{0}\mu(s)\int_{V(B^{c}_{r_{2}})}|\nabla \xi^{tr_{1}}|^{2}{\rm d}x{\rm d}s onumber\\ &\leq &{\rm e}^{\alpha_{3}t_{\varepsilon}}\bigg(\frac{Ct_{\varepsilon}}{(r_{1}+1)^{2}}+\frac{2}{r_{1}+1}\int^{t_{\varepsilon}}_{0}\|\nabla w_{t}^{r_{1}}(y)\|^{2}{\rm d}y\bigg) onumber\\ &\leq &{\rm e}^{\alpha_{3}t_{\varepsilon}}\bigg(\frac{Ct_{\varepsilon}}{(r_{1}+1)^{2}}+\frac{2\Lambda_{0}}{r_{1}+1}\bigg):=\frac{C}{r_{1}+1}.\end{eqnarray*} 此时,固定 $t_\varepsilon$,选取 $r_{1}$ 充分大, 使得 $$\sqrt{{\rm e}^{\alpha_{3}t_{\varepsilon}}\bigg(\frac{Ct_{\varepsilon}}{(r_{1}+1)^{2}} +\frac{2\Lambda_{0}}{r_{1}+1}\bigg)}\leq\frac{\varepsilon}{2}, $$ 从而有 $\|z_{2}^{r_{1}}(t_{\varepsilon})\|_{{\cal M}(B^{c}_{r_{2}})}\leq \frac{\varepsilon}{2}$.

引理 3.8  假设 (1.2)--(1.8)式成立,且 $g\in L^{2}({\Bbb R}^{3})$. 对任意的 $\varepsilon>0$,$z_{0}\in{\Bbb B}_{0}$,固定 $r_{1}$ 和 $r_{2}$ (在引理 3.4 中给出),则存在常数 $K_{\varepsilon,r_{2}}$ 使得 $\|\tilde{z_{2}}^{r_{1}}(t_{\varepsilon})\|_{{\cal M}^{\frac{1}{4}}(B_{r_{2}})}\leq K_{\varepsilon,r_{2}}$ 成立,其中 $\tilde{z_{2}}^{r_{1}}(t_{\varepsilon})=(1-\phi(x))z_{2}^{r_{1}} (t_{\varepsilon})$.

  定义 $\tilde{z_{2}}^{r_{1}}(t)=(1-\phi(x))z_{2}^{r_{1}}(t)$, 当 $|x|\geq 2r_{1}+2$ 时,$\tilde{z}_{2}^{r_{1}}\equiv0$, 所以 $\tilde{z_{2}}^{r_{1}}$ 被限制在 $B_{r_{2}}$ 上,并且对任意的 $t>0$, $\tilde{z_{2}}^{r_{1}}(t)\in {\cal M}_{0}(B_{r_{2}})$.则 $\tilde{z_{2}}^{r_{1}}(t)=(\tilde{w}_{t}^{r_{1}},\tilde{\xi}^{tr_{1}})$ 是对应于 $r=r_{1}$ 时下面初值问题的唯一解 \begin{eqnarray} \left\{\begin{array}{ll} \tilde{w}_{t}^{r_{1}}+A\tilde{w}_{t}^{r_{1}}+A\tilde{w}^{r_{1}}-\int^{\infty}_{0}\mu(s)\triangle\tilde{\xi}^{tr_{1}}(s){\rm d}s+(1-\phi(x)[f_{1}(x,u)-f_{1}(x,v^{r_{1}})] onumber\\ [2mm]=-(1-\phi(x))f_{2}(x,u)+(1-\phi(x))g_{2}+2\nabla\phi(x)(\nabla w^{r_{1}}+\nabla w_{t}^{r_{1}}) onumber\\[2mm] \quad+2\nabla\phi(x)\int^{\infty}_{0}\mu(s)\nabla\xi^{tr_{1}}(s){\rm d}s,\\ [2mm] \tilde{\xi}_{t}^{tr_{1}}=-\tilde{\xi}_{s}^{tr_{1}}+\tilde{w}^{r_{1}},\\ \tilde{w}^{r_{1}}(x,t)\mid_{t=0}=0,\\ \tilde{\xi}^{0r_{1}}(x,s)=0.\end{array} \right.\end{eqnarray}

用 $A^{\frac{1}{4}}\tilde{w}^{r_{1}}$ 在 $L^{2}(B_{r_{2}})$ 作用于上面方程的第一个方程, 用 $\tilde{\xi}^{tr_{1}}$ 在 $L^{2}({\Bbb R}^{+},{\cal H}^{\frac{1}{4}}(B_{r_{2}}))$ 作用于上面方程的和第二个方程,然后作和可得 \begin{eqnarray} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}(\|A^{\frac{1}{8}}\tilde{w}^{r_{1}}\|_{L^{2}(B_{r_{2}})}^{2}+\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|_{L^{2}(B_{r_{2}})}^{2} +\|\tilde{\xi}^{tr_{1}}\|_{\mu,{\cal H}^{\frac{1}{4}}(B_{r_{2}})}) onumber\\ &&+\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|_{L^{2}(B_{r_{2}})}^{2} +\langle\tilde{\xi}^{tr_{1}}(s),\tilde{\xi}_{s}^{tr_{1}}(s)\rangle_{\mu,{\cal H}^{\frac{1}{4}}(B_{r_{2}})} onumber\\ &=&-\langle(1-\phi(x))(f_{1}(x,u)-f_{1}(x,v^{r})),A^{\frac{1}{4}}\tilde{w}^{r_{1}}\rangle-\langle(1-\phi(x))f_{2}(x,u),A^{\frac{1}{4}}\tilde{w}^{r_{1}}\rangle onumber\\ &&+\langle(1-\phi(x))g_{2}(x),A^{\frac{1}{4}}\tilde{w}^{r_{1}}\rangle+\langle2\nabla\phi(x)(\nabla w^{r_{1}} +\nabla w_{t}^{r_{1}}),A^{\frac{1}{4}}\tilde{w}^{r_{1}}\rangle onumber\\ &&+2\int^{\infty}_{0}\mu(s)\int_{B_{r_{2}}}\nabla\phi(x)A^{\frac{1}{4}}\tilde{w}^{r_{1}}\nabla\xi^{tr_{1}}(s){\rm d}x{\rm d}s.\tag{3.32}\end{eqnarray}

现在对 (3.32) 式中的每一项作估计. 首先,由引理 2.1,我们有 \begin{equation} \langle\tilde{\xi}^{tr_{1}}(s),\tilde{\xi}_{s}^{tr_{1}}(s)\rangle_{\mu,{\cal H}^{\frac{1}{4}}(B_{r_{2}})} \geq\frac{\delta}{2}\|\tilde{\xi}^{tr_{1}}\|_{\mu,{\cal H}^{\frac{1}{4}}(B_{r_{2}})}.\tag{3.33} \end{equation}

其次,根据条件 (1.3)--(1.4) 和 (3.15),利用 H\"{o}lder 不等式和连续嵌入 $D(A^{\frac{5}{8}})\hookrightarrow L^{12}(B_{r_{2}})$, $D(A^{\frac{3}{8}})\hookrightarrow L^{4}(B_{r_{2}})$,与 (3.29) 式的估计类似可得 \begin{eqnarray} &&\left|-\langle(1-\phi(x))(f_{1}(x,u)-f_{1}(x,v^{r_{1}})),A^{\frac{1}{4}}\tilde{w}^{r_{1}}\rangle\right| onumber\\ &\leq &c_{2}\int_{B_{r_{2}}}(1+|u|^{4}+|v^{r_{1}}|^{4})|(1-\phi(x))w^{r_{1}}||A^{\frac{1}{4}}\tilde{w}^{r_{1}}|{\rm d}x onumber\\ &\leq& c_{2}(1+\|u\|_{L^{6}}^{4}+\|v^{r_{1}}\|_{L^{6}}^{4})\|\tilde{w}^{r_{1}}\|_{L^{12}(B_{r_{2}})}\|A^{\frac{1}{4}}\tilde{w}^{r_{1}}\|_{L^{4}(B_{r_{2}})} onumber\\ &\leq& C\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|^{2}_{L^{2}(B_{r_{2}})}+\frac{1}{5}\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|^{2}_{L^{2}(B_{r_{2}})},\tag{3.34} \end{eqnarray} 并且注意到 $|\nabla \phi(x)|\leq \frac{1}{r_{2}+1}$,$L^{2}(B_{r_{2}})\hookrightarrow L^{\frac{4}{3}}(B_{r_{2}})$ 可得到 \begin{equation} \langle2\nabla\phi(x)(\nabla w^{r_{1}}+\nabla w_{t}^{r_{1}}),A^{\frac{1}{4}}\tilde{w}^{r_{1}}\rangle\leq C(1+\|\nabla w_{t}^{r_{1}}\|^{2})+\frac{1}{5}\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|^{2}_{L^{2}(B_{r_{2}})}; \tag{3.35} \end{equation} \begin{equation} 2\int^{\infty}_{0}\mu(s)\int_{B_{r_{2}}}\nabla\phi(x)A^{\frac{1}{4}}\tilde{w}^{r_{1}}\nabla\xi^{tr_{1}}(s){\rm d}x{\rm d}s\leq Ck_{0}\|\xi^{tr_{1}}\|^{2}_{\mu,V_{(B_{r_{2}})}}+\frac{1}{5}\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|^{2}_{L^{2}(B_{r_{2}})}.\tag{3.36} \end{equation} 类似地,利用 $f_{2},\;g_{2}$ 的定义,$\phi(x)$ 的性质和 (1.2)式,有 \begin{equation} \left|-\langle(1-\phi(x))f_{2}(x,u),A^{\frac{1}{4}}\tilde{w}^{r_{1}}\rangle\right|\leq C+\frac{1}{5}\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|^{2}_{L^{2}(B_{r_{2}})}, \tag{3.37} \end{equation} \begin{equation} \left|-\langle(1-\phi(x))g_{2},A^{\frac{1}{4}}\tilde{w}^{r_{1}}\rangle\right|\leq C+\frac{1}{5}\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|^{2}_{L^{2}(B_{r_{2}})}.\tag{3.38} \end{equation} 将 (3.33)--(3.38) 式代入 (3.32)式,得到 \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}(\|A^{\frac{1}{8}}\tilde{w}^{r_{1}}\|_{L^{2}(B_{r_{2}})}^{2}+\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|_{L^{2}(B_{r_{2}})}^{2} +\|\tilde{\xi}^{tr_{1}}\|_{\mu,{\cal H}^{\frac{1}{4}}(B_{r_{2}})}) onumber\\ &\leq &C\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|_{L^{2}(B_{r_{2}})}^{2}+C(1+\|\nabla w_{t}^{r_{1}}\|^{2})+Ck_{0}\|\xi^{tr_{1}}\|^{2}_{\mu,V(B_{r_{2}})} onumber\\ &\leq& C(\|A^{\frac{1}{8}}\tilde{w}^{r_{1}}\|_{L^{2}(B_{r_{2}})}^{2}+\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}\|_{L^{2}(B_{r_{2}})}^{2} +\|\tilde{\xi}^{tr_{1}}\|_{\mu,{\cal H}^{\frac{1}{4}}(B_{r_{2}})}) onumber\\ && +C(1+\|\nabla w_{t}^{r_{1}}\|^{2})+Ck_{0}\|\xi^{tr_{1}}\|^{2}_{\mu,V(B_{r_{2}})}\tag{3.39} \end{eqnarray} 取引理 3.3 中的 $t_{\varepsilon}>0$,再利用 (3.15) 式和引理 3.7 (用 $w^{r_{1}}$ 替代 $u$),对上式在 $[0,t_{\varepsilon}]$ 运用 Gronwall 引理, 结合 $\tilde{w}^{r_{1}}(0)=0$,我们得到 $$ \|A^{\frac{1}{8}}\tilde{w}^{r_{1}}(t_{\varepsilon})\|_{L^{2}(B_{r_{2}})}^{2} +\|A^{\frac{5}{8}}\tilde{w}^{r_{1}}(t_{\varepsilon})\|_{L^{2}(B_{r_{2}})}^{2} +\|\tilde{\xi}^{tr_{1}}\|_{\mu,{\cal H}^{\frac{1}{4}}(B_{r_{2}})}\leq C{\rm e}^{Ct_{\varepsilon}}(1+t_{\varepsilon}), $$ 因此,存在 $K_{\varepsilon,r_{2}}$,使得 \begin{equation} \|\tilde{z_{2}}^{r_{1}}(t_{\varepsilon})\|_{{\cal M}^{\frac{1}{4}}(B_{r_{2}})} \leq K_{\varepsilon,r_{2}},~~ \forall \;z_{0}\in{\Bbb B}_{0}.\tag{3.40} \end{equation} 证毕.

下面验证记忆项的紧性.

首先,对于任意的 $\xi^{tr_{1}}_{0}\in L_{\mu}^{2}({\Bbb R}^{+};{\cal H}^{0}(B_{r_{2}}))$, Cauchy 问题 \begin{equation} \left\{\begin{array}{ll} \partial_{t}\xi^{tr_{1}}=-\partial_{s}\xi^{tr_{1}}+w^{r_{1}},~~ t>0,\\ \xi^{0}=\xi_{0} \end{array} \right.\tag{3.41} \end{equation} 有唯一解 $\xi^{tr_{1}}\in{\cal C}({\Bbb R}^{+};L_{\mu}^{2}({\Bbb R}^{+};{\cal H}^{0}(B_{r_{2}})))$,并且有显式表达式 \begin{equation} \xi^{tr_{1}}(x,s)= \left\{\begin{array}{ll} \int_{0}^{s}w^{r_{1}}(t-y){\rm d}y,&0 < s\leq t,\\[3mm] \int_{0}^{t}w^{r_{1}}(t-y){\rm d}y,~~&s>t.\end{array} \right.\tag{3.42} \end{equation} 引理 3.9  在引理 3.8 的条件下,记 $${\cal K}^{r_{1}}_{t_{\varepsilon}}:=\Pi S_{2}(t_{\varepsilon},{\Bbb B}_{0}).$$ 对每一个给定的 $t_{\varepsilon}>0$ 和任意的 $\varepsilon>0$,存在正常数 $N=N(t_{\varepsilon},\varepsilon,r_{2})$,使得

(1) ${\cal K}^{r_{1}}_{t_{\varepsilon}}$ 在 $L^{2}_{\mu}({\Bbb R}^{+}; {\cal H}^{\frac{1}{4}}(B_{r_{2}}))\bigcap H^{1}_{\mu}({\Bbb R}^{+}; {\cal H}^{0}(B_{r_{2}}))$ 有界;

(2) $\sup\limits_{\xi^{t_{\varepsilon}r_{1}}\in{\cal K}^{r_{1}}_{t_{\varepsilon}}} \|\xi^{t_{\varepsilon}r_{1}}(s)\|^{2}_{{\cal H}^{0}(B_{r_{2}})}\leq N,$\\ 其中 $S_{2}(t,\cdot)$ 是方程 $(3.9)(r=r_{1})$ 所对应的解半群, $\Pi:{\cal H}^{0}\times L^{2}_{\mu}({\Bbb R}^{+};{\cal H}^{0}(B_{r_{2}})) \rightarrow L^{2}_{\mu}({\Bbb R}^{+};$ ${\cal H}^{0}(B_{r_{2}}))$ 的投影算子.

  根据 (3.42)式,有 \begin{equation} \partial_{s}\xi^{tr_{1}}(x,s)= \left\{\begin{array}{ll} w^{r_{1}}(t-s),~~& 0 < s\leq t,\\ 0,& s>t.\end{array} \right.\tag{3.43} \end{equation} 从而结合引理 3.8,即知 (1) 成立.

其次,显然有 \begin{equation} \|\xi^{t_{\varepsilon}r_{1}}(x,s)\|_{{\cal H}^{0}(B_{r_{2}})}\leq \left\{\begin{array}{ll} \int_{0}^{s}\|w^{r_{1}}(t-y)\|_{{\cal H}^{0}(B_{r_{2}})}{\rm d}y \leq\int_{0}^{t_{\varepsilon}}\|w^{r_{1}}(t-y)\|_{{\cal H}^{0}(B_{r_{2}})}{\rm d}y, & 0 < s\leq t_{\varepsilon},\\ [3mm] \int_{0}^{t_{\varepsilon}}\|w^{r_{1}}(t-y)\|_{{\cal H}^{0}(B_{r_{2}})}{\rm d}y, & s>t_{\varepsilon}.\end{array} \right.\tag{3.44} \end{equation} 从而结合引理 3.8,即知 (2) 成立.

因此,由引理 3.2 知 ${\cal K}^{r_{1}}_{t_{\varepsilon}}$ 在 $L^{2}_{\mu}({\Bbb R}^{+};{\cal H}^{0}(B_{r_{2}}))$ 中相对紧.

定理1.1的证明  由引理 3.3 和引理 3.4 可知,对任意的 $\varepsilon>0$, 存在 $t_{\varepsilon}$ 和 $r_{2}=r_{2}(\varepsilon)$,使得当 $t\geq t_{\varepsilon}$ 时 $$\|S(t){\Bbb B}_{0}\|_{{\cal M}(B^{c}_{r_{2}})}=\|z(t)\|_{{\cal M}(B^{c}_{r_{2}})}\leq \varepsilon,\quad \forall\;z_{0}\in{\Bbb B}_{0}.$$ 再利用紧嵌入 ${\cal H}^{\frac{1}{4}}(B_{r_{2}})\hookrightarrow {\cal H}^{0}(B_{r_{2}})=V(B_{r_{2}})$ 和引理 3.9,可知 $S(t){\Bbb B}_{0}$ 在 ${\cal M}(B_{r_{2}})$ 中相对紧.

因此,根据 Temam 的经典理论(见文献 [19]),与方程 (1.1) 相对应的半群 $S(t)$ 在$H^{1}({\Bbb R}^{3})\times L^{2}_{\mu}({\Bbb R}^{+},H^{1}({\Bbb R}^{3}))$ 中存在全局吸引子.

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