本文讨论下面的一阶格点系统

(1.1) $ \begin{eqnarray} & \dot{u}_m = \nu (u_{m-1}-2u_{m}+u_{m+1}) -\lambda u_{m} -f(u_{m})+g_m(t), \, m\in {\Bbb Z}, \, t>\tau, \end{eqnarray} $

(1.2) $ \begin{eqnarray} & u_m(\tau) = u_{\tau, m}, \, m\in {\Bbb Z}, \end{eqnarray} $

其中$ {\Bbb Z} = \{\cdots, -m, \cdots, -2, -1, 0, 1, 2, \cdots, m, \cdots\} $表示整数集, $ \nu, \lambda $是正的常数, $ u_m(t) $是未知函数, $ g_m(t) $和$ f(\cdot) $是给定的函数.文中将假设$ g(t) = (g_m(t))_{m\in {\Bbb Z}}\in {\mathcal C}({\mathbb R}; \ell^2) $ (所有定义在$ {\mathbb R} $上取值于$ \ell^2 $中的连续函数全体)且$ (u_{\tau, m})_{m\in {\Bbb Z}}\in \ell^2 $,这里

$ \ell^2 = \Big\{ u = (u_m)_{m\in {\mathbb Z}}\, \big| \, \sum\limits_{m\in{\mathbb Z}}u_m^2<+\infty \Big\}. $

方程(1.1)可以看作是下面反应扩散方程在$ {\mathbb R} $ (实数轴)上的空间离散化近似

(1.3) $ \begin{eqnarray} \frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial x^2} -\lambda u -f(u) +g(t). \end{eqnarray} $

已有一些文献研究了方程(1.3)的解的渐近行为.例如文献[1]研究了有界区域上全局吸引子的存在性,文献[2-4]研究了无界区域上全局吸引子与指数吸引子的存在性,文献[5]研究了该方程的动力学分叉问题.特别地,文献[6]证明了方程(1.3)的离散近似方程,也即方程(1.1)在自治情形全局吸引子的存在性与上半连续性.

格点系统是某些变量离散化的时空系统,包括耦合的常微分方程组、耦合映射格点和细胞自动机^[7].在某些情况下,格点系统表现为偏微分方程的空间变量离散化近似.格点动力系统在现实中有很多应用,如:涉及生物学^[8],化学反应理论^[9],激光理论^[10],材料科学^[11],图象处理与模式识别^[12],等等.

迄今关于格点动力系统的动力学的研究已有很多工作.例如文献[13-14]研究了一阶和二阶耗散格点动力系统整体吸引子的存在性和上半连续性;文献[15-18]研究了非自治格点动力系统存在紧致核截面与一致吸引子的充要条件,以及核截面与一致吸引子的上半连续性和熵的估计,并将理论应用于具体的格点数学物理方程;文献[19]研究了一阶时滞格点系统吸引子的存在性;文献[20-22]研究了随机格点动力系统的随机吸引子;文献[23-25]讨论了格点动力系统指数吸引子与一致指数吸引子的存在性问题.值得指出的是,文献[26]证明了一般非自治格点系统存在拉回吸引子的充要条件,并研究了格点Klein-Gordon-Schrödinger方程组不变测度的存在性.

本文的初衷是研究一般一阶格点系统(1.1)的不变测度的存在性.不变测度的研究来源于统计物理,对于人们理解湍流的运动规律很有帮助^[27],其主要原因是湍流的几个主要物理量(如速度,动能)关于时间平均后表现出很强的规律.如今已有若干文献研究了耗散系统的不变测度和统计解,例如文献[28]研究了耗散系统稳态统计解的上半连续性;文献[29]应用广义Banach极限构造了一般度量空间上的连续动力系统的不变测度;后来,文献[30]改进了文献[28-29]的结果,给出了一类广泛的耗散半群的不变测度和统计解的构造方法.最近,文献[31]把文献[30]的结果推广到非自治耗散系统,证明了完备的度量空间$ X $中的连续过程$ \{{\mathcal U}(t, \tau)\} $$ _{t\geqslant \tau} $存在不变Borel概率测度的充分条件是: (ⅰ)过程$ \{{\mathcal U}(t, \tau)\}_{t\geqslant \tau} $在$ X $中存在拉回吸引子; (ⅱ)对每个$ x\in X $及每个$ t\in {\Bbb R} $, $ X $ -值函数$ \tau\mapsto U(t, \tau)x $在$ (-\infty, t] $上连续且有界.该结果已被应用到全局修正的三维Navier-Stokes方程组^[32], Ladyzhenskaya流体力学方程组^[33],正则化MHD方程组^[34]和格点Klein-Gordon-Schrödinger方程组^[26].另外,文献[35]研究了三维全局修正的Navier-Stokes方程组不变测度与轨道统计解的存在性和渐近正则性.

证得了拉回吸引子上不变测度的存在性之后,本文的第二个目的是证明该不变测度满足Liouville型方程. Liouville方程及其相应的Liouville定理是统计物理中的重要结果.粗略地讲,它是指在哈密顿动力学演化下的点集在相空间中的分布可以随时间演化而变形,但该点集的测度(或相体积)是守恒的(见文献[36, p113]).本文将证明一阶格点系统(1.1)的不变测度也满足Liouville型方程和Liouville定理的结论.

为把问题(1.1)–(1.2)写成一阶常微分方程的初值问题,我们在空间$ \ell^2 $上赋以以下内积和范数

$ (u, v) = \sum\limits_{m\in {\mathbb Z}}u_mv_m, \, \|u\|^2 = (u, u), \, \, \forall\, u = (u_m)_{m\in {\Bbb Z}}, v = (v_m)_{m\in {\Bbb Z}}\in \ell^2, $

则$ \big(\ell^2, (\cdot, \cdot)\big) $是Hilbert空间.在$ \ell^2 $上定义算子$ A, B, B^* $如下

$ \begin{eqnarray*} (Au)_m & = &2u_m-u_{m-1}-u_{m+1}, \, \forall\, u = (u_m)_{m\in {\Bbb Z}}\in \ell^2, \\ (Bu)_m & = &u_{m+1}-u_{m-1}, \, \forall\, u = (u_m)_{m\in {\Bbb Z}}\in \ell^2, \\ (B^*u)_m & = &u_{m-1}-u_{m+1}, \, \forall\, u = (u_m)_{m\in {\Bbb Z}}\in \ell^2. \end{eqnarray*} $

简单计算可得

$ (Au, v) = (B^*Bu, v) = (Bu, Bv), \, \forall\, u = (u_m)_{m\in {\Bbb Z}}, v = (v_m)_{m\in {\Bbb Z}}\in \ell^2, $

$ \|Au\|^2 \leqslant 16\|u\|^2, \|Bu\|^2\leqslant 4\|u\|^2, \, \forall\, u = (u_m)_{m\in {\Bbb Z}}\in \ell^2. $

我们记$ f(u) = (f(u_m))_{m\in {\mathbb Z}} $,则求解的初值问题(1.1)–(1.2)可以写成Hilbert空间$ \ell^2 $中的初值问题

(2.1) $ \begin{eqnarray} && \frac{{\mathrm d} u}{{\mathrm d}t} +\nu Au+\lambda u+f(u) = g(t), \, t>\tau, \end{eqnarray} $

(2.2) $ \begin{eqnarray} && u(\tau) = u_\tau = (u_{\tau, m})_{m\in {\mathbb Z}}\in \ell^2. \end{eqnarray} $

为保证以上初值问题解的存在唯一性和有界性,我们假设函数$ f(s) $和$ g(t) = (g_m(t))_{m\in{\mathbb Z}} $满足下面条件

(2.3) $ \begin{eqnarray} {\rm (H)}: \quad \left\{ \begin{array}{ll} f(s)\in {\mathcal C}^1({\mathbb R}), \mbox {且}\ f(s)s\geqslant 0, & \forall\, s\in {\mathbb R}; \\ g(t) = (g_m(t))_{m\in{\mathbb Z}}\in {\mathcal C}(\ell^2), \mbox {且}\ \int_{-\infty}^t e^{\lambda s}\|g(s)\|^2{\mathrm d}s<+\infty, & \forall\, t\in {\mathbb R}. \end{array} \right. \end{eqnarray} $

引理2.1  设条件$ {\rm (H)}$成立.

(1)对每个$ u_\tau\in \ell^2 $,初值问题(2.1)–(2.2)存在唯一的局部解

(2.4) $ \begin{eqnarray} u\in {\mathcal C}([\tau, T_*); \ell^2)\cap {\mathcal C}^1((\tau, T_*); \ell^2), \end{eqnarray} $

其中$ T_* > \tau $,并且如果$ T_* < +\infty $,则有$ \lim\limits_{t\rightarrow T_*^-}\|u(t)\| = +\infty $.

(2)上述对应于$ u_\tau $的解$ u $满足

(2.5) $ \begin{eqnarray} \|u(t)\|^2 \leqslant \|u_\tau\|^2e^{-\lambda(t-\tau)} +\frac{e^{-\lambda t}}{\lambda}\int_\tau^te^{\lambda s}\|g(s)\|^2{\mathrm d}s, \, \forall\, t\geqslant \tau. \end{eqnarray} $

证   (1)注意到$ u\rightarrow Au $和$ u\rightarrow \lambda u $都是$ \ell^2 $上的有界线性算子,而$ g(t)\in {\mathcal C}(\ell^2) $,我们只需证明$ f(u) $是$ \ell^2 $上的局部Lipschitz映射.事实上,设$ {\mathcal B} $是$ \ell^2 $中的有界闭球,任取$ u, v\in \ell^2 $,由微分中值定理得

(2.6) $ \begin{eqnarray} \|f(u)-f(v)\|^2 = \sum\limits_{m\in {\mathbb Z}}|f(u_m)-f(v_m)|^2 = \sum\limits_{m\in {\mathbb Z}}|f'(\xi_m)||u_m-v_m|^2, \end{eqnarray} $

其中$ \xi_m = u_m+\vartheta_m(v_m-u_m)\in {\mathcal B} $, $ \vartheta_m\in (0, 1) $.因为$ f\in {\mathcal C}^1({\mathbb R}) $, (2.6)式表明

$ \begin{eqnarray*} \|f(u)-f(v)\|^2 \leqslant L({\mathcal B})\|u-v\|^2, \end{eqnarray*} $

其中$ L({\mathcal B}) = \max\limits_{\xi\in {\mathcal B}}|f'(\xi)| $.由经典的常微分方程理论, (1)得证.

(2)用$ u $与方程(2.1)作内积得

$ \frac12\frac{{\mathrm d}}{{\mathrm d}t}\|u(t)\|^2 +\nu \|Bu(t)\|^2 +\lambda\|u(t)\|^2 = -(f(u(t)), u(t))+(g(t), u(t)). $

因为

$ -(f(u(t)), u(t)) = -\sum\limits_{m\in {\mathbb Z}}f(u_m(t))u_m(t)\leqslant 0, $

$ (g(t), u(t))\leqslant \frac{\lambda \|u(t)\|^2}{2}+\frac{\|g(t)\|^2}{2\lambda}, $

所以

$ \frac{{\mathrm d}}{{\mathrm d}t}\|u(t)\|^2 +\lambda\|u(t)\|^2 \leqslant \frac{\|g(t)\|^2}{\lambda}. $

由上式和Gronwall不等式即得(2.5)式.证明完毕.

引理2.1表明若条件(H)成立,则对每个$ u_\tau\in \ell^2 $问题(2.1)–(2.2)相应的解$ u(\cdot) $在$ [0, +\infty) $上全局存在,且

(2.7) $ \begin{eqnarray} u(\cdot) \in {\mathcal C}([\tau, +\infty); \ell^2) \cap {\mathcal C}^1((\tau, +\infty); \ell^2). \end{eqnarray} $

从而问题(2.1)–(2.2)的解算子

$ U(t, \tau): u_\tau\in \ell^2 \longmapsto u(t) = u(t; \tau, u_\tau) = U(t, \tau)u_\tau\in \ell^2, \, \forall\, t\geqslant\tau, $

在$ \ell^2 $上生成连续过程$ \{U(t, \tau)\}_{t\geqslant\tau} $,其中$ u(t) = u(t; \tau, u_\tau) $表示问题(2.1)–(2.2)以$ \tau $时刻$ u_\tau $为初值的解.

记

(2.8) $ \begin{eqnarray} {\mathcal D}_\lambda = \Big\{ {\widehat D} = \{D(s)|s\in {\mathbb R}\} \mid \lim\limits_{s\rightarrow-\infty} e^{\lambda s}\sup\limits_{u\in D(S)}\|u\|^2 = 0 \Big\}. \end{eqnarray} $

显然, $ \ell^2 $中任何与时间无关的有界集都属于$ {\mathcal D}_\lambda $.

引理2.2  设条件${\rm (H)} $成立,则存在$ \widehat{{\mathcal B}}_0 = \{{\mathcal B}_0(s)|s\in {\mathbb R}\} $使得对任意$ {\widehat D} = \{D(s)|s\in {\mathbb R}\}\in {\mathcal D}_\lambda $, $ \exists \tau_0 = \tau_0(t, \widehat{D})\leqslant t $有$ U(t, \tau)D(\tau)\subset {\mathcal B}_0(t) $, $ \forall\, \tau\leqslant \tau_0 $,其中$ {\mathcal B}_0(s) = {\mathcal B}_0(0; R_\lambda(s)) $是$ \ell^2 $中以原点为球心以$ R_\lambda(s) $为半径的球.

证  记

$ R^2_\lambda(s) = 1+\frac{e^{-\lambda t}}{\lambda}\int_{-\infty}^te^{\lambda s}\|g(s)\|^2{\mathrm d}s, \, s\in {\mathbb R}, $

则由(2.5)式知结论成立.

由条件(H)知,对每个给定的时间$ t $,引理2.2中的球$ {\mathcal B}_0(0; R_\lambda(s)) $都是$ \ell^2 $中有界集.集合$ \widehat{{\mathcal B}}_0 = \{{\mathcal B}_0(0; R_\lambda(s))|s\in {\mathbb R}\} $称为过程$ \{U(t, \tau)\}_{t\geqslant\tau} $的有界拉回$ {\mathcal D}_\lambda $ -吸收集.为得到拉回$ {\mathcal D}_\lambda $ -吸引子的存在性,我们下面证明过程$ \{U(t, \tau)\}_{t\geqslant\tau} $具有拉回$ {\mathcal D}_\lambda $渐近零的性质.

引理2.3  设条件$ {\rm (H)} $成立.则对于给定的$ t\in {\mathbb R} $, $ \forall\, \epsilon>0 $和$ \forall\, {\widehat D} = \{D(s)|s\in {\mathbb R}\}\in {\mathcal D}_\lambda $, $ \exists M_* = M_*(t, \epsilon, {\widehat D})\in {\mathbb N} $和$ \tau_* = \tau_*(t, \epsilon, {\widehat D})\leqslant t $,使得

(2.9) $ \begin{eqnarray} \sup\limits_{u_\tau\in D(\tau)} \sum\limits_{|m|\geqslant M_*}|(U(t, \tau)u_\tau)_m|^2 \leqslant \epsilon^2, \quad\forall\, \tau\leqslant \tau_*. \end{eqnarray} $

证  选取适当光滑函数$ \chi(x)\in {\mathcal C}^1({\mathbb R}_+, {\mathbb R}_+) $使得

(2.10) $ \begin{eqnarray} \left\{ \begin{array}{ll} \chi(x) = 0, & 0\leqslant x\leqslant 1, \\ 0\leqslant \chi(x)\leqslant 1, & 1\leqslant x\leqslant 2, \\ \chi(x) = 1, & x\geqslant 2, \\ |\chi'(x)|\leqslant \chi_0 (\mbox{正数}), & x\geqslant 0. \end{array} \right. \end{eqnarray} $

考虑任意的$ {\widehat D} = \{D(s)|s\in {\mathbb R}\}\in {\mathcal D}_\lambda $,对任意$ t, \tau\in {\mathbb R} $, $ t\geqslant \tau $,记$ u(t) = u(t; \tau, u_\tau) = U(t, \tau)u_\tau $为问题(2.1)–(2.2)的以$ u_\tau\in D(\tau) $为初值的解.设$ M $为某正的常数,记$ v_m = \chi(\frac{|m|}{M})u_m $, $ m\in {\mathbb Z} $.用$ v = (v_m)_{m\in {\mathbb Z}} $与方程(2.1)作内积,得

(2.11) $ \begin{eqnarray} (\dot{u}(t), v(t))+\nu(Au(t), v(t))+\lambda(u(t), v(t))+(f(u(t)), v(t)) = (g(t), v(t)). \end{eqnarray} $

经过计算和估计,有

(2.12) $ \begin{equation} (\dot{u}(t), v(t)) = \frac{1}{2}\frac{{\mathrm d}}{{\mathrm d}t} \sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})u^2_m(t), \end{equation} $

(2.13) $ \begin{eqnarray} (Au(t), v(t)) & = &(Bu(t), Bv(t)) = \sum\limits_{m\in{\mathbb Z}} (u_{m+1}-u_{m-1}) \big(\chi(\frac{|m+1|}{M})u_{m+1}-\chi(\frac{|m|}{M})u_{m}\big)\\ & = & \sum\limits_{m\in{\mathbb Z}}\chi(\frac{|m+1|}{M})(u_{m+1}-u_{m})^2 +\sum\limits_{m\in{\mathbb Z}} \chi'(\frac{|\tilde{m}|}{M})\frac{u_m(u_{m+1}-u_{m})}{M}, \end{eqnarray} $

(2.14) $ \begin{equation} (f(u(t)), v(t)) = \sum\limits_{m\in{\mathbb Z}}\chi(\frac{|m|}{M}) f(u_m)u_m\geqslant 0, \end{equation} $

(2.15) $ \begin{equation} (g(t), v(t)) = \sum\limits_{m\in{\mathbb Z}}\chi(\frac{|m|}{M})g_m(t)u_m \leqslant \frac{1}{2\lambda}\sum\limits_{m\in{\mathbb Z}}\chi(\frac{|m|}{M})g^2_m(t) +\frac{\lambda}{2}\sum\limits_{m\in{\mathbb Z}}\chi(\frac{|m|}{M})u^2_m, \end{equation} $

其中(2.13)式中的$ |\tilde{m}| $是介于$ |m| $和$ |m+1| $中的常数.把(2.12)–(2.15)式代入(2.11)式,再应用(2.10)式和拉回吸引性,得到

(2.16) $ \begin{equation} \frac{{\mathrm d}}{{\mathrm d}t} \sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})u^2_m(t) + \lambda\sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})u^2_m(t) \leqslant \frac{1}{\lambda}\sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})g^2_m(t) +\frac{4\chi_0R^2_{\lambda}(t)}{M}, \, \tau\leqslant\tau_0. \end{equation} $

现对任意的$ \epsilon > 0 $,存在$ M_1 = M_1(t, \epsilon)\in {\mathbb N} $,使得

(2.17) $ \begin{equation} \frac{4\chi_0R^2_{\lambda}(t)}{M} \leqslant \frac{\lambda \epsilon^2}{3}, \, \forall\, M>M_1. \end{equation} $

把(2.17)式代入(2.16)式有

$ \frac{{\mathrm d}}{{\mathrm d}t} \sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})u^2_m(t) + \lambda\sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})u^2_m(t) \leqslant \frac{1}{\lambda}\sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})g^2_m(t) +\frac{\lambda \epsilon^2}{3}, \, \tau\leqslant\tau_0, \, M>M_1. $

对上式应用Gronwall不等式,得

(2.18) $ \begin{equation} \sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})u^2_m(t) \leqslant \sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})u^2_m(\tau)e^{-\lambda(t-\tau)} +\frac{e^{-\lambda t}}{\lambda}\int_\tau^t e^{\lambda\theta}\sum\limits_{|m|\geqslant M}g^2_m(\theta){\mathrm d}\theta +\frac{\epsilon^2}{3}. \end{equation} $

由(2.3)式知对上述$ \epsilon > 0 $,存在$ M_2 = M_2(t, \epsilon)\in {\mathbb N} $,使得

(2.19) $ \begin{eqnarray} \int_\tau^te^{\lambda\theta}\sum\limits_{|m|\geqslant M}g^2_m(\theta){\mathrm d}\theta \leqslant \sum\limits_{|m|\geqslant M}\int_{-\infty}^te^{\lambda\theta}g^2_m(\theta){\mathrm d}\theta \leqslant \frac{\epsilon^2}{3}, \, M\geqslant M_2. \end{eqnarray} $

再由(2.8)式知对上述$ \epsilon > 0 $,存在$ \tau_1 = \tau_1(t, \epsilon, {\hat D}) $使得

(2.20) $ \begin{equation} \sum\limits_{m\in {\mathbb Z}} \chi(\frac{|m|}{M})u^2_m(\tau)e^{-\lambda(t-\tau)} \leqslant e^{-\lambda(t-\tau)}\|u_\tau\|^2 \leqslant e^{-\lambda(t-\tau)}\sup\limits_{u_\tau\in D(\tau)}\|u_\tau\|^2 < \frac{\epsilon^2}{3}, \, \tau\leqslant\tau_1. \end{equation} $

从(2.18)–(2.20)式即推得(2.9)式.证明完毕.

应用文献[26,定理2.1]和上面证得的引理2.2、引理2.3,我们得到本节的主要结果如下.

定理2.1  假设条件$ {\rm (H)} $成立.则系统(1.1)–(1.2)的解算子生成的过程$ \{U(t, \tau)\}_{t\geqslant \tau} $在$ \ell^2 $中存在拉回$ {\mathcal D}_\lambda $ -吸引子$ \widehat{{\mathcal A}}_{{\mathcal D}_\lambda} = \{{\mathcal A}_{{\mathcal D}_\lambda}| t\in {\mathbb R}\} $,满足

(1)紧性: $ \forall\, t\in {\mathbb R} $, $ {\mathcal {A}}_{\mathcal {D_{\lambda}}}(t) $是$ \ell^2 $的非空紧子集;

(2)不变性: $ U(t, \tau){\mathcal {A}}_{{\cal D_{\lambda}}}(\tau) = {\mathcal {A}}_{\mathcal {D_{\lambda}}}(t), \, \forall\, t\geqslant \tau $;

(3)拉回吸引性: $ \forall \widehat{D_{\lambda}} = \{ D_{\lambda}(t)| t\in {\mathbb R}\} \in {\mathcal {D_{\lambda}}} $,有

$ \; \lim\limits_{\tau\rightarrow -\infty}{\rm dist}_{\ell^2} \big(U(t, \tau)D_{\lambda}(\tau), {\mathcal {A}}_{\mathcal {D_{\lambda}}}(t) \big) = 0, t\in{\mathbb R}. $

本节我们先证明过程$ \{U(t, \tau)\}_{t\geqslant \tau} $具有$ \tau $ -连续性(指对任意固定的$ t\in {\mathbb R} $和给定的$ u_*\in \ell^2 $, $ \ell^2 $ -值函数$ \tau\longmapsto U(t, \tau)u_* $在$ (-\infty, t] $上连续且有界),进而得到拉回$ {\mathcal D}_\lambda $ -吸引子$ \widehat{{\mathcal A}}_{{\mathcal D}_\lambda} = \{{\mathcal A}_{{\mathcal D}_\lambda}| t\in {\mathbb R}\} $上不变Borel概率测度的存在性.然后我们证明得到的不变测度满足Liouville型方程.

先介绍广义Banach极限的概念.

定义3.1^[27]  记$ \Lambda $为定义在$ [0, +\infty) $上的有界实值函数全体. $ \Lambda $上的任意一个线性泛函(记作) $ {\rm LIM}_{s\rightarrow +\infty} $若满足:

(ⅰ)对任何$ h\in \Lambda $,有$ {\rm LIM}_{s\rightarrow +\infty}h(s)\geqslant 0 $;

(ⅱ)如果通常意义下的极限$ \lim\limits_{s\rightarrow +\infty}h(s) $存在,则LIM$ _{s\rightarrow +\infty}h(s) = \lim\limits_{s\rightarrow +\infty}h(s) $,

则称$ {\rm LIM}_{s\rightarrow +\infty} $为广义Banach极限.

注3.1  我们讨论"拉回"吸引子时需要考虑初始时间$ \tau\rightarrow-\infty $.为此,对于给定的定义在$ (-\infty, 0] $上的实值函数$ \varphi $和给定的广义Banach极限$ {\rm LIM}_{s\rightarrow +\infty} $,我们定义

$ {\rm LIM}_{s\rightarrow-\infty}\varphi(s) = \lim\limits_{s\rightarrow +\infty}\varphi(-s). $

同时,广义Banach极限有下面的性质(见文献[27, (1.38)式]或[30, (2.3)式])

(3.1) $ \begin{eqnarray} |{\rm LIM}_{s\rightarrow +\infty}h(t)| \leqslant \limsup\limits_{s\rightarrow +\infty}|h(t)|, \forall\, h(\cdot)\in\Lambda. \end{eqnarray} $

引理3.1  设条件$ {\rm (H)} $成立.则对任意给定的$ t\in {\mathbb R} $和$ u_*\in \ell^2 $, $ \ell^2 $ -值函数$ \tau\longmapsto U(t, \tau)u_* $在$ (-\infty, t] $上连续且有界.

证  考虑给定的$ u_*\in \ell^2 $和$ t\in {\mathbb R} $.由(2.5)式得

(3.2) $ \begin{eqnarray} \|u(s)\|^2 &\leqslant& \|u_*\|^2e^{-\lambda(s-\tau)} +\frac{e^{-\lambda s}}{\lambda}\int_\tau^s e^{\lambda \theta}\|g(\theta)\|^2{\mathrm d}\theta \\ &\leqslant& \|u_*\|^2 +\frac{e^{-\lambda t}}{\lambda}\int_\tau^t e^{\lambda \theta}\|g(\theta)\|^2{\mathrm d}\theta\\ &\leqslant& \|u_*\|^2 +\frac{e^{-\lambda t}}{\lambda}\int_{-\infty}^t e^{\lambda \theta}\|g(\theta)\|^2{\mathrm d}\theta, \, \forall\, t\geqslant s\geqslant \tau. \end{eqnarray} $

由条件(H)知(3.2)式的右端的上界与$ s\in (-\infty, t] $无关.因此, $ U(t, \cdot)u_* $在$ (-\infty, t] $上有界.

任取$ s_*\in (-\infty, t] $,我们下面证明$ U(t, \tau)u_* $在$ \tau = s_* $处连续.为此,我们需要证明$ \forall\, \epsilon > 0 $, $ \exists \delta = \delta(\epsilon, s_*, t, u_*) > 0 $,使得$ |r-s_*| < \delta $时就有$ \|U(t, s_*)u_*-U(t, r)u_*\| < \epsilon $.不妨设$ r < s_* < t $.记

$ U(\cdot, s_*)U(s_*, r)u_* = u^{(1)}(\cdot), \quad U(\cdot, s_*)U(r, r)u_* = u^{(2)}(\cdot), \quad \tilde{u}(\cdot) = u^{(1)}(\cdot)-u^{(2)}(\cdot). $

则$ \tilde{u}(\cdot) $满足下面的初值问题

(3.3) $ \begin{equation} \frac{{\mathrm d}\tilde{u}}{{\mathrm d}t} +\nu A\tilde{u} +\lambda \tilde{u} +f(u^{(1)})-f(u^{(2)}) = 0, \, t>s_*, \end{equation} $

(3.4) $ \begin{equation} \tilde{u}(s_*) = U(s_*, r)u_*-U(r, r)u_*. \end{equation} $

用$ \tilde{u} $与(3.3)式作内积得

(3.5) $ \begin{eqnarray} \frac12\frac{{\mathrm d}}{{\mathrm d}t}\|\tilde{u}(t)\|^2 +\nu \|B\tilde{u}(t)\|^2 +\lambda\|\tilde{u}(t)\|^2 +\sum\limits_{m\in {\mathbb Z}}(f(u_{m}^{(1)})-f(u_{m}^{(2)}))\tilde{u}_m = 0. \end{eqnarray} $

而由微分中值定理知存在常数$ \vartheta_m\in (0, 1) $,使得

$ f(u_{m}^{(1)})-f(u_{m}^{(2)}) = f'(u_{m}^{(1)}+\vartheta_m(u_{m}^{(1)}-u_{m}^{(2)}))\tilde{u}_m, \, m\in {\Bbb Z}. $

对任意$ m\in {\Bbb Z} $,由(3.2)式知

$ |u_{m}^{(1)}+\vartheta_m(u_{m}^{(1)}-u_{m}^{(2)})| \leqslant 2\|u^{(1)}\|+\|u^{(2)}\| \leqslant 3(\|u_*\|^2 +\frac{e^{-\lambda t}}{\lambda}\int_{-\infty}^t e^{\lambda \theta}\|g(\theta)\|^2{\mathrm d}\theta). $

记

$ L = L(u_*, t) = 3(\|u_*\|^2+\frac{e^{-\lambda t}}{\lambda}\int_{-\infty}^t e^{\lambda \theta}\|g(\theta)\|^2{\mathrm d}\theta), \quad {\mathcal M} = {\mathcal M}(u_*, t) = 2\max\limits_{\eta\in [0, L]}|f'(\eta)|, $

则由(3.5)式得

$ \frac{{\mathrm d}}{{\mathrm d}t}\|\tilde{u}(t)\|^2 \leqslant {\mathcal M}(u_*, t)\|\tilde{u}(t)\|^2. $

对上式用Gronwall不等式且注意到$ {\mathcal M}(u_*, t) $关于$ t $的单调递增性,得

$ \|\tilde{u}(t)\|^2 \leqslant \|\tilde{u}(s_*)\|^2\exp\bigg\{\int_{s_*}^t{\mathcal M}(u_*, \theta){\mathrm d}\theta\bigg\} \leqslant \|\tilde{u}(s_*)\|^2\exp\{(t-s_*){\mathcal M}(u_*, t)\}, $

也即是

$ \|U(t, s_*)u_*-U(t, r)u_*\|^2 \leqslant \|U(s_*, r)u_*-U(r, r)u_*\|^2\exp\{(t-s_*){\mathcal M}(u_*, t)\}. $

由(2.7)式知当$ |r-s_*| $足够小时,上不等式右端的$ \|U(s_*, r)u_*-U(r, r)u_*\|^2 $就能充分地小.证明完毕.

结合文献[31,定理3.1、定理4.1]和本文证得的定理2.1和引理3.1,我们得到下面结果.

定理3.1  设条件$ {\rm (H)} $成立. $ {\rm LIM}_{s\rightarrow +\infty} $是给定的广义$ {\rm Banach }$极限, $ \psi(\cdot): {\mathbb R}\longmapsto \ell^2 $是连续映射且$ \psi(\cdot)\in {\mathcal D}_{\lambda} $.则存在$ \ell^2 $上一族Borel概率测度$ \{\mu_t\}_{t\in {\mathbb R}} $, $ \mu_t $的支集包含于$ {\mathcal A}_{{\mathcal D}_{\lambda}}(t) $中,并有

(3.6) $ \begin{eqnarray} {\rm LIM}_{\tau\rightarrow-\infty} \frac{1}{t-\tau}\int_\tau^t\Phi(U(t, s)\psi(s)){\mathrm d}s = \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(t)} \Phi(u){\mathrm d}\mu_t(u) = \int_{\ell^2} \Phi(u){\mathrm d}\mu_t(u), \end{eqnarray} $

其中$ \Phi $是$ \ell^2 $上连续有界实值函数.同时, $ \mu_t $满足如下不变性质和平均性质

(3.7) $ \begin{equation} \int_{{\mathcal A}_{{\mathcal D}_\lambda}(t)} \Phi(u){\mathrm d}\mu_t(u) = \int_{{\mathcal A}_{{\mathcal D}_\lambda}(\tau)} \Phi(U(t, \tau)u){\mathrm d}\mu_\tau(u), \quad t\geqslant \tau. \end{equation} $

(3.8) $ \begin{equation} {\rm LIM}_{\tau\rightarrow-\infty} \frac{1}{t-\tau}\int_\tau^t\int_{\ell^2} \Phi(U(t, s)u){\mathrm d}\mu_s(u){\mathrm d}s = \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(t)} \Phi(u){\mathrm d}\mu_t(u). \end{equation} $

下面证明定理3.1中的不变Borel测度$ \{\mu_t\}_{t\in {\mathbb R}} $满足Liouville型方程.我们先把方程(2.1)写成下面形式

(3.9) $ \begin{eqnarray} \frac{{\mathrm d}u}{{\mathrm d}t} = F(u, t) = -\nu A u-\lambda u-f(u)+g(t), \quad t>\tau. \end{eqnarray} $

设$ \Phi $是定义在$ \ell^2 $上的实值函数,在$ \ell^2 $的有界子集上有界,且满足

(1)对任意$ u\in \ell^2 $, $ \Phi(u) $的Frechét导数(记作$ \Phi'(u) $)存在,也即对每个$ u\in \ell^2 $,存在$ \ell^2 $中元素$ \Phi'(u) $,使得

$ \lim\limits_{\|v\|\rightarrow 0} \frac{|\Phi(u+v)-\Phi(u)-(\Phi'(u), v)|}{\|v\|} = 0, \quad v\in \ell^2. $

(2)映射$ u\longmapsto \Phi'(u) $是从$ \ell^2 $到$ \ell^2 $的连续有界函数.

我们把满足上述条件(1)–(2)的全体函数的集合记作$ {\cal T} $(称为试验函数类),并称$ {\cal T} $中的函数为试验函数.上述条件(1)–(2)保证了方程(3.9)的解满足下面的等式

(3.10) $ \begin{eqnarray} \frac{{\mathrm d}}{{\mathrm d}t}\Phi(u) = (\Phi'(u), F(u, t)). \end{eqnarray} $

对于一般Banach空间中的试验函数类的定义和常见选取方法可以参考文献[27, p178,定义1.2].

定理3.2  设条件$ {\rm (H)} $成立.则对任意的$ \Phi\in {\mathcal T} $,测度$ \{\mu_t\}_{t\in {\mathbb R}} $满足下面的${\rm Liouville }$型方程

(3.11) $ \begin{equation} \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(t)}\Phi(u){\mathrm d}\mu_t(u) - \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(\tau)}\Phi(u){\mathrm d}\mu_\tau(u) = \int_\tau^t \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(\eta)} (\Phi'(u), F(u, \eta)){\mathrm d}\mu_\eta(u){\mathrm d}\eta, \, \forall\, t\geqslant \tau. \end{equation} $

证  为简洁起见,我们取$ \Phi(u) = \|u\|^2 $进行证明,对于一般的试验函数$ \Phi\in {\mathcal T} $,证明完全类似.

设$ u $是方程(3.9)的解.由于$ \Phi(u) = \|u\|^2 $,经简单计算得$ \Phi'(u) = 2u $.用$ u $与方程(3.9)作内积,然后在$ [\tau, t] $上积分,得

(3.12) $ \begin{eqnarray} \|u(t)\|^2-\|u(\tau)\|^2 = \int_\tau^t(2u(\theta), F(u(\theta), \theta)){\mathrm d}\theta. \end{eqnarray} $

对任意$ s < \tau $,设$ u_0\in\ell^2 $且记$ u(\eta) = U(\eta, s)u_0 $, $ \eta\geqslant s $.由(3.12)式得

(3.13) $ \begin{eqnarray} \|U(t, s)u_0\|^2-\|U(\tau, s)u_0\|^2 = \int_\tau^t(2U(\eta, s)u_0, F(U(\eta, s)u_0, \eta)){\mathrm d}\eta. \end{eqnarray} $

因为测度$ \{\mu_t\}_{t\in {\mathbb R}} $满足(3.8)式,所以

(3.14) $ \begin{eqnarray} && \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(t)} \|u\|^2{\mathrm d}\mu_t(u) - \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(\tau)} \|u\|^2{\mathrm d}\mu_\tau(u) \\ & = & {\rm LIM}_{M\rightarrow-\infty} \frac{1}{t-M}\int_M^t\int_{\ell^2} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0){\mathrm d}s \\ && - {\rm LIM}_{M\rightarrow-\infty} \frac{1}{\tau-M}\int_M^\tau\int_{\ell^2} \|U(\tau, s)u_0\|^2{\mathrm d}\mu_s(u_0){\mathrm d}s\\ & = & {\rm LIM}_{M\rightarrow-\infty} \frac{1}{\tau-(M-t+\tau)}\int_M^\tau\int_{\ell^2} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0){\mathrm d}s \\ && +{\rm LIM}_{M\rightarrow-\infty} \frac{1}{t-M}\int_\tau^t\int_{\ell^2} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0){\mathrm d}s \\ && - {\rm LIM}_{M\rightarrow-\infty} \frac{1}{\tau-M}\int_M^\tau\int_{\ell^2} \|U(\tau, s)u_0\|^2{\mathrm d}\mu_s(u_0){\mathrm d}s. \end{eqnarray} $

由(3.6)和(3.7)式知积分

$ \int_{\ell^2} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0) = \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(s)} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0) = \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(t)} \|u_0\|^2{\mathrm d}\mu_t(u_0) $

与$ s $无关.同时,对每个$ t\in {\mathbb R} $, $ {\mathcal A}_{{\mathcal D}_{\lambda}}(t) $是$ \ell^2 $中紧集.由(2.3)和(2.5)式推得

$ \begin{eqnarray*} \int_{\ell^2} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0) = \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(t)} \|u_0\|^2{\mathrm d}\mu_t(u_0) <+\infty. \end{eqnarray*} $

因此

(3.15) $ \begin{eqnarray} && {\rm LIM}_{M\rightarrow-\infty} \frac{1}{t-M}\int_\tau^t\int_{\ell^2} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0){\mathrm d}s \\ & = & \lim\limits_{M\rightarrow-\infty} \frac{1}{t-M}\int_\tau^t\int_{\ell^2} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0){\mathrm d}s = 0. \end{eqnarray} $

同时,经过简单变换和计算我们也有

(3.16) $ \begin{eqnarray} && {\rm LIM}_{M\rightarrow-\infty} \frac{1}{\tau-(M-t+\tau)}\int_M^\tau\int_{\ell^2} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0){\mathrm d}s\\ & = & {\rm LIM}_{M\rightarrow-\infty} \frac{1}{\tau-M}\int_M^\tau\int_{\ell^2} \|U(t, s)u_0\|^2{\mathrm d}\mu_s(u_0){\mathrm d}s. \end{eqnarray} $

结合(3.13)–(3.16)式并应用Fubini定理,就有

(3.17) $ \begin{eqnarray} & & \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(t)} \|u\|^2{\mathrm d}\mu_t(u) - \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(\tau)} \|u\|^2{\mathrm d}\mu_\tau(u) \\ & = & {\rm LIM}_{M\rightarrow-\infty} \frac{1}{\tau-M}\int_M^\tau\int_{\ell^2} (\|U(t, s)u_0\|^2-\|U(\tau, s)u_0\|^2){\mathrm d}\mu_s(u_0){\mathrm d}s \\ & = & \lim\limits_{M\rightarrow-\infty} \frac{1}{\tau-M}\int_M^\tau\int_{\ell^2} (\|U(t, s)u_0\|^2-\|U(\tau, s)u_0\|^2){\mathrm d}\mu_s(u_0){\mathrm d}s \\ & = & \lim\limits_{M\rightarrow-\infty} \frac{1}{\tau-M}\int_M^\tau\int_{\ell^2}\int_\tau^t (2U(\eta, s)u_0, F(U(\eta, s)u_0, \eta)) {\mathrm d}\eta{\mathrm d}\mu_s(u_0){\mathrm d}s\\ & = & \lim\limits_{M\rightarrow-\infty} \frac{1}{\tau-M}\int_M^\tau\int_\tau^t\int_{\ell^2} (2U(\eta, s)u_0, F(U(\eta, s)u_0, \eta)){\mathrm d}\mu_s(u_0){\mathrm d}\eta{\mathrm d}s. \end{eqnarray} $

再由$ \{\mu_t\}_{t\in {\mathbb R}} $的不变性(见(3.7)式)得

(3.18) $ \begin{eqnarray} && \int_{\ell^2}(2U(\eta, s)u_0, F(U(\eta, s)u_0, \eta)){\mathrm d}\mu_s(u_0) \\ & = & \int_{\ell^2} (2U(\eta, \theta)U(\theta, s)u_0, F(U(\eta, \theta)U(\theta, s)u_0, \eta)){\mathrm d}\mu_s(u_0) \\ & = & \int_{\ell^2}(2U(\eta, \theta)u_0, F(U(\eta, \theta)u_0, \eta)){\mathrm d}\mu_\theta(u_0). \end{eqnarray} $

(3.18)式的右端与$ s $无关.因此,由$ \{\mu_t\}_{t\in {\mathbb R}} $的不变性,且注意到$ \mu_\eta $的支集包含于$ {\mathcal A}_{{\mathcal D}_\lambda}(\eta) $中,得

$ \begin{eqnarray*} && \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(t)}\|u\|^2{\mathrm d}\mu_t(u) - \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(\tau)}\|u\|^2{\mathrm d}\mu_\tau(u) \nonumber\\ & = & \int_\tau^t\int_{\ell^2} (2U(\eta, \theta)u_0, F(U(\eta, \theta)u_0, \eta)){\mathrm d}\mu_\theta(u_0)\nonumber\\ & = & \int_\tau^t \int_{\ell^2} (2u, F(u, \eta)){\mathrm d}\mu_\eta(u){\mathrm d}\eta \nonumber\\ & = & \int_\tau^t \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(\eta)} (2u, F(u, \eta)){\mathrm d}\mu_\eta(u){\mathrm d}\eta, \, \forall\, t\geqslant \tau. \end{eqnarray*} $

证明完毕.

据定理3.2,我们易得下面推论.

推论3.1  设条件$ {\rm (H)} $成立,且$ \Phi\in {\mathcal T} $满足$ \Phi'(u) = 0 $, $ u\in \ell^2 $.则有

(3.19) $ \begin{eqnarray} \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(t)}\Phi(u){\mathrm d}\mu_t(u) = \int_{{\mathcal A}_{{\mathcal D}_{\lambda}}(\tau)} \Phi(u) {\mathrm d}\mu_\tau(u), \, \forall\, t\geqslant \tau. \end{eqnarray} $

在统计物理中,满足$ \Phi'(u) = 0 $ ($ u\in \ell^2 $)的$ \Phi\in {\mathcal T} $称为系统的统计平衡点(见文献[27, p177]),它意味着系统的统计信息不再随时间演化而变化,此时, (3.19)式也可直接由(3.7)式推得,它表明系统的的拉回吸引子可以随时间演化而变形,但拉回吸引子的相体积(测度)是守恒的.