著名的界面扩散模型Cahn-Hilliard-Navier-Stokes(CHNS)系统是由Cahn-Hilliard方程和Navier-Stokes方程耦合而成, 可用来模拟两个不相溶和不可压缩流体的等温混合物的相分离, 参见文献[1-6]及其中的参考文献.过去的几十年里, 在区域$ \Omega $、系数、非线性项$ f $的各种假设下, Cahn-Hilliard-Navier-Stokes系统已经被广泛研究^[7-11], 但耦合的均为低阶的Cahn-Hilliard方程.

关于高阶Cahn-Hilliard方程以及相应耦合的CHNS系统的研究也有很多成果^[12-19]. Cherfils等^[12]得到了高阶各向同性的Cahn-Hilliard方程的适定性结果和全局吸引子.文献[13]证明了具有动态边界条件的高阶Cahn-Hilliard方程的适定性和耗散性.文献[14]证明了高阶各向同性的Cahn-Hilliard-Navier-Stokes系统弱解的存在唯一性.在此基础上, Pan等在文献[15]中也从各向同性的角度提升了高阶CHNS系统的弱解关于时间的正则性结果.另一方面, Caginalp和Esenturk在文献[16]中提出了高阶Cahn-Hilliard系统来考虑各向异性界面.关于高阶各向异性的Cahn-Hilliard方程, 文献[17]中得到了其适定性结果以及全局吸引子的存在性.一般高阶的Cahn-Hilliard方程和修正高阶各向异性的Cahn-Hilliard方程的适定性结果分别在文献[18-19]中得到.

虽然高阶Cahn-Hilliard方程和高阶各向同性的CHNS系统已经被广泛研究, 但是关于高阶各向异性的CHNS系统的研究很少.基于此, 本文研究由高阶各向异性的Cahn-Hilliard方程和不可压缩的Navier-Stokes方程耦合而成的CHNS系统.我们假设两种流体有不可忽略的温差, 相同的常数密度(等于$ 1 $), 两相之间的扩散界面的厚度很小但不为$ 0 $, 即考虑如下的初边值问题

(1.1) $ \left\{ {\begin{array}{*{20}{l}}{{\partial _t}\boldsymbol{u} + \boldsymbol{u} \cdot \nabla \boldsymbol{u} - \nu \Delta \boldsymbol{u} + \nabla p = K\mu \nabla \phi + \boldsymbol{g}, }&{(x, t) \in \Omega \times (0, + \infty ), }\\{\nabla \cdot \boldsymbol{u} = 0, }&{(x, t) \in \Omega \times (0, + \infty ), }\\{{\partial _t}\phi + u \cdot \nabla \phi - {{\varrho}_0}\Delta \mu = 0, }&{(x, t) \in \Omega \times (0, + \infty ), }\\{\mu = \sum\limits_{i = 1}^k {{{( - 1)}^i}} \sum\limits_{|\alpha | = i} {{a_\alpha }} {D^{2\alpha }}\phi + f(\phi ), }&{(x, t) \in \Omega \times (0, + \infty ), }\\{\boldsymbol{u} = 0, \;{D^\alpha }\phi = 0, \;|\alpha | \le k, }&{(x, t) \in \partial \Omega \times (0, + \infty ), }\\{\boldsymbol{u}(0) = {\boldsymbol{u}_0}, \;\phi (0) = {\phi _0}, }&{x \in \Omega , }\end{array}} \right. $

其中$ \Omega\subset{\Bbb R}^2 $是一个有界区域且边界$ \Gamma $光滑, $ \boldsymbol{g} $是与时间无关的外力.序参数$ \phi $代表两种液体的相对浓度, $ \boldsymbol{u} $代表流体速度, 正常数$ \nu $, $ \varrho_0 $和$ {\cal K} $相应代表流体的粘性系数, 流动系数, 应力系数. $ \mu $为以下自由能的变分导数

$ {\cal F}(\phi) = \int_{\Omega} \bigg(\frac{1}{2}\sum\limits_{i = 1}^k\sum\limits_{|\alpha| = i}a_\alpha|{\cal D}^{\alpha}\phi|^2+F(\phi)\bigg){\rm d}x, \ k\in {\Bbb N}, $

其中$ \alpha = (k_1, k_2)\in({\Bbb N}\cup{0})^2 $, $ |\alpha| = k_1+k_2 $, 当$ \alpha\neq(0, 0) $时, 有

$ {\cal D}^\alpha = \frac{\partial^{|\alpha|}}{\partial x_1^{k_1}\partial x_2^{k_2}}. $

特别的, $ {\cal D}^{(0, 0)}\phi = \phi $. $ F(r) = \int_0^rf(\xi){\rm d}\xi $, 通常取多项式近似$ F(r) = C_1r^4-C_2r^2 $, 其中$ C_1 $, $ C_2 $为正常数^[20].

本文我们将讨论高阶各向异性的CHNS系统(1.1)的弱解.第2节给出相应的假设, 介绍所需的空间和算子, 同时给出系统(1.1)弱解的定义; 第3节给出了解的能量估计, 通过Galerkin方法证明弱解的存在性, 并得到弱解的唯一性.

为了本文研究, 假设$ \varrho_0 = 1 $, $ f\in C^2({\Bbb R}) $并且满足

(2.1) $ \begin{equation} \left\{\begin{array}{ll} \liminf\limits_{\mid r \mid\rightarrow +\infty}f'(r)>0, \ f(0) = 0, \\ | f''(r)| \leq c_f(1+| r |^{m-1}), \ \forall r\in {\Bbb R}, \\ F(r)\geq c_1r^4-c_2, \ \exists c_1>0, \ \exists c_2\geq0, \ \forall r\in {\Bbb R}, \end{array}\right. \end{equation} $

其中$ c_f $为正常数, $ m\geq2 $.由(2.1)式可得

(2.2) $ \begin{equation} | f'(r)|\leq c_f(1+| r |^m), \ \ | f(r)|\leq c_f(1+| r |^{m+1}), \ \forall r\in {\Bbb R}, \end{equation} $

其中$ f $为$ F $的导数.特别的, $ f(r) = r^3-r $满足(2.1)式.

假设$ X $为一个实的Hilbert空间, 内积和范数分别为$ (\cdot, \cdot)_X $和$ \mid\cdot\mid_X $, $ X^* $表示$ X $的对偶空间.空间$ {\Bbb X} $表示$ X \times X $, 现定义以下两个Hilbert空间

$ {\rm{\mathbb{H}}} = {\overline {\{ \boldsymbol{u} \in C_c^\infty (\Omega ):{\rm{div}}\boldsymbol{u} = 0, \;x \in \Omega \} } ^{{{\rm{\mathbb{L}}}^2}}}, {\rm{\mathbb{V}}} = {\overline {\{ \boldsymbol{u} \in C_c^\infty (\Omega ):{\rm{div}}\boldsymbol{u} = 0, \;x \in \Omega \} } ^{{\rm{\mathbb{H}}}_0^1}} $

其中$ {\Bbb L}^2(\Omega) = (L^2(\Omega))^2 $, $ {\Bbb H}_0^1(\Omega) = (H_0^1(\Omega))^2 $.空间$ {\Bbb H} $的内积和范数分别为$ (\cdot, \cdot) $和$ \mid\cdot\mid $.空间$ {\Bbb V} $的内积和范数表示如下

$ ((\boldsymbol{u}, \boldsymbol{v})) = \sum\limits_{i = 1}^2 {({\partial _{{x_i}}}\boldsymbol{u}, {\partial _{{x_i}}}\boldsymbol{v})} , \;\;\left\| \boldsymbol{u} \right\| = {((\boldsymbol{u}, \boldsymbol{u}))^{1/2}}. $

由Poincaré不等式可知, $ {\Bbb V} $的范数等价于$ {\Bbb H}_0^1(\Omega) $-范数.

此外, 假设$ k\in{\Bbb N} $, $ k\geq2 $, 当$ |\alpha| = k $时$ a_\alpha>0 $, 则映射

$ (\upsilon, \omega)\in H_0^k(\Omega)\times H_0^k(\Omega)\mapsto\sum\limits_{|\alpha| = k}a_\alpha({\cal D}^\alpha\upsilon, {\cal D}^\alpha\omega) $

是对称连续强制的双线性形式, 因此定义一个从$ H_0^k $到$ H^{-k} $的椭圆算子$ A_k $

$ \langle A_k\upsilon, \omega\rangle_{(H^{-k}, H_0^{k})} = \sum\limits_{| \alpha| = k}a_\alpha({\cal D}^\alpha\upsilon, {\cal D}^\alpha\omega), \ \ \forall\upsilon, \ \omega\in H_0^{k}(\Omega), $

其中$ H^{-k}(\Omega) = (H^k(\Omega))^* $.由$ 2k $阶线性椭圆算子的正则性结果^[21]知, $ A_k $是一个无界线性自伴的正算子且有紧逆.它的定义域如下

$ D(A_k) = H^{2k}(\Omega)\cap H_0^k(\Omega). $

对$ \forall\upsilon\in D(A_k) $, 有

$ A_k\upsilon = (-1)^k\sum\limits_{|\alpha| = k}a_\alpha{\cal D}^{2\alpha}\upsilon. $

由Poincaré不等式易得$ D(A_k^{1/2}) = H_0^k(\Omega) $且对$ \forall(\upsilon, \omega) \in D(A_k^{1/2})\times D(A_k^{1/2}) $, 有

$ (A_k^{1/2}\upsilon, A_k^{1/2}\omega) = \sum\limits_{|\alpha| = k}a_\alpha({\cal D}^\alpha\upsilon, {\cal D}^\alpha\omega), $

则$ | A_k\cdot| $等价于$ H^{2k} $范数, 相应的, $ | A_k^{1/2}\cdot| $等价于$ H^k $范数.

记

$ \sum\limits_{i = 1}^k(-1)^i\sum\limits_{\mid\alpha\mid = i}a_\alpha{\cal D}^{2\alpha}\phi = A_k\phi+B_k\phi, $

其中$ B_k\phi = \sum\limits_{i = 1}^{k-1} (-1)^{i} \sum\limits_{|\alpha| = i}a_\alpha{\cal D}^{2\alpha}\phi $.

定义相空间如下

$ {\Bbb Y} = {\Bbb H}\times H^k(\Omega). $

$ {\Bbb Y} $关于如下范数是一个完备的度量空间

$ \left\| {\left( {\boldsymbol{u}, \phi } \right)} \right\|_{\rm{\mathbb{Y}}}^2: = \frac{1}{\mathcal{K}}{\left| \boldsymbol{u} \right|^2} + \sum\limits_{|\alpha | = k} {{a_\alpha }} |{\mathcal{D}^\alpha }\phi {|^2}. $

由于对$ \forall \boldsymbol{\upsilon} \in {\Bbb V} $, $ \left( {\nabla p, \boldsymbol{\upsilon} } \right) = 0 $, 故系统(1.1)的变分形式如下

(2.3) $ \left\{ {\begin{array}{*{20}{l}}{({\partial _t}\boldsymbol{u}, \boldsymbol{\upsilon}) + \nu (\nabla \boldsymbol{u}, \nabla \boldsymbol{\upsilon}) + (\boldsymbol{u} \cdot \nabla \boldsymbol{u}, \boldsymbol{\upsilon}) - \mathcal{K}(\mu \nabla \phi , \boldsymbol{\upsilon}) = (\boldsymbol{g}, \boldsymbol{\upsilon}), \;\;}\\{({\partial _t}\phi , \varphi ) + (\boldsymbol{u} \cdot \nabla \phi , \varphi ) + (\nabla \mu , \nabla \varphi ) = 0, \;\;}\\{\sum\limits_{i = 1}^k {\sum\limits_{\mid \alpha \mid = i} {{a_\alpha }} } ({\mathcal{D}^\alpha }\phi , {\mathcal{D}^\alpha }\xi ) + (f(\phi ), \xi ) - (\mu , \xi ) = 0, \;}\end{array}} \right. $

其中$ \boldsymbol{\upsilon} \in {\Bbb V} $, $ \varphi\in H^1(\Omega) $, $ \xi\in H_0^k(\Omega) $, 且满足$ \boldsymbol{u}\left( 0 \right) = {\boldsymbol{u}_0} $$ \phi \left( 0 \right) = {\phi _0} $.

定义2.1   令$ \boldsymbol{g} \in {\Bbb V}^* $, $ f\in C^2({\Bbb R}) $且满足(2.1)式.若初值$ \left( {{\boldsymbol{u}_0}, {\phi _0}} \right) \in {\Bbb Y} $, 并且

$ \left( {{\boldsymbol{u}_0}, {\phi _0}} \right)\in C([0, +\infty);{\Bbb Y})\cap L_{loc}^2([0, +\infty);{\Bbb V}\times (D(A_k)\cap H^{2k+1}(\Omega))), $

$ (\partial _{t} \boldsymbol{u}, \partial _{t}\phi)\in L_{loc}^2([0, +\infty);{\Bbb V}^*\times H^{-1}(\Omega)) $

满足(2.3)式, 则称$ \left( {{\boldsymbol{u}_0}, {\phi _0}} \right) $为系统(1.1)的弱解.

为了得到弱解的存在性, 先进行先验估计得到以下结论.

定理3.1   令$ \boldsymbol{g} \in {\Bbb V}^* $, $ f\in C^2({\Bbb R}) $满足(2.1)式.如果$ \left( {{\boldsymbol{u}_0}, {\phi _0}} \right) $是系统(1.1)的弱解, 那么以下估计成立

(3.1) $ \begin{eqnarray} &&\|(\boldsymbol{u}(t), \phi(t))\|_{{\Bbb Y}}^2+\int_t^{t+1}(\frac{\nu}{{\cal K}}\| \boldsymbol{u}(s)\|^2+\|\mu(s)\|^2+| F(\phi(s))|_{L^1}){\rm d}s\\ &&+\int_t^{t+1}(\| \partial_t \boldsymbol{u}(s)\|_{{\Bbb V}^*}^2+\|\phi(s)\|_{H^{2k+1}}^2 +| \partial_t \phi(s)|_{H^{-1}}^2){\rm d}s\\ &\leq & Q(\|(\boldsymbol{u}(0), \phi(0))\|_{{\Bbb Y}}^2) e^{-\kappa t}+C_0(\nu, {\cal K}, \| \boldsymbol{g}\|_{{\Bbb V}^*}), \ \forall t\geq0, \end{eqnarray} $

其中单增非减函数$ Q $, 正常数$ \kappa $和$ C_0 $都与时间和初值无关.

证   在方程(2.3)₁中取$ \boldsymbol{\upsilon} = {\cal K}^{-1}\boldsymbol{u} $, 由$ \left( {\boldsymbol{u} \cdot \nabla \boldsymbol{u}, \boldsymbol{u}} \right) = 0 $可得

(3.2) $ \begin{equation} \frac{1}{2{\cal K}}\frac{\rm d}{{\rm d}t}|\boldsymbol{u}(t)|^2+\frac{\nu}{{\cal K}}\|\boldsymbol{u}(t)\|^2-(\mu\nabla\phi, \boldsymbol{u}(t)) = \frac{1}{{\cal K}}(\boldsymbol{u}(t), \boldsymbol{g}). \end{equation} $

在方程(2.3)₂中取$ \varphi = \mu $可得

(3.3) $ \begin{equation} (\partial_t\phi(t), \mu(t))+(\boldsymbol{u}(t)\cdot\nabla\phi(t), \mu(t))+|\nabla\mu(t)|_{L^2}^2 = 0. \end{equation} $

在方程(2.3)₃中取$ \xi = \partial_t\phi(t) $得到

(3.4) $ \begin{equation} \frac{1}{2}\frac{\rm d}{{\rm d}t}(|A_k^{1/2}\phi(t)|^2+B_k^{1/2}[\phi(t)]+2(F(\phi(t)), 1))-(\mu(t), \partial_t\phi(t)) = 0, \end{equation} $

其中$ B_k^{1/2}[\phi(t)] = \sum\limits_{i = 1}^{k-1} \sum\limits_{|\alpha| = i}a_\alpha| {\cal D}^\alpha\phi(t)|^2 $.由Gagliardo-Nirenberg不等式^[22]知, 对$ i\in\{1, \cdots, k-1\} $, $ k\in {\Bbb N} $, $ k\geq 2 $, 有如下不等式

$ \| \phi(t)\|_{H^i(\Omega)}\leq c(i)\| \phi(t)\|_{H^k(\Omega)}^{\frac{i}{k}}| \phi(t)|_{L^2}^{1-\frac{i}{k}}, \ \phi \in H^k(\Omega). $

由Young不等式得, 对$ \forall i\in\{1, \cdots, k-1\} $, 有

$ \label{eq:12} \| \phi(t)\|_{H^i(\Omega)}^2\leq \epsilon_i\| \phi(t)\|_{H^k(\Omega)}^2+c(\epsilon_i)| \phi(t)|_{L^2}^2, \ \phi \in H^k(\Omega), \ \forall\epsilon_i>0. $

故

$ \begin{eqnarray*} | B_k^{1/2}[\phi(t)]|&\leq&\sum\limits_{i = 1}^{k-1} \sum\limits_{|\alpha| = i}|a_\alpha|| {\cal D}^\alpha\phi(t)|^2\\ &\leq &\epsilon_1\| \phi(t)\|_{H^k(\Omega)}^2+c(\epsilon_1)| \phi(t)|_{L^2}^2+\cdots+\epsilon_{k-1}\| \phi(t)\|_{H^k(\Omega)}^2+c(\epsilon_{k-1})| \phi(t)|_{L^2}^2\\ &\leq&(\epsilon_1+\cdots+\epsilon_{k-1})\| \phi(t)\|_{H^k(\Omega)}^2+(c(\epsilon_1)+\cdots+c(\epsilon_{k-1}))| \phi(t)|_{L^2}^2.\\ \end{eqnarray*} $

当$ \epsilon_i $取合适的值时, 可得

(3.5) $ \begin{equation} | B_k^{1/2}[\phi(t)]|\leq\frac{1}{2}| A_k^{1/2}\phi(t)|^2+c|\phi(t)|_{L^2}^2. \end{equation} $

由(3.2)–(3.4)式得到

(3.6) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}[\frac{1}{{\cal K}}| \boldsymbol{u}(t) |^2+|A_k^{1/2}\phi(t)|^2+B_k^{1/2}[\phi(t)]+2(F(\phi(t)), 1)]\\ &&+\frac{2\nu }{{\cal K}}\| \boldsymbol{u}(t)\|^2+2|\nabla\mu(t)|_{L^2}^2-\frac{2}{{\cal K}}(\boldsymbol{u}(t), \boldsymbol{g}) = 0, \end{eqnarray} $

再在方程(2.3)₃中取$ \xi = 2\zeta\phi(t) $, $ \zeta>0 $, 得到

(3.7) $ \begin{equation} 2\zeta(\mu(t), \phi(t))_{L^2} = 2\zeta| A_k^{1/2}\phi(t)|_{L^2}^2+2\zeta B_k^{1/2}[\phi(t)]+2\zeta(\phi(t), f(\phi(t)))_{L^2}. \end{equation} $

令

$ E(t) = \| (\boldsymbol{u}(t), \phi(t))\|_{{\Bbb Y}}^2+B_k^{1/2}[\phi(t)]+2(F(\phi(t)), 1)+c_E, $

其中$ c_E $为足够大的常数确保$ E(t) $非负.此外, 存在一个单增非减的函数$ Q $使得

(3.8) $ \begin{equation} \|(\boldsymbol{u}(t), \phi(t))\|_{{\Bbb Y}}^2\leq E(t) \leq Q(\|(\boldsymbol{u}(t), \phi(t))\|_{{\Bbb Y}}^2). \end{equation} $

由(3.6)式和(3.7)式得到

(3.9) $ \begin{equation} \frac{\rm d}{{\rm d}t}E(t)+\kappa E(t) = \Lambda_1(t), \end{equation} $

其中$ \kappa\in(0, \zeta) $,

(3.10) $ \begin{eqnarray} \Lambda_1(t):& = &-2\frac{\nu}{{\cal K}}\|\boldsymbol{u}(t)\|_{\Bbb V}^2+\frac{\kappa}{{\cal K}}|\boldsymbol{u}(t)|^2-2|\nabla\mu(t)|_{L^2}^2-(2\zeta-k)| A_k^{1/2}\phi(t)|^2+\frac{2}{{\cal K}}(\boldsymbol{u}(t), \boldsymbol{g})\\ & &+2[\kappa(F(\phi(t))-f(\phi(t))\phi(t), 1)_{L^2}-(\zeta-\kappa)(f(\phi(t))\phi(t), 1)_{L^2}]\\ & &+2\zeta(\mu(t), \phi(t))-(2\zeta-\kappa)B_k^{1/2}[\phi(t)]+\kappa c_E. \end{eqnarray} $

现对(3.10)式的右边进行估计, 由Hölder, Poincaré和Young不等式有

(3.11) $ \begin{eqnarray} 2\zeta(\mu(t), \phi(t))_{L^2}&\leq& 2\zeta |\mu(t)|_{L^2}|\phi(t)|_{L^2}\leq 2\zeta c_\Omega^{1/2}|\Omega|^{1/2}|\nabla\mu(t)|_{L^2}|\phi(t)|_{L^2}{}\\ &\leq&|\nabla\mu(t)|_{L^2}^2+\zeta^2c_\Omega|\Omega| |\phi(t)|_{L^2}^2, \end{eqnarray} $

其中$ c_\Omega $与$ \Omega $有关.

由Hölder, Young不等式可得

(3.12) $ \begin{equation} \frac{2}{{\cal K}}(\boldsymbol{u}(t), \boldsymbol{g})\leq\frac{2}{{\cal K}}\|\boldsymbol{u}(t)\|_{\Bbb V}\|\boldsymbol{g}\|_{{\Bbb V^*}}\leq\frac{\nu}{{\cal K}}\|\boldsymbol{u}(t)\|_{\Bbb V}^2+\frac{1}{{\cal K}\nu}\|\boldsymbol{g}\|_{{\Bbb V^*}}^2. \end{equation} $

另一方面, 通过(2.1)式, 对$ \forall r\in{\Bbb R} $, 有

(3.13) $ \begin{equation} c_*| f(r)|(1+|r|)\leq 2f(r)r + c_f, \end{equation} $

(3.14) $ \begin{equation} F(r)-f(r)r\leq c_f^{'}r^2+c_f^{''}, \end{equation} $

(3.15) $ \begin{equation} |F(r)|-c_0\leq | f(r)|(1+| r|), \end{equation} $

其中$ c_* $, $ c_f $, $ c_f^{'} $, $ c_f^{''} $和$ c_0 $为正的足够大的常数.

由(3.5)式, (3.11)–(3.14)式和Poincaré不等式, 得到

(3.16) $ \begin{eqnarray} \Lambda_1(t)&\leq &-\frac{1}{{\cal K}}(\nu-\kappa c_\Omega|\Omega|)\|\boldsymbol{u}(t)\|_{\Bbb V}^2-|\nabla\mu(t)|_{L^2}^2-\frac{1}{2}(2\zeta-k)| A_k^{1/2}\phi(t)|^2{}\\ & &+\frac{1}{{\cal K}\nu}\|\boldsymbol{g}\|_{{\Bbb V^*}}^2+(2\kappa c_f^{'}+\zeta^2c_\Omega|\Omega|+(2\zeta-k)c)|\phi(t)|_{L^2}^2{}\\ & &-c_*(\zeta-\kappa)(| f(\phi(t))|, 1+| \phi(t)|)+c_1, \end{eqnarray} $

其中$ c_1 = \kappa c_E+2\kappa c_f^{''}|\Omega|+c_f(\zeta-\kappa)|\Omega| $.

由Young不等式, (2.1)$ _3 $式和(3.15)式有

$ \begin{eqnarray*} \label{eq:3.74} |\phi(t)|_{L^2}^2&\leq& \epsilon|\phi(t)|_{L^4}^4+c(\epsilon)\leq\epsilon(F(\phi(t)), 1)+c(\epsilon)\\ &\leq&\epsilon(| f(\phi(t))|, 1+| \phi(t)|)+c(\epsilon), \ \forall\epsilon>0. \end{eqnarray*} $

当$ \epsilon $取适当的值, 使下式成立

(3.17) $ \begin{equation} (2\kappa c_f^{'}+\xi^2c_\Omega|\Omega|+(2\zeta-k)c)|\phi(t)|_{L^2}^2\leq\frac{1}{2}c_*(\zeta-\kappa)(| f(\phi(t))|, 1+| \phi(t)|)+c. \end{equation} $

因此, 由(3.16)式和(3.17)式可得

(3.18) $ \begin{eqnarray} \Lambda_1(t)&\leq &-\frac{1}{{\cal K}}(\nu-\kappa c_\Omega|\Omega|)\|\boldsymbol{u}(t)\|_{\Bbb V}^2-|\nabla\mu(t)|_{L^2}^2-\frac{1}{2}(2\zeta-k)| A_k^{1/2}\phi(t)|^2+\frac{1}{{\cal K}\nu}\|\boldsymbol{g}\|_{{\Bbb V^*}}^2{}\\ & &-\frac{1}{2}c_*(\zeta-\kappa)(| f(\phi(t))|, 1+| \phi(t)|)+c_1. \end{eqnarray} $

设$ \zeta = 1/c_\Omega|\Omega| $且$ \kappa\in(0, \zeta) $, 并令$ \kappa = \min\{\nu/2c_\Omega|\Omega|, 1/c_\Omega|\Omega|\} $.通过(3.9)式和(3.18)式可得

(3.19) $ \begin{eqnarray} && \frac{\rm d}{{\rm d}t}E(t)+\kappa E(t)+\kappa_1(\frac{\nu}{{\cal K}}\|\boldsymbol{u}(t)\|_{\Bbb V}^2 +| A_k^{1/2}\phi(t)|^2)+|\nabla\mu(t)|_{L^2}^2 +\kappa_2(| f(\phi(t))|, 1+| \phi(t)|)_{L^2} {}\\ &\leq& \frac{1}{{\cal K}\nu}\|\boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1. \end{eqnarray} $

由Gronwall不等式(参见文献[23, 引理2.5]), 可得

(3.20) $ \begin{eqnarray} E(t)&\leq&2E(0)e^{-kt}+2\int_0^t(\frac{1}{{\cal K}\nu}\| \boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1)e^{-k(t-y)}{\rm d}y\\ &\leq&2E(0)e^{-kt}+2\kappa^{-1}(\frac{1}{{\cal K}\nu}\| \boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1), \ \forall t\geq 0. \end{eqnarray} $

对(3.19)式在$ [t, t+1] $上积分可得

(3.21) $ \begin{eqnarray} && E(t+1)+\int_t^{t+1}\kappa E(s)+\kappa_1(\frac{\nu}{{\cal K}}\|\boldsymbol{u}(s)\|_{{\Bbb V}}^2 \\ &&+| A_k^{1/2}\phi(s)|^2)+|\nabla\mu(s)|_{L^2}^2+\kappa_2(| f(\phi(s))|, 1+| \phi(s)|)_{L^2}{\rm d}s{}\\ &\leq& E(t)+\frac{1}{{\cal K}\nu}\| \boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1, \ \forall t\geq0. \end{eqnarray} $

通过(3.20)式和(3.21)式有

(3.22) $ \begin{eqnarray} && E(t)+\int_t^{t+1}(\kappa_1(\frac{\nu}{{\cal K}}\|\boldsymbol{u}(s)\|_{{\Bbb V}}^2+| A_k^{1/2}\phi(s)|^2)+|\nabla\mu(s)|_{L^2}^2+\kappa_2(| f(\phi(s))|, 1+| \phi(s)|)_{L^2}){\rm d}s\\ &\leq&4E(0)e^{-\kappa t}+(4\kappa^{-1}+1)(\frac{1}{{\cal K}\nu}\|\boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1), \ \forall t\geq0. \end{eqnarray} $

于是从(3.8)式, (3.15)式和(3.22)式, 可得

(3.23) $ \begin{eqnarray} && \|(\boldsymbol{u}(t), \phi(t)\|_{{\Bbb Y}}^2+\int_t^{t+1}\frac{\nu}{{\cal K}}\|\boldsymbol{u}(s)\|_{\Bbb V}^2+| A_k^{1/2}\phi(s)| ^2+|\nabla\mu(s)|_{L^2}^2+| F(\phi(s))|_{L^1}){\rm d}s\\ &\leq& Q(\|(\boldsymbol{u}(0), \phi(0)\|_{{\Bbb Y}}^2)e^{-\kappa t}+c_2, \ \forall t\geq0, \end{eqnarray} $

其中$ c_2 = (4\kappa^{-1}+1)(\frac{1}{{\cal K}\nu}\| \boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1) $.

接下来证明(3.1)式中的其余项.显然, 从(3.23)式可得

(3.24) $ \begin{equation} \int_t^{t+1}\|\mu(s)\|_{H^1}^2{\rm d}s\leq Q(\|(\boldsymbol{u}(0), \phi(0))\|_{{\Bbb Y}}^2)e^{-\kappa t}+c_3, \ \forall t \geq 0. \end{equation} $

从(1.1)₄式, (2.2)式, (3.23)式, (3.24)式以及$ H^k\hookrightarrow L^\beta $, $ \beta\in[1, +\infty) $, 可得

$ \begin{eqnarray*} \int_t^{t+1}|A_k\phi(s)+B_k\phi(s)|_{L^2}^2{\rm d}s&\leq&\int_t^{t+1}|\mu(s)|_{L^2}^2+|f(\phi(s))|_{L^2}^2{\rm d}s\\ &\leq&\int_t^{t+1}|\mu(s)|_{L^2}^2+c_f(1+|\phi(s)|_{L^{2m+2}}^{2m+2}){\rm d}s\\ &\leq&\int_t^{t+1}c\|\mu(s)\|^2+c_f(1+\|\phi(s)\|_{H^k}^{2m+2}){\rm d}s\\ &\leq& Q(\|(\boldsymbol{u}(0), \phi(0))\|_{{\Bbb Y}}^2)e^{-\kappa t}+c_4, \ \forall t\geq 0. \end{eqnarray*} $

通过(1.1)₄式, (3.24)式和正则性结果, 有

(3.25) $ \begin{equation} \int_t^{t+1}\|\phi(s)\|_{H^{2k+1}}^2{\rm d}s\leq Q(\|(\boldsymbol{u}(0), \phi(0))\|_{{\Bbb Y}}^2)e^{-\kappa t}+c_5, \ \forall t \geq 0. \end{equation} $

此外, 从(1.1)₃式, (3.23)式和(3.24)式, 可得

(3.26) $ \begin{eqnarray} \int_t^{t+1}\|\partial_t\phi(s)\|_{H^{-1}}^2{\rm d}s &\leq &\int_t^{t+1}(\| \Delta\mu(s)\|_{H^{-1}}^2+\| \boldsymbol{u}(s)\cdot\nabla\phi(s)\|_{H^{-1}}^2){\rm d}s\\ &\leq &\int_t^{t+1}(\|\mu(s)\|^2+c_\Omega \|\boldsymbol{u}(s)\|_{\Bbb V}^2\|\phi(s)\|^2){\rm d}s\\ &\leq & Q(\|(\boldsymbol{u}(0), \phi(0))\|_{{\Bbb Y}}^2)e^{-\kappa t}+c_6, \quad\forall t \geq 0. \end{eqnarray} $

因此, $ \partial_t\phi\in L^2([t, t+1];H^{-1}(\Omega)) $.

为了得到$ \partial_t\boldsymbol{u} $在$ L^2([t, t+1];{\Bbb V}^*) $中一致有界, 由

(3.27) $ \begin{equation} \|\boldsymbol{u}\cdot\nabla\boldsymbol{u}\|_{{\Bbb V}^*}^2 \leq c_\Omega |\boldsymbol{u}|^2\|\boldsymbol{u}\|_{{\Bbb V}}^2, \quad\forall \boldsymbol{u} \in {\Bbb V} \end{equation} $

和

$ |(\mu\nabla\phi, \boldsymbol{v})| \leq c_7\|\boldsymbol{v}\|\|\phi\|| \mu|_{L^2}^{1/2}\|\mu\|_{H^1}^{1/2}, \ \forall \boldsymbol{v} \in {\Bbb V}, \ \phi\in D(A_k)\cap H^{2k+1}(\Omega), $

可以得到

(3.28) $ \begin{equation} \| \mu\nabla\phi \|_{\Bbb V^*}^2 \leq c_8\|\phi\|^2| \mu|_{L^2}\|\mu\|_{H^{1}}. \end{equation} $

从(1.1)$ _1 $式, (3.23)式, (3.27)式和(3.28)式, 可得

(3.29) $ \begin{eqnarray} \int_t^{t+1}\|\partial_t\boldsymbol{u}\|_{{\Bbb V^*}}^2{\rm d}s&\leq &\int_t^{t+1}\nu\|\Delta\boldsymbol{u}\|_{{\Bbb V^*}}^2+\|\boldsymbol{u}\cdot\nabla\boldsymbol{u}\|_{{\Bbb V}^*}^2+{\cal K} \| \mu\nabla\phi \|_{\Bbb V^*}^2+\|\boldsymbol{g}\|_{{\Bbb V}^*}^2{\rm d}s\\ &\leq&\int_t^{t+1}\nu\|\boldsymbol{u}\|_{{\Bbb V}}^2+c_\Omega |\boldsymbol{u}|^2\|\boldsymbol{u}\|_{{\Bbb V}}^2+c_8\|\phi\|^2| \mu|_{L^2}\|\mu\|_{H^{1}}+\|\boldsymbol{g}\|_{{\Bbb V}^*}^2{\rm d}s\\ &\leq& Q(\|(\boldsymbol{u}(0), \phi(0))\|_{{\Bbb Y}}^2)e^{-\kappa t}+c_9, \quad\forall t \geq 0. \end{eqnarray} $

结合(3.23)式, (3.25)式, (3.26)式和(3.29)式, 可得(3.1)式.定理3.1证毕.

定理3.2   令$ \boldsymbol{g} \in {\Bbb V}^* $, $ f\in C^2({\Bbb R}) $且满足(2.1)式.若初值$ (\boldsymbol{u}_0, \phi_0)\in {\Bbb Y} $, 系统(1.1)存在弱解$ (\boldsymbol{u}, \phi) $.

证   将用经典的Galerkin方法来证明系统(1.1)弱解的存在性, 参见文献[24-26].

由于$ {\Bbb V}\times H^{2k}\hookrightarrow{\Bbb Y} $是紧嵌入, 设$ \{(\boldsymbol{w}_i, e_i), i = 1, 2, 3 , \cdots \}\subset{\Bbb V}\times H^{2k} $是$ {\Bbb Y} $的正交基.取定一个正整数$ n $, 构造系统(1.1)的近似解$ \{ \boldsymbol{u}^n\} $, $ \{\phi^n\} $, $ \{\mu^n\} $:

$ \begin{eqnarray*} \label{eq:102} \boldsymbol{u}_n(t) = \sum^n_{i = 1}\alpha_n^i(t)\boldsymbol{w}_i, \ \ \phi_n(t) = \sum^n_{i = 1}\beta_n^i(t)e_i, \ \ \mu_n(t) = \sum^n_{i = 1}\gamma_n^i(t)e_i \end{eqnarray*} $

满足对$ i = 1, 2, \cdots, n $, 有

(3.30) $ \begin{eqnarray} \left\{ \begin{array}{ll} (\partial_t\boldsymbol{u}_n, \boldsymbol{w}_i)+\nu( \nabla\boldsymbol{u}_n, \nabla \boldsymbol{w}_i)+(\boldsymbol{u}_n\cdot\nabla\boldsymbol{u}_n, \boldsymbol{w}_i)-{\cal K}(\mu_n\nabla\phi_n, \boldsymbol{w}_i) = (\boldsymbol{g}_n, \boldsymbol{w}_i), \\ (\partial_t\phi_n, e_i) +(\boldsymbol{u}_n\cdot\nabla\phi_n, e_i)+(\nabla\mu_n, \nabla e_i) = 0, \\ { } (\mu_n, e_i) = \sum\limits_{i = 1}^k\sum\limits_{\mid\alpha\mid = i}a_\alpha({\cal D}^{\alpha}\phi_n, {\cal D}^{\alpha}e_i)+(f(\phi_n), e_i), \\ { } \boldsymbol{u}_n(0) = \sum\limits_{i = 1}^n(\boldsymbol{u}_0, \boldsymbol{w}_i)\boldsymbol{w}_i, \ \phi_n(0) = \sum\limits_{i = 1}^n(\phi_0, e_i)e_i. \end{array} \right. \end{eqnarray} $

其中当$ n\rightarrow\infty $时, 在$ {\Bbb Y} $中$ (\boldsymbol{u}_n(0), \phi_n(0))\rightarrow(\boldsymbol{u}_0, \phi_0) $.

实际上, (3.30)式是关于函数$ \alpha_n^i(t) $, $ \beta_n^i(t) $, $ \gamma_n^i(t) $, $ (i = 1, 2, \cdots , n) $的非线性常微分方程组的柯西问题.由常微分方程组解的存在性基本理论知, 方程(3.30)在$ [0, t_n] $上存在唯一解$ \boldsymbol{u}_n(t) $, $ \phi_n(t) $, $ t_n>0 $.

由定理3.1知, 存在与$ n $无关的$ T $, 解可以延拓到整个区间$ [0, T] $.还可得序列$ \{(\boldsymbol{u}_n, \phi_n)\} $在$ L^\infty([0, +\infty);{\Bbb Y})\cap L^2([t, t+1];{\Bbb V}\times (D(A_k)\cap H^{2k+1})) $中有界; $ \{(\partial_t\boldsymbol{u}_n, \partial_t\phi_n)\} $在$ L^2([t, t+1];{\Bbb V^*}\times H^{-1}(\Omega)) $中有界; $ \{\mu_n\} $在$ L^2([t, t+1];H^1(\Omega)) $中有界.于是可取出子序列依然用$ \{\boldsymbol{u}_n\}, \{\phi_n\} $, $ \{\mu_n\} $表示, 使得

$ \begin{eqnarray*} \label{eq:103} \left\{ \begin{array}{ll} (\boldsymbol{u}_n, \phi_n)\rightarrow(\boldsymbol{u}, \phi) \ \ \mbox{在$ L^\infty([0, +\infty);{\Bbb Y})$中弱*收敛}, \\ (\boldsymbol{u}_n, \phi_n)\rightarrow(\boldsymbol{u}, \phi)\ \mbox{在$ L^2([t, t+1];{\Bbb V}\times (D(A_k)\cap H^{2k+1}(\Omega))$中弱收敛}, \\ (\partial_t\boldsymbol{u}_n, \partial_t\phi_n)\rightarrow(\partial_t\boldsymbol{u}, \partial_t\phi)\ \mbox{在$ L^2([t, t+1];{\Bbb V^*}\times H^{-1}(\Omega))$中弱收敛}, \\ \mu_n\rightarrow\mu\ \mbox{在$ L^2([t, t+1]; H^1(\Omega))$中弱收敛}. \end{array} \right. \end{eqnarray*} $

固定$ N(n\geq N) $, 选取$ \boldsymbol{v}_N\in{\Bbb V} $, $ \psi_N\in H^1 $和 $ \xi_N\in H^k $如下

$ \begin{eqnarray*} \boldsymbol{v}_N(t) = \sum^N_{i = 1}\alpha_N^i(t)\boldsymbol{w}_i, \ \ \psi_N(t) = \sum^N_{i = 1}\beta_N^i(t)e_i, \ \ \xi_N(t) = \sum^N_{i = 1}\gamma_N^i(t)e_i. \end{eqnarray*} $

用$ \alpha_N^i(t) $, $ \beta_N^i(t) $, $ \gamma_N^i(t) $分别与(3.30)$ _1 $式, (3.30)$ _2 $式和(3.30)$ _3 $式相乘, 并从$ 1 $到$ N $求和, 可得

$ \begin{eqnarray*} \label{eq:301} \left\{ \begin{array}{ll} (\partial_t\boldsymbol{u}_n, \boldsymbol{v}_N)+\nu( \nabla\boldsymbol{u}_n, \nabla\boldsymbol{v}_N)+(\boldsymbol{u}_n\cdot\nabla\boldsymbol{u}_n, \boldsymbol{v}_N)-{\cal K}(\mu_n\nabla\phi_n, \boldsymbol{v}_N) = (\boldsymbol{g}_n, \boldsymbol{v}_N), \\ (\partial_t\phi_n, \varphi_N) +(\nabla\mu_n, \nabla\varphi_N) +(\boldsymbol{u}_n\cdot\nabla\phi_n, \varphi_N) = 0, \\ { } (\mu_n, \xi_N) = \sum\limits_{i = 1}^k\sum\limits_{\mid\alpha\mid = i}a_\alpha({\cal D}^{\alpha}\phi_n, {\cal D}^{\alpha}\xi_N)+(f(\phi_n), \xi_N). \end{array} \right. \end{eqnarray*} $

令$ n\rightarrow\infty $可得

$ \begin{eqnarray*} \label{eq:402} \left\{ \begin{array}{ll} (\partial_t\boldsymbol{u}, \boldsymbol{v}_N)+\nu( \nabla\boldsymbol{u}, \nabla\boldsymbol{v}_N)+(\boldsymbol{u}\cdot\nabla\boldsymbol{u}, \boldsymbol{v}_N)-{\cal K}(\mu\nabla\phi, \boldsymbol{v}_N) = (\boldsymbol{g}, \boldsymbol{v}_N), \\ (\partial_t\phi, \varphi_N) +(\nabla\mu, \nabla\varphi_N) +(\boldsymbol{u}\cdot\nabla\phi, \varphi_N) = 0, \\ { } (\mu, \xi_N) = \sum\limits_{i = 1}^k\sum\limits_{\mid\alpha\mid = i}a_\alpha({\cal D}^{\alpha}\phi, {\cal D}^{\alpha}\xi_N)+(f(\phi), \xi_N). \end{array} \right. \end{eqnarray*} $

令$ N\rightarrow\infty $, 可得$ \boldsymbol{v} \in{\Bbb V} $, $ \varphi\in H^1 $和$ \xi\in H_0^k $满足

$ \begin{eqnarray*} \label{eq:800} \left\{ \begin{array}{ll} (\partial_t\boldsymbol{u}, \boldsymbol{v})+(\nu \nabla\boldsymbol{u}, \nabla\boldsymbol{v})+(\boldsymbol{u}\cdot\nabla\boldsymbol{u}, \boldsymbol{v})-{\cal K}(\mu\nabla\phi, \boldsymbol{v}) = (\boldsymbol{g}, \boldsymbol{v}), \\ (\partial_t\phi, \varphi) +(\nabla\mu, \nabla\varphi)+(\boldsymbol{u}\cdot\nabla\phi, \varphi) = 0, \\ { } (\mu, \xi) = \sum\limits_{i = 1}^k\sum\limits_{\mid\alpha\mid = i}a_\alpha({\cal D}^{\alpha}\phi, {\cal D}^{\alpha}\xi)+(f(\phi), \xi). \end{array} \right. \end{eqnarray*} $

因此, 系统(1.1)存在弱解.定理3.2证毕.

通过以下定理, 可以得到系统(1.1)的弱解关于初值的连续依赖性.

定理3.3   令$ \boldsymbol{g} \in {\Bbb V}^* $, $ f\in C^2({\Bbb R}) $且满足(2.1)式.令$ (\boldsymbol{u}_i, \phi_i) $是系统(1.1)的两个解, 分别对应的初值为$ (\boldsymbol{u}_i(0), \phi_i(0))\in{\Bbb Y} $, $ i = 1, 2 $, 则有如下估计

(3.31) $ \begin{equation} \|((\boldsymbol{u}_1-\boldsymbol{u}_2)(t), (\phi_1-\phi_2)(t))\|_{\Bbb Y}^2\leq Ce^{Lt}\|((\boldsymbol{u}_1-\boldsymbol{u}_2)(0), (\phi_1-\phi_2)(0))\|_{\Bbb Y}^2, \ \forall t\geq0, \end{equation} $

其中$ C $和$ L $是与时间无关的正常数.

证   令$ \boldsymbol{w}: = \boldsymbol{u}_1-\boldsymbol{u}_2 $, $ \psi: = \phi_1-\phi_2 $, $ \eta: = \mu_1-\mu_2 $, $ \widetilde{p}: = p_1-p_2 $, 由系统(1.1)容易得到

(3.32) $ \begin{equation} \left\{\begin{array}{ll} \partial_t\boldsymbol{w} -\nu\Delta\boldsymbol{w}+\nabla\widetilde{p}+\boldsymbol{w}\cdot\nabla\boldsymbol{u}_1+\boldsymbol{u}_2\cdot\nabla\boldsymbol{w} = {\cal K}\mu_1\nabla\psi+{\cal K}\eta\nabla\phi_2, \\ \nabla\cdot \boldsymbol{w} = 0, \\ \partial_t\psi-\Delta\eta+\boldsymbol{w}\cdot\nabla\phi_1+\boldsymbol{u}_2\cdot\nabla\psi = 0, \\ \eta = A_k\psi+B_k\psi+f(\phi_1)-f(\phi_2), \\ \end{array}\right. \end{equation} $

将(3.32)$ _1 $式与$ {\cal K}^{-1}\boldsymbol{w}(t) $做内积, 由$ (\boldsymbol{u}_2\cdot\nabla\boldsymbol{w}, \boldsymbol{w}) = 0 $, 得到

(3.33) $ \begin{equation} \frac{1}{2{\cal K}}\frac{\rm d}{{\rm d}t}|\boldsymbol{w}(t)|^2+\frac{\nu}{{\cal K}}\|\boldsymbol{w}(t)\|^2 = -\frac{1}{{\cal K}}(\boldsymbol{w}\cdot\nabla\boldsymbol{u}_1, \boldsymbol{w})+(\mu_1\nabla\psi, \boldsymbol{w})+(\eta\nabla\phi_2, \boldsymbol{w}). \end{equation} $

将(3.32)$ _2 $式在$ L^2(\Omega) $中与$ A_k\psi(t) $做内积, 得到

(3.34) $ \begin{equation} \frac{1}{2}\frac{\rm d}{{\rm d}t}| A_k^{1/2}\psi(t)|^2-(A_k\psi, \Delta\eta) = -(\boldsymbol{w}\cdot\nabla\phi_1, A_k\psi)-(\boldsymbol{u}_2\cdot\nabla\psi, A_k\psi). \end{equation} $

再将(3.32)$ _4 $式在$ L^2(\Omega) $中与$ -\Delta\eta(t)-\varsigma A_k\psi(t) $ ($ \varsigma>0 $且足够小)做内积, 得到

(3.35) $ \begin{eqnarray} |\nabla\eta(t)|_{L^2}^2-\varsigma(\eta, A_k\psi)_{L^2}& = &-(A_k\psi, \Delta\eta)-\varsigma(A_k\psi, B_k\psi)-\varsigma| A_k\psi(t)|_{L^2}^2-(B_k\psi, \Delta\eta)\\ & &-(f(\phi_1)-f(\phi_2), \Delta\eta)_{L^2}-\varsigma(f(\phi_1)-f(\phi_2), A_k\psi)_{L^2}. \end{eqnarray} $

从(3.33)–(3.35)式可得

(3.36) $ \begin{eqnarray} & &\frac{1}{2}\frac{\rm d}{{\rm d}t}(\frac{1}{{\cal K}}|\boldsymbol{w}(t)|^2+| A_k^{1/2}\psi(t)|^2)+\frac{\nu}{{\cal K}}\|\boldsymbol{w}(t)\|^2+|\nabla\eta(t)|_{L^2}^2+\varsigma| A_k\psi(t)|_{L^2}^2{}\\ & = &-\frac{1}{{\cal K}}(\boldsymbol{w}\cdot\nabla\boldsymbol{u}_1, \boldsymbol{w}) +(\mu_1\nabla\psi, \boldsymbol{w})+(\eta\nabla\phi_2, \boldsymbol{w})-(\boldsymbol{w}\cdot\nabla\phi_1, A_k\psi){}\\ & &-(\boldsymbol{u}_2\cdot\nabla\psi, A_k\psi)+\varsigma(\eta, A_k\psi)_{L^2}-(B_k\psi, \Delta\eta)-\varsigma(A_k\psi, B_k\psi){}\\ &&-(f(\phi_1)-f(\phi_2), \Delta\eta)_{L^2}-\varsigma(f(\phi_1)-f(\phi_2), A_k\psi)_{L^2}, \end{eqnarray} $

接下来将对(3.36)式的右边进行估计, 通过Hölder, Ladyzhenskaya, Young不等式和(3.5)式, 前五项的估计如下

(3.37) $ \begin{eqnarray} -\frac{1}{{\cal K}}|(\boldsymbol{w}\cdot\nabla\boldsymbol{u}_1, \boldsymbol{w})|&\leq &c|\boldsymbol{w}|_{L^4}|\nabla\boldsymbol{u}_1||\boldsymbol{w}|_{L^4}\\ &\leq& c|\boldsymbol{w}|^{1/2}\|\boldsymbol{w}\|^{1/2}\|\boldsymbol{u}_1\||\bf w|^{1/2}\|\boldsymbol{w}\|^{1/2}\\ &\leq&\frac{\nu}{4{\cal K}}\|\boldsymbol{w}\|^2+c\|\boldsymbol{u}_1\|^2|\boldsymbol{w}|^2, \end{eqnarray} $

(3.38) $ \begin{eqnarray} |(\boldsymbol{w}\cdot\nabla\phi_1, A_k\psi)|&\leq &|\boldsymbol{w}|_{L^4}|\nabla\phi_1|_{L^4}| A_k\psi|_{L^2}\\ &\leq &c|\boldsymbol{w}|^{1/2}\|\boldsymbol{w}\|^{1/2}\|\phi_1\|_{H^1}^{1/2}\|\phi_1\|_{H^2}^{1/2}| A_k\psi|_{L^2}\\ &\leq& c|\boldsymbol{w}|\|\boldsymbol{w}\|\|\phi_1\|_{H^1}\|\phi_1\|_{H^2}+\frac{\varsigma}{16}| A_k\psi|_{L^2}^2\\ &\leq &\frac{\nu}{4{\cal K}}\|\boldsymbol{w}\|^2+\frac{\varsigma}{16}| A_k\psi|_{L^2}^2+c\|\phi_1\|_{H^1}^2\|\phi_1\|_{H^2}^2|\boldsymbol{w}|^2, \end{eqnarray} $

以及

(3.39) $ \begin{eqnarray} | (\boldsymbol{u}_2\cdot\nabla\psi, A_k\psi)|&\leq &c|\boldsymbol{u}_2|^{1/2}\|\boldsymbol{u}_2\|^{1/2}\|\psi\|_{H^1}^{1/2}\|\psi\|_{H^2}^{1/2}| A_k\psi|_{L^2}\\ &\leq &\frac{\varsigma}{16}| A_k\psi|_{L^2}^2+c|\boldsymbol{u}_2|\|\boldsymbol{u}_2\|\|\psi\|_{H^1}\|\psi\|_{H^2}\\ &\leq &\frac{\varsigma}{16}| A_k\psi|_{L^2}^2+c\|\boldsymbol{u}_2\|^{2}| A_k^{1/2}\psi|_{L^2}^2, \end{eqnarray} $

(3.40) $ \begin{eqnarray} |(\mu_1\nabla\psi, \boldsymbol{w})|&\leq &c|\boldsymbol{w}|^{1/2}\|\boldsymbol{w}\|^{1/2}\|\psi\|_{H^1}^{1/2}\|\psi\|_{H^2}^{1/2}| \mu_1|_{L^2}{}\\ &\leq& \frac{\nu}{4{\cal K}}\|\boldsymbol{w}\|^2+c|\boldsymbol{w}|^{2/3}\|\psi\|_{H^1}^{2/3}\|\psi\|_{H^2}^{2/3}| \mu_1|_{L^2}^{4/3}\\ &\leq&\frac{\nu}{4{\cal K}}\|\boldsymbol{w}\|^2+\frac{\varsigma}{16}| A_k\psi|_{L^2}^2+c|\psi|_{L^2}^2+c|\boldsymbol{w}|\|\psi\|_{H^1}| \mu_1|_{L^2}^2\\ &\leq&\frac{\nu}{4{\cal K}}\|\boldsymbol{w}\|^2+\frac{\varsigma}{16}| A_k\psi|_{L^2}^2+c|\psi|_{L^2}^2+c(|\boldsymbol{w}|^2+\epsilon\|\psi\|_{H^1}^2)(| A_k\phi_1|^2+|\phi_1|^2) {}\\ \end{eqnarray} $

和

(3.41) $ \begin{eqnarray} |(\eta\nabla\phi_2, \boldsymbol{w})|&\leq &c|\boldsymbol{w}|^{1/2}\|\boldsymbol{w}\|^{1/2}\|\phi_1\|_{H^1}^{1/2}\|\phi_1\|_{H^2}^{1/2}| \eta|_{L^2}\\ &\leq&\frac{\nu}{4{\cal K}}\|\boldsymbol{w}\|^2+\frac{\varsigma}{16}| A_k\psi|_{L^2}^2+c|\psi|_{L^2}^2+c|\boldsymbol{w}|^2\|\phi_1\|_{H^1}^2\|\phi_1\|_{H^2}^2. \end{eqnarray} $

由Hölder和Sobolev不等式, (3.36)式的最后两项估计如下

(3.42) $ \begin{eqnarray} |(-(f(\phi_1)-f(\phi_2)), \Delta\eta)_{L^2}|& = &|(\nabla(f(\phi_1)-f(\phi_2)), \nabla\eta)_{L^2}|{}\\ &\leq& Q(\|\phi_1\|_{H^1}, \|\phi_2\|_{H^1})(\|\phi_1\|_{H^2}^2+\|\phi_2\|_{H^2}^2)\|\psi\|_{H^k}^2+\frac{1}{2}|\nabla\eta|_{L^2}^2 {}\\ \end{eqnarray} $

和

(3.43) $ \begin{equation} \varsigma|(f(\phi_1)-f(\phi_2), A_k\psi)_{L^2}|\leq Q_\varsigma(\|\phi_1\|_{H^1}, \|\phi_2\|_{H^1})\|\psi\|_{H^k}^2+\frac{\varsigma}{16}| A_k\psi|_{L^2}^2, \end{equation} $

其中$ Q $, $ Q_\zeta $是与时间无关的单增非减的函数.

此外, 通过Hölder, Young, Poincaré和(3.5)式, (3.36)式的余下项估计如下

(3.44) $ \begin{eqnarray} \varsigma|(\eta, A_k\psi)_{L^2}|&\leq&\varsigma|\eta|_{L^2}|A_k\psi|_{L^2}\leq \varsigma c_\Omega^{1/2}|\Omega|^{1/2}|\nabla\eta|_{L^2}|A_k\psi|_{L^2}\\ &\leq&\frac{\varsigma}{2}|A_k\psi|_{L^2}^2+\frac{\varsigma c_\Omega|\Omega|}{2}|\nabla\eta|_{L^2}^2, \end{eqnarray} $

(3.45) $ \begin{equation} |-(B_k\psi, \Delta\eta)_{L^2}|\leq|(\nabla B_k\psi, \nabla\eta)|\leq\frac{1}{4}|\nabla\eta|_{L^2}^2+\frac{\varsigma}{16}| A_k\psi|_{L^2}^2+c|\psi|_{L^2}^2 \end{equation} $

和

(3.46) $ \begin{eqnarray} \varsigma|(A_k\psi, B_k\psi)|&\leq&\varsigma|A_k\psi|_{L^2}|B_k\psi|_{L^2}\leq\varsigma\epsilon|A_k\psi|_{L^2}^2+ c(\epsilon)|B_k\psi|_{L^2}^2{}\\ &\leq& \frac{\varsigma}{8}| A_k\psi|_{L^2}^2+c|\psi|_{L^2}^2. \end{eqnarray} $

由(3.37)–(3.46)式, 可得

(3.47) $ \begin{eqnarray} && \frac{\rm d}{{\rm d}t}(\frac{1}{{\cal K}}|\boldsymbol{w}(t)|^2+| A_k^{1/2}\psi(t)|^2)+(\frac{1}{2}-\varsigma c_\Omega|\Omega|) |\nabla\eta(t)|_{L^2}^2{}\\ &\leq&{\cal J}(t)(\frac{1}{{\cal K}}|\boldsymbol{w}(t)|^2+| A_k^{1/2}\psi(t)|^2), \end{eqnarray} $

其中

$ \begin{eqnarray*} {\cal J}(t):& = &c(\|\boldsymbol{u}_1\|^2+\|\phi_1\|_{H^1}^2\|\phi_1\|_{H^2}^2+\|\boldsymbol{u}_2\|^2+\|\phi_1\|_{H^{2k}}^2+\|\phi_1\|^2)\\ & &+Q(\|\phi_1\|_{H^1}, \|\phi_2\|_{H^1})(1+\|\phi_1\|_{H^2}^2+\|\phi_2\|_{H^2}^2). \end{eqnarray*} $

另一方面, 将(3.19)式在$ (0, t) $上积分并由(3.8)式可得

(3.48) $ \begin{eqnarray} && E(t)+\int_0^t\kappa E(s){\rm d}s+\int_0^t\kappa_1(\frac{\nu}{{\cal K}}\|\boldsymbol{u}(s)\|_{\Bbb V}^2 \\ &&+| A_k^{1/2}\phi(s)|^2)+|\nabla\mu(s)|_{L^2}^2+\kappa_2(| f(\phi(s))|, 1+| \phi(s)|)_{L^2}{\rm d}s{}\\ &\leq &Q(\|(\boldsymbol{u}(0), \phi(0)\|_{{\Bbb Y}}^2)+ (\frac{1}{{\cal K}\nu}\| \boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1)t. \end{eqnarray} $

由(3.1)式和(3.48)式, 容易得到

(3.49) $ \begin{equation} \int_0^t{\cal J}(s){\rm d}s\leq Q(\|(\boldsymbol{u}_1(0), \phi_1(0))\|_{{\Bbb Y}}+\|(\boldsymbol{u}_2(0), \phi_2(0))\|_{{\Bbb Y}})+Lt, \end{equation} $

其中$ L $代表正常数且与初值有关.

因此, 在(3.47)式中取$ \varsigma $足够小并运用Gronwall不等式, 由(3.49)式可得

$ \begin{eqnarray*} \frac{1}{{\cal K}}|\boldsymbol{w}(t)|^2+| A_k^{1/2}\psi(t)|^2 &\leq& e^{\int_0^t{\cal J}(s){\rm d}s}(\frac{1}{{\cal K}}|\boldsymbol{w}(0)|^2+| A_k^{1/2}\psi(0)|^2){\nonumber}\\ &\leq& Ce^{Lt}\|((\boldsymbol{u}_1-\boldsymbol{u}_2)(0), (\phi_1-\phi_2)(0))\|_{\Bbb Y}^2, \end{eqnarray*} $

其中$ C $是与时间无关的正常数.定理3.3证毕.

注3.1   由定理3.3可得系统(1.1)弱解的唯一性.