## Weak Solutions to Higher-Order Anisotropic Cahn-Hilliard-Navier-Stokes Systems

Luo Jiao,, Qi Qian,, Luo Hong,

 基金资助: 国家自然科学基金.  10771306国家自然科学基金.  11701399

 Fund supported: the NSFC.  10771306the NSFC.  11701399

Abstract

In this paper, we are concerned with weak solutions to higher-order anisotropic Cahn-Hilliard-Navier-Stokes systems in a two-dimensional bounded domain. The system consists of an incompressible Navier-Stokes equation coupled with a higher-order anisotropic Cahn-Hilliard equation. First some functional spaces and the definition of weak solutions are introduced. Then energy estimates of the solutions are given, and weak solutions of the system are obtained by Galerkin approximation scheme. Furthermore, the uniqueness of the solution to the system is proved.

Keywords： Anisotropic ; Cahn-Hilliard-Navier-Stokes system ; Weak solutions ; Existence ; Uniqueness

Luo Jiao, Qi Qian, Luo Hong. Weak Solutions to Higher-Order Anisotropic Cahn-Hilliard-Navier-Stokes Systems. Acta Mathematica Scientia[J], 2020, 40(6): 1599-1611 doi:

## 1 引言

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

$\forall\upsilon\in D(A_k)$, 有

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

${\Bbb Y}$关于如下范数是一个完备的度量空间

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

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

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

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

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

$\epsilon_i$取合适的值时, 可得

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

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

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

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

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

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

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

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

$\epsilon$取适当的值, 使下式成立

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

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

$\zeta = 1/c_\Omega|\Omega| $$\kappa\in(0, \zeta) , 并令 \kappa = \min\{\nu/2c_\Omega|\Omega|, 1/c_\Omega|\Omega|\} .通过(3.9)式和(3.18)式可得 \begin{eqnarray} && \frac{\rm d}{{\rm d}t}E(t)+\kappa E(t)+\kappa_1(\frac{\nu}{{\cal K}}\|\boldsymbol{u}(t)\|_{\Bbb V}^2 +| A_k^{1/2}\phi(t)|^2)+|\nabla\mu(t)|_{L^2}^2 +\kappa_2(| f(\phi(t))|, 1+| \phi(t)|)_{L^2} {}\\ &\leq& \frac{1}{{\cal K}\nu}\|\boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1. \end{eqnarray} 由Gronwall不等式(参见文献[23, 引理2.5]), 可得 \begin{eqnarray} E(t)&\leq&2E(0)e^{-kt}+2\int_0^t(\frac{1}{{\cal K}\nu}\| \boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1)e^{-k(t-y)}{\rm d}y\\ &\leq&2E(0)e^{-kt}+2\kappa^{-1}(\frac{1}{{\cal K}\nu}\| \boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1), \ \forall t\geq 0. \end{eqnarray} 对(3.19)式在 [t, t+1] 上积分可得 \begin{eqnarray} && E(t+1)+\int_t^{t+1}\kappa E(s)+\kappa_1(\frac{\nu}{{\cal K}}\|\boldsymbol{u}(s)\|_{{\Bbb V}}^2 \\ &&+| A_k^{1/2}\phi(s)|^2)+|\nabla\mu(s)|_{L^2}^2+\kappa_2(| f(\phi(s))|, 1+| \phi(s)|)_{L^2}{\rm d}s{}\\ &\leq& E(t)+\frac{1}{{\cal K}\nu}\| \boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1, \ \forall t\geq0. \end{eqnarray} 通过(3.20)式和(3.21)式有 \begin{eqnarray} && E(t)+\int_t^{t+1}(\kappa_1(\frac{\nu}{{\cal K}}\|\boldsymbol{u}(s)\|_{{\Bbb V}}^2+| A_k^{1/2}\phi(s)|^2)+|\nabla\mu(s)|_{L^2}^2+\kappa_2(| f(\phi(s))|, 1+| \phi(s)|)_{L^2}){\rm d}s\\ &\leq&4E(0)e^{-\kappa t}+(4\kappa^{-1}+1)(\frac{1}{{\cal K}\nu}\|\boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1), \ \forall t\geq0. \end{eqnarray} 于是从(3.8)式, (3.15)式和(3.22)式, 可得 \begin{eqnarray} && \|(\boldsymbol{u}(t), \phi(t)\|_{{\Bbb Y}}^2+\int_t^{t+1}\frac{\nu}{{\cal K}}\|\boldsymbol{u}(s)\|_{\Bbb V}^2+| A_k^{1/2}\phi(s)| ^2+|\nabla\mu(s)|_{L^2}^2+| F(\phi(s))|_{L^1}){\rm d}s\\ &\leq& Q(\|(\boldsymbol{u}(0), \phi(0)\|_{{\Bbb Y}}^2)e^{-\kappa t}+c_2, \ \forall t\geq0, \end{eqnarray} 其中 c_2 = (4\kappa^{-1}+1)(\frac{1}{{\cal K}\nu}\| \boldsymbol{g}\|_{{\Bbb V^*}}^2+c_1) . 接下来证明(3.1)式中的其余项.显然, 从(3.23)式可得 $$\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.$$ 从(1.1)4式, (2.2)式, (3.23)式, (3.24)式以及 H^k\hookrightarrow L^\beta , \beta\in[1, +\infty) , 可得 通过(1.1)4式, (3.24)式和正则性结果, 有 $$\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.$$ 此外, 从(1.1)3式, (3.23)式和(3.24)式, 可得 \begin{eqnarray} \int_t^{t+1}\|\partial_t\phi(s)\|_{H^{-1}}^2{\rm d}s &\leq &\int_t^{t+1}(\| \Delta\mu(s)\|_{H^{-1}}^2+\| \boldsymbol{u}(s)\cdot\nabla\phi(s)\|_{H^{-1}}^2){\rm d}s\\ &\leq &\int_t^{t+1}(\|\mu(s)\|^2+c_\Omega \|\boldsymbol{u}(s)\|_{\Bbb V}^2\|\phi(s)\|^2){\rm d}s\\ &\leq & Q(\|(\boldsymbol{u}(0), \phi(0))\|_{{\Bbb Y}}^2)e^{-\kappa t}+c_6, \quad\forall t \geq 0. \end{eqnarray} 因此, \partial_t\phi\in L^2([t, t+1];H^{-1}(\Omega)) . 为了得到 \partial_t\boldsymbol{u}$$ L^2([t, t+1];{\Bbb V}^*)$中一致有界, 由

$$$\|\boldsymbol{u}\cdot\nabla\boldsymbol{u}\|_{{\Bbb V}^*}^2 \leq c_\Omega |\boldsymbol{u}|^2\|\boldsymbol{u}\|_{{\Bbb V}}^2, \quad\forall \boldsymbol{u} \in {\Bbb V}$$$

$$$\| \mu\nabla\phi \|_{\Bbb V^*}^2 \leq c_8\|\phi\|^2| \mu|_{L^2}\|\mu\|_{H^{1}}.$$$

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

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].

$\alpha_N^i(t)$, $\beta_N^i(t)$, $\gamma_N^i(t)$分别与(3.30)$_1$式, (3.30)$_2$式和(3.30)$_3$式相乘, 并从$1 $$N 求和, 可得 n\rightarrow\infty 可得 N\rightarrow\infty , 可得 \boldsymbol{v} \in{\Bbb V} , \varphi\in H^1$$ \xi\in H_0^k$满足

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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