数学物理学报, 2023, 43(3): 754-770

部分耗散的三维Boussinesq方程在静力平衡附近的稳定性和指数衰减

黎小丽,, 陈晓莉,

江西师范大学数学与统计学院 南昌 330022

Stability and Exponential Decay of the 3D Boussinesq Equations with Partial Dissipation

Li Xiaoli,, Chen Xiaoli,

School of Mathematics and Statistics, Jiangxi Normal University, Nanchan 330022

收稿日期: 2022-06-29   修回日期: 2023-02-6  

基金资助: 国家自然科学基金(11971209)
国家自然科学基金(11961032)
江西省教育厅基金

Received: 2022-06-29   Revised: 2023-02-6  

Fund supported: NSFC(11971209)
NSFC(11961032)
Foundation of Jiangxi Teacher's Division

作者简介 About authors

黎小丽,E-mail:Lixli629@163.com;

陈晓莉,E-mail:littleli_chen@163.com

摘要

该文研究了速度具有水平部分耗散而温度具有垂直耗散的三维Boussinesq方程在静力平衡附近的稳定性及大时间行为问题.在$R^2\times T$上不仅建立了Boussinesq方程解的稳定性且该解有初值一样的对称性,还证明了速度和温度的振荡部分$(\tilde{u},\tilde{\theta})$是指数衰减的.

关键词： Boussinesq方程; 稳定性; 静力平衡; 大时间行为

Abstract

This paper is devoted to solving the stability and large time behavior problem on three dimensional Boussinesq equations with anisotropic dissipation and vertical thermal diffusion near the hydrostatic equilibrium. The stability of the solution with certain symmetries to the Boussinesq euations is established on the spatial domain $R^2\times \mathrm{T}$ with the periodic box $\mathrm{T}=[-\frac12,\frac12]$. In addition, the oscillators of the velocity $u$ and the temperature $\theta$ admit the exponential decay in time variable $t$.

Keywords： Boussinesq equations; Stability; Hydrostatic equilibrium; Large-time behavior

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本文引用格式

黎小丽, 陈晓莉. 部分耗散的三维Boussinesq方程在静力平衡附近的稳定性和指数衰减[J]. 数学物理学报, 2023, 43(3): 754-770

Li Xiaoli, Chen Xiaoli. Stability and Exponential Decay of the 3D Boussinesq Equations with Partial Dissipation[J]. Acta Mathematica Scientia, 2023, 43(3): 754-770

1 引言

该文研究如下三维Boussinesq方程

$\begin{equation}\label{eq:Bouss} \left\{ \begin{array}{ll} &\partial_t U_h+(U\cdot\nabla)U_h-\nu\Delta_hU_h+\nabla_h \Pi=0,~(t,x)\in R_+\times\Omega,\\ &\partial_t U_3+(U\cdot\nabla)U_3+\partial_3 \Pi=\Theta,\\ &\partial_t \Theta+(U\cdot\nabla)\Theta-\eta\partial_{33}\Theta=0,\\ &\nabla\cdot U=0, \\ &U(x,0)=U_0(x), \Theta(x,0)=\Theta_0(x), \end{array} \right. \end{equation}$

其中$U=(U_h,U_3)$是流体速度, $\Pi$和$\Theta$分别表示压力和温度, $\nu>0,~\eta>0$分别为粘性系数和阻尼系数, 且$\nabla_h=(\partial_1,\partial_2)$, $\Delta_h=\partial_{11}+\partial_{22}$. 假设$\Omega=R^2\times\mathrm{T}$, $\mathrm{T}=[-\frac12,\frac12]$是一个一维的周期盒子.

众所周知, Boussinesq方程是模拟地球物理流体以及各种由浮力驱动的Rayleigh-Bénard对流的模型(见文献[8-9]). 显然

$\begin{eqnarray*} U_{he}=0,\quad \Theta_{he}=x_3, \quad \Pi_{he}=\frac12x_3^2 \end{eqnarray*}$

是方程(1.1)的稳定解, 人们通常称这样的稳定解为静力平衡. 令

$\begin{eqnarray*} u=U-U_{he},\quad \theta=\Theta-\Theta_{he},\quad \pi=\Pi-\Pi_{he}. \end{eqnarray*}$

则方程(1.1)变为

$\begin{equation}\label{eq1:Boussinesq} \left\{ \begin{array}{ll} &\partial_t u_h+(u\cdot\nabla)u_h-\nu\Delta_hu_h+\nabla_h\pi=0,\\ &\partial_t u_3+(u\cdot\nabla)u_3+\partial_3 \pi=\theta,\\ &\partial_t \theta+(u\cdot\nabla)\theta+u_3-\eta\partial_{33}\theta=0,\\ &\nabla\cdot u=0, \\ &u(x,0)=u_0(x), \theta(x,0)=\theta_0(x). \end{array} \right. \end{equation} $

鉴于Boussinesq方程在数学和物理中具有重要的意义, 其解的稳定性和大时间行为问题引起了人们的广泛关注, 并得到了很多不错的结论 (见文献[1,4-5,7,11] 等). Doering, Wu, Zhao和Zheng[5]在有界域上研究了速度全耗散, 温度没有耗散的二维Boussinesq方程的稳定性问题. 之后, Tao, Wu, Zhao和Zheng[11]建立了文献[5] 中稳定解的精确大时间行为. Castro, Cordoba和Lear[3] 考虑了二维Boussinesq方程在速度带阻尼项时的稳定性问题并得到了其在带状区域上的渐近稳定性. 最近, Wu和Zhang[12] 在$R^2\times \mathrm{T}$中利用具有对称性的初值得到了具有水平耗散和垂直热扩散的三维Boussinesq方程解的稳定性. 此外, 他们还得到了具有速度全耗散和垂直热扩散的三维Boussinesq 方程速度和温度振荡的指数衰减. 受文献[6,11-12]的启发, 本文考虑三维Boussinesq方程(1.2)解的稳定性和大时间行为.

本文的主要结论如下.

${\bf定理1.1}$ 设$(u_0,\theta_0)\in H^2(\Omega)$, $\nabla\cdot u_0=0$ 且

$\begin{equation}\label{estimate-01} u_{01},u_{02}\ \mbox{关于$x_3$是偶函数,$u_{03},\theta_0$关于$x_3$是奇函数}, \end{equation}$

其中$u_0=(u_{01},u_{02},u_{03})$. 若存在适当小的$\epsilon(\nu,\eta)>0$使得

$\begin{equation}\label{estimate-2} \|u_0\|_{H^2}+\|\theta_0\|_{H^2}\le\epsilon(\nu,\eta), \end{equation}$

则(1.2)式存在唯一的整体解$(u,\theta)$使得对任意的$t>0$,有

$\begin{equation}\label{estimate-3} \|u(t)\|^2_{H^2}+\|\theta(t)\|^2_{H^2}+\int_0^T(\nu\|\nabla_hu_h(t)\|^2_{H^2}+\eta\|\partial_3\theta(t)\|^2_{H^2}){\rm d}t\le C\epsilon^2, \end{equation} $
$\begin{equation} u_{1},u_{2}\ \mbox{关于$x_3$是偶函数,$u_{3},\theta$关于$x_3$是奇函数} \end{equation}$.

${\bf注1.1}$ 利用文献[12]中的方法易得解的唯一性和对称性. 然而, 由于第三个速度分量缺乏水平耗散$\Delta_{h}u_3$, 所以在建立先验界估计时会比文献[12]更困难. 基于此, 此文的结论改进了文献[12]中解的稳定性. 此外, 在小初值条件下该文还建立了方程(1.2)解的$H^3$稳定性.

接下来考虑方程(1.2)解的大时间行为. 但若想要得到$u$和$\theta$振荡部分的指数衰减, Boussinesq方程需要更多的速度耗散, 故考虑以下三维Boussinesq 方程

$\begin{equation}\label{eq2:Boussinesq} \left\{ \begin{array}{ll} \partial_t u_h+(u\cdot\nabla)u_h-\nu\Delta u_h+\nabla_h P=0,~x\in\Omega,\\ \partial_t u_3+(u\cdot\nabla)u_3-\nu\partial_{33}u_3+\partial_3 P=\theta,\\ \partial_t \theta+(u\cdot\nabla)\theta+u_3-\eta\partial_{33}\theta=0,\\ \nabla\cdot u=0,\\ u(x,0)=u_0(x), \theta(x,0)=\theta_0(x), \end{array} \right. \end{equation}$

并得到以下结论.

${\bf定理1.2}$ 设$(u_0,\theta_0)\in H^3(\Omega)$, $\nabla\cdot u_0=0$且

$\begin{equation}\label{estimate-5} u_{01},u_{02}\ \mbox{关于$x_3$是偶函数,$u_{03},\theta$关于$x_3$是奇函数}, \end{equation} $

其中$u_0=(u_{01},u_{02},u_{03})$. 若存在适当小的$\epsilon(\nu,\eta)>0$使得

$\begin{equation}\label{estimate-6} \|u_0\|_{H^3}+\|\theta_0\|_{H^3}\le\epsilon(\nu,\eta), \end{equation}$

则方程(1.7)存在唯一整体解$(u,\theta)$且满足

(i) 对任意的$t>0$,有

$\begin{equation}\label{estimate-7} \|u(t)\|^2_{H^3}+\|\theta(t)\|^2_{H^3}+\int_0^T(\nu\|\nabla u_h(t)\|^2_{H^3}+\|\partial_{33}u_3\|_{H^3}+\eta\|\partial_3\theta(t)\|^2_{H^3}){\rm d}t\le C\epsilon^2, \end{equation}$
$\begin{equation}\label{estimate-8} u_{1},u_{2}\ \mbox{关于$x_3$是偶函数,$u_{3},\theta$关于$x_3$是奇函数}. \end{equation}$

(ii)若$(\tilde{u},\tilde{\theta})$为$(u,\theta)$的振荡部分, 则对任意$t>0$,有

$\begin{equation}\label{estimate-decay} \|(\tilde{u},\tilde{\theta})(t)\|_{H^1}\le\|(u_0,\theta_0)(t)\|_{H^1}e^{-ct}, \end{equation}$

其中$c=\tilde{C}\min\{\nu,\eta\}$. 同时, 还得到了(1.7)式的极限方程

$\begin{equation}\label{eq3:Boussinesq} \left\{ \begin{array}{ll} \partial_t \overline{u_1}+\overline{u\cdot\nabla \widetilde{u_1}}-\nu\Delta_h \overline{u_1}+\overline{u_1}\partial_1\overline{u_1}+\overline{u_2}\partial_2\overline{u_1}+\partial_1 \overline{P}=0,~x\in\Omega,\\ \partial_t \overline{u_2}+\overline{u\cdot\nabla \widetilde{u_2}}-\nu\Delta_h \overline{u_2}+\overline{u_1}\partial_1\overline{u_2}+\overline{u_2}\partial_2\overline{u_2}+\partial_2 \overline{P}=0,\\ \partial_1\overline{u_1}+\partial_1\overline{u_2}=0. \end{array} \right. \end{equation}$

${\bf注1.2}$ 定理1.2的第一个结论将$H^3$换成$H^2$也是成立的.由于定理1.2研究的方程比文献[12]中对应方程粘性要少,因此该定理在一定程度上改进了文献[12]的结论.

本文的其他部分安排如下: 第2节包含重要的引理和一些符号. 第3节利用靴代理论证明定理1.1. 定理1.2的证明在第4节.

2 关键的引理和符号

回顾平均部分$\bar{f}$和振荡部分$\tilde{f}$的定义,对一般的函数$f:\Omega\rightarrow R^2$, 定义

$\begin{equation}\label{definition-averge} \bar{f}(x,y)=\int_{\mathrm{T}}f(x,y,z')dz',~ \tilde{f}(x,y,z)=f(x,y,z)-\bar{f}(x,y). \end{equation}$

接下来给出在证明过程中需要用到的关键引理.

${\bf引理2.1}$[12] 假设$k\ge0$是整数.

(1) 若定义在$\Omega$上的函数$f$具有足够高的正则性, 即$f\in H^k(\Omega)$, 则

$ (\bar{f},\tilde{f})_{H^k}=0, $
$ \|f\|^2_{H^k}=\|\bar{f}\|^2_{H^k}+\|\tilde{f}\|^2_{H^k}, \|\bar{f}\|^2_{H^k}\le\|f\|^2_{H^k},~\|\tilde{f}\|^2_{H^k}\le\|f\|^2_{H^k}. $

值得注意的是, 由于$\bar{\tilde{f}}=0$, 所以$(\bar{f},\tilde{f})=0$且对任意微分算子$D^{\alpha}:=\partial_1^{\alpha_1}\partial_2^{\alpha_2}\partial_3^{\alpha_3}$,有

$ (D^{\alpha}\bar{f},D^{\alpha}\tilde{f})=0. $

(2) 若$\partial_3\tilde{f}\in{H^k}$, 则$\tilde{f}\in{H^k}$且$\tilde{f}$在$x_3$方向上满足Poincaré等式, 即

$\begin{eqnarray*}\|\tilde{f}\|_{H^k}\le C\|\partial_3\tilde{f}\|_{H^k},\end{eqnarray*} $

其中$C>0$是一个仅与$\Omega$和$k$有关的常数.

(3) 若$i=1,2,3$, 则$\overline{\partial_if}=\partial_i\bar{f}$, $\widetilde{\partial_if}=\partial_i\tilde{f}$且$\partial_3\bar{f}=0$.

(4) 若$\nabla\cdot f=0$, 则$\nabla\cdot\bar{f}=0,~\nabla\cdot\tilde{f}=0$.

${\bf引理2.2}$[12] 设$f\in H^1(R)$, 则

$ \|f\|_{L^\infty(R)}\le \sqrt{2}\|f\|^{\frac12}_{L^2(R)}\|f'\|^{\frac12}_{L^2(R)}. $

若$f\in H^1(\mathrm{T})$, $\tilde{f}$为其振荡, 则

$\begin{eqnarray*} &&\|f\|_{L^\infty(\mathrm{T})}\le \sqrt{2}\|f\|^{\frac12}_{L^2(\mathrm{T})}(\|f\|_{L^2(\mathrm{T})}+\|f'\|_{L^2(\mathrm{T})})^{\frac12},\\ &&\|\tilde{f}\|_{L^\infty(\mathrm{T})}\le \sqrt{2}\|\tilde{f}\|^{\frac12}_{L^2(\mathrm{T})}\|\tilde{f}'\|_{L^2(\mathrm{T})}^{\frac12}. \end{eqnarray*} $

${\bf引理2.3}$[2] 设$\Omega=R^2\times\mathrm{T}$. 对任意$f,g,h\in L^2(\Omega)$, 若$\partial_1f\in L^2(\Omega)$以及$\partial_2g\in L^2(\Omega)$, 则

$\begin{equation}\label{estimate-10} \begin{array}{ll} \Big|\int_{\Omega}fgh{\rm d}x\Big|\le C\|f\|^{\frac12}_{L^2}\|\partial_1f\|_{L^2}^{\frac12}\|g\|^{\frac12}_{L^2}\|\partial_2g\|^{\frac12}_{L^2}\|h\|^{\frac12}_{L^2}(\|h\|_{L^2}+\|\partial_3h\|_{L^2})^{\frac12},\\ \Big|\int_{\Omega}fg\tilde{h}{\rm d}x\Big|\le C\|f\|^{\frac12}_{L^2}\|\partial_1f\|^{\frac12}_{L^2}\|g\|^{\frac12}_{L^2}\|\partial_2g\|^{\frac12}_{L^2}\|\tilde{h}\|^{\frac12}_{L^2}\|\partial_3\tilde{h}\|_{L^2}^{\frac12}. \end{array} \end{equation}$

3 定理1.1的证明

解的局部存在性可以由标准的方法得到[9], 而解的唯一性以及与初值具有相应的对称性也可以利用文献[12]中的方法得到. 为了得到方程(1.2)解的稳定性, 关键的一步是建立其解的整体一致先验界估计. 为了方便起见, 定义以下能量泛函

$\begin{equation}\label{definition-Et} {\cal E}(t)=\sup_{0\le t\le T}(\|u(t)\|^2_{H^2}+\|\theta(t)\|^2_{H^2})+\int_0^T(\nu\|\nabla_hu_h(t)\|^2_{H^2}+\eta\|\partial_3\theta(t)\|^2_{H^2}){\rm d}t. \end{equation} $

利用文献[12]中的方法可以得到(1.2)式解的$H^2$唯一性.

${\bf命题3.1}$ 设$(u^{(1)},\theta^{(1)}),~(u^{(2)},\theta^{(2)})\in L^\infty(0,T;H^2)$且满足方程(1.2). 则对任意$t\in(0,T]$,有

$ (u^{(1)},\theta^{(1)})=(u^{(2)},\theta^{(2)}). $

利用唯一性可证得解有和初值相应的对称性[12].

${\bf推论3.1}$ 设$(u_0,\theta_0)\in H^2$满足对称性条件(1.3). 若$(u,\theta)$是方程(1.2)的解, 则对任意的$t\in[T]$, $(u(t),\theta(t))$ 有相应的对称性(1.6).

进而可得以下先验界估计.

${\bf命题3.2}$ 设$(u_0,\theta_0)\in H^2$满足对称性条件(1.3). 若$(u,\theta)$是方程(1.2)的解, 则对任意的$t\in[T]$, 存在常数$C>0$使得

$\begin{equation}\label{estimate-Et} \begin{array}{ll} {\cal E}(t)\le{\cal E}_0+C{\cal E}(t)^{\frac32}. \end{array} \end{equation}$

${\bf证}$ 由定义(2.1)可知$\tilde{u}=u-\bar{u}$, $\tilde{\theta}=\theta-\bar{\theta}$. 利用推论以及对称性(1.6)易得

$\begin{eqnarray*} \overline{u_3}=\int_{\mathrm{T}}u_3(x_1,x_2,x_3){\rm d}x_3=0,~\overline{\theta}=\int_{\mathrm{T}}\theta(x_1,x_2,x_3){\rm d}x_3=0, \end{eqnarray*}$

因此

$\begin{equation}\label{estimate-u3tildeu3thetatildetheta} \begin{array}{ll} u_3=\widetilde{u_3},\quad \theta=\widetilde{\theta}. \end{array} \end{equation}$

下面将证明命题3.2的过程分为两步.

第一步 $(u,\theta)$的$L^2$ -能量估计.

在$(1.2)_1, (1.2)_2, (1.2)_3$式两边分别与$u_h, u_3, \theta$作$L^2$内积, 利用$\nabla\cdot u=0$, 并分部积分得

$\begin{equation}\label{estimate-L2} \frac12\frac{\rm d}{{\rm d}t}(\|u_h\|^2_{L^2}+\|u_3\|^2_{L^2}+\|\theta\|^2_{L^2})+\nu\|\nabla_hu_h\|^2_{L^2}+\eta\|\partial_3\theta\|^2_{L^2}\le0. \end{equation}$

第二步 $(u,\theta)$的$\dot{H}^2$ -能量估计.

对$(1.2)_1, (1.2)_2, (1.2)_3$式求导$\partial_i^2(i=1,2,3)$后分别与$\partial_i^2 u_h,~ \partial_i^2 u_3,~\partial_i^2\theta$作$L^2$内积得

$\begin{aligned} & \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(\left\|u_{h}\right\|_{\dot{H}^{2}}^{2}+\left\|u_{3}\right\|_{\dot{H}^{2}}^{2}+\|\theta\|_{\dot{H}^{2}}^{2}\right)+\nu\left\|\nabla_{h} u_{h}\right\|_{\dot{H}^{2}}^{2}+\eta\left\|\partial_{3} \theta\right\|_{\dot{H}^{2}}^{2} \\ = & -\sum_{i=1}^{3} \int \partial_{i}^{2}\left(u \cdot \nabla u_{h}\right) \cdot \partial_{i}^{2} u_{h}-\sum_{i=1}^{3} \int \partial_{i}^{2}\left(u \cdot \nabla u_{3}\right) \cdot \partial_{i}^{2} u_{3}-\sum_{i=1}^{3} \int \partial_{i}^{2}(u \cdot \nabla \theta) \cdot \partial_{i}^{2} \theta \\ = & A_{1}+A_{2}+A_{3}. \end{aligned}$

为了估计$A_1$, 将其做如下分解

$\begin{eqnarray*} A_{1}&=&-\sum_{i=1}^2\int\partial_i^2(u\cdot\nabla u_h)\cdot\partial_i^2u_h-\int\partial_3^2(u\cdot\nabla u_h)\cdot\partial_3^2u_h\\ &=&-\sum_{i=1}^2\sum_{j=1}^3\int\partial_i^2(u_j\partial_j u_h)\cdot\partial_i^2u_h-\sum_{j=1}^3\int\partial_3^2(u_j\partial_ju_h)\cdot\partial_3^2u_h\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2(u_j\partial_j u_h)\cdot\partial_i^2u_h-\sum_{i=1}^2\int\partial_i^2(u_3\partial_3 u_h)\cdot\partial_i^2u_h\\ &&-\sum_{j=1}^2\int\partial_3^2(u_j\partial_ju_h)\cdot\partial_3^2u_h-\int\partial_3^2(u_3\partial_3u_h)\cdot\partial_3^2u_h\\ &=&A_{11}+A_{12}+A_{13}+A_{14}. \end{eqnarray*}$

利用$\nabla\cdot u=0$, Hölder不等式, Sobolev不等式和引理2.1(1)-(4)得

$\begin{eqnarray*}\label{estimate-A11A12} A_{11}+A_{12}&=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2(u_j\partial_j u_h)\cdot\partial_i^2u_h-\sum_{i=1}^2\int\partial_i^2(u_3\partial_3 u_h)\cdot\partial_i^2u_h\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2u_j\partial_ju_h\cdot\partial_i^2u_h+2\partial_iu_j\partial_i\partial_ju_h\cdot\partial_i^2u_h\\ &&-\sum_{i=1}^2\int\partial_i^2u_3\partial_3 u_h\cdot\partial_i^2u_h+2\partial_iu_3\partial_i\partial_3u_h\cdot\partial_i^2u_h\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2u_j\partial_ju_h\cdot\partial_i^2u_h+2\partial_iu_j\partial_i\partial_ju_h\cdot\partial_i^2u_h\\ &&-\sum_{i=1}^2\int\partial_i^2\widetilde{u_3}\partial_3 u_h\cdot\partial_i^2u_h+2\partial_i\widetilde{u_3}\partial_i\partial_3u_h\cdot\partial_i^2u_h\\ &\le &C\sum_{i=1}^2\sum_{j=1}^2\|\partial_i^2u_j\|_{L^2}\|\partial_ju_h\|_{L^4}\|\partial_i^2u_h\|_{L^4}+\|\partial_iu_j\|_{L^4}\|\partial_i\partial_ju_h\|_{L^4}\|\partial_i^2u_h\|_{L^2}\\ &&+C\sum_{i=1}^2\|\partial_i^2\widetilde{u_3}\|_{L^2}\|\partial_3 u_h\|_{L^4}\|\partial_i^2u_h\|_{L^4}+\|\partial_i\widetilde{u_3}\|_{L^4}\|\partial_i\partial_3u_h\|_{L^2}\|\partial_i^2u_h\|_{L^4}\\ &\le& C\|u\|_{H^2}\|\nabla_hu_h\|^2_{H^2}, \end{eqnarray*}$

其中用到了$\overline{u_3}=0$及对$i=1,2,3$, $k=1,2$,有

$\begin{matrix}\label{estimate-pa3u3tonablahuh} \|\partial_i^k\widetilde{u_3}\|_{L^2}&\le &C\|\partial_3\partial_i^k\widetilde{u_3}\|_{L^2}\le C\|\partial_i^k\partial_3u_3\|_{L^2}\\ &\le& C\|\partial_i^k(\partial_1u_1+\partial_2u_2)\|_{L^2}\le C\|\nabla_hu_h\|_{H^2}. \end{matrix}$

利用Hölder不等式, Sobolev不等式, 引理2.1(1)-(4), 引理2.3以及(3.7)式得

$\begin{eqnarray*} A_{13}+A_{14}&=&-\sum_{j=1}^2\int\partial_3^2(u_j\partial_ju_h)\cdot\partial_3^2u_h-\int\partial_3^2(u_3\partial_3u_h)\cdot\partial_3^2u_h\\ &=&-\sum_{j=1}^2\int\partial_3^2u_j\partial_ju_h\cdot\partial_3^2u_h+2\partial_3u_j\partial_3\partial_j\widetilde{u_h}\cdot\partial_3^2u_h\\ &&-\int\partial_3^2\widetilde{u_3}\partial_3u_h\cdot\partial_3^2u_h+2\partial_3\widetilde{u_3}\partial_3^2u_h\cdot\partial_3^2u_h\\ &\le& C\sum_{j=1}^2\|\partial_3^2u_j\|^{\frac12}_{L^2}\|\partial_1\partial_3^2u_j\|^{\frac12}_{L^2}\|\partial_ju_h\|^{\frac12}_{L^2}\\ &&\times(\|\partial_ju_h\|_{L^2}+\|\partial_3\partial_ju_h\|_{L^2})^{\frac12}\|\partial_3^2u_h\|^{\frac12}_{L^2}\|\partial_2\partial_3^2u_h\|^{\frac12}_{L^2}\\ && +C\sum_{j=1}^2\|\partial_3u_j\|^{\frac12}_{L^2}\|\partial_1\partial_3u_j\|^{\frac12}_{L^2}\|\partial_3\partial_ju_h\|^{\frac12}_{L^2}\\ &&\times\|\partial_3\partial_3\partial_j\widetilde{u_h}\|_{L^2}^{\frac12}\|\partial_3^2u_h\|^{\frac12}_{L^2}\|\partial_2\partial_3^2u_h\|^{\frac12}_{L^2}\\ &&+C\|\partial_3u_h\|^{\frac12}_{L^2}\|\partial_1\partial_3u_h\|^{\frac12}_{L^2}\|\partial_3^2\widetilde{u_3}\|^{\frac12}_{L^2} \|\partial_3^3\widetilde{u_3}\|_{L^2}^{\frac12} \|\partial_3^2u_h\|^{\frac12}_{L^2}\|\partial_2\partial_3^2u_h\|^{\frac12}_{L^2}\\ &&+C\|\partial_3^2u_h\|^{\frac12}_{L^2}\|\partial_1\partial_3^2u_h\|^{\frac12}_{L^2}\|\partial_3\widetilde{u_3}\|^{\frac12}_{L^2} \|\partial_3^2\widetilde{u_3}\|_{L^2}^{\frac12}\|\partial_3^2u_h\|^{\frac12}_{L^2}\|\partial_2\partial_3^2u_h\|^{\frac12}_{L^2}\\ &\lesssim&\|u\|_{H^2}\|\nabla_hu_h\|^2_{H^2}. \end{eqnarray*}$

因此有

$\begin{equation}\label{estimate-A1} A_{1}\le C\|u\|_{H^2}\|\nabla_hu_h\|^2_{H^2}. \end{equation} $

同理, 将$A_2$做如下分解

$\begin{eqnarray*} A_{2}&=&-\sum_{i=1}^2\int\partial_i^2(u\cdot\nabla u_3)\cdot\partial_i^2u_3-\int\partial_3^2(u\cdot\nabla u_3)\cdot\partial_3^2u_3\\ &=&-\sum_{i=1}^2\sum_{j=1}^3\int\partial_i^2(u_j\partial_j u_3)\cdot\partial_i^2u_3-\sum_{j=1}^3\int\partial_3^2(u_j\partial_ju_3)\cdot\partial_3^2u_3\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2(u_j\partial_j u_3)\cdot\partial_i^2u_3-\sum_{i=1}^2\int\partial_i^2(u_3\partial_3 u_3)\cdot\partial_i^2u_3\\ &&-\sum_{j=1}^2\int\partial_3^2(u_j\partial_ju_3)\cdot\partial_3^2u_3-\int\partial_3^2(u_3\partial_3u_3)\cdot\partial_3^2u_3\\ &=&A_{21}+A_{22}+A_{23}+A_{24}. \end{eqnarray*}$

由$\nabla\cdot u=0$, Hölder不等式, Sobolev不等式, 引理2.1(1)-(4)和(3.7)式得

$\begin{eqnarray*} A_{21}+A_{22}&=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2(u_j\partial_j u_3)\cdot\partial_i^2u_3-\sum_{i=1}^2\int\partial_i^2(u_3\partial_3 u_3)\cdot\partial_i^2u_3\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2u_j\partial_ju_3\cdot\partial_i^2u_3+2\partial_iu_j\partial_i\partial_ju_3\cdot\partial_i^2u_3\\ &&-\sum_{i=1}^2\int\partial_i^2u_3\partial_3 u_3\cdot\partial_i^2u_3+2\partial_iu_3\partial_i\partial_3u_3\cdot\partial_i^2u_3\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2u_j\partial_j\widetilde{u_3}\cdot\partial_i^2\widetilde{u_3}+2\partial_iu_j\partial_i\partial_j\widetilde{u_3}\cdot\partial_i^2\widetilde{u_3}\\ &&-\sum_{i=1}^2\int\partial_i^2\widetilde{u_3}\partial_3 \widetilde{u_3}\cdot\partial_i^2\widetilde{u_3}+2\partial_i\widetilde{u_3}\partial_i\partial_3\widetilde{u_3}\cdot\partial_i^2\widetilde{u_3}\\ &\le& C\sum_{i=1}^2\sum_{j=1}^2\|\partial_i^2u_j\|_{L^4}\|\partial_j\widetilde{u_3}\|_{L^4}\|\partial_i^2\widetilde{u_3}\|_{L^2}+\|\partial_iu_j\|_{L^\infty}\|\partial_i\partial_j\widetilde{u_3}\|_{L^2}\|\partial_i^2\widetilde{u_3}\|_{L^2}\\ &&+C\sum_{i=1}^2\|\partial_i^2\widetilde{u_3}\|_{L^2}\|\partial_3 \widetilde{u_3}\|_{L^\infty}\|\partial_i^2\widetilde{u_3}\|_{L^2}+\|\partial_i\widetilde{u_3}\|_{L^4}\|\partial_i\partial_3\widetilde{u_3}\|_{L^4}\|\partial_i^2\widetilde{u_3}\|_{L^2}\\ &\le& C\|u\|_{H^2}\|\nabla_hu_h\|^2_{H^2}, \end{eqnarray*}$

利用(3.3)式, Hölder不等式, Sobolev不等式, 引理2.1(1)-(4), 引理2.3和$\nabla\cdot u=0$得

$\begin{eqnarray*} A_{23}+A_{24}&=&-\sum_{j=1}^2\int\partial_3^2(u_j\partial_ju_3)\cdot\partial_3^2u_3-\int\partial_3^2(u_3\partial_3u_3)\cdot\partial_3^2u_3\\ &=&-\sum_{j=1}^2\int\partial_3^2u_j\partial_j\widetilde{u_3}\cdot\partial_3^2\widetilde{u_3}+2\partial_3u_j\partial_3\partial_ju_3\cdot\partial_3^2u_3\\ &&-\int\partial_3^2u_3\partial_3u_3\cdot\partial_3^2u_3+2\partial_3u_3\partial_3^2u_3\cdot\partial_3^2u_3\\ &\le& C\sum_{j=1}^2\|\partial_3^2u_j\|^{\frac12}_{L^2}\|\partial_1\partial_3^2u_j\|^{\frac12}_{L^2}\|\partial_j\widetilde{u_3}\|^{\frac12}_{L^2} \|\partial_3\partial_j\widetilde{u_3}\|_{L^2}^{\frac12}\|\partial_3^2u_3\|^{\frac12}_{L^2}\|\partial_2\partial_3^2u_3\|^{\frac12}_{L^2}\\ && +C\sum_{j=1}^2\|\partial_3u_j\|_{L^4}\|\partial_3\partial_ju_3\|_{L^4}\|\partial_3^2u_3\|_{L^2}+C\|\partial_3u_3\|_{L^4}\|\partial_3^2u_3\|_{L^4}\|\partial_3^2u_3\|_{L^2}\\ &\lesssim&\|u\|_{H^2}\|\nabla_hu_h\|^2_{H^2}, \end{eqnarray*}$

所以有

$\begin{equation}\label{estimate-A2} A_{2}\le C\|u\|_{H^2}\|\nabla_hu_h\|^2_{H^2}. \end{equation}$

类似于$A_2$,有

$\begin{eqnarray*} A_{3}&=&-\sum_{i=1}^3\int\partial_i^2(u\cdot\nabla\theta)\cdot\partial_i^2\theta\\ &=&-\sum_{i=1}^2\int\partial_i^2(u\cdot\nabla\theta)\cdot\partial_i^2\theta-\int\partial_3^2(u\cdot\nabla\theta)\cdot\partial_3^2\theta\\ &=&-\sum_{i=1}^2\sum_{j=1}^3\int\partial_i^2(u_j\partial_j\theta)\cdot\partial_i^2\theta-\sum_{j=1}^3\int\partial_3^2(u_j\partial_j\theta)\cdot\partial_3^2\theta\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2(u_j\partial_j\theta)\cdot\partial_i^2\theta-\sum_{i=1}^2\int\partial_i^2(u_3\partial_3\theta)\cdot\partial_i^2\theta\\ &&-\sum_{j=1}^2\int\partial_3^2(u_j\partial_j\theta)\cdot\partial_3^2\theta-\int\partial_3^2(u_3\partial_3\theta)\cdot\partial_3^2\theta\\ &=&A_{31}+A_{32}+A_{33}+A_{34}. \end{eqnarray*}$

利用Hölder不等式, Sobolev不等式, $\overline{u_3}=\overline{\theta}=0$, 引理2.1(1)-(3)和引理2.3有

$\begin{eqnarray*} A_{31}+A_{32}&=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2(u_j\partial_j \theta)\cdot\partial_i^2\theta-\sum_{i=1}^2\int\partial_i^2(u_3\partial_3 \theta)\cdot\partial_i^2\theta\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2u_j\partial_j\theta\cdot\partial_i^2\theta+2\partial_iu_j\partial_i\partial_j\theta\cdot\partial_i^2\theta\\ &&-\sum_{i=1}^2\int\partial_i^2u_3\partial_3 \theta\cdot\partial_i^2\theta+2\partial_iu_3\partial_i\partial_3\theta\cdot\partial_i^2\theta\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^2u_j\partial_j\theta\cdot\partial_i^2\widetilde{\theta}+2\partial_iu_j\partial_i\partial_j\theta\cdot\partial_i^2\widetilde{\theta}\\ &&-\sum_{i=1}^2\int\partial_i^2u_3\partial_3\theta\cdot\partial_i^2\widetilde{\theta}+2\partial_iu_3\partial_i\partial_3\theta\cdot\partial_i^2\widetilde{\theta}\\ &\le &C\sum_{i=1}^2\sum_{j=1}^2(\|\partial_i^2u_j\|_{L^4}\|\partial_j\theta\|_{L^4}\|\partial_i^2\widetilde{\theta}\|_{L^2}+\|\partial_iu_j\|_{L^\infty}\|\partial_i\partial_j\theta\|_{L^2}\|\partial_i^2\widetilde{\theta}\|_{L^2})\\ &&+C\sum_{i=1}^2(\|\partial_i^2u_3\|_{L^2}\|\partial_3 \theta\|_{L^\infty}\|\partial_i^2\widetilde{\theta}\|_{L^2}+\|\partial_iu_3\|_{L^4}\|\partial_i\partial_3\theta\|_{L^4}\|\partial_i^2\widetilde{\theta}\|_{L^2})\\ &\le& C\|u\|_{H^2}\|\partial_3\theta\|^2_{H^2}, \end{eqnarray*}$

最后一个不等式用到了$\|\partial_i^2\tilde{\theta}\|_{L^2}\le C\|\partial_3\partial_i^2\tilde{\theta}\|_{L^2}\le C\|\partial_3\theta\|_{H^2}$. 使用Hölder不等式, Sobolev不等式, $\bar{\theta}=0$, 引理2.1(1)-(3), 引理2.3和$\nabla\cdot u=0$得

$\begin{eqnarray*} A_{33}+A_{34}&=&-\sum_{j=1}^2\int\partial_3^2(u_j\partial_j\theta)\cdot\partial_3^2\theta-\int\partial_3^2(u_3\partial_3\theta)\cdot\partial_3^2\theta\\ &=&-\sum_{j=1}^2\int\partial_3^2u_j\partial_j\widetilde{\theta}\cdot\partial_3^2\widetilde{\theta}+2\partial_3u_j\partial_3\partial_j\theta\cdot\partial_3^2\theta\\ &&-\int\partial_3^2u_3\partial_3\theta\cdot\partial_3^2\theta+2\partial_3u_3\partial_3^2\theta\cdot\partial_3^2\theta\\ &\le &C\sum_{j=1}^2\|\partial_3^2u_j\|^{\frac12}_{L^2}\|\partial_1\partial_3^2u_j\|^{\frac12}_{L^2} \|\partial_j\widetilde{\theta}\|^{\frac12}_{L^2}\|\partial_3\partial_j\widetilde{\theta}\|_{L^2}^{\frac12}\|\partial_3^2\theta\|^{\frac12}_{L^2} \|\partial_2\partial_3^2\theta\|^{\frac12}_{L^2}\\ && +C\sum_{j=1}^2\|\partial_3u_j\|_{L^4}\|\partial_3\partial_j\theta\|_{L^4}\|\partial_3^2\theta\|_{L^2}+C\|\partial_3^2u_3\|_{L^2} \|\partial_3\theta\|_{L^4}\|\partial_3^2\theta\|_{L^4}\\ &&+C\|\partial_3u_3\|_{L^4}\|\partial_3^2\theta\|_{L^4}\|\partial_3^2\theta\|_{L^2}\\ &\le& C\|u\|^{\frac12}_{H^2}\|\nabla_hu_h\|^{\frac12}_{H^2}\|\partial_3\theta\|_{H^2}^{\frac32}\|\theta\|_{H^2}^{\frac12} +C\|u\|_{H^2}\|\partial_3\theta\|^2_{H^2}\\ &\le& C(\|u\|_{H^2}+\|\theta\|_{H^2})(\|\nabla_hu_h\|^2_{H^2}+\|\partial_3\theta\|^2_{H^2}). \end{eqnarray*}$

故

$\begin{equation}\label{estimate-A3} A_3\le C(\|u\|_{H^2}+\|\theta\|_{H^2})(\|\nabla_hu_h\|^2_{H^2}+\|\partial_3\theta\|^2_{H^2}). \end{equation}$

结合(3.8), (3.9)和(3.10)式得

$\begin{eqnarray*}\label{estimate-Hcdot2} &&\frac12\frac{\rm d}{{\rm d}t}(\|u_h\|^2_{\dot{H}^2}+\| u_3\|^2_{\dot{H}^2}+\|\theta\|^2_{\dot{H}^2})+\nu\|\nabla_h u_h\|^2_{\dot{H}^2}+\eta\|\partial_3\theta\|^2_{\dot{H}^2}\\ &\le& C(\|u\|_{H^2}+\|\theta\|_{H^2})(\|\nabla_hu_h\|^2_{H^2}+\|\partial_3\theta\|^2_{H^2}). \end{eqnarray*}$

将(3.4)和(3.11)式相加后两边关于时间积分可得(3.2)式. 即完成命题3.2的证明.

最后给出定理1.1的证明.

${\bf证}$ 类似文献[12,定理1.1]的证明,利用靴代证明法并结合(3.2)式可证定理1.1.具体细节省略.

4 定理1.2的证明

本节将给出定理1.2的证明. 首先建立(1.7)式解的$H^3$整体有界性, 然后证明(1.7)式解的大时间行为, 同时得到(1.7)式的极限方程. 令

$\begin{equation}\label{definition-Ft} {\cal F}(t)=\sup_{0\le t\le T}(\|u(t)\|^2_{H^3}+\|\theta(t)\|^2_{H^3})+\int_0^t(\nu\|\nabla u_h(\tau)\|^2_{H^3}+\nu\|\partial_3u_3(\tau)\|^2_{H^3} +\eta\|\partial_3\theta(\tau)\|^2_{H^3}){\rm d}\tau. \end{equation} $

下面给出方程(1.7)解的$H^3$能量估计.

${\bf命题4.1}$ 设$(u_0,\theta_0)\in H^3$满足对条件(1.8). 若$(u,\theta)$是方程(1.7)的解, 则对任意$t\in[T]$, 存在常数 $C>0$使得

$\begin{equation}\label{estimate-Ft} {\cal F}(t)\le{\cal F}_0+C{\cal F}(t)^{\frac32}. \end{equation}$

${\bf证}$ 与命题3.2的证明一样可得(3.3)式. 将命题4.1的证明过程分成两步.

第一步 $(u,\theta)$的$L^2-$能量估计.

$(1.7)_1,~(1.7)_2,~(1.7)_3$式分别与$u_h,~u_3,~\theta$作$L^2$内积,

利用$\nabla\cdot u=0$, 分部积分得

$\begin{equation}\label{L2} \frac12\frac{\rm d}{{\rm d}t}(\|u_h\|^2_{L^2}+\|u_3\|^2_{L^2}+\|\theta\|^2_{L^2})+\nu\|\nabla u_h\|^2_{L^2}+\nu\|\partial_3u_3\|^2_{L^2}+\eta\|\partial_3\theta\|^2_{L^2}\le0. \end{equation}$

第二步 $(u,\theta)$的$\dot{H}^3$ -能量估计.

$(1.7)_1, (1.7)_2,(1.7)_3$式作用$\partial_i^2(i=1,2,3)$)后分别与$\partial_i^3 u_h,~ \partial_i^3 u_3,~\partial_i^3\theta$作$L^2$内积可得

$\begin{eqnarray*}\label{Hcdot3} &&\frac12\frac{\rm d}{{\rm d}t}(\|u_h\|^2_{\dot{H}^3}+\| u_3\|^2_{\dot{H}^3}+\|\theta\|^2_{\dot{H}^3})+\nu\|\nabla u_h\|^2_{\dot{H}^3}+\nu\|\partial_3u_3\|^2_{\dot{H}^3}+\eta\|\partial_3\theta\|^2_{\dot{H}^3}\\ &=&-\sum_{i=1}^3\int\partial_i^3(u\cdot\nabla u_h)\cdot\partial_i^3u_h-\sum_{i=1}^3\int\partial_i^3(u\cdot\nabla u_3)\cdot\partial_i^3u_3-\sum_{i=1}^3\int\partial_i^3(u\cdot\nabla\theta)\cdot\partial_i^3\theta\\ &=&B_1+B_2+B_3. \end{eqnarray*}$

利用$\nabla\cdot u=0$, 将$B_1$分解为

$\begin{eqnarray*} B_1&=&-\sum_{i=1}^3\int\partial_i^3(u\cdot\nabla u_h)\cdot\partial_i^3u_h\\ &=&-\sum_{i=1}^2\int\partial_i^3(u\cdot\nabla u_h)\cdot\partial_i^3u_h-\int\partial_3^3(u\cdot\nabla u_h)\cdot\partial_3^3u_h\\ &=&-\sum_{i=1}^2\sum_{j=1}^3\int\partial_i^3(u_j\partial_j u_h)\cdot\partial_i^3u_h-\sum_{j=1}^3\int\partial_3^3(u_j\partial_ju_h)\cdot\partial_3^3u_h\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^3(u_j\partial_j u_h)\cdot\partial_i^3u_h-\sum_{i=1}^2\int\partial_i^3(u_3\partial_3 u_h)\cdot\partial_i^3u_h\\ &&-\sum_{j=1}^2\int\partial_3^3(u_j\partial_ju_h)\cdot\partial_3^3u_h-\int\partial_3^3(u_3\partial_3u_h)\cdot\partial_3^3u_h\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_j\partial_j\partial_i^{3-\alpha}u_h\cdot\partial_i^3u_h-\sum_{i=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_3\partial_3\partial_i^{3-\alpha}u_h\cdot\partial_i^3u_h\\ &&-\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_j\partial_j\partial_3^{3-\alpha}u_h\cdot\partial_3^3u_h-\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_3\partial_3\partial_3^{3-\alpha}u_h\cdot\partial_3^3u_h\\ &=&B_{11}+B_{12}+B_{13}+B_{14}. \end{eqnarray*} $

由Hölder不等式和Sobolev不等式可知

$\begin{eqnarray*} B_{11}&=&-\sum_{i=1}^2\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_j\partial_j\partial_i^{3-\alpha}u_h\cdot\partial_i^3u_h\\ &\le & C\sum_{i=1}^2\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\|\partial_i^\alpha u_j\|_{L^4}\|\partial_j\partial_i^{3-\alpha}u_h\|_{L^2}\|\partial_i^3u_h\|_{L^4}\le C\|u\|_{H^3}\|\nabla_hu_h\|^2_{H^3}. \end{eqnarray*}$

类似地有

$\begin{eqnarray*} B_{12}&=&-\sum_{i=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_3\partial_3\partial_i^{3-\alpha}u_h\cdot\partial_i^3u_h\\ &\le & C\sum_{i=1}^2\sum_{\alpha=1}^3C_3^\alpha\|\partial_i^\alpha u_3\|_{L^2}\|\partial_3\partial_i^{3-\alpha}u_h\|_{L^4}\|\partial_i^3u_h\|_{L^4}\le C\|u\|_{H^3}\|\nabla u_h\|^2_{H^3} \end{eqnarray*}$

和

$\begin{eqnarray*} B_{13}&=&-\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_j\partial_j\partial_3^{3-\alpha}u_h\cdot\partial_3^3u_h\\ &\le & C\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\|\partial_3^\alpha u_j\|_{L^4}\|\partial_j\partial_3^{3-\alpha}u_h\|_{L^4}\|\partial_3^3u_h\|_{L^2}\le C\|u\|_{H^3}\|\nabla u_h\|^2_{H^3}. \end{eqnarray*}$

根据Hölder不等式, Sobolev不等式以及$\nabla\cdot u=0$可得

$\begin{eqnarray*} B_{14}&=&-\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_3\partial_3\partial_3^{3-\alpha}u_h\cdot\partial_3^3u_h\\ &\le & C\sum_{\alpha=1}^3C_3^\alpha\|\partial_3^\alpha u_3\|_{L^2}\|\partial_3\partial_3^{3-\alpha}u_h\|_{L^4}\|\partial_3^3u_h\|_{L^4}\le C\|u\|_{H^3}\|\nabla u_h\|^2_{H^3}. \end{eqnarray*} $

因此

$\begin{eqnarray*}\label{B1} B_{1}\le C\|u\|_{H^3}\|\nabla u_h\|^2_{H^3}. \end{eqnarray*}$

利用$\nabla\cdot u=0$将$B_2$分解为下四部分

$\begin{eqnarray*} B_2&=&-\sum_{i=1}^3\int\partial_i^3(u\cdot\nabla u_3)\cdot\partial_i^3u_3\\ &=&-\sum_{i=1}^2\int\partial_i^3(u\cdot\nabla u_3)\cdot\partial_i^3u_3-\int\partial_3^3(u\cdot\nabla u_3)\cdot\partial_3^3u_3\\ &=&-\sum_{i=1}^2\sum_{j=1}^3\int\partial_i^3(u_j\partial_j u_3)\cdot\partial_i^3u_3-\sum_{j=1}^3\int\partial_3^3(u_j\partial_ju_3)\cdot\partial_3^3u_3\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^3(u_j\partial_j u_3)\cdot\partial_i^3u_3-\sum_{i=1}^2\int\partial_i^3(u_3\partial_3 u_3)\cdot\partial_i^3u_3\\ &&-\sum_{j=1}^2\int\partial_3^3(u_j\partial_ju_3)\cdot\partial_3^3u_3-\int\partial_3^3(u_3\partial_3u_3)\cdot\partial_3^3u_3\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_j\partial_j\partial_i^{3-\alpha}u_3\cdot\partial_i^3u_3-\sum_{i=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_3\partial_3\partial_i^{3-\alpha}u_3\cdot\partial_i^3u_3\\ &&-\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_j\partial_j\partial_3^{3-\alpha}u_3\cdot\partial_3^3u_3-\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_3\partial_3\partial_3^{3-\alpha}u_3\cdot\partial_3^3u_3\\ &=&B_{21}+B_{22}+B_{23}+B_{24}. \end{eqnarray*}$

利用$\overline{u_3}=0$, Hölder不等式, Sobolev不等式和引理2.1(1)-(3)得

$\begin{eqnarray*} B_{21}&=&-\sum_{i=1}^2\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_j\partial_j\partial_i^{3-\alpha}u_3\cdot\partial_i^3u_3\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int3\partial_i u_j\partial_j\partial_i^{2}u_3\cdot\partial_i^3\widetilde{u_3}+3\partial_i^2 u_j\partial_j\partial_iu_3\cdot\partial_i^3\widetilde{u_3}+\partial_i^3 u_j\partial_ju_3\cdot\partial_i^3\widetilde{u_3}\\ &\le & C\sum_{i=1}^2\sum_{j=1}^2\|\partial_i u_j\|_{L^\infty}\|\partial_j\partial_i^{2}u_3\|_{L^2}\|\partial_i^3\widetilde{u_3}\|_{L^2}+\|\partial_i^2 u_j\|_{L^\infty}\|\partial_j\partial_iu_3\|_{L^2}\|\partial_i^3\widetilde{u_3}\|_{L^2}\\ &&+\|\partial_i^3 u_j\|_{L^2}\|u_3\|_{L^\infty}\|\partial_i^3\widetilde{u_3}\|_{L^2}\\ &\le& C\|u\|_{H^3}(\|\nabla u_h\|^2_{H^3}+\|\partial_3 u_3\|^2_{H^3}). \end{eqnarray*}$

类似地

$\begin{eqnarray*} B_{22}&=&-\sum_{i=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_3\partial_3\partial_i^{3-\alpha}u_3\cdot\partial_i^3u_3\\ &=&-\sum_{i=1}^2\int3\partial_i\widetilde{u_3}\partial_3\partial_i^{2}u_3\cdot\partial_i^3u_3+3\partial_i^2 \widetilde{u_3}\partial_3\partial_iu_3\cdot\partial_i^3u_3+\partial_i^3 \widetilde{u_3}\partial_3u_3\cdot\partial_i^3u_3\\ &\le & C\|u\|_{H^3}\|\partial_3 u_3\|^2_{H^3}. \end{eqnarray*} $

由Hölder不等式和Sobolev不等式可知

$\begin{eqnarray*} B_{23}&=&-\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_j\partial_j\partial_3^{3-\alpha}u_3\cdot\partial_3^3u_3\\ &\le & C\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\|\partial_3^\alpha u_j\|_{L^2}\|\partial_j\partial_3^{3-\alpha}u_3\|_{L^4}\|\partial_3^3u_3\|_{L^4} \le C\|u\|_{H^3}\|\partial_3 u_3\|^2_{H^3} \end{eqnarray*}$

以及

$\begin{eqnarray*} B_{24}&=&-\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_3\partial_3\partial_3^{3-\alpha}u_3\cdot\partial_3^3u_3\\ &\le & C\sum_{\alpha=1}^3C_3^\alpha\|\partial_3^\alpha u_3\|_{L^4}\|\partial_3\partial_3^{3-\alpha}u_3\|_{L^2}\|\partial_3^3u_3\|_{L^4} \le C\|u\|_{H^3}\|\partial_3 u_3\|^2_{H^3}. \end{eqnarray*}$

故

$\begin{equation}\label{B2} B_{2}\le C\|u\|_{H^3}(\|\nabla u_h\|^2_{H^3}+\|\partial_3u_3\|^2_{H^3}). \end{equation}$

类似于$B_1$或$B_2$有

$\begin{eqnarray*} B_3&=&-\sum_{i=1}^3\int\partial_i^3(u\cdot\nabla \theta)\cdot\partial_i^3\theta\\ &=&-\sum_{i=1}^2\int\partial_i^3(u\cdot\nabla \theta)\cdot\partial_i^3\theta-\int\partial_3^3(u\cdot\nabla \theta)\cdot\partial_3^3\theta\\ &=&-\sum_{i=1}^2\sum_{j=1}^3\int\partial_i^3(u_j\partial_j \theta)\cdot\partial_i^3\theta-\sum_{j=1}^3\int\partial_3^3(u_j\partial_j\theta)\cdot\partial_3^3\theta\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int\partial_i^3(u_j\partial_j \theta)\cdot\partial_i^3\theta-\sum_{i=1}^2\int\partial_i^3(u_3\partial_3 \theta)\cdot\partial_i^3\theta\\ &&-\sum_{j=1}^2\int\partial_3^3(u_j\partial_j\theta)\cdot\partial_3^3\theta-\int\partial_3^3(u_3\partial_3\theta)\cdot\partial_3^3\theta\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_j\partial_j\partial_i^{3-\alpha}\theta\cdot\partial_i^3\theta-\sum_{i=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_3\partial_3\partial_i^{3-\alpha}\theta\cdot\partial_i^3\theta\\ &&-\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_j\partial_j\partial_3^{3-\alpha}\theta\cdot\partial_3^3\theta-\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_3\partial_3\partial_3^{3-\alpha}\theta\cdot\partial_3^3\theta\\ &=&B_{31}+B_{32}+B_{33}+B_{34}. \end{eqnarray*} $

由$\bar{\theta}=0$, Hölder不等式, Sobolev不等式和引理2.1(1)-(3)可知

$\begin{eqnarray*} B_{31}&=&-\sum_{i=1}^2\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_j\partial_j\partial_i^{3-\alpha}\theta\cdot\partial_i^3\theta\\ &=&-\sum_{i=1}^2\sum_{j=1}^2\int3\partial_i u_j\partial_j\partial_i^{2}\theta\cdot\partial_i^3\widetilde{\theta}+3\partial_i^2 u_j\partial_j\partial_i\theta\cdot\partial_i^3\widetilde{\theta}+\partial_i^3 u_j\partial_j\theta\cdot\partial_i^3\widetilde{\theta}\\ &\le & C\sum_{i=1}^2\sum_{j=1}^2\|\partial_i u_j\|_{L^\infty}\|\partial_j\partial_i^{2}\theta\|_{L^2}\|\partial_i^3\widetilde{\theta}\|_{L^2}+\|\partial_i^2 u_j\|_{L^\infty}\|\partial_j\partial_i\theta\|_{L^2}\|\partial_i^3\widetilde{\theta}\|_{L^2}\\ &&+\|\partial_i^3 u_j\|_{L^2}\|\partial_j\theta\|_{L^\infty}\|\partial_i^3\widetilde{\theta}\|_{L^2}\\ &\le& C\|\theta\|_{H^3}(\|\nabla u_h\|^2_{H^3}+\|\partial_3\theta\|^2_{H^3}) \end{eqnarray*}$

和

$\begin{eqnarray*} B_{32}&=&-\sum_{i=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_i^\alpha u_3\partial_3\partial_i^{3-\alpha}\theta\cdot\partial_i^3\theta\\ &=&-\sum_{i=1}^2\int3\partial_iu_3\partial_3\partial_i^{2}\theta\cdot\partial_i^3\widetilde{\theta}+3\partial_i^2 u_3\partial_3\partial_i\theta\cdot\partial_i^3\widetilde{\theta}+\partial_i^3 u_3\partial_3\theta\cdot\partial_i^3\widetilde{\theta}\\ &\le & C\sum_{i=1}^2\|\partial_iu_3\|_{L^\infty}\|\partial_3\partial_i^{2}\theta\|_{L^2}\|\partial_i^3\widetilde{\theta}\|_{L^2}+\|\partial_i^2 u_3\|_{L^4}\|\partial_3\partial_i\theta\|_{L^4}\|\partial_i^3\widetilde{\theta}\|_{L^2}\\ &&+\|\partial_i^3 u_3\|_{L^2}\|\partial_3\theta\|_{L^\infty}\|\partial_i^3\widetilde{\theta}\|_{L^2}\\ &\le &C\|u\|_{H^3}\|\partial_3\theta\|^2_{H^3}. \end{eqnarray*}$

利用Hölder不等式和 Sobolev不等式可得

$\begin{eqnarray*} B_{33}&=&-\sum_{j=1}^2\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_j\partial_j\partial_3^{3-\alpha}\theta\cdot\partial_3^3\theta\\ &=&-\sum_{j=1}^2\int3\partial_3u_j\partial_j\partial_3^{2}\theta\cdot\partial_3^3\theta+3\partial_3^2u_j\partial_j\partial_3\theta\cdot\partial_3^3\theta +\partial_3^3u_j\partial_j\theta\cdot\partial_3^3\theta\\ &\le & C\sum_{j=1}^2\|\partial_3u_j\|_{L^\infty}\|\partial_3^2\partial_j\theta\|_{L^2}\|\partial_3^3\theta\|_{L^2}+\|\partial_3^2u_j\|_{L^4}\|\partial_3\partial_j\theta\|_{L^4}\|\partial_3^3\theta\|_{L^2}\\ &&+\|\partial^3_3u_j\|_{L^2}\|\partial_j\theta\|_{L^4}\|\partial_3^3\theta\|_{L^4}\\ &\le& C(\|u\|_{H^3}+\|\theta\|_{H^3})(\|\nabla u_h\|^2_{H^3}+\|\partial_3\theta\|^2_{H^3}) \end{eqnarray*}$

和

$ B_{34}=-\sum_{\alpha=1}^3C_3^\alpha\int\partial_3^\alpha u_3\partial_3\partial_3^{3-\alpha}\theta\cdot\partial_3^3\theta {\rm d}x\le C\|u\|_{H^3}\|\partial_3\theta\|^2_{H^3}. $

所以有

$\begin{equation}\label{B3} B_{3}\le C(\|u\|_{H^3}+\|\theta\|_{H^3})(\|\nabla u_h\|^2_{H^3}+\|\partial_3\theta\|^2_{H^3}). \end{equation} $

结合(4.1), (4.3), (4.4), (4.5), (4.6)和(4.7)式可完成命题4.1的证明.

下面给出定理1.2的证明.

${\bf证}$ 利用命题4.1中的(4.2)式可得到解的$H^3$ 整体存在性和稳定性, 即可得到定理1.2(i).

对于方程组(1.7)的解$(u,\theta)$, (3.3)式仍然成立. 且通过取方程组(1.7)的垂直平均, 以及利用引理2.1和(3.3)式可得到极限方程组

$\begin{equation}\label{eq3:Boussinesq-average} \left\{ \begin{array}{ll} \partial_t \overline{u_1}+\overline{u\cdot\nabla \widetilde{u_1}}-\nu\Delta_h \overline{u_1}+\overline{u_1}\partial_1\overline{u_1}+\overline{u_2}\partial_2\overline{u_1}+\partial_1 \overline{P}=0,~x\in\Omega,\\ \partial_t \overline{u_2}+\overline{u\cdot\nabla \widetilde{u_2}}-\nu\Delta_h \overline{u_2}+\overline{u_1}\partial_1\overline{u_2}+\overline{u_2}\partial_2\overline{u_2}+\partial_2 \overline{P}=0,\\ \partial_1\overline{u_1}+\partial_1\overline{u_2}=0. \\ \end{array} \right. \end{equation}$

而(1.7)与(4.8)式作差可得$(\tilde{u},\tilde{\theta})$的方程组

$\begin{equation}\label{eq4:Boussinesq-tilde} \left\{ \begin{array}{ll} \partial_t \widetilde{u_1}+\widetilde{u\cdot\nabla \widetilde{u_1}}-\nu\Delta \widetilde{u_1}+\widetilde{u_1}\partial_1\overline{u_1}+\widetilde{u_2}\partial_2\overline{u_1}+\partial_1 \widetilde{P}=0,~x\in\Omega,\\ \partial_t \widetilde{u_2}+\widetilde{u\cdot\nabla \widetilde{u_2}}-\nu\Delta \widetilde{u_2}+\widetilde{u_1}\partial_1\overline{u_2}+\widetilde{u_2}\partial_2\overline{u_2}+\partial_2 \widetilde{P}=0,\\ \partial_t \widetilde{u_3}+u\cdot\nabla \widetilde{u_3}-\nu\partial_{33} \widetilde{u_3}+\partial_3 \widetilde{P}=\widetilde{\theta},\\ \partial_t \widetilde{\theta}+u\cdot\nabla \widetilde{\theta}-\eta\partial_{33} \widetilde{\theta}+\widetilde{u_3}=0,\\ \nabla\cdot\tilde{u}=0. \\ \end{array} \right. \end{equation}$

为了得到$(\tilde{u},\tilde{\theta})$的衰减估计, 需要得到$(\tilde{u},\tilde{\theta})$和$(\nabla\tilde{u},\nabla\tilde{\theta})$的$L^2$能量估计. (49)分别与$(\widetilde{u_1},\widetilde{u_2},\widetilde{u_3},\widetilde{\theta})$ 作$L^2$内积后分部积分得

$\begin{eqnarray*} &&\frac12\frac{\rm d}{{\rm d}t}(\|\tilde{u}\|^2_{L^2}+\|\tilde{\theta}\|^2_{L^2})+\nu\|\nabla \widetilde{u_1}\|^2_{L^2}+\nu\|\nabla \widetilde{u_2}\|^2_{L^2}+\nu\|\partial_3 \widetilde{u_3}\|^2_{L^2}+\eta\|\partial_3\tilde{\theta}\|^2_{L^2}\\ &=&-\int\widetilde{u\cdot\nabla\widetilde{u_1}}\cdot\widetilde{u_1}-\int\widetilde{u_1}\partial_1\overline{u_1}\cdot\widetilde{u_1}-\int\widetilde{u_2}\partial_2\overline{u_1}\cdot\widetilde{u_1}\\ &&-\int\widetilde{u\cdot\nabla\widetilde{u_2}}\cdot\widetilde{u_2}-\int\widetilde{u_1}\partial_1\overline{u_2}\cdot\widetilde{u_2}-\int\widetilde{u_2}\partial_2\overline{u_2}\cdot\widetilde{u_2}\\ &=&D_1+D_2+D_3+D_4+D_5+D_6. \end{eqnarray*}$

由$\nabla\cdot u=0$, $(2.1)$和引理2.1(2)知

$\begin{equation} D_1=-\int\widetilde{u\cdot\nabla\widetilde{u_1}}\cdot\widetilde{u_1}=-\int u\cdot\nabla\widetilde{u_1}\cdot\widetilde{u_1}+\int_{\mathbb{R}^2}\overline{u\cdot\nabla\widetilde{u_1}}\int_{\mathbb{T}}\widetilde{u_1}=0. \end{equation}$

类似地有$D_4=0$. 利用Hölder不等式, Sobolev不等式和引理2.1(2)可得

$\begin{eqnarray*} D_2+D_3+D_5+D_6 &=&-\int\widetilde{u_1}\partial_1\overline{u_1}\cdot\widetilde{u_1}-\int\widetilde{u_2}\partial_2\overline{u_1}\cdot\widetilde{u_1} -\int\widetilde{u_1}\partial_1\overline{u_2}\cdot\widetilde{u_2}-\int\widetilde{u_2}\partial_2\overline{u_2}\cdot\widetilde{u_2}\\ &\le&\|\partial_1\overline{u_1}\|_{L^\infty}\|\widetilde{u_1}\|^2_{L^2}+\|\partial_2\overline{u_1}\|_{L^\infty} \|\widetilde{u_1}\|_{L^2}\|\widetilde{u_2}\|_{L^2}\\ &&+\|\partial_1\overline{u_2}\|_{L^\infty}\|\widetilde{u_1}\|_{L^2}\|\widetilde{u_2}\|_{L^2} +\|\partial_2\overline{u_2}\|_{L^\infty}\|\widetilde{u_2}\|^2_{L^2}\\ &\le & C\|u\|_{H^3}(\|\partial_3\widetilde{u_1}\|_{L^2}^2+\|\partial_3\widetilde{u_2}\|_{L^2}^2). \end{eqnarray*} $

这意味着

$\begin{eqnarray*} &&\frac{\rm d}{{\rm d}t}(\|\tilde{u}\|^2_{L^2}+\|\tilde{\theta}\|^2_{L^2})+2\nu\|\nabla \widetilde{u_1}\|^2_{L^2}+2\nu\|\nabla \widetilde{u_2}\|^2_{L^2}+2\nu\|\partial_3 \widetilde{u_3}\|^2_{L^2}+2\eta\|\partial_3\tilde{\theta}\|^2_{L^2}\\ &\le & C\|u\|_{H^3}(\|\partial_3\widetilde{u_1}\|_{L^2}^2+\|\partial_3\widetilde{u_2}\|_{L^2}^2), \end{eqnarray*}$

利用$\|\partial_3 \widetilde{f}\|_{L^2}\le\|\nabla \widetilde{f}\|_{L^2}$进一步可得

$\begin{eqnarray*} &&\frac{\rm d}{{\rm d}t}(\|\tilde{u}\|^2_{L^2}+\|\tilde{\theta}\|^2_{L^2})+2\nu\|\partial_3 \widetilde{u_1}\|^2_{L^2}+2\nu\|\partial_3 \widetilde{u_2}\|^2_{L^2}+2\nu\|\partial_3 \widetilde{u_3}\|^2_{L^2}+2\eta\|\partial_3\tilde{\theta}\|^2_{L^2}\\ &\le & C\|u\|_{H^3}(\|\partial_3\widetilde{u_1}\|_{L^2}^2+\|\partial_3\widetilde{u_2}\|_{L^2}^2). \end{eqnarray*}$

由引理2.1(2)中的Poincaré不等式可知

$\begin{matrix} &&\frac{\rm d}{{\rm d}t}(\|\tilde{u}\|^2_{L^2}+\|\tilde{\theta}\|^2_{L^2})+\tilde{C}(2\nu-C\|u\|_{H^3})\| \widetilde{u_1}\|^2_{L^2}+\tilde{C}(2\nu-C\|u\|_{H^3})\| \widetilde{u_2}\|^2_{L^2}\\ &&+2\tilde{C}\nu\| \widetilde{u_3}\|^2_{L^2}+2\eta\|\tilde{\theta}\|^2_{L^2}\le 0. \end{matrix}$

而定理1.2(i)给出, 若对足够小的$\epsilon>0$有$\|u_0\|_{H^3}+\|\theta_0\|_{H^3}\le\epsilon$, 则$\|u(t)\|_{H^3}+\|\theta(t)\|_{H^3}\le C\epsilon$且

$ 2\nu-C\|u(t)\|_{H^3}\ge\nu, $

故有

$\begin{equation}\label{decaytildeutheta} \frac{\rm d}{{\rm d}t}(\|\tilde{u}\|^2_{L^2}+\|\tilde{\theta}\|^2_{L^2})+\tilde{C}\min\{\nu,\eta\}(\| \widetilde{u}\|^2_{L^2}+\|\tilde{\theta}\|^2_{L^2})\le 0. \end{equation}$

$(4.9)_1, (4.9)_2, (4.9)_3, (4.9)_4$式作用$\nabla$后分别与$(\nabla \widetilde{u_1},\nabla \widetilde{u_2},\nabla\widetilde{u_3},\nabla\tilde{\theta})$作$L^2$内积可得

$\begin{eqnarray*} &&\frac12\frac{\rm d}{{\rm d}t}(\|\nabla\tilde{u}\|^2_{L^2}+\|\nabla\tilde{\theta}\|^2_{L^2})+\nu\|\nabla^2 \widetilde{u_1}\|^2_{L^2}+\nu\|\nabla^2 \widetilde{u_2}\|^2_{L^2}+\nu\|\partial_3\nabla \widetilde{u_3}\|^2_{L^2}+\eta\|\partial_3\nabla\tilde{\theta}\|^2_{L^2}\\ &=&-\int\nabla(\widetilde{u\cdot\nabla\widetilde{u_1}})\cdot\nabla\widetilde{u_1}-\int\nabla(\widetilde{u_1}\partial_1\overline{u_1})\cdot\nabla\widetilde{u_1}-\int\nabla(\widetilde{u_2}\partial_2\overline{u_1})\cdot\nabla\widetilde{u_1}\\ &&-\int\nabla(\widetilde{u\cdot\nabla\widetilde{u_2}})\cdot\nabla\widetilde{u_2}-\int\nabla(\widetilde{u_1}\partial_1\overline{u_2})\cdot\nabla\widetilde{u_2}-\int\nabla(\widetilde{u_2}\partial_2\overline{u_2})\cdot\nabla\widetilde{u_2}\\ &&-\int\nabla(u\cdot\nabla\widetilde{u_3})\cdot\nabla\widetilde{u_3}-\int\nabla(u\cdot\nabla\widetilde{\theta})\cdot\nabla\widetilde{\theta}=\sum_{i=1}^{8}E_i. \end{eqnarray*} $

由$\nabla\cdot u=0$, (2.1), Hölder不等式, Sobolev不等式和引理2.1(2)可得

$\begin{eqnarray*} E_1&=&-\int\nabla(\widetilde{u\cdot\nabla\widetilde{u_1}})\cdot\nabla\widetilde{u_1}\\ &=&-\int\nabla( u\cdot\nabla\widetilde{u_1})\cdot\nabla\widetilde{u_1}+\int\nabla( \overline{u\cdot\nabla\widetilde{u_1}})\cdot\nabla\widetilde{u_1}\\ &=&-\int\nabla u\cdot\nabla\widetilde{u_1}\cdot\nabla\widetilde{u_1}\le C\|u\|_{H^3}\|\partial_3\nabla\widetilde{u_1}\|^2_{L^2}. \end{eqnarray*} $

类似地有

$ E_4\le C\|u\|_{H^3}\|\partial_3\nabla\widetilde{u_2}\|^2_{L^2}. $

利用Hölder不等式, Sobolev不等式, 引理2.1(2)和引理2.3得

$\begin{eqnarray*} E_2&=&-\int\nabla(\widetilde{u_1}\partial_1\overline{u_1})\cdot\nabla\widetilde{u_1}\\ &=&-\int\nabla\widetilde{u_1}\partial_1\overline{u_1}\cdot\nabla\widetilde{u_1}-\int\widetilde{u_1}\partial_1\nabla\overline{u_1} \cdot\nabla\widetilde{u_1}\\ &\le & C\|\partial_1\overline{u_1}\|_{L^\infty}\|\nabla\widetilde{u_1}\|^2_{L^2}+C\|\nabla\partial_1\overline{u_1}\|^{\frac12}_{L^2} \|\partial_1\nabla\partial_1\overline{u_1}\|^{\frac12}_{L^2} \|\widetilde{u_1}\|^{\frac12}_{L^2}\|\partial_2\widetilde{u_1}\|^{\frac12}_{L^2} \|\nabla\widetilde{u_1}\|^{\frac12}_{L^2}\|\partial_3\nabla\widetilde{u_1}\|^{\frac12}_{L^2}\\ &\le& C\|u\|_{H^3}\|\partial_3\nabla\widetilde{u_1}\|^2_{L^2}. \end{eqnarray*}$

同理

$\begin{eqnarray*} E_3&=&-\int\nabla(\widetilde{u_2}\partial_2\overline{u_1})\cdot\nabla\widetilde{u_1}\\ &=&-\int\nabla\widetilde{u_2}\partial_2\overline{u_1}\cdot\nabla\widetilde{u_1}-\int\widetilde{u_2}\partial_2\nabla\overline{u_1}\cdot\nabla\widetilde{u_1}\\ &\le & C\|\partial_2\overline{u_1}\|_{L^\infty}\|\nabla\widetilde{u_1}\|_{L^2}\|\nabla\widetilde{u_2}\|_{L^2}+C\|\nabla\partial_2\overline{u_1}\|^{\frac12}_{L^2}\|\partial_1\nabla\partial_2\overline{u_1}\|^{\frac12}_{L^2}\\ &&\times\|\widetilde{u_2}\|^{\frac12}_{L^2}\|\partial_2\widetilde{u_2}\|^{\frac12}_{L^2}\|\nabla\widetilde{u_1}\|^{\frac12}_{L^2}\|\partial_3\nabla\widetilde{u_1}\|^{\frac12}_{L^2}\\ &\le & C\|u\|_{H^3}(\|\partial_3\nabla\widetilde{u_1}\|^2_{L^2}+\|\partial_3\nabla\widetilde{u_2}\|^2_{L^2}). \end{eqnarray*}$

采取估计$E_2$和$E_3$时用的方法可得

$\begin{eqnarray*} &&E_5=-\int\nabla(\widetilde{u_1}\partial_1\overline{u_2})\cdot\widetilde{u_2}\le C\|u\|_{H^3}(\|\partial_3\nabla\widetilde{u_1}\|^2_{L^2}+\|\partial_3\nabla\widetilde{u_2}\|^2_{L^2}),\\ && E_6=-\int\nabla(\widetilde{u_2}\partial_2\overline{u_2})\cdot\widetilde{u_2}\le C\|u\|_{H^3}\|\partial_3\nabla\widetilde{u_2}\|^2_{L^2}. \end{eqnarray*}$

根据$\nabla\cdot u=0$, Hölder不等式, Sobolev不等式和引理2.1(2)可得

$\begin{eqnarray*} &&E_7=-\int\nabla(u\cdot\widetilde{u_3})\cdot\nabla\widetilde{u_3} =-\int\nabla u\cdot\widetilde{u_3}\cdot\nabla\widetilde{u_3} \le C\|u\|_{H^3}\|\partial_3\nabla\widetilde{u_3}\|^2_{L^2},\\ &&E_8=-\int\nabla(u\cdot\widetilde{\theta})\cdot\nabla\widetilde{\theta}=-\int\nabla u\cdot\widetilde{\theta}\cdot\nabla\widetilde{\theta} \le C\|u\|_{H^3}\|\partial_3\nabla\widetilde{\theta}\|^2_{L^2}. \end{eqnarray*}$

综合$E_i(i=1,2,\cdots,8)$的估计得

$\begin{eqnarray*} &&\frac{\rm d}{{\rm d}t}(\|\nabla\tilde{u}\|^2_{L^2}+\|\nabla\tilde{\theta}\|^2_{L^2})+2\nu\|\partial_3\nabla \widetilde{u_1}\|^2_{L^2}+2\nu\|\partial_3\nabla \widetilde{u_2}\|^2_{L^2}+2\nu\|\partial_3\nabla \widetilde{u_3}\|^2_{L^2}+2\eta\|\partial_3\nabla\tilde{\theta}\|^2_{L^2}\\ &\le& C\|u\|_{H^3}(\|\partial_3\nabla\widetilde{u_1}\|^2_{L^2}+\|\partial_3\nabla\widetilde{u_2}\|^2_{L^2}+\|\partial_3\nabla\widetilde{u_3}\|^2_{L^2}+\|\partial_3\nabla\widetilde{\theta}\|^2_{L^2}). \end{eqnarray*}$

即

$\begin{matrix} &&\frac{\rm d}{{\rm d}t}(\|\nabla\tilde{u}\|^2_{L^2}+\|\nabla\tilde{\theta}\|^2_{L^2})+(2\nu-C\|u\|_{H^3})\|\partial_3\nabla \widetilde{u_1}\|^2_{L^2}+(2\nu-C\|u\|_{H^3})\|\partial_3\nabla \widetilde{u_2}\|^2_{L^2} \\ &&+(2\nu-C\|u\|_{H^3})\|\partial_3\nabla \widetilde{u_3}\|^2_{L^2}+(2\nu-C\|u\|_{H^3})\|\partial_3\nabla\tilde{\theta}\|^2_{L^2}\le 0. \end{matrix}$

利用Poincaré不等式以及(4.16)式得

$\begin{equation}\label{decaynablatildeutheta} \frac{\rm d}{{\rm d}t}(\|\nabla\tilde{u}\|^2_{L^2}+\|\nabla\tilde{\theta}\|^2_{L^2})+\tilde{C}\min(\nu,\eta)(\|\nabla \tilde{u}\|^2_{L^2}+\|\nabla\tilde{\theta}\|^2_{L^2})\le 0. \end{equation}$

结合(4.16), (4.19)式和Gronwall不等式可得(1.12)式. 即完成了定理1.2的证明.

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