一类二维MHD-Boussinesq方程组整体解的存在性

Existence of Global Solution to a Class of Two-Dimensional MHD-Boussinesq Equations

摘要: 本文证明了一类二维不可压缩MHD-Boussinesq方程组的初值问题在H^s(R²),s>2空间中存在唯一的整体强解。

Abstract: In this paper, we prove that there exists a unique global strong solution to the initial-value problem of a class of two-dimensional incompressible MHD-Boussinesq equations in H^s(R²),s>2.

1. 引言

磁流体力学(Magnetohydrodynamics，简称MHD)是研究等离子体和磁场相互作用的物理学分支。MHD-Boussinesq系统模拟由热场或密度场的浮力效应和磁场产生的洛伦兹力驱动的不可压缩流体的对流，这种对流发生在存在磁场的导电流体的水平层，它与Rayleigh-Benard对流的自然类型密切相关。本文研究如下形式的MHD-Boussinesq方程组的初值问题：

$\{\begin{array}{l}{\partial }_{t}u-\Delta u+u\cdot \nabla u+\nabla \pi =b\cdot \nabla b+\theta e_2,\\ {\partial }_{t}b-\Delta b+u\cdot \nabla b=b\cdot \nabla u,\\ {\partial }_t\theta -\Delta \theta +u\cdot \nabla \theta =0,\\ \nabla \cdot u=0=\nabla \cdot b,\\ \left(u,b,\theta \right)|{}_{t=0}=\left(u_0,b_0,{\theta }_0\right),\end{array}$ (1)

其中， $t\ge 0,x\in R^2,u=u\left(x,t\right),b=b\left(x,t\right),\theta =\theta \left(x,t\right)$ 分别表示速度，磁场和温度， $\pi =\pi \left(x,t\right)$ 表示压力， $e_2$ 是单位矢量 $\left(0,1\right)$。

2009年，Hmidi T.和Keraani S. [1] 证明了当 $u_0\in B_{p,1}^{1+\frac{2}{p}},{\theta }^0\in L^r,2<r\le p<\infty$ 时，Boussinesq方程组的初值问题存在整体解 $u\in C\left(R_{+};B_{p,1}^{1+\frac{2}{p}}\right),\theta \in L^{\infty }\left(R_{+};L^r\right)\cap {\tilde{L}}_{loc}^1\left(R_{+};B_{r,\infty }^2\right)$。2017年，Larios A.和Pei Y. [2] 证明了当 $m>3,u_0,b_0\in H_x^m\cap V,{\theta }_0\in H_x^m$ 时，初值问题(1)在 $H_x^m\left(\Omega \right)$ 空间中存在局部解。2018年，Zhai X.和Chen Z. [3] 证明了当 $u_0\in B_{2,1}^0\left(R^d\right),b_0\in B_{2,1}^s\left(R^d\right)\cap L^{\infty }\left(R^d\right),{\theta }_0\in B_{2,1}^{s+1}\left(R^d\right)$ 且运动粘度依赖温度时，初值问题(1)

在 $B_{2,1}^s\left(R^d\right)$ 空间中存在整体解。2019年，Li Z., Liu P.和Niu P. [4] 证明了当 $\left(u_0,b_0\right)\in H^s\left(R^2\right)$ 时，三维MHD方程组在 $H^s\left(R^2\right)$ 空间中存在整体解。

本文研究初值问题(1)在 $H^s\left(R^2\right)$ 空间中整体解的存在性，主要结果如下：

定理1.1设 $\left(u_0,b_0,{\theta }_0\right)\in H^s\left(R^2\right),s>2$ 且 $\nabla \cdot u_0=\nabla \cdot b_0=0$，则初值问题(1)存在唯一的强解 $\left(u,b,\theta \right)$，满足

$\left(u,b,\theta \right)\in C\left(\left[0,\infty \right);H^s\left(R^2\right)\right)\cap L^2\left(\left[0,\infty \right);H^{s+1}\left(R^2\right)\right).$

本文中用到的符号：

· $C=C\left({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n\right)$ 表示C是仅依赖于 ${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n$ 的常数。

· 如 $C=C\left({\alpha }_0,{\beta }_0,{\gamma }_0\right)$ 表示C是仅依赖于 ${\alpha }_0,{\beta }_0,{\gamma }_0$ 的常数。

· ${\wedge }^s={\left(1-\Delta \right)}^{\frac{s}{2}}$ 表示具有傅里叶符号 ${\left(1+{\left|\xi \right|}^2\right)}^{\frac{s}{2}}$ 的伪微分算子，其中 $\Delta$ 为Laplace算子。

· P表示Leray投影算子，被定义为 [4] $P\left(u\right)=u-\nabla {\Delta }^{-1}\left(\nabla \cdot u\right),u\in H^s\left(R^{{}^2}\right)$。

· $\left(\cdot ,\cdot \right)$ 表示 $L^2$ 空间中的内积。

· 设X和Y是Banach空间， $X\subset Y$。表示X紧嵌入到Y。

2. 初值问题(1)局部解的存在唯一性

为了研究初值问题(1)局部解的存在性，我们需要引用以下引理。

引理2.1 [5] (Aubin-Lions紧性引理)假定 $X\subset E\subset Y$ 是Banach空间且，则

1) 当 $1\le q\le \infty$ 时，。

2) 当 $1\le r\le \infty$ 时，。

引理2.2 [6] 设 $0<\alpha <1,1<p<\infty$ 且 $f,g\in L^p\left(R^2\right)$，则

${\Vert D^{\alpha }\left(fg\right)-f{D}^{\alpha }g-g{D}^{\alpha }f\Vert }_p\le C\left(s\right){\Vert g\Vert }_{\infty }{\Vert D^{\alpha }f\Vert }_p.$

为了证明初值问题(1)存在局部解，先引入初值问题(1)的磨光近似方程组的初值问题，任给 $\epsilon >0$，

考虑

$\{\begin{array}{l}{\partial }_t{u}^{\epsilon }=P{J}_{\epsilon }\left(\Delta u^{\epsilon }\right)+P{J}_{\epsilon }\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{b}^{\epsilon }\right)\right]-P{J}_{\epsilon }\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{u}^{\epsilon }\right)\right]+P\left(J_{\epsilon }{\theta }^{\epsilon }{e}_2\right),\\ {\partial }_t{b}^{\epsilon }=P{J}_{\epsilon }\left(\Delta b^{\epsilon }\right)+P{J}_{\epsilon }\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{u}^{\epsilon }\right)\right]-P{J}_{\epsilon }\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{b}^{\epsilon }\right)\right],\\ {\partial }_t{\theta }^{\epsilon }=P{J}_{\epsilon }\left(\Delta {\theta }^{\epsilon }\right)-P{J}_{\epsilon }\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{\theta }^{\epsilon }\right)\right],\\ u^{\epsilon }\left(0,x\right)=u_0^{\epsilon }\left(x\right),b^{\epsilon }\left(0,x\right)=b_0^{\epsilon }\left(x\right),{\theta }^{\epsilon }\left(0,x\right)={\theta }_0^{\epsilon }\left(x\right),\end{array}$ (2)

其中， $J_{\epsilon }$ 表示磨光算子。

将初值问题(2)视为 $H^s\left(R^2\right)$ 空间中的抽象常微分方程组的初值问题

$\{\begin{array}{l}\frac{\text{d}{u}^{\epsilon }}{\text{d}t}=P{J}_{\epsilon }\left(\Delta u^{\epsilon }\right)+P{J}_{\epsilon }\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{b}^{\epsilon }\right)\right]-P{J}_{\epsilon }\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{u}^{\epsilon }\right)\right]+P\left(J_{\epsilon }{\theta }^{\epsilon }{e}_2\right),\\ \frac{\text{d}{b}^{\epsilon }}{\text{d}t}=P{J}_{\epsilon }\left(\Delta b^{\epsilon }\right)+P{J}_{\epsilon }\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{u}^{\epsilon }\right)\right]-P{J}_{\epsilon }\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{b}^{\epsilon }\right)\right],\\ \frac{\text{d}{\theta }^{\epsilon }}{\text{d}t}=P{J}_{\epsilon }\left(\Delta {\theta }^{\epsilon }\right)-P{J}_{\epsilon }\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{\theta }^{\epsilon }\right)\right],\\ u^{\epsilon }\left(0,x\right)=u_0^{\epsilon }\left(x\right),b^{\epsilon }\left(0,x\right)=b_0^{\epsilon }\left(x\right),{\theta }^{\epsilon }\left(0,x\right)={\theta }_0^{\epsilon }\left(x\right),\end{array}$ (3)

由经典的Picard定理知，存在 $T^{\epsilon }>0$，使得初值问题(3)有唯一的光滑解

$\left(u^{\epsilon },b^{\epsilon },{\theta }^{\epsilon }\right)\in C\left(\left[0,T^{\epsilon }\right);H^s\left(R^2\right)\right).$

引理2.3假设且 $\nabla \cdot u_0=\nabla \cdot b_0=0$，则存在常数 $T>0$，使得初值问题(1)存在唯一的局部强解 $\left(u,b,\theta \right)$，满足

$\left(u,b,\theta \right)\in C\left(\left[0,T\right);H^s\left(R^2\right)\right)\cap L^2\left(\left[0,T\right);H^{s+1}\left(R^2\right)\right).$

证明：在(2)的三个方程两边作用 ${\wedge }^s$，并分别与 ${\wedge }^{s}u,{\wedge }^{s}b,{\wedge }^s\theta$ 作内积，再将左右两边分别相加，得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert {\wedge }^s{u}^{\epsilon }\Vert }_{L^2}^2+{\Vert {\wedge }^s{b}^{\epsilon }\Vert }_{L^2}^2+{\Vert {\wedge }^s{\theta }^{\epsilon }\Vert }_{L^2}^2\right)\\ =\left({\wedge }^s{J}_{\epsilon }\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{b}^{\epsilon }\right)\right],{\wedge }^s{u}^{\epsilon }\right)-\left({\wedge }^s{J}_{\epsilon }\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{u}^{\epsilon }\right)\right],{\wedge }^s{u}^{\epsilon }\right)\\ +\left({\wedge }^s{J}_{\epsilon }\left(\Delta u^{\epsilon }\right),{\wedge }^s{u}^{\epsilon }\right)+\left({\wedge }^s{J}_{\epsilon }\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{u}^{\epsilon }\right)\right],{\wedge }^s{b}^{\epsilon }\right)\\ -\left({\wedge }^s{J}_{\epsilon }\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{b}^{\epsilon }\right)\right],{\wedge }^s{b}^{\epsilon }\right)+\left({\wedge }^s{J}_{\epsilon }\left(\Delta b^{\epsilon }\right),{\wedge }^s{b}^{\epsilon }\right)\\ -\left({\wedge }^s{J}_{\epsilon }\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \nabla \left(J_{\epsilon }{\theta }^{\epsilon }\right)\right],{\wedge }^s{\theta }^{\epsilon }\right)+\left({\wedge }^s{J}_{\epsilon }\left(\Delta {\theta }^{\epsilon }\right),{\wedge }^s{\theta }^{\epsilon }\right)+\left(\left({\wedge }^s{J}_{\epsilon }{\theta }^{\epsilon }{e}_2\right),{\wedge }^s{u}^{\epsilon }\right)\\ \triangleq {\displaystyle \underset{n=1}{\overset{9}{\sum }}{I}_n}.\end{array}$ (4)

由Green公式，得

$\begin{array}{l}{I}_1=-\left(\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot {\wedge }^s\left(J_{\epsilon }{b}^{\epsilon }\right),{\wedge }^s\nabla J_{\epsilon }{u}^{\epsilon }\right)\\ +\left(\left[{\wedge }^s\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \left(\nabla J_{\epsilon }{b}^{\epsilon }\right)\right]-\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot {\wedge }^s\left(\nabla J_{\epsilon }{b}^{\epsilon }\right)\right],{\wedge }^s{J}_{\epsilon }{u}^{\epsilon }\right)\end{array}$ (5)

$\begin{array}{l}{I}_4=\left(\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot {\wedge }^s\left(\nabla J_{\epsilon }{u}^{\epsilon }\right),{\wedge }^s{J}_{\epsilon }{b}^{\epsilon }\right)\\ +\left(\left[{\wedge }^s\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \left(\nabla J_{\epsilon }{u}^{\epsilon }\right)\right]-\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot {\wedge }^s\left(\nabla J_{\epsilon }{u}^{\epsilon }\right)\right],{\wedge }^s{J}_{\epsilon }{b}^{\epsilon }\right)\end{array}$ (6)

由Hölder不等式和引理2.2，得

$\begin{array}{c}\left|I_1+I_4\right|\le \left|\left(\left[{\wedge }^s\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \left(\nabla J_{\epsilon }{b}^{\epsilon }\right)\right]-\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot {\wedge }^s\left(\nabla J_{\epsilon }{b}^{\epsilon }\right)\right],{\wedge }^s{J}_{\epsilon }{u}^{\epsilon }\right)\right|\\ +\left|\left(\left[{\wedge }^s\left[\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot \left(\nabla J_{\epsilon }{u}^{\epsilon }\right)\right]-\left(J_{\epsilon }{b}^{\epsilon }\right)\cdot {\wedge }^s\left(\nabla J_{\epsilon }{u}^{\epsilon }\right)\right],{\wedge }^s{J}_{\epsilon }{b}^{\epsilon }\right)\right|\\ \le C\left(s\right){\Vert {\wedge }^s{J}_{\epsilon }{b}^{\epsilon }\Vert }_{L^2}{\Vert J_{\epsilon }\nabla b^{\epsilon }\Vert }_{L^{\infty }}{\Vert {\wedge }^s{J}_{\epsilon }{u}^{\epsilon }\Vert }_{L^2}+{\Vert {\wedge }^s{J}_{\epsilon }{b}^{\epsilon }\Vert }_{L^2}{\Vert J_{\epsilon }\nabla u^{\epsilon }\Vert }_{L^{\infty }}{\Vert {\wedge }^s{J}_{\epsilon }{b}^{\epsilon }\Vert }_{L^2}\\ \le C\left(s\right)\left({\Vert J_{\epsilon }\nabla u^{\epsilon }\Vert }_{L^{\infty }}+{\Vert \nabla J_{\epsilon }{b}^{\epsilon }\Vert }_{L^{\infty }}\right)\left({\Vert J_{\epsilon }{u}^{\epsilon }\Vert }_{H^s}^2+{\Vert J_{\epsilon }{b}^{\epsilon }\Vert }_{H^s}^2\right),\end{array}$ (7)

类似地，有

$\begin{array}{c}\left|I_5\right|=\left|\left(\left[{\wedge }^s\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \left(\nabla J_{\epsilon }{b}^{\epsilon }\right)\right]-\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot {\wedge }^s\left(\nabla J_{\epsilon }{b}^{\epsilon }\right)+\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot {\wedge }^s\left(\nabla J_{\epsilon }{b}^{\epsilon }\right)\right],{\wedge }^s{J}_{\epsilon }{b}^{\epsilon }\right)\right|\\ \le \left|\left(\left[{\wedge }^s\left[\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot \left(\nabla J_{\epsilon }{b}^{\epsilon }\right)\right]-\left(J_{\epsilon }{u}^{\epsilon }\right)\cdot {\wedge }^s\left(\nabla J_{\epsilon }{b}^{\epsilon }\right)\right],{\wedge }^s{J}_{\epsilon }{b}^{\epsilon }\right)\right|\\ \le C\left(s\right){\Vert {\wedge }^s{J}_{\epsilon }{u}^{\epsilon }\Vert }_{L^2}{\Vert J_{\epsilon }\nabla b^{\epsilon }\Vert }_{L^{\infty }}{\Vert {\wedge }^s{J}_{\epsilon }{b}^{\epsilon }\Vert }_{L^2}\\ \le C\left(s\right){\Vert \nabla J_{\epsilon }{b}^{\epsilon }\Vert }_{L^{\infty }}\left({\Vert J_{\epsilon }{b}^{\epsilon }\Vert }_{H^s}^2+{\Vert J_{\epsilon }{u}^{\epsilon }\Vert }_{H^s}^2\right),\end{array}$ (8)

$\left|I_7\right|\le C\left(s\right){\Vert \nabla J_{\epsilon }{\theta }^{\epsilon }\Vert }_{L^{\infty }}\left({\Vert J_{\epsilon }{\theta }^{\epsilon }\Vert }_{H^s}^2+{\Vert J_{\epsilon }{u}^{\epsilon }\Vert }_{H^s}^2\right),$ (9)

$\left|I_2\right|\le C\left(s\right){\Vert \nabla J_{\epsilon }{u}^{\epsilon }\Vert }_{L^{\infty }}{\Vert J_{\epsilon }{u}^{\epsilon }\Vert }_{H^s}^2.$ (10)

由Green公式，得

$I_3=\left({\wedge }^s{J}_{\epsilon }\left(\Delta u^{\epsilon }\right),{\wedge }^s{u}^{\epsilon }\right)=-{\displaystyle {\int }_{R^2}{\left|{\wedge }^s{J}_{\epsilon }\left(\nabla u^{\epsilon }\right)\right|}^2}\text{d}x=-{\Vert J_{\epsilon }\nabla u^{\epsilon }\Vert }_{H^s}^2.$ (11)

类似地，有

(12)

由Sobolev不等式，得

$\left|I_9\right|\le {\Vert {\wedge }^s{\theta }^{\epsilon }\Vert }_{L^2}{\Vert {\wedge }^s{J}_{\epsilon }{u}^{\epsilon }\Vert }_{L^2}\le C\left(s\right){\Vert {\theta }^{\epsilon }\Vert }_{H^s}{\Vert u^{\epsilon }\Vert }_{H^s}.$ (13)

结合(4)~(13)，得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \left(u^{\epsilon },b^{\epsilon },{\theta }^{\epsilon }\right)\Vert }_{H^s}^2+{\Vert J_{\epsilon }\nabla u^{\epsilon }\Vert }_{H^s}^2+{\Vert J_{\epsilon }\nabla b^{\epsilon }\Vert }_{H^s}^2+{\Vert J_{\epsilon }\nabla {\theta }^{\epsilon }\Vert }_{H^s}^2\le C\left(s\right){\Vert \left(u^{\epsilon },b^{\epsilon },{\theta }^{\epsilon }\right)\Vert }_{H^s}^3.$ (14)

由(14)和 $\left(u^{\epsilon }\left(x,0\right),b^{\epsilon }\left(x,0\right),{\theta }^{\epsilon }\left(x,0\right)\right)=\left(u_0^{\epsilon }\left(x\right),b_0^{\epsilon }\left(x\right),{\theta }_0^{\epsilon }\left(x\right)\right)$ 及 ${\Vert \left(u_0^{\epsilon }\left(x\right),b_0^{\epsilon }\left(x\right),{\theta }_0^{\epsilon }\left(x\right)\right)\Vert }_{H^s}\le {\Vert \left(u_0\left(x\right),b_0\left(x\right),{\theta }_0\left(x\right)\right)\Vert }_{H^s}$ 知，存在一个不依赖于 $\epsilon$ 的 $T=T\left(u_0,b_0,{\theta }_0\right)\left(\le T^{\epsilon }\right)>0$，使得

$\underset{0\le t\le T}{\sup }{\Vert \left(u^{\epsilon },b^{\epsilon },{\theta }^{\epsilon }\right)\Vert }_{H^s}\le C\left(s\right){\Vert \left(u_0,b_0,{\theta }_0\right)\Vert }_{H^s}.$ (15)

由(15)和引理2.1知，存在 $\left\{{\epsilon }_n\right\}$，满足 $\underset{n\to \infty }{\lim }{\epsilon }_n=0$。当 $n\to \infty$ 时，解族 ${\left\{\left(u^{\epsilon },b^{\epsilon },{\theta }^{\epsilon }\right)\right\}}_{\epsilon >0}$ 的子列 $\left\{\left(u^{{\epsilon }_n},b^{{\epsilon }_n},{\theta }^{{\epsilon }_n}\right)\right\}$ 收敛，其极限 $\left(u,b,\theta \right)$ 为初值问题(1)的解，且

$\left(u,b,\theta \right)\in C\left(\left[0,T\right);H^s\left(R^2\right)\right)\cap L^2\left(\left[0,T\right);H^{s+1}\left(R^2\right)\right).$

下证初值问题(1)的局部解的唯一性。

设 $\left(u_1,b_1,{\theta }_1\right)$ 和 $\left(u_2,b_2,{\theta }_2\right)$ 是初值问题(1)的两个解。令和 $\pi ={\pi }_1-{\pi }_2$，则 $\left(u,b,\theta \right)$ 满足

$\{\begin{array}{l}{\partial }_{t}u-\Delta u+u_1\cdot \nabla u+u\cdot \nabla u_2+\nabla \pi =b_1\cdot \nabla b+b\cdot \nabla b_2+\theta e_2,\\ {\partial }_{t}b-\Delta b+u_1\cdot \nabla b+u\cdot \nabla b_2=b_1\cdot \nabla u+b\cdot \nabla u_2,\\ {\partial }_t\theta -\Delta \theta +u_1\cdot \nabla \theta +u\cdot \nabla {\theta }_2=0,\\ \nabla \cdot u=0=\nabla \cdot b,\\ {\left(u,b,\theta \right)}_{t=0}=\left(0,0,0\right).\end{array}$ (16)

在初值问题(16)的三个方程两边分别与 $u,b$ 和 $\theta$ 作内积，再将左右两边分别相加，得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert u\Vert }_{L^2}^2+{\Vert b\Vert }_{L^2}^2+{\Vert \theta \Vert }_{L^2}^2\right)\\ =-\left(u_1\cdot \nabla u,u\right)-\left(u\cdot \nabla u_2,u\right)-\left(u_1\cdot \nabla b,b\right)-\left(u\cdot \nabla b_2,b\right)\\ +\left(b_1\cdot \nabla b,u\right)+\left(b\cdot \nabla b_2,u\right)+\left(b_1\cdot \nabla u,b\right)+\left(b\cdot \nabla u_2,b\right)\\ +\left(\Delta u,u\right)+\left(\Delta b,b\right)+\left(\Delta \theta ,\theta \right)-\left(u_1\cdot \nabla \theta ,\theta \right)-\left(u\cdot \nabla {\theta }_2,\theta \right)+\left(\theta e_2,u\right)\\ \triangleq {\displaystyle \underset{n=1}{\overset{14}{\sum }}{M}_n}.\end{array}$ (17)

由Green公式和 $\mathrm{div}u=\mathrm{div}b=0$，得

$M_1=\left(u_1\cdot \nabla u,u\right)=0,M_3=\left(u_1\cdot \nabla b,b\right)=0.$ (18)

由Hölder不等式和Sobolev嵌入定理，得

$M_2=\left(u\cdot \nabla u_2,u\right)\le {\Vert u\Vert }_{L^2}^2{\Vert \nabla u_2\Vert }_{L^{\infty }}\le C\left(s\right){\Vert u_2\Vert }_{H^s}{\Vert u\Vert }_{L^2}^2,$ (19)

$M_4=\left(u\cdot \nabla b_2,b\right)\le {\Vert u\Vert }_{L^2}{\Vert b\Vert }_{L^2}{\Vert \nabla b_2\Vert }_{L^{\infty }}\le C\left(s\right){\Vert b_2\Vert }_{H^s}\left({\Vert u\Vert }_{L^2}^2+{\Vert b\Vert }_{L^2}^2\right),$ (20)

$M_6=\left(b\cdot \nabla b_2,u\right)\le C\left(s\right){\Vert b_2\Vert }_{H^s}\left({\Vert u\Vert }_{L^2}^2+{\Vert b\Vert }_{L^2}^2\right),$ (21)

$M_8=\left(b\cdot \nabla u_2,b\right)\le C\left(s\right){\Vert u_2\Vert }_{H^s}{\Vert b\Vert }_{L^2}^2,$ (22)

$M_{13}=\left(u\cdot \nabla {\theta }_2,\theta \right)\le {\Vert u\Vert }_{L^2}{\Vert \theta \Vert }_{L^2}{\Vert \nabla {\theta }_2\Vert }_{L^{\infty }}\le C\left(s\right){\Vert {\theta }_2\Vert }_{H^s}\left({\Vert u\Vert }_{L^2}^2+{\Vert \theta \Vert }_{L^2}^2\right),$ (23)

$M_5=\left(b_1\cdot \nabla b,u\right)\le {\Vert \nabla b\Vert }_{L^2}{\Vert u\Vert }_{L^2}{\Vert b_1\Vert }_{L^{\infty }}\le C\left(s\right){\Vert \nabla b\Vert }_{L^2}{\Vert u\Vert }_{L^2}{\Vert b_1\Vert }_{H^s}\le \frac{1}{2}{\Vert \nabla b\Vert }_{L^2}^2+C\left(s\right){\Vert b_1\Vert }_{H^s}^2{\Vert u\Vert }_{L^2}^2,$ (24)

$M_7=\left(b_1\cdot \nabla u,b\right)\le \frac{1}{2}{\Vert \nabla u\Vert }_{L^2}^2+C\left(s\right){\Vert b_1\Vert }_{H^s}^2{\Vert b\Vert }_{L^2}^2,$ (25)

$M_{12}=\left(u_1\cdot \nabla \theta ,u\right)\le \frac{1}{2}{\Vert \nabla \theta \Vert }_{L^2}^2+C\left(s\right){\Vert u_1\Vert }_{H^s}^2{\Vert u\Vert }_{L^2}^2.$ (26)

$M_{14}=\left(\theta e_2,u\right)\le \frac{1}{2}\left({\Vert \theta \Vert }_{L^2}^2+{\Vert u\Vert }_{L^2}^2\right).$ (27)

由Green公式，得

$M_9=\left(\Delta u,u\right)=-{\Vert \nabla u\Vert }_{L^2}^2.$ (28)

类似地，有

$M_{10}=-{\Vert \nabla b\Vert }_{L^2}^2和M_{11}=-{\Vert \nabla \theta \Vert }_{L^2}^2.$ (29)

结合(17)~(29)知，存在 $C=C\left(s\right)$，使得

$\frac{\text{d}}{\text{d}t}\left({\Vert u\Vert }_{L^2}^2+{\Vert b\Vert }_{L^2}^2+{\Vert \theta \Vert }_{L^2}^2\right)+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \nabla b\Vert }_{L^2}^2+{\Vert \nabla \theta \Vert }_{L^2}^2\le C\left({\Vert u\Vert }_{L^2}^2+{\Vert b\Vert }_{L^2}^2+{\Vert \theta \Vert }_{L^2}^2\right).$

因此，对任意 $t\in \left[0,T\right]$，有

$\frac{\text{d}}{\text{d}t}\left({\Vert u\Vert }_{L^2}^2+{\Vert b\Vert }_{L^2}^2+{\Vert \theta \Vert }_{L^2}^2\right)\le C\left({\Vert u\Vert }_{L^2}^2+{\Vert b\Vert }_{L^2}^2+{\Vert \theta \Vert }_{L^2}^2\right).$

由Gronwall不等式及 ${\left(u,b,\theta \right)|}_{t=0}=\left(0,0,0\right)$，可得在 $\left[0,T\right]$ 上恒有 ${\Vert u\Vert }_{L^2}^2+{\Vert b\Vert }_{L^2}^2+{\Vert \theta \Vert }_{L^2}^2=0$，

即 ${\Vert u\Vert }_{L^2}=0,{\Vert b\Vert }_{L^2}=0,{\Vert \theta \Vert }_{L^2}=0$。因此初值问题(1)的局部解唯一。引理2.3证毕。

3. 定理1.1的证明

为了研究初值问题(1)整体解的存在性，需用到如下引理。

引理3.1 [7] (Sobolev对数不等式)设 $s>\frac{d}{2},f\in H^s\left(R^d\right)$，则以下不等式成立：

${\Vert f\Vert }_{L^{\infty }}\le C\left(s\right)\left(1+{\Vert f\Vert }_{H^{\frac{d}{2}}}\right){\log }^{\frac{1}{2}}\left(e+{\Vert f\Vert }_{H^s}\right).$

为了证明初值问题(1)存在整体解，设 $\left[0,T^{*}\right)$ 是初值问题(1)的解存在的最大时间区间，只需证明 $T^{*}=+\infty$。用反证法，假设 $T^{*}<+\infty$。在初值问题(1)中的方程两边分别与 $u,b,\theta$ 作内积，用类似于引理2.3的证明方法，得

${\Vert \left(u,b,\theta \right)\Vert }_{L^{\infty }\left(0,T^{\ast };L^2\right)}+{\Vert \left(\nabla u,\nabla b,\nabla \theta \right)\Vert }_{L^2\left(0,T^{\ast };L^2\right)}\le C\left(s\right){\Vert \left(u_0,b_0,{\theta }_0\right)\Vert }_{L^2}.$ (30)

在初值问题(1)中的方程两边分别与 $\Delta u,\Delta b$ 和 $\Delta \theta$ 作内积，再将左右两边分别相加，得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \nabla b\Vert }_{L^2}^2+{\Vert \nabla \theta \Vert }_{L^2}^2\right)+{\Vert \Delta u\Vert }_{L^2}^2+{\Vert \Delta b\Vert }_{L^2}^2+{\Vert \Delta \theta \Vert }_{L^2}^2\\ =\left(-u\cdot \nabla u,\Delta u\right)+\left(-u\cdot \nabla b,\Delta b\right)+\left(-u\cdot \nabla \theta ,\Delta \theta \right)+\left(\theta e_2,\Delta u\right)\\ \triangleq {\displaystyle \underset{n=1}{\overset{4}{\sum }}{G}_n}.\end{array}$ (31)

由Gagliardo-Nirenberg不等式及Young不等式，得

$\left|G_1\right|\le {\Vert u\cdot \nabla u\Vert }_{L^2}{\Vert \Delta u\Vert }_{L^2}\le C\left(s\right){\Vert u\Vert }_{L^4}{\Vert \nabla u\Vert }_{L^4}{\Vert \Delta u\Vert }_{L^2}\le \frac{1}{4}{\Vert \Delta u\Vert }_{L^2}^2+C\left(s\right)\left({\Vert u\Vert }_{L^2}^2{\Vert \nabla u\Vert }_{L^2}^4\right),$ (32)

类似地，有

$\left|G_2\right|\le {\Vert u\cdot \nabla b\Vert }_{L^2}{\Vert \Delta b\Vert }_{L^2}\le \frac{1}{2}{\Vert \Delta b\Vert }_{L^2}^2+C\left(s\right)\left({\Vert u\Vert }_{L^2}^2{\Vert \nabla u\Vert }_{L^2}^2{\Vert \nabla b\Vert }_{L^2}^2\right),$ (33)

$\left|G_3\right|\le {\Vert u\cdot \nabla \theta \Vert }_{L^2}{\Vert \Delta \theta \Vert }_{L^2}\le \frac{1}{2}{\Vert \Delta \theta \Vert }_{L^2}^2+C\left(s\right)\left({\Vert u\Vert }_{L^2}^2{\Vert \nabla u\Vert }_{L^2}^2{\Vert \nabla \theta \Vert }_{L^2}^2\right),$ (34)

$\left|G_4\right|\le {\Vert \theta e_2\Vert }_{L^2}{\Vert \Delta u\Vert }_{L^2}\le \frac{1}{4}{\Vert \Delta u\Vert }_{L^2}^2+C\left(s\right){\Vert \theta \Vert }_{L^2}^2,$ (35)

结合(31)~(35)知，存在 $C=C\left(s\right)$，使得

$\begin{array}{l}\frac{\text{d}}{\text{d}t}\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \nabla b\Vert }_{L^2}^2+{\Vert \nabla \theta \Vert }_{L^2}^2\right)+{\Vert \Delta u\Vert }_{L^2}^2+{\Vert \Delta b\Vert }_{L^2}^2+{\Vert \Delta \theta \Vert }_{L^2}^2\\ \le C\left({\Vert u\Vert }_{L^2}^2{\Vert \nabla u\Vert }_{L^2}^4+{\Vert u\Vert }_{L^2}^2{\Vert \nabla u\Vert }_{L^2}^2{\Vert \nabla b\Vert }_{L^2}^2+{\Vert u\Vert }_{L^2}^2{\Vert \nabla u\Vert }_{L^2}^2{\Vert \nabla \theta \Vert }_{L^2}^2+{\Vert \theta \Vert }_{L^2}^2\right)\\ \le C{\Vert \nabla u\Vert }_{L^2}^2\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \nabla b\Vert }_{L^2}^2+{\Vert \nabla \theta \Vert }_{L^2}^2\right)+{\Vert \theta \Vert }_{L^2}^2.\end{array}$

由Gronwall不等式及(30)得， $\left(u,b,\theta \right)$ 在 $L^2\left(0,T;H^2\left(R^2\right)\right)$ 中有界。

在方程组(1)的两边作用 ${\wedge }^s$ 并分别与和 ${\wedge }^s\theta$ 作内积，再将左右两边分别相加，得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert {\wedge }^{s}u\Vert }_{L^2}^2+{\Vert {\wedge }^{s}b\Vert }_{L^2}^2+{\Vert {\wedge }^s\theta \Vert }_{L^2}^2\right)\\ =\left({\wedge }^s\left(b\cdot \nabla b\right),{\wedge }^{s}u\right)-\left({\wedge }^s\left(u\cdot \nabla u\right),{\wedge }^{s}u\right)+\left({\wedge }^s\left(\Delta u\right),{\wedge }^{s}u\right)\\ +\left({\wedge }^s\left(b\cdot \nabla u\right),{\wedge }^{s}b\right)-\left({\wedge }^s\left(u\cdot \nabla b\right),{\wedge }^{s}b\right)+\left({\wedge }^s\left(\Delta b\right),{\wedge }^{s}b\right)\\ -\left({\wedge }^s\left(u\cdot \nabla \theta \right),{\wedge }^s\theta \right)+\left({\wedge }^s\left(\Delta \theta \right),{\wedge }^s\theta \right)+\left({\wedge }^s\theta e_2,{\wedge }^{s}u\right)\\ \triangleq {\displaystyle \underset{n=1}{\overset{9}{\sum }}{H}_n}.\end{array}$ (36)

类似 $I_1,I_2,I_4,I_5$ 和 $I_7$ 的估计方法，可得关于 $H_1,H_2,H_4,H_5$ 和 $H_7$ 的如下估计，

$H_1+H_4\le C\left(s\right)\left({\Vert \nabla u\Vert }_{L^{\infty }}+{\Vert \nabla b\Vert }_{L^{\infty }}\right)\left({\Vert b\Vert }_{H^s}^2+{\Vert u\Vert }_{H^s}^2\right),$ (37)

$H_5\le C\left(s\right){\Vert \nabla b\Vert }_{L^{\infty }}\left({\Vert b\Vert }_{H^s}^2+{\Vert u\Vert }_{H^s}^2\right),$ (38)

$H_7\le C\left(s\right){\Vert \nabla \theta \Vert }_{L^{\infty }}\left({\Vert \theta \Vert }_{H^s}^2+{\Vert u\Vert }_{H^s}^2\right),$ (39)

$H_2\le C\left(s\right){\Vert \nabla u\Vert }_{L^{\infty }}{\Vert u\Vert }_{H^s}^2,$ (40)

由Sobolev不等式，得

$H_9\le {\Vert {\wedge }^s\theta e_2\Vert }_{L^2}{\Vert {\wedge }^{s}u\Vert }_{L^2}\le C\left(s\right){\Vert \theta \Vert }_{L^2}{\Vert u\Vert }_{H^s}\le C\left(s\right){\Vert \theta \Vert }_{H^s}{\Vert u\Vert }_{H^s}.$ (41)

由Green公式，得

(42)

结合(36)~(42)知，存在 $C=C\left(s\right)$，使得，

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert u\Vert }_{H^s}^2+{\Vert b\Vert }_{H^s}^2+{\Vert \theta \Vert }_{H^s}^2\right)+\left({\Vert \nabla u\Vert }_{H^s}^2+{\Vert \nabla b\Vert }_{H^s}^2+{\Vert \nabla \theta \Vert }_{H^s}^2\right)\\ \le C\left({\Vert \nabla u\Vert }_{L^{\infty }}+{\Vert \nabla b\Vert }_{L^{\infty }}+{\Vert \nabla \theta \Vert }_{L^{\infty }}\right)\left({\Vert u\Vert }_{H^s}^2+{\Vert b\Vert }_{H^s}^2+{\Vert \theta \Vert }_{H^s}^2\right).\end{array}$

由Gronwall不等式，得

${\Vert u\Vert }_{H^s}^2+{\Vert b\Vert }_{H^s}^2+{\Vert \theta \Vert }_{H^s}^2\le C{\text{e}}^{{\displaystyle {\int }_0^t\left({\Vert \nabla u\Vert }_{L^{\infty }}+{\Vert \nabla b\Vert }_{L^{\infty }}+{\Vert \nabla \theta \Vert }_{L^{\infty }}\right)\text{d}\tau }}\left({\Vert u_0\Vert }_{H^s}^2+{\Vert b_0\Vert }_{H^s}^2+{\Vert {\theta }_0\Vert }_{H^s}^2\right).$

由引理3.1可证，对任意 $t\in \left[0,T^{*}\right]$，有

$\begin{array}{l}{\displaystyle {\int }_0^t\left({\Vert \nabla u\Vert }_{L^{\infty }}+{\Vert \nabla b\Vert }_{L^{\infty }}+{\Vert \nabla \theta \Vert }_{L^{\infty }}\right)\text{d}\tau }\\ \le C{\displaystyle {\int }_0^t\left[1+\log \left(e+{\Vert u\left(\tau \right)\Vert }_{H^s}^2+{\Vert b\left(\tau \right)\Vert }_{H^s}^2+{\Vert \theta \left(\tau \right)\Vert }_{H^s}^2\right)\right]\text{d}\tau }+C\left(u_0,b_0,{\theta }_0,s,T^{*}\right),\end{array}$

因此，

${\Vert u\Vert }_{H^s}^2+{\Vert b\Vert }_{H^s}^2+{\Vert \theta \Vert }_{H^s}^2\le C\left({\Vert u_0\Vert }_{H^s}^2+{\Vert b_0\Vert }_{H^s}^2+{\Vert {\theta }_0\Vert }_{H^s}^2\right){\text{e}}^{{\displaystyle {\int }_0^t\left(1+\log \left(e+{\Vert u\left(\tau \right)\Vert }_{H^s}^2+{\Vert b\left(\tau \right)\Vert }_{H^s}^2+{\Vert \theta \left(\tau \right)\Vert }_{H^s}^2\right)\right)\text{d}\tau }},$

从而，

$\begin{array}{l}\log \left(e+{\Vert u\left(t\right)\Vert }_{H^s}^2+{\Vert b\left(t\right)\Vert }_{H^s}^2+{\Vert \theta \left(t\right)\Vert }_{H^s}^2\right)\\ \le C\left(u_0,b_0,{\theta }_0,s,T^{*}\right)+C{\displaystyle {\int }_0^t\log \left(e+{\Vert u\left(\tau \right)\Vert }_{H^s}^2+{\Vert b\left(\tau \right)\Vert }_{H^s}^2+{\Vert \theta \left(\tau \right)\Vert }_{H^s}^2\right)\text{d}\tau }.\end{array}$

由Gronwall不等式得 $\left(u,b,\theta \right)\in L^{\infty }\left(0,T^{*};H^s\left(R^2\right)\right)$。又由 $\left(u,b,\theta \right)\in L^2\left(0,T^{*};H^2\left(R^2\right)\right)$ 知，解可被延拓到 $t=T^{*}$ 后，这与 $\left(0,T^{*}\right]$ 为初值问题(1)的解的最大存在区间的假设相矛盾，因此 $T^{*}=\infty$。定理1.1证毕。