一类各向异性非牛顿Boussinesq方程组弱解的存在性研究

On the Existence of Weak Solutions for a Class of Anisotropic Non-Newtonian Boussinesq Equations

摘要: 本文拟在三维空间中讨论一类各向异性非牛顿Boussinesq方程组的初边值问题，通过结合使用Galerkin方法、紧性方法与单调性方法证明了该问题弱解的存在性。

Abstract: In this paper, the initial boundary value problem for a class of anisotropic non-Newtonian Boussinesq equations is discussed in three-dimensional space. The existence of weak solutions of the problem is proved by using Galerkin method, compactness method and monotonicity method.

1. 引言

在三维空间中考虑下述一类各向异性非牛顿Boussinesq方程组的初边值问题

$\{\begin{array}{l}\frac{\partial u}{\partial t}+\left(u\cdot \nabla \right)u-{\displaystyle \underset{i=1}{\overset{N}{\sum }}{D}_i\left({\left|D_{i}u\right|}^{q_i-2}{D}_{i}u\right)+\nabla \pi =f,\left(x,t\right)\in Q_T,}\\ \frac{\partial \theta }{\partial t}+\left(u\cdot \nabla \right)\theta -\Delta \theta =g,\left(x,t\right)\in Q_T,\\ divu=0,\left(x,t\right)\in Q_T,\\ u=0,\theta =0,\left(x,t\right)\in {\Gamma }_T,\\ {u|}_{t=0}=u_0\left(x\right),{\theta |}_{t=0}={\theta }_0\left(x\right),x\in \Omega ,\end{array}$ (1)

其中 $\Omega \subset R^3$ 为一边界足够光滑的有界区域， $Q_T=\Omega \times \left[0,T\right]$， ${\Gamma }_T=\partial \Omega \times \left[0,T\right]$。未知向量函数 $u=\left(u_1,u_2,u_3\right)$ 表示流体的速度， $\theta =\theta \left(x,t\right)$ 表示温度， $\pi$ 表示压力， $f=\left(f_1,f_2,f_3\right)$， $g=g\left(x\right)$ 为已给定的外力项， $D_{i}u=\left({\partial }_i{u}_1,{\partial }_i{u}_2,{\partial }_i{u}_3\right)$。指数 $q_i$ 为给定的常数，满足 $1<q_i<\infty \left(i=1,2,3\right)$。

非牛顿流体广泛存在于我们的生产、生活和自然界中，因此越来越多不同领域的学者开始对其展开研究，以求对其有更深刻的了解与利用。在化工、石油、水利、生物工程以及食品材料甚至航空航天领域都对非牛顿流体有着极大的关注，我们主要通过对非牛顿流体方程的研究来增加对非牛顿流体的了解。

粘性导热流体在重力作用下的流动由动量守恒、质量守恒和能量守恒方程组描述(参见 [1] )，在忽略粘性的加热效应及应力张量与应变速率之间呈线性本构关系的假设下就得到一个粘性、不可压缩浮力驱动的流体流动与热对流耦合的运动模型，即所谓的牛顿型Boussinesq近似(参见 [2] 或 [3] )。上述模型在大气科学中具有重要作用 [4]，同时也是许多地球物理应用中的模型 [5]。其在预测观察到的现象中是非常成功的，少有近似模型能与之匹敌，其在天体物理与地球物理流体动力学领域有着重要的影响，基于此，数学上人们对上述模型开展了广泛的研究，取得了大量的研究成果。在该类模型中，若应力张量与应变速率之间显现出非线性关系，即得到所谓的非牛顿Boussinesq模型，目前关于这类模型的研究结果还不多。郭柏灵等在 [6] 中研究了一类修正的双极粘性流体模型，证明了其弱解的存在性和唯一性；随后他们又在 [7] 中研究了周期性边值问题，证明了其弱解的存在性，并在 $p>\left(1+2n\right)/\left(n+2\right)$ 时证明了唯一性及高阶正则性结果。文献 [8] 在三维空间中对周期初边值问题证明了强解的存在唯一性。文献 [9] 研究了一类稳态不可压缩的非牛顿Boussinesq方程组的第一初边值问题，在外力项的某一范数适当小的条件下，证明了方程组正则解的存在唯一性。本文拟在三维空间中讨论一类各向异性非牛顿Boussinesq方程组的初边值问题，通过结合使用Galerkin方法、紧性方法与单调性方法证明了该问题弱解的存在性。

2. 预备知识与主要结果

我们首先介绍本文用到的各向异性Sobolev空间的基本知识。令

$\stackrel{\to }{q}=\left(q_1,\cdots ,q_N\right),1<q_i<\infty ,i=1,2,\cdots ,N,$

定义

$L^{\stackrel{\to }{q}}\left(\Omega \right)=\left\{u|u\in L^{q_i}\left(\Omega \right),i=1,2,\cdots ,N\right\},$

$W^{1,\stackrel{\to }{q}}\left(\Omega \right)=\left\{u|u\in W^{1,1}\left(\Omega \right),D_{i}u\in L^{q_i}\left(\Omega \right),i=1,2,\cdots ,N\right\},$

在其中分别赋予范数

${\Vert u\Vert }_{L^{\stackrel{\to }{q}}\left(\Omega \right)}={\displaystyle \underset{i=1}{\overset{N}{\sum }}{\Vert u\Vert }_{L^{q_i}\left(\Omega \right)}},{\Vert u\Vert }_{W^{1,\stackrel{\to }{q}}\left(\Omega \right)}={\Vert u\Vert }_{L^1\left(\Omega \right)}+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\Vert D_{i}u\Vert }_{L^{q_i}\left(\Omega \right)},}$

所得的线性空间为完备的赋范线性空间，即Banach空间。

令

$\beta =\underset{i=1,\cdots ,N}{\max }{q}_i,\alpha =\underset{i=1,\cdots ,N}{\min }{q}_i,$

假设 $q_i$ 满足

$\alpha =q_1\le q_2\le \cdots \le q_N=\beta .$

本文用到的引理如下：

引理1 [10] 令 $\Omega \subset R^N$ 为一边界充分光滑的有界开集，如果 ${\displaystyle {\sum }_{j=1}^N\frac{1}{q_j}>1}$，则有以下嵌入关系成立

$W^{1,\stackrel{\to }{q}}\left(\Omega \right)$ ↪ $L^s\left(\Omega \right),\forall s:1\le s\le q_a^{\ast },$ (2)

$W^{1,\stackrel{\to }{q}}\left(\Omega \right)$ ↪↪ $L^s\left(\Omega \right),\forall s:1\le s<q_a^{\ast },$ (3)

其中 $q_a^{\ast }=\max \left\{{\bar{q}}^{\ast },\beta \right\}$， ${\bar{q}}^{\ast }=\frac{N}{{\displaystyle {\sum }_{j=1}^N\frac{1}{q_j}-1}}$。

注1由于引理1，本文中总假设指标 $q_i\left(i=1,2,3\right)$ 满足条件 $\frac{1}{q_1}+\frac{1}{q_2}+\frac{1}{q_3}>1$。

引理2 [11] 对 $u,v,\omega \in H_0^1\left(\Omega \right)$，定义

$b\left(u,v,\omega \right)={\displaystyle \underset{i,j=1}{\overset{3}{\sum }}{\displaystyle {\int }_{\Omega }{u}_i\left(x\right)}}\frac{\partial v_j}{\partial x_i}\left(x\right){\omega }_j\left(x\right)\text{d}x,$

若 $divu=0$，则有

$b\left(u,v,\omega \right)=-b\left(u,\omega ,v\right),b\left(u,v,v\right)=0.$

引理3 [12] 设 $X,Y$ 为两个Banach空间，X连续嵌入到Y。倘若函数 $u\in L^{\infty }\left(0,T;X\right)$ 且 $u:\left[0,T\right]\to Y$ 是弱连续的，则 $u:\left[0,T\right]\to X$ 亦是弱连续的。

记 $\mathcal{V}=\left\{u|u\in C_0^{\infty }\left(\Omega \right),divu=0\right\}$，定义如下函数空间

$H:=\mathcal{V}$ 在 $L^2\left(\Omega \right)$ 范数意义下的闭包；

$V_q:=\mathcal{V}$ 在 $W^{1,q}\left(\Omega \right)$ 范数意义下的闭包；

$V_{\stackrel{\to }{q}}:=\mathcal{V}$ 在 $W^{1,\stackrel{\to }{q}}\left(\Omega \right)$ 范数意义下的闭包；

下面考虑抛物型各向异性空间。令

$L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)=\left\{u:\left[0,T\right]\to V_{\stackrel{\to }{q}},u,{\left|D_{i}u\right|}^{q_i}\in L^1\left(Q_T\right),i=1,\cdots ,N\right\};$

赋予范数

${\Vert u\Vert }_{L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)}={\Vert u\Vert }_{L^1\left(Q_T\right)}+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\Vert D_{i}u\Vert }_{L^{q_i}\left(Q_T\right)}};$

对于有界域 $\Omega$ 和有限T，以下嵌入成立 [13]

$L^{\beta }\left(0,T;V_{\beta }\right)$ ↪ $L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$ ↪ $L^{\alpha }\left(0,T;V_{\alpha }\right)$ (4)

且作为 $L^{\alpha }\left(0,T;V_{\alpha }\right)$ 的闭子空间， $L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$ 是自反的且是可分的，用 $L^{{q^{\prime }}_i}\left(0,T;{V^{\prime }}_{q_i}\right)$ 和 $L^{{\stackrel{\to }{q}}^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right)$ 分别来表示 $L^{q_i}\left(0,T;V_{q_i}\right)$ 和 $L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$ 的对偶空间，其中 ${V^{\prime }}_{q_i}$ 和 ${V^{\prime }}_{\stackrel{\to }{q}}$ 表示 $V_{q_i}$ 和 $V_{\stackrel{\to }{q}}$ 的对偶空间。

下面给出问题(1)弱解的定义

定义2.1设 $f\in L^{{\stackrel{\to }{q}}^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right)$， $g\in L^2\left(0,T;H^{-1}\left(\Omega \right)\right)$， $q_a^{\ast }\ge 2$， $\alpha \ge 2$，称 $\left(u,\theta \right)$ 是问题(1)的弱解，如果

1) $u\in L^{\infty }\left(0,T;H\right)\cap L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$， $\theta \in L^{\infty }\left(0,T;L^2\left(\Omega \right)\right)\cap L^2\left(0,T;H_0^1\left(\Omega \right)\right)$ ；

2) $u\left(0\right)=u_0$， $\theta \left(0\right)={\theta }_0$ ；

3) 对 $\forall \phi \in V_{\stackrel{\to }{q}}\cap L^{\theta }\left(\Omega \right)$， $\varphi \in L^2\left(\Omega \right)$ 以及 $a.e.t\in \left[0,T\right]$ 有

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }u\phi \text{d}x}+{\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)u\right]\cdot \phi \text{d}x}+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\displaystyle {\int }_{\Omega }{\left|D_{i}u\right|}^{q_i-2}{D}_{i}u\left(t\right)\cdot D_i\phi \text{d}x}}={\displaystyle {\int }_{\Omega }f\phi \text{d}x},$

$\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }\theta \varphi \text{d}x}+{\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)\theta \right]\cdot \varphi \text{d}x}+{\displaystyle {\int }_{\Omega }\nabla \theta \cdot \nabla \varphi \text{d}x}={\displaystyle {\int }_{\Omega }g\cdot \varphi \text{d}x},$

其中 $\theta$ 由关系式 $\frac{1}{q_a^{\ast }}+\frac{1}{\alpha }+\frac{1}{\theta }=1$ 确定。

注2在上述弱解满足的等式中没有压力项 $\pi$，事实上，如果1) 2) 3) 成立，则利用De Rham定理 [14] 知，有函数 $\pi \in L^2\left(\Omega \right)$ 在广义函数意义下满足上述积分等式。

注3定义中 $u\left(0\right)=u_0$， $\theta \left(0\right)={\theta }_0$ 在下述意义下成立

$\begin{array}{l}\underset{t\to 0^{+}}{\lim }{\displaystyle {\int }_{\Omega }u\left(t\right)\cdot \phi \text{d}x}={\displaystyle {\int }_{\Omega }{u}_0\cdot \phi \text{d}x}\text{,}\forall \phi \in V_{\stackrel{\to }{q}}\cap L^{\theta }\left(\Omega \right),\\ \underset{t\to 0^{+}}{\lim }{\displaystyle {\int }_{\Omega }\theta \left(t\right)\cdot \varphi \text{d}x}={\displaystyle {\int }_{\Omega }{\theta }_0\cdot \varphi \text{d}x}\text{,}\forall \varphi \in L^2\left(\Omega \right)\text{.}\end{array}$

注4当 $u\in L^{\infty }\left(0,T;H\right)\cap L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$ 时，为了使定义中对流项 ${\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)u\right]\phi \text{d}x}$ 有意义，需要 $\phi \in L^{\theta }\left(\Omega \right)$ [15]，但是若 $\theta \le q_a^{\ast }$，则只需 $\phi \in V_{\stackrel{\to }{q}}$，因为此时嵌入 $V_{\stackrel{\to }{q}}$ ↪ $L^{\theta }\left(\Omega \right)$ 成立。

注5若记 $V:=V_{\stackrel{\to }{q}}\cap L^{\theta }\left(\Omega \right)$，则当 $q_a^{\ast }\ge 2$ 时，显然有下列嵌入关系成立

$V$ ↪ $V_{\stackrel{\to }{q}}$ ↪ $H\cong H^{\prime }$ ↪ ${V^{\prime }}_{\stackrel{\to }{q}}$ ↪ $V^{\prime }$. (5)

本文的主要结果如下：

定理1设 $\Omega \subset R^3$ 为一边界充分光滑的有界域， $u_0\in H$， ${\theta }_0\in L^2\left(\Omega \right)$， $f\in L^{{\stackrel{\to }{q}}^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right)$， $g\in L^2\left(0,T;H^{-1}\right)$。若 $\alpha >2$， $q_a^{\ast }\ge q^{\ast }$，其中

$q^{\ast }=\{\begin{array}{l}\frac{2\alpha \left(\alpha -1\right)}{\left(\alpha +1\right)\left(\alpha -2\right)},2<\alpha <3,\\ \frac{2\alpha }{\alpha -1},\alpha \ge 3,\end{array}$ (6)

则问题(1)至少存在一对弱解 $\left(u,\theta \right)$，满足

$u\in C_W\left(\left[0,T\right];H\right)\cap L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right),$

$\theta \in C_W\left(\left[0,T\right];L^2\left(\Omega \right)\right)\cap L^2\left(0,T;H_0^1\left(\Omega \right)\right).$

其中 $u\in C_W\left(\left[0,T\right];X\right)$ 表示 $u:\left[0,T\right]\to X$ 是弱连续的。

3. 定理1的证明

我们将通过以下几步进行定理的证明，

第一步：近似解的构造

定义空间 $\tilde{V}:=\mathcal{V}$ 在 $W^{3,2}\left(\Omega \right)$ 中的闭包，设 ${\left\{{\phi }_r\right\}}_{r\in N}$ 是如下特征值问题的一组非平凡解

${\displaystyle \underset{\left|k\right|=3}{\sum }\left(D^k{\phi }_r,D^{k}v\right)={\lambda }_j\left({\phi }_j,v\right),}\forall v\in \tilde{V}.$ (7)

则 ${\left\{{\phi }_r\right\}}_{r\in N}$ 构成空间 $\tilde{V}$ 的一组正交基 [16]，且在H中是规范正交的；对给定的 $m\in N$，记 $V^m=span\left({\phi }_1,{\phi }_2,\cdots ,{\phi }_m\right)$ 为应由 $\left\{{\phi }_1,{\phi }_2,\cdots ,{\phi }_m\right\}$ 张成的子空间；其次，取Lamé算子的特征函数族 ${\left\{{\varphi }_r\right\}}_{r\in W}$ 为 $H_0^1\left(\Omega \right)$ 中的一组基，并记 $W^m=span\left\{{\varphi }_1,{\varphi }_2,\cdots ,{\varphi }_m\right\}$ 为应由 $\left\{{\varphi }_1,{\varphi }_2,\cdots ,{\varphi }_m\right\}$ 张成的子空间。我们谋求问题(1)类似于以下结构的近似解

$\begin{array}{l}{u}_m\left(x,t\right)={\displaystyle \underset{k=1}{\overset{m}{\sum }}{c}_k^m\left(t\right){\phi }_k}\left(x\right),u_{0m}={\displaystyle \underset{k=1}{\overset{m}{\sum }}{c}_k^m\left(0\right){\phi }_k}\left(x\right),\\ {\theta }_m\left(x,t\right)={\displaystyle \underset{r=1}{\overset{m}{\sum }}{h}_r^m\left(t\right){\varphi }_r}\left(x\right),{\theta }_{0m}={\displaystyle \underset{r=1}{\overset{m}{\sum }}{h}_r^m\left(0\right){\varphi }_r}\left(x\right),\end{array}$

其中对 $k=1,\cdots ,m$， $r=1,\cdots ,m$， $\left(u_m,{\theta }_m\right)$ 满足

$\{\begin{array}{l}{\displaystyle {\int }_{\Omega }\frac{\partial u_m}{\partial t}\cdot {\phi }_k\text{d}x}+{\displaystyle {\int }_{\Omega }\left[\left(u_m\cdot \nabla \right)u_m\right]\cdot {\phi }_k\text{d}x}+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\displaystyle {\int }_{\Omega }{\left|D_i{u}_m\right|}^{q_i-2}{D}_i{u}_m\cdot D_i{\phi }_k\text{d}x}}={\displaystyle {\int }_{\Omega }f\cdot {\phi }_k\text{d}x},\\ {\displaystyle {\int }_{\Omega }\frac{\partial {\theta }_m}{\partial t}\cdot {\varphi }_r\text{d}x+{\displaystyle {\int }_{\Omega }\left[\left(u_m\cdot \nabla \right){\theta }_m\right]\cdot {\varphi }_r\text{d}x}+{\displaystyle {\int }_{\Omega }\nabla {\theta }_m\cdot \nabla {\varphi }_r\text{d}x}={\displaystyle {\int }_{\Omega }g\cdot {\varphi }_r\text{d}x},}\\ u_m\left(0\right)=u_{0m}\text{,}{\theta }_m\left(0\right)={\theta }_{0m}\end{array}$ (8)

并且当 $m\to \infty$ 时

$u_{0m}\to u_0于H;{\theta }_{0m}\to {\theta }_0于L^2\left(\Omega \right).$ (9)

由常微分方程组的相关知识 [16] (定理3.4)知，存在一区间 $\left[0,T_m\right]$， $0<T_m<T$，使(8) (9)存在古典意义下的解 $c_k^m\left(t\right),h_r^m\left(t\right)\in C^1\left[0,T_m\right]$， $k,r=1,\cdots ,m$。结合接下来的一致估计和解的整体存在唯一性定理可得 $T_m=T$。

第二步：推导一致性先验估计

将等式(8)₁两端同时乘以 $c_k^m\left(t\right)$，再关于k求和，得

${\displaystyle {\int }_{\Omega }\frac{\partial u_m}{\partial t}\cdot u_m\text{d}x}+{\displaystyle {\int }_{\Omega }\left[\left(u_m\cdot \nabla \right)u_m\right]\cdot u_m\text{d}x}+{\displaystyle {\int }_{\Omega }{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\left|D_i{u}_m\right|}^{q_i-2}{D}_i{u}_m\cdot D_i{u}_m\text{d}x}}={\displaystyle {\int }_{\Omega }f\cdot u_m\text{d}x}\text{,}$

将上式在0到t， $t\in \left(0,T_m\right)$ 积分，可得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert u_m\Vert }_H^2+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\displaystyle {\int }_{\Omega }{\left|D_i{u}_m\right|}^{q_i}\text{d}x}}+{\displaystyle \underset{i,j=1}{\overset{3}{\sum }}{\displaystyle {\int }_{\Omega }{u}_{m_i}\frac{\partial u_{m_j}}{\partial x_i}{u}_{m_j}}\text{d}x}={\displaystyle {\int }_{\Omega }f\cdot u_m\text{d}x},$

由引理2可知于

$b\left(u,u,u\right)={\displaystyle \underset{i,j=1}{\overset{3}{\sum }}{\displaystyle {\int }_{\Omega }{u}_{m_i}\frac{\partial u_{m_j}}{\partial x_i}{u}_{m_j}\text{d}x}}=0$,

所以由Hölder不等式以及Young不等式，可将上式整理得

$\begin{array}{c}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert u_m\Vert }_H^2+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\displaystyle {\int }_{\Omega }{\left|D_i{u}_m\right|}^{q_i}\text{d}x}}={\displaystyle {\int }_{\Omega }f\cdot u_m\text{d}x}\le {\Vert f\Vert }_{L^{{\stackrel{\to }{q}}^{\prime }}}\cdot {\Vert u_m\Vert }_{L^{\stackrel{\to }{q}}\left(\Omega \right)}\\ \le \frac{v_1}{2}{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\Vert D_i{u}_m\Vert }_{L^{q_i}\left(\Omega \right)}^{q_i}}+C{v}_1{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\Vert D_{i}f\Vert }_{L^{{q^{\prime }}_i}\left(\Omega \right)}^{{q^{\prime }}_i}}\end{array}$ (10)

同理将等式(8)₂两端同时乘以 $h_r^m\left(t\right)$，再关于r求和，得

${\displaystyle {\int }_{\Omega }\frac{\partial {\theta }_m}{\partial t}\cdot {\theta }_m\text{d}x}+{\displaystyle {\int }_{\Omega }\left[\left(u_m\cdot \nabla \right){\theta }_m\right]\cdot {\theta }_m\text{d}x}+{\displaystyle {\int }_{\Omega }\nabla {\theta }_m\cdot \nabla {\theta }_m\text{d}x}={\displaystyle {\int }_{\Omega }g\cdot {\theta }_m\text{d}x}.$

将上式在0到t， $t\in \left(0,T_m\right)$ 积分可得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2+{\Vert \nabla {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2+{\displaystyle \underset{i,j=1}{\overset{3}{\sum }}{\displaystyle {\int }_{\Omega }{u}_{m_i}\frac{\partial {\theta }_{m_j}}{\partial x_i}{\theta }_{m_j}\text{d}x}}={\displaystyle {\int }_{\Omega }g\cdot {\theta }_m\text{d}x},$

由引理2可知

$b\left(u,\theta ,\theta \right)={\displaystyle \underset{i,j=1}{\overset{3}{\sum }}{\displaystyle {\int }_{\Omega }{u}_{m_i}\frac{\partial {\theta }_{m_j}}{\partial x_i}{\theta }_{m_j}\text{d}x}}=0$,

所以由Hölder不等式以及Young不等式，可将上式整理得

$\begin{array}{c}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2+{\Vert \nabla {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2={\displaystyle {\int }_{\Omega }g\cdot {\theta }_m\text{d}x}\\ \le {\Vert g\Vert }_{H^{-1}\left(\Omega \right)}\cdot {\Vert {\theta }_m\Vert }_{W^{1,2}\left(\Omega \right)}\\ \le C{\Vert g\Vert }_{H^{-1}}\cdot {\Vert \nabla {\theta }_m\Vert }_{L^2\left(\Omega \right)}\\ \le \frac{v_2}{2}{\Vert \nabla {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2+C{v}_2{\Vert g\Vert }_{H^{-1}\left(\Omega \right)}^2.\end{array}$ (11)

将(10)式和(11)式相加，整理得

$\begin{array}{l}\frac{\text{d}}{\text{d}t}\left({\Vert u_m\Vert }_H^2+{\Vert {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2\right)+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\displaystyle {\int }_{\Omega }{\left|D_i{u}_m\right|}^{q_i}\text{d}x}}+v_2{\Vert \nabla {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2\\ \le \left(v_1+v_2\right)\left({\Vert u_m\Vert }_H^2+{\Vert {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2\right)+\left(C{v}_1+C{v}_2\right)\left({\Vert D_{i}f\Vert }_{L^{{q^{\prime }}_i}\left(\Omega \right)}^{{q^{\prime }}_i}+{\Vert g\Vert }_{H^{-1}}^2\right),\end{array}$ (12)

将(12)式在 $\left(0,t\right),\left(0\le t\le T\right)$ 上积分可得

$\begin{array}{l}\left({\Vert u_m\left(t\right)\Vert }_H^2+{\Vert {\theta }_m\left(t\right)\Vert }_{L^2\left(\Omega \right)}^2\right)+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\Vert D_i{u}_m\Vert }_{L^{q_i}\left(0,t;L^{q_i}\left(\Omega \right)\right)}^{q_i}}+v_2{\displaystyle {\int }_0^t{\Vert \nabla {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2\text{d}s}\\ \le {\Vert u_m\left(0\right)\Vert }_H^2+{\Vert {\theta }_m\left(0\right)\Vert }_{L^2\left(\Omega \right)}^2+C{\displaystyle {\int }_0^t\left({\Vert u_m\left(s\right)\Vert }_H^2+{\Vert {\theta }_m\left(s\right)\Vert }_{L^2\left(\Omega \right)}^2\right)\text{d}s}\\ +\left(C{v}_1+C{v}_2\right)\left({\displaystyle \underset{i=1}{\overset{3}{\sum }}{\Vert D_{i}f\Vert }_{L^{{q^{\prime }}_i}\left(Q_t\right)}^{{q^{\prime }}_i}}+{\Vert g\Vert }_{L^2\left(0,t;H^{-1}\left(\Omega \right)\right)}^2\right)\\ \le {\Vert u_m\left(0\right)\Vert }_H^2+{\Vert {\theta }_m\left(0\right)\Vert }_{L^2\left(\Omega \right)}^2+C{\displaystyle {\int }_0^t\left({\Vert u_m\left(s\right)\Vert }_H^2+{\Vert {\theta }_m\left(s\right)\Vert }_{L^2\left(\Omega \right)}^2\right)\text{d}s}\\ +\left(C{v}_1+C{v}_2\right)\left({\displaystyle \underset{i=1}{\overset{3}{\sum }}{\Vert D_{i}f\Vert }_{L^{{q^{\prime }}_i}\left(Q_T\right)}^{{q^{\prime }}_i}}+{\Vert g\Vert }_{L^2\left(0,T;H^{-1}\left(\Omega \right)\right)}^2\right)\\ \le C_1+C{\displaystyle {\int }_0^t\left({\Vert u_m\left(s\right)\Vert }_H^2+{\Vert {\theta }_m\left(s\right)\Vert }_{L^2\left(\Omega \right)}^2\right)\text{d}s}\text{.}\end{array}$ (13)

其中 $C_1,C$ 是与m无关的常数，使用Grownwall不等式，设

$y\left(t\right)={\Vert u_m\left(t\right)\Vert }_H^2+{\Vert {\theta }_m\left(t\right)\Vert }_{L^2\left(\Omega \right)}^2,g\left(s\right)=C,$

则

${\Vert u_m\left(t\right)\Vert }_H^2+{\Vert {\theta }_m\left(t\right)\Vert }_{L^2\left(\Omega \right)}^2\le C\cdot {\text{e}}^{{\displaystyle {\int }_0^{t}C\text{d}s}}=C\cdot {\text{e}}^{Ct}\le C\cdot {\text{e}}^{CT}\le C\left(T\right).$ (14)

进而利用(13)可得

$\underset{t\in \left(0,T\right)}{\sup }\left({\Vert u_m\left(t\right)\Vert }_{L^2\left(\Omega \right)}^2\text{+}{\Vert {\theta }_m\left(t\right)\Vert }_{L^2\left(\Omega \right)}^2\right)+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\Vert D_i{u}_m\Vert }_{L^{q_i}\left(0,T;L^{q_i}\left(\Omega \right)\right)}^{q_i}}+{\displaystyle {\int }_0^T{\Vert \nabla {\theta }_m\Vert }_{L^2\left(\Omega \right)}^2\text{d}s}\le C\left(T\right),$

又对 $\forall \phi \in L^{q_i}\left(0,T;V_{q_i}\left(\Omega \right)\right)$ 有

$\begin{array}{c}-{\displaystyle {\iint }_{Q_T}{D}_i\left({\left|D_i{u}_m\right|}^{q_i-2}{D}_i{u}_m\right)\phi \text{d}x\text{d}t}={\displaystyle {\iint }_{Q_T}\left({\left|D_i{u}_m\right|}^{q_i-2}{D}_i{u}_m\right)\cdot D_i\phi \text{d}x\text{d}t}\\ \le {\displaystyle {\iint }_{Q_T}{\left|D_i{u}_m\right|}^{q_i-1}\cdot \left|D_i\phi \right|\text{d}x\text{d}t}\\ \le {\displaystyle {\int }_0^T{\Vert D_i{u}_m\Vert }_{L^{q_i}\left(\Omega \right)}^{q_i-1}}{\Vert D_i\phi \Vert }_{L^{q_i}\left(\Omega \right)}\text{d}t\\ \le {\Vert D_i{u}_m\Vert }_{L^{q_i}\left(0,T;L^{q_i}\left(\Omega \right)\right)}^{q_i-1}{\Vert \phi \Vert }_{L^{q_i}\left(0,T;V_{q_i}\right)},\end{array}$

从而得到

${\Vert D_i\left({\left|D_i{u}_m\right|}^{q_i-2}{D}_i{u}_m\right)\Vert }_{L^{{q^{\prime }}_i}\left(0,T;{V^{\prime }}_{q_i}\left(\Omega \right)\right)}\le C.$ (15)

下面推导 $\frac{\partial u_m}{\partial t}$ 和 $\frac{\partial {\theta }_m}{\partial t}$ 的先验估计。

设 $P_m$ 为 $V_{\stackrel{\to }{q}}$ 到 $V^m$ 上的正交投影算子，即

$P_{m}u={\displaystyle \underset{j=1}{\overset{m}{\sum }}\left(u,{\phi }_j\right){\phi }_j},\forall u\in V_{\stackrel{\to }{q}},$

则 $P_m$ 的伴随算子 $P_m^{\ast }:\widetilde{V^{\prime }}\to \widetilde{V^{\prime }}$，有 $P_m^{\ast }{u^{\prime }}_m={u^{\prime }}_m$，且

${\Vert P_m\Vert }_{L\left(\tilde{V},\tilde{V}\right)}\le 1,{\Vert P_m^{\ast }\Vert }_{L\left({\tilde{V}}^{\prime },{\tilde{V}}^{\prime }\right)}\le 1.$

重写等式 ${\left(\text{8}\right)}_{\text{1}}$ 如下

$\frac{\partial u_m}{\partial t}={\displaystyle \underset{i=1}{\overset{N}{\sum }}{P}_m^{\ast }{D}_i\left({\left|D_i{u}_m\right|}^{q_i-2}{D}_i{u}_m\right)}-P_m^{\ast }\left(u_m\cdot \nabla \right)u_m+P_m^{\ast }f,$ (16)

设 $A_i\left(u_m\right)=D_i\left({\left|D_i{u}_m\right|}^{q_i-2}{D}_i{u}_m\right)$， $A\left(u_m\right)={\displaystyle \underset{i=1}{\overset{N}{\sum }}{A}_i\left(u_m\right)}={\displaystyle \underset{i=1}{\overset{N}{\sum }}{D}_i\left({\left|D_i{u}_m\right|}^{q_i-2}{D}_i{u}_m\right)}$，

对 $\forall \phi \in L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$，有

$\begin{array}{c}{\displaystyle {\iint }_{Q_T}{P}_m^{\ast }}A\left(u_m\right)\phi \text{d}x\text{d}t={\displaystyle {\iint }_{Q_T}A\left(u_m\right)\cdot \left(P_m\phi \right)\text{d}x\text{d}t}\\ \le {\displaystyle \underset{i=1}{\overset{N}{\sum }}\left|{\displaystyle {\iint }_{Q_T}-D_i\left({\left|D_i{u}_m\right|}^{q_i-2}{D}_i{u}_m\right)}\cdot \left(P_m\phi \right)\text{d}x\text{d}t\right|}\\ \le {\displaystyle \underset{i=1}{\overset{N}{\sum }}{\displaystyle {\int }_0^T{\Vert D_i{u}_m\Vert }_{L^{q_i}\left(\Omega \right)}^{q_i-1}\cdot {\Vert P_m\phi \Vert }_{V_{\stackrel{\to }{q}}}\text{d}t}}\\ \le {\displaystyle \underset{i=1}{\overset{N}{\sum }}{\Vert D_i{u}_m\Vert }_{L^{q_i}\left(0,T;L^{q_i}\left(\Omega \right)\right)}^{q_i-1}}\cdot {\Vert \phi \Vert }_{L^{q_i}\left(0,T;V_{\stackrel{\to }{q}}\right)}\\ \le {\Vert u_m\Vert }_{L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)}^{\stackrel{\to }{q}-1}\cdot {\Vert \phi \Vert }_{L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)};\end{array}$ (17)

对于 $-P_m^{\ast }\left(u_m\cdot \nabla \right)u_m$，利用引理2得

$\begin{array}{c}-{\displaystyle {\iint }_{Q_T}{P}_m^{\ast }\left(u_m\cdot \nabla \right)u_m}\cdot \phi \text{d}x\text{d}t=-{\displaystyle {\iint }_{Q_T}\left(u_m\cdot \nabla \right)u_m\cdot \left(P_m\phi \right)\text{d}x\text{d}t}\\ \le {\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\iint }_{Q_T}\left|u_{m_i}\frac{\partial u_{m_j}}{\partial x_i}\cdot P_m{\phi }_j\right|\text{d}x\text{d}t}}\\ ={\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\iint }_{Q_T}\left|u_{m_i}{u}_{m_j}\cdot \frac{\partial \left(P_m{\phi }_j\right)}{\partial x_i}\right|\text{d}x\text{d}t}}\\ \le {\displaystyle {\int }_0^T{\Vert u_m\Vert }_{L^{2{\alpha }^{\prime }}\left(\Omega \right)}^2}{\Vert \nabla \left(P_m\phi \right)\Vert }_{L^{\alpha }\left(\Omega \right)}\text{d}t\text{,}\end{array}$

由于当 $q_a^{\ast }\ge 2{\alpha }^{\prime }=\frac{2\alpha }{\alpha -1}>2$ 时，利用引理1，知 $V_{\stackrel{\to }{q}}$ ↪ $L^{2{\alpha }^{\prime }}\left(\Omega \right)$ 成立，由 $\alpha$ 的定义知 $V_{\stackrel{\to }{q}}$ ↪ $V_{\alpha }$，从而有

$\begin{array}{c}-{\displaystyle {\iint }_{Q_T}{P}_m^{\ast }}\left(u_m\cdot \nabla \right)u_m\cdot \phi \text{d}x\text{d}t\le C{\displaystyle {\int }_0^T{\Vert u_m\Vert }_{V_{\stackrel{\to }{q}}}}{\Vert \phi \Vert }_{V_{\stackrel{\to }{q}}}\text{d}t\\ \le C{\Vert u_m\Vert }_{L^2\left(0,T;V_{\stackrel{\to }{q}}\right)}{\Vert \phi \Vert }_{L^2\left(0,T;V_{\stackrel{\to }{q}}\right)}\\ \le C{\Vert u_m\Vert }_{L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)}{\Vert \phi \Vert }_{L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)};\end{array}$ (18)

对于 ${\displaystyle {\iint }_{Q_T}{P}_m^{\ast }}f\phi \text{d}x\text{d}t$，有

$\begin{array}{c}{\displaystyle {\iint }_{Q_T}{P}_m^{\ast }}f\phi \text{d}x\text{d}t\le {\displaystyle {\iint }_{Q_T}\left|f\cdot \left(P_m\varphi \right)\right|\text{d}x\text{d}t}\\ \le {\Vert f\Vert }_{L^{{\stackrel{\to }{q}}^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right)}\cdot {\Vert \phi \Vert }_{L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)};\end{array}$ (19)

综合(17)~(19)，并利用(14) (15)得

$\frac{\partial u_m}{\partial t}\in L^{{\stackrel{\to }{q}}^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right),$

由于 $L^{{\stackrel{\to }{q}}^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right)$ ↪ $L^{{\beta }^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right)$，所以

$\frac{\partial u_m}{\partial t}\in L^{{\beta }^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right).$ (20)

同理，设 $R_m$ 为 $H_0^1\left(\Omega \right)$ 到 $W^m$ 上的正交投影算子，满足

$R_m\theta ={\displaystyle \underset{j=1}{\overset{m}{\sum }}\left({\theta }_m,{\varphi }_j\right){\varphi }_j},\forall {\theta }_m\in H_0^1\left(\Omega \right).$

对于 $R_m$ 的伴随算子 $R_m^{\ast }:H^{-1}\left(\Omega \right)\to H^{-1}\left(\Omega \right)$，有

$R_m^{\ast }{{\theta }^{\prime }}_m={{\theta }^{\prime }}_m,$

且

${\Vert R_m\Vert }_{L\left(H_0^1\left(\Omega \right),H_0^1\left(\Omega \right)\right)}\le 1,{\Vert R_m\Vert }_{L\left(H^{-1}\left(\Omega \right),H^{-1}\left(\Omega \right)\right)}\le 1.$

重写等式(8)₂如下

$\frac{\partial {\theta }_m}{\partial t}=R_m^{\ast }\Delta {\theta }_m-R_m^{\ast }\left(u_m\cdot \nabla \right){\theta }_m+R_m^{\ast }g,$ (21)

对 $\forall \varphi \in L^2\left(0,T;H_0^1\left(\Omega \right)\right)$，有

$\begin{array}{c}{\displaystyle {\iint }_{Q_T}{R}_m^{\ast }\Delta {\theta }_m\cdot \varphi \text{d}x\text{d}t}={\displaystyle {\iint }_{Q_T}\Delta {\theta }_m\cdot \left(R_m\varphi \right)\text{d}x\text{d}t}\\ \le {\displaystyle {\int }_0^T{\displaystyle {\int }_{\Omega }\left|\nabla {\theta }_m\cdot \nabla \left(R_m\varphi \right)\right|\text{d}x\text{d}t}}\\ \le {\displaystyle {\int }_0^T{\Vert \nabla {\theta }_m\Vert }_{L^2\left(\Omega \right)}\cdot {\Vert \nabla \left(R_m\varphi \right)\Vert }_{L^2\left(\Omega \right)}\text{d}t}\\ \le {\displaystyle {\int }_0^T{\Vert \nabla {\theta }_m\Vert }_{L^2\left(\Omega \right)}\cdot {\Vert \nabla \varphi \Vert }_{L^2\left(\Omega \right)}\text{d}t}\\ \le C{\displaystyle {\int }_0^T{\Vert {\theta }_m\Vert }_{H_0^1\left(\Omega \right)}\cdot {\Vert \varphi \Vert }_{H_0^1\left(\Omega \right)}}\\ \le C{\Vert {\theta }_m\Vert }_{L^2\left(0,T;H_0^1\left(\Omega \right)\right)}\cdot {\Vert \varphi \Vert }_{L^2\left(0,T;H_0^1\left(\Omega \right)\right)}\end{array}$ (22)

对于 $-{\displaystyle {\iint }_{Q_T}{R}_m^{\ast }\left(u_m\cdot \nabla \right){\theta }_m\cdot \varphi \text{d}x\text{d}t}$，有

$\begin{array}{c}-{\displaystyle {\iint }_{Q_T}{R}_m^{\ast }\left(u_m\cdot \nabla \right){\theta }_m\cdot \varphi \text{d}x\text{d}t}=-{\displaystyle {\iint }_{Q_T}\left(u_m\cdot \nabla \right){\theta }_m\cdot \left(R_m\varphi \right)\text{d}x\text{d}t}\\ =-{\displaystyle \underset{i,j=1}{\overset{3}{\sum }}{\displaystyle {\iint }_{Q_T}{u}_{m_i}\frac{\partial {\theta }_{m_j}}{\partial x_i}{R}_m{\varphi }_j\text{d}x\text{d}t}}\\ ={\displaystyle \underset{i,j=1}{\overset{3}{\sum }}{\displaystyle {\iint }_{Q_T}{u}_{m_i}{\theta }_{m_j}}\frac{\partial \left(R_m{\varphi }_j\right)}{\partial x_i}\text{d}x\text{d}t}\\ \le {\displaystyle {\int }_0^T{\Vert u_m\Vert }_{L^2\left(\Omega \right)}{\Vert {\theta }_m\Vert }_{L^3\left(\Omega \right)}{\Vert \nabla R_m\varphi \Vert }_{L^6\left(\Omega \right)}\text{d}t}\\ \le {\displaystyle {\int }_0^T{\Vert u_m\Vert }_{L^2\left(\Omega \right)}{\Vert {\theta }_m\Vert }_{L^3\left(\Omega \right)}{\Vert \varphi \Vert }_{H_0^1\left(\Omega \right)}\text{d}t}\\ \le C{\Vert u_m\Vert }_{L^4\left(0,T;L^2\left(\Omega \right)\right)}{\Vert {\theta }_m\Vert }_{L^4\left(0,T;L^3\left(\Omega \right)\right)}{\Vert \varphi \Vert }_{L^2\left(0,T;H_0^1\left(\Omega \right)\right)},\end{array}$

由(14)知 ${\theta }_m\in L^2\left(0,T;W^{1,2}\left(\Omega \right)\right)\cap L^{\infty }\left(0,T;L^2\left(\Omega \right)\right)$，从而知 [17] (引理2.3.3) ${\theta }_m\in L^4\left(0,T;L^3\left(\Omega \right)\right)$，可得

$-{\displaystyle {\iint }_{Q_T}{R}_m^{\ast }\left(u_m\cdot \nabla \right){\theta }_m\cdot \varphi \text{d}x\text{d}t}\le C{\Vert \varphi \Vert }_{L^2\left(0,T;H_0^1\left(\Omega \right)\right)};$ (23)

对于 ${\displaystyle {\iint }_{Q_T}{R}_m^{\ast }g\cdot \varphi \text{d}x\text{d}t}$，有

${\displaystyle {\iint }_{Q_T}{R}_m^{\ast }g\cdot \varphi \text{d}x\text{d}t}\le {\displaystyle {\iint }_{Q_T}\left|g{R}_m\varphi \right|\text{d}x\text{d}t}\le {\Vert g\Vert }_{L^2\left(0,T;H^{-1}\left(\Omega \right)\right)}\cdot {\Vert \varphi \Vert }_{L^2\left(0,T;H_0^1\left(\Omega \right)\right)};$ (24)

综合(22)~(24)式并利用(14)得

$\frac{\partial {\theta }_m}{\partial t}\in L^2\left(0,T;H^{-1}\left(\Omega \right)\right).$ (25)

第三步：近似解的收敛与解的存在性证明

由(14) (15) (20) (25)可以得到下列收敛性

$\begin{array}{l}{u}_m\rightharpoonup u弱\ast 收敛于L^{\infty }\left(0,T;H\right);\\ u_m\rightharpoonup u于L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right);\\ {u^{\prime }}_m\rightharpoonup u^{\prime }于L^{{\beta }^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right);\\ A\left(u_m\right)\rightharpoonup S于L^{{\stackrel{\to }{q}}^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right);\\ {\theta }_m\rightharpoonup \theta 于L^2\left(0,T;H_0^1\left(\Omega \right)\right);\\ {\theta }_m\rightharpoonup \theta 弱\ast 收敛于L^{\infty }\left(0,T;L^2\left(\Omega \right)\right);\\ {{\theta }^{\prime }}_m\rightharpoonup {\theta }^{\prime }于L^2\left(0,T;H^{-1}\left(\Omega \right)\right);\end{array}$ (26)

由于 $H$ ↪ ${V^{\prime }}_{\stackrel{\to }{q}}$ 是连续的，且当 $\alpha >2$ 时， $q_a^{\ast }>2$ 恒成立，故 $V_{\stackrel{\to }{q}}$ ↪↪ $H$，利用Aubin-Lions引理可得

$u_m\to u于L^{\alpha }\left(0,T;H\right);$ (27)

利用(13)和(27)并利用插值定理可得

$u_m\to u于L^{\alpha }\left(0,T;H\right),\forall \alpha \ge 1$ (28)

另一方面，由于 $H_0^1\left(\Omega \right)$ ↪↪ $L^2\left(\Omega \right)$，利用Aubin-Lions引理可得

${\theta }_m\to \theta 于L^2\left(0,T;L^2\left(\Omega \right)\right).$ (29)

由于上述的收敛性，在(8)₁式中令 $m\to \infty$，可以得到

${\displaystyle {\int }_{\Omega }\frac{\partial u}{\partial t}\cdot {\phi }_k\text{d}x}+{\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)u\right]\cdot {\phi }_k}\text{d}x+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\displaystyle {\int }_{\Omega }{S}_i\cdot D_i{\phi }_k\text{d}x}}={\displaystyle {\int }_{\Omega }f\cdot {\phi }_k}\text{d}x$ (30)

在(8)₂式中令 $m\to \infty$，可以得到

${\displaystyle {\int }_{\Omega }\frac{\partial \theta }{\partial t}\cdot {\varphi }_r\text{d}x}+{\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)\theta \right]\cdot {\varphi }_r\text{d}x}+{\displaystyle {\int }_{\Omega }\nabla \theta \cdot \nabla {\varphi }_r\text{d}x}={\displaystyle {\int }_{\Omega }g\cdot {\varphi }_r\text{d}x}$ (31)

当 $\phi \in V_{\stackrel{\to }{q}}$， $u\in L^{\infty }\left(0,T;H\right)\cap L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$ 时，有

${\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)u\right]\phi \text{d}x}=-{\displaystyle {\int }_{\Omega }u\left(t\right)\otimes u:\nabla \phi \text{d}x}\le {\Vert u\Vert }_{L^{2{\alpha }^{\prime }}\left(\Omega \right)}^2{\Vert \nabla \phi \Vert }_{L^{\alpha }\left(\Omega \right)},$

因为 $q_a^{\ast }\ge \frac{2\alpha }{\alpha -1}$，嵌入 $V_{\stackrel{\to }{q}}$ ↪ $V_{\alpha }$ 成立，并利用引理1可得

${\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)u\right]\phi \text{d}x}\le {\Vert u\Vert }_{V_{\stackrel{\to }{q}}}^2{\Vert \nabla \phi \Vert }_{V_{\stackrel{\to }{q}}}.$

由于 $\tilde{V}={\displaystyle {\cup }_{m=1}^{\infty }{V}^m}$ 以及 $\tilde{V}$ 在 $V_{\stackrel{\to }{q}}$ 中的稠密性可知对 $\forall \phi \in V_{\stackrel{\to }{q}}$， $\varphi \in L^2\left(\Omega \right)$ 均有

${\displaystyle {\int }_{\Omega }\frac{\partial u}{\partial t}\cdot \phi \text{d}x}+{\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)u\right]\cdot \phi \text{d}x}+{\displaystyle \underset{i=1}{\overset{N}{\sum }}{\displaystyle {\int }_{\Omega }{S}_i\cdot D_i\phi \text{d}x}}={\displaystyle {\int }_{\Omega }f\cdot \phi \text{d}x},$ (32)

${\displaystyle {\int }_{\Omega }\frac{\partial \theta }{\partial t}\cdot \varphi \text{d}x}+{\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)\theta \right]\cdot \varphi \text{d}x}+{\displaystyle {\int }_{\Omega }\nabla \theta \cdot \nabla \varphi \text{d}x}={\displaystyle {\int }_{\Omega }g\cdot \varphi \text{d}x},$ (33)

利用(32) (33)可以得到

$\frac{\partial u}{\partial t}\in L^1\left(0,T;V^{\prime }\right),\frac{\partial \theta }{\partial t}\in L^1\left(0,T;H^{-1}\right),$

所以 $u:\left[0,T\right]\to V^{\prime }$， $\theta :\left[0,T\right]\to H^{-1}$ 是几乎处处绝对连续的。又因为 $u\in L^{\infty }\left(0,T;H\right)$， $\theta \in L^{\infty }\left(0,T;L^2\left(\Omega \right)\right)$ 以及 $H$ ↪ $V^{\prime }$， $L^2\left(\Omega \right)$ ↪ $H^{-1}\left(\Omega \right)$，并利用引理3可得 $u\in C_W\left(\left[0,T\right];H\right)$， $\theta \in C_W\left(\left[0,T\right];L^2\left(\Omega \right)\right)$。

为完成定理证明，只需证明

$A_i\left(u\right)=S_i,A\left(u\right)={\displaystyle \underset{i=1}{\overset{N}{\sum }}{A}_i}\left(u\right)=S.$ (34)

由于对 $\forall \xi ,\eta \in L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$，且 $\xi \ne \eta$ 有

${\displaystyle \underset{i=1}{\overset{N}{\sum }}{\langle {\left|D_i\xi \right|}^{q_i-2}{D}_i\xi -{\left|D_i\eta \right|}^{q_i-2}{D}_i\eta ,D_i\xi -D_i\eta \rangle }_{L^{q_i}\left(Q_T\right)\times L^{q_i}\left(Q_T\right)}}>0.$

即 $A_i,A$ 为单调算子，下面运用单调性方法来证明(34)。

首先要证明对 $\forall u,v\in L^{\infty }\left(0,T;H\right)\cap L^{{\stackrel{\to }{q}}^{\prime }}\left(0,T;{V^{\prime }}_{\stackrel{\to }{q}}\right)$，有

${\displaystyle {\int }_{\Omega }\left[\left(u\cdot \nabla \right)u\right]\cdot v\text{d}x}=-{\displaystyle {\int }_{\Omega }u\otimes u:\nabla v\text{d}x}\in L^1\left(0,T\right).$ (35)

通过 $\alpha$ 的定义并由引理1得

$L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$ ↪ $L^{\alpha }\left(0,T;V_{\stackrel{\to }{q}}\right)$ ↪ $L^{\alpha }\left(0,T;L^{q_a^{\ast }}\right),$

由插值公式有

$L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)\cap L^{\infty }\left(0,T;H\right)\subset L^{\alpha }\left(0,T;L^{q_a^{\ast }}\right)\cap L^{\infty }\left(0,T;L^2\left(\Omega \right)\right)\subset L^{\rho }\left(Q_T\right),$

其中 $\rho =2+\alpha -\frac{2\alpha }{q_a^{\ast }}$，当(6)成立时，通过计算可知 $\frac{2}{\rho }+\frac{1}{\rho }\le 1$，所以由Hölder不等式知(35)成立。

类似于文献 [11] 中的方法可得，对 $a.e.t\in \left(0,T\right)$，有

$\frac{1}{2}{\Vert u\left(t\right)\Vert }_{L^2\left(\Omega \right)}^2+{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }S:u\text{d}x\text{d}s}}\ge \frac{1}{2}{\Vert u\left(0\right)\Vert }_{L^2\left(\Omega \right)}^2+{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }fu\text{d}x\text{d}s}}.$ (36)

可任取 $\phi \in L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$，设

$X_{\mu }={\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }\left(A\left(u_{\mu }\right)-A\left(\phi \right):\left(u_{\mu }-\phi \right)\right)\text{d}x\text{d}s}+\frac{1}{2}{\Vert u_{\mu }\left(t\right)\Vert }_{L^2\left(\Omega \right)}^2}\text{,}$ (37)

由A的单调性知

$\underset{\mu \to \infty }{\lim }\inf X_{\mu }\ge \frac{1}{2}{\Vert u\left(t\right)\Vert }_{L^2\left(\Omega \right)}^2,$ (38)

由于(8)₁可知

$X_{\mu }=\frac{1}{2}{\Vert u_{\mu }\left(0\right)\Vert }_{L^2\left(\Omega \right)}^2-{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }A\left(u_{\mu }\right):\phi \text{d}x\text{d}s}}-{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }A\left(\phi \right):\left(u_{\mu }-\phi \right)\text{d}x\text{d}s}}+{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }f{u}_{\mu }\text{d}x\text{d}s}}$

其中令 $m\to \infty$，得

$X=\frac{1}{2}{\Vert u\left(0\right)\Vert }_{L^2\left(\Omega \right)}^2-{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }S:\phi \text{d}x\text{d}s}}-{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }A\left(\phi \right):\left(u-\phi \right)\text{d}x\text{d}s}}+{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }fu\text{d}x\text{d}s}}\text{,}$

综合(37) (38)可得

${\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }\left(S-A\left(\phi \right)\right):\left(u-\phi \right)\text{d}x\text{d}t}}\ge 0,a.e.t\in \left(0,T\right).$ (39)

令 $\phi =u-\lambda \omega$， $\lambda >0$， $\omega \in L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)$，代入到(39)式得

${\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }\left(S-A\left(u-\lambda \omega \right)\text{,}\omega \right)\text{d}x\text{d}t}}\ge 0\text{,}$

令 $\lambda \to 0$，取极限得

${\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }\left(S-A\left(u\right)\omega \right)\text{d}x\text{d}t}}\ge 0,\forall \omega \in L^{\stackrel{\to }{q}}\left(0,T;V_{\stackrel{\to }{q}}\right)\text{,}$

所以

$S=A\left(u\right)\text{,}a.e.\left(x,t\right)\in Q_T.$

定理证得。

4. 总结

本文在三维空间中考虑一类各向异性非牛顿Boussinesq方程组的初边值问题，给出了证明解的存在性的详细步骤，采用Galerkin方法构造近似解序列 $\left\{u_m,{\theta }_m\right\}$，通过对近似解序列 $\left\{u_m,{\theta }_m\right\}$ 进行一直估计以及收敛性的证明并利用单调性方法证明其解的存在性。

基金项目

吉林省教育厅“十三五”科学技术项目(批准号：JJKH20200727KJ)。