三维微极Rayleigh-Bénard对流方程在Triebel-Lizorkin空间中的正则性准则

Regularity Criteria for the Three-Dimensional Micropolar Rayleigh-Bénard Convection Equations in Triebel-Lizorkin Spaces

摘要: 本文考虑三维微极Rayleigh-Bénard对流方程，设$\left(u,\omega ,\theta \right)\left(t,x\right)$ 是方程在$\left(0,T\right)$ 上的Leray-Hopf弱解。证明当压力满足$q>\frac{12}{5}$ ，${\displaystyle {\int }_0^T{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}^{\frac{10q}{5q-6}}}d\tau <\infty$ ，或者压力的梯度满足$\frac{12}{11}<q<4$ ，${\displaystyle {\int }_0^T{\Vert \nabla \pi \Vert }_{{\dot{F}}_{q,\frac{8q}{12-3q}\left({ℝ}^3\right)}^0}^{\frac{8q}{11q-12}}}d\tau <\infty$ 时，解$(u,\omega ,\theta )$ 可以延拓到$t=T$ 。

Abstract: This paper considers the three-dimensional micropolar Rayleigh-Bénard convection equations. We establish regularity criteria for weak solutions, based on the following conditions concerning the pressure: $q>\frac{12}{5}$ , ${\displaystyle {\int }_0^T{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}^{\frac{10q}{5q-6}}}d\tau <\infty$ , or$\frac{12}{11}<q<4$ , ${\displaystyle {\int }_0^T{\Vert \nabla \pi \Vert }_{{\dot{F}}_{q,\frac{8q}{12-3q}\left({ℝ}^3\right)}^0}^{\frac{8q}{11q-12}}}d\tau <\infty$ then the solution can extended beyond $t=T$ .

1. 引言

研究${ℝ}^3$ 上的三维微极Rayleigh-Bénard对流方程

$\{\begin{array}{l}{\partial }_{t}u+\left(u\cdot \nabla \right)u-\left(\nu +\kappa \right)\Delta u+\nabla \pi =2\kappa \nabla \times \omega +\theta e_3,\hfill \\ {\partial }_t\omega +\left(u\cdot \nabla \right)\omega -\left(\alpha +\beta \right)\nabla \nabla \cdot \omega +4\kappa \omega -\gamma \Delta \omega =2\kappa \nabla \times u,\hfill \\ {\partial }_t\theta +u\cdot \nabla \theta -\mu \Delta \theta =u\cdot e_3,\hfill \\ \begin{array}{l}\nabla \cdot u=0,\hfill \\ {\left(u,\omega ,\theta \right)\left(x,t\right)|}_{t=0}=\left(u_0,{\omega }_0,{\theta }_0\right)\left(x\right),\hfill \end{array}\hfill \end{array}$ (1)

其中$u$ 是流体速度场，$\omega$ 为流体粒子旋转角速度的微旋转场，$\pi$ 为标量压力，$\theta$ 为温度，$\nu$ 为牛顿运动粘度系数，$\kappa$ 是微旋转粘度系数，$\alpha$ 和$\beta$ 为角粘度系数，$\mu$ 是热扩散系数，$e_3=\left(0,0,1\right)$ 为垂直单位矢量，$\theta e_3$ 项描述浮力对流体运动的作用，$u\cdot e_3$ 模拟热的无粘性流体中的Rayleigh-Bénard对流运动。由于$\nu ,\mu ,\kappa ,\alpha ,\beta ,\gamma$ 的具体值在本文的讨论中没有特殊作用，为简单起见，取$\mu =\gamma =1$ ，$\nu =\kappa =\frac{1}{2}$ ，$\alpha =\beta =\frac{1}{2}$ ，则方程(1)可以化为

$\{\begin{array}{l}{\partial }_{t}u+\left(u\cdot \nabla \right)u-\Delta u+\nabla \pi =\nabla \times \omega +\theta e_3,\hfill \\ {\partial }_t\omega +\left(u\cdot \nabla \right)\omega -\nabla \nabla \cdot \omega +2\omega -\Delta \omega =\nabla \times u,\hfill \\ {\partial }_t\theta +u\cdot \nabla \theta -\Delta \theta =u\cdot e_3,\hfill \\ \nabla \cdot u=0,\hfill \\ {\left(u,\omega ,\theta \right)\left(x,t\right)|}_{t=0}=\left(u_0,{\omega }_0,{\theta }_0\right)\left(x\right).\hfill \end{array}$ (2)

当$\theta =0,\omega =0$ 时，方程(2)是经典的Navier-Stokes方程

$\{\begin{array}{l}{\partial }_{t}u+\left(u\cdot \nabla \right)u-\Delta u+\nabla \pi =0,\hfill \\ \nabla \cdot u=0,\hfill \\ {u\left(x,t\right)|}_{t=0}=u_0\left(x\right).\hfill \end{array}$ (3)

许多数学家对方程(3)中压力提条件得到解的正则性准则。2001年，Chae和Lee [1]证明当压力$\pi$ 满足

$\pi \in L^r\left(0,T;L^s\left({ℝ}^3\right)\right),\frac{2}{r}+\frac{3}{s}<2,s>\frac{3}{2}$

时，$u$ 是正则解。2002年Berselli和Galdi [2]证明当压力$\pi$ 满足

$\nabla \pi \in L^r\left(0,T;L^s\left({ℝ}^3\right)\right),\frac{2}{r}+\frac{3}{s}=3,s\in \left[\frac{9}{7},3\right]$

时，$u$ 也是正则解。2006年Zhou [3]证明当压力$\pi$ 满足

$\nabla \pi \in L^r\left(0,T;L^s\left({ℝ}^3\right)\right),\frac{2}{r}+\frac{3}{s}\le 3,s\in \left[1,\infty \right]$

时，弱解$u$ 是正则的。2007年Chen和Zhang [4]证明若压力$\pi$ 满足

${\displaystyle {\int }_0^T{\Vert \pi \Vert }_{B_{\infty ,\infty }^0}}\text{d}t<\infty ,$

则$u$ 在$\left(0,T\right]$ 正则。2021年Wan和Chen [5]证明若压力$\pi$ 满足

$\pi \in L^p\left(0,T;{\dot{F}}_{q,\frac{10q}{5q+6}}^0\left({ℝ}^3\right)\right),\frac{2}{p}+\frac{3}{q}<\frac{7}{4},\frac{12}{5}<q\le \infty ,$

或

$\nabla \pi \in L^p\left(0,T;{\dot{F}}_{q,\frac{8q}{12-3q}}^0\left({ℝ}^3\right)\right),\frac{2}{p}+\frac{3}{q}=\frac{11}{4},\frac{12}{11}<q<4,$

则弱解$u$ 能光滑地延拓出$t=T$ 。

当$\theta =0$ 时，方程(2)变为微极流方程

$\{\begin{array}{l}{\partial }_{t}u+\left(u\cdot \nabla \right)u-\left(\nu +\kappa \right)\Delta u+\nabla \pi =2\kappa \nabla \times \omega ,\hfill \\ {\partial }_t\omega +\left(u\cdot \nabla \right)\omega -\left(\alpha +\beta \right)\nabla \nabla \cdot \omega +4\kappa \omega -\gamma \Delta \omega =2\kappa \nabla \times u,\hfill \\ \nabla \cdot u=0,\hfill \\ {\left(u,\omega \right)\left(x,t\right)|}_{t=0}=\left(u_0,{\omega }_0\right)\left(x\right).\hfill \end{array}$ (4)

1996年，Eringen [6]引入微极流方程来模拟微极流体，最先提出有关微极流方程的理论。1997年，Galdi和Rionero [7]得到微极流方程弱解的存在性。2005年，Yamaguchi在[8]得到微极流方程强解的全局存在性。2010年，Yuan [9]证明若速度$u$ 满足

$u\in L^q\left(0,T;L^{p,\infty }\left({ℝ}^3\right)\right),\frac{2}{q}+\frac{3}{p}\le 1,3<p\le \infty ,$

或

$\nabla u\in L^q\left(0,T;L^{p,\infty }\left({ℝ}^3\right)\right),\frac{2}{q}+\frac{3}{p}\le 2,\frac{3}{2}<p\le \infty ,$

则微极流方程的强解$\left(u,\omega \right)$ 可以光滑地延拓到$\left(0,T\right]$ ，从而得到微极流方程经典的Serrin型正则性准则。在此之后大量微极流方程的正则性准则的结论被证明。同年Dong和Zhang [10]证明当速度$u$ 满足

$u\in L^{\frac{2}{1+r}}\left(0,T;B_{\infty ,\infty }^r\left({ℝ}^3\right)\right),-1<r<1$

时微极流方程的弱解$\left(u,\omega \right)$ 在$\left(0,T\right]$ 正则。2017年Gala [11]证明若压力$\pi$ 满足

${\displaystyle {\int }_0^T\frac{{\Vert \pi \left(\cdot ,t\right)\Vert }_{{\dot{B}}_{\infty ,\infty }^{-1}}^2}{1+\ln \left(e+{\Vert \pi \left(\cdot ,t\right)\Vert }_{L^2}\right)}\text{d}t}<\infty ,$

则三维微极流方程的弱解$\left(u,\omega \right)$ 在$\left(0,T\right]$ 上正则。

微极Rayleigh-Bénard对流方程描述的是一种特殊的热对流现象。当研究填充两个刚性表面之间区域的流体层时，流体从下面加热，通常会在Boussinesq方程的近似框架内描述这种热对流现象(见[12]-[15])，这种现象被称作微极Rayleigh-Bénard对流问题。受文献[5]、[10]和[11]的启发，考虑微极Rayleigh-Bénard 对流方程在Triebel-Lizorkin空间上的正则性准则。下面给出本文的主要结论。

定理1.1 设$\left(u_0,{\omega }_0,{\theta }_0\right)\in L^2\left({ℝ}^3\right)\cap L^4\left({ℝ}^3\right)$ ，$\nabla \cdot u_0=\nabla \cdot {\omega }_0=0$ ，且$\left(u,\omega ,\theta \right)\left(t,x\right)$ 是方程组(2)在$\left(0,T\right)$ 上的Leray-hopf弱解。若压力$\pi$ 满足

$$ ${\displaystyle {\int }_0^T{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}^{\frac{10q}{5q-6}}}\text{d}\tau <\infty ,q>\frac{12}{5},$(5)

则(2)的解$\left(u,\omega ,\theta \right)$ 可以延拓到$t=T$ 。

定理1.2 设$\left(u_0,{\omega }_0,{\theta }_0\right)\in L^2\left({ℝ}^3\right)\cap L^4\left({ℝ}^3\right)$ ，$\nabla \cdot u_0=\nabla \cdot {\omega }_0=0$ ，且$\left(u,\omega ,\theta \right)\left(t,x\right)$ 是方程组(2)在$\left(0,T\right)$ 上的Leray-hopf弱解，若$\nabla \pi$ 满足

${\displaystyle {\int }_0^T{\Vert \nabla \pi \Vert }_{{\dot{F}}_{q,\frac{8q}{12-3q}\left({ℝ}^3\right)}^0}^{\frac{8q}{11q-12}}}\text{d}\tau <\infty ,\frac{12}{11}<q<4,$(6)

则(2)的解$\left(u,\omega ,\theta \right)$ 可以延拓到$t=T$ 。

注：定理1.1包含了[5]中定理1.1的结果，定理1.2包含了[5]中定理1.2的结果。

2. 预备知识与重要引理

本节介绍一些常用的函数空间及一些基本性质。

定义2.1 [16] 设$\chi ,\phi \in \mathcal{S}\left({ℝ}^n\right)$ 是非负径向函数，满足$supp\chi \subset B=\left\{\xi \in {ℝ}^n:\left|\xi \right|\le \frac{3}{4}\right\}$ 及$supp\phi \subset \mathcal{C}=\left\{\xi \in {ℝ}^n:\frac{3}{4}\le \left|\xi \right|\le \frac{8}{3}\right\}$ ，使得

${\displaystyle \sum _{j\in Z}\phi }\left(2^{-j}\xi \right)=1,\forall \xi \in {ℝ}^n\backslash \left\{0\right\}$

和

$\chi \left(\xi \right)+{\displaystyle \sum _{j\ge 0}\phi }\left(2^{-j}\xi \right)=1,\forall \xi \in {ℝ}^n.$

$\forall j\in Z$ ，定义如下齐次频投算子${\dot{\Delta }}_j$ 和${\dot{S}}_j$ ：

$\begin{array}{c}{\dot{\Delta }}_{j}u={ℱ}^{-1}\left(\phi \left(2^{-j}\xi \right)\widehat{u}\left(\xi \right)\right)\left(x\right)={ℱ}^{-1}\left(\phi \left(2^{-j}\xi \right)\right)\star u\\ =2^{jn}\left({ℱ}^{-1}\phi \right)\left(2^j\cdot \right)\star u=2^{jn}{\displaystyle {\int }_{{ℝ}^n}\left({ℱ}^{-1}\phi \right)\left(2^{j}y\right)u\left(x-y\right)\text{d}y},\\ {\dot{S}}_{j}u={ℱ}^{-1}\left(\chi \left(2^{-j}\xi \right)\widehat{u}\left(\xi \right)\right)\left(x\right)={ℱ}^{-1}\left(\chi \left(2^{-j}\xi \right)\right)\star u\\ =2^{jn}\left({ℱ}^{-1}\chi \right)\left(2^j\cdot \right)\star u=2^{jn}{\displaystyle {\int }_{{ℝ}^n}\left({ℱ}^{-1}\chi \right)\left(2^{j}y\right)u\left(x-y\right)\text{d}y}.\end{array}$

定义2.2 设${{\mathcal{S}}^{\prime }}_h$ 表示无穷远处趋于零的缓增广义函数，$s\in ℝ$ ，$p,q\in \left[1,\infty \right]$ 。对$\forall f\in {{\mathcal{S}}^{\prime }}_h$ ，使得

${\Vert {\left({\displaystyle \sum _{j\in Z}{2}^{jsq}}{\left|{\dot{\Delta }}_{j}f\right|}^q\right)}^{\frac{1}{q}}\Vert }_{L^p}<\infty$

的这样的$f$ 的全体构成的空间称为Triebel-Lizorkin空间，记为${\dot{F}}_{p,q}^s\left({ℝ}^n\right)$ 。

引理1.3.14 设$u$ ，$\pi$ 分别是方程(2)中的速度和压力，$1<s<\infty$ 。则存在常数$C$ 使得

${\Vert \pi \Vert }_{L^s}\le C\left({\Vert u\Vert }_{L^{2s}}^2+{\Vert {\nabla }^{-1}\theta \Vert }_{L^s}\right),$ (7)

${\Vert \nabla \pi \Vert }_{L^s}\le C{\Vert u\cdot \nabla u\Vert }_{L^s}+{\Vert \theta \Vert }_{L^s}.$ (8)

引理1.3.14的证明 用$\nabla \cdot$ 作用于${\left(\mathrm{1.2.2}\right)}_1$ 两边，结合$\nabla \cdot u=0$ 得

$\begin{array}{l}\pi ={\left(-\Delta \right)}^{-1}\left[\nabla \cdot \left(u\cdot \nabla u\right)-\nabla \cdot \left(\theta e_3\right)\right],\\ \nabla \pi =\nabla {\left(-\Delta \right)}^{-1}\nabla \cdot \left(u\cdot \nabla u\right)-\nabla {\left(-\Delta \right)}^{-1}\nabla \cdot \left(\theta e_3\right).\end{array}$

所以由Riesz变换的$L^p$ 有界性得到

${\Vert \pi \Vert }_{L^s}\le C\left({\Vert u\Vert }_{L^{2s}}^2+{\Vert {\nabla }^{-1}\theta \Vert }_{L^s}\right),$ (9)

${\Vert \nabla \pi \Vert }_{L^s}\le C{\Vert u\cdot \nabla u\Vert }_{L^s}+{\Vert \theta \Vert }_{L^s}.$ (10)

即完成了引理1.3.14的证明。

Riesz位势算子$I_{s}f\left(x\right)={\left(-\Delta \right)}^{-\frac{s}{2}}f\left(x\right)={ℱ}^{-1}\left({\left|\xi \right|}^s\right)\star f\left(x\right)={\displaystyle {\int }_{{ℝ}^n}\frac{f\left(y\right)}{|x-y{|}^{n-s}}\text{d}y}$ ，它是从$L^p$ 到$L^q$ 有界的。

引理1.3.15 [17] 设$f\in L^p\left({ℝ}^n\right)$ ，$0<s<n$ 。则存在$C=C\left(p,q,n,s\right)>0$ ，使得

${\Vert I_{s}f\Vert }_{L^q({ℝ}^3)}\le C\left(p,q,n,s\right){\Vert f\Vert }_{L^p({ℝ}^n)},$

其中p，q满足$\frac{1}{p}-\frac{1}{q}=\frac{s}{n}$ 。

3. Triebel-Lizorkin空间中的正则性准则的证明

本节将给出三维微极Rayleigh-Bénard对流方程在Triebel-Lizorkin空间中的正则性准则的证明。

3.1. 定理1.1的证明

第一步：L²-能量估计

将方程(2)₁，(2)₂和(2)₃分别与$u,\omega$ 和$\theta$ 做$L^2$ 内积后将三式相加，再利用$\nabla \cdot u=0$ 、Hölder不等式和Young不等式得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \left(\omega ,u,\theta \right)\Vert }_2^2+{\Vert \left(\nabla u,\nabla \omega ,\nabla \theta \right)\Vert }_2^2+{\Vert \nabla \cdot \omega \Vert }_2^2+2{\Vert \omega \Vert }_2^2\\ ={\displaystyle {\int }_{{ℝ}^3}\nabla }\times \omega \cdot u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\nabla }\times u\cdot \omega \text{d}x+{\displaystyle {\int }_{{ℝ}^3}\theta }{e}_3\cdot u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}u}{e}_3\cdot \theta \text{d}x\\ =2\left({\displaystyle {\int }_{{ℝ}^3}\nabla }\times u\cdot \omega \text{d}x+{\displaystyle {\int }_{{ℝ}^3}\theta }{e}_3\cdot u\text{d}x\right)\\ \le 2{\Vert \nabla u\Vert }_2{\Vert \omega \Vert }_2+2{\Vert \theta \Vert }_2{\Vert u\Vert }_2\\ \le \frac{1}{2}{\Vert \nabla u\Vert }_2^2+2{\Vert \omega \Vert }_2^2+{\Vert \theta \Vert }_2^2+{\Vert u\Vert }_2^2.\end{array}$

整理得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \left(u,\omega ,\theta \right)\Vert }_2^2+\frac{1}{2}{\Vert \left(\nabla u,\nabla \omega ,\nabla \theta \right)\Vert }_2^2+{\Vert \nabla \cdot \omega \Vert }_2^2\le {\Vert \theta \Vert }_2^2+{\Vert u\Vert }_2^2.$

从而有

$\frac{\text{d}}{\text{d}t}{\Vert \left(u,\omega ,\theta \right)\Vert }_2^2+{\Vert \left(\nabla u,\nabla \omega ,\nabla \theta \right)\Vert }_2^2\le C{\Vert \left(u,\omega ,\theta \right)\Vert }_2^2.$ (11)

对(11)式利用Gronwall不等式可得

$\begin{array}{l}{\Vert u\Vert }_2^2+{\Vert \omega \Vert }_2^2+{\Vert \theta \Vert }_2^2+{\displaystyle {\int }_0^t{\Vert \left(\nabla u,\nabla \omega ,\nabla \theta \right)\Vert }_2^2\text{d}\tau }\\ \le C\left({\Vert u_0\Vert }_2^2+{\Vert {\omega }_0\Vert }_2^2+{\Vert {\theta }_0\Vert }_2^2\right)\le C.\end{array}$

于是可以得出

$u,\omega ,\theta \in L^{\infty }\left(0,T;L^2\right)\cap L^2\left(0,T;H^1\right).$ (12)

第二步：L⁴-能量估计

方程(2)₁两边乘以${\left|u\right|}^{2}u$ 并在${ℝ}^3$ 上积分可得

$\begin{array}{l}\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert u\Vert }_4^4+{\displaystyle {\int }_{{ℝ}^3}\left(u\cdot \nabla \right)u\cdot {\left|u\right|}^{2}u\text{d}x}\\ ={\displaystyle {\int }_{{ℝ}^3}\Delta }u\cdot {\left|u\right|}^{2}u\text{d}x-{\displaystyle {\int }_{{ℝ}^3}\nabla }\pi \cdot {\left|u\right|}^{2}u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\nabla }\times \omega \cdot {\left|u\right|}^{2}u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\theta }{e}_3\cdot {\left|u\right|}^{2}u\text{d}x.\end{array}$ (13)

由分部积分可以得出

$\begin{array}{c}{\displaystyle {\int }_{{ℝ}^3}\Delta }u\cdot {\left|u\right|}^{2}u\text{d}x={\displaystyle \sum _{i,k=1}^3{\displaystyle {\int }_{{ℝ}^3}{\partial }_k^2}}{u}_i{\left|u\right|}^2{u}_i\text{d}x=-{\displaystyle \sum _{i,k=1}^3{\displaystyle {\int }_{{ℝ}^3}{\partial }_k}}{u}_i\cdot {\partial }_k\left({\left|u\right|}^2{u}_i\right)\text{d}x\\ =-{\displaystyle \sum _{i,k=1}^3{\displaystyle {\int }_{{ℝ}^3}{\partial }_k}}{u}_i\cdot {\partial }_k\left({\left|u\right|}^2\right)u_i\text{d}x-{\displaystyle \sum _{i,k=1}^3{\displaystyle {\int }_{{ℝ}^3}{\partial }_k}}{u}_i\cdot {\left|u\right|}^2\cdot {\partial }_k{u}_i\text{d}x\\ =-\frac{1}{2}{\displaystyle \sum _{k=1}^3{\displaystyle {\int }_{{ℝ}^3}{\left({\partial }_k{\left|u\right|}^2\right)}^2}}\text{d}x-{\displaystyle \sum _{i,k=1}^3{\displaystyle {\int }_{{ℝ}^3}{\left({\partial }_k{u}_i\right)}^2}}{\left|u\right|}^2\text{d}x.\end{array}$

由分部积分和$\nabla \cdot u=0$ 得

${\displaystyle {\int }_{{ℝ}^3}\nabla }\pi \cdot {\left|u\right|}^{2}u\mathrm{d}x=-{\displaystyle {\int }_{{ℝ}^3}\pi }\nabla \cdot \left({\left|u\right|}^{2}u\right)\text{d}x=-{\displaystyle {\int }_{{ℝ}^3}\pi }u\cdot \nabla {\left|u\right|}^2\text{d}x$

以及

${\displaystyle {\int }_{{ℝ}^3}\left(u\cdot \nabla \right)u\cdot {\left|u\right|}^{2}u\text{d}x}=\frac{1}{4}{\displaystyle {\int }_{{ℝ}^3}u}\cdot \nabla {\left|u\right|}^4\text{d}x=-\frac{1}{4}{\displaystyle {\int }_{{ℝ}^3}\nabla }\cdot u\cdot {\left|u\right|}^4\text{d}x=0.$

因此(13)式可化简为

$\begin{array}{l}\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert u\Vert }_4^4+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla u\right|}^2{\left|u\right|}^2\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|u\right|}^2\right|}^2\text{d}x}\\ ={\displaystyle {\int }_{{ℝ}^3}\pi }u\cdot \nabla {\left|u\right|}^2\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\nabla }\times \omega \cdot {\left|u\right|}^{2}u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\theta }{e}_3\cdot {\left|u\right|}^{2}u\text{d}x\\ :=J_1+J_2+J_3.\end{array}$ (14)

接下来先估计$J_2$ 利用Hölder不等式和Young不等式可得

$\begin{array}{c}{J}_2={\displaystyle {\int }_{{ℝ}^3}\nabla }\times \omega \cdot {\left|u\right|}^{2}u\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}\omega }\nabla \times \left({\left|u\right|}^{2}u\right)\text{d}x\\ \le 3{\displaystyle {\int }_{{ℝ}^3}\omega {\left|u\right|}^2\left|\nabla u\right|\text{d}x}\\ =3{\displaystyle {\int }_{{ℝ}^3}\left(\omega \left|u\right|\right)\left(\left|u\right|\left|\nabla u\right|\right)\text{d}x}\\ \le 3{\left({\displaystyle {\int }_{{ℝ}^3}|}\omega {|}^2{\left|u\right|}^2\text{d}x\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_{{ℝ}^3}|}u{|}^2{\left|\nabla u\right|}^2\text{d}x\right)}^{\frac{1}{2}}\\ \le 3\left(C{\displaystyle {\int }_{{ℝ}^3}|}\omega {|}^2{\left|u\right|}^2\text{d}x+\frac{1}{24}{\displaystyle {\int }_{{ℝ}^3}|}u{|}^2{\left|\nabla u\right|}^2\text{d}x\right)\\ \le C{\Vert \omega \Vert }_4^2{\Vert u\Vert }_4^2+\frac{1}{8}{\displaystyle {\int }_{{ℝ}^3}|}u{|}^2{\left|\nabla u\right|}^2\text{d}x\\ \le \frac{1}{8}{\displaystyle {\int }_{{ℝ}^3}|}u{|}^2{\left|\nabla u\right|}^2\text{d}x+C{\Vert \omega \Vert }_4^4+C{\Vert u\Vert }_4^4.\end{array}$ (15)

下面估计$J_3$ 利用Hölder不等式和Young不等式可得

$\begin{array}{c}{J}_3={\displaystyle {\int }_{{ℝ}^3}\theta }{e}_3\cdot {\left|u\right|}^{2}u\text{d}x\\ \le {\Vert \theta \Vert }_4{\Vert u\Vert }_4^3\\ \le C{\Vert \theta \Vert }_4^4+C{\Vert u\Vert }_4^4.\end{array}$ (16)

最后估计$J_1$ 根据Littlewood-Paley分解，将$\pi$ 分解为

$\pi ={\displaystyle \sum _{j=-\infty }^{+\infty }{\dot{\Delta }}_j}\pi ={\displaystyle \sum _{j<-N}{\dot{\Delta }}_j}\pi +{\displaystyle \sum _{j=-N}^N{\dot{\Delta }}_j}\pi +{\displaystyle \sum _{j>N}{\dot{\Delta }}_j}\pi ,$ (17)

其中N是待定的正整数。根据(17)式可得

$\begin{array}{c}{J}_1={\displaystyle {\int }_{{ℝ}^3}\pi }u\cdot \nabla {\left|u\right|}^2\text{d}x\\ \le {\displaystyle {\int }_{{ℝ}^3}{\displaystyle \sum _{j<-N}\left|{\dot{\Delta }}_j\pi \right|\left|u\right|\left|\nabla {\left|u\right|}^2\right|\text{d}x}}+{\displaystyle {\int }_{{ℝ}^3}{\displaystyle \sum _{j=-N}^N\left|{\dot{\Delta }}_j\pi \right|\left|u\right|\left|\nabla {\left|u\right|}^2\right|\text{d}x}}\\ +{\displaystyle {\int }_{{ℝ}^3}{\displaystyle \sum _{j>N}\left|{\dot{\Delta }}_j\pi \right|\left|u\right|\left|\nabla {\left|u\right|}^2\right|\text{d}x}}\\ :=J_{11}+J_{12}+J_{13}.\end{array}$(18)

下面对$J_{11}$ 进行估计。利用Hölder不等式，Bernstein不等式，Gagliardo-Nirenberg不等式，引理1.3.14，引理1.3.15和Young不等式可得

$\begin{array}{c}{J}_{11}={\displaystyle {\int }_{{ℝ}^3}{\displaystyle \sum _{j<-N}\left|{\dot{\Delta }}_j\pi \right|\left|u\right|\left|\nabla {\left|u\right|}^2\right|\text{d}x}}\\ \le C{\displaystyle \sum _{j<-N}{\Vert {\dot{\Delta }}_j\pi \Vert }_8}{\Vert u\Vert }_{\frac{8}{3}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{\displaystyle \sum _{j<-N}{2}^{\frac{j}{8}}}{\Vert {\dot{\Delta }}_j\pi \Vert }_6{\Vert u\Vert }_2^{\frac{1}{2}}{\Vert u\Vert }_4^{\frac{1}{2}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{2}^{-\frac{N}{8}}{\Vert \pi \Vert }_6{\Vert u\Vert }_4^{\frac{1}{2}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{2}^{-\frac{N}{8}}\left({\Vert u\Vert }_{12}^2+{\Vert {\nabla }^{-1}\theta \Vert }_6\right){\Vert u\Vert }_4^{\frac{1}{2}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{2}^{-\frac{N}{8}}{\Vert u\Vert }_4^{\frac{1}{2}}{\Vert \nabla {\left|u\right|}^2\Vert }_2^2+2^{-\frac{N}{8}}{\Vert \theta \Vert }_2{\Vert u\Vert }_4^{\frac{1}{2}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{2}^{-\frac{N}{8}}{\Vert u\Vert }_4^{\frac{1}{2}}{\Vert \nabla {\left|u\right|}^2\Vert }_2^2+\frac{1}{16}{\Vert \nabla {\left|u\right|}^2\Vert }_2^2+C{2}^{-\frac{N}{4}}{\Vert u\Vert }_4,\end{array}$ (19)

其中利用了不等式

$\begin{array}{l}{\Vert u\Vert }_{\frac{8}{3}}\le C{\Vert u\Vert }_2^{\frac{1}{2}}{\Vert u\Vert }_4^{\frac{1}{2}},{\Vert \pi \Vert }_6\le C\left({\Vert u\Vert }_{12}^2+{\Vert {\nabla }^{-1}\theta \Vert }_6\right),\\ {\Vert {\nabla }^{-1}\theta \Vert }_6\le C{\Vert \theta \Vert }_2.\end{array}$

对于$J_{12}$ 利用Hölder不等式，Gagliardo-Nirenberg不等式，Sobolev嵌入，(12)式以及Young不等式可得

$\begin{array}{c}{J}_{12}={\displaystyle {\int }_{{ℝ}^3}{\displaystyle \sum _{j=-N}^N\left|{\dot{\Delta }}_j\pi \right|\left|u\right|\left|\nabla {\left|u\right|}^2\right|\text{d}x}}\\ \le C{\displaystyle {\int }_{{ℝ}^3}{N}^{\frac{5q-6}{10q}}}{\left({\displaystyle \sum _{j=-N}^N{\left|{\dot{\Delta }}_j\pi \right|}^{\frac{10q}{5q+6}}}\right)}^{\frac{5q+6}{10q}}\left|u\right|\left|\nabla {\left|u\right|}^2\right|\text{d}x\\ \le C{N}^{\frac{5q-6}{10q}}{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}{\Vert u\Vert }_{\frac{2q}{q-2}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{N}^{\frac{5q-6}{10q}}{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}{\Vert u\Vert }_2^{1-\frac{12}{5q}}{\Vert u\Vert }_{12}^{\frac{12}{5q}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{N}^{\frac{5q-6}{10q}}{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}{\Vert \nabla {\left|u\right|}^2\Vert }_2^{1+\frac{6}{5q}}\\ \le CN{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}^{\frac{10q}{5q-6}}+\frac{1}{16}{\Vert \nabla {\left|u\right|}^2\Vert }_2^2,\end{array}$ (20)

其中利用了不等式

${\Vert u\Vert }_{\frac{2q}{q-2}}\le C{\Vert u\Vert }_2^{1-\frac{12}{5q}}{\Vert u\Vert }_{12}^{\frac{12}{5q}},q>\frac{12}{5}.$

对于$J_{13}$ 利用Hölder不等式，Bernstein不等式，Gagliardo-Nirenberg不等式，引理1.3.14和Young不等式得

$\begin{array}{c}{J}_{13}={\displaystyle {\int }_{{ℝ}^3}{\displaystyle \sum _{j>N}\left|{\dot{\Delta }}_j\pi \right|\left|u\right|\left|\nabla {\left|u\right|}^2\right|\text{d}x}}\\ \le {\displaystyle \sum _{j>N}{\Vert {\dot{\Delta }}_j\pi \Vert }_5}{\Vert u\Vert }_{\frac{10}{3}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le {\displaystyle \sum _{j>N}{2}^{-j}}{\Vert {\dot{\Delta }}_j\nabla \pi \Vert }_5{\Vert u\Vert }_2^{\frac{1}{5}}{\Vert u\Vert }_4^{\frac{4}{5}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{\displaystyle \sum _{j>N}{2}^{-\frac{j}{10}}}{\Vert {\dot{\Delta }}_j\nabla \pi \Vert }_2{\Vert u\Vert }_4^{\frac{4}{5}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{2}^{-\frac{N}{10}}{\Vert \nabla \pi \Vert }_2{\Vert u\Vert }_4^{\frac{4}{5}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{2}^{-\frac{N}{10}}\left({\Vert u\cdot \nabla u\Vert }_2+{\Vert \theta \Vert }_2\right){\Vert u\Vert }_4^{\frac{4}{5}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{2}^{-\frac{N}{10}}{\Vert u\Vert }_4^{\frac{4}{5}}{\Vert \left|u\right|\cdot \left|\nabla u\right|\Vert }_2^2+C{2}^{-\frac{N}{10}}{\Vert \theta \Vert }_2{\Vert u\Vert }_4^{\frac{4}{5}}{\Vert \nabla {\left|u\right|}^2\Vert }_2\\ \le C{2}^{-\frac{N}{10}}{\Vert u\Vert }_4^{\frac{4}{5}}{\Vert \left|u\right|\cdot \left|\nabla u\right|\Vert }_2^2+\frac{1}{16}{\Vert \nabla {\left|u\right|}^2\Vert }_2^2+C{2}^{-\frac{N}{5}}{\Vert u\Vert }_4^{\frac{8}{5}},\end{array}$ (21)

其中利用了不等式

${\Vert u\Vert }_{\frac{10}{3}}\le C{\Vert u\Vert }_2^{\frac{1}{5}}{\Vert u\Vert }_4^{\frac{4}{5}}.$

类似地，将(2)₂式两边乘以${\left|\omega \right|}^2\omega$ 后在${ℝ}^3$ 上积分，利用分部积分和$\nabla \cdot u=0$ 得

$\begin{array}{c}\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert \omega \Vert }_4^4+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \omega \right|}^2{\left|\omega \right|}^2\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|\omega \right|}^2\right|}^2\text{d}x+2{\Vert \omega \Vert }_4^4}\\ ={\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla \cdot \omega {\left|\omega \right|}^2\omega \text{d}x+{\displaystyle {\int }_{{ℝ}^3}\nabla }\times u\cdot {\left|\omega \right|}^2\omega \text{d}x.\end{array}$ (22)

通过分布积分再利用Hölder不等式和Young不等式得

$\begin{array}{l}\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert \omega \Vert }_4^4+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \omega \right|}^2{\left|\omega \right|}^2\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|\omega \right|}^2\right|}^2\text{d}x+2{\Vert \omega \Vert }_4^4}+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \cdot \omega \right|}^2{\left|\omega \right|}^2\text{d}x}\\ ={\displaystyle {\int }_{{ℝ}^3}\nabla }\times u\cdot {\left|\omega \right|}^2\omega \text{d}x-{\displaystyle {\int }_{{ℝ}^3}\nabla }\cdot \omega \omega \cdot \nabla {\left|\omega \right|}^2\text{d}x\\ \le \frac{3}{4}{\Vert \left|\omega \right|\left|\nabla \omega \right|\Vert }_2^2+C{\Vert u\Vert }_4^2{\Vert \omega \Vert }_4^2+\frac{1}{8}{\Vert \nabla {\left|\omega \right|}^2\Vert }_2^2+{\Vert \left|\nabla \cdot \omega \right|\left|\omega \right|\Vert }_2^2+\frac{1}{4}{\Vert \nabla {\left|\omega \right|}^2\Vert }_2^2\\ \le \frac{3}{4}{\Vert \left|\omega \right|\left|\nabla \omega \right|\Vert }_2^2+C{\Vert u\Vert }_4^4+2{\Vert \omega \Vert }_4^4+\frac{1}{8}{\Vert \nabla {\left|\omega \right|}^2\Vert }_2^2+{\Vert \left|\nabla \cdot \omega \right|\left|\omega \right|\Vert }_2^2+\frac{1}{4}{\Vert \nabla {\left|\omega \right|}^2\Vert }_2^2.\end{array}$

即可得到

$\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert \omega \Vert }_4^4+\frac{1}{4}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \omega \right|}^2{\left|\omega \right|}^2\text{d}x}+\frac{1}{8}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|\omega \right|}^2\right|}^2\text{d}x}\le C{\Vert u\Vert }_4^4$ (23)

同样地对(2)₃式两边乘以${\left|\theta \right|}^2\theta$ 后在${ℝ}^3$ 上积分，利用分部积分，Hölder不等式，Young不等式和$\nabla \cdot u=0$ 得出

$\begin{array}{l}\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert \theta \Vert }_4^4+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \theta \right|}^2{\left|\theta \right|}^2\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|\theta \right|}^2\right|}^2\text{d}x}\\ ={\displaystyle {\int }_{{ℝ}^3}u}\cdot e_3{\left|\theta \right|}^2\theta \text{d}x\\ \le {\displaystyle {\int }_{{ℝ}^3}|}u|\cdot {\left|\theta \right|}^3\text{d}x\\ \le {\Vert u\Vert }_4{\Vert \theta \Vert }_4^3\\ \le C\left({\Vert u\Vert }_4^4+{\Vert \theta \Vert }_4^4\right).\end{array}$ (24)

综合(19)式，(20)式，(21)式，(23)式和(24)式可得

$\begin{array}{c}\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert \left(u,\omega ,\theta \right)\Vert }_4^4+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla u\right|}^2{\left|u\right|}^2\text{d}x}+\frac{1}{4}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \omega \right|}^2{\left|\omega \right|}^2\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \theta \right|}^2{\left|\theta \right|}^2\text{d}x}\\ +\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|u\right|}^2\right|}^2\text{d}x}+\frac{1}{8}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|\omega \right|}^2\right|}^2\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|\theta \right|}^2\right|}^2\text{d}x}\\ \le \frac{1}{8}{\displaystyle {\int }_{{ℝ}^3}{\left|u\right|}^2{\left|\nabla u\right|}^2\text{d}x}+C\left({\Vert \omega \Vert }_4^4+{\Vert u\Vert }_4^4+{\Vert \theta \Vert }_4^4\right)+C{2}^{-\frac{N}{8}}{\Vert u\Vert }_4^{\frac{1}{2}}{\Vert \nabla {\left|u\right|}^2\Vert }_2^2\\ +\frac{1}{16}{\Vert \nabla {\left|u\right|}^2\Vert }_2^2+C{2}^{-\frac{N}{4}}{\Vert u\Vert }_4+CN{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}^{\frac{10q}{5q-6}}+\frac{1}{16}{\Vert \nabla {\left|u\right|}^2\Vert }_2^2\\ +C{2}^{-\frac{N}{10}}{\Vert u\Vert }_4^{\frac{4}{5}}{\Vert \left|u\right|\cdot \left|\nabla u\right|\Vert }_2^2+\frac{1}{16}{\Vert \nabla {\left|u\right|}^2\Vert }_2^2+C{2}^{-\frac{N}{5}}{\Vert u\Vert }_4^{\frac{8}{5}}.\end{array}$ (25)

选取(25)式中的N使得

$C{\left(2^{-\frac{N}{8}}{\Vert u\Vert }_4\right)}^{\frac{4}{5}}\le \frac{1}{16},$

具体地说

$N\ge 8\left(\frac{\log \left(C{\Vert u\Vert }_4\right)}{\log 2}+4\right)$

另一方面易得

$C{\left(2^{-\frac{N}{5}}{\Vert u\Vert }_4\right)}^{\frac{4}{5}}\le C{\left(2^{-\frac{N}{8}}{\Vert u\Vert }_4\right)}^{\frac{4}{5}}\le \frac{1}{16}.$

则有

$C{2}^{-\frac{N}{8}}{\Vert u\Vert }_4^{\frac{1}{2}}\le \frac{1}{4}.$

同时还得到

$C{2}^{-\frac{N}{4}}{\Vert u\Vert }_4\le C,C{2}^{-\frac{N}{5}}{\Vert u\Vert }_4^{\frac{8}{5}}\le C.$

因此(25)式变成

$\begin{array}{c}\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert \left(u,\omega ,\theta \right)\Vert }_4^4+\frac{7}{8}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla u\right|}^2{\left|u\right|}^2\text{d}x}+\frac{1}{4}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \omega \right|}^2{\left|\omega \right|}^2\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \theta \right|}^2{\left|\theta \right|}^2\text{d}x}\\ +\frac{1}{8}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|u\right|}^2\right|}^2\text{d}x}+\frac{1}{8}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|\omega \right|}^2\right|}^2\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|\theta \right|}^2\right|}^2\text{d}x}\\ \le C\left({\Vert \omega \Vert }_4^4+{\Vert u\Vert }_4^4+{\Vert \theta \Vert }_4^4\right)+C{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}^{\frac{10q}{5q-6}}\log \left({\Vert u\Vert }_4+e\right)+C,\end{array}$

于是我们有

$\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert \left(u,\omega ,\theta \right)\Vert }_4^4\le C{\Vert \left(u,\omega ,\theta \right)\Vert }_4^4+C{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}^{\frac{10q}{5q-6}}\log \left({\Vert \left(u,\omega ,\theta \right)\Vert }_4+e\right).$ (26)

因此对(26)式利用Gronwall不等式可得

$\begin{array}{c}{\Vert \left(u,\omega ,\theta \right)\Vert }_4^4\le C{\Vert \left(u_0,{\omega }_0,{\theta }_0\right)\Vert }_4^4+{\displaystyle {\int }_0^T{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}^{\frac{10q}{5q-6}}}\log \left({\Vert \left(u,\omega ,\theta \right)\Vert }_4+e\right)\text{d}\tau \\ \le C{\Vert \left(u_0,{\omega }_0,{\theta }_0\right)\Vert }_4^4+{\displaystyle {\int }_0^T{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left({ℝ}^3\right)}^0}^{\frac{10q}{5q-6}}}\left({\Vert \left(u,\omega ,\theta \right)\Vert }_4^4+e\right)\text{d}\tau .\end{array}$

令$M={\Vert \left(u_0,{\omega }_0,{\theta }_0\right)\Vert }_4^4+e$ ，再次利用Gronwall不等式可得

${\Vert \left(u,\omega ,\theta \right)\Vert }_4^4+e\le CM\exp {\displaystyle {\int }_0^T{\Vert \pi \Vert }_{{\dot{F}}_{q,\frac{10q}{5q+6}\left(R^3\right)}^0}^{\frac{10q}{5q-6}}}\text{d}\tau .$

即完成定理1.1的证明。

3.2. 定理1.2的证明

类似定理1.1，考虑如下L⁴能量估计

$\begin{array}{l}\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert u\Vert }_4^4+{\displaystyle {\int }_{{ℝ}^3}u}\cdot \nabla u\cdot {\left|u\right|}^{2}u\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}\Delta }u\cdot {\left|u\right|}^{2}u\text{d}x-{\displaystyle {\int }_{{ℝ}^3}\nabla }\pi \cdot {\left|u\right|}^{2}u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\nabla }\times \omega \cdot {\left|u\right|}^{2}u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\theta }{e}_3\cdot {\left|u\right|}^{2}u\text{d}x\\ =-\frac{1}{2}{\displaystyle \sum _{i,k=1}^3{\displaystyle {\int }_{{ℝ}^3}{\left({\partial }_k{\left|u\right|}^2\right)}^2}}\text{d}x-{\displaystyle \sum _{i,k=1}^3{\displaystyle {\int }_{{ℝ}^3}{\left({\partial }_k{u}_i\right)}^2}}{\left|u\right|}^2\text{d}x-{\displaystyle {\int }_{{ℝ}^3}\nabla }\pi \cdot {\left|u\right|}^{2}u\text{d}x\\ +{\displaystyle {\int }_{{ℝ}^3}\nabla \times \omega \cdot {\left|u\right|}^2}u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\theta e_3\cdot {\left|u\right|}^2}u\text{d}x.\end{array}$ (27)

由(27)式即可得

$\begin{array}{l}\frac{1}{4}\frac{\text{d}}{\text{d}t}{\Vert u\Vert }_4^4+{\displaystyle {\int }_{{ℝ}^3}|}\nabla u{|}^2{\left|u\right|}^2\text{d}x+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla {\left|u\right|}^2\right|}^2\text{d}x}\\ \le -{\displaystyle {\int }_{{ℝ}^3}\nabla }\pi \cdot {\left|u\right|}^{2}u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\nabla }\times \omega \cdot {\left|u\right|}^{2}u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\theta }{e}_3\cdot {\left|u\right|}^{2}u\text{d}x\\ :=K_1+K_2+K_3.\end{array}$

综合文献[5]和定理1.1的证明过程即可证得定理1.2。

基金项目

江西省自然科学基金(20232BAB201014)。