Navier-Stokes-Cahn-Hillard方程在Morrey-Campanato空间中的正则性准则

Regularity Criteria for the Navier-Stokes-Cahn-Hillard Equation in the Morrey-Campanato Space

摘要: 本文给出了Navier-Stokes-Cahn-Hillard方程在三维Morrey-Campanato空间中解的一个正则性准则，利用了Gagliardo-Nirenberg不等式、Sobolev嵌入定理、插值不等式以及先验估计等。

Abstract: This paper gives a regularity criterion for the Navier-Stokes-Cahn-Hillard equation in the 3D Morrey-Campanato space which uses Gagliardo-Nirenberg inequality, Sobolev embedding theorem, interpolation inequality, and prior estimates.

1. 介绍

在本文中考虑在 ${ℝ}^3$ 上带有初值问题的Navier-Stokes-Cahn-Hillard方程

${\partial }_{t}u+\left(u\cdot \nabla \right)u+\nabla \pi -\epsilon \Delta u=\mu \nabla \varphi$, (1.1)

${\partial }_t\varphi +\left(u\cdot \nabla \right)\varphi =\Delta \mu$, (1.2)

$-\Delta \varphi +f^{\prime }\left(\varphi \right)=\mu$, (1.3)

$\nabla \cdot u=0$, (1.4)

$u\left(x,0\right)=u_0,\varphi \left(x,0\right)={\varphi }_0$, (1.5)

其中 $u=\left(u_1\left(x,t\right),u_2\left(x,t\right),u_3\left(x,t\right)\right)$ 表示速度场， $\pi =\pi \left(x,t\right)$ 表示压强， $\varphi =\varphi \left(x,t\right)$ 表示序参量， $\epsilon >0$ 表示粘性系数， $\mu$ 表示化学势， $f\left(\varphi \right)=\frac{1}{4}{\left({\varphi }^2-1\right)}^2$ 表示双井位势。

当 $\varphi =0$ 时，上述方程成为著名的Navier-Stokes方程。Navier-Stokes方程已经被进行了广泛的研究，其中最著名代表性的开创性成果来自于Leray [1] 和Hopf [2]，他们对任意给定的初始速度 $u_0\in L^2\left({ℝ}^3\right)$ 构建了方程的整体弱解，通常称为Leray-Hopf弱解。从20世纪70年代末开始，Scheffer便对Navier-Stokes方程弱解的部分正则性进行了研究，较深刻的结果由Caffarelli，Kohn，Nirenberg [3] 在1982年得到，他们给出了一个适当弱解的定义，并证明了这种解的奇异集的Hausdorff测度为零。

方程(1.1)~(1.5)是由Cahn-Hillard系统和不可压缩Navier-Stokes系统耦合而成。 [4] 中得到了二维Navier-Stokes-Cahn-Hillard系统弱解的H¹正则性和H²时间正则性。关于更多Navier-Stokes-Cahn-Hillard系统的研究，请读者参考 [5] [6] [7] [8] [9]。

在 [10] 中，Montgomery-Smith证明了Navier-Stokes方程的对数改进正则性准则，即解u满足

${\displaystyle {\int }_0^T\frac{{\Vert u\Vert }_{L^q}^p}{1+\ln \left(e+{\Vert u\Vert }_{L^q}\right)}\text{d}t}<\infty ,\frac{2}{p}+\frac{3}{q}\le 1,2<q<\infty$

那么u在 $\left[0,T\right]\times {ℝ}^n$ 上是正则的。 [11] 中建立了Navier-Stokes方程的对数改进正则性准则。

本文给出了Navier-Stokes-Cahn-Hillard方程在三维Morrey-Campanato空间中解的一个正则性准则，主要结果如下，为方便起见，设 $\epsilon =1$。

定理1.1 假设 $\left(u_0,{\varphi }_0\right)\in H^1\left({ℝ}^3\right)\times H^2\left({ℝ}^3\right)$ 且 $\nabla \cdot u_0=0$，对于某个 $T\left(0<T<\infty \right)$， $\left(u,\varphi \right)$ 是方程(1.1)~(1.5)在 $\left[0,T\right)$ 上的局部光滑解。如果u满足：

${\displaystyle {\int }_0^T{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}}\text{d}t<\infty ,0<r<1,2\le p\le \frac{3}{r}$ (1.6)

或

${\displaystyle {\int }_0^T\frac{{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}+{\Vert \Delta \varphi \Vert }_{L^2}^2}{1+\ln \left(e+{\Vert u\Vert }_{L^6}\right)}\text{d}t}<\infty ,0<r<1,2\le p\le \frac{3}{r}$ (1.7)

那么 $\left(u,\varphi \right)\left(x,t\right)$ 可以延拓超过T。

定理1.2 假设 $\left(u_0,{\varphi }_0\right)\in H^1\left({ℝ}^3\right)\times H^2\left({ℝ}^3\right)$ 且 $\nabla \cdot u_0=0$，对于某个 $T\left(0<T<\infty \right)$， $\left(u,\varphi \right)$ 是方程(1.1)~(1.5)在 $\left[0,T\right)$ 上的局部光滑解。如果 $\nabla u$ 满足：

${\displaystyle {\int }_0^T{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}}\text{d}t<\infty ,0<\gamma \le 1,2\le p\le \frac{3}{\gamma }$ (1.8)

或

${\displaystyle {\int }_0^T\frac{{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}+{\Vert \Delta \varphi \Vert }_{L^2}^2}{1+\ln \left(e+{\Vert u\Vert }_{L^6}\right)}\text{d}t}<\infty ,0<\gamma \le 1,2\le p\le \frac{3}{\gamma }$ (1.9)

那么 $\left(u,\varphi \right)\left(x,t\right)$ 可以延拓超过T。

下面我们将介绍一些对定理证明有用的定义和引理。

2. 预备知识

以下定义和定理参考 [12] [13]。

定义2.1 对于 $1<p\le q\le \infty$，Morrey-Campanato空间 ${\stackrel{\dot{}}{M}}^{p,q}$ 定义为

${\stackrel{\dot{}}{M}}^{p,q}=\left\{f\in L_{loc}^p\left({ℝ}^3\right):{\Vert f\Vert }_{{\stackrel{\dot{}}{M}}^{p,q}}=\underset{x\in {ℝ}^3}{\sup }\underset{R>0}{\sup }{R}^{3/q-3/p}{\Vert f\Vert }_{L^p\left(B\left(x,R\right)\right)}<\infty \right\}$,

其中 $B\left(x,R\right)$ 表示 ${ℝ}^3$ 上以x为圆心，R为半径的闭球。

定义2.2 对于 $1\le q^{\prime }\le p^{\prime }<\infty$，定义齐次空间 $N^{p^{\prime },q^{\prime }}$ 是关于函数f的空间且是 $L^{q^{\prime }}\left({ℝ}^3\right)$ 的子空间，其中f可以分解成一个原子序列 $f={\displaystyle \underset{k\in \mathbb{N}}{\sum }{g}_k}$，其中函数 ${\left(g_k\right)}_k\subset L_{comp}^{p^{\prime }}\left({ℝ}^3\right)$ 且满足关于 $g_k$ 支集的直径 $d_k$ 的不等式： $d_k\le 1$ 和 ${\displaystyle \underset{k\in \mathbb{N}}{\sum }{d}_k^{\frac{3}{q^{\prime }}-\frac{3}{p^{\prime }}}}{\Vert g_k\Vert }_{L^{p^{\prime }}}<\infty$。 $N^{p^{\prime },q^{\prime }}$ 是一个Banach空间，其范数为 ${\Vert f\Vert }_{N^{p^{\prime },q^{\prime }}}=\inf \left\{{\displaystyle \underset{k\in \mathbb{N}}{\sum }{d}_k^{\frac{3}{q^{\prime }}-\frac{3}{p^{\prime }}}}{\Vert g_k\Vert }_{L^{p^{\prime }}},f={\displaystyle \underset{k\in \mathbb{N}}{\sum }{g}_k}\right\}$，其中下确界inf是指f分解成一个原子序列的所有可能性。

引理2.3 对于 $1\le q^{\prime }\le p^{\prime }<\infty$，如果 $p,q$ 满足 $\frac{1}{p}+\frac{1}{p^{\prime }}=1$， $\frac{1}{q}+\frac{1}{q^{\prime }}=1$，那么有

${\left({\stackrel{\dot{}}{M}}^{p,q}\right)}^{*}=N^{p^{\prime },q^{\prime }}$.

即对于 $f\in {\stackrel{\dot{}}{M}}^{p,q}$， $g\in N^{p^{\prime },q^{\prime }}$，有 ${\displaystyle {\int }_{{ℝ}^3}fg\text{d}x}\le {\Vert f\Vert }_{{\stackrel{\dot{}}{M}}^{p,q}}{\Vert g\Vert }_{N^{p^{\prime },q^{\prime }}}$。

引理2.4 对于 $1\le q^{\prime }\le p^{\prime }<2$， $\frac{1}{q^{\prime }}=1-\frac{r}{3}$，相应点乘是一个从 $L^2\left({ℝ}^3\right)\times {\stackrel{\dot{}}{H}}^r\left({ℝ}^3\right)$ 到 $N^{p^{\prime },q^{\prime }}\left({ℝ}^3\right)$ 的有界双线性算子，那么存在常数 $C>0$，使得对于任意的 $u\in L^2\left({ℝ}^3\right)$， $v\in {\stackrel{\dot{}}{H}}^r\left({ℝ}^3\right)$，有 ${\Vert uv\Vert }_{N^{p^{\prime },q^{\prime }}}\le {\Vert u\Vert }_{L^2\left({ℝ}^3\right)}{\Vert v\Vert }_{{\stackrel{\dot{}}{H}}^r\left({ℝ}^3\right)}$。

引理2.5 对于 $0\le r\le \frac{3}{2}$，设空间 $M\left({\stackrel{\dot{}}{B}}_{r,1}^2\mapsto L^2\right)$ 是 ${ℝ}^3$ 中局部平方可积函数空间，相应点乘是Besov空间 ${\stackrel{\dot{}}{B}}_2^{r,1}\left({ℝ}^3\right)$ 到 $L^2\left({ℝ}^3\right)$ 的有界双线性映射。 $M\left({\stackrel{\dot{}}{B}}_{r,1}^2\mapsto L^2\right)$ 上的范数由点乘算子的范数决定

${\Vert f\Vert }_{M\left({\stackrel{\dot{}}{B}}_{r,1}^2\mapsto L^2\right)}=\sup \left\{{\Vert fg\Vert }_{L^2}:{\Vert g\Vert }_{{\stackrel{\dot{}}{B}}_2^{r,1}}\le 1\right\}$,

那么， $f\in M\left({\stackrel{\dot{}}{B}}_{r,1}^2\mapsto L^2\right)$ 当且仅当 $f\in {\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}\left({ℝ}^3\right)$。

引理2.6 (Gagliardo-Nirenberg不等式)对于 $0<q$， $r\le \infty$， $0<\alpha <1$， $u\in W^{m,r}\left({ℝ}^n\right)$，有

${\Vert D^{j}u\Vert }_{L^p}\le C{\Vert D^{m}u\Vert }_{L^r}^{\alpha }{\Vert u\Vert }_{L^q}^{1-\alpha }$,

其中 $\frac{j}{m}\le \alpha <1$， $\frac{1}{p}=\frac{j}{n}+\alpha \left(\frac{1}{r}-\frac{m}{n}\right)+\left(1-\alpha \right)\frac{1}{q}$。

3. 定理证明

在这部分，我们将给出定理1.1和定理1.2的证明，利用先验估计和Gronwall不等式解决局部光滑解的正则性准则，即证明下式成立

$\underset{t\to T^{-}}{\lim \sup }\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right)\le C$. (3.1)

将方程(1.1)两端同时乘u，在 ${ℝ}^3$ 上积分并利用分部积分，得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}{\left|u\right|}^2}\text{d}x+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla u\right|}^2}\text{d}x={\displaystyle {\int }_{{ℝ}^3}\mu \nabla \varphi u}\text{d}x$, (3.2)

将方程(1.2)两端同时乘 $\mu$，在 ${ℝ}^3$ 上积分并利用分部积分，得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\left({\left|\nabla \varphi \right|}^2+2f\left(\varphi \right)\right)\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\mu \nabla \varphi u}\text{d}x=-{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \mu \right|}^2}\text{d}x$, (3.3)

将(3.2)和(3.3)两式相加，得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\left({\left|u\right|}^2+{\left|\nabla \varphi \right|}^2+2f\left(\varphi \right)\right)\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\left({\left|\nabla u\right|}^2+{\left|\nabla \mu \right|}^2\right)\text{d}x}=0$,

我们获得基本能量估计

${\Vert u\Vert }_{L^{\infty }\left(0,T;L^2\right)}+{\Vert u\Vert }_{L^2\left(0,T;H^1\right)}\le C$, (3.4)

${\Vert \varphi \Vert }_{L^{\infty }\left(0,T;H^1\right)}\le C$, (3.5)

${\Vert \nabla \mu \Vert }_{L^2\left(0,T;L^2\right)}\le C$. (3.6)

将方程(1.2)两端同时乘 $\varphi$，在 ${ℝ}^3$ 上积分并利用分部积分，结合(1.3)，(1.4)，Young’s不等式，得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}{\left|\varphi \right|}^2}\text{d}x+\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}{\left|\Delta \varphi \right|}^2}\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}\left({\varphi }^3-\varphi \right)\Delta \varphi \text{d}x}=-3{\displaystyle {\int }_{{ℝ}^3}{\varphi }^2{\left|\nabla \varphi \right|}^2}\text{d}x-{\displaystyle {\int }_{{ℝ}^3}\varphi \Delta \varphi \text{d}x}\\ \le -{\displaystyle {\int }_{{ℝ}^3}\varphi \Delta \varphi \text{d}x}\le \frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\Delta \varphi \right|}^2}\text{d}x+\frac{1}{2}{\displaystyle {\int }_{{ℝ}^3}{\left|\varphi \right|}^2}\text{d}x,\end{array}$

利用Gronwall不等式，获得估计

${\Vert \varphi \Vert }_{L^2\left(0,T;H^2\right)}\le C$. (3.7)

下面将利用Gagliardo-Nirenberg不等式( $n=3$ )

${\Vert \varphi \Vert }_{L^{\infty }}^2\le C{\Vert \varphi \Vert }_{L^6}{\Vert \varphi \Vert }_{H^2}$, (3.8)

以及Sobolev嵌入定理

${\Vert \varphi \Vert }_{L^6}\le C{\Vert \varphi \Vert }_{H^1}$, (3.9)

结合(1.3)，(3.5)~(3.7)，(3.8)，(3.9)，Hölder不等式，得

$\begin{array}{c}{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \Delta \varphi \right|}^2}\text{d}x}\text{d}t={\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \left(f^{\prime }\left(\varphi \right)-\mu \right)\right|}^2}\text{d}x}\text{d}t\\ \le C{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \mu \right|}^2}\text{d}x}\text{d}t+C{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla \left({\varphi }^3-\varphi \right)\right|}^2}\text{d}x}\text{d}t\\ \le C+C{\displaystyle {\int }_0^T{\displaystyle {\int }_{{ℝ}^3}{\varphi }^4{\left|\nabla \varphi \right|}^2}\text{d}x}\text{d}t\le C+C{\Vert \nabla \varphi \Vert }_{L^{\infty }\left(0,T;L^2\right)}^2{\displaystyle {\int }_0^T{\Vert \varphi \Vert }_{L^{\infty }}^4}\text{d}t\\ \le C+C{\displaystyle {\int }_0^T{\Vert \varphi \Vert }_{L^6}^2{\Vert \varphi \Vert }_{H^2}^2}\text{d}t\le C+C{\Vert \varphi \Vert }_{L^{\infty }\left(0,T;H^1\right)}^2{\Vert \varphi \Vert }_{L^2\left(0,T;H^2\right)}^2\le C,\end{array}$

获得估计

${\Vert \varphi \Vert }_{L^2\left(0,T;H^3\right)}\le C$, (3.10)

${\Vert \varphi \Vert }_{L^4\left(0,T;L^{\infty }\right)}\le C$, (3.11)

${\Vert \nabla \varphi \Vert }_{L^2\left(0,T;L^{\infty }\right)}\le C{\Vert \varphi \Vert }_{L^2\left(0,T;H^3\right)}\le C$. (3.12)

接下来我们将进行高阶估计，改写方程(1.1)为分量形式( $i=1,2,3$ )

${\partial }_t{u}_i+\left(u\cdot \nabla \right)u_i+{\partial }_i\pi -\Delta u_i=\mu {\partial }_i\varphi$,

将上式左右两端同时乘 $-\Delta u_i$，对i求和，结合(1.3)，(1.4)，利用分部积分，有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla u\right|}^2}\text{d}x+{\displaystyle {\int }_{{ℝ}^3}{\left|\Delta u\right|}^2}\text{d}x\\ ={\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}\left(u\cdot \nabla \right)u_i\Delta u_i}\text{d}x}+{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}\Delta \varphi {\partial }_i\varphi \Delta u_i}\text{d}x},\end{array}$ (3.13)

将(1.2)先作用 $\Delta$，再乘 $\Delta \varphi$，结合(1.3)，(1.4)，利用分部积分，得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}{\left|\Delta \varphi \right|}^2}\text{d}x+{\displaystyle {\int }_{{ℝ}^3}{\left|{\Delta }^2\varphi \right|}^2}\text{d}x\\ ={\displaystyle {\int }_{R^3}\Delta f^{\prime }\left(\varphi \right)}{\Delta }^2\varphi \text{d}x-{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}\Delta \varphi {\partial }_i\varphi \Delta u_i}\text{d}x}-{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}\nabla u_i{\partial }_i\nabla \varphi \Delta \varphi }\text{d}x},\end{array}$ (3.14)

将(3.13)和(3.14)相加，得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\left({\left|\nabla u\right|}^2+{\left|\Delta \varphi \right|}^2\right)\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\left({\left|\Delta u\right|}^2+{\left|{\Delta }^2\varphi \right|}^2\right)\text{d}x}\\ ={\displaystyle {\int }_{{ℝ}^3}\left(u\cdot \nabla \right)u\Delta u}\text{d}x-2{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}\nabla u_i{\partial }_i\nabla \varphi \Delta \varphi }\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\Delta f^{\prime }\left(\varphi \right){\Delta }^2\varphi }\text{d}x\\ =I_1+I_2+I_3.\end{array}$ (3.15)

首先考虑u满足定理1.1中条件的情况，利用Hölder不等式，Young’s不等式，Sobolev嵌入定理(3.9)，估计 $I_3$

$\begin{array}{c}{I}_3={\displaystyle {\int }_{{ℝ}^3}\Delta f^{\prime }\left(\varphi \right){\Delta }^2\varphi }\text{d}x={\displaystyle {\int }_{{ℝ}^3}\Delta \left({\varphi }^3-\varphi \right){\Delta }^2\varphi }\text{d}x={\displaystyle {\int }_{{ℝ}^3}\left(6\varphi {\left|\nabla \varphi \right|}^2+3{\varphi }^2\Delta \varphi -\Delta \varphi \right){\Delta }^2\varphi }\text{d}x\\ \le C\left({\Vert \varphi \Vert }_{L^6}{\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^2{\Vert \Delta \varphi \Vert }_{L^2}+{\Vert \Delta \varphi \Vert }_{L^2}\right){\Vert {\Delta }^2\varphi \Vert }_{L^2}\\ \le C\left({\Vert \varphi \Vert }_{H^1}{\Vert \nabla \varphi \Vert }_{L^6}{\Vert \Delta \varphi \Vert }_{L^2}+{\Vert \varphi \Vert }_{L^{\infty }}^2{\Vert \Delta \varphi \Vert }_{L^2}+{\Vert \Delta \varphi \Vert }_{L^2}\right){\Vert {\Delta }^2\varphi \Vert }_{L^2}\\ \le \frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2+C\left({\Vert \nabla \varphi \Vert }_{L^6}^2{\Vert \Delta \varphi \Vert }_{L^2}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4{\Vert \Delta \varphi \Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right)\\ =\frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2+C\left({\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4+1\right){\Vert \Delta \varphi \Vert }_{L^2}^2.\end{array}$ (3.16)

利用Gagliardo-Nirenberg不等式(3.8)，Sobolev嵌入定理(3.9)和(3.5)， $I_3$ 又可写成估计

$\begin{array}{c}{I}_3\le \frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2++C\left({\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4+1\right){\Vert \Delta \varphi \Vert }_{L^2}^2\\ \le \frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2++C\left({\Vert \nabla \varphi \Vert }_{H^1}^2+{\Vert \varphi \Vert }_{L^6}^2{\Vert \varphi \Vert }_{H^2}^2+1\right){\Vert \Delta \varphi \Vert }_{L^2}^2\\ \le \frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2+C\left({\Vert \Delta \varphi \Vert }_{L^2}^2+1\right){\Vert \Delta \varphi \Vert }_{L^2}^2.\end{array}$ (3.17)

对于 $I_1$ 和 $I_2$ 的估计，分为两种情况。

第一种情况：当 $2<p\le \frac{3}{r}$ 时，利用引理2.3，引理2.4，Young’s不等式和插值不等式 ${\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^r}\le {\Vert u\Vert }_{L^2}^{1-r}{\Vert \nabla u\Vert }_{L^2}^r$， $0<r<1$，估计 $I_1$

$\begin{array}{c}{I}_1={\displaystyle {\int }_{{ℝ}^3}\left(u\cdot \nabla \right)u\Delta u}\text{d}x\le {\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}{\Vert \nabla u\Delta u\Vert }_{N^{p^{\prime },\frac{3}{3-r}}}\le {\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{H}}^r}{\Vert \Delta u\Vert }_{L^2}\\ \le {\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}{\Vert \nabla u\Vert }_{L^2}^{1-r}{\Vert \Delta u\Vert }_{L^2}^r{\Vert \Delta u\Vert }_{L^2}^1={\left({\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \nabla u\Vert }_{L^2}^2\right)}^{\frac{1-r}{2}}{\left({\Vert \Delta u\Vert }_{L^2}^2\right)}^{\frac{1+r}{2}}\\ \le \frac{1}{2}{\Vert \Delta u\Vert }_{L^2}^2+C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \nabla u\Vert }_{L^2}^2.\end{array}$ (3.18)

类似地，我们继续估计 $I_2$，得

$\begin{array}{c}{I}_2=-2{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}\nabla u_i{\partial }_i\nabla \varphi \Delta \varphi }\text{d}x}=2{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}{u}_i{\partial }_i\nabla \varphi \nabla \Delta \varphi }\text{d}x}\le C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}{\Vert \Delta \varphi \nabla \Delta \varphi \Vert }_{N^{p^{\prime },\frac{3}{3-r}}}\\ \le C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}{\Vert \Delta \varphi \Vert }_{{\stackrel{\dot{}}{H}}^r}{\Vert \nabla \Delta \varphi \Vert }_{L^2}\le C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}{\Vert \Delta \varphi \Vert }_{L^2}^{1-r}{\Vert \nabla \Delta \varphi \Vert }_{L^2}^{1+r}\\ =C{\left({\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{1-r}{2}}{\left({\Vert \nabla \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{1+r}{2}}\le C{\left({\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{1-r}{2}}{\left({\Vert \Delta \varphi \Vert }_{L^2}{\Vert {\Delta }^2\varphi \Vert }_{L^2}\right)}^{\frac{1+r}{2}}\\ \le \frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2+C{\Vert \Delta \varphi \Vert }_{L^2}^2+C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \Delta \varphi \Vert }_{L^2}^2.\end{array}$ (3.19)

第二种情况：当 $p=2$ 时，利用引理2.5，Hölder不等式，Young’s不等式和插值不等式 ${\Vert u\Vert }_{{\stackrel{\dot{}}{B}}_2^{r,1}}\le {\Vert u\Vert }_{L^2}^{1-r}{\Vert \nabla u\Vert }_{L^2}^r$， $0<r<1$，估计 $I_1$

$\begin{array}{c}{I}_1={\displaystyle {\int }_{{ℝ}^3}\left(u\cdot \nabla \right)u\Delta u}\text{d}x\le {\Vert u\nabla u\Vert }_{L^2}{\Vert \Delta u\Vert }_{L^2}\le {\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{B}}_2^{r,1}}{\Vert \Delta u\Vert }_{L^2}\le {\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}{\Vert \nabla u\Vert }_{L^2}^{1-r}{\Vert \Delta u\Vert }_{L^2}^r{\Vert \Delta u\Vert }_{L^2}^1\\ ={\left({\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \nabla u\Vert }_{L^2}^2\right)}^{\frac{1-r}{2}}{\left({\Vert \Delta u\Vert }_{L^2}^2\right)}^{\frac{1+r}{2}}\le \frac{1}{2}{\Vert \Delta u\Vert }_{L^2}^2+C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \nabla u\Vert }_{L^2}^2.\end{array}$ (3.20)

同样估计 $I_2$，得

$\begin{array}{c}{I}_2=2{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}{u}_i{\partial }_i\nabla \varphi \nabla \Delta \varphi }\text{d}x}\le C{\Vert u\Delta \varphi \Vert }_{L^2}{\Vert \nabla \Delta \varphi \Vert }_{L^2}\le C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}{\Vert \Delta \varphi \Vert }_{{\stackrel{\dot{}}{B}}_2^{r,1}}{\Vert \nabla \Delta \varphi \Vert }_{L^2}\\ \le C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}{\Vert \Delta \varphi \Vert }_{L^2}^{1-r}{\Vert \nabla \Delta \varphi \Vert }_{L^2}^{1+r}=C{\left({\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{1-r}{2}}{\left({\Vert \nabla \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{1+r}{2}}\\ \le C{\left({\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{1-r}{2}}{\left({\Vert \Delta \varphi \Vert }_{L^2}{\Vert {\Delta }^2\varphi \Vert }_{L^2}\right)}^{\frac{1+r}{2}}\\ \le \frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2+C{\Vert \Delta \varphi \Vert }_{L^2}^2+C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \Delta \varphi \Vert }_{L^2}^2.\\ \le \frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2+C{\Vert \Delta \varphi \Vert }_{L^2}^2+C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \Delta \varphi \Vert }_{L^2}^2.\end{array}$ (3.21)

将上述估计(3.16)，(3.18)~(3.21)代入(3.15)中，得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\left({\left|\nabla u\right|}^2+{\left|\Delta \varphi \right|}^2\right)\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\left({\left|\Delta u\right|}^2+{\left|{\Delta }^2\varphi \right|}^2\right)\text{d}x}\\ \le C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \nabla u\Vert }_{L^2}^2+C{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}{\Vert \Delta \varphi \Vert }_{L^2}^2+C{\Vert \Delta \varphi \Vert }_{L^2}^2+C\left({\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4+1\right){\Vert \Delta \varphi \Vert }_{L^2}^2\\ \le C\left({\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}+{\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4+1\right)\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right),\end{array}$

由(3.7)，(3.11)可知，

${\displaystyle {\int }_0^T\left({\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4+1\right)\text{d}t}\le C.$ (3.22)

利用Gronwall不等式，上式和定理假设条件(1.6)，得

$\begin{array}{c}\sup \left\{{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right\}\le \left({\Vert \nabla u_0\Vert }_{L^2}^2+{\Vert \Delta {\varphi }_0\Vert }_{L^2}^2\right)\exp \left\{{\displaystyle {\int }_0^T\left({\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}+{\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4+1\right)\text{d}t}\right\}\\ =\left({\Vert \nabla u_0\Vert }_{L^2}^2+{\Vert \Delta {\varphi }_0\Vert }_{L^2}^2\right)\exp \left\{T+{\displaystyle {\int }_0^T\left({\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4\right)\text{d}t}+{\displaystyle {\int }_0^T{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}}\text{d}t\right\}\\ <\infty .\end{array}$ (3.23)

或将上述估计(3.16)~(3.20)代入(3.14)中，得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\left({\left|\nabla u\right|}^2+{\left|\Delta \varphi \right|}^2\right)\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\left({\left|\Delta u\right|}^2+{\left|{\Delta }^2\varphi \right|}^2\right)\text{d}x}\\ \le C\left({\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}+{\Vert \Delta \varphi \Vert }_{L^2}^2+1\right)\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right)\\ \le C\left(\frac{{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}+{\Vert \Delta \varphi \Vert }_{L^2}^2}{1+\ln \left(e+{\Vert u\Vert }_{L^6}\right)}+1\right)\ln \left(e+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right)\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right),\end{array}$

因此

$\frac{\text{d}}{\text{d}t}\ln \left(e+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right)\le C\left(\frac{{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}+{\Vert \Delta \varphi \Vert }_{L^2}^2}{1+\ln \left(e+{\Vert u\Vert }_{L^6}\right)}+1\right)\ln \left(e+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right)$,

利用Gronwall不等式，定理假设条件(1.7)，得

$\sup \left\{{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right\}\le \left(e+{\Vert \nabla u_0\Vert }_{L^2}^2+{\Vert \Delta {\varphi }_0\Vert }_{L^2}^2\right)\exp \left\{\exp \left\{T+{\displaystyle {\int }_0^T\frac{{\Vert u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{r}}}^{\frac{2}{1-r}}+{\Vert \Delta \varphi \Vert }_{L^2}^2}{1+\ln \left(e+{\Vert u\Vert }_{L^6}\right)}\text{d}t}\right\}\right\}<\infty$. (3.24)

由(3.23)，(3.24)可知，(3.1)成立，定理1.1得证。

最后考虑 $\nabla u$ 满足定理1.2中条件的情况，对于 $I_1$ 和 $I_2$ 的估计，与之前类似，分为两种情况。

第一种情况：当 $2<p\le \frac{3}{\gamma }$ 时，利用(1.4)，引理2.3，引理2.4，Young’s不等式和插值不等式 ${\Vert u\Vert }_{{\stackrel{\dot{}}{H}}^{\gamma }}\le {\Vert u\Vert }_{L^2}^{1-\gamma }{\Vert \nabla u\Vert }_{L^2}^{\gamma }$， $0<\gamma \le 1$，估计 $I_1$

$\begin{array}{c}{I}_1={\displaystyle {\int }_{{ℝ}^3}\left(u\cdot \nabla \right)u\Delta u}\text{d}x=-{\displaystyle {\int }_{{ℝ}^3}\nabla \cdot \left(\left(u\cdot \nabla \right)u\right)\nabla u}\text{d}x=-{\displaystyle {\int }_{{ℝ}^3}\nabla u\nabla u\nabla u}+u\cdot \nabla \left(\frac{{\left|\nabla u\right|}^2}{2}\right)\text{d}x\\ =-{\displaystyle {\int }_{{ℝ}^3}\nabla u\nabla u\nabla u}\text{d}x\le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}{\Vert \nabla u\nabla u\Vert }_{N^{p^{\prime },\frac{3}{3-\gamma }}}\le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{H}}^{\gamma }}{\Vert \Delta u\Vert }_{L^2}\\ \le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}{\Vert \nabla u\Vert }_{L^2}{\Vert \nabla u\Vert }_{L^2}^{1-\gamma }{\Vert \Delta u\Vert }_{L^2}^{\gamma }=C{\left({\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \nabla u\Vert }_{L^2}^2\right)}^{\frac{2-\gamma }{2}}{\left({\Vert \Delta u\Vert }_{L^2}^2\right)}^{\frac{\gamma }{2}}\\ \le \frac{1}{2}{\Vert \Delta u\Vert }_{L^2}^2+C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \nabla u\Vert }_{L^2}^2.\end{array}$ (3.25)

估计 $I_2$，得

$\begin{array}{c}{I}_2=-2{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}\nabla u_i{\partial }_i\nabla \varphi \Delta \varphi }\text{d}x}\le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}{\Vert {\nabla }^2\varphi \Delta \varphi \Vert }_{N^{p^{\prime },\frac{3}{3-\gamma }}}\le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}{\Vert {\nabla }^2\varphi \Vert }_{{\stackrel{\dot{}}{H}}^{\gamma }}{\Vert \Delta \varphi \Vert }_{L^2}\\ \le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}{\Vert \Delta \varphi \Vert }_{L^2}{\Vert \Delta \varphi \Vert }_{L^2}^{1-\gamma }{\Vert \nabla \Delta \varphi \Vert }_{L^2}^{\gamma }=C{\left({\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{2-\gamma }{2}}{\left({\Vert \nabla \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{\gamma }{2}}\\ \le C{\left({\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{2-\gamma }{2}}{\left({\Vert \Delta \varphi \Vert }_{L^2}{\Vert {\Delta }^2\varphi \Vert }_{L^2}\right)}^{\frac{\gamma }{2}}\le \frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2+C{\Vert \Delta \varphi \Vert }_{L^2}^2+C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \Delta \varphi \Vert }_{L^2}^2.\end{array}$ (3.26)

第二种情况：当 $p=2$ 时，利用引理2.5，Hölder不等式，Young’s不等式和插值不等式 ${\Vert u\Vert }_{{\stackrel{\dot{}}{B}}_2^{\gamma ,1}}\le {\Vert u\Vert }_{L^2}^{1-\gamma }{\Vert \nabla u\Vert }_{L^2}^{\gamma }$， $0<\gamma \le 1$，估计 $I_1$

$\begin{array}{c}{I}_1={\displaystyle {\int }_{{ℝ}^3}\left(u\cdot \nabla \right)u\Delta u}\text{d}x\le C{\Vert \nabla u\Vert }_{L^2}{\Vert \nabla u\nabla u\Vert }_{L^2}\le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{\gamma }}}{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{B}}_2^{\gamma ,1}}{\Vert \nabla u\Vert }_{L^2}\\ \le {\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{\gamma }}}{\Vert \nabla u\Vert }_{L^2}^{2-\gamma }{\Vert \Delta u\Vert }_{L^2}^{\gamma }={\left({\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \nabla u\Vert }_{L^2}^2\right)}^{\frac{2-\gamma }{2}}{\left({\Vert \Delta u\Vert }_{L^2}^2\right)}^{\frac{\gamma }{2}}\\ \le \frac{1}{2}{\Vert \Delta u\Vert }_{L^2}^2+C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \nabla u\Vert }_{L^2}^2.\end{array}$ (3.27)

用同样的方法估计 $I_2$，得

$\begin{array}{c}{I}_2=-2{\displaystyle \underset{i=1}{\overset{3}{\sum }}{\displaystyle {\int }_{{ℝ}^3}\nabla u_i{\partial }_i\nabla \varphi \Delta \varphi }\text{d}x}\le C{\Vert \nabla u\Vert }_{L^2}{\Vert {\nabla }^2\varphi \Delta \varphi \Vert }_{L^2}\le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{\gamma }}}{\Vert {\nabla }^2\varphi \Vert }_{{\stackrel{\dot{}}{B}}_2^{\gamma ,1}}{\Vert \Delta \varphi \Vert }_{L^2}\\ \le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{\gamma }}}{\Vert \Delta \varphi \Vert }_{L^2}^{2-\gamma }{\Vert \nabla \Delta \varphi \Vert }_{L^2}^{\gamma }=C{\left({\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{2-\gamma }{2}}{\left({\Vert \nabla \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{\gamma }{2}}\\ \le C{\left({\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \Delta \varphi \Vert }_{L^2}^2\right)}^{\frac{2-\gamma }{2}}{\left({\Vert \Delta \varphi \Vert }_{L^2}{\Vert {\Delta }^2\varphi \Vert }_{L^2}\right)}^{\frac{\gamma }{2}}\\ \le \frac{1}{4}{\Vert {\Delta }^2\varphi \Vert }_{L^2}^2+C{\Vert \Delta \varphi \Vert }_{L^2}^2+C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{2,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \Delta \varphi \Vert }_{L^2}^2.\end{array}$ (3.28)

将上述估计(3.16)，(3.25)~(3.28)代入(3.15)中，得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\left({\left|\nabla u\right|}^2+{\left|\Delta \varphi \right|}^2\right)\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\left({\left|\Delta u\right|}^2+{\left|{\Delta }^2\varphi \right|}^2\right)\text{d}x}\\ \le C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \nabla u\Vert }_{L^2}^2+C{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}{\Vert \Delta \varphi \Vert }_{L^2}^2+C{\Vert \Delta \varphi \Vert }_{L^2}^2+C\left({\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4+1\right){\Vert \Delta \varphi \Vert }_{L^2}^2\\ \le C\left({\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}+{\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4+1\right)\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right),\end{array}$

利用Gronwall不等式，(3.22)和定理假设条件(1.8)，得

$\begin{array}{c}\sup \left\{{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right\}\le \left({\Vert \nabla u_0\Vert }_{L^2}^2+{\Vert \Delta {\varphi }_0\Vert }_{L^2}^2\right)\exp \left\{{\displaystyle {\int }_0^T\left({\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}+{\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4+1\right)\text{d}t}\right\}\\ =\left({\Vert \nabla u_0\Vert }_{L^2}^2+{\Vert \Delta {\varphi }_0\Vert }_{L^2}^2\right)\exp \left\{T+{\displaystyle {\int }_0^T\left({\Vert \nabla \varphi \Vert }_{L^6}^2+{\Vert \varphi \Vert }_{L^{\infty }}^4\right)\text{d}t}+{\displaystyle {\int }_0^T{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}}\text{d}t\right\}\\ <\infty .\end{array}$ (3.29)

或将上述估计(3.17)，(3.25)~(3.28)代入(3.15)中，得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\left({\left|\nabla u\right|}^2+{\left|\Delta \varphi \right|}^2\right)\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\left({\left|\Delta u\right|}^2+{\left|{\Delta }^2\varphi \right|}^2\right)\text{d}x}\\ \le C\left({\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}+{\Vert \Delta \varphi \Vert }_{L^2}^2+1\right)\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right)\\ \le C\left(\frac{{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}+{\Vert \Delta \varphi \Vert }_{L^2}^2}{1+\ln \left(e+{\Vert u\Vert }_{L^6}\right)}+1\right)\ln \left(e+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right)\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right),\end{array}$

因此

$\frac{\text{d}}{\text{d}t}\ln \left(e+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right)\le C\left(\frac{{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}+{\Vert \Delta \varphi \Vert }_{L^2}^2}{1+\ln \left(e+{\Vert u\Vert }_{L^6}\right)}+1\right)\ln \left(e+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right),$

利用Gronwall不等式，定理假设条件(1.9)，得

$\sup \left\{{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \Delta \varphi \Vert }_{L^2}^2\right\}\le \left(e+{\Vert \nabla u_0\Vert }_{L^2}^2+{\Vert \Delta {\varphi }_0\Vert }_{L^2}^2\right)\exp \left\{\exp \left\{T+{\displaystyle {\int }_0^T\frac{{\Vert \nabla u\Vert }_{{\stackrel{\dot{}}{M}}^{p,\frac{3}{\gamma }}}^{\frac{2}{2-\gamma }}+{\Vert \Delta \varphi \Vert }_{L^2}^2}{1+\ln \left(e+{\Vert u\Vert }_{L^6}\right)}\text{d}t}\right\}\right\}<\infty$. (3.30)

由(3.29)，(3.30)可知，(3.1)成立，定理1.2得证。