三维可压缩液晶流模型解的整体存在唯一性

Global Existence and Uniqueness of a 3D Compressible Nematic Liquid Crystal Flow

摘要: 本文主要研究三维可压缩液晶流方程的解，建立了在H2(R3)中关于整体解的存在性理论。主要利用能量方法，推导出了解的先验估计，再利用连续性技巧将局部解延拓到整体。

Abstract: This paper mainly studies the solution of compressible nematic liquid crystal flow in R³, the existence theory of the global solution to the system is established in H²-framework. The energy method is used to derive the desired a priori estimates and hence the global existence by using the standard continuity method.

1. 引言

液晶相介于液相和晶相之间，它的性质涉及到化学、物理、生物学、电子学、材料学等多学科，其在基础理论得到了广泛的研究，尤其是在显示应用中获得了巨大的成功。液晶是一门方兴未艾的交叉前沿学科 [1] 。关于液晶流的理论最初是由Erickse [2] 和Leslie [3] 提出的。当流体是可压缩时，液晶系统会变得更加复杂，研究结果相对来说比较少，因此本文考虑三维可压缩的简化的液晶流模型，其形式如下：

$\{\begin{array}{l}{\rho }_t+\mathrm{div}\left(\rho u\right)=0,\hfill \\ {\left(\rho u\right)}_t+\mathrm{div}\left(\rho u\otimes u\right)+\nabla P\left(\rho \right)=\mathcal{L}u-\nabla d\cdot \Delta d,\hfill \\ d_t+u\cdot \nabla d=\Delta d+{\left|\nabla d\right|}^{2}d,\hfill \end{array}$ (1)

模型的初始值为

$\rho \left(x,0\right)={\rho }_0\left(x\right),u\left(x,0\right)=u_0\left(x\right),d\left(x,0\right)=d_0\left(x\right),x\in {ℝ}^3,$ (2)

其中 $\rho \left(x,t\right):{ℝ}^3\times \left[0,\infty \right)\to {ℝ}^1$ 和 $u\left(x,t\right):{ℝ}^3\times \left[0,\infty \right)\to {ℝ}^3$ 分别为流体的密度和速度。 $m=\rho u$ 为动量， $P=P\left(\rho \right)$ 是压力函数的表达式。液晶流的光轴矢量是一个单位向量，表达式为 $d\left(x,t\right):{ℝ}^3\times \left[0,\infty \right)\to {\text{S}}^2$ ，即 $d=1$ 。 $\mathcal{L}$ 为Lamé算子，

$\mathcal{L}u=\mu \Delta u+\left(\mu +\lambda \right)\nabla \mathrm{div}u.$

$\mu$ 和 $\lambda$ 为剪切黏度和体积黏度，满足

$\mu >0,2\mu +\text{3}\lambda \ge 0.$

我们现在只回顾一些以前密切相关的结果。许多作者对不可压缩液晶流动进行了研究，如 [4] - [9] 和其中的参考文献。对于可压缩模型(1)，最近也有很多重要的进展。Hu和Wu [10] 研究了小初值条件下模型(1)强解的整体适定性。Ding et al. [11] 和Huang et al. [12] 分别在1维和3维上获得了非负初始密度的初值问题和初值问题的局部时强解。在 [13] [14] 中，作者建立了强解的爆破判据。对于R³上任意有界光滑区域，Wang [15] 建立了以真空为远场密度的二维或三维强解的全局适定性。Li et al. [16] 在初始数据足够光滑且在某能量范数上适当小的条件下，得到了三维经典解的全局适定性。Huang et al. [17] 证明了强解的全局适定性，并得到了初始数据为H²范数稳态附近的小扰动时的 $L^p\left(p\in \left[1,6\right]\right)$ 估计。

Hieber和Prȕss [18] 证明了一个强解的局部适定性，当初始数据接近平衡点时，该强解扩展为全局解。在 [19] 中得到了非等温模式 [20] 的正则性判据。对于具有真空的可压缩非等温模型，Zhong [21] 在二维无热传导的情况下得到了(1)的局部强解，而Liu和Zhong [22] 研究了三维在小条件下强解的全局适定性。Wu [23] 利用标准的能量估计，研究了高维(3维及以上)的液晶流方程组小初值经典解的整体存在性以及运用Green函数方法，得到奇数维情形(3维及以上)该解的逐点估计。本文基于上述基础上，主要考虑三维液晶流模型(1)解在Sobolev空间 $H^2\left(R^3\right)$ 中的整体存在性，由于在H²框架下，二阶导数的估计需要更加精细的能量估计。下面是本文的符号说明和一些引理。

2. 符号说明和一些引理

在本文中，C表示一般的正常数。对于整数 $m\ge 0$ ，Sobolev空间 $H^m\left(R^3\right)$ 中的范数表示为 ${\Vert \cdot \Vert }_{H^m}$ 。特别地，当 $m=0$ 时，我们将简单地使用 $\Vert ·\Vert$ 。同通常一样， $\langle ·,·\rangle$ 表示 $L^2\left(R{}^3\right)$ 中的内积。梯度表示为 $\nabla =\left({\partial }_1,{\partial }_2,{\partial }_3\right)$ ， ${\partial }_i={\partial }_{x_i}$ ， $i=1,2,3$ 。对于任意整数 $l\ge 0$ ， ${\nabla }^{l}f$ 表示函数f的所有l阶导数。

为了后续证明主要结论的需要，下面介绍两个相关引理。

首先，列出一些Sobolev空间中的基本不等式。

引理2.1 令 $f\in H^2\left({ℝ}^3\right)$ ，有

(i) ${\Vert f\Vert }_{L^{\infty }}\le C{\Vert \nabla f\Vert }_{L^2}^{1/2}{\Vert \nabla f\Vert }_{H^1}^{1/2}\le C{\Vert \nabla f\Vert }_{H^1}$ ；

(ii) ${\Vert f\Vert }_{L^6}\le C{\Vert \nabla f\Vert }_{L^2}$ ；

(iii) ${\Vert f\Vert }_{L^q}\le C{\Vert f\Vert }_{H^1},2\le q\le 6$ 。

引理2.2 令 $s>0$ 和 $m\ge 1$ 是整数，有

${\Vert {\nabla }^s\left(fg\right)\Vert }_{L^p}\le C{\Vert f\Vert }_{L^{p_1}}{\Vert {\nabla }^{s}g\Vert }_{L^{p_2}}+C{\Vert {\nabla }^{s}f\Vert }_{L^{p_3}}{\Vert g\Vert }_{L^{p_4}},$

和

${\Vert {\nabla }^m\left(fg\right)-f{\nabla }^{m}g\Vert }_{L^p}\le C{\Vert \nabla f\Vert }_{L^{p_1}}{\Vert {\nabla }^{m-1}g\Vert }_{L^{p_2}}+C{\Vert {\nabla }^{m}f\Vert }_{L^{p_3}}{\Vert g\Vert }_{L^{p_4}},$

这里 $p,p_1,p_2,p_3,p_4\in \left[1,\infty \right]$ ，且

$\frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{p_3}+\frac{1}{p_4}.$

3. 主要结果

在本节，我们给出了本文主要结果。

定理3.1 假设 $\left({\rho }_0-\bar{\rho },u_0\right)\in H^2\left({ℝ}^3\right)$ ， $\nabla d_0\in H^1\left({ℝ}^3\right)$ ， $\bar{\rho }>0$ ，则存在一个足够小的常数 $\epsilon >0$ ，使得如果

${\Vert \left({\rho }_0-\bar{\rho },u_0\right)\Vert }_{H^2\left({ℝ}^3\right)}+{\Vert \nabla d_0\Vert }_{H^1\left({ℝ}^3\right)}\le \epsilon ,$

那么整体存在唯一的光滑解满足

$\left(\rho -\bar{\rho },u,\nabla d\right)\in C\left(\left[0,\infty \right);H^2\left({ℝ}^3\right)\right).$

4. 整体存在性

在本节中，我们研究模型(1)解的局部存在性理论和一些能量估计。将局部存在性结果与一些先验估计相结合，然后使用标准连续性论证，得到模型解的整体存在性。

定理4.1 假设 $\left({\rho }_0-\bar{\rho },u_0\right)\in H^2\left({ℝ}^3\right)$ ， $\nabla d_0\in H^1\left({ℝ}^3\right)$ ， $\bar{\rho }>0$ ，则存在一个大于0的常数 $T_1$ ，使得系统(8)在 ${ℝ}^3\times \left[0,T_1\right]$ 上有一个唯一的整体解 $\left(\rho ,u,d\right)$ 满足

$\left(\rho ,u,d\right)\in C\left(\left[0,T_1\right];H^2\left({ℝ}^3\right)\right),$

证明：利用收缩映射定理的标准论证可以证明这一点。例如，参考文献 [24] [25] 对局部存在结果的研究。我们在这里省略了证明。

首先，我们用能量法给出了一些解的先验估计。在本节中，我们用能量法给出了一些解的先验估计。由定理4.1可知，存在时间 $T^{\prime }>0$ ，解存在于 ${ℝ}^3\times \left[0,T^{\prime }\right]$ 中。如定理4.1所述，假设对于任意 $T\in \left(0,T^{\prime }\right]$ ，解存在于 ${ℝ}^3\times \left[0,T\right]$ 上，我们需要一些关于时间的一致估计来证明定理3.1。为此，我们首先做一个先验假设，当 $\delta >0,{\delta }_1>0$ 足够小的时候，有

$\{\begin{array}{l}\underset{0\le t\le T}{\sup }\left\{{\Vert \left(\rho -\bar{\rho },u\right)\left(\cdot ,t\right)\Vert }_{H^2}\right\}\le \delta ,\\ \underset{0\le t\le T}{\sup }\left\{{\Vert \nabla d\left(\cdot ,t\right)\Vert }_{H^1}\right\}\le {\delta }_1.\end{array}$ (3)

那么，模型(1)的解 $\left(\rho ,u,d\right)$ 存在并且满足

$\begin{array}{l}\left\{{\Vert \left(\rho -\bar{\rho },u\right)\Vert }_{H^2}^2+{\Vert \nabla d\Vert }_{H^1}^2\right\}+{\displaystyle {\int }_0^t\left[{\Vert \nabla \rho \Vert }_{H^1}^2+{\Vert \left(\nabla u,\nabla d\right)\Vert }_{H^2}^2\right]}\left(\cdot ,s\right)\text{d}s\\ \le C_1\left\{{\Vert \left({\rho }_0-\bar{\rho },u_0\right)\Vert }_{H^2}^2+{\Vert \nabla d_0\Vert }_{H^1}^2\right\}.\end{array}$ (4)

通过(3)和sobolev不等式，我们得到

$\underset{x\in {ℝ}^3}{\sup }\left|\left(\rho -\bar{\rho },u,\nabla d\right)\left(t\right)\right|\le C\left(\delta +{\delta }_1\right),$ (5)

其次，我们对方程做一个变形。我们定义函数 $h\left(\rho \right)$ 满足 $h^{\prime }\left(\rho \right)=\frac{1}{\rho }{P}^{\prime }\left(\rho \right)$ ，通过先验假设(3)和Sobolev

不等式可知，存在正常数C₀，C₁和C₂使得

$0<C_0\le h^{\prime }\left(\rho \right)\le C_1\le \infty ,\left|h^{″}\left(\rho \right)\right|\le C_2.$

做变换 $\rho \to \bar{\rho }+\rho$ ，令 $\bar{\rho }=1,\mu =1,\lambda =0$ 。则方程可以写成

$\{\begin{array}{l}{\rho }_t+\text{div}\left[\left(1+\rho \right)u\right]=0,\\ u_t+u\cdot \nabla u+\nabla \left[h\left(1+\rho \right)-h\left(1\right)\right]=\frac{1}{1+\rho }\left(\Delta u+\nabla \text{div}u\right)-\frac{1}{1+\rho }\nabla d\cdot \Delta d,\\ d_t-\Delta d=-u\cdot \nabla d+{\left|\nabla d\right|}^{2}d,\end{array}$ (6)

初值条件满足

$\left(\rho ,u,d\right)\left(x,0\right)=\left({\rho }_0,u_0,d_0\right)\left(x\right).$ (7)

同时为了计算方便，我们对(6)做进一步的简化，可以得到

$\{\begin{array}{l}{\rho }_t+\mathrm{div}u=-\mathrm{div}\left(\rho u\right),\hfill \\ u_t-\Delta u-\nabla \mathrm{div}u+\nabla \rho =-u\cdot \nabla u-l\left(\rho \right)\left[\Delta u+\nabla \mathrm{div}u\right]-f\left(\rho \right)\nabla \rho -\frac{1}{1+\rho }\nabla d\cdot \Delta d,\hfill \\ d_t-\Delta d=-u\cdot \nabla d+{\left|\nabla d\right|}^{2}d,\hfill \end{array}$ (8)

其中 $l\left(\rho \right)=\frac{\rho }{1+\rho }$ ， $f\left(\rho \right)=\frac{P^{\prime }\left(1+\rho \right)}{1+\rho }-1$ 。通过先验假设(3)直接计算可以得到

$\left|l\left(\rho \right)\right|,\left|f\left(\rho \right)\right|\le C\left|\rho \right|,\left|l^{\left(k\right)}\left(\rho \right)\right|,\left|f^{\left(k\right)}\left(\rho \right)\right|\le C,k\ge 1.$ (9)

这些不等式在后面将会用到。

能量估计

接下来，我们将给出解 $\left(\rho ,u,d\right)$ 的能量估计。类似于 [23] 的做法，我们给出 $L^2$ 和一阶导数的 $L^2$ 估计，但由于我们在 $H^2$ 框架下，二阶导数的估计需要更加精细的能量估计。

引理4.1 有

$\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{{\Vert \left(\rho \nabla \rho ,u,\nabla d\right)\Vert }_{L^2}^2\right\}+C\left({\gamma }_0,{\gamma }_1\right){\Vert \left(\nabla \rho ,\nabla u,{\nabla }^{2}d\right)\Vert }^2\le 0.$

证明：将 ${\nabla }^k$ 分别作用于(8)₁和(8)₂，同时用 ${\nabla }^k\rho ,{\nabla }^{k}u$ 分别乘以(8)₁和(8)₂，在 $R^3$ 上积分得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert {\nabla }^k\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{k}u\Vert }_{L^2}^2\right)+{\Vert {\nabla }^{k+1}u\Vert }_{L^2}^2+{\Vert {\nabla }^k\text{div}u\Vert }_{L^2}^2\\ =-\langle {\nabla }^k\rho ,{\nabla }^k\mathrm{div}\left(\rho u\right)\rangle -\langle {\nabla }^{k}u,{\nabla }^k\left(u\cdot \nabla u\right)\rangle -\langle {\nabla }^{k}u,{\nabla }^k\left(l\left(\rho \right)\Delta u+l\left(\rho \right)\nabla \mathrm{div}u\right)\rangle \\ -\langle {\nabla }^{k}u,{\nabla }^k\left(f\left(\rho \right)\nabla \rho \right)\rangle -\langle {\nabla }^{k}u,{\nabla }^k\left(\frac{1}{1+\rho }\nabla d\cdot \Delta d\right)\rangle \\ :=I_1+I_2+I_3+I_4+I_5.\end{array}$ (10)

现在对(10)右边的项进行估计，对于 $k=0$ ，可以得到

$\begin{array}{c}{I}_1=-{\displaystyle {\int }_{{ℝ}^3}\rho }\mathrm{div}\left(\rho u\right)\text{d}x\\ \le C{\Vert \rho \Vert }_{L^3}{\Vert \rho \Vert }_{L^6}{\Vert \nabla u\Vert }_{L^2}+{\Vert \nabla \rho \Vert }_{L^2}{\Vert u\Vert }_{L^3}{\Vert \rho \Vert }_{L^6}\\ \le C\delta \left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2\right),\end{array}$

通过Sobolev不等式和(9)，我们有

$I_3={\displaystyle {\int }_{{ℝ}^3}u\left(l\left(\rho \right){\left|\nabla u\right|}^2+l\left(\rho \right){\left|\nabla \cdot u\right|}^2\right)\text{d}x}\le C\delta \left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2\right),$

$I_4=-{\displaystyle {\int }_{{ℝ}^3}u}\cdot \left(f\left(\rho \right)\nabla \rho \right)\text{d}x\le C\delta \left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2\right),$

其中对向量 $u,v\in {ℝ}^3$ ， $u\otimes v={\left(u_i{v}_j\right)}_{3\times 3}$ ，对于矩阵 $U,V\in {ℝ}^{3\times 3}$ ， $U:V=u_{ij}{v}_{ij}$ 。那么对于 $I_2$ 和 $I_5$ 可以得到

$I_2={\displaystyle \int \left(u\cdot \nabla u\right)}\cdot u\text{d}x\le C{\Vert u\Vert }_{L^3}{\Vert \nabla u\Vert }_{L^2}{\Vert u\Vert }_{L^6}\le C\delta {\Vert \nabla u\Vert }_{L^2}^2,$

$\begin{array}{c}{I}_5=-{\displaystyle {\int }_{{ℝ}^3}\frac{1}{\rho +1}}\left[\text{div}\left(\frac{1}{2}{\left|\nabla d\right|}_{3\times 3}^2-\nabla d\otimes \nabla d\right)\right]\cdot u\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}\nabla }\left(\frac{1}{\rho +1}\right)\cdot \left[\frac{1}{2}{\left|\nabla d\right|}^2{I}_{3\times 3}-\nabla d\otimes \nabla d\right)]\cdot u\text{d}x\\ +{\displaystyle {\int }_{{ℝ}^3}\frac{1}{\rho +1}}\left[\left(\frac{1}{2}{\left|\nabla d\right|}^2{I}_{3\times 3}-\nabla d\otimes \nabla d\right)\cdot \nabla \right]\cdot u\text{d}x\\ \le C{\Vert \nabla \Vert }_{L^2}{\Vert \nabla d\Vert }_{L^6}^2{\Vert u\Vert }_{L^6}+C{\Vert \nabla d\Vert }_{L^3}{\Vert \nabla d\Vert }_{L^6}{\Vert \nabla u\Vert }_{L^2}\\ \le C\delta \left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2\right),\end{array}$

那么，可以得到

$\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \rho \Vert }_{L^2}^2+{\Vert u\Vert }_{L^2}^2\right)+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert \text{div}u\Vert }_{L^2}^2\le C\delta \left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2\right).$ (11)

接下来，对 ${\Vert {\nabla }^{2}d\Vert }_{L^2}^2$ 和 ${\Vert \nabla \rho \Vert }_{L^2}$ 做估计，将 $\nabla$ 作用到(8)₃，将所得的等式乘以 ${\Vert \nabla d\Vert }_{L^2}$ 在 $R^3$ 上积分，利用分部积分和Young不等式可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \nabla d\Vert }^2+{\Vert {\nabla }^{2}d\Vert }^2\\ =-{\displaystyle {\int }_{{ℝ}^3}\nabla }\left(u\cdot \nabla d\right)\cdot \nabla d\text{d}x-{\displaystyle {\int }_{{ℝ}^3}\nabla }\left({\left|\nabla d\right|}^{2}d\right)\cdot \nabla d\text{d}x\\ \le {\displaystyle {\int }_{{ℝ}^3}\left|\nabla d\right|}\left|\nabla u\right|\left|\nabla d\right|\text{d}x+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla d\right|}^2\left|{\nabla }^{2}d\right|\left|d\right|\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}{\left|\nabla d\right|}^4\text{d}x}\\ \le \left({\Vert \nabla d\Vert }_{L^6}{\Vert \nabla u\Vert }_{L^2}+{\Vert {\nabla }^{2}d\Vert }_{L^2}{\Vert u\Vert }_{L^6}+{\Vert {\nabla }^{2}d\Vert }_{L^2}{\Vert \nabla d\Vert }_{L^6}\right){\Vert \nabla d\Vert }_{L^3}+{\Vert \nabla d\Vert }_{L^2}{\Vert \nabla d\Vert }_{L^6}^3\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla u\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2\right).\end{array}$ (12)

将(11)和(12)相加，可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \rho \Vert }_{L^2}^2+{\Vert u\Vert }_{L^2}^2+{\Vert \nabla d\Vert }_{L^2}^2\right)+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2\right).\end{array}$ (13)

为了得到 ${\Vert \nabla \rho \Vert }_{L^2}$ 的估计，我们先引入两个关系式，

$\nabla \rho \cdot \left({\nabla }^2\rho \cdot u\right)=\nabla \frac{{\left|\nabla \rho \right|}^2}{2}\cdot u,{\displaystyle {\int }_{{ℝ}^3}\left(\rho +1\right)\nabla \rho \cdot \left(\Delta u-\nabla \text{div}u\right)\text{d}x}=0,$

定义以下运算

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\left[\frac{1}{2}{\left|\nabla \rho \right|}^2+\frac{{\left(\rho +1\right)}^2}{2}\nabla \rho \cdot u\right]\text{d}x}\\ ={\displaystyle {\int }_{{ℝ}^3}\left[\nabla \rho \cdot \nabla {\rho }_t+\left(\rho +1\right){\rho }_t\nabla \rho \cdot u+\frac{{\left(\rho +1\right)}^2}{2}\nabla {\rho }_t\cdot u+\frac{{\left(\rho +1\right)}^2}{2}\nabla \rho \cdot u_t\right]\text{d}x}.\end{array}$ (14)

那么通过Hölder不等式和Cauchy不等式，则可以得到

$\begin{array}{l}{\displaystyle {\int }_{{ℝ}^3}\left[\nabla \rho \cdot \nabla {\rho }_t+\frac{{\left(\rho +1\right)}^2}{2}\nabla \rho \cdot u_t\right]\text{d}x}\\ =-{\displaystyle {\int }_{{ℝ}^3}\left[\nabla \rho \cdot \nabla \mathrm{div}\left[\left(\rho +1\right)u\right]-\frac{{\left(\rho +1\right)}^2}{2}\nabla \rho \cdot u_t\right]\text{d}x}\\ =-{\displaystyle {\int }_{{ℝ}^3}\left[\nabla \rho \cdot \left({\nabla }^2\rho \cdot u\right)+\nabla \rho \cdot \left(\nabla \rho \cdot \nabla u\right)+{\left|\nabla \rho \right|}^2\mathrm{div}u+\left(\rho +1\right)\nabla \rho \cdot \nabla \mathrm{div}u\right]\text{d}x}\\ +{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^2}{2}\nabla \rho \left\{-\frac{1}{\rho +1}\nabla d\cdot \Delta d-u\cdot \nabla u-\nabla \left[h\left(\rho +1\right)-h\left(1\right)\right]+\frac{1}{\rho +1}\Delta u+\frac{1}{\rho +1}\nabla \mathrm{div}u\right\}\text{d}x}\end{array}$

$\begin{array}{l}\le C\left(\delta +{\delta }_1\right){\Vert \left(\nabla \rho ,\nabla u,{\nabla }^{2}d\right)\Vert }^2-{\displaystyle {\int }_{{ℝ}^3}\left(\rho +1\right)\nabla \rho \cdot \nabla \mathrm{div}u\text{d}x}\\ -{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^2}{2}{h}^{\prime }\left(\rho +1\right){\left|\nabla \rho \right|}^2\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\frac{\rho +1}{2}\nabla \rho \cdot \left(\Delta u+\nabla \mathrm{div}u\right)\text{d}x}\\ =C\left(\delta +{\delta }_1\right){\Vert \left(\nabla \rho ,\nabla u,{\nabla }^{2}d\right)\Vert }^2-{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^2}{2}{h}^{\prime }\left(\rho +1\right){\left|\nabla \rho \right|}^2\text{d}x},\end{array}$

剩下的两项估计如下：

$\begin{array}{l}{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^2}{2}\nabla {\rho }_t\cdot u\text{d}x}=-{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^2}{2}}u\cdot \nabla \mathrm{div}\left[\left(\rho +1\right)u\right]\text{d}x\\ =-{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^2}{2}u\cdot \nabla \left(\nabla \rho \cdot u\right)}+\frac{{\left(\rho +1\right)}^2}{2}u\cdot \left[\mathrm{div}u\nabla \rho +\left(\rho +1\right)\nabla \mathrm{div}u\right]\text{d}x\\ \le C\delta {\Vert \left(\nabla \rho ,\nabla u\right)\Vert }_{L^2}^2+{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^3}{2}}{\left(\mathrm{div}u\right)}^2\text{d}x.\end{array}$

因此，我们可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \nabla \rho \Vert }_{L^2}^2+C{\Vert \nabla \rho \Vert }_{L^2}^2+{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^2}{2}\nabla \rho \cdot u\text{d}x}\\ \le {\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^3}{2}}{\left(\mathrm{div}u\right)}^2\text{d}x+C\left(\delta +{\delta }_1\right){\Vert \left(\nabla u,{\nabla }^{2}d\right)\Vert }_{L^2}^2.\end{array}$ (15)

用两个足够小的数 ${\gamma }_0,{\gamma }_1$ 分别与(13)和(15)相乘之后相加，则可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \left(\rho \nabla \rho ,u,\nabla d\right)\Vert }_{L^2}^2+{\gamma }_0{\Vert \left(\nabla u,{\nabla }^{2}d\right)\Vert }_{L^2}^2+C{\gamma }_1{\Vert \nabla \rho \Vert }_{L^2}^2+{\gamma }_1{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^2}{2}\nabla \rho \cdot u\text{d}x}\\ \le C{\gamma }_0\left(\delta +{\delta }_1\right){\Vert \nabla \rho \Vert }_{L^2}^2+C\left(\delta +{\delta }_1\right)\left({\gamma }_0+{\gamma }_1\right){\Vert \left(\nabla u,{\nabla }^{2}d\right)\Vert }_{L^2}^2+{\gamma }_1{\displaystyle {\int }_{{ℝ}^3}\frac{{\left(\rho +1\right)}^3}{2}{\left(\mathrm{div}u\right)}^2\text{d}x}.\end{array}$ (16)

则有

$\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{{\Vert \left(\rho \nabla \rho ,u,\nabla d\right)\Vert }_{L^2}^2\right\}+C\left({\gamma }_0,{\gamma }_1\right){\Vert \left(\nabla \rho ,\nabla u,{\nabla }^{2}d\right)\Vert }^2\le 0.$ (17)

引理4.1证毕。

引理4.2 有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2\right)+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert \nabla \text{div}u\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2\right)+C\delta \left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right).\end{array}$ (18)

证明：将 ${\nabla }^k$ 分别作用于(8)₁和(8)₂，同时用 ${\nabla }^k\rho ,{\nabla }^{k}u$ 分别乘以(8)₁和(8)₂，在 $R^3$ 上积分得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert {\nabla }^k\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{k}u\Vert }_{L^2}^2\right)+{\Vert {\nabla }^{k+1}u\Vert }_{L^2}^2+{\Vert {\nabla }^k\text{div}u\Vert }_{L^2}^2\\ =-\langle {\nabla }^k\rho ,{\nabla }^k\mathrm{div}\left(\rho u\right)\rangle -\langle {\nabla }^{k}u,{\nabla }^k\left(u\cdot \nabla u\right)\rangle -\langle {\nabla }^{k}u,{\nabla }^k\left(l\left(\rho \right)\Delta u+l\left(\rho \right)\nabla \mathrm{div}u\right)\rangle \\ -\langle {\nabla }^{k}u,{\nabla }^k\left(f\left(\rho \right)\nabla \rho \right)\rangle -\langle {\nabla }^{k}u,{\nabla }^k\left(\frac{1}{1+\rho }\nabla d\cdot \Delta d\right)\rangle \\ :=J_1+J_2+J_3+J_4+J_5.\end{array}$

对于 $k=1$ ，则有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2\right)+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert \nabla \text{div}u\Vert }_{L^2}^2\\ =-\langle \nabla \rho ,\nabla \mathrm{div}\left(\rho u\right)\rangle -\langle \nabla u,\nabla \left(u\cdot \nabla u\right)\rangle -\langle \nabla u,\nabla \left(l\left(\rho \right)\Delta u+l\left(\rho \right)\nabla \mathrm{div}u\right)\rangle \\ -\langle \nabla u,\nabla \left(f\left(\rho \right)\nabla \rho \right)\rangle -\langle \nabla u,\nabla \left(\frac{1}{1+\rho }\nabla d\cdot \Delta d\right)\rangle \\ :=J_1+J_2+J_3+J_4+J_5.\end{array}$

对于上述的项，易得

$\begin{array}{c}{J}_1={\displaystyle {\int }_{{ℝ}^3}{\nabla }^2}\rho \mathrm{div}\left(\rho u\right)\text{d}x\\ \le C{\Vert {\nabla }^2\rho \Vert }_{L^2}\left[{\Vert \nabla \rho u\Vert }_{L^2}+{\Vert \rho \nabla u\Vert }_{L^2}\right]\\ \le C{\Vert {\nabla }^2\rho \Vert }_{L^2}\left[{\Vert \nabla \rho \Vert }_{L^3}{\Vert u\Vert }_{L^6}+{\Vert \rho \Vert }_{L^{\infty }}{\Vert \nabla u\Vert }_{L^2}\right]\\ \le C\delta \left({\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2\right),\end{array}$

$\begin{array}{c}{J}_2=-{\displaystyle {\int }_{{ℝ}^3}\nabla u}\cdot \nabla \left(u\cdot \nabla u\right)\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}\left(u\cdot \nabla \right)\nabla u\cdot \nabla u\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}\left[\nabla \left(u\cdot \nabla u\right)-\left(u\cdot \nabla \right)\nabla u\right]\cdot \nabla u\text{d}x}\\ \le C\delta {\Vert \nabla u\Vert }_{H^1}^2.\end{array}$

和

$\begin{array}{c}{J}_3=-{\displaystyle {\int }_{{ℝ}^3}\nabla u}\cdot \nabla \left(l\left(\rho \right)\Delta u+l\left(\rho \right)\nabla \mathrm{div}u\right)\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}{\nabla }^{2}u}\cdot \left(l\left(\rho \right)\Delta u+l\left(\rho \right)\nabla \mathrm{div}u\right)\text{d}x\le C\delta {\Vert {\nabla }^{2}u\Vert }_{L^2}^2,\end{array}$

$J_4={\displaystyle {\int }_{{ℝ}^3}{\nabla }^{2}u}\cdot \left(f\left(\rho \right)\nabla \rho \right)\text{d}x\le C\delta \left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right).$

同样的，对于 $J_5$ ，

$\begin{array}{c}{J}_5=-{\displaystyle {\int }_{{ℝ}^3}\nabla u}\cdot \nabla \left(\frac{1}{1+\rho }\nabla d\cdot \Delta d\right)\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}\nabla u}\cdot \left[\nabla \left(\frac{1}{1+\rho }\right)\left(\nabla d\cdot \Delta d\right)+\frac{1}{1+\rho }\nabla \left(\nabla d\cdot \Delta d\right)\right]\text{d}x\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2\right).\end{array}$

综上可得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2\right)+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert \nabla \text{div}u\Vert }_{L^2}^2\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2\right)\\ +C\delta \left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right).\end{array}$ (19)

接下来，我们要给出 ${\Vert {\nabla }^{2}d\Vert }_{L^2}^2$ 的估计，将 ${\nabla }^2$ 作用到(18)₃，再对得到的等式乘以 ${\nabla }^{2}d$ 并在 $R^3$ 上积分可得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\nabla }^{2}d\Vert }^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\\ =-{\displaystyle {\int }_{{ℝ}^3}{\nabla }^2}\left(u\cdot \nabla d\right)\cdot {\nabla }^{2}d\text{d}x+{\displaystyle {\int }_{{ℝ}^3}{\nabla }^2}\left({\left|\nabla d\right|}^{2}d\right)\cdot {\nabla }^{2}d\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}\nabla }\left(u\cdot \nabla d\right)\cdot {\nabla }^{3}d\text{d}x-{\displaystyle {\int }_{{ℝ}^3}\nabla }\left({\left|\nabla d\right|}^{2}d\right)\cdot {\nabla }^{3}d\text{d}x\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\right).\end{array}$ (20)

将(19)和(20)相加，得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2\right)+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert \nabla \text{div}u\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert \nabla u\Vert }_{L^2}^2\right)\\ +C\delta \left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right).\end{array}$ (21)

引理4.2证毕。

引理4.3 对于整数 $k=2$ ，有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right)+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\text{div}u\Vert }_{L^2}^2\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2\right)+C\delta \left({\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right).\end{array}$ (22)

证明：将 ${\nabla }^k$ 分别作用于(8)₁和(8)₂，同时用 ${\nabla }^k\rho ,{\nabla }^{k}u$ 分别乘以(8)₁和(8)₂，在 $R^3$ 上积分得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right)+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\text{div}u\Vert }_{L^2}^2\\ =-\langle {\nabla }^2\rho ,{\nabla }^2\mathrm{div}\left(\rho u\right)\rangle -\langle {\nabla }^{2}u,{\nabla }^2\left(u\cdot \nabla u\right)\rangle -\langle {\nabla }^{2}u,{\nabla }^2\left(l\left(\rho \right)\Delta u+l\left(\rho \right)\nabla \mathrm{div}u\right)\rangle \\ -\langle {\nabla }^{2}u,{\nabla }^2\left(f\left(\rho \right)\nabla \rho \right)\rangle -\langle {\nabla }^{2}u,{\nabla }^2\left(\frac{1}{1+\rho }\nabla d\cdot \Delta d\right)\rangle \\ :=K_1+K_2+K_3+K_4+K_5.\end{array}$ (23)

对于 $K_1$ ，由分部积分、Sobolev不等式和先验假设可知

$\begin{array}{c}{K}_1={\displaystyle {\int }_{{ℝ}^3}{\nabla }^2\rho {\nabla }^2\mathrm{div}\left(\rho u\right)\text{d}x}\\ \le C{\Vert {\nabla }^2\rho \Vert }_{L^2}\left[{\Vert {\nabla }^2\rho \nabla u\Vert }_{L^2}+{\Vert \nabla \rho {\nabla }^{2}u\Vert }_{L^2}+{\Vert \rho {\nabla }^{3}u\Vert }_{L^2}\right]\\ \le C{\Vert {\nabla }^2\rho \Vert }_{L^2}\left[{\Vert {\nabla }^2\rho \Vert }_{L^2}{\Vert \nabla u\Vert }_{L^{\infty }}+{\Vert \nabla \rho \Vert }_{L^3}{\Vert {\nabla }^{2}u\Vert }_{L^6}+{\Vert \rho \Vert }_{L^{\infty }}{\Vert {\nabla }^{3}u\Vert }_{L^2}\right]\\ \le C\delta \left({\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2\right),\end{array}$

$\begin{array}{c}{K}_2=\left|\langle {\nabla }^{2}u,{\nabla }^2(u\cdot \nabla u)\rangle \right|\\ ={\displaystyle {\int }_{{ℝ}^3}u}\cdot \nabla {\nabla }^{2}u{\nabla }^{2}u\text{d}x+{\displaystyle {\int }_{{ℝ}^3}\left[{\nabla }^2\left(u\cdot \nabla u\right)-u\cdot \nabla {\nabla }^{2}u\right]{\nabla }^{2}u\text{d}x}\\ \le C{\Vert {\nabla }^{2}u\Vert }_{L^2}^2{\Vert \text{div}u\Vert }_{L^{\infty }}+C{\Vert {\nabla }^{2}u\Vert }_{L^2}\left({\Vert {\nabla }^{2}u\Vert }_{L^2}{\Vert \nabla u\Vert }_{L^{\infty }}+{\Vert \nabla u\Vert }_{L^{\infty }}{\Vert {\nabla }^{2}u\Vert }_{L^2}\right)\\ \le C\delta \left({\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2\right).\end{array}$

和

$\begin{array}{c}{K}_3=-{\displaystyle {\int }_{{ℝ}^3}{\nabla }^{2}u\cdot {\nabla }^2\left(l\left(\rho \right)\Delta u+l\left(\rho \right)\nabla \mathrm{div}u\right)\text{d}x}\\ ={\displaystyle {\int }_{{ℝ}^3}{\nabla }^{3}u\cdot \left[l^{\prime }\left(\rho \right)\nabla \rho \Delta u+l\left(\rho \right)\nabla \Delta u\right]\text{d}x}+{\displaystyle {\int }_{{ℝ}^3}{\nabla }^{3}u\cdot \left[l^{\prime }\left(\rho \right)\nabla \rho \nabla \mathrm{div}u+l\left(\rho \right){\nabla }^2\mathrm{div}u\right]\text{d}x}\\ \le C\delta \left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2\right),\end{array}$

$K_4={\displaystyle {\int }_{{ℝ}^3}{\nabla }^{2}u\cdot {\nabla }^2\left(f\left(\rho \right)\nabla \rho \right)\text{d}x}\le C\delta \left({\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2\right).$

同样的，对于 $K_5$ ，

$\begin{array}{c}{K}_5=-{{\displaystyle {\int }_{{ℝ}^3}\nabla }}^{2}u\cdot {\nabla }^2\left(\frac{1}{1+\rho }\nabla d\cdot \Delta d\right)\text{d}x\\ ={{\displaystyle {\int }_{{ℝ}^3}\nabla }}^{3}u\cdot \left[\nabla \left(\frac{1}{1+\rho }\right)\left(\nabla d\cdot \Delta d\right)+\frac{1}{1+\rho }\nabla \left(\nabla d\cdot \Delta d\right)\right]\text{d}x\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2\right).\end{array}$

最后，综上所得，

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right)+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\text{div}u\Vert }_{L^2}^2\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2\right)+C\delta \left({\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right).\end{array}$ (24)

由引理4.1和引理4.2可得，我们需要给出 ${\nabla }^2\rho$ 的L²估计。由知道

$\nabla \rho =-\frac{1}{\rho +1}\nabla d\cdot \Delta d-u\cdot \nabla u+\frac{1}{\rho +1}\Delta u+\frac{1}{\rho +1}\nabla \mathrm{div}u-u_t,$ (25)

将 $\nabla$ 作用到上式，再对得到的等式乘以 ${\nabla }^2\rho$ ，然后在 $R^3$ 上积分可得

$\begin{array}{l}{\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\displaystyle \int {\nabla }^2\rho }\cdot \nabla u_t\text{d}x\\ =-{\displaystyle \int {\nabla }^2\rho }\cdot \nabla \left(u\cdot \nabla u\right)\text{d}x-{\displaystyle \int {\nabla }^2\rho }\cdot \nabla \left(\frac{1}{1+\rho }\nabla d\cdot \Delta d\right)\text{d}x\\ +{\displaystyle \int \left[{\nabla }^2\rho \cdot \nabla \left(\frac{1}{1+\rho }\Delta u\right)+{\nabla }^2\rho \cdot \nabla \left(\frac{1}{1+\rho }\nabla \mathrm{div}u\right)\right]\text{d}x}\\ :=Q_1+Q_2+Q_3.\end{array}$ (26)

利用Hölder不等式、Sobolev不等式(5)和(9)可得

$Q_1=-{\displaystyle \int {\nabla }^2\rho }\cdot \nabla \left(u\cdot \nabla u\right)\le C\delta \left({\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right),$

和

$Q_3={\displaystyle \int \left[{\nabla }^2\rho \cdot \nabla \left(\frac{1}{1+\rho }\Delta u\right)+{\nabla }^2\rho \cdot \nabla \left(\frac{1}{1+\rho }\nabla \mathrm{div}u\right)\right]\text{d}x}\le C\delta {\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2),$

类似地，有

$\begin{array}{c}{Q}_2=-{\displaystyle \int {\nabla }^2\rho }\cdot \nabla \left(\frac{1}{1+\rho }\nabla d\cdot \Delta d\right)\text{d}x\\ \le {\Vert {\nabla }^2\rho \Vert }_{L^2}{\displaystyle \int \left(\nabla \left(\frac{1}{1+\rho }\right)\nabla d\cdot \Delta d+\frac{1}{1+\rho }\nabla \left(\nabla d\cdot \Delta d\right)\right)\text{d}x}\\ \le {\Vert {\nabla }^2\rho \Vert }_{L^2}{\Vert \nabla \left(\frac{1}{1+\rho }\right)\Vert }_{L^6}{\Vert \nabla d\Vert }_{L^6}{\Vert \Delta d\Vert }_{L^6}+C\left({\Vert \Delta d\Vert }_{L^4}^2+{\Vert \nabla d\Vert }_{L^{\infty }}{\Vert \nabla \Delta d\Vert }_{L^2}\right)\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2\right).\end{array}$

下面对(25)左边的式子进行估计

$\begin{array}{c}{\displaystyle \int {\nabla }^2\rho }\cdot \nabla u_t=\frac{\text{d}}{\text{d}t}{\displaystyle \int \nabla }u\cdot {\nabla }^2\rho \text{d}x-{\displaystyle \int \nabla }u\cdot {\nabla }^2{\rho }_t\text{d}x\\ =\frac{\text{d}}{\text{d}t}{\displaystyle \int \nabla }u\cdot {\nabla }^2\rho \text{d}x-{\displaystyle \int {\left(\nabla \mathrm{div}u\right)}^2\text{d}x}-{\displaystyle \int \nabla }u{\nabla }^2\mathrm{div}\left(\rho u\right)\text{d}x\\ \ge \frac{\text{d}}{\text{d}t}{\displaystyle \int \nabla }u\cdot {\nabla }^2\rho \text{d}x-{\displaystyle \int {\left(\nabla \mathrm{div}u\right)}^2\text{d}x}-C\delta \left({\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2\right).\end{array}$

现在利用插值不等式，可以得到

$\begin{array}{l}\frac{\text{d}}{\text{d}t}\left\{{\Vert {\nabla }^2\left(\rho ,u\right)\Vert }_{L^2}^2-{\gamma }_2{\displaystyle \int \nabla }u\cdot {\nabla }^2\rho \text{d}x\right\}+\left({\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2\right)\\ \le C\left(\delta +{\delta }_1\right)\left({\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2\right).\end{array}$ (27)

通过利用 ${\gamma }_2$ 和 $\delta ,{\delta }_1$ 的小性、引理4.1和引理4.2以及(28)可得

$\frac{\text{d}}{\text{d}t}\left\{{\Vert \left(\rho ,u\right)\Vert }_{H^2}^2+{\Vert \nabla d\Vert }_{H^1}^2\right\}+{\Vert \nabla \rho \Vert }_{H^1}^2+{\Vert \left(\nabla u,\nabla d\right)\Vert }_{H^2}^2\le 0.$ (28)

最后对(28)关于时间t积分得到了定理3.1的结果。证毕。