可压缩液晶流方程解的衰减估计

Decay Estimation of the Solutions to the Compressible Nematic Liquid Crystal Flow Model

摘要: 本文主要研究了可压缩液晶流系统在ℝ3中柯西问题解的高阶导数的衰减估计。本文利用傅立叶变换和中低频分解的方法，完成了证明。

Abstract: This paper primarily studies the decay of higher-order derivatives of the solution to the Cauchy problem on the compressible nematic liquid crystal flow system inℝ3, The proof is accomplished by virtue of Fourier theory and a new observation for cancellation of alow-medium-frequency quantity.

1. 引言

我们下面考虑的系统满足下列方程：

$\{\begin{array}{l}{\rho }_t+\mathrm{div}\left(\rho u\right)=0,\hfill \\ {\left(\rho u\right)}_t+\mathrm{div}\left(\rho u\otimes u\right)+\nabla P\left(\rho \right)=ℒu-\nabla d\cdot \Delta d,\hfill \\ d_t+u\cdot \nabla d=\Delta d+{\left|\nabla d\right|}^{2}d,\hfill \end{array}$ (1.1)

模型的初始值为

$\rho \left(x,0\right)={\rho }_0\left(x\right),u\left(x,0\right)=u_0\left(x\right),d\left(x,0\right)=d_0\left(x\right),x\in {ℝ}^3,$ (1.2)

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$d=1$ 。$ℒ$ 为Lamé算子，

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

$\mu$ 和$\lambda$ 为剪切黏度和体积黏度，满足$\mu >0$ ，$2\mu +\text{3}\lambda \ge 0$ 。

这个系统被称为可压缩液晶流模型。液晶流被认为是缓慢运动的粒子，其中流体速度和粒子的排列相互影响。液晶可以形成并保持在液体和固体之间的中间状态。向列相是液晶中最常见的一种。Erickse [1] [2]和Leslie [3] [4]建立了液晶的流体力学理论。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，由于液晶流系统在数学和物理学的重要性，有非常多的文献研究了它的数学理论。不可压缩液晶流动的数学分析是由Lin和Liu在[5] [6]中提出的。对不可压缩的流体，我们可以参考文献[7]-[13]。

当流体被允许可压缩时，Ericksen-Leslie系统变得更加复杂。据我们所知，目前可用的分析著作似乎很少。Ding-Lin-Wang-Wen [11]和Huang-Wang-Wen [12]研究了具有非负初始密度的系统(1.1)的初值或初边值问题的局部时强解。基于[13] [14]，在[15] [16]中得到了强解的爆破判据。Hu和Wu [17]研究了临界空间强解的全局存在唯一性。在[18]的前提下，当初始数据在某能量范数上足够光滑且适当小时，在[19]中证明了经典解的全局适定性。特别是[20] [21] [22]，在一些较小的条件或几何角度条件下建立了全局弱解(见[23])。Wu [24]研究了高维(3维及以上)的小初值经典解的整体存在性。对于该模型解的局部存在性、整体存在性和正则性，许多学者做了一系列的研究，可以参考[25]-[30]。在定理3.1的假设下，本文致力于建立定理3.2中解的时间衰减率。通过傅立叶变换和低高频分解[31]的方法，完成了证明。

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$\Vert \cdot \Vert$ 。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。其证明可参考文献[32] [33]。

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

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

引理2.3 ([31])对于$f\left(x\right)\in H^m\left({ℝ}^3\right)$ 和任意给定的整数$k,k_0$ 和$k_1$ ($k_0\le k\le k_1\le m$ )，下列式子成立

${\Vert {\nabla }^k{f}^l\Vert }_{L^2}\le r_0^{k-k_0}{\Vert {\nabla }^{k_0}{f}^l\Vert }_{L^2},{\Vert {\nabla }^k{f}^l\Vert }_{L^2}\le {\Vert {\nabla }^{k_1}f\Vert }_{L^2},$ (2.1)

${\Vert {\nabla }^k{f}^h\Vert }_{L^2}\le \frac{1}{R_0^{k_1-k}}{\Vert {\nabla }^{k_1}{f}^h\Vert }_{L^2},{\Vert {\nabla }^k{f}^h\Vert }_{L^2}\le {\Vert {\nabla }^{k_1}f\Vert }_{L^2}.$ (2.2)

和

$r_0^k{\Vert f^m\Vert }_{L^2}\le {\Vert {\nabla }^k{f}^m\Vert }_{L^2}\le R_0^k{\Vert f^m\Vert }_{L^2}.$ (2.3)

3. 主要结果

通过对变量进行换元

$\tilde{\rho }=\rho -{\rho }_{\infty },\tilde{u}=u,\tilde{d}=d-d_{\infty },$

则方程可以写成

$\{\begin{array}{l}{\tilde{\rho }}_t+{\rho }_{\infty }\mathrm{div}\tilde{u}=S_1,\hfill \\ {\tilde{u}}_t+\frac{P_{\rho }\left({\rho }_{\infty }\right)}{{\rho }_{\infty }}\nabla \tilde{\rho }-\frac{\mu }{{\rho }_{\infty }}\Delta \tilde{u}-\frac{\mu +\lambda }{{\rho }_{\infty }}\nabla \mathrm{div}\tilde{u}=S_2,\hfill \\ {\tilde{d}}_t-\Delta \tilde{d}=S_3,\hfill \end{array}$ (3.1)

初值条件满足

$\left(\tilde{\rho },\tilde{u},\tilde{d}\right)\left(x,0\right)=\left({\tilde{\rho }}_0,{\tilde{u}}_0,{\tilde{d}}_0\right)\left(x\right).$ (3.2)

非线性项满足

$\{\begin{array}{l}{S}_1=-\mathrm{div}\left(\tilde{\rho }\tilde{u}\right),\\ S_2=-\tilde{u}\cdot \nabla \tilde{u}+g_1\left(\tilde{\rho }\right)\nabla \tilde{\rho }+\mu g_2\left(\tilde{\rho }\right)\Delta \tilde{u}+\left(\mu +\lambda \right)g_2\left(\tilde{\rho }\right)\nabla \mathrm{div}\tilde{u}-\frac{1}{\tilde{\rho }+{\rho }_{\infty }}\nabla \tilde{d}\cdot \Delta \tilde{d},\\ S_3=-\tilde{u}\cdot \nabla \tilde{d}+{\left|\nabla \tilde{d}\right|}^2\tilde{d}.\end{array}$ (3.3)

其中

$\{\begin{array}{l}{g}_1\left(\tilde{\rho }\right)=\frac{P_{\rho }\left({\rho }_{\infty }\right)}{{\rho }_{\infty }}-\frac{P_{\rho }\left(\tilde{\rho }+{\rho }_{\infty }\right)}{\tilde{\rho }+{\rho }_{\infty }},\\ g_2\left(\tilde{\rho }\right)=\frac{1}{\tilde{\rho }+{\rho }_{\infty }}-\frac{1}{{\rho }_{\infty }},\end{array}$

再做一次变量变换

$\tilde{\rho }\to \tilde{\rho },\tilde{u}\to \alpha u,\tilde{d}\to d,$

令$\alpha =\frac{{\rho }_{\infty }}{\sqrt{P_{\rho }\left({\rho }_{\infty }\right)}},{\lambda }_1=\sqrt{{\rho }_{\infty }},{\mu }_1=\frac{\mu }{{\rho }_{\infty }},{\mu }_2=\frac{\mu +\lambda }{{\rho }_{\infty }}$ ，

则方程可以写成

$\{\begin{array}{l}{\tilde{\rho }}_t+{\lambda }_1\mathrm{div}\tilde{u}={\tilde{S}}_1,\hfill \\ {\tilde{u}}_t+{\lambda }_1\nabla \tilde{\rho }-{\mu }_1\Delta \tilde{u}-{\mu }_2\nabla \mathrm{div}\tilde{u}={\tilde{S}}_2,\hfill \\ {\tilde{d}}_t-\Delta \tilde{d}={\tilde{S}}_3,\hfill \end{array}$ (3.4)

初值条件满足

$\begin{array}{l}{\left(\tilde{\rho },\tilde{u},\tilde{d}\right)|}_{t=0}=\left({\tilde{\rho }}_0,{\tilde{u}}_0,{\tilde{d}}_0\right)\left(x\right)\\ :=\left({\rho }_0-{\rho }_{\infty },u_0,d_0\right)\left(x\right)\to \left(0,0,0\right),\left|x\right|\to +\infty \end{array}$ (3.5)

其中源项

$\left({\tilde{S}}_1,{\tilde{S}}_2,{\tilde{S}}_3\right):=\left(S_1,\alpha S_2,S_3\right)\left(\tilde{\rho },\frac{1}{\alpha }\tilde{u},\tilde{d}\right).$ (3.6)

通过先验假设直接计算可以得到

$\left|g_1\left(\tilde{\rho }\right)\right|,\left|g_2\left(\tilde{\rho }\right)\right|\le C\left|\tilde{\rho }\right|,$ (3.7)

这些不等式在后面将会用到。

命题3.1 (局部存在性)假设$\left({\tilde{\rho }}_0,{\tilde{u}}_0\right)\in H^2\left({ℝ}^3\right)$ ，$\nabla {\tilde{d}}_0\in H^1\left({ℝ}^3\right)$ ，则存在一个大于0的常数$T_1$ ，使得系统(1.8)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

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

证明：利用标准迭代论证、不动点定理和极大值原理进行证明。也可以参考[34] [35]。为了表述简单，我们省略了细节。

命题3.2 (先验估计)由命题3.1可知，存在时间$T^{\prime }>0$ ，解存在于${ℝ}^3\times \left[0,T^{\prime }\right]$ 中。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，则存在一个足够小的数$\delta >0$ ，使得

$\underset{0\le t\le T}{\sup }\left\{{\Vert \left(\tilde{\rho },\tilde{u}\right)\left(\cdot ,t\right)\Vert }_{H^2}+{\Vert \nabla \tilde{d}\left(\cdot ,t\right)\Vert }_{H^1}\right\}\le \delta ,$ (3.8)

那么有以下的估计：

$\left\{{\Vert \left(\tilde{\rho },\tilde{u}\right)\Vert }_{H^2}^2+{\Vert \nabla \tilde{d}\Vert }_{H^1}^2\right\}+{\displaystyle {\int }_0^t\left[{\Vert \nabla \tilde{\rho }\Vert }_{H^1}^2+{\Vert \left(\nabla \tilde{u},\nabla \tilde{d}\right)\Vert }_{H^2}^2\right]}\left(\cdot ,s\right)\text{d}s\le C_1\left\{{\Vert \left({\tilde{\rho }}_0,{\tilde{u}}_0\right)\Vert }_{H^2}^2+{\Vert \nabla {\tilde{d}}_0\Vert }_{H^1}^2\right\}.$ (3.9)

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

${\Vert \left({\tilde{\rho }}_0,{\tilde{u}}_0\right)\Vert }_{H^2\left({ℝ}^3\right)}+{\Vert \nabla {\tilde{d}}_0\Vert }_{H^1\left({ℝ}^3\right)}\le \delta ,$

那么整体存在唯一的光滑解满足

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

在定理3.1的假设下，我们用下面的定理来陈述我们的结果。

定理3.2假设初始值$\left({\tilde{\rho }}_0,{\tilde{u}}_0\right)\in H^2\left({ℝ}^3\right)$ ，$\nabla {\tilde{d}}_0\in H^1\left({ℝ}^3\right)$ ，${\rho }_{\infty }>0$ ，此外，如果初始数据$\left({\tilde{\rho }}_0,{\tilde{u}}_0,\nabla {\tilde{d}}_0\right)$ 在$L^1\left(R^3\right)$ 中有界，则存在一个正常数$C_0$ ，当$t\ge 0$ 时，使得解$\left(\tilde{\rho },\tilde{u},\nabla \tilde{d}\right)$ 具有如下的时间衰减估计

${\Vert {\nabla }^k\left(\tilde{\rho },\tilde{u}\right)\left(t\right)\Vert }_{L^2\left({ℝ}^3\right)}\le C_0{\left(1+t\right)}_{}^{-\frac{3}{4}-\frac{k}{2}}\text{,}k=0,1,2$

${\Vert \nabla \tilde{d}\left(t\right)\Vert }_{H^1}\le {\tilde{C}}_0{\text{e}}^{-Ct}.$

首先，我们先给出高阶导的估计结果。

引理3.1 对于$0\le t\le T$ ，有

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\mathscr{H}}_h\left(t\right)+{\mu }_1{\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\mu }_2{\Vert {\nabla }^2\mathrm{div}u\Vert }_{L^2}^2+{\lambda }_1{\beta }_1{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\\ \le C\delta \left({\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2\right)\\ +{\lambda }_1{\beta }_1{\Vert \nabla \mathrm{div}u\Vert }_{L^2}^2+C\delta \left(1+{\beta }_1\right)\left({\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2\right).\end{array}$ (3.10)

其中${\mathscr{H}}_h\left(t\right)={\Vert {\nabla }^2\left(\rho ,u,d\right)\Vert }_{L^2}^2+2{\beta }_1{\displaystyle \int \nabla }u\cdot {\nabla }^2\rho \text{d}x$ 。

证明：将${\nabla }^2$ 分别作用于(3.4)₁和(3.4)₂，同时用${\nabla }^2\rho ,{\nabla }^{2}u$ 分别乘以(3.4)₁和(3.4)₂，在$R^3$ 上积分，并且利用基本不等式，得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left({\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2\right)+{\mu }_1{\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\mu }_2{\Vert {\nabla }^2\mathrm{div}u\Vert }_{L^2}^2\\ \le C\delta \left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2+{\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert {\nabla }^{3}u\Vert }_{L^2}^2\right).\end{array}$ (3.11)

将$\nabla$ 作用到(3.4)₂, 再对得到的等式乘以${\nabla }^2\tilde{\rho }$ ,利用Hölder不等式、Sobolev不等式，然后在$R^3$ 上积分可得

$\begin{array}{l}{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}^2+{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla \tilde{\rho }\cdot \nabla {\tilde{u}}_t\text{d}x\\ ={{\displaystyle {\int }_{{ℝ}^3}\nabla }}^2\tilde{\rho }\cdot \nabla S_2\text{d}x+{\mu }_{\text{1}}{{\displaystyle {\int }_{{ℝ}^3}\nabla }}^2\tilde{\rho }\cdot \nabla \Delta u\text{d}x+{\mu }_2{{\displaystyle {\int }_{{ℝ}^3}\nabla }}^2\tilde{\rho }\cdot \nabla \nabla \mathrm{div}u\text{d}x\\ \le C\delta \left({\Vert \nabla \tilde{d}\Vert }_{L^2}+{\Vert {\nabla }^2\left(\tilde{\rho },\tilde{u},d\right)\Vert }_{L^2}+{\Vert {\nabla }^3\tilde{u}\Vert }_{L^2}\right),\end{array}$ (3.12)

这里我们利用了

$\begin{array}{l}{\Vert \nabla S_2\Vert }_{L^2}\le C{\Vert \nabla \tilde{u}\Vert }_{L^6}{\Vert \nabla \tilde{u}\Vert }_{L^3}+C{\Vert \tilde{u}\Vert }_{L^{\infty }}{\Vert {\nabla }^2\tilde{u}\Vert }_{L^2}+C{\Vert \nabla g_1\left(\tilde{\rho }\right)\Vert }_{L^6}{\Vert \nabla \tilde{\rho }\Vert }_{L^3}\\ +C{\Vert g_1\left(\tilde{\rho }\right)\Vert }_{L^{\infty }}{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}+C{\Vert \nabla g_2\left(\tilde{\rho }\right)\Vert }_{L^6}{\Vert \Delta \tilde{u}\Vert }_{L^3}+C{\Vert g_2\left(\tilde{\rho }\right)\Vert }_{L^{\infty }}{\Vert {\nabla }^3\tilde{u}\Vert }_{L^2}\\ +C{\Vert \nabla g_3\left(\tilde{\rho }\right)\Vert }_{L^6}{\Vert {\nabla }^2\tilde{u}\Vert }_{L^3}+C{\Vert g_3\left(\tilde{\rho }\right)\Vert }_{L^{\infty }}{\Vert {\nabla }^3\tilde{u}\Vert }_{L^2}+{\Vert \frac{1}{\tilde{\rho }}\Vert }_{L^{\infty }}{\Vert \nabla \tilde{d}\Vert }_{L^2}{\Vert \Delta \tilde{d}\Vert }_{L^2}\\ \le C\delta \left({\Vert \nabla \tilde{d}\Vert }_{L^2}+{\Vert {\nabla }^2\left(\tilde{\rho },\tilde{u},d\right)\Vert }_{L^2}+{\Vert {\nabla }^3\tilde{u}\Vert }_{L^2}\right).\end{array}$ (3.13)

下面对(3.12)左边的式子进行估计，并乘以一个足够小的数${\beta }_1$

$\begin{array}{l}{\beta }_1{\lambda }_1{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}^2+\frac{\text{d}}{\text{d}t}{\beta }_1{\displaystyle {\int }_{{ℝ}^3}{\nabla }^2\tilde{\rho }\cdot \nabla u\text{d}x}\\ \le {\lambda }_1{\beta }_1{\displaystyle {\int }_{{ℝ}^3}{\left(\nabla \mathrm{div}u\right)}^2\text{d}x}+C\delta {\beta }_1\left({\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2\right),\end{array}$ (3.14)

将(3.11)和(3.14)相加，利用插值不等式，可以得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{{\Vert {\nabla }^2\left(\rho ,u\right)\Vert }_{L^2}^2+2{\beta }_1{\displaystyle \int \nabla }u\cdot {\nabla }^2\rho \text{d}x\right\}+{\mu }_1{\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\mu }_2{\Vert {\nabla }^2\mathrm{div}u\Vert }_{L^2}^2+{\beta }_1{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}^2\\ \le C\delta \left({\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2\right)\\ +{\lambda }_1{\beta }_1{\Vert \nabla \mathrm{div}u\Vert }_{L^2}^2+C\delta \left(1+{\beta }_1\right)\left({\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2\right).\end{array}$ (3.15)

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(3.4)₃，再对得到的等式乘以${\nabla }^{2}d$ 并在$R^3$ 上积分可得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert {\nabla }^{2}d\Vert }^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\\ =-{\displaystyle {\int }_{{ℝ}^3}{\nabla }^2}\left(u\cdot \nabla d\right)\cdot {\nabla }^{2}d\text{d}x+{\displaystyle {\int }_{{ℝ}^3}{\nabla }^2}\left({\left|\nabla d\right|}^{2}d\right)\cdot {\nabla }^{2}d\text{d}x\\ ={\displaystyle {\int }_{{ℝ}^3}\nabla }\left(u\cdot \nabla d\right)\cdot {\nabla }^{3}d\text{d}x-{\displaystyle {\int }_{{ℝ}^3}\nabla }\left({\left|\nabla d\right|}^{2}d\right)\cdot {\nabla }^{3}d\text{d}x\\ \le C\delta \left({\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\right).\end{array}$ (3.16)

将(3.15)和(3.16)相加，得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{{\Vert {\nabla }^2\left(\rho ,u,d\right)\Vert }_{L^2}^2+2{\beta }_1{\displaystyle \int \nabla }u\cdot {\nabla }^2\rho \text{d}x\right\}\\ +{\mu }_1{\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\mu }_2{\Vert {\nabla }^2\mathrm{div}u\Vert }_{L^2}^2+{\lambda }_1{\beta }_1{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\\ \le C\delta \left({\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2\right)\\ +{\lambda }_1{\beta }_1{\Vert \nabla \mathrm{div}u\Vert }_{L^2}^2+C\delta \left(1+{\beta }_1\right)\left({\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2\right).\end{array}$ (3.17)

证明结束。

4. 衰减估计

接下来，我们将给出解$\left(\rho ,u,d\right)$ 的衰减估计。

命题4.1在定理3.1和定理3.2的假设下，存在一个正常数C₀，当$t\ge 0$ 时，使得由命题3.1和命题3.2得到的整体解$\left(\tilde{\rho },\tilde{u},\nabla \tilde{d}\right)$ 具有如下的时间衰减估计

${\Vert {\nabla }^k\left(\tilde{\rho },\tilde{u}\right)\left(t\right)\Vert }_{L^2\left({ℝ}^3\right)}\le C_0{\left(1+t\right)}_{}^{-\frac{3}{4}-\frac{k}{2}}\text{,}k=0,1,2$

${\Vert \nabla \tilde{d}\left(t\right)\Vert }_{H^1}\le {\tilde{C}}_0{\text{e}}^{-Ct}.$

引理4.1有

$\begin{array}{l}{\Vert {\nabla }^2\left(\tilde{\rho },\tilde{u},\tilde{d}\right)\left(t\right)\Vert }_{L^2}^2\le C{\text{e}}^{-C_{2}t}{\Vert {\nabla }^2\left({\tilde{\rho }}_0,{\tilde{u}}_0,d_0\right)\Vert }_{L^2}^2+C\delta {\displaystyle {\int }_0^l{\text{e}}^{-C_2\left(t-\tau \right)}}{\Vert \nabla \tilde{d}\left(\tau \right)\Vert }_{H^1}^2\text{d}\tau \\ +C{\displaystyle {\int }_0^t{\text{e}}^{-C_2\left(t-\tau \right)}}{\Vert {\nabla }^2\left({\tilde{\rho }}^L,{\tilde{u}}^L\right)\left(\tau \right)\Vert }_{L^2}^2\text{d}\tau .\end{array}$ (4.1)

其中正常数C₂与$\delta$ 无关。

证明：将$\nabla$ (3.4)₂乘以$\nabla \nabla {\tilde{\rho }}^L$ ，并使用(3.4)₁和分部积分法，我们得到

${\lambda }_1\langle \nabla \nabla \tilde{\rho },\nabla \nabla {\tilde{\rho }}^L\rangle =\langle \nabla S_2,\nabla \nabla {\tilde{\rho }}^L\rangle +{\mu }_1\langle \nabla \Delta \tilde{u},\nabla \nabla {\tilde{\rho }}^L\rangle +{\mu }_2\langle {\nabla }^2\mathrm{div}\tilde{u},\nabla \nabla {\tilde{\rho }}^L\rangle -\langle \nabla {\tilde{u}}_t,\nabla \nabla {\tilde{\rho }}^L\rangle ,$

通过分部积分，可以得到

$\begin{array}{l}\langle \nabla {\tilde{u}}_t,\nabla \nabla {\tilde{\rho }}^L\rangle =\frac{\text{d}}{\text{d}t}\langle \nabla \tilde{u},\nabla \nabla {\tilde{\rho }}^L\rangle -\langle \nabla \tilde{u},\nabla \nabla {\tilde{\rho }}_t{}^L\rangle \\ =\frac{\text{d}}{\text{d}t}\langle \nabla \tilde{u},\nabla \nabla {\tilde{\rho }}^L\rangle -\langle \nabla \tilde{u},\nabla \nabla S_1{}^L\rangle +{\lambda }_1\langle \nabla \tilde{u},\nabla \nabla \mathrm{div}{\tilde{u}}^L\rangle ,\end{array}$

和

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x=-{\lambda }_1{\displaystyle {\int }_{{ℝ}^3}\nabla \nabla \tilde{\rho }\cdot \nabla \nabla {\tilde{\rho }}^L}\text{d}x+{\lambda }_1{\displaystyle {\int }_{{ℝ}^3}\nabla \mathrm{div}\tilde{u}}\nabla \mathrm{div}{\tilde{u}}^L\text{d}x\\ +{\mu }_1{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \Delta \tilde{u}\text{d}x+{\mu }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \nabla \mathrm{div}\tilde{u}\text{d}x\\ -{\displaystyle {\int }_{{ℝ}^3}\left(\nabla \mathrm{div}\tilde{u}\nabla S_1^L-\nabla \nabla {\tilde{\rho }}^L\cdot \nabla S_2\right)\text{d}x}.\end{array}$ (4.2)

然后，通过使用Young不等式，我们有

$\begin{array}{l}-\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x\le \frac{{\lambda }_1}{2}{\Vert \nabla \mathrm{div}\tilde{u}\Vert }_{L^2}^2+\frac{{\lambda }_1}{8}{\Vert \nabla \nabla \tilde{\rho }\Vert }_{L^2}^2+\frac{{\mu }_1}{2}{\Vert \nabla \Delta \tilde{u}\Vert }_{L^2}^2\\ +\frac{{\mu }_2}{2}{\Vert \nabla \nabla \mathrm{div}\tilde{u}\Vert }_{L^2}^2+\frac{1}{2}{\Vert \nabla S_1^L\Vert }_{L^2}^2+\frac{1}{2}{\Vert \nabla S_2\Vert }_{L^2}^2\\ +\left(1+{\lambda }_1+\frac{1+{\mu }_1+{\mu }_2}{2}\right){\Vert \nabla \nabla {\tilde{\rho }}^L\Vert }_{L^2}^2+\frac{1+{\lambda }_1}{2}{\Vert \nabla \mathrm{div}{\tilde{u}}^L\Vert }_{L^2}^2.\end{array}$ (4.3)

根据Plancherel定理，引理2.2和(3.13)我们有

${\Vert \nabla {\tilde{S}}_1^L\Vert }_{L^2}^2+{\Vert \nabla {\tilde{S}}_2\Vert }_{L^2}^2\le C\delta \left({\Vert \nabla \tilde{d}\Vert }_{L^2}+{\Vert {\nabla }^2\left(\tilde{\rho },\tilde{u},d\right)\Vert }_{L^2}+{\Vert {\nabla }^3\tilde{u}\Vert }_{L^2}\right),$ (4.4)

将(3.10)和${\beta }_2\times$ (4.3)相加，利用(3.13)和(2.2)，得到

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}\left\{{\mathscr{H}}_h\left(t\right)-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x\right\}+\frac{{\mu }_1}{4}{R}_0^2{\Vert {\nabla }^2{\tilde{u}}^h\Vert }_{L^2}^2+\frac{{\mu }_1}{4}{\Vert {\nabla }^3\tilde{u}\Vert }_{L^2}^2\\ +\frac{{\mu }_2}{4}{\Vert {\nabla }^2\mathrm{div}\tilde{u}\Vert }_{L^2}^2+\frac{{\beta }_1{\lambda }_1}{8}{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}^2+\frac{1}{4}{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\\ \le \frac{{\lambda }_1{\beta }_2}{2}{\Vert \nabla \mathrm{div}\tilde{u}\Vert }_{L^2}^2+\frac{{\lambda }_1{\beta }_2}{8}{\Vert \nabla \nabla \tilde{\rho }\Vert }_{L^2}^2+\frac{{\mu }_1{\beta }_2}{2}{\Vert \nabla \Delta \tilde{u}\Vert }_{L^2}^2\\ +\frac{{\mu }_2{\beta }_2}{2}{\Vert \nabla \nabla \mathrm{div}\tilde{u}\Vert }_{L^2}^2+\frac{C\delta {\beta }_2}{2}\left({\Vert \nabla \tilde{d}\Vert }_{L^2}+{\Vert {\nabla }^2\left(\tilde{\rho },\tilde{u},d\right)\Vert }_{L^2}+{\Vert {\nabla }^3\tilde{u}\Vert }_{L^2}\right)\end{array}$

$\begin{array}{l}+\left(1+{\lambda }_1+\frac{1+{\mu }_1+{\mu }_2}{2}\right){\beta }_2{\Vert \nabla \nabla {\tilde{\rho }}^L\Vert }_{L^2}^2+\frac{\left(1+{\lambda }_1\right){\beta }_2}{2}{\Vert \nabla \mathrm{div}{\tilde{u}}^L\Vert }_{L^2}^2\\ +C\delta \left({\Vert {\nabla }^{2}u\Vert }_{L^2}^2+{\Vert \nabla d\Vert }_{L^2}^2+{\Vert {\nabla }^{2}d\Vert }_{L^2}^2+{\Vert {\nabla }^{3}d\Vert }_{L^2}^2+{\Vert \nabla \rho \Vert }_{L^2}^2\right)\\ +{\lambda }_1{\beta }_1{\Vert \nabla \mathrm{div}u\Vert }_{L^2}^2+C\delta \left(1+{\beta }_1\right)\left({\Vert {\nabla }^{3}u\Vert }_{L^2}^2+{\Vert {\nabla }^2\rho \Vert }_{L^2}^2\right)\\ \le \left(\frac{{\mu }_2}{8}+\frac{{\lambda }_1\left({\beta }_1+{\beta }_2\right)}{2}\right){\Vert \nabla \mathrm{div}\tilde{u}\Vert }_{L^2}^2+\frac{{\beta }_2{\mu }_1}{2}{\Vert \nabla \Delta \tilde{u}\Vert }_{L^2}^2+\frac{{\beta }_2{\mu }_2}{2}{\Vert {\nabla }^2\mathrm{div}\tilde{u}\Vert }_{L^2}^2\\ +C{\beta }_2{\Vert \nabla \nabla {\tilde{\rho }}^L\Vert }_{L^2}^2+C{\beta }_2{\Vert \nabla \mathrm{div}{\tilde{u}}^L\Vert }_{L^2}^2+C\delta \left(1+{\beta }_1+{\beta }_2\right){\Vert {\nabla }^2\left(\tilde{\rho },\tilde{u},d\right)\Vert }_{L^2}^2.\end{array}$ (4.5)

在(4.5)的两边加$\frac{{\mu }_1}{4}{R}_0^2{\Vert {\nabla }^2{\tilde{u}}^L\Vert }_{L^2}^2$ ，使用分解式(5.2)，我们得到

$\begin{array}{l}\frac{\text{d}}{\text{d}t}\left({\mathscr{H}}_h\left(t\right)-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x\right)+\frac{{\beta }_2{\lambda }_1}{8}{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}^2\\ +\frac{{\mu }_1}{8}{R}_0^2{\Vert {\nabla }^2\tilde{u}\Vert }_{L^2}^2+\frac{{\mu }_1}{4}{\Vert {\nabla }^3\tilde{u}\Vert }_{L^2}^2+\frac{{\mu }_2}{4}{\Vert {\nabla }^2\mathrm{div}\tilde{u}\Vert }_{L^2}^2+\frac{1}{4}{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\\ \le C{\beta }_2{\Vert {\nabla }^2{\tilde{\rho }}^L\Vert }_{L^2}^2+\left(C{\beta }_2+\frac{{\mu }_1}{4}{R}_0^2\right){\Vert {\nabla }^2{\tilde{u}}^L\Vert }_{L^2}^2+\left(\frac{{\mu }_2}{8}+\frac{{\lambda }_1({\beta }_1+{\beta }_2)}{2}\right){\Vert \nabla \mathrm{div}\tilde{u}\Vert }_{L^2}^2\\ +\frac{{\beta }_2{\mu }_1}{2}{\Vert \nabla \Delta \tilde{u}\Vert }_{L^2}^2+\frac{{\beta }_2{\mu }_2}{2}{\Vert {\nabla }^2\mathrm{div}\tilde{u}\Vert }_{L^2}^2+C\delta \left(1+{\beta }_1+{\beta }_2\right){\Vert {\nabla }^2\left(\tilde{\rho },\tilde{u},d\right)\Vert }_{L^2}^2.\end{array}$ (4.6)

注意到

${\beta }_2<\max \left\{\frac{1}{4}\right\}\text{, }{R}_0^2>\max \left\{\frac{4\left({\mu }_2+{\lambda }_1\right)}{{\mu }_1}\right\},$ (4.7)

利用$\delta$ 的小性，我们有

$\begin{array}{l}\frac{\text{d}}{\text{d}t}\left({\mathscr{H}}_h(t)-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x\right)+\frac{{\beta }_2{\lambda }_1}{16}{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}^2+\frac{{\mu }_1}{16}{R}_0^2{\Vert {\nabla }^2\tilde{u}\Vert }_{L^2}^2\\ +\frac{{\mu }_1}{8}{\Vert {\nabla }^3\tilde{u}\Vert }_{L^2}^2+\frac{{\mu }_2}{8}{\Vert {\nabla }^2\mathrm{div}\tilde{u}\Vert }_{L^2}^2+\frac{1}{4}{\Vert {\nabla }^{3}d\Vert }_{L^2}^2\\ \le C{\Vert {\nabla }^2\left({\tilde{\rho }}^L,{\tilde{u}}^L\right)\Vert }_{L^2}^2+C\delta {\Vert \nabla \tilde{d}\Vert }_{H^1}^2.\end{array}$ (4.8)

根据分解(5.2)，我们有

$\begin{array}{l}{\mathscr{H}}_h\left(t\right)-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x\\ =\frac{1}{2}{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}+{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^h\cdot \nabla \tilde{u}\text{d}x+\frac{1}{2}{\Vert {\nabla }^2\tilde{u}\Vert }_{L^2}^2+\frac{1}{2}{\Vert {\nabla }^2\tilde{d}\Vert }_{L^2}^2.\end{array}$

由分部积分法、Young不等式和(2.2)可知

$\begin{array}{l}{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^h\cdot \nabla \tilde{u}\text{d}x=-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }{\tilde{\rho }}^h\nabla \mathrm{div}\tilde{u}\text{d}x\\ \le \frac{{\beta }_2}{2}{\Vert \nabla {\tilde{\rho }}^h\Vert }_{L^2}^2+\frac{{\beta }_2}{2}{\Vert \nabla \mathrm{div}\tilde{u}\Vert }_{L^2}^2\\ \le \frac{{\beta }_2}{2}{\Vert {\nabla }^2\tilde{\rho }\Vert }_{L^2}^2+\frac{{\beta }_2}{2}{\Vert {\nabla }^2\tilde{u}\Vert }_{L^2}^2.\end{array}$

这意味着

${\mathscr{H}}_h\left(t\right)-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x~{\Vert {\nabla }^2\left(\tilde{\rho },\tilde{u},\tilde{d}\right)\Vert }_{L^2}^2,$ (4.9)

我们用到了$0<{\beta }_2<\frac{1}{4}$ ，由式(4.8)和式(4.9)可知，存在一个常数C₂，使得

$\begin{array}{l}\frac{\text{d}}{\text{d}t}\left({\mathscr{H}}_h\left(t\right)-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x\right)+C_2\left({\mathscr{H}}_h\left(t\right)-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x\right)\\ \le C{\Vert {\nabla }^2\left({\tilde{\rho }}^L,{\tilde{u}}^L\right)\Vert }_{L^2}^2+C\delta {\Vert \nabla \tilde{d}\Vert }_{H^1}^2.\end{array}$ (4.10)

用(4.10)乘以${\text{e}}^{C_{2}t}$ ，对$\left[0,t\right]$ 积分，得到

$\begin{array}{l}{\mathscr{H}}_h\left(t\right)-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}^L\cdot \nabla \tilde{u}\text{d}x\le {\text{e}}^{-C_{2}t}\left({\mathscr{H}}_h\left(0\right)-{\beta }_2{\displaystyle {\int }_{{ℝ}^3}\nabla }\nabla {\tilde{\rho }}_0^L\cdot \nabla {\tilde{u}}_0\text{d}x\right)\\ +C{\displaystyle {\int }_0^t{\text{e}}^{-C_2\left(t-\tau \right)}}{\Vert {\nabla }^2\left({\tilde{\rho }}^L,{\tilde{u}}^L\right)\left(\tau \right)\Vert }_{L^2}^2\text{d}\tau \\ +C\delta {\displaystyle {\int }_0^t{\text{e}}^{-C_2\left(t-\tau \right)}}{\Vert \nabla \tilde{d}\Vert }_{H^1}^2\text{d}\tau .\end{array}$ (4.11)

因此我们可以得到引理4.1。

4.1. 中低频部分的衰减估计

在本小节中，设$\mathbb{G}$ 是以下形式的微分算子的矩阵：

$\mathbb{G}=\left(\begin{array}{ccc}0& {\lambda }_1\mathrm{div}& 0\\ {\lambda }_1\nabla & -{\mu }_1\Delta -{\mu }_2\nabla \mathrm{div}& 0\\ 0& 0& -\Delta \end{array}\right)$ (4.12)

和

$\overline{\mathbb{U}}\left(t\right):={\left(\overline{\rho }\left(t\right),\overline{u}\left(t\right),\overline{d}\left(t\right)\right)}^{\text{T}}\text{,}\mathbb{U}\left(0\right)={\left({\tilde{\rho }}_0,{\tilde{u}}_0,{\tilde{d}}_0\right)}^{\text{T}}$ (4.13)

我们有以下相应的线性化问题：

$\{\begin{array}{l}{\partial }_t\overline{\mathbb{U}}+\mathbb{G}\overline{\mathbb{U}}=0\text{, }t>0\hfill \\ {\overline{\mathbb{U}}|}_{t=0}=\mathbb{U}\left(0\right)\hfill \end{array}$ (4.14)

将傅立叶变换运用到(4.14) 中x变量，求解关于t的常微分方程，我们得到

$\overline{\mathbb{U}}\left(t\right)=\mathcal{G}\left(t\right)\mathbb{U}\left(0\right),$

式中半群$\mathcal{G}\left(t\right)={\text{e}}^{-t\mathbb{G}}\left(t\ge 0\right)$ 是由线性算子$\mathbb{G}$ 生成的，并且

$\mathcal{G}\left(t\right)f:={ℱ}^{-1}\left({\text{e}}^{-t{\mathbb{G}}_{\xi }}\widehat{f}\left(\xi \right)\right),$

其中

${\mathbb{G}}_{\xi }=\left(\begin{array}{ccc}0& \text{i}{\xi }^{\text{T}}& 0\\ \text{i}\xi & {\mu }_1{\left|\xi \right|}^2{\delta }_{ij}+{\mu }_2{\xi }_i{\xi }_j& 0\\ 0& 0& {\left|\xi \right|}^2\end{array}\right).$

接下来，我们引入一个重要引理。

引理4.2 ([31])假设$1\le p\le 2$ ，对于任何整数k，有

${\Vert {\nabla }^k\left(\mathcal{G}\left(t\right){\mathbb{U}}^L\left(0\right)\right)\Vert }_{L^2}\le C{\left(1+t\right)}_{}^{-\frac{3}{2}\left(\frac{1}{p}-\frac{1}{2}\right)-\frac{k}{2}}{\Vert \mathbb{U}\left(0\right)\Vert }_{L^p}.$ (4.15)

首先，我们建立非线性问题(3.4)和(3.6)的解的时间衰减估计。表示

$\mathbb{U}\left(t\right):={\left(\tilde{\rho }\left(t\right),\tilde{u}\left(t\right),\tilde{d}\left(t\right)\right)}^{\text{T}}.$

由(3.4)可以得到

$\{\begin{array}{l}{\partial }_t\mathbb{U}+\mathbb{G}\mathbb{U}=S\left(\mathbb{U}\right),t>0\hfill \\ {\mathbb{U}|}_{t=0}=\mathbb{U}\left(0\right),\hfill \end{array}$ (4.16)

其中$S\left(\mathbb{U}\right)={\left({\tilde{S}}_1,{\tilde{S}}_2,{\tilde{S}}_3\right)}^{\text{T}}$，根据Duhamel原理，我们将式(4.16)的解改写为：

$\mathbb{U}\left(t\right)=\mathcal{G}\left(t\right)\mathbb{U}\left(0\right)+{\displaystyle {\int }_0^t\mathcal{G}}\left(t-\tau \right)S\left(\mathbb{U}\right)\left(\tau \right)\text{d}\tau .$ (4.17)

基于引理4.2，我们建立了非线性问题(3.4)~(3.6)解的中低频部分的时间衰减估计。

引理4.3 ([31])假设$1\le p\le 2$ ，对于任何整数k，存在一个正常数C₃，使得

$\begin{array}{l}{\Vert {\nabla }^k{\mathbb{U}}^L\left(t\right)\Vert }_{L^2}\le C_3{\left(1+t\right)}_{}^{-\frac{3}{4}-\frac{k}{2}}{\Vert \mathbb{U}\left(0\right)\Vert }_{L^1}+C_3{\displaystyle {\int }_0^{\frac{t}{2}}{\left(1+t-\tau \right)}_{}^{-\frac{3}{4}-\frac{k}{2}}}{\Vert S\left(\mathbb{U}\right)\left(\tau \right)\Vert }_{L^1}\text{d}\tau \\ +C_3{\displaystyle {\int }_{\frac{t}{2}}^t{\left(1+t-\tau \right)}_{}^{-\frac{k}{2}}{}^{}}{\Vert S\left(\mathbb{U}\right)\left(\tau \right)\Vert }_{L^2}\text{d}\tau .\end{array}$ (4.18)

4.2. 非线性系统的衰减率

在本节中，结合引理3.1和引理4.3，我们得到了非线性问题(3.4)解的时间衰减率。

引理4.4 在命题3.1和3.2的假设下，有

${\Vert {\nabla }^k\left(\tilde{\rho },\tilde{u}\right)\left(t\right)\Vert }_{L^2}\le C{\left(1+t\right)}_{}^{-\frac{3}{4}-\frac{k}{2}},k=0,1,2$ (4.19)

${\Vert \nabla \tilde{d}\left(t\right)\Vert }_{H^1}\le C{\text{e}}^{-Ct}.$ (4.20)

证明：对(3.4)作用k阶导数$\left(0\le k\le 2\right)$ ，在$R^3$ 上积分之后相加，并且利用$\delta$ 的小性可以得到

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \nabla \tilde{d}\Vert }_{H^1}^2+{\Vert \nabla \tilde{d}\Vert }_{H^2}^2\le 0.$ (4.21)

将(4.21)与${\text{e}}^{\frac{c}{2}t}$ 相乘，然后在$\left[0,t\right]$ 积分，得到

${\Vert \nabla \tilde{d}\left(t\right)\Vert }_{H^1}^2\le {\text{e}}^{-\frac{c}{2}t}{\Vert \nabla {\tilde{d}}_0\Vert }_{H^2}^2$ (4.22)

因此，我们可以得到(4.20)。

定义

$M\left(t\right):=\underset{0\le \tau \le t}{\sup }{\displaystyle \sum _{m=0}^2{\left(1+\tau \right)}_{}^{\frac{3}{4}+\frac{m}{2}}}{\Vert {\nabla }^m\left(\tilde{\rho },\tilde{u}\right)\left(\tau \right)\Vert }_{L^2}$ (4.23)

$M\left(t\right)$ 是非递减的，对于$0\le m\le 2$ ，

${\Vert {\nabla }^m\left(\tilde{\rho },\tilde{u}\right)\left(\tau \right)\Vert }_{L^2}\le C_4{\left(1+\tau \right)}_{}^{-\frac{3}{4}-\frac{m}{2}}M\left(t\right),0\le \tau \le t$ (4.24)

C₄是与$\delta$ 无关的正常数。利用Hölder不等式，(4.24)和先验假设，我们有

${\Vert S\left(\mathbb{U}\right)\left(\tau \right)\Vert }_{L^1}\le \delta M\left(t\right){\left(1+\tau \right)}_{}^{-\frac{5}{4}}+\delta {\Vert \nabla {\tilde{d}}_0\Vert }_{H^2}{\text{e}}^{-\frac{c}{4}\tau }.$ (4.25)

和

${\Vert S\left(\mathbb{U}\right)\left(\tau \right)\Vert }_{L^2}\lesssim {\delta }^{1-s_1}{M}^{1+s_1}\left(t\right){\left(1+\tau \right)}_{}^{-\frac{7}{4}-\frac{3}{4}{s}_1}+\delta {\Vert \nabla {\tilde{d}}_0\Vert }_{H^2}{\text{e}}^{-\frac{c}{4}\tau }.$ (4.26)

其中，$s_1\in \left(0,\frac{1}{2}\right)$ 是一个小的固定常数。根据引理4.4，(4.25)和(4.26)，对于$0\le k\le 2$ ，可以得到

(4.27)

由式(4.1)和(4.27)，我们得到

$\begin{array}{c}{\Vert {\nabla }^2\mathbb{U}\left(t\right)\Vert }_{L^2}^2\le C{\text{e}}^{-C_{2}t}{\Vert {\nabla }^2\mathbb{U}\left(0\right)\Vert }_{L^2}^2+C\delta {\displaystyle {\int }_0^t{\text{e}}^{-C_2\left(t-\tau \right)}}{\Vert \nabla \tilde{d}\left(\tau \right)\Vert }_{H^1}^2\text{d}\tau \\ +C\left({\Vert \mathbb{U}\left(0\right)\Vert }_{L^1}^2+{\Vert \nabla {\tilde{d}}_0\Vert }_{H^2}^2+{\delta }^2{M}^2\left(t\right)+{\delta }^{2-2s_1}M{\left(t\right)}^{2+2s_1}\right){\displaystyle {\int }_0^t{\text{e}}^{-C_2\left(t-\tau \right)}}{\left(1+\tau \right)}_{}^{-\frac{7}{2}}\text{d}\tau .\end{array}$ (4.28)

将(4.22)代入式(4.28)，得

${\Vert {\nabla }^2\mathbb{U}\left(t\right)\Vert }_{L^2}^2\le C\left({\Vert \mathbb{U}\left(0\right)\Vert }_{H^2\cap L^1}^2+{\delta }^2{M}^2\left(t\right)+{\delta }^{2-2s_1}{M}^{2+2s_1}\left(t\right)\right){\left(1+t\right)}_{}^{-\frac{7}{2}}$ (4.29)

此外，利用分解式(5.2)和引理2.3，我们得到

${\Vert {\nabla }^k\mathbb{U}\left(t\right)\Vert }_{L^2}^2\le C{\Vert {\nabla }^k{\mathbb{U}}^L\left(t\right)\Vert }_{L^2}^2+C{\Vert {\nabla }^k{\mathbb{U}}^h\left(t\right)\Vert }_{L^2}^2\le C{\Vert {\nabla }^k{\mathbb{U}}^L\Vert }_{L^2}^2+C{\Vert {\nabla }^2\mathbb{U}\Vert }_{L^2}^2.$ (4.30)

由(4.27)，(4.29)和(4.30)得到

${\Vert {\nabla }^k\mathbb{U}\left(t\right)\Vert }_{L^2}^2\le C\left({\Vert \mathbb{U}\left(0\right)\Vert }_{H^2\cap L^1}^2+{\delta }^2{M}^2\left(t\right)+{\delta }^{2-2s_1}{M}^{2+2s_1}\left(t\right)\right){\left(1+t\right)}_{}^{-\frac{3}{2}-k}$ (4.31)

由M(t)的定义并利用式(4.31)中δ的小性，存在一个独立于δ的正常数C₅，使得

$M^2\left(t\right)\le \frac{C_5}{2}\left\{{\Vert \left(\tilde{\rho },\tilde{u},\tilde{d}\right)\left(0\right)\Vert }_{H^2\cap L^1}^2+{\delta }^2{M}^2\left(t\right)+{\delta }^{2-2s_1}M{\left(t\right)}^{2+2s_1}\right\}$ (4.32)

利用Young不等式，我们得到

$C_5{\delta }^{2-2s_1}M{\left(t\right)}^{2+2s_1}\le \frac{1-s_1}{2}{C}_5^{\frac{2}{1-s_1}}+\frac{1+s_1}{2}{\delta }^{\frac{4\left(1-s_1\right)}{1+s_1}}{M}^4\left(t\right).$ (4.33)

定义$K_0:=C_5{\Vert \left(\tilde{\rho },\tilde{u},\tilde{d}\right)\left(0\right)\Vert }_{H^2\cap L^1}^2+\frac{1-s_1}{2}{C}_5^{\frac{2}{1-s_1}}$ 和$C_{\delta }:=\frac{1+s_1}{2}{\delta }^{\frac{4\left(1-s_1\right)}{1+s_1}}$ 。由式(4.32)和δ的小性，我们得到

$M^2\left(t\right)\le K_0+C_{\delta }{M}^4\left(t\right).$ (4.34)

其中$M\left(t\right)\le C$ ，假设$M^2\left(t\right)>2K_0$ ，其中$t\in \left[\overline{t},+\infty \right)$ ，$\overline{t}>0$ 。因为$M^2\left(0\right)={\Vert \left({\tilde{\rho }}_0,{\tilde{u}}_0,{\tilde{d}}_0\right)\Vert }_{H^2}$ 很小，并且$M\left(t\right)\in C^0\left[0,+\infty \right)$ ，存在$t_0\in \left(0,\overline{t}\right)$ 使得$M^2\left(t_0\right)=2K_0$ 。

由(4.34)可以得到

$M^2\left(t_0\right)\le K_0+C_{\delta }{M}^4\left(t_0\right).$

通过直接计算，有

$M^2\left(t_0\right)\le \frac{K_0}{1-C_{\delta }{M}^2\left(t_0\right)},$ (4.35)

设δ是一个很小的常数，使得$C_{\delta }<\frac{1}{4K_0}$ ，即

$C_{\delta }{M}^2\left(t_0\right)<\frac{1}{2},$ (4.36)

然后，由式(4.35)得到$M^2\left(t_0\right)<2K_0$ 。这就与假设$M^2\left(t_0\right)=2K_0$ 相矛盾了。因此，对于任意$t\in \left[\overline{t},+\infty \right)$ 我们有$M^2\left(t\right)\le 2K_0$ ，注意M(t)是非递减的。对于任意$t\in \left[0,+\infty \right)$ 我们有$M\left(t\right)\le C$ 。由式(4.23)中M(t)的定义，我们证明了式(4.19)。

附录

我们介绍关于高低频分解的一些相关知识。

在傅立叶变换的基础上，建立函数$f\left(x\right)\in L^2\left({ℝ}^3\right)$ 的频率分$\left(f^l\left(x\right),f^m\left(x\right),f^h\left(x\right)\right)$ ，如下所示：

$\begin{array}{l}{f}^l\left(x\right)={\chi }_0\left(D_x\right)f\left(x\right),\\ f^m\left(x\right)=\left(I-{\chi }_0\left(D_x\right)-{\chi }_1\left(D_x\right)\right)f\left(x\right),\\ f^h\left(x\right)={\chi }_1\left(D_x\right)f\left(x\right).\end{array}$ (5.1)

这里${\chi }_0\left(D_x\right)$ ，${\chi }_1\left(D_x\right)$ ，$D_x=\frac{1}{\sqrt{-1}}\nabla =\frac{1}{\sqrt{-1}}\left({\partial }_{x_1},{\partial }_{x_2},{\partial }_{x_3}\right)$ 分别为带有特征$\chi \left(\xi \right)$ 和$1-\chi \left(\xi \right)$ 的伪微分算子。这里$\chi \left(\xi \right)$ 为满足$0\le \chi \left(\xi \right)\le 1$ 的光滑截断函数，满足

$0\le {\chi }_0\left(\xi \right),{\chi }_1\left(\xi \right)\le 1\left(\xi \in {ℝ}^3\right)$

和

${\chi }_0\left(\xi \right)=\{\begin{array}{ll}1,\hfill & \left|\xi \right|<\frac{r_0}{2},\hfill \\ 0,\hfill & \left|\xi \right|>r_0,\hfill \end{array}{\chi }_1\left(\xi \right)=\{\begin{array}{ll}0,\hfill & \left|\xi \right|<R_0,\hfill \\ 1,\hfill & \left|\xi \right|>R_0+1,\hfill \end{array}$

对于固定常数$r_0$ 和$R_0$ 满足

$0<r_0\le \min \left\{\frac{1}{2}\sqrt{\frac{{\lambda }_1}{\nu }},\sqrt{\frac{{\gamma }_1{\left({\gamma }_1-d_0\right)}^2}{2d_0}},\frac{1}{2}\right\}$

和

$R_0>\max \left\{2\sqrt{\frac{{\mu }_2+{\lambda }_1}{{\mu }_1}},2\sqrt{\frac{2\left({\gamma }_1+{\lambda }_2\right)}{{\gamma }_0}},1\right\}.$

易得

$\begin{array}{l}f\left(x\right)=f^l\left(x\right)+f^m\left(x\right)+f^h\left(x\right)\\ =:f^L\left(x\right)+f^h\left(x\right)\\ =:f^l\left(x\right)+f^H\left(x\right),\end{array}$ (5.2)

我们定义

$f^L\left(x\right)=f^l\left(x\right)+f^m\left(x\right)\text{,}{f}^H\left(x\right)=f^m\left(x\right)+f^h\left(x\right)\text{.}$ (5.3)