数学物理学报, 2024, 44(6): 1445-1475

变粘可压缩轴对称 Navier-Stokes 方程组全局强解的存在性

龚思梦,2, 张学耀,1,*, 郭真华,1,2

1西北大学数学学院, 非线性科学研究中心 西安 710127

2广西大 学数学与信息科学学院 南宁 530004

The Existence of Global Strong Solution to the Compressible Axisymmetric Navier-Stokes Equations with Density-Dependent Viscosities

Gong Simeng,2, Zhang Xueyao,1,*, Guo Zhenhua,1,2

1School of Mathematics and CNS, Northwest University, Xi'an 710127

2School of Mathematics and Information Science, Guangxi University, Nan'ning 530004

通讯作者: *张学耀,Email：xyzhang05@163.com

收稿日期: 2024-01-9   修回日期: 2024-07-31  

基金资助: 国家自然科学基金(11931013)
广西自然科学基金(2022GXNSFDA035078)

Received: 2024-01-9   Revised: 2024-07-31  

Fund supported: NSFC(11931013)
GXNSF(2022GXNSFDA035078)

作者简介 About authors

龚思梦,Email：1095522546@qq.com;

郭真华,Email：zhguo@gxu.edu.cn

摘要

该文考虑三维空间中粘性依赖密度的可压缩 Navier-Stokes 方程组, 得到了具有小能量大振荡初值的全局轴对称强解的存在唯一性, 其中流体区域为周期域 $\Omega=\{(r,z)\vert r=\sqrt{x^2+y^2},(x,y,z)\in\mathbb{R}^3,r\in I\subset(0,+\infty),z\in(-\infty,+\infty)\}$. 当 $z\rightarrow\pm\infty$ 时, 初始密度保持非真空状态.结果还表明,只要初始密度远离真空, 解在任何时间内都不会发展成真空状态; 并且该文给出了解的精确的衰减速率.

关键词： Navier-Stokes 方程组; 轴对称; 粘性依赖密度; 强解

Abstract

In this paper, we consider the compressible Navier-Stokes equations with viscous-dependent density in 3D space, and obtain a global axisymmetric strong solution with small energy and large initial oscillations in a periodic domain $\Omega=\{(r,z)\vert r=\sqrt{x^2+y^2},(x,y,z)\in\mathbb{R}^3,r\in I\subset(0,+\infty),z\in(-\infty,+\infty)\}$. When $z\rightarrow\pm\infty$, the initial density remains in a non-vacuum state. The results also show that as long as the initial density is far away from the vacuum, the solution will not develop the vacuum state in any time. And the exact decay rates of the solution is obtained.

Keywords： Navier-Stokes equations; Axisymmetric; Density-dependent; Strong solution

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本文引用格式

龚思梦, 张学耀, 郭真华. 变粘可压缩轴对称 Navier-Stokes 方程组全局强解的存在性[J]. 数学物理学报, 2024, 44(6): 1445-1475

Gong Simeng, Zhang Xueyao, Guo Zhenhua. The Existence of Global Strong Solution to the Compressible Axisymmetric Navier-Stokes Equations with Density-Dependent Viscosities[J]. Acta Mathematica Scientia, 2024, 44(6): 1445-1475

1 引言

考虑 $(x,y,z)\in \mathbb{R}^3$ 中的粘性依赖于密度的可压等熵 Navier-Stokes 方程组如下

$\begin{equation} \left\{\begin{array}{cl} &\rho_{t}+\mathrm{div}(\rho U)=0,\\ &(\rho U)_{t}+\mathrm{div}(\rho U\otimes U)+\nabla P-\mathrm{div}(\mu(\rho)D(U))-\nabla (\lambda(\rho)\mathrm{div}U)=0,\\ \end{array}\right. \end{equation}$

其中 $\rho>0,U=(U_1,U_2,U_3),P(\rho)=\rho^\gamma (\gamma>1)$ 分别代表流体的密度, 速度和压力. $D(U)=\frac{\nabla U+{\nabla U}^T}{2}$ 是形变张量. $\mu(\rho)=a\rho^{\delta_1}$ 是剪切粘度, $\lambda(\rho)=b\rho^{\delta_2}$ 是体积粘度, $a$ 和 $b$ 都是常数并且满足

$\begin{equation} a>0, 2a\rho^{\delta_1}+3b\rho^{\delta_2}\geq0. \end{equation}$

考虑轴对称初值

$\begin{equation} \rho|_{t=0}=\rho_0(r,z),\ U|_{t=0}=(\frac{x}{r}u_{1 0}(r,z),\frac{y}{r}u_{1 0}(r,z),u_{2 0}(r,z)), \end{equation}$

其中 $r=\sqrt{x^2+y^2}$.

用轴对称变换

$\begin{array}{c} \rho(x, y, z, t)=\rho(r, z, t) \\ U(x, y, z, t)=\left(U_{1}, U_{2}, U_{3}\right)=\left(\frac{x}{r} u_{1}(r, z, t), \frac{y}{r} u_{1}(r, z, t), u_{2}(r, z, t)\right), \end{array}$

将方程化成轴对称的形式如下

$\begin{equation} \left\{\begin{array}{cl} & \partial_t\rho+\frac{1}{r}\partial_r(r\rho u_1)+\partial_z(\rho u_2)=0,\\[2mm] & \partial_t(\rho u_1)+\frac{1}{r}\partial_r(r\rho u_1^2)+\partial_z(\rho u_1u_2)+\partial_rP\\[2mm] & =\partial_r(\mu(\rho)\partial_ru_1)+\mu(\rho)\partial_r(\frac{u_1}{r})+\partial_z(\mu(\rho)\partial_zu_1)+\partial_r(\lambda(\rho)(\frac{u_1}{r}+\partial_ru_1+\partial_zu_2)),\\[2mm] & \partial_t(\rho u_2)+\frac{1}{r}\partial_r(r\rho u_1u_2)+\partial_z(\rho u_2^2)+\partial_zP\\[2mm] & =\partial_r(\mu(\rho)\partial_ru_2)+\mu(\rho)(\frac{\partial_ru_2}{r})+\partial_z(\mu(\rho)\partial_zu_2)+\partial_z(\lambda(\rho)(\frac{u_1}{r}+\partial_ru_1+\partial_zu_2)), \end{array}\right. \end{equation}$

且有初边值条件

$\begin{equation} (\rho,\rho u)(r,z,0)=(\rho_0,m_0)(r,z)\rightarrow(\widetilde{\rho},0), \vert z\vert\rightarrow\infty. \end{equation}$

引入符号: $\Delta=\partial_r^2+\partial_z^2,\nabla=(\partial_r, \partial_z),\mathrm{div}=\nabla\cdot,\dot{u}=u_t+u\cdot\nabla u,u=(u_1,u_2)$, 则方程可以改写成如下形式

$\begin{equation} \left\{\begin{array}{cl} &(\rho r)_{t}+\mathrm{div}(\rho ur)=0,\\ &\rho \dot {u}+\nabla P =\mathrm{div}(\mu(\rho)\nabla u)+\nabla (\lambda(\rho)\mathrm{div}u)+Q, \end{array}\right. \end{equation}$

其中 $ Q=(\mu(\rho)\partial_r(\frac{u_1}{r})+\partial_r(\lambda(\rho)\frac{u_1}{r}), \frac{1}{r}\mu(\rho)\partial_ru_2+\partial_z(\lambda(\rho)\frac{u_1}{r})).$

在方程 $(1.1)_2$ 中 $\delta_1=\delta_2=0$ 的情况下, 可压缩 Navier-Stokes 方程组 (下称CNS) 的全局适定性得到了广泛的研究. 特别是一维的理论是比较完善的, 见文献 [1⇓⇓-4] 和其中的参考文献. 在多维情况下, Nash[5], Itaya[6] 和 Tani[7]在无真空条件下建立了初值问题和初边值问题经典解的局部适定性理论. Cho 等[8] 和 Luo[9] 分别在三维和二维情况下研究了真空下的强解和经典解的局部适定性. 其中一个重要问题是, 这些对于局部解的解决方法能否扩展到全局. 沿着这条思路的第一个开创性工作是 Matsumura 等[10] 的结果, 基于 CNS 耗散结构的精细能量估计和线性化的谱分析, 他们在 $H^s(\mathbb{R}^3)(s\geq3)$ 中得到了初值接近其远场状态 (非真空平衡状态时) 的全局经典解. 该理论已由 Hoff[11] 推广到不连续解, 由 Danchin[12] 推广到 Besov 空间. 需要注意的是, 文献 [10] 的理论要求解在非真空远场状态下具有较小的振荡, 从而使密度严格远离真空. 一个很自然并且重要的问题是, 是否有适当的理论适用于包含真空的初始数据. 在这个问题的研究上, 做出主要突破的是 Lions[13], 他用弱收敛的方法获得了具有有限能量和大初值的重整化弱解的存在性, 其中 $\gamma\geq9/5$, 这个弱解可以包含真空情形 (对于三维球对称情形, Jiang 等[14] 将其推广到 $\gamma>1$; 后来, Feireisl 等[15] 将这一结果推广到 $\gamma>3/2$). 然而, 对这种重整化弱解的正则性和唯一性还知之甚少. 近来, Huang 等[16] 得到了包含真空的具有小能量大振荡的全局经典解.

当粘性系数依赖密度时, 方程 (1.1) 也得到了很多关注. Liu 等[17] 首先提出了用一些具有粘性依赖密度的可压缩 Navier-Stokes 方程模型来研究空气动力学. 并且我们知道通过 Chapman-Enskog 展开可以从 Boltzmann 方程中推导出 Navier-Stokes 方程[18,19], 此时粘性系数依赖温度. 如果将气体流动限制为等熵的情况, 则这种依赖关系可以通过 Boyle 定律和 Gay-Lussac 定律继承, 此时粘性系数依赖密度. 然而, 在出现真空的存在下, 处理这类系统会遇到较大困难. 一方面，注意到动量方程中 $u_t+u\cdot\nabla u$ 的系数在真空中消失, 这种退化导致了在真空存在时确定速度的一个本质困难. 另一方面, 当密度函数连接到真空时, 粘性项消失, 这给解的正则性分析带来了很大困难, 使得常粘情况下的方法难以适用于当前情况.

针对变粘情形, 仍然有一些重要的工作来克服这些困难. 首先, 当 $\delta_1=0,\delta_2>3$ 时, 在二维情形, Vaigant 等[20]的全局适定性结果是对于一般大初值的第一个重要结果, 其唯一约束是初始密度远离真空. 之后，Jiu 等[21] 证明了包含真空的情形. Huang 等[22] 将这一结果推广到 $\delta_2>\frac43$. Wang 等[23,24] 进一步将此模型的结果推广到三维球对称和轴对称情形, 并分别得到了大初值的全局弱解以及强解的适定性. 其次, 当 $\delta_1>0,\delta_2>0$ 时, 特别地, 粘性系数满足 BD 熵关系 $\lambda(\rho)=2(\rho\mu'(\rho)-\mu(\rho))$ 时, 一个新的熵估计由 Bresch等[25,26] 得到. 进一步, Mellet 等[27] 得到了解的紧性和稳定性分析. 基于这些结果, Guo 等[28,29] 首先得到了球对称大初值的全局弱解. 通过构造合适的逼近系统, Li 等[30]得到了一般大初值的全局弱解. Vasseur 等[31,32] 利用不同的方法得到了类似的结果. 一般地, 对于粘性系数不依赖于 BD 熵关系的情形, 也有一些解的适定性的结果. Li 等[33] 在 $\delta_1=\delta_2=1$ 的情况下, 发现时间演化和粘性项的退化可以转化为特殊项的奇性问题. 基于这一发现, 在假设 $\rho_0(x)\rightarrow0$ (当 $\vert x\vert\rightarrow\infty$), 通过在 $L^6\cap D^1\cap D^2$ 空间中对 $\frac{\nabla\rho}{\rho}$ 建立统一的先验估计, 得到了二维空间 (三维情况见 Zhu[34]) 中正则解的局部存在唯一性. 然而, 这一结果只允许在远场存在真空, 并当真空出现在某些开集上, 甚至出现在单点上都会带来相应的问题. 然后, 通过引入一类合适的解空间, 对高阶项 $\rho^{\frac{\delta_1-1}{2}}\nabla^4u$ 建立一致的先验加权估计, 相同的作者在文献 [35,36] 中给出了 $1<\delta_1=\delta_2\leq\min\{3,\frac{\gamma+1}{2}\}$ 时的局部正则解的存在性. Xin 等[37] 得到了 $0<\delta_1=\delta_2<1$ 时的局部正则解. 特别地, 当 $\delta_{1}=\delta_{2}>1$ 时, 在对初始密度的 $\Vert\rho_0^{\frac{\gamma-1}{2}}\Vert_{H^3}+\Vert\rho_0^{\frac{\delta_{1,2}-1}{2}}\Vert_{H^3}$ 的小性要求下, Xin 等[38]首先得到了正则解的全局存在唯一性. 同时, 当初始密度远离真空, 假设 $\Vert u_0\Vert_{H^1}+\Vert\rho_0-\widetilde{\rho}\Vert_{L^2}$ 小, Guo 等[39] 得到了全局经典解的存在唯一性.

本文研究了三维空间中具有轴对称初值的变粘等熵 CNS, 在初始密度远离真空下, 得到了具有任意小能量大振荡初值的全局轴对称强解, 流体区域为周期域 $\Omega=\{(r,z)\vert r=\sqrt{x^2+y^2},(x,y,z)\in\mathbb{R}^3,r\in I\subset(0,+\infty),z\in(-\infty,+\infty)\}$. 注意到, 对比文献 [23] 中考虑的轴对称初值下的 Vaigant-Kazhikhov 模型 ($\delta_1=0,\delta_2>0$), 此时 $\delta_1>0$ 将带来新的困难, 使得我们需要得到密度导数的可积性估计. 本文证明的关键在于得到 $\int_0^{\infty}\Vert\nabla u\Vert_{L^{\infty}}{\rm d}t$ 和 $\sup\limits_{t\in[0,\infty)}\Vert\nabla\rho\Vert_{L^{q}}(2\leq q<\infty)$, 进一步得到 $\rho$ 的一致的上下界. 通过适用文献 [16,39,40] 中的方法和结构分析，我们从能量估计和初始层分析出发, 得到了新的 $\Vert u\Vert_{H^1}$, $\Vert \dot{u}\Vert_{H^1}$ 以及 $\Vert\nabla\rho\Vert_{L^{q}}$ 的时间加权估计.利用这些关键的估计, 结合动量方程, 可以得到

$\Vert\nabla u\Vert_{L^\infty}\leq \Vert\nabla u\Vert_{L^2}^{\frac{q-2}{2(q-1)}} (\Vert\rho\dot u\Vert_{L^q} +\Vert\nabla\rho^\gamma\Vert_{L^q} +\Vert\nabla\rho\cdot\nabla u\Vert_{L^q} +\Vert\nabla u\Vert_{L^q} +\Vert\nabla\rho \cdot u\Vert_{L^q})^{\frac{q}{2(q-1)}}$

的关于时间的一致可积性,从而得到密度的上下界, 以获得期望的结果. 注意到, 通过新的时间加权估计, 我们可以将文献 [39] 中 $\Vert\nabla u_0\Vert_{L^2}$ 的小性假设去掉, 得到具有小能量大震荡的全局强解; 并且也得到了解的更好的衰减性结果, 具体表现为: 只要初始能量足够小, 解的衰减速率将足够快.

在陈述主要结果之前, 我们首先解释本文中使用的符号和约定.

表示周期区域 $\Omega=\{(r,z)\vert r\in I\subset(0,\infty),z\in(-\infty,+\infty)\}$, 其中 $I$ 为 $(0,\infty)$ 的闭子集, 如 $I=[r_0,R]$( $ R>r_0>0$).

对于 $1\leq m\leq\infty$ 和 $k>0$, 表示标准的齐次和非齐次的 Sobolev 空间如下

$\begin{equation*} \left\{\begin{array}{cl} L^m=L^m(\Omega), D^{k,m}=\{u\in L_{\rm loc}^1(\Omega)\vert\Vert\nabla^ku\Vert_{L^m}<\infty\}, \Vert u\Vert_{D^{k,m}}\triangleq\Vert\nabla^k u\Vert_{L^m},\\ W^{k,m}=L^m\cap D^{k,m}, H^k=W^{k,2}, D^k=D^{k,2}, D^1=\{u\in L^6\vert\Vert\nabla u\Vert_{L^2}<\infty\}.\\ \end{array}\right. \end{equation*}$

初始能量定义为

$\begin{equation*} C_0=\int_\Omega(\frac{1}{2}\rho_0\vert u_0\vert^2+G(\rho_0) ) r\mathrm{d}r\mathrm{d}z, \end{equation*}$

其中 $G$ 的定义如下

$\begin{equation*} G(\rho)=\rho\int^\rho_{\widetilde{\rho}}\frac{P(s)- P(\widetilde{\rho})}{s^2}\mathrm{d}s, \end{equation*}$

由 $G(\rho)$ 的表达式可知

$\begin{equation*} \left\{\begin{array}{cl} G(\rho)=\frac{1}{\gamma-1}\rho^\gamma,&\widetilde{\rho}=0,\\ c_1(\overline{\rho},\widetilde{\rho})(\rho-\widetilde{\rho})^2\leq G(\rho)\leq c_2(\overline{\rho},\widetilde{\rho})(\rho-\widetilde{\rho})^2, &\widetilde{\rho}>0,0\leq\rho\leq\overline{\rho}. \end{array}\right. \end{equation*}$

定义 1.1 对任意 $T>0$, 如果问题 (1.4),(1.5) 的解 $(\rho,u)$ 满足如下条件 ($0<\tau <T$)

$\begin{array}{l}(\rho-\widetilde{\rho}) \in C\left([0, T] ; H^{3}\right), \rho_{t} \in C\left([0, T] ; H^{2}\right), \rho_{t t} \in L^{\infty}\left([0, T] ; L^{2}\right) \cap L^{2}\left([0, T] ; H^{1}\right), \\u \in C\left([0, T] ; D^{1} \cap D^{3}\right) \cap L^{2}\left([0, T] ; D^{4}\right) \cap L^{\infty}\left([\tau, T] ; D^{4}\right), \\u_{t} \in L^{\infty}\left([0, T] ; H^{1}\right) \cap L^{2}\left([0, T] ; D^{2}\right) \cap L^{\infty}\left([\tau, T] ; D^{2}\right) \cap H^{1}\left([\tau, T] ; D^{1}\right), \\u_{t t} \in L^{2}\left([0, T] ; L^{2}\right) \cap L^{\infty}\left([\tau, T] ; D^{1}\right),\end{array}$

那么我们就称$(\rho,u)$ 是问题 (1.4),(1.5) 的全局强解.

本文的主要结果如下

定理 1.1 对于 $\gamma>1,\ \delta_1\geq0,\ \delta_2\geq0$和给定的正常数 $M$, $\underline{\rho}$, $ \overline{\rho}, \widetilde{\rho}$, 初始数据 $(\rho_0,u_0)$ 满足

$\begin{equation} \begin{split} \rho_0\vert u_0\vert^2+G(\rho_0)\in L^1,\ \Vert\nabla\rho_0\Vert_{L^2\cap L^4} \leq M,\\ 0<\underline{\rho}\leq\rho_0\leq\overline{\rho},\ (\rho_0-\widetilde{\rho},u_0)\in H^3, \end{split} \end{equation}$

则存在一个正常数 $\varepsilon$ 依赖于 $\gamma, \delta_1,\delta_2,\underline{\rho},\overline{\rho},\widetilde{\rho},M$, 如果

$\begin{equation} C_0\leq\varepsilon, \end{equation}$

问题(1.4),(1.5) 存在唯一的全局强解 $(\rho,u)(r,z,t)$, 且满足

$\begin{equation} \frac12\underline{\rho}\leq\rho(r,z,t)\leq2\overline{\rho}. \end{equation}$

进一步, 对于任意给定的常数 $1\leq\delta_0<\infty$, 存在正常数$\varepsilon$ 和 $C$ 依赖于 $\delta_0$, 使得当(1.9) 式成立时, 有

$\begin{equation} \vert\rho-\widetilde{\rho}\vert\leq Ct^{-\frac{3(1+\delta_0)}{8}}. \end{equation}$

注 1.1 定理 1.1 包含了粘性系数满足 BD 熵关系 $\lambda(\rho)=2(\rho\mu'(\rho)-\mu(\rho))$ 的情形, 此时,对比已有的弱解的存在性结果[28⇓⇓⇓-32], 得到了无真空时的全局强解.

注 1.2 当 $\delta_1=0,\delta_2=0$, 定理 1.1 得到了常粘情形的具有小能量的全局强解. 在初始远离真空的情形, 这与文献 [16]中得到的结果一致.

2 预备知识

在本章中, 我们给出在后面证明中常用的已知事实和基础的不等式.

引理 2.1 （局部适定性文献 [8,33]） 对 $\widetilde{\rho}>0$, 以及 $\gamma>1, \delta_1\geq0, \delta_2\geq0$, 假设初始数据 $(\rho_0,u_0)$ 满足正则性条件 (1.8), 则存在有限时间 $T^*>0$, 使得问题 （1.4）,（1.5） 在 $\Omega\times(0,T^*]$ 上存在唯一的强解 $(\rho,u)$.

下面是 Gagliardo-Nirenberg 不等式.

引理 2.2 对于 $\forall h\in W^{1,m}(\mathbb{R}^2)\cap L^r(\mathbb{R}^2)$, 有如下不等式

$\begin{equation*} \Vert h\Vert_q\leq C\Vert\nabla h\Vert_m^\theta\Vert h\Vert_r^{1-\theta}, \end{equation*}$

其中 $\theta=(\frac{1}{r}-\frac{1}{q})(\frac{1}{r}-\frac{1}{m}+\frac{1}{2})^{-1}$.

我们现在陈述由引理2.2 得到的一些初等估计: 对于任意 $p'>2$,

$\begin{equation} \begin{split} \Vert u\Vert_{L^\infty} &\leq\Vert u\Vert_{L^2}^\theta \Vert\nabla u\Vert_{L^{p'}}^{1-\theta}\\ &\leq C \Vert\nabla u\Vert_{L^2}^{\theta} \Vert\nabla u\Vert_{L^{p'}}^{1-\theta}\\ &\leq C \Vert\nabla u\Vert_{L^2}^{\theta} (\Vert\nabla u\Vert_{L^2}^{1-\theta} +\Vert\nabla u\Vert_{L^\infty}^{1-\theta})\\ &\leq C (\Vert\nabla u\Vert_{L^2} +\Vert\nabla u\Vert_{L^\infty}). \end{split} \end{equation}$

记 $F\triangleq(1+\frac{\lambda(\rho)}{\mu(\rho)})\mathrm{div}u-\int_{\widetilde{\rho}}^\rho\frac{\gamma s^{\gamma-1}}{\mu(s)}{\rm d}s$. 用 $\mathrm{div}$ 作用$(1.6)_2$ 式可以得到

$\begin{equation*} \begin{split} \Delta F=\mathrm{div}\Big(\frac{1}{a}\rho^{1-\delta_1}\dot u -\delta_1\rho^{-1}\nabla\rho\cdot\nabla u -\frac{b\delta_1}{a}\rho^{\delta_2-\delta_1-1}\nabla\rho \mathrm{div}u -\frac{Q}{\mu(\rho)}\Big), \end{split} \end{equation*}$

由文献 [41,引理 4.27] 对于任意 $p\geq2$, 有

$\begin{aligned}\|\nabla F\|_{L^{p}} \leq & C\left(\left\|\rho^{1-\delta_{1}} \dot{u}\right\|_{L^{p}}+\left\|\rho^{-1} \nabla \rho \cdot \nabla u\right\|_{L^{p}}+\left\|\rho^{\delta_{2}-\delta_{1}-1} \nabla \rho \cdot u\right\|_{L^{p}}\right. \\& \left.+\|\nabla u\|_{L^{p}}+\left\|\rho^{\delta_{2}-\delta_{1}} \nabla u\right\|_{L^{p}}\right).\end{aligned}$

由$(1.6)_2$ 式可知

$\begin{equation*} \begin{split} \mu(\rho)\Delta u+\lambda(\rho)\nabla \mathrm{div}u= \rho\dot{u}+\nabla P-\nabla\mu(\rho)\nabla u-\nabla \lambda(\rho)\mathrm{div}u-Q, \end{split} \end{equation*}$

进一步由椭圆方程 $L^p$ 估计可得

$\begin{equation} \begin{split} \Vert\nabla^2u\Vert_{L^p} \leq C(\Vert\rho\dot u\Vert_{L^p} +\Vert\nabla\rho\Vert_{L^p} +\Vert\nabla\rho\cdot\nabla u\Vert_{L^p} +\Vert\nabla\rho\cdot u\Vert_{L^p} +\Vert\nabla u\Vert_{L^p}). \end{split} \end{equation}$

当 $p=2$ 时, 由 Cauchy-Schwarz 不等式, Poincare 不等式和引理 2.2 (参数为 $m=2,r=2,q=4,\theta=\frac{1}{2}$), 得到

$\begin{equation} \begin{split} \Vert\nabla^2u\Vert_{L^2} &\leq C(\Vert\rho\dot u\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^2} +\Vert\nabla\rho\cdot\nabla u\Vert_{L^2} +\Vert\nabla\rho\cdot u\Vert_{L^2} +\Vert\nabla u\Vert_{L^2})\\ &\leq C(\Vert\rho\dot u\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^4}\Vert\nabla u\Vert_{L^4} +\Vert\nabla\rho\Vert_{L^4}\Vert u\Vert_{L^4} +\Vert\nabla u\Vert_{L^2})\\ &\leq C(\Vert\rho\dot u\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^4}\Vert\nabla u\Vert_{L^2}^{\frac12}\Vert\nabla^2 u\Vert_{L^2}^{\frac12} +\Vert\nabla u\Vert_{L^2}), \end{split} \end{equation}$

对于 $\Vert\nabla\rho\Vert_{L^4}\Vert^2\nabla u\Vert_{L^2}^{\frac12}\Vert\nabla^2 u\Vert_{L^2}^{\frac12}$, 再由 Young 不等式

$\begin{equation*} \begin{split} \Vert\nabla\rho\Vert_{L^4}\Vert\nabla u\Vert_{L^2}^{\frac12}\Vert\nabla^2 u\Vert_{L^2}^{\frac12} \leq C(\varepsilon)\Vert\nabla\rho\Vert_{L^4}\Vert\nabla u\Vert_{L^2} +\varepsilon\Vert\nabla^2 u\Vert_{L^2}, \end{split} \end{equation*}$

因此

$\begin{equation} \begin{split} \Vert\nabla^2u\Vert_{L^2} &\leq C(\Vert\rho\dot u\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^4}^2\Vert\nabla u\Vert_{L^2} +\Vert\nabla u\Vert_{L^2}). \end{split} \end{equation}$

接下来, 引入以下 Zlotnik 不等式来得到密度的一致上下界.

引理 2.3[42,43] 函数 $y$ 满足

$\begin{equation*} y'(t)=g(y)+b'(t), t\in[T],y(0)=y^0, \end{equation*}$

其中$g\in C(R)$ 并且 $y,b\in W^{1,1}(0,T)$. 如果 $g(\infty)=-\infty$ 和

$\begin{equation} b(t_2)-b(t_1)\leq N_0+N_1(t_2-t_1), \end{equation}$

对于所有$0\leq t_1\leq t_2\leq T,N_0\geq0$ 和 $N_1\geq0$ 成立, 则

$\begin{equation*} y(t)\leq\max{\{y^0,\overline{\zeta}\}}+N_0<\infty, t\in[T], \end{equation*}$

其中$\overline{\zeta}$ 是一个常数, 并且下式成立

$g(\zeta) \leq-N_{1}, \quad \zeta \geq \bar{\zeta}.$

3 先验估计

在本章中, 我们将建立必要的先验估计去得到 $\rho$ 的一致上下界. 对于任意的 $T\geq 0$, 假设$(\rho,u)$ 是(1.4),(1.5) 式的全局光滑解. 令 $\sigma(t)\triangleq\min\{1,t\}$, 且对于给定的正常数 $1\leq\delta_0<\infty$, 记

$\begin{equation} A_1(T)\triangleq\sup\limits_{t\in[T]}\Vert\nabla u\Vert_{L^2}^2+\int_0^T\Vert\dot u\Vert_{L^2}^2\mathrm{d}t, \end{equation}$
$\begin{equation} A_2(T)\triangleq\sup\limits_{t\in[T]}\Vert\dot u\Vert_{L^2}^2+\int_0^T\Vert\nabla\dot u\Vert_{L^2}^2\mathrm{d}t, \end{equation}$
$\begin{equation} A_3(T)\triangleq\sup\limits_{t\in[T]}(\sigma(t)^{\frac{3}{4}}\Vert\nabla u\Vert_{L^2}^2)+\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\dot u\Vert_{L^2}^2\mathrm{d}t, \end{equation}$
$\begin{equation} A_4(T)\triangleq\sup\limits_{t\in[T]}(\sigma(t)^{\frac{8}{9}}\Vert\dot u\Vert_{L^2}^2)+\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla\dot u\Vert_{L^2}^2\mathrm{d}t, \end{equation}$
$\begin{equation} A_5(T)\triangleq\sup\limits_{t\in[T]}(t^{1+\delta_0}\Vert u\Vert_{L^2}^2+t^{1+\delta_0}\Vert\rho-\widetilde{\rho}\Vert_{L^2}^2)+\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t, \end{equation}$
$\begin{equation} A_6(T)\triangleq\sup\limits_{t\in[T]}(t^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2)+\int_0^Tt^{1+\delta_0}\Vert\dot u\Vert_{L^2}^2\mathrm{d}t, \end{equation}$
$\begin{equation} A_7(T)\triangleq\sup\limits_{t\in[T]}(t^{1+\delta_0}\Vert\dot u\Vert_{L^2}^2)+\int_0^Tt^{1+\delta_0}\Vert\nabla\dot u\Vert_{L^2}^2\mathrm{d}t. \end{equation}$

下面给出先验估计中的一个重要命题, 这也是得到全局解的关键.

命题 3.1 在定理 1.1 的条件下, 如果 $(\rho,u)$ 是 (1.4),(1.5) 式在 $\Omega\times(0,T]$ 中的光滑解且对于 $2\leq q<\infty,1\leq\delta_0<\infty$ 满足

$\begin{equation} \begin{split} &\frac{1}{2}\underline{\rho}\leq\inf_{\Omega\times[T]}\rho\leq\sup\limits_{\Omega\times[T]}\rho\leq 2\overline{\rho}, \sup\limits_{t\in[T]}\Vert\nabla\rho\Vert_{L^q}^q +\int_0^T\Vert\nabla\rho\Vert^q_{L^q}\mathrm{d}t\leq 4M^q,\\ &\int_{\sigma(T)}^T\Vert\nabla\rho\Vert_{L^q}\mathrm{d}t\leq 4M, \int_0^T\Vert\nabla\rho\Vert_{L^2}^{2} t^{1+\delta_0}\mathrm{d}t \leq 4^{2+\delta_0}(1+\delta_0)^{1+\delta_0}M^2, \end{split} \end{equation}$

则如下的估计成立

$\begin{equation} \begin{split} &\frac{2}{3}\underline{\rho}\leq\inf_{\Omega\times[T]}\rho\leq\sup\limits_{\Omega\times[T]}\rho\leq \frac{3}{2}\overline{\rho}, \sup\limits_{t\in[T]}\Vert\nabla\rho\Vert_{L^q}^q +\int_0^T\Vert\nabla\rho\Vert^q_{L^q}\mathrm{d}t\leq 2M^q,\\ &\int_{\sigma(T)}^T\Vert\nabla\rho\Vert_{L^q}\mathrm{d}t\leq 2M, \int_0^T\Vert\nabla\rho\Vert_{L^2}^{2} t^{1+\delta_0}\mathrm{d}t \leq 4^{1+\delta_0}(1+\delta_0)^{1+\delta_0}M^2. \end{split} \end{equation}$

证 命题3.1 的证明是结合引理 3.1-3.9 得到的.

接下来的引理将给出一些必要的解的时间加权估计.

引理 3.1 （能量估计) 在定理 1.1 的条件下, 如果 $(\rho,u)$ 是 (1.4),(1.5) 式在 $\Omega\times(0,T]$ 中的光滑解, 则存在正常数 $C$, 有如下不等式

$\begin{equation} \begin{split} &\sup\limits_{t\in[T]}\int_\Omega(\rho u^2+G(\rho))r\mathrm{d}r\mathrm{d}z +\int_0^T\int_\Omega\mu(\rho)(|\nabla u|^2+\frac{u_1^2}{r^2})r\mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &+\int_0^T\int_\Omega\lambda(\rho)(\mathrm{div}u+\frac{u_1}{r})^2r\mathrm{d}r\mathrm{d}z\mathrm{d}t\leq CC_0. \end{split} \end{equation}$

证 对 $(1.6)_1,(1.6)_2$ 式分别乘 $G'(\rho),ur$ 并且在 $\Omega\times(0,T]$ 上积分, 将两式相加, 通过远场条件 (1.5) 就可以得到能量不等式 (3.10).

引理 3.2 (时间加权能量估计) 在定理1.1 的条件下, 如果$(\rho,u)$ 是(1.4),(1.5) 式在$\Omega\times(0,T]$ 中的光滑解且满足条件(3.8), 则存在正常数$C$ 依赖于 $\underline{\rho},\overline{\rho}, M, \delta_0$, 有如下不等式

$\begin{equation}A_5(T)\leq C(\underline{\rho},\overline{\rho},M)C_0^{\frac{1}{2(1+\delta_0)}}.\end{equation}$

证 对 $(1.6)_1,(1.6)_2$ 式分别乘 $t^{1+\delta_0}G'(\rho),t^{1+\delta_0}ur$, 将两式相加然后在$(0,T]\times\Omega$ 上积分可以得到

$\begin{equation} \begin{split} &t^{1+\delta_0}\int_\Omega(\frac{1}{2}\rho u^2r+G(\rho)r) \mathrm{d}r\mathrm{d}z +\int_0^T\int_\Omega t^{1+\delta_0}\mu(\rho)(|\nabla u|^2+\frac{u_1^2}{r^2})r\mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &+\int_0^T\int_\Omega t^{1+\delta_0}\lambda(\rho)(\mathrm{div}u+\frac{u_1}{r})^2r\mathrm{d}r\mathrm{d}z\mathrm{d}t\\ =\,&\int_0^T\int_\Omega t^{\delta_0}\frac{1+\delta_0}{2}\rho u^2r\mathrm{d}r\mathrm{d}z\mathrm{d}t +\int_0^T\int_\Omega t^{\delta_0}(1+\delta_0)G(\rho)r \mathrm{d}r\mathrm{d}z\mathrm{d}t \triangleq\sum_{i=1}^2I_i. \end{split} \end{equation}$

下面估计$I_1,I_2$. 由(3.10) 和(3.8) 式, 取 $T_1=C_0^{-\frac{1}{2(1+\delta_0)}}$, 有

$\begin{aligned}I_{1} & \leq C \int_{0}^{T_{1}} t^{\delta_{0}}\|u\|_{L^{2}}^{2} \mathrm{~d} t+C \int_{T_{1}}^{T} t^{\delta_{0}}\|\nabla u\|_{L^{2}}^{2} \mathrm{~d} t \\& \leq C C_{0} T_{1}^{1+\delta_{0}}+C \int_{T_{1}}^{T} \int_{\Omega} t^{1+\delta_{0}} \mu(\rho)|\nabla u|^{2} r \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t \cdot T_{1}^{-1} \\& \leq C C_{0}^{\frac{1}{2}}+\frac{1}{2} \int_{T_{1}}^{T} \int_{\Omega} t^{1+\delta_{0}} \mu(\rho)|\nabla u|^{2} r \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t,\end{aligned}$

以及

$\begin{array}{l}I_{2} \leq C \int_{0}^{T} t^{\delta_{0}}\|\rho-\widetilde{\rho}\|_{L^{2}}^{2} \mathrm{~d} t\\\begin{array}{l}\leq C \int_{0}^{T_{1}} t^{\delta_{0}}\|\rho-\tilde{\rho}\|_{L^{2}}^{2} \mathrm{~d} t+C \int_{T_{1}}^{T} t^{\delta_{0}}\|\nabla \rho\|_{L^{2}}^{2} \mathrm{~d} t \\\leq C T_{1}^{1+\delta_{0}} C_{0}+C \int_{T_{1}}^{T} t^{1+\delta_{0}}\|\nabla \rho\|_{L^{2}}^{2} \mathrm{~d} t \cdot T_{1}^{-1} \\\leq C C_{0}^{\frac{1}{2}}+C C_{0}^{\frac{1}{2\left(1+\delta_{0}\right)}}.\end{array}\end{array}$

将(3.13),(3.14) 式代入(3.12) 式, 利用 (1.9) 式和先验假设(3.8), 得到

$\begin{equation} \begin{split} &t^{1+\delta_0}\int_\Omega(\frac{1}{2}\rho u^2r+G(\rho)r) \mathrm{d}r\mathrm{d}z +\frac12\int_0^T\int_\Omega t^{1+\delta_0}\mu(\rho)(|\nabla u|^2+\frac{u_1^2}{r^2})r\mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &\leq CC_0^{\frac12}+CC_0^{\frac{1}{2(1+\delta_0)}} \leq CC_0^{\frac{1}{2(1+\delta_0)}}, \end{split} \end{equation}$

因此

$A_{5}(T) \leq C C_{0}^{\frac{1}{2\left(1+\delta_{0}\right)}},$

即证明引理成立.

引理 3.3 在定理1.1 的条件下, 如果$(\rho,u)$ 是(1.4),(1.5)式在$\Omega\times(0,T]$ 中的光滑解且满足(3.8) 式, 则存在正常数$C$ 依赖于 $\underline{\rho},\overline{\rho},M$, 有如下不等式

$\begin{equation} A_1(T)+A_2(T)\leq C(\underline{\rho},\overline{\rho},M). \end{equation}$

证 对 $(1.6)_2$ 式乘 $\dot u$, 然后在 $(0,T]\times\Omega$ 上积分有

$\begin{equation} \begin{split} \int_0^T\int_\Omega\rho|\dot u|^2\mathrm{d}r\mathrm{d}z\mathrm{d}t &\;=\int_0^T\int_\Omega-\nabla P\cdot\dot u +\mathrm{div}(\mu(\rho)\nabla u)\cdot\dot u +\mu(\rho)\partial_r(\frac{u_1}{r})\cdot\dot u_1\\ &\quad+\frac{1}{r}\mu(\rho)\partial_ru_2\cdot\dot u_2 +\nabla(\lambda(\rho)\mathrm{div}u)\cdot\dot u +\partial_r(\lambda(\rho)\frac{u_1}{r})\cdot\dot u_1\\ &\quad+\partial_z(\lambda(\rho)\frac{u_1}{r})\cdot\dot u_2\mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &\;\triangleq\sum_{i=1}^7L_i. \end{split} \end{equation}$

下面我们来分别处理这 $7$ 项, 由$(1.4)_1,(3.10)$ 式, 分部积分和 H$\ddot{\mathrm{o}}$lder 不等式可得

$\begin{aligned}L_{1}= & \int_{\Omega}(P-\tilde{P}) \operatorname{div} u \mathrm{~d} r \mathrm{~d} z-\int_{\Omega}\left(P_{0}-\tilde{P}\right) \operatorname{div} u_{0} \mathrm{~d} r \mathrm{~d} z+\int_{0}^{T} \int_{\Omega} \operatorname{div} u P^{\prime} \frac{1}{r} \operatorname{div}(\rho u r) \mathrm{d} r \mathrm{~d} z \mathrm{~d} t \\& +\int_{0}^{T} \int_{\Omega} P\left(\partial_{j} u_{i} \partial_{i} u_{j}\right)+P u_{i} \partial_{i} \operatorname{div} u \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t \\= & \int_{\Omega}(P-\tilde{P}) \operatorname{div} u \mathrm{~d} r \mathrm{~d} z-\int_{\Omega}\left(P_{0}-\tilde{P}\right) \operatorname{div} u_{0} \mathrm{~d} r \mathrm{~d} z+\int_{0}^{T} \int_{\Omega} P\left(\partial_{j} u_{i} \partial_{i} u_{j}\right) \mathrm{d} r \mathrm{~d} z \mathrm{~d} t \\& +\int_{0}^{T} \int_{\Omega}|\operatorname{div} u|^{2}\left(P^{\prime} \rho+P\right)+\operatorname{div} u P^{\prime} \rho u_{1} \frac{1}{r} \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t \\\leq & C \sup _{0 \leq t \leq T}\|\nabla u\|_{L^{2}}\|\rho-\tilde{\rho}\|_{L^{2}}+C+C \int_{0}^{T}\|\nabla u\|_{L^{2}}^{2} \mathrm{~d} t \\\leq & C A_{1}(T)^{\frac{1}{2}} C_{0}^{\frac{1}{2}}+C+C C_{0},\end{aligned}$

类似地, 得到

$\begin{aligned}L_{2}= & -\frac{1}{2} \int_{0}^{T} \int_{\Omega}\left(\mu(\rho)|\nabla u|^{2}\right)_{t}-\mu^{\prime}(\rho) \partial_{t} \rho|\nabla u|^{2} \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t-\int_{0}^{T} \int_{\Omega} \mu(\rho) \nabla u \nabla(u \cdot \nabla u) \mathrm{d} r \mathrm{~d} z \mathrm{~d} t \\= & -\frac{1}{2} \int_{\Omega} \mu(\rho)|\nabla u|^{2} \mathrm{~d} r \mathrm{~d} z+\frac{1}{2} \int_{\Omega} \mu\left(\rho_{0}\right)\left|\nabla u_{0}\right|^{2} \mathrm{~d} r \mathrm{~d} z \\& -\frac{1}{2} \int_{0}^{T} \int_{\Omega} \partial_{k} \mu(\rho) u_{k}|\nabla u|^{2}+\mu^{\prime}(\rho) \rho \operatorname{div} u|\nabla u|^{2}+\mu^{\prime}(\rho) \rho u_{1} \frac{1}{r}|\nabla u|^{2} \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t \\& -\int_{0}^{T} \int_{\Omega} \mu(\rho) \partial_{i} u_{k} \partial_{i} u_{j} \partial_{j} u_{k}+\mu(\rho) \partial_{i} u_{k} u_{j} \partial_{i} \partial_{j} u_{k} \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t \\\leq & -\frac{1}{2} \int_{\Omega} \mu(\rho)|\nabla u|^{2} \mathrm{~d} r \mathrm{~d} z+C+C \int_{0}^{T}\|\nabla u\|_{L^{3}}^{3} \mathrm{~d} t+C\left(\int_{0}^{T}\left\|u_{1}\right\|_{L^{3}}^{3} \mathrm{~d} t\right)^{\frac{1}{3}}\left(\int_{0}^{T}\|\nabla u\|_{L^{3}}^{3} \mathrm{~d} t\right)^{\frac{2}{3}} \\\leq & -\frac{1}{2} \int_{\Omega} \mu(\rho)|\nabla u|^{2} \mathrm{~d} r \mathrm{~d} z+C+C \int_{0}^{T}\|\nabla u\|_{L^{3}}^{3} \mathrm{~d} t\end{aligned}$

和

$\begin{aligned}L_{3}+L_{4} & =\int_{0}^{T} \int_{\Omega}-\mu(\rho) \frac{1}{r^{2}} u_{1} \dot{u}_{1}+\mu(\rho) \frac{1}{r} \partial_{r} u_{1} \dot{u}_{1}+\frac{1}{r} \mu(\rho) \partial_{r} u_{2} \dot{u}_{2} \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t \\& \leq C\left(\int_{0}^{T}\|\dot{u}\|_{L^{2}}^{2} \mathrm{~d} t\right)^{\frac{1}{2}}\left[\left(\int_{0}^{T}\|\nabla u\|_{L^{2}}^{2} \mathrm{~d} t\right)^{\frac{1}{2}}+\left(\int_{0}^{T}\left\|u_{1}\right\|_{L^{2}}^{2} \mathrm{~d} t\right)^{\frac{1}{2}}\right] \\& \leq C C_{0}^{\frac{1}{2}} A_{1}(T)^{\frac{1}{2}}.\end{aligned}$

对于 $\lambda(\rho)$ 项的处理类似于 $\mu(\rho)$ 项的估计, 有

$\begin{aligned} L_5 &=-\frac{1}{2}\int_\Omega(\lambda(\rho)|\mathrm{div}u|^2)\mathrm{d}r\mathrm{d}z +\frac{1}{2}\int_\Omega(\lambda(\rho_0)|\mathrm{div}u_0|^2)\mathrm{d}r\mathrm{d}z\nonumber\\ &\quad+\frac{1}{2}\int_0^T\int_\Omega\lambda(\rho)|\mathrm{div}u|^3 -\lambda'(\rho)\rho|\mathrm{div}u|^3 -\lambda'(\rho)\rho\frac{u_1}{r}|\mathrm{div}u|^2\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad-\int_0^T\int_\Omega\lambda(\rho) \mathrm{div}u\partial_ju_i\partial_iu_j\mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &\leq-\frac{1}{2}\int_\Omega(\lambda(\rho)|\mathrm{div}u|^2)\mathrm{d}r\mathrm{d}z+C +C\int_0^T\Vert\nabla u\Vert_{L^3}^3\mathrm{d}t.\nonumber \end{aligned}$

利用先验假设(3.8) 式和 H$\ddot{\mathrm{o}}$lder 不等式得

$\begin{aligned} L_6+L_7 &=\int_0^T\int_\Omega(\lambda'(\rho)\nabla\rho\frac{u_1}{r} -\lambda(\rho)\frac{u_1}{r^2} +\lambda(\rho)\frac{\nabla u_1}{r})\dot u\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq C\int_0^T\Vert\dot u\Vert_{L^2}\Vert u_1\Vert_{L^4}\Vert\nabla \rho\Vert_{L^4}\mathrm{d}t+C\Big(\int_0^T\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\Big(\int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\leq C(1+\sup\limits_{t\in[T]}\Vert\nabla\rho\Vert_{L^4})\Big(\int_0^T\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\Big(\int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\\ &\leq CC_0^{\frac{1}{2}}A_1(T)^{\frac{1}{2}}.\nonumber \end{aligned}$

进一步, 由引理2.2(参数为$m=2,r=2,q=3,\theta=\frac{1}{3}$), 先验假设(3.8) 和(3.10) 式, 有

$\begin{aligned} \int_0^T\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t &\leq C\int_0^T\Vert\nabla u\Vert^2_{L^2} \Vert\nabla^2u\Vert_{L^2} \mathrm{d}t\nonumber\\ &\leq C\int_0^T\Vert\nabla u\Vert^2_{L^2}(\Vert\rho\dot u\Vert_{L^2}+\Vert\nabla\rho\Vert_{L^2}+ \Vert\nabla\rho\Vert_{L^4}^{2}\Vert\nabla u\Vert_{L^2}\nonumber\\ &\quad+\Vert\nabla u\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^4}\Vert\nabla u\Vert_{L^2}) \mathrm{d}t\nonumber\\ &\leq C(\sup\limits_{t\in[T]}\Vert\nabla u\Vert_{L^2}) \Big(\int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\quad+C(\sup\limits_{t\in[T]}\Vert\nabla\rho\Vert_{L^2}) \int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\nonumber\\ &\quad+C(\sup\limits_{t\in[T]}\Vert\nabla u\Vert_{L^2}) (\sup\limits_{t\in[T]}\Vert\nabla\rho\Vert_{L^4})^{2} \int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\\ &\quad+C(\sup\limits_{t\in[T]}\Vert\nabla u\Vert_{L^2}) \int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\nonumber\\ &\quad+C(\sup\limits_{t\in[T]}\Vert\nabla\rho\Vert_{L^4}) (\sup\limits_{t\in[T]}\Vert\nabla u\Vert_{L^2}) \int_0^T\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\nonumber\\ &\leq C A_1(T)C_0^{\frac{1}{2}}+CC_0+CC_0A_1(T)^{\frac{1}{2}}.\nonumber \end{aligned}$

将(3.18)-(3.23) 式代入 (3.17) 式, 可得

$\begin{equation*} \begin{split} A_1(T)&\leq CC_0^{\frac{1}{2}} A_1(T)^{\frac{1}{2}}+C+CC_0+CC_0^{\frac{1}{2}}A_1(T)+CC_0A_1(T)^{\frac{1}{2}} +CC_0^{\frac{1}{2}}A_1(T)^{\frac{1}{2}}\\ &\leq (\frac12+CC_0^{\frac12})A_1(T)+C(C_0^{2}+C_0+1), \end{split} \end{equation*}$

因此当 $CC_0^{\frac12}<\frac12$ 时, 有

$\begin{equation} \begin{split} A_1(T)\leq C. \end{split} \end{equation}$

对 $(1.6)_2^i$ 作用$[\frac{\partial}{\partial t}+\mathrm{div}(u\cdot)+\frac{u_1}{r}]\cdot\dot u_i$ 关于 $i$ 求和, 并且在 $[T]\times\Omega$ 上积分可知

$\begin{aligned} & \int_0^T\int_\Omega[\partial_t(\rho\dot u_i)+\mathrm{div}(u\cdot\rho\dot u_i) +\frac{u_1}{r}\rho\dot u_i]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &=-\int_0^T\int_\Omega[\partial_t(\partial_iP)+\mathrm{div}(u\cdot\partial_iP)+\frac{u_1}{r}\partial_iP] \cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega[\mathrm{div}(\mu(\rho)\nabla u_i)_t+\mathrm{div}(u\cdot \mathrm{div}(\mu(\rho)\nabla u_i)) +\frac{u_1}{r}\mathrm{div}(\mu(\rho)\nabla u_i)]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega[(\mu(\rho)\partial_r(\frac{u_1}{r}))_t +\mathrm{div}(u\cdot \mu(\rho)\partial_r(\frac{u_1}{r})) +\frac{u_1}{r}\mu(\rho)\partial_r(\frac{u_1}{r})]\cdot\dot u_1\\ &\quad+[\frac{1}{r}(\mu(\rho)\partial_ru_2)_t +\mathrm{div}(u\cdot\frac{1}{r}\mu(\rho)\partial_ru_2) +\frac{u_1}{r}\mu(\rho)\frac{1}{r}\partial_ru_2]\cdot\dot u_2 \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega[\partial_i(\lambda(\rho)\mathrm{div}u)_t +\mathrm{div}(u\cdot\partial_i(\lambda(\rho)\mathrm{div}u)) +\frac{u_1}{r}\partial_i(\lambda(\rho)\mathrm{div}u)]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega[\partial_i(\lambda(\rho)\frac{u_1}{r})_t +\mathrm{div}(u\cdot\partial_i(\lambda(\rho)\frac{u_1}{r})) +\frac{u_1}{r}\partial_i(\lambda(\rho)\frac{u_1}{r})]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\triangleq\sum_{i=1}^6N_i.\nonumber \end{aligned}$

由$(1.4)_1$ 式和分部积分可得

$\begin{aligned} & \int_0^T\int_\Omega[\partial_t(\rho\dot u_i)+\mathrm{div}(u\cdot\rho\dot u_i) +\frac{u_1}{r}\rho\dot u_i]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &=\int_0^T\int_\Omega\frac{1}{2}\partial_t(\rho\vert\dot u\vert^2) +\frac{1}{2}\rho_t\vert\dot u\vert^2 -\frac{1}{2}\rho u_k\partial_k(\vert\dot u\vert^2) +\frac{u_1}{r}\rho\vert\dot u\vert^2\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &=\int_\Omega\frac{1}{2}\rho\vert\dot u\vert^2\mathrm{d}r\mathrm{d}z -\int_\Omega\frac{1}{2}\rho_0\vert\dot u_0\vert^2\mathrm{d}r\mathrm{d}z -\int_0^T\int_\Omega\frac{1}{2r}\mathrm{div}(\rho ur)\vert\dot u\vert^2\mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &\quad-\int_0^T\int_\Omega\frac{1}{2}\rho u\nabla\vert\dot u\vert^2 -\frac{u_1}{r}\rho\vert\dot u\vert^2\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &=\int_\Omega\frac{1}{2}\rho\vert\dot u\vert^2\mathrm{d}r\mathrm{d}z -\int_\Omega\frac{1}{2}\rho_0\vert\dot u_0\vert^2\mathrm{d}r\mathrm{d}z+\int_0^T \int_\Omega\frac{1}{2}\rho\vert\dot u\vert^2\frac{u_1}{r}\mathrm{d}r\mathrm{d}z{\rm d}t,\nonumber \end{aligned}$

其中由(3.10) 式和引理2.2(参数为 $m=2,r=q=4,\theta=0$) 有

$\begin{aligned} \int_0^T \int_\Omega\frac{1}{2}\rho\vert\dot u\vert^2\frac{u_1}{r}\mathrm{d}r\mathrm{d}z{\rm d}t &\leq C\int_0^T \Vert u\Vert_{L^2}^2\Vert \dot u\Vert_{L^4}^2{\rm d}t\nonumber\\ &\leq C C_0^{\frac12}\int_0^T \Vert u\Vert_{L^2}^2\Vert \nabla\dot u\Vert_{L^2}^2{\rm d}t\leq C C_0^{\frac12}A_2(T)^{\frac{1}{2}}. \end{aligned}$

类似地, 我们也可得

$\begin{aligned} N_1&=-\int_0^T\int_\Omega P'\frac{1}{r}\mathrm{div}(\rho ur)\partial_r\dot u_1 +P\partial_r(\partial_k\dot u_1\cdot u_k) +\frac{1}{r^2}u_1\dot u_1P -\frac{1}{r}\partial_r(u_1\dot u_1)P\nonumber\\ &\quad+P'\frac{1}{r}\mathrm{div}(\rho ur)\partial_z\dot u_2 +P\partial_z(\partial_k\dot u_2\cdot u_k) -\frac{1}{r}\partial_z(u_1\dot u_2)P \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &=\int_0^T\int_\Omega-P\partial_r\dot u_{1}\partial_zu_{2}-P\partial_z\dot u_{2}\partial_ru_{1} +P \partial_z\dot u_{1}\partial_ru_{2} +P\partial_r \dot u_{2}\partial_zu_{1}+P'\rho \mathrm{div}u\mathrm{div}\dot u\\ &\quad+\frac{1}{r}P'\rho u_1\mathrm{div}\dot u-\frac{1}{r}Pu_1\mathrm{div}\dot u-\frac{1}{r}P\dot u\cdot\nabla u_1-\frac{1}{r}u_1\dot u_1P \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq C(\int_0^T\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t)^{\frac{1}{2}}(\int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t)^{\frac{1}{2}}\nonumber\\ &\leq CC_0^{\frac{1}{2}}A_2(T)^{\frac{1}{2}},\nonumber \end{aligned}$

同理得到

$\begin{aligned} N_2&=\int_0^T\int_\Omega\partial_j\dot u_i\partial_ku_j\mu(\rho)\partial_ku_i +(\mathrm{div}u+\frac{u_1}{r})\mu'(\rho)\rho\partial_j u_i\partial_j\dot u_i -\mu(\rho)\vert\nabla\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega \frac{1}{r^2}u_1\dot u_i\mu(\rho)\partial_ru_i -\frac{1}{r}\partial_ku_1\dot u_1\partial_k u_i\mu(\rho) -\frac{1}{r}u_1\partial_k\dot u_i\mu(\rho)\partial_ku_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq-\int_0^T\int_\Omega\mu(\rho)\vert\nabla\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t+C\Big(\int_0^T\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2} \nonumber\\ &\leq-\int_0^T\int_\Omega\mu(\rho)\vert\nabla\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t+CA_2(T)^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2}, \end{aligned}$

由$(1.4)_1$ 式和 (3.24) 式, 有

$\begin{aligned} N_3+N_4 &=\int_0^T\int_\Omega (\mu(\rho)-\mu'(\rho))\dot u_1\mathrm{div}u(\frac{\partial_ru_{1}}{r}-\frac{u_1}{r^2})\nonumber\\ &\quad-\mu'(\rho)\rho\dot u_1u_1\frac{\partial_ru_1}{r^2} +(\mu(\rho)+\mu'(\rho))\dot u_1u_1\frac{u_1}{r^3}\nonumber\\ &\quad+\mu(\rho)\frac{1}{r}\dot u\cdot\partial_r\dot u -\frac{1}{r}\mu(\rho)\dot u_1\partial_ru_k\partial_ku_1 -\mu(\rho)\frac{1}{r^2}\vert\dot u_1\vert^2\nonumber\\ &\quad+(\mu(\rho)-\mu'(\rho))\frac{1}{r}\dot u_2\mathrm{div}u\partial_ru_{2} -\dot u_2\frac{1}{r^2}\mu'(\rho)\rho u_1\partial_ru_{2} -\frac{1}{r}\mu(\rho)\dot u_2\partial_ru_k\partial_ku_2 \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq C{A_1(T)}^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2} +C{A_1(T)}^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^\frac{1}{2}\nonumber\\ &\quad+C{A_1(T)}^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^{\frac{1}{4}} \Big(\int_0^T\Vert u\Vert^4_{L^4}\mathrm{d}t\Big)^{\frac{1}{4}} -\int_0^T\int_\Omega\mu(\rho)\frac{\vert\dot u_1\vert^2}{r^2}\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq C\Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2}+ CA_2(T)^{\frac{1}{2}} -\int_0^T\int_\Omega\mu(\rho)\frac{\vert\dot u_1\vert^2}{r^2}\mathrm{d}r\mathrm{d}z\mathrm{d}t. \end{aligned}$

下面处理关于 $\lambda(\rho)$ 的项, 同理可得

$\begin{aligned} N_5 &=-\int_0^T\int_\Omega\lambda(\rho)\vert \mathrm{div}\dot u \vert^2\mathrm{d}r\mathrm{d}z\mathrm{d}t+\int_0^T\int_\Omega \mathrm{div}\dot u(\lambda'(\rho)\rho-\lambda(\rho))\vert \mathrm{div}u\vert^2\nonumber\\ &\quad +\mathrm{div}\dot u\lambda'(\rho)\rho\frac{1}{r}u_1\mathrm{div}u +\mathrm{div}\dot u\lambda(\rho)\partial_ju_k\partial_ku_j +\partial_j\dot u_1\partial_iu_j\lambda(\rho)\mathrm{div}u -\frac{1}{r^2}u_1\dot u_1\lambda(\rho)\mathrm{div}u\nonumber\\ &\quad-\frac{1}{r}\partial_iu_1\dot u_i\lambda(\rho)\mathrm{div}u -\frac{1}{r}u_1\partial_i\dot u_i\lambda(\rho)\mathrm{div}u \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq-\int_0^T\int_\Omega\lambda(\rho)\vert\mathrm{div}\dot u \vert^2\mathrm{d}r\mathrm{d}z\mathrm{d}t +C\Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2} \Big(\int_0^T\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^\frac{1}{2}\nonumber\\ &\leq-\int_0^T\int_\Omega\lambda(\rho)\vert\mathrm{div}\dot u \vert^2\mathrm{d}r\mathrm{d}z\mathrm{d}t+CA_2(T)^{\frac{1}{2}}\Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2}, \end{aligned}$
$\begin{aligned} N_6 &=\int_0^T\int_\Omega\partial_i \dot u_i\lambda'(\rho)\rho \mathrm{div}u u_1\frac{1}{r} +\partial_i \dot u_i\lambda'(\rho)\rho\frac{1}{r^2}u_1^2 -\partial_i \dot u_i\dot u_1\lambda(\rho)\frac{1}{r}\nonumber\\ &\quad+\partial_i\dot u_i\lambda(\rho)\partial_ku_k\frac{u_1}{r} +\dot u_1\lambda(\rho)\frac{u_1^2}{r^3} +\partial_j\dot u_i\partial_i u_j\lambda(\rho)\frac{u_1}{r} -\partial_iu_1\dot u_i\lambda(\rho)\frac{u_1}{r^2} \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq C{A_1(T)}^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2}+C\Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2} \Big(\int_0^T\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^\frac{1}{2}\nonumber \\ &\quad+C{A_1(T)}^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^\frac{1}{2}\nonumber\\ &\leq C\Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2} +CA_2(T)^{\frac{1}{2}}+CA_2(T)^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2}. \end{aligned}$

并且由于引理2.2 (参数为$m=2,r=2,q=4,\theta=\frac{1}{2}$), (2.4)式和先验假设条件 (3.8), 有

$\begin{aligned} \int_0^T\Vert\nabla u\Vert_{L^4}^4\mathrm{d}t &\leq C\int_0^T\Vert\nabla u\Vert^2_{L^2} \Vert\nabla^2u\Vert_{L^2}^2 \mathrm{d}t\nonumber\\ &\leq C\int_0^T\Vert\nabla u\Vert^2_{L^2}(\Vert\dot u\Vert_{L^2}^2+\Vert\nabla\rho\Vert_{L^2}^2+ \Vert\nabla\rho\Vert_{L^4}^{4} \Vert\nabla u\Vert_{L^2}^2 +\Vert\nabla u\Vert_{L^2}^2) \mathrm{d}t\nonumber\\ &\leq C\sup\limits_{t\in[T]}(\Vert\dot u\Vert^2_{L^2}) \int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t +C\sup\limits_{t\in[T]}(\Vert\nabla\rho\Vert_{L^2}^2) \int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\nonumber\\ &\quad+C\sup\limits_{t\in[T]}(\Vert\nabla u\Vert^2_{L^2}) \sup\limits_{t\in[T]}(\Vert\nabla\rho\Vert_{L^4}^4) \int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\nonumber\\ &\quad+C\sup\limits_{t\in[T]}(\Vert\nabla u\Vert^2_{L^2}) \int_0^T\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\nonumber\\ &\leq CC_0A_2(T)+CC_0. \end{aligned}$

将 (3.26)-(3.33) 式代入 (3.25) 式, 并且由于(3.24) 式可得

$\begin{equation*} \begin{split} A_2(T)&\leq C +CA_2(T)^{\frac{1}{2}}C_0^{\frac{1}{2}} +CA_2(T)^{\frac{1}{2}}(C_0A _2(T)+CC_0A_1(T))^{\frac{1}{2}}\\ & +C(C_0A_2(T)+CC_0A_1(T))^{\frac{1}{2}}\\ &\leq (\frac12+CC_0^{\frac12})A_2(T)+C(C_0^{\frac12}+C_0+C_0^{2})+C, \end{split} \end{equation*}$

因此当 $CC_0^{\frac12}<\frac12$ 时, 则有

$\begin{equation} A_2(T)\leq C. \end{equation}$

结合(3.24) 和(3.34) 式可以得到引理的结论.

引理 3.4 在定理 1.1 的条件下, 如果$(\rho,u)$ 是(1.4), (1.5) 式在$\Omega\times(0,T]$ 中的光滑解且满足(3.8) 式, 则存在正常数$C$ 依赖于 $\underline{\rho},\overline{\rho},M$, 有如下不等式

$\begin{equation} A_3(T)+A_4(T)\leq C(\underline{\rho},\overline{\rho},M)C_0^{\frac{1}{4}}. \end{equation}$

证 对 $(1.6)_2$ 式乘 $\sigma(t)^{\frac{3}{4}}\dot u$ 并且在 $[T]\times\Omega$ 上积分得到

$\begin{aligned} \int_0^T\int_\Omega\sigma(t)^{\frac{3}{4}}\rho\vert\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t &=\int_0^T\int_\Omega-\nabla P\cdot\sigma(t)^{\frac{3}{4}}\dot u +\mathrm{div}(\mu(\rho)\nabla u)\cdot\sigma(t)^{\frac{3}{4}}\dot u\nonumber\\ &\quad +\mu(\rho)\partial_r(\frac{u_1}{r})\cdot\sigma(t)^{\frac{3}{4}}\dot u_1 +\frac{1}{r}\mu(\rho)\partial_ru_2\cdot\sigma(t)^{\frac{3}{4}}\dot u_2\nonumber\\ &\quad+\nabla(\lambda(\rho)\mathrm{div}u)\cdot\sigma(t)^{\frac{3}{4}}\dot u\\ &\quad+\partial_r(\lambda(\rho)\frac{u_1}{r})\cdot\sigma(t)^{\frac{3}{4}}\dot u_1 +\partial_z(\lambda(\rho)\frac{u_1}{r})\cdot\sigma(t)^{\frac{3}{4}}\dot u_2\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\;\triangleq\sum_{i=1}^7M_i.\nonumber \end{aligned}$

下面来估计这7项, 证明过程类似于引理 3.3 中关于 $A_1(T)$ 的证明. 我们可以由$(1.4)_1$ 式, $\sigma(t)$ 的定义和(3.10) 式得到

$\begin{aligned} M_1 &\leq\Big(\int_\Omega\sigma(t)^{\frac{3}{4}}\vert\nabla u\vert^2 \mathrm{d}r\mathrm{d}z\Big)^{\frac{1}{2}} \Big(\int_\Omega\sigma(t)^{\frac{3}{4}}\vert p-\tilde p\vert^2 \mathrm{d}r\mathrm{d}z\Big)^{\frac{1}{2}}\nonumber\\ &\quad+\sup\limits_{t\in[T]}(\Vert P-\tilde P\Vert_{L^2}) \Big(\int_0^1\Vert\nabla u\Vert^2_{L^2} \mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^1\sigma(t)^{-\frac{1}{2}} \mathrm{d}t\Big)^{\frac{1}{2}}\\ &\quad+\int_0^T\sigma(t)^{\frac{3}{4}}(\Vert\nabla u\Vert_{L^2}^2+\Vert u\Vert_{L^2}^2)\mathrm{d}t\nonumber\\ &\leq A_3(T)^{\frac{1}{2}}C_0^{\frac{1}{2}}+CC_0,\nonumber \end{aligned}$

类似的证明过程, 有

$\begin{aligned} M_2&\leq -\frac{1}{2}\int_\Omega\mu(\rho)\vert\nabla u\vert^2\sigma(t)^{\frac{3}{4}} \mathrm{d}r\mathrm{d}z +\int_0^1\int_\Omega\frac{3}{8}\mu(\rho)\vert\nabla u\vert^2\sigma(t)^{-\frac{1}{4}}\sigma'(t) \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+C\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t +\Big(\int_0^T\sigma(t)^{\frac{3}{2}}\Vert\nabla\dot u\Vert^2_{L^2} \mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla u\Vert^2_{L^2} \mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\leq -\frac{1}{2}\int_\Omega\mu(\rho)\vert\nabla u\vert^2\sigma(t)^{\frac{3}{4}} \mathrm{d}r\mathrm{d}z +\Big(\int_0^1\sigma(t)^{-\frac{3}{4}}\mathrm{d}t\Big)^{\frac{1}{3}} \Big(\int_0^1\Vert\nabla u\Vert^3_{L^2}\mathrm{d}t\Big)^{\frac{2}{3}}\nonumber\\ &\quad +C\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t +CA_2(T)^{\frac{1}{2}}C_0^{\frac{1}{2}}\\ &\leq -\frac{1}{2}\int_\Omega\mu(\rho)\vert\nabla u\vert^2\sigma(t)^{\frac{3}{4}} \mathrm{d}r\mathrm{d}z +CC_0^{\frac{2}{3}} +C\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t +CC_0^{\frac{1}{2}}\nonumber \end{aligned}$

和

$\begin{aligned} M_3+M_4&=\int_0^T\int_\Omega-\mu(\rho)\frac{1}{r^2}u_1\dot u_1\sigma(t)^{\frac{3}{4}}+\mu(\rho)\frac{1}{r}\partial_ru_1\dot u_1\sigma(t)^{\frac{3}{4}}+\frac{1}{r}\mu(\rho)\partial_ru_2\dot u_2\sigma(t)^{\frac{3}{4}}\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq C\Big(\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}}\Big(\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\dot u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}}\leq CC_0^{\frac{1}{2}}. \end{aligned}$

同理, 对于 $\lambda(\rho)$ 的项, 有

$\begin{equation} \begin{split} M_5+M_6+M_7&\leq -\frac{1}{2}\int_\Omega\lambda(\rho)\vert \mathrm{div}u\vert^2\sigma(t)^{\frac{3}{4}}\mathrm{d}r\mathrm{d}z +C\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t\\ &\quad+CC_0^{\frac{2}{3}} +CC_0^{\frac{1}{2}}. \end{split} \end{equation}$

并且由于 $\sigma(t)$ 的定义和 (3.23) 式, 可得

$\begin{aligned} \int_0^T\sigma(t)^{\frac{3}{4}}\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t &\leq \int_0^T\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t \leq CC_0+CC_0^{\frac{1}{2}}. \end{aligned}$

因此将(3.37)-(3.41) 式代入(3.36) 式就有

$\begin{equation} \begin{split} A_3(T)\leq CC_0^{\frac{1}{2}}+CC_0^{\frac23}\leq CC_0^{\frac{1}{2}}. \end{split} \end{equation}$

对 $(1.6)_2^i$ 乘 $\sigma(t)^{\frac{8}{9}}[\frac{\partial}{\partial t}+\mathrm{div}(u\cdot)+\frac{u_1}{r}]\cdot\dot u_i$ 关于 $i$ 求和并且在 $[T]\times\Omega$ 上积分得到

$\begin{aligned} & \int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}[\partial_t(\rho\dot u_i)+\mathrm{div}(u\cdot\rho\dot u_i) +\frac{u_1}{r}\rho\dot u_i]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &=-\int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}[\partial_t(\partial_iP)+\mathrm{div}(u\cdot\partial_iP)+\frac{u_1}{r}\partial_iP] \cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}[\mathrm{div}(\mu(\rho)\nabla u_i)_t+\mathrm{div}(u\cdot \mathrm{div}(\mu(\rho)\nabla u_i)) +\frac{u_1}{r}\mathrm{div}(\mu(\rho)\nabla u_i)]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}[(\mu(\rho)\partial_r(\frac{u_1}{r}))_t +\mathrm{div}(u\cdot \mu(\rho)\partial_r(\frac{u_1}{r})) +\frac{u_1}{r}\mu(\rho)\partial_r(\frac{u_1}{r})]\cdot\dot u_1\nonumber\\ &\quad+\sigma(t)^{\frac{8}{9}}[\frac{1}{r}(\mu(\rho)\partial_ru_2)_t +\mathrm{div}(u\cdot\frac{1}{r}\mu(\rho)\partial_ru_2) +\frac{u_1}{r}\mu(\rho)\frac{1}{r}\partial_ru_2]\cdot\dot u_2 \mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &\quad+\int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}[\partial_i(\lambda(\rho)\mathrm{div}u)_t +\mathrm{div}(u\cdot\partial_i(\lambda(\rho)\mathrm{div}u)) +\frac{u_1}{r}\partial_r(\lambda(\rho)\mathrm{div}u)]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}[\partial_i(\lambda(\rho)\frac{u_1}{r})_t +\mathrm{div}(u\cdot\partial_i(\lambda(\rho)\frac{u_1}{r})) +\frac{u_1}{r}\partial_i(\lambda(\rho)\frac{u_1}{r})]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\triangleq\sum_{i=1}^6H_i.\nonumber \end{aligned}$

下面来估计等式两端这6 项, 证明过程类似于引理 3.3 中关于$A_2(T)$ 的证明. 由$(1.4)_1$ 式,

$\begin{aligned} & \int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}[\partial_t(\rho\dot u_i)+\mathrm{div}(u\cdot\rho\dot u_i) +\frac{u_1}{r}\rho\dot u_i]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &=\int_\Omega\frac{1}{2}\rho\vert\dot u\vert^2\sigma(t)^{\frac{8}{9}} \mathrm{d}r\mathrm{d}z -\int_0^1\int_\Omega\frac{4}{9}\sigma(t)^{-\frac{1}{9}}\sigma'(t)\rho\vert\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t+\int_0^T \int_\Omega\frac{1}{2}\sigma(t)^{\frac{8}{9}}\rho\vert\dot u\vert^2\frac{u_1}{r}\mathrm{d}r\mathrm{d}z{\rm d}t.\nonumber \end{aligned}$

并且由(3.10), (3.16) 和 (3.42) 式得到

$\begin{aligned} &-\int_0^1\int_\Omega\frac{4}{9}\sigma(t)^{-\frac{1}{9}}\sigma'(t)\rho\vert\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t+\int_0^T \int_\Omega\frac{1}{2}\sigma(t)^{\frac{8}{9}}\rho\vert\dot u\vert^2\frac{u_1}{r}\mathrm{d}r\mathrm{d}z{\rm d}t\nonumber\\ &\leq \sup\limits_{t\in[0,1]}(\Vert\dot u\Vert_{L^2}) \Big(\int_0^1\sigma(t)^{\frac{3}{4}}\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^1t^{-\frac{35}{36}}\mathrm{d}t\Big)^{\frac{1}{2}}+CC_0^{\frac12}\\ &\leq CA_3(T)^{\frac{1}{2}}+CC_0^{\frac12}\leq CC_0^{\frac{1}{4}}.\nonumber \end{aligned}$

类似地, 可得

$\begin{aligned} H_1&\leq C\Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^\frac{1}{2} \Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\Big)^\frac{1}{2}\leq CA_4(T)^{\frac{1}{2}}C_0^{\frac{1}{2}},\end{aligned}$
$\begin{aligned} H_2 &\leq-\int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}\mu(\rho)\vert\nabla\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t+C\Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^\frac{1}{2} \Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2}\nonumber\\ &\leq-\int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}\mu(\rho)\vert\nabla\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t+CA_4(T)^{\frac{1}{2}} \Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2} \end{aligned}$

和

$\begin{aligned} H_3+H_4 &\leq C\Big(\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^{\frac{1}{2}}\\ &\quad+\!C\Big(\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\!-\!\int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}\mu(\rho)\frac{\vert\dot u_1\vert^2}{r^2}\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq CC_0^{\frac{1}{4}}\Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2} +CC_0^{\frac{1}{4}}A_4(T)^{\frac{1}{2}}-\int_0^T\int_\Omega\sigma(t)^{\frac{8}{9}}\mu(\rho)\frac{\vert\dot u_1\vert^2}{r^2}\mathrm{d}r\mathrm{d}z\mathrm{d}t.\nonumber \end{aligned}$

那么对于 $\lambda(\rho)$ 的项, 同理有

$\begin{aligned}H_{5} \leq & -\int_{0}^{T} \int_{\Omega} \sigma(t)^{\frac{8}{9}} \lambda(\rho)|\operatorname{div} \dot{u}|^{2} \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t+C\left(\int_{0}^{T} \sigma(t)^{\frac{3}{4}}\|\dot{u}\|_{L^{2}}^{2} \mathrm{~d} t\right)^{\frac{1}{2}}\left(\int_{0}^{T} \sigma(t)^{\frac{8}{9}}\|\nabla u\|_{L^{4}}^{4} \mathrm{~d} t\right)^{\frac{1}{2}} \\& +C\left(\int_{0}^{T} \sigma(t)^{\frac{8}{9}}\|\nabla u\|_{L^{4}}^{4} \mathrm{~d} t\right)^{\frac{1}{2}}\left(\int_{0}^{T} \sigma(t)^{\frac{8}{9}}\|\nabla \dot{u}\|_{L^{2}}^{2} \mathrm{~d} t\right)^{\frac{1}{2}} \\\leq & -\int_{0}^{T} \int_{\Omega} \sigma(t)^{\frac{8}{9}} \lambda(\rho)|\nabla \dot{u}|^{2} \mathrm{~d} r \mathrm{~d} z \mathrm{~d} t+C A_{3}(T)^{\frac{1}{2}}\left(\int_{0}^{T} \sigma(t)^{\frac{8}{9}}\|\nabla u\|_{L^{4}}^{4} \mathrm{~d} t\right)^{\frac{1}{2}} \\& +C A_{4}(T)^{\frac{1}{2}}\left(\int_{0}^{T} \sigma(t)^{\frac{8}{9}}\|\nabla u\|_{L^{4}}^{4} \mathrm{~d} t\right)^{\frac{1}{2}}\end{aligned}$

和

$\begin{equation} \begin{split} H_6 &\leq C(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\dot u\Vert^2_{L^2}\mathrm{d}t)^{\frac{1}{2}} \Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert u\Vert^4_{L^4}\mathrm{d}t\Big)^{\frac{1}{2}}\\ &\quad+C\Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{4} \Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla\dot u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert u\Vert_{L^4}^4\mathrm{d}t\Big)^{\frac{1}{4}}\\ &\quad+C\Big(\int_0^T\sigma(t)^{\frac{3}{4}}\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla\dot u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}}\\ &\leq CA_4(T)^{\frac{1}{2}}\Big(\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{4} +CA_3(T)^{\frac{1}{2}}A_4(T)^{\frac{1}{2}}. \end{split} \end{equation}$

并且由于 $\sigma(t)$ 的定义和 (3.33) 式, 可得

$\begin{aligned} &\int_0^T\sigma(t)^{\frac{8}{9}}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\leq \int_0^T\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\leq CC_0. \end{aligned}$

因此将 (3.44)-(3.50) 式代入(3.43) 式, 我们得到 $A_4(T)\leq CC_0^{\frac{1}{4}}$, 再结合 (3.42) 式得到引理的证明.

引理 3.5 在定理1.1 的条件和先验假设(3.8) 下, 如果$(\rho,u)$ 是(1.4), (1.5) 式在$\Omega\times(0,T]$ 中的光滑解, 则存在正常数 $C$ 依赖于 $\underline{\rho},\overline{\rho},M,\delta_0$, 有如下不等式

$A_{6}(T)+A_{7}(T) \leq C(\underline{\rho}, \bar{\rho}, M) C_{0}^{\frac{1}{2\left(1+\delta_{0}\right)}}.$

证 对 $(1.6)_2$ 乘 $t^{1+\delta_0}\dot u$ 并且在 $[T]\times\Omega$ 上积分得到

$\begin{aligned} \int_0^T\int_\Omega t^{1+\delta_0}\rho|\dot u|^2\mathrm{d}r\mathrm{d}z\mathrm{d}t &\;=\int_0^T\int_\Omega t^{1+\delta_0}[-\nabla P\cdot \dot u +\mathrm{div}(\mu(\rho)\nabla u)\cdot \dot u+\mu(\rho)\partial_r(\frac{u_1}{r})\cdot\dot u_1\nonumber\\ &\quad +\frac{1}{r}\mu(\rho)\partial_ru_2\cdot \dot u_2+\nabla(\lambda(\rho)\mathrm{div}u)\cdot\dot u +\partial_r(\lambda(\rho)\frac{u_1}{r})\cdot \dot u_1\nonumber\\ &\quad+\partial_z(\lambda(\rho)\frac{u_1}{r})\cdot \dot u_2]\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\;=\sum_{i=1}^7K_i. \end{aligned}$

由于 (3.8), (3.10),(3.11) 式和 (3.16)式, 取$T_2=C_0^{-\frac{1}{4(1+\delta_0)}}$, 有

$\begin{aligned} K_1 &\leq C(\sup\limits_{t\in[T]}t^{1+\delta_0}\Vert \rho-\widetilde{\rho}\Vert^2 )^{\frac{1}{2}} (\sup\limits_{t\in[T]}t^{1+\delta_0}\Vert\nabla u\Vert^2 )^{\frac{1}{2}}\nonumber\\ &\quad+C\int_{0}^{T_2}t^{\delta_0}\Vert\rho-\widetilde{\rho}\Vert_{L^2}\Vert\nabla u\Vert_{L^2} \mathrm{d}t +\int_{T_2}^Tt^{\delta_0}\Vert\nabla\rho\Vert_{L^2}\Vert \nabla u\Vert_{L^2} \mathrm{d}t +C\int_{0}^Tt^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\nonumber\\ &\leq CA_6(T)^{\frac{1}{2}}C_0^{\frac{1}{4(1+\delta_0)}} +C\Big(\int_{0}^{T_2}t^{\delta_0}\Vert\rho-\widetilde{\rho}\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac12} \Big(\int_{0}^{T_2}t^{\delta_0}\Vert\nabla u\Vert^2_{L^2} \mathrm{d}t\Big)^{\frac12}\nonumber\\ &\quad+\Big(\int_{T_2}^Tt^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac12} \Big(\int_{T_2}^Tt^{1+\delta_0}\Vert \nabla u\Vert^2_{L^2} \mathrm{d}t\Big)^{\frac12}\cdot T_2^{-1}+CC_0^{\frac{1}{2(1+\delta_0)}}\nonumber\\ &\leq CA_6(T)^{\frac{1}{2}}C_0^{\frac{1}{4(1+\delta_0)}}+CC_0T_2^{\frac12+\delta_0} +CC_0^{\frac{1}{4(1+\delta_0)}}T_2^{-1} +CC_0^{\frac{1}{2(1+\delta_0)}}\nonumber\\ &\leq CA_6(T)^{\frac{1}{2}}C_0^{\frac{1}{4(1+\delta_0)}} +CC_0^{\frac{1}{2(1+\delta_0)}}. \end{aligned}$
$\begin{aligned} K_2&\leq -\frac{1}{2}\int_\Omega t^{1+\delta_0}\mu(\rho)\vert\nabla u\vert^2 \mathrm{d}r\mathrm{d}z +\int_0^T\int_\Omega\frac{(1+\delta_0)}{2}t^{\delta_0}\mu(\rho)\vert\nabla u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+C\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t +C\Big(\int_0^Tt^{1+\delta_0}\Vert\dot u\Vert^2_{L^2} \mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert^2_{L^2} \mathrm{d}t\Big)^{\frac{1}{2}}\\ &\leq -\frac{1}{2}\int_\Omega t^{1+\delta_0}\mu(\rho)\vert\nabla u\vert^2 \mathrm{d}r\mathrm{d}z +CC_0^{\frac{1}{2(1+\delta_0)}} +C\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t +CA_6(T)^{\frac{1}{2}}C_0^{\frac{1}{4(1+\delta_0)}}.\nonumber \end{aligned}$

同理, 可得

$\begin{aligned} K_3+K_4&=\int_0^T\int_\Omega t^{1+\delta_0}(-\mu(\rho)\frac{1}{r^2}u_1\dot u_1+\mu(\rho)\frac{1}{r}\partial_ru_1\dot u_1+\frac{1}{r}\mu(\rho)\partial_ru_2\dot u_2)\mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq C\Big(\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}}\Big(\int_0^Tt^{1+\delta_0}\Vert\dot u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\leq CC_0^{\frac{1}{4(1+\delta_0)}}A_6(T)^{\frac{1}{2}}. \end{aligned}$

对于$\lambda(\rho)$ 的项同理可得

$\begin{aligned} K_5&\leq -\frac{1}{2}\int_\Omega t^{1+\delta_0}\lambda(\rho)\vert \mathrm{div}u\vert^2\mathrm{d}r\mathrm{d}z +CC_0^{\frac{1}{2(1+\delta_0)}}\nonumber\\ &\quad+C\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t +CA_6(T)^{\frac{1}{2}}C_0^{\frac{1}{4(1+\delta_0)}} \end{aligned}$

和

$\begin{aligned} K_6+K_7 &=\int_0^T\int_\Omega t^{1+\delta_0}(\lambda'(\rho)\nabla\rho\frac{u_1}{r} -\lambda(\rho)\frac{u_1}{r^2} +\lambda(\rho)\frac{\nabla u_1}{r})\dot u \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq C\int_0^Tt^{1+\delta_0}\Vert\dot u\Vert_{L^2}\Vert u_1\Vert_{L^4}\Vert\nabla \rho\Vert_{L^4}\mathrm{d}t +C\Big(\int_0^Tt^{1+\delta_0}\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\Big(\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\leq C\Big(\int_0^Tt^{1+\delta_0}\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}(\sup\limits_{t\in[T]}\Vert \nabla\rho\Vert_{L^4})+C_0^{\frac{1}{4(1+\delta_0)}}A_6(T)^{\frac{1}{2}}\nonumber\\ &\leq CC_0^{\frac{1}{4(1+\delta_0)}}A_6(T)^{\frac{1}{2}}. \end{aligned}$

并由引理2.2 (参数为 $m=2,r=2,q=3,\theta=\frac{1}{3}$), (2.5) 式, 先验假设(3.8), (3.10) 和 (3.11) 式, 有

$\begin{aligned} & \int_0^T t^{1+\delta_0}\Vert\nabla u\Vert^3_{L^3}\mathrm{d}t\nonumber\\ &\leq C\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2 (\Vert\dot u\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^2} +\Vert\nabla\rho\Vert_{L^4}^{2} \Vert\nabla u\Vert_{L^2} +\Vert\nabla u\Vert_{L^2})\mathrm{d}t\nonumber\\ &\leq C\sup\limits_{t\in[T]}(t^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2)^{\frac{1}{2}} \Big(\int_0^Tt^{1+\delta_0}\Vert\dot u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\quad+C \sup\limits_{t\in[T]}(\Vert\nabla \rho\Vert_{L^2}) \int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\nonumber\\ &\quad+ C\sup\limits_{t\in[T]}(t^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2)^{\frac{1}{2}} \int_0^Tt^{\frac{1+\delta_0}{2}}\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t \sup\limits_{t\in[T]}(\Vert\nabla \rho\Vert_{L^4}^{2})\nonumber\\ &\quad+ C\sup\limits_{t\in[T]}(t^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2)^{\frac{1}{2}} \int_0^Tt^{\frac{1+\delta_0}{2}}\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\nonumber\\ &\leq CA_6(T)C_0^{\frac12}+CC_0^{\frac{1}{2(1+\delta_0)}} +CA_6(T)^{\frac{1}{2}}C_0^{\frac{1}{2(1+\delta_0)}}. \end{aligned}$

将(3.53)-(3.58) 式代入(3.52) 式得到

$\begin{equation*} \begin{split} A_6(T)&\leq CC_0^\frac12A_6(T) +CC_0^\frac{1}{4(1+\delta_0)}A_6(T)^\frac12 +CC_0^{\frac{1}{2(1+\delta_0)}}\\ &\leq (\frac12+CC_0^\frac12)A_6(T) +CC_0^\frac{1}{2(1+\delta_0)}, \end{split} \end{equation*}$

因此当 $CC_0^\frac12<\frac12$ 时, 有

$\begin{equation} \begin{split} A_6(T)\leq CC_0^{\frac{1}{2(1+\delta_0)}}. \end{split} \end{equation}$

对 $(1.6)_2^i$ 乘 $t^{1+\delta_0}[\frac{\partial}{\partial t}+\mathrm{div}(u\cdot)+\frac{u_1}{r}]\cdot\dot u_i$ 关于 $i$ 求和并且在 $[T]\times\Omega$ 上积分得到

$\begin{aligned} & \int_0^T\int_\Omega t^{1+\delta_0}[\partial_t(\rho\dot u_i)+\mathrm{div}(u\cdot\rho\dot u_i) +\frac{u_1}{r}\rho\dot u_i]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega t^{1+\delta_0}[\partial_t(\partial_iP)+\mathrm{div}(u\cdot\partial_iP)+\frac{u_1}{r}\partial_iP] \cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &=\int_0^T\int_\Omega t^{1+\delta_0}[\mathrm{div}(\mu(\rho)\nabla u_i)_t+\mathrm{div}(u\cdot \mathrm{div}(\mu(\rho)\nabla u_i)) +\frac{u_1}{r}\mathrm{div}(\mu(\rho)\nabla u_i)]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega t^{1+\delta_0}[(\mu(\rho)\partial_r(\frac{u_1}{r}))_t +\mathrm{div}(u\cdot \mu(\rho)\partial_r(\frac{u_1}{r})) +\frac{u_1}{r}\mu(\rho)\partial_r(\frac{u_1}{r})]\cdot\dot u_1\nonumber\\ &\quad+t^{1+\delta_0}[\frac{1}{r}(\mu(\rho)\partial_ru_2)_t +\mathrm{div}(u\cdot\frac{1}{r}\mu(\rho)\partial_ru_2) +\frac{u_1}{r}\mu(\rho)\frac{1}{r}\partial_ru_2]\cdot\dot u_2 \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega t^{1+\delta_0}[\partial_i(\lambda(\rho)\mathrm{div}u)_t +\mathrm{div}(u\cdot\partial_i(\lambda(\rho)\mathrm{div}u)) +\frac{u_1}{r}\partial_r(\lambda(\rho)\mathrm{div}u)]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\quad+\int_0^T\int_\Omega t^{1+\delta_0}[\partial_i(\lambda(\rho)\frac{u_1}{r})_t +\mathrm{div}(u\cdot\partial_i(\lambda(\rho)\frac{u_1}{r})) +\frac{u_1}{r}\partial_i(\lambda(\rho)\frac{u_1}{r})]\cdot\dot u_i\mathrm{d}r\mathrm{d}z\mathrm{d}t. \end{aligned}$

由(3.26) 式, 有

$\begin{aligned} &\int_0^T\int_\Omega t^{1+\delta_0}[\partial_t(\rho\dot u_i)+\mathrm{div}(u\cdot\rho\dot u_i) +\frac{u_1}{r}\rho\dot u_i]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &=\int_\Omega t^{1+\delta_0}\frac{1}{2}\rho\vert\dot u\vert^2 \mathrm{d}r\mathrm{d}z -\int_0^T\int_\Omega(1+\delta_0) t^{\delta_0}\frac{1}{2}\rho\vert\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t+\int_0^T \int_\Omega\frac{1}{2}t^{1+\delta_0}\rho\vert\dot u\vert^2\frac{u_1}{r}\mathrm{d}r\mathrm{d}z\mathrm{d}t,\nonumber \end{aligned}$

其中由 (3.16) 和 (3.35) 式, 对于 $1\leq\delta_0<\infty$, 有

$\begin{aligned} &\int_0^T\int_\Omega\frac{(1+\delta_0)}{2} t^{\delta_0}\rho\vert\dot u\vert^2 \mathrm{d}r\mathrm{d}z\mathrm{d}t+\int_0^T \int_\Omega\frac{1}{2}t^{1+\delta_0}\rho\vert\dot u\vert^2\frac{u_1}{r}\mathrm{d}r\mathrm{d}z{\rm d}t\nonumber\\ &\leq C\int_0^1t^{\delta_0}\Vert\dot u\Vert^2_{L^2}\mathrm{d}t+A_6(T)+C\sup\limits_{0\leq t\leq T}(\Vert u\Vert_{L^2})\int_0^Tt^{1+\delta_0}\Vert \nabla\dot u\Vert_{L^2}^2{\rm d}t\\ &\leq CC_0^{\frac{1}{4}}+A_6(T)+CC_0^{\frac12}A_7(T).\nonumber \end{aligned}$

与引理3.3 中 $A_2(T)$ 的证明类似, 并且结合(3.11) 式得到

$\begin{equation} \begin{split} &\int_0^T\int_\Omega t^{1+\delta_0}[\partial_t(\partial_iP)+\mathrm{div}(u\cdot\partial_iP)+\frac{u_1}{r}\partial_iP] \cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &\leq C(\int_0^T t^{1+\delta_0}\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t)^\frac{1}{2} (\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert^2_{L^2}\mathrm{d}t)^\frac{1}{2}\\ &\leq CA_7(T)^{\frac{1}{2}}C_0^{\frac{1}{4(1+\delta_0)}}. \end{split} \end{equation}$

由于 $(1.4)_1$, (3.8) 和 (3.11) 式, 可知

$\begin{aligned} &\int_0^T\int_\Omega t^{1+\delta_0}[\mathrm{div}(\mu(\rho)\nabla u_i)_t+\mathrm{div}(u\cdot \mathrm{div}(\mu(\rho)\nabla u_i)) +\frac{u_1}{r}\mathrm{div}(\mu(\rho)\nabla u_i)]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq-\int_0^T t^{1+\delta_0}\mu(\rho)\Vert\nabla\dot u\Vert_{L^2}^2\mathrm{d}t +C\Big(\int_0^T t^{1+\delta_0}\Vert\nabla\dot u\Vert_{L^2}^2\mathrm{d}t\Big)^\frac{1}{2} \Big(\int_0^T t^{1+\delta_0}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2}\nonumber\\ &\leq-\int_0^Tt^{1+\delta_0}\mu(\rho)\Vert\nabla\dot u\Vert_{L^2}^2\mathrm{d}t+CA_7(T)^{\frac{1}{2}} \Big(\int_0^Tt^{1+\delta_0}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^\frac{1}{2} \end{aligned}$

和

$\begin{aligned} &\int_0^T\int_\Omega t^{1+\delta_0}[(\mu(\rho)\partial_r(\frac{u_1}{r}))_t +\mathrm{div}(u\cdot \mu(\rho)\partial_r(\frac{u_1}{r})) +\frac{u_1}{r}\mu(\rho)\partial_r(\frac{u_1}{r})]\cdot\dot u_1\nonumber\\ &\quad+t^{1+\delta_0}[\frac{1}{r}(\mu(\rho)\partial_ru_2)_t +\mathrm{div}(u\cdot\frac{1}{r}\mu(\rho)\partial_ru_2) +\frac{u_1}{r}\mu(\rho)\frac{1}{r}\partial_ru_2]\cdot\dot u_2 \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq C(\int_0^T t^{1+\delta_0}\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t)^{\frac{1}{2}} \Big(\int_0^T t^{1+\delta_0}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^{\frac{1}{2}}-\int_0^T\int_\Omega t^{1+\delta_0}\mu(\rho)\frac{\vert\dot u_1\vert^2}{r^2}\mathrm{d}r\mathrm{d}z\mathrm{d}t\\ &\quad+C\Big(\int_0^T t^{1+\delta_0}\Vert\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T t^{1+\delta_0}\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\leq CA_7(T)^{\frac{1}{2}}\Big(\int_0^T t^{1+\delta_0}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^{\frac{1}{2}} +CA_6(T)^{\frac{1}{2}}A_7(T)^{\frac{1}{2}} -\int_0^T\int_\Omega t^{1+\delta_0}\mu(\rho)\frac{\vert\dot u_1\vert^2}{r^2}\mathrm{d}r\mathrm{d}z\mathrm{d}t.\nonumber \end{aligned}$

对于 $\lambda(\rho)$ 同理, 有

$\begin{aligned} &\int_0^T\int_\Omega t^{1+\delta_0}[\partial_i(\lambda(\rho)\mathrm{div}u)_t +\mathrm{div}(u\cdot\partial_i(\lambda(\rho)\mathrm{div}u)) +\frac{u_1}{r}\partial_i(\lambda(\rho)\mathrm{div}u)]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq-\int_0^T\int_\Omega t^{1+\delta_0}\lambda(\rho)\vert \mathrm{div}\dot u\vert^2\mathrm{d}r\mathrm{d}z\mathrm{d}t +CA_7(T)^{\frac{1}{2}}\Big(\int_0^T t^{1+\delta_0}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^{\frac{1}{2}} \end{aligned}$

和

$\begin{aligned} &\int_0^T\int_\Omega t^{1+\delta_0}[\partial_i(\lambda(\rho)\frac{u_1}{r})_t +\mathrm{\mathrm{div}}(u\cdot\partial_i(\lambda(\rho)\frac{u_1}{r})) +\frac{u_1}{r}\partial_i(\lambda(\rho)\frac{u_1}{r})]\cdot\dot u_i \mathrm{d}r\mathrm{d}z\mathrm{d}t\nonumber\\ &\leq CA_7(T)^{\frac{1}{2}}\Big(\int_0^T t^{1+\delta_0}\Vert\nabla u\Vert^4_{L^4}\mathrm{d}t\Big)^{\frac{1}{4}} +CA_6(T)^{\frac{1}{2}}A_7(T)^{\frac{1}{2}}. \end{aligned}$

并且由于引理 2.2(参数为 $m=2,r=2,q=4,\theta=\frac{1}{2}$), (2.5), 先验假设条件 (3.8), (3.10), (3.11) 和 (3.16) 式可得

$\begin{array}{l}\begin{aligned}& \int_{0}^{T} t^{1+\delta_{0}}\|\nabla u\|_{L^{4}}^{4} \mathrm{~d} t \\\leq & C \int_{0}^{T} t^{1+\delta_{0}}\|\nabla u\|_{L^{2}}^{2}\left\|\nabla^{2} u\right\|_{L^{2}}^{2} \mathrm{~d} t \\\leq & C \int_{0}^{T} t^{1+\delta_{0}}\|\nabla u\|_{L^{2}}^{2}\left(\|\dot{u}\|_{L^{2}}^{2}+\|\nabla \rho\|_{L^{2}}^{2}+\|\nabla \rho\|_{L^{4}}^{4}\|\nabla u\|_{L^{2}}^{2}+\|\nabla u\|_{L^{2}}^{2}\right) \mathrm{d} t \\\leq & C \sup _{t \in[0, T]}\left(t^{1+\delta_{0}}\|\dot{u}\|_{L^{2}}^{2}\right) \int_{0}^{T}\|\nabla u\|_{L^{2}}^{2} \mathrm{~d} t+\sup _{t \in[0, T]}\left(\|\nabla \rho\|_{L^{2}}^{2}\right) \int_{0}^{T} t^{1+\delta_{0}}\|\nabla u\|_{L^{2}}^{2} \mathrm{~d} t \\& +C \sup _{t \in[0, T]}\left(\|\nabla u\|_{L^{2}}^{2}\right) \sup _{t \in[0, T]}\left(\|\nabla \rho\|_{L^{4}}^{4}\right) \int_{0}^{T} t^{1+\delta_{0}}\|\nabla u\|_{L^{2}}^{2} \mathrm{~d} t\end{aligned}\\\begin{aligned}& +C \sup _{t \in[0, T]}\left(\|\nabla u\|_{L^{2}}^{2}\right) \int_{0}^{T} t^{1+\delta_{0}}\|\nabla u\|_{L^{2}}^{2} \mathrm{~d} t \\\leq & C C_{0} A_{7}(T)+C C_{0}^{\frac{1}{2\left(1+\delta_{0}\right)}}\end{aligned}\end{array}$

将(3.61)-(3.68) 式代入(3.60) 式, 由于(3.59) 式, 得到

$\begin{equation*} \begin{split} A_7(T)&\leq CC_0^\frac12A_7(T) +CC_0^\frac{1}{4(1+\delta_0)}A_7(T)^\frac12 +CA_6(T)^\frac12A_7(T)^\frac12 +CC_0^{\frac{1}{4}} +CC_0^{\frac{1}{2(1+\delta_0)}}\\ & \leq (\frac12+CC_0^\frac12)A_7(T) +CC_0^\frac{1}{2(1+\delta_0)}+CC_0^{\frac{1}{4}}, \end{split} \end{equation*}$

因此当 $CC_0<\frac12$ 适当小时, 得到$A_7(T)\leq CC_0^{\frac{1}{2(1+\delta_0)}}$,即完成了引理的证明.

接下来的引理3.6-3.9 将封闭 (3.8) 式中提出的先验假设, 从而完成命题 3.1的证明.

引理 3.6 在定理1.1 的条件和先验假设（3.8） 式下, 如果$(\rho,u)$ 是（1.4）, （1.5） 式在$\Omega\times(0,T]$ 中的光滑解, 对于$2\leq q<\infty$, 有如下不等式

$\begin{equation} \sup\limits_{t\in[T]}\Vert\nabla\rho\Vert_{L^q}^q +\int_0^T\Vert\nabla\rho\Vert^q_{L^q}\mathrm{d}t\leq 2M^q. \end{equation}$

证 将$\partial_i$ 作用到$(1.6)_1$ 式上后乘 $q\partial_i\rho\vert\nabla\rho\vert^{q-2}$ 可得

$\begin{equation} \begin{split} &(\partial_i\rho)_trq\partial_i\rho\vert\nabla\rho\vert^{q-2} +\rho_tq\partial_r\rho\vert\nabla\rho\vert^{q-2} +\partial_i\partial_k\rho u_krq\partial_i\rho\vert\nabla\rho\vert^{q-2} +\partial_k\rho\partial_iu_krq\partial_i\rho\vert\nabla\rho\vert^{q-2}\\ &+\partial_r\rho u_1q\partial_r\rho\vert\nabla\rho\vert^{q-2} +\partial_i\rho \mathrm{div}urq\partial_i\rho\vert\nabla\rho\vert^{q-2} +\rho\partial_i\mathrm{div}urq\partial_i\rho\vert\nabla\rho\vert^{q-2}\\ & +\rho \mathrm{div}uq\partial_r\rho\vert\nabla\rho\vert^{q-2}+\partial_i\rho u_1q\partial_i\rho\vert\nabla\rho\vert^{q-2} +\rho\partial_iu_1q\partial_i\rho\vert\nabla\rho\vert^{q-2} =0. \end{split} \end{equation}$

将(3.70) 式在$\Omega$ 上积分, 通过简单计算可得

$\begin{aligned} & \frac{\mathrm{d}}{\mathrm{d}t}\Big(\int_\Omega\vert\nabla\rho\vert^qr\mathrm{d}r\mathrm{d}z\Big) +\int_\Omega \rho\partial_i\mathrm{div}urq\partial_i\rho\vert\nabla\rho\vert^{q-2} \mathrm{d}r\mathrm{d}z\nonumber\\ &\leq \int_\Omega (1-q)\vert\nabla\rho\vert^{q}(\mathrm{div}ur+u_1) +\rho\frac{u_1}{r}q\partial_r\rho\vert\nabla\rho\vert^{q-2} \mathrm{d}r\mathrm{d}z\\ & -\int_\Omega \partial_k\rho\partial_iu_krq\partial_i\rho\vert\nabla\rho^\zeta\vert^{q-2} +\rho\partial_iu_1q\partial_i\rho\vert\nabla\rho\vert^{q-2} \mathrm{d}r\mathrm{d}z.\nonumber \end{aligned}$

又由于$F$ 的定义, 有

$\begin{equation} \begin{split} \partial_i\mathrm{div}u=(1+\frac{\lambda(\rho)}{\mu(\rho)})^{-1} [\partial_iF-\frac{\nabla\lambda(\rho)}{\mu(\rho)}\mathrm{\mathrm{div}}u +\frac{\nabla\mu(\rho)\lambda(\rho)}{\mu(\rho)^2}\mathrm{\mathrm{div}}u +\frac{\gamma\rho^{\gamma-1}\partial_i\rho}{\mu(\rho)}]. \end{split} \end{equation}$

从而由(2.1), (3.71) 式和 (3.72) 式得到

$\begin{aligned} & \frac{\mathrm{d}}{\mathrm{d}t}\Vert\nabla\rho\Vert^q_{L^q} +\Vert\nabla\rho\Vert^q_{L^q}\nonumber\\ &\leq C\Vert\nabla\rho\Vert_{L^q}^q (\Vert\nabla u\Vert_{L^\infty}+\Vert u\Vert_{L^\infty}) +C\int_\Omega\vert\nabla\rho\vert^{q-1} (\vert\nabla F\vert+\vert\nabla u\vert)\mathrm{d}r\mathrm{d}z\\ &\leq C\Vert\nabla\rho\Vert_{L^q}^q (\Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2}) +C\Vert\nabla\rho\Vert_{L^q}^{q-1}(\Vert\nabla F\Vert_{L^q}+\Vert\nabla u\Vert_{L^q}).\nonumber \end{aligned}$

由(2.3) 式和引理2.2 (参数为 $m=3,r=2,q=\infty,\theta=\frac{3}{4}$) 可知

$\begin{aligned} \int_0^T\Vert\nabla u\Vert_{L^\infty}\mathrm{d}t &\leq \int_0^T\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\nabla^2u\Vert_{L^3}^{\frac34} \mathrm{d}t\\ &\leq \int_0^T\Vert\nabla u\Vert_{L^2}^{\frac14} (\Vert\rho\dot u\Vert_{L^3} +\Vert\nabla\rho^\gamma\Vert_{L^3} +\Vert\nabla\mu(\rho)\nabla u\Vert_{L^3}\nonumber\\ &\quad+\Vert\nabla\lambda(\rho)\mathrm{\mathrm{div}}u\Vert_{L^3} +\Vert\nabla u\Vert_{L^3} +\Vert\nabla\lambda(\rho)u\Vert_{L^3})^{\frac34} \mathrm{d}t.\nonumber \end{aligned}$

结合(3.8), (3.35), (3.51) 式和 Sobolev 嵌入($W^{1,2}(\Omega)\hookrightarrow L^q(\Omega), 2\leq q < \infty)$ 得到, 对任意 $\delta_0>\frac{q-2}{2q-2}(<\frac12)$,

$\begin{aligned} & \int_0^T\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\rho\dot u\Vert_{L^3}^{\frac34}\mathrm{d}t\nonumber\\ &\leq C\int_0^T\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\nabla\dot u\Vert_{L^2}^{\frac34}\mathrm{d}t\nonumber\\ &\leq C\int_0^{\sigma(T)}\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\nabla\dot u\Vert_{L^2}^{\frac34}\mathrm{d}t +C\int_{\sigma(T)}^T\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\nabla\dot u\Vert_{L^2}^{\frac34}\mathrm{d}t\nonumber\\ &\leq C\int_0^{\sigma(T)}\Vert\nabla\dot u\Vert_{L^2}^{\frac34} t^{\frac13}\cdot t^{-\frac13} \Vert\nabla u\Vert_{L^2}^{\frac14} \mathrm{d}t +C\int_{\sigma(T)}^T (t^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2)^{\frac{1}{8}} \Vert\nabla\dot u\Vert_{L^2}^{\frac{3}{4}}t^{\frac{3(1+\delta_0)}{8}} \cdot t^{-\frac{(1+\delta_0)}{2}} \mathrm{d}t\nonumber\\ &\leq C\Big(\int_0^{\sigma(T)}\Vert\nabla\dot u\Vert_{L^2}^2t^{\frac{8}{9}}\mathrm{d}t\Big)^{\frac{3}{8}} \Big(\int_0^{\sigma(T)}\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{8}} \Big(\int_0^{\sigma(T)}t^{-\frac23}\mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\quad+ C\sup\limits_{\sigma(T)\leq t\leq T}(t^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2)^{\frac{1}{8}}\Big(\int_{\sigma(T)}^Tt^{1+\delta_0}\Vert\nabla\dot u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{3}{8}} \Big(\int_{\sigma(T)}^Tt^{-\frac{4(1+\delta_0)}{5}}\mathrm{d}t\Big)^{\frac{5}{8}}\nonumber\\ &\leq CC_0^{\frac{7}{32}}+CC_0^{\frac{1}{4(1+\delta_0)}}. \end{aligned}$

再结合(3.8), (3.11) 和 (3.35) 式有

$\begin{aligned} & \int_0^T\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\nabla\rho^\gamma\Vert_{L^3}^{\frac{3}{4}}\mathrm{d}t\nonumber\\ &\leq C\sup\limits_{0\leq t\leq\sigma(T)}(t^{\frac{3}{4}}\Vert\nabla u\Vert_{L^2}^2)^{\frac{1}{8}} \Big(\int_0^{\sigma(T)}\Vert\nabla\rho\Vert_{L^3}^{\frac{3}{4}} t^{-\frac{3}{32}}\mathrm{d}t\Big)\nonumber\\ &\quad+C\int_{\sigma(T)}^T \Vert\nabla\rho\Vert_{L^3}^{\frac{3}{4}} \Vert\nabla u\Vert_{L^3}^{\frac34} t^{\frac{(1+\delta_0)}{8}}\cdot t^{-\frac{(1+\delta_0)}{8}}\mathrm{d}t\nonumber\\ &\leq CC_0^{\frac{1}{32}} \Big(\int_0^{\sigma(T)}\Vert\nabla\rho\Vert_{L^3}\mathrm{d}t\Big) ^{\frac34} \Big(\int_0^{\sigma(T)}t^{-\frac{3}{8}}\mathrm{d}t\Big)^{\frac{1}{4}}\\ &\quad+C\Big(\int_{\sigma(T)}^T \Vert\nabla\rho\Vert_{L^3}\mathrm{d}t\Big) ^{\frac34} \Big(\int_{\sigma(T)}^T\Vert\nabla u\Vert_{L^2}^2 t^{1+\delta_0}\mathrm{d}t\Big)^{\frac{1}{8}} \Big(\int_{\sigma(T)}^Tt^{-(1+\delta_0)}\mathrm{d}t\Big)^{\frac18}\nonumber\\ &\leq CC_0^{\frac{1}{32}}+CC_0^{\frac{1}{16(1+\delta_0)}},\nonumber \end{aligned}$

由(3.8) 式, Young 不等式和 H$\ddot{\mathrm{o}}$lder 不等式可得

$\begin{aligned} & \int_0^T\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\nabla\mu(\rho)\nabla u\Vert_{L^3} ^{\frac14}\mathrm{d}t\nonumber\\ &\leq C\int_0^T\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\nabla u\Vert_{L^\infty}^{\frac34} \Vert\nabla\rho\Vert_{L^3}^{\frac34} \mathrm{d}t\nonumber\\ &\leq \frac18\int_0^T\Vert\nabla u\Vert_{L^\infty} \mathrm{d}t +C\int_0^T\Vert\nabla u\Vert_{L^2} \Vert\nabla\rho\Vert_{L^3}^{3} \mathrm{d}t\\ &\leq \frac18\int_0^T\Vert\nabla u\Vert_{L^\infty} \mathrm{d}t +C\Big(\int_0^T\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\sup\limits_{t\in[T]}\Vert\nabla\rho\Vert_{L^3}^{\frac{5}{2}}\Big)\Big(\int_0^T\Vert\nabla\rho\Vert_{L^3}\mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\leq \frac18\int_0^T\Vert\nabla u\Vert_{L^\infty} \mathrm{d}t +CC_0^{\frac{1}{2}}.\nonumber \end{aligned}$

同理，可得

$\begin{aligned} & \int_0^T\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\nabla\lambda(\rho)\mathrm{div} u\Vert_{L^3} ^{\frac34}\mathrm{d}t\nonumber\\ &\leq \frac18\int_0^T\Vert\nabla u\Vert_{L^\infty} \mathrm{d}t +C\Big(\int_0^T\Vert\nabla u\Vert_{L^2}^2\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_0^T\Vert\nabla\rho\Vert_{L^3}^{6}\mathrm{d}t\Big)^{\frac{1}{2}}\\ &\leq \frac18\int_0^T\Vert\nabla u\Vert_{L^\infty} \mathrm{d}t +CC_0^{\frac{1}{2}}\nonumber \end{aligned}$

和

$\begin{aligned} & \int_0^T\Vert\nabla u\Vert_{L^2}^{\frac14} \Vert\nabla u\Vert_{L^3} ^{\frac34}\mathrm{d}t\nonumber\\ &\leq \frac18\int_0^T\Vert\nabla u\Vert_{L^\infty} \mathrm{d}t+C(\sup\limits_{t\in[\sigma(T)]}t^{\frac34}\Vert\nabla u\Vert_{L^2}^2)^{\frac12}\int_0^{\sigma(T)} t^{-\frac38} \mathrm{d}t\nonumber\\ &\ \ \ +C\Big(\int_{\sigma(T)}^Tt^{1+\delta_0}\Vert \nabla u\Vert_{L^2}^2 \mathrm{d}t\Big)^{\frac12}\Big(\int_{\sigma(T)}^T t^{-(1+\delta_0)} \mathrm{d}t\Big)^{\frac12}\\ &\leq \frac18\int_0^T\Vert\nabla u\Vert_{L^\infty} \mathrm{d}t +CC_0^{\frac18}+CC_0^{\frac{1}{4(1+\delta_0)}}.\nonumber \end{aligned}$

类似地, 结合(2.1), (3.8) 和(3.10) 式可知

$\begin{aligned}& \int_{0}^{T}\|\nabla u\|_{L^{2}}^{\frac{1}{4}}\|\nabla \lambda(\rho) u\|_{L^{3}}^{\frac{3}{4}} \mathrm{~d} t \\\leq & C \int_{0}^{T}\|\nabla u\|_{L^{2}}^{\frac{1}{4}}\|\nabla \rho\|_{L^{3}}^{\frac{3}{4}}\|u\|_{L^{\infty}}^{\frac{3}{4}} \mathrm{~d} t \\\leq & \frac{1}{8} \int_{0}^{T}\|\nabla u\|_{L^{\infty}} \mathrm{d} t+C\left(\int_{0}^{T}\|\nabla u\|_{L^{2}}^{2} \mathrm{~d} t\right)^{\frac{1}{2}}\left(\int_{0}^{T}\|\nabla \rho\|_{L^{3}}^{6}+\|\nabla \rho\|_{L^{3}}^{\frac{3}{2}} \mathrm{~d} t\right)^{\frac{1}{2}} \\\leq & \frac{1}{8} \int_{0}^{T}\|\nabla u\|_{L^{\infty}} \mathrm{d} t+C C_{0}^{\frac{1}{2}}.\end{aligned}$

将(3.75)-(3.80) 式代入(3.74) 式可以得到以下关键估计

$\begin{equation} \begin{split} \int_0^T\Vert\nabla u\Vert_{L^\infty}\mathrm{d}t\leq CC_0^{\frac{1}{16(1+\delta_0)}}. \end{split} \end{equation}$

因此对(3.73) 式在$(0,T]$ 上积分有如下形式

$\begin{aligned} & \Vert\nabla\rho\Vert^q_{L^q} +\int_0^T\Vert\nabla\rho\Vert^q_{L^q}\mathrm{d}t\nonumber\\ &\leq \Vert\nabla\rho_0\Vert^q_{L^q} +C\int_0^T\Vert\nabla\rho\Vert^q_{L^q} (\Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^2})\mathrm{d}t\\ &\quad+C\int_0^T\Vert\nabla\rho\Vert^q_{L^q} (\Vert\nabla F\Vert_{L^q} +\Vert\nabla u\Vert_{L^q})\mathrm{d}t +C\int_0^T(\Vert\nabla F\Vert_{L^q} +\Vert\nabla u\Vert_{L^q})\mathrm{d}t.\nonumber \end{aligned}$

由(2.2), (3.8), (3.11), (3.35), (3.51) 和 (3.81) 式有

$\begin{aligned} \int_0^T\Vert\nabla F\Vert_{L^q}\mathrm{d}t &\leq C\int_0^T[(\Vert\nabla\rho\Vert_{L^q}+1)( \Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2}) +\Vert\dot u\Vert_{L^q}]\mathrm{d}t\nonumber\\ &\leq C\int_0^T(\Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^2} +\Vert\nabla\dot u\Vert_{L^2})\mathrm{d}t\\ &\leq CC_0^{\frac{1}{16(1+\delta_0)}}+C(\sup\limits_{t\in[\sigma(T)]}t^{\frac34}\Vert\nabla u\Vert_{L^2}^2)^{\frac12}\int_0^{\sigma(T)} t^{-\frac38}{\rm d}t\nonumber\\ &\quad +\Big(\int_0^{\sigma(T)}t^{\frac89}\Vert\nabla\dot u\Vert^2_{L^2}\mathrm{d}t\Big)^{\frac{1}{2}}\Big(\int_0^{\sigma(T)}t^{-\frac89} \mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\quad +\Big(\int_{\sigma(T)}^Tt^{1+\delta_0}(\Vert\nabla u\Vert^2_{L^2} +\Vert\nabla\dot u\Vert^2_{L^2})\mathrm{d}t\Big)^{\frac{1}{2}} \Big(\int_{\sigma(T)}^Tt^{-(1+\delta_0)}\mathrm{d}t\Big)^{\frac{1}{2}}\nonumber\\ &\leq CC_0^{\frac{1}{16(1+\delta_0)}}.\nonumber \end{aligned}$

将(3.83) 式代入(3.82) 式后再结合(3.16), (3.8) 式和 Gronwall 不等式可以得到

$\begin{aligned} & \Vert\nabla\rho\Vert^q_{L^q} +\int_0^T\Vert\nabla\rho\Vert^q_{L^q}\mathrm{d}t\nonumber\\ &\leq (\Vert\nabla\rho_0\Vert^q_{L^q} +C\int_0^T(\Vert\nabla F\Vert_{L^q} +\Vert\nabla u\Vert_{L^2}+\Vert\nabla u\Vert_{L^\infty})\mathrm{d}t) {\rm e}^{C\int_0^T(\Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^2}+\Vert\nabla F\Vert_{L^q})\mathrm{d}t}\nonumber\\ &\leq (M^q +C_1C_0^{\frac{1}{16(1+\delta_0)}}){\rm e}^{C_2C_0^{\frac{1}{16(1+\delta_0)}}}\\ &\leq 2M^q,\nonumber \end{aligned}$

其中存在$C_0$ 适当小使得$C_1C_0^{\frac{1}{16(1+\delta_0)}}<\frac12M^q$, 并且 ${\rm e}^{C_2C_0^{\frac{1}{16(1+\delta_0)}}}<\frac43$. 即完成引理3.6 的证明.

引理 3.7 在定理1.1 的条件和先验假设(3.8) 下, 如果$(\rho,u)$ 是(1.4),(1.5) 式在$\Omega\times(0,T]$ 中的光滑解, 对于$2\leq q<\infty$, 有如下不等式

$\begin{equation} \int_0^T\Vert\nabla\rho\Vert_{L^q}\mathrm{d}t\leq 2M. \end{equation}$

证 对(3.73) 式两端同时除$\Vert\nabla\rho\Vert_{L^q}^{q-1}$, 可得

$\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t}\Vert\nabla\rho\Vert_{L^q} +\Vert\nabla\rho\Vert_{L^q} \leq C\Vert\nabla\rho\Vert_{L^q} (\Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2}) +C(\Vert\nabla F\Vert_{L^q}+\Vert\nabla u\Vert_{L^q}), \end{equation}$

再继续对(3.86) 式在$(0,T]$ 上积分有

$\begin{aligned} & \Vert\nabla\rho\Vert_{L^q} +\int_0^T\Vert\nabla\rho\Vert_{L^q}\mathrm{d}t\nonumber\\ &\leq \Vert\nabla\rho_0\Vert_{L^q} +C\int_0^T\Vert\nabla\rho\Vert_{L^q} (\Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^2})\mathrm{d}t +C\int_0^T(\Vert\nabla F\Vert_{L^q} +\Vert\nabla u\Vert_{L^q})\mathrm{d}t. \end{aligned}$

结合 Gronwall 不等式,(3.81) 和(3.83) 式得到

$\begin{aligned} & \Vert\nabla\rho\Vert_{L^q} +\int_0^T\Vert\nabla\rho\Vert_{L^q}\mathrm{d}t\nonumber\\ &\leq (\Vert\nabla\rho_0\Vert_{L^q} +C\int_0^T(\Vert\nabla F\Vert_{L^q} +\Vert\nabla u\Vert_{L^q})\mathrm{d}t) {\rm e}^{\int_0^T(\Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^2})\mathrm{d}t}\\ &\leq (M +C_3C_0^{\frac{1}{16(1+\delta_0)}}){\rm e}^{C_4C_0^{\frac{1}{16(1+\delta_0)}}}\nonumber\\ &\leq 2M,\nonumber \end{aligned}$

其中存在$C_0$ 适当小使得$C_3C_0^{\frac{1}{16(1+\delta_0)}}<\frac12 M$, 并且${\rm e}^{C_4C_0^{\frac{1}{16(1+\delta_0)}}}<\frac43$. 就可以得到引理的证明.

引理 3.8 在定理1.1 的条件和先验假设(3.8) 下, 如果$(\rho,u)$ 是(1.4),(1.5) 式在$\Omega\times(0,T]$ 中的光滑解, 有如下不等式

$\begin{equation} \int_0^T\Vert\nabla\rho\Vert_{L^2}^{2} t^{1+\delta_0}\mathrm{d}t \leq 4^{1+\delta_0}(1+\delta_0)^{1+\delta_0}M^2. \end{equation}$

证 对(3.73) 式中取$q=2$ 并且对于不等式两端同乘 $t^{1+\delta_0}$ 可得

$\begin{align*} & \frac{\mathrm{d}}{\mathrm{d}t} (t^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2}) +\int_\Omega t^{1+\delta_0}(\mu(\rho)+\lambda(\rho))^{-1}\rho^\gamma\vert\nabla\rho\vert^2\mathrm{d}r\mathrm{d}z\nonumber\\ &\leq Ct^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2} (\Vert\nabla u\Vert_{L^\infty} +\Vert u\Vert_{L^\infty}) +(1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert^2_{L^2}\\ &\quad+C t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^2}(\Vert\nabla F\Vert_{L^2}+\Vert\nabla u\Vert_{L^2}),\nonumber \end{align*}$

对上式再关于$(0,T]$ 积分和先验假设(3.8) 有

$\begin{aligned} & t^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2} +\int_0^T t^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2}\mathrm{d}t\nonumber\\ &\leq C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2} (\Vert\nabla u\Vert_{L^\infty} +\Vert u\Vert_{L^\infty})\mathrm{d}t +\int_0^T(1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert^2_{L^2} \mathrm{d}t\\ &\quad+C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^2}(\Vert\nabla F\Vert_{L^2}+\Vert\nabla u\Vert_{L^2})\mathrm{d}t.\nonumber \end{aligned}$

由(3.8) 式和 (3.69) 式, 取$T_3=4(1+\delta_0)$, 得

$\begin{aligned} \int_0^T(1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert^2_{L^2} \mathrm{d}t &=\int_0^{T_3}(1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert^2_{L^2} \mathrm{d}t +\int^T_{T_3}(1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert^2_{L^2} \mathrm{d}t\nonumber\\ &\leq (1+\delta_0)T_3^{\delta_0}\int_0^{T_3}\Vert\nabla\rho\Vert^2_{L^2} \mathrm{d}t +\frac{(1+\delta_0)}{T_3}\int^T_{T_3}t^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2} \mathrm{d}t\\ &\leq 2^{1+2\delta_0}(1+\delta_0)^{1+\delta_0}M^2 +\frac{1}{4}\int^T_{T_3}t^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2} \mathrm{d}t.\nonumber \end{aligned}$

结合(3.11),(3.51) 式, 有

$\begin{aligned} & \int_0^T t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^2} (\Vert\nabla F\Vert_{L^2}+\Vert\nabla u\Vert_{L^2})\mathrm{d}t\nonumber\\ &\leq \frac{1}{4} \int_0^T t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^2}^2\mathrm{d}t \!+\!C\int_0^T t^{1+\delta_0}(\Vert\nabla\dot u\Vert^2_{L^2} +\Vert\nabla u\Vert^2_{L^2})\mathrm{d}t\!+\!C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^2}^2 \Vert\nabla u\Vert_{L^\infty}\mathrm{d}t\nonumber\\ &\leq CC_0^{\frac{1}{2(1+\delta_0)}}+ \frac{1}{4} \int_0^T t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^2}^2\mathrm{d}t +C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^2}^2 \Vert\nabla u\Vert_{L^\infty}\mathrm{d}t, \end{aligned}$

其中

$\begin{equation*} \begin{split} \Vert\nabla F\Vert_{L^2}\leq (\Vert\nabla\rho\Vert_{L^2}+1)( \Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^2}) +\Vert\nabla\dot u\Vert_{L^2}. \end{split} \end{equation*}$

将(3.91), (3.92) 式代入(3.90) 式得到

$\begin{equation} \begin{split} & t^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2} +\frac{1}{2}\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^2}^2\mathrm{d}t\\ &\leq C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2} (\Vert\nabla u\Vert_{L^\infty} +\Vert \nabla u\Vert_{L^2})\mathrm{d}t +2^{1+2\delta_0}(1+\delta_0)^{1+\delta_0}M^2+CC_0^{\frac{1}{2(1+\delta_0)}}, \end{split} \end{equation}$

再结合(2.1),(3.81) 式和 Gronwall 不等式可知

$\begin{aligned} & t^{1+\delta_0}\Vert\nabla\rho\Vert^2_{L^2} +\frac{1}{2}\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^2}^2\mathrm{d}t\nonumber\\ &\leq (2^{1+2\delta_0}(1+\delta_0)^{1+\delta_0}M^2+CC_0^{\frac{1}{2(1+\delta_0)}}) {\rm e}^{\int_0^T\Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^2}\mathrm{d}t}\\ &\leq (2^{1+2\delta_0}(1+\delta_0)^{1+\delta_0}M^2+C_5C_0^{\frac{1}{2(1+\delta_0)}}){\rm e}^{C_6C_0^{\frac{1}{16(1+\delta_0)}}}\nonumber\\ &\leq 2^{3+2\delta_0}(1+\delta_0)^{1+\delta_0}M^2,\nonumber \end{aligned}$

其中存在 $C_0$ 适当小使得 ${\rm e}^{C_6C_0^{\frac{1}{16(1+\delta_0)}}}<2$, 且 $C_5C_0^{\frac{1}{2(1+\delta_0)}}<2^{1+2\delta_0}(1+\delta_0)^{1+\delta_0}M^2$. 即得到引理的证明.

现在我们继续推导密度的一致上下界, 这是获得所有高阶估计的关键, 进而将强解扩展到全局. 我们使用的方法主要是引理2.3 并结合前面已有的估计.

引理 3.9 在定理1.1 和命题3.1 的条件下, 如果$(\rho,u)$ 是(1.4),(1.5) 式在$\Omega\times(0,T]$ 中的光滑解, 有如下不等式

$\begin{equation} \frac{2}{3}\underline{\rho}\leq\inf_{\Omega\times(0,T]}\rho\leq\sup\limits_{\Omega\times(0,T]}\rho\leq \frac{3}{2}\overline{\rho}. \end{equation}$

证 重写方程 $(1.6)_1$ 如下

$\begin{equation*} \begin{split} D_t(\rho r)\triangleq g_1(\rho)+b_1'(t), \end{split} \end{equation*}$

其中

$\begin{equation*} \begin{split} D_t(\rho r)\triangleq \rho_tr+u\cdot\nabla(\rho r), g_1(\rho)\triangleq-\rho r\frac{\int_{\widetilde{\rho}}^\rho\frac{\gamma s^{\gamma-1}}{\mu(s)}{\rm d}s}{1+\frac{\lambda(\rho)}{\mu(\rho)}}, b_1(t)\triangleq-\int_0^t\frac{\rho rF}{1+\frac{\lambda(\rho)}{\mu(\rho)}}\mathrm{d}t. \end{split} \end{equation*}$

对于$t\in[T]$ 和所有 $0\leq t_1<t_2\leq T$, 由引理 (3.16),(3.81),(3.69) 式有

$\begin{aligned} \vert b_1(t_2)-b_1(t_1)\vert &\leq C\int_{t_1}^{t_2}\Vert\frac{\rho rF}{1+\frac{\lambda(\rho)}{\mu(\rho)}}\Vert_{L^\infty}\mathrm{d}t\nonumber\\ &\leq C \int_0^T\Vert F\Vert^{\frac{q-2}{2(q-1)}}_{L^2}\Vert\nabla F\Vert^{\frac{q}{2(q-1)}}_{L^q}\mathrm{d}t\nonumber\\ &\leq C \int_{t_1}^{t_2}(\Vert\nabla u\Vert_{L^2} +\Vert\int_{\widetilde{\rho}}^\rho\frac{\gamma s^{\gamma-1}}{\mu(s)}\mathrm{d}s\Vert_{L^2})^{\frac{q-2}{2{q-1}}}\\ &\ \ \ \ \times(\Vert\nabla\rho\Vert_{L^q} \Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2} +\Vert\dot u\Vert_{L^q})^{\frac{q}{2{q-1}}}\mathrm{d}t\nonumber\\ &\leq C\int_{t_1}^{t_2} \Vert\nabla\rho\Vert_{L^q} \Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2} +\Vert\nabla\dot u\Vert_{L^2}\mathrm{d}t+(t_2-t_1)\sup\limits_{t\in[t_1,t_2]}\Vert\rho-{\widetilde{\rho}}\Vert_{L^2}\nonumber\\ &\leq C(\underline{\rho},\overline{\rho},{\widetilde{\rho}},M)C_0^\beta+(t_2-t_1)C_0^{\frac12},\nonumber \end{aligned}$

这里$\beta={\frac{1}{16(1+\delta_0)}}$. 我们可以选择 (2.3) 式中的$N_0$ 和$N_1$ 如下

$\begin{equation*} \begin{split} N_1=C_0^{\frac12}, N_0=C(\underline{\rho},\overline{\rho},{\widetilde{\rho}},M)C_0^\beta, \end{split} \end{equation*}$

并且选择(2.7)式中的$\overline{\zeta}=\widetilde{\rho}$, 则对于所有$\zeta\geq2\widetilde{\rho}$ 有

$\begin{equation*} \begin{split} g_1(\zeta)=-\zeta r\frac{\int_{\widetilde{\rho}}^\zeta\frac{\gamma s^{\gamma-1}}{\mu(s)}{\rm d}s}{1+\frac{\lambda(\zeta)}{\mu(\zeta)}} \leq-C_0^{\frac12}. \end{split} \end{equation*}$

根据引理2.3 可以得到

$\begin{equation} \begin{split} \sup\limits_{t\in[T]}\Vert\rho\Vert_{L^\infty}\leq\max\{\overline{\rho},2\widetilde{\rho}\}+N_0 \leq\overline{\rho}+C(\underline{\rho},\overline{\rho},{\widetilde{\rho}},M)C_0^\beta \leq2\overline{\rho}, \end{split} \end{equation}$

其中$C_0\leq(\frac{\overline{\rho}}{2C(\underline{\rho},\overline{\rho},{\widetilde{\rho}},M)} )^{\frac{1}{\beta}}$.

对于$(\rho r)^{-1}$ 进行类似的处理

$\begin{equation*} \begin{split} D_t(\rho r)^{-1}=(\rho r)^{-1}\mathrm{div}u\triangleq g_2(\rho)+b_2'(t), \end{split} \end{equation*}$

其中

$\begin{equation*} \begin{split} g_2(\rho^{-1})\triangleq(\rho r)^{-1}\frac{\int_{\widetilde{\rho}}^\rho\frac{\gamma s^{\gamma-1}}{\mu(s)}{\rm d}s}{1+\frac{\lambda(\rho)}{\mu(\rho)}}, b_2(t)\triangleq\int_0^t\frac{(\rho r)^{-1}F}{1+\frac{\lambda(\rho)}{\mu(\rho)}}\mathrm{d}t. \end{split} \end{equation*}$

对于$t\in[T]$ 对于所有 $0\leq t_1<t_2\leq T$, 由(3.96) 式可知

$\begin{aligned} \vert b_2(t_2)-b_2(t_1)\vert &\leq C\int_{t_1}^{t_2}\Vert\frac{(\rho r)^{-1}F}{1+\frac{\lambda(\rho)}{\mu(\rho)}}\Vert_{L^\infty}\mathrm{d}t\nonumber\\ &\leq C(\underline{\rho},\overline{\rho},{\widetilde{\rho}},M)C_0^\beta+(t_2-t_1)C_0^{\frac12},\nonumber \end{aligned}$

其中的$\beta={\frac{1}{16(1+\delta_0)}}$. 我们可以选择(2.3) 式中的 $N_0$ 和 $N_1$ 如下

$\begin{equation*} \begin{split} N_1=C_0^{\frac12}, N_0=C(\underline{\rho},\overline{\rho},{\widetilde{\rho}},M)C_0^\beta, \end{split} \end{equation*}$

并且选择(2.7) 式中的 $\overline{\zeta}=\widetilde{\rho}^{-1}$. 则对于所有 $\zeta\geq2\widetilde{\rho}^{-1}$ 有

$\begin{equation*} \begin{split} g_2(\zeta)=\zeta r^{-1}\frac{\int_{\widetilde{\rho}}^{\zeta^{-1}}\frac{\gamma s^{\gamma-1}}{\mu(s)}{\rm d}s}{1+\frac{\lambda(\zeta)}{\mu(\zeta)}} \leq-C_0^{\frac12}. \end{split} \end{equation*}$

同样根据引理2.3 可得

$\begin{equation} \begin{split} \sup\limits_{t\in[T]}\Vert\rho^{-1}\Vert_{L^\infty}\leq\max\{{\underline{\rho}}^{-1},2\widetilde{\rho}^{-1}\}+N_0 \leq{\underline{\rho}}^{-1}+C(\underline{\rho},\overline{\rho},{\widetilde{\rho}},M)C_0^\beta \leq2{\underline{\rho}}^{-1}, \end{split} \end{equation}$

其中$C_0\leq(\frac{1}{2C(\underline{\rho},\overline{\rho},{\widetilde{\rho}},M)\underline{\rho}})^{\frac{1}{\beta}} $.

最后, 结合 (3.97) 式和 (3.98) 式就完成了引理3.9的证明.

下面我们可以得到解的重要的高阶导数估计.

引理 3.10 在定1.1 的条件下, 对于 $\tau\in(0,T)$, 存在正常数 $C(\tau)$, 使得 $(\rho,u)$ 有如下不等式成立

$\begin{equation} \begin{split} \sup\limits_{t\in[T]}\int_\Omega\rho\vert u_t\vert^2\mathrm{d}r\mathrm{d}z +\int_0^T\int_\Omega\vert\nabla u_t\vert^2\mathrm{d}r\mathrm{d}z\mathrm{d}t\leq C, \end{split} \end{equation}$
$\begin{equation} \begin{split} \sup\limits_{t\in[T]}\int_\Omega\vert\nabla u_t\vert^2\mathrm{d}r\mathrm{d}z +\int_0^T\int_\Omega\rho u_{tt}^2\mathrm{d}r\mathrm{d}z\mathrm{d}t\leq C, \end{split} \end{equation}$
$\begin{equation} \begin{split} \sup\limits_{t\in[T]}(\Vert\rho-\widetilde{\rho}\Vert_{H^3} +(\Vert P(\rho)-P(\widetilde{\rho})\Vert_{H^3})\leq C, \end{split} \end{equation}$
$\begin{equation} \begin{split} \sup\limits_{t\in[T]}(\Vert\nabla u_t\Vert_{L^2} +\Vert\nabla u\Vert_{H^2}) +\int_0^T(\Vert\nabla u\Vert^2_{H^3}+\Vert\nabla u_t\Vert^2_{H^1})\mathrm{d}t \leq C, \end{split} \end{equation}$
$\begin{equation} \begin{split} \sup\limits_{t\in[\tau,T]}(\Vert\nabla u_t\Vert_{H^1} +\Vert\nabla^4u\Vert_{L^2}) +\int_\tau^T\int_\Omega\vert\nabla u_{tt}\vert^2\mathrm{d}r\mathrm{d}z\mathrm{d}t \leq C(\tau). \end{split} \end{equation}$

证 引理3.10 的证明与文献 [16] 中证明高阶估计类似, 在此省略.

4 定理1.1 证明

有了第 3 节中的所有先验估计, 就可以证明本文的主要结果.

定理 1.1 的证明 根据引理2.1, 存在$T'$ 使得(1.4),(1.5) 式在 $\mathbb{R}^2\times(0,T']$ 有一个唯一的强解$(\rho,u)$. 我们将用先验估计,命题2.1 和引理3.10 来扩展局部强解到全局.

由于 (1.8) 式可知

$\begin{align*} 0<\underline{\rho}\leq\rho_0\leq\overline{\rho}, \Vert\nabla\rho_0\Vert_{L^p} \leq M, \end{align*}$

由于 $C_0\leq\varepsilon$, 则存在 $T_1\in(0,T']$ 使得对于 $T=T_1$, (3.8) 式成立.

接下来, 令

$\begin{aligned} T^*=\sup\{ T \vert (3.8) \mbox{式成立}\}, \end{aligned}$

则 $T^*\geq T_1>0$. 因此对于有限的 $T$ ($0<\tau<T\leq T^*$), 根据引理 3.10 有

$\begin{equation} \begin{split} \nabla u_t,\nabla^3u\in C([\tau,T];L^2\cap L^4), \nabla u,\nabla^2u\in C([\tau,T];L^2\cap C(\overline{\Omega})), \end{split} \end{equation}$

这里我们用到了下面的嵌入

$\begin{equation*} \begin{split} L^{\infty}(\tau,T;H^1)\cap H^1(\tau,T;H^{-1})\hookrightarrow C([\tau,T];L^q) \mbox{对} q\in[1,\infty). \end{split} \end{equation*}$

由于 (3.99),(3.100),(3.103) 式可得

$\begin{equation*} \begin{split} &\int^T_\tau\Vert(\rho\vert u_t\vert^2)_t\Vert_{L^1}\mathrm{d}t\\ &\leq\int^T_\tau\Vert\rho_t\vert u_t\vert^2\Vert_{L^1} +2\Vert\rho u_t\cdot u_{tt}\Vert_{L^1})\mathrm{d}t\\ &\leq C\int^T_\tau(\Vert\rho \vert \mathrm{div}u\vert\vert u_t\vert^2\Vert_{L^1} +\Vert\vert u\vert\vert\nabla\rho\vert\vert u_t\vert^2\Vert_{L^1} +\Vert\rho^{\frac{1}{2}}u_t\Vert_{L^2}\Vert\rho^{\frac{1}{2}}u_{tt}\Vert_{L^2})\mathrm{d}t\\ &\leq C\int^T_\tau(\Vert\rho\vert u_t\vert^2\Vert_{L^1}\Vert\nabla u\Vert_{L^\infty} +\Vert u\Vert_{L^6}\Vert\nabla\rho\Vert_{L^2} \Vert u_t\Vert_{L^6}^2 +\Vert\rho^{\frac{1}{2}}u_{tt}\Vert_{L^2})\mathrm{d}t\\ &\leq C. \end{split} \end{equation*}$

于是

$\rho^{\frac{1}{2}}u_t\in C([\tau,T];L^2).$

再结合 (4.2) 式, 就得到

$\begin{equation} \rho^{\frac{1}{2}}\dot u,\nabla\dot u\in C([\tau,T];L^2). \end{equation}$

下面我们断言

$\begin{equation} T^*=\infty. \end{equation}$

否则如果 $T^*<\infty$, 那么根据命题 3.1, (3.8) 式对于 $T=T^*$ 成立. 由引理 3.10 和 (4.3) 式可得 $(\rho(r,z,T^*),u(r,z,T^*))$ 满足 (1.8) 式. 因此, 可将 $(\rho(r,z,T^*),u(r,z,T^*))$ 看作引理 2.1 中的初值 $(\rho_0,u_0)$, 而 $(\rho(r,z,T^*),u(r,z,T^*))$ 满足正则性条件(1.8), 那么由引理 2.1 可知存在有限时间 $T^{**}>T^*>0$, 使得问题(1.4), (1.5) 在 $\Omega\times(T^*,T^{**}]$ 上存在唯一的强解 $(\rho,u)$, 即存在 $T^{**}>T^*$, 使得 (3.8) 式对 $T=T^{**}$ 成立, 这与 (4.1) 式相矛盾, 这就说明了 (4.4) 式成立. 要完成定理 1.1 的证明, 现在我们只需证 (1.11) 式成立.

对 (3.73) 式两端同乘 $t^{1+\delta_0}$ 并且取 $q=3$, 可以得到

$\begin{equation} \begin{split} & \frac{\mathrm{d}}{\mathrm{d}t}t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 +t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3\\ &\leq (1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert_{L^3}^3+Ct^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 (\Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2})\\ &\quad+Ct^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^{2}(\Vert\nabla F\Vert_{L^3}+\Vert\nabla u\Vert_{L^3}), \end{split} \end{equation}$

再继续对 (4.5) 式在 $(0,T]$ 上积分有

$\begin{aligned} & t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 +\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3\mathrm{d}t\nonumber\\ &\leq \int_0^T(1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert_{L^3}^3\mathrm{d}t\nonumber +C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 (\Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^2})\mathrm{d}t\\ &\quad+C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^{2}(\Vert\nabla F\Vert_{L^3} +\Vert\nabla u\Vert_{L^3})\mathrm{d}t. \end{aligned}$

接下来的证明跟引理 3.8 类似, 由 (3.69) 式, 取 $T_4=4(1+\delta_0)$, 得

$\begin{aligned} \int_0^T(1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 \mathrm{d}t &=\int_0^{T_4}(1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 \mathrm{d}t +\int^T_{T_4}(1+\delta_0)t^{\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 \mathrm{d}t\nonumber\\ &\leq (1+\delta_0)T_4^{\delta_0}\int_0^{T_4}\Vert\nabla\rho\Vert_{L^3}^3 \mathrm{d}t +\frac{(1+\delta_0)}{T_4}\int^T_{T_4}t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 \mathrm{d}t\\ &\leq 2^{1+2\delta_0}(1+\delta_0)^{1+\delta_0}M^3 +\frac{1}{4}\int^T_{T_4}t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 \mathrm{d}t.\nonumber \end{aligned}$

结合 (3.16),(3.51),(3.81) 式, Poincare 不等式和引理 2.2 (参数为 $m=2,r=2,q=3,\theta=\frac{1}{3}$) 有

$\begin{aligned} & \int_0^T t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^{2} (\Vert\nabla F\Vert_{L^3}+\Vert\nabla u\Vert_{L^3})\mathrm{d}t\nonumber\\ &\leq C\int_0^T t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^{2}(\Vert\dot u\Vert_{L^3} +\Vert\nabla u\Vert_{L^3})\mathrm{d}t+C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert^3_{L^3} (\Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^3})\mathrm{d}t\nonumber\\ &\leq C(\int_0^T t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^{3}\mathrm{d}t)^{\frac{2}{3}}(\int_0^T t^{1+\delta_0}(\Vert\dot u\Vert_{L^3}^3 +\Vert\nabla u\Vert_{L^3}^3)\mathrm{d}t)^{\frac{1}{3}}\nonumber\\ &\quad+C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 (\Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2})\mathrm{d}t\\ &\leq\frac{1}{4}\int^T_{0}t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3\mathrm{d}t +C\int_0^T t^{1+\delta_0}(\Vert\dot u\Vert_{L^2}\Vert\nabla\dot u\Vert_{L^2}^2 +\Vert\nabla u\Vert_{L^2}^2\Vert\nabla u\Vert_{L^\infty}) \mathrm{d}t \nonumber\\ &\quad+C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 (\Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2})\mathrm{d}t\nonumber\\ &\leq\frac{1}{4}\int^T_{0}t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3\mathrm{d}t +C\sup\limits_{t\in[t]}(\Vert\dot u\Vert_{L^2})\int_0^Tt^{1+\delta_0}\Vert\nabla\dot u\Vert_{L^2}^2 \mathrm{d}t\nonumber\\ &\quad+C\sup\limits_{t\in[t]}(t^{1+\delta_0}\Vert\nabla u\Vert_{L^2}^2)\int_0^T \Vert\nabla u\Vert_{L^\infty}\mathrm{d}t +C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 (\Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2})\mathrm{d}t\nonumber\\ &\leq\frac{1}{4}\int^T_{0}t^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3\mathrm{d}t +CC_0^{\frac{1}{2(1+\delta_0)}} +CC_0^{\frac{9}{16(1+\delta_0)}} \nonumber\\ &\quad +C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3 (\Vert\nabla u\Vert_{L^\infty}+\Vert\nabla u\Vert_{L^2})\mathrm{d}t,\nonumber \end{aligned}$

其中

$\begin{equation*} \Vert\nabla F\Vert_{L^3}\leq C[\Vert\dot u\Vert_{L^3}+ (\Vert\nabla\rho\Vert_{L^3}+1) (\Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^3})]. \end{equation*}$

将 (4.7),(4.8) 式代入 (4.6) 式得到

$\begin{aligned} & t^{1+\delta_0}\Vert\nabla\rho\Vert^3_{L^3} +\frac{1}{2}\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3\mathrm{d}t\\ &\leq C\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert^3_{L^3} (\Vert\nabla u\Vert_{L^\infty} +\Vert \nabla u\Vert_{L^2})\mathrm{d}t +2^{1+2\delta_0}(1+\delta_0)^{1+\delta_0}M^3+CC_0^{\frac{1}{2(1+\delta_0)}},\nonumber \end{aligned}$

再结合 (2.1),(3.81) 式和 Gronwall 不等式可知

$\begin{aligned} & t^{1+\delta_0}\Vert\nabla\rho\Vert^3_{L^3} +\frac{1}{2}\int_0^Tt^{1+\delta_0}\Vert\nabla\rho\Vert_{L^3}^3\mathrm{d}t\nonumber\\ &\leq (2^{1+2\delta_0}(1+\delta_0)^{1+\delta_0}M^3+CC_0^{\frac{1}{2(1+\delta_0)^2}}) {\rm e}^{\int_0^T\Vert\nabla u\Vert_{L^\infty} +\Vert\nabla u\Vert_{L^2}\mathrm{d}t}\\ &\leq (2^{1+2\delta_0}(1+\delta_0)^{1+\delta_0}M^3+CC_0^{\frac{1}{2(1+\delta_0)}}) {\rm e}^{CC_0^{\frac{1}{16(1+\delta_0)}}}\nonumber\\ &\leq C.\nonumber \end{aligned}$

这说明

$\begin{equation} \Vert\nabla\rho\Vert_{L^3}^3<Ct^{-(1+\delta_0)}. \end{equation}$

由引理 3.2 和 (4.11) 式得到

$\begin{equation} \vert\rho-\widetilde{\rho}\vert\leq\Vert\rho-\widetilde{\rho}\Vert_{L^2}^{\frac{1}{4}} \Vert\nabla\rho\Vert_{L^3}^{\frac{3}{4}} \leq Ct^{-\frac{3(1+\delta_0)}{8}}. \end{equation}$

这就完成了定理 1.1 的证明.

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