数学物理学报, 2026, 46(4): 1374-1392

球对称非等熵可压缩 Navier-Stokes 方程的自由边界问题——献给邓引斌教授 70 寿辰

董文超,1,2, 郭真华,1,2,*, 李振甲,1,2

1 广西大学数学学院 南宁 530004

2 广西应用数学中心 (广西大学) 南宁 530004

Free Boundary Problem for the Spherically Symmetric Non-Isentropic Compressible Navier-Stokes Equations

Dong Wenchao,1,2, Guo Zhenhua,1,2,*, Li Zhenjia,1,2

1 School of Mathematics, Guangxi University, Nanning 530004

2 Center for Applied Mathematics of Guangxi (Guangxi University), Nanning 530004

通讯作者: * 郭真华,E-mail: zhguo@gxu.edu.cn

收稿日期: 2025-12-19   修回日期: 2026-04-30  

基金资助: 国家自然科学基金(12501298)
国家自然科学基金(11931013)
广西自然科学基金(2026GXNSFBA00640111)
广西自然科学基金(2022GXNSFDA035078)

Received: 2025-12-19   Revised: 2026-04-30  

Fund supported: NSFC(12501298)
NSFC(11931013)
GXNSF(2026GXNSFBA00640111)
GXNSF(2022GXNSFDA035078)

作者简介 About authors

董文超,E-mail:wcdong@gxu.edu.cn;

李振甲,E-mail:1467663039@qq.com

摘要

关于输运系数依赖于温度的非等熵可压缩 Navier-Stokes 方程大初值全局适定性问题的研究, 目前主要集中于一维情形, 高维情形的结果相对较少. 针对黏性系数为常数、热传导系数依赖于温度和密度的三维球对称非等熵可压缩 Navier-Stokes 方程的自由边界问题, 在初值属于 $H^1$-空间的条件下, 建立了全局强解的存在唯一性.

关键词： 球对称 Navier-Stokes 方程; 自由边界问题; 适定性

Abstract

Research on the global well-posedness for the non-isentropic compressible Navier-Stokes equations with large initial data, where the transport coefficients depend on temperature, has primarily focused on the one-dimensional case, while results in higher dimensions remain relatively scarce. This paper studies a free boundary problem for the three-dimensional spherically symmetric non-isentropic compressible Navier-Stokes equations, assuming constant viscosity and heat conductivity depending on both temperature and density. Under the condition that the initial data belong to the $H^1$ space, we establish the existence and uniqueness of global strong solution.

Keywords： spherically symmetric Navier-Stokes equations; free boundary problem; well-posedness

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董文超, 郭真华, 李振甲. 球对称非等熵可压缩 Navier-Stokes 方程的自由边界问题——献给邓引斌教授 70 寿辰[J]. 数学物理学报, 2026, 46(4): 1374-1392

Dong Wenchao, Guo Zhenhua, Li Zhenjia. Free Boundary Problem for the Spherically Symmetric Non-Isentropic Compressible Navier-Stokes Equations[J]. Acta Mathematica Scientia, 2026, 46(4): 1374-1392

1 引言

考虑三维非等熵可压缩 Navier-Stokes 方程组

$\begin{cases}\partial_\tau \rho + \operatorname{div}(\rho U) = 0, \\\partial_\tau (\rho U) + \operatorname{div}(\rho U \otimes U) + \nabla p - \operatorname{div}\mathbb{S} = 0, \\\partial_\tau (\rho E) + \operatorname{div}\big((\rho E + p) U\big) = \operatorname{div}(\mathbb{S} U) + \operatorname{div}(\kappa(\rho,\theta) \nabla \theta),\end{cases}$

其中空间 $x\in\Omega_{\tau}\subset\mathbb{R}^3$, 时间 $\tau\geq0$, $\rho$ 是密度, $U$ 是速度场, $\theta$ 是温度, 压力 $p$ 与黏性应力张量 $\mathbb{S}$ 满足

$\begin{align*} p = a \rho^{\gamma} + R \rho \theta, \qquad \mathbb{S} = \lambda(\rho,\theta) (\operatorname{div} U) \mathbb{I} + \mu(\rho,\theta)\left(\frac{\nabla U + (\nabla U)^T}{2}\right), \end{align*}$

总能量 $\rho E$ 为

$\begin{align*} \rho E = \frac{1}{2}\rho|U|^2+\frac{a}{\gamma - 1} \rho^{\gamma} + c_v \rho\theta, \end{align*}$

$a>0, R>0, \gamma>1, c_v>0$ 均为给定常数, 黏性系数 $\lambda(\rho,\theta), \mu(\rho,\theta)$ 与热传导系数 $\kappa(\rho,\theta)$ 为依赖于 $\rho, \theta$ 的光滑函数. 初值条件

$\begin{matrix} (\rho, U, \theta)(x,0)=(\rho_0, U_0, \theta_0)(x), \qquad x\in\Omega_{0}:=\big\{x\in\mathbb{R}^3 \, \big| \, a_0\leq|x|\leq a_1\big\}, \end{matrix}$

边界条件为

$\begin{matrix} \left\{ \begin{array}{ll} u=0, \qquad &x\in\partial\Omega_{a_0}, \\ (p\mathbb{I}-\mathbb{S})\cdot n =P(\tau)n, \qquad &x\in\partial\Omega_{a_1(\tau)}, \\ \nabla \theta\cdot n =0, \qquad &x\in\partial\Omega_{\tau}, \end{array} \right. \end{matrix}$

其中常数 $0<a_0< a_1$, $\partial\Omega_{a_1(\tau)}=\psi(\partial\Omega_{a_1},\tau)$ 是自由边界, $\partial\Omega_{a_1}$ 表示自由边界的初始位置,

$\partial\Omega_ {a_i}:=\big\{x\in\mathbb{R}^3 \, \big| \, |x|= a_i\big\}, i=0,1, \qquad \partial\Omega_{\tau}:= \partial\Omega_{a_0}\cup\partial\Omega_{a_1(\tau)},$

$n$ 是 $\partial\Omega_{a_1(\tau)}$ 的单位外法向量, $P(\tau)\geq0$ 是给定的外压力 (依赖于时间 $\tau$ 的 $C^1$-函数), 流函数 $\psi$ 满足

$\begin{align*} \left\{ \begin{array}{ll} \partial_{\tau}\psi(x,\tau)=U(\psi(x,\tau), \tau), \\ \psi(x,0)=x. \end{array} \right. \end{align*}$

可压缩 Navier-Stokes 方程组 (1.1) 的各类初边值问题一直是偏微分方程领域的重点研究方向之一. 对于常系数情形, 已有大量研究成果. Kazhikhov 与 Shelukhin[25] 利用能量方法, 首次证明了一维有界区域上初边值问题大初值 $H^1$-整体经典解的存在唯一性. 随后 Kanel[22] 于 1979 年建立了 Cauchy 问题在小初始扰动情形下经典解的全局适定性. Kawashima 与 Nishida[23] 则建立了一类 Nishida-Smoller[33] 型稳定性结果, 表明当绝热指数 $\gamma$ 充分接近 $1$ 时, 一维 Cauchy 问题存在大初值整体强解. 至 1982 年, Kazhikhov[24] 去掉了上述关于 $\gamma-1$ 的小性条件. 此后, 一系列文献 [19],[20],[21],[27] 等进一步分析了此类强解的大时间行为.

在高维一般区域中, 文献 [35] 在初始能量充分小的条件下, 建立了初边值问题的整体光滑解. 随后, Hoff[14],[15] 将该类结果推广至间断初值的情形. Danchin[6]则在 Besov 空间中建立了解的存在唯一性. 对于含真空的问题, Lions[28]运用有效黏性通量方法, 证明了在 $\gamma>\frac{9}{5}$ 时等熵情形有限能量弱解的整体存在性. 该结果随后被 Feireisl 等[13]改进至 $\gamma>\frac{3}{2}$. Huang 等[16],[18]则研究了具小能量大振荡的经典解的整体存在性.

在非等熵情形下, Matsumura 与 Nishida 的系列工作 [30],[31],[32] 研究了三维空间中强解与经典解的全局适定性, 首次建立了平衡态附近小扰动解的整体存在性. 1998 年, Xin[37] 对初始密度具紧支集的情形, 给出了光滑解的爆破准则. 上述结果随后被 Xin 与 Yan[38] 推广至初始密度无紧支集的情形. Huang 与 Li[17] 以及 Lai 等[26]分别证明了 Cauchy 问题中小能量大振荡经典解的整体存在性. 最近, Chen 等[4]针对球对称问题进行研究, 获得了无真空情形下大初值整体解的存在性.

然而, 若从 Boltzmann 方程出发, 基于 Chapman-Enskog 二阶展开理论[3]并结合若干物理假设, 可导出可压缩 Navier-Stokes 方程组 (1.1), 其输运系数 $\mu$, $\kappa$ 满足关系

$\begin{matrix} \mu,\kappa\propto\theta^\alpha, \ \ \ \alpha\geq\frac{1}{2}. \end{matrix}$

其中, $\alpha=\frac{1}{2}$ 对应于弹性球体模型, $\alpha=1$ 对应于 Maxwell 分子模型, 而 $\alpha=\frac{5}{2}$ 则对应于电离气体情形. 因此, 研究满足 (1.4) 的输运系数更具物理现实意义. 已有文献 [5],[7],[8],[34] 分别针对满足 (1.4) 的不同类型输运系数, 研究了一维初边值问题. 对于一维 Cauchy 问题, Liu 等[29]首次获得了一类大初值下的 Nishida-Smoller 型稳定性结果. 随后, Wang 与 Zhao[36] 研究了输运系数形如 (其中 $v=1/\rho$ 表示比容)

$\begin{array}{c} \mu, \kappa \propto h(v) \theta^{\alpha}, \quad|\alpha| \ll 1, \\ C^{-1}\left(v^{l_{1}}+v^{-l_{2}}\right) \leq h(v) \leq C\left(v^{l_{3}}+v^{-l_{4}}\right), h^{\prime}(v)^{2} v \leq C h(v)^{3}, \\ \frac{37\left(1-l_{1}\right)_{+}}{4\left(1+2 l_{1}\right)}+\frac{9\left(1-l_{2}\right)_{+}}{2 l_{2}}+\frac{1}{4} \max \left\{\frac{\left(l_{3}-1\right)_{+}}{1+2 l_{1}}, \frac{\left(l_{4}+1\right)_{+}}{2 l_{2}}\right\}<2, \quad l_{i}>0, \end{array}$

在 $H^3 \times H^3 \times H^3$ 范数意义下大扰动平衡态解的稳定性. 近期, Dong 与 Guo[9] 进一步讨论了 $\mu = \theta^\alpha,\ \kappa = \theta^\beta$ 且 $|\alpha| \ll 1,\ 0 \leq \beta < +\infty$ 时, $H^2 \times H^1 \times H^1$ 大扰动下平衡态解的稳定性. 在高维问题方面, Feireisl 等[10],[11],[12]围绕黏性系数非退化情形开展了一系列关于弱解的研究. Cao 与 Li 在文献 [2] 中则研究了输运系数依赖于温度的初边值问题局部强解的存在性.

针对问题 (1.1)-(1.3) 在 $P(\tau) \equiv 0$ 的情形, Bian 等在文献 [1] 中研究了: 当

$\begin{matrix} \mu={\rm const.}, \qquad \lambda=0, \\ C^{-1}(1+\theta^q)\leq \kappa=\kappa(\theta)\leq C(1+\theta^q), \ q\geq1, \\ |\kappa'(\theta)|\leq C(1+\theta^q), \end{matrix}$

且初值满足

$\begin{matrix} \rho_0\in W^{1,\infty}(\Omega_0), \ (U_0,\theta_0)\in H^2(\Omega_0) \end{matrix}$

时, 球对称强解的整体存在唯一性问题. 然而, 对于 $P(\tau)\not\equiv 0$ 的情形, 相关研究尚属空白. 因此, 本文旨在对问题 (1.1)-(1.3) 的球对称强解进行系统研究, 重点聚焦于 $P(\tau) \geq 0$ 的情形, 同时放宽对热传导系数及初值正则性的要求.

2 主要结论

关注三维球对称问题, 即取 $r = |x|$,

$\rho(x,\tau) = \rho(r,\tau), \quad U(x,\tau) = u(r,\tau)\frac{x}{r},\quad \theta(x,\tau) = \theta(r,\tau).$

假设黏性系数 $\mu>0$, $\lambda\geq0$ 均为常数, 存在 $q \geq 0$ 与常数 $C>0$ 使得

$\begin{matrix} C^{-1}\theta^q\leq\kappa(\rho, \theta)\leq C(1+\theta^{q+1-\epsilon}), \quad 0<\epsilon\ll1. \end{matrix}$

自由边界问题 (1.1)-(1.3) 可转化为

$\begin{cases}\left(r^2\rho\right)_\tau + \left(r^2\rho u\right)_r = 0, \\\displaystyle\left(r^2\rho u\right)_\tau + \left(r^2\rho u^2\right)_r + r^2 p_r-r^2\left(\lambda u_r +\mu u_r + \frac{2\lambda}{r}u + \frac{2\mu}{r}u\right)_r = 0, \\\displaystyle \left(r^2\rho E\right)_\tau + \left(r^2(\rho E + p)u\right)_r= \left(r^2\lambda u u_r + 2r\lambda u^2+r^2\mu u u_r \right)_r+ \left(r^2\kappa(\rho,\theta)\theta_r\right)_r,\end{cases}$

$a_0\leq r\leq a_1(\tau)$, $\tau\geq0$, 初值为

$(\rho, u, \theta)(r,0) = (\rho_0, u_0, \theta_0)(r), \quad r \in [a_0, a_1],$

边界条件如下

$\begin{cases}\displaystyle u(a_0,\tau) = 0,\\\displaystyle\left(p - \lambda u_r -\frac{2\lambda}{r} u -\mu u_r \right)(a_1(\tau),\tau) = P(\tau), \\\displaystyle \theta_r(a_0,\tau) =\theta_r(a_1(\tau),\tau) = 0,\end{cases}$

其中边界 $a_1(\tau)$ 满足

$\begin{matrix} a_1'(\tau) = u(a_1(\tau), \tau). \end{matrix}$

那么, 我们有如下的本文主要定理.

定理2.1 设边界压力 $P(\tau)\geq0$ 满足

$\begin{matrix} \begin{array}{cc} \frac{P'(\tau)_+}{P(\tau)}:=\frac{\max\{P'(\tau),0\}}{P(\tau)}\in L_{\text{loc}}^1(0,+\infty);\\ P(\tau)>0, \quad \text{当} 0\leq q<1 \text{时}, \end{array} \end{matrix}$

初值 $(\rho_0, u_0, \theta_0)(r)$ 与边界条件 (2.5) 相容, 且

$\begin{matrix} \min_{r \in [a_0, a_1]} \rho_0(r) > 0, \quad \min_{r \in [a_0, a_1]}\theta_0(r) > 0, \quad (\rho_0, u_0, \theta_0) \in H^1(a_0, a_1), \end{matrix}$

热传导系数满足 (2.2) 式, 则自由边界问题 (2.3)-(2.5) 存在唯一全局强解 $(\rho, u, \theta)(r, \tau)$. 此外, 对任意 $T>0$, $(r, \tau)\in \left\{(r,\tau) \, \big| \, a_0 \leq r \leq a_1(\tau), 0 \leq \tau \leq T \right\}$, 有

$\begin{align*} C^{-1} \leq \rho(r, \tau) \leq C, \quad C^{-1} \leq \theta(r, \tau) \leq C \end{align*}$

且

$\begin{align*} \sup_{\tau\in[T]}\|(\rho, u, \theta)\|_{H^1(a_0, a_1(\tau))}^2 +\int_0^T \|(\rho_r, u_r, \theta_r, u_{rr}, \theta_{rr}, u_\tau, \theta_\tau )\|_{L^2(a_0, a_1(\tau))}^2 \mathrm{d}\tau \leq C, \end{align*}$

其中 $C$ 为依赖于初值和时间 $T$ 的正常数.

注2.1 定理 2.1 中的条件 (2.7) 涵盖多种常见情形, 例如 $P(\tau)\equiv C$ ($C$ 为非负常数) 以及 $P(\tau)=(1+\tau)^{c}$ ($c$ 为任意实数). 事实上, 任意正的光滑函数 $P(\tau)$ 均满足条件 (2.7); 特别地, 当 $q \geq 1$ 时, 任意单调递减的非负光滑函数 $P(\tau)$ 亦满足这一条件.

注2.2 由不等式

$\begin{align*} C^{-1}\theta^q\leq C^{-1}(1+\theta^q)\leq \kappa(\rho,\theta)\leq C(1+\theta^q) \leq C(1+\theta^{q+1-\epsilon}) \end{align*}$

可知, 定理 2.1 中条件 (2.2) 弱于条件 (1.5). 值得注意的是 (2.2) 式包含了更具物理意义的 (1.4) 情形.

注2.3 定理 2.1 的另一创新之处在于显著降低了对初值正则性的要求. 我们不再要求初始密度 $\rho_0\in W^{1,\infty}(a_0,a_1)$, 亦无需初始速度与温度满足 $(u_0,\theta_0)\in H^2(a_0,a_1)$, 而仅需较弱的条件 (2.8).

3 先验估计

本节旨在通过建立先验估计, 进而完成定理 2.1 的证明. 首先, 我们对原系统作 Lagrange 变换. 质量守恒方程 $ (3)_1$ 关于空间变量 $r$ 积分可得

$\begin{align*} \frac{{\rm d}}{{\rm d}\tau}\int_{a_0}^{a_1(\tau)} \rho r^2 \mathrm{d}r=0, \end{align*}$

不失一般性, 假设 $\int_{a_0}^{a_1} \rho_0 r^2 \mathrm{d}r = 1$, 所以

$\int_{a_0}^{a_1(\tau)} \rho r^2 \mathrm{d}r\equiv 1.$

定义 Lagrange 坐标变换 $(r,\tau)\rightarrow(x,t)$:

$x(r,\tau) = \int_{a_0}^r \rho(y,\tau) y^2 \mathrm{d}y = 1 - \int_r^{a_1(\tau)} \rho(y,\tau) y^2 \mathrm{d}y,\qquad t = \tau,$

其将定义域由 $ \left\{(r,\tau) \, \big| \, a_0 \leq r \leq a_1(\tau), 0 \leq \tau \leq T \right\}$ 映射为 $[0,1] \times [T]$ 且满足

$\frac{\partial x}{\partial r} = \rho r^2, \quad\frac{\partial x}{\partial \tau} = -\rho u r^2, \quad\frac{\partial t}{\partial r} = 0, \quad\frac{\partial t}{\partial \tau} = 1, \quad\frac{\partial r}{\partial t} = u.$

因此, 自由边界问题 (2.3)-(2.5) 可转换为

$\begin{cases}\rho_t + \rho^2 (r^2 u)_x = 0, \\u_t + r^2 p_x -(\mu + \lambda) r^2 \left( \rho (r^2 u)_x \right)_x = 0, \\E_t + (r^2 p u)_x = \left( \rho r^4 \lambda u u_x + 2 r \lambda u^2 + \rho r^4 \mu u u_x \right)_x+ \left( \rho r^4 \kappa(\rho, \theta) \theta_x \right)_x,\end{cases}$

初边值条件为

$\begin{cases}(\rho, u, \theta)(x, 0) = (\rho_0, u_0, \theta_0)(x), \qquad x \in [0,1],\\u(0, t) = 0, \\\left( p - \rho r^2\lambda u_x -\frac{2}{r}\lambda u- \rho r^2\mu u_x \right)(1, t) = P(t), \\\theta_x(0, t) = \theta_x(1, t) = 0.\end{cases}$

此外, 固定边界 $x = 1$ 对应原 Euler 坐标下的自由边界 $a_1(t) = r(1, t)$:

$a_1'(t) = u(1, t), \qquad a_1(0) = a_1.$

符号说明 记 $L^m:=L^m(0,1)$ ($1 \leq m \leq +\infty$) 与 $W^{k,m}:=W^{k,m}(0,1)$ 分别为定义在区间 $[0,1]$ 上的 Lebesgue 空间和 Sobolev 空间, 其相应范数记为 $\|\cdot\|_{L^m}$ 与 $\|\cdot\|_{W^{k,m}}$; 简记 $H^k:=W^{k,2}$, $\|(f_1, \cdots, f_n)\|_{W^{k,m}}:=\sum\limits_{i=1}^{n}\|f_i\|_{W^{k,m}}$. 文中 $C=C(T)$ 表示依赖于初值及时间 $T$ 的正常数, 不同行中其具体取值可能不同.

因此, 为证明定理 2.1, 只需建立如下命题

命题3.1 设边界压力 $P(t)\geq0$ 满足

$\begin{matrix} \begin{array}{cc} \frac{P'(t)_+}{P(t)}\in L_{\text{loc}}^1(0,+\infty);\\ P(t)>0, \quad \text{当} 0\leq q<1 \text{时}, \end{array} \end{matrix}$

初值 $(\rho_0, u_0, \theta_0)(x)$ 与边界条件 (3.5) 相容, 且

$\begin{matrix} \min_{x \in [0,1]} \rho_0(x) > 0, \quad \min_{x \in [0,1]}\theta_0(x) > 0, \quad (\rho_0, u_0, \theta_0) \in H^1, \end{matrix}$

热传导系数满足 (2.2) 式, 则初边值问题 (3.4)-(3.5) 存在唯一全局强解 $(\rho, u, \theta)(x, t)$. 此外, 对任意 $T>0$, $(x, t)\in [0,1]\times[T]$, 有

$\begin{align*} C^{-1} \leq \rho(x, t) \leq C, \quad C^{-1} \leq \theta(x, t) \leq C \end{align*}$

且

$\begin{align*} \sup_{t\in[T]}\|(\rho, u, \theta)\|_{H^1}^2 +\int_0^T \|(\rho_x, u_x, \theta_x, u_{xx}, \theta_{xx}, u_t, \theta_t )\|_{L^2}^2 \mathrm{d}t \leq C. \end{align*}$

命题 3.1 的证明可基于以下一系列引理逐步完成.

引理3.1 在命题 3.1 的条件下, 有

$\begin{matrix} \sup_{t\in[T]}\left(\int_0^1 E\mathrm{d}x + a_1^3(t) P(t)\right) \leq C, \end{matrix}$

其中 $ E = \frac{1}{2}u^2+\frac{a}{\gamma - 1} \rho^{\gamma-1} + c_v \theta.$

证 $ (3.4)_3$ 关于 $x$ 在 $[0,1]$ 上积分

$\begin{matrix} \frac{\mathrm{d}}{\mathrm{d}t} \int_0^1 E \mathrm{d}x &= -P(t)(r^2 u)(1, t) \\ &= -P(t)a_1^2(t)a_1'(t) \\ &= -\frac{1}{3} \frac{\mathrm{d}}{\mathrm{d}t}\left( a_1^3(t) P(t) \right) + \frac{1}{3} a_1^3(t) P'(t). \end{matrix}$

所以

$\begin{matrix} \int_0^1 E\mathrm{d}x + \frac{1}{3} a_1^3(t) P(t) &= \int_0^1 E_0\mathrm{d}x + \frac{1}{3} a_1^3(0) P(0) + \frac{1}{3} \int_0^t a_1^3(s) P'(s)\mathrm{d}s \\ &\leq C + \frac{1}{3} \int_0^t \frac{P'(s)_+}{P(s)} a_1^3(s) P(s)\mathrm{d}s, \end{matrix}$

其中 $ E_0 = \frac{1}{2}u_0^2+\frac{a}{\gamma - 1} \rho_0^{\gamma-1} + c_v \theta_0$. 结合 (3.7) 式: $\frac{P'(s)_+}{P(s)}\in L^1(0,T)$ 以及 Grönwall 不等式可得

$\int_0^1 E\mathrm{d}x + a_1^3(t) P(t) \leq C, \qquad \forall \, t\in[T]. $

故有 (3.9) 式成立.

引理3.2 在命题 3.1 的条件下, 对任意 $t\in[T]$, 有

$\begin{matrix} a_1(t) \geq a_0 + C^{-1}. \end{matrix}$

证 利用 (3.2) 式, Hölder 不等式, 引理 3.1 以及

$\begin{align*} \int_{a_0}^r \rho^\gamma(y,\tau) r^2 \mathrm{d}y \leq\int_{a_0}^{a_1(\tau)} \rho^\gamma(y,\tau) r^2 \mathrm{d}y =\int_{0}^{1} \rho^\gamma(x,t) r^2 \frac{1}{\rho(x,t) r^2}\mathrm{d}x =\int_{0}^{1} \rho^{\gamma-1}(x,t) \mathrm{d}x \leq C, \end{align*}$

可得

$\begin{align*} x &= \int_{a_0}^r \rho(y,\tau) y^2\mathrm{d}y \leq \left( \int_{a_0}^r \left( \rho(y,\tau) y^{\frac{2}{\gamma}} \right)^\gamma \mathrm{d}y \right)^{\frac{1}{\gamma}} \left( \int_{a_0}^r y^{\frac{2\gamma - 2}{\gamma} \cdot \frac{\gamma}{\gamma - 1}} \mathrm{d}y \right)^{\frac{\gamma - 1}{\gamma}} \\ &\leq \left( \int_{a_0}^r \rho^\gamma(y,\tau) r^2 \mathrm{d}y \right)^{\frac{1}{\gamma}} \left( \frac{1}{3}r^3 - \frac{1}{3}a_0^3 \right)^{\frac{\gamma - 1}{\gamma}} \\ &\leq C\left( r^3 - a_0^3 \right)^{\frac{\gamma - 1}{\gamma}}, \end{align*}$

所以

$\begin{align*} r^3 \geq a_0^3 + C^{-1} x^{\frac{\gamma}{\gamma - 1}}. \end{align*}$

因此, 取 $x=1$, 结合 $a_1(t)=r(1,t)$ 有

$\begin{align*} a_1(t)\geq \left( a_0^3 + C^{-1} \right)^{\frac{1}{3}} \geq a_0 + C^{-1}. \end{align*}$

则可得到 (3.12) 式.

引理3.3 在命题 3.1 的条件下, 有

$\begin{matrix} \int_0^T \int_0^1 \left(\frac{ \rho r^4}{\theta} u_x^2 + \frac{1}{\rho \theta r^2} u^2 + \frac{\rho \kappa( \rho,\theta) r^4}{\theta^2} \theta_x^2\right) \mathrm{d}x\mathrm{d}t \leq C. \end{matrix}$

证 首先, 利用 (3.4) 式可得如下温度方程

$c_v \theta_t + R \rho \theta (r^2 u)_x = \mu \rho r^4 u_x^2 + \frac{2\mu}{\rho r^2}u^2 + \lambda \rho (r^2 u)_x^2 + (\rho r^4 \kappa(\rho,\theta) \theta_x)_x.$

定义截断函数 $\chi(\eta) \in C^\infty(0, \infty)$ 满足

$\begin{align*} \chi(\eta) = \begin{cases} 1, & 0 < \eta \leq 1, \\ 0, & \eta \geq 2, \end{cases} \qquad \text{且} \quad 0 \leq \chi(\eta) \leq 1. \end{align*}$

(3.14) 式乘以 $ \theta^{-1} $ 后在 $ [0,1] \times [0,t] $ 上积分, 通过引理 3.1, (3.3) 与 $ (3.4)_1$ 式可得

$\begin{align*} & \int_0^t \int_0^1 \left(\frac{\mu \rho r^4}{\theta} u_x^2 + \frac{2\mu}{\rho \theta r^2} u^2 +\frac{\lambda \rho}{\theta} (r^2 u)_x^2 + \frac{\rho \kappa( \rho,\theta) r^4}{\theta^2} \theta_x^2\right) \mathrm{d}x\mathrm{d}s\\ &=c_v \int_0^1 \ln\theta \mathrm{d}x - c_v \int_0^1 \ln\theta_0 \mathrm{d}x + \int_0^t \int_0^1 R \rho (r^2 u)_x \mathrm{d}x\mathrm{d}s\\ &\leq C\int_0^1 \theta \mathrm{d}x + C + \int_0^t \int_0^1 R \rho (r^2 u)_x \big( \chi(\rho) + 1 - \chi(\rho) \big) \mathrm{d}x\mathrm{d}s \\ &\leq C + \int_0^t \int_0^1 \left(R \rho r^2 u_x \chi(\rho) + 2R \rho r u \frac{1}{\rho r^2} \chi(\rho) - R \rho \frac{\rho_s}{\rho^2} (1 - \chi(\rho))\right) \mathrm{d}x\mathrm{d}s \\ &\leq C + \frac{\mu}{4} \int_0^t \int_0^1 \frac{\rho r^4}{\theta} u_x^2 \mathrm{d}x\mathrm{d}s + C \int_0^t \int_0^1 \rho \theta \chi^2(\rho) \mathrm{d}x\mathrm{d}s + \frac{\mu}{4} \int_0^t \int_0^1 \frac{1}{\rho \theta r^2 }u^2 \mathrm{d}x\mathrm{d}s \\ &\quad - R \int_0^t \int_0^1 \frac{\mathrm{d}}{\mathrm{d}s} \left(\ln \rho - \int_{1}^{\rho} \eta^{-1} \chi(\eta) \mathrm{d}\eta \right) \mathrm{d}x\mathrm{d}s \\ &\leq C + \frac{\mu}{4} \int_0^t \int_0^1 \left(\frac{\rho r^4}{\theta} u_x^2 + \frac{1}{\rho \theta r^2 }u^2\right) \mathrm{d}x\mathrm{d}s + C \int_0^t \int_0^1 \theta \mathrm{d}x\mathrm{d}s \\ &\quad - R \int_0^1 \left(\ln \rho - \int_{1}^{\rho} \eta^{-1} \chi(\eta) \mathrm{d}\eta \right) \cdot \boldsymbol{1}_{\{x | \rho(x,t)\geq1\}} \mathrm{d}x \\ &\leq C + \frac{\mu}{4} \int_0^t \int_0^1 \left(\frac{\rho r^4}{\theta} u_x^2 + \frac{1}{\rho \theta r^2 }u^2\right) \mathrm{d}x\mathrm{d}s, \end{align*}$

其中 $\boldsymbol{1}_{\{x | \rho(x,t)\geq1\}}$ 表示区域 $\{x | \rho(x,t)\geq1\}$ 上的特征函数. 故引理结论成立.

引理3.4 在命题 3.1 的条件下, 若 $q\geq1$, 则对任意 $(x, t)\in[0,1]\times[T]$, 有

$\begin{align*} C^{-1}\leq \rho(x,t)\leq C, \qquad \int_0^T \|\theta\|_{L^\infty}^{1+q} \mathrm{d}t \leq C. \end{align*}$

证 本证明分以下四步.

第一步 证明 $\rho(x,t)$ 有上界. 利用 (3.3) 和 (3.4) 式计算

$\begin{align*} \left(\frac{u}{r^2} \right)_t + \frac{2u^2}{r^3} + p_x =(\mu + \lambda) \left( \rho (r^2 u)_x \right)_x =-(\mu +\lambda)(\ln\rho )_{xt}. \end{align*}$

对上式在 $[x,1] \times [0, t]$ 上积分

$\begin{matrix} (\mu +\lambda)\ln \rho(x,t) &= (\mu +\lambda)\ln \rho_0(x) + (\mu +\lambda)\ln \rho(1,t) - (\mu +\lambda)\ln \rho_0(1) \\ &\quad + \int_x^1 \frac{u}{r^2}(y,t) \mathrm{d}y - \int_x^1 \frac{u}{r^2}(y,0)\mathrm{d}y + \int_0^t \int_x^1 \frac{2u^2}{r^3}(y,s) \mathrm{d}y\mathrm{d}s \\ &\quad + \int_0^t \left(a\rho^\gamma + R\rho \theta\right)(1,s)\mathrm{d}s - \int_0^t \left(a\rho^\gamma + R\rho \theta\right)(x,s)\mathrm{d}s. \end{matrix}$

再由边界条件 (3.5), $a_1(t) = r(1, t)$ 与 (3.6) 式可得

$\begin{matrix} (a\rho^\gamma + R\rho \theta)(1,t) &= (\lambda + \mu )\left(\rho (r^2 u)_x\right)(1,t) - \frac{2}{a_1(t)}\mu u(1,t) + P(t)\\ &=-\frac{\mathrm{d}}{\mathrm{d}t}\Big((\mu +\lambda)\ln \rho(1,t) +2\mu\ln a_1(t)\Big)+ P(t). \end{matrix}$

所以

$\begin{matrix} & \int_0^t (a\rho^\gamma + R\rho \theta)(1,s)\mathrm{d}s \\ &=-(\mu +\lambda)\ln \rho(1,t) -2\mu\ln a_1(t)+(\mu +\lambda)\ln \rho_0(1)+2\mu\ln a_1 + \int_0^t P(s)\mathrm{d}s. \end{matrix}$

将 (3.17) 式带入 (3.15) 式得

$\begin{matrix} (\mu +\lambda)\ln \rho(x,t) &= (\mu+\lambda)\ln\rho_0(x) - 2\mu \ln a_1(t) + 2\mu \ln a_1 + \int_0^t P(s) \mathrm{d}s\\ & + \int_x^1 \frac{u}{r^2}(y,t) \mathrm{d}y - \int_x^1 \frac{u}{r^2}(y,0)\mathrm{d}y + \int_0^t \int_x^1 \frac{2u^2}{r^3}(y,s) \mathrm{d}y\mathrm{d}s\\ & - \int_0^t \left(a\rho^\gamma + R\rho \theta\right)(x,s)\mathrm{d}s, \end{matrix}$

利用引理 3.1-3.2 与 $r\geq a_0$, 有

$\begin{align*} \ln \rho(x,t) \leq C + C \int_0^1 \frac{u^2}{r^2} \mathrm{d}x + C\int_0^t \int_0^1 \frac{u^2}{r^3} \mathrm{d}x\mathrm{d}s \leq C, \end{align*}$

因此

$\begin{matrix} \rho(x,t)\leq C. \end{matrix}$

第二步 做温度 $\theta$ 的 $L^{q}(0,T; L^{\infty})$-估计. 由引理 3.1, (2.2) 式, 以及

$\begin{matrix} \int_0^1 \frac{1}{\rho r^4 } \mathrm{d}x =\int_{a_0}^{a_1(t)} \frac{1}{ r^2 } \mathrm{d}r =\frac{1}{a_0}-\frac{1}{ a_1(t)}\leq C \end{matrix}$

得

$\begin{matrix} \max_{x\in[0,1]}\theta^q &\leq \left(\int_0^1 \theta \mathrm{d}x\right)^q + q\int_0^1 \theta^{q-1} |\theta_x| \mathrm{d}x \\ &\leq C + q\left( \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x\right)^{\frac{1}{2}} \left( \int_0^1 \frac{\theta^{2q}}{\rho \kappa(\rho, \theta)r^4 } \mathrm{d}x \right)^{\frac{1}{2}} \\ &\leq C + q\left( \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x\right)^{\frac{1}{2}} \left( \int_0^1 \frac{1}{\rho r^4 } \mathrm{d}x \right)^{\frac{1}{2}}\max_{x\in[0,1]}\theta^{\frac{q}{2}} \\ &\leq\frac{1}{2}\max_{x\in[0,1]}\theta^q+C +C \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x. \end{matrix}$

结合引理 3.3, 则有

$\begin{matrix} \int_0^T \|\theta\|_{L^\infty}^q \mathrm{d}t \leq C. \end{matrix}$

第三步 证明 $\rho(x,t)$ 有非负下界. 重新回到 (3.18) 式, 利用引理 3.1, (3.19), (3.22) 式以及 $q\geq1$, 可得

$\begin{align*} \ln \rho(x,t)&\geq -C- 2\mu \int_0^t \frac{u(1,s)}{a_1(s)}\mathrm{d}s +\int_x^1 \frac{u}{r^2}(y,t) \mathrm{d}y - \int_0^t \left(a\rho^\gamma + R\rho \theta\right)(x,s)\mathrm{d}s\\ &\geq -C- C \int_0^t |u(1,s)|\mathrm{d}s - C\int_0^t\|\theta\|_{L^\infty}\mathrm{d}s\\ &\geq -C, \end{align*}$

这里我们还用到了引理 3.3, (3.20) 式和

$\begin{align*} \int_0^t |u(1,s)|\mathrm{d}s &\leq \int_0^t \left|u(1,s)-\int_{0}^{1}u \mathrm{d}x\right|\mathrm{d}s +\int_0^t \int_{0}^{1}|u| \mathrm{d}x \mathrm{d}s \leq \int_0^t\int_{0}^{1}|u_x| \mathrm{d}x \mathrm{d}s +C\\ &\leq \left(\int_0^t\int_{0}^{1}\frac{ \rho r^4}{\theta} u_x^2 \mathrm{d}x \mathrm{d}s\right)^{\frac{1}{2}} \left(\int_0^t\int_{0}^{1}\frac{\theta}{ \rho r^4} \mathrm{d}x \mathrm{d}s\right)^{\frac{1}{2}}+C\\ &\leq C\left(\int_0^t\|\theta\|_{L^\infty}\mathrm{d}s\right)^{\frac{1}{2}}+C\leq C. \end{align*}$

故可得

$\begin{matrix} \rho(x,t) \geq C^{-1}. \end{matrix}$

第四步 最后, 重新估计温度 $\theta$ 的 $L^{1+q}(0,T; L^{\infty})$-范数. 回顾 (3.21) 式

$\begin{matrix} \max_{x\in[0,1]}\theta^{1+q} &\leq C + C\left( \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x\right)^{\frac{1}{2}} \left( \int_0^1 \frac{\theta^{2q+2}}{\rho \kappa(\rho, \theta)r^4 } \mathrm{d}x \right)^{\frac{1}{2}} \\ &\leq C + C\left( \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x\right)^{\frac{1}{2}} \left( \int_0^1 \theta \mathrm{d}x \right)^{\frac{1}{2}} \left( \max_{x\in[0,1]}\frac{1}{\rho } \right)^{\frac{1}{2}} \max_{x\in[0,1]}\theta^{\frac{1+q}{2}} \\ &\leq\frac{1}{2}\max_{x\in[0,1]}\theta^{1+q}+C +C \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x. \end{matrix}$

所以,

$\begin{matrix} \int_0^T \|\theta\|_{L^\infty}^{1+q} \mathrm{d}t \leq C. \end{matrix}$

综上所述, 该引理得证.

引理3.5 在命题 3.1 的条件下, 若 $0\leq q<1$, $P(t)>0$, 则对任意 $(x, t)\in[0,1]\times[T]$, 有

$\begin{align*} C^{-1}\leq \rho(x,t)\leq C, \qquad \int_0^T \|\theta\|_{L^\infty}^{1+q} \mathrm{d}t \leq C. \end{align*}$

证 密度 $\rho$ 的上界估计与引理 3.4 的第一步相同, 故在此省略. 接下来, 我们将重点研究 $\rho$ 的下界估计. 首先, 类似于 (3.24) 式, 可得

$\begin{align*} \max_{x\in[0,1]}\theta^{1+q} &\leq C + C\left( \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x\right)^{\frac{1}{2}} \left( \int_0^1 \theta \mathrm{d}x \right)^{\frac{1}{2}} \left( \max_{x\in[0,1]}\frac{1}{\rho } \right)^{\frac{1}{2}} \max_{x\in[0,1]}\theta^{\frac{1+q}{2}} \\ &\leq\frac{1}{2}\max_{x\in[0,1]}\theta^{1+q}+C +C \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x \left( \max_{x\in[0,1]}\frac{1}{\rho } \right), \end{align*}$

则

$\begin{matrix} \max_{x\in[0,1]}\theta^{1+q} \leq C+C \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x \left( \max_{x\in[0,1]}\frac{1}{\rho } \right). \end{matrix}$

重新回到 (3.18) 式, 两端取指数得

$\begin{matrix} \rho(x,t)=\mathcal{A}(x,t)\exp\left(- \int_0^t R(\rho \theta)(x,s)\mathrm{d}s\right), \end{matrix}$

其中

$\begin{matrix} \mathcal{A}(x,t)&:=\rho_0(x)\exp\left(\frac{1}{\mu+\lambda}\int_0^t\left( P(s)-2\mu\frac{a_1'(s)}{a_1(s)} -a\rho^\gamma (x,s)\right)\mathrm{d}s \right.\\ &\quad\left.+\frac{1}{\mu+\lambda}\int_x^1 \frac{u}{r^2}(y,t) \mathrm{d}y - \frac{1}{\mu+\lambda}\int_x^1 \frac{u}{r^2}(y,0)\mathrm{d}y + \frac{1}{\mu+\lambda}\int_0^t \int_x^1 \frac{2u^2}{r^3}(y,s) \mathrm{d}y\mathrm{d}s\right)\\ &=:a_1(t)^{-\frac{2\mu}{\mu+\lambda}}\mathcal{B}(x,t). \end{matrix}$

又因为

$\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\exp\left( \int_0^t R(\rho \theta)(x,s)\mathrm{d}s\right) =R\mathcal{A}(x,t)\theta(x,t), \end{align*}$

所以

$\begin{align*} \exp\left( \int_0^t R(\rho \theta)(x,s)\mathrm{d}s\right) =R\int_{0}^{t}\mathcal{A}(x,s)\theta(x,s)\mathrm{d}s+1. \end{align*}$

将上式带入 (3.27) 式可得

$\begin{align*} \rho^{-1}(x,t)=\mathcal{A}^{-1}(x,t)+R\int_{0}^{t}\frac{\mathcal{A}(x,s)}{\mathcal{A}(x,t)}\theta(x,s)\mathrm{d}s. \end{align*}$

利用 $\mathcal{A}(x,t), \mathcal{B}(x,t)$ 的定义 (3.28), 引理 3.1-3.2 和 (3.19) 式, 则有

$\begin{matrix} \left(a_1(t)^{\frac{2\mu}{\mu+\lambda}}\rho(x,t)\right)^{-1} &=\mathcal{B}^{-1}(x,t)+R\int_{0}^{t}\frac{\mathcal{B}(x,s)}{\mathcal{B}(x,t)} a_1(s)^{-\frac{2\mu}{\mu+\lambda}}\theta(x,s)\mathrm{d}s \\ &\leq C+C\int_{0}^{t} a_1(s)^{-\frac{2\mu}{\mu+\lambda}}\theta(x,s)\mathrm{d}s. \end{matrix}$

再结合 (3.26) 式得

$\begin{align*} & a_1(t)^{-\frac{2\mu}{\mu+\lambda}}\max_{x\in[0,1]}\frac{1}{\rho } \\ &\leq C+C\int_{0}^{t} a_1(s)^{-\frac{2\mu}{\mu+\lambda}} \mathrm{d}s +C\int_{0}^{t} \int_0^1 \frac{\rho\kappa(\rho, \theta) r^4 }{\theta^2} \theta_x^2\mathrm{d}x \left(a_1(s)^{-\frac{2\mu}{\mu+\lambda}} \max_{x\in[0,1]}\frac{1}{\rho } \right)\mathrm{d}s, \end{align*}$

由 Grönwall 不等式, $a_{1}(t)\geq C^{-1}$ 与引理 3.3 得

$\begin{matrix} a_1(t)^{-\frac{2\mu}{\mu+\lambda}}\max_{x\in[0,1]}\frac{1}{\rho }\leq C. \end{matrix}$

因此, 如果 $P(t)>0$, 通过引理 3.1 可知

$\begin{matrix} a_1(t) \leq C. \end{matrix}$

合并 (3.30) 与 (3.26) 式即完成了该引理证明.

引理3.6 在命题 3.1 的条件下, 对任意 $(x, t)\in[0,1]\times[T]$, $k\geq0$, 有

$\begin{align*} \theta(x,t)\geq C^{-1}, \qquad \int_0^T \int_0^1 \left(\frac{ r^4}{\theta^k} u_x^2 + \frac{1}{\theta^k r^2} u^2 + k\frac{ \kappa( \rho,\theta) r^4}{\theta^{k+1}} \theta_x^2\right) \mathrm{d}x\mathrm{d}t \leq C. \end{align*}$

证 温度方程 (3.14) 乘以 $\theta^{-k} (k\geq2)$ 后关于 $(x,t)$ 积分得

$\begin{matrix} & \frac{1}{k-1} \int_0^1 \theta^{1-k} \mathrm{d}x + \int_0^t \int_0^1 \left(\frac{ r^4}{\theta^k} u_x^2 + \frac{1}{\theta^k r^2} u^2 + k\frac{ \kappa( \rho,\theta) r^4}{\theta^{k+1}} \theta_x^2\right) \mathrm{d}x\mathrm{d}s \\ &\leq C^{k-1} + C\int_0^t \int_0^1 \theta^{2-k} \mathrm{d}x\mathrm{d}s, \end{matrix}$

此处的常数 $C$ 仅依赖于初值及时间 $T$, 而与参数 $k$ 无关. 接着计算

$\begin{align*} & \int_0^1 \theta^{1-k} \mathrm{d}x +(k-1)\int_0^t \int_0^1 \left(\frac{ r^4}{\theta^k} u_x^2 + \frac{1}{\theta^k r^2} u^2 + k\frac{ \kappa( \rho,\theta) r^4}{\theta^{k+1}} \theta_x^2\right) \mathrm{d}x\mathrm{d}s \\ &\leq C^{k-1} + C(k-1)\int_0^t\|\theta\|_{L^\infty} \int_0^1 \theta^{1-k} \mathrm{d}x\mathrm{d}s, \end{align*}$

结合 Grönwall 不等式和引理 3.4-3.5 得

$\begin{align*} \int_0^1 \theta^{1-k} \mathrm{d}x &\leq C^{k-1} {\rm e}^{C(k-1) \int_0^t \|\theta\|_{L^\infty} \mathrm{d}s} \leq C^{k-1}, \end{align*}$

即

$\begin{align*} \left\Vert \theta^{-1}\right\Vert_{L^{k-1}} \leq C. \end{align*}$

令 $k \to +\infty$, 则

$\begin{matrix} \left\Vert\theta^{-1}\right\Vert_{L^{\infty}} \leq C. \end{matrix}$

当 $0<k<1$ 时, 重回 (3.32) 式, 借助引理 3.1, 知

$\begin{matrix} & \int_0^t \int_0^1 \left(\frac{ r^4}{\theta^k} u_x^2 + \frac{1}{\theta^k r^2} u^2 + k\frac{ \kappa( \rho,\theta) r^4}{\theta^{k+1}} \theta_x^2\right) \mathrm{d}x\mathrm{d}s \leq C +C\int_0^1 \theta^{1-k} \mathrm{d}x + C\int_0^t \int_0^1 \theta^{2-k} \mathrm{d}x\mathrm{d}s \\ &\leq C +C \int_0^t\|\theta\|_{L^\infty}\mathrm{d}s \leq C. \end{matrix}$

最后, 直接对温度方程 (3.14) 积分发现

$\begin{matrix} & \int_0^t \int_0^1 \left( r^4 u_x^2 + \frac{1}{ r^2} u^2 \right) \mathrm{d}x\mathrm{d}s \leq C +C\int_0^1 \theta \mathrm{d}x + C\int_0^t \int_0^1 \theta^{2} \mathrm{d}x\mathrm{d}s \leq C. \end{matrix}$

合并 (3.33)-(3.35) 式可得该引理结论.

引理3.7 在命题 3.1 的条件下, 对任意 $(x, t)\in[0,1]\times[T]$, 有

$\begin{align*} \int_0^1 \rho_x^2 \mathrm{d}x\leq c, \qquad a_0\leq r\leq a_1(t) = r(1,t) \leq c. \end{align*}$

证 利用 (3.4) 式, 可知

$\begin{align*} u_t + r^2 p_x+(\mu + \lambda) r^2 (\ln\rho )_{xt}=0, \end{align*}$

乘以 $u +(\mu+\lambda)r^2(\ln \rho)_x$ 后关于 $x$ 积分得

$\begin{matrix} & \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d}t} \int_0^1 \left( u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2 \mathrm{d}x \\ &= 2(\mu+\lambda)\int_0^1 \left(u +(\mu+\lambda)r^2(\ln \rho)_x\right) r u \left(\ln \rho\right)_x\mathrm{d}x \\ &\quad - \int_0^1 r^2 p_x \left(u +(\mu+\lambda)r^2(\ln \rho)_x \right)\mathrm{d}x \\ &= 2(\mu+\lambda)\int_0^1r u^2(\ln\rho)_x \mathrm{d}x + 2(\mu+\lambda)^2\int_0^1r^3 u (\ln\rho)_x^2 \mathrm{d}x \\ &\quad - a\int_0^1 r^2 (\rho^\gamma)_x u \mathrm{d}x - a(\mu+\lambda)\int_0^1 r^2 (\rho^\gamma)_x r^2 (\ln\rho)_x \mathrm{d}x \\ &\quad - R\int_0^1 r^2 (\rho \theta)_x \left(u +(\mu+\lambda)r^2(\ln \rho)_x \right)\mathrm{d}x =: \sum_{i=1}^5 \mathcal{J}_i. \end{matrix}$

随后对上式右端的五项 $\mathcal{J}_i(i=1,\cdots,5)$ 逐一进行估计.

利用引理 3.1-3.2 以及 $\|u\|_{L^\infty}\leq C\|u_x\|_{L^2}$ 有

$\begin{align*} \mathcal{J}_1 &\leq \int_0^1 \left( u +(\mu+\lambda)r^2(\ln \rho)_x - u \right)^2 \mathrm{d}x + C \int_0^1 \frac{u^4}{r^2} \mathrm{d}x \\ &\leq C \int_0^1 \left( u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2 \mathrm{d}x + C \int_0^1 u^2 \mathrm{d}x + C \|u\|_{L^\infty}^2 \int_0^1 u^2 \mathrm{d}x\\ &\leq C \int_0^1 \left( u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2 \mathrm{d}x + C + C\|u_x\|_{L^2}^2. \end{align*}$

直接计算 $\mathcal{J}_2$ 可得

$\begin{align*} \mathcal{J}_2 &\leq 2\left\|\frac{u}{r}\right\|_{L^\infty} \int_0^1 \left( u +(\mu+\lambda)r^2(\ln \rho)_x - u \right)^2 \mathrm{d}x \\ &\leq C\|u_x\|_{L^2}\int_0^1 \left( u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2 \mathrm{d}x + C \|u_x\|_{L^2}. \end{align*}$

对于 $\mathcal{J}_3$ 与 $\mathcal{J}_4$, 通过分部积分和 $ (8)_1$ 式计算得

$\begin{align*} &\mathcal{J}_3 = -a\left( r^2 \rho^\gamma u \right)(1, t) + \int_0^1 a \rho^\gamma (r^2 u)_{x} \mathrm{d}x = -a\left( r^2 \rho^\gamma u \right)(1, t) - \frac{\mathrm{d}}{\mathrm{d}t} \int_0^1 \frac{a}{\gamma-1}\rho^{\gamma-1} \mathrm{d}x,\\ &\mathcal{J}_4 =-a\gamma(\mu+\lambda)\int_0^1 r^4 \rho^{\gamma-2} \rho_x^2 \mathrm{d}x. \end{align*}$

对于最后一项 $\mathcal{J}_5$, 我们有

$\begin{align*} \mathcal{J}_5 &= - R\int_0^1 r^2 (\rho_x \theta + \rho \theta_x) \left(u +(\mu+\lambda)r^2(\ln \rho)_x \right)\mathrm{d}x \\ &\leq \int_0^1 \theta \left(u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2 \mathrm{d}x + C\int_0^1 r^4 \rho_x^2 \theta \mathrm{d}x \\ &\quad + C\int_0^1 \frac{ \kappa(\rho, \theta)r^4}{\theta^{k+1}}\theta_x^2\mathrm{d}x +\int_0^1 \frac{ \theta^q\theta^{k+1-q}}{\kappa(\rho, \theta)}\left(u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2 \mathrm{d}x \\ &\leq C\left(\|\theta\|_{L^{\infty}}\!+\!\left\|\theta^{k+1-q}\right\|_{L^{\infty}}\right) \left(\int_0^1 \left(u \!+\!(\mu+\lambda)r^2(\ln \rho)_x \right)^2 \mathrm{d}x \!+\! 1\right) \!+\! C\int_0^1 \frac{ \kappa(\rho, \theta)r^4}{\theta^{k+1}}\theta_x^2\mathrm{d}x. \end{align*}$

进行后续估计之前, 我们先利用方程 $ (3.4)_1$ 与边界条件 $ (3.5)_3$ 发现

$\begin{align*} \rho_t(1,t)=-\left(\rho^2(r^2u)_x\right)(1,t) = \left(\frac{\rho}{\mu+\lambda}\left( P(t)-a \rho^{\gamma}- R \rho \theta-\frac{2}{r}\mu u\right)\right)(1,t). \end{align*}$

现在, 将上述 $\mathcal{J}_i(i=1,\cdots,5)$ 的估计代入 (3.36) 式并关于时间积分, 借助引理 3.4-3.6 以及

$\begin{align*} & -\int_0^t \left( r^2 \rho^\gamma u \right)(1, s)\mathrm{d}s =-\int_0^t \rho^\gamma(1, s)a^2_1(s)a'_1(s)\mathrm{d}s\\ &=-\frac{1}{3}\rho^\gamma(1, t)a^3_1(t) +\frac{1}{3}\rho_0^\gamma(1)a^3_1 +\frac{\gamma}{3}\int_0^t\left(\rho^{\gamma-1}\rho_s\right)(1, s) a^3_1(s)\mathrm{d}s\\ &=-\frac{1}{3}\rho^\gamma(1, t)a^3_1(t) +\frac{1}{3}\rho_0^\gamma(1)a^3_1 +\frac{\gamma}{3(\mu+\lambda)}\int_0^t\left(\rho^{\gamma}\left( P(s)-a \rho^{\gamma}\!-\! R \rho \theta-\frac{2}{r}\mu u\right) \right)(1, s) a^3_1(s)\mathrm{d}s, \end{align*}$

可得 $\forall \, k>0$,

$\begin{matrix} & \int_0^1 \left( u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2\mathrm{d}x+a^3_1(t) +\int_0^t\int_0^1 r^4 \rho_x^2 \mathrm{d}x\mathrm{d}s\\ &\leq C\int_0^t\left(1+\|u_x\|_{L^2}+\|\theta\|_{L^{\infty}}+\left\|\theta^{k+1-q}\right\|_{L^{\infty}}\right) \left(\int_0^1 \left(u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2 \mathrm{d}x + 1\right)\mathrm{d}s\\ &\quad + C + C \int_0^t \left(P(s)+\|u_x\|_{L^2}\right)a^3_1(s)\mathrm{d}s. \end{matrix}$

因此, 对任意 $q>0$, 取 $k=q$, 结合 Grönwall 不等式和引理 3.4-3.6 有

$\begin{matrix} \int_0^1 \left( u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2\mathrm{d}x+a^3_1(t) +\int_0^t\int_0^1 r^4 \rho_x^2 \mathrm{d}x\mathrm{d}s \leq C. \end{matrix}$

当 $q=0$ 时, 我们需要重新做温度的 $L^{2-k}(0,T; L^{\infty})$-估计. 重回 (3.24) 式

$\begin{matrix} \max_{x\in[0,1]}\theta^{2-k} &\leq C + \left( \int_0^1 \frac{ \kappa(\rho, \theta) r^4 }{\theta^{k+1}} \theta_x^2\mathrm{d}x\right)^{\frac{1}{2}} \left( \int_0^1 \frac{\theta^{3-k}}{ \kappa(\rho, \theta)r^4 } \mathrm{d}x \right)^{\frac{1}{2}} \\ &\leq C + \left( \int_0^1 \frac{ \kappa(\rho, \theta) r^4 }{\theta^{k+1}} \theta_x^2\mathrm{d}x\right)^{\frac{1}{2}} \max_{x\in[0,1]}\theta^{\frac{2-k}{2}} \\ &\leq\frac{1}{2}\max_{x\in[0,1]}\theta^{2-k}+C +C \int_0^1 \frac{ \kappa(\rho, \theta) r^4 }{\theta^{k+1}} \theta_x^2\mathrm{d}x, \qquad \forall \, k\in(0,1), \end{matrix}$

结合引理 3.6 可得

$\begin{matrix} \int_0^T \|\theta\|_{L^\infty}^{ 2-k } \mathrm{d}t \leq C, \qquad \forall \, k\in(0,1). \end{matrix}$

故, 当 $q=0$ 时, 取 (3.37) 式中 $k=\frac{1}{2}$ 即可得到 (3.38) 式.

(3.38) 式中 $a_1(t)\leq C$ 与引理 3.2 意味着

$\begin{matrix} a_0\leq r\leq a_1(t)\leq C. \end{matrix}$

此外, 由不等式

$\begin{align*} \int_0^1 \rho_x^2\mathrm{d}x \leq C\int_0^1 \left( u +(\mu+\lambda)r^2(\ln \rho)_x \right)^2\mathrm{d}x+C, \end{align*}$

结合 (3.38) 式即可完成引理 3.7 的证明.

引理3.8 在命题 3.1 的条件下, 对任意 $(x, t)\in[0,1]\times[T]$, 有

$\begin{align*} \theta(x,t)\leq C, \qquad \sup_{t\in[T]}\Vert u_x\Vert_{L^2}^2 +\int_0^T \Vert(\theta_x, u_{xx})\Vert_{L^2}^2\mathrm{d}t \leq C. \end{align*}$

证 本证明分以下三步.

第一步 做速度场的 $H^{1}$-估计. 重写 $ (8)_2$ 式 $u_t = r^2 \left( (\mu+\lambda)\rho (r^2 u)_x - p \right)_x, $ 乘以 $\left( \lambda \rho (r^2 u)_x + \mu \rho r^2 u_x - p + P(t) \right)_x$ 后关于 $x$ 积分可得

$\begin{align*} 0 &= \int_0^1 u_{xt} \left( (\mu+\lambda) \rho r^2 u_x + \frac{2\lambda}{r}u - p + P(t) \right) \mathrm{d}x \\ &\quad + \int_0^1 r^2 \left( (\mu+\lambda) \rho r^2 u_x + \frac{2(\lambda + \mu)}{r} u - p \right)_x \left( (\mu+\lambda) \rho r^2 u_x + \frac{2\lambda}{r} u - p + P(t) \right)_x \mathrm{d}x \\ &=\frac{\mathrm{d}}{\mathrm{d}t} \int_0^1 \left(\frac{\mu+\lambda}{2} \rho r^2 u_x^2 + \frac{2\lambda }{r}u u_x - p u_x+ P(t) u_x\right) \mathrm{d}x - \frac{\mu+\lambda}{2}\int_0^1 (\rho r^2)_t u_x^2\mathrm{d}x \\ &\quad -\int_0^1 \left( 2\lambda \left(\frac{u}{r}\right)_t - p_t + P'(t) \right) u_x \mathrm{d}x\\ &\quad + \int_0^1 r^2 \left( (\mu+\lambda) \rho r^2 u_x + \frac{2(\lambda + \mu)}{r} u - p \right)_x \left( (\mu+\lambda) \rho r^2 u_x + \frac{2\lambda}{r} u - p + P(t) \right)_x \mathrm{d}x. \end{align*}$

对上式关于时间积分后, 有

$\begin{matrix} & \int_0^1 u_x^2 \mathrm{d}x + \int_0^t\int_0^1 u_{xx}^2 \mathrm{d}x\mathrm{d}s\\ &\leq C+C \int_0^1\left(u^2 + p^2 + P^2(t) \right) \mathrm{d}x +C\int_0^t\int_0^1 \left(| u |+| u_x|\right) u_{ x}^2 \mathrm{d}x\mathrm{d}s\\ &\quad +C\int_0^t\int_0^1 \left(| u_s |+u^2+|\rho_s|+|\rho_s|\theta+|P'(s)|\right) |u_{ x}| \mathrm{d}x\mathrm{d}s\\ &\quad +C\int_0^t\int_0^1 \left(\theta |u u_x|+\theta u_x^2+|u_x|^3+u^2|u_x| +\rho_x^2u_x^2+\kappa^2(\rho,\theta) \theta^2_x+\rho^2_x +\rho^2_x\theta^2+\theta_x^2\right) \mathrm{d}x\mathrm{d}s\\ &\leq C+C \int_0^1 \theta^2 \mathrm{d}x+\frac{1}{4}\int_0^t\int_0^1 u_{xx}^2 \mathrm{d}x\mathrm{d}s+C\int_0^t\int_0^1 \left(| u |u_{ x}^2+| u_x|^3+ \theta^2+\theta_x^2 +\theta |u u_x|+\theta u_x^2 \right.\\ &\quad\left.+u^2|u_x| +\rho_x^2u_x^2+\kappa^2(\rho,\theta) \theta^2_x+\rho^2_x\theta^2\right) \mathrm{d}x\mathrm{d}s \\ &\leq C+C \int_0^1 \theta^2 \mathrm{d}x+\frac{1}{4}\int_0^t\int_0^1 u_{xx}^2 \mathrm{d}x\mathrm{d}s +C\int_0^t\left(\|u \|_{L^2}^{\frac{1}{2}}\|u_{x}\|_{L^2}^{\frac{1}{2}} +\|u_{x}\|_{L^2}^{\frac{1}{2}}\|u_{x}\|_{H^1}^{\frac{1}{2}}\right)\|u_{x}\|_{L^2}^2\mathrm{d}s\\ &\quad+C\int_0^t\left(\|u \|_{H^1} \|u_{x}\|_{H^1} +\|\theta\|_{L^\infty}^2\right) \int_0^1(\theta+u^2+\rho_x^2)\mathrm{d}x\mathrm{d}s+C\int_0^t\int_0^1 \kappa^2(\rho,\theta) \theta_x^2 \mathrm{d}x\mathrm{d}s\\ &\leq C+C \int_0^1 \theta^2 \mathrm{d}x+\frac{1}{2}\int_0^t\int_0^1 u_{xx}^2 \mathrm{d}x\mathrm{d}s +C\int_0^t \|u_{x}\|_{L^2}^4\mathrm{d}s +C\int_0^t\int_0^1 \kappa^2(\rho,\theta) \theta_x^2 \mathrm{d}x\mathrm{d}s, \end{matrix}$

其中用到引理 3.1, 3.4-3.7, Cauchy 不等式, $\kappa(\rho,\theta)\geq C^{-1}\theta^q\geq C^{-1}$, $\|\theta\|_{L^\infty}^2\leq C+C\int_0^1\theta_x^2\mathrm{d}x$, (3.14) 式,

$\begin{align*} & R\int_0^1 \rho\theta_s u_x \mathrm{d}x\\ &=\frac{R}{c_v}\int_0^1 \rho\left(-R \rho \theta (r^2 u)_x+\mu \rho r^4 u_x^2 + \frac{2\mu}{\rho r^2}u^2 + \lambda \rho (r^2 u)_x^2 + (\rho r^4 \kappa(\rho,\theta) \theta_x)_x\right) u_x \mathrm{d}x\\ &\leq C \int_0^1 \left(\theta |u u_x|+\theta u_x^2+|u_x|^3+u^2|u_x|\right) \mathrm{d}x -\frac{R}{c_v}\int_0^1 (\rho_xu_x+\rho u_{xx})\rho r^4 \kappa(\rho,\theta) \theta_x \mathrm{d}x \end{align*}$

以及 $ |u_s|\leq C\left(|\rho_x|+|\rho_x|\theta+|\theta_x|+|\rho_xu|+|\rho_xu_x|+|u|+|u_x|+|u_{xx}|\right).$ 利用 (3.42) 式, Grönwall 不等式和引理 3.6 得

$\begin{matrix} \int_0^1 u_x^2 \mathrm{d}x + \int_0^t\int_0^1 u_{xx}^2 \mathrm{d}x\mathrm{d}s \leq C+C \int_0^1 \theta^2 \mathrm{d}x +C\int_0^t\int_0^1 \kappa^2(\rho,\theta) \theta_x^2 \mathrm{d}x\mathrm{d}s. \end{matrix}$

第二步 做温度 $\theta$ 的 $L^{k}$-估计($k\geq 2$). 温度方程 (3.14) 乘以 $\theta^{k-1}$ 后关于 $(x,t)$ 积分得

$\begin{matrix} & \frac{1}{k} \int_0^1 \theta^{k} \mathrm{d}x +(k-1)\int_0^t \int_0^1 \kappa( \rho,\theta) \theta^{k-2} \theta_x^2 \mathrm{d}x\mathrm{d}s \\ &\leq C^{k} + C\int_0^t\int_0^1\Big(\theta^{k}|(r^2u)_x|+\theta^{k-1} u_x^2+\theta^{k-1} u^2 +\theta^{k-1}(r^2u)_x^2\Big) \mathrm{d}x\mathrm{d}s \\ &\leq C^{k} + C\int_0^t\int_0^1\Big(\theta^{k+1} +\theta^{k-1} u_x^2+\theta^{k-1} u^2 \Big) \mathrm{d}x\mathrm{d}s \\ &\leq C^{k} + C\int_0^t \|\theta\|_{L^{\infty}}\int_0^1 \theta^{k} \mathrm{d}x\mathrm{d}s + C\int_0^t \|\theta\|_{L^{\infty}}^{k-1} \int_0^1(u_x^2+ u^2)\mathrm{d}x\mathrm{d}s, \end{matrix}$

这里我们用到了 (3.3) 式和引理 3.4, 3.5, 3.7. 又因为

$\begin{matrix} \|\theta\|_{L^{\infty}}^{k-1} &\leq \left(\int_0^1 \theta \mathrm{d}x\right)^{k-1}+(k-1)\|\theta^{k-2}\theta_x\|_{L^1}\\ &\leq C^{k-1}+(k-1)\left(\int_0^1\kappa( \rho,\theta) \theta^{k-2} \theta_x^2 \mathrm{d}x\right)^{\frac{1}{2}} \left(\int_0^1 \frac{\theta^{k-2}}{\kappa( \rho,\theta)}\mathrm{d}x\right)^{\frac{1}{2}}, \end{matrix}$

所以, 取 $k=q+3$, 合并 (3.44)-(3.45) 式, 借助 Grönwall 不等式与 (2.2) 式有

$\begin{matrix} & \int_0^1 \theta^{q+3} \mathrm{d}x +\int_0^t \int_0^1 \kappa( \rho,\theta) \theta^{q+1} \theta_x^2 \mathrm{d}x\mathrm{d}s \\ &\leq C + C\int_0^t\left(\int_0^1\kappa( \rho,\theta) \theta^{q+1} \theta_x^2 \mathrm{d}x\right)^{\frac{1}{2}}\|(u,u_x)\|_{L^2}^2\mathrm{d}s \\ &\leq C +\frac{1}{2}\int_0^t \int_0^1 \kappa( \rho,\theta) \theta^{q+1} \theta_x^2 \mathrm{d}x\mathrm{d}s +C\Big(\sup_{s\in[0,t]}\|u_x\|_{L^2}^2\Big)\int_0^t\|u_x\|_{L^2}^2\mathrm{d}s. \end{matrix}$

因此, 结合 (3.43) 和 (3.46) 式, 由 (2.2) 式得

$\begin{align*} & \sup_{s\in[0,t]}\int_0^1 (u_x^2+\theta^{q+3}) \mathrm{d}x + \int_0^t\int_0^1\left( u_{xx}^2+\kappa( \rho,\theta) \theta^{q+1} \theta_x^2\right) \mathrm{d}x\mathrm{d}s \\ &\leq C+C \sup_{s\in[0,t]}\int_0^1 \theta^2 \mathrm{d}x +C\int_0^t\int_0^1 \kappa(\rho,\theta)(1+\theta^{q+1-\epsilon}) \theta_x^2 \mathrm{d}x\mathrm{d}s \\ &\leq C+\frac{1}{2}\sup_{s\in[0,t]}\int_0^1 \theta^{q+3} \mathrm{d}x +\frac{1}{2}\int_0^t\int_0^1\kappa( \rho,\theta) \theta^{q+1}\theta_x^2\mathrm{d}x\mathrm{d}s +C\int_0^t\int_0^1 \frac{\kappa(\rho,\theta)}{\theta^2}\theta_x^2 \mathrm{d}x\mathrm{d}s. \end{align*}$

利用引理3.3 可知

$\begin{matrix} \sup_{t\in[T]}\int_0^1 (u_x^2+\theta^{q+3}) \mathrm{d}x + \int_0^T\int_0^1\left( u_{xx}^2+\kappa( \rho,\theta) \theta^{q+1} \theta_x^2\right) \mathrm{d}x\mathrm{d}t\leq C. \end{matrix}$

第三步 最后, 做温度 $\theta$ 的 $L^{\infty}$-估计. 重回 (3.44)-(3.45) 式

$\begin{align*} & \frac{1}{k} \int_0^1 \theta^{k} \mathrm{d}x +(k-1)\int_0^t \int_0^1 \kappa( \rho,\theta) \theta^{k-2} \theta_x^2 \mathrm{d}x\mathrm{d}s \\ &\leq C^{k}+ C(k-1)\int_0^t\left(\int_0^1\kappa( \rho,\theta) \theta^{k-2} \theta_x^2 \mathrm{d}x\right)^{\frac{1}{2}}\mathrm{d}s \\ &\leq C^{k}+\frac{k-1}{2}\int_0^t \int_0^1 \kappa( \rho,\theta) \theta^{k-2} \theta_x^2 \mathrm{d}x\mathrm{d}s, \end{align*}$

那么

$\begin{align*} \|\theta\|_{L^k}\leq C, \end{align*}$

其中常数 $C$ 不依赖于 $k$. 取 $k\rightarrow+\infty$, 则

$\begin{matrix} \|\theta\|_{L^\infty}\leq C. \end{matrix}$

综上所述, (3.47)-(3.48) 式即证该引理.

引理3.9 在命题 3.1 的条件下, 有

$\begin{align*} \sup_{t\in[T]}\|\theta_x\|_{L^2}^2 +\int_0^T\|(\theta_t, \theta_{xx})\|_{L^2}^2\mathrm{d}t \leq C. \end{align*}$

证 温度方程 (3.14) 乘以 $\rho^{-1} r^{-4} \kappa(\rho, \theta) \theta_t$ 后关于 $x$ 积分得

$\begin{matrix} & \int_0^1\rho r^4 \kappa(\rho, \theta) \theta_x \left( \rho^{-1} r^{-4} \kappa(\rho, \theta) \theta_t \right)_x\mathrm{d}x + \int_0^1 c_v \rho^{-1} r^{-4}\kappa(\rho, \theta) \theta_t^2 \mathrm{d}x \\ &=\int_0^1\left( \mu \rho r^4 u_x^2 + \frac{2\mu}{\rho r^2}u^2 + \lambda \rho (r^2 u)_x^2-R \rho \theta (r^2 u)_x\right)\rho^{-1} r^{-4} \kappa(\rho, \theta) \theta_t\mathrm{d}x. \end{matrix}$

因为

$\begin{align*} & \rho r^4 \kappa(\rho, \theta) \theta_x \left( \rho^{-1} r^{-4} \kappa(\rho, \theta) \theta_t \right)_x\\ &=\rho r^4 \kappa(\rho, \theta) \theta_x\kappa(\rho, \theta) \theta_t \left( -\rho^{-2}\rho_x r^{-4}- 4\rho^{-1} r^{-5}\rho^{-1} r^{-2}\right)\\ &\quad+\rho r^4 \kappa(\rho, \theta)\theta_x\rho^{-1} r^{-4}\left(\kappa_{\rho}(\rho, \theta)\rho_x \theta_t+\kappa_{\theta}(\rho, \theta)\theta_x \theta_t+\kappa(\rho, \theta) \theta_{xt} \right)\\ &=\kappa^2(\rho, \theta) \theta_x \theta_t \left( -\rho^{-1}\rho_x- 4\rho^{-1} r^{-3}\right)+\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left(\kappa^2(\rho, \theta)\theta_{x}^2 \right) + \kappa(\rho, \theta)\kappa_{\rho}(\rho, \theta)\theta_x\left(\rho_x \theta_t-\rho_t \theta_x \right), \end{align*}$

所以有

$\begin{matrix} & \frac{\mathrm{d}}{\mathrm{d}t}\int_0^1\kappa^2(\rho, \theta)\theta_{x}^2\mathrm{d}x + \int_0^1 c_v \rho^{-1} r^{-4}\kappa(\rho, \theta) \theta_t^2 \mathrm{d}x \\ &\leq C\int_0^1\Big((\rho_x^2+1)\theta_{x}^2+|\rho_t|\theta_x^2+u_x^4+u^4+1\Big)\mathrm{d}x \\ &\leq C \left(\|\rho_x\|_{L^2}^2 +1\right)\|\theta_x\|_{L^\infty}^2 + C(\|u_x\|_{L^\infty} + 1) \int_0^1 \theta_x^2\mathrm{d}x + C\|u_x\|_{L^\infty}^2\|u_x\|_{L^2}^2+ C \\ &\leq C \|\theta_x\|_{L^2}^2+C \|\theta_x\|_{L^2}\|\theta_{xx}\|_{L^2} + C\|u_{xx}\|_{L^2}^2\int_0^1\kappa^2(\rho, \theta)\theta_x^2\mathrm{d}x + C\|u_{xx}\|_{L^2}^2+ C. \end{matrix}$

回顾温度方程 (3.14) 发现 $ |\theta_{xx}| \leq C(1 + |\theta_t| + u_x^2 + u^2 + |\rho_x \theta_x| + \theta_x^2), $ 因此, 由引理 3.7-3.8 得

$\begin{matrix} \|\theta_{xx}\|_{L^2} &\leq C\left(1 + \|\theta_t\|_{L^2} + \|u_x\|_{L^2}^{\frac{1}{2}} \|u_{x}\|_{H^1}^{\frac{1}{2}}\|u_x\|_{L^2} + \|u\|_{L^2}^{\frac{1}{2}} \|u\|_{H^1}^{\frac{1}{2}}\|u\|_{L^2}\right. \\ &\quad \left. +\|\theta_x\|_{L^2}^{\frac{1}{2}} \|\theta_x\|_{H^1}^{\frac{1}{2}}(\|\rho_x\|_{L^2}+\|\theta_x\|_{L^2})\right)\\ &\leq C\left(1 + \|\theta_t\|_{L^2} + \|u_{xx}\|_{L^2}+\|\theta_x\|_{L^2}^3\right) +\frac{1}{2}\|\theta_{xx}\|_{L^2}. \end{matrix}$

最后, 将 (3.51) 式带入 (3.50) 式可得

$\begin{matrix} & \frac{\mathrm{d}}{\mathrm{d}t}\int_0^1\kappa^2(\rho, \theta)\theta_{x}^2\mathrm{d}x + \frac{1}{2}\int_0^1 c_v \rho^{-1} r^{-4}\kappa(\rho, \theta) \theta_t^2 \mathrm{d}x \\ &\leq C \left(\|\theta_x\|_{L^2}^2 + \|u_{xx}\|_{L^2}^2+1\right) \left(\int_0^1\kappa^2(\rho, \theta)\theta_x^2\mathrm{d}x+1\right), \end{matrix}$

结合 Grönwall 不等式和引理 3.8, 则

$\begin{matrix} \sup_{t\in[T]}\|\theta_x\|_{L^2}^2 + \int_0^T\|\theta_t\|_{L^2}^2\mathrm{d}t \leq C. \end{matrix}$

再由 (3.51) 式可知

$\begin{matrix} \int_0^T\|\theta_{xx}\|_{L^2}^2\mathrm{d}t \leq C. \end{matrix}$

故, 合并 (3.53)-(3.54) 式即可得到引理的结论.

基于引理 3.1-3.9 所建立的一系列先验估计, 命题 3.1 得证, 由此完成了定理 2.1 的证明.

参考文献

Bian D F, Guo B L, Zhang J J.

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In this paper, we investigate the Cauchy problem for the one-dimensional compressible Navier–Stokes equations with temperature-dependent transport coefficients and large initial data. Specifically, we consider the case where the viscosity coefficient is characterized by [Formula: see text] and the heat conductivity coefficient is given by [Formula: see text], for [Formula: see text]. Our main result establishes the existence and asymptotic stability of a global strong solution provided that the initial data belongs to the [Formula: see text]-space and [Formula: see text] is sufficiently small. This finding provides a positive response to the question posed in [Remarks 1.3 and 1.4 in T. Wang and H. J. Zhao, One-dimensional compressible heat-conducting gas with temperature-dependent viscosity, Math. Models Methods Appl. Sci. 26 (2016) 2237–2275].

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We consider the one-dimensional compressible Navier–Stokes system for a viscous and heat-conducting ideal polytropic gas when the viscosity [Formula: see text] and the heat conductivity [Formula: see text] depend on the specific volume [Formula: see text] and the temperature [Formula: see text] and are both proportional to [Formula: see text] for certain non-degenerate smooth function [Formula: see text]. We prove the existence and uniqueness of a global-in-time non-vacuum solution to its Cauchy problem under certain assumptions on the parameter [Formula: see text] and initial data, which imply that the initial data can be large if [Formula: see text] is sufficiently small. Such a result appears to be the first global existence result for general adiabatic exponent and large initial data when the viscosity coefficient depends on both the density and the temperature.

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