Vlasov-Poisson-Boltzmann 方程组的 Navier-Stokes-Poisson 方程组逼近——献给陈化教授 70 寿辰

Vlasov-Poisson-Boltzmann 方程组的 Navier-Stokes-Poisson 方程组逼近——献给陈化教授 70 寿辰

董丽娜^,, 刘双乾^,, 马璇^,, 马袁园^,^*

华中师范大学数学与统计学学院武汉 430079

Compressible Navier-Stokes-Poisson System Approximation to the Vlasov-Poisson-Boltzmann System

Dong Lina^,, Liu Shuangqian^,, Ma Xuan^,, Ma Yuanyuan^,^*

School of Mathematics and Statistics, Central China Normal University, Wuhan 430079

通讯作者: ^*马袁园, Email：yyma@mails.ccnu.edu.cn

收稿日期: 2025-12-31 修回日期: 2026-01-27

基金资助:

国家自然科学基金(12325107)

Received: 2025-12-31 Revised: 2026-01-27

Fund supported:

NSFC(12325107)

作者简介 About authors

董丽娜,Email：2163835253@mails.ccnu.edu.cn

刘双乾,Email：sqliu@ccnu.edu.cn

马璇,Email：maxuan@ccnu.edu.cn

摘要

当 Knudsen 数趋于 $0$ 时, 可压缩 Navier-Stokes-Poisson 方程组并不是无量纲化 Vlasov-Poisson-Boltzmann 方程组的极限, 但通过 Chapman-Enskog 展开, 它是 Vlasov-Poisson-Boltz-mann 方程组的二阶近似. 该文的目的是严格证明若可压缩 Navier-Stokes-Poisson 方程组对应的局部 Maxwellian 与 Vlasov-Poisson-Boltzmann 方程组的初值之差是 Knudsen 数的二阶小量, 则两者解在所有时间内的差也保持该量级. 该文的分析基于宏观方程的能量估计以及宏观-微观分解框架下的精细能量方法.

关键词： Vlasov-Poisson-Boltzmann 方程组; 宏观-微观分解; 可压缩 Navier-Stokes-Poisson 方程组

Abstract

When the Knudsen number approaches zero, the compressible Navier-Stokes-Poisson (NSP) system is not the limit of the dimensionless Vlasov-Poisson-Boltzmann (VPB) system; however, via the Chapman-Enskog expansion, it constitutes a second-order approximation to the VPB system. The purpose of this paper is to rigorously prove that if the difference between the local Maxwellian corresponding to the compressible NSP system and the initial value of the VPB system is a second-order small quantity in the Knudsen number, then the solutions of the two systems will maintain this order of magnitude difference for all times. The analysis in this paper is based on the energy estimates of the macroscopic equations, as well as the refined energy method within the macroscopic-microscopic decomposition framework.

Keywords： Vlasov-Poisson-Boltzmann system; macroscopic-microscopic decomposition; compressible Navier-Stokes-Poisson system

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本文引用格式

董丽娜, 刘双乾, 马璇, 马袁园. Vlasov-Poisson-Boltzmann 方程组的 Navier-Stokes-Poisson 方程组逼近——献给陈化教授 70 寿辰[J]. 数学物理学报, 2026, 46(2): 683-708

Dong Lina, Liu Shuangqian, Ma Xuan, Ma Yuanyuan. Compressible Navier-Stokes-Poisson System Approximation to the Vlasov-Poisson-Boltzmann System[J]. Acta Mathematica Scientia, 2026, 46(2): 683-708

1 引言

1.1 问题阐述

在无磁场的情况下, 等离子体在动理学层面的演化可由 Vlasov-Poisson-Boltzmann (VPB) 系统描述

(1.1) $ \begin{equation} \begin{cases} \partial_tF+\xi\cdot\nabla_xF-\nabla_x\phi\cdot\nabla_\xi F=\frac{1}{\epsilon}Q(F,F),\\ -\Delta_x\phi=\int_{\mathbb{R}^3}F\ \mathrm{d}\xi-\rho_1,\quad \text{当 }|x|\to \infty \text{ 时}, |\phi|\to 0, \end{cases}\end{equation}$

这里, $ F=F(t,x,v)\geq 0 $ 是电子在时刻 $ t\geq 0 $, 位置 $ x=(x_1,x_2,x_3)\in \mathbb{R}^3 $, 速度 $ \xi=(\xi_1,\xi_2,\xi_3)\in \mathbb{R}^3 $ 处的密度分布函数. 常数 $ \epsilon>0 $ 是与平均自由程成正比的 Knudsen 数. 自洽电势 $ \phi(t,x) $ 通过泊松方程与分布函数 $ F $ 耦合, 且 $ \rho_1>0 $ 是给定的背景电荷密度. (1.1) 式中的 Boltzmann 碰撞算子 $ Q(F,G) $, 对于硬球模型具有如下形式

$ \begin{align*} Q(F,G)=&\frac{1}{2}\int_{\mathbb{R}^3\times\mathbb{S}_+^2} |(\xi-\xi_*)\cdot\sigma|[F(\xi')G(\xi_*^\prime)+F(\xi_*')G(\xi')\\ &-F(\xi)G(\xi_*)-F(\xi_*)G(\xi)]\ \mathrm{d}\xi_*\mathrm{d}\sigma,\end{align*}$

其中 $ \mathbb{S}_+^2=\left\{\sigma\in\mathbb{S}^2:(\xi-\xi_*)\cdot\sigma\geq 0\right\} $, $ (\xi,\xi_*) $ 与 $ (\xi',\xi_*') $ 分别表示碰撞前后的速度,它们满足

$ \xi'=\xi-[(\xi-\xi_*)\cdot\sigma]\sigma,\quad \xi_*'=\xi_*+[(\xi-\xi_*)\cdot\sigma]\sigma. $

该文的目标是讨论当 Navier-Stokes-Poisson 系统决定的局部 Maxwellian (平衡态) 与 Vlasov-Poisson-Boltzman 系统初值之差为 Knudsen 数的二阶小量时, 两个方程组的解在 Sobolev 范数的意义下也是 Knudsen 数的二阶小量.

1.2 文献综述

现在, 我们来回顾有关 Boltzmann 型方程或方程组的既有研究. 众多研究探究了动理学方程与流体动力学系统 (如 Euler 方程组或 Navier-Stokes 方程组) 之间的关联. 首先, Bardos-Golse-Levermore^[1,2] 分别于 1991 年和 1993 年探讨了动理学理论与宏观流体动力学系统之间的关联, 并且系统地推导出相关形式极限. 关于 Boltzmann 方程收敛于可压缩流体方程组的问题, 我们列举了如下几个经典结果. Huang-Xin-Yang^[15] 于 2008 年得到了 Navier-Stokes 方程组与 Boltzmann 方程接触间断的大时间稳定性, 并给出了收敛速率. Guo-Jang-Jiang^[11] 于 2009 年采用 $ L^2\cap L^\infty $ 方法, 得到了 Boltzmann 方程的可压缩 Euler 极限. Xin-Zeng^[26] 于 2010 年致力于研究当 $ \epsilon \rightarrow 0 $ 时, Boltzmann 方程与可压缩 Euler 方程组的稀疏波逼近, 针对 Navier-Stokes 方程组, 同样得到了相似的结果. 针对复合波的 Euler 方程组黎曼解框架, Huang-Wang-Wang-Yang^[14] 证明了当 $ \epsilon \rightarrow 0 $ 时, Boltzmann 方程向可压缩 Euler 方程组的收敛性问题. Duan-Liu^[5] 于 2021 年在有界区域中的初边值问题中, 对 Boltzmann 方程的可压缩 Navier-Stokes-Fourier 近似进行了严格的数学分析.

接下来, 我们来考虑关于 Boltzmann 方程收敛于不可压缩流体方程组的情况. Golse 和 Saint-Raymond^[8] 于 2004 年在证明了适当缩放的 DiPerna-Lions 解序列的极限是 Leray 解, 这是首个将 Boltzmann 方程与不可压缩的 Navier-Stokes 方程组联系起来的渐近定理. Guo^[9] 于 2006 年证明了 Boltzmann 方程扩散展开的极限是不可压缩的 Navier-Stokes-Fourier (NSF) 方程组. Guo-Liu^[12] 于 2017年在周期区域的 $ L^2\cap L^\infty $ 空间框架下, 证明了不可压缩的 NSF 方程组近似对 Boltzmann 的适用性. Jiang-Masmoudi^[16] 于 2017 年探究了有界区域中, Boltzmann 方程的边界层及不可压缩的 NSF 极限的相关问题.

最后, 我们来讨论与 Vlasov-Poisson-Boltzmann 方程组相关的流体动力学极限. 首先, Guo-Jang^[10] 于 2010 年证明了 Vlasov-Poisson-Boltzmann 系统中任何接近光滑局部 Maxwellian 且具有小无旋速度的解, 当 $ \epsilon \rightarrow 0 $ 时, 会全局地随时间收敛到 Euler-Poisson 系统对应的解. Duan-Liu^[4] 于 2015 年证明了 Vlasov-Poisson-Boltzmann 方程组的稀疏波的稳定性. Li-Yang-Zhong^[19]于 2016 年研究了双极 Vlasov-Poisson-Boltzmann 方程组的谱分析和最佳衰减率. 基于谱分析的方法, Li-Yang-Zhong^[20] 于 2021 年证明了单极 Vlasov-Poisson-Boltzmann 方程组通过扩散展开收敛到不可压缩的 Navier-Stokes-Poisson-Fourier 方程组. Guo-Xiao^[13] 于 2021 年证明了当 $ \epsilon \rightarrow 0 $ 时, 相对论 Vlasov-Maxwell-Boltzmann 系统的解在时间上全局收敛于相对论 Euler-Maxwell 系统的解. Daun-Yang-Yu^[6] 于 2023 年证明了从 Vlasov-Maxwell-Boltzmann 系统到可压缩 Euler-Maxwell 系统的流体力学极限. 带有角截断的软势情形下, Li-Wang^[18] 于 2023 年证明了 Vlasov-Poisson-Boltzmann 系统全局收敛于可压缩 Euler-Poisson 极限. Lei-Liu-Xiao-Zhao^[17] 于 2024 年严格证明了在 Hilbert 展开的框架下, 当 $ \epsilon \rightarrow 0 $ 时, Vlasov-Maxwell-Landau (VML) 或非截段 Vlasov-Maxwell-Boltzmann (VMB) 系统的唯一经典解随时间全局收敛于 Euler-Maxwell 系统的光滑全局解. 在局部 Maxwellian 附近的能量框架下, 从 Vlasov-Poisson-Boltzmann 方程组到可压缩 Navier-Stokes-Poisson 方程组的结果是未知的, 本文开始准备研究这个问题.

1.3 符号说明

设 $ F $ 是 Vlasov-Poisson-Boltzmann 方程组 (1.1) 的解. 我们考虑宏观--微观分解, 其中宏观部分由局部 Maxwellian $ \mathbf{M} $ 表示, 微观部分记为 $ \mathbf{G} $, 即

$ F(t,x,\xi)=\mathbf{M}(t,x,\xi)+\mathbf{G}(t,x,\xi). $

这里, $ \mathbf{M}(t,x,\xi) $ 由 $ F $ 及五个守恒量来定义, 即质量密度 $ \rho(t,x) $, 动量密度 $ m(t,x)=\rho(t,x)u(t,x) $ 和能量密度 $ \mathbf{E}(t,x)+\frac{1}{2}|u(t,x)|^2 $:

(1.2) $ \begin{equation}\begin{cases} \rho(t,x)\equiv \displaystyle\int_{\mathbb{R}^3} F(t,x,\xi)\ \mathrm{d}\xi,\\[6pt] m^i(t,x)\equiv \displaystyle\int_{\mathbb{R}^3} \psi_i(\xi)F(t,x,\xi)\ \mathrm{d}\xi,\quad i=1,2,3,\\[6pt] \left[ \rho\left(\mathbf{E}(t,x)+\frac{1}{2}\lvert u(t,x) \rvert^2\right) \right]\equiv \displaystyle\int_{\mathbb{R}^3}\psi_4(\xi)F(t,x,\xi)\ \mathrm{d}\xi,\end{cases}\end{equation}$

其形式为

(1.3) $ \begin{equation} \mathbf{M}\equiv\mathbf{M}_{[\rho,u,\theta]}(\xi)\equiv \frac{\rho}{(2\pi R \theta)^\frac{3}{2}}\exp\left(-\frac{\lvert \xi-u\rvert^2}{2R\theta}\right),\end{equation}$

其中 $ \theta=\theta(t,x) $ 是温度, 与内能 $ \mathbf{E}(t,x) $ 相关, 满足 $ \mathbf{E}=\frac{3}{2}R\theta $ (为方便计算, 气体常数 $ R $ 取 $ \frac{2}{3} $, 此时可简化为 $ \mathbf{E}=\theta $); $ u(t,x)=[u^1(t,x),u^2(t,x),u^3(t,x)] $ 是流体速度. 此外, $ \psi_i,i=0,1,2,3,4 $ 是五个碰撞不变量

$ \psi_0=1,\quad \psi_i=\xi^i\ (i=1,2,3),\quad \psi_4=\frac{1}{2}|\xi|^2, $

满足

$ \int_{\mathbb{R}^3} \psi_\ell Q(f,g)\ \mathrm{d}\xi=0\quad \text{其中}\ \ell=0,1,2,3,4. $

对于任意给定的 Maxwellian $ \mathbf{\widetilde{M}}=\mathbf{\widetilde{M}}_{[\widetilde{\rho},\widetilde{u},\widetilde{\theta}]} $, 在 $ \xi \in \mathbb{R}^3 $ 上定义内积为

$ \langle f,g\rangle_{\mathbf{\widetilde{M}}}\equiv \int_{\mathbb{R}^3}\frac{f(\xi)g(\xi)}{\mathbf{\widetilde{M}}}\ \mathrm{d}\xi. $

利用上述关于 Maxwellian $ \widetilde{\mathbf{M}} $ 定义的内积, 张成宏观子空间的以下函数两两正交

$ \begin{align*} & \chi_0^{\mathbf{\widetilde{M}}}(\xi;\widetilde{\rho},\widetilde{u}, \widetilde{\theta})\equiv\frac{1}{\sqrt{\widetilde{\rho}}} \mathbf{\widetilde{M}},\\ & \chi_i^{\mathbf{\widetilde{M}}}(\xi;\widetilde{\rho},\widetilde{u}, \widetilde{\theta})\equiv\frac{\xi^i-\widetilde{u}^i} {\sqrt{\frac{2}{3}\widetilde{\rho}\widetilde{\theta}}} \widetilde{\mathbf{M}},\quad i=1,2,3,\\ & \chi_4^{\mathbf{\widetilde{M}}}(\xi;\widetilde{\rho},\widetilde{u}, \widetilde{\theta})\equiv\frac{1}{\sqrt{6\widetilde{\rho}}} \left(\frac{3\lvert \xi-\widetilde{u}\rvert^2}{2\widetilde{\theta}}-3 \right) \mathbf{\widetilde{M}},\\ & \langle \chi_i^{\mathbf{\widetilde{M}}},\chi_j^{\mathbf{\widetilde{M}}}\rangle =\delta^{i,j},\quad \ i,j=0,1,2,3,4,\end{align*}$

其中 $ \delta^{i,j} $ 是克罗内克函数. 利用上述基, 宏观投影 $ \mathbf{P}_0^{\mathbf{\widetilde{M}}} $ 和微观投影 $ \mathbf{P}_1^{\mathbf{\widetilde{M}}} $ 可定义为

$ \begin{cases} \mathbf{P}_0^{\mathbf{\widetilde{M}}} h \equiv \displaystyle \sum_{j=0}^{4} \langle h, \chi_j^{\mathbf{\widetilde{M}}} \rangle_{\mathbf{\widetilde{M}}} \chi_j^{\mathbf{\widetilde{M}}}, \\ \mathbf{P}_1^{\mathbf{\widetilde{M}}} h \equiv h - \mathbf{P}_0^{\mathbf{\widetilde{M}}} h,\end{cases} $

注意, 算子 $ \mathbf{P}_0^{\mathbf{\widetilde{M}}} $ 和 $ \mathbf{P}_1^{\mathbf{\widetilde{M}}} $ 是关于内积 $ \langle \cdot,\cdot\rangle_{\mathbf{\widetilde{M}}} $ 的正交且自伴投影算子.

采用这些符号约定, 对应于 (1.1) 式的解 $ F(t,x,\xi) $ 满足

$\mathbf{P}_0^{\mathbf{M}} F = \mathbf{M}, \quad \mathbf{P}_1^{\mathbf{M}} F = \mathbf{G}.$

于是, (1.1) 式可改写为

(1.4) $ \begin{equation}\label{1.4}\partial_t (\mathbf{M}+\mathbf{G}) +\xi\cdot\nabla_x (\mathbf{M}+\mathbf{G})-\nabla_x\phi \cdot \nabla_\xi (\mathbf{M}+\mathbf{G})=\frac{1}{\epsilon} L_\mathbf{M} \mathbf{G} + \frac{1}{\epsilon} Q(\mathbf{G},\mathbf{G}),\end{equation}$

这里 $ L_\mathbf{M} \mathbf{G}=2Q (\mathbf{G},\mathbf{M}) $ 是关于局部 Maxwellian $ \mathbf{M} $ 的线性化玻尔兹曼碰撞算子.

分别对 (1.4) 式应用于 $ \mathbf{P}_0^{\mathbf{M}} $ 和 $ \mathbf{P}_1^{\mathbf{M}} $, 可得

(1.5) $ \begin{equation}\mathbf{M}_t+\mathbf{P}_0^{\mathbf{M}} (\xi\cdot\nabla_x \mathbf{M})+\mathbf{P}_0^{\mathbf{M}} (\xi\cdot\nabla_x \mathbf{G})-\mathbf{P}_0^{\mathbf{M}} (\nabla_x\phi \cdot \nabla_\xi \mathbf{M})=0\end{equation}$

和

(1.6) $ \begin{equation}\label{1.6}\mathbf{G}_t+\mathbf{P}_1^{\mathbf{M}} (\xi\cdot\nabla_x \mathbf{M})+\mathbf{P}_1^{\mathbf{M}} (\xi\cdot\nabla_x \mathbf{G})-\mathbf{P}_1^{\mathbf{M}} (\nabla_x\phi \cdot \nabla_\xi \mathbf{G})=\frac{1}{\epsilon} L_\mathbf{M} \mathbf{G} + \frac{1}{\epsilon} Q(\mathbf{G},\mathbf{G}).\end{equation}$

现在, 守恒律方程组为

$\int_{\mathbb{R}^{3}} \psi_{\ell} (F_{t} + \xi \cdot \nabla_{x} F - \nabla_x\phi \cdot \nabla_\xi F) \ d\xi = 0, \quad \ell = 0, 1, 2, 3, 4. $

由 (1.6) 式可知

$\mathbf{G} = \epsilon L_{\mathbf{M}}^{-1}\left[ \mathbf{P}_1^{\mathbf{M}} (\xi \cdot \nabla_x \mathbf{M}) \right] =: \epsilon L_{\mathbf{M}}^{-1}\left[ \mathbf{P}_1^{\mathbf{M}} (\xi \cdot \nabla_x \mathbf{M}) \right] + \varTheta,$

其中 $ \varTheta= \epsilon L_{\mathbf{M}}^{-1}\left[ \mathbf{G}_t + \mathbf{P}_1^{\mathbf{M}} (\xi \cdot \nabla_x \mathbf{G})- \mathbf{P}_1^{\mathbf{M}}(\nabla_x\phi \cdot\nabla_\xi\mathbf{G}) \right] - L_{\mathbf{M}}^{-1}\left[ Q(\mathbf{G}, \mathbf{G}) \right] $.

定义黏性系数 $ \mu(\theta) $ 和热传导系数 $ \kappa(\theta) $ 为

$ \begin{cases}A^j(\xi) = \dfrac{|\xi|^2 - 5}{2} \xi^j, \quad B^{ij}(\xi) = \xi^i \xi^j - \dfrac{1}{3} \delta^{ij}|\xi|^2,& i, j = 1, 2, 3, \\\mu(\theta) = -R\theta \displaystyle\int_{\mathbb{R}^3} B^{ij}\left( \frac{\xi - u}{\sqrt{R\theta}} \right) L_{\mathbf{M}_{[1, u, \theta]}}^{-1}\left( B^{ij}\left( \frac{\xi - u}{\sqrt{\frac{2}{3}\theta}} \right) \mathbf{M}_{[1, u, \theta]} \right) > 0, & i \neq j, \\\kappa(\theta) = -R^2\theta \displaystyle\int_{\mathbb{R}^3} A^j\left( \frac{\xi - u}{\sqrt{\frac{2}{3}\theta}} \right) L_{\mathbf{M}_{[1, u, \theta]}}^{-1}\left( A^j\left( \frac{\xi - u}{\sqrt{\frac{2}{3}\theta}} \right) \mathbf{M}_{[1, u, \theta]} \right) > 0,\end{cases}$

其中 $ A=(A^j),B=(B^{i,j}) $ 为 Burnett 函数,参见文献 [1,2,4].

宏观-微观分解^[27], 我们可以得到

(1.7) $ \begin{equation}\label{1.7}\begin{cases}\rho_t + \nabla_x \cdot m = 0, \\\quad m_t^i + \displaystyle\sum_{j=1}^3 (u^i m^j)_{x^j} + \frac{2}{3} (\rho \theta)_{x^i}+\rho\phi_{x^i} \\ = \epsilon \displaystyle\sum_{j=1}^3 \left[ \mu(\theta) \left( u_{x^j}^i + u_{x^i}^j - \frac{2}{3} \delta^{ij} \nabla_x \cdot u \right) \right]_{x^j} - \displaystyle\int_{\mathbb{R}^3} \psi_i(\xi) (\xi \cdot \nabla_x \varTheta){\rm d}\xi, \quad i = 1, 2, 3, \\\quad \left[ \bar{\rho} \left( \dfrac{1}{2} |\bar{u}|^2 + \bar{\theta} \right) \right]_t \!+\! \displaystyle\sum_{j=1}^3 \left[ u^j \rho \left( \frac{1}{2} |u|^2 \!+\! \frac{5}{3} \theta \right) \right]_{x^j}+\rho u \cdot \nabla_x \phi +\displaystyle\int_{\mathbb{R}^3} \psi_4(\xi) (\xi \cdot \nabla_x \varTheta){\rm d}\xi\\ = \epsilon \displaystyle\sum_{j=1}^3 (\kappa(\theta) \theta_{x^j})_{x^j} + \epsilon \displaystyle\sum_{i,j=1}^3 \left\{ \mu(\theta) u^i \left( u_{x^j}^i + u_{x^i}^j - \frac{2}{3} \delta^{ij} \nabla_x \cdot u \right) \right\}_{x^j},\\-\Delta_x\phi=\rho-\rho_1.\end{cases}\end{equation}$

由方程组 (1.7) 可形式上推出: 若令 $ \epsilon=0 $ 并舍弃所有含 $ \varTheta $ 的项, 则 (1.7) 式成为可压缩 Euler-Poisson (EP) 方程组

$ \begin{cases}\bar{\rho}_t + \nabla_x \cdot \overline{m} = 0, \\\overline{m}_t^i + \displaystyle\sum_{j=1}^3 (\bar{u}^i \overline{m}^j)_{x^j} + \frac{2}{3} (\bar{\rho} \bar{\theta})_{x^i} +\bar{\rho}\bar{\phi}_{x^i}= 0, \quad i = 1, 2, 3, \\\left[ \bar{\rho} \left( \dfrac{1}{2} |\bar{u}|^2 + \bar{\theta} \right) \right]_t + \displaystyle\sum_{j=1}^3 \left[ \bar{u}^j \bar{\rho} \left( \frac{1}{2} |\bar{u}|^2 + \frac{5}{3} \bar{\theta} \right) \right]_{x^j} +\bar{\rho}\bar{u}\cdot\nabla_x \bar{\phi} = 0,\\-\Delta_x\bar{\phi}=\bar{\rho}-\rho_1.\end{cases} $

而若仅舍弃所有含 $ \varTheta $ 的项, 则 (1.7) 式为可压缩 Navier-Stokes-Poisson (NSP) 方程组

(1.8) $ \begin{equation}\label{1.8} \begin{cases} \bar{\rho}_t + \nabla_x \cdot \overline{m} = 0, \\ \overline{m}_t^i + \displaystyle\sum_{j=1}^3 (\bar{u}^i \overline{m}^j)_{x^j} + \dfrac{2}{3} (\bar{\rho} \bar{\theta})_{x^i}+\bar{\rho}\bar{\phi}_{x^i} = \epsilon \displaystyle\sum_{j=1}^3 \left[ \mu(\bar{\theta}) \left( \bar{u}_{x^j}^i + \bar{u}_{x^i}^j - \dfrac{2}{3} \delta^{ij} \nabla_x \cdot \bar{u} \right) \right]_{x^j}, & i = 1, 2, 3, \\ \quad \left[ \bar{\rho} \left( \dfrac{1}{2} |\bar{u}|^2 + \bar{\theta} \right) \right]_t + \displaystyle\sum_{j=1}^3 \left[ \bar{u}^j \bar{\rho} \left( \dfrac{1}{2} |\bar{u}|^2 + \dfrac{5}{3} \bar{\theta} \right) \right]_{x^j}+\bar{\rho}\bar{u}\cdot\nabla_x \bar{\phi} \\ = \epsilon \displaystyle\sum_{j=1}^3 (\kappa(\bar{\theta}) \bar{\theta}_{x^j})_{x^j} + \epsilon \displaystyle\sum_{i,j=1}^3 \left\{ \mu(\bar{\theta}) \bar{u}^i \left( \bar{u}_{x^j}^i + \bar{u}_{x^i}^j - \dfrac{2}{3} \delta^{ij} \nabla_x \cdot \bar{u} \right) \right\}_{x^j},\\ -\Delta_x\bar{\phi}=\bar{\rho}-\rho_1. \end{cases}\end{equation}$

本文致力于在扰动框架下研究 Vlasov-Poisson-Boltzmann 方程组的柯西问题. 此时,方程组 (1.1) 需补充初值条件

(1.9) $ \begin{equation}\label{1.9} F(0,x,\xi)=F_0(x,\xi)= \mathbf{M}_{[\rho_0(x),u_0(x),\theta_0(x)]}(\xi)+\mathbf{G}_0(x,\xi),\end{equation}$

而方程组 (1.7) 对应的初值条件为

$ \rho(0,x)=\rho_0(x),\quad u(0,x)=u_0(x),\quad \theta(0,x)=\theta_0(x). $

记 $ [\bar{\rho}, \bar{u}, \bar{\theta}] = [\bar{\rho}(t, x), \bar{u}(t, x), \bar{\theta}(t, x)] $ 为可压缩 Navier-Stokes-Poisson 方程组 (1.8) 在平凡稳态解 $ [\rho, u, \theta] = [1,0,1] $ 附近的整体经典解. 给定 (1.8) 式的初值

(1.10) $ \begin{equation}\label{1.10} \bar{\rho}(0, x) = \bar{\rho}_0(x), \quad \bar{u}(0, x) = \bar{u}_0(x), \quad \bar{\theta}(0, x) = \bar{\theta}_0(x),\end{equation}$

满足

(1.11) $ \begin{equation}\label{1.11} \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \| \partial^\gamma [\bar{\rho}_0 - 1, \bar{u}_0, \bar{\theta}_0 - 1, \Delta^{-1}\nabla\bar{\rho}_0] \| + \epsilon \sum_{|\alpha| = N_1 - 1} \| \partial^\alpha \nabla_x [\bar{\rho}_0 - 1] \| \leqslant \bar{\lambda} \epsilon.\end{equation}$

基于可压缩 Navier-Stokes-Poisson 方程组 (1.8) 的整体解 $ [\bar{\rho}, \bar{u}, \bar{\theta}] $, Vlasov-Poisson-Boltzmann 方程组的近似解可被定义为一种局部热平衡态

(1.12) $ \begin{equation}\label{1.12} \overline{\mathbf{M}} = \mathbf{M}_{[\bar{\rho}(t, x), \bar{u}(t, x), \bar{\theta}(t, x)]}(\xi) = \frac{\bar{\rho}(t, x)}{\sqrt{\left( \frac{4}{3} \pi \bar{\theta}(t, x) \right)^3}} \exp \left( - \frac{|\xi - \bar{u}(t, x)|^2}{\frac{4}{3} \bar{\theta}(t, x)} \right).\end{equation}$

记 $ \overline{\mathbf{M}} $ 附近的碰撞频率为

$ \nu_{\overline{\mathbf{M}}}(\xi) = \int_{\mathbb{R}^3} \int_{\mathbb{S}_+^2} |(\xi - \xi_*) \cdot \sigma| \overline{\mathbf{M}}(\xi_*){\rm d}\xi {\rm d}\sigma, $

参见[3,25] 可知, 若 $ [\bar{\rho}, \bar{u}, \bar{\theta}] $ 是 $ [1,0,1] $ 的充分小扰动, 则 $ \nu_{\overline{\mathbf{M}}}(\xi) \sim 1 + |\xi| $.

注记在本文中, $ C $ 表示某个一般的正 (通常为较大的) 常数. $ \mathcal{A} \lesssim \mathcal{B} $ 表示存在常数 $ C > 0 $, 使得 $ \mathcal{A} \leqslant C \mathcal{B} $. $ \mathcal{A} \sim \mathcal{B} $ 意为 $ \mathcal{A} \lesssim \mathcal{B} $ 且 $ \mathcal{B} \lesssim \mathcal{A} $. 我们用 $ L^2 $ 表示常用的 Hilbert 空间 $ L^2 = L_{x, \xi}^2 $ 或 $ L_x^2 $, 其范数记为 $ \| \cdot \| $, 用 $ \langle \cdot, \cdot \rangle $ 表示 $ L_\xi^2 $ 上的内积, $ (\cdot, \cdot) $ 表示 $ L_{x, \xi}^2 $ 上的内积.

对于多重指标 $ \gamma = (\gamma_0, \alpha) = (\gamma_0, \alpha_1, \alpha_2, \alpha_3) $ 和 $ \beta = (\beta_1, \beta_2, \beta_3) $, 有 $ \partial^\gamma = \partial_t^{\gamma_0} \partial_x^\alpha = \partial_t^{\gamma_0} \partial_{x_1}^{\alpha_1} \partial_{x_2}^{\alpha_2} \partial_{x_3}^{\alpha_3} $, $ \partial^\beta = \partial_\xi^\beta = \partial_{\xi_1}^{\beta_1} \partial_{\xi_2}^{\beta_2} \partial_{\xi_3}^{\beta_3} $. $ \gamma $ 的长度为 $ |\gamma| = \gamma_0 + |\alpha| = \gamma_0 + \alpha_1 + \alpha_2 + \alpha_3 $, $ |\beta| = \beta_1 + \beta_2 + \beta_3 $. 我们也用 $ \partial_t^{\gamma_0} $ 和 $ \partial_x^\alpha $ 分别表示纯 $ t $ 导数和纯 $ x $ 导数.

现在我们定义 Vlasov-Poisson-Boltzmann 方程组 (1.1) 的解所在的函数空间. 对任意 $ T \in (0, +\infty] $, 定义

$ \mathbf{H}_{x, \xi}^N([T]) = \left\{ h(t, x, \xi) \,\bigg|\, \begin{aligned}& \frac{\partial^\gamma \partial^\beta h(t, x, \xi)}{\sqrt{\mathbf{M}_-(\xi)}} \in \mathbf{BC}\left( [T], L_{x, \xi}^2\left( \mathbb{R}^3 \times \mathbb{R}^3 \right) \right) \\& \frac{1}{\sqrt{\epsilon}} \frac{\sqrt{\nu_{\overline{\mathbf{M}}}(\xi)} \partial^\gamma \partial^\beta \mathbf{P} _1^{\overline{\mathbf{M}}} h(t, x, \xi)}{\sqrt{\mathbf{M}_-(\xi)}} \in L_{t, x, \xi}^2\left( (0, T] \times \mathbb{R}^3 \times \mathbb{R}^3 \right) \\& \text{其中 } |\gamma| + |\beta| \leqslant N\end{aligned} \right\} $

对每个 $ t \in [T] $, 对应的范数 $ \| h(t) \|_{\mathbf{H}_{x, \xi}^N} $ 定义为

$ \| h(t) \|_{\mathbf{H}_{x, \xi}^N} \equiv \sum_{|\gamma| + |\beta| \leqslant N} \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{|\partial^\gamma \partial^\beta h(t, x, \xi)|^2}{\mathbf{M}_-} {\rm d}\xi {\rm d}x, $

这里 $ \mathbf{M}_- = \mathbf{M}_{[\rho_-, u_-, \theta_-]} $ 是给定的全局 Maxwellian, 满足

(1.13) $ \begin{equation}\label{1.13}\begin{cases}\dfrac{1}{2} \bar{\theta}(t, x) < \theta_- < \bar{\theta}(t, x), \\|\bar{\rho}(t, x) - \rho_-| + |\bar{u}(t, x) - u_-| + |\bar{\theta}(t, x) - \theta_-| < C \epsilon,\end{cases}\end{equation}$

对所有 $ (t, x) \in \mathbb{R}^+ \times \mathbb{R}^3 $ 成立.

1.4 问题难点以及证明思路

我们研究的问题主要需要克服以下几个困难

(1) 得到电场项的能量和耗散,使得估计封闭;

(2) 处理外力项 $ \nabla_x\phi \cdot \nabla_\xi F $;

我们针对上述问题给予了如下的解决措施

• 首先, $ \bar{\rho}\bar{u} \cdot \nabla_x\bar{\phi} $ 关于 $ \tau $ 和 $ x $ 在 $ [0, t] \times \mathbb{R}^3 $ 上积分, 我们借助质量守恒 $ \bar{\rho}_t + \nabla_x \cdot \overline{m} = 0 $ 和分部积分, 可得 $ \|\nabla_x \bar{\phi}(t)\|_{L_x^2}^2 $, $ (\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\rho-\bar{\rho})_t) $ 利用泊松方程和分部积分,便有 $ \|\partial^\gamma \nabla_x(\phi-\bar{\phi})(t)\|^2 $. 我们通过 $ (\bar{\rho},\bar{u},\bar{\theta}) $ 动量守恒方程与 $ \frac{\epsilon\bar{\rho}^2\partial^\gamma\bar{\rho}_{x^i}} {\frac{4}{3}\mu(\bar{\theta})} $ 做内积, 有

$ \begin{align*}&\sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} (\partial^\gamma\bar{\phi}_{x^i},\ \frac{\epsilon\bar{\rho}^2}{\frac{4}{3}\mu(\bar{\theta})}\partial^\gamma\bar{\rho}_{x^i}) \\=&\sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} (\partial^\gamma\bar{\phi}_{x^i},\ \nabla_x\frac{\epsilon\bar{\rho}^2}{\frac{4}{3}\mu(\bar{\theta})}\partial^\gamma\nabla_x\bar{\phi}_{x^i})+\sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3}\frac{\bar{\rho}^2}{\frac{4}{3}\mu(\bar{\theta})}|\partial^\gamma\nabla_x\bar{\phi}_{x^i}|^2\ \mathrm{d}x.\end{align*}$

刚好就是 $ \bar{\phi} $ 的二阶耗散. 在做宏观部分的估计的过程中, 类似地, 我们让 $ u-\bar{u} $ 的方程与 $ \epsilon\partial^\gamma(\rho-\bar{\rho})_{x^i} $ 做内积得到了 $ \phi-\bar{\phi} $ 的二阶耗散.

• 至于外力项 $ \nabla_x\phi \cdot \nabla_\xi F $, 我们采用宏观--微观分解, 将 $ F $ 拆解为 $ \mathbf{M}+\mathbf{G} $, $ g $ 作为 (1.7) 与 (1.8) 式的纽带. 由于宏观量的 $ L^2 $ 范数是没有耗散的, $ (\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta}) $ 的耗散是从一阶导数开始的. 所以我们将 $ g $ 的表达式与 $ \frac{\mathbf{P}_1^{\overline{\mathbf{M}}} g}{\mathbf{M}_-} $ 做内积, 再结合索伯列夫不等式进行估计外力项.

• 关于高阶混合导数项 $ \left(\partial^\gamma\partial^\beta (\nabla_x\phi \cdot \nabla_\xi F),\ \partial^\gamma\partial^\beta F \right) $, 其中 ($ |\beta|\geq1,|\gamma|+|\beta|\leq N $), 关键在于利用莱布尼茨求导法则和索伯列夫不等式来进行估计, 而且注意到 $ \mathbf{P}_0^{\overline{\mathbf{M}}}g=\mathbf{P}_0^{\overline{\mathbf{M}}} (\mathbf{M}-\overline{\mathbf{M}}) $, 我们可以将 $ g $ 拆解成 $ \mathbf{P}_0^{\overline{\mathbf{M}}} g $ 和 $ \mathbf{P}_1^{\overline{\mathbf{M}}} g $ 分别来处理.

1.5 主要结果

借助上述符号, 主要结果可表述如下

定理 1.1 取 $ N \geqslant 4 $, $ N_1 \geqslant N + 2 $, $ 0 < \epsilon \leqslant 1 $, 再取两个充分小且与 $ \epsilon $ 无关的常数 $ \lambda_0 > \bar{\lambda} > 0 $. 设

(1.14) $ \begin{equation}\| F_0(x, \xi) - \mathbf{M}_{[\bar{\rho}(0, x), \bar{u}(0, x), \bar{\theta}(0, x)]} \|_{\mathbf{H}_{x, \xi}^N}+\|\Delta^{-1}\nabla_x (\rho(0,x)-\bar{\rho}(0,x))\|_{H_x^N} \lesssim \lambda_0 \epsilon^2,\end{equation}$

且 $ [\bar{\rho}(0, x), \bar{u}(0, x), \bar{\theta}(0, x), \Delta^{-1}\nabla_x \bar{\rho}(0,x)] $

满足 (1.11) 式, 使得可压缩 Navier-Stokes-Poisson方程组 (1.8) 的全局光滑解 $ [\bar{\rho}(t, x), \bar{u}(t, x), \bar{\theta}(t, x), \bar{\phi}(t,x)] $, 满足

(1.15) $\begin{equation}\sup_{t \in \mathbb{R}^+}\left\{\| \overline{\mathbf{M}} - \mathbf{M}_{[1,0,1]} \|_{\mathbf{H}_{x, \xi}^{N_1 - 1}}+\|\nabla_x \bar{\phi}\|_{H_x^{N_1-1}}\right\}\lesssim \bar{\lambda} \epsilon,\end{equation}$

其中 $ \overline{\mathbf{M}} $ 如 (1.12) 式定义, $ [\rho_-, u_-, \theta_-] $ 在范数 $ \mathbf{H}_{x, \xi}^N $ 与 $ \mathbf{H}_{x, \xi}^{N_1 - 1} $ 中满足 (1.13) 式.则初值问题 (1.1), (1.9) 存在唯一的全局经典解 $ [F(t, x, \xi), \phi(t,x)] $ 满足 $ F(t, x, \xi) \geqslant 0 $, $ F(t, x, \xi)-\overline{\mathbf{M}} \in \mathbf{H}_{x, \xi}^N(\mathbb{R}^+) $ 且

(1.16) $ \begin{equation}\sup_{t \in \mathbb{R}^+} \left\{\| F(t, x, \xi) - \overline{\mathbf{M}} \|_{\mathbf{H}_{x, \xi}^N}+ \|\nabla_x (\phi(t,x)-\bar{\phi}(t,x))\|_{H_x^N}\right\}\lesssim \lambda_0 \epsilon^2. \end{equation}$

2 完全可压缩 Navier-Stokes-Poisson 方程组

本节旨在探究可压缩 Navier-Stokes-Poisson 方程组解的整体存在性, 关键在于推导柯西问题 (1.8), (1.10) 的一致先验估计, 我们有如下定理

定理 2.1 设 $ N_1\geqslant N + 2 $, $ N\geqslant 4 $, 且假设 (1.11) 成立. 则对任意给定的 $ 0< \epsilon \leqslant 1 $, 柯西问题 (1.8), (1.10) 存在唯一的整体经典解 $ [\bar{\rho}, \bar{u}, \bar{\theta}] = [\bar{\rho}(t, x), \bar{u}(t, x), \bar{\theta}(t, x)] $, 满足

(2.1) $ \begin{aligned}\label{2.1} & E(\bar{\rho}, \bar{u}, \bar{\theta},\bar{\phi})(t) + \epsilon \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_0^t \| \partial^\gamma \nabla_x [\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1] \|^2 {\rm d}\tau \nonumber\\ & +\epsilon\sum_{|\gamma| \leqslant N_1 }\int_0^t\int_{\mathbb{R}^3}|\partial^\gamma\nabla_x^2\bar{\phi}|^2\ \mathrm{d}x\mathrm{d}\tau \lesssim E(\bar{\rho}, \bar{u}, \bar{\theta})(0),\end{aligned}$

其中

$E(\bar{\rho}, \bar{u}, \bar{\theta},\bar{\phi})(t) \sim \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \| \partial^\gamma [\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1] \|^2 + \epsilon^2 \sum_{|\alpha| = N_1 - 1} \| \partial^\alpha \nabla_x [\bar{\rho} - 1] \|^2+\epsilon \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \| \partial^\gamma\nabla_x\bar{\phi}(t)\|^2.$

注这里以及后文的 $ \phi_0(x) $ 和 $ \bar{\phi}_0(x) $ 是指: $ \phi_0(x)=-\Delta^{-1}\nabla_x \rho_0(x),\ \bar{\phi}_0(x)=-\Delta^{-1}\nabla_x \bar{\rho}_0(x) $.

证设 $ [\bar{\rho}(t, x), \bar{u}(t, x), \bar{\theta}(t, x)] $ 是柯西问题 (1.8), (1.10) 在时间区间 $ 0\leqslant t \leqslant T $ 上定义的经典解, 其初值 $ [\bar{\rho}_0, \bar{u}_0, \bar{\theta}_0] $ 满足 (1.11) 式 (其中 $ 0 < T \leqslant \infty $), 进一步假设 $ [\bar{\rho}(t, x), \bar{u}(t, x), \bar{\theta}(t, x)] $ 满足

(2.2) $ \begin{equation}\label{2.2}\max_{0\leqslant t \leqslant T}\left\{\sum_{\substack{|\gamma| \leqslant N_1 -1 \\ \gamma_0 \leqslant N_1 - 2}} \|\partial^\gamma[\bar{\rho}-1, \bar{u}, \bar{\theta}-1, \Delta^{-1}\nabla_x\bar{\rho}]\| + \epsilon \sum_{|\alpha|=N_1-1} \|\partial^\alpha \nabla_x [\bar{\rho}-1]\| \right\}\leqslant\bar{\lambda} \epsilon.\end{equation}$

其中 $ N_1 \geqslant N + 2 $ 且 $ \bar{\lambda} $ 足够小. 接下来, 我们旨在基于先验假设 (2.1) 式, 得到柯西问题 (1.8), (1.10) 解的先验估计, 以探究 NSP 方程组经典解的存在性与唯一性. 首先, 我们采用熵-熵流对方法, 对 Vlasov-Poisson-Boltzmann 方程组进行低阶能量估计. 为此, 构造熵 $ \bar{S} $

$ \bar{S}= -\frac{2}{3} \ln \bar{\rho} + \ln \left( \frac{4}{3} \pi \bar{\theta} \right) + 1. $

在平凡稳定解 $ [\bar{\rho}_s, \bar{u}_s, \bar{\theta}_s] = [1,0,1] $ 附近, 构造如下熵-熵流对 $ (\bar{\eta}, \bar{q}) $

$ \begin{cases}\bar{\eta} = \bar{\theta}_s \left\{ -\dfrac{3}{2} \bar{\rho} \bar{S} + \dfrac{3}{2} \bar{\rho}_s \bar{S}_s + \left. \dfrac{3}{2} \nabla_{\overline{\mathbf{m}}} (\bar{\rho} \bar{S}) \right|_{\overline{\mathbf{m}} = \overline{\mathbf{m}}_s} \cdot (\overline{\mathbf{m}} - \overline{\mathbf{m}}_s) \right\}, \\\bar{q}^j = \bar{\theta}_s \left\{ -\dfrac{3}{2} \bar{\rho} \bar{u}^j \bar{S} + \left. \dfrac{3}{2} \nabla_{\overline{\mathbf{m}}} (\bar{\rho} \bar{S}) \right|_{\overline{\mathbf{m}} = \overline{\mathbf{m}}_s} \cdot (\bar{\mathbf{n}}^j - \bar{\mathbf{n}}_s^j) \right\}, \quad j = 1, 2, 3,\end{cases} $

其中

$ \begin{cases}\displaystyle \overline{\mathbf{m}} = (\bar{m}^0, \bar{m}^1, \bar{m}^2, \bar{m}^3, \bar{m}^4)^t = \left( \bar{\rho}, \bar{\rho} \bar{u}^1, \bar{\rho} \bar{u}^2, \bar{\rho} \bar{u}^3, \bar{\rho} \left( \frac{1}{2} |\bar{u}|^2 + \bar{\theta} \right) \right)^t, \\\displaystyle \overline{\mathbf{n}} = (\bar{\mathbf{n}}^1, \bar{\mathbf{n}}^2, \bar{\mathbf{n}}^3), \\\displaystyle \overline{\mathbf{n}}^j = (\bar{n}_0^j, \bar{n}_1^j, \bar{n}_2^j, \bar{n}_3^j, \bar{n}_4^j)^t \\\quad\,\, \displaystyle = \left( \bar{\rho} \bar{u}^j, \bar{u}^1 \bar{m}^j + \delta^{1j} \frac{2}{3} \bar{\rho} \bar{\theta}, \bar{u}^2 \bar{m}^j + \delta^{2j} \frac{2}{3} \bar{\rho} \bar{\theta}, \bar{u}^3 \bar{m}^j \right. \\\qquad \displaystyle \left. + \delta^{3j} \frac{2}{3} \bar{\rho} \bar{\theta}, \bar{\rho} \bar{u}^j \left( \frac{1}{2} |\bar{u}|^2 + \frac{5}{3} \bar{\theta} \right) \right)^t, \quad j = 1, 2, 3,\end{cases} $

且 $ \overline{\mathbf{m}}_s = \overline{\mathbf{m}}(\bar{\rho}_s, \bar{u}_s, \bar{\theta}_s) $, $ \overline{\mathbf{n}}_s^j = \overline{\mathbf{n}}^j(\bar{\rho}_s, \bar{u}_s, \bar{\theta}_s) $, $ \bar{S}_s = \bar{S}(\bar{\rho}_s, \bar{u}_s, \bar{\theta}_s) $.

由于

$ \begin{cases}\displaystyle (\bar{\rho} \bar{S})_{\bar{m}^0} = \bar{S} + \frac{|\bar{u}|^2}{2 \bar{\theta}} - \frac{5}{3}, \\\displaystyle (\bar{\rho} \bar{S})_{\bar{m}^i} = -\frac{\bar{u}^i}{\bar{\theta}}, \quad i = 1, 2, 3, \\\displaystyle (\bar{\rho} \bar{S})_{\bar{m}^4} = \frac{1}{\bar{\theta}},\end{cases} $

我们得到

$ \begin{cases}\displaystyle \bar{\eta} = \frac{3}{2} \left\{ \bar{\rho} \bar{\theta} - \bar{\theta}_s \bar{\rho} \bar{S} + \bar{\rho} \left[ \left( \bar{S}_s - \frac{5}{3} \right) \bar{\theta}_s + \frac{|\bar{u}|^2}{2} \right] + \frac{2}{3} \bar{\rho}_s \bar{\theta}_s \right\}, \\\displaystyle \bar{q}^j = \bar{u}^i \bar{\eta} + \bar{u}^j \left( \bar{\rho} \bar{\theta} - \bar{\rho}_s \bar{\theta}_s \right), \quad j = 1, 2, 3.\end{cases} $

注意, 对于任意有界闭区域 $ \mathcal{D} \subset \Sigma=\{\overline{\mathbf{m}}:\bar{\rho}>0,\bar{\theta}>0\} $ 内的 $ \overline{\mathbf{m}} $, 存在仅依赖于 $ \mathcal{D} $ 的正常数 $ C $, 使得如此构造的熵-熵流满足 (参见[22,23])

(2.3) $ \begin{equation}\label{2.3}C^{-1} |\overline{\mathbf{m}} - \overline{\mathbf{m}}_s|^2 \leqslant \bar{\eta} \leqslant C |\overline{\mathbf{m}} - \overline{\mathbf{m}}_s|^2.\end{equation}$

而 $ (\bar{\eta}, \bar{q}_1, \bar{q}_2, \bar{q}_3) $ 是下述偏微分方程的解

(2.4) $ \begin{aligned}\label{2.4}\bar{\eta}_t + \text{div}_x \bar{\mathbf{q}} = & -\sum_{i,j=1}^3 \frac{3}{2} \epsilon \frac{\bar{\theta}_s}{\bar{\theta}} \left[ \frac{\mu(\bar{\theta})}{2} \left( \bar{u}_{x^j}^i + \bar{u}_{x^i}^j \right)^2 - \frac{2}{3} \mu(\bar{\theta}) \left( \text{div}_x \bar{u} \right)^2 \right] - \sum_{j=1}^3 \frac{3}{2} \epsilon \frac{\bar{\theta}_s}{\bar{\theta}} \left( \kappa(\bar{\theta}) \bar{\theta}_{x^j} \right)_{x^j} \nonumber\\& + \sum_{i,j=1}^3 \frac{3}{2} \epsilon \left[ \mu(\bar{\theta}) \bar{u}^i \left( \bar{u}_{x^j}^i + \bar{u}_{x^i}^j - \frac{2}{3} \delta^{ij} \text{div}_x \bar{u} \right) \right]_{x^j} + \sum_{j=1}^3 \frac{3}{2} \epsilon \left( \kappa(\bar{\theta}) \bar{\theta}_{x^j} \right)_{x^j} \nonumber\\&-\frac{3}{2}\bar{\rho}\bar{u}\cdot\nabla_x\bar{\phi},\end{aligned}$

回顾先验假设 (2.2), 易知

$ \max_{0 \leqslant t \leqslant T} \| \bar{\theta} - 1 \|_{L_x^\infty} \lesssim \bar{\lambda} \epsilon. $

下面我们估计含电场 $ \bar{\phi} $ 的能量. 由 (1.8) 式和分部积分, 可得

(2.5) $ \begin{aligned}\label{2.5}&-\int_0^t \int_{\mathbb{R}^3} \bar{\rho} \bar{u} \cdot \nabla_x \bar{\phi} \,\mathrm{d}x\mathrm{d}\tau= \int_0^t \int_{\mathbb{R}^3} \nabla_x (\bar{\rho} \bar{u}) \cdot \bar{\phi} \,\mathrm{d}x\mathrm{d}\tau=-\int_0^t \int_{\mathbb{R}^3} \bar{\rho}_t \bar{\phi} \,\mathrm{d}x\mathrm{d}\tau \nonumber\\=&\int_0^t \int_{\mathbb{R}^3} \Delta_x \bar{\phi}_t \bar{\phi} \,\mathrm{d}x\mathrm{d}\tau=-\int_0^t \int_{\mathbb{R}^3} \nabla_x \bar{\phi}_t \cdot \nabla_x \bar{\phi} \,\mathrm{d}x\mathrm{d}\tau \\=&-\int_0^t\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\|\nabla_x\bar{\phi}\|_{L_x^2}^2\ \mathrm{d}\tau =-\frac{1}{2}\|\nabla_x\bar{\phi}(t)\|_{L_x^2}^2+\frac{1}{2}\|\nabla_x\bar{\phi}(0)\|_{L_x^2}^2, \nonumber\end{aligned}$

因此, 对 (2.4) 式关于 $ \tau $ 和 $ x $ 在 $ [0, t] \times \mathbb{R}^3 $ 上积分 (其中 $ t \in [T] $), 并结合柯西-施瓦茨不等式, (2.3) 和 (2.5) 式, 我们得到如下估计式

(2.6) $ \begin{equation}\label{2.6}\|\nabla_x\bar{\phi}(t)\|^2+\| (\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1)(t) \|^2 + \epsilon \int_0^t \| \nabla_x (\bar{u}, \bar{\theta})(\tau) \|^2{\rm d}\tau\leqslant C \bar{\lambda}^2\epsilon^2.\end{equation}$

为了得到 $ (\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1) $ 的高阶估计, 我们将重写方程组 (1.8), 结果如下

(2.7) $ \begin{equation}\label{2.7}\begin{cases}\displaystyle \bar{\rho}_t = -(\bar{\rho} - 1) \text{div}_x \bar{u} - \nabla_x (\bar{\rho} - 1) \cdot \bar{u} - \text{div}_x \bar{u}, \\\quad \displaystyle \bar{u}_t^i + \sum_{j=1}^3 \bar{u}^j \bar{u}_{x^j}^i + \frac{2}{3 \bar{\rho}} (\bar{\rho} \bar{\theta} - 1)_{x^i}+\bar{\phi}_{x^i} \\ \displaystyle = \frac{\epsilon}{\bar{\rho}} \sum_{j=1}^3 \left\{ \mu(\bar{\theta}) \left( \bar{u}_{x^j}^i + \bar{u}_{x^i}^j - \frac{2}{3} \delta^{ij} \text{div}_x \bar{u} \right) \right\}_{x^j}, \quad i = 1, 2, 3, \\\quad \displaystyle \bar{\theta}_t + \sum_{j=1}^3 \left( \bar{u}^j \bar{\theta}_{x^j} + \frac{2}{3} \bar{\theta} \bar{u}_{x^j}^j \right) \\ \displaystyle = \frac{\epsilon}{\bar{\rho}} \left\{ \sum_{j=1}^3 \left( \kappa(\bar{\theta}) \bar{\theta}_{x^j} \right)_{x^j} + \frac{1}{2} \mu(\bar{\theta}) \sum_{i,j=1}^3 \left( \bar{u}_{x^j}^i + \bar{u}_{x^i}^j \right)^2 - \frac{2}{3} \mu(\bar{\theta}) \left( \text{div}_x \bar{u} \right)^2 \right\}.\end{cases}\end{equation}$

接下来, 我们对 (2.7) 式作用 $ \partial^\gamma=\partial_t^{\gamma_0}\partial_x^\alpha (1 \leq |\gamma| \leq N_1 - 1 $, $ \gamma_0 \leqslant N_1 - 2 ) $, 将得到的恒等式分别乘以 $ \frac{2}{3} \partial^\gamma \bar{\rho},\ \bar{\rho} \partial^\gamma \bar{u}^i $, 和 $ \partial^\gamma\bar{\theta} $, 对 $ i $ 从 $ 1 $ 到 $ 3 $ 求和, 再将所得方程关于 $ \tau $ 和 $ x $ 在 $ [0,t]\times\mathbb{R}^3 $ 上积分

(2.7) $ \begin{aligned}\label{2.8} &{\sum_{1\leq|\gamma|\leq N_1-1}\|\partial^\gamma\nabla_x\bar{\phi}(t)\|^2+\sum_{\substack{1 \leqslant |\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} | \partial^\gamma (\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1) |^2{\rm d}x} \nonumber\\ &+ \sum_{\substack{1 \leqslant |\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \epsilon \int_0^t \int_{\mathbb{R}^3} | \nabla_x \partial^\gamma (\bar{u}, \bar{\theta}) |^2{\rm d}x{\rm d}\tau \nonumber \\ \leqslant & C \sum_{\substack{1 \leqslant |\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \| \partial^\gamma (\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1)(0) \|^2 + C\sum_{1\leq|\gamma|\leq N_1-1}\|\partial^\gamma\nabla_x\bar{\phi}(0)\|^2 \nonumber\\ &+ C \bar{\lambda} \epsilon \sum_{\substack{1 \leqslant |\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_0^t \int_{\mathbb{R}^3} | \partial^\gamma (\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1) |^2{\rm d}x{\rm d}\tau \nonumber\\ &+C\max_{0\leq\tau\leq t}\|\bar{u}(\tau)\|_{H^{N_1-1}}\sum_{\substack{1 \leqslant |\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \epsilon \int_0^t \int_{\mathbb{R}^3}|\partial^\gamma (\bar{\rho}-1,\nabla_x\bar{\phi})|^2\ \mathrm{d}\tau.\end{aligned}$

这是因为

$ (\bar{\rho}\partial^\gamma\bar{u}^i, \partial^{\gamma}\bar{\phi}_{x^i})=-(\partial_{x^i}(\bar{\rho}\partial^\gamma \bar{u}^i), \partial^\gamma\bar{\phi})=-\left(\partial_{x^i}\partial^\gamma(\bar{\rho}\bar{u}^i)-\sum_{0<\gamma_1\leq \gamma}C_\gamma^{\gamma_1}\partial_{x^i}(\partial^{\gamma_1}\bar{\rho}\partial^{\gamma-\gamma_1} \bar{u}^i),\partial^\gamma\bar{\phi}\right). $

首先, 我们分析上式右端的第一项. 与 (2.5) 式类似, 有

$ (\partial_{x^i}\partial^\gamma(\bar{\rho}\bar{u}^i),\ \partial^\gamma\bar{\phi})=-\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\|\partial^\gamma\nabla_x\bar{\phi}\|_{L_x^2}^2, $

而且

$ \begin{align*}&-(\sum_{0<\gamma_1\leq\gamma}C_\gamma^{\gamma_1}\partial_{x^i}(\partial^{\gamma_1}\bar{\rho}\partial^{\gamma-\gamma_1}\bar{u}^i),\ \partial^\gamma\bar{\phi)}=(\sum_{0<\gamma_1\leq \gamma}C_\gamma^{\gamma_1}\partial^{\gamma_1}\bar{\rho}\partial^{\gamma-\gamma_1}\bar{u}^i,\ \partial_{x^i}\partial^\gamma\bar{\phi})\\\lesssim&\|\nabla_x\bar{\rho}\|_{H^{N_1-1}}\|\bar{u}\|_{H^{N_1-1}}\sum_{1\leq|\gamma|\leq N_1-1}\|\partial_{x^i}\partial^\gamma\bar{\phi}\|_{L_x^2}.\end{align*}$

现在只需推导当 $ |\gamma| \leqslant N_1 - 1 $ 且 $ \gamma_0 \leqslant N_1 - 2 $ 时, $ \partial^\gamma \nabla_x (\bar{\rho} - 1) $ 的 $ L_{t,x}^2 $ 估计. 为此, 对 $ (2.7)_1 $ 式作用 $ \partial_{x^i}\partial^\gamma $, 对 $ (2.7)_2 $ 式作用 $ \partial^\gamma $, 将所得的恒等式分别乘以 $ \epsilon^2\partial^\gamma \bar{\rho}_{x^i} $ 和 $ \frac{\epsilon\bar{\rho}^2\partial^\gamma\bar{\rho}_{x^i}}{\frac{4}{3}\mu(\bar{\theta})} $, 对 $ i $ 从 $ 1 $ 到 $ 3 $ 求和, 再将所得式子关于 $ \tau $ 和 $ x $ 在 $ [0,t]\times\mathbb{R}^3 $ 上积分, 即可得到

(2.9) $ \begin{aligned}\label{2.9}& \epsilon^2 \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} | \partial^\gamma \nabla_x \bar{\rho} |^2{\rm d}x + \epsilon \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} (\partial^\gamma \bar{u}) \cdot \left( \frac{\bar{\rho}^2}{\frac{4}{3} \mu(\bar{\theta})} \partial^\gamma \nabla_x \bar{\rho} \right){\rm d}x \nonumber \\& \quad + \epsilon \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_0^t \int_{\mathbb{R}^3} | \partial^\gamma \nabla_x \bar{\rho} |^2{\rm d}x{\rm d}\tau+\epsilon \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_0^t\int_{\mathbb{R}^3}\frac{\bar{\rho}^2}{\frac{4}{3}\mu(\bar{\theta})}|\partial^\gamma\nabla_x\bar{\phi}_{x^i}|^2\ \mathrm{d}x\mathrm{d}\tau \nonumber \\& \leqslant C \epsilon^2 \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} | \partial^\gamma \nabla_x \bar{\rho}_0 |^2{\rm d}x + C \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} | \partial^\gamma \bar{u}_0 |^2{\rm d}x \\& \quad + C \epsilon \sum_{\substack{1 \leqslant |\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_0^t \int_{\mathbb{R}^3} | \partial^\gamma (\bar{u}, \bar{\theta}) |^2{\rm d}x{\rm d}\tau \nonumber\\&\quad + C \max_{0 \leq \tau \leq t} \| \nabla_x \Big( \frac{\epsilon \bar{\rho}^2}{\frac{4}{3} \mu(\bar{\theta})} \Big)(\tau) \|_{H^1} \sum_{\substack{|\gamma| \leq N_1 - 1 \\ \gamma_0 \leq N_1 - 2}} \int_0^t \int_{\mathbb{R}^3} \vert \partial^\gamma \nabla_x \bar{\phi}_{x^i} \vert^2 \mathrm{d}x \mathrm{d}\tau, \nonumber\end{aligned}$

与 (2.5) 式类似, 这是因为

$ (\partial^\gamma\bar{\phi}_{x^i},\ \frac{\epsilon\bar{\rho}^2}{\frac{4}{3}\mu(\bar{\theta})}\partial^\gamma\bar{\rho}_{x^i})=(\partial^\gamma\bar{\phi}_{x^i},\ \nabla_x\left(\frac{\epsilon\bar{\rho}^2}{\frac{4}{3}\mu(\bar{\theta})}\right)\partial^\gamma\nabla_x\bar{\phi}_{x^i})+(\partial^\gamma\nabla_x\bar{\phi}_{x^i},\ \frac{\epsilon\bar{\rho}^2}{\frac{4}{3}\mu(\bar{\theta})}\partial^\gamma\nabla_x\bar{\phi}_{x^i}). $

此外, 由 (2.7) 式可推出

(2.10) $ \begin{equation}\label{2.10}\sum_{|\gamma_0'| \leqslant N_1 - 1 } \int_0^t \int_{\mathbb{R}^3} | \partial_t^{\gamma_0'} (\bar{\rho}, \bar{u}, \bar{\theta}) |^2{\rm d}x{\rm d}\tau \lesssim \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_0^t \int_{\mathbb{R}^3} | \nabla_x \partial^\gamma (\bar{\rho}, \bar{u}, \bar{\theta}, \nabla_x \bar{\phi}) |^2{\rm d}x{\rm d}\tau.\end{equation}$

令 $ \lambda_1 > 0 $ 且 $ \bar{\lambda} $ 为充分小的正数. 将 (2.10) 式代入 (2.8) 和 (2.9) 式, 再由 $(2.6)+ (2.8)+(2.9) \times \lambda_1 $ 可得

(2.11) $ \begin{aligned}\label{2.11}& \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}}\|\partial^\gamma\nabla_x\bar{\phi}\|^2+ \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} | \partial^\gamma (\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1) |^2{\rm d}x + \lambda_1 \epsilon^2 \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} | \partial^\gamma \nabla_x \bar{\rho} |^2{\rm d}x \nonumber\\& \quad + \lambda_1 \epsilon \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} (\partial^\gamma \bar{u}) \cdot \left( \frac{\bar{\rho}^2}{\frac{4}{3} \mu(\bar{\theta})} \partial^\gamma \nabla_x \bar{\rho} \right){\rm d}x \\& \quad + \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} C \epsilon \int_0^t \int_{\mathbb{R}^3} | \nabla_x \partial^\gamma (\bar{\rho}, \bar{u}, \bar{\theta}) |^2{\rm d}x{\rm d}\tau+\lambda_1\epsilon \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_0^t\int_{\mathbb{R}^3} \frac{\bar{\rho}^2}{\frac{4}{3}\mu(\bar{\theta})}|\partial^\gamma\nabla_x\bar{\phi}_{x^i}|^2\ \mathrm{d}x\mathrm{d}\tau \nonumber\\& \lesssim \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \| \partial^\gamma (\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1)(0) \|^2 + \epsilon^2 \sum_{|\alpha| = N_1 - 1} \int_{\mathbb{R}^3} | \partial_x^\alpha \nabla_x \bar{\rho}_0 |^2 + \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \|\partial^\gamma\nabla_x\bar{\phi}(0)\|^2.\nonumber\end{aligned}$

然后, 我们定义

$ \begin{aligned}E(\bar{\rho}, \bar{u}, \bar{\theta},\bar{\phi})(t) = & \frac{1}{\lambda_1} \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} | \partial^\gamma (\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1) |^2{\rm d}x + \epsilon^2 \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} | \partial^\gamma \nabla_x \bar{\rho} |^2{\rm d}x \\& + \epsilon \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} \int_{\mathbb{R}^3} (\partial^\gamma \bar{u}) \cdot \left( \frac{\bar{\rho}^2}{\frac{4}{3} \mu(\bar{\theta})} \partial^\gamma \nabla_x \bar{\rho} \right){\rm d}x+\frac{1}{\lambda_1}\sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}}\|\partial^\gamma\nabla_x\bar{\phi}(t)\|^2,\end{aligned} $

显然, 当 $ \lambda_1 > 0 $ 充分小时, 有

$ E(\bar{\rho}, \bar{u}, \bar{\theta},\bar{\phi})(t) \sim \sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}} | \partial^\gamma [\bar{\rho} - 1, \bar{u}, \bar{\theta} - 1] \|^2 + \epsilon^2 \!\!\sum_{|\alpha| = N_1 - 1}\!\! \| \partial_x^\alpha \nabla_x [\bar{\rho} - 1] \|^2 +\sum_{\substack{|\gamma| \leqslant N_1 - 1 \\ \gamma_0 \leqslant N_1 - 2}}\|\partial^\gamma\nabla_x\bar{\phi}(t)\|^2.$

进一步, 由 (2.11) 式可得

(2.12) $ \begin{aligned}\label{2.12}&E(\bar{\rho}, \bar{u}, \bar{\theta},\bar{\phi})(t) + \sum_{\substack{|\gamma| \leqslant N_1 - 1}} \epsilon \int_0^t \int_{\mathbb{R}^3} | \nabla_x \partial^\gamma (\bar{\rho}, \bar{u}, \bar{\theta}) |^2{\rm d}x{\rm d}\tau \nonumber\\&\quad +\epsilon\sum_{|\gamma| \leq N_1 } \int_0^t\int_{\mathbb{R}^3} |\partial^\gamma\nabla_x^2\bar{\phi}|\ \mathrm{d}x\mathrm{d}\tau\lesssim \ E(\bar{\rho}, \bar{u}, \bar{\theta},\bar{\phi})(0).\end{aligned}$

综上, 我们得到了估计 (2.12), 从而得到了 (1.8), (1.10) 式解的全局存在性. 详情参见文献 [24]. 至此, 我们完成了定理 2.1 的证明.

3 Vlasov-Poisson-Boltzmann 方程组的解

在本节中, 我们将基于下述先验假设, 对宏观部分 $ \rho(t, x) - \bar{\rho}(t, x) $, $ u(t, x) - \bar{u}(t, x) $, $ \theta(t, x) - \bar{\theta}(t, x) $ 与微观部分 $ \mathbf{G} $ 建立能量估计

(3.1) $ \begin{aligned}\label{3.1}N^2(t) \equiv & \sup_{0 \leqslant \tau \leqslant t} \sum_{|\gamma| \leqslant N}\int_{\mathbb{R}^3} | \partial^\gamma (\rho(\tau, x) - \bar{\rho}, u(\tau, x) - \bar{u}, \theta(\tau, x) - \bar{\theta}) |^2 \mathrm{d}x \nonumber\\&+\sup_{0 \leqslant \tau \leqslant t}\sum_{|\gamma|\leq N}\int_{\mathbb{R}^3} |\partial^\gamma\nabla_x(\phi(\tau,x)-\bar{\phi})|^2\ \mathrm{d}x \nonumber\\& + \sup_{0 \leqslant \tau \leqslant t} \sum_{|\gamma| + |\beta| \leqslant N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{ | \partial^\gamma \partial^\beta \mathbf{G}(\tau, x, \xi) |^2 }{ \mathbf{M}_- } \mathrm{d}\xi \mathrm{d}x\lesssim \lambda_0^2 \epsilon^4.\end{aligned}$

此外, 定理 2.1 所构造的唯一整体光滑解还满足

(3.2) $ \begin{aligned}\label{3.2} &\quad\sum_{\substack{|\gamma| \leq N_1 -1\\ \gamma_0 \leq N_1-2 }}\|\partial^\gamma[\bar{\rho}-1,\bar{u},\bar{\theta}-1]\|^2+\epsilon^2\sum_{|\alpha|=N_1-1}\|\partial_x^\alpha\nabla_x[\bar{\rho}-1](t)\|^2+\sum_{\substack{|\gamma|\leq N_1-1\\ \gamma_0\leq N_1-2}}\|\partial^\gamma\nabla_x\bar{\phi}(t)\|^2 \nonumber\\ &\quad +\epsilon\sum_{\substack{|\gamma|\leq N_1-1\\ \gamma_0\leq N_1-2}} \int_0^t\int_{\mathbb{R}^3} |\partial^\gamma\nabla_x(\bar{\rho},\bar{u},\bar{\theta})|^2\ \mathrm{d}x\mathrm{d}\tau +\epsilon \sum_{\substack{|\gamma|\leq N_1-1\\ \gamma_0\leq N_1-2}} \int_0^t\int_{\mathbb{R}^3} \frac{\bar{\rho}^2}{\frac{4}{3}\mu(\bar{\theta})}|\partial^\gamma \nabla_x\bar{\phi}_{x^i}|^2\ \mathrm{d}x\mathrm{d}\tau \nonumber\\ & \lesssim \bar{\lambda}^2\epsilon^2.\end{aligned}$

后续小节将致力于基于先验假设 (3.1) 以及关于 $ [\bar{\rho}(t.x),\bar{u}(t.x),\bar{\theta}(t.x)] $ 的估计 (3.2), 推导所需的能量型估计,即完成定理 1.1 的证明. 第一个小节将聚焦于宏观部分的能量估计.

3.1 宏观部分的能量估计

在本节中, 我们主要讨论 $ [\rho-\bar{\rho}, u-\bar{u}, \theta-\bar{\theta}] $ 的能量估计. 有如下引理

引理 3.1 设 $ F(t,x,\xi) $ 是满足柯西问题 (1.1), (1.9) 的经典解, 且 $ [\bar{\rho}(t,x),\bar{u}(t,x),\bar{\theta}(t,x)] $ 是由柯西问题 (1.8), (1.10) 所对应的定理 2.1 得到的经典解, 则以下估计式成立

(3.3) $ \begin{aligned}\label{3.3}& \sum_{|\gamma|\leq N-1}\|\partial^\gamma \nabla_x(\phi-\bar{\phi})(t)\|^2 \nonumber\\&\quad + \sum_{|\gamma| \leq N-1} \left\{ \int_{\mathbb{R}^3} |\partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})|^2{\rm d}x + \epsilon \int_{\mathbb{R}^3} (\partial^\gamma (u - \bar{u})) \cdot (\partial^\gamma \nabla_x (\rho - \bar{\rho})){\rm d}x \right\} \nonumber\\&\quad + \sum_{1 \leq |\gamma| \leq N} \epsilon \int_0^t \int_{\mathbb{R}^3} |\partial^\gamma (u - \bar{u}, \theta - \bar{\theta}, \rho - \bar{\rho})|^2{\rm d}x{\rm d}\tau+\epsilon\sum_{|\gamma|\leq N}\int_0^t\|\partial^\gamma \nabla_x^2(\phi-\bar{\phi})\|^2\ \mathrm{d}\tau \nonumber\\&\lesssim \sum_{|\gamma| \leq N} \|\partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})(0)\|^2 + \epsilon \sum_{|\gamma| \leq N} \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi)}{\mathbf{M}_-} |\partial^\gamma \mathbf{G}|^2 {\rm d}\xi {\rm d}x {\rm d}\tau \nonumber\\&\quad +\sum_{|\gamma|\leq N-1}\|\partial^\gamma \nabla_x(\phi-\bar{\phi})(0)\|^2 + \epsilon^3 \sum_{|\gamma| \leq N-1} \int_0^t \|\partial^\gamma \nabla_x (\bar{u}, \bar{\theta})\|_{H^1}^2{\rm d}\tau\\&\quad +\|\nabla_x \phi\|_{H^{N}}^2 \sum_{|\gamma'| \leq N-1} \int_0^t \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi)}{\mathbf{M}_-} |\nabla_\xi\partial^{\gamma'} \mathbf{G}|^2 \mathrm{d}\xi \mathrm{d}x \mathrm{d}\tau \nonumber.\end{aligned}$

证由 (1.7) 和 (1.8) 式容易看出

(3.4) $ \begin{equation}\label{3.4}(\rho - \bar{\rho})_t = - (\rho - \bar{\rho}) \mathrm{div}_x u - \nabla_x (\rho - \bar{\rho}) \cdot u - \nabla_x \bar{\rho} \cdot (u - \bar{u}) - \bar{\rho} (\mathrm{div}_x u - \mathrm{div}_x \bar{u}),\end{equation}$

(3.5) $\begin{aligned}\label{3.5} &\quad (u^i - \bar{u}^i)_t + \sum_{j=1}^3 (u^j - \bar{u}^j) u^i_{x^j} + \sum_{j=1}^3 (u_{x^j}^i - \bar{u}_{x^j}^i) \bar{u}^j \nonumber\\ &\quad + \frac{2}{3} (\theta_{x^i} - \bar{\theta}_{x^i}) + \frac{2 \bar{\theta}}{3 \bar{\rho}} (\rho_{x^i} - \bar{\rho}_{x^i}) + \frac{2}{3} \rho_{x^i} \left( \frac{\theta}{\rho} - \frac{\bar{\theta}}{\bar{\rho}} \right) + \phi_{x^i}-\bar{\phi}_{x^i} \nonumber \\ &= -\int_{\mathbb{R}^3} \frac{\psi_i (\xi \cdot \nabla_x \varTheta)}{\rho}{\rm d}\xi + \frac{\epsilon}{\rho} \sum_{j=1}^3 \left\{ \mu(\theta) \left( u^i_{x^j} + u^j_{x^i} - \frac{2}{3} \delta^{ij} \mathrm{div}_x u \right) \right\}_{x^j} \\ &\quad - \frac{\epsilon}{\bar{\rho}} \sum_{j=1}^3 \left\{ \mu(\bar{\theta}) \left( \bar{u}^i_{x^j} + \bar{u}^j_{x^i} - \frac{2}{3} \delta^{ij} \mathrm{div}_x \bar{u} \right) \right\}_{x^j}, \quad i = 1,2,3,\nonumber \end{aligned}$

(3.6) $ \begin{aligned}\label{H3.6}& (\theta - \bar{\theta})_t + \sum_{j=1}^3 \left( (u^j - \bar{u}^j) \theta_{x^j} + \bar{u}^j (\theta_{x^j} - \bar{\theta}_{x^j}) \right) + \frac{2}{3} \sum_{j=1}^3 \left( (u^j - \bar{u}^j)_{x^j} \bar{\theta} + u^j_{x^j} (\theta - \bar{\theta}) \right) \nonumber\\&=-\int_{\mathbb{R}^3} \frac{\psi_4 -\xi \cdot u}{\rho} (\xi \cdot \nabla_x \varTheta){\rm d}\xi \nonumber\\&\quad+ \frac{\epsilon}{\rho} \left\{ \sum_{j=1}^3 \left( \kappa(\theta) \theta_{x^j} \right)_{x^j} + \frac{1}{2} \mu(\theta) \sum_{i,j=1}^3 \left( u^i_{x^j} + u^j_{x^i} \right)^2 - \frac{2}{3} \mu(\theta) (\mathrm{div}_x u)^2 \right\} \\&\quad- \frac{\epsilon}{\bar{\rho}} \left\{ \sum_{j=1}^3 \left( \kappa(\bar{\theta}) \bar{\theta}_{x^j} \right)_{x^j} + \frac{1}{2} \mu(\bar{\theta}) \sum_{i,j=1}^3 \left( \bar{u}^i_{x^j} + \bar{u}^j_{x^i} \right)^2 - \frac{2}{3} \mu(\bar{\theta}) (\mathrm{div}_x \bar{u})^2 \right\}.\nonumber\end{aligned}$

与参考文献 [21,引理 4.1] 的证明类似, 我们可以得到如下估计

(3.7) $ \begin{aligned}\label{3.7}& \sum_{|\gamma|\leq j}\|\partial^\gamma\nabla_x(\phi-\bar{\phi})(t)\|^2+ \sum_{|\gamma|=j} \int_{\mathbb{R}^3} |\partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})|^2 \mathrm{d}x \nonumber \\& + \sum_{|\gamma|=j} \epsilon \int_0^t \int_{\mathbb{R}^3} |\nabla_x \partial^\gamma (u - \bar{u}, \theta - \bar{\theta})|^2 \mathrm{d}x \mathrm{d}\tau \nonumber \\&\leq C\sum_{|\gamma|\leq j}\|\partial^\gamma\nabla_x(\phi -\bar{\phi})(0)\|^2 +C \sum_{|\gamma|=j} \|\partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})(0)\|^2 \nonumber \\&\quad + C \epsilon \sum_{|\gamma| \leq j + 1} \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_\mathbf{M}(\xi)}{\mathbf{M}_-}|\partial^\gamma \mathbf{G}|^2 \mathrm{d}\xi \mathrm{d}x \mathrm{d}\tau \\&\quad + C \|\nabla_x \phi\|_{H^{N}}^2 \sum_{|\gamma'| \leq j} \int_0^t \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi)}{\mathbf{M}_-} |\nabla_\xi\partial^{\gamma'} \mathbf{G}|^2 \mathrm{d}\xi \mathrm{d}x \mathrm{d}\tau \nonumber\\&\quad + \lambda_0 \epsilon \sum_{|\gamma| \leq j} \int_0^t \int_{\mathbb{R}^3} |\partial^\gamma \nabla_x (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})|^2 \mathrm{d}x \mathrm{d}\tau \nonumber\\&\quad +C \sup_{0\leq \tau\leq t}[\|u-\bar{u}(\tau)\|_{H^{N-1}} +\|\bar{u}(\tau)\|_{H^{N-1}} ]\sum_{|\gamma|\leq j}\int_0^t \left[\|\partial^\gamma \nabla_x^2(\phi-\bar{\phi)}\|_2^2+\|\partial^\gamma \nabla_x\bar{\rho}\|_2^2\right] \mathrm{d}\tau \nonumber \\&\quad +C\sup_{0\leq \tau \leq t}\|\nabla_x(\phi -\bar{\phi})\|_{H^{N-1}}\sum_{|\gamma|\leq j}\int_0^t \left[\|\partial\nabla_x^2 \bar{\phi}\|^2+\|\partial^\gamma \nabla_x(u-\bar{u})\|^2\right] \mathrm{d}\tau, \nonumber\end{aligned}$

其中 $ j\leq N-1 $.

我们只推导 (3.7) 式中电场项的估计. 为此,我们对 (3.5), (3.6) 式作用 $ \partial^\gamma (|\gamma|\leq N-1) $, 再将所得的恒等式分别与 $ \partial^\gamma (u^i-\bar{u}^i),\partial^\gamma (\theta-\bar{\theta}) $ 在 $ \mathbb{R}^3 $ 上作内积

$ \begin{align*}\quad& \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\sum_{i=1}^3 \left( \partial^\gamma (u^i - \bar{u}^i), \partial^\gamma (u^i - \bar{u}^i) \right)\quad \\= & -\sum_{i,j=1}^3 \sum_{\gamma_1 \leq \gamma} C_\gamma^{\gamma_1} \left( \partial^{\gamma_2}(u^i - \bar{u}^i)\partial^{\gamma_1}u^i_{x^j} + \partial^{\gamma_2}(u^i_{x^j} - \bar{u}^i_{x^j})\partial^{\gamma_1}\bar{u}^i, \partial^\gamma (u^i - \bar{u}^i) \right) \\& -\frac{2}{3}\sum_{i=1}^3 \sum_{\gamma_1 \leq \gamma} C_\gamma^{\gamma_1} \left( \partial^{\gamma_2}\rho_{x^i}\partial^{\gamma_1}( \frac{\theta}{\rho} - \frac{\bar{\theta}}{\bar{\rho}} ), \partial^\gamma (u^i - \bar{u}^i) \right) \\& -\frac{2}{3}\sum_{i=1}^3 \sum_{\gamma_1 \leq \gamma} C_\gamma^{\gamma_1} \left( \partial^{\gamma_2}\left(\frac{\bar{\theta}}{\bar{\rho }}\right)\partial^{\gamma_1}( \rho_{x^i} - \bar{\rho}_{x^i}), \partial^\gamma (u^i - \bar{u}^i) \right)\\& -\frac{2}{3} \sum_{i=1}^3\left( \partial^{\gamma}(\theta_{x^i}-\bar{\theta}_{x^i}), \partial^\gamma (u^i - \overline{u}^i) \right) -\underbrace{\sum_{i=1}^3 (\partial^\gamma(\phi_{x^i}-\bar{\phi}_{x^i}), \partial^\gamma(u^i-\bar{u}^i))}_{I_1}.\\& - \left( \int_{\mathbb{R}^3} \frac{\psi_i (\xi \cdot \nabla_x \varTheta)}{\rho} {\rm d}\xi + \frac{\epsilon}{\rho} \sum_{j=1}^3 \left\{ \mu(\theta) \left( u^i_{x^j} + u^j_{x^i} - \frac{2}{3} \delta^{ij} \mathrm{div}_x u \right) \right\}_{x^j}\, \partial^\gamma (u^i - \bar{u}^i) \right) \\& - \left( \frac{\epsilon}{\bar{\rho}} \sum_{j=1}^3 \left\{ \mu(\bar{\theta}) \left( \bar{u}^i_{x^j} + \bar{u}^j_{x^i} - \frac{2}{3} \delta^{ij} \mathrm{div}_x \bar{u} \right) \right\}_{x^j}\, \partial^\gamma (u^i - \bar{u}^i) \right).\end{align*}$

接下来, 我们只考察 $ I_1 $ 这项, 其余项的估计参见文献 [21], 方法类似, 这里便不再赘述. 由恒等式可知

$ (\partial^\gamma(\phi_{x^i}-\bar{\phi}_{x^i}), \partial^\gamma (u^i-\bar{u}^i))+(\partial^\gamma (\phi-\bar{\phi}),\partial^\gamma(u^i-\bar{u}^i)_{x^i})=0, $

由分部积分得

$ \begin{align*} &(\partial^\gamma(\phi_{x^i}-\bar{\phi}_{x^i}),\ \partial^\gamma(u^i-\bar{u}^i)) =\ (\partial^\gamma(\phi-\bar{\phi}),\ -\partial_{x^i}\partial^\gamma(u^i-\bar{u}^i))\\ =&\ (\partial^\gamma(\phi-\bar{\phi}),\ (\bar{\rho}-1)\partial_{x^i}\partial^\gamma(u^i-\bar{u}^i)) - (\partial^\gamma(\phi-\bar{\phi}),\ \bar{\rho}\partial^\gamma(\nabla_x \cdot u - \nabla_x \cdot\bar{u})).\end{align*}$

现在我们来分析

$ \begin{align*} &(\partial^\gamma(\phi-\bar{\phi}),\ \bar{\rho}\partial^\gamma(\nabla_x \cdot u - \nabla_x \cdot \bar{u}))\\ =&\ (\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\bar{\rho}(\nabla_x \cdot u- \nabla_x \cdot \bar{u}))) - \underbrace{\sum_{\gamma_1<\gamma}C_\gamma^{\gamma_1}(\partial^\gamma(\phi-\bar{\phi}),\ \partial^{\gamma-\gamma_1}\bar{\rho}\partial^{\gamma_1}(\nabla_x\cdot u-\nabla_x\cdot\bar{u}))}_{J_1}.\end{align*}$

将 (3.4) 式代入上式, 可得

$ \begin{align*} &(\partial^\gamma(\phi-\bar{\phi}),\ \bar{\rho}\partial^\gamma (\nabla_x\cdot u-\nabla_x\cdot\bar{u}))\\ =&\ (\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma\left[-(\rho-\bar{\rho})_t-(\rho-\bar{\rho})\text{div}_xu-\nabla_x(\rho-\bar{\rho})\cdot u-\nabla_x\bar{\rho}\cdot(u-\bar{u})\right])-J_1\\ =&\ \underbrace{(\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\rho-\bar{\rho})_t)}_{J_2}-\underbrace{(\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\rho-\bar{\rho})\text{div}_xu)}_{J_3}\\ &\ -\underbrace{(\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\nabla_x(\rho-\bar{\rho})\cdot u))}_{J_4}-\underbrace{(\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\nabla_x\bar{\rho}\cdot (u-\bar{u})))}_{J_5}-J_1.\end{align*}$

首先, 由 (1.7), (1.8) 式和分部积分, 可得 $ J_2 $ 的表达式

$J_2=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\|\partial^\gamma \nabla_x(\phi-\bar{\phi})\|^2. $

接下来, 让我们估计 $ J_3+J_4 $

$ \begin{align*} J_3+J_4=&(\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\nabla_x\cdot(\rho-\bar{\rho})u))\\ =&\underbrace{(\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\nabla_x\cdot((\rho-\bar{\rho})(u-\bar{u})))}_{J_1^\gamma}+(\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\nabla_x\cdot((\rho-\bar{\rho})\bar{u}))).\end{align*}$

通过应用索不列夫不等式, 我们容易得到 $ J_1^\gamma $ 的估计式

$ \begin{align*} J_1^\gamma=&(\partial^\gamma(\phi-\bar{\phi}),\ \sum_{\gamma_1\leq\gamma}C_\gamma^{\gamma_1}\nabla_x\cdot(\partial^{\gamma-\gamma_1}(\rho-\bar{\rho})\partial^{\gamma_1}(u-\bar{u})))\\ =&C_\gamma^{\gamma_1}(\partial^\gamma\nabla_x(\phi-\bar{\phi}),\ \sum_{\gamma_1\leq \gamma}\partial^{\gamma-\gamma_1}\Delta_x(\phi-\bar{\phi})\partial^{\gamma_1}(u-\bar{u}))\\ \lesssim&\|u-\bar{u}\|_{H_x^{N-1}}\|\nabla_x^2(\phi-\bar{\phi})\|_{H_x^{N-1}}^2.\end{align*}$

类似地, 可得

$ (\partial^\gamma(\phi-\bar{\phi}),\ \partial^\gamma(\nabla_x\cdot((\rho-\bar{\rho})\bar{u})))\lesssim\|\bar{u}\|_{H_x^{N-1}}\|\nabla_x^2(\phi-\bar{\phi})\|_{H_x^{N-1}}^2. $

$ J_1 $ 和 $ J_5 $ 的估计与 $ J_3+J_4 $ 的估计类似, 因此有

$\begin{align*} &|J_1|+|J_5|\\ \lesssim&\|\nabla_x(\phi-\bar{\phi})\|_{H^1}\|\nabla_x^2\bar{\phi}\|_{L_x^2}\|\nabla_x(u-\bar{u})\|_{L_x^2}+\|\nabla_x(\phi-\bar{\phi})\|_2\|\nabla_x^2\bar{\phi}\|_2\|\nabla_x(u-\bar{u})\|_{H^1}\\ &+\|u-\bar{u}\|_{H_x^{N-1}}\|\nabla_x^2(\phi-\bar{\phi})\|_{H_x^{N-1}}\|\nabla_x\bar{\rho}\|_{H_x^{N-1}}. \end{align*}$

注意到, 由于 $ \varTheta $ 的表达式中, 含有与电场有关的项 $ -\epsilon L_{\mathbf{M}}^{-1}\left[\mathbf{P}_1^{\mathbf{M}}(\nabla_x\phi \cdot\nabla_\xi\mathbf{G}) \right] $, 我们仍需估计下面的式子. 借助 $ \mathbf{P}_1^{\mathbf{M}} $的定义和引理 3.4, 可以得到

$ \begin{align*}& \left|\left( \partial^\gamma \int_{\mathbb{R}^3} \frac{\psi_4 - \xi \cdot u}{\rho} (\xi \cdot \nabla_x \varTheta){\rm d}\xi,\ \partial^\gamma (\theta - \bar{\theta}) \right) \right|+\left|\sum_i \left(\partial^\gamma \int_{\mathbb{R}^3} \frac{\psi_i (\xi \cdot \nabla_x \varTheta)}{\rho}{\rm d}\xi,\ \partial^\gamma (u^i - \bar{u}^i) \right) \right| \\&\lesssim \sum_{|\gamma_1| \leq |\gamma|/2} \left\| \partial^{\gamma_1} \nabla_x \left[ \frac{1}{\rho}, \frac{u}{\rho} \right] \right\|_{L^3} \left\| \partial^\gamma [(u - \bar{u}, \theta - \bar{\theta})] \right\|_{L^6} \left\| |\frac{\nu_{\mathbf{M}}^{1/2} \partial^{\gamma - \gamma_1} \varTheta}{\sqrt{\mathbf{M}}} |_{L_{\xi}^2} \right\|_{L^2} \\&\quad + \sum_{\substack{|\gamma_1| \geq |\gamma|/2 \geq 1}} \left\| \partial^{\gamma_1} \left[ \frac{1}{\rho}, \frac{u}{\rho} \right] \right\|_{L^2} \left\| \partial^\gamma [(u - \bar{u}, \theta - \bar{\theta})] \right\|_{L^6} \left\| |\frac{\nu_{\mathbf{M}}^{1/2} \partial^{\gamma - \gamma_1} \nabla_x \varTheta}{\sqrt{\mathbf{M}}} |_{L_{\xi}^2} \right\|_{L^3} \\&\lesssim \lambda_0 \epsilon \sum_{|\gamma'| \leq |\gamma|} \left\| \partial^{\gamma'} \nabla_x (u - \bar{u}, \theta - \bar{\theta}) \right\|^2 + \epsilon \sum_{|\gamma'| \leq |\gamma|+1} \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi)}{\mathbf{M}_-} |\partial^{\gamma'} \mathbf{G}|^2 {\rm d}\xi{\rm d}x \\&\quad +\|\nabla_x \phi\|_{H^{N}}^2 \sum_{|\gamma'| \leq |\gamma|} \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi)}{\mathbf{M}_-} |\nabla_\xi\partial^{\gamma'} \mathbf{G}|^2 {\rm d}\xi{\rm d}x.\end{align*}$

将上述估计结合起来并关于 $ \tau $ 从 $ 0 $ 到 $ t $ 上积分, 我们就得到了含电场项的估计. 参见文献 [21,引理 4.1], 我们便得到了估计式 (3.7).

接下来, 我们来估计 $ (\rho-\bar{\rho}) $ 和 $ (\phi_{x^i}-\bar{\phi}_{x^i}) $. 对 (3.5) 式作用 $ \partial^\gamma(|\gamma|\leq N-1) $, 再将所得的结果与 $ \epsilon\partial^\gamma(\rho-\bar{\rho})_{x^i} $ 在 $ \mathbb{R}^3 $ 上取内积, 得到

(3.8) $ \begin{aligned}\label{3.8}&\sum_{i} \frac{\rm d}{{\rm d} t}\left(\partial^{\gamma}\left(u^{i}-\bar{u}^{i}\right), \epsilon \partial^{\gamma}(\rho-\bar{\rho})_{x^{i}}\right)+\sum_{i}\left(\frac{2\bar{\theta}}{3\bar{\rho}} \partial^{\gamma}(\rho-\bar{\rho})_{x^{i}}, \epsilon \partial^{\gamma}(\rho-\bar{\rho})_{x^{i}}\right) \nonumber\\=&-(\partial^\gamma(\phi-\bar{\phi})_{x^i},\epsilon\partial^\gamma (\rho-\bar{\rho})_{x^i})-\sum_{i}\left(\partial^{\gamma}\left(u^{i}-\bar{u}^{i}\right)_{x^{i}}, \epsilon \partial_t\partial^{\gamma}(\rho-\bar{\rho})\right)\nonumber \\&-\sum_{i, j=1}^{3} \sum_{\gamma_1\leq \gamma} C_{\gamma}^{\gamma_{1}}\left(\partial^{\gamma_{2}}\left(u^{j}-\bar{u}^{j}\right) \partial^{\gamma_{1}} u_{x^{j}}^{i}+\partial^{\gamma_{2}}\left(u_{x^{j}}^{i}-\bar{u}_{x^{j}}^{i}\right) \partial^{\gamma_{1}} \bar{u}^{j}, \epsilon \partial^{\gamma}(\rho-\bar{\rho})_{x^{i}}\right)\nonumber \\&-\frac{2}{3} \sum_{i=1}^{3} \sum_{\gamma_1 \leq \gamma} C_{\gamma}^{\gamma_{1}}\left(\partial^{\gamma_{2}} \rho_{x^{i}} \partial^{\gamma_{1}}\left(\frac{\theta}{\rho}-\frac{\bar{\theta}}{\bar{\rho}}\right), \epsilon \partial^{\gamma}(\rho-\bar{\rho})_{x^{i}}\right) \\&-\frac{2}{3} \sum_{i}\left(\partial^{\gamma}\left(\theta_{x^{i}}-\bar{\theta}_{x^{i}}\right), \epsilon \partial^{\gamma}(\rho-\bar{\rho})_{x^{i}}\right) -\sum_{i}\left(\partial^{\gamma}\left[\int_{\mathbb{R}^{3}} \frac{\psi_i(\xi \cdot \nabla_{x} \mathbf{G})}{\rho} {\rm d} \xi\right], \epsilon \partial^{\gamma}(\rho-\bar{\rho})_{x^{i}}\right) \nonumber\\&-\sum_{i} \sum_{0<\gamma_{1} \leq \gamma} C_{\gamma}^{\gamma_{1}}\left(\partial^{\gamma_{1}}\left(\frac{2 \bar{\theta}}{3 \bar{\rho}}\right) \partial^{\gamma_{2}}(\rho-\bar{\rho})_{x^{i}}, \epsilon \partial^{\gamma}(\rho-\bar{\rho})_{x^{i}}\right) \nonumber\\&-\sum_{i}\left(\partial^{\gamma}\left[\frac{\epsilon}{\bar{\rho}} \sum_{j=1}^{3}\left\{\mu(\bar{\theta})\left(\bar{u}_{x^{j}}^{i}+\bar{u}_{x^{i}}^{j}-\frac{2}{3} \delta^{i j} \operatorname{div}_{x} \bar{u}\right)\right\}_{x^{j}}\right], \epsilon \partial^{\gamma}(\rho-\bar{\rho})_{x^{i}}\right).\nonumber\end{aligned}$

此处, 我们用到了以下等式

$ \begin{align*} &\epsilon(\partial^\gamma(\phi-\bar{\phi})_{x^i},\partial^\gamma(\rho-\bar{\rho})_{x^i}) =-\epsilon(\partial^\gamma(\phi_{x^i}-\bar{\phi}_{x^i}),\partial^\gamma(\Delta_x\phi_{x^i}-\Delta_x\bar{\phi}_{x^i}))\\ =&\epsilon(\partial^\gamma\nabla_x(\phi_{x^i}-\bar{\phi}_{x^i}),\partial^\gamma\nabla_x(\phi_{x^i}-\bar{\phi}_{x^i})) =\epsilon \|\partial^\gamma \nabla_x(\phi_{x^i}-\bar{\phi}_{x^i})\|^2.\end{align*}$

由 (3.1) 式, 引理 3.4 以及带 $ \eta_1 $ 的柯西-施瓦茨不等式可得, 对任意 $ \eta_1>0 $, (3.8) 式的右端可由以下表达式控制

(3.9) $ \begin{aligned}\label{3.9}&\max\{\epsilon^2, \lambda_0 \epsilon, \eta_1 \epsilon\} \sum_{|\gamma'| \leq |\gamma|} \|\partial^{\gamma'} \nabla_x (\rho - \bar{\rho})\|^2 + \epsilon^3 \sum_{|\gamma'| \leq |\gamma|} \|\partial^{\gamma'} \nabla_x (\bar{u}, \bar{\theta})\|_{H^1}^2 \nonumber\\&+ C \epsilon \sum_{|\gamma'| \leq |\gamma|} \|\partial^{\gamma'} \nabla_x (u - \bar{u}, \theta - \bar{\theta})\|^2 + C \epsilon \sum_{|\gamma'| \leq |\gamma|} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_\mathbf{M}(\xi)}{\mathbf{M}_-} |\nabla_x \partial^{\gamma'} \mathbf{G}|^2 {\rm d}\xi{\rm d}x.\end{aligned}$

所以, 结合 (3.8) 与 (3.9) 式可容易得到

(3.10) $ \begin{aligned}\label{3.10}&\sum_{i} \sum_{|\gamma| \leq N-1} \left( \partial^{\gamma}(u^i - \bar{u}^i), \epsilon \partial^{\gamma} (\rho - \bar{\rho})_{x^i} \right) + \epsilon \sum_{|\gamma| \leq N-1} \int_0^t \|\partial^{\gamma} \nabla_x (\rho - \bar{\rho})\|^2{\rm d}\tau \nonumber\\&\quad +\epsilon\sum_{|\gamma|\leq N-1}\int_0^t \|\partial^\gamma\nabla_x^2(\phi-\bar{\phi})\|^2\ \mathrm{d}\tau \nonumber\\&\lesssim \sum_{|\gamma| \leq N} \|\partial^{\gamma} (\rho - \bar{\rho}, u - \bar{u})(0)\|^2 + \epsilon^3 \sum_{|\gamma'| \leq N-1} \int_0^t \|\partial^{\gamma'} \nabla_x (\bar{u}, \bar{\theta})\|_{H^1}^2{\rm d}\tau \\&+ \epsilon \sum_{|\gamma'| \leq |\gamma|} \int_0^t \|\partial^{\gamma'} \nabla_x (u - \bar{u}, \theta - \bar{\theta})\|^2{\rm d}\tau+ \epsilon \sum_{|\gamma'| \leq N-1} \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_\mathbf{M}(\xi)}{\mathbf{M}_-} |\nabla_x \partial^{\gamma'} \mathbf{G}|^2 {\rm d}\xi{\rm d}x{\rm d}\tau \nonumber.\end{aligned}$

将 (3.7) 与 (3.10) 式线性组合后, 可得

(3.11) $ \begin{aligned}\label{3.11}&\sum_{|\gamma| \leq N-1} \left\{ \int_{\mathbb{R}^3} |\partial^{\gamma} (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})|^2{\rm d}x + \epsilon \int_{\mathbb{R}^3} (\partial^{\gamma} (u - \bar{u})) \cdot (\partial^{\gamma} \nabla_x(\rho - \bar{\rho})){\rm d}x \right\}\nonumber \\&+ \sum_{|\gamma| \leq N-1} \epsilon \int_0^t \int_{\mathbb{R}^3} |\nabla_x \partial^{\gamma} (u - \bar{u}, \theta - \bar{\theta}, \rho - \bar{\rho})|^2{\rm d}x{\rm d}\tau+\epsilon\sum_{|\gamma|\leqslant N-1}\int_0^t \|\partial^\gamma \nabla_x^2 (\phi-\bar{\phi})\|^2\ \mathrm{d}\tau \nonumber\\&+ \sum_{|\gamma|\leq N-1}\|\partial^\gamma (\phi-\bar{\phi})(t)\|^2 \nonumber\\&\leq C \sum_{|\gamma|\leq N-1}\|\partial^\gamma (\phi-\bar{\phi})(0)\|^2 + C \sum_{|\gamma| \leq N} \|\partial^{\gamma} (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})(0)\|^2 \\&+ C \epsilon \sum_{|\gamma| \leq N} \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_\mathbf{M}(\xi)}{\mathbf{M}_-} |\partial^{\gamma} \mathbf{G}|^2 {\rm d}\xi{\rm d}x{\rm d}\tau + C \epsilon^3 \sum_{|\gamma| \leq N-1} \int_0^t \|\partial^{\gamma} \nabla_x (\bar{u}, \bar{\theta})\|_{H^1}^2{\rm d}\tau.\nonumber\end{aligned}$

一方面, 我们利用先验假设 (2.2) 和 (3.1). 另一方面, 由于 $ N+1\leq N_1-1 $, 再结合 (3.4)-(3.6) 式可知

(3.12) $ \begin{aligned}\label{3.12}&\sum_{1 \leq |\gamma_0| \leq N} \int_0^t \int_{\mathbb{R}^3} |\partial^{\gamma_0}_t (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})|^2{\rm d}x{\rm d}\tau \nonumber\\\lesssim& \sum_{|\gamma_0'| \leq N-1} \int_0^t \int_{\mathbb{R}^3} |\partial^{\gamma_0'}_t \nabla_x (u - \bar{u}, \theta - \bar{\theta}, \rho - \bar{\rho})|^2{\rm d}x{\rm d}\tau \nonumber\\&+ \epsilon^2 \sum_{|\gamma'| \leq N-1} \int_0^t \|\partial^{\gamma'} \nabla_x (\bar{u}, \bar{\theta})\|_{H^1}^2{\rm d}\tau+ \sum_{|\gamma'| \leq N} \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_\mathbf{M}(\xi)}{\mathbf{M}_-} |\partial^{\gamma'} \mathbf{G}|^2 {\rm d}\xi{\rm d}x{\rm d}\tau.\end{aligned}$

则由 (3.11) 与 (3.12) 式可得 (3.3) 式, 至此完成引理 3.1 的证明.

3.2 非流体部分的估计

现在我们推导非流体部分 $ \mathbf{G} $ 的能量估计. 为实现这一目标, 我们做如下分解

$\mathbf{P}_1^{\overline{\mathbf{M}}} \mathbf{G} = \mathbf{G}, \quad \mathbf{P}_1^{\overline{\mathbf{M}}} F = \mathbf{G} + \mathbf{P}_1^{\overline{\mathbf{M}}} \mathbf{M},$

因此, 我们需要估计 $ \mathbf{P}_1^{\overline{\mathbf{M}}} F $ 与 $ \mathbf{P}_1^{\overline{\mathbf{M}}} \mathbf{M} $. 为得到 $ \mathbf{P}_1^{\overline{\mathbf{M}}} \mathbf{M} $ 的估计, 我们给出以下几个引理^[28].

引理 3.2 在 (3.1) 式的假设下, 对任意正整数 $ k>0 $ 以及满足 $ \theta_0>\frac{1}{2}\max \left\{\theta,\bar{\theta}\right\} $ 的 Maxwellian, 有

$\int_{\mathbb{R}^3} \frac{\nu_{\mathbf{M}}^k(\xi) |\mathbf{P}_1^{\overline{\mathbf{M}}} \mathbf{M}|^2}{\mathbf{M}_0}{\rm d}\xi \leqslant C |\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta}|^4. $

因此, 对所有 $ |\gamma| + |\beta| \leqslant N $, 有

$ \begin{align*}\int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi) |\partial^\gamma \partial^\beta \mathbf{G}|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau &\leqslant \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi) |\partial^\gamma \partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} F|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \\&\quad + C \lambda_0^2 \epsilon^2 \sum_{1 \leqslant |\gamma| \leqslant N} \int_0^t \|\partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})\|^2{\rm d}\tau.\end{align*}$

推论 3.1 由定理 2.1, 引理 3.1, 引理 3.2, 以及关系式 $ \nu_\mathbf{M}(\xi) \sim \nu_\mathbf{\overline{\mathbf{M}}}(\xi) \sim 1+|\xi| $, 可以知道

$ \begin{align*}& \sum_{|\gamma| \leqslant N - 1} \left\{ \|\partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})\|^2(t) + \epsilon \int_{\mathbb{R}^3} (\partial^\gamma (u - \bar{u})) \cdot (\partial^\gamma \nabla_x (\rho - \bar{\rho})){\rm d}x \right\} \\&\quad + \epsilon \sum_{1 \leqslant |\gamma| \leqslant N} \int_0^t \|\partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})(\tau)\|^2{\rm d}\tau \\&\quad +\sum_{|\gamma|\leq N-1} \|\partial^\gamma\nabla_x(\phi-\bar{\phi})(t)\|^2+\epsilon \sum_{|\gamma|\leq N-1}\int_0^t \|\partial^\gamma\nabla_x^2(\phi-\bar{\phi})\|^2\ \mathrm{d}\tau\\&\leqslant C \sum_{|\gamma| \leqslant N} \|\partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})(0)\|^2 + C\sum_{|\gamma|\leq N-1}\|\partial^\gamma \nabla_x (\phi-\bar{\phi})(0)\|^2\\&\quad + C \epsilon \sum_{|\gamma| \leqslant N} \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi) |\partial^\gamma \mathbf{P}_1^{\overline{\mathbf{M}}} F|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau + C \bar{\lambda}^2 \epsilon^4.\end{align*}$

现在我们利用原始的 Vlasov-Poisson-Boltzmann 系统 (1.1) 完成 $ \mathbf{P}_1^{\overline{\mathbf{M}}} F $ 的剩余估计. 我们的能量估计可总结如下: 首先, 记

(3.13) $ \begin{equation}\label{3.13}g = F - \overline{\mathbf{M}} - \epsilon L_{{\overline{\mathbf{M}}}}^{-1} \{ \mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}}) \},\end{equation}$

则有

$\mathbf{P}_1^{\overline{\mathbf{M}}} F = \mathbf{P}_1^{\overline{\mathbf{M}}} g + \epsilon L_{{\overline{\mathbf{M}}}}^{-1} \{ \mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}}) \}. $

这意味着, 想要估计 $ \mathbf{P}_1^{\overline{\mathbf{M}}} F $, 只需得到 $ g(t, x, \xi) $ 的足够精确估计即可. 根据先验假设 (3.1), 可观察到 $ g $ 满足如下先验估计

(3.14) $ \begin{equation}\sup_{0 \leqslant \tau \leqslant t} \sum_{|\gamma| + |\beta| \leqslant N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma \partial^\beta g(t, x, \xi)|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x \lesssim \lambda_0^2 \epsilon^2.\end{equation}$

引理 3.3 基于定理 2.1 和引理 3.1 的先验估计, 我们有 $ g $ 的如下估计式

(3.15) $ \begin{aligned}\label{3.15} & \sum_{1 \leqslant |\gamma| \leqslant N} \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{|\partial^\gamma g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x + \sum_{|\gamma| + |\beta| \leqslant N} \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{|\partial^\gamma \partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x \nonumber\\ & \quad + \frac{1}{\epsilon} \sum_{|\gamma| + |\beta| \leqslant N} \int_0^t \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi)|\partial^\gamma \partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \nonumber\\ & \lesssim \sum_{1 \leqslant |\gamma| \leqslant N} \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{|\partial^\gamma g_0|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x + \sum_{|\gamma| + |\beta| \leqslant N} \iint_{\mathbb{R}^3 \times \mathbb{R}^3} \frac{|\partial^\gamma \partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} g_0|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x \nonumber\\ & \quad + \epsilon \sum_{1\leq |\gamma| \leqslant N} \int_0^t \int_{\mathbb{R}^3} |\partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})|^2{\rm d}x{\rm d}\tau \nonumber\\ &\quad + [\|\nabla_x^2(\phi-\bar{\phi})\|^2+\|\nabla_x^2\bar{\phi}\|^2] \left[ \epsilon\|\frac{g}{\sqrt{\mathbf{M}_-}}\|_{H^1}^2+\epsilon^3\|\nabla_x (\bar{\rho},\bar{u},\bar{\theta})\|_{H1}^2 \right] \\ &\quad +\epsilon [\|\nabla_x^2(\phi-\bar{\phi})\|_{H^{N-1}}+\|\nabla_x^2 \bar{\phi}\|_{H^{N-1}}]\|\nabla_x(\bar{\rho},\bar{u},\bar{\theta})\|_{H^N} \|\frac{g}{\sqrt{\mathbf{M}_-}}\|_{H^N}\nonumber\\ &\quad +[\|\nabla_x^2(\phi-\bar{\phi})\|_{H^{N-1}} +\|\nabla_x^2 \bar{\phi}\|_{H^{N-1}}] \|\frac{g}{\sqrt{\mathbf{M}_-}}\|_{H^N} \|\nabla_x (\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})\|_{H^{N-1}}\nonumber\\ &\quad +\|\nabla_x^2\phi\|_{H^{N-1}} \|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N} \left[ \|\frac{g}{\sqrt{\mathbf{M}_-}}\|_{H^N}+\|\nabla_x (\bar{\rho},\bar{u},\bar{\theta})\|_{H^N} +\|(\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})\|_{H^N} \right]\nonumber\\ &\quad +[\|\nabla_x \bar{\phi}\|_{H^1} + \|\nabla_x (\phi-\bar{\phi})\|_{H^1} ]\|\nabla_x (\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})\|_{H^{N-1}}^2\nonumber,\end{aligned}$

其中 $ g_0 $ 表示初始值.

证我们将分为三步完成引理的证明.

步骤一 零阶能量估计. 已知 $ F $ 是 Vlasov-Poisson-Boltzmann 方程组 (1.1) 的解, 结合 (3.13) 式, 可得

(3.16) $ \begin{aligned}\label{3.16}g_t + \xi \cdot \nabla_x g -\nabla_x\phi \cdot\nabla_\xi g &= \frac{1}{\epsilon} L_{\overline{\mathbf{M}}} g + \frac{1}{\epsilon} Q(\mathbf{P}_1^{\overline{\mathbf{M}}} g, \mathbf{P}_1^{\overline{\mathbf{M}}} g) + \frac{1}{\epsilon} Q(\mathbf{P}_0^{\overline{\mathbf{M}}} (\mathbf{M} - \overline{\mathbf{M}}), \mathbf{P}_0^{\overline{\mathbf{M}}} (\mathbf{M} - \overline{\mathbf{M}})) \notag \nonumber\\&\quad + \frac{2}{\epsilon} Q(\mathbf{P}_1^{\overline{\mathbf{M}}} g, \mathbf{P}_0^{\overline{\mathbf{M}}} (\mathbf{M} - \overline{\mathbf{M}})) + 2 Q(\mathbf{P}_1^{\overline{\mathbf{M}}} g, L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}})\}) \notag \nonumber\\&\quad + 2 Q(\mathbf{P}_0^{\overline{\mathbf{M}}} (\mathbf{M} - \overline{\mathbf{M}}), L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}})\}) \nonumber\\&\quad + \epsilon Q(L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}})\}, L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}})\}) \nonumber \\&\quad - \epsilon \partial_t L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}})\} - \epsilon \mathbf{P}_1^{\overline{\mathbf{M}}}\{\xi \cdot \nabla_x L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}})\}\} \nonumber\\&\quad + \epsilon \nabla_x \phi \cdot\nabla_\xi [L_{\overline{\mathbf{M}}}^{-1} \{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot\nabla_x \overline{\mathbf{M}})\}].\end{aligned}$

这里我们用到了如下事实

$ 2 Q\left( \overline{\mathbf{M}}, L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}})\} \right) = \mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}}), $

以及

(3.17) $ \begin{equation}\overline{\mathbf{M}}_t + \mathbf{P}_0^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}})-\nabla_x\phi \cdot\nabla_\xi\overline{\mathbf{M}}= - \epsilon \mathbf{P}_0^{\overline{\mathbf{M}}}\{\xi \cdot \nabla_x L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}})\}\}.\end{equation}$

在 $ [0,t]\times \mathbb{R}^3 \times \mathbb{R}^3 $ 上, 将 (3.16) 式与 $ \frac{\mathbf{P}_1^{\overline{\mathbf{M}}} g}{\mathbf{M}_-} $ 作内积, 可得

(3.18) $ \begin{aligned} &\iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{|\mathbf{P}_1^{\overline{\mathbf{M}}}g|^2}{\mathbf{M}_-}{\rm d}\xi{\rm d}x + \frac{\delta}{\epsilon} \int_0^t\iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi)|\mathbf{P}_1^{\overline{\mathbf{M}}}g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \nonumber\\ \leq & \iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{|\mathbf{P}_1^{\overline{\mathbf{M}}}g_0|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x +C\epsilon^3\sum _{|\gamma|\leq 1}\int_0^t \int_{\mathbb{R}^3} |\partial^\gamma \nabla_x(\bar{\rho},\bar{u},\bar{\theta})|^2{\rm d}x{\rm d}\tau \nonumber\\ &+C\epsilon\int_0^t \int_{\mathbb{R}^3} |\nabla_x (\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta}) |^2{\rm d}x{\rm d}\tau +\int_0^t \iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{\mathbf{P}_1^{\overline{\mathbf{M}}} g\nabla_x\phi \cdot\nabla_\xi g }{\mathbf{M}_-}\ {\rm d}\xi{\rm d}x{\rm d}\tau \nonumber\\&+ \int_0^t \iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{\mathbf{P}_1^{\overline{\mathbf{M}}} g \nabla_x \phi \cdot\epsilon \nabla_\xi [L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}}(\xi\cdot \nabla_x\overline{\mathbf{M}})\}]}{\mathbf{M}_-}{\rm d}\xi{\rm d}x{\rm d}\tau.\end{aligned}$

需要说明的是, 与参考文献 [21] 相比, 仅需估计上式右端的最后两项. 其中一项可分解为

$ \begin{align*} (\nabla_x\phi \cdot\nabla_\xi g, \frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}) =(\nabla_x\phi \cdot\nabla_\xi \mathbf{P}_0^{\overline{\mathbf{M}}}g, \frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-})+(\nabla_x\phi \cdot\nabla_\xi \mathbf{P}_1^{\overline{\mathbf{M}}}g, \frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}),\end{align*}$

而且

$ \begin{align*} (\nabla_x\phi \cdot\nabla_\xi \mathbf{P}_1^{\overline{\mathbf{M}}}g, \frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}) =&\frac{1}{2}(\nabla_x\phi \cdot\nabla_\xi (\mathbf{P}_1^{\overline{\mathbf{M}}} g)^2,\frac{1}{\mathbf{M}_-}) \lesssim\frac{1}{2}(\nabla_x\phi \cdot(\mathbf{P}_1^{\overline{\mathbf{M}}}g)^2, \langle\xi \rangle\frac{1}{\mathbf{M}_-}) \\ \lesssim&\|\nabla_x^2\phi \|_{H^1} \|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}}}\|_2^2,\end{align*}$

此外

$ \begin{align*} (\nabla_x\phi \cdot\nabla_\xi \mathbf{P}_0^{\overline{\mathbf{M}}}g, \frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}) =&(\nabla_x\phi \cdot\nabla_\xi(\sum_{j=0}^4 \langle g,\chi_j^{\overline{\mathbf{M}}}\rangle_{\overline{\mathbf{M}}} \chi_j^{\overline{\mathbf{M}}}),\frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}).\end{align*}$

以 $ j=0 $ 为例做估计, $ j=1,2,3,4 $ 的情形类似

$ \begin{align*} &\left(\nabla_x\phi \cdot\nabla_\xi (\langle g, \frac{1}{\sqrt{\bar{\rho}}}\overline{\mathbf{M}}\rangle_{\overline{\mathbf{M}}}\frac{1}{\sqrt{\bar{\rho}}}\overline{\mathbf{M}}), \frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}\right) \lesssim (\nabla_x\phi \cdot \langle g, \frac{1}{\sqrt{\bar{\rho}}}\overline{\mathbf{M}}\rangle_{\overline{\mathbf{M}}}\frac{\langle \xi \rangle}{\sqrt{\bar{\rho}}}\overline{\mathbf{M}}, \frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-})\\ \lesssim&\eta \frac{1}{\epsilon}\|\frac{\nu^{\frac{1}{2}}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|^2+C_\eta \epsilon \|\nabla_x\phi \cdot\langle g,\frac{1}{\sqrt{\bar{\rho}}}\rangle\frac{\langle \xi \rangle \overline{\mathbf{M}}}{\sqrt{\bar{\rho}}\sqrt{\mathbf{M}_-}}\nu^{-\frac{1}{2}}\|^2\\ \lesssim&\eta \frac{1}{\epsilon}\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|^2+\epsilon \|\nabla_x\phi \cdot\frac{g}{\sqrt{\mathbf{M}_-}}\|^2\\ \lesssim& \eta \frac{1}{\epsilon}\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|^2+\epsilon \left[\|\nabla_x^2(\phi-\bar{\phi})\|^2+\|\nabla_x^2\bar{\phi}\|^2\right]\|\frac{g}{\sqrt{\mathbf{M}_-}}\|_{H^1}^2.\end{align*}$

接下来, 我们估计

$ \begin{align*} &\epsilon \left(\nabla_x\phi \cdot\nabla_\xi (L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}}(\xi \cdot \nabla_x\overline{\mathbf{M}}\}),\frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}\right)\\ \lesssim& \eta \frac{1}{\epsilon} \|\frac{\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\nu^\frac{1}{2}\|^2+C_\eta\epsilon^3\|\nabla_x\phi \cdot \nabla_\xi (L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}}(\xi\cdot\nabla_x\overline{\mathbf{M}})\})\frac{\nu^{-\frac{1}{2}}}{\sqrt{\mathbf{M}_-}}\|^2\\ \lesssim& \eta\frac{1}{\epsilon} \|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|^2+C_\eta\epsilon^3 \|\nabla_x\phi \cdot\nabla_x(\bar{\rho},\bar{u},\bar{\theta})\frac{\langle \xi \rangle^4 \nu^{-\frac{1}{2}} \overline{\mathbf{M}}}{\sqrt{\mathbf{M}_-}}\|^2\\ \lesssim&\eta \frac{1}{\epsilon} \|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|^2+C_\eta \epsilon^3\|\nabla_x\phi \cdot \nabla_x (\bar{\rho},\bar{u},\bar{\theta})\|^2\\ \lesssim&\eta \frac{1}{\epsilon}\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|^2+C_\eta \epsilon^3\left[\|\nabla_x^2(\phi-\bar{\phi})\|_2^2+\|\nabla_x^2\bar{\phi}\|_2^2\right]\|\nabla_x(\bar{\rho},\bar{u},\bar{\theta})\|_{H^1}^2.\end{align*}$

综上所述, 将以上不等式相加并结合参考文献 [21], 有

(3.19) $ \begin{aligned}\label{3.19} &\iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{|\mathbf{P}_1^{\overline{\mathbf{M}}}g|^2}{\mathbf{M}_-}{\rm d}\xi{\rm d}x + \frac{\delta}{\epsilon} \int_0^t\iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi)|\mathbf{P}_1^{\overline{\mathbf{M}}}g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \nonumber\\ \leq & \iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{|\mathbf{P}_1^{\overline{\mathbf{M}}}g_0|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x +C\epsilon^3\sum _{|\gamma|\leq 1}\int_0^t \int_{\mathbb{R}^3} |\partial^\gamma \nabla_x(\bar{\rho},\bar{u},\bar{\theta})|^2{\rm d}x{\rm d}\tau \nonumber\\ &+C\epsilon\int_0^t \int_{\mathbb{R}^3} |\nabla_x (\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta}) |^2{\rm d}x{\rm d}\tau\\ &+C \left[\|\nabla_x^2(\phi-\bar{\phi})\|_{H^1} ^2+\|\nabla_x^2\bar{\phi}\|_{H^1}^2\right]\left[ \|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|^2 +\epsilon\|\frac{g}{\sqrt{\mathbf{M}_-}}\|_{H^1}^2 +\epsilon^3 \|\nabla_x(\bar{\rho},\bar{u},\bar{\theta})\|_{H^1}^2 \right].\nonumber\end{aligned}$

步骤二 不含 $ \xi $ 导数的能量估计. 我们现在推导 $ \partial^\gamma g (1\leq|\gamma|\leq N) $ 的高阶能量估计. 为实现这一目标, 我们对(3.16) 式作用 $ \partial^\gamma $ $ (1\leq |\gamma|\leq N) $, 并将其与 $ \frac{\partial^\gamma g}{\sqrt{\mathbf{M}_-}} $ 在 $ [0,t]\times\mathbb{R}^3\times\mathbb{R}^3 $ 上积分, 类似步骤一, 由索伯列夫不等式和分部积分, 我们可以得到

$ \begin{align*}&\epsilon(\partial^\gamma[\nabla_x\phi\cdot\nabla_\xi (L_{\mathbf{M}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x\overline{\mathbf{M}})\})],\frac{\partial^\gamma g}{\mathbf{M}_-})\\\lesssim& \epsilon \left[\|\nabla_x^2\bar{\phi}\|_{H^{N-1}} +\|\nabla_x^2(\phi-\bar{\phi})\|_{H^{N-1}}\right] \|\nabla_x(\bar{\rho},\bar{u},\bar{\theta})\|_{H^N}\|\frac{g}{\sqrt{\mathbf{M}_-}}\|_{H^N},\end{align*}$

以及

$ \begin{align*} &(\partial^\gamma[\nabla_x\phi \cdot\nabla_\xi g],\frac{\partial^\gamma g}{\mathbf{M}_-})\\ \lesssim&\|\frac{g}{\sqrt{\mathbf{M}}_-}\|_{H_x^N}\|\nabla_x^2\phi\|_{H^{N-1}}\left[\|\nabla_x(\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})\|_{H^{N-1}}+\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\mathbf{M}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N}\right]\\ &+[\|\nabla_x^2\bar{\phi}\|_{H_x^1}+\|\nabla_x^2(\phi-\bar{\phi})\|_{H_x^1}]\left[\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N}^2+\|\nabla_x(\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})\|_{H^{N-1}}^2\right].\end{align*}$

综上所述, 我们有

(3.20) $ \begin{aligned}\label{3.20} &\sum_{1\leq |\gamma|\leq N}\iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{|\partial^{\gamma} g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x +\frac{\delta}{\epsilon}\sum_{1\leq|\gamma|\leq N}\int_0^t\iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi)|\partial^\gamma\mathbf{P}_1^{\overline{\mathbf{M}}}g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \nonumber\\ \lesssim&\sum_{1\leq|\gamma|\leq N}\iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{|\partial^{\gamma} g_0|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x +C\epsilon\int_0^t\int_{\mathbb{R}^3} |\partial^\gamma(\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})|^2{\rm d}x{\rm d}\tau \nonumber\\ &+C\epsilon^3\sum_{|\gamma|\leq N+1} \int_0^t \|\partial^\gamma \nabla_x(\bar{\rho},\bar{u},\bar{\theta})\|^2{\rm d}\tau+\frac{C\epsilon}{\eta_1}\int_0^t\iint_{\mathbb{R}^3\times\mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi)|\mathbf{P}_1^{\overline{\mathbf{M}}}g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \nonumber\\ &+\epsilon [\|\nabla_x^2(\phi-\bar{\phi})\|_{H^{N-1}}+\|\nabla_x^2\bar{\phi}\|_{H^{N-1}}] \|\nabla_x(\bar{\rho},\bar{u},\bar{\theta})\|_{H^N} \|\frac{g}{\sqrt{\mathbf{M}_-}}\|_{H^N}\\ &+\|\frac{g}{\sqrt{\mathbf{M}}_-}\|_{H_x^N}\|\nabla_x^2\phi\|_{H^{N-1}}\left[\|\nabla_x(\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})\|_{H^{N-1}}+\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\mathbf{M}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N}\right] \nonumber\\ &+[\|\nabla_x^2\bar{\phi}\|_{H_x^1}+\|\nabla_x^2(\phi-\bar{\phi})\|_{H_x^1}]\left[\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N}^2+\|\nabla_x(\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})\|_{H^{N-1}}^2\right] \nonumber.\end{aligned}$

步骤三 混合导数的能量估计. 接下来, 我们推导含混合导数项 $ \partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g $ 的能量估计.令 $ |\beta|\geq1 $ 且 $ |\gamma|+|\beta|\leq N $, 将 $ \partial^\gamma\partial^\beta $ 作用于 (3.16) 式, 并将所得的结果与 $ \frac{\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-} $ 在 $ [0,t]\times\mathbb{R}^3\times\mathbb{R}^3 $ 上作内积, 同样地, 我们只分析

$ \epsilon(\partial^\gamma\partial^\beta[\nabla_x\phi \cdot\nabla_\xi L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}}(\xi \cdot\nabla_x\overline{\mathbf{M}})\}],\frac{\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}) $

和

$ (\partial^\gamma\partial^\beta [\nabla_x\phi \cdot\nabla_\xi g],\frac{\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}). $

首先, 我们估计

$ \begin{align*} \epsilon(\partial^\gamma\partial^\beta(\nabla_x\phi \cdot\nabla_\xi L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}}(\xi \cdot\nabla_x\overline{\mathbf{M}})\}),\frac{\partial^\gamma\partial^\beta\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}),\end{align*}$

注意到

$ \partial^\beta(\nabla_\xi L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}}(\xi \cdot\nabla_x\overline{\mathbf{M}})\}) \lesssim \partial^\beta(\langle \xi\rangle^m\overline{\mathbf{M}}\cdot\nabla_x(\bar{\rho},\bar{u},\bar{\theta})) \lesssim \langle \xi \rangle^{m+\beta}\overline{\mathbf{M}}\cdot \nabla_x(\bar{\rho},\bar{u},\bar{\theta}). $

故而, 当 $ m $ 足够大时, 有

$ \begin{align*} &\epsilon(\partial^\gamma\partial^\beta(\nabla_x\phi \cdot\nabla_\xi L_{\overline{\mathbf{M}}}^{-1}\{\mathbf{P}_1^{\overline{\mathbf{M}}}(\xi \cdot\nabla_x\overline{\mathbf{M}})\}),\frac{\partial^\gamma\partial^\beta\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-}) \\ \lesssim& \epsilon\sum_{\gamma_1 \leq \gamma} (\nabla_x\partial^{\gamma-\gamma_1}\phi \cdot\partial^{\gamma_1}\nabla_x(\bar{\rho},\bar{u},\bar{\theta}),\frac{\langle \xi \rangle^{m+|\beta|}}{\mathbf{M}_-}\overline{\mathbf{M}}\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} g)\\ \lesssim&\epsilon\|\nabla_x^2\phi \|_{H^{N-1}}\|\nabla_x(\bar{\rho},\bar{u},\bar{\theta})\|_{H^N}\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N}.\end{align*}$

接下来, 由 $ \mathbf{P}_0^{\overline{\mathbf{M}}}g=\mathbf{P}_0^{\overline{\mathbf{M}}}(\mathbf{M}-\overline{\mathbf{M}}) $ 及索伯列夫不等式, 我们来估计

$ \begin{align*}&(\partial^\gamma\partial^\beta(\nabla_x\phi\cdot\nabla_\xi g), \frac{\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-})\\=&\sum_{\gamma_1\leq \gamma} C_\gamma^{\gamma_1} (\nabla_x\partial^{\gamma-\gamma_1}\phi \cdot\nabla_\xi \partial^{\gamma_1}\partial^\beta(\mathbf{P}_1^{\overline{\mathbf{M}}}g+\mathbf{P}_0^{\overline{\mathbf{M}}}g),\frac{\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-})\\=&\sum_{\gamma_1\leq \gamma} C_\gamma^{\gamma_1} (\nabla_x\partial^{\gamma-\gamma_1}\phi \cdot\nabla_\xi \partial^{\gamma_1}\partial^\beta(\mathbf{P}_1^{\overline{\mathbf{M}}}g),\frac{\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-})\\&+\sum_{\gamma_1\leq \gamma} C_\gamma^{\gamma_1} (\nabla_x\partial^{\gamma-\gamma_1}\phi \cdot\nabla_\xi \partial^{\gamma_1}\partial^\beta(\mathbf{P}_0^{\overline{\mathbf{M}}}g),\frac{\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g}{\mathbf{M}_-})\\\lesssim&\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N}^2\|\nabla_x^2\phi\|_{H^{N-1}} +\|\nabla_x^2\phi\|_{H^{N-1}}\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N} \|(\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})\|_{H^N}.\end{align*}$

综上, 在混合导数的情形下, 我们得到如下估计

(3.21) $ \begin{aligned}\label{3.21} &\sum_{\substack{|\gamma|+|\beta|\leq N\\ |\beta|\geq 1}}\iint_{\mathbb{R}^3\times \mathbb{R}^3} \frac{|\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g|^2}{\sqrt{\mathbf{M}_-}} {\rm d}\xi{\rm d}x +\frac{\delta}{\epsilon} \sum_{\substack{|\gamma|+|\beta|\leq N\\ |\beta|\geq 1}} \int_0^t \iint_{\mathbb{R}^3\times\mathbf{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi)|\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \nonumber\\ \lesssim& \sum_{\substack{|\gamma|+|\beta|\leq N\\ |\beta|\geq1}} \int_{\mathbb{R}^6} \frac{|\partial^\gamma\partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}}g_0|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x+ \frac{C}{\epsilon} \sum_{|\gamma|\leq N-1} \int_0^t\iint_{\mathbb{R}^3\times \mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi)|\partial^\gamma \mathbf{P}_1^{\overline{\mathbf{M}}}g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \nonumber\\ &+ C\epsilon\sum_{1\leq |\gamma|\leq N}\int_0^t \int_{\mathbb{R}^3} |\partial^\gamma (\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})|^2{\rm d}xd\tau \nonumber +C\epsilon^3\sum_{|\gamma|\leq N-1}\int_0^t \int_{\mathbb{R}^3} |\partial^\gamma \nabla_x (\bar{\rho},\bar{u},\bar{\theta})|^2{\rm d}x{\rm d}\tau \nonumber\\ &+\epsilon \|\nabla_x^2\phi\|_{H^{N-1}} \|\nabla_x(\bar{\rho},\bar{u},\bar{\theta})\|_{H^N} \|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N} +\|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N}^2 \|\nabla_x^2 \phi\|_{H^{N-1}} \nonumber\\ &+\|\nabla_x^2 \phi \|_{H^{N-1}} \|\frac{\nu^\frac{1}{2}\mathbf{P}_1^{\overline{\mathbf{M}}}g}{\sqrt{\mathbf{M}_-}}\|_{H^N} \|(\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})\|_{H^N}.\end{aligned}$

最终, (3.19)-(3.21) 式即可得到 (3.15) 式. 至此, 引理 3.3 的证明完成.

3.3 定理 1.1 的证明

现在我们可以开始着手定理 1.1 的证明. 为此, 取充分大的 $ K > 0 $ 和足够小的 $ \epsilon > 0 $,结合推论 3.1 和引理 3.3, 我们可以得到

(3.22) $ \begin{aligned}\label{3.22}& \sum_{|\gamma| \leq N-1} K \left\{\left \| \partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta}) \right\|^2(t) + \epsilon \int_{\mathbb{R}^3} (\partial^\gamma (u - \bar{u})) \cdot (\partial^\gamma \nabla_x (\rho - \bar{\rho})){\rm d}x \right\} \nonumber\\&\quad+ \sum_{1 \leq |\gamma| \leq N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x + \sum_{|\gamma| + |\beta| \leq N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma \partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x \nonumber\\&\quad+ \epsilon \sum_{1 \leq |\gamma| \leq N} \int_0^t \left\| \partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})(\tau) \right\|^2{\rm d}\tau \nonumber\\&\quad+ \frac{1}{\epsilon} \sum_{|\gamma| + |\beta| \leq N} \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi) |\partial^\gamma \partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \\&\quad+ \epsilon\sum_{|\gamma|\leq N} \int_0^t \|\partial^\gamma \nabla_x^2(\phi-\bar{\phi})(\tau)\|^2{\rm d}\tau +\sum_{|\gamma|\leq N-1} K \|\partial^\gamma \nabla_x (\phi-\bar{\phi})(t)\|^2 \nonumber\\&\leq C \sum_{|\gamma| \leq N} \left\| \partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})(0) \right\|^2+C \sum_{|\gamma|\leq N-1} \|\partial^\gamma \nabla_x (\phi-\bar{\phi})(0)\|^2 \nonumber \\&\quad+ \sum_{1 \leq |\gamma| \leq N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma g_0|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x+ \sum_{|\gamma| + |\beta| \leq N}\int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma \partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} g_0|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x \nonumber\\&\quad+ C\epsilon^2 \sum_{|\gamma| \leq N+1} \int_0^t \left\| \partial^\gamma \nabla_x (\bar{\rho}, \bar{u}, \bar{\theta}) \right\|^2{\rm d}\tau,\nonumber\end{aligned}$

其中我们利用了以下事实

$ \begin{align*}&\sum_{|\gamma| \leq N} \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi) |\partial^\gamma \mathbf{P}_1^{\overline{\mathbf{M}}} F|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau \\\lesssim& \sum_{|\gamma| \leq N} \int_0^t \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{\nu_{\overline{\mathbf{M}}}(\xi) |\partial^\gamma \mathbf{P}_1^{\overline{\mathbf{M}}} g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x{\rm d}\tau + C\epsilon^2 \sum_{|\gamma| \leq N} \int_0^t \left\| \partial^\gamma \nabla_x (\bar{\rho}, \bar{u}, \bar{\theta}) \right\|^2{\rm d}\tau.\end{align*}$

回顾 $ g $ 的定义

$ \begin{align*}g(t, x, \xi) &= F(t, x, \xi) - \overline{\mathbf{M}}(t, x, \xi) - \epsilon L_{\overline{\mathbf{M}}}^{-1} \left\{ \mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}}) \right\} \\&= \mathbf{G}(t, x, \xi) + \mathbf{M}(t, x, \xi) - \overline{\mathbf{M}}(t, x, \xi) - \epsilon L_{\overline{\mathbf{M}}}^{-1} \left\{ \mathbf{P}_1^{\overline{\mathbf{M}}} (\xi \cdot \nabla_x \overline{\mathbf{M}}) \right\},\end{align*}$

这蕴含了

(3.23) $ \begin{equation}\label{3.23}\mathbf{G} = \mathbf{P}_1^{\overline{\mathbf{M}}} g + \epsilon L_{\overline{\mathbf{M}}}^{-1} \left\{ \mathbf{P}_1^{\overline{\mathbf{M}}} (g \cdot \nabla_x \overline{\mathbf{M}}) \right\} - \mathbf{P}_1^{\overline{\mathbf{M}}} \mathbf{M}, \quad \mathbf{P}_0^{\overline{\mathbf{M}}} g = \mathbf{P}_0^{\overline{\mathbf{M}}} (\mathbf{M} - \overline{\mathbf{M}}).\end{equation}$

从 (3.23) 式可直接得到

$ \begin{align*}&\sum_{|\gamma| = N} \left\| \partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta}(t)) \right\|^2 \\\lesssim& \sum_{|\gamma| = N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x + \sum_{|\gamma| = N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma \mathbf{P}_1^{\overline{\mathbf{M}}} g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x,\end{align*}$

而且

$ \begin{align*}&\sup_{0 \leq \tau \leq t} \sum_{|\gamma| + |\beta| \leq N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma \partial^\beta \mathbf{G}(t, x, \xi)|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x\\\lesssim& \sup_{0 \leq \tau \leq t} \sum_{|\gamma| + |\beta| \leq N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma \partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x \\&+ \sup_{0 \leq \tau \leq t} \sum_{|\gamma| \leq N} \lambda_0^2 \epsilon^2 \left\| \partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})(\tau) \right\|^2 + \bar{\lambda}^2 \epsilon^4.\end{align*}$

因此, 通过选取合适的 $ K $, 可得

(3.24) $ \begin{equation}\label{3.24}\begin{aligned}& \sup_{0 \leq \tau \leq t} \sum_{|\gamma| \leq N-1} K \left\{ \left\| \partial^\gamma (\rho - \bar{\rho}, u - \bar{u}, \theta - \bar{\theta})(\tau) \right\|^2 + \epsilon \int_{\mathbb{R}^3} (\partial^\gamma (u - \bar{u})) \cdot (\partial^\gamma \nabla_x (\rho - \bar{\rho})){\rm d}x \right\} \\&\quad + \sup_{0 \leq \tau \leq t} \sum_{1 \leq |\gamma| \leq N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x + \sup_{0 \leq \tau \leq t} \sum_{|\gamma| + |\beta| \leq N} \int_{\mathbb{R}^3} \int_{\mathbb{R}^3} \frac{|\partial^\gamma \partial^\beta \mathbf{P}_1^{\overline{\mathbf{M}}} g|^2}{\mathbf{M}_-} {\rm d}\xi{\rm d}x \\&\quad + \sup_{0\leq \tau \leq t} \sum_{|\gamma|\leq N-1} K\|\partial^\gamma \nabla_x (\phi-\bar{\phi})(\tau)\|^2\\&\gtrsim N^2(t) - \bar{\lambda}^2 \epsilon^4,\end{aligned}\end{equation}$

其中我们用到了

$ \begin{align*} &\sum_{|\gamma|=N} \|\partial^\gamma \nabla_x (\phi-\bar{\phi})\|^2 =\sum_{|\gamma|=N} \|\partial^\gamma \Delta_x^{-1}\nabla_x (\rho-\bar{\rho})\|^2\\ \lesssim &\sum_{|\gamma|= N-1} \|\partial^\gamma (\rho-\bar{\rho})\|^2\lesssim\sup_{0\leq \tau \leq t}\sum_{|\gamma|\leq N-1} \|\partial^\gamma (\rho-\bar{\rho},u-\bar{u},\theta-\bar{\theta})(\tau)\|^2.\end{align*}$

由 (3.22), (3.24) 式与定理 2.1, 我们可以推断出

$ N^2(t) \lesssim N_0^2 + \bar{\lambda}^2 \epsilon^4. $

附录

在附录中, 我们将列出本文常用的基本不等式. 第一个不等式涉及索不列夫型不等式.

引理 3.4^[7] 对 $ f(x),g(x)\in H^2(\mathbb{R}^3) $ 有

$ \begin{align*} & \| f(x) \|_{L^6(\mathbb{R}^3)} \lesssim \| \nabla_x f(x) \|,\\ & \| f(x) g(x) \|_{L^2(\mathbb{R}^3)} \lesssim \| f(x) \|_{L^6(\mathbb{R}^3)} \| g(x) \|_{L^3(\mathbb{R}^3)},\\ & \| f(x) \|_{L^p(\mathbb{R}^3)} \lesssim \| f(x) \|_{H^1}, \quad \text{当 } 2 \leqslant p \leqslant 6,\\ & \| g(x) \|_{L^\infty(\mathbb{R}^3)} \lesssim \| \nabla_x g(x) \|_{H^1}. \end{align*}$

接下来, 我们给出非线性及线性化碰撞算子 $ Q(f, f) $ 和 $ L_{\mathbf{M}} \mathbf{G} $ 的若干不等式.

引理 3.5^[23] 假设下面的积分均有意义, 则存在正常数 $ C > 0 $ 使得

$ \int_{\mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi)^{-1} Q(f, g)^2}{\mathbf{\widetilde{M}}}{\rm d}\xi \leqslant C \left\{ \int_{\mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi) f^2}{\mathbf{\widetilde{M}}}{\rm d}\xi \cdot \int_{\mathbb{R}^3} \frac{g^2}{\mathbf{\widetilde{M}}}{\rm d}\xi + \int_{\mathbb{R}^3} \frac{f^2}{\mathbf{\widetilde{M}}}{\rm d}\xi \cdot \int_{\mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi) g^2}{\mathbf{\widetilde{M}}}{\rm d}\xi \right\}. $

引理 3.6 若 $ \frac{\theta}{2} < \widetilde{\theta}<\theta $, 则存在正常数$ \delta = \delta(u, \theta; \widetilde{u}, \widetilde{\theta}) $ 和 $ \eta_0 = \eta_0(u, \theta; \widetilde{u}, \widetilde{\theta}) $, 使得当 $ |u - \widetilde{u}| + |\theta - \widetilde{\theta}| < \eta_0 $ 时, 对 $ h(\xi) \in (\operatorname{Ker} L_{\mathbf{\widetilde{M}}})^\perp $, 有

$- \int_{\mathbb{R}^3} \frac{h L_{\widetilde{\mathbf{M}}} h}{\widetilde{\mathbf{M}}}{\rm d}\xi \geqslant \delta \int_{\mathbb{R}^3} \frac{\nu_{\mathbf{M}}(\xi) h^2}{\widetilde{\mathbf{M}}}{\rm d}\xi \geq 0. $

推论 3.2 在引理 3.6 的假设下, 对任意 $ h(\xi)\in (\operatorname{Ker} L_{\mathbf{M}})^\perp$, 有

$ \begin{cases} \int_{\mathbb{R}^3} \frac{\nu(\xi)}{\mathbf{M}}|L_{\mathbf{M}}^{-1}h|^2\ \mathrm{d}\xi \leq \delta^{-2} \int_{\mathbb{R}^3} \frac{\nu(\xi)^{-1}h^2(\xi)}{\mathbf{M}}\ \mathrm{d}\xi,\\[6pt] \int_{\mathbb{R}^3} \frac{\nu(\xi)}{\mathbf{M}_-}|L_{\mathbf{M}}^{-1}h|^2\ \mathrm{d}\xi \leq \delta^{-2} \int_{\mathbb{R}^3} \frac{\nu(\xi)^{-1}h^2(\xi)}{\mathbf{M}_-}\ \mathrm{d}\xi.\end{cases}$

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