本文主要研究有界区域上的量子Navier-Stokes(QNS)方程有限能量弱解的全局存在性, 该系统也称为Navier-Stokes-Korteweg(N-S-K)系统. 在区域$ (0, T)\times \Omega $上我们考虑如下系统

(1.1) $ \begin{equation} \left\{ \begin{array}{lcl} \partial_{t}\rho+{\rm{div}}(\rho u)=0, \\ { } \partial_{t}(\rho u)+{\rm{div}}(\rho u\otimes u)+\nabla P(\rho)-2\nu{\rm{div}}(\rho Du)-2\kappa^{2}\rho\nabla(\frac{\Delta \sqrt{\rho}}{\sqrt{\rho}})=0, \end{array} \right. \end{equation} $

初值为

(1.2) $ \begin{equation} \left\{ \begin{array}{lcl} \rho(0, x)=\rho^{0}(x), \\ (\rho u)(0, x)=\rho^{0}(x)u^{0}(x), \end{array} \right. \end{equation} $

其中, $ \Omega={\Bbb T}^{d}(d=2, 3) $是一个$ d $维的环面, 未知函数$ \rho=\rho(t, x) $和$ u=u(t, x) $分别表示流体粒子的质量密度和速度, 物理参数$ \nu $和$ \kappa $分别表示粘性系数和色散系数, $ Du=\frac{\nabla u+(\nabla u)^{T}}{2} $是应力张量, 压力$ P(\rho) $是关于密度的非单调函数. 对于任意的$ z\geq 0, \gamma>1 $, 压力$ P(\rho) $满足如下条件

(1.3) $ \begin{equation} \left\{ \begin{array}{lcl} P\in C^{1}({\Bbb R}_{+}), P(0)=0, \\ { } \frac{1}{a}z^{\gamma-1}-b\leq P' (z)\leq az^{\gamma-1}+b, \end{array} \right. \end{equation} $

其中, $ a>0, b\geq 0 $都是常数.

量子力学是当代科学研究中最重要的课题之一, 而量子流体建模是量子力学中最经典的模型. 众所周知, 量子修正项$ (\Delta\sqrt{\rho})/\sqrt{\rho} $在量子Navier-Stokes方程中可以解释为量子势能, 即Bohm势能, 它是从单态薛定谔动力学方程中推导出来的. 特别地, 如果$ \nu=0 $, 则量子Navier-Stokes方程(1.1)就是量子流体动力学(QHD)方程. 文献[1-3, 14, 23, 27]得到了QHD方程有限能量弱解的整体存在性. Donatelli, Feireisl和Marcati在文献[14]中证明了Euler-Korteweg-Poisson方程对于包括真空区域在内的任何足够光滑的初值都允许无限多个在时间上的全局弱解.顺便说一下, 可压缩流动的真空问题早在文献[29]中就已经考虑过了. Jüngel和李海梁在文献[23]中研究了具有松弛项的QHD方程解的全局存在性. 李海梁和Marcati在文献[27]中研究了多维QHD方程在空间周期域中电子粒子密度, 电流密度和静电势的存在性和时间渐近性.

量子Navier-Stokes方程产生于20世纪60年代, 由Brull和M$ \acute{\rm e} $hats^[11]首次推导出来. 由于它广泛应用于数学和工程学等方面, 因此吸引了大量科学家来研究.之前的研究都集中于粘性系数是常数的N-S-K方程, 然而粘性系数依赖于密度的N-S-K方程是最近研究的热点领域. Kotschote在文献[25]中证明了强解的局部存在性, 在文献[20]中Haspot得到了强解的全局存在性. 下面我们回顾一下有关弱解的存在性结果.

在一维情况下, Jüngel在文献[21]中通过利用一个有效速度变换得到了带有量子修正项的正压可压缩Navier-Stokes方程的存在性. 在文献[17]中, Gisclon和Lacroix-Violet通过增加冷压力项证明了正压量子Navier-Stokes方程弱解的全局存在性. Germain和LeFloch在文献[18]中结合有效能量估计、一个新的非线性Sobolev不等式以及一些紧性, 证明了Euler系统有限能量弱解的存在性和熵解的收敛性. 对于多维的情况, 在文献[7]中, 边东芬, 姚磊和朱长江得到了初值问题光滑解的消失毛细血管极限. 汪文军和姚磊在文献[36]中基于能量方法证明了Korteweg型可压缩流体模型在时间上存在全局解. Jüngel在文献[22]中通过选择$ \rho\phi $作为试验函数得到了方程(1.1)–(1.2)有限能量弱解的全局存在性. 其他结果见文献[1, 5, 8, 9, 32, 37].

弱解存在性的证明往往通过下列步骤获得: 首先, 构造逼近解系统(通常是用Galerkin方法或有限差分方法得到), 然后通过紧性分析取极限. 此外, 对于具有退化粘性的动力学, 郭真华, 酒全森和辛周平在文献[19]中证明了在大初值和球对称条件下粘性系数依赖于密度的可压缩Navier-Stokes方程弱解的整体存在性. 同时, 在文献[26]中李海梁, 李竞和辛周平得到了上述系统在有界空间区域或周期区域上的初边值问题. 最近, Vasseur和Yu在文献[34]中证明了N-S-K方程弱解的全局存在性, 之后他们利用这个结果在文献[35]中得到了带有退化粘性极限和阻尼项的3维可压缩Navier-Stokes方程弱解的全局存在性. 与经典的Navier-Stokes系统相比, Navier-Stokes-Korteweg系统逼近解的构造是一个复杂而棘手的问题. 一方面, 本文研究的粘性系数依赖于密度, 而依赖于密度的粘性系数可能产生真空, 更多的细节详见文献[13, 24, 28, 31]. 另一方面, 由于动量方程中存在三阶非线性色散项, 要得到Mellet-Vasseur估计和紧性是一个艰难的过程. 为了克服这些困难, Antonelli和Spirito在文献[4]中运用一些技巧构造了逼近解系统, 并通过限制$ \nu $, $ \kappa $和$ \gamma $, 得到了弱解的全局存在性. 他们的主要思想是构造带有冷压力项和阻尼项的逼近解系统, 然后使用有效速度变换将逼近解系统转换成新的系统, 从而使得色散项和冷压力项消失. 最近, 在文献[6]中, 他们通过引入适当的截断得到了一个新的紧性结果. 然而, 这里我们要指出的是文献[4]的工作是受李竞和辛周平[28]工作的启发. 此外, 唐童和张祖锦在文献[33]中利用了一个有趣的恒等式得到了在粘性系数等于色散系数这种临界情况下弱解的全局存在性.

上述结果研究的压力都是单调函数. 然而, 越来越多的科学实验表明, 当压力为非单调函数时更有意义, 参见文献[10, 12, 13, 15, 16]. Feireisl在文献[15, 16]中分别证明了可压缩Navier-Stokes方程在压力为非单调函数的情况下有界解的紧性和弱-强唯一性. Ducomet, Nečasová和Vasseur在文献[12]中研究了当$ \gamma>\frac{4}{3} $时, 在压力为非单调函数的情况下Navier-Stokes-Poisson方程Cauchy问题弱解的全局稳定性, 之后在文献[13]中, 他们将结果延拓到在球对称运动中$ \gamma>1 $的情况. 因此, 研究带有非单调压力的N-S-K方程弱解的存在性是一个有意义的问题.

本文的目的是证明量子Navier-Stokes方程(1.1)–(1.2)有限能量弱解的全局存在性. 值得指出的是, 本文的主要困难在于获得压力是非单调函数时的B-D熵估计和Mellet-Vasseur不等式, 所以我们重新写了色散项, 并使用一个恒等式克服了这个困难. 受文献[4, 12, 13]工作的启发, 我们在$ \kappa\leq\nu $和压力P($ \rho $)是非单调函数的情况下得到了方程(1.1)–(1.2)有限能量弱解的全局存在性.

本文的主要结构如下: 第1节, 我们简要阐述了量子Navier-Stokes方程的物理背景以及相关问题的国内外研究现状; 第2节, 我们陈述了本论文的主要研究成果, 即定理2.1, 并给出了量子Navier-Stokes方程弱解的定义; 第3节, 我们构造了逼近解系统; 第4节, 我们得到了一系列估计, 即能量估计, B-D熵估计和Mellt-Vasseur不等式; 第5节, 我们获得了一些紧性并证明了定理2.1.

本节分为两部分, 首先我们给出了弱解的定义. 在第二部分我们分别在2维和3维情况下陈述了主要结果.

下面我们介绍弱解的定义. 基于Feireisl, Lions等学者关于Navier-Stokes方程弱解的工作, 我们给出了量子Navier-Stokes方程(1.1)–(1.2)弱解的定义.

定义2.1  设$ \rho\geq0 $, $ (\rho, u) $满足如下条件:

● $ \rho\in L^{\infty}(0, T;L^{1}\cap L^{\gamma}({\Bbb T}^{d})), \; \; \; \sqrt{\rho}u\in L^{\infty}(0, T;L^{2}({\Bbb T}^{d})), \; \; \; \sqrt{\rho}\in L^{\infty}(0, T;H^{1}({\Bbb T}^{d})). $

● $ (\rho, \sqrt{\rho}u) $在分布意义$ {\cal D}' ((0, T)\times {\Bbb T}^{d}) $下满足

$ \begin{eqnarray*} \left\{ \begin{array}{lcl} \partial_{t}\rho+{\rm{div}}(\sqrt{\rho}\sqrt{\rho} u)=0, \\ \rho(0, x)=\rho^{0}(x), \end{array} \right. \end{eqnarray*} $

● 对任意试验函数$ \psi\in C_{c}^{\infty}([0, T);C^{\infty}({\Bbb T}^{d})) $, 有

$ \begin{eqnarray*} &&\int_{\Omega}\rho^{0}u^{0}\psi(0)+\int_{0}^{T}\int_{\Omega}\sqrt{\rho}(\sqrt{\rho}u)\psi_{t}+\sqrt{\rho}u\otimes\sqrt{\rho}u\nabla\psi+P(\rho){\rm{div}}\psi\\ &&-2\nu\int_{0}^{T}\int_{\Omega}(\sqrt{\rho}u\otimes\nabla\sqrt{\rho})\nabla\psi-2\nu\int_{0}^{T}\int_{\Omega}(\nabla\sqrt{\rho}\otimes\sqrt{\rho}u)\nabla\psi\\ &&+\nu\int_{0}^{T}\int_{\Omega}\sqrt{\rho}\sqrt{\rho}u\Delta\psi+\nu\int_{0}^{T}\int_{\Omega}\sqrt{\rho}\sqrt{\rho}u\nabla{\rm{div}}\psi\\ &&-4\kappa^{2}\int_{0}^{T}\int_{\Omega}(\nabla\sqrt{\rho}\otimes\nabla\sqrt{\rho})\nabla\psi+2\kappa^{2}\int_{0}^{T}\int_{\Omega}\sqrt{\rho}\nabla\sqrt{\rho}\nabla{\rm{div}}\psi=0, \end{eqnarray*} $

其中

$ 2\rho\nabla(\frac{\Delta\sqrt{\rho}}{\sqrt{\rho}})={\rm{div}}(\rho\nabla^{2}{\rm{log}} \rho)=\nabla\Delta\rho-4{\rm{div}}(\nabla\sqrt{\rho}\otimes\nabla\sqrt{\rho}). $

● 如果$ E(t)=\int_{\Omega}\frac{1}{2}\rho|u|^{2}+\Pi(\rho)+2\kappa^{2}|\nabla\sqrt{\rho}|^{2} $, 其中$ \Pi(\rho)=\rho\int_{1}^{\rho}\frac{P(s)}{s^{2}}{\rm d}s $, 则对几乎处处的$ t\in(0, T) $, 有

$ E(t)\leq E(0), $

则我们称$ (\rho, u) $是方程(1.1)–(1.2)的有限能量弱解.

在陈述主要结果之前我们需要对初值做一些假设. 令$ \nu\geq\kappa $, $ \eta $是固定的正数, 则初值应满足如下条件

(2.1) $ \begin{equation} \begin{array}{l} \rho^{0}\geq 0 \ \ \mbox{in}\ {\Bbb T}^{d}, \\ \rho^{0}\in L^{1}\cap L^{\gamma}({\Bbb T}^{d}), \\ \nabla\sqrt{\rho^{0}}\in L^{2}\cap L^{2+\eta}({\Bbb T}^{d}), \\ u^{0}=0 \ \ \mbox{若} \ \rho^{0}=0, \\ \sqrt{\rho^{0}}u^{0}\in L^{2}\cap L^{2+\eta}({\Bbb T}^{d}), \\ { } \rho^{0}(1+\frac{|\upsilon^{0}|^{2}}{2}){\rm{log}}(1+\frac{|\upsilon^{0}|^{2}}{2})\ \mbox{在}\ L^{1}({\Bbb T}^{d})\ \mbox{中一致有界}, \end{array} \end{equation} $

其中$ \upsilon^{0}=u^{0}+C\nabla{\rm{log}}\rho^{0} $, $ C>0 $. 对于初值$ \rho^{0} $, 我们假设它在本文中是有界的, 即存在$ \overline{\rho}^{0}>0 $, 使得

(2.2) $ \begin{equation} 0<\frac{1}{\overline{\rho}^{0}}\leq\rho^{0}\leq\overline{\rho}^{0}. \end{equation} $

在2维和3维情况下, 我们有如下结果

定理2.1  令$ \nu $, $ \kappa $和$ \gamma $都是正数, 并且初值(1.2)满足条件(2.1)和(2.2), 则我们有

(1) 在2维情况下, 对任意的$ \kappa\leq\nu $, $ \gamma>1 $和$ 0<T<\infty $, 方程(1.1)–(1.2)在区域$ (0, T)\times {\Bbb T}^{2} $上存在全局弱解;

(2) 在2维情况下, 对任意的$ \kappa^{2}\leq\nu^{2}\leq\frac{9}{8}\kappa^{2} $, $ 1<\gamma<3 $和$ 0<T<\infty $, 方程(1.1)–(1.2)在区域$ (0, T)\times {\Bbb T}^{3} $上存在全局弱解.

注2.1  这里我们要特别指出的是为了获得逼近解系统解的全局正则性, 我们需要证明$ \omega_{\varepsilon} $的充分可积性. 值得注意的是, 这种方法依赖于条件$ \kappa^{2}<\nu^{2}<\frac{9}{8}\kappa^{2} $, 更多细节详见文献[4]的引理11, 否则我们不能得到期望的B-D熵估计. 然而, 在文献[4]中作者没有得到粘性系数等于色散系数这种临界情况下弱解的全局存在性. 本文我们重新写了色散项, 并使用一个恒等式(3.13)移除了对$ \kappa $和$ \nu $的限制, 这就意味着在$ \kappa=\nu $这种临界情况下我们可以得到与文献[4]相同的结果.

本节致力于构建逼近解系统. 受到文献[4]工作的启发, 我们采用了相同的方法和框架. 为了方便读者, 我们只阐述重要步骤.

首先, 我们构造了含有冷压力项和阻尼项的如下逼近解系统

(3.1) $ \begin{equation} \left\{ \begin{array}{lcl} \partial_{t}\rho_{\varepsilon}+{\rm{div}}(\rho_{\varepsilon}u_{\varepsilon})=0, \\ \partial_{t}(\rho_{\varepsilon}u_{\varepsilon})+{\rm{div}}(\rho_{\varepsilon}u_{\varepsilon}\otimes u_{\varepsilon})-2\nu{\rm{div}}{\Bbb S}_{\varepsilon}+\nabla(P_{\varepsilon}(\rho_{\varepsilon})+p_{\varepsilon}(\rho_{\varepsilon}))+\widetilde{p}_{\varepsilon}(\rho_{\varepsilon})u_{\varepsilon}=\kappa^{2}{\rm{div}}{\Bbb K}_{\varepsilon}, \end{array} \right. \end{equation} $

初值为

(3.2) $ \begin{equation} \begin{array}{l} \rho_{\varepsilon}|_{t=0}=\rho_{\varepsilon}^{0}(x), \ \rho_{\varepsilon}u_{\varepsilon}|_{t=0}=\rho_{\varepsilon}^{0}(x)u_{\varepsilon}^{0}(x), \end{array} \end{equation} $

逼近解系统(3.1)中$ {\Bbb S}_{\varepsilon}, {\Bbb K}_{\varepsilon}, \widetilde{p}_{\varepsilon}(\rho_{\varepsilon}), p_{\varepsilon}(\rho_{\varepsilon}) $分别表示粘性应力张量、色散项、阻尼项和冷压力项, 具体表达式如下

(3.3) $ \begin{eqnarray} &&{\rm{div}}{\Bbb S}_{\varepsilon}={\rm{div}}(h_{\varepsilon}(\rho_{\varepsilon})Du_{\varepsilon})+\nabla(g_{\varepsilon}(\rho_{\varepsilon}){\rm{div}}u_{\varepsilon}), \end{eqnarray} $

(3.4) $ \begin{eqnarray} &&{\rm{div}}{\Bbb K}_{\varepsilon}=2\rho_{\varepsilon}\nabla \bigg(\frac{h_{\varepsilon}' (\rho_{\varepsilon}){\rm{div}}(h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}})}{\sqrt{\rho_{\varepsilon}}}\bigg), \end{eqnarray} $

(3.5) $ \begin{eqnarray} &&\widetilde{p}_{\varepsilon}(\rho_{\varepsilon})=\lambda(\varepsilon)(\rho_{\varepsilon}^{\frac{1}{\varepsilon^{2}}}+\rho_{\varepsilon}^{-\frac{1}{\varepsilon^{2}}}), \end{eqnarray} $

(3.6) $ \begin{eqnarray} &&p_{\varepsilon}' (\rho_{\varepsilon})=\mu\widetilde{p}_{\varepsilon}(\rho_{\varepsilon})\frac{h_{\varepsilon}' (\rho_{\varepsilon})}{\rho_{\varepsilon}}, \end{eqnarray} $

其中

(3.7) $ \begin{eqnarray} &&h_{\varepsilon}(\rho_{\varepsilon})=\rho_{\varepsilon}+\varepsilon\rho_{\varepsilon}^{\frac{7}{8}}+\varepsilon\rho_{\varepsilon}^{\gamma}, \end{eqnarray} $

(3.8) $ \begin{eqnarray} &&g_{\varepsilon}(\rho_{\varepsilon})=\rho_{\varepsilon}h_{\varepsilon}' (\rho_{\varepsilon})-h_{\varepsilon}(\rho_{\varepsilon}), \end{eqnarray} $

(3.9) $ \begin{eqnarray} &&\lambda(\varepsilon)=e^{-\frac{1}{\varepsilon^{4}}}, \mu=\nu-\sqrt{\nu^{2}-\kappa^{2}}. \end{eqnarray} $

因此, 由$ h_{\varepsilon}(\rho_{\varepsilon}) $和$ \widetilde{p}_{\varepsilon}(\rho_{\varepsilon}) $的定义, 我们可以得到$ p_{\varepsilon}(\rho_{\varepsilon}) $的表达式

(3.10) $ \begin{eqnarray} p_{\varepsilon}(\rho_{\varepsilon})&=&\mu\varepsilon^{2}\lambda(\varepsilon)\rho_{\varepsilon}^{\frac{1}{\varepsilon^{2}}}+\frac{\varepsilon^{3}\mu7\lambda(\varepsilon)}{8-\varepsilon^{2}}\rho_{\varepsilon}^{\frac{1}{\varepsilon^{2}}-\frac{1}{8}} -\mu\varepsilon^{2}\lambda(\varepsilon)\rho_{\varepsilon}^{-\frac{1}{\varepsilon^{2}}}+\frac{\varepsilon^{3}\mu\lambda(\varepsilon)\gamma}{1+\varepsilon^{2}(\gamma-1)}\rho_{\varepsilon}^{\frac{1}{\varepsilon^{2}}+\gamma-1}\\ &&-\frac{\varepsilon^{3}\mu7\lambda(\varepsilon)}{\varepsilon^{2}+8}\rho_{\varepsilon}^{-\frac{1}{\varepsilon^{2}}-\frac{1}{8}}-\frac{\varepsilon^{3}\mu\lambda(\varepsilon)\gamma}{1-\varepsilon^{2}(\gamma-1)}\rho_{\varepsilon}^{-\frac{1}{\varepsilon^{2}}+\gamma-1}. \end{eqnarray} $

令$ f_{\varepsilon}(\rho_{\varepsilon}) $满足$ p_{\varepsilon}(\rho_{\varepsilon})=\rho_{\varepsilon}f_{\varepsilon}' (\rho_{\varepsilon})-f_{\varepsilon}(\rho_{\varepsilon}) $, 通过直接计算可以得到

(3.11) $ \begin{eqnarray} f_{\varepsilon}(\rho_{\varepsilon})&=&\frac{\mu\varepsilon^{4}\lambda(\varepsilon)}{1-\varepsilon^{2}}\rho_{\varepsilon}^{\frac{1}{\varepsilon^{2}}}+\frac{\varepsilon^{5}\mu7\lambda(\varepsilon)}{(8-\varepsilon^{2})(8-9\varepsilon^{2})}\rho_{\varepsilon}^{\frac{1}{\varepsilon^{2}}-\frac{1}{8}}+\frac{\mu\varepsilon^{2}\lambda(\varepsilon)}{\varepsilon^{2}+1}\rho_{\varepsilon}^{-\frac{1}{\varepsilon^{2}}}\\ &&+\frac{\varepsilon^{5}\mu7\lambda(\varepsilon)8}{(\varepsilon^{2}+8)(8\varepsilon^{2}+9)}\rho_{\varepsilon}^{-\frac{1}{\varepsilon^{2}}-\frac{1}{8}}+\frac{\varepsilon^{5}\mu\lambda(\varepsilon)\gamma}{(1+\varepsilon^{2}(\gamma-1))(1+\varepsilon^{2}(\gamma-2))}\rho_{\varepsilon}^{\frac{1}{\varepsilon^{2}}+\gamma-1}\\ &&+\frac{\varepsilon^{5}\mu\lambda(\varepsilon)\gamma}{(1-\varepsilon^{2}(\gamma-1))(1-\varepsilon^{2}(\gamma-2))}\rho_{\varepsilon}^{-\frac{1}{\varepsilon^{2}}+\gamma-1}. \end{eqnarray} $

特别地, 由$ h_{\varepsilon}(\rho_{\varepsilon}) $和$ g_{\varepsilon}(\rho_{\varepsilon}) $的定义, 我们有

(3.12) $ \begin{equation} \begin{array}{l} { } h_{\varepsilon}(\rho_{\varepsilon})\geq0, \ \ \ |g_{\varepsilon}(\rho_{\varepsilon})|\leq\max(\frac{1}{8}, (\gamma-1))h_{\varepsilon}(\rho_{\varepsilon}), \\ h_{\varepsilon}' (\rho_{\varepsilon})\rho_{\varepsilon}\leq\gamma h_{\varepsilon}(\rho_{\varepsilon}), \ \ \ |h_{\varepsilon}''(\rho_{\varepsilon})|\rho_{\varepsilon}\leq(\gamma-1)h_{\varepsilon}' (\rho_{\varepsilon}). \end{array} \end{equation} $

由文献[4, 引理1]的和文献[33]可知

(3.13) $ \begin{eqnarray} {\rm{div}} ({\Bbb K}_{\varepsilon}(\rho_{\varepsilon}, \nabla\rho_{\varepsilon}))&=&2\rho_{\varepsilon}\nabla \bigg(\frac{h_{\varepsilon}' (\rho_{\varepsilon}){\rm{div}}(h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}})}{\sqrt{\rho_{\varepsilon}}} \bigg)\\ &=&{\rm{div}}(h_{\varepsilon}(\rho_{\varepsilon})\nabla^{2}\phi_{\varepsilon}(\rho_{\varepsilon}))+\nabla(g_{\varepsilon}(\rho_{\varepsilon})\Delta \phi_{\varepsilon}(\rho_{\varepsilon}))\\ &=&\nabla(h_{\varepsilon}' (\rho_{\varepsilon})\Delta h_{\varepsilon}(\rho_{\varepsilon}))-4{\rm{div}}((h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}})\otimes(h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}}))\\ &=&\rho_{\varepsilon}\nabla \bigg(\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s) \bigg), \end{eqnarray} $

其中$ \sqrt{K_{\varepsilon}(\rho_{\varepsilon})\rho_{\varepsilon}}=h_{\varepsilon}'(\rho_{\varepsilon})=\rho_{\varepsilon}\phi' _{\varepsilon}(\rho_{\varepsilon}) $. 最后, 初值(3.2)的构造和文献[4]类似, 这里省略.

为了获得在Navier-Stokes方程中起至关重要的B-D熵估计和其他估计, 我们引入有效速度变换$ \omega_{\varepsilon}=u_{\varepsilon}+\mu\nabla\phi_{\varepsilon}(\rho_{\varepsilon}) $. 因此, 我们考虑如下新的系统

(3.14) $ \begin{equation} \left\{ \begin{array}{lcl} \partial_{t}\rho_{\varepsilon}+{\rm{div}}(\rho_{\varepsilon}\omega_{\varepsilon})=\mu\Delta h_{\varepsilon}(\rho_{\varepsilon}), \\ \partial_{t}(\rho_{\varepsilon}\omega_{\varepsilon})+{\rm{div}}(\rho_{\varepsilon}\omega_{\varepsilon}\otimes\omega_{\varepsilon})-\mu\Delta(h_{\varepsilon}(\rho_{\varepsilon})\omega_{\varepsilon})+\nabla P_{\varepsilon}(\rho_{\varepsilon})+\widetilde{p}_{\varepsilon}(\rho_{\varepsilon})\omega_{\varepsilon}\\ \ -2(\nu-\mu){\rm{div}}(h_{\varepsilon}(\rho_{\varepsilon})D\omega_{\varepsilon})-(2\nu-\mu)\nabla(g_{\varepsilon}(\rho_{\varepsilon}){\rm{div}}\omega_{\varepsilon})=0. \end{array} \right. \end{equation} $

注3.1  由文献[4]的引理2知: 如果$ (\rho_{\varepsilon}, u_{\varepsilon}) $是方程(3.1)的光滑解, 则$ (\rho_{\varepsilon}, \omega_{\varepsilon}) $将满足系统(3.14).

在本章, 我们将推导出一系列的先验估计, 即能量估计, B-D熵估计和Mellt-Vasseur不等式, 目的是获得弱解的紧性.

首先, 我们得到经典的能量估计.

引理4.1  令$ (\rho_{\varepsilon}, u_{\varepsilon}) $是方程(3.1)的光滑解, 则有如下等式成立

(4.1) $ \begin{eqnarray} &&\frac{{\rm d}}{{\rm d}t} \bigg(\int_{\Omega}\frac{\rho_{\varepsilon}|u_{\varepsilon}|^{2}}{2}+\Pi(\rho_{\varepsilon}) +f_{\varepsilon}(\rho_{\varepsilon})+2\kappa^{2}|h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}}|^{2} \bigg) \\ &&+2\nu\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|Du_{\varepsilon}|^{2} +2\nu\int_{\Omega} g_{\varepsilon}(\rho_{\varepsilon})|{\rm{div}}u_{\varepsilon}|^{2}+\int_{\Omega}\widetilde{p}_{\varepsilon}(\rho_{\varepsilon})|u_{\varepsilon}|^{2}=0, \end{eqnarray} $

其中$ \Pi(\rho_{\varepsilon})=\rho_{\varepsilon}\int_{1}^{\rho_{\varepsilon}}\frac{P_{\varepsilon}(s)}{s^{2}}{\rm d}s $.

证  首先, 在方程(3.1)的动量方程两边同乘$ u_{\varepsilon} $, 然后分部积分, 利用(3.1)的连续性方程, 得到

(4.2) $ \begin{eqnarray} &&\frac{{\rm d}}{{\rm d}t}\int_{\Omega}\frac{\rho_{\varepsilon}|u_{\varepsilon}|^{2}}{2}+2\nu\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|Du_{\varepsilon}|^{2}+2\nu\int_{\Omega} g_{\varepsilon}(\rho_{\varepsilon})|{\rm{div}}u_{\varepsilon}|^{2}+\int_{\Omega}\widetilde{p}_{\varepsilon}(\rho_{\varepsilon})|u_{\varepsilon}|^{2}\\ &&-\kappa^{2}\int_{\Omega}{\rm{div}}{\Bbb K}_{\varepsilon}u_{\varepsilon}+\int_{\Omega}\nabla P_{\varepsilon}(\rho_{\varepsilon})u_{\varepsilon}+\int_{\Omega}\nabla p_{\varepsilon}(\rho_{\varepsilon})u_{\varepsilon}=0. \end{eqnarray} $

下面我们只考虑压力项, 其他项的处理参见文献[4].

(4.3) $ \begin{eqnarray} \int_{\Omega}\nabla P_{\varepsilon}(\rho_{\varepsilon})u_{\varepsilon}&=&\int_{\Omega} P_{\varepsilon}' (\rho_{\varepsilon})\nabla\rho_{\varepsilon}u_{\varepsilon}=\int_{\Omega}\nabla\int_{1}^{\rho_{\varepsilon}}\frac{P_{\varepsilon}' (s)}{s}{\rm d}s(\rho_{\varepsilon}u_{\varepsilon}){\rm d}x\\ &=&-\int_{\Omega}\int_{1}^{\rho_{\varepsilon}}\frac{P_{\varepsilon}' (s)}{s}{\rm d}s({\rm{div}}(\rho_{\varepsilon}u_{\varepsilon})){\rm d}x =\int_{\Omega}\int_{1}^{\rho_{\varepsilon}}\frac{P_{\varepsilon}' (s)}{s}{\rm d}s(\partial_{t}\rho_{\varepsilon}){\rm d}x\\ &=&\frac{{\rm d}}{{\rm d}t}\int_{\Omega}\rho_{\varepsilon}\int_{1}^{\rho_{\varepsilon}}\frac{P_{\varepsilon}(s)}{s^{2}}{\rm d}s{\rm d}x :=\frac{{\rm d}}{{\rm d}t}\int_{\Omega}\Pi(\rho_{\varepsilon}){\rm d}x, \end{eqnarray} $

其中$ \Pi(\rho_{\varepsilon})=\rho_{\varepsilon}\int_{1}^{\rho_{\varepsilon}}\frac{P_{\varepsilon}(s)}{s^{2}}{\rm d}s $.

将所有的估计带入到(4.2)式中, 我们就得到了能量估计(4.1).

注4.1  根据(1.3)式可知

$ P(z)\geq\frac{1}{a\gamma}z^{\gamma}-bz. $

故由$ \Pi(\rho_{\varepsilon}) $定义可得

(4.4) $ \begin{equation} \Pi(\rho_{\varepsilon})\geq\frac{1}{a\gamma(\gamma-1)}(\rho_{\varepsilon}^{\gamma}-\rho_{\varepsilon})-b\rho_{\varepsilon}{\rm{log}}\rho_{\varepsilon}. \end{equation} $

更多细节可参见文献[12]. 再者, 由于$ \gamma>1 $, 利用$ {\rm{log}}x\leq x-1 $可以得到

(4.5) $ \begin{eqnarray} \int_{\Omega}\rho_{\varepsilon}{\rm{log}}\rho_{\varepsilon}&=&\frac{2}{\gamma-1}\int_{\Omega}\rho_{\varepsilon}{\rm{log}}\rho_{\varepsilon}^{\frac{\gamma-1}{2}}\leq\frac{2}{\gamma-1}\int_{\Omega}\rho_{\varepsilon}(\rho_{\varepsilon}^{\frac{\gamma-1}{2}}-1)\\ &\leq&\frac{2}{\gamma-1}\int_{\Omega}\rho_{\varepsilon}^{\frac{\gamma+1}{2}}=\frac{2}{\gamma-1}\int_{\Omega}\frac{1}{\delta}\rho_{\varepsilon}^{\frac{1}{2}}\delta\rho_{\varepsilon}^{\frac{\gamma}{2}}\\ &\leq&\frac{1}{\gamma-1}\int_{\Omega}\frac{1}{\delta^{2}}\rho_{\varepsilon}+\frac{1}{\gamma-1}\int_{\Omega}\delta^{2}\rho_{\varepsilon}^{\gamma}\\ &\leq& C+\frac{1}{\gamma-1}\int_{\Omega}\delta^{2}\rho_{\varepsilon}^{\gamma}, \end{eqnarray} $

其中$ C>0 $是个常数. 令$ \delta $充分小, 这就意味着不等式(4.4)右边所有的负项都可以被$ \rho^{\gamma} $给控制住. 因此, 有如下不等式成立

(4.6) $ \begin{eqnarray} &&\frac{{\rm d}}{{\rm d}t} \bigg(\int_{\Omega}\frac{\rho_{\varepsilon}|u_{\varepsilon}|^{2}}{2}+\frac{\rho_{\varepsilon}^ {\gamma}}{a\gamma(\gamma-1)}+f_{\varepsilon}(\rho_{\varepsilon})+2\kappa^{2}|h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}}|^{2}\bigg) \\ &&+2\nu\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|Du_{\varepsilon}|^{2} +2\nu\int_{\Omega} g_{\varepsilon}(\rho_{\varepsilon})|{\rm{div}}u_{\varepsilon}|^{2}+\int_{\Omega}\widetilde{p}_{\varepsilon}(\rho_{\varepsilon})|u_{\varepsilon}|^{2}\leq C. \end{eqnarray} $

下面的引理我们证明了方程(3.14)的能量估计.

引理4.2  令$ (\rho_{\varepsilon}, u_{\varepsilon}) $是方程(3.1)的光滑解, 则$ (\rho_{\varepsilon}, \omega_{\varepsilon}) $将满足如下估计

(4.7) $ \begin{eqnarray} &&\frac{{\rm d}}{{\rm d}t} \bigg(\int_{\Omega}\frac{\rho_{\varepsilon}|\omega_{\varepsilon}|^{2}}{2}+\frac{\rho_{\varepsilon}^{\gamma}}{a\gamma(\gamma-1)}+\frac{1}{2}|h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}}|^{2}\bigg) +\mu\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|A\omega_{\varepsilon}|^{2}\\ && +(2\nu-\mu)\int_{\Omega}(h_{\varepsilon}(\rho_{\varepsilon})|D\omega_{\varepsilon}|^{2}+g_{\varepsilon}(\rho_{\varepsilon})|{\rm{div}}\omega_{\varepsilon}|^{2})+\int_{\Omega}\widetilde{p}_{\varepsilon}(\rho_{\varepsilon})|\omega_{\varepsilon}|^{2}\\ && +\frac{\mu}{a}\int_{\Omega}\rho_{\varepsilon}^{\gamma-2}|\nabla\rho_{\varepsilon}|^{2}h_{\varepsilon}' (\rho_{\varepsilon})+\frac{\mu}{2}\int_{\Omega}h_{\varepsilon}(\rho_{\varepsilon})|\nabla^{2}\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}\\ && +\frac{\mu}{2}\int_{\Omega} g_{\varepsilon}(\rho_{\varepsilon})|\Delta\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}\leq C, \end{eqnarray} $

其中$ A\omega_{\varepsilon}=\frac{\nabla\omega_{\varepsilon}-(\nabla\omega_{\varepsilon})^{T}}{2} $是梯度的反对称部分.

证  因为$ (\rho_{\varepsilon}, u_{\varepsilon}) $是方程(3.1)的光滑解, 由注3.1可知$ (\rho_{\varepsilon}, \omega_{\varepsilon}) $一定满足系统(3.14). 因此在系统(3.14)的动量方程两边同乘$ \omega_{\varepsilon} $, 然后分部积分, 利用系统(3.14)的连续性方程和$ |\nabla\omega_{\varepsilon}|^{2}=|D\omega_{\varepsilon}|^{2}+|A\omega_{\varepsilon}|^{2} $, 得到

(4.8) $ \begin{eqnarray} &&\frac{{\rm d}}{{\rm d}t}\int_{\Omega}\frac{\rho_{\varepsilon}|\omega_{\varepsilon}|^{2}}{2}+\int_{\Omega}\nabla P_{\varepsilon}(\rho_{\varepsilon})\omega_{\varepsilon}+\int_{\Omega}\widetilde{p_{\varepsilon}}(\rho_{\varepsilon})|\omega_{\varepsilon}|^{2}+\mu\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|A\omega_{\varepsilon}|^{2}\\ &&+(2\nu-\mu)\int_{\Omega}(h_{\varepsilon}(\rho_{\varepsilon})|D\omega_{\varepsilon}|^{2}+g_{\varepsilon}(\rho_{\varepsilon})|{\rm{div}}\omega_{\varepsilon}|^{2})=0. \end{eqnarray} $

首先, 我们考虑压力项

(4.9) $ \begin{eqnarray} \int_{\Omega}\nabla P_{\varepsilon}(\rho_{\varepsilon})\omega_{\varepsilon}&=&-\int_{\Omega}\int_{1}^{\rho_{\varepsilon}}\frac{P_{\varepsilon}' (s)}{s}{\rm d}s({\rm{div}}(\rho_{\varepsilon}\omega_{\varepsilon})){\rm d}x\\ &=&\int_{\Omega}\int_{1}^{\rho_{\varepsilon}}\frac{P_{\varepsilon}' (s)}{s}{\rm d}s(\partial_{t}\rho_{\varepsilon}-\mu\Delta h_{\varepsilon}(\rho_{\varepsilon})){\rm d}x\\ &=&\frac{{\rm d}}{{\rm d}t}\int_{\Omega}\Pi(\rho_{\varepsilon}){\rm d}x-\mu\int_{\Omega}\int_{1}^{\rho_{\varepsilon}}\frac{P_{\varepsilon}' (s)}{s}{\rm d}s(\Delta h_{\varepsilon}(\rho_{\varepsilon})){\rm d}x\\ &=&\frac{{\rm d}}{{\rm d}t}\int_{\Omega}\Pi(\rho_{\varepsilon}){\rm d}x+\mu\int_{\Omega}\frac{P_{\varepsilon}' (\rho_{\varepsilon})}{\rho_{\varepsilon}}h_{\varepsilon}' (\rho_{\varepsilon})|\nabla\rho_{\varepsilon}|^{2}{\rm d}x. \end{eqnarray} $

由(1.3)式可知

(4.10) $ \begin{equation} \mu\int_{\Omega}\frac{P_{\varepsilon}' (\rho_{\varepsilon})}{\rho_{\varepsilon}}h_{\varepsilon}' (\rho_{\varepsilon})|\nabla\rho_{\varepsilon}|^{2}\geq\frac{\mu}{a}\int_{\Omega}\rho_{\varepsilon}^{\gamma-2}|\nabla\rho_{\varepsilon}|^{2}h_{\varepsilon}' (\rho_{\varepsilon})-b\mu\int_{\Omega}\frac{h_{\varepsilon}' (\rho_{\varepsilon})}{\rho_{\varepsilon}}|\nabla\rho_{\varepsilon}|^{2}. \end{equation} $

根据(4.6)式, 我们有

(4.11) $ \begin{equation} \int_{\Omega}\frac{h_{\varepsilon}' (\rho_{\varepsilon})}{\rho_{\varepsilon}}|\nabla\rho_{\varepsilon}|^{2}=4\int_{\Omega}h_{\varepsilon}' (\rho_{\varepsilon})|\nabla\sqrt{\rho_{\varepsilon}}|^{2}\leq C\int_{\Omega}|h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}}|^{2}\leq C. \end{equation} $

将(4.4), (4.10)–(4.11)式带入(4.9)式, 得到

(4.12) $ \begin{equation} \int_{\Omega}\nabla P_{\varepsilon}(\rho_{\varepsilon})\omega_{\varepsilon}\geq \frac{{\rm d}}{{\rm d}t}\int_{\Omega}\frac{\rho_{\varepsilon}^{\gamma}}{a\gamma(\gamma-1)}+\frac{\mu}{a}\int_{\Omega}\rho_{\varepsilon}^{\gamma-2}|\nabla\rho_{\varepsilon}|^{2}h_{\varepsilon}' (\rho_{\varepsilon})-C. \end{equation} $

将(4.12)式带入(4.8)式, 我们有

(4.13) $ \begin{eqnarray} &&\frac{{\rm d}}{{\rm d}t} \bigg(\int_{\Omega}\frac{\rho_{\varepsilon}|\omega_{\varepsilon}|^{2}}{2}+\frac{\rho_{\varepsilon}^{\gamma}}{a\gamma(\gamma-1)}\bigg) +\frac{\mu}{a}\int_{\Omega}\rho_{\varepsilon}^{\gamma-2}|\nabla\rho_{\varepsilon}|^{2}h_{\varepsilon}' (\rho_{\varepsilon})+\int_{\Omega}\widetilde{p_{\varepsilon}}(\rho_{\varepsilon})|\omega_{\varepsilon}|^{2}\\ &&+\mu\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|A\omega_{\varepsilon}|^{2}+(2\nu-\mu)\int_{\Omega}(h_{\varepsilon}(\rho_{\varepsilon})|D\omega_{\varepsilon}|^{2}+g_{\varepsilon}(\rho_{\varepsilon})|{\rm{div}}\omega_{\varepsilon}|^{2}) \leq C. \end{eqnarray} $

在系统(3.14)的第一个方程两边同乘$ \sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s) $, 即

(4.14) $ \begin{eqnarray} &&\int_{\Omega}\partial_{t}\rho_{\varepsilon}\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s)+\int_{\Omega}{\rm{div}}(\rho_{\varepsilon}\omega_{\varepsilon})\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s)\\ &=&\mu\int_{\Omega}\Delta h_{\varepsilon}(\rho_{\varepsilon})\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s). \end{eqnarray} $

下面我们逐一计算每一项.

(4.14)式左边第一项为

(4.15) $ \begin{eqnarray} \int_{\Omega}\partial_{t}\rho_{\varepsilon}\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s)=-\frac{1}{2}\frac{{\rm d}}{{\rm d}t}\int_{\Omega}K_{\varepsilon}(\rho_{\varepsilon})|\nabla\rho_{\varepsilon}|^{2}. \end{eqnarray} $

计算第二项, 利用Young不等式和Hölder不等式得到

(4.16) $ \begin{eqnarray} &&\int_{\Omega}{\rm{div}}(\rho_{\varepsilon}\omega_{\varepsilon})\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s)\\ & =&-\int_{\Omega}\rho_{\varepsilon}\omega_{\varepsilon}\cdot\nabla(\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s))\\ & =&-\int_{\Omega}\omega_{\varepsilon}\cdot({\rm{div}}(h_{\varepsilon}(\rho_{\varepsilon})\nabla^{2}\phi_{\varepsilon}(\rho_{\varepsilon}))+\nabla(g_{\varepsilon}(\rho_{\varepsilon})\Delta\phi_{\varepsilon}(\rho_{\varepsilon})))\\ & =&\int_{\Omega}\sqrt{h_{\varepsilon} (\rho_{\varepsilon})}\nabla\omega_{\varepsilon}:\sqrt{h_{\varepsilon}(\rho_{\varepsilon})}\nabla^{2}\phi_{\varepsilon}(\rho_{\varepsilon}) +\int_{\Omega}{\rm{div}}\omega_{\varepsilon}g_{\varepsilon}(\rho_{\varepsilon})\Delta\phi_{\varepsilon}(\rho_{\varepsilon})\\ & \leq&\frac{\mu}{2}\int_{\Omega}h_{\varepsilon}(\rho_{\varepsilon})|\nabla^{2}\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}+\frac{1}{2\mu}\int_{\Omega}h_{\varepsilon}(\rho_{\varepsilon})|\nabla\omega_{\varepsilon}|^{2}\\ &&+\frac{\mu}{2}\int_{\Omega}g_{\varepsilon}(\rho_{\varepsilon})|\Delta\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}+\frac{1}{2\mu}\int_{\Omega}g_{\varepsilon}(\rho_{\varepsilon})|{\rm{div}}\omega_{\varepsilon}|^{2}. \end{eqnarray} $

利用(3.13)式, 得到最后一项

(4.17) $ \begin{eqnarray} &&\mu\int_{\Omega}\Delta h_{\varepsilon}(\rho_{\varepsilon})\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s)\\ &=&-\mu\int_{\Omega}\nabla h_{\varepsilon}(\rho_{\varepsilon})\cdot\nabla(\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s))\\ &=&-\mu\int_{\Omega}\frac{\nabla h_{\varepsilon}(\rho_{\varepsilon})}{\rho_{\varepsilon}}\cdot\rho_{\varepsilon}\nabla(\sqrt{K_{\varepsilon}(\rho_{\varepsilon})}\Delta(\int_{0}^{\rho_{\varepsilon}}\sqrt{K_{\varepsilon}(s)}{\rm d}s))\\ &=&-\mu\int_{\Omega}\nabla\phi(\rho_{\varepsilon})\cdot({\rm{div}}(h_{\varepsilon}(\rho_{\varepsilon})\nabla^{2} \phi_{\varepsilon}(\rho_{\varepsilon}))+\nabla(g_{\varepsilon}(\rho_{\varepsilon})\Delta\phi_{\varepsilon}(\rho_{\varepsilon})))\\ &=&\mu\int_{\Omega}h_{\varepsilon}(\rho_{\varepsilon})|\nabla^{2}\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}+g_{\varepsilon}(\rho_{\varepsilon})|\Delta\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}. \end{eqnarray} $

将(4.15)–(4.17)式带入(4.14)式, 得到

(4.18) $ \begin{eqnarray} &&\frac{1}{2}\frac{{\rm d}}{{\rm d}t}\int_{\Omega}K_{\varepsilon}(\rho_{\varepsilon})|\nabla\rho_{\varepsilon}|^{2}+\frac{\mu}{2}\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|\nabla^{2}\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}+\frac{\mu}{2}\int_{\Omega} g_{\varepsilon}(\rho_{\varepsilon})|\Delta\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}\\ & \leq&\frac{1}{2\mu}\int_{\Omega}h_{\varepsilon}(\rho_{\varepsilon})|\nabla\omega_{\varepsilon}|^{2}+\frac{1}{2\mu}\int_{\Omega}g_{\varepsilon}(\rho_{\varepsilon})|{\rm{div}}\omega_{\varepsilon}|^{2}\leq C. \end{eqnarray} $

结合(4.13)和(4.18)式, 得到(4.7)式. 引理4.2得证.

接下来我们的目标是证明以下辅助引理, 它在获取(3.14)的Mellet-Vasseur不等式中起关键作用.

引理4.3  令$ (\rho_{\varepsilon}, u_{\varepsilon}) $是(3.1)的光滑解, 则对任意的$ \beta\in C^{1}({\Bbb R}) $, 我们有

(4.19) $ \begin{eqnarray} &&\frac{{\rm d}}{{\rm d}t}\int_{\Omega}\rho_{\varepsilon}\beta(\frac{|\omega_{\varepsilon}|^{2}}{2})+ \mu\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|A\omega_{\varepsilon}\cdot\omega_{\varepsilon}|^{2}\beta'' (\frac{|\omega_{\varepsilon}|^{2}}{2})+\mu\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|A\omega_{\varepsilon} |^{2}\beta' (\frac{|\omega_{\varepsilon}|^{2}}{2})\\ && +(2\nu-\mu)\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|D\omega_{\varepsilon}|^{2}\beta' (\frac{|\omega_{\varepsilon}|^{2}}{2})+(2\nu-\mu)\int_{\Omega} g_{\varepsilon}(\rho_{\varepsilon}) |{\rm{div}}\omega_{\varepsilon}|^{2}\beta' (\frac{|\omega_{\varepsilon}|^{2}}{2})\\ && +\int_{\Omega} \widetilde{p}_{\varepsilon}(\rho_{\varepsilon})|\omega_{\varepsilon}|^{2}\beta' (\frac{|\omega_{\varepsilon}|^{2}}{2})+(2\nu-\mu)\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon}) |D\omega_{\varepsilon}\cdot\omega_{\varepsilon}|^{2}\beta''(\frac{|\omega_{\varepsilon}|^{2}}{2})\\ &=&-\int_{\Omega} \nabla P_{\varepsilon}(\rho_{\varepsilon})\omega_{\varepsilon}\beta' (\frac{|\omega_{\varepsilon}|^{2}}{2})-2\nu\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon}) (D\omega_{\varepsilon}\cdot\omega_{\varepsilon})(A\omega_{\varepsilon}\cdot\omega_{\varepsilon})\beta'' (\frac{|\omega_{\varepsilon}|^{2}}{2})\\ && -(2\nu-\mu)\int_{\Omega} g_{\varepsilon}(\rho_{\varepsilon}){\rm{div}}\omega_{\varepsilon} \omega_{\varepsilon}(D\omega_{\varepsilon}\cdot\omega_{\varepsilon})\beta''(\frac{|\omega_{\varepsilon}|^{2}}{2}). \end{eqnarray} $

证  证明见文献[4].

有了上述结果, 我们就可以证明Mellet-Vasseur不等式.

引理4.4  令$ (\rho_{\varepsilon}, u_{\varepsilon}) $是(3.1)的光滑解, 则存在一个不依赖于$ \varepsilon $的常数$ C>0 $, 对任意的$ \delta\in(0, 2) $, 使得如下不等式成立

(4.20) $ \begin{eqnarray} &&\sup\limits_t\int_{\Omega}\rho_{\varepsilon}(1+\frac{|\omega_{\varepsilon}|^{2}}{2}){\rm{log}} (1+\frac{|\omega_{\varepsilon}|^{2}}{2}){}\\ &\leq&\int_{0}^{T}\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|\nabla\omega_{\varepsilon}|^{2} +C\int_{0}^{T}\bigg(\int_{\Omega}\bigg(\frac{\rho_{\varepsilon}^{-\frac{\delta}{2}}P_{\varepsilon}^{2} (\rho_{\varepsilon})}{h_{\varepsilon}(\rho_{\varepsilon})} \bigg)^{\frac{2}{2-\delta}}{\rm d}x\bigg)^{\frac{2-\delta}{2}}{}\\ &&\times \bigg(\int_{\Omega}\rho_{\varepsilon} \bigg(1+{\rm{log}}(1+\frac{|\omega_{\varepsilon}|^{2}}{2})\bigg)^{\frac{2}{\delta}}{\rm d}x \bigg)^{\frac{\delta}{2}}{\rm d}t +\int_{\Omega}\rho_{\varepsilon}^{0} (1+\frac{|\omega_{\varepsilon}^{0}|^{2}}{2}){\rm{log}}(1+\frac{|\omega_{\varepsilon}^{0}|^{2}}{2}). \end{eqnarray} $

证  在上述引理4.3中, 令$ \beta(t)=(1+t){\rm{log}}(1+t) $, 然后去掉我们不需要的项, 可以得到

(4.21) $ \begin{eqnarray} &&\frac{{\rm d}}{{\rm d}t}\int_{\Omega}\rho_{\varepsilon}(1+\frac{|\omega_{\varepsilon}|^{2}}{2}){\rm{log}} (1+\frac{|\omega_{\varepsilon}|^{2}}{2})+\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|\nabla\omega_{\varepsilon}|^{2} \bigg(1+\log(1+\frac{|\omega_{\varepsilon}|^{2}}{2})\bigg)\\ &\leq&\bigg|-\int_{\Omega}\nabla P_{\varepsilon}(\rho_{\varepsilon})\omega_{\varepsilon}\beta' (\frac{|\omega_{\varepsilon}|^{2}}{2})\bigg| +C\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|\nabla\omega_{\varepsilon}|^{2}+C\int_{\Omega} |g_{\varepsilon}(\rho_{\varepsilon})||{\rm{div}}\omega_{\varepsilon}||\nabla\omega_{\varepsilon}|. \end{eqnarray} $

下面考虑不等式右边的第一项, 其他项的处理见文献[4].

(4.22) $ \begin{eqnarray} &&\bigg|-\int_{\Omega}\nabla P_{\varepsilon}(\rho_{\varepsilon})\omega_{\varepsilon}\beta' (\frac{|\omega_{\varepsilon}|^{2}}{2})\bigg|\\ &\leq&\bigg|\int_{\Omega} P_{\varepsilon}(\rho_{\varepsilon}){\rm{div}}\omega_{\varepsilon}\beta' (\frac{|\omega_{\varepsilon}|^{2}}{2})\bigg|+ \bigg|\int_{\Omega} P_{\varepsilon}(\rho_{\varepsilon})\omega_{\varepsilon}\beta'' (\frac{|\omega_{\varepsilon}|^{2}}{2})\nabla(\frac{|\omega_{\varepsilon}|^{2}}{2}) \bigg|\\ &\leq&\bigg|\int_{\Omega} P_{\varepsilon}(\rho_{\varepsilon}){\rm{div}}\omega_{\varepsilon}(1+{\rm{log}} (1+\frac{|\omega_{\varepsilon}|^{2}}{2}))\bigg| +2\int_{\Omega} |P_{\varepsilon}(\rho_{\varepsilon})||\nabla \omega_{\varepsilon}|\\ &\leq& C\int_{\Omega}\frac{P_{\varepsilon}^{2}}{h_{\varepsilon}(\rho_{\varepsilon})}(1+{\rm{log}} (1+\frac{|\omega_{\varepsilon}|^{2}}{2}))+\frac{1}{2}\int_{\Omega} h_{\varepsilon} (\rho_{\varepsilon})|\nabla\omega_{\varepsilon}|^{2}(1+{\rm{log}}(1+\frac{|\omega_{\varepsilon}|^{2}}{2}))\\ &&+\int_{\Omega}\frac{P_{\varepsilon}^{2}}{h_{\varepsilon}(\rho_{\varepsilon})}+\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|\nabla\omega_{\varepsilon}|^{2}\\ &\leq& C\int_{\Omega}\frac{P_{\varepsilon}^{2}}{h_{\varepsilon}(\rho_{\varepsilon})}(1+{\rm{log}}(1+\frac{|\omega_{\varepsilon}|^{2}}{2}))+\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|\nabla\omega_{\varepsilon}|^{2}\\ &&+\frac{1}{2}\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|\nabla\omega_{\varepsilon}|^{2}(1+{\rm{log}}(1+\frac{|\omega_{\varepsilon}|^{2}}{2})). \end{eqnarray} $

由Hölder不等式可以得到

(4.23) $ \begin{eqnarray} &&\int_{\Omega}\frac{P_{\varepsilon}^{2}}{h_{\varepsilon}(\rho_{\varepsilon})}(1+{\rm{log}}( 1+\frac{|\omega_{\varepsilon}|^{2}}{2}))\\ &\leq& C\int_{\Omega}(\frac{\rho_{\varepsilon}^{-\frac{\delta}{2}}P_{\varepsilon}^{2}(\rho_{\varepsilon})}{h_{\varepsilon} (\rho_{\varepsilon})})^{\frac{2}{2-\delta}}{\rm d}x)^{\frac{2-\delta}{2}} \bigg(\int_{\Omega}\rho_{\varepsilon}(1+{\rm{log}}(1+\frac{|\omega_{\varepsilon}|^{2}}{2}) )^{\frac{2}{\delta}}{\rm d}x\bigg)^{\frac{\delta}{2}}. \end{eqnarray} $

最后, 通过使用(4.22)和(4.23)式, 我们从(4.21)式得到(4.20)式. 引理4.4得证.

在前一章能量估计的基础之上, 本章节, 我们将推导出主要的一致有界和紧性分析.

考虑引理4.1和(4.6)式, 我们首先得到

(5.1) $ \begin{equation} \begin{array}{ll} { } \sup\limits_t\int_{\Omega}\rho_{\varepsilon}|u_{\varepsilon}|^{2}\leq C, \ \ \ \ \ \ \ \ \ \sup\limits_t\int_{\Omega}|h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}}|^{2}\leq C, \\ { } \sup\limits_t\int_{\Omega}(\rho_{\varepsilon}+\rho_{\varepsilon}^{\gamma})\leq C, \ \ \ \ \ \ \ \ \int_{0}^{T}\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|Du_{\varepsilon}|^{2}\leq C, \\ { } \sup\limits_t\int_{\Omega} f_{\varepsilon}(\rho_{\varepsilon})\leq C, \ \ \ \ \ \ \ \ \ \ \ \ \int_{0}^{T}\int_{\Omega}|\widetilde{p}_{\varepsilon}(\rho_{\varepsilon})||u_{\varepsilon}|^{2}\leq C. \end{array} \end{equation} $

特别地, 利用(3.7)和(5.1)式, 我们可以推断出

(5.2) $ \begin{equation} \sup\limits_t\int_{\Omega}|\nabla\sqrt{\rho_{\varepsilon}}|^{2}\leq C, \ \ \ \ \ \ \ \ \ \int_{0}^{T}\int_{\Omega}\rho_{\varepsilon}|Du_{\varepsilon}|^{2}\leq C. \end{equation} $

通过引理4.2, 我们可以得到

(5.3) $ \begin{equation} \begin{array}{ll} { } \int_{0}^{T}\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|Au_{\varepsilon}|^{2}\leq C, \ \ \ \ \ \ \ \ \ \ \ \ \int_{0}^{T}\int_{\Omega} g_{\varepsilon}(\rho_{\varepsilon})|\Delta\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}\leq C, \\ { } \int_{0}^{T}\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|\nabla^{2}\phi_{\varepsilon}(\rho_{\varepsilon})|^{2}\leq C, \ \ \ \ \ \int_{0}^{T}\int_{\Omega} h_{\varepsilon}' (\rho_{\varepsilon})|\nabla\rho_{\varepsilon}|^{2}\rho^{\gamma-2}\leq C. \end{array} \end{equation} $

结合上述我们得到的(5.1)和(5.3)式, 有

(5.4) $ \begin{equation} \int_{0}^{T}\int_{\Omega} h_{\varepsilon}(\rho_{\varepsilon})|\nabla u_{\varepsilon}|^{2}\leq C, \ \ \ \ \ \ \int_{0}^{T}\int_{\Omega}\rho_{\varepsilon}|\nabla u_{\varepsilon}|^{2}\leq C. \end{equation} $

再次利用(3.7)和(5.3)式, 可以得到

(5.5) $ \begin{equation} \int_{0}^{T}\int_{\Omega}|\nabla\rho_{\varepsilon}^{\frac{\gamma}{2}}|^{2}\leq C. \end{equation} $

为了处理色散项, 我们还需要下列一致有界

(5.6) $ \begin{equation} \int_{0}^{T}\int_{\Omega}|\nabla^{2}\sqrt{\rho_{\varepsilon}}|^{2}+|\nabla\rho_{\varepsilon}^{\frac{1}{4}}|^{4}\leq C. \end{equation} $

注5.1  上述我们得到的一致有界中$ C>0 $是一个常数且不依赖于$ \varepsilon $.

在继续进一步的分析之前, 我们需要推导出主要的收敛性, 这将有助于我们证明我们的结论.

引理5.1  令$ (\rho_{\varepsilon}, u_{\varepsilon}) $是方程(3.1)的解, 则存在一个函数$ \sqrt{\rho} $使得

(5.7) $ \begin{eqnarray} \begin{array}{l} \mbox{在}\ L^{2}(0, T;H^{1}({\Bbb T}^{d}))\ \mbox{中}, \ \sqrt{\rho_{\varepsilon}}\rightarrow \sqrt{\rho}, \\ \mbox{在}\ L^{2}(0, T;H^{1}({\Bbb T}^{d}))\ \mbox{中}, \ h_{\varepsilon}' (\rho_{\varepsilon})\sqrt{\rho_{\varepsilon}}\rightarrow \sqrt{\rho}, \\ \mbox{在}\ L^{2}(0, T;H^{1}({\Bbb T}^{d}))\ \mbox{中}, \ h_{\varepsilon}' (\rho_{\varepsilon})\nabla\sqrt{\rho_{\varepsilon}}\rightarrow \nabla\sqrt{\rho}, \\ \mbox{在}\ L^{2}(0, T;H^{1}({\Bbb T}^{d}))\ \mbox{中}, \ h_{\varepsilon}"(\rho_{\varepsilon})\rho_{\varepsilon}\nabla\sqrt{\rho_{\varepsilon}}\rightarrow 0, \\ \mbox{在}\ L^{2}(0, T;H^{1}({\Bbb T}^{d}))\ \mbox{中}, \ h_{\varepsilon}(\rho_{\varepsilon})-\rho_{\varepsilon}\rightarrow 0, \\ \mbox{在}\ L^{1}((0, T)\times {\Bbb T}^{d})\ \mbox{中}, \ g_{\varepsilon}(\rho_{\varepsilon})\rightarrow 0, \\ \mbox{在}\ L^{1}((0, T)\times {\Bbb T}^{d})\ \mbox{中}, \ p_{\varepsilon}(\rho_{\varepsilon})\rightarrow 0, \\ \mbox{在}\ L^{1}((0, T)\times {\Bbb T}^{d})\ \mbox{中}, \ \widetilde{p}_{\varepsilon}(\rho_{\varepsilon})\rightarrow 0.\\ \end{array} \end{eqnarray} $

证  具体步骤详见文献[4].

现在我们可以证明压力的收敛性了.

引理5.2  令$ (\rho_{\varepsilon}, u_{\varepsilon}) $是方程(3.1)的光滑解, 则对所有的$ r\in[1, 2) $, $ \rho_{\varepsilon}^{\gamma} $在$ L^{\frac{5}{3}}((0, T)\times {\Bbb T}^{3}) $和$ L^{r}((0, T)\times {\Bbb T}^{2}) $中是有界的.

特别地, 在$ L^{1}((0, T)\times {\Bbb T}^{d}) $中, $ P_{\varepsilon}(\rho_{\varepsilon})\rightarrow P(\rho) $.

证  第一步: 证明$ \rho_{\varepsilon}^{\gamma} $的有界性.

不等式(5.5)意味着$ \rho_{\varepsilon}^{\frac{\gamma}{2}}\in L^{2}(0, T;H^{1}({\Bbb T}^{d})) $.

当$ d=2 $时, 由Sobolev嵌入公式可知, 对所有的$ q\in[2, \infty) $, 有$ \rho_{\varepsilon} ^{\frac{\gamma}{2}}\in L^{2}(0, T;L^{q}({\Bbb T}^{2})) $, 即$ \rho_{\varepsilon}^ {\gamma}\in L^{1}(0, T;L^{p}({\Bbb T}^{2})) $, $ p\in[1, \infty) $. 因此, 对所有的$ p\in[1, \infty) $, $ \rho_{\varepsilon}^{\gamma} $在$ L^{1}(0, T;L^{p}({\Bbb T}^{2})) \cap L^{\infty}(0, T;L^{1}({\Bbb T}^{2})) $中是有界的, 故对所有的$ r\in[1, 2) $, $ \rho_{\varepsilon}^{\gamma} $在$ L^{r}((0, T)\times {\Bbb T}^{2}) $中是有界的.

当$ d=3 $时, 使用Sobolev嵌入公式我们可以得到: $ \rho_{\varepsilon}^{\frac{\gamma}{2}}\in L^{2}(0, T;L^{6}({\Bbb T}^{3})) $, 即$ \rho_{\varepsilon}^{\gamma}\in L^{1}(0, T;L^{3} ({\Bbb T}^{3})) $, 由于$ \rho_{\varepsilon}^{\gamma} $在$ L^{\infty}(0, T;L^{1}({\Bbb T}^{3})) $中有界, 并且根据Lebesgue空间的插值公式, 我们有

$ \begin{eqnarray*} ||\rho_{\varepsilon}^{\gamma}||_{L^{\frac{5}{3}}((0, T)\times {\Bbb T}^{3})}\leq|| \rho_{\varepsilon}^{\gamma}||_{L^{\infty}(0, T;L^{1}({\Bbb T}^{3})}^{\frac{2}{5}}||\rho_ {\varepsilon}^{\gamma}||_{L^{1}(0, T;L^{3}({\Bbb T}^{3})}^{\frac{3}{5}}\leq C. \end{eqnarray*} $

第二步: 证明压力的强收敛.

由引理5.1可知$ \rho_{\varepsilon}\mathop{\longrightarrow}\limits^{ \mbox{ a.e. }}\rho $, 因此$ \rho_{\varepsilon}^{\gamma}\mathop{\longrightarrow}\limits^{ \mbox{ a.e. }}\rho^{\gamma} $. 又因为压力满足式(1.3), 积分, 可得: $ |P_{\varepsilon}(\rho_{\varepsilon})|\leq C(\rho_ {\varepsilon}^{\gamma}+\rho_{\varepsilon}) $. 因此, 由引理的第一部分可知: 在$ L^{1}((0, T)\times {\Bbb T}^{d})) $中, $ P_{\varepsilon}(\rho_{\varepsilon})\rightarrow P(\rho) $. 引理5.2得证.

对于动量的收敛, 我们有如下引理.

引理5.3  令$ (\rho_{\varepsilon}, u_{\varepsilon}) $是系统(3.1)的解, 则在$ L^{2}(0, T;L^{p}({\Bbb T}^{d})) $中, $ m_{1, \varepsilon}=\rho_{\varepsilon}u_{ \varepsilon}\rightarrow m_{1} $, 其中$ p\in[1, \frac{3}{2}) $.

证  证明步骤直接参照文献[4].

引理5.4  令$ (\rho_{\varepsilon}, u_{\varepsilon}) $是系统(3.1)的解, 则在$ L^{2}((0, T)\times {\Bbb T}^{d}) $中, $ \sqrt{\rho_{\varepsilon}}u_{\varepsilon}\rightarrow \sqrt{\rho}u $, 这里我们定义$ u $为

$ \begin{eqnarray*} u=\left\{ \begin{array}{ll} { } \frac{m}{\rho}, {\quad} & {\rho>0}, \\ 0, &{\rho=0}. \end{array} \right. \end{eqnarray*} $

证  与文献[4]证明类似, 我们只关注不同的部分.

首先考虑Mellet-Vasseur不等式(4.20), 令$ \delta>0 $充分小, 根据(5.1)和(5.4)式(一致有界性), 我们可以推导出

$ \begin{eqnarray*} \sup\limits_t\int_{\Omega}\rho_{\varepsilon}(1+\frac{|\omega_{\varepsilon}|^{2}}{2}){\rm{log}}(1+\frac{ |\omega_{\varepsilon}|^{2}}{2}) &\leq& C+C\int_{0}^{T} \bigg(\int_{\Omega}(\frac{\rho_{\varepsilon}^{- \frac{\delta}{2}}P_{\varepsilon}^{2}(\rho_{\varepsilon})}{h_{\varepsilon}(\rho_{\varepsilon})} )^{\frac{2}{2-\delta}}{\rm d}x\bigg)^{\frac{2-\delta}{2}}{\rm d}t\nonumber\\ &\leq& C+C\int_{0}^{T} \bigg(\int_{\Omega}(\rho_{\varepsilon}^{2\gamma-\frac{\delta}{2}-1}+\rho_{\varepsilon} ^{1-\frac{\delta}{2}})^{\frac{2}{2-\delta}}{\rm d}x\bigg)^{\frac{2-\delta}{2}}{\rm d}t. \end{eqnarray*} $

然后, 利用引理5.2, 当$ d=2 $时, 不等式右边在没有任何条件限制情况下是属于$ L^{1}(0, T) $的; 当$ d=3 $时, 在条件$ 2\gamma-1<\frac{5}{3}\gamma $, 即$ \gamma<3 $时不等式右边也是属于$ L^{1}(0, T) $的. 因此, 我们得到

(5.8) $ \begin{equation} \sup\limits_t\int_{\Omega}\rho_{\varepsilon}(1+\frac{|\omega_{\varepsilon}|^{2}}{2}){\rm{log}}(1+\frac{| \omega_{\varepsilon}|^{2}}{2})\leq C. \end{equation} $

其余部分的证明见文献[4].

由上面得到的一致有界和紧性分析的结果, 利用引理4.3, 可以充分得到逼近解系统(3.1)的下界和上界, 与文献[4]的引理11和引理12相同.然后利用拟线性抛物型方程的经典方法, 得到逼近系统(3.1)光滑解的存在性.结合逼近解系统光滑解的存在性和紧性, 定理2.1得证.