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数学物理学报, 2021, 41(3): 629-641 doi:

论文

三维Navier-Stokes-Korteweg方程组的整体大解

于洋海,1, 李金禄,2,3, 吴星,4

A Class of Global Large Solutions to 3D Navier-Stokes-Korteweg Equations

Yu Yanghai,1, Li Jinlu,2,3, Wu Xing,4

通讯作者: 吴星, ny2008wx@163.com

收稿日期: 2020-06-11  

基金资助: 安徽省自然科学基金.  1908085QA05
安徽师范大学博士科研启动基金
国家自然科学基金.  11801090
中国博士后科学基金.  2020T130129
中国博士后科学基金.  2020M672565

Received: 2020-06-11  

Fund supported: the NSF of Anhui Province.  1908085QA05
the PhD Scientific Research Start-up Foundation of Anhui Normal University
the NSFC.  11801090
the Postdoctoral Science Foundation of China.  2020T130129
the Postdoctoral Science Foundation of China.  2020M672565

作者简介 About authors

于洋海,E-mail:yuyanghai214@sina.com , E-mail:yuyanghai214@sina.com

李金禄,E-mail:lijinlu@gnnu.cn , E-mail:lijinlu@gnnu.cn

Abstract

In this paper, we consider the Cauchy problem to the tri-dimensional compressible Navier-Stokes-Korteweg system with a specific choice on the Korteweg tensor in R3, and construct the global solutions to the tri-dimensional Navier-Stokes-Korteweg equations with a class of of large initial data, where the L norm of the initial velocity and the third component of initial vorticity could be arbitrarily large.

Keywords: Navier-Stokes-Korteweg ; Large solutions

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本文引用格式

于洋海, 李金禄, 吴星. 三维Navier-Stokes-Korteweg方程组的整体大解. 数学物理学报[J], 2021, 41(3): 629-641 doi:

Yu Yanghai, Li Jinlu, Wu Xing. A Class of Global Large Solutions to 3D Navier-Stokes-Korteweg Equations. Acta Mathematica Scientia[J], 2021, 41(3): 629-641 doi:

1 引言

本文主要研究三维可压缩Navier-Stokes-Korteweg方程组(记为NSK) 的Cauchy问题(该方程组的推导, 参见文献[2-3, 6])

{tρ+div(ρu)=0,t(ρu)+div(ρuu)div(2μ(ρ)Du)(λ(ρ)divu)+P(ρ)=divK,(u,ρ)|t=0=(u0,ρ0),
(1.1)

其中u=(u1(t,x),u2(t,x),u3(t,x))R3ρ=ρ(t,x)R+分别表示流体的速度和密度. 依赖密度的函数μ(ρ)λ(ρ)分别表示流体的剪切(第一) 粘性系数和体积(第二) 粘性系数, 并且满足强抛物性条件

μ>02μ+λ>0.

Korteweg张量divK描述了两相流体(一般指液体与气体混合) 在交接面处密度的变化, 形式为

divK=(ρκ(ρ)Δρ+κ(ρ)+ρκ(ρ)2|ρ|2)div(κ(ρ)ρρ),
(1.2)

其中光滑函数κ表示毛细管系数. 压强P是关于密度ρ的充分光滑的递增函数. 我们特别注意到一类典型的压强P(ρ)服从如下规律

P(ρ)=Aργ,
(1.3)

其中A称为熵常数, γ1称为绝热指数.

不失一般性, 我们假设A=1且选取平衡态

lim

当忽略Korteweg张量 {\mathord{{{\rm{div}}}}}{\Bbb K} , (1.1)退化为经典的Navier-Stokes方程(记为NS). Kazhikhov等[21]研究了具光滑初值的一维NS方程, 首次证明了全局解的存在性. 对于一维情形, 参见Serre[28-29], Hoff[13]和Mellet[24]等的结果. Matsumura等[26]首次考虑了高维情形. 确切地说, 他们在Sobolev空间的框架下证明了三维NS方程存在整体光滑解, 其中初值 (\bar{\varrho}_0, u_0)\in H^3\times H^3 且充分接近非真空平衡态 (\bar{\varrho}, 0) . 对于不连续初值情形, 参见Hoff的系列工作[14-16].

当在粘性可压缩NS方程中考虑毛细管现象, 即Korteweg张量 {\mathord{{{\rm{div}}}}}{\Bbb K} 被引入, 可得到NSK系统(1.1). 该系统描述了丰富且复杂的物理现象, 又具有数学理论上的挑战性, 被大量数学工作者们广泛研究. 接下来, 我们回顾一些与本文研究相关的结果. Hattori等[7-8]建立了方程组(1.1)–(1.2) 在Sobolev空间中解的局部存在性及光滑解的整体存在性. Kotschote[20]证明了有界域上局部强解的存在性. Haspot[9]证明了当粘性系数 \mu(\rho) , \lambda(\rho) 及压强 P(\rho) 线性地依赖于密度且毛细管系数为 \kappa(\rho) = \frac{\kappa}{\rho} 时, 该问题在 L^2 框架下存在整体小解. 随后, Haspot[10]研究在 \big(\mu(\rho), \lambda(\rho), P(\rho) \big) = (\mu{\rho}, 0, \rho) 时, 方程组(1.1)–(1.2)在 L^p 框架下存在整体解. 值得注意的是密度的特殊选取扮演着极其重要的角色. 对于整体弱解及解的大时间行为, 读者可参见文献[1, 12, 19, 27, 32]及其所引文献. 在上述提到的结果中, 第二粘性系数 \lambda(\rho) 至少是零或非零常数. 然而, 当第二粘性系数依赖密度时, 相应的结果却很少. 受Haspot工作[10]的启发, 我们研究具特殊第二粘性系数的方程组(1.1)–(1.2)的Cauchy问题, 即

\lambda(\rho) = \rho^2+({\kappa}-2\mu) \rho,

并期望对某类大初值, 该问题存在整体解.

尽管解的局部适定性及小初值解的整体适定性理论相对丰富, 但是对于一般的初值, 解的整体适定性结果还不尽如人意. 针对这条线, 我们回顾一些相近的大初值时解整体存在的有趣工作. 对于二维等熵的可压NS方程, 值得一提的是Vaigant等[31]的杰出工作, Vaigant及其合作者证明对任意光滑的初值, 在第一粘性系数 \mu(\rho) 是正常数且第二粘性系数 \lambda(\rho) 有形如 \rho^{\beta}(\beta>3) 的增长假设下, 二维等熵的可压NS方程存在整体强解. Huang等[17]和Jiu等[18]将上述指标改进为 \beta>\frac{4}{3} . Haspot[11]研究了方程组(1.1)–(1.2) 的Cauchy问题, 其中粘性系数满足 (\mu(\rho), \lambda(\rho)) = (\mu \rho, 0) , 证明了该方程组存在整体强解, 其中允许初始速度的无旋部分可以任意大. 而当第二粘性系数为常数且充分大时, 对任意大初值速度场和恒常初始密度, Danchin等[4]证明了二维可压NS方程存在整体光滑解. Fang等[5]证明了具特殊初值的等熵的可压NS方程存在整体解. Lei等[22]得到了三维不可压NS方程的一类整体解, 其中初值的 L^{2} 范数允许任意大. Li等[23]构造了三维不可压Hall-MHD方程的一类整体解, 其中初值的 L^{\infty} 范数可以任意大. 本文考虑方程组(1.1)–(1.2) 的Cauchy问题, 在第二粘性系数满足 \lambda(\rho) = \rho^2+({\kappa}-2\mu) \rho 时, 我们期望某类大初值也能产生该问题的整体解.

1.1 原系统的变形

研究允许真空的可压缩流体方程组的主要困难在于动量方程会失去抛物正则效应. 为避免真空出现, 本文假设初始密度 \rho_0 在平衡态 \bar{\rho} = 1 附近作小扰动. 我们指出该平衡态 \bar{\rho} = 1 的假定, 仅仅是为方便起见, 不是本质的.

由系数的特殊选择

\begin{equation} \mu( \rho) = \mu \rho, \quad\kappa(\rho) = \kappa>0\quad\rm{和}\quad\lambda(\rho) = \rho^2+(\kappa-2\mu) \rho, \end{equation}
(1.4)

我们从(1.2) 式得 {\mathord{{{\rm{div}}}}}{\Bbb K} = \kappa\rho\nabla\Delta\rho. \varrho = \rho-1 , 通过有效速度变换 v = u+\nabla \rho , 只要 \rho\neq0 , 我们可将方程组(1.1)–(1.2) 等价转化为(更多细节推导, 参见附录)

\begin{equation} \left\{\begin{array}{ll} \partial_t \varrho -\Delta\varrho+{\mathord{{{\rm{div}}}}} v = {\mathord{{{\rm{div}}}}}(\varrho \nabla \varrho)-{\rm{div}}(\varrho v), \\ \partial_t v-\mu \Delta v-(\kappa-\mu)\nabla{\mathord{{{\rm{div}}}}} v+ \nabla \varrho = -v\cdot\nabla v+K(\varrho)\nabla\varrho\\ \; \; \; \; + \nabla \varrho \cdot (S_1^1v+S_3^2\varrho) + \nabla\ln(1+\varrho) \cdot (S_2^1v-S_4^2 \varrho), \\ (\varrho, v)|_{t = 0} = (\varrho_0, v_{0}), \end{array}\right. \end{equation}
(1.5)

其中, 我们令

S_1^1 = (\nabla- \nabla^{\mathit{T}}+{\Bbb Id}{\mathord{{{\rm{div}}}}}), \quad S_2^1 = 2\mu{\Bbb D}+(\kappa-2\mu){\Bbb I}{\mathord{{{\rm{div}}}}}, \quad S_3^2 = ( \nabla^2-{\Bbb I}\Delta),

S_4^2 = \big(2\mu{\Bbb D}+(\kappa-2\mu){\Bbb I}{\mathord{{{\rm{div}}}}} \big)\nabla, \quad K(s) = 1- \frac{P'(1+s)}{1+s},

这里我们不妨假设 P'(1) = 1 .

a = e^{{\mu} t\Delta}a_0 是如下热方程的解

\begin{equation} \partial_ta-{\mu} \Delta a = 0, \quad a|_{t = 0} = a_0. \end{equation}
(1.6)

U = \big( \partial_2a, \;- \partial_1a, \; ( \partial_2- \partial_1)a \big)^{\mathit{T}}, 我们发现 U 是如下齐次热方程的解

\partial_tU-\mu \Delta U = 0, \quad U|_{t = 0} = U_0.

引入新的变量 {\bf u} = v-U , 方程组(1.5) 可转化为

\begin{eqnarray} \left\{\begin{array}{ll} \partial_t \varrho -\Delta\varrho+{\mathord{{{\rm{div}}}}} {\bf u} = {\mathord{{{\rm{div}}}}}(\varrho \nabla \varrho)-{\rm{div}}(\varrho {\bf u})-{\rm{div}}(\varrho U)-{\rm{div}} U, \\ \partial_t {\bf u}-\mu \Delta {\bf u}-(\kappa-\mu) \nabla {\mathord{{{\rm{div}}}}} {\bf u}+ \nabla \varrho = {\bf F}, \\ (\varrho, {\bf u})|_{t = 0} = (\rho_0-1, v_0-U_0), \end{array}\right. \end{eqnarray}
(1.7)

其中

\begin{eqnarray*} {\bf F}& = &- {\bf u} \cdot \nabla {\bf u}- {\bf u} \cdot \nabla U-U \cdot \nabla {\bf u}-U \cdot \nabla U+K(\varrho) \nabla \varrho+ \nabla \varrho \cdot S_3^2\varrho\\ &&- \nabla\ln(1+\varrho) \cdot S_4^2 \varrho+ \nabla \varrho \cdot S_1^1( {\bf u}+U)+ \nabla\ln(1+\varrho) \cdot S_2^1( {\bf u}+U). \end{eqnarray*}

1.2 主要结果

我们的主要目标是构造具某类大初值的方程组(1.1)–(1.2) 的整体光滑解. 从现在开始, 我们考虑新的等价方程组(1.7) 而不再关注原方程组(1.1)–(1.2).

该文的主要结果陈述如下.

定理1.1   假设初值 (\varrho_0, v_0) 满足

\varrho_0 = \rho_0-1 = 0, \quad\quad v_0 = U_0 = \big( \partial_2a_0, \;- \partial_1a_0, \; ( \partial_2- \partial_1)a_0 \big)^{\mathit{T}},

其中

{\rm supp} \ \hat{a}_0(\xi)\subset{\cal C}\triangleq\Big\{\xi = (\xi_1, \xi_2, \xi_3): \ |\xi_1-\xi_2|\leq \varepsilon, \; 1\leq |\xi|^2\leq2\Big\}, \quad 0< \varepsilon\ll1.

若存在充分小的正常数 \delta 和一致常数 C 使得

\begin{eqnarray} \varepsilon\|a_0\|_{L^2} \big(1+\|\hat{a}_0\|_{L^1} \big)\exp \Big(C \big(1+\|\hat{a}_0\|_{L^1}+ \varepsilon\|{a}_0\|_{L^2} \big)\|\hat{a}_0\|_{L^1} \Big)\leq \delta, \end{eqnarray}
(1.8)

则方程组(1.7) 存在唯一整体解.

注1.1   定理1.1表明具某类特殊大初值的方程组(1.7) 存在整体解. 我们可以构造出该类初值. 事实上, 令

\hat{a}_0(\xi_1, \xi_2, \xi_3) = \varepsilon^{-1} \Big(\log\log\frac1 \varepsilon \Big)^{\frac12} \hat{\phi}(\xi_1, \xi_2)\hat{\psi}(\xi_3),

其中偶函数 \hat{\phi}\in {\cal C}^\infty_0({\Bbb R}^2), \hat{\psi}\in {\cal C}^\infty_0({\Bbb R}) 均取值于 [0, 1] 且满足

{\rm supp} \hat{\phi}\subset \Big\{\xi_h = (\xi_1, \xi_2): \ |\xi_1-\xi_2|\leq \varepsilon, \ \frac{\sqrt{2}}2\leq|\xi_h|\leq 1\Big\},

{\rm supp} \hat{\psi}\subset\Big\{\xi_3: \ \frac{\sqrt{2}}2\leq|\xi_3|\leq1\Big\},

\hat{\phi}(\xi_1, \xi_2) = \hat{\psi}(\xi_3)\equiv1, \quad \big(|\xi_h|, |\xi_3| \big)\in \Big[\frac{\sqrt{3}}2, \frac{2\sqrt{2}}3 \Big]^2.

直接计算得

\int_{ {\Bbb R}^2}\hat{\phi}(\xi_1, \xi_2) {{\rm{d}}}\xi\approx \varepsilon \quad\rm{和}\quad\int_{ {\Bbb R}} \hat{\psi}(\xi_3) {{\rm{d}}}\xi_3 \approx1.

从而我们有

\|\hat{a}_0\|_{L^1}\approx \Big(\log\log\frac1 \varepsilon\Big)^\frac12\quad\rm{和}\quad\|{a}_0\|_{L^2}\approx \varepsilon^{- \frac12}\Big(\log\log\frac1 \varepsilon\Big)^\frac12,

这意味着

(1.8){\rm{ 式左端}} = C \varepsilon^{\frac12} \Big(\log\log \frac1 \varepsilon \Big)\exp \Big(C\log\log \frac1 \varepsilon \Big).

因此, 只要选择 \varepsilon 足够小, 我们可推出方程组(1.7) 存在唯一整体解.

然而, 我们注意到初始涡度的第三分量 w_0 = \nabla\times U_0 满足

w_0^3 = \partial_1U^2_0- \partial_2U^1_0 = -( \partial_1^2+ \partial_2^2)a_0,

取其 \rm Fourier 变换得

\widehat{w_0^3}(\xi) = (\xi^2_1+\xi^2_2)\hat{a}_0(\xi)\geq0.

所以我们推出

\|w_0^3\|_{L^\infty}\approx \big\|\widehat{w_0^3} \big\|_{L^1} = \big\|(\xi^2_1+\xi^2_2)\hat{a}_0 \big\|_{L^1}\approx \Big(\log\log \frac1 \varepsilon \Big)^\frac12.

进一步, 我们还有

\|U_0\|_{L^\infty}\gtrsim\|\nabla\times U_0\|_{L^\infty}\gtrsim \Big(\log\log \frac1 \varepsilon \Big)^\frac12.

即说明初始速度和初始涡度的第三分量的 L^{\infty} 范数可以任意大, 方程组(1.7) 存在相应的唯一整体解.

2 预备知识

首先, 我们引入一些后文需要的符号.

\bullet 本文中的常数 C 是随估计变化的, 与时间无关.

\bullet 对某个Banach空间 X , 我们简记 \|f_1, \cdots, f_n\|_{X}\triangleq\|f_1\|_{X}+\cdots+\|f_n\|_{X} .

\bullet 给定多重指标 \alpha = (\alpha_1, \alpha_2, \alpha_3)\in {\Bbb N}^3 , 微分算子 D^{\alpha}\triangleq\frac{ \partial^{|\alpha|}}{ \partial^{\alpha_1}_{x_1} \partial^{\alpha_2}_{x_2} \partial^{\alpha_3}_{x_3}}, \;|\alpha| = \alpha_1+\alpha_2+\alpha_3 .

\bullet \langle f, g \rangle 表示 L^2( {\Bbb R}^3) 内积, 即 \langle f, g \rangle\triangleq\int_{ {\Bbb R}^3}fg {{\rm{d}}} x .

\bullet 函数 f 的Fourier变换定义为

\hat{f}(\xi)\triangleq\int_{ {\Bbb R}^3}e^{-{\rm i}x \cdot\xi}f(x) {{\rm{d}}} x.

\bullet 整数阶Sobolev空间 H^m( {\Bbb R}^3) ( m\in {\Bbb N} ) 的范数定义为

\|f\|_{H^m( {\Bbb R}^3)}\triangleq \Big(\sum\limits_{|\alpha|\leq m}\|D^{\alpha}f\|^2_{L^2( {\Bbb R}^3)} \Big)^{ \frac12} = \|f\|_{\dot{H}^m( {\Bbb R}^3)}+\|f\|_{L^2( {\Bbb R}^3)}.

下面的估计会在定理的证明中多次用到.

引理2.1[25] (交换子估计)  

\sum\limits_{0<|\alpha|\leq 3}\|D^{\alpha}(fg)-(D^\alpha f)g\|_{L^2}\leq C \big(\|f\|_{{H}^{2}}\| \nabla g\|_{L^\infty}+\|f\|_{L^\infty}\|g\|_{{H}^3} \big),

\sum\limits_{0<|\alpha|\leq 3}\|D^{\alpha}(fg)-(D^\alpha f)g\|_{L^2}\leq C \big\| \nabla g, \nabla^3 g \big\|_{L^\infty}\|f\|_{H^2}.

引理2.2[25] (乘积估计)   设 m\in {\Bbb Z}^+ m\geq2 , 成立

\sum\limits_{|\alpha|\leq m}\|D^{\alpha}(fg)\|_{L^2}\leq C\|f, \nabla^mf \big\|_{L^\infty}\|g\|_{H^m}, \; \; \sum\limits_{|\alpha|\leq m}\|D^{\alpha}(fg)\|_{L^2}\leq C\|f\|_{H^m}\|g\|_{H^m}.

引理2.3[30] (复合估计)   设 m\in {\Bbb Z}^+ f\in \dot{H}^m\cap L^{\infty} . F\in W_{loc}^{m+2, \infty} F(0) = 0 , 则

\|F(f)\|_{\dot{H}^m}\leq C(M)\|f\|_{\dot{H}^m},

其中常数 C(M) 依赖 M\triangleq\sup\limits_{k\leq m+2, |t|\leq \|f\|_{L^{\infty}}}\|F^{(k)}(t)\|_{L^{\infty}}.

引理2.4   在定理1.1的假设下, 对任意的 m\in{\Bbb Z}^+\cup\{0\} , 我们有

\|\nabla^m U\|_{L^{\infty}}\leq Ce^{-\mu t}\|\hat{a}_0\|_{L^1},

\|{\mathord{{{\rm{div}}}}} U\|_{H^m}\leq Ce^{-\mu t} \varepsilon\|a_0\|_{L^2},

\|U \cdot \nabla U\|_{H^m}\leq Ce^{-\mu t} \varepsilon\|a_0\|_{L^2}\|\hat{a}_0\|_{L^1}.

  注意到 a = e^{\mu t\Delta}a_0 , 由条件 {\rm supp}\ \hat{a}_0(\xi)\subset{\cal C} , 我们得

\begin{eqnarray} \|( \partial_2- \partial_1)a\|_{H^m}\leq Ce^{-\mu t} \varepsilon\|a_0\|_{L^2}, \end{eqnarray}
(2.1)

这就给出

\|{\mathord{{{\rm{div}}}}} U\|_{H^m} = \|( \partial_2- \partial_1) \partial_3a\|_{H^m}\leq Ce^{-\mu t} \varepsilon\|a_0\|_{L^2}.

\|{\bf f}\|_{L^\infty}\leq C\|\hat{{\bf f}}\|_{L^{1}} , 我们有

\begin{eqnarray} \|\nabla^m a\|_{L^\infty}\leq Ce^{-\mu t} \big\||\xi|^m\hat{a}_0 \big\|_{L^{1}}\leq Ce^{-\mu t}\|\hat{a}_0\|_{L^{1}}. \end{eqnarray}
(2.2)

上式意味着

\|\nabla^m U\|_{L^\infty}\leq C\|\nabla^{m+1} a\|_{L^\infty}\leq Ce^{-{\mu}t}\|\hat{a}_0\|_{L^{1}}.

直接计算得

\begin{eqnarray*} U \cdot \nabla U^1& = &(U^1+U^2) \partial_1U^1-U^2( \partial_1- \partial_2)U^1+U^3 \partial_3U^1 \\& = &( \partial_2- \partial_1)a\cdot \partial_1 \partial_2a- \partial_1a\cdot( \partial_2- \partial_1) \partial_2a+( \partial_2- \partial_1)a\cdot \partial_3 \partial_2a \\& = &( \partial_2- \partial_1)a\cdot( \partial_1+ \partial_3) \partial_2a- \partial_1a\cdot( \partial_2- \partial_1) \partial_2a, \\U \cdot \nabla U^2& = &(U^1+U^2) \partial_1U^2-U^2( \partial_1- \partial_2)U^2+U^3 \partial_3U^2 \\& = &-( \partial_2- \partial_1)a\cdot \partial_1 \partial_1a+ \partial_1a\cdot( \partial_2- \partial_1) \partial_1a-( \partial_2- \partial_1)a\cdot \partial_3 \partial_1a \\& = &-( \partial_2- \partial_1)a\cdot( \partial_1+ \partial_3) \partial_1a+ \partial_1a\cdot( \partial_2- \partial_1) \partial_1a, \\U \cdot \nabla U^3& = &U^1 \cdot \partial_1U^3+U^2 \cdot \partial_2U^3+U^3 \cdot \partial_3U^3 \\& = & \partial_2a\cdot( \partial_2- \partial_1) \partial_1a- \partial_1a\cdot( \partial_2- \partial_1) \partial_2a+( \partial_2- \partial_1)a\cdot( \partial_2- \partial_1) \partial_3a \\& = & \partial_2a\cdot( \partial_2- \partial_1)( \partial_1+ \partial_3)a- \partial_1a\cdot( \partial_2- \partial_1)( \partial_2+ \partial_3)a. \end{eqnarray*}

由引理2.2 (2.1)式和(2.2)式, 我们有

\|U \cdot \nabla U^1\|_{H^3}\leq C\|( \partial_2- \partial_1)a\|_{H^{4}}\|a, \nabla^5a\|_{L^{\infty}} \leq Ce^{-\mu t} \varepsilon\|a_0\|_{L^2}\|\hat{a}_0\|_{L^1}.

类似地, 我们还有

\|U \cdot \nabla U^2, \;U \cdot \nabla U^3\|_{H^3}\leq Ce^{-\mu t} \varepsilon\|a_0\|_{L^2}\|\hat{a}_0\|_{L^1}.

于是我们完成引理2.4的证明.

3 定理1.1的证明

定理1.1证明的核心是建立解的先验估计 \|\varrho, {\bf u}\|_{H^3( {\Bbb R}^3)} , 这是本节的主要目标.

我们定义

\begin{equation} \Gamma\triangleq\sup \Big\{t\in[0, T^*):\sup\limits_{\tau\in[0, t]}\|(\varrho, {\bf u})(\tau)\|^2_{H^3}\leq \eta\ll1 \Big\}. \end{equation}
(3.1)

由Sobolev不等式, 得

\begin{equation} \sup\limits_{\tau\in[0, t]}\|\varrho(\tau)\|_{L^\infty}\leq C\eta\leq\frac12, \quad t\in[0, \Gamma], \end{equation}
(3.2)

其中 \eta 是一个待定的充分小的正常数.

接下来, 我们给出定理1.1的证明.

   首先, 将算子 D^\alpha 作用到方程 (1.7)_1 两边, 取其与 D^\alpha \varrho 的内积, 然后将 \sum\limits_{|\alpha|\leq 3} 作用到所得的式子, 合并后我们有

\begin{eqnarray} &&\frac12\frac{ {{\rm{d}}}}{ {{\rm{d}}} t}\|\varrho\|^2_{H^3}+\|\nabla\varrho\|^2_{H^3}+\sum\limits_{|\alpha|\leq 3} \langle D^\alpha{\mathord{{{\rm{div}}}}} {\bf u}, D^\alpha \varrho \rangle \\ & = &-\sum\limits_{|\alpha|\leq 3} \langle D^\alpha(\varrho \nabla \varrho), D^\alpha \nabla\varrho \rangle+\sum\limits_{|\alpha|\leq 3} \langle D^\alpha(\varrho {\bf u}), D^\alpha \nabla\varrho \rangle\\ &&+\sum\limits_{|\alpha|\leq 3} \langle D^\alpha(\varrho U), D^\alpha \nabla\varrho \rangle-\sum\limits_{|\alpha|\leq 3} \langle D^\alpha{\mathord{{{\rm{div}}}}} U, D^\alpha \varrho \rangle \triangleq\sum\limits_{i = 1}^4I_i. \end{eqnarray}
(3.3)

对于 I_1 , 我们可改写为

I_1 = \sum\limits_{|\alpha|\leq 3} \langle D^\alpha(\varrho \nabla \varrho)- D^{\alpha}\varrho\nabla\varrho, D^\alpha \nabla\varrho \rangle+\sum\limits_{|\alpha|\leq 3} \langle D^{\alpha}\varrho\nabla\varrho, D^\alpha \nabla\varrho \rangle\triangleq I_{1, 1}+I_{1, 2}.

由Sobolev嵌入 H^2( {\Bbb R}^3)\hookrightarrow L^\infty( {\Bbb R}^3) , Hölder不等式和引理2.1得

|I_{1, 1}|\leq C \big(\|\varrho\|_{H^2}\| \nabla^2 \varrho\|_{L^\infty}+\| \nabla \varrho\|_{H^3}\|\varrho\|_{L^\infty} \big)\| \nabla \varrho\|_{H^3}\leq C\|\varrho\|_{H^2}\| \nabla \varrho\|_{H^3}^2,

|I_{1, 2}|\leq C \|\nabla\varrho\|_{L^\infty}\|\varrho\|_{H^3}\| \nabla \varrho\|_{H^3}\leq C\|\varrho\|_{H^3}\| \nabla \varrho\|_{H^3}^2,

这就给出

\begin{eqnarray} |I_1|\leq C\|\varrho\|_{H^3}\| \nabla \varrho\|_{H^3}^2. \end{eqnarray}
(3.4)

对于 I_2 , 由Hölder不等式和引理2.2得

\begin{eqnarray} |I_2|\leq \big(\|\varrho {\bf u}\|_{L^2}+\|\varrho {\bf u}\|_{\dot{H}^3} \big)\|\nabla\varrho\|_{H^3} \leq C\|\varrho, {\bf u}\|_{H^3}\| \nabla \varrho\|_{H^3}^2. \end{eqnarray}
(3.5)

对于 I_3 , 由Hölder不等式和引理2.2得

\begin{eqnarray} |I_3|\leq\|\varrho U\|_{H^3}\|\nabla\varrho\|_{H^3} \leq C\|U, \nabla^3 U\|^2_{L^\infty}\|\varrho\|^2_{H^3}+ \frac12\| \nabla \varrho\|_{H^3}^2. \end{eqnarray}
(3.6)

对于 I_4 , 由Hölder不等式得

\begin{eqnarray} |I_4|\leq\|{\mathord{{{\rm{div}}}}} U\|_{H^3}\|\varrho\|_{H^3}. \end{eqnarray}
(3.7)

将上述估计(3.4)–(3.7) 代入(3.3) 得

\begin{eqnarray} &&\frac{ {{\rm{d}}}}{ {{\rm{d}}} t}\|\varrho\|^2_{H^3}+\|\nabla\varrho\|^2_{H^3}+2\sum\limits_{|\alpha|\leq 3} \langle D^\alpha {\mathord{{{\rm{div}}}}} {\bf u}, D^\alpha\varrho \rangle\\ &\leq &C\|\varrho, {\bf u}\|_{H^3}\| \nabla \varrho\|_{H^3}^2+C\|U, \nabla^3 U\|^2_{L^\infty}\|\varrho\|^2_{H^3}+\|{\mathord{{{\rm{div}}}}} U\|_{H^3}\|\varrho\|_{H^3}. \end{eqnarray}
(3.8)

其次, 注意到如下事实

\mu_1\int_{ {\Bbb R}^3}|\nabla f|^2 {{\rm{d}}} x+\mu_2\int_{ {\Bbb R}^3}|{\mathord{{{\rm{div}}}}} f|^2 {{\rm{d}}} x\geq \min(\mu_1, \mu_1+\mu_2 \big)\int_{ {\Bbb R}^3}|\nabla f|^2 {{\rm{d}}} x,

再将算子 D^\alpha 作用到方程 (1.7)_2 两端, 取其与 D^\alpha {\bf u} 的内积, 然后将 \sum\limits_{|\alpha|\leq 3} 作用于所得的式子, 合并后我们有

\begin{eqnarray} \frac12\frac{ {{\rm{d}}}}{ {{\rm{d}}} t}\| {\bf u}\|^2_{H^3}+\min(\mu, \kappa \big) \| \nabla {\bf u}\|^2_{H^3}+\sum\limits_{|\alpha|\leq 3} \langle D^\alpha \nabla \varrho, D^\alpha {\bf u} \rangle\triangleq\sum^{8}_{i = 1}J_i, \end{eqnarray}
(3.9)

其中

J_1 = -\sum\limits_{|\alpha|\leq 3} \langle D^\alpha( {\bf u} \cdot \nabla {\bf u}), D^\alpha {\bf u} \rangle, \quad J_2 = -\sum\limits_{|\alpha|\leq 3} \langle D^\alpha(U \cdot \nabla {\bf u}), D^\alpha {\bf u} \rangle,

J_3 = -\sum\limits_{|\alpha|\leq 3} \langle D^\alpha( {\bf u} \cdot \nabla U), D^\alpha {\bf u} \rangle, \quad J_4 = -\sum\limits_{|\alpha|\leq 3} \langle D^\alpha(U \cdot \nabla U), D^\alpha {\bf u} \rangle,

J_5 = \sum\limits_{|\alpha|\leq 3} \big \langle D^\alpha \big( \nabla \varrho \cdot S_3^2\varrho \big), D^\alpha {\bf u} \big \rangle-\sum\limits_{|\alpha|\leq 3} \big \langle D^\alpha \big( \nabla\ln(1+\varrho) \cdot S_4^2 \varrho \big), D^\alpha {\bf u} \big \rangle,

J_6 = \sum\limits_{|\alpha|\leq 3} \big \langle D^\alpha \big( \nabla \varrho \cdot S_1^1 {\bf u}+ \nabla\ln(1+\varrho) \cdot S_2^1 {\bf u} \big), D^\alpha {\bf u} \big \rangle,

J_7 = \sum\limits_{|\alpha|\leq 3} \big \langle D^\alpha \big( \nabla \varrho \cdot S_1^1U+ \nabla\ln(1+\varrho) \cdot S_2^1U \big), D^\alpha {\bf u} \big \rangle,

J_8 = \sum\limits_{|\alpha|\leq 3} \big \langle D^\alpha \big(K(\varrho) \nabla \varrho \big), D^\alpha {\bf u} \big \rangle.

对于 J_1 , 我们将其改写为

\begin{eqnarray*} J_1& = &-\sum\limits_{0<|\alpha|\leq 3} \langle D^\alpha( {\bf u} \cdot \nabla {\bf u})- {\bf u} \cdot \nabla D^{\alpha} {\bf u}, D^\alpha {\bf u} \rangle -\sum\limits_{0<|\alpha|\leq 3} \langle {\bf u} \cdot \nabla D^{\alpha} {\bf u}, D^{\alpha} {\bf u} \rangle- \langle {\bf u} \cdot \nabla {\bf u}, {\bf u} \rangle\\ & = &\sum\limits_{i = 1}^3J_{1, i}. \end{eqnarray*}

类似于 I_1 的推导, 我们有

|J_{1, 1}|\leq C \big(\| \nabla {\bf u}\|_{H^2}\| \nabla {\bf u}\|_{L^\infty}+\| {\bf u}\|_{H^3}\| \nabla {\bf u}\|_{L^\infty} \big)\| \nabla {\bf u}\|_{H^2}\leq C\| {\bf u}\|_{H^3}\| \nabla {\bf u}\|_{H^2}^2,

|J_{1, 2}|\leq \|{\mathord{{{\rm{div}}}}} {\bf u}\|_{L^\infty}\| \nabla {\bf u}\|^2_{H^2}\leq C\| {\bf u}\|_{H^3}\| \nabla {\bf u}\|_{H^2}^2,

|J_{1, 3}|\leq \| \nabla {\bf u}\|_{L^2}\| {\bf u}\|_{L^6}\| {\bf u}\|_{L^3}\leq C\| {\bf u}\|_{H^3}\| \nabla {\bf u}\|_{L^2}^2,

这给出

\begin{eqnarray} |J_1|\leq C\| {\bf u}\|_{H^3}\| \nabla {\bf u}\|_{H^2}^2. \end{eqnarray}
(3.10)

同理, 对于 J_2 J_3 , 我们得到

\begin{eqnarray} |J_2|+|J_3|&\leq& \|U \cdot \nabla {\bf u}\|_{H^3}\| {\bf u}\|_{H^3}+\| {\bf u} \cdot \nabla U\|_{H^3}\| {\bf u}\|_{H^3}\\ &\leq& C\| U, \nabla^3U\|_{L^\infty}\| {\bf u}\|_{H^3}\| \nabla {\bf u}\|_{H^3}+C\| \nabla U, \nabla^4U\|_{L^\infty}\| {\bf u}\|^2_{H^3}\\ &\leq& C \big(\|U, \nabla^3U\|^2_{L^\infty}+\| \nabla U, \nabla^4U\|_{L^\infty} \big)\| {\bf u}\|^2_{H^3}+ \frac{\min(\mu, \kappa)}{4}\| \nabla {\bf u}\|_{H^3}^2. \end{eqnarray}
(3.11)

对于 J_4 , 由Hölder不等式得

\begin{eqnarray} |J_4|\leq \|U \cdot \nabla U\|_{H^3}\| {\bf u}\|_{H^3}. \end{eqnarray}
(3.12)

对于 J_5 的第二项, 我们将其改写为

\begin{eqnarray*} J_{5, 2} = -\sum\limits_{0<|\alpha|\leq 3} \big \langle D^\alpha \big( \nabla\ln(1+\varrho) \cdot S_4^2 \varrho \big), D^\alpha {\bf u} \big \rangle- \big \langle \big( \nabla\ln(1+\varrho) \cdot S_4^2 \varrho \big), {\bf u} \big \rangle = \sum\limits_{i = 1}^2J^i_{5, 2}. \end{eqnarray*}

由Hölder不等式和引理2.3得

\begin{eqnarray*} |J^1_{5, 2}|&\leq& \Big|\sum\limits_{0<|\alpha|\leq 3} \big \langle D^{\alpha-e_i} \big( \nabla\ln(1+\varrho) \cdot S_4^2 \varrho \big), \partial_{x_i}D^\alpha {\bf u} \big \rangle \Big|\\ &\leq& \big\| \nabla\ln(1+\varrho) \cdot S_4^2 \varrho \big\|_{{H}^2}\|\nabla {\bf u}\|_{H^3}\\ &\leq&C\|\ln(1+\varrho)\|_{\dot{H}^3}\| \nabla \varrho\|_{H^3}\| \nabla {\bf u}\|_{H^3}\\ &\leq&C\|\varrho\|_{\dot{H}^3}\| \nabla \varrho, \nabla {\bf u}\|^2_{H^3}, \\ |J^2_{5, 2}|&\leq& \big\| \nabla\ln(1+\varrho) \big\|_{L^2} \big\|S_4^2 \varrho \big\|_{L^2} \| {\bf u}\|_{L^\infty} \leq C\| {\bf u}\|_{H^3}\| \nabla \varrho\|_{H^3}^2, \end{eqnarray*}

这给出

\begin{equation} |J_5|\leq C\|\varrho, {\bf u}\|_{{H}^3}\| \nabla \varrho, \nabla {\bf u}\|^2_{H^3}. \end{equation}
(3.13)

类似于 J_5 的推导, 我们有

\begin{equation} |J_6|\leq C\|\varrho, {\bf u}\|_{{H}^3}\| \nabla \varrho, \nabla {\bf u}\|^2_{H^3}. \end{equation}
(3.14)

对于 J_7 , 我们将其改写为

\begin{eqnarray*} J_7& = &\sum\limits_{0<|\alpha|\leq 3} \big \langle D^{\alpha-e_i} \big( \nabla \varrho \cdot S_1^1U+ \nabla\ln(1+\varrho) \cdot S_2^1U \big), D^\alpha \partial_{x_i} {\bf u} \big \rangle\\ &&- \big \langle \big( \nabla \varrho \cdot S_1^1U+ \nabla\ln(1+\varrho) \cdot S_2^1U \big), {\bf u} \big \rangle = \sum\limits_{i = 1}^2J_{7, i}. \end{eqnarray*}

由Hölder不等式和引理2.1–2.3得

\begin{eqnarray*} |J_{7, 1}|&\leq& C \big\| \nabla \varrho \cdot S_1^1U+ \nabla\ln(1+\varrho) \cdot S_2^1U \big\|_{{H}^2}\|\nabla {\bf u}\|_{H^3}\\ &\leq& C\| \nabla U, \nabla^3U\|_{L^\infty}\|\nabla\varrho\|_{H^2}\| \nabla {\bf u}\|_{H^3}, \\ |J_{7, 2}|&\leq& C\| \nabla U\|_{L^\infty}\| {\bf u}\|_{L^2}\| \nabla \varrho\|_{H^2}, \end{eqnarray*}

这给出

\begin{equation} |J_7|\leq C\| \nabla U, \nabla^3U\|^2_{L^\infty}\|\varrho, {\bf u}\|_{H^3}^2+\frac{\min(\mu, \kappa)}4\| \nabla {\bf u}\|^2_{H^3}+\frac{1}4\| \nabla \varrho\|^2_{H^3}. \end{equation}
(3.15)

对于 J_8 , 我们还有

J_8 = \sum\limits_{0<|\alpha|\leq 3} \big \langle D^\alpha \big(K(\varrho) \nabla \varrho \big), D^\alpha {\bf u} \big \rangle+ \big \langle (K(\varrho) \nabla \rho, {\bf u} \big \rangle = \sum\limits_{i = 1}^2J_{8, i}.

由Hölder不等式和引理2.3得

|J_{8, 1}|\leq C\|\varrho\|_{H^2}\| \nabla\varrho\|_{H^2}\| \nabla {\bf u}\|_{H^3},

|J_{8, 2}|\leq C\|K(\varrho)\|_{\dot{H}^1}\| \nabla \varrho\|_{L^2}\| {\bf u}\|_{L^3}\leq C\| \nabla \varrho\|^2_{H^2}\| {\bf u}\|_{H^3},

这给出

\begin{eqnarray} |J_8|\leq C\|\varrho, {\bf u}\|_{{H}^3}\| \nabla \varrho, \nabla {\bf u}\|^2_{H^3}. \end{eqnarray}
(3.16)

将上述估计代入(3.9)式, 我们推出

\begin{eqnarray} &&\frac{ {{\rm{d}}}}{ {{\rm{d}}} t}\| {\bf u}\|^2_{H^3}+\min(\mu, \kappa)\| \nabla {\bf u}\|^2_{H^3}+2\sum\limits_{|\alpha|\leq 3} \langle D^\alpha \nabla \varrho, D^\alpha {\bf u} \rangle\\ &\leq& \frac{1}2\| \nabla \varrho\|^2_{H^3}+C\|\varrho, {\bf u}\|_{{H}^3}\| \nabla \varrho, \nabla {\bf u}\|^2_{H^3}+C \big(\|U, \nabla^3U\|^2_{L^\infty}+\| \nabla U, \nabla^4U\|_{L^\infty} \big)\|\varrho, {\bf u}\|_{H^3}^2\\ &&+C\|{\mathord{{{\rm{div}}}}} U, U \cdot \nabla U\|_{H^3}\|\varrho, {\bf u}\|_{H^3}. \end{eqnarray}
(3.17)

最后, 我们需要封闭上式估计. 联立(3.8)和(3.17)式, 关于时间积分得

\begin{eqnarray} &&\|(\varrho, {\bf u})(t)\|^2_{H^3}+\min \Big(\mu, \kappa, \frac12 \Big) \int_0^t\|( \nabla \varrho, \nabla {\bf u})(s)\|^2_{H^3} {{\rm{d}}} s\\ &\leq& C\int_0^t\|(\varrho, {\bf u})(s)\|_{{H}^3}\|( \nabla \varrho, \nabla {\bf u})(s)\|^2_{H^3} {{\rm{d}}} s +C\int_0^t\|{\mathord{{{\rm{div}}}}} U, U \cdot \nabla U\|_{H^3}\|(\varrho, {\bf u})(s)\|_{H^3} {{\rm{d}}} s\\ &&+C\int_0^t \big(\|U, \nabla^3U\|^2_{L^\infty}+\| \nabla U, \nabla^4U\|_{L^\infty} \big)\|(\varrho, {\bf u})(s)\|_{H^3}^2 {{\rm{d}}} s. \end{eqnarray}
(3.18)

选取 \eta 充分小, 吸收掉(3.18) 式右端的第一项, 则当 t\in[0, \Gamma] 时, (3.18) 式退化为

\begin{eqnarray} &&\|(\varrho, {\bf u})(t)\|^2_{H^3}+\int_0^t\| \nabla \varrho, \nabla {\bf u}\|^2_{H^3} {{\rm{d}}} s{}\\ &\leq& C\int_0^t\|{\mathord{{{\rm{div}}}}} U, U \cdot \nabla U\|_{H^3} {{\rm{d}}} s +C\int_0^t \big(\|U, \nabla^3U\|^2_{L^\infty}+\| \nabla U, \nabla^4U\|_{L^\infty} \big)\|(\varrho, {\bf u})(s)\|_{H^3}^2 {{\rm{d}}} s.{\qquad} \end{eqnarray}
(3.19)

由小性条件(1.8), Gronwall不等式和引理2.3, 当 t\in[0, \Gamma] 时, 我们得到

\begin{eqnarray*} \|(\varrho, {\bf u})(t)\|^2_{H^3}&\leq& C\int_0^t\|{\mathord{{{\rm{div}}}}} U, U \cdot \nabla U\|_{H^3} {{\rm{d}}} s\exp \Big(C\int_0^t \big(\|U, \nabla^3U\|^2_{L^\infty}+\| \nabla U, \nabla^4U\|_{L^\infty} \big) {{\rm{d}}} s \Big)\nonumber\\ &\leq& C \varepsilon\|a_0\|_{L^2} \big(1+\|\hat{a}_0\|_{L^1} \big)\exp \Big(C \big(1+\|\hat{a}_0\|_{L^1}+ \varepsilon\|{a}_0\|_{L^2} \big)\|\hat{a}_0\|_{L^1} \Big)\nonumber\\ &\leq& C \delta. \end{eqnarray*}

选择 \eta = 2C\delta , 我们有

\sup\limits_{\tau\in[0, t]}\|(\varrho, {\bf u})(\tau)\|^2_{H^3}\leq \frac\eta2, \quad t\leq \Gamma,

由此可以断言 \Gamma = T^* . 否则, 若 \Gamma<T^* , 由连续性方法, 必存在 0<\epsilon\ll1 使得

\sup\limits_{\tau\in[0, t]}\|(\varrho, {\bf u})(\tau)\|^2_{H^3}\leq \eta, \quad t\leq \Gamma+\epsilon<T^*,

这与(3.1) 式中 \Gamma 的定义矛盾.

综上, 我们得出 \Gamma = T^*

\sup\limits_{\tau\in[0, t]}\|(\varrho, {\bf u})(\tau)\|^2_{H^3}\leq C<\infty, \; \forall t\in(0, T^*),

这就意味着 T^* = +\infty . 于是, 定理1.1得证.

4 附录

本节我们给出方程组(1.5) 的导出细节.

  由质量方程 \partial_t \rho+u\cdot \nabla\rho+\rho{\mathord{{{\rm{div}}}}} u = 0 , 得

\begin{eqnarray} \partial_t \nabla \rho+u\cdot \nabla \nabla\rho+ \nabla \rho\cdot \nabla^{\mathit{T}} u+ \nabla \rho{\mathord{{{\rm{div}}}}} u+ \rho \nabla{\mathord{{{\rm{div}}}}} u = 0. \end{eqnarray}
(4.1)

由动量方程 (1.1)_2 , 得

\begin{eqnarray*} &&\rho(\partial_t u+ u\cdot\nabla u-\mu \Delta u-\mu \nabla{\mathord{{{\rm{div}}}}} u)-2\mu \nabla\rho\cdot{\Bbb D}u\\ &&-\lambda(\rho)\nabla{\mathord{{{\rm{div}}}}} u- \nabla \lambda(\rho){\mathord{{{\rm{div}}}}} u+\nabla P(\rho) = \kappa \rho \nabla \Delta \rho, \end{eqnarray*}

只要 \rho\neq0 , 上式变为

\begin{eqnarray} &&\partial_t u+ u\cdot\nabla u-\mu \Delta u- \big(\mu+ \rho^{-1}\lambda(\rho) \big)\nabla{\mathord{{{\rm{div}}}}} u\\ &&-2\mu \nabla\ln\rho\cdot{\Bbb D}u- \rho^{-1} \nabla \lambda(\rho){\mathord{{{\rm{div}}}}} u+ \rho^{-1}\nabla P(\rho) = \kappa \nabla \Delta \rho. \end{eqnarray}
(4.2)

引入有效速度 v = u+\nabla\rho, 联立(4.2)与 (4.1) 式, 得

\begin{eqnarray} &&\partial_t v+u\cdot\nabla v-\mu \Delta v+ \big( \rho-\mu- \rho^{-1}\lambda(\rho) \big)\nabla{\mathord{{{\rm{div}}}}} v-2\mu \nabla\ln\rho\cdot{\Bbb D}u+ \nabla \rho\cdot \nabla^{\mathit{T}} u\\ &&+ \big( \nabla \rho- \rho^{-1} \nabla \lambda(\rho) \big){\mathord{{{\rm{div}}}}} u+ \rho^{-1}\nabla P(\rho) = \big(\kappa-(2\mu+ \rho^{-1}\lambda(\rho)- \rho) \big) \nabla \Delta \rho. \end{eqnarray}
(4.3)

为消掉三阶导数项 \kappa \nabla \Delta \rho , 假设如下代数关系

2\mu+ \rho^{-1}\lambda(\rho)- \rho = \kappa\quad\Leftrightarrow\quad\lambda(\rho) = \rho^2+(\kappa-2\mu) \rho,

则我们从(4.3) 式得

\begin{eqnarray} &&\partial_t v+ u\cdot\nabla v-\mu \Delta v-(\kappa-\mu)\nabla{\mathord{{{\rm{div}}}}} v-2\mu \nabla\ln\rho\cdot{\Bbb D}u \\ &&-(\kappa-2\mu) \nabla\ln\rho{\mathord{{{\rm{div}}}}} u+ \nabla \rho\cdot \nabla^{\mathit{T}} u- \nabla \rho{\mathord{{{\rm{div}}}}} u+ \rho^{-1}\nabla P(\rho) = 0. \end{eqnarray}
(4.4)

注意到 u = v-\nabla\rho , 我们有

\begin{eqnarray*} &&\partial_t v+ v\cdot\nabla v-\mu \Delta v-(\kappa-\mu)\nabla{\mathord{{{\rm{div}}}}} v+ \rho^{-1}\nabla P(\rho)\nonumber\\ & = & \nabla \rho\cdot\nabla v+ \nabla \rho{\mathord{{{\rm{div}}}}} u- \nabla \rho\cdot \nabla^{\mathit{T}} u+2\mu \nabla\ln\rho\cdot{\Bbb D}u+(\kappa-2\mu) \nabla\ln\rho{\mathord{{{\rm{div}}}}} u\nonumber\\ & = & \nabla \rho\cdot\nabla v+ \nabla \rho{\mathord{{{\rm{div}}}}} (v- \nabla \rho)- \nabla \rho\cdot \nabla^{\mathit{T}} (v- \nabla \rho)+2\mu \nabla\ln \rho \cdot{\Bbb D}(v- \nabla \rho)\nonumber\\ && +(\kappa-2\mu) \nabla\ln \rho{\mathord{{{\rm{div}}}}} (v- \nabla \rho)\nonumber\\ & = & \nabla \rho\cdot \big[(\nabla- \nabla^{\mathit{T}}+{\Bbb I}{\mathord{{{\rm{div}}}}})v+( \nabla^2 -{\Bbb I}\Delta) \rho \big]\nonumber\\ && + \nabla\ln \rho \cdot \big[(2\mu{\Bbb D}+(\kappa-2\mu){\Bbb I}{\mathord{{{\rm{div}}}}})v-(2\mu{\Bbb D}+(\kappa-2\mu){\Bbb I}{\mathord{{{\rm{div}}}}}) \nabla \rho \big]. \end{eqnarray*}

\varrho = \rho-1 , 我们将原方程组(1.1) 等价地转化为

\begin{eqnarray*} \left\{\begin{array}{ll} \partial_t \varrho -\Delta\varrho+{\mathord{{{\rm{div}}}}} v = {\mathord{{{\rm{div}}}}}(\varrho \nabla \varrho)-{\rm{div}}(\varrho v), \\ \partial_t v+ v\cdot\nabla v-\mu \Delta v-(\kappa-\mu)\nabla{\mathord{{{\rm{div}}}}} v+ \nabla \varrho = \nabla \varrho\cdot \big[(\nabla- \nabla^{\mathit{T}}+{\Bbb I}{\mathord{{{\rm{div}}}}})v+( \nabla^2 -{\Bbb I}\Delta)\varrho \big]\nonumber\\ \; \; \; \; + \nabla\ln(1+\varrho) \cdot \big[(2\mu{\Bbb D}+(\kappa-2\mu){\Bbb I}{\mathord{{{\rm{div}}}}})v-(2\mu{\Bbb D}+(\kappa-2\mu){\Bbb I}{\mathord{{{\rm{div}}}}}) \nabla \varrho \big]+ K(\varrho) \nabla \varrho, \\ (v, \varrho)|_{t = 0} = (v_{0}, \varrho_{0}).\end{array}\right. \end{eqnarray*}

其中 K(s) = 1- \frac{P'(1+s)}{1+s} .

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