二维空间中, 非齐次不可压缩流体的运动可以由如下方程组描述

(1.1) $ \begin{equation} \left\{ \begin{array}{lll} \varrho_t+( {\bf u}\cdot\nabla) \varrho = 0, \\ ( \varrho {\bf u})_t+ {{\rm{div }}}( \varrho {\bf u}\otimes {\bf u})+\nabla \pi = \mu \triangle {\bf u}+( \nabla\times {\bf H})\times {\bf H}, \\ {\bf H}_t+ {\bf u}\cdot \nabla {\bf H} = \nu\triangle {\bf H} + {\bf H}\cdot \nabla {\bf u}, \\ {{\rm{div }}} {\bf u} = 0, \quad {{\rm{div }}} {\bf H} = 0, \end{array} \right. \end{equation} $

这里$ \varrho = \varrho(t, {\bf x}) $, $ {\bf u} = ( {\bf u}_1, {\bf u}_2)(t, {\bf x}) $, $ {\bf H} = ( {\bf H}_1, {\bf H}_2)(t, {\bf x}) $以及$ \pi = \pi(t, {\bf x}) $分别表示流体的密度、速度、磁场强度以及流体受到的压力, 常数$ \mu > 0 $表示流体的粘性系数, $ \nu > 0 $表示磁耗散系数.

磁流体动力学方程组(即MHD方程组)描述了导电流体在磁场中的运动规律, 它具有十分重要的物理意义以及很强的物理应用背景.由于流体的运动深受磁场的影响, 因此流体的速度和磁场产生强耦合, 导致方程具有很强的非线性.关于磁流体动力学方程的物理特性, 很多文献给出了许多详细的介绍^{[3, 22]}.

如果不考虑磁场对流体的影响, 即$ {\bf H} = 0 $, 则MHD方程组(1.1)可简化为大家熟知的Navier-Stokes方程组.关于Navier-Stokes方程组的适定性问题已经有大量的研究结果, 参见文献[1, 8-10, 18, 20-21, 24, 26, 29, 33], 对于MHD方程组同样也有很多研究, 参见文献[2, 7, 27-28, 32].

该文主要考虑当磁场强度随着流体运动而变化, 而非简单的往外扩散, 且流体具有非常强导电性的MHD方程组, 称为无磁扩散MHD方程组

(1.2) $ \begin{equation} \left\{ \begin{array}{lll} \varrho_t+( {\bf u}\cdot\nabla) \varrho = 0, \\ ( \varrho {\bf u})_t+ {{\rm{div }}}( \varrho {\bf u}\otimes {\bf u})+\nabla \pi = \mu \triangle {\bf u}+( \nabla\times {\bf H})\times {\bf H}, \\ {\bf H}_t+ {\bf u}\cdot \nabla {\bf H} = {\bf H}\cdot \nabla {\bf u}, \\ {{\rm{div }}} {\bf u} = 0, \quad {{\rm{div }}} {\bf H} = 0, \end{array} \right. \end{equation} $

初值条件为

(1.3) $ \begin{equation} ( \varrho, {\bf u}, {\bf H})(0, {\bf x}) = ( \varrho_0, {\bf u}_0, {\bf H}_0)( {\bf x}), \quad {\bf x}\in {\mathbb{R}}^2, \end{equation} $

在弱意义下, 满足如下的远场条件

(1.4) $ \begin{equation} {\bf u}(t, {\bf x})\rightarrow {\bf 0}, \quad \varrho(t, {\bf x})\rightarrow 0, \quad {\bf H}(t, {\bf x})\rightarrow 0\quad {\rm \mbox{当}}\ | {\bf x}|\rightarrow\infty, \quad t\geq0. \end{equation} $

关于该方程的物理描述可参见文献[3, 4, 14, 22]的结果.

事实上, 对方程(1.2)同样有很多的结论, 当密度$ \varrho $为常数时, 方程(1.2)称为齐次的.酒全森和牛冬娟^[19]出了初值在$ H^s $ ($ s $为大于等于3的整数)空间时方程组(1.2)解的局部存在性.林芳华等^[25]对于方程组(1.2)给出了一个标志性结果, 当初值充分靠近平衡态解时, 在二维情形下证明了全局解的存在性.任晓霞等^[31]对林芳华的结果进行了改进, 当对初值去掉一些假设之后结论仍然成立, 并证明了MHD方程组的能量以不依赖于电阻的速率衰减.在三维周期情形, 潘荣华等^[30]证明了当初值靠近平衡态, 初始速度满足对称性条件时, 不可压MHD方程全局经典解的存在性. Fefferman等^[12]将潘荣华的结论推广到$ d $维全空间$ d = (2, 3) $, 并给出一个令人惊喜的结论:当$ {\bf H}_0\in H^s( {\mathbb{R}}^d) $且$ {\bf u}_0\in H^{s-1+ \varepsilon}( {\mathbb{R}}^d) $, 其中$ s>d/2 $以及$ 0< \varepsilon<1 $时, 即在靠近临界指标索伯列夫空间附近证明了局部解的存在性和唯一性.由于动量方程含有耗散项, Chemin等^[5]通过对初值提出更少的正则性假设, 当$ {\bf u}_0\in B_{2, 1}^{d/2-1}( {\mathbb{R}}^d) $以及$ {\bf H}_0\in B_{2, 1}^{d/2}( {\mathbb{R}}^d) $时, 在Besov空间中证明了解的局部存在性.

当密度$ \varrho $不为常数时, 方程(1.2)称为非齐次的.在真空远场条件下流体动力学适定性问题一直是个研究的热点.最近, 李竞等^[23]在整个平面$ {\mathbb{R}}^2 $上证明了二维可压Navier-Stokes方程柯西问题强解和经典解的存在性.当初值包含真空时, 黄祥娣等^[17]给出了非齐次不可压MHD全局解存在性的证明.受文献[16-17, 23]的启发, 梁之磊^[24]和吕博强等人^[27-29]得到一系列关于初始真空非齐次不可压Navier-Stokes方程的适定性结论.然而对于非齐次不可压无磁扩散的流体方程的相关结论还非常少, 尤其是在初始真空具有紧支集时.因此, 该文主要讨论真空远场条件下二维零磁耗散MHD方程的适定性问题.下面给出主要结果.

定理1.1  设$ \eta_0 > 0 $为一常数, 定义

(1.5) $ \begin{equation} \bar {\bf x}\triangleq (e+| {\bf x}|^2)^{1/2}\ln^{1+\eta_0}(e+| {\bf x}|^2). \end{equation} $

令$ L^r = L^r( {\mathbb{R}}^2), \ W^{s, r} = W^{s, r}( {\mathbb{R}}^2), \ H^s = W^{s, 2}( {\mathbb{R}}^2), $其中$ 1\leq r\leq \infty $.对于任意的常数$ q>2 $以及$ a>1 $, 假设初值$ ( \varrho_0, {\bf u}_0, {\bf H}_0) $满足

(1.6) $ \begin{equation} \left\{ \begin{array}{lll} \varrho_0\geq0, \ \varrho_0\bar {\bf x}^a\in L^1\cap H^1\cap W^{1, q}, \ {\bf H}_0 \bar {\bf x}^a \in H^1\cap W^{1, q}, \\ \nabla {\bf u}_0\in L^2, \ \sqrt{ \varrho_0} {\bf u}_0\in L^2, \ {{\rm{div }}} {\bf u}_0 = {{\rm{div }}} {\bf H}_0 = 0, \end{array} \right. \end{equation} $

则存在$ T_0>0 $使得方程(1.2)–(1.4)在$ {\mathbb{R}}^2\times (0, T_0) $上有唯一的强解$ ( \varrho, {\bf u}, \pi, {\bf H}) $满足$ {{\rm{div }}} {\bf u} = 0 $以及$ {{\rm{div }}} {\bf H} = 0 $, 并使得

(1.7) $ \begin{equation} \left\{ \begin{array}{lll} \varrho\in C([0, T_0]; L^1\cap H^1\cap W^{1, q}), \ \varrho\bar {\bf x}^a\in L^\infty (0, T_0; L^1\cap H^1\cap W^{1, q}), \\ {\bf H}\in C([0, T_0]; L^2\cap H^1\cap W^{1, q}), \ {\bf H}\bar {\bf x}^a\in L^\infty(0, T_0; H^1\cap W^{1, q}), \\ \sqrt \varrho {\bf u}, \ \nabla {\bf u}, \ \sqrt{t}\sqrt \varrho {\bf u}_t, \ \sqrt{t} \nabla\pi, \ \sqrt t \nabla^2 {\bf u}\in L^\infty (0, T_0; L^2), \\ \nabla {\bf u}\in L^2(0, T_0; H^1)\cap L^{(q+1)/q}(0, T_0; W^{1, q}), \ \sqrt{t} \nabla {\bf u}\in L^2(0, T_0; W^{1, q}), \\ \nabla\pi\in L^2(0, T_0; L^2)\cap L^{(q+1)/q}(0, T_0; L^q), \\ \sqrt \varrho {\bf u}_t, \sqrt t \nabla {\bf u}_t\in L^2( {\mathbb{R}}^2\times (0, T_0)), \end{array} \right. \end{equation} $

以及对适当的常数$ N>0 $和$ B_N\triangleq\{ {\bf x}\in {\mathbb{R}}^2|| {\bf x}|<N\} $有

(1.8) $ \begin{equation} \inf\limits_{0\leq t\leq T_0}\int_{B_N} \varrho(t, {\bf x}){\rm d} {\bf x}\geq \frac14\int_{ {\mathbb{R}}^2} \varrho_0( {\bf x}){\rm d} {\bf x}. \end{equation} $

进一步, 若初值$ ( \varrho_0, {\bf u}_0, {\bf H}_0) $满足额外的正则性和兼容性条件, 则定理1.1中得到的局部解$ ( \varrho, {\bf u}, \pi, {\bf H}) $就是经典解.

定理1.2  除了方程(1.6)的初值假设外, 进一步假设初值满足

(1.9) $ \begin{equation} \nabla^2 \varrho_0, \ \nabla^2 {\bf H}_0\in L^2\cap L^q, \quad\bar {\bf x}^{\delta_0} \nabla^2 \varrho_0, \ \bar {\bf x}^{\delta_0} \nabla^2 {\bf H}_0, \ \nabla^2 {\bf u}_0\in L^2, \end{equation} $

其中$ \delta_0\in (0, 1) $.当$ {\bf g}\in L^2 $时, 若方程在初始时刻满足下面兼容性条件

(1.10) $ \begin{equation} -\mu\triangle {\bf u}_0+ \nabla \pi_0-( \nabla\times {\bf H}_0)\times {\bf H}_0 = \varrho_0^{1/2}{\bf g}, \end{equation} $

则方程组存在唯一的局部强解$ ( \varrho, {\bf u}, \pi, {\bf H}) $, 除了满足(1.7)和(1.8)式之外, 还满足

(1.11) $ \begin{equation} \left\{ \begin{array}{lll} \nabla^2 \varrho, \ \nabla^2 {\bf H}\in C([0, T_0]; L^2\cap L^q), \ \bar {\bf x}^{\delta_0} \nabla^2 \varrho, \ \bar {\bf x}^{\delta_0} \nabla^2 {\bf H} \in L^\infty(0, T_0; L^2), \\ \nabla^2 {\bf u}, \ \sqrt \varrho {\bf u}_t, \ \sqrt t \nabla {\bf u}_t, \ t\sqrt \varrho {\bf u}_{tt}, \ t \nabla^2 {\bf u}_t\in L^\infty(0, T_0; L^2), \\ \nabla \pi, \ \sqrt t\pi_t, \ t\pi_t\in L^\infty(0, T_0; L^2), \nabla {\bf u}_t, \ t \nabla {\bf u}_{tt}\in L^2(0, T_0; L^2), \\ t \nabla^3 {\bf u}, \ t \nabla^2\pi \in L^\infty(0, T_0; L^2\cap L^q), t \nabla^2( \varrho {\bf u})\in L^\infty(0, T_0; L^{(q+2)/2}). \end{array} \right. \end{equation} $

就我们的存在性结果而言, 当$ {\bf H} = 0 $时其实定理1.1的结果与吕博强^[27]的结果类似.然而对于全局存在性, 像非齐次Navier-Stokes方程^[29]和带磁扩散的流体方程^{[17, 28]}, 只有给出比局部存在性更多的初值假设, 才可以得到全局解的存在性, 如文献[25, 31].该文受到文献[6, 29, 34]的启发, 当初值满足(1.6)式, (1.9)式以及兼容性条件(1.10)时, 得到了仅依赖于磁场$ {\bf H} $的爆破准则.

定理1.3  假设初值$ ( \varrho_0, {\bf u}_0, {\bf H}_0) $满足(1.6)式, (1.9)式以及兼容性条件(1.10).令$ ( \varrho, {\bf u}, \pi, {\bf H}) $是方程组(1.2)–(1.4)的局部强解.若$ 0<T^*<+\infty $是局部解的最大存在区间, 则对任意的$ r>2 $有

(1.12) $ \begin{equation} \limsup\limits_{T\rightarrow T^*}\left(\| {\bf H}\|_{L^\infty(0, T^*; L^\infty ( {\mathbb{R}}^2))}+\| \nabla {\bf H}\| _{L^1(0, T_*; L^r( {\mathbb{R}}^2)}\right) = \infty. \end{equation} $

讨论零磁扩散磁流体方程组主要有如下难点:首先对于定理1.1和定理1.2, 在全空间情形没有紧致性, $ {\bf u} $的$ L^p $范数无法被$ \sqrt \varrho {\bf u} $和$ \nabla {\bf u} $的$ L^2 $范数控制.因此, Cho等人^[7-9]的方法无法直接应用于该文情形.同时, 速度场和磁场的强耦合作用导致很强的非线性, 给该问题的研究带来很大的困难.因此, 像文献[23-24, 26]中一样, 对密度和磁场给出一些带权估计.另一方面, 磁扩散项的缺失为高阶导数能量估计带来一些麻烦.令人欣慰的是二维情形, 由于$ {{\rm{div }}} {\bf H} = 0 $, 可以假设存在一个势函数$ \phi $使得$ {\bf H} = (\partial_2 \phi, -\partial_1 \phi) $, 则方程$ (1.2)_3 $可以改写为$ \phi_t+ {\bf u}\cdot\nabla\phi = 0, $很显然, 它具有和方程$ (1.2)_1 $相类似的结构.因此, 估计磁场正则性的方法同样适用于密度的正则性估计.

最后, 值得一提的是该文中初值条件的假设比文献[5, 12]的更弱, 而该文中爆破的结论与文献[6]相比更好.

该文后面的内容作如下安排:第2部分中, 给出了证明过程中会用到的基本引理和不等式; 第3, 4部分, 分别给出了局部强解和经典解存在性所需要的先验估计; 定理1.1–1.3的证明则由第5部分给出.

下面给出该文中会用到的一系列引理和估计.当初值严格远离真空时候, 有界球上零磁扩散磁流体方程组的局部存在性定理可由下面引理给出^[7-9].

引理2.1  令$ B_R = \{ {\bf x}\in {\mathbb{R}}^2|| {\bf x}|<R\} $表示半径小于$ R>0 $的有界球, 设磁流体方程组的初值$ ( \varrho_0, {\bf u}_0, {\bf H}_0) $满足

(2.1) $ \begin{equation} ( \varrho_0, {\bf u}_0, {\bf H}_0)\in H^3(B_R), \ \ {{\rm{div }}} {\bf u}_0 = 0, {{\rm{div }}} {\bf H}_0 = 0, \ \inf\limits_{ {\bf x}\in B_R} \varrho_0( {\bf x})>0, \end{equation} $

则存在一个适当小的$ T_R>0 $, 下面的初边值问题

(2.2) $ \begin{equation} \left\{ \begin{array}{lll} \varrho_t+ {{\rm{div }}}( \varrho {\bf u}) = 0, \\ \varrho {\bf u}_t+ \varrho( {\bf u}\cdot \nabla) {\bf u}+ \nabla \pi = \mu\triangle {\bf u}+(\nabla\times {\bf H})\times {\bf H}, \\ {\bf H}_t+( {\bf u}\cdot \nabla) {\bf H} = ( {\bf H}\cdot \nabla) {\bf u}, \\ {{\rm{div }}} {\bf u} = 0, \quad {{\rm{div }}} {\bf H} = 0, \\ {\bf u} = 0, {\bf x}\in \partial B_R, \quad t>0, \\ ( \varrho, {\bf u}, {\bf H})(0, {\bf x}) = ( \varrho_0, {\bf u}_0, {\bf H}_0)( {\bf x}), \quad {\bf x}\in B_R \end{array} \right. \end{equation} $

在$ B_R\times [0, T_R] $上具有唯一的局部解$ ( \varrho, {\bf u}, \pi, {\bf H}) $, 并且满足正则性

(2.3) $ \begin{equation} \left\{ \begin{array}{lll} \varrho, {\bf H}\in C([0, T_R]; H^3), \ {\bf u}\in C([0, T_R]; H^3)\cap L^2(0, T_R; H^4), \\ \pi\in C([0, T_R]; H^2)\cap L^2(0, T_R; H^3), \end{array} \right. \end{equation} $

这里$ L^2 = L^2(B_R) $, 并且$ H^k = H^k(B_R) $, $ k $为正整数.

当$ \Omega = {\mathbb{R}}^2 $或$ \Omega = B_R (R \geq 1) $时, 下面关于$ \tilde D^{1, 2} (\Omega) \triangleq \{v\in H_{\rm loc}^1(\Omega)| \nabla v\in L^2(\Omega)\} $空间中元素的带权估计在该文的后面的证明中起到关键作用^{[16, 24]}.

引理2.2  若$ \bar{{\bf x}} $是(1.5)式中定义的量, $ \Omega = {\mathbb{R}}^2 $或$ \Omega = B_R, R\geq 1 $.假设$ \varrho\in L^1(\Omega)\cap L^\infty(\Omega) $是一个非负函数, 并且对正的常数$ M_1 $, $ M_2 $满足

(2.4) $ \begin{equation} M_1\leq\int_{B_{N_1}} \varrho {\rm d} {\bf x}, \quad \| \varrho\|_{ L^1(\Omega)\cap L^\infty(\Omega)}\leq M_2, \end{equation} $

其中$ N_1\geq1 $且$ B_{N_1}\subset\Omega $.则对任意的$ \varepsilon>0 $以及$ \eta>0 $, 存在依赖于$ \varepsilon, \eta, M_1, M_2, N_1 $和$ \eta_0 $正的常数$ C $, 使得对任意的$ {\bf v}\in \tilde D^{1, 2}( \Omega) $有

(2.5) $ \begin{equation} \|{\bf v}\bar {\bf x}^{-\eta}\|_{L^{(2+\varepsilon)/\tilde{\eta}}(\Omega)} \leq C\|\sqrt \varrho{\bf v}\|_{L^2(\Omega)}+C\| \nabla {\bf v}\|_{L^2(\Omega)}, \end{equation} $

$ \tilde{\eta} = \min\{1, \eta\} $.

考虑如下的Stokes系统

(2.6) $ \begin{equation} \left\{ \begin{array}{lll} -\triangle {\bf u}+\nabla \pi = F, & \mbox{于 }\ B_R \ \mbox{上}, \\ \nabla\cdot u = 0, & \mbox{于 }\ B_R \ \mbox{上} \\ {\bf u} = 0, & \mbox{于 }\ \partial B_R \ \mbox{上}. \end{array} \right. \end{equation} $

则对上述的Stokes系统有$ L^p $有界估计^{[15, Theorem Ⅳ.6.1]}.

引理2.3  设$ {\bf u}\in W_0^{1, q}(B_R) $是方程(2.6)的一个弱解, 其中$ q> 1 $.若对$ k\geq 0 $, $ F\in W^{k, q}(B_R) $, 则有$ {\bf u}\in W^{k+2, q}(B_R) $以及

(2.7) $ \begin{equation} \|\nabla^{k+2} {\bf u}\|_{L^{q}(B_R)}+\|\nabla^{k+1}\pi\|_{L^q}\leq C\| F\|_{W^{k, q}(B_R)}, \end{equation} $

这里$ C $不依赖于$ R $.

另外, 关于有界均振空间($ BMO( {\mathbb{R}}^2) $)以及哈代空间($ \mathcal{H}^1( {\mathbb{R}}^2) $), 该文需要用到如下的重要引理^[11].

引理2.4  存在常数$ C > 0 $,

● 对任意的$ {\bf E}\in L^2( {\mathbb{R}}^2) $以及$ {\bf B}\in L^2( {\mathbb{R}}^2) $, 有

(2.8) $ \|\mathbf{E} \cdot \mathbf{B}\|_{\mathcal{H}^{\infty}\left(\mathbb{R}^{\in}\right)} \leq C\|\mathbf{E}\|_{L^{2}\left(\mathbb{R}^{2}\right)}\|\mathbf{B}\|_{L^{2}\left(\mathbb{R}^{2}\right)}, $

$ {\rm{div}} {\bf E} = 0, \quad \nabla^\bot\cdot {\bf B} = 0 \ {\rm in}\ \mathcal{D}'( {\mathbb{R}}^2); $

● 对任意的$ {\bf v}\in \mathcal{D}^1( {\mathbb{R}}^2) $, 有

(2.9) $ \begin{equation} \|{\bf v}\|_{BMO}\leq C\| \nabla{\bf v}\|_{L^2( {\mathbb{R}}^2)}, \end{equation} $

从这部分开始, 将给出一些证明局部适定性的先验估计.对任意的$ p\in[1, \infty] $以及$ k\geq0 $, 定义

$ \int f {\rm d} {\bf x} = \int_{B_R} f{\rm d} {\bf x}, \quad L^p = L^p(B_R), \quad W^{k, p} = W^{k, p}(B_R), \quad H^k = W^{k, 2}. $

设

(3.1) $ \begin{equation} \phi(t)\triangleq 1+\|\sqrt \varrho {\bf u}\|_{L^2}+\| \nabla {\bf u}\|_{L^2}+\| {\bf H}\|_{L^2} +\| {\bf H}\bar {\bf x}^a\|_{H^1\cap W^{1, q}}+\| \varrho\bar {\bf x}^a\|_{L^1\cap H^1\cap W^{1, q}}. \end{equation} $

该部分的目的是得到关于$ \phi(t) $的先验估计.

命题3.1  当$ R>4, N_0\geq 4 $, 设初值$ ( \varrho_0, {\bf u}_0, {\bf H}_0) $是光滑的并且满足2.1式以及

(3.2) $ \begin{equation} \frac12\leq \int_{B_{N_0}} \varrho_0( {\bf x}){\rm d} {\bf x}\leq \int_{B_R} \varrho_0( {\bf x}){\rm d} {\bf x}\leq \frac32. \end{equation} $

若$ ( \varrho, {\bf u}, \pi, {\bf H}) $是引理2.1中初边值问题2.2在$ B_R\times (0, T_R] $上的解.则存在依赖于$ \mu, q, a, \eta_0, N_0 $和$ C_0 $的常数$ T_0 > 0 $以及$ M > 0 $使得

(3.3) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}\phi(t)+\int_0^{T_0}\left(\| \nabla^2 {\bf u}\|_{L^q} ^{(q+1)/q}+t\| \nabla^2 {\bf u}\|_{L^q}^2+\| \nabla^2 {\bf u}\|_{L^2}^2\right){\rm d}t\leq M \end{equation} $

成立, 其中$ C_0 = \|\sqrt{ \varrho_0} {\bf u}_0\|_{L^2}+\| \nabla {\bf u}_0\|_{L^2}+\| {\bf H}_0\|_{L^2} +\| {\bf H}_0\bar {\bf x}^a\|_{H^1\cap W^{1, q}}+\| \varrho_0\bar {\bf x}^a\| _{L^1\cap H^1\cap W^{1, q}}\;. $命题3.1是一系列先验估计封闭的结果, 它的证明放在该部分的最后.首先, 需要对方程中未知函数$ ( \varrho, {\bf u}, \pi, {\bf H}) $以及$ \nabla {\bf u} $给出$ L^2 $估计.

引理3.1  令$ ( \varrho, {\bf u}, \pi, {\bf H}) $是初边值问题2.2的光滑解.则存在常数$ \alpha = \alpha( q)>1 $以及$ T_1 = T_1(C_0, N_0)>0 $使得对任意的$ t\in (0, T_1] $, 有

(3.4) $ \begin{equation} \sup\limits_{0\leq s\leq t}\left(\| \nabla {\bf u}\|_{L^2}^2+\| {\bf H}\|_{L^2}^2+ \|\sqrt \varrho {\bf u}\|_{L^2}^2 \right) +\int_0^t\left(\| \nabla {\bf u}\|_{L^2}^2+\|\sqrt \varrho {\bf u}_s\|_{L^2}^2\right){\rm d}s \leq C+C\int_0^t\phi^\alpha(s) {\rm d}s. \end{equation} $

证  通过对方程进行基本能量估计可得

(3.5) $ \begin{equation} \sup\limits_{0\leq s\leq t}\left(\|\sqrt \varrho {\bf u}\|_{L^2}^2+\| {\bf H}\|_{L^2}^2 \right) +\int_0^t\| \nabla {\bf u}\|_{L^2}^2{\rm d}s\leq C, \qquad \sup\limits_{0\le s\le t}\| \varrho\|_{L^1\cap L^\infty}\le C. \end{equation} $

取具有紧支集的光滑函数$ \varphi_N\in C_0^\infty(B_R) $, 其中$ N>1 $, 满足

(3.6) $ \begin{equation} 0\leq \varphi_N\leq 1, \ \varphi_N( {\bf x}) = 1, \ {\rm if}\ | {\bf x}|\leq N/2, \ \ | \nabla^k \varphi_N|\leq CN^{-k} (k = 1, 2), \end{equation} $

则由(3.1)式, (3.5)式和$ \int \varrho {\rm d} {\bf x} = \int \varrho_0 {\rm d} {\bf x}, $可推出

(3.7) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int \varrho \varphi_{2N_0}{\rm d} {\bf x}& = &\int \varrho {\bf u}\cdot \nabla \varphi_{2N_0}{\rm d} {\bf x} \geq -CN_0^{-1}\left(\int \varrho {\rm d} {\bf x}\right)^{1/2}\left( \int \varrho| {\bf u}|^2 \right)^{1/2} \\ &\geq& - \tilde C(C_0, N_0), \end{eqnarray} $

对不等式(3.7)在$ (0, T_1) $上两边积分, 易得

(3.8) $ \begin{equation} \inf\limits_{0\leq t\leq T_1}\int_{B_{2N_0}} \varrho {\rm d} {\bf x}\geq \inf\limits_{0\leq t\leq T_1} \int \varrho \varphi_{2N_0}{\rm d} {\bf x}\geq \int \varrho_0 \varphi_{2N_0}{\rm d} {\bf x}- \tilde CT_1\geq1/4, \end{equation} $

取$ T_1\triangleq\min\{1, (4 \tilde C)^{-1}\} $.从下文开始, 总是假设$ t\leq T_1 $.结合(2.5), (3.5)以及(3.8)式, 对任意的$ \varepsilon>0 $, $ \eta>0 $, $ {\bf v}\in \tilde D^{1, 2}(B_R) $满足

(3.9) $ \begin{equation} \|{\bf v}\bar {\bf x}^{-\eta}\|_{L^{(2+ \varepsilon)/ \tilde\eta}}^2\leq C( \varepsilon, \eta)\|\sqrt \varrho{\bf v}\|_{L^2}^2+C( \varepsilon, \eta)\| \nabla{\bf v}\|_{L^2}^2, \end{equation} $

这里$ \tilde\eta = \min\{1, \eta\} $.进一步, 由(3.9)式可得

(3.10) $ \begin{equation} \| \varrho^\eta {\bf u}\|_{L^{(2+ \varepsilon)/ \tilde\eta}}+\| {\bf u}\bar {\bf x}^{-\eta}\| _{L^{(2+ \varepsilon)/ \tilde\eta}}\leq C( \varepsilon, \eta)\phi^{1+\eta}(t). \end{equation} $

接下来, 对方程(2.2)$ _2 $两边乘以$ {\bf u}_t $, 并积分.及分部积分可得

(3.11) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\left(\mu\| \nabla {\bf u}\|_{L^2}^2\right)+\|\sqrt \varrho {\bf u}_t\|_{L^2}^2 \leq C\int \varrho| {\bf u}|^2| \nabla {\bf u}|^2{\rm d} {\bf x}+2\int ( {\bf H}\cdot \nabla) {\bf H} \cdot {\bf u}_t {\rm d} {\bf x}. \end{eqnarray} $

由Gagliardo-Nirenberg不等式以及$ H^{2} $范数的定义, 有

(3.12) $ \begin{equation} \| \nabla {\bf u}\|_{L^p}\leq C\| \nabla {\bf u}\|_{L^2}^{2/p}\| \nabla {\bf u}\|_{H^1}^{1-2/p} \leq C\phi(t)+C\phi(t)\| \nabla^2 {\bf u}\|_{L^2}^{1-2/p}, \end{equation} $

结合(3.10)式, 对$ \eta>0 $和$ \tilde\eta = \min\{1, \eta\} $, 有

(3.13) $ \begin{eqnarray} \int \varrho^\eta| {\bf u}|^2| \nabla {\bf u}|^2{\rm d} {\bf x}&\leq & C\| \varrho^{\eta/2} {\bf u}\|_{L^{8/ \tilde\eta} }^2\| \nabla {\bf u}\|_{L^{8/(4- \tilde\eta)}}^2 \\ &\leq& C(\eta)\phi^{4+2\eta}(t)\left(1+\| \nabla^2 {\bf u}\|_{L^2}^{ \tilde\eta/2} \right) \\ &\leq&C\phi^{\alpha(\eta)}(t)+ \varepsilon\| \nabla^2 {\bf u}\|_{L^2}^2. \end{eqnarray} $

关于$ x $分部积分, 再结合方程(2.2)$ _3 $, 方程(3.11)右端第二项可得

(3.14) $ \begin{eqnarray} 2\int ( {\bf H}\cdot \nabla) {\bf H} \cdot {\bf u}_t {\rm d} {\bf x}& = &-2\frac{\rm d}{{\rm d}t}\int ( {\bf H}\cdot \nabla ) {\bf u}\cdot {\bf H} {\rm d} {\bf x}+2\int ( {\bf H}_t\cdot \nabla) {\bf u}\cdot {\bf H} {\rm d} {\bf x}\\ &&+2\int ( {\bf H}\cdot \nabla ) {\bf u}\cdot {\bf H}_t {\rm d} {\bf x} \\ & = &-2\frac{\rm d}{{\rm d}t}\int ( {\bf H}\cdot \nabla ) {\bf u}\cdot {\bf H} {\rm d} {\bf x}+2\int (( {\bf H}\cdot \nabla) {\bf u}\cdot \nabla) {\bf u}\cdot {\bf H} {\rm d} {\bf x} \\ & &-2\int (( {\bf u}\cdot \nabla) {\bf H}\cdot \nabla) {\bf u}\cdot {\bf H} {\rm d} {\bf x} \\ & &+2\int ( {\bf H}\cdot \nabla ) {\bf u}\cdot( {\bf H}\cdot \nabla) {\bf u} {\rm d} {\bf x}-2\int ( {\bf H}\cdot \nabla ) {\bf u}\cdot( {\bf u}\cdot \nabla) {\bf H} {\rm d} {\bf x}. \end{eqnarray} $

上式第二, 四项可以得到

$ \begin{eqnarray*} \left|\int (( {\bf H}\cdot \nabla) {\bf u}\cdot \nabla) {\bf u}\cdot {\bf H} {\rm d} {\bf x}\right|+\left|\int ( {\bf H}\cdot \nabla ) {\bf u}\cdot( {\bf H}\cdot \nabla) {\bf u} {\rm d} {\bf x}\right|\leq C\| {\bf H}\|_{L^\infty}^2\| \nabla {\bf u}\|_{L^2}^2. \end{eqnarray*} $

上式三, 五项, 由Hölder不等式以及Young不等式可以得到

$ \begin{eqnarray*} &&\left|\int (( {\bf u}\cdot \nabla) {\bf H}\cdot \nabla) {\bf u}\cdot {\bf H} {\rm d} {\bf x}\right|+\left|\int ( {\bf H}\cdot \nabla ) {\bf u}\cdot( {\bf u}\cdot \nabla) {\bf H} {\rm d} {\bf x}\right| \nonumber\\ & \leq &C\int| {\bf H}|| \nabla {\bf u}|| {\bf u}|| \nabla {\bf H}|{\rm d} {\bf x}\\ & \leq &C\| {\bf H}\|_{L^\infty}\| {\bf u}\bar {\bf x}^{-1}\|_{L^{2a}}\| {\bf H}\bar {\bf x}^a\| _{H^1}\| \nabla {\bf u}\|_{L^{2a/(a-1)}} \nonumber\\ &\leq &C\phi^\alpha(t)+ \varepsilon\phi^{-1}(t)\| \nabla^2 {\bf u}\|_{L^2}^2. \end{eqnarray*} $

把上面两部分的估计代入等式(3.14), 得到

(3.15) $ \begin{equation} 2\int ( {\bf H}\cdot \nabla) {\bf H} \cdot {\bf u}_t {\rm d} {\bf x}\leq-2\frac{\rm d}{{\rm d}t}\int ( {\bf H}\cdot \nabla ) {\bf u}\cdot {\bf H} {\rm d} {\bf x} +C\phi^\alpha(t)+ \varepsilon\phi^{-1}(t)\| \nabla^2 {\bf u}\|_{L^2}^2. \end{equation} $

结合(3.13)和(3.15)式, 有

(3.16) $ \begin{equation} \frac{\rm d}{{\rm d}t}\mu\| \nabla {\bf u}\|_{L^2}^2 +2\frac{\rm d}{{\rm d}t}\int \left( ( {\bf H}\cdot \nabla ) {\bf u}\cdot {\bf H}\right) {\rm d} {\bf x}+\|\sqrt \varrho {\bf u}_t\|_{L^2}^2 \leq C\phi^\alpha(t)+4 \varepsilon\phi^{-1}(t)\| \nabla^2 {\bf u}\|_{L^2}^2. \end{equation} $

因此, 只需估计(3.16)式右端最后一项即可.从而设不等式(2.7)的右边所有项$ F = \varrho {\bf u}_t+ \varrho {\bf u}\cdot \nabla {\bf u}+ {\bf H}\cdot \nabla {\bf H} $, 取$ p\in [2, q] $, 有

(3.17) $ \begin{eqnarray} \| \nabla^2 {\bf u}\|_{L^p}\leq C\left(\| \varrho {\bf u}_t\|_{L^p}+\| \varrho {\bf u}\cdot \nabla {\bf u}\|_{L^p}+\|| {\bf H}|| \nabla {\bf H}|\|_{L^p}\right). \end{eqnarray} $

结合不等式(3.17), (3.12)和(3.13), 可得

(3.18) $ \begin{eqnarray} \| \nabla^2 {\bf u}\|_{L^2}&\leq& C\phi^{1/2}(t)\|\sqrt \varrho {\bf u}_t\|_{L^2} +C\| \varrho {\bf u}\cdot \nabla {\bf u}\|_{L^2} +C\phi^\alpha (t)\\ &\leq &C\phi^{1/2}(t)\|\sqrt \varrho {\bf u}_t\|_{L^2}+\frac12\| \nabla^2 {\bf u}\|_{L^2} +C\phi^\alpha (t). \end{eqnarray} $

将估计(3.18)代入(3.16)式, 对得到的不等式在$ (0, t) $上积分, 取$ \varepsilon $适当小, 从而有

(3.19) $ \begin{eqnarray} \frac\mu2\| \nabla {\bf u}\|_{L^2}^2+\int_0^t \|\sqrt \varrho {\bf u}_s\|_{L^2}^2{\rm d}s \leq C+C\| {\bf H}\|_{L^4}^4+C\int_0^t\phi^\alpha(s){\rm d}s, \end{eqnarray} $

而对于$ \int \left( ( {\bf H}\cdot \nabla ) {\bf u}\cdot {\bf H}\right) {\rm d} {\bf x} $的估计用到如下不等式

$ \begin{eqnarray*} \int \left( ( {\bf H}\cdot \nabla ) {\bf u}\cdot {\bf H}\right) {\rm d} {\bf x}\ge-\frac{\mu}{2}\| \nabla {\bf u}\|_{L^2}^2-C\| {\bf H}\|_{L^4}^4. \end{eqnarray*} $

至于(3.19)式右端第二项的估计, 对(2.2)$ _3 $式两边乘以$ 4| {\bf H}|^2 {\bf H} $, 然后在$ B_R $上积分

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}\| {\bf H}\|_{L^4}^4\leq C\int| \nabla {\bf u}|| {\bf H}|^4{\rm d} {\bf x}\leq C\| \nabla {\bf u}\|_{L^2}\| {\bf H}\|_{L^2}\| {\bf H}\|_{L^\infty}^3. \end{eqnarray*} $

并$ t $在$ (0, t) $积分, 可推出

(3.20) $ \begin{equation} \| {\bf H}\|_{L^4}^4\leq C+\int_0^t\phi^\alpha(s){\rm d}s. \end{equation} $

把(3.20)式代入(3.19)式, 再结合(3.5)式, 即可得(3.4)式.证毕.

引理3.2  设$ ( \varrho, {\bf u}, \pi, {\bf H}) $以及$ T_1 $为引理3.1中相关的量.则对任意$ t\in (0, T_1] $, 有

(3.21) $ \begin{equation} \sup\limits_{0\leq s\leq t}s\|\sqrt \varrho {\bf u}_t\|_{L^2}^2+\int_0^ts\| \nabla {\bf u}_s\| _{L^2}^2{\rm d}s\leq C\exp\left(C\int_0^t\phi^\alpha(s){\rm d}s\right). \end{equation} $

证  对方程(2.2)$ _2 $两端关于$ t $求导

(3.22) $ \begin{equation} \varrho {\bf u}_{tt}+ \varrho {\bf u}\cdot \nabla {\bf u}_t-\mu\triangle {\bf u}_t = - \varrho_t( {\bf u}_t+ {\bf u}\cdot \nabla {\bf u})- \varrho {\bf u}_t\cdot \nabla {\bf u}- \nabla \pi_t+ {\bf H}_t\cdot \nabla {\bf H}+ {\bf H}\cdot \nabla {\bf H}_t. \end{equation} $

对上式两端乘以$ {\bf u}_t $并在$ B_R $上积分, 有

(3.23) $ \begin{eqnarray} &&\frac12\frac{\rm d}{{\rm d}t}\int \varrho| {\bf u}_t|^2{\rm d} {\bf x}+\mu\int| \nabla {\bf u}_t|^2{\rm d} {\bf x} \\ & = &-2\int \varrho {\bf u}\cdot \nabla {\bf u}_t\cdot {\bf u}_t {\rm d} {\bf x}-\int \varrho {\bf u}\cdot \nabla( {\bf u}\cdot \nabla {\bf u}\cdot {\bf u}_t){\rm d} {\bf x} \\ &&-\int \varrho {\bf u}_t\cdot \nabla {\bf u}\cdot {\bf u}_t {\rm d} {\bf x}-\int( {\bf H}\otimes {\bf H})_t: \nabla {\bf u}_t{\rm d} {\bf x} \\ &\leq &C\int \varrho| {\bf u}|| {\bf u}_t|\left(| \nabla {\bf u}_t|+| \nabla {\bf u}|^2+| {\bf u}|| \nabla^2 {\bf u} |\right){\rm d} {\bf x}+C\int \varrho| {\bf u}|^2| \nabla {\bf u}|| \nabla {\bf u}_t|{\rm d} {\bf x} \\ &&+C\int \varrho| {\bf u}_t|^2| \nabla {\bf u}|{\rm d} {\bf x}+C \int | {\bf H}|| {\bf H}_t|| \nabla {\bf u}_t|{\rm d} {\bf x}. \end{eqnarray} $

和引理3.1一样, 下面对上式右端进行逐项估.首先根据(3.1), (3.5), (3.9), (3.10)和(3.12)式, 对任意的$ \varepsilon\in (0, 1) $, (3.23)式右端第一项和第二项的类似部分可以做如下估计

(3.24) $ \begin{eqnarray} &&\int \varrho| {\bf u}|| {\bf u}_t|\left(| \nabla {\bf u}_t|+| \nabla {\bf u}|^2+| {\bf u}|| \nabla^2 {\bf u} |\right){\rm d} {\bf x} \\ & \leq &C\|\sqrt \varrho {\bf u}\|_{L^6}\|\sqrt \varrho {\bf u}_t\|_{L^2}^{1/2} \|\sqrt \varrho {\bf u}_t\|_{L^6}^{1/2}\left(\| \nabla {\bf u}_t\|_{L^2} +\| \nabla {\bf u}\|_{L^4}^2\right) \\ &&+C\| \varrho^{1/4} {\bf u}\|_{L^{12}}^2\|\sqrt \varrho {\bf u}_t\|_{L^2}^{1/2} \|\sqrt \varrho {\bf u}_t\|_{L^6}^{1/2}\| \nabla^2 {\bf u}\|_{L^2} \\ &\leq &C\phi^\alpha(t)\|\sqrt \varrho {\bf u}_t\|_{L^2}^{1/2}\left( \|\sqrt \varrho {\bf u}_t\|_{L^2}^{1/2}+ \| \nabla {\bf u}_t\|_{L^2}^{1/2}\right) \left(\| \nabla {\bf u}_t\|_{L^2}+\| \nabla^2 {\bf u}\|_{L^2}+\phi(t)\right) \\ &\leq& \varepsilon\| \nabla {\bf u}_t\|_{L^2}^2+C\phi^\alpha(t)\left( \| \nabla^2 {\bf u}\|_{L^2}^2+\|\sqrt \varrho {\bf u}_t\|_{L^2}^2+1\right). \end{eqnarray} $

对于(3.23)式右端第二项的剩下部分, 利用Hölder不等式, (3.10)式以及(3.12)式可得

(3.25) $ \begin{eqnarray} \int \varrho| {\bf u}|^2| \nabla {\bf u}|| \nabla {\bf u}_t|{\rm d} {\bf x}&\leq & C\|\sqrt \varrho {\bf u}\|_{L^8}^2\| \nabla {\bf u}\|_{L^4}\| \nabla {\bf u}_t\|_{L^2}\\ & \leq & \varepsilon\| \nabla {\bf u}_t\|_{L^2}^2+C\left(\phi^\alpha (t)+\| \nabla^2 {\bf u}\|_{L^2}^2 \right). \end{eqnarray} $

同样地, 利用Hölder不等式以及(3.9)式可得

(3.26) $ \begin{eqnarray} \int \varrho| {\bf u}_t|^2| \nabla {\bf u}|{\rm d} {\bf x}&\leq &\| \nabla {\bf u}\|_{L^2}\|\sqrt \varrho {\bf u}_t\|_{L^6}^{3/2} \|\sqrt \varrho {\bf u}_t\|_{L^2}^{1/2} \\ & \leq & \varepsilon\| \nabla {\bf u}_t\|_{L^2}^2+C\phi^\alpha(t)\|\sqrt \varrho {\bf u}_t\|_{L^2}^2. \end{eqnarray} $

最后, 借助(2.2)$ _3 $和(3.10)式, 可知

(3.27) $ \begin{eqnarray} \int | {\bf H}|| {\bf H}_t|| \nabla {\bf u}_t|{\rm d} {\bf x} &\leq &C\int \left(| {\bf H}|| \nabla {\bf u}|+| \nabla {\bf H}|| {\bf u}|\right)| \nabla {\bf u}_t|{\rm d} {\bf x} \\ & \leq &C\left(\| {\bf H}\|_{L^\infty}\| \nabla {\bf u}\|_{L^2}+\| {\bf H}\bar {\bf x}^a\|_{W^{1, q}} \| {\bf u}\bar {\bf x}^{-a}\|_{L^{2q/(q-2)}}\right)\| \nabla {\bf u}_t\|_{L^2} \\ &\leq & \varepsilon\| \nabla {\bf u}_t\|_{L^2}^2+C\phi^\alpha(t). \end{eqnarray} $

对上述所有估计进行汇总, 并令$ \varepsilon $适当小, 利用估计(3.18), 有

(3.28) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\int \varrho| {\bf u}_t|^2{\rm d} {\bf x}+\mu\int| \nabla {\bf u}_t|^2{\rm d} {\bf x} & \leq &C\phi^\alpha(t)\left(1+\|\sqrt \varrho {\bf u}_t\|_{L^2}^2 +\| \nabla^2 {\bf u}\|_{L^2}^2 \right) \\ & \leq &C\phi^\alpha(t)\|\sqrt \varrho {\bf u}_t\|_{L^2}^2+C\phi^\alpha(t), \end{eqnarray} $

因此, 利用Gronwall不等式就可以得到(3.21).证毕.

引理3.3  在引理3.1中$ ( \varrho, {\bf u}, \pi, {\bf H}) $以及$ T_1 $的假设下.对任意的$ t\in (0, T_1] $, 有

(3.29) $ \begin{equation} \sup\limits_{0\leq s\leq t}\left(\| \varrho\bar {\bf x}^a\|_{L^1\cap H^1\cap W^{1, q}}+\| {\bf H}\bar {\bf x}^a\|_{H^1\cap W^{1, q}}\right) \leq \exp\left(C\exp\left(C\int_0^t\phi^\alpha(s){\rm d}s\right)\right). \end{equation} $

证  对方程(2.2)$ _1 $两边同时乘以$ \bar {\bf x}^a $, 对任意的$ \bf u\in \tilde D^{1, 2}(B_R) $, 由(3.9)式, 有

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}t}\int \varrho\bar {\bf x}^a{\rm d} {\bf x} &\leq &C\int \varrho | {\bf u}|\bar {\bf x}^{a-1} \ln^{1+\eta_0}(e+| {\bf x}|^2){\rm d} {\bf x}\\ &\leq &C\| \varrho\bar {\bf x}^{a-1+8/(8+a)}\|_{L^{(8+a)/(7+a)}}\| {\bf u} \bar {\bf x}^{-4/(8+a)}\|_{L^{8+a}}\\ &\leq &C\phi^\alpha(t), \end{eqnarray*} $

这表明

(3.30) $ \begin{equation} \sup\limits_{0\leq s\leq t}\| \varrho\bar {\bf x}^a\|_{L^1}\leq C\left(1+ \int_0^t\phi^\alpha(s){\rm d}s\right). \end{equation} $

至于$ \varrho\bar {\bf x}^a $的高阶估计类似于下面的$ {\bf H}\bar {\bf x}^a $估计.首先, 对$ 0<\delta<1 $根据Sobolev不等式以及(3.10)式, 有

(3.31) $ \begin{eqnarray} \| {\bf u}\bar {\bf x}^{-\delta}\|_{L^\infty}&\leq &C\left(\| {\bf u}\bar {\bf x}^{-\delta} \|_{L^{4/\delta}}+\| \nabla( {\bf u}\bar {\bf x}^{-\delta})\|_{L^3} \right) \\ &\leq& C\left(\| {\bf u}\bar {\bf x}^{-\delta} \|_{L^{4/\delta}}+ \| \nabla {\bf u}\|_{L^3}+\| {\bf u}\bar {\bf x}^{-\delta} \|_{L^{4/\delta}}\|\bar {\bf x}^{-1} \nabla\bar {\bf x}\|_{L^{12/(4-3\delta)}} \right) \\ &\leq &C\left(\phi^\alpha (t) +\| \nabla^2 {\bf u}\|_{L^2}\right). \end{eqnarray} $

令$ \tilde {\bf H}\triangleq {\bf H}\bar {\bf x}^a $, 则(2.2)$ _3 $式可改写为

(3.32) $ \begin{equation} \tilde {\bf H}_t+ {\bf u}\cdot \nabla \tilde {\bf H}-a {\bf u}\cdot \nabla\ln\bar {\bf x}\cdot \tilde {\bf H} = \tilde {\bf H}\cdot \nabla {\bf u}. \end{equation} $

由(3.31)式的估计以及Gagliardo-Nirenberg不等式, 上述等式容易得到

(3.33) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\| \tilde {\bf H}\|_{L^2} &\leq &C\left(\| \nabla {\bf u}\|_{L^\infty} +\| {\bf u}\cdot \nabla\ln\bar {\bf x}\|_{L^\infty}\right)\| \tilde {\bf H}\|_{L^2}\\ &\leq &C\left(\phi^\alpha(t)+\| \nabla^2 {\bf u}\|_{L^2\cap W^{1, q}}\right)\| \tilde {\bf H}\|_{L^2}. \end{eqnarray} $

对$ p\in [2, q] $, 进一步可以得到

(3.34) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\| \nabla \tilde {\bf H}\|_{L^p} &\leq & C\left( 1+\| \nabla {\bf u}\|_{L^\infty} +\| {\bf u}\cdot \nabla\ln\bar {\bf x}\|_{L^\infty}\right)\| \nabla \tilde {\bf H}\|_{L^p} \\ &&+C\left(\|| \nabla {\bf u}|| \nabla\ln\bar {\bf x}|\|_{L^p}+\|| {\bf u}|| \nabla^2\ln\bar {\bf x}|\| _{L^p} +\| \nabla^2 {\bf u}\|_{L^p}\right)\| \tilde {\bf H}\|_{L^\infty} \\ &\leq &C\left(\phi^\alpha(t)+\| \nabla {\bf u}\|_{L^2\cap W^{1, q}}\right)\| \nabla \tilde {\bf H}\|_{L^p} \\ &&+C\left(\| \nabla {\bf u}\|_{L^p}+\| {\bf u}\bar {\bf x}^{-1/4}\|_{L^\infty} \|\bar {\bf x}^{-3/2}\|_{L^p}+\| \nabla^2 {\bf u}\|_{L^p}\right)\| \tilde {\bf H}\|_{L^\infty} \\ &\leq& C\left(\phi^\alpha(t)+\| \nabla^2 {\bf u}\|_{L^2\cap L^p}\right) \left(1+\| \nabla \tilde {\bf H}\|_{L^p}+\| \nabla \tilde {\bf H}\|_{L^q}\right). \end{eqnarray} $

结合(3.33)和(3.34)式, 有

(3.35) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\left(\| \tilde {\bf H}\|_{L^2}+\| \nabla \tilde {\bf H}\|_{L^p}\right) \\ &\leq & C\left(\phi^\alpha(t)+\| \nabla^2 {\bf u}\|_{L^2}+\| \nabla^2 {\bf u}\|_{ L^p}\right) \left(1+\| \nabla \tilde {\bf H}\|_{L^p}+\| \tilde {\bf H}\|_{L^2}+ \| \nabla \tilde {\bf H}\|_{L^q}\right). \end{eqnarray} $

对于$ \| \nabla \tilde {\bf H}\|_{L^p} $, 由文献[27]有

(3.36) $ \begin{equation} \begin{array}{ll} { } \int_0^t\left(\| \nabla^2 {\bf u}\|_{ L^p}^{(p+1)/p}+\| \nabla\pi\|_{ L^p}^{(p+1)/p}\right){\rm d}s\le C\exp\left(C\int_0^t\phi^\alpha(s){\rm d}s\right), \\ { } \int_0^t\left(s\| \nabla^2 {\bf u}\|_{L^2\cap L^p}^2+s\| \nabla\pi\|_{L^2\cap L^p}^2\right){\rm d}s\leq C\exp\left(C\int_0^t\phi^\alpha(s){\rm d}s\right). \end{array} \end{equation} $

把(3.36)式代入(3.35)式, 再利用Gronwall不等式可以得到(3.29)式.证毕.

有了引理3.1–3.3的估计之后, 可以直接给出命题3.1的证明.

$ \phi(t)\leq \exp\left(C\exp\left(C\int_0^t \phi^\alpha(s){\rm d}s\right)\right). $

令$ M\triangleq e^{Ce} $以及$ T_0 \triangleq \min\{T_1, (CM^\alpha)^{-1}\} $, $ { }\sup_{0\leq t\leq T_0}\phi(t)\leq M, $再结合(3.4), (3.18)和(3.36)式可以推出(3.3)式.命题3.1证毕.

在这部分的先验估计中, 一般常数$ C $除了依赖于$ \mu, q, a, \eta_0, N_0 $和$ C_0 $之外, 还依赖于一些初值$ \delta_0 $, $ \| \nabla^2 {\bf u}_0\|_{L^2} $, $ \| \nabla^2 \varrho_0\|_{L^q} $, $ \| \nabla^2 {\bf H}_0\|_{L^q} $, $ \|\bar {\bf x}^{\delta_0} \nabla^2 \varrho_0\|_{L^2} $, $ \|\bar {\bf x}^{\delta_0} \nabla^2 {\bf H}_0\|_{L^2} $以及$ \|{\bf g}\|_{L^2} $.

首先, 有下列估计成立

引理4.1  存在常数$ C > 0 $, 使得

(4.1) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}\left(\|\bar {\bf x}^{\delta_0} \nabla^2 \varrho\|_{L^2} +\|\bar {\bf x}^{\delta_0} \nabla^2 {\bf H}\|_{L^2}\right)\leq C. \end{equation} $

证  由兼容性条件(1.10), (2.1)和方程(2.2)$ _2 $, 有$ \sqrt \varrho {\bf u}_t(t = 0, {\bf x})\triangleq -{\bf g}-\sqrt{ \varrho_0} {\bf u}_0\cdot \nabla {\bf u}_0, $对方程(3.28)在$ (0, T_0) $上积分并利用(3.3)和(3.4)式, 可得

(4.2) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}\|\sqrt \varrho {\bf u}_t\|_{L^2}^2+\int_0^{T_0} \| \nabla {\bf u}_t\|_{L^2}^2dt\leq C, \end{equation} $

结合上式和(3.3)式以及(3.18)式可以推出

(4.3) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}\| \nabla {\bf u}\|_{H^1}\leq C. \end{equation} $

另一方面, 结合(3.3)式, (3.31)式以及(4.3)式, 对$ \delta\in (0, 1] $, 显然有

(4.4) $ \begin{equation} \| \varrho^\delta {\bf u}\|_{L^\infty}+\|\bar {\bf x}^{-\delta} {\bf u}\|_{L^\infty}\leq C(\delta). \end{equation} $

而对$ 2\leq r\leq q $, 通过直接计算

(4.5) $ \begin{equation} \| \varrho_t(\bar {\bf x}^{(1+a)/2}+| {\bf u}|)\|_{L^r}+\| {\bf H}_t(\bar {\bf x}^{(1+a)/2} +| {\bf u}|)\|_{L^r}\leq C. \end{equation} $

对$ \delta\in (0, 1] $和$ s>2/\delta $, 由(2.2)$ _1 $, (2.2)$ _3 $, (3.3), (4.3)和(4.4)式, 可以从(3.9), (4.2), (4.3)和(4.4)式得到

(4.6) $ \begin{eqnarray} \|\bar {\bf x}^{-\delta} {\bf u}_t\|_{L^s}+\|\bar {\bf x}^{-\delta} {\bf u}\cdot \nabla {\bf u}\|_{L^s} &\leq & C\|\bar {\bf x}^{-\delta} {\bf u}_t\|_{L^s} +C\|\bar {\bf x}^{-\delta} {\bf u}\|_{L^\infty}\| \nabla {\bf u}\|_{L^s}\\ &\leq &C(\delta, s)+C(\delta, s)\| \nabla {\bf u}_t\|_{L^2}. \end{eqnarray} $

下面定义$ \bar {\bf H}\triangleq \bar {\bf x}^{\delta_0} {\bf H} $和$ \bar \varrho\triangleq \bar {\bf x}^{\delta_0} \varrho $, 则由(3.3)式容易得到

(4.7) $ \begin{equation} \|\bar \varrho\|_{L^\infty}+\| \nabla\bar \varrho\|_{L^2\cap L^q}+\|\bar {\bf H}\|_{L^\infty}+ \| \nabla \bar {\bf H}\|_{L^2\cap L^q}\leq C, \end{equation} $

其中$ \bar \varrho $和$ \bar {\bf H} $分别满足

(4.8) $ \begin{equation} \bar \varrho_t+ {\bf u}\cdot \nabla\bar \varrho-\delta_0 \bar \varrho {\bf u}\cdot \nabla \ln\bar {\bf x} = 0 \end{equation} $

和

(4.9) $ \begin{equation} \bar {\bf H}_t+ {\bf u}\cdot \nabla\bar {\bf H}-\delta_0 {\bf u}\cdot \nabla\ln\bar {\bf x}\cdot \bar {\bf H} = \bar {\bf H}\cdot \nabla {\bf u}, \end{equation} $

因此, 通过能量方法可以得到

(4.10) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\| \nabla^2\bar {\bf H}\|_{L^2}&\leq & C\left(1+\| \nabla {\bf u}\| _{L^\infty} +\| {\bf u}\cdot \nabla\ln\bar {\bf x}\|_{L^\infty}\right)\| \nabla^2\bar {\bf H}\|_{L^2} +C\|| \nabla^2 {\bf u}|| \nabla\bar {\bf H}|\|_{L^2} \\ &&+C\|| \nabla\bar {\bf H}|| \nabla {\bf u}|| \nabla\ln\bar {\bf x}|\|_{L^2}+C\|| \nabla\bar {\bf H}|| {\bf u}| | \nabla^2 \ln\bar {\bf x}|\|_{L^2} \\ &&+C\|\bar {\bf H}\|_{L^\infty}\left(\| \nabla^2\left( {\bf u}\cdot \nabla\ln\bar {\bf x} \right)\|_{L^2}+\| \nabla^3 {\bf u}\|_{L^2}\right) \\ &\leq &C\left(1+\| \nabla {\bf u}\|_{L^\infty}\right)\| \nabla^2\bar {\bf H}\|_{L^2} +C\| \nabla^2 {\bf u}\|_{L^{2q/(q-2)}}\| \nabla\bar {\bf H}\|_{L^q} \\ &&+C\| \nabla\bar {\bf H}\|_{L^2}\| \nabla {\bf u}\|_{L^\infty}+C\| \nabla\bar {\bf H}\|_{L^2} \|| {\bf u}|| \nabla^2\ln\bar {\bf x}|\|_{L^\infty} \\ &&+C\| \nabla^2 {\bf u}\|_{L^2}+C\| \nabla {\bf u}\|_{L^2}+C\|| {\bf u}|| \nabla^3\ln\bar {\bf x}|\|_{L^2} +C\| \nabla^3 {\bf u}\|_{L^2} \\ &\leq & C\left(1+\| \nabla {\bf u}\|_{L^\infty}\right)\| \nabla^2\bar {\bf H}\|_{L^2}+C +C\| \nabla^3 {\bf u}\|_{L^2}, \end{eqnarray} $

在第二个和第三个不等式用到了(4.4)式和(4.7)式.类似的, 可以从(4.8)式得到

(4.11) $ \begin{equation} \frac{\rm d}{{\rm d}t}\| \nabla^2\bar \varrho\|_{L^2}\leq C\left(1+\| \nabla {\bf u}\|_{L^\infty}\right)\| \nabla^2\bar \varrho\|_{L^2}+C +C\| \nabla^3 {\bf u}\|_{L^2}. \end{equation} $

结合(4.10)和(4.11)式, 有

(4.12) $ \begin{equation} \frac{\rm d}{{\rm d}t}\left(\| \nabla^2\bar \varrho\|_{L^2}+\| \nabla^2\bar {\bf H}\|_{L^2}\right) \leq C\left(1+\| \nabla {\bf u}\|_{L^\infty}\right)\left(\| \nabla^2\bar \varrho\|_{L^2}+ \| \nabla^2\bar {\bf H}\|_{L^2}\right) +C+C\| \nabla^3 {\bf u}\|_{L^2}. \end{equation} $

至于$ \| \nabla^3 {\bf u}\|_{L^2} $, 由(2.7)和(4.3)式, 有估计

(4.13) $ \begin{eqnarray} \| \nabla^3 {\bf u}\|_{L^2}&\leq &C \| \nabla( \varrho {\bf u}_t)\|_{L^2}+C\| \nabla( \varrho {\bf u}\cdot \nabla {\bf u})\|_{L^2} +C\| \nabla^2| {\bf H}|^2\|_{L^2}+C \| \nabla( {\bf H}\cdot \nabla {\bf H})\|_{L^2} \\ &\leq &C\|\bar \varrho\|_{L^\infty}\| \nabla {\bf u}_t\|_{L^2}+C\|\bar {\bf x}^a \nabla \varrho\|_{L^q}\|\bar {\bf x}^{-a} {\bf u}_t\|_{L^{2q/(q-2)}} \\ &&+C\|\bar {\bf x}^{-\delta} {\bf u}\|_{L^\infty} \|\bar {\bf x}^\delta \nabla \varrho\|_{L^q} \| \nabla {\bf u}\|_{L^{2q/(q-2)}}+C\|\bar \varrho\|_{L^\infty}\| \nabla {\bf u}\|_{L^4}^2 \\ &&+C\|\bar \varrho\|_{L^\infty}\|\bar {\bf x}^{-\delta} {\bf u}\|_{L^\infty}\| \nabla^2 {\bf u}\| _{L^2}^2 +C\| \nabla\bar {\bf H}\|_{L^4}^2 +C\|\bar {\bf H}\|_{L^\infty}\| \nabla^2 {\bf H}\|_{L^2} \\ &\leq &C\| \nabla {\bf u}_t\|_{L^2}+C\left(\| \nabla^2\bar \varrho\|_{L^2} +\| \nabla^2\bar {\bf H}\|_{L^2}\right)+C, \end{eqnarray} $

上面最后一个不等式用了估计(3.3), (4.3), (4.4), (4.6)以及如下估计

(4.14) $ \begin{eqnarray} \|\bar {\bf x}^{\delta_0} \nabla^2 \varrho\|_{L^2}+\|\bar {\bf x}^{\delta_0} \nabla^2 {\bf H}\|_{L^2} \leq C\| \nabla^2(\bar {\bf x}^{\delta_0} \varrho)\|_{L^2}+\| \nabla^2(\bar {\bf x}^{\delta_0} {\bf H})\|_{L^2}+C. \end{eqnarray} $

把(4.13)式代入(4.12)式, 可知

$ \begin{eqnarray*} &&\frac{\rm d}{{\rm d}t}\left(\| \nabla^2(\bar {\bf x}^{\delta_0} \varrho)\|_{L^2} +\| \nabla^2(\bar {\bf x}^{\delta_0} {\bf H})\|_{L^2}\right) \nonumber\\ &\leq &C(1+\| \nabla^2 {\bf u}\|_{L^q})\left(\| \nabla^2(\bar {\bf x}^{\delta_0} \varrho)\|_{L^2} +\| \nabla^2(\bar {\bf x}^{\delta_0} {\bf H})\|_{L^2}\right) +C\| \nabla {\bf u}_t\|_{L^2}+C, \end{eqnarray*} $

最后通过(3.3), (4.2)和(4.14)式, 以及Gronwall不等式, 可得(4.1)式.引理4.1证毕.

关于$ \nabla {\bf u}_t $, 有如下的正则性估计以及相应的扩散估计.

引理4.2  存在常数$ C > 0 $, 成立

(4.15) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}t\| \nabla {\bf u}_t\|_{L^2}^2+\int_0^{T_0}t\left( \|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2+\| \nabla^2 {\bf u}_{t}\|_{L^2}^2\right){\rm d}t\leq C. \end{equation} $

证  对(3.22)式两端乘以$ {\bf u}_{tt} $并在$ B_R $上积分, 通过分部积分可得

(4.16) $ \begin{eqnarray} &&\frac12\frac{\rm d}{{\rm d}t}\left(\mu\| \nabla {\bf u}_t\|_{L^2}^2\right)+\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2 \\ & = &-\int\left(2 \varrho {\bf u}\cdot \nabla {\bf u}_t\cdot {\bf u}_{tt}+ \varrho {\bf u}_t\cdot \nabla {\bf u} \cdot {\bf u}_{tt}\right){\rm d} {\bf x}-\int \varrho {\bf u}\cdot \nabla( {\bf u}\cdot \nabla {\bf u})\cdot {\bf u}_{tt}{\rm d} {\bf x} \\ &&-\int \varrho {\bf u}\cdot \nabla {\bf u}_{tt}\cdot {\bf u}_t{\rm d} {\bf x}-\int \varrho {\bf u}\cdot \nabla {\bf u}_{tt} \cdot( {\bf u} \cdot \nabla) {\bf u} {\rm d} {\bf x} \\ &&-\int {\bf H}_t\cdot \nabla {\bf u}_{tt}\cdot {\bf H} {\rm d} {\bf x}-\int {\bf H}\cdot \nabla {\bf u}_{tt}\cdot {\bf H}_t{\rm d} {\bf x} \\ & = & \sum\limits_{i = 1}^6{I_i}. \end{eqnarray} $

下面对(4.16)式进行逐项估计.对于$ I_1 $, $ I_2 $项, 由(3.3), (4.2)–(4.4)式以及(4.6)式有

(4.17) $ \begin{eqnarray} &&\left|\int\left(2 \varrho {\bf u}\cdot \nabla {\bf u}_t\cdot {\bf u}_{tt}+ \varrho {\bf u}_t\cdot \nabla {\bf u} \cdot {\bf u}_{tt}\right){\rm d} {\bf x}\right|+\left|\int \varrho {\bf u}\cdot \nabla( {\bf u}\cdot \nabla {\bf u}) \cdot {\bf u}_{tt}{\rm d} {\bf x}\right| \\ &\leq & \varepsilon\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2 +C( \varepsilon)\left(\|\sqrt \varrho {\bf u}\| _{L^\infty}^2 \| \nabla {\bf u}_t\|_{L^2}^2+\|\sqrt \varrho {\bf u}_t\|^2_{L^4}\| \nabla {\bf u}\| _{L^4}^2\right) \\ &&+C( \varepsilon)\left(\|\sqrt \varrho {\bf u}\|_{L^\infty}^2\| \nabla {\bf u}\|_{L^4}^2 +\| \varrho^{1/4} {\bf u}\|_{L^\infty}^4\| \nabla^2 {\bf u}\|_{L^2}^2\right) \\ &\leq & \varepsilon\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2 +C( \varepsilon)\left(1+\| \nabla {\bf u}_t\|_{L^2}^2 \right). \end{eqnarray} $

对于$ I_3 $, $ I_4 $项, 有

(4.18) $ \begin{eqnarray} &&-\int \varrho {\bf u}\cdot \nabla {\bf u}_{tt}\cdot {\bf u}_t{\rm d} {\bf x}-\int \varrho {\bf u}\cdot \nabla {\bf u}_{tt} \cdot( {\bf u} \cdot \nabla) {\bf u} {\rm d} {\bf x} \\ & = &-\frac{\rm d}{{\rm d}t}\int\left( \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot {\bf u}_t+ \varrho {\bf u} \cdot \nabla {\bf u}_{t}\cdot( {\bf u} \cdot \nabla) {\bf u}\right){\rm d} {\bf x} +\int( \varrho {\bf u})_t\cdot \nabla {\bf u}_{t}\cdot {\bf u}_t{\rm d} {\bf x} \\ &&+\int( \varrho {\bf u})_t\cdot \nabla {\bf u}_{t}\cdot( {\bf u} \cdot \nabla) {\bf u} {\rm d} {\bf x}+\int \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot {\bf u}_{tt}{\rm d} {\bf x} \\ &&+\int \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot( {\bf u}_t \cdot \nabla) {\bf u} {\rm d} {\bf x} +\int \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot( {\bf u} \cdot \nabla) {\bf u}_t {\rm d} {\bf x}. \end{eqnarray} $

上式右端不含$ \frac{\rm d}{{\rm d}t} $的项都可以估计.首先, 由Hölder不等式以及(3.3)和(4.4)–(4.6)式有

(4.19) $ \begin{eqnarray} \int( \varrho {\bf u})_t\cdot \nabla {\bf u}_{t}\cdot {\bf u}_t{\rm d} {\bf x} &\leq & C\| \varrho\bar {\bf x}^a\|_{L^\infty}\|\bar {\bf x}^{-a/2} {\bf u}_t\|_{L^4}^2 \| \nabla {\bf u}_t\|_{L^2}\\ &&+C\|\bar {\bf x}^{(1+a)/2} \varrho_t\|_{L^2} \|\bar {\bf x}^{-1/2} {\bf u}\|_{L^\infty}\|\bar {\bf x}^{-a/2} {\bf u}_t\|_{L^4} \| \nabla {\bf u}_t\|_{L^4} \\ &\leq &\delta\| \nabla^2 {\bf u}_t\|_{L^2}^2+C(\delta)\| \nabla {\bf u}_t\|_{L^2}^4 +C(\delta), \end{eqnarray} $

同样不难得到下面估计

(4.20) $ \begin{eqnarray} \int( \varrho {\bf u})_t\cdot \nabla {\bf u}_{t}\cdot( {\bf u} \cdot \nabla) {\bf u} {\rm d} {\bf x} &\leq & C\| \varrho\bar {\bf x}^a\|_{L^\infty} \|\bar {\bf x}^{-a/2} {\bf u}_t\|_{L^4} \|\bar {\bf x}^{-a/2} {\bf u}\cdot \nabla {\bf u}\|_{L^4}\| \nabla {\bf u}_t\|_{L^2} \\ &&+C\|\bar {\bf x}^{(1+a)/2} \varrho_t\|_{L^2}\|\bar {\bf x}^{-1/2} {\bf u}\|_{L^\infty} \|\bar {\bf x}^{-a/2} {\bf u}\|_{L^\infty}\| \nabla {\bf u}\|_{L^4}\| \nabla {\bf u}_t\|_{L^4} \\ &\leq &\delta\| \nabla^2 {\bf u}_t\|_{L^2}^2+C(\delta)\| \nabla {\bf u}_t\|_{L^2}^4 +C(\delta). \end{eqnarray} $

从(4.4)式容易得到

(4.21) $ \begin{eqnarray} \int \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot {\bf u}_{tt}{\rm d} {\bf x}&\leq & C\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}\|\sqrt \varrho {\bf u}\|_{L^\infty}\| \nabla {\bf u}_t\|_{L^2}\\ &\leq & \varepsilon\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2 +C( \varepsilon)\| \nabla {\bf u}_t\|_{L^2}^2. \end{eqnarray} $

由(4.3), (4.4)及(4.6)式, 对于(4.18)式右端的第五项

(4.22) $ \begin{eqnarray} \int \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot( {\bf u}_t \cdot \nabla) {\bf u} {\rm d} {\bf x} &\leq &C\| \varrho\bar {\bf x}^a\|_{L^\infty}\|\bar {\bf x}^{-a/2} {\bf u}\|_{L^\infty} \|\bar {\bf x}^{-a/2} {\bf u}_t\|_{L^4}\| \nabla {\bf u}\|_{L^4}\| \nabla {\bf u}_t\|_{L^2} \\ &\leq & C+C( \varepsilon)\| \nabla {\bf u}_t\|_{L^2}^2, \end{eqnarray} $

同样地, 有

(4.23) $ \begin{eqnarray} \int \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot( {\bf u} \cdot \nabla) {\bf u}_t {\rm d} {\bf x} \leq C\| \varrho\bar {\bf x}^a\|_{L^\infty}\|\bar {\bf x}^{-a/2} {\bf u}\|_{L^\infty}^2\| \nabla {\bf u}_t\|_{L^2}^2 \leq C\| \nabla {\bf u}_t\|_{L^2}^2. \end{eqnarray} $

把上述估计(4.19)–(4.23)代入(4.18)式, 可以得到

(4.24) $ \begin{eqnarray} &&-\int \varrho {\bf u}\cdot \nabla {\bf u}_{tt}\cdot {\bf u}_t{\rm d} {\bf x}-\int \varrho {\bf u}\cdot \nabla {\bf u}_{tt} \cdot( {\bf u} \cdot \nabla) {\bf u} {\rm d} {\bf x}\\ & = &-\frac{\rm d}{{\rm d}t}\int\left( \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot {\bf u}_t+ \varrho {\bf u} \cdot \nabla {\bf u}_{t}\cdot( {\bf u} \cdot \nabla) {\bf u}\right){\rm d} {\bf x}+ \varepsilon\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2 \\ &&+C( \varepsilon, \delta)\| \nabla {\bf u}_t\|_{L^2}^4+\delta\| \nabla^2 {\bf u}_t\|_{L^2}^2 +C( \varepsilon, \delta). \end{eqnarray} $

对于$ I_5 $以及$ I_6 $项, 由(2.2)$ _3 $, (4.1), (4.3)以及(4.5)式, 容易得到

(4.25) $ \begin{eqnarray} &&-\int {\bf H}_t\cdot \nabla {\bf u}_{tt}\cdot {\bf H} {\rm d} {\bf x}-\int {\bf H}\cdot \nabla {\bf u}_{tt}\cdot {\bf H}_t{\rm d} {\bf x} \\ & = &-\frac{\rm d}{{\rm d}t}\int\left( {\bf H}_t\cdot \nabla {\bf u} _t\cdot {\bf H} + {\bf H}\cdot \nabla {\bf u}_t\cdot {\bf H}_t\right){\rm d} {\bf x} \\ &&+\int {\bf H}_{tt}\cdot \nabla {\bf u}_t\cdot {\bf H} {\rm d} {\bf x}+2\int {\bf H}_t\cdot \nabla {\bf u}_t\cdot {\bf H}_t {\rm d} {\bf x}+\int {\bf H}\cdot \nabla {\bf u}_t\cdot {\bf H}_{tt}{\rm d} {\bf x} \\ &\leq &-\frac{\rm d}{{\rm d}t}\int\left( {\bf H}_t\cdot \nabla {\bf u} _t\cdot {\bf H} + {\bf H}\cdot \nabla {\bf u}_t\cdot {\bf H}_t\right){\rm d} {\bf x} \\ &&+C\| {\bf H}_t\bar {\bf x}^{(a+1)/2}\|_{L^q} \| \nabla {\bf u}_t\|_{L^2}\left(\| {\bf u}\bar {\bf x}^{-1/2}\|_{L^\infty} \| \nabla {\bf H}\|_{(q-2)/2q}+\| {\bf H}\|_{L^\infty}\| \nabla {\bf u}\|_{(q-2)/2q}\right) \\ &&+C\|\bar {\bf x}^{a/2} {\bf H}\|_{L^\infty}\|\bar {\bf x}^{-a/2} {\bf u}_t\|_{L^4} \| \nabla {\bf H}\|_{L^4} \| \nabla {\bf u}_t\|_{L^2}+C\| {\bf H}\|_{L^\infty}^2\| \nabla {\bf u}_t\|_{L^2}^2 \\ &&+C\|\bar {\bf x}^{a/2} {\bf H}\|_{L^\infty} \|\bar {\bf x}^{-a/2} {\bf u}\|_{L^\infty} \| \nabla {\bf H}_t\|_{L^2} \| \nabla {\bf u}_t\|_{L^2} \\ &&+C\| {\bf H}\|_{L^\infty}\| {\bf H}_t \bar {\bf x}^{a/2}\|_{L^q}\| \nabla {\bf u}_t\|_{L^2}\| \nabla {\bf u}\|_{(q-2)/2q} \\ &\leq &-\frac{\rm d}{{\rm d}t}\int\left( {\bf H}_t\cdot \nabla {\bf u} _t\cdot {\bf H} + {\bf H}\cdot \nabla {\bf u}_t\cdot {\bf H}_t\right){\rm d} {\bf x}+C \left(\| \nabla {\bf u}_t\|_{L^2}^2+1\right), \end{eqnarray} $

上式最后一个不等式用到了估计(4.1), (4.3), (4.4), (4.6), (4.7)以及下面事实

(4.26) $ \begin{eqnarray} \| \nabla {\bf H}_t\|_{L^2}&\leq& C\|| \nabla {\bf u}|| \nabla {\bf H}|\|_{L^2}+C\|| {\bf u}|| \nabla^2 {\bf H}|\| _{L^2}+C\|| {\bf H}|| \nabla^2 {\bf u}|\|_{L^2} \\ &\leq &C\| \nabla {\bf u}\|_{L^{2q/(q-2)}}\| \nabla {\bf H}\|_{L^q}+C\|\bar {\bf x}^{-\delta_0} {\bf u}\|_{L^\infty} \|\bar {\bf x}^{\delta_0} \nabla^2 {\bf H}\|_{L^2}+C\| {\bf H}\|_{L^\infty} \| \nabla^2 {\bf u}\|_{L^2} \\ &\leq &C. \end{eqnarray} $

令

$ \begin{eqnarray*} \Phi(t)&\triangleq&\mu\| \nabla {\bf u}_t\|_{L^2}^2+\int\left( \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot {\bf u}_t+ \varrho {\bf u} \cdot \nabla {\bf u}_{t}\cdot( {\bf u} \cdot \nabla) {\bf u}\right){\rm d} {\bf x} \nonumber\\ &&+\int\left( {\bf H}_t\cdot \nabla {\bf u} _t\cdot {\bf H} + {\bf H}\cdot \nabla {\bf u}_t\cdot {\bf H}_t\right){\rm d} {\bf x}, \end{eqnarray*} $

把(4.17)式和(4.24)–(4.25)式代入(4.16)式, 取$ \varepsilon $适当小, 可知

(4.27) $ \begin{equation} \Phi'(t)+\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2\leq C\delta\| \nabla^2 {\bf u}_t\|_{L^2}^2 +C\| \nabla {\bf u}_t\|_{L^2}^4+C. \end{equation} $

通过(4.2)–(4.5)式, 有

$ \begin{eqnarray*} &&\left|\int\left( \varrho {\bf u}\cdot \nabla {\bf u}_{t}\cdot {\bf u}_t+ \varrho {\bf u} \cdot \nabla {\bf u}_{t}\cdot( {\bf u} \cdot \nabla) {\bf u}\right){\rm d} {\bf x}\right|+\left|\int\left( {\bf H}_t\cdot \nabla {\bf u} _t\cdot {\bf H} + {\bf H}\cdot \nabla {\bf u}_t\cdot {\bf H}_t\right){\rm d} {\bf x}\right| \nonumber\\ &\leq &C\|\sqrt \varrho {\bf u}\|_{L^\infty}\| \nabla {\bf u}_t\|_{L^2}\|\sqrt \varrho {\bf u}_t\|_{L^2} +C\|\sqrt \varrho {\bf u}\|_{L^\infty}^2\| \nabla {\bf u}_t\|_{L^2}\| \nabla {\bf u}\|_{L^2} \nonumber\\ &&+C\| {\bf H}\|_{L^\infty}\| {\bf H}_t\|_{L^2}\| \nabla {\bf u}_t\|_{L^2} \leq \varepsilon\| \nabla {\bf u}_t\|_{L^2}^2+C( \varepsilon). \end{eqnarray*} $

因此

(4.28) $ \begin{equation} C(\mu)\| \nabla {\bf u}_t\|_{L^2}^2-C\leq \Phi(t)\leq C\| \nabla {\bf u}_t\|_{L^2}^2 +C, \end{equation} $

为了完成证明, 下面只需给出(4.27)式右边第一项$ \| \nabla^2 {\bf u}_t\|_{L^2}^2 $的估计.事实上, 对方程(3.22)改写成下面形式

$ -\mu\triangle {\bf u}_t+\nabla \pi_t = - \varrho {\bf u}_{tt}- \varrho {\bf u}\cdot \nabla {\bf u}_t - \varrho_t( {\bf u}_t+ {\bf u}\cdot \nabla {\bf u})- \varrho {\bf u}_t\cdot \nabla {\bf u}+ {\bf H}_t\cdot \nabla {\bf H} + {\bf H}\cdot \nabla {\bf H}_t. $

由引理2.3, (4.1)– (4.7)式以及(4.26)式, 利用Gagliardo-Nirenberg不等式可知

(4.29) $ \begin{eqnarray} \| \nabla^2 {\bf u}_t\|_{L^2}^2 & \leq&C(\|- \varrho {\bf u}_{tt}- \varrho {\bf u}\cdot \nabla {\bf u}_t- \varrho_t( {\bf u}_t+ {\bf u}\cdot \nabla {\bf u}) - \varrho {\bf u}_t\cdot \nabla {\bf u}+ {\bf H}_t\cdot \nabla {\bf H}+ {\bf H}\cdot \nabla {\bf H}_t\|_{L^2}^2) \\ &\leq & C\| \varrho\|_{L^\infty} \|\sqrt \varrho {\bf u}_{tt} \|_{L^2}^2 +C\| \varrho\| _{L^\infty} \|\sqrt \varrho {\bf u} \|_{L^\infty}^2 \| \nabla {\bf u}_t\|_{L^2}^2 \\ && +C\|\bar {\bf x}^{(a+1)/2} \varrho_t\|_{L^q}^2\|\bar {\bf x}^{-1} {\bf u}_t\|_{L^{2q/(q-2)}}^2+C\|\bar {\bf x}^{(a+1)/2} \varrho_t\|_{L^q}^2\|\bar {\bf x}^{-1} {\bf u}\|_{L^\infty}^2 \| \nabla {\bf u}\|_{L^{2q/(q-2)}}^2 \\ &&+C\|\bar {\bf x}^{(a+1)/2} \varrho\|_{L^q}^2 \|\bar {\bf x}^{-1} {\bf u}\|_{L^\infty}^2\| \nabla {\bf u}\|_{L^{2q/(q-2)}}^2 +C\|\sqrt \varrho\|_{L^\infty}\|\sqrt \varrho {\bf u}_t\|_{L^2}^2\|\nabla {\bf u}\|_{L^2}^2 \\ && +C\|\bar {\bf x}^{(a+1)/2} {\bf H}_t\|_{L^q}^2 \| \nabla {\bf H}\|_{L^{2q/(q-2)}}^2+C\| {\bf H}\|_{L^\infty}^2\| \nabla {\bf H}_t\|_{L^2}^2 \\ &\leq & C\|\sqrt \varrho {\bf u}_{tt} \|_{L^2}^2+C\| \nabla {\bf u}_t\|_{L^2}^4+C, \end{eqnarray} $

把(4.29)式代入(4.27)式, 取$ \delta $适当小, 可得

(4.30) $ \begin{equation} \Phi'(t)+\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2\leq C\| \nabla {\bf u}_t\|_{L^2}^4+C. \end{equation} $

对方程(4.30)两端乘以$ t $并在$ (0, T_0) $上关于$ t $积分, 由(4.2), (4.28)和(4.29)式以及Gronwall不等式可知

$ \sup\limits_{0\leq t\leq T_0}t\| \nabla {\bf u}_t\|_{L^2}^2+\int_0^{T_0}t\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2{\rm d}t\leq C. $

因此, (4.15)式成立, 引理4.2得证.

关于密度和磁场有如下的高阶估计.

引理4.3  存在常数$ C > 0 $, 有

(4.31) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}\left( \| \nabla^2 \varrho\|_{L^q}+\| \nabla^2 {\bf H}\|_{L^q}\right)\leq C. \end{equation} $

证  对方程(4.8)和(4.9)两端分别用微分算子$ \nabla^2 $作用, 并对两个方程分别乘以\linebreak $ q| \nabla^2\bar \varrho| ^{q-2} \nabla^2\bar \varrho $和$ q| \nabla^2\bar {\bf H}|^{q-2} \nabla^2\bar {\bf H} $, 然后在$ B_R $积分可得

(4.32) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\left(\| \nabla^2\bar {\bf H}\|_{L^q}+\| \nabla^2\bar \varrho\|_{L^q}\right) \\ &\leq &C\| \nabla {\bf u}\|_{L^\infty}\left(\| \nabla^2\bar {\bf H}\|_{L^q} +\| \nabla^2 \bar \varrho\|_{L^q}\right)+ \left( \| \nabla\bar {\bf H}\| _{L^\infty} +\| \nabla \bar \varrho\|_{L^\infty}\right)\| \nabla^2 {\bf u}\|_{L^q} \\ &\leq &C\left(1+\| \nabla^2 {\bf u}\|_{L^q}\right)\left(1+\| \nabla^2\bar {\bf H}\|_{L^q} +\| \nabla^2 \bar \varrho\|_{L^q}\right)+C\| \nabla^3 {\bf u} \|_{L^q}. \end{eqnarray} $

根据(2.7)式, 方程(4.32)右边的最后一项可以做如下估计

(4.33) $ \begin{eqnarray} \| \nabla^3 {\bf u}\|_{L^q}+\|\nabla^2\pi\|_{L^q}&\leq &C\| \nabla( \varrho {\bf u}_t)\|_{L^q}+C\| \nabla( \varrho {\bf u}\cdot \nabla {\bf u}) \|_{L^q}+C\| \nabla ( {\bf H}\cdot \nabla {\bf H} )\|_{L^q} \\ &\leq & C\|\bar {\bf x}^{-a} {\bf u}_t\|_{L^\infty}\|\bar {\bf x}^a \nabla \varrho\|_{L^q} +C\|\bar {\bf x}^a \varrho\|_{L^\infty}\|\bar {\bf x}^{-a} \nabla {\bf u}_t\|_{L^q} \\ &&+C\|\bar {\bf x}^a \nabla \varrho\|_{L^q}\|\bar {\bf x}^{-a} {\bf u}\|_{L^\infty} \| \nabla {\bf u}\|_{L^\infty}+C\| \varrho\|_{L^\infty} \| \nabla {\bf u}\|_{L^{2q}}^2 \\ &&+C\|\bar {\bf x}^a \varrho\|_{L^\infty}\|\bar {\bf x}^{-a} {\bf u}\|_{L^\infty} \| \nabla^2 {\bf u}\|_{L^q} \\ &&+C\|\bar {\bf x}^{a} {\bf H}\|_{L^\infty}\|\bar {\bf x}^{-a} \nabla^2 {\bf H}\|_{L^q} +C\| \nabla {\bf H}\|_{L^{2q}}^2 \\ &\leq &C\| \nabla {\bf u}_t\|_{L^q}+C\|\bar {\bf x}^{-a} {\bf u}_t\|_{L^q} +\frac12\| \nabla^3 {\bf u}\|_{L^q}+C\| \nabla^2\bar {\bf H}\|_{L^q}+C \\ &\leq &C\| \nabla {\bf u}_t\|_{L^2}^{2/q}\| \nabla^2 {\bf u}_t\|_{L^2}^{(q-2)/q} +C\| \nabla {\bf u}_t\|_{L^2}+\frac12\| \nabla^3 {\bf u}\|_{L^q} \\ &&+C\| \nabla^2\bar {\bf H}\|_{L^q}+C, \end{eqnarray} $

上述估计还用到了(3.3), (4.3), (4.4), (4.6)以及(4.7)式.

另一方面, 由(4.15)式

(4.34) $ \begin{eqnarray} &&\int_0^{T_0}\left(\| \nabla {\bf u}_t\|_{L^2}^{2/q}\| \nabla^2 {\bf u}_t\|_{L^2} ^{(q-2)/q}\right)^{(q+1)/q}{\rm d}t \\ &\leq &C\sup\limits_{0\leq t\leq T_0}\left(t\| \nabla {\bf u}_t\|_{L^2}^2\right) ^{(q+1)/q^2}\int_0^{T_0}\left(t\| \nabla^2 {\bf u}_t\|_{L^2}^2 +t^{-(q^2+q)/(q^2+q+2)}\right){\rm d}t \\ &\leq &C. \end{eqnarray} $

把(4.33)式代入(4.32)式, 利用Gronwall不等式可得(4.31), (3.3), (4.2)以及(4.34)式.引理4.3得证.

关于速度的三阶导数, 以及$ {\bf u}_t $二阶导数有下面的带权估计.

引理4.4  对常数$ C > 0 $, 有

(4.35) $ \begin{eqnarray} &&\sup\limits_{0\leq t\leq T_0}t\left( \| \nabla^3 {\bf u}\|_{L^2\cap L^q}+\| \nabla {\bf u}_t\|_{H^1} +\| \nabla^2( \varrho {\bf u})\|_{L^{(q+2)/2}}\right) \\ &&+\int_0^{T_0}t^2\left(\| \nabla {\bf u}_{tt}\| _{L^2}^2+\|\bar {\bf x}^{-1} {\bf u}_{tt}\|_{L^2}^2 \right){\rm d}t \leq C. \end{eqnarray} $

证  如果有下面估计成立

(4.36) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}t^2\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2+\int_0^{T_0} t^2\| \nabla {\bf u}_{tt}\|_{L^2}^2 {\rm d}t\leq C, \end{equation} $

则不难得到(4.35)式的证明.事实上, 结合上式与不等式(2.5), (4.15)以及(4.29)可得

(4.37) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}t\| \nabla {\bf u}_t\|_{H^1}+\int_0^{T_0}t^2\|\bar {\bf x}^{-1} {\bf u}_{tt}\|_{L^2}^2dt \leq C. \end{equation} $

进一步, 根据(4.13), (4.31), (4.33)和(4.34)式有

(4.38) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}t\| \nabla^3 {\bf u}\|_{L^2\cap L^q}\leq C, \end{equation} $

最后, 结合(3.3), (4.1)和(4.31)式, 不难得到

(4.39) $ \begin{eqnarray} t\| \nabla^2( \varrho {\bf u})\|_{L^{(q+2)/2}} &\leq &C t\|| \nabla^2 \varrho|| {\bf u}|\|_{L^{(q+2)/2}} +Ct\|| \nabla \varrho|| \nabla {\bf u}|\|_{L^{(q+2)/2}}+Ct\| \varrho| \nabla^2 {\bf u}|\|_{L^{(q+2)/2}} \\ &\leq &Ct\|\bar {\bf x}^{\delta_0} \nabla^2 \varrho\|_{L^2}^{2/(q+2)}\| \nabla^2 \varrho\|_{L^q} ^{q/(q+2)}\|\bar {\bf x}^{-2\delta_0/(q+2)} {\bf u}\|_{L^\infty} \\ &&+Ct\| \nabla \varrho\|_{L^q}\| \nabla {\bf u}\|_{L^{q(q+2)/(q-2)}}+Ct\| \nabla^2 {\bf u}\| _{L^{(q+2)/2}} \\ &\leq &C. \end{eqnarray} $

因此, 从(4.36)–(4.39)式可得不等式(4.35).

下面只需证明(4.36)式成立即可.首先对方程(3.22)关于$ t $求导可得

$ \begin{eqnarray*} && \varrho {\bf u}_{ttt}+ \varrho {\bf u}\cdot \nabla {\bf u}_{tt}-\mu\triangle {\bf u}_{tt}+\nabla\pi_{tt} \nonumber\\ & = &2 {{\rm{div }}}( \varrho {\bf u}) {\bf u}_{tt}+ {{\rm{div }}}( \varrho {\bf u})_t {\bf u}_t-2( \varrho {\bf u})_t\cdot \nabla {\bf u}_t - \varrho_{tt} {\bf u}\cdot \nabla {\bf u}-2 \varrho_t {\bf u}_t\cdot \nabla {\bf u} \nonumber\\ &&- \varrho {\bf u}_{tt}\cdot \nabla {\bf u}+ {\bf H}_{tt}\cdot \nabla {\bf H}+2 {\bf H}_t\cdot \nabla {\bf H}_t+ {\bf H} \cdot \nabla {\bf H}_{tt}, \end{eqnarray*} $

方程两边乘以$ {\bf u}_{tt} $并在$ B_R $分部积分, 得

(4.40) $ \begin{eqnarray} &&\frac12\frac{\rm d}{{\rm d}t}\int \varrho| {\bf u}_{tt}|^2{\rm d} {\bf x}+\int\left(\mu| \nabla {\bf u}_{tt}|^2 \right){\rm d} {\bf x} \\ & = &-4\int \varrho {\bf u}\cdot \nabla {\bf u}_{tt}\cdot {\bf u}_{tt}{\rm d} {\bf x}-\int( \varrho {\bf u})_t\cdot\left( \nabla( {\bf u}_t\cdot {\bf u}_{tt})+2 \nabla {\bf u}_t\cdot {\bf u}_{tt}\right) \\ &&-\int( \varrho {\bf u})_t\cdot \nabla( {\bf u}\cdot \nabla {\bf u}\cdot {\bf u}_{tt}){\rm d} {\bf x}-2\int \varrho_t {\bf u}_t \cdot \nabla {\bf u}\cdot {\bf u}_{tt}{\rm d} {\bf x}-\int \varrho {\bf u}_{tt}\cdot \nabla {\bf u}\cdot {\bf u}_{tt}{\rm d} {\bf x} \\ &&+\int {\bf H}_{tt}\cdot \nabla {\bf H}\cdot {\bf u}_{tt}{\rm d} {\bf x}+2\int {\bf H}_t\cdot \nabla {\bf H}_t\cdot {\bf u}_{tt}{\rm d} {\bf x}+\int {\bf H}\cdot \nabla {\bf H}_{tt}\cdot {\bf u}_{tt}{\rm d} {\bf x} \\ &\triangleq&\sum\limits_{i = 1}^{8}J_i. \end{eqnarray} $

下面只需对方程(4.40)右端每一项进行估计即可.对于$ J_1 $, 利用(4.4)式可以直接估计

(4.41) $ \begin{equation} |J_1|\leq C\|\sqrt \varrho {\bf u}\|_{L^\infty}\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}\| \nabla {\bf u}_{tt}\|_{L^2}\leq \varepsilon\| \nabla {\bf u}_{tt}\|_{L^2}^2+C( \varepsilon) \|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2. \end{equation} $

对于$ J_2 $, 有如下估计

(4.42) $ \begin{eqnarray} |J_2|&\leq &C\|\bar {\bf x}( \varrho {\bf u})_t\|_{L^q}\left(\|\bar {\bf x}^{-1} {\bf u}_{tt}\| _{2q/(q-2)}\| \nabla {\bf u}_t\|_{L^2}+\|\bar {\bf x}^{-1} {\bf u}_{t}\| _{2q/(q-2)}\| \nabla {\bf u}_{tt}\|_{L^2}\right) \\ &\leq &C\left(1+\| \nabla {\bf u}_t\|_{L^2}^2\right)\left(\|\sqrt \varrho {\bf u}_{tt}\| _{L^2}+\| \nabla {\bf u}_{tt}\|_{L^2} \right) \\ &\leq & \varepsilon\left(\|\sqrt \varrho {\bf u}_{tt}\| _{L^2}^2+\| \nabla {\bf u}_{tt}\|_{L^2}^2 \right)+C( \varepsilon)\left(1+\| \nabla {\bf u}_t\|_{L^2}^4\right), \end{eqnarray} $

上述估计用到了(2.6), (3.9)以及(4.5)式.除此之外, 还有用到了如下事实

(4.43) $ \begin{eqnarray} \|\bar {\bf x}( \varrho {\bf u})_t\|_{L^q} &\leq &C\|\bar {\bf x}| \varrho_t|| {\bf u}|\|_{L^q} +C\|\bar {\bf x} \varrho| {\bf u}_t|\|_{L^q} \\ &\leq &C\| \varrho_t\bar {\bf x}^{(1+a)/2}\|_{L^q}\|\bar {\bf x}^{-(a-1)/2} {\bf u}\| _{L^\infty} +C\| \varrho\bar {\bf x}^a\|_{L^{2q/(3- \tilde a)}}\| {\bf u}_t\bar {\bf x}^{1-a}\|_{L^{2q/( \tilde a-1)}} \\ &\leq &C+C\| \nabla {\bf u}_t\|_{L^2}. \end{eqnarray} $

其中$ \tilde a = \min\{2, a\} $, 不等式(4.43)得到用到了(4.4)–(4.6)式.

根据(3.9), (4.3), (4.4)以及(4.43)式可以得到$ J_3 $的估计.

(4.44) $ \begin{eqnarray} |J_3|&\leq &C\int|( \varrho {\bf u})_t|\left(| {\bf u}|| \nabla^2 {\bf u}|| {\bf u}_{tt}|+| {\bf u}|| \nabla {\bf u}| | \nabla {\bf u}_{tt}|+| \nabla {\bf u}|^2| {\bf u}_{tt}|\right){\rm d} {\bf x}\\ &\leq &C\|\bar {\bf x}( \varrho {\bf u})_t\|_{L^q}\|\bar {\bf x}^{-1/q} {\bf u}\|_{L^\infty}\| \nabla^2 {\bf u}\| _{L^2} \|\bar {\bf x}^{-(q-1)/q} {\bf u}_{tt}\|_{L^{2q/(q-2)}} \\ &&+C\|\bar {\bf x}( \varrho {\bf u})_t\|_{L^q}\|\bar {\bf x}^{-1} {\bf u}\|_{L^\infty}\| \nabla {\bf u}\| _{L^{2q/ (q-2)}}\| \nabla {\bf u}_{tt}\|_{L^2} \\ &&+C\|\bar {\bf x}( \varrho {\bf u})_t\|_{L^q}\| \nabla {\bf u}\|_{L^4}^2\|\bar {\bf x}^{-1} {\bf u}_{tt}\| _{L^{2q/(q-2)}} \\ &\leq &C(1+\| \nabla {\bf u}_t\|_{L^2})\left(\|\sqrt \varrho {\bf u}_{tt}\| _{L^2}+\| \nabla {\bf u}_{tt}\|_{L^2} \right) \\ &\leq & \varepsilon\left(\|\sqrt \varrho {\bf u}_{tt}\| _{L^2}^2+\| \nabla {\bf u}_{tt}\|_{L^2}^2 \right)+C( \varepsilon)\left(1+\| \nabla {\bf u}_t\|_{L^2}^2\right). \end{eqnarray} $

由柯西不等式以及(3.9), (4.5)和(4.6)式, 对$ J_4 $有

(4.45) $ \begin{eqnarray} |J_4|&\leq &C\int| \varrho_t|| {\bf u}_t|| \nabla {\bf u}|| {\bf u}_{tt}|{\rm d} {\bf x}\\ &\leq & C\|\bar {\bf x} \varrho_t\|_{L^q}\|\bar {\bf x}^{-1/2} {\bf u}_t\|_{L^{4q/(q-2)}} \| \nabla {\bf u}\|_{L^2}\|\bar {\bf x}^{-1/2} {\bf u}_{tt}\|_{L^{4q/(q-2)}} \\ &\leq & C\left(1+\| \nabla {\bf u}_t\|_{L^2}\right)\left(\|\sqrt \varrho {\bf u}_{tt}\| _{L^2}+\| \nabla {\bf u}_{tt}\|_{L^2} \right) \\ &\leq & \varepsilon\left(\|\sqrt \varrho {\bf u}_{tt}\| _{L^2}^2+\| \nabla {\bf u}_{tt}\|_{L^2}^2 \right)+C( \varepsilon)\left(1+\| \nabla {\bf u}_t\|_{L^2}^2\right). \end{eqnarray} $

利用Gagliardo-Nirenberg不等式和(4.3)式可以得到$ J_5 $的估计

(4.46) $ \begin{eqnarray} |J_5|\leq C\| \nabla {\bf u}\|_{L^\infty}\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2\leq C\left(1+\| \nabla^2 {\bf u}\|_{L^q}\right)\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2. \end{eqnarray} $

对于$ J_6 $–$ J_8 $的估计, 首先由方程(2.2)$ _3 $和不等式(3.3)以及(4.3)–(4.6)式, 有

(4.47) $ \begin{eqnarray} \| {\bf H}_{tt}\|_{L^2}&\leq & C\|\bar {\bf x}^{-\theta_0} {\bf u}_t\| _{2q/[(q-2)\theta_0]} \|\bar {\bf x}^{\theta_0} \nabla {\bf H}\|_{L^{2q/[q-(q-2)\theta_0]}}+C\| \nabla {\bf u}_t\|_{L^2} \\ &&+C\|\bar {\bf x}^{-\delta_0/2} {\bf u}\|_{L^\infty}\|\bar {\bf x}^{\delta_0/2} \nabla {\bf H}_t\|_{L^2}+C\| {\bf H}_t\|_{L^q}\| \nabla {\bf u}\|_{L^{2q/(q-2)}} \\ &\leq &C\left(1+\| \nabla {\bf u}_t\|_{L^2}\right), \end{eqnarray} $

对于上式最后一个不等号, 这里用到了下面估计

$ \begin{eqnarray*} \|\bar {\bf x}^{\delta_0/2} \nabla {\bf H}_t\|_{L^2}&\leq& C\|\bar {\bf x}^{\delta_0/2}| \nabla {\bf u}|| \nabla {\bf H}|\|_{L^2}+C\|\bar {\bf x}^{\delta_0/2} | {\bf u}|| \nabla^2 {\bf H}|\| _{L^2}+C\|\bar {\bf x}^{\delta_0/2} | {\bf H}|| \nabla^2 {\bf u}|\|_{L^2} \nonumber\\ &\leq &C\| \nabla {\bf u}\|_{L^{2q/(q-2)}}\|\bar {\bf x}^{a} \nabla {\bf H}\|_{L^q} +C\|\bar {\bf x}^{-\delta_0/2} {\bf u}\|_{L^\infty} \|\bar {\bf x}^{\delta_0} \nabla^2 {\bf H}\|_{L^2} \nonumber\\ &&+C\|\bar {\bf x}^{\delta_0/2} {\bf H}\|_{L^\infty} \| \nabla^2 {\bf u}\|_{L^2} \nonumber\\ &\leq &C, \end{eqnarray*} $

该不等式在条件(4.1), (4.3), (4.4)和(4.7)下是成立的.有了上述不等式做准备之后, 关于$ J_6 $–$ J_8 $再利用(4.47), (4.5), (4.7)和(4.26)式以及分部积分, 则

(4.48) $ \begin{eqnarray} |J_6|+|J_7|+|J_8|&\leq &C\| {\bf H}_t\|_{L^4}^2\| \nabla {\bf u}_{tt}\|_{L^2} +C\| {\bf H}\|_{L^\infty}\| {\bf H}_{tt}\|_{L^2}\| \nabla {\bf u}_{tt}\|_{L^2} \\ &\leq &C\left(\| \nabla {\bf u}_t\|_{L^2}+\| {\bf H}_t\|_{L^2}^2 +\| \nabla {\bf H}_t\|_{L^2}^2\right) \| \nabla {\bf u}_{tt}\|_{L^2} \\ &\leq & \varepsilon\| \nabla {\bf u}_{tt}\|_{L^2}^2 +C( \varepsilon)\left(1+\| \nabla {\bf u}_t\|_{L^2}^2\right). \end{eqnarray} $

把$ J_1 $–$ J_8 $的估计代入(4.40)式, 令$ \varepsilon $足够小, 对(4.40)式乘以权重$ t^2 $并积分, 则由Gronwall不等式可得(4.36)式.引理4.4证毕.

有了第3部分, 第4部分的一系列先验估计, 下面可以证明该文的主要结论, 即定理1.1, 定理1.2和定理1.3.首先, 给出定理1.1的证明.

定理1.1的证明  设$ ( \varrho_0, {\bf u}_0, {\bf H}_0) $是定理1.1中的初值.不失一般性, 假设初始密度$ \varrho_0 $满足

$ \int_{ {\mathbb{R}}^2} \varrho_0 {\rm d} {\bf x} = 1, $

这意味着存在$ N_0 > 0 $使得

(5.1) $ \begin{equation} \int_{B_{N_0}} \varrho_0{\rm d} {\bf x}\geq\frac34\int_{ {\mathbb{R}}^2} \varrho_0 {\rm d} {\bf x} = \frac34. \end{equation} $

令$ \varrho_0^R = \hat \varrho_0^R+R^{-1}e^{-| {\bf x}|^2} $, 其中$ 0\leq \hat \varrho_0^R\in C_0^\infty( {\mathbb{R}}^2) $并满足

(5.2) $ \begin{equation} \left\{ \begin{array}{lll} { } \int_{B_{N_0}}\hat \varrho_0^R {\rm d} {\bf x}\geq 1/2, \\ \bar {\bf x}^a\hat \varrho_0^R\rightarrow\bar {\bf x}^a \varrho_0, \quad \mbox{在}\ L^1( {\mathbb{R}}^2)\cap H^1( {\mathbb{R}}^2)\cap W^{1, q}( {\mathbb{R}}^2), \end{array} \right.\ \mbox{当} \ R\rightarrow\infty. \end{equation} $

取$ {\bf H}_0^R\in \{{\bf w}\in C_0^\infty(B_R)| {{\rm{div }}} {\bf w} = 0\} $满足

(5.3) $ \begin{equation} {\bf H}_0^R\bar {\bf x}^a\rightarrow {\bf H}_0\bar {\bf x}^a \quad {\rm \mbox{在} \ }H^1( {\mathbb{R}}^2)\cap W^{1, q}( {\mathbb{R}}^2) \ {\rm \mbox{上} \ }, \qquad \mbox{当}\ R \rightarrow\infty. \end{equation} $

因为$ \nabla {\bf u}_0\in L^2( {\mathbb{R}}^2) $, 存在$ {\bf v}_i^R\in C_0^\infty(B_R) $$ (i = 1, 2) $使得

(5.4) $ \begin{equation} \lim\limits_{R\rightarrow\infty}\|{{\bf v}}_i^R-\partial_i {\bf u}_0\|_{L^2( {\mathbb{R}}^2)} = 0, \quad i = 1, 2, \end{equation} $

接下来考虑下面Stokes系统的光滑解$ {\bf u}_0^R $

(5.5) $ \begin{equation} \left\{ \begin{array}{lll} -\triangle {\bf u}_0^R+ \varrho_0^R {\bf u}_0^R+\nabla\pi^R_0 = \sqrt{ \varrho_0^R}{\bf h}^R-\partial_i{\bf v}_i^R, &\ {\rm \mbox{在} \ }B_R \ {\rm \mbox{上} \ }, \\ {{\rm{div }}} {\bf u} = 0, &\ {\rm \mbox{在} \ } B_R \ {\rm \mbox{上} \ }, \\ {\bf u}_0^R = 0, &\ {\rm \mbox{在} \ } \partial B_R \ {\rm \mbox{上} \ }, \end{array} \right. \end{equation} $

这里$ {\bf h}^R = (\sqrt{ \varrho_0} {\bf u}_0)*j_{1/R} $, 其中$ j_\delta $是宽度为$ \delta $的标准磨光函数.通过定义$ B_R $之外的部分为零, 把$ {\bf u}_0^R $延拓到整个平面$ {\mathbb{R}}^2 $, 见文献[28], 则有

(5.6) $ \begin{equation} \lim\limits_{R\rightarrow\infty}\left(\| \nabla( \tilde {\bf u}_0^R- {\bf u}_0)\|_{L^2( {\mathbb{R}}^2)} +\|\sqrt{ \varrho_0^R} \tilde {\bf u}_0^R-\sqrt{ \varrho_0} {\bf u}_0\|_{L^2( {\mathbb{R}}^2)} \right) = 0. \end{equation} $

因此, 由引理2.1, 初值为$ ( \varrho_0^R, {\bf u}_0^R, {\bf H}_0^R) $的初边值问题2.2在$ B_R\times[0, T_R] $上存在经典解$ ( \varrho^R, {\bf u}^R, {\bf H}^R) $.进一步, 由命题3.1可知存在不依赖于$ R $的$ T_0 $使得(3.3)式对$ ( \varrho^R, {\bf u}^R, {\bf H}^R) $成立.取$ ( \varrho^R, {\bf u}^R, {\bf H}^R) $在$ {\mathbb{R}}^2\backslash B_R $上为零, 使得$ ( \varrho^R, {\bf u}^R, {\bf H}^R) $在$ {\mathbb{R}}^2 $上延拓, 表示为

$ \tilde \varrho^R\triangleq\varphi_R \varrho^R, \quad \tilde {\bf u}^R, \quad \tilde {\bf H}^R\triangleq\varphi_R {\bf H}^R, $

$ \varphi_R $为(3.6)中定义的磨光函数, 则由(3.3)可以得到

(5.7) $ \begin{eqnarray} \sup\limits_{0\leq t\leq T_0}\left(\|\sqrt{ \tilde \varrho^R} \tilde {\bf u}^R\|_{L^2} +\| \nabla \tilde {\bf u}^R\|_{L^2}\right) \leq C+C\sup\limits_{0\leq t\leq T_0}\| \nabla {\bf u}^R\|_{L^2} \leq C, \end{eqnarray} $

以及

(5.8) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}\| \tilde \varrho^R\bar {\bf x}^a\|_{L^1\cap L^\infty}\leq C. \end{equation} $

接下来, 对$ p\in [2, q] $, 由(3.3)和(3.29)式可知

(5.9) $ \begin{eqnarray} &&\sup\limits_{0\leq t\leq T_0}\left(\| \nabla(\bar {\bf x}^a \tilde \varrho^R)\|_{L^p( {\mathbb{R}}^2)}+ \|\bar {\bf x}^a \nabla \tilde \varrho^R\|_{L^p( {\mathbb{R}}^2)}+\| \nabla(\bar {\bf x}^a \tilde {\bf H}^R)\| _{L^p( {\mathbb{R}}^2)} +\|\bar {\bf x}^a \nabla \tilde {\bf H}^R\|_{L^p( {\mathbb{R}}^2)}\right) \\ &\leq &C\sup\limits_{0\leq t\leq T_0}\left(\|\bar {\bf x}^a \nabla \varrho^R\|_{L^p(B_R)} +\|\bar {\bf x}^a \varrho^R \nabla\varphi_R\|_{L^p(B_R)}+\| \varrho^R \nabla\bar {\bf x}^a\|_{L^p(B_R)} \right) \\ &&+C\sup\limits_{0\leq t\leq T_0}\left(\|\bar {\bf x}^a \nabla {\bf H}^R\|_{L^p(B_R)} +\|\bar {\bf x}^a {\bf H}^R \nabla\varphi_R\|_{L^p(B_R)}+\| {\bf H}^R \nabla\bar {\bf x}^a\|_{L^p(B_R)} \right) \\ &\leq &C +C\|\bar {\bf x}^a \varrho^R\|_{L^p(B_R)}+C\|\bar {\bf x}^a {\bf H}^R\|_{L^p(B_R)} \\ &\leq &C. \end{eqnarray} $

另外, 由(3.3)以及(3.31)式有

(5.10) $ \begin{equation} \int_0^{T_0}\left(\| \nabla^2 \tilde {\bf u}^R\|_{L^q( {\mathbb{R}}^2)}^{(q+1)/q}+t \| \nabla^2 \tilde {\bf u}^R\|_{L^q( {\mathbb{R}}^2)}^2+\| \nabla^2 \tilde {\bf u}^R\|_{L^2( {\mathbb{R}}^2)} ^2\right){\rm d}t\leq C, \end{equation} $

以及对任意的$ p\in [2, q] $, 有

(5.11) $ \begin{eqnarray} &&\int_0^{T_0}\left(\|\bar {\bf x} \tilde \varrho_t^R\|_{L^p( {\mathbb{R}}^2)}^2 +\|\bar {\bf x} \tilde {\bf H}_t^R\|_{L^p( {\mathbb{R}}^2)}^2\right){\rm d}t \\ &\leq & C\int_0^{T_0}\left(\||\bar {\bf x}|| {\bf u}^R|| \nabla \varrho^R|\|_{L^p(B_R)}^2 +\|\bar {\bf x} \varrho^R {{\rm{div }}} {\bf u}^R\|_{L^p(B_R)}^2\right){\rm d}t \\ &&+C\int_0^{T_0}\left(\||\bar {\bf x}|| {\bf u}^R|| \nabla {\bf H}^R|\|_{L^p(B_R)}^2 +\|\bar {\bf x}| {\bf H}^R|| \nabla {\bf u}^R|\|_{L^p(B_R)}^2\right){\rm d}t \\ &\leq &C\int_0^{T_0}\|\bar {\bf x}^{1-a} {\bf u}^R\|_{L^\infty}^2\left( \|\bar {\bf x}^a \nabla \varrho^R\|_{L^p(B_R)}^2 +\|\bar {\bf x}^a \nabla {\bf H}^R\|_{L^p(B_R)}^2\right){\rm d}t \\ &\leq &C. \end{eqnarray} $

同时, 根据(3.3)和(3.21)式可以推出

(5.12) $ \begin{equation} \sup\limits_{0\leq t\leq T_0}t\|\sqrt{ \tilde \varrho^R} \tilde {\bf u}_t^R\|_{L^2( {\mathbb{R}}^2)}^2 +\int_0^{T_0}t\| \nabla \tilde {\bf u}_t^R\|_{L^2( {\mathbb{R}}^2)}^2{\rm d}t \leq C+C\int_0^{T_0}t\| \nabla {\bf u}_t^R\|_{L^2(Ø)}^2 {\rm d}t \leq C. \end{equation} $

结合估计(5.7)–(5.12), $ R\rightarrow\infty $, 可以得到函数列$ ( \tilde \varrho^R, {\bf u}^R, \tilde{{\bf {\bf H}}}^R) $或者它的子列是弱收敛的, 假设收敛极限为$ ( \varrho, {\bf u}, {\bf H}) $, 因此有

(5.13) $ \begin{equation} \bar {\bf x} \tilde{ \varrho^R}\rightarrow \bar {\bf x} \varrho, \quad \bar {\bf x} \tilde{ {\bf H}}^R\rightarrow \bar {\bf x} {\bf H}, \ \ \mbox{在}\ C(\overline{B_N}\times[0, T_0]) \mbox{ 中强收敛, 对任意的$ N>0, $} \end{equation} $

(5.14) $ \begin{equation} \bar {\bf x}^a \tilde{ \varrho}^R\rightharpoonup \bar {\bf x}^a \varrho, \quad \bar {\bf x}^a \tilde{ {\bf H}}^R\rightharpoonup \bar {\bf x}^a {\bf H}, \ \ \mbox{ 在$ L^\infty(0, T_0; H^1( {\mathbb{R}}^2)\cap W^{1, q}( {\mathbb{R}}^2)) $ 中弱*收敛, } \end{equation} $

(5.15) $ \begin{equation} \sqrt{ \tilde \varrho^R} \tilde {\bf u}^R\rightharpoonup\sqrt \varrho {\bf u}, \quad \nabla \tilde {\bf u}^R\rightharpoonup \nabla {\bf u}, \quad \mbox{在$ L^\infty(0, T_0; L^2( {\mathbb{R}}^2))$ 中弱*收敛, } \end{equation} $

(5.16) $ \begin{equation} \nabla^2 \tilde {\bf u}^R\rightharpoonup \nabla^2 {\bf u}, \ \ \mbox{在 $ L^{(q+1)/q}(0, T_0; L^q( {\mathbb{R}}^2))\cap L^2((0, T_0)\times {\mathbb{R}}^2)$ 中弱收敛, } \end{equation} $

(5.17) $ \begin{equation} t^{1/2} \nabla^2 \tilde {\bf u}^R\rightharpoonup t^{1/2} \nabla^2 {\bf u}, \ \ \mbox{ 在$ L^{2}(0, T_0; L^q( {\mathbb{R}}^2))$ 中弱收敛, } \end{equation} $

(5.18) $ \begin{equation} t^{1/2}\sqrt{ \tilde \varrho^R} \tilde {\bf u}^R_t\rightharpoonup t^{1/2}\sqrt \varrho {\bf u}_t, \quad \nabla \tilde {\bf u}^R\rightharpoonup \nabla {\bf u}, \ \ \mbox{ 在$ L^{\infty}(0, T_0; L^2( {\mathbb{R}}^2))$中弱*收敛, } \end{equation} $

(5.19) $ \begin{equation} t^{1/2} \nabla \tilde {\bf u}^R_t\rightharpoonup t^{1/2} \nabla {\bf u}_t, \ \ \mbox{ 在$ L^{\infty}((0, T_0)\times {\mathbb{R}}^2) $ 中弱收敛} \end{equation} $

和

(5.20) $ \begin{equation} \bar {\bf x}^a \varrho\in L^\infty(0, T_0; L^1( {\mathbb{R}}^2)), \quad\inf\limits_{0\leq t\leq T_0} \int_{B_{2N_0}} \varrho(t, {\bf x}){\rm d} {\bf x}\geq \frac14. \end{equation} $

最后, 令$ R\rightarrow\infty $, 由(5.13)–(5.20)式可知$ ( \varrho, {\bf u}, {\bf H}) $是方程(1.1)–(1.3)定义于$ (0, T_0]\times {\mathbb{R}}^2 $上满足(1.6)和(1.7)式的强解.对于方程(1.1)–(1.3)的唯一性的证明可以参见文献[27].定理1.1证毕.

定理1.2的证明  设$ ( \varrho_0, {\bf u}_0, {\bf H}_0) $是定理1.2中的初值.和定理1.1的证明一样, 设

$ \int_{ {\mathbb{R}}^2} \varrho_0{\rm d} {\bf x} = 1, $

则存在$ N_0 > 0 $使得(5.1)式成立.令$ \varrho_0^R = \hat \varrho_0^R+R^{-1}e^{-| {\bf x}|^2} $, 其中$ 0\leq \hat \varrho_0^R\in C_0^\infty( {\mathbb{R}}^2) $, 并满足(5.2)式以及当$ R\rightarrow\infty $时

(5.21) $ \begin{equation} \left\{ \begin{array}{lll} \nabla^2\hat \varrho_0^R\rightarrow \nabla^2 \varrho_0, &{\rm \mbox{在}\ } L^q( {\mathbb{R}}^2) \ {\rm \mbox{上}\ }, \\ \bar {\bf x}^{\delta_0} \nabla^2\hat \varrho_0^R\rightarrow \bar {\bf x}^{\delta_0} \nabla^2 \varrho_0, &{\rm \mbox{在}\ } L^2( {\mathbb{R}}^2) \ {\rm \mbox{上}\ }, \end{array} \right. \end{equation} $

另一方面, 取$ {\bf H}_0^R\in \{{\bf w}\in C_0^\infty(B_R)| {{\rm{div }}} {\bf w} = 0\} $满足(5.3)式和

(5.22) $ \begin{equation} \left\{ \begin{array}{lll} \nabla^2 {\bf H}^R_0\rightarrow \nabla^2 {\bf H}_0, &{\rm \mbox{在}\ } L^q( {\mathbb{R}}^2) \ {\rm \mbox{上}\ }, \\ \bar {\bf x}^{\delta_0} \nabla^2 {\bf H}_0^R\rightarrow \bar {\bf x}^{\delta_0} \nabla^2 {\bf H}_0, &{\rm \mbox{在}\ } L^2( {\mathbb{R}}^2) \ {\rm \mbox{上}\ }, \end{array} \right.\quad {\rm \mbox{当}\ } R\rightarrow\infty. \end{equation} $

接下来, 考虑Stokes问题的光滑解

(5.23) $ \begin{equation} \left\{ \begin{array}{lll} -\mu\triangle {\bf u}_0^R+ \nabla\pi_0^R = (\nabla\times {\bf H}_0^R)\times {\bf H}_0^R - \varrho_0^R {\bf u}_0^R+\sqrt{ \varrho_0^R} {\bf h}^R, & \ {\rm \mbox{在}\ } B_R \ {\rm \mbox{上}\ }, \\ {{\rm{div }}} {\bf u} = 0, & \ {\rm \mbox{在}\ } {\rm in\ } B_R \ {\rm \mbox{上}\ }, \\ {\bf u}_0^R = 0, & \ {\rm \mbox{在}\ } \partial B_R \ {\rm \mbox{上}\ }, \end{array} \right. \end{equation} $

对(5.23)式两边乘以$ {\bf u}_0^R $并在$ B_R $上积分

$ \begin{eqnarray*} \|\sqrt{ \varrho_0^R} {\bf u}_0^R\|_{L^2(B_R)}^2+\mu\| \nabla {\bf u}_0^R\|_{L^2(B_R)}^2 &\leq &\int_{Ø}| {\bf H}_0^R|| \nabla {\bf H}_0^R|| {\bf u}_0^R|{\rm d} {\bf x}+\|\sqrt{ \varrho_0^R} {\bf u}_0^R\|_{L^2(B_R)} \|{\bf h}^R\|_{L^2(B_R)} \nonumber\\ &\leq & \varepsilon\left(\|\sqrt{ \varrho_0^R} {\bf u}_0^R\|_{L^2(B_R)}^2+\| \nabla {\bf u}_0^R \|_{L^2(B_R)}^2 \right)+C( \varepsilon), \end{eqnarray*} $

可推出

(5.24) $ \begin{equation} \|\sqrt{ \varrho_0^R} {\bf u}_0^R\|_{L^2(B_R)}^2+\| \nabla {\bf u}_0^R \|_{L^2(B_R)}^2 \leq C, \end{equation} $

这里常数$ C $与$ R $无关.从(2.7)式, 可知

(5.25) $ \begin{eqnarray} &&\| \nabla^2 {\bf u}_0^R\|_{L^2(B_R)}+\| \nabla \pi_0^R\|_{L^2(Ø)} \\ &\leq &C\|| {\bf H}_0^R|| \nabla {\bf H}_0^R|\|_{L^2(B_R)}+C\| \varrho_0^R {\bf u}_0^R\|_{L^2(B_R)} +C\|\sqrt{ \varrho_0^R}{\bf h}^R \|_{L^2(B_R)} \\ &\leq &C. \end{eqnarray} $

把$ {\bf u}_0^R $在$ {\mathbb{R}}^2 $上进行延拓, 延拓之后的函数记为$ \tilde {\bf u}_0^R $, 从(5.24)和(5.25)式可推出

$ \| \nabla \tilde {\bf u}_0^R\|_{H^1( {\mathbb{R}}^2)}\leq C, $

结合上述估计以及(5.21)和(5.24)式可知存在子列$ R_j\rightarrow\infty $以及函数$ \tilde {\bf u}_0\in \{ \tilde {\bf u}_0\in H^2_{\rm loc}( {\mathbb{R}}^2)| $$ \sqrt{ \varrho_0} \tilde {\bf u}_0 \in L^2( {\mathbb{R}}^2), \nabla \tilde {\bf u}_0\in H^1( {\mathbb{R}}^2)\} $和$ \nabla \tilde\pi_0\in L^2( {\mathbb{R}}^2) $使得

(5.26) $ \begin{equation} \left\{ \begin{array}{lll} \sqrt{ \varrho_0^{R_j}} \tilde {\bf u}_0^{R_j}\rightharpoonup \sqrt{ \varrho_0} \tilde {\bf u}_0, & {\rm \mbox{在} \ } L^2( {\mathbb{R}}^2) {\rm \mbox{上弱收敛} \ }, \\ \nabla \tilde {\bf u}_0^{R_j}\rightharpoonup \nabla \tilde {\bf u}_0, &{\rm \mbox{在} \ } H^1( {\mathbb{R}}^2) {\rm \mbox{上弱收敛} \ }, \\ \nabla\pi^R_0\rightharpoonup \nabla \tilde\pi_0, &{\rm \mbox{在} \ } L^2( {\mathbb{R}}^2) {\rm \mbox{上弱收敛} \ }. \end{array} \right. \end{equation} $

容易验证$ \tilde {\bf u}_0^R $满足(5.23)式, 则可以从(5.21), (5.22), (5.23)和(5.26)式推出$ ( \tilde {\bf u}_0, \tilde\pi_0) $满足

$ -\mu\triangle \tilde {\bf u}_0+ \nabla \tilde\pi_0+ \varrho_0 \tilde {\bf u}_0 = (\nabla\times {\bf H}_0^R)\times {\bf H}_0^R+ \varrho_0 {\bf u}_0 +\sqrt{ \varrho_0}{\bf g}, $

因此, 由(1.9)式可知

(5.27) $ \begin{equation} \tilde {\bf u}_0 = {\bf u}_0. \end{equation} $

另一方面, 由(5.23)式可知

$ \limsup\limits_{R_j\rightarrow\infty}\int_{ {\mathbb{R}}^2}\left(| \nabla \tilde {\bf u}_0^{R_j}|^2+ \varrho_0^{R_j}| \tilde {\bf u}_0^{R_j}|^2\right){\rm d} {\bf x} \leq \int_{ {\mathbb{R}}^2}\left(| \nabla {\bf u}_0|^2+ \varrho_0| {\bf u}_0|^2\right){\rm d} {\bf x}, $

结合(5.26)式有

$ \lim\limits_{R_j\rightarrow\infty}\int_{ {\mathbb{R}}^2}| \nabla \tilde {\bf u}_0^{R_j}|^2{\rm d} {\bf x} = \int_{ {\mathbb{R}}^2}| \nabla {\bf u}_0|^2{\rm d} {\bf x}, \quad\lim\limits_{R_j\rightarrow\infty}\int_{ {\mathbb{R}}^2} \varrho_0^{R_j}| \tilde {\bf u}_0^{R_j}|^2 {\rm d} {\bf x} = \int_{ {\mathbb{R}}^2} \varrho_0| {\bf u}_0|^2{\rm d} {\bf x}. $

因此, 由(5.26)和(5.27)式表明

(5.28) $ \begin{equation} \lim\limits_{R\rightarrow\infty}\left(\| \nabla( \tilde {\bf u}_0^R- {\bf u}_0)\|_{L^2( {\mathbb{R}}^2)}+ \|\sqrt{ \varrho_0^R} \tilde {\bf u}_0^R-\sqrt{ \varrho_0} {\bf u}_0\|_{L^2( {\mathbb{R}}^2)}\right) = 0. \end{equation} $

和(5.28)式类似, 容易得到$ { }\lim_{R\rightarrow\infty}\| \nabla^2( \tilde {\bf u}_0^R- {\bf u}_0)\|_{L^2( {\mathbb{R}}^2)} = 0. $

最后, 由引理2.1, 初边值问题(2.2)在$ [0, T_R]\times B_R $上有局部经典解$ ( \varrho^R, {\bf u}^R, {\bf H}^R) $.因此, 存在与$ R $无关的常数$ C $使得命题3.1和引理4.1–4.4的估计对$ ( \varrho^R, {\bf u}^R, {\bf H}^R) $都成立.对$ ( \varrho^R, {\bf u}^R, {\bf H}^R) $在$ {\mathbb{R}}^2\setminus B_R $上零延拓, 并令$ \tilde \varrho^R\triangleq\varphi_R \varrho^R, \ \tilde {\bf u}^R, \ \tilde {\bf H}^R\triangleq\varphi_R {\bf H}^R, $$ \varphi_R $和(3.6)式一样定义.可以从(3.3)式以及引理4.1–4.4推出序列$ ( \tilde \varrho^R, \tilde {\bf u}^R, \tilde {\bf H}^R) $或是它的一列子列是弱收敛到$ ( \varrho, {\bf u}, {\bf H}) $, 并满足(1.6), (1.7)和(1.10)式.通过标准的能量估计可知$ ( \varrho, {\bf u}, {\bf H}) $实际上是方程(1.1)–(1.3)的经典解.定理1.2证毕.

定理1.3的证明  下面给出定理1.3的证明, 采用反证法.设

(5.29) $ \begin{equation} \limsup\limits_{T\rightarrow T^*}\left(\| {\bf H}\|_{L^\infty(0, T^*; L^\infty ( {\mathbb{R}}^2))}+\| \nabla {\bf H}\| _{L^1(0, T_*; L^r( {\mathbb{R}}^2)}\right) = M_0<\infty, \end{equation} $

对任意的$ r>2 $成立.

首先, 由于(3.5)式成立, 就有如下引理.

引理5.1  设$ ( \varrho, {\bf u}, \pi, {\bf H}) $是方程组(1.1)–(1.3)的光滑解.在假设(5.29)下, 对$ 0\leq T\leq T^* $, 成立

(5.30) $ \begin{equation} \sup\limits_{t\in [0, T]}\mu\| \nabla {\bf u}\|_{L^2}^2+\int_0^T\|\sqrt \varrho\dot {\bf u}\|_{L^2}^2 {\rm d}t \leq C. \end{equation} $

证  对(1.1)$ _2 $式两边乘以$ \dot {\bf u} $并在$ {\mathbb{R}}^2 $上积分, 可得

(5.31) $ \begin{eqnarray} \int \varrho\dot {\bf u} {\rm d} {\bf x} = \mu\int \triangle {\bf u}\cdot\dot {\bf u} {\rm d} {\bf x} -\int \nabla \pi \cdot \dot {\bf u} {\rm d} {\bf x} -\frac12\int \nabla| {\bf H}|^2\cdot\dot {\bf u} {\rm d} {\bf x} +\int {\bf H}\cdot \nabla {\bf H}\cdot\dot {\bf u} {\rm d} {\bf x}. \end{eqnarray} $

对其每一项可以做估计, 由分部积分以及Gagliardo-Nirenberg不等式有

(5.32) $ \begin{eqnarray} \int \triangle {\bf u}\cdot\dot {\bf u} {\rm d} {\bf x}& = &\int \triangle {\bf u}\cdot ( {\bf u}_t+ {\bf u}\cdot \nabla {\bf u}){\rm d} {\bf x}\\ & = &-\frac12\frac{\rm d}{{\rm d}t}\int | \nabla {\bf u}|^2{\rm d} {\bf x}-\int \partial_i {\bf u}^j \partial_i( {\bf u}^k \partial_k {\bf u}^j){\rm d} {\bf x} \\ &\leq & -\frac12\frac{\rm d}{{\rm d}t}\int | \nabla {\bf u}|^2{\rm d} {\bf x} +C\int | \nabla {\bf u}|^3{\rm d} {\bf x} \\ &\leq & -\frac12\frac{\rm d}{{\rm d}t}\| \nabla {\bf u}\|_{L^2}^2+C\| \nabla {\bf u}\|_{L^2}^2\| \nabla^2 {\bf u}\| _{L^2}, \end{eqnarray} $

对于第二项, 由有界均振空间与哈代空间的对偶性可知

(5.33) $ \begin{eqnarray} -\int \nabla \pi \cdot\dot {\bf u} {\rm d} {\bf x}& = &-\int \nabla\pi\cdot( {\bf u}_t+ {\bf u}\cdot \nabla {\bf u}) {\rm d} {\bf x} = \int\pi \partial_i {\bf u}^j \partial_j {\bf u}^i {\rm d} {\bf x} \\ &\leq & C\|\pi\|_{BMO}\| \partial_i {\bf u}^j \partial_j {\bf u}^i\|_{\mathcal{H}^1} \leq C\| \nabla\pi\|_{L^2}\| \nabla {\bf u}\|_{L^2}^2, \end{eqnarray} $

上式第二个不等号用到了(2.8)和(2.9)式.

第三项关于磁场的估计, 由假设(5.29)有

(5.34) $ \begin{eqnarray} -\frac12\int \nabla| {\bf H}|^2\cdot\dot {\bf u} {\rm d} {\bf x} = \frac12\int | {\bf H}|^2 \partial_i {\bf u}^j \partial_j {\bf u}^i {\rm d} {\bf x} \leq C\| \nabla {\bf u}\|_{L^2}^2, \end{eqnarray} $

最后一项, 利用方程(1.1)$ _3 $, 可得

(5.35) $ \begin{eqnarray} \int( {\bf H}\cdot \nabla) {\bf H}\cdot\dot {\bf u} {\rm d} {\bf x} & = &\int ( {\bf H}\cdot \nabla) {\bf H}\cdot {\bf u}_t {\rm d} {\bf x} +\int ( {\bf H}\cdot \nabla) {\bf H}\cdot( {\bf u}\cdot \nabla) {\bf u} {\rm d} {\bf x} \\ & = &-\frac{\rm d}{{\rm d}t}\int {\bf H}\cdot \nabla {\bf u}\cdot {\bf H} {\rm d} {\bf x} +\int {\bf H}_t\cdot \nabla {\bf u}\cdot {\bf H} {\rm d} {\bf x}+\int {\bf H}\cdot \nabla {\bf u}\cdot {\bf H}_t {\rm d} {\bf x} \\ &&+\int ( {\bf H}\cdot \nabla) {\bf H}\cdot( {\bf u}\cdot \nabla) {\bf u} {\rm d} {\bf x} \\ &\leq& -\frac{\rm d}{{\rm d}t}\int {\bf H}\cdot \nabla {\bf u}\cdot {\bf H} {\rm d} {\bf x} +C\int| {\bf H}|^2| \nabla {\bf u}|^2{\rm d} {\bf x}. \end{eqnarray} $

把上述(5.32)–(5.35)式的估计代入(5.31)式, 由(5.29)式可以推出

(5.36) $ \begin{eqnarray} &&\frac{\mu}2\frac{\rm d}{{\rm d}t}\| \nabla {\bf u}\|_{L^2}^2+\frac{\rm d}{{\rm d}t}\int {\bf H}\cdot \nabla {\bf u}\cdot {\bf H} {\rm d} {\bf x}+\|\sqrt \varrho\dot {\bf u} \|_{L^2}^2 \\ &\leq& C\left(1+\| \nabla^2 {\bf u}\| _{L^2}+\| \nabla\pi\|_{L^2}\right)\| \nabla {\bf u}\|_{L^2}^2 \\ &\leq & C\left(1+\| \nabla {\bf H}\|_{L^r}\right)\| \nabla {\bf u}\|_{L^2}^2+ C\| \nabla {\bf u}\|_{L^2}^4+\frac12\|\sqrt \varrho\dot {\bf u} \|_{L^2}^2, \end{eqnarray} $

结合(3.5)和(5.29)式可知(5.30)式成立.引理5.1证毕.

引理5.2  假设(5.29)式成立, 对$ 0\leq T\leq T^* $以及常数$ C > 0 $, 有

(5.37) $ \begin{equation} \sup\limits_{t\in [0, T]}\left(\|\sqrt \varrho\dot {\bf u}\|_{L^2}^2+\| \nabla {\bf H}\|_{L^2}^2\right)+\int_0^T \mu\| \nabla\dot {\bf u}\|_{L^2}^2 {\rm d}t \leq C. \end{equation} $

证  把算子$ \dot {\bf u}^j [\frac{\partial}{\partial t} + {\bf u}\cdot \nabla] $作用于$ {\bf u} $的第$ j $个分量, 然后对$ j $求和, 并在$ {\mathbb{R}}^2 $上分部积分, 可得

(5.38) $ \begin{eqnarray} \frac12\frac{\rm d}{{\rm d}t}\left(\int \varrho|\dot {\bf u}|^2{\rm d} {\bf x}\right) +\mu\| \nabla\dot {\bf u}\|_{L^2}^2 & = &-\mu\int \partial_i( \partial_i {\bf u}\cdot \nabla {\bf u}^j)\dot {\bf u}^j {\rm d} {\bf x} -\mu\int {\mbox{div}}( \partial_i {\bf u} \partial_i {\bf u}^j)\dot {\bf u}^j {\rm d} {\bf x} \\ &&-\int(\dot {\bf u}^j \partial_t \partial_j\pi+\dot {\bf u}^j( {\bf u}\cdot \nabla) \partial_j\pi){\rm d} {\bf x} \\ &&-\frac12\int\dot {\bf u}^j\left( \partial_t( \partial_j| {\bf H}|^2)+ {\bf u}\cdot \nabla( \partial_j| {\bf H}|^2) \right){\rm d} {\bf x} \\ &&+\int\dot {\bf u}^j\left( \partial_t( {\bf H}\cdot \nabla {\bf H}^j ) + {\bf u}\cdot \nabla( {\bf H}\cdot \nabla {\bf H}^j)\right){\rm d} {\bf x} \\ & = &\sum\limits_{i = 1}^5K_i. \end{eqnarray} $

下面对(5.38)右端的每一项进行分别估计.首先, 由文献[29, Lemma 3.3], 有

(5.39) $ \begin{equation} \sum\limits_{i = 1}^3K_i\leq \frac{\rm d}{{\rm d}t}\int\pi \partial_j {\bf u}^i \partial_i {\bf u}^j {\rm d} {\bf x} +C\left(\|\pi\|_{L^4}^4 +\| \nabla {\bf u}\|_{L^4}^4\right) +\frac{\mu}{6}\| \nabla\dot {\bf u}\|_{L^2}^2. \end{equation} $

根据方程(1.1)$ _3 $和(1.1)$ _4 $, 由分部积分可得

(5.40) $ \begin{eqnarray} K_4& = &\int \partial_j\dot {\bf u}^j {\bf H}\cdot {\bf H}_t {\rm d} {\bf x}-\frac12\int\dot {\bf u}^j {\bf u}^i \partial_i \partial_j| {\bf H}|^2{\rm d} {\bf x} \\ & = &\int \partial_j\dot {\bf u}^j {\bf H} \cdot\left( {\bf H}\cdot \nabla {\bf u}- {\bf u}\cdot \nabla {\bf H} \right){\rm d} {\bf x}+\frac12\int {\bf u}^i \partial_i\dot {\bf u}^j \partial_j| {\bf H}|^2{\rm d} {\bf x} \\ & = & \int {\mbox{div}}\dot {\bf u} {\bf H}\cdot( {\bf H}\cdot \nabla {\bf u}){\rm d} {\bf x}-\frac12 \int \partial_j\dot {\bf u}^j {\bf u}^i \partial_i| {\bf H}|^2{\rm d} {\bf x}+\frac12\int {\bf u}^i \partial_i\dot {\bf u}^j \partial_j| {\bf H}|^2{\rm d} {\bf x} \\ &\leq & C\int | \nabla {\bf u}|^3| {\bf H}|^2{\rm d} {\bf x}+C\int | \nabla {\bf u}|| \nabla\dot {\bf u}|| {\bf H}|^2{\rm d} {\bf x} \\ &\leq & C\| \nabla {\bf u}\|_{L^2}^2+C\| \nabla {\bf u}\|_{L^4}^4 +\frac{\mu}{6}\| \nabla\dot {\bf u}\|_{L^2}^2. \end{eqnarray} $

类似的

(5.41) $ \begin{equation} K_5\leq C\| \nabla {\bf u}\|_{L^2}^2+C\| \nabla {\bf u}\|_{L^4}^4 +\frac{\mu}{6}\| \nabla\dot {\bf u}\|_{L^2}^2. \end{equation} $

把估计(5.39)–(5.41)代入(5.38)式, 则有

(5.42) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\|\sqrt \varrho\dot {\bf u}\|_{L^2}^2 +\mu\| \nabla\dot {\bf u}\|_{L^2}^2 \\ &\leq & 2\frac{\rm d}{{\rm d}t}\int\pi \partial_j {\bf u}^i \partial_i {\bf u}^j {\rm d} {\bf x}+C\left(\|\pi\|_{L^4}^4 +\| \nabla {\bf u}\|_{L^4}^4 +\| \nabla {\bf u}\|_{L^2}^2\right) \\ &\leq &2\frac{\rm d}{{\rm d}t}\int\pi \partial_j {\bf u}^i \partial_i {\bf u}^j {\rm d} {\bf x} +C\left(1+\| \nabla\pi\|_{L^\frac43}^4+\| \nabla^2 {\bf u}\|_{L^\frac43}^4\right) \\ &\leq &2\frac{\rm d}{{\rm d}t}\int\pi \partial_j {\bf u}^i \partial_i {\bf u}^j {\rm d} {\bf x} +C\left(1+ \| \varrho\|_{L^2}^2\|\sqrt \varrho\dot {\bf u}\|_{L^2}^4+\| {\bf H}\|_{L^\infty}^4 \| {\bf H}\|_{L^2}^2\| \nabla {\bf H}\|_{L^2}^2\right), \end{eqnarray} $

上述估计用到了(2.7)式以及嵌入不等式.

下面对方程(1.1)$ _3 $关于空间变量求导, 乘以$ \partial_i {\bf H} $, 在$ {\mathbb{R}}^2 $上积分, 由(5.29), (5.30)和(2.7)式可得

(5.43) $ \begin{eqnarray} \frac{\rm d}{{\rm d}t}\| \nabla {\bf H}\|_{L^2}^2&\leq & C\int | \nabla {\bf u}|| \nabla {\bf H}|^2{\rm d} {\bf x} +C\int | {\bf H}|| \nabla {\bf H}|| \nabla^2 {\bf u}|{\rm d} {\bf x} \\ &\leq &C\| \nabla {\bf H}\|_{L^2}\| \nabla {\bf H}\|_{L^r}\| \nabla {\bf u}\|_{L^\frac{2r}{r-2}} +C\| {\bf H}\|_{L^\infty}\| \nabla {\bf H}\|_{L^2}\| \nabla^2 {\bf u}\|_{L^2} \\ &\leq&C \| \nabla {\bf H}\|_{L^2}\| \nabla {\bf H}\|_{L^r}\| \nabla {\bf u}\|_{L^2}^{1-\frac{2}{r}} \| \nabla^2 {\bf u}\|_{L^2}^{\frac2r}+C\| \nabla {\bf H}\|_{L^2}\left(\| \nabla {\bf H}\|_{L^2} +\|\sqrt \varrho\dot {\bf u}\|_{L^2}\right) \\ &\leq&C\left(1+\| \nabla {\bf H}\|_{L^r}\right)\left(1+\| \nabla {\bf H}\|_{L^2}^2+\|\sqrt \varrho\dot {\bf u}\|_{L^2}^2 \right). \end{eqnarray} $

结合(5.42)和(5.43)式, 由(5.29), (5.30)和(2.7)式, 以及

$ \int\pi \partial_i {\bf u}^j \partial_j {\bf u}^i {\rm d} {\bf x} \leq C\| \nabla\pi\|_{L^2} \| \nabla {\bf u}\|_{L^2}^2 \leq \varepsilon\|\sqrt \varrho\dot {\bf u}\|_{L^2}^2+ \varepsilon\| \nabla {\bf H}\|_{L^2}^2 +C\| \nabla {\bf u}\|_{L^2}^4, $

可得(5.37)式.引理5.2得证.

同文献[6, Lemma 5.3]同样的推导, 有如下引理.

引理5.3  设(5.29)式成立, 对任意的$ 0\leq T\leq T^* $以及常数$ C > 0 $, 有

(5.44) $ \begin{equation} \sup\limits_{t\in [0, T]}\left(\| \varrho\bar {\bf x}^a\|_{L^1\cap H^1\cap W^{1, q}} +\| {\bf H}\bar {\bf x}^a\|_{H^1\cap W^{1, q}}\right) \leq C. \end{equation} $

最后, 由(5.30), (5.37)以及(5.44)式的一系列先验估计, 可以得到引理4.1–4.4的高阶估计, 这些高阶估计说明局部解就是经典解.

引理5.4  设(5.29)式成立, 对任意的$ 0\leq T\leq T^* $以及常数$ C > 0 $, 有

(5.45) $ \begin{eqnarray} &&\sup\limits_{0\leq t\leq T}\Big(\|\bar {\bf x}^{\delta_0} \nabla^2 \varrho\|_{L^2} +\|\bar {\bf x}^{\delta_0} \nabla^2P( \varrho)\|_{L^2}+\|\bar {\bf x}^{\delta_0} \nabla^2 {\bf H}\|_{L^2} +t\| \nabla {\bf u}_t\|_{L^2}^2+\| \nabla^2 \varrho\|_{L^q} \\ &&+\| \nabla^2P( \varrho)\|_{L^q} +\| \nabla^2 {\bf H}\|_{L^q}+t\| \nabla^3 {\bf u}\|_{L^2\cap L^q}+t\| \nabla^2 {\bf u}_t\|_{H^1} +t\| \nabla^2( \varrho {\bf u})\|_{L^{(q+2)/2}}\Big) \\ &&+\int_0^{T}\left( t\|\sqrt \varrho {\bf u}_{tt}\|_{L^2}^2 +t\| \nabla^2 {\bf u}_{t} \|_{L^2}^2+t^2\| \nabla {\bf u}_{tt}\| _{L^2}^2+t^2 \|\bar {\bf x}^{-1} {\bf u}_{tt}\|_{L^2}^2\right){\rm d}t\leq C. \end{eqnarray} $

有了上述一系列的估计, 定理1.3就可以得到证明.事实上, 在引理5.1–5.4的估计下, 可以得到函数

$ ( \varrho, {\bf u}, {\bf H})(T^*, {\bf x}) = \lim\limits_{T\rightarrow T^*}( \varrho, {\bf u}, {\bf H})(T, {\bf x}) $

满足所有(1.5)和(1.8)式中出现的条件.则由定理1.2可以把经典解延拓到$ t> T^* $.这与定理1.3中所说$ T^* $是经典解的最大存在时间相矛盾.