Euler-Poisson方程组通常用来描述半导体材料中带电粒子流的输运现象^{[1, 9, 15, 23]}.它由描述电子流运动的双极非等熵可压缩Euler方程组和描述电场变化的Poisson方程耦合而成,形式如下

$\left\{ \begin{matrix} {{\partial }_{t}}{{n}^{\nu }}+\nabla \cdot ({{n}^{\nu }}{{u}^{\nu }})=0,\\ {{\partial }_{t}}{{u}^{\nu }}+({{u}^{\nu }}\cdot \nabla ){{u}^{\nu }}+\nabla {{\theta }^{\nu }}+\frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}\nabla {{n}^{\nu }}=-{{q}_{\nu }}\nabla \phi -{{u}^{\nu }},\\ {{\partial }_{t}}{{\theta }^{\nu }}+{{u}^{\nu }}\cdot \nabla {{\theta }^{\nu }}+{{\theta }^{\nu }}\nabla \cdot {{u}^{\nu }}-\frac{1}{{{n}^{\nu }}}{{\left| \nabla {{\theta }^{\nu }} \right|}^{2}}=-({{\theta }^{\nu }}-1)+\frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}\Delta {{\theta }^{\nu }},\\ -\Delta \phi ={{n}^{i}}-{{n}^{e}}+b(x),(t,x)\in {{\mathbb{R}}^{+}}\times \mathbb{T},\\ \end{matrix} \right.$

(1.1)

其中 ${\Bbb T}= \left( {\Bbb R}/{\Bbb Z} \right) ^3$表示三维空间的环,指标$\nu = e,i,$分别表示半导体材料中的电子和离子. 未知变量分别为: 密度${n^\nu } > 0,$ 速度${{u}^{\nu }}=\left( u_{1}^{\nu },u_{2}^{\nu },u_{3}^{\nu } \right)$,绝对温度${\theta ^\nu } > 0$,函数$\nabla \phi$表示电场势,函数$b = b(x) \ge{{\rm const}} > 0$表示半导体材料的掺杂浓度,其中 const 表示某个绝对正常数. 这里描述温度变化的方程源自下述能量方程

${{\partial }_{t}}({{n}^{\nu }}{{\varpi }^{\nu }})+\nabla \cdot \left( {{n}^{\nu }}{{\varpi }^{\nu }}{{u}^{\nu }}+{{p}_{\nu }}{{u}^{\nu }} \right)=\nabla \cdot \left( {{\kappa }_{\nu }}\nabla {{\theta }^{\nu }} \right)-{{q}_{\nu }}{{n}^{\nu }}{{u}^{\nu }}\nabla \phi -{{n}^{\nu }}{{\left| {{u}^{\nu }} \right|}^{2}}-({{\theta }^{\nu }}-1),\ \nu =e,i,$

此处总能量函数${\varpi ^\nu } = {\theta ^\nu } + \frac{1}{2}|{u^\nu }{|^2}$,压差函数${p_\nu } = {n^\nu }{\theta ^\nu }$; 再由Boltzmann方程可知热传导系数${\kappa _\nu }$为温度$\theta ^\nu $的函数,简单起见本文只考查$\kappa _\nu ({\theta ^\nu }) = \theta ^\nu $的情形,由此结合方程组(1.1)的前两个方程即得描述温度变化的第三个方程.

本文研究方程组(1.1)的空间周期问题,其初始条件为

$({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }}){{|}_{t=0}}=({{n}^{\nu 0}},{{u}^{\nu 0}},{{\theta }^{\nu 0}}),x\in \mathbb{T},\nu =e,i,$

(1.2)

电场势函数初值$\nabla \phi {|_{t = 0}} = \nabla {\phi ^0}$满足

$-\Delta {{\phi }^{0}}={{n}^{i0}}-{{n}^{e0}}+b,x\in \mathbb{T}.$

(1.3)

设$({n^\nu },{u^\nu },{\theta ^\nu },\phi ) = ({\bar n^\nu }(x),0,1,\bar \phi (x))$为方程组(1.1)的一个稳态解,则其满足

$\left\{ \begin{matrix} \nabla \ln ({{{\bar{n}}}^{e}})=\nabla \bar{\phi }=-\nabla \ln ({{{\bar{n}}}^{i}}),\\ -\Delta \bar{\phi }={{{\bar{n}}}^{i}}-{{{\bar{n}}}^{e}}+b(x),x\in \mathbb{T}. \\ \end{matrix} \right.$

(1.4)

为保证$\phi$的唯一性,附加条件如下

$\int_{\mathbb{T}}{\phi (\cdot ,x)}dx=\int_{\mathbb{T}}{\bar{\phi }(x)}dx.$

(1.5)

由(1.4)式的第一式知:$ {\bar n^e} = {{e}^{\bar \phi }},{\bar n^i} = {{e}^{C - \bar \phi }}$,其中常数C > 0,再由其第二式可知:$ \bar \phi$满足严格单调半线性椭圆方程如下

$ - \Delta \bar \phi = {{e}^{C - \bar \phi }} - C{{e}^{\bar \phi }} + b(x),x \in {\mathbb T} .$

(1.6)

于是借助于经典的Schauder不动点定理或极小化方法可得解$\bar \phi$的存在性^{[3, 6]},唯一性则由函数$f:{\rm{ }}\bar \phi \mapsto {{e}^{C - \bar \phi }} - C{{e}^{\bar \phi }}$的严格单调性保证,进而由最大值原理结合条件$b \ge{{\rm const}} > 0$可知: ${\bar n^\nu } \ge {{\rm const}} > 0.$

众所周知,当${n^\nu },{\theta ^\nu } > 0$时,方程组(1.1)为可对称化的双曲-抛物组. 于是由Kato^[10]的结论以及Matsumura-Nishida^[16-17]的结果可知: 只要初值光滑,周期问题(1.1)--(1.2)就一定存在局部唯一光滑解.

命题 1.1 (局部存在唯一性^{[11, 14]}) 令整数$s\geq 3$,光滑周期函数b=b(x)满足$b \ge {{\rm const}} > 0$. 若对给定常数$\kappa > 0$,初值满足$({n^{\nu 0}}-{\bar n^\nu }, {u^{\nu 0}},{\theta ^{\nu 0}} - 1) \in {H^s}({\Bbb T})$ 且${n^{\nu 0}},{\theta ^{\nu 0}} \ge 2\kappa $, 则存在T>0使得问题(1.1)--(1.2)存在局部唯一光滑解,满足 $ {n^\nu },{\theta ^\nu } \ge \kappa, (t,x) \in \left[{0,T} \right] \times {\Bbb T} $ 及

$\begin{align} & {{\theta }^{\nu }}-1\in {{C}^{1}}\left( [0,T],{{H}^{s-2}}(\mathbb{T}) \right)\cap C\left( [0,T],{{H}^{s}}(\mathbb{T}) \right),\nu =e,i,\\ & ({{n}^{\nu }}-{{{\bar{n}}}^{\nu }},{{u}^{\nu }},\nabla \phi -\nabla \bar{\phi })\in {{C}^{1}}\left( [0,T],{{H}^{s-1}}(\mathbb{T}) \right)\cap C\left( [0,T],{{H}^{s}}(\mathbb{T}) \right),\nu =e,i,\\ \end{align}$

此处${H^m}({\Bbb T})$,表示三维环${\Bbb T}$上的m阶Sobolev空间,其范数记为${\left\| \cdot \right\|_m}$.

目前已有一些关于半导体Euler-Poisson模型的相关研究. 2000年,借助于仿微分算子的工具,Cordier和 Grenier^[2]解决了一维等熵Euler-Poisson模型的拟中性极限问题. 2001年,关于给定的边界值,Slemrod和 Sternberg^[22]获得了一维稳态Euler-Poisson系统的拟中性极限. 2003年Hisao和 Markowich,Wang^[7] 证明了高维Euler-Poisson系统的Cauchy问题存在唯一按指数速率衰减至平衡态的小摄动 光滑解. 2004年,Peng和 Wang^[19]研究了带有初始层的Euler-Poisson高维位势流的拟中性极限问题. 同年,借助于加权能量方法、 迭代技巧以及紧性方法,Wang^[24] 研究了环上黏性与无黏性的Euler-Poisson系统 的拟中性极限问题. 2005年,借助经典能量方法,Loeper^[12]严格证明了无压力的Euler-Poisson系统 收敛到不可压Euler方程组. 2005年,应用渐近展开法、迭代技巧以及对称双曲组的性质,Peng和 Wang^[20]获得了时变Euler-Poisson系统的 拟中性极限就是不可压的Euler方程组以及不可压的Euler方程组存在局部光滑解. 2006年,借助渐近展开的方法,Peng,Wang和 Yong^[21]研究了高维非等熵Euler-Poisson系统的拟中性极限问题. 2008年,Ju,Li和 Wang^[8] 研究了环上及全空间上的Navier-Stokes-Poisson方程组的拟中性极限问题. 2009年,Luo和 Smoller^[13]研究了可压Euler-Poisson系统旋转星型解的存在性及非线性稳定性问题.

上述工作都是围绕常数稳态解开展的. 2005年,Guo和 Strauss^[6]研究了物理边界无耗散项等熵的Euler-Poisson方程组,获得了非常数平衡解附近光滑解的整体存在性. 最近Peng^[18],Feng,Peng和 Wang^[4],Feng和 Wang^[5]运用时空混合导数迭代方法分别研究了单、双极等熵以及单极非等熵Euler-Poisson方程组非常数平衡解的稳定性问题,得到相关解的正则性为

$({{n}^{\nu }}-{{\bar{n}}^{\nu }},{{u}^{\nu }},E-\bar{E},B-\bar{B})\in \bigcap\limits_{k=0}^{s}{{{C}^{k}}}({{\mathbb{R}}^{+}},{{H}^{s-k}}(\mathbb{T})).$

然而目前尚无关于双极非等熵可压缩Euler-Poisson方程组非常数平衡解的稳定性的相关研究. 本文意在研究上述问题.通过研究可以发现双极非等熵模型解的正则性(1.8)--(1.10)式与单极非等熵模型^[5]以及双极等熵模型^[4]的相关结果存在本质的不同。这是由于这里所研究的双极非等熵模型中同时耦合了带正电的离子方程与温度场方程的缘由.

本文的主要结果如下:

定理 1.1 令整数$s \ge 6$,光滑周期函数b=b(x)满足$b \ge {{\rm const} }> 0$. 设初值满足

$\int_{\mathbb{T}}{{{n}^{\nu 0}}(x)dx=}\int_{\mathbb{T}}{{{{\bar{n}}}^{\nu }}(x)dx},\nu =e,i,$

(1.7)

那么存在常数${{\delta }_{0}}$ > 0足够小,不依赖于任何时间t>0,使得若

${{\left\| \left( {{n}^{\nu 0}}-{{{\bar{n}}}^{\nu }},{{u}^{\nu 0}},{{\theta }^{\nu 0}}-1 \right) \right\|}_{s}}\le {{\delta }_{0}},\nu =e,i,$

则周期问题(1.1)--(1.2)存在唯一整体光滑解$({n^\nu },{u^\nu },{\theta ^\nu },\phi )$满足

$({{n}^{\nu }}-{{\bar{n}}^{\nu }},\nabla \phi -\nabla \bar{\phi })\in \bigcap\limits_{k=0}^{2}{{{C}^{k}}}\left( {{\mathbb{R}}^{+}},{{H}^{s-k}}(\mathbb{T}) \right)\bigcap\limits_{k=3}^{\left[s/2 \right]}{{{C}^{k}}}\left( {{\mathbb{R}}^{+}},{{H}^{s-2k+2}}(\mathbb{T}) \right),\nu =e,i,$

(1.8)

${u^\nu } \in C \left( {{{\Bbb R}^ + },{H^{s }}({\Bbb T})} \right) \mathop \bigcap \limits_{k = 1}^{\left[{s/2} \right]} {C^k}\left( {{{\Bbb R}^ + },{H^{s - 2k + 1}}({\Bbb T})} \right), \nu = e,i, $

(1.9)

$ {\theta ^\nu } - 1 \in \mathop \bigcap \limits_{k = 0}^{\left[{s/2} \right]} {C^k}\left( {{{\Bbb R}^ + },{H^{s - 2k}}({\Bbb T})} \right) , \nu = e,i, $

(1.10)

进而有

$\begin{align} & \sum\limits_{\nu =e,i}{|||\left( {{n}^{\nu }}(t)-{{{\bar{n}}}^{\nu }},{{u}^{\nu }}(t),{{\theta }^{\nu }}(t)-1,\nabla \phi (t)-\nabla \bar{\phi } \right)||{{|}_{[s/2]}}} \\ & \le C{{e}^{-\eta t}}\sum\limits_{\nu =e,i}{{{\left\| ({{n}^{\nu 0}}-\bar{n},{{u}^{\nu 0}},{{\theta }^{\nu 0}}-1) \right\|}_{s}}},\\ \end{align}$

(1.11)

此处符号$[\cdot]$表示取整数,范数$|||\cdot|||_m$的定义见第二节.

注1.1 方程组(1.1)中速度耗散项- ${u^\nu }$,温度耗散项 - $({\theta ^\nu } - 1)$以及扩散项 $\frac{{{\theta ^\nu }}}{{{n^\nu }}}\Delta {\theta ^\nu }$在证明定理1.1的过程中起关键作用.

定理1.1的证明基于能量方法、对称子技巧以及时空混合导数迭代方法. 首先指出,该迭代方法由Peng^[18]使用于单极等熵Euler-Poisson方程组周期问题,随后被Feng,Peng和 Wang^[4]推广至双极等熵模型的周期问题,能有效地克服非常数稳态解带来的困难. 其次,双极非等熵Euler-Poisson方程组要比单极非等熵的情形更为复杂,因为它增加了描述带正电离子运动的相关方程; 与双极极等熵Euler-Poisson方程组不同,本文研究的双极非等熵模型增加了描述能量演化的温度方程,这里压差函数不仅依赖密度,而且依赖温度,从而导致在为建立迭代关系式而做的能量估计的过程中出现新的困难,进而解的各个分量的正则性不同。幸运的是,系统(1.1)的能量方程中的温度扩散项克服了这个困难,进而保证了光滑解的整体存在性.

本文其余部分结构如下: 第二节给出准备工作; 第三节给出定理1.1的证明所需的能量估计,进而给出迭代关系式; 第四节,通过迭代给出先验估计,进而证明解的整体存在性.

首先引入一些记号以备后用. 对于常数0 <$ \gamma $ < 1,表达式$f \sim g$意指$\gamma g \le f \le \frac{1}{\gamma }g$. 用$\left\| \cdot \right\|$和${\left\| \cdot \right\|_{{L^\infty }}}$分别表示空间${L^2}\left( {\Bbb T} \right)$和的范数${L^\infty }\left( {\Bbb T} \right). 用\left\langle { \cdot ,\cdot } \right\rangle $表示空间${L^2}\left( {\Bbb T} \right)$上的内积. 对于多重指标$\alpha =\left( {{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}} \right)\in {{\mathbb{N}}^{3}}$,记:${{\partial }^{\alpha }}=\partial _{{{x}_{1}}}^{{{\alpha }_{1}}}\partial _{{{x}_{2}}}^{{{\alpha }_{2}}}\partial _{{{x}_{3}}}^{{{\alpha }_{3}}}=\partial _{1}^{{{\alpha }_{1}}}\partial _{2}^{{{\alpha }_{2}}}\partial _{3}^{{{\alpha }_{3}}}$. 对于两个多重指标$\alpha ,\beta \in {{\Bbb N}}^3,\beta \leqslant \alpha $表示对$1 \le j \le 3$有${\beta _j} \le {\alpha _j}; \beta < \alpha$表示$\beta \leqslant \alpha $且$\beta \neq \alpha.$最后,对于T>0以及$m \ge 1$,引入空间

${{B}_{m,T}}(\mathbb{T})=\bigcap\limits_{k=0}^{m}{{{C}^{k}}}\left( [0,T],{{H}^{m-k}}(\mathbb{T}) \right),$

其范数定义为

$|||\cdot ||{{|}_{m}}={{(\sum\limits_{k+|\alpha |\le m}{{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\left( \cdot \right) \right\|}^{2}}})}^{\frac{1}{2}}}.$

显然有$ || \cdot |{|_m} \le ||| \cdot ||{|_m}$.

接下来回顾定理1.1证明过程中常用的公式及引理.

Leibniz 公式

$\begin{align} & {{\partial }^{\alpha }}\left( fg \right)=f{{\partial }^{\alpha }}g+\sum\limits_{\beta <\alpha }{C_{\alpha }^{\beta }{{\partial }^{\alpha -\beta }}f{{\partial }^{\beta }}g},\forall \alpha \in {{\mathbb{N}}^{3}},\\ & \partial _{t}^{k}\left( fg \right)=f\partial _{t}^{k}g+\sum\limits_{l<k}{C_{k}^{l}\partial _{t}^{k-l}f\partial _{t}^{l}g},\forall k\in \mathbb{N},\\ \end{align}$

其中常数$C_\alpha ^\beta > 0,C_k^l > 0.$

引理 2.1 (Poincaré不等式) 设$1 \le p < \infty ,\Omega \subset {{\Bbb R}}^d$为具有Lipschitz边界的有界连通开区域,于是存在依赖p与$\Omega$的常数使得下式成立

${\left\| {u - {u_\Omega }} \right\|_{{L^p}\left( \Omega \right)}} \le C{\left\| {\nabla u} \right\|_{{L^p}\left( \Omega \right)}}, \forall u \in {W^{1,p}}\left( \Omega \right), $

(2.1)

其中${u_\Omega } = \frac{1}{{\left| \Omega \right|}}\int_\Omega {u(x){d}x}$表示u在$\Omega上$的平均值.

引理 2.2^[18] 设整数$m\ge 3$,若 $u,v \in {B_{m,T}}\left( {\Bbb T} \right)$,则

$|||uv||{|_m} \le C|||u||{|_m}|||v||{|_m}. $

(2.2)

引理 2.3 设整数$m\ge 3$,函数f光滑且f(x,0,0,0) = 0. 若$v \in {B_{m,T}}\left( {\Bbb T} \right)$满足

${{\partial }_{t}}v=f(x,v,{{\partial }_{x}}v,{{\partial }_{xx}}v),$

则对任意$t \in [0,T]$,有

$||\partial _t^k{\partial ^\alpha }v(t,\cdot )|| \le C||v(t,\cdot )|{|_m},\forall k + |\alpha | \le [m/2], $

(2.3)

这里正常数C可以连续的依赖于$||v||_m$.

证 该过程与文献^[18],引理 2.8的证明类似,故略去详细过程. 证毕.

设$\left( {{n^\nu },{u^\nu },{\theta ^\nu },\phi } \right)$ 为周期问题(1.1)--(1.2)的唯一光滑解. 令

${n^\nu } = {\bar n^\nu } + {N^\nu }, {\theta ^\nu } = 1 + {\Theta ^\nu },\phi = \bar \phi + \Phi , $

(3.1)

则系统(1.1)改写为

$ \left\{ {\begin{array}{*{20}{c}} {{\partial _t}{N^\nu } + {u^\nu } \cdot \nabla {N^\nu } + ({{\bar n}^\nu } + {N^\nu })\nabla \cdot {u^\nu } + {u^\nu } \cdot \nabla {{\bar n}^\nu } = 0,} \\ \begin{array}{l} {\partial _t}{u^\nu } + ({u^\nu } \cdot \nabla ){u^\nu } + \nabla {\Theta ^\nu } + \nabla (\ln ({{\bar n}^\nu } + {N^\nu }) - \ln {{\bar n}^\nu }) + {\Theta ^\nu }\nabla (\ln ({{\bar n}^\nu } + {N^\nu }) \\ + {u^\nu } = - {q_\nu }\nabla \Phi ,\\ \end{array} \\ {{\partial _t}{\Theta ^\nu } + {u^\nu } \cdot \nabla {\Theta ^\nu } + (1 + {\Theta ^\nu })\nabla \cdot {u^\nu } + \displaystyle \frac{1}{{{n^\nu }}}{{\left| {\nabla {\Theta ^\nu }} \right|}^2} + {\Theta ^\nu } = \displaystyle \frac{{1 + {\Theta ^\nu }}}{{{{\bar n}^\nu } + {N^\nu }}}\Delta {\Theta ^\nu },} \\ {\Delta \Phi = {N^i} - {N^e},(t,x) \in {{\Bbb R}^ + } \times{\Bbb T} .} \\ \end{array}} \right.$

(3.2)

初始条件为

$({N^\nu },{u^\nu },{\Theta ^\nu }){|_{t = 0}} = ({N^{\nu 0}},{u^{\nu 0}},{\Theta ^{\nu 0}}), x \in {{\Bbb T}},\nu =e,i, $

(3.3)

其中 ${N^{\nu 0}} = {n^{\nu 0}} - {\bar n^\nu },{\Theta ^{\nu 0}} = {\theta ^{\nu 0}} - 1.$

直接计算可得

$\begin{align} & \nabla (\ln ({{{\bar{n}}}^{\nu }}+{{N}^{\nu }}))=\frac{\nabla {{{\bar{n}}}^{\nu }}}{{{{\bar{n}}}^{\nu }}}+\frac{1}{{{n}^{\nu }}}\nabla {{N}^{\nu }}-\frac{\nabla {{{\bar{n}}}^{\nu }}}{{{n}^{\nu }}{{{\bar{n}}}^{\nu }}}{{N}^{\nu }},\\ & \nabla (\ln ({{{\bar{n}}}^{\nu }}+{{N}^{\nu }})-\ln {{{\bar{n}}}^{\nu }})=\frac{1}{{{n}^{\nu }}}\nabla {{N}^{\nu }}-\frac{\nabla {{{\bar{n}}}^{\nu }}}{{{\left( {{{\bar{n}}}^{\nu }} \right)}^{2}}}{{N}^{\nu }}+\frac{1}{{{n}^{\nu }}{{\left( {{{\bar{n}}}^{\nu }} \right)}^{2}}}{{\left( {{N}^{\nu }} \right)}^{2}}. \\ \end{align}$

记

${{U}^{\nu }}=\left( \begin{matrix} {{N}^{\nu }} \\ {{u}^{\nu }} \\ {{\Theta }^{\nu }} \\ \end{matrix} \right),U=\left( \begin{matrix} {{U}^{e}} \\ {{U}^{i}} \\ \end{matrix} \right),W=\left( \begin{matrix} U \\ \nabla \Phi \\ \end{matrix} \right),$

(3.4)

$ {U^{\nu 0}} = \left( {\begin{array}{*{20}{c}} {{n^{\nu 0}} - \bar n} \\ {{u^{\nu 0}}} \\ {{\theta ^{\nu 0}} - 1} \\ \end{array}} \right),{U^0} = \left( {\begin{array}{*{20}{c}} {{U^{e0}}} \\ {{U^{i0}}} \\ \end{array}} \right),{W^0} = \left( {\begin{array}{*{20}{c}} {{U^0}} \\ {\nabla {\Phi ^0}} \\ \end{array}} \right), $

(3.5)

其中${\Phi ^0} = {\phi ^0} - \bar \phi$.

于是(3.2)式中的Euler方程可改写为如下矩阵形式

${\partial _t}{U^\nu } + \sum\limits_{j = 1}^3 {A_j^\nu } ({n^\nu },{u^\nu },{\theta ^\nu }){\partial _j} {U^\nu } + {L^\nu }(x){U^\nu } + {M^\nu }(W) = {f^\nu },\nu =e,i.$

(3.6)

此处

$A_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }})=\left( \begin{matrix} u_{j}^{\nu } & {{n}^{\nu }}e_{j}^{T} & 0 \\ \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}{{e}_{j}} & ~~u_{j}^{\nu }{{I}_{3}}~~ & {{e}_{j}} \\ 0 & {{\theta }^{\nu }}e_{j}^{T} & u_{j}^{\nu } \\ \end{matrix} \right),j=1,2,3,$

(3.7)

${L^\nu }(x) = \left( {\begin{array}{*{20}{c}} 0 & ~~{{{(\nabla {{\bar n}^\nu })}^T}}~~ & 0 \\ { - \displaystyle \frac{{\nabla {{\bar n}^\nu }}}{{{{({{\bar n}^\nu })}^2}}}} & 0 & {\displaystyle\frac{{\nabla {{\bar n}^\nu }}}{{{{\bar n}^\nu }}}} \\ 0 & 0 & 0 \\ \end{array}} \right), $

(3.8)

${M^\nu }(W) = \left( {\begin{array}{*{20}{c}} 0 \\ {{u^\nu } + {q_\nu }\nabla \Phi } \\ {{\Theta ^\nu } - \displaystyle\frac{{{\theta ^\nu }}}{{{n^\nu }}}\Delta {\Theta ^\nu }} \\ \end{array}} \right), $

(3.9)

${{f}^{\nu }}=\left( \begin{matrix} 0 \\ \frac{\nabla {{{\bar{n}}}^{\nu }}}{{{n}^{\nu }}{{{\bar{n}}}^{\nu }}}{{N}^{\nu }}{{\Theta }^{\nu }}-\frac{{{\left( {{N}^{\nu }} \right)}^{2}}}{{{n}^{\nu }}{{({{{\bar{n}}}^{\nu }})}^{2}}} \\ -\frac{1}{{{n}^{\nu }}}|\nabla {{\Theta }^{\nu }}{{|}^{2}} \\ \end{matrix} \right),$

(3.10)

此处$\left( {{e}_{1}},{{e}_{2}},{{e}_{3}} \right)$为${{\Bbb R} }^3$的标准正交基,I₃为$3\times 3$单位矩阵.

当${\bar n^\nu } + {N^\nu },{\rm{ }}1 + {\Theta ^\nu } > 0$时,易知(3.6)式是关于$U^\nu$可对称化的双曲-抛物组. 这是因${\bar n^\nu } \ge {{\rm const}} > 0$且由于${N^\nu },{\rm{ }}{\Theta ^\nu } \to 0,$从而有${\bar n^\nu } + {N^\nu },{\rm{ }}1 + {\Theta ^\nu } \ge {{\rm const} }> 0$.

令

$A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})=\left( \begin{matrix} \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} & 0 & 0 \\ 0 & ~~{{n}^{\nu }}{{I}_{3}}~~ & 0 \\ 0 & 0 & \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}} \\ \end{matrix} \right),$

则

$\begin{align} & \tilde{A}_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }})=A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})A_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }}) \\ & =\left( \begin{matrix} \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}u_{j}^{\nu } & {{\theta }^{\nu }}e_{j}^{T} & 0 \\ {{\theta }^{\nu }}{{e}_{j}} & ~~~{{n}^{\nu }}u_{j}^{\nu }{{I}_{3}}~~ & {{n}^{\nu }}{{e}_{j}} \\ 0 & {{n}^{\nu }}e_{j}^{T} & \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}u_{j}^{\nu } \\ \end{matrix} \right),j=1,2,3, \\ \end{align}$

为一对称矩阵,再由矩阵$A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})$,的正定对称性可知,(3.6)式关于$U^\nu$是可对称化的双曲-抛物组.

从现在开始,令T>0,W为命题1.1给出,定义在区间[0,T]上,初值为W⁰的系统(3.2)的光滑解. 定义

${{\omega }_{T}}=\underset{0\le t\le T}{\mathop{\sup }}\,|||W\left( t \right)||{{|}_{[s/2]}}$

(3.11)

及任意常数C>0不依赖任何时间t,T.

易知,当$m\ge3$时, 嵌入$H^{m-1}({\Bbb T})\hookrightarrow L^\infty({\Bbb T})$连续,于是存在常数C>0使得

${{\left\| h \right\|}_{{{L}^{\infty }}}}\le C{{\left\| h \right\|}_{m-1}},\forall h\in {{H}^{m-1}}(\mathbb{T}).$

因此,若${\omega _T}$充分小,从${\bar n^\nu } \ge {{\rm const}} > 0$即得${\bar n^\nu } + {N^\nu },{\rm{ }}1 + {\Theta ^\nu } \ge{{\rm const} }> 0$.

引理 3.1在定理1.1的条件下,若${\omega _T}$足够小则有

$\begin{align} & \frac{d}{dt}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{U}^{\nu }},{{U}^{\nu }} \right\rangle }+{{\left\| \nabla \Phi \right\|}^{2}})+2\sum\limits_{\nu =e,i}{\left\langle {{n}^{\nu }}{{u}^{\nu }},{{u}^{\nu }} \right\rangle } \\ & +2\sum\limits_{\nu =e,i}{\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}{{\Theta }^{\nu }},{{\Theta }^{\nu }} \right\rangle }+2\sum\limits_{\nu =e,i}{{{\left\| \nabla {{\Theta }^{\nu }} \right\|}^{2}}}\le C|||U|||_{[s/2]}^{2}|||W||{{|}_{[s/2]}}.\forall t\in [0,T]. \\ \end{align}$

(3.12)

证 分以下两个步骤来完成.

步骤 1 首先,断言下面两式成立

${{\left\| {{\partial }_{t}}({{N}^{\nu }},{{\Theta }^{\nu }}) \right\|}_{{{L}^{\infty }}}}\le C|||U||{{|}_{[s/2]}},{{\left\| {{\partial }_{t}}A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}) \right\|}_{{{L}^{\infty }}}}\le C|||U||{{|}_{[s/2]}},$

(3.13)

$\begin{array}{*{35}{l}} \left| \left\langle (\sum\limits_{j=1}^{3}{{{\partial }_{j}}\tilde{A}_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }})-2A_{0}^{O}({{n}^{\nu }},{{\theta }^{\nu }}){{L}^{\nu }}})V,V \right\rangle \right|\le C||U|{{|}_{[s/2]}}||V|{{|}^{2}},\forall V\in {{L}^{2}}(\mathbb{T}). \\ \end{array}$

(3.14)

事实上,由连续性嵌入${H^{[s/2] - 1}}\left( {\Bbb T} \right) \hookrightarrow {L^\infty }\left( {\Bbb T} \right)$可知

${{\left\| {{\partial }_{t}}({{N}^{\nu }},{{\Theta }^{\nu }}) \right\|}_{{{L}^{\infty }}}}\le C{{\left\| {{\partial }_{t}}({{N}^{\nu }},{{\Theta }^{\nu }}) \right\|}_{[s/2]-1}}\le C|||U||{{|}_{[s/2]}}.$

直接计算可得

${{\partial }_{t}}A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})=\left( \begin{matrix} \frac{{{\partial }_{t}}{{\Theta }^{\nu }}}{{{n}^{\nu }}}-\frac{{{\partial }_{t}}{{N}^{\nu }}}{{{({{n}^{\nu }})}^{2}}} & 0 & 0 \\ 0 & ~~{{\partial }_{t}}{{N}^{\nu }}{{I}_{3}}~~ & 0 \\ 0 & 0 & \frac{{{\partial }_{t}}{{N}^{\nu }}}{{{\theta }^{\nu }}}-\frac{{{n}^{\nu }}{{\partial }_{t}}{{\Theta }^{\nu }}}{{{({{\theta }^{\nu }})}^{2}}} \\ \end{matrix} \right),$

进而可知

${{\left\| {{\partial }_{t}}A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}) \right\|}_{{{L}^{\infty }}}}\le C{{\left\| {{\partial }_{t}}({{N}^{\nu }},{{\Theta }^{\nu }}) \right\|}_{{{L}^{\infty }}}}\le C|||U||{{|}_{[s/2]}}.$

最后计算可得

$\begin{align} & \sum\limits_{j=1}^{3}{{{\partial }_{j}}}\tilde{A}_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }})-2A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{L}^{\nu }} \\ & =\left( \begin{matrix} \nabla \cdot \left( \frac{{{\theta }^{\nu }}{{u}^{\nu }}}{{{n}^{\nu }}} \right) & ~{{\left( \nabla {{\Theta }^{\nu }}-2\frac{{{\theta }^{\nu }}\nabla {{{\bar{n}}}^{\nu }}}{{{n}^{\nu }}} \right)}^{T}}~~ & 0 \\ \nabla {{\Theta }^{\nu }}+2\frac{{{n}^{\nu }}\nabla {{{\bar{n}}}^{\nu }}}{{{({{{\bar{n}}}^{\nu }})}^{2}}} & \nabla \cdot ({{n}^{\nu }}{{u}^{\nu }}){{I}_{3}} & \nabla {{n}^{\nu }}-2\frac{{{n}^{\nu }}\nabla {{{\bar{n}}}^{\nu }}}{{{{\bar{n}}}^{\nu }}} \\ 0 & {{\left( \nabla {{{\bar{n}}}^{\nu }} \right)}^{T}} & \nabla \cdot \left( \frac{{{n}^{\nu }}{{u}^{\nu }}}{{{\theta }^{\nu }}} \right) \\ \end{matrix} \right) \\ \end{align}$

在$({n^\nu },{u^\nu },{\theta ^\nu }) = ({\bar n^\nu },0,1)$处为一反对称矩阵,由此可知(3.14)式成立.

步骤2 由(3.6)式与$A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{U}^{\nu }}$在空间$L^2({\Bbb T})$上内积可得

$\begin{align} & \frac{d}{dt}\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{U}^{\nu }},{{U}^{\nu }} \right\rangle } \\ & =\sum\limits_{\nu =e,i}{\left\langle {{\partial }_{t}}A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{U}^{\nu }},{{U}^{\nu }} \right\rangle }+2\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})\left( {{f}^{\nu }}-{{M}^{\nu }}(W) \right),{{U}^{\nu }} \right\rangle } \\ & +\sum\limits_{\nu =e,i}{\left\langle (\sum\limits_{j=1}^{3}{{{\partial }_{j}}\tilde{A}_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }})}-2A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{L}^{\nu }}){{U}^{\nu }},{{U}^{\nu }} \right\rangle }. \\ \end{align}$

由(3.13)--(3.14)式及(3.10)式可知

$\begin{align} & \left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{U}^{\nu }},{{U}^{\nu }} \right\rangle \le {{\left\| {{\partial }_{t}}A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}) \right\|}_{{{L}^{\infty }}}}||{{U}^{\nu }}|{{|}^{2}}\le C|||U|||_{[s/2]}^{3},\\ & \left\langle (\sum\limits_{j=1}^{3}{{{\partial }_{j}}\tilde{A}_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }})}-2A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{L}^{\nu }}){{U}^{\nu }},{{U}^{\nu }} \right\rangle \le C|||U|||_{[s/2]}^{3},\\ & \left| \left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{f}^{\nu }},{{U}^{\nu }} \right\rangle \right|\le C|||U|||_{[s/2]}^{3},\\ & \sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{M}^{\nu }}(W),{{U}^{\nu }} \right\rangle } \\ & =\left\langle {{n}^{e}}{{u}^{e}}-{{n}^{i}}{{u}^{i}},\nabla \Phi \right\rangle +\sum\limits_{\nu =e,i}{\left\langle {{n}^{\nu }}{{u}^{\nu }},{{u}^{\nu }} \right\rangle }++\sum\limits_{\nu =e,i}{\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}{{\Theta }^{\nu }},{{\Theta }^{\nu }} \right\rangle }+\sum\limits_{\nu =e,i}{{{\left\| \nabla {{\Theta }^{\nu }} \right\|}^{2}}}. \\ \end{align}$

于是有

$\begin{align} & \frac{d}{dt}\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{U}^{\nu }},{{U}^{\nu }} \right\rangle }+2\sum\limits_{\nu =e,i}{\left\langle {{n}^{\nu }}{{u}^{\nu }},{{u}^{\nu }} \right\rangle } \\ & +2\sum\limits_{\nu =e,i}{\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}{{\Theta }^{\nu }},{{\Theta }^{\nu }} \right\rangle }+2\sum\limits_{\nu =e,i}{{{\left\| \nabla {{\Theta }^{\nu }} \right\|}^{2}}}\le 2\left\langle {{n}^{e}}{{u}^{e}}-{{n}^{i}}{{u}^{i}},\nabla \Phi \right\rangle +C|||U|||_{[s/2]}^{3}. \\ \end{align}$

(3.15)

然后由方程组(3.2)的第一、四式并运用分部积分可得

$\left\langle {{n}^{e}}{{u}^{e}}-{{n}^{i}}{{u}^{i}},\nabla \Phi \right\rangle =-\frac{1}{2}\frac{d}{dt}{{\left\| \nabla \Phi \right\|}^{2}}.$

再与(3.15)式联立可知(3.12)式成立. 证毕.

对于$k\in {\Bbb N},\alpha \in {{\Bbb N}}^3$且$1 \le k + |\alpha| \le [s / 2],$对(3.6)式求混合导数$\partial _t^k{\partial ^\alpha }$可得

${\partial _t}U_{k,\alpha }^\nu + \sum\limits_{j = 1}^3 {A_j^\nu } ({n^\nu },{u^\nu },{\theta ^\nu }){\partial _j} U_{k,\alpha }^\nu + {L^\nu }U_{k,\alpha }^\nu + M_{k,\alpha }^\nu = f_{k,\alpha }^\nu + g_\nu ^{k,\alpha }, $

(3.16)

其中

$\begin{align} & U_{k,\alpha }^{\nu }=\partial _{t}^{k}{{\partial }^{\alpha }}{{U}^{\nu }},M_{k,\alpha }^{\nu }=\partial _{t}^{k}{{\partial }^{\alpha }}{{M}^{\nu }},f_{k,\alpha }^{\nu }=\partial _{t}^{k}{{\partial }^{\alpha }}{{f}^{\nu }}, \\ & g_{\nu }^{k,\alpha }=\sum\limits_{j=1}^{3}{(A_{j}^{\nu }}({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }}){{\partial }_{j}}U_{k,\alpha }^{\nu }-\partial _{t}^{k}{{\partial }^{\alpha }}(A_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }}){{\partial }_{j}}{{U}^{\nu }})) \\ & +\left( L{{U}_{k,\alpha }}-\partial _{t}^{k}{{\partial }^{\alpha }}(LU) \right). \\ \end{align}$

(3.17)

再由Poisson方程可得

$\partial _t^k{\partial ^\alpha }\Delta \Phi = N_{k,\alpha }^i - N_{k,\alpha }^e, $

(3.18)

这里$N_{k,\alpha }^\nu = \partial _t^k{\partial ^\alpha }{N^\nu },\nu = e,i$.

定义耗散函数${D_m}( \cdot )$形式如下

${D_m}(t) = |||U(t)|||_m^2 + \sum\limits_{\nu = e,i} {|||\nabla {\Theta ^\nu }(t)|||_m^2} ,\forall m \in {\Bbb N}.$

(3.19)

引理 3.2 在定理1.1的条件下,若${\omega _T}$足够小,则对任意$k \in {\Bbb N},\alpha \in {{\Bbb N}}^3$且$|\alpha | \ge 1,k + |\alpha | \le [s/2],$存在正常数C₀使得$\forall t \in [0,T]$,有

$\begin{align} & \frac{d}{dt}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}})+{{C}_{0}}\sum\limits_{\nu =e,i}{{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}({{u}^{\nu }},{{\Theta }^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|}^{2}}} \\ & \le C\sum\limits_{\nu =e,i}{\left\| \partial _{t}^{k}{{N}^{\nu }} \right\|_{|\alpha |}^{2}}+C\sum\limits_{\nu =e,i}{\left\| (\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k}\nabla {{\Theta }^{\nu }},\partial _{t}^{k+1}{{\Theta }^{\nu }},\partial _{t}^{k}\nabla \Phi ) \right\|_{|\alpha |-1}^{2}} \\ & +C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

(3.20)

证 分以下三个步骤来完成.

步骤1 对$k\in {\Bbb N},\alpha \in {{\Bbb N}}^3$且$1 \le k + |\alpha | \le [s/2]$,断言下式成立

$\begin{align} & \frac{d}{dt}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}}) \\ & \le -2\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})M_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }-2\left\langle \partial _{t}^{k}{{\partial }^{\alpha }}({{n}^{e}}{{u}^{e}}-{{n}^{i}}{{u}^{i}}),\partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\rangle \\ & +2\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})g_{\nu }^{k,\alpha },U_{k,\alpha }^{\nu } \right\rangle }+C{{D}_{[s/2]}}|||W||{{|}_{[s/2]}}. \\ \end{align}$

(3.21)

事实上,对于$1 \le k + |\alpha | \le [s/2]$,由(3.16)式与$A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,a}^{\nu }$在空间$L^2({\Bbb T})$上内积可得

$\begin{align} & \frac{d}{dt}\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle = \\ & \left\langle {{\partial }_{t}}A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle +\left\langle (\sum\limits_{j=1}^{3}{{{\partial }_{j}}\tilde{A}_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }})}-2A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{L}^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle \\ & +2\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})f_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle -2\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})M_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle +2\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})g_{\nu }^{k,\alpha },U_{k,\alpha }^{\nu } \right\rangle . \\ \end{align}$

进而由(3.18)式可得

$\begin{array}{*{35}{l}} -\left\langle \partial _{t}^{k}{{\partial }^{\alpha }}({{n}^{e}}{{u}^{e}}-{{n}^{i}}{{u}^{i}}),\partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\rangle =\frac{1}{2}\frac{d}{dt}{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}}. \\ \end{array}$

(3.22)

因而有

$\begin{align} & \frac{d}{dt}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}}) \\ & =\sum\limits_{\nu =e,i}{\left\langle {{\partial }_{t}}A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle } \\ & +\sum\limits_{\nu =e,i}{\left\langle (\sum\limits_{j=1}^{3}{{{\partial }_{j}}\tilde{A}_{j}^{\nu }}({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }})-2A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }}){{L}^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle } \\ & +2\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})f_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }-2\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})M_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle } \\ & +2\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})g_{\nu }^{k,\alpha },U_{k,\alpha }^{\nu } \right\rangle }-2\left\langle \partial _{t}^{k}{{\partial }^{\alpha }}({{n}^{e}}{{u}^{e}}-{{n}^{i}}{{u}^{i}}),\partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\rangle . \\ \end{align}$

显然,不等式右端前三项都能被$C{D_{[s/2]}}|||W||{|_{[s/2]}}$控制,于是(3.21)式成立.

步骤2 对任意$k \in {\Bbb N}$,且 $|\alpha | \ge 1,k + |\alpha | \le [s/2],$断言下面两式成立

$\begin{align} & -2\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})M_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }-2\left\langle \partial _{t}^{k}{{\partial }^{\alpha }}({{n}^{e}}{{u}^{e}}-{{n}^{i}}{{u}^{i}}),\partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\rangle \\ & \le -\sum\limits_{\nu =e,i}{\left( \left\langle {{n}^{\nu }}u_{k,\alpha }^{\nu },u_{k,\alpha }^{\nu } \right\rangle +\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\Theta _{k,\alpha }^{\nu },\Theta _{k,\alpha }^{\nu } \right\rangle +{{\left\| \nabla \Theta _{k,\alpha }^{\nu } \right\|}^{2}} \right)} \\ & +C\sum\limits_{\nu =e,i}{\left\| \left( \partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k+1}{{\Theta }^{\nu }},\partial _{t}^{k}\nabla \Phi \right) \right\|_{|\alpha |-1}^{2}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}},\\ \end{align}$

(3.23)

$\begin{align} & \left| \left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})g_{\nu }^{k,\alpha },U_{k,\alpha }^{\nu } \right\rangle \right| \\ & \le \varepsilon {{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\left( {{u}^{\nu }},\nabla {{\Theta }^{\nu }} \right) \right\|}^{2}}+\left\| \partial _{t}^{k}\left( {{u}^{\nu }},{{\Theta }^{\nu }},\nabla {{\Theta }^{\nu }} \right) \right\|_{\left| \alpha \right|-1}^{2}+C\left\| \partial _{t}^{k}{{N}^{\nu }} \right\|_{\left| \alpha \right|}^{2}+C|||U|||_{[s/2]}^{3},\\ \end{align}$

(3.24)

这里$u_{k,\alpha }^\nu = \partial _t^k{\partial ^\alpha }{u^\nu },\Theta _{k,\alpha }^\nu = \partial _t^k{\partial ^\alpha }{\Theta ^\nu }$,待定系数$\varepsilon $> 0充分小.

事实上,由(3.9)式可知

$M_{k,\alpha }^{\nu }=\left( \begin{matrix} 0 \\ u_{k,\alpha }^{\nu }+{{q}_{\nu }}\partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \\ \Theta _{k,\alpha }^{\nu }-\partial _{t}^{k}{{\partial }^{\alpha }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}\Delta {{\Theta }^{\nu }} \right) \\ \end{matrix} \right).$

进而有

$\begin{align} & \left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})M_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle \\ & =\left\langle {{n}^{\nu }}u_{k,\alpha }^{\nu },u_{k,\alpha }^{\nu } \right\rangle +\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\Theta _{k,\alpha }^{\nu },\Theta _{k,\alpha }^{\nu } \right\rangle -\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k}{{\partial }^{\alpha }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}\Delta {{\Theta }^{\nu }} \right),\Theta _{k,\alpha }^{\nu } \right\rangle \\ & +{{q}_{\nu }}\left\langle {{n}^{\nu }}u_{k,\alpha }^{\nu },\partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\rangle . \\ \end{align}$

(AAA)

由此可得

$\begin{align} & \left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})M_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle -{{q}_{\nu }}\left\langle \partial _{t}^{k}{{\partial }^{\alpha }}({{n}^{\nu }}{{u}^{\nu }}),\partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\rangle \\ & =\left\langle {{n}^{\nu }}u_{k,\alpha }^{\nu },u_{k,\alpha }^{\nu } \right\rangle +\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\Theta _{k,\alpha }^{\nu },\Theta _{k,\alpha }^{\nu } \right\rangle -\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k}{{\partial }^{\alpha }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}\Delta {{\Theta }^{\nu }} \right),\Theta _{k,\alpha }^{\nu } \right\rangle \\ & -{{q}_{\nu }}\left\langle \partial _{t}^{k}{{\partial }^{\alpha }}({{n}^{\nu }}{{u}^{\nu }})-{{n}^{\nu }}u_{k,\alpha }^{\nu },\partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\rangle . \\ \end{align}$

(3.25)

下面处理(3.25)式右端各项. 首先,对于第三项,由Leibniz公式可知

$\begin{align} & \left| \left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k}{{\partial }^{\alpha }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}\Delta {{\Theta }^{\nu }} \right),\Theta _{k,\alpha }^{\nu } \right\rangle \right| \\ & =|I_{1}^{\nu }+\sum\limits_{\begin{matrix} \left| \beta \right|<\left| \alpha \right| \\ l<k \\ \end{matrix}}{C_{\alpha }^{\beta }C_{k}^{l}I_{2l\beta }^{\nu }}+\sum\limits_{\left| \beta \right|<\left| \alpha \right|}{C_{\alpha }^{\beta }I_{3\beta }^{\nu }}+\sum\limits_{l<k}{C_{k}^{l}I_{4l}^{\nu }}|,\\ \end{align}$

(3.26)

其中

$\begin{align} & I_{1}^{\nu }=\left\langle \Delta \Theta _{k,\alpha }^{\nu },\Theta _{k,\alpha }^{\nu } \right\rangle ,I_{2l\beta }^{\nu }=\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k-l}{{\partial }^{\alpha -\beta }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right)\partial _{t}^{k}{{\partial }^{\beta }}\Delta {{\Theta }^{\nu }},\Theta _{k,\alpha }^{\nu } \right\rangle ,\\ & I_{3\beta }^{\nu }=\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}{{\partial }^{\alpha -\beta }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right)\partial _{t}^{k}{{\partial }^{\beta }}\Delta {{\Theta }^{\nu }},\Theta _{k,\alpha }^{\nu } \right\rangle ,I_{4l}^{\nu }=\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k-l}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right)\partial _{t}^{l}{{\partial }^{\alpha }}\Delta {{\Theta }^{\nu }},\Theta _{k,\alpha }^{\nu } \right\rangle . \\ \end{align}$

容易看出,当k=0时${I_{2l\beta }} = {I_{4l}} = 0$. 利用分部积分可得

$I_{1}^{\nu }=\left\langle \Delta \Theta _{k,\alpha }^{\nu },\Theta _{k,\alpha }^{\nu } \right\rangle =-{{\left\| \nabla \Theta _{k,\alpha }^{\nu } \right\|}^{2}}.$

(3.27)

再由Sobolev嵌入定理可得: 当$l = \left| \beta \right| = 0$时有

$\begin{align} & \left| I_{2l\beta }^{\nu } \right|\le C{{\left\| \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}} \right\|}_{{{L}^{\infty }}}}\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right) \right\|{{\left\| \Delta {{\Theta }^{\nu }} \right\|}_{{{L}^{\infty }}}}\left\| \Theta _{k,\alpha }^{\nu } \right\| \\ & \le C\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right) \right\|{{\left\| \Delta {{\Theta }^{\nu }} \right\|}_{2}}\left\| \Theta _{k,\alpha }^{\nu } \right\|\le C{{D}_{[s/2]}}(t){{\left\| \Theta \right\|}_{[s/2]}}; \\ \end{align}$

(3.28)

当$l + \left| \beta \right| = 1$时有

$\begin{align} & \left| I_{2l\beta }^{\nu } \right|\le C{{\left\| \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}} \right\|}_{{{L}^{\infty }}}}{{\left\| \partial _{t}^{k-l}{{\partial }^{\alpha -\beta }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right) \right\|}_{{{L}^{4}}}}{{\left\| \partial _{t}^{l}{{\partial }^{\beta }}\Delta {{\Theta }^{\nu }} \right\|}_{{{L}^{4}}}}\left\| \Theta _{k,\alpha }^{\nu } \right\| \\ & \le C{{\left\| \partial _{t}^{k-l}{{\partial }^{\alpha -\beta }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right) \right\|}_{1}}{{\left\| \partial _{t}^{l}{{\partial }^{\beta }}\Delta {{\Theta }^{\nu }} \right\|}_{1}}\left\| \Theta _{k,\alpha }^{\nu } \right\|\le C{{D}_{[s/2]}}(t){{\left\| \Theta \right\|}_{[s/2]}}; \\ \end{align}$

(3.29)

以及当$l + \left| \beta \right| \ge 2$时有

$\begin{align} & \left| I_{2l\beta }^{\nu } \right|\le C{{\left\| \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}} \right\|}_{{{L}^{\infty }}}}{{\left\| \partial _{t}^{k-l}{{\partial }^{\alpha -\beta }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right) \right\|}_{{{L}^{\infty }}}}\left\| \partial _{t}^{l}{{\partial }^{\beta }}\Delta {{\Theta }^{\nu }} \right\|\left\| \Theta _{k,\alpha }^{\nu } \right\| \\ & \le C{{\left\| \partial _{t}^{k-l}{{\partial }^{\alpha -\beta }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right) \right\|}_{2}}\left\| \partial _{t}^{l}{{\partial }^{\beta }}\Delta {{\Theta }^{\nu }} \right\|\left\| \Theta _{k,\alpha }^{\nu } \right\|\le C{{D}_{[s/2]}}(t){{\left\| \Theta \right\|}_{[s/2]}}. \\ \end{align}$

(3.30)

另一方面,分部积分并结合(3.2)式中的第三个方程可得

$I_{3\beta }^{\nu }\le \varepsilon {{\left\| \left( \Theta _{k,\alpha }^{\nu },\nabla \Theta _{k,\alpha }^{\nu } \right) \right\|}^{2}}+C\left\| \left( \partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k+1}{{\Theta }^{\nu }} \right) \right\|_{\left| \alpha \right|-1}^{2},$

(3.31)

$I_{3\beta }^{\nu }\le \varepsilon {{\left\| \left( \Theta _{k,\alpha }^{\nu },\nabla \Theta _{k,\alpha }^{\nu } \right) \right\|}^{2}}+C\left\| \left( \partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k+1}{{\Theta }^{\nu }} \right) \right\|_{\left| \alpha \right|-1}^{2},$

(3.32)

其次,估计(3.25)式右端最后一项. 由于$|\alpha | \ge 1$以及当$\omega_T$足够小时,$ n = \bar n + N \ge {const} > 0$. 于是分部积分并进行与^[4]中类似的估计可得

$\begin{align} & \left| \left\langle \partial _{t}^{k}{{\partial }^{\alpha }}({{n}^{\nu }}{{u}^{\nu }})-{{n}^{\nu }}u_{k,\alpha }^{\nu },\partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\rangle \right| \\ & \le \varepsilon \left\langle nu_{k,\alpha }^{\nu },u_{k,\alpha }^{\nu } \right\rangle +C||\partial _{t}^{k}{{u}^{\nu }}||_{\alpha |-1}^{2}+C||\partial _{t}^{k}\nabla \Phi ||_{\alpha |-1}^{2}+C|||U|||_{[s/2]}^{2}|||W||{{|}_{[s/2]}},\\ \end{align}$

这里${{\alpha }_{1}}\in {{\mathbb{N}}^{3}}$且$|{{\alpha }_{1}}|=|\alpha |-1$. 因此,取$\varepsilon > 0$足够小,上述估计联合(3.25)--(3.32)式可知(3.23)式成立.

接下来,由(3.17)式可得

$\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})g_{\nu }^{k,\alpha },U_{k,\alpha }^{\nu } \right\rangle =\sum\limits_{j=1}^{3}{K_{1j}^{\nu }}+K_{2}^{\nu },$

(3.33)

其中

$\begin{align} & K_{1j}^{\nu }=\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})\left( A_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }}){{\partial }_{j}}\partial _{t}^{k}{{\partial }^{\alpha }}{{U}^{\nu }}-\partial _{t}^{k}{{\partial }^{\alpha }}\left( A_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }}){{\partial }_{j}}{{U}^{\nu }} \right) \right),U_{k,\alpha }^{\nu } \right\rangle ,\\ & K_{2}^{\nu }=\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})\left( {{L}^{\nu }}\partial _{t}^{k}{{\partial }^{\alpha }}{{U}^{\nu }}-\partial _{t}^{k}{{\partial }^{\alpha }}\left( {{L}^{\nu }}{{U}^{\nu }} \right) \right),U_{k,\alpha }^{\nu } \right\rangle . \\ \end{align}$

由(3.7)--(3.8)式以及$A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})$的定义可得

$\begin{align} & K_{1j}^{\nu }\le \varepsilon {{\left\| \left( \partial _{t}^{k}{{\partial }^{\alpha }}\left( {{u}^{\nu }},\nabla {{\Theta }^{\nu }} \right) \right) \right\|}^{2}}+C\left\| \partial _{t}^{k}({{u}^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|_{|\alpha |-1}^{2} \\ & +C\left\| \partial _{t}^{k}{{N}^{\nu }} \right\|_{|\alpha |}^{2}+C|||U|||_{[s/2]}^{3},\\ \end{align}$

(3.34)

$\begin{array}{*{35}{l}} K_{2}^{\nu }\le \varepsilon {{\left\| u_{k,\alpha }^{\nu } \right\|}^{2}}+C\left\| \partial _{t}^{k}{{N}^{\nu }} \right\|_{|\alpha |}^{2}+C\left\| \partial _{t}^{k}({{u}^{\nu }},{{\Theta }^{\nu }}) \right\|_{|\alpha |-1}^{2} \\ \end{array}$

(3.35)

联合(3.33)--(3.35)式即得(3.24)式.

步骤3 基于上面两个步骤的准备,联合(3.21)式与(3.23)--(3.24)式 并取$\varepsilon>0$足够小即得(3.9)式. 证毕.

引理3.2只对$|\alpha | \ge 1$成立. 下面考察$|\alpha | = 0$时的情形,它为随后第四节的迭代过程提供基础.

引理 3.3 在定理1.1的条件下,若${\omega _T}$足够小,则对任意 $0 \le k \le [s/2],$存在正常数C₀使得$\forall t \in [0,T]$,有

$\begin{align} & \frac{d}{dt}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})\partial _{t}^{k}{{U}^{\nu }},\partial _{t}^{k}{{U}^{\nu }} \right\rangle }+{{\left\| \partial _{t}^{k}\nabla \Phi \right\|}^{2}}) \\ & +{{C}_{0}}\sum\limits_{\nu =e,i}{{{\left\| \partial _{t}^{k}\left( {{u}^{\nu }},{{\Theta }^{\nu }},\nabla {{\Theta }^{\nu }} \right) \right\|}^{2}}}\le C{{D}_{[s/2]}}\left( t \right)|||W||{{|}_{[s/2]}}. \\ \end{align}$

(3.36)

证 注意到$n ^\nu,{\theta ^\nu } \ge {const} > 0$. 当k = 0时,估计式(3.36)由引理3.1给出. 当$1 \le k \le [s/2]$时,(3.36)式由(3.21)式的$\alpha $ = 0情形以及下面两个估计给出.

$\begin{align} & I_{1}^{\nu }=\left\langle \Delta \Theta _{k,\alpha }^{\nu },\Theta _{k,\alpha }^{\nu } \right\rangle ,I_{2l\beta }^{\nu }=\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k-l}{{\partial }^{\alpha -\beta }}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}} \right)\partial _{t}^{k}{{\partial }^{\beta }}\Delta {{\Theta }^{\nu }},\Theta _{k,\alpha }^{\nu } \right\rangle ,\\ & -\left\langle \partial _{t}^{k}({{n}^{e}}{{u}^{e}}-{{n}^{i}}{{u}^{i}}),\partial _{t}^{k}\nabla \Phi \right\rangle -\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})\partial _{t}^{k}{{M}^{\nu }},\partial _{t}^{k}{{U}^{\nu }} \right\rangle } \\ & \le -\sum\limits_{\nu =e,i}{\left( \left\langle {{n}^{\nu }}\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}{{u}^{\nu }} \right\rangle +\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }} \right\rangle +{{\left\| \nabla \partial _{t}^{k}{{\Theta }^{\nu }} \right\|}^{2}} \right)} \\ & +C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}},\\ \end{align}$

(3.37)

$\left| \left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})\partial _{t}^{k}{{g}_{\nu }},\partial _{t}^{k}{{U}^{\nu }} \right\rangle \right|\le C|||U|||_{[s/2]}^{3}.$

(3.38)

事实上,$\alpha $ = 0时,(3.25)式变为

$\begin{align} & {{q}_{\nu }}\left\langle \partial _{t}^{k}({{n}^{\nu }}{{u}^{\nu }}),\partial _{t}^{k}\nabla \Phi \right\rangle -\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})\partial _{t}^{k}{{M}^{\nu }},\partial _{t}^{k}{{U}^{\nu }} \right\rangle \\ & =-\left\langle {{n}^{\nu }}\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}{{u}^{\nu }} \right\rangle -\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }} \right\rangle +\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}\Delta \partial _{t}^{k}{{\Theta }^{\nu }} \right),\partial _{t}^{k}{{\Theta }^{\nu }} \right\rangle \\ & +{{q}_{\nu }}\left\langle \partial _{t}^{k}({{n}^{\nu }}{{u}^{\nu }})-{{n}^{\nu }}\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}\nabla \Phi \right\rangle . \\ \end{align}$

(3.39)

接下来处理上式右端各项. 由Leibniz公式结合分部积分可得

$\left\langle \frac{{{n}^{\nu }}}{{{\theta }^{\nu }}}\partial _{t}^{k}\left( \frac{{{\theta }^{\nu }}}{{{n}^{\nu }}}\Delta {{\Theta }^{\nu }} \right),\partial _{t}^{k}{{\Theta }^{\nu }} \right\rangle \le -{{\left\| \nabla \partial _{t}^{k}{{\Theta }^{\nu }} \right\|}^{2}}+C{{D}_{[s/2]}}(t)|||U||{{|}_{[s/2]}},$

(3.40)

$\left| \left\langle \partial _{t}^{k}({{n}^{\nu }}{{u}^{\nu }})-{{n}^{\nu }}\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}\nabla \Phi \right\rangle \right|\le C|||U|||_{[s/2]}^{2}|||W||{{|}_{[s/2]}}.$

(3.41)

于是,联合(3.39)--(3.41)式可知(3.37)式成立.

最后,当$\alpha $ = 0时,(3.33)式变为

$\begin{align} & \left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})\partial _{t}^{k}{{g}_{\nu }},\partial _{t}^{k}{{U}^{\nu }} \right\rangle = \\ & -\sum\limits_{j=1}^{3}{\sum\limits_{l<k}{C_{k}^{l}}\left. A_{0}^{\nu } \right\rangle }({{n}^{\nu }},{{\theta }^{\nu }})\partial _{t}^{k-l}A_{j}^{\nu }({{n}^{\nu }},{{u}^{\nu }},{{\theta }^{\nu }})\partial _{t}^{l}{{\partial }_{j}}{{U}^{\nu }},\left\langle \partial _{t}^{k}{{U}^{\nu }} \right.\le C|||U|||_{[s/2]}^{3},\\ \end{align}$

于是(3.38)式成立. 证毕.

命题 3.1 在定理1.1的条件下,若${\omega _T}$足够小,则对任意$ k \in {\Bbb N},\alpha \in {{\Bbb N}}^3 且 |\alpha| \ge 1,k + |\alpha| \le [s/2],$存在正常数C₀使得$\forall t \in [0,T]$,有

$\begin{align} & \frac{d}{dt}\sum\limits_{\beta \le \alpha }{(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\beta }^{\nu },U_{k,\beta }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\beta }}\nabla \Phi \right\|}^{2}})}+{{C}_{0}}\sum\limits_{\nu =e,i}{\left\| \partial _{t}^{k}({{u}^{\nu }},{{\Theta }^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|_{|\alpha |}^{2}} \\ & \le C\sum\limits_{\nu =e,i}{\left( \left\| \partial _{t}^{k}{{N}^{\nu }} \right\|_{|\alpha |}^{2}+C\left\| (\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k}\nabla {{\Theta }^{\nu }},\partial _{t}^{k+1}{{\Theta }^{\nu }},\partial _{t}^{k}\nabla \Phi ) \right\|_{|\alpha |-1}^{2} \right)} \\ & +C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

(3.42)

证 固定$k \ge 0$. 注意到

${{\left\| \partial _{t}^{k}({{u}^{\nu }},{{\Theta }^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|}^{2}}+\sum\limits_{1\le |\beta |,\beta \le \alpha }{{{\left\| \partial _{t}^{k}{{\partial }^{\beta }}({{u}^{\nu }},{{\Theta }^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|}^{2}}}\tilde{\ }\left\| \partial _{t}^{k}({{u}^{\nu }},{{\Theta }^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|_{|\alpha |}^{2},$

于是对(3.20)式关于$|\alpha | \le [s/2] - k$求和并结合(3.36)式即可得到(3.42)式.

估计式(3.42)给出了${u^\nu },{\Theta ^\nu }$及$\nabla {\Theta ^\nu }$的耗散的递推关系式. 但该估计式的左端项仍不足以控制所有右端项,因此必须建立密度${N^\nu }$和电势$\nabla \Phi $的耗散估计进而得出迭代关系式.

引理 3.4 在定理1.1的条件下,若${\omega _T}$足够小,则对任意$ k \in {\Bbb N},\alpha \in {{\Bbb N}}^3 $且$ |\alpha| \ge 1,k + |\alpha| \le [s/2],\forall t \in [0,T],$ 有

$\begin{align} & \sum\limits_{\nu =e,i}{\left\| \partial _{t}^{k}{{N}^{\nu }} \right\|_{|\alpha |}^{2}}+\left\| \partial _{t}^{k}({{N}^{e}}-{{N}^{i}}) \right\|_{|\alpha |-1}^{2} \\ & \le C\sum\limits_{\nu =e,i}{\left\| (\partial _{t}^{k}{{N}^{\nu }},\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k}\nabla {{\Theta }^{\nu }}) \right\|_{|\alpha |-1}^{2}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}},\\ \end{align}$

(3.43)

$\begin{align} & \left\| \partial _{t}^{k}\nabla \Phi \right\|_{|\alpha |-1}^{2}\le C\sum\limits_{\nu =e,i}{\left\| (\partial _{t}^{k}{{N}^{\nu }},\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k}\nabla {{\Theta }^{\nu }}) \right\|_{|\alpha |-1}^{2}} \\ & +C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

(3.44)

证 首先,对于$ k \in {\Bbb N},\beta \in {{\Bbb N}}^3, k + |\beta| \le [s/2] - 1$,对(3.2)式第二式求$\partial _t^k{\partial ^\beta }$,并注意到

$\nabla \left( \ln ({{{\bar{n}}}^{\nu }}+{{N}^{\nu }})-\ln {{{\bar{n}}}^{\nu }} \right)=\nabla \left( \frac{{{N}^{\nu }}}{{{{\bar{n}}}^{\nu }}} \right)+\frac{{{({{N}^{\nu }})}^{2}}}{{{n}^{\nu }}{{({{{\bar{n}}}^{\nu }})}^{2}}}-\frac{{{N}^{\nu }}\nabla {{N}^{\nu }}}{{{n}^{\nu }}{{{\bar{n}}}^{\nu }}},$

可得

${{\partial }^{\beta }}\nabla \left( \frac{\partial _{t}^{k}{{N}^{\nu }}}{{{{\bar{n}}}^{\nu }}} \right)+{{q}_{\nu }}\partial _{t}^{k}{{\partial }^{\beta }}\nabla \Phi =\nabla \Theta _{k,\beta }^{\nu }-u_{k,\beta }^{\nu }-u_{k+1,\beta }^{\nu }+{{\partial }^{\beta }}\left( \frac{\nabla {{{\bar{n}}}^{\nu }}}{{{{\bar{n}}}^{\nu }}}\partial _{t}^{k}\Theta \right)-R_{\nu 1}^{k,\beta },$

(3.45)

其中

$\begin{array}{*{35}{l}} \begin{array}{*{35}{l}} R_{\nu 1}^{k,\beta }=\partial _{t}^{k}{{\partial }^{\beta }}\left( \frac{{{({{N}^{\nu }})}^{2}}}{{{n}^{\nu }}{{({{{\bar{n}}}^{\nu }})}^{2}}}+({{u}^{\nu }}\cdot \nabla ){{u}^{\nu }}+\frac{{{\Theta }^{\nu }}\nabla {{N}^{\nu }}}{{{n}^{\nu }}}-\frac{{{N}^{\nu }}{{\Theta }^{\nu }}}{{{n}^{\nu }}{{{\bar{n}}}^{\nu }}}-\frac{{{N}^{\nu }}\nabla {{N}^{\nu }}}{{{n}^{\nu }}{{{\bar{n}}}^{\nu }}} \right). \\ \end{array} \\ \end{array}$

由引理2.2可得$ \left\| {R_{\nu 1}^{k,\beta }} \right\| \le C|||U||{|_{[s/2]}}|||W||{|_{[s/2]}}$.

另外,易知$ {\partial ^\beta }\nabla \left( { \frac{{N_k^\nu }}{{{{\bar n}^\nu }}}} \right) = \frac{1}{{{{\bar n}^\nu }}}\nabla N_{k,\beta }^\nu + R_{\nu 2}^{k,\beta }. $此处$ R_{\nu 2}^{k,\beta } = \sum\limits_{\gamma \le \beta } {{d_{k\beta \gamma }}(x)N_{k,\gamma }^\nu } ,$ 其中${d_{k\beta \gamma }}$为一给定光滑函数. 容易看出$\left\| {R_{2\nu }^{k,\beta }} \right\| \le C{\left\| {\partial _t^k{N^\nu }} \right\|_{|\beta |}}$.

然后,注意到${\bar n^\nu } \ge {const}$ > 0,由(3.45)式与 $ \nabla {N_{k,\beta }} $ 在空间L^2({\Bbb T})上内积可得

$\begin{align} & {{\left\| {{\left( {{{\bar{n}}}^{\nu }} \right)}^{-\frac{1}{2}}}\nabla N_{k,\beta }^{\nu } \right\|}^{2}}+{{q}_{\nu }}\left\langle \partial _{t}^{k}{{\partial }^{\beta }}\nabla \Phi ,\nabla N_{k,\beta }^{\nu } \right\rangle \\ & \le \frac{1}{2}{{\left\| {{\left( {{{\bar{n}}}^{\nu }} \right)}^{-\frac{1}{2}}}\nabla N_{k,\beta }^{\nu } \right\|}^{2}}+C\left\| (\partial _{t}^{k}{{N}^{\nu }},\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\nabla \partial _{t}^{k}{{\Theta }^{\nu }}) \right\|_{|\beta |}^{2} \\ & +C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

再由Poisson方程可得

$\sum\limits_{\nu =e,i}{{{q}_{\nu }}\left\langle \partial _{t}^{k}{{\partial }^{\beta }}\nabla \Phi ,\nabla N_{k,\beta }^{\nu } \right\rangle }={{\left\| N_{k,\beta }^{i}-N_{k,\beta }^{e} \right\|}^{2}}\ge 0.$

故

$\begin{align} & \sum\limits_{\nu =e,i}{{{\left\| \nabla N_{k,\beta }^{\nu } \right\|}^{2}}}+{{\left\| N_{k,\beta }^{e}-N_{k,\beta }^{i} \right\|}^{2}} \\ & \le \sum\limits_{\nu =e,i}{\left\| (\partial _{t}^{k}{{N}^{\nu }},\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\nabla \partial _{t}^{k}{{\Theta }^{\nu }}) \right\|_{|\beta |}^{2}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

对此不等式关于所有指标$\beta$求和可得

$\begin{align} & \sum\limits_{\nu =e,i}{\left\| \nabla \partial _{t}^{k}{{N}^{\nu }} \right\|_{|\beta |}^{2}}+\left\| \partial _{t}^{k}({{N}^{e}}-{{N}^{i}}) \right\|_{|\beta |}^{2} \\ & \le C\sum\limits_{\nu =e,i}{\left\| (\partial _{t}^{k}{{N}^{\nu }},\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\nabla \partial _{t}^{k}{{\Theta }^{\nu }}) \right\|_{|\beta |}^{2}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

注意到(1.7)式,运用引理2.1,然后更换$\beta$为$\alpha$即得(3.43)式,这里$|\alpha | = 1 + |\beta |. $最终,联合(3.45)式可知

${{\left\| \partial _{t}^{k}{{\partial }^{\beta }}\nabla \Phi \right\|}^{2}}\le C\sum\limits_{\nu =e,i}{\left( \left\| \partial _{t}^{k}{{N}^{\nu }} \right\|_{|\beta |+1}^{2}+\left\| (\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\nabla \partial _{t}^{k}{{\Theta }^{\nu }}) \right\|_{|\beta |}^{2} \right)}+{{\left\| R_{1\nu }^{k,\beta } \right\|}^{2}},$

于是,上式关于所有指标$\beta$求和然后再联合(3.43)式可得(3.44)式成立.

注意到${U^\nu } = {({N^\nu },{u^\nu },{\Theta ^\nu })^T}$以及$U = {\left( {{U^e},{U^i}} \right)^T}$并取$\varepsilon>0$充分小可得下面结果.

命题 3.2 在定理1.1的条件下,若${\omega _T}$足够小,则对任意$ k \in {\Bbb N},\alpha \in {{\Bbb N}}^3 $且$ |\alpha| \ge 1,k + |\alpha| \le [s/2],$存在正常数C₀使得$\forall t \in [0,T],$ 有

$\begin{align} & \frac{d}{dt}\sum\limits_{\beta \le \alpha }{(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }U_{k,\beta }^{\nu },U_{k,\beta }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\beta }}\nabla \Phi \right\|}^{2}})}+{{C}_{0}}\sum\limits_{\nu =e,i}{\left\| \partial _{t}^{k}({{U}^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|_{|\alpha |}^{2}} \\ & \le C\sum\limits_{\nu =e,i}{\left\| (\partial _{t}^{k}{{U}^{\nu }},\partial _{t}^{k}\nabla {{\Theta }^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k+1}{{\Theta }^{\nu }}) \right\|_{|\alpha |-1}^{2}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

(3.46)

命题3.2只包含$0 \le k \le [s/2] - 1$的情形. 此外为了运用迭代关系式(3.46),还需估计k = [s/2]的情形作为初值以及$\left\| {\partial _t^k{N^\nu }} \right\|$(详见第四节迭代过程). 下面给出这些估计.

命题 3.3 在定理1.1的条件下,若${\omega _T}$足够小,则对任意$ k \in {\Bbb N},0 \le k \le [s/2] - 1 $有

$\left\| \partial _{t}^{k}{{N}^{\nu }} \right\|_{1}^{2}\le C\sum\limits_{\nu =e,i}{{{\left\| (\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k}\nabla {{\Theta }^{\nu }}) \right\|}^{2}}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}.$

(3.47)

$\begin{align} & \frac{d}{dt}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }\partial _{t}^{[s/2]}{{U}^{\nu }},\partial _{t}^{[s/2]}{{U}^{\nu }} \right\rangle }+{{\left\| \partial _{t}^{[s/2]}\nabla \Phi \right\|}^{2}})+{{c}_{0}}\sum\limits_{\nu =e,i}{{{\left\| \partial _{t}^{[s/2]}({{U}^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|}^{2}}} \\ & \le C\sum\limits_{\nu =e,i}{\left\| \partial _{t}^{[s/2]-1}{{u}^{\nu }} \right\|_{1}^{2}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

(3.48)

证 首先,对于$ k \in {\Bbb N},k \le [s/2] - 1$,对(3.2)式第二式求$\partial _t^k$可得

$\nabla \left( \frac{\partial _{t}^{k}{{N}^{\nu }}}{{{{\bar{n}}}^{\nu }}} \right)+{{q}_{\nu }}\partial _{t}^{k}\nabla \Phi =-\partial _{t}^{k}{{u}^{\nu }}-\partial _{t}^{k+1}{{u}^{\nu }}-\nabla \partial _{t}^{k}{{\Theta }^{\nu }}-\frac{\nabla {{{\bar{n}}}^{\nu }}}{{{{\bar{n}}}^{\nu }}}\partial _{t}^{k}{{\Theta }^{\nu }}-R_{\nu ,1}^{k,0},$

(3.49)

其中

$R_{\nu ,1}^{k,0}=\partial _{t}^{k}\left( \left( {{u}^{\nu }}\cdot \nabla \right){{u}^{\nu }}+\frac{|{{N}^{\nu }}{{|}^{2}}}{{{n}^{\nu }}{{({{{\bar{n}}}^{\nu }})}^{2}}}+\frac{{{\Theta }^{\nu }}\nabla {{N}^{\nu }}}{{{n}^{\nu }}}-\frac{{{N}^{\nu }}{{\Theta }^{\nu }}}{{{n}^{\nu }}{{{\bar{n}}}^{\nu }}}-\frac{{{N}^{\nu }}\nabla {{N}^{\nu }}}{{{n}^{\nu }}{{{\bar{n}}}^{\nu }}} \right).$

定义势函数$\nabla \Psi $满足

$\nabla \cdot \left( \nabla \Psi \right)=\Delta \Psi ={{N}^{e}}-{{N}^{i}},\int_{\mathbb{T}}{\Psi (t,x)dx=0.}$

于是 $ \nabla \cdot \left( { - \nabla \Phi + \nabla \Psi } \right) = - ({N^e} - {N^i}) + {N^e} - {N^i} = 0,$进而有

$\nabla \cdot \left( -\nabla \partial _{t}^{k}\Phi +\nabla \partial _{t}^{k}\Psi \right)=0,0\le k\le [s/2]-1.$

从(3.49)式可知

$\begin{align} & \nabla \eta _{k}^{\nu }-{{q}_{\nu }}\left( -\nabla \partial _{t}^{k}\Phi +\nabla \partial _{t}^{k}\Psi \right) \\ & =-\partial _{t}^{k}{{u}^{\nu }}-\partial _{t}^{k+1}{{u}^{\nu }}-\nabla \partial _{t}^{k}{{\Theta }^{\nu }}-\frac{\nabla {{{\bar{n}}}^{\nu }}}{{{{\bar{n}}}^{\nu }}}\partial _{t}^{k}{{\Theta }^{\nu }}-R_{\nu ,1}^{k,0},\\ \end{align}$

(3.50)

其中

$\eta _{k}^{\nu }=\frac{1}{{{{\bar{n}}}^{\nu }}}\partial _{t}^{k}{{N}^{\nu }}+{{q}_{\nu }}\partial _{t}^{k}\Psi .$

由$\left\langle { - \nabla \partial _t^k\Phi + \nabla \partial _t^k\Psi ,\nabla \eta _k^\nu } \right\rangle = - \left\langle {\nabla \cdot \left( { - \nabla \partial _t^k\Phi + \nabla \partial _t^k\Psi } \right),\eta _k^\nu } \right\rangle = 0,$可得

${{\left\| \nabla \eta _{k}^{\nu } \right\|}^{2}}\le C{{\left\| (\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\nabla \partial _{t}^{k}{{\Theta }^{\nu }}) \right\|}^{2}}+C{{\left\| R_{\nu 1}^{k,0} \right\|}^{2}}.$

(3.51)

因$\partial _t^k{N^\nu } = {\bar n^\nu }\eta _k^\nu - {q_\nu }{\bar n^\nu }\partial _t^k\Psi,\Delta \Psi = {N^e} - {N^i},$可得

$-\Delta \partial _{t}^{k}\Psi =\partial _{t}^{k}{{N}^{i}}-\partial _{t}^{k}{{N}^{e}}={{{\bar{n}}}^{i}}\eta _{k}^{i}+{{{\bar{n}}}^{i}}\partial _{t}^{k}\Psi -{{{\bar{n}}}^{e}}\eta _{k}^{e}-{{{\bar{n}}}^{e}}\partial _{t}^{k}\Psi .$

因此,$ - \Delta \partial _t^k\Psi + \left( {{{\bar n}^e} + {{\bar n}^i}} \right)\partial _t^k\Psi = {\bar n^i}\eta _k^i - {\bar n^e}\eta _k^e.$

因${\bar n^\nu } \ge {const} >$ 0,上式与$ \partial _t^k\Psi$ 在$L^2({\Bbb T})$上做内积可得

${{\left\| \partial _{t}^{k}\nabla \Psi \right\|}^{2}}+{{c}_{0}}{{\left\| \partial _{t}^{k}\Psi \right\|}^{2}}\le C\sum\limits_{\nu =e,i}{{{\left\| \eta _{k}^{\nu } \right\|}^{2}}}\le C\sum\limits_{\nu =e,i}{{{\left\| \nabla \eta _{k}^{\nu } \right\|}^{2}}},$

(3.52)

这里用到了引理2.1.

从(3.49),(3.51)--(3.52)式以及$\eta _k^\nu $的定义可得

$\begin{align} & {{\left\| \nabla \left( \frac{\partial _{t}^{k}{{N}^{\nu }}}{{{{\bar{n}}}^{\nu }}} \right) \right\|}^{2}}\le {{\left\| \nabla \eta _{k}^{\nu } \right\|}^{2}}+{{\left\| \partial _{t}^{k}\nabla \Psi \right\|}^{2}} \\ & \le C{{\left\| (\partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\nabla \partial _{t}^{k}{{\Theta }^{\nu }}) \right\|}^{2}}+C{{\left\| R_{\nu 1}^{k,0} \right\|}^{2}}. \\ \end{align}$

注意到${\bar n^\nu } \ge {const} > 0$时,结合引理2.1可得如下等价关系式

${{\left\| {{\left( {{{\bar{n}}}^{\nu }} \right)}^{-\frac{1}{2}}}\partial _{t}^{k}{{N}^{\nu }} \right\|}^{2}}\tilde{\ }\left\| \partial _{t}^{k}{{N}^{\nu }} \right\|_{1}^{2}.$

此即完成了对(3.47)式的证明.

接下来,由(3.2)式的第一式可得

$\partial _{t}^{[s/2]}{{N}^{\nu }}=-\nabla \cdot (\partial _{t}^{[s/2]-1}({{N}^{\nu }}{{u}^{\nu }}))-\nabla \cdot ({{\bar{n}}^{\nu }}\partial _{t}^{[s/2]-1}{{u}^{\nu }}),$

由此可得${{\left\| \partial _{t}^{[s/2]}{{N}^{\nu }} \right\|}^{2}}\le C\left\| \partial _{t}^{[s/2]-1}{{u}^{\nu }} \right\|_{1}^{2}+C\left\| U \right\|_{[s/2]}^{3}.$

上式再联合k = [s/2]时的(3.36)式即得(3.48)式. 证毕.

定理1.1的证明基于以下结果,它是上节所有估计的推论.

引理 4.1 在定理1.1的条件下,若${\omega _T}$充分小,则对任意$ k \in {\Bbb N},\alpha \in {{\Bbb N}}^3 $且$ k + |\alpha| \le [s/2]$,存在正常数$ {\lambda _{k,\alpha }}$使得$\forall t \in [0,T]$,有

$\frac{d}{dt}\sum\limits_{k+|\alpha |\le [s/2]}{{{\lambda }_{k,\alpha }}}\left( \left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle +{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}} \right)+{{D}_{[s/2]}}(t)\le 0.$

(4.1)

证 在命题3.2的(3.46)式中取$(k,|\alpha |) = ([s/2] - 1,1)$可得

$\begin{align} & \frac{d}{dt}\sum\limits_{\left| \beta \right|\le 1}{(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{[s/2]-1,\beta }^{\nu },U_{[s/2]-1,\beta }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{[s/2]-1}{{\partial }^{\beta }}\nabla \Phi \right\|}^{2}})} \\ & +{{C}_{0}}\sum\limits_{\nu =e,i}{\left\| \partial _{t}^{[s/2]-1}({{U}^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|_{1}^{2}} \\ & \le \text{C}\sum\limits_{\nu =e,i}{{{\left\| (\partial _{t}^{[s/2]-1}{{U}^{\nu }},\partial _{t}^{[s/2]-1}\nabla {{\Theta }^{\nu }},\partial _{t}^{[s/2]}{{u}^{\nu }},\partial _{t}^{[s/2]}{{\Theta }^{\nu }}) \right\|}^{2}}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

(4.2)

容易看出,(3.48)式右端项$\left\| \partial _{t}^{[s/2]-1}{{u}^{\nu }} \right\|_{1}^{2}$可以被(4.2)式的左端项$\left\| \partial _{t}^{[s/2]-1}({{U}^{\nu }},\nabla {{\Theta }^{\nu }}) \right\|_{1}^{2}$控制. 于是在(3.46)式中按照k降低,$|\alpha|$升高的规则依次对指标$(k,|\alpha |)$取遍所有可能的值可知: (3.46)式中的右端项$\left\| (\partial _{t}^{k}{{U}^{\nu }},\partial _{t}^{k}\nabla {{\Theta }^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k+1}{{\Theta }^{\nu }}) \right\|_{|\alpha |-1}^{2}$(在k降低,$|\alpha|$升高后)可间接被原不等式左端项所控制.于是由(3.46)--(3.48)式以及(4.2)式及整个迭代过程可知: 存在正常数${\lambda _{k,\alpha }}$使得下式成立

$\begin{align} & \frac{d}{dt}\sum\limits_{k+|\alpha |\le [s/2]}{{{\lambda }_{k,\alpha }}}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}}) \\ & +\sum\limits_{k+|\alpha |\le [s/2]}{\sum\limits_{\nu =e,i}{||\partial _{t}^{k}({{U}^{\nu }},\nabla {{\Theta }^{\nu }})||_{|\alpha |}^{2}}} \\ & \le C\sum\limits_{k=0}^{[s/2]-1}{\sum\limits_{\nu =e,i}{{{\left\| \left( \partial _{t}^{k}{{U}^{\nu }},\partial _{t}^{k}\nabla {{\Theta }^{\nu }},\partial _{t}^{k+1}{{u}^{\nu }},\partial _{t}^{k+1}{{\Theta }^{\nu }} \right) \right\|}^{2}}}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

再次应用命题3.3并注意到等价关系式$\sum\limits_{k+|\alpha |\le [s/2]}{\sum\limits_{\nu =e,i}{\left\| \partial _{t}^{k}\left( {{U}^{\nu }},\nabla {{\Theta }^{\nu }} \right) \right\|_{|\alpha |}^{2}}}\tilde{\ }{{D}_{[s/2]}}(t),$重整系数(仍然记为${\lambda _{k,\alpha }})$可得

$\begin{align} & \frac{d}{dt}\sum\limits_{k+|\alpha |\le [s/2]}{{{\lambda }_{k,\alpha }}}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}})+{{D}_{[s/2]}}(t) \\ & \le C\sum\limits_{k=0}^{[s/2]}{\sum\limits_{\nu =e,i}{{{\left\| \left( \partial _{t}^{k}{{u}^{\nu }},\partial _{t}^{k}{{\Theta }^{\nu }},\partial _{t}^{k}\nabla {{\Theta }^{\nu }} \right) \right\|}^{2}}}}+C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

接下来,运用命题3.1并再度重整系数${\lambda _{k,\alpha }}$可得

$\begin{align} & \frac{d}{dt}\sum\limits_{k+|\alpha |\le [s/2]}{{{\lambda }_{k,\alpha }}}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}}) \\ & +{{D}_{[s/2]}}(t)\le C{{D}_{[s/2]}}(t)|||W||{{|}_{[s/2]}}. \\ \end{align}$

最后,由${\omega _T}$的小性可知(4.1)式成立. 证毕.

首先给出范数等价性结果以备后用.

引理 4.2 对任意$m\in {\Bbb N}$下述等价关系式成立

$||||U||{{|}_{m}}\tilde{\ }|||W||{{|}_{m}},{{\left\| U \right\|}_{s}}\tilde{\ }{{\left\| W \right\|}_{s}}.$

证 由(1.5)式及表达式(3.4)及引理2.1知结果成立. 证毕.

现在综合上述估计建立小摄动光滑解的整体存在性及渐近稳定性. 由引理4.1--4.2知

$\frac{d}{dt}\sum\limits_{k+|\alpha |\le [s/2]}{{{\lambda }_{k,\alpha }}}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}})+|||W|||_{[s/2]}^{2}\le 0.$

注意到等价关系

$\sum\limits_{k+|\alpha |\le [s/2]}{{{\lambda }_{k,\alpha }}}(\sum\limits_{\nu =e,i}{\left\langle A_{0}^{\nu }({{n}^{\nu }},{{\theta }^{\nu }})U_{k,\alpha }^{\nu },U_{k,\alpha }^{\nu } \right\rangle }+{{\left\| \partial _{t}^{k}{{\partial }^{\alpha }}\nabla \Phi \right\|}^{2}})\tilde{\ }|||W|||_{[s/2]}^{2},$

可知存在常数$\eta $> 0使得

$\frac{d}{dt}|||W|||_{[s/2]}^{2}+\eta |||W|||_{[s/2]}^{2}\le 0.$

(4.3)

在[0,T]上积分(4.3)式并应用引理2.3可得

$|||W(t)|||_{[s/2]}^{2}\le C{{e}^{-\eta t}}|||W(0)|||_{[s/2]}^{2}\le C{{e}^{-\eta t}}||{{W}^{0}}||_{s}^{2}\le C{{e}^{-\eta t}}||{{U}^{0}}||_{s}^{2}.$

于是可知(1.8)--(1.11)式成立.证毕.