微极流体模型, 20世纪60年代由Eringen^[8-9]提出. 该模型研究具有微观结构的材料的微观现象. 微极流体的应用涵盖了非常广泛的领域. 例如, 动物血液, 悬浮液和液晶等. 更多详情可参见文献[7, 24-25]. 在欧拉坐标系中, 一维非等熵可压缩微极流体模型可描述如下

(1.1) $ \begin{equation} \left\{ \begin{array}{rl} &\rho_t+(\rho u)_x = 0, \\ &(\rho u)_t+(\rho u^2+P )_x = ({\lambda u_x})_x, \\ &(\rho w)_t+(\rho u w)_x+A w = ({ A w_x})_x, \\ &E_t+\left(u\big(E+P\big)\right)_x+q_x = \left(\lambda uu_x+ A {ww}_x\right)_x+ A w^2, \end{array}\right. \end{equation} $

其中$ \rho(x, t) $, $ u(x, t) $, $ w(x, t) $, $ P $分别代表密度, 速度, 微旋转速度和压力. $ \lambda $是粘性系数, $ A $是一个正常数. 总能量$ E $由下式给出

$ E = \rho\left(e+\frac{1}{2}\big(u^2+w^2\big)\right). $

我们考虑多方气体, 因此压力函数$ P $和内能函数$ e $分别为

(1.2) $ \begin{equation} P = R\rho T, \quad\quad e = C_v T, \end{equation} $

热流密度为

(1.3) $ \begin{equation} q = -\kappa T_x, \end{equation} $

其中参数$ R>0, \; C_v>0 $, $ \kappa $分别为气体常数, 定容热容, 热传导系数. 为了简单起见, 我们我们假设$ \lambda $以及$ \kappa $是正常数, 并将$ R $和$ C_v $归一化, 即$ R = 1 $和$ C_v = 1 $.

低马赫数近似的目的是要说明压缩对于压力的变化, 可以忽略不计. 这是讨论高亚音速流体力学时常见的假设. 我们先简要回顾一下欧拉方程、纳维-斯托克斯方程、磁流体力学方程和微极流体方程在这方面的研究结果. 对于欧拉方程, 第一个结果来自于Klainerman和Majda^[20-21], 他们证明了等熵可压欧拉方程到不可压缩欧拉方程的局部光滑解的不可压缩极限. Ukai^[34]利用声波的快速衰减, 验证了一般初值的低马赫数极限. Schochet^[30]在有界区域上对准备充分的初值得到了非等熵可压缩欧拉方程到非等熵不可压缩欧拉方程局部光滑解的收敛性. 对于纳维-斯托克斯方程, Alazard^[2]在全空间对准备不足的初值证明了低马赫数极限. Huang, Wang和Wang^[13]在全空间上分别对准备充分的初值和准备不足的初值, 得到了背景不是恒定状态的一维非等熵纳维-斯托克斯方程的低马赫数极限. 在有界区域, 低马赫数极限由Jiang和Ou^[15]以及Dou, Jiang和Ou^[6]所证明. 对于磁流体力学方程, Klainerman和Majda^[20]在周期区域对准备充分的初值研究了可压缩等熵磁流体力学方程的低马赫数极限. Hu和Wang^[12]分别在无界区域, 周期区域和全空间得到了可压缩粘性磁流体力学方程弱解的收敛性. Fan, Gao和Guo^[11]研究了具有零热传导系数的非等熵磁流体力学方程的低马赫数极限. Jiang, Ju, Li和Xin^[18]在三维空间上建立了具有一般初值的完全可压缩磁流体力学方程的低马赫数极限. Su^[32-33]分别对准备充分和准备不足的初值, 得到了三维可压缩微极流体模型的低马赫数极限. 其他有趣的结果, 对欧拉方程, 可参见文献[1, 27-28]. 对纳维-斯托克斯方程, 可参见文献[5, 10, 19, 22, 26-27]. 对磁流体力学方程, 可参见文献[11-12, 14-17, 23]. 对微极流体方程, 可参见文献[4].

本文我们研究背景不是恒定状态的微极流体的低马赫数极限, 即

$ (\rho^\epsilon, T^\epsilon)(x, t) \rightarrow (\rho_\pm, T_\pm), \; x\rightarrow \pm\infty, \;\; \rho_-T_- = \rho_+T_+, $

因为$ T_- $可能不等于$ T_+ $, 所以会造成一些数学上的困难. 为此, 我们引入了一个背景状态函数$ \tilde \theta $来解决这个问题, 详见第2节.

在本文中, 令$ \varepsilon $为可压缩参数, 代表流体的最大马赫数. 和文献[31]一样, 我们设

(1.4) $ \begin{equation} u(x, t) = \varepsilon u^{\varepsilon}(x, \varepsilon t), \quad w(x, t) = \varepsilon w^{\varepsilon}(x, \varepsilon t), \end{equation} $

$ \lambda = \varepsilon\lambda^\varepsilon, \quad A = \varepsilon A ^\varepsilon, \quad\kappa = \varepsilon\kappa^\varepsilon. $

考虑物理状态中的压力$ P $满足

(1.5) $ \begin{equation} P^{\varepsilon} = \overline{P} +O(\varepsilon), \end{equation} $

其中$ \overline{P} $是某个给定的常量, 它被标准化为$ \overline{P} = 1 $. 于是, 系统(1.1)变为

(1.6) $ \begin{equation} \left\{ \begin{array}{ll} \rho^{\varepsilon}_t+(\rho^{\varepsilon} u^{\varepsilon})_x = 0, \\ { } \rho^{\varepsilon}( u^{\varepsilon}_t+u^{\varepsilon}u^{\varepsilon}_x) +\frac{P^{\varepsilon}_x}{\varepsilon^2} = \lambda^\varepsilon u^{\varepsilon}_{xx}, \\ \rho^{\varepsilon}( w^{\varepsilon}_t+u^{\varepsilon}w^{\varepsilon}_x) + A ^\varepsilon w = A ^\varepsilon w^{\varepsilon}_{xx}, \\ \rho^{\varepsilon}( T^{\varepsilon}_t+u^{\varepsilon}T^{\varepsilon}_x)+P^{\varepsilon}u^{\varepsilon}_x = \kappa T^{\varepsilon}_{xx}+\varepsilon^2\left(\lambda^\varepsilon |u^{\varepsilon}_x|^2+ A ^\varepsilon\big( |w^{\varepsilon}_x|^2+ |w^{\varepsilon}|^2\big)\right), \end{array} \right. \end{equation} $

它的极限系统为

(1.7) $ \begin{equation} \left\{ \begin{array}{l} (2\overline u-\overline T_x)_x = 0, \quad \overline\rho = \overline T^{-1}, \\ { } \overline\rho( \overline u_t+\overline u \cdot\overline u_x)+\overline\pi_x = \lambda \overline u_{xx}, \\ \overline\rho( \overline w_t+\overline u\cdot\overline w_x) + A \overline w = A \overline w_{xx}, \\ \overline\rho( \overline T_t+\overline u\overline T_x)+\overline u_x = \kappa \overline T_{xx}, \end{array} \right. \end{equation} $

其中$ \overline\pi = \lambda\overline u_x-\overline\rho{\overline u}^2+\frac{\kappa}{2}\frac{\overline\rho_t}{\overline\rho} $, 为方便起见, 我们仍用$ (\lambda, A, \kappa) $代替$ (\overline{\lambda}, \overline{ A }, \overline\kappa) $. 利用系统(1.7), 我们得到如下非线性扩散方程

(1.8) $ \begin{equation} \rho_t = \left(\frac{\kappa}{2}\frac{\rho_x}{\rho}\right)_x. \end{equation} $

像文献[3]一样, 考虑远场条件$ \lim\limits_{x\rightarrow \pm\infty}\rho = \rho_{\pm} $, 我们得到方程(1.8)的一个唯一的自相似解$ {\mathcal P}(\xi), \;\xi = \frac{x}{\sqrt{1+t}} $满足$ {\mathcal P}(\pm\infty, t) = \rho_{\pm} $. 令$ \delta = |\rho_+-\rho_{-}| $, 则$ {\mathcal P}(x, t) $满足

$ {\mathcal P}_x(x, t) = \frac{O(1)\delta}{\sqrt{1+t}}e^{-\frac{x^2}{4d(\rho_{\pm})(1+t)}}, \quad x\rightarrow \pm\infty, \quad d(\rho) = \frac{\kappa}{2\rho}. $

为了说明本文的主要结果, 我们考虑了压力$ P^\varepsilon $是给定状态1的一个小扰动的情况. 为了理解热力学的作用, 如文献[2]中所述, 我们引入以下变换来确保$ P^\varepsilon $和$ T^\varepsilon $的正性

(1.9) $ \begin{equation} P^\varepsilon(x, t) = e^{\varepsilon p^\varepsilon(x, t)}, \quad T^\varepsilon(x, t) = e^{\theta^\varepsilon(x, t)}. \end{equation} $

从式(1.2) 和(1.9) 得出$ \rho^{\varepsilon} = e^{\varepsilon p^\varepsilon(x, t)-\theta^\varepsilon(x, t)} $. 在这些变量的变化下, 一维非等熵可压缩微极流体模型(1.6)具有以下等价形式

(1.10) $ \begin{equation} \left\{ \begin{array}{ll} { } p^{\varepsilon}_t+u^{\varepsilon}p_x+\frac{1}{\varepsilon}\big(2u^{\varepsilon}-\kappa^\varepsilon e^{-\varepsilon p^{\varepsilon}+\theta^{\varepsilon}}\theta^{\varepsilon}_x\big)_x = \varepsilon e^{-\varepsilon p^{\varepsilon}}\big(\lambda^\varepsilon|u^{\varepsilon}_x|^2+ A ^\varepsilon |w^{\varepsilon}_x|^2+ A ^\varepsilon |w^{\varepsilon}|^2\big)\\ { }{\qquad}{\qquad}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\; \; \; \; \; \; \; { } +\kappa e^{-\varepsilon p^{\varepsilon}+\theta^{\varepsilon}}p_x^\varepsilon \theta^{\varepsilon}_x, \\ { } e^{-\theta^{\varepsilon}}( u^{\varepsilon}_t+u^{\varepsilon}u^{\varepsilon}_x) +\frac{p^{\varepsilon}_x}{\varepsilon} = e^{-\varepsilon p^{\varepsilon}}\lambda^\varepsilon u^{\varepsilon}_{xx}, \\ { } e^{-\theta^{\varepsilon}}( w^{\varepsilon}_t+u^{\varepsilon}w^{\varepsilon}_x) = e^{-\varepsilon p^{\varepsilon}}\big( A ^\varepsilon w^{\varepsilon}_{xx}- A ^\varepsilon w^{\varepsilon}\big), \\ \theta^{\varepsilon}_t+u^{\varepsilon}\theta^{\varepsilon}_x+u^{\varepsilon}_x = \kappa^\varepsilon e^{-\varepsilon p^{\varepsilon}}\big(e^{\theta^{\varepsilon}} \theta^{\varepsilon}_{x}\big)_x+\varepsilon^2e^{-\varepsilon p^{\varepsilon}}\left(\lambda^\varepsilon |u^{\varepsilon}_x|^2+ A ^\varepsilon\big( |w^{\varepsilon}_x|^2+ |w^{\varepsilon}|^2\big)\right). \end{array} \right. \end{equation} $

我们将研究系统(1.10) 解的极限. 形式上, 当$ \varepsilon\rightarrow 0 $时, 如果$ (p^\varepsilon, u^\varepsilon, w^\varepsilon, \theta^\varepsilon) $强收敛到极限$ (1, \overline u, \overline{w}, \overline \theta) $, 并且$ (\lambda^\varepsilon, A ^\varepsilon, \kappa^\varepsilon) $收敛到一个常数向量$ (\overline{\lambda}, \overline{ A }, \overline\kappa) $, 然后对系统(1.10)取极限, 我们有

(1.11) $ \begin{equation} \left\{ \begin{array}{ll} (2\overline u-\kappa e^{\overline \theta}\overline \theta_x)_x = 0, \\ e^{-\overline \theta}( \overline u_t+\overline u \cdot\overline u_x)+\overline\pi_x = \lambda \overline u_{xx}, \\ e^{-\overline \theta}( \overline w_t+\overline u\cdot\overline w_x) + A \overline w = A \overline w_{xx}, \\ \overline \theta_t+\overline u\overline\theta_x+\overline u_x = \kappa \big(e^{\overline \theta}\overline \theta_{x}\big)_x. \end{array} \right. \end{equation} $

本文的目的是用一般的初始数值严格建立上述极限过程. 为此, 我们补充系统(1.10)的初始条件如下

(1.12) $ \begin{equation} \big(p^{\varepsilon}, u^{\varepsilon}, w^{\varepsilon}, \theta^\varepsilon\big)|_{t = 0} = \big(p^{\varepsilon}_{in}, u^{\varepsilon}_{in}, w^{\varepsilon}_{in}, \theta^\varepsilon_{in}\big), \end{equation} $

并且

$ 0<\underline{a}\leq e^{-\varepsilon p^{\varepsilon}_{in}}\leq\overline{a}, \quad 0<\underline{b}\leq e^{ \theta^{\varepsilon}_{in}}\leq\overline{b}, $

其中$ \underline{a}, \;\overline{a}, \;\underline{b}, \;\overline{b} $是给定的不依赖于$ \varepsilon $的常数.

本文的布局如下: 在第2节, 我们给出了本文的主要结果和一些注释. 第3节, 我们建立了先验估计, 并在这些估计的帮助下, 证明定理2.1. 在第4节中, 我们通过修改Métivier和Schochet^[28]提出的论点来证明定理2.2.

在本节中, 我们首先给出一些符号, 然后陈述柯西问题(1.10), (1.12)的主要结果. 符号选择如下

我们用$ ||\cdot||_B $表示空间$ B $中的范数, 特别的, 用$ ||f|| $表示$ ||f||_{L^2({\mathbb R}) } $. 对于多索引数$ \alpha = (\alpha_0, \alpha_1) $, 定义$ \partial^\alpha: = (\varepsilon\partial_t)^{\alpha_0}\partial^{\alpha_1}_x $和$ |\alpha| = |\alpha_0|+|\alpha_1| $, 在此注释下, 我们设$ \|f\|_{{\mathcal H}^{s}} = \sum\limits_{|\alpha|\leq s}\|\partial^\alpha f(t)\| $.

由于$ \theta_{-} = \ln T_- $可能不等于$ \theta_{+} = \ln T_+ $, 我们需要为$ \theta^\varepsilon $引入一个背景状态函数$ \tilde\theta $. 我们选择$ \tilde\theta = -\ln{\mathcal P} $满足, 当$ x\rightarrow \pm\infty $时, $ \tilde\theta\rightarrow \theta_{\pm} $. 当$ s\geq 3 $时, 我们定义如下解空间

(2.1) $ \begin{eqnarray} {{\cal N}}(t):& = &\|(p^{\varepsilon}, u^{\varepsilon}, {w}^{\varepsilon}, \theta^{\varepsilon}-\tilde\theta)(t)\|_{H^{s, \varepsilon}}^2\\ & = &\sum\limits_{|\alpha| = 0}^{s}\|\partial^\alpha(p^{\varepsilon}, u^{\varepsilon}, w^{\varepsilon} )(t)\|^2+\sum\limits_{|\alpha| = 0}^{s+1}\|\partial^\alpha(\varepsilon p^{\varepsilon}, \varepsilon u^{\varepsilon}, \varepsilon w^{\varepsilon} )(t)\|^2\\ &\quad&+\|(\theta^{\varepsilon}-\tilde\theta)(t)\|^2 +\sum\limits_{|\alpha| = 0}^{s}\int_0^t\|\partial^\alpha(p^{\varepsilon}_x, u^{\varepsilon}_x, w^{\varepsilon}_x )(\tau)\|^2{\rm d}\tau \\ &\quad&+\sum\limits_{|\alpha| = 0}^{s+1}\|\partial^\alpha\theta^{\varepsilon}(t)\|^2 +\sum\limits_{|\alpha| = 0}^{s+1}\int_0^t\|\partial^\alpha(\theta^{\varepsilon}_x, \varepsilon u^{\varepsilon}_x, \varepsilon w^{\varepsilon}_x )(\tau)\|^2{\rm d}\tau. \end{eqnarray} $

现在给出我们的主要结果.

定理2.1   令$ s\geq 3 $是一个整数, 假设初始条件(1.12) 满足

(2.2) $ \begin{equation} \|\big(p^{\varepsilon}_{in}, u^{\varepsilon}_{in}, w^{\varepsilon}_{in}, \theta^\varepsilon_{in}\big)\|_{H^{s, \varepsilon}}^2\leq C_0<\infty, \quad \varepsilon \in (0, 1], \end{equation} $

其中$ C_0 $不依赖于$ \varepsilon $. 则存在仅依赖$ C_0 $和$ |\theta_+-\theta_-| $的正常数$ T_0 $和$ \varepsilon_0 $使得柯西问题(1.10), (1.12) 有唯一的光滑解$ (p^{\varepsilon}, u^{\varepsilon}, w^{\varepsilon}, \theta^\varepsilon) $满足

(2.3) $ \begin{equation} \|(p^{\varepsilon}, u^{\varepsilon}, w^{\varepsilon}, \theta^\varepsilon)\|_{H^{s, \varepsilon}}^2\leq \tilde{C_0} , \end{equation} $

其中$ \tilde{C_0}>0 $仅依赖于$ C_0 $和$ |\theta_+-\theta_-| $.

注2.1   波的强度$ |\theta_+-\theta_{-}| $允许较大, 但是$ T_0 $和$ \varepsilon_0 $可能依赖波的强度.

定理2.2(低马赫数极限)   假设初始条件(1.12) 满足

(2.4) $ \begin{eqnarray} &&\big(p^{\varepsilon}_{in}, u^{\varepsilon}_{in}, w^{\varepsilon}_{in}, \theta^\varepsilon_{in}-\tilde\theta\big)\rightarrow \big(p_{in}, u_{in}, w_{in}, \theta_{in}-\tilde\theta\big), \;{\rm in}\quad H^s({ \mathbb R})\quad {\rm as} \quad\varepsilon \rightarrow 0, \end{eqnarray} $

(2.5) $ \begin{eqnarray} &&|\theta_{in}-\theta_+|\leq C x^{-1-\sigma}, \quad x\in [1, +\infty), \end{eqnarray} $

这里$ \sigma $和$ C $是正常数, 则有定理2.1知, 系统(1.10)的解$ (p^{\varepsilon}, u^{\varepsilon}, w^{\varepsilon}, \theta^\varepsilon) $在$ L^2(0, T_0; H_{\rm loc}^{s'}(\mathbb R)) $中对所有的$ s'<s $强收敛到$ (0, u, w, \theta) $, 其中$ ( u, w, \theta) $是具有初始条件$ (u^{\varepsilon}_{in}, w^{\varepsilon}_{in}, \theta^\varepsilon_{in}) $的系统(1.9) 的唯一解.

注2.2  当$ x\in (-\infty, -1] $时, 条件(2.5) 可以被$ |\theta_{in}-\theta_{-}|\leq C x^{-1-\sigma} $替换.

为方便起见, 我们省略了变量的上标$ \varepsilon $, 并令

(3.1) $ \begin{equation} a(\varepsilon p) = e^{-\varepsilon p}, \quad b(\theta) = e^{\theta} \end{equation} $

和

(3.2) $ \begin{equation} \left\{ \begin{array}{rl} &{\mathcal Q}(t): = \sup \limits_{0\leq \tau\leq t}\big(\|(p, u, w)\|_{{\mathcal H}^{s}}+\|(\varepsilon p, \varepsilon u, \varepsilon w)\|_{{\mathcal H}^{s+1}}\\ &\quad\quad\quad\quad +\|\theta-\tilde\theta\|_{L^2} +\|((\varepsilon\partial_t)\theta, \theta_x)\|_{{\mathcal H}^{s}}\big), \\ &{\mathcal S}(t): = \|(p_x, u_x, w_x)\|_{{\mathcal H}^{s}}+\|(\theta_x, \varepsilon u_x, \varepsilon w_x)\|_{{\mathcal H}^{s+1}}. \end{array} \right. \end{equation} $

我们也假设以下的估计成立

(3.3) $ \begin{equation} 0<\frac{1}{2}\underline{a}\leq e^{-\varepsilon p^{\varepsilon}_{in}}\leq 2\overline{a}, \quad 0<\frac{1}{2}\underline{b}\leq e^{ \theta^{\varepsilon}_{in}}\leq 2\overline{b}, \quad {\rm on }\quad t\in [0, T]. \end{equation} $

定理2.1的证明取决于下面的关键命题

命题3.1   对任意给定的$ s \geq 3 $和$ \varepsilon \in (0, 1] $, 令$ (p_x, u_x, w_x, \theta_x) $是柯西问题(1.10), (1.12)的经典解. 若式(3.3)成立, 则

(3.4) $ \begin{equation} {\mathcal N}(t)\leq C[1+\Lambda({\mathcal Q}(0))]+C(t^{\frac{1}{2}}+\varepsilon)\Lambda({\mathcal N}(t)), \end{equation} $

其中$ {\mathcal N}(t) $由式(2.1)定义, 常数$ C>0 $依赖于$ \underline a, \;\overline a, \;\underline b $和$ \overline b $, $ \Lambda(\cdot) $是有限阶多项式.

引理3.1   当$ s\geq 3 $, 成立

(3.5) $ \begin{equation} \|\theta-\tilde \theta\|^2+\sum\limits_{|\alpha| = 1}^{s}\|\partial^\alpha\theta\|^2 +\sum\limits_{|\alpha| = 0}^{s}\int_0^t\|\partial^\alpha\theta_x\|^2{\rm d}\tau \leq C+{\mathcal Q}(0)+\int_0^t\Lambda({\mathcal Q}(\tau)){\rm d}\tau. \end{equation} $

证  方程$ (1.10)_4 $能转化为

(3.6) $ \begin{eqnarray} &&(\theta-\tilde \theta)_t + u(\theta-\tilde \theta)_x+u_x-\kappa e^{-\varepsilon p}(e^\theta(\theta-\tilde \theta)_x) = \kappa e^{-\varepsilon p}(e^\theta \tilde \theta_x)_x\\ &&+\varepsilon^2e^{-\varepsilon p}\left(\lambda|u_x|^2+ A \big(| w_x|^2+|w|^2\big)\right)-u\tilde \theta_x-\tilde \theta_t. \end{eqnarray} $

将式(3.6)两边乘以$ \theta-\tilde \theta $, 然后关于$ x $在$ {\mathbb R} $上积分(以下关于$ x $的积分都在$ {\mathbb R} $上), 我们有

(3.7) $ \begin{eqnarray} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}\int |\theta-\tilde \theta|^2{\rm d}x+\kappa\int a(\varepsilon p)b(\theta)|(\theta-\tilde \theta)_x|^2{\rm d}x +\int\kappa a(\varepsilon p)b(\theta)\varepsilon p_x(\theta-\tilde \theta)(\theta-\tilde \theta)_x{\rm d}x\\ &\leq& C\|u_x\|_{L^\infty}\|\theta-\tilde \theta\|^2+C\|u_x\|^2\|\theta-\tilde \theta\| +C\varepsilon^2\|\theta-\tilde \theta\|_{L^\infty}\big(\|u_x\|^2+\| w_x\|^2+\|w\|^2\big)\\ &&+\|\tilde \theta\|_{L^\infty}\big(\|u_x\|^2+\|p_x\|^2\big)\|\theta-\tilde \theta\|+C\|\theta-\tilde \theta\|^2+C\|\tilde \theta_t\|\\ &\leq& C\Lambda({\mathcal Q}(t)). \end{eqnarray} $

将式(3.7)在$ [0, t] $上积分, 我们得到

(3.8) $ \begin{equation} \|\theta(t)-\tilde \theta\|^2+\int_0^t \|(\theta-\tilde \theta)_x(\tau)\|^2{\rm d}\tau\leq C+\|(\theta-\tilde \theta)(0)\|^2+\int_0^t\Lambda({\mathcal Q}(\tau)){\rm d}\tau. \end{equation} $

令$ \theta_\alpha = \partial^\alpha\theta $, 其中$ 1\leq |\alpha|\leq s $. 将式$ (1.10)_5 $求$ \alpha $阶导, 我们得到

(3.9) $ \begin{eqnarray} &&\partial_t\theta_\alpha+u\partial_x\theta_\alpha+\partial^\alpha u_x-\kappa a(\varepsilon p)(b(\theta)\partial_x\theta_\alpha)_x {}\\ & = &\varepsilon^2\partial^\alpha\big(a(\varepsilon p)(\lambda|u_x|^2+ A | w_x|^2+ A |w|^2)\big)\\ &&-(\partial^\alpha, u)\theta_x+\kappa\Big\{\partial^\alpha\big(b(\theta)\theta_x\big)_x-a(\varepsilon p)(b(\theta) \partial_x\theta_\alpha)_x\Big\}. \end{eqnarray} $

将式(3.9)两边乘以$ \theta_\alpha $, 并关于$ x $求积分, 我们有

(3.10) $ \begin{eqnarray} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}\int |\theta_\alpha|^2{\rm d}x+\kappa\int a(\varepsilon p)b(\theta)|\partial_x\theta_\alpha|^2{\rm d}x +\frac{1}{2}\int u\partial_x(|\theta_\alpha|^2){\rm d}x+\int\theta_\alpha\partial_\alpha u_x{\rm d}x\\ &\leq &C\bigg|\int a(\varepsilon p)b(\theta)\varepsilon p_x\theta_\alpha\partial_x\theta_\alpha {\rm d}x\bigg|+C\|\theta_\alpha\|\|\partial^\alpha\big(a(\varepsilon p)(\lambda|\varepsilon u_x|^2+ A |\varepsilon w_x|^2+ A |\varepsilon w|^2)\big)\\ &&+C\|\theta_\alpha\|\|(\partial^\alpha, u)\theta_x\|^2+C\bigg|\int \theta_\alpha\big[\partial^\alpha(a(\varepsilon p)(b(\theta)\partial_x\theta)_x)-a(\varepsilon p)(b(\theta)\partial_x\theta_\alpha)_x\big]\bigg|{\rm d}x. \end{eqnarray} $

利用柯西不等式, 我们得到

(3.11) $ \begin{eqnarray} &&\int u\partial_x(|\theta_\alpha|^2){\rm d}x+\int\theta_\alpha\partial_\alpha u_x{\rm d}x\\ &\leq& C\big(\|u_x\|_{L^\infty}\|\theta_\alpha\|^2+\|u_\alpha\|^2+\|\partial_x\theta_\alpha\|^2\big)\leq C\big(\|u\|_{H^2}\|\theta_\alpha\|^2+\|u_\alpha\|^2+\|\partial_x\theta_\alpha\|^2\big)\\ &\leq& C\Lambda({\mathcal Q}(t)), \end{eqnarray} $

(3.12) $ \begin{eqnarray} &&\bigg|\int a(\varepsilon p)b(\theta)\varepsilon p_x\theta_\alpha\partial_x\theta_\alpha {\rm d}x\bigg|\\ &\leq &C\big(\|\varepsilon p_x\|_{L^\infty}^2\|\theta_\alpha\|^2+\|\partial_x\theta_\alpha\|^2\big)\leq C\big(\|\varepsilon p\|_{H^2}^2\|\theta_\alpha\|^2+\|\partial_x\theta_\alpha\|^2\big)\\ &\leq &C\Lambda({\mathcal Q}(t)). \end{eqnarray} $

注意到

(3.13) $ \begin{equation} (\partial^\alpha, u)\theta_x = \sum\limits_{1\leq |\beta|\leq \alpha}C_{\alpha, \beta}\partial^\beta u\partial^{\alpha-\beta}\theta_x, \end{equation} $

于是有

(3.14) $ \begin{equation} \|(\partial^\alpha, u)\theta_x\|\leq C\big(\|u\|_{W^{1, \infty}}\|\theta_x\|_{{\mathcal H}^{s-1}}+\|u\|_{{\mathcal H}^{s}}\|\theta_x\|_{L^\infty}\big)\leq \Lambda({\mathcal Q}(t)). \end{equation} $

另一方面, 我们易知

(3.15) $ \begin{eqnarray} &&\|\partial^\alpha\big(a(\varepsilon p)(\lambda|\varepsilon u_x|^2+ A |\varepsilon w_x|^2+ A |\varepsilon w|^2)\big)\|\\ &\leq& C\|\partial^\alpha\big(\lambda|\varepsilon u_x|^2+ A |\varepsilon w_x|^2\big)\|+ C\Lambda(\|\varepsilon p\|_{{\mathcal H}^{s}})\|\big(|\varepsilon u_x|^2, |\varepsilon w_x|^2\big)\|_{{\mathcal H}^{s-1}}\\ &\leq& C\|\big(\varepsilon u_x, \varepsilon w_x\big)\|_{{\mathcal H}^{s}}+ C\Lambda(\|\varepsilon p\|_{{\mathcal H}^{s}})\|\big(|\varepsilon u_x|^2, |\varepsilon w_x|^2\big)\|_{{\mathcal H}^{s-1}}\\ &\leq &C\Lambda({\mathcal Q}(t)). \end{eqnarray} $

(3.16) $ \begin{eqnarray} &&\|\partial^\alpha(a(\varepsilon p)(b(\theta)\partial_x\theta)_x)-a(\varepsilon p)(b(\theta)\partial_x\theta_\alpha)_x\|\\ &\leq& C\sum\limits_{1\leq |\beta|\leq |\alpha|}\Big\{\|\partial^\alpha(a(\varepsilon p))\partial^{\alpha-\beta}\big(b(\theta)\theta_{xx}+b(\theta)\theta_x^2\big)\| +\|\partial^\alpha\big(b(\theta)\theta_x\big)\partial^{\alpha-\beta}\theta_x\|\\ &\quad&+ \|\partial^\alpha\big(b(\theta)\big)\partial^{\alpha-\beta}\theta_{xx}\|\Big\}\\ &\leq& C\Lambda(\|\varepsilon p\|_{{\mathcal H}^{s}})\big(\|\theta_{xx}\|_{{\mathcal H}^{s-1}}+\Lambda(\|\theta_{x}\|_{{\mathcal H}^{s}})\big) +C\Lambda(\|\theta_{x}\|_{{\mathcal H}^{s}})\big(1+\|\theta_{xx}\|_{{\mathcal H}^{s-1}}\big)\\ &\leq& C\Lambda({\mathcal Q}(t)). \end{eqnarray} $

将式(3.11)–(3.12) 和(3.14)–(3.16) 带入式(3.10), 并关于$ t $积分, 我们得到

(3.17) $ \begin{equation} \|\partial^\alpha\theta\|^2 +\int_0^t\|\partial^\alpha\theta_x\|^2{\rm d}\tau \leq C+\int_0^t\Lambda({\mathcal Q}(\tau)){\rm d}\tau. \end{equation} $

由式(3.8) 和(3.17), 我们得到了式(3.5).

引理3.2   当$ s\geq 3 $, 成立

(3.18) $ \begin{equation} \|(\varepsilon p, \varepsilon u, \varepsilon w)\|_{{\mathcal H}^{s}}^2+ \int_0^t\|(\varepsilon u_x, \varepsilon w_x)\|^2_{{\mathcal H}^{s}}{\rm d}\tau \leq C{\mathcal Q}^2(0)+C\int_0^t\Lambda({\mathcal Q}(\tau))\big[1+{\mathcal S(\tau)}\big]{\rm d}\tau. \end{equation} $

证  对满足$ 0\leq |\alpha|\leq s $的$ \alpha $, 令

$ \check p = \varepsilon p, \;\; \check u = \varepsilon u, \;\;\check w = \varepsilon w, \;\; \big(\check p_\alpha, \check u_\alpha, \check w_\alpha\big) = \partial^\alpha\big(\check p, \check u, \check w\big). $

我们易知

(3.19) $ \begin{equation} \left\{ \begin{array}{ll} \partial_t \check p+u\partial_x \check p+\big(2u-\kappa a(\check p)b(\theta)\theta_x\big)_x\\ = \kappa a(\check p)b(\theta)\theta_x\partial_x \check p +a(\check p)\big(\lambda|\partial_x \check u|^2+ A (|\partial_x \check w|^2+ |\check w|^2)\big), \\ b(-\theta)( \partial_t \check u+u\partial_x \check u) +p_x = \lambda a(\check p)\partial_{xx} \check u, \\ b(-\theta)(\partial_t \check w+u\partial_x \check w)+ A a(\check p)\check w = A a(\check p)\partial_{xx} \check w. \end{array} \right. \end{equation} $

对式(3.19)关于$ \alpha $求导, 我们得到

(3.20) $ \begin{equation} \left\{ \begin{array}{ll} \partial_t \check p_{\alpha}+u\partial_x \check p_{\alpha} = h_1+h_2+h_3+h_4, \\ b(-\theta)( \partial_t \check u_{\alpha}+u\partial_x \check u_{\alpha}) +\partial^\alpha p_x-\lambda a(\check p)\partial_{xx} \check u_{\alpha} = h_5+h_6+h_7, \\ b(-\theta)(\partial_t \check w_{\alpha}+u\partial_x \check w_{\alpha})+ A a(\check p)\check w_{\alpha}- A a(\check p)\partial_{xx} \check w_{\alpha} = h_8+h_{9}+h_{10}+h_{11}, \end{array} \right. \end{equation} $

其中

(3.21) $ \begin{equation} \left\{ \begin{array}{ll} h_1 = -[\partial^\alpha, u]\partial_x\check p, \;h_2 = -\partial^\alpha\big(2u-\kappa a(\check p)b(\theta)\theta_x\big)_x, \\ h_3 = \partial^\alpha(\kappa a(\check p)b(\theta)\theta_x\partial_x \check p), \\ h_4 = \partial^\alpha\left[a(\check p)\big(\lambda|\partial_x \check u|^2+ A |\partial_x \check w|^2+ A |\check w|^2\big)\right], \\ h_5 = -[\partial^\alpha, b(-\theta)]\partial_t\check u, \;h_6 = -[\partial^\alpha, b(-\theta)u]\partial_x\check u, \\ h_7 = \lambda[\partial^\alpha, a(\check p)]\partial_{xx}\check u, \ h_8 = -[\partial^\alpha, b(-\theta)]\partial_t\check w, \\ h_{9} = -[\partial^\alpha, b(-\theta)u]\partial_x\check w, \ h_{10} = A [\partial^\alpha, a(\check p)]\partial_{xx}\check w, \\ h_{11} = -A [\partial^\alpha, a(\check p)]\check w. \end{array} \right. \end{equation} $

将式$ (3.20)_1 $乘以$ \check p_\alpha $, 并关于$ x $积分, 我们得到

(3.22) $ \begin{equation} \frac{1}{2}\frac{\rm d}{{\rm d}t}\int |\check p_\alpha|^2{\rm d}x = \frac{1}{2}\int u_x|\check p_\alpha|^2{\rm d}x+\int \big(h_1+h_2+h_3+h_4\big)\check p_\alpha {\rm d}x. \end{equation} $

利用式(3.2), 我们推得

(3.23) $ \begin{equation} \left|\int u_x|\check p_\alpha|^2{\rm d}x\right|\leq C\|u\|_{L^\infty}\|\check p_\alpha\|^2\leq C\Lambda({\mathcal Q}(t)). \end{equation} $

(3.24) $ \begin{equation} \|h_1\| = \|[\partial^\alpha, u]\partial_x\check p\|\leq C\big(\|u\|_{W^{1, \infty}}\|\partial_x\check p\|_{{\mathcal H}^{s-1}}+\|u\|_{{\mathcal H}^{s}}\|\partial_x\check p\|_{L^\infty}\big)\leq C\Lambda({\mathcal Q}(t)). \end{equation} $

类似的, 我们有

$ \|(h_3, h_4)\|\leq C\Lambda({\mathcal Q}(t)), $

结合式(3.24), 我们得到

(3.25) $ \begin{equation} \left|\int \big(h_1+h_3+h_4\big)\check p_\alpha {\rm d}x\right|\leq C\Lambda({\mathcal Q}(t)). \end{equation} $

利用Sobolev不等式, 有

$ \|h_2\|\leq C\left(\|u\|_{{\mathcal H}^{s}}+\Lambda({\mathcal Q}(t))\right)\big(\|\varepsilon p\|_{{\mathcal H}^{s+1}}+\|\theta_x\|_{{\mathcal H}^{s+1}}\big)\leq C\left({\mathcal S}(t)+\Lambda({\mathcal Q}(t))\right), $

即

(3.26) $ \begin{equation} \left|\int h_2\check p_\alpha {\rm d}x\right|\leq C\left({\mathcal S}(t)+\Lambda({\mathcal Q}(t))\right). \end{equation} $

由式(3.22)–(3.23) 和(3.25)–(3.26), 我们得到

(3.27) $ \begin{equation} \|\check p_\alpha(t)\|^2\leq \|\check p_\alpha(0)\|^2+C\int_0^t\Lambda({\mathcal Q}(\tau))\big(1+{\mathcal S}(\tau)\big){\rm d}\tau. \end{equation} $

将式$ (3.20)_2 $乘以$ \check u_\alpha $, 然后关于$ x $积分, 得到

(3.28) $ \begin{equation} \frac{1}{2}\frac{\rm d}{{\rm d}t}\int b(-\theta)|\check u_\alpha|^2{\rm d}x-\lambda\int a(\check p)\check u_{\alpha}\partial_{xx} \check u_{\alpha}{\rm d}x\leq\int \big(h_5+h_6+h_7\big)\check u_\alpha {\rm d}x, \end{equation} $

这里我们使用了下面的两个估计

$ \left|\int \partial_t b(-\theta)|\check u_\alpha|^2{\rm d}x\right|+\left|\int b(-\theta)u\partial_x(\|\check u_\alpha|^2){\rm d}x\right|\leq C\Lambda({\mathcal Q}(t)), \nonumber $

$ \left|-\int \partial^\alpha p_x\check u_\alpha {\rm d}x\right| = \left|\int p_\alpha \partial_x \check u_\alpha {\rm d}x\right|\leq\|p_\alpha\|\|\partial_x \check u_\alpha\|\leq C\Lambda({\mathcal Q}(t)), \nonumber $

由式(3.2), 我们得到

$ \begin{eqnarray*} \|h_5\|& = &\|[\partial^\alpha, b(-\theta)]\partial_t\check u\| = \|[\partial^\alpha, b(-\theta)](\varepsilon\partial)_t u\|\\ &\leq& C\big(\|\theta\|_{H^2}\|(\varepsilon\partial)_t u\|_{{\mathcal H}^{s-1}}+\Lambda({\mathcal Q}(t))\|(\varepsilon\partial)_t u\|_{L^\infty}\big)\leq C\Lambda({\mathcal Q}(t)), \\ \|h_6\|&\leq& C\big(\|b(-\theta)u\|_{W^{1, \infty}}\|\partial_{x} \check u_{\alpha}\|_{{\mathcal H}^{s-1}}+\|b(-\theta)u\|_{{\mathcal H}^{s}}\|\partial_{x} \check u_{\alpha}\|_{L^\infty}\big)\leq C\Lambda({\mathcal Q}(t)), \\ \|h_7\|&\leq& C\big(\|a(\check p)\|_{W^{1, \infty}}\|\partial_{xx} \check u_{\alpha}\|_{{\mathcal H}^{s-1}}+\|a(\check p)\|_{{\mathcal H}^{s}}\|\partial_{xx} \check u_{\alpha}\|_{L^\infty}\big)\leq C\Lambda({\mathcal Q}(t)), \end{eqnarray*} $

于是有

(3.29) $ \begin{equation} \left|\int \big(h_5+h_6+h_7\big)\check u_\alpha {\rm d}x\right|\leq C\Lambda({\mathcal Q}(t)). \end{equation} $

另一方面, 我们有

$ \begin{eqnarray*} -\lambda\int a(\check p)\check u_{\alpha}\partial_{xx} \check u_{\alpha}{\rm d}x & = &\lambda\int a(\check p)|\partial_{x} \check u_{\alpha}|^2{\rm d}x+\lambda\int \partial_{x} a(\check p)\check u_{\alpha}\partial_{x} \check u_{\alpha}{\rm d}x\nonumber\\ &\geq&\frac{3}{4}\lambda\int a(\check p)|\partial_{x} \check u_{\alpha}|^2{\rm d}x-C\Lambda({\mathcal Q}(t)), \end{eqnarray*} $

结合式(3.28)–(3.29), 得到

(3.30) $ \begin{equation} \|\varepsilon u(t)\|^2_{{\mathcal H}^{s}}+\int_0^t\|\varepsilon u_x(t)\|^2_{{\mathcal H}^{s}}(\tau) {\rm d}\tau\leq C\|\varepsilon u(0)\|^2_{{\mathcal H}^{s}}+C\int_0^t\Lambda({\mathcal Q}(\tau)){\rm d}\tau. \end{equation} $

将式$ (3.20)_3 $乘以$ \check w_\alpha $并关于$ x $积分, 我们有

(3.31) $ \begin{eqnarray} && \frac{1}{2}\frac{\rm d}{{\rm d}t}\int b(-\theta)|\check w_\alpha|^2{\rm d}x- A \int a(\check p)\check w_{\alpha}\partial_{xx} \check w_{\alpha}{\rm d}x+ A \int a(\check p)\check w_{\alpha}^2{\rm d}x {}\\ &\leq&\int \big(h_8+h_{9}+h_{10}+h_{11}\big)\check w_\alpha {\rm d}x, \end{eqnarray} $

从式(3.2)知

$ \begin{eqnarray*} \|h_8\|& = &\|[\partial^\alpha, b(-\theta)]\partial_t\check w\| = \|[\partial^\alpha, b(-\theta)](\varepsilon\partial)_t w\|\\ &\leq& C\big(\|\theta\|_{H^2}\|(\varepsilon\partial)_t w\|_{{\mathcal H}^{s-1}}+\Lambda({\mathcal Q}(t))\|(\varepsilon\partial)_t w\|_{L^\infty}\big)\leq C\Lambda({\mathcal Q}(t)), \\ \|h_9\|&\leq& C\big(\|b(-\theta)u\|_{W^{1, \infty}}\|\partial_{x} \check w_{\alpha}\|_{{\mathcal H}^{s-1}}+\|b(-\theta)w\|_{{\mathcal H}^{s}}\|\partial_{x} \check w_{\alpha}\|_{L^\infty}\big)\leq C\Lambda({\mathcal Q}(t)), \\ \|h_{10}\|&\leq &C\big(\|a(\check p)\|_{W^{1, \infty}}\|\partial_{xx} \check w_{\alpha}\|_{{\mathcal H}^{s-1}}+\|a(\check p)\|_{{\mathcal H}^{s}}\|\partial_{xx} \check w_{\alpha}\|_{L^\infty}\big)\leq C\Lambda({\mathcal Q}(t)), \\ \|h_{11}\|&\leq &C\big(\|a(\check p)\|_{W^{1, \infty}}\|\check w_{\alpha}\|_{{\mathcal H}^{s-1}}+\|a(\check p)\|_{{\mathcal H}^{s}}\| \check w_{\alpha}\|_{L^\infty}\big)\leq C\Lambda({\mathcal Q}(t)), \end{eqnarray*} $

于是有

(3.32) $ \begin{equation} \left|\int \big(h_8+h_9+h_{10}+h_{11}\big)\check w_\alpha {\rm d}x\right|\leq C\Lambda({\mathcal Q}(t)), \end{equation} $

类似于式(3.30), 我们得到

(3.33) $ \begin{eqnarray} \|\varepsilon w(t)\|^2_{{\mathcal H}^{s}}+\int_0^t\|\varepsilon w_x(t)\|^2_{{\mathcal H}^{s}}(\tau){\rm d}\tau\leq C\|\varepsilon w(0)\|^2_{{\mathcal H}^{s}}+C\int_0^t\Lambda({\mathcal Q}(\tau)){\rm d}\tau. \end{eqnarray} $

整合式(3.27), (3.30) 和(3.33), 我们得到了式(3.18).

接下来, 我们将推导$ \|(p, u, w)\|_{{\mathcal H}^{s}} $的估计. 为此, 我们需要控制$ \partial^{s+1}(\varepsilon p, \varepsilon u, \varepsilon w, \theta) $和$ (\partial_t)^s(p, u, w) $两项. 我们先对$ \|(\varepsilon p, \varepsilon u, \varepsilon w, \theta)\|_{{\mathcal H}^{s+1}} $进行估计.

引理3.3   当$ s\geq 3 $, 成立

(3.34) $ \begin{eqnarray} &&\sum\limits_{|\alpha| = s+1}\|\partial^\alpha(\varepsilon p, \varepsilon u, \varepsilon w, \theta)(t)\|^2 +\sum\limits_{|\alpha| = s+1}\int_0^t\|\partial^\alpha(\varepsilon p_x, \varepsilon u_x, \varepsilon w_x, \theta_x)(t)\|^2{\rm d}\tau \\ &\leq& C{\mathcal Q}^2(0)+C\int_0^t\big[1+{\mathcal S(\tau)}\big]\Lambda({\mathcal Q}(\tau)){\rm d}\tau. \end{eqnarray} $

证   令$ \hat p = \varepsilon p-\theta $, 则$ (\hat p, \check u, \check w, \theta) $满足

(3.35) $ \begin{equation} \left\{ \begin{array}{ll} { } \partial_t \hat p+u\partial_x \hat p+\frac{1}{\varepsilon}\partial_x \check u = 0, \\ { } b(-\theta)( \partial_t \check u+u\partial_x \check u) +\frac{1}{\varepsilon}(\partial_x \hat p+\theta_x) = \lambda a(\varepsilon p)\partial_{xx} \check u, \\ { } b(-\theta)(\partial_t \check w+u\partial_x \check w)+ A a(\varepsilon p)\check w = A a(\varepsilon p)\partial_{xx} \check w, \\ { } \theta_t+u\theta_x+\frac{1}{\varepsilon}\partial_x \check u = a(\varepsilon p)(b(\theta)\theta_x)_x+\varepsilon^2a(\varepsilon p)\big(\lambda|\partial_x \check u|^2+ A |\partial_x \check w|^2+ A | \check w|^2\big). \end{array} \right. \end{equation} $

当$ \alpha $满足$ |\alpha| = s+1 $时, 令

$ \big(\hat p_\alpha, \check u_\alpha, \check w_\alpha, \theta_\alpha\big) = \partial^\alpha\big(\hat p, \check u, \check w, \theta\big). $

将式(3.35)关于$ \alpha $求导, 我们得到

(3.36) $ \begin{equation} \left\{ \begin{array}{ll} { } \partial_t \hat p_\alpha+u\partial_x \hat p_\alpha+\frac{1}{\varepsilon}\partial_x \check u_\alpha = h_{12}, \\ { } b(-\theta)( \partial_t \check u_\alpha+u\partial_x \check u_\alpha) +\frac{1}{\varepsilon}(\partial_x \hat p_\alpha+\partial_x\theta_\alpha) = \lambda a(\varepsilon p)\partial_{xx} \check u_\alpha+h_{13}, \\ { } b(-\theta)(\partial_t \check w_\alpha+u\partial_x \check w_\alpha)+ A a(\varepsilon p)\check w = A a(\varepsilon p)\partial_{xx} \check w+h_{14}, \\ { } \partial_t\theta_\alpha+u\partial_x\theta_\alpha+\frac{1}{\varepsilon}\partial_x \check u_\alpha = a(\varepsilon p)(b(-\theta)\partial_x\theta_\alpha)_x+h_{15}, \end{array} \right. \end{equation} $

其中

$ \begin{eqnarray*} &&h_{12} = -[\partial^\alpha, u]\partial_x\hat p, \\ &&h_{13} = -[\partial^\alpha, b(-\theta)]\partial_t\check u-[\partial^\alpha, b(-\theta)u]\partial_x\check u+[\partial^\alpha, \lambda a(\varepsilon p)]\partial_{xx}\check u, \\ &&h_{14} = -[\partial^\alpha, b(-\theta)]\partial_t\check w-[\partial^\alpha, b(-\theta)u]\partial_x\check w+[\partial^\alpha, A a(\varepsilon p)]\partial_{xx}\check w-[\partial^\alpha, A a(\varepsilon p)]\check w, \\ &&h_{15} = -[\partial^\alpha, u]\partial_x\theta+\partial^\alpha\big(a(\varepsilon p)\big(\lambda|\partial_x \check u|^2+ A |\partial_x \check w|^2+ A | \check {w}|^2\big)\big)\\ &&\quad\quad\quad+\Big[\partial^\alpha\big(a(\varepsilon p)(b(\theta)\theta_x)_x\big)-a(\varepsilon p)(b(\theta)\partial_x\theta_\alpha)_x\Big]. \end{eqnarray*} $

分别将式$ (3.36)_1 $–$ (3.36)_4 $乘以$ \hat p_\alpha, \check u_\alpha, \check w_\alpha, \theta_\alpha $, 相加并关于$ x $积分, 我们得到

(3.37) $ \begin{eqnarray} &&\frac{1}{2}\frac{\rm d}{{\rm d}t}\int\left(|\hat p_{\alpha}|^2+b(-\theta)|\check u_\alpha|^2+b(-\theta)|\check w_\alpha|^2 +|\theta_\alpha|^2\right){\rm d}x\\ &&+\frac{3}{4}\int\left[a(\varepsilon p)\big(\lambda|\partial_x \check u_\alpha|^2+ A |\partial_x \check w_\alpha|^2+ A | \check w_\alpha|^2\big)+a(\varepsilon p)b(\theta)|\partial_x \theta_\alpha|^2\right]{\rm d}x\\ &\leq& C\Lambda({\mathcal Q}(t))+\int\big(h_{12}\hat p_{\alpha}+h_{13}\check u_{\alpha}+h_{14}\check w_{\alpha}+h_{15}\theta_{\alpha}\big) {\rm d}x, \end{eqnarray} $

这里我们用到了以下估计

$ \begin{eqnarray*} &&\left|\int b(-\theta)u \big(\check u_{\alpha}\partial_x\check u_{\alpha}+\check w_{\alpha}\partial_x\check w_{\alpha}\big){\rm d}x\right|+\left|\int \partial_tb(-\theta)\big(|\check u_{\alpha}|^2+|\check w_{\alpha}|^2\big){\rm d}x\right|\\ &&+\left|\int u \hat p_{\alpha}\partial_x\hat p_{\alpha}{\rm d}x\right|\leq C\Lambda({\mathcal Q}(t)), \end{eqnarray*} $

$ \begin{eqnarray*} \int \lambda a(\varepsilon p)\check u_{\alpha}\partial_{xx}\check u_{\alpha}{\rm d}x& = &-\int \lambda a(\varepsilon p)\big|\partial_{x}\check u_{\alpha}\big|^2{\rm d}x-\int \lambda \partial_{x}a(\varepsilon p)\check u_{\alpha}\partial_{x}\check u_{\alpha}{\rm d}x\\ &\leq& -\frac{3}{4}\int \lambda a(\varepsilon p)\big|\partial_{x}\check u_{\alpha}\big|^2{\rm d}x+C\Lambda({\mathcal Q}(t)), \\ \int A a(\varepsilon p)\check w_{\alpha}\partial_{xx}\check w_{\alpha}{\rm d}x& = &-\int A a(\varepsilon p)\big|\partial_{x}\check w_{\alpha}\big|^2{\rm d}x-\int \lambda \partial_{x}a(\varepsilon p)\check w_{\alpha}\partial_{x}\check w_{\alpha}{\rm d}x\\ &\leq& -\frac{3}{4}\int A a(\varepsilon p)\big|\partial_{x}\check w_{\alpha}\big|^2{\rm d}x+C\Lambda({\mathcal Q}(t)), \\ \int a(\varepsilon p)\theta_{\alpha}\partial_{x}(b(\theta)\partial_x\theta_{\alpha}){\rm d}x& = &-\int a(\varepsilon p)b(\theta)\big|\partial_{x}\theta_{\alpha}\big|^2{\rm d}x-\int \partial_{x}a(\varepsilon p)b(\theta)\theta_{\alpha}\partial_{x}\theta_{\alpha}{\rm d}x\\ &\leq &-\frac{3}{4}\int a(\varepsilon p)b(\theta)\big|\partial_{x}\theta_{\alpha}\big|^2{\rm d}x+C\Lambda({\mathcal Q}(t)), \end{eqnarray*} $

$ \int\left(\check u_{\alpha}\partial_x\hat p_{\alpha}+\hat p_{\alpha}\partial_{x}\check u_{\alpha}+\check u_{\alpha}\partial_x\theta_{\alpha}+\theta_{\alpha}\partial_{x}\check u_{\alpha}\right){\rm d}x = 0. $

从式(1.10)可以得到

$ \varepsilon \partial_tu = -u\partial_x(\varepsilon u)-b(\theta)\partial_xp+\lambda a(\varepsilon p)b(\theta)\partial_{xx}(\varepsilon u), $

$ \varepsilon \partial_t w = -u\partial_x(\varepsilon w)-a(\varepsilon p)b(\theta)(\varepsilon w)+ A a(\varepsilon p)b(\theta)\partial_{xx}(\varepsilon w), $

于是

(3.38) $ \begin{eqnarray} \|\varepsilon \partial_t u(t)\|^2_{{\mathcal H}^{s}}&\leq& C\Lambda({\mathcal Q}(t))\left[1+\|\partial_x(\varepsilon u)(t)\|^2_{{\mathcal H}^{s}}+\|\partial_xp(t)\|^2_{{\mathcal H}^{s}}+\|\partial_{xx}(\varepsilon u)(t)\|^2_{{\mathcal H}^{s}}\right]\\ &\leq &C\Lambda({\mathcal Q}(t))\left[1+{\mathcal S}(t)\right], \end{eqnarray} $

(3.39) $ \begin{equation} \|\varepsilon \partial_t w(t)\|^2_{{\mathcal H}^{s}}\leq C\Lambda({\mathcal Q}(t))\left[1+{\mathcal S}(t)\right], \end{equation} $

利用式(3.2)和(3.39), 我们得到

$ \begin{eqnarray*} \|h_{12}\|&\leq& C\big(\|u\|_{W^{1, \infty}}\|\varepsilon\partial_x p-\partial_x\theta\|_{{\mathcal H}^{s}}+\|u\|_{{\mathcal H}^{s+1}}\|\varepsilon\partial_x p-\partial_x\theta\|_{L^\infty}\big)\\ &\leq& C\Lambda({\mathcal Q}(t))\big(\|\varepsilon\partial_tu\|_{{\mathcal H}^{s}}+\|\partial_xu\|_{{\mathcal H}^{s}}+\|\varepsilon\partial_xp\|_{{\mathcal H}^{s}}\big)\\ &\leq& C\Lambda({\mathcal Q}(t))\left[1+{\mathcal S}(t)\right], \end{eqnarray*} $

利用Hölder不等式, 可得

(3.40) $ \begin{equation} \left|\int h_{12}\hat p_\alpha {\rm d}x\right|\leq C\Lambda({\mathcal Q}(t))\left[1+{\mathcal S}(t)\right]. \end{equation} $

另一方面, 我们得到

$ \begin{eqnarray*} &&\|[\partial^\alpha, b(-\theta)u]\partial_x(\varepsilon u)\|+\|[\partial^\alpha, \lambda a(\varepsilon p)]\partial_{xx}(\varepsilon u)\|\\ &\leq& C\Lambda({\mathcal Q}(t))\left[1+\|\varepsilon u\|_{{\mathcal H}^{s+1}}+\|\varepsilon \partial_{xx}u\|_{{\mathcal H}^{s}}\right]\leq C\Lambda({\mathcal Q}(t))\left[1+{\mathcal S}(t)\right], \end{eqnarray*} $

$ \|[\partial^\alpha, b(-\theta)u]\partial_x(\varepsilon u)\|\leq C\Lambda({\mathcal Q}(t))\left[1+\|\varepsilon \partial_{t}u\|_{{\mathcal H}^{s}}\right]\leq C\Lambda({\mathcal Q}(t))\left[1+{\mathcal S}(t)\right], $

结合式(3.40), 有

(3.41) $ \begin{equation} \left|\int h_{13}\check u_\alpha {\rm d}x\right|\leq C\Lambda({\mathcal Q}(t))\left[1+{\mathcal S}(t)\right]. \end{equation} $

类似的, 我们得到

(3.42) $ \begin{equation} \left|\int \big(h_{14}\check w_\alpha+h_{15}\theta_\alpha\big) {\rm d}x\right|\leq C\Lambda({\mathcal Q}(t))\left[1+{\mathcal S}(t)\right]. \end{equation} $

利用式(3.37)和(3.40)–(3.42), 我们得到了式(3.34).

在本小节, 我们估计$ (\partial_t)^s(p, u, w) $. 为此, 对于给定的状态$ (\underline p, \underline u, \underline{ w}, \underline\theta) $, 我们给出式(1.10)的下述线性系统的$ L^2 $估计

(3.43) $ \begin{equation} \left\{ \begin{array}{ll} { } \partial_t p+\underline u\partial_x p+\frac{1}{\varepsilon}\big(2u-\kappa a(\varepsilon\underline p)b(\underline \theta) \theta_x\big)_x\\ { } = \kappa a(\varepsilon\underline p)b(\underline \theta){\underline p}_x\theta_x +\varepsilon a(\varepsilon\underline p)\big(\lambda {\underline u}_x u_x+ A {\underline { w}}_x w_x+ A {\underline w} w\big)+f_1, \\ { } b(-\underline\theta)( \partial_t u+\underline u\partial_x u) +\frac{1}{\varepsilon}p_x = \lambda a(\varepsilon\underline p)\partial_{xx} u+f_2, \\ { } b(-\underline\theta)(\partial_t w+\underline u\partial_x w)+ A a(\varepsilon\underline p) w = A a(\varepsilon\underline p)\partial_{xx} w+f_3, \\ { } \partial_t\theta+\underline u\partial_x \theta+\partial_x u = \kappa a(\varepsilon\underline p)(b(-\underline\theta)\theta_x)_x+\varepsilon^2 a(\varepsilon\underline p)\big(\lambda {\underline u}_x u_x+ A {\underline { w}}_x w_x+ A {\underline w} w\big)+f_4, \end{array} \right. \end{equation} $

其中$ f_i, \;i = 1, 2, \cdots, 5 $是本源项.

引理3.4   令$ (p, u, w, \theta) $是系统(3.43)的解, 并假设

(3.44) $ \begin{equation} 0<\frac{1}{2}\underline{a}\leq a(\varepsilon \underline p)\leq 2\overline{a}, \quad 0<\frac{1}{2}\underline{b}\leq b(\underline \theta)\leq 2\overline{b}, \quad \quad t\in [0, T], \end{equation} $

于是, 对$ 0<t\leq T $, 成立

(3.45) $ \begin{eqnarray} &&\|(p, u, w)(t)\|^2+\int_0^t \|(u_x, w_x)(\tau)\|^2{\rm d}\tau{}\\ &\leq& \|(p, u, w)(0)\|^2+C\sup\limits_{0\leq \tau\leq t}\|\theta_x(\tau)\|^2\\ &&+ C\Lambda({\mathcal R}_0)\bigg\{\int_0^t \|(\varepsilon u_x, \varepsilon w_x, \theta_{xx})(\tau)\|^2{\rm d}\tau+\int_0^t \|(p, u, w, \theta_x)(\tau)\|^2{\rm d}\tau\\ &&+\int_0^t \|f_4(\tau)\|^2{\rm d}\tau+\left(\int_0^t \|(p, u, w, \theta_x)(\tau)\|^2{\rm d}\tau\right)^{\frac{1}{2}}\left(\int_0^t \|(f_1, f_2, f_3)(\tau)\|^2{\rm d}\tau\right)^{\frac{1}{2}}\bigg\}, \end{eqnarray} $

其中$ {\mathcal R}_0 = \sup\limits_{0\leq \tau\leq t}\left\{\|\partial_t\underline\theta(\tau)\|_{L^{\infty}}+ \|(\underline p, \underline u, \underline{ w}, \underline\theta)(\tau)\|_{W^{1, \infty}}\right\} $.

证  设$ \textbf{u}: = 2u-\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x $, 则式(3.43) 变为

(3.46) $ \begin{equation} \left\{ \begin{array}{ll} { } p_t+\underline up_x+\frac{1}{\varepsilon}{{\bf u}}_x = \kappa a(\varepsilon\underline p)b(\underline \theta){\underline p}_x\theta_x +\frac{\lambda}{2}\varepsilon a(\varepsilon\underline p)\underline u{ {\bf u}}_x+\frac{\lambda}{2}\varepsilon a(\varepsilon\underline p){\underline u}_x\big(\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x\big)_x \\ { } \quad+\varepsilon a(\varepsilon\underline p)\big( A {\underline { w}}_x w_x+ A {\underline w}w\big)+f_1, \\ { } b(-\underline\theta)( { {\bf u}}_t+\underline u{ {\bf u}}_x) +\frac{1}{\varepsilon}p_x-\frac{\lambda}{2} a(\varepsilon\underline p){{\bf u}}_{xx} = \frac{1}{2}b(\underline \theta)\big(\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x\big)_{xx}\\ { } \quad-\frac{1}{2}b(-\underline \theta)\big[(\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x)_t+\underline u(\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x)_x\big] +f_2, \\ { } b(-\underline\theta)( w_t+\underline u w_x)+ A a(\varepsilon\underline p)w = A a(\varepsilon\underline p) w_{xx}+f_3, \\ { } \theta_t+\underline u\theta_x+ \frac{1}{2}{\mathcal {\bf u}}_x +\big[\frac{\kappa}{2} a(\varepsilon\underline p)b(\underline \theta)\theta_x\big]_x -\kappa a(\varepsilon\underline p)\big(b(\underline \theta)\theta_x\big)_x = \frac{\lambda}{2}\varepsilon^2 a(\varepsilon\underline p)\underline u{\mathcal {\bf u}}_x\\ { } \quad+\frac{\lambda}{2}\varepsilon^2 a(\varepsilon\underline p){\underline u}_x\big(\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x\big)_x+\varepsilon^2 a(\varepsilon\underline p)\big( A {\underline { w}}_x w_x+ A {\underline w}w\big)+f_4, \end{array} \right. \end{equation} $

将式$ \partial_x(3.46)_4 $乘以$ \frac{\kappa}{2}a(\varepsilon\underline p) $, 再加上式$ (3.46)_2 $, 我们得到

(3.47) $ \begin{eqnarray} &&\quad\frac{1}{2}b(-\underline\theta)( { {\bf u}}_t+\underline u{ {\bf u}}_x) +\frac{1}{\varepsilon}p_x-\frac{\kappa}{4} a(\varepsilon\underline p){ {\bf u}}_{xx}-\frac{\lambda}{2} a(\varepsilon\underline p){ {\bf u}}_{xx}\\ && = \frac{\lambda}{2} a(\varepsilon\underline p)\big(\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x\big)_{xx}+f_2-\frac{1}{2}b(-\underline \theta)\big[(\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x)_t+\underline u(\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x)_x\big]\\ &&\quad-\frac{\kappa}{2} a(\varepsilon\underline p)\big[\kappa a(\varepsilon\underline p)(b(\underline \theta)\theta_x)_x\big]_x-\frac{\kappa}{4} a(\varepsilon\underline p)\big[\lambda \varepsilon^2 a(\varepsilon\underline p){\underline u}_x{ {\bf u}}_x)_x\big]-\frac{\kappa}{2} a(\varepsilon\underline p)\partial_xf_4\\ &&\quad-\frac{\lambda}{4}\varepsilon^2\kappa a(\varepsilon\underline p)\big[{\underline u}_x(\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x)_x\big]_x +\frac{\kappa}{4} a(\varepsilon\underline p)\big[\kappa a(\varepsilon\underline p)b(\underline \theta)\theta_x\big]_{xx}+\frac{1}{2}a(\varepsilon\underline p){\underline u}_x\theta_x\\ && = g+f_2-\frac{\kappa}{2} a(\varepsilon\underline p)\partial_xf_4 \end{eqnarray} $

将式$ (3.46)_1 $–$ (3.46)_3 $分别乘以$ p, \textbf{u}, w $, 相加, 并关于$ x $积分, 我们得到

(3.48) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\left(\int\big(\frac{1}{2}|p|^2+\frac{1}{4}b(-\theta)|\textbf{u}|^2 +\frac{1}{2}b(-\theta)| w|^2\big){\rm d}x\right)\\ &&+\int\bigg[(\frac{\kappa}{4} +\frac{\lambda}{2}) a(\varepsilon\underline p)|\textbf{u}_x|^2+a(\varepsilon\underline p) \big(| w_x|^2\big)\bigg]{\rm d}x\\ &\leq &\frac{1}{8}\int\big((\frac{\kappa}{4}+\frac{\lambda}{2}\big) a(\varepsilon\underline p)|\textbf{u}_x|^2{\rm d}x+ \frac{1}{2}\int a(\varepsilon\underline p) \big(| w_x|^2+|w|^2\big)\bigg]{\rm d}x+\|f_5\|^2+\int g\textbf{u}{\rm d}x\\ &&+\int\left(f_1p+f_2\textbf{u}+f_3 w\right){\rm d}x+C\Lambda({\mathcal R}_0)\|(p, u, w, \theta_x, \theta_{xx})\|^2. \end{eqnarray} $

直接计算得

$ \int g{\bf u}{\rm d}x\leq \frac{1}{8}\int\big(\frac{\kappa}{4}+\frac{\lambda}{2}\big) a(\varepsilon\underline p)|{ \textbf{u}}_x|^2{\rm d}x+C\Lambda({\mathcal R}_0)\|(p, u, w, \theta_x, \theta_{xx})\|^2, $

结合式(3.48), 我们得到了式(3.45).

接下来我们利用引理3.4去估计$ \|(\varepsilon \partial_t)^k(p, u, w)\|, \;1\leq k\leq s $.

引理3.5   令$ s\geq 3 $当$ 0\leq k\leq s $时, 成立

(3.49) $ \begin{eqnarray} &&\|(\varepsilon\partial_t)^k(p, u, w)(t)\|^2+\int_0^t \|(\varepsilon\partial_t)^k(u_x, w_x)(\tau)\|^2{\rm d}\tau {}\\ &\leq& \|(\varepsilon\partial_t)^k(p, u, w)(0)\|^2 +C\sup\limits_{0\leq \tau\leq t}\|\theta_x(\tau)\|^2_{{\mathcal H}^k} +Ct^{\frac{1}{2}}\Lambda({\mathcal N}_0)+ C\Lambda({\mathcal R})\int_0^t \|\theta_{xx}(\tau)\|^2_{{\mathcal H}^k}{\rm d}\tau, {\qquad}{\quad} \end{eqnarray} $

其中$ {\mathcal R} = \sup\limits_{0\leq \tau\leq t}\left\{\|\partial_t\theta(\tau)\|_{L^{\infty}}+\{\|( p, u, { w}, \theta)(\tau)\|_{W^{1, \infty}}\right\} $.

证  设$ (p_k, u_k, w_k, \theta_k) = (\varepsilon\partial_t)^k(p, u, w, \theta) $, 将$ (\varepsilon\partial_t)^k $应用到式(1.10), 得到

(3.50) $ \begin{equation} \left\{ \begin{array}{ll} { } \partial_t p_k+u\partial_x p_k+\frac{1}{\varepsilon}\big(2u_k-\kappa a(\varepsilon p)b( \theta)\partial_x\theta_k\big)_x \\ { } = \kappa a(\varepsilon p)b( \theta){ p}_x\theta_x +\varepsilon a(\varepsilon p)\big(\lambda \partial_x{u} \partial_x u_k+ A \partial_x{{ w}}\partial_x w_k+ A ww_k\big)+f_{k1}, \\ { } b(-\theta)( \partial_t u_k+ u\partial_x u_k) +\frac{1}{\varepsilon}\partial_x p_k = \lambda a(\varepsilon p)\partial_{xx} u_k+f_{k2}, \\ { } b(-\theta)(\partial_t w_k+ u\partial_x w_k)-a(\varepsilon p) A w_k = A a(\varepsilon p)\partial_{xx} w_k+f_{k3}, \\ { } \partial_t\theta_k+u\partial_x \theta_k+\partial_x u_k = \kappa a(\varepsilon p)(b(-\theta)\partial_{x}\theta_k)_x\\ +\varepsilon^2 a(\varepsilon p)\big(\lambda \partial_{x}{u} \partial_{x}u_k+ A \partial_{x}{{ w}} \partial_{x} w_k+ A {w}w_k\big)+f_{k4}, \end{array} \right. \end{equation} $

利用引理3.4, 得到

(3.51) $ \begin{eqnarray} &&\|(\varepsilon\partial_t)^k(p, u, w)(t)\|^2+\int_0^t \|(\varepsilon\partial_t)^k(u_x, w_x)(\tau)\|^2{\rm d}\tau{}\\ &\leq& \|(\varepsilon\partial_t)^k(p, u, w)(0)\|^2+C\sup\limits_{0\leq \tau\leq t}\|\theta_x(\tau)\|^2_{{\mathcal H}^k}\\&& +C\Lambda({\mathcal R})\bigg[\int_0^t \|(\varepsilon\partial_t)^k\theta_{xx}(\tau)\|^2{\rm d}\tau+\int_0^t \|(p_k, u_k, w_k, \partial_x\theta_k)\|^2{\rm d}\tau\\ &&+\int_0^t \|f_{k4}(\tau)\|^2{\rm d}\tau +t^{\frac{1}{2}}\Lambda({\mathcal Q}(\tau))\left(\int_0^t\|(f_{k1}, f_{k2}, f_{k3})(\tau)\|^2{\rm d}\tau\right)^{\frac{1}{2}}\bigg], \end{eqnarray} $

其中

$ \begin{eqnarray*} f_{k1}& = &-[(\varepsilon\partial_t)^k, u]p_x+\kappa\frac{1}{\varepsilon}\left(\big[(\varepsilon\partial_t)^k, a(\varepsilon p)b(\theta)\big]\theta_x\right)_x +\big[(\varepsilon\partial_t)^k, \kappa a(\varepsilon p)b(\theta)p_x\big]\theta_x\\ &&+\varepsilon\big[(\varepsilon\partial_t)^k, \lambda\varepsilon a(\varepsilon p)u_x\big]u_x+\varepsilon\big[(\varepsilon\partial_t)^k, A \varepsilon a(\varepsilon p) w_x\big] w_x+\varepsilon\big[(\varepsilon\partial_t)^k, A \varepsilon a(\varepsilon p)w\big]w, \\ f_{k2}& = &-[(\varepsilon\partial_t)^k, b(-\theta)]u_t-[(\varepsilon\partial_t)^k, b(-\theta)u]u_x+\lambda\big[(\varepsilon\partial_t)^k, a(\varepsilon p)\big]u_{xx}, \\ f_{k3}& = &-[(\varepsilon\partial_t)^k, b(-\theta)] w_t-[(\varepsilon\partial_t)^k, b(-\theta)u] w_x+ A \big[(\varepsilon\partial_t)^k, a(\varepsilon p)\big] w_{xx}-\big[(\varepsilon\partial_t)^k, A a(\varepsilon p)\big]w, \\ f_{k4}& = &-[(\varepsilon\partial_t)^k, u]\theta_x+\kappa[(\varepsilon\partial_t)^k, a{\varepsilon p}](b(\theta)\theta_x)_x+\kappa a(\varepsilon p)\left([(\varepsilon\partial_t)^k, b{\theta}]\theta_x\right)_x\\ &&+\varepsilon\big[(\varepsilon\partial_t)^k, \lambda\varepsilon a(\varepsilon p)u_x\big]u_x+\varepsilon\big[(\varepsilon\partial_t)^k, A \varepsilon a(\varepsilon p) w_x\big] w_x+\varepsilon\big[(\varepsilon\partial_t)^k, A \varepsilon a(\varepsilon p)w\big]w, \end{eqnarray*} $

接下来我们估计$ (f_{k1}, f_{k2}, f_{k3}, f_{k4}) $. 首先, 对$ f_{k1} $, 我们有

(3.52) $ \begin{equation} \left(\big[(\varepsilon\partial_t)^k, a(\varepsilon p)b(\theta)\big]\theta_x\right)_x = \big[(\varepsilon\partial_t)^k, a(\varepsilon p)b(\theta)\big]\theta_{xx}+\big[(\varepsilon\partial_t)^k, \big(a(\varepsilon p)b(\theta)\big)_x\big]\theta_x, \end{equation} $

其中

(3.53) $ \begin{eqnarray} &&\frac{1}{\varepsilon}\|\big[(\varepsilon\partial_t)^k, a(\varepsilon p)b(\theta)\big]\theta_{xx}\| = \|\big[(\varepsilon\partial_t)^{k-1}\partial_t, a(\varepsilon p)b(\theta)\big]\theta_{xx}\|\\ &\leq &\|\big[(\varepsilon\partial_t)^{k-1}, a(\varepsilon p)b(\theta)\big]\theta_{txx}\|+\|\big[(\varepsilon\partial_t)^{k-1}, \big(a(\varepsilon p)b(\theta)\big)_t\big]\theta_{xx}\|\\ &\leq& C\Lambda({\mathcal Q}(t))\left(1+\|\theta_{tx}\|_{{\mathcal H}^{s-1}}+\|\theta_{x}\|_{{\mathcal H}^{s}}\right), \end{eqnarray} $

(3.54) $ \begin{eqnarray} &&\frac{1}{\varepsilon}\|\big[(\varepsilon\partial_t)^k, \big(a(\varepsilon p)b(\theta)\big)_x\big]\theta_x\| = \|\big[(\varepsilon\partial_t)^{k-1}\partial_t, \big(a(\varepsilon p)b(\theta)\big)_x\big]\theta_{x}\|\\ &\leq &\|\big[(\varepsilon\partial_t)^{k-1}, \big(a(\varepsilon p)b(\theta)\big)_{tx}\big]\theta_{x}\|+\|\big[(\varepsilon\partial_t)^{k-1}, \big(a(\varepsilon p)b(\theta)\big)_x\big]\theta_{tx}\|\\ &\leq& C\Lambda({\mathcal Q}(t))\left(1+\|\theta_{tx}\|_{{\mathcal H}^{s-1}}+\|p_{x}\|_{{\mathcal H}^{s}}\right). \end{eqnarray} $

另一方面, 我们易知

(3.55) $ \begin{eqnarray} &&\|\big[(\varepsilon\partial_t)^{k}, u\big]p_x\|+\|\big[(\varepsilon\partial_t)^k, \kappa a(\varepsilon p)b(\theta)p_x\big]\theta_x\|+\|\varepsilon\big[(\varepsilon\partial_t)^k, \lambda\varepsilon a(\varepsilon p)u_x\big]u_x\|\\ && +\|\varepsilon\big[(\varepsilon\partial_t)^k, A \varepsilon a(\varepsilon p) w_x\big] w_x\|+\|\varepsilon\big[(\varepsilon\partial_t)^k, A \varepsilon a(\varepsilon p)w\big]w\|\\ &\leq& C\Lambda({\mathcal Q}(t))\left(1+\|p_{x}\|_{{\mathcal H}^{s}}\right), \end{eqnarray} $

结合式(3.52)–(3.54), 得到

(3.56) $ \begin{equation} \|f_{k1}\|\leq C\Lambda({\mathcal Q}(t))\big(1+\|\theta_{tx}\|_{{\mathcal H}^{s-1}}+\|p_{x}\|_{{\mathcal H}^{s}}\big). \end{equation} $

对$ \|f_{k2}\| $, 我们注意到

$ \big[(\varepsilon\partial_t)^{k}, b(-\theta)\big]u_t = \sum\limits_{i = 1}^{k}C_{k, i}(\varepsilon\partial_t)^{i}b(-\theta)\cdot(\varepsilon\partial_t)^{k-i}u_t = \sum\limits_{i = 1}^{k}C_{k, i}(\varepsilon\partial_t)^{i-1}\partial_tb(-\theta)\cdot(\varepsilon\partial_t)^{k-i+1}u, $

于是得到

(3.57) $ \begin{equation} \|\big[(\varepsilon\partial_t)^{k}, b(-\theta)\big]u_t\|\leq C\Lambda({\mathcal Q}(t))\big(1+\|\theta_{t}\|_{{\mathcal H}^{k-1}}+\Lambda\left(\|\theta_{t}\|_{{\mathcal H}^{k-2}}\right)\big). \end{equation} $

另外, 我们易知

$ \|\big[(\varepsilon\partial_t)^{k}, b(-\theta)u\big]u_x\|+\|\lambda\big[(\varepsilon\partial_t)^{k}, a(\varepsilon p)\big]u_{xx}\|\leq C\Lambda({\mathcal Q}(t)), $

结合式(3.57), 得到

(3.58) $ \begin{equation} \|f_{k2}\|\leq C\Lambda({\mathcal Q}(t))\big(1+\|\theta_{t}\|_{{\mathcal H}^{k-1}}+\Lambda\left(\|\theta_{t}\|_{{\mathcal H}^{k-2}}\right)\big). \end{equation} $

类似的, 我们得到

(3.59) $ \begin{eqnarray} &&\|f_{k3}\|\leq C\Lambda({\mathcal Q}(t))\big(1+\|\theta_{t}\|_{{\mathcal H}^{k-1}}+\Lambda\left(\|\theta_{t}\|_{{\mathcal H}^{k-2}}\right)\big), \end{eqnarray} $

(3.60) $ \begin{eqnarray} &&\|f_{k4}\|\leq C\Lambda({\mathcal Q}(t)). \end{eqnarray} $

从式$ (1.10)_4 $和(3.2), 我们有

(3.61) $ \begin{equation} \|\theta_{t}\|_{{\mathcal H}^{s-1}}\leq C\Lambda({\mathcal Q}(t)), \quad \|\theta_{tx}\|_{{\mathcal H}^{s-1}}\leq C\Lambda({\mathcal Q}(t))\big(1+\|(u_x, w_x, \theta_{xx})\|_{{\mathcal H}^{s}}\big), \end{equation} $

于是, 利用式(3.56)和(3.58)–(3.61), 我们得到

(3.62) $ \begin{equation} \|(f_{k1}, f_{k2}, f_{k3})\|\leq C\Lambda({\mathcal Q}(t))\big(1+\|(p_x, u_x, w_x, \theta_{xx})\|_{{\mathcal H}^{s}}\big). \end{equation} $

将式(3.62)带入式(3.51), 我们得到了式(3.49).

利用引理3.1, 引理3.3和引理3.5, 我们易知

推论3.1  对$ s\geq 3, \; 0\leq k\leq s-1 $, 成立

(3.63) $ \begin{eqnarray} &&\quad\|(\varepsilon\partial_t)^k(p, u, w)(t)\|^2+\int_0^t \|(\varepsilon\partial_t)^k(u_x, w_x)(\tau)\|^2{\rm d}\tau\\ &&\leq C\left(1+{\mathcal Q}^2(0)\right)+ Ct^{\frac{1}{2}}\Lambda({\mathcal N}(t)) \end{eqnarray} $

和

(3.64) $ \begin{eqnarray} &&\quad\|(\varepsilon\partial_t)^s(p, u, w)(t)\|^2+\int_0^t \|(\varepsilon\partial_t)^s(u_x, w_x)(\tau)\|^2{\rm d}\tau\\ &&\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ Ct^{\frac{1}{2}}\Lambda({\mathcal N}(t))+ C\Lambda({\mathcal R}(t)). \end{eqnarray} $

在本小节, 我们将使用推论3.1估计$ \|(p, u, w)\|_{{\mathcal H}^{s}} $.

引理3.6   对$ s\geq 3 $, 成立

(3.65) $ \begin{equation} \|(p, u, w)\|_{{\mathcal H}^{s}}\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)). \end{equation} $

证  从式$ (1.10)_1 $, 引理3.1–3.3和推论3.1可知

$ \begin{eqnarray*} \|u_x\|&\leq&C\Big\{\|\varepsilon p_t\|+\varepsilon\|u\|_{L^\infty}\|p_x\|+\|a(\varepsilon p)b(\theta)\|_{L^\infty}\|\theta_{xx}\|+\|a(\varepsilon p)b(\theta)\|_{L^\infty}\|\theta_{x}\|^2\\ &\quad&+\|a(\varepsilon p)b(\theta)\|_{L^\infty}\|p_{x}\|\|\theta_{x}\|+\|(\lambda u_x, A w_{x}, A w)\|^2\Big\}\\ &\leq &C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)). \end{eqnarray*} $

对$ 0\leq k\leq s-2 $, 我们有

(3.66) $ \begin{eqnarray} \|(\varepsilon\partial_t)^ku_x\|&\leq&C\Big\{\|(\varepsilon\partial_t)^{k+1} p\|+\Lambda\left(\|(\varepsilon p_x, \varepsilon u_x, \varepsilon w_{x})\|_{{\mathcal H}^k}+\|\theta_{xx}\|_{{\mathcal H}^k}\right)\Big\}\\ &\leq &C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)) \end{eqnarray} $

和

(3.67) $ \begin{eqnarray} \|(\varepsilon\partial_t)^{s-1}u_x\|&\leq&C\Big\{\|(\varepsilon\partial_t)^{s} p\|+\Lambda\left(\|(\varepsilon p, \varepsilon u, \varepsilon w)\|_{{\mathcal H}^{s+1}}+\|(\varepsilon \theta_t, \theta_{xx})\|_{{\mathcal H}^{s}}\right)\Big\}\\ &\leq &C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t))+C\Lambda({\mathcal R}(t)), \end{eqnarray} $

类似的, 对$ 0\leq k\leq s-2 $, 我们有

(3.68) $ \begin{eqnarray} \|(\varepsilon\partial_t)^kp_x\|&\leq&C\Big\{\|(\varepsilon\partial_t)^{k+1} u\|+\Lambda\left(\|(\varepsilon p, \varepsilon u, , \varepsilon \textbf{b})\|_{{\mathcal H}^{k+1}}+\|\varepsilon\theta_t, u_{xx}\|_{{\mathcal H}^k}\right)\Big\}\\ &\leq &C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)) \end{eqnarray} $

和

(3.69) $ \begin{eqnarray} \|(\varepsilon\partial_t)^{s-1}p_x\|&\leq&C\Big\{\|(\varepsilon\partial_t)^{s} u\|+\Lambda\left(\|(\varepsilon p, \varepsilon u, , \varepsilon \textbf{b})\|_{{\mathcal H}^{s}}+\|\varepsilon\theta_t, u_{xx}\|_{{\mathcal H}^{s-1}}\right)\Big\}\\ &\leq &C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t))+C\Lambda({\mathcal R}(t)). \end{eqnarray} $

利用Sobolev不等式, 可得

(3.70) $ \begin{eqnarray} \|(\varepsilon\partial_t)^kw_x\|&\leq&C\Big\{\|(\varepsilon\partial_t)^{k+1} w\|+\Lambda\left(\|(\varepsilon w, \varepsilon w_{xx})\|_{{\mathcal H}^{k}}\right)\Big\}\\ &\leq &C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)) \end{eqnarray} $

和

(3.71) $ \begin{eqnarray} \|(\varepsilon\partial_t)^{s-1}w_x\|&\leq& C\Big\{\|(\varepsilon\partial_t)^{s} w\|+\Lambda\left(\|(\varepsilon w, \varepsilon w_{xx})\|_{{\mathcal H}^{s}}\right)\Big\}\\ &\leq &C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t))+C\Lambda({\mathcal R}(t)). \end{eqnarray} $

对$ 0\leq k\leq s-3 $, 从式(1.10), (3.66), (3.68)和(3.70)我们得到

(3.72) $ \begin{equation} \|(\varepsilon\partial_t)^k(p_{xx}, u_{xx}, w_{xx})\|\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)) \end{equation} $

和

(3.73) $ \begin{equation} \|(p, u, w)\|_{{\mathcal H}^{s-1}}\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)), \end{equation} $

结合式$ (1.10)_4 $和(3.5), 就得到

(3.74) $ \begin{equation} {\mathcal R}(t)\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)). \end{equation} $

利用式(1.10), (3.64), (3.67), (3.69), (3.71)和(3.73)–(3.74), 我们就得到了式(3.65).

引理3.7   对$ s\geq 3 $成立

(3.75) $ \begin{equation} \int_0^t\|(p_x, u_x, w_x)(\tau)\|^2_{{\mathcal H}^{s}}{\rm d}\tau\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)). \end{equation} $

证  首先, 从式(3.64)和(3.74)我们得到

(3.76) $ \begin{equation} \int_0^t\|(\varepsilon\partial_t)^su_x(\tau)\|^2{\rm d}\tau\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)), \end{equation} $

结合式$ (1.10)_1 $, 就有

(3.77) $ \begin{eqnarray} \int_0^t\|(\varepsilon \partial_t)^{s+1}p(\tau)\|^2{\rm d}\tau&\leq& C\int_0^t\|(\partial_t)^su_x(\tau)\|^2{\rm d}\tau \big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t))\\ &\leq& C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)). \end{eqnarray} $

另一方面, 从式$ (1.10)_2 $可知

(3.78) $ \begin{equation} (\varepsilon \partial_t)^{s}p_x = -\varepsilon b(-\theta)(\varepsilon \partial_t)^{s}u_t+(\varepsilon \partial_t)^{s}\Big[-\varepsilon b(-\theta)uu_x+\lambda\varepsilon a(\varepsilon p)u_{xx}\Big]. \end{equation} $

将式(3.78) 乘以$ (\varepsilon \partial_t)^{s}p_x $, 并将结果关于$ x $积分, 得

(3.79) $ \begin{eqnarray} \|(\varepsilon \partial_t)^{s}p_x\|^2& = &\int b(-\theta)(\varepsilon \partial_t)^{s}p_x(\varepsilon \partial_t)^{s}\big(-\varepsilon b(-\theta)uu_x+\lambda\varepsilon a(\varepsilon p)u_{xx}\big)\\ &&-\varepsilon \int b(-\theta)(\varepsilon \partial_t)^{s}u_t(\varepsilon \partial_t)^{s}p_x{\rm d}x \\ &\leq& -\frac{\rm d}{{\rm d}t}\int b(-\theta)(\varepsilon \partial_t)^{s}u(\varepsilon \partial_t)^{s}p_x{\rm d}x+\int b(-\theta)(\varepsilon \partial_t)^{s}u(\varepsilon \partial_t)^{s+1}p_x{\rm d}x\\ &&+\int \partial_tb(-\theta)(\varepsilon \partial_t)^{s}u(\varepsilon \partial_t)^{s}(\varepsilon p)_x{\rm d}x+\frac{1}{8}\|(\varepsilon \partial_t)^{s}p_x\|^2\\ &&+\int \Big|(\varepsilon \partial_t)^{s}\left[-\varepsilon b(-\theta)uu_x+\lambda\varepsilon a(\varepsilon p)u_{xx}\right]\Big|^2{\rm d}x. \end{eqnarray} $

将式(3.79)关于$ t $积分, 利用式(3.76)–(3.77), 引理3.1–3.3和引理3.6, 我们得到

(3.80) $ \begin{equation} \int_0^t\|(\varepsilon\partial_t)^sp_x(\tau)\|^2{\rm d}\tau\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)), \end{equation} $

结合式(3.79), 有

(3.81) $ \begin{eqnarray} \int_0^t\|(\varepsilon\partial_t)^{s-1}p_{xx}(\tau)\|^2{\rm d}\tau&\leq& C\int_0^t\|(\varepsilon\partial_t)^sp_x(\tau)\|^2{\rm d}\tau+C\big(1+\Lambda({\mathcal Q}(0))\big)\\ &\quad&+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t))\\ &\leq& C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)). \end{eqnarray} $

由(3.79)和(3.81)式, 可得

(3.82) $ \begin{equation} \int_0^t\|p_x(\tau)\|^2_{{\mathcal H}^s}{\rm d}\tau\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)). \end{equation} $

类似可得

(3.83) $ \begin{equation} \int_0^t\|(u_x, w_x)(\tau)\|^2_{{\mathcal H}^s}{\rm d}\tau\leq C\big(1+\Lambda({\mathcal Q}(0))\big)+ C\big(t^{\frac{1}{2}}+\varepsilon\big)\Lambda({\mathcal N}(t)). \end{equation} $

将式(3.82)和(3.83)相加就得到了式(3.65).

命题3.1的证明   利用引理3.1–3.3, 引理3.6和引理3.7, 我们立即就得到了命题3.1.

定理2.1的证明   利用命题3.1, 局部存在性定理和拔靴带方法, 我们就得到了定理2.1.

在本节中, 我们将通过修改Métivier和Schochet^[28]提出的论点来证明定理2.2, 也可参见文献[1-2, 13, 22].

从定理2.1, 我们得到

(4.1) $ \begin{equation} \sup\limits_{0\leq \tau\leq T_0}\|(p^\varepsilon, u^\varepsilon, w^\varepsilon)\|_{{\mathcal H}^s}+\sup\limits_{0\leq \tau\leq T_0}\|(\theta^\varepsilon-\tilde\theta)\|_{{\mathcal H}^{s+1}}< \infty. \end{equation} $

因此, 在提取子序列后, 当$ \varepsilon\rightarrow 0 $, 我们得到

(4.2) $ \begin{equation} \left\{ \begin{array}{ll} (p^\varepsilon, u^\varepsilon, w^\varepsilon)\rightharpoonup (\overline p, \overline u, \overline w), \; \; &L^{\infty}\left(0, T_0; {\mathcal H}^{s}(\mathbb R)\right), \\ \theta^\varepsilon-\tilde\theta\rightharpoonup\overline\theta-\tilde\theta, \; \; &L^{\infty}\left(0, T_0; {\mathcal H}^{s+1}(\mathbb R)\right), \end{array} \right. \end{equation} $

其中$ \rightharpoonup $为相应空间上的弱$ \ast $收敛.

从$ w^\varepsilon, \theta^\varepsilon $的方程和式(4.1), 我们得到

(4.3) $ \begin{equation} w^\varepsilon_t, \theta^\varepsilon_t\in C\left(0, T_0; {\mathcal H}^{s-2}(\mathbb R)\right), \end{equation} $

结合Aubin-Lions引理易知, 当$ s'<s $时

(4.4) $ \begin{equation} \left\{ \begin{array}{ll} w^\varepsilon\rightarrow \overline w, \; &L^{\infty}\big(0, T_0; {\mathcal H}^{s'}(\mathbb R)\big), \\ \theta^\varepsilon \rightarrow \overline\theta-\tilde\theta, \; \;&L^{\infty}\big(0, T_0; {\mathcal H}^{s'+1}({\mathbb R})\big), \end{array} \right. \end{equation} $

其中$ \rightarrow $为相应空间上的强收敛, 后文出现的$ \rightarrow $代表相同含义.

为了获得极限系统(1.11), 我们需要$ (p^\varepsilon, u^\varepsilon) $在$ L^{2}\big(0, T_0; {\mathcal H}^{s'}({\mathbb R})\big), s'<s $的强收敛性. 为此, 我们将得到$ p^\varepsilon $和$ \big(2u^{\varepsilon}-\kappa e^{-\varepsilon p^{\varepsilon}+\theta^{\varepsilon}}\theta^{\varepsilon}_x\big)_x $当$ \varepsilon\rightarrow 0 $时强收敛于0. 事实上, 式$ (1.10)_1 $和$ (1.10)_2 $能转化为

(4.5) $ \begin{equation} \left\{ \begin{array}{ll} { }\varepsilon p^{\varepsilon}_t\big(2u^{\varepsilon}-\kappa e^{-\varepsilon p^{\varepsilon}+\theta^{\varepsilon}}\theta^{\varepsilon}_x\big)_x = \varepsilon f^\varepsilon, \\ { } \varepsilon e^{-\theta^{\varepsilon}}u^{\varepsilon}_t +{p^{\varepsilon}_x} = \varepsilon g^\varepsilon. \end{array} \right. \end{equation} $

从式(2.3)可知, $ f^\varepsilon, g^\varepsilon $在$ C\big(0, T_0; {\mathcal H}^{s-1}({\mathbb R})\big) $中有界, $ p^{\varepsilon} $在$ L^\infty\big(0, T_0; L^\infty({\mathbb R})\big) $中一致有界. 在式(4.5)中取弱极限, 我们有$ {\overline p}_x = 0 $, 并且$ (2\overline u-\kappa e^{\overline\theta}{\overline\theta}_x\big)_x = 0 $. 因为$ {\overline p}\in L^\infty\left(0, T_0; {\mathcal H}^{s}({\mathbb R})\right) $, 我们能得到$ {\overline p} = 0 $. 另外, 类似于文献[2, 13], 我们可以得到$ \overline \theta $满足以下估计

(4.6) $ \begin{equation} |\overline \theta-\theta_+|\leq Cx^{-1-\sigma}, \quad x\in [1, +\infty). \end{equation} $

为了获得$ u^\varepsilon $和$ p^\varepsilon $的强收敛性, 我们需要以下命题

命题4.1   令式(2.3) 和(4.6) 成立, 则对$ s'<s $, 有

(4.7) $ \begin{equation} \left\{ \begin{array}{ll} p^\varepsilon\rightarrow 0, \; \;&L^{2}\big(0, T_0; {\mathcal H}^{s'}({\mathbb R})\big), \\ (2u^{\varepsilon}-\kappa e^{-\varepsilon p^{\varepsilon}+\theta^{\varepsilon}}\theta^{\varepsilon}_x\big)_x\rightarrow 0, \; \; &L^{2}\big(0, T_0; {\mathcal H}^{s'-1}({\mathbb R})\big). \end{array} \right. \end{equation} $

命题4.1的证明基于Métivier和Schochet^[28]在波动方程中的色散估计, 并在文献[2] 被重新表述.

引理4.1   令$ T>0 $, $ v^\varepsilon $是$ C([0, T], H^2({\mathbb R}^d)) $上的有界序列, $ \varepsilon\partial_tv^\varepsilon $在$ L^2(0, T; L^2({\mathbb R}^d)) $上有界, 且满足

(4.8) $ \begin{equation} \varepsilon^2\partial_t\big(a^\varepsilon \partial_t v^\varepsilon\big)-\nabla\cdot(b^\varepsilon\nabla v^\varepsilon) = c^\varepsilon, \end{equation} $

其中$ c^\varepsilon $在$ L^2(0, T; L^2({\mathbb R}^d)) $上收敛于0. 假设对$ k>1 +\frac{d}{2} $, $ (a^\varepsilon, b^\varepsilon) $为正, 在$ C\left(0, T; H^{k}_{\rm loc}({\mathbb R}^d)\right) $上有界, 在$ C\left(0, T; H^{k}_{\rm loc}({\mathbb R}^d)\right) $上收敛于$ (a, b) $且满足, 对任意的$ \tau \in {\mathbb R} $, 有

(4.9) $ \begin{equation} a\tau^2 + \nabla \cdot(b\nabla)\rightarrow {0}, \quad {\rm in}\; L^2({\mathbb R}^d), \end{equation} $

则序列$ v^\varepsilon $在$ L^2(0, T; L^2_{\rm loc}({\mathbb R}^d)) $上强收敛于0.

命题4.1的证明   将$ \varepsilon\partial_t $应用到式$ (4.5)_1 $, 将$ \partial_x $应用到$ e^{\theta^\varepsilon}\times (4.5)_2 $, 我们得到

(4.10) $ \begin{equation} \frac{1}{2}\varepsilon^2p_{tt}^\varepsilon-\big(e^{\theta^\varepsilon}p_{x}^\varepsilon\big)_x = \varepsilon F^\varepsilon(p^\varepsilon, u^\varepsilon, w^\varepsilon, \theta^\varepsilon), \end{equation} $

其中$ F^\varepsilon(p^\varepsilon, u^\varepsilon, w^\varepsilon, \theta^\varepsilon) $是一个光滑函数, 并且$ F(\textbf{0}) = 0 $. 从式(2.3)可知, 当$ \varepsilon\rightarrow 0 $时, $ \varepsilon F^\varepsilon $在$ L^2(0, T; L^2({\mathbb R})) $强收敛于0.

利用$ \theta^\varepsilon $的强收敛性, 初始条件(2.5)和文献[2]中第8.1节的结论, 我们可以证明式(4.10)中的系数满足引理4.1的条件. 因此, 应用引理4.1, 我们得到

(4.11) $ \begin{equation} p^\varepsilon\rightarrow 0\;\; L^2(0, T; L^2_{\rm loc}({\mathbb R})). \end{equation} $

因为$ \theta^\varepsilon $在$ C([0, T_0], H^s({\mathbb R})) $上一致有界, 利用插值定理可得

(4.12) $ \begin{equation} p^\varepsilon\rightarrow 0\;\; L^2(0, T; H^{s'}_{\rm loc}({\mathbb R})), \quad s'<s. \end{equation} $

类似的, 我们得到$ (2u^{\varepsilon}-\kappa e^{-\varepsilon p^{\varepsilon}+\theta^{\varepsilon}}\theta^{\varepsilon}_x\big)_x $的收敛性.

定理2.2的证明   如果命题4.1成立, 在方程(1.10)中对$ (p^\varepsilon, u^\varepsilon, w^\varepsilon, \theta^\varepsilon) $求极限, 我们证得, 极限$ (0, \overline u, \overline w, \overline \theta) $是方程(1.11)在分布意义上的解. 另一方面, 根据文献[28]中的参数, 可以得到$ (\overline u, \overline w, \overline \theta) $满足初始条件

(4.13) $ \begin{equation} (\overline u, \overline w, \overline \theta)|_{t = 0} = (u_{in}, w_{in}, \theta_{in}), \end{equation} $

其中$ u_{in} $由$ u_{in} = \frac{1}{2}\kappa e^{\theta_{in}}(\theta_{in})_x $确定. 此外, 可以通过能量法得到具有初始数据(4.13)的极限系统(1.11)的解的唯一性, 以上结论对整个序列$ (\overline u, \overline w, \overline \theta) $成立.