## 一维非等熵可压缩微极流体的低马赫数极限

1 上海对外经贸大学统计与信息学院 上海 201620

2 东华大学信息科学与技术学院 上海 201620

## Low Mach Number Limit to One-Dimensional Non-Isentropic Compressible Viscous Micropolar Fluid Model

Liu Xin,1, Dong Xiaolei,2

1 School of Statistics and Information, Shanghai University of International Business and Economics, Shanghai 201620

2 College of Information Sciences and Technology, Donghua University, Shanghai 201620

 基金资助: 国家自然科学基金.  11801357

 Fund supported: the NSFC.  11801357

Abstract

In this paper, we consider the one dimensional non-isentropic compressible micropolar fluid model with general initial data, and justify rigorously the low Mach number limit of this system. The limit relies on the uniform estimates including weighted time derivatives and an extended convergence lemma. Moreover, the difference between the states at ±∞ can be arbitrary large in this case.

Keywords： Micropolar fluid model ; Non-isentropic ; Low Mach number limit ; Uniform estimates

Liu Xin, Dong Xiaolei. Low Mach Number Limit to One-Dimensional Non-Isentropic Compressible Viscous Micropolar Fluid Model. Acta Mathematica Scientia[J], 2021, 41(5): 1445-1464 doi:

## 1 引言

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

$$$P = R\rho T, \quad\quad e = C_v T,$$$

$$$q = -\kappa T_x,$$$

$$$u(x, t) = \varepsilon u^{\varepsilon}(x, \varepsilon t), \quad w(x, t) = \varepsilon w^{\varepsilon}(x, \varepsilon t),$$$

$$$P^{\varepsilon} = \overline{P} +O(\varepsilon),$$$

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

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

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

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

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

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

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

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

### 3 一致估计

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

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

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

### 3.1 $(\varepsilon p, \varepsilon u, \varepsilon w, \theta)$的${\mathcal H}^{s}$估计

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

方程$(1.10)_4$能转化为

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

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

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

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

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

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

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

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

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

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

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

对满足$0\leq |\alpha|\leq s $$\alpha , 令 我们易知 $$\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.$$ 对式(3.19)关于 \alpha 求导, 我们得到 $$\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.$$ 其中 $$\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.$$ 将式 (3.20)_1 乘以 \check p_\alpha , 并关于 x 积分, 我们得到 $$\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.$$ 利用式(3.2), 我们推得 $$\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)).$$ $$\|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)).$$ 类似的, 我们有 结合式(3.24), 我们得到 $$\left|\int \big(h_1+h_3+h_4\big)\check p_\alpha {\rm d}x\right|\leq C\Lambda({\mathcal Q}(t)).$$ 利用Sobolev不等式, 有 $$\left|\int h_2\check p_\alpha {\rm d}x\right|\leq C\left({\mathcal S}(t)+\Lambda({\mathcal Q}(t))\right).$$ 由式(3.22)–(3.23) 和(3.25)–(3.26), 我们得到 $$\|\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.$$ 将式 (3.20)_2 乘以 \check u_\alpha , 然后关于 x 积分, 得到 $$\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,$$ 这里我们使用了下面的两个估计 由式(3.2), 我们得到 于是有 $$\left|\int \big(h_5+h_6+h_7\big)\check u_\alpha {\rm d}x\right|\leq C\Lambda({\mathcal Q}(t)).$$ 另一方面, 我们有 结合式(3.28)–(3.29), 得到 $$\|\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.$$ 将式 (3.20)_3 乘以 \check w_\alpha 并关于 x 积分, 我们有 \begin{eqnarray} && \frac{1}{2}\frac{\rm d}{{\rm d}t}\int b(-\theta)|\check w_\alpha|^2{\rm d}x- A \int a(\check p)\check w_{\alpha}\partial_{xx} \check w_{\alpha}{\rm d}x+ A \int a(\check p)\check w_{\alpha}^2{\rm d}x {}\\ &\leq&\int \big(h_8+h_{9}+h_{10}+h_{11}\big)\check w_\alpha {\rm d}x, \end{eqnarray} 从式(3.2)知 于是有 $$\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)),$$ 类似于式(3.30), 我们得到 \begin{eqnarray} \|\varepsilon w(t)\|^2_{{\mathcal H}^{s}}+\int_0^t\|\varepsilon w_x(t)\|^2_{{\mathcal H}^{s}}(\tau){\rm d}\tau\leq C\|\varepsilon w(0)\|^2_{{\mathcal H}^{s}}+C\int_0^t\Lambda({\mathcal Q}(\tau)){\rm d}\tau. \end{eqnarray} 整合式(3.27), (3.30) 和(3.33), 我们得到了式(3.18). 接下来, 我们将推导 \|(p, u, w)\|_{{\mathcal H}^{s}} 的估计. 为此, 我们需要控制 \partial^{s+1}(\varepsilon p, \varepsilon u, \varepsilon w, \theta)$$ (\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.

### 3.2 $(\varepsilon p, \varepsilon u, \varepsilon w, \theta)$的${\mathcal H}^{s+1}$估计

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

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

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

$\alpha$满足$|\alpha| = s+1$时, 令

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

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

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

$$$\varepsilon^2\partial_t\big(a^\varepsilon \partial_t v^\varepsilon\big)-\nabla\cdot(b^\varepsilon\nabla v^\varepsilon) = c^\varepsilon,$$$

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

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

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

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