考虑如下一维非等熵的可压缩Navier-Stokes方程组

$ \begin{eqnarray} \left\{\begin{array}{ll} v_t-u_x = 0, \\ u_t+p_x = \mu\left(\frac{u_x}{v}\right)_x, \\ \left(e+\frac{u^2}{2}\right)_t+(pu)_x = \kappa \left(\frac{\theta_x}{v}\right)_x+\mu\left(\frac{uu_x}{v}\right)_x, \end{array}\right. \, \, t>0, \, x\in\mathbb{R}, \end{eqnarray} $

(1.1)

其中$ v $, $ u $, $ \theta $, $ p $, $ e $分别表示流体的比容、速度、温度、压强和内能, $ \mu $, $ \kappa $分别是粘性系数和热传导系数.

本文假设粘性系数$ \mu $是热传导系数$ \kappa $的同阶或高阶函数, 且流体压强$ p $和内能$ e $由下式给出

$ \begin{equation} p = \frac{R\theta}{v}, \; \; \; e = \frac{R\theta}{\gamma-1}, \end{equation} $

(1.2)

其中$ R>0 $是气体常数, $ \gamma>1 $是绝热指数.

对于理想流体, 即$ \mu = 0, \kappa = 0 $, 方程组(1.1)退化为如下可压缩Euler方程组

$ \begin{eqnarray} \left\{\begin{array}{ll} v_t-u_x = 0, \\ u_t+p_x = 0, \\ \left(e+\frac{u^2}{2}\right)_t+(pu)_x = 0, \end{array}\right. \, \, t>0, \, x\in\mathbb{R}. \end{eqnarray} $

(1.3)

本文考虑方程组(1.1)当$ \mu\rightarrow 0, \kappa\rightarrow 0 $时的零耗散极限问题, 并且期望方程组(1.1)的光滑解趋向于相应的Euler方程组(1.3)的Riemann问题的解.假设方程组(1.3)具有如下Riemann初值

$ \begin{equation} (v, u, \theta)(0, x) = \left\{\begin{array}{ll} (v_{-}, u_{-}, \theta_{-}), \; \; \; \; \; \; \; x<0, \\ (v_{+}, u_{+}, \theta_{+}), \; \; \; \; \; \; \; x>0, \end{array}\right. \end{equation} $

(1.4)

其中$ v_{\pm}>0, \theta_{\pm}>0, u_{\pm}\in \mathbb{R} $为给定的常数.

众所周知, 当

$ \begin{equation} u_{-} = u_{+}, \quad p_{-} = \frac{R\theta_{-}}{v_{-}} = \frac{R\theta_{+}}{v_{+}} = p_{+} \end{equation} $

(1.5)

时, Riemann问题(1.3)–(1.4)具有如下接触间断解

$ \begin{equation} (v^{cd}, u^{cd}, \theta^{cd})(0, x) = \left\{\begin{array}{ll} (v_{-}, u_{-}, \theta_{-}), \; \; \; \; \; \; \; x<0, \\ (v_{+}, u_{+}, \theta_{+}), \; \; \; \; \; \; \; x>0. \end{array}\right. \end{equation} $

(1.6)

对于可压缩Navier-Stokes方程组(1.1), 构造接触间断波$ (v^{cd}, u^{cd}, \theta^{cd})(t, x) $的一个光滑逼近$ (\bar{v}, \bar{u}, \bar{\theta})(t, x) $, 称之为“粘性接触波”.类似于文献[3], 我们希望$ (\bar{v}, \bar{u}, \bar{\theta})(t, x) $的压强接近于一个特定常数, 即

$ \begin{equation} \bar{p}\equiv\frac{R\bar{\theta}}{\bar{v}}\approx p_{+}. \end{equation} $

(1.7)

这表明, 当不考虑高阶项时, 能量方程$ (1.1)_3 $可写成

$ \begin{equation} \frac{R\theta_t}{\gamma-1}+p_{+}u_x = \kappa\left(\frac{\theta_x}{v}\right)_x. \end{equation} $

(1.8)

将(1.7)式代入(1.8)式中, 可得到一个非线性扩散方程

$ \begin{equation} \theta_t = m\kappa\left(\frac{\theta_x}{\theta}\right)_x, \; \; \; \; \; \; \theta(t, \pm\infty) = \theta_{\pm}, \; \; \; \; \; \; m = \frac{p_{+}(\gamma-1)}{\gamma R^2}>0. \end{equation} $

(1.9)

由文献[4]知, 问题(1.9)存在一个唯一自相似解$ \Theta(t, x) = \Theta(\xi), \xi = \frac{x}{\sqrt{\kappa(1+t)}} $, 并且$ \Theta(\xi) $是一个单调函数.当$ \theta_{+}>\theta_{-} $时, $ \Theta(\xi) $单调递增; 当$ \theta_{+}<\theta_{-} $时, $ \Theta(\xi) $单调递减.此外，存在常数$ \tilde{\delta}>0 $, 使得当$ \delta = |\theta_{+}-\theta_{-}|\leq\tilde{\delta} $时, $ \Theta(t, x) $满足

$ \begin{equation} \left|(\kappa(1+t))^{\frac{l}{2}}\partial_{x}^{l}\Theta\right|+\left|\Theta-\theta_{\pm}\right|\leq c_{1}\delta e^{-\frac{c_{2}x^2}{\kappa(1+t)}}, \; \; \; \; \; \; |x|\rightarrow\infty, \; \; l\geq1, \end{equation} $

(1.10)

其中$ c_1, c_2 $为两个仅依赖与$ \theta_- $和$ \tilde{\delta} $的正常数.

有了$ \Theta(t, x) $的定义后, 可以定义粘性接触波$ (\bar{v}, \bar{u}, \bar{\theta})(t, x) $如下:

$ \begin{equation} \begin{split} &\bar{v} = \frac{R}{p_{+}}\Theta, \; \; \; \; \bar{u} = u_{-}+\frac{\kappa(\gamma-1)}{\gamma R}(\ln\Theta)_x, \\& \bar{\theta} = \Theta-\frac{(\gamma-1)\kappa}{\gamma Rp_{+}}\Theta_t+\frac{(\gamma-1)\mu\kappa}{\gamma R^2}(\ln\Theta)_{xx}, \end{split} \end{equation} $

(1.11)

从而$ (\bar{v}, \bar{u}, \bar{\theta}) $满足

$ \begin{equation} \left\|\left(\bar{v}-v^{cd}, \bar{u}-u^{cd}, \bar{\theta}-\theta^{cd}\right)(t)\right\|_{L^{p}} = O(\kappa^{\frac{1}{2p}}) (1+t)^{\frac{1}{2p}}, \; \; \; \; \; \; p\geq1, \end{equation} $

(1.12)

且

$ \begin{eqnarray} \left\{\begin{array}{ll} \bar{v}_t-\bar{u}_x = 0, \\ \bar{u}_t+\bar{p}_x = \mu\left(\frac{\bar{u}_x}{\bar{v}}\right)_x, \\ \left(\bar{e}+\frac{\bar{u}^2}{2}\right)_t+(\bar{p}\bar{u})_x = \kappa\left(\frac{\bar{\theta}_x}{\bar{v}}\right)_x+\mu\left(\frac{\bar{u}\bar{u}_x}{\bar{v}}\right)_x+R_{1}, \end{array}\right. \end{eqnarray} $

(1.13)

其中$ \bar{e} = \frac{R\bar{\theta}}{\gamma-1} $,

$ \begin{equation} \begin{split} R_{1}& = \frac{m\kappa^2}{p_{+}}\left\{\frac{1}{\Theta}\Theta_{xxt}+\frac{1}{\Theta^2}\Theta_{xt}\Theta_x-\frac{(\gamma-1)p_{+}}{\gamma\Theta}(\ln\Theta)_{xx}\Theta_t\right\} \\&\; \; \; +\frac{\mu}{\kappa}\left\{\frac{m\gamma\kappa^3}{R\Theta}(\ln\Theta)_{xxxx}-\frac{m\kappa^3}{R\Theta^2}(\ln\Theta)_{xxx}\Theta_x+ \frac{m(\gamma-1)\kappa^3}{\gamma R\Theta}\left((\ln\Theta)_{xx}\right)^2\right.\\&\; \; \; \left.+\frac{\kappa^2}{R}(\ln\Theta)_{xxt}-\frac{m^2\kappa^3}{\Theta^2}\left((\ln\Theta)_{xx}\right)^2\right\} \\& = O(\delta)\kappa(1+t)^{-2} e^{-\frac{c_{2}x^2}{\kappa(1+t)}}, \; \; \; \; \; |x|\rightarrow\infty. \end{split} \end{equation} $

(1.14)

本文的主要结果如下.

定理1.1   假设常数状态$ (v_{\pm}, u_{\pm}, \theta_{\pm}) $满足(1.5), $ (v^{cd}, u^{cd}, \theta^{cd})(t, x) $是Euler方程组的黎曼问题(1.3)–(1.4)的一个接触间断解, 且$ (\bar{v}, \bar{u}, \bar{\theta})(t, x) $是由(1.11)给出的对应于$ (v^{cd}, u^{cd}, \theta^{cd})(t, x) $的粘性接触波.进一步假设粘性系数$ \mu $是热传导系数$ \kappa $的同阶或高阶函数, 并且可压缩Navier-Stokes方程组(1.1)的初始值与$ (\bar{v}, \bar{u}, \bar{\theta})(t, x) $的初始值相同.那么存在常数$ \kappa_{0}>0 $, 使得当$ 0<\kappa\leq\kappa_{0} $时, 方程组(1.1)存在唯一的整体光滑解$ (v, u, \theta)(t, x) $, 满足对任意的常数$ T>0 $及$ h>0 $, 有

$ \begin{equation} \sup\limits_{0\leq t \leq T, |x|\geq h}\left|(v, u, \theta)(t, x)-(v^{cd}, u^{cd}, \theta^{cd})(t, x)\right|\leq C\kappa^{\frac{7}{8}}, \end{equation} $

(1.15)

其中$ C $是一个不依赖于$ \kappa $的正常数.

注1.1   在定理1.1中, 接触间断波$ (v^{cd}, u^{cd}, \theta^{cd})(t, x) $的强度$ \delta = |\theta_+-\theta_-| $只要求是有限的, 而不需要任何小性条件.

注1.2   在文献[1, 2]中, 作者研究了可压缩Navier-Stokes方程组(1.1)的解趋向于接触间断波$ (v^{cd}, u^{cd}, \theta^{cd})(t, x) $的零耗散极限, 并且证明了方程组(1.1)的解趋向于接触间断波的收敛率分别为$ \kappa^{\frac{1}{4}} $和$ \kappa^{\frac{3}{4}} $.在定理1.1中, 得到了一个更快的收敛率$ \kappa^{\frac{7}{8}} $ (见(1.15)式), 因此在这种意义上, 改进了文[1, 2]中的主要结果.

下面简要地回顾已有的相关结果, 并且指出本文证明的主要思想.目前关于可压缩流体力学方程趋向于基本波的零耗散极限问题已有许多结果.在不考虑初始层的情形下, Goodman和Xin^[5]研究了粘性守恒律方程组趋向于无粘激波的粘性消失极限.后来, Yu^[6]考虑了具有初始层的粘性守恒律方程组趋向于无粘激波的粘性消失极限问题.当可压缩Euler方程组存在一个激波解时, Hoff和Liu^[7], Wang^[8]以及Wang^[9]证明了一维可压缩Navier-Stokes方程组趋向于激波的粘性消失极限. Xin^[10]研究了一维等熵的可压缩Navier-Stokes方程组趋向于稀疏波的零耗散极限问题, 并且得到了解在初始时间$ t = 0 $以外的关于粘性系数的收敛率.后来, Jiang^[11], 以及Xin和Zeng^[12]将文[10]中的结果分别推广到非等熵的可压缩Navier-Stokes方程组以及Boltzmann方程的情形. Ma^{[1, 2]}研究了一维非等熵的可压缩Navier-Stokes方程组趋向于强接触间断波的零耗散极限, 并且得到了收敛率.关于可压缩流体力学方程趋向于复合波的零耗散极限问题, 读者可参见文献[3, 13-16]以及其中的参考文献.

本文的主要目的在于改进文[1, 2]中的收敛率.在文[1, 2]中, 作者在先验假设

$ \begin{equation} N(\tau)\equiv\|(\phi, \psi, \zeta)(\tau)\|_{H^2(\mathbb{R})}\leq\varepsilon \end{equation} $

(1.16)

下, 利用基本能量方法证明了一维非等熵的可压缩Navier-Stokes方程组的解趋向于接触间断波的收敛率分别为$ \kappa^{\frac{1}{4}} $和$ \kappa^{\frac{3}{4}} $, 这里$ \varepsilon>0 $是一个充分小的正常数.与文[1,2]中的分析相比, 在本文中, 我们利用一个依赖于热传导系数$ \kappa $的先验假设(2.6), 以及一些更加精细的能量估计得到了一个更快的收敛率$ \kappa^{\frac{7}{8}} $.特别地, 我们证明了扰动函数$ (\phi, \psi, \zeta)(\tau, y) $的$ \|(\phi_y, \psi_y, \zeta_y)(\tau)\| $估计为$ C\kappa $ ($ C>0 $为正常数), 这比文[1, 2]中相应的结果更优(见以下引理2.2).

本文的结构安排如下.在第二节中, 首先将Navier-Stokes方程组(1.1)的解在粘性接触波附近作扰动, 从而得到扰动方程组; 然后证明扰动方程组解的先验估计; 最后给出主要定理1.1的证明.

记号  在本文中, $ {\bf{a}} = (a_{i}) $表示$ \mathbb{R}^{n} $中的向量且$ |{\bf{a}}| = \left(\mathop\sum\limits_{i = 1}^{n}a_{i}^{2}\right)^{\frac{1}{2}} $, 而$ A = (A_{ij})_{n\times n} $是矩阵且$ |A| = \left(\mathop\sum\limits_{i = 1}^{n} \mathop\sum\limits_{j = 1}^{n}A_{ij}^{2}\right)^{\frac{1}{2}} $. $ C $表示某个常数, 且它在不同的能量估计中可能会改变. $ L^p(\mathbb{R}) $ ($ 1\leq p\leq+\infty $)表示通常的Lebesgue空间, $ H^{l}(\mathbb{R}) $是通常的$ l $阶-Sobolve空间, 其范数为$ \|f\|_{l} = \left(\mathop\sum\limits_{i = 0}^{l}\|\partial_{x}^{i}f\|^{2}\right)^{\frac{1}{2}} $, 其中$ \|\cdot\|\triangleq \|\cdot\|_{L^{2}(\mathbb{R})} $.

首先, 假设方程组(1.1)的初始值与粘性接触波$ (\bar{v}, \bar{u}, \bar{\theta})(t, x) $的初始值相同, 即$ (v, u, \theta)(0, x) = (\bar{v}, \bar{u}, \bar{\theta})(0, x) $.定义扰动函数$ (\phi, \psi, \zeta)(t, x) $为

$ \begin{equation} \phi = v-\bar{v}, \; \; \; \; \; \; \psi = u-\bar{u}, \; \; \; \; \; \; \zeta = \theta-\bar{\theta}, \end{equation} $

(2.1)

则由(1.1)和(1.13)式可得

$ \begin{eqnarray} \left\{\begin{array}{ll} \phi_t-\psi_x = 0, \\ \psi_t+\left(\frac{R\zeta-\bar{p}\phi}{v}\right)_x = \mu\left(\frac{u_x}{v}-\frac{\bar{u}_x}{\bar{v}}\right)_x, \\ \frac{R\zeta_t}{\gamma-1}+pu_x-\bar{p}\bar{u}_x = \kappa\left(\frac{\theta_x}{v}-\frac{\bar{\theta}_x}{\bar{v}}\right)_x +\mu\left(\frac{u_x^2}{v}-\frac{\bar{u}_x^2}{\bar{v}}\right)-R_{1}, \\ \phi(0, x) = \psi(0, x) = \zeta(0, x) = 0.\\ \end{array}\right. \end{eqnarray} $

(2.2)

作如下坐标变换

$ \begin{equation} y = \frac{x}{\kappa}, \; \; \; \; \; \; \; \tau = \frac{1+t}{\kappa}, \end{equation} $

(2.3)

则方程组(2.2)可改写成如下形式

$ \begin{eqnarray} \left\{\begin{array}{ll} \phi_\tau-\psi_y = 0, \\ \psi_\tau+\left(\frac{R\zeta-\bar{p}\phi}{v}\right)_y = \frac{\mu}{\kappa}\left(\frac{u_y}{v}-\frac{\bar{u}_y}{\bar{v}}\right)_y, \\ \frac{R\zeta_\tau}{\gamma-1}+pu_y-\bar{p}\bar{u}_y = \left(\frac{\theta_y}{v}-\frac{\bar{\theta}_y}{\bar{v}}\right)_y +\frac{\mu}{\kappa}\left(\frac{u_y^2}{v}-\frac{\bar{u}_y^2}{\bar{v}}\right)-R_{2}, \\ \phi(\tau_{0}, y) = \psi(\tau_{0}, y) = \zeta(\tau_{0}, y) = 0, \end{array}\right. \end{eqnarray} $

(2.4)

其中$ \tau_{0} = \frac{1}{\kappa}, \; R_{2} = \kappa R_{1} $, 而且有

$ \begin{equation} \left|\partial_y^{l}\Theta\right|\leq c_{1}\kappa^{\frac{l}{2}}e^{-\frac{c_{2}y^2}{\tau}}, \; \; \; \; \; \; l\geq1, \; \; \; \; \; \; |R_{2}|\leq c_{1}\kappa^2e^{-\frac{c_{2}y^2}{\tau}}. \end{equation} $

(2.5)

令$ \tau_{1} = \frac{1+T}{\kappa} $, 仅需要证明对充分小的$ \kappa $, 方程组(2.4)在$ \mathbb{R}\times[\tau_{0}, \tau_{1}] $上存在唯一光滑解.为此, 定义方程组(2.4)的解空间如下

$ \begin{align*} \begin{split} X(\tau_0, \hat{\tau}) = &\left\{(\phi, \psi, \zeta)(\tau, y)|(\phi, \psi, \zeta)(\tau, y)\in C^{1}(\tau_{0}, \hat{\tau}: H^{2}(\mathbb{R})), \right. \\&\left.\quad(\phi, \psi)_y(\tau, y)\in L^{2}(\tau_{0}, \hat{\tau}: H^{2}(\mathbb{R})), \quad \zeta_y(\tau, y)\in L^{2}(\tau_{0}, \hat{\tau}: H^{1}(\mathbb{R}))\right\}, \end{split} \end{align*} $

其中$ \tau_0<\hat{\tau}\leq\tau_1 $为某个常数.

Cauchy问题(2.4)的局部解的存在唯一性可类似于文[17]得到.为简洁起见, 在此省略其证明.为了得到Cauchy问题(2.4)的整体解, 需要建立其解的一定形式的能量估计.

在这一小节中, 将证明Cauchy问题(2.4)解的如下先验估计.

命题2.1 (先验估计)   在定理1.1的条件下, 假设$ (\phi, \psi, \zeta)\in X(\tau_0, \tau_2) $为Cauchy问题(2.4)的解, 其中$ \tau_0<\tau_2\leq\tau_1 $, 并且满足如下先验假设

$ \begin{equation} N(\tau_2)\equiv\sup\limits_{\tau\in[\tau_0, \tau_2]}\|(\phi, \psi, \zeta)(\tau)\|_{2}\leq\kappa^{\frac{1}{2}}. \end{equation} $

(2.6)

那么存在两个不依赖于$ \kappa $和$ \tau_{2} $的正常数$ \kappa_{0} $和$ C_0>0 $, 使得当$ 0<\kappa\leq\kappa_0 $时, 对任意的$ \tau\in[\tau_0, \tau_2] $成立

$ \begin{eqnarray} &&\sup\limits_{\tau_0\leq\tau\leq\tau_2}\|(\phi, \psi, \zeta)(\tau)\|^2+\int_{\tau_0}^{\tau_2}\|\zeta_y(s)\|_1^2ds\leq C_0\kappa^{\frac{3}{2}}, \end{eqnarray} $

(2.7)

$ \begin{eqnarray} &&\sup\limits_{\tau_0\leq\tau\leq\tau_2}\|(\phi_y, \psi_y, \zeta_y)(\tau)\|_1^2+\int_{\tau_0}^{\tau_2}(\|(\phi_y, \psi_y)(s)\|_1^{2}+\|\zeta_{yy}(s)\|_1^{2})ds\leq C_0\kappa^{2}. \end{eqnarray} $

(2.8)

在证明命题2.1之前, 由先验假设(2.6)及Sobolev不等式

$ \begin{equation} \|f(\tau)\|_{L^\infty}\leq \|f(\tau)\|^{\frac{1}{2}}\|f_y(\tau)\|^{\frac{1}{2}}, \end{equation} $

(2.9)

得

$ \begin{equation} \|(\phi, \psi, \zeta, \phi_y, \psi_y, \zeta_y)(\tau)\|_{L^\infty}\leq \kappa^{\frac{1}{2}}. \end{equation} $

(2.10)

此外, 利用$ \kappa $的小性有

$ \begin{equation} v(\tau, y) = \phi(\tau, y)+\bar{v}(\tau, y)\geq\bar{v}(\tau, y)-\|\phi(\tau)\|_{L^\infty}\geq\frac{1}{2}\min\{v_-, v_+\}>0, \end{equation} $

(2.11)

及

$ \begin{equation} \begin{array}{ll} \theta(\tau, y)& = \zeta(\tau, y)+\bar{\theta}(\tau, y) = \zeta+\Theta-\frac{(\gamma-1)}{\gamma Rp_{+}}\Theta_\tau+\frac{(\gamma-1)\mu}{\gamma R^2\kappa}(\ln\Theta)_{yy}\\ &\geq\Theta(\tau, y)-\|\zeta(\tau)\|_{L^\infty}-C\kappa e^{-\frac{c_2y^2}{\tau}}\\ &\geq \frac{1}{2}\min\{\theta_-, \theta_+\}>0. \end{array} \end{equation} $

(2.12)

命题2.1可由下面的一系列引理得到.对于低阶估计, 有

引理2.1   在命题2.1的假设下, 存在两个不依赖于$ \kappa $和$ \tau_{2} $的正常数$ \kappa_{1} $和$ C_1 $, 使得当$ 0<\kappa\leq\kappa_{1} $时, 有

$ \begin{equation} \sup\limits_{\tau_{0}\leq\tau\leq\tau_{1}}\|(\phi, \psi, \zeta)(\tau)\|^{2}+\int_{\tau_{0}}^{\tau_{1}}\|\zeta_y(s)\|^{2}ds\leq C_1\kappa^{\frac{3}{2}}. \end{equation} $

(2.13)

引理2.1的证明与文献[1]中引理3.1的证明完全类似, 其详细过程在此省略.下面开始作高阶能量估计.

引理2.2   在命题2.1的假设下, 存在两个不依赖于$ \kappa $和$ \tau_{2} $的正常数$ \kappa_{2} $和$ C_2 $, 使得当$ 0<\kappa\leq\kappa_{2} $时, 对任意的$ \tau\in[\tau_0, \tau_2] $有

$ \begin{equation} \|(\phi_y, \psi_y, \zeta_y)(\tau)\|^{2}+\int_{\tau_{0}}^{\tau}(\|(\phi_y, \psi_y)(s)\|^{2}+\|\zeta_{yy}(s)\|^{2})ds\leq C_2\kappa^{2}. \end{equation} $

(2.14)

证  为方便起见, 记$ M = (v, u, \theta)(x, t) $, $ \bar{M} = (\bar{v}, \bar{u}, \bar{\theta})(x, t) $, 则方程组(1.1)可改写为

$ \begin{equation} A^{0}(M)M_{\tau}+A(M)M_y = B(M)M_{yy}+g(M, M_y), \end{equation} $

(2.15)

其中

$ \begin{eqnarray*} &&g(M, M_y) = \left(0, \, \, \frac{\mu}{\kappa}\left(\frac{1}{v}\right)_y\theta u_y, \, \, \left(\frac{1}{v}\right)_y\theta_y+\frac{\mu}{\kappa}\frac{u_y^2}{v}\right)^{T}, \\ &&{A^{0}(M)} = \left(\begin{array}{lrc} -\theta p_v&0&0\\ 0&\theta&0\\ 0&0&\frac{R}{\gamma-1} \end{array}\right), \, \, {A(M)} = \left(\begin{array}{lrc} 0&\theta p_v&0\\ \theta p_v&0&p\\ 0&p&0 \end{array}\right), \, \, {B(M)} = \left(\begin{array}{lrc} 0&0&0\\ 0&\frac{\mu}{\kappa}\frac{\theta}{v}&0\\ 0&0&\frac{1}{v} \end{array}\right). \end{eqnarray*} $

因此(1.13)式可以改写为

$ \begin{equation} A^{0}(\bar{M})\bar{M}_\tau+A(\bar{M})\bar{M}_y = B(\bar{M})\bar{M}_{yy}+g(\bar{M}, \bar{M}_y)+\bar{F}, \end{equation} $

(2.16)

其中$ \bar{F} = (0, 0, R_{2})^{t} $.现在定义矩阵$ \tilde{A}(M) $为

$ \begin{equation} {\tilde{A}(M)} = \left(\begin{array}{lrc} A_{11}(M)&A_{12}(\bar{M})\\ A_{21}(\bar{M})&0\\ \end{array}\right), \end{equation} $

(2.17)

其中$ {A_{11}(M)} = \left(\begin{array}{lrc} 0&\theta p_v\\ \theta p_v&0\\ \end{array}\right), $ $ {A_{12}(M)} = \left(\begin{array}{lrc} 0\\ p\\ \end{array}\right), $ $ {A_{21}(M)} = \left(\begin{array}{lrc} 0&p\\ \end{array}\right). $

令$ W = M-\bar{M} $, 则由(2.15)–(2.17)式可推出

$ \begin{equation} A^{0}(M)W_{\tau}+\tilde{A}(M)W_y = B(M)W_{yy}+g_{2}(M, M_y)+(\tilde{A}(M)-A(M))W_y, \end{equation} $

(2.18)

其中

$ g_{2}(M, M_y)= \left\{(A^{0}(\bar{M})-A^{0}(M))\bar{M}_\tau+(B(M)-B(\bar{M}))\bar{M}_{yy}\right\}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+(A(\bar{M})-A(M))\bar{M}_y+(g(M, M_y)-g(\bar{M}, \bar{M}_y))-\bar{F}. $

将(2.18)式关于$ y $求导, 再将所得方程乘以$ W_y $, 并在$ \mathbb{R} $上关于$ y $求积分得

$ \begin{equation} \begin{aligned} &\underbrace{\int_\mathbb{R}\langle A^{0}(M)\partial_{y}W_{\tau}, \partial_{y}W\rangle dy}_{I_1}+\underbrace{\int_\mathbb{R}\langle\tilde{A}(M)\partial_y^{2}W, \partial_{y}W\rangle dy}_{I_2}\\ = &\underbrace{\int_\mathbb{R}\langle B(M)\partial_y^{3}W, \partial_{y}W\rangle dy}_{I_3}+\underbrace{\int_\mathbb{R}\langle\tilde{H}, \partial_{y}W\rangle dy}_{I_4}, \end{aligned} \end{equation} $

(2.19)

这里$ \langle\cdot, \cdot\rangle $表示$ \mathbb{R}^3 $上的内积, 且

$ \begin{eqnarray} &&\tilde{H} = A^{0}(M)(A^{0}(M)^{-1}g_{2})_y+A^{0}(M)\left(A^{0}(M)^{-1} B(M)\right)_yW_{yy}\\ &&\quad\; \; \; +A^{0}(M)\left\{A^{0}(M)^{-1}(\tilde{A}(M)-A(M))W_y\right\}_y-A^{0}(M)\left(A^{0}(M)^{-1}\tilde{A}(M)\right)_yW_y. \end{eqnarray} $

现在来逐项估计$ I_i, i = 1, 2, 3, 4 $.首先由(1.11), (2.1), (2.4), (2.5)及(2.10)–(2.12)式, 有

$ \begin{eqnarray} &&I_1 = \frac{1}{2}\frac{d}{d\tau}\int_\mathbb{R} A^{0}(M)W_y^2dy-\frac{1}{2}\int_\mathbb{R} A^{0}(M)_\tau W_y^2dy\\ &&\quad = \frac{1}{2}\frac{d}{d\tau}\int_\mathbb{R} A^{0}(M)W_y^2dy-\frac{1}{2}\int_\mathbb{R}\left[(p^2)_\tau\phi_y^2+\theta_\tau\psi_y^2\right]dy \\ &&\quad\geq\frac{1}{2}\frac{d}{d\tau}\int_\mathbb{R} A^{0}(M)W_y^2dy-C\int_\mathbb{R}\left(\|(\Theta_\tau, \psi_y)(\tau)\|_{L^\infty}(\phi_y^2+\psi_y^2)+\|\phi_y(\tau)\|_{L^\infty}|\zeta_{yy}\phi_y|\right)dy\\ &&\quad\quad-C\int_\mathbb{R}\|pu_y-\bar{p}\bar{u}_y-\frac{\zeta_yv_y}{v^2}+\left(\frac{\bar{\theta}}{v}-\frac{\bar{\theta}}{\bar{v}}\right)_y +\frac{\mu}{\kappa}\left(\frac{\bar{u}_y^2}{v}-\frac{\bar{u}_y^2}{\bar{v}}\right)-R_1\|_{L^\infty}(\phi_y^2+\psi_y^2)dy\\ && \quad\geq\frac{1}{2}\frac{d}{d\tau}\int_\mathbb{R} A^{0}(M)W_y^2dy-C\kappa^{\frac{1}{2}}\int_\mathbb{R}(\phi_y^2+\psi_y^2+\zeta_{yy}^2)dy. \end{eqnarray} $

(2.20)

类似地有

$ \begin{equation} \begin{split} I_2 = \frac{1}{2}\int_\mathbb{R}\tilde{A}(M)_y W_y^2dy &\leq C\int_\mathbb{R}\|(W_y, \bar{M}_y)(\tau)\|_{L^\infty}|W_y|^2dy\\ &\leq C\kappa^{\frac{1}{2}}\int_\mathbb{R}(\phi_y^2+\psi_y^2+\zeta_{y}^2)dy; \end{split} \end{equation} $

(2.21)

$ \begin{equation} \begin{split} I_3& = \int_\mathbb{R}\left(\frac{\mu}{\kappa}\frac{\theta}{v}\partial_y^3\psi\partial_y\psi+\frac{1}{v}\partial_y^3\zeta\partial_y\zeta\right)dy\\& \leq-\int_\mathbb{R}\left(\frac{\mu}{\kappa}\frac{\theta}{v}\psi_{yy}^2+\frac{1}{v}\zeta_{yy}^2\right)dy+C\int_\mathbb{R}(|W_y|+|\bar{M}_y|)\left(\frac{\mu}{\kappa}|\psi_y||\psi_{yy}|+|\zeta_y||\zeta_{yy}|\right)dy\\& \leq-\int_\mathbb{R}\left(\frac{\mu}{\kappa}\frac{\theta}{v}\psi_{yy}^2+\frac{1}{v}\zeta_{yy}^2\right)dy+C\kappa^{\frac{1}{2}}\int_\mathbb{R}\left(\frac{\mu}{\kappa}\psi_{yy}^2+\zeta_{yy}^2+\psi_y^2+\zeta_y^2\right)dy. \end{split} \end{equation} $

(2.22)

对于$ I_4 $, 有

$ \begin{equation} \begin{split} I_4& = \int_\mathbb{R}\langle A^{0}(M)(A^{0}(M)^{-1})_yg_{2}, \partial_yW\rangle dy+ \int_\mathbb{R}\langle(g_{2})_y, \partial_yW\rangle dy\\& \; \; \; +\int_\mathbb{R}\langle A^{0}(M)\left(A^{0}(M)^{-1} B(M)\right)_yW_{yy}, W_y\rangle dy\\& \; \; \; +\int_\mathbb{R}\langle A^{0}(M)\left(A^{0}(M)^{-1}(\tilde{A}(M)-A(M))W_y\right)_y, W_y\rangle dy\\& \; \; \; -\int_\mathbb{R}\langle A^{0}(M)\left(A^{0}(M)^{-1}\tilde{A}(M)\right)_yW_y, W_y\rangle dy\\& = I_{41}+I_{42}+I_{43}+I_{44}+I_{45}. \end{split} \end{equation} $

(2.23)

利用Cauchy不等式, (2.5)式和引理3.1, 有

$ \begin{eqnarray} I_{41}&\leq& C\int_\mathbb{R}(|W_y|+|\bar{M}_y|)\{|W_y|+(|\bar{M}_\tau|+|\bar{M}_{yy}|+|\bar{M}_y|)|W|+|\bar{F}|\}|W_y|dy\\& \leq& C\int_\mathbb{R}\left\{|W_y|^3+(|\bar{M}_\tau|+|\bar{M}_{yy}|+|\bar{M}_y|)|W||W_y|^2+|R_{2}W_y^2|+|\bar{M}_y||W_y|^2\right.\\& &\left.+|\bar{M}_y|(|\bar{M}_\tau|+|\bar{M}_{yy}|+|\bar{M}_y|)|W||W_y|+|\bar{M}_y||W_y||R_{2}|\right \}dy\\& \leq& \kappa^{\frac{1}{2}}\int_\mathbb{R}|W_y|^2dy+(\kappa^{\frac{3}{2}}+\kappa^2)\int_\mathbb{R}|W|^2dy+C\kappa^{\frac{1}{2}}\int_\mathbb{R}|R_{2}|^2dy\\& \leq&\kappa^{\frac{1}{2}}\int_\mathbb{R}|W_y|^2dy+C\kappa^3; \end{eqnarray} $

(2.24)

$ \begin{array}{l} {I_{42}} = \int_{\mathbb{R}} \langle \{ ({A^0}(\bar M) - {A^0}(M)){{\bar M}_\tau } + (B(M) - B(\bar M)){{\bar M}_{yy}} + \\ \;\;\;\;\;\;(A(\bar M) - A(M)){{\bar M}_y}{\} _y},{W_y}\rangle dy\\ \;\; + \int_{\mathbb{R}} {\langle {{\left( {g(M,{M_y}) - g(\bar M,{{\bar M}_y})} \right)}_y},{W_y}\rangle } dy - \int_{\mathbb{R}} {\langle {{\bar F}_y},{W_y}\rangle } dy \equiv I_{42}^1 + I_{42}^2 + I_{42}^3. \end{array} $

(2.25)

由$ \bar{M} $的定义, Cauchy不等式, (2.5)式和引理3.1得

$ \begin{equation} \begin{split} I_{42}^{1}&\leq C\int_\mathbb{R}(|\bar{M}_\tau|+|\bar{M}_{yy}|+|\bar{M}_y|)(|W_y|+|\bar{M}_y||W|)|W_y|dy\\& \; \; \; +C\int_\mathbb{R}(|\bar{M}_{y\tau}|+|\bar{M}_{yyy}|+|\bar{M}_{yy}|)|W||W_y|dy\\& \leq \kappa^{\frac{1}{2}}\int_\mathbb{R}|W_y|^2dy+C\kappa^{\frac{3}{2}}\int_\mathbb{R}|W|^2dy\\& \leq\kappa^{\frac{1}{2}}\int_\mathbb{R}|W_y|^2dy+C\kappa^3; \end{split} \end{equation} $

(2.26)

$ \begin{eqnarray} I_{42}^{2} &\leq& C\int_\mathbb{R}\left\{|W_y|^2+|\bar{M}_y||W_y|+|\bar{M}_y^2||W|\right\}\left(\frac{\mu}{\kappa}|\psi_{yy}|+|\zeta_{yy}|\right)dy \\ &\leq&\kappa^{\frac{1}{2}}\int_\mathbb{R}\left(\phi_y^2+\psi_y^2+\zeta_y^2+\frac{\mu}{\kappa}\psi_{yy}^2+\zeta_{yy}^2\right)dy+\kappa^{\frac{3}{2}}\int_\mathbb{R}|W|^2dy\\ &\leq& \kappa^{\frac{1}{2}}\int_\mathbb{R}\left(\phi_y^2+\psi_y^2+\zeta_y^2+\frac{\mu}{\kappa}\psi_{yy}^2+\zeta_{yy}^2\right)dy+C\kappa^{3}; \end{eqnarray} $

(2.27)

$ \begin{eqnarray} I_{42}^{3}&\leq& C\kappa\int_\mathbb{R}|W_y|^2dy+C\kappa^{-1}\int_\mathbb{R}|(R_{2})_y|^2dy\leq C\kappa\int_\mathbb{R}(\phi_y^2+\psi_y^2+\zeta_y^2)dy+C\kappa^{\frac{7}{2}}. \end{eqnarray} $

(2.28)

联立(2.25)–(2.28)式得

$ \begin{equation} I_{42}\leq \kappa^{\frac{1}{2}}\int_\mathbb{R}\left(\phi_y^2+\psi_y^2+\zeta_y^2+\frac{\mu}{\kappa}\psi_{yy}^2+\zeta_{yy}^2\right)dy+C\kappa^{3}. \end{equation} $

(2.29)

类似于$ I_{42} $的估计有

$ \begin{equation} \begin{split} I_{43}& = \int_\mathbb{R}\langle A^{0}(M)(A^{0}(M)^{-1}B(M))_yW_{yy}, \partial_yW\rangle dy\\& = \int_\mathbb{R}\left(\frac{\mu}{\kappa}\theta\left(\frac{1}{v}\right)_y\psi_y\psi_{yy}+\left(\frac{1}{v}\right)_y\zeta_y\zeta_{yy}\right)dy\\& \leq C\kappa^{\frac{1}{2}}\int_\mathbb{R}\left(\psi_y^2+\zeta_y^2+\frac{\mu}{\kappa}\psi_{yy}^2+\zeta_{yy}^2\right)dy. \end{split} \end{equation} $

(2.30)

利用分部积分, Cauchy不等式以及(2.10)–(2.12)式可得

$ \begin{equation} \begin{split} I_{44}& = \int_\mathbb{R}\langle A^{0}(M)\left(A^{0}(M)^{-1}\right)_y(\tilde{A}(M)-A(M))W_y, W_y\rangle dy\\& \quad+\int_\mathbb{R}\langle\{(\tilde{A}(M)-A(M))W_y\}_y, W_y\rangle dy \\& \leq C\int_\mathbb{R}(|W_y|+|\bar{M}_y|)|W||W_y|^2dy+\int_\mathbb{R}\{\psi_y((\bar{p}-p)\zeta_y)_y+\zeta_y((\bar{p}-p)\psi_y)_y\}dy \\&\leq C\int_\mathbb{R}(|W_y|+|\bar{M}_y|)|W||W_y|^2dy+\int_\mathbb{R}(\bar{p}-p)_y\psi_y\zeta_y dy \\& \leq C\kappa^{\frac{1}{2}}\int_\mathbb{R}(\phi_y^2+\psi_y^2+\zeta_y^2)dy. \end{split} \end{equation} $

(2.31)

同理可得

$ \begin{equation} \begin{split} I_{45}& = -\int_\mathbb{R}\left\{\theta\psi_y\left(\phi_yp_{vy}+\zeta_y\left(\frac{\bar{p}}{\theta}\right)_y\right)+\bar{p}_y\psi_y\zeta_y\right\}dy\\& = -\int_\mathbb{R}\left(p_{vy}\psi_y\phi_y+\left(\frac{\bar{p}}{\theta}\right)_y\psi_y\zeta_y+\bar{p}_y\psi_y^2\zeta_y\right)dy\\& \leq c\kappa^{\frac{1}{2}}\int_\mathbb{R}(\phi_y^2+\psi_y^2+\zeta_y^2)dy. \end{split} \end{equation} $

(2.32)

将(2.24), (2.29)–(2.32)式代入(2.23)式得

$ \begin{equation} I_4\leq \kappa^{\frac{1}{2}}\int_\mathbb{R}\left(\phi_y^2+\psi_y^2+\zeta_y^2+\frac{\mu}{\kappa}\psi_{yy}^2+\zeta_{yy}^2\right)dy+C\kappa^{3}. \end{equation} $

(2.33)

联立(2.19), (2.20)–(2.22)和(2.33)式, 利用$ \kappa $的小性有

$ \frac{d}{d\tau}\int_\mathbb{R} A^{0}(M)W_y^2dy+\int_\mathbb{R}\left(\frac{\mu}{\kappa}\frac{\theta}{v}\psi_{yy}^2+\frac{1}{v}\zeta_{yy}^2\right)dy\leq \kappa^{\frac{1}{2}}\int_\mathbb{R}(\phi_y^2+\psi_y^2+\zeta_y^2)dy+C\kappa^{3}. $

将上式在$ [\tau_{0}, \tau] $上积分, 则有

$ \begin{equation} \|(\phi_y, \psi_y, \zeta_y)(\tau)\|^2+\int_{\tau_{0}}^{\tau}\left(\frac{\mu}{\kappa}\|\psi_{yy}(\tau)\|^2+\|\zeta_{yy}(\tau)\|^2\right)ds\leq \\ \kappa^{\frac{1}{2}}\int_{\tau_{0}}^{\tau}\|(\phi_y, \psi_y)(\tau)\|^2ds+C\kappa^{2}. \end{equation} $

(2.34)

对于$ \int_{\tau_{0}}^{\tau}\|(\phi_y, \psi_y)(\cdot, \tau)\|^2ds $的估计, 在文献[1]中已经证得

$ \begin{equation} \int_{\tau_{0}}^{\tau}\|(\phi_y, \psi_y)(\cdot, \tau)\|^2ds\leq C\int_\mathbb{R}(\phi_y^2+\psi_y^2+\zeta_y^2)dy+C\int_{\tau_{0}}^{\tau}\int_\mathbb{R}\left(\frac{\mu}{\kappa}\psi_{yy}^2+\zeta_{yy}^2\right)dyds+C\kappa^{\frac{3}{2}}. \end{equation} $

(2.35)

将(2.35)式代入(2.34)式中, 且令$ \kappa $充分小, 可以得到(2.13)式.引理2.2证毕.

类似于引理2.2的证明, 可得

引理2.3   在命题2.1的假设下, 存在两个不依赖于$ \kappa $和$ \tau_{2} $的正常数$ \kappa_{3} $和$ C_3 $, 使得当$ 0<\kappa\leq\kappa_{3} $时, 对任意的$ \tau\in[\tau_0, \tau_2] $有

$ \begin{equation} \|(\phi_{yy}, \psi_{yy}, \zeta_{yy})(\cdot, \tau)\|^{2}+\int_{\tau_{0}}^{\tau}(\|(\phi_{yy}, \psi_{yy})(\cdot, \tau)\|^{2}+\|\zeta_{yyy}(\cdot, \tau)\|^{2})ds\leq C_3\kappa^{\frac{5}{2}}. \end{equation} $

(2.36)

令$ \kappa_0 = \min\{\kappa_1, \kappa_2, \kappa_3\} $以及$ C_0 = \max\{C_1, C_2, C_3\} $, 由引理2.1–2.3可得命题2.1.

现在开始证明本文的主要定理.

定理1.1的证明  由于命题2.1已证, 可以选取(2.7)–(2.8)式中的$ \kappa $充分小, 使得$ \sqrt{2C_0}\kappa^{\frac{1}{4}}<1 $, 这样就封闭了先验假设(2.6)式.再由标准的连续性技巧, 可以将局部解延拓到$ \tau = \tau_1 $时刻.此外, 估计式(2.7)–(2.8)对$ \tau_2 = \tau_1 $时刻也成立, 即

$ \begin{eqnarray} &&\sup\limits_{\tau_0\leq\tau\leq\tau_1}\|(\phi, \psi, \zeta)(\tau)\|^2+\int_{\tau_0}^{\tau_1}\|\zeta_y(s)\|_1^2ds\leq C_0\kappa^{\frac{3}{2}}, \end{eqnarray} $

(2.37)

$ \begin{eqnarray} &&\sup\limits_{\tau_0\leq\tau\leq\tau_1}\|(\phi_y, \psi_y, \zeta_y)(\tau)\|_1^2+\int_{\tau_0}^{\tau_1}(\|(\phi_y, \psi_y)(s)\|_1^{2}+\|\zeta_{yy}(s)\|_1^{2})ds \leq C_0\kappa^{2}. \end{eqnarray} $

(2.38)

从而对任意的常数$ T>0 $, 由(2.37)–(2.38)式及Sobolev不等式得

$ \begin{equation} \begin{aligned} \|(v-\bar{v}, u-\bar{u}, \theta-\bar{\theta})(t)\|_{L^{\infty}}&\leq \|(\phi, \psi, \zeta)(\tau)\|^{\frac{1}{2}}\|(\phi_y, \psi_y, \zeta_y)(\tau)\|^{\frac{1}{2}}\\ &\leq C\kappa^{\frac{7}{8}}, \; \; \; \forall\, t\, \in[0, T]. \end{aligned} \end{equation} $

(2.39)

又由(1.10)式可得

$ \begin{equation} \begin{split} \sup\limits_{0\leq t\leq T, \, |x|\geq h}\|(\bar{v}-v^{cd}, \bar{u}-u^{cd}, \bar{\theta}-\theta^{cd})(t, x)\| \leq c_{1}\delta e^{-\frac{c_{2}h^2}{\kappa(1+T)}} \leq C\kappa^{2}, \end{split} \end{equation} $

(2.40)

因此由(2.39)和(2.40)式可推出

$ \begin{eqnarray} &&\sup\limits_{0\leq t \leq T, \, |x|\geq h}\left|(v, u, \theta)(x, t)-(v^{cd}, u^{cd}, \theta^{cd})(x, t)\right|\\ &\leq&\sup\limits_{0\leq t \leq T, \, |x|\geq h}\left|(v-\bar{v}, u-\bar{u}, \theta-\bar{\theta})(t, x)\right|+\sup\limits_{0\leq t \leq T, \, |x|\geq h}\left|(\bar{v}-v^{cd}, \bar{u}-u^{cd}, \bar{\theta}-\theta^{cd})(t, x)\right|\\ &\leq& C\kappa^{\frac{7}{8}}. \end{eqnarray} $

(2.41)

这样就证明了(1.15)式.定理1.1证毕.