数学物理学报, 2023, 43(4): 1123-1132

二维不可压缩 Navier-Stokes-Allen-Cahn 系统的边界层分离

陈敏,, 胡碧艳,, 罗宏,*

四川师范大学数学科学学院 成都 610068

Boundary Layer Separation of 2-D Incompressible Navier-Stokes-Allen-Cahn System

Chen Min,, Hu Biyan,, Luo Hong,*

School of Mathematics Science, Sichuan Normal University, Chengdu 610068

通讯作者: *罗宏，E-mail: lhscnu@163.com

收稿日期: 2022-07-17   修回日期: 2023-02-11  

基金资助: 国家自然科学基金(12171343)
四川省科研厅科研基金(22CXTD0029)

Received: 2022-07-17   Revised: 2023-02-11  

Fund supported: NSFC(12171343)
Scientific Research Fund of the Science and Technology Department of Sichuan Province(22CXTD0029)

作者简介 About authors

陈敏，E-mail:1653637845@qq.com;

胡碧艳，E-mail:2838954298@qq.com

摘要

该文研究了二维不可压缩 Navier-Stokes-Allen-Cahn 系统的边界层分离. 首先利用不可压缩流的几何理论以及泰勒展开式, 得到平直边界下边界层分离的条件; 其次求出边界奇点, 得到弯曲边界下边界层分离的条件. 这些条件由初值和外力决定, 可以预测 Navier-Stokes-Allen-Cahn 系统发生边界层分离的时间和地点.

关键词： 边界层分离; Navier-Stokes-Allen-Cahn 系统; 分歧; 二维不可压缩流

Abstract

In this paper, boundary layer separation of 2-D incompressible Navier-Stokes-Allen-Cahn system is considered. Firstly, the condition of boundary layer separation under flat boundary is obtained with the help of the geometric theory of incompressible flow and Taylor expansion. Secondly, the expression for boundary singularity is presented and the condition of boundary layer separation under curved boundary is discovered. The conditions, determined by initial values and external forces, can predict when and where boundary layer separation for Navier-Stokes-Allen-Cahn system will occur.

Keywords： Boundary layer separation; Navier-Stokes-Allen-Cahn system; Bifurcation; 2-D incompressible flow

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本文引用格式

陈敏, 胡碧艳, 罗宏. 二维不可压缩 Navier-Stokes-Allen-Cahn 系统的边界层分离[J]. 数学物理学报, 2023, 43(4): 1123-1132

Chen Min, Hu Biyan, Luo Hong. Boundary Layer Separation of 2-D Incompressible Navier-Stokes-Allen-Cahn System[J]. Acta Mathematica Scientia, 2023, 43(4): 1123-1132

1 引言

气液两相流广泛应用于航空、海洋、生物等领域, 其研究具有重要的理论和应用价值. 众所周知, Navier-Stokes 方程是描述粘性不可压缩流体的动量守恒方程, Navier-Stokes-Allen-Cahn(简称 NSAC) 系统是通过将 Navier-Stokes 方程与 Allen-Cahn 方程耦合而形成的. 近年来, NSAC 系统已经被广泛的研究[1⇓⇓⇓⇓⇓⇓⇓-9]. Xu 等[1]证明了大粘度和小初始数据条件下的三维不可压缩NSAC系统的全局解的存在性. Gal 和 Grasselli[2]研究了二维 NSAC 系统在有界区域中弱解的存在唯一性、全局吸引子和指数吸引子的存在性, 同时假设外力项为零且 $f$ 为实解析函数时, 证明了解的收敛性, 并对收敛速率进行了估计. Zhao 等[3] 利用 Galerkin 近似方法得到了粘性系数趋于零的 NSAC 系统局部光滑解的存在性, 并证明了在适当小的时间内 NSAC 系统收敛于 Allen-Cahn-Euler 系统. Medjo[4] 利用 Galerkin 近似方法证明了二维 NSAC 系统变分解的存在唯一性. Zhao[5]研究了三维可压缩 NSAC 系统柯西问题解的全局稳定性和时间衰减率, 得到了高阶空间导数的最佳衰减率.

在流体动力学中,物理学家 Prandtl 于 1904 年提出了边界层, 此后, 边界层研究就成为流体力学中的重要课题, 取得了很多成果[3,10⇓⇓⇓-14]. Zhao 等[3] 使用边界层函数来处理 Navier-Stokes 方程和Euler 方程之间边界条件的不匹配问题. Xie[12] 研究了小粘度通道中的 NSAC 系统的边界层, 并证明了在一定的厚度下该系统存在边界层. Fan 和 Hou[13] 研究了具有非粘性的可压缩 Navier-Stokes 模型的初边值问题, 并严格证明了退化边界层的渐近稳定性. 张浩和汪娜[14]用边界层函数法研究了一类带有积分初边值条件的奇异摄动问题, 得到了解的存在唯一性.

在流体力学中有一个重要的分歧现象-边界层分离, 近年来, 关于边界层分离的研究也有许多的结果[15⇓⇓⇓⇓⇓⇓⇓⇓-24]. Chorin 和 Marsden[15] 提出了如何定义边界层分离及如何揭示其机理等问题. Ghil, Ma 和 Wang[16-17]给出了边界层分离的数学定义, 提出了判断边界层分离的标准. Ghil 和 Liu 等[18]研究了 Dirichlet 边界条件下二维不可压缩流的边界层分离. Luo 等[19-20]得到了平直和弯曲边界下二维不可压缩 Navier-Stokes 边界层分离的条件. 姜倩[21]研究了二维阻尼 Navier-Stokes 系统的边界层分离条件. Shen 等[22]证明了 Oleinik 数据下 Prandtl 方程的边界层分离, 并验证了解局部行为的假设.

本文将考虑如下 NSAC 系统的边界层分离

$u_{t}+(u \cdot \nabla) u-\gamma \triangle u+\nabla p=\mathcal{K} \mu \nabla \phi+g, \quad(x, t) \in \Omega \times(0,+\infty)$
$\phi_{t}+(u \cdot \nabla) \phi+\mu=0, \quad(x, t) \in \Omega \times(0,+\infty)$
$\mu=-\varepsilon \Delta \phi+\alpha\left(\phi^{3}-\phi\right), \quad(x, t) \in \Omega \times(0,+\infty)$,
$\operatorname{div} u=0, \quad(x, t) \in \Omega \times(0,+\infty)$
$u(x, t)=0, \phi(x, t)=M(x), \quad(x, t) \in \partial \Omega \times(0,+\infty)$
$u(x, 0)=\theta(x), \phi(x, 0)=b(x), \quad x \in \Omega$,

其中 $\Omega\subset \mathbb{R} ^2$ 为有界开集且边界为 $\partial \Omega$, ${u}$ 代表流体速度, 序参数 $\phi$ 表示两种液体的相对浓度, 正常数 $\gamma$ 和 ${\cal K}$ 是流体的粘性系数和应力系数, ${g}$ 是与时间无关的外力, $\varepsilon$ 和 $\alpha$ 是描述两相相互作用的两个正参数.

在 Ghil, Ma 和 Wang等建立边界层分离理论[16-17,23-24]的基础上, 我们得到了问题1.1a-1.1f边界层分离的条件. 本文的后续安排如下, 在第 $2$ 节中, 给出了边界奇点, 边界层分离的概念以及边界层分离理论. 第 $3$ 节和第 $4$ 节分别给出了平直边界和弯曲边界下 NSAC 系统边界层分离的条件.

2 预备知识

假设 $\Omega$ 为 $\mathbb{R} ^2$ 中的有界开集, $\tau$ 和 $n$ 分别是 $\partial \Omega$ 上的单位切向量和单位法向量, $C^{r}(\Omega)$ 为 $\Omega$ 上 $C^{r}$ 集所组成的集合.

$B_0^r(\Omega)=\{ u\in C^r(\Omega)\mid {\rm div} u=0, u|_{\partial \Omega}=0\}.$

下面将给出边界奇点和边界层分离的定义及边界层分离引理.

定义 2.1[16-17,24] 假设 $ u\in B^r_0(\Omega)(r\geq 2)$. 如果 $\frac{\partial u_\tau(x_0)}{\partial n}=0$, 那么 $x_0\in \partial \Omega$ 称为 $ u$ 的边界奇点.

定义 2.2[16-17,24] 若当 $t<t_0$ 时, $u(x, t)$ 与图1(a) 的结构是拓扑等价的; 当 $t>t_0$ 时, $ u(x, t)$ 与图1(c) 的结构是拓扑等价的, 则我们称在 $ u\in C^1([T];B_0^2(\Omega))$ 在 $t_0$ 时刻发生边界层分离. 即 $t<t_0$ 时, $ u(x,t)$ 拓扑等价于平行流, $t>t_0$ 时, $ u(x,t)$ 分离为一个漩涡. 如果 $x_0$ 是 $t=t_0$ 时刻的孤立边界奇异点, 那么我们称在 $x_0\in \partial \Omega$ 上发生边界层分离.

图1

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图1   边界层分离


引理 2.1[16-17,24] 令 $ u\in C^1([T];B_0^2(\Omega))$ 是一个二维向量场, $x_0 \in \partial \Omega$. 若存在 $0<t_0<T$ 使得

$\begin{equation}uaduad\ \frac{\partial u_\tau}{\partial n}(x, t)\neq 0, \ t<t_0,x\in\partial \Omega, \end{equation} $
$\begin{equation} \frac{\partial u_\tau}{\partial n}(x_0, t_0)=0, \end{equation} $
$\begin{equation} \ \frac{\partial ^2 u_\tau}{\partial t \partial n}(x_0, t_0)\neq 0, \end{equation} $

其中 $x_0$ 是 $ u(\cdot, t_0)$ 在 $\partial \Omega$ 上的孤立边界奇异点, 则 $u$ 会在 $(x_0, t_0)$ 发生边界层分离.

3 平直边界下的边界层分离

设 $\Gamma \subset\partial \Omega$. 不失一般性, 我们取一个坐标系统 $(x_1,x_2)$, 其原点为 $x_0$, 存在$\delta > 0$ 使得 $\Gamma=\{{(x_1,0)\mid0<|x_1|<\delta}\}$. 显然,$\Gamma$ 上的切向量和法向量分别是 $x_1$ -和 $x_2$ -方向上的单位向量.

设 ${\theta}(x)=({\theta}_1,{\theta}_2)\in C^3(\overline{\Omega};\mathbb{R} ^2)$. 由于 ${\theta}(x)|_{\partial \Omega}=0$, 则对 $\theta$ 在 $x_2=0$ 处进行 Taylor 展开可得

$\begin{equation} {\theta}_1=x_2 \theta_{11}(x_1)+x_2^2 \theta_{12}(x_1)+x_2^3 \theta_{13}(x_1)+o(x_2^3).\end{equation}$

因为 ${\rm div} \theta=0$, 所以

$\begin{equation}{\theta}_2=x_2^2 \theta_{21}(x_1)+o(x_2^2),\end{equation}$

其中 $\theta_{21}(x_1)=-\frac{\theta'_{11}(x_1)}{2}$.

同理, 根据 $\phi|_{\partial \Omega}=M(x)$, 在 $ x_2=0$ 处对 $b\in C^3(\overline{\Omega})$ 进行 Taylor 展开

$\begin{equation}b=M(x_1)+x_2 b_1(x_1)+x_2^2b_2(x_1)+x_2^3 b_{3}(x_1)+o(x_2^3). \end{equation}$ (3.3)

设 $ g(x)=( g_1, g_2)\in C^1(\overline{\Omega};\mathbb{R} ^2)$. 考虑 $ g(x)$ 在 $x_2=0$ 处的 Taylor 展开

$\begin{equation} g_1= g_{10}(x_1)+x_2 g_{11}(x_1)+o(x_2),\end{equation}$
$\begin{equation} g_2= g_{20}(x_1)+x_2 g_{21}(x_1)+o(x_2).\end{equation}$

下面的定理给出了平直边界下发生边界层分离的条件.

定理 3.1 设 $ \theta \in C^3(\overline{\Omega},\mathbb{R} ^2)$, $b \in C^3(\overline{\Omega})$, $M\in C^1(\partial \Omega)$ 以及 $g\in C^1(\overline{\Omega},\mathbb{R} ^2)$. 若

$\begin{equation} 0< t_0=\min_{\Gamma} \frac{-\theta_{11}}{2\gamma \theta_{11}^{''}+6 \gamma \theta_{13}+[-{\cal K}\varepsilon(b_1^{''}+6b_3)+{\cal K}\alpha(3M^2b_1-b_1)]M'+ g_{11}- g'_{20}}\ll1, \end{equation} $

则存在 $x_0\in \Gamma$ 使得问题(1.1)-(1.6)的解在$(t_0, x_0)$ 处发生边界层分离. 其中 $\theta_{11}$, $\theta_{13}$, $b_1$, $b_3$, $g_{11}$ 和 $g_{20}$ 分别满足(3.1),(3.3)-(3.5)式.

证 在定理3.1的假设下, 二维 NSAC 系统存在全局正则解. $ u$ 在 $t=0$ 处有 Taylor 展开式

$\begin{equation} u={\theta}+t\lambda+o(t).\end{equation} $

因为 $u|_{\partial \Omega}=0$ 以及 ${\rm div} u=0$, 所以 $\lambda |_{x_2=0}=0$ 和 ${\rm div} \lambda=0$. 令 $\lambda$ 在 $ x_2=0$ 处的 Taylor 展开为

$\begin{equation}\lambda=(\lambda_1,\lambda_2)=(x_2\lambda_{11}+o(x_2), x^2_2\lambda_{21}+o(x_2^2)). \end{equation}$

同理, 将 $\phi$ 在 $t=0$ 处进行 Taylor 展开可得

$\begin{equation} \phi=b+t\psi+o(t). \end{equation}$

根据 $\phi|_{\partial \Omega}=b(x_1,0)+t\psi(x_1,0)+o(t)=M(x_1)$, 可以得到 $\psi(x_1,0)=0$, 所以设

$\begin{equation}\psi=x_2\psi_1+o(x_2). \end{equation} $

对压强 $p$ 进行 Taylor 展开可得

$\begin{equation} p=p_0+tp_1+o(t), \end{equation}$
$\begin{equation}p_0=p_{01}(x_1)+x_2p_{02}(x_1)+o(x_2). \end{equation}$

将(1.1)式的第一个分量方程改写为

$\begin{matrix}&& u_{1t}+u_1\frac{\partial u_1}{\partial x_1}+ u_2\frac{\partial u_1}{\partial x_2}-\gamma\frac{\partial ^2 u_1}{\partial x_1^2}-\gamma\frac{\partial ^2 u_1}{\partial x_2^2}+\frac{\partial p}{\partial x_1}\\&=&{\cal K}\left[-\varepsilon\frac{\partial ^2 \phi}{\partial x_1^2}-\varepsilon\frac{\partial ^2 \phi}{\partial x_2^2}+\alpha\phi\left(\phi^2-1\right)\right]\frac{\partial \phi}{\partial x_1}+ g_1.\end{matrix}$

将(3.7),(3.9),(3.11)代入(3.13)式可得

$\begin{matrix} \lambda_1+\theta_1\frac{\partial \theta_1}{\partial x_1}+ \theta_2\frac{\partial \theta_1}{\partial x_2}-\gamma\frac{\partial ^2 \theta_1}{\partial x_1^2}-\gamma\frac{\partial ^2 \theta_1}{\partial x_2^2}+\frac{\partial p_0}{\partial x_1}={\cal K}\left[-\varepsilon\frac{\partial ^2b}{\partial x_1^2}-\varepsilon\frac{\partial ^2b}{\partial x_2^2}+\alpha (b^3-b)\right]\frac{\partial b}{\partial x_1}+ g_1.\end{matrix}$

将(3.1)-(3.5),(3.8),(3.10)和(3.12)式代入(3.14)式有

$\begin{matrix}\lambda_{11}-\gamma\theta^{''}_{11}-6\gamma\theta_{13}+p'_{02}&= &[-{\cal K}\varepsilon(M^{''}+2b_2)+{\cal K}\alpha(M^3-M)]b'_1\\&&+[-{\cal K}\varepsilon(b_1^{''}+6b_3)+{\cal K}\alpha(3M^2b_1-b_1)]M'+ g_{11}.\end{matrix}$

同理, 将(1.1)式的第二个分量改写为

$\begin{eqnarray*}&& u_{2t}+u_1\frac{\partial u_2}{\partial x_1}+u_2\frac{\partial u_2}{\partial x_2}-\gamma\frac{\partial ^2 u_2}{\partial x_1^2}-\gamma\frac{\partial ^2 u_2}{\partial x_2^2}+\frac{\partial p}{\partial x_2}\\&=&{\cal K}\left[-\varepsilon\frac{\partial ^2 \phi}{\partial x_1^2}-\varepsilon\frac{\partial ^2 \phi}{\partial x_2^2}+\alpha\phi(\phi^2-1)\right]\frac{\partial \phi}{\partial x_2}+ g_2.\end{eqnarray*}$

结合(3.1)-(3.5),(3.7)-(3.12)式可得

$\begin{equation}-2\gamma\theta_{21}+p_{02}=[-{\cal K}\varepsilon (M^{''}+2b_2)+{\cal K}\alpha(M^3-M)]b_1+g_{20}.\end{equation}$

类似, 由(1.2)式可得

$\begin{eqnarray*}\phi_t+ u_1\frac{\partial \phi}{\partial x_1}+u_2\frac{\partial \phi}{\partial x_2}-\varepsilon\frac{\partial ^2\phi}{\partial x_1^2}-\varepsilon\frac{\partial ^2 \phi}{\partial x_2^2}+\alpha\phi(\phi^2-1)=0.\end{eqnarray*}$

结合(3.7)和(3.9)式可得

$\begin{eqnarray*}&&\frac{\partial (b+t\psi)}{\partial t}+( \theta_1+t \lambda_1)\frac{\partial (b+t\psi)}{\partial x_1}+(\theta_2+t\lambda_2)\frac{\partial (b+t\psi)}{\partial x_2}\\&=&\varepsilon\frac{\partial ^2 (b+t\psi)}{\partial x_1^2}+\varepsilon\frac{\partial ^2 (b+t\psi)}{\partial x_2^2}-\alpha(b+t\psi)[(b+t\psi)^2-1],\end{eqnarray*}$

因此

$\psi+\theta_1\frac{\partial b}{\partial x_1}+ \theta_2\frac{\partial b}{\partial x_2}=\varepsilon\frac{\partial ^2b}{\partial x_1^2}+\varepsilon\frac{\partial ^2b}{\partial x_2^2}-\alpha(b^3-b).$

结合(3.1)-(3.5)以及(3.10)式有

$\begin{equation}-\varepsilon M^{''}-2\varepsilon b_2+\alpha(M^3-M)=0.\end{equation}$

根据(3.15)-(3.17)式可以得到

$\begin{equation}\lambda_{11}=2\gamma \theta_{11}^{''}+6 \gamma\theta_{13}+[-{\cal K}\varepsilon(b_1^{''}+6b_3)+{\cal K}\alpha(3M^2b_1-b_1)]M'+ g_{11}-g'_{20}.\end{equation}$

利用(3.7),(3.8)和(3.18)式能够得到

$\begin{eqnarray*}\frac{\partial u_\tau}{\partial n}\bigg|_{\Gamma}&=&\frac{\partial u_1}{\partial x_2}\bigg|_{\Gamma}=\left(\frac{\partial \theta_1}{\partial x_2}+t \lambda_{11}+o(t)\right)\bigg|_{\Gamma}\\&=&\theta_{11}+t \{2\gamma \theta_{11}^{''}+6 \gamma\theta_{13}+[-{\cal K}\varepsilon(b_1^{''}+6b_3)+{\cal K}\alpha(3M^2b_1-b_1)]M'+g_{11}-g'_{20}\}+o(t).number\end{eqnarray*}$

如果

$0<t_0=\min_{x_1 \in \Gamma}\frac{-\theta_{11}}{2\gamma\theta_{11}^{''}+6 \gamma \theta_{13}+[-{\cal K}\varepsilon(b_1^{''}+6b_3)+{\cal K}\alpha(3M^2b_1-b_1)]M'+ g_{11}- g'_{20}}\ll1,$

那么 $\frac{\partial u_\tau}{\partial n}\big|_{\Gamma}=0$,即存在一个点 $x_0 \in \Gamma$ 使得

$\frac{-\theta_{11}}{2\gamma\theta_{11}^{''}+6 \gamma \theta_{13}+[-{\cal K}\varepsilon(b_1^{''}+6b_3)+{\cal K}\alpha(3M^2b_1-b_1)]M'+ g_{11}- g'_{20}}$

取得最小值, 因此, $\frac{\partial u_\tau(x_0,t_0)}{\partial n}=0$, 即引理 2.1中(2.2)式成立.

因为当 $t<t_0$ 时

$\begin{eqnarray*}\frac{\partial u_\tau}{\partial n}\bigg|_{\Gamma}&=&\frac{\partial u_1}{\partial x_2}\bigg|_{\Gamma}=\left(\frac{\partial \theta_1}{\partial x_2}+t \lambda_{11}+o(t)\right)\bigg|_{\Gamma}\\&=& \theta_{11}+t \{2\gamma \theta_{11}^{''}+6 \gamma\theta_{13}+[-{\cal K}\varepsilon(b_1^{''}+6b_3)+{\cal K}\alpha(3M^2b_1-b_1)]M'+g_{11}- g'_{20}\}+o(t)\\&\neq& 0,\end{eqnarray*}$

所以引理 2.1 的第一个条件(2.1)式成立.

由于

$\begin{eqnarray*}\frac{\partial ^2 u_\tau}{\partial n\partial t}\bigg|_{(t_0,x_0)}=\frac{\partial ^2 u_1}{\partial x_2\partial t}\bigg|_{(t_0,x_0)}=\lambda_{11}+o(t)\bigg|_{(t_0,x_0)}\neq 0,\end{eqnarray*}$

故引理2.1的第三个条件(2.3)式成立. 利用引理 2.1可知, 问题(1.1)-(1.6)的解在 $(t_0, x_0)$ 处会发生边界层分离. 定理3.1证毕.

注3.1 如果 $- \theta_{11}$ 和 $ 2\gamma \theta_{11}^{''}+6 \gamma \theta_{13}+[-{\cal K}\varepsilon(b_1^{''}+6b_3) +{\cal K}\alpha(3M^2b_1-b_1)]M'+ g_{11}- g'_{20} $ 是线性相关的, 那么

$ t'=\frac{-\theta_{11}}{2\gamma \theta_{11}^{''}+6 \gamma \theta_{13}+[-{\cal K}\varepsilon(b_1^{''}+6b_3) +{\cal K}\alpha(3M^2b_1-b_1)]M'+ g_{11}-g'_{20}} $

为常数, 故存在多个点$x'\in \Gamma$ 使得 $t'$ 取得最小值, 即 $x'$ 不是孤立奇异点. 事实上, $-\theta_{11}$ 和 $2\gamma \theta_{11}^{''}+6 \gamma \theta_{13}+[-{\cal K}\varepsilon(b_1^{''}+6b_3)+{\cal K}\alpha(3M^2b_1-b_1)]M'+g_{11}-g'_{20}$ 线性相关的可能性很小, 所以没有特别强调它.

注3.2 若相对浓度 $\phi$ 为零, 则NSAC 系统变为 Navier-Stokes 方程, 而系统(1.1)-(1.6)边界层分离的条件(3.6)式变为

$0< \min_{\Gamma} \frac{-\theta_{11}}{2\gamma \theta_{11}^{''}+6 \gamma\theta_{13}+ g_{11}- g'_{20}}\ll1, $

与文献[19] 的结果吻合.

4 弯曲边界下的边界层分离

设 $\Gamma \subset\partial \Omega$. 不失一般性, 我们取 $x_0$ 为原点建立一个坐标系统 $(x_1,x_2)$, 定义$\Gamma= \{{(x_1,q(x_1))\mid0<|x_1|<\delta}\}$, 其中 $\delta>0$. 令 $\tau$ 和 $n$ 分别为 $\Gamma$ 上的单位切向量和单位法向量, 且满足

$\begin{equation}\tau=\left(\frac{1}{\sqrt{1+{q'}^2}},\frac{q'}{\sqrt{1+{q'}^2}}\right),\ n=\left(-\frac{q'}{\sqrt{1+{q'}^2}},\frac{1}{\sqrt{1+{q'}^2}}\right). \end{equation}$

定理 4.1 设 $q\in C^2(\mathbb{R} )$, $\theta \in C^3(\overline{\Omega},\mathbb{R} ^2)$, $b \in C^3(\overline{\Omega})$ 和 $g\in C^1(\overline{\Omega},\mathbb{R} ^2)$. 如果

$\begin{equation} 0< t_1=\min_{\Gamma} \frac{\frac{\partial \theta_2}{\partial x_1}-\frac{\partial \theta_1}{\partial x_2}}{\gamma\frac{\partial \triangle \theta_1}{\partial x_2}-\gamma\frac{\partial \triangle\theta_2}{\partial x_1}+{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_1}\frac{\partial b}{\partial x_2}-{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_2}\frac{\partial b}{\partial x_1}+\frac{\partial g_1}{\partial x_2}-\frac{\partial g_2}{\partial x_1}}\ll1, \end{equation} $

那么存在 $x'_0\in \Gamma$ 使得问题(1.1)-(1.6)的解在 $(t_1, x'_0)$ 处发生边界层分离.

证 在定理 4.1 的假设下, 二维 NSAC 系统存在全局正则解. 根据(1.1)式可知

$\begin{eqnarray*} \frac{\partial }{\partial n}\frac{\partial u_{\tau}}{\partial t}=-\frac{\partial (( u \cdot \nabla) u\cdot\tau)}{\partial n}+\frac{\partial (\gamma\triangle u\cdot\tau)}{\partial n}-\frac{\partial (\nabla p\cdot\tau)}{\partial n}+\frac{\partial ({\cal K}\mu \nabla \phi\cdot\tau)}{\partial n}+\frac{\partial ( g\cdot\tau)}{\partial n}.number \end{eqnarray*} $

在 $(0,t)$ 上进行积分, 可以得到

$\begin{equation}\frac{\partial u_{\tau}}{\partial n}-\frac{\partial \theta_{\tau}}{\partial n}=\int_0^t\bigg[-\frac{\partial (( u \cdot \nabla)u\cdot\tau)}{\partial n}+\frac{\partial (\gamma\triangle u\cdot\tau)}{\partial n}-\frac{\partial (\nabla p\cdot\tau)}{\partial n}+\frac{\partial ({\cal K}\mu \nabla \phi\cdot\tau)}{\partial n}+\frac{\partial ( g\cdot\tau)}{\partial n}\bigg]{\rm d}s.\end{equation} $

因为 $(u \cdot \nabla)u\cdot\tau=u_{\tau}\frac{\partial u_{\tau}}{\partial \tau}-ku_{\tau}u_n+u_{n}\frac{\partial u_{\tau}}{\partial n}-u_{n}u\frac{\partial \tau}{\partial n}$, 其中 $k$ 为曲率, 所以

$\begin{eqnarray*} \frac{\partial (( u \cdot \nabla) u\cdot\tau)}{\partial n}&=&\frac{\partial ( u_{\tau}\frac{\partial u_{\tau}}{\partial \tau})}{\partial n}-\frac{\partial (k u_{\tau}u_n)}{\partial n}+\frac{\partial ( u_{n}\frac{\partial u_{\tau}}{\partial n})}{\partial n}-\frac{\partial ( u_{n} u\frac{\partial \tau}{\partial n})}{\partial n}\\ &=&\frac{\partial u_{\tau}}{\partial n}\frac{\partial u_{\tau}}{\partial \tau}+ u_{\tau}\frac{\partial ^2 u_{\tau}}{\partial \tau\partial n}-\frac{\partial k}{\partial n} u_{\tau} u_n-k u_{n}\frac{\partial u_{\tau}}{\partial n}-k u_{\tau}\frac{\partial u_n}{\partial n}\\ &&+\frac{\partial u_{n}}{\partial n}\frac{\partial u_{\tau}}{\partial n} + u_{n}\frac{\partial ^2 u_{\tau}}{\partial n^2}-\frac{\partial u_{n}}{\partial n} u\frac{\partial \tau}{\partial n}- u_n\frac{\partial u}{\partial n}\frac{\partial \tau}{\partial n}- u_{n} u\frac{\partial ^2 \tau}{\partial n^2}, \end{eqnarray*} $

根据 $u|_{\partial \Omega}=0$ 和 ${\rm{div}} u=0$, 可以得到在 $\Gamma$ 上有

$\begin{equation}\frac{\partial (( u \cdot \nabla) u\cdot\tau)}{\partial n}=0. \end{equation}$ (4.4)

因为

$\begin{eqnarray*} \frac{\partial }{\partial n}\frac{\partial p}{\partial \tau}&=&n_1\frac{\partial }{\partial x_1}\frac{\partial p}{\partial \tau} +n_2\frac{\partial }{\partial x_2}\frac{\partial p}{\partial \tau} \\ &=&n_1\frac{\partial ^2 p}{\partial x_1^2} \tau_1+n_1\frac{\partial p}{\partial x_1}\frac{\partial \tau_1}{\partial x_1}+n_1\tau_2\frac{\partial ^2 p}{\partial x_1\partial x_2}+n_1\frac{\partial p}{\partial x_2}\frac{\partial \tau_2}{\partial x_1} \\ &&+n_2\frac{\partial ^2 p}{\partial x_1\partial x_2}\tau_1+n_2\frac{\partial p}{\partial x_1}\frac{\partial \tau_1}{\partial x_2}+n_2\tau_2\frac{\partial ^2 p}{\partial x_2^2}+n_2\frac{\partial p}{\partial x_2}\frac{\partial \tau_2}{\partial x_2}, \end{eqnarray*} $

以及

$\begin{eqnarray*} \frac{\partial }{\partial \tau}\frac{\partial p}{\partial n}\\&=&\tau_1\frac{\partial }{\partial x_1}\frac{\partial p}{\partial n} +\tau_2\frac{\partial }{\partial x_2}\frac{\partial p}{\partial n} \\& =&\tau_1\frac{\partial ^2 p}{\partial x_1^2} n_1+\tau_1\frac{\partial p}{\partial x_1}\frac{\partial n_1}{\partial x_1}+\tau_1 n_2\frac{\partial ^2 p}{\partial x_1\partial x_2}+\tau_1\frac{\partial p}{\partial x_2}\frac{\partial n_2}{\partial x_1} \\& &+\tau_2\frac{\partial ^2 p}{\partial x_1\partial x_2}n_1+\tau_2\frac{\partial p}{\partial x_1}\frac{\partial n_1}{\partial x_2}+\tau_2n_2\frac{\partial ^2 p}{\partial x_2^2}+\tau_2\frac{\partial p}{\partial x_2}\frac{\partial n_2}{\partial x_2}, \end{eqnarray*} $

所以

$\begin{equation}\frac{\partial }{\partial n}\frac{\partial p}{\partial \tau}-\frac{\partial }{\partial \tau}\frac{\partial p}{\partial n}=\frac{\partial p}{\partial x_1}k\sqrt{1+{q'}^2}. \end{equation}$

显然

$\begin{eqnarray*} \frac{\partial p}{\partial n}\bigg|_{\Gamma}=(\gamma\triangle u+{\cal K}\mu \nabla \phi+g)\cdot n|_{\Gamma}. \end{eqnarray*} $

故在 $x\in \Gamma$ 上, 有

$\begin{equation} \frac{\partial }{\partial \tau}\frac{\partial p}{\partial n}=\frac{\partial (\gamma\triangle u\cdot n)}{\partial \tau}+\frac{\partial ({\cal K}\mu \nabla \phi\cdot n)}{\partial \tau}+\frac{\partial (g\cdot n)}{\partial \tau}. \end{equation}$

根据(4.3)-(4.6)式可以得到, 当 $x\in \Gamma$ 时

$\begin{matrix}\frac{\partial u_{\tau}(x,t)}{\partial n}-\frac{\partial \theta_{\tau}(x)}{\partial n}&=&\int_0^t\bigg[\frac{\partial (\gamma\triangle u\cdot\tau)}{\partial n}-\frac{\partial (\gamma\triangle u\cdot n)}{\partial \tau}-\frac{\partial p}{\partial x_1}k\sqrt{1+{q'}^2}\\&&+\frac{\partial ({\cal K}\mu \nabla \phi\cdot\tau)}{\partial n}-\frac{\partial ({\cal K}\mu \nabla \phi\cdot n)}{\partial \tau}+\frac{\partial (g\cdot\tau)}{\partial n}-\frac{\partial (g\cdot n)}{\partial \tau}\bigg]{\rm d}s.\end{matrix}$

根据(4.1)式有

$\begin{eqnarray*}\frac{\partial (\gamma\triangle u\cdot\tau)}{\partial n}-\frac{\partial (\gamma\triangle u\cdot n)}{\partial \tau}&=&\gamma\bigg(n_1\triangle u_1\frac{\partial \tau_1}{\partial x_1}+n_1\tau_2\frac{\partial \triangle u_2}{\partial x_1}+n_1\triangle u_2\frac{\partial \tau_2}{\partial x_1}+n_2\tau_1\frac{\partial \triangle u_1}{\partial x_2}\\&&-\tau_1\triangle u_1\frac{\partial n_1}{\partial x_1}-\tau_1n_2\frac{\partial \triangle u_2}{\partial x_1}-\tau_1\triangle u_2\frac{\partial n_2}{\partial x_1}-\tau_2n_1\frac{\partial \triangle u_1}{\partial x_2}\bigg),\end{eqnarray*}$

即

$\begin{equation}\frac{\partial (\gamma\triangle u\cdot\tau)}{\partial n}-\frac{\partial (\gamma\triangle u\cdot n)}{\partial \tau}=\gamma\left(\triangle u_1k\sqrt{1+{q'}^2}-\frac{\partial \triangle u_2}{\partial x_1}+\frac{\partial \triangle u_1}{\partial x_2}\right).\end{equation}$

同理

$\begin{equation}\frac{\partial ({\cal K}\mu \nabla \phi\cdot\tau)}{\partial n}-\frac{\partial ({\cal K}\mu \nabla \phi\cdot n)}{\partial \tau}={\cal K}\left(\mu\frac{\partial \phi}{\partial x_1}k\sqrt{1+{q'}^2}-\frac{\partial \mu}{\partial x_1}\frac{\partial \phi}{\partial x_2}+\frac{\partial \mu}{\partial x_2}\frac{\partial \phi}{\partial x_1}\right),\end{equation}$
$\begin{equation}\frac{\partial ( g\cdot\tau)}{\partial n}-\frac{\partial ( g\cdot n)}{\partial \tau}= g_1k\sqrt{1+{q'}^2}-\frac{\partial g_2}{\partial x_1}+\frac{\partial g_1}{\partial x_2}.\end{equation}$

因为

$\begin{array}{c}\frac{\partial p}{\partial \tau}=\tau_{1} \frac{\partial p}{\partial x_{1}}+\tau_{2} \frac{\partial p}{\partial x_{2}}, \\ \frac{\partial p}{\partial \tau}=\left(\gamma \Delta u_{1}+\mathcal{K} \mu \frac{\partial \phi}{\partial x_{1}}+g_{1}\right) \cdot \tau_{1}+\left(\gamma \triangle u_{2}+\mathcal{K} \mu \frac{\partial \phi}{\partial x_{2}}+g_{2}\right) \cdot \tau_{2} \text {, 在 } \Gamma \text { 上, }\end{array}$

所以在 $\Gamma$ 上可以得到

$\begin{equation}\frac{\partial p}{\partial x_1}=\gamma\triangle u_1 +{\cal K}\mu \frac{\partial \phi}{\partial x_1}+g_1.\end{equation}$

利用(4.7)-(4.11)式可以得到

$\begin{equation}\frac{\partial u_{\tau}(x,t)}{\partial n}-\frac{\partial \theta_{\tau}(x)}{\partial n} =\int_0^t\bigg(-\gamma\frac{\partial \triangle u_2}{\partial x_1}+\gamma\frac{\partial \triangle u_1}{\partial x_2}-{\cal K}\frac{\partial \mu}{\partial x_1}\frac{\partial \phi}{\partial x_2}+{\cal K}\frac{\partial \mu}{\partial x_2}\frac{\partial \phi}{\partial x_1}-\frac{\partial g_2}{\partial x_1}+\frac{\partial g_1}{\partial x_2}\bigg){\rm d}s. \end{equation}$

将 $u$ 和 $\phi$ 在 $t=0$ 进行 Taylor 展开可得

$\begin{eqnarray*} u=\theta+t\lambda+o(t),uad \phi=b+t\psi+o(t), \end{eqnarray*} $

将其代入(4.12)式可得

$\begin{eqnarray*} \frac{\partial u_{\tau}(x,t)}{\partial n}&=&\frac{\partial \theta_{\tau}(x)}{\partial n} +\int_0^t\bigg[-\gamma\frac{\partial \triangle (\theta_2+s \lambda_2)}{\partial x_1}+\gamma\frac{\partial \triangle ( \theta_1+s \lambda_1)}{\partial x_2}-\frac{\partial g_2}{\partial x_1}+\frac{\partial g_1}{\partial x_2} \\ &&-{\cal K}\frac{\partial (-\varepsilon\triangle(b+s\psi)+\alpha(b+s\psi)((b+s\psi)^2-1))}{\partial x_1}\frac{\partial (b+s\psi)}{\partial x_2} \\ &&+{\cal K}\frac{\partial (-\varepsilon\triangle(b+s\psi)+\alpha(b+s\psi)((b+s\psi)^2-1))}{\partial x_2}\frac{\partial (b+s\psi)}{\partial x_1}\bigg]{\rm d}s\\ &=&\frac{\partial \theta_{\tau}(x)}{\partial n} +t\bigg[-\gamma\frac{\partial \triangle \theta_2}{\partial x_1}+\gamma\frac{\partial \triangle\theta_1}{\partial x_2}-\frac{\partial g_2}{\partial x_1}+\frac{\partial g_1}{\partial x_2} -{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_2}\frac{\partial b}{\partial x_1}+{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_1}\frac{\partial b}{\partial x_2}\bigg]\\&&+o(t). \end{eqnarray*}$

因为 $u|_{\partial \Omega}=0$, 所以

$ -\frac{\partial \theta_{\tau}}{\partial n}\bigg|_{\Gamma}=\frac{\partial \theta_2}{\partial x_1}-\frac{\partial \theta_1}{\partial x_2}, $

故

$\begin{eqnarray*}\frac{\partial u_{\tau}(x,t)}{\partial n} &=&\frac{\partial \theta_2}{\partial x_1}-\frac{\partial \theta_1}{\partial x_2}\\ &&+t\bigg[-\gamma\frac{\partial \triangle \theta_2}{\partial x_1}+\gamma\frac{\partial \triangle\theta_1}{\partial x_2}-\frac{\partial g_2}{\partial x_1}+\frac{\partial g_1}{\partial x_2} -{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_2}\frac{\partial b}{\partial x_1} +{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_1}\frac{\partial b}{\partial x_2}\bigg]+o(t). \end{eqnarray*}$

显然, 若

$\begin{eqnarray*} t_1=\min_{\Gamma} \frac{\frac{\partial \theta_2}{\partial x_1}-\frac{\partial \theta_1}{\partial x_2}}{\gamma\frac{\partial \triangle \theta_1}{\partial x_2}-\gamma\frac{\partial \triangle \theta_2}{\partial x_1}+{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_1}\frac{\partial b}{\partial x_2}-{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_2}\frac{\partial b}{\partial x_1}+\frac{\partial g_1}{\partial x_2}-\frac{\partial g_2}{\partial x_1}}\ll1,number \end{eqnarray*} $

则

$\begin{eqnarray*} \left\{ \begin{array}{ll} \frac{\partial u_\tau(x,t)}{\partial n}\ne 0, uad &0<t<t_1,\\ [3mm] \frac{\partial u_\tau(x'_0,t)}{\partial n}=0, uad &t=t_1.\end{array} \right.number \end{eqnarray*}$

由于

$\frac{\partial ^2 u_\tau}{\partial t\partial n}\bigg|_{\Gamma}=\gamma\frac{\partial \triangle \theta_1}{\partial x_2}-\gamma\frac{\partial \triangle \theta_2}{\partial x_1}+{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_1}\frac{\partial b}{\partial x_2}-{\cal K}\varepsilon\frac{\partial \triangle b}{\partial x_2}\frac{\partial b}{\partial x_1}+\frac{\partial g_1}{\partial x_2}-\frac{\partial g_2}{\partial x_1}\ne 0, $

所以利用引理2.1可知系统(1.1)-(1.6)的解在$(t_1, x'_0)$ 处发生边界层分离. 证毕.

注4.1 由(4.2)式可知, 可以通过初值 $\theta$, $b$ 和外力 $g$ 判断 NSAC 系统是否发生边界层分离.

注4.2 若相对浓度 $\phi$ 为零, 则NSAC 系统变为 Navier-Stokes 方程, 而系统(1.1)-(1.6)边界层分离的条件(4.2)式变为

$ 0< \min_{\Gamma} \frac{\frac{\partial \theta_2}{\partial x_1}-\frac{\partial \theta_1}{\partial x_2}}{\gamma\frac{\partial \triangle \theta_1}{\partial x_2}-\gamma\frac{\partial \triangle \theta_2}{\partial x_1}+\frac{\partial g_1}{\partial x_2}-\frac{\partial g_2}{\partial x_1}}\ll1, $

与文献[20]的结果相吻合.

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