广东金融学院应用数学系，广东 广州

收稿日期：2022年10月23日；录用日期：2022年11月22日；发布日期：2022年11月30日

摘要

考虑了具有Soret效应的Brinkman方程组的解对方程系数 $\sigma$ 的连续依赖性。首先，运用微分不等式技术，得到温度和盐浓度的相关估计，尤其是获得了盐浓度的四阶范数估计；其次，利用先验估计，推导出能量函数所满足的微分不等式；最后，求解该不等式，建立了解对系数 $\sigma$ 的连续依赖性结果，该结果表明系数 $\sigma$ 的微小变化不会引起解的急剧变化，因此Brinkman方程组对Soret系数具有结构稳定性。

关键词

Brinkman方程组，连续依赖性，Brinkman系数，Soret系数

Structural Stability of a Class of Brinkman Equations with Soret Term

Jianshen Cheng

Department of Applied Mathematics, Guangdong University of Finance, Guangzhou Guangdong

Received: Oct. 23^rd, 2022; accepted: Nov. 22^nd, 2022; published: Nov. 30^th, 2022

ABSTRACT

The continuous dependence of the solution of Brinkman equations with Soret effect on equation coefficient $\sigma$ is considered. Firstly, the correlation estimates of temperature and salt concentration are obtained by using differential inequality technique. Especially, we can get the fourth-order norm estimates for the concentration of the salt. Secondly, the differential inequality satisfied by the energy function is derived by using a priori estimate. Finally, by solving the inequality, the continuous dependence of solution on coefficient $\sigma$ is established, the results show that the small change of coefficient $\sigma$ will not cause the sharp change of solution, so Brinkman equations have structural stability for Soret coefficients.

Keywords:Brinkman Equations, Continuous Dependence, Brinkman Coefficient, Soret Coefficient

Copyright © 2022 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

Nield [1] 和Straughan [2] 详细介质了多孔介质中的具有Soret效应的Brinkman方程，它具有双扩散效应，而且采用了Boussinesq逼近。Straughan等 [2] 在有界区域内，建立了方程组的解对Soret系数的连续依赖性，其结果仅表明Soret系数微小的变化，不会引起解的急剧变化，此时方程组是稳定的，这类稳定性的研究称之为结构稳定性研究。结构稳定性研究主要是考察模型自身的微小变化对模型解的影响，而传统的稳定性主要是考虑初始数据的变化对解的影响，关于结构稳定性系统的介绍见文献 [3] [4]。

近年来，人们越来越关注多孔介质中Brinkman，Darcy和Forchheimer流体方程组解的性态研究，它已经成为是数学与物理交叉学科领域的热点问题。Payne等 [5] 研究了Darcy方程组的空间衰减性。文献 [6] - [18] 研究了这三类方程组的结构稳定性，取得了新的研究成果。盐浓度方程若有含Soret项，导致处理盐浓度的最大值难度大，从而使得研究具有Soret效应的方程组解的结构稳定性较少，本文考虑如下具有Soret效应非齐次边界条件下的Brinkman方程组：

$\{\begin{array}{l}\frac{\partial u_i}{\partial t}=\nu \Delta u_i-u_i-p_i+g_{i}T+h_{i}C,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial u_i}{\partial x_i}=0,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial T}{\partial t}+u_i\frac{\partial T}{\partial x_i}=\Delta T,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial C}{\partial t}+u_i\frac{\partial C}{\partial x_i}=\Delta C+\sigma \Delta T,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\end{array}$ (1)

其中 $u_i$，p，T，C分别为速度，压强，温度和盐浓度。 $g_i\left(x\right)$ 和 $h_i\left(x\right)$ 为重力函数且 $\left|g_i\right|,\left|h_i\right|\le 1$，$\Delta$ 为Laplace算子。 $\sigma$，$\nu$ 分别为Soret和Brinkman系数。在方程组(1)中， $\nu$，$\sigma$ 都是大于零的常数。

方程组(1)在 $\Omega \times \left[0,\tau \right]$ 区域内成立，其中 $\Omega$ 是 $R^3$ 中一个有界连通的凸区域， $\tau$ 是给定的常数且 $0\le \tau <\infty$。我们考虑绝缘材料有损坏，导致流体与外界有热交换，溶质通过边界的通量为零情形，相应的边界条件为

$u_i=0,\frac{\partial T}{\partial n}+k_{1}T=f_1\left(x,t\right),\frac{\partial C}{\partial n}+k_{2}C=f_2\left(x,t\right),\left(x,t\right)\in \partial \Omega \times \left[0,\tau \right]$ (2)

初始条件为

$u_i\left(x,0\right)=u_{i0}\left(x\right),T\left(x,0\right)=T_0\left(x\right),C\left(x,0\right)=C_0\left(x\right),x\in \Omega$ (3)

本文研究了方程组(1)的解对方程系数 $\sigma$ 的连续依赖性。为了得到连续依赖的结果，通常的做法是需要利用温度与盐浓度的最大值估计，去处理交叉项。由于方程组(1)含有Soret项 $\sigma \text{Δ}T$，导致盐浓度的最大值估计难度很大，为了克服这个难题，我们采用给出盐浓度的四阶范数估计。如何构造合适的函数进行盐浓度的四阶范数估计是本文的最大创新之处。

本文用逗号表示求偏导， $i$ 表示对 $x_i$ 求偏导，如： $u_{,i}$ 表示为 $\frac{\partial u}{\partial x_i}$，重复指标表示求和， $u_{i,i}={\displaystyle {\sum }_{i=1}^3\frac{\partial u_i}{\partial x_i}}$，$u_{i,j}{u}_{i,j}={\displaystyle {\sum }_{i,j=1}^3{\left(\frac{\partial u_i}{\partial x_j}\right)}^2}$。

2. 一些有用的估计

为了推导出本文的主要结果，首先给出一些引理。

引理1温度T满足以下最大值估计

${\text{Sup}}_{\left[0,\tau \right]}{\Vert T\Vert }_{\infty }\le T_M,$ (4)

其中 $T_M=\max \left\{{\Vert T_0\Vert }_{\infty },\frac{1}{k_1}\underset{\left[0,\tau \right]}{\text{Sup}}{f}_1{}_{{}_{\infty }}\right\}$。

证明 在方程组(1)第三个方程两边同时乘以 $2r{T}^{2r-1}\left(r\ge 1\right)$，并且在 $\Omega \times \left[0,t\right]\left(t\in \left[0,\tau \right]\right)$ 上积分，可得

$2r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}^{2r-1}{T}_{,\eta }\text{d}x\text{d}\eta }}+2r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}^{2r-1}{u}_i{T}_{,i}\text{d}x\text{d}\eta }}=2r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}^{2r-1}\text{Δ}T\text{d}x\text{d}\eta }}$ . (5)

由散度定理，Hölder不等式和Young不等式，可得

$2r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}^{2r-1}{u}_i{T}_{,i}\text{d}x\text{d}\eta }}={\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\left(T^{2r}\right)}_{,i}{u}_i\text{d}x\text{d}\eta }}=0.$ (6)

$\begin{array}{l}2r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}^{2r-1}\text{Δ}T\text{d}x\text{d}\eta }}\\ =2r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{T}^{2r-1}\frac{\partial T}{\partial n}\text{d}S\text{d}\eta }}-\frac{2\left(2r-1\right)}{r}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\left(T^r\right)}_{,i}{\left(T^r\right)}_{,i}\text{d}x\text{d}\eta }}\\ =2r{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1{T}^{2r-1}\text{d}S\text{d}\eta }}-2r{k}_1{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{T}^{2r}\text{d}S\text{d}\eta }}-\frac{2\left(2r-1\right)}{r}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{\left(T^r\right)}_{,i}{\left(T^r\right)}_{,i}\text{d}x\text{d}\eta }}\\ \le {\left(\frac{2r-1}{2r{k}_1}\right)}^{2r-1}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1{}^{2r}\text{d}S\text{d}\eta }}.\end{array}$ (7)

联合式(5)~式(7)，可得

${\displaystyle {\int }_{\Omega }{T}^{2r}\text{d}x}\le {\displaystyle {\int }_{\Omega }{T}_0^{2r}\text{d}x}+{\left(\frac{2r-1}{2r{k}_1}\right)}^{2r-1}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1{}^{2r}\text{d}S\text{d}\eta }}\text{.}$ (8)

式(8)式从0到t积分，有

${\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}^{2r}\text{d}x\text{d}\eta }}\right)}^{\frac{1}{2r}}\le {\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}_0^{2r}\text{d}x\text{d}\eta }}\right)}^{\frac{1}{2r}}+{\left(\frac{2r-1}{2r{k}_1}\right)}^{\frac{2r-1}{2r}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_0^{\xi }{\displaystyle {\int }_{\partial \Omega }{f}_1{}^{2r}\text{d}S\text{d}\xi \text{d}\eta }}}\right)}^{\frac{1}{2r}}\text{.}$ (9)

在(9)式中，当 $r\to +\infty$，有

${\text{Sup}}_{\left[0,\tau \right]}{\Vert T\Vert }_{\infty }\le \max \left\{{\Vert T_0\Vert }_{\infty },\frac{1}{k_1}\underset{\left[0,\tau \right]}{\text{Sup}}{f}_1{}_{{}_{\infty }}\right\}=T_M.$

证毕！

引理2对于温度T，有如下估计

${\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{T}^4\text{d}s\text{d}\eta }}\le m_1\left(t\right),$ (10)

${\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{T}^2\text{d}s\text{d}\eta }}\le m_2\left(t\right),$ (11)

其中 $m_1\left(t\right)=\frac{1}{2k_1^2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^4\text{d}s\text{d}\eta }}+\frac{1}{2k_1}{\displaystyle {\int }_{\partial \Omega }{T}_0^4\text{d}s}$，$m_2\left(t\right)=\frac{1}{2k_1^2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s\text{d}\eta }}+\frac{1}{2k_1}{\displaystyle {\int }_{\partial \Omega }{T}_0^2\text{d}s}$。

证明 在方程组(1)第三个方程两边同时乘以 $4T^3$，并在 $\Omega$ 上积分得

$\begin{array}{c}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{T}^4\text{d}x}\le 4{\displaystyle {\int }_{\Omega }{T}^3\left(\text{Δ}T-u_i{T}_{,i}\right)\text{d}x\text{d}\eta }\\ =-12{\displaystyle {\int }_{\Omega }{T}^2{\left|\nabla T\right|}^2\text{d}x\text{d}\eta }-4k_1{\displaystyle {\int }_{\partial \Omega }{T}^4\text{d}s}+4{\displaystyle {\int }_{\partial \Omega }{T}^3{f}_1\text{d}s}.\end{array}$ (12)

式(12)从0到 积分，并由Young不等式，可得

${\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{T}^4\text{d}s\text{d}\eta }}\le \frac{1}{2k_1^2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^4\text{d}s\text{d}\eta }}+\frac{1}{2k_1}{\displaystyle {\int }_{\partial \Omega }{T}_0^4\text{d}s}.$

同理可知

${\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{T}^2\text{d}s\text{d}\eta }}\le \frac{1}{2k_1^2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s\text{d}\eta }}+\frac{1}{2k_1}{\displaystyle {\int }_{\partial \Omega }{T}_0^2\text{d}s}.$

证毕！

引理3对于温度T和盐浓度C，有如下估计

${\displaystyle {\int }_{\Omega }{T}^2\text{d}x}+2{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x\text{d}\eta }}\le m_3\left(t\right),$ (13)

${\displaystyle {\int }_{\Omega }{C}^2\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{C}_{,i}{C}_{,i}\text{d}x\text{d}\eta }}\le m_4\left(t\right),$ (14)

其中

$m_3\left(t\right)={\displaystyle {\int }_{\Omega }{T}_0^2\text{d}x}+\frac{1}{2k_1}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s\text{d}\eta }},$

$m_4\left(t\right)=\frac{{\sigma }^2}{2}{m}_3\left(t\right)+{\displaystyle {\int }_{\Omega }{C}_0^2\text{d}x}+\frac{1}{k_2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_2^2\text{d}s\text{d}\eta }}+\frac{2{\sigma }^2}{k_2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s\text{d}\eta }}+\frac{2{\sigma }^2{k}_1^2}{k_2}{m}_2\left(t\right).$

证明 在方程组(1)第三个方程两边乘以2T，并在 $\Omega \times \left[0,t\right]$ 上积分，由Young不等式，可得

${\displaystyle {\int }_{\Omega }{T}^2\text{d}x}+2{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x\text{d}\eta }}\le {\displaystyle {\int }_{\Omega }{T}_0^2\text{d}x}+\frac{1}{2k_1}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s\text{d}\eta }}.$

在方程组(1)第四个方程两边乘以2T，并在 $\Omega \times \left[0,t\right]$ 上积分得

$\begin{array}{c}{\displaystyle {\int }_{\Omega }{C}^2\text{d}x}+2{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{C}_{,i}{C}_{,i}\text{d}x\text{d}\eta }}\le {\displaystyle {\int }_{\Omega }{C}_0^2\text{d}x}-2\sigma {\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{C}_{,i}{T}_{,i}\text{d}x\text{d}\eta }}\\ +2{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }C\left(f_2-k_{2}C\right)\text{d}s\text{d}\eta }}\\ +2{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }C\left(f_1-k_{1}T\right)\text{d}s\text{d}\eta }}.\end{array}$ (15)

式(15)运用Schwarz不等式，可知

$\begin{array}{c}{\displaystyle {\int }_{\Omega }{C}^2\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{C}_{,i}{C}_{,i}\text{d}x\text{d}\eta }}\le {\displaystyle {\int }_{\Omega }{C}_0^2\text{d}x}+{\sigma }^2{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x\text{d}\eta }}+\frac{1}{k_2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_2^2\text{d}s\text{d}\eta }}\\ +\frac{2{\sigma }^2}{k_2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s\text{d}\eta }}+\frac{2{\sigma }^2{k}_1^2}{k_2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{T}^2\text{d}s\text{d}\eta }}.\end{array}$ (16)

将式(13)和式(4)代入式(16)，可得

$\begin{array}{c}{\displaystyle {\int }_{\Omega }{C}^2\text{d}x}+{\displaystyle {\int }_0^t{\displaystyle {\int }_{\Omega }{C}_{,i}{C}_{,i}\text{d}x\text{d}\eta }}\le \frac{{\sigma }^2}{2}{m}_3\left(t\right)+{\displaystyle {\int }_{\Omega }{C}_0^2\text{d}x}+\frac{1}{k_2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_2^2\text{d}s\text{d}\eta }}\\ +\frac{2{\sigma }^2}{k_2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s\text{d}\eta }}+\frac{2{\sigma }^2{k}_1^2}{k_2}{m}_2\left(t\right).\end{array}$

证毕！

引理4对于盐浓度C，有如下的4阶范数估计

${\displaystyle {\int }_{\Omega }{C}^4\text{d}x}\le m_5\left(t\right),$ (17)

其中 $m_5\left(t\right)=n_1{\displaystyle {\int }_{\partial \Omega }{f}_1^4\text{d}s}+n_2{\displaystyle {\int }_{\partial \Omega }{f}_2^4\text{d}s}+n_4{m}_4\left(t\right)+\frac{n_5}{2}{m}_3\left(t\right)+n_3{m}_1\left(t\right)$，$n_1,n_2,n_3,n_4,n_5$ 为大于零的常数。

证明 在方程组(1)第四个方程两边乘以 $C^3$，并在 $\Omega$ 上积分得

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{C}^4\text{d}x}=4{\displaystyle {\int }_{\Omega }{C}^3\left(\Delta C+\sigma \text{Δ}T-u_i{C}_{,i}\right)\text{d}x}\\ =-12{\displaystyle {\int }_{\Omega }{C}^2{\left|\nabla C\right|}^2\text{d}x}-12\sigma {\displaystyle {\int }_{\Omega }{C}^2{C}_{,i}{T}_{,i}\text{d}x}+{\displaystyle {\int }_{\partial \Omega }{C}^3\left(f_2-k_{2}C\right)\text{d}s}+\sigma {\displaystyle {\int }_{\partial \Omega }{C}^3\left(f_1-k_{1}T\right)\text{d}s}\\ =-12{\displaystyle {\int }_{\Omega }{C}^2{\left|\nabla C\right|}^2\text{d}x}-12\sigma {\displaystyle {\int }_{\Omega }C{C}_{,i}\left(C{T}_{,i}+T{C}_{,i}\right)\text{d}x}+12\sigma {\displaystyle {\int }_{\Omega }CT{\left|\nabla C\right|}^2\text{d}x}\\ -4k_2{\displaystyle {\int }_{\partial \Omega }{C}^4\text{d}s}+4{\displaystyle {\int }_{\partial \Omega }{C}^3{f}_2\text{d}s}-4\sigma k_1{\displaystyle {\int }_{\partial \Omega }{C}^{3}T\text{d}s}+4\sigma k_1{\displaystyle {\int }_{\partial \Omega }{C}^3{f}_1\text{d}s}\end{array}$

$\begin{array}{l}\le -12{\displaystyle {\int }_{\Omega }{C}^2{\left|\nabla C\right|}^2\text{d}x}+6\sigma {\lambda }_1{\displaystyle {\int }_{\Omega }}{C}^2{\left|\nabla C\right|}^2\text{d}x+\frac{6\sigma }{{\lambda }_1}{\displaystyle {\int }_{\Omega }\left(C{T}_{,i}+T{C}_{,i}\right)\left(C{T}_{,i}+T{C}_{,i}\right)\text{d}x}\\ +6\sigma {\lambda }_2{\displaystyle {\int }_{\Omega }{C}^2{\left|\nabla C\right|}^2\text{d}x}+\frac{6\sigma T_M^2}{{\lambda }_2}{\displaystyle {\int }_{\Omega }{\left|\nabla C\right|}^2\text{d}x}-k_2{\displaystyle {\int }_{\partial \Omega }{C}^4\text{d}s}\\ +\frac{27}{k_2^3}{\displaystyle {\int }_{\partial \Omega }{f}_2^4\text{d}s}+\frac{27{\sigma }^4{k}_1^4}{k_2^3}{\displaystyle {\int }_{\partial \Omega }{T}^4\text{d}s}+\frac{27{\sigma }^4{k}_1^4}{k_2^3}{\displaystyle {\int }_{\partial \Omega }{f}_1^4\text{d}s},\end{array}$ (18)

其中 ${\lambda }_1$，${\lambda }_2$ 是大于零的任意常数。

运用方程组(1)第三个和第四个方程以及Hölder不等式，可得

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{T}^2{C}^2\text{d}x}=2{\displaystyle {\int }_{\Omega }T{C}^2{T}_{,t}\text{d}x}+2{\displaystyle {\int }_{\Omega }C{T}^2{C}_{,t}\text{d}x}\\ =2{\displaystyle {\int }_{\Omega }T{C}^2\left(\Delta T-u_i{T}_{,i}\right)\text{d}x}+2{\displaystyle {\int }_{\Omega }C{T}^2\left(\Delta C+\sigma \text{Δ}T-u_i{C}_{,i}\right)\text{d}x}\\ =-2{\displaystyle {\int }_{\Omega }\left(C{T}_{,i}+T{C}_{,i}\right)\left(C{T}_{,i}+T{C}_{,i}\right)\text{d}x}-4{\displaystyle {\int }_{\Omega }T{T}_{,i}C{C}_{,i}\text{d}x}\\ -4\sigma {\displaystyle {\int }_{\Omega }TC{\left|\nabla T\right|}^2\text{d}x}-2\sigma {\displaystyle {\int }_{\Omega }{T}^2{T}_{,i}{C}_{,i}\text{d}x}+2{\displaystyle {\int }_{\partial \Omega }T{C}^2\left(f_1-k_{1}T\right)\text{d}s}\\ +2{\displaystyle {\int }_{\partial \Omega }C{T}^2\left(f_2-k_{2}C\right)\text{d}s}+2\sigma {\displaystyle {\int }_{\partial \Omega }C{T}^2\left(f_1-k_{1}T\right)\text{d}s}\end{array}$

$\begin{array}{l}=-2{\displaystyle {\int }_{\Omega }\left(C{T}_{,i}+T{C}_{,i}\right)\left(C{T}_{,i}+T{C}_{,i}\right)\text{d}x}-4{\displaystyle {\int }_{\Omega }T{T}_{,i}C{C}_{,i}\text{d}x}\\ +4\sigma {\displaystyle {\int }_{\Omega }T{T}_{,i}\left(C{T}_{,i}+T{C}_{,i}\right)\text{d}x}+2\sigma {\displaystyle {\int }_{\Omega }{T}^2{T}_{,i}{C}_{,i}\text{d}x}\\ -2\left(k_1+k_2\right){\displaystyle {\int }_{\partial \Omega }{T}^2{C}^2\text{d}s}+2{\displaystyle {\int }_{\partial \Omega }T{C}^2{f}_1\text{d}s}+2{\displaystyle {\int }_{\partial \Omega }C{T}^2{f}_2\text{d}s}\\ +2\sigma {\displaystyle {\int }_{\partial \Omega }C{T}^2{f}_1\text{d}s}-2\sigma k_1{\displaystyle {\int }_{\partial \Omega }C{T}^3\text{d}s}\end{array}$

$\begin{array}{l}\le -2{\displaystyle {\int }_{\Omega }\left(C{T}_{,i}+T{C}_{,i}\right)\left(C{T}_{,i}+T{C}_{,i}\right)\text{d}x}+2{\lambda }_3{\displaystyle {\int }_{\Omega }{C}^2{\left|\nabla C\right|}^2\text{d}x}\\ +\frac{2T_M^2}{{\lambda }_3}{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x}+2\sigma {\lambda }_4{\displaystyle {\int }_{\Omega }\left(C{T}_{,i}+T{C}_{,i}\right)\left(C{T}_{,i}+T{C}_{,i}\right)\text{d}x}\\ +\frac{T_M^2}{2{\lambda }_4}{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x}+\sigma T_M^2{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x}+\sigma T_M^2{\displaystyle {\int }_{\Omega }{C}_{,i}{C}_{,i}\text{d}x}\\ +\frac{k_2}{k}{\displaystyle {\int }_{\partial \Omega }{C}^4\text{d}s}+\left(\frac{k}{4k_1^2}+\frac{{\sigma }^2}{2k_2}\right){\displaystyle {\int }_{\partial \Omega }{f}_1^4\text{d}s}+\left(\frac{1}{2}+\frac{{\sigma }^2}{2k_2}+\frac{{\sigma }^2{k}_1^2}{k_2^2}\right){\displaystyle {\int }_{\partial \Omega }{T}^4\text{d}s}+\frac{1}{2}{\displaystyle {\int }_{\partial \Omega }{f}_2^4\text{d}s},\end{array}$ (19)

其中k， ${\lambda }_3$，${\lambda }_4$ 是大于零的任意常数。

联合式(18)和式(19)得

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{C}^4\text{d}x}+k\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{T}^2{C}^2\text{d}x}\\ \le -\left(12-6\sigma {\lambda }_1-2k{\lambda }_3-6\sigma {\lambda }_2\right){\displaystyle {\int }_{\Omega }{C}^2{\left|\nabla C\right|}^2\text{d}x}\\ -\left(2k-\frac{6\sigma }{{\lambda }_1}-2k\sigma {\lambda }_4\right){\displaystyle {\int }_{\Omega }\left(C{T}_{,i}+T{C}_{,i}\right)\left(C{T}_{,i}+T{C}_{,i}\right)\text{d}x}\\ +\left(\frac{6\sigma T_M^2}{{\lambda }_2}+k\sigma T_M^2\right){\displaystyle {\int }_{\Omega }{C}_{,i}{C}_{,i}\text{d}x}+\left(\frac{2k{T}_M^2}{{\lambda }_3}+\frac{k{T}_M^2}{2{\lambda }_4}+k\sigma T_M^2\right){\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x}\\ +n_1{\displaystyle {\int }_{\partial \Omega }{f}_1^4\text{d}s}+n_2{\displaystyle {\int }_{\partial \Omega }{f}_2^4\text{d}s}+n_3{\displaystyle {\int }_{\partial \Omega }{T}^4\text{d}s},\end{array}$ (20)

其中 $n_1=\frac{27{\sigma }^4{k}_1^4}{k_2^3}+\left(\frac{k}{4k_1^2}+\frac{{\sigma }^2}{2k_2}\right)k$，$n_2=\frac{k}{2}+\frac{27}{k_2^3}$，$n_3=\left(\frac{1}{2}+\frac{{\sigma }^2}{2k_2}+\frac{{\sigma }^2{k}_1^2}{k_2^2}\right)k+\frac{27{\sigma }^4{k}_1^4}{k_2^3}$。

式(20)中，取 ${\lambda }_1={\lambda }_2=\frac{1}{3\sigma }$，${\lambda }_3=\frac{1}{18{\sigma }^2}$，${\lambda }_4=\frac{1}{2\sigma }$，$k=18{\sigma }^2$，可得

$\begin{array}{l}\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{C}^4\text{d}x}+k\frac{\text{d}}{\text{d}t}{\displaystyle {\int }_{\Omega }{T}^2{C}^2\text{d}x}\\ \le n_4{\displaystyle {\int }_{\Omega }{C}_{,i}{C}_{,i}\text{d}x}+n_5{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x}+n_1{\displaystyle {\int }_{\partial \Omega }{f}_1^4\text{d}s}+n_2{\displaystyle {\int }_{\partial \Omega }{f}_2^4\text{d}s}+n_3{\displaystyle {\int }_{\partial \Omega }{T}^4\text{d}s},\end{array}$ (21)

其中 $n_4=18{\sigma }^2{T}_M^2\left(1+\sigma \right)$，$n_5=18{\sigma }^3{T}_M^2\left(1+36\sigma \right)+183\sigma T_M^2$。

积分式(21)并将式(10)，式(14)和(15)的结果代入，可得

${\displaystyle {\int }_{\Omega }{C}^4\text{d}x}\le m_5\left(t\right).$

证毕！

3. 解对Soret系数 $\sigma$ 连续依赖性

在本节我们将建立方程组的解对Soret系数 $\sigma$ 连续依赖性。

假设 $\left(u_i,T,C,p\right)$ 是如下Brinkman方程组的解

$\{\begin{array}{l}\frac{\partial u_i}{\partial t}=\nu \Delta u_i-u_i-p_{,i}+g_{i}T+h_{i}C,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial u_i}{\partial x_i}=0,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial T}{\partial t}+u_i\frac{\partial T}{\partial x_1}=\Delta T,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial C}{\partial t}+u_i\frac{\partial C}{\partial x_1}=\Delta C+{\sigma }_1\Delta T,\left(x,t\right)\in \Omega \times \left[0,\tau \right]\end{array}$ (22)

边界条件为

$u_i=0,\frac{\partial T}{\partial n}+k_{1}T=f_1\left(x,t\right),\frac{\partial C}{\partial n}+k_{2}C=f_2\left(x,t\right),\left(x,t\right)\in \partial \Omega \times \left[0,\tau \right]$ (23)

初始条件为

$u_i\left(x,0\right)=u_{i0}\left(x\right),T\left(x,0\right)=T_0\left(x\right),C\left(x,0\right)=C_0\left(x\right),x\in \Omega .$ (24)

此外，假设 $\left(u_i^{*},T^{*},C^{*},p^{*}\right)$ 也是如下Brinkman方程组的解

$\{\begin{array}{l}\frac{\partial u_i^{*}}{\partial t}=\nu \Delta u_i^{*}-u_i^{*}-p_{,i}^{*}+g_i{T}^{*}+h_i{C}^{*},\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial u_i^{*}}{\partial x_i}=0,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial T^{*}}{\partial t}+u_i^{*}\frac{\partial T^{*}}{\partial x_i}=\Delta T^{*},\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial C^{*}}{\partial t}+u_i^{*}\frac{\partial C^{*}}{\partial x_i}=\Delta C^{*}+{\sigma }_2\Delta T^{*},\left(x,t\right)\in \Omega \times \left[0,\tau \right].\end{array}$ (25)

边界条件为

$u_i^{*}=0,\frac{\partial T^{*}}{\partial n}+k_1{T}^{*}=f_1\left(x,t\right),\frac{\partial C^{*}}{\partial n}+k_2{C}^{*}=f_2\left(x,t\right),\left(x,t\right)\in \partial \Omega \times \left[0,\tau \right].$ (26)

初始条件为

$u_i^{*}\left(x,0\right)=u_{i0}\left(x\right),T^{*}\left(x,0\right)=T_0\left(x\right),C^{*}\left(x,0\right)=C_0\left(x\right),x\in \Omega .$ (27)

定义解的差为： ${\omega }_i=u_i-u_i^{*}$，$\theta =T-T^{*}$，$S=C-C^{*}$，$\pi =p-p^{*}$，$\sigma ={\sigma }_1-{\sigma }_2$ 则 $\left({\omega }_i,\theta ,S,\pi \right)$ 满足如下初边值问题

$\{\begin{array}{l}\frac{\partial {\omega }_i}{\partial t}=\nu \Delta {\omega }_i-{\omega }_i-{\pi }_{,i}+g_i\theta +h_{i}S,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial {\omega }_i}{\partial x_i}=0,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial \theta }{\partial t}+{\omega }_i{T}_{,i}+u_i^{*}{\theta }_{,i}=\Delta \theta ,\left(x,t\right)\in \Omega \times \left[0,\tau \right],\\ \frac{\partial S}{\partial t}+{\omega }_i{C}_{,i}+u_i^{*}{S}_{,i}=\Delta S+\sigma \Delta T+{\sigma }_2\Delta \theta ,\left(x,t\right)\in \Omega \times \left[0,\tau \right].\end{array}$ (28)

边界条件为

${\omega }_i=0,\frac{\partial \theta }{\partial n}+k_1\theta =0,\frac{\partial S}{\partial n}+k_{2}S=0,\left(x,t\right)\in \partial \Omega \times \left[0,\tau \right].$ (29)

初始条件为

${\omega }_i\left(x,0\right)=0,\theta \left(x,0\right)=0,S\left(x,0\right)=0,x\in \Omega .$ (30)

定理1假设 $\left(u_i,T,C,p\right)$ 是式(22)~式(24)初边值问题的古典解， $\left(u_i^{*},T^{*},C^{*},p^{*}\right)$ 是式(25)~式(27)初边值问题的古典解， $\left({\omega }_i,\theta ,S,\pi \right)$ 为这两个解的差，则当方程系数 $\sigma$ 趋于 时，解 $\left(u_i,T,C,p\right)$ 收敛于解 $\left(u_i^{*},T^{*},C^{*},p^{*}\right)$，且有下列不等式成立

${\epsilon }_1{\Vert \omega \Vert }^2+{\epsilon }_2{\Vert \theta \Vert }^2+{\Vert S\Vert }^2\le {\sigma }^2{\text{e}}^{{\epsilon }_{3}t}{m}_6\left(t\right),$ (31)

其中 ${\epsilon }_1,{\epsilon }_2,{\epsilon }_3$ 是大于零的常数， $m_6\left(t\right)=\frac{1}{k_2}{\displaystyle {\int }_0^t{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s\text{d}\eta }}+m_2\left(t\right)+m_3\left(t\right)$。

证明 在方程组(28)第一个方程两边乘以 ${\omega }_i$，并在 $\Omega$ 上积分得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \omega \Vert }^2+\nu {\displaystyle {\int }_{\Omega }{\omega }_{i,j}{\omega }_{i,j}\text{d}x}=-{\displaystyle {\int }_{\Omega }{\omega }_i{\omega }_i\text{d}x}+{\displaystyle {\int }_{\Omega }{\omega }_i{g}_i\theta \text{d}x}+{\displaystyle {\int }_{\Omega }{\omega }_i{h}_{i}S\text{d}x}.$

上式运用Schwarz不等式得

$\frac{\text{d}}{\text{d}t}{\omega }^2+2\nu {\displaystyle {\int }_{\Omega }{\omega }_{i,j}{\omega }_{i,j}\text{d}x}\le {\displaystyle {\int }_{\Omega }{\theta }^2\text{d}x}+{\displaystyle {\int }_{\Omega }{S}^2\text{d}x}$ (32)

在方程组(28)第三个方程两边同时乘以 $\theta$，并在 $\Omega$ 上积分得

$\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert \theta \Vert }^2+{\displaystyle {\int }_{\Omega }{\theta }_{,i}{\theta }_{,i}\text{d}x}=-{\displaystyle {\int }_{\Omega }{\omega }_i{T}_{,i}\theta \text{d}x}-k_1{\displaystyle {\int }_{\partial \Omega }{\theta }^2\text{d}s}={\displaystyle {\int }_{\Omega }{\omega }_{i}T{\theta }_{,i}\text{d}x}-k_1{\displaystyle {\int }_{\partial \Omega }{\theta }^2\text{d}s}.$

上式运用Schwarz不等式得

$\frac{\text{d}}{\text{d}t}{\Vert \theta \Vert }^2+2k_1{\displaystyle {\int }_{\partial \Omega }{\theta }^2\text{d}x}+{\displaystyle {\int }_{\Omega }{\theta }_{,i}{\theta }_{,i}\text{d}x}\le T_M^2{\Vert \omega \Vert }^2.$ (33)

在方程组(28)第四个方程两边同时乘以S，并在 $\Omega$ 上积分得

$\begin{array}{l}\frac{1}{2}\frac{\text{d}}{\text{d}t}{\Vert S\Vert }^2+{\displaystyle {\int }_{\Omega }{S}_{,i}{S}_{,i}\text{d}x}+k_2{\displaystyle {\int }_{\partial \Omega }{S}^2\text{d}s}\\ ={\displaystyle {\int }_{\Omega }{\omega }_{i}C{S}_{,i}\text{d}x}+\sigma {\displaystyle {\int }_{\partial \Omega }S{f}_1\text{d}s}-\sigma k_1{\displaystyle {\int }_{\partial \Omega }ST\text{d}s}-\sigma {\displaystyle {\int }_{\Omega }{S}_{,i}{T}_{,i}\text{d}x}\\ -{\sigma }_2{k}_1{\displaystyle {\int }_{\partial \Omega }S\theta \text{d}s}-{\sigma }_2{\displaystyle {\int }_{\Omega }{S}_{,i}{\theta }_{,i}\text{d}x}.\end{array}$

上式运用Young和Hölder不等式得

$\begin{array}{c}\frac{\text{d}}{\text{d}t}{\Vert S\Vert }^2\le {\left({\displaystyle {\int }_{\Omega }{\left({\omega }_i{\omega }_i\right)}^2\text{d}x}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_{\Omega }{C}^4\text{d}x}\right)}^{\frac{1}{2}}+\frac{{\sigma }^2}{k_2}{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s}+\frac{2{\sigma }^2{k}_1^2}{k_2}{\displaystyle {\int }_{\partial \Omega }{T}^2\text{d}s}\\ +2{\sigma }^2{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x}+\frac{2{\sigma }_2^2{k}_1^2}{k_2}{\displaystyle {\int }_{\partial \Omega }{\theta }^2\text{d}s}+2{\sigma }_2^2{\displaystyle {\int }_{\Omega }{\theta }_{,i}{\theta }_{,i}\text{d}x}.\end{array}$ (34)

对于满足在边界上为零的函数G，由 [17] 的结论，有如下Sobolev不等式成立

${\displaystyle {\int }_{\Omega }{\left|G\right|}^4\text{d}x}\le c_1{\left({\displaystyle {\int }_{\Omega }{\left|G\right|}^2\text{d}x}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_{\Omega }{G}_{i,j}{G}_{i,j}\text{d}x}\right)}^{\frac{3}{2}}\le c_2{\left({\displaystyle {\int }_{\Omega }{G}_{i,j}{G}_{i,j}\text{d}x}\right)}^2,$ (35)

其中 $c_1$，$c_2$ 是大于零的常数。

在式(35)中，取 $G={\omega }_i$，可知

${\displaystyle {\int }_{\Omega }{\left({\omega }_i{\omega }_i\right)}^2\text{d}x}\le c_2{\left({\displaystyle {\int }_{\Omega }{\omega }_{i,j}{\omega }_{i,j}\text{d}x}\right)}^2.$ (36)

联合式(34)和式(36)得

$\begin{array}{c}\frac{\text{d}}{\text{d}t}{S}^2\le \sqrt{c_2{m}_5\left(\tau \right)}{\displaystyle {\int }_{\Omega }{\omega }_{i,j}{\omega }_{i,j}\text{d}x}+\frac{{\sigma }^2}{k_2}{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s}+\frac{2{\sigma }^2{k}_1^2}{k_2}{\displaystyle {\int }_{\partial \Omega }{T}^2\text{d}s}\\ +2{\sigma }^2{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x}+\frac{2{\sigma }_2^2{k}_1^2}{k_2}{\displaystyle {\int }_{\partial \Omega }{\theta }^2\text{d}s}+2{\sigma }_2^2{\displaystyle {\int }_{\Omega }{\theta }_{,i}{\theta }_{,i}\text{d}x}.\end{array}$ (37)

定义

$F\left(t\right)={\epsilon }_1{\Vert \omega \Vert }^2+{\epsilon }_2{\Vert \theta \Vert }^2+{\Vert S\Vert }^2,$

其中 ${\epsilon }_1=\frac{\sqrt{c_2{m}_5\left(\tau \right)}}{\nu }$，${\epsilon }_2=\max \left\{\frac{{\sigma }_2^2{k}_1}{k_2},2{\sigma }_2^2\right\}$。

联合式(32)，(33)和(37)，可得

$\frac{\text{d}F\left(t\right)}{\text{d}t}\le {\epsilon }_{3}F\left(t\right)+{\sigma }^2\left(\frac{1}{k_2}{\displaystyle {\int }_{\partial \Omega }{f}_1^2\text{d}s}+\frac{2k_1^2}{k_2}{\displaystyle {\int }_{\partial \Omega }{T}^2\text{d}s}+2{\displaystyle {\int }_{\Omega }{T}_{,i}{T}_{,i}\text{d}x}\right),$ (38)