一类热扩散方程组的空间衰减估计

Spatial Decay Estimates for a Class of Thermo-Diffusion Equations

摘要: 本文研究了一类热扩散方程组的空间衰减估计，应用一阶微分不等式的方法，建立了半无限管内热扩散线性微分方程组的解的空间指数衰减估计。为了使估计明确，还导出了总能量的界。该结果可看作Saint-Venant原则在热扩散方程组中的应用。

Abstract: In this paper, we study the spatial decay estimates for a class of thermo-diffusion equations. Using the technique of a first-order differential inequality, the exponential decay estimates for the linear differential equations of thermodiffusion in a semi-infinite pipe were established. To make the estimate explicit, the bound for the total energy was also derived.

1. 前言

在20世纪70年代，W. Nowacki在他的论文 [1] 中给出了一维空间中热扩散的微分方程，参考文献中的许多论文研究了这一系统。例如， [2] 和 [3] 利用不同的参数研究了热扩散线性系统的初边界值问题。通过 [4] 得到了相关线性柯西问题解的 $L^p-L^q$ 时间衰减估计。然而，在本文中，我们考虑了三维热扩散的微分方程，并研究了系统解的空间衰减估计。实际上，很多文献都是关于各种微分方程系统的空间衰减估计问题，为了查阅关于圣维南原理的这类工作，可以参考 [5] - [13] 以及其中引用的论文。

我们假设瞬态流体占据了边界 $\partial R$ 的半无限圆柱管R的内部。管道的截面用D表示，截面的边界用 $\partial D$ 表示，管道R平行于 $x_3$ 轴。我们定义：

$R_z=\left\{\left(x_1,x_2,x_3\right)|\left(x_1,x_2\right)\in D,x_3>z\ge 0\right\},$

$D_z=\left\{\left(x_1,x_2,x_3\right)|\left(x_1,x_2\right)\in D,x_3=z\ge 0\right\},$

其中z是沿 $x_3$ 轴的变量。显然， $R_0=R$ 和 $D_0=D$。设 $u_i$ 、T和C分别表示位移、温度和化学势为独立场。这些依赖于空间变量 $\left(x_1,x_2,x_3\right)$ 和时间变量t，并满足以下方程组：

$\rho {\ddot{\mu }}_i-v\Delta u_i-\left(\lambda +v\right)u_{j,ji}+{\gamma }_1{T}_{,i}+{\gamma }_2{C}_{,i}=0,R\times \left\{t\ge 0\right\},$ (1.1)

$c\stackrel{\dot{}}{T}-K\Delta T+{\gamma }_1{\stackrel{\dot{}}{u}}_{i,i}+d\stackrel{\dot{}}{C}=0,R\times \left\{t\ge 0\right\},$ (1.2)

$n\stackrel{\dot{}}{C}-M\Delta C+{\gamma }_2{\stackrel{\dot{}}{u}}_{i,i}+d\stackrel{\dot{}}{T}=0,\text{in}R\times \left\{t\ge 0\right\},$ (1.3)

在初始边界条件下

$u_i=0,T=C=0\text{on}\partial D\times \left\{t\ge 0\right\},$ (1.4)

$u_i=\stackrel{\dot{}}{u}=0,T=C=0\text{in}R\times \left\{t=0\right\}\text{.}$ (1.5)

$u_i=f_i\left(x_1,x_2,t\right),T=F\left(x_1,x_2,t\right),C=G\left(x_1,x_2,t\right)\text{on}{D}_0\times \left\{t\ge 0\right\},$ (1.6)

$u_i,u_{i,j},T,T_i,C,C_{,i},p=o\left(x_3^{-1}\right)x_1,x_2,t\text{as}{x}_3\to \infty \text{.}$ (1.7)

在方程(1.1)~(1.3)中， $\Delta$ 是拉普拉斯算子； $\rho$ 表示密度； ${\gamma }_1$ 和 ${\gamma }_2$ 是热和扩散扩张的系数； $\lambda$ 和 $\nu$ 是材料系数；K是导热系数；M是扩散系数。 $n,c,d$ 是热扩散的系数。以上常数均为正，满足

$cn-d^2>0$ (1.8)

这意味着(1.1)~(1.3)是一个偏微分方程组的双曲分解系统。在下面的几节中，我们可以使用下面的不等式。设D为带t的平面域D，他的边界为 $\partial D$。如果w在 $\partial D$ 上等于0，那么

${\displaystyle {\int }_D{w}_{,a}{w}_{,a}\text{d}A}\ge {\lambda }_1{\displaystyle {\int }_D{w}^2\text{d}x},$ (1.9)

其中 ${\lambda }_1$ 是问题的最小特征值

$\Delta \phi +\lambda \phi =0\text{in}D,$

$\phi =0\text{on}\partial D$

这种不平等现象已经得到了很好的研究(见 [14] [15])。在本文中，通常的求和约定是使用重复的拉丁下标从1到3，并重复希腊字母下标从1到2。逗号用来表示部分区分，例如：

$u_{i,j}=\frac{\partial u_i}{\partial x_i},{\phi }_{\alpha ,\alpha }={\displaystyle {\sum }_{\alpha =1}^2\frac{\partial {\phi }_{\alpha }}{\partial x_{\alpha }}},{\stackrel{\dot{}}{u}}_i=\frac{\partial u_i}{{\partial }_t}.$

2. 能量衰减

在这一部分中，我们导出了问题(1.3)~(1.6)的主要指数衰减结果。应用方程(1.1)和(1.4)~(1.6)，得到

$\begin{array}{c}0={\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}\left[\rho {\ddot{u}}_i-v\Delta u_i-\left(\lambda +v\right)u_{j,ji}+{\gamma }_1{T}_{,i}+{\gamma }_2{C}_{,i}\right]{\stackrel{\dot{}}{u}}_i\text{d}x\text{d}\eta }}\\ =\frac{1}{2}\rho {\displaystyle {\int }_{R_z}{\stackrel{\dot{}}{u}}_i{\stackrel{\dot{}}{u}}_i\text{d}x}+v{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{u}_{i,3}{\stackrel{\dot{}}{u}}_i\text{d}A\text{d}\eta }}+\frac{v}{2}{\displaystyle {\int }_{R_z}{u}_{i,j}{u}_{i,j}\text{d}x}+\left(\lambda +v\right){\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{u}_{j,j}{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}\\ +\frac{\lambda +v}{2}{\displaystyle {\int }_{R_z}{u}_{i,i}^2\text{d}x}+{\gamma }_1{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}{T}_{,i}{\stackrel{\dot{}}{u}}_i\text{d}x\text{d}\eta }}+{\gamma }_2{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}{C}_{,i}{\stackrel{\dot{}}{u}}_i\text{d}x\text{d}\eta }}.\end{array}$ (2.1)

我们用T乘以(1.2)，并积分得

$\begin{array}{c}0={\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}\left[c\stackrel{\dot{}}{T}-K\Delta T+{\gamma }_1{\stackrel{\dot{}}{u}}_{i,i}+d\stackrel{\dot{}}{C}\right]T\text{d}x\text{d}\eta }}\\ =\frac{c}{2}{\displaystyle {\int }_{R_z}{T^2\text{d}x|}_{\eta =t}}+K{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}_{,3}T\text{d}A\text{d}\eta }}+K{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}{T}_{,j}{T}_{,j}\text{d}x\text{d}\eta }}\\ -{\gamma }_1{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}T{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}-{\gamma }_1{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}{T}_{,i}{\stackrel{\dot{}}{u}}_i\text{d}x\text{d}\eta }}+d{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}C\stackrel{\dot{}}{T}\text{d}x\text{d}\eta }}\text{.}\end{array}$ (2.2)

以同样的方式，我们也可得

$\begin{array}{c}0={\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}\left[n\stackrel{\dot{}}{C}-M\Delta C+{\gamma }_2{\stackrel{\dot{}}{u}}_{i,i}+d\stackrel{\dot{}}{T}\right]C\text{d}x\text{d}\eta }}\\ =\frac{n}{2}{\displaystyle {\int }_{R_z}{C^2\text{d}x|}_{\eta =t}}+M{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}_{,3}C\text{d}A\text{d}\eta }}+M{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}{C}_{,j}{C}_{,j}\text{d}x\text{d}\eta }}\\ -{\gamma }_2{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}C{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}-{\gamma }_2{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}{C}_{,i}{\stackrel{\dot{}}{u}}_i\text{d}x\text{d}\eta }}+d{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}C\stackrel{\dot{}}{T}\text{d}x\text{d}\eta }}\end{array}$ (2.3)

现在，我们定义一个函数

$\begin{array}{c}E\left(z,t\right)=\frac{1}{2}\rho {\displaystyle {\int }_{R_z}{\stackrel{\dot{}}{u}}_i{\stackrel{\dot{}}{u}}_i\text{d}x}+\frac{v}{2}{\displaystyle {\int }_{R_z}{u}_{i,j}{u}_{i,j}\text{d}x}+\frac{\lambda +v}{2}{\displaystyle {\int }_{R_z}{u}_{i,i}^2\text{d}x}+K{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}{T}_{,j}{T}_{,j}\text{d}x\text{d}\eta }}\\ +M{\displaystyle {\int }_0^t{\displaystyle {\int }_{R_z}{C}_{,j}{C}_{,j}\text{d}x\text{d}\eta }}+{\displaystyle {\int }_{R_z}\left[\frac{c}{2}{T}^2+dCT+\frac{n}{2}{C}^2\right]\text{d}x}\end{array}$ (2.4)

然后得

$\begin{array}{c}-\frac{\partial E\left(z,t\right)}{\partial z}=\frac{1}{2}\rho {\displaystyle {\int }_{D_z}{\stackrel{\dot{}}{u}}_i{\stackrel{\dot{}}{u}}_i\text{d}A}+\frac{v}{2}{\displaystyle {\int }_{D_z}{u}_{i,j}{u}_{i,j}\text{d}A}+\frac{\lambda +v}{2}{\displaystyle {\int }_{D_z}{u}_{i,i}^2\text{d}A}+K{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}_{,j}{T}_{,j}\text{d}A\text{d}\eta }}\\ +M{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}_{,j}{C}_{,j}\text{d}A\text{d}\eta }}+{\displaystyle {\int }_{D_z}\left[\frac{c}{2}{T}^2+dCT+\frac{n}{2}{C}^2\right]\text{d}A}\text{.}\end{array}$ (2.5)

很明显，(2.5)的最后一项是正的，因为 $cn-d^2>0$。从(2.1)~(2.3)开始，我们有

$\begin{array}{c}E\left(z,t\right)=-v{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{u}_{i,3}{\stackrel{\dot{}}{u}}_i\text{d}A\text{d}\eta }}-\left(\lambda +v\right){\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{u}_{j,j}{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}+{\gamma }_1{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}T{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}\\ +{\gamma }_2{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}C{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}-K{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}_{,3}T\text{d}A\text{d}\eta }}-M{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}_{,3}C\text{d}A\text{d}\eta }}.\end{array}$ (2.6)

利用Schwarz，Poincar’e定理，以及AG的平均不等式，我们得到了

$\begin{array}{c}v{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{u}_{i,3}{\stackrel{\dot{}}{u}}_i\text{d}A\text{d}\eta }}\le v{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{\stackrel{\dot{}}{u}}_i{\stackrel{\dot{}}{u}}_i\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{u}_{i,3}{u}_{i,3}\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}\\ \le v{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{\stackrel{\dot{}}{u}}_i{\stackrel{\dot{}}{u}}_i\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{u}_{i,3}{u}_{i,3}\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}\\ \le \frac{\sqrt{v}t}{\sqrt{\rho }}\left(-\frac{\partial E\left(z,t\right)}{\partial z}\right),\end{array}$ (2.7)

和

$\begin{array}{c}\left(\lambda +v\right){\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{u}_{j,j}\stackrel{\dot{}}{u}\text{d}A\text{d}\eta }}\le \left(\lambda +v\right){\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{\stackrel{\dot{}}{u}}_i{\stackrel{\dot{}}{u}}_i\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{u}_{j,j}^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}\\ \begin{array}{c}\le \frac{\sqrt{\lambda +vt}}{\sqrt{\rho }}\left(-\frac{\partial E\left(z,t\right)}{\partial z}\right),\end{array}\end{array}$ (2.8)

还有

$\begin{array}{l}{\gamma }_1{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}T{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}+{\gamma }_2{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}C{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}\\ \le {\gamma }_1{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{\stackrel{\dot{}}{u}}_3^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}+{\gamma }_2{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{\stackrel{\dot{}}{u}}_3^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}\\ \le \frac{{\gamma }_1}{\sqrt{{\lambda }_1}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}_{,\alpha }{T}_{,\alpha }\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{\stackrel{\dot{}}{u}}_3^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}+\frac{{\gamma }_2}{\sqrt{{\lambda }_1}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}_{,\alpha }{C}_{,\alpha }\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{\stackrel{\dot{}}{u}}_3^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}\\ \le \left[\frac{{\gamma }_1\sqrt{t}}{2\sqrt{\rho {\lambda }_{1}K}}+\frac{{\gamma }_2\sqrt{t}}{2\sqrt{\rho {\lambda }_{1}M}}\right]\left(-\frac{\partial E\left(z,t\right)}{\partial z}\right).\end{array}$ (2.9)

利用Schwarz定理，(1.9)和AG平均不等式，我们得到了

$\begin{array}{l}-K{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}_{,3}T\text{d}A\text{d}\eta }}-M{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}_{,3}C\text{d}A\text{d}\eta }}\\ \le K{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}_{,3}^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}+M{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}_{,3}^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}\\ \le \frac{K}{\sqrt{{\lambda }_1}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}_{,3}^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{T}_{,\alpha }{T}_{,\alpha }\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}+\frac{M}{\sqrt{{\lambda }_1}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}_{,3}^2\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}{\left({\displaystyle {\int }_0^t{\displaystyle {\int }_{D_z}{C}_{,\alpha }{C}_{,\alpha }\text{d}A\text{d}\eta }}\right)}^{\frac{1}{2}}\\ \begin{array}{c}\le \frac{1}{2\sqrt{{\lambda }_1}}\left(-\frac{\partial E\left(z,t\right)}{\partial z}\right).\end{array}\end{array}$ (2.10)

将(2.7)~(2.10)代入(2.6)，我们有

$E\left(z,t\right)\le m_1\left(t\right)\left(-\frac{\partial E\left(z,t\right)}{\partial z}\right),$ (2.11)

其中

$m_1\left(t\right)=\frac{\sqrt{v}t}{\sqrt{\rho }}+\frac{\sqrt{\lambda +vt}}{\sqrt{\rho }}+\frac{{\gamma }_1\sqrt{t}}{2\sqrt{\rho {\lambda }_{1}K}}+\frac{{\gamma }_2\sqrt{t}}{2\sqrt{\rho {\lambda }_{1}M}}+\frac{1}{2\sqrt{{\lambda }_1}}$ (2.12)

我们可以知道，(2.11)得出的结论是：

$E\left(z,t\right)\le E\left(0,t\right){\text{e}}^{-\frac{1}{m_1\left(t\right)}z}.$ (2.13)

不等式(2.13)就是我们所寻求的空间衰减结果。

3. 总能量E(0, t)的边界

为了使我们的衰变结果在第3节中明确化，我们根据已知的数据导出了 $E\left(0,t\right)$ 的界。首先，我们将等式(2.4)和(2.6)代入 $z=0$ 的时候，即，

$\begin{array}{c}E\left(0,t\right)=\frac{1}{2}\rho {\displaystyle {\int }_R{\stackrel{\dot{}}{u}}_i{\stackrel{\dot{}}{u}}_i\text{d}x}+\frac{v}{2}{\displaystyle {\int }_R{u}_{i,j}{u}_{i,j}\text{d}x}+\frac{\lambda +v}{2}{\displaystyle {\int }_R{u}_{i,i}^2\text{d}x}+K{\displaystyle {\int }_0^t{\displaystyle {\int }_R{T}_{,j}{T}_{,j}\text{d}x\text{d}\eta }}\\ +M{\displaystyle {\int }_0^t{\displaystyle {\int }_R{C}_{,j}{C}_{,j}\text{d}x\text{d}\eta }}+{\displaystyle {\int }_R\left[\frac{c}{2}{T}^2+dCT+\frac{n}{2}{C}^2\right]\text{d}x}\\ =-v{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_0}{u}_{i,3}{\stackrel{\dot{}}{u}}_i\text{d}A\text{d}\eta }}-\left(\lambda +v\right){\displaystyle {\int }_0^t{\displaystyle {\int }_{D_0}{u}_{j,j}{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}+{\gamma }_1{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_0}T{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}\\ +{\gamma }_2{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_0}C{\stackrel{\dot{}}{u}}_3\text{d}A\text{d}\eta }}-K{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_0}{T}_{,3}T\text{d}A\text{d}\eta }}-M{\displaystyle {\int }_0^t{\displaystyle {\int }_{D_0}{C}_{,3}C\text{d}A\text{d}\eta }}.\end{array}$ (3.1)

为了得到 $E\left(0,t\right)$ 的界，我们现在介绍函数

(3.2)

其中和是待定的正常数。所以，我们可得

(3.3)

根据散度定理可知

(3.4)

由Schwarz的不等式和AG的平均不等式，从(3.4)我们得到了

(3.5)

对于。挑选，，，，，，，，，，，，，，，，，结合(3.1)和(3.5)，我们有

(3.6)

因此，

(3.7)

其中

(3.8)

回顾(3.2)中和的定义，我们得出结论，我们通过选择适当的和，的界，不等式(2.13)就可以明确表示。

基金项目

广东大学生攀登计划(pdjh2019b0335)。