贵州师范大学数学科学学院，贵州 贵阳

收稿日期：2024年2月8日；录用日期：2024年3月22日；发布日期：2024年3月29日

摘要

本文介绍了Gronwall不等式及傅里叶变换的性质与推论，应用傅里叶变换法分析三维Navier-Stokes-Poisson (NSP)方程与三维可压缩Navier-Stokes-Korteweg方程的格林矩阵，得到NSP方程与Navier-Stokes-Korteweg方程的傅里叶变换。

关键词

傅里叶变换，Green矩阵，Navier-Stokes-Poisson方程

Fourier Transform of Green Matrices in Linear Partial Differential Equations

Mei Liu, Fuqiang Lu

School of Mathematical Sciences, Guizhou Normal University, Guiyang Guizhou

Received: Feb. 8^th, 2024; accepted: Mar. 22^nd, 2024; published: Mar. 29^th, 2024

ABSTRACT

In this paper, the properties and inferences of Gronwall inequality and Fourier transform are introduced, and the Fourier transform of the NSP equation and the Navier-Stokes-Korteweg equation is analyzed by using the Fourier transform method, and the Fourier transform of the NSP equation and the Navier-Stokes-Korteweg equation is obtained.

Keywords:Fourier Transform, Green Matrix, Navier-Stokes-Poisson Equation

Copyright © 2024 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

傅里叶变换是傅里叶级数由周期函数向非周期函数的演变，它通过特定形式的积分建立函数之间的对应关系，且傅里叶变换在许多领域都有着广泛的应用，比如化学、航空、航天、工业技术等，当然在数学领域也有着十分重要的作用，比如红外光谱检测、信号处理、通信系统、量子力学等等。在现代数学中，傅里叶是一种非常重要的变换，为解决现代物理学问题提供了一件必不可少的工具，所以在求解线性偏微分方程的Green矩阵中运用傅里叶变换会比较简单，傅里叶变换是一种重要的积分变换，正是由于其积分的性质，它可以求解微分方程、偏微分方程，积分变换的作用是把初始函数变成另一类比较容易求解的象函数，因此用积分变换求解偏微分方程的方法与我们采用对数来计算数的乘、除、乘方和开方的技巧完全类似。在用傅里叶变换法求解偏微分方程的过程中用到了一个很重要的不等式——Gronwall不等式，该不等式在一阶常微分方程解的存在唯一性定理的证明过程中起核心作用，还有一个很重要的作用是给出相关未知函数的上界估计。

在文献 [1] 中，曹瑞华利用傅里叶变换法研究偏微分方程的应用；在文献 [2] 中，Hoff D等对给出了若干空间维中可压缩流的纳维–斯托克斯方程的柯西问题解的渐近性态的详细点态进行描述；徐小蓉介绍了分离变数法及积分变换法在解数学物理方程的应用参考文献 [3] ，并在分离变数法中对齐次方程及非齐次方程进行了区分；张礼涛从三个方面讨论傅里叶变换方法在求解偏微分方程中的具体运用，参考文献 [4] 。

2. Gronwall不等式及傅里叶变换性质与推论

2.1. 傅里叶变换的基本概念

若函数 $f\left(t\right)$ 在 $\left(-\infty ,+\infty \right)$ 上满足傅里叶积分定理的条件，则函数

$F\left(\omega \right)={\displaystyle {\int }_{-\infty }^{+\infty }f\left(t\right){\text{e}}^{-j\omega t}\text{d}t}$

为 $f\left(t\right)$ 的傅里叶变换，而称函数

$f\left(t\right)=\frac{1}{2\pi }{\displaystyle {\int }_{-\infty }^{+\infty }F\left(\omega \right){\text{e}}^{j\omega t}\text{d}\omega }$

为 $F\left(\omega \right)$ 的傅里叶变换。

2.2. Gronwall不等式

(1) 微分形式的Gronwall不等式 [5]

设 $\eta \left(t\right)$ 是 $\left[0,T\right]$ 上的非负、绝对连续的函数，且对几乎处处的t满足微分不等式

${\eta }^{\prime }\left(t\right)\le \phi \left(t\right)\eta \left(t\right)+\psi \left(t\right),$

这里 $\phi \left(t\right)$ 和 $\psi \left(t\right)$ 是 $\left[0,T\right]$ 上非负、可积的函数，那么对所有的 $0\le t\le T$ 有

$\eta \left(t\right)\le {\text{e}}^{{\displaystyle {\int }_0^t\phi \left(s\right)\text{d}s}}\left[\eta \left(0\right)+{\displaystyle {\int }_0^t\phi \left(s\right)\text{d}s}\right].$

(2) 积分形式的Gronwall不等式

设 $\xi \left(t\right)$ 是 $\left[0,T\right]$ 上的非负、绝对连续的函数，且对几乎处处的t和常数 $C_1,C_2\ge 0$ 满足积分不等式

$\xi \left(t\right)\le C_1{\displaystyle {\int }_0^t\xi \left(s\right)\text{d}s}+C_2,$

那么对几乎处处的 $0\le t\le T$ 有

$\xi \left(t\right)\le C_2\left(1+C_{1}t{\text{e}}^{C_{1}t}\right).$

2.3. 傅里叶变换的性质与推论

性质1. 线性性质

若 $f_i\left(x\right)\in L^1\left(-\infty ,+\infty \right)$ ， $a_i\left(i=1,2\right)$ 为任意常数，则

${\left(a_1{f}_1+a_2{f}_2\right)}^{\wedge }=a_1{\hat{f}}_1+a_2{\hat{f}}_2.$

性质2. 微商性质

用 $C\left(-\infty ,\infty \right)$ 表示 $\left(-\infty ,\infty \right)$ 上连续函数的全体构成的线性空间。若

则

${\left(\frac{\text{d}f}{\text{d}x}\right)}^{\wedge }=i\lambda \hat{f}\left(\xi \right).$

推论1. 若 $f\left(x\right),f^{\prime }\left(x\right),\cdots ,f^{\left(m\right)}\left(x\right)\in C\left(-\infty ,\infty \right)\cap L^1\left(-\infty ,\infty \right)$ ，则

${\left(\frac{{\text{d}}^{m}f}{\text{d}{x}^m}\right)}^{\wedge }={\left(i\lambda \right)}^m\hat{f}\left(\xi \right)\left(m\ge 1\right).$

性质3. 乘多项式性质

若 $f\left(x\right),xf\left(x\right)\in L^1\left(-\infty ,\infty \right)$ ，则

${\left(xf\left(x\right)\right)}^{^}=i\frac{\text{d}}{\text{d}\lambda }\hat{f}\left(\lambda \right).$

推论2. $f\left(x\right),xf\left(x\right),\cdots ,x^{m}f\left(x\right)\in L^1\left(-\infty ,\infty \right)$ ，则

${\left(x^{m}f\left(x\right)\right)}^{^}=i^m\frac{{\text{d}}^m}{\text{d}{\lambda }^m}\hat{f}\left(\lambda \right)\left(m\ge 1\right).$

以上的性质及推论的证明参见文献 [5] 。

3. 傅里叶变换法研究线性偏微分方程的格林矩阵

3.1. Navier-Stokes-Poisson方程中Green矩阵的傅里叶变换

我们考虑线性化的Navier-Stokes-Poisson方程：

$\{\begin{array}{l}{\rho }_t+divm=0,\\ m_t+\nabla \rho -\mu \Delta m-\lambda \nabla divm+2\nabla {\left(-\Delta \right)}^{-1}\rho =0，\\ \left(\rho ,m\right)\left(x,0\right)=\left({\rho }_0,m_0\right)\left(x\right),x\in R^3.\end{array}$ (1)

方程(1)的傅里叶变换 $\hat{G}\left(\xi ,t\right)$ 的格林矩阵为

$\hat{G}\left(\xi ,t\right)=\left(\begin{array}{cc}\frac{{\lambda }_1{\text{e}}^{{\lambda }_{2}t}-{\lambda }_2{\text{e}}^{{\lambda }_{1}t}}{{\lambda }_1-{\lambda }_2}& -i\left(\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\right){\xi }^t\\ -i\left(\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\right)\xi & {\text{e}}^{-\mu {\left|\xi \right|}^{2}t}\left(I-\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\right)+\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\end{array}\right)+\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\left(\begin{array}{cc}0& 0\\ \frac{-i\xi }{\frac{1}{2}{\left|\xi \right|}^2}& 0\end{array}\right),$ (2)

其特征值为

${\lambda }_{1，2}=-\frac{1}{2}\left(\mu +\lambda \right){\left|\xi \right|}^2\pm \frac{1}{2}\sqrt{{\left(\mu +\lambda \right)}^2{\left|\xi \right|}^4-4\left({\left|\xi \right|}^2+2\right)},$ (3)

$\left|\xi \right|\ne 0,2/\left(\mu +\lambda \right)$ ，且 ${\lambda }_1+\left(\mu +\lambda \right){\left|\xi \right|}^2=-{\lambda }_2$ ， ${\lambda }_2+\left(\mu +\lambda \right){\left|\xi \right|}^2=-{\lambda }_1$ ， ${\lambda }_1{\lambda }_2={\left|\xi \right|}^2+2$ 。

证明：设初始数据 $\left({\rho }_0,m_0\right)$ ，将算子 ${\Sigma }_j\frac{\partial }{\partial x_j}$ 应用于(1)的第二个式子，且令 $D=divm$ ，结合 ${\rho }_t=-divm$ ，有 $\Delta D=\Delta divm=-\Delta {\rho }_t$ ，得到

$D_t+\text{Δ}\rho =\left(\mu +\lambda \right)\text{Δ}D+2\rho ,$ (4)

首先结合(1)的第一个式子有 $\left(\mu +\lambda \right)\text{Δ}D=\left(\mu +\lambda \right)\text{Δ}divm=-\left(\mu +\lambda \right)\text{Δ}{\rho }_t$ ，得到：

${\rho }_{tt}=-D_t=\text{Δ}\rho +\left(\mu +\lambda \right)\text{Δ}{\rho }_t-2\rho ,$ (5)

即：

${\rho }_{tt}-\text{Δ}\rho -\left(\mu +\lambda \right)\text{Δ}{\rho }_t+2\rho =0.$ (6)

对(6)与 ${\rho }_t=-divm$ 应用傅里叶变换性质及推论有：

$\{\begin{array}{l}{\hat{\rho }}_{tt}+\left(\mu +\lambda \right){\left|\xi \right|}^2{\hat{\rho }}_t+{\left|\xi \right|}^2{\hat{\rho }}_t+2\hat{\rho }=0,\\ \hat{\rho }\left(\xi ,0\right)={\hat{\rho }}_0\left(\xi \right),\\ {\hat{\rho }}_t\left(\xi ,0\right)=-i\xi \cdot {\hat{m}}_0\left(\xi \right).\end{array}$ (7)

因为 ${\lambda }_1\left(\xi \right),{\lambda }_2\left(\xi \right)$ 是其相应的根，所以可求其通解

$\hat{\rho }\left(\xi ,t\right)=A\left(\xi \right){\text{e}}^{{\lambda }_2\left(\xi \right)t}+B\left(\xi \right){\text{e}}^{{\lambda }_1\left(\xi \right)t},$ (8)

因此，当 ${\lambda }_1\ne {\lambda }_2$ 时，有

$\{\begin{array}{l}\hat{\rho }\left(\xi ,0\right)=A\left(\xi \right)+B\left(\xi \right)={\hat{\rho }}_0\left(\xi \right),\\ {\hat{\rho }}_t\left(\xi ,0\right)={\lambda }_2\left(\xi \right)A\left(\xi \right)+{\lambda }_1\left(\xi \right)B\left(\xi \right)=-i\xi \cdot {\hat{m}}_0\left(\xi \right),\end{array}$ (9)

由 ${(9)}_1\times {\lambda }_2-{(9)}_2$ 、 ${(9)}_1\times {\lambda }_1-{(9)}_2$ 得到

$A\left(\xi \right)=\frac{{\lambda }_1{\hat{\rho }}_0+i\xi \cdot {\hat{m}}_0}{{\lambda }_1-{\lambda }_2},B\left(\xi \right)=\frac{-i\xi \cdot {\hat{m}}_0-{\lambda }_2{\hat{\rho }}_0}{{\lambda }_1-{\lambda }_2},$ (10)

将(10)带回(8)整理有

$\hat{\rho }=\left(\frac{{\lambda }_1{\text{e}}^{{\lambda }_{2}t}-{\lambda }_{-}{\text{e}}^{{\lambda }_{1}t}}{{\lambda }_1-{\lambda }_2}\right){\hat{\rho }}_0\left(\xi \right)-i\left(\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\right)\xi \cdot {\hat{m}}_0\left(\xi \right).$ (11)

其次，为了计算 $\hat{m}$ ，对(1)的第二个式子直接运用傅里叶变换的性质及推论，有

${\hat{m}}_t+\mu {\left|\xi \right|}^2\hat{m}+\lambda \xi {\xi }^{\top }\hat{m}+i\xi \hat{\rho }+2i\xi {\left|\xi \right|}^2\hat{\rho }=0,$

即

${\hat{m}}_t=\left(-\mu {\left|\xi \right|}^2-\lambda \xi {\xi }^{\top }\right)\hat{m}-i\xi \left(1+2i{\left|\xi \right|}^{-2}\right)\hat{\rho },$ (12)

之后我们引入平行于 $\xi$ 的分量 $a=\frac{\xi \cdot \hat{m}}{\left|\xi \right|}$ 和正交于 $\xi$ 的分量b，再将其单位化有

$\hat{m}\left(\xi ,t\right)=a\left(\xi ,t\right)\frac{\xi }{\left|\xi \right|}+b\left(\xi ,t\right).$ (13)

对(13)求导，结合(12)得

$a_t\left(\xi ,t\right)+b_t\left(\xi ,t\right)\frac{\left|\xi \right|}{\xi }=-\left(\mu +\lambda \right){\left|\xi \right|}^{2}a-i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|\hat{\rho }-\mu {\left|\xi \right|}^{2}b\cdot \frac{\left|\xi \right|}{\xi },$

即

$\{\begin{array}{l}{a}_t=-\left(\mu +\lambda \right){\left|\xi \right|}^{2}a-i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|\hat{\rho },\\ b_t=-\mu {\left|\xi \right|}^{2}b,\end{array}$ (14)

由(13)知 $b_0={\hat{m}}_0-a_0\frac{\xi }{\left|\xi \right|}={\hat{m}}_0\left(I-\frac{\xi \cdot {\xi }^t}{{\left|\xi \right|}^2}\right)$ 。

对(14)应用Gronwall不等式可得

$b\left(\xi ,t\right)={\text{e}}^{-\mu {\left|\xi \right|}^{2}t}\left(I-\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\right){\hat{m}}_0,$ (15)

$a\left(\xi ,t\right)={\text{e}}^{-\left(\mu +\lambda \right){\left|\xi \right|}^{2}t}\left[a\left(\xi ,0\right)-i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|{\displaystyle {\int }_0^t{\text{e}}^{\left(\mu +\lambda \right){\left|\xi \right|}^{2}s}\hat{\rho }\left(\xi ,s\right)\text{d}s}\right],$ (16)

将(11)带入(16)，再结合 ${\lambda }_1+\left(\mu +\lambda \right){\left|\xi \right|}^2=-{\lambda }_2$ ， ${\lambda }_2+\left(\mu +\lambda \right){\left|\xi \right|}^2=-{\lambda }_1$ ， ${\lambda }_1{\lambda }_2={\left|\xi \right|}^2$ ， $b\cdot \xi =0$ 有

$\begin{array}{c}a\left(\xi ,t\right)={\text{e}}^{-\left(\mu +\lambda \right){\left|\xi \right|}^{2}t}\left[a\left(\xi ,0\right)-i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|{\displaystyle {\int }_0^t{\text{e}}^{\left(\mu +\lambda \right){\left|\xi \right|}^{2}t}}\left[\frac{{\lambda }_1{\text{e}}^{{\lambda }_{2}t}-{\lambda }_2{\text{e}}^{{\lambda }_{1}t}}{{\lambda }_1-{\lambda }_2}{\hat{\rho }}_0\left(\xi \right)-i\left(\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\right)\xi \cdot {\hat{m}}_0\left(\xi \right)\right]\text{d}t\right]\\ ={\text{e}}^{-\left(\mu +\lambda \right){\left|\xi \right|}^{2}t}a\left(\xi ,0\right)-i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|{\text{e}}^{-\left(\mu +\lambda \right){\left|\xi \right|}^{2}t}{\displaystyle {\int }_0^t\frac{{\lambda }_1{\text{e}}^{-{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{-{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}{\hat{\rho }}_0\left(\xi \right)-i\frac{{\text{e}}^{-{\lambda }_{2}t}-{\text{e}}^{-{\lambda }_{1}t}}{{\lambda }_1-{\lambda }_2}\xi \cdot {\hat{m}}_0\left(\xi \right)\text{d}t}\\ ={\text{e}}^{-\left(\mu +\lambda \right){\left|\xi \right|}^{2}t}a\left(\xi ,0\right)-i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|\frac{-{\text{e}}^{-{\lambda }_{2}t}+{\text{e}}^{-{\lambda }_{1}t}}{{\lambda }_1-{\lambda }_2}{\hat{\rho }}_0-i\frac{-{\lambda }_2^{-1}{\text{e}}^{{\lambda }_{1}t}-\left(-{\lambda }_1^{-1}\right){\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|\cdot \xi \cdot {\hat{m}}_0\left(\xi \right)\\ +i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|\cdot i\cdot \frac{1}{{\lambda }_1{\lambda }_2}\xi \cdot {\hat{m}}_0\left(\xi \right)\cdot {\text{e}}^{-\left(\mu +\lambda \right){\left|\xi \right|}^{2}t}\\ =-i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}{\hat{\rho }}_0+\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot \frac{\xi \cdot {\hat{m}}_0}{\left|\xi \right|},\end{array}$

因此

$a\left(\xi ,t\right)=-i\left(1+2{\left|\xi \right|}^{-2}\right)\left|\xi \right|\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}{\hat{\rho }}_0+\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot \frac{\xi \cdot {\hat{m}}_0}{\left|\xi \right|},$ (17)

最后将(17)、(15)带回(13)得到最终结果：

$\begin{array}{c}\hat{m}\left(\xi ,t\right)=\left[-i\left(1+2{\left|\xi \right|}^{-2}\right)\cdot \frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot {\hat{\rho }}_0\left|\xi \right|+\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot \frac{\xi \cdot {\hat{m}}_0}{\left|\xi \right|}\right]\frac{\xi }{\left|\xi \right|}+{\text{e}}^{-\mu {\left|\xi \right|}^{2}t}\left(I-\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\right){\hat{m}}_0\\ =\left(-i\cdot \frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}-2i{\left|\xi \right|}^{-2}\cdot \xi \cdot \frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\right){\hat{\rho }}_0+\left[\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot \frac{\xi {\xi }^t}{{\left|\xi \right|}^2}+{\text{e}}^{-\mu {\left|\xi \right|}^{2}t}\left(I-\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\right)\right]{\hat{m}}_0,\end{array}$

得证

$\hat{G}\left(\xi ,t\right)=\left(\begin{array}{cc}\frac{{\lambda }_1{\text{e}}^{{\lambda }_{2}t}-{\lambda }_2{\text{e}}^{{\lambda }_{1}t}}{{\lambda }_1-{\lambda }_2}& -i\left(\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\right){\xi }^t\\ -i\left(\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\right)\xi & {\text{e}}^{-\mu {\left|\xi \right|}^{2}t}\left(I-\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\right)+\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\end{array}\right)+\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\left(\begin{array}{cc}0& 0\\ \frac{-i\xi }{\frac{1}{2}{\left|\xi \right|}^2}& 0\end{array}\right).$

求出格林矩阵的傅里叶变换后，就可以得到其半群的表达式，即傅里叶变换的精确表达式，为了能够建立NSP方程中 $L^2$ 时间衰减率，需要对半群分别进行低频和高频的分析，所以我们必须掌握求傅里叶空间中格林函数的公式和对其元素的渐近分析的方法。

3.2. Navier-Stokes-Korteweg方程中Green矩阵的傅里叶变换

我们考虑线性化Navier-Stokes-Korteweg的方程：

$\{\begin{array}{l}{\rho }_t+divm=0,\\ m_t-\mu \Delta m-\left(\mu +\nu \right)\nabla divm+\nabla \rho -\kappa \nabla \Delta \rho =0,\\ \left(\rho ,m\right)\left(x,0\right)=\left({\rho }_0,m_0\right)\left(x\right),x\in R^3.\end{array}$ (18)

方程(18)的傅里叶变换 $\hat{G}\left(\xi ,t\right)$ 的格林矩阵为

$\hat{G}\left(\xi ,t\right)=\left(\begin{array}{cc}\frac{{\lambda }_1{\text{e}}^{{\lambda }_{2}t}-{\lambda }_2{\text{e}}^{{\lambda }_{1}t}}{{\lambda }_1-{\lambda }_2}& -\frac{i{\xi }^t\left({\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}\right)}{{\lambda }_1-{\lambda }_2}\\ -\frac{i\xi \left(1+\kappa {\left|\xi \right|}^2\right)\left({\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}\right)}{{\lambda }_1-{\lambda }_2}& {\text{e}}^{-\mu {\left|\xi \right|}^{2}t}\left(I-\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\right)+\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\end{array}\right),$

其中

${\lambda }_{1，2}=-\frac{1}{2}\left(2\mu +\nu \right){\left|\xi \right|}^2\pm \frac{1}{2}i\sqrt{4\left(1+\kappa {\left|\xi \right|}^2\right){\left|\xi \right|}^2-{\left(2\mu +\nu \right)}^2{\left|\xi \right|}^4},$

同样， $\left|\xi \right|\ne 0,2/\left(2\mu +\nu \right)$ ，且

${\lambda }_1+\left(2\mu +\nu \right){\left|\xi \right|}^2=-{\lambda }_2,{\lambda }_2+\left(2\mu +\nu \right){\left|\xi \right|}^2=-{\lambda }_1,{\lambda }_1{\lambda }_2=\left(1+\kappa {\left|\xi \right|}^2\right){\left|\xi \right|}^2$

证明：与上述矩阵分析方法相同，将算子 ${\Sigma }_j\frac{\partial }{\partial x_j}$ 应用于(18)的第二个式子，得到

$D_t+\text{Δ}\rho =\left(2\mu +\nu \right)\text{Δ}D+\kappa {\text{Δ}}^2\rho ,$ (19)

其中 $D=divm$ ，由(18)的第一个式子有 $\left(2\mu +\nu \right)\text{Δ}D=\left(2\mu +\nu \right)\text{Δ}divm=-\left(2\mu +\nu \right)\text{Δ}{\rho }_t$ ，得到：

${\rho }_{tt}-\Delta \rho -\left(2\mu +\nu \right)\Delta {\rho }_t+\kappa {\Delta }^2\rho =0,$ (20)

由性质及推论有：

$\{\begin{array}{l}{\hat{\rho }}_{tt}+\left(2\mu +\nu \right){\left|\xi \right|}^2{\hat{\rho }}_t+{\left|\xi \right|}^2\hat{\rho }+\kappa {\left|\xi \right|}^4\hat{\rho }=0,\\ \hat{\rho }\left(\xi ,0\right)={\hat{\rho }}_0\left(\xi \right),\\ {\hat{\rho }}_t\left(\xi ,0\right)=-i\xi \cdot {\hat{m}}_0\left(\xi \right).\end{array}$ (21)

${\lambda }_1\left(\xi \right),{\lambda }_2\left(\xi \right)$ 是其相应的根，所以其通解可写为：

$\hat{\rho }\left(\xi ,t\right)=A\left(\xi \right){\text{e}}^{{\lambda }_2\left(\xi \right)t}+B\left(\xi \right){\text{e}}^{{\lambda }_1\left(\xi \right)t},$ (22)

当 ${\lambda }_1\left(\xi \right)\ne {\lambda }_2\left(\xi \right)$ 时，有

$\{\begin{array}{l}\hat{\rho }\left(\xi ,0\right)=A\left(\xi \right)+B\left(\xi \right)={\hat{\rho }}_0\left(\xi \right),\\ {\hat{\rho }}_t\left(\xi ,0\right)={\lambda }_2\left(\xi \right)A\left(\xi \right)+{\lambda }_1\left(\xi \right)B\left(\xi \right)=-i\xi \cdot {\hat{m}}_0\left(\xi \right),\end{array}$ (23)

因此

$A\left(\xi \right)=\frac{{\lambda }_1{\hat{\rho }}_0+i\xi \cdot {\hat{m}}_0}{{\lambda }_1-{\lambda }_2},B\left(\xi \right)=\frac{-i\xi \cdot {\hat{m}}_0-{\lambda }_2{\hat{\rho }}_0}{{\lambda }_1-{\lambda }_2},$ (24)

将(24)带入(22)有

$\hat{\rho }=\frac{{\lambda }_1{\text{e}}^{{\lambda }_{2}t}-{\lambda }_{-}{\text{e}}^{{\lambda }_{1}t}}{{\lambda }_1-{\lambda }_2}{\hat{\rho }}_0\left(\xi \right)-i\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\xi \cdot {\hat{m}}_0\left(\xi \right).$ (25)

同(12)，为了计算 $\hat{m}$ ，对(18)的第二个式子进行傅里叶变换，有

${\hat{m}}_t=\left(-\mu {\left|\xi \right|}^{2}I-\left(\mu +\nu \right)\xi {\xi }^t\right)\hat{m}-i\xi \left(1+\kappa {\left|\xi \right|}^2\right)\hat{\rho },$ (26)

同样，我们引入平行于 $\xi$ 的分量 $a=\frac{\xi \cdot \hat{m}}{\left|\xi \right|}$ 和正交于 $\xi$ 的分量b，再将其单位化有

$\hat{m}\left(\xi ,t\right)=a\left(\xi ,t\right)\frac{\xi }{\left|\xi \right|}+b\left(\xi ,t\right).$ (27)

对(27)求导结合(26)有

$\{\begin{array}{l}{a}_t=-\left(2\mu +\nu \right){\left|\xi \right|}^{2}a-i\xi \left(1+\kappa {\left|\xi \right|}^2\right)\hat{\rho },\\ b_t=-\mu {\left|\xi \right|}^{2}b,\end{array}$ (28)

对(28)应用Gronwall不等式得到

$b\left(\xi ,t\right)={\text{e}}^{-\mu {\left|\xi \right|}^{2}t}\left(I-\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\right){\hat{m}}_0,$ (29)

$a\left(\xi ,t\right)={\text{e}}^{-\left(2\mu +\nu \right){\left|\xi \right|}^{2}t}\left[a\left(\xi ,0\right)-i\left(1+\kappa {\left|\xi \right|}^2\right)\left|\xi \right|{\displaystyle {\int }_0^t{\text{e}}^{\left(2\mu +\nu \right){\left|\xi \right|}^{2}s}\hat{\rho }\left(\xi ,s\right)\text{d}s}\right],$ (30)

将(25)带入(30)，且 ${\lambda }_1+\left(2\mu +\nu \right){\left|\xi \right|}^2=-{\lambda }_2$ ， ${\lambda }_2+\left(2\mu +\nu \right){\left|\xi \right|}^2=-{\lambda }_1$ ， ${\lambda }_1{\lambda }_2=\left(1+\kappa {\left|\xi \right|}^2\right){\left|\xi \right|}^2$ 有：

$a\left(\xi ,t\right)=-i\left(1+\kappa {\left|\xi \right|}^2\right)\left|\xi \right|\frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}{\hat{\rho }}_0+\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot \frac{\xi \cdot {\hat{m}}_0}{\left|\xi \right|},$ (31)

最后将(31)、(29)带回(27)有

$\begin{array}{c}\hat{m}\left(\xi ,t\right)=\left[-i\left(1+\kappa {\left|\xi \right|}^2\right)\cdot \frac{{\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot {\hat{\rho }}_0\left|\xi \right|+\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot \frac{\xi \cdot {\hat{m}}_0}{\left|\xi \right|}\right]\frac{\xi }{\left|\xi \right|}+{\text{e}}^{-\mu {\left|\xi \right|}^{2}t}\left(I-\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\right){\hat{m}}_0\\ =-\frac{i\left(1+\kappa {\left|\xi \right|}^2\right)\left({\text{e}}^{{\lambda }_{1}t}-{\text{e}}^{{\lambda }_{2}t}\right)}{{\lambda }_1-{\lambda }_2}{\hat{\rho }}_0+\left[\frac{{\lambda }_1{\text{e}}^{{\lambda }_{1}t}-{\lambda }_2{\text{e}}^{{\lambda }_{2}t}}{{\lambda }_1-{\lambda }_2}\cdot \frac{\xi {\xi }^t}{{\left|\xi \right|}^2}+{\text{e}}^{-\mu {\left|\xi \right|}^{2}t}\left(I-\frac{\xi {\xi }^t}{{\left|\xi \right|}^2}\right)\right]{\hat{m}}_0.\end{array}$

通过计算Navier-Stokes-Korteweg线性方程格林函数的傅里叶变换可得到 $\rho$ 、m的表达式，为后面计算上、下界衰减估计做好准备。其应用事例参见文献 [6] 。

4. 结论

本文运用Gronwall不等式和这些性质推论求出NSP方程和Navier-Stokes-Korteweg方程的Green矩阵的傅里叶变换，由此我们可以知道，不管是在有静电势的情况下还是有常数毛细管系数的时候，都可以运用此方法来求方程的Green函数的傅里叶变换，事实上，傅里叶变换在偏微分方程的理论研究中还有很多重要的应用，尤其是在求偏微分方程的衰减估计中，接下来我将会运用此方法研究磁流体动力学(MHD)方程在Besove空间中的衰减估计。

基金项目

贵州师范大学学术新苗基金项目(No. 黔师新苗[2022] 03)。

文章引用

刘 梅,陆富强. 线性偏微分方程中Green矩阵的傅里叶变换
Fourier Transform of Green Matrices in Linear Partial Differential Equations[J]. 理论数学, 2024, 14(03): 164-171. https://doi.org/10.12677/pm.2024.143096

参考文献