二阶PDE的分类与总结

4.1.1 二阶常系数偏微分方程的分类

$u = u(x,y), \, n=2 \\ a_11 u_xx + 2a_{12} u_{xy}+ a_{22}u_{yy} + b_1 u_x + b_2 u_y +c_0 u = f$

换一个坐标

$\begin{cases} x = x(\xi,\eta) = a\xi + b \eta \\ y = y(\xi,\eta) = c\xi + d \eta \end{cases}$

得到

$u(x,y) = u(x(\xi,\eta)+y(\xi,\eta)) = u(\xi,\eta)$

并且有

$u_x = u_{\xi} \xi_x + u_{\eta} \eta_x \\ u_y = u_{\xi} \xi_y + u_{\eta} \eta_y \\ u_{xx} = (u_{\xi})_x \xi_x + u_{\xi} \xi_{xx} + (u_{\eta})_{x} \eta_x + u_{\eta} \eta_{xx} \\ = (u_{\xi \xi} \xi_x + u_{\xi \eta} \eta_x) \xi_x + (u_{\eta \xi} \xi_x + u_{\eta \eta} \eta_x) \eta_x \\ u_{yy} = = (u_{\xi \xi} \xi_y + u_{\xi \eta} \eta_y) \xi_y + (u_{\eta \xi} \xi_y + u_{\eta \eta} \eta_y) \eta_y$

写成矩阵形式

$D^2_{x,y} u = \begin{pmatrix} u_{xx} & u_{xy} \\ u_{yx} & u_{xy} \end{pmatrix} \quad D^2_{\xi,\eta} u = \begin{pmatrix} u_{\xi\xi} & u_{\xi\eta} \\ u_{\eta\xi} & u_{\eta\eta} \end{pmatrix}$

那么二阶导数可以写成

$u_{xx} = \begin{pmatrix} \xi_x & \eta_x \end{pmatrix} \begin{pmatrix} u_{\xi\xi} & u_{\xi\eta} \\ u_{\eta\xi} & u_{\eta\eta} \end{pmatrix} \begin{pmatrix} \xi_x \\ \eta_x \end{pmatrix}$

进一步可以得到

$D^2_{(x,y)} u = \begin{pmatrix} \xi_x & \eta_x \\ \xi_y & \eta_y \end{pmatrix} \begin{pmatrix} u_{\xi\xi} & u_{\xi\eta} \\ u_{\eta\xi} & u_{\eta\eta} \end{pmatrix} \begin{pmatrix} \xi_x & \xi_y\\ \eta_x & \eta_y \end{pmatrix} = J_0 D^2_{\xi,\eta} u J_0^{T}$

$Tr(AD^2_{(x,y)}) = Tr(A J_0 D^2_{(x,y)} u J_0^T) = Tr(J_0^T A J_0 D^2_{\xi,\eta} )$

选取正交矩阵 $J_0$ ，使得

$J_0^T J_0 = J_0 J_0^T = I \\ J_0^T A J_0 = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$

故

$Tr(AD^2_{(x,y)}) = Tr\left( \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \begin{pmatrix} u_{\xi\xi} & u_{\xi\eta} \\ u_{\eta\xi} & u_{\eta\eta} \end{pmatrix} \right) = \lambda_1 u_{\xi \xi} + \lambda_2 u_{\eta \eta}$

其中 $\lambda_1, \lambda_2$ 为矩阵的特征值

矩阵的特征值满足以下特征方程

$\lvert \lambda I - A \rvert = \det \left( \begin{pmatrix} \lambda - a_{11} & -a_{12} \\ -a_{12} & \lambda - a_{22} \end{pmatrix} \right) \\ = (\lambda - a_{11})(\lambda - a_{22}) - a^2_{12} \\ = \lambda^2 - (a_{11} + a_{22}) \lambda - (a^2_{12} - a_{11} a_{22}) = 0$

得到特征值的解

$\lambda_{\pm} = \dfrac{T_0 \pm \sqrt{T_0^2 + 4 \Delta_0}}{2}$