MIT 18.03 Differential Equation - 4.3 Decoupling

01 Feb 2020 - Less than 1 minute read

常微分方程组（三）：解耦

对于一个系统 另\(\left\{\begin{matrix}u = mx+ny\\ v = px+qy\end{matrix}\right.\)

使得

\[\left\{\begin{matrix}x'=ax+by\\y'=cx+dy\end{matrix}\right.\ \Rightarrow \left\{\begin{matrix}u'=k_1u\\v'=k_2v\end{matrix}\right.\]

称为解耦 decoupling

注：并不能简化运算 但能给出一些物理意义

一般方法

• 条件

  特征值必须为实数且完备

• 目的

  找到解耦矩阵$D$令\(\begin{pmatrix}u\\v\end{pmatrix} = D\begin{pmatrix}x\\y\end{pmatrix}\)使得方程组被解耦

  所需的代换 \(\begin{pmatrix}x\\y\end{pmatrix} = D^{-1}\begin{pmatrix}u\\v\end{pmatrix}\) 称$D^{-1}$为$E$

• $E$ 需满足的条件

  $E = [\overrightarrow{\alpha_1},\overrightarrow{\alpha_2}]$，$\overrightarrow{\alpha_1},\overrightarrow{\alpha_2}$为系统矩阵的特征值

• 为什么？

  \[AE = A[\overrightarrow{\alpha_1},\overrightarrow{\alpha_2}] = [A\overrightarrow{\alpha_1},A\overrightarrow{\alpha_2}] = [\lambda_1\overrightarrow{\alpha_1},\lambda_2\overrightarrow{\alpha_2}] = E\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}\]

  令\(\begin{pmatrix}x\\y\end{pmatrix} = \textbf{x}\), \(\begin{pmatrix}u\\v\end{pmatrix} = \textbf{u}\), 则$\textbf{x} = E\textbf{u}$

  原方程组为$\textbf{x}’ = A\textbf{x}$

  带入得 \(E\textbf{u}' = AE\textbf{u} = E\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}\textbf{u}\)

  \[\textbf{u}' = \begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}\textbf{u}\] \[\left\{\begin{matrix}u'=\lambda_1u\\v'=\lambda_2v\end{matrix}\right.\]

  方程组被解耦