# 相平面

## 線性系統的例子

{\displaystyle {\begin{aligned}{\frac {dx}{dt}}&=Ax+By\\{\frac {dy}{dt}}&=Cx+Dy\end{aligned}}}

{\displaystyle {\begin{aligned}&{\frac {d}{dt}}{\begin{pmatrix}x\\y\\\end{pmatrix}}={\begin{pmatrix}A&B\\C&D\\\end{pmatrix}}{\begin{pmatrix}x\\y\\\end{pmatrix}}\\&{\frac {d\mathbf {x} }{dt}}=\mathbf {A} \mathbf {x} .\end{aligned}}}

${\displaystyle {\frac {dy}{dx}}={\frac {Cx+Dy}{Ax+By}}}$

### 用特徵值求解

${\displaystyle \det(\mathbf {A} -\lambda \mathbf {I} )=0}$

${\displaystyle \mathbf {A} \mathbf {x} =\lambda \mathbf {x} }$

${\displaystyle x={\begin{bmatrix}k_{1}\\k_{2}\end{bmatrix}}c_{1}e^{\lambda _{1}t}+{\begin{bmatrix}k_{3}\\k_{4}\end{bmatrix}}c_{2}e^{\lambda _{2}t}.}$

${\displaystyle \lambda ^{2}-(A+D)\lambda +(AD-BC)=0}$

${\displaystyle \lambda ^{2}-p\lambda +q=0}$

${\displaystyle p=A+D=\mathrm {tr} (\mathbf {A} )\,,}$

（"tr"表示矩陣的）以及

${\displaystyle q=AD-BC=\det(\mathbf {A} )\,.}$

${\displaystyle \lambda ={\frac {1}{2}}(p\pm {\sqrt {\Delta }})\,}$

${\displaystyle \Delta =p^{2}-4q\,.}$

### 特徵向量及節點

• 若二個符號一正一負，特徵向量的交點為鞍點
• 若二個符號都為正，表示系統會遠離特徵向量，交點是不穩定節點
• 若二個符號都為負，表示系統會趨向特徵向量，交點是穩定節點。

## 參考資料

1. D.W. Jordan; P. Smith. Non-Linear Ordinary Differential Equations: Introduction for Scientists and Engineers 4th. Oxford University Press. 2007. ISBN 978-0-19-920825-8.
2. ^ K.T. Alligood; T.D. Sauer; J.A. Yorke. Chaos: An Introduction to Dynamical Systems. Springer. 1996. ISBN 978-0-38794-677-1.
3. ^ W.E. Boyce; R.C. Diprima. Elementary Differential Equations and Boundary Value Problems 4th. John Wiley & Sons. 1986. ISBN 0-471-83824-1. 已忽略未知参数|url-access= (帮助)