雅可比矩阵及其有关的线性化

1、雅可比矩阵是什么

​ 考虑向量方程对向量的导数：

​ 设：$\vec{f}\left(\vec{y}\right)={\left[\begin{array}{c}\begin{array}{c}{f}_1\left(\vec{y}\right)\\ f_2\left(\vec{y}\right)\end{array}\\ \begin{array}{c}⋮\\ f_n\left(\vec{y}\right)\end{array}\end{array}\right]}_{n\times 1}$ 而 $\vec{y}={\left[\begin{array}{c}\begin{array}{c}{y}_1\\ y_2\end{array}\\ \begin{array}{c}⋮\\ m\end{array}\end{array}\right]}_{m\times 1}$

​

​ 当其按照分子布局展开时即为雅可比矩阵：

​ $\frac{{\partial \vec{f}\left(\vec{y}\right)}_{n\times 1}}{{\partial \vec{y}}_{m\times 1}}={\left[\begin{array}{cc}\begin{array}{cc}\frac{\partial f_1\left(\vec{y}\right)}{\partial y_1}& \frac{\partial f_1\left(\vec{y}\right)}{\partial y_2}\end{array}& \begin{array}{cc}\dots & \frac{\partial f_1\left(\vec{y}\right)}{\partial y_m}\end{array}\\ \begin{array}{cc}\begin{array}{c}⋮\\ \begin{array}{c}⋮\\ \frac{\partial f_n\left(\vec{y}\right)}{\partial y_1}\end{array}\end{array}& \begin{array}{c}\ddots \\ \begin{array}{c}\\ \frac{\partial f_n\left(\vec{y}\right)}{\partial y_2}\end{array}\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}\\ \begin{array}{c}\ddots \\ \dots \end{array}\end{array}& \begin{array}{c}⋮\\ \begin{array}{c}⋮\\ \frac{\partial f_n\left(\vec{y}\right)}{\partial y_m}\end{array}\end{array}\end{array}\end{array}\right]}_{n\times m}$

考虑一个例子：$\vec{f}\left(\vec{y}\right)=\left[\begin{array}{c}{f}_1\left(\vec{y}\right)\\ f_2\left(\vec{y}\right)\end{array}\right]={\left[\begin{array}{c}{y}_1^2+y_2^2+y_3\\ y_0^2+2y_1\end{array}\right]}_{2\times 1}$ 而 $\vec{y}={\left[\begin{array}{c}\begin{array}{c}{y}_1\\ y_2\end{array}\\ y_3\end{array}\right]}_{3\times 1}$

求偏导得：$\vec{f}\left(\vec{y}\right)=\left[\begin{array}{ccc}\frac{\partial f_1\left(\vec{y}\right)}{\partial y_1}& \frac{\partial f_1\left(\vec{y}\right)}{\partial y_2}& \frac{\partial f_1\left(\vec{y}\right)}{\partial y_3}\\ \frac{\partial f_2\left(\vec{y}\right)}{\partial y_1}& \frac{\partial f_2\left(\vec{y}\right)}{\partial y_2}& \frac{\partial f_2\left(\vec{y}\right)}{\partial y_3}\end{array}\right]={\left[\begin{array}{ccc}2y_1& 2y_2& 1\\ 2& 0& 2y_3\end{array}\right]}_{2\times 3}$

---

2、利用雅可比进行线性化

​ 考虑二维情况在平衡点处线性化

​ ${\dot{x}}_1=f_1\left(x_1,x_2\right)\Rightarrow$ $\left[\begin{array}{c}\dot{x_{1d}}\\ \dot{x_{2d}}\end{array}\right]={\left[\begin{array}{cc}\frac{\partial f_1}{\partial x_1}& \frac{\partial f_1}{\partial x_2}\\ \frac{\partial f_2}{\partial x_1}& \frac{\partial f_2}{\partial x_2}\end{array}\right]}_{x=x_0}\left[\begin{array}{c}{x}_{1d}\\ x_{2d}\end{array}\right]$

​ ${\dot{x}}_2=f_2\left(x_1,x_2\right)\Rightarrow$

中间的偏导矩阵为雅可比矩阵

​

​ 来个例子：$\ddot{x}+\dot{x}+\frac{1}{x}=1$

​ 令$x_1=x$ $x_2=\dot{x}$

​ 得到状态空间方程： ${\dot{x}}_1=x_2$

​ ${\dot{x}}_2=\ddot{x}=1-\frac{1}{x}-\dot{x}=1-\frac{1}{x_1}-x_2$

​

​ 寻找其平衡点：令${\dot{x}}_1=0$，${\dot{x}}_2=0$ 可得：

​ $x_{10}=1,x_{20}=0$

​

​ 代入公式得：$\left[\begin{array}{c}\dot{x_{1d}}\\ \dot{x_{2d}}\end{array}\right]={\left[\begin{array}{cc}0& 1\\ -\left(-\frac{1}{x_1^2}\right)& -1\end{array}\right]}_{x_0}\left[\begin{array}{c}{x}_{1d}\\ x_{2d}\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}{x}_{1d}\\ x_{2d}\end{array}\right]$

​

​ 此时，$\dot{x_{2d}}=x_{1d}-x_{2d}$

​ 又因为$\ddot{x_d}=\dot{x_{2d}}$，$x_d=x_{1d}$，$\dot{x_d}=x_{2d}$

​ 所以，可得$\ddot{x_d}+\dot{x_d}-x_d=0$

​ 完成线性化！