极坐标和直角坐标的雅克比矩阵推导

我们经常需要在一些问题中研究坐标系的关系，这里讲讲最常见的极坐标和直角坐标的雅克比矩阵的推导。以二维坐标为例，三维坐标也是同理。

1. 直角坐标和极坐标

直角坐标表示为$(x,y)$，极坐标表示为$(\rho ,\phi )$，它们之间有如下的关系：
$$\begin{array}{rl}{\rho }^2=x^2+y^2,& \phi =\arctan \frac{y}{x};\\ x=\rho \cos \phi ,& y=\rho \sin \phi \end{array}$$

2. 向量之间的雅克比矩阵

向量X和向量Y的微分映射由雅克比矩阵来刻画，给定两个向量$\mathbf{x}=(x_1,x_2,\cdots ,x_n{)}^T$，$\mathbf{y}=(y_1,y_2,\cdots ,y_m{)}^T$

$$\begin{array}{r}\{\begin{array}{l}\mathrm{d}{x}_1=\frac{\partial x_1}{\partial y_1}\mathrm{d}{y}_1+\frac{\partial x_1}{\partial y_2}\mathrm{d}{y}_2+\cdots +\frac{\partial x_1}{\partial y_m}\mathrm{d}{y}_m\\ \mathrm{d}{x}_2=\frac{\partial x_2}{\partial y_1}\mathrm{d}{y}_1+\frac{\partial x_2}{\partial y_2}\mathrm{d}{y}_2+\cdots +\frac{\partial x_2}{\partial y_m}\mathrm{d}{y}_m\\ ⋮\\ \mathrm{d}{x}_n=\frac{\partial x_n}{\partial y_1}\mathrm{d}{y}_1+\frac{\partial x_n}{\partial y_2}\mathrm{d}{y}_2+\cdots +\frac{\partial x_n}{\partial y_m}\mathrm{d}{y}_m\end{array}\end{array}$$

写成矩阵的形式就是：

$$\left(\begin{array}{c}\mathrm{d}{x}_1\\ \mathrm{d}{x}_2\\ ⋮\\ \mathrm{d}{x}_n\end{array}\right)=\left[\begin{array}{cccc}\frac{\partial x_1}{\partial y_1}& \frac{\partial x_1}{\partial y_2}& \cdots & \frac{\partial x_1}{\partial y_m}\\ \frac{\partial x_2}{\partial y_1}& \frac{\partial x_2}{\partial y_2}& \cdots & \frac{\partial x_2}{\partial y_m}\\ ⋮& ⋮& & ⋮\\ \frac{\partial x_n}{\partial y_1}& \frac{\partial x_n}{\partial y_2}& \cdots & \frac{\partial x_n}{\partial y_m}\end{array}\right]\left(\begin{array}{c}\mathrm{d}{y}_1\\ \mathrm{d}{y}_2\\ ⋮\\ \mathrm{d}{y}_m\end{array}\right)$$

其中的矩阵

$$\frac{\partial (x_1,x_2,\cdots ,x_n)}{\partial (y_1,y_2,\cdots ,y_m)}=\left[\begin{array}{cccc}\frac{\partial x_1}{\partial y_1}& \frac{\partial x_1}{\partial y_2}& \cdots & \frac{\partial x_1}{\partial y_m}\\ \frac{\partial x_2}{\partial y_1}& \frac{\partial x_2}{\partial y_2}& \cdots & \frac{\partial x_2}{\partial y_m}\\ ⋮& ⋮& & ⋮\\ \frac{\partial x_n}{\partial y_1}& \frac{\partial x_n}{\partial y_2}& \cdots & \frac{\partial x_n}{\partial y_m}\end{array}\right]$$

就是雅克比矩阵。我们称从坐标$\mathbf{y}$(分母)到$\mathbf{x}$(分子)的雅克比矩阵。

3. 极坐标到直角坐标的雅克比矩阵

这个比较简单，利用关系$x=\rho \cos \phi ,y=\rho \sin \phi$，

$$\begin{array}{rl}\frac{\partial x}{\partial \rho }=\cos \phi ,& \frac{\partial x}{\partial \phi }=-\rho \sin \phi \\ \frac{\partial y}{\partial \rho }=\sin \phi ,& \frac{\partial y}{\partial \phi }=\rho \cos \phi \end{array}$$

我们可以写出雅克比矩阵
$$\frac{\partial (x,y)}{\partial (\rho ,\phi )}=\left[\begin{array}{cc}\frac{\partial x}{\partial \rho }& \frac{\partial x}{\partial \phi }\\ \frac{\partial y}{\partial \rho }& \frac{\partial y}{\partial \phi }\end{array}\right]=\left[\begin{array}{cc}\cos \phi & -\rho \sin \phi \\ \sin \phi & \rho \cos \phi \end{array}\right]$$

4. 直角坐标到极坐标的雅克比矩阵

这里有两种方法。

4.1 直接求解

利用关系${\rho }^2=x^2+y^2,\phi =\arctan \frac{y}{x}$，我们可以对上式直接应用求导

对于第一个式子：$$\rho =\sqrt{x^2+y^2}$$

直接求导有：

$$\begin{gathered} \frac{\partial \rho }{\partial x}=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{\rho }=\cos \phi \\ \frac{\partial \rho }{\partial y}=\frac{2y}{2\sqrt{x^2+y^2}}=\frac{y}{\rho }=\sin \phi \end{gathered}$$

对于第二个式子直接求导有：

$$\begin{gathered} \frac{\partial \phi }{\partial x}=\frac{-\frac{y}{x^2}}{1+\frac{y^2}{x^2}}=\frac{-y}{x^2+y^2}=\frac{-y}{{\rho }^2}=\frac{-\sin \phi }{\rho } \\ \frac{\partial \phi }{\partial y}=\frac{\frac{1}{x}}{1+\frac{y^2}{x^2}}=\frac{x}{x^2+y^2}=\frac{x}{{\rho }^2}=\frac{\cos \phi }{\rho } \end{gathered}$$

当然也可以用全微分的方法来求解，我们对第一个式子全微分：

$$2\rho \mathrm{d}\rho =2x\mathrm{d}x+2y\mathrm{d}y$$

于是得到

$$\mathrm{d}\rho =\frac{x}{\rho }\mathrm{d}x+\frac{y}{\rho }\mathrm{d}y$$

于是有：
$$\frac{\partial \rho }{\partial x}=\frac{x}{\rho }=\cos \phi ,\frac{\partial y}{\partial \rho }=\frac{y}{\rho }=\sin \phi$$

对第二个式子变换一下：

$$\tan \phi =\frac{y}{x}$$

然后我们再求全微分：

$$\frac{1}{{\cos }^2\phi }\mathrm{d}\phi =-\frac{y}{x^2}\mathrm{d}x+\frac{1}{x}\mathrm{d}y$$

于是得到

$$\mathrm{d}\phi =-\frac{y{\cos }^2\phi }{x^2}\mathrm{d}x+\frac{{\cos }^2\phi }{x}\mathrm{d}y=-\frac{y}{{\rho }^2}\mathrm{d}x+\frac{x}{{\rho }^2}\mathrm{d}y=-\frac{\sin \phi }{\rho }\mathrm{d}x+\frac{\cos \phi }{\rho }\mathrm{d}y$$

于是有：
$$\frac{\partial \phi }{\partial x}=\frac{-\sin \phi }{\rho },\frac{\partial \phi }{\partial y}=\frac{\cos \phi }{\rho }$$

$$\frac{\partial (\rho ,\phi )}{\partial (x,y)}=\left[\begin{array}{cc}\frac{\partial \rho }{\partial x}& \frac{\partial \rho }{\partial y}\\ \frac{\partial \phi }{\partial x}& \frac{\partial \phi }{\partial y}\end{array}\right]=\left[\begin{array}{cc}\cos \phi & \sin \phi \\ \frac{-\sin \phi }{\rho }& \frac{\cos \phi }{\rho }\end{array}\right]$$

4.2 求逆

这里刚好是一个二阶方阵，所以可以直接对3中的雅克比矩阵求逆：

$$\frac{\partial (\rho ,\phi )}{\partial (x,y)}={\left(\frac{\partial (x,y)}{\partial (\rho ,\phi )}\right)}^{-1}={\left[\begin{array}{cc}\cos \phi & -\rho \sin \phi \\ \sin \phi & \rho \cos \phi \end{array}\right]}^{-1}=\left[\begin{array}{cc}\cos \phi & \sin \phi \\ \frac{-\sin \phi }{\rho }& \frac{\cos \phi }{\rho }\end{array}\right]$$