习题 An Introduction to Manifold - Loring W. Tu, Problem 8.4

给定 $x,y$ 为 $\mathbb{R}^2$ 上的标准坐标
给定开集 $U = \mathbb{R}^2 \backslash \{(x,0) \big| x \geqslant 0\}$

$U$ 上可唯一地定义极坐标 $$ \bbox[5pt]{ \begin{aligned} x & = r\cos{\theta} \\ y & = r\sin{\theta} & r > 0 , 0 < \theta < 2\pi \end{aligned} } $$

计算 $\displaystyle \frac{\partial}{\partial{r}} , \frac{\partial}{\partial{\theta}}$ 表示为 $\displaystyle \frac{\partial}{\partial{x}} , \frac{\partial}{\partial{y}}$ 的形式

Proof.
给定 $p = (x,y) \in U$
记 $F : \mathbb{R}^2 \to (0,+\infty)\times(0,\frac{\pi}{2})$

$$ \bbox[5pt]{ (r,\theta) = F(x,y) = \left(\sqrt{x^2+y^2},\varphi_{\arctan}(x,y)\right) } $$

其中

$$ \bbox[5pt]{ \varphi_{\arctan}(x,y) = \left\{\begin{aligned} & \arctan{\frac{y}{x}} & x > 0 , y > 0 \\ & \frac{\pi}{2} & x = 0 , y > 0 \\ & \pi + \arctan{\frac{y}{x}} & x < 0 \\ & \frac{3\pi}{2} & x = 0 , y < 0 \\ & 2\pi + \arctan{\frac{y}{x}} & x > 0 , y < 0 \\ \end{aligned}\right. } $$

又记

$$ \bbox[5pt]{ \begin{aligned} F_{\ast,p}\left(\left.\frac{\partial}{\partial{x}}\right|_{p}\right) & = a\left.\frac{\partial}{\partial{r}}\right|_{F(p)} + b\left.\frac{\partial}{\partial{\theta}}\right|_{F(p)} \\ F_{\ast,p}\left(\left.\frac{\partial}{\partial{y}}\right|_{p}\right) & = c\left.\frac{\partial}{\partial{r}}\right|_{F(p)} + d\left.\frac{\partial}{\partial{\theta}}\right|_{F(p)} \end{aligned} } $$

则

$$ \bbox[5pt]{ \begin{aligned} a = & F_{\ast,p}\left(\left.\frac{\partial}{\partial{x}}\right|_{p}\right)r \\ \mathtip{\color{skyblue}{=}}{定义: F_{\ast,p}} & \frac{\partial}{\partial{x}}(r \circ F(p))\big|_{p} \\ = & \frac{\partial}{\partial{x}}(\sqrt{x^2+y^2})\big|_{p} \\ = & \frac{x}{\sqrt{x^2+y^2}} \\\\ b = & F_{\ast,p}\left(\left.\frac{\partial}{\partial{x}}\right|_{p}\right)\theta \\ \mathtip{\color{skyblue}{=}}{定义: F_{\ast,p}} & \frac{\partial}{\partial{x}}(\theta \circ F(p))\big|_{p} \\ = & \frac{\partial}{\partial{x}}\left(\varphi_{\arctan}{\frac{y}{x}}\right)\big|_{p} \\ = & \frac{-y}{x^2+y^2} \\\\ c = & F_{\ast,p}\left(\left.\frac{\partial}{\partial{y}}\right|_{p}\right)r \\ \mathtip{\color{skyblue}{=}}{定义: F_{\ast,p}} & \frac{\partial}{\partial{y}}(r \circ F(p))\big|_{p} \\ = & \frac{\partial}{\partial{y}}(\sqrt{x^2+y^2})\big|_{p} \\ = & \frac{y}{\sqrt{x^2+y^2}} \\\\ d = & F_{\ast,p}\left(\left.\frac{\partial}{\partial{y}}\right|_{p}\right)\theta \\ \mathtip{\color{skyblue}{=}}{定义: F_{\ast,p}} & \frac{\partial}{\partial{y}}(\theta \circ F(p))\big|_{p} \\ = & \frac{\partial}{\partial{y}}\left(\varphi_{\arctan}{\frac{y}{x}}\right)\big|_{p} \\ = & \frac{x}{x^2+y^2} \end{aligned} } $$

即

$$ \bbox[5pt]{ \begin{aligned} F_{\ast,p}\left(\left.\frac{\partial}{\partial{x}}\right|_{p}\right) & = \frac{x}{\sqrt{x^2+y^2}}\left.\frac{\partial}{\partial{r}}\right|_{F(p)} + \frac{-y}{x^2+y^2}\left.\frac{\partial}{\partial{\theta}}\right|_{F(p)} \\ F_{\ast,p}\left(\left.\frac{\partial}{\partial{y}}\right|_{p}\right) & = \frac{y}{\sqrt{x^2+y^2}}\left.\frac{\partial}{\partial{r}}\right|_{F(p)} + \frac{x}{x^2+y^2}\left.\frac{\partial}{\partial{\theta}}\right|_{F(p)} \end{aligned} } $$

或

$$ \bbox[5pt]{ \begin{pmatrix} \displaystyle F_{\ast,p}\left(\left.\frac{\partial}{\partial{x}}\right|_{p}\right) \\ \displaystyle F_{\ast,p}\left(\left.\frac{\partial}{\partial{y}}\right|_{p}\right) \end{pmatrix} = \begin{pmatrix} \displaystyle \frac{x}{\sqrt{x^2+y^2}} & \displaystyle \frac{-y}{x^2+y^2} \\ \displaystyle \frac{y}{\sqrt{x^2+y^2}} & \displaystyle \frac{x}{x^2+y^2} \end{pmatrix} \begin{pmatrix} \displaystyle \left.\frac{\partial}{\partial{r}}\right|_{F(p)} \\ \displaystyle \left.\frac{\partial}{\partial{\theta}}\right|_{F(p)} \end{pmatrix} } $$

那么

$$ \bbox[5pt]{ \begin{pmatrix} \displaystyle \left.\frac{\partial}{\partial{r}}\right|_{F(p)} \\ \displaystyle \left.\frac{\partial}{\partial{\theta}}\right|_{F(p)} \end{pmatrix} = \begin{pmatrix} \displaystyle \frac{x}{\sqrt{x^2+y^2}} & \displaystyle \frac{-y}{x^2+y^2} \\ \displaystyle \frac{y}{\sqrt{x^2+y^2}} & \displaystyle \frac{x}{x^2+y^2} \end{pmatrix}^{-1}\begin{pmatrix} \displaystyle F_{\ast,p}\left(\left.\frac{\partial}{\partial{x}}\right|_{p}\right) \\ \displaystyle F_{\ast,p}\left(\left.\frac{\partial}{\partial{y}}\right|_{p}\right) \end{pmatrix} } $$

故

$$ \bbox[5pt]{ \begin{pmatrix} \displaystyle \left.\frac{\partial}{\partial{r}}\right|_{F(p)} \\ \displaystyle \left.\frac{\partial}{\partial{\theta}}\right|_{F(p)} \end{pmatrix} = \begin{pmatrix} \displaystyle \frac{x}{\sqrt{x^2+y^2}} & \displaystyle \frac{y}{\sqrt{x^2+y^2}} \\ -y & x \end{pmatrix} \begin{pmatrix} \displaystyle F_{\ast,p}\left(\left.\frac{\partial}{\partial{x}}\right|_{p}\right) \\ \displaystyle F_{\ast,p}\left(\left.\frac{\partial}{\partial{y}}\right|_{p}\right) \end{pmatrix} } $$

即

$$ \bbox[5pt]{ \begin{aligned} \left.\frac{\partial}{\partial{r}}\right|_{F(p)} & = \frac{x}{\sqrt{x^2+y^2}}F_{\ast,p}\left(\left.\frac{\partial}{\partial{x}}\right|_{p}\right) + \frac{y}{\sqrt{x^2+y^2}}F_{\ast,p}\left(\left.\frac{\partial}{\partial{y}}\right|_{p}\right) \\ \left.\frac{\partial}{\partial{\theta}}\right|_{F(p)} & = -yF_{\ast,p}\left(\left.\frac{\partial}{\partial{x}}\right|_{p}\right) + xF_{\ast,p}\left(\left.\frac{\partial}{\partial{y}}\right|_{p}\right) \end{aligned} } $$

或简记为

$$ \bbox[5pt]{ \begin{aligned} \frac{\partial}{\partial{r}} & = \frac{x}{\sqrt{x^2+y^2}}\frac{\partial}{\partial{x}} + \frac{y}{\sqrt{x^2+y^2}}\frac{\partial}{\partial{y}} \\ \frac{\partial}{\partial{\theta}} & = -y\frac{\partial}{\partial{x}} + x\frac{\partial}{\partial{y}} \end{aligned} } $$