极坐标求导中的误区（个人向）&C-R条件极坐标表示

大家好，欢迎来到IT知识分享网。

上一篇的方法似乎是错误的，不过倒是挺利于记忆的。

这一篇用于记录一下个人在极坐标求导中经常犯错/不会的地方。

习题：

设 $f(z)=u(r,\phi )+\text{i}v(r,\phi )(z=r{e}^{\mathrm{i}\phi })$ 推导极坐标系下的 $C-R$ 条件。
$$\begin{array}{rl}x& =r\cos \phi ,y=r\sin \phi ,\\ r& =\sqrt{x^2+y^2},\phi =\arctan \frac{y}{x}.\end{array}$$
则有
$$\begin{array}{rl}\frac{\partial r}{\partial x}& =\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}=\cos \phi ,\frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}=\frac{y}{r}=\sin \phi \\ \frac{\partial \phi }{\partial x}& =\frac{-\frac{y}{x^2}}{1+{\left(\frac{y}{x}\right)}^2}=\frac{-y}{r^2}=-\frac{\sin \phi }{r},\frac{\partial \phi }{\partial x}=\frac{\frac{1}{x}}{1+{\left(\frac{y}{x}\right)}^2}=\frac{x}{r^2}=\frac{\cos \phi }{r}.\end{array}$$
上面这里就是误区所在

比如求 $\frac{\partial r}{\partial x}$ 的时候，不能使用关系式 $x=r\cos \theta \to r=\frac{x}{\cos \theta }$ 来计算。因为 $\cos \theta$ 和 $x$ 看似不相干，是相互独立的，但实际上却不是。所以求导的时候，应该回归到 $x$ 和 $y$ 这两个变量上面来，这两变量是完全独立的。

$$\begin{array}{rl}\frac{\partial u}{\partial x}& =\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \phi }\frac{\partial \phi }{\partial x}=\cos \phi \frac{\partial u}{\partial r}-\frac{\sin \phi }{r}\frac{\partial u}{\partial \phi },\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial u}{\partial \phi }=\sin \phi \frac{\partial u}{\partial r}+\frac{\cos \phi }{r}\frac{\partial u}{\partial \phi },\\ \frac{\partial v}{\partial x}& =\frac{\partial v}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial v}{\partial \phi }\frac{\partial \phi }{\partial x}=\cos \phi \frac{\partial v}{\partial r}-\frac{\sin \phi }{r}\frac{\partial v}{\partial \phi },\frac{\partial v}{\partial y}=\frac{\partial v}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial v}{\partial \phi }\frac{\partial \phi }{\partial y}=\sin \phi \frac{\partial v}{\partial r}+\frac{\cos \phi }{r}\frac{\partial v}{\partial \phi }.\end{array}$$
结合直角坐标系中的 $C-R$ 条件：
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y},$$
得
$$\begin{array}{rlr}\cos \phi \frac{\partial u}{\partial r}-\frac{\sin \phi }{r}\frac{\partial u}{\partial \phi }& =\sin \phi \frac{\partial v}{\partial r}+\frac{\cos \phi }{r}\frac{\partial v}{\partial \phi },& \text{(1.1)}\\ \sin \phi \frac{\partial u}{\partial r}+\frac{\cos \phi }{r}\frac{\partial u}{\partial \phi }& =-\cos \phi \frac{\partial v}{\partial r}+\frac{\sin \phi }{r}\frac{\partial v}{\partial \phi }.& \text{(1.2)}\end{array}$$
考虑 $(1.1)\times \cos \phi +(1.2)\times \sin \phi$ 和 $(1.1)\times \sin \phi -(1.2)\times \cos \phi$，得到
$$\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \phi },\frac{\partial u}{\partial \phi }=-r\frac{\partial v}{\partial r}.$$