嘿，我当初啃Ahlfors的复分析教材时，也在拉普拉斯方程转极坐标这步卡了好久！刚好能给你掰扯清楚整个推导过程～

首先得明确直角坐标$(x,y)$和极坐标$(r,\theta)$的转换关系：
$$
x = r\cos\theta, \quad y = r\sin\theta
$$
反过来也可以写成：
$$
r = \sqrt{x^2 + y^2}, \quad \theta = \arctan\left(\frac{y}{x}\right)
$$

接下来我们要用链式法则把直角坐标下的偏导数转换成极坐标的形式，这是整个推导的核心。

第一步：计算一阶偏导数的转换 #

先算出$r$和$\theta$对$x$、$y$的偏导数：

• 对$x$的偏导：$\frac{\partial r}{\partial x} = \frac{x}{r} = \cos\theta$，$\frac{\partial \theta}{\partial x} = -\frac{y}{r^2} = -\frac{\sin\theta}{r}$
• 对$y$的偏导：$\frac{\partial r}{\partial y} = \frac{y}{r} = \sin\theta$，$\frac{\partial \theta}{\partial y} = \frac{x}{r^2} = \frac{\cos\theta}{r}$

根据链式法则，$u$对$x$、$y$的一阶偏导可以表示为：
$$
\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r}\cdot\frac{\partial r}{\partial x} + \frac{\partial u}{\partial \theta}\cdot\frac{\partial \theta}{\partial x} = \cos\theta\frac{\partial u}{\partial r} - \frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}
$$
$$
\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r}\cdot\frac{\partial r}{\partial y} + \frac{\partial u}{\partial \theta}\cdot\frac{\partial \theta}{\partial y} = \sin\theta\frac{\partial u}{\partial r} + \frac{\cos\theta}{r}\frac{\partial u}{\partial \theta}
$$

第二步：计算二阶偏导数并求和 #

接下来我们需要计算$\frac{\partial^2 u}{\partial x^{2}$和$\frac{\partial}2 u}{\partial y^2}$，这里要注意$\frac{\partial u}{\partial r}$和$\frac{\partial u}{\partial \theta}$本身也是$r$和$\theta$的函数，所以求二阶偏导时还要再次用链式法则。

先算$\frac{\partial^2 u}{\partial x^2}$：对$\frac{\partial u}{\partial x}$再次求$x$的偏导，拆分后逐项计算，最后整理得到：
$$
\frac{\partial^2 u}{\partial x^2} = \cos^{2\theta\frac{\partial}2 u}{\partial r^2} - \frac{2\sin\theta\cos\theta}{r}\frac{\partial^2 u}{\partial r\partial\theta} + \frac{\sin^2\theta}{r2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\sin^2\theta}{r}\frac{\partial u}{\partial r} + \frac{2\sin\theta\cos\theta}{r^2}\frac{\partial u}{\partial \theta}
$$

用同样的方法计算$\frac{\partial^2 u}{\partial y^2}$，得到：
$$
\frac{\partial^2 u}{\partial y^2} = \sin^{2\theta\frac{\partial}2 u}{\partial r^2} + \frac{2\sin\theta\cos\theta}{r}\frac{\partial^2 u}{\partial r\partial\theta} + \frac{\cos^2\theta}{r2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cos^2\theta}{r}\frac{\partial u}{\partial r} - \frac{2\sin\theta\cos\theta}{r^2}\frac{\partial u}{\partial \theta}
$$

现在把这两个二阶偏导相加，你会发现很多项会相互抵消：

• 交叉项$\frac{\partial^2 u}{\partial r\partial\theta}$的系数抵消为0
• $\frac{\partial u}{\partial \theta}$的系数也抵消为0
• 剩下的项利用$\sin^2\theta + \cos^2\theta = 1$合并，得到：
  $$
  \Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^{2}\frac{\partial}2 u}{\partial \theta^2}
  $$

第三步：整理成Ahlfors书中的形式 #

把上面的式子两边乘以$r^2$，或者对前两项做变形：
$$
r\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right) = r\left(\frac{\partial u}{\partial r} + r\frac{\partial^2 u}{\partial r^2}\right) = r\frac{\partial u}{\partial r} + r^{2\frac{\partial}2 u}{\partial r^2}
$$

把它和$\frac{\partial^2 u}{\partial \theta^{2}$相加，就等于$r}2\Delta u$，而因为$u$是调和函数，$\Delta u = 0$，所以最终得到：
$$
r\frac{\partial\left(r\frac{\partial u}{\partial r}\right)}{\partial r} + \frac{\partial^2 u}{\partial \theta^2} = 0
$$

这样就和Ahlfors书里的形式完全一致啦！是不是一下子就通了😉

备注：内容来源于stack exchange，提问作者tianhaowu