二维拉普拉斯方程极坐标形式

二维Laplace方程
$$
\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0
$$
令$x=rcos\theta, y=rsin\theta$，则
$$
\left\{\begin{aligned}\frac{\partial u}{\partial r}&=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial u}{\partial x}cos\theta+\frac{\partial u}{\partial y}sin\theta\\\frac{\partial u}{\partial \theta}&=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta}=-\frac{\partial u}{\partial x}rsin\theta+\frac{\partial u}{\partial y}rcos\theta\end{aligned}\right.
$$

$$
\left\{\begin{aligned}\frac{\partial^2u}{\partial r^2}&=\frac{\partial^2 u}{\partial x^2}\frac{\partial x}{\partial r}cos\theta+\frac{\partial^2u}{\partial x\partial y}\frac{\partial y}{\partial r}cos\theta+\frac{\partial^2u}{\partial y\partial x}\frac{\partial x}{\partial r}sin\theta+\frac{\partial^2 u}{\partial y^2}\frac{\partial y}{\partial r}sin\theta\\&=\frac{\partial u}{\partial x^2}cos^2\theta+2\frac{\partial ^2u}{\partial x\partial y}sin\theta cos\theta+\frac{\partial^2u}{\partial y^2}sin^2\theta\\\frac{\partial^2u}{\partial \theta^2}&=-\frac{\partial^2u}{\partial x^2}\frac{\partial x}{\partial \theta}rsin\theta-\frac{\partial^2 u}{\partial x\partial y}rsin\theta-\frac{\partial u}{\partial x}sin^2\theta\\&+\frac{\partial^2u}{\partial y\partial x}\frac{\partial x}{\partial\theta}rcos\theta+\frac{\partial^2 u}{\partial y^2}\frac{\partial y}{\partial \theta}rcos\theta-\frac{\partial u}{\partial y}rsin\theta\\&=\frac{\partial^2 u}{\partial x^2}r^2sin^2\theta-\frac{\partial^2u}{\partial x\partial y}2r^2sin\theta cos\theta+\frac{\partial^2u}{\partial y^2}r^2cos^2\theta-\frac{\partial u}{\partial x}rsin\theta-\frac{\partial u}{\partial y}rsin\theta\end{aligned}\right.
$$

所以有
$$
r^2\frac{\partial^2u}{\partial r^2}+\frac{\partial^2 u}{\partial \theta^2}=r^2\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)-\left(u_xrcos\theta+u_yrsin\theta\right)
$$
右端项采用如下处理
$$
\begin{aligned}\frac{\partial^2u}{\partial r^2}+r^2\frac{\partial^2 u}{\partial \theta^2}=&r^2\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right)-r\left(u_x\frac{\partial x}{\partial r}+u_y\frac{\partial y}{\partial r}\right)\\=&r^2\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}-ru_r\right)\end{aligned}
$$
其中有
$$
r^2\frac{\partial^2u}{\partial r^2}+ru_r+\frac{\partial^2u}{\partial\theta^2}=r^2\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right)=0
$$
由此可得最终形式
$$
\frac{\partial^2u}{\partial r^2}+\frac{1}{r}u_r+\frac{1}{r^2}\frac{\partial^2u}{\partial\theta^2}=0
$$