问题及解答

平面 $\mathbb{R}^2$ 上的 Laplace 方程如下:

\[
\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}
\]

将它写成极坐标的形式.

考虑极坐标 $(r,\theta)$, 它与直角坐标系的转换为

\[
\begin{cases}
x=r\cos\theta\\
y=r\sin\theta
\end{cases}
\]

从而,

\[
\begin{aligned}
r=\sqrt{x^2+y^2}\\
\theta=\arctan\frac{y}{x}+k\pi
\end{aligned}
\]

这里要注意的是角度 $\theta$ 的范围是 $[-\pi,\pi)$, 即 $k$ 的取值可能是 $0,1,-1$ 之一, 根据 $y,x$ 的符号而定.

首先, 我们证明

Claim1:

\[
\begin{cases}
\partial_r=\frac{1}{r}(x\partial_x+y\partial_y)\\
\partial_\theta=-y\partial_x+x\partial_y
\end{cases}
\]

Pf.

\[
\begin{split}
\frac{\partial}{\partial r}&=\frac{\partial x}{\partial r}\cdot\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\cdot\frac{\partial}{\partial y}\\
&=\cos\theta\frac{\partial}{\partial x}+\sin\theta\frac{\partial}{\partial y}\\
&=\frac{x}{r}\frac{\partial}{\partial x}+\frac{y}{r}\frac{\partial}{\partial y}\\
&=\frac{1}{r}(x\partial_x+y\partial_y)
\end{split}
\]

\[
\begin{split}
\frac{\partial}{\partial\theta}&=\frac{\partial x}{\partial\theta}\cdot\frac{\partial}{\partial x}+\frac{\partial y}{\partial\theta}\cdot\frac{\partial}{\partial y}\\
&=-r\sin\theta\frac{\partial}{\partial x}+r\cos\theta\frac{\partial}{\partial y}\\
&=-y\partial_x+x\partial_y
\end{split}
\]

根据 Claim 1, 容易反解出

\[
\begin{cases}
\partial_x=\frac{x}{r}\partial_r-\frac{y}{r^2}\partial_\theta\\
\partial_y=\frac{y}{r}\partial_r+\frac{x}{r^2}\partial_\theta
\end{cases}
\]
 

然后计算 $\partial_x \partial_x+\partial_y \partial_y$.

首先

\[
\begin{split}
\partial_x \partial_x&=(\frac{x}{r}\partial_r-\frac{y}{r^2}\partial_\theta)(\frac{x}{r}\partial_r-\frac{y}{r^2}\partial_\theta)\\
&=(\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_\theta)(\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_\theta)\\
&=\cos^2\theta\partial^2_{rr}-\cos\theta\partial_r(\frac{\sin\theta}{r}\partial_\theta)-\frac{\sin\theta}{r}\partial_\theta(\cos\theta\partial_r)+\frac{\sin\theta}{r}\partial_\theta(\frac{\sin\theta}{r}\partial_\theta)\\
&=\cos^2\theta\partial^2_{rr}-\sin\theta\cos\theta(-\frac{1}{r^2}\partial_\theta+\frac{1}{r}\partial_r\partial_\theta)-\frac{\sin\theta}{r}(-\sin\theta\partial_r+\cos\theta\partial_\theta\partial_r)\\
&\quad+\frac{\sin\theta}{r^2}(\cos\theta\partial_\theta+\sin\theta\partial_{\theta\theta}^2)\\
&=\cos^2\theta\partial^2_{rr}+\frac{\sin^2\theta}{r^2}\partial_{\theta\theta}^2-\frac{\sin 2\theta}{r}\partial_r\partial_\theta+\frac{\sin^2\theta}{r}\partial_r+\frac{\sin 2\theta}{r^2}\partial_\theta,
\end{split}
\]

另一方面

\[
\begin{split}
\partial_y \partial_y&=(\frac{y}{r}\partial_r+\frac{x}{r^2}\partial_\theta)(\frac{y}{r}\partial_r+\frac{x}{r^2}\partial_\theta)\\
&=(\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_\theta)(\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_\theta)\\
&=\sin^2\theta\partial^2_{rr}+\sin\theta\partial_r(\frac{\cos\theta}{r}\partial_\theta)+\frac{\cos\theta}{r}\partial_\theta(\sin\theta\partial_r)+\frac{\cos\theta}{r^2}\partial_\theta(\cos\theta\partial_\theta)\\
&=\sin^2\theta\partial^2_{rr}+\sin\theta\cos\theta(-\frac{1}{r^2}\partial_\theta+\frac{1}{r}\partial_r\partial_\theta)+\frac{\cos\theta}{r}(\cos\theta\partial_r+\sin\theta\partial_\theta\partial_r)\\
&\quad+\frac{\cos\theta}{r^2}(-\sin\theta\partial_\theta+\cos\theta\partial_{\theta\theta}^2)\\
&=\sin^2\theta\partial^2_{rr}+\frac{\cos^2\theta}{r^2}\partial_{\theta\theta}^2+\frac{\sin 2\theta}{r}\partial_\theta\partial_r+\frac{\cos^2\theta}{r}\partial_r-\frac{\sin 2\theta}{r^2}\partial_\theta,
\end{split}
\]

因此,

\[
\partial_{xx}^2+\partial_{yy}^2=\partial_{rr}^2+\frac{1}{r}\partial_r+\frac{1}{r^2}\partial_{\theta\theta}^2.
\]

设 $u=u(x,y)$ 具有二阶连续偏导数, 从而 $u_{xy}=u_{yx}$.  这里 $x=r\cos\theta$, $y=r\sin\theta$.

\[
\begin{split}
\frac{\partial u}{\partial r}&=\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial r}\\
&=\frac{\partial u}{\partial x}\cos\theta+\frac{\partial u}{\partial y}\sin\theta\\
&=\frac{1}{r}\biggl(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\biggr)
\end{split}
\]

此即 $\partial_r=\frac{1}{r}(x\partial_x+y\partial_y)$.

\[
\begin{split}
\frac{\partial u}{\partial\theta}&=\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial\theta}+\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial\theta}\\
&=\frac{\partial u}{\partial x}\cdot(-r\sin\theta)+\frac{\partial u}{\partial y}\cdot r\cos\theta\\
&=-y\frac{\partial u}{\partial x}+x\frac{\partial u}{\partial y},
\end{split}
\]

即 $\partial_\theta=-y\partial_x+x\partial_y$.