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. 

即, 若 $u=f(x,y)\in C^2$, 在极坐标变换 $x=\rho\cos\theta$, $y=\rho\sin\theta$ 下, 二维拉普拉斯方程为

\[
\Delta u=\frac{\partial^2 u}{\partial\rho^2}+\frac{1}{\rho}\frac{\partial u}{\partial\rho}+\frac{1}{\rho^2}\frac{\partial^2 u}{\partial\theta^2}.
\]

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,

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

$u=f(x,y)\in C^2$, 故 $\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y\partial x}$. $u=f(\rho\cos\theta,\rho\sin\theta)$, 由复合函数求导法则,
\[
\begin{aligned}
\frac{\partial u}{\partial\rho}&=\frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial\rho}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial\rho}=\frac{\partial f}{\partial x}\cos\theta+\frac{\partial f}{\partial y}\sin\theta,\\
\frac{\partial u}{\partial\theta}&=\frac{\partial f}{\partial x}\cdot\frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y}\cdot\frac{\partial y}{\partial\theta}=\frac{\partial f}{\partial x}\cdot(-\rho\sin\theta)+\frac{\partial f}{\partial y}\cdot\rho\cos\theta.
\end{aligned}
\]

\[
\begin{split}
\frac{\partial^2 u}{\partial\rho^2}&=\frac{\partial}{\partial\rho}\Bigl(\frac{\partial u}{\partial\rho}\Bigr)=\frac{\partial}{\partial\rho}\Bigl(\frac{\partial f}{\partial x}\cos\theta+\frac{\partial f}{\partial y}\sin\theta\Bigr)\\
&=\cos\theta\cdot\frac{\partial}{\partial\rho}\Bigl(\frac{\partial f}{\partial x}\Bigr)+\sin\theta\cdot\frac{\partial}{\partial\rho}\Bigl(\frac{\partial f}{\partial y}\Bigr)\\
&=\cos\theta\cdot\Bigl(\frac{\partial^2 f}{\partial x^2}\cdot\frac{\partial x}{\partial\rho}+\frac{\partial^2 f}{\partial x\partial y}\cdot\frac{\partial y}{\partial\rho}\Bigr)+\sin\theta\cdot\Bigl(\frac{\partial^2 f}{\partial y\partial x}\cdot\frac{\partial x}{\partial\rho}+\frac{\partial^2 f}{\partial y^2}\cdot\frac{\partial y}{\partial\rho}\Bigr)\\
&=\cos\theta\cdot\Bigl(\frac{\partial^2 f}{\partial x^2}\cdot\cos\theta+\frac{\partial^2 f}{\partial x\partial y}\cdot\sin\theta\Bigr)+\sin\theta\cdot\Bigl(\frac{\partial^2 f}{\partial y\partial x}\cdot\cos\theta+\frac{\partial^2 f}{\partial y^2}\cdot\sin\theta\Bigr)\\
&=\frac{\partial^2 f}{\partial x^2}\cdot\cos^2\theta+\frac{\partial^2 f}{\partial x\partial y}\cdot\sin\theta\cos\theta+\frac{\partial^2 f}{\partial y\partial x}\cdot\sin\theta\cos\theta+\frac{\partial^2 f}{\partial y^2}\cdot\sin^2\theta\\
%-------------------------
&=\frac{\partial^2 f}{\partial x^2}\cdot\cos^2\theta+\frac{\partial^2 f}{\partial x\partial y}\cdot 2\sin\theta\cos\theta+\frac{\partial^2 f}{\partial y^2}\cdot\sin^2\theta.
\end{split}
\]

\[
\begin{split}
\frac{\partial^2 u}{\partial\theta^2}&=\frac{\partial}{\partial\theta}\Bigl(\frac{\partial u}{\partial\theta}\Bigr)=\frac{\partial}{\partial\theta}\Bigl(\frac{\partial f}{\partial x}\cdot(-\rho\sin\theta)+\frac{\partial f}{\partial y}\cdot\rho\cos\theta\Bigr)\\
&=\rho\cdot\frac{\partial}{\partial\theta}\Bigl(-\sin\theta\cdot\frac{\partial f}{\partial x}+\cos\theta\cdot\frac{\partial f}{\partial y}\Bigr)\\
&=\rho\biggl[-\cos\theta\cdot\frac{\partial f}{\partial x}-\sin\theta\cdot\frac{\partial}{\partial\theta}\Bigl(\frac{\partial f}{\partial x}\Bigr)+(-\sin\theta)\frac{\partial f}{\partial y}+\cos\theta\cdot\frac{\partial}{\partial\theta}\Bigl(\frac{\partial f}{\partial y}\Bigr)\biggr],
\end{split}
\]
其中
\[
\begin{aligned}
\frac{\partial}{\partial\theta}\Bigl(\frac{\partial f}{\partial x}\Bigr)&=\frac{\partial^2 f}{\partial x^2}\cdot\frac{\partial x}{\partial\theta}+\frac{\partial^2 f}{\partial x\partial y}\cdot\frac{\partial y}{\partial\theta}=\frac{\partial^2 f}{\partial x^2}\cdot(-\rho\sin\theta)+\frac{\partial^2 f}{\partial x\partial y}\cdot\rho\cos\theta,\\
%--------------------
\frac{\partial}{\partial\theta}\Bigl(\frac{\partial f}{\partial y}\Bigr)&=\frac{\partial^2 f}{\partial y\partial x}\cdot\frac{\partial x}{\partial\theta}+\frac{\partial^2 f}{\partial y^2}\cdot\frac{\partial y}{\partial\theta}=\frac{\partial^2 f}{\partial y\partial x}\cdot(-\rho\sin\theta)+\frac{\partial^2 f}{\partial y^2}\cdot\rho\cos\theta,\\
\end{aligned}
\]

于是
\[
\begin{split}
%-----------------------
\frac{\partial^2 u}{\partial\theta^2}&=\rho\biggl[-\cos\theta\cdot\frac{\partial f}{\partial x}-\sin\theta\cdot\Bigl(\frac{\partial^2 f}{\partial x^2}\cdot(-\rho\sin\theta)+\frac{\partial^2 f}{\partial x\partial y}\rho\cos\theta\Bigr)+(-\sin\theta)\frac{\partial f}{\partial y}\\
&\qquad+\cos\theta\cdot\Bigl(\frac{\partial^2 f}{\partial y\partial x}(-\rho\sin\theta)+\frac{\partial^2 f}{\partial y^2}\cdot\rho\cos\theta\Bigr)\biggr]\\
%--------------------------
&=\rho\biggl[-\cos\theta\cdot\frac{\partial f}{\partial x}+\rho\sin^2\theta\cdot\frac{\partial^2 f}{\partial x^2}-\rho\sin\theta\cos\theta\cdot\frac{\partial^2 f}{\partial x\partial y}-\sin\theta\cdot\frac{\partial f}{\partial y}\\
&\qquad-\rho\sin\theta\cos\theta\cdot\frac{\partial^2 f}{\partial y\partial x}+\rho\cos^2\theta\frac{\partial^2 f}{\partial y^2}\biggr]\\
%--------------------------
&=-\rho\biggl[\cos\theta\cdot\frac{\partial f}{\partial x}+\sin\theta\cdot\frac{\partial f}{\partial y}\biggr]+\rho^2\biggl[\sin^2\theta\cdot\frac{\partial^2 f}{\partial x^2}+\cos^2\theta\cdot\frac{\partial^2 f}{\partial y^2}\\
&\qquad -2\sin\theta\cos\theta\cdot\frac{\partial^2 f}{\partial x\partial y}\biggr]\\
\end{split}
\]

因此
\[
\begin{split}
&\frac{\partial^2 u}{\partial\rho^2}+\frac{1}{\rho}\frac{\partial u}{\partial\rho}+\frac{1}{\rho^2}\frac{\partial^2 u}{\partial\theta^2}\\
=&\biggl[\frac{\partial^2 f}{\partial x^2}\cdot\cos^2\theta+\frac{\partial^2 f}{\partial x\partial y}\cdot 2\sin\theta\cos\theta+\frac{\partial^2 f}{\partial y^2}\cdot\sin^2\theta\biggr]+\frac{1}{\rho}\cdot\biggl[\frac{\partial f}{\partial x}\cos\theta+\frac{\partial f}{\partial y}\sin\theta\biggr]\\
&\quad+\frac{1}{\rho^2}\biggl[-\rho\Bigl(\cos\theta\cdot\frac{\partial f}{\partial x}+\sin\theta\cdot\frac{\partial f}{\partial y}\Bigr)+\rho^2\Bigl(\sin^2\theta\cdot\frac{\partial^2 f}{\partial x^2}+\cos^2\theta\cdot\frac{\partial^2 f}{\partial y^2}\\
&\qquad -2\sin\theta\cos\theta\cdot\frac{\partial^2 f}{\partial x\partial y}\Bigr)\biggr]\\
&=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}.
\end{split}
\]
 

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