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\section{极坐标变换下的Laplace算子}
对于函数$u=u(x,y)$,其中$(x,y)\in D_{xy}\subseteq \mathbb R^2\backslash\{(0,0)\}$,构造极坐标变换
\begin{equation}
x=r \cos \theta ,
\end{equation}
\begin{equation}
y=r\sin\theta,
\end{equation}
其中$(r,\theta)\in D_{r\theta}\subseteq(0,+\infty )\times\left[ 0,2\pi\right),$计算得雅可比行列式
$\dfrac{\partial (x,y)}{\partial (r,\theta)}=r>0$,因此（1）式和（2）式表示一个双射$T: (r,\theta )\mapsto (x,y)$，从而映射（函数）$(r,\theta)\mapsto u$存在。为了方便，我们还假设$u$是二阶连续可微的,使得$u$对$x$,$y$的混合偏导数与求导顺序无关。
由（1）（2）易得$r^2=x^2+y^2$，两边对变量$x$求导得$r\dfrac{\partial r}{\partial x}=x$，所以$\dfrac{\partial r}{\partial x}=\dfrac{x}{r}$，同理可得$\dfrac{\partial r}{\partial y}=\dfrac{y}{r}$。由（1）（2）也易得$x\sin\theta=y\cos\theta$，两边对变量$x$求导得$\sin\theta+x\dfrac{\partial \theta}{\partial x}\cos\theta=-y\dfrac{\partial \theta}{\partial x}\sin\theta$，即$\dfrac{\partial\theta}{\partial x}=-\dfrac{\sin\theta}{x\cos\theta+y\sin\theta}$，为了使得表达式简洁，我们在分子分母都乘以非零的$r$并将（1）（2）分别代入式中的$x$,$y$得$\dfrac{\partial\theta}{\partial x}=-\dfrac{y}{r^2}$，同理可得$\dfrac{\partial\theta}{\partial y}=\dfrac{x}{r^2}$.
为了计算$\nabla^2 u$，用链式法则先求对变量$x$的一阶偏导数并代入上面的结论和化简得
\[\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial x}=\dfrac{\partial u}{\partial r}\dfrac{x}{r}-\dfrac{\partial u}{\partial \theta}\dfrac{y}{r^2},\]
\[\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial y}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial y}=\dfrac{\partial u}{\partial r}\dfrac{y}{r}+\dfrac{\partial u}{\partial \theta}\dfrac{x}{r^2},\]
运用求导的乘积法则和链式法则得
\begin{equation}
\dfrac{\partial^2 u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial x}\right)=\left(\dfrac{\partial^2 u}{\partial r^2}\dfrac{\partial r}{\partial x}+\dfrac{\partial^2 u}{\partial r\partial \theta}\dfrac{\partial \theta}{\partial x}\right)\dfrac{\partial r}{\partial x}+\dfrac{\partial u}{\partial r}\dfrac{\partial^2 r}{\partial x^2}+\left(\dfrac{\partial^2 u}{\partial r\partial \theta}\dfrac{\partial r}{\partial x}+\dfrac{\partial^2 u}{\partial \theta^2}\dfrac{\partial \theta}{\partial x}\right)\dfrac{\partial \theta}{\partial x}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial^2 \theta}{\partial x^2},
\end{equation}
将$x$换成$y$得
\begin{equation}
\dfrac{\partial^2 u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial y}\right)=\left(\dfrac{\partial^2 u}{\partial r^2}\dfrac{\partial r}{\partial y}+\dfrac{\partial^2 u}{\partial r\partial \theta}\dfrac{\partial \theta}{\partial y}\right)\dfrac{\partial r}{\partial y}+\dfrac{\partial u}{\partial r}\dfrac{\partial^2 r}{\partial y^2}+\left(\dfrac{\partial^2 u}{\partial r\partial \theta}\dfrac{\partial r}{\partial y}+\dfrac{\partial^2 u}{\partial \theta^2}\dfrac{\partial \theta}{\partial y}\right)\dfrac{\partial \theta}{\partial y}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial^2 \theta}{\partial y^2}.
\end{equation}
注意由于假设二阶连续可微，所以两个混合偏导数用同一个记号表示。下面计算所需的四个二阶偏导数。$\dfrac{\partial r}{\partial x}=\dfrac{x}{r}$的两边求对变量$x$的偏导数得\[\dfrac{\partial^2 r}{\partial x^2}=\dfrac{r^2-rx\dfrac{\partial r}{\partial x}}{r^3}=\dfrac{y^2}{r^3},\]
这里利用了一个小技巧，为了能够通过关系式$r^2=x^2+y^2$来化简最后的结果，分子分母同时乘以非零的$r$。把$x$和$y$互换即得\[\dfrac{\partial^2 r}{\partial y^2}=\dfrac{x^2}{r^3};\]
$\dfrac{\partial\theta}{\partial x}=-\dfrac{y}{r^2}$的两边求对变量$x$的偏导数得\[\dfrac{\partial^2 \theta}{\partial x^2}=\dfrac{2y}{r^3}\dfrac{\partial r}{\partial x}=\frac{2xy}{r^4}\],同理可得
\[\dfrac{\partial^2 \theta}{\partial y^2}=-\frac{2xy}{r^4},\]
将这些结果代入（3）（4）式得
\[\dfrac{\partial^2 u}{\partial x^2}=\dfrac{1}{r^4}\left[ r^2\dfrac{\partial^2 u}{\partial r^2}x^2+\left(r\dfrac{\partial u}{\partial r}+\dfrac{\partial^2 u}{\partial \theta^2}\right)y^2+2\left(\dfrac{\partial u}{\partial \theta}-r\dfrac{\partial^2 u}{\partial r\partial\theta}\right)xy\right],\]
\[\dfrac{\partial^2 u}{\partial y^2}=\dfrac{1}{r^4}\left[ \left(r\dfrac{\partial u}{\partial r}+\dfrac{\partial^2 u}{\partial \theta^2}\right)x^2+r^2\dfrac{\partial^2 u}{\partial r^2}y^2+2\left(r\dfrac{\partial^2 u}{\partial r\partial \theta}-\dfrac{\partial u}{\partial \theta}\right)xy\right],\]
两式相加得
\[\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=\dfrac{1}{r^4}\left(r^2\dfrac{\partial^2 u}{\partial r^2}+r\dfrac{\partial u}{\partial r}+\dfrac{\partial^2 u}{\partial \theta^2}\right)\left(x^2+y^2\right),\]
注意到$r^2=x^2+y^2$，所以在极坐标变换下二维Laplace算子的表达式为
\[\boxed{\nabla^2 =\dfrac{\partial^2 }{\partial r^2}+\dfrac{1}{r}\dfrac{\partial }{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2 }{\partial \theta^2}}\]
\section{柱面坐标变换下的Laplace算子}
函数$u=u(x,y,z)$的柱面坐标变换是指
\[x=\rho \cos\phi,\]
\[y=\rho \sin\phi,\]
\[z=z,\]
于是由上一节的讨论易得Laplace算子在柱面坐标变换下的表示为
\[\boxed{\nabla^2 =\dfrac{\partial^2 }{\partial \rho^2}+\dfrac{1}{\rho}\dfrac{\partial }{\partial \rho}+\dfrac{1}{\rho^2}\dfrac{\partial^2 }{\partial \phi^2}+\dfrac{\partial^2 }{\partial z^2}}\]
\section{球面坐标变换下的Laplace算子}
对于函数$u=u(x,y,z)$,其中$(x,y,z)\in \Omega_{xyz}\subseteq \mathbb R^3\backslash\{(0,0,0)\}$,构造球面坐标变换
\[x=r\sin\theta\cos\phi,\]
\[y=r\sin\theta\sin\phi,\]
\[z=r\cos\theta,\]
其中$(r,\theta,\phi)\in \Omega_{r\theta\phi}\subseteq(0,+\infty)\times[0,\pi]\times[0,2\pi)$.类似于极坐标变换的情形，以上三式给出了一个双射$S:(r,\theta,\phi)\mapsto (x,y,z)$,从而函数$(r,\theta,\phi)\mapsto u$存在。为了方便我们仍然假设$u$是二阶连续可微的。
为了能够利用第一节的结论，我们令$\rho=r\sin\theta$（这实际上是引入了柱面坐标$(\rho,\phi,z)$），于是$x=\rho \cos\phi$，$y=\rho\sin\phi$,对$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}$可以利用极坐标变换的结论得
\begin{equation}
\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=\dfrac{\partial^2 u }{\partial \rho^2}+\dfrac{1}{\rho}\dfrac{\partial u}{\partial \rho}+\dfrac{1}{\rho^2}\dfrac{\partial^2 u}{\partial \phi^2}
\end{equation}
同理，由$\rho=r\sin\theta$，$z=r\cos\theta$得
\begin{equation}
\dfrac{\partial^2 u}{\partial \rho^2}+\dfrac{\partial^2 u}{\partial z^2}=\dfrac{\partial^2 u }{\partial r^2}+\dfrac{1}{r}\dfrac{\partial u}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2 u}{\partial \theta^2}
\end{equation}
（5）和（6）相加得
\begin{equation}
\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}+\dfrac{\partial^2 u}{\partial z^2}=\dfrac{1}{\rho}\dfrac{\partial u}{\partial \rho}+\dfrac{1}{\rho^2}\dfrac{\partial^2 u}{\partial \phi^2}+\dfrac{\partial^2 u }{\partial r^2}+\dfrac{1}{r}\dfrac{\partial u}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2 u}{\partial \theta^2}
\end{equation}
我们还需计算$\dfrac{\partial u}{\partial \rho}$,注意到$(z,\rho)\mapsto (r,\theta)$是一个极坐标变换，于是由第一节的讨论可知
\[\dfrac{\partial u}{\partial \rho}=\dfrac{\partial u}{\partial r}\dfrac{\rho}{r}+\dfrac{\partial u}{\partial \theta}\dfrac{z}{r^2},\]
将$z=r\cos\theta$和$\rho=r\sin\theta$代入得
\[\dfrac{\partial u}{\partial \rho}=\dfrac{\partial u}{\partial r}\sin\theta+\dfrac{1}{r}\dfrac{\partial u}{\partial \theta}\cos\theta,\]
代入（7）式得到Laplace算子在球面坐标变换下的表示为
\[\boxed{\nabla^2=\dfrac{\partial^2}{\partial r^2}+\dfrac{2}{r}\dfrac{\partial}{\partial r}+\dfrac{1}{r^2}\left(\dfrac{\partial^2}{\partial \theta^2}+\cot\theta\dfrac{\partial}{\partial\theta}+\csc^2 \theta \dfrac{\partial^2}{\partial \phi^2}\right)}\]
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