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\section{極坐標變換下的Laplace算子}
對於函數$u=u(x,y)$,其中$(x,y)\in D_{xy}\subseteq \mathbb R^2\backslash\{(0,0)\}$,構造極坐標變換
\begin{equation}
x=r \cos \theta ,
\end{equation}
\begin{equation}
y=r\sin\theta,
\end{equation}
其中$(r,\theta)\in D_{r\theta}\subseteq(0,+\infty )\times\left[ 0,2\pi\right),$計算得雅可比行列式
$\dfrac{\partial (x,y)}{\partial (r,\theta)}=r>0$,因此（1）式和（2）式表示一個雙射$T: (r,\theta )\mapsto (x,y)$，從而映射（函數）$(r,\theta)\mapsto u$存在。為了方便，我們還假設$u$是二階連續可微的,使得$u$對$x$,$y$的混合偏導數與求導順序無關。
由（1）（2）易得$r^2=x^2+y^2$，兩邊對變量$x$求導得$r\dfrac{\partial r}{\partial x}=x$，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$x\sin\theta=y\cos\theta$，兩邊對變量$x$求導得$\sin\theta+x\dfrac{\partial \theta}{\partial x}\cos\theta=-y\dfrac{\partial \theta}{\partial x}\sin\theta$，即$\dfrac{\partial\theta}{\partial x}=-\dfrac{\sin\theta}{x\cos\theta+y\sin\theta}$，為了使得表達式簡潔，我們在分子分母都乘以非零的$r$並將（1）（2）分別代入式中的$x$,$y$得$\dfrac{\partial\theta}{\partial x}=-\dfrac{y}{r^2}$，同理可得$\dfrac{\partial\theta}{\partial y}=\dfrac{x}{r^2}$.
為了計算$\nabla^2 u$，用鏈式法則先求對變量$x$的一階偏導數並代入上面的結論和化簡得
\[\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial x}=\dfrac{\partial u}{\partial r}\dfrac{x}{r}-\dfrac{\partial u}{\partial \theta}\dfrac{y}{r^2},\]
\[\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial y}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial y}=\dfrac{\partial u}{\partial r}\dfrac{y}{r}+\dfrac{\partial u}{\partial \theta}\dfrac{x}{r^2},\]
運用求導的乘積法則和鏈式法則得
\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}
將$x$換成$y$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}
注意由於假設二階連續可微，所以兩個混合偏導數用同一個記號表示。下面計算所需的四個二階偏導數。$\dfrac{\partial r}{\partial x}=\dfrac{x}{r}$的兩邊求對變量$x$的偏導數得\[\dfrac{\partial^2 r}{\partial x^2}=\dfrac{r^2-rx\dfrac{\partial r}{\partial x}}{r^3}=\dfrac{y^2}{r^3},\]
這里利用了一個小技巧，為了能夠通過關系式$r^2=x^2+y^2$來化簡最后的結果，分子分母同時乘以非零的$r$。把$x$和$y$互換即得\[\dfrac{\partial^2 r}{\partial y^2}=\dfrac{x^2}{r^3};\]
$\dfrac{\partial\theta}{\partial x}=-\dfrac{y}{r^2}$的兩邊求對變量$x$的偏導數得\[\dfrac{\partial^2 \theta}{\partial x^2}=\dfrac{2y}{r^3}\dfrac{\partial r}{\partial x}=\frac{2xy}{r^4}\],同理可得
\[\dfrac{\partial^2 \theta}{\partial y^2}=-\frac{2xy}{r^4},\]
將這些結果代入（3）（4）式得
\[\dfrac{\partial^2 u}{\partial x^2}=\dfrac{1}{r^4}\left[ r^2\dfrac{\partial^2 u}{\partial r^2}x^2+\left(r\dfrac{\partial u}{\partial r}+\dfrac{\partial^2 u}{\partial \theta^2}\right)y^2+2\left(\dfrac{\partial u}{\partial \theta}-r\dfrac{\partial^2 u}{\partial r\partial\theta}\right)xy\right],\]
\[\dfrac{\partial^2 u}{\partial y^2}=\dfrac{1}{r^4}\left[ \left(r\dfrac{\partial u}{\partial r}+\dfrac{\partial^2 u}{\partial \theta^2}\right)x^2+r^2\dfrac{\partial^2 u}{\partial r^2}y^2+2\left(r\dfrac{\partial^2 u}{\partial r\partial \theta}-\dfrac{\partial u}{\partial \theta}\right)xy\right],\]
兩式相加得
\[\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=\dfrac{1}{r^4}\left(r^2\dfrac{\partial^2 u}{\partial r^2}+r\dfrac{\partial u}{\partial r}+\dfrac{\partial^2 u}{\partial \theta^2}\right)\left(x^2+y^2\right),\]
注意到$r^2=x^2+y^2$，所以在極坐標變換下二維Laplace算子的表達式為
\[\boxed{\nabla^2 =\dfrac{\partial^2 }{\partial r^2}+\dfrac{1}{r}\dfrac{\partial }{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2 }{\partial \theta^2}}\]
\section{柱面坐標變換下的Laplace算子}
函數$u=u(x,y,z)$的柱面坐標變換是指
\[x=\rho \cos\phi,\]
\[y=\rho \sin\phi,\]
\[z=z,\]
於是由上一節的討論易得Laplace算子在柱面坐標變換下的表示為
\[\boxed{\nabla^2 =\dfrac{\partial^2 }{\partial \rho^2}+\dfrac{1}{\rho}\dfrac{\partial }{\partial \rho}+\dfrac{1}{\rho^2}\dfrac{\partial^2 }{\partial \phi^2}+\dfrac{\partial^2 }{\partial z^2}}\]
\section{球面坐標變換下的Laplace算子}
對於函數$u=u(x,y,z)$,其中$(x,y,z)\in \Omega_{xyz}\subseteq \mathbb R^3\backslash\{(0,0,0)\}$,構造球面坐標變換
\[x=r\sin\theta\cos\phi,\]
\[y=r\sin\theta\sin\phi,\]
\[z=r\cos\theta,\]
其中$(r,\theta,\phi)\in \Omega_{r\theta\phi}\subseteq(0,+\infty)\times[0,\pi]\times[0,2\pi)$.類似於極坐標變換的情形，以上三式給出了一個雙射$S:(r,\theta,\phi)\mapsto (x,y,z)$,從而函數$(r,\theta,\phi)\mapsto u$存在。為了方便我們仍然假設$u$是二階連續可微的。
為了能夠利用第一節的結論，我們令$\rho=r\sin\theta$（這實際上是引入了柱面坐標$(\rho,\phi,z)$），於是$x=\rho \cos\phi$，$y=\rho\sin\phi$,對$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}$可以利用極坐標變換的結論得
\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}
我們還需計算$\dfrac{\partial u}{\partial \rho}$,注意到$(z,\rho)\mapsto (r,\theta)$是一個極坐標變換，於是由第一節的討論可知
\[\dfrac{\partial u}{\partial \rho}=\dfrac{\partial u}{\partial r}\dfrac{\rho}{r}+\dfrac{\partial u}{\partial \theta}\dfrac{z}{r^2},\]
將$z=r\cos\theta$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算子在球面坐標變換下的表示為
\[\boxed{\nabla^2=\dfrac{\partial^2}{\partial r^2}+\dfrac{2}{r}\dfrac{\partial}{\partial r}+\dfrac{1}{r^2}\left(\dfrac{\partial^2}{\partial \theta^2}+\cot\theta\dfrac{\partial}{\partial\theta}+\csc^2 \theta \dfrac{\partial^2}{\partial \phi^2}\right)}\]
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