Laplace Operator in Spherical Coordinates
在這篇文章中我們主要探討拉普拉斯算子\(\Delta\)如何用三維球坐標表示。我們考慮三維球坐標\((r,\theta,\varphi)\)，其對應於三維歐氏坐標有以下關係
\[x=r\sin\varphi\cos\theta,\quad y=r\sin\varphi\sin\theta,\quad z=r\cos\varphi\]
\begin{array}{l}
\displaystyle r=\sqrt{x^2+y^2+z^2},\\ \\
\displaystyle\frac{\partial r}{\partial x}=\frac{x}{r}=\sin\varphi\cos\theta,\\ \\
\displaystyle\frac{\partial r}{\partial y}=\frac{y}{r}=\sin\varphi\sin\theta,\\ \\
\displaystyle\frac{\partial r}{\partial z}=\frac{z}{r}=\cos\varphi,
\end{array}
\right .\]
\begin{array}{l}
\displaystyle\frac{\partial }{\partial x}=(\sin\varphi\cos\theta)\frac{\partial}{\partial r}-\frac{\sin\theta}{r\sin\varphi}\frac{\partial}{\partial \theta}+\frac{\cos\varphi\cos\theta}{r}\frac{\partial}{\partial\varphi},\\ \\
\displaystyle\frac{\partial \varphi}{\partial y}=(\sin\varphi\sin\theta)\frac{\partial}{\partial r}+\frac{\cos\theta}{r\sin\varphi}\frac{\partial}{\partial\theta}+\frac{\cos\varphi\sin\theta}{r}\frac{\partial}{\partial\varphi},\\ \\
\displaystyle\frac{\partial }{\partial z}=(\cos\varphi)\frac{\partial}{\partial r}-\frac{\sin\varphi}{r}\frac{\partial}{\partial\varphi}.
\end{array}
\right .\]
和
\[\frac{\partial}{\partial\theta}=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\]
以及
\[\frac{\partial}{\partial\varphi}=r\cos\varphi\cos\theta\frac{\partial}{\partial x}+r\cos\varphi\sin\theta\frac{\partial}{\partial y}-r\sin\varphi\frac{\partial}{\partial z}.\]
\left (
\begin{array}{c}
\displaystyle r\frac{\partial}{\partial r} \\
\displaystyle\frac{\partial}{\partial\theta} \\
\displaystyle\tan\varphi\frac{\partial}{\partial\varphi}
\end{array}\right )=
\left (
\begin{array}{ccc}
x & y & z \\ \\
-y & x & 0 \\ \\
\displaystyle x & y & -\frac{x^2+y^2}{z}
\end{array}\right )
\left (
\begin{array}{c}
\displaystyle\frac{\partial}{\partial x} \\
\displaystyle\frac{\partial}{\partial y} \\
\displaystyle\frac{\partial}{\partial z}
\end{array}\right ).
\]
M=\left (\begin{array}{ccc}
x & y & z \\ \\
-y & x & 0 \\ \\
\displaystyle x & y & -\frac{x^2+y^2}{z}
\end{array}\right )
\]
x & y & z \\ \\
-y & x & 0 \\ \\
\displaystyle x & y & -\frac{x^2+y^2}{z}
\end{array}\right )\left (\begin{array}{ccc}
x & -y & x \\ \\
y & x & y \\ \\
\displaystyle z & 0 & -\frac{x^2+y^2}{z}
\end{array}\right )=r^2\left (\begin{array}{ccc}
1 & 0 & 0 \\ \\
0 & \sin^2\varphi & 0 \\ \\
0 & 0 & \tan^2\varphi
\end{array}\right )\]
\Delta u(p)=\sum_{i=1}^n\frac{\partial^2 u}{\partial x_i^2}(p)=\sum_{i=1}^n\frac{d^2}{dt^2}\bigg\vert_{t=0}u(p+tv_i),\quad t\in(-\epsilon,\epsilon)
\]
其中
其中
\begin{aligned}
g_{rr}
&=\mathbf{e}_r\cdot \mathbf{e}_r\\
&=(\sin\varphi\cos\theta,\sin\varphi\sin\theta,\cos\varphi)\cdot (\sin\varphi\cos\theta,\sin\varphi\sin\theta,\cos\varphi)\\
&=1
\end{aligned}
\begin{aligned}
g_{\theta\theta}
&=\mathbf{e}_\theta\cdot \mathbf{e}_\theta\\
&=(-r\sin\varphi\sin\theta,r\sin\varphi\cos\theta,0)\cdot(-r\sin\varphi\sin\theta,r\sin\varphi\cos\theta,0)\\
&=r^2\sin^2\varphi
\end{aligned}
\begin{aligned}
g_{\varphi\varphi}
&=\mathbf{e}_\varphi\cdot \mathbf{e}_\varphi\\
&=(r\cos\varphi\cos\theta,r\cos\varphi\sin\theta,-r\sin\varphi)\cdot(r\cos\varphi\cos\theta,r\cos\varphi\sin\theta,-r\sin\varphi)\\
&=r^2.
\end{aligned}
1 & 0 & 0 \\
0 & \frac{1}{r^2\sin^2\theta} & 0 \\
0 & 0 & \frac{1}{r^2}
\end{array}\right)\]
\begin{array}{ccc}
1 & 0 & 0 \\
0 & r^2\sin^2\theta & 0 \\
0 & 0 & r^2
\end{array}\right\vert=r^4\sin^2\varphi.\]
\Delta
&=\frac{1}{r^2\sin\varphi}\frac{\partial}{\partial r}\left(\frac{r^2\sin\varphi}{1}\frac{\partial}{\partial r}\right)+\frac{1}{r^2\sin\varphi}\frac{\partial}{\partial \theta}\left(\frac{r^2\sin\varphi}{r^2\sin^2\varphi}\frac{\partial}{\partial \theta}\right)+\frac{1}{r^2\sin\varphi}\frac{\partial}{\partial \varphi}\left(\frac{r^2\sin\varphi}{r^2}\frac{\partial}{\partial \varphi}\right)\\
&=\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{1}{r^2\sin^2\varphi}\frac{\partial^2}{\partial\theta^2}+\frac{1}{r^2\sin\varphi}\frac{\partial}{\partial\varphi}\left(\sin\varphi\frac{\partial }{\partial\varphi}\right)
\end{aligned}
我們之所以需要將拉普拉斯算子表示成球坐標的形式，是因為它可以幫助我們得到拉普拉斯方程
\[\Delta u(x)=0\]
在\(n\)維球\(B=B_R(0)\)上的 Poisson Integral Formula
\[u(x)=\frac{R^2-\vert x\vert^2}{\omega_nR}\int_{\partial B}\frac{\varphi(y)}{\vert x-y\vert^n}\,dS_y\]
其中
\[u(x)=\varphi(x),\quad x\in\partial B\]
以及
\[\omega_n=\frac{2\pi^{n/2}}{\Gamma(n/2)}\]
而當\(n=2\)時，我們將\(x\)用極坐標 (Polar Coordinate) 表示可以得到
\[u(r,\theta)=\frac{1}{2\pi}\int_0^{2\pi}\frac{\left(R^2-r^2\right)\varphi(\phi)}{R^2+r^2-2Rr\cos(\theta-\phi)}\,d\phi,\quad r<R\]
\[x=r\sin\varphi\cos\theta,\quad y=r\sin\varphi\sin\theta,\quad z=r\cos\varphi\]
其中
\[r>0,\quad\theta\in(0,2\pi),\quad\varphi\in(0,\pi)\]
在這個球坐標系統之下，拉普拉斯算子的形式可以表示如下
\[\boxed{\displaystyle\Delta u(x,y,z)=\frac{\partial^2U}{\partial r^2}+\frac{2}{r}\frac{\partial U}{\partial r}+\frac{1}{r^2\sin^2\varphi}\frac{\partial^2U}{\partial\theta^2}+\frac{1}{r^2\sin\varphi}\frac{\partial}{\partial\varphi}\left(\sin\varphi\frac{\partial U}{\partial\varphi}\right)}\]
其中函數\(u(x,y,z)\)對應於函數\(U(r,\theta,\varphi)\)。以下我們提供四種方法來說明其由來。
方法一
首先根據球坐標的定義我們有
\[\left\{\begin{array}{l}
\displaystyle r=\sqrt{x^2+y^2+z^2},\\ \\
\displaystyle\frac{\partial r}{\partial x}=\frac{x}{r}=\sin\varphi\cos\theta,\\ \\
\displaystyle\frac{\partial r}{\partial y}=\frac{y}{r}=\sin\varphi\sin\theta,\\ \\
\displaystyle\frac{\partial r}{\partial z}=\frac{z}{r}=\cos\varphi,
\end{array}
\right .\]
和
\[\left\{
\begin{array}{l}
\displaystyle \theta=\tan^{-1}\frac{y}{x},\\ \\
\displaystyle\frac{\partial \theta}{\partial x}=\frac{-y}{x^2+y^2}=-\frac{r\sin\varphi\sin\theta}{r^2\sin^2\varphi}=-\frac{\sin\theta}{r\sin\varphi},\\ \\
\displaystyle\frac{\partial \theta}{\partial y}=\frac{x}{x^2+y^2}=\frac{r\sin\varphi\cos\theta}{r^2\sin^2\varphi}=\frac{\cos\theta}{r\sin\varphi},\\ \\
\displaystyle\frac{\partial \theta}{\partial z}=0,
\end{array}
\right .\]
\[\left\{
\begin{array}{l}
\displaystyle \theta=\tan^{-1}\frac{y}{x},\\ \\
\displaystyle\frac{\partial \theta}{\partial x}=\frac{-y}{x^2+y^2}=-\frac{r\sin\varphi\sin\theta}{r^2\sin^2\varphi}=-\frac{\sin\theta}{r\sin\varphi},\\ \\
\displaystyle\frac{\partial \theta}{\partial y}=\frac{x}{x^2+y^2}=\frac{r\sin\varphi\cos\theta}{r^2\sin^2\varphi}=\frac{\cos\theta}{r\sin\varphi},\\ \\
\displaystyle\frac{\partial \theta}{\partial z}=0,
\end{array}
\right .\]
以及
\[\left\{
\begin{array}{l}
\displaystyle \varphi=\cos^{-1}\frac{z}{r},\\ \\
\displaystyle\frac{\partial \varphi}{\partial x}=\frac{-1}{\sqrt{1-(z/r)^2}}\frac{-z}{r^2}\frac{\partial r}{\partial x}=\frac{z}{r\sqrt{r^2-z^2}}\frac{\partial r}{\partial x}=\frac{r\cos\varphi}{r\sqrt{r^2\sin^2\varphi}}\sin\varphi\cos\theta=\frac{\cos\varphi\cos\theta}{r},\\ \\
\displaystyle\frac{\partial \varphi}{\partial y}=\frac{r\cos\varphi}{r\sqrt{r^2\sin^2\varphi}}\sin\varphi\sin\theta=\frac{\cos\varphi\sin\theta}{r},\\ \\
\displaystyle\frac{\partial \varphi}{\partial z}=\frac{-1}{\sqrt{1-(z/r)^2}}\left(\frac{1}{r}-\frac{z}{r^2}\frac{\partial r}{\partial z}\right)=\frac{-1}{\sqrt{r^2-z^2}}\left(1-\cos^2\varphi\right)=-\frac{\sin^2\varphi}{\sqrt{r^2\sin^2\varphi}}=-\frac{\sin\varphi}{r}.
\end{array}
\right .\]
\[\left\{
\begin{array}{l}
\displaystyle \varphi=\cos^{-1}\frac{z}{r},\\ \\
\displaystyle\frac{\partial \varphi}{\partial x}=\frac{-1}{\sqrt{1-(z/r)^2}}\frac{-z}{r^2}\frac{\partial r}{\partial x}=\frac{z}{r\sqrt{r^2-z^2}}\frac{\partial r}{\partial x}=\frac{r\cos\varphi}{r\sqrt{r^2\sin^2\varphi}}\sin\varphi\cos\theta=\frac{\cos\varphi\cos\theta}{r},\\ \\
\displaystyle\frac{\partial \varphi}{\partial y}=\frac{r\cos\varphi}{r\sqrt{r^2\sin^2\varphi}}\sin\varphi\sin\theta=\frac{\cos\varphi\sin\theta}{r},\\ \\
\displaystyle\frac{\partial \varphi}{\partial z}=\frac{-1}{\sqrt{1-(z/r)^2}}\left(\frac{1}{r}-\frac{z}{r^2}\frac{\partial r}{\partial z}\right)=\frac{-1}{\sqrt{r^2-z^2}}\left(1-\cos^2\varphi\right)=-\frac{\sin^2\varphi}{\sqrt{r^2\sin^2\varphi}}=-\frac{\sin\varphi}{r}.
\end{array}
\right .\]
於是我們得到了以下一階算子的關係
\[\left\{\begin{array}{l}
\displaystyle\frac{\partial }{\partial x}=(\sin\varphi\cos\theta)\frac{\partial}{\partial r}-\frac{\sin\theta}{r\sin\varphi}\frac{\partial}{\partial \theta}+\frac{\cos\varphi\cos\theta}{r}\frac{\partial}{\partial\varphi},\\ \\
\displaystyle\frac{\partial \varphi}{\partial y}=(\sin\varphi\sin\theta)\frac{\partial}{\partial r}+\frac{\cos\theta}{r\sin\varphi}\frac{\partial}{\partial\theta}+\frac{\cos\varphi\sin\theta}{r}\frac{\partial}{\partial\varphi},\\ \\
\displaystyle\frac{\partial }{\partial z}=(\cos\varphi)\frac{\partial}{\partial r}-\frac{\sin\varphi}{r}\frac{\partial}{\partial\varphi}.
\end{array}
\right .\]
接著我們只要繼續去計算二階算子
\[\frac{\partial^2 }{\partial x^2},\quad\frac{\partial^2 }{\partial y^2},\quad\frac{\partial^2 }{\partial z^2}\]
便可得到拉普拉斯算子的球坐標表示。這個方法需要極大的耐心與細心!
方法二
在方法一中我們將算子\(\frac{\partial }{\partial x},\,\frac{\partial }{\partial y},\,\frac{\partial }{\partial z}\)表示成算子\(\frac{\partial }{\partial r},\,\frac{\partial }{\partial \theta},\,\frac{\partial }{\partial \varphi}\)的組合。而在方法二中則剛好相反，我們會有
\[r\frac{\partial}{\partial r}=x\frac{\partial }{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}\]和
\[\frac{\partial}{\partial\theta}=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\]
以及
\[\frac{\partial}{\partial\varphi}=r\cos\varphi\cos\theta\frac{\partial}{\partial x}+r\cos\varphi\sin\theta\frac{\partial}{\partial y}-r\sin\varphi\frac{\partial}{\partial z}.\]
接著我們將\(\frac{\partial}{\partial\varphi}\)乘上\(\tan\varphi\)會得到
\[\tan\varphi\frac{\partial}{\partial\varphi}=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}-\frac{x^2+y^2}{z}\frac{\partial}{\partial z}\]
將上述算子用矩陣來表示我們會得到
\[\left (
\begin{array}{c}
\displaystyle r\frac{\partial}{\partial r} \\
\displaystyle\frac{\partial}{\partial\theta} \\
\displaystyle\tan\varphi\frac{\partial}{\partial\varphi}
\end{array}\right )=
\left (
\begin{array}{ccc}
x & y & z \\ \\
-y & x & 0 \\ \\
\displaystyle x & y & -\frac{x^2+y^2}{z}
\end{array}\right )
\left (
\begin{array}{c}
\displaystyle\frac{\partial}{\partial x} \\
\displaystyle\frac{\partial}{\partial y} \\
\displaystyle\frac{\partial}{\partial z}
\end{array}\right ).
\]
接著我們只要去計算
\[\left(r\frac{\partial}{\partial r}\right)^2+\left(\frac{\partial}{\partial\theta} \right)^2+\left(\tan\varphi\frac{\partial}{\partial\varphi}\right)^2.\]
經過適當整理便可得到拉普拉斯算子的球坐標表示。其中如果我們定義
\[M=\left (\begin{array}{ccc}
x & y & z \\ \\
-y & x & 0 \\ \\
\displaystyle x & y & -\frac{x^2+y^2}{z}
\end{array}\right )
\]
我們會發現到
\[MM^T=\left (\begin{array}{ccc}x & y & z \\ \\
-y & x & 0 \\ \\
\displaystyle x & y & -\frac{x^2+y^2}{z}
\end{array}\right )\left (\begin{array}{ccc}
x & -y & x \\ \\
y & x & y \\ \\
\displaystyle z & 0 & -\frac{x^2+y^2}{z}
\end{array}\right )=r^2\left (\begin{array}{ccc}
1 & 0 & 0 \\ \\
0 & \sin^2\varphi & 0 \\ \\
0 & 0 & \tan^2\varphi
\end{array}\right )\]
這個方法一樣需要龐大的計算!
方法三
對任意定點\(p\in\mathbb{R}^n\)，我們會有
\[\Delta u(p)=\sum_{i=1}^n\frac{\partial^2 u}{\partial x_i^2}(p)=\sum_{i=1}^n\frac{d^2}{dt^2}\bigg\vert_{t=0}u(p+tv_i),\quad t\in(-\epsilon,\epsilon)
\]
其中\(\lbrace v_1,\cdots,v_n\rbrace\subset\mathbb{R}^n\)是一組標準正交基底 (orthonormal basis)。這個性質可參考"Laplace Operator in Spherical Coordinate of \(\mathbb{R}^n\)"這份講議，這個性質幫助我們可以建構在\(n\)維空間中的球坐標拉普拉斯算子。接著我們只要挑選適當的標準正交基底即可。
方法四
這個方法我們會需要微分幾何中黎曼流形 (Riemannian manifold) 上拉普拉斯算子 (Laplace-Beltrami operator) 的定義
\[\Delta =\frac{1}{\sqrt{G}}\sum_{i=1}^n\frac{\partial}{\partial x^i}\left(\sum_{j=1}^m\sqrt{G}g^{ij}\frac{\partial }{\partial x^j}\right)\]其中
\[g_{ij}=\left\langle\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right\rangle ,\quad G=\text{det}\left(g_{ij}\right),\quad\left(g^{ij}\right)=(g_{ij})^{-1}\]
在此我們令
\[x^1=r,\quad x^2=\theta,\quad x^3=\varphi\]
並考慮三維球\((\mathbb{S}^2, g)\)伴隨黎曼度量(Riemannian metric)
\[ds^2=dr^2+r^2\sin^2\theta\,d\theta^2+r^2\,d\varphi^2.\]其中
\begin{aligned}
g_{rr}
&=\mathbf{e}_r\cdot \mathbf{e}_r\\
&=(\sin\varphi\cos\theta,\sin\varphi\sin\theta,\cos\varphi)\cdot (\sin\varphi\cos\theta,\sin\varphi\sin\theta,\cos\varphi)\\
&=1
\end{aligned}
\begin{aligned}
g_{\theta\theta}
&=\mathbf{e}_\theta\cdot \mathbf{e}_\theta\\
&=(-r\sin\varphi\sin\theta,r\sin\varphi\cos\theta,0)\cdot(-r\sin\varphi\sin\theta,r\sin\varphi\cos\theta,0)\\
&=r^2\sin^2\varphi
\end{aligned}
\begin{aligned}
g_{\varphi\varphi}
&=\mathbf{e}_\varphi\cdot \mathbf{e}_\varphi\\
&=(r\cos\varphi\cos\theta,r\cos\varphi\sin\theta,-r\sin\varphi)\cdot(r\cos\varphi\cos\theta,r\cos\varphi\sin\theta,-r\sin\varphi)\\
&=r^2.
\end{aligned}
以及
\[g_{r\theta}=g_{\theta\varphi}=g_{r\varphi}=0\]
接著我們會有
\[(g_{ij})=\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & r^2\sin^2\theta & 0 \\
0 & 0 & r^2
\end{array}\right)\]
1 & 0 & 0 \\
0 & r^2\sin^2\theta & 0 \\
0 & 0 & r^2
\end{array}\right)\]
和
\[(g^{ij})=\left( \begin{array}{ccc}1 & 0 & 0 \\
0 & \frac{1}{r^2\sin^2\theta} & 0 \\
0 & 0 & \frac{1}{r^2}
\end{array}\right)\]
以及
\[G=\left\vert\begin{array}{ccc}
1 & 0 & 0 \\
0 & r^2\sin^2\theta & 0 \\
0 & 0 & r^2
\end{array}\right\vert=r^4\sin^2\varphi.\]
最後我們根據黎曼流形上拉普拉斯算子的定義便會得到
\begin{aligned}\Delta
&=\frac{1}{r^2\sin\varphi}\frac{\partial}{\partial r}\left(\frac{r^2\sin\varphi}{1}\frac{\partial}{\partial r}\right)+\frac{1}{r^2\sin\varphi}\frac{\partial}{\partial \theta}\left(\frac{r^2\sin\varphi}{r^2\sin^2\varphi}\frac{\partial}{\partial \theta}\right)+\frac{1}{r^2\sin\varphi}\frac{\partial}{\partial \varphi}\left(\frac{r^2\sin\varphi}{r^2}\frac{\partial}{\partial \varphi}\right)\\
&=\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{1}{r^2\sin^2\varphi}\frac{\partial^2}{\partial\theta^2}+\frac{1}{r^2\sin\varphi}\frac{\partial}{\partial\varphi}\left(\sin\varphi\frac{\partial }{\partial\varphi}\right)
\end{aligned}
我們之所以需要將拉普拉斯算子表示成球坐標的形式，是因為它可以幫助我們得到拉普拉斯方程
\[\Delta u(x)=0\]
在\(n\)維球\(B=B_R(0)\)上的 Poisson Integral Formula
\[u(x)=\frac{R^2-\vert x\vert^2}{\omega_nR}\int_{\partial B}\frac{\varphi(y)}{\vert x-y\vert^n}\,dS_y\]
其中
\[u(x)=\varphi(x),\quad x\in\partial B\]
以及
\[\omega_n=\frac{2\pi^{n/2}}{\Gamma(n/2)}\]
而當\(n=2\)時，我們將\(x\)用極坐標 (Polar Coordinate) 表示可以得到
\[u(r,\theta)=\frac{1}{2\pi}\int_0^{2\pi}\frac{\left(R^2-r^2\right)\varphi(\phi)}{R^2+r^2-2Rr\cos(\theta-\phi)}\,d\phi,\quad r<R\]
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