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
$$x=r\sin\varphi\cos\theta,\quad y=r\sin\varphi\sin\theta,\quad z=r\cos\varphi$$

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
$$\left\{ \begin{array}{l} \displaystyle \theta=\tan^{-1}\frac{y}{x},\\ \\ \displaystyle\frac{\partial \theta}{\partial x}=\frac{-y}{x^2+y^2}=-\frac{r\sin\varphi\sin\theta}{r^2\sin^2\varphi}=-\frac{\sin\theta}{r\sin\varphi},\\ \\ \displaystyle\frac{\partial \theta}{\partial y}=\frac{x}{x^2+y^2}=\frac{r\sin\varphi\cos\theta}{r^2\sin^2\varphi}=\frac{\cos\theta}{r\sin\varphi},\\ \\ \displaystyle\frac{\partial \theta}{\partial z}=0, \end{array} \right .$$

以及
$$\left\{ \begin{array}{l} \displaystyle \varphi=\cos^{-1}\frac{z}{r},\\ \\ \displaystyle\frac{\partial \varphi}{\partial x}=\frac{-1}{\sqrt{1-(z/r)^2}}\frac{-z}{r^2}\frac{\partial r}{\partial x}=\frac{z}{r\sqrt{r^2-z^2}}\frac{\partial r}{\partial x}=\frac{r\cos\varphi}{r\sqrt{r^2\sin^2\varphi}}\sin\varphi\cos\theta=\frac{\cos\varphi\cos\theta}{r},\\ \\ \displaystyle\frac{\partial \varphi}{\partial y}=\frac{r\cos\varphi}{r\sqrt{r^2\sin^2\varphi}}\sin\varphi\sin\theta=\frac{\cos\varphi\sin\theta}{r},\\ \\ \displaystyle\frac{\partial \varphi}{\partial z}=\frac{-1}{\sqrt{1-(z/r)^2}}\left(\frac{1}{r}-\frac{z}{r^2}\frac{\partial r}{\partial z}\right)=\frac{-1}{\sqrt{r^2-z^2}}\left(1-\cos^2\varphi\right)=-\frac{\sin^2\varphi}{\sqrt{r^2\sin^2\varphi}}=-\frac{\sin\varphi}{r}. \end{array} \right .$$

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

$$(g_{ij})=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & r^2\sin^2\theta & 0 \\ 0 & 0 & r^2 \end{array}\right)$$

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
$$u(r,\theta)=\frac{1}{2\pi}\int_0^{2\pi}\frac{\left(R^2-r^2\right)\varphi(\phi)}{R^2+r^2-2Rr\cos(\theta-\phi)}\,d\phi,\quad r<R$$