3次元ラプラシアンの極座標表示

この記事では、デカルト座標における3次元ラプラシアン

$$\begin{array}{}\text{(1)}& {\mathrm{\Delta }}_{\mathrm{C}}=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}\end{array}$$

の極座標表示を導出します。

方針

デカルト座標系$(x,y,z)$と極座標系$(r,\theta ,\varphi )$は、次の関係式で結ばれます。

$$\begin{array}{}\text{(2)}& x& =& r\sin \theta \cos \varphi \text{(3)}& y& =& r\sin \theta \sin \varphi \text{(4)}& z& =& r\cos \theta \end{array}$$

$(x,y,z)$はそれぞれ$(r,\theta ,\varphi )$の関数なので、以下の微分の連鎖律が成り立ちます。

$$\begin{array}{}\text{(5)}& \frac{\partial }{\partial x}& =& \frac{\partial r}{\partial x}\frac{\partial }{\partial r}+\frac{\partial \theta }{\partial x}\frac{\partial }{\partial \theta }+\frac{\partial \varphi }{\partial x}\frac{\partial }{\partial \varphi }\text{(6)}& \frac{\partial }{\partial y}& =& \frac{\partial r}{\partial y}\frac{\partial }{\partial r}+\frac{\partial \theta }{\partial y}\frac{\partial }{\partial \theta }+\frac{\partial \varphi }{\partial y}\frac{\partial }{\partial \varphi }\text{(7)}& \frac{\partial }{\partial z}& =& \frac{\partial r}{\partial z}\frac{\partial }{\partial r}+\frac{\partial \theta }{\partial z}\frac{\partial }{\partial \theta }+\frac{\partial \varphi }{\partial z}\frac{\partial }{\partial \varphi }\end{array}$$

これら9つの係数を$r,\theta ,\varphi$で表すことができれば、$\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}$の極座標表示が得られるので、それぞれを2乗して足し合わせることで3次元ラプラシアンの極座標表示を計算できます。

一階微分の導出

$\frac{\partial r}{\partial x},\frac{\partial r}{\partial y},\frac{\partial r}{\partial z}$の導出

$r$と$x,y,z$は、以下の関係式で結ばれます。

$$\begin{array}{}\text{(8)}& r^2=x^2+y^2+z^2\end{array}$$

この両辺を$x$で偏微分すると

$$\begin{array}{rcl}2r\frac{\partial r}{\partial x}& =& 2x\\ \text{(9)}& \therefore \frac{\partial r}{\partial x}& =& \frac{x}{r}=\frac{r\sin \theta \cos \varphi }{r}=\sin \theta \cos \varphi \end{array}$$

が得られます。$y,z$についても同様に

$$\begin{array}{rcl}2r\frac{\partial r}{\partial y}& =& 2y\\ \text{(10)}& \therefore \frac{\partial r}{\partial y}& =& \frac{y}{r}=\frac{r\sin \theta \sin \varphi }{r}=\sin \theta \sin \varphi \end{array}$$

$$\begin{array}{rcl}2r\frac{\partial r}{\partial z}& =& 2z\\ \text{(11)}& \therefore \frac{\partial r}{\partial z}& =& \frac{z}{r}=\frac{r\cos \theta }{r}=\cos \theta \end{array}$$

となります。

$\frac{\partial \theta }{\partial x},\frac{\partial \theta }{\partial y},\frac{\partial \theta }{\partial z}$の導出

$\theta$と$x,y,z$は、以下の関係式で結ばれます。

$$\begin{array}{}\text{(12)}& {\tan }^2\theta =\frac{x^2+y^2}{z^2}\end{array}$$

この両辺を$x$で偏微分すると

$$\begin{array}{c}2\tan \theta \frac{1}{{\cos }^2\theta }\frac{\partial \theta }{\partial x}=\frac{2x}{z^2}\\ \text{(13)}& \therefore \frac{\partial \theta }{\partial x}=\frac{x}{z^2}\frac{{\cos }^2\theta }{\tan \theta }=\frac{r\sin \theta \cos \varphi {\cos }^2\theta }{r^2{\cos }^2\theta \tan \theta }=\frac{\cos \theta \cos \varphi }{r}\end{array}$$

となります。$y$についても同様に

$$\begin{array}{c}2\tan \theta \frac{1}{{\cos }^2\theta }\frac{\partial \theta }{\partial y}=\frac{2y}{z^2}\\ \text{(14)}& \therefore \frac{\partial \theta }{\partial y}=\frac{y}{z^2}\frac{{\cos }^2\theta }{\tan \theta }=\frac{r\sin \theta \sin \varphi {\cos }^2\theta }{r^2{\cos }^2\theta \tan \theta }=\frac{\cos \theta \sin \varphi }{r}\end{array}$$

となります。$z$については、両辺の$\ln$をとって$z$で偏微分すると

$$\begin{array}{c}2\ln (\tan \theta )=\ln (x^2+y^2)-\ln (z^2)\\ 2\frac{1}{\tan \theta }\frac{1}{{\cos }^2\theta }\frac{\partial \theta }{\partial z}=-\frac{2}{z}\\ \text{(15)}& \therefore \frac{\partial \theta }{\partial z}=-\frac{\tan \theta {\cos }^2\theta }{z}=-\frac{\tan \theta {\cos }^2\theta }{r\cos \theta }=-\frac{\sin \theta }{r}\end{array}$$

が得られます。

$\frac{\partial \varphi }{\partial x},\frac{\partial \varphi }{\partial y},\frac{\partial \varphi }{\partial z}$の導出

$\varphi$と$x,y,z$は、以下の関係式で結ばれます。

$$\begin{array}{}\text{(16)}& \tan \varphi =\frac{y}{x}\end{array}$$

この両辺を$x$で偏微分すると

$$\begin{array}{c}\frac{1}{{\cos }^2\varphi }\frac{\partial \varphi }{\partial x}=-\frac{y}{x^2}\\ \text{(17)}& \therefore \frac{\partial \varphi }{\partial x}=-\frac{y{\cos }^2\varphi }{x^2}=-\frac{r\sin \theta \sin \varphi {\cos }^2\varphi }{r^2{\sin }^2\theta {\cos }^2\varphi }=-\frac{\sin \varphi }{r\sin \theta }\end{array}$$

となります。$y$についても同様に

$$\begin{array}{c}\frac{1}{{\cos }^2\varphi }\frac{\partial \varphi }{\partial y}=\frac{1}{x}\\ \text{(18)}& \therefore \frac{\partial \varphi }{\partial y}=\frac{{\cos }^2\varphi }{x}=\frac{{\cos }^2\varphi }{r\sin \theta \cos \varphi }=\frac{\cos \varphi }{r\sin \theta }\end{array}$$

となります。$z$については、$\varphi$は$z$に依存しないため

$$\begin{array}{}\text{(19)}& \frac{\partial \varphi }{\partial z}=0\end{array}$$

です。

以上をまとめると、$\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}$の極座標表示は以下のようになります。

$$\begin{array}{}\text{(20)}& \frac{\partial }{\partial x}& =& \sin \theta \cos \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }\text{(21)}& \frac{\partial }{\partial y}& =& \sin \theta \sin \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }\text{(22)}& \frac{\partial }{\partial z}& =& \cos \theta \frac{\partial }{\partial r}-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta }\end{array}$$

二階微分の導出

次に、得られた$\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}$の2乗を計算していきます。

$\frac{{\partial }^2}{\partial x^2}$の導出

$$\begin{array}{rcl}\frac{{\partial }^2}{\partial x^2}& =& (\sin \theta \cos \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }{)}^2\\ & =& \sin \theta \cos \varphi \frac{\partial }{\partial r}(\sin \theta \cos \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ & & +\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }(\sin \theta \cos \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ \text{(23)}& & & -\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }(\sin \theta \cos \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\end{array}$$

$$\begin{array}{rcl}\mathrm{第}1\mathrm{項}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{部}\mathrm{分}& & =\frac{\partial }{\partial r}(\sin \theta \cos \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ & & =\sin \theta \cos \varphi \frac{{\partial }^2}{\partial r^2}\\ & & –\frac{1}{r^2}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\cos \theta \cos \varphi \frac{{\partial }^2}{\partial r\partial \theta }\\ \text{(24)}& & & +\frac{1}{r^2}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{{\partial }^2}{\partial r\partial \varphi }\end{array}$$

$$\begin{array}{rcl}\mathrm{第}2\mathrm{項}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{部}\mathrm{分}& & =\frac{\partial }{\partial \theta }(\sin \theta \cos \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ & & =\cos \theta \cos \varphi \frac{\partial }{\partial r}+\sin \theta \cos \varphi \frac{{\partial }^2}{\partial \theta \partial r}\\ & & –\frac{1}{r}\sin \theta \cos \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\cos \theta \cos \varphi \frac{{\partial }^2}{\partial {\theta }^2}\\ \text{(25)}& & & +\frac{1}{r}\frac{\cos \theta \sin \varphi }{{\sin }^2\theta }\frac{\partial }{\partial \varphi }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{{\partial }^2}{\partial \theta \partial \varphi }\end{array}$$

$$\begin{array}{rcl}\mathrm{第}3\mathrm{項}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{部}\mathrm{分}& & =\frac{\partial }{\partial \varphi }(\sin \theta \cos \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ & & =-\sin \theta \sin \varphi \frac{\partial }{\partial r}+\sin \theta \cos \varphi \frac{{\partial }^2}{\partial \varphi \partial r}\\ & & –\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\cos \theta \cos \varphi \frac{{\partial }^2}{\partial \varphi \partial \theta }\\ \text{(26)}& & & –\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }-\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{{\partial }^2}{\partial {\varphi }^2}\end{array}$$

$\frac{{\partial }^2}{\partial y^2}$の導出

$$\begin{array}{rcl}\frac{{\partial }^2}{\partial y^2}& =& (\sin \theta \sin \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }{)}^2\\ & =& \sin \theta \sin \varphi \frac{\partial }{\partial r}(\sin \theta \sin \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ & & +\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }(\sin \theta \sin \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ \text{(27)}& & & +\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }(\sin \theta \sin \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\end{array}$$

$$\begin{array}{rcl}\mathrm{第}1\mathrm{項}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{部}\mathrm{分}& & =\frac{\partial }{\partial r}(\sin \theta \sin \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ & & =\sin \theta \sin \varphi \frac{{\partial }^2}{\partial r^2}\\ & & –\frac{1}{r^2}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\cos \theta \sin \varphi \frac{{\partial }^2}{\partial r\partial \theta }\\ \text{(28)}& & & –\frac{1}{r^2}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{{\partial }^2}{\partial r\partial \varphi }\end{array}$$

$$\begin{array}{rcl}\mathrm{第}2\mathrm{項}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{部}\mathrm{分}& & =\frac{\partial }{\partial \theta }(\sin \theta \sin \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ & & =\cos \theta \sin \varphi \frac{\partial }{\partial r}+\sin \theta \sin \varphi \frac{{\partial }^2}{\partial \theta \partial r}\\ & & –\frac{1}{r}\sin \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\cos \theta \sin \varphi \frac{{\partial }^2}{\partial {\theta }^2}\\ \text{(29)}& & & –\frac{1}{r}\frac{\cos \theta \cos \varphi }{{\sin }^2\theta }\frac{\partial }{\partial \varphi }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{{\partial }^2}{\partial \theta \partial \varphi }\end{array}$$

$$\begin{array}{rcl}\mathrm{第}3\mathrm{項}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{部}\mathrm{分}& & =\frac{\partial }{\partial \varphi }(\sin \theta \sin \varphi \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi })\\ & & =\sin \theta \cos \varphi \frac{\partial }{\partial r}+\sin \theta \sin \varphi \frac{{\partial }^2}{\partial \varphi \partial r}\\ & & +\frac{1}{r}\cos \theta \cos \varphi \frac{\partial }{\partial \theta }+\frac{1}{r}\cos \theta \sin \varphi \frac{{\partial }^2}{\partial \varphi \partial \theta }\\ \text{(30)}& & & –\frac{1}{r}\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }+\frac{1}{r}\frac{\cos \varphi }{\sin \theta }\frac{{\partial }^2}{\partial {\varphi }^2}\end{array}$$

$\frac{{\partial }^2}{\partial z^2}$の導出

$$\begin{array}{rcl}\frac{{\partial }^2}{\partial z^2}& =& (\cos \theta \frac{\partial }{\partial r}-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta }{)}^2\\ & =& \cos \theta \frac{\partial }{\partial r}(\cos \theta \frac{\partial }{\partial r}-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta })\\ \text{(31)}& & & –\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta }(\cos \theta \frac{\partial }{\partial r}-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta })\end{array}$$

$$\begin{array}{rcl}\mathrm{第}1\mathrm{項}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{部}\mathrm{分}& & =\frac{\partial }{\partial r}(\cos \theta \frac{\partial }{\partial r}-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta })\\ & & =\cos \theta \frac{{\partial }^2}{\partial r^2}\\ \text{(32)}& & & +\frac{1}{r^2}\sin \theta \frac{\partial }{\partial \theta }-\frac{1}{r}\sin \theta \frac{{\partial }^2}{\partial r\partial \theta }\end{array}$$

$$\begin{array}{rcl}\mathrm{第}2\mathrm{項}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{部}\mathrm{分}& & =\frac{\partial }{\partial \theta }(\cos \theta \frac{\partial }{\partial r}-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta })\\ & & =-\sin \theta \frac{\partial }{\partial r}+\cos \theta \frac{{\partial }^2}{\partial \theta \partial r}\\ \text{(33)}& & & –\frac{1}{r}\cos \theta \frac{\partial }{\partial \theta }-\frac{1}{r}\sin \theta \frac{{\partial }^2}{\partial {\theta }^2}\end{array}$$

まとめ

最後に、ここまで計算した$\frac{{\partial }^2}{\partial x^2},\frac{{\partial }^2}{\partial y^2},\frac{{\partial }^2}{\partial z^2}$すべて足し上げます。

$\frac{{\partial }^2}{\partial r^2}$の係数

$$\begin{array}{}\text{(34)}& {\sin }^2\theta {\cos }^2\varphi +{\sin }^2\theta {\cos }^2\varphi +{\cos }^2\varphi =1\end{array}$$

$\frac{\partial }{\partial r}$の係数

$$\begin{array}{}\text{(35)}& \frac{1}{r}({\cos }^2\theta {\cos }^2\varphi +{\sin }^2\varphi +{\cos }^2\theta {\sin }^2\varphi +{\cos }^2\varphi +{\sin }^2\theta )=\frac{2}{r}\end{array}$$

$\frac{{\partial }^2}{\partial {\theta }^2}$の係数

$$\begin{array}{}\text{(36)}& \frac{1}{r^2}({\cos }^2\theta {\cos }^2\varphi +{\cos }^2\theta {\sin }^2\varphi +{\sin }^2\theta )=\frac{1}{r^2}\end{array}$$

$\frac{\partial }{\partial \theta }$の係数

$$\begin{array}{rcl}& & \frac{1}{r^2}(-2\sin \theta \cos \theta {\cos }^2\varphi +\frac{\cos \theta {\sin }^2\varphi }{\sin \theta }\\ & & -2\sin \theta \cos \theta {\sin }^2\varphi +\frac{\cos \theta {\cos }^2\varphi }{\sin \theta }+2\sin \theta \cos \theta )\\ \text{(37)}& & & =\frac{1}{r^2}\frac{\cos \theta }{\sin \theta }\end{array}$$

$\frac{{\partial }^2}{\partial {\varphi }^2}$の係数

$$\begin{array}{}\text{(38)}& \frac{1}{r^2}(\frac{{\sin }^2\varphi }{{\sin }^2\theta }+\frac{{\cos }^2\varphi }{{\sin }^2\theta })=\frac{1}{r^2}\frac{1}{{\sin }^2\theta }\end{array}$$

$\frac{\partial }{\partial \varphi }$の係数

$$\begin{array}{rcl}& & \frac{1}{r^2}(\sin \varphi \cos \varphi +\frac{{\cos }^2\theta \sin \varphi \cos \varphi }{{\sin }^2\theta }+\frac{\sin \varphi \cos \varphi }{{\sin }^2\theta }\\ & & -\sin \varphi \cos \varphi -\frac{{\cos }^2\theta \sin \varphi \cos \varphi }{{\sin }^2\theta }-\frac{\sin \varphi \cos \varphi }{{\sin }^2\theta })\\ \text{(39)}& & & =0\end{array}$$

$\frac{{\partial }^2}{\partial r\partial \theta }$の係数

$$\begin{array}{}\text{(40)}& \frac{1}{r}(2\sin \theta \cos \theta {\cos }^2\varphi +2\sin \theta \cos \theta {\sin }^2\varphi -2\sin \theta \cos \theta )=0\end{array}$$

$\frac{{\partial }^2}{\partial \theta \partial \varphi }$の係数

$$\begin{array}{}\text{(41)}& \frac{1}{r^2}(-\frac{2\cos \theta \sin \varphi \cos \varphi }{\sin \theta }+\frac{2\cos \theta \sin \varphi \cos \varphi }{\sin \theta })=0\end{array}$$

$\frac{{\partial }^2}{\partial \varphi \partial r}$の係数

$$\begin{array}{}\text{(42)}& \frac{1}{r}(-2\sin \varphi \cos \varphi +2\sin \varphi \cos \varphi )=0\end{array}$$

以上をまとめると、3次元ラプラシアンの極座標表示は以下のようになります。

$$\begin{array}{rcl}{\mathrm{\Delta }}_{\mathrm{P}}& =& \frac{{\partial }^2}{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}(\frac{{\partial }^2}{\partial {\theta }^2}+\frac{\cos \theta }{\sin \theta }\frac{\partial }{\partial \theta }+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2})\\ \text{(43)}& & =& \frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{\partial }{\partial r})+\frac{1}{r^2}(\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }(\sin \theta \frac{\partial }{\partial \theta })+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2})\end{array}$$