今回は円柱座標のラプラシアンを求めていきます。
めちゃくちゃめんどくさいですが、一生に一回はやってみてもいいかなと思います。
証明
円柱座標において
$$
\begin{eqnarray}
\begin{cases}
x=r\cos\phi\\
y=r\sin\phi\\
z=z\\
\end{cases}
\end{eqnarray}
$$
と書けます。
下準備
基底ベクトルの整理
\({\bf e_r} = {\bf e_x}\cos\phi +{\bf e_y}\sin\phi,\;\;\;{\bf e_\phi} = {\bf e_x}\cos\phi -{\bf e_y}\sin\phiなので、\)
$$ \frac{\partial {\bf e_r}}{\partial r}=0,\;\;\;\frac{\partial {\bf e_\phi}}{\partial r}=0,\;\;\;\frac{\partial {\bf e_z}}{\partial r}=0,\\ \frac{\partial e_r}{\partial {\phi}}=-\sin{\phi}{\bf e_x}+\cos\phi e_y={\bf e_\phi},\;\;\;\frac{\partial e_{\phi}}{\partial \phi}=-\sin\phi {\bf e_x}-\cos\phi {\bf e_y}={\bf -e_r}, \;\;\;\frac {\partial e_z}{\partial \phi}=0\\ \frac{\partial e_r}{\partial z}=0,\;\;\;\frac{\partial e_\phi}{\partial z}=0,\frac{\partial e_z}{\partial z}=0 $$
また、 $$ {\bf e_r} \cdot {\bf e_r}={\bf e_\phi} \cdot {\bf e_\phi}={\bf e_z} \cdot {\bf e_z}=1, \;\;\;{\bf e_r} \cdot {\bf e_\phi}={\bf e_\phi} \cdot {\bf e_z}={\bf e_z} \cdot {\bf e_r}=0 $$
となります。
円柱座標のラプラシアン
$$ {\bf{\bigtriangledown}} f= {\bf e_r}\frac{\partial f}{\partial r}+{\bf e_\phi}\frac{1}{r}\frac{\partial f}{\partial \phi}+{\bf e_z}\frac{\partial f}{\partial z}＝A_r e_r+A_\phi e_\phi+A_z e_z = {\bf A} $$
とします。
すなわち、
$$
\begin{eqnarray}
\begin{cases}
A_r = \frac{\partial f}{\partial r} \\
A_\phi = \frac{1}{r}\frac{\partial f}{\partial \phi} \\
A_z = \frac{\partial f}{\partial z}
\end{cases}
\end{eqnarray}
$$
となります。
\( \bigtriangledown \cdot {\bf A}=\left({\bf e_r}\frac{\partial}{\partial r}+{\bf e_\phi}\frac{1}{r}\frac{\partial}{\partial \phi}+{\bf e_z}\frac{\partial}{\partial z}\right) \cdot \left(A_r {\bf e_r}+A_{\phi} {\bf e_{\phi}}+A_z {\bf e_z}\right) \\ = {\bf e_r}\frac{\partial A_r}{\partial r}{\bf e_r}+{\bf e_r}\frac{\partial e_r}{\partial r}{\bf A_r}+{\bf e_r}\frac{\partial A_{\phi}}{\partial r}{\bf e_{\phi}}+{\bf e_r}\frac{\partial e_\phi}{\partial r}{\bf A_\phi}+{\bf e_r}\frac{\partial A_z}{\partial r}{\bf e_z}+{\bf e_r}\frac{\partial e_z}{\partial r}{\bf A_z} \\ +{\bf e_\phi}\frac{\partial A_r}{\partial \phi}{\bf e_r}+{\bf e_\phi}\frac{\partial e_r}{\partial \phi}{\bf A_r}+{\bf e_\phi}\frac{\partial A_\phi}{\partial \phi}{\bf e_\phi}+{\bf e_\phi}\frac{\partial e_\phi}{\partial \phi}{\bf A_\phi}+{\bf e_\phi}\frac{\partial A_z}{\partial \phi}{\bf e_z}+{\bf e_\phi} \frac{\partial e_z}{\partial \phi}{\bf A_z}\\ +{\bf e_z}\frac{\partial A_r}{\partial z}{\bf e_r}+{\bf e_z}\frac{\partial e_r}{\partial z}{\bf A_r}+{\bf e_z}\frac{\partial A_{\phi}}{\partial z}{\bf e_\phi}+{\bf e_z}\frac{\partial e_\phi}{\partial z}{\bf A_\phi}+{\bf e_z}\frac{\partial A_z}{\partial z}{\bf e_z}+{\bf e_z}\frac{\partial e_z}{\partial z}{\bf A_z} \\なので、 \) \( {\bf e_x} = {\bf e_r}\cos\phi – {\bf e_{\phi}}\sin\phi,\;\;\;{\bf e_y} = {\bf e_r} \sin\phi +{\bf e_\phi}\cos\phi,\;\;\;{\bf e_z}={\bf e_z} \)より、
$$ \bigtriangledown \cdot {\bf A}=\frac{\partial A_r}{\partial r}+\frac{A_r}{r}+ \frac{1}{r} \frac{\partial A_\phi}{\partial \phi}+\frac{\partial A_z}{\partial z}=\frac{1}{r}\frac{\partial}{\partial r}(rA_r)+\frac{1}{r}\frac{\partial A_\phi}{\partial\phi}+\frac{\partial A_z}{\partial z} $$
よって、
$$\bigtriangledown \cdot (\bigtriangledown f)=\bigtriangleup f= \frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial f}{\partial r})+\frac{1}{r^{2}}\frac{\partial^{2} f}{\partial \phi^{2}}+\frac{\partial^{2} f}{\partial z^{2}} $$
