2次元でのラプラシアンの極座標表示 †
誰もが一生に一度はやっておいた方が良いと言われている(？)，2次元ラプラシアンの極座標表示の導出過程をここに書き散らす．
ゴール †
2次元直交座標系 \( (x,y) \) でのラプラシアン
\[
\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \tag{1}
\]
を，次式のような2次元極座標 \( (r,\theta) \) での表現を導出する．
\[
\nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} \tag{2}
\]
ここで
\[\begin{align}
x = &~r\cos\theta \tag{3}\\
y = &~ r\sin\theta \tag{4}\\
r = &~ \sqrt{x^2+y^2} \tag{5}\\
\tan\theta = &~ \frac{y}{x} \tag{6}\\
\end{align}\]
である．
導出 †
一階微分 †
まず \(\dfrac{\partial}{\partial x}\) と \(\dfrac{\partial}{\partial y}\) を \(r,~\theta\) で表す．
\[\begin{align}
\frac{\partial}{\partial x} = & \frac{\partial r}{\partial x}\frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta} \tag{7}\\
\frac{\partial}{\partial y} = & \frac{\partial r}{\partial y}\frac{\partial}{\partial r} + \frac{\partial \theta}{\partial y}\frac{\partial}{\partial \theta}\tag{8}
\end{align}\]
であるから，\(\dfrac{\partial r}{\partial x}\), \(\dfrac{\partial \theta}{\partial x}\), \(\dfrac{\partial r}{\partial y}\), \(\dfrac{\partial \theta}{\partial y}\) を求める．
式(5)を \(x\) で微分して
\[
\frac{\partial r}{\partial x} = \frac{2x}{2\sqrt{x^2+y^2}} = \frac{x}{r} = \cos\theta \tag{9}
\]
が得られる．次に式(6)
\[\begin{align}
\tan\theta = &~ \frac{y}{x} \tag{6}
\end{align}\]
を \(x\) で微分して
\[\begin{align}
\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial x} = &~ -\frac{y}{x^2} = -\frac{\sin\theta}{r\cos^2\theta} \\
\frac{\partial\theta}{\partial x} = &~ - \frac{\sin\theta}{r} \tag{10}
\end{align}\]
を得る．同様に \(y\) についても行うと
\[
\frac{\partial r}{\partial y} = \frac{2y}{2\sqrt{x^2+y^2}} = \frac{y}{r} = \sin\theta \tag{11}
\]
および
\[\begin{align}
\tan\theta = &~ \frac{y}{x} \tag{6}
\end{align}\]
を \(y\) で微分して
\[\begin{align}
\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial y} = &~ \frac{1}{x} = \frac{1}{r\cos\theta} \\
\frac{\partial\theta}{\partial y} = &~ \frac{\cos\theta}{r} \tag{12}
\end{align}\]
となる．したがって
\[\begin{align}
\frac{\partial}{\partial x} = & \frac{\partial r}{\partial x}\frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta} \tag{7} \\
= &~\cos\theta \frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta} \tag{13}
\end{align}\]
\[\begin{align}
\frac{\partial}{\partial y} = & \frac{\partial r}{\partial y}\frac{\partial}{\partial r} + \frac{\partial \theta}{\partial y}\frac{\partial}{\partial \theta}\tag{8} \\
= &~ \sin\theta \frac{\partial}{\partial r} + \frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\tag{14}
\end{align}\]
となる．これで一階微分が求まった．
二階微分 †
次に \(\dfrac{\partial^2}{\partial x^2}\) と \(\dfrac{\partial^2}{\partial y^2}\) を \(r,~\theta\) で表す．
式(13)より
\[\begin{align}
\frac{\partial}{\partial x} \left( \frac{\partial}{\partial x} \right) = & \frac{\partial r}{\partial x}\frac{\partial}{\partial r}\left( \cos\theta \frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right) +\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}\left( \cos\theta \frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right) \\
= &~ \cos\theta \frac{\partial}{\partial r}\left( \cos\theta \frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right) -\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\left( \cos\theta \frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right) \\
= &~ \cos\theta \left( \cos\theta\frac{\partial^2}{\partial r^2} + \frac{\sin\theta}{r^2}\frac{\partial}{\partial \theta} - \frac{\sin\theta}{r}\frac{\partial^2}{\partial r \partial \theta} \right) \\
& -\frac{\sin\theta}{r}\left( -\sin\theta \frac{\partial}{\partial r} + \cos\theta\frac{\partial^2}{\partial r \partial \theta} - \frac{\cos\theta}{r}\frac{\partial}{\partial \theta} - \frac{\sin\theta}{r} \frac{\partial^2}{\partial \theta^2} \right) \\
\end{align}\]
これをまとめると,
\[\begin{align}
\frac{\partial^2}{\partial x^2} =
\cos^2\theta \frac{\partial^2}{\partial r^2} +
\frac{\sin^2\theta}{r} \frac{\partial}{\partial r}-
\frac{2\sin\theta\cos\theta}{r} \frac{\partial^2}{\partial r \partial \theta} +
\frac{2\sin\theta\cos\theta}{r^2} \frac{\partial}{\partial \theta}+
\frac{\sin^2\theta}{r^2} \frac{\partial^2}{\partial \theta^2}
\tag{15}
\end{align}\]
となる．分母の \(r\) の次数と分子の \(\sin\theta\) または \(\cos\theta\) の次数が各項で揃っていることを確認．
揃っていなかったらどこかで計算を間違えているはず．
同様にして式(14)より
\[\begin{align}
\frac{\partial}{\partial y} \left( \frac{\partial}{\partial y} \right) = &
\frac{\partial r}{\partial y}\frac{\partial}{\partial r}\left( \sin\theta \frac{\partial}{\partial r} + \frac{\cos\theta}{r}\frac{\partial}{\partial \theta} \right) +
\frac{\partial \theta}{\partial y}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial r} + \frac{\cos\theta}{r}\frac{\partial}{\partial \theta} \right) \\
= &~ \sin\theta \frac{\partial}{\partial r}\left( \sin\theta \frac{\partial}{\partial r} + \frac{\cos\theta}{r}\frac{\partial}{\partial \theta} \right) +
\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial r} + \frac{\cos\theta}{r}\frac{\partial}{\partial \theta} \right) \\
= &~ \sin\theta \left( \sin\theta \frac{\partial^2}{\partial r^2} - \frac{\cos\theta}{r^2}\frac{\partial}{\partial \theta} +\frac{\cos\theta}{r}\frac{\partial^2}{\partial r \partial \theta} \right) \\
&+\frac{\cos\theta}{r}\left( \cos\theta \frac{\partial}{\partial r} + \sin\theta\frac{\partial^2}{\partial r \partial \theta} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta} + \frac{\cos\theta}{r}\frac{\partial^2}{\partial \theta^2}\right)
\end{align}\]
これをまとめると
\[\begin{align}
\frac{\partial^2}{\partial y^2} =
\sin^2\theta \frac{\partial^2}{\partial r^2} +
\frac{\cos^2\theta}{r} \frac{\partial}{\partial r} +
\frac{2\sin\theta\cos\theta}{r} \frac{\partial^2}{\partial r \partial \theta} -
\frac{2\sin\theta\cos\theta}{r^2} \frac{\partial}{\partial \theta}+
\frac{\cos^2\theta}{r^2} \frac{\partial^2}{\partial \theta^2}
\tag{16}
\end{align}\]
仕上げ †
式(15)と式(16)よりゴールとして設定した式(2)が導ける．
\[\begin{align}
\nabla^2 = & \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \\
= &~ \left( \sin^2\theta + \cos^2\theta \right) \frac{\partial^2}{\partial r^2} +
\frac{\left( \sin^2\theta + \cos^2\theta \right)}{r} \frac{\partial}{\partial r} +
\frac{\left( \sin^2\theta + \cos^2\theta \right)}{r^2} \frac{\partial^2}{\partial \theta^2} \\
= & \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}
\end{align}\]
おわり．
