#author("2020-12-28T11:03:28+09:00","default:Miyashita","Miyashita") #author("2020-12-28T11:05:55+09:00","default:Miyashita","Miyashita") *2次元でのラプラシアンの極座標表示 [#oeb7ce98] 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．~ ~ #contents **ゴール [#f268ec26] 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 [#w8747832] ***一階微分 [#x1b790f4] 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 [#z80ddc60] 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 \(\theta\) の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．~ 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 [#fdbdaaed] 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．