極座標表示におけるラプラシアン (2次元)

関数${\displaystyle \psi =\xi (x,y)}$ において極座標表示${\displaystyle x=r\cos \theta }$ ，${\displaystyle {\displaystyle {\displaystyle y=r\sin \theta }}}$ におけるラプラシアンは

${\displaystyle \Delta \psi =\left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}\right)\psi }$${\displaystyle =\left(\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}\right)\psi }$

で与えられる．

■導出手順

与式の左辺の${\displaystyle \frac{{\partial }^2\psi }{\partial x^2}}$ ， ${\displaystyle \frac{{\partial }^2\psi }{\partial y^2}}$ を求め，
${\displaystyle \frac{\partial r}{\partial x}}$ ， ${\displaystyle \frac{{\partial }^{2}r}{\partial x^2}}$ ，${\displaystyle \frac{\partial \theta }{\partial x}}$ ，${\displaystyle \frac{{\partial }^2\theta }{\partial x^2}}$ ，${\displaystyle \frac{\partial r}{\partial y}}$ ，${\displaystyle \frac{{\partial }^{2}r}{\partial y^2}}$ ，${\displaystyle \frac{\partial \theta }{\partial y}}$ ，${\displaystyle \frac{{\partial }^2\theta }{\partial y^2}}$ を用いて右辺へ式変形する．

■導出

始めに，関数 ${\displaystyle \psi }$ を${\displaystyle x}$ で偏微分する．

${\displaystyle \frac{\partial \psi }{\partial x}}$ ${\displaystyle =\frac{\partial \psi }{\partial r}\frac{\partial r}{\partial x}+\frac{\partial \psi }{\partial \theta }\frac{\partial \theta }{\partial x}}$　

参考:合成関数の偏導関数

${\displaystyle \frac{{\partial }^2\psi }{\partial x^2}}$ ${\displaystyle =\frac{\partial }{\partial x}\left(\frac{\partial \psi }{\partial x}\right)}$

${\displaystyle {\displaystyle {\displaystyle =\frac{\partial }{\partial x}}\left({\displaystyle \frac{\partial \psi }{\partial r}\frac{\partial r}{\partial x}+\frac{\partial \psi }{\partial \theta }\frac{\partial \theta }{\partial x}}\right)}}$

${\displaystyle =\frac{\partial }{\partial x}\left(\frac{\partial \psi }{\partial r}\frac{\partial r}{\partial x}\right)+\frac{\partial }{\partial x}\left(\frac{\partial \psi }{\partial \theta }\frac{\partial \theta }{\partial x}\right)}$

ここで

より

${\displaystyle =\left\{\frac{\partial }{\partial x}\left(\frac{\partial \psi }{\partial r}\right)\right\}\frac{\partial r}{\partial x}}$${\displaystyle +\frac{\partial \psi }{\partial r}\left\{\frac{\partial }{\partial x}\left(\frac{\partial r}{\partial x}\right)\right\}}$${\displaystyle +\left\{\frac{\partial }{\partial x}\left(\frac{\partial \psi }{\partial \theta }\right)\right\}\frac{\partial \theta }{\partial x}}$${\displaystyle +\frac{\partial \psi }{\partial \theta }\left\{\frac{\partial }{\partial x}\left(\frac{\partial \theta }{\partial x}\right)\right\}}$

さらにここで

${\displaystyle \frac{\partial }{\partial x}\left(\frac{\partial \psi }{\partial r}\right)}$${\displaystyle =\left\{\frac{\partial }{\partial r}\frac{\partial \psi }{\partial r}\right\}\frac{\partial r}{\partial x}+\left\{\frac{\partial }{\partial \theta }\frac{\partial \psi }{\partial r}\right\}\frac{\partial \theta }{\partial x}}$${\displaystyle =\frac{{\partial }^2\psi }{\partial r^2}\frac{\partial r}{\partial x}+\frac{{\partial }^2\psi }{\partial \theta \partial r}\frac{\partial \theta }{\partial x}}$

${\displaystyle \frac{\partial }{\partial x}\left(\frac{\partial \psi }{\partial \theta }\right)}$${\displaystyle =\left\{\frac{\partial }{\partial r}\frac{\partial \psi }{\partial \theta }\right\}\frac{\partial r}{\partial x}+\left\{\frac{\partial }{\partial \theta }\frac{\partial \psi }{\partial \theta }\right\}\frac{\partial \theta }{\partial x}}$${\displaystyle =\frac{{\partial }^2\psi }{\partial r\partial \theta }\frac{\partial r}{\partial x}+\frac{{\partial }^2\psi }{\partial {\theta }^2}\frac{\partial \theta }{\partial x}}$

より

${\displaystyle =\left(\frac{{\partial }^2\psi }{\partial r^2}\frac{\partial r}{\partial x}+\frac{{\partial }^2\psi }{\partial \theta \partial r}\frac{\partial \theta }{\partial x}\right)\frac{\partial r}{\partial x}}$${\displaystyle +\frac{\partial \psi }{\partial r}\frac{{\partial }^{2}r}{{\partial }^{2}x}}$${\displaystyle +\left(\frac{{\partial }^2\psi }{\partial r\partial \theta }\frac{\partial r}{\partial x}+\frac{{\partial }^2\psi }{\partial {\theta }^2}\frac{\partial \theta }{\partial x}\right)\frac{\partial \theta }{\partial x}}$${\displaystyle +\frac{\partial \psi }{\partial \theta }\frac{{\partial }^2\theta }{{\partial }^{2}x}}$

${\displaystyle =\frac{{\partial }^2\psi }{\partial r^2}{\left(\frac{\partial r}{\partial x}\right)}^2+\frac{{\partial }^2\psi }{\partial \theta \partial r}\frac{\partial \theta }{\partial x}\frac{\partial r}{\partial x}}$${\displaystyle +\frac{\partial \psi }{\partial r}\frac{{\partial }^{2}r}{\partial x^2}}$${\displaystyle +\frac{{\partial }^2\psi }{\partial r\partial \theta }\frac{\partial r}{\partial x}\frac{\partial \theta }{\partial x}}$ ${\displaystyle +\frac{{\partial }^2\psi }{\partial {\theta }^2}{\left(\frac{\partial \theta }{\partial x}\right)}^2}$${\displaystyle +\frac{\partial \psi }{\partial \theta }\frac{{\partial }^2\theta }{\partial x^2}}$
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　 ･･････(1)

次に

${\displaystyle \frac{\partial r}{\partial x}}$ ， ${\displaystyle \frac{{\partial }^{2}r}{\partial x^2}}$ ，${\displaystyle \frac{\partial \theta }{\partial x}}$ ，${\displaystyle \frac{{\partial }^2\theta }{\partial x^2}}$

を用いて(1)を式変形する．

始めに

${\displaystyle x=r\cos \theta }$ ， ${\displaystyle {\displaystyle {\displaystyle y=r\sin \theta }}}$

の両辺を2乗して加えると

${\displaystyle x^2+y^2={\left(r\cos \theta \right)}^2+{\left(r\sin \theta \right)}^2}$

となり，これを整理すると

${\displaystyle x^2+y^2=r^2}$ 　　　･･････(2)

また，左辺同士、右辺同士で比をとることにより

${\displaystyle \frac{y}{x}=\frac{r\sin \theta }{r\cos \theta }}$

となり，これを整理して

${\displaystyle \tan \theta =\frac{y}{x}}$ 　　　　　･･････(3)

参考：三角関数の相互関係

が得られる．上記の(2)，(3)を用いて

${\displaystyle \frac{\partial r}{\partial x}}$ ， ${\displaystyle \frac{{\partial }^{2}r}{\partial x^2}}$ ，${\displaystyle \frac{\partial \theta }{\partial x}}$ ，${\displaystyle \frac{{\partial }^2\theta }{\partial x^2}}$

を求める．

(2)の両辺を${\displaystyle x}$ で微分して

${\displaystyle 2x}$ ${\displaystyle =2r\frac{\partial r}{\partial x}}$

${\displaystyle \frac{\partial r}{\partial x}}$ ${\displaystyle =\frac{x}{r}}$ ${\displaystyle =\frac{r\cos \theta }{r}}$ ${\displaystyle =\cos \theta }$　　　･･････(4)

が得られる． (3)の両辺をxで微分して

${\displaystyle \frac{1}{{\cos }^2\theta }\frac{\partial \theta }{\partial x}}$ ${\displaystyle =y\left(-\frac{1}{x^2}\right)}$

${\displaystyle \left({\tan }^2\theta +1\right)\frac{\partial \theta }{\partial x}}$ ${\displaystyle =-\frac{y}{x^2}}$

${\displaystyle \left(\frac{y^2}{x^2}+1\right)\frac{\partial \theta }{\partial x}}$ ${\displaystyle =-\frac{y}{x^2}}$

${\displaystyle \left(\frac{y^2+x^2}{x^2}\right)\frac{\partial \theta }{\partial x}}$ ${\displaystyle =-\frac{y}{x^2}}$

${\displaystyle \left(\frac{r^2}{x^2}\right)\frac{\partial \theta }{\partial x}}$ ${\displaystyle =-\frac{y}{x^2}}$

${\displaystyle \frac{\partial \theta }{\partial x}}$ ${\displaystyle =-\frac{y}{r^2}}$ ${\displaystyle =-\frac{r\sin \theta }{r^2}}$ ${\displaystyle =-\frac{\sin \theta }{r}}$ 　　　･･････(5)

が得られる．

${\displaystyle \frac{{\partial }^{2}r}{\partial x^2}}$${\displaystyle =\frac{\partial }{\partial x}\left(\frac{\partial r}{\partial x}\right)}$

(4)を代入して

${\displaystyle =\frac{\partial }{\partial x}\cos \theta }$ ${\displaystyle =\frac{\partial }{\partial r}\cos \theta \frac{\partial r}{\partial x}+\frac{\partial }{\partial \theta }\cos \theta \frac{\partial \theta }{\partial x}}$ ${\displaystyle =-\sin \theta \frac{\partial \theta }{\partial x}}$

(5)を代入して

${\displaystyle =-\sin \theta \left(-\frac{\sin \theta }{r}\right)}$ ${\displaystyle =\frac{{\sin }^2\theta }{r}}$ 　　　･･････(6)

が得られる．

${\displaystyle \frac{{\partial }^2\theta }{\partial x^2}}$ ${\displaystyle =\frac{\partial }{\partial x}\left(\frac{\partial \theta }{\partial x}\right)}$

(5)を代入して

${\displaystyle =\frac{\partial }{\partial x}\left(-\frac{\sin \theta }{r}\right)}$ ${\displaystyle =\frac{\partial }{\partial r}\left(-\frac{\sin \theta }{r}\right)\frac{\partial r}{\partial x}}$${\displaystyle +\frac{\partial }{\partial \theta }\left(-\frac{\sin \theta }{r}\right)\frac{\partial \theta }{\partial x}}$

(4)，(5)より

${\displaystyle =\frac{\sin \theta \cos \theta }{r^2}+\frac{\cos \theta \sin \theta }{r^2}}$ ${\displaystyle =\frac{2\sin \theta \cos \theta }{r^2}}$ 　　　･･････(7)

(4)〜(7)を用いて(1)を整理すると

${\displaystyle \frac{{\partial }^2\psi }{\partial x^2}}$ ${\displaystyle =\frac{{\partial }^2\psi }{\partial r^2}{\cos }^2\theta +\frac{{\partial }^2\psi }{\partial \theta \partial r}\left(-\frac{\sin \theta }{r}\right)\cos \theta }$ ${\displaystyle +\frac{\partial \psi }{\partial r}\frac{{\sin }^2\theta }{r}+\frac{{\partial }^2\psi }{\partial r\partial \theta }\cos \theta \left(-\frac{\sin \theta }{r}\right)}$ ${\displaystyle +\frac{{\partial }^2\psi }{\partial r^2}\frac{{\sin }^2\theta }{r^2}+\frac{\partial \psi }{\partial \theta }\frac{2\sin \theta \cos \theta }{r^2}}$　･･････(8)

次に，関数 ${\displaystyle \psi }$ を${\displaystyle y}$ で偏微分する．

${\displaystyle \frac{\partial \psi }{\partial y}}$ ${\displaystyle =\frac{\partial \psi }{\partial r}\frac{\partial r}{\partial y}+\frac{\partial \psi }{\partial \theta }\frac{\partial \theta }{\partial y}}$

参考:合成関数の偏導関数

${\displaystyle y}$ に関しても ${\displaystyle x}$ と同様に計算して

${\displaystyle \frac{{\partial }^2\psi }{\partial y^2}}$ ${\displaystyle =\frac{{\partial }^2\psi }{\partial r^2}{\left(\frac{\partial r}{\partial y}\right)}^2+\frac{{\partial }^2\psi }{\partial \theta \partial r}\frac{\partial \theta }{\partial y}\frac{\partial r}{\partial y}}$${\displaystyle +\frac{\partial \psi }{\partial r}\frac{{\partial }^{2}r}{\partial y^2}+\frac{{\partial }^2\psi }{\partial r\partial \theta }\frac{\partial r}{\partial y}\frac{\partial \theta }{\partial y}}$${\displaystyle +\frac{{\partial }^2\psi }{\partial {\theta }^2}{\left(\frac{\partial \theta }{\partial y}\right)}^2+\frac{\partial \psi }{\partial \theta }\frac{{\partial }^2\theta }{\partial y^2}}$　･･････(9)

また，(2)，(3)を用いて

${\displaystyle \frac{\partial r}{\partial y}}$ ， ${\displaystyle \frac{{\partial }^{2}r}{\partial y^2}}$ ， ${\displaystyle \frac{\partial \theta }{\partial y}}$ ， ${\displaystyle \frac{{\partial }^2\theta }{\partial y^2}}$

を求める．

${\displaystyle \frac{\partial r}{\partial y}}$ は (4) と同様に計算して

${\displaystyle \frac{\partial r}{\partial y}}$ ${\displaystyle =\sin \theta }$ 　　･･････(10)

(3)より

${\displaystyle \tan \theta =\frac{y}{x}}$

これを ${\displaystyle y}$ で微分して

${\displaystyle \frac{1}{{\cos }^2\theta }\frac{\partial \theta }{\partial y}}$ ${\displaystyle =\frac{1}{x}}$

${\displaystyle \left({\tan }^2\theta +1\right)\frac{\partial \theta }{\partial y}}$ ${\displaystyle =\frac{1}{x}}$

${\displaystyle \left(\frac{y^2}{x^2}+1\right)\frac{\partial \theta }{\partial y}}$ ${\displaystyle =\frac{1}{x}}$

${\displaystyle \left(\frac{y^2+x^2}{x^2}\right)\frac{\partial \theta }{\partial y}}$ ${\displaystyle =\frac{1}{x}}$

${\displaystyle \left(\frac{r^2}{x^2}\right)\frac{\partial \theta }{\partial y}}$ ${\displaystyle =\frac{1}{x}}$

${\displaystyle \frac{\partial \theta }{\partial y}}$ ${\displaystyle =\frac{x}{r^2}}$ ${\displaystyle =\frac{r\cos \theta }{r^2}}$ ${\displaystyle =\frac{\cos \theta }{r}}$　　･･････(11)

${\displaystyle \frac{{\partial }^{2}r}{\partial y^2}}$${\displaystyle =\frac{\partial }{\partial y}\left(\frac{\partial r}{\partial y}\right)}$

(10)を代入して

${\displaystyle =\frac{\partial }{\partial y}\sin \theta }$ ${\displaystyle =\cos \theta \frac{\partial \theta }{\partial y}}$

(11)を代入して

${\displaystyle =\frac{{\cos }^2\theta }{r}}$ 　　･･････(12)

${\displaystyle \frac{{\partial }^2\theta }{\partial y^2}}$ は(7)と同様に計算して

${\displaystyle \frac{{\partial }^2\theta }{\partial y^2}}$ ${\displaystyle =-\frac{2\sin \theta \cos \theta }{r^2}}$ 　　･･････(13)

(10)〜(13)を用いて(9)を整理すると

${\displaystyle \frac{{\partial }^2\psi }{\partial y^2}}$${\displaystyle =\frac{{\partial }^2\psi }{\partial r^2}{\sin }^2\theta +\frac{{\partial }^2\psi }{\partial \theta \partial r}\left(\frac{\cos \theta }{r}\right)\sin \theta $ {\displaystyle \frac{{\partial }^2\psi }{\partial y^2}}$+\frac{\partial \psi }{\partial r}\frac{{\cos }^2\theta }{r}+\frac{{\partial }^2\psi }{\partial r\partial \theta }\sin \theta \left(\frac{\cos \theta }{r}\right)}$ ${\displaystyle +\frac{{\partial }^2\psi }{\partial {\theta }^2}\frac{{\cos }^2\theta }{r^2}+\frac{\partial \psi }{\partial \theta }\left(-\frac{2\sin \theta \cos \theta }{r^2}\right)}$
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　 ･･････(14)

ここで(8)，(14)より，

${\displaystyle \frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}}$

${\displaystyle =\frac{{\partial }^2\psi }{\partial r^2}+\frac{1}{r}\frac{\partial \psi }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2\psi }{\partial {\theta }^2}}$${\displaystyle =\left(\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}\right)\psi }$

となる．よって示された．

 

ホーム>>カテゴリー別分類>>微分>>偏微分>>極座標表示におけるラプラシアン(2次元)

最終更新日：2023年1月17日