極座標ラプラシアンの割とラクな計算法（高次元）

はじめに

前の記事 では2次元の極座標ラプラシアンを計算しましたが，今度は3次元以上のものも効率的に計算してみます．今回は意地でも微分計算をしたくなかったので，前記事の結果を少し引用します．

2次元の復習

極座標変換$x=r\sin \theta ,y=r\cos \theta$のもとで,
$$r\frac{\partial }{\partial r}=x\frac{\partial }{\partial x}+y\frac{\partial }{\partial y},\frac{\partial }{\partial \theta }=y\frac{\partial }{\partial x}-x\frac{\partial }{\partial y}$$
$$r^2\frac{{\partial }^2}{\partial x^2}+r^2\frac{{\partial }^2}{\partial y^2}=r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }^2}$$
となるのでした．なお，高次元への拡張を意識して$x$と$y$をいれかえていますが，やってることは 前の記事 と同じです．
今後のために，もう1つ式を導出します．上段の2式から$\frac{\partial }{\partial y}$を消去して
$$\begin{array}{rl}xr\frac{\partial }{\partial r}+y\frac{\partial }{\partial \theta }& =x^2\frac{\partial }{\partial x}+y^2\frac{\partial }{\partial x}\\ & =r^2\frac{\partial }{\partial x}(\because x^2+y^2=r^2)\end{array}$$
$$\therefore \frac{r}{\sin \theta }\frac{\partial }{\partial x}=r\frac{\partial }{\partial r}+\frac{1}{\tan \theta }\frac{\partial }{\partial \theta }\cdots (\ast )$$

3次元Ver.

ラプラシアン$\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}$を，極座標変換$x=r\sin \theta \sin \varphi ,y=r\sin \theta \cos \varphi ,z=r\cos \theta$のもとで書きかえます(通常の記法と$x$と$y$が逆なので注意)．
$w=r\sin \theta$とおいてみます．$x=w\sin \varphi ,y=w\cos \varphi$ですが，これは2次元の極座標変換に他ならないので，既知の結果から
$$w^2\frac{{\partial }^2}{\partial x^2}+w^2\frac{{\partial }^2}{\partial y^2}=w^2\frac{{\partial }^2}{\partial w^2}+w\frac{\partial }{\partial w}+\frac{{\partial }^2}{\partial {\varphi }^2}$$
となり，両辺$1/{\sin }^2\theta$倍して
$$r^2\frac{{\partial }^2}{\partial x^2}+r^2\frac{{\partial }^2}{\partial y^2}=r^2\frac{{\partial }^2}{\partial w^2}+\frac{r}{\sin \theta }\frac{\partial }{\partial w}+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}\cdots (1)$$
とします．一方$w=r\sin \theta ,z=r\cos \theta$であり，これも極座標変換なので，
$$r^2\frac{{\partial }^2}{\partial w^2}+r^2\frac{{\partial }^2}{\partial z^2}=r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }^2}\cdots (2)$$
が成り立ちます．(1)(2)から$r^2\frac{{\partial }^2}{\partial w^2}$を消すと，
$$r^2(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2})=r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }^2}+\frac{r}{\sin \theta }\frac{\partial }{\partial w}+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}$$
となります．$\frac{\partial }{\partial w}$の項が残っていますが，序盤で導いた$(\ast )$を$w,z$の極座標変換に適用した
$$\frac{r}{\sin \theta }\frac{\partial }{\partial w}=r\frac{\partial }{\partial r}+\frac{1}{\tan \theta }\frac{\partial }{\partial \theta }$$
を使うと，最終的に
$$r^2(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2})=r^2\frac{{\partial }^2}{\partial r^2}+2r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }^2}+\frac{1}{\tan \theta }\frac{\partial }{\partial \theta }+\frac{1}{{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}$$
$$\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}=\frac{{\partial }^2}{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}+\frac{1}{r^2\tan \theta }\frac{\partial }{\partial \theta }+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}$$
が得られます．
2次元の結果をフルに活用して，3次元ラプラシアンの極座標表示を手際よく導出できました！

4次元Ver.

ここまで低次の結果をうまく使えるケースもなかなかないので，せっかくならより高次のラプラシアンも極座標表示してみようと思います．
まずは小手調べで4次元をやってみます．4次元の極座標変換は
$$x_1=r\sin {\theta }_1\sin {\theta }_2\sin {\theta }_3,x_2=r\sin {\theta }_1\sin {\theta }_2\cos {\theta }_3,$$
$$x_3=r\sin {\theta }_1\cos {\theta }_2,x_4=r\cos {\theta }_1$$
です．3次元のときと同じように，$w=r\sin {\theta }_1$とおいて考えればよさそうです．
$x_1=w\sin {\theta }_2\sin {\theta }_3,x_2=w\sin {\theta }_2\cos {\theta }_3,x_3=w\cos {\theta }_2$なので，3次元の結果から
$$w^2(\frac{{\partial }^2}{\partial x_1^2}+\frac{{\partial }^2}{\partial x_2^2}+\frac{{\partial }^2}{\partial x_3^2})=w^2\frac{{\partial }^2}{\partial w^2}+2w\frac{\partial }{\partial w}+\frac{{\partial }^2}{\partial {\theta }_2^2}+\frac{1}{\tan {\theta }_2}\frac{\partial }{\partial {\theta }_2}+\frac{1}{{\sin }^2{\theta }_2}\frac{{\partial }^2}{\partial {\theta }_3^2}$$
となり，両辺$1/{\sin }^2{\theta }_1$倍して
$$r^2(\frac{{\partial }^2}{\partial x_1^2}+\frac{{\partial }^2}{\partial x_2^2}+\frac{{\partial }^2}{\partial x_3^2})=r^2\frac{{\partial }^2}{\partial w^2}+\frac{2r}{\sin {\theta }_1}\frac{\partial }{\partial w}+\frac{1}{{\sin }^2{\theta }_1}(\frac{{\partial }^2}{\partial {\theta }_2^2}+\frac{1}{\tan {\theta }_2}\frac{\partial }{\partial {\theta }_2})+\frac{1}{{\sin }^2{\theta }_1{\sin }^2{\theta }_2}\frac{{\partial }^2}{\partial {\theta }_3^2}$$
としておきます．一方，$w=r\sin {\theta }_1,x_4=r\cos {\theta }_1$から，2次元の結果を使えて
$$r^2\frac{{\partial }^2}{\partial w^2}+r^2\frac{{\partial }^2}{\partial x_4^2}=r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }_1^2}$$
$$\frac{r}{\sin {\theta }_1}\frac{\partial }{\partial w}=r\frac{\partial }{\partial r}+\frac{1}{\tan {\theta }_1}\frac{\partial }{\partial {\theta }_1}$$
なので，以上をまとめて
$$\begin{array}{rl}{r}^2(\frac{{\partial }^2}{\partial x_1^2}+\frac{{\partial }^2}{\partial x_2^2}+\frac{{\partial }^2}{\partial x_3^2}+\frac{{\partial }^2}{\partial x_4^2})& =r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }_1^2}+2(r\frac{\partial }{\partial r}+\frac{1}{\tan {\theta }_1}\frac{\partial }{\partial {\theta }_1})+\frac{1}{{\sin }^2{\theta }_1}(\frac{{\partial }^2}{\partial {\theta }_2^2}+\frac{1}{\tan {\theta }_2}\frac{\partial }{\partial {\theta }_2})+\frac{1}{{\sin }^2{\theta }_1{\sin }^2{\theta }_2}\frac{{\partial }^2}{\partial {\theta }_3^2}\\ & =r^2\frac{{\partial }^2}{\partial r^2}+3r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }_1^2}+\frac{2}{\tan {\theta }_1}\frac{\partial }{\partial {\theta }_1}+\frac{1}{{\sin }^2{\theta }_1}(\frac{{\partial }^2}{\partial {\theta }_2^2}+\frac{1}{\tan {\theta }_2}\frac{\partial }{\partial {\theta }_2})+\frac{1}{{\sin }^2{\theta }_1{\sin }^2{\theta }_2}\frac{{\partial }^2}{\partial {\theta }_3^2}\end{array}$$
となります．だんだん式が長くなってきましたが，やることは3次元のときとほぼ同じでしたね．

$n$次元Ver.

ここまでくると意地の領域ですが，$n$次元でもラプラシアンを極座標表示してみます．まず，$n$次元の極座標変換は
$$\begin{array}{rl}{x}_1& =r\sin {\theta }_1\sin {\theta }_2\cdots \sin {\theta }_{n-2}\sin {\theta }_{n-1},\\ x_2& =r\sin {\theta }_1\sin {\theta }_2\cdots \sin {\theta }_{n-2}\cos {\theta }_{n-1},\\ x_3& =r\sin {\theta }_1\sin {\theta }_2\cdots \sin {\theta }_{n-3}\cos {\theta }_{n-2},\\ & ⋮\\ x_{n-1}& =r\sin {\theta }_1\cos {\theta }_2,\\ x_n& =r\cos {\theta }_1\end{array}$$
です．例のごとく，$w=r\sin {\theta }_1$とおいて$n-1$次元の結果を使いたいので，帰納法でやることにします．
置き換えによって得られる$n-1$次の極座標変換は
$$\begin{array}{rl}{x}_1& =w\sin {\theta }_2\sin {\theta }_3\cdots \sin {\theta }_{n-2}\sin {\theta }_{n-1},\\ x_2& =w\sin {\theta }_2\sin {\theta }_3\cdots \sin {\theta }_{n-2}\cos {\theta }_{n-1},\\ x_3& =w\sin {\theta }_2\sin {\theta }_3\cdots \sin {\theta }_{n-3}\cos {\theta }_{n-2},\\ & ⋮\\ x_{n-2}& =w\sin {\theta }_2\cos {\theta }_3,\\ x_{n-1}& =w\cos {\theta }_2\end{array}$$
です．2次元の極座標変換$w=r\sin {\theta }_1,x_n=r\cos {\theta }_1$からは
$$r^2\frac{{\partial }^2}{\partial w^2}+r^2\frac{{\partial }^2}{\partial x_n^2}=r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }_1^2}\cdots (3)$$
$$\frac{r}{\sin {\theta }_1}\frac{\partial }{\partial w}=r\frac{\partial }{\partial r}+\frac{1}{\tan {\theta }_1}\frac{\partial }{\partial {\theta }_1}\cdots (4)$$
が出るのでした．
よって，これまでの結果や再帰の流れを考えると，$n$次元の極座標ラプラシアンが
$$r^2\sum _{i=1}^n\frac{{\partial }^2}{\partial x_i^2}=r^2\frac{{\partial }^2}{\partial r^2}+(n-1)r\frac{\partial }{\partial r}+\sum _{i=0}^{n-2}\frac{1}{(\sin {\theta }_0\cdots \sin {\theta }_i{)}^2}(\frac{{\partial }^2}{\partial {\theta }_{i+1}^2}+\frac{n-i-2}{\tan {\theta }_{i+1}}\frac{\partial }{\partial {\theta }_{i+1}})$$
と書けることが予想されます．ここで，${\theta }_0=\pi /2$としています．
さっそくinduction stepに入りましょう．$n-1$次の極座標変換では，$r$が$w$に変わり，${\theta }_1$が抜けていることから，
$$w^2\sum _{i=1}^{n-1}\frac{{\partial }^2}{\partial x_i^2}=w^2\frac{{\partial }^2}{\partial w^2}+(n-2)w\frac{\partial }{\partial w}+\sum _{i=1}^{n-2}\frac{1}{(\sin {\theta }_0\widehat{\sin {\theta }_1}\cdots \sin {\theta }_i{)}^2}(\frac{{\partial }^2}{\partial {\theta }_{i+1}^2}+\frac{n-i-2}{\tan {\theta }_{i+1}}\frac{\partial }{\partial {\theta }_{i+1}})$$
となります．ここで，$t_0,\cdots ,\widehat{t_j},\cdots ,t_N$は$t_j$を除くという意味です．両辺に$1/{\sin }^2{\theta }_1$をかけて
$$r^2\sum _{i=1}^{n-1}\frac{{\partial }^2}{\partial x_i^2}=r^2\frac{{\partial }^2}{\partial w^2}+\frac{(n-2)r}{\sin {\theta }_1}\frac{\partial }{\partial w}+\sum _{i=1}^{n-2}\frac{1}{(\sin {\theta }_0\sin {\theta }_1\cdots \sin {\theta }_i{)}^2}(\frac{{\partial }^2}{\partial {\theta }_{i+1}^2}+\frac{n-i-2}{\tan {\theta }_{i+1}}\frac{\partial }{\partial {\theta }_{i+1}})$$
とし，(3)(4)を使うと
$$\begin{array}{rl}{r}^2\sum _{i=1}^n\frac{{\partial }^2}{\partial x_i^2}& =r^2\frac{{\partial }^2}{\partial r^2}+r\frac{\partial }{\partial r}+\frac{{\partial }^2}{\partial {\theta }_1^2}+(n-2)(r\frac{\partial }{\partial r}+\frac{1}{\tan {\theta }_1}\frac{\partial }{\partial {\theta }_1})+\sum _{i=1}^{n-2}\frac{1}{(\sin {\theta }_0\sin {\theta }_1\cdots \sin {\theta }_i{)}^2}(\frac{{\partial }^2}{\partial {\theta }_{i+1}^2}+\frac{n-i-2}{\tan {\theta }_{i+1}}\frac{\partial }{\partial {\theta }_{i+1}})\\ & =r^2\frac{{\partial }^2}{\partial r^2}+(n-1)r\frac{\partial }{\partial r}+\sum _{i=0}^{n-2}\frac{1}{(\sin {\theta }_0\cdots \sin {\theta }_i{)}^2}(\frac{{\partial }^2}{\partial {\theta }_{i+1}^2}+\frac{n-i-2}{\tan {\theta }_{i+1}}\frac{\partial }{\partial {\theta }_{i+1}})\end{array}$$
となるので，$n$次元でも成り立つことが言えました．

おわりに

当初は思いついた2次元のものだけを書く予定でしたが，歯切れが悪いので，高次元Ver.も書くことにしました（笑）．3次元以降は一切微分の計算をしていないことからもわかる通り，必要な計算をうまく2次元に押し込められているわけですね．