5-30 座標系変換

解答(2階微分の変換)

少し式展開を補いながら進めると

 \displaystyle \left(\frac{\partial^2 f}{\partial x^2}\right)_y = \left[\frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x}\right)_y \right]_y

ここで \displaystyle \left( \frac{\partial f}{\partial x}\right)_yf (演算対象)とみて 偏微分の連鎖規則 (2)式を再び適用

 \displaystyle \\ = \left[\frac{\partial}{\partial r} \left( \frac{\partial f}{\partial x}\right)_y \right]_\theta \left(\frac{\partial r}{\partial x}\right)_y + \left[\frac{\partial}{\partial \theta} \left( \frac{\partial f}{\partial x}\right)_y \right]_r \left(\frac{\partial \theta}{\partial x}\right)_y

第 1 項は r が固定なので 考慮しなくて良い

\displaystyle = \left[\frac{\partial}{\partial \theta} \left( \frac{\partial f}{\partial x}\right)_y \right]_r \left(\frac{\partial \theta}{\partial x}\right)_y

前に求めた (6), (4)式を代入

\displaystyle = \left\{ \frac{\partial}{\partial \theta} \left[ -\frac{\sin{\theta}}{r} \left( \frac{\partial f}{\partial \theta} \right)_r \right] \right\}_r \left( - \frac{\sin{\theta}}{r} \right)

これで rθ だけが含まれている状態になったので、あとは通常の微分を行えばよい。 {}内部に 合成関数の微分
(fg)′ = f′g + fg′ を適用して

\displaystyle = \left\{ -\frac{\cos{\theta}}{r} \left( \frac{\partial f}{\partial \theta} \right)_r -\frac{\sin{\theta}}{r} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r  \right\} \left( - \frac{\sin{\theta}}{r} \right)\\ \\ \\ =\frac{\sin{\theta}\cos{\theta}}{r^2}\left(\frac{\partial f}{\partial \theta} \right)_r + \frac{\sin^2{\theta}}{r^2} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r\\ …(9)

同様に

\displaystyle  \left(\frac{\partial^2 f}{\partial y^2} \right)_x  = \left[\frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y}\right)_x \right]_x\\ \\ \\ = \left[\frac{\partial}{\partial r} \left( \frac{\partial f}{\partial y}\right)_x \right]_\theta \left(\frac{\partial r}{\partial y}\right)_x + \left[\frac{\partial}{\partial \theta} \left( \frac{\partial f}{\partial y}\right)_x \right]_r \left(\frac{\partial \theta}{\partial y}\right)_x\\ \\ \\ = \left[\frac{\partial}{\partial \theta} \left( \frac{\partial f}{\partial y}\right)_x \right]_r \left(\frac{\partial \theta}{\partial y}\right)_x\\ \\ \\ = \left\{ \frac{\partial}{\partial \theta} \left[ \frac{\cos{\theta}}{r} \left( \frac{\partial f}{\partial \theta} \right)_r \right] \right\}_r \left( \frac{\cos{\theta}}{r} \right)\\ \\ \\ = \left\{ -\frac{\sin{\theta}}{r} \left( \frac{\partial f}{\partial \theta} \right)_r +\frac{\cos{\theta}}{r} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r  \right\} \left( \frac{\cos{\theta}}{r} \right)\\ \\ \\ = - \frac{\sin{\theta}\cos{\theta}}{r^2}\left(\frac{\partial f}{\partial \theta} \right)_r + \frac{\cos^2{\theta}}{r^2} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r\\ …(10)

求めた \displaystyle \left(\frac{\partial^2 f}{\partial x^2} \right)_y,\left(\frac{\partial^2 f}{\partial y^2} \right)_x ((9),(10)式)を加えると 

\displaystyle \nabla^2 f = \left( \frac{\partial^2 f}{\partial x^2} \right)_y + \left( \frac{\partial^2 f}{\partial y^2} \right)_x\\ \\ \\ = \frac{\sin{\theta}\cos{\theta}}{r^2}\left(\frac{\partial f}{\partial \theta} \right)_r + \frac{\sin^2{\theta}}{r^2} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r -\frac{\sin{\theta}\cos{\theta}}{r^2}\left(\frac{\partial f}{\partial \theta} \right)_r + \frac{\cos^2{\theta}}{r^2} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r\\ \\ \\ = \left\{\frac{\sin{\theta}\cos{\theta}}{r^2}-\frac{\sin{\theta}\cos{\theta}}{r^2} \right\} \left(\frac{\partial f}{\partial \theta} \right)_r  + \left\{ \frac{\sin^2{\theta}}{r^2} + \frac{\cos^2{\theta}}{r^2}\right\} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r\\ \\ \\ = \frac{1}{r^2} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r …(11)