5-30 座標系変換

問題 (2階微分の変換)

(問題の続き)

式(2)をもう一度適用して、

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

となることを示せ。同じように

\displaystyle  \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{\cos^2{\theta}}{r^2} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r\\ (r 固定)

および

\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{1}{r^2} \left( \frac{\partial^2 f}{\partial \theta^2} \right)_r (r 固定)

であることを示せ。

(問題の続きは第5ページ)