5-30 座標系変換(*) – ページ 4 – Shinshu Univ., Physical Chemistry Lab., Adsorption Group

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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 $\displaystyle \left( \frac{\partial f}{\partial x}\right)_y$ を f (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 r と θ 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)

固定ページ: 123456