$u = \theta + \log r$, $x = r^2 \cos \theta$, $y = \sin (r\theta)$ が与えられたとき、以下の問題を解く。 (1) $(r, \theta) = (1, \frac{\pi}{4})$ における $\frac{\partial r}{\partial x}$, $\frac{\partial \theta}{\partial x}$, $\frac{\partial r}{\partial y}$, $\frac{\partial \theta}{\partial y}$ を求める。 (2) $(r, \theta) = (1, \frac{\pi}{4})$ における $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}$ を求める。

以下に問題の解答を示します。

1. 問題の内容

$u = \theta + \log r$, $x = r^2 \cos \theta$, $y = \sin (r\theta)$ が与えられたとき、以下の問題を解く。

(1) $(r, \theta) = (1, \frac{\pi}{4})$ における $\frac{\partial r}{\partial x}$, $\frac{\partial \theta}{\partial x}$, $\frac{\partial r}{\partial y}$, $\frac{\partial \theta}{\partial y}$ を求める。

(2) $(r, \theta) = (1, \frac{\pi}{4})$ における $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}$ を求める。

2. 解き方の手順

(1)

まず、$x = r^2 \cos \theta$ と $y = \sin (r\theta)$ を $r$ と $\theta$ で偏微分する。

$\frac{\partial x}{\partial r} = 2r \cos \theta$

$\frac{\partial x}{\partial \theta} = -r^2 \sin \theta$

$\frac{\partial y}{\partial r} = \theta \cos (r\theta)$

$\frac{\partial y}{\partial \theta} = r \cos (r\theta)$

次に、逆関数の微分を用いて $\frac{\partial r}{\partial x}$, $\frac{\partial \theta}{\partial x}$, $\frac{\partial r}{\partial y}$, $\frac{\partial \theta}{\partial y}$ を求める。

まず、ヤコビアン $J$ を計算する。

$J = \frac{\partial (x, y)}{\partial (r, \theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} 2r \cos \theta & -r^2 \sin \theta \\ \theta \cos (r\theta) & r \cos (r\theta) \end{vmatrix} = 2r^2 \cos \theta \cos (r\theta) + r^2 \theta \sin \theta \cos (r\theta) = r^2 (2 \cos \theta + \theta \sin \theta) \cos (r\theta)$

$(r, \theta) = (1, \frac{\pi}{4})$ のとき、$x = \frac{\sqrt{2}}{2}$、$y = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$である。また、

$\frac{\partial x}{\partial r} = 2 \cos \frac{\pi}{4} = \sqrt{2}$

$\frac{\partial x}{\partial \theta} = - \sin \frac{\pi}{4} = -\frac{\sqrt{2}}{2}$

$\frac{\partial y}{\partial r} = \frac{\pi}{4} \cos \frac{\pi}{4} = \frac{\pi}{4} \cdot \frac{\sqrt{2}}{2} = \frac{\pi \sqrt{2}}{8}$

$\frac{\partial y}{\partial \theta} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$

$J = (2 \cos \frac{\pi}{4} + \frac{\pi}{4} \sin \frac{\pi}{4}) \cos \frac{\pi}{4} = (2 \cdot \frac{\sqrt{2}}{2} + \frac{\pi}{4} \cdot \frac{\sqrt{2}}{2}) \frac{\sqrt{2}}{2} = (1 + \frac{\pi}{8}) \sqrt{2} \frac{\sqrt{2}}{2} = 1 + \frac{\pi}{8}$

$\frac{\partial (r, \theta)}{\partial (x, y)} = J^{-1} = \frac{1}{1+\frac{\pi}{8}}$

$\frac{\partial r}{\partial x} = \frac{1}{J} \frac{\partial y}{\partial \theta} = \frac{\sqrt{2}/2}{1 + \pi/8} = \frac{4\sqrt{2}}{8+\pi}$

$\frac{\partial \theta}{\partial x} = \frac{1}{J} (-\frac{\partial y}{\partial r}) = \frac{-\pi \sqrt{2}/8}{1 + \pi/8} = \frac{-\pi \sqrt{2}}{8+\pi}$

$\frac{\partial r}{\partial y} = \frac{1}{J} (-\frac{\partial x}{\partial \theta}) = \frac{\sqrt{2}/2}{1 + \pi/8} = \frac{4\sqrt{2}}{8+\pi}$

$\frac{\partial \theta}{\partial y} = \frac{1}{J} \frac{\partial x}{\partial r} = \frac{\sqrt{2}}{1 + \pi/8} = \frac{8\sqrt{2}}{8+\pi}$

(2)

$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial x}$

$\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial y}$

$\frac{\partial u}{\partial r} = \frac{1}{r}$

$\frac{\partial u}{\partial \theta} = 1$

$\frac{\partial u}{\partial x} = \frac{1}{r} \frac{\partial r}{\partial x} + \frac{\partial \theta}{\partial x} = \frac{4\sqrt{2}}{8+\pi} - \frac{\pi \sqrt{2}}{8+\pi} = \frac{(4-\pi)\sqrt{2}}{8+\pi}$

$\frac{\partial u}{\partial y} = \frac{1}{r} \frac{\partial r}{\partial y} + \frac{\partial \theta}{\partial y} = \frac{4\sqrt{2}}{8+\pi} + \frac{8\sqrt{2}}{8+\pi} = \frac{12\sqrt{2}}{8+\pi}$

$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \frac{(4-\pi)\sqrt{2}}{8+\pi} + \frac{12\sqrt{2}}{8+\pi} = \frac{(16-\pi)\sqrt{2}}{8+\pi}$

3. 最終的な答え

(1)

$\frac{\partial r}{\partial x} = \frac{4\sqrt{2}}{8+\pi}$

$\frac{\partial \theta}{\partial x} = \frac{-\pi \sqrt{2}}{8+\pi}$

$\frac{\partial r}{\partial y} = \frac{4\sqrt{2}}{8+\pi}$

$\frac{\partial \theta}{\partial y} = \frac{8\sqrt{2}}{8+\pi}$

(2)

$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \frac{(16-\pi)\sqrt{2}}{8+\pi}$