$u = \theta + \log r$, $x = r^2 \cos \theta$, $y = r \sin \theta$ が与えられている。 (1) $(r, \theta) = (1, \pi/4)$ 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) $(r, \theta) = (1, \pi/4)$ における $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}$ を求める。

1. 問題の内容

$u = \theta + \log r$, $x = r^2 \cos \theta$, $y = r \sin \theta$ が与えられている。

(1) $(r, \theta) = (1, \pi/4)$ 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) $(r, \theta) = (1, \pi/4)$ 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 $y = r \sin \theta$ から $\frac{\partial x}{\partial r}$, $\frac{\partial x}{\partial \theta}$, $\frac{\partial y}{\partial r}$, $\frac{\partial y}{\partial \theta}$ を計算する。

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

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

$\frac{\partial y}{\partial r} = \sin \theta$

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

次に、ヤコビアン $J = \frac{\partial (x, y)}{\partial (r, \theta)}$ を計算する。

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

$(r, \theta) = (1, \pi/4)$ のとき、$\cos (\pi/4) = \sin (\pi/4) = \frac{1}{\sqrt{2}}$ なので、

$J = 1^2 \left( 2 \left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2 \right) = 2 \cdot \frac{1}{2} + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}$

次に、逆ヤコビアンを使って $\frac{\partial r}{\partial x}$, $\frac{\partial \theta}{\partial x}$, $\frac{\partial r}{\partial y}$, $\frac{\partial \theta}{\partial y}$ を計算する。

$\frac{\partial r}{\partial x} = \frac{1}{J} \frac{\partial y}{\partial \theta} = \frac{1}{J} r \cos \theta = \frac{1}{3/2} \cdot 1 \cdot \frac{1}{\sqrt{2}} = \frac{2}{3} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{3}$

$\frac{\partial \theta}{\partial x} = \frac{-1}{J} \frac{\partial y}{\partial r} = \frac{-1}{J} \sin \theta = \frac{-1}{3/2} \cdot \frac{1}{\sqrt{2}} = -\frac{2}{3} \cdot \frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{3}$

$\frac{\partial r}{\partial y} = \frac{-1}{J} \frac{\partial x}{\partial \theta} = \frac{-1}{J} (-r^2 \sin \theta) = \frac{r^2 \sin \theta}{J} = \frac{1^2 \cdot \frac{1}{\sqrt{2}}}{3/2} = \frac{2}{3} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{3}$

$\frac{\partial \theta}{\partial y} = \frac{1}{J} \frac{\partial x}{\partial r} = \frac{1}{J} (2r \cos \theta) = \frac{2r \cos \theta}{J} = \frac{2 \cdot 1 \cdot \frac{1}{\sqrt{2}}}{3/2} = \frac{4}{3} \cdot \frac{1}{\sqrt{2}} = \frac{2\sqrt{2}}{3}$

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

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

$u = \theta + \log r$ より

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

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

したがって、

$\frac{\partial u}{\partial x} = \frac{1}{r} \frac{\partial r}{\partial x} + \frac{\partial \theta}{\partial x} = \frac{1}{1} \frac{\sqrt{2}}{3} + (-\frac{\sqrt{2}}{3}) = \frac{\sqrt{2}}{3} - \frac{\sqrt{2}}{3} = 0$

$\frac{\partial u}{\partial y} = \frac{1}{r} \frac{\partial r}{\partial y} + \frac{\partial \theta}{\partial y} = \frac{1}{1} \frac{\sqrt{2}}{3} + \frac{2\sqrt{2}}{3} = \frac{\sqrt{2}}{3} + \frac{2\sqrt{2}}{3} = \frac{3\sqrt{2}}{3} = \sqrt{2}$

$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0 + \sqrt{2} = \sqrt{2}$

3. 最終的な答え

(1) $\frac{\partial r}{\partial x} = \frac{\sqrt{2}}{3}$, $\frac{\partial \theta}{\partial x} = -\frac{\sqrt{2}}{3}$, $\frac{\partial r}{\partial y} = \frac{\sqrt{2}}{3}$, $\frac{\partial \theta}{\partial y} = \frac{2\sqrt{2}}{3}$

(2) $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \sqrt{2}$