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

1. 問題の内容

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

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

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

2. 解き方の手順

(1) まず、$x$と$y$を$r$と$\theta$で偏微分する。

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

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

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

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

次に、逆関数の偏微分の公式を用いる。

$\frac{\partial r}{\partial x} = \frac{\frac{\partial y}{\partial \theta}}{J}$

$\frac{\partial \theta}{\partial x} = -\frac{\frac{\partial y}{\partial r}}{J}$

$\frac{\partial r}{\partial y} = -\frac{\frac{\partial x}{\partial \theta}}{J}$

$\frac{\partial \theta}{\partial y} = \frac{\frac{\partial x}{\partial r}}{J}$

ここで、$J$はヤコビアンで、

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

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

$\frac{\partial x}{\partial r} = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}$

$\frac{\partial x}{\partial \theta} = -1 \cdot \frac{1}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$

$\frac{\partial y}{\partial r} = \frac{1}{\sqrt{2}}$

$\frac{\partial y}{\partial \theta} = 1 \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$

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

$\frac{\partial r}{\partial x} = \frac{\frac{1}{\sqrt{2}}}{\frac{3}{2}} = \frac{1}{\sqrt{2}} \cdot \frac{2}{3} = \frac{\sqrt{2}}{3}$

$\frac{\partial \theta}{\partial x} = -\frac{\frac{1}{\sqrt{2}}}{\frac{3}{2}} = -\frac{1}{\sqrt{2}} \cdot \frac{2}{3} = -\frac{\sqrt{2}}{3}$

$\frac{\partial r}{\partial y} = -\frac{-\frac{1}{\sqrt{2}}}{\frac{3}{2}} = \frac{1}{\sqrt{2}} \cdot \frac{2}{3} = \frac{\sqrt{2}}{3}$

$\frac{\partial \theta}{\partial y} = \frac{\sqrt{2}}{\frac{3}{2}} = \sqrt{2} \cdot \frac{2}{3} = \frac{2\sqrt{2}}{3}$

(2)

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

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

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

$(r, \theta) = (1, \pi/4)$ のとき、$\frac{\partial u}{\partial r} = 1$

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

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

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

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)

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