$u = \log \sqrt{x^2 + y^2}$ ($x^2 + y^2 \ne 0$), $v = \tan^{-1} \frac{y}{x}$ ($x \ne 0$) が与えられたとき、以下の2つを証明する問題です。 (1) $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$, $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$ (2) $\Delta u = 0$, $\Delta v = 0$ (ここで、$\Delta$ はラプラシアンを表し、$\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ です)

1. 問題の内容

$u = \log \sqrt{x^2 + y^2}$ ($x^2 + y^2 \ne 0$), $v = \tan^{-1} \frac{y}{x}$ ($x \ne 0$) が与えられたとき、以下の2つを証明する問題です。

(1) $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$, $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

(2) $\Delta u = 0$, $\Delta v = 0$ (ここで、$\Delta$ はラプラシアンを表し、$\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ です)

2. 解き方の手順

(1) まず、それぞれの偏微分を計算します。

$u = \log \sqrt{x^2 + y^2} = \frac{1}{2} \log (x^2 + y^2)$なので、

$\frac{\partial u}{\partial x} = \frac{1}{2} \cdot \frac{2x}{x^2 + y^2} = \frac{x}{x^2 + y^2}$

$\frac{\partial u}{\partial y} = \frac{1}{2} \cdot \frac{2y}{x^2 + y^2} = \frac{y}{x^2 + y^2}$

次に、$v = \tan^{-1} \frac{y}{x}$を偏微分します。

$\frac{\partial v}{\partial x} = \frac{1}{1 + (\frac{y}{x})^2} \cdot (-\frac{y}{x^2}) = \frac{x^2}{x^2 + y^2} \cdot (-\frac{y}{x^2}) = -\frac{y}{x^2 + y^2}$

$\frac{\partial v}{\partial y} = \frac{1}{1 + (\frac{y}{x})^2} \cdot (\frac{1}{x}) = \frac{x^2}{x^2 + y^2} \cdot (\frac{1}{x}) = \frac{x}{x^2 + y^2}$

したがって、

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

$\frac{\partial u}{\partial y} = \frac{y}{x^2 + y^2} = - (-\frac{y}{x^2 + y^2}) = -\frac{\partial v}{\partial x}$

(2) ラプラシアンを計算します。

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

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

$\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} + \frac{x^2 - y^2}{(x^2 + y^2)^2} = 0$

$\frac{\partial^2 v}{\partial x^2} = \frac{\partial}{\partial x} (-\frac{y}{x^2 + y^2}) = -y \cdot \frac{-2x}{(x^2 + y^2)^2} = \frac{2xy}{(x^2 + y^2)^2}$

$\frac{\partial^2 v}{\partial y^2} = \frac{\partial}{\partial y} (\frac{x}{x^2 + y^2}) = x \cdot \frac{-2y}{(x^2 + y^2)^2} = -\frac{2xy}{(x^2 + y^2)^2}$

$\Delta v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = \frac{2xy}{(x^2 + y^2)^2} - \frac{2xy}{(x^2 + y^2)^2} = 0$

3. 最終的な答え

(1) $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$, $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$ は成立する。

(2) $\Delta u = 0$, $\Delta v = 0$ は成立する。