$f: \mathbb{R}^2 \rightarrow \mathbb{R}$, $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ が $\mathbb{R}^2$ 上で $C^2$ 級であり、次の関係式を満たすと仮定する。 $\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}$, $\frac{\partial g}{\partial x} = - \frac{\partial f}{\partial y}$ (1) $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$ を証明せよ。 (2) $r > 0$, $\theta \in \mathbb{R}$ に対して、$\varphi(r, \theta) = r \cos \theta$, $\psi(r, \theta) = r \sin \theta$ とし、$F(r, \theta) = f(\varphi(r, \theta), \psi(r, \theta))$, $G(r, \theta) = g(\varphi(r, \theta), \psi(r, \theta))$ とする。このとき、次の関係式が成り立つことを証明せよ。 $r \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}$, $\frac{1}{r} \frac{\partial F}{\partial \theta} = - \frac{\partial G}{\partial r}$

1. 問題の内容

$f: \mathbb{R}^2 \rightarrow \mathbb{R}$, $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ が $\mathbb{R}^2$ 上で $C^2$ 級であり、次の関係式を満たすと仮定する。

$\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}$, $\frac{\partial g}{\partial x} = - \frac{\partial f}{\partial y}$

(1) $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$ を証明せよ。

(2) $r > 0$, $\theta \in \mathbb{R}$ に対して、$\varphi(r, \theta) = r \cos \theta$, $\psi(r, \theta) = r \sin \theta$ とし、$F(r, \theta) = f(\varphi(r, \theta), \psi(r, \theta))$, $G(r, \theta) = g(\varphi(r, \theta), \psi(r, \theta))$ とする。このとき、次の関係式が成り立つことを証明せよ。

$r \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}$, $\frac{1}{r} \frac{\partial F}{\partial \theta} = - \frac{\partial G}{\partial r}$

2. 解き方の手順

(1) 与えられた条件 $\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}$ を $x$ で偏微分すると $\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 g}{\partial x \partial y}$ となる。

また、$\frac{\partial g}{\partial x} = - \frac{\partial f}{\partial y}$ を $y$ で偏微分すると $\frac{\partial^2 g}{\partial y \partial x} = - \frac{\partial^2 f}{\partial y^2}$ となる。

$f, g$ は $C^2$ 級であるから、$\frac{\partial^2 g}{\partial x \partial y} = \frac{\partial^2 g}{\partial y \partial x}$ が成り立つ。したがって、

$\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 g}{\partial x \partial y} = \frac{\partial^2 g}{\partial y \partial x} = - \frac{\partial^2 f}{\partial y^2}$

よって、$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$ が示された。

(2) $F(r, \theta) = f(\varphi(r, \theta), \psi(r, \theta)) = f(r \cos \theta, r \sin \theta)$ であるから、連鎖律を用いて

$\frac{\partial F}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial \varphi}{\partial r} + \frac{\partial f}{\partial y} \frac{\partial \psi}{\partial r} = \frac{\partial f}{\partial x} \cos \theta + \frac{\partial f}{\partial y} \sin \theta$

$\frac{\partial F}{\partial \theta} = \frac{\partial f}{\partial x} \frac{\partial \varphi}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial \psi}{\partial \theta} = \frac{\partial f}{\partial x} (-r \sin \theta) + \frac{\partial f}{\partial y} (r \cos \theta)$

同様に、$G(r, \theta) = g(\varphi(r, \theta), \psi(r, \theta)) = g(r \cos \theta, r \sin \theta)$ であるから、

$\frac{\partial G}{\partial r} = \frac{\partial g}{\partial x} \frac{\partial \varphi}{\partial r} + \frac{\partial g}{\partial y} \frac{\partial \psi}{\partial r} = \frac{\partial g}{\partial x} \cos \theta + \frac{\partial g}{\partial y} \sin \theta$

$\frac{\partial G}{\partial \theta} = \frac{\partial g}{\partial x} \frac{\partial \varphi}{\partial \theta} + \frac{\partial g}{\partial y} \frac{\partial \psi}{\partial \theta} = \frac{\partial g}{\partial x} (-r \sin \theta) + \frac{\partial g}{\partial y} (r \cos \theta)$

ここで、$\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}$, $\frac{\partial g}{\partial x} = - \frac{\partial f}{\partial y}$ を用いると、

$\frac{\partial G}{\partial \theta} = - \frac{\partial f}{\partial y} (-r \sin \theta) + \frac{\partial f}{\partial x} (r \cos \theta) = r \sin \theta \frac{\partial f}{\partial y} + r \cos \theta \frac{\partial f}{\partial x} = r (\cos \theta \frac{\partial f}{\partial x} + \sin \theta \frac{\partial f}{\partial y}) = r \frac{\partial F}{\partial r}$

$\frac{\partial G}{\partial r} = - \frac{\partial f}{\partial y} \cos \theta + \frac{\partial f}{\partial x} \sin \theta$

$- \frac{1}{r} \frac{\partial F}{\partial \theta} = - \frac{1}{r} [\frac{\partial f}{\partial x} (-r \sin \theta) + \frac{\partial f}{\partial y} (r \cos \theta)] = \frac{\partial f}{\partial x} \sin \theta - \frac{\partial f}{\partial y} \cos \theta = \frac{\partial g}{\partial y} \sin \theta - \frac{\partial g}{\partial x} \cos \theta = - \frac{\partial G}{\partial r}$

したがって、$r \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}$, $\frac{1}{r} \frac{\partial F}{\partial \theta} = - \frac{\partial G}{\partial r}$ が示された。

3. 最終的な答え

(1) $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$

(2) $r \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}$, $\frac{1}{r} \frac{\partial F}{\partial \theta} = - \frac{\partial G}{\partial r}$