$f: \mathbb{R}^2 \to \mathbb{R}$ と $g: \mathbb{R}^2 \to \mathbb{R}$ は $\mathbb{R}^2$ 上で $C^2$ 級であり、以下の関係式を満たすとします。 $\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}, \quad \frac{\partial g}{\partial x} = -\frac{\partial f}{\partial y}$ このとき、以下の3つの問題に答えます。 (1) $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$ を証明します。 (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、$F(r, \theta) = f(\varphi(r, \theta), \psi(r, \theta)), G(r, \theta) = g(\varphi(r, \theta), \psi(r, \theta))$ とします。このとき、次の関係式が成り立つことを証明します。 $r \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}, \quad \frac{1}{r} \frac{\partial F}{\partial \theta} = - \frac{\partial G}{\partial r}$ (3) $\frac{\partial^2 F}{\partial r^2} + \frac{1}{r^2} \frac{\partial^2 F}{\partial \theta^2} + \frac{1}{r} \frac{\partial F}{\partial r} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ が $\mathbb{R}^2 \setminus \{(0,0)\}$ 上で成り立つことを証明します。

1. 問題の内容

$f: \mathbb{R}^2 \to \mathbb{R}$ と $g: \mathbb{R}^2 \to \mathbb{R}$ は $\mathbb{R}^2$ 上で $C^2$ 級であり、以下の関係式を満たすとします。

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

このとき、以下の3つの問題に答えます。

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

(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、$F(r, \theta) = f(\varphi(r, \theta), \psi(r, \theta)), G(r, \theta) = g(\varphi(r, \theta), \psi(r, \theta))$ とします。このとき、次の関係式が成り立つことを証明します。

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

(3) $\frac{\partial^2 F}{\partial r^2} + \frac{1}{r^2} \frac{\partial^2 F}{\partial \theta^2} + \frac{1}{r} \frac{\partial F}{\partial r} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ が $\mathbb{R}^2 \setminus \{(0,0)\}$ 上で成り立つことを証明します。

2. 解き方の手順

(1) 与えられた関係式 $\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}$ と $\frac{\partial g}{\partial x} = -\frac{\partial f}{\partial y}$ を使います。

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

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

$f$ と $g$ は $C^2$ 級なので、$\frac{\partial^2 g}{\partial x \partial y} = \frac{\partial^2 g}{\partial y \partial x}$ が成り立ちます。

したがって、$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 g}{\partial x \partial y} - \frac{\partial^2 g}{\partial y \partial x} = 0$ となります。

(2) $F(r, \theta) = f(r \cos \theta, r \sin \theta)$ と $G(r, \theta) = g(r \cos \theta, r \sin \theta)$ に対して、偏微分を計算します。

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

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

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

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

与えられた関係式 $\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}$ と $\frac{\partial g}{\partial x} = -\frac{\partial f}{\partial y}$ を使うと、

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

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

よって、$r \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}$ と $\frac{1}{r} \frac{\partial F}{\partial \theta} = - \frac{\partial G}{\partial r}$ が成り立ちます。

(3) (2)の結果を使い、さらに偏微分を計算して$\frac{\partial^2 F}{\partial r^2}$, $\frac{\partial^2 F}{\partial \theta^2}$を求めます。

それらを$\frac{\partial^2 F}{\partial r^2} + \frac{1}{r^2} \frac{\partial^2 F}{\partial \theta^2} + \frac{1}{r} \frac{\partial F}{\partial r}$に代入して、(1)の結果を使うと、$\frac{\partial^2 F}{\partial r^2} + \frac{1}{r^2} \frac{\partial^2 F}{\partial \theta^2} + \frac{1}{r} \frac{\partial F}{\partial r} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$ が得られます。

3. 最終的な答え

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

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

(3) $\frac{\partial^2 F}{\partial r^2} + \frac{1}{r^2} \frac{\partial^2 F}{\partial \theta^2} + \frac{1}{r} \frac{\partial F}{\partial r} = 0$