SENSEI AI
About App

与えられた条件の下で、以下の3つの命題を証明する。 (1) $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$ (2) $r \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}, \frac{1}{r} \frac{\partial F}{\partial \theta} = -\frac{\partial G}{\partial r}$ ただし、$F(r, \theta) = f(r\cos\theta, r\sin\theta)$, $G(r, \theta) = g(r\cos\theta, r\sin\theta)$, $\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}, \frac{\partial g}{\partial x} = -\frac{\partial f}{\partial y}$ (3) $\frac{\partial^2 F}{\partial r^2} + \frac{1}{r^2} \frac{\partial^2 F}{\partial \theta^2} + \frac{1}{r} \frac{\partial F}{\partial r} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ (ただし、$ (r, \theta) \in \mathbb{R}^2 \setminus \{(0, 0)\} $)

解析学偏微分ラプラス方程式偏微分方程式座標変換
2025/7/16

1. 問題の内容

与えられた条件の下で、以下の3つの命題を証明する。
(1) ∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0
(2) r∂F∂r=∂G∂θ,1r∂F∂θ=−∂G∂rr \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}, \frac{1}{r} \frac{\partial F}{\partial \theta} = -\frac{\partial G}{\partial r}
ただし、F(r,θ)=f(rcos⁡θ,rsin⁡θ)F(r, \theta) = f(r\cos\theta, r\sin\theta), G(r,θ)=g(rcos⁡θ,rsin⁡θ)G(r, \theta) = g(r\cos\theta, r\sin\theta), ∂f∂x=∂g∂y,∂g∂x=−∂f∂y\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}, \frac{\partial g}{\partial x} = -\frac{\partial f}{\partial y}
(3) ∂2F∂r2+1r2∂2F∂θ2+1r∂F∂r=∂2f∂x2+∂2f∂y2\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} (ただし、(r,θ)∈R2∖{(0,0)} (r, \theta) \in \mathbb{R}^2 \setminus \{(0, 0)\} )

2. 解き方の手順

(1)
与えられた条件∂f∂x=∂g∂y,∂g∂x=−∂f∂y\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}, \frac{\partial g}{\partial x} = -\frac{\partial f}{\partial y}をxxとyyでそれぞれ偏微分すると、
∂2f∂x2=∂2g∂x∂y,∂2g∂x∂y=−∂2f∂y2\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 g}{\partial x \partial y}, \frac{\partial^2 g}{\partial x \partial y} = -\frac{\partial^2 f}{\partial y^2}.
したがって、∂2f∂x2=−∂2f∂y2\frac{\partial^2 f}{\partial x^2} = -\frac{\partial^2 f}{\partial y^2}となるため、∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0が成り立つ。
(2)
まず、F(r,θ)=f(rcos⁡θ,rsin⁡θ)F(r, \theta) = f(r\cos\theta, r\sin\theta)とG(r,θ)=g(rcos⁡θ,rsin⁡θ)G(r, \theta) = g(r\cos\theta, r\sin\theta)をrrとθ\thetaで偏微分する。
∂F∂r=∂f∂x∂(rcos⁡θ)∂r+∂f∂y∂(rsin⁡θ)∂r=∂f∂xcos⁡θ+∂f∂ysin⁡θ\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
∂F∂θ=∂f∂x∂(rcos⁡θ)∂θ+∂f∂y∂(rsin⁡θ)∂θ=∂f∂x(−rsin⁡θ)+∂f∂y(rcos⁡θ)\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)
∂G∂r=∂g∂x∂(rcos⁡θ)∂r+∂g∂y∂(rsin⁡θ)∂r=∂g∂xcos⁡θ+∂g∂ysin⁡θ\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
∂G∂θ=∂g∂x∂(rcos⁡θ)∂θ+∂g∂y∂(rsin⁡θ)∂θ=∂g∂x(−rsin⁡θ)+∂g∂y(rcos⁡θ)\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)
与えられた条件∂f∂x=∂g∂y,∂g∂x=−∂f∂y\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、
∂G∂θ=−∂f∂y(−rsin⁡θ)+∂f∂x(rcos⁡θ)=rsin⁡θ∂f∂y+rcos⁡θ∂f∂x=r(∂f∂xcos⁡θ+∂f∂ysin⁡θ)=r∂F∂r\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(\frac{\partial f}{\partial x}\cos\theta + \frac{\partial f}{\partial y}\sin\theta) = r\frac{\partial F}{\partial r}
∂G∂r=−∂f∂ycos⁡θ+∂f∂xsin⁡θ=−1r(∂f∂x(−rsin⁡θ)+∂f∂y(rcos⁡θ))=−1r∂F∂θ\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 x}(-r\sin\theta) + \frac{\partial f}{\partial y}(r\cos\theta)) = -\frac{1}{r}\frac{\partial F}{\partial \theta}
したがって、r∂F∂r=∂G∂θ,1r∂F∂θ=−∂G∂rr \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}, \frac{1}{r} \frac{\partial F}{\partial \theta} = -\frac{\partial G}{\partial r}が成り立つ。
(3)
∂F∂r=∂f∂xcos⁡θ+∂f∂ysin⁡θ\frac{\partial F}{\partial r} = \frac{\partial f}{\partial x}\cos\theta + \frac{\partial f}{\partial y}\sin\theta
∂F∂θ=−rsin⁡θ∂f∂x+rcos⁡θ∂f∂y\frac{\partial F}{\partial \theta} = -r\sin\theta\frac{\partial f}{\partial x} + r\cos\theta\frac{\partial f}{\partial y}
∂2F∂r2=∂∂r(∂f∂xcos⁡θ+∂f∂ysin⁡θ)=cos⁡θ(∂2f∂x2cos⁡θ+∂2f∂y∂xsin⁡θ)+sin⁡θ(∂2f∂x∂ycos⁡θ+∂2f∂y2sin⁡θ)=cos⁡2θ∂2f∂x2+2sin⁡θcos⁡θ∂2f∂x∂y+sin⁡2θ∂2f∂y2\frac{\partial^2 F}{\partial r^2} = \frac{\partial}{\partial r}(\frac{\partial f}{\partial x}\cos\theta + \frac{\partial f}{\partial y}\sin\theta) = \cos\theta(\frac{\partial^2 f}{\partial x^2}\cos\theta + \frac{\partial^2 f}{\partial y \partial x}\sin\theta) + \sin\theta(\frac{\partial^2 f}{\partial x \partial y}\cos\theta + \frac{\partial^2 f}{\partial y^2}\sin\theta) = \cos^2\theta\frac{\partial^2 f}{\partial x^2} + 2\sin\theta\cos\theta\frac{\partial^2 f}{\partial x\partial y} + \sin^2\theta\frac{\partial^2 f}{\partial y^2}
∂2F∂θ2=∂∂θ(−rsin⁡θ∂f∂x+rcos⁡θ∂f∂y)=−rcos⁡θ∂f∂x−rsin⁡θ(∂2f∂x2(−rsin⁡θ)+∂2f∂y∂x(rcos⁡θ))−rsin⁡θ∂f∂y+rcos⁡θ(∂2f∂x∂y(−rsin⁡θ)+∂2f∂y2(rcos⁡θ))=−rcos⁡θ∂f∂x−rsin⁡θ∂f∂y+r2sin⁡2θ∂2f∂x2−2r2sin⁡θcos⁡θ∂2f∂x∂y+r2cos⁡2θ∂2f∂y2\frac{\partial^2 F}{\partial \theta^2} = \frac{\partial}{\partial \theta}(-r\sin\theta\frac{\partial f}{\partial x} + r\cos\theta\frac{\partial f}{\partial y}) = -r\cos\theta\frac{\partial f}{\partial x} - r\sin\theta(\frac{\partial^2 f}{\partial x^2}(-r\sin\theta) + \frac{\partial^2 f}{\partial y\partial x}(r\cos\theta)) -r\sin\theta\frac{\partial f}{\partial y} + r\cos\theta(\frac{\partial^2 f}{\partial x\partial y}(-r\sin\theta) + \frac{\partial^2 f}{\partial y^2}(r\cos\theta)) = -r\cos\theta\frac{\partial f}{\partial x} - r\sin\theta\frac{\partial f}{\partial y} + r^2\sin^2\theta\frac{\partial^2 f}{\partial x^2} - 2r^2\sin\theta\cos\theta\frac{\partial^2 f}{\partial x\partial y} + r^2\cos^2\theta\frac{\partial^2 f}{\partial y^2}
1r2∂2F∂θ2=−cos⁡θr∂f∂x−sin⁡θr∂f∂y+sin⁡2θ∂2f∂x2−2sin⁡θcos⁡θ∂2f∂x∂y+cos⁡2θ∂2f∂y2\frac{1}{r^2}\frac{\partial^2 F}{\partial \theta^2} = -\frac{\cos\theta}{r}\frac{\partial f}{\partial x} - \frac{\sin\theta}{r}\frac{\partial f}{\partial y} + \sin^2\theta\frac{\partial^2 f}{\partial x^2} - 2\sin\theta\cos\theta\frac{\partial^2 f}{\partial x\partial y} + \cos^2\theta\frac{\partial^2 f}{\partial y^2}
1r∂F∂r=cos⁡θr∂f∂x+sin⁡θr∂f∂y\frac{1}{r}\frac{\partial F}{\partial r} = \frac{\cos\theta}{r}\frac{\partial f}{\partial x} + \frac{\sin\theta}{r}\frac{\partial f}{\partial y}
したがって、
∂2F∂r2+1r2∂2F∂θ2+1r∂F∂r=(cos⁡2θ+sin⁡2θ)∂2f∂x2+(sin⁡2θ+cos⁡2θ)∂2f∂y2=∂2f∂x2+∂2f∂y2\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} = (\cos^2\theta + \sin^2\theta)\frac{\partial^2 f}{\partial x^2} + (\sin^2\theta + \cos^2\theta)\frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}

3. 最終的な答え

(1) ∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0
(2) r∂F∂r=∂G∂θ,1r∂F∂θ=−∂G∂rr \frac{\partial F}{\partial r} = \frac{\partial G}{\partial \theta}, \frac{1}{r} \frac{\partial F}{\partial \theta} = -\frac{\partial G}{\partial r}
(3) ∂2F∂r2+1r2∂2F∂θ2+1r∂F∂r=∂2f∂x2+∂2f∂y2\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}

「解析学」の関連問題

与えられたべき級数 $\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} x^{2n-1}$ の収束半径を求める問題です。

べき級数収束半径比判定法極限
2025/7/17

与えられた4つの級数の収束半径を求める問題です。各級数は以下の通りです。 (1) $\sum_{n=1}^{\infty} \frac{n!}{n^n} x^n$ (2) $\sum_{n=1}^{\...

級数収束半径極限
2025/7/17

与えられた2つの級数が条件収束することを示す問題です。条件収束とは、級数自体は収束するが、絶対値をとった級数は発散することを意味します。 (1) $\sum_{n=1}^{\infty} (-1)^{...

級数条件収束交代級数収束判定法
2025/7/17

方程式 $x^2 = ae^x$ が異なる3個の実数解を持つように、定数 $a$ の値の範囲を定める問題です。

指数関数微分グラフ増減実数解方程式
2025/7/17

$x > 0$ のとき、不等式 $x - \frac{x^2}{2} < \log(1+x)$ が成り立つことを証明します。

不等式対数関数導関数単調増加関数証明
2025/7/17

$y = 2\sin2(x + \frac{\pi}{4}) + 2$ のグラフとして正しいものを選択する問題です。

三角関数グラフ振幅周期平行移動
2025/7/17

与えられた微分方程式 $\frac{dy}{dx} = 1 - e^{-x}$ を解く問題です。

微分方程式変数分離形積分積分定数
2025/7/17

## 問題の解答

積分不定積分三角関数指数関数
2025/7/17

問題は3つの部分から構成されています。 1. $z = \tan(2x+y)$ 上の点 $(\frac{\pi}{4}, \frac{\pi}{4}, \boxed{1})$ における接平面の方程式...

偏微分接平面ヤコビアン多変数関数
2025/7/17

与えられた問題は以下の6つの小問から構成されています。 1. $y = \arccos x$ の $x = -\frac{\sqrt{3}}{2}$ における接線の方程式を求める。

微分接線極限テイラー展開導関数
2025/7/17