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$u=u(x, y)$, $x=r\cos\theta$, $y=r\sin\theta$ のとき、以下の問いに答えます。 (1) $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$ を $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$ で表してください。 (2) $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$ を $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$ で表してください。 (3) $(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2$ と $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ を $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$ で表してください。

解析学偏微分連鎖律座標変換ラプラシアン
2025/7/4

1. 問題の内容

u=u(x,y)u=u(x, y), x=rcos⁡θx=r\cos\theta, y=rsin⁡θy=r\sin\theta のとき、以下の問いに答えます。
(1) ∂u∂r\frac{\partial u}{\partial r}, ∂u∂θ\frac{\partial u}{\partial \theta} を ∂u∂x\frac{\partial u}{\partial x}, ∂u∂y\frac{\partial u}{\partial y} で表してください。
(2) ∂u∂x\frac{\partial u}{\partial x}, ∂u∂y\frac{\partial u}{\partial y} を ∂u∂r\frac{\partial u}{\partial r}, ∂u∂θ\frac{\partial u}{\partial \theta} で表してください。
(3) (∂u∂x)2+(∂u∂y)2(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2 と ∂2u∂x2+∂2u∂y2\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} を ∂u∂r\frac{\partial u}{\partial r}, ∂u∂θ\frac{\partial u}{\partial \theta} で表してください。

2. 解き方の手順

(1) 連鎖律を用いると、
∂u∂r=∂u∂x∂x∂r+∂u∂y∂y∂r\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r}
∂u∂θ=∂u∂x∂x∂θ+∂u∂y∂y∂θ\frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta}
x=rcos⁡θx=r\cos\theta, y=rsin⁡θy=r\sin\thetaより、
∂x∂r=cos⁡θ\frac{\partial x}{\partial r} = \cos\theta
∂y∂r=sin⁡θ\frac{\partial y}{\partial r} = \sin\theta
∂x∂θ=−rsin⁡θ\frac{\partial x}{\partial \theta} = -r\sin\theta
∂y∂θ=rcos⁡θ\frac{\partial y}{\partial \theta} = r\cos\theta
よって、
∂u∂r=cos⁡θ∂u∂x+sin⁡θ∂u∂y\frac{\partial u}{\partial r} = \cos\theta \frac{\partial u}{\partial x} + \sin\theta \frac{\partial u}{\partial y}
∂u∂θ=−rsin⁡θ∂u∂x+rcos⁡θ∂u∂y\frac{\partial u}{\partial \theta} = -r\sin\theta \frac{\partial u}{\partial x} + r\cos\theta \frac{\partial u}{\partial y}
(2) (1)の結果から、
∂u∂r=cos⁡θ∂u∂x+sin⁡θ∂u∂y\frac{\partial u}{\partial r} = \cos\theta \frac{\partial u}{\partial x} + \sin\theta \frac{\partial u}{\partial y}
∂u∂θ=−rsin⁡θ∂u∂x+rcos⁡θ∂u∂y\frac{\partial u}{\partial \theta} = -r\sin\theta \frac{\partial u}{\partial x} + r\cos\theta \frac{\partial u}{\partial y}
この連立方程式を∂u∂x\frac{\partial u}{\partial x}と∂u∂y\frac{\partial u}{\partial y}について解きます。
∂u∂r⋅rcos⁡θ=rcos⁡2θ∂u∂x+rcos⁡θsin⁡θ∂u∂y\frac{\partial u}{\partial r}\cdot r\cos\theta = r\cos^2\theta \frac{\partial u}{\partial x} + r\cos\theta\sin\theta \frac{\partial u}{\partial y}
∂u∂θ⋅sin⁡θ=−rsin⁡2θ∂u∂x+rcos⁡θsin⁡θ∂u∂y\frac{\partial u}{\partial \theta}\cdot \sin\theta = -r\sin^2\theta \frac{\partial u}{\partial x} + r\cos\theta\sin\theta \frac{\partial u}{\partial y}
Subtracting the second equation from the first yields:
rcos⁡θ∂u∂r−sin⁡θ∂u∂θ=r(cos⁡2θ+sin⁡2θ)∂u∂x=r∂u∂xr\cos\theta \frac{\partial u}{\partial r} - \sin\theta \frac{\partial u}{\partial \theta} = r(\cos^2\theta + \sin^2\theta)\frac{\partial u}{\partial x} = r \frac{\partial u}{\partial x}
∂u∂x=cos⁡θ∂u∂r−sin⁡θr∂u∂θ\frac{\partial u}{\partial x} = \cos\theta \frac{\partial u}{\partial r} - \frac{\sin\theta}{r} \frac{\partial u}{\partial \theta}
Similarly,
∂u∂r⋅rsin⁡θ=rcos⁡θsin⁡θ∂u∂x+rsin⁡2θ∂u∂y\frac{\partial u}{\partial r}\cdot r\sin\theta = r\cos\theta\sin\theta \frac{\partial u}{\partial x} + r\sin^2\theta \frac{\partial u}{\partial y}
∂u∂θ⋅cos⁡θ=−rsin⁡θcos⁡θ∂u∂x+rcos⁡2θ∂u∂y\frac{\partial u}{\partial \theta}\cdot \cos\theta = -r\sin\theta\cos\theta \frac{\partial u}{\partial x} + r\cos^2\theta \frac{\partial u}{\partial y}
Adding the above two equations,
rsin⁡θ∂u∂r+cos⁡θ∂u∂θ=r(sin⁡2θ+cos⁡2θ)∂u∂y=r∂u∂yr\sin\theta \frac{\partial u}{\partial r} + \cos\theta \frac{\partial u}{\partial \theta} = r(\sin^2\theta + \cos^2\theta)\frac{\partial u}{\partial y} = r \frac{\partial u}{\partial y}
∂u∂y=sin⁡θ∂u∂r+cos⁡θr∂u∂θ\frac{\partial u}{\partial y} = \sin\theta \frac{\partial u}{\partial r} + \frac{\cos\theta}{r} \frac{\partial u}{\partial \theta}
(3) (∂u∂x)2+(∂u∂y)2=(cos⁡θ∂u∂r−sin⁡θr∂u∂θ)2+(sin⁡θ∂u∂r+cos⁡θr∂u∂θ)2(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2 = (\cos\theta \frac{\partial u}{\partial r} - \frac{\sin\theta}{r} \frac{\partial u}{\partial \theta})^2 + (\sin\theta \frac{\partial u}{\partial r} + \frac{\cos\theta}{r} \frac{\partial u}{\partial \theta})^2
=cos⁡2θ(∂u∂r)2−2cos⁡θsin⁡θr∂u∂r∂u∂θ+sin⁡2θr2(∂u∂θ)2+sin⁡2θ(∂u∂r)2+2sin⁡θcos⁡θr∂u∂r∂u∂θ+cos⁡2θr2(∂u∂θ)2= \cos^2\theta(\frac{\partial u}{\partial r})^2 - 2\cos\theta\frac{\sin\theta}{r}\frac{\partial u}{\partial r}\frac{\partial u}{\partial \theta} + \frac{\sin^2\theta}{r^2}(\frac{\partial u}{\partial \theta})^2 + \sin^2\theta(\frac{\partial u}{\partial r})^2 + 2\sin\theta\frac{\cos\theta}{r}\frac{\partial u}{\partial r}\frac{\partial u}{\partial \theta} + \frac{\cos^2\theta}{r^2}(\frac{\partial u}{\partial \theta})^2
=(cos⁡2θ+sin⁡2θ)(∂u∂r)2+(sin⁡2θ+cos⁡2θr2)(∂u∂θ)2= (\cos^2\theta+\sin^2\theta)(\frac{\partial u}{\partial r})^2 + (\frac{\sin^2\theta + \cos^2\theta}{r^2})(\frac{\partial u}{\partial \theta})^2
=(∂u∂r)2+1r2(∂u∂θ)2= (\frac{\partial u}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial u}{\partial \theta})^2
∂2u∂x2+∂2u∂y2=∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}

3. 最終的な答え

(1)
∂u∂r=cos⁡θ∂u∂x+sin⁡θ∂u∂y\frac{\partial u}{\partial r} = \cos\theta \frac{\partial u}{\partial x} + \sin\theta \frac{\partial u}{\partial y}
∂u∂θ=−rsin⁡θ∂u∂x+rcos⁡θ∂u∂y\frac{\partial u}{\partial \theta} = -r\sin\theta \frac{\partial u}{\partial x} + r\cos\theta \frac{\partial u}{\partial y}
(2)
∂u∂x=cos⁡θ∂u∂r−sin⁡θr∂u∂θ\frac{\partial u}{\partial x} = \cos\theta \frac{\partial u}{\partial r} - \frac{\sin\theta}{r} \frac{\partial u}{\partial \theta}
∂u∂y=sin⁡θ∂u∂r+cos⁡θr∂u∂θ\frac{\partial u}{\partial y} = \sin\theta \frac{\partial u}{\partial r} + \frac{\cos\theta}{r} \frac{\partial u}{\partial \theta}
(3)
(∂u∂x)2+(∂u∂y)2=(∂u∂r)2+1r2(∂u∂θ)2(\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2 = (\frac{\partial u}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial u}{\partial \theta})^2
∂2u∂x2+∂2u∂y2=∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}

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