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$u=u(x, y)$ であり、$x=r\cos\theta$, $y=r\sin\theta$ のとき、以下の問いに答える。 (1) $\frac{\partial u}{\partial r}, \frac{\partial u}{\partial \theta}$ を $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}$ で表す。 (2) $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}$ を $\frac{\partial u}{\partial r}, \frac{\partial u}{\partial \theta}$ で表す。 (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 $\frac{\partial u}{\partial r}, \frac{\partial u}{\partial \theta}$ で表す。

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) であり、x=rcos⁡θx=r\cos\theta, y=rsin⁡θy=r\sin\theta のとき、以下の問いに答える。
(1) ∂u∂r,∂u∂θ\frac{\partial u}{\partial r}, \frac{\partial u}{\partial \theta} を ∂u∂x,∂u∂y\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y} で表す。
(2) ∂u∂x,∂u∂y\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y} を ∂u∂r,∂u∂θ\frac{\partial u}{\partial r}, \frac{\partial u}{\partial \theta} で表す。
(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,∂u∂θ\frac{\partial u}{\partial r}, \frac{\partial u}{\partial \theta} で表す。

2. 解き方の手順

(1)
偏微分の連鎖律より、
∂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}
ここで、
∂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
したがって、
∂u∂r=∂u∂xcos⁡θ+∂u∂ysin⁡θ\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\cos\theta + \frac{\partial u}{\partial y}\sin\theta
∂u∂θ=∂u∂x(−rsin⁡θ)+∂u∂y(rcos⁡θ)\frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x}(-r\sin\theta) + \frac{\partial u}{\partial y}(r\cos\theta)
(2)
(1) の結果より、
∂u∂r=∂u∂xcos⁡θ+∂u∂ysin⁡θ\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\cos\theta + \frac{\partial u}{\partial y}\sin\theta
∂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}
これらを ∂u∂x\frac{\partial u}{\partial x} と ∂u∂y\frac{\partial u}{\partial y} について解く。
∂u∂x=A,∂u∂y=B\frac{\partial u}{\partial x} = A, \frac{\partial u}{\partial y} = B とおく。
Acos⁡θ+Bsin⁡θ=∂u∂rA\cos\theta + B\sin\theta = \frac{\partial u}{\partial r}
−Arsin⁡θ+Brcos⁡θ=∂u∂θ-Ar\sin\theta + Br\cos\theta = \frac{\partial u}{\partial \theta}
最初の式に rsin⁡θr\sin\theta をかけ、2番目の式に cos⁡θ\cos\theta をかけると
Arsin⁡θcos⁡θ+Brsin⁡2θ=rsin⁡θ∂u∂rAr\sin\theta\cos\theta + Br\sin^2\theta = r\sin\theta\frac{\partial u}{\partial r}
−Arsin⁡θcos⁡θ+Brcos⁡2θ=cos⁡θ∂u∂θ-Ar\sin\theta\cos\theta + Br\cos^2\theta = \cos\theta\frac{\partial u}{\partial \theta}
足し合わせると、
Br=rsin⁡θ∂u∂r+cos⁡θ∂u∂θBr = r\sin\theta\frac{\partial u}{\partial r} + \cos\theta\frac{\partial u}{\partial \theta}
B=sin⁡θ∂u∂r+cos⁡θr∂u∂θB = \sin\theta\frac{\partial u}{\partial r} + \frac{\cos\theta}{r}\frac{\partial u}{\partial \theta}
同様に、最初の式に rcos⁡θr\cos\theta をかけ、2番目の式に sin⁡θ\sin\theta をかけると
Arcos⁡2θ+Brsin⁡θcos⁡θ=rcos⁡θ∂u∂rAr\cos^2\theta + Br\sin\theta\cos\theta = r\cos\theta\frac{\partial u}{\partial r}
−Arsin⁡2θ+Brsin⁡θcos⁡θ=sin⁡θ∂u∂θ-Ar\sin^2\theta + Br\sin\theta\cos\theta = \sin\theta\frac{\partial u}{\partial \theta}
引き算すると、
Ar=rcos⁡θ∂u∂r−sin⁡θ∂u∂θAr = r\cos\theta\frac{\partial u}{\partial r} - \sin\theta\frac{\partial u}{\partial \theta}
A=cos⁡θ∂u∂r−sin⁡θr∂u∂θA = \cos\theta\frac{\partial u}{\partial r} - \frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}
したがって、
∂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=(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)2−2sin⁡θcos⁡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\frac{\sin\theta\cos\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\frac{\sin\theta\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⁡2θr2+cos⁡2θr2)(∂u∂θ)2= (\cos^2\theta + \sin^2\theta)(\frac{\partial u}{\partial r})^2 + (\frac{\sin^2\theta}{r^2} + \frac{\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=∂∂x(∂u∂x)+∂∂y(∂u∂y)\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial x}(\frac{\partial u}{\partial x}) + \frac{\partial}{\partial y}(\frac{\partial u}{\partial y})
=∂∂x(cos⁡θ∂u∂r−sin⁡θr∂u∂θ)+∂∂y(sin⁡θ∂u∂r+cos⁡θr∂u∂θ)= \frac{\partial}{\partial x}(\cos\theta\frac{\partial u}{\partial r} - \frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}) + \frac{\partial}{\partial y}(\sin\theta\frac{\partial u}{\partial r} + \frac{\cos\theta}{r}\frac{\partial u}{\partial \theta})
=(cos⁡θ∂∂r−sin⁡θr∂∂θ)(cos⁡θ∂u∂r−sin⁡θr∂u∂θ)+(sin⁡θ∂∂r+cos⁡θr∂∂θ)(sin⁡θ∂u∂r+cos⁡θr∂u∂θ)= (\cos\theta\frac{\partial}{\partial r} - \frac{\sin\theta}{r}\frac{\partial}{\partial \theta})(\cos\theta\frac{\partial u}{\partial r} - \frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}) + (\sin\theta\frac{\partial}{\partial r} + \frac{\cos\theta}{r}\frac{\partial}{\partial \theta})(\sin\theta\frac{\partial u}{\partial r} + \frac{\cos\theta}{r}\frac{\partial u}{\partial \theta})
=∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2= \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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