Takuya Miyashita
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*2次元でのラプラシアンの極座標表示 [#oeb7ce98]
誰もが一生に一度はやっておいた方が良いと言われている(?)...
~
#contents
**ゴール [#f268ec26]
2次元直交座標系 \( (x,y) \) でのラプラシアン
\[
\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\parti...
\]
を,次式のような2次元極座標 \( (r,\theta) \) での表現を...
\[
\nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\...
\]
ここで
\[\begin{align}
x = &~r\cos\theta \tag{3}\\
y = &~ r\sin\theta \tag{4}\\
r = &~ \sqrt{x^2+y^2} \tag{5}\\
\tan\theta = &~ \frac{y}{x} \tag{6}\\
\end{align}\]
である.
**導出 [#w8747832]
***一階微分 [#x1b790f4]
まず \(\dfrac{\partial}{\partial x}\) と \(\dfrac{\partia...
\[\begin{align}
\frac{\partial}{\partial x} = & \frac{\partial r}{\partia...
\frac{\partial}{\partial y} = & \frac{\partial r}{\partia...
\end{align}\]
であるから,\(\dfrac{\partial r}{\partial x}\), \(\dfrac{...
式(5)を \(x\) で微分して
\[
\frac{\partial r}{\partial x} = \frac{2x}{2\sqrt{x^2+y^2}...
\]
が得られる.次に式(6)
\[\begin{align}
\tan\theta = &~ \frac{y}{x} \tag{6}
\end{align}\]
を \(x\) で微分して
\[\begin{align}
\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial x} =...
\frac{\partial\theta}{\partial x} = &~ - \frac{\sin\theta...
\end{align}\]
を得る.同様に \(y\) についても行うと
\[
\frac{\partial r}{\partial y} = \frac{2y}{2\sqrt{x^2+y^2}...
\]
および
\[\begin{align}
\tan\theta = &~ \frac{y}{x} \tag{6}
\end{align}\]
を \(y\) で微分して
\[\begin{align}
\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial y} =...
\frac{\partial\theta}{\partial y} = &~ \frac{\cos\theta}{...
\end{align}\]
となる.したがって
\[\begin{align}
\frac{\partial}{\partial x} = & \frac{\partial r}{\partia...
= &~\cos\theta \frac{\partial}{\partial r} - \frac{\sin\...
\end{align}\]
\[\begin{align}
\frac{\partial}{\partial y} = & \frac{\partial r}{\partia...
= &~ \sin\theta \frac{\partial}{\partial r} + \frac{\cos...
\end{align}\]
となる.これで一階微分が求まった.
***二階微分 [#z80ddc60]
次に \(\dfrac{\partial^2}{\partial x^2}\) と \(\dfrac{\pa...
式(13)より
\[\begin{align}
\frac{\partial}{\partial x} \left( \frac{\partial}{\parti...
= &~ \cos\theta \frac{\partial}{\partial r}\left( \cos\th...
= &~ \cos\theta \left( \cos\theta\frac{\partial^2}{\part...
& -\frac{\sin\theta}{r}\left( -\sin\theta \frac{\partial}...
\end{align}\]
これをまとめると,
\[\begin{align}
\frac{\partial^2}{\partial x^2} =
\cos^2\theta \frac{\partial^2}{\partial r^2} +
\frac{\sin^2\theta}{r} \frac{\partial}{\partial r}-
\frac{2\sin\theta\cos\theta}{r} \frac{\partial^2}{\parti...
\frac{2\sin\theta\cos\theta}{r^2} \frac{\partial}{\partia...
\frac{\sin^2\theta}{r^2} \frac{\partial^2}{\partial \the...
\tag{15}
\end{align}\]
となる.分母の \(r\) の次数と分子の \(\sin\theta\) または...
揃っていなかったらどこかで計算を間違えているはず.~
~
同様にして式(14)より
\[\begin{align}
\frac{\partial}{\partial y} \left( \frac{\partial}{\parti...
\frac{\partial r}{\partial y}\frac{\partial}{\partial r}\...
\frac{\partial \theta}{\partial y}\frac{\partial}{\partia...
= &~ \sin\theta \frac{\partial}{\partial r}\left( \sin\th...
\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\left...
= &~ \sin\theta \left( \sin\theta \frac{\partial^2}{\par...
&+\frac{\cos\theta}{r}\left( \cos\theta \frac{\partial}{\...
\end{align}\]
これをまとめると
\[\begin{align}
\frac{\partial^2}{\partial y^2} =
\sin^2\theta \frac{\partial^2}{\partial r^2} +
\frac{\cos^2\theta}{r} \frac{\partial}{\partial r} +
\frac{2\sin\theta\cos\theta}{r} \frac{\partial^2}{\parti...
\frac{2\sin\theta\cos\theta}{r^2} \frac{\partial}{\partia...
\frac{\cos^2\theta}{r^2} \frac{\partial^2}{\partial \the...
\tag{16}
\end{align}\]
***仕上げ [#fdbdaaed]
式(15)と式(16)よりゴールとして設定した式(2)が導ける.
\[\begin{align}
\nabla^2 = & \frac{\partial^2}{\partial x^2} + \frac{\par...
= &~ \left( \sin^2\theta + \cos^2\theta \right) \frac{\pa...
\frac{\left( \sin^2\theta + \cos^2\theta \right)}{r} \fra...
\frac{\left( \sin^2\theta + \cos^2\theta \right)}{r^2} \f...
= & \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\p...
\end{align}\]
おわり.
End:
*2次元でのラプラシアンの極座標表示 [#oeb7ce98]
誰もが一生に一度はやっておいた方が良いと言われている(?)...
~
#contents
**ゴール [#f268ec26]
2次元直交座標系 \( (x,y) \) でのラプラシアン
\[
\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\parti...
\]
を,次式のような2次元極座標 \( (r,\theta) \) での表現を...
\[
\nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\...
\]
ここで
\[\begin{align}
x = &~r\cos\theta \tag{3}\\
y = &~ r\sin\theta \tag{4}\\
r = &~ \sqrt{x^2+y^2} \tag{5}\\
\tan\theta = &~ \frac{y}{x} \tag{6}\\
\end{align}\]
である.
**導出 [#w8747832]
***一階微分 [#x1b790f4]
まず \(\dfrac{\partial}{\partial x}\) と \(\dfrac{\partia...
\[\begin{align}
\frac{\partial}{\partial x} = & \frac{\partial r}{\partia...
\frac{\partial}{\partial y} = & \frac{\partial r}{\partia...
\end{align}\]
であるから,\(\dfrac{\partial r}{\partial x}\), \(\dfrac{...
式(5)を \(x\) で微分して
\[
\frac{\partial r}{\partial x} = \frac{2x}{2\sqrt{x^2+y^2}...
\]
が得られる.次に式(6)
\[\begin{align}
\tan\theta = &~ \frac{y}{x} \tag{6}
\end{align}\]
を \(x\) で微分して
\[\begin{align}
\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial x} =...
\frac{\partial\theta}{\partial x} = &~ - \frac{\sin\theta...
\end{align}\]
を得る.同様に \(y\) についても行うと
\[
\frac{\partial r}{\partial y} = \frac{2y}{2\sqrt{x^2+y^2}...
\]
および
\[\begin{align}
\tan\theta = &~ \frac{y}{x} \tag{6}
\end{align}\]
を \(y\) で微分して
\[\begin{align}
\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial y} =...
\frac{\partial\theta}{\partial y} = &~ \frac{\cos\theta}{...
\end{align}\]
となる.したがって
\[\begin{align}
\frac{\partial}{\partial x} = & \frac{\partial r}{\partia...
= &~\cos\theta \frac{\partial}{\partial r} - \frac{\sin\...
\end{align}\]
\[\begin{align}
\frac{\partial}{\partial y} = & \frac{\partial r}{\partia...
= &~ \sin\theta \frac{\partial}{\partial r} + \frac{\cos...
\end{align}\]
となる.これで一階微分が求まった.
***二階微分 [#z80ddc60]
次に \(\dfrac{\partial^2}{\partial x^2}\) と \(\dfrac{\pa...
式(13)より
\[\begin{align}
\frac{\partial}{\partial x} \left( \frac{\partial}{\parti...
= &~ \cos\theta \frac{\partial}{\partial r}\left( \cos\th...
= &~ \cos\theta \left( \cos\theta\frac{\partial^2}{\part...
& -\frac{\sin\theta}{r}\left( -\sin\theta \frac{\partial}...
\end{align}\]
これをまとめると,
\[\begin{align}
\frac{\partial^2}{\partial x^2} =
\cos^2\theta \frac{\partial^2}{\partial r^2} +
\frac{\sin^2\theta}{r} \frac{\partial}{\partial r}-
\frac{2\sin\theta\cos\theta}{r} \frac{\partial^2}{\parti...
\frac{2\sin\theta\cos\theta}{r^2} \frac{\partial}{\partia...
\frac{\sin^2\theta}{r^2} \frac{\partial^2}{\partial \the...
\tag{15}
\end{align}\]
となる.分母の \(r\) の次数と分子の \(\sin\theta\) または...
揃っていなかったらどこかで計算を間違えているはず.~
~
同様にして式(14)より
\[\begin{align}
\frac{\partial}{\partial y} \left( \frac{\partial}{\parti...
\frac{\partial r}{\partial y}\frac{\partial}{\partial r}\...
\frac{\partial \theta}{\partial y}\frac{\partial}{\partia...
= &~ \sin\theta \frac{\partial}{\partial r}\left( \sin\th...
\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\left...
= &~ \sin\theta \left( \sin\theta \frac{\partial^2}{\par...
&+\frac{\cos\theta}{r}\left( \cos\theta \frac{\partial}{\...
\end{align}\]
これをまとめると
\[\begin{align}
\frac{\partial^2}{\partial y^2} =
\sin^2\theta \frac{\partial^2}{\partial r^2} +
\frac{\cos^2\theta}{r} \frac{\partial}{\partial r} +
\frac{2\sin\theta\cos\theta}{r} \frac{\partial^2}{\parti...
\frac{2\sin\theta\cos\theta}{r^2} \frac{\partial}{\partia...
\frac{\cos^2\theta}{r^2} \frac{\partial^2}{\partial \the...
\tag{16}
\end{align}\]
***仕上げ [#fdbdaaed]
式(15)と式(16)よりゴールとして設定した式(2)が導ける.
\[\begin{align}
\nabla^2 = & \frac{\partial^2}{\partial x^2} + \frac{\par...
= &~ \left( \sin^2\theta + \cos^2\theta \right) \frac{\pa...
\frac{\left( \sin^2\theta + \cos^2\theta \right)}{r} \fra...
\frac{\left( \sin^2\theta + \cos^2\theta \right)}{r^2} \f...
= & \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\p...
\end{align}\]
おわり.
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