円柱座標の勾配、発散、ラプラシアン

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ベクトル解析11　円柱座標の勾配、発散、ラプラシアン

基本事項の整理。直交座標との関係性について。

\(x=r\cos\theta\)、\(y=r\sin\theta\)、\(z=z\)

上から以下二つの式がわかる。

\(r=\sqrt{x^2+y^2}\)、\(\theta=\tan^{-1}\displaystyle\frac{y}{x}\)

勾配

ここから、勾配の成分を一つ一つ計算していく。

\(\nabla=\displaystyle\frac{\partial}{\partial x}\boldsymbol{e_{x}}+\displaystyle\frac{\partial}{\partial y}\boldsymbol{e_{y}}+\displaystyle\frac{\partial}{\partial z}\boldsymbol{e_{z}}\)　のそれぞれの成分を計算していく。

x成分

\(\displaystyle\frac{\partial}{\partial x}=\displaystyle\frac{\partial r}{\partial x}\displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{\partial \theta}{\partial x}\displaystyle\frac{\partial}{\partial \theta}+\displaystyle\frac{\partial z}{\partial x}\displaystyle\frac{\partial}{\partial z}\)を求めたい。

ここで、以下の3式が成立する。

\(\displaystyle\frac{\partial r}{\partial x}=\displaystyle\frac{\partial}{\partial x}\sqrt{x^2+y^2}=\displaystyle\frac{x}{\sqrt{x^2+y^2}}=\displaystyle\frac{x}{r}=\cos\theta\)

\(\displaystyle\frac{\partial \theta}{\partial x}=\displaystyle\frac{\partial}{\partial x}\tan^{-1}\displaystyle\frac{y}{x}=-\displaystyle\frac{y}{x}\cdot\displaystyle\frac{1}{1+(\frac{y}{x})^2}=-\displaystyle\frac{y}{x^2+y^2}=-\displaystyle\frac{\sin\theta}{r}\)

\(\displaystyle\frac{\partial z}{\partial x}=0\)

これらをもとの式に代入すると

\[\displaystyle\frac{\partial}{\partial x}=\cos\theta \displaystyle\frac{\partial}{\partial r}-\displaystyle\frac{\sin\theta}{r}\displaystyle\frac{\partial}{\partial \theta}\]

y成分

\(\displaystyle\frac{\partial}{\partial y}=\displaystyle\frac{\partial r}{\partial y}\displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{\partial \theta}{\partial y}\displaystyle\frac{\partial}{\partial \theta}+\displaystyle\frac{\partial z}{\partial y}\displaystyle\frac{\partial}{\partial z}\)を求める。

ここで、次の3式が成立する。

\(\displaystyle\frac{\partial r}{\partial y}=\displaystyle\frac{\partial}{\partial y}\sqrt{x^2+y^2}=\displaystyle\frac{y}{\sqrt{x^2+y^2}}=\displaystyle\frac{y}{r}=\sin\theta\)

\(\displaystyle\frac{\partial \theta}{\partial y}=\displaystyle\frac{\partial}{\partial y}\tan^{-1}\displaystyle\frac{y}{x}=\displaystyle\frac{1}{x}\cdot\displaystyle\frac{1}{1+(\frac{y}{x})^2}=\displaystyle\frac{x}{x^2+y^2}=\displaystyle\frac{\cos\theta}{r}\)

\(\displaystyle\frac{\partial z}{\partial y}=0\)

これらをもとの式に代入すると
\[\displaystyle\frac{\partial}{\partial y}=\sin\theta \displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{\cos\theta}{r}\displaystyle\frac{\partial}{\partial \theta}\]

z成分

\[\displaystyle\frac{\partial}{\partial z}\]

まとめ

\(\nabla=\displaystyle\frac{\partial}{\partial x}\boldsymbol{e_{x}}+\displaystyle\frac{\partial}{\partial y}\boldsymbol{e_{y}}+\displaystyle\frac{\partial}{\partial z}\boldsymbol{e_{z}}\)

\(=\biggl(\cos\theta\displaystyle\frac{\partial}{\partial r}-\displaystyle\frac{\sin\theta}{r}\displaystyle\frac{\partial}{\partial \theta}\biggr)\boldsymbol{e_{x}}+\biggl(\sin\theta\displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{\cos\theta}{r}\displaystyle\frac{\partial}{\partial \theta}\biggr)\boldsymbol{e_{y}}+\displaystyle\frac{\partial}{\partial z}\boldsymbol{e_{z}}\)

\(=\biggl(\cos\theta\displaystyle\frac{\partial}{\partial r}-\displaystyle\frac{\sin\theta}{r}\displaystyle\frac{\partial}{\partial \theta}\biggr)(\boldsymbol{e_{r}}\cos\theta-\boldsymbol{e_{\theta}}\sin\theta)\)

\(+\biggl(\sin\theta\displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{\cos\theta}{r}\displaystyle\frac{\partial}{\partial \theta}\biggr)(\boldsymbol{e_{r}}\sin\theta+\boldsymbol{e_{\theta}}\cos\theta)+\displaystyle\frac{\partial}{\partial z}\boldsymbol{e_{z}}\)

\(=\displaystyle\frac{\partial}{\partial r}\boldsymbol{e_{r}}+\displaystyle\frac{1}{r}\displaystyle\frac{\partial}{\partial \theta}\boldsymbol{e_{\theta}}+\displaystyle\frac{\partial}{\partial z}\boldsymbol{e_{z}}\)

※\(\boldsymbol{e_{x}}=\boldsymbol{e_{r}}\cos\theta-\boldsymbol{e_{\theta}}\sin\theta\)、\(\boldsymbol{e_{y}}=\boldsymbol{e_{r}}\sin\theta+\boldsymbol{e_{\theta}}\cos\theta\)　という関係式を使った。

結果

\(\nabla f=\displaystyle\frac{\partial f}{\partial r}\boldsymbol{e_{r}}+\displaystyle\frac{1}{r}\displaystyle\frac{\partial f}{\partial \theta}\boldsymbol{e_{\theta}}+\displaystyle\frac{\partial f}{\partial z}\boldsymbol{e_{z}}\)

発散

\(\boldsymbol{A}=A_{r}\boldsymbol{e_{r}}+A_{\theta}\boldsymbol{e_{\theta}}+A_{z}\boldsymbol{e_{z}}\)

ここで発散とは

\[\nabla\cdot\boldsymbol{A}=\displaystyle\frac{\partial A_{x}}{\partial x}+\displaystyle\frac{\partial A_{y}}{\partial y}+\displaystyle\frac{\partial A_{z}}{\partial z}\]

右辺第一項 

\(\displaystyle\frac{\partial A_{x}}{\partial x}=\displaystyle\frac{\partial r}{\partial x}\displaystyle\frac{\partial A_{x}}{\partial r}+\displaystyle\frac{\partial \theta}{\partial x}\displaystyle\frac{\partial A_{x}}{\partial \theta}+\displaystyle\frac{\partial z}{\partial x}\displaystyle\frac{\partial A_{x}}{\partial z}\)

\(=\cos\theta\displaystyle\frac{\partial A_{x}}{\partial r}-\displaystyle\frac{\sin \theta}{r}\displaystyle\frac{\partial A_{x}}{\partial \theta}\)

\(=\cos\theta\displaystyle\frac{\partial}{\partial r}(A_{r}\cos\theta-A_{\theta}\sin\theta)-\displaystyle\frac{\sin\theta}{r}\displaystyle\frac{\partial}{\partial \theta}(A_{r}\cos\theta-A_{\theta}\sin\theta)\)

\(=\cos\theta\biggl(\displaystyle\frac{\partial A_{r}}{\partial r}\cos\theta-\displaystyle\frac{A_{\theta}}{\partial r}\sin\theta\biggr)\)

\(-\displaystyle\frac{\sin \theta}{r}\biggl(-A_{r}\sin\theta+\displaystyle\frac{\partial A_{r}}{\partial \theta}\cos\theta-A_{\theta}\cos\theta-\displaystyle\frac{A_{\theta}}{\partial \theta}\sin\theta\biggr)\)

変形過程で、勾配計算の時に導いた以下の式を利用した。

\(\displaystyle\frac{\partial r}{\partial x}=\cos\theta\)、\(\displaystyle\frac{\partial \theta}{\partial x}=-\displaystyle\frac{\sin\theta}{r}\)、\(\displaystyle\frac{\partial z}{\partial x}=0\)

また、次の式も。

\(A_{x}=A_{r}\cos\theta-A_{\theta}\sin\theta\)

右辺第二項

\(\displaystyle\frac{\partial A_{y}}{\partial y}=\displaystyle\frac{\partial r}{\partial y}\displaystyle\frac{\partial A_{y}}{\partial r}+\displaystyle\frac{\partial \theta}{\partial y}\displaystyle\frac{\partial A_{y}}{\partial \theta}+\displaystyle\frac{\partial z}{\partial y}\displaystyle\frac{\partial A_{y}}{\partial z}\)

\(=\sin\theta\displaystyle\frac{\partial A_{y}}{\partial r}+\displaystyle\frac{\cos \theta}{r}\displaystyle\frac{\partial A_{y}}{\partial \theta}\)

\(=\sin\theta\displaystyle\frac{\partial}{\partial r}(A_{r}\sin\theta+A_{\theta}\cos\theta)+\displaystyle\frac{\cos \theta}{r}\displaystyle\frac{\partial}{\partial \theta}(A_{r}\sin\theta+A_{\theta}\cos\theta)\)

\(=\sin\theta\biggl(\displaystyle\frac{\partial A_{r}}{\partial r}\sin\theta+\displaystyle\frac{A_{\theta}}{\partial r}\cos\theta\biggr)\)

\(+\displaystyle\frac{\cos \theta}{r}\biggl(A_{r}\cos\theta+\displaystyle\frac{\partial A_{r}}{\partial \theta}\sin\theta-A_{\theta}\sin\theta+\displaystyle\frac{A_{\theta}}{\partial \theta}\cos\theta\biggr)\)

変形過程で、勾配計算の時に導いた以下の式を利用した。

\(\displaystyle\frac{\partial r}{\partial y}=\sin\theta\)、\(\displaystyle\frac{\partial \theta}{\partial y}=\displaystyle\frac{\cos\theta}{r}\)、\(\displaystyle\frac{\partial z}{\partial y}=0\)

また、次の式も。

\(A_{y}=A_{r}\sin\theta+A_{\theta}\cos\theta\)

右辺第三項

\[\displaystyle\frac{\partial A_{z}}{\partial z}\]

まとめ

上の結果を発散の式に代入すると

\[\nabla\cdot\boldsymbol{A}=\displaystyle\frac{\partial A_{r}}{\partial r}+\displaystyle\frac{A_{r}}{r}+\displaystyle\frac{1}{r}\displaystyle\frac{\partial A_{\theta}}{\partial \theta}+\displaystyle\frac{\partial A_{z}}{\partial z}\]

\[=\displaystyle\frac{1}{r}\displaystyle\frac{\partial}{\partial r}(rA_{r})+\displaystyle\frac{1}{r}\displaystyle\frac{\partial A_{\theta}}{\partial \theta}+\displaystyle\frac{\partial A_{z}}{\partial z}\]

ラプラシアン

\(\Delta f=\nabla\cdot(\nabla f)=\nabla\cdot\biggl(\displaystyle\frac{\partial f}{\partial r}\boldsymbol{e_{r}}+\displaystyle\frac{1}{r}\displaystyle\frac{\partial f}{\partial \theta}\boldsymbol{e_{\theta}}+\displaystyle\frac{\partial f}{\partial z}\boldsymbol{e_{z}}\biggr)\)

※勾配の式を利用。ここで

\(A_{r}=\displaystyle\frac{\partial f}{\partial r}\)、\(A_{\theta}=\displaystyle\frac{1}{r}\displaystyle\frac{\partial f}{\partial \theta}\)、\(A_{z}=\displaystyle\frac{\partial f}{\partial z}\)として、上の発散の式を適用すると

\[\Delta f=\displaystyle\frac{1}{r}\displaystyle\frac{\partial}{\partial r}\biggl(r\displaystyle\frac{\partial f}{\partial r}\biggr)+\displaystyle\frac{1}{r^2}\displaystyle\frac{\partial^2 f}{\partial \theta ^2}+\displaystyle\frac{\partial^2 f}{\partial z^2}\]