円柱座標の勾配、発散、ラプラシアン

$x=r\cos\theta$、$y=r\sin\theta$、$z=z$

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

勾配

$\nabla=\displaystyle\frac{\partial}{\partial x}\boldsymbol{e_{x}}+\displaystyle\frac{\partial}{\partial y}\boldsymbol{e_{y}}+\displaystyle\frac{\partial}{\partial z}\boldsymbol{e_{z}}$　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。

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

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

$\displaystyle\frac{\partial z}{\partial x}=0$

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

$\displaystyle\frac{\partial}{\partial y}=\displaystyle\frac{\partial r}{\partial y}\displaystyle\frac{\partial}{\partial r}+\displaystyle\frac{\partial \theta}{\partial y}\displaystyle\frac{\partial}{\partial \theta}+\displaystyle\frac{\partial z}{\partial y}\displaystyle\frac{\partial}{\partial z}$

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

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

$\displaystyle\frac{\partial z}{\partial y}=0$

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

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

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。

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

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。

$\displaystyle\frac{\partial r}{\partial y}=\sin\theta$、$\displaystyle\frac{\partial \theta}{\partial y}=\displaystyle\frac{\cos\theta}{r}$、$\displaystyle\frac{\partial z}{\partial y}=0$、$A_{y}=A_{r}\sin\theta+A_{\theta}\cos\theta$

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

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

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

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