本稿では、
\begin{eqnarray}
x &=& r \cos \theta \\
y &=& r \sin \theta \\
z &=& z \tag{1}
\end{eqnarray}で表される3次元の円柱座標系$(r, \theta, z)$の偏微分について述べます。
ここでは、
3次元円柱座標系の偏微分 - 数式で独楽する
とは異なるアプローチで見ていきます。
極座標 - 数式で独楽する
円柱座標系による直交座標系の偏微分
式(1)より、容易に
\begin{eqnarray}
&& \frac{\partial x}{\partial r} = \cos \theta, & \quad \frac{\partial x}{\partial \theta} = -r \sin \theta , & \quad \frac{\partial x}{\partial z} = 0 \\
&& \frac{\partial y}{\partial r} = \sin \theta, & \quad \frac{\partial y}{\partial \theta} = r \cos \theta , & \quad \frac{\partial y}{\partial z} = 0 \\
&& \frac{\partial z}{\partial r} = 0, & \quad \frac{\partial z}{\partial \theta} = 0, & \quad \frac{\partial z}{\partial z} = 1
\end{eqnarray}を得ることができます。
これより、任意の関数$u$を偏微分していきます。
\begin{eqnarray}
\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} + \frac{\partial u}{\partial z} \frac{\partial z}{\partial r} = \cos \theta \, \frac{\partial u}{\partial x} + \sin \theta \, \frac{\partial u}{\partial y}\\
\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} + \frac{\partial u}{\partial z} \frac{\partial z}{\partial \theta} = -r\sin \theta \, \frac{\partial u}{\partial x} + r\cos \theta \, \frac{\partial u}{\partial y} \\
\frac{\partial u}{\partial z} &=& \frac{\partial u}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial z} + \frac{\partial u}{\partial z} \frac{\partial z}{\partial z} = \frac{\partial u}{\partial z}
\end{eqnarray}
この関係を、行列を用いて
\begin{equation}
\left( \begin{array}{c} \frac{\partial u}{\partial r} \\ \frac{1}{r} \frac{\partial u}{\partial \theta} \\ \frac{\partial u}{\partial z} \end{array} \right)
= \left( \begin{array}{ccc} \cos \theta & \sin \theta & 0 \\
- \sin \theta & \cos \theta & 0\\
0 & 0 & 1 \end{array} \right)
\left( \begin{array}{c} \frac{\partial u}{\partial x} \\ \frac{\partial u}{\partial y} \\ \frac{\partial u}{\partial z} \end{array} \right) \tag{2}
\end{equation}と表記します。
\begin{equation}
R = \left( \begin{array}{ccc} \cos \theta & \sin \theta & 0 \\
-\sin \theta & \cos \theta & 0 \\
0 & 0 & 1 \end{array} \right)
\end{equation}を用いて表すと、
\begin{equation}
\left( \begin{array}{c} \frac{\partial u}{\partial r} \\ \frac{1}{r} \frac{\partial u}{\partial \theta} \\ \frac{\partial u}{\partial z} \end{array} \right)
= R \left( \begin{array}{c} \frac{\partial u}{\partial x} \\ \frac{\partial u}{\partial y} \\ \frac{\partial u}{\partial z} \end{array} \right)
\end{equation}となります。
直交座標系による極座標系の偏微分
式(2)の左から逆行列を掛けて、
\begin{equation}
\left( \begin{array}{c} \frac{\partial u}{\partial x} \\ \frac{\partial u}{\partial y} \\ \frac{\partial u}{\partial z} \end{array} \right)
= \left( \begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1 \end{array} \right)
\left( \begin{array}{c} \frac{\partial u}{\partial r} \\ \frac{1}{r} \frac{\partial u}{\partial \theta} \\ \frac{\partial u}{\partial z} \end{array} \right) \tag{3}
\end{equation}を得ます。
なお、
\begin{equation}
\left( \begin{array}{c} \frac{\partial u}{\partial x} \\ \frac{\partial u}{\partial y} \\ \frac{\partial u}{\partial z} \end{array} \right)
= R^{-1} \left( \begin{array}{c} \frac{\partial u}{\partial r} \\ \frac{1}{r} \frac{\partial u}{\partial \theta} \\ \frac{\partial u}{\partial z} \end{array} \right)
\end{equation}となります。
式(3)を成分ごとに展開すると、
\begin{eqnarray}
\frac{\partial u}{\partial x} &=& \cos \theta \, \frac{\partial u}{\partial r} - \frac{\sin \theta}{r} \frac{\partial u}{\partial \theta} \\
\frac{\partial u}{\partial y} &=& \sin \theta \, \frac{\partial u}{\partial r} + \frac{\cos \theta}{r} \frac{\partial u}{\partial \theta} \\
\frac{\partial u}{\partial z} &=& \frac{\partial u}{\partial z} \tag{4}
\end{eqnarray}となります。
式(4)で$u=r, \theta, z$とすれば、
\begin{eqnarray}
&& \frac{\partial r}{\partial x} = \cos \theta ,
& \quad \frac{\partial r}{\partial y} = \sin \theta ,
& \quad \frac{\partial r}{\partial z} = 0 \\
&& \frac{\partial \theta}{\partial x} = -\frac{\sin \theta}{r} ,
& \quad \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r} ,
& \quad \frac{\partial \theta}{\partial z} = 0 \\
&& \frac{\partial z}{\partial x} = 0,
& \quad \frac{\partial z}{\partial y} = 0,
& \quad \frac{\partial z}{\partial z} = 1
\end{eqnarray}を得ます。
