Cauchy-Riemann の微分方程式の極座標表示

(1) $z=re^{i\theta}=r(\cos\theta+i\sin\theta)$としたとき, Cauchy-Riemann の微分方程式は, 実部を$u(r,\ \theta)$, 虚部を$v(r, \theta)$ とすると

$\dfrac{\partial u}{\partial r} = \dfrac{1}{r}\dfrac{\partial v}{\partial \theta},\ \dfrac{\partial v}{\partial r} = -\dfrac{1}{r}\dfrac{\partial u}{\partial \theta}\ (r \neq 0)$

となることを示せ.

(2) 正則関数の実部$u(r, \theta)$, 虚部$v(r, \theta)$ は, 次のラプラス方程式

$\dfrac{\partial ^2 u}{\partial r^2} + \dfrac{1}{r}\dfrac{\partial u}{\partial r} + \dfrac{1}{r^2}\dfrac{\partial ^2 u}{\partial \theta^2} = 0,\ \dfrac{\partial ^2 v}{\partial r^2} + \dfrac{1}{r}\dfrac{\partial v}{\partial r} + \dfrac{1}{r^2}\dfrac{\partial ^2 v}{\partial \theta^2} = 0$

を満たすことを示せ.

(1) Cauchy-Riemann の微分方程式は

$\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y},\ \dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x}$

で与えられたことを思い出しておく.

また, 連鎖律から

$\dfrac{\partial u}{\partial x} = \cos \theta \dfrac{\partial u}{\partial r} - \dfrac{\sin \theta}{r} \dfrac{\partial u}{\partial \theta},\ \dfrac{\partial u}{\partial y} = \sin \theta \dfrac{\partial u}{\partial r} + \dfrac{\cos \theta}{r} \dfrac{\partial u}{\partial \theta}$

$\dfrac{\partial v}{\partial x} = \cos \theta \dfrac{\partial v}{\partial r} - \dfrac{\sin \theta}{r} \dfrac{\partial v}{\partial \theta},\ \dfrac{\partial v}{\partial y} = \sin \theta \dfrac{\partial v}{\partial r} + \dfrac{\cos \theta}{r} \dfrac{\partial v}{\partial \theta}$

と変数変換し, 元の微分方程式にそれぞれの右辺を代入すると

$$
\left\{
\begin{array}{l}
\cos \theta \left ( \dfrac{\partial u}{\partial r} - \dfrac{1}{r}\dfrac{\partial v}{\partial \theta} \right ) - \sin \theta \left ( \dfrac{\partial v}{\partial r} + \dfrac{1}{r}\dfrac{\partial u}{\partial \theta} \right ) = 0 \\
\sin \theta \left ( \dfrac{\partial u}{\partial r} - \dfrac{1}{r}\dfrac{\partial v}{\partial \theta} \right ) + \cos \theta \left ( \dfrac{\partial v}{\partial r} + \dfrac{1}{r}\dfrac{\partial u}{\partial \theta} \right ) = 0
\end{array}
\right.
$$

となる. これを行列を用いて書き直すと

$$
\begin{pmatrix}
\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta
\end{pmatrix}
\begin{pmatrix}
\dfrac{\partial u}{\partial r} - \dfrac{1}{r}\dfrac{\partial v}{\partial \theta} \\
\dfrac{\partial v}{\partial r} + \dfrac{1}{r}\dfrac{\partial u}{\partial \theta}
\end{pmatrix}
=
\begin{pmatrix}
0 \\ 0
\end{pmatrix}
$$

となり, 左辺の回転行列は逆行列をもつから,

$$
\begin{pmatrix}
\dfrac{\partial u}{\partial r} - \dfrac{1}{r}\dfrac{\partial v}{\partial \theta} \\
\dfrac{\partial v}{\partial r} + \dfrac{1}{r}\dfrac{\partial u}{\partial \theta}
\end{pmatrix}
=
\begin{pmatrix}
0 \\ 0
\end{pmatrix}
$$

となり,

$\dfrac{\partial u}{\partial r} = \dfrac{1}{r}\dfrac{\partial v}{\partial \theta},\ \dfrac{\partial v}{\partial r} = -\dfrac{1}{r}\dfrac{\partial u}{\partial \theta}$

が示された. $//$

(2) $u$ のラプラス方程式について示す. (1) で得た式をさらに偏微分することで

$$
\dfrac{\partial ^2 u}{\partial r^2} = -\dfrac{1}{r^2}\dfrac{\partial v}{\partial \theta} + \dfrac{1}{r}\dfrac{\partial ^2 v}{\partial \theta \partial r},\ \dfrac{\partial ^2 u}{\partial \theta^2} = -r \dfrac{\partial ^2 v}{\partial r \partial \theta}
$$

を得る.

与えられた式にこれらを代入すると

$$
\dfrac{\partial ^2 u}{\partial r^2} + \dfrac{1}{r}\dfrac{\partial u}{\partial r} + \dfrac{1}{r^2}\dfrac{\partial ^2 u}{\partial \theta^2} = -\dfrac{1}{r^2} \dfrac{\partial v}{\partial \theta} + \dfrac{1}{r} \dfrac{\partial ^2 v}{\partial \theta \partial r} + \dfrac{1}{r^2} \dfrac{\partial v}{\partial \theta} - \dfrac{1}{r}\dfrac{\partial ^2 v}{\partial r \partial \theta} = 0
$$

を得る. $v$ についてのラプラス方程式も同様にして導かれる. $//$