複素解析をざっとまとめるー15（第３章の章末問題の解答）

目次

(1)

\begin{align}
u_{x} &=3x^{2}+6xy-3y^{2} \nonumber \\
u_{y} &=3x^{2}-6xy+3y^{2} \nonumber
\end{align}
となります. CR関係式より,
\begin{align}
v_{x} &=-u_{y}\nonumber \\
&=-3x^{2}+6xy-3y^{2} \tag{1} \\
v_{y} &=u_{x}\nonumber \\
&= 3x^{2}+6xy-3y^{2}\tag{2} \\
\end{align}
　(1)より
\begin{align}
v &= -x^{3}+3x^{2}y+3xy^{2}+c_{1}(y) \tag{3}
\end{align}
　(2)より
\begin{align}
v &= -y^{3}+3x^{2}y+3xy^{2}+c_{2}(x) \tag{4}
\end{align}
　(3)と(4)を比較しては
\begin{align}
v = -x^{3}-y^{3}+3x^{2}y+3xy^{2}+C\quad(C\mbox{は実定数}) \nonumber
\end{align}
　よっては
\begin{align}
f(z)&=(x-y)(x^{2}+y^{2}+4xy)+i(-x^{3}-y^{3}+3x^{2}y+3xy^{2}+C) \nonumber \\
[&= (1-i)z^{3}+iC] \nonumber
\end{align}

(2)

「にの項が存在しない」ということは
\begin{align}
\frac{\partial f}{\partial \overline{z}} = 0 \nonumber
\end{align}
ということです. に注意すると
\begin{align}
\frac{\partial f}{\partial \overline{z}} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \overline{z}}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \overline{z}}
\end{align}
　ここで,
\begin{align}
x &= \frac{z+\overline{z}}{2} \nonumber \\
y &= \frac{z-\overline{z}}{2i} \nonumber
\end{align}
より,
\begin{align}
\frac{\partial f}{\partial \overline{z}} &=\frac{1}{2}\left( \frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right) \nonumber \\
&= \frac{1}{2}\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}+i\frac{\partial u}{\partial y}-\frac{\partial v}{\partial y} \right)
\end{align}
がCR関係式を満たすならば
\begin{align}
\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}+i\frac{\partial u}{\partial y}-\frac{\partial v}{\partial y} = 0 \nonumber \\
\therefore \quad \frac{\partial f}{\partial \overline{z}} = 0
\end{align}
　よって「複素関数がコーシーリーマン関係式を満たす」ことと「にの項が存在しない」ことは等しいのです.

(3)

　CR関係式より
\begin{align}
\frac{\partial^{2} u}{\partial x^{2}} +\frac{\partial^{2} u}{\partial y^{2}}
&=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x} \right)+\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial y} \right)\nonumber \\
&=\frac{\partial }{\partial x}\left(\frac{\partial v}{\partial y} \right)+\frac{\partial }{\partial y}\left(-\frac{\partial v}{\partial x} \right)\nonumber \\
&= 0
\end{align}
同様にして
\begin{align}
\frac{\partial^{2} v}{\partial x^{2}} +\frac{\partial^{2} v}{\partial y^{2}}
&=\frac{\partial }{\partial x}\left(\frac{\partial v}{\partial x} \right)+\frac{\partial }{\partial y}\left(\frac{\partial v}{\partial y} \right)\nonumber \\
&=\frac{\partial }{\partial x}\left(-\frac{\partial u}{\partial y} \right)+\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x} \right)\nonumber \\
&= 0
\end{align}
　上の解答で,
\begin{align}
\frac{\partial^{2} u}{\partial x\partial y} &= \frac{\partial^{2} u}{\partial y\partial x} \nonumber \\
\frac{\partial^{2} v}{\partial x\partial y} &= \frac{\partial^{2} v}{\partial y\partial x} \nonumber
\end{align}
を普通に使っていますが, これが成立する条件をちゃあんと覚えていますか？シュワルツの定理と呼ばれるものです.

シュワルツの定理[シュヴァルツの定理]
　変数関数を異なる2つの変数で2階偏微分した結果について, が存在し, かつが連続ならば, が存在して
\begin{align}
f_{x_{j}x_{k}}=f_{x_{k}x_{j}} \nonumber
\end{align}
が成立する.

　実用的にを考えるときには, 次の系の形を覚えておけば十分です.

　のすべての2階偏導関数が存在してそれらが連続ならば, 2階偏導関数の値は微分の順序に依存しない.

(4)

　直交座標系(デカルト座標系, あるいはカーテシアンともいう)と極座標系の相互関係
\begin{align}
r = \sqrt{x^{2}+y^{2}},\quad \theta=\tan^{-1}\frac{y}{x} \nonumber
\end{align}
より,
\begin{align}
\frac{\partial r}{\partial x}
&= \frac{\partial}{\partial x}\sqrt{x^2 + y^2} \nonumber \\
&= \frac{2x}{2\sqrt{x^2 + y^2}}\nonumber \\
&= \frac{x}{r} \nonumber \\
&= \cos\theta \nonumber
\end{align}
　また, 逆三角関数の微分
\begin{align}
(\tan^{-1}x)' = \frac{1}{1+x^{2}} \nonumber
\end{align}
を用いて,
\begin{align}
\frac{\partial\theta}{\partial x}
&=\frac{\partial}{\partial x}\tan^{-1}\frac{y}{x}\nonumber \\
&= \frac{1}{1+\left(\frac{y}{x}\right)^2}\left(-\frac{y}{x^2}\right)\nonumber \\
&= -\frac{y}{x^2+y^2}\nonumber \\
&=-\frac{r\sin\theta}{r^{2}} \nonumber \\
&= -\frac{\sin\theta}{r}\nonumber
\end{align}
　同様にして,
\begin{align}
\frac{\partial r}{\partial y}
&= \frac{\partial}{\partial y}\sqrt{x^2 + y^2} \nonumber \\
&= \frac{2y}{2\sqrt{x^2 + y^2}}\nonumber \\
&= \frac{y}{r} \nonumber \\
&= \sin\theta \nonumber
\end{align}
\begin{align}
\frac{\partial\theta}{\partial y}
&=\frac{\partial}{\partial y}\tan^{-1}\frac{y}{x}\nonumber \\
&= \frac{1}{1+\left(\frac{y}{x}\right)^2}\left(\frac{1}{x}\right)\nonumber \\
&= \frac{x}{x^2+y^2}\nonumber \\
&=\frac{r\cos\theta}{r^{2}} \nonumber \\
&= \frac{\cos\theta}{r}\nonumber
\end{align}
　これらを,
\begin{align}
\frac{\partial u}{\partial x} &= \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x} \nonumber \\
\frac{\partial u}{\partial y} &= \frac{\partial u}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial y} \nonumber
\end{align}
に代入すると,
\begin{align}
\frac{\partial u}{\partial x} &=\frac{\partial u}{\partial r}\cos\theta-\frac{\partial u}{\partial \theta}\frac{\sin\theta}{r} \nonumber \\
\frac{\partial u}{\partial y} &=\frac{\partial u}{\partial r}\sin\theta+\frac{\partial u}{\partial \theta}\frac{\cos\theta}{r} \nonumber
\end{align}
となります. は上式のを単にに置き換えただけの式なります. よってCR関係式は
\begin{align}
\begin{cases}
\frac{\partial u}{\partial r}\cos\theta-\frac{\partial u}{\partial \theta}\frac{\sin\theta}{r}&=\frac{\partial v}{\partial r}\sin\theta+\frac{\partial v}{\partial \theta}\frac{\cos\theta}{r} \nonumber \\
\frac{\partial u}{\partial r}\sin\theta+\frac{\partial u}{\partial \theta}\frac{\cos\theta}{r}&=- (\frac{\partial v}{\partial r}\cos\theta-\frac{\partial v}{\partial \theta}\frac{\sin\theta}{r} \nonumber \ ]
\end{cases}
\end{align}
　式(1),(2)のそれぞれの両辺にをかけて整理すると
\begin{align}
\begin{cases}
(r\frac{\partial u}{\partial r}-\frac{\partial v}{\partial \theta})\cos\theta- (\frac{\partial u}{\partial \theta}+r\frac{\partial v}{\partial r})\sin\theta &= 0 \nonumber \\
(r\frac{\partial u}{\partial r}-\frac{\partial v}{\partial \theta})\sin\theta+ (\frac{\partial u}{\partial \theta}+r\frac{\partial v}{\partial r})\cos\theta &= 0 \nonumber
\end{cases}
\end{align}
　連立方程式を行列で表すと
\begin{align}
\begin{pmatrix}
r\frac{\partial u}{\partial r}-\frac{\partial v}{\partial \theta} \\
\frac{\partial u}{\partial \theta}+r\frac{\partial v}{\partial r}
\end{pmatrix}
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0
\end{pmatrix}
\nonumber
\end{align}
となります. 行列
\begin{align}
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
\nonumber
\end{align}
は正則行列なので, 両辺にその逆行列を右側からかけることで,
\begin{align}
\begin{pmatrix}
r\frac{\partial u}{\partial r}-\frac{\partial v}{\partial \theta} \\
\frac{\partial u}{\partial \theta}+r\frac{\partial v}{\partial r}
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0
\end{pmatrix}
\nonumber
\end{align}
　よってCR関係式は極座標系では
\begin{align}
r\frac{\partial u}{\partial r} &= \frac{\partial v}{\partial \theta} \nonumber \\
\frac{\partial u}{\partial \theta} &= -r\frac{\partial v}{\partial r} \nonumber
\end{align}
となります.