九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2019年度 解析学・微積分
Author
Yu, Miyake
Description
微分方程式
\(2\) つの関数 \(x(t),y(t)\) について, 次の連立微分方程式を解け。
\[
\left\{
\begin{aligned}
&\frac{\text{d}x}{\text{d}t} = x - 5y\\
&\frac{\text{d}y}{\text{d}t} = x - 3y\\
&x(0) = 3,y(0) = 1
\end{aligned}
\right.
\]
複素関数論
解析関数 \(f(z) = u + iv\) を考える。ただし, \(z = x + iy\) は複素数, \(x\) と \(y\) は実数, \(u\) と \(v\) は実数値関数, \(i = \sqrt{-1}\) である。 \(x\) と \(y\) が極形式 \(x = r\cos \theta\) と \(y = r\sin \theta\) で表されるとき, 極形式のコーシー \(\cdot\) リーマンの方程式は以下の式で書けることを示せ。
\[
\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta} , \frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta}
\]
Kai
微分方程式
\[
\frac{\text{d}x}{\text{d}t} = x - 5y \Rightarrow y = \frac{1}{5}\big(x - \frac{\text{d}x}{\text{d}t}\big)
\]
\[
\frac{\text{d}y}{\text{d}t} = x - 3y \text{ に代入して,}
\]
\[
\frac{\text{d}^2x}{\text{d}t^2} + 2\frac{\text{d}x}{\text{d}t} + 2x = 0
\]
\[
\lambda^2 + 2\lambda + 2 = 0 \Rightarrow \lambda = -1 \pm i
\]
\[
x = e^{-t}(c_1\cos t + c_2\sin t)
\]
\[
\frac{\text{d}x}{\text{d}t} = -e^{-t}[(c_1 -c_2)\cos t + (c_1 + c_2)\sin t]
\]
\[
y = \frac{1}{5}e^{-t}[(2c_1 - c_2)\cos t + (c_1 + 2c_2)\sin t]
\]
\[
\left\{
\begin{aligned}
&x(0) = c_1 = 3 \\
&y(0) = \frac{1}{5}(2c_1 - c_2) = 1
\end{aligned}
\right.
\Rightarrow
\left\{
\begin{aligned}
c_1 = 3\\
c_2 = 1
\end{aligned}
\right.
\]
\[
\left\{
\begin{aligned}
x &= e^{-t}(3\cos t + \sin t) \\
y &= e^{-t}(\cos t + \sin t)
\end{aligned}
\right.
\]
複素関数論
\(x = r \cos \theta, y = r \sin \theta\) より、
\[
\begin{aligned}
\frac{\partial x}{\partial r} = \cos \theta
, \ \
\frac{\partial x}{\partial \theta} = -r \sin \theta
, \ \
\frac{\partial y}{\partial r} = \sin \theta
, \ \
\frac{\partial y}{\partial \theta} = r \cos \theta
\end{aligned}
\]
なので、
\[
\begin{aligned}
\frac{\partial u}{\partial r}
&= \frac{\partial x}{\partial r} \frac{\partial u}{\partial x}
+ \frac{\partial y}{\partial r} \frac{\partial u}{\partial y}
\\
&= \cos \theta \frac{\partial u}{\partial x}
+ \sin \theta \frac{\partial u}{\partial y}
, \\
\frac{\partial u}{\partial \theta}
&= \frac{\partial x}{\partial \theta} \frac{\partial u}{\partial x}
+ \frac{\partial y}{\partial \theta} \frac{\partial u}{\partial y}
\\
&= -r \sin \theta \frac{\partial u}{\partial x}
+ r \cos \theta \frac{\partial u}{\partial y}
, \\
\frac{\partial v}{\partial r}
&= \frac{\partial x}{\partial r} \frac{\partial u}{\partial x}
+ \frac{\partial y}{\partial r} \frac{\partial u}{\partial y}
\\
&= \cos \theta \frac{\partial v}{\partial x}
+ \sin \theta \frac{\partial v}{\partial y}
, \\
\frac{\partial v}{\partial \theta}
&= \frac{\partial x}{\partial \theta} \frac{\partial u}{\partial x}
+ \frac{\partial y}{\partial \theta} \frac{\partial u}{\partial y}
\\
&= -r \sin \theta \frac{\partial v}{\partial x}
+ r \cos \theta \frac{\partial v}{\partial y}
\end{aligned}
\]
である。
さらに、コーシー・リーマンの方程式
\[
\begin{aligned}
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}
, \ \
\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}
\end{aligned}
\]
を使うと、
\[
\begin{aligned}
\frac{\partial u}{\partial r}
&= \cos \theta \frac{\partial u}{\partial x}
+ \sin \theta \frac{\partial u}{\partial y}
\\
&= \cos \theta \frac{\partial v}{\partial y}
- \sin \theta \frac{\partial v}{\partial x}
\\
&= \frac{1}{r} \frac{\partial v}{\partial \theta}
, \\
\frac{\partial v}{\partial r}
&= \cos \theta \frac{\partial v}{\partial x}
+ \sin \theta \frac{\partial v}{\partial y}
\\
&= - \cos \theta \frac{\partial u}{\partial y}
+ \sin \theta \frac{\partial u}{\partial x}
\\
&= - \frac{1}{r} \frac{\partial u}{\partial \theta}
\end{aligned}
\]
を得る。