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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 2018年8月実施 解析学・微積分

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Author​

Yu, Miyake

Description​

微分方程式​

22 つの関数 x(t),y(t)x(t),y(t) について, 次の連立微分方程式を解け。

{dxdt=x−5ydydt=x−3yx(0)=3,y(0)=1\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+ivf(z) = u + iv を考える。ただし, z=x+iyz = x + iy は複素数, xx と yy は実数, uu と vv は実数値関数, i=−1i = \sqrt{-1} である。 xx と yy が極形式 x=rcos⁡θx = r\cos \theta と y=rsin⁡θy = 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。

∂u∂r=1r∂v∂θ,∂v∂r=−1r∂u∂θ\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​

dxdt=x−5y⇒y=15(x−dxdt)\frac{\text{d}x}{\text{d}t} = x - 5y \Rightarrow y = \frac{1}{5}\big(x - \frac{\text{d}x}{\text{d}t}\big)
dydt=x−3y に代入して,\frac{\text{d}y}{\text{d}t} = x - 3y \text{ に代入して,}
d2xdt2+2dxdt+2x=0\frac{\text{d}^2x}{\text{d}t^2} + 2\frac{\text{d}x}{\text{d}t} + 2x = 0
λ2+2λ+2=0⇒λ=−1±i\lambda^2 + 2\lambda + 2 = 0 \Rightarrow \lambda = -1 \pm i
x=e−t(c1cos⁡t+c2sin⁡t)x = e^{-t}(c_1\cos t + c_2\sin t)
dxdt=−e−t[(c1−c2)cos⁡t+(c1+c2)sin⁡t]\frac{\text{d}x}{\text{d}t} = -e^{-t}[(c_1 -c_2)\cos t + (c_1 + c_2)\sin t]
y=15e−t[(2c1−c2)cos⁡t+(c1+2c2)sin⁡t]y = \frac{1}{5}e^{-t}[(2c_1 - c_2)\cos t + (c_1 + 2c_2)\sin t]
{x(0)=c1=3y(0)=15(2c1−c2)=1⇒{c1=3c2=1\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.
{x=e−t(3cos⁡t+sin⁡t)y=e−t(cos⁡t+sin⁡t)\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=rcos⁡θ,y=rsin⁡θx = r \cos \theta, y = r \sin \theta より、

∂x∂r=cos⁡θ,  ∂x∂θ=−rsin⁡θ,  ∂y∂r=sin⁡θ,  ∂y∂θ=rcos⁡θ \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、

∂u∂r=∂x∂r∂u∂x+∂y∂r∂u∂y=cos⁡θ∂u∂x+sin⁡θ∂u∂y,∂u∂θ=∂x∂θ∂u∂x+∂y∂θ∂u∂y=−rsin⁡θ∂u∂x+rcos⁡θ∂u∂y,∂v∂r=∂x∂r∂u∂x+∂y∂r∂u∂y=cos⁡θ∂v∂x+sin⁡θ∂v∂y,∂v∂θ=∂x∂θ∂u∂x+∂y∂θ∂u∂y=−rsin⁡θ∂v∂x+rcos⁡θ∂v∂y \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

∂u∂x=∂v∂y,  ∂u∂y=−∂v∂x \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、

∂u∂r=cos⁡θ∂u∂x+sin⁡θ∂u∂y=cos⁡θ∂v∂y−sin⁡θ∂v∂x=1r∂v∂θ,∂v∂r=cos⁡θ∂v∂x+sin⁡θ∂v∂y=−cos⁡θ∂u∂y+sin⁡θ∂u∂x=−1r∂u∂θ \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。

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