ラプラス変換

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ラプラス変換

ラプラス変換による微分方程式の解法

振動

\[ f''(t) + 2 \zeta \omega f'(t) + \omega^2 f(t) = g(t) \]

\[ (s^2F(s) - sf(0) - f'(0)) + 2 \zeta \omega (sF(s)-f(0)) + \omega^2 F(s) = G(s) \]

\[ (s^2 + 2 \zeta \omega s + \omega^2) F(s) = G(s) + sf(0) + f'(0) + 2\zeta\omega f(0) \]

減衰なし \(\zeta=0\)

\[ (s^2 + \omega^2) F(s) = G(s) + sf(0) + f'(0) \]

\[ F(s) = \frac{G(s)}{s^2 + \omega^2} + \frac{sf(0)}{s^2 + \omega^2} + \frac{f'(0)}{s^2 + \omega^2} \]

積の逆ラプラス変換は畳み込み

\[ \L^{-1} \left[\frac{G(s)}{s^2 + \omega^2}\right] = \frac{1}{\omega}\sin(\omega t) * g(t) = \frac{1}{\omega} \int_0^t \sin(\omega (\tau - t)) * g(\tau) d\tau \]

\[ f(t) = \frac{1}{\omega} \int_0^t \sin(\omega (\tau - t))g(\tau) d\tau + f(0) \cos\omega t + f'(0) \frac{1}{\omega} \sin\omega t \]

二階線形微分方程式

\[ f''(t) + P(t) f'(t) + Q(t) f(t) + R(t) = 0 \]

固有関数

\[ \dd[u]{x} = -\lambda u \quad (0 < x < 1 , \lambda \geq 0,) \]

自由端（\(u'(0)=u'(1)=0\)）

\[ (s^2 U(s) - su(0) - u'(0)) + \lambda U(s) = 0 \]

\[ U(s) = \frac{s}{s^2+\lambda} u(0) + \frac{1}{s^2+\lambda} u'(0) \]

\[ u(x) = u(0) \cos(\sqrt{\lambda}t) + \frac{u'(0)}{\sqrt{\lambda}} \sin(\sqrt{\lambda}t) \]

境界条件から

\[ u(x) = u(0) \cos(\sqrt{\lambda}x) \]

\[ u'(1) = - u(0) \sqrt{\lambda} \sin(\sqrt{\lambda}) = 0 \]

\[ u(x) = u(0) \cos(n\pi x) \]

例１

\[ f''(t) + P(t) f'(t) + Q(t) f(t) + R(t) = 0 \]