相似定理

${\displaystyle \mathcal{L}\left\{f\left(at\right)\right\}=\frac{1}{a}F\left(\frac{s}{a}\right)}$

${\displaystyle \mathcal{L}\left\{\frac{1}{a}f\left(\frac{t}{a}\right)\right\}=F\left(as\right)}$

■証明

${\displaystyle \mathcal{L}\left\{f\left(at\right)\right\}={\int }_0^{\infty }{e}^{-st}f\left(at\right)dt}$

ここで，${\displaystyle \tau =at}$  とおくと， ${\displaystyle dt=\frac{1}{a}d\tau }$, ${\displaystyle f\left(at\right)=f\left(\tau \right)}$ となり

${\displaystyle \mathcal{L}\left\{f\left(\tau \right)\right\}={\int }_0^{\infty }{e}^{-\frac{s}{a}\tau }f\left(\tau \right)·\frac{1}{a}d\tau }$

${\displaystyle =\frac{1}{a}{\int }_0^{\infty }{e}^{-\frac{s}{a}\tau }f\left(\tau \right)d\tau }$

${\displaystyle =\frac{1}{a}F\left(\frac{s}{a}\right)}$

${\displaystyle \tau =at}$ の関係から

${\displaystyle \mathcal{L}\left\{f\left(at\right)\right\}=\frac{1}{a}F\left(\frac{s}{a}\right)}$

■証明

${\displaystyle \mathcal{L}\left\{f\left(\frac{t}{a}\right)\right\}={\int }_0^{\infty }{e}^{-st}f\left(\frac{t}{a}\right)dt}$

ここで， ${\displaystyle \tau =\frac{t}{a}}$  とおくと， ${\displaystyle dt=ad\tau }$ , ${\displaystyle f\left(at\right)=f\left(\tau \right)}$ となり

${\displaystyle \mathcal{L}\left\{f\left(\tau \right)\right\}={\int }_0^{\infty }{e}^{-sa\tau }f\left(\tau \right)·ad\tau }$

${\displaystyle =a{\int }_0^{\infty }{e}^{-sa\tau }f\left(\tau \right)d\tau }$

${\displaystyle =aF\left(as\right)}$

${\displaystyle \tau =\frac{t}{a}}$  の関係から

${\displaystyle \mathcal{L}\left\{f\left(\frac{t}{a}\right)\right\}}$${\displaystyle =aF\left(as\right)}$

${\displaystyle \frac{1}{a}\mathcal{L}\left\{f\left(\frac{t}{a}\right)\right\}}$${\displaystyle =F\left(as\right)}$

ラプラス変換の線形性より

${\displaystyle \frac{1}{a}\mathcal{L}\left\{f\left(\frac{t}{a}\right)\right\}=\mathcal{L}\left\{\frac{1}{a}f\left(\frac{t}{a}\right)\right\}=F\left(as\right)}$

　

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　最終更新日： 2023年6月6日