22 Jun 2025
Laplace Transforms and Applications
Elementary and inverse Laplace transforms, scaling and shifting, delta and periodic functions, derivatives, integrals, convolution, and differential equations.
For $t\ge0$, the Laplace transform is
\[\boxed{F(s)=\mathcal L\{f(t)\} =\int_0^\infty e^{-st}f(t)\,dt,}\]where $\operatorname{Re}s$ is large enough for the integral to converge.
Elementary functions
For $\operatorname{Re}s>0$,
\[\mathcal L\{1\}=\int_0^\infty e^{-st}dt=\frac1s.\]With $u=st$,
\[\mathcal L\{t^n\} =\frac1{s^{n+1}}\int_0^\infty u^ne^{-u}du =\boxed{\frac{n!}{s^{n+1}}}.\]Direct integration gives
\[\mathcal L\{e^{at}\}=\frac1{s-a}, \qquad \operatorname{Re}s>\operatorname{Re}a.\]Using $e^{i\omega t}=\cos\omega t+i\sin\omega t$,
\[\frac1{s-i\omega}=\frac{s+i\omega}{s^2+\omega^2},\]so comparison of real and imaginary parts gives
\[\boxed{ \mathcal L\{\cos\omega t\}=\frac{s}{s^2+\omega^2}, \qquad \mathcal L\{\sin\omega t\}=\frac{\omega}{s^2+\omega^2}. }\]Change of scale and shifting
For $a>0$, put $u=at$:
\[\boxed{ \mathcal L\{f(at)\}(s) =\frac1aF\!\left(\frac sa\right). }\]Multiplication by an exponential shifts $s$:
\[\boxed{\mathcal L\{e^{at}f(t)\}=F(s-a).}\]For a delay $a>0$, the Heaviside factor is essential. Substituting $u=t-a$ gives
\[\boxed{ \mathcal L\{H(t-a)f(t-a)\}=e^{-as}F(s). }\]Dirac delta and periodic functions
The sampling property gives, for $a\ge0$,
\[\boxed{\mathcal L\{\delta(t-a)\}=e^{-as}.}\]If $f(t+T)=f(t)$, split the transform into periods:
\[\begin{aligned} F(s)&=\sum_{n=0}^\infty \int_{nT}^{(n+1)T}e^{-st}f(t)dt\\ &=\sum_{n=0}^\infty e^{-nsT} \int_0^T e^{-su}f(u)du. \end{aligned}\]The geometric sum yields
\[\boxed{ F(s)=\frac{\int_0^T e^{-st}f(t)dt}{1-e^{-sT}}. }\]Derivatives and integrals
Integration by parts gives
\[\begin{aligned} \mathcal L\{f'(t)\} &=\left[e^{-st}f(t)\right]_0^\infty +s\int_0^\infty e^{-st}f(t)dt\\ &=sF(s)-f(0^+), \end{aligned}\]assuming $e^{-st}f(t)\to0$ at infinity. Repetition gives
\[\boxed{ \mathcal L\{f^{(n)}\} =s^nF-s^{n-1}f(0^+)-s^{n-2}f'(0^+)-\cdots-f^{(n-1)}(0^+). }\]Let $g(t)=\int_0^t f(u)du$. Then $g’=f$ and $g(0)=0$, hence
\[\boxed{\mathcal L\left\{\int_0^t f(u)du\right\}=\frac{F(s)}s.}\]Convolution theorem
For the causal convolution
\[(f*g)(t)=\int_0^t f(u)g(t-u)du,\]change the order of integration over $0\le u\le t<\infty$ and put $v=t-u$:
\[\begin{aligned} \mathcal L\{f*g\} &=\int_0^\infty du\,f(u)e^{-su} \int_0^\infty dv\,g(v)e^{-sv}\\ &=\boxed{F(s)G(s)}. \end{aligned}\]Inverse transform and an application
The inverse transform may be found by decomposing $F(s)$ into known transform pairs. For example,
\[F(s)=\frac{2s+5}{(s+1)(s+2)} =\frac3{s+1}-\frac1{s+2},\]so
\[\mathcal L^{-1}\{F(s)\}=3e^{-t}-e^{-2t}.\]Now consider the initial-value equation
\[y''+\omega_0^2y=\frac{F_0}{m}H(t), \qquad y(0)=0,\quad y'(0)=0.\]Transforming each derivative,
\[[s^2Y(s)-sy(0)-y'(0)]+\omega_0^2Y(s) =\frac{F_0}{ms},\]and therefore
\[Y(s)=\frac{F_0}{m}\frac1{s(s^2+\omega_0^2)} =\frac{F_0}{m\omega_0^2} \left(\frac1s-\frac{s}{s^2+\omega_0^2}\right).\]Taking the inverse transform gives
\[\boxed{ y(t)=\frac{F_0}{m\omega_0^2}[1-\cos(\omega_0t)]H(t). }\]The displacement has units $[F_0/(m\omega_0^2)]=\mathrm m$. Direct substitution gives $y’’+\omega_0^2y=F_0/m$ for $t>0$, and the initial conditions are satisfied.
The transform theorems, partial fractions, initial values, and differential- equation residual are checked in the Maxima worksheet; every printed residual is zero.
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