Ordinary Differential Equations/Laplace Transform

Definition

Let $f(t)$ be a function on $[0,\infty )$. The Laplace transform of $f$ is defined by the integral

$F(s)={\mathcal {L}}\{f\}(s)=\int _{0}^{\infty }e^{-st}f(t)dt\,.$

The domain of $F(s)$ is all values of $s$ such that the integral exists.

Properties

Linearity

Let $f$ and $g$ be functions whose Laplace transforms exist for $s>\alpha$ and let $a$ and $b$ be constants. Then, for $s>\alpha$,

${\mathcal {L}}\{af+bg\}=a{\mathcal {L}}\{f\}+b{\mathcal {L}}\{g\}\,,$

which can be proved using the properties of improper integrals.

Shifting in s

If the Laplace transform ${\mathcal {L}}\{f\}(s)=F(s)$ exists for $s>\alpha$, then

${\mathcal {L}}\{e^{at}f(t)\}(s)=F(s-a)\,$

for $s>\alpha +a$.

Proof.

${\begin{aligned}{\mathcal {L}}\{e^{at}f(t)\}(s)&{}=\int _{0}^{\infty }e^{-st}e^{at}f(t)dt\\&{}=\int _{0}^{\infty }e^{-(s-a)t}f(t)dt\\&{}=F(s-a)\,.\end{aligned}}$

Laplace Transform of Higher-Order Derivatives

If $F(s)={\mathcal {L}}\{f(t)\}$, then ${\mathcal {L}}\{f'(t)\}=sF(s)-f(0)$

Proof:

${\mathcal {L}}\{f'(t)\}=\int _{0}^{\infty }f'(t)e^{-st}dt$

$=\lim _{C\to \infty }\int _{0}^{C}f'(t)e^{-st}dt$

$=\lim _{C\to \infty }\left.e^{-st}f(t)\right|_{0}^{C}-\int _{0}^{C}-sf(t)e^{-st}dt$ (integrating by parts)

$=-f(0)+s\lim _{C\to \infty }\int _{0}^{C}f(t)e^{-st}dt$

$=s{\mathcal {L}}\{f(t)\}-f(0)$

$=sF(s)-f(0)\,$

Using the above and the linearity of Laplace Transforms, it is easy to prove that ${\mathcal {L}}\{f''(t)\}=s^{2}F(s)-sf(0)-f'(0)$

Derivatives of the Laplace Transform

If ${\mathcal {L}}\{f(t)\}=F(s)$, then ${\mathcal {L}}\{tf(t)\}=-F'(s)$

Laplace Transform of Few Simple Functions

1. ${\mathcal {L}}\{1\}={1 \over s}$
2. ${\mathcal {L}}\{e^{at}\}={1 \over s-a}$
3. ${\mathcal {L}}\{\cos \omega t\}={s \over s^{2}+\omega ^{2}}$
4. ${\mathcal {L}}\{\sin \omega t\}={\omega \over s^{2}+\omega ^{2}}$
5. ${\mathcal {L}}\{1\}={1 \over s}$
6. ${\mathcal {L}}\{t^{n}\}={n! \over s^{n+1}}$