Ordinary Differential Equations/Laplace Transform

Definition

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

$F(s)=ℒ\{f\}(s)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}f(t)dt.$

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 $s>\alpha$,

$ℒ\{af+bg\}=aℒ\{f\}+bℒ\{g\},$

which can be proved using the properties of improper integrals.

Shifting in s

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

$ℒ\{e^{at}f(t)\}(s)=F(s-a)$

for $s>\alpha +a$.

Proof.

$\begin{array}{cc}ℒ\{e^{at}f(t)\}(s)& =\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{e}^{at}f(t)dt\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-(s-a)t}f(t)dt\\ & =F(s-a).\end{array}$

Laplace Transform of Higher-Order Derivatives

If $F(s)=ℒ\{f(t)\}$, then $ℒ\{f^{\prime }(t)\}=sF(s)-f(0)$

Proof:

$ℒ\{f^{\prime }(t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{f}^{\prime }(t)e^{-st}dt$

$=\underset{C\to \mathrm{\infty }}{\lim }\underset{0}{\overset{C}{\int }}{f}^{\prime }(t)e^{-st}dt$

$=\underset{C\to \mathrm{\infty }}{\lim }{e^{-st}f(t)|}_0^C-\underset{0}{\overset{C}{\int }}-sf(t)e^{-st}dt$ (integrating by parts)

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

$=sℒ\{f(t)\}-f(0)$

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

Using the above and the linearity of Laplace Transforms, it is easy to prove that $ℒ\{f^{″}(t)\}=s^{2}F(s)-sf(0)-f^{\prime }(0)$

Derivatives of the Laplace Transform

If $ℒ\{f(t)\}=F(s)$, then $ℒ\{tf(t)\}=-F^{\prime }(s)$

Laplace Transform of Few Simple Functions

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