• What is Laplace transform?
• What's the difference between that and Fourier transform?
• When to use it?

Given a function $f(t)$ on the positive real line $t>0$, we can define the following function of $p$: $F(p) = \int_{t=0}^\infty f(t) e^{-pt} dt.$ Again, we require the function $f(t)$ to have moderate growth at $t \to \infty$ for the integral to be well-defined.

• $f(t) = 1$, $F(p) = 1/p,$ valid for $Re(p) > 0$
• $f(t) = e^{a t}$, $F(p) = 1/(p-a),$valid for $Re(p-a) > 0$.
• $f(t) = \cos(at)$, $F(p) = (1/2)[1/(p-ia) + 1/(p+ia)] = p/(p^2 + a^2).$ valid for $Re(p)>0$ if $a$ is real.

That was about function with (linear) exponential decay or growth at infinty

How about Gaussian?

• If $f(t) = e^{-t^2}$, then

$F(p) = \int_0^\infty e^{-t^2} e^{-pt} dt = \int_0^\infty e^{-(t+p/2)^2 + p^2/4} dt = e^{p^2/4} \int_{p/2}^\infty e^{-t^2} dt.$ OK, that's not nice, you can express the result using Gaussian error function, which is about $\int_0^a e^{-t^2} dt$, but let's not worry about it.

How about rational function?

• If $f(t) = 1/(1+t)$, we know $F(p)$ exists, and it is holomorphic (at least) for $Re(p)>0$.

Suppose we know the Laplace transform of $f(t)$, let's denote $F = LT(f)$ (note we just write the name of the function $f$, not including its input variables t). What can we say about $LT(f')$?

We can do integration by part $LT(f') = \int_0^\infty e^{-pt} \frac{df}{dt} dt = \int_{t=0}^{t=\infty} e^{-pt} df = \int_{t=0}^{t=\infty} d[e^{-pt} f] - d[e^{-pt}] f= e^{-pt} f(t)|_0^\infty + \int_0^\infty p e^{-pt} f(t) dt = -f(0) + pF(p).$

$f(t) = \frac{1}{2\pi i} \int_{c - i\infty}^{c+i \infty} e^{pt} F(p) dp. \quad c \gg 0$ We want to take $c$ large enough so that there is no singularity of $F(p)$ for $Re(p) > c$.

For example, if $F(p) = 1/p$, or $1/(p-a)$, we can get $f(t) = 1$ and $f(t)=e^{at}$ respectively.

If you have a differential equation about $f(t)$ on some domain $t > 0$, and you know the initial conditions, say $f(t=0)$ etc, then you can use it to compute the Laplace transform of $f$. We have

Example $f'(t) + f(t) = 3, \quad f(0) = 1$ We apply Laplace transform to the equation, we get $pF(p) - f(0) + F(p) = 3 / p.$ Then, we get $F(p) (p+1) = (3/p + 1) \Rightarrow F(p) = \frac{3+p} {p (p+1)}$

Then, we may apply the inverse Laplace transformation, to get $f(t) = Res_{p=0} (e^{pt} \frac{3+p} {p (p+1)}) + Res_{p=-1} (e^{pt} \frac{3+p} {p (p+1)}) = \frac{e^{0t} (3+0)}{0+1} + e^{1t} \frac{3-1} {-1} = -2 e^{-t} + 3.$ Double check $f'(t) + f(t) = 2 e^{-t} + (-2 e^{-t} + 3)=3, \quad f(0) = 1.$ yeah.