I have recently been learning about Laplace transforms and how to use them for solving ODEs. I understand how to calculate and use both $\mathcal{L}$ and $\mathcal{L}^{-1}$ however I am struggling to understand where the convolution theorem has come from, and what convolution is. My understanding is this:
Say we have two functions $f(t)$ and $g(t)$ (which we will presume are real and continuous) we can say that:
$$\mathcal{L}[f(t)]=F(s),\,\mathcal{L}[g(t)]=G(s)$$
and we wish to find:
$$\mathcal{L}^{-1}[F(s)G(s)]$$
we can use the fact that this is equal to:
$$\mathcal{L}^{-1}[F(s)G(s)]=(f*g)(t)$$
where:
$$f*g=\int_0^tf(\tau)g(t-\tau)d\tau$$
Whilst this defines what convolution actually is, I do not see why this is the case or what the convolution of two functions represents outside of this use.
Best Answer
Let's take the Laplace Transform of the convolution, $(f*g)(t)$. We assume that $f$ and $g$ are causal functions (i.e., ,$f(t)=g(t)=0$ for $t<0$). Proceeding, we find that
$$\begin{align} \mathscr{L}\left(\int_0^t f(\tau)g(t-\tau)\,d\tau\right)&=\int_0^\infty e^{-st}\int_0^t f(\tau)g(t-\tau)\,d\tau\,dt\\\\ &=\int_0^\infty f(\tau) \int_\tau^\infty e^{-st}g(t-\tau)\,dt\,d\tau\\\\ &=\int_0^\infty f(\tau) \int_0^\infty e^{-s(t+\tau)}g(t)\,dt\,d\tau\\\\ &=\left(\int_0^\infty f(\tau) e^{-s\tau}\,d\tau\right)\left(\int_0^\infty g(t) e^{-st}\,dt\right)\\\\ &=\mathscr{L}\{f\}(s) \mathscr{L}\{g\}(s) \\\\ &=F(s)G(s)\tag1 \end{align}$$
whereupon inversion of $(1)$ yields the coveted expression
$$\mathscr{L}^{-1}\{FG\}(t)=(f*g)(t)$$