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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.

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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

$$\mathscr{L}^{-1}\{FG\}(t)=(f*g)(t)$$

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  • $\begingroup$ Great explanation thank you, I struggled to find this when I searched elsewhere and I did not think to simplify the integral using the fact that $f(t)$ and $g(t)$ are non-zero for $t<0$. However could you explain how you got from $4$ to $5$ as I got: $$\int_0^t{f(\tau)e^{-s\tau}d\tau}\int_0^{\infty}{g(t)e^{-st}dt}$$ $\endgroup$
    – Henry Lee
    Apr 16, 2019 at 23:05
  • $\begingroup$ @HenryLee You're welcome. My pleasure. I've edited the typo. $\endgroup$
    – Mark Viola
    Apr 16, 2019 at 23:21
  • $\begingroup$ I'm not sure I understand the ending still $\endgroup$
    – Henry Lee
    Apr 16, 2019 at 23:23
  • $\begingroup$ How does $$\int_0^t g(t)e^{-st}dt=G(s)$$ and how can a limit contain the same variable as the integral $\endgroup$
    – Henry Lee
    Apr 17, 2019 at 1:51
  • $\begingroup$ @HenryLee That was obviously another typo. I've edited it accordingly. $\endgroup$
    – Mark Viola
    Apr 17, 2019 at 2:36

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