Monday Math 107

The Laplace Transform
Part 8: Convolution

Let’s consider two functions, f(t) and g(t), defined for t≥0. Next, we denote their Laplace transforms by F(s) and G(s), respectively. What, then, might we say about the product of those transforms, F(s)G(s)?

From the definition of the Laplace transform, we see
$F(s)=\mathcal{L}\left[f(t)\right](s)=\int_0^{\infty}e^{-st}f(t)\,dt$
and
$G(s)=\mathcal{L}\left[g(t)\right](s)=\int_0^{\infty}e^{-st}g(t)\,dt$.
So then the product is
$F(s)G(s)=\left(\int_0^{\infty}e^{-st}f(t)\,dt\right)\left(\int_0^{\infty}e^{-st}g(t)\,dt\right)$.
Renaming the variables of integration from t to u and v, we can combine the two into one double integral:
$\begin{eqnarray}F(s)G(s)&=&\left(\int_0^{\infty}e^{-st}f(u)\,du\right)\left(\int_0^{\infty}e^{-sv}g(v)\,dv\right)\\&=&\int_0^{\infty}\int_0^{\infty}e^{-su}f(u)\cdot{e^{-sv}g(v)}\,dv\,du\\&=&\int_0^{\infty}\int_0^{\infty}e^{-s(u+v)}f(u)g(v)\,dv\,du\end{eqnarray}$.
Now, we perform a change of variable substitution on the inner integral, defining new variable w=v+u, dw=dv; note that the lower limit v=0 becomes w=u, while the infinite limit remains infinite. We then get
$\begin{eqnarray}F(s)G(s)&=&\int_0^{\infty}\int_0^{\infty}e^{-s(u+v)}f(u)g(v)\,dv\,du\\&=&\int_0^{\infty}\int_u^{\infty}e^{-sw}f(u)g(w-u)\,dw\,du\end{eqnarray}$.
Now, reversing the order of integration over the infinite triangular region, we get:
$\begin{eqnarray}F(s)G(s)&=&\int_0^{\infty}\int_u^{\infty}e^{-sw}f(u)g(w-u)\,dw\,du\\&=&\int_0^{\infty}\int_0^we^{-sw}f(u)g(w-u)\,du\,dw\\&=&\int_0^{\infty}e^{-sw}\left(\int_0^wf(u)g(w-u)\,du\right)\,dw\end{eqnarray}$.
Now, to make this more clear, let us rename u with τ, and w with t:
$F(s)G(s)=\int_0^{\infty}e^{-st}\left(\int_0^tf(\tau)g(t-\tau)\,d\tau\right)\,dt$

Now, note that if we denote the term of the inner integral by defining the function $h(t)=\int_0^tf(\tau)g(t-\tau)\,d\tau$, we see the above is
$F(s)G(s)=\int_0^{\infty}e^{-st}h(t)\,dt=\mathcal{L}\left[h(t)\right](s)$,
the Laplace transform of h(t). But what is this function? Note that we said that f(t) and g(t) were defined for t≥0, which means that f(τ)g(τ-t) is clearly defined only over the range of the integral, 0≤τt. However, if we extend the functions into negative t by zero; that is, define f(t)=g(t)=0 for t≤0, then the product is defined for all τ, but is zero for τ outside the range 0≤τt. With this extension in mind, we see that $\int_{-\infty}^{\infty}f(\tau)g(t-\tau)\,d\tau=\int_0^tf(\tau)g(t-\tau)\,d\tau$, and we can see that
$\begin{eqnarray}h(t)&=&\int_0^tf(\tau)g(t-\tau)\,d\tau\\&=&\int_{-\infty}^{\infty}f(\tau)g(t-\tau)\,d\tau\\&=&f*g(t)\end{eqnarray}$,
where f*g denotes the convolution of f and g; thus the Laplace transform of the convolution of two functions is the product of the Laplace transforms of the functions, with the caveat that the functions be zero for negative time. Compare this to the very similar convolution theorem for the Fourier transform.

To demonstrate an example, let us try to find the function f(t) with Laplace transform $F(s)=\frac{\sqrt{s}}{s^2+4}$. Using
$\sqrt{s}=\frac{s}{\sqrt{s}}$, we see
$\begin{eqnarray}F(s)&=&\frac{\sqrt{s}}{s^2+4}\\&=&\frac{1}{\sqrt{s}}\frac{s}{s^2+2^2}\\&=&\frac{1}{\sqrt{\pi}}\sqrt{\frac{\pi}{s}}\frac{s}{s^2+2^2}\end{eqnarray}$.
Now, we recall from here and here that
$\mathcal{L}\left[\frac1{\sqrt{t}}\right](s)=\sqrt{\frac{\pi}{s}}$
and
$\mathcal{L}\left[\cos(\omega{t})\right](s)=\frac{s}{s^2+\omega^2}$, so we see that
$F(s)=\frac{1}{\sqrt{\pi}}\mathcal{L}\left[\frac1{\sqrt{t}}\right](s)\mathcal{L}\left[\cos(2t)\right](s)$,
and so we see that f(t) is $\frac{1}{\sqrt{\pi}}$ times the convolution of $g(t)=\frac1{\sqrt{t}}H(t)$ and $h(t)=\cos(2t)H(t)$, where H(t) is the Heaviside step function, needed to make the functions zero for negative t:
$\begin{eqnarray}f(t)&=&\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{\tau}}H(\tau)\cdot\cos(2(t-\tau))H(t-\tau)\,d\tau\\&=&\frac{1}{\sqrt{\pi}}\int_0^t\frac{\cos(2(t-\tau))}{\sqrt{\tau}}\,d\tau\end{eqnarray}$,
which can be expressed in terms of Fresnel integrals.