Laplace transform IV - Convolution redux

Recall that for the Fourier transform we had

where $f\ast g$ was the convolution of $f$ and $g$, defined by

Something very similar holds for the Laplace transform, but we have to make a few small adjustments.

Definition of convolution

First note that, when dealing with the Laplace transform, we are now dealing with functions defined on $[0,\mathrm{\infty })$. In particular, in the integral above the function $g(x)$ would not be defined for $x<0$. Similar, the function $f(t-x)$ would not be defined for $t-x<0$, i.e., for $x>t$. So we need to restrict the bounds on the definite integral in order for both terms in the integrand to even be defined.

Fortunately, it turns out that is the only thing we need to do:

Example

Define functions $f,g$ on $[0,\mathrm{\infty })$ by $f(t)=t$ and $g(t)=e^t$. We then compute

Convolution and the Laplace transform

Just as with the Fourier transform, it turns out that the Laplace transform converts convolution product into usual product:

Read "in reverse", this same property tells us

and this is the situation in which we'll most often find ourselves running into the convolution.

Example

Suppose we wished/needed to compute the inverse Laplace transform of the function $F(s)=\frac{1}{s^2(s-1)}$.

One option is to first use partial fractions to rewrite $F(s)$ as

The above calculation took a few minutes behind the scenes, but the payoff is that we can now instantly conclude

A second (new) option is to view $F(s)$ as a product of two functions with known inverse transforms, namely

Then

Although very different, these two methods (at least in this example) take roughly the same amount of time. In the first method, we needed to compute a partial fraction decomposition, while in the second we needed to compute a convolution integral.

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Laplace transform V - Solving differential equations