Definition 5.8.1.
Let \(f,g\) be integrable functions on \([0,\infty)\text{.}\) Their convoluiton is defined by this integral.
\begin{equation*}
f \star g (t) = \int_0^t f(\tau) g(t-\tau) d \tau\text{.}
\end{equation*}
Section 5.8 Algebraic Properties of the Laplace Transform
The Laplace transform is linear, but it is not multiplicative. How does it interact with product of function. To start, it is easier to think in reverse: if there is a product \(FG\) in the \(s\)-domain, where did it come from in the \(t\)-domain? Here is the relevant integration calculation, using a substitution. (The change in the order of integration near the end is jutified by theorems in real analysis.)
\begin{align*}
FG \amp = \int_0^\infty e^{-st} f(t) df \int_0^\infty e^{-su}
g(t) du\\
\amp = \int_0^\infty \int_0^\infty e^{-s(u+t)} f(t)g(u) dt du\\
\amp = \int_0^\infty f(t) \left[ \int_0^\infty e^{-s(u+t)}
g(u) du \right] du\\
v \amp = u + t\\
\amp = \int_0^\infty f(t) \left[ \int_t^\infty e^{-sv} g(v-t)
dv \right) dt\\
\amp = \int_0^\infty e^{-sv} \left[\int_0^t f(t) g(v-t) dt
\right] dv\\
\amp = \calL \left\{ \int_0^t f(t) g(v-t) dt \right\}
\end{align*}
The product \(FG\) in the \(s\)-domain turns into this strange integral-based combination of the function \(f\) and \(g\text{.}\) This is a new ‘product’; it is called a convolution.
The convolution takes two functions and produces a new function, so it is a product. However, it is a strange product with new properties.
Proposition 5.8.2.
- The convolution is commutative: \(f \star g = g \star f\text{.}\)
- The convolution is associative: \((f \star g) \star h = f \star (g \star h)\text{.}\)
- The convolution is distributive: \(f \star (g \pm h) = f \star g \pm f \star h\text{.}\)
- The convolution respect constants. \(a (f \star g) = (af) \star g = f \star (ag)\)
- The Laplace transform turns convolutions into products:\begin{equation*} \calL \left\{ f \star g \right\} = F(s) G(s) \end{equation*}
Here is an interesting question: if this is a product, what is the identity? That is, what function \(g\) satisfies
\begin{equation*}
f \star g = \int_0^t f(\tau) g(t-\tau) d \tau = f(t)\text{?}
\end{equation*}
In this question, the integral needs to evaluate \(f(\tau)\) at \(\tau = t\text{.}\) I know a ‘function’ that does this: \(\delta_0\text{.}\)
\begin{equation*}
f \star \delta_0 = \int_0^t f(\tau) \delta_0 (t-\tau) d \tau = f(t)
\end{equation*}
Proposition 5.8.3.
The \(\delta\)-function at 0 is the identity for the convolution.
The convolution behaves well with differentiation.
\begin{equation*}
\frac{d}{dt} (f \star g) = \frac{df}{dt} \star g + f \star
\frac{dg}{dt}
\end{equation*}
It also behaves well with integration.
\begin{equation*}
\int_0^\infty f \star g dt = \int_0^\infty f(t) dt \int_0^\infty
g(t) dt
\end{equation*}
Finally, it lets me understand inverse Lapalce transforms of products. I’ll show an example.
Example 5.8.4.
In this example, I’ll look at the inverse transform of a two rational functions that transform back to sine functions.
\begin{align*}
\calL^{-t} \left\{ \frac{k^2}{(s^2+k^2)^2} \right\} \amp =
\calL^{-1} \left\{ \frac{k}{s^2+k^2} \frac{k}{s^2+k^2}
\right\}\\
\amp = \sin kt \star \sin kt
\end{align*}
The inverse transformation goes to the convolution. What does this convolution actually look like? I’ll need a trig identity to approach the integral.
\begin{align*}
\sin kt \star \sin kt \amp = \int_0^t \sin ku \sin (kt-ku) du\\
\sin A \sin B \amp = \frac{1}{2} \cos
(A-B) - \cos (A+B) \\
\sin kt \star \sin kt \amp = \frac{1}{2} \int_0^t \cos (ku
- kt + ku) - \cos (ku + kt - ku) du\\
\amp = \frac{1}{2} \int_0^t \cos (2ku - 2t) - \cos (kt) du\\
\amp = \frac{1}{2} \left. \frac{\sin (2ku - kt)}{2k}
\right|_0^t - \left. \frac{1}{2} u \cos (kt) \right|_0^t\\
\amp = \frac{\sin (2kt-kt) - \sin (-kt)}{4k} - \frac{t\cos
kt}{2}\\
\amp = \frac{\sin (kt)}{2k} - \frac{t\cos kt}{2} = \frac{\sin
kt - kt \cos kt}{2k}
\end{align*}
This is the value of the convolution, hence the inverse transform of the starting rational function.
