2702 Definition and Properties

Definition

The convolution of two functions $f$ and $g$ is a third function which we denote $f*g$. It is defined as the following integral $$(f*g)(t)=\int_{0^-}^{t^+}f(\tau)g(t-\tau)d\tau\tag{1}$$ We will leave this unmotivated until the next note, and for now just learn how to work with it.
There are a few things to point out about the formula. * The variable of integration is $\tau$. We can't use $t$ because that is already used in the limits and in the integrand. We can choose any symbol we want for the variable of integration – it is just a dummy variable. * The limits of integration are $0^-$ and $t^+$. This is important, particularly when we work with delta functions. If $f$ and $g$ are continuous or have at worst jump discontinuities then we can use 0 and $t$ for the limits. You will often see convolution written like this: $$(f*g)(t)=\int_{0}^{t}f(\tau)g(t-\tau)d\tau$$ * We are considering one-sided convolution. There is also a two-sided convolution where the limits of integration are $\pm \infty$. * (Important) One-sided convolution is only concerned with functions on the interval $(0^-, \infty)$. When using convolution we never look at t < 0.

Examples

Example 1 below calculates two useful convolutions from the definition $(1)$. As you can see, the form of $f*g$ is not very predictable from the form of $f$ and $g$.
Example 1. Show that $$e^{at}*e^{bt}=\frac{e^{at}-e^{bt}}{a-b}, a\neq b$$ $$e^{at}*e^{at}=te^{at}$$ Solution. We show the first; the second calculation is similar.
If $a = b$, $$\begin{aligned} e^{at}*e^{bt}&=\int_0^t e^{a\tau}e^{b(t-\tau)}d\tau\\ &=e^{bt}\int_0^t e^{(a-b)\tau}d\tau\\ &=e^{bt}\frac{e^{(a-b)\tau}}{a-b}\bigg|_0^t\\ &=e^{bt}\frac{e^{(a-b)t}-1}{a-b}\\ &=\frac{e^{at}-e^{bt}}{a-b} \end{aligned}$$ If $a=b$, $$\begin{aligned} e^{at}*e^{at}&=\int_0^t e^{a\tau}e^{a(t-\tau)}d\tau\\ &=\int_0^t e^{at}d\tau\\ &=e^{at}\int_0^t 1d\tau\\ &=te^{at} \end{aligned}$$ Note that because the functions are continuous we could safely integrate just from 0 to $t$ instead of having to specify precisely $0^-$ to $t^+$.

The convolution gives us a formula for a particular solution $y_p$ to an inhomogeneous linear ODE. The next example illustrates this for a first order equation.
Example 2. Express as a convolution the solution to the first order constantcoefficient linear IVP. $$y'+ky=q(t);y(0)=0\tag{2}$$ Solution. The integrating factor is $e^{kt}$; multiplying both sides by it gives $$(ye^{kt})'=q(t)e^{kt}$$ Integrate both sides from 0 to $t$, and apply the Fundamental Theorem of Calculus to the left side; since we have $y(0) = 0$, the solution we seek satisfies $$ye^{kt}=\int_0^t q(\tau)e^{k\tau}d\tau$$ Moving the $e^{kt}$ to the right side and placing it under the integral sign gives $$y_p=\int_0^t q(\tau)e^{-k(t-\tau)}d\tau$$ $$y_p=q(t)*e^{-kt}$$ Now we observe that the solution is the convolution of the input $q(t)$ with $e^{-kt}$, which is the solution to the corresponding homogeneous DE $y'+ky=0$, but with IC $y(0) = 1$. This is the simplest case of Green's formula, which is the analogous result for higher order linear ODE's, as we will see shortly.

Properties

1. Linearity: Convolution is linear. That is, for functions $f_1$, $f_2$, $g$ and constants $c_1$, $c_2$ we have $$(c_1f_1+c_2f_2)*g=c_1(f_1*g)+c_2(f_2*g)$$ This follows from the exact same property for integration. This might also be called the distributive law.

2. Commutivity: $f*g = g*f$.
   Proof: This follows from the change of variable $v=t-\tau$.
   Limits: $\tau=0^- \rArr t-\tau=t^+$, and $\tau=t^+ \rArr t-\tau=0^-$
   $dv = -d\tau$
   Integral: $$\begin{aligned} (f*g)(t)&=\int_{0^-}^{t^+} f(\tau)g(t-\tau)d\tau\\ &=\int_{t^+}^{0^-} f(t-v)g(v) (-dv)\\ &=\int_{0^-}^{t^+} g(v)f(t-v)dv\\ &=(g*f)(t) \end{aligned}$$

3. Associativity: $f * (g * h) = (f * g) * h$. The proof just amounts to changing the order of integration in a double integral. $$\begin{aligned} (f*(g*h))(t)&=\int_0^t f(u)(g*h)(t-u)du\\ &=\int_{u=0}^tf(u)(\int_{s=0}^{t-u}g(s)h(t-u-s)ds)du\\ &=\int_{u=0}^t\int_{s=0}^{t-u}f(u)g(s)h(t-s-u)dsdu\\ &=\int_{u=0}^t\int_{s=u}^{t}f(u)g(s-u)h(t-s)dsdu\\ &=\int_{s=0}^t\int_{u=0}^sf(u)g(s-u)h(t-s)duds\\ &=\int_{s=0}^t(\int_{u=0}^sf(u)g(s-u)du)h(t-s)ds\\ &=\int_0^t(f*g)(s)h(t-s)ds\\ &=((f*g)*h)(t) \end{aligned}$$

Delta Functions

We have $$(\delta*f)(t)=f(t)\text{ and }(\delta(t-a)*f)(t)=f(t-a)\tag{3}$$ The notation for the second equation is ugly, but its meaning is clear.

We prove these formulas by direct computation. First, remember the rules of integration with delta functions: for $b>0$ $$\int_{0^-}^b \delta(\tau)f(\tau)d\tau=f(0)$$ The formulas follow easily for $t\geq 0$ $$\begin{aligned} (\delta*f)(t)&=\int_{0^-}^{t^+}\delta(\tau)f(t-\tau)d\tau\\ &=\int_{0^-}^{t^+}\delta(\tau)f(t-0)d\tau\\ &=f(t-0)\int_{0^-}^{t^+}\delta(\tau)d\tau\\ &=f(t) \end{aligned}$$ $$\begin{aligned} (\delta(t-a)*f)(t)&=\int_{0^-}^{t^+} \delta(\tau-a)f(t-\tau)d\tau\\ &=\int_{0^-}^{t^+} \delta(\tau-a)f(t-a)d\tau\\ &=f(t-a)\int_{0^-}^{t^+} \delta(\tau-a)d\tau\\ &=f(t-a) \end{aligned}$$

Convolution is a Type of Multiplication

You should think of convolution as a type of multiplication of functions. In fact, it is often referred to as the convolution product. In fact, it has the properties we associate with multiplication: * It is commutative. * It is associative. * It is distributive over addition. * It has a multiplicative identity. For ordinary multiplication, 1 is the multiplicative identity. Formula $(3)$ shows that $\delta(t)$ is the multiplicative identity for the convolution product.