MATH 334 CONVOLUTIONS
“Convolution” is an operation involving two functions that turns out to be rather useful in many
applications. We have two
... [Show More] reasons for introducing it here. First of all, convolution will give
us a way to deal with inverse transforms of fairly arbitrary products of functions. Secondly, it
will be a major element in some relatively simple formulas for solving a number of differential
equations.
Let us start with just seeing what “convolution” is. After that, we’ll discuss using it with the
Laplace transform and in solving differential equations.
27.1 Convolution, the Basics
Definition and Notation
Let f (t) and g(t) be two functions. The convolution of f and g , denoted by f ∗ g , is the
function on t ≥ 0 given by
f ∗ g(t) =
Z t
x=0
f (x)g(t − x) dx .
!◮Example 27.1: Let
f (t) = e
3t
and g(t) = e
7t
.
Since we will use f (x) and g(t − x) in computing the convolution, let us note that
f (x) = e
3x
and g(t − x) = e
7(t−x)
.
So,
f ∗ g(t) =
Z t
x=0
f (x)g(t − x) dx
=
Z t
x=0
e
3x
e
7(t−x)
dx
=
Z t
x=0
e
3x
e
7t
e
−7x
dx
539
540 Convolution and Laplace Transforms
= e
7t
Z t
x=0
e
−4x
dx
= e
7t
·
−1
4
e
−4x
t
x=0
=
−1
4
e
7t
e
−4t −
−1
4
e
7t
e
−4·0 =
−1
4
e
3t +
1
4
e
7t
.
Simplifying this slightly, we have
f ∗ g(t) =
1
4
e
7t − e
3t
when f (t) = e
3t
and g(t) = e
7t
.
It is common practice to also denote the convolution f ∗ g(t) by f (t) ∗ g(t) where, here,
f (t) and g(t) denote the formulas for f and g . Thus, instead of writing
f ∗ g(t) =
1
4
e
7t − e
3t
when f (t) = e
3t
and g(t) = e
7t
,
we may just write
e
3t
∗ e
7t =
1
4
e
7t − e
3t
.
This simplifies notation a little, but be careful — t is being used for two different things in this
equation: On the left side, t is used to describe f and g ; on the right side, t is the variable in
the formula for the convolution. By convention, if we assign t a value, say, t = 2 , then we are
setting t = 2 in the final formula for the convolution. That is,
e
3t
∗ e
7t with t = 2
means compute the convolution and replace the t in the resulting formula with 2 , which, by the
above computations, is
1
4
e
7·2 − e
3·2
=
1
4
e
14 − e
6
.
It does NOT mean to compute
e
3·2
∗ e
7·2
,
which would give you a completely different result, namely,
e
6
∗ e
14 =
Z t
x=0
e
6
e
14 dt = e
20t .
!◮Example 27.2: Let us find
1
√
t
∗ t
2 when t = 4 .
Here,
f (t) =
1
√
t
and g(t) = t
2
.
So
f (x) =
1
√
x
and g(t − x) = (t − x)
2
,
and
1
√
t
∗ t
2 = f ∗ g(t) =
Z t
x=0
1
√
x
(t − x)
2
=
Z t
x=0
x
−1/2
t
2 − 2tx + x
2
dx
Convolution, the Basics 541
=
Z t
x=0
t
2
x
−1/2 − 2tx
1/2 + x
3/2
dx
= t
2
2x
1/2 − 2t
2
3
x
3/2 +
2
5
x
5/2
t
x=0
= 2t
2
· t
1/2 −
4
3
t · t
3/2 +
2
5
t
5/2 .
After a little algebra and arithmetic, this reduces to
1
√
t
∗ t
2 =
16
15
t
5/2 . (27.1)
Thus, to compute
1
√
t
∗ t
2 when t = 4 ,
we actually compute
16
15
t
5/2 with t = 4 ,
obtaining
16
15
4
5/2 =
16
15
2
5 =
512
15
.
Basic Identities
Let us quickly note a few easily verified identities that can simplify the computation of some
convolutions.
The first identity is trivial to derive. Let α be a constant, and let f and g be two functions.
Then, of course,
Z t
x=0
[αf (x)]g(t − x) dx =
Z t
x=0
f (x)[αg(t − x)] dx = α
Z t
x=0
f (x)g(t − x) dx ,
which we can rewrite as
[αf ] ∗ g = f ∗ [αg] = α[ f ∗ g] .
In other words, we can “factor out constants”.
A more substantial identity comes from looking at how switching the roles of f and g
changes the convolution. That is, how does the result of computing
g ∗ f (t) =
Z t
x=0
g(x) f (t − x) dx
compare to what we get by computing
f ∗ g(t) =
Z t
x=0
f (x)g(t − x) dx ?
Well, in the last integral, let’s use the substitution y = t − x . Then x = t − y , dx = −dy and
f ∗ g(t) =
Z t
x=0
f (x)g(t − x) dx
=
Z t−t
y=t−0
f (t − y)g(y)(−1) dy
= − Z 0
y=t
g(y) f (t − y) dy =
Z t
y=0
g(y) f (t − y) dy .
542 Convolution and Laplace Transforms
The last integral is exactly the same as the integral for computing g∗ f (t), except for the cosmetic
change of denoting the variable of integration by y instead of x . So that integral is the formula
for formula for g ∗ f (t), and our computations just above reduce to
f ∗ g(t) = g ∗ f (t) . (27.2)
Thus we see that convolution is “commutative”.
!◮Example 27.3: Let’s consider the convolution
t
2
∗
1
√
t
.
Since we just showed that convolution is commutative, we know that
t
2
∗
1
√
t
=
1
√
t
∗ t
2
.
What an incredible stroke of luck! We’ve already computed the convolution on the right in
example 27.2. Checking back to equation (27.1), we find
1
√
t
∗ t
2 =
16
15
t
5/2 .
Hence,
t
2
∗
1
√
t
=
1
√
t
∗ t
2 =
16
15
t
5/2 .
In addition to being commutative, convolution is “distributive” and “associative”. That is,
given three functions f , g and h ,
[ f + g] ∗ h = [ f ∗ h] + [g ∗ h] , (27.3)
f ∗ [g + h] = [ f ∗ g] + [ f ∗ h] (27.4)
and
f ∗ [g ∗ h] = [ f ∗ g] ∗ h . (27.5)
The first and second equations are that “addition distributes over convolution”. They are easily
confirmed using the basic definition of convolution. For the first:
[ f + g] ∗ h(t) =
Z t
x=0
[ f (x) + g(x)]h(t − x) dx
=
Z t
x=0
[ f (x)h(t − x) + g(x)h(t − x)] dx
=
Z t
x=0
f (x)h(t − x) dx +
Z t
x=0
g(x)h(t − x)] dx
= [ f ∗ g] + [g ∗ h] .
The second, equation (27.4) follows in a similar manner or by combining (27.3) with the commutativity of the convolution. The last equation in the list, equation (27.5), states that convolution [Show Less]