Section 7.4
Derivatives, Integrals and Products of Transforms
As noted previously, the most difficult part of solving initial value problems using Laplace
transforms is recognizing the inverse transform. In this section, we will develop a few more tools
that can help us recognize inverse transforms more easily.
We will often come across transforms that we recognize as products of known transforms; as a
very simple example, consider the transform
F (s) =
1
,
s3
which we can think of as
1 1
· .
s s2
Of course, we already know the inverse transform of F (s), but if we did not know from our
table that
1
1
L−1 { 3 } = t2 ,
s
2
then we might be tempted to think of F (s) as the product of the transforms 1/s and 1/s2 , whose
inverse transforms we know:
1
1
L−1 { } = 1 and L−1 { 2 } = t.
s
s
While it is tempting to guess that
F (s) =
L{1} · L{t} = L{1 · t},
it is clear that this is not the case:
L{1} · L{t} =
1
,
s3
while
1
.
s2
Clearly the product of a transform is not the transform of the product.
Regardless, it seems as if there should be some tie between the inverse transforms of the factors
of 1/s3 and the inverse transforms of its factors 1/s and 1/s2 –can we use the inverse transforms 1
and t to build the inverse transform 12 t2 of 1/s3 ?
It turns out that we can. Thinking of g(t) = t as g(τ ) = τ , Notice that
∫ t
1
1 2 t
τ · 1 dτ = τ = t2 ,
2
2
0
0
L{1 · t} = L{t} =
whose transform is
So
{
}
1 2
1 2
1
L t = · 3 = 3.
2
2 s
s
{∫ t
}
L
τ · 1 dτ = L{1} · L{t}.
0
In this case at least, the transform of the integral of a product is the product of the transforms!
1
Section 7.4
While the general case is a bit more complicated, the idea that we have seen here leads to a
general theme in Laplace transforms: integrals of certain products of functions have transforms that
are products of the transforms of the original functions. To help us codify this idea, we record the
following definition:
Definition. The convolution f ∗ g of the piecewise continuous functions f and g is the function
∫ t
f (t) ∗ g(t) =
f (τ )g(t − τ ) dτ.
0
While it is not apparent from the definition, the product is commutative; in other words, the
functions f ∗ g and g ∗ f are exactly the same,
f ∗ g = g ∗ f.
Example. Calculate the convolution of f (t) = t and g(t) = et .
Using the definition (and being careful to integrate with respect to τ ), we see that
∫
f (t) ∗ g(t) =
t
f (τ )g(t − τ ) dτ
0
∫
0
We’ll need to use integration by parts here:
u=τ
du = dτ
t
τ et−τ dτ.
=
dv = et−τ dτ
v = −et−τ
2
Section 7.4
so that
∫
f (t) ∗ g(t) =
t
f (τ )g(t − τ ) dτ
0
∫
t
τ et−τ dτ
=
0
= −τ e
t
t−τ ∫
0
0
= −τ e
t
t−τ −e
t
et−τ dτ
+
t
t−τ 0
0
= −tet−t + 0 − et−t + et−0
= −t − 1 + et .
Thus
f (t) ∗ g(t) = et − t − 1.
As indicated above, transforms of convolutions have a special property, as indicated in the
following theorem:
Theorem 1. If f (t) and g(t) are piecewise continuous for t ≥ 0 and f and g are of exponential
order for the same c, then the Laplace transform of the convolution function f (t) ∗ g(t) exists, and
is given by
L{f (t) ∗ g(t)} = L{f (t)}L{g(t)};
equivalently, if F (s) is the transform of f (t) and G(s) is the transform of g(t), then
L−1 {F (s)G(s)} = f (t) ∗ g(t).
In other words, the transform of a convolution of functions is the product of the transforms of
the functions; on the other hand, if we wish to find the inverse transform of a function H(s) and
realize that H(s) is a product of functions F (s) and G(s) whose transforms f (t) and g(t) we know,
H(s) = F (s)G(s),
then the inverse transform h(t) of H(s) is just the convolution of f and g,
h(t) = f (t) ∗ g(t).
3
Section 7.4
Example. Given f (t) = t and g(t) = et , calculate
L{f (t) ∗ g(t)}
in two different ways.
1. Since f (t) has Laplace transform F (s) = 1/s2 and g(t) has transform G(s) = 1/(s − 1),
Theorem 1 tells us immediately that
L{f (t) ∗ g(t)} = F (s)G(s)
1 1
= 2
s s−1
1
= 2
.
s (s − 1)
2. On the other hand, we computed f (t) ∗ g(t) earlier;
f (t) ∗ g(t) = et − t − 1.
Let’s use this formula to calculate the transform of f (t) ∗ g(t):
L{f (t) ∗ g(t)} = L{et − t − 1}
1
1
1
=
− 2−
s−1 s
s
s2 − (s − 1) − s(s − 1)
=
s2 (s − 1)
s2 − s + 1 − s2 + s)
=
s2 (s − 1)
1
,
= 2
s (s − 1)
confirming the answer we arrived at in the previous computation.
Example. Find
−1
L
{
}
3s
.
(s2 + 9)2
While the function
(s2
3s
+ 9)2
is not on our table of transforms, we recognize that it can be factored:
(s2
3s
3
s
= 2
;
2
2
+ 9)
s +9s +9
4
Section 7.4
since we know the inverse transforms of each of the factors, we can use Theorem 1 to answer the
original question.
Setting
s
3
F (s) = 2
and G(s) = 2
,
s +9
s +9
we know that
f (t) = cos 3t and g(t) = sin 3t.
Thus
L
−1
{
3s
2
(s + 9)2
}
∫
t
=
cos 3τ sin(3t − 3τ ) dτ.
0
We’ll need to use a few trig identities to answer the question; focusing on the indefinite integral,
we have
∫
∫
cos 3τ sin(3t − 3τ ) dτ =
cos 3τ (sin 3t cos 3τ − cos 3t sin 3τ ) dτ
∫
cos 3τ sin 3t cos 3τ − cos 3τ cos 3t sin 3τ ) dτ
=
∫
=
cos2 3τ sin 3t − cos 3t cos 3τ sin 3τ dτ
∫
= sin 3t
∫
cos2 3τ dτ − cos 3t
cos 3τ sin 3τ dτ.
Let’s evaluate the integrals separately. The first requires us to use the half angle identity:
∫
∫
sin 3t
2
sin 3t cos 3τ dτ =
1 + cos 6τ dτ
2
(
)
1
sin 3t
τ + sin 6τ .
=
2
6
The second integral can be solved with the u-substitution u = sin 3τ :
∫
∫
cos 3t
cos 3t cos 3τ sin 3τ dτ =
u du
3
cos 3t 2
=
u
6
cos 3t
=
sin2 3τ.
6
5
Section 7.4
∫
Thus
cos 3τ sin(3t − 3τ ) dτ =
(
)
sin 3t
1
cos 3t
τ + sin 6τ −
sin2 3τ.
2
6
6
Evaluating the definite integral, we have
∫
t
cos 3τ sin(3t − 3τ ) dτ
=
0
=
=
=
=
=
=
=
=
=
=
Thus
L
−1
t
(
)
sin 3t
1
cos 3t
2
τ + sin 6τ −
sin 3τ 2
6
6
0
(
)
sin 3t
1
cos 3t
t + sin 6t −
sin2 3t − 0 − 0
2
6
6
(
)
sin 3t
1
cos 3t
t + sin 6t −
sin2 3t
2
6
6
(
)
sin 3t
1
6t + sin 6t − cos 3t sin2 3t
12
6
(
)
sin 3t
1
6t + sin 6t −
cos 3t(1 − cos 6t)
12
12
(
)
1
6t sin 3t + sin 6t sin 3t − cos 3t + cos 6t cos 3t
12
(
)
1
6t sin 3t + sin 6t sin 3t + cos 6t cos 3t − cos 3t
12
(
)
1
6t sin 3t + cos 6t cos 3t + sin 6t sin 3t − cos 3t
12
(
)
1
6t sin 3t + cos(6t − 3t) − cos 3t
12
(
)
1
6t sin 3t + cos 3t − cos 3t
12
1
t sin 3t
2
{
3s
2
(s + 9)2
6
}
1
= t sin 3t.
2