Lecture Notes in Differential Equations - Bruce E. Shapiro

350 LESSON 32. THE LAPLACE TRANSFORM
Using trig substitution,
sin t ∗ cos t = 1 2 cos t sin2 t + 1 sin t (t − sin t cos t) (32.166)
2
= 1 t sin t (32.167)
2
Theorem 32.13 (Convolution Theorem). Let f(t) and g(t) be piecewise
continuous functions on [0, ∞) of expnonential order with Laplace Transforms
F (s) and G(s). Then
L[f ∗ g] = F (s)G(s) (32.168)
The Laplace transform of the convolution is the product of the transforms.
In its inverse form, the inverse transform of the product is the convolution:
Proof.
(∫ ∞
F (s)G(s) =
=
=
0
∫ ∞ ∫ ∞
0
∫ ∞
0
L −1 [F (s)G(s)] = f ∗ g (32.169)
) (∫ ∞
f(t)e −ts dt
0
)
g(x)e −sx dx
(32.170)
f(t)g(x)e −ts−xs dtdx (32.171)
0
(∫ ∞
)
f(t) g(x)e −s(t+x) dx dt (32.172)
0
In the inner integral let u = t + x. Then
∫ ∞
(∫ ∞
)
F (s)G(s) = f(t) g(u − t)e −su du dt (32.173)
0
0
∫ ∞
t
Interchanging the order of integration:
∫ ∞
(∫ t
)
F (s)G(s) = e −su g(u − t)f(t)dt du (32.174)
as expected.
=
0
0
e −su (f ∗ g)(u)du (32.175)
= L[f ∗ g] (32.176)
Example 32.16. Find a function f(t) whose Laplace Transform is
F (s) =
1
s(s + 1)
(32.177)