596 CHAPTER 30. LAPLACE TRANSFORM METHODS
Theorem 30.3.6 Suppose F (s) is the Laplace transform of f (t) and G(s) is the Laplacetransform of g(t). Then F (s)G(s) is the Laplace transform of∫ t
0f (u)g(t−u)du =
∫ t
0f (t−u)g(u)du≡ f ∗g(t)
Proof: To be rigorous, you really need to replace improper integrals with integrals overa finite interval and then take a limit, but the idea is essentially as follows:∫
∞
0e−st
∫ t
0f (t−u)g(u)dudt =
∫∞
0
∫∞
ue−st f (t−u)g(u)dtdu
=∫
∞
0
∫∞
ue−s(t−u) f (t−u)e−sug(u)dtdu
=∫
∞
0
∫∞
0e−sr f (r)e−sug(u)drdu
=∫
∞
0e−sug(u)
(∫∞
0e−sr f (r)dr
)du
=∫
∞
0e−sr f (r)dr
∫∞
0e−sug(u)du
= F (s)G(s)
The other formula follows from changing the variable. ■Note that F (s) could be a matrix and G(s) could be a vector. You simply need the
multiplication to make sense.
Example 30.3.7 Find a particular solution to
x′ (t) = Ax(t)+f (t)
where
f (t) =
tt
ln(t2 +1
)
and
A =
−1 0 −6−2 1 −51 0 4
There is no way you will find a decent closed form solution to this in terms of elemen-
tary functions because of the horrible ln(t2 +1
)but this is not really a problem because
you can find a particular solution in terms of a convolution. You just need to find the fun-damental matrix which is not hard. I will use the following to find the fundamental matrix.
syms s t; b=eye(3); c=[-1 0 -6;-2 1 -5;1 0 4];simplify(ilaplace(inv(s*b-c)))This yields for Φ(t) −et (2et −3) 0 −6et (et −1)
−et (t + et −1) et −et (2t +3et −3)et (et −1) 0 et (3et −2)