Page 118 - Advanced engineering mathematics

98 CHAPTER 3 The Laplace Transform

Convolution is commutative:
f ∗ g = g ∗ f.

This can be proved by a straightforward change of variables in the integral defining the convol-
ution.
In addition to its use in computing the inverse transform of products, convolution allows us
to solve certain general initial value problems.

EXAMPLE 3.15
Solve the initial value problem

y − 2y − 8y = f (t); y(0) = 1, y (0) = 0.
We want a formula for the solution that will hold for any “reasonable” forcing function f . Apply
the Laplace transform to the differential equation in the usual way, obtaining
2
s Y(s) − s − 2(sY(s) − 1) − 8Y(s) = F(s).
Then

2
(s − 2s − 8)Y(s) = s − 2 + F(s).
Then
s − 2 1
Y(s) = + F(s).
2
s − 2s − 8 s − 2s − 8
2
2
Factor s − 2s − 8 = (s − 4)(s + 2), and use a partial fractions decomposition to write
1 1 2 1 1 1 1 1
Y(s) = + + F(s) − F(s).
3 s − 4 3 s + 2 6 s − 4 6 s + 2
Now apply the inverse transform to obtain the solution
1 2 1 1
4t
4t
y(t) = e + e −2t + e ∗ f (t) − e −2t ∗ f (t),
3 3 6 6
which is valid for any function f for which these convolutions are defined.
Convolution also enables us to solve some kinds of integral equations, which are equations
in which the unknown function appears in an integral.

EXAMPLE 3.16
Solve for f (t) in the integral equation
t

2 −τ
f (t) = 2t + f (t − τ)e dτ.
0
Recognize the integral on the right as the convolution of f (t) with e . Therefore, the integral
−t
equation has the form
2
−t
f (t) = 2t + f (t) ∗ e .
Apply the Laplace transform and the convolution theorem to this equation to get
4 1
F(s) = + F(s).
s 3 s + 1

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October 14, 2010 14:14 THM/NEIL Page-98 27410_03_ch03_p77-120