DIFFERENTIAL EQUATIONS WITH DISCONTINUOUS
FORCING FUNCTIONS
KIAM HEONG KWA
Consider the initial value problem
(1)
y 00 (t) + αy 0 (t) + βy(t) = ua (t), y(0) = y0 , y 0 (0) = y00 ,
where α, β ∈ R and ua is the Heaviside function
(
0 if t < a,
(2)
ua (t) =
1 if t ≥ a
for some a ≥ 0. Note that the non-homogeneous term ua in (1) is
discontinuous at t = a. It is an instance of the class of discontinuous
forcing functions. In general, the discontinuity of ua at t = a is manifested in a corresponding discontinuity of the solution y(t) to (1) at
the same point.
Without the aid of the Laplace transform, this initial value problem
(1) has to be solved in two steps as follows. First of all, we note that
for 0 ≤ t < a, the solution y(t) satisfies the initial value problem
(3)
y 00 (t) + αy 0 (t) + βy(t) = 0, y(0) = y0 , y 0 (0) = y00 .
Let yI (t) be the solution to (3). Next, we note that for t ≥ a, the
solution y(t) to (1) fulfills the other initial value problem, i.e.,
(4)
y 00 (t) + αy 0 (t) + βy(t) = 1, y(a) = yI (a), y 0 (a) = yI0 (a).
Let yII (t) be the solution to (4). Then the solution y(t) to (1) is given
by
(
yI (t) if 0 ≤ t < a,
(5)
y(t) =
yII (t) if t ≥ a.
On the other hand, the application of the Laplace transform to the
initial value problem (1) has the advantage of solving the problem in
Date: February 22, 2011.
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KIAM HEONG KWA
a more direct manner. Let Y (s) = Ly(s). Then taking the Laplace
transform of (1) yields
(s2 + αs + β)Y (s) − (s + α)y0 − y00 =
e−as
,
s
from which it follows that
(s + α)y0 + y00
e−as
(6)
Y (s) = 2
+
.
s + αs + β
s(s2 + αs + β)
The solution y(t) to (1) is then obtained by inverting the transform:
(7)
0
e−as
−1
−1 (s + α)y0 + y0
y(t) = L [Y (s)](t) = L
(t).
+
s2 + αs + β
s(s2 + αs + β)