Chapter 6.4: Discontinuous forcing
functions
We now consider non-homogeneous (forced)
equations
ay 00 + by 0 + cy = g
where g is a discontinuous forcing function.
Such equations are everywhere in engineering,
where the forcing function might be a voltage
which is switched on and then turned off.
The only method we know so far which can
solve such an equation is the Laplace transform
method.
1
Table
We often use:
L−1(e−csF (s)) = uc(t)f (t − c).
Also
L[ectf (t)] = F (s − c).
2
An example
For the first example, let’s drive an oscillatory
system with the force


 1, t ≤ 1,
g(t) = 1 − u1(t) =

 0, t > 1
We thus solve:
y 00 + y = g,
y(0) = 0 = y 0(0).
Taking Laplace transforms, we get:
(s2
1
+ 1)Y (s) = (1 − e−s).
s
Then,
1
−s ).
(1
−
e
Y (s) =
s(s2 + 1)
3
An example
We undo the first factor by partial fractions:
1
s(s2 +1)
Bs+C
=A
+
s
s2 +1
⇐⇒ As2 + A + Bs2 + Cs + A = 1
⇐⇒ A = 1, B = −1, C = 0.
Thus,
1
1
s
=
−
.
2
2
s(s + 1)
s s +1
So we need:
1
−1
−s
L {(1 − e )[
s
− 2
]}.
s s +1
Using the translation formula,
y = 1 − cos t − u1(t) + u1(t) cos(t − 1).
4
Checking the solution
This is hard, because we cannot differentiate
u1!! It is not even continuous. It is easy to
check that the initial condition holds.
Let’s check it formally as if u1 had derivatives.
The term cos t is a homogeneous solution so
we can ignore it. The factor cos(t − 1) is also
a homogeneous solution. All told, we get:
y 00+y = 1−u001(t)−u1(t)+u001(t) cos(t−1)+2u01 sin(t−1).
The trick is that u0(t), u00(t) = 0 unless t = 1.
So we can set t = 1 in cos(t − 1) (getting 1)
and in sin(t − 1) (getting 0). We thus have:
y 00 + y = 1 − u001(t) − u1(t) + u001(t) = 1 − u1.
5
Second example
Solve the initial value problem

00
0

 L[y] = 2y + y + 2y = g(t),

 y(0) = 0 = y 0(0),
where


 1, 5 ≤ t ≤ 20,
g(t) = u5(t) = u20(t) =

 0 ≤ t < 5, t ≥ 20.
Solution: The equation transforms to:
(2s2
e−5s e−20s
+ s + 2)Y (s) =
−
.
s
s
6
Solution
Thus,
1
−5s − e−20s].
[e
s(2s2 + s + 2)
Let us write this as
Y (s) =
Y (s) = H(s)[e−5s−e−20s], H(s) =
1
.
2
s(2s + s + 2)
Let h(t) = L−1[H]. Then the solution is:
u5(t)h(t − 5) − u20(t)h(t − 20).
To find h we use partial fractions:
1
a
bs + c
= + 2
.
2
s(2s + s + 2)
s
2s + s + 2
We get:
bs2 + cs + 2as2 + as + 2a = 1
⇐⇒ 2a = 1, 2a + b = 0, a + c = 0.
Thus, a = 1/2, b = −1, c = −1/2.
7
Solution
Thus,
s+1
1
−s − 1/2
1 1
2
H(s) =
+ 2
=
−
2 + 15
2s 2s + s + 2
2s 2 (s + 1
)
4
16
From the table of inverse Laplace transforms,
we get:
√
√
√
−t/4
1 e
15
15
15
[cos(
t) +
sin(
t)].
h(t) = −
2
2
4
15
4
Recall that the solution is:
u5(t)h(t − 5) − u20(t)h(t − 20).
8
Third example
This time we use non-trivial initial conditions:
Solve the initial value problem:
y 00 + 2y 0 + 5y = 10u1(t), y(0) = 3y 0(0) = 0
We have
Ly 00 + 2Ly 0 + 5Ly = 10L[u1(t)]
=⇒ (s2 + 2s + 5)Y (s) − 3s − 6 = 10e−s/s
=⇒ Y (s) =
3s+6
s2+2s+5
+
10e−s
.
s(s2+2s+5)
9
Solution
We use partial fractions on the second term
10
A
Bs + C
= +
s(s2 + 2s + 4)
s
s2 + 2s + 5
or equivalently
A(s2 + 2s + 5) + (Bs + C)s
= 10(A + B)s2 + (2A + C)s + 5A = 10.
The constant term gives A = 2.Thus B =
−2, C = −4 We c complete the square
s2 + 2s + 5 = (s + 1)2 + 4
We then have:
2
−2s − 4
3s + 6
−s
−s
+e
+e
.
Y (s) =
(s + 1)2 + 4
s
(s + 1)2 + 4
10
Solution
Simplifying, we get:
s+1+1
−s 2 − 2e−s (s+1)+1
Y (s) = 3 (s+1)
2 +4 + e
s
(s+1)2 +4
s+1
3
2
−s 2
+
+
e
= 3 (s+1)
2 +4
2 (s+1)2 +4
s
s+1
2
−s
−2e−s (s+1)
−
e
.
2 +4
(s+1)2 +4
Now use the table of inverse Laplace transforms to get
y = 3e−t cos(2t) + 3/2e−t−1 sin(2t) + 2u1
−2u1(t)e−t+1 cos(2t − 2) − u1(t)e−t+1 sin(2t − 2).
11
Fourth example
The forcing function is now a ramp function
(because of its graph):
y 00 + 4y = g(t),
with
y(0) = y 0(0) = 0,


0, 0 ≤ t < 5,









 (t − 5)/5, 5 ≤ t < 10
g(t) =



1,








t ≥ 10.
Thus,
g(t) = [u5(t)(t − 5) − u10(t)(t − 10)]/5.
12
Fourth example
Taking the Laplace transform, we get:
Y (s) = (e−5s − e−10s)H(s)/5,
with
1
1 1
1
H(s) = 2 2
= [ 2− 2
].
s (s + 4)
4 s
s +4
Then
1
1
h(t) = t − sin(2t)
4
8
and
y = u5(t)h(t − 5) − u10(t)h(t − 10).
13