8.7: Constant Coefficient Equations with Impulses
So far in this chapter, we’ve considered initial value problems for the constant coefficient equation
ay
′′
′
+ b y + cy = f (t),
where f is continuous or piecewise continuous on [0, ∞). In this section we consider initial value problems where f
represents a force that’s very large for a short time and zero otherwise. We say that such forces are impulsive. Impulsive
forces occur, for example, when two objects collide. Since it isn’t feasible to represent such forces as continuous or
piecewise continuous functions, we must construct a different mathematical model to deal with them.
t0 +h
If f is an integrable function and f (t) = 0 for t outside of the interval [t , t + h] , then ∫
f (t) dt is called the total
impulse of f . We’re interested in the idealized situation where h is so small that the total impulse can be assumed to be
applied instantaneously at t = t . We say in this case that f is an impulse function. In particular, we denote by δ(t − t )
the impulse function with total impulse equal to one, applied at t = t . (The impulse function δ(t) obtained by setting
t = 0 is the Dirac δ function.) It must be understood, however, that δ(t − t ) isn’t a function in the standard sense, since
our “definition” implies that δ(t − t ) = 0 if t ≠ t , while
0
0
t0
0
0
0
0
0
0
0
t0
∫
δ(t − t0 ) dt = 1.
t0
From calculus we know that no function can have these properties; nevertheless, there’s a branch of mathematics known as
the theory of distributions where the definition can be made rigorous. Since the theory of distributions is beyond the scope
of this book, we’ll take an intuitive approach to impulse functions.
Our first task is to define what we mean by the solution of the initial value problem
ay
′′
′
+ b y + cy = δ(t − t0 ),
y(0) = 0,
′
y (0) = 0,
where t is a fixed nonnegative number. The next theorem will motivate our definition.
0
 Theorem 8.7.1
Suppose t
0
≥ 0.
For each positive number h, let y be the solution of the initial value problem
h
ay
′′
h
+ by
′
h
+ c yh = fh (t),
yh (0) = 0,
′
y (0) = 0,
h
(8.7.1)
where
⎧
⎪
⎪
0,
0 ≤ t < t0 ,
fh (t) = ⎨ 1/h,
t0 ≤ t < t0 + h,
(8.7.2)
⎪
⎩
⎪
0,
t ≥ t0 + h,
so f has unit total impulse equal to the area of the shaded rectangle in Figure 8.7.1 . Then
h
lim yh (t) = u(t − t0 )w(t − t0 ),
(8.7.3)
h→0+
where
−1
w =L
1
(
2
as
).
+ bs + c
William F. Trench 8.7.1 2/2/2022
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Figure 8.7.1 : y = f
h (t)
Proof
Taking Laplace transforms in Equation 8.7.1yields
2
(as
+ bs + c)Yh (s) = Fh (s),
so
Fh (s)
Yh (s) =
2
as
.
+ bs + c
The convolution theorem implies that
t
yh (t) = ∫
w(t − τ )fh (τ ) dτ .
0
Therefore, Equation 8.7.2implies that
0,
⎧
⎪
⎪
⎪
yh (t) = ⎨
⎪
⎪
⎩
⎪
h
Since y
h (t)
=0
∫
t0
h
1
0 ≤ t < t0 ,
t
1
w(t − τ )dτ ,
t0 +h
∫
t0
t0 ≤ t ≤ t0 + h,
w(t − τ )dτ ,
(8.7.4)
t > t0 + h.
for all h if 0 ≤ t ≤ t , it follows that
0
lim yh (t) = 0
if
h→0+
0 ≤ t ≤ t0 .
(8.7.5)
We’ll now show that
lim yh (t) = w(t − t0 )
if
t > t0 .
(8.7.6)
h→0+
Suppose t is fixed and t > t . From Equation 8.7.4,
0
t0 +h
1
yh (t) =
∫
h
w(t − τ )dτ
if
h < t − t0 .
(8.7.7)
t0
Since
t0 +h
1
∫
h
dτ = 1,
(8.7.8)
t0
we can write
t0 +h
1
w(t − t0 ) =
h
w(t − t0 ) ∫
t0
t0 +h
1
dτ =
∫
h
William F. Trench 8.7.2 2/2/2022
w(t − t0 ) dτ .
t0
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From this and Equation 8.7.7,
t0 +h
1
yh (t) − w(t − t0 ) =
∫
h
(w(t − τ ) − w(t − t0 )) dτ .
t0
Therefore
t0 +h
1
| yh (t) − w(t − t0 )| ≤
∫
h
|w(t − τ ) − w(t − t0 )| dτ .
(8.7.9)
t0
Now let M be the maximum value of |w(t − τ ) − w(t − t )| as τ varies over the interval
(Remember that t and t are fixed.) Then Equation 8.7.8and Equation 8.7.9imply that
h
0
[ t0 , t0 + h]
.
0
t0 +h
1
| yh (t) − w(t − t0 )| ≤
h
Mh ∫
dτ = Mh .
(8.7.10)
t0
But lim
M = 0 , since w is continuous. Therefore Equation 8.7.10implies Equation 8.7.6. This and Equation
8.7.5 imply Equation 8.7.3.
h→0+
h
Theorem 8.7.1 motivates the next definition.
 Definition 8.7.2
If t
0
>0
, then the solution of the initial value problem
ay
′′
′
+ b y + cy = δ(t − t0 ),
′
y(0) = 0,
y (0) = 0,
(8.7.11)
is defined to be
y = u(t − t0 )w(t − t0 ),
where
1
−1
w =L
(
2
as
).
+ bs + c
In physical applications where the input f and the output y of a device are related by the differential equation
ay
w
′′
′
+ b y + cy = f (t),
is called the impulse response of the device. Note that w is the solution of the initial value problem
aw
′′
′
+ b w + cw = 0,
′
w(0) = 0,
w (0) = 1/a,
(8.7.12)
as can be seen by using the Laplace transform to solve this problem. (Verify.) On the other hand, we can solve Equation
8.7.12 by the methods of Section 5.2 and show that w is defined on (−∞, ∞) by
e
r2 t
−e
r1 t
w =
1
,
w =
te
r1 t
1
,
or
w =
a
a(r2 − r1 )
e
λt
sin ωt,
(8.7.13)
aω
depending upon whether the polynomial p(r) = ar + br + c has distinct real zeros r and r , a repeated zero r , or
complex conjugate zeros λ ± iω . (In most physical applications, the zeros of the characteristic polynomial have negative
real parts, so lim
w(t) = 0 .) This means that y = u(t − t )w(t − t ) is defined on (−∞, ∞) and has the following
properties:
2
1
t→∞
0
y(t) = 0,
ay
′′
′
+ b y + cy = 0
on
2
1
0
t < t0 ,
(−∞, t0 )
and
(t0 , ∞),
and
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y
′
−
(t0 ) = 0,
y
′
(t0 ) = 1/a
+
(8.7.14)
(remember that y (t ) and y (t ) are derivatives from the right and left, respectively) and y (t ) does not exist. Thus,
even though we defined y = u(t − t )w(t − t ) to be the solution of Equation 8.7.11, this function doesn’t satisfy the
differential equation in Equation 8.7.11 at t , since it isn’t differentiable there; in fact Equation 8.7.14 indicates that an
impulse causes a jump discontinuity in velocity. (To see that this is reasonable, think of what happens when you hit a ball
with a bat.) This means that the initial value problem Equation 8.7.11 doesn’t make sense if t = 0 , since y (0) doesn’t
exist in this case. However y = u(t)w(t) can be defined to be the solution of the modified initial value problem
′
−
′
0
+
′
0
0
0
0
0
′
0
ay
′′
′
+ b y + cy = δ(t),
y(0) = 0,
′
y− (0) = 0,
where the condition on the derivative at t = 0 has been replaced by a condition on the derivative from the left.
Figure 8.7.2 illustrates Theorem 8.7.1 for the case where the impulse response w is the first expression in Equation 8.7.13
and r and r are distinct and both negative. The solid curve in the figure is the graph of w. The dashed curves are
solutions of Equation 8.7.1 for various values of h . As h decreases the graph of y moves to the left toward the graph of
w.
1
2
h
Figure 8.7.2 : An illustration of Theorem 8.7.1
 Example 8.7.1 :
Find the solution of the initial value problem
y
where t
0
>0
′′
′
− 2 y + y = δ(t − t0 ),
′
y(0) = 0,
. Then interpret the solution for the case where t
=0
0
y (0) = 0,
(8.7.15)
.
Solution
Here
−1
w =L
1
(
2
s
−1
) =L
1
(
− 2s + 1
(s − 1)2
) = te
−t
,
so Theorem 8.7.2 yields
y = u(t − t0 )(t − t0 )e
−(t−t0 )
as the solution of Equation 8.7.15 if t > 0 . If t = 0 , then Equation 8.7.15 doesn’t have a solution; however,
y = u(t)te
(which we would usually write simply as y = te ) is the solution of the modified initial value problem
0
0
−t
−t
y
The graph of y = u(t − t
0 )(t
− t0 )e
′′
′
− 2 y + y = δ(t),
−(t−t0 )
y(0) = 0,
′
y− (0) = 0.
is shown in Figure 8.7.3
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Figure 8.7.3 : y = u(t − t
0 )(t
−(t− t0 )
− t0 ) e
Definition 8.7.2 and the principle of superposition motivate the next definition.
 Definition 8.7.3
Suppose α is a nonzero constant and
value problem
ay
′′
f
is piecewise continuous on
′
+ b y + cy = f (t) + αδ(t − t0 ),
. If
[0, ∞)
t0 > 0
, then the solution of the initial
′
y(0) = k0 ,
y (0) = k1
is defined to be
^(t) + αu(t − t0 )w(t − t0 ),
y(t) = y
where y^ is the solution of
ay
This definition also applies if t
0
′′
=0
′
+ b y + cy = f (t),
′
y(0) = k0 ,
y (0) = k1 .
, provided that the initial condition y
′
(0) = k1
is replaced by y
′
−
(0) = k1
.
 Example 8.7.2
Solve the initial value problem
y
′′
′
+ 6 y + 5y = 3 e
−2t
+ 2δ(t − 1),
′
y(0) = −3,
y (0) = 2.
(8.7.16)
Solution
We leave it to you to show that the solution of
y
′′
′
+ 6 y + 5y = 3 e
−2t
′
,
y(0) = −3, y (0) = 2
is
^ = −e
y
−2t
1
+
e
−5t
5
−
2
e
−t
.
2
Since
w(t)
−1
=
L
1
=
4
−1
L
(
(
1
2
s +6s+5
1
s+1
−
)
1
s+5
=
)
−1
L
e
=
1
(
(s+1)(s+5)
−t
−e
)
−5t
4
the solution of Equation 8.7.16 is
y = −e
−2t
1
+
e
2
−5t
5
−
e
−t
e
−(t−1)
−e
+ u(t − 1)
2
William F. Trench 8.7.5 2/2/2022
−5(t−1)
(8.7.17)
2
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(Figure 8.7.4 ).
Figure 8.7.4 : Graph of Equation 8.7.17
Definition 8.7.3 can be extended in the obvious way to cover the case where the forcing function contains more than one
impulse.
 Example 8.7.3
Solve the initial value problem
y
′′
+ y = 1 + 2δ(t − π) − 3δ(t − 2π),
′
y(0) = −1, y (0) = 2.
(8.7.18)
Solution
We leave it to you to show that
^ = 1 − 2 cos t + 2 sin t
y
is the solution of
y
′′
+ y = 1,
y(0) = −1,
′
y (0) = 2.
Since
−1
w =L
1
(
2
s
) = sin t,
+1
the solution of Equation 8.7.18 is
y
= 1 − 2 cos t + 2 sin t + 2u(t − π) sin(t − π) − 3u(t − 2π) sin(t − 2π)
= 1 − 2 cos t + 2 sin t − 2u(t − π) sin t − 3u(t − 2π) sin t,
or
⎧ 1 − 2 cos t + 2 sin t,
⎪
⎪
y =⎨
1 − 2 cos t,
0 ≤ t < π,
π ≤ t < 2π,
(8.7.19)
⎪
⎩
⎪
1 − 2 cos t − 3 sin t,
t ≥ 2π
(Figure 8.7.5 ).
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Figure 8.7.5 : Graph of Equation 8.7.19
William F. Trench 8.7.7 2/2/2022
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