LECTURE 18: THE LAPLACE TRANSFORM AND TRANSFORMS OF DERIVATIVES AND
ODES
MINGFENG ZHAO
October 23, 2015
Example 1. A mass of 4 kg on a spring with k = 4 and a damping constant c = 1. Suppose F0 = 2. Using forcing
function F0 cos(ωt), find the ω that causes practical resonance and find the practical amplitude.
By the assumption, m = 4, c = 1, k = 4, and F0 = 2, we have
4x00 + x0 + 4x = 2 cos(ωt).
The characteristic equation of 4x00 + x0 + 4x = 0 is 4r2 + r + 4 = 0, then
√
−1 ± 1 − 4 · 4 · 4
r1,2 =
8
Let the stead periodic solution be:
xsp (t) = A cos(ωt) + B sin(ωt).
That is, xsp is a particular solution to 4x00 + x0 + 4x = 2 cos(ωt). Then
x0sp (t)
= −Aω sin(ωt) + Bω cos(ωt)
x00sp (t)
= −Aω 2 cos(ωt) − Bω 2 sin(ωt).
Then
4x00sp + x0sp + 4xsp
=
4 −Aω 2 cos(ωt) − Bω 2 sin(ωt) + [−Aω sin(ωt) + Bω cos(ωt)] + 4 [A cos(ωt) + B sin(ωt)]
=
=
2 cos(ωt).
−4Aω 2 + Bω + 4A cos(ωt) + −4Bω 2 − Aω + 4B sin(ωt)
Then
−4Aω 2 + Bω + 4A = 2,
and
− 4Bω 2 − Aω + 4B = 0.
Solve A and B, we get
A=
ω2
8 − 8ω 2
,
+ (4 − ω 2 )2
and B =
1
2ω
.
ω 2 + (4 − ω 2 )2
2
MINGFENG ZHAO
Then
xsp (t) = A cos(ωt) + B sin(ωt) =
8 − 8ω 2
2ω
cos(ωt) + 2
sin(ωt).
2
2
2
ω + (4 − 4ω )
ω + (4 − 4ω 2 )2
So the amplitude is:
C(ω)
=
=
=
=
=
=
=
p
A2 + B 2
s
2 2
2ω
8 − 8ω 2
+
ω 2 + (4 − 4ω 2 )2
ω 2 + (4 − 4ω 2 )2
p
(8 − 8ω 2 )2 + 4ω 2
ω 2 + (4 − 4ω 2 )2
p
2 (4 − 4ω 2 )2 + ω 2
ω 2 + (4 − 4ω 2 )2
2
p
ω 2 + (4 − 4ω 2 )2
2
+ 16 − 32ω 2 + 16ω 4
2
√
.
4
16ω − 31ω 2 + 16
√
ω2
So we get
C 0 (ω) = −
64ω 3 − 62ω
=
2ω(31 − 32ω 2 )
3
3 .
(16ω 4 − 31ω 2 + 16) 2
(16ω 4 − 31ω 2 + 16) 2
r
31
So C(ω) will achieve its maximum at ω =
, that is, the practical resonance frequency is:
32
r
31
ωpr =
≈ 0.984 .
32
The practical resonance amplitude is:
r
max C(ω) = C(ωpr ) = C
ω>0
31
32
!
2
=q
63
64
16
= √ ≈ 2.016 .
63
The Laplace transform
Definition 1. Let f (t) be a function on [0, ∞), then
I. The Laplace transform of f , denoted by L[f ](s), is defined as:
Z ∞
L[f (t)](s) =
f (t)e−st dt, for all s > 0.
0
LECTURE 18: THE LAPLACE TRANSFORM AND TRANSFORMS OF DERIVATIVES AND ODES
3
II. If F (s) = L[f ](s), the inverse Laplace transform of F , denoted by L−1 [F ](t), is defined as:
L−1 [F (s)](t) = f (t),
for all t > 0.
t2
Z
Remark 1. Not for any function f (t) on [0, ∞), the Laplace transform L[f ] exists, e.g., f (t) = e , then
0
does not exist.
Example 2. Find L[1].
In fact, we have
Z
L[1](s) =
∞
e
0
−st
∞
1 −st 1
dt = − e = ,
s
s
0
for all s > 0.
So
L[1] =
Also we know that L−1
1
.
s
1
=1.
s

 1, if t ≥ a,
Example 3 (Heaviside Function). Let a be a real constant, and ua (t) =
, find L[ua ].
 0, if t < a.
If a ≤ 0, then ua (t) = 1 for all t ≥ 0, then
L[ua (t)](s) = L[1](s) =
1
,
s
for all s > 0.
If a > 0, then ua (t) = 0 for all 0 ≤ t ≤ a, which implies that
∞
Z ∞
1
e−as
L[ua (t)](s) =
e−st dt = − e−st =
,
s
s
a
a
for all s > 0.
∞
2
et e−st dt
4
MINGFENG ZHAO
In summary, we have
 1

if a ≤ 0,
 ,
s
L[ua (t)] =
−as

 e
, if a > 0.
s
for all s > 0 .
Proposition 1. Let a, b be two constants, then
I. Linearity:
L[af (t) + bg(t)] = aL[f (t)] + bL[g(t)],
and
L−1 [aF (s) + bG(s)] = aL−1 [F (s)] + bL−1 [G(s)].
II. First Shifting Property:
L e−at f (t) (s) = L[f ](s + a),
and
L−1 [F (s + a)](t) = e−at L−1 [F (s)](t).
III. Transforms of derivatives:
L[f 0 (t)](s)
=
sL[f (t)](s) − f (0)
L[f 00 (t)](s)
=
s2 L[f (t)](s) − sf (0) − f 0 (0).
Proof. I. The linearity of L and L−1 follow directly from the linearity of integration.
II. In fact, we have
L e−at f (t) (s) =
Z
∞
e−at f (t)e−st dt =
0
Z
∞
f (t)e−(s+a)t dt = L[f ](s + a).
0
III. Use the integration by parts, we have
L[f 0 ](s)
∞
Z
f 0 (t)e−st dt
=
0
∞
Z
e−st df (t)
=
0
=
=
e
−st
∞
f (t)0 + s
Z
∞
f (t)e−st dt
0
−f (0) + sL[f ](s).
Let g(t) = f 0 (t), then
L[f 00 ](s)
= L[g 0 ](s)
= −g(0) + sL[g](s)
= −f 0 (0) + sL[f 0 ](s)
By the result for the first derivative
LECTURE 18: THE LAPLACE TRANSFORM AND TRANSFORMS OF DERIVATIVES AND ODES
= −f 0 (0) + s [−f (0) + sL[f ](s)]
5
By the result for the first derivative
= s2 L[f ](s) − sf (0) − f 0 (0).

 1, if 1 ≤ t < 2,
Example 4. Let f (t) =
, find L[f (t)].
 0, otherwise.
It’s easy to see that f (t) = u1 (t) − u2 (t), then
L[f (t)] = L[u1 (t)] − L[u2 (t)].
By looking up the table, we have
L[u1 (t)] =
e−s
,
s
and L[u2 (t)] =
Then
L[f (t)] =
Example 5. Find L
Since L[1] =
−1
e−s
e−2s
−
.
s
s
1
.
s+1
1
, by the First Shifting Property, then
s
L[e−t ] =
1
,
s+1
which implies that
−1
L
1
= e−t .
s+1
e−2s
.
s
6
MINGFENG ZHAO
Example 6. Solve x0 = 1, x(0) = 0.
Let X(s) = L[x](s), apply the Laplace transform on the both sides of x0 = t, then
L[x0 ](s) = L[1](s) =
1
.
s
Since x(0) = 0, then
L[x0 ](s) = sL[x](s) − x(0) = sX(s).
Then we have sX(s) = 1s . So
X(s) =
1
.
s2
By looking up the table, we have
−1
x(t) = L
[X(s)](t) = L
−1
1
(t) = t
s2
So the solution to x0 = 1, x(0) = 0 is:
x(t) = t .
Example 7. Solve x00 + x = cos(2t), x(0) = 0 and x0 (0) = 1.
Let X(s) = L[x(t)](s), apply the Laplace transform on the both sides of x00 + x = cos(2t), then
L[x00 ] + L[x] = L[cos(2t)] =
s
s
= 2
.
s2 + 22
s +4
Since x(0) = 0 and x0 (0) = 1, then
L[x00 ] = s2 L[x] − sx(0) − x0 (0) = s2 F (s) − 1.
So we get
s2 X(s) − 1 + X(s) =
s
.
s2 + 4
So
X(s) =
1
s
+
.
(s2 + 1)(s2 + 4) s2 + 1
Use the method of partial fractions, we have
(s2
s
As + B
Ds + E
= 2
+ 2
.
2
+ 1)(s + 4)
s +1
s +4
That is, we have
s =
=
(As + B)(s2 + 4) + (Ds + E)(s2 + 1)
As3 + 4As + Bs2 + 4B + Ds3 + Ds + Es2 + E
LECTURE 18: THE LAPLACE TRANSFORM AND TRANSFORMS OF DERIVATIVES AND ODES
=
7
(A + D)s3 + (B + E)s2 + (4A + D)s + 4B + E.
Then
A + D = 0,
B + E = 0,
4A + D = 1,
1
,
3
1
D=− ,
3
and
4B + E = 0.
So
A=
B = 0,
and E = 0.
That is,
(s2
s
1
s
1
s
= · 2
− · 2
.
2
+ 1)(s + 4)
3 s +1 3 s +4
So
X(s) =
1
s
1
s
1
·
− ·
+
.
3 s2 + 1 3 s2 + 4 s2 + 1
Then
x(t)
−1
= L
=
s
1 −1
s
1
1 −1
−1
− L
+L
[X(s)] = L
3
s2 + 1
3
s2 + 4
s2 + 1
1
1
cos(t) − cos(2t) + sin(t),
3
3
By looking up the table.
Therefore, the solution to x00 + x = cos(2t), x(0) = 0 and x0 (0) = 1 is:
x(t) =
1
1
cos(t) − cos(2t) + sin(t) .
3
3
Problems you can do:
Lebl’s Book [2]: All exercises on Page 255.
Braun’s Book [1]: All exercises on Page 232 and Page 233. Read all materials in Section 2.9.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca