LECTURE 19: THE LAPLACE TRANSFORM AND TRANSFORMS OF DERIVATIVES AND
ODES
MINGFENG ZHAO
October 26, 2015
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
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.
Proposition 1. The followings hold:
I. Transforms of derivatives:
L[f 0 ](s)
= sL[f ](s) − f (0)
L[f 00 ](s)
= s2 L[f ](s) − sf (0) − f 0 (0).
II. First Shifting Property:
L e−at f (t) (s)
=
L[f ](s + a)
L−1 [F (s + a)](t)
=
e−at L−1 [F (s)](t).
s2 + s + 1
, find the inverse Laplace transform L−1 [F (s)].
s3 + s
Use the method of partial fractions, we have
Example 1. Take F (s) =
F (s) =
s2 + s + 1
A Bs + C
s2 + s + 1
=
= + 2
.
3
s +s
s(s2 + 1)
s
s +1
Then we get
s2 + s + 1 = A(s2 + 1) + s(Bs + C) = (A + B)s2 + Cs + A.
1
2
MINGFENG ZHAO
So
A + B = 1,
C = 1,
and A = 1.
That is,
A = 1,
B = 0,
and C = 1.
Then
F (s) =
s2 + s + 1
s2 + s + 1
1
1
=
= + 2
.
3
s +s
s(s2 + 1)
s s +1
So we have
L
−1
[F ](t)
−1
= L
=
1
1
−1
+L
s
s2 + 1
1 + sin(t),
By looking up the table.
Therefore, we get
L−1 [F ](t) = 1 + sin(t) .
Example 2. Find L−1
1
.
s2 + 4s + 8
Notice that
s2
1
1
=
.
+ 4s + 8
(s + 2)2 + 4
By looking up the table, we have
L[sin(2t)](s) =
2
.
s2 + 4
So we get
sin(2t)
1
L
(s) = 2
.
2
s +4
So we have
−1
L
1
2
s + 4s + 8
−1
1
(s + 2)2 + 4
=
L
=
1 −2t
e
sin(2t),
2
By the First Shifting Property.
Therefore, we get
L
−1
1
1
= e−2t sin(2t) .
s2 + 4s + 8
2
LECTURE 19: THE LAPLACE TRANSFORM AND TRANSFORMS OF DERIVATIVES AND ODES
3
Using the Heaviside function
Recall that for any real constant a, and the Heaviside function is defined as:

 1, if t ≥ a,
u(t − a) := ua (t) =
 0, if t < a.
Remark 1. When a = 0, we know that u(t − 0) = 1 for all t ≥ 0. When a = ∞, we define u(t − ∞) := 0 for all t ≥ 0.
Proposition 2 (Second Shifting Property). For any constant a, then
L[u(t − a)f (t − a)](s) = e−as L[f (t)](s),
for all s > 0.
That is,
L−1 e−as F (s) = u(t − a)L−1 [F (s)](t − a).
Proof. By the definition of Laplace transform, we have
Z ∞
L[(t − a)f (t − a)] =
u(t − a)f (t − a)e−st dt
0
Z
∞
f (t − a)e−st dt
=
By the definition of u(t − a)
a
Z
∞
=
0
f (t0 )e−s(t +a) dt0
Let t − a = t0
0
=
e−as
=
−as
Z
∞
0
f (t0 )e−st dt0
0
e
L[f (t)],
for all s > 0.
Remark 2. In practice, we can use the second shifting property in the following way:
L[u(t − a)f (t)](s) = e−as L[f (t + a)](s).
Example 3. Solve x00 + x = f (t), x(0) = x0 (0) = 0, where

 1, if 1 ≤ t < 5
f (t) =
 0, otherwise.
Let X(s) = L[x] and F (s) = L[f (t)], apply the Laplace transform on the both sides of x00 + x = f (t), then
L[x00 ] + L[x] = L[f (t)] = F (s).
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MINGFENG ZHAO
Since x(0) = x0 (0) = 0, then
L[x00 ] = s2 L[x] − sx(0) − x00 (0) = s2 X(s).
So we get
s2 X(s) + X(s) = F (s).
That is,
X(s) =
F (s)
.
s2 + 1
For F (s) = L[f (t)], by the definition of f (t), we know that
f (t) = u(t − 1) − u(t − 5),
∀t ≥ 0.
Then
F (s)
=
L[f (t)]
=
L[u(t − 1)] − L[u(t − 5)]
=
e−5s
e−s
−
.
s
s
So
X(s) =
F (s)
e−s
e−5s
=
−
.
s2 + 1
s(s2 + 1) s(s2 + 1)
In order to compute L−1 [X(s)], by the Second Shifting Property, we need to compute L−1
of partial fractions, we have
1
A Bs + C
= + 2
.
+ 1)
s
s +1
s(s2
Then
1 = A(s2 + 1) + (Bs + C)s = (A + B)s2 + Cs + A.
So
A + B = 0,
C = 0,
and A = 1.
Then
A = 1,
B = −1. and C = 0.
Then
1
1
s
= − 2
.
s(s2 + 1)
s s +1
h
1
s(s2 +1)
i
. Use the method
LECTURE 19: THE LAPLACE TRANSFORM AND TRANSFORMS OF DERIVATIVES AND ODES
5
So we have
−1
L
1
s(s2 + 1)
−1
1
s
−1
−L
s
s2 + 1
=
L
=
1 − cos(t)
By looking the table.
So we know that
x(t)
= L−1 [X(s)]
e−5s
e−s
−1
−
L
= L−1
s(s2 + 1)
s(s2 + 1)
= u1 (t)[1 − cos(t − 1)] − u5 (t)[1 − cos(t − 5)],
By the Second Shifting Property.
Therefore, the solution to x00 + x = f (t), x(0) = x0 (0) = 0 is:
x(t) = u(t − 1)[1 − cos(t − 1)] − u(t − 5)[1 − cos(t − 5)] .

 w(t), if a ≤ t < b
Remark 3. For any 0 ≤ a < b ≤ ∞, let f (t) =
, then f (t) = w(t) · [u(t − a) − u(t − b)] for all
 0,
otherwise.
t ≥ 0.
Example 4. Find the Laplace transform of
f (t) =




1




t






0
if t < 1
if 1 ≤ t < 2
if t ≥ 2.
We can rewrite the piecewise definition of f using Heaviside functions:
f (t) = 1 · (u(t − 0) − u(t − 1)) + t · u(t − 1) − u(t − 2) + 0 · [u(t − 2) − u(t − ∞)]
= 1 − u(t − 1) + tu(t − 1) − tu(t − 2)
= 1 + (t − 1)u(t − 1) − tu(t − 2).
This last expression makes it easier to apply the second shifting formula:
F (s)
= L[1] + L[(t − 1)u(t − 1)] − L[tu(t − 2)]
=
=
1
+ e−s L[t + 1 − 1] − e−2s L[t + 2]
s
1
+ e−s L[t] − e−2s (L[t] + 2L[1])
s
6
MINGFENG ZHAO
=
1 e−s
+ 2 − e−2s
s
s
1
2
+
2
s
s
Remark 4. In general, let a0 = 0 < a1 < a2 < a3 < · · · < an < ∞, if



f1 (t),
if 0 ≤ t < a1 ,







f (t),
if a1 ≤ t < a2 ,

 2



 f3 (t),
if a2 ≤ t < a3 ,
f (t) =
.
..

..

.







fn (t),
if an−1 ≤ t < an ,






fn+1 (t), if t ≥ an ,
then,
f (t)
=
n
X
fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 [u(t − an ) − u(t − ∞)]
k=1
= f1 (t)[1 − u(t − a1 )] +
n
X
fk (t)[u(t − ak−1 ) − u(t − ak )] + fn+1 (t)u(t − an ).
k=2
Problems you can do:
Lebl’s Book [2]: All exercises on Page 255, Page 261, Page 262 and Page 263.
Braun’s Book [1]: All exercises on Page 232, Page 233, Page 237, Page 238 and Page 243. Read all materials
in Section 2.9, Section 2.10 and Section 2.11.
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