CHAPTER 6: THE LAPLACE TRANSFORM
MINGFENG ZHAO
October 25, 2014
Definition 1 (Definitions of the Laplace transform and the inverse Laplace transform). Let f (t) be a function on [0, ∞),
then
I. The Laplace transform of f , denoted by L[f ](s), is defined as:
∞
Z
f (t)e−st dt,
L[f ](s) =
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 ](t) = f (t),
∀for all t > 0.
Example 1. Let f (t) = cos(t), find L[f (t)].
Let L[f (t)](s) = F (s), by the definition of the Laplace transform, we have
∞
Z
F (s)
cos(t)e−st dt
=
0
=
∞
sin(t)e−st 0
Z
= s
∞
Z
sin(t)e−st dt
+s
Use the integration by parts
0
∞
sin(t)e−st dt
0
= s − cos(t)e−st
∞
0
Z
−s
∞
cos(t)e−st dt
0
= s [1 − sF (s)]
= s − s2 F (s).
So we have
F (s) = s − s2 F (s).
Then
F (s) =
1
s2
s
.
+1
Use the integration by parts
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MINGFENG ZHAO
Therefore, we have
L[f (t)](s) =
s
.
s2 + 1
Definition 2 (Definition of the convolution). Let f (t) and g(t) be two functions on [0, ∞), the convolution of f and g
is defined as:
t
Z
(f ∗ g)(t) =
f (τ )g(t − τ ) dτ.
0
Example 2. Let f (t) = 2 sin(t) and g(t) = cos(t), find (f ∗ g)(t).
By the definition of the convolution, we have
t
Z
(f ∗ g)(t)
f (τ )g(t − τ ) dτ
=
0
t
Z
=
sin(τ ) cos(t − τ ) dτ.
2
0
Recall the identity
2 sin(α) cos(β) = sin(α + β) + sin(α − β).
Then
Z
(f ∗ g)(t)
t
[sin(t) + sin(2τ − t)] dτ
=
0
Z
=
t
sin(t)
Z
0
= t sin(t) −
t
sin(2τ − t) dτ
dτ +
0
1
cos(2τ − t)
2
t
0
1
1
= t sin(t) − cos(t) − cos(t)
2
2
=
t sin(t) − cos(t).
Therefore, we get
(f ∗ g)(t) = t sin(t) − cos(t) .
Definition 3 (Definition of Heaviside function). Let a be a constant, then

 0, if t < a,
ua (t) =
 1, if t ≥ a.
CHAPTER 6: THE LAPLACE TRANSFORM
3



a(t), if t < 1,



Example 3. Let a(t), b(t) and c(t) be continuous functions on [0, ∞). Define f (t) =
b(t), if 1 ≤ t < 2, . Represent




 c(t), if t ≥ 2.
f (t) in terms of the Heaviside functions.
By the definitions of the Heaviside functions, it’s easy to see that

 1, if t < 1,
1 − u1 (t) =
 0, if t ≥ 1,

 1, if 1 ≤ t < 2,
u1 (t) − u2 (t) =
 0, if t ≥ 2,

 1, if t < 2,
u2 (t) =
 0, if t ≥ 2.
Therefore, we have
f (t) = a(t)[1 − u1 (t)] + b(t)[u1 (t) − u2 (t)] + c(t)u2 (t) .
Definition 4 (Definition of Dirac delta function). For any continuous function f (t) on (−∞, ∞), we have
Z
∞
δ(t)f (t) dt = f (0).
−∞
Example 4. For any constant a > 0, find L[δ(t − a)]
By the definition of the Laplace transform, we have
∞
Z
L[δ(t − a)]
e−st δ(t − a) dt
=
0
Z
∞
0
t−s(t +a) dt0
=
Let t0 = t − a
−a
= e
−as
,
Since a > 0, by the definition of δ.
Therefore, we have
L[δ(t − a)](s) = e−as .
Froposition 1. There hold that
I. Linearity:
L[af (t) + bg(t)](s) = aL[f (t)](s) + bL[g(t)](s).
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MINGFENG ZHAO
That is,
L−1 [aF (s) + bG(s)](t) = aL−1 [F (s)](t) + bL−1 [G(s)](t).
II. First Shifting Property:
L e−at f (t) (s) = L[f (t)](s + a).
That is,
L−1 [F (s + a)] (t) = e−at L−1 [F (s)](t).
III. 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).
IV. Second Shifting Property:
L[ua (t)f (t − a)] = e−as L[f (t)](s),
for all s > 0.
That is,
L−1 e−as G(s) (t) = ua (t)L−1 [G(s)](t − a).
V. Transform of Integrals:
t
Z
L
f (τ ) dτ (s) =
0
L[f (t)](s)
.
s
That is,
−1
L
Z t
F (s)
=
L−1 [G(s)](τ ) dτ.
s
0
VI. Transform of Convolution:
L[(f ∗ g)(t)] = L[f (t)] · L[g(t)].
That is,
L−1 [L[f (t)] · L[g(t)]] = f ∗ g(t)
Example 5. Find L−1
h
s
s2 −6s+8
i
.
First, we know that s2 − 6s + 8 = (s − 2)(s − 4). By the partial fractions, we have
s
s
A
B
=
=
+
.
s2 − 6s + 8
(s − 2)(s − 4)
s−2 s−4
Then
s = A(s − 4) + B(s − 2) = (A + B)s − 4A − 2B.
CHAPTER 6: THE LAPLACE TRANSFORM
That is,
A + B = 1,
and
0 = −4A − 2B.
So
A = −1,
and B = 2.
So we get
s
1
2
=−
+
.
− 6s + 8
s−2 s−4
s2
Then
L
−1
s
1
1
−1
−1
= −L
+ 2L
.
s2 − 6s + 8
s−2
s−4
By looking up the table, we have
L
−1
1
= e2t ,
s−2
−1
and L
1
= e4t .
s−4
Therefore, we have
−1
L
s
= −e2t + 2e4t .
s2 − 6s + 8
1
.
s2 + 2s + 3
Since s2 + 2s + 3 = (s + 1)2 + 2, then
Example 6. Find L−1
1
1
=
.
s2 + 2s + 3
(s + 1)2 + 2
By looking up the table, we have
L−1
√
e−t
1
= √ sin( 2t) .
2
s + 2s + 3
2
Example 7. Solve y 00 + 4y 0 + 5y = δ(t − 3), y(0) = 1, y 0 (0) = 0.
Let Y (s) = L[y(t)], apply the Laplace transform on the both sides of y 00 + 4y 0 + 5y = δ(t − 3), then
L[y 00 ] + 4L[y 0 ] + 5L[y] = L[δ(t − 3)].
By the transform of derivatives, we have
L[y 0 ] = sL[y] − y(0) = sY (s) − 1,
and L[y 00 ] = s2 L[y] − sy(0) − y 0 (0) = s2 Y (s) − s.
By looking the table, we have
L[δ(t − 3)] = e−3s .
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MINGFENG ZHAO
So we get
s2 Y (s) − s + 4[sY (s) − 1] + 5Y (s) = e−3s .
Then
Y (s)
=
=
=
=
e−3s + s + 4
s2 + 4s + 5
e−3s
s+4
+ 2
2
s + 4s + 5 s + 4s + 5
s+4
e−3s
+
2
(s + 2) + 1 (s + 2)2 + 1
s+2
2
e−3s
+
+
.
(s + 2)2 + 1 (s + 2)2 + 1 (s + 2)2 + 1
By looking up the table, we have
s+2
L−1
= e−2t cos(t),
(s + 2)2 + 1
and L−1
1
= e−2t sin(t).
(s + 2)2 + 1
By the Second Shifting Property, we have
e−3s
1
−1
−1
L
= µ3 (t) · L
(t − 3) = µ3 (t)e−2(t−3) sin(t − 3).
(s + 2)2 + 1
(s + 2)2 + 1
Then we get
y(t) = L−1 [Y (s)] = e−2t cos(t) + 2e−2t sin(t) + µ3 (t)e−2(t−3) sin(t − 3).
Therefore, the solution to y 00 + 4y 0 + 5y = δ(t − 3), y(0) = 1, y 0 (0) = 0 is:
y(t) = e−2t cos(t) + 2e−2t sin(t) + µ3 (t)e−2(t−3) sin(t − 3) .
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