LECTURE 21: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC
DELTA
MINGFENG ZHAO
October 30, 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.
Definition 2. Let a be a real constant, and the Heaviside function is defined as:

 1, if t ≥ a,
u(t − a) =
 0, if x < a.
Remark 1. Notice that 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.
Definition 3. Let f (t) and g(t) be two functions on [0, ∞), the convolution of f and g is defined as:
Z
(f ∗ g)(t) =
t
f (τ )g(t − τ ) dτ.
0
Definition 4 (Dirac’s delta function). For any continuous function f (t) on (−∞, ∞), we have
Z
∞
δ(t)f (t) dt = f (0).
−∞
Proposition 1. The followings hold:
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2
MINGFENG ZHAO
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).
III. Second Shifting Property:
= e−as L[f (t)](s)
L[u(t − a)f (t − a)](s)
L−1 e−as F (s) = u(t − a)L−1 [F (s)](t − a).
IV. Transform of Integrals:
Z
L
t
f (τ ) dτ (s)
0
L−1
F (s)
(t)
s
L[f (t)](s)
s
Z t
=
L−1 [F (s)](τ ) dτ
=
0
V. Transform of Convolution:
L[(f ∗ g)(t)](s)
=
L[f (t)](s) · L[g(t)](s)
L−1 [F (s) · G(s)] (t)
=
L−1 [F (s)] ∗ L−1 [G(s)](t).
VI. Transform of Dirac delta:
L[δ(t − a)](s)
L−1 [e−as ](t)
= e−as
= δ(t − a).
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 1. Find the Laplace transform Y (s) = L[y(t)] of the solution of y 00 + y = g(t), y(0) = 1 and y 0 (0) = 2, where

 1,
if 0 ≤ t < 2,
g(t) =
 cos(t − 2), if t ≥ 2.
LECTURE 21: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA
Apply the Laplace transform on the both sides of y 00 + y = g(t), then
L[y 00 ] + L[y] = L[g(t)].
By transform of derivatives, we have
L[y 00 ] = s2 L[y] − sy(0) − y 0 (0) = s2 Y (s) − s − 2.
Then
s2 Y (s) − s − 2 + Y (s) = L[g(t)].
So we get
Y (s) =
L[g(t)] + s + 2
s2 + 1
For L[g(t)], it’s easy to see that
g(t)
=
1 · [u(t − 0) − u(t − 2)] + cos(t − 2)[u(t − 2) − u(t − ∞)]
=
1 − u(t − 2) + cos(t − 2)u(t − 2)
=
1 + [cos(t − 2) − 1]u(t − 2).
By the Second Shifting Property, we have
L[g(t)]
= L[1] + L[(cos(t − 2) − 1)u(t − 2)]
=
L[1] + e−2s L[cos(t + 2 − 2) − 1](s)
=
L[1] + e−2s L[cos(t)](s) − e−2s L[1](s).
By looking up the table, we know that
L[1] =
1
,
s
and L[cos(t)] =
s
.
s2 + 1
Then we get
L[g(t)] =
1 e−2s
se−2s
−
+ 2
.
s
s
s +1
So
Y (s)
=
=
1
s
−
e−2s
s
+
s2
se−2s
s2 +1
+s+2
+1
s+2
1
e−2s
se−2s
+
−
+ 2
.
2
2
2
s + 1 s(s + 1) s(s + 1) (s + 1)2
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MINGFENG ZHAO
Therefore, we know that
Y (s) =
s+2
1
e−2s
se−2s
.
+
−
+
s2 + 1 s(s2 + 1) s(s2 + 1) (s2 + 1)2
Remark 3. In general, let a0 = 0 < a1 < a2 < a3 < · · · < an < ∞, if



f1 (t),
if 0 ≤ t < a1 ,







f2 (t),
if a1 ≤ t < a2 ,





 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
Example 2. Let ω > 0, f (t) = sin(ωt) and g(t) = cos(ωt) for all t ≥ 0, find (f ∗ g)(t).
In fact, we have
t
Z
(f ∗ g)(t)
sin(ωτ ) cos(ω(t − τ )) dtτ
=
0
=
=
=
1
2
Z
t
[sin(ωt) − sin(ωt − 2ωτ )] dτ
0
t
1
1
sin(ωt)τ +
cos(2ωτ − ωt) 2
4ω
0
1
t sin(ωt).
2
Example 3. Let f (t) be any nice function on [0, ∞), find the solution to x00 + ω02 x = f (t), x(0) = x0 (0) = 0.
Let X(s) = L[x(t)](s) and F (s) = L[f (t)](s), apply the Laplace transform on the both sides of x00 + ω02 x = f (t), then
L[x00 ] + ω02 L[x] = L[f (t)].
By the transform of derivatives, we have
L[x00 ] = s2 L[x] − sx(0) − x00 (0) = s2 X(s).
LECTURE 21: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA
Then we have
s2 X(s) + ω02 X(s) = F (s).
So we get
X(s) =
F (s)
1
= F (s) · 2
.
s2 + ω02
s + ω02
By the Laplace transform of convolution, we know that
= L−1 [X(s)](t)
=
L−1 [F (s)] ∗ L−1
x(t)
1
2
s + ω02
.
By looking up the table, we get
L−1
1
sin(ω0 t)
=
.
s2 + ω02
ω0
So we have
t
Z
x(t)
f (t − τ ) ·
=
0
=
Example 4. Find L−1
1
ω0
Z
sin(ω0 τ )
dτ
ω0
t
sin(ω0 τ )f (t − τ ) dτ.
0
s+1
.
s
In fact, we have
L−1
s+1
1
1
= L−1 1 +
= L−1 [1] + L−1
.
s
s
s
By looking up the table, we have
−1
L
[1] = δ(t),
and L
−1
1
= 1.
s
Then we know that
L−1
s+1
= δ(t) + 1. .
s
Example 5. Let ω0 > 0, solve x00 + ω02 x = δ(t), x(0) = x0 (0) = 0.
Let X(s) = L[x(t)], apply the Laplace transform on the both sides of x00 + ω02 x = δ(t), then
L[x00 ] + ω02 L[x] = L[δ(t)].
By the transform of derivatives, we have
L[x00 ] = s2 L[x] − sx(0) − x00 (0) = s2 X(s).
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MINGFENG ZHAO
By looking up the table, we have
L[δ(t)] = 1.
So we have
s2 X(s) + ω02 X(s) = 1.
That is,
X(s) =
s2
1
.
+ ω02
By looking up the table, we get
x(t) = L
−1
[X(s)] = L
−1
1
sin(ω0 t)
=
.
2
2
s + ω0
ω0
Therefore, the solution to x00 + ω02 x = δ(t), x(0) = x0 (0) = 0 is:
x(t) =
sin(ω0 t)
.
ω0
Example 6. 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 .
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
s+4
e−3s
+
s2 + 4s + 5 s2 + 4s + 5
e−3s
s+4
+
(s + 2)2 + 1 (s + 2)2 + 1
s+2
2
e−3s
+
+
.
2
2
(s + 2) + 1 (s + 2) + 1 (s + 2)2 + 1
LECTURE 21: THE LAPLACE TRANSFORMS OF INTEGRAL, CONVOLUTION AND DIRAC DELTA
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By the first shifting property and looking up the table, we have
s+2
1
−2t
−1
L−1
(t)
=
e
(t) = e−2t sin(t).
cos(t),
and
L
(s + 2)2 + 1
(s + 2)2 + 1
By the Second Shifting Property, we have
e−3s
−1
=
L
(s + 2)2 + 1
=
u(t − 3) · L
−1
1
(t − 3)
(s + 2)2 + 1
u(t − 3)e−2(t−3) sin(t − 3).
Then we get
y(t) = L−1 [Y (s)] = e−2t cos(t) + 2e−2t sin(t) + u(t − 3)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) + u(t − 3)e−2(t−3) sin(t − 3) .
Problems you can do:
Lebl’s Book [2]: All exercises on Page 267, Page 268, Page 274 and Page 275.
Braun’s Book [1]: All exercises on Page 250, Page 251, Page 256 and Page 257. Read all materials in Section
2.12 and Section 2.13.
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