LECTURE 20: TRANSFORMS OF DERIVATIVES AND ODES
MINGFENG ZHAO
October 22, 2014
Proposition 1. There holds that
I. Linearity:
L[af (t) + bg(t)](s) = aL[f (t)](s) + bL[g(t)](s).
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).
Using the Heaviside function
The Heaviside function: Let a be a constant, then

 0, if t < a,
ua (t) =
 1, if t ≥ a.
Proposition 2 (Second Shifting Property). For any constant a, then
L[ua (t)f (t − a)](s) = e−as L[f (t)](s),
1
for all s > 0.
2
MINGFENG ZHAO
That is,
L−1 e−as F (s) = ua (t)L−1 [F (s)](t − a).
Proof. By the definition of Laplace transform, we have
Z ∞
L[ua (t)f (t − a)] =
ua (t)f (t − a)e−st dt
0
Z
∞
f (t − a)e−st dt
=
By the definition of ua (t)
a
Z
∞
=
0
f (t0 )e−s(t +a) dt0
Let t − a = t0
0
= e
−as
= e
−as
Z
∞
0
f (t0 )e−st dt0
0
L[f (t)],
for all s > 0.
Example 1. 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).
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) = u1 (t) − u5 (t),
Then
F (s)
= L[f (t)]
∀t ≥ 0.
LECTURE 20: TRANSFORMS OF DERIVATIVES AND ODES
3
= L[u1 (t)] − L[u5 (t)]
e−s
e−5s
−
.
s
s
=
So
X(s) =
F (s)
e−s
e−5s
=
−
.
s2 + 1
s(s2 + 1) s(s2 + 1)
In orde to compute L−1 [X(s)], by the Second Shifting Property, we need to compute L−1
h
1
s(s2 +1)
i
. Use the method
of partial fractions, we have
1
A Bs + C
= + 2
.
s(s2 + 1)
s
s +1
Then
1 = A(s2 + 1) + (Bs + C)s = (A + B)s2 + Cs + A.
So
A + B = 0,
C = 0,
and A = 1.
Then
B = −1. and C = 0.
A = 1,
Then
1
s
1
= − 2
.
s(s2 + 1)
s s +1
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−s
e−5s
−1
= L−1
−
L
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) = u1 (t)[1 − cos(t − 1)] − u5 (t)[1 − cos(t − 5)] .
4
MINGFENG ZHAO
Transform of integrals
Proposition 3 (Transform of integrals). There holds that
Z t
L[f (t)](s)
f (τ ) dτ (s) =
L
.
s
0
That is,
L
−1
Z t
F (s)
=
L−1 [F (s)](τ ) dτ.
s
0
t
Z
Proof. Let g(t) =
f (τ ) dτ , by the Fundamental Theorem of Calculus, we know that
0
g(0) = 0,
and g 0 (t) = f (t).
Then we have
Z
L
t
f (τ ) dτ
Z
∞
g(t) · e−st dt
= L[g(t)] =
0
0
=
=
=
∞
Z
1 ∞ 0
1 −st
g (t)e−st dt
− e g(t) +
s
s
0
0
Z
1 ∞ 0
g (t)e−st dt
s 0
Use the integration by parts
L[f (t)]
.
s
1
.
s(s2 + 1)
By looking up the table, we have
Example 2. Find L−1
L
By the transform of integrals, we have
−1
L
−1
1
s(s2 + 1)
1
= sin(t).
s2 + 1
"
=
−1
1
s2 +1
t
s
L
Z
L−1
=
0
Z
=
1
(τ ) dτ
s2 + 1
t
sin(τ ) dτ
0
=
#
1 − cos(t).
LECTURE 20: TRANSFORMS OF DERIVATIVES AND ODES
So
L−1
Example 3. Solve t2 =
Z
1
= 1 − cos(t) .
s(s2 + 1)
t
eτ x(τ ) dτ .
0
Let X(s) = L[x(t)], apply the Laplace transform on the both sides of t2 =
Z
t
eτ x(τ ) dτ , we have
0
L[t2 ] = L
t
Z
eτ x(τ ) dτ .
0
By the transform of integrals, then
Z t
τ
L
e x(τ ) dτ
=
0
=
=
L [et x(t)]
s
L[x(t)](s − 1)
s
X(s − 1)
.
s
By the First Shifting Property
By looking up the table, we have
L[t2 ] =
2
.
s3
Then
2
X(s − 1)
=
.
s3
s
Then
X(s − 1) =
2
.
s2
That is,
X(s) =
2
.
(s + 1)2
By the First Shifting Property, we have
x(t) = L−1 [X(s)] = 2e−t t.
Therefore, the solution to t2 =
Z
t
eτ x(τ ) dτ is:
0
x(t) = 2e−t t .
Convolution
5
6
MINGFENG ZHAO
Definition 1. Let f (t) and g(t) be two functions on [0, ∞), the convolution of f and g is defined as:
Z t
(f ∗ g)(t) =
f (τ )g(t − τ ) dτ.
0
Transform of Convolution
Proposition 4. There holds that
L[(f ∗ g)(t)] = L[f (t)] · L[g(t)].
That is,
L−1 [F (s) · G(s)] = L−1 [F (s)] ∗ L−1 [G(s)].
Proof. In fact, we have
Z
L[f ∗ g]
∞
(f ∗ g)(t)e−st dt
=
0
Z
∞
=
e
−st
∞
0
Z
∞
e−st f (τ )g(t − τ ) dtdτ
τ
∞
=
∞
Z
f (τ )
0
Z
∞
Z
0
f (τ )
g(t )e
0
Z
g(t − τ )e−st dt dτ
τ
∞
=
=
dt
0
=
Z
f (τ )g(t − τ ) dτ
0
Z
t
Z
−s(t0 +τ )
0
dt
dτ
Let t0 = t − τ
0
∞
f (τ )e−sτ
Z
0
∞
0
g(t0 )e−st dt0
dτ
0
Z
= L[g(t)] ·
∞
f (τ )e−sτ dτ
0
= L[f (t)] · L[g(t)].
Example 4. Let f (t) = et and g(t) = t for t ≥ 0, find f ∗ g and L[f ∗ g].
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