LECTURE 19: TRANSFORMS OF DERIVATIVES AND ODES
MINGFENG ZHAO
October 20, 2014
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 ](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. 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).
1
2
MINGFENG ZHAO
Example 1. 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) = sF (s).
Then we have
sX(s) =
1
.
s
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 .

 1, if t ≥ a,
Example 2 (Heaviside Function). Let a be a real constant, and ua (t) =
, find L[ua ].
 0, if x < a.
If a ≤ 0, then ua (t) ≡ 1 for all t ≥ 0, then
L[ua ](s) = L[1](s) =
1
,
s
for all s > 0.
LECTURE 19: TRANSFORMS OF DERIVATIVES AND ODES
If a > 0, then ua (t) = 0 for all 0 ≤ t ≤ a, which implies that
Z ∞
e−st dt
L[ua ](s) =
a
=
∞
1
− e−st s
a
=
e−at
,
s
for all s > 0.
In summary, we have
 1

if a ≤ 0,
 ,
s
L[ua ] =
−at

 e
, if a > 0.
s
for all s > 0 .

 1, if 1 ≤ t < 2,
Example 3. 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)] =
e−s
e−2s
−
.
s
s
e−2s
.
s
3
4
MINGFENG ZHAO
Example 4. Solve x00 + x = cos(2t), x(0) = 0 and x0 (0) = 1.
Let X(s) = L[x](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
.
s + 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) =
s2
s
.
+4
So
X(s) =
(s2
1
s
+ 2
.
2
+ 1)(s + 4) s + 1
Use the method of partial fractions, we have
s
As + B
Ds + E
= 2
+ 2
.
(s2 + 1)(s2 + 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
=
(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,
s
1
s
1
s
= · 2
− ·
.
(s2 + 1)(s2 + 4)
3 s + 1 3 s2 + 4
So
X(s) =
Then
x(t)
= L−1 [X(s)]
1
s
1
s
1
·
− ·
+
.
3 s2 + 1 3 s2 + 4 s2 + 1
LECTURE 19: TRANSFORMS OF DERIVATIVES AND ODES
=
1 −1
s
1
s
1
L
− L−1 2
+ L−1 2
2
3
s +1
3
s +4
s +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
s2 + s + 1
, find the inverse Laplace transform L−1 [F ].
s3 + s
Use the method of partial fractions, we have
Example 5. Take F (s) =
F (s) =
s2 + s + 1
s2 + s + 1
A Bs + C
=
= + 2
.
3
2
s +s
s(s + 1)
s
s +1
Then we get
s2 + s + 1 = A(s2 + 1) + s(Bs + C) = (A + B)s2 + Cs + A.
So
A + B = 1,
C = 1,
and A = 1.
That is,
A = 1,
B = 0,
and C = 1.
Then
F (s) =
s2 + s + 1
1
1
s2 + s + 1
=
= + 2
.
3
2
s +s
s(s + 1)
s s +1
So we have
L−1 [F ](t)
= L−1
=
1
1
+ L−1 2
s
s +1
1 + sin(t),
By looking up the table.
Therefore, we get
L−1 [F ](t) = 1 + sin(t) .
Example 6. Find L−1
1
.
s2 + 4s + 8
Notice that
1
1
=
.
s2 + 4s + 8
(s + 2)2 + 4
5
6
MINGFENG ZHAO
By looking up the table, we have
L[sin(2t)](s) =
s2
2
.
+4
So we get
1
sin(2t)
(s) = 2
.
L
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) .
2
s + 4s + 8
2
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