5Transform theory
IB Complex Methods
5.3 Elementary properties of the Laplace transform
We will come up with seven elementary properties of the Laplace transform. The
first 4 properties are easily proved by direct substitution
Proposition.
(i) Linearity:
L(αf + βg) = αL(f ) + βL(g).
(ii) Translation:
L(f(t − t
0
)H(t − t
0
)) = e
−pt
0
ˆ
f(p).
(iii) Scaling:
L(f(λt)) =
1
λ
ˆ
f
p
λ
,
where we require λ > 0 so that f(λt) vanishes for t < 0.
(iv) Shifting:
L(e
p
0
t
f(t)) =
ˆ
f(p − p
0
).
(v) Transform of a derivative:
L(f
0
(t)) = p
ˆ
f(p) − f (0).
Repeating the process,
L(f
00
(t)) = pL(f
0
(t)) − f
0
(0) = p
2
ˆ
f(p) − pf (0) − f
0
(0),
and so on. This is the key fact for solving ODEs using Laplace transforms.
(vi) Derivative of a transform:
ˆ
f
0
(p) = L(−tf(t)).
Of course, the point of this is not that we know what the derivative of
ˆ
f
is. It is we know how to find the Laplace transform of
tf
(
t
)! For example,
this lets us find the derivative of t
2
with ease.
In general,
ˆ
f
(n)
(p) = L((−t)
n
f(t)).
(vii) Asymptotic limits
p
ˆ
f(p) →
(
f(0) as p → ∞
f(∞) as p → 0
,
where the second case requires f to have a limit at ∞.
Proof.
(v) We have
Z
∞
0
f
0
(t)e
−pt
dt = [f(t)e
−pt
]
∞
0
+ p
Z
∞
0
f(t)e
−pt
dt = p
ˆ
f(p) − f (0).
(vi) We have
ˆ
f(p) =
Z
∞
0
f(t)e
−pt
dt.
Differentiating with respect to p, we have
ˆ
f
0
(p) = −
Z
∞
0
tf(t)e
−pt
dt.
(vii) Using (v), we know
p
ˆ
f(p) = f(0) +
Z
∞
0
f
0
(t)e
−pt
dt.
As
p → ∞
, we know
e
−pt
→
0 for all
t
. So
p
ˆ
f
(
p
)
→ f
(0). This proof looks
dodgy, but is actually valid since
f
0
grows no more than exponentially fast.
Similarly, as p → 0, then e
−pt
→ 1. So
p
ˆ
f(p) → f(0) +
Z
∞
0
f
0
(t) dt = f(∞).
Example. We can compute
L(t sin t) = −
d
dp
L(sin t) = −
d
dp
1
p
2
+ 1
=
2p
(p
2
+ 1)
2
.