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

.