• Status: Solved
• Priority: Medium
• Security: Public
• Views: 1557

# Derivation of Laplace Transforms

I'm trying to derive a few laplace transforms, I'll use L for the symbol for Laplace Transform...

L{ e^(at) * f(t) } = F(s - a)
L{ u_a(t) } = e^(-sa) / s
L{ t * f(t) } = -F`(s)

u_a(t) is the heaviside step function.

I've started with the first and gotten stuck:

L{ e^(at) * f(t) } =

/ inf
|      e^-(s - a)t * f(t) dt
/0

I'm not sure how to show that's F(s - a)....

Thanks,
Brian
0
BrianGEFF719
• 5
• 3
3 Solutions

Author Commented:
The heaviside derivation isn't too hard...

L{ u_a(t) } = int_0->a ( 0 dt ) + int_a->inf( e^(-st) )
= 0 - (e^(-sa) / -s) = e^(-sa) / s
0

Commented:
L{ e^(at) * f(t) } = F(s - a)

This is the First Shifting Theorem.

Start from the definition of the Laplace Transform:

F(s) = L{f} = Int_0->inf (e^(-st).f(t) dt)

And replace s by (s-a):

F(s-a) = Int_0->inf (e^-(s-a)t.f(t) dt) = Int_o->inf (e^(-st)[e^(at).f(t)] dt) = L{e^(at).f(t)}.

0

Commented:
L{ t * f(t) } = -F`(s)

If f(t) satisfies the conditions of the existence theorem, then the derivative F'(s) of its transform with respect to s can be obtained by differentiating under the integral sign w.r.t s, thus

F'(s) = - Int_0->inf (e^-(st).[t.f(t)] dt)

There is proof of this in a book at my uni's library, so I'm going to quickly check if it's available....
0

Commented:
Right, well, I found the book - but couldn't find the proof I was after.

So, a basic way of looking at it, is this:

F(s) = L{f} = Int_0->inf (e^-(st).f(t) dt)

F'(s) = d/ds L{f} = d/ds Int_0->inf (e^-(st).f(t) dt)

Because you're integrating with respect to t, you can just take the d/ds inside the integral:

F '(s) = Int_o->inf (d/ds e^-(st).f(t) dt) = Int_o->inf (-t.e^-(st).f(t) dt) = L{-t.f(t)}

Therefore  F '(s) = -L{t.f(t)}

If you want a rigorous proof of this though, then it's SOMEWHERE in:

Differential and Integral Calculus, vol.2. By Courant.

:-P
0

Author Commented:
Is it acceptable in a proof to work backwards, or to work forward half way and work the end of the problem back to the same form? For example, in the proof of e^(at) f(t), if we work to where we have the integral with -(s - a), and then work backwards with F(s - a)?
0

Commented:
It's not always acceptable to work backwards. (A=>B does not necessarily mean that B=>A).
(For example, solve: sqrt(x+3)=sqrt(1-x)-sqrt(1+x), and check your answers ;))

In the case of starting from F(s-a) and working backwards: it's perfectly valid.

Effectively, all you're doing is rearranging the Laplace transform; it's not a case of showing/proving that the Laplace transform _implies_ that F(s-a)=L{e^(at).f(t)}; instead, it's a case of showing that they're _identical_. And in such a case, the order of 'proof' does not matter.
0

Author Commented:
>> (A=>B does not necessarily mean that B=>A).

The only situation when I think that would be true is with even powers, right?
0

Commented:
My word is certainly not gospel, so I will say "I don't know" to that one.

I mean, A and B could be any mathematical statements; for example,

A: f(x,y) satisfies the Uniqueness Theorm.
B: f(x,y) is continuous.

A => B, but B =/> A.

But, maybe this is be an invalid example, as perhaps a square does pop up in the proof that A=>B.

So, I will shrug at this one. Sorry.
0

Commented:
Brian: Interactive's proof of the Shifting Theorem is the standard proof. He is not working "backwards" but simply rearranging terms.

The derivation of the third rule requires differentiation under the integral sign. This requires the functions and variables involved to be continuous in the classic sense. Most of the use of such transforms in practical work involve such continuous functions. Problems generally arise when one is transforming infinite series involving trigonometrical function.
0

## Featured Post

• 5
• 3
Tackle projects and never again get stuck behind a technical roadblock.