From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Newsgroups: sci.math
Subject: Re: Intuition on the Laplace Transform?
Date: 11 Oct 1998 01:36:29 -0400

In article <6vmjma$591@news1.panix.com>, k y n n  <kj0@mailcity.com> wrote:
>
>
>Is it possible to understand intuitively what the Laplace transform
>does to a function, beyond rendering it more tractable in DE problems?
[...]

I have seen the following: represent s=u+iv, so that Laplace Transform
is the Fourier Transform (the variable being v) of the function 

  f(t) * exp(-u*t)

where the factor exp(-u*t) is there to damp the anticipated growth of
f(t), so that the product is indeed Fourier transformable (for s
sufficiently large, if |f| does not grow faster than some exponential).
 Remark: As you know, f can be taken as defined on all of R but flat zero
for negative t.

 This makes it possible to handle "large" functions by a method similar to
Fourier Transform (by first damping them in a way which can be reversed).
 
 If your intuition includes linear algebra relationships, you can think of
both Fourier and Laplace Transforms as analogues to expansions into
eigenvectors of the operator of the derivative:

 (d/dt) exp(-s*t) = (-s) * exp(-s*t)

(here exp(-s*t) is an "eigenvector" of the operator d/dt, corresponding
to the eigenvalue (-s)).

 It then looks like "diagonalization" of d/dt, and handling polynomials of
s  replaces the manipulations with polynomials of (d/dt).

Hope some of it helps, ZVK(Slavek).