From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Meaning of Laplace transform
Date: 7 Sep 1999 22:52:52 -0400
Newsgroups: sci.math

In article <7r434c$8ij$1@nnrp1.deja.com>,  <hursun@my-deja.com> wrote:
>I'm sorry I don't know how to edit my post, so I add my another
>question about Laplace transform to another posting.
>
>I know where Laplace transform is useful (solving Diff.equ) ,,But
>I can't hardly grasp what intrgral transform (in general) does tell
>about the original function.(Any geometrical meaning?)
>
[...]

There is a "geometric", actually linear-algebraic meaning, the one that
makes it useful in differential equations:

Take the operator of the derivative D, defined on the differentiable
functions on [0, inf} that vanish at 0 and are dominated by an exponential
(such functions are called Laplace transformable). Call the Laplace
transform operator L. Then (integration by parts)
   (LDf)(s) = s * (Lf)(s)
(If f does not vanish at 0, the correction is well-known.)

One more piece of notation, plus the fact that L is one-to-one:
  S will be the operator of multiplication by s; it turns a function
 s |-> F(s) into a function s |-> s*F(s).

 Then L*D = S*L, that is, D = L^(-1) * S * L

If you imagine S as a "diagonal" operator (acting similarly to diagonal
matrices on vectors), then the above equation says that D (a complicated
operator) has been "diagonalized" to become S, a simple-looking operator.

The eigenvectors of D are the exponentials 

(of course, Dexp(a*t)=a*exp(a*t)), so that L sort of expands D in its
system of eigenvectors.

D has no claim to be symmetric - that is, on the set of transformable and
differentiable functions on [0,inf), which does not even have a reasonable
inner product, so one cannot expect anything near orthogonality of
eigenvectors.

If you are somehow familiar with Fourier Transform, there is another
"apology" for L:

Take a (piecewise continuous) function f, vanishing on (-inf, 0), and try
to find its Fourier Transform (i.e. determine the frequency image of f).
An obstacle could be that |f| does not have a finite integral. In an
attempt to extract some frequency information anyway, apply a damping
factor to f, i.e. consider a function
    x |-> exp(-u*x)*f(x)
Now if f is not too fast growing, there will be values of u for which the
damped function will be Fourier transformable. Then the Fourier Transform
will be

   integral[0 to inf) exp(-i*v*x) * exp(-u*x) * f(x) dx

All you have to do now is to wrap up u and v into one complex variable

    u + i*v = s

and Laplace Transform pops out. Summary: it is a coded version of Fourier
Transform applied to an exponentially damped function.

Hope some of it helps, ZVK(Slavek).
==============================================================================

From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: LaPlace Invariant
Date: 6 Mar 1999 02:58:57 -0500
Newsgroups: sci.math

In article <36E07877.3627@rochester.rr.com>,
Jonathan W. Hoyle <jhoyle1@rochester.rr.com> wrote:
>As my Differential Equation knowledge has gotten a bit rusty lately, I
>have been recently thinking about Laplace Transforms, particularly, if
>f(x) and g(x) are two real valued functions over the positive reals, and
>L{f(x)} = L{g(x)}, what can we say about f(x) and g(x)?
>
>It's certainly not true that f(x) must equal g(x), as they can differ
>over a set of measure 0 without affecting the integral.  But more than
>that, I have the following questions I was wondering if are already
>known:
>
>1. If L{f(x)} = L{g(x)}, and f and g are both continuous, must f = g?
>
 Yes; it boils down (subtract right side from left side) to: If
L{h(x)} = 0, and if h is continuous, must h = 0? And this follows, after a
change of variable, from Weierstrass's Approximation Theorem.

>2. When does L^-1{L{f(x)}} = f(x)?  (Same question with L{L^-1{f(x)}}.)
>
 Every time it's defined.

>3. If L{f(x)} exists, does there always exist a continuous g(x) such
>that L{f(x)} = L{g(x)}?
>

 No. Recall that for the step function b: b(x)=1 in (0,A) and 0 otherwise,
L{b(x)}(s) = (1-e^(A*s))/s, and no continuous function can equal b almost
everywhere.

>4. Finally, given arbitrary f(x) such that L{f(x)} exists, can one
>determine the family of functions {fn(x)} such that L{fn(x)} = L{f(x)}
>for all n?

Too easy as stated: simply put f_n = f for all f. I wonder what the
apparently more difficult question could have been. (Provided n was a
subscript. If it was something else, description in words will help.)

Hope it helps, ZVK(Slavek).