Why does the Laplace Transform work? [closed]

I would like to know why the Laplace transform works.

Is it that when multiplying a function by exp(-st) that the area captured beneath the curve during integration in the time domain, gives a transform of the function to a unique function in another linear vector space, the frequency domain, and that combinations in that other space uniquely transform back to functions in the time domain?

John

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$\exp(-st)$ is an eigenfunction of $d/dt$. – Terry Tao Mar 7 2012 at 18:17
As a cartoon, you can view the Laplace transform as a mapping between time and energy representations (this is what the Gibbs distribution is about). Meanwhile, the Fourier transform is a mapping between time and complex frequency representations. – Steve Huntsman Mar 7 2012 at 18:28
"why the Laplace transform works" - do you mean "why are the usual formulas for Laplace transforms correct"? or "what is the conceptual reason why Laplace transforms turn convolutions into products"? – Yemon Choi Mar 7 2012 at 18:46
maybe it's also worth mentioning Stieltjes integrals at this point? – S. Sra Mar 8 2012 at 0:38
The main question is not well posed. Works for doing what? – Michael Mar 8 2012 at 9:20

3 Answers

As other posters indicated, the Laplace transform is closely related to the Fourier transform. It is easier to explain the versatility of the Fourier transform.

If you are interested in differential equations, you wish that all functions were linear combinations of exponentials

$$e_\xi(x)= e^{ i \xi x}.$$

The reason is the following simple identity

$$\frac{d}{dx}\left(\sum_k A(\xi_k) e_{\xi_k}(x)\right) = i\sum_k \xi_k A(\xi_k) e_{\xi_k}(x)$$

which shows that for linear combinations of exponentials the transcendental operation of derivation is replaced with a much simpler algebraic operation. It is natural to ask if any function f(x) can be described as a linear combination of exponentials

$$f(x) = \sum_\xi A(\xi) e_\xi(x).$$

The answer is yes, if we allow for infinte superpositions of exponentials

$$f(x) "= " \sum_{\xi\in\mathbb{R}} A(\xi) e_\xi(x) :=\int_{\mathbb{R}} A(\xi) e_\xi(x) d\xi.$$

More precisely, the above function $\xi\mapsto A(\xi)$ is the Fourier transfrom of $f(x)$

$$A(\xi)=\frac{1}{2\pi} \int_{\mathbb{R}} f(x) e_{-\xi}(x) dx.$$

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 I like this a lot. – Jeff Strom Mar 8 2012 at 15:32 I feel that this comment, and the eigenfunction comment, are most likely to lead me to a better understanding of why circuit problems can be represented and solved when transformed to another space. Thanks to all. – John Mar 9 2012 at 18:19

I am particularly interested in 'why' it Works for ac theory. I say 'why' because I know 'how' it works - ie. how to do the transform.

I am puzzled that combinations of voltage waveforms, together with integrated and differentiated waveforms, can be combined, transformed to another space, manipulated and transformed back and that they successfully solve a circuit problem. The maths works, but why does the Physics? I was hoping the maths would help me understand the Physics.

On comments: I can see exp(-st) as an eigenfunction - it doesn't change with d/dt, its job seems to be to cause the transformed values to be finite.

The Gibbs distribution may be along the lines I was thinking - which is that perhaps that the transformed function is an encapsulation of the energy beneath the decaying exp(-st) voltage.

Stieltjes didn't take me anywhere towards an understanding.

"Laplace transform is closely related to the Fourier transform." Yes I understand that. Laplace just adds a real part to s and that causes a decay - the transformed function then occupies 2D in the W plane with both real and imaginary axes.

You have given me more to think about, for which I am grateful, but it doesn't seem to be leading me to a physical reason for it's success. Some of your answers are going to take me quite a while to digest, but thank you for them.

After being on MathFlow the other day I was led to this which helped quite a bit:

http://www.codee.org/library/articles/the-laplace-transform-motivating-the-definition

and file itself:

http://www.codee.org/library/articles/the-laplace-transform-motivating-the-definition/at_download/files|files:000

John

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 mathoverflow.net/faq#whatnot – Yemon Choi Mar 8 2012 at 17:03 Answer boxes should not be used for comments or updates. The software/format here is not set up like a forum thread – Yemon Choi Mar 8 2012 at 17:04

Consider these relations for the Fourier, Mellin, and Laplace transforms:

$\int^{\infty}_{-\infty}{exp(2 \pi ifx)exp(-2 \pi ify)df} = \delta(x-y)$

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= y \delta(x-y)$

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{-xp} e^{yp} dp=\delta(x-y)$.

For me, the delta fct. results seem intuitive (also analogous to the orthogonality relationship for characters of character groups), and the eqns. encapsulate the properties of the transforms (try deriving the transform pairs and other relations from them, e.g., Plancherel, convolution, Poisson summation) and illustrate the transformations from one transform to another.

(Tried this as a comment initially, but had formatting problems.)

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 Per Fourier's original motivation, apply F(d/dy) to the LPT/FT relations to obtain an operational calculus. For the Mellin relation, apply G(xd/dx). A good part of operational calculus deals with the issue of for what kinds of functions this is valid. – Tom Copeland Mar 8 2012 at 23:06