Fourier transform of a functional

  • Thread starter Irid
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  • #1
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Main Question or Discussion Point

Hello,
I was wondering if such a thing even exists, so here it goes... Let's say I have a function x(s) (it is real, smooth, differentiable, etc.) defined on (0,1). In addition, dx/ds = 0 on the boundary (s=0 and s=1). I can compute its Fourier transform (?) as
[tex] a_p = \int_0^1 x(s) \cos(sp\pi)\, ds [/tex]
and now I have a set of numbers [itex]a_p[/itex] which contain the same information as the original function x(s).

The good news is that if I compute the same Fourier transform on the derivatives of x, i.e.
[tex] \int_0^1 \frac{d^2 x}{ds^2} \cos(sp\pi)\, ds = -p^2 \pi^2 a_p[/tex]
I get an answer in terms of the [itex]a_p[/itex] which I already know.

So here's my question:

What if I want to find the Fourier transform of higher powers of x?

[tex] \int_0^1 x^2 \cos(sp\pi)\, ds =\, ? [/tex]

Can it be expressed in terms of, let's say, a power series in [itex]a_p[/itex]?

[tex] \int_0^1 x^2 \cos(sp\pi)\, ds =\, ?\, \sum_{n=0}^{\infty} c_n a_p^n[/tex]

And what if I want to find the Fourier transform of a generic functional f[x(s)] (smooth, differentiable, etc.)? Is it related somehow to the Fourier transform of the original function x(s)?
 

Answers and Replies

  • #2
mathman
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Have you tried integrating by parts? Note that a Fourier series has sine terms as well as cosine.
 
  • #3
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Have you tried integrating by parts? Note that a Fourier series has sine terms as well as cosine.
Well I'm trying to keep things simple here by requiring dx/ds = 0 at the boundaries, hence all the sine terms are zero.

I don't know what to make of the integration by parts. Here's what I get

[tex] \int_0^1 x^2 cos(ps\pi)\, ds = -2\int_0^1 x\frac{dx}{ds}\sin(ps\pi)\, ds [/tex]

I don't see how can this bring me back to the a_p...
 
  • #4
mathman
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I am not sure where you are going. However if you integrate by parts again you will get the second derivative and a cos.
 
  • #5
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I'm trying to solve a non-linear stochastic PDE of the type
dx/dt = d2x/ds2 + F[x] + noise(t)

and I would really benefit if it could be done with the Fourier transform because then I only need the first few terms of the Fourier expansion to have enough information about the x(s).. of course, all of this is to be done numerically
 
  • #6
strangerep
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Irid,

I'm not sure if this helps, but... the Fourier transform of a product ##f(x)g(x)## is the convolution of their respective FTs.
 

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