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Fourier transform of a functional

  1. Jan 18, 2014 #1
    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)?
     
  2. jcsd
  3. Jan 18, 2014 #2

    mathman

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    Have you tried integrating by parts? Note that a Fourier series has sine terms as well as cosine.
     
  4. Jan 19, 2014 #3
    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...
     
  5. Jan 19, 2014 #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.
     
  6. Jan 19, 2014 #5
    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
     
  7. Jan 19, 2014 #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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