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Homework Help: How to invert an integral equation

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose we have a physical quantity [tex] f(r) [/tex] depending on another quantity [tex] q(r). [/tex] [tex] f(r) [/tex] is known at all points.
    If the following relationship holds:

    2. Relevant equations
    [tex] f(r)=\int_{\Omega}q(r-r')dr' [/tex]
    where [tex] \Omega [/tex] is a bounded volume,
    is there any possibility to invert somehow such relationship
    in order to have informations on [tex]q(r)[/tex]?
    Something like (but not necessarily):
    where [tex] L [/tex] is a linear operator.

    3. The attempt at a solution
    It is a problem similar to that of the Poisson equation, but I should procede
    in the opposite way, starting from the integral relationship to get the differential form.
    I have already tried to do that but with no success.
    This is a textbook like example but I have to say I have no idea whether a solution exists.
    (I have not taken it from a book)
    Thank you very much to all
    Last edited: Jun 10, 2010
  2. jcsd
  3. Jun 10, 2010 #2


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    Homework Helper

    not an answer but maybe an insight, not totally sure on the generalistion to multiple dimensions, as I have only really played with convolutions in 1D...

    here's some stuff on convolutions if you haven't seen them

    define an indicator function for [itex] \Omega[/itex]:
    [tex] s(r) = 1, \ r \ in \ \Omega, \ 0 \ otherwise [/tex]

    Now using the indicator function, you can change the integral to a convolution over all space
    [tex] f(r)=s*q = \int s(r') q(r-r')dr' [/tex]

    Using the properties of convolution q*s = s*q
    [tex] f(r)=q*s = \int q(r') s(r-r')dr' [/tex]

    so the integral is summing up q(r) over the volume Omega, shifted to center r
  4. Jun 10, 2010 #3


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    Homework Helper

    then, the convolution theorem says, if you take the fourier transform a convolution, then:
    [tex] F=\mathbb{F} \{ f(r) \} = \mathbb{F} \{ q*s (r) \} =\mathbb{F} \{ q(r) \} \mathbb{F} \{ s(r) \} = QS[/tex]

    then rearranging for q
    [tex] q(r) = \mathbb{F}^{-1} \{ \frac{F}{S} \} [/tex]
  5. Jun 11, 2010 #4
    I do not see anything I could complain about this derivation.
    I guessed you solved the problem.
    What else to say ...
    Many thanks lanedance!
    Hope you have a nice day,
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