How to invert an integral equation

[nista]

Homework Statement

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

Homework Equations

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

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

 

[lanedance]

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
http://en.wikipedia.org/wiki/Convolution

define an indicator function for $\Omega$:
$$s(r) = 1, \ r \ in \ \Omega, \ 0 \ otherwise$$

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

Using the properties of convolution q*s = s*q
$$f(r)=q*s = \int q(r') s(r-r')dr'$$

so the integral is summing up q(r) over the volume Omega, shifted to center r

 

[lanedance]

then, the convolution theorem says, if you take the Fourier transform a convolution, then:
$$F=\mathbb{F} \{ f(r) \} = \mathbb{F} \{ q*s (r) \} =\mathbb{F} \{ q(r) \} \mathbb{F} \{ s(r) \} = QS$$

then rearranging for q
$$q(r) = \mathbb{F}^{-1} \{ \frac{F}{S} \}$$

 

[nista]

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,
Cheers