How to invert an integral equation

In summary, the author is trying to solve a problem from a textbook example, but they are not sure if a solution exists.
  • #1
nista
11
0

Homework Statement



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:

Homework 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):
[tex]q(r)=Lf(r)[/tex]
where [tex] L [/tex] 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
 
Last edited:
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  • #2
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 [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
 
  • #3
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]
 
  • #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,
Cheers
 

1. How do you solve an integral equation?

To solve an integral equation, you need to use an inverse operation called integration. This involves finding the anti-derivative of the function inside the integral symbol and evaluating it at the upper and lower limits of integration.

2. What is the purpose of inverting an integral equation?

Inverting an integral equation allows you to find the original function from its integral. This can be useful in solving problems in physics, engineering, and other fields where the integral equation represents a physical phenomenon.

3. What are the steps involved in inverting an integral equation?

The first step is to take the derivative of both sides of the equation. Then, use algebraic manipulation to isolate the original function on one side of the equation. Finally, integrate both sides of the equation to find the original function.

4. Are there any special cases when inverting an integral equation?

Yes, there are some special cases where the integral equation cannot be inverted using traditional methods. These include improper integrals, oscillating integrals, and singular integrals. In these cases, advanced techniques such as contour integration or Laplace transforms may be used.

5. Can you provide an example of inverting an integral equation?

Sure, let's say we have the integral equation ∫0x f(t) dt = x² + 2x. To invert this equation, we first take the derivative of both sides, which gives us f(x) = 2x + 2. Then, we integrate both sides to find the original function: ∫0x (2x + 2) dx = x² + 2x. Therefore, the original function is f(x) = x² + 2x + C, where C is the integration constant.

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