Group Abstract Group Abstract

Message Boards Message Boards

Difficult convolution problem: can I ask for a solution in terms of elliptic integrals?

I have a convolution problem that is difficult. I want to convolve a function with itself. The function is as follows: f(t_) = 2 * (2/Pi)*Sqrt[1 - t^2], on the -1 <= t <= 1 interval. This is a semi-circle distribution with a radius of 1. It is also called a Wigner Distribution. I expect the resultant convolved function to exist on the interval 0 <= t <= 2.

My Mathematica code is attached. My question is as follows: can I make an assumption such that Mathematica returns a solution with Elliptic integrals? Presently, Mathematica just sort of hangs up after running for a while.
POSTED BY: Patrick McMullen
4 Replies
Sort By:

Same solution as Henrik's one, a little more simplified :

f[t_] := If[-1 <= t <= 1, 2*(2/Pi)*Sqrt[1 - t^2], 0];
g[z_] := FourierTransform[f[x], x, z];
Carré[u_] := g[u] g[u];
FullSimplify@
 FunctionExpand[
  InverseFourierTransform[Carré[\[Omega]], \[Omega], t], -2 <= t <= 2]

giving :

(8 Sqrt[2] ((4 + t^2) EllipticE[1 - t^2/4] - 
   2 t^2 EllipticK[1 - t^2/4]))/(3 \[Pi]^(5/2))
POSTED BY: Claude Mante

Thank you -- this is helpful.
POSTED BY: Patrick McMullen

Strange that MMA has problems doing it ...

Another option would be to use the convolution theorem:

f[t_] := HeavisidePi[t/2] Sqrt[1 - t^2]
ft = FourierTransform[f[t], t, k];
convol = InverseFourierTransform[ft^2, k, t] // FullSimplify

enter image description here

Plotting this strange function took me quite some time - it seems not easy to calculate the respective values.
POSTED BY: Henrik Schachner

With Maple is easy to do: enter image description here

UPDATE:
POSTED BY: Mariusz Iwaniuk
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard