Polar Fourier transform of derivatives

Click For Summary

Discussion Overview

The discussion revolves around the polar Fourier transform, specifically its application to derivatives of functions. Participants explore the differences between Cartesian and polar coordinates in the context of Fourier transforms, with a focus on how to compute these transforms in polar coordinates, particularly for derivatives.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the polar Fourier transform and expresses confusion about how to relate the Fourier transform of a derivative to the transform of the original function.
  • Another participant suggests searching for the "Hankel transform" and mentions that differentiating the inverse transform and using Bessel function identities might be a viable approach.
  • A participant reflects on their background in pure mathematics and mathematical physics, indicating a personal context for their inquiry related to solving PDEs in a water wave problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method for computing the polar Fourier transform of derivatives, and multiple viewpoints and approaches are presented without resolution.

Contextual Notes

There are limitations regarding the assumptions made about the functions involved and the specific conditions under which the polar Fourier transform is applied. The discussion also hints at unresolved mathematical steps related to the application of Bessel functions.

hunt_mat
Homework Helper
Messages
1,816
Reaction score
33
TL;DR
Computing Fourier transforms is simple in Cartesian co-ordinates but how do you do it for polar co-ordinates?
The 2D Fourier transform is given by: \hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy
In terms of polar co-ordinates: \hat{f}(\rho,\phi)=\int_{0}^{\infty}\int_{-\pi}^{\pi}rf(r,\theta)e^{-ir\rho\cos(\theta-\phi)}drd\theta

For Fourier transforms in cartesian co-ordinates, relating the Fourier transform of a derivative of a function to the Fourier transform of the function. However, what happens with the polar Fourier transform? I've done a simple calculation for the r-derivative and I get:
\widehat{\frac{\partial f}{\partial r}}=-\rho\sum_{n\in\mathbb{Z}}\int_{0}^{\infty}rf_{n}(r)J_{n-1}(r\rho)dr
This is after I expanded f(r,\theta) as a Fourier series:f(r,\theta)=\sum_{n\in\mathbb{Z}}f_{n}(r)e^{in\theta}
I feel a little lost at this point. Can anyone suggest anything?
 
Physics news on Phys.org
hunt_mat said:
TL;DR Summary: Computing Fourier transforms is simple in Cartesian co-ordinates but how do you do it for polar co-ordinates?

The 2D Fourier transform is given by: \hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy[\tex]<br />
<br /> Did you mean to type this? <br /> The 2D Fourier transform is given by: ##\hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy##
 
DaveE said:
Did you mean to type this?
The 2D Fourier transform is given by: ##\hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy##
I was correcting my memory on how to include LaTeX here, you'll see that it's corrected now.
 
  • Like
Likes   Reactions: DaveE
Search for "Hankel transform". The Wikipedia article doesn't give an expression for the transform of a derivative, but differentiating the inverse transform and using Bessel function identities should work.
 
  • Like
Likes   Reactions: vanhees71 and hunt_mat
I had a feeling that this would be the case. A lot of my undergraduate was pure maths, mainly analysis and mathematical physics. Solving PDEs was something I had to teach myself about.

This is about a water wave problem, looking for the free surface from a vorticity distribution. The azimuthal component will be the most important.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K