Polar Fourier transform of derivatives

In summary: The wave height at a point is proportional to the square of the distance from the center of the wave, and the phase speed is proportional to the inverse of the distance from the center.In summary, the water wave problem is to find the free surface from a vorticity distribution.
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hunt_mat

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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: [tex]\hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy[/tex]
In terms of polar co-ordinates: [tex]\hat{f}(\rho,\phi)=\int_{0}^{\infty}\int_{-\pi}^{\pi}rf(r,\theta)e^{-ir\rho\cos(\theta-\phi)}drd\theta[/tex]

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 [itex]r-[/itex]derivative and I get:
[tex]\widehat{\frac{\partial f}{\partial r}}=-\rho\sum_{n\in\mathbb{Z}}\int_{0}^{\infty}rf_{n}(r)J_{n-1}(r\rho)dr[/tex]
This is after I expanded [itex]f(r,\theta)[/itex] as a Fourier series:[tex]f(r,\theta)=\sum_{n\in\mathbb{Z}}f_{n}(r)e^{in\theta}[/tex]
I feel a little lost at this point. Can anyone suggest anything?
 
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  • #2
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: [tex]\hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy[\tex]
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##
 
  • #3
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.
 
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  • #4
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.
 
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  • #5
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.
 

What is the Polar Fourier transform of derivatives?

The Polar Fourier transform of derivatives is a mathematical operation that transforms a function's derivatives from the Cartesian coordinate system to the polar coordinate system. It is used to analyze functions that are expressed in polar coordinates, such as those found in polar graphs.

How is the Polar Fourier transform of derivatives calculated?

The Polar Fourier transform of derivatives is calculated by taking the derivative of a function in the Cartesian coordinate system, converting it to polar coordinates, and then applying the Fourier transform to the resulting function. This process is repeated for each derivative of the function.

What is the significance of the Polar Fourier transform of derivatives?

The Polar Fourier transform of derivatives is significant because it allows for the analysis of functions expressed in polar coordinates, which cannot be easily analyzed using traditional methods. It also provides a way to study the frequency and amplitude content of a function in the polar coordinate system.

What are some applications of the Polar Fourier transform of derivatives?

The Polar Fourier transform of derivatives has various applications in fields such as signal processing, image processing, and pattern recognition. It is also used in the study of waves and oscillations in physics and engineering.

Are there any limitations to the Polar Fourier transform of derivatives?

Like any mathematical tool, the Polar Fourier transform of derivatives has its limitations. It may not be suitable for functions that do not have well-defined derivatives, and it may not be able to accurately analyze functions with sharp discontinuities or singularities. Additionally, it may be computationally intensive for complex functions.

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