Problem with Fourier bessel transform of Yukawa potential

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SUMMARY

The discussion focuses on the challenges faced in calculating the Fourier Bessel Transform of the Yukawa potential using the FFTW library's discrete sine transform (DST) routine. The analytical result for the transform is given as f^hat(k) = -4π * e1 * e2 / (kappa^2 + k^2). However, significant discrepancies between numerical and analytical results were observed, even with an increased number of points (up to 16,384). The participant, Pradipta, is advised that a sine transform is inappropriate due to the function's behavior at r=0, and a cosine transform may yield better results.

PREREQUISITES
  • Understanding of Fourier Bessel Transform and Hankel Transform of order zero
  • Familiarity with the Yukawa potential and its mathematical representation
  • Experience with FFTW library, specifically the discrete sine transform (DST) routine
  • Knowledge of numerical analysis techniques for potential functions
NEXT STEPS
  • Explore the implementation of the Fourier Bessel Transform using cosine transforms
  • Investigate numerical techniques for handling singularities in potential functions
  • Learn about the properties and applications of the Yukawa potential in physics
  • Review advanced FFTW usage and optimization strategies for numerical transforms
USEFUL FOR

Physicists, computational scientists, and numerical analysts working with potential functions, particularly those involved in quantum mechanics and electrostatics, will benefit from this discussion.

praban
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Hello,

I am trying to find Fourier Bessel Transform (i.e. Hankel transform of order zero) for Yukuwa potential of the form
f(r) = - e1*e2*exp(-kappa*r)/(r) (e1, e2 and kappa are constants). I am using the discrete sine transform routine from FFTW ( with dst routine). For this potential there is analytical result - f^hat (k) = - 4*pi *e1*e2/(kappa^2 + k^2).

I was comparing the results from numerical and analytical transform. However, I see that there is a significant difference (delR =0.3, 1st point is at 0.1, 4096 points are used for the numerical transform but the error remains even if I increase it 16384). Is there any trick to get better numerical transform for (screened) coulomb potential?

analytical numerical r
-711437635.18197799 -748996019.05573177 0.1
-275116156.66050136 -261467385.57794687 0.4
-136050696.75080600 -143418942.44284841 0.7
-79670334.886979684 -75837581.595151573 1.0
-51976687.575621709 -54872935.399959348 1.3

thanks,

Pradipta
 
A sine transform is a poor choice since your function is not zero at r=0. You might have better luck with a cosine transform. In any case you will have a problem with a numerical transform since the function is infinite at the origin.
 

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