Solving the 3D Poisson equation

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SUMMARY

This discussion focuses on solving the 3D Poisson equation using spherical coordinates and spherical harmonics. The equation addressed is ##d^2\phi/dr^2 + 1/rd\phi/dr - l*(l+1)/r^2 = \rho (r)##, where ##\phi## represents the gravitational potential and ##\rho## the mass density. The user employs LU-decomposition for solving the equation and utilizes a hyperbolic transformation to manage the infinite grid. The discussion highlights challenges in setting the potential to zero at infinity and concludes with a successful solution, prompting others to share their insights.

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  • Understanding of the 3D Poisson equation
  • Familiarity with spherical coordinates and spherical harmonics
  • Knowledge of LU-decomposition methods
  • Experience with hyperbolic transformations in mathematical modeling
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  • Explore numerical methods for boundary value problems
  • Learn about potential theory in gravitational fields
  • Investigate alternative coordinate transformations for infinite domains
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Mathematicians, physicists, and engineers working on gravitational modeling, numerical analysis, or anyone interested in solving complex PDEs in spherical coordinates.

debilo
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TL;DR
I try to solve 3D Poisson equation with boundaries condition at infinity.
Hello !

I want to solve the 3D Poisson equation using spherical coordinates and spherical harmonics.
First I must solve this : ##d^2\phi/dr^2 + 1/rd\phi/dr - l*(l+1)/r^2 = \rho (r)## with ##\phi (\infty ) = 0## (here ##\phi## is the gravitationnal potential and ##\rho## is the mass density).
To deal with the infinite grid I tried to use the hyperbolic transformation ##r = X*arctanh(R)## with ##X## a constant and ##R## between 0 and 1.
I use a LU-decomposition method to solve this. For ##\phi (0)## I apply symmetrical conditions.
I tried to solve the problem in the case of a spherical mass distribution. I compared the calculated potential with the theoretical potential.
I find a pretty good result when I fix the value in rmax.
But for ##\phi (\infty )## I have a problem. In fact I obviously get in trouble when I let R go to 1 (or 0) ; so I take ##Rmin = 0+\varepsilon## and ##Rmax = 1-\varepsilon##.

My question is : do you know a trick to really set the value of the potential to zero at infinity ?
 
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I finally found a solution it's ok !
 
Are you going to post it here? I'm sure others would be interested and they may see something you can to make it even better.
 
debilo said:
I finally found a solution it's ok !
Hi, can you please post the solution?
 

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