I don't know if it's me, but there are a lot of things I don't understand. Among these things :

1) The geometry : " torque that a razor thin disc applies onto a razor thin ring" ???

2) "The outer box imposes a symmetry which is not in the sought solution" For, me the problem is symetric, but the boundary conditions at Infinity are not respected. So it's another problem

@andre point 2 is a typo. it now reads 0 <= x <= 1 && -1 <= y <= 1. But its not at infinity.

3) "This is imposing fixed potential on the disk not fixed density" I Don't understand. My background in mechanics is low, but not in electrostatics, and the problem should be the same. Is potential imposed at X<x<1/2 or is somthing else imposed ? In this caes what ?

@andre I was not sure about this: is it really equivalent to fix the potential or the density? If so I ll remove this point from my question.

Hello and thanks for your time

do you think I should remove the mention of torquing?

My implicit question behind this question is how to impose boundary condition at infinity and use FEM?

@chris AhAh !!! Gravitational Potential ? For me Razor = "rasoir" (in french) !!

razor fin =disque infiniment fin

OK thanks. Concerning 1) I think you should write : "that I am interested in the gravitational potential generated by a ring and a disc?"

ok

I have put a cartoon to make it clearer.

Concerning "remove the mention of torquing" : For me "torquing " is related to "rotation", so it's not related to your question ???

I'm discovering the photos you have just added.

... I trying to understand ... (suspens !)

Can you confirm that the disk and the ring are not coplanar ?

Well that's the idea but I think I know how to deal with this once I have the potential everywhere.

Two misaligned discs or ring will torque each other (ils exercent un couple l un sur l autre)

for instance the alt azimuthal torque for a disc has this structure !Mathematica graphics

I'm still trying to understant, but I must go.

I'll be back in 1 hour.

Here is a reminder (graphic 1):

thanks: if they are co planar there is no torque;

but say in this configuration yes. !Mathematica graphics

or may be more clear here !Mathematica graphics

OK. Now my ideas are relatively clear. I'm thinking how to explain them concisely. But the conclusion is a little bit desappointing. It is : you don't need FEM. To obtain the gravity, it suffices to integrate the matter quantity over the space with the ponderation 1/distance^2

(cf Newton, who is "new" only in his name)

While I agree in principle (and I do give the analytical result in my post), I am still interested technically in FEM being able to reproduce this result. Thanks for your help anyway.