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Re: evaluate_basis update

 

k.b.oelgaard@xxxxxxxxx wrote:
Hi,

I'm running some tests on evaluate_basis with different elements and here are some results.

The following elements are no longer supported by FIAT:
Crouzeix-Raviart and Brezzi-Douglas-Fortin-Marini, will they be supported again?



I would say that BDFM is not important, but that CR is.


The following elements could not be compiled by gcc because of a problem with and undefined variable 'copy' in evaluate_dof() and evaluate_dofs():
 FiniteElement('Brezzi-Douglas-Marini', Cell('triangle', 1, Space(2)), 2)
 FiniteElement('Brezzi-Douglas-Marini', Cell('triangle', 1, Space(2)), 3)
FiniteElement('Brezzi-Douglas-Marini', Cell('tetrahedron', 1, Space(3)), 2) FiniteElement('Brezzi-Douglas-Marini', Cell('tetrahedron', 1, Space(3)), 3) FiniteElement('Nedelec 1st kind H(curl)', Cell('triangle', 1, Space(2)), 2) FiniteElement('Nedelec 1st kind H(curl)', Cell('triangle', 1, Space(2)), 3) FiniteElement('Nedelec 1st kind H(curl)', Cell('tetrahedron', 1, Space(3)), 3)


Known potential issue. Will fix.


A funny thing to note is that
FiniteElement('Nedelec 1st kind H(curl)', Cell('tetrahedron', 1, Space(3)), 2)
compiles fine.

Values differ for the following DG element:
FiniteElement('Discontinuous Lagrange', Cell('interval', 1, Space(1)), 2) FiniteElement('Discontinuous Lagrange', Cell('interval', 1, Space(1)), 3) FiniteElement('Discontinuous Lagrange', Cell('interval', 1, Space(1)), 4) FiniteElement('Discontinuous Lagrange', Cell('tetrahedron', 1, Space(3)), 2) FiniteElement('Discontinuous Lagrange', Cell('tetrahedron', 1, Space(3)), 3) FiniteElement('Discontinuous Lagrange', Cell('tetrahedron', 1, Space(3)), 4) FiniteElement('Discontinuous Lagrange', Cell('triangle', 1, Space(2)), 2) FiniteElement('Discontinuous Lagrange', Cell('triangle', 1, Space(2)), 3) FiniteElement('Discontinuous Lagrange', Cell('triangle', 1, Space(2)), 4)
but Marie changed the ordering of dofs, so that is not big mystery.


I refuse to take the blame for changing the ordering of the dofs! ;)

I checked some values manually and it looks like the values has just been shifted around so the result should be OK.



Good!


Values also differ for the following elements:
FiniteElement('Nedelec 1st kind H(curl)', Cell('tetrahedron', 1, Space(3)), 2)
 FiniteElement('Raviart-Thomas', Cell('tetrahedron', 1, Space(3)), 2)
 FiniteElement('Raviart-Thomas', Cell('tetrahedron', 1, Space(3)), 3)
 FiniteElement('Raviart-Thomas', Cell('triangle', 1, Space(2)), 2)
 FiniteElement('Raviart-Thomas', Cell('triangle', 1, Space(2)), 3)
For the Nedelec element values look the same, but they appear in different places.

If I call tabulate on the FIAT element (reference) I get matching results as when I call evaluate_basis(), so I guess it is OK. Marie, has the ordering of dofs changed for Nedelec?


Don't know actually.  I can check evaluate_dof.



For the Raviart-Thomas elements the story is almost the same. The only difference is that for e.g.,
 FiniteElement('Raviart-Thomas', Cell('triangle', 1, Space(2)), 2)
I get values like '0.44444 0.55556 0.22222 -0.22222 -0.33333' in the 'old' version of FFC and FIAT, but in the new FFC (and the new FIAT) I get values like: 0.66667 0.33333 -0.00000 -0.00000 -0.33333' when I tabulate the basis at the same point. Marie, has something changed for RT elements?


Yes. Rob changed the interior degrees of freedom for polynomial degrees > 1 from integral moments to point values. Hence the basis
functions have changed.

(Note that there is a unit test for RT2 on a triangle that passes :) )

--
Marie


For both old and new FFC FIAT.tabulate() and evaluate_basis() agrees so I guess the values are OK.

If the above issues can be confirmed and addressed It looks like evaluate_basis can be trusted.

Kristian
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