ffc team mailing list archive

[Anders Logg <logg@xxxxxxxxx>]

On Tue, Dec 01, 2009 at 10:36:42AM +0000, Garth N. Wells wrote:
>
>
> Anders Logg wrote:
> > On Tue, Dec 01, 2009 at 09:41:04AM +0100, Marie Rognes wrote:
> >> Anders Logg wrote:
> >>> On Tue, Dec 01, 2009 at 09:15:53AM +0100, Marie Rognes wrote:
> >>>> Anders Logg wrote:
> >>>>> On Mon, Nov 30, 2009 at 11:01:59PM +0000, Garth N. Wells wrote:
> >>>>>> Anders Logg wrote:
> >>>>>>> We still haven't decided on the correct strategy for choosing the
> >>>>>>> degree of an unspecified element.
> >>>>>>>
> >>>>>>> What we have now looks at the total degree of the form and then sets
> >>>>>>> the degree accordingly. This doesn't really work weoull and the reason
> >>>>>>> is quite simple: We can't figure out the total degree correctly if we
> >>>>>>> don't know the degree of the coefficient.
> >>>>>>>
> >>>>>>> So my new suggestion is the following. We simply scan all elements in
> >>>>>>> the form with specified degrees and set the degree to the maximum
> >>>>>>> degree among the elements.
> >>>>>>>
> >>>>>> Or should it be the maximum degree of the test and trial functions?
> >>>>> Yes, that's basically what happens now. It looks at everything that
> >>>>> has a degree so it looks at the test and trial functions but also at
> >>>>> any coefficients that may happen to have a degree.
> >>>>>
> >>>>> That's useful for say v*f*g*dx if one of f and g happen to have a
> >>>>> degree specified. Then the degree for the other needs to be the same.
> >>>>>
> >>>> Why?
> >>> Since that's how the generated quadrature code works. It loops over
> >>> quadrature points for an integrand and then it's useful if everything
> >>> in the integrand is evaluated at those points.
> >>>
> >> Ok.
> >
> > I have made a new change of the selection strategy.
> >
>
> Does the change in UFL affect the polynomial degree of the integrand as
> computed by UFL?  I see that it's no longer taking into account
> derivatives, so if v and u1 are P1, will UFL tell the form compiler that
>
>    dot(grad(v), grad(u))
>
> is zero-order or second-order?

It does now.

For the above case, the quadrature degree is chosen to

  2*(q - 1) = 2q - 2 = 0 for q = 1

There are still some problems when "Quadrature" is selected instead of
"Lagrange" for coefficients, but I need Kristian's help to sort that
out.

I expect we will need to modify the degree selection further when
people start using it.

--
Anders


> > For quadrature, FFC now looks at the maximum degree of anything in the
> > form (including test and trial functions and coefficients).
> >
> > For tensor representation, FFC first looks at the maximum degree to
> > select the degree but then looks at the total degree to select the
> > quadrature it uses at compile time.
> >
> > I think this looks ok now and both FFC and UFL are ready for a new
> > release. There's still the bug fix Kristian is working on for using
> > quadrature elements for coefficients instead of Lagrange, but we
> > don't need to wait with the release.
> >
>
> OK.
>
> > FFC is still verbose about its selection of degrees so keep an eye on
> > what FFC prints during compilation and report any strange behavior.
> >
>
> Yes, good to keep it on initially.
>
> Garth
>
>
>

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