ffc team mailing list archive

[Anders Logg <logg@xxxxxxxxx>]

On Sun, May 23, 2010 at 03:32:51PM +0200, Kristian Oelgaard wrote:
> On 23 May 2010 15:08, Garth N. Wells <gnw20@xxxxxxxxx> wrote:
> >
> >
> > On 23/05/10 13:58, Kristian Oelgaard wrote:
> >>
> >> On 23 May 2010 12:51, Kristian Oelgaard<k.b.oelgaard@xxxxxxxxx>  wrote:
> >>>
> >>> On 23 May 2010 12:14, Garth N. Wells<gnw20@xxxxxxxxx>  wrote:
> >>>>
> >>>> Where are we at with attaching data to integrals? The quadrature order
> >>>> can
> >>>> be attached, but I's like to do:
> >>>>
> >>>>    dS({"representation": "tensor"})
> >>>>
> >>>> Is this supported by FFC?
> >>
> >> It looks like this is already supported in FFC.
> >> Note that the representation must be the same on each subdomain (this
> >> is also a requirement for quadrature_degree).
> >>
> >>  dS(0, {"representation": "tensor"}) + dS(0, {"representation":
> >> "quadrature"})
> >>
> >> is not possible, but
> >>
> >>  dS(1, {"representation": "tensor"}) + dS(0, {"representation":
> >> "quadrature"})
> >>
> >
> > OK.
> >
> >    dS(0, {"representation": "tensor"}) \
> >  + dS(0, {"representation": "tensor"})
> >
> > works, but
> >
> >    dS({"representation": "tensor"}) \
> >  + dS({"representation": "tensor"})
> >
> > doesn't. Seems like a bug.
>
> It's because domain_id is not a keyword argument in the __call__
> function of Measure.
> I changed this such that you can do:
>
>  dS(metadata={"representation": "tensor"}) \
>  + dS(metadata={"representation": "tensor"})
>
> > Turns out I can't use this anyway because facet normals are not supported by
> > FFC with the tensor representation,
> >
> >    ffc.tensor.monomialextraction.MonomialException: No handler defined for
> > terminal FacetNormal
>
> I don't know if this is just because it is not implemented yet, or
> because it is not possible to support this with
> tensor representation.

It's possible to support (and has been before), but I removed some
"extra features" of the tensor representation since those were
supported by quadrature anyway.

--
Anders

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