dolfin team mailing list archive

["Robert C. Kirby" <kirby@xxxxxxxxxxxx>]

There is a "trace_tabulate" method in FIAT that computes the values  
of the basis functions on a boundary facet (or the interior, if you  
like), FFC should make use of that and then throw away things that  
are zero.


On Apr 12, 2006, at 10:22 AM, Johan Hoffman wrote:

> Ok, good to know. Any ideas on what would be the best solution?
>
> /Johan
>
>
> > You most definitely don't want to use the d-1 dimensional element of
> > the same type.  Works great for Lagrange (and actually Hermite, for
> > what it's worth), but is disastrously catastrophic for everything  
> > else.
> >
> >
> > On Apr 12, 2006, at 8:29 AM, Garth N. Wells wrote:
> >
> > > On Wed, 2006-04-12 at 15:09 +0200, Garth N. Wells wrote:
> > > > On Wed, 2006-04-12 at 14:30 +0200, Johan Hoffman wrote:
> > > >
> > > > > Apart from the issue of duplicating code with or without
> > > > > templates, what
> > > > > do we need to figure out to get the assembly of boundary integrals
> > > > > correct?
> > > > >
> > > > > Do we have a clear way to obtain the FE basis for the boundary of
> > > > > a cell
> > > > > given the basis of the cell? Do we use the corresponding basis
> > > > > for for d-1
> > > > > dimensions?
> > > > >
> > > > > /Johan
> > > >
> > > > This will have to come from FFC won't it? It might be less  
> > > > efficient,
> > > > but we could work with the finite element of the whole cell and
> > > > perhaps
> > > > the edge/face number is enough for FFC to do the rest.
> > >
> > > Thinking about this some more, it might be better not to restrict  
> > > the
> > > finite element basis to the boundary in order to work with problems
> > > where derivatives of the basis functions appear in the boundary
> > > integrals, and hence involve all nodes of the element. Similar
> > > story for
> > > non-conforming elements.
> > >
> > > Garth
> > >
> > >
> > > > I noticed that
> > > > the FFC-produced function eval takes "unsigned int boundary" as an
> > > > argument, but doesn't do anything with it yet.
> > > >
> > > > Garth
> > > >
> > > >
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