ffc team mailing list archive

[Kristian Oelgaard <k.b.oelgaard@xxxxxxxxxx>]

Quoting Anders Logg <logg@xxxxxxxxx>:

> On Tue, Apr 29, 2008 at 08:29:18AM +0100, Garth N. Wells wrote:
> >
> >
> > Kristian Oelgaard wrote:
> >> Quoting Anders Logg <logg@xxxxxxxxx>:
> >>
> >>> On Mon, Apr 28, 2008 at 08:25:53PM +0200, Kristian Oelgaard wrote:
> >>>
> >>>> # 2D plasticity, cases
> >>>> 1) 1st order elements, mesh(1000, 1000)
> >>>> 2) 2nd order elements, mesh(500, 500)
> >>>> 3) 3rd order elements, mesh(250, 250)
> >>>> 4) 4th order elements*, mesh(125, 125)
> >>>>
> >>>> *Note that because the bilinear form in this case is a 9th order form
> we
> >>> take
> >>>> the number of quadrature points equal to 5 when declaring the
> quadrature
> >>> element.
> >>>
> >>> Another optimization would be to reduce the number of quadrature
> >>> points. If your method is order p, then you only need to integrate the
> >>> form with quadrature exact for degree p polynomials. So in particular,
> >>> you don't necessarily need many quadrature points just because the
> >>> integrand has many factors.
> >>
> >> I don't quite follow. If the displacement field is 4th order, the gradient
> is
> >> 3rd order and the bilinear form is 9th order? And since 5 integration
> points
> >> can integrate 2*5-1 = 9th order polynomials exactly I would say we should
> use
> >> 5 integration points in each direction. This is also what the
> >> __init_quadrature() and __compute_degree() functions in
> monomialintegration.py
> >> will return.
> >>
> >
> > I think the point Anders is making is that you don't need exact
> quadrature.
> >
> > Garth
> 
> Yes. The quadrature order should be chosen to retain the convergence
> of the numerical scheme. Integrating exactly may be overkill. That
> will speedup quadrature even further.

OK. If FIAT could return only 4 quadrature points instead of 2x2x2 = 8, the
integration is still exact and it would make life a lot easier for both tensor
and quadrature representation.

Kristian

> -- 
> Anders
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