ffc team mailing list archive

[Jake Ostien <tostien@xxxxxxxxx>]

Jake Ostien wrote:
> Anders Logg wrote:
>   
> > On Thu, Jan 10, 2008 at 09:29:56AM -0500, Jake Ostien wrote:
> >   
> >     
> > > Hi,
> > >
> > > I'd like to write a form based on the lifting operator.  If the lifting 
> > > operator (R) is defined as
> > >
> > >     -\int_Gamma avg(A):jump(B) dS = \int_Omega A:R(B) dV
> > >
> > > Is there any way I can define a form such as
> > >
> > >     \int_Omega R(A):R(B) dV
> > >
> > > Where e.g. A and B are BasisFunctions?
> > >
> > > I can do this already for a known Function, say H, where I say
> > >
> > >     V = TestFunction
> > >     U = TrialFunction
> > >
> > >     H = Function
> > >
> > >     a = dot(V,U)*dx
> > >     L = dot(avg(V),jump(H))*dS
> > >
> > > Then U is the lifted H, projected onto the basis of V.
> > >
> > > Should this same approach work for BasisFunctions?
> > >
> > > Jake
> > >     
> > >       
> > I don't understand what you want to do. The only way I see to compute
> > a projection (or lifting) is to define a bilinear form and linear form
> > as you do and solve a linear system.
> >
> >   
> >     
> Yes, I see.  Here's the problem.  The lifting operator formulation looks 
> like
>     \int_Omega R(V):R(H) dV
>
> For the R(V) term, I can lift the BasisFunction V and project it to get 
> a DiscreteFunction in DOLFIN, but then when I pass in those coefficients 
> as arguments to the lifting operator form, how can I tell FFC that those 
> coefficients are for the TestFunction?  If I could do this then I could 
> just do the projections as a separate step and use the results.
>
>   
Actually, this isn't right.  What I suggested was the projection

  -\int_gamma avg(V):jump(V) dS = \int_Omega V:U dV

where U = R(V), which is pretty much nonsense.

I'll have to figure out some other way.

Thanks,
Jake


> Jake
>
>
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