dolfin team mailing list archive

[Johan Hake <johan.hake@xxxxxxxxx>]

On Tuesday September 28 2010 04:46:30 Marie Rognes wrote:
> On 25. sep. 2010 22:02, Johan Hake wrote:
> > On Saturday September 25 2010 00:24:12 Marie E. Rognes wrote:
> >> On 25. sep. 2010, at 02:37, Johan Hake <johan.hake@xxxxxxxxx> wrote:
> >>> Hello!
> >>> 
> >>> I have been experimenting with forms including spaces of Real. I am
> >>> using the Real space to calculate average values of other species
> >>> across the boundaries. These values are then used to cacluate derived
> >>> values such as fluxes across a boundary. In this way I manage to
> >>> connect two sepatated meshes (stored in the same mesh) via the
> >>> calculated flux. As everything is contained in one form I can
> >>> linearize it an use it in a Newton solver, which converges nicely.
> >>> 
> >>> In this procedure I have used unknowns of Reals to caluclate other
> >>> Reals. This is somewhat combersome when I have to express this using
> >>> an integral.
> >>> 
> >>> Consider:
> >>>  V = MixedFunctionSpace([FunctionSpace(mesh, "CG", 1)]+
> >>>  
> >>>                         [FunctionSpace(mesh, "R", 0)]*3)
> >>>  
> >>>  u, r0, r1, r2 = split(Function(V))
> >>>  v, v0, v1, v2 = TestFunctions(V)
> >>>  ...
> >>>  L0 = (r0-(r2-r1))*v0*ds(1)
> >>> 
> >>> Here I express r0 in terms of r1 and r2. However in the form I need to
> >>> integrate the expression over an arbitrary domain in the mesh. I am not
> >>> interested in this, as the value is not spatialy varying. Do you have
> >>> any suggestions of how to avoid this or to do it in some other manner?
> >> 
> >> It is not entirely clear to me which values that are unknown above, but
> >> if everything is known or you are just operating on real numbers, I
> >> imagine that extracting the value of the Function using values() could
> >> be simpler?
> > 
> > r0, r1, and r2 are unknowns together with the field u. r1 and r2 are the
> > mean of the field across a boundary, and r0 are the flux between the two
> > distinct domains.
> > 
> > A more complete form would be:
> >   # Boundary forms
> >   L1 = (r1-u)/area[1]*v1*ds(1)
> >   L2 = (r2-u)/area[2]*v2*ds(2)
> >   L0 = (r0-(r2-r1))*v0*ds(1)
> >   Lu = r0*v*ds(1) - r0*v*ds(2)
> >   
> >   # Stiffnes + Boundary
> >   L = inner(grad(u),grad(v))*dx + L0 + L1 + L2 + Lu
> > 
> > My question relates to the algebraic relation of the reals in:
> >   L0 = (r0-(r2-r1))*v0*ds(1)
> > 
> > which really does not need an integral.
> 
> Do you really want:
> 
>     r0 = (r2 - r1)

Yes.

> If yes, why not replace r0 by (r2 - r1)?

That is true but I figure I want to compute that value for book keeping :P

On a related topic:
The Real space can be used to compute ODEs coupled to PDEs by expressing them 
using a form. I know several applications that this might be usefull. But then 
we get the same "problem" when a scalar value are computed by integrating over 
a domain.

Johan





> 
> --
> Marie