dolfin team mailing list archive

[Johan Hake <johan.hake@xxxxxxxxx>]

On Tuesday September 28 2010 09:08:01 Marie Rognes wrote:
> On 28. sep. 2010 17:16, Johan Hake wrote:
> > 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
> 
> So, you want a simpler way of doing it, but not a really simple way of
> doing it ;) ?

The point is that the flux quantity is nice to have computed, as I use it for 
further computation. The above form works fine.

> I don't think I have any more suggestions at the moment -- could you
> remind me why the integration is a problem again? Or, what would you
> like to write?

No problem. Just overkill (I assume...). I am calculating a scalar quantity 
(r0) from other scalar quantities (r1, r2) using a form. But the only way I 
can use a form to evaluate these are by an integral. An integral is over a 
spatial domain. r0, r1 and r2 are just values on R not on the domain. Why 
should I then have to integrate over the domain to calculate r0?
 
Because nothing else is implemented in UFL/UFC/DOLFIN. I am fine with this, 
but it is a unintutive and it might be usefull to introduce some other measure 
for these guys.

> > 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.
> 
> Sounds like a bit of an overhead there, yes ...

Yes and the ODE solver can probably not be used to solve particular stiff 
ODEs. BUT, and here comes the raisin in the hot dog, it can be included in the 
automated linearization so you can get quadratic convergence in a coupled 
ODE/PDE system.

Johan