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.h...@gmail.com> 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 ;) ? 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? > 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 ... -- Marie _______________________________________________ Mailing list: https://launchpad.net/~dolfin Post to : dolfin@lists.launchpad.net Unsubscribe : https://launchpad.net/~dolfin More help : https://help.launchpad.net/ListHelp
