On Wed, 22 Jan 2014 11:39:30 -0800 Nikolaus Rath <[email protected]> wrote:
> Hi Jan, > > It was my impression from the other thread that I could handle the > problem of multiple solutions by giving the nullspace to the solver. > But I'll try the Laplace constraint as well. > > But then, even with u constrained, I still need to somehow put the > same constraint on the test functions, don't I? That's the part I'm > struggling with. Sure. This is why it seems logical to me to constraint both trial and test space by Laplace equation. Nevertheless I did not think it over a much. Jan > > > Best, > Nikolaus > > > On 01/22/2014 11:22 AM, Jan Blechta wrote: > > Actually there is a very good reason to constraint also u, say by > > Laplace equation in the interior, as your original problem is > > singular on a usual H^1 space - there are infinintely many > > solutions to the problem. So you need to pick some. > > > > Jan > > ------------------------------------------------------------------------ > > From: Nikolaus Rath <mailto:[email protected]> > > Sent: 22/01/2014 19:27 > > To: Jan Blechta <mailto:[email protected]> > > Cc: [email protected] <mailto:[email protected]> > > Subject: Re: [FEniCS] How to impose constraints on test functions? > > > > On 01/21/2014 01:57 AM, Jan Blechta wrote: > >>>>>> However, I am still struggling to combine the finite element > >>>>>> method with Lagrange multipliers. I think I have a good handle > >>>>>> on Lagrange multipliers for constrained optimization of a > >>>>>> scalar function or integral, but I fail to transfer this to FE. > >>>>> > >>>>> Ok, potential for Poisson problem is > >>>>> > >>>>> \Psi(u) = 1/2 \int |\nabla u|^2 - L(u) > >>>> > >>>> Ah, the potential is the starting point. Thanks! > >>>> > >>>>> So if you want to minimize \Psi on V = H^1(\Omega) subject to > >>>>> constraint \int u = 0, you do can try to find a minimum (u, c) > >>>>> \in (V \times R) of > >>>>> > >>>>> \Psi(u) - c \int u > >>>>> > >>>> > >>>> My apologies if I'm slow, but why would I want to find a minimum > >>>> (u,c) \in (V * R)? It seems to me that I don't want to find a > >>>> specific value c -- I want a minimum u \in V \forall c. > >>> > >>> Nevermind that question. Since c is the Lagrange multiplier of > >>> course we need to solve for it. I got confused because I didn't > >>> see any additional constraint equations being used to actually > >>> determine the value. But that is just because of the special case > >>> \int u = 0, correct? > >>> > >>> For the more general case \int u = u0, am I right that we'd need > >>> to set up an extra term in the linear form? > >>> > >>> (u, c) = TrialFunction(W) > >>> (v, d) = TestFunctions(W) > >>> g = Expression("-sin(5*x[0])") > >>> a = (inner(grad(u), grad(v)) + c*v + u*d)*dx > >>> L = g*v*ds + Constant(u0) * d * dx > >> > >> Correct. > > > > Ah, good :-). Thanks! > > > > > > Could you also give me a similar hint for my original problem? > > > > To recap: I want to find u on the boundary such that with > > > > L = dot(f, grad(v)) * dx > > a = u * dot(grad(v), FacetNormal(mesh)) * ds > > > > L(v) == a(u, v) holds for all v that satisfy div(grad(v)) = 0 > > (without any boundary conditions on v). > > > > > > I tried to start setting up the potential problem: > > > > \Psi(u) = 1/2 a(u,u) - L(u) > > > > but in contrast to the Poisson-Neumann example, now the constraint > > isn't on the solution u but on the test functions v, so I'm not > > sure how I can add the Lagrange multiplier here... As far as I can > > see, there's no reason to constrain the variation of u to satisfy > > Laplace's equation when looking for a stationary point of \Psi. > > > > What am I missing this time? > > > > > > Thanks so much for all your help! > > Nikolaus > > _______________________________________________ fenics mailing list [email protected] http://fenicsproject.org/mailman/listinfo/fenics
