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. 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 -- Nikolaus Rath, Ph.D. Senior Scientist Tri Alpha Energy, Inc. +1 949 830 2117 ext 211 _______________________________________________ fenics mailing list [email protected] http://fenicsproject.org/mailman/listinfo/fenics
