Hi Kramer, Thank you for the reply. The only problem I see with the solution you suggest, is that both I and lambda will be trial functions, so both terms should be on the bilinear part of the variational form. I could, of course, define a residual and solve everything using nonlinear solvers, but I don't know how efficient that would be. Best, Anna. Dr Anna Kalogirou Research Fellow School of Mathematics University of Leeds http://www1.maths.leeds.ac.uk/~matak/ On 04/08/16 17:29, Stephan Kramer wrote:
On 27/07/16 11:01, Anna Kalogirou wrote:
Dear all,
I am trying to solve the system found in the attached pdf as a mixed system. I want to solve the same problem as the one found here <https://bitbucket.org/annakalog/buoy2d/src/14334d3c20b9f10ed7c1246cde9e3cb60b1c75e4/Inequality%20constraint/?at=master>,
but with the use of Schur complements and not linear algebra.
In a previous discussion about this problem, Lawrence mentioned that the test function for integral(lambda*Theta(x-Lp)dx) needs to be considered as coming from the real space of constant functions. Could you please elaborate on this?
My guess at what Lawrence means is (it's how I would probably approach it), is that you add an additional equation that says
(1) I = integral(lambda*Theta(x-Lp)dx)
and then substitute I in your third equation. Equation (1) can be turned into a finite element weak formulation by considering the function space of functions that are constant over the entire domain This function space is just 1-dimensional and therefore equivalent to the space of reals R and hence we denote this function space of constants simply as R. Then we can consider I to be a trial function in R, and with a test function v4 in R, we can write:
v4*lambda*Theta(x-Lp)*dx == v4*I*dx/area(domain)
This is a proper finite element equation that you could solve in conjuction with the other equations. I believe however that this function space R is not currently implemented in Firedrake. It is available in fenics, and is on the wishlist.
Cheers Stephan
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