Hi Matt, your derivation is spot on. However, the problem is not linear, which is why I am using SNES. So you need to replace
U = A^{-1} f - A^{-1} B L with dU = A^{-1} f - A^{-1} B dL On Thu, Nov 30, 2023 at 1:47 PM Matthew Knepley <knep...@gmail.com> wrote: > On Thu, Nov 30, 2023 at 4:23 PM Alexander Lindsay < > alexlindsay...@gmail.com> wrote: > >> If someone passes me just L, where L represents the "global" degrees of >> freedom, in this case they represent unknowns on the trace of the mesh, >> this is insufficient information for me to evaluate my function. Because in >> truth my degrees of freedom are the sum of the trace unknowns (the unknowns >> in the global solution vector) and the eliminated unknowns which are >> entirely local to each element. So I will say my dofs are L + U. >> > > I want to try and reduce this to the simplest possible thing so that I can > understand. We have some system which has two parts to the solution, L and > U. If this problem is linear, we have > > / A B \ / U \ = / f \ > \ C D / \ L / \ g / > > and we assume that A is easily invertible, so that > > U + A^{-1} B L = f > U = f - A^{-1} B L > > C U + D L = g > C (f - A^{-1} B L) + D L = g > (D - C A^{-1} B) L = g - C f > > where I have reproduced the Schur complement derivation. Here, given any > L, I can construct the corresponding U by inverting A. I know your system > may be different, but if you are only solving for L, > it should have this property I think. > > Thus, if the line search generates a new L, say L_1, I should be able to > get U_1 by just plugging in. If this is not so, can you write out the > equations so we can see why this is not true? > > Thanks, > > Matt > > >> I start with some initial guess L0 and U0. I perform a finite element >> assembly procedure on each element which gives me things like K_LL, K_UL, >> K_LU, K_UU, F_U, and F_L. I can do some math: >> >> K_LL = -K_LU * K_UU^-1 * K_UL + K_LL >> F_L = -K_LU * K_UU^-1 * F_U + F_L >> >> And then I feed K_LL and F_L into the global system matrix and vector >> respectively. I do something (like a linear solve) which gives me an >> increment to L, I'll call it dL. I loop back through and do a finite >> element assembly again using **L0 and U0** (or one could in theory save off >> the previous assemblies) to once again obtain the same K_LL, K_UL, K_LU, >> K_UU, F_U, F_L. And now I can compute the increment for U, dU, according to >> >> dU = K_UU^-1 * (-F_U - K_UL * dL) >> >> Armed now with both dL and dU, I am ready to perform a new residual >> evaluation with (L0 + dL, U0 + dU) = (L1, U1). >> >> The key part is that I cannot get U1 (or more generally an arbitrary U) >> just given L1 (or more generally an arbitrary L). In order to get U1, I >> must know both L0 and dL (and U0 of course). This is because at its core U >> is not some auxiliary vector; it represents true degrees of freedom. >> >> On Thu, Nov 30, 2023 at 12:32 PM Barry Smith <bsm...@petsc.dev> wrote: >> >>> >>> Why is this all not part of the function evaluation? >>> >>> >>> > On Nov 30, 2023, at 3:25 PM, Alexander Lindsay < >>> alexlindsay...@gmail.com> wrote: >>> > >>> > Hi I'm looking at the sources, and I believe the answer is no, but is >>> there a dedicated callback that is akin to SNESLineSearchPrecheck but is >>> called before *each* function evaluation in a line search method? I am >>> using a Hybridized Discontinuous Galerkin method in which most of the >>> degrees of freedom are eliminated from the global system. However, an >>> accurate function evaluation requires that an update to the "global" dofs >>> also trigger an update to the eliminated dofs. >>> > >>> > I can almost certainly do this manually but I believe it would be more >>> prone to error than a dedicated callback. >>> >>> > > -- > What most experimenters take for granted before they begin their > experiments is infinitely more interesting than any results to which their > experiments lead. > -- Norbert Wiener > > https://www.cse.buffalo.edu/~knepley/ > <http://www.cse.buffalo.edu/~knepley/> >