Thanks Matt. For the immediate present I will probably use a basic line
search with a precheck, but if I want true line searches in the future I
will pursue option 2
On Thu, Nov 30, 2023 at 2:27 PM Matthew Knepley <knep...@gmail.com> wrote:
> On Thu, Nov 30, 2023 at 5:08 PM Alexander Lindsay <
> alexlindsay...@gmail.com> wrote:
>
>> 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
>>
>
> I see two cases:
>
> 1) There is an easy nonlinear elimination for U. In this case, you do
> this to get U_1.
>
> 2) There is only a linear elimination. In this case, I don't think the
> nonlinear system should be phrased
> only on L, but rather on (U, L) itself. The linear elimination can
> be used as an excellent preconditioner
> for the Newton system.
>
> Thanks,
>
> Matt
>
>
>> 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/>
>>>
>>
>
> --
> 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/>
>
