Hi Robert,
Thank you for taking the time to review these bits of code.

> In general, I am not too happy about LCBCs, as they work only for the nodal
> (Lagrange) basis. Remind me what kind of constraints do you use? Enforcing
> those by some penalty or Lagrange multipliers might be a more future-proof way
> to go.

Basically I have two domains with a different PDE defined on each and I want the solution
to match at a frontier shared by the two domains (a bit like ). To do so, I use something like

```
def zero_shift(ts, coors, region):
return np.zeros_like(coors[:, 0], dtype=np.float64)
match_coors = Function('match_coors', per.match_coors)
shift_fun = Function('shift_fun', zero_shift)
lcbcs = Conditions([LinearCombinationBC('lc12', [gamma1, gamma2],
{'u2.all' : 'u1.all'}, match_coors, 'shifted_periodic', arguments=(shift_fun,))])

```
I tend to find it quite convenient as Sfepy automatically builds the full stiffness matrix,
with DOFs from both domains. Could this be easily achieved with Lagrange multipliers?

> Yes, but it is not your case - you solve a different linear system in each
> iteration, right?

Yes indeed, so no need to pre-compute the factorization beforehand.

I have an additional question. Say I have assembled both the matrix and rhs (in the reduced form,
just as detailed in the previous messages) and I want to modify the essential boundary conditions
of my problem. According to , the stiffness matrix stays the same while the rhs vector is modified.
Is there a way to update the rhs vector without having to make a full `ev.eval_residual()` call?

Either way, you have answered my initial questions and my implementation seems to be working.

Cheers!

Hugo

Le lun. 2 mai 2022 à 20:01, Robert Cimrman <cimrman3@ntc.zcu.cz> a écrit :
Hi Hugo,

On 29/04/2022 15:57, Hugo Lévy wrote:
> Hi Robert,
>
> Thank you for your swift reply, very useful as always!
>
> I have dug into the code (`Problem.solve`, `Newton` class from
> solvers\nls.py) in order
> to see what's happening with the residual vector and the tangent matrix.
> Especially, since
> I only solve linear systems at each iteration of my own Newton algorithm, I
> could by-pass the
> whole Sfepy's Newton call and keep only the lines
>
> vec_dx = lin_solver(vec_r, x0=vec_x, eps_a=eps_a, eps_r=eps_r, mtx=mtx_a,
> status=ls_status)
>
> where I would have pre-computed vec_r and mtx_a the way you proposed.

Yes.

> There are a few catches, though (at least for me):
>
> - I am using LCBCs. As you pointed out in , LCBSs are not compatible
> with setting
> 'active_only' to False at the creation of the Problem instance. This means
> the matrices &
> vectors I assemble "by hand" will be reduced in size (because of the
> constrained DOFs), and so
> will be my solution vector. I will then recover the full solution with
>
> variables.set_state(vec_dx, pb.active_only)

In general, I am not too happy about LCBCs, as they work only for the nodal
(Lagrange) basis. Remind me what kind of constraints do you use? Enforcing
those by some penalty or Lagrange multipliers might be a more future-proof way
to go.

> - LCBCs modify the residual vector as well as the tangent matrix (class
> `Evaluator` methods
> `eval_residual` and `eval_tangent_matrix`), which would not be accounted
> for if I do
>
> mtx = pb.mtx_a.copy()
> mtx_k = eqs['K'].evaluate(mode='weak', dw_mode='matrix', asm_obj=mtx)
>
> So I believe creating two Problem instances (each with the same BCs) as you
> suggested is better
> as I wouldn't have to worry about it.
>
> - When pb.is_linear() is True, there is a presolve step which computes a LU
> matrix factorization
> from what I understand. Does this mean I have to call ls.presolve(mtx)
> before calling the lin_solver()
> function? Just like in  for instance.

Yes, but it is not your case - you solve a different linear system in each
iteration, right?

> So far, the code could look something like this
>
> ...
>
> pb_constant = Problem('pb_constant', equations=eqs_constant,
> active_only=True)
> pb_constant.set_bcs(ebcs=my_ebcs, epbcs=my_epbcs, lcbcs=my_lcbcs)
>
> # Compute matrix & vector of linear terms
> pb_constant.set_solver(nls)
> tss = pb_constant.get_solver()
> pb_constant.equations.set_data(None, ignore_unknown=True)
> variables = pb_constant.get_initial_state(vec=None)
> pb_constant.time_update(tss.ts)
> variables.apply_ebc(force_values=None)
> pb_constant.update_materials()
> ev_constant = pb_constant.get_evaluator()
> mtx_constant = ev.eval_tangent_matrix(variables(), is_full=True)
> rhs_constant = ev.eval_residual(variables(), is_full=True)
>
> pb_update = Problem('pb_update', equations=eqs_update, active_only=True)
> pb_update.set_bcs(ebcs=my_ebcs, epbcs=my_epbcs, lcbcs=my_lcbcs)
>
> while 1: # my Newton iteration loop
>
> # Compute matrix & vector of nonlinear terms
> pb_update.set_solver(nls)
> tss = pb_constant.get_solver()
> pb_update.equations.set_data(None, ignore_unknown=True)
> variables = pb_update.get_initial_state(vec=None)
> pb_update.time_update(tss.ts)
> variables.apply_ebc(force_values=None)
> pb_update.update_materials()
> ev_constant = pb_update.get_evaluator()
> mtx_update = ev.eval_tangent_matrix(variables(), is_full=True)
> rhs_update = ev.eval_residual(variables(), is_full=True)
>
> # Construct the full matrix / rhs
> mtx = mtx_update + mtx_constant
> rhs = rhs_update + rhs_constant
>
> # Solve
> ls = nls.lin_solver
> ls.presolve(mtx)
> vec_x = ls(rhs, x0=None, eps_a=eps_a, eps_r=eps_r, mtx=mtx, status={})
> sol = variables.set_state(vec_x, pb_update.active_only) # This might not be
> the correct syntax with the last release!
>
> if stop_condition:
> break
>
> ...
> ...
>
> I do not quite get what the `is_full=True` option does exactly. Should it
> be set to True when
> active_only is True?

The `is_full` just says whether the passed-in vector is full or reduced - in
the latter case the evaluator function calls make_full_vec(). So yes, it
interplays with active_only - when active_only is True, is_full in the Newton
solver is False, as it works with the reduced linear system.

> I would be very glad if you could give me a quick feedback on the code /
> remarks. If you think this
> looks fine (i.e. the computational cost is actually reduced by proceeding
> in this manner), I will implement
> this solution for both of my initial questions.

It looks OK at a glance.

r.

> Thanks again,
>
> Hugo
>
>  https://github.com/sfepy/sfepy/issues/785
> 
> https://sfepy.org/doc-devel/examples/diffusion-time_poisson_interactive.html
>
> Le jeu. 28 avr. 2022 à 11:08, Robert Cimrman <cimrman3@ntc.zcu.cz> a écrit :
>
>> Hi Hugo,
>>
>> On 27/04/2022 09:29, hugo.levy@gmail.com wrote:
>>> Hi Robert,
>>>
>>> I am currently facing a new difficulty using Sfepy for solving nonlinear
>> PDEs. I have coded my own
>>> Newton’s iterative method that simply consists in linearizing the PDE
>> F(u)=0 around the previous
>>> iteration u_k. I have two questions:
>>>
>>> 1) At each iteration, I create a `Problem` instance for the linearized
>> PDE and solve it the usual
>>> way in Sfepy. However, only the “linearized-nonlinear” terms need to be
>> recomputed from one iteration
>>> to another. For the sake of computational efficiency, I wish not to
>> reconstruct all matrices / rhs vectors
>>> but only the ones that need to be updated. How would you achieve that?
>> In other words, how can you
>>> store some assembled terms while re-assembling some others?
>>
>> There is no built-in support for this, but because you implement your own
>> solver, it should be possible to get without too much hassle. I suggest
>> you to
>> split your equations to linear and nonlinear parts and evaluate those
>> separately. There are several ways how to do that, e.g. (pb is a Problem
>> instance):
>>
>> eqs = {
>> 'K' : '...',
>> }
>>
>> ...
>> pb.time_update()
>> pb.update_materials()
>>
>> mtx = pb.mtx_a.copy()
>> mtx_k = eqs['K'].evaluate(mode='weak', dw_mode='matrix', asm_obj=mtx)
>>
>> etc. See also assemble_matrices() in .
>>
>> Another way would be to have two problem instance for the linear and
>> nonlinear
>> parts and evaluate those.
>>
>> 
>> https://sfepy.org/doc-devel/examples/linear_elasticity-dispersion_analysis.html
>>
>>> 2) For some reason, I need to evaluate another function, say G (a
>> residual), at each iteration in
>>> the weak form. Again, some terms in G are constant while others are
>> iteration-dependent (through
>>> their material). How would you efficiently evaluate G(u_k) at each
>> iteration? Currently, the only way
>>> I found is to do
>>>
>>> dummy_ts = TimeStepper(0, 100) # ficticious time-stepper
>>>           eqs = Equations([my_equation]) # equation associated with
>> function G
>>>           eqs.init_state()
>>>           ebcs = Conditions(my_ebcs)
>>>           epbcs = Conditions(my_epbcs)
>>>           eqs.time_update(dummy_ts, ebcs=ebcs, epbcs=epbcs)
>>>           res_vec = eqs.eval_residuals(sol) # sol is the solution at
>> current iteration
>>>
>>> which I guess this is not Sfepy’s philosophy…
>>
>> It looks OK. Note that you probably do not need to use the dummy_ts. Maybe
>> you
>> could use Problem.evaluate(), or Problem.create_evaluable(),
>> .eval_equations()
>> combo?
>>
>>> I hope my questions are clear enough. Thank you very much in advance for