Should d2udt2 be set to 0 when Dirichlet boundary condition is applied?

[Wayne901]

I have a little suggestion/question about ex23. When the dirichlet boundary conditions are applied, dudt and d2udt2 should be 0. However when I run the wave euqation example ex23 and output the d2udt2 vector in ImplicitSolver function, d2udt2 is not zero at the boundary DoFs. I set d2udt2's boundary entries to 0, and the final result is slightly different to the original code (./ex23 -m ../../data/disc-nurbs.mesh -r 3 -o 4 -tf 5 --dirichlet)
Result from original code:

Result when boundary entris of d2udt2 set to 0:

I think the nonzero boundary DoFs in d2udt2 comes from solving the linear system
$[M+\beta (\Delta t)^2K]\frac{d^2u(t+\Delta t)}{dt^2}=-Ku_{ImplicitSolve}$? But why would the solution on boundary entries be nonzero when $u_{ImplicitSolve}$'s are 0 there?

Please let me know if my choice of setting d2udt2's boundary DoFs 0 is correct.

Thank

[Wayne901]

May someone give some hint on this problem? I'm still confused about it. The different result is not observed when t_final <= 4. I set d2udt2 to 0 after T_solver with

T_solver.Mult(z, d2udt2);
for (int i = 0; i < ess_tdof_list.Size(); i++)
  d2udt2[ess_tdof_list[i]] = 0.0;

[Wayne901]

I wrote another wave equation code based on the coupled first-order equations, like ex10.
The result at t=5 is same as the ex23 with d2udt2 set to 0.

[IdoAkkerman]

I have tried to address the issue in a small PR #4233. I added as step after solving for the acceleration where the acceleration is hardwired set to zero.

I think, this should technically not have been necessary, as the solve should have resulted in zero accelerations on those places. But apparently round-off and/or the approximate nature of the linear solves create for excessive slip of the solution.

[IdoAkkerman]

I have checked the BC values and after zeroing the accelerations the solution values at the BC were of the order e-105 which is effectively zero.