Newton Solver max value

Question

Hi!

I am sorry if my question is too simple or if the answer is in another topic, but I couldn't find it.

Is there a way of setting up a maximum value for tolerance when solving a nonlinear problem with Newton solver?

For example, if the problem is diverging:

...
Newton iteration 15: r (abs) = 1.836e+09 (tol = 1.000e-12) r (rel) = 1.292e+06 (tol = 1.000e-07)
Newton iteration 16: r (abs) = 7.548e+08 (tol = 1.000e-12) r (rel) = 5.312e+05 (tol = 1.000e-07)
Newton iteration 17: r (abs) = 1.952e+09 (tol = 1.000e-12) r (rel) = 1.374e+06 (tol = 1.000e-07)
Newton iteration 18: r (abs) = 1.811e+10 (tol = 1.000e-12) r (rel) = 1.274e+07 (tol = 1.000e-07)
Newton iteration 19: r (abs) = 7.576e+09 (tol = 1.000e-12) r (rel) = 5.332e+06 (tol = 1.000e-07)
Newton iteration 20: r (abs) = 2.295e+09 (tol = 1.000e-12) r (rel) = 1.615e+06 (tol = 1.000e-07)
...

I already know that this solution won't converge, so setting up some maximum value (1e+05, for example) to stop this attempt would really save some time!

Thanks!

Answer 1

Hello,

I suppose you are using the Newton method available in FEniCS, as we can see in http://fenicsproject.org/documentation/tutorial/nonlinear.html

On the solver, you can define some parameters:

absolute_tolerance
convergence_criterion
error_on_nonconvergence
linear_solver
maximum_iterations
method
preconditioner
relative_tolerance
relaxation_parameter
report
krylov_solver
lu_solver

From what I see, you can define the tolerance for the newton solver by using the following:

problem = NonlinearVariationalProblem(F, u_, bcs, J)
solver  = NonlinearVariationalSolver(problem)

prm = solver.parameters
prm['newton_solver']['absolute_tolerance'] = 1E-8
prm['newton_solver']['relative_tolerance'] = 1E-7

Also, you might want to take a look at the "convergence_criterion", which by can be 'residual' (default) or 'incremental'.

If you still can not find your solution, I recommend you develop your netwon solver "manually". Also demonstrated on the same link.

du = Function(V)
u  = Function(V)  # u = u_k + omega*du
omega = 1.0       # relaxation parameter
eps = 1.0
tol = 1.0E-5
iter = 0
maxiter = 25
while eps > tol and iter < maxiter:
    iter += 1
    A, b = assemble_system(a, L, bcs_du)
    solve(A, du.vector(), b)
    eps = numpy.linalg.norm(du.vector().array(), ord=numpy.Inf)
    print 'Norm:', eps
    u.vector()[:] = u_k.vector() + omega*du.vector()
    u_k.assign(u)

Where you can evaluate the value of the variable "eps" from within the newton iteration.

Comments

Thanks for the answer!!