Newtoniteration output / see solution / initial data

[Noya]

Hello,
can someone explain me the output from the newton iteration??
I want to solve a nonlinear problem and use

solve(F == 0, m, bc)
yh, ph = m.split()

And I have the following output. Can someone explain to me what this means?

No Jacobian form specified for nonlinear variational problem.
Differentiating residual form F to obtain Jacobian J = F’.
Solving nonlinear variational problem.
Newton iteration 0: r (abs) = 8.011e+01 (tol = 1.000e-10) r (rel) = 1.000e+00 (tol = 1.000e-09)
Newton iteration 1: r (abs) = 3.742e-01 (tol = 1.000e-10) r (rel) = 4.671e-03 (tol = 1.000e-09)
Newton iteration 2: r (abs) = 1.308e-03 (tol = 1.000e-10) r (rel) = 1.633e-05 (tol = 1.000e-09)
Newton iteration 3: r (abs) = 3.430e-08 (tol = 1.000e-10) r (rel) = 4.282e-10 (tol = 1.000e-09)
Newton solver finished in 3 iterations and 3 linear solver iterations.

Can i get the solutions output at each step?
And how is the start value chosen?

Thanks
Noya

[nate]

See for example:
Newton’s method, The nonlinear Poisson demo and

Set krylov linear solver paramters in newton solver

I recommend designing your own NewtonSolver and setting the options using PETSc. Here’s a quick example based on the nonlinear poisson demo. import matplotlib.pyplot as plt from dolfin import * class Problem(NonlinearProblem): def __init__(self, J, F, bcs): self.bilinear_form = J self.linear_form = F self.bcs = bcs NonlinearProblem.__init__(self) def F(self, b, x): assemble(self.linear_form, tensor=b) for bc in self.bcs: bc.…

The initial residual is constructed from whatever you pass for the solution vector. In the example you’ve provided m.vector().