On 12/28/18 1:14 AM, llf m wrote:
> Yes, you are right. At first in Step-15
> pseudo-timestepping(Lecture 31.7),
> it linearized the nonlinear term that makes me confused, linearized a
> nonlinear term
> is a general approach? Is it acceptable? But finally I get the idea, yes, we
> need
> linearization in nonlinear iteration.
> In a nonlinear time-dependent problem F[u(t)]=0:
> Time iteration:
> Newton-Raphson iteration:
> solve linearized equation F'[u(t_n)^k]delta_u =
> - F(u(t_n)^k)
> (while k is the index of newton
> iteration,u(t_n)^0=u(t_{n-1}))
> update u(t_n)^{k+1} = u(t_n)^{k} + delta_u
> check convergence e.g., norm(F(u(t_n)^k))<1e-10
> if satisfied go to next time iteration
> else go to next newton iteration
> The above algorithm is the idea I get, maybe I still have some
> misunderstanding.
That's exactly the idea. In each time step, you would want to start with
something like
u(t_n)^0 = u(t_{n-1})^*
where the right hand side is the final Newton iterate of the previous time
step. This gives you a reasonably good starting guess and you will not need a
lot of Newton iterations in time step n. An even better starting guess would
be to *extrapolate* from the previous time steps to the current time step,
e.g., using
u(t_n)^0 = u(t_{n-1})^* + [u(t_{n-1})^* - u(t_{n-2})^*]/dt * dt
= 2*u(t_{n-1})^* - u(t_{n-2})^*
where in the first line on the right, the term [...]/dt is an approximation of
the time derivative, and consequently the right hand side of the first line is
simply a linear extrapolation from the previous two time steps. Of course, if
you wanted to go back in time, you can come up with even better starting
guesses. With such a starting guess, you typically only need two or three
Newton iterations per time step.
Best
W.
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Wolfgang Bangerth email: bange...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/
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