Dear Wolfgang,
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.
Thanks.
Best,
M
Wolfgang Bangerth <bange...@colostate.edu> 于2018年12月21日周五 下午2:20写道：
> On 12/20/18 1:15 AM, miffy....@gmail.com wrote:
> > I'm sorry to reply you in a old question mail, I want to know when
> rhs
> > depend on the solution i.e., rhs is a polynomials of the solution,
> should we
> > handle it with test function to get a bilinear form while we construct
> the
> > weak form?
> > The description may be confused, for example consider the rhs as
> f(u,x,t)
> > in Step-26 (the heat euqation,the original rhs is f(x,t)), f(u,t) may be
> > u^2+u^3+u^4, how to deal with such a problem?
> > I have read the toturials to find the answer, but the rhs always
> only
> > depend on x,y,z and Step-15 can't answer my question(maybe) .
>
> No, step-15 is exactly what you need :-) There, the right hand side does
> depend on u in some (more complicated) way. The only different is that in
> step-15, the iterations we do are Newton steps, whereas what you want to
> do is
> a time iteration; but the principle is still the same.
>
> Can you explain why step-15 doesn't answer your questions?
>
> Best
> W
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: bange...@colostate.edu
> www: http://www.math.colostate.edu/~bangerth/
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