On Sun, Nov 6, 2011 at 9:56 PM, Dominik Szczerba <dominik at itis.ethz.ch>wrote:
> On Sun, Nov 6, 2011 at 10:40 PM, Matthew Knepley <knepley at gmail.com> > wrote: > > On Sun, Nov 6, 2011 at 9:34 PM, Dominik Szczerba <dominik at itis.ethz.ch> > > wrote: > >> > >> On Sun, Nov 6, 2011 at 6:59 PM, Matthew Knepley <knepley at gmail.com> > wrote: > >> > On Sun, Nov 6, 2011 at 5:52 PM, Dominik Szczerba < > dominik at itis.ethz.ch> > >> > wrote: > >> >> > >> >> >>> I want to start small by porting a very simple code using fixed > >> >> >>> point > >> >> >>> iterations as follows: A(x)x = b(x) is approximated as A(x0)x = > >> >> >>> b(x0), > >> >> >>> then solved by KSP for x, then x0 is updated to x, then repeat > >> >> >>> until > >> >> >>> convergence. > >> >> > > >> >> > Run the usual "Newton" methods with A(x) in place of the true > >> >> > Jacobian. > >> >> > >> >> When I substitute A(x) into eq. 5.2 I get: > >> >> > >> >> A(x) dx = -F(x) (1) > >> >> A(x) dx = -A(x) x + b(x) (2) > >> >> A(x) dx + A(x) x = b(x) (3) > >> >> A(x) (x+dx) = b(x) (4) > >> >> > >> >> My questions: > >> >> > >> >> * Will the procedure somehow optimally group the two A(x) terms into > >> >> one, as in 3-4? This requires knowledge, will this be efficiently > >> >> handled? > >> > > >> > There is no grouping. You solve for dx and do a vector addition. > >> > > >> >> > >> >> * I am solving for x+dx, while eq. 5.3 solves for dx. Is this, and > >> >> how, correctly handled? Should I somehow disable the update myself? > >> > > >> > Do not do any update yourself, just give the correct A at each > iteration > >> > in > >> > your FormJacobian routine. > >> > Matt > >> > >> OK, no manual update, this is clear now. What is still not clear is > >> that by substituting A for F' I arrive at an equation in x+dx (my eq. > >> 4), and not dx (Petsc eq. 5.3)... > > > > Newton's equation is for dx. Then you add that to x to get the next > guess. > > This > > is described in any book on numerical analysis, e.g. Henrici. > > Matt > > I understand that, but I am asking something else... when taking F' = > A I arrive at an equation in x+dx, which is not Newton equation. What > is wrong in this picture? Keep the residual on the rhs, not b Matt > > Dominik -- What most experimenters take for granted before they begin their experiments is infinitely more interesting than any results to which their experiments lead. -- Norbert Wiener -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.mcs.anl.gov/pipermail/petsc-users/attachments/20111106/32b9c15a/attachment.htm>
