I think I get it now. Thanks a lot. On Nov 6, 2011 11:06 PM, "Matthew Knepley" <knepley at gmail.com> wrote:
> 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/e344873f/attachment-0001.htm>
