Dear Jed, What if my F has the form(the full version):
F(f) = fxx+fyy+fxx*fyy-fxy*fxy-g = 0, how should my Newton iteration look like? J(f)h = -F(f) = -(fxx+fyy+fxx*hyy+fyy*hxx-fxy*hxy-fxy*hxy-g)? I know h is the update, but what are hxx,hyy? Any more references on this? Thanks a lot! Rebecca Quoting Jed Brown <jed at 59A2.org>: > On Wed, 10 Feb 2010 15:35:14 -0500, "(Rebecca) Xuefei YUAN" > <xy2102 at columbia.edu> wrote: >> Hi,all, >> >> I was writing analytic Jacobian to my FormJacobian(). However, I found >> that there are quite big differences between finite difference >> Jacobian and hand coded Jacobian. To simply the problem, assume that I >> am solving a very simple equation as the residual function has the >> form of >> F = fxx*fyy - g. >> Where f is the unknown, g is the given r.h.s. > > It looks like you have coded the Picard linearization. You want the > Newton linearization > > J(f)h = fxx*hyy + hxx*fyy > > Jed > > -- (Rebecca) Xuefei YUAN Department of Applied Physics and Applied Mathematics Columbia University Tel:917-399-8032 www.columbia.edu/~xy2102
