On Fri, 15 Mar 2013, Jens Lohne Eftang wrote: > What do you mean by that exactly. My Jacobian is A and the residual is Ax-b. > The linear solver seems to successfully solve x=A\b so I don't see how it's > possible that something is inconsistent (independent of whether my assembly > loops are actually correct ... )
Sure it's possible. A common example: your residual is r(x):=Ax-b; if you instead just always return "b", then (assuming your A and b are correct and you don't set any initial guess) the first residual evaluation r(0)=A*0-b=b will be correct, the first solve will be correct, you'll get approximately x=inv(A)*b after this step.. but then when the solver asks for a nonlinear residual to test, you'll be handing it "b" instead of "A*x-b"="0", and it'll think you haven't converged on the nonlinear problem at all. So it will try a line search, but that search will never find a residual reduction no matter how many intermediate values of "x" it tries. --- Roy ------------------------------------------------------------------------------ Everyone hates slow websites. So do we. Make your web apps faster with AppDynamics Download AppDynamics Lite for free today: http://p.sf.net/sfu/appdyn_d2d_mar _______________________________________________ Libmesh-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/libmesh-users
