[petsc-users] KSPBuildSolution

[Matthew Knepley]

On Thu, Feb 17, 2011 at 8:08 AM, Juha Jäykkä <juhaj at iki.fi> wrote:

> > > [0]PETSC ERROR: Zero pivot row 1 value 1.90611e-13 tolerance 1e-12!
> > You have a singular Jacobian, which leads me to believe that your
> boundary
> > conditions are incorrect.
>
> Thanks for the tip. I also found the following in the -info output:
>
> [0] SNESLSCheckResidual_Private(): ||J^T(F-Ax)||/||F-AX|| 27.9594 near zero
> implies inconsistent rhs
>
> which is strange at first sight: my RHS is my equation, how can that be
> inconsistent? It is what defines the problem. BUT your comment about the
> boundary conditions led me to look into them in more detail.
>
> It seems one of them may be trivially satisfied, leaving me with an
> infinite
> number of solutions satisfying the other one too. Could this be the reason
> for
> my problems? If so, I need to start thinking of another boundary
> condition...
> not nice, since the problem I am solving does not really give them! =(
>
> Take an example:
>
> r f''/f - r (f'/f)^2 + f'/f = 0.
>
> This has the general solution
>
> f(r) = a r^b,
>
> but if the boundary conditions are f(0)=0 and f(1)=1 (like I had in my real
> problem), we have (if b>0):
>
> a*0 = 0
> a*1 = 1
>
> and b is left undetermined or (if b<0):
>
> lim(a*r^b) = 0 as r->0 => a=0
> 0*1^b = 1 => no solution.
>
> I am unsure how to interpret b=0 since then we have 0^0 at the boundary.
> But
> even disregarding the undefined value at the boundary, there cannot be a
> continuous solution with b=0 since then f(r>0)=1, but f(0)=0. Looks like
> Heaviside step to me.
>
> In summary, I cannot even solve this simpler problem with SNES, but I think
> I
> understand the reason now: I have not specified boundary conditions which
> would give a unique solution - they probably do not even give a finite
> number
> of solutions, but an inifinite (quite likely uncountable, like in my
> example)
> number of solutions.
>
> Any thoughts on this? Is there any sense in my analysis? If so, I will need
> to
> go think about Neumann boundaries...


Yes, if your BC do not give at least a locally unique solution, then your
Jacobian will
be rank deficient and Newton breaks down. You can still try Picard, but I
recommend
understanding what you mean by a solution first.

   Matt


> Cheers,
> Juha
>
> --
>                 -----------------------------------------------
>                | Juha Jäykkä, juhaj at iki.fi                     |
>                | http://www.maths.leeds.ac.uk/~juhaj           |
>                 -----------------------------------------------
>



-- 
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/20110217/3cd32b89/attachment.htm>