[petsc-users] KSPBuildResidual and KSPType compatibility

[Jed Brown]

Please try with -pc_type none.

There is always a small difference due to using the recurrence, but it should be small so long as the Krylov basis is close to orthogonal.

I'd note that if you're using this expensive convergence test, then IBCGS probably isn't helping over BCGS (the I part is to amortize some vector and reduction costs). It'd be worth comparing when you use normal BCGS.

"Fischer, Greg A." <fischega at westinghouse.com> writes:

> I tried your suggestion, but the values are still quite different.
>
> -----Original Message-----
> From: Jed Brown <jed at jedbrown.org>
> Sent: Monday, December 6, 2021 12:01 PM
> To: Fischer, Greg A. via petsc-users <petsc-users at mcs.anl.gov>; petsc-users at mcs.anl.gov
> Cc: Fischer, Greg A. <fischega at westinghouse.com>
> Subject: Re: [petsc-users] KSPBuildResidual and KSPType compatibility
>
> [External Email]
>
> "Fischer, Greg A. via petsc-users" <petsc-users at mcs.anl.gov> writes:
>
>> Hello petsc-users,
>>
>> I would like to check convergence against the infinity norm, so I defined my own convergence test routine with KSPSetConvergenceTest. (I understand that it may be computationally expensive.)
>>
>> I would like to do this with the "ibcgs" method. When I use KSPBuildResidual and calculate the NORM_2 against the output vector, I get a value that differs from the 2-norm that gets passed into the convergence test function. However, when I switch to the "gcr" method, the value I calculate matches the function input value.
>
> IBCGS uses the preconditioned norm by default while GCR uses the unpreconditioned norm. You can use -ksp_norm_type unpreconditioned or KSPSetNormType() to make IBCGS use unpreconditioned.
>
>> Is the KSPBuildResidual function only compatible with a subset of the KSPType methods? If I want to evaluate convergence against the infinity norm, do I need to set KSPSetInitialGuessNonzero and continually re-start the solver with a lower tolerance values until I get a satisfactory value of the infinity norm?
>>
>> Thanks,
>> Greg
>>
>>
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