Professor Bangerth -
Thank you very much for your very helpful reply! I was thinking that the
addition of large vector of numbers at or below machine precision would
accumulate a lot of numerical roundoff error but I see what you mean. The
Eisenstat Walker algorithm looks like exactly what I want. Thank you very
much again.
Jonathan
On Saturday, September 14, 2019 at 12:25:15 AM UTC-4, Wolfgang Bangerth
wrote:
>
>
> Jonathan,
>
> > I have a simple question I was hoping to get advice on related to
> setting the
> > linear solver tolerance for an iterative solver. In the tutorial
> examples (and
> > in most of the codes I have seen and written) the solver tolerance is
> some
> > fraction of the initial right hand side 2-norm (e.g. tol = 1.0e-8 *
> > rhs.l2_norm() ). However, I'm questioning this strategy for a nonlinear
> > problem using Newton's method to solve the nonlinear system. As the
> iterations
> > progress, the initial norm of the right hand side (i.e. the norm of the
> > residual) is decreasing substantially. For a number of problems that I
> am
> > trying to solve, this causes my tolerance to drop below machine
> precision at
> > some point (if this happens I take 1e-15 to be the tolerance).
>
> This isn't quite the right way to see it. I believe you are thinking that
> 1e-16 is the machine precision, but that doesn't mean that 1e-16 is the
> *smallest number a machine can represent*. It's not an *absolute*
> precision,
> but a *relative* precision. A different way to see this is to say that in
> double precision, the processor stores numbers with approximately 16
> decimal
> digits.
>
> So if the norm of your right hand side (the residual of your nonlinear
> problem) is already at 1e-10, then there is nothing wrong with requiring
> that
> the linear solver gets the residual down to 1e-18=1e-8*1e-10. All you're
> requiring is that the individual entries of the (linear) residual b-Ax
> are, on
> average, 10^8 times smaller than the individual entries of the right hand
> side
> b. That's fine.
>
>
> > I would like to have a tolerance setting strategy that allows me to have
> a
> > reasonable and consistent number of iterations for each linear solve
> required
> > in a Newton increment.
>
> No, that's not actually what you want. What you really want (or *should*
> want
> ;-) ) is a strategy in which you solve the first few Newton steps
> inexactly
> (you're still far from the solution, so who cares whether you compute the
> Newton update with great accuracy!) and gradually tighten the tolerance of
> the
> linear solver as your Newton iteration gets you closer and closer to the
> solution. The key word to search for in the literature is
> "Eisenstat-Walker
> algorithm". The canonical reference for this method is this:
>
> @article{eiwa96,
> author = {Stanley C. Eisenstat and Homer F. Walker},
> title = {Choosing the Forcing Terms in an Inexact {N}ewton Method},
> journal = {SIAM Journal on Scientific Computing},
> volume = {17},
> number = {1},
> pages = {16-32},
> year = {1996},
> doi = {10.1137/0917003},
> URL = {http://dx.doi.org/10.1137/0917003},
> eprint = {http://dx.doi.org/10.1137/0917003}
> }
>
> The price you will pay is that the number of linear iterations goes up as
> you
> get closer to the solution. But the way you should really see this is that
> you
> are saving yourself iterations in the early Newton steps.
>
> Best
> Wolfgang
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: bang...@colostate.edu
> <javascript:>
> www: http://www.math.colostate.edu/~bangerth/
>
>
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