On Tue, Sep 15, 2009 at 6:04 PM, Ryan Yan <span dir="ltr"><<a href="mailto:vyan2000@gmail.com">vyan2000@gmail.com</a>></span> wrote:<br><div class="gmail_quote"><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
<div>Thanks Jed. </div>
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<div>And in the term ||Ae||/||Ax||, PETSc impilictly assume that we use zero as initial guess and here the x really means exact solution, right?</div></blockquote><div><br>This has nothing to do with the initial guess. Ae is the residual, and x is the current guess. We do not have the exact solution,<br>
but I guess A x^* = b, so that could have been used.<br><br> Matt<br> </div><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"><div> <br></div>
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<div>Yan</div></font><div><div></div><div class="h5">
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<div class="gmail_quote">On Tue, Sep 15, 2009 at 6:41 PM, Jed Brown <span dir="ltr"><<a href="mailto:jed@59a2.org" target="_blank">jed@59a2.org</a>></span> wrote:<br>
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<div>Ryan Yan wrote:<br>> But can you say a little more about what happened in the first scenario, I<br>> did not quite fellow you. Let's say, If I set -ksp_rtol 1e-2, what does this<br>> mean in the left PC case. More specifically, which term will be considered<br>
> as a stop creteria for the KSP solve.<br><br></div>Look at the first column (preconditioned residual norm). Notice that<br>this is decreasing by 1e-2 and 1e-3 respectively in your examples.<br><font color="#888888"><br>
Jed<br><br></font></blockquote></div><br>
</div></div></blockquote></div><br><br clear="all"><br>-- <br>What most experimenters take for granted before they begin their experiments is infinitely more interesting than any results to which their experiments lead.<br>
-- Norbert Wiener<br>