Hi Roy,
I am sorry about mixing terminology in my previous communication.
What I was referring to as the residual is actually the "|F|" output from
the time solver, so essentially the \dot{x}. And the "iterations" are
actually the time steps.
I use only two Newton iterations per time step.
So, the TwoStepTimeSolver reduces the |F| to 10^-8 in about three time
steps, as opposed to about 100 that I would expect (with two Newton
iterations per time step).
The code modification that I mentioned in my previous email is all that I
changed for use of the TwostepTimeSolver class.
Manav
On Thu, Apr 4, 2013 at 12:18 PM, Roy Stogner <[email protected]>wrote:
>
> On Thu, 4 Apr 2013, Manav Bhatia wrote:
>
> I am attempting to use the TwostepTimeSolver in place of the
>> EulerSolver
>> for time marching in my code.
>>
>> This, however, is leading to unexpected behavior: the nonlinear
>> residuals drop down to 10^-8 in 3 iterations even though the time step has
>> changed from 0.05 to 0.051 (should take about 100 iterations).
>>
>
> By "iterations" do you mean Newton iterations? That's the context
> where "nonlinear residuals" really make sense, but there's no way you
> should be running a solve that normally takes 100 inexact-Newton
> steps.
>
>
> I am in the process of looking into the code, but wanted to run this by
>> the developers to see if there is something I may be missing to setup the
>> solver.
>>
>
> Maybe something you're missing in the reporting? For each time step
> of deltat you're now really doing one deltat step and two deltat/2
> steps; depending on how your problems nonlinearities behave it's quite
> possible that the deltat/2 steps are much easier to solve.
> ---
> Roy
>
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