[petsc-users] TimeStepper norm problems.

[Matthew Knepley]

This sounds like a problem in your calculation function where a Vec or Mat
does not get reset to 0, but it does in your by hand code.

   Matt
On Mar 20, 2015 2:52 PM, "Andrew Spott" <ansp6066 at colorado.edu> wrote:

>  I have a fairly simple problem that I’m trying to timestep:
>
> u’ = A(t) u
>
> I’m using the crank-nicholson method, which I understand (for this
> problem) to be:
>
> u(t + h) = u(t) + h/2[A(t+h)*u(t+h) + A(t)*u(t)]
> or
> [1 - h/2 * A(t+1)] u(t+1) = [1 + h/2 * A(t)] u(t)
>
> When I attempt to timestep using PETSc, the norm of `u` blows up.  When I
> do it directly (using the above), the norm of `u` doesn’t blow up.
>
> It is important to note that the solution generated after the first step
> is identical for both, but the second step for Petsc has a norm of ~2,
> while for the directly calculated version it is ~1.  The third step for
> petsc has a norm of ~4, while the directly calculated version it is still
> ~1.
>
> I’m not sure what I’m doing wrong.
>
> PETSc code is taken out of the manual and is pretty simple:
>
>          TSCreate( comm, &ts );
>         TSSetProblemType( ts, TS_LINEAR);
>         TSSetType( ts, TSCN );
>         TSSetInitialTimeStep( ts, 0, 0.01 );
>         TSSetDuration( ts, 5, 0.03 );
>         TSSetFromOptions( ts );
>         TSSetRHSFunction( ts, NULL, TSComputeRHSFunctionLinear, NULL );
>         TSSetRHSJacobian( ts, A, A, func, &cntx );
>         TSSolve( ts, psi0 );
>
> `func` just constructs A(t) at the time given.  The same code for
> calculating A(t) is used in both calculations, along with the same initial
> vector psi0, and the same time steps.
>
> Let me know what other information is needed.  I’m not sure what could be
> the problem.  `func` doesn’t touch U at all (should it?).
>
> -Andrew
>
>
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