[petsc-users] TimeStepper norm problems.

[Barry Smith]

Data files are needed

 PetscViewerBinaryOpen( PETSC_COMM_WORLD, "hamiltonian/energy_eigenvalues_vector.dat", FILE_MODE_READ, &view );
    VecLoad( H0, view );
    PetscViewerBinaryOpen( PETSC_COMM_WORLD, "hamiltonian/dipole_matrix.dat", FILE_MODE_READ, &view );

BTW: You do not need to call Mat/VecAssembly on Mats and Vecs after they have been loaded.

  Barry


> On Mar 20, 2015, at 6:39 PM, Andrew Spott <ansp6066 at colorado.edu> wrote:
> 
> Sorry it took so long, I wanted to create a “reduced” case (without all my parameter handling and other stuff…)
> 
> https://gist.github.com/spott/aea8070f35e79e7249e6
> 
> The first section does it using the time stepper.  The second section does it by explicitly doing the steps.  The output is:
> 
> //first section, using TimeStepper:
> t: 0 step: 0 norm-1: 0
> t: 0.01 step: 1 norm-1: 0
> t: 0.02 step: 2 norm-1: 0.999995
> t: 0.03 step: 3 norm-1: 2.99998
> 
> //Second section, using explicit code.
> t: 0.01 norm-1: 0
> t: 0.02 norm-1: 0
> t: 0.02 norm-1: 2.22045e-16
> 
> 
> 
> On Fri, Mar 20, 2015 at 4:45 PM, Barry Smith <bsmith at mcs.anl.gov> wrote:
> 
> Andrew, 
> 
> Send your entire code. It will be easier and faster than talking past each other. 
> 
> Barry 
> 
> > On Mar 20, 2015, at 5:00 PM, Andrew Spott <ansp6066 at colorado.edu> wrote: 
> > 
> > I’m sorry, I’m not trying to be difficult, but I’m not following. 
> > 
> > The manual states (for my special case): 
> > • u ̇ = A(t)u. Use 
> > 
> > TSSetProblemType(ts,TS LINEAR); TSSetRHSFunction(ts,NULL,TSComputeRHSFunctionLinear,NULL); TSSetRHSJacobian(ts,A,A,YourComputeRHSJacobian,&appctx); 
> > 
> > where YourComputeRHSJacobian() is a function you provide that computes A as a function of time. Or use ... 
> > My `func` does this. It is 7 lines: 
> > 
> > context* c = static_cast<context*>( G_u ); 
> > PetscScalar e = c->E( t_ ); 
> > MatCopy( c->D, A, SAME_NONZERO_PATTERN ); 
> > MatShift( A, e ); 
> > MatDiagonalSet( A, c->H0, INSERT_VALUES); 
> > MatShift( A, std::complex<double>( 0, -1 ) ); 
> > return 0; 
> > 
> > SHOULD `func` touch U? If so, what should `func` do to U? I thought that the RHSJacobian function was only meant to create A, since dG/du = A(t) (for this special case). 
> > 
> > -Andrew 
> > 
> > 
> > 
> > On Fri, Mar 20, 2015 at 3:26 PM, Matthew Knepley <knepley at gmail.com> wrote: 
> > 
> > On Fri, Mar 20, 2015 at 3:09 PM, Andrew Spott <ansp6066 at colorado.edu> wrote: 
> > So, it doesn’t seem that zeroing the given vector in the function passed to TSSetRHSJacobian is the problem. When I do that, it just zeros out the solution. 
> > 
> > I would think you would zero the residual vector (if you add to it to construct the residual, as in FEM methods), not the solution. 
> > 
> > The function that is passed to TSSetRHSJacobian has only one responsibility — to create the jacobian — correct? In my case this is A(t). The solution vector is given for when you are solving nonlinear problems (A(t) also depends on U(t)). In my case, I don’t even look at the solution vector (because my A(t) doesn’t depend on it). 
> > 
> > Are you initializing the Jacobian to 0 first? 
> > 
> > Thanks, 
> > 
> > Matt 
> > 
> > Is this the case? or is there some other responsibility of said function? 
> > 
> > -Andrew 
> > 
> > >Ah ha! 
> > > 
> > >The function passed to TSSetRHSJacobian needs to zero the solution vector? 
> > > 
> > >As a point, this isn’t mentioned in any documentation that I can find. 
> > > 
> > >-Andrew 
> > 
> > On Friday, Mar 20, 2015 at 2:17 PM, Matthew Knepley <knepley at gmail.com>, wrote: 
> > 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 
> > 
> > 
> > 
> > 
> > -- 
> > What most experimenters take for granted before they begin their experiments is infinitely more interesting than any results to which their experiments lead. 
> > -- Norbert Wiener 
> > 
> 
> 
>