So, I’m now even more confused.
I’m attempting to solve an equation that looks like this:
u’ = -i(H0 + e(t) D) u
Where H0 is a purely real diagonal matrix, D is an off-diagonal block
matrix, and e(t) is a function of time (The schrödinger equation in the
energy basis).
I’ve rewritten the e(t) function in my code to just return 0.0. So the
new equation is just u’ = -iH0 u. The matrix is time independent and
diagonal (I’ve checked this). H0[0] ~= -.5 (with no imaginary
component). and u(t=0) = [1,0,0,0,..]
This problem SHOULD be incredibly simple: u’ = i (0.5) u.
However, I’m still getting the same blowup with the TS.:
//with e(t) == 0
//TS
t: 0 step: 0 norm-1: 0
t: 0.01 step: 1 norm-1: 0
t: 0.02 step: 2 norm-1: 0.9999953125635765
t: 0.03 step: 3 norm-1: 2.999981250276277
//Hand rolled
t: 0.01 norm-1: 0 ef 0
t: 0.02 norm-1: 0 ef 0
t: 0.03 norm-1: -1.110223024625157e-16 ef 0
——————————————————————————————
//with e(t) != 0
//TS
t: 0 step: 0 norm-1: 0
t: 0.01 step: 1 norm-1: 0
t: 0.02 step: 2 norm-1: 0.9999953125635765
t: 0.03 step: 3 norm-1: 2.999981250276277
//Hand rolled
t: 0.01 norm-1: 0 ef 9.474814647559372e-11
t: 0.02 norm-1: 0 ef 7.57983838406065e-10
t: 0.03 norm-1: -1.110223024625157e-16 ef 2.558187954267552e-09
I’ve updated the gist.
-Andrew
On Fri, Mar 20, 2015 at 9:57 PM, Barry Smith <[email protected]
<mailto:[email protected]>> wrote:
Andrew,
I'm afraid Emil will have to take a look at this and explain it. The
-ts_type beuler and -ts_type theta -ts_theta_theta .5 are stable but
the -ts_type cn is not stable. It turns out that -ts_type cn is
equivalent to -ts_type theta -ts_theta_theta .5 -ts_theta_endpoint
and somehow this endpoint business (which I don't understand) is
causing a problem. Meanwhile if I add -ts_theta_adapt to the
endpoint one it becomes stable ? Anyways all cases are displayed below.
Emil,
What's up with this? Does the endpoint business have a bug or can it
not be used for this problem (the matrix A is a function of t.)
Barry
$ ./ex2 -ts_type cn
t: 0 step: 0 norm-1: 0
t: 0.01 step: 1 norm-1: 0
t: 0.02 step: 2 norm-1: 1
t: 0.03 step: 3 norm-1: 3
~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
$ ./ex2 -ts_type theta
t: 0 step: 0 norm-1: 0
t: 0.01 step: 1 norm-1: 0
t: 0.02 step: 2 norm-1: 0
t: 0.03 step: 3 norm-1: 0
~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
$ ./ex2 -ts_type theta -ts_theta_theta .5
t: 0 step: 0 norm-1: 0
t: 0.01 step: 1 norm-1: 0
t: 0.02 step: 2 norm-1: 0
t: 0.03 step: 3 norm-1: 0
~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
$ ./ex2 -ts_type theta -ts_theta_theta .5 -ts_theta_endpoint
t: 0 step: 0 norm-1: 0
t: 0.01 step: 1 norm-1: 0
t: 0.02 step: 2 norm-1: 1
t: 0.03 step: 3 norm-1: 3
~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
$ ./ex2 -ts_type theta -ts_theta_theta .5 -ts_theta_endpoint
-ts_theta_adapt
t: 0 step: 0 norm-1: 0
t: 0.01 step: 1 norm-1: 0
t: 0.02 step: 2 norm-1: 0
t: 0.03 step: 3 norm-1: 0
~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
$ ./ex2 -ts_type theta -ts_theta_theta .5 -ts_theta_endpoint
-ts_theta_adapt -ts_monitor
0 TS dt 0.01 time 0
t: 0 step: 0 norm-1: 0
0 TS dt 0.01 time 0
1 TS dt 0.1 time 0.01
t: 0.01 step: 1 norm-1: 0
1 TS dt 0.1 time 0.01
2 TS dt 0.1 time 0.02
t: 0.02 step: 2 norm-1: 0
2 TS dt 0.1 time 0.02
3 TS dt 0.1 time 0.03
t: 0.03 step: 3 norm-1: 0
3 TS dt 0.1 time 0.03
~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
$ ./ex2 -ts_type theta -ts_theta_theta .5 -ts_theta_endpoint
-ts_theta_adapt -ts_monitor -ts_adapt_monitor
0 TS dt 0.01 time 0
t: 0 step: 0 norm-1: 0
0 TS dt 0.01 time 0
TSAdapt 'basic': step 0 accepted t=0 + 1.000e-02 wlte= 0
family='theta' scheme=0:'(null)' dt=1.000e-01
1 TS dt 0.1 time 0.01
t: 0.01 step: 1 norm-1: 0
1 TS dt 0.1 time 0.01
TSAdapt 'basic': step 1 rejected t=0.01 + 1.000e-01 wlte=1.24e+03
family='theta' scheme=0:'(null)' dt=1.000e-02
TSAdapt 'basic': step 1 accepted t=0.01 + 1.000e-02 wlte= 0
family='theta' scheme=0:'(null)' dt=1.000e-01
2 TS dt 0.1 time 0.02
t: 0.02 step: 2 norm-1: 0
2 TS dt 0.1 time 0.02
TSAdapt 'basic': step 2 rejected t=0.02 + 1.000e-01 wlte=1.24e+03
family='theta' scheme=0:'(null)' dt=1.000e-02
TSAdapt 'basic': step 2 accepted t=0.02 + 1.000e-02 wlte= 0
family='theta' scheme=0:'(null)' dt=1.000e-01
3 TS dt 0.1 time 0.03
t: 0.03 step: 3 norm-1: 0
3 TS dt 0.1 time 0.03
~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
$ ./ex2 -ts_type beuler
t: 0 step: 0 norm-1: 0
t: 0.01 step: 1 norm-1: 0
t: 0.02 step: 2 norm-1: 0
t: 0.03 step: 3 norm-1: 0
~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
$ ./ex2 -ts_type euler
t: 0 step: 0 norm-1: 0
t: 0.01 step: 1 norm-1: 0
t: 0.02 step: 2 norm-1: 0
t: 0.03 step: 3 norm-1: 0
~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
> On Mar 20, 2015, at 10:18 PM, Andrew Spott
<[email protected]> wrote:
>
> here are the data files.
>
> dipole_matrix.dat:
> https://www.dropbox.com/s/2ahkljzt6oo9bdr/dipole_matrix.dat?dl=0
>
> energy_eigenvalues_vector.dat
>
https://www.dropbox.com/s/sb59q38vqvjoypk/energy_eigenvalues_vector.dat?dl=0
>
> -Andrew
>
>
>
> On Fri, Mar 20, 2015 at 7:25 PM, Barry Smith <[email protected]>
wrote:
>
> 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
<[email protected]> 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
<[email protected]> 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
<[email protected]> 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
<[email protected]> wrote:
> > >
> > > On Fri, Mar 20, 2015 at 3:09 PM, Andrew Spott
<[email protected]> 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
<[email protected]>, 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"
<[email protected]> 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
> > >
> >
> >
> >
>
>
>
