[petsc-users] TimeStepper norm problems. EMIL Please read this
Emil Constantinescu
emconsta at mcs.anl.gov
Sun Mar 22 21:21:04 CDT 2015
Hi Andrew,
I can reproduce this issue and I agree that there is something wrong.
I'll look into it.
Emil
On 3/22/15 3:29 PM, Andrew Spott wrote:
> 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 <bsmith at mcs.anl.gov
> <mailto:bsmith at mcs.anl.gov>> 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
> <ansp6066 at colorado.edu> 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 <bsmith at mcs.anl.gov>
> 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
> <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
> > > >
> > >
> > >
> > >
> >
> >
> >
>
>
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