<span id="mailbox-conversation"><div>So, I’m now even more confused.</div>
<div><br></div>
<div>I’m attempting to solve an equation that looks like this:</div>
<div><br></div>
<div>u’ = -i(H0 + e(t) D) u</div>
<div><br></div>
<div>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).</div>
<div><br></div>
<div>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,..]</div>
<div><br></div>
<div>This problem SHOULD be incredibly simple: u’ = i (0.5) u.</div>
<div><br></div>
<div>However, I’m still getting the same blowup with the TS.:</div>
<div><br></div>
<div>//with e(t) == 0</div>
<div>
<div>
<div id="mb-reply">//TS</div>
<div id="mb-reply">t: 0 step: 0 norm-1: 0</div>
<div>t: 0.01 step: 1 norm-1: 0</div>
<div>t: 0.02 step: 2 norm-1: 0.9999953125635765</div>
<div>t: 0.03 step: 3 norm-1: 2.999981250276277</div>
<div id="mb-reply">//Hand rolled</div>
<div id="mb-reply">t: 0.01 norm-1: 0 ef 0</div>
<div>t: 0.02 norm-1: 0 ef 0</div>
<div>t: 0.03 norm-1: -1.110223024625157e-16 ef 0</div>
</div>
<div id="mb-reply">——————————————————————————————</div>
<div id="mb-reply">//with e(t) != 0</div>
<div id="mb-reply">//TS</div>
<div>
<div>t: 0 step: 0 norm-1: 0</div>
<div>t: 0.01 step: 1 norm-1: 0</div>
<div>t: 0.02 step: 2 norm-1: 0.9999953125635765</div>
<div id="mb-reply">t: 0.03 step: 3 norm-1: 2.999981250276277</div>
<div id="mb-reply">//Hand rolled</div>
<div>t: 0.01 norm-1: 0 ef 9.474814647559372e-11</div>
<div>t: 0.02 norm-1: 0 ef 7.57983838406065e-10</div>
<div id="mb-reply">t: 0.03 norm-1: -1.110223024625157e-16 ef 2.558187954267552e-09</div>
</div>
<div id="mb-reply"><br></div>
<div id="mb-reply">I’ve updated the gist.</div>
<div id="mb-reply"><br></div>
<div id="mb-reply">-Andrew</div>
</div></span><div class="mailbox_signature"><br></div>
<br><br><div class="gmail_quote"><p>On Fri, Mar 20, 2015 at 9:57 PM, Barry Smith <span dir="ltr"><<a href="mailto:bsmith@mcs.anl.gov" target="_blank">bsmith@mcs.anl.gov</a>></span> wrote:<br></p><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><br> Andrew,
<br><br> 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.
<br><br> Emil,
<br><br> 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.)
<br><br> Barry
<br><br><br>$ ./ex2 -ts_type cn
<br>t: 0 step: 0 norm-1: 0
<br>t: 0.01 step: 1 norm-1: 0
<br>t: 0.02 step: 2 norm-1: 1
<br>t: 0.03 step: 3 norm-1: 3
<br>~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
<br>$ ./ex2 -ts_type theta
<br>t: 0 step: 0 norm-1: 0
<br>t: 0.01 step: 1 norm-1: 0
<br>t: 0.02 step: 2 norm-1: 0
<br>t: 0.03 step: 3 norm-1: 0
<br>~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
<br>$ ./ex2 -ts_type theta -ts_theta_theta .5
<br>t: 0 step: 0 norm-1: 0
<br>t: 0.01 step: 1 norm-1: 0
<br>t: 0.02 step: 2 norm-1: 0
<br>t: 0.03 step: 3 norm-1: 0
<br>~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
<br>$ ./ex2 -ts_type theta -ts_theta_theta .5 -ts_theta_endpoint
<br>t: 0 step: 0 norm-1: 0
<br>t: 0.01 step: 1 norm-1: 0
<br>t: 0.02 step: 2 norm-1: 1
<br>t: 0.03 step: 3 norm-1: 3
<br>~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
<br>$ ./ex2 -ts_type theta -ts_theta_theta .5 -ts_theta_endpoint -ts_theta_adapt
<br>t: 0 step: 0 norm-1: 0
<br>t: 0.01 step: 1 norm-1: 0
<br>t: 0.02 step: 2 norm-1: 0
<br>t: 0.03 step: 3 norm-1: 0
<br>~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
<br>$ ./ex2 -ts_type theta -ts_theta_theta .5 -ts_theta_endpoint -ts_theta_adapt -ts_monitor
<br>0 TS dt 0.01 time 0
<br>t: 0 step: 0 norm-1: 0
<br>0 TS dt 0.01 time 0
<br>1 TS dt 0.1 time 0.01
<br>t: 0.01 step: 1 norm-1: 0
<br>1 TS dt 0.1 time 0.01
<br>2 TS dt 0.1 time 0.02
<br>t: 0.02 step: 2 norm-1: 0
<br>2 TS dt 0.1 time 0.02
<br>3 TS dt 0.1 time 0.03
<br>t: 0.03 step: 3 norm-1: 0
<br>3 TS dt 0.1 time 0.03
<br>~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
<br>$ ./ex2 -ts_type theta -ts_theta_theta .5 -ts_theta_endpoint -ts_theta_adapt -ts_monitor -ts_adapt_monitor
<br>0 TS dt 0.01 time 0
<br>t: 0 step: 0 norm-1: 0
<br>0 TS dt 0.01 time 0
<br> TSAdapt 'basic': step 0 accepted t=0 + 1.000e-02 wlte= 0 family='theta' scheme=0:'(null)' dt=1.000e-01
<br>1 TS dt 0.1 time 0.01
<br>t: 0.01 step: 1 norm-1: 0
<br>1 TS dt 0.1 time 0.01
<br> TSAdapt 'basic': step 1 rejected t=0.01 + 1.000e-01 wlte=1.24e+03 family='theta' scheme=0:'(null)' dt=1.000e-02
<br> TSAdapt 'basic': step 1 accepted t=0.01 + 1.000e-02 wlte= 0 family='theta' scheme=0:'(null)' dt=1.000e-01
<br>2 TS dt 0.1 time 0.02
<br>t: 0.02 step: 2 norm-1: 0
<br>2 TS dt 0.1 time 0.02
<br> TSAdapt 'basic': step 2 rejected t=0.02 + 1.000e-01 wlte=1.24e+03 family='theta' scheme=0:'(null)' dt=1.000e-02
<br> TSAdapt 'basic': step 2 accepted t=0.02 + 1.000e-02 wlte= 0 family='theta' scheme=0:'(null)' dt=1.000e-01
<br>3 TS dt 0.1 time 0.03
<br>t: 0.03 step: 3 norm-1: 0
<br>3 TS dt 0.1 time 0.03
<br>~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
<br>$ ./ex2 -ts_type beuler
<br>t: 0 step: 0 norm-1: 0
<br>t: 0.01 step: 1 norm-1: 0
<br>t: 0.02 step: 2 norm-1: 0
<br>t: 0.03 step: 3 norm-1: 0
<br>~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
<br>$ ./ex2 -ts_type euler
<br>t: 0 step: 0 norm-1: 0
<br>t: 0.01 step: 1 norm-1: 0
<br>t: 0.02 step: 2 norm-1: 0
<br>t: 0.03 step: 3 norm-1: 0
<br>~/Src/petsc/test-dir (barry/more-tchem-work *=) arch-debug
<br><br><br>> On Mar 20, 2015, at 10:18 PM, Andrew Spott <ansp6066@colorado.edu> wrote:
<br>>
<br>> here are the data files.
<br>>
<br>> dipole_matrix.dat:
<br>> https://www.dropbox.com/s/2ahkljzt6oo9bdr/dipole_matrix.dat?dl=0
<br>>
<br>> energy_eigenvalues_vector.dat
<br>> https://www.dropbox.com/s/sb59q38vqvjoypk/energy_eigenvalues_vector.dat?dl=0
<br>>
<br>> -Andrew
<br>>
<br>>
<br>>
<br>> On Fri, Mar 20, 2015 at 7:25 PM, Barry Smith <bsmith@mcs.anl.gov> wrote:
<br>>
<br>> Data files are needed
<br>>
<br>> PetscViewerBinaryOpen( PETSC_COMM_WORLD, "hamiltonian/energy_eigenvalues_vector.dat", FILE_MODE_READ, &view );
<br>> VecLoad( H0, view );
<br>> PetscViewerBinaryOpen( PETSC_COMM_WORLD, "hamiltonian/dipole_matrix.dat", FILE_MODE_READ, &view );
<br>>
<br>> BTW: You do not need to call Mat/VecAssembly on Mats and Vecs after they have been loaded.
<br>>
<br>> Barry
<br>>
<br>>
<br>> > On Mar 20, 2015, at 6:39 PM, Andrew Spott <ansp6066@colorado.edu> wrote:
<br>> >
<br>> > Sorry it took so long, I wanted to create a “reduced” case (without all my parameter handling and other stuff…)
<br>> >
<br>> > https://gist.github.com/spott/aea8070f35e79e7249e6
<br>> >
<br>> > The first section does it using the time stepper. The second section does it by explicitly doing the steps. The output is:
<br>> >
<br>> > //first section, using TimeStepper:
<br>> > t: 0 step: 0 norm-1: 0
<br>> > t: 0.01 step: 1 norm-1: 0
<br>> > t: 0.02 step: 2 norm-1: 0.999995
<br>> > t: 0.03 step: 3 norm-1: 2.99998
<br>> >
<br>> > //Second section, using explicit code.
<br>> > t: 0.01 norm-1: 0
<br>> > t: 0.02 norm-1: 0
<br>> > t: 0.02 norm-1: 2.22045e-16
<br>> >
<br>> >
<br>> >
<br>> > On Fri, Mar 20, 2015 at 4:45 PM, Barry Smith <bsmith@mcs.anl.gov> wrote:
<br>> >
<br>> > Andrew,
<br>> >
<br>> > Send your entire code. It will be easier and faster than talking past each other.
<br>> >
<br>> > Barry
<br>> >
<br>> > > On Mar 20, 2015, at 5:00 PM, Andrew Spott <ansp6066@colorado.edu> wrote:
<br>> > >
<br>> > > I’m sorry, I’m not trying to be difficult, but I’m not following.
<br>> > >
<br>> > > The manual states (for my special case):
<br>> > > • u ̇ = A(t)u. Use
<br>> > >
<br>> > > TSSetProblemType(ts,TS LINEAR); TSSetRHSFunction(ts,NULL,TSComputeRHSFunctionLinear,NULL); TSSetRHSJacobian(ts,A,A,YourComputeRHSJacobian,&appctx);
<br>> > >
<br>> > > where YourComputeRHSJacobian() is a function you provide that computes A as a function of time. Or use ...
<br>> > > My `func` does this. It is 7 lines:
<br>> > >
<br>> > > context* c = static_cast<context*>( G_u );
<br>> > > PetscScalar e = c->E( t_ );
<br>> > > MatCopy( c->D, A, SAME_NONZERO_PATTERN );
<br>> > > MatShift( A, e );
<br>> > > MatDiagonalSet( A, c->H0, INSERT_VALUES);
<br>> > > MatShift( A, std::complex<double>( 0, -1 ) );
<br>> > > return 0;
<br>> > >
<br>> > > 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).
<br>> > >
<br>> > > -Andrew
<br>> > >
<br>> > >
<br>> > >
<br>> > > On Fri, Mar 20, 2015 at 3:26 PM, Matthew Knepley <knepley@gmail.com> wrote:
<br>> > >
<br>> > > On Fri, Mar 20, 2015 at 3:09 PM, Andrew Spott <ansp6066@colorado.edu> wrote:
<br>> > > 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.
<br>> > >
<br>> > > 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.
<br>> > >
<br>> > > 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).
<br>> > >
<br>> > > Are you initializing the Jacobian to 0 first?
<br>> > >
<br>> > > Thanks,
<br>> > >
<br>> > > Matt
<br>> > >
<br>> > > Is this the case? or is there some other responsibility of said function?
<br>> > >
<br>> > > -Andrew
<br>> > >
<br>> > > >Ah ha!
<br>> > > >
<br>> > > >The function passed to TSSetRHSJacobian needs to zero the solution vector?
<br>> > > >
<br>> > > >As a point, this isn’t mentioned in any documentation that I can find.
<br>> > > >
<br>> > > >-Andrew
<br>> > >
<br>> > > On Friday, Mar 20, 2015 at 2:17 PM, Matthew Knepley <knepley@gmail.com>, wrote:
<br>> > > 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.
<br>> > >
<br>> > > Matt
<br>> > >
<br>> > > On Mar 20, 2015 2:52 PM, "Andrew Spott" <ansp6066@colorado.edu> wrote:
<br>> > > I have a fairly simple problem that I’m trying to timestep:
<br>> > >
<br>> > > u’ = A(t) u
<br>> > >
<br>> > > I’m using the crank-nicholson method, which I understand (for this problem) to be:
<br>> > >
<br>> > > u(t + h) = u(t) + h/2[A(t+h)*u(t+h) + A(t)*u(t)]
<br>> > > or
<br>> > > [1 - h/2 * A(t+1)] u(t+1) = [1 + h/2 * A(t)] u(t)
<br>> > >
<br>> > > 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.
<br>> > >
<br>> > > 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.
<br>> > >
<br>> > > I’m not sure what I’m doing wrong.
<br>> > >
<br>> > > PETSc code is taken out of the manual and is pretty simple:
<br>> > >
<br>> > > TSCreate( comm, &ts );
<br>> > > TSSetProblemType( ts, TS_LINEAR);
<br>> > > TSSetType( ts, TSCN );
<br>> > > TSSetInitialTimeStep( ts, 0, 0.01 );
<br>> > > TSSetDuration( ts, 5, 0.03 );
<br>> > > TSSetFromOptions( ts );
<br>> > > TSSetRHSFunction( ts, NULL, TSComputeRHSFunctionLinear, NULL );
<br>> > > TSSetRHSJacobian( ts, A, A, func, &cntx );
<br>> > > TSSolve( ts, psi0 );
<br>> > >
<br>> > > `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.
<br>> > >
<br>> > > 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?).
<br>> > >
<br>> > > -Andrew
<br>> > >
<br>> > >
<br>> > >
<br>> > >
<br>> > > --
<br>> > > What most experimenters take for granted before they begin their experiments is infinitely more interesting than any results to which their experiments lead.
<br>> > > -- Norbert Wiener
<br>> > >
<br>> >
<br>> >
<br>> >
<br>>
<br>>
<br>>
<br><br></blockquote></div><br>