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]> 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 >> > > >> > >> > >> > >> >> >>
