Re: User supplied jacobian makes ode23s very slow

[Alexander Bessman]

On Fri, 2015-11-06 at 16:48 +0000, Carlo De Falco wrote:
> 
> On 6 Nov 2015, at 13:01, Alexander Bessman <address@hidden> wrote:
> 
> > Hello,
> > 
> > I'm using ode23s from odepkg to solve a simple 1D PDE, ∂u/∂t = ∂²u/dx².
> > I have discretized it via the forward-time central-space scheme,
> > yielding a tridiagonal system of ODEs, which ode23s has no problem
> > solving. However, when I supply the system's Jacobian via odeset, the
> > solution takes much longer to complete. The solution is still correct,
> > but I would have expected it to be faster, not slower. I've included my
> > code below. The result of running it is thus:
> > 
> > |Timesteps             |CPU-time            |Max stepsize        |
> > |N  |with Jac |w/o Jac |with Jac  |w/o Jac  |with Jac  |w/o Jac  |
> > |10 |432      |79      |1.3801 s  |1.1953 s |1.9914e-2 |2.0000e-1|
> > |20 |1633     |98      |5.4225 s  |2.6037 s |5.2711e-3 |1.9447e-1|
> > |40 |6442     |119     |26.7256 s |5.9159 s |8.4338e-4 |1.7272e-1|
> > 
> > I'd be grateful of someone could help me understand this phenomenon.
> > //Alexander Bessman
> > 
> > -----------------------------------------------------------------------
> > 
> > pkg load odepkg
> > 
> > cpu = zeros(2,3);
> > steps = zeros(2,3);
> > htmax = zeros(2,3);
> > iter = 0;
> > for N = [10, 20, 40]    
> >    % Jacobian
> >    e = ones (N, 1);
> >    jac = spdiags ([e, -2*e, e], [-1:1], N, N);
> >    jac(end-1, end) = 2;
> >    conf = odeset ('Jacobian', jac);
> > 
> >    u0 = zeros(N,1);
> >    iter = iter + 1;
> > 
> >    tic
> >    [t, y] = ode23s (@heat_eq, [0, 2], u0);
> >    cpu(1, iter) = toc;
> >    steps(1, iter) = length(t);
> >    htmax(1, iter) = max(diff(t));
> > 
> >    tic
> >    [t, y] = ode23s (@heat_eq, [0, 2], u0, conf);
> >    cpu(2, iter) = toc;
> >    steps(2, iter) = length(t);
> >    htmax(2, iter) = max(diff(t));
> > end
> > 
> > function dudt = heat_eq (t, u, kappa=1)
> >    N = length(u);
> >    e = ones (N, 1);
> >    A = spdiags ([e, -2*e, e], [-1:1], N, N);
> > 
> >    % Boundary conditions
> >    A(end-1, end) = 2;
> >    b = zeros(N, 1);
> >    if t <= 1
> >        b(1) = 1;
> >    end
> > 
> >    dudt = kappa * N^2 * (A*u + b);
> > 
> > 
> > _______________________________________________
> > Help-octave mailing list
> > address@hidden
> > https://lists.gnu.org/mailman/listinfo/help-octave
> 
> 
> 
> You have an incorrect jacobian which causes an unnecessary high number of 
> iterations.
> You have to multiply jac by N^2 * kappa, by doing so I get:
> 
> steps =
> 
>     79    98   119
>     79    98   119
> 
> cpu =
> 
>    1.02229   2.04385   4.87316
>    0.20588   0.24211   0.29443
> 
> cpu (1,:) ./cpu (2,:)
> ans =
> 
>     4.9654    8.4419   16.5513
> 
> so solution with jacobian is up to 16x faster.
> For larger problems the gain is even more impressive.
> 
> c.
> 

Thank you. Such a silly mistake, but I suppose I should have expected as
much. Very impressive speed gains indeed!

//Alexander