Re: Trying to solve du/dt = i*u

[Thomas Shores]

The routine lsode expects a real system.  The difficulty is easily 
overcome.  Simply separate arguments into real and imaginary parts thusly:

octave:2> u0 = [1,0] % u = 1 in real and imaginary parts
octave:3> t = linspace(0,5,10);
octave:4> function udot = f(u,t)
> udot = [-u(2),u(1)]; % this is I*u in real and imaginary parts
> end
octave:5> u = lsode('f',u0,t)
u =

  1.00000   0.00000
  0.84961   0.52742
  0.44367   0.89619
 -0.09572   0.99541
 -0.60632   0.79522
 -0.93455   0.35584
 -0.98167  -0.19057
 -0.73353  -0.67966
 -0.26475  -0.96432
  0.28366  -0.95892

octave:6> plot(t,u(:,1))
octave:7>hold on
octave:8>plot(t,u:,2))

This gives exactly what you expect, the real and imaginary parts of 
u(t)=exp(i*t).

Thomas Shores


John B. Thoo wrote:
> Hi, everyone.
>
> As a toy problem, I'm trying to solve  du/dt = i*u,  where i^2 = -1.   
> This is what I've tried and gotten:
>
> octave-3.0.5:1> function udot = f (u, t)
>  > udot = I*u;
>  > endfunction
> octave-3.0.5:2> u0 = 1;
> octave-3.0.5:3> t = linspace (0, 5, 10);
> octave-3.0.5:4> u = lsode ("f", u0, t);
> warning: lsode: ignoring imaginary part returned from user-supplied  
> function
> octave-3.0.5:5> plot (t, u)
> octave-3.0.5:6>
>
> Gnuplot gives the constant graph  u = 1  for  t = 0..5.
>
> What am I doing wrongly?
>
> Eventually, I'd like to solve a system like this:
>
> udot(1) = -i*u(3)*u(2)' - i*u(2)*u(1)';
> udot(2) = -2*i*u(3)*u(1)' - i*u(1)*u(1);
> udot(3) = -3*i*u(2)*u(1);
>
> (I'm using  '  for complex conjugate.  Is that correct?)
>
> Thanks for any help you can give.
>
> ---John.
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