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

[John B. Thoo]

That's very neat.  I tried it and it works! :-)  But now how do I  
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?)

I don't see how to index the real and imaginary parts of the  
functions u(1), u(2), and u(3).

Thank again.

---John.


On Apr 30, 2009, at 9:09 PM, Thomas Shores wrote:

> 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.