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

[James Sherman Jr.]

I don't believe that lsode can handle imaginary components (hence the warning).  There may be a function that can handle complex functions, but I don't know of any myself.  There maybe something in the odepkg found at http://octave.sourceforge.net/packages.html.

Sorry I can't be of more help.

James

On Thu, Apr 30, 2009 at 11:25 PM, John B. Thoo <address@hidden> 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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