Is there a numerical ODE solver facility in Maxima

Ok, thanks everyone, this has been useful.  I am still unsuccessful, but
have a direction to work in.  Jaime, thanks for the text, I don't know
Portuguese but the examples still help.  I talked with a friend who had some
experience with Runge-Kutta stuff and he helped me out a bit.

Here is the equation:

diff( f(x), x, 2) = -k * sin( f(x) ) * cos( f(x) ) /* k is just some
positive constant */

I think the correct approach here is to introduce the function g(x) that is
the first derivative.  Looking at the docs and examples, I think I should
have something like...

f(x)=diff(g(x), x)
diff(g(x), x) = -k * sin*(f(x)) * cos(f(x))

rk(['diff(g, x), -1*sin(f)*cos(f)], [f,g], [0,0]/*initial value*/, [x, -.5,
.5, .01]); /*for k=1*/

Which should return a list of lists of form [xval, fval, gval].  However, if
I run that, the program hangs.

Also, after some playing around, I see that I cannot pass something like
'diff(g, x) since x is substituted before evaluation (I am told that
diff(g,-0.5) makes no sense).  Any idea how to get this into the right form?

Lastly, the boundary conditions are a bit harder than an initial value and
slope at x=-0.5.  I have mixed boundary conditions:

f'(-0.5) + K*sin(f(-0.5))*cos(f(-0.5)) = 0; /* K is just some other constant
*/
-f'(0.5) + K*sin(f(0.5))*cos(f(0.5)) = 0;

This leads me to believe that I can still use a shooting method, (provide an
initial guess of f(-.5) which gives me f'(-.5), then play around with my
guess of f(-0.5) until I get a solution that satisfies the BC at f(0.5)).

Of course, a small angle approximation (sin(f(x)) * cos(f(x)) -> f(x)) can
be made which makes this easy to solve (f(x) ~ cos(x)) and this can be used
for my initial conditions for the shooting method.

So, I think I know a way to proceed except I am not sure how to do it with
rk.  Maybe it can't be done?  I guess I could always do exactly what I
describe and just use an Euler integrator and take small steps in x... maybe
the error won't be that bad.

(I hope I made no mistakes above, I haven't gotten a result yet, so I am
unsure that my thinking is 100% correct)

Zach

2009/4/2 Martin Sch?necker <ms_usenet at gmx.de>

> (%i1) ? rk;
>  -- Function: rk (<ODE>, <var>, <initial>, <domain>)
>  -- Function: rk ([<ODE1>,...,<ODEm>], [<v1>,...,<vm>],
>           [<init1>,...,<initm>], <domain>)
>      The first form solves numerically one first-order ordinary
>      differential equation, and the second form solves a system of m of
>      those equations, using the 4th order Runge-Kutta method. <var>
>      represents the dependent variable. <ODE> must be an expression
>      that depends only on the independent and dependent variables and
>      defines the derivative of the dependent variable with respect to
>      the independent variable.
>      The independent variable is specified with `domain', which must be
>      a list of four elements as, for instance:
>           [t, 0, 10, 0.1]
>      the first element of the list identifies the independent variable,
>      the second and third elements are the initial and final values for
>      that variable, and the last element sets the increments that
>      should be used within that interval.
>      If <m> equations are going to be solved, there should be <m>
>      dependent variables <v1>, <v2>, ..., <vm>. The initial values for
>      those variables will be <init1>, <init2>, ..., <initm>.  There
>      will still be just one independent variable defined by `domain',
>      as in the previous case. <ODE1>, ..., <ODEm> are the expressions
>      that define the derivatives of each dependent variable in terms of
>      the independent variable. The only variables that may appear in
>      those expressions are the independent variable and any of the
>      dependent variables. It is important to give the derivatives
>      <ODE1>, ..., <ODEm> in the list in exactly the same order used for
>      the dependent variables; for instance, the third element in the
>      list will be interpreted as the derivative of the third dependent
>      variable.
>      The program will try to integrate the equations from the initial
>      value of the independent variable until its last value, using
>      constant increments. If at some step one of the dependent
>      variables takes an absolute value too large, the integration will
>      be interrupted at that point. The result will be a list with as
>      many elements as the number of iterations made. Each element in
>      the results list is itself another list with <m>+1 elements: the
>      value of the independent variable, followed by the values of the
>      dependent variables corresponding to that point.
>   There are also some inexact matches for `rk'.
>   Try `?? rk' to see them.
> (%o1) true
>
> Best regards,
> Martin
>
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