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ode
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1995-05-30
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ode:
Syntax: ode ( rhsf, tstart, tend, ystart, dtout, relerr, abserr, uout )
Synopsis: Integrate a system of first order differential
equations.
Description:
ode integrates a system of N first order ordinary
differential equations of the form:
dy(i)/dt = f(t,y(1),y(2),...,y(N))
y(i) given at t .
The arguments to ode are:
rhsf A function that evaluates dy(i)/dt at t. The function
takes two arguments and returns dy/dt. An example that
generates dy/dt for Van der Pol's equation is shown
below.
vdpol = function ( t , x )
{
local (xp)
xp = zeros(2,1);
xp[1] = x[1] * (1 - x[2]^2) - x[2];
xp[2] = x[1];
return xp;
};
ystart The initial values of y, y(tstart).
tstart The initial value of the independent variable.
tend The final value of the independent variable.
dtout The output interval. The vector y will be saved at
tstart, increments of tstart + dtout, and tend. If
dtout is not specified, then the default is to store
output at 101 values of the independent variable.
relerr The relative error tolerance. Default value is 1.e-6.
abserr The absolute error tolerance. At each step, ode
requires that:
abs(local error) <= abs(y)*relerr + abserr
For each component of the local error and solution
vectors. The default value is 1.e-6.
uout Optional. A user-supplied function that computes an
arbitrary output during the integration. uout must
return a row-matrix at each dtout during the
integration. It is entirely up to the user what to put
in the matrix. The matrix is used to build up a larger
matrix of the output, with one row for each dtout. The
resulting matrix is returned by ode when the
integration is complete.
The Fortran source code for ode() is completely explained and
documented in the text, "Computer Solution of Ordinary
Differential Equations: The Initial Value Problem" by
L. F. Shampine and M. K. Gordon.
Example:
//-----------------------------------------------------------//
// Integrate the Van der Pol equation, and measure the effect
// of relerr and abserr on the solution.
//
vdpol = function ( t , x )
{
local (xp)
xp = zeros(2,1);
xp[1] = x[1] * (1 - x[2]^2) - x[2];
xp[2] = x[1];
return xp;
};
t0 = 0;
tf = 10;
x0 = [0; 0.25];
dtout = 0.05;
relerr = [1e-6, 1e-5, 1e-4, 1e-3, 1e-2, 1e-1];
abserr = relerr;
//
// Baseline
//
xbase = ode( vdpol, x0, 0, 20, 0.05, 1e-9, 1e-9);
results = zeros (relerr.n, abserr.n);
elapse = zeros (relerr.n, abserr.n);
//
// Now loop through the combinations of relerr
// and abserr, saving the results, and computing
// the maximum difference.
//
"start testing loop"
for (i in 1:abserr.n)
{
xode.[i] = <<>>;
for (j in 1:relerr.n)
{
printf("\t%i %i\n", i, j);
tic();
xode.[i].[j] = ode( vdpol, 0, 20, x0, 0.05, relerr[j], abserr[i]);
elapse[i;j] = toc();
// Save results
results[i;j] = max (max (abs (xode.[i].[j] - xbase)));
}
}
//-----------------------------------------------------------//
> results
results =
matrix columns 1 thru 6
1.97e-05 0.000297 0.000634 0.00815 0.078 1.44
0.000128 7.89e-05 0.000632 0.00924 0.0732 1.61
0.000647 0.000625 0.00112 0.0147 0.0995 1.46
0.00355 0.00352 0.00271 0.0118 0.0883 0.862
0.0254 0.0254 0.0254 0.104 0.218 1.72
0.513 0.513 0.513 0.589 0.467 1.82
Each row of results is a function of the absolute error
(abserr) and each column is a function of the relative error
(relerr).
See Also: ode4