I have an ODE of type:
x'(t) = a(t)x+g(t)
Which I am trying to solve. The only GSL ODE example isn't very helpful because the only coefficient (\mu) is not time dependent.
This question has been answered on the GSL mailing list however the answer is very unclear - g(t)
is ignored and it has not been explained how to incorporate a(t)
into func
( should it be passed in *params
?).
Is there any example I can see where such an ODE is solved using GSL?
UPDATE: As has been pointed out below, this has been answered on the GSL mailing list. Here is a full example program of how this is done:
#include <stdio.h>
#include <math.h>
#include "gsl/gsl_errno.h"
#include "gsl/gsl_matrix.h"
#include "gsl/gsl_odeiv2.h"
int func(double t, const double y[], double f[], void *params) {
f[0] = -t* y[0];
return GSL_SUCCESS;
}
int jac(double t, const double y[], double *dfdy, double dfdt[], void
*params) {
gsl_matrix_view dfdy_mat = gsl_matrix_view_array(dfdy, 1, 1);
gsl_matrix * m = &dfdy_mat.matrix;
gsl_matrix_set(m, 0, 0, -t);
dfdt[0] = -1;
return GSL_SUCCESS;
}
int main(void) {
double mu = 0;
gsl_odeiv2_system sys = { func, jac, 1, &mu };
gsl_odeiv2_driver * d = gsl_odeiv2_driver_alloc_y_new(&sys,
gsl_odeiv2_step_rk1imp, 1e-7, 1e-7, 0.0);
int i;
double t = 0.0, t1 = 2.0;
const double x0 = 1.0;
double y[1] = {x0};
const int N = 100;
printf("time\t \tapprox solution \t exact solution\n");
for (i = 0; i <= N; i++) {
double ti = i * (t1 / N);
int status = gsl_odeiv2_driver_apply(d, &t, ti, y);
if (status != GSL_SUCCESS) {
printf("error, return value=%d\n", status);
break;
}
printf("%.5e\t%.5e\t\t%.5e\n", t, y[0], x0*exp(-0.5*t*t));
}
gsl_odeiv2_driver_free(d);
printf("\n...and done!\n");
return 0;
}