Prince-Dormand's 4th and 5th Order Embedded Runge-Kutta Method Version 1
Function List
Function(s):
- int Embedded_Prince_Dormand_v1_4_5( double (*f)(double, double), double y[ ], double x0, double h, double xmax, double *h_next, double tolerance )
Solve the differential equation y' = f(x,y) from x0 to xmax with initial condition y(x0) = y[0] using the initial step size h. The result at x = xmax is returned in y[1]. Upon returning h_next contains the estimated step size so that the final answer is within tolerance of the actual solution at x = xmax, this value of h_next can be used as the initial step size h in the subsequent call to this function. This function returns a 0 if a solution was found, -1 if a solution could not be found, and -2 if xmax < x0 or if h < = 0.
C Source
////////////////////////////////////////////////////////////////////////////////
// File: embedded_prince_dormand_v1_4_5.c //
// Routines: //
// Embedded_Prince_Dormand_v1_4_5 //
////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////
// //
// Description: //
// This Runge-Kutta-Prince-Dormand method is an adaptive procedure for //
// approximating the solution of the differential equation y'(x) = f(x,y) //
// with initial condition y(x0) = c. This implementation evaluates //
// f(x,y) six times per step using embedded fourth order and fifth order //
// Runge-Kutta estimates to estimate the not only the solution but also //
// the error. //
// The next step size is then calculated using the preassigned tolerance //
// and error estimate. //
// For step i+1, //
// y[i+1] = y[i] + h * ( 31 / 540 * k1 + 190 / 297 * k3 //
// - 145 / 108 * k4 + 351 / 220 * k5 + 1/20 * k6 ) //
// where //
// k1 = f( x[i],y[i] ), //
// k2 = f( x[i]+h/5, y[i] + h*k1/5 ), //
// k3 = f( x[i]+3h/10, y[i]+(h/40)*(3 k1 + 9 k2) ), //
// k4 = f( x[i]+3h/5, y[i]+(h/10)*(3 k1 - 9 k2 + 12 k3) ), //
// k5 = f( x[i]+2h/3, y[i]+(h/729)*(226 k1 - 675 k2 + 880 k3 + 55 k4) ) //
// k6 = f( x[i]+h, y[i]+(h/2970)*(-1991 k1 + 7425 k2 - 2660 k3 //
// - 10010 k4 + 10206 k5) ) //
// x[i+1] = x[i] + h. //
// //
// The error is estimated to be //
// err = h*( 77 k1 - 400 k3 + 1925 k4 - 1701 k5 + 99 k6 ) / 2520 //
// The step size h is then scaled by the scale factor //
// scale = 0.8 * | epsilon * y[i] / [err * (xmax - x[0])] | ^ 1/4 //
// The scale factor is further constrained 0.125 < scale < 4.0. //
// The new step size is h := scale * h. //
////////////////////////////////////////////////////////////////////////////////
#include < math.h >
#define max(x,y) ( (x) < (y) ? (y) : (x) )
#define min(x,y) ( (x) < (y) ? (x) : (y) )
#define ATTEMPTS 12
#define MIN_SCALE_FACTOR 0.125
#define MAX_SCALE_FACTOR 4.0
static double Runge_Kutta(double (*f)(double,double), double *y, double x,
double h);
////////////////////////////////////////////////////////////////////////////////
// int Embedded_Prince_Dormand_v1_4_5( double (*f)(double, double), //
// double y[], double x, double h, double xmax, double *h_next, //
// double tolerance ) //
// //
// Description: //
// This function solves the differential equation y'=f(x,y) with the //
// initial condition y(x) = y[0]. The value at xmax is returned in y[1]. //
// The function returns 0 if successful or -1 if it fails. //
// //
// Arguments: //
// double *f Pointer to the function which returns the slope at (x,y) of //
// integral curve of the differential equation y' = f(x,y) //
// which passes through the point (x0,y0) corresponding to the //
// initial condition y(x0) = y0. //
// double y[] On input y[0] is the initial value of y at x, on output //
// y[1] is the solution at xmax. //
// double x The initial value of x. //
// double h Initial step size. //
// double xmax The endpoint of x. //
// double *h_next A pointer to the estimated step size for successive //
// calls to Runge_Kutta_Prince_Dormand_v1_4_5. //
// double tolerance The tolerance of y(xmax), i.e. a solution is sought //
// so that the relative error < tolerance. //
// //
// Return Values: //
// 0 The solution of y' = f(x,y) from x to xmax is stored y[1] and //
// h_next has the value to the next size to try. //
// -1 The solution of y' = f(x,y) from x to xmax failed. //
// -2 Failed because either xmax < x or the step size h <= 0. //
// //
////////////////////////////////////////////////////////////////////////////////
// //
int Embedded_Prince_Dormand_v1_4_5( double (*f)(double, double), double y[],
double x, double h, double xmax, double *h_next, double tolerance ) {
double scale;
double temp_y[2];
double err;
double yy;
int i;
int last_interval = 0;
// Verify that the step size is positive and that the upper endpoint //
// of integration is greater than the initial enpoint. //
if (xmax < x || h <= 0.0) return -2;
// If the upper endpoint of the independent variable agrees with the //
// initial value of the independent variable. Set the value of the //
// dependent variable and return success. //
*h_next = h;
y[1] = y[0];
if (xmax == x) return 0;
// Insure that the step size h is not larger than the length of the //
// integration interval. //
h = min(h, xmax - x);
// Redefine the error tolerance to an error tolerance per unit //
// length of the integration interval. //
tolerance /= (xmax - x);
// Integrate the diff eq y'=f(x,y) from x=x to x=xmax trying to //
// maintain an error less than tolerance * (xmax-x) using an //
// initial step size of h and initial value: y = y[0] //
temp_y[0] = y[0];
while ( x < xmax ) {
scale = 1.0;
for (i = 0; i < ATTEMPTS; i++) {
err = fabs( Runge_Kutta(f, temp_y, x, h) );
if (err == 0.0) { scale = MAX_SCALE_FACTOR; break; }
yy = (temp_y[0] == 0.0) ? tolerance : fabs(temp_y[0]);
scale = 0.8 * sqrt( sqrt ( tolerance * yy / err ) );
scale = min( max(scale,MIN_SCALE_FACTOR), MAX_SCALE_FACTOR);
if ( err < ( tolerance * yy ) ) break;
h *= scale;
if ( x + h > xmax ) h = xmax - x;
else if ( x + h + 0.5 * h > xmax ) h = 0.5 * h;
}
if ( i >= ATTEMPTS ) { *h_next = h * scale; return -1; };
temp_y[0] = temp_y[1];
x += h;
h *= scale;
*h_next = h;
if ( last_interval ) break;
if ( x + h > xmax ) { last_interval = 1; h = xmax - x; }
else if ( x + h + 0.5 * h > xmax ) h = 0.5 * h;
}
y[1] = temp_y[1];
return 0;
}
////////////////////////////////////////////////////////////////////////////////
// static double Runge_Kutta(double (*f)(double,double), double *y, //
// double x0, double h) //
// //
// Description: //
// This routine uses Prince-Dormand's embedded 4th and 5th order methods //
// to approximate the solution of the differential equation y'=f(x,y) //
// with the initial condition y = y[0] at x = x0. The value at x + h is //
// returned in y[1]. The function returns err / h ( the absolute error //
// per step size ). //
// //
// Arguments: //
// double *f Pointer to the function which returns the slope at (x,y) of //
// integral curve of the differential equation y' = f(x,y) //
// which passes through the point (x0,y[0]). //
// double y[] On input y[0] is the initial value of y at x, on output //
// y[1] is the solution at x + h. //
// double x Initial value of x. //
// double h Step size //
// //
// Return Values: //
// This routine returns the err / h. The solution of y(x) at x + h is //
// returned in y[1]. //
// //
////////////////////////////////////////////////////////////////////////////////
static double Runge_Kutta(double (*f)(double,double), double *y, double x0,
double h) {
static const double two_thirds = 2.0 / 3.0;
static const double one_seventwoninths = 1.0 / 729.0;
static const double one_twoninesevenzero = 1.0 / 2970.0;
static const double one_twofivetwozero = 1.0 / 2520.0;
static const double one_fiveninefourzero = 1.0 / 5940.0;
double k1, k2, k3, k4, k5, k6;
double h5 = 0.2 * h;
k1 = (*f)(x0, *y);
k2 = (*f)(x0+h5, *y + h5 * k1);
k3 = (*f)(x0+0.3*h, *y + h * ( 0.075 * k1 + 0.225 * k2) );
k4 = (*f)(x0+0.6*h, *y + h * ( 0.3 * k1 - 0.9 * k2 + 1.2 * k3) );
k5 = (*f)(x0+two_thirds * h, *y + one_seventwoninths * h * ( 226.0 * k1
- 675.0 * k2 + 880.0 * k3 + 55.0 * k4) );
k6 = (*f)(x0+h, *y + one_twoninesevenzero * h * ( - 1991 * k1 + 7425.0 * k2
- 2660.0 * k3 - 10010.0 * k4 + 10206.0 * k5) );
*(y+1) = *y + one_fiveninefourzero * h * ( 341.0 * k1 + 3800.0 * k3
- 7975.0 * k4 + 9477.0 * k5 + 297.0 * k6 );
return one_twofivetwozero * (77.0 * k1 - 400.0 * k3 + 1925.0 * k4
- 1701.0 * k5 + 99.0 * k6);
}