Verner's 5th and 6th Order Embedded Runge-Kutta Method
Function List
Function(s):
- int Embedded_Verner_5_6( 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_verner_5_6.c //
// Routines: //
// Embedded_Verner_5_6 //
////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////
// //
// Description: //
// The Runge-Kutta-Verner method is an adaptive procedure for approxi- //
// mating the solution of the differential equation y'(x) = f(x,y) with //
// initial condition y(x0) = c. This implementation evaluates f(x,y) //
// eight times per step using embedded fifth order and sixth 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 / 5600 * ( 210 * k1 + 896 * k3 + 1215 * k4 //
// + 2695 * k5 + 584 * k6 ) //
// where //
// k1 = f( x[i],y[i] ), //
// k2 = f( x[i]+h/18, y[i] + h*k1/18), //
// k3 = f( x[i]+h/6, y[i]+h/12*(- k1 + 3 k2) ), //
// k4 = f( x[i]+2h/9, y[i]+h/81*(-2 k1 + 12 k2 + 8 k3) ), //
// k5 = f( x[i]+2h/3, y[i]+h/33*(40 k1 - 12 k2 - 168 k3 + 162 k4)) //
// k6 = f( x[i]+h, y[i]+h/1752*(-8856 k1 + 1728 k2 + 43040 k3 //
// - 36855 k4 + 2695 k5) ) //
// k7 = f( x[i]+8h/9, y[i]+h/891*( -8716 k1 + 1968 k2 + 39520 k3 //
// - 33696 k4 + 1716 k5) ) //
// k8 = f( x[i]+h, y[i]+h/9984*( 117585 k1 - 22464 k2 - 540032 k3 //
// + 466830 k4 - 14014 k5 + 2079 K7) ) //
// x[i+1] = x[i] + h. //
// //
// The error is estimated to be //
// err = h*( 15015 k1 - 118272 k3 + 115830 k4 - 30030 k5 - 30368 k6 //
// + 31185 k7 + 16640 k8 ) / 291200 //
// The step size h is then scaled by the scale factor //
// scale = 0.8 * | epsilon * y[i] / [err * (xmax - x[0])] | ^ 1/5 //
// 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_Verner_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 Embedded_Verner_5_6. //
// 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_Verner_5_6( 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 * pow( tolerance * yy / err , 0.2 );
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 Verner's embedded 5th and 6th 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 c1 = 210.0 / 5600.0;
static const double c3 = 896.0 / 5600.0;
static const double c4 = 1215.0 / 5600.0;
static const double c5 = 2695.0 / 5600.0;
static const double c6 = 584.0 / 5600.0;
static const double a2 = 1.0 / 18.0;
static const double a3 = 1.0 / 6.0;
static const double a4 = 2.0 / 9.0;
static const double a5 = 2.0 / 3.0;
static const double a7 = 8.0 / 9.0;
static const double b31 = -1.0 / 12.0;
static const double b32 = 3.0 / 12.0;
static const double b41 = -2.0 / 81.0;
static const double b42 = 12.0 / 81.0;
static const double b43 = 8.0 / 81.0;
static const double b51 = 40.0 / 33.0;
static const double b52 = -12.0 / 33.0;
static const double b53 = -168.0 / 33.0;
static const double b54 = 162.0 / 33.0;
static const double b61 = -8856.0 / 1752.0;
static const double b62 = 1728.0 / 1752.0;
static const double b63 = 43040.0 / 1752.0;
static const double b64 = -36855.0 / 1752.0;
static const double b65 = 2695.0 / 1752.0;
static const double b71 = -8716.0 / 891.0;
static const double b72 = 1968.0 / 891.0;
static const double b73 = 39520.0 / 891.0;
static const double b74 = -33696.0 / 891.0;
static const double b75 = 1716.0 / 891.0;
static const double b81 = 117585.0 / 9984.0;
static const double b82 = -22464.0 / 9984.0;
static const double b83 = -540032.0 / 9984.0;
static const double b84 = 466830.0 / 9984.0;
static const double b85 = -14014.0 / 9984.0;
static const double b87 = 2079.0 / 9984.0;
static const double e1 = 15015.0 / 291200.0;
static const double e3 = - 118272.0 / 291200.0;
static const double e4 = 115830.0 / 291200.0;
static const double e5 = - 30030.0 / 291200.0;
static const double e6 = - 30368.0 / 291200.0;
static const double e7 = 31185.0 / 291200.0;
static const double e8 = 16640.0 / 291200.0;
double k1, k2, k3, k4, k5, k6, k7, k8;
double h18 = a2 * h;
k1 = (*f)(x0, *y);
k2 = (*f)(x0+h18, *y + h18 * k1);
k3 = (*f)(x0+a3*h, *y + h * ( b31*k1 + b32*k2) );
k4 = (*f)(x0+a4*h, *y + h * ( b41*k1 + b42*k2 + b43*k3) );
k5 = (*f)(x0+a5*h, *y + h * ( b51*k1 + b52*k2 + b53*k3 + b54*k4) );
k6 = (*f)(x0+h, *y + h * ( b61*k1 + b62*k2 + b63*k3 + b64*k4 + b65*k5) );
k7 = (*f)(x0+a7*h, *y + h * ( b71*k1 + b72*k2 + b73*k3 + b74*k4 + b75*k5) );
k8 = (*f)(x0+h, *y + h * ( b81*k1 + b82*k2 + b83*k3 + b84*k4
+ b85*k5 + b87*k7) );
*(y+1) = *y + h * (c1 * k1 + c3 * k3 + c4 * k4 + c5 * k5 + c6 * k6);
return e1*k1 + e3*k3 + e4*k4 + e5*k5 + e6*k6 + e7*k7 + e8*k8;
}