Solver uses Runge Kutta fifth order algorithm with adaptive step size, it is quite fast and convenient to use.

Example Problem

In order to demonstrate how to use the solver, consider the example problem from Mathematica.com. This is second order ODE or system of two first order equations.

In order to solve this equation, we need to specify

• right hand side function (rhs)
• initial condition
• solvers parameters (optional)

RHS Function:

Inputs: First parameter double x; is value of independent variable (time in this example), second parameter double *y is array of dependent variables, size of array is specified by fourth parameter int nSize.

Outputs: function calculate derivatives double *ydot; of each dependent variable.

void rhs(double x, double *y, double *ydot, int nSize)
{
	static double  f = 700000;
	static double  R = 60;
	static double  L = 1e-4;
	static double  C1 = 1e-7;
	static double  C2 = 4e-10;
	static double  E0 = 0.1;
	static double  E  = 0.2; //0.2 0.3
	static double  pi = 3.1415926535897932384626433832795;

	ydot[0] = y[1];
   ydot[1] = ( -R*y[1]-( (C2-C1)*fabs(y[0])/(2*C1*C2)
       +(C1+C2)*y[0]/(2*C1*C2)+E0)+E*sin(2*pi*f*x))/L;
}

Solver initialization:

const int nSize = 2; //size of ODE system
CRungeKutta solver(nSize); //Create solver 
double y[nSize] = {0, 0}; //Specify initial conditions
double t0 = 0; //specify initial value of independent variable
double t1 = 1.4e-4;//specify final value of independent variable
double t, dt, h;
double nSteps = 10000; //number of output points 

solver.SetEps(1e-8);

t = t0;
dt = (t1-t0)/nSteps;
h = dt/100;

try
{
    for(int i = 0; i<nSteps; i++)
    {
       solver.Solve(rhs, y, t0+i*dt, t0+(i+1)*dt,h);
       printf("%e\t %e\t %e\n",t0+(i+1)*dt,y[0], y[1]);
   }
}
catch(CError error)
{
   printf("%s\n", error.ShowReason());
}
catch(...)
{
   printf("Unknow Error\n");
};
return 0; 

Solution: