Demo: Class ODESolver usage: A linear ODE given by
numerical data
We present the same problem as the previous example but the
differential equation is given as linear one by its coefficients.
We consider the linear differential equation
c_{1}y'(t) + c_{0}y(t) = f(t)
We test the code with the data:
f(t)=2*exp(t)1,
c_{0}=c_{1}=1, y(0)=0.
With these data we obtain the solution
y(t)=exp(t)1.
 The code stars in the same way as for the
previous demo code.
#include "OFELI.h"
using namespace OFELI;
int main(int argc, char *argv[])
{
theFinalTime = 1.;
theTimeStep = atof(argv[1]);


The ODE is defined by declaring an instance of class
ODESolver.
The constructor defined the numerical scheme to solve the
equation. Note that we use the enum variable TimeScheme that
contains all implemented schemes. Other possiblities are FORWARD_EULER,
FORWARD_EULER, HEUN and
AB2. Note that only explicit methods are
used. Implicit methods lead to a nonlinear equation at each time
step, which we avoid here.

Once the instance created, we set the initial solution, and define
then the ODE by the member function setF.
We note that the used variables are t and y
since we attempt to solve the equation
y'(t) = F(t,y(t))
It is important that the initial solution is to be given before
defining the equation. We eventually run the time marching procedure.
ODESolver ode(HEUN);
ode.setInitial(0.);
ode.setInitialRHS(1.);
TimeLoop {
ode.setCoef(1,1,0,2*exp(theTime)1);
ode.runOneTimeStep();
}
cout << ode << endl;
cout << "Error: " << fabs(exp(theFinalTime)1ode.get()) << endl;
return 0;
}


We can now display the ode data and the solution. We also dispolay the
error for this case where the exact solution is
y(t) = exp(t)1
.
cout << ode << endl;
cout << "Solution: " << ode.get() << endl;
cout << "Error: " << fabs(exp(theFinalTime)1ode.get()) < endl;
return 0;
}
