Runge Kutta in Fortran - fortran

Question

I'm trying to implement the Runge Kutta method in Fortran and am facing a convergence problem. I don't know how much of the code I should show, so I'll describe the problem in detail, and please guide me as to what I should add/remove to/from the post to make it answerable.
I have a 6-dimensional vector of position and velocity of a ball, and a corresponding system of diff. eqs. that describe the equations of motions, from which I want to calculate the trajectory of the ball, and compare results for different orders of the RK method.
Let's focus on 3rd order RK. The model I use is implemented as follows:
k1 = h * f(vec_old,omega,phi)
k2 = h * f(vec_old + 0.5d0 * k1,omega,phi)
k3 = h * f(vec_old + 2d0 * k2 - k1,omega,phi)
vec = vec_old + (k1 + 4d0 * k2 + k3) / 6d0
Where f is the function that constitutes the equations of motion (or equivalently the RHS of my system of diff. eqs). Note that f is time independent, therefore has only 1 argument. h takes the role of a small time step dt.
If we wish to calculate the trajectory of the ball for a finite time total_time, and allow for a total error of epsilon, then we need to ensure each step takes a proportional fraction of the error. For the first step, I then did the following:
vec1 = solve(3,vec_old,h,omega,phi)
vec2 = solve(3,vec_old,h/2d0,omega,phi)
do while (maxval((/(abs(vec1(i) - vec2(i)),i=1,6)/)) > eps * h / (tot_time - current_time))
h = h / 2d0
vec1 = solve(3,vec_old,h,omega,phi)
vec2 = solve(3,vec_old,h/2d0,omega,phi)
end do
vec = (8d0/7d0) * vec2 - (1d0/7d0) * vec1
Where solve(3,vec_old,h,omega,phi) is the function that calculates the single RK step described above. 3 denotes the RK order we are using, vec_old is the current state of the position-velocity vector, h, h/2d0 both represent the time step being used, and omega,phi are just some extra parameters for f. Finally, for the first step we set current_time = 0d0.
The point is that if we use a 3rd order RK, we should have an error in $O(h^3)$, and thus fall off faster than linearly in h. Therefore, we should expect the while loop to eventually come to a halt for small enough h.
My problem is that the loop doesn't converge, and not even close - the ratio
maxval(...) / eps * (...)
remains pretty much constant, all the way until eps * h / (tot_time - current_time)) becomes zero due to finite precision.
For completeness, this is my definition for f:
function f(vec_old,omega,phi) result(vec)
real(8),intent(in) :: vec_old(6),omega,phi
real(8) :: vec(6)
real(8) :: v,Fv
v = sqrt(vec_old(4)**2+vec_old(5)**2+vec_old(6)**2)
Fv = 0.0039d0 + 0.0058d0 / (1d0 + exp((v-35d0)/5d0))
vec(1) = vec_old(4)
vec(2) = vec_old(5)
vec(3) = vec_old(6)
vec(4) = -Fv * v * vec_old(4) + 4.1d-4 * omega * (vec_old(6)*sin(phi) - vec_old(5)*cos(phi))
vec(5) = -Fv * v * vec_old(5) + 4.1d-4 * omega * vec_old(4)*cos(phi)
vec(6) = -Fv * v * vec_old(6) - 4.1d-4 * omega * vec_old(4)*sin(phi) - 9.8d0
end function f
Does anyone have any idea as to why the while loop doesn't converge?
If anything else is needed (output, other pieces of code etc.) please tell me and I'll add it. Also, if trimming is required, I'll cut whatever would be considered unnecessary. Thanks!

Answer 1

To compute the step error using the half step method, you need to compute the approximation at t+h in both cases, which means two steps with step size h/2. As it is now you compare the approximation at t+h to the approximation at t+h/2 which gives you an error of size f(vec(t+h/2))*h/2.
Thus change to a 3-step procedure
vec1 = solve(3,vec_old,h,omega,phi)
vec2 = solve(3,vec_old,h/2d0,omega,phi)
vec2 = solve(3,vec2 ,h/2d0,omega,phi)
in both locations, the difference of vec2-vec1 should then be of order h^4.