Consider that I am solving a second order ODE using RK2/RK4. The ODE represents
simple equations of motion:
Equations of motion I am trying to solve:
\begin{align}
\frac{dx}{dt} &= v
\\[.3em]
m·\frac{dv}{dt}&= f_{1}(x)+f_{2}(x,v)
\end{align}
RK2 method:
\begin{align}
s_{x1}&=h f_x (t_i, v_i)
\\
s_{v1}&=h f_v (t_i, x_i, v_i)
\\[.5em]
s_{x2}&=h f_x(t_i+\tfrac{1}{2} h, v_i+\tfrac{1}{2}s_{v1})
\\
s_{v2}&=h f_v(t_i+\tfrac{1}{2}h, x_i+\tfrac{1}{2}s_{x1}, v_i +\tfrac{1}{2} s_{v1})
\\[1em]
x_{i+1}&=x_i+s_{x2}
\\
v_{i+1}&=v_{i}+s_{v2}
\end{align}
where $f_x(t,x,v)=v$ and $f_v(t,x,v)=\frac{1}{m}(f_1(x)+f_2(x,v))$
Now along with $x$ and $v$, I also require to compute the $f_{1}(x)+f_{2}(x,v)$ at each time-step $h$. In such case what should I take velocity and position pair at that particular time-step?
Method Runge Kutta $4^{th}$ order
Basic Formulae
\begin{align}
x^{'}&=x_{0}+ \frac{1}{6}(k_{0}+2k_{1}+2k_{2}+k_{3})
\\
v^{'}&=v_{0}+ \frac{1}{6}(l_{0}+2l_{1}+2l_{2}+l_{3})
\end{align}
Calculation of coefficients
\begin{align}
k_0 &= h v_0
\\
l_0 &= \frac {h (F_{p}(x_0) +F_g(x_0,v_0)) }{ m}
\\[.5em]
k_1 &= h (v_0+ \frac{l_0}{2})
\\
l_1 &= \frac {h (F_{p}(x_0 + \frac {k_0}{2}) +F_g(x_0 + \frac {k_0}{2}, v_0 + \frac {l_0}{2})) }{ m}
\\[.5em]
k_2 &= h (v_0+ \frac{ l_1}{2})
\\
l_2 &= \frac {h (F_{p}(x_0 + \frac {k_1}{2}) +F_g(x_0 + \frac {k_1}{2}, v_0 + \frac {l_1}{2})) }{ m}
\\[.5em]
k_3 &= h (v_0+ {l_2})
\\
l_3 &= \frac {h (F_{p}(x_0 + {k_2}) +F_g(x_0 + {k_2},v_0 + {l_2})) }{ m}
\end{align}
In the above method, what will the value of $F_p$ and $F_g$. Should I take it the ones at the before applying runge-kutta simply at the $x_0$ and $v_0$ at that time-step. But this may seem incorrect as velocity and position are not computed using these force values.
Best Answer
You have to evaluate them exactly as the method prescribes. You already did this correctly for the RK2 method.
This might seem to be a lot more effort for RK4. But consider that RK4 is $O(h^4)$. Roughly, to get a accuracy of e.g. about $10^{-4}$ for $t=1$ you need $h=0.1$ and $10$ steps netting $40$ function evaluations. To get the same accuracy for the $O(h^2)$ RK2 method you need a step size $h=0.01$ and $100$ steps netting $200$ function evaluations.
A detailed example for using equal amounts of function evaluations in Euler, Heun, RK2, RK3 and RK4 (with one out of 3 in each method aiming for $10^{-4}$ accuracy) can be found in this answer: https://math.stackexchange.com/a/1239002/115115