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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.

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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

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  • $\begingroup$ Thanks! But I am still a bit confused . Can you check the above edit please. $\endgroup$ – Abhishek Bhatia Apr 23 '15 at 11:42
  • $\begingroup$ Looks correct. I do not understand your final question. All the input arguments for the functions are perfectly defined, the functions can be evaluated without further preparation. $\endgroup$ – Dr. Lutz Lehmann Apr 23 '15 at 12:37
  • $\begingroup$ Using RK4 now at each time-step I can compute the velocity and position. But I tend to measure the forces too at each time-step. RK4 computes forces several times. Which value of force should I take? $\endgroup$ – Abhishek Bhatia Apr 23 '15 at 12:41
  • $\begingroup$ This is a different question. It depends. Is the function evaluation an experiment with freely selectable inputs or are you following a trajectory and trying to reconstruct it from its acceleration values measured in regular intervals? Runge-Kutta methods might be unusable in this situation, explore multi-step methods. $\endgroup$ – Dr. Lutz Lehmann Apr 23 '15 at 12:48
  • $\begingroup$ If I understand correctly the forces are computed directly and not using the acceleration values. So I wish to simplify measure the force at regular intervals of the timestep itself. $\endgroup$ – Abhishek Bhatia Apr 23 '15 at 13:05

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