I would like to use Boost C++ odeint Runge-Kutta integrator on a system that looks like this :

$$\ddot x = - \frac A{||x||^3} * x $$

$ x $ is a vector in 3D space, so basicaly $ x(i, j, k) $
$ \ddot x $ is its second derivative
$ {||x||^3} $ is magnitude cubed of $ x $
$ A $ is a constant

I know the initial conditions of the problem, namely $ \dot x(t=0) $ and $ x(t=0) $ .

I have checked this example in the odeint documentation, as well as the full code here. Examples show use of odeint with a single ODE. But my problem would have to be split into 6 ODE.

Can I use odeint Runge Kutta method for such a system ( 6 ODEs ) and if yes, is there any example that I can follow to help me implement my problem ?


1 Answer 1


In theory, sure, although I'd be wary of your particular ODE system for a couple reasons: (1) the right-hand side is undefined when $x = 0$, which will trip up any integrator in the Boost C++ suite, (2) based on its similarity to functions known to be non-Lipschitz in neighborhoods of the origin, even existence of solutions is unclear.

Implementing your system should be straightforward. Transform your system into first-order form, and then model your implementation after the Lorenz system attractor example on the odeint website.

  • $\begingroup$ Thank you for your answer, especially the warnings. I wouldn't worry about x=0 too much, because given what I know about the system I am solving, I believe this state cannot occur, at least not in a reasonable interval of integration that I want to observe. I would like to ask a thing about Lorenz system example. Are dxdt[0], dxdt[1] and dxdt[2] the "equivalent" for the IJK coordinates in my system ? Thank you. $\endgroup$
    – James C
    Nov 18, 2014 at 3:12
  • 1
    $\begingroup$ @JamesC: You would introduce an auxiliary variable, say, $y$ and set $\dot{x} = y$, and $\ddot{x} = \dot{y} = -Ax/\|x\|^{3}$. At that point, the abstraction is basically that you have a vector of six state variables $x[i]$, $x[j]$, $x[k]$, $y[i]$, $y[j]$, $y[k]$, so if you want to think of these as being x[0] through x[5], you're representing the same quantities, just relabeled so that they're in a form convenient for use in an ODE solver. $\endgroup$ Nov 18, 2014 at 3:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.