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A system of differential equations can be solved using a numerical integrator with adaptive stepsize $ h$, e.g. Runge Kutta Fehlberg

Now suppose I want to integrate a set of differential equations $\{\dot{x}, \dot{y}, \dot{z} \}$ numerically (e.g. the 3D trajectory of some particle). Each differential equation uses the same adaptive stepsize $h$.

I want to evaluate $x,y$ at some specific target value of $z = z_t$

Now, due to the adaptive stepsize, it is unlikely that $z$ will ever be exactly $z_t$. That is to say, I might get for integration step $n$ and step $n+1$, $z_n=z_t + \epsilon$, $z_{n+1}=z_t - \delta$ for small values $\epsilon, \delta$

Obviously, by setting a more strict tolerance on the integrator, I can get close to $z_t$, but this would be at the expense of speed.

Given that I know a priori that I want to evaluate $x,y$ at $z=z_t$, is there a method which permits both an adaptive stepsize, and an exact $z=z_t$ result?

Thanks

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  • $\begingroup$ See for example stackoverflow.com/q/48892663/3088138 where zero-crossings of an oscillatory process are computed. The last attempt uses Newton's method for finding the best t. Note that odeint uses lsoda which uses adaptive step sizes. $\endgroup$ Feb 21 '18 at 16:59
  • $\begingroup$ Why do you want to control the position of the last step indirectly via the tolerances and not directly by cutting it to the desired length? $\endgroup$ Feb 21 '18 at 17:01
  • $\begingroup$ @LutzL Thanks for your reply. Can you explain what you mean by 'cutting it to the desired length'? $\endgroup$ Feb 21 '18 at 17:02
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    $\begingroup$ something like if t+1.01*dt > tf then dt = tf-t. The factor 1.01 is to avoid/preempt to cut to a ridiculously small step. $\endgroup$ Feb 21 '18 at 17:06
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Once you choose a method, it has an order which approximates how the error depends on stepsize. The popular fourth order Runge-Kutta has an error that scales as $h^4$. You can get an approximation to the accuracy by doing your step once as a full step and once as two half steps. You would expect the error to be reduced by a factor $16$ so the difference between them is about $15$ times the error of the two step approach. You can use this to update your stepsize. There is a discussion in chapter 17.2 of Numerical Recipes and probably in many other numerical analysis texts.

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  • $\begingroup$ The question was IMO not about implementing the adaptive step size, but how to implement what is sometimes called "events" on top of an adaptive scheme. $\endgroup$ Feb 22 '18 at 9:02

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