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I came up with the following problem while using Euler-midpoint method and the classical fourth-order Runge-Kutta method to solve ordinary differential equations.

As step size $h$ decreases, which method is more efficient in estimation? Euler-midpoint method and the classical fourth-order Runge-Kutta method?

By solving several questions and comparing them to actual value given by exact solution, I realize that for the same step size $h$, the classical fourth-order Runge-Kutta method gives a more accurate estimation compared to the Euler-midpoint method.

However, does the same hold true as $h$ tends to $0$ from positive?

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  • $\begingroup$ This comes down to the errors of each method. The midpoint rule is known for $\mathcal O(h^2)$ local error (and is in fact a 2nd order RK method), while the classical RK 4 method has $\mathcal O(h^4)$ local error, which decays much faster. $\endgroup$ Commented Oct 17, 2020 at 13:42
  • $\begingroup$ From the speed of decaying, as $h$ tends to $0$, one can conclude that RK4 method converges much faster than Euler midpoint method? $\endgroup$
    – Idonknow
    Commented Oct 17, 2020 at 13:43
  • $\begingroup$ Yes, $h^4$ vanishes faster than $h^2$. $\endgroup$ Commented Oct 17, 2020 at 13:45
  • $\begingroup$ I see. Perhaps you can post an answer so that I can accept your answer? $\endgroup$
    – Idonknow
    Commented Oct 17, 2020 at 13:45
  • $\begingroup$ See math.stackexchange.com/a/1239002/115115 for an example. RK2 there is the explicit midpoint method. Note also the practical limitations of using floating-point arithmetic, there is no convergence for $h\to 0$ in the experiments, or rather one would have to switch to a variable-precision numerical data type. $\endgroup$ Commented Oct 17, 2020 at 14:07

1 Answer 1

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This comes down to the errors of each method. The midpoint rule is known for $\mathcal O(h^3)$ local error or $\mathcal O(h^2)$ global error (and is in fact a 2nd order RK method), while the classical RK 4 method has $\mathcal O(h^5)$ local error or $\mathcal O(h^4)$ global error, which decay much faster.

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