0

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?

Idonknow
  • 15,643
  • 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. – Simply Beautiful Art Oct 17 '20 at 13:42
  • From the speed of decaying, as $h$ tends to $0$, one can conclude that RK4 method converges much faster than Euler midpoint method? – Idonknow Oct 17 '20 at 13:43
  • Yes, $h^4$ vanishes faster than $h^2$. – Simply Beautiful Art Oct 17 '20 at 13:45
  • I see. Perhaps you can post an answer so that I can accept your answer? – Idonknow Oct 17 '20 at 13:45
  • See https://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. – Lutz Lehmann Oct 17 '20 at 14:07
  • How do you measure efficiency? Are you also considering the cost of getting a reliable error estimate? You may well get a sufficiently accurate result and a reliable error estimate at a lower computational cost with a low order method than with a high order method. – Carl Christian Oct 17 '20 at 15:55

1 Answers1

1

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.

Simply Beautiful Art
  • 74,685