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

1k Views Asked by At

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?

1

There are 1 best solutions below

0
On

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.