Midpoint method

Remarks

Proof idea: by Taylor's theorem, $y(t + h) = y + h f + \frac{h^2}{2}(f_t + f_y f) + O(h^3)$, with $f$ and its partial derivatives taken at $(t, y(t))$. Expanding the increment gives $h f\big(t + \frac h2, y + \frac h2 f\big) = h f + \frac{h^2}{2}(f_t + f_y f) + O(h^3)$, and the two expansions agree up to $h^2$. In Butcher form it has $c = (0, \tfrac12)$, $a_{21} = \tfrac12$ and $b = (0, 1)$. Example: for $y' = \lambda y$ it gives $y_{n+1} = \big(1 + z + \frac{z^2}{2}\big) y_n$ with $z = h\lambda$, the Taylor polynomial of $e^z$ of degree $2$. The implicit midpoint method $y_{n+1} = y_n + h f\big(t_n + \frac h2, \frac{y_n + y_{n+1}}{2}\big)$ also has order $2$; it is the one-stage Gauss–Legendre method.

Going deeper

For a student

The explicit midpoint method is one of the simplest second-order Runge–Kutta methods, together with Heun's method. Both satisfy the order conditions $\sum b_i = 1$ and $\sum b_i c_i = \frac12$, and they differ in the choice of the second stage. The implicit midpoint method is symplectic and preserves quadratic invariants, which makes it useful for Hamiltonian systems.