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This explicit numerical method is a two-stage Runge-Kutta method and a one-step method. For a sufficiently smooth ordinary differential equation, the chain rule gives $y^{\prime \prime }=f_x+f_{y}f$, and Taylor's theorem shows that its local truncation error is $O(h^3)$. Uniform Lipschitz continuity of $f$ in its solution argument gives an increment-function Lipschitz constant at most $L(1+hL/2)$. The Lipschitz one-step error bound then proves second-order numerical convergence on a fixed time interval. This method uses an intermediate prediction from $y_n$; the explicit midpoint linear multistep method instead requires a separate previous value $y_{n-1}$.

By the chain rule, $y^{\prime \prime }=f_x+f_{y}f$ along the exact solution. Applying Taylor's theorem to the exact increment and the two arguments of $f$ givesThe assumed differentiability makes the remainders uniform on $0\le x\le 1-h$. Subtracting gives the consistency errorHere $O$ denotes Big O notation. The general order of a numerical method is therefore one for $\beta \ne 1/2$ and at least two for $\beta =1/2$. For example, on $y^{\prime }=y$ with unit initial value, $y_N=(1+h+\beta h^2{)}^{1/h}$. Expanding its logarithm gives $y_N=e[1+(\beta -1/2)h+O(h^2)]$, so the first-order term is actually nonzero when $\beta \ne 1/2$.

To control accumulation, the mean value theorem and the bound on $f_y$ givePut $K_h=L(1+\beta hL)$, $q_h=1+h{K}_h$ and $e_n=y(x_n)-y_n$. The exact and numerical updates yield $|e_{n+1}|\le q_h|e_n|+h|T_n|$, with $e_0=0$. Iterating this recurrence, as in the Lipschitz one-step error bound, givesBecause $K_h\le L(1+L)$ for $h\le 1$, the prefactor is bounded independently of the mesh. When $\beta =1/2$, the consistency error is $O(h^2)$, hence the maximum global error is $O(h^2)$. This proves at least second-order numerical convergence of the explicit midpoint method.

Put $e_n=y(x_n)-y_n$. Subtracting the numerical update from the exact update with its local truncation error givesThe Lipschitz condition in the solution variable and the triangle inequality implyThe assumed positive denominator permits division. With $K=[1+L(1-\theta )h]/(1-L\theta h)$ and $e_0=0$, repeated substitution of this Lipschitz one-step error bound yieldsHere $K>1$ because $L,h>0$, so the displayed geometric series quotient is well defined. For implicit steps, the same condition makes the scalar update map a contraction mapping, ensuring a unique numerical next value.