= Implicit midpoint consistency bound
{title2=$\|\tau_n\|\leq(M_3/24+LM_2/8)h^2$}
For an exact solution $y'=f(t,y)$, define the <normalized local truncation error> of the <implicit midpoint rule> by
$$
\tau_n=\frac{y(t_n+h)-y(t_n)}h-
f\left(t_n+\frac h2,\frac{y(t_n+h)+y(t_n)}2\right).
$$
If $\|y''\|\leq M_2$, $\|y'''\|\leq M_3$ and the state <Jacobian matrix> of $f$ has <operator norm> at most $L$ on the intervening state segments, then the displayed bound holds uniformly on a fixed finite interval. To prove it, put $t=t_n+h/2$. Symmetric <Taylor expansions> give
$$
\left\|\frac{y(t+h/2)-y(t-h/2)}h-y'(t)\right\|
\leq\frac{M_3h^2}{24},\qquad
\left\|\frac{y(t+h/2)+y(t-h/2)}2-y(t)\right\|
\leq\frac{M_2h^2}{8}.
$$
For the first bound, integrate the second-order remainder of $y'(t+s)$ over the symmetric step; for the second, average the second-order remainders of $y$ at the endpoints. The odd terms cancel. The <Lipschitz condition> for $f$ and $y'(t)=f(t,y(t))$ finish the proof by the <triangle inequality>. The unnormalized step residual is $h\tau_n=O(h^3)$.
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