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For a uniformly Lipschitz continuous state field $f$ with constant $L$, an implicit midpoint rule step is uniquely solvable when $hL/2<1$, by the contraction mapping theorem. Write its step map as ${\Psi }_h(u)=u+h{\Phi }_h(u)$, where ${\Phi }_h(u)=f(t+h/2,u+h{\Phi }_h(u)/2)$. Subtracting the increment equations givesTherefore $\mathrm{Lip}({\Phi }_h)\le L/(1-hL/2)$, and the triangle inequality gives the bound for ${\Psi }_h$. Combining it with the implicit midpoint consistency bound yieldsfor $L>0$. This follows by summing the geometric error recurrence with forcing $K{h}^3/(1-hL/2)$. For $hL\le 1$ and $nh\le T$, $R_h^n\le e^{2LT}$, proving second-order global error with exact starting data. If $L=0$, the direct bound is $\Vert e_n\Vert \le \Vert e_0\Vert +KT{h}^2$.

Work on a fixed finite interval $[a,a+T]$. Write $Y_n=y(x_n)$, $t=x_n+h/2$, and use the paper's normalized local truncation errorAssume $y$ has bounded continuous derivatives through order three on this interval, with $\Vert y^{\prime \prime }\Vert \le M_2$ and $\Vert y^{\prime \prime \prime }\Vert \le M_3$. Assume also that the Jacobian matrix $f_y$ has operator norm at most $L$ on the segments joining $y(t)$ to the endpoint averages. For a scalar equation this last hypothesis is simply $|\partial f/\partial y|\le L$ there.

Two symmetric estimates separate the errors. By Taylor's theorem applied to $y^{\prime }$ about $t$, the linear term integrates to zero on $[-h/2,h/2]$, givingLikewise the two linear terms in the endpoint expansions cancel, and their second-order remainders giveThe derivative bound on $f$ gives a Lipschitz condition in its second variable. Since $y^{\prime }(t)=f(t,y(t))$, the triangle inequality now yieldsThe constant is independent of $n$ and $h$ as long as the steps remain in the fixed interval. This implicit midpoint consistency bound proves numerical consistency of order two. The corresponding unnormalized residual is $h{\tau }_n=O(h^3)$; the two powers conventionally called local truncation error differ because the paper divides by the step size. If local error instead means the error of an exactly solved numerical step started at $Y_n$, the small-step Lipschitz condition gives the bound $h\Vert {\tau }_n\Vert /(1-hL/2)=O(h^3)$ as well.