One-node Gauss collocation is implicit midpoint of order two

Substituting [F1] under [A2] into [F2] gives $U=u+(h\mathrm{/}2)f(t+h\mathrm{/}2,U)$ and ${\mathrm{\Psi }}_h(u)=u+hf(t+h\mathrm{/}2,U)$, which is precisely implicit midpoint. Since $f$ is bounded on the compact tube, the stage equation gives $U-u=O(h)$. Taylor expansion of each coordinate at $(t,u)$ then gives $U-u=(h\mathrm{/}2)f(t,u)+O(h^2)$: first use $U-u=O(h)$ in $f(t+h\mathrm{/}2,U)=f(t,u)+O(h)$ and substitute that estimate back into the stage equation.

Write $f_0=f(t,u)$. Applying the first-order scalar Taylor formula with quadratic remainder [F3] to each coordinate at $(t+h\mathrm{/}2,U)$ and using step 1.1 yields $f(t+h\mathrm{/}2,U)=f_0+\frac{h}{2}(f_t(t,u)+D_{y}f(t,u)f_0)+O(h^2)$ in Euclidean norm. The exact path has $y^{\mathrm{\prime }}(t)=f_0$ by [F7] and $y^{\mathrm{\prime }\mathrm{\prime }}(t)=f_t(t,u)+D_{y}f(t,u)f_0$ by [F3]. Coordinatewise Taylor's formula for $y$ and the endpoint expression in step 1.1 therefore give $y(t+h)-{\mathrm{\Psi }}_h(u)=O(h^3)$ uniformly.

Let $\mathcal{K}$ be the compact convex time-state tube in the Statement, and let $V$ be its open neighborhood where $f$ is componentwise $C^2$. By [F8] choose $r>0$ such that the closed $4r$-neighborhood ${\mathcal{K}}^{+}$ lies in $V\subseteq D$. This neighborhood is compact and convex, and [F8] gives uniform bounds $B_{+}$ for $f$, $L_{+}$ for $D_{y}f$, and bounds for the second partial derivatives there. If $\mathrm{\parallel }u-y(t_n){\mathrm{\parallel }}_2\le r$ and $h\le 2r$, every stage point $(t_n+h\mathrm{/}2,Y)$ with $\mathrm{\parallel }Y-u{\mathrm{\parallel }}_2\le r$ is within $3r$ of $(t_n,y(t_n))\in \mathcal{K}$, hence lies in ${\mathcal{K}}^{+}$. For $h{B}_{+}\mathrm{/}2\le r$ and $h{L}_{+}\mathrm{/}2\le 1\mathrm{/}2$, [F9] supplies the unique midpoint stage in that ball. Any other midpoint stage in ${\mathcal{K}}^{+}$ equals it by subtracting the equations and using $h{L}_{+}\mathrm{/}2<1$; thus it agrees with the branch assumed in the Statement whenever that branch lies in $\mathcal{K}$. The Taylor estimates of steps 1.1–2.1 and stability [F5] apply uniformly on ${\mathcal{K}}^{+}$, with endpoint Lipschitz bound $1+2L_{+}h$.

Set $y_0=y(0)$ and compute on $h=T\mathrm{/}N$. As long as $e_n:=\mathrm{\parallel }{y}_n-y(t_n){\mathrm{\parallel }}_2\le r$, step 3.1 defines the next stage and endpoint. Inserting the exact-start endpoint and using steps 2.1 and 3.1 gives $e_{n+1}\le (1+2L_{+}h)e_n+C{h}^3$. On every such mesh prefix [F6] bounds all errors, including the newly computed endpoint error, by $C_T{h}^2$, with $C_T$ independent of the prefix length because that length is at most $T$. Choose $h$ small enough that $C_T{h}^2<r$. Starting with $e_0=0$, induction now continues every step and keeps every grid state within $r$ of the exact path. All grid states and stages therefore lie in ${\mathcal{K}}^{+}$, and the same recurrence gives ${\max }_{0\le n\le N}{e}_n\le C_T{h}^2$. This proves global order two without assuming endpoint containment in the original tube.