Each SDC sweep raises the order up to the collocation order

The stage points of the contraction limit $U^{\ast }$ lie in the compact set $\mathcal{T}$, since all iterates do and they converge by [F3]. The constant predictor gives $\mathrm{\parallel }{U}^0-U^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty },2}=\mathrm{\parallel }hQF(U^{\ast }){\mathrm{\parallel }}_{\mathrm{\infty },2}\le h\mathrm{\parallel }Q{\mathrm{\parallel }}_{\mathrm{\infty }}B$ by [F1] and the row-sum norm. Since $q\le C_{q}h$, [F3] gives $\mathrm{\parallel }{U}^k-U^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty },2}\le C_q^k\mathrm{\parallel }Q{\mathrm{\parallel }}_{\mathrm{\infty }}B{h}^{k+1}$ for fixed $k$ (with the same $O(h)$ bound when $k=0$).

To bound dependence on the input, let $D_j=\mathrm{\parallel }{U}^j(u)-U^j(v){\mathrm{\parallel }}_{\mathrm{\infty },2}$ for two starts $u,v\in {\mathcal{U}}_n$, and put $\delta =hL\mathrm{\parallel }Q-Q_{\mathrm{\Delta }}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Subtracting the stage equations [F2] and using the row-sum norm gives $(1-a)D_{j+1}\le \mathrm{\parallel }u-v{\mathrm{\parallel }}_2+\delta D_j$. Since $a\le 1\mathrm{/}2$ and $\delta =O(h)$, $D_0=\mathrm{\parallel }u-v{\mathrm{\parallel }}_2$ and induction for the fixed number $k$ gives $D_k\le C_k\mathrm{\parallel }u-v{\mathrm{\parallel }}_2$ with $C_k$ independent of $h$.

Let $\mathrm{\Omega }$ be the open neighborhood of $\mathcal{T}$ on which the stated smoothness holds, within the IVP domain. There is $\rho >0$ such that the closed $3\rho$-neighborhood $N$ of $\mathcal{T}$ lies in $\mathrm{\Omega }$: cover $\mathcal{T}$ by finitely many balls whose concentric doubled balls lie in $\mathrm{\Omega }$, and take $3\rho$ smaller than the minimum of their radii. The set $N$ is compact, so all partial derivatives through order $M$ are bounded there. In particular, since $M\ge 1$, the state Jacobian has a uniform bound $\widehat{L}$; applying the scalar mean-value estimate along state segments gives a state Lipschitz bound $\widehat{L}$ on each convex cylinder contained in $N$. For an exact start $u=y(t)$, the fixed cylinder $K_t=[t,t+\rho ]\times \bar{B}(u,\rho )$ is compact, convex, and contained in $N$, because its points have distance at most $\sqrt{2}\rho$ from $(t,u)\in \mathcal{T}$. These bounds are uniform in $t$ and independent of $h$.

For an exact start, [F1] and step 1.1 give ${\sup }_{0\le \tau \le 1}\mathrm{\parallel }{u}_h(\tau )-u{\mathrm{\parallel }}_2\le h{\mathrm{\Lambda }}_{M}B$. The exact path has speed at most $B$, hence ${\sup }_{0\le \tau \le 1}\mathrm{\parallel }y(t+\tau h)-u{\mathrm{\parallel }}_2\le hB$. Thus for $h\le \rho$ and $hB\max \{1,{\mathrm{\Lambda }}_M\}\le \rho$, both paths, the stages, and their comparison segments lie in $K_t$. Any two collocation solutions $V,W$ with stage points in $K_t$ satisfy $\mathrm{\parallel }V-W{\mathrm{\parallel }}_{\mathrm{\infty },2}\le h\widehat{L}\mathrm{\parallel }Q{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }V-W{\mathrm{\parallel }}_{\mathrm{\infty },2}$ by subtracting [F1]. Taking $h\widehat{L}\mathrm{\parallel }Q{\mathrm{\parallel }}_{\mathrm{\infty }}<1$ proves uniqueness there; existence follows from [F3] and the containment of $U^{\ast }$. Together with the assumed exact-path derivative bound, step 1.3 now verifies every tube, smoothness, and branch hypothesis of [F4], with uniform constants.

Applying the endpoint formula [F2] to the two stage vectors gives $\mathrm{\parallel }{\mathrm{\Psi }}_h^{(k)}(u)-{\mathrm{\Psi }}_h^{(k)}(v){\mathrm{\parallel }}_2\le \mathrm{\parallel }u-v{\mathrm{\parallel }}_2+hL\mathrm{\parallel }b{\mathrm{\parallel }}_1{D}_k\le (1+C_{s}h)\mathrm{\parallel }u-v{\mathrm{\parallel }}_2$ for $u,v\in {\mathcal{U}}_n$.

The endpoint formula in [F2], the state Lipschitz bound, and step 1.1 give a difference from the exact collocation endpoint of at most $hL\mathrm{\parallel }b{\mathrm{\parallel }}_1\mathrm{\parallel }{U}^k-U^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty },2}=O(h^{k+2})$. Adding the exact collocation defect [F4], whose hypotheses were checked in step 2.1, gives an exact-start defect $O(h^{M+1})+O(h^{k+2})=O(h^{p+1})$, since $p+1=\min (M+1,k+2)$ and $h\le 1$.

Let $e_n=\mathrm{\parallel }{y}_n-y(t_n){\mathrm{\parallel }}_2$. Both starts lie in ${\mathcal{U}}_n$ by the Statement. Insert and subtract the exact-start one-step output to obtain, using step 2.2 and step 3.1, $e_{n+1}\le (1+C_{s}h)e_n+C{h}^{p+1}$ with $p=\min (M,k+1)$ and a uniform $C$. The initial error is $e_0=0$, so [F5] gives global order $p$.