The validated Euler tube encloses the exact periodic flow and monodromy matrix

Fix any exact state $y_n\in Y_n$ at time $nh$ and take its maximal solution from [F3]. If $M=0$, [F1] gives $H(y_n)=0$, so the constant solution and uniqueness in [F3] give the claim for this cell. Suppose $M>0$. Then $y_n$ is interior to $Q_n$. While the solution remains in $K$, each coordinate derivative $H_i(y(t))$ is continuous; [F5] and [F6] therefore give, coordinatewise, $-M(t-nh)\le y_i(t)-y_{n,i}\le M(t-nh)$ and hence $\mathrm{\parallel }y(t)-y_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le M(t-nh)$. A first exit from $K$ at $\tau <nh+h$ would lie strictly inside $y_n+[-Mh,Mh{]}^m\subseteq Q_n\subseteq K$, hence in the interior of $K$, contradicting first exit. Nor can the maximal solution end at a finite $b\le nh+h$ while remaining in $K$: the same estimate makes $y(t)$ Cauchy as $t\uparrow b$, with limit $y_b\in K\subset \mathrm{\Omega }$; local existence from $(b,y_b)$ in [F3] extends it, since continuity of $H$ gives the matching derivative at the join. This contradicts maximality. Thus the solution exists through the cell endpoint and, by the bound and continuity, stays in $Q_n$ throughout the closed cell.

By [F5], the exact Euler defect on this cell has coordinate integrals ${\delta }_i={\int }_{nh}^{nh+h}(H_i(y(t))-H_i(y_n))dt$. By [F1] and step 1.1, each integrand lies between $-LM(t-nh)$ and $LM(t-nh)$. Applying [F6] on $[nh,nh+h]$ gives $\mathrm{\mid }{\delta }_i\mathrm{\mid }\le {\int }_{nh}^{nh+h}LM(t-nh)dt=LM{h}^2\mathrm{/}2$ for every coordinate. Also $H(y_n)\in H^{\mathrm{\#}}(Y_n)$; hence the exact next value lies in $Y_n+h{H}^{\mathrm{\#}}(Y_n)+[-LM{h}^2\mathrm{/}2,LM{h}^2\mathrm{/}2{]}^m=Y_{n+1}$.

Initially the exact state lies in $Y_0$. Apply steps 1.1 and 2.1 successively for $n=0,\dots ,N-1$. Induction places each exact cell in $Q_n$ and its endpoint in $Y_{n+1}$, in particular the final endpoint in $Y_N$. The no-exit argument and maximal-solution property [F3] establish existence and uniqueness through all cells.

For $H(x,Z,\theta )=(\theta f(x),\theta Df(x)Z,0)$, the state block solves $x^{\mathrm{\prime }}=\theta f(x)$ and the matrix block solves $Z^{\mathrm{\prime }}=\theta Df(x)Z$ with $Z(0)=I_d$. Rescaling physical time $t=\theta s$ makes the state block ${\varphi }_{\theta s}(x_0)$. Step 3.1 establishes existence through that time. Since $f$ is $C^2$ on open $U$, [F4] gives the state derivative there on its open maximal-flow domain, even if $X_0$ is degenerate or the point is on its boundary. Its variational equation and uniqueness identify $Z(s)$ with $D_x{\varphi }_{\theta s}(x_0)$. The terminal block inclusions from step 3.1 are therefore the claimed flow and monodromy enclosures.