\section{Setting and main results}
\label{q:sec:statements}

\subsection{Autonomous weakly reversible networks}
We use the reaction indices $E$, vectors $\nu_e$, field $f_k$, stoichiometric
subspace $S$, classes $P_c$ and boxes $B_h$ of Sections~\ref{u:sec:setting}--\ref{u:sec:affine}.
In this part complexes lie in $\N^d$, $d\ge1$, and the rates are constant in time;
their prefactors may depend on $\eps$. Thus the positive Carath\'eodory solutions
of Part~\ref{part:uniform} are $C^1$ solutions here.

A network is \emph{weakly reversible} when every reaction lies on a directed
cycle. Such a network is endotactic: if a reaction has negative projection in a
direction $r$, a return path crosses upward from below its source level,
providing the witness in Definition~\ref{u:def:endo}. Consequently
Theorem~\ref{u:thm:main}, applied to a singleton rate box, supplies global positive
solutions for every positive constant rate vector. The task below is to quantify
their eventual containment as that rate vector degenerates.

\subsection{The general theorem}

\begin{theorem}[General rate-dependent containment]\label{q:thm:general}
For every finite weakly reversible network with $d\ge1$ species there is $H>0$, depending on the network alone, with the following property. Fix a positive class representative $c$ and prefactor bounds $0<\kappa_-\le\kappa_+$. There is $\alpha\in(0,1)$, depending on the network, $c$, $\kappa_\pm$ only, such that for all
\[
0\le a_-\le a_+,\qquad a_e\in[a_-,a_+],\qquad 0<\eps\le1,\qquad \bar k_e\in[\kappa_-,\kappa_+],
\]
the rates $k_e=\bar k_e\eps^{a_e}$ admit a global forward solution from every $x_0\in P_c$, and every global forward solution from $x_0$ has a finite $T\ge0$ such that for all $t\ge T$ and every coordinate $i$,
\begin{equation}\label{q:eq:general-bound}
 \sqrt\alpha\,\eps^{\Delta/(2H)}\le x_i(t)\le\alpha^{-1/2}\eps^{-\Delta/(2H)},\qquad \Delta=a_+-a_-.
\end{equation}
The entry time $T$ may depend on $x_0$ and on the rate data.
\end{theorem}

The bound depends on the \emph{spread} of the exponents, not on their common level. Multiplying every rate by $\eps^{a}$ changes the clock but not the phase portrait (Corollary~\ref{q:cor:clock}), so a common offset cannot affect concentration geometry; it does affect physical time. An empty reaction set gives constant solutions and singleton classes and is covered trivially. The constants $H$ and $\alpha$ are existence constants: $H$ is a finite minimum over the slope data of the affine construction, and $\alpha$ is chosen from the class threshold and rate-margin inequality; Section~\ref{q:sec:effective-H} discusses what blocks their numerical extraction.

\subsection{The coupled network and the prepared basin}
Consider the four reversible pairs
\begin{equation}\label{q:eq:reactions}
 2Y+Z\ \rightleftharpoons\ 2X,\qquad X+2Z\ \rightleftharpoons\ 2Y,\qquad 2X+Y\ \rightleftharpoons\ 2Z,\qquad 3Z\ \rightleftharpoons\ X,
\end{equation}
with forward rate constants $a_1,a_2,a_3,K$ and reverse rate constants $b_1p^2,\ b_2p^2,\ b_3ph^6,\ b_4p^2$, where $p=uh$. The admitted parameter rectangle is
\begin{equation}\label{q:eq:parameters}
 0<h\le\tfrac1{600},\qquad 0<u\le\tfrac1{20},\qquad a_1,a_2,a_3,b_1,b_2,b_3,b_4\in\big[\tfrac{99}{100},\tfrac{101}{100}\big],\qquad K\in\big[\tfrac{99}{25},\tfrac{101}{25}\big].
\end{equation}
The net currents and the literal mass-action field are
\begin{equation}\label{q:eq:currents}
\begin{aligned}
 J_1&=a_1y^2z-b_1p^2x^2, & J_2&=a_2xz^2-b_2p^2y^2,\\
 J_3&=a_3x^2y-b_3ph^6z^2, & J_4&=Kz^3-b_4p^2x,\\
 \dot x&=2J_1-J_2-2J_3+J_4, & \dot y&=-2J_1+2J_2-J_3,\qquad \dot z=-J_1-2J_2+2J_3-3J_4.
\end{aligned}
\end{equation}
For positive states define the inventory, the inventory coordinate and two shape coordinates
\begin{equation}\label{q:eq:coordinates}
\begin{gathered}
 W=6x+5y+2z,\qquad Q=\frac{W}{2ph},\qquad A=\frac{x}{ph^3Q^3},\qquad B=\frac{y}{ph^2Q^2},\\
 C=1-3h^2AQ^2-\tfrac52hBQ,
\end{gathered}
\end{equation}
so that $x=ph^3AQ^3$, $y=ph^2BQ^2$ and $z=phQC$.

\paragraph{Why these coordinates and boxes?}
The vector $(6,5,2)$ annihilates the reaction vectors of pairs $1,2,4$, so these
fast reactions preserve $W$; only pair $3$ changes it. In a $Z$-dominated state,
$z\approx W/2=phQ$. Balancing the leading fast $X$ terms gives
\[
 Kz^3\approx b_4p^2x\quad\Longrightarrow\quad
 x\approx\frac{K}{b_4}\,ph^3Q^3.
\]
The leading $Y$ balance is $J_1\approx J_2$, or
$a_1y^2z\approx a_2xz^2$, and therefore
\[
 y\approx\sqrt{\frac{a_2K}{a_1b_4}}\,ph^2Q^2.
\]
At nominal prefactors these shape coordinates are centered at $A=4$, $B=2$.
The positive inventory input is $13b_3ph^6z^2$, whose scale after division by
$2ph$ is $p^2h^7Q^2$; this identifies the slow clock $\lambda=p^2h^7$.
At the same leading balance, the slow residual is $1-32Q^6$, with zero
$Q=2^{-5/6}\in(1/2,3/4)$. The inner intervals below surround these balances,
while the wider outer box provides a preparation margin. This scaling argument
motivates the choices; Section~\ref{q:sec:coupled-proof} verifies them for the
full coupled field by exact differential identities and interval inequalities.
The prepared outer basin and the inner target are the preimages under $(A,B,Q)$ of
\begin{equation}\label{q:eq:regions}
 \calO=[3,5]\times[1,3]\times[\tfrac14,1],\qquad \calI=[\tfrac72,\tfrac92]\times[\tfrac32,\tfrac52]\times[\tfrac12,\tfrac34].
\end{equation}
Membership in $\calO$ is a condition on the initial state only; no future behaviour is assumed.

\begin{theorem}[Prepared-basin concentration scales and recovery]\label{q:thm:coupled}
Assume~\eqref{q:eq:parameters} and write $\lambda=p^2h^7$. Every positive initial state with $(A(0),B(0),Q(0))\in\calO$ admits a global positive solution of~\eqref{q:eq:currents}. Every such solution remains in the preimage of $\calO$ and satisfies, for all $t\ge0$,
\begin{equation}\label{q:eq:sharp-bounds}
 \tfrac3{64}\,ph^3\le x(t)\le5\,ph^3,\qquad \tfrac1{16}\,ph^2\le y(t)\le3\,ph^2,\qquad \tfrac18\,ph\le z(t)\le ph.
\end{equation}
It lies in the preimage of $\calI$ for every $t\ge(31/10)\lambda^{-1}$. For the prepared initial state
\begin{equation}\label{q:eq:witness}
 x_0=\frac{ph^3}{16},\qquad y_0=\frac{ph^2}{8},\qquad z_0=\frac{ph}{4}\Big(1-\frac54h-\frac34h^2\Big),
\end{equation}
which has $(A,B,Q)(0)=(4,2,\tfrac14)$, every time $T\ge0$ at which the solution is in the target satisfies
\begin{equation}\label{q:eq:time-bounds}
 T\ \ge\ \frac{400}{1313}\,\lambda^{-1}\ \approx\ 0.3046\,\lambda^{-1}.
\end{equation}
\end{theorem}

Thus the worst-case prepared recovery time is pinned between $0.3046\,\lambda^{-1}$ and $3.1\,\lambda^{-1}$, a ratio of $40703/4000\approx10.18$, uniformly over~\eqref{q:eq:parameters}. The three concentration orders in~\eqref{q:eq:sharp-bounds} hold along every prepared trajectory with the displayed parameter-independent constants. The witness trajectory supplies the matching recovery order.

\begin{remark}[Reciprocal inventory comparison]
The constants $400/1313$ and $31/10$ result from comparing $1/Q$: its derivative
cancels the factor $Q^2$ in $\dot Q$ (Section~\ref{q:sec:capture}).
The resulting certified window can be compared with the sampled entry near
$0.358\,\lambda^{-1}$ in Figure~\ref{q:fig:trajectory}.
\end{remark}

\section{Proof of the general theorem}
\label{q:sec:general-proof}

\subsection{The common affine geometry}\label{q:sec:imported}
Use the family $I$, slopes $r_\ell$, offsets $b_\ell(h)$, tolerance $\vartheta$
and threshold $h_g$ from Lemma~\ref{u:lem:affine}. The comparison set $V$ already
contains every complex difference, so the gap $H\in(0,1]$ in~\eqref{u:eq:gap}
is fixed by the network before its rates are chosen. For an active label at
$x=h^q\in B_h$, the sign and ordering estimates
\eqref{u:eq:sign-close}--\eqref{u:eq:ratio} apply. Weak reversibility supplies
the additional return paths needed to turn these geometric estimates into an
explicit margin in terms of the rate spread.

\subsection{The quantitative cut-drift lemma}
Use the maximum source degree $D_*$ defined in Section~\ref{u:sec:drift}. For $x\in B_h$ every source monomial satisfies $x^{y_e}\ge h^{D_*}$. Let $k_0>0$ be a lower bound for all rates and assume the \emph{rate margin}
\begin{equation}\label{q:eq:rate-margin}
 h^{H}\sum_e k_e\ \le\ \frac{k_0}{2}.
\end{equation}

\begin{lemma}[Quantitative cut drift]\label{q:lem:drift}
Let $0<h<h_g$, $x=h^q\in B_h$, and let $\ell$ be active at $x$ with slope $r=r_\ell$. Then $r\cdot f_k(x)\ge0$. If moreover $r\notin S^\perp$, then
\begin{equation}\label{q:eq:drift}
 r\cdot f_k(x)\ \ge\ H\,k_0\,h^{D_*}.
\end{equation}
\end{lemma}

\begin{proof}
Order the complexes by the level $r\cdot y$ and consider the finitely many distinct levels $L_1<\dots<L_m$. For two complexes $y,z$ with $r\cdot z>r\cdot y$ the difference $z-y\in V$ has $r\cdot(z-y)\ge2H$ by~\eqref{u:eq:gap}, so by~\eqref{u:eq:sign-close} $q\cdot(z-y)>H$ and
\[
 x^{z}=h^{q\cdot z}=h^{q\cdot(z-y)}x^{y}\le h^{H}x^{y}.
\]
Fix a cut between consecutive levels $L_j<L_{j+1}$. Each reaction crossing the cut downward (source above, product below) has, by weak reversibility, a directed return path, which must cross the cut upward somewhere; hence if any reaction crosses the cut downward, some reaction $e^*$ crosses it upward. Let $w=y_{e^*}$, which lies at or below $L_j$. Every downward source $v$ lies at or above $L_{j+1}$, so $x^{v}\le h^Hx^{w}$. The total downward flux through the cut is therefore at most $h^H(\sum_ek_e)x^w\le(k_0/2)x^w$ by~\eqref{q:eq:rate-margin}, while the upward flux is at least $k_{e^*}x^w\ge k_0x^w$. The net upward flux across a crossed cut is thus at least $(k_0/2)x^w\ge(k_0/2)h^{D_*}$; a cut with upward crossings only has positive net flux, and a cut with no crossing has zero net flux.

Writing $r\cdot f_k(x)=\sum_ek_ex^{y_e}(r\cdot y'_e-r\cdot y_e)$ and telescoping each level difference over the cuts it spans gives
\[
 r\cdot f_k(x)=\sum_{j=1}^{m-1}(L_{j+1}-L_j)\cdot(\text{net upward flux across cut }j)\ \ge0 .
\]
Each gap $L_{j+1}-L_j$ is at least $2H$ by~\eqref{u:eq:gap}. If $r\notin S^\perp$, some reaction has $r\cdot y'_e\neq r\cdot y_e$, so at least one cut is crossed, and the sum is at least $2H\cdot(k_0/2)h^{D_*}$. If $r\in S^\perp$ all level differences vanish and the work is zero.
\end{proof}

\subsection{Applying the shared plateau and entry argument}
For the chosen representative $c$, take the class threshold $h_c<h_g$ as in
Section~\ref{u:sec:drift}, so that
\begin{equation}\label{q:eq:class-threshold}
 \norm{\log c/\log h}_\infty<H\le1\qquad(0<h\le h_c).
\end{equation}
In particular $c\in\interior B_h$. Write $M_h=F_h(c)$ and use the same plateau
$C_h=\{x:F_h(x)=M_h\}$ as in~\eqref{u:eq:plateau}.

Lemma~\ref{u:lem:plateau} gives compact convex containment
$C_h\subset\interior B_{\sqrt h}$ directly from the common affine family.

\begin{lemma}[Invariance and entry at one scale]\label{q:lem:entry}
Let $0<h\le h_c$, let $k$ satisfy~\eqref{q:eq:rate-margin} with lower bound $k_0$,
and let $x$ be a global forward solution in $P_c$.
\begin{enumerate}[label=(\alph*)]
\item If $x(a)\in C_h$ then $x(t)\in C_h$ for all $t\ge a$.
\item If $x(t)\in B_h$ for all $t\ge a$, then $x(t)\in C_h$ for every
\[
 t\ge a+\frac{M_h-F_h(x(a))}{Hk_0h^{D_*}}.
\]
\end{enumerate}
\end{lemma}
\begin{proof}
The neutral-branch argument of Lemma~\ref{u:lem:plateau} excludes active slopes
in $S^\perp$ off $C_h$. Lemma~\ref{q:lem:drift} therefore supplies the common
margin $\delta_h=Hk_0h^{D_*}$. Apply the trapping argument of
Section~\ref{u:sec:trapping} and the height-deficit deadline
\eqref{u:eq:exact-delay}; autonomous $C^1$ solutions satisfy the same
absolutely continuous envelope calculation.
\end{proof}

\subsection{Rate margin and the dyadic scale chain}
Let $m\ge1$ be the number of reactions, $k_e=\bar k_e\eps^{a_e}$, $\Delta=a_+-a_-$. Then $k_e\ge\kappa_-\eps^{a_+}=:k_0$ and $\sum_ek_e\le m\kappa_+\eps^{a_-}$. Choose
\begin{equation}\label{q:eq:alpha}
 0<\alpha<h_c\qquad\text{with}\qquad \alpha^{H}\,m\,\kappa_+\le\frac{\kappa_-}{2},
\end{equation}
and put $h_*=\alpha\,\eps^{\Delta/H}\le\alpha$. For every $0<h\le h_*$,
\[
 h^H\sum_ek_e\le\alpha^H\eps^{\Delta}\,m\kappa_+\eps^{a_-}\le\frac{\kappa_-}{2}\eps^{a_+}=\frac{k_0}2 ,
\]
which is~\eqref{q:eq:rate-margin}. Note that $\alpha$ depends on the network (through $H$, $m$, $h_c$), on $c$ (through $h_c$) and on $\kappa_\pm$, and on nothing else.

\begin{proof}[Proof of Theorem~\ref{q:thm:general}]
Theorem~\ref{u:thm:main} gives global positive solutions for these fixed rates,
and every solution remains in $P_c$. At every scale $h\le h_*$ the preceding
rate-margin estimate and Lemma~\ref{q:lem:entry} give invariance and finite entry.
Apply the dyadic descent of Section~\ref{u:sec:iteration} with base $h_0=h_*$:
$h_n=h_0^{2^n}$ and $C_{h_{n+1}}\subset\interior B_{h_n}$.
For any initial state, $\norm{\log x_0/\log h_n}_\infty<H$ for sufficiently large
$n$, so the zero-slope argument places it in $C_{h_n}$. The finite chain of
entry estimates then gives eventual residence in
$C_{h_*}\subset\interior B_{\sqrt{h_*}}$.
Since $\sqrt{h_*}=\sqrt\alpha\,\eps^{\Delta/(2H)}$, this is
\eqref{q:eq:general-bound}. The finite number of steps and their deadlines
may depend on the seed and rate data.
\end{proof}

\begin{remark}[Explicit single-scale deadline]
Slopes have coordinates in $[-1,1]$, so $F_h$ is $1$-Lipschitz in $\ell^1$ and $0\le M_h-F_h(x)\le\|c-x\|_1\le2d/h$ on $B_h$. Lemma~\ref{q:lem:entry}(b) therefore gives the explicit deadline $a+2d\,\eps^{-a_+}h^{-(D_*+1)}/(H\kappa_-)$ for entering $C_h$ from a box-contained trajectory. Summing over the chain gives an explicit, seed-dependent total deadline. The formal statement keeps the sharper height-deficit form.
\end{remark}

\subsection{On the effectiveness of \texorpdfstring{$H$ and $\alpha$}{H and alpha}}
\label{q:sec:effective-H}
Given an explicit affine family, the remaining choices are finite:
$H$ is the minimum in~\eqref{u:eq:gap}, $h_c$ satisfies
\eqref{q:eq:class-threshold}, and $\alpha$ satisfies~\eqref{q:eq:alpha}.
The construction used here supplies the family existentially. Its induction on
dimension chooses layer thresholds through limiting comparisons of offsets known
to be $o(h^{-1})$. Extracting a network-specific numerical $H$ therefore requires
making those choices and the uniform activity threshold explicit. This is the
constructive step left open by the present proof.

The inventory clocks of Theorem~\ref{q:thm:global-clocks} and
Section~\ref{q:sec:bottleneck} instead use directly available data: a left null
vector of the fast stoichiometry, the slow rate constants and regional bounds on
the slow source monomials. Corollary~\ref{q:cor:scaling} also applies these clocks
after entry into the general reciprocal box. Thus geometric containment and
inventory replenishment have separate quantitative inputs.

\section{The coupled network}
\label{q:sec:coupled-proof}

\subsection{Stoichiometry and the neutral inventory}
The four forward reaction vectors of~\eqref{q:eq:reactions} are the columns of
\begin{equation}\label{q:eq:stoich}
 \nu=\begin{pmatrix}2&-1&-2&1\\-2&2&-1&0\\-1&-2&2&-3\end{pmatrix},
\end{equation}
and the first three columns have determinant $-13$. Hence $S=\R^3$: every positive stoichiometric class is the whole open orthant, and there is no nonzero linear conservation law. In particular no positive mass assignment makes~\eqref{q:eq:reactions} a closed system; it is an effective open network. Nevertheless $\ell=(6,5,2)$ annihilates columns $1$, $2$ and $4$, and the full field satisfies the exact identity
\begin{equation}\label{q:eq:neutral}
 \dot W=-13\,J_3,\qquad W=6x+5y+2z ,
\end{equation}
for every choice of prefactors. Only the reaction pair $2X+Y\rightleftharpoons2Z$ changes $W$, and its reverse direction, with rate constant $b_3ph^6$, is the only process that increases it. This is the structural source of slow recovery.

\subsection{Normalised coordinates and three clocks}
For reference, $p=uh$, $W=6x+5y+2z$ and $\lambda=p^2h^7$. The coordinate
definitions and characteristic inverse speeds are
\begin{center}\small
\begin{tabular}{lll}
\toprule
Coordinate & Definition & Characteristic time\\
\midrule
$A$ & $x/(ph^3Q^3)$ & $p^{-2}$\\
$B$ & $y/(ph^2Q^2)$ & $(p^2h^3)^{-1}$\\
$Q$ & $W/(2ph)$ & $\lambda^{-1}=(p^2h^7)^{-1}$\\
\bottomrule
\end{tabular}
\end{center}
\nopagebreak[4]
The exact equations below retain the additional powers of $Q$ in each derivative.
On the preimage of $\calO$ under~\eqref{q:eq:coordinates}, the admitted rectangle gives
\begin{equation}\label{q:eq:C-bound}
 \tfrac12\ \le\ 1-15h^2-\tfrac{15}{2}h\ \le\ C\ \le\ 1,
\end{equation}
so the map $(A,B,Q)\mapsto(x,y,z)=(ph^3AQ^3,\,ph^2BQ^2,\,phQC)$ is positive and nonsingular there. Differentiating the full literal field~\eqref{q:eq:currents}, including the derivatives of $Q$ inside $A$ and $B$, gives exact equations with three separated characteristic speeds,
\begin{equation}\label{q:eq:three-fields}
 \dot A=p^2R_A,\qquad \dot B=p^2h^3Q^3R_B,\qquad \dot Q=\tfrac{13}{2}\,p^2h^7\,Q^2R_Q ,
\end{equation}
where, with $G=a_3A^2BQ^6-b_3C^2$,
\begin{align}
 R_A&=KC^3-b_4A+h^2Q^2\big(2a_1B^2C-a_2AC^2\big)-2b_1uh^4A^2Q^3+b_2uh^2B^2Q\notag\\
 &\qquad-2a_3h^5A^2BQ^5+\frac{2b_3h^5C^2}{Q}+\frac{39}{2}Ah^7QG,\label{q:eq:RA}\\
 R_B&=2\big(a_2AC^2-a_1B^2C\big)+2b_1uh^2A^2Q-\frac{2b_2uB^2}{Q}-a_3h^3A^2BQ^3+\frac{b_3h^3C^2}{Q^3}+\frac{13Bh^4G}{Q^2},\label{q:eq:RB}\\
 R_Q&=b_3C^2-a_3A^2BQ^6 .\label{q:eq:RQ}
\end{align}
Equations~\eqref{q:eq:three-fields} express the full coupled mass-action system exactly in the new coordinates. The transformation separates characteristic clocks $p^2$, $p^2h^3$, $p^2h^7$ and exposes useful signs; on the reference path $h=t$, $u=t^5$, $\eps=t^{12}$ these are $t^{12},t^{15},t^{19}$.

\subsection{The corridor certificate}
\begin{lemma}[Continuous corridor certificate]\label{q:lem:corridor}
Under~\eqref{q:eq:parameters} the following inequalities hold for all $(A,B,Q)$ in the indicated sets, simultaneously at face intersections and uniformly over the parameter and prefactor boxes.
\begin{enumerate}[label=(\alph*)]
\item \emph{Outer faces.} $R_A>0$ on $\{A=3\}\cap\calO$ and $R_A<0$ on $\{A=5\}\cap\calO$; $R_B>0$ on $\{B=1\}\cap\calO$ and $R_B<0$ on $\{B=3\}\cap\calO$; $R_Q>0$ on $\{Q=\tfrac14\}\cap\calO$ and $R_Q<0$ on $\{Q=1\}\cap\calO$.
\item \emph{Inner faces.} The same six strict sign statements on the faces of $\calI$.
\item \emph{Entry slabs with margins.}
\begin{center}\small\renewcommand{\arraystretch}{1.35}
\begin{tabular}{lll}
\toprule
slab (other coordinates) & lower slab & upper slab\\
\midrule
$A$ ($B,Q$ in outer ranges) & $R_A\ge\tfrac14$ for $3\le A\le\tfrac72$ & $R_A\le-\tfrac14$ for $\tfrac92\le A\le5$\\
$B$ ($A$ inner, $Q$ outer) & $R_B\ge1$ for $1\le B\le\tfrac32$ & $R_B\le-1$ for $\tfrac52\le B\le3$\\
$Q$ ($A,B$ inner) & $R_Q\ge\tfrac1{10}$ for $\tfrac14\le Q\le\tfrac12$ & $R_Q\le-\tfrac1{10}$ for $\tfrac34\le Q\le1$\\
\bottomrule
\end{tabular}
\end{center}
\end{enumerate}
\end{lemma}

\begin{proof}
Each of the eighteen statements is an inclusion of the range of a rational function of thirteen variables ($h,u,Q,A,B$ and the eight prefactors) over a rational box in a rational interval with the required sign or margin. We verify it by interval arithmetic~\cite{MooreKearfottCloud2009,Tucker2011}: constants and variable boxes initialise an induction over the expression tree; addition adds endpoints, negation swaps them, multiplication takes the extreme endpoint products, nonnegative integer powers and reciprocals of positive intervals preserve inclusion. Every reciprocal in~\eqref{q:eq:RA}--\eqref{q:eq:RB} has base $Q\ge\tfrac14$, and every power base is nonnegative on the boxes. The checker is a small expression language whose evaluation is proved sound for every real point of the box; its rational endpoint computations are verified by kernel arithmetic, and separately proved polynomial identities equate the checked expressions with~\eqref{q:eq:RA}--\eqref{q:eq:RQ}. Appendix~\ref{q:app:certificate} lists the eighteen boxes and the certified enclosures. The inclusions establish the face signs and slab margins on the complete declared boxes.
\end{proof}

\subsection{Invariance, global existence and the concentration bounds}
\begin{proposition}\label{q:prop:invariance}
Assume~\eqref{q:eq:parameters}. Every positive initial state in the preimage of $\calO$ has a global positive solution of~\eqref{q:eq:currents}; every such solution stays in the preimage of $\calO$ for all $t\ge0$ and satisfies~\eqref{q:eq:sharp-bounds}.
\end{proposition}

\begin{proof}
On any time interval on which a positive solution exists, $(A,B,Q)$ is $C^1$ and obeys~\eqref{q:eq:three-fields}. Consider the six margins $A-3,\,5-A,\,B-1,\,3-B,\,Q-\tfrac14,\,1-Q$. At a time when all are nonnegative and one of them vanishes, Lemma~\ref{q:lem:corridor}(a) gives that margin a strictly positive derivative (using $p^2>0$, $Q^3>0$, $Q^2>0$); corner ties are included since each face statement holds on the closed face. A standard first-exit argument for closed boxes then shows that no margin can become negative, so the solution remains in the preimage of $\calO$. There, \eqref{q:eq:C-bound} and $A\in[3,5]$, $B\in[1,3]$, $Q\in[\tfrac14,1]$ give~\eqref{q:eq:sharp-bounds}, a compact subset of the open orthant. Local existence for the polynomial field together with this a priori compact bound yields continuation to all $t\ge0$ by the usual argument; the same reasoning applies to every solution with the stated initial data.
\end{proof}

\subsection{Sequential capture by reciprocal comparison}
\label{q:sec:capture}
The following comparison gives a deadline for permanent residence in the target interval.

\begin{lemma}[Slab entry]\label{q:lem:slab}
Let $f:[s,t]\to\R$ be $C^1$ with $f\ge L$ on $[s,t]$ and let $\delta>0$. Suppose $f'(\tau)\ge\delta$ whenever $f(\tau)\le l$. If $t-s\ge(l-L)/\delta$ then $f(t)\ge l$. Symmetrically, if $f\le U$, $f'\le-\delta$ whenever $f\ge l$, and $t-s\ge(U-l)/\delta$, then $f(t)\le l$.
\end{lemma}

\begin{proof}
If $f(t)<l$, then $f\le l$ on all of $[s,t]$: otherwise take the last time $\tau_1<t$ with $f(\tau_1)\ge l$; just after $\tau_1$ we have $f<l$ and hence $f'\ge\delta>0$, so $f$ cannot drop below $f(\tau_1)\ge l$, a contradiction. Then $f'\ge\delta$ on $[s,t]$ and $f(t)\ge f(s)+\delta(t-s)\ge L+(l-L)=l$, again a contradiction. The second statement follows by replacing $f$ with $-f$.
\end{proof}

\begin{proposition}[Entry deadline]\label{q:prop:deadline}
Assume~\eqref{q:eq:parameters}, let $\lambda=p^2h^7$, and let $(x,y,z)$ be a solution from the preimage of $\calO$. Put
\begin{equation}\label{q:eq:phases}
 T_A=\frac{2}{p^2},\qquad T_B=T_A+\frac{32}{p^2h^3},\qquad T_Q=T_B+\frac{40}{13\lambda}.
\end{equation}
Then $A(t)\in[\tfrac72,\tfrac92]$ for $t\ge T_A$, $B(t)\in[\tfrac32,\tfrac52]$ for $t\ge T_B$, $Q(t)\in[\tfrac12,\tfrac34]$ for $t\ge T_Q$, and
\begin{equation}\label{q:eq:sequential-deadline}
 T_Q=\frac{2h^7+32h^4+40/13}{\lambda}\ \le\ \frac{31}{10}\,\lambda^{-1}\qquad(0<h\le\tfrac1{600}).
\end{equation}
\end{proposition}

\begin{proof}
Throughout, Proposition~\ref{q:prop:invariance} keeps $(A,B,Q)\in\calO$.

\emph{Phase $A$.} On $[0,T_A]$ apply Lemma~\ref{q:lem:slab} to $f=A$ with $L=3$, $l=\tfrac72$, $\delta=p^2/4$: whenever $A\le\tfrac72$, the entry slab of Lemma~\ref{q:lem:corridor}(c) gives $\dot A=p^2R_A\ge p^2/4$, and $(l-L)/\delta=2/p^2=T_A$. Hence $A(T_A)\ge\tfrac72$, and since the same inequality holds at later times whenever $A\le\tfrac72$, the first-exit argument of Lemma~\ref{q:lem:slab} keeps $A\ge\tfrac72$ for all $t\ge T_A$. The upper slab gives $A\le\tfrac92$ in the same way.

\emph{Phase $B$.} On $[T_A,T_B]$, $A$ is inner, so whenever $B\le\tfrac32$ we have $R_B\ge1$ and $\dot B=p^2h^3Q^3R_B\ge p^2h^3/64$ since $Q\ge\tfrac14$. With $L=1$, $l=\tfrac32$, $\delta=p^2h^3/64$ the deadline is $32/(p^2h^3)$. The upper slab is symmetric.

\emph{Phase $Q$ by reciprocal comparison.} On $[T_B,T_Q]$, $A$ and $B$ are inner. Put $f=1/Q$, which is $C^1$ with $f'=-\dot Q/Q^2=-\tfrac{13}{2}\lambda R_Q$ by~\eqref{q:eq:three-fields}; the factor $Q^2$ cancels exactly. Whenever $Q\le\tfrac12$, i.e.\ $f\ge2$, the slab gives $R_Q\ge\tfrac1{10}$ and hence $f'\le-\tfrac{13}{20}\lambda$. Since $Q\ge\tfrac14$ we have $f\le4$. Lemma~\ref{q:lem:slab} (second form) with $U=4$, $l=2$, $\delta=\tfrac{13}{20}\lambda$ gives $f\le2$, i.e.\ $Q\ge\tfrac12$, after $(4-2)/\delta=40/(13\lambda)$. Whenever $Q\ge\tfrac34$, i.e.\ $f\le\tfrac43$, we have $R_Q\le-\tfrac1{10}$ and $f'\ge\tfrac{13}{20}\lambda$; with $f\ge1$ the first form gives $f\ge\tfrac43$, i.e.\ $Q\le\tfrac34$, after $(\tfrac43-1)/\delta=20/(39\lambda)\le40/(13\lambda)$. Residence thereafter follows as in phase $A$.

Finally $T_Q\lambda=2h^7+32h^4+40/13$, and for $h\le\tfrac1{600}$ the first two terms are below $3\cdot10^{-10}$ while $40/13<3.0770$, so $T_Q\le3.1\lambda^{-1}$.
\end{proof}

Comparing $1/Q$ retains the exact cancellation of $Q^2$. A direct comparison of $Q$ using $Q^2\ge1/16$ on the slab would lose a factor of $16$ in the $Q$-phase estimate.

\subsection{The witness}
\begin{proposition}[Witness lower bound]\label{q:prop:witness}
Assume~\eqref{q:eq:parameters}. Let $(x,y,z)$ be any solution from a prepared state with $Q(0)=\tfrac14$ (for instance~\eqref{q:eq:witness}). If $Q(T)\ge\tfrac12$ for some $T\ge0$, then $T\ge\frac{400}{1313}\lambda^{-1}$.
\end{proposition}

\begin{proof}
On the preimage of $\calO$ the forward term of $R_Q$ is nonnegative and $b_3C^2\le\tfrac{101}{100}$ by~\eqref{q:eq:C-bound}, so
\[
 \dot Q=\tfrac{13}{2}\lambda Q^2R_Q\ \le\ \tfrac{13}{2}\cdot\tfrac{101}{100}\,\lambda\,Q^2=\tfrac{1313}{200}\,\lambda\,Q^2 .
\]
Hence $f=1/Q$ satisfies $f'=-\dot Q/Q^2\ge-\tfrac{1313}{200}\lambda$ for all $t\ge0$. Integrating from $f(0)=4$ gives $f(T)\ge4-\tfrac{1313}{200}\lambda T$. Target membership gives $Q(T)\ge\tfrac12$, i.e.\ $f(T)\le2$, so $\tfrac{1313}{200}\lambda T\ge2$ and $T\ge\tfrac{400}{1313}\lambda^{-1}$. The differential inequality holds throughout the trajectory, so the bound applies at every target time.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{q:thm:coupled}]
Global existence, outer invariance and~\eqref{q:eq:sharp-bounds} are Proposition~\ref{q:prop:invariance}; the deadline is Proposition~\ref{q:prop:deadline} together with~\eqref{q:eq:sequential-deadline}; the state~\eqref{q:eq:witness} is positive, has $(A,B,Q)(0)=(4,2,\tfrac14)\in\calO$, hence has a global solution, and Proposition~\ref{q:prop:witness} applies to it since the target requires $Q\ge\tfrac12$.
\end{proof}

\subsection{A numerical illustration}
Figure~\ref{q:fig:trajectory} shows the solution from~\eqref{q:eq:witness} at the corner $h=\tfrac1{600}$, $u=\tfrac1{20}$ with nominal prefactors, integrated with the stiff Radau method~\cite{HairerWanner1996} on the complete transformed system~\eqref{q:eq:three-fields} in the exact slow time $\tau=\lambda t$ (solver parameters are recorded in the reproducibility material). The first sampled time at which all three coordinates are inside $\calI$ is $\tau\approx0.358$, inside the certified window $[400/1313,\,31/10]\approx[0.3046,\,3.1]$. The fast coordinates $A,B$ settle on the $p^2$ and $p^2h^3$ clocks, invisible on the slow axis and shown on a logarithmic inset; the slow coordinate $Q$ rises monotonically here, although no such monotonicity is used in the proofs.

\begin{figure}[tb]
\centering
\includegraphics[width=\linewidth]{figures/recovery_trajectory.pdf}
\caption{Numerical illustration of the complete system~\eqref{q:eq:currents} at $h=1/600$, $u=1/20$, nominal prefactors, from the witness~\eqref{q:eq:witness}, plotted against the exact slow time $\tau=p^2h^7t$ (left) and on a logarithmic time axis for the fast coordinates (right). Grey bands are the outer intervals of $\calO$, green bands the target intervals of $\calI$. The dashed vertical line marks the first sampled time at which all three coordinates lie in $\calI$, $\tau\approx0.358$; the shaded region $\tau<400/1313$ is excluded for the witness by Proposition~\ref{q:prop:witness}, and the certified deadline $\tau=31/10$ lies beyond the plotted window. One slow unit is $4.03\times10^{27}$ seconds in the calibration of Section~\ref{q:sec:examples}.}
\label{q:fig:trajectory}
\end{figure}

\subsection{Global inventory clocks for arbitrary seeds}
\label{q:sec:global-clocks}
Theorem~\ref{q:thm:coupled} assumes a prepared seed. The inventory identity, however, carries information on the whole positive orthant, because positivity alone gives
\begin{equation}\label{q:eq:orthant-bounds}
 z\le\tfrac12W,\qquad x\le\tfrac16W,\qquad y\le\tfrac15W\qquad\text{for every positive state}.
\end{equation}
Dropping the nonnegative term of the appropriate sign in $\dot W=-13J_3=-13a_3x^2y+13b_3ph^6z^2$ and inserting~\eqref{q:eq:orthant-bounds} gives the two-sided differential inequality
\begin{equation}\label{q:eq:W-inequality}
 -\tfrac{13}{180}\,a_3\,W^3\ \le\ \dot W\ \le\ \tfrac{13}{4}\,b_3ph^6\,W^2 ,
\end{equation}
valid along every positive solution at every time.

\begin{theorem}[Global inventory clocks]\label{q:thm:global-clocks}
Assume~\eqref{q:eq:parameters}. Every positive initial state has a global positive solution of~\eqref{q:eq:currents}, and every global positive solution satisfies, for all $T\ge0$,
\begin{align}
 T&\ \ge\ \frac{4}{13\,b_3\,ph^6}\Big(\frac1{W(0)}-\frac1{W(T)}\Big)
 &&\text{(restoration clock)},\label{q:eq:restoration-clock}\\
 T&\ \ge\ \frac{90}{13\,a_3}\Big(\frac1{W(T)^2}-\frac1{W(0)^2}\Big)
 &&\text{(depletion clock)}.\label{q:eq:depletion-clock}
\end{align}
In particular, any positive solution that raises its inventory from $W_0$ to $W_1>W_0$ needs time at least $\frac{4}{13b_3ph^6}(W_0^{-1}-W_1^{-1})$, and any solution that lowers it from $W_0$ to $W_1<W_0$ needs at least $\frac{90}{13a_3}(W_1^{-2}-W_0^{-2})$, whatever the shape coordinates $A,B$ do.
\end{theorem}

\begin{proof}
Global existence from every positive seed follows from Theorem~\ref{u:thm:main} applied to the weakly reversible network~\eqref{q:eq:reactions} with fixed positive rates. For~\eqref{q:eq:restoration-clock}, $f=1/W$ is $C^1$ with $f'=-\dot W/W^2\ge-\tfrac{13}{4}b_3ph^6$ by~\eqref{q:eq:W-inequality}; integrating, $f(0)-f(T)\le\tfrac{13}{4}b_3ph^6\,T$. For~\eqref{q:eq:depletion-clock}, $g=1/W^2$ has $g'=-2\dot W/W^3\le\tfrac{13}{90}a_3$, so $g(T)-g(0)\le\tfrac{13}{90}a_3T$. Neither bound uses monotonicity of $W$, so both hold at every time, not only at first passage.
\end{proof}

The restoration clock with $W(0)=\tfrac{ph}2$ and $W(T)\ge ph$ gives $T\ge\frac{4}{13b_3}\lambda^{-1}\ge\frac{400}{1313}\lambda^{-1}$: Proposition~\ref{q:prop:witness} is the special case of Theorem~\ref{q:thm:global-clocks} at the witness, and the prepared basin was never needed for the lower bound. The depletion clock quantifies the opposite transient: from a macroscopic seed such as $x=y=z=1$ ($W=13$), reaching the corridor inventory $W\le ph$ needs at least $\frac{90}{13a_3}\big((ph)^{-2}-13^{-2}\big)\approx6.9\,(ph)^{-2}$ time units at nominal $a_3$, which is $h^{5}$ times the restoration scale $\lambda^{-1}$: collapse of a surplus is governed by the forward reaction $2X+Y\to2Z$ and is at least $(ph)^{-2}$ slow, while rebuilding a deficit is $(p^2h^7)^{-1}$ slow. What the theorem does not give is an upper bound on the time for an arbitrary seed to enter the sharp corridor; see Section~\ref{q:sec:discussion}.

\section{Corollaries and the inventory mechanism}
\label{q:sec:corollaries}

\subsection{Common clocks}
\begin{corollary}[Exact common-clock transport]\label{q:cor:clock}
For autonomous mass action and $b>0$, $\tilde x$ is a global forward solution for rates $\tilde k$ if and only if $x(t)=\tilde x(bt)$ is a global forward solution for rates $k_e=b\tilde k_e$ from the same initial state. Phase-space paths, invariant sets and residence statements correspond exactly; all times are divided by $b$.
\end{corollary}

\begin{proof}
$f_{b\tilde k}=bf_{\tilde k}$, and the chain rule gives $\dot x(t)=b\dot{\tilde x}(bt)=bf_{\tilde k}(\tilde x(bt))=f_{b\tilde k}(x(t))$; the converse uses $b^{-1}$.
\end{proof}

This is why~\eqref{q:eq:general-bound} depends only on the spread $\Delta$: a common factor $\eps^{a_-}$ leaves concentration geometry unchanged and multiplies all times by $\eps^{-a_-}$. Different exponent assignments with the same spread can nevertheless have different sharp species scales, as the coupled network shows.

\subsection{A continuous reverse-exponent family}
Let $\tfrac12<\gamma<\tfrac72$ and set
\begin{equation}\label{q:eq:gamma-path}
 p=\eps^{1/2},\qquad h=\eps^{(2\gamma-1)/12},\qquad u=\eps^{(7-2\gamma)/12},
\end{equation}
so that $p=uh$, $p^2=\eps$ and $ph^6=\eps^\gamma$: the reverse rates on pairs $1,2,4$ are proportional to $\eps$ and pair $3$ has reverse exponent $\gamma$.

\begin{corollary}[Certified exponent family]\label{q:cor:gamma}
Define
\begin{equation}\label{q:eq:improved-cutoff}
 \eps_{\max}(\gamma)=\min\big\{600^{-12/(2\gamma-1)},\ 20^{-12/(7-2\gamma)}\big\}.
\end{equation}
For $0<\eps\le\eps_{\max}(\gamma)$ and all admitted prefactors, Theorem~\ref{q:thm:coupled} applies to~\eqref{q:eq:gamma-path}. Along every prepared trajectory,
\begin{equation}\label{q:eq:gamma-orders}
 x\asymp\eps^{(2\gamma+1)/4},\qquad y\asymp\eps^{(\gamma+1)/3},\qquad z\asymp\eps^{(2\gamma+5)/12},
\end{equation}
with the explicit constants of~\eqref{q:eq:sharp-bounds}; every prepared start recovers by $\tfrac{31}{10}\eps^{-(14\gamma+5)/12}$ and the witness needs at least $\tfrac{400}{1313}\eps^{-(14\gamma+5)/12}$.
\end{corollary}

\begin{proof}
$h\le\tfrac1{600}$ and $u\le\tfrac1{20}$ are exactly the two real-power inequalities in~\eqref{q:eq:improved-cutoff}, and $ph^3=\eps^{(2\gamma+1)/4}$, $ph^2=\eps^{(\gamma+1)/3}$, $ph=\eps^{(2\gamma+5)/12}$, $p^2h^7=\eps^{(14\gamma+5)/12}$ by exponent arithmetic. Substitute into Theorem~\ref{q:thm:coupled}.
\end{proof}

The symmetric reference sub-rectangle $h,u\le10^{-3}$ corresponds to the cutoff
\begin{equation}\label{q:eq:original-cutoff}
 \eps\le10^{-36/\min\{2\gamma-1,\,7-2\gamma\}};
\end{equation}
at $\gamma=1$ it gives $10^{-36}$, while~\eqref{q:eq:improved-cutoff} gives $600^{-12}\approx4.59\times10^{-34}$. At $\gamma=2$ the concentration exponents are $\tfrac54,1,\tfrac34$ and the recovery exponent $\tfrac{11}4$. The endpoints $\gamma=\tfrac12$ and $\tfrac72$ are not covered: there one of $h,u$ stops tending to zero. Figure~\ref{q:fig:exponents} displays the exponent laws and the cutoffs.

\begin{figure}[tb]
\centering
\includegraphics[width=\linewidth]{figures/exponent_family.pdf}
\caption{Left: exponent laws~\eqref{q:eq:gamma-orders} and the recovery exponent along the family~\eqref{q:eq:gamma-path}. Right: the certified cutoff $\log_{10}\eps_{\max}(\gamma)$ for the admitted rectangle (solid) and the symmetric reference sub-rectangle (dashed). The marked point is the corner $\gamma_*$ of Proposition~\ref{q:prop:corner}, where the two branches of~\eqref{q:eq:improved-cutoff} balance and $\eps_{\max}$ attains its largest value $1/144{,}000{,}000$.}
\label{q:fig:exponents}
\end{figure}

\begin{proposition}[Best corner of the certificate]\label{q:prop:corner}
Over the rectangle~\eqref{q:eq:parameters}, $p=uh\le\tfrac1{12000}$, so the baseline reverse parameter satisfies $\eps=p^2\le\tfrac1{144{,}000{,}000}\approx6.94\times10^{-9}$. The family~\eqref{q:eq:gamma-path} reaches this corner exactly when
\begin{equation}\label{q:eq:gamma-star}
 \gamma_*=\frac12+\frac{3\ln600}{\ln12000}\approx2.5432 ,
\end{equation}
which maximises $\eps_{\max}(\gamma)$. For fixed $p\le\tfrac1{12000}$ the admissible $h$ range is $[20p,\tfrac1{600}]$, and $\lambda=p^2h^7$ is increasing in $h$ and in $p$, so the fastest certified recovery scale over the whole rectangle is attained at $h=\tfrac1{600}$, $u=\tfrac1{20}$:
\begin{equation}\label{q:eq:corner-scale}
 \lambda^{-1}=12000^2\cdot600^7=4.0310784\times10^{27}.
\end{equation}
\end{proposition}

\begin{proof}
The two branches of~\eqref{q:eq:improved-cutoff} are respectively increasing and decreasing in $\gamma$ and balance when $(7-2\gamma)\ln600=(2\gamma-1)\ln20$, i.e.\ at~\eqref{q:eq:gamma-star}; substituting $h=\tfrac1{600}$, $p=\tfrac1{12000}$ into~\eqref{q:eq:gamma-path} gives the same value. The remaining statements are monotonicity of $p^2h^7$ in each variable.
\end{proof}

Figure~\ref{q:fig:rectangle} shows $\log_{10}\lambda^{-1}$ over the rectangle together with two family paths. The corner minimises the certified recovery scale over this rectangle. Enlarging the rectangle requires new face and entry estimates (Section~\ref{q:sec:discussion}).

\begin{figure}[tb]
\centering
\includegraphics[width=.55\linewidth]{figures/parameter_rectangle.pdf}
\caption{The certified recovery scale $\log_{10}(p^2h^7)^{-1}$ over the admitted rectangle (logarithmic axes). Dotted: the family path $\gamma=2$. Solid: the path $\gamma_*\approx2.543$, which passes through the corner $(h,u)=(1/600,1/20)$ (dot), the point of fastest certified recovery.}
\label{q:fig:rectangle}
\end{figure}

\subsection{Scale identities and necessary reaction extent}
\begin{corollary}[Preparation--recovery trade-off]\label{q:cor:tradeoff}
With the preparation scales $s_x=ph^3$, $s_y=ph^2$, $s_z=ph$,
\begin{equation}\label{q:eq:tradeoff}
 s_y^2=s_xs_z,\qquad \lambda^{-1}=\frac{s_y}{s_x^3}=\frac{s_z^3}{s_y^5}.
\end{equation}
Hence the witness lower bound and the universal prepared deadline are $\tfrac{400}{1313}s_y/s_x^3$ and $\tfrac{31}{10}s_y/s_x^3$.
\end{corollary}

\begin{proof}
Substitute and cancel positive powers of $p$ and $h$.
\end{proof}

These dimensionless preparation-scale identities explain why extreme separation of concentration scales in this family is accompanied by slow recovery.

\begin{corollary}[Necessary cumulative reverse extent]\label{q:cor:extent}
For any solution from~\eqref{q:eq:witness} and any $T\ge0$ at which it is in the target,
\begin{equation}\label{q:eq:extent}
 -\int_0^TJ_3(t)\dd t=\frac{W(T)-W(0)}{13}\ \ge\ \frac{ph}{26},\qquad \int_0^Tb_3ph^6z(t)^2\dd t\ \ge\ \frac{ph}{26}.
\end{equation}
\end{corollary}

\begin{proof}
Integrate~\eqref{q:eq:neutral}; $W=2phQ$ rises from $\tfrac{ph}2$ to at least $ph$. Since $-J_3=b_3ph^6z^2-a_3x^2y\le b_3ph^6z^2$, the second bound follows.
\end{proof}

\subsection{A general inventory-bottleneck principle}
\label{q:sec:bottleneck}
The witness argument is an instance of a general principle that uses stoichiometry and kinetics rather than the reaction graph alone. Write a mass-action system as $\dot x=\sum_e\nu_ev_e(x)$ with $v_e\ge0$, each direction of a reversible pair counted separately.

\begin{proposition}[Inventory bottleneck, constant capacity]\label{q:prop:bottleneck}
Partition the reactions into a fast set $F$ and a remainder $R$, and let $\ell\in\R^d$ satisfy $\ell\cdot\nu_e=0$ for all $e\in F$. Put $U=\ell\cdot x$. Suppose that, up to the target time $T$, the trajectory stays in a region $\calD$ on which
\[
 \sum_{e\in R}(\ell\cdot\nu_e)_+\,v_e(x)\le M .
\]
If reaching the target requires $U(T)\ge U(0)+\Delta U$ with $\Delta U>0$, then $T\ge\Delta U/M$ when $M>0$, and the target is unreachable along such a trajectory when $M=0$.
\end{proposition}

\begin{proof}
$\dot U=\sum_{e\in R}(\ell\cdot\nu_e)v_e\le\sum_{e\in R}(\ell\cdot\nu_e)_+v_e\le M$ on $\calD$; integrate.
\end{proof}

\begin{proposition}[Inventory bottleneck, state-dependent capacity]\label{q:prop:bottleneck-state}
In the setting of Proposition~\ref{q:prop:bottleneck} suppose instead that $\dot U\le g(U)$ on $\calD$ for a continuous $g>0$ on an interval containing $[U(0),U_{\rm target}]$ and all values taken by $U$ before $T$. Then
\begin{equation}\label{q:eq:state-capacity}
 T\ \ge\ \int_{U(0)}^{U_{\rm target}}\frac{\dd s}{g(s)} .
\end{equation}
\end{proposition}

\begin{proof}
Let $\Phi'=1/g$; then $(\Phi\circ U)'=\dot U/g(U)\le1$, so $\Phi(U(T))-\Phi(U(0))\le T$, and $\Phi$ is increasing.
\end{proof}

\begin{corollary}[Explicit $\eps$-scaling]\label{q:cor:scaling}
In the setting of Proposition~\ref{q:prop:bottleneck}, suppose every reaction $e\in R$ with $\ell\cdot\nu_e>0$ has rate $k_e=\bar k_e\eps^{a_e}$ with $a_e\ge a$ and $\bar k_e\le\kappa_+$, and that $x^{y_e}\le m_e$ on $\calD$. Then any target requiring $U$ to rise by $\Delta U>0$ satisfies
\begin{equation}\label{q:eq:scaling}
 T\ \ge\ \frac{\Delta U}{\kappa_+\sum_{e\in R,\ \ell\cdot\nu_e>0}(\ell\cdot\nu_e)\,m_e}\ \eps^{-a}\ =\ \Omega(\eps^{-a}),
\end{equation}
uniformly over the admitted prefactors. In the general setting of Theorem~\ref{q:thm:general}, after the entry time the trajectory lies in the box~\eqref{q:eq:general-bound}, so $m_e=(\alpha^{-1/2}\eps^{-\Delta/(2H)})^{|y_e|}$ is admissible and~\eqref{q:eq:scaling} holds from that time on.
\end{corollary}

\begin{proof}
$\dot U\le\sum(\ell\cdot\nu_e)_+k_ex^{y_e}\le\eps^a\kappa_+\sum(\ell\cdot\nu_e)_+m_e=:C\eps^a$ on $\calD$, and Proposition~\ref{q:prop:bottleneck} with $M=C\eps^a$ gives $T\ge(\Delta U/C)\eps^{-a}$.
\end{proof}

This is the universal form of the slow-channel obstruction: it uses a left null vector of the fast reaction vectors, the slow rate scalings and a region bound. It is one-sided; a matching upper bound requires network-specific progress estimates, which is what Theorem~\ref{q:thm:coupled} supplies for~\eqref{q:eq:reactions}. For the coupled network take $\ell=(6,5,2)$, $F$ the six directed reactions of pairs $1,2,4$, and $R$ the two directions of pair $3$. The only positive contribution to $\dot W$ is $13\,b_3ph^6z^2$, and on the preimage of $\calO$ we have $z=phQC\le phQ=W/2$, so $\dot W\le g(W):=\tfrac{13}{4}b_3ph^6W^2$. For the witness $W$ must rise from $\tfrac{ph}{2}$ to $ph$, and~\eqref{q:eq:state-capacity} gives
\[
 T\ \ge\ \frac{4}{13\,b_3\,ph^6}\Big(\frac{2}{ph}-\frac1{ph}\Big)=\frac{4}{13\,b_3}\,\lambda^{-1}\ \ge\ \frac{400}{1313}\,\lambda^{-1},
\]
which is Proposition~\ref{q:prop:witness} and, since~\eqref{q:eq:orthant-bounds} needs no basin, Theorem~\ref{q:thm:global-clocks}: the reciprocal comparison \emph{is} the state-dependent capacity bound for $g(W)\propto W^2$. The constant-capacity version with $z\le ph$ gives the weaker $T\ge\lambda^{-1}/(26\cdot\tfrac{101}{100})$, and Corollary~\ref{q:cor:scaling} with $a=\gamma$ along the family~\eqref{q:eq:gamma-path} gives the recovery exponent up to the region-dependent prefactor. A matching upper deadline additionally uses forward progress ($R_Q\ge\tfrac1{10}$ on the slab) and invariance of the comparison region.

Fast-subsystem conservation laws as slow variables have an established literature; see~\cite{Desoeuvres2024} for approximate conservation laws and reduction, and~\cite{Fenichel1979} for the geometric theory. The propositions above are elementary; their value is that they turn such an inventory into an explicit, regionally justified lower bound on recovery time. The productive search pattern for related mechanisms is: candidate fast subnetwork $\to$ exact left null vector $\ell$ $\to$ regional bound on the increasing slow fluxes $\to$ lower time bound $\to$ a separate progress mechanism for an upper bound. The reaction graph alone does not determine $\ell$, the flux bound or a useful region.

\section{Stochastic remarks}
\label{q:sec:stochastic}
Consider the standard stochastic mass-action model of~\eqref{q:eq:reactions}, a continuous-time Markov chain on molecule counts $n=(n_X,n_Y,n_Z)\in\N^3$ with combinatorial propensities~\cite{Kurtz1972,Gillespie1977,AndersonKurtz2015}, in which every one of the eight directed reactions has a positive rate constant.

\begin{proposition}[No jump into an absorbing state]\label{q:prop:absorbing}
\begin{enumerate}[label=(\alph*)]
\item For any reversible network with all rate constants positive, no reaction event can lead from a state in which some reaction is enabled to a state in which no reaction is enabled.
\item For~\eqref{q:eq:reactions} the absorbing count states are exactly $(0,0,0)$, $(0,1,0)$, $(0,0,1)$ and $(0,1,1)$, and none of them is reachable from any other state.
\end{enumerate}
\end{proposition}

\begin{proof}
(a) A jump consuming the complex $y$ and producing $y'$ leaves a state containing at least $y'$ molecules of the product complex, so the reverse reaction $y'\to y$, which has positive rate, is enabled at the destination. (b) $n_X\ge1$ enables $X\to3Z$; $n_Y\ge2$ enables $2Y\to X+2Z$; $n_Z\ge2$ enables $2Z\to2X+Y$. If none holds then $n_X=0$, $n_Y\le1$, $n_Z\le1$, and inspection shows that none of the eight reactions is enabled. Unreachability is (a).
\end{proof}

Individual species counts can reach zero and return: for instance, $n_X=0$ is left by $3Z\to X$. Species absence, loss of a functional threshold and absorption are therefore distinct events. Recurrence and stationary distributions for this four-linkage-class network require a separate analysis; the single-linkage-class results of~\cite{AndersonCappellettiKim2019} provide related context.

The deterministic identity~\eqref{q:eq:neutral} has an exact discrete counterpart. Let $N_3^{+}(t)$ and $N_3^{-}(t)$ count the forward and reverse events of pair $3$ and put $W_n=6n_X+5n_Y+2n_Z$. Before any explosion,
\begin{equation}\label{q:eq:discrete-inventory}
 W_n(t)=W_n(0)+13\big(N_3^-(t)-N_3^+(t)\big),
\end{equation}
so $W_n\bmod13$ is invariant, and a target requiring $W_n$ to rise by at least $L$ requires at least $m=\lceil L/13\rceil$ reverse events of pair $3$.

The reverse propensity of pair $3$ is $\lambda_3^-(n)=\kappa\,n_Z(n_Z-1)$ with $\kappa=b_3ph^6/(VN_{\!A})$ in the units of Section~\ref{q:sec:examples}, and $2n_Z\le W_n$ for every count state. Hence $\lambda_3^-\le\tfrac{\kappa}{4}W_n^2$ everywhere, which is the discrete form of~\eqref{q:eq:W-inequality}. This turns the inventory bottleneck into a pathwise statement with no localisation.

\begin{proposition}[Pure-birth comparison]\label{q:prop:birth}
Let $W_0=W_n(0)$ and let $T$ be any random time at which $W_n(T)\ge W_0+13m$ for an integer $m\ge1$ (with $T=\infty$ if this never happens before explosion). Put $g_k=\tfrac{\kappa}{4}(W_0+13k)^2$. Then there are i.i.d.\ unit exponential random variables $E_0,\dots,E_{m-1}$ with
\begin{equation}\label{q:eq:birth-pathwise}
 T\ \ge\ \sum_{k=0}^{m-1}\frac{E_k}{g_k}\qquad\text{pathwise}.
\end{equation}
Consequently
\begin{equation}\label{q:eq:birth-moments}
 \mathbb E[T]\ \ge\ \sum_{k=0}^{m-1}\frac1{g_k}\ \ge\ \frac{4}{13\kappa}\Big(\frac1{W_0}-\frac1{W_0+13m}\Big),\qquad
 \Pr\{T\le t\}\ \le\ e^{\theta t}\prod_{k=0}^{m-1}\Big(1+\frac\theta{g_k}\Big)^{-1}\quad(\theta>0),
\end{equation}
and the lower bound in~\eqref{q:eq:birth-pathwise} has variance $\sum_kg_k^{-2}$.
\end{proposition}

\begin{proof}
By the random time-change representation~\cite{Kurtz1972,AndersonKurtz2015}, $N_3^-(t)=Y\big(\int_0^t\lambda_3^-(n(s))\dd s\big)$ for a unit-rate Poisson process $Y$, up to explosion. Let $\sigma_k$ be the time of the $k$-th reverse pair-$3$ event and $E_k$ the $k$-th interarrival time of $Y$, so the $E_k$ are i.i.d.\ unit exponentials. Forward pair-$3$ events lower $W_n$ and all other events preserve it, so $W_n(s)\le W_0+13N_3^-(s)$ by~\eqref{q:eq:discrete-inventory}; on $(\sigma_k,\sigma_{k+1}]$ we have $N_3^-=k$ and therefore $\lambda_3^-\le\tfrac\kappa4(W_0+13k)^2=g_k$. Hence $E_k=\int_{\sigma_k}^{\sigma_{k+1}}\lambda_3^-\le g_k(\sigma_{k+1}-\sigma_k)$. Reaching $W_n\ge W_0+13m$ requires $N_3^-\ge m$, so $T\ge\sigma_m=\sum_{k<m}(\sigma_{k+1}-\sigma_k)\ge\sum_{k<m}E_k/g_k$. The expectation, the integral comparison $\sum_{k<m}(W_0+13k)^{-2}\ge\int_0^m(W_0+13s)^{-2}\dd s$, the variance and the Chernoff bound $\Pr\{S\le t\}\le e^{\theta t}\mathbb E e^{-\theta S}$ follow from independence.
\end{proof}

Writing $W_n=WVN_{\!A}$ and $\kappa=b_3ph^6/(VN_{\!A})$, the mean bound in~\eqref{q:eq:birth-moments} is exactly the deterministic restoration clock~\eqref{q:eq:restoration-clock}, up to the discreteness correction $W_0+13m$ versus $W_1$: the deterministic clock is the expectation of a pathwise stochastic lower bound. At the corner of Section~\ref{q:sec:examples} ($V=1\,\mathrm L$), $W_0=4.18\times10^{16}$ counts, $m=3.2\times10^{15}$, and the bound has mean $1.228\times10^{27}\,\mathrm s$ with relative standard deviation $1.9\times10^{-8}$; the stochastic recovery time is at least this, essentially deterministically. At the reference endpoint $\eps=10^{-36}$, where the deterministic $X$ concentration corresponds to $0.003$ molecules, the inventory sits almost entirely in $Z$: $W_0\approx301$ counts, $m=24$, mean bound $3.2\times10^{56}\,\mathrm s$ with relative standard deviation $0.22$. The bottleneck therefore survives the breakdown of the deterministic description; what changes at low counts is the width of the distribution, not the order of the delay. The comparison is global up to explosion and uses only the inventory itself.

For completeness, a localised Poisson bound also holds: if the reverse propensity is at most $\Lambda$ on a finite set $\calD$ of count states, then $\Pr\{T_{\rm target}\le t,\ T_{\rm target}\le\tau_{\rm exit}\}\le\Pr\{\mathrm{Poisson}(\Lambda t)\ge m\}$. It is weaker than Proposition~\ref{q:prop:birth}: at the corner, bounding $n_Z$ by its outer-box maximum gives $m/\Lambda\approx1.5\times10^{26}\,\mathrm s$, a factor eight below the birth-chain mean, because it ignores that the propensity is smallest exactly when the deficit is largest.

\section{Dimensional calibration and operational limitations}
\label{q:sec:examples}
Fix a concentration unit $C_0$, a time unit $\tau_0$ and a volume $V$, independently of the parameters. If $\hat k$ is a dimensionless rate constant for a source of molecularity $|y|$, then
\begin{equation}\label{q:eq:dimensional}
 k_{\rm dim}=\frac{\hat kC_0^{1-|y|}}{\tau_0},\qquad c_i=C_0x_i,\qquad t_{\rm dim}=\tau_0t,\qquad N_i=c_iVN_{\!A},
\end{equation}
with $N_{\!A}=6.02214076\times10^{23}\,\mathrm{mol^{-1}}$ exactly~\cite{BIPM2019}. All forward sources in~\eqref{q:eq:reactions} are cubic; the reverse sources of pairs $1$--$3$ are quadratic and that of pair $4$ is linear. Take $C_0=1\,\mathrm{mol\,L^{-1}}$, $\tau_0=1\,\mathrm s$, $V=1\,\mathrm L$ and the corner $h=\tfrac1{600}$, $u=\tfrac1{20}$ ($p=\tfrac1{12000}$) with nominal prefactors. The forward constants are $1,1,1,4\ \mathrm{L^2mol^{-2}s^{-1}}$; the reverse constants are $6.944\times10^{-9}\ \mathrm{L\,mol^{-1}s^{-1}}$ for pairs $1$ and $2$, $1.786\times10^{-21}\ \mathrm{L\,mol^{-1}s^{-1}}$ for pair $3$ and $6.944\times10^{-9}\ \mathrm{s^{-1}}$ for pair $4$. The certified all-time bounds, rounded outward, are
\begin{center}
\begin{tabular}{lrrr}
\toprule
species & lower bound (mol/L) & upper bound (mol/L) & upper count in $1\,\mathrm L$\\
\midrule
$X$ & $1.8084\times10^{-14}$ & $1.9291\times10^{-12}$ & $1.1617\times10^{12}$\\
$Y$ & $1.4467\times10^{-11}$ & $6.9445\times10^{-10}$ & $4.1821\times10^{14}$\\
$Z$ & $1.7361\times10^{-8}$ & $1.3889\times10^{-7}$ & $8.3641\times10^{16}$\\
\bottomrule
\end{tabular}
\end{center}
and the certified recovery window is
\[
 1.2281\times10^{27}\,\mathrm s\ \le\ T_{\rm witness},\qquad T_{\rm all}\ \le\ 1.2497\times10^{28}\,\mathrm s .
\]
Copy numbers are macroscopic, yet the certified recovery time exceeds the age of the universe by nine orders of magnitude.

Other admissible points have the opposite limitation. On the reference path $\gamma=1$ at the symmetric reference endpoint $\eps=10^{-36}$, the upper bound for $X$ is $5\times10^{-27}\,\mathrm{mol/L}$, fewer than $0.004$ expected molecules in one litre, with $3.05\times10^{56}\,\mathrm s\le T_{\rm witness}$ and $T_{\rm all}\le3.1\times10^{57}\,\mathrm s$; at the admitted-rectangle endpoint $\eps=600^{-12}$ the $X$ upper count is still below $0.30$, with $1.86\times10^{52}\,\mathrm s\le T_{\rm witness}$ and $T_{\rm all}\le1.89\times10^{53}\,\mathrm s$. These expected counts lie below one molecule, outside the macroscopic regime of a deterministic concentration description.

There is also a thermodynamic distinction. With $r_i$ the forward-to-reverse ratio of pair $i$ on the reference path, the rate ratio around the independent cycle formed by pairs $1$, $2$ and $4$ is
\begin{equation}\label{q:eq:cycle-ratio}
 \frac{r_1r_2}{r_4}=\frac1{4\eps},
\end{equation}
whose logarithm is about $81.5$ at $\eps=10^{-36}$ and $17.4$ at the corner. Graph reversibility therefore does not imply equilibrium thermodynamic reversibility; reading such a cycle logarithm as an affinity or chemical work~\cite{Schnakenberg1976} requires a thermodynamically specified reservoir realisation that this three-species effective model does not supply. For practitioners the model-specific conclusion is: adding tiny reverse reactions can restore mathematical permanence while the certified concentrations or recovery times remain operationally unacceptable. The resulting diagnostic is to identify nearly conserved inventories, bound their replenishment capacity, compare with the operating horizon and check copy numbers.


\section{Discussion and open problems}
\label{q:sec:discussion}
Weak reversibility protects against concentration loss quantitatively as rates degenerate, but concentration geometry, preparation and recovery obey different laws. The general bound is governed by the exponent spread and an existence constant $H$. In the coupled network the preparation scales obey exact algebraic relations, the slow reverse reaction must supply a definite cumulative inventory restoration, and the state-dependent capacity bound reproduces the certified witness constant exactly. These facts explain the recovery exponent $(14\gamma+5)/12$ without any convergence or equilibrium argument.

\begin{enumerate}[label=(O\arabic*)]
\item \emph{Whole-class capture.} Theorem~\ref{q:thm:coupled} is restricted to the prepared basin. Theorem~\ref{q:thm:global-clocks} removes the basin hypothesis from the \emph{lower} bounds: from any positive seed, restoring a deficit costs at least the restoration clock and shedding a surplus at least the depletion clock. The general theorem absorbs every positive seed into a reciprocal box, but neither result gives a seed-independent deadline for entering the sharp corridor. A capture theorem would need to control $A$ and $B$ before corridor entry; the inventory identity alone does not, since $\dot Q$ depends on $A^2BQ^6$ and the identity gives no bound on $A,B$.
\item \emph{Limiting scaled entry time.} Matching powers do not give a limit for $\lambda T_{\rm entry}$ as $h,u\to0$. A promising route is a reduced slow equation after fast-layer control, with a quantified residual.
\item \emph{Optimal constants.} The certified window $[0.3046,3.1]$ leaves room for sharper constants. Subdividing the entry slabs would sharpen the $A$ and $B$ phases, which are negligible here, and the $Q$ phase through a better bound on $R_Q$ along the actual corridor.
\item \emph{Larger rectangles.} At $h\le\tfrac1{500}$ with $u\le10^{-3}$ only the sufficient lower $A$ entry margin falls slightly below $\tfrac14$ (a failure of the estimate, not a counterexample), and at $u\le\tfrac1{10}$ the term $-2b_2uB^2/Q$ pushes the lower $B$ entry margin below $1$. There is also a genuine adverse face: at $h=\tfrac1{100}$, $u=\tfrac1{20}$, $(A,B,Q)=(\tfrac72,\tfrac52,\tfrac34)$, $a_i=b_1=b_2=b_3=1$, $K=\tfrac{99}{25}$, $b_4=\tfrac{101}{100}$ the exact value of $R_A$ is negative, so the inner box of~\eqref{q:eq:regions} is not invariant there; the mechanism is $KC^3-b_4A$, where inventory stored in $Y$ reduces $C$ and weakens the cubic restoring source. Enlarging $h$ substantially requires different boxes, not merely more interval subdivision.
\item \emph{Effective affine data and network-specific clocks.} The barrier delays of Part~\ref{part:uniform} depend on the finite affine data. Section~\ref{q:sec:effective-H} identifies the limiting choices that prevent their numerical extraction in the construction used here. Explicit slopes and activity thresholds, a worked $H$ for a small network, or bounds in terms of the complexes would make Theorem~\ref{q:thm:general} numerically usable. Direct network structure can also yield sharper recovery constants, as the staged scalar example shows.
\item \emph{A general staged-recovery principle.} In Section~\ref{u:sec:example}, a recovered component accelerates the guaranteed recovery of the components it drives. A general result should specify the dependency order and the clock inequalities needed to propagate these improved bounds through a network.
\item \emph{Periodic entrainment.} Corollary~\ref{u:cor:periodic} gives periodic operation under periodic rate histories. Conditions for stability, uniqueness and entrainment of that periodic regime, including state-dependent feedback under the hypotheses of Section~\ref{u:sec:consequences}, remain to be determined.
\item \emph{Stochastic viability.} Proposition~\ref{q:prop:birth} bounds the stochastic recovery time from below, pathwise, with no localisation. Matching stochastic upper bounds, positive recurrence, the time spent below functional thresholds and the full distribution of inventory events for this four-linkage-class network are open.
\end{enumerate}