\section{Setting and main theorem}\label{u:sec:setting}

Let $E$ be a finite index set of reactions. For $e\in E$ let $y_e,y'_e\in\R^d_{\ge0}$
be its source and target complex and $\nu_e=y'_e-y_e$ its reaction vector. Parallel
reactions, self-loops ($\nu_e=0$) and the empty family are permitted. Put
\[
 S=\spanR_{\R}\{\nu_e:e\in E\},\qquad P_c=(c+S)\cap\U\quad(c\in\U).
\]
For $x\in\U$ and $y\in\R^d_{\ge0}$ write $x^{y}=\exp(\sum_i y_i\log x_i)$, the
usual monomial when $y$ has natural coordinates.

\begin{definition}[Ordinary endotacticity]\label{u:def:endo}
The network is \emph{endotactic} if for every $r\in\R^d$ and every $e\in E$,
\begin{equation}\label{u:eq:endo}
 r\cdot\nu_e<0\quad\Longrightarrow\quad
 \exists\,j\in E:\quad r\cdot\nu_j>0\ \text{ and }\ r\cdot y_j<r\cdot y_e .
\end{equation}
\end{definition}
All directions $r\in\R^d$ are tested, not only nonnegative ones (the latter would
give the weaker ``lower endotactic'' condition of~\cite{Pantea2012}). Nothing is
required of neutral reactions ($r\cdot\nu_e=0$), and the witness need not start at
a globally lowest source; requiring that defines strong
endotacticity~\cite{GopalkrishnanMillerShiu2014,JohnstonPanteaDonnell2016}.

Fix numbers $0<\underline k_e\le\overline k_e<\infty$ and put
\[
 \B=\prod_{e\in E}[\underline k_e,\overline k_e],\qquad
 f_k(x)=\sum_{e\in E}k_e\,x^{y_e}\,\nu_e\qquad(k\in\B,\ x\in\U).
\]
An \emph{admissible history} is a function $k:[0,\infty)\to\B$ each of whose
coordinates is Lebesgue measurable. A \emph{solution} from $x_0\in\U$ under $k$ is a
function $x:[0,\infty)\to\U$ that is absolutely continuous on every compact
interval, satisfies $x(0)=x_0$, and satisfies $x'(t)=f_{k(t)}(x(t))$ for almost
every $t$. Modifying $k$ on a null set changes neither the class of admissible
histories nor the solutions, so histories that are merely essentially bounded in
$\B$ are covered after a modification on one common null set ($E$ is finite).

\begin{theorem}[Uniform endotactic permanence]\label{u:thm:main}
Assume the network is endotactic. For every rate box $\B$ and every $c\in\U$ there
is a nonempty compact convex set $K\subset P_c$ such that:
\begin{enumerate}
\item every $x_0\in P_c$ and every admissible history have a unique global
 solution, and it stays in $P_c$; uniqueness is among all solutions in the above
 sense, with no a priori boundedness assumption;
\item $K$ is invariant: a solution starting in $K$ remains in $K$ for all $t\ge0$,
 under every admissible history;
\item for every compact $Q\subset P_c$ there is $T_Q\ge0$ such that every solution
 from every $x_0\in Q$ under every admissible history lies in $K$ for every
 $t\ge T_Q$.
\end{enumerate}
The set $K$ is chosen before $Q$, and $T_Q$ before $x_0$ and the history.
\end{theorem}

Compact containment in $P_c$ gives one positive lower bound and one finite upper
bound for every coordinate, valid after $T_Q$ for all histories; these constants
depend on the network, $\B$ and $c$, not on the history, and need not be uniform
across classes or boxes. If $d=0$ the state space is a point and there is nothing
to prove; if $S=\{0\}$ (empty family or only self-loops) then $P_c=\{c\}$ and
$K=\{c\}$ works with $T_Q=0$. The proof for $d\ge1$ occupies
Sections~\ref{u:sec:affine}--\ref{u:sec:solutions}. The argument handles a proper
stoichiometric subspace without modification; the class enters through the
neutral normals discussed in Section~\ref{u:sec:drift}.

\section{The finite affine geometric input}\label{u:sec:affine}

For $0<h<1$ put $B_h=[h,h^{-1}]^d$. Every $x\in B_h$ can be written uniquely as
$x=h^{q}$ coordinatewise, with $q=\log x/\log h\in[-1,1]^d$. Let $V\subset\R^d$ consist of the coordinate vectors $e_i$ and all nonzero
ordered differences of complexes occurring as sources or targets. This common
comparison set includes differences between different connected components and
will also be used for the cut estimates in Part~\ref{part:reverse}. For
$r\in[-1,1]^d$ define the tolerance
\begin{equation}\label{u:eq:tolerance}
 \vartheta(r)=\min\Bigl(\{1/2\}\cup
 \Bigl\{\frac{|r\cdot v|}{2\norm{v}_1}:v\in V,\ r\cdot v\ne0\Bigr\}\Bigr).
\end{equation}
Zero scalar products are omitted, so $\vartheta(r)>0$ for every $r$ and
$\vartheta(0)=1/2$; no continuity of $\vartheta$ is assumed or true.

\begin{lemma}[Affine approximation with discontinuous tolerance]\label{u:lem:affine}
There are a nonempty finite label set $I$, slopes $r_\ell\in[-1,1]^d$ ($\ell\in I$)
independent of $h$, offsets $b_\ell(h)\in\R$ defined for small $h>0$, and
$h_g\in(0,1)$ such that, whenever $0<h<h_g$, $q\in[-1,1]^d$, and $\ell$ attains the
minimum in
\[
 F_h(h^{q})=\min_{m\in I}\bigl\{r_m\cdot h^{q}+b_m(h)\bigr\},
\]
one has $\norm{q-r_\ell}_\infty<\vartheta(r_\ell)$.
\end{lemma}

This is the uniform form, Corollary~2.2, of Lemma~2.1 in~\cite{OpenAI2026Boundedness}
(reproved in~\cite[Section~2]{OpenAI2026Uniform}) applied with $m=d$, $[a,b]=[-1,1]$
and the tolerance~\eqref{u:eq:tolerance}. The lemma is graph independent: it concerns
only the finite set $V$ of comparison vectors. Its sequential statement covers every
fixed label, including ties; the uniform statement follows because a failure
sequence $h_\nu\to0$, $q_\nu\in[-1,1]^d$ with a violating active label would, by
finiteness and compactness, have a subsequence with one fixed label and convergent
exponents, contradicting the sequential bound for that label, the tolerance being
evaluated at the fixed slope throughout. The family, hence its finite set of slopes,
is fixed before the threshold $h_g$. The Lean port of this construction is
discussed in Section~\ref{sec:formal}.

Fix the family and define the positive separation
\begin{equation}\label{u:eq:gap}
 H=\min\bigl(\{1\}\cup\{|r_\ell\cdot v|/2:\ \ell\in I,\ v\in V,\ r_\ell\cdot v\ne0\}\bigr)>0 .
\end{equation}
Let $x=h^{q}\in B_h$ with $0<h<h_g$, let $\ell$ be active at $x$ and write
$r=r_\ell$. For $v\in V$ with $r\cdot v\ne0$,
\[
 |(q-r)\cdot v|\le\norm{q-r}_\infty\norm{v}_1<\vartheta(r)\norm{v}_1\le|r\cdot v|/2 ,
\]
and hence
\begin{equation}\label{u:eq:sign-close}
 \operatorname{sign}(q\cdot v)=\operatorname{sign}(r\cdot v),\qquad
 |q\cdot v|>|r\cdot v|/2\ge H .
\end{equation}
Since $x^{y}=h^{q\cdot y}$ and $0<h<1$, this gives the \emph{complex-ordering estimate}
\begin{equation}\label{u:eq:ratio}
 r\cdot y<r\cdot z\quad\Longrightarrow\quad x^{z}\le h^{H}x^{y}
 \qquad(x=h^{q}\in B_h,\ r=r_\ell\text{ active at }x),
\end{equation}
and the \emph{coordinate-sign estimate}: if $r_i\ne0$ then $q_i$ has the sign of
$r_i$ and $|q_i|>H$. Here $y,z$ are any complexes of the network. The positive gap $H$ belongs to
the fixed finite slope family; exact zero comparisons are omitted from its definition.

\section{Endotactic drift and the convex plateau}\label{u:sec:drift}

Call a slope $r$ \emph{neutral} if $r\cdot\nu_e=0$ for every $e$, that is,
$r\perp S$.

\begin{lemma}[One witness below all adverse sources]\label{u:lem:witness}
Let $r$ be nonneutral. There is $j\in E$ with $r\cdot\nu_j>0$ and
$r\cdot y_j<r\cdot y_e$ for every $e$ with $r\cdot\nu_e<0$.
\end{lemma}
\begin{proof}
If adverse reactions exist, pick $e_-$ minimizing $r\cdot y_e$ among them.
Condition~\eqref{u:eq:endo} gives $j$ with $r\cdot\nu_j>0$ and
$r\cdot y_j<r\cdot y_{e_-}\le r\cdot y_e$ for every adverse $e$. If no adverse
reaction exists, nonneutrality gives some $j$ with $r\cdot\nu_j>0$, and there is no
inequality to check.
\end{proof}

For each nonneutral slope $r=r_\ell$ of the fixed family choose such a witness
$j=j(\ell)$ and set
\[
 \alpha_r=\underline k_{j}\,(r\cdot\nu_{j})>0,\qquad
 \beta_r=\sum_{e:\,r\cdot\nu_e<0}\overline k_e\,|r\cdot\nu_e|\ \ge0 .
\]
Let $k\in\B$, $0<h<h_g$, $x=h^{q}\in B_h$, and let $\ell$ be active at $x$. Keeping
the witness term, discarding the other nonnegative terms, bounding the coefficients
by the box, and applying~\eqref{u:eq:ratio} to every adverse source gives
\begin{equation}\label{u:eq:drift}
 r\cdot f_k(x)\ \ge\ \alpha_r x^{y_j}-\sum_{r\cdot\nu_e<0}\overline k_e|r\cdot\nu_e|\,x^{y_e}
 \ \ge\ x^{y_j}\bigl(\alpha_r-h^{H}\beta_r\bigr).
\end{equation}
The coefficients retain the projection lengths $|r\cdot\nu_e|$ of each reaction. Since the family
is finite, there is $h_*\in(0,h_g)$ such that $h^{H}\beta_r\le\alpha_r/2$ for all
nonneutral slopes whenever $0<h<h_*$. Let $D_*=\max(\{0\}\cup\{\sum_iy_{e,i}:e\in E\})$.
Sources are nonnegative, so $x\in B_h$ implies $x^{y}\ge h^{D_*}$ for every source
$y$. If nonneutral slopes exist let $\alpha_*>0$ be the least of their $\alpha_r$.
Then for $0<h<h_*$, every $k\in\B$, and every $x\in B_h$,
\begin{equation}\label{u:eq:margin}
 r_\ell\cdot f_k(x)\ \ge\ \delta_h:=\tfrac12\alpha_*h^{D_*}>0
 \qquad\text{for every nonneutral label $\ell$ active at $x$.}
\end{equation}
A neutral slope has $r\cdot f_k(x)=0$ identically. If every slope is neutral the
strict statements below are vacuous and any $\delta_h>0$ may be used. All choices
so far depend on the network and $\B$ only.

\paragraph{The plateau at a class representative.}
Now fix $c\in\U$. Choose $h_c\in(0,h_g)$, depending on $c$, so that
$\norm{\log c/\log h}_\infty<H\le1$ whenever $0<h\le h_c$
(hence $c\in\interior B_h$), and reduce the drift threshold $h_*$ to be at most $h_c$. If a label $\ell$ is active
at $c$ and $r_{\ell,i}\ne0$ for some $i$, the coordinate-sign estimate would give
$|\log c_i/\log h|>H$; hence every slope active at $c$ is zero. Pick such a label
and put $M_h=F_h(c)$. Its branch is the constant $M_h$, so $F_h\le M_h$ on all of
$\R^d$, and every other constant branch is $\ge M_h$. Define the \emph{plateau}
\begin{equation}\label{u:eq:plateau}
 C_h=\{x\in\R^d:F_h(x)=M_h\}=\bigcap_{\ell\in I}\{x:r_\ell\cdot x+b_\ell(h)\ge M_h\},
\end{equation}
a closed convex polyhedron containing $c$.

\begin{lemma}[Compact plateau and classwise strict drift]\label{u:lem:plateau}
For every $0<h\le h_c$,
\[
 C_h\subset\interior B_{\sqrt h},\qquad C_h\ \text{is compact}.
\]
If in addition $h<h_*$ and $x\in(P_c\cap B_h)\setminus C_h$, every label active at $x$ is nonneutral, so
$r_\ell\cdot f_k(x)\ge\delta_h$ for every active $\ell$ and every $k\in\B$.
\end{lemma}
\begin{proof}
Let $x=h^{q}\in C_h\cap B_h$. The chosen zero-slope branch equals $M_h=F_h(x)$, so it
is active at $x$, and Lemma~\ref{u:lem:affine} with $\vartheta(0)=1/2$ gives
$|q_i|<1/2$, that is $\sqrt h<x_i<h^{-1/2}$: $C_h\cap B_h\subset\interior B_{\sqrt h}$.
Suppose some $z\in C_h$ lay outside $B_h$. Since $c\in\interior B_h$ and $C_h$ is
convex, the segment from $c$ to $z$ lies in $C_h$; its first point on the boundary
of $B_h$ would belong to $C_h\cap B_h$ yet not to $\interior B_{\sqrt h}$, a
contradiction. Hence $C_h\subset\interior B_{\sqrt h}$; it is closed and bounded,
hence compact.

If $r_\ell$ is neutral then $r_\ell\cdot x=r_\ell\cdot c$ on $c+S$, so the branch
$\ell$ is constant on the class with value $\ge F_h(c)=M_h$. At $x\in P_c$ with
$F_h(x)<M_h$ this branch is not minimal, so $\ell$ is not active. The remaining
active labels are nonneutral and~\eqref{u:eq:margin} applies.
\end{proof}

The classwise argument uses neutrality with respect to the whole subspace $S$;
individual reactions may have zero projection even for a nonneutral slope.

\section{Solutions, trapping, and uniform absorption}\label{u:sec:solutions}

\subsection{Carath\'eodory solutions for measurable histories}\label{u:sec:bridge}

On any compact positive box the finitely many monomials $x^{y_e}$ and their state
derivatives $\partial_i x^{y}=y_ix^{y}/x_i$ are bounded. Hence
$(t,x)\mapsto f_{k(t)}(x)$ is measurable in $t$, smooth in $x$ on $\U$, and bounded
and Lipschitz in $x$ on each compact positive box uniformly over all histories
with values in $\B$. These are the standard Carath\'eodory conditions
\cite[Chapter~2]{CoddingtonLevinson1955}, \cite[\S1]{Filippov1988}, and we record the
argument in the form used by the formal proof. Fix $x_0\in\U$, $s\ge0$, a closed
ball of radius $\rho$ about $x_0$ inside $\U$, a bound $A$ for $\norm{f_k}$ and a
Lipschitz constant $L$ for $f_k$ on that ball, uniform in $k\in\B$. For $\eta>0$
with $A\eta\le\rho$ and $L\eta<1$, the Picard map
\[
 (\mathcal Tu)(t)=x_0+\int_s^t f_{k(v)}(u(v))\,dv\qquad(t\in[s,s+\eta])
\]
sends continuous paths in the ball to continuous paths in the ball and is a
contraction; its fixed point is an integral solution, hence absolutely continuous
and a solution of the equation almost everywhere, and any two solutions agree on
$[s,s+\eta]$ by the same estimate. Repeating the construction yields a unique
maximal positive solution, and all its integral increments lie in the closed
subspace $S$, so it stays in the affine class $c+S$ as long as it exists.

Let $x$ be a solution segment in $B_h$ and put $g(t)=F_h(x(t))$. The minimum of
finitely many affine functions is Lipschitz with constant at most $d$ in the
maximum norm, so $g$ is absolutely continuous on every compact interval. At almost
every $t$ the curve, $g$, and each branch composition $t\mapsto r_\ell\cdot x(t)+b_\ell(h)$
are differentiable and the equation holds. If $\ell$ is active at such a $t$, the
nonnegative differentiable function (branch minus envelope) vanishes at $t$, so
its derivative there is zero and $g'(t)=r_\ell\cdot f_{k(t)}(x(t))$. This handles
ties among active branches without selecting one. Hence, while $x(t)\in P_c\cap B_h$,
\begin{equation}\label{u:eq:acdrift}
 g'(t)\ge0\quad\text{a.e.},\qquad g'(t)\ge\delta_h\quad\text{a.e. on }\{t:g(t)<M_h\},
\end{equation}
by Lemma~\ref{u:lem:plateau} and the neutral-slope identity. These are
almost-everywhere statements; nothing is assumed at switching times of the
active set.

\subsection{Trapping and global existence}\label{u:sec:trapping}

Let $x_0\in C_h\cap P_c$. Then $x_0\in\interior B_h$. While the maximal solution
stays in $B_h$, \eqref{u:eq:acdrift} and the fundamental theorem for absolutely
continuous functions make $g$ nondecreasing; since $g(0)=M_h$ and $g\le M_h$, the
solution stays in $C_h$. A first exit from $\interior B_h$ would be a point of
$C_h$ on the boundary of $B_h$, contradicting Lemma~\ref{u:lem:plateau}. So the
maximal solution stays in the compact positive set $C_h\cap(c+S)$.

If its maximal interval of existence ended at a finite $b$, the field bound $A$ on
a positive box containing $C_h$ would give $\norm{x(t)-x(s)}\le A|t-s|$, so $x(t)$
would converge as $t\uparrow b$ to a point $x_b$ of that compact set; local
existence from $x_b$ under the shifted history and concatenation of the integral
equations would extend the solution beyond $b$. Hence the solution is global and
positive. The same argument shows that $C_h\cap P_c$ is invariant for every
admissible history, and uniqueness among all solutions follows from local
Lipschitz uniqueness on overlapping compact intervals. For natural-number complexes the field is polynomial on all of $\R^d$. The registered real-complex uniqueness statement concerns positive competitors; no extension of the real-power field outside $\U$ is used.

\subsection{Finite entry from a common drift margin}\label{u:sec:entry}
The trapping and entry arguments use only compact plateau geometry, zero drift
for neutral slopes and a common positive drift margin for active nonneutral
slopes. Thus they apply with any such margin $\delta_h$; the endotactic estimate
\eqref{u:eq:margin} is one choice, and Part~\ref{part:reverse} will supply another.

Suppose a global positive solution stays in $P_c\cap B_h$ for all $t\ge a$.
Write $g(t)=F_h(x(t))$ and set
\begin{equation}\label{u:eq:exact-delay}
 b=a+\frac{M_h-F_h(x(a))}{\delta_h}.
\end{equation}
If $g(a)=M_h$, invariance applies immediately. Otherwise, until the first entry
into $C_h$, integration of $g'\ge\delta_h$ gives
$g(t)\ge g(a)+\delta_h(t-a)$. Continuity at $b$ then gives $g(b)\ge M_h$ if entry
has not occurred earlier. Since $F_h\le M_h$, entry occurs by $b$ and invariance
gives residence thereafter. With a known lower bound $A_h\le F_h(x(a))$ we may,
in particular, use the convenient deadline
\begin{equation}\label{u:eq:delay}
 a+\frac{M_h-A_h}{\delta_h}+1.
\end{equation}

\subsection{The dyadic scale iteration}\label{u:sec:iteration}

Choose one $h_0\in(0,1)$ below all thresholds above; it depends only on the network,
$\B$ and $c$. Define
\[
 h_n=h_0^{2^n},\qquad C_n=C_{h_n},\qquad F_n=F_{h_n},\qquad M_n=F_n(c)\qquad(n\ge0).
\]
These scales decrease to $0$ and satisfy $\sqrt{h_{n+1}}=h_n$, so
Lemma~\ref{u:lem:plateau} gives
\begin{equation}\label{u:eq:boxes}
 C_{n+1}\subset\interior B_{h_n}.
\end{equation}
The plateaux need not be nested; only~\eqref{u:eq:boxes} is used. By compactness
$A_n=\min_{x\in C_{n+1}}F_n(x)$ exists, and $A_n\le M_n$ because $c\in C_{n+1}$.
A solution trapped in $C_{n+1}\cap P_c$ from time $a$ on stays in $B_{h_n}$, so by
\eqref{u:eq:delay} it enters, and then stays in, $C_n$ by time $a+D_n$ where
\begin{equation}\label{u:eq:totaldelay}
 D_n=\frac{M_n-A_n}{\delta_{h_n}}+1,\qquad T_N=\sum_{n=0}^{N-1}D_n .
\end{equation}
By induction, a solution trapped in $C_N\cap P_c$ from time $0$ lies in $C_0$ for
all $t\ge T_N$, under every admissible history.

\paragraph{Proof of Theorem~\ref{u:thm:main}.}
Let $x_0\in P_c$. The numbers $|\log x_{0,i}|$ are finite and $|\log h_N|\to\infty$,
so there is $N$ with $\norm{\log x_0/\log h_N}_\infty<H$; by the coordinate-sign
estimate every slope active at $x_0$ at scale $h_N$ is zero, so $F_N(x_0)$ equals a
constant branch, which is $\ge M_N$, while $F_N\le M_N$; thus $x_0\in C_N$. Trapping
gives the unique global positive solution in $C_N\cap P_c$, and the iteration
places it in $C_0$ by time $T_N$. Put
\[
 K=C_0\cap(c+S).
\]
It contains $c$, is compact and convex, and lies in $\U$ by
Lemma~\ref{u:lem:plateau}, so $K\subset P_c$; it is invariant because $C_0\cap P_c$ and
$c+S$ are. This proves (1) and (2) and the absorption of each individual initial
state.

For (3) let $Q\subset P_c$ be compact. The coordinate logarithms are continuous on
$Q$, so $|\log x_i|\le A_Q$ on $Q$ for some $A_Q\ge0$, and one index $N$ with
$A_Q/|\log h_N|<H$ places all of $Q$ in $C_N$. The deadline $T_Q=T_N$ then works
for every $x_0\in Q$ and every admissible history. This common starting plateau is
what gives uniformity; the pointwise existence of entry times together with
compactness of $Q$ would not. \qed

Formula~\eqref{u:eq:totaldelay} is a rigorous bound in terms of the barrier data
$(M_n,A_n,\delta_{h_n})$. The affine construction does not make these data
efficiently computable, and Section~\ref{u:sec:example} shows that direct scalar
structure, when available, yields far better constants.

\subsection{Nonnegative real complexes}\label{u:sec:real}

The argument above requires regularity only on compact positive boxes. The polynomial extension available for natural complexes is not needed for the real-complex conclusion. For $y_e,y'_e\in\R^d_{\ge0}$ the monomials $x^{y}$ are smooth on $\U$,
bounded with bounded derivatives on compact positive boxes, and satisfy
$x^{y}\ge h^{D_*}$ on $B_h$ because the exponents are nonnegative; the comparison
set $V$ is still finite; and Lemma~\ref{u:lem:affine} concerns exponent vectors
$q\in[-1,1]^d$ only. Hence Theorem~\ref{u:thm:main} holds verbatim for nonnegative
real complexes, with uniqueness understood among solutions in $\U$, the domain
where the real-power field is defined. The argument gives this extension directly. The registered statement also covers real complexes; Section~\ref{sec:formal} records its exact scope.

\section{Immediate consequences}\label{u:sec:consequences}

\begin{proposition}[Uniform continuous dependence]\label{u:prop:dependence}
Let $C\subset\U$ be compact. There is $L\ge0$ depending on the network, $\B$ and
$C$ only, such that any two solutions $x,y$ under the same admissible history that
both remain in $C$ satisfy
\begin{equation}\label{u:eq:gronwall}
 \norm{x(t)-y(t)}\le\norm{x(0)-y(0)}\,e^{Lt}\qquad(t\ge0).
\end{equation}
In particular this applies on the invariant set $K$ of Theorem~\ref{u:thm:main}.
\end{proposition}
\begin{proof}
Let $L$ be a Lipschitz constant of $f_k$ on $C$, uniform over $k\in\B$. The two
integral equations give $q(t)\le q(0)+L\int_0^t q(s)\,ds$ for the continuous
function $q(t)=\norm{x(t)-y(t)}$. Put $U(t)=q(0)+L\int_0^tq(s)\,ds$; then $q\le U$,
$U$ is $C^1$ with $U'=Lq\le LU$, and $(e^{-Lt}U)'\le0$ gives $q\le U\le q(0)e^{Lt}$.
Only the auxiliary primitive $U$ is differentiated pointwise; the measurable field
enters in integral form.
\end{proof}

\begin{corollary}[Periodic operation]\label{u:cor:periodic}
Let $\tau>0$ and let an admissible history satisfy $k(t+\tau)=k(t)$ for almost
every $t\ge0$. Then there is a positive solution with $x(t+\tau)=x(t)$ for all
$t\ge0$, taking all its values in $K$.
\end{corollary}
\begin{proof}
The time-$\tau$ map $\Phi_\tau(z)=x_z(\tau)$ sends $K$ to itself by invariance and
is continuous by Proposition~\ref{u:prop:dependence}. Brouwer's fixed-point theorem,
applied in the affine hull of the nonempty compact convex set $K$, gives
$z\in K$ with $\Phi_\tau(z)=z$. The shifted curve $t\mapsto x_z(t+\tau)$ is
absolutely continuous on compact intervals, starts at $z$, and satisfies the
equation almost everywhere because translation preserves null sets and the
coefficients agree almost everywhere; by uniqueness it equals $x_z$.
\end{proof}

\begin{corollary}[Autonomous equilibria]\label{u:cor:equilibrium}
For every constant $k\in\B$, every positive stoichiometric class of an endotactic
network contains a positive equilibrium of $f_k$.
\end{corollary}
\begin{proof}
Write $f=f_k$ and $\Phi_s$ for its flow. For $\tau_n=1/(n+1)$, Brouwer gives
$z_n\in K$ with $\Phi_{\tau_n}(z_n)=z_n$; pass to a subsequence $z_n\to z_*\in K$.
With $A$ a bound for $\norm f$ on $K$, $\norm{\Phi_s(z_n)-z_n}\le As$ for
$0\le s\le\tau_n$, so these short arcs converge uniformly to $z_*$, and uniform
continuity of $f$ on $K$ gives
\[
 0=\frac{\Phi_{\tau_n}(z_n)-z_n}{\tau_n}=\frac1{\tau_n}\int_0^{\tau_n}f(\Phi_s(z_n))\,ds
 \longrightarrow f(z_*).
\]
\end{proof}
A fixed point of a single time-$\tau$ map could lie on a periodic orbit; the
shrinking times are needed. The permanence-plus-Brouwer route to equilibria has a
two-species antecedent in \cite[Remark~6.2]{CraciunNazarovPantea2013}; here it is
available at the general endotactic scope.

\begin{corollary}[Bounded multiplicative modulation]\label{u:cor:modulation}
If $k_e(t)=\bar k_e\theta_e(t)$ with $\bar k_e>0$ and measurable
$0<\underline\theta_e\le\theta_e(t)\le\overline\theta_e<\infty$, then
Theorem~\ref{u:thm:main} applies with
$\underline k_e=\bar k_e\underline\theta_e$ and
$\overline k_e=\bar k_e\overline\theta_e$. No slow-variation, switching-rate or
independence assumption is involved.
\end{corollary}

\begin{proposition}[State feedback with local hypotheses]\label{u:prop:feedback}
Let $k:[0,\infty)\times\U\to\B$ be jointly measurable and such that the closed-loop
field $(t,x)\mapsto f_{k(t,x)}(x)$ is bounded and Lipschitz in $x$ on every compact
positive box, uniformly in $t$. Then every positive initial state has a unique
global positive closed-loop solution, $K$ is invariant for it, and the deadline
$T_Q$ of Theorem~\ref{u:thm:main} applies to closed-loop solutions from $Q$.
\end{proposition}
\begin{proof}
The closed-loop field satisfies the Carath\'eodory conditions, so a local solution
exists and is unique. Along it, the evaluated history $t\mapsto k(t,x(t))$ is
measurable with values in $\B$, so the trapping argument of
Section~\ref{u:sec:trapping} applies to the local solution and confines it to a
compact positive plateau; the field bound then gives a limit at any finite
maximal time and local existence continues the solution. Only after global
existence is established is the evaluated history a global admissible history,
to which Theorem~\ref{u:thm:main} applies; its bounds and the starting plateau do not
depend on the feedback law.
\end{proof}

\section{A quantitative three-species example}\label{u:sec:example}

Consider the nondimensional system
\begin{equation}\label{u:eq:fixture}
 \dot a=c\,(k_1-k_2a^2),\qquad
 \dot b=ab\,(k_3-k_4b^2),\qquad
 \dot c=bc\,(k_5-k_6c^2),\qquad \tfrac12\le k_i(t)\le2,
\end{equation}
with six independent measurable coefficients. It is the mass-action system of the
six reactions of Figure~\ref{u:fig:network}.
\begin{figure}[ht]
\centering
\renewcommand{\arraystretch}{1.45}
\begin{tabular}{ccc}
$C\xrightarrow{k_1}A+C$ & $A+B\xrightarrow{k_3}A+2B$ & $B+C\xrightarrow{k_5}B+2C$\\
$2A+C\xrightarrow{k_2}A+C$ & $A+3B\xrightarrow{k_4}A+2B$ & $B+3C\xrightarrow{k_6}B+2C$
\end{tabular}
\caption{Three inward reaction pairs with terminal targets. Each column has
sources $u$ and $u+2v$ and common target $u+v$, with $v=e_A,e_B,e_C$
respectively. For the direction $r=(0,1,0)$ the globally lowest sources are the
neutral ones in the first column.}
\label{u:fig:network}
\end{figure}

\subsection{Structural classification}\label{u:sec:classification}
For a pair $u\to u+v$, $u+2v\to u+v$ and a direction $r$: if $r\cdot v>0$ the first
reaction is a forward witness with source $u$ below the adverse source $u+2v$; if
$r\cdot v<0$ the roles are reversed; if $r\cdot v=0$ neither reaction is adverse.
Hence each pair, and their union, is endotactic in every real direction. The
reaction vectors are $\pm e_A,\pm e_B,\pm e_C$, so $S=\R^3$ and the positive class
is the whole open orthant. At $r=(0,1,0)$ the globally lowest sources $C$ and
$2A+C$ are neutral while the forward witness $A+B$ starts at level $1$, so the
network is not strongly endotactic. Each target is terminal, so the network is not
weakly reversible. It has nine complexes, three linkage classes, rank $3$ and
deficiency $9-3-3=3$; the displayed reaction graph is acyclic and cannot be complex
balanced at positive rates, since a terminal target receives positive influx and
has no outflux (this concerns the displayed realization; dynamically equivalent
realizations are not excluded here). Full rank excludes any nonzero linear
conservation law.

\subsection{Invariant boxes}\label{u:sec:boxes}
For a positive initial state put
\begin{equation}\label{u:eq:startup}
 m=\min(a_0,b_0,c_0,1/2),\qquad M=\max(a_0,b_0,c_0,2).
\end{equation}
The box $[m,M]^3$ is invariant for every admissible history: on a lower face of a
coordinate the bracket in its equation is at least $1/2-2m^2\ge0$, on an upper face
at most $2-M^2/2\le0$, and the remaining factors are positive. To make this a
rigorous almost-everywhere argument even with weak face signs, clamp the state to
the box, construct the bounded Lipschitz integral solution of the clamped field,
and note that the positive part of the distance of each coordinate to its interval
is absolutely continuous with nonpositive derivative a.e. where positive and zero
derivative a.e. where zero; starting from zero it stays zero, so the clamped
solution solves~\eqref{u:eq:fixture} and uniqueness proves invariance. The same face
signs, now strict ($9/50>0$ and $-9/8<0$), make the operating box
$[2/5,5/2]^3$ invariant. Enlarging the operating box beyond $[1/2,2]^3$ matters:
with $k_1=1/2$, $k_2=2$ and $a_0<1/2$ the first coordinate approaches $1/2$ from
below without reaching it in finite time.

\subsection{Clocks and scalar comparisons}\label{u:sec:comparisons}
Assume $(a_0,b_0,c_0)\in[1/10,10]^3$; by invariance all three coordinates stay in
$[1/10,10]$. Each coordinate has an \emph{operational clock}
\[
 \sigma_a(t)=\int_0^t c(s)\,ds,\qquad \sigma_b(t)=\int_0^t a(s)\,ds,\qquad
 \sigma_c(t)=\int_0^t b(s)\,ds,
\]
an absolutely continuous nondecreasing function whose derivative is the prefactor
of the corresponding equation. The startup box bounds every clock speed below by
$1/10$: $\sigma(t)\ge t/10$ for each clock.

\paragraph{The first coordinate (conventional).}
Since $\sigma_a$ is locally bi-Lipschitz with $\sigma_a'\ge1/10$, the substitution
$t\mapsto\sigma_a$ is valid for absolutely continuous equations and preserves the
almost-everywhere statements. In the clock $\sigma=\sigma_a$,
\[
 \tfrac12-2a^2\ \le\ \frac{da}{d\sigma}\ \le\ 2-\tfrac12a^2\qquad\text{a.e.}
\]
The Riccati comparison equations $a'=\frac12-2a^2$ from $a(0)=1/10$ and
$a'=2-\frac12a^2$ from $a(0)=10$ have the solutions
$\frac12\tanh(\sigma+\atanh\frac15)$ and $2\cothh(\sigma+\atanh\frac15)$, and the
positive-part argument with a local Lipschitz bound (valid for measurable
coefficients) gives
\begin{equation}\label{u:eq:aenvelopes}
 \tfrac12\tanh\bigl(\sigma+\atanh\tfrac15\bigr)\le a\le2\cothh\bigl(\sigma+\atanh\tfrac15\bigr).
\end{equation}
Both $a\ge2/5$ and $a\le5/2$ hold once $\sigma\ge\atanh\frac45-\atanh\frac15=\frac12\log6$,
hence for all
\begin{equation}\label{u:eq:Ta}
 t\ \ge\ T_a:=5\log6\ <\ 9,
\end{equation}
and they persist because the comparison envelopes move inward. The rounding
$5\log6<9$ is the exact inequality $e^{9/5}>\sum_{j=0}^{6}(9/5)^j/j!=7542629/1250000>6$.

\paragraph{The second and third coordinates (clocked linear comparison).}
Let $u\in\{b,c\}$ with clock $\sigma\in\{\sigma_b,\sigma_c\}$ and coefficients
$(k_+,k_-)\in\{(k_3,k_4),(k_5,k_6)\}$, so that $u'=\sigma'u(k_+-k_-u^2)$ almost
everywhere. Because $u$ is confined to $[1/10,10]$ and $v\mapsto v^{-2}$ is
Lipschitz there, $z=u^{-2}$ is absolutely continuous, and by the chain rule, for
almost every $t$,
\begin{equation}\label{u:eq:zode}
 z'=-2u^{-3}u'=\sigma'\bigl(-2k_+z+2k_-\bigr),\qquad\text{so}\qquad
 \sigma'(1-4z)\ \le\ z'\ \le\ \sigma'(4-z),
\end{equation}
using $k_+\ge1/2$, $k_-\le2$, $z>0$ and $\sigma'\ge0$. The relevant comparison is
stated in the clock directly, without changing the time variable.

\begin{lemma}[Clocked linear comparison]\label{u:lem:clocked}
Let $z,\sigma:[0,T]\to\R$ be absolutely continuous, $\beta\in\R$, and suppose
$z'(t)\le\sigma'(t)\bigl(\beta-z(t)\bigr)$ for almost every $t\in[0,T]$. Then
\[
 z(T)\ \le\ \beta+\bigl(z(0)-\beta\bigr)e^{-(\sigma(T)-\sigma(0))}.
\]
The symmetric statement with $\ge$ holds for a lower bound.
\end{lemma}
\begin{proof}
$\sigma$ is continuous, hence bounded by some $M$ on $[0,T]$, and $\exp$ is
Lipschitz on $(-\infty,M]$; so $w=e^{\sigma}(z-\beta)$ is absolutely continuous as
a product of absolutely continuous functions. Almost everywhere
$w'=e^{\sigma}\bigl(\sigma'(z-\beta)+z'\bigr)\le0$. Integrating,
$w(T)\le w(0)$, which is the claim after division by $e^{\sigma(T)}>0$.
\end{proof}
This lemma, including the composition step and the integration, is verified in
Lean (\lean{ac_clocked_linear_upper}; the constant-speed case is
\lean{ac_linear_upper}). Applied to~\eqref{u:eq:zode} with $z(0)=u_0^{-2}\in[1/100,100]$:
\begin{equation}\label{u:eq:zenvelopes}
 \tfrac14-\tfrac{6}{25}e^{-4\sigma(t)}\ \le\ z(t)\ \le\ 4+96\,e^{-\sigma(t)}\qquad(t\ge0).
\end{equation}
The upper bound on $z$ is a \emph{lower} bound on $u$: $u\ge2/5$ as soon as
$\sigma(t)\ge\log(128/3)$, and persistently so because $\sigma$ is nondecreasing.
The lower bound gives $u\le5/2$ as soon as $\sigma(t)\ge\frac14\log(8/3)$.

\subsection{Two certified deadlines}\label{u:sec:deadlines}
With the uniform clock floor $\sigma(t)\ge t/10$ alone,
\eqref{u:eq:zenvelopes} gives $u\ge2/5$ for $t\ge T_*:=10\log(128/3)$ and $u\le5/2$
for $t\ge\frac52\log(8/3)$; together with~\eqref{u:eq:Ta} this is the one-stage
certificate.

\begin{proposition}[One-stage certificate]\label{u:prop:38}
Every solution of~\eqref{u:eq:fixture} from $[1/10,10]^3$ lies in $[2/5,5/2]^3$ for
every $t\ge T_*=10\log(128/3)$, and $T_*<38$.
\end{proposition}
\begin{proof}
All deadlines in the preceding paragraph are at most $T_*$ since $5\log6<T_*$ and
$\frac52\log(8/3)<T_*$. The rounding is the exact rational inequality
\begin{equation}\label{u:eq:rational38}
 e^{19/5}\ \ge\ \sum_{j=0}^{9}\frac{(19/5)^j}{j!}=\frac{1968639501739}{44296875000}\ >\ \frac{128}{3},
\end{equation}
so $\log(128/3)<19/5$ and $T_*<38$.
\end{proof}

The one-stage bound of Proposition~\ref{u:prop:38} wastes the fact that the first
coordinate recovers early. Once
$a\ge2/5$, the clock $\sigma_b$ runs at speed at least $2/5$, and once $b\ge2/5$,
so does $\sigma_c$. Feeding these floors back in sequence gives a staged
certificate; the order matters, since assuming all three floors from the start
would be circular.

\begin{theorem}[Staged certificate]\label{u:thm:staged}
Every solution of~\eqref{u:eq:fixture} from $[1/10,10]^3$ under every admissible
history satisfies
\[
 a(t)\in[\tfrac25,\tfrac52]\ \ (t\ge9),\qquad
 b(t)\in[\tfrac25,\tfrac52]\ \ (t\ge\tfrac{65}{4}),\qquad
 c(t)\in[\tfrac25,\tfrac52]\ \ (t\ge\tfrac{347}{16}).
\]
In particular every such solution lies in $[0.4,2.5]^3$ for all $t\ge347/16=21.6875$,
hence for all $t\ge22$.
\end{theorem}
\begin{proof}
\emph{Upper bounds.} By~\eqref{u:eq:Ta}, $a\le5/2$ for $t\ge9$. For $u\in\{b,c\}$
the lower half of~\eqref{u:eq:zenvelopes} with $\sigma\ge t/10$ gives $z\ge4/25$,
i.e.\ $u\le5/2$, for $t\ge\frac52\log(8/3)$, and $\frac52\log(8/3)<\frac52<9$ since
$8/3<e$. So all upper bounds hold from $t=9$ on, and persist.

\emph{Stage $a$.} By~\eqref{u:eq:Ta}, $a(t)\ge2/5$ for all $t\ge9$; also $a\ge1/10$
for all $t\ge0$.

\emph{Stage $b$.} The clock $\sigma_b$ has $\sigma_b'=a\ge1/10$ a.e.\ on $[0,9]$ and
$\sigma_b'\ge2/5$ a.e.\ on $[9,\infty)$, so for $t\ge9$
\[
 \sigma_b(t)\ \ge\ \frac{9}{10}+\frac25\,(t-9),
\]
and the right side equals $19/5$ at $t=65/4$. By~\eqref{u:eq:zenvelopes} and
\eqref{u:eq:rational38}, $z_b(t)\le4+96e^{-19/5}<4+96\cdot\frac{3}{128}=\frac{25}{4}$
for $t\ge65/4$, that is $b(t)>2/5$ for all $t\ge65/4$.

\emph{Stage $c$.} Now $\sigma_c'=b\ge1/10$ a.e.\ on $[0,65/4]$ and $\sigma_c'\ge2/5$
a.e.\ on $[65/4,\infty)$, so for $t\ge65/4$
\[
 \sigma_c(t)\ \ge\ \frac{65}{40}+\frac25\Bigl(t-\frac{65}{4}\Bigr),
\]
which equals $19/5$ at $t=347/16$. The same comparison gives $c(t)>2/5$ for all
$t\ge347/16$. Finally $347/16<22$.
\end{proof}

The two clock identities $\frac9{10}+\frac25(\frac{65}4-9)=\frac{19}5$ and
$\frac{65}{40}+\frac25(\frac{347}{16}-\frac{65}4)=\frac{19}5$ are exact. Keeping the
exact stage times instead of rounding gives $T_a=5\log6$,
$T_b=\frac34T_a+\frac52\log\frac{128}{3}$ and $T_c=\frac34T_b+\frac52\log\frac{128}3$,
approximately $8.959$, $16.103$ and $21.461$; neither this nor $347/16$ is claimed
optimal. Table~\ref{u:tab:times} summarizes the deadlines.

\begin{table}[ht]
\centering
\begin{tabular}{llll}
\toprule
Requirement & Deadline (exact) & Decimal & Rounded, certified\\
\midrule
$a\in[2/5,5/2]$ & $5\log6$ & $8.95880$ & $9$\\
$b,c\le5/2$ & $\tfrac52\log(8/3)$ & $2.45207$ & $9$\\
$b,c\ge2/5$, one stage & $10\log(128/3)$ & $37.53418$ & $38$\\
$b\ge2/5$, staged & $\tfrac34\,5\log6+\tfrac52\log(128/3)$ & $16.10264$ & $65/4=16.25$\\
$c\ge2/5$, staged & $\tfrac34T_b+\tfrac52\log(128/3)$ & $21.46053$ & $347/16=21.6875$\\
\bottomrule
\end{tabular}
\caption{Sufficient recovery deadlines in nondimensional time, uniform over all
measurable histories in $[1/2,2]^6$ and all startups in $[1/10,10]^3$. The decimal
column only evaluates the exact formulas; the certified column uses the rational
roundings proved in Lean.}
\label{u:tab:times}
\end{table}

Figure~\ref{u:fig:trajectories} shows illustrative trajectories under three declared
rate schedules; every sampled state after $347/16$ lies in the operating box, as
Theorem~\ref{u:thm:staged} requires, and the pictures show how far from tight the
deadline is for these particular histories.

\begin{figure}[ht]
\centering
\includegraphics[width=\linewidth]{figures/operating_trajectories.pdf}
\caption{Illustrative trajectories under three declared schedules. The shaded
strip is the proved operating interval $[0.4,2.5]$ for each coordinate and the
dashed line is the staged deadline $347/16$. Colours distinguish $a,b,c$; line
styles distinguish the four startups $(0.1,0.1,0.1)$ (solid), $(10,10,10)$
(dashed), $(0.1,10,0.1)$ (dotted) and $(10,0.1,10)$ (dash-dotted). Constant rates
are $(0.5,2,0.5,2,0.5,2)$; periodic rates are
$k_i(t)=1.25+0.75\sin(2\pi t/8+(i-1)\pi/3)$; switching rates are $0.5$ when
$(\lfloor t/2\rfloor+i-1)\bmod3=0$ and $2$ otherwise, integrated segment by segment
at the exact switching times (SciPy LSODA, relative tolerance $10^{-9}$, absolute
tolerance $10^{-11}$). }
\label{u:fig:trajectories}
\end{figure}

\subsection{Dimensions and physical reading}\label{u:sec:dimensions}
Let $A=X_*a$, $B=X_*b$, $C=X_*c$ and $t_{\rm phys}=t_*t$ for declared positive
reference scales. The corresponding physical effective model is
\[
 \dot A=C(\kappa_1-\kappa_2A^2),\qquad \dot B=AB(\kappa_3-\kappa_4B^2),\qquad \dot C=BC(\kappa_5-\kappa_6C^2),
\]
with $k_1=t_*\kappa_1$, $k_2=t_*X_*^2\kappa_2$, $(k_3,k_5)=t_*X_*(\kappa_3,\kappa_5)$ and
$(k_4,k_6)=t_*X_*^3(\kappa_4,\kappa_6)$. The $\kappa_i$ carry different units
($\kappa_1$: time$^{-1}$; $\kappa_2$: concentration$^{-2}$time$^{-1}$; $\kappa_3,\kappa_5$:
concentration$^{-1}$time$^{-1}$; $\kappa_4,\kappa_6$: concentration$^{-3}$time$^{-1}$), so the
common dimensionless box $[1/2,2]^6$ is an assumption on scaled coefficients, not a
statement that unlike physical constants share a numerical range; whether useful
$X_*,t_*$ exist for given measured ranges must be checked case by case. The
physical startup and operating boxes are $[0.1X_*,10X_*]^3$ and $[0.4X_*,2.5X_*]^3$,
and the deadline is $22t_*$. Calibrating $t_*$ requires physical rate data.

The reduced full-rank system is an open, driven effective model. One mass-balanced
bookkeeping realization replaces the forward reactions by
$F_A+C\to A+C$, $F_B+A+B\to A+2B$, $F_C+B+C\to B+2C$ and the backward ones by
$2A+C\to A+C+W_A$, $A+3B\to A+2B+W_B$, $B+3C\to B+2C+W_C$, with maintained feed
reservoirs $F_\bullet$ and removed waste $W_\bullet$ absorbed into the effective
coefficients. This realization supplies open-system bookkeeping for effective cooperative
kinetics; its physical use requires specifying the maintained reservoirs and
the detailed reaction mechanisms.

\subsection{Limits of the fixed-box hypothesis and transition to Part II}\label{u:sec:limits}
The positive rate box fixes both a lower and an upper bound. For the endotactic
pair $A\rightleftarrows2A$ with $k_+(t)=k_-(t)=e^{-t}$, the operational clock has
finite total length and the logistic solution satisfies $a(t)\le e\,a_0$;
arbitrarily small startups therefore never reach a common positive floor.
For $0\rightleftarrows A$ with birth rate $1+t$ and death rate $1$, the equation
$a'=1+t-a$ has unbounded solutions.

Part~\ref{part:reverse} keeps the rates autonomous and strictly positive for each
$\eps>0$, while allowing their lower bounds to degenerate across a parameter
family. The resulting questions are quantitative: how the absorbing concentration
scales change, and how long an inventory can take to recover. Both use the shared geometry and absorption argument.