\subsection{A growing wall away from Taub}
\label{sec:late-wall-entrance}

We now bring a physical orbit into the core coordinates of the preceding
subsection. Throughout this step the orbit is assumed to stay at least
$\epsilon>0$ from every Taub point. Decay of the pairwise wall products
will make the transverse core amplitudes as small as needed.

For $\Delta\geq0$, write
\begin{equation}
 \mathcal P_\Delta(q):\quad
        4N_i(q)N_j(q)\leq\Delta^2\quad(i\ne j).
 \label{eq:small-pairwise-products}
\end{equation}
The constraint immediately gives, when $\Delta\leq1/4$,
\begin{equation}
 p^2+s^2+N_1^2+N_2^2+N_3^2
       \leq1+\tfrac32\Delta^2\leq35/32,
       \qquad |p|<17/16.
 \label{eq:pair-product-state-bound}
\end{equation}

\begin{lemma}[Leaving a dominant wall]
\label{lem:dominant-wall-exit}
Let $0<\nu\leq1/4$ and $0\leq\Delta\leq\nu/8$.
If~\eqref{eq:small-pairwise-products} holds on an interval of length
$5/\nu^2$, then that interval contains a point at which all three
walls are less than $\nu$.
\end{lemma}

\begin{proof}
If the walls are initially below $\nu$, there is nothing to prove.
Otherwise permute axes so that $n=N_1\geq\nu$ initially, and write
$b=N_2$, $c=N_3$. As long as $n\geq\nu$,
\[
 b+c\leq\frac{\Delta^2}{2n}\leq\frac{\nu}{128}
                                      \leq\frac n{16}.
\]
The first shear equation can be written exactly as
\[
 p'=(p-2)n^2+(1+p)(b-c)^2+(1-2p)n(b+c).
\]
Using~\eqref{eq:pair-product-state-bound}, the three terms are bounded
above by $-15n^2/16$, $3n^2/256$, and $n^2/4$, respectively.
Their sum is at most $-n^2/2\leq-\nu^2/2$.
The bound $|p|<17/16$ makes a stay of length $5/\nu^2$ impossible.
There is therefore a time $T$ in the interval with $n(T)<\nu$.

By continuity, between the initial time and $T$ the wall takes the
value $(\nu+n(T))/2\in[\nu/2,\nu)$. At this time the product bounds
give $b+c\leq\Delta^2/\nu<\nu$. All three walls are then strictly
below $\nu$.
\end{proof}

\begin{lemma}[One growing direction]
\label{lem:away-wall-growth-gap}
For each $\epsilon>0$, there are $0<\nu_0\leq1/4$ and $0<c\leq1$
such that every constrained state with $0\leq N_i\leq\nu_0$ and
$\dist(q,T_i)\geq\epsilon$ has exactly one wall rate at least $c$;
the other two are at most $-c$.
\end{lemma}

\begin{proof}
Let $\lambda_i=N_i'/N_i$, interpreted by their polynomial formulas
also when a wall vanishes. On the Kasner circle the rates are
$-1+2p$, $-1-p+\sqrt3s$, and $-1-p-\sqrt3s$.
The circle identity $p^2+s^2=1$ shows that exactly one is positive
and two are negative away from the three Taub points. For example,
if $p>1/2$ the first is positive and
$3s^2<(1+p)^2$ makes the other two negative. If $-1<p<1/2$, the
sign of $s$ distinguishes the other two possibilities.

Define the continuous margin
\[
 m(q)=\max_i\min\bigl(\lambda_i(q),\{-\lambda_j(q):j\ne i\}\bigr).
\]
On the compact constrained set with $0\leq N_i\leq1$ and all Taub
distances at least $\epsilon$, the function
$\max(m,N_1+N_2+N_3)$ is strictly positive. At a point where the
wall sum is zero this follows from the Kasner calculation; elsewhere
it follows from the sum itself. Choose a positive lower bound $d$,
and put $\nu_0=\min(1/4,d/6)$ and $c=\min(1,d)$.
When $N_i\leq\nu_0$, their sum is at most $d/2$, so $m\geq d$.
This proves the assertion. If the compact set is empty, the assertion
is vacuous and any positive sufficiently small constants suffice.
\end{proof}

\begin{proposition}[A transverse fixed-height entrance]
\label{prop:late-fixed-wall-entrance}
Choose $\nu_0,c$ as above. Let $0<\nu\leq\nu_0$ and
$0\leq\Delta\leq\nu/8$. If $q\in\M$ satisfies uniform Taub avoidance
and $\mathcal P_\Delta(\phi_tq)$ for all $t\geq0$, then there are
$s\in[0,5/\nu^2]$, $T>s$, and an axis $i$ such that all walls are
less than $\nu$ at $s$, and $T$ is their first subsequent arrival
at height $\nu$. Only the $i$th wall arrives, and
\[
 N_i'(T)=\lambda_i(T)\nu\geq c\nu>0.
\]
\end{proposition}

\begin{proof}
Lemma~\ref{lem:dominant-wall-exit} supplies $s$. Until the first
height-$\nu$ arrival, Lemma~\ref{lem:away-wall-growth-gap} applies.
Its growing index is constant on every connected time interval:
a change would force one continuous rate to pass through zero,
contrary to the gap. Hence one wall grows at least exponentially
with rate $c$, and the other two decrease strictly.

There must be an arrival in finite time. Indeed, if no wall reached
$\nu$, the same index would remain growing for all future time and
$N_i(t)\geq N_i(s)e^{c(t-s)}$ would contradict $N_i(t)<\nu$.
Continuity of the maximum of the walls gives a first arrival, and
the two decreasing walls remain strictly below their values at $s$.
The rate gap persists at the endpoint, proving transversality.
\end{proof}