\subsection{Passage across the cyclic change of axes}
\label{sec:terminal-passage}

Write the terminal exit radius as
\[
 R_0=2+2\alpha,\qquad 0<\alpha\leq1.
\]
The new projective denominator is of order $\alpha^2$. To enter the
next core with small transverse walls, we follow the departing wall
down to a height
\begin{equation}
 h=\eta\alpha^J,\qquad
 L=\log(1/h)=\log(1/\eta)+J\log(1/\alpha).
 \label{eq:terminal-height}
\end{equation}
Here $\eta$ and the integer $J$ will be chosen once. The outgoing
segment can become long as $\alpha$ tends to zero, but its wall
rates have size $\alpha$ apart from integrable terms. This permits
uniform control before changing frames.

\subsubsection{Descent to the auxiliary section}

Start at the downward core exit
\begin{equation}
 y_-(R_0,u,w)=(1-\sqrt{R_0^2-1},u,1,w).
 \label{eq:terminal-entry}
\end{equation}
We allow signed $u,w$ in this first passage. This will give ordinary
smooth dependence at points where a wall vanishes. Denote the first
arrival at $V=h$ by $S_h$, and the state there by $y_h$.

\begin{lemma}[Uniform outgoing passage]
\label{lem:terminal-outgoing}
Fix $0<\eta<1$ and an integer $J\geq1$. There are constants
$B\geq2$ and $\varepsilon>0$ such that, if
\begin{equation}
 0<\alpha\leq1,\qquad 0<b\leq\varepsilon\alpha^J,\qquad
 |u|+|w|\leq b,
 \label{eq:terminal-outgoing-domain}
\end{equation}
the actual orbit from~\eqref{eq:terminal-entry} reaches $V=h$
for the first time at $S_h>0$. Throughout this segment,
\begin{gather}
 A\geq\frac18,\qquad M<0,\qquad
 \frac74<\rho<\frac{17}{4},\qquad h\leq V\leq1,
 \quad \rho=H(M-1,V),
 \label{eq:terminal-outgoing-region}\\
 |U|+|W|\leq\frac{Bb}{2},\qquad
 |\rho-R_0|\leq40Bb,\qquad
 \rho>2+\alpha,\qquad |U|\leq\frac V2.
 \label{eq:terminal-outgoing-bounds}
\end{gather}
There are positive wall multipliers $B_U(t),B_W(t)$ with
\begin{equation}
 U(t)=uB_U(t),\qquad W(t)=wB_W(t),\qquad
 \frac2B\leq B_U(t),B_W(t)\leq\frac B2.
 \label{eq:terminal-outgoing-multipliers}
\end{equation}
The departing wall is strictly decreasing and
\begin{equation}
 0<S_h\leq16L.
 \label{eq:terminal-outgoing-time}
\end{equation}
For each such passage, holding its numerical height $h$ fixed,
the arrival time and state are $C^2$ on a neighborhood of
$(R_0,u,w)$.
\end{lemma}

\begin{proof}
Use the decreasing wall as a clock, $\sigma=-\log V$. On the
downward branch set
\[
 g=\sqrt{\rho^2-V^2}=1-M,\qquad q_-=g+C,\qquad C=U(V+W-U).
\]
Then $d\sigma/d\tau=q_-$ and $V=e^{-\sigma}$. The energy identity
and the two transverse equations give
\begin{align}
 \frac{d\rho}{d\sigma}
   &=\frac{-C\rho-g(U-W)^2/\rho}{q_-},\notag\\
 \frac{dU}{d\sigma}&=a_U U,\qquad
 a_U=\frac{2g-\rho^2+2VW-W^2+U^2}{2q_-},\notag\\
 \frac{dW}{d\sigma}&=a_W W,\qquad
 a_W=\frac{g-2-C}{q_-}.
 \label{eq:terminal-outgoing-field}
\end{align}
On the region
\begin{equation}
 \frac74\leq\rho\leq\frac{17}{4},\qquad
 |U|,|W|\leq\frac1{100},\qquad 0\leq V\leq1,
 \label{eq:terminal-clock-region}
\end{equation}
one has $1<g\leq\rho$ and $q_-\geq1/2$. The latter follows from
$|C|\leq |U|V+(|U|+|W|)^2$. At zero transverse walls the rates
are
\begin{align}
 a_U^0&=-\frac{\rho-2}{2}
             -\frac{\rho V^2}{2g(\rho+g)},\notag\\
 a_W^0&=\frac{\rho-2}{\rho}
             -\frac{2V^2}{\rho g(\rho+g)}.
 \label{eq:terminal-boundary-rates}
\end{align}
These identities use $\rho-g=V^2/(\rho+g)$. In particular,
\begin{equation}
 |a_U^0|+|a_W^0|
       \leq10\bigl(\alpha+|\rho-R_0|+V^2\bigr).
 \label{eq:terminal-boundary-rate-bound}
\end{equation}
For example, the two fractions involving $V^2$ in
\eqref{eq:terminal-boundary-rates} are at most $5V^2$ and
$2V^2$, respectively, and
$|\rho-2|\leq|\rho-R_0|+2\alpha$.

Put $\beta=|U|+|W|$. The differences from the boundary rates are
\[
 a_U-a_U^0=\frac{2VW-W^2+U^2-2Ca_U^0}{2q_-},
 \qquad
 a_W-a_W^0=-\frac{C(1+a_W^0)}{q_-}.
\]
On~\eqref{eq:terminal-clock-region},
$|a_U^0|\leq20$, $|a_W^0|\leq7$, and hence
\begin{align}
 \left|\frac{d\rho}{d\sigma}\right|
       &\leq20\bigl(|U|V+\beta^2\bigr),\notag\\
 |a_j-a_j^0|&\leq100(\beta V+\beta^2),
                       \qquad j=U,W.
 \label{eq:terminal-perturbation-rates}
\end{align}
For the first inequality, use $g/\rho\leq1$,
$(U-W)^2\leq\beta^2$, $\rho\leq5$ and $q_-\geq1/2$.
The other two follow directly from the displayed difference formulas.

We now close these estimates uniformly over the full clock interval.
The elementary inequality $x e^{-x}\leq e^{-1}$ for $x\geq0$
gives
\begin{align}
 \alpha L&\leq K:=\log(1/\eta)+J e^{-1},\notag\\
 \alpha^J(1+L)&\leq Q:=1+\log(1/\eta)+e^{-1}.
 \label{eq:terminal-clock-budgets}
\end{align}
Set $B=2\exp(10K+6)$. On any existing prefix $0\leq\sigma\leq T\leq L$
assume, in addition to~\eqref{eq:terminal-clock-region},
\begin{equation}
 \beta\leq Bb,\qquad |\rho-R_0|\leq80Bb.
 \label{eq:terminal-bootstrap-region}
\end{equation}
Integrating~\eqref{eq:terminal-boundary-rate-bound} and
\eqref{eq:terminal-perturbation-rates}, using
$\int_0^T e^{-\sigma}\dd\sigma\leq1$ and
$\int_0^T e^{-2\sigma}\dd\sigma\leq1/2$, yields
\begin{align*}
 \left|\int_0^T a_j\dd\sigma\right|
   &\leq10\alpha T+800BbT+100(Bb)^2T+100Bb+5,\\
 |\rho(T)-R_0|&\leq20\bigl[Bb+(Bb)^2T\bigr].
\end{align*}
Choose a constant $\delta>0$, depending on $B$, so small that
$b(1+T)\leq\delta$ makes the first right-hand side at most
$10K+6$ and the second at most $40Bb$. For instance,
\[
 \delta\leq\frac1{900B+100B^2+1}
\]
suffices. Define
$\Lambda_j(T)=\exp(\int_0^T a_j\dd\sigma)$.
The scalar equations give $U(T)=u\Lambda_U(T)$ and
$W(T)=w\Lambda_W(T)$ in this clock, including for signed or zero
inputs. Thus~\eqref{eq:terminal-bootstrap-region} improves to
\[
 \frac2B\leq\Lambda_j(T)\leq\frac B2,\qquad
 \beta(T)\leq\frac{Bb}{2},\qquad
 |\rho(T)-R_0|\leq40Bb.
\]
Take $\varepsilon Q\leq\delta$ and
$B\varepsilon\leq\min(1/1000,\eta)$. Equations
\eqref{eq:terminal-outgoing-domain} and
\eqref{eq:terminal-clock-budgets} ensure all the preceding smallness
conditions. They also give
\begin{equation*}
 Bb\leq\frac{\alpha}{1000},\qquad Bb\leq h.
\end{equation*}
The improved bounds put $\rho$ strictly between $7/4$ and $17/4$
and both walls strictly below $1/100$. Continuity therefore extends
the improved bounds over every existing prefix. Their last two
consequences in~\eqref{eq:terminal-outgoing-bounds} follow from
$\rho\geq2+2\alpha-40Bb$ and $V\geq h\geq Bb$.

To prove arrival at the lower section, return to the normalized flow.
It exists for all positive time also for signed initial walls on $G=1$:
the proof of Lemma~\ref{lem:forward} bounds $|N_i(t)|$ by
$|N_i(0)|e^t$ and then bounds the shear using absolute values of
the curvature products. Continue the orbit in the closed region
given by $M\leq0$, \eqref{eq:terminal-clock-region},
$h\leq V\leq1$ and $A\geq1/16$. In this region,
\begin{equation*}
 0<D\leq48,\qquad A=\frac6D\geq\frac18.
\end{equation*}
Indeed, $-5\leq M<0$ follows from the radius and active-wall bounds.
In~\eqref{eq:projective-inverse} the cross term has absolute
value at most one, while $U^2\leq1$ and $(V-W)^2\leq4$.
These bounds give both positivity of $D$ and its stated upper bound.

On every prefix of this region, the clock is strictly increasing and
its inverse converts the actual orbit into
\eqref{eq:terminal-outgoing-field}. Thus all the improved estimates
already proved apply. Before $V=h$, each boundary condition has a
strict margin, except $V=1$, where $dV/dt=-Aq_-V<0$ points inward.
The region can therefore be continued until the lower section is
reached. Moreover,
\[
 \frac{d\sigma}{dt}=Aq_-\geq\frac1{16}.
\]
As long as $V>h$ one has $\sigma<L$, so the segment cannot last
longer than $16L$. This proves existence, strict decrease, first
arrival and~\eqref{eq:terminal-outgoing-time}.
The multipliers in original time are
$B_j(t)=\Lambda_j(-\log V(t))$, giving
\eqref{eq:terminal-outgoing-multipliers}.

Finally fix $h$ at its value for the given input. On this finite
segment the chart is positive and $dV/dt<0$. These properties
persist on a neighborhood of the segment and its moving endpoint.
The implicit function theorem gives a $C^2$ arrival time, and
strict decrease identifies it with the first arrival. Composing
with the flow gives the $C^2$ arrival state, including for zero or
signed transverse walls.
\end{proof}

\subsubsection{The cyclic change of frame}

The normalized permutation $\mathsf C$ in~\eqref{eq:axis-actions}
has the following projective expression:
\begin{align}
 \mathcal H&=(M+1)^2+U^2+V^2+W^2-2UV-2UW-2VW,\notag\\
 \mathsf C_p(M,U,V,W)
   &=\left(1-\frac{4(M+1)}{\mathcal H},
             \frac{4V}{\mathcal H},
             \frac{4W}{\mathcal H},
             \frac{4U}{\mathcal H}\right).
 \label{eq:terminal-projective-cycle}
\end{align}
Substitution in~\eqref{eq:projective-coordinates} gives
\begin{equation}
 A_{\rm new}=\frac{A\mathcal H}{4}.
 \label{eq:terminal-cycle-clock}
\end{equation}
Thus $\mathcal H>0$ ensures that both frames describe the same
orbit in valid charts. The permutation commutes with the flow in
original time; the two projective clocks have the ratio displayed
in~\eqref{eq:terminal-cycle-clock}.

The small denominator can be estimated directly along the whole
outgoing segment. For $M=1-g$ and $g=\sqrt{\rho^2-V^2}$,
\begin{equation}
 \mathcal H=(\rho-2)^2+\frac{4V^2}{\rho+g}
                         +(U-W)^2-2V(U+W).
 \label{eq:terminal-cycle-split}
\end{equation}
This identity separates the radial scale from the wall error.

\begin{lemma}[Cyclic entrance estimates]
\label{lem:terminal-cycle-entrance}
Fix $0<\eta\leq1/100$ and an integer $J\geq2$. There are
$C_*>0$ and $\varepsilon>0$ such that, if
\begin{equation}
 0<\alpha\leq1,\qquad
 0<b\leq\varepsilon\alpha^{2J},\qquad |u|+|w|\leq b,
 \label{eq:terminal-cycle-domain}
\end{equation}
the outgoing segment of Lemma~\ref{lem:terminal-outgoing} satisfies
\begin{equation}
 \mathcal H>\frac{39}{10}\alpha^2+\frac25V^2,\qquad
 |V_{\rm new}|\leq\frac{2C_*b}{\alpha^2}\leq\frac1{5000}.
 \label{eq:terminal-cycle-prefix}
\end{equation}
At its endpoint write
$z=\mathsf C_p y_h=(m_e,u_e,v_e,w_e)$ and
$k=H(m_e-1,v_e)$. Then
\begin{gather}
 \frac{399}{100}\alpha^2<\mathcal H(y_h)
                         <\frac{401}{100}\alpha^2,
 \label{eq:terminal-cycle-end-denom}\\
 \frac{19}{10\alpha}<m_e-1<\frac{21}{10\alpha},
 \qquad \frac{19}{10\alpha}<k<\frac{11}{5\alpha},
 \label{eq:terminal-cycle-end-radius}\\
 \frac{\eta}{2}\alpha^{J-2}\leq u_e\leq2\eta\alpha^{J-2},
 \qquad |v_e|,|w_e|\leq\frac{2C_*b}{\alpha^2}.
 \label{eq:terminal-cycle-end-walls}
\end{gather}
For positive input walls, $v_e$ and $w_e$ are positive.
The cyclic chart remains positive throughout the outgoing segment.
\end{lemma}

\begin{proof}
Use Lemma~\ref{lem:terminal-outgoing} and take $C_*=40B$.
After decreasing $\varepsilon$, its estimates apply and, with
$E=C_*b$, give
\[
 |\rho-R_0|\leq E,\qquad |U|+|W|\leq E,\qquad
 E\leq\frac{\alpha^2}{10000}\leq\frac{\alpha}{10000}.
\]
Here $\alpha^{2J}\leq\alpha^J$ ensures the previous smallness
condition, and $\alpha^{2J}\leq\alpha^2$ gives the new one.
In~\eqref{eq:terminal-cycle-split},
$\rho+g\leq9$ and $\rho-2\geq(199/100)\alpha$.
The inequality $2VE\leq V^2/100+100E^2$ therefore yields
\[
 \mathcal H\geq
  \left(\frac{199}{100}\right)^2\alpha^2
       -100\left(\frac{\alpha}{10000}\right)^2
       +\left(\frac49-\frac1{100}\right)V^2
  >\frac{39}{10}\alpha^2+\frac25V^2.
\]
The wall formula in~\eqref{eq:terminal-projective-cycle} implies
$|4W/\mathcal H|\leq2E/\alpha^2\leq1/5000$.
Together with~\eqref{eq:terminal-cycle-clock}, this proves
the assertions along the segment.

At $V=h$ we have $V\leq\alpha/100$. Now
$\rho+g\geq3$ and
\[
 \frac{19999}{10000}\alpha\leq\rho-2
                        \leq\frac{20001}{10000}\alpha.
\]
The part of $\mathcal H$ involving the two transverse walls
has absolute value at most
\[
 E^2+2VE\leq
       \left(\frac{\alpha}{10000}\right)^2
          +2\frac{\alpha}{100}\frac{\alpha}{10000}.
\]
The remaining correction $4V^2/(\rho+g)$ lies between zero
and $4\alpha^2/(3\cdot10000)$. Substitution in
\eqref{eq:terminal-cycle-split} proves
\eqref{eq:terminal-cycle-end-denom}.

At the same endpoint,
\begin{equation}
 \frac{199}{100}\alpha<g-2<\frac{201}{100}\alpha.
 \label{eq:terminal-cycle-gap}
\end{equation}
The upper bound follows from $g\leq\rho$. For the lower bound,
the preceding lower bound for $\rho$ and $V\leq\alpha/100$
give
$\rho^2-V^2>(2+(199/100)\alpha)^2$ by expansion.
Since $g\geq0$, taking square roots proves it.
As $m_e-1=4(g-2)/\mathcal H$, equations
\eqref{eq:terminal-cycle-end-denom} and
\eqref{eq:terminal-cycle-gap} give the first pair of bounds in
\eqref{eq:terminal-cycle-end-radius}. The second follows from
\[
 m_e-1\leq H(m_e-1,v_e)\leq m_e-1+|v_e|,
 \qquad |v_e|\leq1/5000,\quad \alpha\leq1.
\]
Finally $u_e=4h/\mathcal H$ gives its two bounds, and
$v_e=4W_h/\mathcal H$, $w_e=4U_h/\mathcal H$ give the others.
Their positivity follows from
\eqref{eq:terminal-outgoing-multipliers}.
\end{proof}

The purpose of the exponent $J$ is now explicit. Suppose the
incoming estimates require its two decreasing walls to be small
even after multiplication by $k^q$, where $q\in\N$. Choose
\begin{equation}
 J\geq q+2,\qquad 0<\varepsilon_{\rm in}\leq10^{-4},\qquad
 0<\eta\leq\min\left(\frac1{100},
                    \frac{\varepsilon_{\rm in}}{4(11/5)^q}\right).
 \label{eq:terminal-entrance-choices}
\end{equation}
The next result records both this smallness and the actual event
that it makes possible.

\subsubsection{Arrival at the next core}

\begin{proposition}[The actual terminal passage]
\label{prop:terminal-first-arrival}
For the choices in~\eqref{eq:terminal-entrance-choices}, there is
$\varepsilon>0$ such that every positive pair $u,w$ satisfying
\eqref{eq:terminal-cycle-domain} has the following properties.
The cyclic endpoint $z=(m_e,u_e,v_e,w_e)$ is an incoming point with
\begin{gather}
 m_e>2,\qquad k>\frac74,\qquad 0<v_e<\frac12,\notag\\
 0<u_e,w_e\leq\varepsilon_{\rm in},\qquad
 u_e k^q,\ w_e k^q\leq\frac{\varepsilon_{\rm in}}2.
 \label{eq:terminal-entrance-smallness}
\end{gather}
For the exact flow starting at the cyclic permutation of the
normalized state associated with~\eqref{eq:terminal-entry}, the time
and state
\begin{equation}
 T_h=S_h+T_{\rm in}(z),\qquad y_+=\mathcal P(z)
 \label{eq:terminal-composition}
\end{equation}
give its first positive arrival at the new $V=1$. The new chart
is positive up to that time, the new $V$ is less than one before
arrival, and the final crossing is upward with $M_+>1$.
For each input, the two expressions in
\eqref{eq:terminal-composition}, with $h$ held fixed, are locally
$C^2$ functions of $(R,u,w)$ at $(R_0,u,w)$.

If $\eta$ is replaced by another value satisfying
\eqref{eq:terminal-entrance-choices}, then on a common input domain
the resulting total time and final state are unchanged.
\end{proposition}

\begin{proof}
Apply Lemma~\ref{lem:terminal-cycle-entrance} and put
\[
 d=\frac{\varepsilon_{\rm in}}{4(11/5)^q}.
\]
Decrease the input threshold so that $C_*\varepsilon\leq d$.
The endpoint bounds give
\begin{align*}
 u_e k^q&\leq2\eta(11/5)^q\alpha^{J-q-2}
                                  \leq\frac{\varepsilon_{\rm in}}2,\\
 w_e k^q&\leq2d(11/5)^q\alpha^{2J-q-2}
                                  \leq\frac{\varepsilon_{\rm in}}2.
\end{align*}
Both exponents are nonnegative. Since $k>1$, these inequalities
also imply the unweighted wall bounds. The remaining entrance
conditions follow from
\eqref{eq:terminal-cycle-prefix} and
\eqref{eq:terminal-cycle-end-radius}.
Thus Proposition~\ref{prop:incoming-passage} applies to $z$.

The original-time flow commutes with $\mathsf C$.
Consequently its cyclic state at time $S_h$ is exactly $z$, and
flow uniqueness identifies the continuation from $z$ with the
same normalized orbit. It reaches $y_+$ at time $T_h$.
During $[0,S_h]$, the new chart is positive and the new active wall
is at most $1/5000$ by~\eqref{eq:terminal-cycle-prefix}.
During the incoming segment, that wall is strictly increasing
and stays below one until its first arrival. This proves that
$T_h$ is the first positive event for the whole orbit. Its
transversality and $M_+>1$ follow from the incoming passage.

For fixed $h$, Lemma~\ref{lem:terminal-outgoing} gives $C^2$
dependence of $S_h$ and $y_h$. The rational map $\mathsf C_p$ is
smooth wherever $\mathcal H\ne0$, and
$\mathcal H(y_h)>0$ here. The incoming arrival time and state
are $C^2$ on an ordinary neighborhood of $z$. Composition proves
the stated local regularity.

Finally, for two admissible values $\eta_1,\eta_2$, choose the
smaller of their input thresholds. Both constructions give the
first positive arrival of the same cyclic orbit at the same
section. Uniqueness of this first time gives equality of the
total times, and flow uniqueness gives equality of the final states.
\end{proof}

The terminal passage therefore admits a decomposition adapted to
any admissible auxiliary height. The radius bounds show how an
exit with $R_0-2=2\alpha$ feeds an incoming radius of order
$\alpha^{-1}$. To determine the era map quantitatively, we next
estimate the phase and derivatives accumulated in this passage,
retaining the wall factors through the cyclic denominator.