\subsection{Passage to the growing wall}
\label{sec:connection-passage}

We use the growing wall as a clock throughout the connection. A phase
function relating the radii of the adjacent cores on their single-wall
boundaries then makes the accumulated error proportional to the small
wall amplitudes, uniformly in the waiting time.

\subsubsection{The growing wall as a clock}

Keep the frame of the preceding core during the connection and set
$m=-M$. Consider the initial state
\begin{equation}
 (M,U,V,W)(0)=(-\mu,u,1,w),\qquad
 \mu\geq4,\quad 0\leq u\leq\frac{\varepsilon_0}{\mu},
 \quad 0<w\leq\varepsilon_0,
 \label{eq:connection-entry}
\end{equation}
where $\varepsilon_0=10^{-4}$ as before. The next entrance occurs
when $W$ first reaches one. After the reflection
$(M,U,V,W)\mapsto(-M,U,W,V)$, this becomes an upward crossing of
the new active wall. Equivariance under this fixed reflection makes
the result identical to changing the frame at the preceding core exit.

Write
\begin{equation}
 C=U(V+W-U),\qquad q=m-1-C,\qquad
 a=\frac{1-m^2-(V-W)^2+U^2}{2}.
 \label{eq:connection-rates}
\end{equation}
On an interval where $W>0$ and $q>0$, the variable
$\ell=\log(W/w)$ has derivatives
\begin{equation}
 \frac{d\ell}{d\tau}=q,\qquad
 \frac{d\ell}{dt}=Aq.
 \label{eq:connection-clock}
\end{equation}
The projective equations become
\begin{equation}
 \begin{aligned}
 m'&=\frac{V^2-W^2-U(V-W)-mC}{q},&
 U'&=\frac aq U,\\
 V'&=\frac{-m-1-C}{q}V,& W'&=W,
 \end{aligned}
 \label{eq:connection-clock-system}
\end{equation}
where primes now mean $d/d\ell$. In particular $W=we^\ell$, so the
target $W=1$ is reached at the prescribed clock value
$L=\log(1/w)$ whenever the passage exists.

We control every initial segment in the closed region
\begin{equation}
 |m-\mu|\leq\frac12,\qquad
 0\leq U\leq\varepsilon_0,\qquad 0\leq V,W\leq1.
 \label{eq:connection-region}
\end{equation}
The elementary bounds needed below are
\begin{gather}
 -U^2\leq C\leq2U,\qquad
 \frac\mu2\leq q\leq\mu,\qquad
 a\leq-\frac{\mu^2}{4},\qquad a+m+1+C<0,
 \label{eq:connection-elementary-rates}\\
 0<3+m^2-4U\leq D\leq2\mu^2,
 \qquad A\geq\frac3{\mu^2},\qquad Aq\geq\frac1\mu.
 \label{eq:connection-chart-bounds}
\end{gather}
For the first line, use $\mu-1/2\leq m\leq\mu+1/2$ and
$U\leq10^{-4}$. In particular $m\geq7\mu/8$ gives the stated
bound on $a$, and $m\geq7/2$ gives the negative ratio rate.
For the second line, expand $D$ in~\eqref{eq:projective-inverse}:
the term $(V-W)^2$ is nonnegative, while
$-2U(V+W)\geq-4U$. The upper bound follows from
$m^2\leq(\mu+1/2)^2$ and $|V-W|\leq1$.
The two clock bounds then follow from $A=6/D$ and $q\geq\mu/2$.

\begin{lemma}[Decay and finite integrals before arrival]
\label{lem:connection-prefix}
Suppose an initial segment from~\eqref{eq:connection-entry} remains
in~\eqref{eq:connection-region}, with $V,W>0$. At every clock value
$s\geq0$ in this segment,
\begin{align}
 U&\leq V,& C&\geq0,\notag\\
 U(s)&\leq u e^{-\mu s/2},&
 V(s)&\leq e^{-(1+2/\mu)s},\notag\\
 V(s)W(s)&\leq w e^{-2s/\mu},& W(s)&=we^s.
 \label{eq:connection-prefix-decay}
\end{align}
Moreover,
\begin{align}
 \int_0^s\frac{V^2}{q}\dd\ell&\leq\frac1{2\mu},&
 \int_0^s\frac{W^2}{q}\dd\ell&\leq\frac1\mu,\notag\\
 \int_0^s\frac Uq\dd\ell&\leq\frac{4u}{\mu^2},&
 \int_0^s\frac{(VW)^2}{q}\dd\ell&\leq\frac{w^2}{4},
 \label{eq:connection-integral-bounds}
\end{align}
and the phase band improves to
\begin{equation}
 |m(s)-\mu|\leq\frac{3/2+12u}{\mu}<\frac12.
 \label{eq:connection-band-improvement}
\end{equation}
\end{lemma}

\begin{proof}
The ratio equation is
\[
 \left(\frac UV\right)'
       =\frac{a+m+1+C}{q}\frac UV.
\]
Its coefficient is negative by~\eqref{eq:connection-elementary-rates}.
Since $U(0)/V(0)=u\leq1$, it follows that $U\leq V$, and hence
$C\geq0$.
With this ordering the rates satisfy
\begin{equation*}
 \frac aq\leq-\frac\mu2,\qquad
 \frac{-m-1-C}{q}=-1-\frac{2(1+C)}q
                      \leq-1-\frac2\mu.
\end{equation*}
To check the first inequality, multiply by $2q>0$. The required
numerator is
\[
 1-m^2-(V-W)^2+U^2+\mu(m-1-C)
 \leq1-\mu+U^2+\frac m2<0.
\]
Here $\mu-m\leq1/2$, $C\geq0$, and $m\leq\mu+1/2$ were used.
The scalar differential inequalities give the two wall bounds in
\eqref{eq:connection-prefix-decay}. The product equation is
\[
 (VW)'=-\frac{2(1+C)}q VW\leq-\frac2\mu VW,
\]
which gives the product bound as well.

For the integrals, retain the exact clock denominator in the
dissipation inequalities:
\begin{align*}
 (V^2)'&\leq-\frac{2\mu V^2}{q},&
 (1-W^2)'&\leq-\frac{\mu W^2}{q},\\
 U'&\leq-\frac{\mu^2U}{4q},&
 ((VW)^2)'&\leq-\frac{4(VW)^2}{q}.
\end{align*}
The first uses $-m-1-C\leq-\mu$, the second uses $2q\geq\mu$,
and the last uses $C\geq0$. Integrate each inequality and discard
its nonnegative terminal value. The initial values are $1$,
$1-w^2\leq1$, $u$ and $w^2$, respectively. This proves
\eqref{eq:connection-integral-bounds}.

Finally, the numerator of $m'$ has absolute value at most
$V^2+W^2+3\mu U$. Indeed, $|V-W|\leq1$, $0\leq C\leq2U$, and
$1+2m\leq3\mu$. Integrating and using the first three bounds in
\eqref{eq:connection-integral-bounds} gives
\eqref{eq:connection-band-improvement}. Its strict inequality follows
from $\mu\geq4$ and $u\leq10^{-4}$.
\end{proof}

The integrals in~\eqref{eq:connection-integral-bounds} are projective
time integrals, written using $d\tau=d\ell/q$. Their bounds have no
factor $L$. This is essential when $w$ is much smaller than the other
initial quantities.

\begin{proposition}[First arrival at the growing wall]
\label{prop:long-connection-arrival}
For every input~\eqref{eq:connection-entry}, the original flow has a
first positive time $T_c$ at which $W=1$. Up to that time it remains
in~\eqref{eq:connection-region}, with $A\geq3/\mu^2$.
The wall $V$ is strictly decreasing and $W$ strictly increasing;
there is exactly one time in $(0,T_c)$ at which $V=W$.
The exit is transverse, with $dW/dt>0$.

Put $L=\log(1/w)$, and let $\tau_c$ be the elapsed projective time.
Then
\begin{equation}
 \frac L\mu\leq\tau_c\leq\frac{2L}\mu,
 \qquad 0<T_c\leq\mu L.
 \label{eq:connection-duration}
\end{equation}
After reflection, the two transverse walls are
$u_+=U(T_c)$ and $w_+=V(T_c)$, and they satisfy
\begin{equation}
 0\leq u_+\leq u e^{-\mu L/2},\qquad
 0<w_+\leq w e^{-2L/\mu}.
 \label{eq:connection-output-decay}
\end{equation}
Here $u_+>0$ if and only if $u>0$. The time $T_c$ and the returned
state are locally $C^2$ functions of $(\mu,u,w)$, also at $u=0$.
\end{proposition}

\begin{proof}
The denominator is positive at the entrance by
\eqref{eq:connection-chart-bounds}. Lift the initial state to the
normalized constraint surface and use its forward flow.
At any unfinished segment in the proposed region, positivity of $V,W$
persists by their multiplicative equations. The strong wall cannot
cross its upper boundary because $a<0$ there; its zero boundary is
invariant. The old active wall points strictly into $V<1$, since
$-m-1-C<0$ throughout the region. Initially $W<1$, so continuity
gives a first short segment with the required bounds.

For continuation in the original coordinates, include the closed
condition $A\geq3/(2\mu^2)$. Within the region it improves to
$A\geq3/\mu^2$ by~\eqref{eq:connection-chart-bounds}. The phase
condition improves strictly by Lemma~\ref{lem:connection-prefix}.
Thus, as long as $W<1$, none of these boundaries can terminate the
passage. The original constrained flow exists at every finite forward
time by Lemma~\ref{lem:forward}; the positive lower bound on $A$
keeps the projective coordinates defined. Applying the same local
continuation argument at a finite endpoint extends the segment unless
$W=1$ there.

There must be such a finite endpoint. Otherwise
\eqref{eq:connection-chart-bounds} and
\eqref{eq:connection-clock} give
\[
 \frac t\mu\leq\log\frac{W(t)}w\leq L
\]
for arbitrarily large $t$, a contradiction. This also proves
$T_c\leq\mu L$. Since $q$ lies between $\mu/2$ and $\mu$,
integrating $d\tau=d\ell/q$ from $0$ to $L$ proves the other
duration bounds.

The rates of $V$ and $W$ have strict opposite signs. Hence the hit is
the first one, and $V-W$ decreases from $1-w>0$ to $V(T_c)-1<0$,
giving one and only one equality $V=W$. At the hit,
$dW/dt=Aq>0$. Substitution of $s=L$ in
\eqref{eq:connection-prefix-decay} gives
\eqref{eq:connection-output-decay}; the sign statements follow from
wall sign preservation.

To obtain local regularity, apply the implicit function theorem to
the actual equation $W(t;\mu,u,w)=1$ at its transverse hit.
The flow and the event are $C^2$ on a neighborhood of the whole
finite segment. The derivative $dW/dt$ is positive on this segment
and remains positive on a neighborhood of it, while the initial
value remains below one. The nearby implicit hit is therefore still
the first hit, with the intervening flow in the same chart. Composition
with the flow gives the $C^2$ returned state. This argument also
provides a local extension through $u=0$.
\end{proof}

\subsubsection{Phase change across the connection}

The band estimate~\eqref{eq:connection-band-improvement} bounds $m$
throughout the passage, but it does not yet give an error that vanishes
with both incoming walls. The appropriate quantity combines the
radii of the two adjacent cores. Set
\begin{equation}
 H(a,b)=\sqrt{a^2+b^2},\qquad
 J(m,V,W)=H(m+1,V)+H(m-1,W)-m.
 \label{eq:connection-matching-phase}
\end{equation}
For $m>1$,
\begin{equation}
 J(m,V,0)=H(m+1,V)-1,\qquad
 J(m,0,W)=H(m-1,W)+1.
 \label{eq:connection-phase-boundaries}
\end{equation}
The first expression is the preceding core radius minus one; the
second is the reflected core radius plus one. Thus a small change of
$J$ gives a decrement close to two between the physical core radii.

Write $\rho_+=H(m+1,V)$, $\rho_-=H(m-1,W)$, and define
\begin{align}
 S&=\frac{m+1}{\rho_+}+\frac{m-1}{\rho_-}-1,\notag\\
 B&=mS+\frac{V^2}{\rho_+}+\frac{W^2}{\rho_-},\qquad
 K_\pm=\frac1{\rho_\pm(\rho_\pm+m\pm1)}.
 \label{eq:connection-phase-coefficients}
\end{align}
Direct differentiation in projective time gives the exact identity
\begin{equation}
 \dot J=(VW)^2(K_+-K_-)-US(V-W)-CB.
 \label{eq:connection-phase-cancellation}
\end{equation}
To verify its cancellation, use
\[
 H(a,b)-a=\frac{b^2}{H(a,b)+a},\qquad
 1-\frac a{H(a,b)}=\frac{b^2}{H(a,b)(H(a,b)+a)}
 \quad(a>0).
\]
Before this substitution, the terms in $\dot J$ independent of $U$
are
\[
 V^2\left(\frac{m-1}{\rho_-}-1\right)
       +W^2\left(1-\frac{m+1}{\rho_+}\right).
\]
The two identities turn their sum into $(VW)^2(K_+-K_-)$.
The remaining terms are precisely $-US(V-W)-CB$.

In~\eqref{eq:connection-region}, these coefficients satisfy
\begin{equation}
 \frac34\leq S\leq1,\qquad 0\leq B\leq2\mu,
 \qquad 0\leq K_\pm\leq\frac2{\mu^2},
 \quad |K_+-K_-|\leq\frac4{\mu^2}.
 \label{eq:connection-phase-coefficient-bounds}
\end{equation}
Indeed, if $a\geq2$ and $0\leq b\leq1$, then
$7/8\leq a/H(a,b)\leq1$, which gives the bounds on $S$.
Also $V^2/\rho_+\leq1$ and $W^2/\rho_-\leq1$, so
$B\leq m+2\leq2\mu$. For the kernels use
$m\pm1\geq\mu/2$ and $H(a,b)\geq a$; their denominators are at
least $\mu^2/2$.

Since $|C|\leq2U$ and $|V-W|\leq1$, the derivative identity yields
\begin{equation}
 |\dot J|\leq5\mu U+\frac4{\mu^2}(VW)^2.
 \label{eq:connection-matching-rate-bound}
\end{equation}
Consequently the last two integrals in
\eqref{eq:connection-integral-bounds} give, for every completed
initial segment,
\begin{equation}
 |J(m(s),V(s),W(s))-J(\mu,1,w)|
       \leq\frac{20u}{\mu}+\frac{w^2}{\mu^2}.
 \label{eq:connection-matching-drift}
\end{equation}
The interaction of the two walls is controlled by the integral of
$(VW)^2$, whose initial value is $w^2$. Its contribution stays small
even when the clock interval becomes long.

\begin{proposition}[The radius decrement]
\label{prop:connection-radius-decrement}
For the passage in Proposition~\ref{prop:long-connection-arrival},
put $m_*=m(T_c)$ and
\[
 r_-=H(\mu+1,1),\qquad r_+=H(m_*-1,1).
\]
These are the preceding core exit radius and the next core entrance
radius after reflection. They satisfy
\begin{equation}
 |r_+-(r_--2)|\leq\frac{24(u+w^2)}\mu.
 \label{eq:connection-radius-decrement}
\end{equation}
\end{proposition}

\begin{proof}
Let $v_*=V(T_c)$. By~\eqref{eq:connection-output-decay},
$0<v_*\leq w$. The two endpoint defects
\[
 e_-=H(\mu-1,w)-(\mu-1),\qquad
 e_+=H(m_*+1,v_*)-(m_*+1)
\]
belong to $[0,w^2/\mu]$. This follows from the identity for
$H(a,b)-a$, since $\mu-1$ and $m_*+1$ are at least $\mu/2$.
Expanding~\eqref{eq:connection-matching-phase} at the two ends gives
\[
 r_+-r_-+2
       =J(m_*,v_*,1)-J(\mu,1,w)+e_--e_+.
\]
Apply~\eqref{eq:connection-matching-drift} and the defect bounds.
The absolute value is at most
$20u/\mu+w^2/\mu^2+2w^2/\mu$, which is bounded by the right-hand
side of~\eqref{eq:connection-radius-decrement}.
\end{proof}

To express the result directly in core coordinates, take a downward
exit with radius $R\geq11/2$ and walls $u,w$. Its state is
\[
 (M,U,V,W)=(1-\sqrt{R^2-1},u,1,w),\qquad
 \mu=\sqrt{R^2-1}-1.
\]
Then $\mu\geq4$, $R/2\leq\mu<R$, and $r_-=R$.
If $0\leq u\leq\varepsilon_0/R$ and
$0<w\leq\varepsilon_0/R$, all the entrance hypotheses above hold.
Define the reflection map by
\begin{equation}
 \mathcal F(R,u,w)=(r_+,u_+,w_+).
 \label{eq:long-reflection-map}
\end{equation}
It is locally $C^2$, and its core entrance embedding is exactly
$(m_*,U(T_c),1,V(T_c))$. Here $m_*>1$, so the positive entrance
branch of the radius coordinates is the correct one. The estimates give
\begin{equation}
 |\mathcal F_R(R,u,w)-(R-2)|\leq48\frac{u+w}{R},\qquad
 0\leq u_+\leq u,\qquad 0<w_+\leq w.
 \label{eq:long-reflection-values}
\end{equation}
The value error is now measured by the same wall amplitudes as in the
core estimate. To control repeated reflections, we also need its
derivative along the transverse cone. The next estimates keep the
cancellation in~\eqref{eq:connection-phase-cancellation} when the
entrance and the logarithmic clock interval vary.