\subsection{Radial phase through the slow decay}
\label{sec:small-section-slow-radius}

The coupling between the shear and the difference of the
weak walls cancels from the evolution of their Euclidean
radius. The remaining
terms carry the small projection height or the wall
amplitudes, and the same cancellation survives ordinary
differentiation in the initial shear. This gives a quadratic
value error and a linear derivative error over the entire
interval of slow decay.

Use the actual family $z(t,\rho)=(p,s,n,x,y)(t,\rho)$ of
Section~\ref{sec:small-section-slow-variation}, with
$\zeta,\nu>0$, and put
\begin{equation}
 d=x-y,\qquad R=\sqrt{s^2+d^2},\qquad
 A=1+p,\qquad E=n+x^2+y^2.
 \label{eq:slow-radius-quantities}
\end{equation}
All derivatives in $\rho$ keep the original parameters and
both times $t_0,T$ fixed. We first estimate the rate of $R$
and its ordinary differential, then integrate them using
the two time moments of $E$.

On the Einstein constraint, the equations for $s$ and $d$
take the form
\[
 \Dt s=Qs+\sqrt3(n-x-y)d,\qquad
 \Dt d=(Q-A)d+\sqrt3s(x+y).
\]
The terms containing $s(x+y)d$ cancel in
$s\Dt s+d\Dt d$. Since $s>0$ on the region
\eqref{eq:slow-decay-region}, we obtain the smooth identity
\begin{equation}
 \Dt R=\frac{\mathcal N}{R},\qquad
 \mathcal N=QR^2-Ad^2+\sqrt3\,nsd.
 \label{eq:slow-radius-rate}
\end{equation}
In particular, the small height $A$ appears before any
estimate is made on the weak-wall contribution.

\begin{lemma}
\label{lem:slow-radius-rate}
There is an absolute constant $C_R\geq1$ with the following
property. Let $0<a\leq1/256$ and let a state belong to
\eqref{eq:slow-decay-region}. Let $\delta$ be a tangent
direction to $G=1$ such that, for some $L\geq0$,
\[
 |\delta s|\leq5,\qquad
 |\delta n|\leq nL,\qquad |\delta x|\leq xL,
       \qquad |\delta y|\leq yL.
\]
Then
\begin{equation}
 \frac a2\leq R\leq4a,\qquad
 |\Dt R|\leq C_R aE,\qquad
 \left|\delta\!\left(\frac{\mathcal N}{R}\right)\right|
                  \leq C_R E(1+aL).
 \label{eq:slow-radius-rate-bounds}
\end{equation}
Here $\Dt R$ denotes the radius rate in
\eqref{eq:slow-radius-rate}.
\end{lemma}

\begin{proof}
The regional bounds give $a/2\leq s\leq3a$,
$x,y\leq a$, and $|d|\leq2a$. Thus $a/2\leq R\leq4a$.
Also
\begin{equation*}
 0\leq E\leq3a^2,\qquad n\leq E,\qquad
 b\leq3E,
\end{equation*}
where $b$ is defined in \eqref{eq:slow-variation-moments}.
Indeed, $n\leq a^2\leq1$ and $x+y\leq2$, so
$n^2+n(x+y)\leq3n$.
The polynomial estimates
\eqref{eq:slow-variation-polynomial-bounds} consequently give
\[
 |Q|\leq6E,\qquad |\delta Q|\leq12EL
                                      \leq36a^2L.
\]

Write $B=1+aL$. The ordinary differential of the radius is
\begin{equation*}
 \delta R=\frac{s\delta s+d\delta d}{R}.
\end{equation*}
Since $|\delta d|\leq(x+y)L\leq2aL$, its numerator has
absolute value at most $15a+4a^2L$. Division by $R\geq a/2$
therefore gives $|\delta R|\leq40B$.
The tangent constraint gives
\[
 (-p)\delta p=s\delta s+\frac12\delta Q.
\]
Using $-p\geq1/2$ yields
\begin{equation}
 |\delta R|\leq40B,\qquad
 |\delta p|\leq40aB.
 \label{eq:slow-radius-tangent-bounds}
\end{equation}

The numerator in \eqref{eq:slow-radius-rate} retains the
factor $E$. In fact,
\[
 d^2\leq2E,\qquad
 |d\delta d|\leq(x+y)^2L\leq2EL,
 \qquad 0\leq A\leq6a^2.
\]
Together with $R\leq4a$, $n\leq E$, and $|s|\leq3a$,
these inequalities bound $|\mathcal N|$ by an absolute
constant times $a^2E$.
Its differential is
\begin{equation}
 \begin{split}
 \delta\mathcal N={}&(\delta Q)R^2
       +2Q(s\delta s+d\delta d)
       -(\delta p)d^2-2Ad\delta d\\
     &+\sqrt3\bigl((\delta n)sd+n(\delta s)d
                                      +ns\delta d\bigr).
 \end{split}
 \label{eq:slow-radius-numerator-differential}
\end{equation}
Each term is bounded by an absolute constant times $aEB$.
For the second term, use
$|s\delta s+d\delta d|=|R\delta R|\leq160aB$.
For the terms containing wall differentials, retain their
wall factors and use $a^2EL\leq aEB$; the term containing
$\delta p$ is controlled by
\eqref{eq:slow-radius-tangent-bounds} and $d^2\leq2E$.
These estimates account for all terms in
\eqref{eq:slow-radius-numerator-differential}.

Finally,
\[
 \delta\!\left(\frac{\mathcal N}{R}\right)
       =\frac{\delta\mathcal N}{R}
           -\frac{\mathcal N\delta R}{R^2}.
\]
The lower bound $R\geq a/2$ and
\eqref{eq:slow-radius-tangent-bounds} prove both rate
estimates after choosing one absolute constant $C_R$.
\end{proof}

The energy $E$ differs from $b$ only in how it treats the
strong wall. Since $E\leq b+n$ and $\Dt n\leq-2n$,
the dissipative moment estimate used in
Lemma~\ref{lem:slow-variation-moments} gives
\begin{equation}
 \begin{split}
 I_0:=\int_0^T E(t)\dd t&\leq J_0+\frac{n(0)}2\leq5a,\\
 I_1:=\int_0^T tE(t)\dd t&\leq J_1+\frac{n(0)}4
                            \leq\frac1{256}+(3+D)a.
 \end{split}
 \label{eq:slow-radius-moments}
\end{equation}
Here we used \eqref{eq:slow-variation-linear-moments} and
$n(0)\leq a^2$. These estimates hold uniformly for the
actual family and are independent of how small either
positive initial wall is relative to the other.

\begin{proposition}
\label{prop:small-section-slow-radius-transfer}
There exists $h_*>0$ such that, for every $q>2$ and
$D\geq0$, one can choose $K_{q,D}>0$ with the following
property. For every $0<h\leq h_*$ and $P>6D+2$, and for
all sufficiently small $a>0$, uniformly over
\eqref{eq:slow-variation-label-domain}, the function
\[
 F(\rho)=R(T,\rho)-R(0,\rho)
\]
has an ordinary derivative at each $\rho\in[a,2a]$ and
\begin{equation}
 |F(\rho)|+a|F'(\rho)|\leq K_{q,D}a^2.
 \label{eq:slow-radius-transfer}
\end{equation}
The threshold for $a$ may depend on $h,q,D,P$.
\end{proposition}

\begin{proof}
Use the thresholds of
Theorem~\ref{thm:small-section-slow-variation}, and fix a
member of the actual family. With $\delta=\partial_\rho$,
that theorem gives
\[
 |\delta s|\leq5,\qquad
 L(t)\leq2L_0+20t,\qquad
 L_0\leq2+4Cq|\log a|.
\]
Its differential is tangent to $G=1$, since the initial
chart lies on the constraint and the flow preserves it.
Lemma~\ref{lem:slow-radius-rate} therefore yields
\begin{equation}
 |\partial_tR|\leq C_R aE,\qquad
 |\partial_\rho(\mathcal N/R)|
                \leq C_R E(1+2aL_0+20at).
 \label{eq:slow-radius-family-rates}
\end{equation}

These are derivatives of one ordinary smooth family.
Indeed, the initial chart and the normalized flow are
jointly $C^2$ near each $(\rho,t)$ under consideration,
and $s(t,\rho)>0$ keeps the square root defining $R$
smooth there. The constraint and
\eqref{eq:slow-radius-rate} hold on the same ordinary
neighborhood. Equality of mixed derivatives thus gives
\begin{equation}
 \partial_t\partial_\rho R
                  =\partial_\rho(\mathcal N/R).
 \label{eq:slow-radius-mixed-derivative}
\end{equation}
In particular, the time derivative on the left is that of
the actual initial response, including when $\rho$ is an
endpoint of $[a,2a]$.

Integrate the first inequality in
\eqref{eq:slow-radius-family-rates}. By
\eqref{eq:slow-radius-moments},
\[
 |F(\rho)|\leq C_R aI_0\leq5C_Ra^2.
\]
Using \eqref{eq:slow-radius-mixed-derivative} for the second
inequality gives
\begin{equation*}
 |F'(\rho)|\leq C_R\bigl((1+2aL_0)I_0+20aI_1\bigr).
\end{equation*}
For $0<a\leq1$, the inequality $a|\log a|\leq1$ implies
$aL_0\leq2+4Cq$. Also
$I_1\leq4+D$ by \eqref{eq:slow-radius-moments}.
Consequently,
\[
 (1+2aL_0)I_0+20aI_1
                   \leq(105+40Cq+20D)a.
\]
Thus \eqref{eq:slow-radius-transfer} holds, for example,
with $K_{q,D}=C_R(110+40Cq+20D)$.
The constants $C$ and $C_R$ are absolute; in particular
this choice precedes $h$ and $P$.
\end{proof}

Define the physical phase at the end of the slow interval by
\begin{equation}
 f(\rho)=\frac{R(T,\rho)}{\sqrt3},\qquad
 f_0(\rho)=\Phi_h(\rho,\zeta,\nu),
 \label{eq:slow-radius-phase}
\end{equation}
where $\Phi_h$ is the reference phase
\eqref{eq:small-section-reference-phase} on the original
section. At the beginning of the slow interval,
$R(0,\rho)/\sqrt3$ is exactly the physical phase after
the initial decay that was estimated in
Corollary~\ref{cor:small-section-shear-monotonicity}.

\begin{corollary}
\label{cor:small-section-slow-physical-phase}
There exists $h_*>0$ such that, for every $0<h\leq h_*$,
$q>2$, and $D\geq0$, there is $K_{h,q,D}>0$ with the
following property. For every $P>6D+2$ and all
sufficiently small $a>0$, uniformly over
\eqref{eq:slow-variation-label-domain}, the physical phase
$f$ has an ordinary derivative and satisfies
\begin{equation}
 \begin{split}
 |f-f_0|&\leq K_{h,q,D}a^2,\qquad
       |f'-f_0'|\leq K_{h,q,D}a,\\
 \frac1{4\sqrt3}&\leq f'\leq\frac2{\sqrt3}+1.
 \end{split}
 \label{eq:slow-radius-phase-comparison}
\end{equation}
The constant $K_{h,q,D}$ is chosen before $P$.
\end{corollary}

\begin{proof}
Let $g(\rho)=R(0,\rho)/\sqrt3$. Then the exact identity
\[
 f=\frac{F}{\sqrt3}+g
\]
joins the two portions of the same orbit.
Proposition~\ref{prop:small-section-slow-radius-transfer}
bounds the value and derivative of $F/\sqrt3$ by
$K_{q,D}a^2$ and $K_{q,D}a$, respectively.
Corollary~\ref{cor:small-section-shear-monotonicity} bounds
$g-f_0$ and its derivative by $K_ha^2$ and $K_ha$,
with $K_h$ chosen before $q$.
The first two assertions follow with
$K_{h,q,D}=K_{q,D}+K_h$.

Decrease $h_*$ to be at most $1/16$ and then decrease
the threshold for $a$ so that
$K_{h,q,D}a\leq1/(4\sqrt3)$.
The original reference slope satisfies
\[
 \frac1{2\sqrt3}\leq f_0'\leq\frac2{\sqrt3}
\]
by Proposition~\ref{prop:small-section-reference-slope}.
Combining this with the derivative error proves the
last two bounds in \eqref{eq:slow-radius-phase-comparison}.
\end{proof}

The radius has therefore retained the original phase and
its positive response to the initial shear throughout
the additional waiting time. The endpoint wall bounds
\eqref{eq:slow-decay-projective-endpoint} and this phase
control provide the physical data for the subsequent
projection and upward core entrance.