\subsection{Phase along the stable boundary}
\label{sec:stable-boundary-phase}

During the incoming passage, the two decaying walls move the shear
coordinate even before the growing wall becomes appreciable. The
endpoint of this boundary motion provides a useful phase. Its level
sets collect boundary orbits approaching the same Kasner state.
We construct this phase directly from the flow, prove its smoothness,
and use it to match the two parts of a bounded-radius connection.
It is an auxiliary coordinate for the connection estimates. Its
comparison with the radius at the next entrance will recover the
observable era phase.

\subsubsection{The limiting Kasner coordinate}

On the invariant boundary $V=0$, write $y=(m,u,w)$. The vector field
in projective time is
\begin{equation}
 f(y)=\left(
 mu^2-(m+1)uw+w^2,\quad
 \frac{1-m^2-w^2+u^2}{2}u,\quad
 -(m+1+uw-u^2)w\right).
 \label{eq:stable-boundary-field}
\end{equation}
Its expression in the original normalized time is
\begin{equation}
 g(y)=\frac6{D(y)}f(y),\qquad
 D(y)=3+m^2+(u-w)^2.
 \label{eq:stable-original-field}
\end{equation}
In particular, the projective denominator is positive throughout this
boundary. Signed walls will be allowed in the construction, so that
the resulting differentiability statements hold in open sets.

\begin{proposition}[The stable phase]
\label{prop:stable-phase-limit}
Fix $c\geq2$, put $K=c+1$, and suppose that
\begin{equation}
 |m-c|\leq\frac1{40},\qquad |u|,|w|\leq\frac1{200}.
 \label{eq:stable-initial-box}
\end{equation}
The solution $(M(t),U(t),W(t))$ of~\eqref{eq:stable-original-field}
exists for all $t\geq0$ and remains in
\begin{equation}
 |M-c|\leq\frac1{10},\qquad |U|,|W|\leq\frac1{100}.
 \label{eq:stable-band}
\end{equation}
There is a limit
\[
 P(m,u,w)=\lim_{t\to\infty}M(t),
 \qquad U(t),W(t)\longrightarrow0.
\]
For every $t\geq0$,
\begin{equation}
 |P(m,u,w)-M(t)|
 \leq\frac8K\bigl(|U(t)|+|W(t)|\bigr)^2.
 \label{eq:stable-phase-tail}
\end{equation}
The value of $P$ is unchanged by restarting this boundary orbit.
In particular, $P(m,0,0)=m$.
\end{proposition}

\begin{proof}
Let $a=(1-M^2-W^2+U^2)/2$ and $b=-(M+1+UW-U^2)$ be the
two wall rates in projective time. Throughout~\eqref{eq:stable-band},
\begin{equation*}
 \frac{5K}{8}\leq M\leq K,\qquad
 \frac aK\leq-\frac K8,\qquad \frac bK\leq-\frac14,
 \qquad D\leq2K^2.
\end{equation*}
These follow by substitution, using $K\geq3$ and the two wall bounds.
Set
\[
 \xi(t)=K\int_0^t\frac6{D(M(r),U(r),W(r))}\dd r.
\]
The multiplicative wall equations give, on any initial interval in
the band,
\begin{equation}
 |U(t)|\leq |u|e^{-K\xi(t)/8},\qquad
 |W(t)|\leq |w|e^{-\xi(t)/4}.
 \label{eq:stable-boundary-decay}
\end{equation}
The first equation of~\eqref{eq:stable-boundary-field} gives
\[
 \left|\frac{dM}{d\xi}\right|
 \leq U^2+2|U||W|+\frac{W^2}K
 \leq (u^2+2|u||w|)e^{-K\xi/8}
       +\frac{w^2}K e^{-\xi/2}.
\]
Integrating proves
\begin{equation*}
 |M(t)-m|\leq\frac8K(|u|+|w|)^2<\frac1{40}.
\end{equation*}
Thus $|M-c|<1/20$ and the wall bounds improve to $1/200$.
Also $A=6/D\geq3/K^2$. To apply continuation in the original
normalized variables, include $A\geq3/(2K^2)$ in the closed
conditions. All conditions improve strictly. The proof of
Lemma~\ref{lem:forward} also covers signed curvature variables:
$|N_i(t)|\leq |N_i(0)|e^t$, and the constraint bounds the shear
using the absolute values of the products $N_iN_j$. Hence the
normalized solution continues for all positive time; the lower
bound for $A$ keeps its projective coordinates regular. The band
therefore holds globally.

We now have $\xi(t)\geq3t/K$, so both walls tend to zero.
The same integral estimate, begun at any time $t_0$, yields
\[
 |M(t)-M(t_0)|
 \leq\frac8K\bigl(|U(t_0)|+|W(t_0)|\bigr)^2,
 \qquad t\geq t_0.
\]
It makes $M(t)$ Cauchy at infinity and proves
\eqref{eq:stable-phase-tail}. Flow uniqueness identifies a
restarted orbit with the time translate of the original one.
Their first coordinates have the same limit. The zero-wall orbit
is constant, giving the last assertion.
\end{proof}

\subsubsection{Smooth dependence of the limiting phase}

The limit must be differentiated with respect to its initial state.
For this purpose we keep $c$, and therefore $K$, fixed and use
\begin{equation}
 Y(s;y)=(M(Ks;y),U(Ks;y),W(Ks;y)),\qquad
 \frac{dY}{ds}=K g(Y).
 \label{eq:stable-fixed-rescaling}
\end{equation}
This is a fixed rescaling of the original time. In particular, varying
$y$ leaves the elapsed original time $Ks$ unchanged.
The preceding decay estimates become
\begin{equation}
 \begin{gathered}
 |Y_u(s)|\leq|u|e^{-3Ks/8},\qquad
 |Y_w(s)|\leq|w|e^{-3s/4},\\
 B(s):=|Y_u(s)|+|Y_w(s)|\leq b_0e^{-s/2},
 \end{gathered}
 \label{eq:stable-rescaled-decay}
\end{equation}
where $b_0=|u|+|w|\leq1/100$.
All norms below are supremum norms and their induced operator norms.
For a direction $h=(h_m,h_u,h_w)$, write $h_s=(h_u,h_w)$.

\begin{lemma}[Differential estimates on the boundary]
\label{lem:stable-differential-coefficients}
On~\eqref{eq:stable-band}, put $\beta=|u|+|w|$ and
$\mathcal A=K Dg(y)$. Then
\begin{align}
 |\mathcal A_{mm}|&\leq64\beta^2,&
 |\mathcal A_{mu}|,|\mathcal A_{mw}|&\leq64\beta,
 \notag\\
 |\mathcal A_{um}|&\leq64|u|,&
 |\mathcal A_{wm}|&\leq64|w|,
 \label{eq:stable-center-coupling}\\
 \mathcal A_{uu}+|\mathcal A_{uw}|&\leq-3K/16,&
 \mathcal A_{ww}+|\mathcal A_{wu}|&\leq-3/8.
 \label{eq:stable-dissipative-rows}
\end{align}
Every full signed row sum of $\mathcal A$ is at most $256\beta$.
Moreover,
\begin{equation}
 \|K D^2g(y)[h,z]\|
 \leq1024\left(\beta\|h\|\|z\|
       +\|h_s\|\|z\|+\|h\|\|z_s\|\right).
 \label{eq:stable-hessian-bound}
\end{equation}
\end{lemma}

\begin{proof}
The polynomial Jacobian is
\[
 Df=\begin{pmatrix}
 u^2-uw&2mu-(m+1)w&-(m+1)u+2w\\
 -mu&(1-m^2-w^2+3u^2)/2&-uw\\
 -w&(2u-w)w&-(m+1+2uw-u^2)
 \end{pmatrix}.
\]
The denominator satisfies $K^2\leq3D$ and $K\leq D$, while
\[
 |f_m/D|\leq\beta^2,\qquad
 |f_u/D|\leq|u|,\qquad |f_w/D|\leq|w|,
 \qquad dD=(2m,2(u-w),2(w-u)).
\]
Insert these bounds into
\begin{equation}
 \mathcal A_{ij}
 =\frac{6K}{D}\left((Df)_{ij}-\frac{f_i}{D}\partial_jD\right).
 \label{eq:stable-jacobian-factor}
\end{equation}
The center row of $Df$ is bounded by
$(\beta^2,2K\beta,2K\beta)$, and its center column below the
diagonal by $(K|u|,|w|)$. This gives
\eqref{eq:stable-center-coupling}.
For the stable rows, direct substitution in the displayed matrix gives
\[
 (Df)_{uu}+|(Df)_{uw}|+4\beta^2\leq-K^2/16,
 \qquad
 (Df)_{ww}+|(Df)_{wu}|+4\beta^2\leq-K/8.
\]
The two rational corrections in each stable row of
\eqref{eq:stable-jacobian-factor} contribute at most $4\beta^2$
inside the parentheses. Since $D\leq2K^2$, this proves
\eqref{eq:stable-dissipative-rows}. Both stable row sums are at most
$-1/4$. Adding the center column, and using $\beta\leq1$, proves
the bound for the full rows.

To estimate the Hessian, retain the distinction between center and
stable directions. Differentiating the matrix above gives
\[
 \|D^2f[h,z]\|\leq12K\|h\|\|z\|,
 \qquad
 D^2f[(h_m,0,0),(z_m,0,0)]=(0,-u h_mz_m,0).
\]
For the first bound, the sums of the absolute coefficient bounds in
the three rows are at most $12K$, $9K$, and $10K$, respectively.
Splitting each direction into its center and stable parts therefore
gives, with
\[
 R_0=\beta\|h\|\|z\|,\qquad
 S_0=\|h_s\|\|z\|+\|h\|\|z_s\|,
\]
the sharper bound $\|D^2f[h,z]\|\leq R_0+12K S_0$.
We also have
\begin{gather*}
 \|g\|\leq6\beta,\qquad
 |dD[h]|\leq6K\|h\|,\qquad
 |D^2D[h,z]|\leq10\|h\|\|z\|,\\
 \|\mathcal A h\|\leq64\beta\|h\|+36K\|h_s\|.
\end{gather*}
Here $D^2D[h,z]=2h_mz_m+2(h_u-h_w)(z_u-z_w)$.
For the last bound, the center column is controlled by
\eqref{eq:stable-center-coupling}; each wall column of $Dg$ is
bounded by $18$, using $|(Df)_{ij}|\leq2D$ for $j=u,w$ in
\eqref{eq:stable-jacobian-factor}.

Twice differentiating the product identity $D(y)g(y)=6f(y)$ gives
\begin{equation}
\begin{split}
 K D^2g[h,z]={}&\frac{6K}{D}D^2f[h,z]
 -\frac{dD[h]}D\mathcal A z
 -\frac{dD[z]}D\mathcal A h
 -\frac{D^2D[h,z]}D K g.
\end{split}
\label{eq:stable-hessian-factor}
\end{equation}
Using $K/D\leq1$ and $K^2/D\leq3$, the four terms are bounded by
\begin{gather*}
 6R_0+216S_0,\qquad
 384R_0+648\|h\|\|z_s\|,\\
 384R_0+648\|h_s\|\|z\|,\qquad 60R_0.
\end{gather*}
Their sum is at most $834R_0+864S_0$, which proves
\eqref{eq:stable-hessian-bound}.
\end{proof}

\begin{proposition}[Regularity of the stable phase]
\label{prop:stable-phase-regularity}
The function $P$ is $C^2$ on each open box
\[
 \Omega_c=\{(m,u,w):|m-c|<1/40,\ |u|,|w|<1/200\},
 \qquad c\geq2.
\]
Set $b_0=|u|+|w|$ and $E=e^{512b_0}$. On this box,
\begin{align}
 |DP(y)h-h_m|
 &\leq256b_0\|h_s\|+66000b_0^2E\|h\|,
 \label{eq:stable-phase-differential}\\
 \|D^2P(y)\|&\leq40000E^3.
 \label{eq:stable-phase-hessian}
\end{align}
For $m\geq2$, we also have $DP(y)f(y)=0$ and
$DP(m,0,0)h=h_m$.
\end{proposition}

\begin{proof}
Every finite-time map in~\eqref{eq:stable-fixed-rescaling} is $C^2$.
Write $J_h(s)=D_yY(s;y)h$. Its variational equation is
\[
 J_h'=\mathcal A(Y)J_h,\qquad J_h(0)=h.
\]
The signed row comparison used in
Lemma~\ref{lem:core-variation-growth}, together with
\eqref{eq:stable-rescaled-decay}, yields
\begin{equation}
 \|J_h(s)\|\leq
 \exp\left(\int_0^s256b_0e^{-r/2}\dd r\right)\|h\|
 \leq E\|h\|.
 \label{eq:stable-first-growth}
\end{equation}
The stable block has row sums at most $-1/4$. Its forcing by the
center component has norm at most $64b_0E\|h\|e^{-s/2}$.
Multiplying by $e^{s/4}$ and integrating gives
\begin{equation}
 \|(J_h(s))_s\|
 \leq\bigl(\|h_s\|+256b_0E\|h\|\bigr)e^{-s/4}
 \leq4E\|h\|e^{-s/4}.
 \label{eq:stable-first-decay}
\end{equation}
In the center row, keep the amplitudes in
\eqref{eq:stable-center-coupling}. They give
\[
 |(J_h)_m'|\leq64B^2|(J_h)_m|+128B\|(J_h)_s\|
 \leq Q_1(y,h)e^{-s/2},
\]
where
\[
 Q_1(y,h)=128b_0\|h_s\|+33000b_0^2E\|h\|.
\]
Consequently, for $t\geq a\geq0$,
\begin{equation}
 |(J_h(t))_m-(J_h(a))_m|
 \leq2Q_1(y,h)e^{-a/2}.
 \label{eq:stable-first-tail}
\end{equation}
Since $b_0\leq1/100$ and $E<e^6$, the linear maps
$D_yY_m(s;y)$ converge uniformly on $\Omega_c$ in operator norm.

We next control their derivatives. Put
$H_{h,z}(s)=D_y^2Y(s;y)[h,z]$. Differentiating the flow equation
once more gives
\[
 H_{h,z}'=\mathcal A(Y)H_{h,z}
             +K D^2g(Y)[J_h,J_z],\qquad H_{h,z}(0)=0.
\]
By~\eqref{eq:stable-hessian-bound},
\eqref{eq:stable-first-growth}, and
\eqref{eq:stable-first-decay}, its forcing obeys
\begin{equation}
 \|K D^2g(Y)[J_h,J_z]\|
 \leq10000E^2\|h\|\|z\|e^{-s/4}.
 \label{eq:stable-second-forcing}
\end{equation}
The three terms in~\eqref{eq:stable-hessian-bound} contribute at most
$E^2$, $4E^2$, and $4E^2$, respectively, after removal of the common
factor $\|h\|\|z\|e^{-s/4}$. Since $9\cdot1024<10000$, this proves
the forcing bound.
The homogeneous growth on every subinterval is at most $E$.
The forced signed row estimate therefore gives
\begin{equation}
 \|H_{h,z}(s)\|\leq40000E^3\|h\|\|z\|.
 \label{eq:stable-second-growth}
\end{equation}
The center row has absolute row sum at most $192B$. Combining
this with~\eqref{eq:stable-second-forcing} yields
\[
 |(H_{h,z})_m'|
 \leq Q_2(y)\|h\|\|z\|e^{-s/4},\qquad
 Q_2(y)=192b_0(40000E^3)+10000E^2.
\]
Thus
\begin{equation*}
 |(H_{h,z}(t))_m-(H_{h,z}(a))_m|
 \leq4Q_2(y)\|h\|\|z\|e^{-a/4}.
\end{equation*}
The second derivatives of $Y_m(s;\cdot)$ also converge uniformly
on $\Omega_c$.

These uniform limits identify the derivatives of the limiting phase.
Take the finite-time functions
$F_n(y)=Y_m(n;y)$ and let $L_1$ be the uniform limit of $DF_n$.
For a line segment in the convex box $\Omega_c$,
\[
 F_n(z)-F_n(y)=\int_0^1
 DF_n(y+\theta(z-y))(z-y)\dd\theta.
\]
Pass to the limit using $F_n\to P$ and uniform convergence of
$DF_n$. The resulting identity and continuity of $L_1$ imply
$DP=L_1$. Apply the same argument to the operator-valued functions
$DF_n$ and their uniformly convergent derivatives $D^2F_n$.
It follows that $DP$ is $C^1$ and that its derivative is the second
limit. Taking $a=0$ in~\eqref{eq:stable-first-tail} proves
\eqref{eq:stable-phase-differential};
\eqref{eq:stable-second-growth} proves
\eqref{eq:stable-phase-hessian}.

At an interior point the boundary orbit remains in $\Omega_c$
for sufficiently small positive time. Differentiate the invariance
from Proposition~\ref{prop:stable-phase-limit} at time zero.
This gives $DP(y)g(y)=0$, and hence $DP(y)f(y)=0$.
The zero-wall differential follows from
\eqref{eq:stable-phase-differential} with $b_0=0$.
\end{proof}

\subsubsection{The phase formula and its wall dependence}

The stable phase has a useful approximation in which the weak wall
is kept exactly. Define
\begin{equation}
 P_0(m,u,w)=H(m+1,w)-1
       +\frac{m}{m^2-1}u^2-\frac2{m+1}uw.
 \label{eq:stable-corrected-phase}
\end{equation}
The first term is the endpoint phase on the plane $u=0$.
The two quadratic corrections cancel the leading motion produced
by the strong wall.

\begin{proposition}[Boundary phase expansion]
\label{prop:stable-phase-expansion}
Under~\eqref{eq:stable-initial-box},
\begin{equation}
 |P(m,u,w)-P_0(m,u,w)|
 \leq\frac{1024}{K^3}|u|(|u|+|w|)^3.
 \label{eq:stable-quartic-remainder}
\end{equation}
In particular,
\begin{equation}
 P(m,0,w)=H(m+1,w)-1.
 \label{eq:stable-weak-phase}
\end{equation}
\end{proposition}

\begin{proof}
Put $\ell=m+1$, $d=m^2-1$, and $\sigma=H(m+1,w)$.
Differentiation along~\eqref{eq:stable-boundary-field} gives
\begin{equation}
 DP_0 f=r_4u^4+r_3u^3w+r_2u^2w^2+r_1uw^3,
 \label{eq:stable-quartic-identity}
\end{equation}
where
\begin{align*}
 r_4&=-\frac{2m}{d^2},&
 r_3&=\frac{2(3m-1)}{d^2},\\
 r_2&=-\frac{m^2+1}{d^2}-\frac m d
              +\frac{\sigma+1}{\sigma(\sigma+\ell)},&
 r_1&=\frac2{\ell^2}+\frac1\ell-\frac1{\sigma+\ell}.
\end{align*}
To see the cancellation, first differentiate $\sigma$:
\[
 D\sigma f=m u^2-(m+1)uw
       +\frac{\sigma+1}{\sigma(\sigma+\ell)}u^2w^2
       -\frac1{\sigma+\ell}uw^3.
\]
This uses $\sigma-\ell=w^2/(\sigma+\ell)$ and
$1-\ell/\sigma=w^2/[\sigma(\sigma+\ell)]$.
The derivatives of the two rational quadratic terms in
\eqref{eq:stable-corrected-phase} cancel the first two terms
and give~\eqref{eq:stable-quartic-identity}.

For $m\geq3/2$, the inequalities
\[
 \frac1d\leq\frac2\ell,\qquad
 \frac m d\leq\frac3\ell,\qquad
 \frac{m^2+1}d\leq3,\qquad \sigma\geq\ell
\]
bound $|r_4|,|r_3|,|r_2|,|r_1|$ by
$12/\ell,36/\ell,11/\ell,4/\ell$, respectively.
All four are therefore at most $64/\ell$.
On~\eqref{eq:stable-band}, $\ell\geq K/2$, and hence
\[
 |DP_0 f|\leq\frac{128}K |u|(|u|+|w|)^3.
\]
Along the boundary orbit, the sum of the wall magnitudes does not
increase, while~\eqref{eq:stable-boundary-decay} retains the
faster factor $|U|\leq|u|e^{-K\xi/8}$. Since
$d\xi=K A\dd t$, integration gives
\begin{align*}
 &|P_0(M(t),U(t),W(t))-P_0(m,u,w)|\\
 &\qquad\leq\frac{128}{K^2}|u|(|u|+|w|)^3
             \int_0^{\xi(t)}e^{-Kx/8}\dd x
 \leq\frac{1024}{K^3}|u|(|u|+|w|)^3.
\end{align*}
As $t\to\infty$, the left-hand phase converges to
$P_0(P(m,u,w),0,0)=P(m,u,w)$. This proves
\eqref{eq:stable-quartic-remainder}; setting $u=0$ proves
\eqref{eq:stable-weak-phase}.
\end{proof}

Extend $P$ to four coordinates by ignoring $V$, and define
\begin{equation}
 \widehat\rho(M,U,V,W)=H(P(M,U,W)-1,V).
 \label{eq:stable-entrance-radius}
\end{equation}
For $M\geq2$ and $|U|,|W|<1/200$, the stable phase is $C^2$.
Its boundary invariance removes all terms without an active wall
from its derivative along the full projective field:
\begin{equation}
 \dot P
 =UV\bigl((1-M)P_M+W(P_U-P_W)\bigr)
       -V^2\bigl(P_M+(U/2)P_U\bigr).
 \label{eq:stable-active-phase-rate}
\end{equation}
Indeed, subtract the boundary field~\eqref{eq:stable-boundary-field}
from the $M,U,W$ components of~\eqref{eq:projective-flow}.
The differences are
\[
 (-V^2+(1-M)UV,\quad UVW-UV^2/2,\quad-UVW).
\]
Applying $DP$ and using $DPf=0$ proves
\eqref{eq:stable-active-phase-rate}. For $V>0$ the radial differential is
\begin{equation}
 D\widehat\rho[h]
 =\frac{(P-1)DP[h_M,h_U,h_W]+Vh_V}{\widehat\rho}.
 \label{eq:stable-entrance-differential}
\end{equation}
The same formula applies whenever the defining radicand is positive.

Recall the finite outgoing endpoint
$y_\eta=(M_\eta,U_\eta,\eta,W_\eta)$ from
Lemma~\ref{lem:compact-outgoing-tail}. Its reflected entrance is
$(-M_\eta,U_\eta,W_\eta,\eta)$. We associate to this endpoint
the phase
\begin{equation}
 F_\eta(R,u,w)
 =H\bigl(P(-M_\eta,U_\eta,\eta)-1,W_\eta\bigr).
 \label{eq:compact-finite-phase}
\end{equation}

\begin{proposition}[Cancellation of the fixed section height]
\label{prop:compact-reference-phase}
Fix $0<\eta\leq10^{-4}$. On a signed neighborhood of
$[31/8,6]\times\{(0,0)\}$, the function $F_\eta$ is $C^2$.
There are $\delta>0$ and $C_\eta<\infty$ such that, for
$R\in[31/8,6]$ and $|u|,|w|\leq\delta$,
\begin{equation}
 |F_\eta|+\|DF_\eta\|+\|D^2F_\eta\|\leq C_\eta.
 \label{eq:compact-finite-phase-bounds}
\end{equation}
On the reference line,
\begin{equation}
 F_\eta(R,0,0)=R-2,\qquad
 \partial_R F_\eta(R,0,0)=1.
 \label{eq:compact-reference-phase}
\end{equation}
\end{proposition}

\begin{proof}
For $u=w=0$, the reflected finite endpoint is
\[
 (\sqrt{R^2-\eta^2}-1,0,0,\eta).
\]
Its first coordinate is greater than two, even for $R\geq7/2$.
By~\eqref{eq:stable-weak-phase},
\[
 P(\sqrt{R^2-\eta^2}-1,0,\eta)
 =H(\sqrt{R^2-\eta^2},\eta)-1=R-1.
\]
Substitution in~\eqref{eq:compact-finite-phase} gives $F_\eta=R-2$.
The finite endpoint is $C^2$ in a full neighborhood, as established
in Lemma~\ref{lem:compact-outgoing-tail}. At each reference point
the composition defining $P$ lies in its $C^2$ domain, and the
radicand defining $F_\eta$ equals $(R-2)^2>0$. The composition is
therefore $C^2$ locally. Compactness of the reference segment
provides a uniform signed neighborhood and a bound for its first
two derivatives, proving~\eqref{eq:compact-finite-phase-bounds}.
The reference identity holds on an open interval around every
$R\in[31/8,6]$, so it may be differentiated there.
\end{proof}

The ordinary reflected radius at the finite endpoint is
$\sqrt{R^2-\eta^2}-2$ on the reference line.
Following the stable boundary to its Kasner endpoint replaces this
quantity by exactly $R-2$. Thus the fixed section height has been
absorbed into the phase before the incoming passage is estimated.
The factors $UV$ and $V^2$ in~\eqref{eq:stable-active-phase-rate}
will allow the remaining phase change to be integrated as the
active wall grows.

We first compute the differential of $P$ on the weak-wall plane.
The transverse derivative on this plane contributes to the linear
response of the outgoing phase.

\begin{lemma}[The weak-wall differential]
\label{lem:stable-weak-differential}
If $m\geq2$ and $|w|<1/200$, then, for every direction
$h=(h_m,h_u,h_w)$,
\begin{equation}
 DP(m,0,w)h
 =\frac{(m+1)h_m+w(h_w-2h_u)}{H(m+1,w)}.
 \label{eq:stable-weak-differential}
\end{equation}
\end{lemma}

\begin{proof}
Use the fixed rescaling~\eqref{eq:stable-fixed-rescaling}, with
$c=m$, and let $Y(s)=(M(s),0,W(s))$ be the orbit from $(m,0,w)$.
For its initial variation $J(s)=D_yY(s;y)h$, consider
\begin{equation*}
 L(s)=(M(s)+1)J_m(s)+W(s)\bigl(J_w(s)-2J_u(s)\bigr).
\end{equation*}
This quantity is constant. To verify it, put
$\ell=(m+1,-2w,w)$ on $u=0$. The boundary field and its
Jacobian there are
\[
 f=(w^2,0,-(m+1)w),\qquad
 Df=\begin{pmatrix}
 0&-(m+1)w&2w\\
 0&(1-m^2-w^2)/2&0\\
 -w&-w^2&-(m+1)
 \end{pmatrix}.
\]
Along this field,
\[
 \dot\ell=(w^2,2(m+1)w,-(m+1)w),\qquad
 \dot\ell+\ell Df=0,\qquad \ell f=0.
\]
The derivative of $g=Af$ is $Dg=A Df+f\,dA$.
The last identity removes the $dA$ term, so the same conservation
holds for the original flow and for its fixed rescaling. Thus
\[
 L(s)=L(0)=(m+1)h_m+w(h_w-2h_u).
\]
As $s\to\infty$, $M(s)\to P(m,0,w)$, $W(s)\to0$,
and $J_m(s)\to DP(m,0,w)h$. The whole variation is bounded by
\eqref{eq:stable-first-growth}, so both products with $W(s)$
vanish. Consequently
\[
 (P(m,0,w)+1)DP(m,0,w)h=(m+1)h_m+w(h_w-2h_u).
\]
Now use~\eqref{eq:stable-weak-phase}, whose right-hand norm is
strictly positive.
\end{proof}

The term $-2wh_u$ in~\eqref{eq:stable-weak-differential} measures
the response to a strong wall introduced transversely to the
weak-wall plane. A related linear invariant carries this response
through the finite outgoing segment.

\begin{proposition}[Strong-wall response of the finite phase]
\label{prop:finite-phase-strong-response}
For $R\in[31/8,6]$ and $0<\eta\leq10^{-4}$,
\begin{equation}
 \partial_u F_\eta(R,0,0)=-\frac2R.
 \label{eq:finite-phase-strong-response}
\end{equation}
\end{proposition}

\begin{proof}
Write $X$ for the full projective vector field in
\eqref{eq:projective-flow}, and let
$Z(\tau)=(M(\tau),0,V(\tau),0)$ be the reference pulse
\eqref{eq:reference-core-pulse} with radius $R$.
An initial-state variation of the original flow, evaluated at the
original times corresponding to this pulse, satisfies
\begin{equation}
 \dot\xi=DX(Z)\xi+a(\tau)X(Z),\qquad
 a(\tau)=-\frac{dD(Z)\xi}{D(Z)}.
 \label{eq:finite-reference-clock-variation}
\end{equation}
Indeed, the original field is $AX$, and multiplication of its
variational equation by $dt/d\tau=1/A$ gives
$DX\xi+(dA\xi/A)X$. Since $A=6/D$, this is
\eqref{eq:finite-reference-clock-variation}.

For every such variation,
\begin{equation*}
 I(\tau)=(M-1)\xi_M+V\xi_V-2V\xi_U
\end{equation*}
is constant. The coefficient row is
$\lambda=(M-1,-2V,V,0)$. On the reference pulse,
\[
 \dot M=-V^2,\qquad \dot V=(M-1)V,
 \qquad \dot\lambda+\lambda DX=0,\qquad \lambda X=0.
\]
The two row identities follow by differentiating
\eqref{eq:projective-flow} at $U=W=0$. They imply $\dot I=0$
even with the tangent forcing in
\eqref{eq:finite-reference-clock-variation}.

Vary $u$ in the downward entry
$y_-(R,u,0)=(1-\sqrt{R^2-1},u,1,0)$, keeping $R$ fixed.
The initial variation is $e_U=(0,1,0,0)$, so $I=-2$.
Let
\[
 z=y_\eta(R,0,0),\qquad
 v=\partial_u y_\eta(R,0,0)
\]
be the actual endpoint and its derivative. The latter is the
fixed-time variation plus $\partial_u S_\eta$ times the original
vector field at $z$. Since $\lambda X=0$, this endpoint correction
also disappears from the invariant. We obtain
\begin{equation*}
 (z_M-1)v_M+\eta(v_V-2v_U)=-2.
\end{equation*}

The stable point after reflection is $(-z_M,0,\eta)$.
Its variation is $(-v_M,v_U,v_V)$.
Let $q$ be the derivative of $P$ in this direction.
Lemma~\ref{lem:stable-weak-differential} gives
\[
 \bigl(P(-z_M,0,\eta)+1\bigr)q
 =(z_M-1)v_M+\eta(v_V-2v_U)=-2.
\]
The reference value is $P=R-1$, so $q=-2/R$.
Finally $F_\eta=H(P-1,W_\eta)$, with $W_\eta=0$ and
$P-1=R-2>0$ at this endpoint. Differentiating this norm gives
$\partial_u F_\eta=q$, as asserted.
\end{proof}

On the plane $u=0$, the energy of the outgoing segment changes
quadratically in the initial weak wall. Retaining this factor through
the two radial norms will also control the radial derivative.

For $Y=(R,u,w)$ in a neighborhood of the reference segment, let
$b_W(Y,t)$ be the positive wall factor from
\eqref{eq:compact-tail-factor-formula}, with $S_\eta$ replaced by
$t$. Thus
\[
 W(t;Y)=w b_W(Y,t),\qquad b_W(Y,S_\eta(Y))=B_W(Y).
\]
The factor is jointly $C^2$ in $Y$ and $t$ around the finite
orbit segment. Define
\begin{align}
 r_\eta(Y)&=H(M_\eta(Y)-1,\eta),\notag\\
 I_\eta(Y)&=\int_0^{S_\eta(Y)}
       2A(t;Y)(M(t;Y)-1)b_W(Y,t)^2\dd t,\notag\\
 Q_\eta(Y)&=
 \frac{I_\eta(Y)\left(1-\dfrac4{r_\eta(Y)+R}\right)+B_W(Y)^2}
      {F_\eta(Y)+R-2}.
 \label{eq:finite-weak-coefficient}
\end{align}
All three functions are $C^2$ on a sufficiently small signed
neighborhood. For $r_\eta$ the radicand is positive because
$\eta>0$. For $I_\eta$, write the moving integral over $[0,1]$
and differentiate the finite flow and wall factor twice. The two
denominators in $Q_\eta$ are positive near the reference line,
where $r_\eta=R$ and $F_\eta=R-2$.

\begin{lemma}[Quadratic weak-wall factorization]
\label{lem:finite-weak-factorization}
After reducing the signed neighborhood if necessary,
\begin{equation}
 F_\eta(R,0,w)=R-2+w^2Q_\eta(R,0,w).
 \label{eq:finite-weak-factorization}
\end{equation}
In particular, $\partial_w F_\eta(R,0,0)=0$. There are
$\delta>0$ and $B_\eta>0$ such that
\begin{equation}
 |\partial_R F_\eta(R,0,w)-1|\leq B_\eta w^2
 \quad\text{for }31/8\leq R\leq6,\quad |w|\leq\delta.
 \label{eq:finite-weak-radial-bound}
\end{equation}
\end{lemma}

\begin{proof}
The plane $U=0$ is invariant. On it,
\eqref{eq:energy-identities} in original time gives
\[
 \frac d{dt}\bigl((M-1)^2+V^2\bigr)
       =2A(M-1)W^2.
\]
Substitute $W=w b_W$ and integrate from the downward core exit
to $V=\eta$. The initial energy is $R^2$, so
\begin{equation}
 r_\eta^2=R^2+w^2I_\eta.
 \label{eq:finite-weak-energy}
\end{equation}
At the reflected endpoint, the exact weak-wall formula
\eqref{eq:stable-weak-phase} gives
\[
 P(-M_\eta,0,\eta)=r_\eta-1,
 \qquad F_\eta^2=(r_\eta-2)^2+w^2B_W^2.
\]
Here the neighborhood has been chosen so that $-M_\eta>2$.
Rationalizing~\eqref{eq:finite-weak-energy} first yields
$r_\eta-R=w^2I_\eta/(r_\eta+R)$. Hence
\begin{align*}
 \bigl(F_\eta-(R-2)\bigr)\bigl(F_\eta+R-2\bigr)
 &=r_\eta^2-R^2-4(r_\eta-R)+w^2B_W^2\\
 &=w^2\left[I_\eta\left(1-\frac4{r_\eta+R}\right)+B_W^2\right].
\end{align*}
Division by $F_\eta+R-2$ proves
\eqref{eq:finite-weak-factorization}.

This identity holds on an open neighborhood in the two variables
$(R,w)$, so both of its partial derivatives can be taken there.
The $w$ derivative vanishes at $w=0$, and
\[
 \partial_R F_\eta(R,0,w)
       =1+w^2\partial_R Q_\eta(R,0,w).
\]
The derivative of $Q_\eta$ is continuous. Compactness of the
reference segment provides a common closed strip on which it is
bounded, giving~\eqref{eq:finite-weak-radial-bound}.
\end{proof}

We can now gather the finite phase estimates in the form needed for
the incoming passage. In particular, the weak-wall square survives
radial differentiation.

\begin{proposition}[Expansion of the finite outgoing phase]
\label{prop:finite-phase-expansion}
Fix $0<\eta\leq10^{-4}$. There are $\delta>0$ and $C_\eta>0$
such that, for
\[
 31/8\leq R\leq6,\qquad |u|,|w|\leq\delta,
\]
the same finite outgoing map is defined and $F_\eta$ is $C^2$.
With $b=|u|+|w|$,
\begin{align}
 \left|F_\eta(R,u,w)-\left(R-2-\frac{2u}{R}\right)\right|
       &\leq C_\eta b^2,\label{eq:finite-phase-expansion}\\
 |\partial_R F_\eta(R,u,w)-1|
       &\leq C_\eta(|u|+w^2),\label{eq:finite-phase-radial-bound}\\
 \left|\partial_u F_\eta(R,u,w)+\frac2R\right|
       +|\partial_w F_\eta(R,u,w)|
       &\leq C_\eta b.\label{eq:finite-phase-wall-bound}
\end{align}
\end{proposition}

\begin{proof}
Choose a closed convex box contained in the common signed
neighborhood of the preceding results. Let $A_\eta$ bound
$\|D^2F_\eta\|$ on this box. The derivative is then Lipschitz:
\begin{equation}
 \|DF_\eta(Y)-DF_\eta(Z)\|\leq A_\eta\|Y-Z\|.
 \label{eq:finite-phase-gradient-lipschitz}
\end{equation}
The reference value and all three partial derivatives have now
been computed:
\[
 F_\eta(R,0,0)=R-2,\qquad
 DF_\eta(R,0,0)=(1,-2/R,0).
\]
Set $Z=(R,0,0)$ and $Y=(R,u,w)$. Applying
\eqref{eq:finite-phase-gradient-lipschitz} to the unit wall
directions proves~\eqref{eq:finite-phase-wall-bound} with
constant $2A_\eta$. To obtain the value estimate, subtract the
linear functional $DF_\eta(Z)$ from $F_\eta$. Its derivative
at any point of the segment from $Z$ to $Y$ has norm at most
$A_\eta\|Y-Z\|$. The mean-value bound therefore gives
\[
 |F_\eta(Y)-F_\eta(Z)-DF_\eta(Z)(Y-Z)|
       \leq A_\eta\|Y-Z\|^2\leq A_\eta b^2,
\]
which is~\eqref{eq:finite-phase-expansion}.

For the radial derivative, compare instead with
$Z_w=(R,0,w)$. The distance to this point is $|u|$, so
\eqref{eq:finite-phase-gradient-lipschitz} and
\eqref{eq:finite-weak-radial-bound} give
\[
 |\partial_R F_\eta(Y)-1|
       \leq A_\eta|u|+B_\eta w^2.
\]
Taking $C_\eta=2A_\eta+B_\eta+1$ proves all three estimates.
\end{proof}

The coefficient $-2/R$ is independent of the section height.
On $u=0$, the weak wall contributes only at quadratic order to the
phase and its radial derivative. These are properties of the finite
outgoing phase. Its comparison with the radius at the next core
will use~\eqref{eq:stable-active-phase-rate} during the intervening
growth of $V$.