\subsection{Radial variation of the terminal boundary phase}
\label{sec:terminal-single-phase}

The cyclic coordinates have derivatives of order $\alpha^{-2}$,
but their contribution to the bounded phase is much smaller.
The single-wall invariant makes this cancellation explicit.
To preserve it along the outgoing passage, we first refine the
radial derivative of that passage: its radius changes almost by
translation, while each wall response retains its own amplitude.

Using $I$ and $\lambda_+$ from~\eqref{eq:terminal-strong-invariant},
define
\begin{equation}
 \vartheta(m,u)=\frac{2}{\lambda_+(I(m,u))-1},\qquad m>1.
 \label{eq:terminal-single-bounded-phase}
\end{equation}
In the domain of Lemma~\ref{lem:terminal-strong-limit}, this is
$2/(P(m,u,0)-1)$. At the actual cyclic entrance
$Z_h(Y)=(m_e,u_e,v_e,w_e)$, put
\begin{equation}
 \psi_h(Y)=\vartheta(m_e,u_e),\qquad Y=(R,u,w).
 \label{eq:terminal-single-composed-phase}
\end{equation}
Thus $\psi_h$ retains the actual entrance shear and strong wall,
and evaluates their single-wall phase. The weak wall and the
subsequent incoming passage will be included in the phase
correction below.

\begin{proposition}
\label{prop:terminal-single-composition}
Fix $0<\eta_0\leq1/100$ and an integer $J\geq2$.
There are $C\geq1$ and $\varepsilon>0$, depending only on
$\eta_0,J$, with the following property. If
\begin{equation}
 \eta_0\leq\eta\leq1/100,\quad 0<\alpha\leq1,\quad
 0<b\leq\varepsilon\alpha^{2J},\quad |u|+|w|\leq b,
 \label{eq:terminal-single-composition-domain}
\end{equation}
then, at $Y_0=(2+2\alpha,u,w)$ and the fixed numerical height
$h=\eta\alpha^J$, the function $\psi_h$ is $C^2$ near $Y_0$ and
\begin{equation}
 |\psi_h(Y_0)-\alpha|
   +|\partial_R\psi_h(Y_0)-1/2|\leq Cb,
 \qquad \|D\psi_h(Y_0)\|\leq C.
 \label{eq:terminal-single-composition-bounds}
\end{equation}
The estimates allow signed and zero old walls.
\end{proposition}

\subsubsection{The outgoing radial response}

Recall the radius map
$E_h(Y)=(\rho_h,U_h,W_h)$ from
\eqref{eq:terminal-outgoing-radius-map} and set $L=\log(1/h)$.
All norms below are maximum norms and their induced operator
norms.

\begin{lemma}
\label{lem:terminal-fine-outgoing}
Fix $0<\eta_0<1$ and an integer $J\geq1$. There are
$C\geq1$ and $\varepsilon>0$ such that, whenever
\[
 \eta_0\leq\eta<1,\quad 0<\alpha\leq1,\quad
 0<b\leq\varepsilon\alpha^J,\quad |u|+|w|\leq b,
\]
the fixed-height map $E_h$ at $Y_0=(2+2\alpha,u,w)$ is $C^2$,
and
\begin{align}
 \|DE_h\|&\leq C,&
 |\partial_R\rho_h-1|
     &\leq C\bigl(|u|+b^2(1+L)^2\bigr),
       \label{eq:terminal-fine-radial}\\
 |U_h|&\leq C|u|,& |W_h|&\leq C|w|,
       \label{eq:terminal-fine-wall-values}\\
 |\partial_R U_h|&\leq C|u|/\alpha,&
 |\partial_R W_h|&\leq C|w|/\alpha.
       \label{eq:terminal-fine-wall-columns}
\end{align}
All derivatives are evaluated at $Y_0$, with $h=\eta\alpha^J$
held fixed.
\end{lemma}

\begin{proof}
Use the actual clock family $\Phi(Y,s)$ of
\eqref{eq:terminal-actual-clock-family}, constructed first up
to the lowest height $\eta_0\alpha^J$. Every height in the
statement corresponds to a shorter interval $0\leq s\leq L$.
For the two wall rates in~\eqref{eq:terminal-outgoing-field},
define
\begin{equation}
 \ell_j(Y,s)=\int_0^s
       a_j(e^{-r},\Phi(Y,r))\dd r,
       \qquad j\in\{U,W\}.
 \label{eq:terminal-wall-rate-integrals}
\end{equation}
On a common initial neighborhood, the scalar wall equations
give
\begin{equation*}
 \Phi_j(Y,s)=Y_j e^{\ell_j(Y,s)},\qquad
 D_Y\Phi_j[d]=e^{\ell_j}d_j+\Phi_j D_Y\ell_j[d].
\end{equation*}
Here $Y_U=u$ and $Y_W=w$. The integral defining $\ell_j$ is
$C^1$ in $Y$: the joint clock family is $C^2$, and
differentiation is under a finite integral with continuous
integrand and derivative. Consequently these identities also
hold at zero walls.

The absolute rate integrals
\eqref{eq:terminal-absolute-rate-integral}, the first-variation
bound in Proposition~\ref{prop:terminal-outgoing-derivatives},
and the rate gradients~\eqref{eq:terminal-gradient-bounds}
give one $K\geq1$ such that, along the whole reference segment,
\begin{equation}
 \|D_Y\Phi\|\leq K,\qquad
 K^{-1}\leq e^{\ell_j}\leq K,\qquad
 \|D_Y\ell_U\|+\|D_Y\ell_W\|\leq Ks.
 \label{eq:terminal-fine-multiplier-bounds}
\end{equation}
For the last inequality, each differentiated integrand has
norm at most $1536\|D_Y\Phi\|$; enlarge $K$ to absorb the
two resulting constants.

Let $\xi(s)=D_Y\Phi(Y_0,s)[e_R]$, where $e_R=(1,0,0)$.
The factor identities and~\eqref{eq:terminal-fine-multiplier-bounds}
imply
\begin{equation}
 \begin{gathered}
 |U(s)|\leq K|u|,\qquad |U(s)|+|W(s)|\leq Kb,
 \qquad |\xi_\rho(s)|\leq K,\\
 |\xi_U(s)|\leq K^2|u|s,\qquad
 |\xi_W(s)|\leq K^2|w|s.
 \end{gathered}
 \label{eq:terminal-radial-wall-response}
\end{equation}
The radial row of the variational equation has no constant
term in the old walls. In fact,
\eqref{eq:terminal-gradient-bounds}, with $v=e^{-s}$ and
$\beta=|U|+|W|$, yields
\[
 |\xi_\rho'|
 \leq256(|U|v+\beta^2)|\xi_\rho|
       +32(v+\beta)|\xi_U|+16\beta|\xi_W|.
\]
Substitution of~\eqref{eq:terminal-radial-wall-response} gives
\begin{equation*}
 |\xi_\rho'(s)|\leq512K^3
       \bigl(|u|(1+s)e^{-s}+b^2(1+s)\bigr).
\end{equation*}
Since $\xi_\rho(0)=1$,
\[
 |\xi_\rho(s)-1|
   \leq512K^3\bigl(2|u|+b^2s(1+s)\bigr).
\]
We used $\int_0^s(1+r)e^{-r}\dd r\leq2$ and
$\int_0^s(1+r)\dd r\leq s(1+s)$.

At the endpoint, $E_h(Y)=\Phi(Y,L)$ on an ordinary
neighborhood of $Y_0$, by
\eqref{eq:terminal-outgoing-clock-identification}. Thus the
last bound proves~\eqref{eq:terminal-fine-radial}; the factor
identity proves~\eqref{eq:terminal-fine-wall-values}.
Finally,
\[
 \alpha L\leq \log(1/\eta_0)+J/e
\]
turns the last two bounds in
\eqref{eq:terminal-radial-wall-response} into
\eqref{eq:terminal-fine-wall-columns}. Increasing $C$ gives
all conclusions with the same constants.
\end{proof}

\subsubsection{The invariant in outgoing coordinates}

Consider an outgoing point
$(1-g,U,v,W)$, where $g=\sqrt{\rho^2-v^2}$. Introduce
\begin{equation}
 \begin{aligned}
 H_0&=(2-g)^2+v^2,&
 \Delta&=(U-W)^2-2v(U+W),\\
 \mathcal H&=H_0+\Delta,&
 N&=\rho^2-4+\Delta,\\
 e&=\Delta\left(1+\frac{N}{H_0}\right),&
 X&=\sqrt{\rho^2+e}.
 \end{aligned}
 \label{eq:terminal-single-radical-data}
\end{equation}
The cyclic denominator is $\mathcal H$, and its shear and
strong-wall coordinates are
\begin{equation}
 m_e=N/\mathcal H,\qquad u_e=4v/\mathcal H.
 \label{eq:terminal-single-cyclic-pair}
\end{equation}
The following identities express their invariant without
separately differentiating these two singular quotients:
\begin{align}
 \mathcal H^2+N^2+16v^2
     &=2\mathcal H N+16H_0,
       \label{eq:terminal-single-polynomial}\\
 \rho^2+e&=4+\frac{\mathcal H N}{H_0}.
       \label{eq:terminal-single-radical-identity}
\end{align}
To check the first, use $N=\mathcal H-4(2-g)$; the second
follows by expanding $(H_0+\Delta)(\rho^2-4+\Delta)$.
If $g>2$ and $H_0,\mathcal H$ are positive, then
$N=\mathcal H+4(g-2)>0$, $m_e>1$, and $X>2$. Moreover,
\[
 I(m_e,u_e)=2+\frac{16H_0}{\mathcal H N}
          =2+\frac{16}{X^2-4}.
\]
The upper root is $(X+2)/(X-2)$. Therefore
\begin{equation}
 \vartheta(m_e,u_e)=s_v(\rho,U,W),\qquad
 s_v(\rho,U,W):=\frac{\sqrt{\rho^2+e}-2}{2}.
 \label{eq:terminal-single-radical-phase}
\end{equation}
At $U=W=0$ the defect $\Delta$ vanishes and
$s_v=(\rho-2)/2$, independently of $v$.

\begin{lemma}
\label{lem:terminal-single-radical-bounds}
Suppose
\begin{equation}
 \begin{gathered}
 0<\alpha\leq1,\qquad
 |\rho-(2+2\alpha)|\leq\alpha/10000,\\
 |U|+|W|\leq\beta\leq\alpha/10000,
 \qquad 0\leq v\leq\alpha/100.
 \end{gathered}
 \label{eq:terminal-single-radical-region}
\end{equation}
Then $s_v$ is $C^2$ near $(\rho,U,W)$, with $v$ held fixed,
and
\begin{align}
 |s_v-(\rho-2)/2|&\leq23\beta(v+\beta)/\alpha,
       \label{eq:terminal-single-radical-value}\\
 |\partial_\rho s_v-1/2|
     &\leq1024\beta(v+\beta)/\alpha^2,
       \label{eq:terminal-single-radical-radial}\\
 |\partial_U s_v|+|\partial_W s_v|
     &\leq1024(v+\beta)/\alpha.
       \label{eq:terminal-single-radical-walls}
\end{align}
\end{lemma}

\begin{proof}
The algebraic gap and denominator estimates used in the proof
of Lemma~\ref{lem:terminal-cycle-entrance}, with the bounds in
\eqref{eq:terminal-single-radical-region}, give
\begin{equation}
 \begin{gathered}
 2<\rho\leq5,\qquad 2<g\leq5,\qquad H_0\geq\alpha^2,\\
 |\Delta|\leq2\beta(v+\beta)\leq\alpha^2/1000,
 \qquad |N|\leq22\alpha,\qquad \mathcal H>0.
 \end{gathered}
 \label{eq:terminal-single-scalar-geometry}
\end{equation}
In particular, $1.99\alpha<g-2<2.01\alpha$ and
$H_0=(g-2)^2+v^2\geq\alpha^2$.
The estimate for $\Delta$ follows from
$|U-W|,|U+W|\leq\beta$, and
$\rho^2-4\leq(3\alpha)7$ gives the bound on $N$.
Also $N=\mathcal H+4(g-2)>0$, so
\eqref{eq:terminal-single-radical-identity} gives $X>2$.
All square roots and denominators are consequently smooth
on a neighborhood.

For the derivative in $\rho$, let
\[
 H_{0,\rho}=2(g-2)\rho/g,\qquad |H_{0,\rho}|\leq20\alpha.
\]
Differentiating~\eqref{eq:terminal-single-radical-data}, while
keeping $v$ fixed, gives
\begin{equation}
 \begin{aligned}
 e_\rho&=\Delta\left(\frac{2\rho}{H_0}
                     -\frac{NH_{0,\rho}}{H_0^2}\right),\\
 e_\Delta&=1+\frac{\rho^2-4+2\Delta}{H_0},\\
 d\Delta&=(2(U-W)-2v)dU+(-2(U-W)-2v)dW,\\
 ds_v&=\frac{(2\rho+e_\rho)d\rho+e_\Delta d\Delta}{4X}.
 \end{aligned}
 \label{eq:terminal-single-radical-differential}
\end{equation}
The scalar estimates in
\eqref{eq:terminal-single-scalar-geometry} imply
\begin{equation}
 |e|\leq\frac{46\beta(v+\beta)}\alpha,\qquad
 |e_\rho|\leq\frac{900\beta(v+\beta)}{\alpha^2},\qquad
 |e_\Delta|\leq\frac{24}\alpha.
 \label{eq:terminal-single-error-coefficients}
\end{equation}
Indeed, the factors multiplying $\Delta$ in $e$ and
$e_\rho$ are at most $23/\alpha$ and $450/\alpha^2$:
\[
 1+\frac{22\alpha}{\alpha^2}\leq\frac{23}\alpha,
 \qquad
 \frac{10}{\alpha^2}
       +\frac{22\alpha\,20\alpha}{\alpha^4}
          =\frac{450}{\alpha^2}.
\]
Adding one more $\Delta$ gives
$|\rho^2-4+2\Delta|\leq23\alpha$, which proves the
last bound in~\eqref{eq:terminal-single-error-coefficients}.

Since $(X-\rho)(X+\rho)=e$ and $X,\rho>2$, we have
$|X-\rho|\leq|e|$. This proves
\eqref{eq:terminal-single-radical-value}. For the radial
derivative, retain the subtraction before estimating:
\[
 \partial_\rho s_v-\frac12
       =\frac{\rho-X}{2X}+\frac{e_\rho}{4X}.
\]
Equation~\eqref{eq:terminal-single-error-coefficients} and
$\alpha\leq1$ bound this by
$1024\beta(v+\beta)/\alpha^2$.
Finally $|\Delta_U|+|\Delta_W|\leq4(v+\beta)$.
The last line of~\eqref{eq:terminal-single-radical-differential}
then proves~\eqref{eq:terminal-single-radical-walls}.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:terminal-single-composition}]
Choose the constants in Lemma~\ref{lem:terminal-fine-outgoing}
and in the uniform terminal geometry. Reduce $\varepsilon$
so that both apply and $\varepsilon\leq1$.
Write
\[
 z=E_h(Y_0)=(\rho,U,W),\qquad \beta=|U|+|W|.
\]
For a constant $B\geq1$ depending only on $\eta_0,J$,
\begin{equation*}
 |\rho-(2+2\alpha)|\leq Bb,\qquad
 \beta\leq Bb,\qquad Bb\leq\alpha/10000.
\end{equation*}
Also $h\leq\alpha/100$, so
Lemma~\ref{lem:terminal-single-radical-bounds} applies at $z$.
The strict inequalities used in
\eqref{eq:terminal-single-radical-phase} persist on an initial
neighborhood. There the actual outgoing state satisfies
$M_h<1$ and is reconstructed from $E_h$ with $V=h$.
Hence
\begin{equation}
 \psi_h=s_h\circ E_h
 \quad\hbox{on a neighborhood of }Y_0.
 \label{eq:terminal-single-local-composition}
\end{equation}
This identity proves $C^2$ regularity and identifies the
ordinary derivatives of the two sides.

Because $J\geq2$, we have $b\leq\alpha^2$ and
$h\leq\alpha^2$. Thus
$h+\beta\leq(1+B)\alpha^2$. The radical estimates give
constants $C_1,C_2$ with
\begin{equation}
 |s_{h,\rho}(z)-1/2|\leq C_1b,\qquad
 |s_{h,U}(z)|+|s_{h,W}(z)|\leq C_2\alpha.
 \label{eq:terminal-single-composition-coefficients}
\end{equation}
They also give
\[
 |\psi_h(Y_0)-\alpha|
 \leq\frac{23\beta(h+\beta)}\alpha
           +\frac12|\rho-(2+2\alpha)|\leq C_3b.
\]

For the radial derivative put $d=DE_h(Y_0)[e_R]$ and
$\overline L=1+\log(1/\eta_0)+J/e$. The fine outgoing bounds
give
\[
 |d_\rho|\leq A,\qquad
 |d_U|+|d_W|\leq Ab/\alpha,
\]
with a uniform $A\geq1$. Moreover,
$\alpha(1+L)\leq\overline L$, and therefore
\[
 b^2(1+L)^2\leq\overline L^{\,2}b,\qquad
 |d_\rho-1|\leq A(1+\overline L^{\,2})b.
\]
The first inequality uses $b\leq\alpha^2$.
Apply the chain rule in~\eqref{eq:terminal-single-local-composition}
and subtract the reference derivative:
\begin{align*}
 |\partial_R\psi_h-1/2|
 &\leq |s_{h,\rho}-1/2|\,|d_\rho|
       +\tfrac12|d_\rho-1|\\
 &\quad+(|s_{h,U}|+|s_{h,W}|)(|d_U|+|d_W|)
 \leq C_4b.
\end{align*}
In particular, the factor $\alpha$ in the wall derivative
of the radical compensates for the factor $1/\alpha$ in
the outgoing wall response.

Finally~\eqref{eq:terminal-single-composition-coefficients}
bounds $\|Ds_h(z)\|$ by a uniform constant, since
$b,\alpha\leq1$. Combining this with $\|DE_h\|\leq A$
proves the operator bound in
\eqref{eq:terminal-single-composition-bounds}. Increase $C$
to cover the sum of the value and radial errors.
\end{proof}

The cancellation has now been retained through the actual
outgoing segment. To obtain the bounded phase of the next
core, the remaining comparison is between this single-wall
phase, the full stable projection at $Z_h$, and the subsequent
incoming arrival.