\subsection{Slow wall decay near a Taub point}
\label{sec:small-section-slow-decay}

We now consider positive shear when $n_2$ is very small and
$n_3$ is allowed to be of order $\varepsilon a$.
After the initial decay of $N_1$, the third wall can still be
too large in projective coordinates. Positive shear makes
this wall decrease at a rate of order $a$. Waiting for a time
of order $a^{-1}\log(1/a)$ therefore supplies the additional
decay. The initial size of $n_2$ must allow for its possible
growth during the same interval.

The estimates below keep the three wall rates separate.
Their integrals control the shear displacement and preserve
the quadratic size of the projection height $1+p$ throughout
this longer interval. All times in this subsection are the
original normalized time $t$ of \eqref{eq:normalized-flow}.

Write a state as $(p,s,n,x,y)=(p,s,N_1,N_2,N_3)$.
For $0<a\leq1/256$, consider the region
\begin{equation}
 \begin{gathered}
 G=1,\qquad p\leq-\frac12,\qquad
       \frac a2\leq s\leq3a,\\
 0\leq n\leq a^2,\qquad 0\leq x\leq a^2,
       \qquad 0\leq y\leq\frac a{16}.
 \end{gathered}
 \label{eq:slow-decay-region}
\end{equation}
The different bounds on $x$ and $y$ reflect their different
roles: the wall that can grow starts at the smaller scale.

\begin{lemma}
\label{lem:slow-decay-rates}
On \eqref{eq:slow-decay-region},
\begin{equation}
 \begin{gathered}
 |Q(n,x,y)|\leq\frac{a^2}{128},\qquad
       \frac{a^2}{5}\leq1-p^2\leq10a^2,\\
 -1<p\leq-\frac{15}{16},\qquad
       \frac{a^2}{10}\leq1+p\leq6a^2.
 \end{gathered}
 \label{eq:slow-decay-constraint}
\end{equation}
Writing the wall equations as
$\Dt n=\lambda_1n$, $\Dt x=\lambda_2x$, and
$\Dt y=\lambda_3y$, their rates satisfy
\begin{equation}
 \lambda_1\leq-2,\qquad |\lambda_2|\leq6a,
       \qquad \lambda_3\leq-\frac a2.
 \label{eq:slow-decay-wall-rates}
\end{equation}
The shear velocity satisfies
\begin{equation}
 |\Dt s|\leq3(x^2+y^2)+\frac a4 n.
 \label{eq:slow-decay-shear-rate}
\end{equation}
\end{lemma}

\begin{proof}
Nonnegativity of the walls gives
\[
 |Q(n,x,y)|\leq(n+x+y)^2,
 \qquad n+x+y\leq2a^2+\frac a{16}
                         \leq\frac{9a}{128}.
\]
This proves the first bound in
\eqref{eq:slow-decay-constraint}. The identity
$1-p^2=s^2+Q$ gives the next two bounds. Since $p<0$ and
$a\leq1/256$, these imply $-1<p\leq-15/16$.
Rationalizing
\[
 1+p=\frac{1-p^2}{1-p},\qquad
       \frac53\leq1-p\leq2,
\]
proves the height bounds. This argument also proves
\eqref{eq:slow-decay-constraint} with the weaker initial
assumption $p<0$ in place of $p\leq-1/2$.

On $G=1$, the three rates are
\[
 \lambda_1=-p^2-s^2+2p,\qquad
 \lambda_2=Q-(1+p)+\sqrt3s,\qquad
 \lambda_3=Q-(1+p)-\sqrt3s.
\]
The bounds just obtained, together with
$3/2\leq\sqrt3\leq7/4$, yield
\eqref{eq:slow-decay-wall-rates}.

For the shear equation, retain the factor $s$ multiplying
the strong-wall square. Directly from the curvature polynomial,
\begin{align*}
 |Q(n,x,y)|&\leq n^2+x^2+y^2+2n(x+y),\\
 |-x^2+y^2+nx-ny|&\leq x^2+y^2+n(x+y).
\end{align*}
Using $|s|\leq3a$, $\sqrt3\leq2$, and $6a\leq1$ in
\eqref{eq:normalized-flow} gives
\[
 |\Dt s|\leq3\bigl(x^2+y^2+n(x+y)+an^2\bigr).
\]
Finally, the wall bounds in
\eqref{eq:slow-decay-region} imply
\[
 3(x+y+an)\leq3\left(a^2+\frac a{16}+a^3\right)
                      \leq\frac a4,
\]
which proves \eqref{eq:slow-decay-shear-rate}.
\end{proof}

Suppose an orbit remains in \eqref{eq:slow-decay-region}
for $0\leq t\leq T$ and its initial walls are bounded by
nonnegative numbers $B_n,B_x,B_y$. Scalar comparison gives
\begin{equation}
 n(t)\leq B_ne^{-2t},\qquad
 x(t)\leq B_xe^{6at},\qquad
 y(t)\leq B_ye^{-at/2}.
 \label{eq:slow-decay-envelopes}
\end{equation}
These comparisons also hold when a wall vanishes, by its
multiplicative evolution equation. Inserting them into
\eqref{eq:slow-decay-shear-rate} and integrating yields
\begin{equation}
 \begin{split}
 |s(t)-s(0)|\leq E(t)
   :={}&\frac{3B_y^2}{a}(1-e^{-at})
       +\frac{B_x^2}{4a}(e^{12at}-1)\\
     &+\frac{aB_n}{8}(1-e^{-2t}).
 \end{split}
 \label{eq:slow-decay-drift}
\end{equation}
In particular,
\begin{equation}
 E(t)\leq\frac{3B_y^2}{a}
       +\frac{B_x^2}{4a}e^{12aT}+\frac{aB_n}{8}
       \quad(0\leq t\leq T).
 \label{eq:slow-decay-drift-budget}
\end{equation}
The decaying wall contributes at most $3B_y^2/a$ even when
$T$ is large. This is the bound that keeps the shear in
its original annulus.

\begin{proposition}
\label{prop:slow-decay-persistence}
Let $0<a\leq1/256$, $T\geq0$, and suppose
\begin{equation}
 0\leq B_n<a^2,\qquad
 0\leq B_x,\quad B_xe^{6aT}<a^2,\qquad
 0\leq B_y<\frac a{16}.
 \label{eq:slow-decay-initial-bounds}
\end{equation}
Let $z_0=(p_0,s_0,n_0,x_0,y_0)$ satisfy $G(z_0)=1$,
$p_0<0$, and
\begin{equation}
 \frac{3a}{4}\leq s_0\leq\frac{9a}{4},\qquad
 0\leq n_0\leq B_n,\quad
 0\leq x_0\leq B_x,\quad
 0\leq y_0\leq B_y.
 \label{eq:slow-decay-initial-state}
\end{equation}
Then the actual orbit $\phi_tz_0$ remains in
\eqref{eq:slow-decay-region} for $0\leq t\leq T$.
Throughout this interval it satisfies
\eqref{eq:slow-decay-constraint},
\eqref{eq:slow-decay-envelopes}, and
\eqref{eq:slow-decay-drift}.
\end{proposition}

\begin{proof}
The forward existence and wall-sign argument of
Section~\ref{sec:small-section-nonlinear} applies also to
zero walls. Thus the actual solution exists throughout
$[0,T]$, preserves the constraint, and has nonnegative walls.
The assumptions imply $B_x<a^2$. The constraint calculation
in Lemma~\ref{lem:slow-decay-rates}, using $p_0<0$, gives
$p_0\leq-15/16$. Hence the initial point satisfies strictly
the six inequalities
\[
 p<-\frac12,\qquad \frac a2<s<3a,\qquad
 n<a^2,\qquad x<a^2,\qquad y<\frac a{16}.
\]

On any initial interval where the corresponding non-strict
inequalities hold, the preceding comparison estimates apply.
The bound in \eqref{eq:slow-decay-drift-budget} is at most
\begin{equation*}
 \frac{3a}{256}+\frac{a^3}{4}+\frac{a^3}{8}
       \leq\frac a8.
\end{equation*}
Indeed, $B_y^2\leq a^2/256$ and
$B_x^2e^{12aT}\leq a^4$, while $B_n\leq a^2$.
Consequently,
\[
 \frac{5a}{8}\leq s(t)\leq\frac{19a}{8},
\]
strictly inside the prescribed shear interval.
Equations \eqref{eq:slow-decay-envelopes} and
\eqref{eq:slow-decay-initial-bounds} give strict upper
bounds for all three walls, and the constraint gives
$p(t)\leq-15/16<-1/2$.
Every defining inequality has therefore improved strictly.

The times up to which the closed inequalities hold form
a closed initial interval. If its endpoint were smaller
than $T$, the strict improvements and continuity would
extend it. The interval is thus all of $[0,T]$.
Applying the comparison estimates there proves the result.
\end{proof}

Return to the small section and write its data as
\begin{equation}
 z_\rho=\Xi_h(\rho,\zeta,\nu),\qquad a\leq\rho\leq2a,
 \qquad
 t_0=\frac q2\log(1/a),\qquad
 T=\frac D a\log(1/a).
 \label{eq:slow-decay-section-data}
\end{equation}
Here $\zeta$ is the initially small second wall, whereas
$\nu$ is the third wall that will decay during $[t_0,t_0+T]$.
The duration $T$ is chosen in advance for the whole family.
The identity
\begin{equation}
 e^{caT}=a^{-cD}\qquad(c\in\R)
 \label{eq:slow-decay-time-powers}
\end{equation}
converts the wall estimates into powers of the radial scale.

\begin{theorem}
\label{thm:small-section-slow-decay}
There exists $h_*>0$ such that, for every $0<h\leq h_*$,
$q>2$, $D\geq0$, and $P>6D+2$, the following holds for
all sufficiently small $a>0$, uniformly over
\begin{equation}
 0\leq\varepsilon\leq\frac1{128},\qquad
 0\leq\zeta\leq\min\{a^P,2\varepsilon a\},\qquad
 0\leq\nu\leq2\varepsilon a,\qquad
 a\leq\rho\leq2a.
 \label{eq:slow-decay-label-domain}
\end{equation}
Let
\[
 z(t)=\phi_{t_0+t}z_\rho=(p(t),s(t),n(t),x(t),y(t)),
       \qquad 0\leq t\leq T.
\]
This is the actual orbit $\phi_t(\phi_{t_0}z_\rho)$.
It lies in \eqref{eq:slow-decay-region}, and
\begin{equation}
 \begin{aligned}
 n(t)&\leq ha^q e^{-2t},
       &x(t)&\leq2\zeta e^{6at},
       &y(t)&\leq4\varepsilon a e^{-at/2}.
 \end{aligned}
 \label{eq:slow-decay-actual-walls}
\end{equation}
Moreover,
\begin{equation}
 \begin{split}
 |s(t)-s(0)|&\leq48\varepsilon^2a
          +\zeta^2a^{-12D-1}+\frac h8a^{q+1},\\
 \frac{a^2}{10}&\leq1+p(t)\leq6a^2,\qquad
       0\leq V(t)<1.
 \end{split}
 \label{eq:slow-decay-actual-height}
\end{equation}
At the endpoint, the projective walls satisfy
\begin{equation}
 \begin{split}
 0\leq U(T)&\leq20h a^{q-2},\\
 0\leq V(T)&\leq40\zeta a^{-6D-2},\\
 0\leq W(T)&\leq80\varepsilon a^{D/2-1}.
 \end{split}
 \label{eq:slow-decay-projective-endpoint}
\end{equation}
\end{theorem}

\begin{proof}
Choose $h_*$ as in
Proposition~\ref{prop:small-section-core-coordinates}.
For sufficiently small $a$, that proposition places
$z(0)=\phi_{t_0}z_\rho$ on $G=1$ with $p(0)<0$ and
\[
 \frac{3a}{4}\leq s(0)\leq\frac{9a}{4},\qquad
 0\leq n(0)\leq ha^q,\qquad
 0\leq x(0)\leq2\zeta,\qquad
 0\leq y(0)\leq2\nu\leq4\varepsilon a.
\]
Decrease the common threshold for $a$ so that
\begin{equation*}
 a\leq\frac1{256},\qquad
 ha^{q-2}<\frac12,\qquad
 40a^{P-6D-2}<\frac12.
\end{equation*}
Both exponents are positive. With
\[
 B_n=ha^q,\qquad B_x=2\zeta,
       \qquad B_y=4\varepsilon a,
\]
these inequalities give
\[
 B_n<a^2,\qquad
 B_xe^{6aT}=2\zeta a^{-6D}<\frac{a^2}{40},\qquad
 B_y\leq\frac a{32}<\frac a{16}.
\]
Also $T\geq0$. Proposition~\ref{prop:slow-decay-persistence}
therefore applies to $z(0)$ on the entire interval $[0,T]$.
Uniqueness of the normalized flow gives
$\phi_tz(0)=\phi_{t+t_0}z_\rho$, so all conclusions concern
the original small-section orbit.

The wall and height estimates follow immediately.
Substitution in \eqref{eq:slow-decay-drift-budget}, using
\eqref{eq:slow-decay-time-powers} with $c=12$, gives the
stated shear displacement. Throughout the interval,
$x(t)<a^2/40$; hence
\[
 0\leq V(t)=\frac{\sqrt3x(t)}{1+p(t)}
            \leq\frac{20x(t)}{a^2}<1.
\]

More generally, each nonnegative physical wall $v$ with
upper bound $B$ satisfies
\[
 0\leq\frac{\sqrt3v}{1+p}\leq\frac{20B}{a^2}.
\]
At $t=T$, use $e^{-2T}\leq1$ for the strong wall and
\eqref{eq:slow-decay-time-powers} with $c=6$ and $c=-1/2$
for the other two walls. This proves
\eqref{eq:slow-decay-projective-endpoint}.
\end{proof}

For $D>2$, the last endpoint bound tends to zero, and the
condition $P>6D+2$ makes the second one tend to zero as well.
Thus the longer interval reduces both projective weak walls
while keeping the shear of order $a$ and the height of order
$a^2$. The bound $V<1$ also places this entire interval
before any subsequent upward crossing of $V=1$. To use the
endpoint as a statistical entrance, we next control its
dependence on the original shear coordinate $\rho$.