\subsection{The reference flow and its radial phase}
\label{sec:small-section-transport}

Fix the section $N_1=h$ near $T_1=(-1,0)$.
The orbit with vanishing transverse coordinates provides an explicit
linear transport for the shear and the two weak walls. Its transported
radial phase has a positive derivative on the initial families used below.

\subsubsection{Section coordinates and the reference orbit}

Write $Z=(s,n_2,n_3)\in\R^3$ and define
\begin{equation}
 \begin{split}
 \mathcal D_h(Z)&=1-s^2-Q(h,n_2,n_3),\\
 \Xi_h(Z)&=\bigl(-\sqrt{\mathcal D_h(Z)},s,h,n_2,n_3\bigr),
 \end{split}
 \label{eq:small-section-chart}
\end{equation}
where $Q$ is the curvature polynomial in
\eqref{eq:phase-space}. Wherever $\mathcal D_h>0$, this is a smooth
chart on the negative shear branch of the constraint with
$N_1=h$. The projection
\[
 \Pi(p,s,N_1,N_2,N_3)=(s,N_2,N_3)
\]
satisfies $\Pi\Xi_h(Z)=Z$. We use the maximum norm on these
three coordinates and its induced operator norm.

For $|h|\leq1/16$ and $\|Z\|\leq1/16$, direct estimates of the
three squares and three products in $Q$ give
\begin{equation}
 |Q(h,n_2,n_3)|\leq\frac9{256},\qquad
 \frac{246}{256}\leq\mathcal D_h(Z)\leq\frac{265}{256},
 \qquad
 \frac{15}{16}\leq\sqrt{\mathcal D_h(Z)}\leq\frac{17}{16}.
 \label{eq:small-section-box}
\end{equation}
On the constraint the logarithmic rate of $N_1$ is
$Q-1+2p$. Thus, at $\Xi_h(Z)$,
\begin{equation}
 -4<Q(h,n_2,n_3)-1-2\sqrt{\mathcal D_h(Z)}<-2.
 \label{eq:small-section-transverse}
\end{equation}
For $h>0$, the flow crosses the section towards decreasing $N_1$.
The signed coordinate neighborhood in
\eqref{eq:small-section-box} also provides ordinary derivatives
at points where either of the two small walls vanishes.

Set
\begin{equation}
 \begin{aligned}
 z(n)&=\sqrt{1-n^2},\\
 F(n)&=n^2-1-2z(n)=-z(n)(2+z(n)),\\
 b(n)&=n^2-1+z(n)=z(n)(1-z(n)).
 \end{aligned}
 \label{eq:small-section-reference-rates}
\end{equation}
The identities on the right hold for $n^2<1$.
The reference solution is
$q_h(t)=\phi_t\Xi_h(0)$. Denote its strong wall by $n_h(t)$.
Here this is also the dominant wall in amplitude. The word
``strong'' refers to decay, while ``dominant'' refers to size.

\begin{lemma}
\label{lem:small-section-reference-orbit}
For $0<h\leq1/16$, the reference solution exists for every
$t\geq0$ and has the form
\begin{equation}
 q_h(t)=\bigl(-z(n_h(t)),0,n_h(t),0,0\bigr),\qquad
 \Dt n_h=n_hF(n_h),\qquad
 0<n_h(t)\leq h e^{-2t}.
 \label{eq:small-section-reference-orbit}
\end{equation}
\end{lemma}

\begin{proof}
The wall equations preserve $N_2=N_3=0$. On this invariant set
the shear equation is $\Dt s=N_1^2s$, so $s=0$ is preserved as
well. The constraint reduces to $p^2+N_1^2=1$, which bounds the
solution and gives forward continuation. Moreover,
\[
 \Dt p=N_1^2(p-2)\leq0,
 \qquad p(t)\leq p(0)=-z(h)<0.
\]
The constraint therefore selects $p(t)=-z(N_1(t))$ throughout.
The multiplicative wall equation preserves $N_1>0$ and gives
$\Dt N_1=N_1F(N_1)$.

Since $z(h)\geq15/16$, we have $p(t)\leq-15/16$.
The original wall rate $-p^2+2p$ is consequently at most $-2$.
Multiplication by $e^{2t}$ shows that
$N_1(t)\leq h e^{-2t}$, proving the assertion.
\end{proof}

Define three smooth coefficients for $n^2<1$ by
\begin{equation}
 A(n)=\frac3{2+z(n)},\qquad
 C(n)=\frac{9(1+z(n))}{2(2+z(n))^2},\qquad
 K(n)=\frac{3\sqrt3\,n}{(2+z(n))^2}.
 \label{eq:small-section-coefficients}
\end{equation}
The linear map
\begin{equation}
 \mathcal T(n)(s,n_2,n_3)
   =\bigl(A(n)s+K(n)(n_2-n_3),C(n)n_2,C(n)n_3\bigr)
 \label{eq:small-section-transfer}
\end{equation}
will express the transverse variation in coordinates conserved
along the reference orbit. In particular, $\mathcal T(0)$ is the
identity.

To see why these coefficients arise, differentiate the exact vector
field at $(-z(n),0,n,0,0)$. Its three transverse components form
the closed system
\begin{equation}
 \Dt\xi=n^2\xi+\sqrt3\,n(\alpha-\beta),\qquad
 \Dt\alpha=b(n)\alpha,\qquad
 \Dt\beta=b(n)\beta.
 \label{eq:small-section-reference-variation}
\end{equation}
The variations of $p$ and $N_1$ do not enter these equations.
Thus \eqref{eq:small-section-reference-variation} is the
transverse part of the actual flow differential.

\begin{lemma}
\label{lem:small-section-conservation}
Along any solution of
$\Dt n=nF(n)$ and
\eqref{eq:small-section-reference-variation} with $n^2<1$,
the three components of
$\mathcal T(n)(\xi,\alpha,\beta)$ are constant.
\end{lemma}

\begin{proof}
Since $z'=-n/z$, differentiation of
\eqref{eq:small-section-coefficients} gives
\[
 A'=\frac{3n}{z(2+z)^2},\qquad
 C'=\frac{9n}{2(2+z)^3},\qquad
 K'=\frac{3\sqrt3\{z(2+z)+2n^2\}}{z(2+z)^3}.
\]
Using $n^2=1-z^2$, these formulas yield
\begin{equation}
 nFA'+n^2A=0,\qquad
 nFC'+bC=0,\qquad
 nFK'+\sqrt3\,nA+bK=0.
 \label{eq:small-section-cancellations}
\end{equation}
For example, the derivative of $A\xi+K(\alpha-\beta)$ is
\[
 (nFA'+n^2A)\xi
   +(nFK'+\sqrt3\,nA+bK)(\alpha-\beta)=0.
\]
The other two conserved quantities follow from the middle
identity in \eqref{eq:small-section-cancellations}.
\end{proof}

For a fixed time, let
\begin{equation}
 Y_h(t)=D_Z\bigl(\Pi\phi_t\Xi_h\bigr)(0).
 \label{eq:small-section-flow-differential}
\end{equation}
The chart is smooth in an ordinary neighborhood of zero, and the
flow is differentiable in its initial state. Differentiating
$\Pi\Xi_h=\mathrm{id}$ gives $Y_h(0)=\mathrm{id}$.
Equation~\eqref{eq:small-section-reference-variation} and
Lemma~\ref{lem:small-section-conservation} therefore give the
finite identity
\begin{equation}
 \mathcal T(n_h(t))Y_h(t)=\mathcal T(h),
       \qquad t\geq0.
 \label{eq:small-section-finite-transfer}
\end{equation}

\begin{proposition}
\label{prop:small-section-reference-transfer}
For $0<h\leq1/16$ and $t\geq0$,
\begin{equation}
 \|Y_h(t)\|\leq2,\qquad
 \|Y_h(t)-\mathcal T(h)\|\leq6h e^{-2t}.
 \label{eq:small-section-reference-tail}
\end{equation}
In particular, the transverse differential converges to
$\mathcal T(h)$ as $t\to\infty$.
For $0<a\leq1$, $q\geq0$ and
$T=(q/2)\log(1/a)$,
\begin{equation}
 0<n_h(T)\leq h a^q,\qquad
 \|Y_h(T)-\mathcal T(h)\|\leq6h a^q.
 \label{eq:small-section-reference-cutoff}
\end{equation}
\end{proposition}

\begin{proof}
For $0\leq n\leq1/16$, one has $1/2\leq z\leq1$ and
$1-z\leq n$. The identities
\[
 A-1=\frac{1-z}{2+z},\qquad
 C-1=\frac{(1-z)(2z+1)}{2(2+z)^2}
\]
and $3\sqrt3\leq(2+z)^2$ imply
\begin{equation}
 0\leq A(n)-1\leq n,\qquad
 0\leq C(n)-1\leq n,\qquad
 0\leq K(n)\leq n.
 \label{eq:small-section-coefficient-bounds}
\end{equation}
For the maximum norm, the first coordinate of
$(\mathcal T(n)-\mathrm{id})Z$ is bounded by
$n|s|+n|n_2-n_3|\leq3n\|Z\|$, and each remaining
coordinate is bounded by $n\|Z\|$. Hence
\begin{equation*}
 \|\mathcal T(n)-\mathrm{id}\|\leq3n,\qquad
 \|\mathcal T(n)\|\leq1+3n.
\end{equation*}

Put $n=n_h(t)$. Equation~\eqref{eq:small-section-finite-transfer}
can be rewritten as
\[
 Y_h(t)-\mathcal T(h)
      =(\mathrm{id}-\mathcal T(n))Y_h(t).
\]
It follows that
$\|Y_h(t)\|\leq1+3h+3n\|Y_h(t)\|$.
Since $h,n\leq1/16$, this implies $\|Y_h(t)\|\leq2$.
The same identity now gives
$\|Y_h(t)-\mathcal T(h)\|\leq6n$.
Apply \eqref{eq:small-section-reference-orbit} to obtain
\eqref{eq:small-section-reference-tail}. Finally
$e^{-2T}=a^q$ proves \eqref{eq:small-section-reference-cutoff}.
\end{proof}

\subsubsection{The transported radius along the nonlinear flow}

The limiting transverse transfer suggests the quadratic form
\begin{equation}
 \mathcal Q(n;s,v)=(A(n)s+K(n)v)^2+C(n)^2v^2.
 \label{eq:small-section-quadratic}
\end{equation}
For $n^2<1$, both $A(n)$ and $C(n)$ are positive, so
$\mathcal Q(n;s,v)>0$ unless $s=v=0$. At $n=0$ it is
$s^2+v^2$. For $0\leq n\leq1/16$, the coefficient bounds
also imply
\begin{equation}
 s^2+v^2\leq2\mathcal Q(n;s,v).
 \label{eq:small-section-quadratic-lower}
\end{equation}
Indeed, $A,C\geq1$ and $2K^2\leq1$, and
\[
 s^2+v^2
 \leq2(As+Kv)^2+(1+2K^2)v^2
 \leq2(As+Kv)^2+2C^2v^2.
\]

We now compute its drift on the exact flow. At a constrained state
$x=(p,s,n,N_2,N_3)$, set
\begin{equation}
 \begin{gathered}
 v=N_2-N_3,\qquad w=N_2+N_3,\qquad
 \Delta_Q=v^2-2nw,\\
 \Delta_z=-p-z(n),\qquad
 \Delta_F=\Delta_Q-2\Delta_z.
 \end{gathered}
 \label{eq:small-section-differences}
\end{equation}
Here $Q=n^2+\Delta_Q$. If $p<0$ and $n^2<1$, the constraint
gives the useful rationalized expression
\begin{equation}
 \Delta_z=-\frac{s^2+\Delta_Q}{-p+z(n)}.
 \label{eq:small-section-root-difference}
\end{equation}
The denominator is positive. The factor $n$ in the term $-2nw$
remains explicit.

Substitution into \eqref{eq:normalized-flow} yields
\begin{equation}
 \begin{aligned}
  \Dt n&=n(F+\Delta_F),\\
  \Dt s&=n^2s+\sqrt3\,nv+e_s,
      &e_s&=\Delta_Qs-\sqrt3\,wv,\\
  \Dt v&=bv+e_v,
      &e_v&=(\Delta_Q+\Delta_z)v+\sqrt3\,ws.
 \end{aligned}
 \label{eq:small-section-perturbed-equations}
\end{equation}
All rates are in the original normalized time.

\begin{proposition}
\label{prop:small-section-quadratic-drift}
Along an exact constrained solution with $n^2<1$, define
\[
 \mathcal R(n;s,v)
     =AK(s^2-v^2)+(K^2+C^2-A^2)sv.
\]
Then
\begin{equation}
 \begin{split}
 \frac{\dd}{\dd t}\mathcal Q(n;s,v)
  ={}&n\Delta_F\,\partial_n\mathcal Q(n;s,v)
       +2\Delta_Q\,\mathcal Q(n;s,v)\\
    &+2\Delta_zv\{AKs+(K^2+C^2)v\}
       +2\sqrt3\,w\mathcal R(n;s,v).
 \end{split}
 \label{eq:small-section-quadratic-drift}
\end{equation}
Moreover, $\mathcal R(0;s,v)=0$.
\end{proposition}

\begin{proof}
Write $U=As+Kv$ and $V=Cv$. The cancellations
\eqref{eq:small-section-cancellations}, applied to
\eqref{eq:small-section-perturbed-equations}, give
\begin{equation*}
 \Dt U=n\Delta_F(A's+K'v)+Ae_s+Ke_v,\qquad
 \Dt V=n\Delta_F C'v+Ce_v.
\end{equation*}
Differentiate $\mathcal Q=U^2+V^2$ using these two identities.
The terms involving $\Delta_F$ give
$n\Delta_F\partial_n\mathcal Q$. Those involving $\Delta_Q$
give $2\Delta_Q(U^2+C^2v^2)$, and the terms involving
$\Delta_z$ give $2\Delta_zv(KU+C^2v)$.
The remaining terms are
\[
 2\sqrt3\,w\{U(-Av+Ks)+C^2vs\}
       =2\sqrt3\,w\mathcal R(n;s,v).
\]
This proves \eqref{eq:small-section-quadratic-drift}.
At zero, $A=C=1$ and $K=0$, so $\mathcal R=0$.
\end{proof}

In particular, the terms $-\sqrt3\,wv$ and $\sqrt3\,ws$
in \eqref{eq:small-section-perturbed-equations} act as a
rotation of $(s,v)$. Their contribution to the transported
radius vanishes when $n=0$. Keeping this cancellation before
estimating the drift will preserve the small factor supplied
by the decaying wall.

Define the reference radial phase on the section by
\begin{equation}
 \Phi_h(Z)=\frac1{\sqrt3}
       \sqrt{\mathcal Q(h;s,n_2-n_3)}.
 \label{eq:small-section-reference-phase}
\end{equation}
This is the Euclidean radius formed from the shear and wall
difference after the limiting linear transfer, divided by
$\sqrt3$. We will compare it with the phase of the actual
entrance.

Fix $\varsigma\in\{-1,1\}$ and vary the shear while keeping both
small walls fixed:
\[
 Z_\varsigma(\rho)=(\varsigma\rho,n_2,n_3).
\]
For $a>0$, $0\leq\varepsilon\leq1/64$ and
$0\leq n_2,n_3\leq2\varepsilon a$, one has, for
$a\leq\rho\leq2a$,
\begin{equation}
 \|Z_\varsigma(\rho)\|\leq2a,\qquad
 \sqrt{\rho^2+(n_2-n_3)^2}\geq a,\qquad
 |n_2-n_3|\leq2\varepsilon a.
 \label{eq:small-section-shear-annulus}
\end{equation}

\begin{proposition}
\label{prop:small-section-reference-slope}
If $0\leq h\leq1/16$ and the preceding conditions hold, then
$\rho\mapsto\Phi_h(Z_\varsigma(\rho))$ is ordinarily
differentiable at every $\rho\in[a,2a]$, and
\begin{equation}
 \frac1{2\sqrt3}
    \leq\frac{\dd}{\dd\rho}\Phi_h(Z_\varsigma(\rho))
    \leq\frac2{\sqrt3}.
 \label{eq:small-section-reference-slope}
\end{equation}
\end{proposition}

\begin{proof}
Let $d=n_2-n_3$, and use the Euclidean norm for vectors in
$\R^2$. Put
\[
 P=(A(h)\varsigma,0),\qquad
 W=(K(h)d,C(h)d).
\]
Then $\sqrt3\,\Phi_h(Z_\varsigma(\rho))=\|\rho P+W\|_2$.
By \eqref{eq:small-section-coefficient-bounds},
$\|P\|_2=A(h)\in[1,2]$ and $\|W\|_2\leq3|d|$.
Consequently,
\[
 3\|W\|_2\leq9|d|
       \leq18\varepsilon a
       \leq a
       \leq\rho\|P\|_2.
\]
The reverse triangle inequality gives
$\|\rho P+W\|_2\geq\rho\|P\|_2-\|W\|_2>0$,
so the norm is differentiable at the point in question.
Cauchy--Schwarz and the triangle inequality show that
\[
 \frac{\dd}{\dd\rho}\|\rho P+W\|_2
   =\frac{P\cdot(\rho P+W)}{\|\rho P+W\|_2}
   \leq\|P\|_2,
\]
whereas
\[
 \frac{P\cdot(\rho P+W)}{\|\rho P+W\|_2}
 \geq
 \frac{\rho\|P\|_2^2-\|P\|_2\|W\|_2}
      {\rho\|P\|_2+\|W\|_2}
 \geq\frac12\|P\|_2.
\]
Division by $\sqrt3$ proves
\eqref{eq:small-section-reference-slope}.
\end{proof}

The reference derivative is thus controlled for both signs of
the shear and for every relative size of the two small walls
in \eqref{eq:small-section-shear-annulus}. The next estimates
control the nonlinear drift over the initial interval of
length $(q/2)\log(1/a)$ and its ordinary derivatives. They will
allow this reference phase to be compared with an actual
core entrance.