\subsection{Variation of the remaining core passage}
\label{sec:partial-core-variation}

The initial state now moves in both radius and angle. To follow
its phase to the outgoing section, we keep these two variations
separate until the last crossing. The weighted variational equation
from Section~\ref{sec:core-variation} still has uniformly bounded
growth. Its initial condition acquires an angular correction, and
the integral of its strong component determines the resulting
change in the radial derivative.

For a projective state $y=(M,U,V,W)$ with $V>0$, write
\[
 \mathcal C(y)=(r(y),U,VW),\qquad
 r(y)=\sqrt{(M-1)^2+V^2},\qquad
 z(y)=\operatorname{arsinh}\frac{1-M}{V}.
\]
Let $\mathcal R_\downarrow$ be the actual downward return
defined in \eqref{eq:partial-core-oriented-return}. Denote its
radius component by $\mathcal R_{\downarrow,R}$, and put
\[
 K_v=\exp(2\pi+48\varepsilon_0),\qquad C_p=4096K_v.
\]
Thus $K_v$ is the growth constant in
\eqref{eq:core-variation-growth}.

\begin{theorem}
\label{thm:partial-core-radius-response}
Let $y_0$ satisfy \eqref{eq:partial-core-initial-state}, and set
$k=r(y_0)$, $u_0=U_0$, $P_0=V_0W_0$. For any ordinary
initial direction $w\in\R^4$, define
\[
 v=(v_R,v_U,v_P)=D\mathcal C(y_0)w,\qquad
 \delta z=Dz(y_0)w.
\]
Then
\begin{equation}
 \begin{split}
 |D\mathcal R_{\downarrow,R}(y_0)w-v_R|
 \leq C_p\biggl\{
    |v_U|+\frac{u_0+P_0}{k}(|v_R|+|v_P|)
       +\left(ku_0+\frac{P_0^2}{k}\right)|\delta z|
                       \biggr\}.
 \end{split}
 \label{eq:partial-core-radius-response-general}
\end{equation}
The estimate holds at interior entrances and at the incoming
boundary $V_0=1$, including states with a zero transverse wall.
\end{theorem}

The first two terms measure changes in the initial radius and
wall amplitudes. The last term accounts for moving the starting
point along the core. Its strong-wall coefficient is $ku_0$,
whereas its weak-wall coefficient is quadratic in $P_0$.
We retain this distinction when passing to the reciprocal radius.

\subsubsection{The moving initial angle}

Let $\mathcal Y=(R,U,P)$ along the reference passage, and let
$F(z,\mathcal Y)$ be the angular vector field in
\eqref{eq:core-angle-system}. Write $\alpha=z(y_0)$ and
$\beta=z(t_*)$ for its initial and final angles.
For the initial direction $w$, the variation at fixed
angle is $\xi=\delta\mathcal Y-\delta z\,F$, as in
\eqref{eq:fixed-angle-variation}. At the initial point,
\begin{equation}
 \xi(\alpha)=v-\delta z\,F(\alpha,k,u_0,P_0).
 \label{eq:partial-core-moving-initial}
\end{equation}
Here $v$ and $\delta z$ are derivatives of the actual initial
coordinates. The identity follows by differentiating the smooth
coordinate changes and the actual flow. In particular the
variational equation \eqref{eq:core-variational-equation}
applies to this $\xi$.
In the original time it reads $\dd\xi/\dd t=ARd\,J\xi$;
the angular integrals below use the positive weight
$\dd z=ARd\,\dd t$ along the actual orbit.

At a core point with $R=k$, the field satisfies
\begin{equation}
 |F_R|\leq2\left(ku_0+\frac{P_0^2}{k}\right),\qquad
 |F_U|\leq2ku_0,\qquad |F_P|\leq\frac{4P_0}{k}.
 \label{eq:partial-core-initial-field}
\end{equation}
For the first bound, use the radial integrand estimate
\eqref{eq:core-radius-integrand} at this point.
Since $1\leq V\leq k$ and $W=P_0/V\leq P_0$,
\[
 UV\leq ku_0,\quad UW\leq u_0/60000,\quad
 U^2/k\leq u_0/60000,\quad W^2/k\leq P_0^2/k.
\]
Division by $d\geq999/1000$ gives the stated bound.
The other two follow from the wall factors
$F_U=UL_U$, $F_P=PL_P$ and
\eqref{eq:core-log-rate-bounds}.

Recall $S_k\xi=(\xi_R,k\xi_U,\xi_P)$. As $P_0\leq1$ and
$k\geq6$, \eqref{eq:partial-core-initial-field} implies
\[
 \|S_kF(\alpha,k,u_0,P_0)\|_\infty
       \leq4(k^2u_0+P_0/k).
\]
Introduce the quantity
\begin{equation}
 Q_0=|v_R|+k|v_U|+|v_P|
                     +(k^2u_0+P_0/k)|\delta z|.
 \label{eq:partial-core-variation-size}
\end{equation}
The triangle inequality in
\eqref{eq:partial-core-moving-initial} gives
$\|S_k\xi(\alpha)\|_\infty\leq4Q_0$.
Lemma~\ref{lem:core-variation-growth}, applied on every
initial part of the passage, therefore gives
\begin{equation}
 \|S_k\xi(z)\|_\infty\leq Q,\qquad
 Q=4K_vQ_0,\qquad \alpha\leq z\leq\beta.
 \label{eq:partial-core-variation-growth}
\end{equation}
The radius $k$ in this weighting is fixed at the reference
initial state; it is not differentiated inside the linear
equation.

We first work with any bound $\|S_k\xi\|_\infty\leq Q$
on a part of the core. Set
\[
 x=k\xi_U,\qquad x_0=x(\alpha).
\]
The strong row of \eqref{eq:core-transverse-rows} and the
decay of $U$ give
\begin{equation}
 |x(z)|\leq e^{-k(z-\alpha)/4}
             \bigl(|x_0|+3ku_0Q(z-\alpha)\bigr).
 \label{eq:partial-core-strong-envelope}
\end{equation}
Indeed $x'=B_{UU}x+h$, with $B_{UU}\leq-k/4$ and
$|h|\leq3kUQ$. Multiplication by $e^{k(z-\alpha)/4}$
leaves a nonpositive damping coefficient and forcing at
most $3ku_0Q$. This is the scalar comparison used in
\eqref{eq:core-strong-variation}; it uses only the angle
difference and applies to the present initial point.
Its integral is
\begin{equation}
 \int_\alpha^\beta|x(z)|\dd z
       \leq\frac{4|x_0|+48u_0Q}{k}.
 \label{eq:partial-core-strong-integral}
\end{equation}
The constants follow by integrating $e^{-ks/4}$ and
$s e^{-ks/4}$ over $s\geq0$.

Put $\gamma=(101/99)^2$. The amplitude integrals needed
for the radius row are
\begin{equation}
 \int_\alpha^\beta U\dd z\leq\frac{4u_0}{k},\qquad
 \int_\alpha^\beta W^2\dd z\leq2\gamma P_0^2,\qquad
 \int_\alpha^\beta\frac{W}{kV}\dd z
                  \leq\frac{2\gamma P_0}{k}.
 \label{eq:partial-core-variation-integrals}
\end{equation}
These are the estimates in
\eqref{eq:core-amplitude-integrals}, with the initial
product $P_0$ in place of the entrance weak wall.
For clarity, $P\leq P_0$ and $R\geq99k/100$, while
$V\geq1$ gives
$\cosh\alpha,\cosh\beta\leq101k/100$. Thus the same
endpoint estimate for $\int\cosh^2z\dd z$ yields
$\int V^{-2}\dd z\leq2\gamma$. The last two integrals
follow from $W=P/V$.

\begin{lemma}
\label{lem:partial-core-radial-variation}
On every part of the passage on which
$\|S_k\xi\|_\infty\leq Q$,
\begin{equation}
 |\xi_R(\beta)-\xi_R(\alpha)|
       \leq256\left\{\frac{|x_0|}{k}
                         +\frac{(u_0+P_0)Q}{k}\right\}.
 \label{eq:partial-core-radial-variation}
\end{equation}
\end{lemma}

\begin{proof}
The radius row estimate
\eqref{eq:core-radial-variation-integrand} gives
\[
 \left|\frac{\dd\xi_R}{\dd z}\right|
 \leq
 \left(16U+\frac{8W^2}{k^2}+\frac{8W}{kV}\right)Q
       +3|x|.
\]
Here $2/\cosh z+1/k\leq3$.
Integrating with
\eqref{eq:partial-core-strong-integral} and
\eqref{eq:partial-core-variation-integrals} gives
\[
 \frac{12|x_0|}{k}
 +\frac{208u_0Q}{k}
 +\frac{16\gamma P_0^2Q}{k^2}
 +\frac{16\gamma P_0Q}{k}.
\]
Since $P_0\leq k$ and $\gamma<2$, this is bounded
by the right-hand side of
\eqref{eq:partial-core-radial-variation}.
\end{proof}

The strong contribution here is $|x_0|/k$.
For a complete core passage the incoming condition
$\cosh\alpha=k$ supplies the further factor $1/k$
in \eqref{eq:core-radial-variation}. Keeping the
present estimate allows the starting angle to vary
through the core.

\subsubsection{Projection onto the moving exit}

At the exit set
\[
 r_*=R(\beta),\qquad c_*=-\tanh\beta,\qquad
 f_*=F_R(\beta,\mathcal Y(\beta)).
\]
The endpoint
variation is tangent to $V=1$. The same elimination
as in \eqref{eq:core-exit-projection} gives
\begin{equation}
 D\mathcal R_{\downarrow,R}(y_0)w
       =s_*\xi_R(\beta),\qquad
 s_*=\frac{c_*}{c_*+f_*/r_*}.
 \label{eq:partial-core-exit-projection}
\end{equation}
This identity includes the derivative of the true
exit time. Its contribution is a multiple of the
flow vector, which is removed by the fixed-angle
projection.

At the downward exit, $c_*\leq-1/2$ and $f_*\leq0$.
The exit factor bounds proved in
Section~\ref{sec:core-variation}, together with
$U(\beta)\leq u_0$ and $W(\beta)=P(\beta)\leq P_0$,
give
\begin{equation}
 0<s_*\leq1,\qquad
 |s_*-1|\leq8\left(\frac{u_0}{k}
                              +\frac{P_0^2}{k^2}\right).
 \label{eq:partial-core-exit-factor}
\end{equation}
The use of the initial product follows from $V=1$
at the exit.

\begin{proof}[Proof of Theorem~\ref{thm:partial-core-radius-response}]
The actual return is $C^2$ by
Proposition~\ref{prop:partial-core-smooth-exit}.
Equations \eqref{eq:partial-core-moving-initial} and
\eqref{eq:partial-core-initial-field} imply
\begin{align}
 |\xi_R(\alpha)-v_R|
       &\leq2(ku_0+P_0^2/k)|\delta z|,\notag\\
 |x_0|&\leq k|v_U|+2k^2u_0|\delta z|.
 \label{eq:partial-core-initial-component-bounds}
\end{align}
Write
\[
 N=|v_U|+\frac{u_0+P_0}{k}(|v_R|+|v_P|)
                      +(ku_0+P_0^2/k)|\delta z|.
\]
The core bounds imply $u_0+P_0\leq1$ and $P_0\leq k$.
They yield
\begin{equation}
 \frac{(u_0+P_0)Q_0}{k}\leq2N,\qquad
 \frac{|x_0|}{k}\leq2N,\qquad
 |\xi_R(\alpha)-v_R|\leq2N.
 \label{eq:partial-core-budget-absorption}
\end{equation}
To verify the first inequality, the $k|v_U|$ term
contributes at most $|v_U|$, and
\[
 \frac{u_0+P_0}{k}(k^2u_0+P_0/k)
                  \leq2(ku_0+P_0^2/k).
\]
For this last bound, use $u_0+P_0\leq1$ in the first
product, then $u_0P_0/k^2\leq ku_0$ and
$P_0^2/k^2\leq P_0^2/k$.
The other two inequalities follow from
\eqref{eq:partial-core-initial-component-bounds}.
Also \eqref{eq:partial-core-exit-factor} gives
\[
 |s_*-1|\,|v_R|
       \leq8\frac{u_0+P_0}{k}|v_R|\leq8N.
\]

Use $Q=4K_vQ_0$ in
\eqref{eq:partial-core-radial-variation}.
The first two inequalities in
\eqref{eq:partial-core-budget-absorption} give
\[
 |\xi_R(\beta)-\xi_R(\alpha)|
                    \leq(512+2048K_v)N.
\]
Because $0<s_*\leq1$, the exact exit identity implies
\begin{align*}
 |s_*\xi_R(\beta)-v_R|
 &\leq|\xi_R(\beta)-\xi_R(\alpha)|
       +|\xi_R(\alpha)-v_R|+|s_*-1|\,|v_R|\\
 &\leq(522+2048K_v)N\leq4096K_vN.
\end{align*}
This proves the theorem.
\end{proof}

Let $Y(\rho)$ be an ordinarily differentiable curve of
projective initial states, and set
\[
 k(\rho)=r(Y(\rho)),\quad U(\rho)=Y_U(\rho),\quad
 \Pin(\rho)=Y_V(\rho)Y_W(\rho),\quad z(\rho)=z(Y(\rho)).
\]
Its outgoing radius is
\[
 f(\rho)=\mathcal R_{\downarrow,R}(Y(\rho)).
\]
The following statement is local at the parameter in
question; nearby points need only lie in the ordinary
domain of the downward return.

\begin{proposition}
\label{prop:partial-core-radial-family}
Suppose $Y(\rho)$ satisfies
\eqref{eq:partial-core-initial-state}. Let
$0<a\leq1$ and $K\geq1$, and assume at this parameter
that
\begin{equation}
 \begin{gathered}
 \frac2{Ka}\leq k\leq\frac K a,\qquad
 |k'|\leq\frac K{a^2},\\
 |U'|\leq\frac{KU}{a},\qquad
 |\Pin'|\leq\frac{K\Pin}{a},\qquad
 |z'|\leq\frac K a.
 \end{gathered}
 \label{eq:partial-core-radial-family-input}
\end{equation}
Then $f$ is ordinarily differentiable,
$k/2\leq f\leq2k$, and
\begin{align}
 |f-k|&\leq16K(U+a\Pin^2),\notag\\
 |(f-k)'|&\leq4C_pK^2\,\frac{U+a\Pin}{a^2}.
 \label{eq:partial-core-family-radius-error}
\end{align}
For the reciprocal phase,
\begin{align}
 \left|\frac2f-\frac2k\right|
       &\leq32K^3a^2(U+a\Pin^2),\notag\\
 \left|\left(\frac2f-\frac2k\right)'\right|
       &\leq(8C_pK^4+64K^5)(U+a\Pin).
 \label{eq:partial-core-family-phase-error}
\end{align}
\end{proposition}

\begin{proof}
The chain rule and
Proposition~\ref{prop:partial-core-smooth-exit}
give ordinary differentiability.
The radius bounds and drift in
\eqref{eq:partial-core-prefix-estimates} give
$k/2\leq f\leq2k$ and
\[
 |f-k|\leq16(U+\Pin^2/k)\leq16K(U+a\Pin^2),
\]
where $1/k\leq Ka$.
Apply Theorem~\ref{thm:partial-core-radius-response}
with $w=Y'(\rho)$. Its four input derivatives become
$k'$, $U'$, $\Pin'$ and $z'$ by the chain rule.
Put $B=(U+a\Pin)/a^2$. Since $a,\Pin\leq1$,
\[
 \frac Ua\leq B,\qquad
 \frac{U+\Pin}{a}\leq B,\qquad
 (U+\Pin)\Pin\leq B,\qquad
 \frac U{a^2}+\Pin^2\leq B.
\]
These inequalities and
\eqref{eq:partial-core-radial-family-input} bound each
of the four terms
\[
 |U'|,\quad \frac{U+\Pin}{k}|k'|,\quad
 \frac{U+\Pin}{k}|\Pin'|,\quad
 (kU+\Pin^2/k)|z'|
\]
by $K^2B$. Their sum proves the derivative estimate
in \eqref{eq:partial-core-family-radius-error}.

For the reciprocal error set $e=f-k$ and
$\psi=2/f-2/k$. The exact identities are
\begin{equation*}
 \psi=-\frac{2e}{kf},\qquad
 \psi'=-\frac{2e'}{f^2}
              +\frac{2e(f+k)k'}{k^2f^2}.
\end{equation*}
Both $f$ and $k$ are at least $1/(Ka)$.
Consequently
\[
 |\psi|\leq2K^2a^2|e|,\qquad
 \frac{f+k}{k^2f^2}
       =\frac1{k^2f}+\frac1{kf^2}\leq2(Ka)^3.
\]
The value estimate follows at once. For the derivative,
\[
 |\psi'|\leq2K^2a^2|e'|+4K^4a|e|.
\]
Insert \eqref{eq:partial-core-family-radius-error}
and use
$a(U+a\Pin^2)\leq U+a\Pin$, which follows from $a\leq1$
and $a\Pin\leq1$. This proves
\eqref{eq:partial-core-family-phase-error}.
\end{proof}

\subsubsection{Application to the small section}

\begin{corollary}
\label{cor:small-section-partial-core-response}
There are absolute constants $h_*>0$ and $K\geq1$
such that for every $0<h\leq h_*$, $q>3$ and
$\ell>4$, all sufficiently small scales $a>0$
have the following property.
Consider the positive-shear family
\eqref{eq:radial-shear-family} with $n_3\leq a^\ell$,
at the fixed time $T=(q/2)\log(1/a)$.
At every $\rho\in[a,2a]$ for which its actual core
state $Y(\rho)$ has $V\geq1$, the outgoing radius
$f=\mathcal R_{\downarrow,R}\circ Y$ satisfies
\eqref{eq:partial-core-family-radius-error} and
\eqref{eq:partial-core-family-phase-error}.
These are ordinary derivatives along the original
shear curve, with $h,a,q,n_2,n_3$ fixed.
\end{corollary}

\begin{proof}
Take the common small-$h$ threshold of
Propositions~\ref{prop:small-section-coordinate-response}
and \ref{prop:small-section-core-data}.
Choose
\[
 K=2000+K_0+K_1
\]
before $h,q,\ell$, and take the small-$a$ thresholds
from those propositions, also requiring $a\leq1$.
The radius bounds give
\[
 \frac2{Ka}\leq\frac1{16a}\leq k
      \leq\frac{32}{a}\leq\frac Ka,\qquad
 |k'|\leq\frac{1536}{a^2}\leq\frac K{a^2}.
\]
The wall and angle bounds give the other three
inequalities in
\eqref{eq:partial-core-radial-family-input}.
Choose the entry bound in
\eqref{eq:partial-core-entry-smallness} with
$\eta=\varepsilon_0$. When $V\geq1$, the state
satisfies \eqref{eq:partial-core-initial-state}.
Proposition~\ref{prop:partial-core-radial-family}
now applies, including at $V=1$.
The flow identity at the end of
Section~\ref{sec:small-section-partial-core}
identifies this endpoint with the outgoing event on
the original orbit, at time
$T+\tau_\downarrow(Y(\rho))$.
\end{proof}

Thus the radial and angular derivatives supplied by
the initial decay control the phase at the actual
outgoing section. The remaining derivative estimates
concern the two wall amplitudes. Their multiplier
form will retain the small factors needed to pass
from this exit to the next incoming section.