\subsection{Variations across long connections}
\label{sec:connection-variation}

The matching phase also controls nearby orbits. Its usefulness at the
level of derivatives comes from two facts: its rate retains a small
wall factor, and the transverse linear equation has bounded growth
throughout a connection, however long the waiting time. We first
establish these facts in coordinates adapted to the phase. We then
treat variations of the entrance phase and of the two incoming walls.
The growing wall requires separate attention because it changes the
length of the logarithmic clock interval.

Throughout this subsection the entrance satisfies
\eqref{eq:connection-entry}, with $u>0$ as well as $w>0$. All three
walls are therefore positive on the finite passage. We use $s$ for
the clock $\ell$, write $L=\log(1/w)$, and retain $P=VW$.
The quantities $S,B,\rho_\pm$ are those in
\eqref{eq:connection-phase-coefficients}.

\subsubsection{The phase coordinates and the actual variation}

Since $\partial_mJ=S\geq3/4$, the variables
$(J,\log U,\log V,\log W)$ form local coordinates. In these
coordinates the equation is
\begin{equation}
 J'=G,\qquad (\log U)'=f,\qquad (\log V)'=g,\qquad
 (\log W)'=1,
 \label{eq:connection-log-coordinate-system}
\end{equation}
where
\begin{equation}
 f=\frac aq,\qquad g=-1-\frac{2(1+C)}q,\qquad
 G=U\alpha+P^2\beta,
 \label{eq:connection-log-rates}
\end{equation}
and the cancellation in~\eqref{eq:connection-phase-cancellation} gives
\begin{equation}
 \alpha=-\frac{S(V-W)+(V+W-U)B}{q},\qquad
 \beta=\frac{K_+-K_-}{q}.
 \label{eq:connection-factored-rate}
\end{equation}

We identify the variations in this equation with derivatives of the
original flow. Let $Y=(m,U,V,W)$, let $F(Y)$ denote the right-hand
side of~\eqref{eq:connection-clock-system}, and let $\eta$ be a
variation of $Y$ at fixed original time. For an entrance direction
$v=(v_\mu,v_u,v_w)$, put $\zeta=v_w/w$ and
\begin{equation}
 \xi=\eta-\left(\frac{\eta_W}{W}-\zeta\right)F(Y).
 \label{eq:connection-relative-clock-variation}
\end{equation}
This compares orbits at the same value of $\log(W/w)$. In particular,
$\xi_W=\zeta W$, and its initial value is
$(v_\mu,v_u,0,v_w)$.

For completeness, set $h=Aq=d\ell/dt$. The original equation is
$dY/dt=hF$. Its variation gives
$d\eta/dt=\delta h\,F+hDF\eta$. Also
$d(\eta_W/W)/dt=\delta h$, by varying $d\log W/dt=h$.
Differentiating~\eqref{eq:connection-relative-clock-variation}
therefore cancels the $\delta h$ term and yields
$d\xi/dt=hDF\xi$. The strictly increasing clock gives
$\xi'=DF\xi$ on the entire passage.

Write the induced variation in the phase coordinates as
\begin{equation}
 Z=\delta J=S\xi_m+\frac V{\rho_+}\xi_V
                        +\frac W{\rho_-}\xi_W,
 \qquad x=\frac{\xi_U}{U},\qquad y=\frac{\xi_V}{V}.
 \label{eq:connection-phase-variation}
\end{equation}
Its initial values are
\begin{equation}
 (Z,x,y,\zeta)(0)=
 \left(S(\mu,1,w)v_\mu+\frac w{H(\mu-1,w)}v_w,
              \frac{v_u}{u},0,\frac{v_w}{w}\right).
 \label{eq:connection-variation-entry}
\end{equation}
Thus the logarithmic wall variations may be large at the entrance.
Their coefficients in the phase equation will retain the corresponding
small wall amplitudes.

To specify these coefficients, introduce differentiation at fixed
values of the other phase coordinates:
\begin{equation}
 \begin{aligned}
 \mathcal D_J&=S^{-1}\partial_m,&
 \mathcal D_U&=U\partial_U,\\
 \mathcal D_V&=V\partial_V+e_V\partial_m,&
 \mathcal D_W&=W\partial_W+e_W\partial_m,
 \end{aligned}
 \qquad
 e_V=-\frac{V^2}{\rho_+S},\quad
 e_W=-\frac{W^2}{\rho_-S}.
 \label{eq:connection-coordinate-derivatives}
\end{equation}
For any function $h$, write $h_J=\mathcal D_Jh$ and similarly for
the other three subscripts. The exact variational equation is
\begin{equation}
 \begin{pmatrix}Z\\x\\y\end{pmatrix}'=
 \begin{pmatrix}
 G_J&G_U&G_V&G_W\\ f_J&f_U&f_V&f_W\\ g_J&g_U&g_V&g_W
 \end{pmatrix}
 \begin{pmatrix}Z\\x\\y\\\zeta\end{pmatrix},
 \qquad \zeta'=0.
 \label{eq:connection-variational-system}
\end{equation}

\begin{lemma}[Coefficients retaining the wall amplitudes]
\label{lem:connection-variation-coefficients}
In the region~\eqref{eq:connection-region}, the phase row satisfies
\begin{align}
 |G_J|&\leq64\left(\frac U{\mu^2}+\frac{P^2}{\mu^4}\right),&
 |G_U|&\leq32U,\notag\\
 |G_V|&\leq32\left(UV+\frac{P^2}{\mu^3}\right),&
 |G_W|&\leq32\left(UW+\frac{P^2}{\mu^3}\right).
 \label{eq:connection-phase-row}
\end{align}
The two amplitude rows satisfy
\begin{align}
 |f_J|&\leq32,& |f_U|&\leq32U,\notag\\
 |f_V|&\leq32\left(\frac{V^2+P}{\mu}+UV\right),&
 |f_W|&\leq32\left(\frac{W^2+P}{\mu}+UW\right),
 \label{eq:connection-strong-row}\\
 |g_J|&\leq\frac{64}{\mu^2},&
 |g_U|&\leq\frac{64U}{\mu},\notag\\
 |g_V|&\leq64\left(\frac{V^2}{\mu^3}+\frac{UV}{\mu}\right),&
 |g_W|&\leq64\left(\frac{W^2}{\mu^3}+\frac{UW}{\mu}\right).
 \label{eq:connection-weak-row}
\end{align}
\end{lemma}

\begin{proof}
The coordinate derivatives have
\begin{equation}
 S^{-1}\leq\frac43,\qquad
 |e_V|\leq\frac{3V^2}{\mu},\qquad
 |e_W|\leq\frac{3W^2}{\mu}.
 \label{eq:connection-inverse-phase-bounds}
\end{equation}
Here $\rho_+\geq\mu$ and $\rho_-\geq5\mu/8$.
We also have $q\geq\mu/2$, $|1+C|\leq2$, $|f|\leq2\mu$,
$|m+f|\leq4\mu$, and $|V+W-2U|\leq3$.
For a variation $(Z,x,y,\zeta)$, put
\[
 \delta m=S^{-1}Z+e_Vy+e_W\zeta,\qquad
 \delta C=U(V+W-2U)x+UVy+UW\zeta.
\]
The quotient rule gives
\begin{align*}
 \delta f&=\frac{-(m+f)\delta m-(V-W)(Vy-W\zeta)
                         +U^2x+f\delta C}{q},\\
 \delta g&=\frac{2(1+C)\delta m-2m\delta C}{q^2}.
\end{align*}
Substitution of~\eqref{eq:connection-inverse-phase-bounds} gives
\eqref{eq:connection-strong-row}--\eqref{eq:connection-weak-row}.
For example, the numerator of $f_V$ has absolute value at most
$12V^2+V(V+W)+2\mu UV$; division by $q$ gives
$26V^2/\mu+2P/\mu+4UV$, within the stated bound.
The weak wall columns are bounded in the same calculation by
$48V^2/\mu^3+16UV/\mu$ and its $W$ counterpart.

For the phase row, differentiating the factored expression
\eqref{eq:connection-factored-rate} preserves the cancellation.
Let $\mathcal K=K_+-K_-$. The elementary kernel
$k(a,b)=[H(a,b)(H(a,b)+a)]^{-1}$ has derivatives
\[
 k_a=-H^{-3},\qquad
 k_b=-\frac{b(2H+a)}{H^3(H+a)^2}.
\]
For $a\geq d>0$, these imply
$0\leq k\leq1/(2d^2)$, $|k_a|\leq d^{-3}$ and
$|bk_b|\leq3b^2/(4d^4)$. Applying these bounds at the two
arguments, and then using~\eqref{eq:connection-coordinate-derivatives},
gives
\[
 |\mathcal K|\leq\frac3{\mu^2},\quad
 |\partial_m\mathcal K|\leq\frac9{\mu^3},\quad
 |\mathcal K_J|\leq\frac{12}{\mu^3},\quad
 |\mathcal K_V|\leq\frac{32V^2}{\mu^4},\quad
 |\mathcal K_W|\leq\frac{32W^2}{\mu^4}.
\]
Indeed, each wall derivative consists of
$e_V\partial_m\mathcal K$ or $e_W\partial_m\mathcal K$ and one
kernel wall derivative, bounded respectively by $27r^2/\mu^4$
and $5r^2/\mu^4$. Since
\[
 (q_J,q_U,q_V,q_W)
  =\bigl(S^{-1},-U(V+W-2U),e_V-UV,e_W-UW\bigr),
\]
the quotient $\beta=\mathcal K/q$ satisfies
\begin{align*}
 |\beta|&\leq\frac6{\mu^3},&
 |\beta_J|&\leq\frac{40}{\mu^4},&
 |\beta_U|&\leq\frac{36U}{\mu^4},\\
 |\beta_V|&\leq\frac{100V^2}{\mu^5}+\frac{12UV}{\mu^4},&&
 |\beta_W|&\leq\frac{100W^2}{\mu^5}+\frac{12UW}{\mu^4}.
\end{align*}

The phase derivative of $\alpha$ needs one further cancellation.
The identity
\[
 B=J-\frac{m+1}{\rho_+}+\frac{m-1}{\rho_-}
\]
gives
\begin{gather*}
 S_m=\frac{V^2}{\rho_+^3}+\frac{W^2}{\rho_-^3},\qquad
 B_m=S-\frac{V^2}{\rho_+^3}+\frac{W^2}{\rho_-^3},\\
 0\leq S_m\leq\frac{9(V^2+W^2)}{2\mu^3},\qquad
 |B_m-1|\leq\frac{3(V^2+W^2)}{\mu^2}.
\end{gather*}
Moreover,
\[
 B-m=\frac{V^2(\rho_++1)}{\rho_+(\rho_++m+1)}
       +\frac{W^2(\rho_--1)}{\rho_-(\rho_-+m-1)}.
\]
Both terms are nonnegative, and their sum is at most
$8(V^2+W^2)/(5\mu)\leq4/5$. It follows that $0\leq B-q\leq2$.
Set
\[
 N_0=S(V-W)+(V+W-U)(B-q).
\]
Then $|N_0|\leq6$, $|(N_0)_m/S|\leq20/\mu^2$, and
\[
 \alpha=-(V+W-U)-\frac{N_0}{q},\qquad
 \alpha_J=-\frac{(N_0)_m}{qS}+\frac{N_0}{q^2S}.
\]
Consequently $|\alpha_J|\leq40/\mu^3+32/\mu^2\leq48/\mu^2$.

For the other columns, let $N=S(V-W)+(V+W-U)B$, so
$\alpha=-N/q$. Differentiation of the displayed formulas for $S,B$
gives, for $r=V,W$,
\[
 |S_r|,|B_r|\leq\frac{5r^2}{\mu^2},\qquad
 |N|\leq4\mu,\quad |N_U|\leq\frac32\mu U,\quad
 |N_r|\leq3\mu r.
\]
For clarity, the wall derivatives of $B$ are
\[
 B_V=\frac{(m+1)V^2}{\rho_+^3}+e_V(B_m-S),\qquad
 B_W=-\frac{(m-1)W^2}{\rho_-^3}+e_W(B_m-S).
\]
The quotient rule now yields
\begin{equation}
 |\alpha|\leq8,\qquad |\alpha_U|\leq16U,
 \qquad |\alpha_V|\leq16V,\qquad |\alpha_W|\leq16W.
 \label{eq:connection-strong-phase-factor-bounds}
\end{equation}
Finally,
\[
 G_J=U\alpha_J+P^2\beta_J,\qquad
 G_U=U\alpha_U+U\alpha+P^2\beta_U,
\]
and $G_r=U\alpha_r+P^2\beta_r+2P^2\beta$ for $r=V,W$.
The phase column is bounded by $48U/\mu^2+40P^2/\mu^4$.
For a wall column, the last two terms have sum at most
$32P^2/\mu^3$, using $\mu\geq4$ and $U,V,W\leq1$.
Together with~\eqref{eq:connection-strong-phase-factor-bounds},
these prove~\eqref{eq:connection-phase-row}.
\end{proof}

\subsubsection{Bounded transverse growth}

The phase column of the weak amplitude equation is of order $\mu^{-2}$,
whereas that of the strong equation is bounded independently of $\mu$.
Set $X=(x,\mu^2y)$, with the maximum norm on $\R^2$, and write
\begin{equation}
 X'=TX+Zc+\zeta d,\qquad
 T=\begin{pmatrix}f_U&f_V/\mu^2\\ \mu^2g_U&g_V\end{pmatrix},
 \quad c=\binom{f_J}{\mu^2g_J},\quad
 d=\binom{f_W}{\mu^2g_W}.
 \label{eq:connection-weighted-system}
\end{equation}
The coefficient bounds give $\|c\|\leq64$ and bound each
absolute row sum of $T$ by
\[
 r(s)=64\left(2\mu U+\frac{V^2}{\mu^3}+\frac P{\mu^3}\right).
\]
Its integral is controlled by the decreasing quantity
\begin{equation}
 \mathcal E(s)=64\left(4U+\frac{V^2}{2\mu^3}
                               +\frac P{2\mu^2}\right).
 \label{eq:connection-transverse-potential}
\end{equation}
In fact $U'\leq-\mu U/2$, $V'\leq-V$, and
$P'\leq-2P/\mu$ imply $\mathcal E'\leq-r$. At the entrance,
\[
 0\leq\mathcal E(0)\leq\frac{2533}{5000}<\log2.
\]
The last comparison follows, for example, from
$2533/5000<2/3$ and $e^{2/3}<2$.

\begin{lemma}[Growth on every subinterval]
\label{lem:connection-transverse-growth}
If $X'=TX+b$ on $[s,t]\subset[0,L]$, with continuous forcing
$b$, then
\begin{equation}
 \|X(t)\|\leq2\left(\|X(s)\|+\int_s^t\|b(r)\|\dd r\right).
 \label{eq:connection-transverse-growth}
\end{equation}
\end{lemma}

\begin{proof}
The right upper derivative of the maximum norm is at most
$r\|X\|+\|b\|$. This follows by an Euler step in each component,
as in the core variation estimate; the diagonal term can retain its
sign, and its signed row sum is bounded by the absolute row sum.
Multiplication by $e^{\mathcal E}$, using
$\mathcal E'\leq-r$, gives
\[
 \|X(t)\|\leq e^{\mathcal E(s)-\mathcal E(t)}\|X(s)\|
       +\int_s^t e^{\mathcal E(r)-\mathcal E(t)}\|b(r)\|\dd r.
\]
The nonnegative function $\mathcal E$ is decreasing and
$e^{\mathcal E(0)}<2$, proving the result.
\end{proof}

The bound is independent of $L$. The phase equation couples back to
$X$, but its total effect is small. To quantify it, the zero and first
clock moments of the amplitudes suffice:
\begin{equation}
 \begin{aligned}
 \int_0^s U\dd r&\leq\frac{2u}{\mu},&
 \int_0^s rU\dd r&\leq\frac{4u}{\mu^2},\\
 \int_0^s P^2\dd r&\leq\frac{\mu w^2}{4},&
 \int_0^s rP^2\dd r&\leq\frac{\mu^2w^2}{16}.
 \end{aligned}
 \label{eq:connection-clock-moments}
\end{equation}
Indeed, if $h'\leq-ch$ and $h\geq0$, integration gives
$\int_0^s h\leq h(0)/c$. Differentiating $(1+cr)h(r)$ gives
an upper bound $-c^2rh(r)$, so
$\int_0^s rh(r)\dd r\leq h(0)/c^2$.
Apply this to $U$ with $c=\mu/2$, and to $P^2$ with $c=4/\mu$.

For $\zeta=0$, the phase row in
Lemma~\ref{lem:connection-variation-coefficients} gives
\begin{equation}
 |Z'|\leq A_0|Z|+B_0\|X\|,
 \quad A_0=64\left(\frac U{\mu^2}+\frac{P^2}{\mu^4}\right),
 \quad B_0=64\left(U+\frac{P^2}{\mu^5}\right).
 \label{eq:connection-phase-feedback-inequality}
\end{equation}
Here the transverse row bound is
$|G_U|+|G_V|/\mu^2\leq B_0$; keeping the factor $32$ in the
two wall columns proves this bound directly.
Define
\begin{equation}
 k(s)=A_0(s)+128sB_0(s),\qquad
 \Lambda=\frac{128u}{\mu^3}+\frac{32768u}{\mu^2}
                         +\frac{528w^2}{\mu^3}.
 \label{eq:connection-feedback-integral}
\end{equation}
The four moments give $\int_0^s k\leq\Lambda$ for every $s\leq L$.
Moreover $\Lambda\leq1/16$, by $\mu\geq4$,
$u\leq1/40000$ and $w^2\leq10^{-8}$.

We will repeatedly use the following elementary absorption argument.
Suppose that, on a compact interval, every bound $|Z|\leq Q$ implies
$|Z'|\leq e+kQ$, where $e,k\geq0$ and $\int k\leq\Lambda<1$.
Choose $Q=\max|Z|$ on this interval and integrate up to a point
where the maximum is attained. Then
\begin{equation}
 \max|Z|\leq\frac{|Z(0)|+\int e}{1-\Lambda},\qquad
 |Z(s)-Z(0)|\leq\int e+
       \Lambda\frac{|Z(0)|+\int e}{1-\Lambda}.
 \label{eq:connection-feedback-absorption}
\end{equation}
In our application $(1-\Lambda)^{-1}\leq16/15$.

\subsubsection{Entrance derivatives and the moving arrival}

\begin{proposition}[Two parameter responses]
\label{prop:connection-phase-strong-responses}
Let $(Z_\mu,x_\mu,y_\mu)$ and $(Z_u,x_u,y_u)$ denote the
fixed-clock variations induced respectively by the ordinary entrance
directions $(1,0,0)$ and $(0,1,0)$. Put $S_0=S(\mu,1,w)$.
For every $0\leq s\leq L$,
\begin{equation}
 \begin{gathered}
 |Z_\mu(s)|\leq\frac{16}{15},\qquad
 |Z_\mu(s)-S_0|\leq\frac{16}{15}\Lambda,\\
 \|(x_\mu,\mu^2y_\mu)(s)\|\leq\frac{2048}{15}s,
 \end{gathered}
 \label{eq:connection-phase-response}
\end{equation}
\begin{equation}
 |Z_u(s)|\leq\frac{70}{\mu},\qquad
 \left\|(x_u,\mu^2y_u)(s)-(u^{-1},0)\right\|
       \leq256+\frac{8960s}{\mu}.
 \label{eq:connection-strong-response}
\end{equation}
\end{proposition}

\begin{proof}
In the phase direction $X(0)=0$ and $\zeta=0$.
Lemma~\ref{lem:connection-transverse-growth} gives
\[
 \|X(s)\|\leq128\int_0^s|Z(r)|\dd r.
\]
If $|Z|\leq Q$ on the interval, this is at most $128sQ$.
Equation~\eqref{eq:connection-phase-feedback-inequality} now gives
$|Z'|\leq kQ$. Apply~\eqref{eq:connection-feedback-absorption}
with $e=0$, and use $0<S_0\leq1$. This proves
\eqref{eq:connection-phase-response}.

For the strong direction $Z(0)=0$, $\zeta=0$ and
$X(0)=(u^{-1},0)$. Subtract this constant vector before estimating
the transverse equation. Its extra forcing satisfies
\[
 \|T(u^{-1},0)\|\leq64\mu U/u.
\]
By~\eqref{eq:connection-clock-moments} and the factor two in
\eqref{eq:connection-transverse-growth},
\begin{equation}
 \|X(s)-(u^{-1},0)\|
      \leq256+128\int_0^s|Z(r)|\dd r.
 \label{eq:connection-shifted-strong-response}
\end{equation}
The constant entry vector contributes only $G_U/u$ to $Z'$, so
\[
 |Z'|\leq32U/u+A_0|Z|+B_0\|X-(u^{-1},0)\|.
\]
For $|Z|\leq Q$, the right-hand side is bounded by $e_u+kQ$,
where
\[
 e_u=32U/u+16384\left(U+P^2/\mu^5\right).
\]
Its integral is at most
\[
 \frac{64}{\mu}+\frac{32768u}{\mu}
                    +\frac{4096w^2}{\mu^4}\leq\frac{65}{\mu}.
\]
Absorption gives $|Z|\leq(16/15)(65/\mu)\leq70/\mu$.
Substitution into~\eqref{eq:connection-shifted-strong-response}
proves the second assertion.
\end{proof}

The singular entry derivative $u^{-1}$ has thus remained in the
strong logarithmic component. Its influence on the phase was
integrated against the decaying wall $U$.

The direction $(0,0,1)$ has $\zeta=1/w$ and
$Z(0)=w/H(\mu-1,w)\leq2w/\mu$. In this direction a large
direct forcing appears in $x$. We integrate that forcing explicitly
before estimating how it influences $y$ and $Z$.

We first record the needed mixed integrals, all in the logarithmic
clock:
\begin{equation}
 \begin{aligned}
 \int_0^s W^2\dd r&\leq\frac{W(s)^2}{2},&
 \int_0^s UW\dd r&\leq\frac{4uw}{\mu},\\
 \int_0^s UW^2\dd r&\leq uw,&
 \int_0^s P\dd r&\leq\frac{\mu w}{2},\\
 \int_0^s P^2W^2\dd r&\leq\frac{w^2W(s)^2}{2}.&&
 \end{aligned}
 \label{eq:connection-growing-moments}
\end{equation}
The first follows from $(W^2)'=2W^2$. The second follows from
$(UW)'\leq-\mu UW/4$, and the fourth from $P'\leq-2P/\mu$.
For the third, $(UW^2)'\leq0$, so its integral is at most $uw^2s$.
Since $W$ is increasing and $W'=W$,
$ws\leq\int_0^sW\dd r=W(s)-w\leq1$.
For the last inequality, use $P\leq w$ and the first bound.
These arguments include the endpoint $\mu=4$, where $UW^2$ need
not have a strictly negative decay rate.

Define the primitive
\begin{equation}
 F_w(s)=\frac1w\int_0^s f_W(r)\dd r.
 \label{eq:connection-growing-primitive}
\end{equation}
The strong row and~\eqref{eq:connection-growing-moments} give
\begin{equation}
 |F_w(s)|\leq32\left(1+\frac{W(s)^2}{\mu w}\right),\qquad
 \int_0^s64\mu U|F_w|\dd r\leq6144u.
 \label{eq:connection-growing-primitive-bounds}
\end{equation}
For the first inequality, the integral of $|f_W|/w$ is at most
$16W(s)^2/(\mu w)+16+128u/\mu$.
For the second, insert the first bound and use
$\int U\leq2u/\mu$ and $\int UW^2\leq uw$.

The residual $\widetilde X=X-(F_w,0)$ starts at zero and solves
\[
 \widetilde X'=T\widetilde X+T(F_w,0)
                 +\left(0,\frac{\mu^2g_W}{w}\right)+Zc.
\]
Its three forcing terms have norm at most
$64\mu U|F_w|$, $64(W^2/\mu+\mu UW)/w$, and $64|Z|$.
Their integrals, together with
Lemma~\ref{lem:connection-transverse-growth}, give
\begin{equation}
 \|\widetilde X(s)\|\leq12800u+
       \frac{64W(s)^2}{\mu w}+128\int_0^s|Z(r)|\dd r.
 \label{eq:connection-growing-residual}
\end{equation}
Indeed, the middle forcing has integral at most
$256u+32W(s)^2/(\mu w)$, while the first contributes at most
$6144u$ before the factor two.

\begin{proposition}[Response to the growing wall]
\label{prop:connection-growing-response}
For the fixed-clock variation $(Z_w,x_w,y_w)$ in the ordinary
direction $(0,0,1)$, and every $0\leq s\leq L$,
\begin{align}
 |Z_w(s)|&\leq\frac{8000(u+w)}{\mu},\notag\\
 |x_w(s)|&\leq32+12800u+\frac{96W(s)^2}{\mu w}
                             +\frac{1024000s(u+w)}{\mu},\notag\\
 |y_w(s)|&\leq\frac1{\mu^2}
       \left(12800u+\frac{64W(s)^2}{\mu w}
                             +\frac{1024000s(u+w)}{\mu}\right).
 \label{eq:connection-growing-response}
\end{align}
\end{proposition}

\begin{proof}
For a bound $|Z|\leq Q$, put $q_w=W^2/(\mu w)$.
Equations~\eqref{eq:connection-growing-primitive-bounds} and
\eqref{eq:connection-growing-residual} give
\[
 |x|\leq32(1+q_w)+(12800u+64q_w+128sQ),\qquad
 |y|\leq\frac{12800u+64q_w+128sQ}{\mu^2}.
\]
Apply the phase row separately to these two contributions to $x$.
The result is $|Z'|\leq e_w+kQ$, with the same $k$ as before and
\begin{align*}
 e_w={}&1024U(1+q_w)
       +64\left(U+\frac{P^2}{\mu^5}\right)(12800u+64q_w)\\
       &+\frac{32}{w}\left(UW+\frac{P^2}{\mu^3}\right).
\end{align*}
The last term is the direct contribution $G_W/w$.
Expansion and the moments in
\eqref{eq:connection-clock-moments} and
\eqref{eq:connection-growing-moments} yield
\begin{equation*}
 \int_0^s e_w\dd r\leq
 \frac{u(7296+1638400u+204800w^2/\mu^3)
               +w(2048/\mu^5+8/\mu)}{\mu}.
\end{equation*}
In detail, the four expanded terms before the direct contribution
are bounded by
\[
 \frac{2048u+1638400u^2}{\mu},\quad
 \frac{5120u}{\mu},\quad
 \frac{204800uw^2}{\mu^4},\quad
 \frac{2048w}{\mu^6}.
\]
The direct contribution has integral at most
$128u/\mu+8w/\mu^2$. Adding $|Z(0)|\leq2w/\mu$ gives a
total at most $7500(u+w)/\mu$: here $u\leq1/40000$,
$w\leq10^{-4}$ and $\mu\geq4$ suffice.
Equation~\eqref{eq:connection-feedback-absorption} proves the
phase bound $|Z|\leq8000(u+w)/\mu$.
Substituting it into the primitive and residual bounds proves the
two amplitude assertions.
\end{proof}

The arrival is at $W=1$, or $s=L(p)$ for $p=(\mu,u,w)$.
Since $DL(p)v=-v_w/w=-\zeta$, differentiation at this moving
endpoint gives
\begin{equation}
 D(J_*,\log u_+,\log w_+)(p)v
       =(Z,x,y)(L)-\zeta(G,f,g)_*.
 \label{eq:connection-moving-arrival}
\end{equation}
Here $J_*=J(m_*,V_*,1)$, $u_+=U_*$, and $w_+=V_*$.
This identity also follows directly from the original-time event:
varying $W(T_c;p)=1$ removes $(\eta_W/W)F$ from the fixed-time
variation. Comparing with
\eqref{eq:connection-relative-clock-variation} leaves exactly
$-\zeta F$. Thus~\eqref{eq:connection-moving-arrival} is the
derivative of the actual first arrival established in
Proposition~\ref{prop:long-connection-arrival}.

\begin{corollary}[Derivatives of the returned amplitudes]
\label{cor:connection-log-amplitude-derivatives}
The ordinary parameter derivatives of the two returned logarithmic
amplitudes satisfy
\begin{align}
 |\partial_\mu\log u_+|&\leq137L,&
 |\partial_\mu\log w_+|&\leq\frac{137L}{\mu^2},\notag\\
 |\partial_u\log u_+|&\leq9000\left(\frac1u+\frac L\mu\right),&
 |\partial_u\log w_+|&\leq\frac{9000(1+L)}{\mu^2},\notag\\
 |\partial_w\log u_+|&\leq\frac{32\mu}{w},&
 |\partial_w\log w_+|&\leq\frac{1+32/\mu}{w}.
 \label{eq:connection-log-amplitude-derivatives}
\end{align}
\end{corollary}

\begin{proof}
For the first two parameter directions $\zeta=0$, so
Proposition~\ref{prop:connection-phase-strong-responses} applies
directly at $s=L$.
Use $2048/15<137$, $u\leq1$ and $\mu\geq1$ to obtain the
first four bounds.

For the growing direction, $W(L)=1$ and $wL\leq1$.
The amplitude bounds in~\eqref{eq:connection-growing-response}
then imply
\[
 |x_w(L)|\leq\frac{16\mu}{w},\qquad
 |y_w(L)|\leq\frac{200}{\mu^3w}.
\]
To check the constants, multiply the first proposed bound by $\mu w$
and the second by $\mu^3w$. The resulting numerators are
\begin{gather*}
 32\mu w+12800\mu uw+96+1024000wL(u+w),\\
 12800\mu uw+64+1024000wL(u+w).
\end{gather*}
Use $\mu uw\leq10^{-8}$, $u+w\leq1/8000$ and $wL\leq1$;
these are at most $16\mu^2$ and $200$, respectively.
Finally $|f_*|\leq2\mu$ and $|g_*|\leq1+8/\mu$.
The endpoint correction in~\eqref{eq:connection-moving-arrival}
therefore gives
\[
 |\partial_w\log u_+|\leq\frac{18\mu}{w}
                                    \leq\frac{32\mu}{w},\qquad
 |\partial_w\log w_+|\leq
           \frac{1+8/\mu+200/\mu^3}{w}
                                    \leq\frac{1+32/\mu}{w}.
\]
\end{proof}

The large logarithmic derivatives are paired with the exponential
decay of the returned walls in~\eqref{eq:connection-output-decay}.
For the radial derivative, we next convert the matching-phase
variation back to the radius of the next core. This conversion must
retain the small endpoint defects, just as in the value estimate.