\section{Closing the fluid, rigid-velocity, and configuration errors}
\label{sec:convergence}

The theorem is conditional on a sufficiently regular physical solution,
not on convergence of numerical forces or trajectories. Geometry is
an unknown of the discrete scheme. The comparison below is constructed
on that computed geometry, and its discrepancy from the exact geometry
is estimated together with the velocity error.
The comparison and residual strategy has precedents in coupled
geometry analyses \citep{legendre2008,vonwahlrichter2023} and
changing-space Stokes estimates \citep{brenner2014}. The constructions
and bounds below specify the regularized volume and surface
consistency terms and the retained-coarse transfer argument needed
for the present method.

\subsection{Explicit assumptions on the exact branch and loads}
Assume that \eqref{eq:physical-stokes}--\eqref{eq:newton} has a
solution on $[0,T]$ with positive wall clearance, and extend its
velocity rigidly inside the body. We assume
\begin{equation}\label{eq:coupled-strong-regularity}
 \begin{gathered}
 u\in H^1(0,T;L^2(\Omega)^2),\qquad X(t)\in\mathbb K(q(t)),
 \qquad \xi\in H^1(0,T;\R^3),\\
 M_0:=\esssup_t\left(\|u(t)\|_{W^{4,\infty}(B(q(t)))}
                  +\|u(t)\|_{W^{4,\infty}(F(q(t)))}\right)<\infty,\\
 M_1(t):=\|u_t(t)\|_{W^{3,\infty}(B(q(t)))}
                  +\|u_t(t)\|_{W^{3,\infty}(F(q(t)))}
                   \in L^2(0,T).
 \end{gathered}
\end{equation}
The time derivative is Eulerian and may have different side traces.
The extra spatial derivative compared with Part~II is used when
differentiating the configuration alignment: that derivative contains
$\nabla u$ multiplied by the speed of the alignment map.
Assume also
\begin{equation}\label{eq:coupled-load-reaction-regularity}
 \begin{gathered}
 f\in L^2(0,T;W^{1,\infty}(\Omega)^2),\qquad
 F_e,T_e\in L^2(0,T),\\
 \Lambda_B^2:=\int_0^T\|\lambda_B(t)\|_{L^2(B(q(t)))}^2dt<\infty,
 \qquad
 \Lambda_\Gamma^2:=\int_0^T
        \|\lambda_\Gamma(t)\|_{L^2(\Gamma(q(t)))}^2dt<\infty.
 \end{gathered}
\end{equation}
Here $F_e$ is vector valued. The volume reaction in
\eqref{eq:reaction-representatives} already has this regularity under
\eqref{eq:coupled-strong-regularity}. The surface-traction bound is
a separate assumption on the exact solution. A mere ambient
$L^2_tU'$ force would not justify the Lipschitz-in-configuration
load estimate below; the stated spatial extension assumption is used.

The primary load choice is the exact ambient time average in
\eqref{eq:discrete-load}. More generally, let a bar denote each
ambient time average and define the data error before computing the
trajectory by
\begin{equation}\label{eq:coupled-load-consistency}
 E_L^2=\sum_n\tau_n\left(
       \|f^n-\bar f^{\,n}\|_{L^2(\Omega)}^2
       +|F_e^n-\bar F_e^{\,n}|^2
       +|T_e^n-\bar T_e^{\,n}|^2\right).
\end{equation}
If $\overline{\mathcal L}_n$ is the exact-average functional evaluated
on $F(q_n)$, Lemma~\ref{lem:joint-coercivity} and Poincare's inequality
bound the squared time sum of
$\sup_{Y_h\ne0}|\mathcal L^n(Y_h)-\overline{\mathcal L}_n(Y_h)|/
\|v_h\|_U$ by $CE_L^2$, uniformly on every admissible prefix.
The data error is thus independent of the unknown computed geometry.
Bounded $E_L$ also implies the uniform data bounds used in
Theorem~\ref{thm:coupled-energy}. Exact time averages have $E_L=0$.
Endpoint loads are also covered: if
$f\in H^1(0,T;L^2(\Omega)^2)$ and $F_e,T_e\in H^1(0,T)$, then
the choices $f^n=f(t_n)$, $F_e^n=F_e(t_n)$, and $T_e^n=T_e(t_n)$
give
\[
 E_L\le C\tau_{\max}
       (\|f_t\|_{L^2_tL^2_x}+\|F_e'\|_{L^2_t}+\|T_e'\|_{L^2_t}).
\]
This follows from the fundamental theorem of calculus, Jensen's
inequality, and Lemma~\ref{lem:joint-coercivity}; the fluid integral
in both compared loads is on the same current domain $F(q_n)$.
Write
\begin{equation}\label{eq:coupled-resolution}
 r_h^2=h+\bar\eps+\bar\ell+\bar\kappa^2+\tau_{\max}+E_L^2,
 \qquad \tau_{\max}=\max_n\tau_n.
\end{equation}
All constants below may depend on the exact branch and the fixed
physical parameters, including the strictly positive excess inertias,
but not on a refinement history or on inverse regularization widths.

\subsection{The mapped identity on all ambient tests}
On an admissible prefix of the computation, align the exact pose to
the piecewise affine computed pose using Lemma~\ref{lem:alignment}:
\begin{equation}\label{eq:computed-alignment}
 \begin{gathered}
 \Psi(t)=\Psi(q(t),q_h(t)),\quad
 O(t)=Q(\theta_h(t)-\theta(t)),\quad A=D\Psi,\\
 \widetilde u=\mathcal T_\Psi u,
 \qquad \widetilde\xi=(OV,\omega),
 \qquad \widetilde X=(\widetilde u,\widetilde\xi),\quad
 d(t)=|q_h(t)-q(t)|.
 \end{gathered}
\end{equation}
The norms used for poses are equivalent finite-dimensional norms.
The comparison belongs to $\mathbb K(q_h(t))$, but its rigid velocity
is not the speed of $q_h$. This distinction is retained in the
following calculation.

\begin{lemma}[An explicit geometric commutator]
\label{lem:mapped-coupled-identity}
For every $Y=(v,W,\eta)\in\mathbb V$, the mapped state satisfies
\begin{equation}\label{eq:mapped-full-identity}
 \begin{split}
 (\partial_t\widetilde X,Y)_M+a(\widetilde u,v)
       +\mathcal G(\omega;\widetilde X,Y)
 &+\int_{B(q_h)}\widetilde\lambda_B
                   \cdot(v-\mathcal R_{q_h}\zeta)\,dx\\
 &+\int_{\Gamma(q_h)}\widetilde\lambda_\Gamma
                   \cdot(\gamma_{q_h}v-\mathcal R_{q_h}\zeta)\,ds\\
 &=\mathcal L(t,q_h;Y)+\mathcal E(t;Y),
 \end{split}
\end{equation}
where $\zeta=(W,\eta)$ and the two reactions are rigid pushforwards
of the exact reactions. Their respective $L^2$ norms are unchanged.
There is a known $a_*(t)\in L^2(0,T)$, independent of the numerical
errors, such that
\begin{equation}\label{eq:geometric-residual-bound}
 |\mathcal E(t;Y)|
 \le C\bigl(a_*(t)d(t)+|q_h'(t)-q'(t)|\bigr)
                       \|Y\|_{\mathbb V}.
\end{equation}
One may take a constant multiple of
$1+\|u_t(t)\|_0+\|f(t)\|_{W^{1,\infty}}+|F_e(t)|$ for $a_*$.
\end{lemma}
\begin{proof}
In \eqref{eq:full-reaction}, use the inverse Piola fluid test
$(\mathcal T_\Psi^{-1}v)(x)=A(x)^{-1}v(\Psi(x))$ and rigid test
$(O^{\mathsf T}W,\eta)$. These are admissible ambient tests;
they need not satisfy any exact or regularized rigidity constraint.
On the body and its boundary $A=O$, and $\Psi$ is a rigid map.
The test mismatch therefore becomes
$O^{\mathsf T}(v-\mathcal R_{q_h}\zeta)\circ\Psi$.
Since area and arclength are preserved, both reaction integrals
transform exactly, with
$\widetilde\lambda\circ\Psi=O\lambda$. There is no difference of
two regularized kernels in this calculation.

Put $z=\Psi_t\circ\Psi^{-1}$. Differentiating the Piola formula
at a fixed physical point gives the exact fluid identity
\begin{equation}\label{eq:piola-time-commutator}
 \widetilde u_t-
       (A^{-\mathsf T}u_t)\circ\Psi^{-1}
 =\bigl((A-A^{-\mathsf T})u_t+A_tu\bigr)\circ\Psi^{-1}
       -\nabla\widetilde u\,z.
\end{equation}
The first term is bounded in $L^2$ by $Cd\|u_t\|_0$.
The others are bounded by
$C(d+|q_h'-q'|)\|u\|_{H^1}$, by
\eqref{eq:alignment-estimates}--\eqref{eq:alignment-time}.
For the body mass, $(OV)_t=OV_t+O_tV$, and
$O_t=(\theta_h'-\omega)JO$.
The transformed excess-mass term thus differs by
$m_e(O_tV)\cdot W$, which has the same bound.
The angular mass term is unchanged. The gyroscopic term transforms
without a remainder because $OJ=JO$:
\[
 (JV)\cdot O^{\mathsf T}W=(JOV)\cdot W.
\]
In particular, the interior acceleration is not replaced by the
acceleration of a body moving with $\widetilde\xi$. Directly there,
\begin{equation}\label{eq:mapped-interior-mismatch}
 \partial_t\widetilde b-O\partial_tb\circ\Psi^{-1}
       =O_tV-\omega J(c_h'-OV).
\end{equation}
This is part of \eqref{eq:piola-time-commutator}, and is controlled
by the pose and pose-speed errors; it has not been omitted.

For the viscous form, change variables using $\det A=1$ and
differentiate $Av$ and $A^{-1}v\circ\Psi$. Each difference from
the identity-map expression contains either $A-I$, $A^{-1}-I$,
or $D A$. Their $L^\infty$ norms are $O(d)$. Poincare's inequality
therefore gives
\[
 |a(\mathcal T_\Psi u,v)-a(u,\mathcal T_\Psi^{-1}v)|
       \le Cd\|u\|_{H^1}\|v\|_{H^1}.
\]
This also follows componentwise for the symmetric-gradient form;
no assumption of a global $H^2$ velocity is needed for this bound.

The transformed fluid load is integrated on precisely $F(q_h)$,
with density $(A^{-\mathsf T}f)\circ\Psi^{-1}$. Its difference
from $f$ is bounded in $L^2$ by $Cd\|f\|_{W^{1,\infty}}$.
The transformed body force is $OF_e$, whose difference from $F_e$
is at most $Cd|F_e|$; the torque is unchanged. Combining these
identities proves \eqref{eq:mapped-full-identity} and
\eqref{eq:geometric-residual-bound}. In particular no convergence
or bounded inverse estimate for a discrete multiplier was used.
\end{proof}

The numerical speed is uniformly bounded by
Theorem~\ref{thm:coupled-energy}. On a sufficiently close admissible
prefix, the preceding alignment consequently has bounded derivatives
independent of all resolutions. Formula
\eqref{eq:piola-time-commutator}, the side regularity in
\eqref{eq:coupled-strong-regularity}, and the chain rule show that
$\widetilde u$ is piecewise uniformly $W^{3,\infty}$ and its
Eulerian derivative is piecewise $W^{3,\infty}$ in $L^2_t$.
The fourth spatial derivative of $u$ is used here in
$\nabla\widetilde u\,z$. Moreover,
$\widetilde\xi\in H^1_t$ with a uniform bound, since its derivative
uses $\xi'$ and $\theta_h'-\omega$, not a derivative of the numerical
velocity. Lemma~\ref{lem:joint-recovery} therefore constructs
\begin{equation}\label{eq:joint-comparison}
 Z_h(t)=(z_h(t),\widetilde\xi(t)),\quad
 H_h(t)=\widetilde X(t)-Z_h(t)=(\eta_h(t),0,0),
 \quad Z_h(t_n)\in\mathbb K_h^n,
\end{equation}
with the uniform estimates \eqref{eq:joint-recovery-estimate}.
No numerical acceleration bound enters their constants.

\subsection{Consistency on the current joint space}
Let $\overline{b}^{,n}=\tau_n^{-1}\int_{t_{n-1}}^{t_n}b(s)\,ds$.
The following estimates compare the time-averaged identity
\eqref{eq:mapped-full-identity} with the current geometry and the
lagged gyroscopic coefficient.

\begin{lemma}[Current-geometry residual bounds]
\label{lem:joint-time-residuals}
On every sufficiently close admissible prefix there exist nonnegative
$k_n,s_n$, independent of any desired trajectory-convergence assertion,
such that
\begin{equation}\label{eq:residual-sums}
 \sum_n\tau_n k_n\le C,\qquad
 \sum_n\tau_n s_n^2\le Cr_h^2.
\end{equation}
For
$e^n=X_h^n-Z_h(t_n)$, $E_n=\|e^n\|_M^2$,
and $P_n=|q_n-q(t_n)|^2$, subtraction of the comparison gives
\begin{equation}\label{eq:joint-error-equation}
 \left(\frac{e^n-P_M^ne^{n-1}}{\tau_n},Y_h\right)_M
  +a(e_u^n,v_h)+\mathcal G(\omega_h^{n-1};e^n,Y_h)
       =\mathcal R_n(Y_h),
\end{equation}
where, for all $Y_h=(v_h,\zeta)\in\mathbb K_h^n$,
\begin{equation}\label{eq:joint-residual-closure}
 |\mathcal R_n(Y_h)|
       \le C\bigl(\sqrt{k_n}\sqrt{E_{n-1}+P_{n-1}}+s_n\bigr)
                         \|v_h\|_U.
\end{equation}
\end{lemma}
\begin{proof}
Orthogonality removes the current projection inside a current test,
including for the previous comparison $Z_h(t_{n-1})$. Thus its tested
time difference is the ordinary difference of $Z_h$, with no
mesh-change term. The exact residual is the sum of:
the load difference; the averaged two reaction integrals;
$-\overline{\mathcal E}^{,n}$;
$(\overline{\partial_tH_h}^{,n},Y_h)_M$;
$a(\eta_h(t_n),v_h)$;
$-a(\widetilde u(t_n)-\overline{\widetilde u}^{,n},v_h)$;
and the gyroscopic difference
\begin{equation}\label{eq:gyro-residual}
 \overline{\mathcal G(\omega;\widetilde X,Y_h)}^{,n}
       -\mathcal G(\omega_h^{n-1};\widetilde X(t_n),Y_h).
\end{equation}
Here the averaged load in the mapped equation uses $q_h(s)$, whereas
the discrete load uses $q_n$.

We bound these terms explicitly. The configuration speed is uniformly
bounded. A rigid shift over one interval has displacement $O(\tau_n)$.
The $L^2$ translation estimate for $H^1$ fields on a body gives
\begin{equation}\label{eq:shifted-volume-defect}
 \|v_h-\mathcal R_{q_h(s)}\zeta\|_{L^2(B(q_h(s)))}
 \le C(\eps_n+\kappa_n+\tau_n)\|Y_h\|_{\mathbb V}.
\end{equation}
Indeed, pull the integral back by the rigid map from $B(q_n)$:
the difference of the pulled field and $v_h$ is bounded by
$C\tau_n\|v_h\|_{H^1}$ by integration along the rigid path, and
the difference of the two affine rigid test fields is
$C\tau_n|\zeta|$. The remaining term is
\eqref{eq:volume-defect}. The analogous surface estimate is
\begin{equation}\label{eq:shifted-surface-defect}
 \|\gamma_{q_h(s)}v_h-\mathcal R_{q_h(s)}\zeta\|_{L^2(\Gamma(q_h(s)))}
 \le C(\sqrt{\eps_n}+\sqrt{\ell_n}+\sqrt{\tau_n})
                              \|Y_h\|_{\mathbb V}.
\end{equation}
For the shifted trace, integration on normal fibers gives a
$C\sqrt{\tau_n}\|v_h\|_{H^1}$ bound, and the tangential change of
parametrization is controlled by its $H^{1/2}$ trace seminorm.
Equivalently one may use the rigid-pullback trace estimate proved in
Part~II \citep{partii2026}. The normal and tangential maps have uniform
Jacobians on this admissible family. Applying
\eqref{eq:surface-defect} proves \eqref{eq:shifted-surface-defect}.
Pairing with the two mapped reactions and Jensen's inequality gives
total squared residual at most
\[
 C(\bar\eps+\bar\kappa+\tau_{\max})^2\Lambda_B^2
 +C(\bar\eps+\bar\ell+\tau_{\max})\Lambda_\Gamma^2.
\]
Thus time averaging never replaces a reaction on one configuration
by a reaction on another without a displacement estimate.

The symmetric difference of $F(q_h(s))$ and $F(q_n)$ has area
$O(\tau_n)$. Therefore the current-domain load discrepancy is at
most $C\sqrt{\tau_n}\|f(s)\|_{L^\infty}\|v_h\|_0$.
Its squared time sum is bounded by
$C\tau_{\max}\|f\|_{L^2_tL^\infty_x}^2$, in addition to $E_L^2$.
The comparison derivative and spatial errors contribute at most
$C\delta$ to that sum by \eqref{eq:joint-recovery-estimate}.
For the viscous endpoint/average term, the swept-strip proof of
Part~II gives
\[
 \|\widetilde u(t)-\widetilde u(s)\|_U
 \le C\sqrt{|t-s|}+C\int_s^t
       \|\widetilde u_r\|_{W^{3,\infty}(B(q_h(r)))\oplus
                                  W^{3,\infty}(F(q_h(r)))}\,dr.
\]
The strip has area $O(|t-s|)$ and gradients are bounded; away from it,
integrate the side derivative. Jensen's inequality then gives a
squared time sum bounded by $C\tau_{\max}$.

To close the geometric residual rather than assume it small, on
$s\in[t_{n-1},t_n]$ use
\begin{align}
 d(s)&\le\sqrt{P_{n-1}}+C\tau_n,\nonumber\\
 |q_h'(s)-q'(s)|
 &\le C\sqrt{E_{n-1}}+C\sqrt{P_{n-1}}
                    +|\xi(s)-\xi(t_{n-1})|.
                    \label{eq:pose-speed-error}
\end{align}
The second line uses
$\widetilde\xi(t_{n-1})=(O(t_{n-1})V(t_{n-1}),\omega(t_{n-1}))$
and $q_h'=\xi_h^{n-1}$. It does not use a derivative of $\xi_h$.
Equations \eqref{eq:geometric-residual-bound} and
\eqref{eq:pose-speed-error} imply the past-error contribution in
\eqref{eq:joint-residual-closure} with
$k_n=C(1+\overline{a_*^2}^{,n})$.
The remaining terms have squared time sum
$C\tau_{\max}^2(\|a_*\|_{L^2}^2+\|\xi'\|_{L^2}^2)$.

Finally, split \eqref{eq:gyro-residual} at
$\mathcal G(\omega(t_n);\widetilde X(t_n),Y_h)$.
Its time-variation part has squared time sum $C\tau_{\max}^2$
because $\omega$ and $\widetilde\xi$ have uniform $H^1_t$ bounds.
The remaining coefficient obeys
\[
 |\omega(t_n)-\omega_h^{n-1}|
 \le C\sqrt{E_{n-1}}+|\omega(t_n)-\omega(t_{n-1})|.
\]
The last term again contributes at most
$C\tau_{\max}^2\|\omega'\|_{L^2}^2$ to the squared time sum.
Throughout, Lemma~\ref{lem:joint-coercivity} replaces
$\|Y_h\|_{\mathbb V}$ by $C\|v_h\|_U$ for small $\bar\kappa$.
Collecting the displayed bounds proves
\eqref{eq:residual-sums}--\eqref{eq:joint-residual-closure}.
\end{proof}

\subsection{The coupled recursion and its consequence}
\begin{theorem}[Conditional convergence with an unknown trajectory]
\label{thm:fluid-driven-convergence}
Assume \eqref{eq:excess-inertia}, the stated body and partition
regularity, and \eqref{eq:coupled-strong-regularity}--
\eqref{eq:coupled-load-reaction-regularity}. Initialize with the exact
pose $q_0=q(0)$ and $X_h^0=P_M^0X(0)$.
For all sufficiently small $r_h$, the numerical configurations remain
in a collision-free neighborhood of the exact branch through time
$T$, and the scheme is defined there. One has
\begin{equation}\label{eq:coupled-convergence-aligned}
 \begin{split}
 \max_n\bigl(&|q_n-q(t_n)|+|\xi_h^n-\xi(t_n)|
                   +\|u_h^n-u(t_n)\|_0\bigr)\\
 &+\left(\sum_n\tau_n
            \|u_h^n-\mathcal T_{\Psi(t_n)}u(t_n)\|_U^2\right)^{1/2}
       \le Cr_h.
 \end{split}
\end{equation}
The same-point physical energy seminorm satisfies the generally
weaker estimate
\begin{equation}\label{eq:coupled-convergence-physical}
 \left(\sum_n\tau_n\|u_h^n-u(t_n)\|_U^2\right)^{1/2}
       \le C(r_h+\sqrt{r_h}).
\end{equation}
In particular it converges strongly. No ratio between the time step,
mesh size, and regularization widths is required, and there is no
factor counting enrichment changes. The theorem gives no multiplier
or force convergence estimate.
\end{theorem}
\begin{proof}
Test \eqref{eq:joint-error-equation} by $e^n$. Its skew term is
zero. Projection orthogonality and polarization give
\[
 E_n-E_{n-1}+\|e^n-e^{n-1}\|_M^2
                  +2\mu\tau_n\|e_u^n\|_U^2
       =2\tau_n\mathcal R_n(e^n).
\]
Use \eqref{eq:joint-residual-closure} and Young's inequality to absorb
one viscous term. After changing constants,
\begin{equation}\label{eq:velocity-error-recursion}
 E_n-E_{n-1}+\|e^n-e^{n-1}\|_M^2
               +\mu\tau_n\|e_u^n\|_U^2
       \le C\tau_n k_n(E_{n-1}+P_{n-1})+C\tau_n s_n^2.
\end{equation}
All error terms on the right are from the previous step.
In particular the lagged gyroscopic coefficient leaves no current
error term: $\mathcal G(\omega_h^{n-1};e^n,e^n)=0$.

The exact kinematic truncation defect is
\[
 r_q^n=q(t_n)-q(t_{n-1})-\tau_n\xi(t_{n-1}),\qquad
 \sum_n\frac{|r_q^n|^2}{\tau_n}
      \le \tau_{\max}^2\|\xi'\|_{L^2(0,T)}^2.
\]
Subtracting the exact pose increment from \eqref{eq:pose-update}
and using
$|\xi_h^{n-1}-\xi(t_{n-1})|
\le C\sqrt{E_{n-1}}+C\sqrt{P_{n-1}}$
gives, for $\tau_n\le1$,
\begin{equation}\label{eq:pose-error-recursion}
 P_n\le(1+C\tau_n)P_{n-1}+C\tau_n E_{n-1}
                           +C\tau_n^{-1}|r_q^n|^2.
\end{equation}
This is simply Young's inequality applied to the pose increment;
it does not assume that the computed path is already convergent.

Set $A_n=E_n+P_n$ and enlarge $k_n$ by a fixed constant if needed.
Combining the two recursions yields
\begin{equation}\label{eq:closed-error-recursion}
 A_n+\|e^n-e^{n-1}\|_M^2+\mu\tau_n\|e_u^n\|_U^2
 \le (1+C\tau_n k_n)A_{n-1}
       +C\tau_n s_n^2+C\tau_n^{-1}|r_q^n|^2.
\end{equation}
Iteration uses
$\prod_j(1+C\tau_jk_j)\le\exp(C\sum_j\tau_jk_j)$.
The sum in the exponent is uniformly bounded by
\eqref{eq:residual-sums}, even though its individual coefficients
need not be uniformly bounded. Applying the same argument after
adding the accumulated nonnegative increment and viscous terms gives
\begin{equation}\label{eq:joint-error-closed}
 \max_n A_n+\sum_n\|e^n-e^{n-1}\|_M^2
                    +\mu\sum_n\tau_n\|e_u^n\|_U^2
       \le C(A_0+r_h^2).
\end{equation}
At initialization the poses agree and $\Psi(0)=\mathrm{id}$.
Since $Z_h(0)\in\mathbb K_h^0$,
$e^0=P_M^0(X(0)-Z_h(0))$.
The best-approximation property and the recovery bound therefore
give $A_0\le C\delta^3\le Cr_h^2$.

The argument has so far been made on a sufficiently close admissible
prefix, to justify the alignment. Choose two nested pose
neighborhoods of the exact branch, with positive clearance on the
larger one. The energy bound supplies a fixed bound for each
numerical pose increment. For small $\tau_{\max}$, a first exit
from the smaller neighborhood still lies in the larger one.
All preceding constants are uniform there, so
\eqref{eq:joint-error-closed} makes the pose error smaller than half
the exit threshold when $r_h$ is sufficiently small, a contradiction.
This continuation argument extends the estimate through $T$ and
proves that the linear solve always uses an admissible geometry.
Its smallness threshold depends on the exact collision-free branch,
not on a ratio of numerical parameters.

The recovery bound now gives the aligned fluid energy estimate.
The rigid error satisfies
$|\xi_h^n-\xi(t_n)|\le C\|e^n\|_M+C\sqrt{P_n}$.
Lemma~\ref{lem:alignment-physical-norms} gives the same-point bulk
$L^2$ estimate, proving \eqref{eq:coupled-convergence-aligned}.
Its $U$ estimate gives
\[
 \left(\sum_n\tau_n
       \|\mathcal T_{\Psi(t_n)}u(t_n)-u(t_n)\|_U^2\right)^{1/2}
       \le C\bigl(\max_n|q_n-q(t_n)|\bigr)^{1/2},
\]
which proves \eqref{eq:coupled-convergence-physical}.
\end{proof}

\begin{corollary}[Recovery of physical interior inertia and dissipation]
\label{cor:physical-interior-limit}
Under Theorem~\ref{thm:fluid-driven-convergence},
\[
 \max_n\|u_h^n-\mathcal R_{q_n}\xi_h^n\|_{L^2(B(q_n))}
      \le Cr_h,\qquad
 \sum_n\tau_n\|D(u_h^n)\|_{L^2(B(q_n))}^2\le Cr_h^2.
\]
The nodal surrogate-to-physical kinetic-energy difference is at most
$Cr_h$ uniformly in time.
\end{corollary}
\begin{proof}
The aligned exact velocity equals
$\mathcal R_{q_n}\widetilde\xi(t_n)$ in $B(q_n)$ and has zero
strain there. The mass error and recovery bounds control the first
displayed difference, including
$|\xi_h^n-\widetilde\xi(t_n)|$; the aligned energy estimate controls
the interior strain. Finally apply the factorization
\eqref{eq:energy-discrepancy} and the uniform discrete mass bound.
Unlike the time-integrated statement obtained from constraints alone,
this nodal assertion uses the convergence theorem.
\end{proof}

\begin{corollary}[Vanishing joint transfer losses]
\label{cor:joint-transfer-loss}
Under Theorem~\ref{thm:fluid-driven-convergence},
\begin{equation}\label{eq:vanishing-joint-transfer}
 \begin{split}
 \sum_n\bigl(&\|X_h^n-P_M^nX_h^{n-1}\|_M^2
             +\|(I-P_M^n)X_h^{n-1}\|_M^2\bigr)\\
 &=\sum_n\|X_h^n-X_h^{n-1}\|_M^2\le Cr_h^2.
 \end{split}
\end{equation}
Thus neither the accumulated transfer loss nor the accumulated
implicit-step increment loss survives the convergence limit.
\end{corollary}
\begin{proof}
The increment term retained in \eqref{eq:joint-error-closed}
controls $\sum_n\|e^n-e^{n-1}\|_M^2$ by $Cr_h^2$.
The mapped state has a uniform $H^1_t\mathbb H$ bound, and
\eqref{eq:joint-recovery-estimate} gives the same bound for $Z_h$.
Consequently
\[
 \sum_n\|Z_h(t_n)-Z_h(t_{n-1})\|_M^2
       \le\tau_{\max}\|\partial_tZ_h\|_{L^2_tM}^2
       \le C\tau_{\max}.
\]
Use $X_h^n=e^n+Z_h(t_n)$ and $\tau_{\max}\le r_h^2$.
The equality is the orthogonal splitting
\eqref{eq:coupled-loss-splitting}. No bound on the number of
changes of the joint constraint space is used.
\end{proof}

For the piecewise constant velocity reconstruction and the piecewise
affine pose, the same convergence conclusions hold on $[0,T]$.
The exact velocity is $H^1_tL^2_x$, and the swept-strip argument gives
its $U$ time modulus of order $\sqrt{|t-s|}$ plus the integral of
its side time-derivative norm. These add at most $C\sqrt{\tau_{\max}}$
to the displayed bounds. Thus the theorem includes strong convergence
of the physical Eulerian velocity in $L^2(0,T;H^1(\Omega)^2)$.
If an additional global $L^\infty_tH^2_x$ bound is available, the
second alignment bound is Lipschitz in the pose, and the physical
energy rate improves to $Cr_h$. Without that extra regularity,
claiming the aligned rate for the same-point $H^1$ error would ignore
the displaced jump in the velocity gradient.