\section{Convergence for a moving prescribed rigid body}
\label{sec:convergence}

The estimate in this section is conditional on explicitly stated
regularity of the exact solution. It does not assert that arbitrary
finite-energy data have that regularity. All transfer operators act in
the original Eulerian bulk inner product. No transported mesh or
material time derivative is introduced in the numerical scheme.

\subsection{Regularity and the comparison trajectory}
Use the piecewise norm $\mathcal M_t$ defined in Section~\ref{sec:moving-approximation}.
Let $u$ satisfy the Eulerian weak equation and reaction identity
\eqref{eq:traction}, and set $w=u-g$. We assume that
\begin{align}
 &w\in H^1(0,T;L^2(\Omega)^2),\qquad
       w(t)\in V(t),\nonumber\\
 &M_0:=\sup_{0\le t\le T}\mathcal M_t(w(t))<\infty,
 \qquad
 M_1:=\left(\int_0^T\mathcal M_t(\partial_t w(t))^2\,dt\right)^{1/2}
       <\infty.\label{eq:strong-regularity}
\end{align}
The side restrictions and their derivatives are interpreted in the
Sobolev sense on the smooth moving space--time phases. The trace
differentiation justified in Lemma~\ref{lem:moving-recovery} uses these
explicit assumptions. In particular,
$\partial_t w$ in \eqref{eq:strong-regularity} is the \emph{Eulerian}
derivative. Bounds for a material derivative alone do not imply the
displayed spatial regularity. The Eulerian derivative may jump across
$\Gamma(t)$ and need not belong to $H^1(\Omega)^2$ or $V(t)$.
The trace compatibility is $w|_{\Gamma(t)}=0$, not
$\partial_t w|_{\Gamma(t)}=0$.

For the reaction in \eqref{eq:traction}, assume
\begin{equation}\label{eq:reaction-norm}
 \Lambda:=\left(\int_0^T
              \norm{\lambda(t)}_{L^2(\Gamma(t))}^2\,dt\right)^{1/2}
             <\infty.
\end{equation}
Surface norms are defined by the rigid parametrization of $\Gamma(t)$,
which preserves arclength. Define also
\[
 \mathfrak G=\norm g_{W^{1,\infty}(0,T;H^3(\Omega)^2)}.
\]
This quantity is finite for the smooth fixed-cutoff lifting constructed
in Lemma~\ref{lem:moving-lift} under the stated motion assumptions.
The constants below may depend on the fixed body, its clearance from
the outer boundary, the prescribed motion, the fixed cutoff, $T$, and
$\nu$. They are independent of the choices and number of local
enrichments, surface partitions, and regularization widths.

Write $\bar\eps=\max_n\eps_n$, $\bar\ell=\max_n\ell_n$, and
$\tau_{\max}=\max_n\tau_n$. Lemma~\ref{lem:moving-recovery} constructs,
on the retained coarse Cartesian grid, a trajectory
\begin{equation}\label{eq:comparison-trajectory}
 z_h(t)=Q_h(t)w(t),\qquad \eta(t)=w(t)-z_h(t),
 \qquad z_h(t_n)\in Z^n.
\end{equation}
With $r_h=\sqrt{h+\bar\eps}$, its estimates imply
\begin{align}
 \sup_t\norm{\eta(t)}_U&\le C r_hM_0,\nonumber\\
 \norm{\partial_t\eta}_{L^2(0,T;L^2(\Omega))}
   &\le C r_h(M_0+M_1).
 \label{eq:comparison-estimates}
\end{align}
These are the two different norms needed by the energy argument. An
$H^1$-small Eulerian derivative of the comparison error is neither
asserted nor used. Lemma~\ref{lem:moving-time-modulus} also gives
\begin{equation}\label{eq:time-modulus-used}
 \norm{w(t)-w(s)}_U
 \le C\sqrt{|t-s|}\,M_0
       +C\int_{\min(s,t)}^{\max(s,t)}
                 \mathcal M_r(\partial_r w(r))\,dr.
\end{equation}

\subsection{Consistency on the current constraint space}
For any time-dependent quantity $q$, let
$\bar q^n=\tau_n^{-1}\int_{t_{n-1}}^{t_n}q(s)\,ds$.
The selected load $f^n$ need not equal $\bar f^n$ exactly; measure its
error by
\begin{equation}\label{eq:load-error}
 E_f=\left(\sum_n\tau_n
                   \norm{f^n-\bar f^n}_{U'}^2\right)^{1/2}.
\end{equation}
The primary theoretical choice $f^n=\bar f^n$ gives $E_f=0$.
For the centered exact and discrete loads, write
\begin{align}
 \dual{\mathcal F(t)}v
    &=\dual{f(t)}v-\ip{\partial_tg(t)}v-a(g(t),v),\nonumber\\
 \dual{\mathcal F_h^n}v
    &=\dual{f^n}v-\ip{\delta g_h^n}v-a(g_h(t_n),v),
 \qquad
 \delta g_h^n=\frac{g_h(t_n)-g_h(t_{n-1})}{\tau_n}.
 \label{eq:loads-comparison}
\end{align}
The centered reaction identity is consequently
\begin{equation}\label{eq:centered-reaction}
 \ip{\partial_t w(t)}v+a(w(t),v)
       +\int_{\Gamma(t)}\lambda(t)\cdot\gamma_t v\,ds
       =\dual{\mathcal F(t)}v,
 \qquad v\in U,
\end{equation}
for almost every $t$.

\begin{lemma}[Load and moving-interface residuals]
\label{lem:consistency}
Under the preceding assumptions,
\begin{equation}\label{eq:lift-load-defect}
 \left(\sum_n\tau_n
   \norm{\mathcal F_h^n-\overline{\mathcal F}^{\,n}}_{U'}^2\right)^{1/2}
       \le E_f+C(h^2+\tau_{\max})\mathfrak G.
\end{equation}
For $v_h\in Z^n$, define
\[
 L_n(v_h)=\frac1{\tau_n}\int_{t_{n-1}}^{t_n}
             \int_{\Gamma(s)}\lambda(s)\cdot\gamma_s v_h\,ds_\Gamma\,ds.
\]
Then
\begin{align}
 |L_n(v_h)|&\le C\bigl(\sqrt{\eps_n}+\sqrt{\ell_n}
                                  +\sqrt{\tau_n}\bigr)
       \overline{\norm\lambda}_{\Gamma}^{\,n}\norm{v_h}_U,
       \label{eq:reaction-residual}\\
 \overline{\norm\lambda}_{\Gamma}^{\,n}
       &:=\frac1{\tau_n}\int_{t_{n-1}}^{t_n}
                        \norm{\lambda(s)}_{L^2(\Gamma(s))}\,ds.
       \nonumber
\end{align}
Finally,
\begin{equation}\label{eq:viscous-time-defect}
 \left(\sum_n\tau_n
             \norm{w(t_n)-\bar w^n}_U^2\right)^{1/2}
       \le C\sqrt{\tau_{\max}}(M_0+M_1).
\end{equation}
\end{lemma}
\begin{proof}
Let $d_h=g_h-g$. Subtraction of the two loads gives the exact identity
\[
 \dual{\mathcal F_h^n-\overline{\mathcal F}^{\,n}}v
  =\dual{f^n-\bar f^n}v-\ip{\overline{\partial_td_h}^{\,n}}v
       -a(d_h(t_n),v)-a(g(t_n)-\bar g^n,v).
\]
The fixed-cutoff Hermite lifting estimate gives
\[
 \norm{\partial_td_h}_{L^2(0,T;L^2)}
     +\sup_t\norm{d_h(t)}_U\le Ch^2\mathfrak G.
\]
Here the supremum follows also from the one-dimensional
$H^1$-in-time embedding for the finite family of lifting coefficients.
Jensen's inequality controls the averaged derivative. The fundamental
theorem of calculus and Cauchy--Schwarz give
\[
 \sum_n\tau_n\norm{g(t_n)-\bar g^n}_U^2
       \le\tau_{\max}^2\norm{\partial_tg}_{L^2(0,T;U)}^2.
\]
Poincare's inequality and the triangle inequality in the weighted
sequence norm prove \eqref{eq:lift-load-defect}.

Let $R_{s,n}$ denote the exact rigid map from $\Gamma(t_n)$ to
$\Gamma(s)$. It preserves surface measure. By the moving-trace estimate
of Lemma~\ref{lem:moving-trace} and the current-space trace defect,
\begin{align*}
 \norm{\gamma_s v_h}_{L^2(\Gamma(s))}
 &\le \norm{\gamma_s v_h\circ R_{s,n}-\gamma_{t_n}v_h}
                                  _{L^2(\Gamma(t_n))}
             +\norm{\gamma_{t_n}v_h}_{L^2(\Gamma(t_n))}\\
 &\le C\bigl(\sqrt{|s-t_n|}+\sqrt{\eps_n}
                      +\sqrt{\ell_n}\bigr)\norm{v_h}_U.
\end{align*}
Integration against $\lambda(s)$ proves
\eqref{eq:reaction-residual}. In particular, the time-averaged
reaction is not incorrectly evaluated on the current interface.
Jensen's inequality gives
$\sum_n\tau_n(\overline{\norm\lambda}_{\Gamma}^{\,n})^2\le\Lambda^2$.

Apply \eqref{eq:time-modulus-used} for $s\in(t_{n-1},t_n)$,
average in $s$, and then use Cauchy--Schwarz. This bounds
$\norm{w(t_n)-\bar w^n}_U$ by
$C\sqrt{\tau_n}(M_0+
\norm{\mathcal M_{\cdot}(\partial_tw)}_{L^2(t_{n-1},t_n)})$.
Summation proves \eqref{eq:viscous-time-defect}, with constants allowed
to depend on $T$.
\end{proof}

\subsection{The energy error estimate}
Define the explicit residual size
\begin{equation}\label{eq:residual-size}
 \begin{split}
 \mathfrak R_h={}&\sqrt{h+\bar\eps}\,(M_0+M_1)
          +\sqrt{\tau_{\max}}(M_0+M_1)\\
       &+(\sqrt{\bar\eps}+\sqrt{\bar\ell}
                       +\sqrt{\tau_{\max}})\Lambda
          +(h^2+\tau_{\max})\mathfrak G+E_f.
 \end{split}
\end{equation}
\begin{theorem}[Moving-geometry velocity convergence]
\label{thm:convergence}
Assume the collision-free rigid motion and Cartesian spaces specified
above, sufficiently small $h+\bar\eps$ for the collar construction,
and \eqref{eq:strong-regularity}--\eqref{eq:reaction-norm}.
Use the true affine initialization
\[
 u_h^0=g_h(0)+P^0\bigl(u(0)-g_h(0)\bigr)
\]
and the scheme \eqref{eq:scheme}, equivalently
\eqref{eq:centered-scheme}. Then
\begin{equation}\label{eq:moving-energy-error}
 \max_{0\le n\le N}\norm{u_h^n-u(t_n)}_0
 +\left(\sum_{n=1}^N\tau_n
                  \norm{u_h^n-u(t_n)}_U^2\right)^{1/2}
       \le C\mathfrak R_h.
\end{equation}
The constant is independent of the number and choices of mesh changes.
There is no restriction coupling $\tau_n$ to $h$, $\eps_n$, or a
remeshing frequency.

For the piecewise constant reconstruction $U_h(t)=u_h^n$ on
$(t_{n-1},t_n]$, set $U_h(0)=u_h^0$. The same right-hand side, with a
different constant, bounds
\begin{equation}\label{eq:moving-reconstruction-error}
 \sup_{0\le t\le T}\norm{U_h(t)-u(t)}_0
                      +\norm{U_h-u}_{L^2(0,T;U)}.
\end{equation}
Consequently, for a fixed exact solution in this class, simultaneous
$h,\bar\eps,\bar\ell,\tau_{\max},E_f\to0$ gives strong convergence in
both norms. In particular the rate for exact time-averaged loads is
$C_{\rm data}(\sqrt h+\sqrt{\bar\eps}
                       +\sqrt{\bar\ell}+\sqrt{\tau_{\max}})$.
\end{theorem}
\begin{proof}
Put $z^n=z_h(t_n)$ and $e^n=y_h^n-z^n$, where
$y_h^n=u_h^n-g_h(t_n)\in Z^n$. The previous comparison $z^{n-1}$
need not belong to the current space. Nevertheless, orthogonality
gives
\begin{equation}\label{eq:comparison-pairing}
 \ip{P^nz^{n-1}}{v_h}=\ip{z^{n-1}}{v_h}
                      \qquad(v_h\in Z^n).
\end{equation}
Thus subtraction of the comparison produces its ordinary Eulerian
time difference, with no additional projection-change term.
Average \eqref{eq:centered-reaction} in time and subtract it from the
discrete centered equation. Since
$\delta w^n=\overline{\partial_tw}^{\,n}$ and
$\delta z^n=\delta w^n-\delta\eta^n$, the result is
\begin{align}
 \ip{(e^n-P^ne^{n-1})/\tau_n}{v_h}+a(e^n,v_h)
     &=\mathcal R_n(v_h),\label{eq:error-equation}\\
 \mathcal R_n(v_h)
     &=\dual{\mathcal F_h^n-\overline{\mathcal F}^{\,n}}{v_h}
       +L_n(v_h)+\ip{\delta\eta^n}{v_h}\nonumber\\
     &\quad +a(\eta(t_n),v_h)
       -a(w(t_n)-\bar w^n,v_h).
       \nonumber
\end{align}
This is the usual residual-energy argument for backward Euler; compare
the dynamic-space argument in \citet[Section 3.1]{brenner2014}.
The moving comparison estimates and the moving-interface reaction
bound above are the specific ingredients that close it here.

Let
$r_n=\sup_{0\ne v_h\in Z^n}|\mathcal R_n(v_h)|/\norm{v_h}_U$,
with $r_n=0$ if $Z^n=\{0\}$. From
\eqref{eq:comparison-estimates}, Jensen's inequality, and
Lemma~\ref{lem:consistency},
\begin{equation}\label{eq:residual-sequence-bound}
 \left(\sum_n\tau_nr_n^2\right)^{1/2}\le C\mathfrak R_h.
\end{equation}
In particular,
$\sum_n\tau_n\norm{\delta\eta^n}_0^2
\le\norm{\partial_t\eta}_{L^2(0,T;L^2)}^2$;
no global $H^1$ regularity of $\partial_tw$ is used.

Test \eqref{eq:error-equation} with $e^n$ and multiply by $2\tau_n$.
By \eqref{eq:comparison-pairing} and polarization,
\[
 \norm{e^n}_0^2-\norm{e^{n-1}}_0^2
 +\norm{e^n-e^{n-1}}_0^2+2\nu\tau_n\norm{e^n}_U^2
       =2\tau_n\mathcal R_n(e^n).
\]
Young's inequality and summation therefore give, for every $k$,
\begin{equation}\label{eq:error-energy}
 \norm{e^k}_0^2+\sum_{n=1}^k\norm{e^n-e^{n-1}}_0^2
       +\nu\sum_{n=1}^k\tau_n\norm{e^n}_U^2
 \le\norm{e^0}_0^2+\nu^{-1}\sum_{n=1}^k\tau_nr_n^2.
\end{equation}
The affine initialization and $P^0z^0=z^0$ give
$e^0=P^0(\eta(0)+g(0)-g_h(0))$. Contractivity bounds its norm
by $Cr_hM_0+Ch^2\mathfrak G$.
Combining \eqref{eq:error-energy} with
\eqref{eq:residual-sequence-bound} proves the desired bounds for $e$.
Finally,
\[
 u_h^n-u(t_n)=e^n-\eta(t_n)+g_h(t_n)-g(t_n).
\]
The spatial recovery and lifting estimates prove
\eqref{eq:moving-energy-error}.

For the reconstructed energy norm, use
\eqref{eq:time-modulus-used} interval by interval and
$g\in H^1(0,T;U)$; the extra error is bounded by
$C\sqrt{\tau_{\max}}(M_0+M_1)
 +C\tau_{\max}\mathfrak G$.
For the reconstructed supremum bulk norm, apply the fundamental
theorem of calculus separately to $w$ and the smooth prescribed lift:
\[
 \sup_n\sup_{t\in[t_{n-1},t_n]}
       \norm{w(t_n)-w(t)}_0
       \le\sqrt{\tau_{\max}}
                  \norm{\partial_tw}_{L^2(0,T;L^2)},
\]
while the corresponding lift difference is at most
$\tau_{\max}\norm{\partial_tg}_{L^\infty(0,T;L^2)}$.
The first derivative norm is bounded by $CM_1$. These two terms are
already included in \eqref{eq:residual-size}, proving
\eqref{eq:moving-reconstruction-error}.
\end{proof}

\begin{corollary}[Vanishing centered transfer losses]
\label{cor:vanishing-loss}
Under the convergent parameter sequence in
Theorem~\ref{thm:convergence},
\[
 \sum_n\norm{y_h^n-y_h^{n-1}}_0^2\longrightarrow0.
\]
Hence both the accumulated centered projection loss and the
accumulated centered time-step loss in the orthogonal decomposition
of the energy identity tend to zero.
\end{corollary}
\begin{proof}
Equation~\eqref{eq:error-energy} implies
$\sum_n\norm{e^n-e^{n-1}}_0^2\to0$.
Moreover, $z_h=w-\eta\in H^1(0,T;L^2)$ and
\eqref{eq:comparison-estimates} bounds its derivative uniformly.
Thus
\[
 \sum_n\norm{z_h(t_n)-z_h(t_{n-1})}_0^2
       \le\tau_{\max}\norm{\partial_tz_h}_{L^2(0,T;L^2)}^2
       \longrightarrow0.
\]
The triangle inequality squared proves the assertion for $y_h=e+z_h$.
The two nonnegative orthogonal losses sum to that full increment loss.
\end{proof}

\begin{remark}[Endpoint forcing and the time rate]
\label{rem:endpoint-load}
The load term \eqref{eq:load-error} permits the independently evaluated
endpoint forcing used in a manufactured-solution test. For example,
suppose the bulk force has uniformly bounded values on each side of
the moving interface, and its sidewise Eulerian time derivative has
an $L^2$-in-time $L^\infty$ bound. Splitting into the region swept by
the interface between $s$ and $t$, of area at most $C|t-s|$, and its
complement gives
\[
 \norm{f(t)-f(s)}_{L^2(\Omega)}
       \le C\sqrt{|t-s|}\,M_{f,0}
             +C\int_{\min(s,t)}^{\max(s,t)}m_{f,1}(r)\,dr.
\]
The same averaging argument as in
\eqref{eq:viscous-time-defect}, followed by Poincare's inequality,
gives $E_f\le C\sqrt{\tau_{\max}}
(M_{f,0}+\norm{m_{f,1}}_{L^2(0,T)})$ for $f^n=f(t_n)$.
This additional regularity is required only for that choice of load.
The half-order time bound in the theorem allows a gradient jump to
move with the interface. A first-order energy estimate is not inferred
from the formal order of backward Euler.
\end{remark}