\section{A quantitative estimate for piecewise smooth solutions}\label{sec:rates}
We next give an a priori bound for a more regular solution. The estimate
is in the retained coarse mesh size; it is not an adaptive optimality
statement. The comparison-function and residual-energy strategy is standard;
compare the dynamic-mesh Stokes error analysis in
\citet[Section 3.1]{brenner2014}. The specific additional ingredient here is
one constructed comparison operator shared by all admissible spaces, so
that its mesh-change residual vanishes exactly. Write
\[
 Y=W^{3,\infty}(B)^2\times W^{3,\infty}(F)^2,
 \qquad
 \norm v_Y=\norm{v|_B}_{W^{3,\infty}(B)}
               +\norm{v|_F}_{W^{3,\infty}(F)}.
\]
The notation $u\in H^1(0,T;Y)$ below means a Bochner absolutely
continuous function whose value and time derivative have square-integrable
$Y$ norms. In particular $u$ has a continuous representative with values
in $Y$.

\begin{theorem}[Coarse-grid energy error bound]\label{thm:rate}
Assume the geometry and exact-integration discretization of the preceding
sections. Suppose the target velocity satisfies
\[
 u\in H^1(0,T;Y)\cap H^1(0,T;V).
\]
Suppose also that there is
$\lambda\in L^2(0,T;L^2(\Gamma)^2)$ such that, for almost every $t$,
\begin{equation}\label{eq:traction-identity}
 \ip{\partial_tu(t)}v+a(u(t),v)
       +\int_\Gamma\lambda(t)\cdot\gamma_\Gamma v\,ds
 =\dual{f(t)}v\qquad(v\in U).
\end{equation}
This assumption specifies the regularity of the continuous reaction; it
does not impose an inf--sup condition on the discrete multiplier.
For sufficiently small $h_m+\bar\eps_m$, the numerical velocity satisfies
\begin{align}
 &\max_{0\le n\le N_m}\norm{u_m^n-u(t_n)}_0
 +\left(\sum_{n=1}^{N_m}\tau_m^n
                   \norm{u_m^n-u(t_n)}_U^2\right)^{1/2}
 \nonumber\\
 &\quad\le C_{\nu,T,\Gamma,\rho}
 \left[
   \sqrt{h_m+\bar\eps_m}\,
       \bigl(\norm u_{L^\infty(0,T;Y)}
                    +\norm{\partial_tu}_{L^2(0,T;Y)}\bigr)
 \right.\nonumber\\
 &\qquad\left.
  +(\sqrt{\bar\eps_m}+\sqrt{\bar\ell_m})
                     \norm\lambda_{L^2(0,T;L^2(\Gamma))}
  +\tau_{\max,m}\norm{\partial_tu}_{L^2(0,T;U)}
 \right].\label{eq:rate}
\end{align}
The constant is independent of the selected refined cells, their relative
positions to $\Gamma$, the number of time steps, and the number of
refinement or coarsening events. For fixed solution norms, the right-hand
side is $O(\sqrt{h_m}+\sqrt{\bar\eps_m}
+\sqrt{\bar\ell_m}+\tau_{\max,m})$.
\end{theorem}

\begin{proof}
We suppress the computation index $m$. Let
$L_h=Q_{h,\bar\eps}:V\cap Y\to K_{h,\bar\eps}$ be the fixed linear comparison operator constructed
in Lemma~\ref{lem:collar}, with $K_{h,\bar\eps}\subset Z^n$ for every $n$. Put
\[
 z(t)=L_hu(t),\qquad \eta(t)=u(t)-z(t),\qquad
 e^n=u^n-z(t_n),\qquad d=h+\bar\eps.
\]
The same, time-independent operator is used throughout this computation.
The quantitative collar bound and its application to $\partial_tu$ give
\begin{equation}\label{eq:comparison-errors}
 \norm\eta_{L^\infty(0,T;U)}
       \le C\sqrt d\,\norm u_{L^\infty(0,T;Y)},\qquad
 \norm{\partial_t\eta}_{L^2(0,T;U)}
       \le C\sqrt d\,\norm{\partial_tu}_{L^2(0,T;Y)}.
\end{equation}
Here differentiation commutes with $L_h$ because it is a fixed linear
map. This is precisely where using a different comparison function after
each mesh change would introduce an additional term.

Let $\ubar^n$ and $\bar\lambda^n$ denote the exact time averages on
$(t_{n-1},t_n)$. Time averaging \eqref{eq:traction-identity} and
subtracting the resulting identity from \eqref{eq:scheme}, using
$P^nz(t_{n-1})=z(t_{n-1})$, gives
\begin{equation}\label{eq:error-equation}
 \ip{(e^n-P^ne^{n-1})/\tau_n}{v_h}+a(e^n,v_h)
    =\mathcal R_n(v_h)\qquad(v_h\in Z^n),
\end{equation}
where
\begin{align}
 \mathcal R_n(v_h)
  ={}&\ip{\tau_n^{-1}\int_{t_{n-1}}^{t_n}
                    \partial_t\eta(t)\,dt}{v_h}
       +a(\eta(t_n),v_h)\nonumber\\
    &+a(\ubar^n-u(t_n),v_h)
       +\int_\Gamma\bar\lambda^n\cdot\gamma_\Gamma v_h\,ds.
       \label{eq:residual}
\end{align}
All four residuals have a specified origin: comparison differentiation,
spatial comparison, time averaging, and the physical boundary reaction.
There is no remeshing residual.

Use $\norm v_a=\sqrt{a(v,v)}=\sqrt\nu\norm v_U$, and let $C_P$ be a
Poincare constant for $\Omega$. Lemma~\ref{lem:trace} gives
$|\mathcal R_n(v_h)|\le r_n\norm{v_h}_a$, with
\begin{align}
 r_n={}&\frac{C_P}{\sqrt\nu}
       \norm{\tau_n^{-1}\int_{t_{n-1}}^{t_n}
                    \partial_t\eta\,dt}_0
       +\sqrt\nu\norm{\eta(t_n)}_U\nonumber\\
    &+\sqrt\nu\norm{\ubar^n-u(t_n)}_U
       +\frac{C_\Gamma}{\sqrt\nu}
          (\sqrt{\eps_n}+\sqrt{\ell_n})
                           \norm{\bar\lambda^n}_{L^2(\Gamma)}.
                           \label{eq:residual-bound}
\end{align}
Test \eqref{eq:error-equation} with $e^n$, use orthogonal-projection
contractivity, and apply Young's inequality. This yields
\[
 \norm{e^n}_0^2-\norm{e^{n-1}}_0^2
       +\tau_n\norm{e^n}_a^2\le\tau_nr_n^2.
\]
Therefore
\begin{equation}\label{eq:error-stability}
 \max_n\norm{e^n}_0^2+\sum_n\tau_n\norm{e^n}_a^2
 \le2\left(\norm{e^0}_0^2+\sum_n\tau_nr_n^2\right).
\end{equation}
The initialization is compatible with the common comparison:
$e^0=P^0u_0-z(0)=P^0\eta(0)$, so
$\norm{e^0}_0\le\norm{\eta(0)}_0$.

Jensen's inequality controls the averaged $\partial_t\eta$ and
$\bar\lambda^n$ terms. For the time residual, the fundamental theorem of
calculus and Cauchy--Schwarz give
\begin{equation}\label{eq:time-residual}
 \sum_n\tau_n\norm{\ubar^n-u(t_n)}_U^2
 \le\tau_{\max}^2\norm{\partial_tu}_{L^2(0,T;U)}^2.
\end{equation}
Indeed, on a single interval the norm of
$u(t_n)-\ubar^n$ is at most
$\sqrt{\tau_n}\norm{\partial_tu}_{L^2(t_{n-1},t_n;U)}$.
Squaring the four terms of \eqref{eq:residual-bound}, using
$(x_1+\cdots+x_4)^2\le4\sum_i x_i^2$, and then applying
\eqref{eq:comparison-errors} proves
\begin{align*}
 \sum_n\tau_nr_n^2\le C_{\nu,\Gamma}\bigl[
 &d\{T\norm u_{L^\infty(0,T;Y)}^2
                      +\norm{\partial_tu}_{L^2(0,T;Y)}^2\}
  +\tau_{\max}^2\norm{\partial_tu}_{L^2(0,T;U)}^2\\
 &+(\sqrt{\bar\eps}+\sqrt{\bar\ell})^2
                   \norm\lambda_{L^2(0,T;L^2(\Gamma))}^2\bigr].
\end{align*}
Insert this into \eqref{eq:error-stability}, take square roots, and use
$u^n-u(t_n)=e^n-\eta(t_n)$ and Poincare's inequality. This proves
\eqref{eq:rate}.
\end{proof}

\begin{remark}[Nodal and continuous-time error norms]
The same right-hand side as in \eqref{eq:rate}, with a changed constant,
controls $\norm{U_m-u}_{L^2(0,T;U)}$, since
\[
 \sum_n\int_{t_{n-1}}^{t_n}\norm{u(t_n)-u(t)}_U^2\,dt
 \le\tau_{\max,m}^2\norm{\partial_tu}_{L^2(0,T;U)}^2.
\]
The first term in \eqref{eq:rate} is a maximum over time nodes. For the
piecewise constant $U_m$, an $L^\infty(0,T;L^2)$ error estimate includes
the additional bound
$\sqrt{\tau_{\max,m}}\norm{\partial_tu}_{L^2(0,T;L^2)}$.
For a first-order bound on that interpolation term, one can assume
$\partial_tu\in L^\infty(0,T;L^2)$.
The qualitative uniform convergence in Theorem~\ref{thm:convergence}
does not require this stronger regularity.
\end{remark}