\section{A physical rigid-body model and its whole-box extension}
\label{sec:model}

\subsection{Exterior Stokes flow coupled to Newton's equations}
Let $\Omega=(0,1)^2$ and let $B_*\Subset\R^2$ be a fixed simply
connected body with connected $C^6$ boundary and centroid zero:
$\int_{B_*}x\,dx=0$. The body mass center is assumed to coincide with
this geometric centroid. Write
\[
 J=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad
 Q(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\
                           \sin\theta&\cos\theta\end{pmatrix}.
\]
For a pose $q=(c,\theta)$, define
$B(q)=c+Q(\theta)B_*$, $F(q)=\Omega\setminus\overline{B(q)}$,
and $\Gamma(q)=\partial B(q)$. The normal $n_q$ points out of the
body into the fluid. The analysis concerns a smooth solution branch
on $[0,T]$ for which the body has a positive clearance from the outer
wall. Quantitative regularity assumptions used for convergence are
stated below; no convergence of numerical geometries is assumed.

Let the fluid density and viscosity be $\rho_f>0$ and $\mu>0$.
The physical body mass and scalar moment of inertia about $c$ are
$m_s>0$ and $I_s>0$. With $\xi=(V,\omega)$, set
\[
 \mathcal R_q\xi(x)=V+\omega J(x-c),\qquad b(t,x)=\mathcal R_{q(t)}\xi(t)(x).
\]
The pose satisfies
\begin{equation}\label{eq:kinematics}
 \dot c=V,\qquad \dot\theta=\omega.
\end{equation}
The exterior fluid equations, with an \emph{Eulerian} time derivative,
are
\begin{equation}\label{eq:physical-stokes}
 \begin{aligned}
 \rho_f\partial_tu-\divg\sigma&=f
                      &&\text{in }F(q(t)),\\
 \divg u&=0,
 &\sigma&=2\mu D(u)-pI,\qquad
 D(u)=\tfrac12(\nabla u+\nabla u^{\mathsf T}),\\
 u&=0&&\text{on }\partial\Omega,\\
 u&=b&&\text{on }\Gamma(q(t)).
 \end{aligned}
\end{equation}
Here $f(t,x)$ is a given ambient extension of the exterior force
density, whose restriction to $F(q(t))$ is used in the equation.
The external force and torque on the body are denoted by $F_e(t)$
and $T_e(t)$. Newton's equations are
\begin{equation}\label{eq:newton}
 \begin{aligned}
 m_s\dot V&=F_e+\int_{\Gamma(q)}\sigma n_q\,ds,\\
 I_s\dot\omega&=T_e+
         \int_{\Gamma(q)}J(x-c)\cdot\sigma n_q\,ds.
 \end{aligned}
\end{equation}
The fluid traction in these equations is the exterior stress acting
on the body. We do not replace it by an uncorrected force spread over
a fictitious interior.

Define the mass and inertia of the fluid displaced by the body,
and their excess-body counterparts, by
\begin{equation}\label{eq:excess-inertia}
 \begin{aligned}
 m_f&=\rho_f|B_*|,&
 I_f&=\rho_f\int_{B_*}|x|^2\,dx,\\
 m_e&=m_s-m_f>0,& I_e&=I_s-I_f>0.
 \end{aligned}
\end{equation}
The strict inequalities are assumptions of the method analyzed here.
For a homogeneous body they follow from a solid density larger than
the fluid density. They make the discrete background-domain mass
form positive even before interior rigidity is accurately enforced.
The result does not cover zero or negative excess inertia.

\subsection{The fixed positive product space}
Let
\[
 U=\{v\in H^1_0(\Omega)^2:\divg v=0\},
 \qquad \norm v_U=\norm{\nabla v}_{L^2(\Omega)},
 \qquad \mathbb V=U\times\R^2\times\R.
\]
The ambient kinetic-energy space is
$\mathbb H=L^2(\Omega)^2\times\R^2\times\R$.
For $X=(u,V,\omega)$ and $Y=(v,W,\eta)$, define
\begin{equation}\label{eq:mass-inner-product}
 (X,Y)_M=\rho_f\ip uv+m_e V\cdot W+I_e\omega\eta,
 \qquad \norm X_M^2=(X,X)_M.
\end{equation}
The inner product is independent of the pose and of every mesh and
moment partition. The viscous and gyroscopic forms are
\begin{equation}\label{eq:viscous-gyro}
 a(u,v)=2\mu\ip{D(u)}{D(v)},
 \qquad
 \mathcal G(\varpi;X,Y)=m_f\varpi(JV)\cdot W.
\end{equation}
For $u,v\in U$, integration by parts gives
$a(u,v)=\mu\ip{\nabla u}{\nabla v}$ on the whole box.
Also $\mathcal G(\varpi;X,Y)=-\mathcal G(\varpi;Y,X)$, and
$\mathcal G(\varpi;X,X)=0$.

Extend the physical fluid velocity into the body by $u=b$ and define
the continuous joint constraint space
\begin{equation}\label{eq:joint-continuous-space}
 \mathbb K(q)=\{(v,W,\eta)\in\mathbb V:
                v=\mathcal R_q(W,\eta)\text{ a.e. in }B(q)\}.
\end{equation}
The extension belongs to $U$ because the exterior and interior traces
match. Its distributional time derivative contains no surface delta:
the coefficient of such a term would be the velocity jump, which is
zero. The exact joint state lies in $\mathbb K(q(t))$.
For any joint test $Y$, let
\begin{equation}\label{eq:external-load}
 \mathcal L(t,q;Y)=\int_{F(q)}f(t,x)\cdot v(x)\,dx
                         +F_e(t)\cdot W+T_e(t)\eta.
\end{equation}

\begin{proposition}[The Eulerian fictitious-inertia correction]
\label{prop:physical-model}
For a sufficiently regular collision-free solution of
\eqref{eq:physical-stokes}--\eqref{eq:newton}, the rigid extension
satisfies
\begin{equation}\label{eq:wholebox-coupled}
 (\partial_tX,Y)_M+a(u,v)+\mathcal G(\omega;X,Y)
         =\mathcal L(t,q;Y),\qquad Y\in\mathbb K(q(t)).
\end{equation}
Conversely, the sufficiently regular constrained formulation
\eqref{eq:kinematics}, \eqref{eq:wholebox-coupled} gives the exterior
Stokes and physical Newton equations, with the exterior pressure
and traction understood in their usual weak sense.
\end{proposition}
\begin{proof}
Since $c'=V$, the Eulerian interior derivative is
\begin{equation}\label{eq:eulerian-rigid-derivative}
 \partial_tb=\dot V+\dot\omega J(x-c)-\omega JV.
\end{equation}
For a rigid interior test $v=W+\eta J(x-c)$, the zero-centroid
condition and rigid invariance of area and polar moment imply
\begin{equation}\label{eq:interior-inertia-calculation}
 \rho_f\int_{B(q)}\partial_tb\cdot v\,dx
       =m_f(\dot V-\omega JV)\cdot W+I_f\dot\omega\eta.
\end{equation}
Thus the whole-box fluid inertia, the excess body inertia, and
$\mathcal G(\omega;X,Y)$ sum to the exterior fluid inertia plus
$m_s\dot V\cdot W+I_s\dot\omega\eta$.
The strain of the rigid extension is zero in the body, so the
whole-box viscous form equals the exterior viscous form for these
states and tests. Integration of the exterior fluid equation by parts
produces the traction work
$\int_\Gamma\sigma n_q\cdot\mathcal R_q(W,\eta)$ on the left.
Substituting \eqref{eq:newton} proves \eqref{eq:wholebox-coupled}.

Conversely, tests with zero rigid component and fluid support in the
exterior give the Stokes equation. Any rigid test velocity admits a
solenoidal, outer-zero extension: multiply its affine/quadratic
streamfunction by a cutoff equal to one near the body and take its
curl. Subtracting the exterior fluid weak identity for these tests
from \eqref{eq:wholebox-coupled}, and using
\eqref{eq:interior-inertia-calculation}, gives both equations in
\eqref{eq:newton}.
\end{proof}

The extra term $m_f\omega JV$ is necessary for the Eulerian Stokes
model. Excess-inertia formulas derived with a material acceleration,
as in the classical distributed-multiplier formulation
\citep[Section 4]{glowinski2001}, cannot be transferred unchanged
after deleting advection. For the present planar rigid motion, this
additional term is gyroscopic and contributes no kinetic-energy work.

\subsection{Volume and surface reactions}
For consistency estimates we need an identity on the full test space,
not only on exactly rigid tests. Under the same regularity, choose
the interior pressure to be zero and set
\begin{equation}\label{eq:reaction-representatives}
 \lambda_B=-\rho_f\partial_tb\quad\text{in }B(q),
 \qquad \lambda_\Gamma=\sigma n_q\quad\text{on }\Gamma(q).
\end{equation}
Then, for every $Y=(v,W,\eta)\in\mathbb V$, with $\zeta=(W,\eta)$,
\begin{equation}\label{eq:full-reaction}
 \begin{split}
 (\partial_tX,Y)_M+a(u,v)+\mathcal G(\omega;X,Y)
 &+\int_{B(q)}\lambda_B\cdot(v-\mathcal R_q\zeta)\,dx\\
 &+\int_{\Gamma(q)}\lambda_\Gamma\cdot
                  (\gamma_qv-\mathcal R_q\zeta)\,ds
       =\mathcal L(t,q;Y).
 \end{split}
\end{equation}
Indeed, the volume reaction cancels the fictitious interior fluid
inertia in the fluid equation. In the rigid equation its opposite
work restores the displaced-fluid inertia in
\eqref{eq:interior-inertia-calculation}; the gyroscopic term restores
the remaining $m_f\omega JV$. The surface reaction supplies the
physical exterior traction. This proves the identity by direct
substitution, without an inf--sup or multiplier-convergence theorem.
The displayed pair is one convenient continuous representative;
neither uniqueness of all reaction representations nor convergence
of discrete multipliers is needed below.

Surface velocity constraints alone do not ensure that an unsteady
fictitious interior has the rigid angular momentum used in
\eqref{eq:interior-inertia-calculation}; this issue is also discussed
by \citet{kallemov2016}. The discrete method therefore uses both
volume and surface moments. At a finite resolution its interior
rigidity is approximate, and its positive energy must be interpreted
as a surrogate, as quantified in
Proposition~\ref{prop:surrogate-physical-energy}.

\subsection{Physical energy of the continuous model}
For an exactly rigid extended state, the mass energy is precisely
the physical kinetic energy:
\begin{equation}\label{eq:continuous-physical-energy}
 \tfrac12\norm X_M^2
   =\frac{\rho_f}{2}\int_{F(q)}|u|^2\,dx
                       +\frac{m_s}{2}|V|^2+\frac{I_s}{2}\omega^2.
\end{equation}
Testing \eqref{eq:wholebox-coupled} by $X$ gives
\begin{equation}\label{eq:continuous-coupled-energy}
 \frac d{dt}\frac12\norm X_M^2+a(u,u)
                        =\mathcal L(t,q;X).
\end{equation}
The moving exterior does not leave an unaccounted kinetic-energy
flux in this identity. On the interface $u=b$ and
\[
 \int_{\Gamma(q)}|b|^2b\cdot n_q\,ds
       =\int_{B(q)}\divg(|b|^2b)\,dx=0,
\]
because $\divg b=0$ and
$b\cdot(\nabla b)b=b\cdot(\omega J)b=0$ pointwise.
The Reynolds-transport boundary term therefore vanishes after
integration. Finally $D(u)=0$ in the rigid interior, so
$a(u,u)=2\mu\int_{F(q)}|D(u)|^2$ is the physical fluid dissipation.