text:
\begin{document}
\title[Existence and uniqueness for an FSI problem]
{On the existence and the uniqueness of the solution to a
fluid-structure interaction problem}
\author{Daniele Boffi}
\address{King Abdullah University of Science and Technology (KAUST), Saudi
Arabia and University of Pavia, Italy}
\email{daniele.boffi@kaust.edu.sa}
\urladdr{https://cemse.kaust.edu.sa/people/person/daniele-boffi}
\author{Lucia Gastaldi}
\address{DICATAM, Universit\`a di Brescia, Italy}
\email{lucia.gastaldi@unibs.it}
\urladdr{http://lucia-gastaldi.unibs.it}
\subjclass{65N30, 65N12, 74F10}

\begin{abstract}
In this paper we consider the linearized version of a system of partial
differential equations arising from a fluid-structure interaction model.
We prove the existence and the uniqueness of the solution under natural
regularity assumptions.
\end{abstract}

\maketitle

\section{Introduction}
\label{se:intro}

The mathematical analysis and the numerical approximation of problems
involving the interaction of fluids and solids are essential for the modeling
and simulation of a variety of applications related to engineering, physics,
and biology.

We consider a model presented in~\cite{BCG,BG} based on a fictitious domain
approach and the use of a distributed Lagrange multiplier. The considered
formulation is the evolution of a model originated from a finite element
approach of the immersed boundary method~\cite{BG_Bathe,BGHP,BCG_rho}. The
immersed boundary method has been introduced by Peskin and his collaborators
in several seminal papers~\cite{Peskin77,PeMcQ89,MQPe2000,PeAN} where the
interaction between the fluid and the solid was modeled by a suitably defined
Dirac delta function and the numerical approximation was performed by finite
differences.
One of the differences of the immersed boundary method with respect to other
possible approaches is that the discretization is performed by using two fixed
meshes: one for the fluid domain (artificially extended to include the
immersed solid) and one for the reference configuration of the solid. This
choice has the advantage that the computational meshes need not be updated at
each time step.  On other hand, the intersection between the fluid mesh and
the image of the solid mesh into the actual solid configuration needs to be
evaluated; in our approach the Lagrange multiplier is responsible for such
coupling.
The numerical analysis of the problem shows appealing properties related to
its stability~\cite{newren,BCG} and the numerical investigations demonstrate
the superiority of the finite element approach with respect to the original
finite difference scheme in terms of mass conservation.
Higher order time discretization has been investigated in~\cite{wolf}.

In this paper we address the study of the existence and uniqueness of the
continuous solution. The solution has four components: fluid velocity $\u$ and
pressure $p$ (extended into the solid region in the spirit of the fictitious
domain), the position of the solid domain inside the fluid, seen as a mapping
$\X$ from a reference configuration, and the Lagrange multiplier $\llambda$
supported in the solid reference domain that is used to enforce the coupling
between the solid and the fluid.
The problem is highly non linear; in particular the unknown $\X$ defines
mathematically the region occupied by the solid at a given time.
We consider a linearization of the problem with respect to the variable $\X$
and, for simplicity, we neglect the convective term of the Navier--Stokes
equations.

Our existence and uniqueness proof is based on a Faedo--Galerkin
approximation as done in~\cite{T} for the study of the Navier--Stokes
equations. We extend the results of~\cite{DGHL} where the coupling of the
incompressible Navier--Stokes equations with a linear elasticity model in a
fixed domain is considered.

Existence and uniqueness results for models related to fluid-structure
interactions have a limited but not empty occurrence in the literature.
In particular, some authors discussed the existence of weak solutions in the
case of a fluid containing rigid solids or elastic bodies whose behavior is
described by a finite number of
modes~\cite{CSMT,DE1999,DE2000,DEGLT,F,GM,GLS,HS,Serre,TaTu,Ta}.
Other results are available for the existence of weak solutions in the case of
a fluid enclosed in a solid
membrane~\cite{Beirao,CDEG,MC2013a,MC2013b,MC2014,MC2015,MC2016} or interacting
with a plate~\cite{FO};
the typical example of application is the blood flowing in a vessel~\cite{QTV}.
The existence of the solution in the case of viscoelastic particles immersed
in a Newtonian fluid is discussed in~\cite{LW} using the Eulerian description
for both fluid and solid.
Local-in-time existence and uniqueness of strong solutions for a model
involving an elastic structure immersed in a fluid is analyzed
in~\cite{Coutand2005,Coutand2006,RaymondVanni2014,Boulakia2017,Boulakia2019}.

In the next section we recall the strong formulation of our model.
Section~\ref{se:FD_DLM} presents the fictitious domain approach together with
its variational formulation. The linearized problem is described in
Section~\ref{se:linearized} and the main existence and uniqueness result for
the velocity $\u$ and the position of the solid $\X$ is stated and proved in
Section~\ref{se:main}. Finally, Section~\ref{se:pl} is devoted to the
existence and uniqueness of the pressure $p$ and the multiplier $\llambda$.

\section{Setting of the problem}

In this section we recall the formulation of the fluid-structure interaction
problem presented in~\cite{BG}. We assume that we are given a two or three
dimensional Lipschitz and convex domain $\Omega$ which is occupied by a fluid
and a solid.  We denote by $\Of_t$ and $\Os_t$ the regions where the fluid and
the solid are respectively located at time $t$, so that $\Omega$ is the
interior of $\overline\Omega^f_t\cup\overline\Omega^s_t$. The regularity of
the two subdomains will be made more precise later on as a consequence of
Assumption~\ref{ass}. For simplicity we assume that
$\partial\Os_t\cup\partial\Omega$ is empty, that is the solid is immersed in
the fluid, and the moving interface $\partial\Of_t\cap\partial\Os_t$ is
denoted by $\Gamma_t$.

We denote by $\u$, $\ssigma$, and $\rho$ the velocity, stress tensor, and mass
density, respectively, and we use subscripts $f$ or $s$ to refer to the fluid
or to the solid. We assume that the densities $\rho_f$ and $\rho_s$ are
positive constants.

The following equations represent the strong form of the problem we are
interested in, corresponding to the interaction of an incompressible fluid and
an incompressible immersed elastic structure.

\begin{equation}
\label{eq:general}
\aligned
&\rho_f\dot\u_f=\div\ssigma_f  &&\text{in }\Of_t\\
&\div\u_f=0 &&\text{in }\Of_t\\
&\rho_s\dot\u_s=\div\ssigma_s &&\text{in }\Os_t\\
&\div\u_s=0 &&\text{in }\Os_t\\
&\u_f=\u_s &&\text{on }\Gamma_t\\
& \ssigma_f\n_f=-\ssigma_s\n_s &&\text{on }\Gamma_t.
\endaligned
\end{equation}
The following initial and boundary conditions are imposed on $\partial\Omega$.

\begin{equation}
\label{eq:ibc}
\aligned
&\u_f(0)=\u_{f0} && \text{on }\Of_0,\\
&\u_s(0)=\u_{s0} && \text{on }\Os_0,\\
&\u_f(t)=0 && \text{on }\partial\Omega.
\endaligned
\end{equation}

The fluid stress tensor is defined by the Navier--Stokes law as it is common
for Newtonian fluids

\begin{equation}
\label{eq:sigmaf}
\ssigma_f=-p_f\mathbb{I}+\nu_f\Grads{\u_f},
\end{equation}
where $\Grads\u=(1/2)\left(\Grad\u+(\Grad\u)^\top\right)$ is the symmetric
gradient and $\nu_f$ represents the viscosity of the fluid.

The solid domain $\Os_t$ is the image of a reference domain $\B=\Os_0$.
Let $\X(t):\B\to\Os_t$ be the mapping that associates to each point $\s\in\B$ a
point $\x\in\Os_t$. When it is needed in order to avoid confusion we will use
the notation $\X(\s,t)$ to denote the dependence on both space and time.
We assume that $\X$ is one to one and that, for all $t\in[0,T]$,
$\|\X(\s_1,t)-\X(\s_2,t)\|\ge\gamma\|\s_1-\s_2\|$ for all $\s_1,\s_2\in\B$ for
a positive constant $\gamma$.
In particular, $\X(t)$ is invertible with Lipschitz inverse.

We denote by $\F=\Grad_s\X$ the deformation gradient; its determinant is
denoted by $|\F|$. We have that $|\F|$ is constant in time since the fluid and
the solid are incompressible; it is not restrictive to assume that
$\X_0(\s)=\s$ for $\s\in\B$, so that $|\F|=1$ for all $t$.

When dealing with moving domains it is essential to be precise with respect to
the Eulerian and Lagrangian descriptions of the involved quantities.
In~\eqref{eq:general}, we used the dot over the velocity in order to denote
the material time derivative. The Eulerian description of the fluid gives
$\dot\u_f=\partial\u_f/\partial t+\u_f\cdot\Grad\u_f$.
The spatial description of the material velocity in the solid, where the
Lagrangian representation is used, reads

\begin{equation}
\label{eq:materialvelocity}
\u_s(\x,t)=\frac{\partial\X(\s,t)}{\partial t}\Big|_{\x=\X(\s,t)}
\end{equation}

so that $\dot\u_s(\x,t)=\partial^2\X(\s,t)/\partial t^2|_{\x=\X(\s,t)}$.

Following what we did in~\cite{BGHP}, we consider a viscous-hyperelastic solid
structure and we define the Cauchy stress tensor as the sum of two
contributions $\ssigma_s=\ssigma_s^f+\ssigma_s^s$.
There is a fluid-like part of the stress

\begin{equation}
\label{eq:sigmas}
\ssigma_s^f=-p_s\mathbb{I}+\nu_s\Grads\u_s
\end{equation}

for a positive constant viscosity $\nu_s$ and an elastic part $\ssigma_s^s$.
The elastic part of the stress $\ssigma_s^s$ can be written in terms of the
first Piola--Kirchhoff stress tensor $\P$ with a change of variables from the
Eulerian to the Lagrangian framework:

\begin{equation}
\label{eq:Piola}
\P(\F(\s,t))=
|\F(\s,t)|\ssigma_s^s(\x,t)\F^{-\top}(\s,t)\qquad\text{for }\x=\X(\s,t).
\end{equation}

Following~\cite{BG} we are going to consider a linear dependence of $\P$ on
$\Grad_s\F$, namely

\begin{equation}
\label{eq:linearelasticity}
\P(\F)=\kappa\Grad_s\F.
\end{equation}

The following notation will be used throughout the paper.

If $D$ is a domain in $\RE^d$, we denote by
$W^{s,p}(D)$ the Sobolev space on $D$ ($s\in\RE$, $1\le p\le\infty$), and by
$\|\cdot\|_{s,p,D}$ its norm (see, for example,~\cite{Adams}). As usual we write
$H^s(D)=W^{s,2}(D)$ and omit $p$ in the norm and seminorm when
it is equal to $2$. Moreover, bold characters denote vector valued
functions and the corresponding functional spaces. The dual space of a Hilbert
space $X$ will be denoted with $X'$. The notation
$(\cdot,\cdot)_D$ stands for the scalar product in $L^2(D)$ and the duality
pairing is denoted by brackets $\langle\cdot,\cdot\rangle$.
The subscript indicating the domain is omitted if the domain is
$\Omega$, while we shall always use it for the reference domain
$\B$.
We will make use of the space $H^1_0(D)$ of functions in $H^1(D)$ with zero
trace on the boundary of $D$ and of its dual $H^{-1}(D)$. Moreover, $L^2_0(D)$
denotes the subspace of $L^2(D)$ of functions with zero mean value on $D$.

We denote by $\mathscr{D}(D)$ the space of $C^\infty$ functions
with compact support in $D$.
When $X$ is a Banach space, we denote by $L^p(0,T;X)$ ($1\le p\le\infty$) the
space of $L^p$-integrable functions from $(0,T)$ into $X$, which is a Banach
space with the norm

\[
\|v\|_{L^p(X)}=\left(\int_0^T\|v(t)\|_X^p\,dt\right)^{1/p}.
\]

Analogously, the space $C^m([0,T];X)$ denotes the space of functions from
$[0,T]$ to $X$ which are continuous up to the $m$-th derivative in $t$.

Finally, we are going to use the following spaces:

\begin{equation}
\label{eq:divfreespaces}
\aligned
&\mathscr{V}_0=\{\v\in\mathscr{D}(\Omega)^d: \div\v=0\}\\
&\H=\text{ the closure of }\mathscr{V}_0\text{ in }\Ldo\\
&\V=\text{ the closure of }\mathscr{V}_0\text{ in }\Huo.
\endaligned
\end{equation}

\section{Fictitious domain approach and Lagrange multiplier}
\label{se:FD_DLM}

We extend the fluid velocity and the pressure into the solid domain
by introducing new unknowns with the following meaning:

\begin{equation}
\label{eq:FDunknowns}
\u=\left\{
\begin{array}{ll}
\u_f&\text{ in } \Of_t\\
\u_s&\text{ in } \Os_t
\end{array}
\right.\qquad
p=\left\{
\begin{array}{ll}
p_f&\text{ in } \Of_t\\
p_s&\text{ in } \Os_t
\end{array}.
\right.
\end{equation}

The condition that the material velocity of the solid is equal to the
velocity of the fictitious fluid is expressed by

\begin{equation}
\label{eq:constraint}
\frac{\partial\X(\s,t)}{\partial t}=\u(\X(\s,t),t) \quad\text{for }\s\in\B.
\end{equation}

We introduce the following bilinear form:

\begin{equation}
\label{eq:defc}
\c(\mmu,\z)=\left(\Grad_s\mmu,\Grad_s\z\right)_\B+\left(\mmu,\z\right)_\B
\quad\forall\mmu,\ \z\in\Hub.
\end{equation}

It is obvious that for all $\mmu,\z\in\Hub$

\[
\aligned
&\c(\z,\z)=\|\z\|^2_{1,B}=\|\z\|^2_{0,\B}+\|\Grad_s\z\|^2_{0,B}\\
&\c(\mmu,\z)\le\|\mmu\|_{1,B}\|\z\|_{1,B}\\
&\c(\mmu,\z)=0 \text{ for all } \mmu\in\Hub \text{ implies } \z=0.
\endaligned
\]

System~\eqref{eq:general} can be formulated as follows.

\begin{problem}
\label{pb:pbvar}
Given $\u_0\in\Huo$, $\u_{s0}\in\mathbf{H}^1(\Os_0)$, and $\X_0(\s)=\s$ for
$\s\in\B$,
for almost every $t\in]0,T]$, find
$(\u(t),p(t))\in\Huo\times\Ldo$, $\X(t)\in\Hub$,
and $\llambda(t)\in\Hub$ such that it holds
\begin{subequations}
\begin{alignat}{2}
&\rho_f\frac d {dt}(\u(t),\v)+b(\u(t),\u(t),\v)+a(\u(t),\v)\qquad\notag&&\\
&\qquad-(\div\v,p(t))+\c(\llambda(t),\vcX)=0
   &&\quad\forall\v\in\Huo
     \label{eq:NS1_DLM}\\
&(\div\u(t),q)=0&&\quad\forall q\in\Ldo
     \label{eq:NS2_DLM}\\
&\dr\left(\frac{\partial^2\X}{\partial t^2}(t),\z\right)_{\B}
  +\kappa(\Grad_s\X(t),\Grad_s\z)_{\B}-\c(\llambda(t),\z)=0
  &&\quad\forall\z\in\Hub
     \label{eq:solid_DLM}\\
&\c\left(\mmu,\ucX-\frac{\partial\X}{\partial t}(t)\right)=0
  &&\quad\forall\mmu\in \Hub
     \label{eq:ode_DLM}\\
&\u(0)=\u_0\quad\mbox{\rm in }\Omega,
     \label{eq:ciu_DLM}\\
&\X(0)=\X_0\quad\mbox{\rm in }\B,\qquad \frac{\partial\X}{\partial t}(0)=\u_{s0}
\quad\mbox{\rm in }\B.
     \label{eq:ciX_DLM}
\end{alignat}
\end{subequations}

\end{problem}

Here $\dr=\rho_s-\rho_f$ and

\[
\aligned
&a(\u,\v)=(\nu\Grads{\u},\Grads{\v})\quad \text{with }
\nu=\left\{\begin{array}{ll}
\nu_f&\text{in }\Of_t\\
\nu_s&\text{in }\Os_t
\end{array}\right.
\\
&b(\u,\v,\w)=
\frac{\rho_f}2\left((\u\cdot\Grad\v,\w)-(\u\cdot\Grad\w,\v)\right).
\endaligned
\]

We assume that $\nu\in L^\infty(\Omega)$ and that there exists a positive
constant $\nu_0>0$ such that $\nu\ge\nu_0>0$ in $\Omega$, hence the following
Korn's inequality holds true for all $\u\in\Huo$

\begin{equation}
\label{eq:Korn}
a(\u,\u)\ge \mathbf{k}\|\Grad\u\|^2_\OO.
\end{equation}

We add the following compatibility conditions for the initial velocity
\begin{equation}
\label{eq:compatibility}
\div\u_0=0,\quad\text{and}\quad \u_0|_{\Os_0}=\u_{s0}.
\end{equation}

The second condition is related to the fact that we are assuming $\B=\Os_0$.

\section{Linearized problem}
\label{se:linearized}

We fix a function $\Xb$ which satisfies the following assumption.

\begin{ass}
\label{ass}
Let $\Xb\in C^{1}([0,T];\W^{1,\infty}(\B))$ be invertible with Lipschitz
inverse for all $t\in[0,T]$, with $\Xb(\s,0)=\s$ for $\s\in\B$.
In addition, we assume that

\begin{equation}
\label{eq:assX}
\Jb(t)=\textrm{det}(\Grad_s\Xb(t))=1 \quad\text{for all }t.
\end{equation}

\end{ass}

From now one we are going to neglect the convective term so that our problem
will read as follows.

\begin{problem}
\label{pb:pblin}
Given $\u_0\in\Huo$ and $\u_{s0}\in\Hub$,
for almost every $t\in]0,T]$ find
$(\u(t),p(t))\in\Huo\times\Ldo$, $\X(t)\in\Hub$,
and $\llambda(t)\in\Hub$ such that it holds
\begin{subequations}
\begin{alignat}{2}
&\rho_f\frac d {dt}(\u(t),\v)+a(\u(t),\v)-(\div\v,p(t))\qquad\notag&&\\
&\qquad+\c(\llambda(t),\vcXb)=0
   &&\quad\forall\v\in\Huo
     \label{eq:NS1_lin}\\
&(\div\u(t),q)=0&&\quad\forall q\in\Ldo
     \label{eq:NS2_lin}\\
&\dr\left(\frac{\partial^2\X}{\partial t^2}(t),\z\right)_{\B}
  +\kappa(\Grad_s\X(t),\Grad_s\z)_{\B}-\c(\llambda(t),\z)=0
  &&\quad\forall\z\in\Hub
     \label{eq:solid_lin}\\
&\c\left(\mmu,\ucXb-\frac{\partial\X}{\partial t}(t)\right)=0
  &&\quad\forall\mmu\in \Hub
     \label{eq:ode_lin}\\
&\u(0)=\u_0\quad\mbox{\rm in }\Omega,
     \label{eq:ciu_lin}\\
&\X(0)=\X_0\quad\mbox{\rm in }\B,\qquad \frac{\partial\X}{\partial t}(0)=\u_{s0}
\quad\mbox{\rm in }\B.
     \label{eq:ciX_lin}
\end{alignat}
\end{subequations}

\end{problem}

In the previous equations we used the notation $\vcXb=\v(\Xb(\cdot,t))$ and
$\ucXb=\u(\Xb(\cdot,t),t)$.

Let us split the second order in time Equation~\eqref{eq:solid_lin} into a
system of two differential equations of first order in time by introducing a
new unknown $\w=\frac{\partial\X}{\partial t}$. Then Problem~\ref{pb:pblin}
becomes:

\begin{problem}
\label{pb:linI}
Given $\u_0\in\Huo$ and $\u_{s0}\in \Hub$, for almost every $t\in]0,T]$ find
$(\u(t),p(t))\in\Huo\times\Ldo$, $(\X(t),\w(t))\in\Hub\times\Hub$,
and $\llambda(t)\in\Hub$ such that it holds
\begin{subequations}
\begin{alignat}{2}
&\rho_f\frac d {dt}(\u(t),\v)+a(\u(t),\v)-(\div\v,p(t))\qquad\notag&&\\
&\qquad+\c(\llambda(t),\vcXb)=0
   &&\quad\forall\v\in\Huo
     \label{eq:NS1_linI}\\
&(\div\u(t),q)=0&&\quad\forall q\in\Ldo
     \label{eq:NS2_linI}\\
&\dr\left(\frac{\partial\w}{\partial t}(t),\z\right)_{\B}
  +\kappa(\Grad_s\X(t),\Grad_s\z)_{\B}-\c(\llambda(t),\z)=0
  &&\quad\forall\z\in\Hub
     \label{eq:solid_linI}\\
&\left(\frac{\partial\X}{\partial t}(t),\y\right)_{\B}=(\w(t),\y)_{\B}
  &&\quad\forall\y\in\Ldb
     \label{eq:vel_linI}\\
&\c\left(\mmu,\ucXb-\w(t)\right)=0 &&\quad\forall\mmu\in \Hub
     \label{eq:ode_linI}\\
&\u(0)=\u_0\quad\mbox{\rm in }\Omega,
     \label{eq:ciu_linI}\\
&\X(0)=\X_0\quad\mbox{\rm in }\B,\qquad \w(0)=\u_{s0}
\quad\mbox{\rm in }\B.
     \label{eq:ciX_linI}
\end{alignat}
\end{subequations}

\end{problem}

We set

\begin{equation}
\label{eq:kernel}
\Kt=\{(\v,\z(t))\in\V\times\Hub: \c(\mmu,\vcXb(t)-\z(t))=0\ \forall\mmu\in\Hub\}.
\end{equation}

We observe that~\eqref{eq:compatibility} implies that
$(\u_0,\u_{s0})\in\mathbb{K}_0$.

Problem~\ref{pb:linI} is equivalent to the following one.

\begin{problem}
\label{pb:inK}
Given $(\u_0,\u_{s0})\in\K_0$, for almost every $t\in]0,T]$, find
$(\u(t),\w(t))\in\Kt$ and $\X(t)\in\Hub$ such that

\begin{equation}
\label{eq:inK}
\aligned
&\rho_f\frac d {dt}(\u(t),\v)+a(\u(t),\v)
+\dr\left(\frac{\partial\w}{\partial t}(t),\z(t)\right)_{\B}
\\
&\qquad
+\kappa(\Grad_s\X(t),\Grad_s\z(t))_{\B}=0&&\quad \forall(\v,\z(t))\in\Kt\\
&\left(\frac{\partial\X}{\partial t}(t),\y\right)_{\B}=(\w(t),\y)_{\B}
  &&\quad\forall\y\in\Ldb\\
&\u(0)=\u_0\quad\mbox{\rm in }\Omega,\qquad \w(0)=\u_{s0} \quad\mbox{\rm in }\B,
\\
&\X(0)=\X_0\quad\mbox{\rm in }\B,
\endaligned
\end{equation}

\end{problem}

In the following section we are going to prove existence and uniqueness of the
solution to Problem~\ref{pb:inK}. In the next section we will show
existence and existence for the pressure $p$ and the multipier $\lambda$ as
well.

\section{Existence and uniqueness}
\label{se:main}

We start this section by showing existence and uniqueness of the solution to
Problem~\ref{pb:inK} by following the Galerkin approximation technique used
in~\cite[Chapt.III.1]{T}. The proof of the next theorem will be obtained in
several steps.

\begin{thm}
\label{th:existence}
We set $\X_0(\s)=\s$ for $\s\in\B$.
Let $\Xb\in C^1([0,T];W^{1,\infty}(\B))$ be such that Assumption~\ref{ass} is
satisfied. Then, given $\u_0\in\V$ and $\u_{s0}\in\Hub$
satisfying the compatibility condition~\eqref{eq:compatibility}, for a.e.
$t\in(0,T)$ there exist $(\u(t),\w(t))\in\Kt$ and $\X(t)\in\Hub$ satisfying
Problem~\ref{pb:inK} and

\[
\aligned
&\u\in L^\infty(0,T;\H)\cap L^2(0,T;\V)\\
&\w\in L^\infty(0,T;\Ldb)\cap L^2(0,T;\Hub)\\
&\X\in L^\infty(0,T;\Hub)\quad \text{with }
\frac{\partial\X}{\partial t}\in L^\infty(0,T;\Ldb)\cap L^2(0,T;\Hub).
\endaligned
\]

\end{thm}

\subsection{Basis in $\Kt$}
We introduce a basis in $\Kt$ that will be used for the Galerkin approximation
of our problem.
Let $\ppsi_j$ ($j\in\NA$) be the complete set of eigenfunctions for the
eigenvalue problem: find $\lf\in\RE$ and $\ppsi\in\V$ with $\ppsi\ne0$
such that

\begin{equation}
\label{eq:eigpsi}
a(\ppsi,\v)=\lf(\ppsi,\v)\qquad\forall \v\in\V.
\end{equation}

It is well known that the eigenvalues are positive and can be enumerated
in an increasing sequence going to $+\infty$. The associated eigenfunctions
$\{\ppsi_j\}_{j=1}^\infty$ are orthogonal with respect to
the scalar product in $\mathbf{L}^2(\Omega)$ and to the bilinear form
$a(\cdot,\cdot)$. We normalize them with respect to the $\mathbf{L}^2(\Omega)$
norm, so that $\|\ppsi_j\|_{0,\Omega}=1$ for all $j\in\NA$.

Moreover, let $\cchi_j$ ($j\in\NA$) be the complete set of
eigenfunctions for the eigenvalue problem: find $\ls\in\RE$ and $\cchi\in\Hub$
with $\cchi\ne0$ such that

\begin{equation}
\label{eq:eigchi}
\c(\mmu,\cchi)=\ls(\cchi,\mmu)_{\B}\qquad\forall\mmu\in\Hub.
\end{equation}

Also the eigenvalues of~\eqref{eq:eigchi} are positive and can be enumerated
in increasing sequence going to $+\infty$.
We have that $\{\cchi_j\}_{j=1}^\infty$ are orthogonal we respect to
the scalar product in $\Ldb$ and to the bilinear form $\c(\cdot,\cdot)$. We
normalize them with respect to $\c$ so that $\c(\cchi_j,\cchi_j)=1$ for all
$j\in\NA$.

\begin{proposition}
\label{pr:base_phi}
For $j\in\NA$ and $t\in[0,T]$, let us set $\pphi_j(t)=\ppsi_j\circ\Xb(t)\in\Hub$.
Then, for each $t\in[0,T]$, $\{\pphi_j(t)\}_{j=1}^\infty$ is a basis of
$\Hub$.
\end{proposition}
\begin{proof}
Given $\z\in\Hub$ we will show that it can be written as a combination of
the $\{\pphi_j(t)\}$'s. Thanks to the assumptions on $\Xb$, we have that
$\v_z(t)=\z\circ\Xb(t)^{-1}\in\HH^1(\overline{\Os_t})$ where
$\overline{\Os_t}=\Xb(\B,t)$. Let $\tilde\v_z(t)\in\Huo$ be an extension of
$\v_z(t)$ to $\Omega$, so that $\tilde\v_z(t)|_{\overline{\Os_t}}=\v_z(t)$. Then
we can write $\tilde\v_z(t)$ in terms of the basis functions $\ppsi_j$, that is

\[
\tilde\v_z(t)=\sum_{j=1}^\infty\alpha_j(t)\ppsi_j.
\]

By construction we have that $\tilde\v_z(\Xb(\cdot,t),t)=\z\in\Hub$, hence we
obtain

\[
\z=\tilde\v_z(\Xb(\cdot,t),t)=\sum_{j=1}^\infty\alpha_j(t)\ppsi_j\circ\Xb
=\sum_{j=1}^\infty\alpha_j(t)\pphi_j(t).
\]

\end{proof}

As a consequence of the previous proposition, a basis in $\Kt$ is given by
$\{(\ppsi_j,\pphi_j(t))\}_{j=1}^\infty\}$.

\subsection{Galerkin approximation}
We introduce a Galerkin approximation of the solution of Problem~\ref{pb:inK}.
Let us consider $\V^m=\Span(\ppsi_1,\dots,\ppsi_m)$,
$\W^m(t)=\Span(\pphi_1(t),\dots,\pphi_m(t))$, and
$\HH^m=\Span(\cchi_1,\dots,\cchi_m)$.
We define a subspace $\Kt^m$ of $\Kt$ generated by the first $m$
basis functions in $\Kt$ as follows

\begin{equation}
\label{eq:Km}
\Kt^m=\{(\v,\z(t))\in\V^m\times\W^m: \c(\pphi_i(t),\vcXb(t)-\z(t))=0\text{ for }
i=1,\dots,m\}
\end{equation}

It is clear that if $(\v,\z(t))\in\Kt^m$ then

\[
\v=\sum_{j=1}^m\alpha_j\ppsi_j\qquad \z(t)=\sum_{j=1}^m\alpha_j\pphi_j(t)
\]

for the same coefficients $\{\alpha_j\}$.
The Galerkin approximation of the solution of Problem~\ref{pb:inK} is given by

\begin{equation}
\label{eq:Galapprox}
\aligned
&\u^m(t)=\sum_{j=1}^m\alm_j(t)\ppsi_j,\quad
\w^m(t)=\sum_{j=1}^m\alm_j(t)\pphi_j(t), \quad\\
&\X^m(t)=\sum_{j=1}^m\bem_j(t)\cchi_j
\endaligned
\end{equation}

such that
\begin{equation}
\label{eq:inKm}
\aligned
&\rho_f\frac d {dt}(\u^m(t),\v)+a(\u^m(t),\v)
+\dr\left(\frac{\partial\w^m}{\partial t}(t),\z(t)\right)_{\B}\\
&\qquad
+\kappa(\Grad_s\X^m(t),\Grad_s\z(t))_{\B}=0&& \forall(\v,\z(t))\in\Kt^m\\
&\left(\frac{\partial\X^m}{\partial t}(t),\y\right)_{\B}=(\w^m(t),\y)_{\B}
  &&\forall\y\in\HH^m\\
&\u^m(0)=\u^m_0\ \text{ in }\Omega,\qquad
\w^m(0)=\u^m_{s0} \ \text{ in }\B\\
&\X^m(\s,0)=\X^m_0\ \text{ for }\s\in\B.
\endaligned
\end{equation}

The initial conditions in~\eqref{eq:inKm} are obtained by projecting the
initial data, that is

\begin{equation}
\label{eq:inKm_ini}
\aligned
&\u^m_0=\sum_{j=1}^m\alm_{0j}\ppsi_j\quad \text{with }
\alm_{0j}=\frac{(\u_0,\ppsi_j)}{(\ppsi_j,\ppsi_j)}\\
&\u^m_{s0}=\sum_{j=1}^m\alm_{0j}\pphi_j(0)\\
&\X^m_0=\sum_{j=1}^m\bem_{0j}\cchi_j\quad\text{with }
\bem_{0j}=\frac{(\s,\cchi_j)}{(\cchi_j,\cchi_j)},
\endaligned
\end{equation}

where we have taken into account the compatibility
assumption~\eqref{eq:compatibility}.

Using~\eqref{eq:Galapprox} in~\eqref{eq:inKm} we obtain for
$(\v,\z)=(\ppsi_i,\pphi_i(t))$ and $\y=\cchi_i$ the following system:

\begin{equation}
\label{eq:systemm}
\aligned
&\rho_f\sum_{j=1}^m\alpha_j'(t)(\ppsi_j,\ppsi_i)
+\sum_{j=1}^m\alpha_j(t)a(\ppsi_j,\ppsi_i)\\
&\quad
+\dr\left(\sum_{j=1}^m\left(\alpha_j'(t)\pphi_j(t)
+\alpha_j(t)\pphi'_j(t)\right),\pphi_i(t)\right)_{\B}\\
&\quad
+\kappa\sum_{j=1}^m\beta_j(t)(\Grad_s\cchi_j,\Grad_s\pphi_i(t))_{\B}=0\\
& \sum_{j=1}^m\beta_j'(t)(\cchi_j,\cchi_i)_{\B}=
\sum_{j=1}^m\alpha_j(t)(\pphi_j(t),\cchi_i)_{\B},
\endaligned
\end{equation}

where we omitted the superscript $m$ in order to simplify the notation.

Thanks Proposition~\ref{pr:base_phi}, $\pphi_j(t)\in\Hub$
can be written in terms of the basis $\{\cchi_i\}_{i=1}^\infty$ as follows

\begin{equation}
\label{eq:phi_chi}
\pphi_j(t)=\sum_{r=1}^\infty\delta_{jr}(t)\cchi_r\qquad\text{with }
\delta_{jr}(t)=\c(\pphi_j(t),\cchi_r),
\end{equation}

therefore $\delta_{jr}(t)$ inherits the regularity in time of $\pphi_j(t)$.

\begin{lemma}
\label{le:serie}
Under Assumption~\ref{ass}, we have for $j\in\NA$

\begin{equation}
\label{eq:serie}
\aligned
&\bigg\|\sum_{r=1}^\infty \delta_{jr}^2(t)c_r\bigg\|_{L^\infty(0,T)}\le C
\|\Xb\|^2_{L^\infty(L^\infty(\B))}\|\ppsi_j\|^2_\OO\\
&\bigg\|\sum_{r=1}^\infty \delta_{jr}^2(t)\bigg\|_{L^\infty(0,T)}
\le C
\|\Xb\|^2_{L^\infty(\W^{1,\infty}(\B))}\|\ppsi_j\|^2_{1,\Omega}
\\
&\bigg\|\sum_{r=1}^\infty (\delta_{jr}'(t))^2c_r\bigg\|_{L^\infty(0,T)}\le C
\|\Xb\|^2_{W^{1,\infty}(L^\infty(\B))}\|\ppsi_j\|^2_{1,\Omega},
\endaligned
\end{equation}

where $c_r=\|\cchi_r\|^2_{\OB}=\frac1{\lambda_{sr}}$ (see~\eqref{eq:eigchi}).
\end{lemma}

\begin{proof}
For each $j\in\NA$, $\ppsi_j\in\V$ is an eigenfunction of~\eqref{eq:eigpsi} with
$\|\ppsi_j\|^2_\OO=1$ and $\|\Grads{\ppsi_j}\|^2_\OO=\lambda_{fj}$.
Hence $\pphi_j$ is continuous from $[0,T]$ into $\Hub$ with the time
derivative in $L^\infty(0,T;\Ldb)$.
Taking into account the properties of the eigensolutions
of~\eqref{eq:eigchi} we have

\[
\|\pphi_j(t)\|^2_\OB=
\sum_{r=1}^\infty\delta_{jr}^2(t)\|\cchi_r\|^2_\OB
=\sum_{r=1}^\infty\delta_{jr}^2(t)c_r.
\]
Hence we have:

\[
\bigg\|\sum_{r=1}^\infty \delta_{jr}^2(t)c_r\bigg\|_{L^\infty(0,T)}=
\|\pphi_j(t)\|^2_{L^\infty(\Ldb)}
\le
C\|\Xb\|^2_{L^\infty(\L^\infty(\B))}\|\ppsi_j\|^2_{\OO}.
\]

Similarly, we set
$d_r=\|\Grad_s\cchi_r\|^2_\OB=\frac{\lambda_{sr}-1}{\lambda_{sr}}$ where
$\lambda_{sr}$ are the eigenvalues of~\eqref{eq:eigchi}. It is easy to see that
$0<\frac{\lambda_{s1}-1}{\lambda_{s1}}\le d_r<1$. Then we have
\[
\|\Grad_s\pphi_j(t)\|^2_\OB
=\sum_{r=1}^\infty\delta_{jr}^2(t)\|\Grad_s\cchi_r\|^2_\OB
=\sum_{r=1}^\infty\delta_{jr}^2(t)d_r.
\]
The above equations imply that for each $t\in[0,T]$ the series on the right
hand side is convergent and that the first two inequalities in~\eqref{eq:serie}
hold true. Let us now show the third one. We have that the time
derivative of $\pphi_j(t)$ is given by $\frac{\partial\pphi_j}{\partial t}(t)
=\Grad\ppsi_j\frac{\partial\Xb}{\partial t}(t)$, hence it belongs to
$L^\infty(0,T;\Ldb)$; moreover we have

\[
\frac{\partial\pphi_j}{\partial t}(t)=\sum_{r=1}^\infty \delta'_{jr}(t)\cchi_r,
\]

from which we obtain

\[
\|\pphi'_j(t)\|^2_\OB=\sum_{r=1}^\infty (\delta'_{jr})^2c_r
\]

and we conclude again that the series on the right hand side is convergent and
that the estimate in the second inequality of~\eqref{eq:serie} is verified.
\end{proof}

Using the expression~\eqref{eq:phi_chi} into~\eqref{eq:systemm}, we arrive at

\[
\aligned
&\rho_f\alpha_i'(t)+a(\ppsi_i,\ppsi_i)\alpha_i(t)\\
&\qquad+\dr\sum_{j=1}^m\left(C_{ij}(t)\alpha_j'(t)+D_{ij}(t)\alpha_j(t)\right)
+\kappa\sum_{j=1}^m \delta_{ij}(t)d_j\beta_j(t)=0\\
&\beta_i'(t)=\sum_{j=1}^m\alpha_j(t)B_{ji}(t),
\endaligned
\]
where $B(t)$, $C(t)$, $D(t)$, and $E(t)$ are real matrices in
$\RE^{m\times m}$ with elements

\begin{equation}
\label{eq:matrici}
\aligned
&B_{ji}(t)=\delta_{ji}(t)&&\quad
C_{ij}(t)=\sum_{r=1}^\infty \delta_{jr}(t)\delta_{ir}(t)c_r\\
&D_{ij}(t)=\sum_{r=1}^\infty \delta'_{jr}(t)\delta_{ir}(t)c_r&&\quad
E_{ij}(t)=\delta_{ij}(t)d_j.
\endaligned
\end{equation}

Let $\aalm(t)$ and $\bbem(t)$ be the vector valued functions with components
$\alm_j(t)$ and $\bem_j(t)$, respectively.
We have obtained the following system of linear ordinary differential equations

\begin{equation}
\label{eq:ode}
\aligned
&\left(\rho_f\I_m+\dr C(t)\right)(\aalm(t))'
+\left(a(\ppsi_i,\ppsi_i)\I_m+D(t)\right)\aalm(t)\\
&\qquad+E(t)\bbem(t)=0\\
&(\bbem(t))'=B^T\aalm(t)\\
&\aalm(0)=\aalm_0\\
&\bbem(0)=\bbem_0,
\endaligned
\end{equation}
where $\aalm_0$ and $\bbem_0$ are the vectors with components $\alm_{0j}$
and $\bem_{0j}$ (see~\eqref{eq:inKm_ini}.

\begin{lemma}
\label{le:matrici}
Under Assumption~\ref{ass},
the matrices $B(t)$, $C(t)$, $D(t)$, and $E(t)$ given by ~\eqref{eq:matrici}
are well defined and continuous in $[0,T]$. Moreover, the matrix
$\rho_f\I_m+\dr C(t)$ is invertible with continuous inverse.
\end{lemma}

\begin{proof}
By definition~\eqref{eq:phi_chi}, it is clear that the elements of $B(t)$ and
$E(t)$ are continuous in $[0,T]$.
Let us consider the elements of $C(t)$. Thanks to the Cauchy--Schwarz inequality
we have

\[
\aligned
\|C_{ij}\|_{L^\infty(0,T)}
&=\left|\sum_{r=1}^\infty \delta_{jr}(t)\delta_{ir}(t)c_r\right|\\
&\le\left(\sum_{r=1}^\infty \delta_{jr}^2(t)c_r\right)^{1/2}
\left(\sum_{r=1}^\infty \delta_{ir}^2(t)c_r\right)^{1/2}\\
&\le\|\pphi_j\|_{L^\infty(\Ldb)}\|\pphi_i\|_{L^\infty(\Ldb)}.
\endaligned
\]

Since $\pphi_j(t)$ for $j\in\NA$ is continuous in $[0,T]$ with values in
$\Hub$, the series $\sum_{r=1}^\infty \delta_{jr}^2(t)c_r$ is continuous in
$[0,T]$.
This implies that the elements of $C(t)$ are also continuous in $[0,T]$.

A similar argument shows that $\|D_{ij}\|_{L^\infty(0,T)}$ is bounded.

Now we show that $\rho_f\I_m+\dr C(t)$ is invertible.
Since $C(t)$ is symmetric, it is enough to show that $C$ is also positive
semidefinite that is $x^TCx\ge0$ for all $x\in\RE^m$. This can be obtained by
direct computation as follows

\[
\aligned
x^TCx&=
\sum_{i,j=1}^mx_i\left(\sum_{r=1}^\infty \delta_{ir}(t)c_r\delta_{jr}(t)\right)x_j
\\
&=\sum_{r=1}^\infty c_r \left(\sum_{i=1}^mx_i\delta_{ir}(t)\right)
\left(\sum_{j=1}^mx_j\delta_{jr}(t)\right)\\
&=\sum_{r=1}^\infty c_r \left(\sum_{i=1}^mx_i\delta_{ir}(t)\right)^2\ge0.
\endaligned
\]

\end{proof}

\begin{proposition}
\label{pr:ode_existence}
The system of ordinary differential
equations~\eqref{eq:ode} has a unique solution $\aalm\in C^1([0,T])$ and
$\bbem\in C^1([0,T])$.
\end{proposition}
\begin{proof}
As a consequence of Lemma~\ref{le:matrici}, the matrix $\rho_f\I_m+\dr C(t)$
is invertible with continuous inverse, hence the standard theory for systems
of linear first order ordinary differential equations gives
that~\eqref{eq:ode} has a unique solution in $C^1([0,T])$.
\end{proof}

The above proposition yields the existence of the solution of~\eqref{eq:inKm},
stated in the following theorem

\begin{thm}
\label{th:gal_existence}
There exists a unique solution $(\u^m(t),\w^m(t))\in\Kt^m$ and $\X^m(t)\in\HH^m$
of~\eqref{eq:inKm} and~\eqref{eq:inKm_ini} with

\begin{equation}
\label{eq:Galregu}
(\u^m,\w^m)\in C^1([0,T];\Kt^m), \quad \X^m\in C^1([0,T];\HH^m).
\end{equation}

\end{thm}

\subsection{A priori estimates}

We have the following a priori estimates for the solution of~\eqref{eq:inKm}
and~\eqref{eq:inKm_ini}.
\begin{proposition}
\label{pr:apriori}
The following bounds hold true with $C>0$ independent of $m$:

\begin{subequations}
\begin{alignat}{1}
&\|\u^m\|_{L^\infty(\Ld)}+\|\u^m\|_{L^2(\Huo)}\le
C\left(\|\u^m_0\|_\OO+\|\u^m_{s0}\|_\OB+|\B|^{1/2}\right)
\label{eq:aprioriu}\\
&\|\w^m\|_{L^\infty(\Ldb)}
+\|\w^m\|_{L^2(\Hub)}\le
C\left(\|\u^m_0\|_\OO+\|\u^m_{s0}\|_\OB+|\B|^{1/2}\right)
\label{eq:aprioriw}\\
&\|\X^m\|_{L^\infty(\Hub)}\le
C\left(\|\u^m_0\|_\OO+\|\u^m_{s0}\|_\OB+|\B|^{1/2}\right)
\label{eq:aprioriX}\\
&\left\|\frac{\partial\X^m}{\partial t}\right\|_{L^\infty(\Ldb)}
+\left\|\frac{\partial\X^m}{\partial t}\right\|_{L^\infty(\Hub)}\le
C\big(\|\u^m_0\|_\OO+\|\u^m_{s0}\|_\OB+|\B|^{1/2}\big),
\label{eq:aprioriXt}
\end{alignat}
\end{subequations}

where $|\B|$ stands for the measure of $\B$.
\end{proposition}

\begin{proof}
By definition~\eqref{eq:Galapprox} we have that

\[
\frac{\partial\u^m}{\partial t}\in L^2(0,T;\V)\qquad
\frac{\partial\w^m}{\partial t}\in L^2(0,T;\Hub)\qquad
\frac{\partial\X^m}{\partial t}\in L^2(0,T;\Hub)
\]

implying that

\[
\aligned
&2\left(\frac{\partial\u^m}{\partial t}(t),\u^m(t)\right)
=\frac{d}{dt}\|\u^m(t)\|^2_\OO\\
&2\left(\frac{\partial\w^m}{\partial t}(t),\w^m(t)\right)_{\B}
=\frac{d}{dt}\|\w^m(t)\|^2_\OB\\
&2\left(\Grad_s\frac{\partial\X^m}{\partial t}(t),\Grad_s\X^m(t)\right)_{\B}
=\frac{d}{dt}\|\Grad_s\X^m(t)\|^2_\OB.
\endaligned
\]

Let us take $(\v,\z(t))=(\u^m(t),\w^m(t))$ in the first equation
in~\eqref{eq:inKm}, then

\[
\aligned
&\frac{\rho_f}2\frac{d}{dt}\|\u^m(t)\|^2_\OO+\nu_0\|\Grads{\u^m(t)}\|^2_\OO\\
&\qquad
+\frac{\dr}2\frac{d}{dt}\|\w^m(t)\|^2_\OB
+\kappa(\Grad_s\X^m(t),\Grad_s\w^m(t))_{\B}\le0.
\endaligned
\]

Thanks to the fact that $(\u^m(t),\w^m(t))\in\Kt^m$, we have that
$\w^m(t)$ belongs to $\Hub$ and the second equation in~\eqref{eq:inKm} implies
that it is equal to the time derivative of $\X^m(t)$, so that the last
inequality can be rewritten as

\[
\aligned
&\frac{\rho_f}2\frac{d}{dt}\|\u^m(t)\|^2_\OO+\nu_0\|\Grads{\u^m(t)}\|^2_\OO\\
&\qquad
+\frac{\dr}2\frac{d}{dt}\|\w^m(t)\|^2_\OB
+\frac{\kappa}2\frac{d}{dt}\|\Grad_s\X^m(t)\|^2_\OB\le0.
\endaligned
\]

Integrating on $(0,t)$ with $t\in(0,T]$ and taking into
account~\eqref{eq:Korn}, we arrive at

\begin{equation}
\label{eq:final}
\aligned
\rho_f\|\u^m(t)\|^2_\OO&+2\mathbf{k}\int_0^t\|\Grad\u^m(\tau)\|^2_\OO\,d\tau

+\dr\|\w^m(t)\|^2_\OB+\kappa\|\Grad_s\X^m(t)\|^2_\OB\\
&\le\rho_f\|\u^m_0\|^2_\OO+\dr\|\u^m_{s0}\|^2_\OB+\kappa\|\X^m_0\|^2_\OB.
\endaligned
\end{equation}

Thanks to~\eqref{eq:inKm_ini}, the last inequality
implies~\eqref{eq:aprioriu}. In order to
obtain~\eqref{eq:aprioriw}, we observe that a.e. in $t$
$\w^m(t)=(\u^m\circ\Xb)(t)$ in $\Hub$. Therefore
\[
\Grad_s\w^m(t)=(\Grad\u^m\circ\Xb)(t)\Grad_s\Xb(t)
\]

and

\[
\aligned
\|\Grad_s\w^m\|^2_{L^2(\Ldb)}&=\int_0^T\|\Grad_s\w^m(t)\|^2_\OB\,dt\\
&=\int_0^T\|(\Grad\u^m\circ\Xb)(t)\Grad_s\Xb(t)\|^2_\OB\,dt\\
&\le \|\Grad_s\Xb(t)\|^2_{L^\infty(\L^\infty(\B))}
\int_0^T\|(\Grad\u^m\circ\Xb)(t)\|^2_\OB\,dt\\
&\le C\int_0^T\|\Grad\u^m\|^2_{0,\Os_t}\,dt\le C\|\Grad\u^m\|^2_{L^2(\Ld)}
\endaligned
\]

which together with~\eqref{eq:final} gives~\eqref{eq:aprioriw}.
It remains to bound $\X^m$. Since

\[
\frac{\partial\X^m(t)}{\partial t}=\w^m(t),
\]

we obtain~\eqref{eq:aprioriXt} directly.
Moreover, the inequality~\eqref{eq:final} gives the estimate for
$\|\Grad_s\X^m\|_{L^\infty(\Ldb)}$. Let us now estimate $\X^m(t)$ in the
$\Ld$-norm. Integrating in time the last equation, we obtain

\begin{equation}
\label{eq:Xm}
\X^m(t)=\X_0^m+\int_0^t\w^m(\tau)d\tau.
\end{equation}

After some computations, we get

\[
\aligned
\|\X^m(t)\|^2_\OB&\le\|\X_0^m\|^2_\OB
+\left\|\int_0^t\w^m(\tau)d\tau\right\|^2_\OB\\
&\le\|\X_0^m\|^2_\OB+\left\|t\int_0^t|\w^m(\tau)|^2d\tau\right\|^2_\OB\\
&\le\|\X_0^m\|^2_\OB+t\int_0^t\|\w^m(\tau)\|^2_\OB d\tau.
\endaligned
\]

It follows

\[
\aligned
&\|\X^m\|^2_{L^2(\Ldb)}\le\|\X_0^m\|^2_\OB+T\|\w^m\|^2_{L^2(\Ldb)}\\
&\|\X^m\|^2_{L^\infty(\Ldb)}\le\|\X_0^m\|^2_\OB
+T^2\|\w^m\|^2_{L^\infty(\Ldb)}.
\endaligned
\]

\end{proof}

\subsection{Passing to the limit}\leavevmode

\medskip
\noindent
{\bf Step 1}: $\u^m$ converges to $\u\in L^\infty(0,T;\H)\cap L^2(0,T;\V)$.\\
The a priori estimate~\eqref{eq:aprioriu} shows the existence of an element
$\u\in L^\infty(0,T;\H)$ and of a subsequence $m'\to\infty$ such that

\[
\u^{m'}\weakstar \u\quad\text{in }L^\infty(0,T;\H).
\]

This means that for each $\v\in L^1(0,T;\H)$

\begin{equation}
\label{eq:ws}
\int_0^T(\u^{m'}(t)-\u(t),\v(t))\,dt\to 0\qquad\text{as } m'\to\infty.
\end{equation}

Since $\u^{m'}$ is also bounded in $L^2(0,T;\V)$, we can extract another
subsequence (still denoted $\u^{m'}$) that converges weakly to
$\u^*\in L^2(0,T;\V)$, that is

\[
\u^{m'}\rightharpoonup\u^* \quad\text{in }L^2(0,T;\V).
\]

The above convergence means that

\begin{equation}
\label{eq:weak}
\int_0^T \langle\u^{m'}(t)-\u^*(t),\v(t)\rangle\,dt\to 0\qquad\text{as }
m'\to\infty\quad\forall\v\in L^2(0,T;\V').
\end{equation}

By the Riesz representation theorem, we can identify $\H$ with $\H'$, so that

\[
\V\subset\H=\H'\subset\V'.
\]

Moreover the duality pairing between $\V'$ and $\V$ can be identified to the
scalar product in $\H$ for $\u\in\V$ and $\v\in\H$, that is

\[
\langle\v,\u\rangle=(\v,\u)\quad\forall\v\in\H,\ \forall\u\in\V.
\]

Comparing~\eqref{eq:ws} and~\eqref{eq:weak} with $\v\in L^2(0,T;\H)$,
we obtain that

\begin{equation}
\label{eq:solu}
\u=\u^*\in L^\infty(0,T;\H)\cap L^2(0,T;\V).
\end{equation}

{\bf Step 2}: $\w^m$ converges to $\w$ in
$L^\infty(0,T;\Ldb)\cap L^2(0,T;\Hub)$.\\
With arguments similar to those used above, we obtain from~\eqref{eq:aprioriw}
the following convergence for some subsequences of $\w^{m'}$:

\begin{equation}
\label{eq:convw}
\aligned
&\w^{m'}\weakstar \w&&\text{in }L^\infty(0,T;\Ldb)\\
&\w^{m'}\rightharpoonup \w^*&&\text{in }L^2(0,T;\Hub).
\endaligned
\end{equation}

Using again the Riesz representation theorem, we have that
$\Hub\subset\Ldb\subset\Hub'$ and we can conclude that

\begin{equation}
\label{eq:solw}
\w=\w^* \in L^\infty(0,T;\Ldb)\cap L^2(0,T;\Hub).
\end{equation}

{\bf Step 3}: The limit $(\u(t),\w(t))$ is contained in $\Kt$.\\
By construction $(\u^m(t),\w^m(t))\in\Kt^m$, that is

\[
\c(\pphi_i(t),(\u^m\circ\Xb)(t)-\w^m(t))=0\quad \text{for }i=1,\dots,m.
\]

Since $\u^m\rightharpoonup\u$ in $L^2(0,T;\V)$, we have that
$\u^m\circ\Xb\rightharpoonup\u\circ\Xb$ in $L^2(0,T;\Hub)$.

Let us consider a scalar function $\phi\in C^\infty(0,T)$. Then we have that
$\phi\pphi_i$ belongs to $L^2(0,T;\Hub)$ and that

\[
\aligned
0&=\int_0^T\phi(t)\c(\pphi_i(t),(\u^m\circ\Xb)(t)-\w^m(t))\,dt\\
&=\int_0^T\c(\phi(t)\pphi_i(t),(\u^m\circ\Xb)(t)-\w^m(t))\,dt.
\endaligned
\]

Recalling that the bilinear form $\c(\cdot,\cdot)$ is the scalar product in
$\Hub$, we can pass to the limit as $m\to\infty$. The weak convergence of
$\u^m\circ\Xb$ and of $\w^m$ in $L^2(0,T;\Hub)$ implies

\begin{equation}
\label{eq:cuno}
\int_0^T\phi(t)\c(\pphi_i,\ucXb-\w(t))\,dt=0
\quad\text{for }i=1,\dots,m.
\end{equation}

The last equality is valid for each $i$, and by linearity for all
finite linear combinations of $\pphi_i(t)$. Using
Proposition~\ref{pr:base_phi}, by continuity, Equation~\eqref{eq:cuno} is
still valid for all $\mmu\in\Hub$ and implies that

\[
\c(\mmu,\ucXb-\w(t))=0\quad \forall\mmu\in\Hub
\]
holds true in the sense of distributions on $(0,T)$, so that we conclude that
$(\u(t),\w(t))$ belongs to $\Kt$.

\medskip\noindent
{\bf Step 4}: Limit of $\X^m$ and ${\partial\X^m}/{\partial t}$.\\
Since $\X^m$ and $\|{\partial\X^m}/{\partial t}\|$ are bounded in
$L^\infty(0,T;\Hub)$ and $L^\infty(0,T;\Ldb)$, respectively, there exist
$\X\in L^\infty(0,T;\Hub)$, $\Y\in L^\infty(0,T;\Ldb)$, and subsequences $m'$
such that

\begin{equation}
\label{eq:convX}
\aligned
&\X^{m'}\weakstar\X&&\text{in }L^\infty(0,T;\Hub)\\
&\frac{\partial\X^{m'}}{\partial t}\weakstar \Y&&
\text{in }L^\infty(0,T;\Ldb)
\endaligned
\end{equation}

in the sense that

\[
\aligned
&\int_0^T\langle\X^{m'}(t)-\X(t),\y(t)\rangle_{\B}\,dt\to 0
&&\text{as }m\to\infty\quad\forall\y\in L^1(0,T;\Hub')\\
&\int_0^T\left(\frac{\partial\X^{m'}}{\partial t}(t)-\Y(t),\y(t)\right)_{\B}\to0\
&&\text{as }m\to\infty \quad\forall\y\in L^1(0,T;\Ldb).
\endaligned
\]

Let us consider a scalar function $\phi(t)$ which is continuously
differentiable in $[0,T]$ and $\phi(T)=0$, and let us denote by $\phi'(t)$ its
derivative. Then for $j=1,\dots,m$

\[
\int_0^T\left(\frac{\partial\X^m}{\partial t}(t),\cchi_j\right)_{\B}\phi(t)\,dt
=-\int_0^T\left(\X^m(t),\phi'(t)\cchi_j\right)_{\B}\,dt
-(\X_0^m,\cchi_j)_{\B}\phi(0).
\]

From~\eqref{eq:inKm_ini} we have that $\X_0^m\to\X_0$ strongly in $\Hub$, hence
we can pass to the limit and obtain

\[
\int_0^T\left(\Y(t),\cchi_j\right)_{\B}\phi(t)\,dt
=-\int_0^T\left(\X(t),\phi'(t)\cchi_j\right)_{\B}\,dt-(\X_0,\cchi_j)_{\B}\phi(0).
\]

The above relation is valid for all finite linear combinations $\y$ of
$\cchi_j$ with $j=1,\dots,m$. Moreover, it depends linearly and continuously
on $\y\in\Ldb$; hence, it is valid for all $\y\in\Ldb$.
Taking $\phi\in{\mathscr D}(0,T)$ and integrating by parts,
we get the following equation in the sense of distributions:

\[
\left(\frac{\partial\X}{\partial t}(t),\y\right)_{\B}=(\Y(t),\y)_{\B}
\quad\forall\y\in\Ldb.
\]

{\bf Step 5}: ${\partial\X}/{\partial t}(t)=\w(t)$.\\
We have (see~\eqref{eq:inKm})

\[
\left(\frac{\partial\X^m}{\partial t}(t)-\w^m(t),\cchi_i\right)_{\B}=0
\qquad\forall\cchi_i\ i=1,\dots,m.
\]

The convergence of $\w^m$ and of $\X^m$ obtained in~\eqref{eq:convw}
and~\eqref{eq:convX} implies that

\[
\Y=\w \in L^\infty(0,T;\Ldb),
\]
therefore the limits $\X$ and $\w$ satisfy the second equation
in~\eqref{eq:inK}.

\noindent
{\bf Step 6}: Passing to the limit in Equation~\eqref{eq:inKm}.\\
Let $\phi(t)$ be defined as before. We have:

\[
\int_0^T\left(\frac{\partial\u^m}{\partial t},\ppsi_j\right)\phi(t)\,dt=
-\int_0^T(\u^m(t),\ppsi_j\phi'(t))\,dt-(\u^m(0),\ppsi_j)\phi(0)
\]

and

\[
\aligned
&\int_0^T\left(\frac{\partial\w^m}{\partial t},\pphi_j(t)\right)_{\B}\phi(t)\,dt=
-\int_0^T(\w^m(t),\pphi_j(t)\phi'(t))_{\B}\,dt\\
&\qquad-\int_0^T(\w^m(t),\pphi_j'(t)\phi(t))_{\B}\,dt
-(\w^m(0),\pphi_j(0))_{\B}\phi(0).
\endaligned
\]

Using these relations in~\eqref{eq:inKm} we obtain

\[
\aligned
&-\rho_f\int_0^T(\u^m(t),\ppsi_j\phi'(t))\,dt+\int_0^Ta(\u^m(t),\ppsi_j\phi(t))\,dt\\
&\qquad
-\dr\int_0^T(\w^m(t),\pphi_j(t)\phi'(t))_{\B}\,dt
-\dr\int_0^T(\w^m(t),\frac{\partial\pphi_j}{\partial t}(t)\phi(t))_{\B}\,dt\\
&\qquad
+\kappa\int_0^T(\Grad_s\X^m(t),\Grad_s\pphi_j(t)\phi(t))_{\B}\,dt\\
&\quad=\rho_f(\u^m(0),\ppsi_j)\phi(0)+\dr(\w^m(0),\pphi_j(0))_{\B}\phi(0).
\endaligned
\]

For $j$ fixed, passing to the limit yields

\begin{equation}
\label{eq:i}
\aligned
&-\rho_f\int_0^T(\u(t),\ppsi_j\phi'(t))\,dt+\int_0^Ta(\u(t),\ppsi_j\phi(t))\,dt\\
&\qquad
-\dr\int_0^T(\w(t),\pphi_j(t)\phi'(t))_{\B}\,dt
-\dr\int_0^T(\w(t),\frac{\partial\pphi_j}{\partial t}(t)\phi(t))_{\B}\,dt\\
&\qquad
+\kappa\int_0^T(\Grad_s\X(t),\Grad_s\pphi_j(t)\phi(t))_{\B}\,dt\\
&\quad=\rho_f(\u_0,\ppsi_j)\phi(0)+\dr(\u_{s0},\pphi_j(0))_{\B}\phi(0).
\endaligned
\end{equation}

Each element $(\v,\z(t))\in\Kt^m$ can be written as

\[
\aligned
&\v=\sum_{j=1}^m a_j\ppsi_j\\
&\z(t)=\sum_{j=1}^m a_j\pphi_j(t)=\sum_{j=1}^m a_j\ppsi_j\circ\Xb(t).
\endaligned
\]

Let us denote by $\z'(t)$ the time derivative of $\z(t)$. We have

\[
\z'(t)=\sum_{j=1}^m a_j\Grad\ppsi_j\circ\Xb(t)\frac{\partial\Xb}{\partial t}(t)
\]
which, due to the regularity of $\Xb$, is continuous from $[0,T]$ into $\Ldb$.

We write~\eqref{eq:i} as follows

\begin{equation}
\label{eq:ognivz}
\aligned
&-\rho_f\int_0^T(\u(t),\v\phi'(t))\,dt+\int_0^Ta(\u(t),\v\phi(t))\,dt\\
&\qquad
-\dr\int_0^T(\w(t),\z(t)\phi'(t))_{\B}\,dt
-\dr\int_0^T(\w(t),\z'(t)\phi(t))_{\B}\,dt\\
&\qquad
+\kappa\int_0^T(\Grad_s\X(t),\Grad_s\z(t)\phi(t))_{\B}\,dt\\
&\quad=\rho_f(\u_0,\v)\phi(0)+\dr(\u_{s0},\z(0))_{\B}\phi(0).
\endaligned
\end{equation}

All the terms depend linearly and continuously on $(\v,\z)\in C^1([0,T];\Kt^m)$,
hence for each $t\in[0,T]$ Equation~\eqref{eq:ognivz} holds true for all
$(\v,\z)\in C^1([0,T];\Kt)$.

Taking $\phi\in{\mathscr D}(0,T)$ and integrating by parts with respect
to $t$, we arrive at

\[
\aligned
&\rho_f\int_0^T\left(\frac{\partial\u}{\partial t}(t),\v\right)\phi(t)\,dt
+\int_0^Ta(\u(t),\v)\phi(t)\,dt\\
&\qquad
+\dr\int_0^T\left(\frac{\partial\w}{\partial t}(t),\z(t)\right)_{\B}\phi(t)\,dt
+\kappa\int_0^T(\Grad_s\X(t),\Grad_s\z(t))_{\B}\phi(t)\,dt=0,
\endaligned
\]

which implies that the first equation in~\eqref{eq:inK} holds true in the sense
of distributions on $(0,T)$.

\medskip\noindent
{\bf Step 7}: Initial conditions.\\
It remains to check that $\u(0)=\u_0$, $\w(0)=\u_{s,0}$ and
$\X(0)=\X_0$.

Since $\frac{\partial\X}{\partial t}=\w\in L^2(0,T;\Hub)$, then $\X$ is
continuous from $[0,T]$ to $\Hub$, and we can pass to the limit
in~\eqref{eq:Xm} for $t=0$ arriving at $\X(0)=\X_0$.

We recall that

\[
(\u,\w)\in L^2(0,T;\V\times\Hub).
\]

Moreover, since $(\u(t),\w(t))\in\Kt$ and $\Kt\subset\V\times\Hub$, we have
that

\[
(\u,\w)\in L^2(0,T;\Kt).
\]

In order to prove the continuity of $\u$ and $\w$ in the correct spaces
for the initial conditions, we can use a general interpolation theorem.
If we can show

\begin{equation}
\left(\frac{\partial\u}{\partial t},\frac{\partial\w}{\partial t}\right)\in
L^2(0,T;\V'\times\Hub')
\label{eq:HB}
\end{equation}

then it follows

\[
(\u,\w)\in C^0([0,T];\H\times\Ldb).
\]

Given $(\v,\z(t))\in\Kt$ the first equation in~\eqref{eq:inK} can be written
as

\[
\rho_f\left(\frac{\partial\u}{\partial t}(t),\v\right)+
\dr\left(\frac{\partial\w}{\partial t}(t),\z(t)\right)_{\B}
=-\langle A\u(t),\v\rangle-\kappa(\Grad_s\X(t),\Grad_s(\z(t)))_{\B},
\]

where $A:\V\to \V'$ is the linear and continuous operator associated with the
bilinear form $a$.
Since $\u\in L^2(0,T;\V)$, the function $A\u$ belongs to $L^2(0,T;\V')$.
Taking into account that $\X\in L^2(0,T;\Hub)$ we obtain also

\begin{equation}
\label{eq:secondo}
\int_0^T(\Grad_s\X(t),\Grad_s(\vcXb(t)))_{\B}\,dt\le
C \|\Grad_s\X(t)\|_{L^2(\Ldb)}\|\v\|_{\V}.
\end{equation}

It follows that

\[
\left(\frac{\partial\u}{\partial t},\frac{\partial\w}{\partial t}\right)\in
L^2(0,T;\Kt'),
\]

which, from Hahn--Banach theorem implies~\eqref{eq:HB}.

The general interpolation theory of Lions--Magenes~\cite{LM}
and~\cite[Lemma~1.2]{T} implies that $\u$ is continuous form $[0,T]$ to $\H$
and $\w$ is continuous from $[0,T]$ to $\Ldb$.

It remains to check that $\u(0)=\u_0$ and $\w(0)=\u_{s0}$.
We multiply the first equation in~\eqref{eq:inK} by a scalar function
$\phi(t)$, continuously differentiable on $[0,T]$ with $\phi(T)=0$, and
integrate with respect to $t$

\[
\aligned
&\int_0^T\rho_f\frac d {dt}(\u(t),\v)\phi(t)\,dt
+\int_0^T\dr\left(\frac{\partial\w}{\partial t}(t),\z(t)\right)_{\B}\phi(t)\,dt
\\
&+\int_0^Ta(\u(t),\v)\phi(t)\,dt
+\int_0^T\kappa(\Grad_s\X(t),\Grad_s\z(t))_{\B}\phi(t)\,dt=0.
\endaligned
\]

Integration by parts in the first two integrals gives for $(\v,\z(t))\in\Kt$

\[
\aligned
&\rho_f\int_0^T\frac d {dt}(\u(t),\v)\phi(t)\,dt
+\dr\int_0^T
\left(\frac{\partial\w}{\partial t}(t),\z(t)\right)_{\B}\phi(t)\,dt\\
&\quad =-\rho_f\int_0^T(\u(t),\v)\phi'(t)\,dt
-\dr\int_0^T\left(\w(t),\z(t)\phi'(t)
+\frac{\partial\z}{\partial t}(t)\phi(t)\right)_{\B}\,dt\\
&\qquad -\rho_f(\u(0),\v)\phi(0)-\dr(\w(0),\z(0))_{\B}\phi(0),
\endaligned
\]

which inserted in the previous equation gives

\[
\aligned
&-\rho_f\int_0^T(\u(t),\v)\phi'(t)\,dt
-\dr\int_0^T\left(\w(t),\z(t)\phi'(t)
+\frac{\partial\z}{\partial t}(t)\phi(t)\right)_{\B}\,dt\\
&\qquad
+\int_0^Ta(\u(t),\v)\phi(t)\,dt
+\int_0^T\kappa(\Grad_s\X(t),\Grad_s\z(t))_{\B}\phi(t)\,dt\\
&\quad
=\rho_f(\u(0),\v)\phi(0)+\dr(\w(0),\z(0))_{\B}\phi(0).
\endaligned
\]

Comparing the last equality with~\eqref{eq:ognivz} gives for $\phi(0)\ne0$

\begin{equation}
\label{eq:initialcond}
\rho_f(\u(0)-\u_0,\v)+\dr(\w(0)-\u_{s0},\z(0))_{\B}=0
\end{equation}

for all $(\v,\z(0))\in\mathbb{K}_0$. Recalling that $\B=\Os_0$,
$\w(0)=\ucXb(0)$, and the
compatibility assumption~\eqref{eq:compatibility}, we can write the last
equation as

\[
\rho_f(\u(0)-\u_0,\v)_{\Of_0}+\rho_s(\u(0)-\u_{0},\v)_{\Os_0}=0,
\]

that is, with obvious notation,

$(\rho(\u(0)-\u_0),\v)_{\Omega}=0$
which gives that $\u(0)=\u_0$ in
$\Omega$. Substituting in~\eqref{eq:initialcond} we obtain also that
$\w(0)=\u_{s0}$ in $\B$.

\subsection{Uniqueness}
Let us assume that $(\u_1,\w_1,\X_1)$ and $(\u_2,\w_2,\X_2)$ are two solutions
of~\eqref{eq:inK}. Since the problem is linear the differences
$\hat\u=\u_1-\u_2$, $\hat\w=\w_1-\w_2$ and $\hat\X=\X^1-\X^2$ satisfy the
same equations as $\u,\w,\X$ with vanishing initial conditions, that is

\[
\aligned
&\rho_f\left(\frac{\partial\hat\u}{\partial t}(t),\v\right)+a(\hat\u(t),\v)
+\dr\left(\frac{\partial\hat\w}{\partial t}(t),\z(t)\right)_{\B}
\\
&\qquad
+\kappa(\Grad_s\hat\X(t),\Grad_s\z(t))_{\B}=0&&\quad \forall(\v,\z(t))\in\Kt\\
&\left(\frac{\partial\hat\X}{\partial t}(t),\y\right)_{\B}=(\hat\w(t),\y)_{\B}
  &&\quad\forall\y\in\Ldb\\
&\hat\u(0)=0\quad\mbox{\rm in }\Omega,\quad \hat\w(0)=0 \quad\mbox{\rm in }\B,
\quad
\hat\X(0)=0\quad\mbox{\rm in }\B,
\endaligned
\]

We take $(\v,\z(t))=(\hat\u(t),\hat\w(t))$ in the first equation and use the
fact that $\hat\w(t)=\frac{\partial\hat\X}{\partial t}(t)$, so that we get

\begin{equation}
\label{eq:uniq1}
\aligned
&\rho_f\left(\frac{\partial\hat\u}{\partial t}(t),\hat\u(t)\right)
+a(\hat\u(t),\hat\u(t))
+\dr\left(\frac{\partial\hat\w}{\partial t}(t),\hat\w(t)\right)_{\B}
\\
&\qquad
+\kappa
\left(\Grad_s\hat\X(t),\Grad_s\frac{\partial\hat\X}{\partial t}(t)\right)_{\B}=0.
\endaligned
\end{equation}

Thanks to~\eqref{eq:solu}, \eqref{eq:solw}, \eqref{eq:convX}, \eqref{eq:HB},
Step 5, and~\cite[Lemma III.1.2]{T}, we can write

\[
\aligned
&\left(\frac{\partial\hat\u}{\partial t}(t),\hat\u(t)\right)
=\frac12\frac{d}{dt}\|\hat\u(t)\|^2_\OO\\
&\left(\frac{\partial\hat\w}{\partial t}(t),\hat\w(t)\right)_{\B}
=\frac12\frac{d}{dt}\|\hat\w(t)\|^2_\OB\\
&\left(\Grad_s\hat\X(t),\Grad_s\frac{\partial\hat\X}{\partial t}(t)\right)_{\B}
=\frac12\frac{d}{dt}\|\Grad_s\hat\X(t)\|^2_\OB.
\endaligned
\]

Inserting these equalities into~\eqref{eq:uniq1} gives

\[
\frac{\rho_f}2\frac{d}{dt}\|\hat\u(t)\|^2_\OO+\mathbf{k}\|\Grad\hat\u(t)\|^2_\OO
+\frac{\dr}2\frac{d}{dt}\|\hat\w(t)\|^2_\OB
+\frac{\kappa}2\frac{d}{dt}\|\Grad_s\hat\X(t)\|^2_\OB\le0,
\]

which integrated from $0$ to $t$ implies

\[
\frac{\rho_f}2\|\hat\u(t)\|^2_\OO
+\frac{\dr}2\|\hat\w(t)\|^2_\OB
+\frac{\kappa}2\|\Grad_s\hat\X(t)\|^2_\OB\le0.
\]

Therefore $\u_1(t)=\u_2(t)$, $\w_1(t)=\w_2(t)$, and $\X_1(t)=\X_2(t)$ for all
$t$.

\section{Recovery of the pressure and of the Lagrange multiplier}
\label{se:pl}

In order to obtain existence and uniqueness of the solution of
Problem~\ref{pb:linI}, we need to show that starting from the solution
$(\u,\w,\X)$ of Problem~\ref{pb:inK}, we can define a Lagrange multiplier
$\llambda$ and a pressure $p$ so that $(\u,p,\X,\w,\llambda)$
satisfies~\eqref{eq:NS1_linI}-\eqref{eq:ciX_linI}.

\begin{proposition}
\label{pr:lambda}
Let $(\u,\w,\X)$ be the solution of Problem~\ref{pb:inK}, then there exists
$\llambda\in L^2(0,T;\Hub)$ such that for all $t\in(0,T)$

\begin{equation}
\label{eq:lambda}
\c(\llambda(t),\z)=
\dr\left(\frac{\partial\w}{\partial t}(t),\z\right)_{\B}
  +\kappa(\Grad_s\X(t),\Grad_s\z)_{\B}\qquad\forall\z\in\Hub.
\end{equation}

\end{proposition}

\begin{proof}
Since $\c$ is equal to the scalar product in $\Hub$, it is enough to show that
the right hand side is a linear functional on $\Hub$. The linearity is obvious.
We check now that both terms are continuous.
Since $\frac{\partial\w}{\partial t}\in L^2(0,T;\Hub')$ and
$\X\in L^2(0,T;\Hub)$ we have

\[
\aligned
&\int_0^T\left(\frac{\partial\w}{\partial t}(t),\z\right)_{\B}\,dt\le
\left\|\frac{\partial\w}{\partial t}\right\|_{L^2(\Hub')}\|\z\|_{\Hub}
\\
&\int_0^T (\Grad_s\X(t),\Grad_s\z)_{\B}\,dt\le\|\X\|_{L^2(\Hub)}\|\z\|_{\Hub}.
\endaligned
\]

These inequalities  imply that
the right hand side of~\eqref{eq:lambda} is a continuous functional on
$L^2(0,T;\Hub)$.
Therefore, from the Lax--Milgram lemma, we obtain existence and uniqueness of
the solution $\llambda\in L^2(0,T;\Hub)$.
\end{proof}

The above proposition allows us to split the first equation in~\eqref{eq:inK}
into two equations as follows:

\begin{equation}
\label{eq:split}
\aligned
&\rho_f\frac d {dt}(\u(t),\v)+a(\u(t),\v)+\c(\llambda(t),\vcXb(t))=0
&&\quad\forall\v\in\V\\
&\dr\left(\frac{\partial\w}{\partial t}(t),\z\right)_{\B}
+\kappa(\Grad_s\X(t),\Grad_s\z)_{\B}-\c(\llambda(t),\z)=0
&&\quad \forall\z\in\Hub.
\endaligned
\end{equation}

In order to obtain the solution of Problem~\ref{pb:linI}, it remains to show
the existence of $p$.

\begin{proposition}
\label{pr:p}
Let $(\u,\w,\X)$ and $\llambda$ be the solutions of Problem~\ref{pb:inK} and
of~\eqref{eq:lambda}. Then there exists a unique $p\in L^2(0,T;\Ldo)$ such that
$(\u,p,\X,\w,\llambda)$ is the solution of Problem~\ref{pb:linI}.
\end{proposition}

\begin{proof}
The existence and uniqueness of $(\u,\w,\X)$ and $\llambda$ are stated in
Theorem~\ref{th:existence} and in Proposition~\ref{pr:lambda}, respectively.
The pressure $p$ can be obtained as the solution of the following equation

\begin{equation}
\label{eq:defp}
(p(t),\div\v)=\rho_f\frac d {dt}(\u(t),\v)+a(\u(t),\v)+\c(\llambda(t),\vcXb(t))
\quad\forall\v\in\Huo.
\end{equation}

In order to see that this problem defines a function $p(t)$ satisfying the
required regularity, we can use standard arguments originating from the Banach
closed range theorem (see, for instance~\cite[Theorem~4.1.4]{bbf}). We need to
show that the right-hand side of~\eqref{eq:defp} is a linear and continuous
functional on $\Huo$ belonging to the polar set of the kernel of the
divergence operator in $\Huo$. Let us denote the right-hand side
of~\eqref{eq:defp} by $\ell(\v)$; the continuity of $\ell$ can be shown as
follows

\[
\aligned
&\left\vert\int_0^T\ell(\v)\,dt\right\vert\le
\int_0^T|\ell(\v)|\,dt\\
&\qquad\le
C\left(\left\|\frac{\partial\u}{\partial t}\right\|_{L^2(H^{-1}(\Omega))}
+\|\u\|_{L^2(\Huo)}+\|\llambda\|_{L^2(\Hub)}\right)\|\v\|_{\Huo}.
\endaligned
\]

Moreover, it is clear that $\ell$ belongs to the polar set of the kernel of the
divergence operator in $\Huo$: this is exactly what is stated in the first
equation of~\eqref{eq:split}.

From the closed range theorem, it follows that there exists $p(t)$
satisfying~\eqref{eq:defp} such that

\[
\|p\|_{L^2(\Ldo)}\le(1/\beta)\|\ell\|_{L^2(H^{-1}(\Omega))},
\]

where $\beta$ is the inf-sup constant associated with the divergence operator
in $\Huo$ (see~\cite[(I.1.51) and Prop.~I.1.2]{T}).

\end{proof}

\section*{Acknowledgments}
The authors are members of the INdAM Research group GNCS and their research
is partially supported by IMATI/CNR and by PRIN/MIUR.

\bibliographystyle{plain}
\bibliography{ref}

\end{document}


TITLE: Relationships between $\det(A+B)$ and $A+B$
QUESTION [4 upvotes]: When computing $\det(A+B)$ we notice that there is no relation between $\det A + \det B$.
However does the $\det(A+B)$ have any relation to the matrices $A+B$ as they stand?

REPLY [0 votes]: I can provide the following insight:
for every $A,B \in F^{n\times n}$
if we define $W=\left\{ M\in \left( \mathbb{Z} _{2}\right) ^{n\times n}| M \ is \ diagonal \right\} $ then the formal way of describing it is:
$$ det\left( A+B\right) =\sum _{M\in W}det\left( A\cdot M+B\cdot \left( I-M\right) \right) $$
in words, the sum of determinants above is the sum of all the determinants of the matrices that result from either removing the column from $B$ or from $A$ for each column in $A+B$ , the following is an example where $n = 2$ to better understand the idea:
$$
\left| A+B\right| =\begin{vmatrix}
a_{11}+b_{11} & a_{12}+b_{12} \\
a_{21}+b_{21} & a_{22}+b_{22}
\end{vmatrix}=\begin{vmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{vmatrix}+\begin{vmatrix}
b_{11} & a_{12} \\
b_{21} & a_{22}
\end{vmatrix}+\begin{vmatrix}
a_{11} & b_{12} \\
a_{21} & b_{22}
\end{vmatrix}+\begin{vmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{vmatrix}
$$
we can notice that:
$$
\left| A+B\right| =det(A)+\begin{vmatrix}
\vert & \vert \\
B_{1} & A_{2}  \\
\vert & \vert
\end{vmatrix}+\begin{vmatrix}
\vert & \vert \\
A_{1} & B_{2} \\
\vert & \vert
\end{vmatrix}+det(B)
$$
so in a way there is a relation between $det(A+B)$ and $det(A)+det(B)$ .


TITLE: tiling a rectangle with the smallest number of squares
QUESTION [33 upvotes]: This is based on another thread. For $m,n\in \mathbb N$, let $f(m,n)$ be the minimum number of squares with integer sides needed to tile a $m\times n$ rectangle. Recently, a table of values for $n\le m\le 85$, obtained by what seems to be a brute force search, has been put online here.
The table looks quite fuzzy, but if we restrict it to values of coprime $m,n$ such that $2n\ge m\ge n$, there is surprisingly little fluctuation. For convenience, the following table gives $f(m,n)$ in reverse order, showing in row $n$ the values for $m=n-1,n-2,...,\lbrace n/2\rbrace$ but putting "o" wherever $(m,n)>1$.
(The number following $m$ is $ g(m):=\frac{\log(m\sqrt{5})}{\log(\phi)}$ where $\phi=\frac{\sqrt{5}+1}2$, see below.)
 3 :  3.955  [ 3]
 4 :  4.553  [ 4]
 5 :  5.016  [ 5, 4]
 6 :  5.395  [ 5, o]
 7 :  5.716  [ 5, 5, 5]
 8 :  5.993  [ 7, o, 5]
 9 :  6.238  [ 7, 6, o, 6]
10 :  6.457  [ 6, o, 6, o]
11 :  6.655  [ 6, 7, 6, 6, 6]
12 :  6.836  [ 7, o, o, o, 6]
13 :  7.002  [ 7, 6, 7, 7, 6, 6]
14 :  7.156  [ 7, o, 7, o, 7, o]
15 :  7.299  [ 7, 8, o, 7, o, o, 8]
16 :  7.433  [ 7, o, 8, o, 7, o, 7]
17 :  7.559  [ 8, 8, 7, 8, 7, 7, 7, 8]
18 :  7.678  [ 8, o, o, o, 7, o, 7, o]
19 :  7.791  [ 7, 9, 7, 7, 7, 7, 7, 7, 7]
20 :  7.897  [ 9, o, 7, o, o, o, 7, o, 8]
21 :  7.999  [ 8, 7, o, 9, 8, o, o, 7, o, 7]
22 :  8.095  [ 8, o, 8, o, 8, o, 8, o, 8, o]
23 :  8.188  [ 8, 8, 8, 9, 8, 8, 8, 8, 8, 8, 8]
24 :  8.276  [ 8, o, o, o, 9, o, 8, o, o, o, 7]
25 :  8.361  [ 8, 8, 8, 8, o, 8, 8, 9, 8, o, 8, 8]
26 :  8.442  [ 8, o, 8, o, 8, o, 8, o, 9, o, 8, o]
27 :  8.521  [ 8,10, o, 8, 8, o, 8, 8, o, 8, 8, o, 8]
28 :  8.596  [ 8, o,10, o, 9, o, o, o, 8, o, 8, o, 8]
29 :  8.669  [ 9, 8, 8,10,10, 9, 9, 8, 9, 8, 8, 8, 9, 8]
30 :  8.740  [ 9, o, o, o, o, o, 9, o, o, o, 8, o, 9, o]
31 :  8.808  [ 8, 8, 8,10, 8, 8, 8, 8, 8, 8, 9, 8, 8, 8, 8]
32 :  8.874  [ 9, o, 8, o, 8, o, 9, o, 9, o, 8, o, 8, o, 9]
33 :  8.938  [ 9, 9, o, 9, 8, o, 8,10, o, 9, o, o, 8, 8, o, 9]
34 :  9.000  [ 9, o, 9, o, 9, o, 9, o, 8, o, 9, o, 8, o, 8, o]
35 :  9.060  [ 8, 9,10, 8, o, 9, o, 9, 8, o, 8, 9, 9, o, o, 8, 9]
36 :  9.119  [ 9, o, o, o,10, o,10, o, o, o, 9, o, 9, o, o, o,10]
37 :  9.176  [ 9, 9, 9, 9, 8,10, 9, 8, 9, 9, 9, 9, 8, 9, 8, 9, 8, 8]
38 :  9.231  [ 9, o, 9, o, 9, o, 9, o,10, o, 9, o, 9, o, 9, o,10, o]
39 :  9.285  [ 9, 9, o, 9, 9, o,10, 9, o, 9, 9, o, o, 9, o, 9, 9, o, 9]
40 :  9.338  [ 9, o, 9, o, o, o, 9, o, 9, o, 8, o, 9, o, o, o, 9, o, 8]
41 :  9.389  [ 9, 9, 9, 9,11, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
42 :  9.439  [ 9, o, o, o, 9, o, o, o, o, o,10, o,10, o, o, o,10, o,10, o]
43 :  9.488  [ 9, 9, 9, 9, 9, 9,10,10, 9, 9, 9, 9, 9, 9, 9, 9,10, 9, 9, 9, 9]
44 :  9.536  [ 9, o, 9, o, 9, o, 9, o, 9, o, o, o, 9, o, 9, o, 9, o, 9, o, 9]
45 :  9.582  [10, 9, o,10, o, o, 9, 9, o, o, 9, o, 9, 9, o,10, 9, o, 9, o, o, 9]
46 :  9.628  [ 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o]
47 :  9.673  [ 9, 9, 9,11, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9,10, 9, 9, 9, 9, 9, 9, 9, 9]
48 :  9.716  [10, o, o, o, 9, o, 9, o, o, o,10, o,10, o, o, o, 9, o, 9, o, o, o, 9]
49 :  9.759  [ 9, 9, 9, 9,10,10, o, 9, 9, 9, 9, 9, 9, o, 9,10,10, 9, 9,10, o, 9, 9, 9]
50 :  9.801  [10, o, 9, o, o, o, 9, o, 9, o, 9, o, 9, o, o, o, 9, o, 9, o, 9, o, 9, o]
51 :  9.842  [ 9,10, o, 9,10, o,10, 9, o,10,10, o, 9,10, o, 9, o, o, 9,10, o,10,10, o, 9]
52 :  9.883  [10, o,11, o, 9, o,10, o,10, o, 9, o, o, o, 9, o,10, o, 9, o, 9, o,10, o,11]
53 :  9.922  [10, 9, 9,11, 9,10,10, 9,10,11,10, 9,10, 9,10,10,11,10, 9, 9, 9, 9, 9,11,11, 9]
54 :  9.961  [10, o, o, o, 9, o,10, o, o, o, 9, o, 9, o, o, o, 9, o, 9, o, o, o, 9, o,10, o]
55 :  9.999  [10,10,10,10, o, 9, 9, 9, 9, o, o, 9,10, 9, o, 9, 9, 9,10, o, 9, o,10, 9, o, 9, 9]
56 :  10.03  [ 9, o, 9, o,10, o, o, o, 9, o,10, o,11, o,10, o,10, o, 9, o, o, o,10, o, 9, o, 9]
57 :  10.07  [10,10, o,10, 9, o,10,10, o,10, 9, o, 9,10, o,10,10, o, o, 9, o,10, 9, o,10, 9, o,10]
58 :  10.11  [10, o,10, o,10, o,10, o,10, o,11, o,10, o,10, o,10, o,11, o,10, o,10, o,10, o,10, o]
59 :  10.14  [10,10, 9, 9,10,11, 9, 9, 9,10,10, 9, 9,10, 9, 9, 9,10, 9, 9,10, 9,10, 9, 9, 9, 9, 9,10]
60 :  10.18  [10, o, o, o, o, o,11, o, o, o,10, o,10, o, o, o,11, o, 9, o, o, o,10, o, o, o, o, o, 9]
61 :  10.21  [10, 9,10,10,10,10, 9, 9,10,10, 9,11,10,10,10,10, 9, 9,10,10, 9,10, 9, 9,10, 9,10, 9, 9, 9]
62 :  10.24  [10, o,10, o,10, o,10, o,10, o,10, o,11, o,11, o,10, o,11, o,10, o,10, o,10, o,10, o,10, o]
63 :  10.28  [10,10, o,10,10, o, o,10, o,10,10, o,10, o, o,10,10, o,10,10, o,10,10, o,10,10, o, o,10, o,10]
64 :  10.31  [10, o,10, o,10, o,10, o,11, o,11, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10]
65 :  10.34  [10,10,10,10, o,10,10,10,10, o,10,10, o,10, o,11,10,10,10, o,10,10,10,10, o, o,10,10,11, o,10,10]
66 :  10.37  [10, o, o, o,10, o,10, o, o, o, o, o,10, o, o, o, 9, o,10, o, o, o,10, o,10, o, o, o, 9, o, 9, o]
67 :  10.40  [10,10,10,10,10,10,11,10,10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,10,10,10,10]
68 :  10.44  [10, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o, o, o,10, o,10, o,10, o,10, o,10, o,10, o,11, o,10]
69 :  10.47  [10,10, o,10,11, o,10, 9, o,10,10, o,10,11, o,11,11, o,10,11, o,10, o, o,10,10, o, 9, 9, o,10, 9, o, 9]
70 :  10.50  [11, o,10, o, o, o, o, o,10, o,10, o,10, o, o, o,11, o,10, o, o, o,10, o, o, o,10, o,10, o,10, o,10, o]
71 :  10.53  [10,10,10,12,10,10,10,10,10,11,10,12,10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,10,10,10]
72 :  10.55  [10, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,10]
73 :  10.58  [10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]
74 :  10.61  [10, o,10, o,10, o,10, o,10, o,10, o,11, o,11, o,10, o,11, o,10, o,10, o,10, o,10, o,11, o,10, o,10, o,10, o]
75 :  10.64  [10,11, o,10, o, o,10,10, o, o,10, o,10,10, o,10,10, o,10, o, o,10,11, o, o,10, o,10,10, o,10,10, o,10, o, o,10]
76 :  10.67  [10, o,10, o,10, o,11, o,10, o,10, o,10, o,10, o,10, o, o, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10]
77 :  10.69  [10,10,10,10,10,11, o,10,10,12, o,10,11, o,10,11,12,10,10,10, o, o,10,12,12,10,10, o,10,10,10,10, o,10, o,12,10,10]
78 :  10.72  [11, o, o, o,10, o,10, o, o, o,10, o, o, o, o, o,11, o,10, o, o, o,11, o,11, o, o, o,11, o,10, o, o, o,11, o,10, o]
79 :  10.75  [11,10,10,11,11,10,10,10,10,11,11,10,10,10,10,10,10,11,10,10,10,10,10,10,10,10,11,11,10,10,10,11,10,10,10,11,10,10,10]
80 :  10.77  [10, o,10, o, o, o,10, o,10, o,12, o,10, o, o, o,10, o,11, o,11, o,10, o, o, o,10, o,11, o,10, o,10, o, o, o,10, o,10]
81 :  10.80  [10,10, o,12,11, o,10,10, o,10,10, o,10,12, o,10,10, o,10,10, o,10,10, o,10,10, o,11,11, o,10,11, o,10,10, o,10,10, o,10]
82 :  10.82  [10, o,11, o,11, o,11, o,10, o,11, o,11, o,10, o,10, o,11, o,10, o,10, o,10, o,10, o,11, o,10, o,11, o,10, o,10, o,10, o]
83 :  10.85  [10,10,10,10,10,11,10,10,10,10,10,10,10,10,11,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,10,10,10,10,10,10]
84 :  10.87  [10, o, o, o,10, o, o, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,11, o,10, o, o, o,10, o,10, o, o, o, o, o,10, o, o, o,10]
85 :  10.90  [10,10,11,10, o,12,11,10,10, o,10,12,10,10, o,11, o,10,10, o,10,11,11,10, o,10,11,11,10, o,10,10,10, o, o,10,11,10,10, o,10,10]

For a Fibonacci rectangle, we obviously have $f(F_{k+1},F_k)\le k$, and it seems straightforward to show that this bound is sharp. But is this really trivial?

It looks like under the above restrictions on $m$ and $n$, the values of $f(m,n)$ are very close to $g(m)$, more precisely $$\boxed{\lfloor g(m)\rfloor -1\le f(m,n)\le \lceil g(m)\rceil +1}.$$ Is it possible that in the minimal tilings, patterns like in a 'Fibonacci rectangle tiling' occur frequently?
Note that it is already known or at least plausible from the article quoted in the first thread that $f(m,n)\sim g(m)$.
What about $f(m,n)$ if the rectangle sides are not coprime? Obviously $f(km,kn)\le f(m,n)$ for $k\in \mathbb N$. In the range of the table, there is equality everywhere.

Is anything known concerning the conjecture $f(km,kn)= f(m,n)$?
Moreover, does there even exist a minimal tiling of a $km\times kn$ rectangle such that not all square sides are multiples of $k$?

For $m> n$, let’s call a $m\times n$ rectangle reducible if $f(m,n)=f(n,m-n)+1$. This means there is a minimal tiling such that the biggest square has side $n$. (There may exist other minimal tilings also, though I'd rather doubt it).
Using the same order as above, I have kept in row $m$ only those $n$’s for which the $m\times n$ rectangle is reducible and put "o" where it is not:
 3 : [ 2]
 4 : [ 3]
 5 : [ 4, 3]
 6 : [ o, 4]
 7 : [ o, 5, 4]
 8 : [ o, 6, 5]
 9 : [ o, 7, 6, 5]
10 : [ o, 8, 7, 6]
11 : [ o, 9, 8, 7, 6]
12 : [ o, o, 9, 8, 7]
13 : [ o, o,10, 9, 8, 7]
14 : [ o, o,11,10, 9, 8]
15 : [ o, o,12,11,10, 9, 8]
16 : [ o, o,13,12,11,10, 9]
17 : [ o, o, o,13,12,11,10, 9]
18 : [ o, o, o,14,13,12,11,10]
19 : [ o, o, o, o, o,13,12,11,10]
20 : [ o, o, o,16,15,14,13,12,11]
21 : [ o, o, o,17,16,15,14,13,12,11]
22 : [ o, o, o,18,17,16, o,14,13,12]
23 : [ o, o, o,19,18,17,16, o,14,13,12]
24 : [ o, o, o, o,19,18,17,16,15,14,13]
25 : [ o, o, o, o,20,19,18,17,16,15,14,13]
26 : [ o, o, o, o, o,20,19,18,17,16,15,14]
27 : [ o, o, o, o, o,21,20,19,18,17,16,15,14]
28 : [ o, o, o, o,23,22,21,20,19,18,17,16, o]
29 : [ o, o, o, o,24,23, o,21,20,19,18,17,16,15]
30 : [ o, o, o, o, o,24,23,22,21,20,19,18,17,16]
31 : [ o, o, o, o, o, o, o, o, o,21,20,19,18,17,16]
32 : [ o, o, o, o, o,26,25,24,23,22,21,20,19,18,17]
33 : [ o, o, o, o, o,27, o,25,24,23,22,21,20,19,18,17]
34 : [ o, o, o, o, o, o,27,26, o,24,23,22,21,20,19,18]
35 : [ o, o, o, o, o, o,28,27, o,25,24,23,22,21,20,19,18]
36 : [ o, o, o, o, o, o, o,28,27,26,25,24,23,22,21,20,19]
37 : [ o, o, o, o, o,31, o, o,28,27,26,25,24,23, o,21,20,19]
38 : [ o, o, o, o, o, o, o, o,29, o,27,26,25,24,23,22,21,20]
39 : [ o, o, o, o, o, o,32, o,30,29,28,27,26,25,24,23,22,21, o]
40 : [ o, o, o, o, o, o, o,32, o,30, o,28,27,26,25,24,23,22,21]
41 : [ o, o, o, o, o, o, o, o, o,31,30,29, o,27,26,25,24,23,22,21]
42 : [ o, o, o, o, o, o, o,34,33,32,31,30,29,28,27,26,25,24,23,22]
43 : [ o, o, o, o, o, o, o,35, o, o,32,31, o,29,28,27,26,25, o,23,22]
44 : [ o, o, o, o, o, o, o,36, o,34,33,32,31, o,29,28,27,26,25,24,23]
45 : [ o, o, o, o, o, o, o, o,36,35, o,33,32,31,30,29,28,27,26,25,24,23]
46 : [ o, o, o, o, o, o, o,38, o,36,35,34,33,32,31, o,29,28,27,26,25,24]
47 : [ o, o, o, o, o, o, o, o, o, o, o, o,34,33,32,31, o,29,28,27,26,25,24]
48 : [ o, o, o, o, o, o, o, o,39,38,37,36,35,34,33,32,31,30,29,28,27,26,25]
49 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,35,34,33,32,31,30,29,28,27,26,25]
50 : [ o, o, o, o, o, o, o, o, o,40, o,38,37,36,35,34, o,32,31,30,29,28,27,26]
51 : [ o, o, o, o, o, o, o, o, o,41,40,39, o,37,36,35,34,33,32,31,30,29,28,27,26]
52 : [ o, o, o, o, o, o, o, o, o, o, o,40,39,38, o,36,35,34,33,32,31,30,29,28,27]
53 : [ o, o, o, o, o, o, o, o, o,43, o, o,40, o,38,37,36,35,34,33,32,31, o,29,28,27]
54 : [ o, o, o, o, o, o, o, o, o, o, o,42, o,40,39,38,37,36,35,34,33,32,31,30, o,28]
55 : [ o, o, o, o, o, o, o, o, o,45,44, o, o, o,40, o, o,37, o,35,34,33,32,31,30,29,28]
56 : [ o, o, o, o, o, o, o, o, o,46, o,44,43,42,41,40,39,38,37,36,35,34,33,32,31, o,29]
57 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,43, o,41,40,39,38,37,36,35, o,33,32,31,30,29]
58 : [ o, o, o, o, o, o, o, o, o,48,47,46,45, o,43,42,41,40,39,38,37,36,35,34, o,32, o,30]
59 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,45, o, o, o,41,40, o, o,37,36,35,34,33,32,31,30]
60 : [ o, o, o, o, o, o, o, o, o, o, o,48,47,46,45,44,43,42, o,40,39,38,37,36,35,34,33,32,31]
61 : [ o, o, o, o, o, o, o, o, o, o, o,49, o,47,46, o, o, o, o,41,40,39, o,37,36,35,34,33,32,31]
62 : [ o, o, o, o, o, o, o, o, o, o, o, o,49, o,47, o,45, o,43,42,41,40,39,38,37,36,35,34,33,32]
63 : [ o, o, o, o, o, o, o, o, o, o, o,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37,36,35,34,33,32]
64 : [ o, o, o, o, o, o, o, o, o, o, o,52, o,50,49,48, o,46,45,44,43,42,41,40,39,38,37,36,35,34,33]
65 : [ o, o, o, o, o, o, o, o, o, o, o, o,52, o,50,49,48,47,46,45,44,43, o,41,40,39,38,37,36,35,34,33]
66 : [ o, o, o, o, o, o, o, o, o, o, o,54, o, o,51,50, o,48,47,46, o,44,43,42,41,40,39,38,37,36,35,34]
67 : [ o, o, o, o, o, o, o, o, o, o, o,55, o, o, o,51, o,49,48,47,46,45,44,43, o,41,40,39,38,37, o, o,34]
68 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,54, o,52,51, o,49,48,47,46,45,44,43,42,41,40,39,38,37,36,35]
69 : [ o, o, o, o, o, o, o, o, o, o, o,57, o,55,54,53,52,51,50,49,48,47,46, o,44, o,42, o,40,39,38,37,36,35]
70 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,56,55,54, o, o,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37,36]
71 : [ o, o, o, o, o, o, o, o, o, o, o,59, o, o, o,55,54, o,52, o,50,49,48,47,46,45,44,43, o,41,40,39,38,37,36]
72 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o,57,56, o,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37]
73 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,55,54,53,52, o,50,49,48,47,46,45,44,43, o,41,40,39,38,37]
74 : [ o, o, o, o, o, o, o, o, o, o, o,62, o, o,59, o, o,56, o,54,53,52, o,50,49,48,47,46,45, o,43,42,41,40,39,38]
75 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o,60, o, o,57,56,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39,38]
76 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,58,57, o,55,54, o,52,51,50,49,48,47,46,45,44,43,42,41,40,39]
77 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,63, o, o,60, o, o,57,56,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39]
78 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,64, o, o,61,60,59,58,57,56,55,54, o,52,51,50,49,48,47,46,45,44,43,42,41, o]
79 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,61, o, o, o, o, o,55, o,53, o,51,50,49,48,47,46,45,44,43,42,41,40]
80 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o,65,64, o, o, o,60, o, o, o,56,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41]
81 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,67, o, o, o,63, o, o,60,59, o,57,56,55,54, o,52,51,50,49,48,47,46,45,44,43,42,41]
82 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,63,62,61,60, o,58, o, o,55,54, o,52, o,50,49,48,47,46,45,44,43,42]
83 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,59, o,57,56,55, o,53,52,51,50,49,48,47,46,45,44,43,42]
84 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o,69,68, o,66, o,64,63,62,61,60,59,58,57,56,55,54,53,52,51,50,49,48,47,46, o,44,43]
85 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,68, o, o,65, o,63,62,61,60,59, o,57,56,55,54,53,52,51,50,49,48,47,46,45,44,43]

The general tendency is clear, but the overall situation looks rather irregular. For some values of $m$, there are much more 'holes' than for others. Any ideas why?

REPLY [10 votes]: For rectangles with maximum size 760 or less, there is one remaining possible counterexample. The 17-square tiling of the 697x611 is not proven minimal.

16 1394 1222 723 671 120 551 499 155 69 1 32 87 39 31 8 55 47 344
17 697 611 51 51 119 119 119 119 119 34 68 34 41 82 82 492 41 205 205
I've put 4944 more at Possible Counterexamples to the Minimal Squaring Conjecture.
My original example shows two known ways to divide a 2(7125×7081) rectangle into 20 squares.

The smallest known dissection of a (7125×7081) rectangle needs 21 squares.


TITLE: Uncountable Pre-Image
QUESTION [4 upvotes]: I've been reading about space filling curves, and been asking myself this question.
If $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ is a continuous open map, is it true that $\forall x \in$ range$(f)$ , $f^{-1}(x)$ is always uncountable?

REPLY [14 votes]: I think so, yes. Let $x$ be in the range of $f$ and define $U = f^{-1}((-\infty, x))$, $V = f^{-1}((x,\infty))$. Since $f$ is open, $U$ and $V$ are both nonempty. So they are disjoint nonempty open sets, which means that the complement of $f^{-1}(x)$ is disconnected. But the complement of any countable subset of ${\bf R}^2$ is connected. (Even path connected, by a simple cardinality argument: for any distinct $p$ and $q$ one can find uncountably many paths from $p$ to $q$, any two of which are disjoint except for the points $p$ and $q$. So a countable set can obstruct only countably many of them.)


text:
\begin{appendix}

\subsection{Proof of Lemma~\ref{lem:degree_iso_bounds}}
\begin{IEEEproof}
The degree bounds are readily obtained from classical concentration inequalities.  Specifically, let $d$ denote the degree of a given vertex.  Note that $d$ is a Binomial random variable with parameters $n-1$ and $p$, with expected value $\mathbb{E}[d] = (n-1)p$.  Now, for any $0 < \beta \le \sqrt{3}$ we have\footnote{The statement of this concentration inequality in \cite{Mitzenmacher} has $0 < \beta \le 1$, but the improved upper bound of $\sqrt{3}$ can be obtained from the same proof {\it mutatis mutandis}.} \cite{Mitzenmacher}
$$
\mathbf{Pr}(d \geq (1+\beta)\mathbb{E}[d])\leq e^{\frac{-\mathbb{E}[d] \beta^2}{3}}.
$$
Choose $\beta = \sqrt{3}\left(\frac{\ln{n}}{np}\right)^{\frac{1}{2}-\epsilon}$, which is at most $\sqrt{3}$ for probability functions satisfying the conditions in the lemma and for sufficiently large $n$.   Substituting into the above expression, we have
\begin{equation*}
\mathbf{Pr}(d \geq (1+\beta)\mathbb{E}[d]) \leq e^{-(n-1)p \left(\frac{\ln{n}}{np}\right)^{1-2\epsilon}} =  O\left(e^{-\ln{n}\left(\frac{\ln{n}}{np}\right)^{-2\epsilon}}\right).
\end{equation*}
To show that the maximum degree is smaller than the given bound asymptotically almost surely, we show that all vertices have degree less than the given bound with probability tending to $1$.  By the union bound, the probability that at least one vertex has degree larger than $(1+\beta)\mathbb{E}[d]$ is upper bounded by
$$
n\mathbf{Pr}(d \geq (1+\beta)\mathbb{E}[d]) = O\left(e^{\ln{n}-\ln{n}\left(\frac{\ln{n}}{np}\right)^{-2\epsilon}}\right).
$$
Since $\limsup_{n\rightarrow\infty}\frac{\ln{n}}{np} < 1$, the above expression goes to zero as $n \rightarrow \infty$, proving the upper bound on the maximum degree.

We now show the lower bound for $i(\mathcal{G})$.  Specifically, we will show that for $p$ satisfying the properties in the lemma, almost every graph has the property that  all sets of vertices of size $s$, $1 \le s  \le \lfloor\frac{n}{2}\rfloor$, have at least $\alpha{snp}$ edges leaving that set, for some constant $\alpha$ that we will specify later.  For any specific set $\mathcal{S}$ of vertices of size $s$, the probability that $\mathcal{S}$ has $\lfloor\alpha{snp}\rfloor$ or fewer edges leaving the set is $\sum_{j = 0}^{\lfloor\alpha{snp}\rfloor}\binom{s(n-s)}{j}p^j(1-p)^{s(n-s)-j}$.  Let $E_s$ denote the event that at least one set of vertices of size $s$ has $\lfloor\alpha{snp}\rfloor$ or fewer edges leaving the set.  Then
\begin{equation}
\mathbf{Pr}\left[E_s\right] \le \binom{n}{s}\sum_{j = 0}^{\lfloor\alpha{snp}\rfloor}\binom{s(n-s)}{j}p^j(1-p)^{s(n-s)-j}.
\label{eqn:prob_Es}
\end{equation}
Note that for $1 \le j \le \lfloor\alpha{snp}\rfloor$,
\begin{align*}
\frac{\binom{s(n-s)}{j}p^j(1-p)^{s(n-s)-j}}{\binom{s(n-s)}{j-1}p^{j-1}(1-p)^{s(n-s)-j+1}} = \frac{s(n-s) - j + 1}{j}\frac{p}{1-p} &\ge \frac{s(n-s) -\alpha{s}{np}}{\alpha{s}{np}}\frac{p}{1-p} \\
&\ge \frac{1-2\alpha{p}}{2\alpha}\frac{1}{1-p} \ge \frac{1}{2\alpha},
\end{align*}
for $s \le \lfloor\frac{n}{2}\rfloor$ and $2\alpha < 1$ (which will be satisfied by our eventual choice of $\alpha$).  Thus, there exists some constant $r > 0$ such that
\begin{equation*}
\sum_{j = 0}^{\lfloor\alpha{snp}\rfloor}\binom{s(n-s)}{j}p^j(1-p)^{s(n-s)-j} \le r\binom{s(n-s)}{\lfloor\alpha{snp}\rfloor}p^{\lfloor\alpha{snp}\rfloor}(1-p)^{s(n-s)-\lfloor\alpha{snp}\rfloor}.
\end{equation*}
Substituting into \eqref{eqn:prob_Es} and using the fact that $\binom{n}{k} \le \left(\frac{ne}{k}\right)^{k}$, we have
\begin{align}
\mathbf{Pr}\left[E_s\right] &\le r\left(\frac{ne}{s}\right)^{s}\left(\frac{s(n-s)ep}{\alpha{snp}}\right)^{\alpha{snp}}e^{-p(s(n-s)-\alpha{snp})} \nonumber \\
&\le re^{s\ln\frac{ne}{s}}\left(\frac{e}{\alpha}\right)^{\alpha{snp}}e^{-p(s(n-s)-\alpha{snp})} \nonumber \\
&= re^{sh(s)}, \label{eqn:prob_e_bound}
\end{align}
where
\begin{equation}
h(s) = 1 + np\underbrace{\left(\frac{\ln{n}}{np} + \alpha - \alpha\ln{\alpha} + \alpha{p} -1\right)}_{\Gamma(\alpha)} + ps - \ln{s}.
\label{eqn:hs}
\end{equation}
Noting that $h(s)$ is decreasing in $s$ until $s = \frac{1}{p}$ and increasing afterwards, we have
\begin{align*}
h(s) &\le \max\left\{h(1), h\left(\frac{n}{2}\right)\right\} \\
&= \max\left\{1 +p + np\Gamma(\alpha), 1 + \ln{2} + np\left(\Gamma(\alpha) - \frac{\ln{n}}{np} + \frac{1}{2}\right)\right\}.
\end{align*}
From \eqref{eqn:hs},  $\Gamma(\alpha)$ is increasing in $\alpha$ for $0 \le \alpha < 1$, with $\Gamma(0) = \frac{\ln{n}}{np} - 1$ being negative and bounded away from $0$ for sufficiently large $n$ (by the assumption on $p$ from the statement of the lemma).   Thus, there exists some sufficiently small positive constant $\alpha$  such that $h(s) \le -\bar{\alpha}np$ for some constant $\bar{\alpha} > 0$ and for sufficiently large $n$.  Thus \eqref{eqn:prob_e_bound} becomes $\mathbf{Pr}\left[E_s\right] \le re^{-s\bar{\alpha}np}$ for sufficiently large $n$.

By the union bound, the probability that $i(\mathcal{G}) < \alpha{np}$ is upper bounded by the sum of the probabilities of the events $E_s$ for $1 \le s \le \lfloor\frac{n}{2}\rfloor$.  Using the above expression, we have
$$
\sum_{s =1}^{\lfloor\frac{n}{2}\rfloor}\mathbf{Pr}[E_s] \le r\sum_{s = 1}^{\lfloor\frac{n}{2}\rfloor}e^{-s\bar{\alpha}np} \le r\sum_{s = 1}^{\infty}e^{-s\bar{\alpha}np} = r\frac{e^{-\bar{\alpha}np}}{1-e^{-\bar{\alpha}np}}
$$
which goes to $0$ as $n \rightarrow \infty$.  Thus, we have $i(\mathcal{G}) \ge \alpha{np}$ asymptotically almost surely.
\end{IEEEproof}

\end{appendix}

config: arxiv


TITLE: Force creates the motion or Motion creates the force?
QUESTION [1 upvotes]: Motion Creates Force ?:

When you move your hand (to push a box for example), it creates a force.
When electron moves through a wire it creates electrical force.

Or Force Creates Motion ?:

Gravity force caused an apple to fall.
We pushed a ball and it started to move, but friction force made it stop.

So which one is the correct one?
Actually it feels weird to me if you say : "the earth is static and nothing is moving, but it attract everything to itself (gravity).".
I think like: maybe there is some moving thing on the earth which causes the gravity, or maybe the movement of the earth itself causes the gravity. so when I see a force, I search for a motion that causes it. Is it correct?
Ps: My major is not physics, so please feel free to edit the question.
Ps2: I know the earth is not static. just wanted to simplify the question.

REPLY [1 votes]: If you want the short answer, then motion does not create a force. The force is what changes the velocity of a object.
The longer answer:
The classical equation to calculate a force acting on a object (excluding special and general relativity) is: $$F = m \frac{d^2x}{dt^2}$$
I think the confusion lies in this equation. If you substitute the acceleration of the object into this equation then it gives you a force. However, this does not mean that the acceleration creates a force, all it means is that if we know the acceleration then we know how strong the force has to be to create that acceleration.
However, forces are not fundamental. It is a way of measuring the strength of a interaction. The reason we need interactions in physics is because, even though objects moving at a constant velocity will always move at a constant velocity, the same does not apply to objects that are accelerating. Acceleration is the result of an external influence, unlike inertia. Fundamentally, interactions can be described as a field that fill up all of space. These fields cause objects with certain properties to change their motion. These fields are how attractive and repulsive forces mediate their influence, such as the electric field.
With electric fields, the strength of the force is determined not only by the charge of the particle attracting the second particle, but is also determined by the charge of the second particle as shown in the equation below:
$$F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$$
Solving for acceleration we can find that:
$$a = \frac{1}{4 \pi \epsilon_0 m_2} \frac{q_1 q_2}{r^2}$$
With this we can deduce that the acceleration (or change in steady motion) is determined by both of the charges. This is where the rule for like charges repel and opposite charges attract. Because of the fact that $1 \times 1 = -1 \times -1 = 1 $ and $ -1 \times 1 = 1 \times -1 = -1$ with positive accelleration being repulsive and negative being attractive.
With gravity, it's a bit different. The equation for gravity is:
$$F = - G \frac{m_1 m_2}{r^2}$$
solving for accelleration we find that:
$$a = - G \frac{m_1}{r^2}$$
This means that all objects attracted by gravity get accelerated by the same amount and is completely independent from the particle getting attracted. This is why things attracted by gravity fall at the same rate
There is a way to seem that motion of one object can cause another object to change its motion, however, the way to describe this is through a field. When the first object moves, it interacts with the field and is in a different configuration than it was before, which changes how the second object moves. If the first object did not move, then the second object would be affected differently, which makes it seem that motion is what generated these forces. Objects are constantly interacting with these fields, which means the field tells objects how to move; objects tells the field how it should change. Which sounds suspiciously like John Wheeler's quote about general relativity.
Even in electromagnetism, where the magnetic field is dependent on the velocity of the object generating the field, the same concept can be applied. The velocity of the object changes the field configuration of the magnetic field which changes how charged objects will act in its vicinity.
There are four fundamental fields that permeate space: Gravity, Electromagnetism, Strong force, Weak force. These fields are the reason anything happens in the universe, without them there would be no change in motion. So in short: matter interacts with fields and fields interact with the matter, however movement is not the reason for why things have a force, the force is why things move.
Hope this helps :)


TITLE: Isomorphism between the Tangent Space and Euclidean Space
QUESTION [0 upvotes]: I attempt to understand the invariance of dimension of topological manifolds from Loring Tu's An Introduction to Manifolds (Second Edition, page no. 89). The corresponding section of the book is given below.

My Question
In the proof of Corollary 8.7, it is claimed that the tangent space $T_p(U)$ at point $p \in U \subset \mathbb{R}^n$ is isomorphic to the Euclidean $n$-space $\mathbb{R}^n$. How do we know that this is the case? In other words, what is the proof of this claim?

REPLY [3 votes]: Let $A$ be an affine space with translation space $V$. That is, $A$ is a set whose elements are points, and we have an operation $+:A\times V \to A$ such that $p+0=p$ for all $p \in A$, we have $(p+v)+w=p+(v+w)$ (where the last $+$ is addition in $V$) and given $p,q \in A$, there is a unique $v\in V$ such that $q=p+v$ (people write $v = q-p$).
Once $O \in A$ has been chosen, we have a bijection $V \to A$ given by $v\mapsto O+v$, and so every structure $V$ has (vector space structure, topology, manifold structure, etc) is transferred to $A$. A different choice of $O$ gives an isomorphic vector space structure, same topology, and same manifold structure.
For every $p \in A$, we have $T_pA =V$ in a natural way. For example, if $$T_pA = \{\alpha'(0)\mid \alpha:(-\epsilon,\epsilon)\to A \mbox{ with }\alpha(0)=p\},$$one can write $$v = \frac{\rm d}{{\rm d}t}\bigg|_{t=0}p+tv.$$If you define $T_pA$ as the space of derivations of germs of smooth functions at $p$, then the isomorphism is $$v\mapsto \left([f]_p \mapsto \frac{\rm d}{{\rm d}t}\bigg|_{t=0}f(p+tv)\right),$$and so on.