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     {\huge СИБИРСКИЕ \ ЭЛЕКТРОННЫЕ} \\[2mm]
     {\huge МАТЕМАТИЧЕСКИЕ ИЗВЕСТИЯ} \\[2mm]
     {\large Siberian Electronic Mathematical Reports} \\[1mm]
     {\LARGE\tt{http://semr.math.nsc.ru}}\\[0.5mm]
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     {\flushleft\it Том , стр.  \hspace{65mm}{\rm\small УДК 517.95, 531 }}
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        {\rm\small DOI~10.33048/semi.2019.16.xxx}\hphantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}
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\title[Rotationally symmetric solutions to isothermal compressible Navier-Stokes equations]{
Rotationally symmetric solutions to compressible Navier-Stokes
equations}
\author{{P.I. Plotnikov }}%
\address{Plotnikov P.I.  % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Voronej State University
\newline\hphantom{iii} University square, 1,
\newline\hphantom{iii} 394018, Voronej, Russia}%
\email{piplotnikov@mail.ru}%

\thanks{\sc Plotnikov P.I.}
\thanks{\copyright \ 2021 Плотников П.И.}
\thanks{\rm Исследования проводились за счет  Российского Научного Фонда, грант  22-11-00103}\thanks{\it Поступила  г., г.}%


\semrtop \vspace{1cm} \maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } We consider  a boundary value problem for
compressible Navier-Stokes equations with the pressure function
$p=\varrho$, where $\varrho$ is the density of fluid. It is assumed
the given data and a flow domain are invariant with respect to
rotations around the vertical axis.  The existence of weak
rotationally symmetric solutions is proved. \medskip

\noindent{\bf Keywords:} viscous gas, concentrations problem,
isothermal fluid.
 \end{quote}
}

\tableofcontents
\section{Problem formulation. Results}\label{s1}
\renewcommand{\theequation}{1.\arabic{equation}}
\setcounter{equation}{0}


 The paper is devoted to the study of boundary value problems for  3D compressible
Navier-Stokes equations.  Suppose a viscous compressible fluid
occupies a bounded domain $\Omega\subset \mathbb R^3$ with
$C^\infty$ boundary $\partial\Omega$. The state of the fluid is
characterized by the macroscopic quantities: the {density}
 $\varrho(x,t)$ and the {velocity}
 $\mathbf{u}(x,t)$.
For isentropic fluid, the pressure $p(\varrho)$ and internal energy
$e(\varrho)$ are given by the formulae
\begin{equation}\label{constitute1}
    p=\varrho^\gamma, \quad
    e=\frac{1}{\gamma-1}\varrho^\gamma,
\end{equation}
where $\gamma>1$ is the adiabatic constant. In the end-point
isothermal case the pressure and internal energy are defined by the
equalities
\begin{equation}\label{constitute2}
    p=\varrho, \quad
    e=\varrho\,\ln\varrho-\varrho.
\end{equation}

  The problem is to  find
$\mathbf u(x,t)$ and $\varrho(x,t)$ satisfying the following
equations and boundary conditions in the cylinder $Q_T=\Omega\times
(0,T)$.


\begin{gather}
\nonumber \partial_t (\varrho \mathbf{u})+\text{~div~}(\varrho
\mathbf u \otimes\mathbf{u})+\nabla
p(\varrho)=\text{~ div~}\mathbb S(\mathbf u)+\varrho\mathbf f\quad\text{in~~}Q_T,\\
 \label{stokes1}\partial_t\varrho+\text{~ div~}(\varrho \mathbf{u})=0,\quad \varrho\geq
 0\text{~~in~~}Q_T,\\\nonumber
\mathbf u=0\quad\text{on~} \partial\Omega\times (0,T),\\
\nonumber \mathbf u(x,0)=\mathbf u_0(x),\quad
\varrho(x,0)=\varrho_0(x)> 0\quad\text{in~} \Omega.
\end{gather}
Here, the vector field $\mathbf f\in L^\infty (Q_T)$  denotes the
density of external mass forces,
 the {viscous stress
tensor}
 $\mathbb S(\mathbf u)$ is defined by the equality
\begin{equation}
\label{stokes4a}
 \mathbb S(\mathbf{u})= \big(\nabla
 \mathbf{u}+\nabla\mathbf{u}^\top\big)-
 \frac{2}{3}\,\, \text{div~}
 \mathbf u\,\,\mathbf{I}\,.
\end{equation}
This is a typical form of the viscous stress tensor with shear
viscosity equals 1 and bulk viscosity equals 0. We assume that given
initial data satisfy the conditions
$$\mathbf u_0\in  L^2(\Omega), \quad \varrho_0\in L^\infty(\Omega),\quad \rho_0>0.
$$
Regular solutions to problem \eqref{stokes1} satisfy the energy
diffusion equality
\begin{equation}\label{stokes101}\begin{split}
\int_{\Omega}\big(\, \frac{1}{2}\varrho(t)|\mathbf
u(t)|^2+e(t)\,\big)\,dx+\int_0^t\int_\Omega \mathbb S(\mathbf
u(s)):\nabla \mathbf u(s)\, dxds\\= \int_{\Omega}\big(\,
\frac{1}{2}\varrho_0|\mathbf
u_0|^2+e(\varrho_0)\,\big)\,dx\text{~~for~~} t\in (0,T).
\end{split}\end{equation}
In particular, they admit the energy estimates:
\begin{equation}\label{stokes102}\begin{split}
\|\varrho|\mathbf u|^2+ p(\varrho)\|_{L^\infty(0,T;
L^1(\Omega))}+\|\mathbf u\|_{L^2(0,T; W^{1,2}(\Omega))}\leq E
\text{~for~}\gamma>1,\\
\|\varrho|\mathbf u|^2+ |\varrho\, \ln \varrho|\,\|_{L^\infty(0,T;
L^1(\Omega))}+\|\mathbf u\|_{L^2(0,T; W^{1,2}(\Omega))}\leq
E\text{~for~}\gamma=1,
\end{split}\end{equation}
where the constant $E$ depends only on $\Omega$, $T$,
$\|\varrho_0\|_{L^\infty(\Omega)}$, $\|\mathbf u_0\|_{L^2(\Omega)}$,
and $\|f\|_{L^\infty(Q_T)}$. For the macroscopic variables
satisfying estimates \eqref{stokes102}, we can define weak solutions
to problem \eqref{stokes1}.
\begin{definition}\label{stokes9} A weak solution to problem
\eqref{stokes1} is a couple $(\varrho, \mathbf u): Q_T\to \mathbb
R^+\times\mathbb R^3$ satisfying the following conditions
\begin{itemize}
\item

\begin{equation}\label{stokes103}
\varrho, p(\varrho), \varrho|\mathbf{u}|^2\in L^\infty(0,T;
L^1(\Omega)), \quad \mathbf{u}\in L^2(0,T; W^{1,2}_0(\Omega)).
\end{equation}

\item The integral identity
\begin{multline}\label{stokes10}
\int_{Q_T}\big (\varrho\mathbf{u}\cdot
\partial_t\boldsymbol{\xi}+
\varrho\mathbf{u}\otimes\mathbf{u}:\nabla\boldsymbol{\xi}+p(\varrho)\,\text{\rm
div~} \boldsymbol{\xi}-\mathbb
S(\mathbf{u}):\nabla\boldsymbol{\xi}\big)\,dxdt \\+\int_{Q_T}
\varrho\mathbf f\cdot\boldsymbol{\xi}\, dxdt+ \int_{\Omega
}\varrho_0 \mathbf u_0(x)\cdot \boldsymbol{\xi}(x,0)\,dx=0
\end{multline}
holds for every vector fields $\boldsymbol{\xi}\in C^\infty(Q_T)$
vanishing in a neighborhood of the lateral boundary
$\partial\Omega\times[0,T]$ and the lid $\Omega\times\{t=T\}$.
\item The integral identity
\begin{equation}\label{stokes11}
\int_{Q_T}\big (\varrho
\partial_t\psi+\varrho\mathbf{u}\cdot\nabla\psi
\big)\,dxdt+\int_{\Omega } \varrho_0(x)\psi(x,0)\, dx=0
\end{equation}
holds for every  $\psi\in C^\infty(Q_T)$ vanishing in a neighborhood
of ~ $\Omega\times\{t=T\}$.
\end{itemize}
\end{definition}

 The first nonlocal results
concerning the mathematical theory of compressible Navier-Stokes
equations are due to P.-L. Lions. In monograph \cite{PLL} he
established the existence of a weak solution to nonstationary
boundary value problem for the Navier-Stokes equations with the
pressure function $p\sim\varrho^\gamma$ for all $\gamma>5/3$.
 More recently, Feireisl, Novotn\'{y},  and Petzeltov\'{a}, see
  \cite{FNP},  proved the existence result for all
 $\gamma>3/2$,
 see also monographs \cite{FEIRBOOK},  \cite{NovotnyStra}
 for references and details. For   $\gamma\leq 3/2$,
the question on solvability of problem \eqref{stokes1}  is still
open. The main difficulty is the
 so-called concentration problem,
which was widely discussed in the mathematical literature in
connection with vortex sheets dynamics.\\[1ex]
\emph{In this work, we focus on the end-point case of the isothermal
flows.  Hereinafter we will assume that $\gamma=1$.}
\\[1ex]
In this case the concentration problem can be formulated as follows.
Let us consider a sequence $\varrho_\varepsilon$, $\mathbf
u_\varepsilon$ of  approximate solutions to problem \eqref{stokes1}
satisfying energy estimate \eqref{stokes102}. Here $\varepsilon$ is
the approximation parameter. Such solutions can be obtained by the
regularization method.

We may assume, after passing to a subsequence, that
\begin{gather*}
 \mathbf u_\varepsilon\to\mathbf u\text{~~weakly in~~}L^2(0,T;
W^{1,2}_0(\Omega)), \\\varrho_\varepsilon\to\varrho\text{~~star
weakly in~~}L^\infty(0,T; L^1(\Omega)),
\\
\varrho_\varepsilon \mathbf u_\varepsilon\otimes\mathbf
u_\varepsilon\to \mathcal{M} \text{~~star weakly in the space of
Radon measures as~~} \varepsilon \to 0,
\end{gather*}
where $\mathcal{M}=(\mathcal{M}_{ij})_{3\times 3}$ denotes a
matrix-valued Radon measure in $\Omega$. In the general case the
weak star defect measure
$$\boldsymbol\mu= \mathcal{M}- \varrho\mathbf u\otimes \mathbf u\neq 0.$$
In particular, the momentum balance equation for the limits
$\varrho$ and $\mathbf u$ reads
$$
\partial_t(\varrho\mathbf u)+\text{div~}(\varrho \mathbf
u\otimes\mathbf u)-\text{div~}\mathbf S(\mathbf u)+\nabla\varrho
+\text{div~}\boldsymbol\mu=\varrho f\text{~~in~~} Q_T.
$$
The concentrations are canceled if $\text{~div~}\boldsymbol\mu=0$.
 Hence the problem is to describe
the structure of the defect measure and to find conditions under
which it  is canceled.

 For the compressible Navier-Stokes equations, the concentration problem is poorly
 investigated. Recently, the Hausdorff and parabolic dimensions of the support of
the defect measure were estimated in \cite{Xi} and \cite{Xi1}. The
absence of concentrations in the critical case  $\gamma=3/2$ have
been proved in \cite{plotn1}.  The structure of the concentrations
set may be very complicated.  Therefore,  we restrict our
considerations by the case of rotationally symmetric flows.

Denote by $O(\alpha)$ the orthogonal matrix defined by the
equalities
\begin{equation}\label{rotor2}
O(\alpha)= \left(
\begin{array}{cc}
\displaystyle {O_{12}(\alpha)}&\displaystyle { 0}
\\ \displaystyle {0 } &  \displaystyle {1
 }
  \end{array}
\right), \quad O_{12}(\alpha)=\left(
\begin{array}{cc}\displaystyle
{\cos\alpha}&\displaystyle { \sin\alpha}
\\ \displaystyle {-\sin\alpha } &  \displaystyle {\cos\alpha
 }
  \end{array}
\right).
\end{equation}


\begin{definition}\label{rita3}
We say that a solution to equations \eqref{stokes1} is rotationally
symmetric  if the equalities
\begin{equation}\label{rotor1}
\Omega=O(\alpha)\Omega, \quad    \mathbf
u\big(O(\alpha)x,t\big)=O(\alpha)\mathbf u(x,t), \quad
    \varrho(x,t)=\varrho\big(O(\alpha) x,t\big)
\end{equation}
holds for every $\alpha\in \mathbb R$.
\end{definition}

We are now in a position to formulate the main results of this
paper. Introduce the regularized internal energy
\begin{eqnarray}\label{stokes105}
e&=&\varrho\ln\varrho-\varrho + e_1(\varrho)+ e_2(\varrho), \\\notag
e_1&=&\varepsilon
\varrho^{1+\varepsilon}\ln(1+\varepsilon\varrho),\,\,\,\,
 e_2=\frac{1}{2}\delta^3\varrho^3,
\end{eqnarray}
depending on the small parameters $\varepsilon, \delta\in (0,1/2)$.
The corresponding pressure function $p=e'\varrho-e$ is defined by
the equalities
\begin{eqnarray}\label{stokes106}
p&=&\varrho + p_1(\varrho)+ p_2(\varrho), \\\notag
 p_1&=&\varepsilon
\varrho^{1+\varepsilon}\big(\varepsilon\ln(1+\varepsilon\varrho)+
\frac{\varepsilon\varrho}{1+\varepsilon\varrho}\big), \,\,\,\,
p_2=\delta^3\varrho^3. \end{eqnarray} For positive $\varepsilon,
\delta$, the function $p$ meet all requirements of existence theorem
proved in \cite{FNP}, which leads to the following

\begin{proposition}\label{stokes107}
Let $\varrho_0>0$ and
\begin{equation}\label{stokes108a}
    \|\varrho_0\|_{ L^\infty(\Omega)}+ \|\mathbf
    u_0\|_{L^2(\Omega)}+\|f\|_{L^\infty(Q_T)}\leq c_E.
\end{equation}
Then the boundary value problem
\begin{gather}
\notag \partial_t (\varrho \mathbf{u})+\text{\rm~div~}(\varrho
\mathbf u \otimes\mathbf{u})+\nabla
p=\text{\rm~ div~}\mathbb S(\mathbf u)+\varrho\mathbf f\quad\text{in~~}Q_T,\\
 \label{stokes108}\partial_t\varrho+\text{\rm~ div~}(\varrho \mathbf{u})=0,\quad \varrho\geq
 0\text{~~in~~}Q_T,\\\notag
\mathbf u=0\quad\text{on~} \partial\Omega\times (0,T),\\
\notag \mathbf u(x,0)=\mathbf u_0(x),\quad
\varrho(x,0)=\varrho_0(x)> 0\quad\text{in~} \Omega
\end{gather}
has a weak solution, which admits the estimate
\begin{equation}\label{stokes113}
 \|\varrho|\mathbf u|^2\|_{L^\infty(0,T;
L^1(\Omega))}+ \|e(\varrho)\|_{L^\infty(0,T; L^1(\Omega))}+\|\mathbf
u\|_{L^2(0,T; W^{1,2}(\Omega))}\leq E. \end{equation} Here the
constant $E$ depends only on $\Omega$, $T$, and $c_E$.
\end{proposition}
Note that
$$
\varrho\ln (2+\varrho)\leq 2(\varrho\ln\varrho-\rho)+c,
$$
where $ c $ is some absolute constant. It follows from this and the
expressions for $e$ and $p$ that
\begin{eqnarray}\label{stokes113cor}
\|\varrho\ln(2+\varrho)\|_{L^\infty(0,T;
L^1(\Omega))}&+&\|e_1(\varrho)\|_{L^\infty(0,T;
L^1(\Omega))}\\\notag+ \|e_2(\varrho)\|_{L^\infty(0,T;
L^1(\Omega))}&\leq &c E+c.
\end{eqnarray}
and
\begin{equation}\label{stokes114cor}
\|p\|_{L^\infty(0,T; L^1(\Omega))}+\|p_1(\varrho)\|_{L^\infty(0,T;
L^1(\Omega))}+ \|p_2(\varrho)\|_{L^\infty(0,T; L^1(\Omega))}\leq c
E+c.
\end{equation}



The following remark is important.
\begin{remark}\label{stokes114}
Recall definition \eqref{rotor2} of the matrix $O(\alpha)$. If the
domain $\Omega$, initial data $(\varrho_0, \mathbf u_0)$ and mass
force $\mathbf f$ satisfy the symmetry conditions
\begin{equation}\label{stokes115}\begin{split}
\Omega=O(\alpha)\Omega, \quad   \mathbf
u_0\big(O(\alpha)x,t\big)=O(\alpha)\mathbf u_0(x,t), \\
    \varrho_0(x,t)=\varrho_0\big(O(\alpha) x,t\big), \quad
    \mathbf f\big(O(\alpha)x,t\big)=O(\alpha)\mathbf f(x,t),
\end{split}\end{equation} for all $\alpha\in\mathbb R$, then problem
\eqref{stokes108} has a rotationally -symmetric solution which meets
all requirements of Definition \ref{rita3} and Proposition
\ref{stokes107}. We do not claim that all weak solutions to problem
\eqref{stokes108} given by Proposition \ref{stokes107} are
symmetric. We mean that at least one of such solutions has symmetry
properties.
\end{remark}
The particular cases of rotationally symmetric solutions are the
axially symmetric solutions with zero azimuth velocity component,
and radially symmetric solutions with $\varrho$ and $\mathbf u$
depending only on  $|x|$. In the radially symmetric case, the
existence of global weak solutions for all adiabatic exponents
$\gamma>1$ was proved in
 \cite{jang1},  see also \cite{sarka}. The similar results for axially symmetric and
 helically symmetric solutions were proved in
\cite{ping2} and \cite{sunjiang}. However, the obtained solutions
are defined only in the flow domain with punctured axis of symmetry.
In paper \cite{WPR} the existence of rotationally symmetric
solutions was proved for all $\gamma> (7+\sqrt{73})/12$. The
end-point case of isothermal flows with $\gamma=1$ is still poorly
studied. Our goal is to fill this gap.





The following theorem is the main result of this paper.  Assume that
the flow domain $\Omega$ with the boundary of the class $C^\infty$
satisfies the symmetry condition
$$
O(\alpha)\Omega=\Omega\text{~~for all ~~}\alpha\in\mathbb R,
$$ and the intersection of $\Omega$ with the symmetry axis $Ox_3$
is an interval
$$
\mathcal I=\{x:\,\,x_1=x_2=0, \,\, x_3\in (-a,a)\}.
$$
Furthermore assume that the initial data $(\varrho_0, \mathbf u_0)$
and mass force $f$ satisfy condition \eqref{stokes108a} and symmetry
condition \eqref{stokes115}. Denote by $(\varrho_{\varepsilon,
\delta}, \mathbf u_{\varepsilon,\delta})$ the rotationally-symmetric
solutions to problem \eqref{stokes108} satisfying conditions of
Proposition \ref{stokes107}.

\begin{theorem}\label{thm1} Under the above assumptions, there is a subsequence
of the sequence
$$
\varrho_\varepsilon=\varrho_{\varepsilon, \kappa(\varepsilon)},
\quad \mathbf u_\varepsilon=\mathbf u_{\varepsilon,
\kappa(\varepsilon)}, \quad
\kappa(\varepsilon)=e^{-\frac{1}{\varepsilon^9}},
$$
still denoted by $\varrho_\varepsilon$, $\mathbf u_\varepsilon$,
with the following properties.
\begin{eqnarray}\label{stokes116}
\varrho_\varepsilon\to \varrho,\,\,\,\varrho_\varepsilon\mathbf
u_\varepsilon\to \varrho\mathbf u&\text{~~weakly in ~~}&
L^1(Q_T),\\\notag  \mathbf u_\varepsilon\to \mathbf u
&\text{~~weakly in ~~}& L^2(0,T; W^{1,2}_0(\Omega)).
\end{eqnarray}
The functions $\varrho$, $\mathbf u$ satisfy symmetry condition
\eqref{rotor1} and admit energy estimate \eqref{stokes113},
\eqref{stokes113cor}, \eqref{stokes114cor}.

\begin{equation}\label{stokes117}
\varrho_\varepsilon\mathbf u_\varepsilon\otimes \mathbf
u_\varepsilon\to \varrho\mathbf u\otimes \mathbf u+\boldsymbol\mu,
\quad p(\varrho_\varepsilon)\to \varrho
\end{equation}
star weakly in the space of parametrized measures $L^r_w(0,T;
C_0(\Omega)')$ for every $r\in (1,\infty)$. (See Section
\ref{thm1proof} and Lemma \ref{lita9} for  details.)


The parametrized  Radon measure
\begin{equation*}
\boldsymbol\mu(t)= \left(
\begin{array}{ccc}
\displaystyle {0}&\displaystyle { 0}&\displaystyle { 0}
\\
\displaystyle {0}&\displaystyle { 0}&\displaystyle { 0}\\
\displaystyle {0 } &  \displaystyle {0
 }&\displaystyle{\mu_{33}(t)}
  \end{array}
\right)
\end{equation*}
is supported on the segment $\mathcal I$ of the symmetry axis for
a.e. $t\in (0,T)$.

The functions $\varrho$, $\mathbf u$ serve as  a rotationally
-symmetric weak solution to the boundary value problem
\begin{gather}\nonumber
 \partial_t (\varrho \mathbf{u})+\text{\rm~div~}(\varrho \mathbf u
\otimes\mathbf{u})+\nabla \varrho +\text{\rm ~div~}\boldsymbol \mu=
\text{\rm~ div~}\mathbb S(\mathbf u)+\varrho\mathbf
f\quad\text{in~~}Q_T,\\\label{isothermal}
\partial_t\varrho+\text{\rm~ div~}(\varrho \mathbf{u})=0,\quad \varrho\geq
 0\text{~~in~~}Q_T,\\\nonumber
\mathbf u=0\quad\text{on~~} \partial\Omega\times (0,T),\\\nonumber
 \mathbf u(x,0)=\mathbf u_0(x),\quad \varrho(x,0)=\varrho_0(x)>
0\quad\text{in~} \Omega.
\end{gather}


\end{theorem}
The rest of the paper is devoted to the proof of this theorem. In
order to obtain a priori estimates of rotationally symmetric weak
solutions to the regularized problem,  we rewrite equations
\eqref{stokes108} in the cylinder coordinate.  The next section is
devoted to this procedure.



\section{Cylindrical coordinates. Integral identities}\label{s2}
\renewcommand{\theequation}{2.\arabic{equation}}
\setcounter{equation}{0}



Denote by  $r$, $\vartheta$, $x_3$ the cylindrical coordinates in
the space $\mathbb R^3$ of points $x=(x_1,x_2,x_3)$,
\begin{equation}\label{rita1}
    x_1=r\cos \vartheta, \quad x_2=r \sin\vartheta,\quad x_3=x_3.
\end{equation}
It is convenient to use the notation
\begin{equation}\label{rita1y}
y=(y_1,y_3)=(r,x_3), \quad z=(z_1,z_2,z_3)= (r,\theta, x_3)=(y_1,
\theta, y_3).
\end{equation}
Let $G\subset\mathbb R^2$ be an intersection of the rotationally
symmetric flow domain $\Omega$ with the half-plane
$$
\Pi^+=\{x_1>0,x_2=0\},
$$
i.e.,
$$
G=\{x\in \Omega: x_1>0, x_2=0\}.
$$
It can be regarded as the domain in the plane of variable $y$.
Assume that $\Omega$ and $G$ are simply-connected. We also assume
that the intersection $\Omega$ and the symmetry axis $Ox_3$ is the
interval
$$
\mathcal I=\big\{\, x:\,\, x_3\in (-a, a), \quad x_1=x_2=0\,\big\}.
$$
We will consider $\mathcal I$ as the interval $-a<y_3<a$, $y_1=0$ in
the plane of variable $y$. It follows that the boundary $G$ consists
of the interval $\mathcal I$, the arc $\partial\Omega\cap \Pi^+$,
and the end-points of $\mathcal I$.  In the cylindrical coordinates,
we have
$$
\Omega= (G\times \mathbb S^1)\cup\mathcal I.
$$
Denote by $v_1$, $v_2$, and $v_3$ the radial, azimuth, and axial
components of the velocity $\mathbf u$, i.e.,
\begin{equation}\label{rita2}\begin{split}
\mathbf u=\mathbf U\,\mathbf v,\quad\mathbf v=(v_1,v_2, v_3), \quad
\mathbf U=\left(
\begin{array}{ccc}
\displaystyle {\cos\theta}&\displaystyle {
-\sin\theta}&\displaystyle { 0}
\\
\displaystyle {\sin\theta}&\displaystyle { \cos\theta}&\displaystyle { 0}\\
\displaystyle {0 } &  \displaystyle {0
 }&\displaystyle{1}
  \end{array}
\right). \end{split}\end{equation} The flow is rotationally
symmetric if and only if $\varrho$ и $\mathbf v$ are independent of
$\theta$,
$$
\varrho=\varrho(y,t), \quad \mathbf v=\mathbf v(y,t).
$$
The main energy dissipation estimate \eqref{stokes113} has analogy
in the cylindrical coordinates. This results from the following
technical lemma.
\begin{lemma}\label{sveta9m}
Let $\mathbf u:\Omega\to \mathbb R^3$ be a rotationally symmetric
vector field such that $\mathbf u=\mathbf U(\theta)\, \mathbf v(y)$,
$y=(r,x_3)=(y_1,y_3)$. Then
\begin{equation}\label{rita20}
\int_\Omega |\nabla_x \mathbf u|^2\, dx=2\pi\int_G |\nabla_y \mathbf
v|^2\, y_1 dy+2\pi\int_G (v_1^2+v_2^2)\frac{1}{y_1} dy.
\end{equation}
\end{lemma}
\begin{proof} The proof given in Section \ref{cylinder}, Lemma \ref{sveta9}. \end{proof} It follows from this lemma and
\eqref{stokes113} that the  rotationally symmetric solution $(\rho,
\mathbf u)$ defined by Proposition \ref{stokes10} in the cylindrical
coordinates admits the estimate
\begin{equation}\label{rita21}\begin{split}
\int_0^T\int_G \Big(|\nabla\mathbf
v|^2+\frac{1}{y_1^2}(v_1^2+v_2^2)\Big)\, y_1dydt\leq c,\\
\text{\rm ess}\sup\limits_{t\in (0,T)}\int_G \Big(\varrho|\mathbf
v|^2+e(\varrho)\Big)\, y_1dy\leq c,
\end{split}\end{equation}






 Here the constant $E$ depends only on
$\Omega$, $T$, and $c_E$. Further we will use the special integral
identity for functions $v$ and $\rho$.
\begin{proposition}\label{rita22}
Let $\varrho (x,t)$, $\mathbf u(x,t)$ be a rotationally symmetric
solution to problem \eqref{stokes108} given by Proposition
\ref{stokes107}.  Furthermore, assume that  the vector fields
$\mathbf v(y,t)$ and $\overline{\mathbf v}(y,t)$ are defined by the
equalities
$$
\mathbf u=\mathbf U(\theta)\mathbf v(y,t),\quad \overline{\mathbf
v}(y,t)=(v_1(y,t),\, v_3(y,t)).
$$
Then the integral identities
\begin{multline}\label{rita23}
\int_{G\times(0,T)}\Big (\varrho\overline{\mathbf{v}}\cdot
\partial_t\overline{\boldsymbol{\varsigma}}+
\varrho\overline{\mathbf{v}}\otimes\overline{\mathbf{v}}:\nabla
\overline{\boldsymbol{\varsigma}}+p(\varrho)\,\,\text{\rm div~}
\overline{\boldsymbol{\varsigma}}-\mathbf B(\overline{\mathbf v}):
\nabla\overline{\boldsymbol\varsigma}\\+\mathbf b(\overline{\mathbf
v}, \varrho) \varsigma_1+\varrho \overline{\mathbf f}\cdot\overline
{\boldsymbol\varsigma}\Big)\,y_1 dy\,dt=0,
\end{multline}


\begin{equation}\label{rita24}
\int_{G\times (0,T)}\Big(\,\,\varrho \partial_t \psi +\varrho
\overline{\mathbf v}\cdot \nabla\psi\,\,\Big )\,y_1\, dy\, dt=0.
\end{equation}
hold for every continuously differentiable, compactly supported in
$G\times (0,T)$ vector field
$$\overline{\boldsymbol\varsigma}=(\varsigma_1, \varsigma_3):G\times
(0,T)\to\mathbb R^2.
$$
and  continuously differentiable, compactly supported in $G\times
(0,T)$ function $\psi:G\times (0,T)\to\mathbb R$.

Here $\nabla \mathbf a=\nabla_y \mathbf a$ for each vector field
$\mathbf a(y,t)$. The matrix  $\mathbf B$ and vector field  $\mathbf
b$ are defined by the equalities
\begin{equation}\label{rita25}\begin{split}
    \mathbf B=(\nabla\overline{\mathbf v}+\nabla\overline{\mathbf
v}^\top)- \frac{2}{3}(\text{\rm div~}\overline{\mathbf
v}+\frac{1}{r} v_1)\, \mathbf I,\\
\mathbf b=\frac{1}{y_1}\varrho v_2^2+\frac{1}{y_1}p-\frac{4}{3y_1^2}
v_1+\frac{2}{3y_1}\text{\rm div~}\overline{\mathbf v}.
\end{split}\end{equation}
For every compact set $K\Subset G$, the matrix $\mathbf B$ and
vector field $\mathbf b$ admit the estimate
\begin{equation}\label{rita26}
   \|\mathbf B\|_{L^2(K\times (0,T))}+\|\mathbf b\|_{L^1(K\times
   (0,T))}\leq c,
\end{equation}
where the constant $c$ depends only on $K$ and the constant in
estimate \eqref{rita21}.
\end{proposition}
\begin{proof} The proof given in Section \ref{cylinder}, Proposition \ref{sveta24}. \end{proof}






\section{ Localization. Potential estimate}\label{inna}
\renewcommand{\theequation}{3.\arabic{equation}}
\setcounter{equation}{0} The main difficulty of our problem is the
proof that the kinetic energy tensors $\varrho\mathbf u\times\mathbf
u$ of rotationally symmetric solutions to problem \ref{stokes108}
given by Proposition \ref{stokes107} are equiintegrable on ever
compact set $K\subset\Omega\setminus \mathcal I$. In order to prove
this result we developed the approach based on estimates of the
density potentials. The main ingredient of the method is the $L^2$
estimate of the function $(-\Delta)^{1/4}\varrho$. In the
cylindrical coordinates, the vector field $v$ demonstrates a
complicated behavior near the boundary of $G$. Since we are
interesting in interior estimate, it is convenient to localize the
problem and work with compactly supported  functions defined in the
whole space.

\subsection{Localization}
 Fix a compact set  $K\Subset G\times (0,T)$, and choose two functions
 $\zeta, \eta\in C^\infty_0(G_T)$ such that
\begin{equation}\label{rita10}
    \zeta=\eta=1 \text{~on~} K,  \quad  \eta=1  \text{~on~}  \text{\rm supp} \,\,
    \zeta, \quad 0\leq \zeta,\eta\leq 1,
\end{equation}
Assume that they are extended by zero to the whole space $\mathbb
R^2\times\mathbb R$.   Set
\begin{equation}\label{rita7}
\rho(y,t)=\zeta(y,t)\varrho(y,t), \quad \mathbf w(y,
t)=\eta(y,t)\mathbf v(y,t), \quad \overline{\mathbf
w}=(w_1,w_3)=\eta \overline{\mathbf v}.
\end{equation}
Extend these functions by zero to $\mathbb R^3$. The following Lemma
characterizes the properties of the extended functions.
\begin{proposition}\label{rita12}
The functions $\rho$ and $\mathbf w$ are compactly supported in
$G\times (0,T)$. There is $R>0$, depending only on $K$, such that
$$
0<R^{-1}\leq y_1\leq R\text{~~on the supports of ~~}
\rho\text{~~and~~}\zeta.
$$
There is a constant $c$, depending only on $K$ and the constants in
estimates \eqref{stokes113}-\eqref{stokes114cor}, such that
\begin{equation}\label{rita13}\begin{split}
\|\zeta\varrho\ln(2+\varrho)\|_{L^\infty(\mathbb R; L^1( \mathbb
R^2))}+\|\zeta\, e_1\|_{L^\infty(\mathbb R; L^1( \mathbb
R^2))}+\|\zeta\, e_2\|_{L^\infty(\mathbb R; L^1(
\mathbb R^2))}\leq c,\\
\|\rho|\mathbf w|^2 \|_{L^\infty(\mathbb R;L^1(\mathbb R^2))}+
\|{\mathbf w}\|_{L^2(\mathbb R; W^{1,2}(\mathbb R^2))}\leq c.
\end{split}\end{equation}
 Moreover, the integral identities
\begin{equation}\label{rita16}\begin{split}
\int_{\mathbb R^3}\Big(\,\rho \overline{\mathbf{w}}
\cdot\partial_t\overline{\boldsymbol{\varsigma}} + \rho
\overline{\mathbf w}\otimes \overline{\mathbf
w}:\nabla\overline{\boldsymbol\varsigma}+ \zeta p(\varrho)\text{\rm
~div~}\overline{\boldsymbol\varsigma}\\ -\zeta\mathbf
B:\nabla\overline{\boldsymbol\varsigma}
 +\mathbf{F}\cdot \overline{\boldsymbol{\varsigma}}\, \Big )\, y_1dydt
 =0,
\end{split}\end{equation}
\begin{equation}\label{rita16a}\begin{split}
\int_{\mathbb R^3}\big(\rho \partial_t \psi+\rho \overline{\mathbf
w}\cdot \nabla \psi +\varrho\,\mathbf{h}\,\psi\,\big)\, y_1 dydt
 =0,
\end{split}\end{equation}
hold true for all $\overline{\boldsymbol\varsigma}, \psi\in
C^1_{\text{\rm loc}}(\mathbb R^3\setminus\{y_1=0\})$. The vector
field $\mathbf F$ and function $\mathbf h$ are compactly supported
in $G\times (0,T)$ and admit the estimates
\begin{equation}\label{rita17}
 \|\mathbf F\|_{L^1(\mathbb
    R^3)}+\|\mathbf h\|_{L^2(\mathbb R^3)}+
\|\varrho \mathbf h\|_{L^\infty(\mathbb R; L^1(\mathbb R^2))}\leq c,
\end{equation} where  $c$  depends on $K$ and does not depend on
the regularization parameters  $\varepsilon$, $\delta$.
\end{proposition}
\textbf{Proof.} Compactness of the support of  $(\rho, \mathbf w)$
obviously follows from \eqref{rita7}.

For the proof of integral identities  \eqref{rita16} and
\eqref{rita16a} choose arbitrary functions
$\overline{\boldsymbol{\varsigma}}, \psi\in C^1_{\text{loc}}(\mathbb
R^3)$. Next, replace $\overline{\boldsymbol{\varsigma}}$ and  $\psi$
in integral identities \eqref{rita23} and \eqref{rita24} by
 $\zeta\overline{\boldsymbol{\varsigma}}$ and  $\zeta\psi$. By doing
 so we arrive at integral identities \eqref{rita16}, \eqref{rita16a}
 with
\begin{eqnarray}\notag
\mathbf F&=&\partial_t\zeta\varrho\overline{\mathbf
w}+\varrho\overline{\mathbf w}\otimes\overline{\mathbf
w}\nabla\zeta-\mathbf B\nabla\zeta+p\nabla\zeta+\\\notag
&&\zeta\mathbf b\, \mathbf e_1+\rho\overline{\mathbf f},\\\notag
\mathbf h&=&\overline{\mathbf w}\cdot \nabla\zeta.
\end{eqnarray}
It remains to note that estimate \eqref{rita17} follows from
estimates \eqref{rita13} and \eqref{rita25}. \qed



\subsection{Potential estimate}

In  this section, we prove the estimate of the Riesz potential for
the function $\rho$. Recall that for every integrable, compactly
supported in $\mathbb R^2$ function $\rho$ the fractional Laplacian
$(-\Delta)^{-1/2}\rho$ is defined by the equality, \cite{adams},
\begin{equation}\label{rita27}
    (-\Delta)^{-1/2}\rho(y)= c_{1/2}\,\int_{\mathbb
    R^2}\frac{\rho(z)}{|y-z|}\, dz.
\end{equation}
We have $(-\Delta)^{-1/2}=(-\Delta)^{-1/4}(-\Delta)^{-1/4}$, where
\begin{equation}\label{rita28}
    (-\Delta)^{-1/4}\rho(y)= c_{1/4}\,\int_{\mathbb
    R^2}\frac{\rho(z)}{|y-z|^{3/2}}\, dz.
\end{equation}
Here $c_{1/2}$, $c_{1/4}$ are some unessential absolute constants.
The following proposition is the main result of this section.



\begin{proposition}\label{inna1}
Under the assumptions of Proposition \ref{rita12}, the truncated
density $\rho$ admits the estimate
\begin{equation}\label{inna2}
    \|(-\Delta)^{-1/4}\rho\|_{L^2(\mathbb R^3)}\leq c.
\end{equation}
Here the constant $c$ depends on the compact set $K\subset G\times
(0,T)$ and does not depend on the regularization parameters
$\varepsilon$, $\delta$ in the expressions for the internal energy.
\end{proposition}
The proof falls into a sequence of lemmas.  Introduce the kernels
\begin{eqnarray}\label{inna3}
K_0(y)&=&\frac{y}{|y|}, \,\, K_{0i}=\frac{y_i}{|y|}\\
\notag
K_{ij}(y)&=&\partial_jK_{0i}(y)=\frac{|y|^2\delta_{ij}-y_iy_j}{|y|^3}\text{~~for~~}
y\in \mathbb R^2\setminus \{0\}, \quad i,j=1,3.
\end{eqnarray}
We adapt the convention that these kernels equal zero at $y=0$. Set
\begin{equation}\label{inna4a}
\overline{\varsigma}_i(y,t)=K_{0i}* \rho(y,t)\equiv \int_{\mathbb
R^2} K_{0i}(y-z)\rho(t,z)\, dz, \quad i=1,3.
\end{equation}
\begin{lemma}\label{inna4} Under the above assumptions,
the vector field $\overline{\boldsymbol\varsigma}$ belongs to the
class  $L^\infty(\mathbb R; C^1(\mathbb R^2))$. Its time derivative
 $\partial_t\overline{\boldsymbol\varsigma}$ belongs to the class $L^2(\mathbb R;
C(\mathbb R^2))$ and is defined by the equality
\begin{equation}\label{inna5}
    \partial_t\overline{\varsigma}_i=-\sum_{j=1,3} K_{ij}*(\rho w_j)+\sigma_i.
\end{equation}
The vector field $\boldsymbol\sigma=(\sigma_1, \sigma_3)$ admits the
estimate
\begin{equation}\label{inna5s}
\|\sigma_i\|_{L^\infty(\mathbb R^3)}\leq c(K).
\end{equation}
\end{lemma}
\begin{proof}
  We begin with the observation that
 estimate  \eqref{rita13} and formula
 \eqref{stokes105} imply
\begin{equation}\label{inna60}
\delta^3\|\rho^3\|_{L^\infty(\mathbb R; L^1(\mathbb R^2))}\leq c
\end{equation}
From this and the H\"{o}lder inequality we obtain the estimate
\begin{equation}\label{inna6}\begin{split}
    \|\varrho\mathbf w\|_{L^2(\mathbb R; L^s(\mathbb
    R^2))}^2=\int_\mathbb R \|\rho\mathbf w\|_{L^s(\mathbb
    R^2)}^2dt\leq\int_\mathbb R \|\rho\|_{L^3(\mathbb R^2)}^2\,
    \|\mathbf w\|_{L^{3s/(3-s)}(\mathbb R^2)}^2\,dt\\
    \leq c(s)\delta^{-2}\int_\mathbb R
    \|\nabla \mathbf w\|_{L^{2}(\mathbb R^2)}^2dt\leq
    c(\delta,s),
\end{split}\end{equation}
 which holds true for all  $1\leq s<3$.   From this and \eqref{rita16a}
 we conclude that the integral identity
\begin{equation}\label{inna7}\begin{split}
\int_{\mathbb R^3}\big(\rho(z,t)
\partial_t{\psi}(z,t)\big)\,z_1 dydt +
\sum_{j=1,3}\int_{\mathbb R^3}\big(\rho(z,t) w_j(z,t)
\partial_j\psi(z,t)\big)z_1 dydt\\
+\int_{\mathbb R^3}\varrho(z,t)\,\mathbf{h}(z,t)\,\psi(z,t)\,z_1 ydt
 =0
\end{split}\end{equation}
holds for all $\psi(z,t)$ satisfying the conditions
\begin{equation}\label{rita30}
\psi\in C^1_{loc}(\mathbb R; L^\infty_{loc}(\mathbb
R^2\setminus\{z_1=0\}))\,\cap C_{loc} (\mathbb R;
W^{1,\sigma}_{loc}(\mathbb R^2\setminus\{z_1=0\})), \quad \sigma>
3/2.
\end{equation}
Now choose an arbitrary function $\chi\in C^\infty_0(\mathbb R)$. It
is easily seen that  for an arbitrary fixed $y$, the function $\psi=
z_1^{-1}\chi(t)K_{0i} (y-z)$ satisfies condition \eqref{rita30}.
Substituting $\psi$ into \eqref{inna7} leads to the integral
identity
\begin{equation*}\begin{split}
\int_{\mathbb R}\chi'(t)\Big\{\int_{\mathbb R^2}
K_{0i}(y-z)\rho(z,t)\, dz\,\Big\} dt -\sum_{j=1,3}
\int_{\mathbb R} \chi(t)\Big\{\int_{\mathbb R^2}K_{ij}(y-z)(\rho w_j(z,t))\,dz\Big\} dt\\
+\int_{\mathbb R}\chi(t)\Big\{\int_{\mathbb R^2}
K_{0i}(y-z)\big(\varrho \mathbf h(z,t)-\frac{1}{z_1}\rho
w_1\big)\,dz\Big\} \,dt=0.
\end{split}\end{equation*}
Since $\chi$ is an arbitrary function of the class $ C^\infty_0$, it
follows that the vector field
\begin{equation*}
\overline{\boldsymbol\varsigma}=(\overline{\varsigma}_1,
\overline{\varsigma}_3), \quad \overline{\varsigma}_i= K_{0i}*
\varrho
\end{equation*}
satisfies the desired equality
\begin{equation}\label{inna20}
    \partial_t\overline{\varsigma}_i= -\sum_{j=1,3}K_{ij}*(\rho
    w_j(z,t))+ \sigma_i,
\end{equation}
where
\begin{equation}\label{inna21}
 \sigma_i=K_{0i}*(\varrho \mathbf h(z,t)-\frac{1}{z_1}\rho
w_1\big).
\end{equation}
Note that $1/|z_1|\leq R$ on the support of $\rho$ and $\mathbf h$.
On the other hand,  estimates \eqref{rita13} and \eqref{rita17}
imply the inequality
\begin{equation*}
    \|\varrho \mathbf h\|_{L^\infty(\mathbb R;L^1(\mathbb R^2))}+
    \|\varrho |\mathbf w|\|_{L^\infty(\mathbb R;L^1(\mathbb
    R^2))}\leq c(K).
\end{equation*}
Since the the kernel $K_0(y)$ is bounded it follows that
\begin{equation}\label{inna22}
\|\sigma_i\|_{L^\infty(\mathbb R; C(\mathbb R^2))}\leq c(K).
\end{equation}
Next, the boundedness and continuity of
$\overline{\boldsymbol\varsigma}$ obviously follows from equality
\eqref{inna4} and estimate \eqref{inna60}. We have
\begin{equation*}
    \partial_j\overline{\varsigma}_i= K_{ij}*\rho=\int_{\mathbb
    R^2}\frac{|y-z|^2\delta_{ij}-(y_i-z_i)(y_j-z_j)}{|y-z|^3}\rho(z,t)dz.
\end{equation*}
The kernel in the right hand side is smooth for $y\neq z$ and has
the weak singularity of order $\sim |y-z|^{-1}$ at $y=z$. Since the
function $\rho\in L^\infty(\mathbb R; L^3(\mathbb R^2))$ is
compactly supported and $3>2$, it follows from the standard
estimates of potential operators, see \cite{adams}, that $
\partial_j\overline{\varsigma}_i$ belongs to the class
$L^\infty(\mathbb R; C(\mathbb R^2))$.  Hence
$\overline{\boldsymbol\varsigma}$ belongs to the class
$L^\infty(\mathbb R; C(\mathbb R^2))$



Repeating these arguments and using estimate \eqref{inna6} we
conclude that $K_{ij}*(\rho w_j)$ belongs to the class $L^2(\mathbb
R; C(\mathbb R^2))$. This result along with equality \eqref{inna20}
and estimate \eqref{inna22} imply the inclusion
$\partial_t\overline{\boldsymbol\varsigma}\in L^2(\mathbb R;
C(\mathbb R^2))$.  These completes the proof of the lemma.
\end{proof}

Lemma \ref{inna4}  implies the following important identity.
\begin{lemma}\label{inna8} Under the above assumptions, we have
\begin{equation}\label{inna9}\begin{split}
\frac{1}{2}\int_{\mathbb
R^5}\rho(y,t)\rho(z,t)\frac{|(y-z)^\bot\cdot(\overline{\mathbf
w}(y,t)-\overline{\mathbf w}(z,t))|^2}{|y-z|^3}\,y_1 dydzdt+\\
\int_{\mathbb
R^5}\zeta(y,t)p(y,t)\rho(z,t)\,\frac{y_1dydzdt}{|z-y|}+\Xi=0,
\end{split}\end{equation}
where
$$
(y-z)^\bot=(y_2-z_2, z_1-y_1),\quad \overline{\mathbf w}=(w_1, w_3),
\quad |\Xi|\leq c,
$$
 the constant  $c$  does not depend on the regularization parameters.
\end{lemma}
\begin{proof} Substituting the vector field   $\overline{\boldsymbol\varsigma}$ given by formula
 \eqref{inna4a} into integral identity   \eqref{rita16} and using
 formula \eqref{inna5} for
 $\partial_t\overline{\boldsymbol\varsigma}$ we arrive at the
 equality
\begin{eqnarray}\label{inna10}
\int_{\mathbb R^3}\Big((\rho w_iw_j)( K_{ij}*\rho)-(\rho
w_i)(K_{ij}*(\rho w_j))\Big)\, y_1dydt&+&\\\notag \int_{\mathbb
R^3}\zeta p(K_{ii}*\rho)\,y_1dy dt+\Xi&=&0,
\end{eqnarray}
where
\begin{equation}\label{inna11}
\Xi=\int_{\mathbb R^3}\Big(\rho w_i \sigma_i-\zeta
B_{ij}K_{ij}*\rho+ F_iK_{0i}*\rho\Big)\, y_1 dydt.
\end{equation}
Let us prove that   $|\Xi|$  is bounded by the constant independent
of the regularization parameters. Since  $|K_0|$ does not exceed
$1$, and $\varrho$ is compactly supported in $G\times (0,T)$, we
have
$$
|K_0*\rho|\leq c\|\rho\|_{L^\infty(\mathbb R, L^1(\mathbb R^2))}\leq
c.
$$
From this and estimates  \eqref{rita13} and \eqref{inna5s} we
conclude that
\begin{equation}\label{inna12}
\Big|\int_{\mathbb R^3}\Big(\rho w_i \sigma_i+
F_iK_{0i}*\rho\Big)y_1dydt\Big|\leq c.
\end{equation}
It remains to estimate the product  of the matrix $\mathbf B$ and
the convolution $K_{ij}*\rho$. Formula
 \eqref{inna3} for $K_{ij}$ implies the inequality
\begin{equation}\label{inna31}
\Big|\int_{\mathbb R^3}\zeta B_{ij}K_{ij}*\rho y_1dydt\Big|\leq
c\int_{(0,T)\times G}
\Big(\int_G\frac{\rho(z,t)dz}{|y-z|}\Big\}|\zeta\mathbf B(y,t)|\,y_1
dydt.
\end{equation}
Recall that there is a constant $R$, depending only on the compact
set $K$ such that the inequalities
\begin{equation}\label{inna36}
0< R^{-1}\leq y_1\leq R\text{~~on the support of~~}\zeta.
\end{equation}
 Since the supports of $\zeta \mathbf B$ and $\rho$ are contained
in the support of $\zeta$, it follows from this and \eqref{inna31}
that
\begin{eqnarray}\label{inna32}
\Big|\int_{\mathbb R^3}\zeta B_{ij}K_{ij}*\rho y_1dydt\Big|&\leq&
c\int_{(0,T)\times G}
\Big(\int_G\frac{\rho(z,t)dz}{|y-z|}\Big\}|\zeta\mathbf B(y,t)|
dydt\\\notag &=&c\int_{(0,T)\times G} \Big(\int_G\frac{|\zeta\mathbf
B(y,t)|dy}{|y-z|}\Big\}\rho(z,t) dzdt.
\end{eqnarray}
Introduce the functions
$$
M(t)=\|\zeta\mathbf B(\cdot,t)\|_{L^2(\mathbb R^2)}
$$
$$
\mathbf S(y,t)=\frac{1}{M(t)}\zeta \mathbf B(y,t)\text{~~for
~~}M(t)> 0 \text{~~and~~} \mathbf S(y,t)=0\text{~~ otherwise~~},
$$
$$
\mathbf V(y,t)=\int_{G}\frac{|\mathbf S(t,z)|dz}{|y-z|}
$$
It is clear that
\begin{equation}\label{inna14}
    \|\mathbf V( \cdot,t)\|_{W^{1,2}(G)}\leq c.
\end{equation}
With this notation inequality \eqref{inna32} becomes
\begin{eqnarray}\notag
\Big|\int_{\mathbb R^3}\zeta B_{ij}K_{ij}*\rho y_1 dydt\Big|&\leq&
c\int_0^T M(t) \Big\{\int_G \Big(\int_G\frac{|S(z,t)|
dz}{|y-z|}\Big)\rho(y,t)\,dy\Big\}\,dt
\\\label{inna13}&\leq&c\int_0^T M(t) \Big\{\int_G \mathbf
V(y,t)\rho(y,t)\,dy\Big\}\,dt.
\end{eqnarray}
Now our task is to estimate the integral in the right hand side of
this inequality. Applying the Young inequality we obtain
\begin{equation}\label{inna34}
     \mathbf
V\, \rho\leq e^{\mathbf V}\, -\mathbf V-1+(1+\rho)\ln(1+\rho)-\rho.
\end{equation}
We have
$$
\ln(1+\rho)\leq \rho\ln(2+\rho)\text{~~for~~}\rho\geq
1\text{~~and~~}\ln(1+\rho)\leq \ln 2\text{~~for~~} \rho\leq 1,
$$
which yields
$$
(1+\rho)\ln(1+\rho)-\rho\leq 2\rho\ln(2+\rho)+\ln 2.
$$
It follows from this and  estimate \eqref{rita13}in Proposition
\ref{rita12}, that
\begin{eqnarray}\label{inna37}
\int_G\big((1+\rho(t))\ln(1+\rho(t))-\rho(t)\big)\, dy
&\leq&\\\notag \int_G\big(2\rho(t)\ln(2+\rho(t))+\ln 2\big)\, dy
&=&\\\notag\int_G\big(2\zeta\varrho(t)\ln(2+\zeta\varrho(t))+\ln
2\big)\, dy&\leq &\\\notag 2\int_G\zeta\varrho(t)\ln(2+\varrho(t))\,
dy +c\leq c
\end{eqnarray}
for a.e. $t\in \mathbb R$. On the other hand the Pokhozhaev-
Trudinger inequality along with estimate \eqref{inna14} implies
\begin{equation*}
\int_G\big(e^{\mathbf V(y,t)}\, -\mathbf V(y,t)-1\big)\,dy\leq
c(\|\mathbf V(\cdot,t)\|_{W^{1,2}(G)})\leq c\text{~~for a.e. ~~}
t\in \mathbb R. \end{equation*} It follows from this estimate,
estimate \eqref{inna37} and the Young inequality \eqref{inna34} that
\begin{equation}\begin{split}\label{inna16}
\int_G V(t)\rho(t)\,dy\leq  c\text{~~for a.e. ~~} t\in \mathbb R.
\end{split}\end{equation}
Combining this result with inequality \eqref{inna13} we arrive at
the estimate
\begin{equation*}
\Big|\int_{\mathbb R^3}\zeta B_{ij}\,K_{ij}*\rho dtdy\Big|\leq
c\int_0^T M(t)dt.
\end{equation*}
Since $\zeta \mathbf B$ is compactly supported in $G\times (0,T)$,
there is a compact $K'\subset G$ such that $\text{spt~}\zeta\subset
K'\times (0,T)$. In fact, $K'$ is the orthogonal projection of the
support of  $\zeta$ onto $G$. Next, estimate \eqref{rita26} in
Proposition \ref{rita22} implies the inequality
$$
\int_0^T M(t)^2dt\leq \int_{K'\times (0,T)}|\mathbf B(y,t)|^2
dydt\leq c.
$$
Thus we get
\begin{equation}\label{inna39}
\Big|\int_{\mathbb R^3}\zeta B_{ij}\,K_{ij}*\rho y_1dydt\Big|\leq c.
\end{equation}
It follows from the expression \eqref{inna11} for the quantity $\Xi$
that
\begin{equation*}
|\Xi|\leq \int_{\mathbb R^3}\big|\,\rho w_i \sigma_i+
F_iK_{0i}*\rho\big|\, y_1 dydt+\Big|\int_{\mathbb R^3}\zeta
B_{ij}K_{ij}*\rho\, y_1 dydt\Big|.
\end{equation*}
It remains to note that desired estimate
 $|\Xi|\leq c$ is an obviously follows from this and
 estimates \eqref{inna12}  \eqref{inna39}.
 Finally notice that the main identity \eqref{inna9}is a straightforward consequence
 of relation  \eqref{inna10} and the identities
\begin{equation*}\begin{split}
\int_{\mathbb R^3}\Big((\rho w_iw_j)( K_{ij}*\rho)-(\rho
w_i)(K_{ij}*(\rho w_j))\Big)y_1\,dydt=\\\frac{1}{2}\int_{\mathbb
R^5}\rho(t,y)\rho(t,z)\frac{|(y-z)^\bot\cdot(\overline{\mathbf
w}(t,y)-\overline{\mathbf w}(z,t))|^2}{|y-z|^3}y_1\, dydzdt,\\
\int_{\mathbb R^3}\zeta p(K_{ii}*\rho)y_1\, dy dt=\int_{\mathbb
R^5}\zeta(y,t)p(y,t)\rho(z,t)\,\frac{y_1dydzt}{|z-y|}.
\end{split}\end{equation*}
This completes the proof of Lemma \ref{inna8}

\end{proof}



We are now in a position to complete the proof of Proposition
\ref{inna1}. To this end, note that the first integral in the left
hand side of identity \eqref{inna9} is nonnegative. From this and
estimate for $\Xi$ in Lemma \ref{inna8} we get the inequality
\begin{equation*}\begin{split}
 \int_{\mathbb
R^3}\rho(y,t)\Big\{\int_{\mathbb
R^2}\zeta(z,t)p(z,t)\,\frac{dz}{|z-y|}\Big\}y_1 dydt\leq c.
\end{split}\end{equation*}
Since  $\rho$  and   $\zeta p$   are compactly supported in $G\times
(0,T)$, and
$$
0<R^{-1}<y_1<R
$$ on the supports $\zeta$ and $\rho$, we have
\begin{equation}\label{inna17}\begin{split}
 \int_{\mathbb
R^3}\rho(y,t)\Big\{\int_{\mathbb
R^2}\zeta(z,t)p(z,t)\,\frac{dz}{|z-y|}\Big\}dydt=
\\\int_{(0,T)\times
G}\rho(y,t)\Big\{\int_{G}\zeta(z,t)p(z,t)\,\frac{dz}{|z-y|}\Big\}dydt\leq
c.
\end{split}\end{equation}
Formula  \eqref{stokes106} for $p$  and relation $\rho=\zeta\varrho$
imply the inequality
$$
\zeta p=\zeta\varrho+\zeta\varepsilon
\varrho^{1+\varepsilon}\big(\varepsilon\ln(1+\varepsilon\varrho)+
\frac{\varepsilon\varrho}{1+\varepsilon\varrho}\big)+
\zeta\delta^3\varrho^3\geq \rho,
$$
which yields the estimate
\begin{equation*}\begin{split}
 \int_{G}\zeta(z,t)p(z,t)\,\frac{dz}{|z-y|}\geq
 \int_{G}\,\frac{\rho(t,z)dz}{|z-y|}.
\end{split}\end{equation*}
From this and \eqref{inna17} we obtain
\begin{equation*}\begin{split}
\int_{G\times
(0,T)}\rho(y,t)\Big\{\int_{G}\frac{\rho(z,t)dz}{|z-y|}\Big\}dydt\leq
c.
\end{split}\end{equation*}
Noting that    $\rho$ is compactly supported in $
 G\times(0,T)$ we arrive at the inequality
\begin{equation*}\begin{split}
 \int_{\mathbb
R^3}\rho\, (-\Delta)^{-1/2}\rho\, dydt= c_{1/2}\int_{\mathbb
R^3}\rho(y,t)\Big\{\int_{\mathbb
R^2}\frac{\rho(z,t)dz}{|z-y|}\Big\}\,dydt\leq c.
\end{split}\end{equation*}
It remains to note
\begin{equation*}\begin{split}
\int_{\mathbb R^3}(-\Delta)^{-1/4}\rho\,\,
(-\Delta)^{-1/4}\rho\,\,dydt = \int_{\mathbb R^3}\rho\,
(-\Delta)^{-1/2}\rho\,\, dydt,
\end{split}\end{equation*}
and the proposition follows\qed

\section{Estimates of rotationally symmetric functions in Orlicz spaces }\label{lena}
\renewcommand{\theequation}{4.\arabic{equation}}
\setcounter{equation}{0} In this section we prove the Orlicz type
estimates for the density $\varrho$ and the velocity field $\mathbf
u$.  The result is of general character and does not related with
the theory of Navier-Stokes equations. Suppose  that the bounded
flow domain $\Omega\subset \mathbb R^3$ is rotationally symmetric
and has a smooth boundary $\partial\Omega$. The intersection of
$\Omega$ and the symmetry axis $Ox_3$ consists of the only interval
$\mathcal I$.

Recall that the cylindrical coordinates $(y,\vartheta)$,
$y=(y_1,y_3)$ are defined by the equalities.
\begin{equation}\label{lena13}
    x_1=y_1\cos \vartheta, \quad x_2=y_1 \sin\vartheta,\quad x_3=y_3.
\end{equation}
In particular,  $\Omega$ is defined by the relation
\begin{equation}\label{lena14}
  \Omega\setminus \mathcal I=\big\{\,y,\vartheta:\,\,
  y\in G,\,\, \theta\in [0,2\pi]\,\big\}.
\end{equation}
Here the domain  $G\subset\mathbb R^2$ is the   intersection of the
rotationally symmetric flow domain $\Omega$ and  the half-plane
$\Pi^+$:$x_1>0$, $x_2=0$. Since $y_1=x_1$ and $y_3=x_3$, $G$ can be
regarded as a domain in the plane of the variable $y$.

Let us consider the function $\varrho: \Omega\times(0,T)\to\mathbb
R^+$ and the vector field $\mathbf u:\Omega\times (0,T)\to\mathbb
R^3$ satisfying the following conditions.

%\end{document}
\begin{itemize}
\item[\textbf{C.1}]
The functions $\varrho\in L^\infty(0,T; L^1(\Omega))$ and $\mathbf
u\in L^2(0,T; W^{1,2}_0(\Omega))$ admit the estimate
\begin{equation}\label{lena50}
\text{ess}\sup\limits_{t\in (0,T)}\int_\Omega \big(\,\varrho\ln (2+
\varrho)+\varrho|\mathbf u|^2\,\big)\, dx+\|\nabla\mathbf
u\|_{L^2(\Omega\times (0,T))}\leq c_0.
\end{equation}

\item[\textbf{C.2}]
The functions $\varrho$ and $\mathbf u$ are rotationally symmetric,
i.e.,
\begin{equation}\label{lena51}
\varrho=\varrho (y,t), \quad \mathbf u(y,\theta,t)=\mathbf
U(\theta)\,\mathbf v(y,t),
\end{equation}
where the orthogonal matrix $\mathbf U$ is defined by \eqref{rita2}.


\item[\textbf{C.3}] For every compact set $K\Subset G\times (0,T)$,
the function $\varrho(y, t)\boldsymbol{1}_K$ admits the estimate
\begin{equation}\label{lena52}
\int_K\, \big(\, (-\Delta)^{-1/4}(\varrho \boldsymbol{1}_K)
\big)^2dydt\leq c(K).
\end{equation}
Here
\begin{equation*}
   (-\Delta)^{-1/4}\varrho(y,t)=c_{1/4}\int_G\frac{1}{|y-z|^{3/2}}\,
    \varrho(z,t)\, dz.
\end{equation*}






\end{itemize}













The main result of this section is the following theorem.

\begin{theorem}\label{lena11}
Let $\varrho:\Omega\times (0,T)\to\mathbb R^+$ and $\mathbf
u:\Omega\times (0,T)\to \mathbb R^3$ satisfy conditions
 (\textbf{C.1})-(\textbf{C.1}). Then the
inequality
\begin{equation}\label{lena12}
    \int_{K'}\varrho |\mathbf u|^2\, \ln(2+\varrho|\mathbf
    u|^2)^\alpha\,dxdt \leq c
\end{equation}
holds for all compact sets $K'\subset (\Omega\setminus \mathcal
I)\times (0,T)$ and for all $0<\alpha<1/4$. Here the constant $c$
depends only on $K'$, $\alpha$, and the constant $c _0$ in condition
(\textbf{C.1}).
\end{theorem}

The proof falls  into three steps. First, we reformulate the theorem
in the cylindrical coordinates.


\subsection{Step 1.
Cylindrical coordinates. Localization}



 Without loss of generality we may assume that
$K'$ is a toroidal compact set in the form
$$
K'=\big\{(y,\theta,t): (y,t)\in K, \quad \theta\in [0,2\pi]\big\},
$$
where $K$ is a compact subset $G\times (0,T)$. In other words, $K'$
is obtained by rotation of $K$ around the symmetry axis $Ox_3$. By
virtue of Condition (\textbf{C.2}), it suffices to prove that the
inequality
\begin{equation}\label{lena16}
    \int_{K}\varrho |\mathbf v|^2\, \ln(2+\varrho|\mathbf
    v|^2)^\alpha\,y_1dydt \leq c
\end{equation}
holds for all compact sets $K\subset G\times (0,T)$ and for all
$0<\alpha<1/4$. In order to simplify the proof, we  use the
localization procedure defined in Section \ref{inna}.

Fix a compact set  $K\Subset G\times (0,T)$, and choose two
functions
 $\zeta, \eta\in C^\infty_0(G_T)$ such that
\begin{equation}\label{lena17}
    \zeta=\eta=1 \text{~на~} K,  \quad  \eta=1  \text{~на~}  \text{\rm supp} \,\,
    \zeta, \quad 0\leq \zeta,\eta\leq 1.
\end{equation}
Assume that they are extended by zero to the whole space $\mathbb
R^2\times\mathbb R$.   Set
\begin{equation}\label{lena18}
\rho(y,t)=\zeta(y,t)\varrho(y,t), \quad \mathbf w(y,
t)=\eta(y,t)\mathbf v(y,t), \quad \overline{\mathbf
w}=(w_1,w_3)=\eta \overline{\mathbf v}.
\end{equation}
Extend these functions by zero to $\mathbb R^3$. Note that the
function $\rho$ and  vector field $\mathbf w$ meet all requirements
of Proposition \ref{rita12}. Recall that the is $R>0$, depending on
$K$, such that $R^{-1}\leq y_1\leq R$ on the support of $\zeta$.
Therefore, in order to prove Theorem \eqref{lena11}, it suffices to
show that the estimate
\begin{equation}\label{lena19}
    \int_{\mathbb R^3}\rho |\mathbf w|^2\, \ln(2+\rho|\mathbf
    w|^2)^\alpha\,dydt \leq c
\end{equation}
holds for all $0<\alpha<1/4$.



\subsection{Step 2. Weak estimates of the kinetic energy}
Now we prove the weak type estimates for the localized kinetic
energy $\rho|\mathbf w|^2$.  Our considerations are based on the
following proposition.
\begin{proposition}\label{lena53}
Let the vector field $\mathbf W:\mathbb R^3\to\mathbb R^3$ compactly
supported in $G\times (0,T)$ and
\begin{equation}\label{lena54}
    \|\mathbf W(t)\|_{W^{1,2}(\mathbb R^2)}\leq 1\text{~~for
    a.e.~~}t\in \mathbb R.
\end{equation}
Furthermore assume that $\varrho$ satisfies conditions
\textbf{C.1}-\textbf{C.3} of Theorem \ref{lena11}. Then,  there is a
constant $c$, depending on $K$ and the constant $c_0$ in condition
\textbf{C.1}, such that
\begin{equation}\label{lena21}
    \int_{\mathbb R^3}\rho |\mathbf W|^2 \, \ln(2+\rho|\mathbf W|^2)^{1/2}
    \, dydt\leq c,
\end{equation}
where $c$ depends on $K$.
\end{proposition}
\begin{proof}
The proof is given in Appendix \ref{app9}.
\end{proof}


The main goal of this section is the weak-type estimate for the
kinetic energy tensor. The result is given by the following
proposition. Fix an arbitrary $N\geq 0$ and denote by $\mathcal
A_N\subset G\times (0,T)$ the set
\begin{equation}\label{lena31}
    \mathcal A_N=\big\{\, (y,t)\in G\times (0,T):\,\, \rho(y,t)|\mathbf
    w(y,t)|^2\geq N\,\big\}.
\end{equation}
\begin{proposition}\label{lena32}
Under the above assumptions,  we have
\begin{equation}\label{lena33}
    \int_{\mathcal A_N}\rho|\mathbf w|^2\, dydt\leq
    c\,\ln(2+N)^{-1/4}.
\end{equation}
Here the constant $c$ depends on the compact set  $K$, and does not
depend on the regularization parameters $\varepsilon$, $\delta$ in
the expressions for the internal energy $e$ and the pressure
function $p$.
\end{proposition}

\begin{proof}
Since the function $\rho |\mathbf w|^2$ is integrable, it suffices
to prove the lemma for $N\geq 3$. In this case, we may replace
$\ln(2+N)$ by $\ln N$. Therefore, it suffices to prove the
inequality
\begin{equation}\label{lena33a}
    \int_{\mathcal A_N}\rho|\mathbf w|^2\, dydt\leq
    c\,\ln N^{-1/4} \text{~~for all~~} N\geq 3.
\end{equation}
Our  considerations are based on the following construction. Denote
by $M(t)$ the function
$$
M(t)=\|\mathbf w(\cdot,t)\|_{W^{1,2}(\mathbb R^2)}.
$$
Note that $M(t)$ is compactly supported in $(0,T)$. Choose an
arbitrary $\sigma\geq 1$ and set
\begin{eqnarray}\notag
    \mathcal B&=&\big\{ (y,t)\in G\times (0,T):\,\,
     (y,t)\in \mathcal A_N,\, M(t)\leq \sigma\,\big\},
    \\\notag \mathcal C&=&\big\{ (y,t)\in G\times (0,T):\,\,
     (y,t)\in \mathcal A_N,\, M(t)> \sigma\,\big\}.
\end{eqnarray}
It is clear that
\begin{equation}\label{lena34}
  \int_{\mathcal A_N}\rho|\mathbf w|^2\, dydt=  \int_{\mathcal B}\rho|\mathbf w|^2\, dydt
  +  \int_{\mathcal C}\rho|\mathbf w|^2\, dydt.
\end{equation}
Let us estimate the second integral in the right hand side. It
follows from the energy estimate \eqref{rita13} that
\begin{eqnarray*}
\int_{\mathcal C}\rho|\mathbf w|^2\, dydt&\leq &
\int_{t:M(t)>\sigma}\Big(\int_G \rho|\mathbf w|^2\, dy\,\Big)\, dt\\
&\leq& c\text{~~meas~~}\{M(t)>\sigma\}
\end{eqnarray*}
Next, energy estimate \eqref{rita13} implies
\begin{equation*}
\int_{\mathbb R} M(t)^2\, dt= \int_{\mathbb R^2}\|\mathbf
w(t)\|_{W^{1,2}(\mathbb R^2)}^2\, dt\leq c,
\end{equation*}
which gives
\begin{equation}
\text{~~meas~~}\{M(t)>\sigma\}\leq c/\sigma^2.
\end{equation}
Combining the obtained inequalities we arrive at the estimate
\begin{equation}\label{lena35}
   \int_{\mathcal C}\rho|\mathbf w|^2\, dydt\leq c/\sigma^2.
\end{equation}
Now our task is to estimate the first integral in the right hand
side of \eqref{lena34}. Introduce the function
$$
\mathbf W(y,t)=M(t)^{-1}\mathbf w(y,t)\text{~when~}
M(t)>0,\text{~and~} \mathbf W(y,t)=0\text{~~otherwise}.
$$
It is clear that $(y,t)\in \mathcal B$ if and only if
\begin{equation*}
M(t)\leq \sigma,  \quad \rho(y,t)|\mathbf W(y,t)|^2\geq
\frac{N}{M(t)^2}
\end{equation*}
It follows from this that
\begin{equation}\label{lena36a}
\mathcal B\subset \mathcal B^*=\big\{\,(y,t)\in G\times
(0,T):M(t)\leq \sigma, \,\,\, \rho(y,t)|\mathbf W(y,t)|^2\geq
N/\sigma^2\,\big\}.
\end{equation}
Thus we get
\begin{eqnarray}\label{lena36}
\int_{\mathcal B}\rho|\mathbf w|^2\, dydt&\leq&\int_{\mathcal
B^*}\rho|\mathbf w|^2\, dydt\\\notag &=& \int_{\mathcal
B^*}M(t)^2\rho|\mathbf W|^2\, dydt\\\notag &\leq& \sigma^2
\int_{\mathcal B^*}\rho|\mathbf W|^2\, dydt. \end{eqnarray} On the
other hand, definition \eqref{lena36a} of the set $\mathcal B^*$
implies
\begin{eqnarray*}
 \ln (2+N/\sigma^2)^{1/2}\int_{\mathcal B^*}\rho|\mathbf W|^2\, dydt
 &\leq&\int_{\mathcal B^*}\rho|\mathbf W|^2\ln(2+\rho|\mathbf W|^2)^{1/2}dydt\\
 &\leq&\int_{G\times (0,T)}\rho|\mathbf W|^2\ln(2+\rho|\mathbf
 W|^2)^{1/2}dydt.
\end{eqnarray*}
From this and estimate \eqref{lena21} in Proposition \ref{lena53} we
obtain
\begin{eqnarray*}\notag
 \int_{\mathcal B^*}\rho|\mathbf W|^2\, dydt
 &\leq&\\ &&\ln (2+N/\sigma^2)^{-1/2}\int_{G\times (0,T)}\rho|\mathbf W|^2\ln(2+\rho|\mathbf
 W|^2)^{1/2}dydt\\&\leq& c\ln (2+N/\sigma^2)^{-1/2}\leq c\ln(N/\sigma^2)^{-1/2}.
\end{eqnarray*}
Substituting this inequality into \eqref{lena36} we obtain the
estimate
\begin{equation}\label{lena37}
    \int_{\mathcal B}\rho|\mathbf w|^2\, dydt\leq c\sigma^2
    \ln(N/\sigma^2)^{-1/2}.
\end{equation}
Next, substituting inequalities \eqref{lena35} and \eqref{lena37}
into identity \eqref{lena34} we arrive at the estimate
\begin{equation}\label{lena38}
    \int_{\mathcal A_N}\rho|\mathbf w|^2\, dydt\leq c/\sigma^{-2}
+c\sigma^2
    \ln(N/\sigma^2)^{-1/2}.
\end{equation}
Now set $\sigma=\ln N^{1/8}$. We have
$$
\ln\big(N/(\ln N)^{1/4}\big)\geq \ln\big(N/
N^{1/4}\big)=\frac{3}{4}\ln N.
$$
Substituting these relations into \eqref{lena38} we finally obtain
\begin{equation*}
    \int_{\mathcal A_N}\rho|\mathbf w|^2\, dydt\leq c\ln N^{-1/4}+c
    \ln N^{1/4}/(\ln N)^{1/2}\leq c \ln N^{-1/4},
\end{equation*}
and the proposition follows.
\end{proof}

\subsection{Step 3. The proof of Theorem \ref{lena11}} We are now in a
position to complete the proof of Theorem \ref{lena11}. To this end,
we use the weak-type estimate \eqref{lena33} in Proposition
\ref{lena32}. Denote by $E(\lambda)$, $\lambda\geq 0$ the monotone
function defined by the relation
\begin{equation*}
E(\lambda)=\text{meas~}\,\big\{\,(y,t)\in G\times
(0,T):\,\,\rho(y,t)|\mathbf w(y,t)|^2\geq \lambda\,\big\}.
\end{equation*}
It is the standard fact of calculus, that the equality
\begin{equation*}
    \int_{G\times(0,T)} f(\rho|\mathbf w|^2)\,dydt
    =-\int_{[0,\infty)}\, f(\lambda)\,\, d E(\lambda)
\end{equation*}
holds for every piecewise continuous function $f$ such that
$f(\rho|\mathbf w|^2)\in L^1(G\times (0,T)).$ Here $-dE(\lambda)$ is
the Borel-Stieltjes measure associated with the function
$-E(\lambda)$. In particular, we have
\begin{equation}\label{lena39}
\int_{\mathcal A_N}\rho|\mathbf w|^2\, dydt=-\int_{[N, \infty)}
\lambda \, d E(\lambda).
\end{equation}
With this notation the desired inequality \eqref{lena19} reads
\begin{equation}\label{lena40}
\int_{G\times (0,T)}\rho|\mathbf w|^2\ln(2+\rho|\mathbf
w|^2)^\alpha\, dydt\equiv -\int_{[0, \infty)}
\lambda\ln(2+\lambda)^\alpha \, d E(\lambda)\leq c.
\end{equation}
Introduce the monotone function $F$ and the corresponding
 Borel-Stieltjes measure $-dF$ defined by the equalities
$$
F(s)=-\int_{[s,\infty)}\lambda\, dE(\lambda), \quad
dF(\lambda)=\lambda\, dE(\lambda).
$$
Note that identity \eqref{lena39} and estimate \eqref{lena33} in
Proposition \ref{lena32} imply
\begin{equation}\label{lena41}
    F(s)\leq c \ln(2+s)^{-1/4}\text{~~for~~} s\geq 0.
\end{equation}
It is clear that
\begin{equation*}
-\int_{[0, \infty)} \lambda\ln(2+\lambda)^\alpha \, d E(\lambda)=
-\int_{[0, \infty)} \ln(2+\lambda)^\alpha \, d F(\lambda)
\end{equation*}
Now our task is to prove that the integral in the right hand side
exists and admits the effective estimate. Let us consider the
regularized integral
\begin{equation*}
    I(\epsilon,
    N)=-\int_{[0,N)}\,\,
  g(\lambda)  d F(\lambda), \quad g(\lambda)=\frac{\ln(2+\lambda)^\alpha}{1+\epsilon^2\lambda^2}
\end{equation*}
Note that  $g$ is bounded and continuously differentiable. On the
other hand, the total variation of $F$ is finite. Hence, we may
integrate by parts to obtain
\begin{equation}\label{lena42}
    I(\epsilon,
    N)=\int_{[0,N)}\,\,g'(\lambda)\, F(\lambda)\,d\lambda+ g(0)
    F(0)-g(N)\lim\limits_{\lambda\to N-0}F(\lambda).
\end{equation}
We have
\begin{eqnarray*}
g'(\lambda)&=&g_0(\lambda)-\mathbf R(\lambda),\\\notag
    g_0(\lambda)&=&\alpha
    \ln(2+\lambda)^{\alpha-1}(2+\lambda)^{-1}(1+\epsilon^2\lambda^2)^{-1}\\\notag
\mathbf R(\lambda)&=&2\epsilon^2
\lambda\ln(2+\lambda)^{\alpha}(1+\epsilon^2\lambda^2)^{-2}
\end{eqnarray*}
Since $\alpha\in (0,1/4)$, estimate \eqref{lena41} implies
$$
|g'(\lambda) F(\lambda)|\leq c (1+\epsilon^2\lambda^2)^{-1}
$$
which yields $g' F\in L^1(\mathbb R^+)$. On the other hand, we have
\begin{equation*}
0\leq g(N) F(N)\leq c \ln(2+N)^{\alpha-1/4}(1+\epsilon^2
N^2)^{-1}\to 0\text{~~as~~} N\to\infty.
\end{equation*}
Letting $N\to\infty$ in equality \eqref{lena42} we conclude that
there exists the integral
\begin{eqnarray}\notag
-\int_{[0,\infty)}g(\lambda)\,
dF(\lambda)&=&\int_0^\infty\,\,g_0(\lambda)\,
F(\lambda)\,d\lambda\\\notag &-&\int_0^\infty\,\,\mathbf
R(\lambda)\, F(\lambda)\,d\lambda+ g(0) F(0).
\end{eqnarray}

Since $\mathbf R$ and $F$ are nonnegative, we have
\begin{equation}\label{lena44}
-\int_{[0,\infty)}g(\lambda)\,
dF(\lambda)\leq\int_0^\infty\,\,g_0(\lambda)\, F(\lambda)\,d\lambda+
g(0) F(0).
\end{equation}
It follows from \eqref{lena31} and \eqref{lena39} that
\begin{eqnarray}\label{lena440}
 F(0)&=&-\int_{[0,\infty)}d
F(\lambda)=-\int_{[0,\infty)} \lambda d E(\lambda)\\\notag
&=&\int_{\mathcal A_0}\rho|\mathbf w|^2\, dydt=\int_{G\times
(0,T)}\rho|\mathbf w|^2\, dydt\leq c.
\end{eqnarray}
Next, estimate \eqref{lena41} yields
\begin{eqnarray}\notag
\int_0^\infty g_0 F\, d\lambda&\leq &\alpha\int_0^\infty
\ln(2+\lambda)^{\alpha-1}(2+\lambda)^{-1}\,F(\lambda)d\lambda\\\notag
&\leq &\int_0^\infty
\ln(2+\lambda)^{\alpha-1-1/4}(2+\lambda)^{-1}\,d\lambda
 \leq    c(\alpha).
\end{eqnarray}
It follows from this and \eqref{lena440} that
\begin{equation}\label{lena45}
-\int_{[0,\infty)}g_0(\lambda)\, dF(\lambda)\leq c(\alpha)+\ln
2^\alpha c\leq c.
\end{equation}
Recall that $-dF$ is a nonnegative Borel measure. Notice that $g_0$
is an increasing function of $\epsilon$ and
$$
g_0(\lambda)\,\nearrow \,\ln(2+\lambda)^\alpha \text{~~as~~}
\epsilon\, \searrow\,  0.
$$
Letting $\epsilon\to 0$ in \eqref{lena45} and applying the Fatou
theorem we conclude that there exists the integral
\begin{equation*}
-\int_{[0,\infty)}\ln(2+\lambda)^\alpha\, dF(\lambda)\equiv
-\int_{[0,\infty)}\lambda \ln(2+\lambda)d E(\lambda)\leq c.
\end{equation*}
Combining this inequality with relation \eqref{lena40} we obtain
desired estimate \eqref{lena19}. This completes the proof of Theorem
\ref{lena11}.

\section{ Singular limits }\label{olga}
\renewcommand{\theequation}{5.\arabic{equation}}
\setcounter{equation}{0}

Until now, we have been dealing with solutions of regularized
equations with internal energy $e$ and pressure functions $p$ given
by the equalities
\begin{eqnarray}\label{olga10}
e&=&\varrho\ln\varrho-\varrho + e_1(\varrho)+ e_2(\varrho), \\\notag
e_1&=&\varepsilon
\varrho^{1+\varepsilon}\ln(1+\varepsilon\varrho),\,\,\,\,
 e_2=\frac{1}{2}\delta^3\varrho^3,
\end{eqnarray}
\begin{eqnarray}\label{olga11}
p&=&\varrho + p_1(\varrho)+ p_2(\varrho), \\\notag
 p_1&=&\varepsilon
\varrho^{1+\varepsilon}\big(\varepsilon\ln(1+\varepsilon\varrho)+
\frac{\varepsilon\varrho}{1+\varepsilon\varrho}\big), \,\,\,\,
p_2=\delta^3\varrho^3. \end{eqnarray} Here  $\varepsilon, \delta\in
(0,1/2)$ are the regularization parameters.


Proposition \ref{stokes107} guarantees the existence of weak
rotationally symmetric solutions of regularized equations. These
solutions depend on the regularization parameters $\varepsilon$ and
$\delta$. In this section, we will show this dependence and use the
following notation for solutions of regularized equations
\eqref{stokes108}
\begin{equation*}
\varrho=\varrho_{\varepsilon,\delta}, \quad \mathbf u=\mathbf
u_{\varepsilon,\delta}
\end{equation*}
Recall that the main goal of the article is to prove the existence
od weak rotationally symmetric solutions to the isothermal
Navier-Stokes equations \eqref{isothermal}. The main Theorem
\ref{thm1} gives a precise formulation of this result. We obtain the
solution of  equations \eqref{isothermal}  as the limit of the
solutions to the regularized equations \eqref{stokes108}. The first
step on this way is to prove that the pressure functions $p_1$, $p_2
$ converge to zero on compact subsets of  $\Omega\setminus \mathcal
I$ as the regularization parameters $\varepsilon$, $\delta$ tend to
zero.






  The main idea is the following. We assume that the
  regularization parameters $\varepsilon$
  and $\delta$ are dependent,
\begin{equation}\label{olga12}
    \delta=\kappa(\varepsilon)\ll \varepsilon, \quad
    \kappa(\varepsilon)= e^{-\frac{1}{\varepsilon^9}},\quad\varepsilon\in
    (0,1/2).
\end{equation}
  In this case, we
 use the denotations
\begin{equation}\label{olga12den}
   \varrho_{\varepsilon}=\varrho_{\varepsilon, \kappa(\varepsilon)}, \quad
    \mathbf u_\varepsilon=\mathbf u_{\varepsilon, \kappa(\varepsilon)}.
\end{equation}
The following theorem is the main result of this section.
\begin{theorem}\label{olga13}
Under the above assumptions,
\begin{equation}\label{olga15}
    \int_{K'}(p_1(\varrho_\varepsilon)+p_2(\varrho_\varepsilon))\,dxdt
    \to 0\text{~~as~~}\varepsilon\to 0
\end{equation}
for every compact set $K'\subset (\Omega\setminus \mathcal I)\times
(0,T)$
\end{theorem}
We reformulate this theorem in the cylindrical coordinates arguing
as in the beginning of Section \ref{lena}. Recall that the
cylindrical coordinates $(y,\vartheta)$, $y=(y_1,y_3)$ are defined
by the equalities.
\begin{equation*}
    x_1=y_1\cos \vartheta, \quad x_2=y_1 \sin\vartheta,\quad x_3=y_3.
\end{equation*}
In particular, we have
\begin{equation*}
  \Omega\setminus \mathcal I=\big\{\,y,\vartheta,t:\,\,
  y\in G,\,\, \theta\in [0,2\pi]\,\big\}.
\end{equation*}
Here the domain  $G\subset\mathbb R^2$ is the   intersection of the
rotationally symmetric flow domain $\Omega$ and  the half-plane
$\{x_1>0, x_2=0\}$.  It can be regarded as the domain in the plane
of variable $y$. Without loss of generality we may assume that $K'$
is a toroidal compact set
$$
K'=\big\{(y,\theta,t): (y,t)\in K, \quad \theta\in [0,2\pi]\big\}.
$$
Here $K$ is a compact subset of $G\times (0,T)$. In other words,
$K'$ is obtained by rotation of $K$ around the symmetry axis $Ox_3$.
A rotationally symmetric weak solution to problem \eqref{stokes108}
admits the representation, see Section \ref{s2},
\begin{equation*}
\mathbf u_{\varepsilon,\delta}(x,t)=\mathbf U(\theta)\, \mathbf
v_{\varepsilon, \delta}(y,t), \quad \varrho_{\varepsilon,
\delta}=\varrho_{\varepsilon, \delta}(y,t), \quad (y,t)\in G\times
(0,T),
\end{equation*}
where the orthogonal matrix $\mathbf U$ is given by \eqref{rita2}.
Hence Theorem \ref{olga13} can be  reformulated as follows
\begin{theorem}\label{olga13cyl}
Under the above assumptions,
\begin{equation}\label{olga16}
    \int_{K}p_1(\varrho_\varepsilon)\,dydt+\int_{K}p_2(\varrho_\varepsilon)\,dydt
    \to 0\text{~~as~~}\varepsilon\to 0
\end{equation}
for every compact set $K\subset G\times (0,T)$
\end{theorem}
The rest of the section is devoted to the proof of Theorems
\ref{olga13} and \ref{olga13cyl}. We start with the analysis of the
behavior of the function $p_2$ for small $\delta\ll \varepsilon$.

\subsection{Estimates of $p_2$. Auxiliary theorem}
The proofs of Theorems \ref{olga13} and \ref{olga13cyl} are based on
the intermediate technical result. In order to formulate it, recall
the localization procedure, see Section \ref{inna} for details.

 Fix a compact
set  $K\Subset G\times (0,T)$, and choose two functions
 $\zeta, \eta\in C^\infty_0(G_T)$ such that
\begin{equation*}
    \zeta=\eta=1 \text{~on~} K,  \quad  \eta=1  \text{~on~}  \text{\rm supp} \,\,
    \zeta, \quad 0\leq \zeta,\eta\leq 1.
\end{equation*}
Assume that they are extended by zero to the whole space $\mathbb
R^3$.   Set
\begin{equation*}
\rho_{\varepsilon, \delta}(y,t)=\zeta(y,t)\varrho_{\varepsilon,
\delta}(y,t), \quad \mathbf w_{\varepsilon, \delta}(y,
t)=\eta(y,t)\mathbf v_{\varepsilon, \delta}(y,t).
\end{equation*}
Extend these functions by zero to $\mathbb R^3$. Note that the
function $\rho=\rho_{\varepsilon,\delta}$ and and the vector field
$\mathbf w=\mathbf w_{\varepsilon, \delta}$ meet all requirements of
Proposition \ref{rita12}. We also  notice that
$\varrho_{\varepsilon,\delta}=\rho_{\varepsilon, \delta}$ in $K$.
The following auxiliary theorem is the main result of this
subsection.






\begin{theorem}\label{olga100}
For every compact set $K\in G\times (0,T)$, there is a constant $c$
such that the inequality
\begin{equation}\label{olga101}
\int_{\mathbb R^3}p_2(\rho_{\varepsilon, \delta})\, dydt=
\int_{\mathbb R^3}\delta^3\,\rho_{\varepsilon, \delta}^3\, dydt\leq
c\varepsilon
\end{equation}
holds for all $(\varepsilon, \delta)$ satisfying the inequalities
\begin{equation}\label{olga102}
    0<\varepsilon\leq 1/2, \quad 0<\delta\leq \kappa(\varepsilon)=
    e^{-\frac{1}{\varepsilon^9}}.
\end{equation}
\end{theorem}
The proof falls into three steps. By abuse of notation, we will
write simply  $\rho$ and $\mathbf w$ instead of $\rho_{\varepsilon,
\delta}$ and $\mathbf w_{\varepsilon, \delta}$.
\paragraph{ \textbf{Step 1. The basic inequality.}}
Introduce the function
\begin{equation}\label{olga26}
    \varphi(\rho)=\ln(1+\delta \rho)\text{~~such
    that~~}p_2\varphi=\delta^3\rho^3\, \ln(1+\delta \rho)
\end{equation}
Recall that the potential $\nabla\Delta^{-1}$ admits the
representation
\begin{equation*}
\nabla \Delta^{-1}u=c\int_{\mathbb R^2} \frac{y-z}{|y-z|^2}\, u(z)\,
dz.
\end{equation*}
Our first goal is the derivation of  the special integral inequality
for $p_2\varphi$. The result is   given by the following lemma.
\begin{lemma}\label{olga27}
Let $\mathbf B$ and  $\mathbf F$  be defined by Propositions
\ref{rita22} and \ref{rita12}. Then we have
\begin{equation}\label{olga28}
\int_{\mathbb R^3} p_2(\rho)\,\varphi(\rho)\, y_1
dydt\leq\sum_{i=1}^{5} I_i,
\end{equation}
where
\begin{eqnarray}\notag
I_1&=&\int_{\mathbb R^3}\big\{\nabla\Delta^{-1}\text{~\rm
div~}(\varphi\overline{\mathbf w})\big\}\,\cdot
(\rho\overline{\mathbf w})\,y_1dydt
\\\label{olga29}I_2&=&-\int_{\mathbb R^3}
(\rho \overline{\mathbf w}\otimes \overline{\mathbf w}):
(\nabla^2\Delta^{-1}\varphi)\,y_1dydt,
\\\notag I_3&=& \int_{\mathbb R^3}\zeta \mathbf
B:(\nabla^2\Delta^{-1}\varphi)\,y_1dydt,\\\notag I_4&=&
-\int_{\mathbb R^3}\mathbf F\cdot (\nabla\Delta^{-1}\varphi)\,
y_1dydt,\\\notag I_5&=& -\int_{\mathbb R^3}
(\nabla\Delta^{-1}\boldsymbol\pi)\cdot (\rho\overline{\mathbf w})\,
y_1dydt,
\end{eqnarray}
Here $\overline{\mathbf w}= (w_1,w_3)$, $\zeta$ is a smooth cutoff
function such that $\rho=\zeta\varrho$. The function $\boldsymbol
\pi$ is defined by the equality
\begin{equation}\label{olga30}
    \boldsymbol{\pi}=\frac{\delta\rho}{1+\delta\rho}\big(\,
    \nabla\zeta\cdot \overline{\mathbf w}-\frac{1}{y_1}
    w_1-\text{\rm div~}\overline{\mathbf w} \,\big)+
    \varphi\text{\rm ~div~}\overline{\mathbf w}.
\end{equation}

\end{lemma}

\begin{proof} The proof is given in Appendix \ref{hurma}.\end{proof}


\paragraph{ \textbf{Step 2. Estimates of $I_i$}}Now our task is to estimate the quantities $I_i$ in the right hand
side of inequality \eqref{olga28}. The derivation of these estimate
is based on the following auxiliary proposition, which is not
related directly to Navier-Stokes equations.  Let us consider the
function $\rho:\mathbb R^2\to\mathbb R^+$ and vector field $\mathbf
W=(W_1,W_3):\mathbb R^2\to \mathbb R^2$ satisfying the following
condition.

\begin{condition}\label{olga38ab}
The functions $\rho$ and $\mathbf W$ are compactly supported in $G$.
They admit the estimate
\begin{equation}\label{olga39}\begin{split}
\int_{\mathbb R^2}\rho\ln(2+\rho)\, dy+    \int_{\mathbb R^2}
p_1(\rho)\, dy\leq c_1<\infty, \quad
    \|\mathbf W\|_{W^{1,2}(\mathbb R^3)}\leq 1,\\
0<c_1^{-1}\leq y_1\leq c_1\text{~~ on the supports of~~} \rho
\text{~~and~~} \mathbf W.
\end{split}\end{equation}
\end{condition}
\begin{proposition}\label{olga43ab} Let $\rho$ and $\mathbf W$ satisfy
Condition \ref{olga38ab}. Then there is a constant $c$, depending on
$c_1$ and $G$ such that
\begin{eqnarray}\label{olga44ab}
\|\rho \mathbf W\|_{L^{1+\frac{\varepsilon}{2}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^2}, & & \|\rho |\mathbf
W|^2\|_{L^{1+\frac{\varepsilon}{2}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^3},\\
\label{olga45ab} \|\nabla\Delta^{-1}\text{\rm ~div~}(\varphi \mathbf
W)\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^6},&&\|\nabla^2\Delta^{-1}\varphi
\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^5},\\\label{olga46ab}
\|\nabla\Delta^{-1}\boldsymbol\pi^*\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}\leq \frac{c}{\varepsilon^{3}}, &&
\|\nabla\Delta^{-1}\varphi\|_{L^\infty(\mathbb R^2)}\leq {c}.
\end{eqnarray}
Here

\begin{equation}\label{olga46piab}
\varphi=\ln(1+\delta \rho), \quad
\boldsymbol\pi^*=\frac{\delta\rho}{1+\delta\rho}\big(\,
    \nabla\zeta\cdot \mathbf W-\frac{1}{y_1}
    W_1-\text{\rm ~div~}\mathbf W\,\big)+\varphi\text{\rm ~div~}\mathbf W.
\end{equation}
\end{proposition}

\begin{proof} The proof is given in Appendix \ref{sliva}.\end{proof}

We are now in a position to estimate the quantities $I_i$. The
result is given by the following lemma.
\begin{lemma}\label{olga110}
Under the assumptions of Theorem \ref{olga13cyl}, the quantities
$I_i$ admit the estimates
\begin{equation}\label{olga111}
|I_i|\leq c \varepsilon^{-8}, \quad \varepsilon, \delta\in (0,1/2).
\end{equation}
Here the constant $c$ does not depend on $\varepsilon$ and $\delta$
\end{lemma}
\begin{proof}
We estimate $I_i$ step by step. Introduce the functions
\begin{eqnarray}\notag
M(t)&=&\|\overline{\mathbf w}(\cdot,t)\|_{W^{1,2}(\mathbb
R^2)},\quad t\in \mathbb R,\\\label{olga112} \mathbf W(y,t)&=&
M(t)^{-1} \overline{\mathbf w}(y,t)\text{~for~} M(t)\neq 0,\\\notag
\mathbf W(y,t)&=& 0 \text{~~otherwise}.
\end{eqnarray}
Recall that the cutoff function $\zeta:\mathbb R^3\to [0,1]$ is
compactly supported in $G\times (0,T)$. In particular, we have
\begin{equation*}
\rho\ln(2+\rho)=\zeta\varrho\ln(2+\zeta\varrho)\leq
\zeta\varrho\ln(2+\varrho).
\end{equation*}
From this and estimate \eqref{rita13} in Proposition \ref{rita12} we
conclude that the inequality
\begin{equation}\label{olga115}
\int_{\mathbb R^2}\rho(t)\ln(2+\rho(t))\, dydt \leq c.
\end{equation}
holds for a.e. $t\in\mathbb R$. Here $c$ does not depend on $t$.
Therefore, the functions $\rho$ and $\mathbf W$ meet al requirements
of Proposition \ref{olga43ab} and satisfy inequalities
\eqref{olga44ab}-\eqref{olga46ab}. Now we may estimate $I_i$.

\emph{Estimate of $I_1$.} We have
\begin{equation*}
\big\{\nabla\Delta^{-1}\text{~\rm div~}(\varphi\overline{\mathbf
w})\big\}\,\cdot (\rho\overline{\mathbf w})= M(t)^2 \rho\mathbf W
\cdot \big\{\nabla\Delta^{-1}\text{~\rm div~}(\varphi\mathbf
W)\big\}.
\end{equation*}
Recall that
$$
0<R^{-1}<y_1< R\text{~~on the support of ~~}\rho.
$$
From this and expression \eqref{olga29} for $I_1$ we conclude that
\begin{equation*}
|I_1|\leq c\int_{\mathbf R}M(t)^2\Big(\int_{\mathbf R^2}
|\rho(t)\mathbf W(t) |\,|\nabla\Delta^{-1}\text{~\rm
div~}(\varphi(t)\mathbf W(t))|\, dy\Big)\,dt.
\end{equation*}
Note that $\rho(t)$ and $\mathbf W(t)$ meet all requirements of
Proposition \ref{olga43ab} and admit estimates \eqref{olga44ab},
\eqref{olga45ab} with the constant $c$ independent of $t$. It
follows from these estimate and the H\"{o}lder inequality that
\begin{equation*}\begin{split}
\int_{\mathbf R^2} |\rho(t)\mathbf W(t)
|\,|\nabla\Delta^{-1}\text{~\rm div~}(\varphi(t)\mathbf W(t))|\, dy
\leq\\ \|\rho(t)\mathbf W(t)\|_{L^{1+\frac{\varepsilon}{2}}(\mathbb
R^2)}\, \|\nabla\Delta^{-1}\text{~\rm div~}(\varphi(t)\mathbf
W(t)\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^8}. \end{split}\end{equation*} This gives the
desired estimate
\begin{equation*}
|I_1|\leq \frac{c}{\varepsilon^6}\int_{\mathbf R}M(t)^2\,dt\leq
\frac{c}{\varepsilon^6}\|\overline{\mathbf w}\|_{L^2(\mathbb
R;W^{1,2}(\mathbb R^2))}^2\leq \frac{c}{\varepsilon^8}.
\end{equation*}
\emph{Estimate of $I_2$.} Arguing as before we obtain
\begin{equation*}
|I_2|\leq c\int_{\mathbf R}M(t)^2\Big(\int_{\mathbf R^2}
\rho(t)|\mathbf W(t) |^2\,|\nabla^2\Delta^{-1}\varphi(t)|\,
dy\Big)\,dt.
\end{equation*}
Estimates \eqref{olga44ab} and \eqref{olga45ab} in Proposition
\ref{olga43ab} imply the inequality
\begin{equation*}\begin{split}
\int_{\mathbf R^2} \rho(t)|\mathbf W(t)
|^2\,|\nabla^2\Delta^{-1}\varphi(t)|\, dy\leq\\
\|\rho(t)|\mathbf W(t)^2\|_{L^{1+\frac{\varepsilon}{2}}(\mathbb
R^2)}\,
\|\nabla^2\Delta^{-1}\varphi(t)\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}\leq \frac{c}{\varepsilon^8}. \end{split}\end{equation*} Thus
we get
\begin{equation*}
|I_2|\leq \frac{c}{\varepsilon^6}\int_{\mathbf R}M(t)^2\,dt\leq
\frac{c}{\varepsilon^6}\|\overline{\mathbf w}\|_{L^2(\mathbb
R;W^{1,2}(\mathbb R^2))}^2\leq \frac{c}{\varepsilon^8}.
\end{equation*}
\emph{Estimate of $I_3$.}Note that the function $\zeta$ is compactly
supported in a compact subset of $G\times (0,T)$. Moreover, this
subset depends only on the compact set $K$ in Proposition
\ref{olga100}. It follows from this and estimate \eqref{rita26} in
this proposition  that
\begin{equation}\label{olga114}
\|\zeta\mathbf B\|_{L^2(\mathbb R^3)}\leq c.
\end{equation}
On the other hand the inequality
$$
\varphi(\rho(t))^2=\ln(1+\delta\rho(t))^2\leq c  \rho(t)
$$
along with estimate \eqref{olga115} implies
\begin{equation*}
\|\varphi(t)\|_{L^2(\mathbb R^2)}\leq c.
\end{equation*}
From this and  estimate \eqref{uruk7} for the Riesz singular
integral in Appendix \ref{uruk} we obtain
\begin{equation*}
\|\nabla^2\Delta^{-1}\varphi(t)\|_{L^2(\mathbb R^2)}\leq c,
\end{equation*}
where $c$ is independent on $t$. Since $\varphi$ is compactly
supported in $G\times (0,T)$ it follows that
\begin{equation}\label{olga116}
\|\nabla^2\Delta^{-1}\varphi\|_{L^2(\mathbb R^3)}\leq c.
\end{equation}
Combining estimates \eqref{olga114} and \eqref{olga116} we arrive at
desired estimate for $I_3$,
\begin{equation*}
    |I_3|\leq \|\zeta\mathbf B\|_{L^2(\mathbb R^3)}\,
    \|\nabla^2\Delta^{-1}\varphi\|_{L^2(\mathbb R^3)}\leq c.
\end{equation*}
\emph{Estimate of $I_4$.} Since $\varphi(t)$ meet all requirements
of Proposition \ref{olga43ab} and is compactly supported in $G\times
(0,T)$, it follows from estimate \eqref{olga46ab} in  Proposition
\ref{olga43ab} that
\begin{equation}
    \|\nabla\Delta^{-1}\varphi\|_{L^\infty(\mathbb R^3)}\leq c
\end{equation}
From this, estimate \eqref{rita17}  for $\mathbf F$, and expression
\eqref{olga29} for $I_4$ we obtain the needed estimate
\begin{equation*}
    |I_4|\leq \|\mathbf F\|_{L^1(\mathbb R^3)}\,
    \|\nabla\Delta^{-1}\varphi\|_{L^\infty(\mathbb R^3)}\leq c.
\end{equation*}
\emph{Estimate of $I_5$.} Rewrite formula \eqref{olga29} for $I_5$
in the equivalent form
\begin{equation*}
I_5=-\int_{\mathbb R} M(t)^2\Big(\int_{\mathbb
R^2}(\nabla\Delta^{-1}\boldsymbol\pi^*)\cdot(\rho\mathbf W) y_1
dy\Big)\, dt.
\end{equation*}
It follows that
\begin{equation*}
|I_5|\leq\int_{\mathbb R} M(t)^2\Big(\int_{\mathbb
R^2}|\nabla\Delta^{-1}\boldsymbol\pi^*|\,|\rho\mathbf W| dy\Big)\,
dt.
\end{equation*}
It follows from estimates \eqref{olga44ab} and \eqref{olga46ab} that
\begin{equation*}\begin{split}
\int_{\mathbb
R^2}|\nabla\Delta^{-1}\boldsymbol\pi^*(t)|\,|\rho(t)\mathbf
W(t)| dy\leq\\
\|\rho(t)\mathbf W(t)\|_{L^{1+\frac{\varepsilon}{2}}(\mathbb R^2)}\,
\|\nabla\Delta^{-1}\boldsymbol\pi^*\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}\leq \frac{c}{\varepsilon^6}. \end{split}\end{equation*}
 This
gives the desired estimate
\begin{equation*}
|I_5|\leq \frac{c}{\varepsilon^6}\int_{\mathbf R}M(t)^2\,dt\leq
\frac{c}{\varepsilon^6}\|\overline{\mathbf w}\|_{L^2(\mathbb
R;W^{1,2}(\mathbb R^2))}^2\leq \frac{c}{\varepsilon^6}.
\end{equation*}

\end{proof}


\paragraph{ \textbf{Step 3. Proof of Theorem \ref{olga100}}}
We are now in a position to complete the proof of the auxiliary
Theorem \ref{olga100}. We begin with the observation that
$$
0< R^{-1}\leq y_1\leq R\text{~~in the support of ~~}\rho.
$$
Here the constant $R$ depends only on the compact set $K$ in Theorem
\ref{olga100}. It follows from this, inequality \eqref{olga28} in
Proposition \ref{olga27}, and estimates \eqref{olga111} in Lemma
\ref{olga110} that
\begin{equation}\label{olga201}
\int_{\mathbb R^3}\delta^3\rho^3\ln(1+\delta \varrho)\, dydt =
\int_{\mathbb R^3}p_2\varphi\, dydt\leq c\sum_{i=1}^5 I_i\leq
\frac{c}{\varepsilon^8},
\end{equation}
where $c$ depends only on $K$. Let us prove that there is a constant
$c$, independent of $\varepsilon$ and $\delta$, such that
\begin{equation}\label{olga202}
\int_{\mathbb R^3}\delta^3\rho^3\, dydt\leq
c\Big(\,\delta^{3/8}+\frac{1}{\ln(1+\delta^{-1/6})}\,
\frac{1}{\varepsilon^6}\, \Big).
\end{equation}
To this end, choose two arbitrary numbers $m$ and $N$ satisfying the
following conditions
\begin{equation*}
m,N\geq 1, \quad \frac{N}{\delta}\geq m.
\end{equation*}
Introduce two sets $A_N\subset A_m\subset G\times (0,T)$ defined by
the equalities
\begin{eqnarray*}
A_m&=&\big\{\, (y,t):\,\,\rho(y,t)\geq m\,\big\},\\
A_N&=&\big\{\, (y,t):\,\,\delta\rho(y,t)\geq N\,\big\}.
\end{eqnarray*}
We have
\begin{eqnarray}\label{olga203}
\int_{\mathbb R^3}(\delta\rho)^3\,dydt&=&
\int_{(G\times(0,T))\setminus A_m}(\delta\rho)^3\,dydt\\\notag & + &
\int_{A_m\setminus A_N}(\delta\rho)^3\,dydt+
\int_{A_N}(\delta\rho)^3\,dydt.
\end{eqnarray}
It is easy to see that
\begin{equation}\label{olga204}
\int_{(G\times(0,T))\setminus A_m}(\delta\rho)^3\,dydt\leq \delta^3
m^3\, T\, \text{meas~} G\leq c\delta^3 m^3.
\end{equation}
Obviously we have
\begin{equation*}
(\delta\rho)^3\leq
\frac{1}{\ln(1+N)}\,(\delta\rho)^3\,\ln(1+\delta\rho)\text{~~in~~}
A_N.
\end{equation*}
From this and estimate \eqref{olga201} we conclude that
\begin{equation}\label{olga205}
\int_{A_N}(\delta\rho)^3\,dydt\leq
c\,\frac{1}{\ln(1+N)}\,\frac{1}{\varepsilon^8}.
\end{equation}
Next, the Chebyshev inequality and the energy estimate imply
\begin{equation*}
    \text{meas~} A_m \leq \frac{1}{m}\int_{\mathbb R^3}\rho(y,t)\,
    dydt\leq \frac{c}{m}.
\end{equation*}
Noting that $\delta\rho\leq N$ in $A_m\setminus A_N$ we obtain
\begin{equation}\label{olga206}
\int_{A_m\setminus A_N}(\delta\rho)^3\,dydt\leq c\,\frac{N^3}{m}.
\end{equation}
Estimates \eqref{olga204}-\eqref{olga206} along with equality
\eqref{olga203} imply
\begin{equation*}
\int_{\mathbb R^3}(\delta\rho)^3\,dydt\leq c\, \Big(\,\delta^3
m^3+\frac{N^3}{m}+\frac{1}{\ln(1+N)}\frac{1}{\varepsilon^8}\,\Big).
\end{equation*}
Substituting $m=\delta^{-5/6}$, $N=\delta^{-1/9}$ in this inequality
we obtain  desired estimate \eqref{olga202}. It remains to note that
for $\delta\leq \exp(-1/\varepsilon^9)$, inequality \eqref{olga202}
implies the estimate
\begin{eqnarray*}
\int_{\mathbb R^3}\delta^3\rho^3\, dydt&\leq&
c\Big(\,\delta^{1/2}+\frac{1}{\ln(1+\delta^{-1/9})}\,
\frac{1}{\varepsilon^8}\, \Big)\\ &\leq & c
e^{-\frac{1}{2\varepsilon^9}}+\frac{1}{
\ln(\,e^{\frac{1}{9\varepsilon^9}}\,)} \frac{c}{\varepsilon^6}\leq
c\varepsilon.
\end{eqnarray*}
This completes the proof of Theorem \ref{olga100}.


\subsection{Proof of Theorems \ref{olga13} and \ref{olga13cyl}.}
\paragraph{\textbf{Proof of Theorem \ref{olga13cyl}.}}
Introduce the sequence of compact sets $K_n\subset G\times (0,T)$,
$n\geq 1$ such that
\begin{equation*}
    K_n\subset K_{n+1}, \quad \bigcup_n K_n= G\times (0,T).
\end{equation*}
For every $n$, choose a function $\zeta_n\in C^\infty_0(\mathbb
R^3)$ with the properties
\begin{equation*}
    \text{~the support of ~}\zeta_n\Subset G\times (0,T), \quad 0\leq
    \zeta_n\leq 1, \quad \zeta_n=1\text{~~on ~~ }K_n.
\end{equation*}
Set
$$
\rho_{n,\varepsilon}\,:=\,\zeta_n\varrho_\varepsilon,\text{~~where~~}
\varrho_\varepsilon= \varrho_{\varepsilon,\kappa(\varepsilon)},
\quad \kappa(\varepsilon)=e^{-\frac{1}{\varepsilon^9}}.
$$
It is easy to see that the function
$$
\zeta_n(y,t)\varrho_{\varepsilon, \delta}(y,t)
$$
meets all requirements of Theorem \ref{olga100} with $\zeta$ and $K$
replaced by $\zeta_n$ and $K_n$. It follows from this and estimate
\eqref{olga101} in Theorem \ref{olga100} that
\begin{equation}\label{olga207}
\int_{\mathbb R^3}p_2(\rho_{n,\varepsilon}(y,t))\, dydt\leq
c_n\varepsilon\text{~~for all~~} \varepsilon\in (0,1/2].
\end{equation}
for all
$$
\varepsilon\in (0,1/2], \quad 0<\delta\leq \kappa(\varepsilon).
$$
Here the constant $c_n$ is independent of $\varepsilon$ and
$\delta$.

Now choose an arbitrary compact set $K\subset G\times (0,T)$. It is
clear that $K\subset K_n$ for some $n\geq 1$. Hence
$$
\varrho_\varepsilon=\rho_{n, \varepsilon}\text{~~on~~} K.
$$
Since the functions $p_1$ and $p_2$ are nonnegative, we have
\begin{equation*}
\int_{K}p_1(\varrho_{\varepsilon})\,
dydt+\int_{K}p_2(\varrho_{\varepsilon})\, dydt \leq\int_{\mathbb
R^3}p_1(\rho_{n,\varepsilon})\, dydt+\int_{\mathbb
R^3}p_2(\rho_{n,\varepsilon})\, dydt.
\end{equation*}
Therefore, it suffices to prove that
\begin{equation}\label{olga208}
\int_{\mathbb R^3}p_1(\rho_{n,\varepsilon})\, dydt+\int_{\mathbb
R^3}p_2(\rho_{n,\varepsilon})\, dydt\to 0\text{~~as~~ } \varepsilon
\to 0.
\end{equation}
By virtue of \eqref{olga207}, we have
\begin{equation}\label{olga209}
\int_{\mathbb R^3}p_1(\rho_{n,\varepsilon})\, dydt+\int_{\mathbb
R^3}p_2(\rho_{n,\varepsilon})\, dydt\leq \int_{\mathbb
R^3}p_1(\rho_{n,\varepsilon})\, dydt+ c_n\varepsilon.
\end{equation}
Consider the integral in the right hand side of this inequality in
more details. \emph{By abuse of notation, we will write $\varrho$,
$\zeta$, and  $\rho$ instead of $\varrho_\varepsilon$, $\zeta_n$,
and $\rho_{n,\varepsilon}$}.


Note that  the function $\varrho(y,t)$ meets all requirements of
Proposition \ref{rita12}. In particular, it admits the estimate
\begin{equation*}
    \text{~~ess}\sup\limits_{t\in \mathbb R}\int_{\mathbb
    R^2}\zeta(y,t)\,\, e_1(\varrho)\,dydt\leq C(K).
\end{equation*}
Recall that $\zeta$ is compactly supported in the bounded domain
$G\times(0,T)$. Since
$$
\rho=\zeta\varrho, \quad 0\leq \zeta\leq 1,
$$
we have
$$
\zeta e_1(\varrho)=\zeta\varepsilon
\varrho^{1+\varepsilon}\ln(1+\varepsilon\varrho)\geq \varepsilon
\rho^{1+\varepsilon}\ln(1+\varepsilon\rho)=e_1(\rho),
$$
which yields the estimate
\begin{equation}\label{olga20}
    \text{~~ess~~}\sup\limits_{t\in \mathbb R}\int_{\mathbb
    R^2} e_1(\rho)\,dydt\leq C(K).
\end{equation}
Introduce the function $\lambda(y,t)$ defined by the equalities
\begin{equation*}
\lambda=\varepsilon\,  \rho, \quad \rho=\lambda/\varepsilon.
\end{equation*}
Formulae \eqref{olga10} and \eqref{olga11} implies
\begin{equation*}
p_1(\rho)=\varepsilon\, e_1(\rho)+\varepsilon\,
\Big(\frac{\lambda}{\varepsilon}\Big)^{1+\varepsilon}\,\frac{\lambda}{1+\lambda}=
\varepsilon\,
e_1(\rho)+\varepsilon^{-\varepsilon}\,\frac{\lambda^{2+\varepsilon}}{1+\lambda}.
\end{equation*}
From this, estimate \eqref{olga20}, and the inequality
$\varepsilon^{-\varepsilon}\leq c$ we conclude that the estimate
\begin{equation}\label{olga21}
    \int_{\mathbb R^2} p_1(\rho(t))\, dy\leq c
    \varepsilon+c\int_{
    G}\frac{\lambda(t)^{2+\varepsilon}}{1+\lambda(t)}\, dy
\end{equation}
holds for a.e. $t\in\mathbb R$. Recall that $\rho$ and $\lambda$ are
compactly supported in $G\times (0,T)$. Let us evaluate the integral
in the right hand side.  We split it into three parts
\begin{equation}\label{olga22}
    \int_{\mathbb
    G}\frac{\lambda(y,t)^{2+\varepsilon}}{1+\lambda(y,t)}\, dy =
    J_1(t)+J_2(t)+ J_3(t),
\end{equation}
where
\begin{eqnarray}\notag
J_1(t)&=&\int_{G^-(t)}\frac{\lambda(y,t)^{2+\varepsilon}}{1+\lambda(y,t)}dy,
\\\notag J_2(t)&=&\int_{G^+(t)}\frac{\lambda(y,t)^{2+\varepsilon}}{1+\lambda(y,t)}dy\\\notag
J_3(t)&=&\int_{G\setminus(G^-(t)\cup
G^+(t))}\frac{\lambda(y,t)^{2+\varepsilon}}{1+\lambda(y,t)}dy,
\end{eqnarray}
\begin{equation*}
  G^-(t)=\big\{\, y\in G:\,\, \lambda (y,t)\leq m^{-1}\,\big\}, \quad
 G^+(t)=\big\{\, y\in G:\,\, \lambda (y,t)\geq m\,\big\},
\end{equation*}
$m>2$ is an arbitrary number. It is easily seen that
\begin{equation}\label{olga23}
J_1(t)\leq c m^{-2}\int_G\,dy= c\, m^{-2}.
\end{equation}
For every $\lambda\geq m$, we have
$$
\frac{\lambda^{2+\varepsilon}}{1+\lambda}\leq
\lambda^{1+\varepsilon}\ln(1+\lambda)\frac{1}{\ln(1+m)}.
$$
Next, we have
$$
\lambda^{1+\varepsilon}\ln(1+\lambda)=\varepsilon^\varepsilon
\varepsilon\rho^{1+\varepsilon}\ln(1+\varepsilon\rho)=\varepsilon^\varepsilon
e_1(\rho)\leq c\,e_1(\rho).
$$
From this we conclude that
\begin{equation}\label{olga24}
    J_2(t)=\int_{G^+(t)}\frac{\lambda(y,t)^{2+\varepsilon}}{1+\lambda(y,t)}dy\,\leq\,
    \frac{c}{\ln(1+m)}\int_G e_1 (\rho(y,t))\,dy\,\leq\,
    \frac{c}{\ln(1+m)}.
\end{equation}
It remains to estimate $J_3(t)$. To this end, note that the estimate
$$
\int_G \rho(y,t)\, dy\,\leq \,c
$$
and the Chebyshev inequality imply
\begin{eqnarray}\label{olga25}
J_3(t)&\leq &\int_{m^{-1}\leq \lambda\leq
m}\frac{\lambda(y,t)^{2+\varepsilon}}{1+\lambda(y,t)}dy \leq
m^{1+\varepsilon}\int_{m^{-1}\leq \lambda}dy\\\notag
&=&m^{1+\varepsilon}\int_{(\varepsilon m)^{-1}\leq \rho}\,dy\leq c\,
m^{1+\varepsilon}(\varepsilon m)\,\leq \,c\,\varepsilon\,
m^{2+\varepsilon}.
\end{eqnarray}
Substituting \eqref{olga23}-\eqref{olga24} in \eqref{olga22} leads
to the inequality
\begin{equation*}
    \int_{
    G}\frac{\lambda(y,t)^{2+\varepsilon}}{1+\lambda(y,t)}\, dy\leq
c\, m^{-2}+ \frac{c}{\ln(1+m)}+c\,\varepsilon\, m^{2+\varepsilon}.
\end{equation*}
Note that the constant $c$ does not depend on $t$, $m$, $\delta$,
and $\varepsilon$. Since $\lambda(y,t)$ is compactly supported in
$G\times (0,T)$, we conclude that
\begin{equation*}
    \int_{\mathbb
    R^2}\frac{\lambda(y,t)^{2+\varepsilon}}{1+\lambda(y,t)}\, dy\leq
c\, m^{-2}+ \frac{c}{\ln(1+m)}+c\,\varepsilon\, m^{2+\varepsilon}.
\end{equation*}
Since $p_1(\rho)$ is compactly supported in $G\times (0,T)$, itt
follows from this and \eqref{olga21} that
\begin{equation*}
    \int_{\mathbb R^2} p_1(\rho(t))\, dy\leq c
    \varepsilon+c\, m^{-2}+ \frac{c}{\ln(1+m)}+c\,\varepsilon\,
    m^{2+\varepsilon}.
\end{equation*}
Combining this inequality with \eqref{olga209} and recalling that
$\rho=\rho_{n,\varepsilon}$ we arrive at the estimate
\begin{equation*}
\int_{\mathbb R^3}p_1(\rho_{n,\varepsilon})\, dydt+\int_{\mathbb
R^3}p_2(\rho_{n,\varepsilon})\, dydt\leq c
    \varepsilon+c\, m^{-2}+ \frac{c}{\ln(1+m)}+c\,\varepsilon\,
    m^{2+\varepsilon}.
\end{equation*}
Letting $\varepsilon\to 0$ we obtain
\begin{equation*}
\limsup\limits_{\varepsilon\to 0}\, \int_{\mathbb R^3}(
p_1(\rho_{n,\varepsilon})+p_2(\rho_{n,\varepsilon})\, dydt\leq c\,
m^{-2}+ \frac{c}{\ln(1+m)}.
\end{equation*}
Letting $m\to \infty$ we arrive at desired relation \eqref{olga208}.
This completes the proof of Theorem \ref{olga13cyl}.
\paragraph{\textbf{Proof of Theorem \ref{olga13}.}}
Theorem \ref{olga13} is a straightforward consequence of Theorem
\ref{olga13cyl}.

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\section{ Proof of Theorem \ref{thm1}}\label{thm1proof}
\renewcommand{\theequation}{6.\arabic{equation}}
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\subsection{Part 1. Weak limits}\label{part1} We obtain a solution to isothermal
compressible Navier- Stokes equations \eqref{isothermal} as a weak
limit of solutions $\varrho_{\varepsilon, \delta}=\varrho$, $\mathbf
u_{\varepsilon, \delta}=\mathbf u$ to regularized equations
\eqref{stokes108}.  Note that these solutions are defined by
Proposition \ref{stokes107}.  Set
\begin{equation}\label{lita1}
    \varrho_\varepsilon=\varrho_{\varepsilon,\kappa(\varepsilon)},
    \quad \mathbf u_\varepsilon=\mathbf
    u_{\varepsilon,\kappa(\varepsilon)}, \quad \kappa(\varepsilon)=
    e^{-\frac{1}{\varepsilon ^9}}.
\end{equation}
In this subsection, we study the weak compactness properties of
sequences  $\varrho_\varepsilon$ and $\mathbf u_\varepsilon$. It
follows from estimates \eqref{stokes113}-\eqref{stokes114cor} in
Proposition \ref{stokes107} that
\begin{equation}\label{lita2}
 \|\varrho_\varepsilon|\mathbf u_\varepsilon|^2\|_{L^\infty(0,T;L^1(\Omega))}
 +\|\mathbf u_\varepsilon\|_{L^2(0,T; W^{1,2}(\Omega))}\leq c.
\end{equation}

\begin{eqnarray}\label{lita3}
\|\varrho_\varepsilon\ln(2+\varrho_\varepsilon)\|_{L^\infty(0,T;
L^1(\Omega))}&+&\|e_1(\varrho_\varepsilon)\|_{L^\infty(0,T;
L^1(\Omega))}\\\notag+ \|e_2(\varrho_\varepsilon)\|_{L^\infty(0,T;
L^1(\Omega))}&\leq &c.
\end{eqnarray}

\begin{equation}\label{lita4}
\|p(\varrho_\varepsilon)\|_{L^\infty(0,T;
L^1(\Omega))}+\|p_1(\varrho_\varepsilon)\|_{L^\infty(0,T;
L^1(\Omega))}+ \|p_2(\varrho_\varepsilon)\|_{L^\infty(0,T;
L^1(\Omega))}\leq c.
\end{equation}
Here the constant $c$ depends only on $\Omega$, $T$, and the
constant $c_E$ in condition \eqref{stokes108a} of Proposition
\ref{stokes107}. Our further considerations are based on the
following lemma.

\begin{lemma}\label{lita5} The estimate
\begin{equation}\label{lita6}
    \|\varrho_\varepsilon |\mathbf u _\varepsilon|
    \ln (2+\varrho_\varepsilon |\mathbf u
    _\varepsilon|)^{\alpha}\|_{L^\infty(0,T; L^1(\Omega))}\leq c
\end{equation}
holds for every $\alpha<1/2$. Here the constant $c$ depends only on
$\alpha$ and the constants in estimates \eqref{lita2}-\eqref{lita4}.
\end{lemma}
\begin{proof}
The proof is given in Appendix \ref{persik}.


\end{proof}

Estimates \eqref{lita3} and \eqref{lita6} imply that the sequences
$\varrho_\varepsilon$ and $\varrho_\varepsilon\mathbf u_\varepsilon$
are equi-integrable. It follows from this and the Dunford-Pettis
Theorem (see \cite{fonseca}, thm. 2.54) that there are subsequences,
still denoted by $\varrho_\varepsilon$, $\varrho_\varepsilon\mathbf
u_\varepsilon$, and functions $\varrho$, $\overline{\varrho\mathbf
u}$ such that
\begin{equation}\label{lita7}
\varrho_\varepsilon\to \varrho, \quad \varrho_\varepsilon\mathbf
u_\varepsilon\to \overline{\varrho\mathbf u}\text{~~weakly in~~}
L^1(Q_T)\text{~~as~~} \varepsilon\to 0.
\end{equation}
In particular, for every measurable set $\mathcal T\subset (0,T)$,
we have
\begin{eqnarray*}
\int_{\Omega\times \mathcal T}\varrho\,
dxdt&=&\lim\limits_{\varepsilon\to 0}\int_{\Omega\times \mathcal
T}\varrho_\varepsilon\, dxdt\\ \leq \text{~meas~}\mathcal T
\limsup\limits_{\varepsilon\to
0}\|\varrho_\varepsilon\|_{L^\infty(0,T; L^1(\Omega))}&\leq&
c\text{~meas~}\mathcal T,
\end{eqnarray*}
\begin{eqnarray*}
\Big|\int_{\Omega\times \mathcal T}\overline{\varrho\mathbf u}\,
dxdt\Big|&=&\lim\limits_{\varepsilon\to 0}\Big|\int_{\Omega\times
\mathcal T}\varrho_\varepsilon\mathbf u_\varepsilon\, dxdt\Big|\\
\leq \text{~meas~}\mathcal T \limsup\limits_{\varepsilon\to
0}\|\varrho_\varepsilon\mathbf u_\varepsilon\|_{L^\infty(0,T;
L^1(\Omega))}&\leq& c\text{~meas~}\mathcal T.
\end{eqnarray*}
It follows from this that $\varrho, \overline{\varrho\mathbf u}\in
L^\infty(0,T; L^1(\Omega))$.  By virtue of estimate \eqref{lita2},
there is a subsequence, still denoted by $\mathbf u_\varepsilon$,
and a vector field $\mathbf u\in L^2(0,T; W^{1,2}_0(\Omega))$ such
that
\begin{equation}\label{lita8}
\mathbf u_\varepsilon\to \mathbf u\text{~~weakly in~~} L^2(0,T;
W^{1,2}(\Omega))\text{~~as~~} \varepsilon\to 0.
\end{equation}
Let us consider the sequence $K_n$, $n\geq 1$, of compact subsets of
$(\Omega\setminus \mathcal I)\times (0,T)$ such that
\begin{equation*}
    K_1\subset\dots \subset K_n\subset K_{n+1}\subset\dots\subset
(\Omega\setminus \mathcal I)\times (0,T) , \quad
    \bigcup_{n\geq 1} K_n=(\Omega\setminus \mathcal I)\times (0,T).
\end{equation*}
By virtue of Proposition \ref{inna1}, the sequences
$\varrho_\varepsilon$ and $\mathbf u_\varepsilon$ satisfy conditions
(\textbf{C.1})-(\textbf{C.3}) of Theorem \ref{lena11}. It follows
from this theorem that for every $n\geq 1$ and $\alpha\in (0,1/4)$,
\begin{equation*}
    \int_{K_n}\varrho_\varepsilon |\mathbf u_\varepsilon|^2\,
    \ln(2+\varrho_\varepsilon |\mathbf u_\varepsilon|^2)\, dxdt\leq
    c_n,
\end{equation*}
where $c_n$ is independent of $\varepsilon$. It follows that the
sequence $\varrho_\varepsilon |\mathbf u_\varepsilon|^2$ is
equi-integrable on every compact set $K\subset(\Omega\setminus
\mathcal I)\times (0,T)$. Using the Dunford-Pettis Theorem and  a
diagonal procedure  we may extract  a subsequuence of
$\varrho_\varepsilon \mathbf u_\varepsilon\otimes \mathbf
u_\varepsilon$ (not relabeled)   such that
\begin{equation}\label{lita30}
\varrho_\varepsilon \mathbf u_\varepsilon\otimes \mathbf
u_\varepsilon\to\overline{\varrho \mathbf u\otimes \mathbf u} \text
{~~ weakly in~~} L^1(K)\text{~~as~~}\varepsilon\to 0
\end{equation}
for every compact set $K\subset (\Omega\setminus \mathcal I)\times
(0,T)$. Here $\overline{\varrho\mathbf u\otimes \mathbf u}\in
L^1(Q_T)$  is a matrix  valued function.



Let us consider the weak compactness properties of the sequences of
the pressure functions $p(\varrho_\varepsilon)$ and kinetic energy
tensors $\varrho_\varepsilon\mathbf u_\varepsilon\otimes \mathbf
u_\varepsilon$. To this end, recall briefly the basic facts
concerning $L^p$ spaces on Banach spaces.

Let  $Y$ be  a Banach space and let $Y'$ be a Banach space dual to
$Y$. A function $\nu:(0,T)\to Y'$ is said to be  weakly star
measurable if for any $L\in Y$, the scalar function
$$
t\in (0,T)\mapsto \langle\, \nu(t)\,,\, L\,\rangle \text{ ~~is
measurable~}.
$$
Further the notation $L^r_w(0,T; Y')$, $1\leq r <\infty$, stands for
the Banach space of weakly star measurable functions with the finite
norm
$$
\|\nu\|_{L^r_w(0,T;Y')}=\Big(\int_0^T
\|\nu(t)\|_{Y'}^r\,dt\Big)^{1/r}.
$$
If a Banach space  $Y$ is separable, then every weakly measurable
mapping $\varphi:(0,T)\to Y$ is strongly measurable. In this case,
$L^r(0,T;Y)$ is the standard space of functions integrable in the
sense of Bochner. This leads to the following lemma  (see
\cite{fonseca}, thm. 2.112)

\begin{lemma}\label{lita9}
Assume that $Y$ is separable. If $L\in \big(\, L^r(0,T; Y)\big)'$,
$1<r<\infty$, then there exists a unique $\nu\in
L^{\frac{r}{r-1}}_w(0,T; Y')$ such that
\begin{equation}\label{lita10}
    \langle\, L\,,\, \varphi\,\rangle= \int_0^T \langle\,
    \nu(t),\,\varphi(t)\,\rangle\, dt
\end{equation}
for every  $\varphi\in L^r(0,T; Y)$. Moreover,
$$
\|L\|_{\big(\, L^r(0,T; Y)\big)'}=\|\nu\|_{L^{\frac{r}{r-1}}_w(0,T;
Y')}.
$$
In particular, we have
\begin{equation*}
    L^r_w(0,T; Y')\,=\, \big(\, L^{\frac{r}{r-1}}(0,T;
    C_0(\Omega))\, \big)' \text{~~for~~}1\leq r<\infty.
\end{equation*}

\end{lemma}
Denote by $\mathbb M(\Omega)$ the Banach space of all Radon measures
in $\Omega$. Note that $\mathbb M(\Omega) =C_0(\Omega)'$ is dual to
the space $C_0(\Omega)$ of continuous functions
$u:\overline{\Omega}\to \mathbb R$ vanishing at $\partial\Omega$.
 By the Riesz representation theorem (see \cite{fonseca}, thm. 1.195), for every $\nu\in\mathbb M(\Omega)$
there is a unique signed Borel measure $\nu^*$ defined on the
algebra of Borel subsets of $\Omega$ such that
$$
\langle \,\nu\,,\, \varphi\,\rangle\,=\, \int_\Omega \varphi\,
d\nu^*.
$$
for all  $\varphi\in C_0$. The norm of $\nu$ admits the
representation
$$
\|\nu\|_{\mathbb M(\Omega)}= \|\nu^*\|(\Omega),
$$
where $\|\nu^*\|(\Omega)$ is the total variation of a signed measure
$\nu^*$  over $\Omega$. Further we will identify $\nu$ and $\nu^*$.

Every function $g\in L^1(\Omega)$ determines a unique Radon measure
\begin{equation}\label{lita11}
d\nu =g\, dx, \quad \langle \,\nu\,,\, \varphi\,\rangle\,=\,
\int_\Omega \varphi\,g dx\text{~~for all~~} \varphi\in C_0(\Omega).
\end{equation}
It is clear that $\|\nu\|_{\mathbb M(\Omega)}=\|g\|_{L^1(\Omega)}$.
Relation \eqref{lita11} defines the isometry
$$
L^r(0,T; L^1(\Omega))\hookrightarrow L^r_w(0,T; \mathbb M(\Omega)).
$$
In particular, $L^1$-functions $p(\varrho_\varepsilon(t))$ and
$\varrho_\varepsilon(t) u_{\varepsilon,i}(t)u_{\varepsilon,j}(t)$
determine the Radon measures $\sigma_\varepsilon(t)$ and
$\nu_{\varepsilon, ij}$ given by the equalities
\begin{equation*}
\int_\Omega \varphi\, d\sigma_\varepsilon(t)=\int_\Omega \varphi\,
p(\varrho_\varepsilon)(t)\,d x, \quad \int_\Omega \varphi\,
d\nu_{\varepsilon, ij}=\int_\Omega \varphi\,\varrho_\varepsilon
u_{\varepsilon,i}u_{\varepsilon,j}\,dx
\end{equation*}
for all $\varphi\in C_0(\Omega)$. It follows from estimates
\eqref{lita2} and \eqref{lita3} that the sequences
$\sigma_\varepsilon(t)$ and $\nu_{\varepsilon, ij}$ are bounded in
every space $L^r_w(0,T; \mathbb M(\Omega))$, $1\leq r\leq \infty$.

Now fix an arbitrary $r\in (1,\infty)$ and note that the space
$L^{\frac{r}{r-1}}(0,T; C_0(\Omega))$ is separable. From this and
Lemma \ref{lita9} we conclude that
\begin{equation*}
L^r_w(0,T; \mathbb M(\Omega))=\big(\,L^{\frac{r}{r-1}}(0,T;
C_0(\Omega))\,\big)'
\end{equation*}
By virtue of the Alaoglu theorem, there are subsequences of the
sequences $\sigma_\varepsilon(t)$, $\nu_{\varepsilon, ij}$, still
denoted by $\sigma_\varepsilon(t)$, $\nu_{\varepsilon, ij}$, and the
parametrized measures $\sigma, \nu_{ij}\in L^r_w(0,T;\mathbb
M(\Omega))$ such that
\begin{equation*}
\sigma_\varepsilon\to \sigma, \quad \nu_{\varepsilon, ij}\to\nu_{
ij}\text{~~weakly star in~~} L^r_w(0,T; \mathbb
M(\Omega))\text{~~as~~}\varepsilon\to 0.
\end{equation*}
This means that
\begin{eqnarray}\label{lita20}
\int_{Q_T} p(\varrho_\varepsilon) \varphi\, dxdt&\to&
\int_0^T\Big(\,\int_\Omega \varphi\, d\sigma(t)\,\Big)dt,\\\notag
\int_{Q_T}\varrho_\varepsilon u_{\varepsilon,i}
u_{\varepsilon,j}\,\varphi\, dxdt&\to& \int_0^T\Big(\,\int_\Omega
\varphi\, d\nu_{ij}(t)\,\Big)dt
\end{eqnarray}
for all $\varphi\in L^{\frac{r}{r-1}}(0,T; C_0(\Omega))$.


Recall that $\varrho_\varepsilon$ and $\mathbf u_\varepsilon$ are
weak solutions to boundary value problem \eqref{stokes1}. They
satisfy integral identities \eqref{stokes10} and \eqref{stokes11}
with the pressure function
\begin{eqnarray*}
p(\varrho)&=&\varrho + p_1(\varrho)+ p_2(\varrho), \\
 p_1(\varrho)&=&\varepsilon
\varrho^{1+\varepsilon}\big(\varepsilon\ln(1+\varepsilon\varrho)+
\frac{\varepsilon\varrho}{1+\varepsilon\varrho}\big), \,\,\,\,
p_2(\varrho)=\kappa(\varepsilon)^3\varrho^3. \end{eqnarray*}

Substituting $\varrho_\varepsilon$, $\mathbf u_\varepsilon$ into
\eqref{stokes10} and \eqref{stokes11}, letting $\varepsilon\to 0$,
and using relations \eqref{lita7}, \eqref{lita8}, and \eqref{lita20}
we arrive at the integral identities
\begin{equation}\label{lita21}\begin{split}
\int_{Q_T}\overline{\varrho\mathbf{u}}\cdot
\partial_t\boldsymbol{\xi}\,dxdt+\int_0^T \Big(\,\int_\Omega
\nabla\boldsymbol{\xi}: d\boldsymbol\nu(t)\,\Big)\, dt \\+\int_0^T
\Big(\,\int_\Omega \text{\rm div~} \boldsymbol{\xi}\,
d\sigma(t)\,\Big)\,dt -\int_{Q_T}\mathbb
S(\mathbf{u}):\nabla\boldsymbol{\xi}\,dxdt
\\+\int_{Q_T} \varrho\mathbf f\cdot\boldsymbol{\xi}\, dxdt+
\int_{\Omega }\varrho_0 \mathbf u_0(x)\cdot
\boldsymbol{\xi}(x,0)\,dx=0,
\end{split}\end{equation}
\begin{equation}\label{lita22}
\int_{Q_T}\big (\varrho
\partial_t\psi+\overline{\varrho\mathbf{u}}\cdot\nabla\psi
\big)\,dxdt+\int_{\Omega } \varrho_0(x)\psi(x,0)\, dx=0.
\end{equation}
Here $\boldsymbol\nu(t)=\big(\nu_{ij}(t)\big)_{1\leq i,j\leq 3}$ is
a symmetric matrix-valued Radon measure in $\Omega$.

Identities \eqref{lita21} and \eqref{lita22} hold for all vector
fields $\boldsymbol{\xi}\in C^\infty(Q_T)$ vanishing in a
neighborhood of  $\partial\Omega\times[0,T]$ and
$\Omega\times\{t=T\}$, and for all functions $\psi\in C^\infty(Q_T)$
vanishing in a neighborhood of ~ $\Omega\times\{t=T\}$.



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\subsection{Part 2. Weak limits of the  momentum and kinetic energy
tensor}\label{part2}

 In this subsection we study in details the
vector field $\overline{\varrho\mathbf u}$ and the measures
$\sigma$, $\boldsymbol \nu$.  The following lemma is the first
result in this direction.

\begin{proposition}\label{lita23} Under the assumptions of Theorem
\ref{thm1}, we have
\begin{equation}\label{lita24}
\overline{\varrho\mathbf u}=\varrho\mathbf u\text{~~in~~}Q_T,
\end{equation}
\begin{equation}\label{lita25}
\overline{\varrho\mathbf u\otimes\mathbf u}=\varrho\mathbf
u\otimes\mathbf u \text{~~in~~}Q_T.
\end{equation}



\end{proposition}


In order to avoid repetition, we prove the relation \eqref{lita24}.
The proof of \eqref{lita25} is similar. The proof of \eqref{lita24}
falls into three steps.

\paragraph{\textbf{Step1}.} By virtue of the mass balance equation \eqref{stokes108}, the
density function $\varrho_\varepsilon$ satisfies the integral
identity
$$
\int_{Q_T}\varrho_\varepsilon\,\partial_t\psi\, dxdt=-\int_{Q_T}
\varrho_\varepsilon \mathbf u_\varepsilon\cdot \nabla \psi\,dxdt
$$
for every $\psi\in C^\infty_0(Q_T)$. It follows from the energy
estimates \eqref{lita2}, \eqref{lita3}, and the embedding $
W^{3,2}_0(\Omega) \hookrightarrow C^1(\Omega)$ that
$$
\Big|\int_{Q_T} \varrho_\varepsilon \mathbf u_\varepsilon\cdot
\nabla \psi\,dxdt\Big|\leq c \int_0^T
\|\psi(t)\|_{W^{3,2}_0(\Omega)}\, dt.
$$
This means that
\begin{equation*}
    \|\partial_t\varrho_\varepsilon\|_{L^\infty(0,T;
    W^{-3,2}(\Omega))}\leq c \text{~~and~~}
    \|\varrho_\varepsilon\|_{Lip(0,T; W^{-3,2}(\Omega))}\leq c.
\end{equation*}
On the other hand, we have
$$
\|\varrho_\varepsilon(t)\|_{W^{-2,2}(\Omega)}\leq c.
$$
Note that the embedding $W^{-2,2}(\Omega)\hookrightarrow W^{-3,
2}(\Omega)$ is compact. Hence the sequence
$\varrho_\varepsilon:[0,T]\to W^{-3,2}(\Omega)$ meets all
requirements of the Arzel\`{a}-Ascoli Theorem (see \cite{deudonne},
thm. 7.5.7). Therefore,  after passing to a subsequence  we may
assume that
\begin{equation}\label{lita26}
    \varrho_\varepsilon\to \varrho \text{~~in~~} C(0,T; W^{-3,2}(\Omega)).
\end{equation}
\paragraph{\textbf{Step 2}.} Choose an arbitrary integer $N\geq 3$. Introduce the
truncated vector field
\begin{eqnarray*}
\mathbf u_{N,\varepsilon}(x,t)&=&\mathbf u_\varepsilon
(x,t)\text{~~if~~} |\mathbf u_\varepsilon (x,t)|\leq N,\\
\mathbf u_{N,\varepsilon}(x,t)&=&\frac{N}{|\mathbf
u_\varepsilon(x,t)|}\, \mathbf u_\varepsilon (x,t)\text{~~if~~}
|\mathbf u_\varepsilon (x,t)|\geq N.
\end{eqnarray*}
It is clear that $\mathbf u_{N,\varepsilon}\in L^2(0,T;
W^{1,2}_0(\Omega))$  and
\begin{equation}\label{lita27}
\|\mathbf u_{N, \varepsilon}\|_{L^2(0,T;W^{1,2}(\Omega))}\leq c,
\quad |\mathbf u_{N, \varepsilon}|\leq N.
\end{equation}
Using the diagonal procedure we may extract a subsequence (nor
relabeled) such that
\begin{equation}\label{lita28}
\mathbf u_{N, \varepsilon}\to \overline{\mathbf u}_N\text{~~weakly
in ~~} L^2(0,T; W^{1,2}(\Omega))\text{~~and star weakly in ~~}
L^\infty(Q_T).
\end{equation}
The limiting vector field $\overline{\mathbf u}_N$ admits estimates
\eqref{lita28}. The following lemma is the main result of this
paragraph
\begin{lemma}\label{lita29a}
\begin{equation}\label{lita29}
\int_{Q_T}\varphi\, \varrho_\varepsilon \,\mathbf u_{N,
\varepsilon}\, dxdt\to \int_{Q_T}\varphi\,
\varrho\,\overline{\mathbf u}_N\, dxdt \text{~~as~~} \varepsilon\to
0\text{~~for all ~~}\varphi\in C^\infty_0(Q_T).
\end{equation}
\end{lemma}


\begin{proof} Introduce the mollifier
\begin{equation}
    \mathbf T_\omega u(x)=\frac{1}{\omega^3}\int_{\mathbb
    R^3}\vartheta\big(\, \frac{x-y}{\omega}\,\big)\, u(y)\, dy
\end{equation}
where the kernel
$$\theta\in C^\infty_0(\mathbb R^3), \quad
\text{support~}\theta\subset\{|x|\leq 1\}, \quad \int_{\mathbb
R^3}\theta(x)\, dx=1.
$$
Note that
\begin{equation}\label{lita41}
\int_{Q_T}\varphi\, \big(\,\varrho_\varepsilon \,\mathbf u_{N,
\varepsilon}- \varrho\,\overline{\mathbf u}_N\,\big)\, dxdt=
\Pi_1+\Pi_2+\Pi_3,
\end{equation}
where
\begin{eqnarray*}
\Pi_1&=& \int_{Q_T}\varphi \varrho_\varepsilon\,\big(\mathbf
u_{N,\varepsilon}- \mathbf T_\omega\mathbf
u_{N,\varepsilon}\,\big)\, dxdt,\\
\Pi_2&=& \int_{Q_T}\varphi \varrho\,\big(\mathbf T_\omega
\overline{\mathbf u}_{N}- \overline{\mathbf
u}_{N}\,\big)\, dxdt,\\
\Pi_3&=&\int_{Q_T}\varphi\big( \varrho_\varepsilon\mathbf
T_\omega\mathbf u_{N,\varepsilon}-\varrho \mathbf
T_\omega\overline{\mathbf u}_{N}\,\big)\, dxdt.
\end{eqnarray*}
Note that the embedding $W^{1,2}(\Omega)\hookrightarrow L^2(\Omega)$
is compact. It follows that
\begin{eqnarray}\label{lita42}
\|\mathbf u_{N,\varepsilon}(t)- \mathbf T_\omega\mathbf
u_{N,\varepsilon}(t)\|_{L^2(\Omega)}&\leq& \delta(\omega)\,
\|\mathbf u_{N,\varepsilon}(t)\|_{W^{1,2}(\Omega)},\\\notag
\|\overline{\mathbf u}_{N}(t)- \mathbf T_\omega\overline{\mathbf
u}_{N}(t)\|_{L^2(\Omega)}&\leq& \delta(\omega)\, \|\overline{\mathbf
u}_{N}(t)\|_{W^{1,2}(\Omega)},
\end{eqnarray}
where $\delta(\omega)\sim c\,\omega^{1/2}\to 0$ as $\omega \to 0$.
We also have
\begin{eqnarray}\label{lita43}
\|\mathbf u_{N,\varepsilon}(t)- \mathbf T_\omega\mathbf
u_{N,\varepsilon}(t)\|_{L^\infty(\Omega)}&\leq& 2N ,\\\notag
\|\overline{\mathbf u}_{N}(t)- \mathbf T_\omega\overline{\mathbf
u}_{N}(t)\|_{L^\infty(\Omega)}&\leq& 2N.
\end{eqnarray}
Fix an arbitrary $m>3$ and set
$$
A_m=\{(x,t)\in Q_T:\,\, \rho_\varepsilon >m\}\subset Q_T.
$$
We have
\begin{eqnarray}\label{lita44}
   | \Pi_1|&\leq& \Big|\int_{Q_T\setminus A_m}\varphi \varrho_\varepsilon\,\big(\mathbf
u_{N,\varepsilon}- \mathbf T_\omega\mathbf
u_{N,\varepsilon}\,\big)\, dxdt\Big|\\\notag &+&\Big|\int_{
A_m}\varphi \varrho_\varepsilon\,\big(\mathbf u_{N,\varepsilon}-
\mathbf T_\omega\mathbf u_{N,\varepsilon}\,\big)\, dxdt\Big|.
\end{eqnarray}
By virtue of \eqref{lita42}, the first term in the right hand side
admits the estimate
\begin{eqnarray*}
\Big|\int_{Q_T\setminus A_m}\varphi
\varrho_\varepsilon\,\big(\mathbf u_{N,\varepsilon}- \mathbf
T_\omega\mathbf u_{N,\varepsilon}\,\big)\, dxdt\Big|&\leq&
cm\int_{Q_T} |\mathbf u_{N,\varepsilon}- \mathbf T_\omega\mathbf
u_{N,\varepsilon}|\, dxdt\\&\leq& cm\int_0^T \|\mathbf
u_{N,\varepsilon}(t)- \mathbf T_\omega\mathbf
u_{N,\varepsilon}(t)\|_{L^2(\Omega)}\,dt\\&\leq&
cm\delta(\omega)\int_0^T\|\mathbf
u_{N,\varepsilon}(t)\|_{W^{1,2}(\Omega)}\,dt\leq cm\delta(\omega).
\end{eqnarray*}
On the other hand, energy estimate \eqref{lita3} and estimate
\eqref{lita43} imply the inequality
\begin{equation*}
\Big|\int_{ A_m}\varphi \varrho_\varepsilon\,\big(\mathbf
u_{N,\varepsilon}- \mathbf T_\omega\mathbf
u_{N,\varepsilon}\,\big)\, dxdt\Big|\leq \frac{cN}{\ln m}.
\end{equation*}
Substituting the obtained results  into  \eqref{lita44} we arrive at
the estimate
\begin{equation*}
|\Pi_1|\leq c \,\big(m\delta(\omega)+\frac{N}{\ln m}\,\big).
\end{equation*}
Now set $m=1/(\delta|\ln\delta|)$ and choose $\omega$ so small that
$\delta(\omega)<1$. We thus get
$$
m\delta(\omega)+\frac{N}{\ln
m}=\frac{1}{|\ln\delta(\omega)|}+\frac{N}{|\ln(\delta(\omega)|\ln\delta(\omega)|)|}\leq
\frac{c N}{|\ln(\delta(\omega)|\ln\delta(\omega)|)|},
$$
which gives
\begin{equation}\label{lita45}
    |\Pi_1|\leq
\frac{c N}{|\ln(\delta(\omega)|\ln\delta(\omega)|)|}.
\end{equation}
Repeating these arguments and using estimates \eqref{lita42} and
\eqref{lita43} for $\overline{\mathbf u}_N$ we obtain the similar
estimate for $\Pi_2$,
\begin{equation}\label{lita46}
    |\Pi_2|\leq
\frac{c N}{|\ln(\delta(\omega)|\ln\delta(\omega)|)|}.
\end{equation}
Note that $c$ is independent on $\omega$. Let us consider the
quantity $\Pi_3$. We have
\begin{equation}\label{lita47}
    \Pi_3=\int_{Q_T}(\varrho_\varepsilon-\varrho)\big(\varphi\mathbf
T_\omega\mathbf u_{N,\varepsilon}\big)\, dxdt+
\int_{Q_T}\varrho\varphi \mathbf T_\omega\big( \mathbf
u_{N,\varepsilon}-\overline{\mathbf u}_{N}\,\big)\, dxdt.
\end{equation}
Recall that the function $\mathbf u_{N,\varepsilon}(t)\in
W^{1,2}_0(\Omega)$ equals zero outside of $\Omega$. It follows from
the properties of the mollifiers that
$$
\|\mathbf T_\omega \mathbf
u_{N,\varepsilon}(t)\|_{W^{3,2}(\Omega)}\leq c(\omega)\|\mathbf
u_{N,\varepsilon}\|_{L^2(\Omega)}\leq c(\omega)\|\mathbf
u_\varepsilon(t)\|_{L^2(\Omega)}.
$$
Since $\varphi\in C^\infty(Q_T)$ is compactly supported in the
cylinder $Q_T=\Omega\times (0,T)$, the vector field
$\varphi(t)\mathbf u_{N,\varepsilon}(t)\in W^{3,2}_0(\Omega)$ admits
the estimate
$$
\|\varphi(t)\mathbf T_\omega \mathbf
u_{N,\varepsilon}(t)\|_{W^{3,2}(\Omega)}\leq c(\omega)\|\mathbf
u_\varepsilon(t)\|_{L^2(\Omega)}.
$$
Here $c$ depends on $\omega$, $ \varphi$ and is independent on $t$.
We thus get
\begin{eqnarray*}
   \Big| \int_{Q_T}(\varrho_\varepsilon-\varrho)\big(\varphi\mathbf
T_\omega\mathbf u_{N,\varepsilon}\big)\, dxdt\Big|&\leq&
\|\varrho_\varepsilon-\varrho\|_{C(0,T;
W^{-3,2}(\Omega))}\,\int_0^T\|\varphi\mathbf T_\omega\mathbf
u_{N,\varepsilon}(t)\|_{W^{3,2}(\Omega)}\,dt\\\notag &\leq &
c(\omega)\|\varrho_\varepsilon-\varrho\|_{C(0,T;
W^{-3,2}(\Omega))}\,\int_0^T\|\mathbf
u_{\varepsilon}(t)\|_{L^2(\Omega)}\,dt\\\notag &\leq&
c(\omega)\|\varrho_\varepsilon-\varrho\|_{C(0,T; W^{-3,2}(\Omega))}.
\end{eqnarray*}
From this and \eqref{lita26} we conclude that
\begin{equation}\label{lita49}
 \int_{Q_T}(\varrho_\varepsilon-\varrho)\big(\varphi\mathbf
T_\omega\mathbf u_{N,\varepsilon}\big)\, dxdt\to
0\text{~~as~~}\varepsilon\to 0.
\end{equation}
Next we have
\begin{equation*}
\int_{Q_T}\varrho\varphi \mathbf T_\omega\big( \mathbf
u_{N,\varepsilon}-\overline{\mathbf u}_{N}\,\big)\, dxdt=
\int_{Q_T}\mathbf T_\omega(\varrho\varphi) \big( \mathbf
u_{N,\varepsilon}-\overline{\mathbf u}_{N}\,\big)\, dxdt.
\end{equation*}
Since the mapping $\mathbf T_\omega: L^1(\Omega)\to
L^\infty(\Omega)$ is bounded, we have
\begin{equation*}
    \|\mathbf T_\omega(\varrho\varphi)\|_{L^\infty(Q_T)}\leq
    c(\omega).
\end{equation*}
On the other hand,
\begin{equation}
\mathbf u_{N,\varepsilon}\to\overline{\mathbf u}_N\text{~~weakly
in~~} L^2(Q_T)\text{~~as~~}\varepsilon\to 0.
\end{equation}
It follows that
\begin{equation*}
\int_{Q_T}\varrho\varphi \mathbf T_\omega\big( \mathbf
u_{N,\varepsilon}-\overline{\mathbf u}_{N}\,\big)\, dxdt\to
0\text{~~as~~}\varepsilon\to 0.
\end{equation*}
Combining this relation with \eqref{lita47} and \eqref{lita49} we
conclude that
\begin{equation*}
\Pi_3\to 0\text{~~as~~}\varepsilon\to 0
\end{equation*}
for fixed $\omega$ and $N$. From this, estimates
\eqref{lita45}-\eqref{lita46}, and identity \eqref{lita41} we obtain
\begin{equation*}
\limsup\limits_{\varepsilon\to 0} \Big|\int_{Q_T}\varphi\,
\big(\,\varrho_\varepsilon \,\mathbf u_{N, \varepsilon}-
\varrho\,\overline{\mathbf u}_N\,\big)\, dxdt\Big|\leq \frac{c
N}{|\ln(\delta(\omega)|\ln\delta(\omega)|)|}.
\end{equation*}
Letting $\omega \to 0$ we arrive at the desired relation
\eqref{lita29}.
\end{proof}
\paragraph{\textbf{Step 3.}}
We are now in a position to complete the proof of Proposition
\ref{lita23}.  The main idea is to consider the truncated sequence
$\varrho_{\varepsilon}\mathbf u_{N,\varepsilon}$ as an approximation
of the sequence $\varrho_{\varepsilon}\mathbf u_{\varepsilon}$.  Let
us estimate the discrepancy
\begin{equation}\label{lita60}
    \mathbf D_1(N, \varepsilon)=\int_{Q_T} \varrho_\varepsilon|\mathbf
    u_\varepsilon-\mathbf u_{N,\varepsilon}|\, dxdt.
\end{equation}
 It follows from estimate \eqref{lita6} in Lemma \ref{lita5} and De
la Vall\`{e}e Poussin criterion that the sequence
$\varrho_\varepsilon\mathbf u_\varepsilon$ is equi-integrable. This
means that there is a function $\delta_1:\mathbb R^+\to \mathbb
R^+$, independent of $\varepsilon$, such that
\begin{equation}\label{lita53}
\lim\limits_{\omega\to 0}\delta_1(\omega)=0, \quad \int_C
\varrho_\varepsilon |\mathbf u_\varepsilon|\,dxdt\leq
\delta_1(\text{~meas~} C)\text{~~for every~~} C\subset Q_T.
\end{equation}
Note that
$$
\varrho_\varepsilon|\mathbf
    u_\varepsilon-\mathbf u_{N,\varepsilon}|\leq
    2\varrho_\varepsilon|\mathbf u_\varepsilon|\text{~~in~~}
    Q_T,\quad \varrho_\varepsilon|\mathbf
    u_\varepsilon-\mathbf u_{N,\varepsilon}|=0\text{~~in~~}
    Q_T\setminus C_{N,\varepsilon},
$$
where
$$
C_{N,\varepsilon}=\big\{\, (x,t)\in Q_T:  |\mathbf
u_\varepsilon(x,t)|>N\,\big\}.
$$
We thus get
\begin{equation}\label{lita54}\begin{split}
\mathbf D_1(N, \varepsilon)\leq
    c\int_{C_{N,\varepsilon}}\varrho_\varepsilon|\mathbf u_\varepsilon|dxdt\leq
    c\delta_1(\text{~meas~} C_{N, \varepsilon}).
\end{split}\end{equation}
The Chebyshev inequality and estimate \eqref{lita2} imply
\begin{equation}\label{lita54a}
\text{meas~} C_{N,\varepsilon}\leq \frac{1}{N^2}\int_{Q_T}|\mathbf
u_\varepsilon|^2\, dxdt\leq \frac{c}{N^2}.
\end{equation}
From this and \eqref{lita54} we conclude that
\begin{equation}\label{lita55}\begin{split}
\mathbf D_1(N,\varepsilon)\leq c\delta_1(c N^{-2 }).
\end{split}\end{equation}
Note that we have no information about the limiting vector field
$\varrho \mathbf u$. Now our task is to prove that it is integrable
and
\begin{equation}\label{lita62}
    \mathbf D_2(N):= \int_{Q_T} |\varrho \mathbf
    u-\varrho\overline{\mathbf u}_N|\,dxdt\to 0 \text{~~as~~} N\to
    \infty.
\end{equation}
By virtue of the estimate \eqref{lita6} in Lemma \ref{lita5}, we
have
$$
\int_{Q_T} \varrho_\varepsilon|\mathbf u_{N,\varepsilon}| \ln(2+
\varrho_\varepsilon|\mathbf u_{N,\varepsilon}|)^\alpha\,dxdt\leq
\int_{Q_T} \varrho_\varepsilon|\mathbf u_{\varepsilon}| \ln(2+
\varrho_\varepsilon|\mathbf u_{\varepsilon}|)^\alpha\,dxdt \leq c
$$
for all $0<\alpha<1/2$.  Therefore,  the sequence
$\varrho_{\varepsilon}\mathbf u_{N,\varepsilon}$ is equi-integrable.
From this and \eqref{lita29} we conclude that
$\varrho_\varepsilon\mathbf u_{N,\varepsilon}$ converge weakly in
$L^1(Q_T)$ to $\varrho\overline{\mathbf u}_N$. Choose an arbitrary
measurable  $C\subset Q_T$ and set
$$
\boldsymbol\varsigma= \frac{\overline{\mathbf
u}_N(x,t)}{|\overline{\mathbf u}_N(x,t)|}\, \boldsymbol{1_C}
\text{~~for~~}|\overline{\mathbf u}_N(x,t)|>0\text{~~and~~}
\boldsymbol\varsigma=0\text{~~otherwise}.
$$
It is clear that the measurable function $\boldsymbol\varsigma$
satisfies the inequality $|\boldsymbol\varsigma|\leq 1$ and is
supported in $C$. It follows from this and inequality \eqref{lita53}
that
\begin{eqnarray*}
\delta_1(\text{meas C})&\geq &\int_C \varrho_\varepsilon|\mathbf
u_{\varepsilon}|dxdt\geq \int_C \varrho_\varepsilon|\mathbf
u_{N,\varepsilon}|dxdt\\
&\geq &\int_{Q_T} \varrho_\varepsilon\mathbf u_{N,\varepsilon}\cdot
\boldsymbol\varsigma dxdt \to \int_{Q_T} \varrho \overline{\mathbf
u}_N\cdot \boldsymbol\varsigma dxdt\\ &=&\int_{C}
\varrho|\overline{\mathbf u}_N| dxdt.
\end{eqnarray*}
It follows that
$$
\int_{C} \varrho|\overline{\mathbf u}_N| dxdt\leq
\delta_1(\text{~~meas~~}C)\text{~~for all measurable~~} C\subset
Q_T.
$$
Hence the sequence $\varrho\overline{\mathbf u}_N$, $N\geq 3$, is
equi-interable. Let us prove that $\overline{\mathbf u}_N$ converges
to $\mathbf u$ in $L^1(Q_T)$. Recalling estimate \eqref{lita54a} and
applying the Cauchy inequality we  obtain
\begin{eqnarray*}
\int_{Q_T}|\mathbf u_{N,\varepsilon}-\mathbf u_\varepsilon|\,dxdt
&=&\int_{C_{N,\varepsilon}}|\mathbf u_{N,\varepsilon}-\mathbf
u_\varepsilon|\,dxdt\\
\leq c (\text{~meas ~} C_{N,\varepsilon})^{1/2}&\leq& c N^{-1}.
\end{eqnarray*}
Since
$$
\mathbf u_\varepsilon \to \mathbf u, \quad \mathbf
u_{N,\varepsilon}\to \overline{\mathbf u}_N\text{~~weakly in~~}
L^2(Q_T) \text{~~as~~} \varepsilon\to 0,
$$
we have
$$
\int_{Q_T}|\mathbf u-\overline{\mathbf u}_N|dxdt\leq
\liminf\limits_{\varepsilon\to 0}\int_{Q_T}|\mathbf
u_\varepsilon-\mathbf u_{N, \varepsilon}|dxdt \leq \frac{c}{N}.
$$
It follows that
$$
\|\mathbf u-\overline{\mathbf u}_N\|_{L^1(Q_T)}\to 0\text{~~as~~}
N\to\infty.
$$
In particular, the the vector fields $\overline{\mathbf u}_N$
converges to $\mathbf u$ in measure. Therefore, the sequence
$\varrho\overline{\mathbf u}_N$ is equi-integrable in $Q_T$ and
converges to $\varrho\mathbf u$ in measure. It follows that
$$
\varrho\overline{\mathbf u}_N\to  \varrho \mathbf u\text{~~in~~}
L^1(Q_T)\text{~~as~~} N\to\infty.
$$
This gives the desired relation \eqref{lita62} In particular, we
prove that the vector field  $\varrho\mathbf u$ is integrable. For
every fixed $\varphi\in L^\infty(Q_T)$, we have
\begin{eqnarray*}
    \int_{Q_T}\varphi (\varrho_\varepsilon\mathbf
    u_\varepsilon-\varrho\mathbf u)\, dxdt&=&
     \int_{Q_T}\varphi \varrho_\varepsilon(\mathbf
    u_\varepsilon-\mathbf u_{N,\varepsilon})\, dxdt\\\notag
    + \int_{Q_T}\varphi \varrho(\overline{\mathbf
    u}_N-\mathbf u)\, dxdt &+&\int_{Q_T}\varphi (\varrho_\varepsilon\mathbf u_{N,\varepsilon}
    -\varrho\overline{\mathbf
    u}_N)\, dxdt.
\end{eqnarray*}
It follows that
\begin{equation*}
\Big|\int_{Q_T}\varphi (\varrho_\varepsilon\mathbf
    u_\varepsilon-\varrho\mathbf u)\, dxdt\Big|\leq c \mathbf D_1(N,
    \varepsilon)+c\mathbf D_2(N)+\mathbf D_3(N, \varepsilon),
\end{equation*}
where
$$
\mathbf D_3(N,\varepsilon)=\Big |\int_{Q_T}\varphi
(\varrho_\varepsilon\mathbf u_{N,\varepsilon}
    -\varrho\overline{\mathbf
    u}_N)\, dxdt\Big |\to 0\text{~~as~~}\varepsilon\to 0
$$
The latter follows from relation \eqref{lita29} in Lemma
\ref{lita29a}.  From this and estimate \eqref{lita55} we obtain
\begin{equation*}
\Big|\int_{Q_T}\varphi (\varrho_\varepsilon\mathbf
    u_\varepsilon-\varrho\mathbf u)\, dxdt\Big|\leq c
    \delta_1(cN)+c\mathbf D_2(N)+\mathbf D_3(N, \varepsilon).
\end{equation*}
Letting $\varepsilon\to 0$ we arrive at the inequality
\begin{equation*}
\limsup\limits_{\varepsilon\to 0}\Big|\int_{Q_T}\varphi
(\varrho_\varepsilon\mathbf
    u_\varepsilon-\varrho\mathbf u)\, dxdt\Big|\leq c \delta_1(cN)+c\mathbf
    D_2(N).
\end{equation*}
Note that the left hand side of this inequality is independent on
$N$. Letting $N\to\infty$ and using \eqref{lita62} we arrive at the
desired relation
\begin{equation*}
\lim\limits_{\varepsilon\to 0}\int_{Q_T}\varphi
(\varrho_\varepsilon\mathbf
    u_\varepsilon-\varrho\mathbf u)\, dxdt=0.
\end{equation*}
It remains to note that $\varrho_\varepsilon\mathbf u_\varepsilon$
converge weakly to $\overline{\varrho \mathbf u}$ as $\varepsilon\to
0$. This completes the proof of Proposition \ref{lita23}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Part  3. Specification of the measures $\boldsymbol\nu$ and
$\sigma$}\label{part3} In this subsection we study in details the
structure of the measures  $\boldsymbol \nu(t)$ and $\sigma (t)$
introduced in Subsection \ref{part1}. Note that $\boldsymbol\nu(t)$
and $\sigma(t)$ are Borel measures defined on the $\sigma$- algebra
of Borel sets $B\subset\Omega$. By the Lebesgue dominant convergence
theorem,
$$
\int_\Omega \varphi_n\, d\boldsymbol\nu(t)\to \int_\Omega\varphi\ \,
d\boldsymbol\nu(t), \quad \int_\Omega \varphi_n\,d\sigma(t)\to
\int_\Omega\varphi\ \, d\sigma(t)\text{~~as~~} n\to\infty
$$
for every sequence of Borel functions $\varphi_n$, $n\geq 1$,
satisfying the conditions
$$
\varphi_n:\Omega\to\mathbb R, \quad |\varphi_n|\leq c,\quad
\varphi_n(x)\to\varphi(x)\text{~~for every ~~} x\in\Omega.
$$
Denote by $\boldsymbol\mu(t)$ and $\varpi (t)$ the restrictions of
the measures $\boldsymbol\nu(t)$ and $\sigma(t)$ on $ \mathcal I$,
$$
\boldsymbol\mu(t)=\lfloor_{\mathcal I}\,\,\boldsymbol\nu(t),\quad
\varpi(t)=\lfloor_{\mathcal I}\,\,\sigma(t).
$$
Recall  that the interval $\mathcal I$ is the intersection of
$\Omega$ and the symmetry axis.  The following proposition is the
main result of this subsection
\begin{proposition}\label{lita65}
Assume that the function $\varphi: Q_T\to \mathbb R$ satisfies the
conditions
\begin{equation*}
    \varphi\in C^\infty(Q_T), \quad \varphi(\cdot,t)\in
    C^\infty_0(\Omega)\text{~~for every~~} t\in [0,T].
\end{equation*}
Then
\begin{equation}\label{lita66}
    \int_0^T\Big(\,\int_\Omega \varphi\, d\boldsymbol
    \nu(t)\,\Big)\,dt
    =\int_{Q_T} \varphi \varrho\mathbf u\otimes\mathbf u\, dxdt+
\int_0^T\Big(\,\int_{\mathcal I} \varphi\, d\boldsymbol
    \mu(t)\,\Big)\,dt,
\end{equation}
\begin{equation}\label{lita66a}
    \int_0^T\Big(\,\int_\Omega \varphi\, d\sigma(t)\,\Big)\,dt
    =\int_{Q_T} \varphi \varrho\, dxdt+
\int_0^T\Big(\,\int_{\mathcal I} \varphi\, d\varpi(t)\,\Big)\,dt.
\end{equation}

\end{proposition}



\begin{proof}
The proof is   based on the following construction. Fix an arbitrary
$C^\infty$ function $\Theta:\mathbb R^+\to \mathbb R^+$ satisfying
the conditions
\begin{equation*}
    0\leq \Theta\leq 1, \quad \Theta(s)=1\text{~~for~~}0\leq s\leq
    1, \quad \Theta(s)=0\text{~~for~~}s\geq 2.
\end{equation*}
Set
\begin{equation}\label{lita67}
    \theta_r(x)=\Theta\big(\frac{1}{r}\sqrt{x_1^2+ x_2^2}\big), \quad
    r>0.
\end{equation}
Let us consider the integral
\begin{equation*}
\int_{Q_T}\varphi\varrho_\varepsilon\mathbf u_\varepsilon\otimes
\mathbf u_\varepsilon\,
dxdt=\int_{Q_T}(1-\theta_r)\varphi\varrho_\varepsilon\mathbf
u_\varepsilon\otimes \mathbf u_\varepsilon\,
dxdt+\int_{Q_T}\theta_r\varphi\varrho_\varepsilon\mathbf
u_\varepsilon\otimes \mathbf u_\varepsilon\, dxdt.
\end{equation*}
Note that $(1-\theta_r)\varphi$ is compactly supported in
$\Omega\setminus\mathcal I$. It follows from relation \eqref{lita30}
and equality \eqref{lita25} that
\begin{equation*}
\lim\limits_{\varepsilon\to
0}\int_{Q_T}(1-\theta_r)\varphi\varrho_\varepsilon\mathbf
u_\varepsilon\otimes \mathbf u_\varepsilon\,
dxdt=\int_{Q_T}(1-\theta_r)\varphi\varrho\mathbf u\otimes \mathbf
u\, dxdt.
\end{equation*}
On the other hand, relation \eqref{lita20} implies
\begin{equation*}
\lim\limits_{\varepsilon\to
0}\int_{Q_T}\theta_r\varphi\varrho_\varepsilon\mathbf
u_\varepsilon\otimes \mathbf u_\varepsilon\, dxdt=\int_0^T\Big(\,
\int_\Omega\theta_r\varphi d\boldsymbol\nu(t)\,\Big)\, dt.
\end{equation*}
We thus get
\begin{equation}\label{lita68}
\lim\limits_{\varepsilon\to
0}\int_{Q_T}\varphi\varrho_\varepsilon\mathbf u_\varepsilon\otimes
\mathbf u_\varepsilon\, dxdt=
\int_{Q_T}(1-\theta_r)\varphi\varrho\mathbf u\otimes \mathbf u\,
dxdt+\int_0^T\Big(\, \int_\Omega\theta_r\varphi
d\boldsymbol\nu(t)\,\Big) dt.
\end{equation}
Note that the integral in the left hand side is independent of $r$.
We have for fixed $t\in [0,T]$,
$$
\theta_r(x)\varphi(x,t)\to 0\text{~~as~~}r\to 0\text{~~for~~}
x\in\Omega\setminus \mathcal I, \quad \theta_r(x)\varphi(x,t)=
\varphi(x,t)\text{~~for~~} x\in \mathcal I.
$$
For every $t\in (0,T)$, the limiting function equals $0$ outside of
$\mathcal I\times (0,T)$ and equals $\varphi$ on $\mathcal I$. It is
a Borel function as a pointwise limit of Borel functions. Obviously,
the sequence $\theta_r\varphi$ is uniformly bounded. Applying the
Lebesgue dominating convergence Theorem we conclude that
$$
J_r(t):=\int_\Omega \theta_r(x)\varphi(x,t)d\boldsymbol\nu(t) \to
\int_{\mathcal I}\varphi(x,t)\, d\boldsymbol\mu(t)=: J(t).
$$
It is clear that
$$
|J_r(t)|\leq c\|\boldsymbol \nu(t)\|_{\mathbb M(\Omega)}.
$$
Recall that $\|\boldsymbol \nu(t)\|_{\mathbb M(\Omega)}\in L^r(0,T)$
for every $r\in [1,\infty)$. Hence the sequence  $J_r(t)$ meets all
requirements of the Lebesgue dominated convergence theorem on the
interval $(0,T)$. Applying this theorem we obtain
$$
\int_0^T J_r(t)\, dt \to \int_0^T J(t)\,dt,
$$
which gives
\begin{equation}\label{lita69}
  \lim\limits_{r\to 0}\int_0^T\Big(\,  \int_\Omega
  \theta_r(x,t)\varphi(x,t)\,d\boldsymbol\nu(t)\,\Big) dt =
\int_0^T\Big (\,\int_{\mathcal I}\varphi(x,t)\,
d\boldsymbol\mu(t)\,\Big)\,dt.
\end{equation}
Since $\varrho\mathbf u\otimes\mathbf u\in L^1(Q_T)$, we have
\begin{equation}\label{lita70}
  \lim\limits_{r\to 0}
  \int_{Q_T}(1-\theta_r)\varphi\varrho\mathbf u\otimes \mathbf u\,
dxdt=\int_{Q_T}\varphi\varrho\mathbf u\otimes \mathbf u\, dxdt.
\end{equation}
Letting $r\to 0$ in \eqref{lita68} and using \eqref{lita69},
\eqref{lita70} we obtain
\begin{equation*}
\lim\limits_{\varepsilon\to
0}\int_{Q_T}\varphi\varrho_\varepsilon\mathbf u_\varepsilon\otimes
\mathbf u_\varepsilon\, dxdt=\int_{Q_T}\varphi\varrho\mathbf
u\otimes \mathbf u\, dxdt+\int_0^T\Big (\,\int_{\mathcal
I}\varphi(x,t)\, d\boldsymbol\mu(t)\,\Big)\,dt.
\end{equation*}
It remains to note that by virtue of \eqref{lita20},
$$
\lim\limits_{\varepsilon\to
0}\int_{Q_T}\varphi\varrho_\varepsilon\mathbf u_\varepsilon\otimes
\mathbf u_\varepsilon\, dxdt= \int_0^T\Big(\, \int_\Omega\varphi
d\boldsymbol\nu(t)\,\Big)\, dt.
$$
This completes the proof of \eqref{lita66}. Let us prove
\eqref{lita66a}. We have
\begin{equation*}
\int_{Q_T}\varphi p(\varrho_\varepsilon)\,
dxdt=\int_{Q_T}(1-\theta_r)\varphi p(\varrho_\varepsilon)\,
dxdt+\int_{Q_T}\theta_r\varphi p(\varrho_\varepsilon)\, dxdt.
\end{equation*}
Recall the equality
$$
p(\varrho_\varepsilon)= \varrho_\varepsilon
+p_1(\varrho_\varepsilon)+ p_2(\varrho_\varepsilon).
$$
Note that $(1-\theta_r)\varphi$ is compactly supported in
$\Omega\setminus\mathcal I$. From this and relation \eqref{olga15}
in Theorem \ref{olga13} we conclude that
\begin{equation*}
\lim\limits_{\varepsilon\to
0}\int_{Q_T}(1-\theta_r)\varphi(p_1(\varrho_\varepsilon)+p_2(\varrho_\varepsilon))\,
dxdt=0.
\end{equation*}
Since $\varrho_\varepsilon\to\varrho$ weakly in $L^1(Q_T)$, we have
\begin{equation}\label{lita73}
\lim\limits_{\varepsilon\to
0}\int_{Q_T}(1-\theta_r)\varphi\,p(\varrho_\varepsilon)\,
dxdt=\int_{Q_T}(1-\theta_r)\varphi\,\varrho\, dxdt.
\end{equation}
It follows from relation \eqref{lita20} that
\begin{equation*}\lim\limits_{\varepsilon\to 0}
 \int_{Q_T}\theta_r \varphi p(\varrho_\varepsilon) \, dxdt=
\int_0^T\Big(\,\int_\Omega \theta_r\varphi\, d\sigma(t)\,\Big)dt.
\end{equation*}
Combining this relation with \eqref{lita73} we arrive at the
equality
\begin{equation*}\begin{split}
\lim\limits_{\varepsilon\to
0}\int_{Q_T}\varphi\,p(\varrho_\varepsilon)\,
dxdt=\int_{Q_T}(1-\theta_r)\varphi\,\varrho\, dxdt\\
+\int_0^T\Big(\,\int_\Omega \theta_r \varphi\, d\sigma(t)\,\Big)dt.
\end{split}\end{equation*}
Letting $r\to 0$ and arguing as before we finally obtain
\begin{equation*}\begin{split}
\lim\limits_{\varepsilon\to
0}\int_{Q_T}\varphi\,p(\varrho_\varepsilon)\,
dxdt=\int_{Q_T}\varphi\,\varrho\, dxdt
+\int_0^T\Big(\,\int_{\mathcal I} \varphi\, d\varpi(t)\,\Big)dt.
\end{split}\end{equation*}
It remains to note that
\begin{equation*}\begin{split}
\lim\limits_{\varepsilon\to
0}\int_{Q_T}\varphi\,p(\varrho_\varepsilon)\, dxdt=
\int_0^T\Big(\,\int_{\Omega} \varphi\, d\sigma(t)(t)\,\Big)dt.
\end{split}\end{equation*}
This completes the proof of \eqref{lita66a}.

\end{proof}

\begin{corollary}\label{lita80}
Under the above assumptions, the integral identities
\begin{equation}\label{lita81}\begin{split}
\int_{Q_T}\big(\,\varrho\mathbf{u}\cdot
\partial_t\boldsymbol{\xi}+\varrho\mathbf u\otimes\mathbf
u:\nabla\boldsymbol\xi+\varrho\text{\rm ~div~}\boldsymbol\xi\,\big)
\,dxdt+\\
\int_0^T \Big(\,\int_{\mathcal I} \nabla\boldsymbol{\xi}:
d\boldsymbol\mu(t)\,\Big)\, dt +\int_0^T \Big(\,\int_{\mathcal I}
\text{\rm div~} \boldsymbol{\xi}\, d\varpi(t)\,\Big)\,dt\\
-\int_{Q_T}\mathbb S(\mathbf{u}):\nabla\boldsymbol{\xi}\,dxdt
+\int_{Q_T} \varrho\mathbf f\cdot\boldsymbol{\xi}\, dxdt\\+
\int_{\Omega }\varrho_0 \mathbf u_0(x)\cdot
\boldsymbol{\xi}(x,0)\,dx=0,
\end{split}\end{equation}
\begin{equation}\label{lita82}
\int_{Q_T}\big (\varrho
\partial_t\psi+\varrho\mathbf{u}\cdot\nabla\psi
\big)\,dxdt+\int_{\Omega } \varrho_0(x)\psi(x,0)\, dx=0
\end{equation}
 hold for all vector fields $\boldsymbol{\xi}\in C^\infty(Q_T)$
vanishing in a neighborhood of $\partial\Omega\times[0,T]$ and
$\Omega\times\{t=T\}$, and for all functions $\psi\in C^\infty(Q_T)$
vanishing in a neighborhood of ~ $\Omega\times\{t=T\}$.

\end{corollary}
\begin{proof} The proof obviously follows from
integral identities \eqref{lita21}- \eqref{lita22} and equalities
\eqref{lita24}-\eqref{lita25}, \eqref{lita66}-\eqref{lita66a}.
\end{proof}


\subsection{Part  4. Proof of Theorem \ref{thm1} }\label{part4}
We are now in a position to complete the proof of Theorem
\ref{thm1}. It remains to show that matrix value parametrized
measure $\boldsymbol\mu=(\mu_{ij})_{i,j=1,2,3}$ has the only nonzero
entry $\mu_{33}$, and the parametrized measure $\varpi=0$. To this
and choose an arbitrary functions
$$
\eta\in C^\infty_0(0,T), \quad \chi\in C^\infty_0(\mathcal I).
$$
Introduce the vector field $\boldsymbol\xi: Q_T\to\mathbb R^3$
defined by the equalities
$$
\boldsymbol \xi=(\xi_1, 0, 0), \quad
\xi_1=x_1\theta_r(x)\chi(x_3)\eta(t),
$$
where $\theta_r$ is defined by \eqref{lita67}. For sufficiently
small $r$, the smooth vector field $\boldsymbol\xi$ is compactly
supported in $Q_T$. Denote by $Q_{T,r}$ the cylinder
$$
Q_{T,r}=\{\, (x,t)\in Q_T:\,\, \sqrt{x_1^2+x_2^2}\leq 2r\,\}.
$$
Obviously, $\boldsymbol\xi$ is supported in $Q_{T,r}$. Substituting
$\boldsymbol\xi$ in integral identity \eqref{lita81} we obtain
\begin{equation}\label{lita91}\begin{split}
\int_{Q_{T,r}}\big(\,\varrho u_1
\partial_t\xi_1+\varrho u_1^2\partial_1\xi_1+
\varrho u_1 u_2\partial_2\xi_1 +\varrho u_1 u_3\partial_3\xi_1
\,\big)
\,dxdt+\\
\int_{Q_{T,r}}\big(\,\varrho\partial_1\xi_1
-S_{11}\partial_1\xi_1-S_{12}\partial_2\xi_1 -S_{13}\partial_3\xi_1+
\varrho\mathbf
f_1\xi_1\big)\, dxdt\\
+\int_0^T \eta(t) \Big(\,\int_{\mathcal I}\chi(x_3)
d\mu_{11}(t)\,\Big)\, dt +\int_0^T \eta(t)\Big(\,\int_{\mathcal
I}\chi(x_3)\, d\varpi(t)\,\Big)\,dt .
\end{split}\end{equation}
Note that
\begin{equation}\label{lita92a}
|\nabla\xi_1|+|\partial_t\xi_1|\leq c\text{~~in~~}Q_T, \quad
\varrho, \varrho |\mathbf u|^2, S_{ij}\in L^1(Q_T).
\end{equation}
Moreover, we have
\begin{equation}\label{lita92b}
\text{meas~} Q_{T,r}\to 0\text{~~as~~}r\to 0.
\end{equation}
It follows that
\begin{equation*}\begin{split}
\int_{Q_{T,r}}\big(\,\varrho u_1
\partial_t\xi_1+\varrho u_1^2\partial_1\xi_1+
\varrho u_1 u_2\partial_2\xi_1 +\varrho u_1 u_3\partial_3\xi_1
\,\big)
\,dxdt+\\
\int_{Q_{T,r}}\big(\,\varrho\partial_1\xi_1
-S_{11}\partial_1\xi_1-S_{12}\partial_2\xi_1 -S_{13}\partial_3\xi_1+
\varrho\mathbf f_1\xi_1\big)\, dxdt\to 0
\end{split}\end{equation*}
as $r\to 0$. Letting $r\to 0$ in \eqref{lita91} we obtain
$$
\int_0^T \eta(t) \Big(\,\int_{\mathcal I}\chi(x_3)
d\mu_{11}(t)\,\Big)\, dt +\int_0^T \eta(t)\Big(\,\int_{\mathcal
I}\chi(x_3)\, d\varpi(t)\,\Big)\,dt=0
$$
for all $\chi\in C^\infty_0(\mathcal I)$ and $\eta\in
C^\infty_0(0,T).$ Recalling that $\mu_{11}\geq 0$ and $\varpi\geq 0$
we conclude that
\begin{equation}\label{92}
    \mu_{11}=\varpi=0.
\end{equation}
In order to prove the equality $\mu_{12}=0$, introduce the vector
field
$$
\boldsymbol \xi=(\xi_1, 0, 0), \quad
\xi_1=x_2\theta_r(x)\chi(x_3)\eta(t).
$$
Substituting $\boldsymbol\xi$ in  \eqref{lita81} and noting that
$\varpi=0$ we arrive at the equality
\begin{equation}\label{lita91a}\begin{split}
\int_{Q_{T,r}}\big(\,\varrho u_1\partial_t\xi_1+\varrho
u_1^2\partial_1\xi_1+ \varrho u_1 u_2\partial_2\xi_1 +\varrho u_1
u_3\partial_3\xi_1 \,\big)
\,dxdt+\\
\int_{Q_{T,r}}\big(\,\varrho\partial_1\xi_1
-S_{11}\partial_1\xi_1-S_{12}\partial_2\xi_1 -S_{13}\partial_3\xi_1+
\varrho\mathbf
f_1\xi_1\big)\, dxdt\\
+\int_0^T \eta(t) \Big(\,\int_{\mathcal I}\chi(x_3)
d\mu_{12}(t)\,\Big)\, dt =0 .
\end{split}\end{equation}
Recalling relations \eqref{lita92a} and \eqref{lita92b} we obtain
\begin{equation*}\begin{split}
\int_{Q_{T,r}}\big(\,\varrho u_1
\partial_t\xi_1+\varrho u_1^2\partial_1\xi_1+
\varrho u_1 u_2\partial_2\xi_1 +\varrho u_1 u_3\partial_3\xi_1
\,\big)
\,dxdt+\\
\int_{Q_{T,r}}\big(\,\varrho\partial_1\xi_1
-S_{11}\partial_1\xi_1-S_{12}\partial_2\xi_1 -S_{13}\partial_3\xi_1+
\varrho\mathbf f_1\xi_1\big)\, dxdt\to 0
\end{split}\end{equation*}
as $r\to 0$. Letting $r\to 0$ in \eqref{lita91a} we get
$$
\int_0^T \eta(t) \Big(\,\int_{\mathcal I}\chi(x_3)
d\mu_{12}(t)\,\Big)\, dt =0
$$
for all $\chi\in C^\infty_0(\mathcal I)$ and $\eta\in
C^\infty_0(0,T).$ We thus get
\begin{equation}\label{93}
    \mu_{12}\equiv \mu_{21}=0.
\end{equation}
Let us prove that $\mu_{22}=0$. To this end, take the vector field
$\boldsymbol \xi$ in the form
$$
\boldsymbol \xi=(0, \xi_2, 0), \quad
\xi_2=x_2\theta_r(x)\chi(x_3)\eta(t).
$$
Substituting $\boldsymbol\xi$ in  \eqref{lita81} and noting that
$\varpi=0$ we obtain
\begin{equation*}\begin{split}
\int_{Q_{T,r}}\big(\,\varrho u_2\partial_t\xi_2+\varrho u_1
u_2\partial_1\xi_1+ \varrho u_2^2\partial_2\xi_2 +\varrho u_2
u_3\partial_3\xi_1 \,\big)
\,dxdt+\\
\int_{Q_{T,r}}\big(\,\varrho\partial_2\xi_2
-S_{12}\partial_1\xi_2-S_{22}\partial_2\xi_2 -S_{23}\partial_3\xi_2+
\varrho\mathbf
f_2\xi_2\big)\, dxdt\\
+\int_0^T \eta(t) \Big(\,\int_{\mathcal I}\chi(x_3)
d\mu_{22}(t)\,\Big)\, dt =0 .
\end{split}\end{equation*}
Arguing as before we conclude that $\mu_{22}=0$.  Combining this
result with \eqref{92} and \eqref{93} we arrive at the equalities
\begin{equation}\label{94}
    \mu_{ij}=0\text{~~for~~} 1\leq i,j\leq 2, \quad \varpi=0.
\end{equation}
It remains to prove that
\begin{equation}\label{95}
\mu_{13}\equiv \mu_{31}=\mu_{23}\equiv \mu_{32}=0.
\end{equation}
We begin with the observation that
$$
\varrho_\varepsilon u_{\varepsilon, i}
u_{\varepsilon,j}\varphi_i\varphi_j=\varrho_\varepsilon (\mathbf
u_\varepsilon\cdot\boldsymbol\varphi)^2\geq 0\text{~~for
all~~}\boldsymbol\varphi \in \mathbb R^3.
$$
Choose an arbitrary functions
$$\boldsymbol \varphi:\mathcal I\to
\mathbb R^3, \quad \varphi_i\in C_0(\mathcal I), \quad \eta\in
C^+_0(0,T). $$
 and set
$$
\boldsymbol \xi= \theta_r(x) \eta(t)\, \boldsymbol\varphi(x_3).
$$
It follows from this and \eqref{lita20}, \eqref{lita26} that
\begin{eqnarray*}\notag
0\leq \int_{Q_T}\theta_r\eta \varrho_\varepsilon u_{\varepsilon,i}
u_{\varepsilon_j}\varphi_i\varphi_j \, dxdt
&\to&\int_{Q_T}\theta_r\eta \varrho u_{i} u_j\varphi_i\varphi_j \,
dxdt\\ &+& \int_0^T \eta(t)\Big (\int_{\mathcal
I}\varphi_i(x_3)\varphi_j(x_3)\, d\mu_{i,j}(t)\,\Big)\, dt.
\end{eqnarray*}
Note that the functions $\theta_r \eta\boldsymbol \varphi$ are
uniformly bounded and are compactly supported in the small cylinder
$Q_{T,r}$, which measure tends to $0$ as $r\to 0$. Since the
function $\varrho|\mathbf u|^2$ is integrable, we conclude that
$$
\lim\limits_{r\to 0}\int_{Q_T}\theta_r\eta \varrho u_{i}
u_j\varphi_i\varphi_j \, dxdt=0.
$$
We thus get
$$
\int_0^T \eta(t)\Big (\int_{\mathcal
I}\varphi_i(x_3)\varphi_j(x_3)\, d\mu_{i,j}(t)\,\Big)\, dt\geq
0\text{~~for all~~}\eta\in C^+_0(0,T).
$$
It follows that
$$
\int_{\mathcal I}\varphi_i(x_3)\varphi_j(x_3)\, d\mu_{i,j}(t)\, \geq
0\text{~~for all~~}\boldsymbol\varphi \in C_0(\mathcal I)\text{~~and
for a.e.~} t\in (0,T).
$$
Recalling \eqref{94} we conclude that the inequality
$$
0\leq\int_{\mathcal I}\big(2\varphi_1\varphi_3d\mu_{13}(t)+
2\varphi_2\varphi_3 d\mu_{23}(t)+\varphi_3^2 d\mu_{33}(t)\,\big)
$$
holds for all $\varphi_j \in C_0(\mathcal I)$ and for a.e.
$t\in(0,T)$. It is possible if and only if $\mu_{13}=\mu_{23}=0$.
Therefore, the matrix $\mu_{ij}$ may have the only nonzero entry
$\mu_{33}$. This completes the proof of Theorem \ref{thm1}.





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\section{Appendix. Navier-Stokes equations in the cylindrical coordinates }\label{cylinder}
\renewcommand{\theequation}{7.\arabic{equation}}
\setcounter{equation}{0}
 In this section we give the proof of the lemma and proposition
 formulated in Section \ref{s2}. To this end we recall the
 denotations for the cylindrical coordinates
Denote by  $r$, $\vartheta$, $x_3$ the cylindrical coordinates in
the space $\mathbb R^3$ of points $x=(x_1,x_2,x_3)$,
\begin{equation}\label{sveta1}
    x_1=r\cos \vartheta, \quad x_2=r \sin\vartheta,\quad x_3=x_3,
\end{equation}
\begin{equation}\label{sveta2}
y=(y_1,y_3)=(r,x_3), \quad z=(z_1,z_2,z_3)= (r,\theta, x_3)=(y_1,
\theta, y_3).
\end{equation}
and the  domain
$$
G=\{x\in \Omega: x_1>0, x_2=0\}.
$$
We will consider $G$ as a domain in the plane of variable $y$. The
radial, azimuth and axial components $v_i$ of the velocity field
$\mathbf u$ form the vector field $\mathbf v$ such that
\begin{equation}\label{sveta3}\begin{split}
\mathbf u=\mathbf U\,\mathbf v,\quad\mathbf v=(v_1,v_2, v_3), \quad
\mathbf U=\left(
\begin{array}{ccc}
\displaystyle {\cos\theta}&\displaystyle {
-\sin\theta}&\displaystyle { 0}
\\
\displaystyle {\sin\theta}&\displaystyle { \cos\theta}&\displaystyle { 0}\\
\displaystyle {0 } &  \displaystyle {0
 }&\displaystyle{1}
  \end{array}
\right). \end{split}\end{equation}

We start with the proof of the following technical lemma. Throughout
of this section the denotation
$$ \mathbf A\,=\,\big[\, A_1, A_2, A_3\,\big]
$$
stands for a matrix $\mathbf A$ with the columns $A_i$. Next, we
denote by $\mathbf A:\mathbf B$ the scalar product
$$
\mathbf A:\mathbf B\, =\text{~tr~} (\mathbf A^\top\, \mathbf B).
$$
Note that $|\mathbf A|^2=\mathbf A:\mathbf A$ is the squared
Hilbert-Schmidt norm of $\mathbf A$.

\begin{lemma}\label{sveta4} Let $\Omega$ be rotationally symmetric
and $\mathbf u\in W^{1,2}(\Omega)$. Then
\begin{equation}\label{sveta5}
    \nabla_x\mathbf u(x)=\mathbf U(\theta)\,\big (\boldsymbol
    \lambda(\mathbf v)+\boldsymbol \nu(\mathbf v)\,\big)\, \mathbf U^\top.
\end{equation}
Here the matrices $\boldsymbol \lambda$ and $\boldsymbol\nu$ are
defined by the equalities
\begin{equation}\label{sveta6}\begin{split}
\boldsymbol \lambda(\mathbf v)=\left(
\begin{array}{ccc}
\displaystyle {\partial_r v_1}&\displaystyle {
\frac{1}{r}\partial_\theta v_1}&\displaystyle { \partial_{x_3} v_1}
\\
\displaystyle {\partial_r v_2}&\displaystyle
{\frac{1}{r}\partial_\theta v_2 }
&\displaystyle { \partial_{x_3} v_2 }\\
\displaystyle {\partial_r v_3} &  \displaystyle
{\frac{1}{r}\partial_\theta v_3
 }&\displaystyle{\partial_{x_3} v_3}
  \end{array}\right),\quad
\boldsymbol \nu(\mathbf v)=\left(
\begin{array}{ccc}
\displaystyle {0}&\displaystyle { -\frac{1}{r} v_2}&\displaystyle {
0}
\\
\displaystyle {0}&\displaystyle {\frac{1}{r} v_1 }
&\displaystyle { 0}\\
\displaystyle {0} &  \displaystyle {0
 }&\displaystyle{0}
  \end{array}\right). \end{split}\end{equation}
\end{lemma}
\begin{proof} By virtue of the chain rule, the equality
\begin{equation*}
    \nabla_x\mathbf u=\nabla_z\mathbf u\,\nabla_x z
\end{equation*}
holds true for almost all $z\in G\times \mathbb S^1$. We have
$$
z=(\sqrt{x_1^2+x_2^2},\, \arctan\frac{x_2}{x_1},\, x_3)
$$
Which gives
$$
\nabla_x z=\left(
\begin{array}{ccc}
\displaystyle {\cos\theta}&\displaystyle {\sin\theta }&\displaystyle
{ 0}
\\
\displaystyle {-\frac{1}{r}\sin\theta}&\displaystyle {\frac{1}{r}
\cos\theta}
&\displaystyle { 0}\\
\displaystyle {0} &  \displaystyle {0
 }&\displaystyle{1}
  \end{array}\right)=\mathbf R\mathbf U^\top, \quad \mathbf
  R=\text{~diag~}\{1,\frac{1}{r},1\}.
$$
Thus we get
\begin{equation}\label{sveta7}
    \nabla_x\mathbf u= \nabla_z\mathbf u\, \mathbf R\, \mathbf
    U^\top
\end{equation}
On the other hand, we have
\begin{equation}\label{sveta8}
\nabla_z\mathbf u=\big[\, \partial_r\mathbf u,\partial_\theta\mathbf
u,\partial_{x_3}\mathbf u\,\big]= \big[\, \mathbf U\partial_r\mathbf
v,\mathbf U\partial_\theta\mathbf v,\mathbf U\partial_{x_3}\mathbf
v\,\big]+\big[\, 0,\partial_\theta\mathbf U\mathbf v,0\,\big]
\end{equation}
Notice that
$$
 \mathbf
U^\top\partial_\theta\mathbf U=\left(
\begin{array}{ccc}
\displaystyle {\cos\theta}&\displaystyle {\sin\theta }&\displaystyle
{ 0}
\\
\displaystyle {-\sin\theta}&\displaystyle { \cos\theta}
&\displaystyle { 0}\\
\displaystyle {0} &  \displaystyle {0
 }&\displaystyle{1}
  \end{array}\right)\left(
\begin{array}{ccc}
\displaystyle {-\sin\theta}&\displaystyle {-\cos\theta
}&\displaystyle { 0}
\\
\displaystyle {\cos\theta}&\displaystyle { -\sin\theta}
&\displaystyle { 0}\\
\displaystyle {0} &  \displaystyle {0
 }&\displaystyle{0}
  \end{array}\right)=\left(
\begin{array}{ccc}
\displaystyle {0}&\displaystyle {-1}&\displaystyle { 0}
\\
\displaystyle {1}&\displaystyle {0}
&\displaystyle { 0}\\
\displaystyle {0} &  \displaystyle {0
 }&\displaystyle{0}
  \end{array}\right)=\mathbf S.
$$
Since $\mathbf U\mathbf U^\top=\mathbf I$, we can rewrite identity
\eqref{sveta8} in the equivalent form
\begin{equation*}
\nabla_z\mathbf u=\big[\, \mathbf U\partial_r\mathbf v,\mathbf
U\partial_\theta\mathbf v,\mathbf U\partial_{x_3}\mathbf
v\,\big]+\big[\, 0,\mathbf U(\mathbf S\mathbf v),0\,\big].
\end{equation*}
Note that for every matrix $\mathbf A$ and columns $a$, $b$, $c$, we
have
\begin{equation*}
\big[\, \mathbf A a,\mathbf A b,\mathbf A c\,\big]= \mathbf
A\big[\,a, b, c\,\big],
\end{equation*}
which yields
\begin{equation*}
\nabla_z\mathbf u=\mathbf U\big[\, \partial_r\mathbf
v,\partial_\theta\mathbf v,\partial_{x_3}\mathbf v\,\big]+\mathbf U
\big[\, 0,\mathbf S\mathbf  v,0\big].
\end{equation*}
Substituting this identity into \eqref{sveta7} we obtain
\begin{equation*}
\nabla_x\mathbf u=\mathbf U\,\Big(\,\big[\, \partial_r\mathbf
v,\partial_\theta\mathbf v,\partial_{x_3}\mathbf v\,\big]\mathbf R+
\big[\, 0,\mathbf S\mathbf  v,0\big]\mathbf R\,\Big)\, \mathbf
U^\top.
\end{equation*}
It remains to note that
$$
\big[\, \partial_r\mathbf v,\partial_\theta\mathbf
v,\partial_{x_3}\mathbf v\,\big]\mathbf R=\boldsymbol\lambda(\mathbf
v), \quad \big[\, 0,\mathbf S\mathbf v,0\big]\mathbf
R=\boldsymbol\nu(\mathbf v)
$$
and the lemma follows.
\end{proof}
The next technical lemma gives the representation of $\mathbf
u\otimes\mathbf u$ and $\text{div}\mathbf u$.
\begin{lemma}\label{sveta11}
Let $\mathbf u\in W^{1,2}(\Omega)$ admits the representation
$\mathbf u=\mathbf U(\theta) \mathbf v(z)$. Then
\begin{equation}\label{sveta12}
    \mathbf u\otimes\mathbf u=\mathbf U(\mathbf v\otimes \mathbf
    v)\mathbf U^\top, \quad \text{\rm div~} \mathbf u=\text{\rm
    tr~}\big(\,\boldsymbol\lambda(\mathbf v)+\boldsymbol\nu(\mathbf
    v)\,\big).
\end{equation}
\end{lemma}
\begin{proof}
The first equality in \eqref{sveta12} easily follows from direct
calculations. In order to prove the second, it suffices to note that
by virtue of Lemma \ref{sveta4} we have
$$
\text{div~}\mathbf u=\text{tr~}\nabla_x\mathbf u=
\text{tr~}\big(\,\mathbf U(\boldsymbol\lambda+\boldsymbol\nu)\mathbf
U^T\,\big)=\text{tr~}\big(\,\boldsymbol\lambda+\boldsymbol\nu\,\big).
$$
\end{proof}


The following lemma is a straightforward consequence of Lemma
\ref{sveta4}.
\begin{lemma}\label{sveta9}
Let $\mathbf u:\Omega\to \mathbb R^3$ be a rotationally symmetric
vector field such that $\mathbf u=\mathbf U(\theta)\, \mathbf v(y)$,
$y=(r,x_3)$. Then
\begin{equation}\label{sveta10}
\int_\Omega |\nabla_x \mathbf u|^2\, dx=2\pi\int_G |\nabla_y \mathbf
v|^2\, r drdx_3+2\pi\int_G (v_1^2+v_2^2)\frac{1}{r} drdx_3.
\end{equation}

\end{lemma}
\begin{proof}
By virtue of Lemma \ref{sveta4}, we have
\begin{equation*}\begin{split}
|\nabla_x\mathbf u|^2=|\mathbf U(\boldsymbol\lambda(\mathbf
v)+\boldsymbol\nu(\mathbf v))\mathbf
U^\top|^2=|\boldsymbol\lambda(\mathbf v)+\boldsymbol\nu(\mathbf
v)|^2=\\
(\boldsymbol\lambda(\mathbf v)+\boldsymbol\nu(\mathbf
v)):(\boldsymbol\lambda(\mathbf v)+\boldsymbol\nu(\mathbf v))= |
\boldsymbol\lambda(\mathbf v)|^2+| \boldsymbol\nu(\mathbf v)|^2 +2
\boldsymbol\lambda(\mathbf v): \boldsymbol\nu(\mathbf v).
\end{split}\end{equation*}
Note that
\begin{equation*}\begin{split}
\boldsymbol \lambda(\mathbf v)=\left(
\begin{array}{ccc}
\displaystyle {\partial_r v_1}&\displaystyle {0}&\displaystyle {
\partial_{x_3} v_1}
\\
\displaystyle {\partial_r v_2}&\displaystyle {0}
&\displaystyle { \partial_{x_3} v_2 }\\
\displaystyle {\partial_r v_3} &  \displaystyle {0
 }&\displaystyle{\partial_{x_3} v_3}
  \end{array}\right),\quad
\boldsymbol \nu(\mathbf v)=\left(
\begin{array}{ccc}
\displaystyle {0}&\displaystyle { -\frac{1}{r} v_2}&\displaystyle {
0}
\\
\displaystyle {0}&\displaystyle {\frac{1}{r} v_1 }
&\displaystyle { 0}\\
\displaystyle {0} &  \displaystyle {0
 }&\displaystyle{0}
  \end{array}\right). \end{split}\end{equation*}
It follows that $\boldsymbol\lambda(\mathbf v)$ and
$\boldsymbol\nu(\mathbf v)$ are orthogonal which yields
$$
|\nabla_x\mathbf u|^2=|\boldsymbol\lambda(\mathbf
v)|^2+|\boldsymbol\nu(v)|^2.
$$
Since $\mathbf v$ is independent of $\theta$, we have
\begin{equation*}
    \int_\Omega |\nabla_x\mathbf u|^2\, dx=2\pi\int_G(|\boldsymbol\lambda(\mathbf
v)|^2+|\boldsymbol\nu(\mathbf v)|^2)\, r\,drdx_3.
\end{equation*}
It remains to note that
$$
|\boldsymbol\lambda(\mathbf v)|^2=|\nabla_y\mathbf v|^2, \quad
|\boldsymbol\nu(\mathbf v)|^2=\frac{1}{r^2}(v_1^2+v_2^2).
$$
\end{proof}


We are now in a position to formulate   the compressible
Navier-Stokes equations in the cylindrical coordinates. Let us
consider the totality of test vector fields $\boldsymbol \xi:Q_T\to
\mathbb R^3$ satisfying the following conditions.

\begin{condition}\label{svetac}
\begin{equation}\label{sveta13}
    \boldsymbol \xi\in C^1_0(Q_T), \quad \boldsymbol \xi=\mathbf
    U(\theta)\boldsymbol \varsigma(z,t).
\end{equation}
 The vector field $\boldsymbol\varsigma\in
 C^1(G\times \mathbb S^1\times (0,T))$ is compactly supported in
 $G\times\mathbb S^1\times (0,T)$.
\end{condition}
\begin{proposition}\label{sveta14}
Let $\varrho (x,t)$, $\mathbf u(x,t)$ be a rotationally symmetric
solution to problem \ref{stokes108} given by Proposition
\ref{stokes107}.  Then the vector field $\mathbf v(y,t)$ defined by
the equality $\mathbf u=\mathbf U(\theta)\mathbf v(y,t)$, and the
density $\varrho=\varrho(y,t)$ admit the estimate
\begin{equation}\label{sveta15}\begin{split}
\int_0^T\int_G \Big(|\nabla\mathbf
v|^2+\frac{1}{y_1^2}(v_1^2+v_2^2)\Big)\, y_1dydt\leq c,\\
\text{\rm ess}\sup\limits_{t\in (0,T)}\int_G \Big(\varrho\mathbf
v|^2+e(\varrho)\Big)\, y_1dy\leq c,
\end{split}\end{equation}
where the internal energy $e$ is defined by \eqref{stokes105}, the
constant $c$ depends only on the constant $E$ in \eqref{stokes113}.
The integral identity
\begin{equation}\label{sveta16}\begin{split}
\int_0^T\int_{G\times \mathbb S^1}\Big(\,\, \varrho\mathbf
v\cdot\partial_t\boldsymbol\varsigma+\varrho \mathbf v\otimes\mathbf
v:\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big)
+p(\varrho)\text{\rm tr~}\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big)\,\,\Big)\,y_1\,
dyd\theta dt\\
=\int_0^T\int_{G\times \mathbb S^1}\Big(\,\,\big(\,\boldsymbol
\lambda(\mathbf v)+\boldsymbol\nu(\mathbf v)+ \boldsymbol
\lambda(\mathbf v)^\top+\boldsymbol\nu(\mathbf v)^\top\,\big
):\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big
)\,\Big)\,y_1\, dy\,d\theta\,dt\\
 -\frac{2}{3}\int_0^T\int_{G\times
\mathbb S^1}\Big(\,\big(\,\text{\rm tr~}\big(\,\boldsymbol
\lambda(\mathbf v)+\boldsymbol\nu(\mathbf v)\,\big )\,\text{\rm
tr~}\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big
)\,\,\Big)\,y_1\, dyd\theta dt\\-\int_0^T\int_{G\times \mathbb
S^1}\varrho\mathbf f^*\cdot\boldsymbol\varsigma\, y_1\, dyd\theta dt
\end{split}\end{equation}
holds for all test functions $\boldsymbol\varsigma$ satisfying
Condition \ref{svetac}.  The integral identity
\begin{equation}\label{sveta17}\begin{split}
\int_0^T\int_{G\times \mathbb S^1}\Big(\,\,\varrho \partial_t \psi
+\varrho \mathbf v\cdot (\mathbf R\nabla_z\psi)\,\,\Big )\,y_1\,
dyd\theta dt=0
\end{split}\end{equation}
holds for all functions $\psi\in C^1(G\times \mathbb S^1\times
(0,T))$ compactly supported in $G\times \mathbb S^1\times (0,T)$.
Here $\boldsymbol \lambda(\mathbf v)$ and $\boldsymbol \nu(\mathbf
v)$ are defined by equalities \eqref{sveta6} with
$\partial_\theta\mathbf v=0$, the matrices $\boldsymbol
\lambda(\boldsymbol\varsigma)$ and $\boldsymbol
\nu(\boldsymbol\varsigma)$ are defined by equalities \eqref{sveta6}
with $\mathbf v$ replaced by $\boldsymbol\varsigma$, the vector
field $\mathbf f^*(y,t)$ is defined by the equality $\mathbf
f=\mathbf U(\theta)\mathbf f^*$.

\end{proposition}

\begin{proof}
Estimate \eqref{sveta15} obviously follows from estimate
\eqref{stokes113} in Proposition \ref{stokes107} and
Lemma\ref{sveta9}.


Let us turn to the proof if identities \eqref{sveta16} and
\eqref{sveta17}. It follows from Proposition \ref{stokes107} that
the functions $\varrho$, $\mathbf u=\mathbf U\mathbf v$, and
$\boldsymbol\xi=\mathbf U\boldsymbol\varsigma$ satisfy the integral
identity
\begin{equation}\label{sveta18}
\int_{Q_T}\big (\varrho\mathbf{u}\cdot
\partial_t\boldsymbol{\xi}+
\varrho\mathbf{u}\otimes\mathbf{u}:\nabla\boldsymbol{\xi}+p(\varrho)\text{\rm
div~} \boldsymbol{\xi}-\mathbb S(\mathbf{u}):\nabla\boldsymbol{\xi}+
\varrho\mathbf f\cdot\boldsymbol{\xi}\big)\, dxdt=0.
\end{equation}
Note that
\begin{equation}\label{sveta19}
\mathbf{u}\cdot
\partial_t\boldsymbol{\xi}=\mathbf U\mathbf v\cdot\mathbf
U\partial_t\boldsymbol \varsigma=\mathbf v\cdot\partial_t\boldsymbol
\varsigma, \quad \mathbf{f}\cdot \boldsymbol{\xi}=\mathbf U\mathbf
f^*\cdot\mathbf U\boldsymbol \varsigma=\mathbf f^*\cdot\boldsymbol
\varsigma.
\end{equation}
Next, we have
\begin{equation*}
\mathbf{u}\otimes \mathbf u=(\mathbf U\mathbf{v})\otimes (\mathbf
U\mathbf v)=\mathbf U(\mathbf v\otimes\mathbf v)\mathbf U^\top,
\end{equation*}
which yields
\begin{equation*}
\mathbf{u}\otimes \mathbf u:\nabla\boldsymbol\xi=\mathbf U(\mathbf
v\otimes\mathbf v)\mathbf U^\top\,:\, \nabla\boldsymbol\xi.
\end{equation*}
It follows from Lemma \ref{sveta4} that
\begin{equation*}
\nabla_x\boldsymbol \xi= \mathbf U\,\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big)\mathbf
U^\top.
\end{equation*}
Combining the obtained results we arrive at the identity
\begin{equation}\label{sveta20}\begin{split}
\mathbf{u}\otimes \mathbf u:\nabla\boldsymbol\xi=\big(\,\mathbf U(
\mathbf v\otimes\mathbf v)\mathbf U^\top\,\big):\big(\,\mathbf
U\,\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big)\mathbf
U^\top\,\big)\\= \big(\, \mathbf v\otimes\mathbf
v\,\big):\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big).
\end{split}\end{equation}
Note that
\begin{equation*}\begin{split}
\mathbb S(\mathbf{u})=(\nabla_x \mathbf u+\nabla_x  \mathbf
u^\top)-\frac{2}{3}\text{div~} \mathbf u\,\,\,\mathbf I,
\end{split}\end{equation*}
which yields
\begin{equation}\label{sveta21}\begin{split}
\mathbb S(\mathbf{u}):\nabla\boldsymbol\xi=(\nabla_x \mathbf
u+\nabla_x \mathbf
u^\top):\nabla\boldsymbol\xi-\frac{2}{3}\text{div~} \mathbf
u\,\text{div~} \boldsymbol\xi.
\end{split}\end{equation}
It follows from Lemma \ref{sveta4} and Lemma \ref{sveta11} that
\begin{equation*}\begin{split}
(\nabla_x \mathbf u+\nabla_x \mathbf
u^\top):\nabla\boldsymbol\xi=\\\big(\mathbf U\,\big(\,\boldsymbol
\lambda(\mathbf v)+\boldsymbol\nu(\mathbf v)+\lambda(\mathbf
v)^\top+\boldsymbol\nu(\mathbf v)^\top\,\big)\mathbf
U^\top\big):\big(\mathbf U\,\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big)\mathbf
U^\top\big)\\
=\big(\,\boldsymbol \lambda(\mathbf v)+\boldsymbol\nu(\mathbf
v)+\boldsymbol \lambda(\mathbf v)^\top+\boldsymbol\nu(\mathbf
v)^\top\,\big):\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big)
\end{split}\end{equation*}
and
\begin{equation*}\begin{split}
\text{div~} \mathbf u\,\text{div~} \boldsymbol\xi=\text{tr~}
\big(\,\boldsymbol \lambda(\mathbf v)+\boldsymbol\nu(\mathbf
v)\,\big)\,\text{tr~}\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\big)
\end{split}\end{equation*}
Substituting these results into \eqref{sveta21} we arrive at the
identity
\begin{equation}\label{sveta22}\begin{split}
\mathbb S(\mathbf{u}):\nabla\boldsymbol\xi=(\nabla_x \mathbf
u+\nabla_x \mathbf
u^\top):\nabla\boldsymbol\xi-\frac{2}{3}\text{div~} \mathbf
u\,\text{div~} \boldsymbol\xi=\\
\big(\,\boldsymbol \lambda(\mathbf v)+\boldsymbol\nu(\mathbf
v)+\boldsymbol \lambda(\mathbf v)^\top+\boldsymbol\nu(\mathbf
v)^\top\,\big):\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big)-\\
\frac{2}{3}\text{tr~} \big(\,\boldsymbol \lambda(\mathbf
v)+\boldsymbol\nu(\mathbf v)\,\big)\,\text{tr~}\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\big).
\end{split}\end{equation}
Finally, substituting \eqref{sveta19}, \eqref{sveta20}, and
\eqref{sveta22} into \eqref{sveta18} and noting that $dxdt=y_1
dy\,d\theta\, dt$ we arrive at the desired identity \eqref{sveta16}.

In order to prove \eqref{sveta16}, notice that
\begin{equation}\label{sveta23}
\int_0^T\int_\Omega (\varrho\partial_t\psi+\varrho\mathbf
u\cdot\nabla_x\psi)\,dxdt=0.
\end{equation}
It is easy to see that
\begin{equation*}
\varrho\mathbf u\cdot\nabla_x\psi=\varrho\mathbf U\mathbf v\cdot
\mathbf U(\mathbf R\nabla_z\psi)=\varrho\mathbf v\cdot(\mathbf
R\nabla_z\psi).
\end{equation*}
Substituting these equalities into \eqref{sveta23} we obtain
\eqref{sveta17}.
\end{proof}
In the paper we use the simplified version of identities
\eqref{sveta16} and \eqref{sveta17} with  special test function
$\boldsymbol\varsigma$. The result is given by the following
proposition.


\begin{proposition}\label{sveta24}
Let $\varrho$, $\mathbf v$,  $\boldsymbol\varsigma$, and $\psi$ meet
all requirements of Proposition \ref{sveta14}.  Furthermore assume
that
\begin{equation}\label{sveta25}
\partial_\theta\boldsymbol\varsigma=0,\quad
\varsigma_2=0,\quad\partial_\theta\psi=0.
\end{equation}
Then the vector fields
\begin{equation}\label{sveta26}
\overline{\mathbf v}=(v_1(y,t), v_3(y,t)), \quad
\overline{\boldsymbol\varsigma}=(v_1(y,t), v_3(y,t)), \quad
\overline{\mathbf f}= (f_1^*(y,t), f_3^*(y,t))
\end{equation}
satisfy the integral identities
\begin{multline}\label{sveta27}
\int_{G\times(0,T)}\Big (\varrho\overline{\mathbf{v}}\cdot
\partial_t\overline{\boldsymbol{\varsigma}}+
\varrho\overline{\mathbf{v}}\otimes\overline{\mathbf{v}}:\nabla
\overline{\boldsymbol{\varsigma}}+p(\varrho)\text{\rm div~}
\overline{\boldsymbol{\varsigma}}-\mathbf B(\overline{\mathbf v}):
\nabla\overline{\boldsymbol\varsigma}\\+\mathbf b(\overline{\mathbf
v}, \varrho) \,\varsigma_1+\varrho \overline{\mathbf
f}\cdot\overline {\boldsymbol\varsigma}\Big)\,y_1 dy\,dt=0,
\end{multline}
\begin{equation}\label{sveta28}
\int_{G\times (0,T)}\Big(\,\,\varrho \partial_t \psi +\varrho
\overline{\mathbf v}\cdot \nabla\psi\,\,\Big )\,y_1\, dy\, dt=0.
\end{equation}
Here $\nabla \mathbf a=\nabla_y \mathbf a$ for each vector field
$\mathbf a(y,t)$. The matrix  $\mathbf B$ and vector field  $\mathbf
b$ are defined by the equalities
\begin{equation}\label{sveta29}\begin{split}
    \mathbf B=(\nabla\overline{\mathbf v}+\nabla\overline{\mathbf
v}^\top)- \frac{2}{3}(\text{\rm div~}\overline{\mathbf
v}+\frac{1}{r} v_1),\\
\mathbf b=\frac{1}{y_1}\varrho v_2^2+\frac{1}{y_1}p-\frac{4}{3y_1^2}
v_1+\frac{2}{3y_1}\text{\rm div~}\overline{\mathbf v}
\end{split}\end{equation}
For every compact set $K\Subset G$, the functions $\mathbb B$ and
$\mathbf b$ admit the estimate
\begin{equation}\label{sveta29a}
   \|\mathbf B\|_{L^2(K\times (0,T))}+\|\mathbf b\|_{L^1(K\times
   (0,T))}\leq c,
\end{equation}
where the constant $c$ depends only on $K$ and the constant in
estimate \eqref{sveta15}.
\end{proposition}
\begin{proof}
The proof is based on the identities \eqref{sveta16} and
\eqref{sveta17}. It follows from the conditions of the proposition
and relation \eqref{sveta6} that
\begin{equation}\label{sveta30}\begin{split}
\boldsymbol \lambda(\mathbf v)=\left(
\begin{array}{ccc}
\displaystyle {\partial_r v_1}&\displaystyle { 0}&\displaystyle {
\partial_{x_3} v_1}
\\
\displaystyle {\partial_r v_2}&\displaystyle {0 }
&\displaystyle { \partial_{x_3} v_2 }\\
\displaystyle {\partial_r v_3} &  \displaystyle
{0}&\displaystyle{\partial_{x_3} v_3}
  \end{array}\right),\quad
\boldsymbol \nu(\mathbf v)=\left(
\begin{array}{ccc}
\displaystyle {0}&\displaystyle { -\frac{1}{r} v_2}&\displaystyle {
0}
\\
\displaystyle {0}&\displaystyle {\frac{1}{r} v_1 }
&\displaystyle { 0}\\
\displaystyle {0} &  \displaystyle {0
 }&\displaystyle{0}
  \end{array}\right). \end{split}\end{equation}
and
\begin{equation}\label{sveta31}\begin{split}
\boldsymbol \lambda(\boldsymbol\varsigma)=\left(
\begin{array}{ccc}
\displaystyle {\partial_r \varsigma_1}&\displaystyle {
0}&\displaystyle { \partial_{x_3} \varsigma_1}
\\
\displaystyle {0}&\displaystyle {0}
&\displaystyle { 0}\\
\displaystyle {\partial_r \varsigma_3} &  \displaystyle {0
 }&\displaystyle{\partial_{x_3} \varsigma_3}
  \end{array}\right),\quad
\boldsymbol \nu(\boldsymbol{\varsigma})=\left(
\begin{array}{ccc}
\displaystyle {0}&\displaystyle { 0}&\displaystyle { 0}
\\
\displaystyle {0}&\displaystyle {\frac{1}{r} \varsigma_1 }
&\displaystyle { 0}\\
\displaystyle {0} &  \displaystyle {0
 }&\displaystyle{0}
  \end{array}\right). \end{split}\end{equation}
In particular, $\boldsymbol \lambda(\boldsymbol\varsigma)$ is
orthogonal to $\boldsymbol\nu(\mathbf v)$ and to
$\boldsymbol\nu(\mathbf v)^\top$. It follows that
\begin{equation*}\begin{split}
\big(\,\boldsymbol \lambda(\mathbf v)+\boldsymbol\nu(\mathbf v)+
\boldsymbol \lambda(\mathbf v)^\top+\boldsymbol\nu(\mathbf
v)^\top\,\big ):\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big
)=\\\big(\,\boldsymbol \lambda(\mathbf v)+\boldsymbol
\lambda(\mathbf v)^\top\big):\boldsymbol
\lambda(\boldsymbol\varsigma)+ \big(\, \boldsymbol\nu(\mathbf v)
+\boldsymbol\nu(\mathbf v)^\top\,\big
):\boldsymbol\nu(\boldsymbol\varsigma).
\end{split}\end{equation*}
It is easy to check that
\begin{equation*}
\big(\,\boldsymbol \lambda(\mathbf v)+\boldsymbol \lambda(\mathbf
v)^\top\big):\boldsymbol
\lambda(\boldsymbol\varsigma)=(\nabla\overline{\mathbf
v}+\nabla\overline{\mathbf
v}^\top):\nabla\overline{\boldsymbol\varsigma}
\end{equation*}
and
\begin{equation*}
\big(\,\boldsymbol\nu(\mathbf v)  +  \boldsymbol\nu(\mathbf
v)^\top\,\big):\boldsymbol\nu(\boldsymbol\varsigma)\big)=\frac{2}{r^2}\,v_1\,\varsigma_1.
\end{equation*}
We thus get
\begin{equation}\label{sveta32}\begin{split}
\big(\,\boldsymbol \lambda(\mathbf v)+\boldsymbol\nu(\mathbf v)+
\boldsymbol \lambda(\mathbf v)^\top+\boldsymbol\nu(\mathbf
v)^\top\,\big ):\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big
)=\\(\nabla\overline{\mathbf v}+\nabla\overline{\mathbf
v}^\top):\nabla\overline{\boldsymbol\varsigma}+\frac{2}{r^2}\,v_1\,\varsigma_1.
\end{split}\end{equation}
Next, we have
\begin{equation*}
\text{tr~}\big(\,\boldsymbol \lambda(\mathbf
v)+\boldsymbol\nu(\mathbf v)\big)=\text{div~}\overline{\mathbf
v}+\frac{1}{r} v_1, \quad \text{tr~}\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\big)=
\text{div~}\overline{\boldsymbol\varsigma}+\frac{1}{r} \varsigma_1,
\end{equation*}
which yields
\begin{equation}\label{sveta33}\begin{split}
\frac{2}{3}\text{tr~}\big(\,\boldsymbol \lambda(\mathbf
v)+\boldsymbol\nu(\mathbf v)\big)\text{tr~}\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\big)=\\
\frac{2}{3}(\text{div~}\overline{\mathbf v}+\frac{1}{r} v_1)
(\text{div~}\overline{\boldsymbol\varsigma}+\frac{1}{r}
\varsigma_1).
\end{split}\end{equation}
Combining \eqref{sveta32} and \eqref{sveta33} we arrive at the
identity
\begin{equation*}\begin{split}
\big(\,\boldsymbol \lambda(\mathbf v)+\boldsymbol\nu(\mathbf v)+
\boldsymbol \lambda(\mathbf v)^\top+\boldsymbol\nu(\mathbf
v)^\top\,\big ):\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big
)-\\
\frac{2}{3}\text{tr~}\big(\,\boldsymbol \lambda(\mathbf
v)+\boldsymbol\nu(\mathbf v)\big)\text{tr~}\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\big)=\\
(\nabla\overline{\mathbf v}+\nabla\overline{\mathbf
v}^\top):\nabla\overline{\boldsymbol\varsigma}+\frac{2}{r^2}\,v_1\,\varsigma_1-\\
\frac{2}{3}(\text{div~}\overline{\mathbf v}+\frac{1}{r} v_1)
(\text{div~}\overline{\boldsymbol\varsigma}+\frac{1}{r}
\varsigma_1)=\\
\big(\nabla\overline{\mathbf v}+\nabla\overline{\mathbf v}^\top -
\frac{2}{3}(\text{div~}\overline{\mathbf v}+\frac{1}{r}
v_1)\,\mathbf
I):\nabla\overline{\boldsymbol\varsigma}+\Big(-\frac{4}{3r^2}\,v_1+
\frac{2}{3r}\text{div~}\overline{\mathbf v}\Big)\varsigma_1.
\end{split}\end{equation*}
From this and expression \eqref{sveta29} for $\mathbf B$, we
conclude that
\begin{equation}\label{sveta34}\begin{split}
\big(\,\boldsymbol \lambda(\mathbf v)+\boldsymbol\nu(\mathbf v)+
\boldsymbol \lambda(\mathbf v)^\top+\boldsymbol\nu(\mathbf
v)^\top\,\big ):\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\,\big
)-\\
\frac{2}{3}\text{tr~}\big(\,\boldsymbol \lambda(\mathbf
v)+\boldsymbol\nu(\mathbf v)\big)\text{tr~}\big(\,\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol\nu(\boldsymbol\varsigma)\big)=\\
\mathbf
B:\nabla\overline{\boldsymbol\varsigma}+\Big(-\frac{4}{3r^2}\,v_1+
\frac{2}{3r}\text{div~}\overline{\mathbf v}\Big)\,\varsigma_1.
\end{split}\end{equation}
Next, notice that relations \eqref{sveta31} imply
\begin{equation}\label{sveta35}\begin{split}
\varrho \mathbf v\otimes\mathbf v:(\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol
\nu(\boldsymbol\varsigma))= \varrho\overline{\mathbf
v}\otimes\overline{\mathbf
v}:\nabla\overline{\boldsymbol\varsigma}+\frac{1}{r}\varrho
v_2^2\,\varsigma_1.
\end{split}\end{equation}
Relation \eqref{sveta31} also implies
\begin{equation}\label{sveta36}\begin{split}
p(\varrho)\text{tr~}(\boldsymbol
\lambda(\boldsymbol\varsigma)+\boldsymbol
\nu(\boldsymbol\varsigma))=
p(\varrho)\text{div~}\overline{\boldsymbol\varsigma}+p(\varrho)\frac{1}{r}\varsigma_1.
\end{split}\end{equation}
Obviously we have
\begin{equation}\label{sveta37}
\varrho \mathbf v\cdot \partial_t\boldsymbol\varsigma =\varrho
\overline{\mathbf v}\cdot \partial_t\overline{\boldsymbol\varsigma}
, \quad \varrho \mathbf f^*\cdot \boldsymbol\varsigma = \varrho
\overline{\mathbf f}\cdot \overline{\boldsymbol\varsigma}.
\end{equation}
Substituting \eqref{sveta34}-\eqref{sveta37} into integral identity
\eqref{sveta16}, recalling formula \eqref{sveta29} for $\mathbf b$,
and noting that the integrand in \eqref{sveta16} is independent of
$\theta$ we arrive at the desired identity \eqref{sveta27}.

Estimate \eqref{sveta29a} for $\mathbf B$ and $\mathbf b$ is a
straightforward consequence of estimate  \eqref{sveta15}.

It remains to prove integral identity \eqref{sveta29}. To this end,
it suffices to note that the test function $\psi$ is independent of
$\theta$ and hence
$$
\varrho\mathbf v\cdot (\mathbf
R\nabla_z\psi)=\varrho\overline{\mathbf v}\cdot\nabla_y\psi.
$$
Substituting this equality into integral identity \eqref{sveta17} we
arrive at \eqref{sveta29}.

\end{proof}


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\section{Appendix. Potential estimates. Singular operators}\label{uruk}
\renewcommand{\theequation}{8.\arabic{equation}}
\setcounter{equation}{0} In this section we collect the basic facts
from the theory of potential operators and singular integrals.

\subsection{Estimates of potentials}
We begin with the representation of the Sobolev functions via their
gradients.

\begin{lemma}\label{uruk1}
Let $\Omega\subset \mathbb R^n$, $n\geq  2$, be a bounded domain
with a Lipschitz boundary. Then the equality
\begin{equation}\label{uruk2}
    u(x)=\frac{1}{n\omega_n}\int_\Omega \frac{(x_i-y_i) \partial_i
    u(y)}{|x-y|^n}\, dy \text{~~a.e. in ~~}\mathbb R^n
\end{equation}
holds for ever function $u\in W^{1,1}(\mathbb R^n)$ compactly
supported in $\Omega$. Here $\omega_n$ is a volume of $\mathbb
S^{n-1}$
\end{lemma}
\begin{proof}
This statement is a particular case of Lemma 7.14 in
\cite{trudinger}.
\end{proof}

Let $\Omega\subset \mathbb R^n$ be a bounded measurable set of
positive measure. For every $\mu\in (0,1)$, define the integral
operator
\begin{equation}\label{uruk3}
    \mathcal V_\mu u(x)=\int_\Omega |x-y|^{-n(1-\mu)}u(y)\, dy.
\end{equation}
\begin{lemma}\label{uruk4}
Let $1<s\leq r<\infty$ and
$$
0\leq \kappa= s^{-1}-r^{-1}<\mu.
$$
Then the operator $\mathcal V_\mu: L^s(\Omega)\to L^r(\Omega)$ is
bounded and
\begin{equation}\label{uruk5}
    \|\mathcal V_\mu u\|_{L^r(\Omega)}\leq
    \Big(\,\frac{1-\kappa}{\mu-\kappa}\,\Big)^{1-\kappa}\,\omega_n^{1-\mu}
    (\text{\rm meas~} \Omega)^{\mu-\kappa}\,\,
    \|u\|_{L^s(\Omega)}
\end{equation}
for all $u\in L^s(\Omega)$.
\end{lemma}
\begin{proof}
This statement is a particular case of Lemma 7.12 in
\cite{trudinger}.
\end{proof}

\subsection{Singular integrals}
The Calderon-Zygmund singular integral operators are defined as
follows
$$
\mathbf S u(x)=\lim\limits_{\epsilon\to 0}\int_{|x-y|\geq \epsilon}
K(x, x-y) u(y)\, dy, \quad x,y\in \mathbb R^d, \quad d\geq 1.
$$
Here the Calderon-Zygmund kernel $K$ satisfies the conditions
$$
K(x,z)=\frac{1}{|z|^d}\, \omega\big(x, \frac{z}{|z|}\,\big), \quad
\omega\in L^\infty (\mathbb R^d\times \mathbb S^{d-1}),\quad
\int_{\mathbb S^{d-1}}\omega(x, \xi)\, d S^{d-1}(\xi)=0.
$$
It is known that the the operator $\mathbf S: L^r(\mathbb R^d)\to
L^r(\mathbb R^d)$ is bounded for all $1<r<\infty$. Moreover, it
admits the estimate, see \cite{calderon},
\begin{equation*}
    \|\mathbf S u\|_{L^r(\mathbb R^d)}\leq c\frac{r^2}{r-1}\| u\|_{L^r(\mathbb
    R^d)}, \quad  u\in L^r(\mathbb
    R^d),
\end{equation*}
where $c$ is independent of $r$ and $u$. The typical example of the
Calderon-Zygmund singular integral is the Riesz operator
$$
\mathbf R_j u=\partial_j (-\Delta)^{-1/2} u.
$$
In particular it satisfies the inequality
\begin{equation}\label{uruk6}
    \|\mathbf R_i u\|_{L^r(\mathbb R^d)}\leq c\frac{r^2}{r-1}\| u\|_{L^r(\mathbb
    R^d)}, \quad  u\in L^r(\mathbb
    R^d).
\end{equation}
The operators $\mathbf R_{ij}=\mathbf R_i\mathbf R_j$ are not
Calderon-Zygmund operators. However, they are products of such
operators and admit the estimates
\begin{equation}\label{uruk7}
    \|\mathbf R_{ij}u\|_{L^r(\mathbb R^d)}\leq c\frac{r^4}{(r-1)^2}\| u\|_{L^r(\mathbb
    R^d)}, \quad  u\in L^r(\mathbb
    R^d), \quad 1\leq i,j\leq d.
\end{equation}




\section{Appendix. Proof of Proposition \ref{lena53}}\label{app9}
\renewcommand{\theequation}{9.\arabic{equation}}
Note that conditions \textbf{C.1}-\textbf{C.3} and equality
$\rho=\zeta\varrho$ imply the estimates

\begin{equation}\label{lena55}
\text{ess}\sup\limits_{t\in (0,T)}\int_{\mathbb R^2} \rho\ln (2+
\rho)\, dy\leq c_0.
\end{equation}

\begin{equation}\label{lena56}
\int_{G\times (0,T)}\, \big(\, (-\Delta)^{-1/4}\rho \big)^2dydt\leq
c.
\end{equation}
Here
\begin{equation*}
   (-\Delta)^{-1/4}\rho(y,t)=c_{1/4}\int_G\frac{1}{|y-z|^{3/2}}\,
    \rho(z,t)\, dz.
\end{equation*}

The further considerations are based on the following lemma.
\begin{lemma}\label{lena20}
Let $\rho$ and $\mathbf W$ satisfy all conditions of Proposition
\ref{lena53}. Then for every integer $n\geq 2$, we have
\begin{equation}\label{lena60}
    \int_{\mathbb R^3}\rho |\mathbf W|^n\, dydt\leq c,
\end{equation}
where $c$ depends on $K$, $n$, and constants in \eqref{lena55},
\eqref{lena56}.
\end{lemma}

\begin{proof} Since $\mathbf W(\cdot,t)$ is compactly supported in $G$, it follows from
the embedding theorem that
\begin{equation}\label{lena22}
    \int_{\mathbb R^2}|\mathbf W(y,t)|^r\, dy=\int_{G}|\mathbf W(y,t)|^r\,
    dy\leq
c(r)\|\mathbf W(\cdot,t)\|_{W^{1,2}(G)}^r\leq c(r)
\end{equation}
for every $r\in (1,\infty)$ and a.e. $t\in\mathbb R$.  Now set
\begin{equation}\label{lena23}
\varphi=|\mathbf W|^{n-1}|\nabla\mathbf W|, \quad |\nabla |\mathbf
W|^n|=n|\mathbf W|^{n-2}|\nabla \mathbf W^\top \mathbf W|\leq
n\varphi.
\end{equation}
Estimate \eqref{lena22} and the H\"{o}lder inequality imply the
estimate
\begin{equation}\label{lena24}\begin{split}
\|\varphi(t)\|_{L^s(\mathbb R^2)}=\||\mathbf W|^{n-1}
(t)|\nabla\mathbf W|(t)\|_{L^s(\mathbb R^2)}\leq\\
\||\mathbf W(t)|^{n-1}\|_{L^{\frac{2s}{2-s}}(\mathbb R^2)}\,
\||\nabla\mathbf W(t)|\|_{L^2(\mathbb R^2)}\leq c,
\end{split}\end{equation}
which holds for every $s\in [1,2)$ and a.e. $t\in \mathbb R$. It
follows from identity \eqref{uruk2} in Lemma \ref{uruk1} that
\begin{equation*}
|\mathbf W(y,t)|^n=\frac{1}{4\pi}\int_ G\frac{\nabla|\mathbf
W|^n(z,t)\cdot (y-z)}{|y-z|^2}\,dz, \quad y\in \mathbb R^2
\end{equation*}
We thus get
\begin{equation}\label{lena25}
|\mathbf W(y,t)|^n\leq c\int_{\mathbb
R^2}\frac{1}{|y-z|}\,\varphi(z,t)\,dz= c
(-\Delta)^{-1/2}\varphi(y,t).
\end{equation}
Recall that $\varphi$ is compactly supported in $G\times (0,T)$. We
have
$$
(-\Delta)^{-1/2}\varphi=(-\Delta)^{-1/4}(-\Delta)^{-1/4}\varphi,
\quad (-\Delta)^{-1/4}\varphi=c_{1/4}\int_{\mathbb
R^2}\frac{\varphi(z,t)}{|y-z|^{3/2}}\, dz.
$$
Since $G$ is bounded, there is a disc $B_R=\{|y|\leq R\}$ such that
$G\Subset B_R$.  Note that the integral operator
$$
(-\Delta)^{-1/4}\varphi=c_{1/4}\int_{\mathbb
R^2}\frac{\varphi(z,t)}{|y-z|^{3/2}}\,
dz=c_{1/4}\int_{B_{2R}}\frac{\varphi(z,t)}{|y-z|^{3/2}}\, dz
$$
meets all requirements of Lemma \ref{uruk4} with $\Omega=B_{2R}$,
$\mu=1/4$. Now set $r=3$, $s=2$. We have $\kappa=2^{-1}-3^{-1} <
4^{-1}$ Applying  inequality \eqref{uruk5} in Lemma \ref{uruk4} we
arrive at the estimate
\begin{equation}\label{inna26}
    \|(-\Delta)^{-1/4}\varphi(t)\|_{L^3(B_{2R})}\leq c\text{~~for
    a.e.~~} t\in \mathbb R.
\end{equation}
Since $\varphi(t)$ is supported in $G\subset B_R$, we have
\begin{equation*}
|(-\Delta)^{-1/4}\varphi(t)|\leq c|y|^{-3/2}\|\varphi\|_{L^1(G)}\leq
c|y|^{-3/2} \text{~~in~~} \mathbb R^2\setminus B_{2R}.
\end{equation*}
It follows from this and \eqref{inna26} that
\begin{equation*}
\|(-\Delta)^{-1/4}\varphi(t)\|_{L^2(\mathbb R^2)}\leq c\text{~~for
    a.e.~~} t\in \mathbb R.
\end{equation*}
Since $\varphi(t)$ vanishes outside of the interval $(0,T)$, we have
\begin{equation*}
\|(-\Delta)^{-1/4}\varphi\|_{L^2(\mathbb R^3)}\leq c.
\end{equation*}
Combining this result with estimates \eqref{lena25} and estimate
\eqref{lena56}     we finally obtain the desired estimate
\begin{eqnarray*}
\int_{\mathbb R^3}\rho |\mathbf W|^n\, dydt &\leq & c\int_{\mathbb
R^3}\rho  (-\Delta)^{-1/2}\varphi\, dydt\\ &=& c\int_{\mathbb
R^3}\rho (-\Delta)^{-1/4}(-\Delta)^{-1/4}\varphi\, dydt\\
&=&c\int_{\mathbb R^3}(-\Delta)^{-1/4}\rho (-\Delta)^{-1/4}\varphi\,
dydt\leq c.
\end{eqnarray*}
\end{proof}

We are now in a position to complete the proof of Proposition
\ref{lena53}. We have
\begin{eqnarray}\label{lena30}
\rho |\mathbf W|^2\ln(2+\rho |\mathbf W|^2)^{1/2}&\leq & c\rho
\ln(2+\rho )^{1/2}|\mathbf W|^2+\\\notag &&c \rho|\mathbf
W|^2\ln(2+|\mathbf W|^2)^{1/2}.
\end{eqnarray}
It is easily seen that
\begin{eqnarray*}
\rho \ln(2+\rho )^{1/2}|\mathbf W|^2&\leq&\rho \ln(2+\rho
)+\rho|\mathbf W|^4,\\
\rho|\mathbf W|^2\ln(2+|\mathbf W|^2)&\leq& c\rho |\mathbf
W|^2+c\rho |\mathbf W|^3. \end{eqnarray*} Substituting these
inequalities into \eqref{lena30} we arrive at the estimate
\begin{equation*}
\rho |\mathbf W|^2\ln(2+\rho |\mathbf W|^2)^{1/2}\leq c\rho
\ln(2+\rho )+c\rho\sum_{k=2}^4 |\mathbf W|^3.
\end{equation*}
Integrating both the  sides of this inequality over $\mathbb R^3$
and applying estimates \eqref{lena55} and \eqref{lena60} we obtain
the desired estimate \eqref{lena21}. This completes the proof of
Proposition \ref{lena53}.













\section{Appendix. Proof of Lemma \ref{olga27}}\label{hurma}
\renewcommand{\theequation}{10.\arabic{equation}}
 By virtue of  \eqref{rita16a}, the function $\rho$ satisfies the
integral identity
\begin{equation*}
\int_{\mathbb R^3}\big(\rho \partial_t \psi+\rho \overline{\mathbf
w}\cdot \nabla \psi +\varrho\,\mathbf{h}\,\psi\,\big)\, y_1 dydt
 =0
\end{equation*} for every $\psi\in C^1_{\rm loc}(\mathbb R^3)$. It
follows that it satisfies the first order differential equation
\begin{equation*}
  y_1  \partial_t\rho +\text{~div~}(\rho y_1\overline{\mathbf
  w})=y_1\rho \mathbf h,
\end{equation*}
which is understood in the sense of distribution. Note that $\rho$
and $\mathbf w$ are compactly supported in $G\times (0,T)$.
Moreover, Proposition \ref{rita12} implies that
\begin{equation*}
    0<R^{-1}\leq  y_1\leq R,
\end{equation*}
where $R$ depends only on the compact set $K$. Hence we may rewrite
equation for $\rho$ in the form
\begin{equation}\label{olga31}
    \partial_t\rho +\text{~div~}(\rho \overline{\mathbf
  w})=\rho\big(\, \nabla\zeta \cdot \overline{ \mathbf w}-\frac{1}{y_1} w_1\big).
\end{equation}
Here we use the relation $\mathbf
h=\nabla\zeta\cdot\overline{\mathbf w}$. ~Since $\rho\in
L^\infty(\mathbb R; L^3(\mathbb R^2))$ and $\overline{\mathbf w}\in
L^2(\mathbb R;L^r(\mathbb R^2))$, the right hand side of this
equation belongs to the class $L^2(\mathbb R; L^q(\mathbb R^2))$ for
every $q\in [1,3)$. Next, the vector field $\overline{\mathbf
w}$~belongs to the class $L^2(\mathbb R; W^{1,2}(\mathbb R^2))$.
Moreover, the functions $\rho$ and $\overline{\mathbf w}$ are
compactly supported in $\mathbb R^3$. Hence we may apply the
renormalization procedure to to obtain the equation for
$\varphi(\rho)$.
\begin{equation}\label{olga32}
    \partial_t\varphi  +\text{~div~}(\varphi \overline{\mathbf
  w})=\boldsymbol\pi.
\end{equation}
where
\begin{equation*}
    \boldsymbol{\pi}=\frac{\delta\rho}{1+\delta\rho}\big(\,
    \nabla\zeta\cdot \overline{\mathbf w}-\frac{1}{y_1}
    w_1-\text{\rm ~div~}\overline{\mathbf w}\,\big)+\varphi\text{\rm ~div~}\overline{\mathbf
    w}.
\end{equation*}

Equation \eqref{olga32} is equivalent to the system of integral
identities
\begin{equation}\label{olga33}
   \int_{\mathbb R^3}\big(\,\varphi \partial_t\psi +\varphi \overline{\mathbf
  w})\cdot\nabla\psi+\boldsymbol\pi\,\psi\,\big)\, dydt,
\end{equation}
where $\psi\in C^1_{\rm loc}(\mathbb R^3)$ is an arbitrary function.
Now choose an arbitrary vector field $\boldsymbol{\upsilon}\in
C^\infty_0(\mathbb R^3)$. Substituting the test function
$\psi=\Delta^{-1} \text{~div~}\boldsymbol{\upsilon}$ into identity
\eqref{olga33} and noting that
$$
(\Delta^{-1} \text{~div~})^\top=-\nabla\Delta^{-1}
$$
we arrive at the integral identity
\begin{equation*}
\int_{\mathbb R^3}\big(\,
(\nabla\Delta^{-1}\varphi)\cdot\partial_t\boldsymbol{\upsilon}+
\nabla \Delta^{-1}\text{~div}(\varphi\overline{\mathbf
w})\cdot\boldsymbol\upsilon +\nabla\Delta^{-1}\boldsymbol\pi\cdot
\boldsymbol\upsilon)\, dydt=0.
\end{equation*}
Since $\boldsymbol{\upsilon}$ is an arbitrary smooth compactly
supported vector field, it follows from this that
\begin{equation}\label{olga37}
\partial_t(\nabla\Delta^{-1}\varphi)=-\nabla \Delta^{-1}\text{~div}(\varphi\overline{\mathbf
w})+\nabla\Delta^{-1}\boldsymbol\pi.
\end{equation}
Notice that
$$
\varphi\in L^\infty(\mathbb R; L^r(\mathbb R^2)), \quad
\overline{\mathbf w}\in L^2(\mathbb R;L^r(\mathbb R^2)),\quad
1<r<\infty,
$$
which yields
$$
\varphi\overline{\mathbf w}\in  L^2(\mathbb R;L^r(\mathbb R^2)),
\quad \boldsymbol\pi\in  L^2(\mathbb R;L^q(\mathbb R^2)).
$$
for all $r\in (1,\infty)$ and $q\in (1,2)$.  From this and estimates
\eqref{uruk5}, \eqref{uruk7} for potential and singular integrals in
Section \ref{uruk} we conclude that
\begin{eqnarray}\label{olga34}
\nabla \Delta^{-1}\text{~div}(\varphi\overline{\mathbf w})\in
L^2(\mathbb R; L^r(\mathbb R^2)), && \nabla^2\Delta^{-1}\varphi\in
L^\infty(\mathbb R;L^2(\mathbb R^2)), \\\notag \nabla\Delta^{-1}
\varphi\in L^\infty(\mathbb R; C(\mathbb R^2)),&&
\nabla\Delta^{-1}\boldsymbol\pi\in L^2(\mathbb R; L^r(\mathbb R^2)),
\end{eqnarray}
for all $r\in (1,\infty)$. In particular, we have
\begin{equation}\label{olga35}
\partial_t(\nabla\Delta^{-1}\varphi)\in  L^2(\mathbb R;L^r(\mathbb
R^2))\text{~~for all~~} r\in (1,\infty).
\end{equation}
On the other hand, we have
\begin{equation}\label{olga36}\begin{split}
\rho\in L^\infty(\mathbb R, L^3(\mathbb R^2)), \quad
\rho\overline{\mathbf w}\in L^2(\mathbb R; L^q(\mathbb R^2)), \\
\varrho \overline{\mathbf w}\otimes \overline{\mathbf w}\in
L^1(\mathbb R; L^q(\mathbb R^2))
\end{split}\end{equation}
for all $q\in (1,3)$.  Hence we may substitute the test function
$\overline{\boldsymbol\varsigma}=\nabla \Delta^{-1}\varphi$ into
integral identity \eqref{rita16}. Using formula \eqref{olga37} for
$\partial_t\overline{\boldsymbol\varsigma}=\partial_t\nabla\Delta^{-1}\varphi$
we arrive at the identity
\begin{equation}
\sum_{i=1}^{5} I_i=\int_{\mathbf R^3}\zeta p(\varrho)\varphi\,
y_1dydt.
\end{equation}
It remains to note that desired estimate \eqref{olga28} follows from
this and the obvious relations
$$
p_2(\rho)\equiv \zeta p_2(\zeta\varrho)\leq \zeta p(\varrho).
$$
This completes the proof of Lemma \ref{olga27}







\section{Appendix. Proof of Proposition \ref{olga43ab}}\label{sliva}
\renewcommand{\theequation}{11.\arabic{equation}}
\setcounter{equation}{0} Throughout of this section we deal with the
functions $\rho:\mathbb R^2\to\mathbb R^+$ and vector fields
$\mathbf W=(W_1,W_3):\mathbb R^2\to \mathbb R^2$ satisfying the
following condition.

\begin{condition}\label{sliva38}
The functions $\rho$ and $\mathbf W$ are compactly supported in $G$.
They admit the estimate
\begin{equation}\label{sliva39}\begin{split}
\int_{\mathbb R^2}\rho|\ln\rho|\, dy+    \int_{\mathbb R^2}
e_1(\rho)\, dy\leq c_1<\infty, \quad
    \|\mathbf W\|_{W^{1,2}(\mathbb R^3)}\leq 1,\\
0<c_1^{-1}\leq y_1\leq c_1\text{~~ on the supports of~~} \rho
\text{~~and~~} \mathbf W.
\end{split}\end{equation}
\end{condition}

\begin{remark}\label{sliva40}
Under the above assumptions,
\begin{equation}\label{sliva41}
\int_{\mathbb R^2}\rho^{1+\varepsilon}\, dy\leq c
\,\varepsilon^{-1}.
\end{equation}
Here $c$ depends only on $c_1$ and $G$.  Indeed, set
$\lambda=\varepsilon \rho$. We have
$$
\lambda ^{1+\varepsilon}\ln(1+\lambda)=\varepsilon^{1+\varepsilon}
\rho^{1+\varepsilon}\ln(1+\varepsilon\rho)\leq c e_1(\rho),
$$
which yields
\begin{equation}\label{sliva42}
\int_{\mathbb R^2} \lambda^{1+\varepsilon}\ln(1+\lambda)\, dy\leq c.
\end{equation}
Since $\rho$ and $\lambda$ are supported in $G$, we also have
$$
\lambda^{1+\varepsilon}\leq c\big(\,
\lambda^{1+\varepsilon}\ln(1+\lambda)+\boldsymbol{1}_G\,\big).
$$
From this and \eqref{sliva42} we obtain the inequality
$$
\int_{\mathbb R^2}\lambda^{1+\varepsilon}\, dy\leq c.
$$
which obviously yields \eqref{sliva41}
\end{remark}
Recall the formulation  of Proposition \ref{olga43ab}.
\begin{proposition}\label{sliva43} Let $\rho$ and $\mathbf W$ satisfy
Condition \ref{sliva38}. Then there is a constant $c$, depending on
$c_1$ and $G$ such that
\begin{eqnarray}\label{sliva44}
\|\rho \mathbf W\|_{L^{1+\frac{\varepsilon}{2}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^2}, & & \|\rho |\mathbf
W|^2\|_{L^{1+\frac{\varepsilon}{2}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^3},\\
\label{sliva45} \|\nabla\Delta^{-1}\text{\rm ~div~}(\varphi \mathbf
W)\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^6},&&\|\nabla^2\Delta^{-1}\varphi
\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^5},\\\label{sliva46}
\|\nabla\Delta^{-1}\boldsymbol\pi^*\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}\leq \frac{c}{\varepsilon^{3/2}}, &&
\|\nabla\Delta^{-1}\varphi\|_{L^\infty(\mathbb R^2)}\leq {c}.
\end{eqnarray}
Here
\begin{equation}\label{sliva46pi}
\varphi=\ln(1+\delta \rho), \quad
\boldsymbol\pi^*=\frac{\delta\rho}{1+\delta\rho}\big(\,
    \nabla\zeta\cdot \mathbf W-\frac{1}{y_1}
    W_1-\text{\rm div~}\mathbf W\,\big)+\varphi\text{\rm ~div~}\mathbf W.
\end{equation}
\end{proposition}
 The proof is based on the two technical lemmas.

\begin{lemma}\label{sliva49} Under the assumptions of Proposition
\eqref{sliva43}, the function $\varphi$ admits the estimate
\begin{equation}\label{sliva50}
    \|\varphi\|_{L^\gamma(\mathbb R^2)}\leq c \gamma\text{~~for
    all~~} \gamma\in [2,\infty).
\end{equation}
Here $c$ depends only on $G$ and the constant $c_1$ in Condition
\ref{sliva38}.
\end{lemma}

\begin{proof}
Note that the function $\ln(1+s)^\gamma/(1+s)$, $s\in [0,\infty)$,
takes the maximum at the point $s=e^\gamma-1$, where it equals
$\gamma^\gamma\, e^{-\gamma}$. Since $\varphi(\rho)$ and $\rho$ are
supported in $G$, we have
\begin{eqnarray*}
\int_{\mathbb R^2}\varphi^\gamma\, dy
&=&\int_{G}\varphi^\gamma\,dy\leq \int_G \ln(1+\rho)^\gamma\,dy\\
&\leq &\gamma^\gamma\, e^{-\gamma}\int_G (1+\rho)\, dy \leq c\,
\gamma^\gamma\, e^{-\gamma},
\end{eqnarray*}
which obviously yields \eqref{sliva50}
\end{proof}

\begin{lemma}\label{sliva52}
Under the assumptions of Proposition \eqref{sliva43}, the vector
$\mathbf W$ admits the estimate
\begin{equation}\label{sliva53}
    \|\mathbf W\|_{L^\gamma(\mathbb R^2)}\leq c \gamma\text{~~for
    all~~} \gamma\in [2,\infty).
\end{equation}
Here $c$ depends only on $G$.
\end{lemma}
\begin{proof}
It follows from identity \eqref{uruk2} in Lemma \ref{uruk1} that
\begin{equation*}
    \mathbf W=\mathcal V_i\partial_i \mathbf W,\text{~~where~~}\mathcal V_i
    f=
    c\int_{\mathbb R^2}\frac{y_i-z_i}{|y-z|^2} f(z)\, dz.
\end{equation*}
The potential operator $\mathcal V_i$ meets all requirements of
Lemma \ref{uruk4} with $\Omega=G$, $n=2$, and $\mu=1/2$. It follows
from estimate \eqref{uruk5} in Lemma \ref{uruk4} with $s=2$ and
$r=\gamma$ that
\begin{equation*}
\|\mathcal V_i \nabla\mathbf W\|_{L^\gamma(G)}\leq
c\Big(\frac{1-\kappa}{\frac{1}{2}-\kappa}\Big)^{1-\kappa}\,
\|\nabla\mathbf W\|_{L^2(G)}\leq
c\Big(\frac{1-\kappa}{\frac{1}{2}-\kappa}\Big)^{1-\kappa}, \quad
\kappa=\frac{1}{2}-\frac{1}{\gamma}.
\end{equation*}
Thus we get
\begin{eqnarray}\label{sliva48}
\|\mathbf W\|_{L^\gamma(G)}&\leq&\|\mathcal V_i \nabla\mathbf
W\|_{L^\gamma(G)}\\\notag&\leq&
c\Big(\frac{1-\kappa}{\frac{1}{2}-\kappa}\Big)^{1-\kappa} =
c\Big(\frac{2+\gamma}{2}\Big)^{\frac{1}{2}+\frac{1}{\gamma}}\leq
c\,\gamma,
\end{eqnarray}
and the lemma follows.
\end{proof}
 We split the rest of proof of Proposition \ref{sliva43} into three steps.
 \\[2ex]\noindent \textbf{ Step 1.
Proof of estimates \eqref{sliva44}.} It follows from the  H\"{o}lder
inequality that the estimate
\begin{equation*}
\int_{\mathbb R^2} \rho^{1+\frac{\varepsilon}{2}}\, |\mathbf
W|^{1+\frac{\varepsilon}{2}}\, dy \leq\Big(\,\int_{\mathbb R^2}
\rho^{\alpha(1+\frac{\varepsilon}{2})}\,
dy\,\Big)^{\frac{1}{\alpha}}\,\Big(\,\int_{\mathbb R^2} |\mathbf
W|^{\frac{\alpha}{\alpha-1}(1+\frac{\varepsilon}{2})}\,
dy\,\Big)^{\frac{\alpha-1}{\alpha}}
\end{equation*}
holds for every $1<\alpha<\infty$. Now set
\begin{equation*}
\alpha=\frac{1+\varepsilon}{1+\frac{\varepsilon}{2}}=\frac{2+2\varepsilon}{2+\varepsilon},
\quad \frac{\alpha}{\alpha-1}=\frac{2+2\varepsilon}{\varepsilon}.
\end{equation*}
By virtue of Remark \eqref{sliva40}, we have
\begin{equation*}
\Big(\,\int_{\mathbb R^2}
\rho^{\alpha(1+\frac{\varepsilon}{2})}\,dy\Big)^{\frac{1}{\alpha}}
=\Big(\,\int_{\mathbb R^2}
\rho^{1+\varepsilon}\,dydy\,\Big)^{\frac{1}{\alpha}}\leq
c\Big(\frac{1}{\varepsilon}\Big)^{\frac{2+\varepsilon}{2+2\varepsilon}}.
\end{equation*}
Noting that
\begin{equation*}
\frac{\alpha}{\alpha-1}(1+\frac{\varepsilon}{2})=(2+\varepsilon)
\frac{1+\varepsilon}{\varepsilon} :=\gamma>2,
\end{equation*}
we obtain
\begin{equation*}
    \|\rho\mathbf W\|_{L^{1+\frac{\varepsilon}{2}}(\mathbb R^2)}\leq
    c\Big(\frac{1}{\varepsilon}\Big)^{\frac{2+\varepsilon}{2+2\varepsilon}}\,\,\|\mathbf
    W\|_{L^\gamma(\mathbb R^2)}.
\end{equation*}
From this and inequality \eqref{sliva53} in Lemma \ref{sliva52} we
get desired estimate \eqref{sliva44} for $\rho\mathbf W$. Repeating
these arguments we obtain
\begin{equation*}
    \|\rho|\mathbf W|^2\|_{L^{1+\frac{\varepsilon}{2}}(\mathbb R^2)}\leq
    c\Big(\frac{1}{\varepsilon}\Big)^{\frac{2+\varepsilon}{2+2\varepsilon}}\,\,\||\mathbf
    W|^2\|_{L^\gamma(\mathbb R^2)}\leq
    \frac{c}{\varepsilon}\,\|\mathbf W\|_{L^{2\gamma}(G)}^2.
\end{equation*}
Applying estimate \eqref{sliva53} with $\gamma$ replaced by
$2\gamma$ we arrive at the estimate \eqref{sliva44} for
$\rho|\mathbf W|^2$. This completes the proof of inequalities
\eqref{sliva44}.
\\[2ex]\noindent \textbf{ Step 2.
Proof of estimates \eqref{sliva45}.} Note that the matrix-value
operator $\nabla\Delta^{-1}\text{~div~}$ admits the representation
$$
\big(\,\nabla\Delta^{-1}\text{~div~}\,\big)_{ij}=\mathbf R_i\mathbf
R_j=\mathbf R_{ij},
$$
where $\mathbf R_i$ is the Riesz operator. It follows from the
Calderon-Zygmund estimate \eqref{uruk7} that
\begin{equation}\label{sliva65}
    \|\mathbf R_{ij}f\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^4}\,\|f\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}\text{~~for all~~}  f\in
L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2).
\end{equation}
In particular, we have
\begin{equation}\label{sliva60}
    \|\nabla\Delta^{-1}\text{~div~}(\varphi \mathbf W)\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2)}\leq
\frac{c}{\varepsilon^4}\,\|\varphi\mathbf
W\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2)}.
\end{equation}
The Cauchy inequality implies
\begin{equation*}
\|\varphi\mathbf W\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}\leq
\|\varphi\|_{L^{2\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}\,\,\|\mathbf
W\|_{L^{2\frac{2+\varepsilon}{\varepsilon}}(\mathbb R^2)}.
\end{equation*}
From this and  estimates \eqref{sliva50}, \eqref{sliva53} with
$\gamma=2(2+\varepsilon)\varepsilon^{-1}$ we obtain
\begin{equation*}
\|\varphi\mathbf W\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}\leq \frac{c}{\varepsilon^2}
\end{equation*}
Substituting this inequality into \eqref{sliva60} we arrive at the
first estimate in \eqref{sliva45}. In order to prove the second,
note that $\nabla^2\Delta^{-1}\varphi$ is the matrix valued func
tion with entries
$$
(\nabla^2\Delta^{-1}\varphi)_{ij}= \mathbf R_j\mathbf R_i \varphi.
$$
Applying inequality \eqref{sliva65} we arrive at the estimate
\begin{equation*}
\|\nabla^2\Delta^{-1}\varphi\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}\leq
\frac{c}{\varepsilon^4}\|\varphi\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(\mathbb
R^2)}.
\end{equation*}
From this and estimate \eqref{sliva50} with
$\gamma=(2+\varepsilon)\varepsilon^{-1}$ we obtain the second
estimate  \eqref{sliva45}.\\[2ex]
\textbf{Step 3. The proof of  estimates \eqref{sliva46}.} We begin
with the observation that
\begin{eqnarray*}
|\nabla\Delta^{-1}\boldsymbol\pi^*|&=&
c\Big|\int_G\frac{y-z}{|y-z|^2}\boldsymbol\pi^*(z)\, dz\Big|\\
&\leq& c \int_G\frac{1}{|y-z|}|\boldsymbol\pi^*(z)|\, dz.
\end{eqnarray*}
On the other hand, relation \eqref{sliva46pi} implies
\begin{equation*}
|\boldsymbol\pi^*|\leq c |\mathbf W|+|1+\varphi|\,
|\text{~div~}\mathbf W|.
\end{equation*}
We get
\begin{eqnarray}\label{sliva66}
|\nabla\Delta^{-1}\boldsymbol\pi^*|&\leq& c(-\Delta)^{-1/2}(|\mathbf
W|)+ c(-\Delta)^{-1/2}\big(|1+\varphi|\, |\text{~div~}\mathbf
W|\,\big)=\\\notag c\int_G\frac{1}{|y-z|}|\mathbf W|\, dz&+& c
\int_G\frac{1}{|y-z|}|1+\varphi|\, |\text{~div~}\mathbf W|\, dz.
\end{eqnarray}
It is easy to see that the potential $(-\Delta)^{-1/2}$ meets all
requirements of Lemma \ref{uruk4} with $n=2$, $\mu=1/2$, and
$\Omega=G$. Now set
$$
s=2, \quad r=\frac{2+\varepsilon}{\varepsilon}, \quad \kappa=
\frac{1}{2}-\frac{\varepsilon}{2+\varepsilon}.
$$
Applying estimate \eqref{uruk5} in Lemma \ref{uruk4} to the
potential $\Delta^{-1/2}(|\mathbf W|)$ and noting that
$$
\Big(\,
\frac{1-\kappa}{\mu-\kappa}\,\Big)^{1-\kappa}=\Big(\frac{2+3\varepsilon}{2\varepsilon}\,\Big)
^{\frac{2+3\varepsilon}{2\varepsilon}}\leq \frac{c}{\varepsilon},
$$
we arrive at the estimate for the first integral in the right hand
side of \eqref{sliva66}.
\begin{equation}\label{sliva68}
    \|(-\Delta)^{-1/2}(|\mathbf
    W|)\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(G)}\,\leq\,
    \frac{c}{\varepsilon}\,\|\mathbf W\|_{L^2(G)}\leq
    \frac{c}{\varepsilon}.
\end{equation}
The estimate of the second integral in \eqref{sliva66} is more
complicated. Choose the number $s$ satisfying the equalities
\begin{equation}\label{sliva69s}
\frac{2s}{2-s}=\frac{2+\varepsilon}{\varepsilon(1-\varepsilon)},
\quad
s=\frac{2(2+\varepsilon)}{2+3\varepsilon-2\varepsilon^2}\in(1,2).
\end{equation}
 Now set
$$
r=\frac{2+\varepsilon}{\varepsilon}, \quad \kappa
=\frac{1}{s}-\frac{1}{r}=\frac{2+\varepsilon-2\varepsilon^2}{2(2+\varepsilon)}<\frac{1}{2}.
$$
Recall that the potential operator $(-\Delta)^{-1/2}$ satisfies all
conditions of Lemma \ref{uruk4} with $n=2$, $\mu=1/2$, and
$\Omega=G$. For our choice $s$ and $r$, the quantity $\kappa<1/2$.
Hence we may apply estimate \eqref{uruk5} to obtain
\begin{equation*}\begin{split}
\|(-\Delta)^{-1/2}\big(|1+\varphi|\, |\text{~div~}\mathbf
W|\,\big)\|_{L^r(G)}\leq\\
c\Big(\frac{1-\kappa}{1/2-\kappa}\Big)^{1-\kappa}\,\||1+\varphi|\,
|\text{~div~}\mathbf W|\|_{L^s(G)}. \end{split}\end{equation*} It
follows from the expression for $\kappa$ that
\begin{equation*}
\Big(\frac{1-\kappa}{1/2-\kappa}\Big)^{1-\kappa}=
\Big(\frac{2+\varepsilon+2\varepsilon^2}{\varepsilon^2}
\,\Big)^{\frac{2+\varepsilon+2\varepsilon^2}{2(2+\varepsilon)}}\leq
\frac{c}{\varepsilon^2}.
\end{equation*}
Thus we get
\begin{equation}\label{sliva69}
\|(-\Delta)^{-1/2}\big(|1+\varphi|\, |\text{~div~}\mathbf
W|\,\big)\|_{L^r(G)}\leq\\
\frac{c}{\varepsilon^2}\,\|1+\varphi|\, |\text{~div~}\mathbf
W|\|_{L^s(G)}.
\end{equation}
Let us estimate the right hand side of this inequality. Set
\begin{equation*}
    \alpha=\frac{2}{s}, \quad \frac{\alpha}{\alpha-1}=\frac{2}{2-s}
\end{equation*}
It follows from the H\"{o}lder inequality that
\begin{eqnarray*}
\||1+\varphi|\, |\text{~div~}\mathbf W|\|_{L^s(G)}&\leq&
\|1+\varphi\|_{L^{\frac{\alpha s}{\alpha-1}}(G)}
\|\text{~div~}\mathbf W|\|_{L^{\alpha s}(G)},
\end{eqnarray*}
or equivalently
\begin{eqnarray}\label{sliva70}
\||1+\varphi|\, |\text{~div~}\mathbf W|\|_{L^s(G)}&\leq&
\|1+\varphi\|_{L^{\frac{2 s}{2-s}}(G)} \|\text{~div~}\mathbf
W|\|_{L^{2}(G)}\\\notag &\leq& c \|1+\varphi\|_{L^{\frac{2
s}{2-s}}(G)}.
\end{eqnarray}
Estimate \eqref{sliva69s} yields
\begin{eqnarray*}
\|1+\varphi\|_{L^{\frac{2 s}{2-s}}(G)}&\leq& c\big(\,1+\|
\varphi\|_{L^{\frac{2 s}{2-s}}(G)}\,\big)\\
&=& c\big(\,1+
\|\varphi\|_{L^{\frac{2+\varepsilon}{\varepsilon(1-\varepsilon)}}(G)}\,\big).
\end{eqnarray*}
Applying Lemma \ref{sliva49} with
$\gamma=(2+\varepsilon)/(\varepsilon(1-\varepsilon))$ we arrive at
the estimate
\begin{equation*}
\|1+\varphi\|_{L^{\frac{2 s}{2-s}}(G)}\leq\frac{c}{\varepsilon}.
\end{equation*}
From this and inequality \eqref{sliva70} we get the estimate
\begin{equation*}
\||1+\varphi|\, |\text{~div~}\mathbf W|\|_{L^s(G)}\leq
\frac{c}{\varepsilon}.
\end{equation*}
Substituting this inequality into \eqref{sliva69} and noting that
$r=(2+\varepsilon)/\varepsilon$ we  get the estimate
\begin{equation*}
\|(-\Delta)^{-1/2}\big(|1+\varphi|\, |\text{~div~}\mathbf
W|\,\big)\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(G)}\leq\\
\frac{c}{\varepsilon^3}.
\end{equation*}
Combining this result with estimate \eqref{sliva68} and recalling
inequality \eqref{sliva66} we finally obtain the desired estimate
\begin{eqnarray*}
\|\nabla\Delta^{-1}\boldsymbol\pi^*\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(G)}&\leq&
c\|(-\Delta)^{-1/2}\big(|(1+\varphi)\, \text{~div~}\mathbf
W|\,\big)\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(G)}\\&+&
c\|(-\Delta)^{-1/2}\big(\,|\mathbf
W|\,\big)\|_{L^{\frac{2+\varepsilon}{\varepsilon}}(G)} \,\,\leq
\,\,\frac{c}{\varepsilon ^3}.
\end{eqnarray*}
It remains to prove the second inequality in \eqref{sliva46}.
Arguing as before we obtain
\begin{equation*}
|\nabla\Delta^{-1}\varphi|\leq c(-\Delta)^{-1/2}\varphi=
c\int_G\frac{1}{|y-z|}\,\varphi(z)\, dz.
\end{equation*}
Obviously we have
\begin{equation*}
    (-\Delta)^{-1/2}\varphi\leq c\|\varphi\|_{L^3(G)}
\end{equation*}
Applying Lemma \ref{sliva49} with $\gamma=3$ we arrive at the
desired estimate
\begin{equation*}
|\nabla\Delta^{-1}\varphi|\leq c(-\Delta)^{-1/2}\varphi\leq
c\|\varphi\|_{L^3(G)}\leq c.
\end{equation*}
This completes the proof of estimate \eqref{sliva46} and Proposition
\ref{sliva43}.



\section{Appendix. Proof of Lemma  \ref{lita5} }\label{persik}
\renewcommand{\theequation}{12.\arabic{equation}}
\setcounter{equation}{0} First we prove the equi-integrability of
the sequence $\varrho_\varepsilon\mathbf u_\varepsilon$. To this
end, choose an arbitrary set $A\subset\Omega$ and $N$ such that
\begin{equation}\label{persik1}
    |A|:=\text{~meas~} A<1/3, \quad N>3.
\end{equation}
Fix $t\in (0,T)$ and set
$$
A_N=\{x\in A:\, \varrho_\varepsilon(x,t)\geq N\}\subset A.
$$
For every $\epsilon>0$ we have
\begin{eqnarray*}
\int_A \varrho_\varepsilon(t)|\mathbf u_\varepsilon(t)|\, dx&\leq&
\frac{1}{2\epsilon}\int_A \varrho_\varepsilon(t)\,
dx+\frac{\epsilon}{2}\int_A \varrho_\varepsilon(t)|\mathbf
u_\varepsilon(t)|^2\, dx\\\notag &=&
\frac{1}{2\epsilon}\Big(\int_{A\setminus A_N}
\varrho_\varepsilon(t)\, dx +\int_{ A_N} \varrho_\varepsilon(t)\,
dx\Big) +\frac{\epsilon}{2}\int_A \varrho_\varepsilon(t)|\mathbf
u_\varepsilon(t)|^2\, dx.
\end{eqnarray*}
It follows from estimates \eqref{lita2} and \eqref{lita3} that
$$
\int_{ A_N} \varrho_\varepsilon(t)\, dx\leq \frac{c}{\ln N}, \quad
\int_A \varrho_\varepsilon(t)|\mathbf u_\varepsilon(t)|^2\, dx \leq
c.
$$
Thus we get
\begin{equation}\label{persik2}
\int_A \varrho_\varepsilon(t)|\mathbf u_\varepsilon(t)|\, dx\leq
c\frac{1}{2\epsilon}\big (N|A|+ \frac{1}{\ln N}\big)+ c\epsilon.
\end{equation}
Introduce the function $\lambda: [3,\infty)\to \mathbb R^+$ defined
by the equality $\lambda(s)=s\ln s$. It has the smooth inverse
$\lambda^{-1}:[3\ln3,\infty)\to\mathbb R$ with the properties
$$
\frac{\ln s}{s}\,\lambda^{-1}(s)\to 1,\quad \ln(\lambda^{-1}(s))/\ln
s\to 1\text{~~as~~} s\to\infty.
$$
Set
$$
N=\lambda^{-1}\big(\,\frac{1}{|A|}\,\big),\quad
\epsilon=\frac{1}{\sqrt{\ln N}}.
$$
From this and inequality \eqref{persik2} we obtain the estimate
\begin{equation}\label{persik3}
\int_A \varrho_\varepsilon(t)|\mathbf u_\varepsilon(t)|\, dx\leq c
\ln \big(\,|A|^{-1}\,\big)^{-1/2}
\end{equation}
for $|A|\leq 1/3$ and for a.e. $t\in (0,T)$. Note that the constant
$c$ is independent of $t$. Inequality \eqref{persik3} implies the
weak type estimate for the momentum $\varrho_\varepsilon \mathbf
u_\varepsilon$. By virtue of estimates \eqref{lita2} and
\eqref{lita3}, we have
$$
\int_\Omega \varrho_\varepsilon(t) |\mathbf u_\varepsilon(t)|\, dx
\leq \frac{1}{2}\int_\Omega \varrho_\varepsilon(t)\,
dx+\frac{1}{2}\int_\Omega \varrho_\varepsilon(t) |\mathbf
u_\varepsilon(t)|^2\, dx\leq c_1,
$$
where $c_1\geq 1$ depends only of the constants in \eqref{lita2},
\eqref{lita3} and is independent of $t$.  From this and the
Chebyshev inequality we obtain that for every $m> 0$,
\begin{equation*}
\text{meas~}\{\varrho_\varepsilon(t)|\mathbf u_\varepsilon(t)|\geq
m\}\leq \frac{c_1}{m}.
\end{equation*}
Now set
\begin{equation}\label{persik4a}
B_m=\{\varrho_\varepsilon(t)|\mathbf u_\varepsilon(t)|\geq m\}.
\end{equation}
Inequality \eqref{persik3} with $A= B_m$ implies the weak type
estimate
\begin{equation}\label{persik4}
\int_{B_m}\varrho_\varepsilon(t)|\mathbf u_\varepsilon(t)|\,dx\leq c
\ln(m/c_1)^{-1/2}\text{~~for~~} m\geq 3c_1.
\end{equation}
Now we proceed as in the proof of Theorem \ref{lena11}. Fix $t\in
(0,T)$. Denote by $E(\lambda)$, $\lambda\geq 0$ the monotone
function defined by the relation
\begin{equation*}
E(\lambda)=\text{meas~}\,\big\{\,x\in
\Omega:\,\,\varrho_\varepsilon(x,t)|\mathbf u_\varepsilon(x,t)|\geq
\lambda\,\big\}.
\end{equation*}
It is the standard fact of calculus, that the equality
\begin{equation*}
    \int_{\Omega} f(\varrho_\varepsilon(t)| \mathbf
    u_\varepsilon(t)|)\,dx
    =-\int_{[0,\infty)}\, f(\lambda)\,\, d E(\lambda)
\end{equation*}
holds for every piecewise continuous function $f$ such that
$f(\varrho_\varepsilon(t)|\mathbf u_\varepsilon|)\in L^1(\Omega).$
Here $-dE(\lambda)$ is the Borel-Stieltjes measure associated with
the function $-E(\lambda)$. In particular, we have
\begin{equation}\label{persik5}
\int_{B_m}\varrho_\varepsilon|\mathbf u_\varepsilon|\,
dydt=-\int_{[m, \infty)} \lambda \, d E(\lambda).
\end{equation}
With this notation, the desired inequality
\begin{equation}\label{persik6}
    \|\varrho_\varepsilon |\mathbf u _\varepsilon|
    \ln (2+\varrho_\varepsilon |\mathbf u
    _\varepsilon|)^{\alpha}\|_{L^\infty(0,T; L^1(\Omega))}\leq c
\end{equation}
reads
\begin{equation}\label{persik7}
\int_{\Omega}\varrho_\varepsilon(t)|\mathbf
u_\varepsilon(t)|\ln(2+\varrho_\varepsilon(t)|\mathbf
u_\varepsilon(t)|)^\alpha\, dydt\equiv -\int_{[0, \infty)}
\lambda\ln(2+\lambda)^\alpha \, d E(\lambda)\leq c.
\end{equation}
Here $\alpha\in [0,1/4)$, $c$ is independent of $t$.

Introduce the monotone function $F$ and the corresponding
 Borel-Stieltjes measure $-dF$ defined by the equalities
$$
F(s)=-\int_{[s,\infty)}\lambda\, dE(\lambda), \quad
dF(\lambda)=\lambda\, dE(\lambda).
$$
Identity \eqref{persik7} and estimate \eqref{persik4}
 imply
\begin{equation}\label{persik8}
    F(s)\leq c \ln(s/c_1)^{-1/2}\text{~~for~~} s\geq 3c_1.
\end{equation}
Next note that
$$
\ln(1+\frac{2}{s})\leq \frac{2}{s}\leq \frac{2}{3}\leq \ln3\leq
\ln\big(\, \frac{s}{c_1^2}\,\big)=\ln s-\ln c_1\text{~~for~~} s\geq
3c_1^2,
$$
which yields
$$
\ln (2+s)=\ln s+\ln(1+\frac{2}{s})\leq 2\ln s-2\ln c_1= 2\ln\big(\,
\frac{s}{c_1}\,\big) \text{~~for~~} s\geq 3c_1.
$$
From this and \eqref{persik8} we conclude that
\begin{equation}\label{persik9}
    F(s)\leq c \ln(2+s)^{-1/2}\text{~~for~~} s\geq 3c_1.
\end{equation}
Since
$$
F(s)= -\int_{0,\infty}\lambda d E(\lambda)=\int_\Omega
\varrho_\varepsilon(t)|\mathbf u_\varepsilon(t)|\,dx\leq c,
$$
we obviously have
\begin{equation*}
    F(s)\leq c \ln(2+s)^{-1/2}\text{~~for~~} s\leq 3c_1^2.
\end{equation*}
Combining this result with \eqref{persik9} we finally obtain
\begin{equation}\label{persik10}
    F(s)\leq c \ln(2+s)^{-1/2}\text{~~for~~} s\geq 0.
\end{equation}
Here the constant $c$ depends only on the constant in energy
estimates \eqref{lita2}, \eqref{lita3} and is independent of $t$. It
is easy to see that
\begin{equation*}
-\int_{[0, \infty)} \lambda\ln(2+\lambda)^\alpha \, d E(\lambda)=
-\int_{[0, \infty)} \ln(2+\lambda)^\alpha \, d F(\lambda)
\end{equation*}
Therefore,  our task is to prove that the integral in the right hand
side exists and admits the effective estimate. Let us consider the
regularized integral
\begin{equation*}
    I(\epsilon,
    N)=-\int_{[0,N)}\,\,
  g(\lambda)  d F(\lambda), \quad g(\lambda)=\frac{\ln(2+\lambda)^\alpha}{1+\epsilon^2\lambda^2}
\end{equation*}
Note that  $g$ is bounded and continuously differentiable. On the
other hand, the total variation of $F$ is finite. Hence, we may
integrate by parts to obtain
\begin{equation}\label{lena42a}
    I(\epsilon,
    N)=\int_{[0,N)}\,\,g'(\lambda)\, F(\lambda)\,d\lambda+ g(0)
    F(0)-g(N)\lim\limits_{\lambda\to N-0}F(\lambda).
\end{equation}
We have
\begin{eqnarray}\label{persik11}
g'(\lambda)&=&g_0(\lambda)-\mathbf R(\lambda),\\\notag
    g_0(\lambda)&=&\alpha
    \ln(2+\lambda)^{\alpha-1}(2+\lambda)^{-1}(1+\epsilon^2\lambda^2)^{-1}\\\notag
\mathbf R(\lambda)&=&2\epsilon^2
\lambda\ln(2+\lambda)^{\alpha}(1+\epsilon^2\lambda^2)^{-2}
\end{eqnarray}
Since $\alpha\in (0,1/2)$, estimate \eqref{persik10} implies that
$$
|g'(\lambda) F(\lambda)|\leq c (1+\epsilon^2\lambda^2)^{-1}
$$
and hence $g' F\in L^1(\mathbb R^+)$. On the other hand, we have
\begin{equation*}
0\leq g(N) F(N)\leq c \ln(2+N)^{\alpha-1}(1+\epsilon N)^{-1}\to
0\text{~~as~~} N\to\infty.
\end{equation*}
Letting $N\to\infty$ in equality \eqref{lena42a} we conclude that
there exists the integral
\begin{eqnarray}\notag
-\int_{[0,\infty)}g(\lambda)\,
dF(\lambda)&=&\int_0^\infty\,\,g_0(\lambda)\,
F(\lambda)\,d\lambda\\\notag &-&\int_0^\infty\,\,\mathbf
R(\lambda)\, F(\lambda)\,d\lambda+ g(0) F(0).
\end{eqnarray}

Since $\mathbf R$ and $F$ are nonnegative, we have
\begin{equation}\label{persik12}
-\int_{[0,\infty)}g(\lambda)\,
dF(\lambda)\leq\int_0^\infty\,\,g_0(\lambda)\, F(\lambda)\,d\lambda+
g(0) F(0).
\end{equation}
It follows from \eqref{persik10}  that $F(0)\leq c$. Next, estimate
\eqref{persik10} yields
\begin{eqnarray}\notag
\int_0^\infty g_0 F\, d\lambda&\leq &\alpha\int_0^\infty
\ln(2+\lambda)^{\alpha-1}(2+\lambda)^{-1}\,F(\lambda)d\lambda\\\notag
&\leq &\int_0^\infty
\ln(2+\lambda)^{\alpha-1-1/2}(2+\lambda)^{-1}\,d\lambda
 \leq    c(\alpha).
\end{eqnarray}
It follows from this and \eqref{persik12} that
\begin{equation}\label{persik13}
-\int_{[0,\infty)}g_0(\lambda)\, dF(\lambda)\leq c(\alpha)+\ln
2^\alpha c\leq c.
\end{equation}
Recall that $-dF$ is a nonnegative Borel measure. Note that $g_0$ is
an increasing function of $\epsilon$ and
$$
g_0(\lambda)\,\nearrow \,\ln(2+\lambda)^\alpha \text{~~as~~}
\epsilon\, \searrow\,  0.
$$
Letting $\epsilon\to 0$ in \eqref{lena45} and applying the Fatou
theorem we conclude that there exists the integral
\begin{equation*}
-\int_{[0,\infty)}\ln(2+\lambda)^\alpha\, dF(\lambda)\equiv
-\int_{[0,\infty)}\lambda \ln(2+\lambda)d E(\lambda)\leq c.
\end{equation*}
This completes the proof of Lemma \ref{lita5}.












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