text:
\section{Introduction}

A spray is a process in which drops are dispersed in a gas. Sprays can help people to distribute material over a cross-section and to generate liquid surface area.
 There are various applications in different fields, the examples include but are not  limited to  fuel injectors for gasoline and Diesel engines, atomizers for jet engines, atomizers for injecting heavy fuel oil into combustion air in steam boiler injectors, and rocket engine injectors.
 It also has a big impact on crop yields, plant health, efficiency of pest control and of course, profitability. Here we will focus on a model of a moderately thick spray. In general, we refer the reader to \cite{Caflisch} for the physical background.

For a moderately thick spray, we can assume that the volume fraction occupied by the droplets is small enough to be
neglected. Thus, we are able to apply the Vlasov-Boltzmann equation to model the liquid phase, which can be performed by the use of a particle density function. In particular, a function
$f(t,x,\xi)$ denotes a number density of droplets of which at time $t$ and physical position $x$ with velocity $\xi$, which is a solution of the following Vlasov-Boltzmann equation
\begin{equation}
\label{vlasov equation}
f_t+\xi\cdot\nabla_x f+\nabla(f\mathfrak{F})=Q(f,f),\end{equation}
where $\mathfrak{F}$ is an acceleration resulting from the drag force exerted by the gas,
and $Q(f,f)$ is an operator taking into account complex phenomena happening at the level of the droplets
(collisions, coalescences, breakup).

The incompressible Navier-Stokes equations could be used to describe the motion of gas
when instigating  the transport of sprays in the upper airways of the human lungs:
\begin{equation}
\begin{split}&
\label{NS equation}
u_t+u\cdot\nabla u+\nabla P-\mu\Delta u=F_e,
\\&\Dv u=0
\end{split}
\end{equation}
where $u$ is the velocity of the  gas, and $F_e$ denotes the external force with \begin{equation}
\label{external force}
F_e=-c\int_{\R^3}f\mathfrak{F}\,d\xi,
\end{equation}
$\mathfrak{F}$ denotes the acceleration, it is given by
\begin{equation*}
\label{accelation}
\mathfrak{F}=-\frac{9\mu}{2\rho}\frac{\xi-u}{r^2},
\end{equation*} $\mu$ is the viscosity constant of the incompressible Navier-Stokes equations, $\rho$ is the density of gas, $r$ is the radius of the droplet. We can assume that
$\frac{9\mu}{2\rho r^2}=1$ in the whole paper, thus
\begin{equation}
\label{accelation-}
\mathfrak{F}=-(\xi-u).
\end{equation}

\noindent One of the typical forms of the collision kernel is given by
\begin{equation}
\label{operator of collision}
Q(f,f)=-\lambda f(t,x,\xi)+\lambda\int_{\R^3} T(\xi,\xi')f(t,x,\xi')\,d\xi'.
\end{equation}
The constant $\lambda>0$ is  the breakup frequency.
 The kernel $T(\xi,\xi^\prime)$ is the probability of a change with respect to velocity from $\xi^\prime$ to $\xi$, and $\xi$ is the velocity of individuals before the collision while $\xi^\prime$ is the velocity immediately after the collision.
  Given that a reorientation occurs, the probability function $T(\xi,\xi^\prime)$ is a non-negative function and after normalization we have
\begin{equation}\label{E1.4}
\int_{\mathbb{R}^3}T(\xi,\xi^\prime)d\xi=1.
\end{equation}

By the acceleration of the Vlasov-Boltzmann equations and the external force of the incompressible Navier-Stokes equations, the above equations \eqref{vlasov equation}-\eqref{operator of collision} can be coupled with each other. As a result, we arrive at
a  model for a moderately thick spray, namely the Navier-Stokes-Vlasov-Boltzmann equations:
\begin{equation}\label{E1.1}
\left\{\begin{array}{l}
\partial_t f
  +
  \xi\cdot \nabla_x f+{\rm div}_\xi((u-\xi)f)=-\lambda f+\lambda\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f(t,x,\xi^\prime)d\xi^\prime,\\[2mm]
\partial_t u
 +u\cdot \nabla_x u +\nabla_x P-\mu\Delta_x u=-\int_{\mathbb{R}^3}(u-\xi)fd\xi,   \\[2mm]
 {\rm div}u=0,
 \end{array}
        \right.
\end{equation}
with $$\int_{\R^3} T(\xi,\xi')\,d\xi=1.$$
The main goal of this paper is to investigate the existence of weak solutions globally in time $t$ for the equations \eqref{E1.1} with the following initial data
\begin{equation}\label{E1.2}
(f,u)|_{t=0}=(f_0(x,\xi),u_0(x)), \ \ \  x\in \mathbb{T}^3,\ \xi\in \mathbb{R}^3,
\end{equation}
and the initial data $f_0(x,\xi)$ is suitable decay condition as $|\xi|\to\infty.$
\bigbreak

The mathematical analysis for spray models is very challengingly because the coupled term for unknowns that does not depend on the same set of variables.
  The existence theory of global weak solutions for such models,  dates back to the late of 1990's. The first work in this field was  \cite{Hamdache} where the author proved the global  existence of weak solution and their large-time behavior for the Vlasov-Stokes equations. The existence theorem for weak solutions was extended in \cite{Ano, Boudin, Mellet1, Yu2, Yu1}, where the authors did not neglect the convection term and considered the Navier-Stokes equations, including incompressible and compressible ones. In \cite{Goudon1}, the existence and uniqueness of global smooth solutions near an equilibrium was proved under smallness conditions for the Navier-Stokes system coupled with the Vlasov-Fokker-Planck equation in 3D. In the meantime,  there has been  a lot of work  in hydrodynamic limits; we refer the reader to \cite{Goudon2, Goudon3, Mellet2}. In these works, the authors rely on convergence methods, such as the  compactness and relative entropy methods, to investigate hydrodynamic limits. It is natural to consider the mathematical analysis of  models with
  effects of collision or the effects of breakup, are but more challenging.
   In \cite{Legar}, Legar-Vasseur established the existence theory for  weak solutions for a system involving the coupling of  the Navier-Stokes equations and the Vlasov-Boltzmann equation where they restricted the velocity $\xi$ to  a bounded domain.  Benjelloun-Desvillettes-Moussa in \cite{Benjelloun}
   introduced  the Navier-Stokes-Vlasov-Boltzman model for spray theory, and a typical operator is given when the droplets after breakup have the same velocities as before breakup.
   Finally, they
   established an existence result of weak solutions for a simple model which was derived from the Navier-Stokes-Vlasov-Boltzman model. In particular, they
assumed that the aerosol is bidispersed, in other words,  there are only two possible radius $r_1 > r_2$ exist for the droplets, and the result
for  the breakup of particles of radii $r_1$ are particles of radius $r_2$. Under this assumption, the density function will be split up as follows $$ f(t, x, \xi, r) = f_1(t, x, \xi)\delta_{r=r_1} + f_2(t, x, \xi)\delta_{r=r_2},$$
therefore, the  Navier-Stokes-Vlasov-Boltzman model will reduce to the Navier-Stokes-Vlasov equations with damping terms. Motivated by the work of Yu \cite{Yu}, of particular interests in this paper is to establish the existence of weak solutions of
 Navier-Stokes-Vlasov-Boltzman equations with a typical breakup operator given in \cite{Benjelloun}, for the velocity $\xi\in \R^3.$ However, the restriction of the same velocities is not necessary in our paper.
  \\

It is not necessary to assume that the droplets after breakup have the same velocities as before breakup in this paper. In fact,  we deduce the following  from the conservation of kinetic energy,
\begin{equation*}
|\xi|^2=|\xi^\prime|^2,
\end{equation*}
which implies
\begin{equation}\label{E1.3}
|\xi|=|\xi^\prime|.
\end{equation}
 We will assume \eqref{E1.3} in the whole paper. In other words, different from the work of \cite{Benjelloun}, we only need to assume that the droplets after breakup have the same speeds as before breakup, not the same velocities.

\noindent As \cite{Legar}, we assume that $T(\xi,\xi^\prime)$ satisfies a self-similarity property, namely,
\begin{equation}\label{E1.5}
T(\xi,\xi^\prime)=H(|\xi^\prime|)T(\frac{\xi}{|\xi^\prime|},\frac{\xi^\prime}{|\xi^\prime|}), \textrm{\ for\  some\  function}\  H(\cdot).
\end{equation}

\noindent Note that, smooth solutions of the   problem \eqref{E1.1}-\eqref{E1.2} satisfy an  energy equality. In particular, we have the energy inequality
\begin{equation}\label{E1.6}
\begin{split}
&\frac{1}{2}\int_{\mathbb{T}^3}|u|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(1+\frac{1}{2}|\xi|^2)d\xi dx +\int_0^T \int_{\mathbb{T}^3}|\nabla u|^2dx dt+\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f|u-\xi|^2d\xi dx  dt\\
=&\frac{1}{2}\int_{\mathbb{T}^3}|u_0|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0(1+\frac{1}{2}|\xi|^2)d\xi dx ,
\end{split}
\end{equation}
for any $T>0$.   Thus, it is natural to suppose that the initial data satisfy
\begin{equation}\label{E1.7}
\int_{\mathbb{T}^3}|u_0|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0(1+\frac{1}{2}|\xi|^2)d\xi dx <\infty.
\end{equation}

\bigbreak

Based on the energy equality \eqref{E1.6}, we define the concept of weak solution for the   problem \eqref{E1.1}-\eqref{E1.2} as follows:\\
\begin{definition}\label{Def1.1}
A pair $(u,f)$ is called a  global weak solution to the problem \eqref{E1.1}-\eqref{E1.2} if, for any $T>0$, the following properties hold:
\begin{itemize}
\item $u\in L^\infty(0,T; L^2(\mathbb{T}^3))\cap L^2(0,T; H^1(\mathbb{T}^3))$;
\item $f(t,x,\xi)\geq 0$, for any $(t,x,\xi)\in(0,T)\times\mathbb{T}^3\times\mathbb{R}^3$;
\item $f\in L^\infty(0,T; L^\infty(\mathbb{T}^3\times\mathbb{R}^3)\cap L^1(\mathbb{T}^3\times\mathbb{R}^3) )$;
\item $|\xi|^3f \in L^\infty(0,T; L^1(\mathbb{T}^3\times\mathbb{R}^3))$;
\item for any test function $\varphi\in C^\infty([0,T]\times \mathbb{T}^3)$ with ${\rm div}\varphi=0$, we have
\begin{equation}\label{E1.8}
\begin{split}&
\int_0^T\int_{\mathbb{T}^3}(-u\cdot\varphi_t+(u\cdot \nabla) u\cdot\varphi+\nabla u: \nabla \varphi)dxd
\\&\quad\quad\quad\quad=-\int_0^T\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(u-\xi)\cdot \varphi d\xi dx  dt+\int_{\mathbb{T}^3}u_0\cdot\varphi(0,x)dx;
\end{split}
\end{equation}
\item for any test function $\phi\in C^\infty([0,T]\times\mathbb{T}^3\times\mathbb{R}^3)$, we have
\begin{equation}\label{E1.9}
\begin{split}
&-\int_0^T\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(\phi_t+\xi\cdot \nabla_x \phi +(u-\xi)\cdot \nabla_{\xi}\phi)d\xi dx dt=\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0\phi(0,x,\xi)d\xi dx\\
&+\lambda\int_0^T\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f\phi d\xi dx  dt
-\lambda\int_0^T\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f(t,x,\xi^\prime)\phi d\xi^\prime d\xi  dx dt;
\end{split}
\end{equation}
\item the energy inequality
\begin{equation*}
\begin{split}
&\frac{1}{2}\int_{\mathbb{T}^3}|u|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^T \int_{\mathbb{T}^3}|\nabla_x u|^2dx dt\\
&+\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f|u-\xi|^2d\xi dx  dt\\
\leq &\frac{1}{2}\int_{\mathbb{T}^3}|u_0|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0(1+\frac{1}{2}|\xi|^2)d\xi dx,
\end{split}
\end{equation*}
for any $t\in [0,T].$
\end{itemize}
\end{definition}

Our main results in this paper is as follows:
\begin{theorem}\label{Thm 1}
If the initial data satisfy $\Dv u_0=0$, and \eqref{E1.7}, the probability function $T(\xi,\xi^\prime)$ satisfies \eqref{E1.4} and \eqref{E1.5},
then there exists a global weak solution to the problem \eqref{E1.1}-\eqref{E1.2}.
\end{theorem}

\begin{remark}\label{Rem 1.1}
The same result as in Theorem \ref{Thm 1} holds for the initial boundary value problem \eqref{E1.1}-\eqref{E1.2}, with the boundary condition
$u=0$ on $\partial\Omega$ and $f(t,x,\xi)=f(t,x,\xi^*)$ for $x\in \partial\Omega$, $\xi\cdot n(x)<0$, where $\xi^*=\xi-2(\xi\cdot n(x))n(x)$ is the specular
velocity, $n(x)$ is the outward normal to $\Omega,$ and $\Omega\subset\mathbb{R}^3$ is a bounded domain.
Meanwhile,
the incompressible Navier-Stokes equations in the system \eqref{E1.1} can be replaced by the inhomogeneous ones; the extension of our result, in this
context, is considered in the forthcoming paper \cite{Yao}.
\end{remark}

 An interesting aspect of our main result is that it  handles the model with some breakup effects while the velocity of particles is in the whole space. As we mentioned before, we need to assume that the droplets after breakup have the same speeds as before breakup, but not the same velocities. This is different from the work of  \cite{Benjelloun}.
A key observation of our proof is Lemma \ref{Lem2.2}, which gives us some uniform control on the density function $f(t,x,\xi)$. By  Fubini's Theorem, we are able to show the following {\it a priori} estimate
\begin{equation}
\label{operator 0}
\int_{\mathbb{T}^3}\int_{\R^{3}}|\xi|^p Q(f) \,d\xi\; dx=0\;\text{ for any } p\geq1,
\end{equation}
which allows us to obtain further bounds and compactness of smooth solutions.  Thus, with a suitable approximation, a weak solution could be recovered. Our idea of the approximation is to construct an iteration for the kinetic part, and to adopt
a  Galerkin  method for the fluid part. However, the breakup operator in the  iteration is given by
\begin{equation}
\label{operator iteration}
-\lambda f^n+\lambda\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n-1}(t,x,\xi^\prime)d\xi^\prime,
\end{equation}
for any integer $n\geq 1.$
Note that, $\{f^n\}$ is an increasing sequence. This allows us to obtain the bounds on $f^n$ and $\int|\xi|^kf ^n\,d\xi$, which yields   weak stability. By the weak convergence method, the existence of weak solutions can be  deduced.

{\bf Notations}: In the following, $C$ from line to line denote a constant  depending   on the initial data, $T$ and the physical coefficients; $C(E,B)$    denotes  a generic   positive constant  depending   on the initial data, $T$, the physical coefficients and $E$, $B$.

We organize the rest of the paper as follows. In Section 2, we deduce a priori estimates, state some useful lemmas. In Section 3, we construct a smooth solution of an approximation scheme for  \eqref{E1.1}-\eqref{E1.2}. In Section 4, we recover   weak solutions from the approximations  by a  weak convergence method.

\bigskip

\section{{\it A  priori} estimates and some useful lemmas}\label{S2}

In this section, we derive a {\it priori} estimates for   the problem \eqref{E1.1}-\eqref{E1.2}, which will help us to derive the weak stability of the solutions. Firstly, we derive an  energy inequality for any smooth solution  of \eqref{E1.1}-\eqref{E1.2}.

\begin{lemma}\label{Lem 2.1}
For any smooth solution $(u,f)$ to the  problem \eqref{E1.1}-\eqref{E1.2},  the following equality holds
\begin{equation}
\begin{split}
\label{energy equality}
&\frac{1}{2}\int_{\mathbb{T}^3}|u|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^T \int_{\mathbb{T}^3}|\nabla u|^2dx dt+\int_0^T\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f|u-\xi|^2d\xi dx dt\\
=& \frac{1}{2}\int_{\mathbb{T}^3}|u_0|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0(1+\frac{1}{2}|\xi|^2)d\xi dx.
\end{split}
\end{equation}
\end{lemma}
\begin{proof}
Taking the scalar product with  $u$ on both sides of \eqref{E1.1}$_2$, and integrating over $\mathbb{T}^3$, we have
\begin{equation}\label{E2.1}
\frac{1}{2}\frac{d}{dt}\int_{\mathbb{T}^3}|u|^2dx +\int_{\mathbb{T}^3}|\nabla u|^2dx=-\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(u-\xi)\cdot ud\xi dx.
\end{equation}
Multiplying by $1+\frac{1}{2}|\xi|^2$ on both sides of \eqref{E1.1}$_1$, and integrating over $\mathbb{T}^3\times \mathbb{R}^3$, we have
\begin{equation}\label{E2.2}
\begin{split}
&\frac{d}{dt}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f|u-\xi|^2d\xi dx\\
=&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(u-\xi)\cdot ud\xi dx-\lambda\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(1+\frac{1}{2}|\xi|^2)d\xi dx\\
&+ \lambda\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f(t,x,\xi^\prime)d\xi^\prime(1+\frac{1}{2}|\xi|^2)d\xi dx.
\end{split}
\end{equation}
By using  Fubini's theorem, \eqref{E1.3} and \eqref{E1.4}, the last term on the right-hand side of \eqref{E2.2} can be estimated as follows
\begin{equation}
\begin{split}
\label{aaaaa}
& \lambda\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f(t,x,\xi^\prime)d\xi^\prime(1+\frac{1}{2}|\xi|^2)d\xi dx\\
 =&\lambda\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3} T(\xi,\xi^\prime)f(t,x,\xi^\prime)(1+\frac{1}{2}|\xi^\prime|^2)d\xi d\xi^\prime dx\\
= &\lambda\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}  f(t,x,\xi)(1+\frac{1}{2}|\xi|^2)d\xi dx.
\end{split}
\end{equation}
Adding \eqref{E2.1} to \eqref{E2.2} and using \eqref{aaaaa},  \eqref{energy equality} follows.
\end{proof}

\bigbreak

To develop further estimates, we will rely on the following lemma.
\begin{lemma}[\cite{Yu}]\label{Lem2.2}
Assume that $T(\xi,\xi^\prime)$ satisfies \eqref{E1.4} and \eqref{E1.5}. Then  there exists a constant $K>0$, such that
\begin{equation}\label{E2.4}
\int_{\mathbb{R}^3}T(\xi,\xi^\prime)d\xi^\prime\leq K<\infty.
\end{equation}
\end{lemma}

\bigbreak

\noindent In what follows, we denote
\begin{equation*}
m_\alpha f(t,x)=\int_{\mathbb{R}^3}|\xi|^\alpha fd\xi,\ \ and\   M_\alpha f(t)=\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}|\xi|^\alpha fdxd \xi,
\end{equation*}
here $\alpha\geq 0$ is a constant.
Clearly,
\begin{equation*}
M_\alpha f(t)=\int_{\mathbb{T}^3}m_\alpha f dx.
\end{equation*}

\begin{lemma}[\cite{Hamdache}]\label{Lem2.3}
Let $\beta>0$ and let $f$ be a nonnegative function in $L^\infty((0,T)\times \mathbb{T}^3\times\mathbb{R}^3),$  such that $m_\beta f(t,x)<+\infty,$ for a.e. $(t,x)$. Then the following estimate  holds for any $\alpha<\beta$:
\begin{equation}
m_\alpha f(t,x)\leq C(\|f(t,x,\cdot)\|_{L^\infty(\mathbb{R}^3)}+1)m_\beta f(t,x)^{\frac{\alpha+3}{\beta+3}}, \ a.e. \ (t,x).
\end{equation}
\end{lemma}

\section{An approximation scheme}\label{S3}
In this  section, we construct  smooth solutions of an approximation system.   For that purpose, we define a finite-dimensional space
$
X_m=\textrm{span}\{\phi_i\}_{i=1}^{m},
$
where  $\{\phi_i\}_{i=1}^{m}\subset C_0^\infty(\T^3)$ is an orthonormal  basis of $\{v\in L^2(\T^3): \textrm{div}v=0 \  \textrm{in}\  \mathcal{D}^\prime\}.$ Define $Y_m=C([0,T];X_m). $

We propose the following approximation scheme associated with  the  Navier-Stokes-Boltzmann equations \eqref{E1.1}
\begin{equation}\label{E3.1}
\left\{\begin{array}{l}
\partial_t f_m
  +
  \xi\cdot \nabla_x f_m+{\rm div}_\xi((\tilde{u}-\xi)f_m)=-\lambda f_m+\lambda\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)d\xi^\prime,\\[2mm]
\partial_t u_m
 +(\tilde{u}\cdot \nabla_x)u_m+\nabla_x P_m-  \Delta_x u_m=-\int_{\mathbb{R}^3}(\tilde{u}-\xi)f_m  d\xi,    \\[2mm]
 {\rm div}u_m=0,  \ \ \ \   \qquad x\in \mathbb{T}^3,\ \xi\in \mathbb{R}^3, \  t\in(0,T), \\[2mm]
 \end{array}
        \right.
\end{equation}
with initial data
\begin{equation}\label{E3.2}
(f_m,u_m)|_{t=0}=(f_0^\epsilon(x,\xi),u_0^\epsilon(x)), \ \ \  x\in \mathbb{T}^3,\ \xi\in \mathbb{R}^3,
\end{equation}
where $\tilde{u}$ is given in $Y_m$. The initial data  $f_0^\epsilon$ and $u_0^\epsilon$ are $C^\infty$ functions  such that  $f_0^\epsilon\rightarrow f_0$ strongly in $L^p(\mathbb{T}^3\times \mathbb{R}^3)$, for all $p<\infty$, and weakly in $weak\ast$-$L^\infty(\mathbb{T}^3\times \mathbb{R}^3)$;$f_0^\epsilon$ has a compact support with respect to $\xi$ in  $\R^3$,  $M_3 f_0^\epsilon$ is uniformly bounded with respect to $\epi,$  and $u_0^\epsilon\rightarrow u_0$ strongly in $L^2(\mathbb{T}^3)$.

\subsection{The weak solutions of kinetic part}
\noindent The first step of solving \eqref{E3.1}-\eqref{E3.2} is to investigate the global existence of weak solutions of  the following problem:
\begin{equation}\label{E3.3-1}
\left\{\begin{array}{l}
\partial_t f_m
  +
  \xi\cdot \nabla_x f_m+{\rm div}_\xi((\tilde{u}-\xi)f_m)=-\lambda f_m+\lambda\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)d\xi^\prime.\\[2mm]
f_m|_{t=0}=f_0^\epsilon. \\[2mm]
 \end{array}
        \right.
\end{equation}
To this end,
we construct a sequence of solutions in $n$ verifying
\begin{equation}\label{E3.4-1}
\left\{\begin{array}{l}
\partial_t f_m^n
  +
  \xi\cdot \nabla_x f_m^n+{\rm div}_\xi((\tilde{u}-\xi)f_m^n)=-\lambda f_m^n+\lambda\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m^{n-1}(t,x,\xi^\prime)d\xi^\prime,\\[2mm]
f_m^n|_{t=0}=f_0^\epsilon, \\[2mm]
f_m^0=0.
 \end{array}
        \right.
\end{equation}
The method of characteristics gives us a solution of \eqref{E3.4-1} for any given $f_m^{n-1}$. Indeed, when  $f_m^{n-1}$ is given, we are able to define the trajectories $x(\tau)=x(\tau,t,x,\xi)$ and $\xi(\tau)=\xi(\tau,t,x,\xi)$ with the following ODE system
\begin{equation}\notag
\left\{\begin{split}
&
\frac{d x}{d\tau}(\tau)=\xi(\tau),\\
&
\frac{d\xi}{d\tau}(\tau)= \tilde{u}(t, x(\tau))-\xi(\tau),\\
&
x(t,t,x,\xi)=x,\\
&
\xi(t,t,x,\xi)=\xi.
\end{split}
\right.
\end{equation}
Along the trajectories above, the solutions of \eqref{E3.4-1} satisfy
\begin{equation}\label{E3.3}
\frac{d}{d\tau}f^n_m(\tau,x(\tau),\xi(\tau))=(3-\lambda)f^n_m(\tau,x(\tau),\xi(\tau))+\lambda\int_{\mathbb{R}^3}T(\xi(\tau),\xi^\prime)f^{n-1}_m(\tau,x(\tau),\xi^\prime)d\xi^\prime.
\end{equation}
By   standard theory of ODES,  there exists a smooth solution of \eqref{E3.3} as follows
\begin{equation}\label{E3.4}
\begin{split}
&f^n_m(t,x,\xi)
=e^{(3-\lambda)t}f_0^\epsilon(x(0,t,x,\xi),\xi(0,t,x,\xi))\\
&+\lambda\int_0^t\int_{\mathbb{R}^3}e^{(3-\lambda)(t-\tau)}T(\xi(\tau,t,x,\xi),\xi^\prime)f^{n-1}_m(\tau,x(\tau,t,x,\xi),\xi^\prime)d\xi^\prime d\tau.
\end{split}
\end{equation}
It is also a smooth solution of \eqref{E3.4-1} for any given $f_m^{n-1}$.
Thanks to \eqref{E3.4}, we find
\beq\notag
f^n_m\geq 0, \ {\rm for}\ {\rm all} \  n\geq 0,
\eeq
and $\{f^n_m\}_{n=1}^{\infty} $ is an increasing sequence of measure functions with respect to $n$. For this proof, we refer the reader  to \cite{Legar}.\\

Next we shall apply a compactness argument to recover  weak solutions of problem \eqref{E3.3-1} by passing to the limit in $f_m^n$ as $n$ goes to infinity.
To this end, we need to derive uniform bounds on $f^{n}_m$ with respect to $n$. Note that, $\{f^n_m\}_{n=1}^{\infty} $ is   increasing sequence of nonnegative functions. The Gronwall inequality yields a uniform estimates of $f_m^n$ with respect to $n$.\\

The following Lemma \ref{Lem3.1} provides that $f_m^n$ is bounded in $ L^\infty(0,T;L^p(\T^3\times\R^3)),\text{ for any } p\geq 1,$ with respect to $n>0.$
\begin{lemma}\label{Lem3.1}
For any $n\geq 0$ and fixed $m>0$, $f^n_m(t,x,\xi)$ is bounded in
\beq\notag
L^\infty(0,T;L^\infty(\T^3\times\R^3))\cap L^\infty(0,T;L^1(\T^3\times\R^3)),
\eeq
and hence
\beq\label{E3.6}
f^n_m(t,x,\xi)\ is \ bounded\ in\ L^\infty(0,T;L^p(\T^3\times\R^3)),\text{ for any } p\geq 1.
\eeq
\end{lemma}
\bp
Thanks to Lemma \ref{Lem2.2}, we can deduce the following bound  from  \eqref{E3.4}
\beq\label{E3.7}
\begin{split}
\|f^n_m(t,x,\xi)\|_{L^\infty}\leq & e^{|3-\lambda |T}\|f^\epi_0\|_{L^\infty}+\lambda e^{|3-\lambda |T}K\int_0^t\|f^{n-1}_m(\tau,x,\xi)\|_{L^\infty}d\tau\\
\leq & e^{|3-\lambda |T}\|f_0\|_{L^\infty}+\lambda e^{|3-\lambda |T}K\int_0^t\|f^{n}_m(\tau,x,\xi)\|_{L^\infty}d\tau,\\
\end{split}
\eeq
where we used that $\{f_m^n\}$ is an increasing sequence.

\noindent Applying the Gronwall inequality to \eqref{E3.7}, one obtains
\beq\notag
\|f^n_m\|_{L^\infty}\leq e^{|3-\lambda |T}\|f_0\|_{L^\infty}(1+\lambda e^{|3-\lambda |T}Kte^{\lambda e^{|3-\lambda |T}Kt}),\ for\ any\  0\leq t\leq T.
\eeq
Next, integrating \eqref{E3.3} over $(0,t)\times \T^3\times \R^3$, we find
\beq\notag
\begin{split}
\int_{\T^3}\int_{\R^3}f^n_m(t,x,\xi)d\xi dx=&\int_{\T^3}\int_{\R^3}f^\epi_0(x(0,t,x,\xi),\xi(0,t,x,\xi))d\xi dx\\
 &+(3-\lambda) \int_0^t\int_{\T^3}\int_{\R^3}f^n_m(\tau,x(\tau),\xi(\tau))d\xi dx d\tau\\
&+\lambda   \int_0^t\int_{\T^3}\int_{\R^3}\int_{\mathbb{R}^3}T(\xi(\tau),\xi^\prime)f^{n-1}_m(t,x(\tau),\xi^\prime)d\xi^\prime d\xi dx d\tau,
\end{split}
\eeq
which implies
\beq \label{E3.8}
\begin{split}
\int_{\T^3}\int_{\R^3}f^n_m(t,x,\xi)d\xi dx\leq &\int_{\T^3}\int_{\R^3}f^\epi_0(x(0,t,x,\xi),\xi(0,t,x,\xi))d\xi dx\\
 &+\lambda   \int_0^t\int_{\T^3}\int_{\R^3}\int_{\mathbb{R}^3}T(\xi(\tau),\xi^\prime)f^{n-1}_m(\tau,x(\tau),\xi^\prime)d\xi^\prime d\xi dx d\tau\\
 &+ 3\int_0^t\int_{\T^3}\int_{\R^3}f^n_m(\tau,x(\tau),\xi(\tau))d\xi dx d\tau\\
\leq &\int_{\T^3}\int_{\R^3}f_0(x,\xi)d\xi dx +\lambda   \int_0^t\int_{\T^3}\int_{\R^3}\int_{\mathbb{R}^3}T(\xi(\tau),\xi^\prime)f^{n-1}_m(\tau,x(\tau),\xi^\prime)d\xi^\prime d\xi dxd\tau \\
&+ 3\int_0^t\int_{\T^3}\int_{\R^3}f^n_m(\tau,x(\tau),\xi(\tau))d\xi dx d\tau,
\end{split}
\eeq
where we have used the facts that $T(\xi,\xi^\prime)\geq 0 $, $
f^n_m\geq 0 \ {\rm for}\ {\rm all} \  n\geq 0,$
 and $\lambda>0.$\\

\noindent Thanks to \eqref{E1.4} and   Fubini's Theorem, we have
\beq \notag
\begin{split}
\int_{\T^3}\int_{\R^3}f^n_m(t,x,\xi)d\xi dx\leq &\int_{\T^3}\int_{\R^3}f_0(x,\xi)d\xi dx +(\lambda +3)  \int_0^t\int_{\T^3}\int_{\R^3}f^{n}_m(\tau,x,\xi) d\xi dx d\tau.
\end{split}
\eeq
Applying  the Gronwall inequality  again, this yields
\beq\notag
\|f^n_m\|_{L^\infty(0,T;L^1(\T^3\times \R^3))}\leq \lambda e^{|3-\lambda |T}K(1+(\lambda+3)t e^{(\lambda+3)t}),\ for\ any\  0\leq t\leq T.
\eeq
\ep
\noindent Next we shall show that $\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n}_m(t,x,\xi^\prime)d\xi^\prime$ is bounded in $
L^\infty(0,T;L^p(\T^3\times\R^3)),\ for\ any\ p\geq 1$ in the following lemma.
\begin{lemma}\label{Lem4.2}
For any $n\geq 0$, $\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n}_m(t,x,\xi^\prime)d\xi^\prime$ is bounded in
\beq\notag
L^\infty(0,T;L^p(\T^3\times\R^3)),\ for\ any\ p\geq 1.
\eeq
\end{lemma}
\bp
We derive the following bound from \eqref{E2.4} and \eqref{E3.6}:
\beq
\left\|\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n}_m(t,x,\xi^\prime)d\xi^\prime\right\|_{L^\infty}\leq  \|f^{n}_m\|_{L^\infty}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)d\xi^\prime\leq C.
\eeq
Also, by \eqref{E1.4}, \eqref{E3.6} and  Fubini's theorem, we find
\beq
\begin{split}
& \int_{\T^3}\int_{\R^3}\left|\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n}_m(t,x,\xi^\prime)d\xi^\prime\right|d\xi dx\\
=& \int_{\T^3}\int_{\R^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n}_m(t,x,\xi^\prime)d\xi^\prime d\xi dx\\
=& \int_{\T^3}\int_{\R^3}f^{n}_m d\xi dx\leq C,
\end{split}
\eeq
here $C>0$  depends only on the initial data, $\lambda$ and $T$.
This completes the proof of Lemma \ref{Lem4.2}.
\ep

In order to pass to the limit as $n$ goes to infinity,  we need   bounds on $\int_{\R^3} f^n_m d\xi$,  $\int_{\R^3}\xi f^n_m d\xi$  and $\int_{\R^3}|\xi|^2f^n_m d\xi$ stated in the following lemma.
\begin{lemma}\label{Lem4.3-1}
For any $n\geq 0$, if
\beq\notag
M_k f^\epi_0 =\int_{\T^3}\int_{\R^3}|\xi|^kf^\epi_0d\xi dx<+\infty,
\eeq
for some $k\geq 1$,  then the following estimates hold
\beq\label{E4.10-}
\left\|\int_{\R^3} f^n_m d \xi\right\|_{L^\infty(0,T;L^\frac{3+k}{3}(\T^3))}\leq C(\lambda,T,M_k f^\epi_0,\|f^\epi_0\|_{L^\infty},\tilde{u}),
\eeq
\beq\label{E4.10}
\left\|\int_{\R^3}\xi f^n_m d \xi\right\|_{L^\infty(0,T;L^\frac{3+k}{4}(\T^3))}\leq C(\lambda,T,M_k f^\epi_0,\|f^\epi_0\|_{L^\infty},\tilde{u}),
\eeq
and
\beq\label{E4.11}
\left\|\int_{\R^3}|\xi|^2 f^n_m d \xi\right\|_{L^\infty(0,T;L^\frac{3+k}{5}(\T^3))}\leq C(\lambda,T,M_k f^\epi_0,\|f^\epi_0\|_{L^\infty},\tilde{u}).
\eeq
\end{lemma}
\bp

Multiplying by $|\xi|^k$  on both sides of \eqref{E3.3}, then integrating over $\T^3\times\R^3$,     for $k\geq 1$, we have
\beq\label{E4.12-1}
\begin{split}
&\frac{d}{d\tau}\int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(\tau,x(\tau),\xi(\tau)) d\xi dx+\lambda\int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(\tau,x(\tau),\xi(\tau)) d\xi dx\\
=& \lambda \int_{\T^3}\int_{\R^3}\int_{\R^3}T(\xi(\tau),\xi')|\xi|^kf^{n-1}_m(\tau,x(\tau),\xi')\,d\xi' d\xi dx+3 \int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(\tau,x(\tau),\xi(\tau)) d\xi dx\\
\leq &\lambda \int_{\T^3}\int_{\R^3}|\xi|^kf^{n-1}_m(\tau,x(\tau),\xi) d\xi dx +3\int_{\T^3}\int_{\R^3}|\xi|^{k}f^n_m(\tau,x(\tau),\xi(\tau)) d\xi dx.
\end{split}
\eeq
Integrating with respect to time $\tau$ over $(0,t)$, we have
\begin{equation}
\begin{split}
\label{inequality on k order}
&\int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(t,x,\xi) d\xi dx
\\&\leq
\lambda \int_0^T\int_{\T^3}\int_{\R^3}|\xi|^kf^{n}_m d\xi dx\,dt +3\int_0^T\int_{\T^3}\int_{\R^3}|\xi|^{k}f^n_m d\xi dx\,dt
\\&+\int_{\T^3}\int_{\R^3}|\xi|^kf_0^\eps d\xi dx.
\end{split}
\end{equation}
Applying the Gronwall inequality to  \eqref{inequality on k order},
there exists constant $K>0$ such that,
$$\int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(t,x,\xi) d\xi dx\leq Ke ^{Kt},$$
for any $m,n>0,$ and $k\geq 1.$ Here  $K$ depends on the initial data.
This estimate, together  with Lemma \ref{Lem2.3},  yields
\eqref{E4.10-}, \eqref{E4.11} and \eqref{E4.10}.

\ep

Similarly, we can show the following lemma:
\begin{lemma}\label{Lem2.4-1}
 Assume that $f_0\in L^\infty(\T^3\times\R^3)\cap L^1(\T^3\times\R^3)$ and $|\xi|^kf_0\in L^1(\T^3\times\R^3)$. If  $f_m^n\in L^\infty((0,T)\times\T^3\times\R^3)$, then
\beq\notag
M_kf_m^n(t)\leq C_{N,T}\left((M_kf_0)^{\frac{1}{3+k}}+(\|f_m^n\|_{L^\infty}+1)\|\tilde{u}\|_{L^p(0,T;L^{3+k}(\T^3))}\right)^{3+k},
\eeq
for all $0\leq t\leq T.$
\end{lemma}
\begin{proof}
Multiplying by $|\xi|^k$  on both sides of \eqref{E3.4-1}, one finds
\begin{equation*}
\begin{split}
&\frac{d}{d\tau}\int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(t,x,\xi) d\xi dx- k\int_{\T^3}\int_{\R^3}(\tilde{u}-\xi)f_m^n|\xi|^{k-1}\,d\xi\,dx
\\&=-\lambda\int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(t,x,\xi) d\xi dx+ \lambda \int_{\T^3}\int_{\R^3}\int_{\R^3}T(\xi,\xi')|\xi|^kf^{n-1}_m(t,x,\xi')\,d\xi' d\xi dx,\\
\end{split}
\end{equation*}
which yields

\begin{equation*}
\begin{split}
&\frac{d}{d\tau}\int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(t,x,\xi) d\xi dx+ k\int_{\T^3}\int_{\R^3}|\xi|^kf_m^n\,d\xi\,dx+\lambda\int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(t,x,\xi) d\xi dx
\\&\leq k\int_{\T^3}\int_{\R^3}|\tilde{u}||\xi|^{k-1}f_m^n\,d\xi\,dx+ \lambda \int_{\T^3}\int_{\R^3}\int_{\R^3}T(\xi,\xi')|\xi|^kf^{n-1}_m(t,x,\xi')\,d\xi' d\xi dx,\\
&\leq  k\int_{\T^3}\int_{\R^3}|\tilde{u}||\xi|^{k-1}f_m^n\,d\xi\,dx+ \lambda \int_{\T^3}\int_{\R^3}\int_{\R^3}|\xi|^kf^{n-1}_m(t,x,\xi)\,d\xi dx
\\
&\leq  k\int_{\T^3}\int_{\R^3}|\tilde{u}||\xi|^{k-1}f_m^n\,d\xi\,dx+ \lambda \int_{\T^3}\int_{\R^3}\int_{\R^3}|\xi|^kf^n_m(t,x,\xi)\,d\xi dx,
\end{split}
\end{equation*}
where  we have used that  $f_m^n$ is an increasing sequence with respect to $n$. Thus, we have
\begin{equation}
\label{E3.26}
\begin{split}
&\frac{d}{d\tau}\int_{\T^3}\int_{\R^3}|\xi|^kf^n_m(t,x,\xi) d\xi dx+ k\int_{\T^3}\int_{\R^3}|\xi|^kf_m^n\,d\xi\,dx\\
&\leq  k\int_{\T^3}\int_{\R^3}|\tilde{u}||\xi|^{k-1}f_m^n\,d\xi\,dx.
\end{split}
\end{equation}
Using H\"{o}lder's inequality,  the right-hand side of \eqref{E3.26} can be estimated as follows:
\beq\notag
k\int_{\T^3}\int_{\R^3}|\xi|^{k-1}f_m^{n}|\tilde{u}|d\xi dx\leq k\|\tilde{u}\|_{L^q(\T^3)}\|m_{k-1}f_m^{n}\|_{L^{q^\prime}(\T^3)},
\eeq
here $\frac{1}{q}+\frac{1}{q^\prime}=1$.  Let $R>0$ be fixed, then we have
\beq\notag
m_{k-1}f_m^n(t,x)\leq C\|f_m^n(t)\|_{L^\infty}R^{k+2}+\frac{1}{R}\int_{|\xi|>R} |\xi|^kf_m^nd\xi,
\eeq
taking $R=(\int_{\R^3} |\xi|^kf_m^nd\xi)^{\frac{1}{k+3}}$ and $q=k+3$,   we get
\beq\notag
k\int_{\T^3}\int_{\R^3}|\xi|^{k-1}f_m^n|\tilde{u}|d\xi dx\leq  Ck \|w(t)\|_{L^{k+3}(\T^3)}(\|f_m^n(t)\|_{L^\infty}+1)\left(\int_{\T^3}\int_{\R^3} |\xi|^kf_m^nd\xi dx\right)^{\frac{k+2}{k+3}},
\eeq
substituting this into \eqref{E3.26},  we complete the proof of Lemma \ref{Lem2.4-1}.
\end{proof}

Thus, we are ready to show the following existence of weak solutions of approximation \eqref{E3.3-1}.
\begin{lemma}
For any $T>0$ and fixed $m>0,$ there exists a weak solution of the  approximation \eqref{E3.3-1} in the following sense:
\begin{equation}\label{E3.17}
\begin{split}
&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(t)\phi(t,x,\xi)d\xi dx-\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(\phi_t+\xi\cdot \nabla_x \phi +(\tilde{u}-\xi)\cdot \nabla_{\xi}\phi)d\xi dx ds\\
=&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f^\epi_0\phi(0,x,\xi)d\xi dx
+\lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m\phi d\xi dx ds\\
&
-\lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)\phi d\xi^\prime d\xi dx ds.
\end{split}
\end{equation}
In particular, the solution satisfies the following bounds:
\beq\label{E3.27}
\|f_m(t,x,\xi)\|_{ L^\infty(0,T;L^p(\T^3\times\R^3))}\leq C(\lambda,T,\|f^\epi_0\|_{L^\infty\cap L^1(\T^3\times \R^3)}),\ for\ any\ p\geq 1,
\eeq
\beq\label{E3.28}
\left\|\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)d\xi^\prime\right\|_{\ L^\infty(0,T;L^p(\T^3\times\R^3))}\leq C(\lambda,T,\|f^\epi_0\|_{L^\infty\cap L^1(\T^3\times \R^3)}),\ for\ any\ p\geq 1,
\eeq
\beq\label{E3.29}
\left\|\int_{\R^3} f_m d \xi\right\|_{L^\infty(0,T;L^2(\T^3))}\leq C(\lambda,T,M_3f_0^\eps,\|f^\epi_0\|_{L^\infty},\tilde{u}),
\eeq
\beq\label{E3.30-}
\left\|\int_{\R^3}\xi f_m d \xi\right\|_{L^\infty(0,T;L^\frac{3}{2}(\T^3))}\leq C(\lambda,T,M_3 f_0^\eps,\|f^\epi_0\|_{L^\infty},\tilde{u}),
\eeq
and
\beq\label{E3.31}
\left\|\int_{\R^3}|\xi|^2 f_m d \xi\right\|_{L^\infty(0,T;L^\frac{6}{5}(\T^3))}\leq C(\lambda,T,M_3 f_0^\eps,\|f^\epi_0\|_{L^\infty},\tilde{u}).
\eeq

\end{lemma}
\begin{proof}
First,   from  \eqref{E4.10-}, \eqref{E4.10} and \eqref{E4.11}, we deduce the following convergence results
\beq\label{E4.17-}
\int_{\R^3} f^n_m d \xi \rightharpoonup  \int_{\R^3} f_m d\xi \ weakly(*) \ in\ L^\infty(0,T;L^2(\T^3)),
\eeq
\beq\label{E4.17}
\int_{\R^3}\xi f^n_m d \xi \rightharpoonup  \int_{\R^3}\xi f_m d\xi \ weakly(*) \ in\ L^\infty(0,T;L^\frac{3}{2}(\T^3)),
\eeq
and
\beq\label{E4.18}
\int_{\R^3}|\xi|^2 f^n_m d \xi \rightharpoonup  \int_{\R^3}|\xi|^2 f_m d\xi \ weakly(*) \ in\ L^\infty(0,T;L^\frac{6}{5}(\T^3)),
\eeq
as $n\rightarrow \infty.$

In order to pass to the limits as $n\to \infty,$ we investigate  the convergence of the operator $\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n}_m(t,x,\xi^\prime)d\xi'$ in a  suitable space.
To this end, for any function $\psi(t,x)$ in $L^2(0,T;L^2(\R^3))$,  we find
\begin{equation}
\begin{split}
&\int_{\T^3}\int_{\R^3}\left(\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n-1}_m(t,x,\xi^\prime)d\xi^\prime-\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)d\xi^\prime\right)\psi(t,x)d\xi dx\\
=& \int_{\T^3}\int_{\R^3}\int_{\R^3}T(\xi,\xi^\prime)(f^{n-1}_m(t,x,\xi^\prime)-f_m(t,x,\xi^\prime))\psi(t,x)d\xi^\prime d\xi dx\\
=&  \int_{\T^3}\int_{\R^3}(f^{n-1}_m(t,x,\xi)-f_m(t,x,\xi))\psi(t,x) d\xi dx
\rightarrow 0
\end{split}
\end{equation}
as $n\rightarrow \infty$, thanks to \eqref{E4.17-}.    By Lemma \ref{Lem4.2}, this yields
\beq\label{E4.12}
\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n}_m(t,x,\xi^\prime)d\xi^\prime \rightharpoonup \int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)d\xi^\prime \ weakly(*)\ in\  L^\infty(0,T;L^2(\T^3)),\\
\eeq
 for any $q>1$.

\bigbreak
Since $f_m^n$ given by \eqref{E3.4} is a smooth solution of problem \eqref{E3.4-1}, it satisfies the following weak formulation
\begin{equation}\label{E3.15}
\begin{split}
&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f^n_m(t,x,\xi)\phi(t,x,\xi)d\xi dx-\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f^n_m(\phi_t+\xi\cdot \nabla_x \phi +(\tilde{u}-\xi)\cdot \nabla_{\xi}\phi)d\xi dx ds\\
=&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f^\epi_0\phi(0,x,\xi)d\xi dx
+\lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f^n_m\phi d\xi dx ds\\
&
-\lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f^{n-1}_m(t,x,\xi^\prime)\phi d\xi^\prime d\xi dx ds,
\end{split}
\end{equation}
for any test function $\phi\in C^\infty([0,\infty)\times\mathbb{T}^3\times\mathbb{R}^3)$.\\
  Letting $n$ tend to infinity in  \eqref{E3.15}, and using \eqref{E4.12}, \eqref{E4.17-} and \eqref{E4.17}, one obtains
\begin{equation*}
\begin{split}
&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(t)\phi(t,x,\xi)d\xi dx-\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(\phi_t+\xi\cdot \nabla_x \phi +(\tilde{u}-\xi)\cdot \nabla_{\xi}\phi)d\xi dx ds\\
=&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f^\epi_0\phi(0,x,\xi)d\xi dx
+\lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m\phi d\xi dx ds\\
&
-\lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)\phi d\xi^\prime d\xi dx ds.
\end{split}
\end{equation*}

\noindent  Here we should  remark that the solution $f_m$ satisfies the following bounds
\beq\label{E3.27!}
\|f_m\|_{ L^\infty(0,T;L^p(\T^3\times\R^3))}\leq C,\ for\ any\ p\geq 1,
\eeq
\beq\label{E3.28!}
\left\|\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(d\xi^\prime\right\|_{\ L^\infty(0,T;L^p(\T^3\times\R^3))}\leq C,\ for\ any\ p\geq 1,
\eeq
\beq\label{E3.29!}
\left\|\int_{\R^3} f_m d \xi\right\|_{L^\infty(0,T;L^2(\T^3))}\leq C,
\eeq
\beq\label{E3.30-!}
\left\|\int_{\R^3}\xi f_m d \xi\right\|_{L^\infty(0,T;L^\frac{3}{2}(\T^3))}\leq C,
\eeq
and
\beq\label{E3.31!}
\left\|\int_{\R^3}|\xi|^2 f_m d \xi\right\|_{L^\infty(0,T;L^\frac{6}{5}(\T^3))}\leq C,
\eeq
where $C$ are positive constants depend only on $\lambda,T,M_3 f_0^\eps$ and $\|f^\epi_0\|_{L^\infty\cap L^1(\T^3\times \R^3)}$.
\end{proof}
\bigbreak
\subsection{ The Navier-Stokes part}
\noindent In this subsection, we shall study the solution of the Navier-Stokes part, and the energy inequality for the whole approximation \eqref{E3.1}.  First, we are able to view the right-hand side of  \eqref{E3.1}$_2$ as a external force of the Navier-Stokes equations.   From \eqref{E3.29} and \eqref{E3.30-}, the right-hand side of \eqref{E3.1}$_2$ can be estimated as follows:
 \beq\label{E3.33--}
\begin{split}
& \left\|-\int_{\mathbb{R}^3}f_m(\tilde{u}-\xi)d\xi\right\|_{L^\infty(0,T;L^{\frac{3}{2}}(\T^3))}\\
&=\left\| -\tilde{u}\int_{\mathbb{R}^3}f_md\xi +\int_{\mathbb{R}^3}\xi f_md\xi\right\|_{L^\infty(0,T;L^{\frac{3}{2}}(\T^3))}\\
& \leq C(m,\lambda,T,M_3 f^\epi_0,\|f^\epi_0\|_{L^\infty\cap L^1(\T^3\times \R^3)},\|\tilde{u}\|_{L^\infty(0,T;L^2(\T^3))}).
\end{split}
\eeq
Thus, we can apply the classical theory of Navier-Stokes equations to solve  \eqref{E3.1}$_2$ when $f_{m}$ is a solution of \eqref{E3.3-1}.
\bigbreak

\noindent
 Next we consider  the following weak formulation of the Navier-Stokes equation \eqref{E3.1}$_2$:
\begin{equation}\label{E3.14}
\begin{split}
& \int_{\mathbb{T}^3}[\partial_tu_m\cdot \varphi+(\tilde{u}\cdot \nabla) u_m\cdot \varphi+\nabla u_m: \nabla \varphi]dx\\
=&-\int_{\mathbb{T}^3}\left(\int_{\mathbb{R}^3}f_m(\tilde{u}-\xi)d\xi\right) \cdot \varphi dx,
\end{split}
\end{equation}
where  $\varphi\in X_m$. Since  $X_m$ is a finite-dimensional space, we can write $u_m$ as follows
\beq\notag
u_m=\sum\limits_{i=1}^{m}\alpha_{im}(t) \varphi_i.
\eeq
By the standard Galerkin method, the approximation \eqref{E3.1}$_2$  yields the following ODE:
\beq\label{E3.34}
\begin{split}
\frac{d}{dt}\alpha_{im}(t)=&-\int_{\T^3}(\tilde{u}\cdot \nabla \varphi_j)\cdot\varphi_idx   \alpha_{jm}(t)-\int_{\T^3} \nabla \varphi_i:\nabla\varphi_j dx \alpha_{jm}(t)\\
-& \int_{\T^3}\left(\int_{\mathbb{R}^3}f_m(\tilde{u}-\xi)d\xi \right)\cdot \varphi_idx.
\end{split}
\eeq
From \eqref{E3.33--}, then by using the classical ODE theory, there exists a unique solution  $\alpha_{im}(t)$$(i=1,\cdots,m)$ of \eqref{E3.34} for any $0\leq t\leq  T_m$, where $0<T_m\leq T.$ This gives us a unique solution $u_m\in Y_m$ of the weak formulation \eqref{E3.14} for any $0\leq t\leq T_m$.

 Next, we can derive the energy inequality for $u_m$. Indeed, taking $\varphi_i$ for \eqref{E3.14} and multiplying the equation by $\alpha_{im}$, then summing  the resulting equality from $i=1$ to $m$,  we have
\beq\notag
\begin{split}
&\frac{1}{2}\frac{d}{dt}\int_{\T^3}|u_m|^2dx+\int_{\T^3}|\nabla u_m|^2dx\\
=& -\int_{\T^3}\int_{\R^3}f_m(\tilde{u}-\xi)\cdot u_md\xi dx\\
\leq & \|u_m\|_{L^\infty(\T^3)}\cdot\left\|\int_{\R^3}f_m(\tilde{u}-\xi) d\xi\right\|_{L^1(\T^3)}\\
\leq & \frac{1}{2}\int_{\T^3}|\nabla u_m|^2dx+ C(m)\left\|\int_{\R^3}f_m(\tilde{u}-\xi) d\xi\right\|_{L^{\frac{3}{2}}(\T^3)}^2,
\end{split}
\eeq
where we have used the  fact that all norms in $X_m$ are equivalent. This yields
\beq\label{E3.35}
\begin{split}
&\int_{\T^3}|u_m|^2dx+\int_0^t\int_{\T^3}|\nabla u_m|^2dxds\leq \int_{\T^3}|u_0^\epi|^2dx
\\&\quad\quad\quad\quad\quad\quad\quad\quad\quad+C(m)\int_0^t \left\|\int_{\R^3}f_m(\tilde{u}-\xi) d\xi\right\|_{L^{\frac{3}{2}}(\T^3)}^2ds.
\end{split}
\eeq
 With the help of \eqref{E3.33--}, this gives us the following estimates
 \begin{equation}
 \begin{split}&
 \label{local estimate for NS equations}\|u_m\|_{L^2(\T^3)}\leq M<\infty
 \quad\text{ for any  } t\in[0,T_m],\\
 & \|\nabla u_m\|_{L^2(0,T;L^2(\T^3))}\leq M.
 \end{split}
 \end{equation}

Now we define a convex set
\beq\notag
\mathbf{A}:=\{\tilde{u}\in C([0,T_m];X_m):  \sup\limits_{0\leq t\leq T_m}\|\tilde{u}\|_{L^2(\T^3)}\leq M,\;\;\Dv\tilde{u}=0\},
\eeq
and a map $S:\mathbf{A}\to \mathbf{A}$ such that $u_m:=S(\tilde{u})$.
We shall apply   Schauder's fixed point theorem to show that the operator $S$ has a fixed point in the following lemma.

\begin{lemma}\label{lemma3.4}
The operator $S$ has a fixed point in $\mathbf{A}$, that is, there is  a point $u_m\in \mathbf{A}$ such that $Su_m=u_m=\tilde{u}$.
\end{lemma}
\begin{proof} Thanks to \eqref{local estimate for NS equations}, $u_m$ is bounded in the set $\mathbf{A}.$
Meanwhile, taking $\varphi$ in \eqref{E3.14} to be $\varphi_i$, multiplying the equation by $\alpha_{im}^\prime(t)$,  then summing the resulting equality   from $i=1$ to $m$,    we have
\beq\notag
\begin{split}
&\int_{\T^3}|\partial_t u_m|^2dx +\frac{1}{2}\frac{d}{dt}\int_{\T^3} |\nabla u_m|^2 dx =-\int_{\mathbb{T}^3}\left(\int_{\mathbb{R}^3}f_m(\tilde{u}-\xi)d\xi\right) \cdot \partial_t u_m dx\\
&\quad\quad\quad \quad\quad\quad \quad\quad\quad \quad\quad\quad \quad\quad\quad -\int_{\T^3} ((\tilde{u}\cdot \nabla)u_m)\cdot \partial_t u_m dx\\
&\leq \|\partial_t u_m \|_{L^\infty(\T^3)}\cdot\|\tilde{u}\|_{L^2(\T^3)}\cdot\|\nabla u_m\|_{L^2(\T^3)}
+\|\partial_t u_m \|_{L^\infty(\T^3)}\cdot \left\|\int_{\R^3}f_m(\tilde{u}-\xi) d\xi\right\|_{L^{1}(\T^3)}\\
&\leq \frac{1}{2}\int_{\T^3}|\partial_t u_m|^2dx+C(m) \left\|\int_{\R^3}f_m(\tilde{u}-\xi) d\xi\right\|_{L^{1}(\T^3)}^2 +C(m)\|\tilde{u}\|_{L^2(\T^3)}^2\|\nabla u_m\|_{L^2(\T^3)}^2.
\end{split}
\eeq
Note that all norms in $X_m$ are equivalent to each other,  so by  \eqref{E3.33--} and \eqref{local estimate for NS equations}  one obtains that
\beq
\begin{split}
\label{local estimate for Aubin's lemma}
\int_0^t\int_{\T^3}|\partial_t u_m|^2dxds \leq C(m,\lambda,T, M_3 f^\epi_0,\|f^\epi_0\|_{L^\infty\cap L^1(\T^3\times \R^3)},\|\tilde{u}\|_{L^\infty(0,T;L^2(\T^3))}),
\end{split}
\eeq
for any $t\in[0,T_m].$
  Thanks to \eqref{local estimate for NS equations}$_2$ and \eqref{local estimate for Aubin's lemma}, the Aubin-Lions  Lemma,  $u_m=S(\tilde{u})$ is compact in $\mathbf{A}$.  On the other hand, it is easy to verify that $S$ is sequentially continuous, see \cite{Legar,Yu2} for the details.  Thus  Schauder's  fixed point   theorem gives that  $S$ has a fixed point $u_m$ in $\mathbf{A}.$
\end{proof}

We have shown that there exists   a pair $(f_m,u_m)$ on a short time interval $[0,T_m].$ In an effort to extend $T_m$ to $T$, we rely on the following  uniform bounds on $u_m$.
\begin{lemma}\label{lemma3.5}
For any $t\in [0,T_m],$  we have
\begin{equation}\label{E3.37-}
\begin{split}
&\frac{1}{2}\int_{\mathbb{T}^3}|u_m|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^t \int_{\mathbb{T}^3}|\nabla_x u_m|^2dx ds\\
&+\int_0^t \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m|u_m-\xi|^2d\xi dx  ds\\
\leq &\frac{1}{2}\int_{\mathbb{T}^3}|u_0^\epi|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx.
\end{split}
\end{equation}
\end{lemma}
\begin{proof}
Taking $\varphi$ in \eqref{E3.14} to be $\varphi_i$ and multiplying the equation by $\alpha_{im}$,   summing   from $i=1$ to $m$,  we have
\begin{equation}
\label{ssss}
\begin{split}
& \frac{1}{2}\int_{\T^3}|u_m|^2dx+\int_0^t\int_{\T^3}|\nabla u_m|^2dxds\\
=& \frac{1}{2}\int_{\T^3}|u_0^\epi|^2dx-\int_0^t\int_{\T^3}\int_{\R^3}f_m(\tilde{u}-\xi)\cdot u_md\xi dxds.\\
\end{split}
\end{equation}
Note that $\tilde{u}=u_m$ from Lemma \ref{lemma3.4}.  We are able to use $u_m$ to replace $\tilde{u}$ in the energy equality of the Navier-Stokes part \eqref{ssss} on a short  time $[0,T_m]$,  thus
\begin{equation}
\begin{split}
\label{E3.38}
& \frac{1}{2}\int_{\T^3}|u_m|^2dx+\int_0^t\int_{\T^3}|\nabla u_m|^2dxds\\
=& \frac{1}{2}\int_{\T^3}|u_0^\epi|^2dx-\int_0^t\int_{\T^3}\int_{\R^3}f_m(u_m-\xi)\cdot u_md\xi dxds.
\end{split}
\end{equation}

   Note that, $f_m^n$ given by \eqref{E3.4} is a smooth solution to the problem \eqref{E3.4-1}. We can multiply $1+\frac{1}{2}|\xi|^2$ on both sides of \eqref{E3.4-1} and integrate to deduce the  following equality
\begin{equation}
\begin{split}
\label{energy for kinetic part}
& \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m^n(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m^n|\tilde{u}-\xi|^2d\xi dxds\\
=&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m^n(\tilde{u}-\xi)\cdot \tilde{u}d\xi dxds-\lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m^n(1+\frac{1}{2}|\xi|^2)d\xi dxds\\
&+ \lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime) f_m^{n-1}(t,x,\xi^\prime)d\xi^\prime(1+\frac{1}{2}|\xi|^2)d\xi dxds\\
=&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m^n(\tilde{u}-\xi)\cdot \tilde{u}d\xi dxds-\lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m^n(1+\frac{1}{2}|\xi|^2)d\xi dxds\\
&+ \lambda\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3} f_m^{n-1}(t,x,\xi)(1+\frac{1}{2}|\xi|^2)d\xi dxds,
\end{split}
\end{equation}
 where we have used the Fubini's theorem, \eqref{E1.3} and \eqref{E1.4} in the last equality of \eqref{energy for kinetic part}. Next,  by the convergence \eqref{E4.17-}-\eqref{E4.12}, we are able to pass to the limits in \eqref{energy for kinetic part} as $n\to\infty$. In fact, we can recover the following inequality
\begin{equation}\label{TTTT}
\begin{split}
& \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m|\tilde{u}-\xi|^2d\xi dxds\\
\leq &\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(\tilde{u}-\xi)\cdot \tilde{u}d\xi dxds.
\end{split}
\end{equation}
Applying Lemma \ref{lemma3.4} to \eqref{TTTT}, one obtains
 \begin{equation}\label{E3.39}
\begin{split}
& \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m|u_m-\xi|^2d\xi dxds\\
\leq &\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^t\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(u_m-\xi)\cdot u_m d\xi dxds.
\end{split}
\end{equation}
  Combining \eqref{E3.38} and \eqref{E3.39}, we have \eqref{E3.37-} for any $t\in[0,T_m].$

\end{proof}

Note that all norms in $X_m$ are equivalent, Lemma \ref{lemma3.5} yields a uniform estimate on $u_m$ as follows $$\sup\limits_{0\leq t\leq T_m}\|u_m\|_{X_m}\leq C(m)<\infty.$$
 It allows us to have $T_m=T.$ Thus, the solution $(u_m,f_m)$ exists on $[0,T]$. Hence, we have the following result on the global existence of weak solutions of the approximation system:
\begin{proposition}\label{Pro3.1}
For any  $T>0$, under the assumptions of Theorem \ref{Thm 1}, there exists a  weak solution $(u_m, f_m)$  of the following problem
\begin{equation}\label{E3.18}
\left\{\begin{array}{l}
\partial_t f_m
  +
  \xi\cdot \nabla_x f_m+{\rm div}_\xi((u_m-\xi)f_\epsilon)=-\lambda f_m+\lambda\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)d\xi^\prime,\\[2mm]
\partial_t u_m
 +(u_m\cdot \nabla_x)u_m +\nabla_x P_m- \Delta_x u_m=-\int_{\mathbb{R}^3}(u_m-\xi)f_m  d\xi,    \\[2mm]
 {\rm div}u_m=0,  \ \ \ \   \qquad x\in \mathbb{T}^3,\ \xi\in \mathbb{R}^3, \  t\in(0,T), \\[2mm]
 \end{array}
        \right.
\end{equation}
with initial data
\begin{equation}\label{E3.19-}
(f_m,u_m)|_{t=0}=(f_0^\epsilon(x,\xi),u_0^\epsilon(x)), \ \ \  x\in \mathbb{T}^3,\ \xi\in \mathbb{R}^3.
\end{equation}
Moreover, for any $0\leq t\le T$, the weak solution   $(u_m, f_m)$ satisfies the following energy inequality
\begin{equation}\label{E3.19}
\begin{split}
&\frac{1}{2}\int_{\mathbb{T}^3}|u_m|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^T \int_{\mathbb{T}^3}|\nabla_x u_m|^2dx ds\\
&+\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m|u_m-\xi|^2d\xi dx  ds\\
\leq & \frac{1}{2}\int_{\mathbb{T}^3}|u_0^\epi|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx,
\end{split}
\end{equation}
 and
\beq\label{E3.20}
\|f_m\|_{ L^\infty(0,T;L^p(\R^3\times\T^3))}\leq C,\ for\ any\ 1\leq p\leq \infty,
\eeq
where $C>0$ depends only on the initial data, $\lambda$ and $T$.
\end{proposition}

\bigbreak
\section{Recover the weak solutions}
The main goal of this section is to recover the weak solutions of problem \eqref{E1.1}-\eqref{E1.2} by passing to the limit in the sequence  $(u_m,f_m)$ which was constructed in
Proposition \ref{Pro3.1}.
In particular,
we shall pass to the limits as $m$ goes to infinity and $\epi$ tends to zero, and show that
the limit function is a weak solution of problem \eqref{E1.1}-\eqref{E1.2}.
In the following, we will investigate the weak limit
with respect to $m$ in Step 1 and pass to the limit with respect to $\epi$ in
Step 2.

\bigbreak

\textbf{Step 1. Passing to the limit as $m\rightarrow \infty$.}

In this step, we keep $\epi>0$ fixed, deducing from Proposition \ref{Pro3.1}, we have the following estimates independent of $m$:
\beq\label{E4.1}
\|f_m\|_{ L^\infty(0,T;L^p(\T^3\times\R^3))}\leq C,\text{ for any}\ 1\leq p\leq  \infty,
\eeq
\beq\label{E4.4}
M_2f_m(t)\leq C,\text{ for any}\ 0\leq t\leq T,
\eeq
\beq\label{E4.2-}
\|u_m\|_{ L^\infty(0,T;L^2(\T^3))}\leq C,
\eeq
\beq\label{E4.3}
\|u_m\|_{ L^2(0,T;H^1(\T^3))}\leq C.
\eeq
Meanwhile, as the same in Section \ref{S3}, the solution satisfies the following estimate:
\beq\label{E3.28--}
\left\|\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)d\xi^\prime\right\|_{\ L^\infty(0,T;L^p(\T^3\times\R^3))}\leq C,\text{ for any } 1\leq p\leq \infty.
\eeq
Furthermore, by \eqref{E4.1}-\eqref{E4.3} and Lemmas \ref{Lem2.3} and Lemma \ref{Lem2.4-1}, any solution satisfies the following uniform bounds
\beq\label{E4.39--}
M_3f_m(t)\leq C,\text{ for any}\ 0\leq t\leq T,
\eeq
\beq\label{E4.39-}
\left\|\int_{\R^3} f_m d \xi\right\|_{L^\infty(0,T;L^2(\T^3))}\leq C,
\eeq
\beq\label{E4.40-}
\left\|\int_{\R^3}\xi f_m d \xi\right\|_{L^\infty(0,T;L^\frac{3}{2}(\T^3))}\leq C,
\eeq
and
\beq\label{E4.41-}
\left\|\int_{\R^3}|\xi|^2 f_m d \xi\right\|_{L^\infty(0,T;L^\frac{6}{5}(\T^3))}\leq C.
\eeq

\bigbreak

With the above estimates \eqref{E4.1}-\eqref{E4.41-}, we are ready to investigate the limit  as $m$ goes to infinity. To this end, we shall rely on the Aubin-Lions lemma for the Navier-Stokes part and the $L^p$ average velocity lemma for the kinetic part.
Using the same arguments as that in \cite{Boudin}, we prove that $\partial_t u_m$ is bounded in ${L^{\frac{4}{3}}(0,T;H^{-1}(\mathbb{T}^3))}$. For   completeness, we give the proof as follows.
\begin{lemma}\label{Lem4.4}
For any $m$, it holds that
\beq\label{E4.29}
\|\partial_t u_m\|_{L^{\frac{4}{3}}(0,T;H^{-1}(\mathbb{T}^3))}\leq C.
\eeq
\end{lemma}
\bp Note that $$\partial_t u_m=-
 (u_m\cdot \nabla_x)u_m-\nabla_x P_m+\Delta_x u_m+\int_{\mathbb{R}^3}(u_m-\xi)f_m  d\xi. $$
 We control the first term
\beq\label{E4.30}
\int_0^{T}\int_{\mathbb{T}^3} (u_m\cdot \nabla) u_m\cdot\varphi dxds=-\int_0^{T}\int_{\mathbb{T}^3} (u_m\cdot \nabla)\varphi\cdot u_m dxds.
\eeq
By \eqref{E4.2-}-\eqref{E4.3} and an interpolation inequality, one obtains
\beq\label{E4.31}
\|u_m\|_{L^{\frac{8}{3}}(0,T;L^4(\T^3))}\leq C.
\eeq
From \eqref{E4.30} and \eqref{E4.31}, we deduce that
\beq
\left|\int_0^{T}\int_{\mathbb{T}^3}( u_m\cdot \nabla) u_m\cdot\varphi dxds\right|\leq C \|\nabla \varphi\|_{L^{4}(0,T;L^2(\T^3))}.
\eeq
The second term vanishes if $\Dv\varphi=0,$ and the third term is controlled by
$$\int_0^T\int_{\mathbb{T}^3}\Delta u_m \varphi\,dx\,dt=-\int_0^T\int_{\mathbb{T}^3}\nabla u_m :\nabla\varphi\,dx\,dt,$$
which is bounded by
$C\|\nabla\varphi\|_{L^2(0,T;L^2(\T^3))}.$

Next, for the last term, by H\"{o}lder inequality, we have
\beq
\begin{split}
&\left|\int_0^{T}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(u_m-\xi)\cdot \varphi d\xi dx ds\right|\\
\leq & C\|u_m\|_{L^{2}(0,T;L^6(\T^3))}\|\varphi\|_{L^{2}(0,T;L^6(\T^3))}\|m_0f_m\|_{L^{\infty}(0,T;L^{\frac{3}{2}}(\T^3))}\\
& +  C\|\varphi\|_{L^{2}(0,T;L^5(\T^3))}\|m_1f_m\|_{L^{2}(0,T;L^{\frac{5}{4}}(\T^3))},
\end{split}
\eeq
which implies that
$
\int_{ \mathbb{R}^3}f_m(u_m-\xi)d\xi
$
is uniformly bounded in  $L^{2}(0,T;H^{-1}(\T^3))$.  This completes the proof of Lemma  \ref{Lem4.4}.

\ep

\noindent  With \eqref{E4.3} and Lemma \ref{Lem4.4}, the Aubin-Lions Lemma yields
\begin{equation}
\label{strong convergence of velocity}
u_m\to u\quad\text{ strongly in } L^{\infty}(0,T;L^r(\mathbb{T}^3))
\end{equation}
for any $1<r\leq 6.$

 By \eqref{E4.1}, \eqref{E4.2-}-\eqref{E4.3},  as $m\rightarrow \infty,$ we have
\beq\label{E3.37}
\begin{split}
& f_m \rightharpoonup f, \ weakly(*)\ in\  L^\infty(0,T;L^q(\T^3\times\R^3),\ for\ any\ q>1;\\
& u_m \rightharpoonup u, \ \ weakly(*)\ in\   L^\infty(0,T;L^2(\T^3))\cap  L^2(0,T;H^1(\T^3)).
\end{split}
\eeq
Thanks to \eqref{E3.28--} and \eqref{E3.37}$_1$, we employ the same arguments as in \eqref{E4.12}, to have,  for any $q>1$, as $m\rightarrow \infty, $
\beq\label{E4.43-}
\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)d\xi^\prime \rightharpoonup \int_{\mathbb{R}^3}T(\xi,\xi^\prime)f(t,x,\xi^\prime)d\xi^\prime, \ weakly(*)\ in\  L^\infty(0,T;L^q(\T^3\times\R^3)).\\
\eeq

\noindent Furthermore,  by \eqref{E4.39-}-\eqref{E4.41-} and \eqref{E3.37}$_1$,  we have
\beq \label{E4.52}
\int_{\R^3}f_m d \xi \rightharpoonup  \int_{\R^3} f d\xi,\ weakly(*) \ in\ L^\infty(0,T;L^2(\T^3)),
\eeq
\beq\label{E4.53}
\int_{\R^3}\xi f_m d \xi \rightharpoonup  \int_{\R^3}\xi fd\xi,\ weakly(*) \ in\ L^\infty(0,T;L^\frac{3}{2}(\T^3)),
\eeq
and
\beq\label{E4.54}
\int_{\R^3}|\xi|^2 f_m d  \xi \rightharpoonup  \int_{\R^3}|\xi|^2 fd\xi,\ weakly(*) \ in\ L^\infty(0,T;L^\frac{6}{5}(\T^3)),
\eeq
as $m\rightarrow \infty.$

  \noindent Thus, we are ready to investigate the limits as $m$ goes to infinity. In particular, with \eqref{strong convergence of velocity}-\eqref{E3.37},  as $m\to\infty,$ from the weak formulation:
\begin{equation}\label{E4.37}
\begin{split}
&\int_0^{T}\int_{\mathbb{T}^3}(-u_m\cdot \partial_t\varphi+(u_m\cdot \nabla) u_m\cdot \varphi+\nabla u_m: \nabla \varphi)dxds\\
&=-\int_0^{T}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(u_m-\xi)\cdot \varphi d\xi dx ds+\int_{\T^3} u^\epi_0\cdot \varphi(0,\cdot)dx,
\end{split}
\end{equation}we can recover the following ones:
\begin{equation}
\begin{split}
\label{weak formulation for NS part}
&\int_0^{T}\int_{\mathbb{T}^3}(-u\cdot \partial_t\varphi+(u\cdot \nabla) u\cdot \varphi+\nabla u: \nabla \varphi)dxds\\
&=-\int_0^{T}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(u-\xi)\cdot \varphi d\xi dx ds+\int_{\T^3} u^\epi_0\cdot \varphi(0,\cdot)dx,
\end{split}
\end{equation}
for any test function $\varphi\in C^\infty([0,T]\times \mathbb{T}^3)$ and ${\rm div}\varphi=0$.

 Similarly, thanks to \eqref{strong convergence of velocity}-\eqref{E4.53}, as $m\to\infty,$ from the weak formulation of the  kinetic equation
\begin{equation}\label{E4.38-}
\begin{split}
&-\int_0^{T}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(\phi_t+\xi\cdot \nabla_x \phi +(u_m-\xi)\cdot \nabla_{\xi}\phi)d\xi dx ds\\
=&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f^\epi_0\phi(0,x,\xi)d\xi dx
+\lambda\int_0^{T}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m\phi d\xi dx ds\\
&
-\lambda\int_0^{T}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_m(t,x,\xi^\prime)\phi d\xi d\xi^\prime dx ds,
\end{split}
\end{equation}
we can recover
\begin{equation}\label{E4.38}
\begin{split}
&-\int_0^{T}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(\phi_t+\xi\cdot \nabla_x \phi +(u-\xi)\cdot \nabla_{\xi}\phi)d\xi dx ds\\
=&\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f^\epi_0\phi(0,x,\xi)d\xi dx
+\lambda\int_0^{T}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f\phi d\xi dx ds\\
&
-\lambda\int_0^{T}\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f(t,x,\xi^\prime)\phi d\xi d\xi^\prime dx ds,
\end{split}
\end{equation}

\noindent for any test function  $\phi\in C^\infty([0,T]\times\mathbb{T}^3\times\mathbb{R}^3)$.\\

\noindent The last task of this step is to study the limit of the energy inequality as  $m$ goes to infinity. In particular, we can state this limit in the following lemma.
 \begin{lemma}
 \label{lemma of energy inequality}
 The following energy inequality holds as $m\to\infty$:
\begin{equation}\label{E3.19--}
\begin{split}
&\frac{1}{2}\int_{\mathbb{T}^3}|u|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^T \int_{\mathbb{T}^3}|\nabla_x u|^2dx ds\\
&+\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f|u-\xi|^2d\xi dx  ds\\
\leq &\frac{1}{2}\int_{\mathbb{T}^3}|u_0^\epi|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx.
\end{split}
\end{equation}
\end{lemma}
\begin{proof}
Here,
 we only focus on the most challenging term \begin{equation}
 \label{most difficulty term}\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m|u_m-\xi|^2d\xi dx  ds,
 \end{equation}
when
  studying the limit of \begin{equation*}
\begin{split}
&\frac{1}{2}\int_{\mathbb{T}^3}|u_m|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^T \int_{\mathbb{T}^3}|\nabla_x u_m|^2dx ds\\
&+\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m|u_m-\xi|^2d\xi dx  ds\\
\leq & \frac{1}{2}\int_{\mathbb{T}^3}|u_0^\epi|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx,
\end{split}
\end{equation*}
as $m\to\infty.$
For that purpose,   as in \cite{Hamdache}, we rewrite \eqref{most difficulty term} as follows:
\beq
\begin{split}
\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m|u_m-\xi|^2d\xi dx  ds&=\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m |u_m|^2d\xi dxds\\
-& 2\int_0^T \int_{\mathbb{T}^3}u_m\cdot\int_{\mathbb{R}^3}f_m\xi d\xi dxds +\int_0^T \int_{\mathbb{T}^3} m_2f_mdxds.\\
\end{split}
\eeq
By \eqref{E3.37}$_1$($ f_m \rightharpoonup f,  weakly*\ in\  L^\infty(0,T;L^\infty(\T^3\times\R^3)$) and \eqref{strong convergence of velocity},  Fatou's Lemma yields
\beq\notag
 \int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}|u|^2 f \mathbf{1}_{|\xi|\leq l} dxd\xi ds\leq  \liminf\limits_{m\rightarrow \infty}\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m |u_m|^2d\xi dxds,
\eeq
where $l>0$ is any positive number.
Letting $l\to\infty$, we can apply the monotone convergence theorem to obtain
\begin{equation}
\label{11}
 \int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}|u|^2 f  dxd\xi ds\leq  \liminf\limits_{m\rightarrow \infty}\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m |u_m|^2d\xi dxds.
\end{equation}
Thanks to  \eqref{strong convergence of velocity} and \eqref{E4.53}, we have
\begin{equation}
\label{convergence}
\int_0^T \int_{\mathbb{T}^3}u_m\cdot \int_{\mathbb{R}^3}f_m\xi d\xi dxds \rightarrow \int_0^T \int_{\mathbb{T}^3}u\cdot\int_{\mathbb{R}^3}f\xi d\xi dxds,\  as\   m\rightarrow \infty.
\end{equation}
Note that by \eqref{E4.54},   Fatou's lemma yields
\begin{equation}
\label{22}
\int_0^T \int_{\mathbb{T}^3} m_2fdxds \leq \liminf\limits_{m\rightarrow \infty} \int_0^T \int_{\mathbb{T}^3} m_2f_mdxds.
\end{equation}
By \eqref{11}-\eqref{22}, we have
\beq
\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f|u-\xi|^2d\xi dx  ds \leq \liminf\limits_{m\rightarrow \infty} \int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_m|u_m-\xi|^2d\xi dx  ds,\  as\   m\rightarrow \infty.
\eeq
Thus, we are able to show \eqref{E3.19--}.
\end{proof}
\bigbreak

By \eqref{weak formulation for NS part}, \eqref{E4.38}, and Lemma \ref{lemma of energy inequality},  we have the following existence result of weak solutions for all time $t>0$:
\begin{proposition}\label{Pro3.2}
For any  $T>0$, under the assumption of Theorem \ref{Thm 1}, there exists a  weak solution $(u_\epi, f_\epi)$   of the following problem
\begin{equation}\label{E3.181}
\left\{\begin{array}{l}
\partial_t f_\epi
  +
  \xi\cdot \nabla_x f_\epi+{\rm div}_\xi((u_\epi-\xi)f_\epsilon)=-\lambda f_\epi+\lambda\int_{\mathbb{R}^3}T(\xi,\xi^\prime)f_\epi(t,x,\xi^\prime)d\xi^\prime,\\[2mm]
\partial_t u_\epi
 +(u_\epi\cdot \nabla_x)u_\epsilon +\nabla_x P_\epi- \Delta_x u_\epi=-\int_{\mathbb{R}^3}(u_\epi-\xi)f_\epsilon  d\xi,    \\[2mm]
 {\rm div}u_\epi=0,  \ \ \ \   \qquad x\in \mathbb{T}^3,\ \xi\in \mathbb{R}^3, \  t\in(0,T), \\[2mm]
 \end{array}
        \right.
\end{equation}
with initial data
\begin{equation}\label{E3.19----}
(f_\epi,u_\epi)|_{t=0}=(f_0^\epsilon(x,\xi),u_0^\epsilon(x)), \ \ \  x\in \mathbb{T}^3,\ \xi\in \mathbb{R}^3.
\end{equation}
Moreover, for any $0\leq t\le T$, the weak solution   $(u_\epi, f_\epi)$ satisfies the following energy inequality:
\begin{equation}\label{E3.19-------}
\begin{split}
&\frac{1}{2}\int_{\mathbb{T}^3}|u_\epi|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_\epi(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^T \int_{\mathbb{T}^3}|\nabla_x u_\epi|^2dx ds\\
&+\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_\epi|u_\epi-\xi|^2d\xi dx  ds\\
\leq &\left(\frac{1}{2}\int_{\mathbb{T}^3}|u_0^\epi|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx\right),
\end{split}
\end{equation}
 and
\beq\label{E3.201}
\|f_\epi(t,x,\xi)\|_{ L^\infty(0,T;L^p(\R^3\times\T^3))}\leq C,\ for\ any\ 1\leq p\leq  \infty,
\eeq
here $C>0$ depends only on the initial data, $\lambda$ and $T$.
\end{proposition}

\bigbreak

\textbf{Step 2. Passing to the limit as $\epsilon \rightarrow 0$.}\\
Next, we aim to pass to the limit to  recover the weak solution to the problem \eqref{E1.1}-\eqref{E1.2} as $\epsilon\rightarrow 0.$
\begin{remark}
\label{remark on last level}
Although we have fixed $\epsilon>0$ in Step 1, estimates \eqref{E4.1}-\eqref{E4.41-} are uniform  with respect to $\epsilon$. Thus, we have the same convergence results for   $(u_{\epsilon},f_{\epsilon})$ to the same  $(u_m,f_m)$.
\end{remark}

To recover the weak solutions of  \eqref{E1.1}-\eqref{E1.2}, we only need to pass to the limit for  $(u_{\epsilon},f_{\epsilon})$ as $\epsilon\rightarrow 0.$ Thanks to Remark \ref{remark on last level}, we can take the same limit process as $m\to \infty$ to handle the limit  in the weak formulation. It allows us to deduce \eqref{E1.8}, \eqref{E1.9} by letting $\epsilon$ go  to zero.

Note that by the restriction on $u_0^\epi$ and $f_0^\epi$ at the beginning of Section 3, we have
$$\frac{1}{2}\int_{\mathbb{T}^3}|u_0^\epi|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0^\epi(1+\frac{1}{2}|\xi|^2)d\xi dx
\to \frac{1}{2}\int_{\mathbb{T}^3}|u_0|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0(1+\frac{1}{2}|\xi|^2)d\xi dx
$$
as $\epsilon\to 0.$ With the help of the convergence of $(u_{\epsilon},f_{\epsilon})$, this allows us to deduce, as $\epsilon\to 0$,
\begin{equation}\label{E4.36}
\begin{split}
&\frac{1}{2}\int_{\mathbb{T}^3}|u|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f(1+\frac{1}{2}|\xi|^2)d\xi dx+\int_0^T \int_{\mathbb{T}^3}|\nabla_x u|^2dx ds\\
&+\int_0^T \int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f|u-\xi|^2d\xi dx  ds\\
\leq & \frac{1}{2}\int_{\mathbb{T}^3}|u_0|^2dx+\int_{\mathbb{T}^3}\int_{\mathbb{R}^3}f_0(1+\frac{1}{2}|\xi|^2)d\xi dx.
\end{split}
\end{equation}
Thus, we have completed the proof of our main result.

\section*{Acknowledgement}
 Yao   was supported by the National
Natural Science Foundation of China $\#$11571280, 11331005, FANEDD $\#$201315  and China Scholarship Council.

\bibliographystyle{plain}


Another week, another seven days of cycling. Commuting, taking the long way home, the weekly coffee-outside ride, a Sunday road ride in the sun (with its attendant early season sunburn), and more porteur-ing to work and a conference. Also, the Necro-Pug is sorta for sale. Kinda.
Last night's ride the long way home (LWH) sent me westward on a coffee-outside reconnaissance mission. If it's not already obvious, I like to drink my coffee near the water. The ride wasn't as long as I would have liked, but the fall scenery more than made up for it–and recon was a definite success.
Read More Time or Miles?


text:
By David B. Kopel and Vincent Harinam
This article appeared on National Review (Online) on August 28, 2016.
Despite very severe anti-knife laws, Great Britain has been suffering from a surge in knife crime. Some Britons propose making the laws even harsher. Others are offering more constructive solutions to get to the root causes of the problem.
Britain's experience demonstrates the importance of the Second Amendment. Under the logic of the Supreme Court's District of Columbia v. Heller decision, knives are certainly among the "arms" protected by the Second Amendment. Courts in Connecticut, Wisconsin, and Washington are among those that have recognized as much, with courts in the first two states finding that particular knife controls went too far and were unconstitutional.
Although England's 1689 Bill of Rights recognized the right to possess defensive arms, that right is now a dead letter, as are many of the others enumerated in that document. So today, Great Britain has trapped itself in a vicious cycle of rising crime and intensifying repression.
The slippery slope doesn't end with the confiscation of guns, but with the destruction of the right to self-defense itself.
By the government's count, knife crime in Britain rose 36 percent between 2013 and 2017. Some of the statistical increase can be attributed to changes in the recording practices of police departments, which have long underreported all sorts of crime. But the Home Office, whose functions include collecting crime statistics, acknowledges that knife crime is up sharply.
National Health Service hospitals reported a 13 percent increase in admissions of victims of knife-related assaults between 2015 and 2016. The next year, between 2016 and 2017, there was a further 7 percent increase. London mayor Sadiq Khan tweeted, "No excuses: there is never a reason to carry a knife. Anyone who does will be caught, and they will feel the full force of the law."
The problem certainly isn't a lack of laws against carrying knives. As Joyce Malcolm details in her book Guns and Violence: The English Experience, since the 1950s, the British have banned carrying anything with the intent to use it for self-defense. This even includes a hatpin, if a woman were to use it against an attempted rapist. In the Orwellian language of British law, the willingness to use something for self-defense makes it an "offensive weapon."
According to a British police website, it is illegal to carry any "product which is made or adapted to cause a person injury." Britons are allowed, for example, to carry colored dye spray to mark an attacker, but if they spray the dye in the attacker's eyes, it "would become an offensive weapon because it would be used in a way that was intended to cause injury."
An American tourist was even convicted of carrying an "offensive weapon" after she used a pen knife to stab some men who were attacking her. Then, in the 1996 Offensive Weapons Act, carrying a knife was made presumptively illegal, even without the "offensive" intent to use the knife defensively. A person accused of the crime must "prove that he had a good reason or lawful authority for having" it to avoid punishment.
And even then, in practice, having a good reason is no protection. The first victim of the anti-knife law was Dean Payne, a man whose job at a distribution plant was to cut the straps on newspaper bundles. During what a local newspaper called "a routine search of his car," the police found a lock knife, a small printer's knife, and a Stanley knife. The magistrate readily accepted Payne's testimony that he had no intention of using the knife for "offensive" purposes, but nevertheless sent him to jail for two weeks.
The persecution of crime victims and laborers seems to have emboldened rather than deterred violent criminals. So in 2016, the government banned the sale of so-called zombie knives, horror-film-inspired blades that are marketed as collectors' items. Furthermore, online knife purchases cannot legally be delivered to residential addresses, and all sales to persons under 18 are prohibited.
Earlier this year, Poundland, a British chain of discount stores, terminated the sale of kitchen knives at its 850 locations in the U.K. and Ireland. The company expressed hope that "other retailers will join us." Dr. John Crichton, chairman of the Royal College of Psychiatrists, has urged lawmakers to prohibit the sale of pointed kitchen knives. Luton Crown Court judge Nic Madge has proposed a national program to file down the points of kitchen knives.
In the late nineteenth and early twentieth centuries, Great Britain had very low homicide rates, and knife controls were close to nil. So what's driving the present surge in knife attacks? According to the British Home Office, gang and drug activity are significant, interrelated contributors. Between 2014 and 2017, the proportion of homicides involving drugs increased from 50 percent to 57 percent. Conversely, non-drug-related homicides decreased.
Burgeoning crack-cocaine markets have mainstreamed the use of knives (and guns) among British youth. As in the U.S., illegal crack markets incentivize weapons possession and violence. For example, British gangs routinely engage in "taxing" - a new term for old-fashioned violence in territorial battles between gangs. Reporting on a particular method of drug distribution favored by British gangs, U.K. police forces recorded increased knife and firearm possession.
Meanwhile, weapons possession by gangsters has prompted non-gang-affiliated youth to arm themselves for protection.
In short, the U.K. has a drug and gang problem masquerading as a knife problem. Knife control is, by itself, a shallow solution. The futile effort to restrict the supply of knives and anything else that could possibly be used as a weapon ignores the root causes of criminal activity: As is the case everywhere else, crime in the U.K. is strongly associated with broken homes and poverty.
Making things worse, the number of police officers was reduced from 143,734 in 2010 to 123,142 in 2017. Leaked Home Office documents acknowledge that the police cuts "likely contributed" to rising violence, notwithstanding public denials from the Conservative government. Meanwhile, between 2010 and 2016, youth services were cut by £387 million, and 603 youth clubs were closed. Idle youth, broken families, and police cutbacks are a deadly combination.
Fortunately, the Home Office's recent Serious Violence Strategy offers some sensible ideas, including early intervention and prevention with youth and community partnerships. Somewhat belatedly, there is now also a greater emphasis on hot-spots policing, which allocates scarce policing resources to the areas most affected by violent crime.
But there is more to be done. The U.K. might consider Cure Violence's violence-interruption program, in which ex-convicts are trained to work in the community to prevent homicides. In the U.S. the program lowered shootings in seven Chicago neighborhoods (reductions of 41 percent to 73 percent), four in Baltimore (reductions of 34 percent to 56 percent), and two in New York (reductions of 37 percent and 50 percent). Perhaps this might help in Britain too.
Arms rights in England were never as robust as in the United States. The U.S. Second Amendment, ratified in 1791, declares that the right to keep and bear arms "shall not be infringed." The more limited 1689 English Bill of Rights allowed "subjects" to "have arms for their defence suitable to their conditions and as allowed by law." Today, there are no conditions under which English subjects may possess a suitable defensive arm in public. The English government prioritizes the safety of criminals over the safety of their victims. As England shows, the slippery slope of gun control doesn't end with the confiscation of handguns, but with destruction of the right to self-defense itself.
David Kopel is an associate policy analyst at the Cato Institute in Washington, D.C. Vincent Harinam is a law-enforcement consultant and incoming Ph.D candidate at the University of Cambridge.

redpajama_set_name: RedPajamaCommonCrawl


Celebrity diets have the ultimate power, but diets can be dangerous for cats, said the Western doctor. Veterinarian in New York "Diets without cereals are linked to a deficiency of amino acids." Cardiac, "she also once discussed meals, especially for cats," "Male cats have a lot of urinary system but How to find some do not have one. with regard to annoying "components by elements"? Gayer, they are also excellent. "The remnants of the most needed diets might be better but moderately.
. Chaton's snack and food businesses are also divided according to the shipping center, which mainly comprises convenience stores. dog food and groceries, including Hypermarkets, online retailers and many others. The other people section of the submission station is made up of grocery stores, pharmacists, etc. Food marketsOrhypermarkets disperse snacks and kitty meals with a selection of brands and cat-food.website brands categories of items and, therefore, the best income reveals more than 35% in 2018. This station is generally desired in developed economic climates and displays reasonable transmission in climates in development because it offers a common market, where buyers can find dog food with their monthly purchases. These channels, grocery stores or hypermarkets that submit are the best sellers, which may be accompanied by the foods in question. look at a significant growth rate around modern times and is expected to increase significantly in the perspective period. Online retailers offering kitten meals and snack foods present a selection of snacks and kitten feed through online channels at a very competitive cost. Worldwide, International Market Observations, Inc. estimates that its turnover exceeds US $ 38 billion by 2025. The size of the international market is strongly influenced by the growing interest in humanizing homeowners. of dogs. The global pet-care company generated revenues close to US $ 90 billion in 2013 and exceeded US $ 115 billion in 2018, with growth close to 5% over the last 5 years.
An icon Cat Food and of the world. world .


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At IBC2018 (14-18 September 2018, Amsterdam), Imagine Communications is unveiling a range of business-driving enhancements to Versio™ Platform, the latest evolution of the company's groundbreaking and industry-proven playout software solutions, on their stand at the Amtrium, #4.A01.
Imagine has added MPEG-DASH support to Versio Platform, enabling extremely high-quality and low-latency proxies for quick turnaround review of media that simply is not possible using other systems. Ubiquitous across browsers and player plugins, this industry-standard streaming format simplifies implementation by significantly reducing the need for custom code or licensed codecs for any customer-specific UI or API. The use of MPEG-DASH was pioneered for OTT delivery, but has now been proven as a scalable, cost-optimized solution for a wide range of high-quality video applications. The MPEG-DASH format also easily carries multiple audio programs and caption/subtitle tracks in industry-standardized formats. Other available solutions require either additional proprietary equipment to perform this function, or vendor-specific proprietary implementations.
Media companies using Imagine's widely adopted ADC™ automation can now continue on their platform of choice, while layering in new features and capabilities that the Versio playout engine offers. At IBC2018, Imagine will demonstrate how customers can use Versio Redundancy to synchronize ADC and Versio playlists and integrate content metadata to existing environments. This unique capability enables media organizations to operate hybrid environments of legacy and next-gen systems, as well as enabling a seamless and staged migration to Versio Automation for operations transitioning to full next-generation architectures.
Versio Master Control offers a complete software-based solution with innovative, user-friendly tools that enable media companies to maximize operational efficiency. At IBC2018, Imagine will demonstrate tight integration of Adobe® After Effects® (AE) with Versio Master Control, enabling export from AE to Versio with sophisticated, embedded metadata and triggers for interactive, real-time playback. Direct operator control of customizable graphics assets, as defined by the broadcaster's graphics design team, ensures a consistent look and feel beyond simple text-based templates. Other competing solutions use a proprietary graphics environment, which requires specialized, trained designers and operators. The integration of industry-standard Adobe AE creation tools within Versio Master Control enables media companies to open the door to a broad pool of graphic designers and advanced capabilities from Adobe.
At IBC2018, Imagine is introducing "Container" functionality in Versio Automation. Containers are high-level packages of primary and secondary events, which can be treated as a single object. These objects can be created manually or generated from a Traffic system and are easily expanded, collapsed and re-ordered. This innovative approach to schedule management streamlines media workflows and significantly improves operational efficiency compared to traditional, cluttered systems.
In Versio Ingest, Imagine will demonstrate new pooling and redundancy functionality. Rather than "fixed channels," Versio Ingest will distribute record bookings across available channels of all chassis to ensure that main and redundant records are never on the same server. Pools can reserve channels for scheduled records, secondary events, crash records or "unassigned." This industry-first capability offers media organizations an intelligent way of leveraging fewer hardware devices, simultaneously supporting contemporary and legacy hardware such as Imagine Nexio® media servers, and reducing overall cost.
Visit Imagine Communications at IBC2018 at the Amtrium, Stand #4.A01. For more information about Imagine Communications' products and solutions, please visit www.imaginecommunications.com.
Imagine Communications empowers the media and entertainment industry through transformative innovation. Broadcasters, networks, video service providers and enterprises around the world rely on our optimized, futureproof, multiscreen video and revenue enablement solutions every day to support their mission-critical operations. Today, nearly half of the world's video channels traverse our products, and our software solutions drive close to a third of global ad revenue. Through continuous innovation, we are delivering the most advanced IP, cloud-enabled, software-defined network and workflow solutions in the industry. Visit www.imaginecommunications.com for more information, and follow us on Twitter @imagine_comms.


text:
The Onion: An interview with 'America's Finest News Source'
Wikinews Shorts: June 4, 2007
Arson charge for man who cleaned home with gasoline
By MhVgPyTE | December 5, 2018 - 2:10 am | Uncategorized
Ernest Krajniak from Chilton, Wisconsin in the United States has been charged with arson after a lit cigarette ignited gasoline soaked clothes, setting his apartment ablaze.
On Friday April 3, Krajniak, 47, cleaned his entire apartment with about five gallons of gasoline, wiping everything down with the soaked clothes. After he was finished, he piled the soaked clothes in the center of his bedroom, lit a cigarette and then threw what was left of the still lit cigarette, into the pile.
Krajniak never called the fire department and never pulled the alarm. Instead he yelled 'fire' a few times then walked to the police station where an ambulance took him to a local hospital for the treatment of minor burns. The fire department later arrived to put out the blaze and his apartment was extensively smoke damaged. 11 other apartments were also damaged, leaving the occupants without a place to stay for at least a week.
"I should have never used that," said Krajniak during a court appearance on Monday. He admitted to knowing that gasoline was highly flammable. He was arrested and his bond has been set a US$2,500. Krajniak's next court appearance is scheduled for Monday, April 13. According to WISinfo.com, Krajniak has no prior criminal record.
The careless smoking of cigarettes has been blamed for thousands of fires across the U.S. In January 2008, an unnamed elderly woman in Buffalo, New York was receiving oxygen for medical problems in her home and lit a cigarette and began to smoke it. The oxygen coming from her mask then facilitated the ignition of her clothing, setting her on fire.
In the U.S. in 2002, only 4% of all residential fires were reportedly caused by smoking materials. These fires, however, were responsible for 19% of residential fire fatalities and 9% of injuries. The fatality rate due to smoking is nearly four times higher than the overall residential fire rate; injuries are more than twice as likely. Forty percent of all smoking fires start in the bedroom or living room/family room; in 35% of these fires, bedding or upholstered furniture are the items first ignited.


text:
LAHS Senior Sarah Criscuolo Represents Los Alamos Key Club
Los Alamos Key Club member Sarah Criscuolo speaks at the March 14 mid-year conference for the Southwestern District of the Kiwanis International at the Courtyard by Marriott in Farmington. Courtesy photo
KIWANIS News:
Los Alamos High School Key Club member Sarah Criscuolo spoke March 14 at the annual mid-year conference for the Southwestern District of Kiwanis International at the Marriott in Farmington.
Carl Mueller also spoke at the event. He is the father of Kayla Mueller, 26, of Prescott, Ariz., who was killed by the Islamic State in Syria last month. She had spoken to the Prescott Kiwanis Club in May 2013 during a vacation from her work in Turkey helping Syrian refugees. Since her death, Mueller's family has received an outpouring of support from Kiwanis and her father traveled to speak at the mid-year conference to show his appreciation to the group.
The Southwest District Key Club includes divisions in Arizona, New Mexico and El Paso, Texas. With 18 divisions, the Southwest board lead close to 6,000 members through a year filled with fundraising, bringing awareness to various major emphasis projects, and producing hundreds of community service hours.
Criscuolo is serving on this year's Southwest District Key Club Board as lieutenant governor for the Sangre de Cristo and Petroglyphs divisions and also is chairing this year's District Convention March 27-29 at the Embassy Suites in Albuquerque.


MECKLENBURG COUNTY, NC (WBTV) - Government officials in Mecklenburg County filed a lawsuit against several drug manufacturers and distributors Thursday.
The lawsuit states that Mecklenburg County officials are seeking to recover damages as a result of the opioid epidemic they claim the manufacturers caused.
Mecklenburg County is joining several other states and counties including Gaston County and Burke County who have also filed lawsuits against the manufacturers of prescription opioids.
The 88-page lawsuit claims that "county governments and the services they provide their citizens have been strained to the breaking point by this public health crisis."
Officials say North Carolina and Mecklenburg County are in the midst of this crisis according to reported statistics.
According to the lawsuit, "in 2016 112 people died from opioid-related death, nearly doubling the number of deaths the year before."
There are currently more than 300 lawsuits filed by cities and counties nationwide against opioid manufacturers.


Did you reunite with your long-lost childhood best friend while waiting in the lift line? Did you see a moose running down the hill while sitting safely in a chair above? Or maybe you crashed into the person who got off the lift before you only to end up marrying him or her five years later. Tell us about it.
The team here at The Know Outdoors wants to hear your favorite tales from riding Colorado ski lifts for our new project: Stories From the Lift.
It doesn't matter if your story is from the lift line, the chair, or from when you ate it while trying to get off at the top of the mountain. It can be funny, it can be romantic, it can be shocking, it can be just plain weird. It can be anything! Just tell us about it.
I was traveling in Colorado alone on a work trip and decided to take an adventure to Breckenridge, my first time snowboarding outside of New England. My phone must have fallen out of my pocket while I was up on the lift because when I got off, my phone was nowhere to be found. I was panicked. I was alone, traveling for work in a state where I didn't know anyone and now I was without my phone. I did a very horrible, distracted run down the mountain, took the lift back up and walked over to an Epic photo taker with frozen snot on his face.
It was my phone. My white iPhone in a white case that fell face down into the white powder that someone happened to find, ski down with and take the lift back up to get it to its owner.
Back in 2016-17, I had more free time than I had friends. Translation: skiing alone on weekdays. So, I showed up at Mary Jane early one morning and hopped on the Super Gauge lift.
On the way up, I fell into conversation with my chairmates. They were newly married and in their early 20s — from Nebraska, a place unfamiliar to me. We talked about farm economics and politics, and before we got off I offered to show them one of my favorite little tree runs. Turns out the flatlanders could rip.
And then the really funny thing happened: We kept riding the lift and talking, and we kept accumulating more of their family members. Soon, I was leading seven or eight Nebraskans from multiple generations through all my favorite spots. They even invited me to come visit them for the solar eclipse, which was passing over their town that summer. I wasn't able to make the trip, but the offer still stood when I got in touch six-plus months later. It was a great introduction to the joy of skiing with strangers.
We'll be collecting your stories and choosing some to be featured both online at The Know Outdoors and in the paper in The Denver Post.
So what are you waiting for? Tell us your stories.
Last hit: Just in time for Colorado Craft Beer Week, here's the latest on Denver's brewing scene Up next: What do royalty, dogs, God and the sea have in common? Denver Art Museum's newest show, of course.
3Celebrating 4/20 this year? Check out these events, deals around Denver for the annual pot holiday.


text:
namespace SimpleBlogSystem.Services
{
    using System;
    using System.Linq;
    using Common.Constants;
    using Data;
    using Models;
    using Microsoft.AspNet.Identity.EntityFramework;

    public class UsersService : IUsersService
    {
        private readonly EfGenericRepository<User> users = new EfGenericRepository<User>(new SimpleBlogSystemDbContext());

        public IQueryable<User> All(int page = 1, int pageSize = GlobalConstants.DefaultPageSize)
        {
            var result = this.users
                .All()
                .OrderByDescending(u => u.UserName)
                .Skip((page - 1) * pageSize)
                .Take(pageSize);

            return result;
        }

        public IQueryable<User> GetById(int userId)
        {
            var result = this.users
                .All()
                .Where(u => int.Parse(u.Id) == userId);

            return result;
        }

        public void Update(string userId, string email, string username)
        {
            var user = this.users
                .All()
                .FirstOrDefault(u => u.Id == userId);

            if (email != null)
            {
                user.Email = email;
            }

            if (username != null)
            {
                user.UserName = username;
            }

            users.SaveChanges();
        }

        public void Remove(int userId)
        {
            var user  = this.users
                    .All()
                    .FirstOrDefault(u => int.Parse(u.Id) == userId);

            this.users.Delete(user);
            this.users.SaveChanges();
        }

        public int Register()
        {
            return 0;
        }
    }
}

redpajama_set_name: RedPajamaGithub


The Lab, FiveThirtyEight's new NBA podcast, goes inside the Western Conference.
We're starting a new podcast! It's called The Lab, and it's a weekly conversation about the NBA that will come to you all season long. In this week's episode (Oct. 4, 2017), FiveThirtyEight writers Neil Paine, Chris Herring and Kyle Wagner break down the NBA's draft lottery reform and discuss how it might affect tanking. Then they review every team in the Western Conference — the good, the bad and the ugly. How long can the Golden State Warriors' dynasty last — and would it withstand a serious injury to one of its key players? Will the Oklahoma City Thunder find a way to utilize its new star power? Can anything be done to improve the Sacramento Kings? We answer these questions and more. Next week, we're on to the Eastern Conference.
The Ringer's Haley O'Shaughnessy wrote about how the NBA's draft reform changes tanking strategies.
ESPN's Kevin Pelton takes a look at why Vegas and ESPN projections differ on eight NBA teams.
Chris argues that the Timberwolves might have too many stars.
The Ringer compiled the important, very serious questions facing each Western Conference team.


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Cleaning performance - dust pickup on carpet boasts a 'C' rating and on hard flooring an 'A' rating. This performance is achieved through optimization throughout the whole machine, without compromising on energy use and performance.
Smaller for easy storage – we've made the Nilfisk Select small enough to store easily whilst still retaining all the latest features you would expect from a larger vacuum cleaner. A remote control handle with adjustable suction levels, and on/off functionality makes using the Select a breeze.


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text:
Dr. René Carlson, Former President of AVMA and WVA, Dies
Press release: American Veterinary Medical Association
Dr. René Carlson, who served as president of the American Veterinary Medical Association (AVMA) and the World Veterinary Association (WVA), has died.
"René was an exceptional leader and voice for our profession," said Dr. John de Jong, president of the AVMA. "She had been incredibly strong these past few years as she battled her illness and fought with the optimism and dignity that we will always remember her for. Personally, I will always remember her positive energy, her humility and her love of her husband, the AVMA and the profession."
A resident of Chetek, Wisconsin, Dr. Carlson graduated from the University of Minnesota College of Veterinary Medicine in 1978 and served as a small animal practitioner in Wisconsin for several years before taking on an active role in AVMA leadership. In addition to serving as AVMA president from 2011-2012, Dr. Carlson took on several other leadership roles at the AVMA as a member of the AVMA House of Delegates, the AVMA Council on Education, and as AVMA vice president. She also served as the AVMA director of international affairs and as chair of the association's Committee on International Veterinary Affairs.
In 2014, Dr. Carlson was elected president of the WVA. "It's amazing. I'm this person from a town of 2,000 people in Wisconsin, and I'm president of an international organization representing fellow veterinarians around the world," she said in a 2017 interview with JAVMA News.
At the 2018 Convention in Denver, Dr. Carlson was honored with a lifetime honorary membership in the WVA and the AVMA awarded her with the Global Veterinary Service Award for promoting veterinary medicine worldwide.
Dr. Carlson is survived by her husband, Dr. Mark Carlson, who was also her veterinary school classmate.
"Dr. Carlson's accomplishments are reflective of her lifelong dedication and commitment to the veterinary profession," said Dr. Janet Donlin, AVMA chief executive officer. "We will all miss René so much, as she had such a profound impact on all of us and our profession for so long. Let us all keep her husband, Mark, and her family in our prayers and thoughts."
The family suggests sending memorial contributions in Dr. Carlson's name to the American Veterinary Medical Foundation, the Chetek Lutheran Church or the University of Minnesota, where plans are underway to establish a scholarship in her name.
For more information, contact Michael San Filippo, AVMA senior media relations specialist, at 847-285-6687 (office), 847-732-6194 (cell) or [email protected].