\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 167, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/167\hfil Streaming semigroups]
{New approach to streaming semigroups
with dissipative boundary conditions}

\author[M. Boulanouar\hfil EJDE-2007/167\hfilneg]
{Mohamed Boulanouar}

\address{Mohamed Boulanouar \newline
LMCM, University of Poitiers,
22 rue des Canadiens, Poitiers, France}
\email{boulanouar@free.fr}

\thanks{Submitted September 6, 2007. Published December 3, 2007.}
\subjclass[2000]{47D06}
\keywords{Streaming operator; general boundary conditions;
semigroup}

\begin{abstract}
 This paper concerns the generation of a $C_0$-semigroup by a
 streaming operator with general dissipative boundary
 conditions. Here, we give a third approach based on the
 construction of the generated semigroup without using the
 Hille-Yosida's Theorem.  The first approach, based on the
 Hille-Yosida's Theorem, was given by Dautray \cite{Dautray},
 Protopopescu \cite{Protopopescu} and Voigt \cite{Voigt}.
 The second approach, based  on the characteristic method,
 was given by Beals \cite{Beals} and  Protopopescu \cite{Protopopescu}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{proposition}[theorem]{Proposition}
\newcommand{\abs}[1]{|#1|}
\newcommand{\norm}[1]{\|#1\|}

\section{Introduction}

In this paper, we are concerned by generation Theorem
and the explicit expression of the
generated semigroup of the streaming operator $T_K$ defined by
\begin{equation}\label{EQUATION:a}
\begin{gathered}
T_K\varphi(x,v)=-v\cdot\nabla_x\varphi(x,v),
\quad \text{on the domain}\\
D(T_K)=\{\varphi\in W_-^p(\Omega):
\gamma_-\varphi=K\gamma_+\varphi\}
\end{gathered}
\end{equation}
where $(x,v)\in \Omega=X\times V$ with
$X\subset \mathbb{R}^n$  is a smoothly bounded open subset and
$d\mu$ is a Radon measure on $\mathbb{R}^n$ with support $V$.
The traces $\gamma_+\varphi=\varphi_{\vert\Gamma_+}$ and
$\gamma_-\varphi=\varphi_{\vert\Gamma_-}$ present respectively
the outgoing and the incoming particles fluxes and
$K$ is a bounded linear operator between  the traces spaces
$L^p(\Gamma_+)$ and $L^p(\Gamma_-)$
(see the next section for more explanations).

In the phase space $\Omega=X\times V$, the function
$\varphi(x,v)$ presents
the density of all particles (neutrons, photons, molecules of gas,\dots)
having, at the time $t=0$, the position $x\in X$
with the directional velocity $v\in V$.
The boundary conditions $\gamma_-\varphi=K\gamma_+\varphi$
included in the domain $D(T_K)$ generalize naturally
all well-known boundary conditions
such as (vacuum, reflection, specular, periodic,\dots).
For the convenience of reader
and more explanations, we refer for instance
to \cite{Beals}, \cite[Chapter XI and XII]{Protopopescu},
\cite[Chapter 21]{Dautray} and \cite{Voigt}.

The existence of a strongly continuous semigroup
generated by the streaming operator has been
investigated by several authors and
several important results have been cleared.
When $\norm{K}<1$, the first approach,
based on the characteristic method, has been used
in \cite{Beals} and \cite[Theorem 4.3, p.386]{Protopopescu}.
For the same case (i.e. $\norm{K}<1$),
the second approach, based on the Hille-Yosida's Theorem,
has been used in \cite[Theorem 2.2, p.410]{Protopopescu}
\cite[Theorem 3, p.1118]{Dautray} and \cite[Theorem 4.3, p.66]{Voigt}.

The motivation, of the present work, is to give the third approach
when $\|K\|<1$  without using the Hille-Yosida's Theorem
or characteristic method.
This approach is concerned by two steps.
The first one is devoted to the construction of
a $C_0$-semigroup given in the Proposition \ref{Proposition3.1}.
In the second step, we show that $T_K$ given by the relation \eqref{EQUATION:a}
is the infinitesimal generator of this semigroup in the Theorem \ref{THEOREM4.1}.

To obtain our objective, we use our technics successfully applied in
\cite{Boulanouar3}\cite{Boulanouar4}.
We point out that the present work is new and gives the explicit expression
of the generated semigroup and all of this result doesn't hold for
the case $\|K\|\geq1$.
In \cite{Boulanouar2}, we give another and different treatment for the last case
(i.e., $\|K\|\geq1$).

\section{Setting of the problem}

We consider the Banach space
$L^p(\Omega)$ ($1\leq p<\infty$) with its natural norm
\begin{equation}\label{e:NORME1}
\norm{\varphi}_p=
\Big[\int_{\Omega}\abs{\varphi(x,v)}^pdxd\mu\Big]^{1/p},
\end{equation}
where $\Omega=X\times V$ with
$X\subset \mathbb{R}^n$  be a smoothly bounded open subset and
$d\mu$ be a Radon measure on $\mathbb{R}^n$ with support $V$.
We also consider the partial Sobolev space
$$
W^p(\Omega)
=\big\{\varphi\in L^p(\Omega),\;\; v\cdot\nabla_x\varphi\in L^p(\Omega)
\big\},
$$
with the norm
$\norm{\varphi}_{W^p(\Omega )}=
\big[\norm{\varphi}_p^p+\norm{v\cdot\nabla_x\varphi}_p^p
\big]^{1/p}$.
We set $n(x)$ the outer unit normal at
$x\in\partial X$,
where $\partial X$ is the boundary of $X$
equipped with the measure of surface $d\gamma$.
We denote
\begin{align*}
\Gamma\;&=\partial X\times V,\\
\Gamma_0&=\{(x,v)\in\Gamma,\;\; v\cdot n(x)=0\},\\
\Gamma_+&=\{(x,v)\in\Gamma,\;\; v\cdot n(x)>0\},\\
\Gamma_-&=\{(x,v)\in\Gamma,\;\; v\cdot n(x)<0\},\\
\end{align*}
and suppose that $d\gamma(\Gamma_0)=0$.
For $(x,v)\in \Omega $,
the time which a particle starting at $x$
with velocity $-v$ needs
until it reaches the boundary $\partial X$ of $X$
is denoted by
$$
t(x,v)=\inf\{t>0: x-tv\not \in X\}.
$$
Similarly, if
$(x,v)\in \Gamma_+$ we set
$$
\tau(x,v)=\inf\{t>0: x-tv\not \in X\}.
$$
We also consider the trace spaces
$L^p(\Gamma_{\pm})$ equipped with the norms
\begin{equation*}\label{}
\norm{\varphi}_{L^p(\Gamma_{\pm})}=
\Big[\int_{\Gamma_{\pm}}\abs{\varphi(x,v)}^pd\xi\Big]^{1/p},
\end{equation*}
where $d\xi=\abs{v\cdot n(x)}d\gamma d\mu$.
In this context we define the trace applications by
$$\gamma_\pm\;\; :\;\;
\varphi\to \varphi\vert_{\Gamma_\pm},
$$
and the Banach spaces
\begin{gather*}
W^p_-(\Omega)=\{
\varphi\in W^p(\Omega),\;\;\gamma_-\varphi\in L^p(\Gamma_-)\}\\
W^p_+(\Omega)=\{
\varphi\in W^p(\Omega),\;\;\gamma_+\varphi\in L^p(\Gamma_+)\}.
\end{gather*}
Note that, by \cite{Cessenat1}, \cite{Cessenat2}, we have
$W^p_-(\Omega)=W^p_+(\Omega)$.
Finally, we consider the boundary operator
\begin{equation}\label{e:K}
K\in\mathcal{L}(L^p(\Gamma_{+}),\;\;
L^p(\Gamma_{-})),
\end{equation}
and we set $\norm{K}:=\norm{K}_{\mathcal{L}(L^p(\Gamma_{+}),L^p(\Gamma_{-}))}$
for the rest of this article.

\begin{lemma}[{\cite[Theorem 2, pp.1087]{Dautray}}]\label{LEMMA2.1}
The operator $T_0$ defined by
\begin{equation}\label{T0:a}
\begin{gathered}
T_0\varphi(x,v)=-v\cdot\nabla_x\varphi(x,v),
\quad \text{on the domain}\\
D(T_0)=\{\varphi\in W^p(\Omega),\;\;\gamma_-\varphi=0\}
\end{gathered}
\end{equation}
generates, on $L^p(\Omega )$, the $C_0$-semigroup
$\{U_0(t)\}_{t\geq0}$ of contractions given by
\begin{equation*}
U_0(t)\varphi(x,v)=
\chi(t-t(x,v))\varphi(x-tv,v),
\end{equation*}
where
\begin{equation}\label{LEMMA2.1:a}
\chi(t-t(x,v))
=\begin{cases}
1&\text{if } t-t(x,v)\leq0,\\
0& \text{otherwise}.
\end{cases}
\end{equation}
\end{lemma}

We complete this section by the following Lemma that we will need later.

\begin{lemma}\label{LEM2.2}
The following applications are contunous:
\begin{enumerate}
\item $\gamma_+ : D(T_0)\to L^p(\Gamma_+)$;
\item $\gamma_+(\lambda-T_0)^{-1} : L^p(\Omega)\to
L^p(\Gamma_+)$, for all $\lambda>0$;
\item $t\geq0\to\gamma_+U_0(t)\varphi\in
L^p(\Gamma_+)$, for all $\varphi\in D(T_0)$.
\end{enumerate}
\end{lemma}

\begin{proof}
(1). For all $\varphi\in D(T_0)$, we have $\gamma_-\varphi=0\in L^p(\Gamma_-)$
and thus the Green's formula holds on $D(T_0)$.
Using the  relation
\begin{equation}\label{LEMMA3.1:xx}
\mathop{\rm sgn} u\abs{u}^{p-1}v\cdot\nabla_xu=\frac{1}{p}v\cdot\nabla_x\abs{u}^p,
\end{equation}
we obtain
\begin{equation}\label{LEM2.2:aa}
\begin{aligned}
-\int_\Omega
\left[\mathop{\rm sgn}\varphi\abs{\varphi}^{p-1}T_0\varphi\right](x,v)dxd\mu
&=\frac{1}{p}\int_{ X\times V}
v\cdot\nabla_x(\abs{\varphi}^p)(x,v)dxd\mu\\
&=\frac{1}{p}\int_{\Gamma_+}
\abs{\gamma_+\varphi(x,v)}^pd\xi
-\frac{1}{p}\int_{\Gamma-}
\abs{\gamma_-\varphi(x,v)}^pd\xi\\
&=\frac{1}{p}\norm{\gamma_+\varphi}_{L^p(\Gamma_+)}^p,
\end{aligned}
\end{equation}
which implies, by H\"{o}lder's inequality, that
\begin{align*}
\norm{\gamma_+\varphi}_{L^p(\Gamma_+)}^p
&\leq p\int_\Omega\abs{\varphi(x,v)}^{p-1}\abs{T_0\varphi(x,v)}dxd\mu\\
&\leq p\Big[\int_\Omega\abs{\varphi(x,v)}^{q(p-1)}dxd\mu\Big]^{\frac{1}{q}}
\Big[\int_\Omega\abs{T_0\varphi(x,v)}^pdxd\mu\Big]^{1/p}\\
&\leq p\norm{\varphi}_p^{p/q}\norm{T_0\varphi}_p
\end{align*}
where $q\geq1$ is the conjugate of $p\geq1$ (i.e. $p^{-1}+q^{-1}=1$).
Now, the Young's formula gives us
\begin{equation*}
\norm{\gamma_+\varphi}_{L^p(\Gamma_+)}^p\leq
p\big[\frac{1}{q}\norm{\varphi}_p^p
+\frac{1}{p}\norm{T_0\varphi}_p^p\big]
\leq \max\{\frac{p}{q},1\}\norm{\varphi}_{D(T_0)}^p
\end{equation*}
which prove the continuity.

\noindent (2).
Let $\lambda>0$. We know from the previous Lemma that,
for all $g\in L^p(\Omega)$, the function
$\varphi=(\lambda-T_0)^{-1}g\in D(T_0)$ is the unique solution of
the equation
$\lambda\varphi=T_0\varphi+g$. Multiplying this last equation by
$\mathop{\rm sgn}\varphi\abs{\varphi}^{p-1}$ and using the relation
\eqref{LEMMA3.1:xx} we obtain
\begin{align*}
\lambda\int_\Omega\abs{\varphi}^p(x,v)dxd\mu
&=\int_\Omega \big[\mathop{\rm sgn}\varphi\abs{\varphi}^{p-1}T_0\varphi\big](x,v)dxd\mu\\
&\quad +\int_\Omega
\big[\mathop{\rm sgn}\varphi\abs{\varphi}^{p-1}\big](x,v)g(x,v)dxd\mu
\end{align*}
which implies, by the relation \eqref{LEM2.2:aa}, that
\begin{equation*}
\lambda\norm{\varphi}_p^p=
-\frac{1}{p}\norm{\gamma_+\varphi}_{L^p(\Gamma_+)}^p
+\int_\Omega
\big[\mathop{\rm sgn}\varphi\abs{\varphi}^{p-1}g\big](x,v)g(x,v)dxd\mu
\end{equation*}
and therefore
\begin{equation*}
\norm{\gamma_+\varphi}_{L^p(\Gamma_+)}^p
\leq p\int_\Omega
\abs{\varphi}^{p-1}(x,v)\abs{g}(x,v)dxd\mu.
\end{equation*}
The H\"{o}der's inequality and
$\norm{\varphi}_p=\norm{(\lambda-T_0)^{-1}g}_p\leq(\norm{g}_p/\lambda)$
which follows from the contractiveness of the semigroup
$\{U_0(t)\}_{t\geq0}$ in the previous Lemma, infer that
\begin{equation*}
\norm{\gamma_+\varphi}_{L^p(\Gamma_+)}^p
\leq p\norm{\varphi}_p^\frac{p}{q}\norm{g}_p
\leq \frac{p}{\lambda^\frac{p}{q}}\norm{g}_p^\frac{p}{q}\norm{g}_p
=\frac{p}{\lambda^\frac{p}{q}}\norm{g}_p^p.
\end{equation*}
Thus
\begin{equation*}
\norm{\gamma_+(\lambda-T_0)^{-1}g}_{L^p(\Gamma_+)}
\leq\big[\frac{p}{\lambda^\frac{p}{q}}\big]^{1/p}\norm{g}_p
\end{equation*}
for all $g\in L^p(\Omega)$. The second statement is proved.

\noindent (3).
Let $h>0$. For all $\varphi\in D(T_0)$ we have
\begin{align*}
\norm{U_0(h)\varphi-\varphi}_{D(T_0)}&=
\left[\norm{U_0(h)\varphi-\varphi}_p^p+
\norm{T_0\left[U_0(h)\varphi-\varphi\right]}_p^p\right]^{1/p}\\
&=\left[\norm{U_0(h)\varphi-\varphi}_p^p+
\norm{U_0(h)T_0\varphi-T_0\varphi}_p^p\right]^{1/p}\\
\end{align*}
which implies
$$
\lim_{h\searrow0}\norm{U_0(h)\varphi-\varphi}_{D(T_0)}=0
$$
and therefore the continuity at $t=0_+$ follows.
Now, the continuity at $t>0$ follows from the previous relation and
the fact that $\{U_0(t)\}_{t\geq0}$ is a semigroup
(i.e., $U_0(t+s)=U_0(t)U_0(s)$).
\end{proof}

\section{Construction of the semigroup}

In this section we give the expression of the generated semigroup
$\{U_K(t)\}_{t\geq0}$ only for the case $\|K\|<1$.
In order to show the proposition \ref{Proposition3.1} which is the main
result of this section, we are going to show some preparatory Lemmas.

\begin{lemma}
The  Cauchy's problem
\begin{equation}
\begin{gathered}
\frac{du}{dt}+v\cdot\nabla_x u=0,\quad (t,x,v)\in(0,\infty)\times \Omega ;\\
\gamma_-u=f_-\in L^p(\mathbb{R}_+,L^p(\Gamma_-));\\
u(0)=f_0\in L^p(\Omega),
\end{gathered}
\label{Pff0} %\leqno\text{\rm P($f_-,f_0$)}
\end{equation}
admits an unique solution $u=u(t,x,v)=u(t)(x,v)$.
Furthermore, for all $t\geq0$, we have
\begin{equation}\label{LEM3.1:a}
\norm{u(t)}_p^p+
\int_0^t\norm{\gamma_+u(s)}_{L^p(\Gamma_+)}^pds
=\int_0^t\norm{f_-(s)}_{L^p(\Gamma_-)}^pds+\norm{f_0}_p^p.
\end{equation}
\end{lemma}

\begin{proof}
Let $f_-\in L^p(\mathbb{R}_+,L^p(\Gamma_-))$
and $f_0\in L^p(\Omega)$.
First. The existence of the solutions $u$ of the Cauchy's problem
\eqref{Pff0} is guaranteed by \cite[pp.1124]{Dautray} and it is given by
\begin{equation}\label{LEMMA3.2:dd}
u(t,x,v)=f_-(t-t(x,v),x-t(x,v)v,v)+U_0(t)\varphi(x,v).
\end{equation}
Next. If $u$ and $u'$ are two solution
of the Cauchy's problem \eqref{Pff0},
then $w=u-u'$ is solution of the
Cauchy's problem \eqref{Pff0} with $f_-=0,f_0=0$
which implies that $w=0$,
by the relation \eqref{LEM3.1:a}, and therefore $u=u'$.

Multiplying the first equation of the Cauchy's problem
\eqref{Pff0}
by $\mathop{\rm sgn} u\abs{u}^{p-1}$, using the relation
\eqref{LEMMA3.1:xx}
and integrating on $\Omega$, we obtain
\begin{align*}
\frac{1}{p}\frac{d\norm{u(t)}_p^p}{dt}
&=\frac{1}{p}\int_{\Gamma_-}\abs{\gamma_-u(t,x,v)}^pd\xi
-\frac{1}{p}\int_{\Gamma_+}\abs{\gamma_+u(t,x,v)}^pd\xi\\
&=\frac{1}{p}\int_{\Gamma_-}\abs{f_-(t,x,v)}^pd\xi
-\frac{1}{p}\int_{\Gamma_+}\abs{\gamma_+u(t,x,v)}^pd\xi\\
&=\frac{1}{p}\norm{f_-(t)}_{L^p(\Gamma_-)}^p-
\frac{1}{p}\norm{\gamma_+u(t)}_{L^p(\Gamma_+)}^p
\end{align*}
which implies, by integration with respect to $t$, that
\begin{equation*}
\norm{u(t)}_p^p-\norm{f_0}_p^p=
\int_0^t\norm{f_-(s)}_{L^p(\Gamma_-)}^pds-
\int_0^t\norm{\gamma_+u(s)}_{L^p(\Gamma_+)}^pds
\end{equation*}
and completes the proof.
\end{proof}

\begin{remark} \rm
In the sequel, we use the fact that all expression on the form of
the relation \eqref{LEMMA3.2:dd} is automatically solution of the
Cauchy's problem \eqref{Pff0}.

In the sequel, when it is necessary, we implicitly define by zero
all function outside their domain (for instance $f_\varphi(\cdot)$
in the next Lemma or the the operator-value functions $A_K(\cdot)$
in the Lemma \ref{LEMMA3.3})
\end{remark}

\begin{lemma}\label{LEMMA3.2}
Let $\|K\|<1$. For all $\varphi\in L^p(\Omega)$, the  equation
\begin{equation}\label{LEMMA3.2:a}
f(t)=V_K(t)\varphi+H_Kf(t),
\end{equation}
has an unique solution $f_\varphi\in L^p(\mathbb{R}_+, L^p(\Gamma_-))$, where
\begin{gather}\label{LEMMA3.2:b}
V_K(t)\varphi=K\left[\gamma_+U_0(t)\varphi\right],\\
\label{LEMMA3.2:c}
H_Kf(t,x,v)= K\left[\gamma_+u(t)\right](x,v),
\end{gather}
with
\begin{gather}\label{LEMMA3.2:d}
u(t,x,v)=\xi(t-t(x,v))f(t-t(x,v),x-t(x,v)v,v), \\
\xi(t-t(x,v))=1-\chi(t-t(x,v))
=\begin{cases}
1& \text{if } t-t(x,v)\geq0,\\
0 &\text{otherwise}.
\end{cases}
\end{gather}
where $\chi$ is given by  \eqref{LEMMA2.1:a}.
Furthermore, the application
\begin{equation}\label{LEMMA3.2:f}
\varphi\in L^p(\Omega)\to
f_\varphi\in L^p(\mathbb{R}_+,L^p(\Gamma_-))
\end{equation}
is linear and bounded satisfying
\begin{equation}\label{LEMMA3.2:e}
\norm{f_\varphi}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}\leq
\frac{\norm{K}}{(1-\norm{K})}\norm{\varphi}_p.
\end{equation}
\end{lemma}

\begin{proof}
Let $\varphi\in L^p(\Omega)$. Using the boundedness of $K$ we get that
\begin{align*}
\norm{V_K(\cdot)\varphi}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}^p&=
\lim_{t\to\infty}\int_0^t\norm{V_K(s)\varphi}_{L^p(\Gamma_-)}^pds\\
&=\lim_{t\to\infty}\int_0^t
\norm{K\left[\gamma_+U_0(s)\varphi\right]}_{L^p(\Gamma_-)}^pds\\
&\leq\norm{K}^p\lim_{t\to\infty}\int_0^t
\norm{\gamma_+U_0(s)\varphi}_{L^p(\Gamma_+)}^pds.
\end{align*}
As the function $u(t,x,v)=U_0(t)\varphi(x,v)$ is solution of Cauchy's problem
\eqref{Pff0} with $f_-=0,f_0=\varphi$, then the relation
\eqref{LEM3.1:a} infers that
\begin{equation}\label{LEMMA3.2:aa}
\norm{V_K(\cdot)\varphi}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}
\leq\norm{K}\norm{\varphi}_p
\end{equation}
and therefore $V_K(\cdot)\varphi\in L^p(\mathbb{R}_+, L^p(\Gamma_-))$.

Let $f\in L^p(\mathbb{R}_+, L^p(\Gamma_-))$.
Using the boundedness of $K$, we get
\begin{align*}
\norm{H_Kf}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}^p&=
\lim_{t\to\infty}\int_0^t\norm{H_Kf(s)}_p^pds\\
&=\lim_{t\to\infty}\int_0^t
\norm{K\left[\gamma_+u(s)\right]}_{L^p(\Gamma_-)}^pds\\
&\leq\norm{K}^p\lim_{t\to\infty}\int_0^t\norm{\gamma_+u(s)}_{L^p(\Gamma_+)}^pds.
\end{align*}
As $u=u(t,x,v)$ given by the relation \eqref{LEMMA3.2:d}
is solution of Cauchy's problem
\eqref{Pff0} with $f_-=f,f_0=0$, then the relation \eqref{LEM3.1:a}
infers that
\[
\norm{H_Kf}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}^p
\leq\norm{K}^p\lim_{t\to\infty}\int_0^t\norm{f(s)}_{L^p(\Gamma_-)}^pds
\leq\norm{K}^p\norm{f}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}^p.
\]
Thus we have
\begin{equation}\label{LEMMA3.2:bb}
\norm{H_Kf}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}
\leq\norm{K}\norm{f}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}
\end{equation}
and therefore $H_Kf\in L^p(\mathbb{R}_+, L^p(\Gamma_-))$.

Since $\norm{K}<1$, for all $\varphi\in L^p(\Omega)$,
the equation \eqref{LEMMA3.2:a} admits an unique
solution $f_\varphi\in L^p(\mathbb{R}_+, L^p(\Gamma_-))$ given by
\begin{equation*}
f_\varphi=(I-H_K)^{-1}V_K(\cdot)\varphi
\end{equation*}
which implies the linearity of the application
\eqref{LEMMA3.2:f}.
Finally, using the relations \eqref{LEMMA3.2:aa} and \eqref{LEMMA3.2:bb}
we get
\begin{equation*}
\norm{f}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}\leq
\norm{K}\norm{\varphi}_p
+\norm{K}\norm{f}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}
\end{equation*}
Now, the relation \eqref{LEMMA3.2:e} follows.
\end{proof}

In the following Lemma, we give the second part
of the semigroup given in the Proposition \ref{Proposition3.1}.

\begin{lemma}\label{LEMMA3.3}
Let $\|K\|<1$. For all $t\geq0$, the operator $A_K(t)$ given  by
\begin{equation*}
A_K(t)\varphi(x,v)=
\xi(t-t(x,v))f_\varphi(t-t(x,v),x-t(x,v)v,v)
\end{equation*}
is a linear and bounded from $L^p(\Omega)$ into itself,
where $f_\varphi$ is given in the previous Lemma.
Furthermore, for all $\varphi\in L^p(\Omega)$, we have
\begin{enumerate}
\item $A_K(0)=0$ and $\lim_{t\searrow0}\norm{A_K(t)\varphi}_p=0$;
\item $t\geq0\to A_K(t)\varphi$ is continuous.
\end{enumerate}
\end{lemma}

\begin{proof}
(1). Let $t\geq0$ and $\varphi\in L^p(\Omega)$.
As $u(t,x,v)=A_K(t)\varphi(x,v)$ is solution of the Cauchy's problem
${\rm P}(f_-=f_\varphi, f_0=0)$
with $f_\varphi\in L^p(\mathbb{R}_+, L^p(\Gamma_-))$,
then the relation \eqref{LEM3.1:a}
infers that
\begin{equation*}
\norm{A_K(t)\varphi}_p^p\leq\int_0^t\norm{f_\varphi(s)}_{L^p(\Gamma_-)}^pds
\leq\norm{f_\varphi}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}^p,
\end{equation*}
which implies $A(0)=0$, $A_K(t)\varphi\in L^p(\Omega)$
and $\lim_{t\searrow0}\norm{A_K(t)\varphi}_p=0$.
Furthermore, the previous relation together the relation \eqref{LEMMA3.2:e}
imply the boundedness of the operator $A_K(t)$ from $L^p(\Omega)$
into itself.

(2).
Let $t\geq0$ and  $\varphi\in L^p(\Omega)$.
For all $h>0$, defining the function
$u_h(t)=A_K(t+h)\varphi-A_K(t)\varphi$.
As $u_h$ is the solution of the Cauchy's problem
\eqref{Pff0} with $f_-=f_\varphi(\cdot+h)-f_\varphi,f_0=A_K(h)\varphi$
and $f_\varphi(\cdot+h)-f_\varphi\in L^p(\mathbb{R}_+, L^p(\Omega))$
and $f_0=A_K(h)\varphi\in L^p(\Omega)$, the relation
\eqref{LEM3.1:a} infers
\begin{equation*}
\norm{A_K(t+h)\varphi-A_K(t)\varphi}_p^p\leq
\int_0^t\norm{f_\varphi(s+h)-f_\varphi(s)}_{L^p(\Gamma_-)}^pds
+\norm{A_K(h)\varphi}_p^p.
\end{equation*}
However, the operator $V_K$ defined by the relation \eqref{LEMMA3.2:b}
satisfies
$$
V_K(t+h)\varphi=K\left[\gamma_+U_0(t+h)\varphi\right]=
K\left[\gamma_+U_0(t)U_0(h)\varphi\right]=V_K(t)\left[U_0(h)\varphi\right].
$$
Thus, by uniqueness of the solution of the equation
\eqref{LEMMA3.2:a} we get that
$f_\varphi(t+h)=f_{U_0(h)\varphi}(t)$ and therefore
\begin{equation*}
\norm{A_K(t+h)\varphi-A_K(t)\varphi}_p^p\leq
\int_0^t\norm{f_{U_0(h)\varphi}(s)-f_\varphi(s)}_{L^p(\Gamma_-)}^pds
+\norm{A_K(h)\varphi}_p^p.
\end{equation*}
Using the linearity of the application $\varphi\to f_\varphi$
in the previous Lemma and the relation \eqref{LEMMA3.2:e}
we finally obtain
\begin{align*}
\norm{A_K(t+h)\varphi-A_K(t)\varphi}_p^p&\leq
\int_0^t\norm{f_{U_0(h)\varphi-\varphi}(s)}_{L^p(\Gamma_-)}^pds
+\norm{A_K(h)\varphi}_p^p\\
&\leq\norm{f_{U_0(h)\varphi-\varphi}}_{L^p(\mathbb{R}_+, L^p(\Gamma_-))}^p
+\norm{A_K(h)\varphi}_p^p\\
&\leq\big[\frac{\norm{K}}{(1-\norm{K})}\big]^p\norm{U_0(h)\varphi-\varphi}_p^p+
\norm{A_K(h)\varphi}_p^p.
\end{align*}
Now the required continuity follows from
the first part and the fact that $\{U_0(t)\}_{t\geq0}$
is a semigroup.
\end{proof}

The following Proposition
is devoted to the explicit expression of a semigroup
which will be, in the Theorem \ref{THEOREM4.1},
the generated semigroup by the streaming
operator given by the relation \eqref{EQUATION:a}.

\begin{proposition}\label{Proposition3.1}
If $\norm{K}<1$, then the family $\{U_K(t)\}_{t\geq0}$ defined by
\begin{equation}\label{Proposition3.1:a}
U_K(t)=U_0(t)+A_K(t),\quad t\geq0,
\end{equation}
is a $C_0$-semigroup on $L^p(\Omega )$.
Furthermore, for all $\varphi\in L^p(\Omega)$, we have
\begin{equation}\label{Proposition3.1:b}
U_K(t)\varphi(x,v)=U_0(t)(x,v)+
\xi(t-t(x,v))K\left[\gamma_+U_K(t-t(x,v))\varphi\right](x-t(x,v)v,v)
\end{equation}
for all $t\geq0$ and a.e. $(x,v)\in \Omega$.
\end{proposition}

\begin{proof}
Let $t\geq0$ and $\varphi\in L^p(\Omega)$.
Note that from the lemmas \ref{LEMMA2.1} and \ref{LEMMA3.3}
the operator $U_K(t)$ is linear and bounded  from
$L^p(\Omega)$ into itself, $U_K(0)=U_0(0)+A_K(0)=0+I=I$
($I$ is the identity operator of $L^p(\Omega)$)
and
\begin{equation*}
\lim_{t\searrow0}\norm{U_K(t)\varphi-\varphi}_p \leq
\lim_{t\searrow0}\norm{U_0(t)\varphi-\varphi}_p +
\lim_{t\searrow0}\norm{A_K(t)\varphi}_p =0.
\end{equation*}
Now, let us shown the formula \eqref{Proposition3.1:b}
and let $\varphi\in D(T_0)$.
First. Using the definition of the operator $A_K(t)$
in the previous Lemma and all of the notations
of the Lemma \ref{LEMMA3.2}
we get that
\begin{equation*}
\gamma_-A_K(t)\varphi-K\left[\gamma_+A_K(t)\varphi\right]=
f_\varphi(t)-K\left[\gamma_+A_K(t)\varphi\right]
=K\left[\gamma_+U_0(t)\varphi\right]
\end{equation*}
a.e. $t\geq0$. As, the first point of the Lemma
\ref{LEM2.2} and the boundedness of the operator $K$
imply the continuity of the application
$t\geq0\to
K\left[\gamma_+U_0(t)\varphi\right]\in L^p(\Gamma_-)$,
then the previous equality holds for all $t\geq0$.
\\
Next. The previous relation and the fact that
$U_0(t)\varphi\in D(T_0)$ imply
\begin{align*}
\gamma_-U_K(t)\varphi-K\gamma_+U_K(t)\varphi=&
\gamma_-A_K(t)\varphi-K\gamma_+U_K(t)\varphi\\
&=\gamma_-A_K(t)\varphi-K\gamma_+\left[U_0(t)\varphi+A_K(t)\varphi\right]\\
&=\gamma_-A_K(t)\varphi-K\gamma_+\left[U_0(t)\varphi\right]-
K\gamma_+\left[A_K(t)\varphi\right]\\
&=K\gamma_+\left[U_0(t)\varphi\right]-K\gamma_+\left[U_0(t)\varphi\right]\\
&=0
\end{align*}
for all $t\geq0$, which implies
$\gamma_-U_K(t)\varphi=K\gamma_+U_K(t)\varphi$
and therefore
\begin{equation*}
\gamma_-A_K(t-t(x,v))\varphi(x-t(x,v)v,v)
=K\left[\gamma_+U_K(t-t(x,v))\varphi\right](x-t(x,v)v,v)
\end{equation*}
for all $t\geq0$ and a.e. $(x,v)\in\Omega$.
From other hand, the definition of $A_K(t)$ in the Lemma
\ref{LEMMA3.3} infers that $\gamma_-A_K(t)\varphi(x,v)=f_\varphi(t,x,v)$
which implies
\begin{align*}
\gamma_-A_K(t-t(x,v))\varphi(x-t(x,v)v,v)&=
f_\varphi(t-t(x,v),x-t(x,v)v,v)\\
&=A_K(t)\varphi(x,v)
\end{align*}
a.e. $(x,v)\in\Omega$ and for all $t\geq0$
because of the second point of the previous Lemma.
Now, the two previous relation gives us
\begin{equation*}
A_K(t)\varphi(x,v)
=K\left[\gamma_+U_K(t-t(x,v))\varphi\right](x-t(x,v)v,v)
\end{equation*}
for all $t\geq0$ and a.e. $(x,v)\in\Omega$.
Replacing this relation in the relation \eqref{Proposition3.1:a},
we obtain the relation \eqref{Proposition3.1:b} which holds
on $L^p(\Omega)$ because of the density of
$D(T_0)$ in $L^p(\Omega)$.

Now, in order to order to finish the proof,  we have to show that
$$G_K(t,t'):=U_K(t)U_K(t')-U_K(t+t')=0$$
for all $t,t'\geq0$.
Using the relation \eqref{Proposition3.1:b}, a simple calculation
shows
$$
G_K(t,t')=A_K(t)U_K(t')+(U_0(t)A_K(t')-A_K(t+t'))
$$
Let $\varphi\in D(T_0)$. Using the definition of $A_K(t)$ and
$U_K(t)$ we easily get
\begin{align*}
G_K(t,t')\varphi(x,v)
&=\xi(t-t(x,v))K\left[\gamma_+U_K(t-t(x,v))U_K(t')\varphi\right]
(x-t(x,v)v,v)\\
&\quad +\left[\chi(t-t(x,v))\xi(t+t'-t(x,v))-\xi(t+t'-t(x,v))\right]\\
&\quad \times K\left[\gamma_+U_K(t+t'-t(x,v),t')\varphi\right](x-t(x,v)v,v).
\end{align*}
Now the definition of $\chi$ and $\xi$ implies
\begin{equation*}
G_K(t,t')\varphi(x,v)=\xi(t-t(x,v))
K\left[\gamma_+G_K(t-t(x,v),t')\varphi\right](x-t(x,v)v,v).
\end{equation*}
As $u_{t'}(t)=G_K(t,t')\varphi$
is solution of the following Cauchy's problem
\eqref{Pff0} with $f_-=K[\gamma_+G_K(\cdot,t')\varphi], f_0=0$ and
$f_-\in\mathcal{R}(K)\subset L^p(\Gamma_-)$,
then the relation \eqref{LEM3.1:a} and the boundedness of $K$
infer that
\begin{align*}
\int_0^t\norm{\gamma_+G(s,t')\varphi}_{L^p(\Gamma_+)}^pds
&\leq\int_0^t\norm{K\left[\gamma_+G_K(s,t')\varphi\right]}_{L^p(\Gamma_-)}^pds\\
&\leq\norm{K}^p\int_0^t\norm{\gamma_+G_K(s,t')\varphi}_{L^p(\Gamma_+)}^pds
\end{align*}
which implies
\begin{equation*}
\int_0^t\norm{\gamma_+G(s,t')\varphi}_{L^p(\Gamma_+)}ds=0
\end{equation*}
because of $\norm{K}<1$. From other hand, the relation \eqref{LEM3.1:a}
 gives us
\begin{align*}
\norm{G_K(t,t')\varphi}_p^p
&\leq\int_0^t\norm{K\left[\gamma_+G_K(s,t')\varphi\right]}_{L^p(\Gamma_-)}^pds\\
&\leq\norm{K}^p\int_0^t\norm{\gamma_+G_K(s,t')\varphi}_{L^p(\Gamma_+)}^pds.
\end{align*}
Now the two previous relations and the density of $D(T_0)$ in $L^p(\Omega)$
imply that $G_K(t,t')=0$ for all $t,t'\geq0$
and thus $\{U_K(t)\}_{t\geq0}$ is a strongly continuous semigroup
on $L^p(\Omega)$. The proof is complete.
\end{proof}

Now, let us calculate the resolvent operator of the generator of
the semigroup $\{U_K(t)\}_{t\geq0}$ given in the previous Proposition.
But, before to state this result, recall that the Albedo operator $A$
associate to the following problem
\begin{gather*}
v\cdot\nabla_x u=0,\quad\text{on }\Omega\\
\gamma_-u=\psi\in L^p(\Gamma_-),
\end{gather*}
is defined by $A\psi(x,v)=\psi(x-\tau(x,v)v,v)$. Furthermore,
a simple calculation shows that
$\norm{A\psi}_{L^p(\Gamma_+)}=\norm{\psi}_{L^p(\Gamma_-)}$
for all $\psi\in L^p(\Gamma_-)$
and thus $\norm{A}_{\mathcal{L}(L^p(\Gamma_-),L^p(\Gamma_+))}=1$.

\begin{proposition}\label{LEM10}
Let $\norm{K}<1$ and suppose that $(B_K,D(B_K))$ is the generator of
the semigroup $\{U_K(t)\}_{t\geq0}$.
Then, for all $\lambda>0$, the resolvent of $B_K$ is
linear and bounded operator from $L^p(\Omega)$ into itself
given by
\begin{equation}\label{LEM10:a}
\begin{aligned}
(\lambda-B_K)^{-1}g(x,v)
&=(\lambda-T_0)^{-1}g(x,v)+ \epsilon_\lambda(x,v)\\
&\quad\times\left[K(I-K_\lambda)^{-1}\gamma_+(\lambda-T_0)^{-1}g\right]
 (x-t(x,v)v,v)
\end{aligned}
\end{equation}
where $I$ is the identity operator of $L^p(\Gamma_+)$,
$\epsilon_\lambda(x,v)=e^{-\lambda t(x,v)}$ and
$K_\lambda=\left[\gamma_+\epsilon_\lambda\right]AK$
with $A$ is the Albedo operator.
\end{proposition}

\begin{proof}
For all $\lambda>0$ and  all $\varphi\in D(T_0)$, we have
\begin{equation}\label{LEM10:aa}
\norm{K_\lambda}_{\mathcal{L}(L^p(\Gamma_+))}\leq
\norm{A}_{\mathcal{L}(L^p(\Gamma_-),L^p(\Gamma_+))}
\norm{K}_{\mathcal{L}(L^p(\Gamma_+),L^p(\Gamma_-))}<1
\end{equation}
which implies that the operators $K_\lambda$
and $(I-K_\lambda)^{-1}$ belong to $\mathcal{L}(L^p(\Gamma_+))$
and therefore the relation \eqref{LEM10:a} admits a sense
because of the second point of the Lemma \ref{LEM2.2}.

Let $\lambda>0$ and $\varphi\in D(T_0)$.
First, for  a.e. $(x,v)\in \Gamma_+$, the relation \eqref{Proposition3.1:b}
gives us
\begin{align*}
\gamma_+U_0(t)\varphi(x,v)=&\gamma_+U_K(t)\varphi(x,v)\\
&-\xi(t-\tau(x,v))
K\left[\gamma_+U_K(t-\tau(x,v))\varphi\right](x-\tau(x,v)v,v)\\
=&\gamma_+U_K(t)\varphi(x,v)\\
&-\xi(t-\tau(x,v))
AK\left[\gamma_+U_K(t-\tau(x,v))\varphi\right](x,v)\\
\end{align*}
for all $t\geq0$ because the second point of the Lemma
\ref{LEM2.2}.
Next, the first point of the Lemma \ref{LEM2.2} and the previous relation
imply that
\begin{align*}
&\gamma_+(\lambda-T_0)^{-1}\varphi(x,v)\\
&= \gamma_+\left[\int_0^\infty e^{-\lambda t}U_0(t)\varphi dt\right](x,v)\\
&=\int_0^\infty e^{-\lambda t}\gamma_+U_0(t)\varphi(x,v)dt\\
&=\int_0^\infty e^{-\lambda t}\gamma_+U_K(t)\varphi(x,v)dt\\
&\quad -\int_0^\infty e^{-\lambda t}\xi(t-\tau(x,v))
K\big[\gamma_+U_K(t-\tau(x,v))\varphi\big](x-\tau(x,v)v,v)dt.
\end{align*}
The change of variable $s=t-\tau(x,v)$ infers
\begin{align*}
&\gamma_+(\lambda-T_0)^{-1}\varphi(x,v)\\
&= \int_0^\infty e^{-\lambda t}\gamma_+U_K(t)\varphi(x,v)dt\\
&\quad -\int_0^\infty e^{-\lambda s}
\left[(\gamma_+\epsilon_\lambda)K
\left[\gamma_+U_K(s)\varphi\right]\right](x-\tau(x,v)v,v)ds\\
&=\int_0^\infty e^{-\lambda t}\gamma_+U_K(t)\varphi(x,v)dt\\
&\quad -(\gamma_+\epsilon_\lambda)A
\Big[\int_0^\infty e^{-\lambda s}K
\left[\gamma_+U_K(s)\varphi\right]ds\Big](x,v)\\
&=\int_0^\infty e^{-\lambda t}\gamma_+U_K(t)\varphi(x,v)dt
  - (\gamma_+\epsilon_\lambda)AK
\left[\int_0^\infty e^{-\lambda s}
\left[\gamma_+U_K(s)\varphi\right]ds\right](x,v)\\
&=\int_0^\infty e^{-\lambda t}\gamma_+U_K(t)\varphi(x,v)dt
-K_\lambda\int_0^\infty e^{-\lambda s}
\gamma_+U_K(s)\varphi(x,v)ds,
\end{align*}
where $A$ is the Albedo operator.
Using the boundedness of $K_\lambda$ we get
\begin{align*}
\gamma_+(\lambda-T_0)^{-1}\varphi
&=\int_0^\infty e^{-\lambda t}\gamma_+U_K(t)\varphi dt
-K_\lambda\int_0^\infty e^{-\lambda s}
\gamma_+U_K(s)\varphi ds\\
&=(I-K_\lambda)
\int_0^\infty e^{-\lambda t}\gamma_+U_K(t)\varphi dt.
\end{align*}
The relation \eqref{LEM10:aa}, implies the invertibility of $(I-K_\lambda)$
and thus
\begin{equation}\label{LEM10:cc}
\int_0^\infty e^{-\lambda t}\gamma_+U_K(t)\varphi dt=
(I-K_\lambda)^{-1}\gamma_+(\lambda-T_0)^{-1}\varphi.
\end{equation}
 On the other hand, using the relation \eqref{Proposition3.1:b}, we get
\begin{align*}
&\left[(\lambda-B_K)^{-1}-(\lambda-T_0)^{-1}\right]\varphi(x,v)\\
&=\int_0^\infty e^{-\lambda t}\left[U_K(t)\varphi-U_0(t)\varphi\right](x,v)dt\\
&=\int_0^\infty
\xi(t-t(x,v))e^{-\lambda t}
K\left[\gamma_+U_K(t-t(x,v))\varphi\right](x-t(x,v)v,v)dt.
\end{align*}
The change of variable $s=t-t(x,v)$ and the boundedness of $K$ infer
\begin{align*}
&\left[(\lambda-B_K)^{-1}-(\lambda-T_0)^{-1}\right]\varphi(x,v)\\
&=\epsilon_\lambda(x,v)\int_0^\infty e^{-\lambda s}
K\left[\gamma_+U_K(s)\varphi\right](x-t(x,v)v,v)ds\\
&=\epsilon_\lambda
K\Big[\int_0^\infty e^{-\lambda s}\gamma_+U_K(s)\varphi ds\Big]
(x-t(x,v)v,v).
\end{align*}
Combining, the last relation with the relation \eqref{LEM10:cc}
and the density of $D(T_0)$ in $L^p(\Omega)$, we obtain
\begin{equation*}
(\lambda-B_K)^{-1}\varphi-(\lambda-T_0)^{-1}\varphi=
\epsilon_\lambda K(I-K_\lambda)^{-1}\gamma_+(\lambda-T_0)^{-1}\varphi
\end{equation*}
for all $\varphi\in L^p(\Omega)$. The proof is complete.
\end{proof}

\section{Generation Theorem}

In this section, we state the main generation Theorem
where we prove that operator $T_K$ given by the relation
\eqref{EQUATION:a} is well the generator of the semigroup
$\{U_K(t)\}_{t\geq0}$.

\begin{lemma}
Suppose that $\norm{K}<1$. If $\lambda>0$, then we have
$\lambda\in\rho(T_K)$ and
\begin{equation}\label{LEM4.1:a}
(\lambda-B_K)^{-1}=(\lambda-T_K)^{-1}.
\end{equation}
\end{lemma}

\begin{proof}
Let $\lambda>0$, $g\in L^p(\Omega)$ and
$\varphi=(\lambda-B_K)^{-1}g\in D(B_K)\subset L^p(\Omega)$.
Using the relation \eqref{LEM10:a}, a simple calculation of derivative
give us
\begin{align*}
v\cdot\nabla_x\varphi(x,v)
&=v\cdot\nabla_x(\lambda-T_0)^{-1}g(x,v)\\
&\quad + v\cdot\nabla_x\left[\epsilon_\lambda(x,v)
\left[K(I-K_\lambda)^{-1}\gamma_+(\lambda-T_0)^{-1}g\right](x-t(x,v)v,v)\right]\\
&=g(x,v)-\lambda(\lambda-T_0)^{-1}g(x,v)\\
&\quad -\lambda\epsilon_\lambda(x,v)
\left[K(I-K_\lambda)^{-1}\gamma_+(\lambda-T_0)^{-1}g\right](x-t(x,v)v,v)\\
&\quad+\epsilon_\lambda(x,v)v\cdot\nabla_x \left[K(I-K_\lambda)^{-1}
\gamma_+(\lambda-T_0)^{-1}g\right](x-t(x,v)v,v).
\end{align*}
The last term vanishes and we can write
$v\cdot\nabla_x\varphi=g-\lambda\varphi$
which implies
$$
\norm{v\cdot\nabla_x\varphi}_p=\norm{g-\lambda\varphi}_p
\leq\norm{g}_p+\lambda\norm{\varphi}_p<\infty
$$
and therefore $\varphi=(\lambda-B_K)^{-1}g\in W^p(\Omega)$.
Furthermore, we trivially have
$$\gamma_-(\lambda-B_K)^{-1}g=
K(I-K_\lambda)^{-1}\gamma_+(\lambda-T_0)^{-1}g\in L^p(\Gamma_-)
$$
because the rang of $K$ is such that $\mathcal{R}(K)\subset L^p(\Gamma_-)$.
Thus we have $(\lambda-B_K)^{-1}g\in W_-^p(\Omega)$.
But, using all of the notations of the Proposition \ref{LEM10},
we get
\begin{align*}
\gamma_-(\lambda-B_K)^{-1}g&=K(I-K_\lambda)^{-1}
\gamma_+(\lambda-T_0)^{-1}g\\
&=K\left[K_\lambda(I-K_\lambda)^{-1}+I\right]\gamma_+(\lambda-T_0)^{-1}g\\
&=K\left[K_\lambda(I-K_\lambda)^{-1}\gamma_+(\lambda-T_0)^{-1}g
+\gamma_+(\lambda-T_0)^{-1}g\right]\\
&=K\left[(\gamma_+\epsilon_\lambda)AK(I-K_\lambda)^{-1}\gamma_+(\lambda-T_0)^{-1}g
+\gamma_+(\lambda-T_0)^{-1}g\right]\\
&=K\gamma_+(\lambda-B_K)^{-1}g
\end{align*}
which implies $(\lambda-B_K)^{-1}g\in D(T_K)$ and therefore
$\varphi=(\lambda-B_K)^{-1}g$ is the solution of
$(\lambda-T_K)\varphi=g$.
The arbitrary of $g\in L^p(\Omega)$
infers that $\lambda\in\rho(T_K)$ and the invertibility of
the operator $(\lambda-T_K)$.
\end{proof}

Now, we are able to state the main result of this work
as follows.

\begin{theorem}\label{THEOREM4.1}
If $\norm{K}<1$, then the operator $T_K$ defined by
the relation \eqref{EQUATION:a}, i.e,
\begin{gather*}
T_K\varphi(x,v)=-v\cdot\nabla_x\varphi(x,v), \quad \text{on the domain}\\
D(T_K)=\{\varphi\in W^p(\Omega),\;\gamma_\pm\varphi\in L^p(\Gamma_\pm),\quad
\gamma_-\varphi=K\gamma_+\varphi\}
\end{gather*}
generates, on $L^p(\Omega)$, the strongly continuous
semigroup $\{U_K(t)\}_{t\geq0}$ satisfying
\begin{equation}\label{THEOREM4.1:a}
\begin{aligned}
U_K(t)\varphi(x,v)&=U_0(t)(x,v)\\
&\quad+\xi(t-t(x,v))
K\left[\gamma_+U_K(t-t(x,v))\varphi\right](x-t(x,v)v,v)
\end{aligned}
\end{equation}
for all $t\geq0$ and a.e. $(x,v)\in \Omega$ and all $\varphi\in L^p(\Omega)$.
Furthermore,
\begin{equation}\label{THEOREM4.1:b}
\norm{U_K(t)}_{\mathcal{L}(L^p(\Omega))}\leq1,\quad t\geq0.
\end{equation}
\end{theorem}

\begin{proof}
The existence of the semigroup $\{U_K(t)\}_{t\geq0}$
and the relation \eqref{THEOREM4.1:a}
are guaranteed by the Proposition \ref{Proposition3.1}.
Now, let us show the identity $B_K=T_K$.

First. If $\varphi\in D(T_K)$, then the relation \eqref{LEM4.1:a}
infers
\begin{equation*}
\varphi=(\lambda-T_K)^{-1}(\lambda-T_K)\varphi=
(\lambda-B_K)^{-1}(\lambda-T_K)\varphi
\end{equation*}
which implies that
$\varphi\in D(B_K)= \mathcal{R}((\lambda-B_K)^{-1}(\lambda-T_K))$
and therefore $D(T_K)\subset D(B_K)$.
Inversely, if $\varphi\in D(B_K)$, the relation \eqref{LEM4.1:a}
also infers
\begin{equation*}
\varphi=(\lambda-B_K)^{-1}(\lambda-B_K)\varphi=
(\lambda-T_K)^{-1}(\lambda-B_K)\varphi
\end{equation*}
which implies that $\varphi\in D(T_K)=
\mathcal{R}((\lambda-T_K)^{-1}(\lambda-B_K))$
and therefore $D(B_K)\subset D(T_K)$.
Thus, $D(T_K)=D(B_K)$.

Next. Since the relation \eqref{LEM4.1:a}
we get, for all $\varphi\in D(T_K)=D(B_K)$, that
$(\lambda-T_K)\varphi=(\lambda-B_K)\varphi$
which implies that $B_K\varphi=T_K\varphi$.
Thus we have $B_K=T_K$.

To complete the proof, let us show the relation \eqref{THEOREM4.1:b}.
Let $\varphi\in D(B_K)=D(T_K)\subset L^p(\Omega)$.
As $u(t)=U_K(t)\varphi$ is the solution of the following Cauchy's problem
\eqref{Pff0} with $f_-=K[\gamma_+U_K(\cdot)\varphi],f_0=\varphi$,
applying the relation \eqref{LEM3.1:a}
together with the boundedness of the operator $K$
we get, for all $t\geq0$, that
\begin{align*}
\norm{U_K(t)\varphi}_p^p-\norm{\varphi}_p^p
&=\int_0^t\norm{K\gamma_+U_K(s)\varphi}_{L^p(\Gamma_-)}^pds
 -\int_0^t\norm{\gamma_+U_K(s)\varphi(s)}_{L^p(\Gamma_+)}^pds\\
&\leq [\norm{K}^p-1]
\int_0^t\norm{\gamma_+U_K(s)\varphi(s)}_{L^p(\Gamma_+)}^pds,
\end{align*}
which implies
$\norm{U_K(t)\varphi}_p\leq\norm{\varphi}_p$ for all $t\geq0$ because of
$\norm{K}<1$.
Now, the density of $D(B_K)$ in $L^p(\Omega)$ archives the proof.
\end{proof}

\begin{remark} \rm
First, we can positively close the conjecture \cite[p 103]{Voigt}.
Next, it is clear that the previous Theorem and all of result of this work
are based on the fact that $\norm{K}<1$
and we cannot apply these results for the case $\norm{K}\geq1$.
In \cite{Boulanouar2}, we give others and different proofs
to obtain the same main objective of the present work, but for the case
$\norm{K}\geq1$.
\end{remark}

\subsection*{Acknowledgements}
I would like to express my gratitude to the Professor S. Noui-Mehidi
for his valuable help.

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\end{document}