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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 155, pp. 1--11.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/155\hfil Streaming semigroup]
{New approach to streaming semigroups with multiplying boundary conditions}

\author[M. Boulanouar\hfil EJDE-2008/155\hfilneg]
{Mohamed Boulanouar}

\address{Mohamed Boulanouar\newline
LMCM, Universit\'e de Poitiers,
86000, Poitiers, France}
\email{boulanouar@free.fr}

\thanks{Submitted May 15, 2008. Published November 12, 2008.}
\subjclass[2000]{47D06}
\keywords{Streaming operator; general boundary conditions;
semigroup}

\begin{abstract}
 This paper concerns the generation of a $C_0$-semigroup by the
 streaming operator with general multiplying boundary conditions.
 A first approach, presented in \cite{Boulanouar1}, is based on
 the Hille-Yosida's Theorem.
 Here, we present a second approach based on the construction of the
 generated semigroup, without using the Hille-Yosida's Theorem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

 \newcommand{\abs}[1]{|#1|}
 \newcommand{\norm}[1]{\|#1\|}

\section{Introduction}

Let us consider a particle population (neutrons, photons, molecules
of gas,\dots ) in some domain of $\mathbb{R}^n$. Each particle is
distinguished by its position $x\in X\subset\mathbb{R}^n$ and its
directional velocity $v\in V\subset\mathbb{R}^n$. If we  denote by
$f(t,x,v)$ the density of particles having, at the time $t$, the
position $x$ with the directional velocity $v$, then particle
population is governed by the following evolution equation
\begin{equation}\label{e:E1}
\frac{\partial f}{\partial t}(t)=-v\cdot\nabla_xf(t)=:T_Kf(t),
\end{equation}
where $(x,y)\in\Omega=X\times V$ and $t\geq0$.
The operator $T_K$ is called the streaming operator
describing the transport of particles and it is
equipped with following general boundary conditions
\begin{equation}\label{e:E2}
f(t)\big|_{\Gamma_-}=K\big(f(t)\big|_{\Gamma_+}\big)
\end{equation}
where $f(t)|_{\Gamma_-}$ (resp. $f(t)|_{\Gamma_+}$)
is the incoming (resp. outgoing) particle flux which is the restriction
of the density $f(t)$ on the subset $\Gamma_-$ (resp. $\Gamma_+$)
of $\partial X\times V $.
The boundary operator
$K$ is linear and bounded on suitable function spaces.
All of known boundary conditions (vacuum, specular reflections,
periodic,\dots ) are special examples of our general context.
(see the next section for more explanations).

When $\|K\|\leq1$, the existence of a strongly continuous semigroup
has been investigated by several authors and
important results have been cleared
in \cite{Beals,Protopopescu,Voigt}.

However, the case $\|K\|\geq1$ has been rarely studied and the first
approach, based on Hille-Yosida's Theorem, is given  in
\cite{Boulanouar1} according to some geometrical restrictions
on $X$ and $V$ that we have expressed in the definition \ref{DEFINITION}.
Namely, the difficulty regarding the case $\|K\|>1$
is linked to the increasing number of incoming particles.
In this case, the time sojourn of particles in $X$
may be arbitrary small and intuitively the boundary operator $K$
does not take too much into account such as particles.

The motivation, of this present work, is to give a second approach
when $\|K\|\geq1$ without using the Hille-Yosida's Theorem.
This approach is concerned by two steps.
The first one is devoted to the construction of
a $C_0$-semigroup.
In the second one, we show that $T_K$
is the infinitesimal generator of this semigroup.

To obtain our objective,
we use our technics successfully applied in
\cite{Boulanouar4,Boulanouar5}.
We point out that this work is new and gives the explicit
expression of the generated semigroup.

\section{Estatement of the problem}

We consider Banach space
$L^p(\Omega)$ ($1\leq p<\infty$) with its natural norm
\begin{equation}\label{e:NORME1}
\|\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
$$
W^p(\Omega)=\{\varphi\in L^p(\Omega),\; v\cdot\nabla_x\varphi\in L^p(\Omega)\},
$$
with the norm
$\|\varphi\|_{W^p(\Omega )}=
[\|\varphi\|_p^p+\norm{v\cdot\nabla_x\varphi}_p^p]^{1/p}$.
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{gather*}
\Gamma=\partial X\times V,\quad
\Gamma_0=\{(x,v)\in\Gamma,\; v\cdot n(x)=0\},\\
\Gamma_+=\{(x,v)\in\Gamma,\; v\cdot n(x)>0\},\quad
\Gamma_-=\{(x,v)\in\Gamma,\; v\cdot n(x)<0\},
\end{gather*}
and suppose that $d\gamma d\mu(\Gamma_0)=0$.
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
$$
\tau(x,v)=\inf\{t>0,\; x-tv\not \in X\}.
$$
Now,  we use the context of \cite{Boulanouar1} as follows

\begin{definition}\label{DEFINITION} \rm
The pair $(X,V)$ is regular if
$$
\tau_0:= \inf_{(x,v)\in \Gamma_+}\tau(x,v)>0.
$$
\end{definition}

We also consider the trace spaces
$L^p(\Gamma_{\pm})$ equipped with the norm
\begin{equation*}
\|\varphi\|_{L^p(\Gamma_{\pm})}=
\Big[\int_{\Gamma_{\pm}}\abs{\varphi(x,v)}^pd\xi\Big]^{1/p},
\end{equation*}
where $d\xi=\abs{v\cdot n(x)}d\gamma d\mu$.
The first consequence of the regularity of the pair $(X,V)$ is
as follows.

\begin{lemma}[\cite{Boulanouar1}]\label{TRACES}
If the pair $(X,V)$ is regular, then the trace applications
$$
\gamma_+\;:W^p(\Omega)\; \longrightarrow L^p(\Gamma_+),\quad
\gamma_-\;:W^p(\Omega)\; \longrightarrow L^p(\Gamma_-),
$$
are linear and continuous.
\end{lemma}


Finally, if we consider the boundary operator
\begin{equation}\label{e:K}
K\in\mathcal{L}\left(L^p(\Gamma_{+}),L^p(\Gamma_{-}) \right),
\end{equation}
then the previous Lemma gives a sense to the operator
\begin{gather*}
T_K\varphi=-v\cdot\nabla_x\varphi\quad
\text{defined on the domain}\\
D(T_K)=\{\varphi\in W^p(X\times V),\;
\gamma_-\varphi=K\gamma_+\varphi\}.
\end{gather*}
We set $\|K\|:=\|K\|_{\mathcal{L}(L^p(\Gamma_{+}),L^p(\Gamma_{-}))}$
for the rest of this article.
If $K=0$, the operator $T_0$ has  properties that we summarize as
follows.

\begin{lemma}\label{LEMMA2.2}
The operator $T_0$, on $L^p(\Omega )$ $(p\geq1)$, generates
a contraction $C_0$-semi\-group
$\{U_0(t)\}_{t\geq0}$ given by
\begin{equation}\label{LEMMA2.2:b}
U_0(t)\varphi(x,v)=
\chi\left(t-t(x,v)\right)\varphi\left(x-tv,v\right),
\end{equation}
where
\begin{equation}\label{LEMMA2.2:c}
\chi\left(t-t(x,v)\right)=
\begin{cases}
1&\text{if } t(x,v)-t\geq0,\\
0&\text{otherwise.}
\end{cases}
\end{equation}
\end{lemma}

We conclude this section with the following lemma that we will need later.

\begin{lemma}\label{LEM2.2}
Suppose that the pair $(X,V)$ is regular and let
$\varphi\in W^p(\Omega)$ and $\lambda>0$.
If  we set
\begin{equation}\label{e:DECOMPOSITION}
\begin{gathered}
\Psi(x,v)=\epsilon_\lambda(x,v)\gamma_-\varphi(x-t(x,v)v,v),\\
\Phi=\varphi-\Psi,
\end{gathered}
\end{equation}
where $\epsilon_\lambda(x,v)=e^{-\lambda t(x,v)}$,
then the following statements hold
\begin{enumerate}
\item $\Psi\in W^p(\Omega)$ and $\Phi\in D(T_0)$;
\item the application
$t\geq0\rightarrow \gamma_+[U_0(t)\varphi]\in L^p(\Gamma_+)$
is continuous.
\end{enumerate}
\end{lemma}

\begin{proof}
(1) Let $\varphi\in W^p(\Omega)$ and $\lambda>0$. As we have
$v\cdot\nabla_x\Psi+\lambda\Psi=0$ with
$\gamma_-\Psi=\gamma_-\varphi\in L^p(\Gamma_-)$, then a simple
calculation gives us
\begin{equation*}
\norm{v\cdot\nabla_x\Psi}_p^p=\lambda\|\Psi\|_p^p
\leq\lambda\big[\frac{1}{p\lambda}\big]^{1/p}
\norm{\gamma_-\varphi}_{L^p(\Gamma_-)}^p<\infty
\end{equation*}
which implies
\begin{gather*}
\|\Psi\|_{W^p(\Omega)}=[\|\Psi\|_p^p+\norm{v\cdot\nabla_x\Psi}_p^p]^{1/p}
<\infty,\\
\norm{\Phi}_{W^p(\Omega)}=\norm{\varphi-\Psi}_{W^p(\Omega)}
\leq\|\varphi\|_{W^p(\Omega)}+\|\Psi\|_{W^p(\Omega)}<\infty,
\end{gather*}
and therefore $\Psi$ and $\Phi$ belong to $W^p(\Omega)$. Furthermore,
we trivially have
$\gamma_-\Phi=\gamma_-(\varphi-\Psi)=\gamma_-\varphi-\gamma_-\varphi=0$
and thus $\Phi\in D(T_0)$.

(2) Let $\varphi\in W^p(\Omega)$ and $\lambda>0$.
For all $h>0$ and all $t\geq0$ we have
\begin{equation}\label{e:az}
\begin{aligned}
&\norm{\gamma_+U_0(t+h)\varphi-\gamma_+U_0(t)\varphi}_{L^p(\Gamma_+)}\\
&=\norm{\gamma_+U_0(t+h)\Psi-\gamma_+U_0(t)\Psi+\gamma_+
U_0(t+h)\Phi-\gamma_+U_0(t)\Phi}_{L^p(\Gamma_+)}\\
&\leq
\norm{\gamma_+U_0(t+h)\Psi-\gamma_+U_0(t)\Psi}_{L^p(\Gamma_+)}
+\norm{\gamma_+U_0(t+h)\Phi-\gamma_+U_0(t)\Phi}_{L^p(\Gamma_+)}\\
&=:I_1(h)+I_2(h).
\end{aligned}
\end{equation}
As $\Phi\in D(T_0)$,
Lemmas \ref{TRACES} and \ref{LEMMA2.2},
imply
\begin{equation*}
\begin{aligned}
\lim_{h\to0}I_2(h)
&=
\lim_{h\to0}\norm{\gamma_+U_0(t+h)\Phi-\gamma_+U_0(t)\Phi}_{L^p(\Gamma_+)}\\
&\leq\norm{\gamma_+}_{\mathcal{L}(D(T_0),L^p(\Gamma_+))}
\lim_{h\to0}\norm{U_0(t+h)\Phi-U_0(t)\Phi}_{D(T_0)}\\
&=0.
\end{aligned}
\end{equation*}
Next, a simple calculation shows that
\begin{align*}
\lim_{h\to0}I_1(h)^p
&=\lim_{h\to0}\norm{\gamma_+U_0(t+h)\Psi
 -\gamma_+U_0(t)\Psi}_{L^p(\Gamma_+)}^p\\
&=\lim_{h\to0}\int_{\Gamma_+}
\abs{\chi(t+h-t(x,v))e^{\lambda(t+h)}-\chi(t-t(x,v))e^{\lambda
t}}^p
\abs{\Psi(x,v)}^pd\xi\\
&=0
\end{align*}
This completes the proof.
\end{proof}

\section{Construction of the semigroup}

In this section, we construct the semigroup $\{U_K(t)\}_{t\geq0}$ when
$\|K\|\geq1$. In order to show Theorem \ref{THEOREM3.1} which is
the main result, we begin by
\begin{lemma}\label{LEMMA3.1}
The following 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\left(\mathbb{R}_+,L^p(\Gamma_-)\right);\\
u(0)=f_0\in L^p(\Omega),
\end{gathered}
\label{Pf-f0}
\end{equation}
admits a unique solution $u=u(t,x,v)=u(t)(x,v)$. Furthermore, for
all $t\geq0$, we have
\begin{equation}\label{LEMMA3.1: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 $f_-\in L^p\left(\mathbb{R}_+,L^p(\Gamma_-)\right)$
and $f_0\in L^p(\Omega)$.
First. Using \cite[pp.1124]{Dautray} it follows that
Cauchy's problem ${\rm P(f_-,f_0)}$ has a unique solution
given by
\begin{equation}\label{LEMMA3.2:dd}
u(t,x,v)=\xi\left(t-t(x,v)\right)f_-(t-t(x,v),x-t(x,v)v,v)+U_0(t)f_0(x,v).
\end{equation}
where $\xi$ is given in Lemma \ref{LEMMA3.2}.
Next. Multiplying
first equation of Cauchy's problem ${\rm (P)}(f_-,f_0)$ by $\mathop{\rm sgn}
u\abs{u}^{p-1}$ and using
\begin{equation*}
\mathop{\rm sgn} u\abs{u}^{p-1}v\cdot\nabla_xu=\frac{1}{p}v\cdot\nabla_x\abs{u}^p,
\end{equation*}
with an integrating over $\Omega$, 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 achieves the proof.
\end{proof}

\begin{remark} \label{rmk3.1} \rm
In the sequel, we use the fact that all expression on the form of
\eqref{LEMMA3.2:dd} is automatically solution of
Cauchy's problem ${\rm P(f_-,f_0)}$.
\end{remark}

The second consequence of the regularity of the pair
$(X,V)$ is as follows.

\begin{lemma}\label{LEMMA3.2}
Suppose that the pair $(X,V)$ is regular and let
\begin{equation*}
\xi\left(t-t(x,v)\right)=
\begin{cases}
1&\text{if } t(x,v)-t\leq0,\\
0&\text{otherwise.}
\end{cases}
\end{equation*}
If $0\leq t\leq \tau_0$, then we have
$\gamma_+\xi\left(t-t(\cdot,\cdot)\right)=0$.
\end{lemma}

\begin{proof}
By the regularity of the pair $(X,V)$, we have
$$
0\leq t\leq\tau_0=
\inf_{(x,v)\in \Gamma_+}\tau(x,v)\leq \tau(x,v)
$$
a.e. $(x,v)\in\Gamma_+$, and therefore
\begin{equation*}
\gamma_+\left[\xi(t-t(\cdot,\cdot))\right](x,v)=
\xi(t-\tau(x,v))=0,
\end{equation*}
a.e. $(x,v)\in\Gamma_+$.
\end{proof}

\begin{lemma}\label{LEMMA3.3}
Suppose that the pair $(X,V)$ is regular.
For all $0\leq t\leq \tau_0$, the operator $A_K(t)$ given  by
\begin{equation*}
A_K(t)\varphi(x,v)=
\xi\left(t-t(x,v)\right)K\left[\gamma_+U_0\left(t-t(x,v)\right)
\varphi\right]\left(x-t(x,v)v,v\right)
\end{equation*}
is a linear and bounded from $L^p(\Omega)$ into itself.
Furthermore, we have
\begin{enumerate}
\item $A_K(0)=0$;
\item $\lim_{t\searrow0}\norm{A_K(t)\varphi}_p=0$
 for all $\varphi\in L^p(\Omega)$;
\item $\gamma_+A_K(t)=0$ for $0\leq t\leq\tau_0$;
\item $A_K(t)A_K(s)=0$ for all
$0\leq t,s\leq\tau_0$ such that $0\leq t+s\leq\tau_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $0\leq t\leq \tau_0$ and $\varphi\in L^p(\Omega)$.
As $u(t)=A_K(t)\varphi$ is the solution
of Cauchy's problem
${\rm P}(f_-=K\left[\gamma_+U_0(\cdot)\varphi\right], f_0=0)$
then \eqref{LEMMA3.1:a} and the boundedness of
$K$ implies
\begin{equation*}
\norm{A_K(t)\varphi}_p^p\leq\int_0^t
\norm{K\left[\gamma_+U_0(s)\varphi\right]}_{L^p(\Gamma_-)}^pds
\leq\|K\|^p\int_0^t
\norm{\gamma_+U_0(s)\varphi}_{L^p(\Gamma_+)}^p.
\end{equation*}
However, $u(t)=U_0(t)\varphi$ is solution
of Cauchy's problem \eqref{Pf-f0} with $f_-=0, f_0=\varphi$,
and therefore \eqref{LEMMA3.1:a} implies
\begin{equation}\label{LEMMA3.3:aa}
\int_0^t\norm{\gamma_+U_0(s)\varphi}_{L^p(\Gamma_+)}^p
=\|\varphi\|_p^p-\norm{U_0(t)\varphi}_p^p.
\end{equation}
 From the previous two relations we obtain
\begin{equation*}
\norm{A_K(t)\varphi}_p^p
\leq\|K\|^p\left[\|\varphi\|_p^p-\norm{U_0(t)\varphi}_p^p\right]
\leq\|K\|^p\|\varphi\|_p^p
\end{equation*}
which implies that $A_K(t)\varphi\in L^p(\Omega)$ and
the boundedness of the operator $A_K(t)$ follows. Points (1) and (2)
follow from the fact that $\{U_0(t)\}_{t\geq0}$ is a $C_0$-semigroup.

(3) This point obviously follows from previous Lemma.

(4) Let $0\leq t,s\leq\tau_0$ such that $0\leq t+s\leq\tau_0$ and
$\varphi\in L^p(\Omega)$.
A simple calculation shows
that the expression of $A_K(t)A_K(s)\varphi$ contains the following
function
\begin{equation*}
\alpha(x,v,x',v'):=
\xi\Bigl(s-t\Bigl(x'-\bigr(t-t(x,v)\bigr)v',v'\Bigr)\Bigr)
\end{equation*}
for a.e $(x,v)\in\Omega$ and a.e $(x',v')\in\Gamma_+$.
Using the definition of $\xi$ in previous Lemma,
we get that
\begin{align*}
\alpha(x,v,x',v')=0&\Longleftrightarrow
s<t\Bigl(x'-\bigr(t-t(x,v)\bigr)v',v'\Bigr)\\
&\Longleftrightarrow
s<\tau(x',v')-(t-t(x,v))\\
&\Longleftrightarrow
s+t<\tau(x',v')+t(x,v)
\end{align*}
for a.e $(x,v)\in\Omega$ and a.e $(x',v')\in\Gamma_+$.
But, the regularity of the pair $(X,V)$ in the sense of Definition
\ref{DEFINITION} gives us
$$
t+s\leq\tau_0=\inf_{(x',v')\in \Gamma_+}\tau(x',v')
\leq\tau(x',v')<\tau(x',v')+t(x,v)
$$
for a.e. $(x,v)\in\Omega$ and a.e. $(x',v')\in\Gamma_+$
which implies that $\alpha(\cdot,\cdot,\cdot,\cdot)=0$
and therefore $A_K(t)A_K(s)=0$. The fourth point is proved.
\end{proof}

The main result of this section is given as follows.

\begin{theorem}\label{THEOREM3.1}
Suppose that the pair $(X,V)$ is regular.
The family of operators $\{U_K(t)\}_{t\geq0}$ defined by
\begin{equation}\label{THEOREM3.1:a}
\begin{gathered}
U_K(t)=\left[U_0(\tau_0)+ A_K(\tau_0)\right]^n\left[U_0(r)+ A_K(r)\right],\\
\text{if $t=n\tau_0+r$ with $0\leq r<\tau_0$ and $n\in\mathbb{N}$},
\end{gathered}
\end{equation}
is a $C_0$-semigroup on $L^p(\Omega)$.
Furthermore, we have
\begin{equation}\label{THEOREM3.1:b}
\begin{gathered}
U_K(t)\varphi(x,v)=U_0(t)(x,v)+\\
\xi\left(t-t(x,v)\right)
K\left[\gamma_+U_K\left(t-t(x,v)\right)\varphi\right]\left(x-t(x,v)v,v\right)
\end{gathered}
\end{equation}
for all $t\geq0$, a.e. $(x,v)\in \Omega$ and all $\varphi\in L^p(\Omega)$,
where $\xi$ is given in Lemma \eqref{LEMMA3.2}.
\end{theorem}

\begin{proof}
Note that from previous Lemma and Lemma \ref{LEMMA2.2}
the operator $U_0(t)+ A_K(t)$
($0\leq t\leq\tau_0$) is a linear bounded from
$L^p(\Omega)$ into itself. Thus $U_K(t)$ is also
linear bounded  for all $t\geq0$,
$U_K(0)=U_0(0)+A_K(0)=I$. Furthermore, if $t\leq\tau_0$
then we trivially have
\begin{equation*}
\lim_{t\searrow0}\|U_K(t)\varphi-\varphi\|_p
=\lim_{t\searrow0}\|U_0(t)\varphi-\varphi\|_p +
\lim_{t\searrow0}\|A_K(t)\varphi t\|_p=0,
\end{equation*}
for all $\varphi\in L^p(X\times V)$.
Now, let us show that $U_K(t)U_K(s) =U_K(t+s)$
for all $t,s\geq0$.

First, note that if $0\leq t,s\leq\tau_0$ such that
$0\leq t+s\leq\tau_0$, a simple calculation shows that
$U_0(t)A_K(s)+A_K(t)U_0(s)=A_K(t+s)$ and therefore
\begin{equation}\label{LEM3.3:BK}
\begin{aligned}
U_K(t)U_K(s)
&= \left[U_0(t)+A_K(t)\right]\left[U_0(s)+A_K(s)\right]\\
&=U_0(t+s)+U_0(t)A_K(s)+A_K(t)U_0(s) +A_K(t)A_K(s)\\
& =U_0(t+s)+A_K(t+s)\\
&=U_K(t+s),
\end{aligned}
\end{equation}
where we have used the relation $A_K(t)A_K(s)=0$ in previous lemma.
Thus
\begin{equation}\label{LEM3.1:ff}
U_K(t)U_K(s)=U_K(t+s),
\end{equation}
for all $0\leq t, s\leq\tau_0$ such that $0\leq t+s\leq\tau_0$.

Next, for all $t,s\geq0$, there exists
$n_t,n_s\in\mathbb{N}$ and $0\leq r_t,r_s<\tau_0$ such that
$t=n_t\tau_0+r_t$ and $t=n_s\tau_0+r_s$. In this case
\eqref{LEM3.1:ff} implies
\begin{align*}
U_K(t)U_K(s)
&=[U_K(\tau_0)]^{n_t}U_K(r_t)
[U_K(\tau_0)]^{n_s}U_K(r_t)\\
&=[U_K(\tau_0/2)]^{2n_t}[U_K(r_t/2)]^2
[U_K(\tau_0/2)]^{2n_s}[U_K(r_t/2)]^2\\
&=[U_K(\tau_0/2)]^{2n_t}
[U_K(\tau_0/2)]^{2n_s}
[U_K(r_t/2)]^2[U_K(r_t/2)]^2\\
&=[U_K(\tau_0)]^{n_t}
[U_K(\tau_0)]^{n_s}
[U_K((r_t+r_t)/2)]^2\\
&=[U_K(\tau_0)]^{n_t+n_s}
[U_K((r_t+r_t)/2)]^2
\end{align*}
because
$$
0\leq (\tau_0/2), (r_t/2), (r_s/2), (r_t+\tau_0)/2,
(r_s+\tau_0)/2, (r_t+r_s)/2<\tau_0$$
Now, if $r_t+r_s<\tau_0$ then $(((r_t+r_s)/2)+((r_t+r_s)/2))<\tau_0$
which implies, by \eqref{LEM3.1:ff},
that $[U_K((r_t+r_t)/2)]^2=U_K(r_t+r_t)$
and therefore $U_K(t)U_K(s)=U_K(t+s)$
because $t+s=(n_t+n_s)\tau_0 +(r_t+r_s)$
with $0\leq (r_t+r_s)<\tau_0$.

If $\tau_0\leq r_t+r_s<2\tau_0$ then
$r_t+r_s=\tau_0+r$ with $0\leq r<\tau_0$.
As we have,
$$
0\leq (r_t/2), (r_s/2), (r_t+r_s)/2, (r/2)<\tau_0
$$
then \eqref{LEM3.1:ff} implies
\begin{align*}
U_K(t)U_K(s)&=[U_K(\tau_0)]^{n_t+n_s}
[U_K((r_t+r_t)/2)]^2\\
&=[U_K(\tau_0)]^{n_t+n_s}
[U_K((\tau_0/2)+(r/2))]^2\\
&=[U_K(\tau_0)]^{n_t+n_s}
[U_K(\tau_0/2)]^2[U_K(r/2)]^2\\
&=[U_K(\tau_0)]^{n_t+n_s}
U_K(\tau_0)U_K(r)\\
&=[U_K(\tau_0)]^{n_t+n_s+1}U_K(r)\\
&=U_K(t+s)
\end{align*}
because $(t+s)=(n_t+n_s)\tau_0+(r_t+r_s)=(n_t+n_s)\tau_0+(\tau_0+r)=
(n_t+n_s+1)\tau_0+r$ with $0\leq r<\tau_0$.

Now, let us show \eqref{THEOREM3.1:b}.
If we denote
\begin{equation*}
\overline A_K(t)\varphi(x,v)=\xi\left(t-t(x,v)\right)
K\left[\gamma_+U_K\left(t-t(x,v)\right)\varphi\right]\left(x-t(x,v)v,v\right),
\end{equation*}
then, we have to show the following formula
\begin{equation}\label{LEM44:aa}
U_K(t)=U_0(t)+\overline A_K(t),\quad t\geq0.
\end{equation}
Let $\varphi\in W^p(\Omega)$.
If $0\leq t<\tau_0$, the third point of Lemma
\ref{LEMMA3.3} infers that
$\gamma_+U_K(t)\varphi=\gamma_+U_0(t)\varphi+\gamma_+A_K(t)\varphi=
\gamma_+U_0(t)\varphi$
which implies
\begin{equation*}
\gamma_+U_K(t-t(x,v))\varphi=\gamma_+U_0(t-t(x,v))\varphi
\end{equation*}
for a.e. $(x,v)\in\Omega$ such that $t(x,v)\leq t$
and therefore
\begin{align*}
\overline A_K(t)\varphi(x,v)&=\xi(t-t(x,v))K\left[\gamma_+U_K(t-t(x,v))\varphi\right](x-t(x,v)v,v)\\
&=\xi(t-t(x,v))K\left[\gamma_+U_0(t-t(x,v))\varphi\right](x-t(x,v)v,v)\\
&=A_K(t)\varphi(x,v)
\end{align*}
for a.e. $(x,v)\in\Omega$ and all $0\leq t<\tau_0$.
Thus, we have
\begin{equation*}
U_K(t)\varphi=U_0(t)\varphi+A_K(t)\varphi=
U_0(t)\varphi+\overline A_K(t)\varphi
\end{equation*}
for all $0\leq t<\tau_0$ and all $\varphi\in W^p(\Omega)$.
Now the density of $ W^p(\Omega)$ in $L^p(\Omega)$
implies that \eqref{LEM44:aa} holds for all $0\leq t<\tau_0$.

Next. Suppose that \eqref{LEM44:aa} holds for
$(n-1)\tau_0\leq t<n\tau_0$ and all $\varphi\in L^p(\Omega)$. If
$n\tau_0\leq t<(n+1)\tau_0$ then
$(n-1)\tau_0\leq t-\tau_0<n\tau_0$
and therefore
\begin{align*}
U_K(t)\varphi&=U_K(t-\tau_0)U_K(\tau_0)\varphi\\
&=\big[U_0(t-\tau_0)+\overline A_K(t-\tau_0)\big]
\big[U_0(\tau_0)+\overline A_K(\tau_0)\big]\varphi\\
&=U_0(t)\varphi+U_0(t-\tau_0)\overline A_K(\tau_0)\varphi
+\overline A_K(t-\tau_0)U_K(\tau_0)\varphi.
\end{align*}
Using the definition of $\xi$ given Lemma \ref{LEMMA3.2},
a simple calculation implies
$U_0(t-\tau_0)\overline A_K(\tau_0)\varphi=0$ and
$\overline A_K(t-\tau_0)U_K(\tau_0)\varphi=\overline A_K(t)\varphi$.
Thus
$$
U_K(t)\varphi=
U_0(t)\varphi+\overline A_K(t)\varphi
$$
which prove \eqref{LEM44:aa} for
$n\tau_0\leq t<(n+1)\tau_0$ and therefore for all $t\geq0$.
\end{proof}

\section{Generation Theorem}

To show the main result of this work, we  needed the following result.

\begin{lemma}\label{LEMMA4.1}
Suppose that the pair $(X,V)$ is regular.
If $\varphi\in W^p(\Omega)$ and $\lambda>0$, then we have
\begin{equation*}
\lim_{t\searrow0}
\big\|\frac{A_K(t)\varphi+U_0(t)\Psi-\Psi}{t}-\lambda\Psi\big\|_p=
\norm{K\gamma_+\varphi-\gamma_-\varphi}_{L^p(\Gamma_-)}
\end{equation*}
where $\Psi$ is given by  \eqref{e:DECOMPOSITION}.
\end{lemma}

\begin{proof}
Let $0<t\leq\tau_0$ and $\varphi\in W^p(\Omega)$.
Using \eqref{e:DECOMPOSITION},
a simple calculation gives us
\begin{align*}
&\big[\frac{A_K(t)\varphi+U_0(t)\Psi-\Psi}{t}-\lambda\Psi\big](x,v)\\
&= \Bigl(\frac{e^{\lambda t}-1}{t}-\lambda\Bigr)\Psi(x,v)\\
&\quad +\frac{1}{t}
\Bigl(A_K(t)\varphi(x,v)
-\xi(t-t(x,v))e^{\lambda t}\epsilon_\lambda(x,v)\gamma_-\varphi\left(x-t(x,v)v,v\right)\Bigr)\\
&=:I_1(t)\varphi+I_2(t)\varphi\\
\end{align*}
a.e. $(x,v)\in\Omega$, which implies
\begin{equation*}
\lim_{t\searrow0}\norm{I_2(t)\varphi}_p-\lim_{t\searrow0}\norm{I_1(t)\varphi}_p
\leq\lim_{t\searrow0}
\big\|\frac{A_K(t)\varphi+U_0(t)\Psi-\Psi}{t}-\lambda\Psi\big\|_p
\end{equation*}
and
\begin{equation*}
\lim_{t\searrow0}
\big\|\frac{A_K(t)\varphi+U_0(t)\Psi-\Psi}{t}-\lambda\Psi\big\|_p
\leq
\lim_{t\searrow0}\norm{I_2(t)\varphi}_p
+\lim_{t\searrow0}\norm{I_1(t)\varphi}_p.
\end{equation*}
As we obviously have $\lim_{t\to0}I_1(t)=0$, then we get
\begin{equation}\label{LEMMA4.1:cc}
\lim_{t\searrow0}
\norm{\frac{A_K(t)\varphi+U_0(t)\Psi-\Psi}{t}-\lambda\Psi}_p
=\lim_{t\searrow0}\norm{I_2(t)\varphi}_p.
\end{equation}
But $u(t)=tI_2(t)\varphi$ is the solution of Cauchy's problem
${\rm P}(f_-=K\left[\gamma_+U_0(t)\varphi\right]-e^{\lambda t}\gamma_-\varphi, f_0=0)$,
thus \eqref{LEMMA3.1:a} implies
\[
\norm{I_2(t)\varphi}_p^p
=\frac{1}{t}\int_0^t
\norm{K\left[\gamma_+U_0(s)\varphi\right]-
e^{\lambda s}\gamma_-\varphi}_{L^p(\Gamma_-)}^pds
-\frac{1}{t}\int_0^t \norm{\gamma_+I_2(t)\varphi}_{L^p(\Gamma_-)}^pds.
\]
 From the regularity of the pair $(X,V)$ we get, by Lemma \ref{LEMMA3.2}
and the third point of Lemma \ref{LEMMA3.3}, that
$\gamma_+I_2(t)\varphi=0$ and therefore
the previous relation becomes
\begin{equation}\label{LEMMA4.1:dd}
\norm{I_2(t)\varphi}_p^p=
\frac{1}{t}\int_0^t
\norm{K\left[\gamma_+U_0(s)\varphi\right]-
e^{\lambda s}\gamma_-\varphi}_{L^p(\Gamma_-)}^pds.
\end{equation}
Using the boundedness of $K$ and the second point of Lemma
\ref{LEM2.2} we obtain the continuity of the application
$$
0\leq t\leq\tau_0\longrightarrow K\left[\gamma_+U_0(s)\varphi\right]-
e^{\lambda s}\gamma_-\varphi\in L^p(\Gamma_-)
$$
for all $\varphi\in W^p(\Omega)$ and therefore
\eqref{LEMMA4.1:dd} becomes
\begin{equation}\label{LEMMA4.1:ee}
\lim_{t\searrow0}\norm{I_2(t)\varphi}_p^p=
\norm{K\left[\gamma_+\varphi\right]-\gamma_-\varphi}_{L^p(\Gamma_-)}^p.
\end{equation}
which achieves the proof by \eqref{LEMMA4.1:cc}.
\end{proof}

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

\begin{theorem}\label{THEOREM4.1}
Suppose that the pair $(X,V)$ is regular.
Then the operator $T_K$ given by
\begin{equation}\label{THEOREM4.1:a}
\begin{aligned}
&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{aligned}
\end{equation}
is the infinitesimal generator of the $C_0$-semigroup $\{U_K(t)\}_{t\geq0}$
satisfying
\begin{equation}\label{THEOREM4.1:b}
U_K(t)\varphi(x,v)=U_0(t)(x,v)
+\xi\left(t-t(x,v)\right) K\left[\gamma_+U_K\left(t-t(x,v)\right)
\varphi\right]\left(x-t(x,v)v,v\right)
\end{equation}
for all $t\geq0$ and a.e. $(x,v)\in \Omega$ and all
$\varphi\in L^p(\Omega)$,
where $\xi$ is given the Lemma \ref{LEMMA3.2}.
Furthermore, if $\|K\|\geq1$, then we have
\begin{equation}\label{THEOREM4.1:c}
\norm{U_K(t)}_{\mathcal{L}(L^p(\Omega))}
\leq\|K\|\exp\bigl(\frac{t}{\tau_0}\ln\|K\|\bigr),\quad
t\geq0.
\end{equation}
\end{theorem}

\begin{proof}
Note that the existence of the $C_0$-semigroup
$\{U_K(t)\}_{t\geq0}$ and \eqref{THEOREM4.1:b}
are already proved in Theorem \ref{THEOREM3.1}.
Let us shown that the operator $T_K$ is the generator of our semigroup.

Let $0<t<\tau_0$ and $\varphi\in W^p(\Omega)$.
Using \eqref{e:DECOMPOSITION} we easily get
\begin{equation}\label{THEOREM1:aa}
\big[\frac{U_K(t)\varphi-\varphi}{t}+v\cdot\nabla_x\varphi\big]
=\big[\frac{U_0(t)\Phi-\Phi}{t}-T_0\Phi\big]
+ \frac{A_K(t)\varphi+U_0(t)\Psi-\Psi}{t}-\lambda\Psi
\end{equation}
which implies
\begin{align*}
&\big\|\frac{U_K(t)\varphi+\varphi}{t}+v\cdot\nabla_x\varphi\big\|_p\\
&\leq\big\|\frac{U_0(t)\Phi-\Phi}{t}-T_0\Phi\big\|_p
+\big\|\frac{A_K(t)\varphi+U_0(t)\Psi-\Psi}{t}-\lambda\Psi\big\|_p
\end{align*}
and
\begin{align*}
&\big\|\frac{A_K(t)\varphi+U_0(t)\Psi-\Psi}{t}-\lambda\Psi\big\|_p
-\norm{\frac{U_0(t)\Phi-\Phi}{t}-T_0\Phi}_p\\
&\leq\big\|\frac{U_K(t)\varphi-\varphi}{t}+v\cdot\nabla_x\varphi\big\|_p.
\end{align*}
Lemmas \ref{LEMMA2.2} and \ref{LEMMA4.1} imply
\begin{equation}\label{e:EQUIVALENCE}
\lim_{t\searrow0}
\big\|\frac{U_K(t)\varphi-\varphi}{t}+v\cdot\nabla_x\varphi\big\|_p
=\lim_{t\searrow0}
\norm{K\gamma_+\varphi-\gamma_-\varphi}_{L^p(\Gamma_-)}^p
\end{equation}
which implies
\begin{equation*}
\varphi\in D(T_K)\Longleftrightarrow
K\gamma_+\varphi-\gamma_-\varphi=0
\Longleftrightarrow
\lim_{t\searrow0}
\big\|\frac{U_K(t)\varphi-\varphi}{t}+v\cdot\nabla_x\varphi\big\|_p=0
\end{equation*}
and therefore \eqref{e:EQUIVALENCE} gives us
\begin{equation*}
\lim_{t\searrow0}
\big\|\frac{U_K(t)\varphi-\varphi}{t}+T_K\varphi\big\|_p=0.
\end{equation*}
Thus $T_K$ is the generator of the semigroup
$\{U_K(t)\}_{t\geq0}$.
Now, let us show \eqref{THEOREM4.1:c}.

Let $0\leq t\leq\tau_0$ and $\varphi\in D(T_K)$.
As, $u(t)=U_K(t)\varphi=U_0(t)\varphi+A_K(t)\varphi$
is the solution of Cauchy's problem
${\rm P}(f_-=K\left[\gamma_+U_0(t)\varphi\right], f_0=\varphi)$,
then \eqref{LEMMA3.1:a} and the boundedness of the operator $K$
infer that
\begin{align*}
\norm{U_K(t)\varphi}_p^p
&=\int_0^t\norm{K\left[\gamma_+U_0(s)\varphi\right]}_{L^p(\Gamma_-)}^pds+
\|\varphi\|_p^p
-\int_0^t\norm{\gamma_+U_0(s)\varphi}_{L^p(\Gamma_+)}^pds\\
&\leq\left[\|K\|^p-1\right]\int_0^t\norm{\gamma_+U_0(s)\varphi}_{L^p(\Gamma_+)}^pds+
\|\varphi\|_p^p
\end{align*}
where we have used the third point of Lemma \ref{LEMMA3.3}.
Using \eqref{LEMMA3.3:aa} and the fact that $\|K\|\geq1$,
the previous relation becomes
\begin{equation*}
\norm{U_K(t)\varphi}_p^p\leq
\left[\|K\|^p-1\right]\|\varphi\|_p^p+\|\varphi\|_p^p
=\|K\|^p\|\varphi\|_p^p
\end{equation*}
and therefore
\begin{equation*}
\norm{U_K(t)}_{\mathcal{L}(L^p(\Omega))}\leq\|K\|,
\quad\text{for all $0\leq t\leq\tau_0$,}
\end{equation*}
because of the density of $D(T_K)$ in
$L^p(\Omega)$.
Now, for all $t\geq0$, there exists $n\in\mathbb{N}$ and
$0\leq r<\tau_0$ such that
$t=n\tau_0+r$. Using previous relation and
\eqref{THEOREM3.1:a}  we get
\begin{align*}
\norm{U_K(t)}_{\mathcal{L}(L^p(\Omega))}
&=\norm{\left[U_K(\tau_0)\right]^nU_K(r)}_{\mathcal{L}(L^p(\Omega))}\\
&\leq\norm{U_K(\tau_0)}_{\mathcal{L}(L^p(\Omega))}^n\norm{U_K(r)}_{\mathcal{L}(L^p(\Omega))}\\
&\leq\|K\|^n\|K\|\\
&\leq\|K\|^{t/\tau_0}\|K\|.
\end{align*}
which prove \eqref{THEOREM4.1:c} and completes the proof.
\end{proof}

\begin{remark} \label{rmk4.1} \rm
Recall that the case $\|K\|<1$ is already studied in \cite{Boulanouar6}
without the hypothesis : $(X,V)$ is regular.
\end{remark}

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\end{document}