\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small {\em 
Electronic Journal of Differential Equations}, 
Vol. 2004(2004), No. 140, pp. 1--10.\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 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/140\hfil A linear model]
{A linear model for the dynamics of fish larvae}

\author[N. Ghouali \& T. M. Touaoula \hfil EJDE-2004/140\hfilneg]
{Noureddine Ghouali, Tarik Mohamed Touaoula} % in alphabetical order

\address{Noureddine Ghouali  \hfill\break
Department of Mathematics, University of Tlemcen \\
BP 119, Tlemcen 13000, Algeria}
\email{n\_ghouali@mail.univ-tlemcen.dz}

\address{Tarik Mohamed Touaoula \hfill\break
Department of Mathematics, University of Tlemcen \\
BP 119, Tlemcen 13000, Algeria} 
\email{touaoulatarik@yahoo.fr}

\date{}
\thanks{Submitted July 10, 2004. Published November 26, 2004.}
\subjclass[2000]{35G10} 
\keywords{Parabolic equations; characteristic curve; dynamics of the fish larvae}

\begin{abstract}
 We consider a linear model for the growth and the dispersion
 of fish larvae of certain species. Dispersion is modeled as
 entailed by the combination of transport and vertical diffusion.
 We generalize the work of Boushaba, Arino and Boussouar
 \cite{Bous}, \cite{Bous1} in the sense that horizontal
 velocities are uniform throughout the water column;
 but we deal with vertical component velocity and vertical diffusion
 depending on the space variables and on time, which was not the
 case in \cite{Bous}, \cite{Bous1}. This new vision leads us
 to non-autonomous problems, the aim of this work is to show
 the existence, uniqueness, and positivity of solutions.
\end{abstract}

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

\section{Introduction.}\label{sec:s0}

In this paper, we introduce a mathematical model for the dynamics
of the fish larvae of certain species. This model takes into
account both the physical and biological effects. For the physical
part, the model considered here stresses two main factors: 1)
Transport entailed by the currents: the currents are computed
using Navier-Stokes equations and are introduced in the equations
of the larvae as functions of space and time with sufficient
regularity to allow existence and uniqueness of stream lines.
 2) Vertical diffusion induced by vertical mixing in the upper part of
the water column. For the biological part the main parameters are
a function which gives the instantaneous rate of progression
within the stages from the egg fertilization to the end of the
yolk-sac period.

The model is expressed in a generality which encompasses a large
variety of situations. The motivation at the origin of this work
is the study of the dynamics of the Bay of Biscay anchovy
\cite{Arino}, that is to say, a region of the Atlantic ocean
close to the French coast, bordered eastward by the continental
shelf. The Bay of Biscay goes from the Northern Spanish coast up
to about 46$^{0}$ in ``latitude''. In this region at the end of
May , a thermocline establishes itself: the top of the
thermocline is roughly at the same distance $z_{therm}$ from the
surface. The thermocline divides the
water column into three regions: the upper part, from the surface to $%
z_{therm}$ deep, the so called mixed layer. This is where the
larvae grow. Below is the thermocline, a rather thin layer where
the temperature loses rapidly a few degrees and the vertical
mixing coefficient is negligibly small. Below the thermocline is
another well mixed layer where the temperature is only slowly
changing with depth. This region is of no concern to us for the
rest of the study, which will be confined to the mathematical
issues related to the above mentioned model.

The domain of study has notably been restricted to the upper
layer, the so called mixed layer of the water column.

The purpose of this work is to perform a mathematical analysis of
the model, notably, show existence, uniqueness and positivity of
solutions. In a previous work coauthored by  Boushaba, Boussouar;
and  Arino \cite {Bous1}, a simplified version of the model of
the phytoplankton had been investigated. It was assumed that the
diffusion rate and the vertical current does not depend on time
and the horizontal current is uniform throughout the water
column. Under this assumption, it was possible to uncouple the
vertical and the horizontal components in the following sense:
the study was restricted to each of the horizontal streamlines:
the restriction to such a line reduces the functions of time
horizontal components to functions of time so that the full model
reduces on such a line to a diffusion equation in the vertical
variable coupled with a first order growth equation. Our purpose
in this work is to extend this method to the more realistic
situation where the diffusion rate and the vertical current
depends also on time and the horizontal current is uniform
throughout the water column. The idea we exploit here is the same
in the first time, that is to uncouple the vertical and the
horizontal components but the restriction on such a line gives
equations of parabolic type with time dependent coefficients. The
study of such equations takes up the main part in this work.

Time dependence is dealt with using results on time-dependent
evolution equations by Acquistapace \cite{Ac1,Ac,Ac2} and several
other authors (Lunardi \cite{Lunardi}, Tanabe \cite{Tana}). A
valuable source of information of this work was a monograph by
Tanabe \cite{Tana}. The main result of this part, stated in
theorem \ref{th:harnack1}, ensures that, under some conditions on
the coefficients of the equation, the Cauchy problem associated
with the equation has a unique classical solution, which moreover
is nonnegative if the initial value is non negative. The paper is
organized as follows. Section 2 is devoted to recall some
important theorems. Section 3 is devoted to a detailed
presentation of the model. Section 4 is devoted to proving
existence, uniqueness and positivity of solution of our problem.

\section{Notation and preliminary results}\label{sec:s1}

Let $Y$ be a Banach space and $[a,b]$ a finite interval of the
real line, then we define the space
$$
C([a,b];Y)=\{f:[a,b]\to Y: f \mbox{ is continuous}\}.
$$
Note that $C([a,b];Y)$ is a Banach space with the norm
$$ \|f\|_{C([a,b];Y)}=\sup_{s\in [a,b]}\|f(s)\|_Y.
$$
We also consider the space $C^1([a,b];Y)$ consisting of functions
$f\in C([a,b];Y)$ such that $f$ is strongly differentiable in
$[a,b]$ and $f'\in C([a,b];Y)$, with the norm
$$
\|f\|_{C^1([a,b];Y)}=\|f\|_{C([a,b];Y)}+\|f'\|_{C([a,b];Y)}.
$$
Let ${\theta} \in ]0,1[$ then we define the following Holder type
spaces
\begin{align*}
C^{\theta}([a,b];Y) &=\{f\in C([a,b];Y): [f]_{C^{\theta}([a,b];Y)}
=\sup\{\frac{\|f(s)-f(t)\|_Y}{|t-s|^{\theta}}\\
&\quad t,s \in [a,b], t\neq s \} <\infty \},
\end{align*}
which is equipped with the norm
$$
\|f\|_{C^{\theta}([a,b];Y)}=\|f\|_{C([a,b];Y)}+[f]_{C^{\theta}([a,b];Y)}\,.
$$
Also let
$$
C^{1,\theta}([a,b];Y)=\{f \in C^1([a,b];Y): f' \in
C^{\theta}([a,b];Y)\},
$$
with norm
$$
\|f\|_{C^{1,\theta}([a,b];Y)}=\|f\|_{C([a,b];Y)}+\|f'\|_{C^{\theta}([a,b];Y)}.
$$
We consider the  problem
\begin{equation}
\begin{gathered}
u_{t}(t,x)-a(t,x)u_{xx}(t,x)-b(t,x)u_{x}(t,x)-c(t,x)u(t,x)=0, \\
(t,x)\in [0,T]\times[0,1] \\
\alpha_0(t)u(t,0)-\beta_0(t)u_x(t,0)=\alpha_1(t)u(t,1)+\beta_1(t)u_x(t,1)=0,
\quad  t\in [0,T],\\
u(0,x)=\Phi(x), \quad x\in [0,1],
\end{gathered}   \label{10}
\end{equation}
under the following assumptions:
\begin{equation}
\begin{gathered}
a,b,c \in C([0,T]\times[0,1]), \\
a(.,x),b(.,x),c(.,x)\in C^{1,\delta}([0,T];\mathbb{R}) \\
\mbox {with norms indpendent of $x\in [0,1]$, for some $\delta \in
]0,1[$,} \\
a>0, c\leq 0 \quad \mbox{in } [0,T]\times[0,1],
\end{gathered}  \label{l1}
\end{equation}
To recall some propositions, we set $E=C([0,1])$,
$\|u\|_E=\sup_{x\in [0,1]}|u(x)|$, and define for each $t
\in(0,T)$,
\begin{equation}
\begin{gathered} \label{ll1}
D(A(t))=\{u\in C^2([0,1]): \alpha_0(t)u(0)-\beta_0(t)u'(0)
=\alpha_1(t)u(1)+\beta_1(t)u'(1)=0,\} \\
A(t)u=a(t,.)u''+b(t,.)u'+c(t,.)u\,.
\end{gathered}
\end{equation}

\begin{proposition}[\cite{Ac}] \label{th:harnack}
 Let $a,b,c$ be as in \eqref{10}, \eqref{l1}, and
suppose that $u\in C^2([0,1])$ is a solution of
\begin{equation}
\begin{gathered}
\lambda u-a(t,.)u''-b(t,.)u'-c(t,.)u=f \in C([0,1]),\\
\alpha_0(t)u(0)-\beta_0(t)u'(0)=z_0 \in \mathbb{C}, \\
\alpha_1(t)u(1)+\beta_1(t)u'(1)=z_1 \in \mathbb{C},
\end{gathered}  \label{ll11}
\end{equation}
where $t\in [0,T]$ is fixed and $\lambda$ is a complex number
lying in the sector
$$
\Sigma_K:=\{z\in \mathbb{C}: \mathop{\rm Re} z\geq 0\}\cup \{z
\in \mathbb{C}: |\mathop{\rm Im} z|>K|\mathop{\rm Re} z| \}\quad
(K>0.)
$$ Then
there exists $M>0$, depending on $K,a,b,c$, but independent of
$t$ such that
\begin{equation}
(1+|\lambda|)\|u\|_E+(1+|\lambda|^{1/2})\|u'\|_E+\|u''\|_E \leq
M(\|f\|_E+(1+|\lambda|^{1/2})(|z_0|+|z_1|)).
\end{equation}
\end{proposition}

As a consequence of the above proposition we have the following
result.

\begin{proposition}[\cite{Ac}]\label{th:harna1}
Let $a,b,c$ be as in \eqref{10}, \eqref{l1}; let $\{A(t)\}_{t\in
[0,T]}$ be defined by (\ref{ll1}). Then we have:
\begin{itemize}
\item[(i)] $[0,\infty[ \subseteq \rho(A(t))$ for all $t\in [0,T]$;
where $\rho (A(t))$ is the resolvent set of $A(t)$ and
 $R(\lambda ,A(t))=(\lambda I-A(t))^{-1}$.
\item[(ii)] $\Sigma_K\subseteq \rho(A(t))$ and for each $K>0$ there
exists $M(K)>0$ (depending also on $a,b,c$) such that
$$
\|R(\lambda,A(t))\|_{\mathcal{L}(E)}\leq
\frac{M(K)}{1+|\lambda|}\quad \forall \lambda \in \Sigma_K,
\forall t \in [0,T],
$$
where $\Sigma_K$ is defined above.
\end{itemize}
\end{proposition}

\begin{definition}[\cite{Tana}] \rm
A classical solution of (\ref{10}) is a function
\[
u\in C([0,T],E)\cap C((0,T],D(A(t)))\cap C^{1}((0,T],E),
\]
such that $u(0)=x$, $u'(t)-A(t)u(t)=0$ in $(0,T]$.
\end{definition}

Let us assume the following hypotheses:
\begin{itemize}
\item[(AT1)] For each $t\in [ 0,T]$,
 $A(t):D(A(t))\subseteq E\to E$ is a closed linear operator
and there exists $M>0$ and $\theta \in ( \frac{\pi }{2},\pi) $
such that
\begin{gather*}
\rho (A(t)) \supseteq S_{\theta}: =\{ \lambda \in
\mathbb{C}:\lambda \neq 0 ,| \arg \lambda | <\theta
\}\cup \{0\}, \\
\| R(\lambda ,A(t))\| \leq \frac{M}{1+| \lambda | } \quad \forall
\lambda \in S_{\theta}\cup \{0\},\;\forall  t\in [ 0,T],
\end{gather*}

\item[(AT2)] There exist $B>0$ and
$\delta_{1},\dots ,\delta _{k},\nu _{1},\dots ,\nu _{k}$ with
$0\leq \nu _{i}<\delta _{i}\leq 2$ such that
\[
\| A(t)R(\lambda ,A(t))( ( A(s)) ^{-1}-( A(t)) ^{-1}) \| \leq
B\sum_{i=1}^{k}| t-s| ^{\delta _{i}}| \lambda | ^{\nu _{i}-1},
\]
for all $\lambda \in S_{\theta }-\{0\},\;0\leq s<t\leq T$.
\end{itemize}
It is obvious that $\Sigma_K$ and $S_{\theta}$ are the same sets.


\begin{theorem}[\cite{Tana}]\label{th:2}
Assume that (AT1) and (AT2) hold. Then, if $x\in \bar{D}(A(0))$,
problem \eqref{10} has a unique classical solution.
\end{theorem}

\begin{remark}\label{rm:3} \rm
In general the function $c$ in the problem (\ref{10})is not
negative. Moreover by setting $u=ve^{\omega t}$ with $\omega \in
\mathbb{R}$, the function $v$ is solution of
\begin{equation}
\begin{gathered}
v'(t)-( A(t)-\omega I) v(t)=0, \quad t\in (0,T] \\
v(0)=x\,.
\end{gathered} \label{111}
\end{equation}
Hence existence, uniqueness and positivity of solutions of problem
(\ref{111}) is equivalent to the same properties of problem
(\ref{10}).
\end{remark}

\section{The model}\label{sec:s2}

The domain under consideration is $\Omega =D\times (0,z^{\ast })$,
where $D $ is an open subset of the surface, that is $D$ is a
portion of the plane and $z^{\ast }$ is the distance from the
surface to a region above the thermocline .

 The state variable for the dynamics of the larvae is
the density of larvae. For the part of the larval cycle which
goes from fertilization to the end of stage, the density
$l=l(t,s,P),\;$where $s$ denotes the position within the stages,
which we take specifically of the Bay of Biscay anchovy in
$[1,12)$ \cite{Arino} and $P=(x,y,z)$ represents a generic point
in the physical space.

 The region of observation is assimilated to the product of the
horizontal plane and a vertical line. The origin is a point of the
surface in the sea, the $x$ axis is oriented westward, the $y$
axis is oriented northward, and the $z$ axis is oriented
downward. Of course $t$ is the chronological time. $l$ is a
density with respect to the stage and the position. The larvae
are characterized by their density, that is to say, at each time
$t\in [ 0,T] $, where $T$ is the maximal time of observation,
$l(t,s,P)$ can be thought of as the larvae biomass per unit of
volume evaluated at the point $P$, at that time. The full model
is as follows
\begin{equation}
\begin{gathered}
\frac{\partial l}{\partial t}+\frac{\partial ( fl) }{\partial s}
+div\,(Vl)-\frac{\partial }{\partial z}( h\frac{\partial
l}{\partial z})
+\mu l=0, \\
l(t,1,x,y,z)=B(t,P), \\
h\frac{\partial l}{\partial z}=0,\quad z=0, \\
h\frac{\partial l}{\partial z}=0\quad z=z^{\ast },
\end{gathered} \label{1}
\end{equation}
We now discuss in detail the parameters and functions of the
model.

\noindent \textbf{The velocity.} The velocity vector
$V(t,P)=(V_{1}(t,P),V_{2}(t,P),V_{3}(t,P)) $ describes the sea
current which is supposed to be known. We assume that the sea
water is incompressible, which yields:
\begin{equation}
\mathop{\rm div}(V)=0,  \label{98}
\end{equation}
with $V_{1}(t,x,y,z)=V_{1}(t,x,y)$, $V_{2}(t,x,y,z)=V_{2}(t,x,y)$.


\noindent\textbf{The mixing coefficient.} The mixing coefficient
$h=h(t,P)$ gives the diffusion rate, supposed to be essentially
vertical.

\noindent \textbf{The growth function.} The main biological
parameters are functions $f(t,s)$, which gives the instantaneous
rate of progression within the stages from the egg fertilization
to the end of the yolk-sac period. For the principle of
determination see \cite{Motos,Arino}.

\noindent \textbf{The mortality of larvae.} The mortality is
modelled by the expression $\mu =\mu(t,s,P)$.

\noindent \textbf{Demographic boundary conditions.} Demographic
boundary conditions are given at $s=1$, at any time during the
spawning period, the variable $s$ takes its values in the
interval $[1,12)$, where $s=1$ corresponds to the newly fertilized
eggs, and $s=12$, to the end of the yolk sac period.

\noindent\textbf{Horizontal boundary conditions.} Model (\ref{1})
does not show any lateral boundary conditions. Choosing the right
boundary in the $x$ and $y$ directions is a difficult issue that
we mainly avoid here by assuming that the initial value has a
compact support contained  in the interior of the domain and we
consider the solution within a time interval $[ 0,T] $ during
which the horizontal projection  of the support is contained  in
the interior of the domain $D$.

\noindent \textbf{Vertical boundary conditions.} Vertical
boundary conditions are imposed at the surface and at $z^{\ast
}$, here we are assuming a no flux conditions.

\noindent \textbf{Initial conditions}
 Initial conditions are given at $t=0$ (beginning of the
year). The standing assumption is that there is no larva alive at
this period of the year, so that $l(0,s,x,y,z)=0$.


\begin{remark} \rm
What we call an initial value in the present context is not the
value of the solution at a given time or rather, the only relevant
information would be that at $t=0$ (that is $1^{st}$ January)
there is no larva in the sea. What we consider as an initial
value is the distribution of newly fertilized eggs, that is the
larvae at stage $s=1$ all over the reproduction season.
\end{remark}

\section{Existence, uniqueness and positivity of the solution of the
Cauchy problem} \label{sec:s4}

 The aim of this section is to show that  model
(\ref{1}) possesses a positive, unique solution. For this we use
an approach by the method of characteristics to build a one
dimensional time dependent parabolic equation whose solution will
yield the solution of equation (\ref {1}). We assume that
\begin{itemize}
\item[(H1)] $V_{1},V_{2}$ are functions in $C^{1}((0,T)\times D)$ and
$f$ $\in $ $C^{1}((0,T)\times (1,12))$.
\end{itemize}
We introduce the flow generated by the horizontal current and the
size growth, that is
\[
\phi:=\phi (\tau ,t_{0},1,x_{0},y_{0}),
\]
and for each initial value
 $\tilde{\zeta}\equiv(t_{0},1,x_{0},y_{0})$, $\phi (\tau
,\tilde{\zeta})$ is the solution of the equation
\begin{equation}
\big( \frac{dt}{d\tau },\frac{ds}{d\tau },\frac{dx}{d\tau
},\frac{dy}{d\tau }\big) =( 1,\;f(t,s),V_{1}(t,x,y),V_{2}(t,x,y))
, \label{caracte}
\end{equation}
satisfying $t(0)=t_{0}$, $s(0)=1$, $x(0)=x_{0}$, $y(0)=y_{0}$,
since the theory of ordinary differential equations guarantees
that a unique characteristic curve passes through each point
$\tilde{\zeta}$.

 We denote $\bar{l}(\tau ,z)\equiv \bar{l}(\tau ,\tilde{\zeta},z)
=l(\phi (\tau ,\tilde{\zeta}),z)$ the restriction of $l$ along the
characteristic line. The equation verified by $\bar{l}$ reads
\[
\frac{\partial \bar{l}(\tau ,z)}{\partial \tau
}+\bar{V}_{3}\frac{\partial \bar{l}(\tau ,z)}{\partial
z}-\frac{\partial }{\partial z}( \bar{h} \frac{\partial
\bar{l}(\tau ,z)}{\partial z}) +\bar{\gamma}\bar{l} (\tau ,z)=0,
\]
where $\bar{V}_{3}:=\bar{V}_{3}(\tau
,\tilde{\zeta},z)$,$\;\bar{h}:=\bar{h}%
(\tau ,\tilde{\zeta},z)$,$\;\bar{\gamma}:=\bar{\gamma}(\tau
,\tilde{\zeta},z)$ are the restrictions of
$V_{3}$,$\;h$,$\;B$,$\;\gamma $ respectively along the
characteristic line and $\gamma $ is equation of order $0$.
 So to each $\tilde{\zeta}$, we have associated the
following problem
\begin{equation}
\begin{gathered}
\frac{\partial \bar{l}}{\partial \tau }+\bar{V}_{3}\frac{\partial
\bar{l}}{\partial z}-\frac{\partial}{\partial z}( \bar{h}
\frac{\partial \bar{l}}{\partial z}) + \bar{\gamma}\bar{l}=0, \\
\bar{l}(0,z)=\bar{B}(z),\\
\bar{h}(\tau,0)\frac{\partial \bar{l}}{\partial z}(\tau,0)=0, \\
\bar{h}(\tau,z^*)\frac{\partial \bar{l}}{\partial z}(\tau,z^*)=0,
\end{gathered} \label{3}
\end{equation}
where $\bar{B}(z)$ is the restriction of $B$ along the
characteristic line.  We consider the operator
$A(\tau):D(A(\tau))\subseteq C([0,z^{\ast }])\to C([0,z^{\ast }])$
defined by
\begin{gather*}
A(\tau)u=\bar{V}_{3}(\tau,.)u'-(\bar{h}(\tau,.)u')'+\bar{\gamma}(\tau,.)u,
\\
D(A(\tau))=\{u\in C^{2}([0,z^{\ast }]),\;\
\bar{h}(\tau,0)u'(0)=\bar{h}(\tau,z^*)u'(z^{\ast })=0\}.
\end{gather*}
We now state the assumptions of this section.
\begin{itemize}
\item[(H2)] $h \in C^1([0,T]\times \bar{\Omega})$,
${V}_{3}\in C([0,T]\times \bar{\Omega})$, ${\gamma}\in
C([0,T]\times[1,12]\times \bar{\Omega)}$.

\item[(H3)]  $h, \frac{\partial h}{\partial z}$,
$V_3 \in C^{1,\delta}([0,T];C(\bar{\Omega}))$, and $ \gamma \in
C^{1,\delta}([0,T];C([1,12]\times \bar{\Omega}))$.

\item[(H4)] $h\geq c_0$ in $[0,T]\times \bar{\Omega}$ where $c_0 >0$.
\end{itemize}


\begin{theorem}\label{th:harnack1}
Assume (H2)--(H4) hold. If the positive function $\bar{B}$ is in
$C([0,z^{\ast}])$, then problem (\ref{3}) has a unique
non-negative classical solution.
\end{theorem}

\begin{proof}
 Without loss of generality we can assume that
$\gamma\geq 0$, otherwise we can replace $\gamma$ by
$\gamma+\omega\geq 0$ see Remark \ref{rm:3}. The main idea is to
use theorem \ref{th:2}. The first assertion (AT1) follows from the
proposition \ref{th:harna1}. Concerning the second assertion
(AT2), for $f\in C([0,z^{\ast }])$, $t,s\in G_{1}$, where $G_1$
is some neighborhood of $\tau=0$, $\lambda \in S_{\theta }-\{0\}$,
we set $v=\nolinebreak  (A(s))^{-1}f$  and $u=R(\lambda
,A(t))(\lambda -A(s))v$, then we have to estimate the
$C([0,z^*])$-norm of
\[
u-v=(A(t)) R(\lambda ,A(t))(A(t))^{-1}-(A(s))^{-1})f.
\]
Now $u-v\in C^{2}([0,z^{\ast }])$ and $u$ and $v$ solve
\begin{equation}
\begin{gathered}
\lambda u-A(t,.)u=\lambda v-f,  \\
\bar{h}(t,0)u'(0)=\bar{h}(t,z^*)u'(z^*)=0,
\end{gathered} \label{610}
\end{equation}
and
\begin{equation}
\begin{gathered}
A(s,.) v=f, \\
\bar{h}(s,0)v'(0)=\bar{h}(s,z^*)v'(z^*)=0,
\end{gathered} \label{611}
\end{equation}
respectively. This shows that
\begin{equation}
\begin{gathered}
\lambda (u-v)-A(t,.)(u-v)=(A(t,.)-A(s,.))v,\;\;\; \\
\bar{h}(t,0)(u'-v')(0)= (\bar{h}(s,0)-\bar{h}(t,0))v'(0), \\
\bar{h}(t,z^*)(u'-v')(z^*)=(\bar{h}(s,z^*)-\bar{h}(t,z^*))v'(z^*).
\end{gathered}  \label{612}
\end{equation}
Applying  Proposition \ref{th:harnack} to (\ref{611}) with
$\lambda=0$, we have
\begin{equation}
\| v\|_E\leq c\| f\| _E. \label{A15}
\end{equation}
Using again the proposition \ref{th:harnack} to (\ref{612}), we
get
\begin{align*}
| \lambda | \| u-v\| _E
&\leq M(\|(A(s,.)-A(t,.))v\|_E+(1+|\lambda|^{1/2})\\
&\quad \times
\big(|(\bar{h}(s,0)-\bar{h}(t,0))v'(0)|+|(\bar{h}(s,z^*)
-\bar{h}(t,z^*))v'(z^*)|\big))\,.
\end{align*}
Using  hypothesis (H3) and by virtue of (\ref{A15}),
\[
\| u-v\| _E\leq c(|t-s|| \lambda|^{-1}|t-s| ^{\delta } | \lambda
|^{-1}+|t-s||\lambda |^{-1/2})\|f\| _E.
\]
Hence the hypothesis (AT2) holds.

Since $D(A(0))$ is dense in $C([0,z^{* }])$ see \cite{Ac}, then
according to Theorem \ref{th:2}, for $\bar{B}\in $ $C([0,z^{*}])$
we have existence and uniqueness of a classical solution of
problem (\ref{3}). It remain to see that the solution is positive
which can be proved by the standard argument. If $u$ is solution
of problem (\ref{3}), we set $u=u^{+}-u^{-}$ where $u^{+}$ and
$u^{-}$ are respectively the positive and negative part of $u$,
so multiplying the equation (\ref{3}) by $u^{-}$, and integrating
over $(0,z^*)$ we have
$$
\int_{0}^{z^{\ast}}(\frac{\partial u}{\partial \tau} u^{-} +
\bar{h}(\tau,z)\frac{\partial u}{\partial z} \frac{\partial
u^-}{\partial z}+\bar{V}_{3}(\tau,z) \frac{\partial u}{\partial
z}{u^{-}}+ \bar{\gamma}(\tau,z)uu^-) dz=0,
$$
hence
$$
-\frac{1}{2}\frac{d}{d\tau}\|u^{-}(\tau)\|_{L^2(0,z^*)}^2 \geq
c_0\int_{0}^{z^{*}}| \frac{\partial u^{-}}{\partial
z}|^2dz-m_3\int_{0}^{z^*} \frac{\partial u^{-}}{\partial z}u^{-}dz
+m_4\int_{0}^{z^*}|u^{-}|^2dz,
$$
with
\[
m_{3}=\sup_{\tau,z}| \bar{V}_{3}(\tau,z)| ,\quad
m_{4}=\inf_{\tau,z}| \bar{\gamma}(\tau,z)|.
\]
Since
\[ \int_{0}^{z^{\ast }}\frac{\partial
u^{-}}{\partial z}u^{-}dz\leq \int_{0}^{z^{\ast }}\big( \rho |
\frac{\partial u^{-}}{\partial z}|^{2} +\frac{1}{\rho }| u^{-}|
^{2}\big) dz,\quad \forall \rho >0,
\]
it follows that
\begin{align*}
&\frac{1}{2}\frac{d}{d\tau}\|u^{-}(\tau)\|_{L^2(0,z^*)}^2+(c_0-m_3\rho)\|
\frac{\partial u^{-}(\tau)}{\partial z}\|_{L^2(0,z^*)}^2\\
&+(m_4-\frac{m_3}{\rho}+\omega)\|u^{-}(\tau)\|_{L^2(0,z^*)}^2\\
& \leq \omega \|u^{-}(\tau)\|_{L^2(0,z^*)}^2,
\end{align*}
choosing $\rho$ and $\omega$  such that
$$
c_0-m_3\rho >0 \quad\mbox{and}\quad m_4-\frac{m_3}{\rho}+\omega
>0,
$$
so
\[
\frac{1}{2}\frac{d}{d\tau}\|u^{-}(\tau)\|_{L^2(0,z^*)}^2\leq
\omega \|u^{-}(\tau)\|_{L^2(0,z^*)}^2,
\]
then
\[
\|u^{-}(\tau)\|_{L^2(0,z^*)}^2\leq \|u^{-}(0)\|_{L^2(0,z^*)}^2 e^{
2\omega \tau},
\]
 which gives $u^{-}(\tau)=0$ provided $B\geq 0$, then the solution is
positive.
\end{proof}

Recall that for $z\in [ 0,z^{\ast }] $ the system
\begin{gather*}
t=T(\tau ,t_{0},x_{0},y_{0}), \quad
s=S(\tau ,t_{0},x_{0},y_{0}), \\
x=X(\tau ,t_{0},x_{0},y_{0}), \quad y=Y(\tau ,t_{0},x_{0},y_{0}),
\end{gather*}
is a solution of the characteristic system (\ref{caracte})
emanating from the point $\tilde{\zeta}$. We have also
\begin{gather*}
t_{0}=T(0,t_{0},x_{0},y_{0}), \quad
1=S(0,t_{0},x_{0},y_{0}), \\
x_{0}=X(0,t_{0},x_{0},y_{0}), \quad
y_{0}=Y(0,t_{0},x_{0},y_{0})\,.
\end{gather*}
If
\[
\mathop{\rm Jac}(T,S,X,Y):=\left|
\begin{matrix}
\frac{\partial T}{\partial \tau } & \frac{\partial T}{\partial
t_{0}} & \frac{\partial T}{\partial x_{0}} & \frac{\partial
T}{\partial y_{0}}
\\
\frac{\partial S}{\partial \tau } & \frac{\partial S}{\partial
t_{0}} & \frac{\partial S}{\partial x_{0}} & \frac{\partial
S}{\partial y_{0}}
\\
\frac{\partial X}{\partial \tau } & \frac{\partial X}{\partial
t_{0}} & \frac{\partial X}{\partial x_{0}} & \frac{\partial
X}{\partial y_{0}}
\\
\frac{\partial Y}{\partial \tau } & \frac{\partial Y}{\partial
t_{0}} & \frac{\partial Y}{\partial x_{0}} & \frac{\partial
Y}{\partial y_{0}}
\end{matrix}
\right| _{\tau =0}\neq 0,
\]
then the Jacobian does not vanish in a neighborhood of the initial
curve. Therefore, the local inversion theorem guarantees that we
can solve for $(\tau ,t_{0},x_{0},y_{0})$ as function of
$(t,s,x,y)$ near the initial curve; that is, there exists a
neighborhood $G_{1}$ of $(0,t_{0},x_{0},y_{0})$ and a
neighborhood $G$ of  $(t,s,x,y)$ such that
\[
(T,S,X,Y):G_{1}\to G
\]
is a diffeomorphism. Then
\begin{gather*}
\tau =\psi _{1}(t,s,x,y), \quad
t_{0}=\psi _{2}(t,s,x,y), \\
x_{0}=\psi _{3}(t,s,x,y), \quad y_{0}=\psi _{4}(t,s,x,y),
\end{gather*}
and for initial data
\begin{gather*}
0=\psi _{1}(t_{0},1,x_{0},y_{0}), \quad
t_{0}=\psi _{2}(t_{0},1,x_{0},y_{0}), \\
x_{0}=\psi _{3}(t_{0},1,x_{0},y_{0}), \quad y_{0}=\psi
_{4}(t_{0},1,x_{0},y_{0}).
\end{gather*}
Once problem (\ref{3}) is solved, we have
\[
l(t,s,x,y,z)=\bar{l}(\psi _{1},\psi _{2},\psi _{3},\psi _{4},z),\;
\]
in a neighborhood of $G$. Indeed, by differentiation we obtain
that
\begin{gather*}
\frac{\partial l}{\partial t} =\frac{\partial \bar{l}}{\partial
\tau } \frac{\partial \psi _{1}}{\partial t}+\frac{\partial
\bar{l}}{\partial t_{0}} \frac{\partial \psi _{2}}{\partial
t}+\frac{\partial \bar{l}}{\partial x_{0}} \frac{\partial \psi
_{3}}{\partial t}+\frac{\partial \bar{l}}{\partial y_{0}}
\frac{\partial \psi _{4}}{\partial t}, \\
f\frac{\partial l}{\partial s} =f( \frac{\partial
\bar{l}}{\partial \tau }\frac{\partial \psi _{1}}{\partial
s}+\frac{\partial \bar{l}}{\partial t_{0}}\frac{\partial \psi
_{2}}{\partial s}+\frac{\partial \bar{l}}{\partial
x_{0}}\frac{\partial \psi _{3}}{\partial s}+\frac{\partial
\bar{l}}{\partial
y_{0}}\frac{\partial \psi _{4}}{\partial s}) , \\
V_{1}\frac{\partial l}{\partial x} = V_{1}( \frac{\partial
\bar{l}}{
\partial \tau }\frac{\partial \psi _{1}}{\partial x}+\frac{\partial
\bar{l}}{
\partial t_{0}}\frac{\partial \psi _{2}}{\partial x}+\frac{\partial
\bar{l}}{
\partial x_{0}}\frac{\partial \psi _{3}}{\partial x}+\frac{\partial
\bar{l}}{
\partial y_{0}}\frac{\partial \psi _{4}}{\partial x}) , \\
V_{2}\frac{\partial l}{\partial y} = V_{2}( \frac{\partial
\bar{l}}{
\partial \tau }\frac{\partial \psi _{1}}{\partial y}+\frac{\partial
\bar{l}}{
\partial t_{0}}\frac{\partial \psi _{2}}{\partial y}+\frac{\partial
\bar{l}}{
\partial x_{0}}\frac{\partial \psi _{3}}{\partial y}+\frac{\partial
\bar{l}}{
\partial y_{0}}\frac{\partial \psi _{4}}{\partial y})\,.
\end{gather*}
Thus
\begin{align*}
\frac{\partial l}{\partial t}+f\frac{\partial l}{\partial
s}+V_{1}\frac{
\partial l}{\partial x}+V_{2}\frac{\partial l}{\partial y}
&= \frac{\partial \bar{l}}{\partial \tau } \big(\frac{\partial
\psi _{1}}{\partial t}+f\frac{\partial \psi _{1}}{\partial s}
+V_{1}\frac{\partial \psi _{1}}{\partial x}
+V_{2}\frac{\partial \psi _{1}}{\partial y}\big)\\
&\quad+ \frac{\partial \bar{l}}{\partial t_{0}}
 \big(\frac{\partial \psi _{2}}{\partial t}+f\frac{\partial \psi
_{2}}{\partial s}
 +V_{1}\frac{\partial \psi _{2}}{\partial x}
 +V_{2}\frac{\partial \psi _{2}}{\partial y}\big)\\
&\quad + \frac{\partial \bar{l}}{\partial
x_{0}}\big(\frac{\partial \psi _{3}}{\partial t}+f\frac{\partial
\psi _{3}}{\partial s}+V_{1}\frac{\partial \psi _{3}}{
\partial x}+V_{2}\frac{\partial \psi _{3}}{\partial y}\big)\\
&\quad+ \frac{\partial \bar{l}}{\partial y_{0}}
\big(\frac{\partial \psi _{4}}{\partial t} +f\frac{\partial \psi
_{4}}{\partial s} +V_{1}\frac{\partial \psi _{4}}{\partial x}
+V_{2}\frac{\partial \psi _{4}}{\partial y}\big).
\end{align*}
Then
\begin{align*}
\frac{\partial l}{\partial t}+f\frac{\partial l}{\partial
s}+V_{1}\frac{
\partial l}{\partial x}+V_{2}\frac{\partial l}{\partial y}
&=\frac{\partial \bar{l}}{\partial \tau } \big(\frac{\partial
T}{\partial \tau }\frac{\partial \psi _{1}}{\partial
t}+\frac{\partial S}{\partial t_{0}}\frac{\partial \psi
_{1}}{\partial s}+\frac{\partial X}{\partial x_{0}}\frac{\partial
\psi _{1}}{
\partial x}+\frac{\partial Y}{\partial y_{0}}\frac{\partial \psi _{1}}{
\partial y}\big)\\
&\quad + \frac{\partial \bar{l}}{\partial t_{0}}
\big(\frac{\partial T}{\partial \tau } \frac{\partial \psi
_{2}}{\partial t}+\frac{\partial S}{\partial t_{0}}\frac{
\partial \psi _{2}}{\partial s}+\frac{\partial X}{\partial x_{0}}\frac{
\partial \psi _{2}}{\partial x}+\frac{\partial Y}{\partial y_{0}}\frac{
\partial \psi _{2}}{\partial y}\big)\\
&\quad + \frac{\partial \bar{l}}{\partial x_{0}}
\big(\frac{\partial T}{\partial \tau } \frac{\partial \psi
_{3}}{\partial t}+\frac{\partial S}{\partial t_{0}}\frac{
\partial \psi _{3}}{\partial s}+\frac{\partial X}{\partial x_{0}}\frac{
\partial \psi _{3}}{\partial x}+\frac{\partial Y}{\partial y_{0}}\frac{
\partial \psi _{3}}{\partial y}\big)\\
&\quad +\frac{\partial \bar{l}}{\partial y_{0}}
\big(\frac{\partial T}{\partial \tau } \frac{\partial \psi
_{4}}{\partial t}+\frac{\partial S}{\partial t_{0}}\frac{
\partial \psi _{4}}{\partial s}+\frac{\partial X}{\partial x_{0}}\frac{
\partial \psi _{4}}{\partial x}+\frac{\partial Y}{\partial y_{0}}\frac{
\partial \psi _{4}}{\partial y}\big).
\end{align*}
For $Z=(T,S,X,Y)^{T}$ and $\psi =(\psi _{1},\psi _{2}, \psi _{3},
\psi _{4})^{T}$, we have $(Z\circ \psi )(t,s,x,y)=(t,s,x,y)$
which implies
\begin{equation}
\mathop{\rm Jac}(Z).\mathop{\rm Jac}(\psi )=\mathop{\rm
Id}{}_{4}\,. \label{Id4}
\end{equation}
By identification in (\ref{Id4}), we find
\begin{gather*}
\frac{\partial T}{\partial \tau }\frac{\partial \psi
_{1}}{\partial t}+ \frac{\partial S}{\partial
t_{0}}\frac{\partial \psi _{1}}{\partial s}+ \frac{\partial
X}{\partial x_{0}}\frac{\partial \psi _{1}}{\partial x}+
\frac{\partial Y}{\partial y_{0}}\frac{\partial \psi
_{1}}{\partial y}=1,
\\
\frac{\partial T}{\partial \tau }\frac{\partial \psi
_{2}}{\partial t}+ \frac{\partial S}{\partial
t_{0}}\frac{\partial \psi _{2}}{\partial s}+ \frac{\partial
X}{\partial x_{0}}\frac{\partial \psi _{2}}{\partial x}+
\frac{\partial Y}{\partial y_{0}}\frac{\partial \psi
_{2}}{\partial y}=0,
\\
\frac{\partial \psi _{3}}{\partial t}+f\frac{\partial \psi
_{3}}{\partial s }+V_{1}\frac{\partial \psi _{3}}{\partial
x}+V_{2}\frac{\partial \psi _{3}
}{\partial y}=0, \\
\frac{\partial T}{\partial \tau }\frac{\partial \psi
_{4}}{\partial t}+ \frac{\partial S}{\partial
t_{0}}\frac{\partial \psi _{4}}{\partial s}+ \frac{\partial
X}{\partial x_{0}}\frac{\partial \psi _{4}}{\partial x}+
\frac{\partial Y}{\partial y_{0}}\frac{\partial \psi
_{4}}{\partial y}=0\,.
\end{gather*}
Therefore,
\[
\frac{\partial l}{\partial t}+f\frac{\partial l}{\partial
s}+V_{1}\frac{%
\partial l}{\partial x}+V_{2}\frac{\partial l}{\partial
y}=\frac{\partial \bar{l}}{\partial \tau }\,.
\]
In addition,
\[
\frac{\partial l}{\partial z}=\frac{\partial \bar{l}}{\partial z},
\]
for the initial data
\begin{align*}
l(t,1,x,y,z) &=\bar{l}(\psi _{1}(t,1,x,y),\psi _{2}(t,1,x,y),\psi
_{3}(t,1,x,y),\psi _{4}(t,1,x,y),z), \\
&=\bar{l}(0,t,x,y,z), \\
&=\bar{B}(t,x,y,z), \\
&=B(T(0,t,x,y),X(0,t,x,y),Y(0,t,x,y),z), \\
&=B(t,x,y,z)\,.
\end{align*}
Then $l$ is a unique solution of (\ref{1}). So the solution $l$
of  (\ref{1}) can be determined in terms of the solution
$\bar{l}$ of (\ref{3}).

\subsection*{Acknowledgments}
The authors wish to thank the anonymous referee for his/her
helpful suggestions and comments.

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\end{document}