\documentclass[twoside]{article}
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\markboth{\hfil High-order mixed-type differential equations\hfil EJDE--2000/60}
{EJDE--2000/60\hfil M. Denche \& A. L. Marhoune \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.~{\bf 2000}(2000), No.~60, pp.~1--10. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
%
High-order mixed-type differential equations with weighted integral 
boundary conditions 
% 
\thanks{ {\em Mathematics Subject Classifications:} 35B45, 35G10, 35M10.
\hfil\break\indent
{\em Key words:} Integral boundary condition, energy inequalities, 
equation of mixed type.
\hfil\break\indent
\copyright 2000 Southwest Texas State University. \hfil\break\indent
Submitted March 27, 2000. Published September 21, 2000.} }
\date{}
%
\author{ M. Denche \& A. L. Marhoune }
\maketitle

\begin{abstract} 
In this paper, we prove the existence and uniqueness of strong solutions 
for high-order mixed-type problems with weighted integral boundary 
conditions. The proof uses energy inequalities and the 
density of the range of the operator generated.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

Let $\alpha $ be a positive integer and $Q$ be the set
$(0,1)\times ( 0,T)$.
We consider the equation
\begin{equation}
\mathcal{L}u:=\frac{\partial ^{2}u}{\partial t^{2}}+(-1)^{\alpha }a(t)\frac{%
\partial ^{2\alpha +1}u}{\partial x^{2\alpha }\partial t}=f(x,t),  
\label{1}
\end{equation}
where the function $a(t)$ and its derivative are bounded on the 
interval $[ 0,T]$:
\begin{eqnarray}
0<a_{0}\leq a(t)\leq a_1\,,\label{1'} \\
\frac{da(t)}{dt}\leq a_2\,.\quad \label{1''}
\end{eqnarray}
To equation (\ref{1}) we attach the initial conditions
\begin{equation}
l_{1}u=u(x,0)=\varphi (x),\quad l_{2}u=\frac{\partial u}{\partial t}(
x,0)=\psi ( x)\mbox{\quad }x\in (0,1),  \label{2}
\end{equation}
the boundary conditions
\begin{equation}
\frac{\partial ^{i}}{\partial x^{i}}u(0,t)=\frac{\partial ^{i}}{\partial
x^{i}}u(1,t)=0,\quad \mbox{for \ }0\leq i\leq \alpha -k-1,\mbox{ }t\in 
(0,T),
\label{3}
\end{equation}
and integral conditions
\begin{equation}
\int_{0}^{1}x^{i}u(x,t)dx=0,\mbox{ for \ }0\leq i\leq 2k-1,\mbox{\quad }%
1\leq k\leq \alpha \quad t\in (0,T).  \label{5}
\end{equation}
where $\varphi $ and $\psi $ are known functions which satisfy the
compatibility conditions given in (\ref{3})-(\ref{5}).

Various problems arising in heat conduction \cite{2', 3, 8, 9}, 
chemical engineering \cite{6}, underground water flow \cite{7}, 
thermo-elasticity \cite{13}, and plasma physics \cite{11'}
 can be reduced to the nonlocal problems with integral boundary conditions.
 This type of boundary value problems has been investigated in 
\cite{1, 2, 2', 3, 4, 6, 8, 9, 10, 12, 14} for
parabolic equations and in \cite{10', 13'} for hyperbolic equations.
The basic tool in \cite{2, 10, 14} is the energy inequality method which, 
of course, requires appropriate multipliers and functional spaces. 
In this paper, we extend this method to the study of a high-order 
mixed-type partial differential equations.

\section{Preliminaries}

In this paper, we prove existence and uniqueness of a strong solution of
problem (\ref{1})-(\ref{5}). For this, we consider the problem (\ref{1})-(%
\ref{5}) as a solution of the operator equation
\begin{equation}
Lu=\mathcal{F},  \label{6}
\end{equation}
where $L=(\mathcal{L},l_{1},l_{2})$, with domain of definition $D(L)$
consisting of functions $u\in W_{2}^{2\alpha ,2}(Q)$ such that $\frac{%
\partial ^{i}}{\partial x^{i}}(\frac{\partial ^{\alpha -k+1}u}{\partial
x^{\alpha -k}\partial t})\in L_{2}(Q)$, $i=\overline{0,\alpha +k-1}$ 
and $u$ satisfies conditions (\ref{3})-(\ref{5}); the operator $L$ is 
considered from $E$ to $F$, where $E$ is the Banach space consisting of 
 functions 
$u\in L_{2}(Q)$, satisfying (\ref{3})-(\ref{5}), with the finite norm
\begin{eqnarray}
\left\| u\right\| _{E}^{2}&=&\int_{Q} \big|J^{k}\frac{\partial ^{2}u}{
\partial t^{2}}\big| ^{2}+\sum_{i=0}^{\alpha -k}
\big|\frac{\partial ^{i+1}u}{\partial x^{i}\partial t}\big| ^{2}
\,dx\,dt    \label{6'} \\
&&+\sup_{0\leq t\leq T} \int_{0}^{1} \sum_{i=0}^{\alpha -k}
\big|\frac{\partial ^{i}u}{\partial x^{i}}\big|
^{2}+\big|\frac{\partial ^{\alpha -k+1}u}{\partial x^{\alpha -k}
\partial t} \big| ^{2}\,dx\,, \nonumber
\end{eqnarray}
where $J^{k}u=\int_{0}^{x}\int_{0}^{\xi _{1}}\dots\int_{0}^{\xi _{k-1}}u(\xi
,t)d\xi $.
Here $F$ is the Hilbert space of vector-valued functions $\mathcal{F}
=(f,\varphi ,\psi )$ obtained by completing of the space $L_{2}(Q)\times
W_{2}^{2\alpha }(0,1)\times W_{2}^{2\alpha }(0,1)$ with respect to the norm
\begin{equation}
\left\| \mathcal{F}\right\| _{F}^{2}=\int_{Q}\big|f\big|
^{2}\,dx\,dt+\int_{0}^{1}\sum_{i=0}^{\alpha -k}
\big|\frac{\partial ^{i}\varphi }{\partial x^{i}}\big| ^{2}+\big|
\frac{\partial ^{\alpha -k}\psi }{\partial x^{\alpha -k}}\big| ^{2}\,dx\,.
\label{8}
\end{equation}
Then we establish an energy inequality
\begin{equation}
\left\| u\right\| _{E}\leq C_{1}\left\| Lu\right\| _{F},  \label{9}
\end{equation}
and we show that the operator $L$ has the closure $\overline{L}$.

\paragraph{Definition}
A solution of the operator equation \ $\overline{L}u=\mathcal{F}$  is
called a strong solution of the problem (\ref{1})-(\ref{5}).
\smallskip

Inequality (\ref{9}) can be extended to $u\in D(\overline{L})$, i.e.,
\[
\left\| u\right\| _{E}\leq C_{1}\left\| \overline{L}u\right\| _{F},
\quad \forall u\in D(\overline{L}).
\]
From this inequality we obtain the uniqueness of a strong solution if it
exists, and the equality of sets $R(\overline{L})$ and $\overline{R(L)}$.
Thus, to prove the existence of a strong solution of the problem 
(\ref{1})-(\ref{5}) for any $\mathcal{F}\in F$, it remains to prove that 
the set $R(L)$ is dense in $F$.

\begin{lemma}
\label{le2} For any function $u\in E$, we have
\begin{equation}
\int_{0}^{1}\big|J^{2k}\frac{\partial ^{2}u}{\partial t^{2}}\big|
^{2}dx\leq 4^{k}\int_{0}^{1}\big|J^{k}\frac{\partial ^{2}u}{\partial 
t^{2}}\big| ^{2}\,dx\,.  \label{10}
\end{equation}
\end{lemma}

\paragraph{Proof}
Integrating $-\int_{0}^{1}xJ^{k}\frac{\partial ^{2}u}{\partial t^{2}}
J^{k+1}\frac{\partial ^{2}\overline{u}}{\partial t^{2}}dx$ by parts, 
and using elementary inequalities we obtain (\ref{10}). 
\hfill$\diamondsuit$

\begin{lemma}
\label{le3} For $u\in E$ and $0\leq i\leq \alpha -k$, we have
\begin{equation}
\int_{0}^{1}\big|\frac{\partial ^{i+1}u}{\partial x^{i}\partial t}\big|
^{2}dx\leq 4^{(\alpha -k)-i}\int_{0}^{1}\big|\frac{\partial ^{\alpha }}{%
\partial x^{\alpha }}(J^{k}\frac{\partial u}{\partial t})\big| 
^{2}dx\,.  \label{11}
\end{equation}
\end{lemma}

\paragraph{Proof} Integrating by parts  $-\int_{0}^{1}x\frac{\partial
^{\alpha -i}}{\partial x^{\alpha -i}}(J^{k}\frac{\partial u}{\partial t})
\frac{\partial ^{\alpha -i-1}}{\partial x^{\alpha -i-1}}
(J^{k}\frac{\partial\overline{u}}{\partial t})\,dx$ and using elementary 
inequalities yield (\ref {11}). \hfill$\diamondsuit$

\begin{lemma}
\label{le4} For $u\in E$ satisfying the condition (\ref{2}) we have
\begin{eqnarray} 
\lefteqn{ \sum_{i=0}^{\alpha -k} \int_{0}^{1}\exp (-c\tau
)\big|\frac{\partial ^{i}u(x,\tau )}{\partial x^{i}}\big| ^{2}\,dx }
\nonumber \\
&\leq&
\sum_{i=0}^{\alpha -k} \int_{0}^{1}(1-x)\big|\frac{
\partial ^{i}\varphi }{\partial x^{i}}\big| ^{2}\,dx \label{12} \\
&& +\frac{1}{3}(4^{\alpha -k+1}-1)\int_{0}^{\tau }\int_{0}^{1}\exp (-ct)
\big|\frac{\partial ^{\alpha }}{\partial x^{\alpha }}(J^{k}
\frac{\partial u}{\partial t})\big| ^{2}\,dx\,dt\,,  \nonumber 
\end{eqnarray}
where $c\geq 1$ and $0\leq \tau \leq T$.
\end{lemma}

\paragraph{Proof}
Integrating by parts $\int_{0}^{\tau }\exp (-ct)\frac{
\partial ^{i+1}u}{\partial x^{i}\partial t}\frac{\partial ^{i}u}{\partial
x^{i}}dt$ for $i=\overline{0,\alpha -k-1}$, using elementary 
inequalities, and lemma \ref{le3}, we obtain (\ref{12}).

\section{An energy inequality and its applications}

\begin{theorem}
\label{th1} For any function $u\in D(L)$, we have
\begin{equation}
\left\| u\right\| _{E}\leq C_{1}\left\| Lu\right\| _{F},  \label{13}
\end{equation}
where  $C_{1}=\exp (cT)\max (8.4^{k},\frac{a_{1}}{2})/
\min (\frac{a_{0}}{2},\frac{7}{8})$ , with the constant $c$ satisfying
\begin{equation}
c\geq 1\quad \mbox{and}\quad 3(ca_{0}-a_{2})\geq 2(4^{\alpha -k-1}-1). 
 \label{14}
\end{equation}
\end{theorem}

\paragraph{Proof} Let 
\[
Mu=(-1)^{k}J^{2k}\frac{\partial ^{2}u}{\partial t^{2}}\,.
\]
For a  constant $c$ satisfying (\ref{14}), we consider the quadratic 
form
\begin{equation}
\mathop{\rm Re}\int_{0}^{\tau }\int_{0}^{1}\exp 
(-ct)\mathcal{L}u\overline{Mu}\,dx\,dt ,
\label{15}
\end{equation}
which is obtained by multiplying (\ref{1}) by $\exp (-ct)\overline{Mu}$,
then integrating over $Q^{\tau }$, with $Q^{\tau }=(0,1)\times (0,\tau )$,
$0\leq \tau \leq T$, and then taking the real part. 
Integrating by parts in (\ref{15}) with the
use of boundary conditions (\ref{3}) and (\ref{5}), we obtain
\begin{eqnarray}
\lefteqn{\mathop{\rm Re}\int_{0}^{\tau }\int_{0}^{1}\exp (-ct)
\mathcal{L}u\overline{Mu}\,dx\,dt} \nonumber\\
&=&\int_{0}^{\tau }\int_{0}^{1}\exp (-ct)\big|J^{k}\frac{\partial 
^{2}u}{\partial t^{2}}\big| ^{2}\,dx\,dt  \label{16} \\
&&+\mathop{\rm Re}\int_{0}^{\tau }\int_{0}^{1}\exp (-ct)a(t)\frac{\partial ^{\alpha
-k+1}u}{\partial x^{\alpha -k}\partial t}\frac{\partial ^{\alpha -k+2}
\overline{u}}{\partial x^{\alpha -k}\partial t^{2}}\,dx\,dt\,. \nonumber 
\end{eqnarray}
By substituting the expression of $Mu$ in (\ref{15}), using elementary
inequalities and Lemma \ref{le2} we obtain
\begin{eqnarray}
\mathop{\rm Re}\int_{0}^{\tau }\int_{0}^{1}\exp (-ct)\mathcal{L}u\overline{Mu}
\,dx\,dt &\leq& 8.4^{k}\int_{0}^{\tau }\int_{0}^{1}\exp (-ct)\big|\mathcal{L}
u\big| ^{2}\,dx\,dt \label{17} \\
&&+\frac{1}{8}\int_{0}^{\tau }\int_{0}^{1}\exp (-ct)\big|J^{k}\frac{\partial
^{2}u}{\partial t^{2}}\big| ^{2}\,dx\,dt\,. \nonumber 
\end{eqnarray}
By integrating the last term on the right-hand side of (\ref{16}) and
combining the obtained results with the inequalities (\ref{14}), (\ref{17})
and lemmas \ref{le3}, \ref{le4} we obtain
\begin{eqnarray}
\lefteqn{
8.4^{k}\int_{0}^{\tau }\int_{0}^{1}\exp (-ct)\big|\mathcal{L}u\big|
^{2}\,dx\,dt +\frac{1}{2}\int_{0}^{1}a(0)\big|\frac{\partial ^{\alpha -k}\psi 
}{\partial x^{\alpha -k}}\big| ^{2}dx
+\sum_{i=0}^{\alpha -k} \int_{0}^{1}\big|\frac{\partial ^{i}\varphi }{\partial x^{i}}%
\big| ^{2}dx }\nonumber \\
&\geq& \frac{7}{8}\int_{0}^{\tau }\int_{0}^{1}\exp (-ct)\big|J^{k}
\frac{\partial ^{2}u}{\partial t^{2}}\big| ^{2}\,dx\,dt +
\sum_{i=0}^{\alpha -k} \int_{0}^{1}\exp (-c\tau )\big|\frac{\partial
^{i}u(x,\tau )}{\partial x^{i}}\big| ^{2}dx \nonumber \\
&&+\sum_{i=0}^{\alpha -k} \int_{0}^{\tau
}\int_{0}^{1}\exp (-ct)\big|\frac{\partial ^{i+1}u}{\partial x^{i}
\partial t}\big| ^{2}\,dx\,dt  \label{20} \\
&&+\frac{1}{2}\int_{0}^{1}\exp (-c\tau )a(\tau )\big|
\frac{\partial ^{\alpha -k+1}u(x,\tau )}{\partial x^{\alpha -k}
\partial t}\big|^{2}dx  \nonumber
\end{eqnarray}
Using elementary inequalities and (\ref{1'}) we obtain
\begin{eqnarray}
\lefteqn{ 8.4^{k}\int_{Q}\big|\mathcal{L}u\big| ^{2}\,dx\,dt 
+\frac{a_{1}}{2}\int_{0}^{1}\big|\frac{\partial ^{\alpha -k}\psi }
{\partial x^{\alpha -k}}\big| ^{2}dx+\sum_{i=0}^{\alpha -k} 
\int_{0}^{1}\big|\frac{\partial ^{i}\varphi }{\partial x^{i}}\big|
^{2}dx }\nonumber \\
&\geq& \exp (-cT)\big[ \frac{7}{8}\int_{0}^{\tau }\int_{0}^{1}\big|J^{k}%
\frac{\partial ^{2}u}{\partial t^{2}}\big| ^{2}\,dx\,dt +
\sum_{i=0}^{\alpha -k} \int_{0}^{1}\big|\frac{\partial ^{i}u(x,\tau )}{%
\partial x^{i}}\big| ^{2}\,dx \label{21} \\
&&+\sum_{i=0}^{\alpha -k} \int_{0}^{\tau
}\int_{0}^{1}\big|\frac{\partial ^{i+1}u}{\partial x^{i}\partial t}
\big|^{2}\,dx\,dt +\frac{a_{0}}{2}\int_{0}^{1}
\big|\frac{\partial ^{\alpha-k+1}u(x,\tau )}{\partial x^{\alpha -k}
\partial t}\big| ^{2}dx\big]
\nonumber
\end{eqnarray}
As the left hand side of (\ref{21}) is independent of $\tau $, by replacing
the right hand side by its upper bound with respect to $\tau $ in the
interval $[ 0,T]$, we obtain the desired inequality. 
\hfill$\diamondsuit$


\begin{lemma}
The operator $L$ from $E$ to $F$ admits a closure.
\end{lemma}

\paragraph{Proof} Suppose that $(u_{n})\in D(L)$ is a sequence such that
\begin{equation}
u_{n}\to 0\quad \mbox{in }E  \label{22}
\end{equation}
and
\begin{equation}
Lu_{n}\to \mathcal{F}\quad \mbox{in }F\,.  \label{23}
\end{equation}
We need to show that $\mathcal{F}=(f,\varphi _{1},\varphi _{2})=0$.
The fact that $\varphi _{i}=0;\ i=1,2$; results directly from the continuity
of the trace operators $l_{i}$.

Introduce the operator
\[
\mathcal{L}_{0}v=\frac{\partial ^{2}((1-x)^{\alpha +k}J^{k}v)}{\partial 
t^{2}}+(-1)^{\alpha +1}\frac{\partial ^{\alpha +1}}{\partial x^{\alpha }
\partial t}(a(t)\frac{\partial ^{\alpha }((1-x)^{\alpha +k}J^{k}v)}
{\partial x^{\alpha}}),
\]
defined on the domain $D(\mathcal{L}_{0})$ of functions 
$v\in W_{2}^{2\alpha,2}(Q)$ satisfying
\[
v|_{t=T}=0\quad \frac{\partial v}{\partial t}|_{t=T}=0
\quad \frac{\partial ^{i}v}{\partial x^{i}}|_{x=0}=\frac{\partial 
^{i}v}{\partial x^{i}}|_{x=1}=0\,, \quad i=\overline{0,\alpha -1}\,.
\]
we note that $D(\mathcal{L}_{0})$ is dense in the Hilbert space obtained
by completing $L_{2}(Q)$ with respect to the norm
\[
\left\| v\right\| ^{2}=\int_{Q}(1-x)^{2(\alpha +k)}\big|J^{k}v\big|
^{2}\,dx\,dt .
\]
Since
\begin{eqnarray*}
\int_{Q}f\overline{(1-x)^{\alpha +k}J^{k}v}\,dx\,dt 
&=&\lim_{n\to\infty} \int_{Q}\mathcal{L}u_{n}
\overline{(1-x)^{\alpha +k}J^{k}v}\,dx\,dt \\
&=&\lim_{n\to \infty} \int_{Q}u_{n}\mathcal{L}_{0}\overline{v}\,dx\,dt =0\,,
\end{eqnarray*}
holds for every function $v\in D(\mathcal{L}_{0})$, it follows that 
$f=0$. \hfill$\diamondsuit$\smallskip

Theorem \ref{th1} is valid for strong solutions, i.e., we have the
inequality
\[
\left\| u\right\| _{E}\leq C_{1}\left\| \overline{L}u\right\|_{F}\,,
\ \forall \, u\in D(\overline{L}),
\]
hence we obtain the following.

\begin{corollary}
A strong solution of (\ref{1})-(\ref{5}) is unique if it exists,
and depends continuously on $\mathcal{F}=(f,\varphi ,\psi )\in F$.
\end{corollary}

\begin{corollary}
The range $R(L)$ of the operator $\overline{L}$ is closed in $F$, and 
$R(\overline{L})=\overline{R(L)}$.
\end{corollary}

\section{Solvability of Problem (\ref{1})-(\ref{5})}

To proof solvability of (\ref{1})-(\ref{5}), it is
sufficient to show that $R( L)$ is dense in $F$. The proof is
based on the following lemma

\begin{lemma}
\label{le9} Let $D_{0}( L)=\left\{ u\in D(L):l_{1}u=0,l_{2}u=0\right\}$.
 If for $u\in D_{0}( L)$ and some $\omega \in L_{2}( Q)$, we have
\begin{equation}
\int_{Q}( 1-x)^{2k}\mathcal{L}u\varpi \,dx\,dt =0\,,  \label{24}
\end{equation}
then $\omega =0$.
\end{lemma}

\paragraph{Proof} The equality (\ref{24}) can be written as follows
\begin{equation}
-\int_{Q}( 1-x)^{2k}\frac{\partial ^{2}u}{\partial t^{2}}\varpi
\,dx\,dt =( -1)^{\alpha }\int_{Q}( 1-x)^{2k}\frac{%
\partial ^{\alpha }}{\partial x^{\alpha }}( a\frac{\partial ^{\alpha
+1}u}{\partial x^{\alpha }\partial t})\varpi \,dx\,dt .  \label{25}
\end{equation}
For $\omega (x,t)$ given, we introduce the function 
$v(x,t)=(-1)^{k}\partial ^{2k}((1-x)^{2k}\omega )/ \partial x^{2k}$, 
then we have $\int_{0}^{1}x^{i}v(x,t)dx=0$  for $i=\overline{0,2k-1}$. 
Then from equality (\ref{25}) we have
\begin{equation}
-\int_{Q}\frac{\partial ^{2}u}{\partial t^{2}}J^{2k}\overline{v}\,dx\,dt 
=(-1)^{\alpha }\int_{Q}\frac{\partial ^{\alpha }}{\partial x^{\alpha }}
( a\frac{\partial ^{\alpha +1}u}{\partial x^{\alpha }\partial t})
J^{2k}\overline{v}\,dx\,dt .  \label{25'}
\end{equation}
Integrating by parts the right hand side of (\ref{25'}) $2k$ times, 
we get
\begin{equation}
-\int_{Q}\frac{\partial ^{2}u}{\partial t^{2}}J^{2k}\overline{v}%
\,dx\,dt =\int_{Q}A(t)\frac{\partial u}{\partial t}\overline{v}\,dx\,dt ,  
\label{25''}
\end{equation}
where $A(t)u=(-1)^{\alpha }\frac{\partial ^{\alpha -k}}{\partial x^{\alpha
-k}}( a\frac{\partial ^{\alpha -k}u}{\partial x^{\alpha -k}})$.

When we introduce the smoothing operators 
$J_{\varepsilon }^{-1}=( I+\varepsilon \frac{\partial }{\partial t})^{-1}$
 and $( J_{\varepsilon }^{-1})^{\ast }$ with respect to $t$ \cite{14},
 then these operators provide solutions of the problems
\begin{eqnarray}
\varepsilon \frac{dg_{_{\varepsilon }}( t)}{dt}+g_{\varepsilon
}( t)&=&g(t),  \label{26} \\
g_{\varepsilon }( t)|_{t=0} &=&0,  \nonumber
\end{eqnarray}
and
\begin{eqnarray}
-\varepsilon \frac{dg_{\varepsilon }^{\ast }( t)}{dt}%
+g_{\varepsilon }^{\ast }( t)&=&g( t),  \label{27} \\
g_{\varepsilon }^{\ast }( t)|_{t=T} &=&0\,.  \nonumber
\end{eqnarray}
The solutions have the following properties: for any $g\in L_{2}(
0,T)$, the functions $g_{\varepsilon }=( J_{\varepsilon
}^{-1})g$ and $\ g_{\varepsilon }^{\ast }=( J_{\varepsilon
}^{-1})^{\ast }g$ are in $W_{2}^{1}( 0,T)$ such that $%
g_{\varepsilon }\mid _{t=0}=0$ and $g_{\varepsilon }^{\ast }\mid _{t=T}=0$.
Moreover, $J_{\varepsilon }^{-1}$ commutes with $\frac{\partial }{\partial 
t}
$, so $\int_{0}^{T}\big|g_{\varepsilon }-g\big| ^{2}dt\to 0$
and $\int_{0}^{T}\big|g_{\varepsilon }^{\ast }-g\big|
^{2}dt\to 0$, for $\varepsilon \to 0$.

Replacing in (\ref{25''}), $\frac{\partial u}{\partial t}$ by the smoothed
function $J_{\varepsilon }^{-1}\frac{\partial u}{\partial t}$,
 using the relation 
$A( t)J_{\varepsilon }^{-1}=J_{\varepsilon }^{-1}A( \tau
)+\varepsilon J_{\varepsilon }^{-1}\frac{\partial A( \tau
)}{\partial \tau }J_{\varepsilon }^{-1}$, and using properties of
the smoothing operators we obtain
\begin{equation}
\int_{Q}\frac{\partial u}{\partial t}\overline{J^{2k}(\frac{\partial
v_{\varepsilon }^{\ast }}{\partial t})}\,dx\,dt =\int_{Q}A( t)\frac{%
\partial u}{\partial t}\overline{v_{\varepsilon }^{\ast }}\,dx\,dt +\varepsilon
\int_{Q}\frac{\partial A}{\partial t}( \frac{\partial u}{\partial t}%
)_{\varepsilon }\overline{v_{\varepsilon }^{\ast }}\,dx\,dt .  \label{30}
\end{equation}
Passing to the limit,(\ref{30}) is satisfied for all
functions satisfying the conditions (\ref{2})-(\ref{5}) such that 
$\frac{\partial ^{i}}{\partial x^{i}}( a\frac{\partial ^{\alpha +1}u}
{\partial x^{\alpha }\partial t})\in L_{2}(Q)$ for $0\leq i\leq \alpha $.

The operator $A( t)$ has a continuous inverse 
on $L_{2}( 0,1)$ defined by
\begin{eqnarray}
\lefteqn{ A^{-1}( t)g} \label{31}\\
&=&(-1)^{\alpha }\int_{0}^{x}\int_{0}^{\eta _{\alpha
-k-1}}\dots\int_{0}^{\eta _{1}}  
\big[ \int_{0}^{\eta }\int_{0}^{\xi _{\alpha-k-1}}\dots
\int_{0}^{\xi _{1}} \frac{1}{a}g( \xi )d\xi d\xi_{1}\dots 
\,d\xi _{\alpha -k-1}  \nonumber \\
&&
 +\sum_{i=1}^{\alpha -k} C_{i}(t)\frac{\eta^{i-1}}{(i-1)!}\big]\,
d\eta d\eta _{1}\dots d\eta _{\alpha -k-1}\,. \nonumber 
\end{eqnarray}
Then we have
\begin{equation}
\int_{0}^{1}A^{-1}( t)g\,dx=0\,.  \label{32}
\end{equation}
Hence the function $( \frac{\partial u}{\partial t})
_{\varepsilon }$ can be represented in the form $( \frac{\partial u}{%
\partial t})_{\varepsilon }=J_{\varepsilon }^{-1}A^{-1}A\frac{%
\partial u}{\partial t}$. Then $\frac{\partial A}{\partial t}( \frac{%
\partial u}{\partial t})_{\varepsilon }=A_{\varepsilon }(
t)A( t)\frac{\partial u}{\partial t}$, where
\begin{equation}
A_{\varepsilon }( t)g=(-1)^{\alpha }\big[ a'(t)J_{\varepsilon }^{-1}
\frac{g}{a}+\sum_{i=1}^{\alpha -k} \frac{\partial ^{\alpha -k}}
{\partial x^{\alpha -k}}
\big\{ \frac{x^{i-1}}{(i-1)!}J_{\varepsilon }^{-1}C_{i}\big\} \big]. 
 \label{32'}
\end{equation}
Consequently, (\ref{30}) becomes
\begin{equation}
\int_{Q}\frac{\partial u}{\partial t}\overline{J^{2k}(\frac{\partial
v_{\varepsilon }^{\ast }}{\partial t})}\,dx\,dt =\int_{Q}A( t)\frac{%
\partial u}{\partial t}( \overline{v_{\varepsilon }^{\ast }+\varepsilon
A_{\varepsilon }^{\ast }v_{\varepsilon }^{\ast }})\,dx\,dt ,  \label{34}
\end{equation}
in which $A_{\varepsilon }^{\ast }( t)$ is the adjoint of the
operator $A_{\varepsilon }( t)$.
The left-hand side of (\ref{34}) is a continuous linear functional of 
$\frac{\partial u}{\partial t}$. Hence the function $h_{\varepsilon
}=v_{\varepsilon }^{\ast }+\varepsilon A_{\varepsilon }^{\ast
}v_{\varepsilon }^{\ast }$ has the derivatives $\frac{\partial
^{i}h_{\varepsilon }}{\partial x^{i}}\in L_{2}( Q)$, $\frac{%
\partial ^{i}}{\partial x^{i}}( a\frac{\partial ^{\alpha
-k}h_{\varepsilon }}{\partial x^{\alpha -k}})\in L_{2}( Q)
$, $i=\overline{0,\alpha -k}$, and the following conditions are satisfied
\begin{equation}
\frac{\partial ^{i}h_{_{\varepsilon }}}{\partial x^{i}}\big| _{x=0}
=\frac{\partial ^{i}h_{_{\varepsilon }}}{\partial x^{i}}\big| _{x=1}=0\,,
\ i=\overline{0,\alpha -k-1}.  \label{35}
\end{equation}
The operators \ $A_{\varepsilon }^{\ast }(t)$ are bounded in $L_{2}(Q)$,
for $\varepsilon $ sufficiently small we have
$\left\| \varepsilon A_{\varepsilon }^{\ast }(t)\right\| _{L_{2}(
Q)}<1$; hence the operator $I+\varepsilon A_{\varepsilon }^{\ast
}(t) $ has a bounded inverse in $L_{2}(Q)$. In addition,
the operators $\frac{\partial ^{i}A_{\varepsilon }^{\ast }(t)}
{\partial x^{i}}$, \ $i=\overline{0,\alpha -k}$ are bounded in $L_{2}(Q)$.
 From the equality
\begin{equation}
\frac{\partial ^{i}h_{\varepsilon }}{\partial x^{i}}=(I+\varepsilon
A_{\varepsilon }^{\ast }(t))\frac{\partial ^{i}v_{\varepsilon }^{\ast }}{%
\partial x^{i}}+\varepsilon 
\sum_{k=1}^i \mathcal{C}_{i}^{k}\frac{\partial ^{k}A_{\varepsilon }^{\ast }(t)}
{\partial x^{k}}\frac{\partial ^{i-k}v_{\varepsilon }^{\ast }}
{\partial x^{i-k}},\quad i=\overline{0,\alpha -k-1}\,  \label{36}
\end{equation}
we conclude that  $v_{\varepsilon }^{\ast }$ has 
derivatives
$\frac{\partial ^{i}v_{\varepsilon }^{\ast }}{\partial x^{i}}$ in 
$L_{2}(Q)$, $i=\overline{0,\alpha -k-1}$. 
Taking into account (\ref{35}) and (\ref{36}), for
 $i=\overline{0,\alpha -k-1}$, we have
\begin{eqnarray}
\big[ (I+\varepsilon A_{\varepsilon }^{\ast }(t))\frac{\partial
^{i}v_{\varepsilon }^{\ast }}{\partial x^{i}}+\varepsilon 
\sum_{k=1}^i \mathcal{C}_{i}^{k}\frac{\partial ^{k}A_{\varepsilon
}^{\ast }(t)}{\partial x^{k}}\frac{\partial ^{i-k}v_{\varepsilon }^{\ast 
}}{\partial x^{i-k}}\big]_{x=0}&=&0\,, \label{37} \\
\big[ (I+\varepsilon A_{\varepsilon }^{\ast }(t))\frac{\partial
^{i}v_{\varepsilon }^{\ast }}{\partial x^{i}}+\varepsilon
 \sum_{k=1}^i \mathcal{C}_{i}^{k}\frac{\partial ^{k}A_{\varepsilon
}^{\ast }(t)}{\partial x^{k}}\frac{\partial ^{i-k}v_{\varepsilon }^{\ast 
}}{\partial x^{i-k}}\big]_{x=1}&=&0\,. \label{38}
\end{eqnarray}
Similarly, for $\varepsilon $ sufficiently small, and each fixed 
$x\in [ 0,1]$ the operators $\frac{\partial ^{i}A_{\varepsilon }^{\ast
}(t)}{\partial x^{i}}$, $i=\overline{0,\alpha -k}$ are bounded in $L_{2}(Q)$
and the operator $I+\varepsilon A_{\varepsilon }^{\ast }(t)$ is 
continuously invertible in $L_{2}(Q)$.
From (\ref{37}) and (\ref{38}) result that $v_{\varepsilon }^{\ast }$
satisfies the conditions
\[
\frac{\partial ^{i}v_{\varepsilon }^{\ast }}{\partial x^{i}}\big|_{x=0}
=\frac{\partial ^{i}v_{\varepsilon }^{\ast }}{\partial x^{i}}\big|_{x=1}=0,
\quad i=\overline{0,\alpha -k-1},
\]
So, for $\varepsilon $ sufficiently small, the function $v_{\varepsilon
}^{\ast }$ has the same properties as $h_{\varepsilon }$. In addition $%
v_{\varepsilon }^{\ast }$ satisfies the integral conditions (\ref{5}).

Putting $u=\int_{0}^{t}\int_{0}^{\tau }\exp (c\eta )v_{\varepsilon }
^{\ast}( \eta ,\tau )d\eta d\tau $ in (\ref{25''}), with the constant
$c$ satisfying
$ca_{0}-a_{2}-\frac{a_{2}^{2}}{a_{0}}\geq 0$, and using (\ref{27}), we
obtain
\begin{eqnarray} 
\int_{Q}(-1)^{k}\exp (ct)v_{\varepsilon }^{\ast }\overline{J^{2k}v}%
\,dx\,dt &=&-\int_{Q}(-1)^{k}A( t)\frac{\partial u}{\partial t}\exp
(-ct)\frac{\partial ^{2}\overline{u}}{\partial t^{2}}\,dx\,dt \nonumber\\
&&+\varepsilon \int_{Q}(-1)^{k}A( t)\frac{\partial u}{\partial t}
\frac{\partial \overline{v_{\varepsilon }^{\ast }}}{\partial t}\,dx\,dt 
\,. \label{40'}
\end{eqnarray}
Integrating by parts each term in the right-hand side of (\ref{40'}), 
we have
\begin{eqnarray}
\lefteqn{ \mathop{\rm Re}\int_{Q}(-1)^{k}A( t)\frac{\partial u}{\partial t}\exp
(-ct)\frac{\partial ^{2}\overline{u}}{\partial t^{2}}\,dx\,dt  }
\label{41}\\
&\geq & \frac{c}{2}\int_{Q}a( t)e^{-ct}\big|
\frac{\partial ^{\alpha -k+1}u}{\partial x^{\alpha -k}\partial t}
\big| ^{2}\,dx\,dt  
-\frac{1}{2}\int_{Q}\frac{\partial a}{\partial t}e^{-ct}\big|\frac{%
\partial ^{\alpha -k+1}u}{\partial x^{\alpha -k}\partial t}
\big| ^{2}\,dx\,dt \,. \nonumber
\end{eqnarray}
\begin{equation}
\mathop{\rm Re}( -\varepsilon \int_{Q}(-1)^{k}A( t)\frac{\partial
u}{\partial t}\frac{\partial \overline{v_{\varepsilon }^{\ast }}}{\partial 
t}\,dx\,dt )\geq \frac{-\varepsilon a_{2}^{2}}{2a_{0}}\int_{Q}\exp
(-ct)\big|\frac{\partial ^{\alpha -k+1}u}{\partial x^{\alpha -k}\partial 
t}\big| ^{2}\,dx\,dt .  \label{42}
\end{equation}
Now, using (\ref{41}) and (\ref{42}) in (\ref{40'}),
 with the choice of $c$ indicated above, we have
$2\mathop{\rm Re}\int_{Q}\exp (ct)v_{\varepsilon }^{\ast }\overline{j^{2k}v}%
\,dx\,dt \leq 0$, then  $2\mathop{\rm Re}\int_{Q}
\exp (ct)v\overline{J^{2k}v}\,dx\,dt \leq 0$ as $\varepsilon$ 
approaches zero.
Since $\mathop{\rm Re}\int_{Q}\exp (ct)\big|J^{k}\overline{v}\big| 
^{2}\,dx\,dt =0$, we conclude that $J^{2k}v=0$, hence $\omega =0$, 
which completes the present proof. \hfill$\diamondsuit$

\begin{theorem}
The range $R(L)$ of $L$ coincides with $F.$
\end{theorem}

\paragraph{Proof} Since $F$ is a Hilbert space, we have $R(L)=F$ if
 and only if the following implication is satisfied:
\begin{equation}
\int_{Q}\mathcal{L}u\overline{f}\,dx\,dt +\int_{0}^{1}( 
\sum_{i=0}^{\alpha -k} \frac{\partial ^{i}l_{1}u}{\partial x^{i}}
\frac{\partial ^{i}\overline{\varphi }}{\partial x^{i}}+\frac{\partial 
^{\alpha-k}l_{2}u}{\partial x^{\alpha -k}}\frac{\partial ^{\alpha -k}
\overline{\psi }}{\partial x^{\alpha -k}})dx=0,  \label{43}
\end{equation}
for arbitrary $u\in E$ and $\mathcal{F}=( f,\varphi ,\psi )\in F$, 
implies  that $f$, $\varphi$, and $\psi$ are zero.

Putting $u\in D_{0}(L)$ in (\ref{43}), we obtain $\int_{Q}\mathcal{L}u
\overline{f}\,dx\,dt =0$. Taking $\omega =f/(1-x)^{2k}$, and using lemma
\ref{le9} we obtain that $f/(1-x)^{2k}=0$, then $f=0$.
Consequently, $\forall u\in D(L)$ we have
\begin{equation}
\int_{0}^{1} \sum_{i=0}^{\alpha -k} \frac{
\partial ^{i}l_{1}u}{\partial x^{i}}\frac{\partial ^{i}\overline{\varphi 
}}{\partial x^{i}}+\frac{\partial ^{\alpha -k}l_{2}u}{\partial x^{\alpha -k}}%
\frac{\partial ^{\alpha -k}\overline{\psi }}{\partial x^{\alpha 
-k}}\, dx=0\,.  \label{44}
\end{equation}
The range of the trace operator $(l_{1,}l_{2})$ is everywhere dense 
in a Hilbert space with norm
\[
\big[ \int_{0}^{1} \sum_{i=0}^{\alpha -k} 
\big|\frac{\partial ^{i}\varphi }{\partial x^{i}}\big| ^{2}+\big|
\frac{\partial ^{\alpha -k}\psi }{\partial x^{\alpha -k}}\big| ^{2}\,dx
\big]^{1/2}\,.
\]
Therefore, $(\varphi ,\psi )=(0,0)$ and the present proof is complete. 

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\end{thebibliography} \medskip

\noindent{\sc Mohamed Denche} (e-mail: m\_denche@hotmail.com)\\
{\sc A. L. Marhoune}\\
 Institut de Mathematiques\\ 
Universit\'{e} Mentouri Constantine\\ 
25000 Constantine, Algeria


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