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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 17, 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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/17\hfil Continuous dependence of solutions]
{Continuous dependence of solutions to mixed boundary value
problems for a parabolic equation}

\author[M. Z. Djibibe, K. Tcharie, N. I. Yurchuk\hfil EJDE-2008/17\hfilneg]
{Moussa Zakari Djibibe, Kokou Tcharie, Nikolay Iossifovich Yurchuk }  % in alphabetical order

\address{Moussa Zakari Djibibe\newline
University of Lom\'e - Togo\\
Department of Mathematics  \\
PO Box 1515 Lom\'e, Togo}
\email{mdjibibe@tg.refer.org\; zakari.djibibe@gmail.com\; tel +228 924 45 21}

\address{Kokou Tcharie \newline
University of Lom\'e - Togo\\
Department of Mathematics\\
PO Box  1515 Lom\'e, Togo}
\email{tkokou@yahoo.fr}

\address{Nikolay Iossifovich Yurchuk \newline
 Dept. of Mechanics and Mathematics, Belarussian State University,
220050,  Minsk, Belarus}
\email{yurchuk@bsu.by}

\thanks{Submitted October 16, 2007. Published February 5, 2008.}
\subjclass[2000]{35K20, 35K25, 35K30}
\keywords{Priori estimate; mixed problem; continuous dependence;
\hfill\break\indent boundary conditions}

\begin{abstract}
 We prove the continuous dependence, upon
 the data, of solutions to second-order parabolic equations.
 We study two boundary-value problems: One has a nonlocal (integral)
 condition and the another has a local boundary condition.
 The proofs are based on a priori estimate for the difference
 of solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
%\allowdisplaybreaks

\section{Introduction}

This paper is devoted to the proof of the continuous dependence, upon
the data, of generalized solutions of a second order parabolic equation.
The boundary conditions are of mixed type.
This article contributes to the development of  the a priori estimates
method for solving such problems.
The questions related to these problems are so miscellaneous that the
elaboration of a general theory is still premature. Therefore,
the investigation of these problems requires at every time a
separate study.

The importance of problems with integral condition has been pointed out
by Samarskii \cite{ref24}. Mathematical modelling by evolution
problems with a nonlocal constraint of the form
${\frac{1}{1 -\alpha}\int_{\alpha}^1 u(x, t)\,dx = E(t)}$
is encountered in heat transmission theory, thermoelasticity,
chemical engineering, underground water flow, and plasma physic.
See for instance   Benouar-Yurchuk \cite{ref1}, Benouar-Bouziani
 \cite{ref2}-\cite{ref3}, Bouziani \cite{ref4}-\cite{ref7},
Cannon et al \cite{ref14}-\cite{ref16}, Ionkin
\cite{ref18}-\cite{ref19}, Kamynin \cite{ref20} and Yurchuk
\cite{ref26}-\cite{ref28}.
 Mixed problems with nonlocal boundary
conditions or with nonlocal initial conditions were studied in
Bouziani \cite{ref7}-\cite{ref9}, Byszewski et al
\cite{ref11}-\cite{ref13}, Gasymov \cite{ref17}, Ionkin
\cite{ref18}-\cite{ref19}, Lazhar \cite{ref22},
Mouravey-Philipovski  \cite{ref23} and Said-Nadia
\cite{ref25}.
The results and the method used here are a further elaboration of
those in \cite{ref1}. We should
mention here that the presence of integral term in the boundary
condition can greatly complicate the application of standard
functional and numerical techniques.
 This work can be considered as a continuation of the results
in \cite{ref7}, \cite{ref26} and \cite{ref28}.

The remainder of the paper is divided into four section.
In Section 2, we give the statement of the problem.
Then in Section 3, we establish a priori estimate. Finally, in section
4, we show the continuous dependence of a solution upon the data.

\section{Statement of the problem}

In the rectangle $G = \{(x,t) : 0<x<1,\; 0<t<T  \}$, we consider
following two mixed boundary value problems for parabolic equations:

 The mixed problem with boundary integral condition
\begin{gather}
\frac{\partial u_{\alpha}}{\partial t} -\frac{\partial}{\partial x}
\big(a(x,t)\frac{\partial u_\alpha}{\partial x}\big)
=f(x,t),  \label{c1}\\
u_\alpha(x,0) =\varphi_\alpha(x),\quad
\frac{\partial u_\alpha (0, t)}{\partial x} =0\label{c2}\\
\frac{1}{1-\alpha}\int_\alpha^1 u_\alpha (x, t)\,dx = h(t),   \label{c3}
\end{gather}
 where $ 0\leq \alpha < 1$.

 The mixed problem with  local boundary conditions
\begin{gather}
\frac{\partial u_1}{\partial t} - \frac{\partial}{\partial x}(a(x,t)
\frac{\partial u_1}{\partial x}) = f(x,t)\label{c4} \\
u_1 (x,0) = \varphi_1(x), \quad  \frac{\partial u_1 (0, t)}{\partial x} =0
 \label{c5}\\
u_1(1,t) = h(t). \label{c6}
\end{gather}

We assume the following conditions:
\begin{itemize}
\item[(A1)]
The coefficient $a(x,t)$ in  \eqref{c1} and \eqref{c4} is
a continuous differentiable function, and  $0<a_0 \leq a(x,t)\leq a_1$.

\item[(A2)]
$f \in L_2(G)$, $h \in W_2^1(0,T )$, $\varphi_\alpha$,
$\varphi_1 \in W_2^1(0,T )$,
$\varphi_{\alpha}'(0) = \varphi '_1(0) =0$,
$ \varphi_1(1) = h(0)$,
$\frac{1}{1-\alpha}\int_\alpha^1 \varphi_\alpha(x)\,dx = h(0)$.

\end{itemize}

In  \cite{ref26}, it is shown that if the conditions (A1)--(A2) are satisfied,
then there exists a unique solution to  \eqref{c1}--\eqref{c3} and
\eqref{c4}--\eqref{c6}, and that the solution is  differentiable almost
everywhere on $G$. It is clear that for every function
$g\in C([0,1]):{g(1) = \lim_{\alpha\to 1}
\frac{1}{1 - \alpha}\int_\alpha^1 g(x)\,dx}$
Therefore, the problem \eqref{c4}--\eqref{c6} is the limit
when $\alpha\to 1$ of the  problem  \eqref{c1}--\eqref{c3}.
 In this paper,  we  establish a priori estimate for the difference
$u_\alpha - u_1$ and  use it  to prove that if
 $\alpha \to 1$ and $\varphi_\alpha\to \varphi_1$ then
$u_\alpha \to u_1$.  In this way we prove a new important property;
 a continuous dependence of solutions of mixed problems for parabolic
 equations of the form of boundary conditions.

\section{A priori estimate}

 \begin{theorem} \label{thm1}
Under assumptions {\rm (A1)--(A2)}, there exists a positive constant
$c$  independent of  $u_\alpha, u_1$ and $\alpha$ such that
\begin{equation} \label{c7}
\begin{aligned}
& \int_0^1 \int_0^T(1-x)[ |\frac{\partial u_\alpha}{\partial t}
 - \frac{\partial u_1}{\partial t} |^2
 + |\frac{\partial^2 u_\alpha}{\partial x^2}
 - \frac{\partial^2 u_1}{\partial x^2} |^2 ]\,dx\,dt  \\
& + \sup_{0\leq t \leq T}\int_0^1[ |u_\alpha -u_1|^2
 +(1-x)|\frac{\partial u_\alpha}{\partial x}
 - \frac{\partial u_1}{\partial x} |^2 ]\,dx \\
&\leq c\Big\{\int_0^1 |\varphi_\alpha ' (x)
 - \varphi'_1 (x)|^2\,dx
 + \big|h(0) - \frac{1}{1-\alpha} \int_\alpha^1 u_1(x , 0)\,dx \big|^2  \\
&\quad+ \int_0^T \big[\big|h' (t) - \frac{1}{1-\alpha}\int_\alpha^1
 \frac{\partial u_1(x , t)}{\partial t}\,dx  \big|^2
 + \big|h(t)-\frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,t)\,dx  \big|^2 \big]\,dt
\Big\}
\end{aligned}
\end{equation}
\end{theorem}

\begin{proof}
Consider the problem \eqref{c1}--\eqref{c3}, with
\begin{equation}
u_\alpha = v_\alpha + \frac{3x^2}{1+\alpha + \alpha^2}h(t)  \label{c8}
\end{equation}
where $v_\alpha$ is a solution of the problem
\begin{gather*}
 \frac{\partial v_\alpha}{\partial t}
 -\frac{\partial}{\partial x}(a(x,t)\frac{\partial v_\alpha}{\partial x} )
 =  g_{\alpha}(x, t) \\
 v_\alpha (x,0) =\psi_{\alpha}(x),\quad
 \frac{\partial v_\alpha (0, t)}{\partial x} = 0, \quad
 \frac{1}{1-\alpha}\int_\alpha^1 v_\alpha (x,t)\,dx = 0\,,
\end{gather*}
where
\begin{gather*}
 g_{\alpha}(x, t) =  f(x,t) -  \frac{3x^2}{1+\alpha +\alpha^2}h'(t)
 + \frac{6 h(t)}{1 +\alpha + \alpha^2}(a(x, t)
 + x\frac{\partial a(x, t)}{\partial x})\\
\psi_{\alpha}(x) = \varphi_\alpha (x) - \frac{3x^2}{1+\alpha
 + \alpha^2}h(0)
\end{gather*}
In  \eqref{c4}-\eqref{c6} we also pose
\begin{equation}
 u_1 = v_1 + \frac{3x^2}{1-\alpha^3}\int_\alpha^1 u_1(x , t)\,dx \label{c9}
\end{equation}
where  $v_1$ is a solution of the problem
\begin{gather*}
 \frac{\partial v_1}{\partial t} -\frac{\partial}{\partial x}
\big(a(x,t)\frac{\partial v_1}{\partial x} \big) = g_1(x,t)\\
 v_1(x,0) = \psi_1(x),\quad  \frac{\partial v(0, t)}{\partial x} = 0,\\
v_1(1,t) = h(t) - \frac{3}{1-\alpha^3}\int_\alpha^1 u_1(x, t)\,dx\,
\end{gather*}
where
\begin{align*}
 &g_1(x) \\
&= f(x,t) -\frac{3x^2}{1-\alpha^3}\int_\alpha^1
  \frac{\partial u_1(x ,t)}{\partial t}\,dx
 + \frac{6}{1-\alpha^3} (a(x,t) + x\frac{\partial a(x,t)}{\partial x})
 \int_\alpha^1 u_1(x ,t )\,dx,
\end{align*}
\[
 \psi_(x) =  \varphi_1(x) - \frac{3x^2}{1-\alpha^3}
 \int_\alpha^1 u_1(x , 0)\,dx\,.
\]
Then the function $w_\alpha = v_1 - v_\alpha$ is a solution of the
 problem
\begin{gather}
\frac{\partial w_\alpha}{\partial t}
 -\frac{\partial}{\partial x}(a(x,t)\frac{\partial w_\alpha}{\partial x} )
 = F_\alpha (x,t)\label{c10} \\
w_\alpha(x,0) = \phi_\alpha(x),\quad\frac{\partial w_\alpha(0,t)}{\partial x}
= 0,\quad \frac{1}{1-\alpha}\int_\alpha^1 w_\alpha(x,t)\,dx = 0\,,\label{c11}
\end{gather}
where
\begin{gather}
\begin{aligned}
 F_\alpha(x,t)& = -\frac{6}{1+\alpha +\alpha^2}\big(a(x,t)
   + x\frac{\partial a(x, t)}{\partial x}\big)\big[ h(t)
   - \frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,t)\,dx  \big] \\
  &\quad +   \frac{3x^2}{1+\alpha + \alpha^2}
  \big[ h'(t) - \frac{1}{1-\alpha}\int_\alpha^1
  \frac{\partial u_1(x, t)}{\partial t}\,dx   \big]
\end{aligned}\label{c12}
\\
 \phi_\alpha(x) = \varphi_1(x) - \varphi_\alpha(x)
  + \frac{3x^2}{1+\alpha + \alpha^2}
  \big(  h(0) - \frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,0)\,dx \big)\,.
\label{c13}
\end{gather}
For this purpose we need the following result.

\begin{lemma} \label{lem1}
For a function $w_\alpha$ satisfying {\rm \eqref{c10})--\eqref{c11}},
there exists a positive constant $c$ independent of $w_\alpha, \phi_\alpha$
and  $F_\alpha$ such that
\begin{equation}
\begin{aligned}
&\int_G \psi_\alpha(x)\big[|\frac{\partial w_\alpha}{\partial t} |^2
 +  |\frac{\partial^2 w_\alpha}{\partial x^2} |^2\big]\,dx\,dt
 +  \sup_{0 \leq t\leq T}\int_0^1\big[ \psi_\alpha(x)|
  \frac{\partial w_\alpha}{\partial x} |^2 + |w_\alpha |^2\big]\,dx \\
&\leq  c_1\big\{\int_0^1\psi_\alpha(x)|\phi_\alpha'(x)|^2\,dx
+ \int_G |F_\alpha(x,t)|^2\,dx\,dt \big\}
\end{aligned}\label{c14}
\end{equation}
where
$\psi_\alpha (x) = \begin{cases}
1&\text{if } 0\leq x \leq \alpha \\
\frac{1 -x}{1 - \alpha} &\text{if } \alpha \leq x\leq 1\,.
\end{cases}$
\end{lemma}

Suppose, for the moment, that Lemma \ref{lem1} has been proved,
and turn to the proof of Theorem \ref{thm1}.
In accordance with the equality
\[
 u_1 - u_\alpha = w_\alpha - \frac{3x^2}{1+\alpha + \alpha^2}
\big[ h(t) - \frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,t)\,dx  \big],
\]
and the estimates
\begin{align*}
&\int_G (1-x)|\frac{\partial u_1}{\partial t}
- \frac{\partial u_\alpha}{\partial t}|^2\,dx\,dt \\
&\leq   \frac{3}{5}\int_0^T|h'(t)-\frac{1}{1-\alpha}\int_\alpha^1
  \frac{\partial u_1(x ,t)}{\partial t}\,dx|^2\,dt
  +  2\int_G(1-x)|\frac{\partial w_\alpha}{\partial t}|^2\,dx\,dt\,,
\end{align*}
\begin{align*}
& \int_G(1-x)|\frac{\partial^2 u_1}{\partial x^2}
  - \frac{\partial^2 u_\alpha}{\partial x^2}|^2\,dx\,dt \\
& \leq   36\int_0^T |h(t) - \frac{1}{1 - \alpha}
  \int_\alpha^1 u_1(x,t)\,dx|^2\,dt
 + 2\int_G(1-x)|\frac{\partial^2 w_\alpha}{\partial x^2}|^2\,dx\,dt\,,
\end{align*}
\begin{align*}
&\sup_{0\leq t\leq T}\int_0^1(1-x)|\frac{\partial u_1}{\partial x}
  - \frac{\partial u_\alpha}{\partial x}|^2\,dx \\
&\leq   6\sup_{0\leq t\leq T}|h(t)
  - \frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,t)\,dx |^2
  + 2\sup_{0\leq t\leq T}\int_0^1(1-x)
  |\frac{\partial w_\alpha}{\partial x}|^2\,dx\,,
\end{align*}
\begin{align*}
&\sup_{0\leq t\leq T}\int_0^1 |u_1 - u_\alpha |^2\,dx\\
&\leq   \frac{18}{5}\sup_{0\leq t\leq T}|h(t)
  - \frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,t)\,dx|^2
  +  2\sup_{0\leq t\leq T}\int_0^1 |w_\alpha |^2\,dx\,,
\end{align*}
\[
\int_0^1  |\phi_\alpha'|^2\,dx
\leq  2\int_0^1 |\varphi'_1(x) - \varphi_\alpha'|^2\,dx
  + 24|h(0) - \frac{1}{1-\alpha}\int_\alpha^1 u_1(x,  0)\,dx|^2,
\]
\begin{align*}
 \int_G |F_\alpha(x,t)|^2\,dx\,dt
 &\leq  72\int_G|a(x,t) +x \frac{\partial a(x,t)}{\partial x}|^2\,dx\,dt \\
 &\quad \times \int_0^T |h(t)- \frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,t)\,dx|^2\,dt  \\
 &\quad +  \frac{18}{5}\int_0^T |h'(t)
  - \frac{1}{1-\alpha}\int_\alpha^1
  \frac{\partial u_1(x ,t)}{\partial t}\,dx|^2\,dt\,,
\end{align*}
\begin{align*}
 &\sup_{0\leq t\leq T}|h(t)
  - \frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,t)\,dx|^2 \\
&\leq  2|h(0) - \frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,0)\,dx|^2
 +  2T\int_0^T |h'(t)
  - \frac{1}{1-\alpha}\int_\alpha^1
  \frac{\partial u_1(x ,t)}{\partial t}\,dx|^2\,dt\,.
\end{align*}
We have the estimate
\begin{equation}
 \int_0^1 |\varphi_1(x)- \varphi_\alpha(x)|^2\,dx
 \leq 2\int_0^1 |\varphi'_1(x) - \varphi_\alpha'(x)|^2\,dx
 + 2|\varphi_1 (0) - \varphi_\alpha (0)|^2\,.
\end{equation}
Then we obtain
\begin{align*}
&\int_G (1 - x)\big[|\frac{\partial u_1}{\partial t}
 - \frac{\partial u_\alpha}{\partial t}|^2
 + |\frac{\partial^2 u_1}{\partial x^2}
 - \frac{\partial^2 u_\alpha}{\partial x^2}|^2\big]\,dx\,dt  \\
&+\sup_{0\leq t\leq T} \int_0^1\big[(1 - x)|\frac{\partial u_1}{\partial x}
 - \frac{\partial u_\alpha}{\partial x}|^2
 +   |u_1 - u_\alpha|^2\big]\,dx\,dt  \\
&\leq 4c_1\int_0^1|\varphi'_1(x) - \varphi'_{\alpha}(x)|^2\,dx
 + 48( \frac{5c_1 + 2}{5}) \big| h(0)
 - \frac{1}{1 -\alpha}\int_{\alpha}^1 u_1(x, 0)\,dx\big|^2 \\
&\quad + 3(\frac{12c_1 + 1 + 32T}{5})
  \int_0^T \big|h'(t) - \frac{1}{1-\alpha}\int_{\alpha}^1
  \frac{\partial u_1(x, t)}{\partial t}\,dx\big|^2\,dt \\
&\quad + \big(36 + 144c_1\max_{0\leq t\leq T}\int_0^1|a(x,t)
  + x\frac{\partial a(x,t)}{\partial x}|^2\,dx\big)
  \int_0^T \big|h(t)\\
&\quad - \frac{1}{1 - \alpha}\int_\alpha^1 u_1(x,t)\,dx\big|^2\,dt\,.
\end{align*} % \label{c16}
 From the above inequality, it follows that \eqref{c7} holds with
\begin{equation}
 c = \max\Big(4c_1,\; 48(\frac{5c_1 + 2}{5}),\;
\max \big(3(\frac{12c_1 + 1 + 32T}{5}),\; N\big)  \Big)\,,
\end{equation}
where
\[
N =36 + 144c_1\max_{0\leq t\leq T}\int_0^1|a(x,t)
+ x\frac{\partial a(x,t)}{\partial x}|^2\,dx\,.
\]

To complete the proof of Theorem \ref{thm1}, it remains to prove the Lemma.


\begin{proof}[Proof of Lemma \ref{lem1}]
 Let
\[
 M_{\alpha} w_{\alpha} = \begin{cases}
\frac{\partial w_{\alpha}}{\partial t} &\text{if } 0\leq x \leq \alpha \\
\frac{1 -x}{1 - \alpha}\frac{\partial w_{\alpha}}{\partial t}
+ \frac{1}{1 - \alpha} I_{\alpha} \frac{\partial w_{\alpha}}{\partial t}
&\text{if } \alpha \leq x\leq 1\,,
\end{cases}
\]
where ${I_\alpha w_\alpha = \int_\alpha^x w_\alpha(\xi ,t)\,d\xi}$.
Multiplying both sides of  (\ref{c10}) by $M_\alpha w_\alpha$ and
integrating the result  with respect to  $x\in [0; 1]$, we obtain
\begin{equation}
 \int_0^1 \frac{\partial w_{\alpha}}{\partial t}M_{\alpha} w_{\alpha} \,dx
- \int_0^1 \frac{\partial}{\partial x}(a(x,t)
 \frac{\partial w_\alpha}{\partial x})M_\alpha w_\alpha\,dx
 = \int_0^1 F_\alpha (x; t)M_\alpha w_\alpha\,dx\,dt\,. \label{c17}
\end{equation}
 Integrating by parts the first two integrals on the left-hand
of (\ref{c17}), and using the boundary conditions (\ref{c11}), we obtain
\begin{equation}
\int_0^1 \frac{\partial w_{\alpha}}{\partial t}M_{\alpha} w_{\alpha} \,dx
= \int_0^1 \psi_{\alpha}(x)(\frac{\partial w_{\alpha}}{\partial t})^2\,dx;
 \label{c18}
\end{equation}
and
\begin{equation}
\begin{aligned}
&-\int_0^1 \frac{\partial}{\partial x}\big(a(x,t)
  \frac{\partial w_\alpha}{\partial x}\big)M_\alpha w_\alpha\,dx \\
&=  \frac{1}{2}\frac{\partial}{\partial t}
\Big(\int_0^1\psi_{\alpha} (x) a(x; t)|\frac{\partial w_\alpha}{\partial x}
|^2\,dx\Big)
 -  \frac{1}{2}\int_0^1 \psi_{\alpha} (x)
 \frac{\partial a(x, t)}{\partial t}|
 \frac{\partial w_\alpha}{\partial x}|^2\,dx\,.
\end{aligned}\label{c19}
\end{equation}
 Substituting (\ref{c18}) and (\ref{c19}) into (\ref{c17}), we obtain
\begin{align*}
  &\int_0^1 \psi_{\alpha}(x)|\frac{\partial w_{\alpha}}{\partial t}|^2\,dx
  + \frac{1}{2}\frac{\partial}{\partial t}
 \Big(\int_0^1\psi_{\alpha} (x) a(x; t)|\frac{\partial w_
  \alpha}{\partial x}|^2\,dx\Big)  \\
&= \frac{1}{2}\int_0^1 \psi_{\alpha} (x)
  \frac{\partial a(x, t)}{\partial t}|
  \frac{\partial w_\alpha}{\partial x}|^2\,dx
+ \int_0^1 F_\alpha (x; t)M_\alpha w_\alpha\,dx\,dt\,.
%\label{c20}
\end{align*}
 Multiplying both sides of  the above equality by  $e^{c(\tau - t)}$,
and   integrating  the result  with  respect  $t\in [0; \tau]$, we obtain
\begin{equation}
\begin{aligned}
 &\int_0^{\tau}\int_0^1 e^{c(\tau - t)}\psi_{\alpha}(x)|
  \frac{\partial w_\alpha}{\partial t}|^2\, dx\,dt
  + \frac{1}{2}\int_0^1 a(x,t)\psi_{\alpha} (x)|
  \frac{\partial w_\alpha}{\partial x}|^2\,dx \mid_{t = \tau}  \\
 &= \int_0^{\tau}\int_0^1 e^{c(\tau - t)}
 F_{\alpha} (x, t)M_{\alpha}w_{\alpha}\,dx\,dt
 + \frac{1}{2}\int_0^{\tau}\int_0^1 e^{c(\tau - t)}
 \frac{\partial a(x, t)}{\partial t}\psi_{\alpha} (x)|
 \frac{\partial w_\alpha}{\partial x}|^2\,dx\,dt  \\
 &\quad+ \frac{1}{2}\int_0^1 e^{c\tau}a(x,0)\psi_{\alpha} (x)|\phi'(x) |^2\,dx
  -  \frac{1}{2}\int_0^{\tau}\int_0^1 ca(x,t)
 e^{c(\tau - t)}\psi_\alpha(x)|
 \frac{\partial w_\alpha}{\partial x}|^2\,dx\,dt\,.
\end{aligned} \label{c21}
\end{equation}
Using the following estimates
\begin{align*}
&\int_0^1 F_\alpha(x,t)M_\alpha w_\alpha \,dx \\
&\leq  \int_0^1 |F_\alpha(x,t)|^2\,dx
 + \frac{1}{4}\int_0^1\psi_\alpha(x)|
 \frac{\partial w_\alpha}{\partial t}|^2\,dx
 + \frac{1}{1-\alpha}\int_\alpha^1 F_\alpha(x,t)I_\alpha
 \frac{\partial w_\alpha}{\partial t}\,dx\,,
\end{align*}
\[
 \frac{1}{1-\alpha}\int_\alpha^1 F_\alpha(x,t)I_\alpha
 \frac{\partial w_\alpha}{\partial t}\,dx \leq 2\int_\alpha^1
 |F_\alpha(x,t)|^2\,dx + \frac{1}{8(1-\alpha)^2}
 \int_\alpha^1|I_\alpha\frac{\partial w_\alpha}{\partial t}|^2\,dx\,,
\]
\[
 \frac{1}{8(1-\alpha)^2}\int_\alpha^1 |I_\alpha
 \frac{\partial w_\alpha}{\partial t}|^2\,dx
 \leq \frac{1}{2}\int_\alpha^1\frac{(1-x)^2}{(1-\alpha)^2}|
 \frac{\partial w_\alpha}{\partial t}|^2\,dx \leq \frac{1}{2}
 \int_0^1 \psi_\alpha(x)|\frac{\partial w_\alpha}{\partial t}|^2\,dx;
\]
and choosing a constant $c$ in (\ref{c21}) such that
 ${ca_0 \geq |\frac{\partial a(x,t)}{\partial t}|}$, we deduce the
inequality
\begin{align*}
& \frac{1}{4}\int_0^{\tau}\int_0^1 e^{c(\tau - t)}\psi_{\alpha}(x)|
 \frac{\partial w_\alpha}{\partial t}|^2\, dx\,dt
 + \frac{1}{2}a_0 \int_0^1 \psi_{\alpha} (x)|
  \frac{\partial w_\alpha}{\partial x}|^2\,dx \mid_{t = \tau} \\
&\leq 3  \int_0^{\tau}\int_0^1 e^{c(\tau - t)}|F_{\alpha} (x, t)|^2\,dx\,dt
 +\int_0^1 e^{c\tau}a(x,t)\psi_{\alpha} (x)|\phi'(x) |^2\,dx
\end{align*} %\label{c22}
which can be written as
\begin{equation}
\begin{aligned}
 &\frac{1}{4}\int_0^{\tau}\int_0^1 \psi_\alpha(x)|
 \frac{\partial w_\alpha}{\partial t}|^2\,dx\,dt
 + a_0 \int_0^1 \psi_\alpha(x)|
 \frac{\partial w_\alpha}{\partial t}|^2\mid_{t =\tau}\,dx  \\
&\leq 3\int_G e^{cT}|F_\alpha(x,t)|^2\,dx\,dt +\int_0^1
 e^{cT}\psi_\alpha(x)a(x,t) |\phi'_\alpha(x)|^2\,dx\,.
\end{aligned}\label{c23}
\end{equation}
For the function $w_\alpha(x,t)$, there hold the relation
\begin{equation}
 \int_0^1 |w_\alpha(x,t)|^2\,dx \leq 4\int_0^1 (1-x)^2|
 \frac{\partial w_\alpha(x,t)}{\partial x}|^2\,dx
 \leq 4\int_0^1 \psi_\alpha(x)|
 \frac{\partial w_\alpha(x,t)}{\partial x}|^2\,dx\,. \label{c24}
\end{equation}
It follows from (\ref{c23}) and (\ref{c24}) that
\begin{equation}
\begin{aligned}
&\frac{1}{4}\int_0^{\tau}\int_0^1 \psi_\alpha(x)|
 \frac{\partial w_\alpha}{\partial t}|^2\,dx\,dt
 +  \frac{a_0}{2}\int_0^1 \psi_\alpha(x)|
 \frac{\partial w_\alpha}{\partial x}|^2\mid_{t = \tau}\,dx
 + \frac{a_0}{8}\int_0^1 |w_\alpha|^2\mid_{t = \tau}\,dx \\
&\leq  e^{cT}\int_0^1 a(x,t)\psi_\alpha(x)|\phi'_\alpha(x)|^2\,dx
 + 3e^{cT}\int_G |F_\alpha(x,t)|^2\,dx\,dt\,.
\end{aligned}\label{c25}
\end{equation}
Since the right-hand side  of (\ref{c25}) does not depend on
$\tau$, we take the  supremum of the left-hand side of
(\ref{c25}) with respect to $0\leq t \leq T$; we obtain
\begin{equation}
\begin{aligned}
 &\frac{1}{4}\int_G \psi_\alpha (x)|
 \frac{\partial w_\alpha}{\partial t}|^2\,dx\,dt
 +\frac{1}{2}a_0\sup_{0\leq t\leq T}\int_0^1[\psi_\alpha(x)|
 \frac{\partial w_\alpha}{\partial x}|^2 + \frac{1}{4}|w_\alpha |^2]\,dx \\
 &\leq  e^{cT}\int_0^1 a(x,t)\psi_\alpha(x)|\phi'_\alpha(x)|^2\,dx
 +  3e^{cT}\int_G |F_\alpha(x,t)|^2\,dx\,dt\,.
\end{aligned} \label{c26}
\end{equation}
It follows from (\ref{c10}) that
\begin{align*}
 & a_0\int_G \psi_\alpha(x)|\frac{\partial w_\alpha ^2}{\partial x^2}
 |^2\,dx\,dt  \\
&\leq  2\int_G \psi_\alpha(x)|\frac{\partial w_\alpha}{\partial t}|^2\,dx\,dt
  + 2T\max_{0\leq t\leq t}|\frac{\partial a(x,t)}{\partial t}|^2\\
 &\times \sup_{0\leq t\leq T}\int_0^1\psi_\alpha(x)|
 \frac{\partial w_\alpha}{\partial x}|^2\,dx
 + 2\int_G \psi_\alpha(x)|F_\alpha(x,t)|^2\,dx\,dt\,.
\end{align*} % \label{c27}
Combining this inequality and (\ref{c26}), we obtain (\ref{c14}), with
\begin{equation}
 c_1 = \frac{\frac{48e^{cT}\max(a_0,3)\max(1,T|
\frac{\partial a}{\partial t}|^2)}{\min(a_0,2)} + 3}{\min(a_0, 3)
\max(1, T|\frac{\partial a}{\partial t}|^2)}\,.
\end{equation}
Thus the proof is complete, which also proves Theorem \ref{thm1}.
\end{proof}

\begin{corollary} \label{coro1}
Since  $ 1-x\leq 1-\alpha < 1$ for $\alpha \leq x\leq 1 $ it follows that
$1-x\leq \psi_\alpha(x)\leq 1$ and the inequality \eqref{c14} yields
\begin{align*}
&\int_G (1-x)\big[ |\frac{\partial w_\alpha}{\partial t}|^2
+ |\frac{\partial^2 w_\alpha}{\partial x^2}|^2\big]\,dx\,dt
+\sup_{0\leq t \leq T}\int_0^1[(1-x)|\frac{\partial w_\alpha}{\partial x}|^2
+|w_\alpha |^2]\,dx \\
&\leq  c_1\{\int_0^1|\phi_\alpha'(x)|^2\,dx
 + \int_G |F_\alpha(x,t)|^2\,dx\,dt \}\,.
\end{align*}% \label{c15}
\end{corollary}

\section{Continuous dependence of solutions}

Using a priori estimate (\ref{c7}) for the  difference
$(u_\alpha - u_1)$ of solution $u_\alpha$ of the mixed problem
\eqref{c1}--\eqref{c3} with integral boundary condition,
and the solution $u_1$ of the the mixed problem \eqref{c4}--\eqref{c6}
 with the local condition, we come to the following result.
We remark that this important property has not been
established prior to this work.

\begin{theorem} \label{thm2}
Assume that {\rm (A1)--(A2)} hold. If
\begin{equation}
 \lim_{\alpha \to 1}\int_0^1 |\varphi_1'(x) - \varphi_{\alpha}'(x)|\,dx
= 0, \label{c28}
\end{equation}
then
\begin{equation}
\begin{aligned}
 &\lim_{\alpha \to 1}\Big\{\int_0^1\int_0^T (1-x)
\big[|\frac{\partial u_\alpha}{\partial t}
 - \frac{\partial u_1}{\partial t}|^2
 + |\frac{\partial^2 u_\alpha}{\partial x^2}
 - \frac{\partial u_1}{\partial t}|^2\big]\,dx\,dt  \\
 &+  \sup_{0\leq t\leq T}\int_0^1
 \big[(1-x)|\frac{\partial u_\alpha}{\partial x}
  - \frac{\partial u_1}{\partial t}|^2 + |u_\alpha -u_1|^2\big]\,dx \Big\}
= 0\,.
\end{aligned}\label{c29}
\end{equation}
\end{theorem}

\begin{proof}
Since (A1) and (A2) are satisfied, the  a priori estimate  (\ref{c7})
holds for the difference $u_\alpha - u_1$. Moreover, the equality
$u_1(1,t) = h(t)$ yields
\begin{gather*} %\albel{c30}
\lim_{\alpha\to 1}\big|\frac{1}{1-\alpha}\int_\alpha^1 u_1(x, t)\,dx
 - h(t)\big|
= \lim_{\alpha\to 1}\big|\frac{1}{1 - \alpha}\int_\alpha^1 u_1(x, t)\,dx
  - u_1(1,t)\big|^2 = 0 \\
\lim_{\alpha\to 1}\big|\frac{1}{1-\alpha}\int_\alpha^1
 \frac{\partial u_1(x,t)}{\partial t} - h'(t)\big|^2
= \lim_{\alpha\to 1}\big|\frac{1}{1-\alpha}\int_\alpha^1
 \frac{\partial u_1(x ,t)}{\partial t}\,dx
 - \frac{\partial u_1(1, t)}{\partial t}\big|^2 = 0 \\ %\label{c31}
\lim_{\alpha\to 1}\big|\frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,0)\,dx
 - h(0)|^2
= \lim_{\alpha\to 1}\big|\frac{1}{1-\alpha}\int_\alpha^1 u_1(x ,0)\,dx
- u_1(1,0)|^2 = 0\,. %\label{c32}
\end{gather*}
 From these three equalities, (\ref{c28}),  and (\ref{c7}),
we obtain (\ref{c33}). Thus theorem is
proved.
\end{proof}

To complete our investigation we show that for any
function $\varphi_1\in W_2^1(0,T)$ satisfying the conditions
$\varphi'_1(0) = 0$ and $\varphi_1(1) = h(0)$, there exist
functions $\varphi_\alpha\in W_2^1(0,T)$  such that
\begin{equation}
\varphi_\alpha'(0) = 0, \quad
\frac{1}{1-\alpha}\int_\alpha^1 \varphi_\alpha(x)\,dx = h(0) \label{c33}
\end{equation}
and the equality (\ref{c28}) is valid.
Set
\[
\varphi_\alpha(x) = \varphi_1(x) + \frac{3x^2}{1+\alpha +\alpha^2}
\big( h(0) - \frac{1}{1-\alpha}\int_\alpha^1 \varphi_1(x)\,dx \big)\,.
\]
Then $\varphi_\alpha\in W^1_2(0,T)$, $\varphi_\alpha'(0)
= \varphi_1'(0) = 0$,
\begin{align*}
 \frac{1}{1-\alpha}\int_\alpha^1 \varphi_\alpha(x)\,dx
 &=\frac{1}{1-\alpha}\int_\alpha^1 \varphi_1(x)\,dx\\
 &\quad+ \frac{3}{1-\alpha^3}\int_\alpha^1 x^2\,dx
   \Big( h(0) - \frac{1}{1-\alpha}\int_\alpha^1 \varphi_1(x)\,dx \Big)\\
 &= \frac{1}{1-\alpha}\int_\alpha^1 \varphi_1(x)\,dx
 + \Big( h(0) - \frac{1}{1-\alpha}\int_\alpha^1 \varphi_1(x)\,dx\Big) \\
 &= h(0)
\end{align*}
and
\begin{align*}
\lim_{\alpha\to 1}\int_0^1 |\varphi'_\alpha(x) - \varphi_1'(x)|\,dx
 &= \lim_{\alpha\to 1}\frac{3}{1+\alpha + \alpha^2}\big|h(0)
   - \frac{1}{1-\alpha}\int_\alpha^1 \varphi_1(x) \big| \\
 &= |\varphi_1(1) - h(0)|
  = 0\,,
\end{align*}
\[
 \lim_{\alpha\to 1}\frac{1}{1 - \alpha}\int_{\alpha}^1 |\varphi_1(x)\,dx
= \varphi_1(1) = h(0)\,.
\]
This completes the proof.
\end{proof}

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\end{document}