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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 140, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/140\hfil Initial boundary value problem]
{Initial boundary value problem for a system in elastodynamics
with viscosity}
\author[K. T. Joseph\hfil EJDE-2005/140\hfilneg]
{Kayyunnapara Thomas Joseph }

\address{Kayyunnapara Thomas Joseph \hfill\break
School of Mathematics\\
Tata Institute of Fundamental Research\\
Homi Bhabha Road\\
Mumbai 400005, India}
\email{ktj@math.tifr.res.in}

\date{}
\thanks{Submitted June 10, 2005. Published December 5, 2005.}
\subjclass[2000]{35B40, 35L65}
\keywords{Elastodynamics equation; viscosity; initial
boundary value problem}

\begin{abstract}
 In this paper we prove existence of  global solutions to
 boundary-value problems for two systems with a small
 viscosity coefficient and derive estimates uniform in
 the viscosity parameter. We do not assume any smallness
 conditions on the data.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

In this paper first we consider the boundary-value problem, for a
system of nonlinear ordinary differential equations,
\begin{equation}
\begin{gathered}
 -\xi \frac {du}{d\xi} + u \frac {du}{d\xi} -\frac {d\sigma}{d\xi} =
\epsilon \frac {d^2u}{d\xi^2},  \\
-\xi  \frac {d\sigma}{d \xi} +u \frac {d\sigma}{d \xi} - k^2
\frac {du}{d \xi} = \epsilon
\frac {d^2\sigma}{d \xi^2}
\end{gathered}\label{e1.1}
\end{equation}
for $\xi \in [0,\infty)$ with boundary
conditions
\begin{equation}
\begin{gathered}
u(0) = u_B, u(\infty)=u_R, \\
\sigma(0) = \sigma_B,
\sigma(\infty)=\sigma_R.
\end{gathered}
\label{e1.2}
\end{equation}
Next we consider the  initial boundary
value problem, for a system of parabolic equations in $x>0$ $t>0$,
\begin{equation}
\begin{gathered}
u_t + u u_x - \sigma_x = \epsilon u_{xx}, \\
\sigma_t + u \sigma_x - k^2  u_x = \epsilon \sigma_{xx}
\end{gathered}
\label{e1.3}
\end{equation}
in $\Omega ={(x,t) : x>0, t>0}$, with the initial condition at $t=0$
\begin{equation}
u(x,0) = u_0(x) , \sigma(x,0) = \sigma_0(x) \quad x>0,
\label{e1.4}
\end{equation}
and boundary condition, at $x =0$,
\begin{equation}
u(0,t) = u_B(t) , \sigma(0,t) = \sigma_B(t) \quad t>0.
\label{e1.5}
\end{equation}
In both of these problems, $\epsilon>0$ is a small
parameter. The system of equations \eqref{e1.1} and \eqref{e1.3} are
 approximations of initial boundary value problem for the system of
 equations which comes in elastodynamics:
\begin{equation}
\begin{gathered}
u_t + u u_x - \sigma_x = 0,\\
\sigma_t + u \sigma_x - k^2  u_x = 0,
\end{gathered}
\label{e1.6}
\end{equation}
where $u$
is the velocity, $\sigma$ is the stress and $k>0$ is the speed
of propagation of the elastic waves. This equation has been studied by
many authors \cite{c1,j1,j2,j3} for the case when there is no
boundary. The system \eqref{e1.6} is nonconservative, strictly
hyperbolic system with characteristic speeds
\begin{equation}
\lambda_1(u,\sigma) = u - k,   \lambda_2(u,\sigma) = u + k
\label{e1.7}
\end{equation}
with Riemann invariants
\begin{equation}
r(u,\sigma)=\sigma + k u, s(u,\sigma)= \sigma - k u \label{e1.8}
\end{equation}
respectively.
The problem \eqref{e1.1}-\eqref{e1.2} is the
vanishing self-similar approximations to study the boundary-Riemann
problem for \eqref{e1.6} and the problem \eqref{e1.3}-\eqref{e1.5} is the vanishing
diffusion approximations for \eqref{e1.6} with general initial-boundary data.
 Our aim is to show the existence of
smooth solutions of these problems and derive
estimates in the space of bounded variation, uniformly
in $\epsilon>0$. We do not give any restrictions on the
size of the initial data.

In the study of $(u^\epsilon,\sigma^\epsilon)$ as
$\epsilon$ tends to $0$, there are
two difficulties. The first is the nonconservative product which appear
in the equation \eqref{e1.6}. For the self-similar case this difficulty can be
overcome by the work of LeFloch and Tzavaras \cite{l1} on nonconservative
products. The second is the study of the behaviour of
$(u^\epsilon,\sigma^\epsilon)$ near the boundary $x=0$. Since the
characteristic speeds may change sign, the boundary may be characterestic
at some points. This makes the study of the behaviour of
$(u^\epsilon,\sigma^\epsilon)$ near $x=0$, as $\epsilon$ goes to $0$
difficult. This aspects are under investigation and
will be taken up in a subsequent paper.

\section{ Self-similar vanishing diffusion approximation}

 In this section, we consider the system \eqref{e1.1} and \eqref{e1.2} and prove the
existence of smooth solutions.
Given the data $(u_B,\sigma_B), (u_R,\sigma_R)$, we define
\begin{equation}
r_B = \sigma_B + k u_B, r_R = \sigma_R + k u_R, s_B = \sigma_B - k
u_B, s_R = \sigma_R - k u_R \label{e2.1}
\end{equation}
The characteristic speeds \eqref{e1.7} in terms of the Riemann invariants
 take the form
\[
\lambda_1(r,s) = \frac{r-s}{2k}-k,\quad
\lambda_2(r,s) = \frac{r-s}{2k}+k.
\]
Consider the square
\[
D = [\min(r_{B},r_{R}),\max(r_{B},r_{R})]\times
[\min(s_{B},s_{R}),\max(s_{B},s_{R})],
\]
and consider the minimum and maximum of the eigenvalues on this square
\[
\lambda_j^m = \min_{D}\lambda_j(r,s), \lambda_j^M =
\max_{D}\lambda_j(r,s),\quad j=1,2. \label{e2.2}
\]
We shall prove the following result.

\begin{theorem} \label{thm2.1}
  For each fixed $\epsilon>0$
there exits a smooth solution $(u^\epsilon(\xi),\sigma^\epsilon(\xi))$
for \eqref{e1.1} and \eqref{e1.2} satisfying the estimates
\begin{equation}
|u^\epsilon(\xi)| + |\sigma^\epsilon(\xi)| \leq C,
\int_{0}^{\infty}| \frac {du^\epsilon}{d \xi}|d \xi +
\int_{0}^{\infty}| \frac {d\sigma^\epsilon}{d \xi}(\xi)|d \xi \leq
C, \label{e2.3}
\end{equation}
If $\lambda_1^m >0$, then
\begin{equation}
|u^\epsilon(\xi) - u_B| + |\sigma^\epsilon(\xi) -\sigma_B| \leq
\frac{C}{\delta} e^\frac{-(\xi -\lambda_1^m)^2}{2\epsilon} , \quad
0\leq \xi\leq\lambda_1^m-\delta \label{e2.4)}
\end{equation}
If $\lambda_2^M>0$, then
\begin{equation}
|u^\epsilon(\xi) - u_R| + |\sigma^\epsilon(\xi) -\sigma_R| \leq
\frac{C}{\delta} e^\frac{-(\xi -\lambda_2^M)^2}{2\epsilon} , \quad
\xi\geq\lambda_2^M+\delta, \label{e2.5}
\end{equation}
for some constant $C>0$ independent of $\epsilon>0$ and for
$\delta>0$, small.
\end{theorem}

\begin{proof} To prove the theorem it is easier to work with Riemann
invariants \eqref{e1.8}. The problem \eqref{e1.1} and \eqref{e1.2}
takes the form
\begin{equation}
-\xi \frac {dr}{d \xi} + \lambda_1(r,s) \frac {dr}{d \xi}
 = \epsilon \frac {d^2 r}{d \xi^2}, \quad
 -\xi \frac {ds}{d \xi} + \lambda_2(r,s) \frac {ds}{d \xi} =
\epsilon \frac {d^2 s}{d \xi^2} \label{e2.6}
\end{equation}
on $[0,\infty)$ with boundary conditions
\begin{equation}
r(0) = r_{B} ,\quad  r(\infty) = r_{R} ,\quad s(0) = s_{B},\quad
s(\infty) = s_{R} \label{e2.7}
\end{equation}
where $r_{B}$, $r_{R}$, $s_{B}$ and $s_{R}$ are given by \eqref{e2.1}.

 From the definition \eqref{e1.8} of $r , s$,
$u=\frac{r - s}{2k}, \sigma = \frac{r + s}{2}$. Then to prove
\eqref{e2.3}-\eqref{e2.5}, it is sufficient to prove the
following  estimates
\begin{equation}
\begin{gathered}
r^\epsilon(\xi) \in [\min(r_{B},\quad
r_{R}),\max(r_{B},r_{R})], \quad  \xi \in [0,\infty),\\
s^\epsilon(\xi) \in [\min(s_{B}, \quad
s_{R}),\max(s_{B},s_{R})], \quad \xi \in [0,\infty);
\end{gathered}
\label{e2.8}
\end{equation}
\begin{equation}
\begin{gathered}
|r^\epsilon(\xi) - r_{B}| \leq \frac{C}{\delta} e^\frac{-(\xi
-\lambda_1^m)^2}{2\epsilon} ,\quad \xi \leq \lambda_1^m - \delta,\\
|s^\epsilon(\xi) - s_{B}| \leq \frac{C}{\delta} e^\frac{-(\xi
-\lambda_2^m)^2}{2\epsilon} ,\quad  \xi \leq \lambda_2^m - \delta;
\end{gathered}
\label{e2.9}
\end{equation}
\begin{equation}
\begin{gathered}
|r^\epsilon(\xi) - r_{R}| \leq \frac{C}{\delta} e^\frac{-(\xi
-\lambda_1^M)^2}{2\epsilon} , \quad \xi \geq \lambda_1^M + \delta,\\
|s^\epsilon(\xi) - s_{R}| \leq \frac{C}{\delta} e^\frac{-(\xi
-\lambda_2^M)^2}{2\epsilon} , \quad \xi \geq \lambda_2^M + \delta;
\end{gathered}
\label{e2.10}
\end{equation}
\begin{equation}
\int_{0}^{\infty}| \frac {dr^\epsilon}{d \xi}|d \xi \leq
|r_{R}-r_{B}|,\quad \int_{0}^{\infty}| \frac {ds^\epsilon}{d \xi}|d
\xi \leq |s_{R}-s_{B}|. \label{e2.11}
\end{equation}
To prove these estimates  we reduce \eqref{e2.6} and \eqref{e2.7} to an
integral equation and use some ideas of Tzavaras \cite{t1} and
Joseph and LeFloch \cite{j4}.
Note that \eqref{e2.1} can be
written in the form
\begin{equation}
\begin{gathered}
\frac{d^2r}{d\xi^2} = (\frac{\lambda_1(r,s) - \xi}{\epsilon})
\frac{dr}{d\xi},\\
\frac{d^2s}{d\xi^2} = (\frac{\lambda_2(r,s) - \xi}{\epsilon})
\frac{ds}{d\xi}.
\end{gathered}
\label{e2.12}
\end{equation}
 For $j=1,2$, let
\begin{equation}
g^{\epsilon}{_j}(\xi)= \int_{\alpha_j}^\xi (y - \lambda_j(r,s)(y))
dy \label{e2.13}
\end{equation}
Integrating the equation \eqref{e2.12} once leads to
\begin{equation}
\begin{gathered}
\frac {dr^\epsilon}{d\xi} = (r_{R} - r_{B})\frac{e^\frac{-g_1(\xi)}{\epsilon}}
{\int_{0}^{\infty} e^\frac{-g_1(y)}{\epsilon} dy},\\
\frac {ds^\epsilon}{d\xi} = (s_{R} - s_{B})\frac{e^{-g_2(\xi)}{\epsilon}}
{\int_{0}^{\infty} e^\frac{-g_2(y)}{\epsilon} dy}.
\end{gathered}
\label{e2.14}
\end{equation}
On integrating \eqref{e2.14} using the boundary condition \eqref{e2.7}
we get,
\begin{equation}
\begin{gathered}
r^\epsilon(\xi) = r_{B} + (r_{R} -
r_{B})\frac{\int_{0}^{\xi}e^\frac{-g_1(y)}{\epsilon} dy}
{\int_{0}^{\infty} e^\frac{-g_1(y)}{\epsilon} dy},\\
s^\epsilon(\xi) = s_{B} + (s_{R} -
s_{B})\frac{\int_{0}^{\xi}e^\frac{-g_2(y)}{\epsilon} dy}
{\int_{0}^{\infty} e^\frac{-g_2(y)}{\epsilon} dy}.
\end{gathered}
\label{e2.15}
\end{equation}
It follows that to solve \eqref{e2.6} and \eqref{e2.7} with estimates
 \eqref{e2.8}--\eqref{e2.11},
it is enough to solve \eqref{e2.15}. To solve \eqref{e2.15}, we use the
Schauder fixed point theorem applied to the function
\[
F(r,s)(\xi) = (F_1(r,s)(\xi),F_2(r,s)(\xi))
\]
where
\begin{equation}
\begin{gathered}
F_1(r,s)(\xi) = r_{B} +
(r_{R} - r_{B})\frac{\int_{0}^{\xi}e^\frac{-g_1(y)}{\epsilon} dy}
{\int_{0}^{\infty} e^\frac{-g_1(y)}{\epsilon} dy},\\
F_2(r,s)(\xi) = s_{B} +
(s_{R} - s_{B})\frac{\int_{0}^{\xi}e^\frac{-g_2(y)}{\epsilon} dy}
{\int_{0}^{\infty} e^\frac{-g_2(y)}{\epsilon} dy}
\end{gathered}
\label{e2.16}
\end{equation}
and $g_j$, $j=1,2$ are given by \eqref{e2.13}. From \eqref{e2.16}
 it is clear that
$F_1(r,s)$ is a convex combination of $r_{B}$ and $r_{R}$ and
$F_2(r,s)$ is a convex combination of $s_B$ and $s_R$. So the estimate
\begin{equation}
\begin{gathered}
F_1(r,s)(\xi) \in  [\min(r_{B},r_{R}),\max(r_{B},r_{R})],\\
F_2(r,s)(\xi) \in  [\min(s_{B},s_{R}),\max(s_{B},s_{R})]
\end{gathered}
\label{e2.17}
\end{equation}
easily follows. Next we note that the expression on the right
of \eqref{e2.16} is independent
of the choice of $\alpha_j$ because adding a constant to $g_j$ does not
change the value of the right hand side of \eqref{e2.16}.
Take $\rho_j$ as the
point $\xi$ where minimum of
\[
\min {\int_{\alpha_j}^{\xi}(y-\lambda_j(r,s)(y)) dy}
\]
is achieved. This minimum is achieved because $\lambda_j(r,s)$ is
bounded  by the estimate \eqref{e2.17} and so the term
$\int_{\alpha_j}^{\xi}\lambda_j(r,s)(y) dy$ has at most linear
growth as $\xi \to \infty$ where as the first term is ${\xi^2}/2
- {\alpha_j^2}/2$ has quadratic growth.
Now take $\alpha_j = \rho_j$ in the definition of $g_j$, we have
\begin{equation}
g_j(\xi) \geq 0 , \xi \in [0,\infty). \label{e2.18}
\end{equation}
Suppose $\lambda_j^M>0$, then because of the choice of $\rho_j$,
\begin{align*}
g_j(\xi) &= \int_{\rho_j}^\xi (y-\lambda_j(r,s)(y))dy\\
&\geq \int_{\lambda_j^M}^\xi (y-\lambda_j(r,s)(y)) dy\\
&\geq \int_{\lambda_j^M}^\xi (y-\lambda_j^M) dy\\
&= \frac{(\xi - \lambda_j^M)^2}{2} ,\quad\mbox{if }\xi \geq \lambda_j^M.
\end{align*}
So we have, for $\lambda_j^M>0$,
\begin{equation}
g_j(\xi)\geq \frac{(\xi - \lambda_j^M)^2}{2} , \text{ if } \xi
\geq \lambda_j^M. \label{e2.19}
\end{equation}
Similarly, for $\lambda_j^m>0$, we have
\begin{equation}
g_j(\xi) \geq \frac{(\xi - \lambda_j^m)^2}{2} , if \xi \leq
\lambda_j^m. \label{e2.20}
\end{equation}
Further,
\begin{equation}
\int_{0}^{\infty} e^\frac{-{g_j(\xi)}{\epsilon}} d\xi \geq
\epsilon^{1/2}\int_{0}^{\infty} e^\frac{-{g_j(\rho_j+
{\epsilon}^{1/2}\xi)} {\epsilon}} d\xi \,.\label{e2.21}
\end{equation}
Now
\begin{equation}
\begin{aligned}
g_j(\rho_j + \epsilon^{1/2}\xi) &= \int_{\rho_j}^{\rho_j+\epsilon^{1/2}
\xi} (y-\lambda_j(y)) dy\\
&= \int_0^{{\epsilon}^{1/2}\xi}(y+\rho_j
-\lambda_j(\rho_j + y)) dy\\
&\leq  \epsilon \frac{\xi^2}{2} + (\lambda_j^M
-\lambda_j^m) \epsilon^{1/2}\xi.
\end{aligned}
\label{e2.22}
\end{equation}
 From \eqref{e2.21} and \eqref{e2.22}. we get for $j=1,2$
\begin{equation}
\begin{aligned}
\int_{0}^{\infty} e^\frac{-{g_j(\xi)}{\epsilon}} d\xi & \geq
\epsilon^{1/2} \int_{0}^{\infty} e^{\frac{-y^2}{2} -
(\lambda_j^M-\lambda_j^m)\frac{y}{\epsilon^{1/2}}} dy\\
&= \epsilon \int_{0}^{\infty} e^{\frac{-\epsilon y^2}{2} -
(\lambda_j^M-\lambda_j^m) y} dy\\
&\geq \epsilon   \int_{0}^{\infty} e^{\frac{- y^2}{2} -
(\lambda_j^M-\lambda_j^m) y} dy
\end{aligned}
\label{e2.23}
\end{equation}
 From \eqref{e2.16} and \eqref{e2.23} we get for $j=1,2$
\begin{equation}
|\frac {d F_j(r,s)}{d \xi}(\xi)| \leq \frac{C}{\epsilon}.
\label{e2.24}
\end{equation}
Further, from \eqref{e2.16}, \eqref{e2.19}, \eqref{e2.20} and \eqref{e2.23},
we get: For $\lambda_1^m>0$,
\[
|F_1(r,s)(\xi) - r_{B}| \leq \frac{C}{\epsilon}\int_{0}^\xi
e^\frac{-(s -\lambda_1^m)^2}{2\epsilon} ds = \frac{C \sqrt{2 \epsilon}}
{\epsilon}
\int_\frac{-\lambda_1^m}{\sqrt{2\epsilon}}^\frac{(\xi-\lambda_1^m)}
{\sqrt{2 \epsilon}}
e^{-s^2 } ds, 0\leq \xi \leq \lambda_1^m.
\]
For the case $\lambda_2^m>0$,
\[
|F_2(r,s)(\xi) - s_{B}| \leq \frac{C}{\epsilon}\int_{0}^\xi
e^\frac{-(s -\lambda_2^m)^2}{2\epsilon} ds = \frac{C \sqrt{2 \epsilon}}
{\epsilon}\int_\frac{-\lambda_2^m}{\sqrt{2 \epsilon}}^{
\frac{(\xi-\lambda_2^m)}{\sqrt{2 \epsilon}}}
e^{-s^2} ds, 0\leq\xi\leq\lambda_2^m.
\]
 From \eqref{e2.16}, \eqref{e2.19}, \eqref{e2.20} and \eqref{e2.23},
 we have for the case $\lambda_1^M>0$,
\[
|F_1(r,s)(\xi) - r_{R}| \leq \frac{C}{\epsilon}\int_\xi^{\infty}
e^\frac{-(s -\lambda_k^M)^2}{2\epsilon} ds = \frac{C \sqrt{2 \epsilon}}
{\epsilon}\int_{\frac{(\xi-\lambda_1^M)}{\sqrt{2 \epsilon}}}^{\infty}
e^{-s^2} ds, \xi\geq\lambda_1^M.
\]
For the case $\lambda_2^M>0$
\[
|F_2(r,s)(\xi) - s_{R}| \leq \frac{C}{\epsilon}\int_\xi^{\infty}
e^\frac{-(s -\lambda_k^M)^2}{2\epsilon} ds
= \frac{C \sqrt{2 \epsilon}}
{\epsilon}\int_{\frac{(\xi-\lambda_2^M)}{\sqrt{2 \epsilon}}}^{\infty}
e^{-s^2} ds, \xi>\lambda_2^M
\]
Now using the asymptotic expansion
\[
\int_y^\infty e^{-y^2} dy = (\frac{1}{2y} -O(\frac{1}{y^2}))e^{-y^2}, \quad
y \to \infty
\]
in the above two inequalities, we get
\begin{equation}
\begin{gathered}
|F_1(r,s)(\xi) - r_{B}| \leq \frac{C}{\delta} e^\frac{-(\xi
-\lambda_1^m)^2}{2\epsilon} ,\quad \xi \leq \lambda_1^m - \delta,\\
|F_2(r,s)(\xi) - s_{B}| \leq \frac{C}{\delta} e^\frac{-(\xi
-\lambda_2^m)^2}{2\epsilon} , \quad \xi \leq \lambda_2^m - \delta;
\end{gathered}
\label{e2.25a}
\end{equation}
\begin{equation}
\begin{gathered}
|F_1(r,s)(\xi) - r_{R}| \leq \frac{C}{\delta} e^\frac{-(\xi
-\lambda_1^M)^2}{2\epsilon} , \quad\xi \geq \lambda_1^M + \delta,\\
|F_2(r,s)(\xi) - s_{R}| \leq \frac{C}{\delta} e^\frac{-(\xi
-\lambda_2^M)^2}{2\epsilon} ,\quad  \xi \geq \lambda_2^M + \delta.
\end{gathered}
\label{e2.25b)}
\end{equation}
If $\lambda_j^M<0$, it can be easily seen that
$g_j(\xi)\geq\frac{\xi^2}{2}$
and an analysis similar to the one given earlier gives
\begin{gather}
|F_1(r,s)(x)-r_R|\leq \frac{C}{\delta} e^\frac{-\xi^2}{2\epsilon},
\quad \xi>0 \label{e2.26a}
\\
|F_2(r,s)(x)-s_R|\leq \frac{C}{\delta} e^\frac{-\xi^2}{2\epsilon},
\quad \xi>0 .\label{e2.26b}
\end{gather}
The estimates \eqref{e2.17}, \eqref{e2.24}--\eqref{e2.26b}
show that $F$ is  compact and maps the convex set
\[
[\min(r_{B},r_{R}),\max(r_{B},r_{R})]
\times [\min(s_{B},s_{R}),\max(s_{B},s_{R})]
\]
into itself. So by Schauder fixed point
theorem $F$ has a fixed point and hence \eqref{e2.11} has a solution.
Further it  satisfies the
estimates \eqref{e2.3}-\eqref{e2.5}. The proof of the theorem is complete.
\end{proof}


\section{Vanishing diffusion approximation}

In this section we consider \eqref{e1.3} in the domain
$\Omega_T = [{x>0,0 \leq t \leq T}]$, for $T>0$,
with initial condition \eqref{e1.4} and boundary condition \eqref{e1.5}
and prove the following result.

\begin{theorem} \label{thm3.1}  Assume that
$u^{\epsilon}_0(x), \sigma^{\epsilon}_0(x) \in W^{1,1}(0,\infty)$
and $u^{\epsilon}_B, \sigma^{\epsilon}_B \in W^{1,1}(0,T)$ for every
$T>0$. Further assume that
$(u^{\epsilon}_0(0),\sigma^{\epsilon}_0(0))=(u^{\epsilon}_B(0),
\sigma^{\epsilon}_B(0)$.
Then there exists a classical solution $(u^\epsilon,\sigma^\epsilon)$ of
the problem \eqref{e1.3}--\eqref{e1.5} in $\Omega_T$ with the
following estimates:
\begin{equation}
\begin{gathered}
\|u^\epsilon\|_{L^\infty(\Omega_T)}
\leq \frac{1}{k}
\max\big[\|\sigma^{\epsilon}_0\|_{L^\infty}+k\|u^{\epsilon}_0\|_{L^\infty},
\|\sigma^{\epsilon}_B\|_{L^\infty(0,T)}+k\|u^{\epsilon}_B\|_{L^\infty(0,T)}\big]\\
\|\sigma^\epsilon\|_{L^\infty(\Omega_T)} \leq
\max\big[\|\sigma^{\epsilon}_0\|_{L^\infty}+k\|u^{\epsilon}_0\|_{L^\infty},
\|\sigma^{\epsilon}_B\|_{L^\infty(0,T)}+k\|u^{\epsilon}_B\|_{L^\infty(0,T)}
\big]\\
\end{gathered}
\label{e3.1}
\end{equation}
\begin{equation}
\begin{aligned}
\int_0^\infty(|\partial_x u^\epsilon(x,t)\| dx
&\leq \frac{1}{k}\int_0^\infty(|\partial_x u^{\epsilon}_0(x)|+
k|\partial_x \sigma^{\epsilon}_0(x)|) \, dx\\
&\quad  +\frac{1}{k}\int_0^T(|\partial_t u^{\epsilon}_B(t)|+k
|\partial_t \sigma^{\epsilon}_B(t)|) \, dt , \\
\int_0^\infty|\partial_x \sigma^\epsilon(x,t)|) \, dx
&\leq \int_0^\infty(\partial_x|u^{\epsilon}_0(x)|+k
|\partial_x \sigma^{\epsilon}_0(x)|) \, dx\\
&\quad +\int_0^T(|\partial_t u^{\epsilon}_B(t)|+k
|\partial_t \sigma^{\epsilon}_B(t)|) \,dt . \\
\end{aligned}
\label{e3.2}
\end{equation}
\end{theorem}

We prove this theorem in several steps. Since we are dealing with
the case $\epsilon >0$ fixed in this theorem we suppress the dependence
of $\epsilon$ and write $u,\sigma,r,s$
for$u^\epsilon,\sigma^\epsilon,r^\epsilon, s^\epsilon$.
We rewrite the problem \eqref{e1.1} - \eqref{e1.3} in terms of the Riemann
invariants $(r,s)$ as
\begin{equation}
\begin{gathered}
r_t + (\frac{{s-r}}{{2k}} - k) r_x = \epsilon r_{xx}, \\
s_t + (\frac{{s-r}}{{2k}} + k) s_x = \epsilon s_{xx}. \\
\end{gathered}
\label{e3.3}
\end{equation}
with initial conditions
\begin{equation}
r(x,0) = r_0(x) = \sigma_0(x) + k u_0(x), s(x,0) = s_0(x)
 = \sigma_0(x) - k u_0(x)
\label{e3.4}
\end{equation}
and the boundary conditions
\begin{equation}
r(0,t) = r_B(t) = \sigma_B(t) + k u_B(t), s(0,t) = s_B(t)
 = \sigma_B(t) - k u_B(t).
\label{e3.5}
\end{equation}
First we assume that $r_0$ and $s_0$ are $C^\infty$ functions
on$[0,\infty)$ which are in $W^{1,1}(0,\infty)$ and boundary data
$r_B $ and $s_B$ are $C^\infty$ which are in $W^{1,1}(0,T)$.
The general result then follows from a simple density arguments.
To prove the theorem we define
a sequence of functions $(r_n(x,t),s_n(x,t))$,$n=0,1,2,\dots $ ,
iteratively,
\[
(r_0(x,t),s_0(x,t)) = (r_0(x),s_0(x)), \label{e3.60}
\]
and for $n=1,2,\dots ,(r_n(x,t),s_n(x,t))$ is defined by the
solution of linear problems
\begin{equation}
\begin{gathered}
(r_n)_t + (\frac{s_{n-1}-r_{n-1}}{2k} - k) (r_n)_x =
\epsilon (r_n)_{xx}, \\
(s_n)_t + (\frac{s_{n-1}-r_{n-1}}{2k} + k) (s_n)_x =
 \epsilon (s_n)_{xx}. \\
\end{gathered}
\label{e3.6n}
\end{equation}
with initial conditions
\begin{equation}
r_n(x,0) = r_0(x) ,  s_n(x,0) = s_0(x) \label{e3.7n}
\end{equation}
and the boundary conditions
\begin{equation}
r_n(0,t) = r_B(t) , s_n(0,t) = s_B(t). \label{e3.8n}
\end{equation}

Fix $T>0$, then by linear theory of parabolic equations,
see Friedman \cite{f1},
there exists a unique $C^\infty$
solution $(r_1,s_1)$ to \eqref{e3.6n}--\eqref{e3.8n}. Further,
the solution
decay to $0$ as $x$ tends to $\infty$ and by maximum
principle
\begin{equation}
\begin{gathered}
\|r_1(x,t)\|_{L^\infty(\Omega_T)} =
     \max\big[\|r_0\|_{L^\infty[0,\infty)},\|r_B\|_{L^\infty[0,T]}\big], \\
\|s_1(x,t)\|_{L^\infty(\Omega_T)} =
     \max\big[\|s_0\|_{L^\infty[0,\infty)},\|s_B\|_{L^\infty[0,T]}\big]. \\
\end{gathered}
\label{e3.9}
\end{equation}
Iteratively we get unique solution $(r_n,s_n)$ of the problem
\eqref{e3.6n}--\eqref{e3.8n} in $C^\infty(\Omega_T)$ and
\begin{equation}
\begin{gathered}
\|r_n(x,t)\|_{L^\infty(\Omega_T)} =
        \max\big[\|r_0\|_{L^\infty[0,\infty)},\|r_B\|_{L^\infty[0,T]}\big], \\
\|s_n(x,t)\|_{L^\infty(\Omega_T)} =
        \max\big[\|s_0\|_{L^\infty[0,\infty)},\|s_B\|_{L^\infty[0,T]}\big]. \\
\end{gathered}
\label{e3.10}
\end{equation}
Note that
\begin{equation}
\begin{gathered}
\lambda_{1n}(x,t) = \frac{s_n(x,t) - r_n(x,t)}{2k} - k, \\
\lambda_{2n}(x,t) = \frac{s_n(x,t) - r_n(x,t)}{2k} + k. \\
\end{gathered}
\label{e3.11}
\end{equation}
By \eqref{e3.9} and \eqref{e3.10}, we have there exists a constant
$\lambda \geq 1$ such that
\begin{equation}
\sup_{\Omega_T}|\lambda_{in}(x,t)| \leq \lambda ,\quad \text{ for }
 i=1,2,\; n=0,1,2,\dots
\label{e3.12}
\end{equation}
For future use we write \eqref{e3.6n}--\eqref{e3.8n} in the integral
formulation.
For this we introduce the standard boundary heat kernels
\begin{gather*}
p_\epsilon(x,y,t) = \frac{1}{\sqrt{4\pi t
\epsilon}}[e^\frac{-(x-y)^2} {4 t \epsilon} - e^\frac{-(x+y)^2}{4
t \epsilon}], \label{e3.13}
\\
q_\epsilon(x,t,s) = \frac{-2}{\sqrt
\pi}\partial_s[\int_\frac{x}{{2 \sqrt{\epsilon(t-s)}}}^\infty e^{-
y^2} \,dy]. \label{e3.14}
\end{gather*}
Then \eqref{e3.6n}--\eqref{e3.8n} is equivalent to
\begin{equation}
\begin{aligned}
r_n(x,t) =& \int_0^\infty r_0(y) p_\epsilon(x,y,t) \, dy + \int_0^t r_B(s)
 q_\epsilon(x,t,s) \, ds \\
 &-\int_0^t \int_0^\infty p_\epsilon (x,y,t-s)
\lambda_{1n-1}(y,s)\partial_y r_n(y,s) \, dy \, ds \\
s_n(x,t) =& \int_0^\infty s_0(y) p_\epsilon(x,y,t) \, dy + \int_0^t s_B(s)
 q_\epsilon(x,t,s) \, ds \\
 &- \int_0^t \int_0^\infty p_\epsilon (x,y,t-s)
\lambda_{2,n-1}(y,s)\partial_y s_n(y,s) \, dy \, ds.
\end{aligned}
\label{e3.15}
\end{equation}

With these preliminaries we start the proof of the theorem.
First we show that the map $(r_{n-1},s_{n-1}) \to (r_n,s_n)$
is a contraction in $L^\infty(\Omega_{T_0})$, where $T_0$ is given by
\begin{equation} \label{e3.16}
T_0 = \frac{1}{{9 C_{0}^2}}
\end{equation}
where
\[
C_{0}= \frac{1}{{(\pi \epsilon)^{1/2}}} {[2 \lambda + \frac{1}{2k}
 (\int_0^\infty (|v'_0(x)| + |w'_0(x)|) \, dx
 + \int_0^T (|v'_B(t)| + |w'_B(t)|) \, dt)]}
\]
With this notation we shall prove the following lemma.

\begin{lemma} \label{lem3.2}
(a) Let $T>0$ be fixed. Then for $n=1,2,\dots $ and $0 \leq t \leq T$,
\begin{equation}
\begin{gathered}
\int_0^\infty|\partial_x r_n(x,t)| \, dx \leq \int_0^\infty|r'_0| \, dx +
\int_0^T|r'_B(t)| \, dt , \\
\int_0^\infty|\partial_x s_n(x,t)| \, dx \leq \int_0^\infty|s'_0| \, dx +
\int_0^T|s'_B(t)| \, dt . \\
\end{gathered}
\label{e3.17}
\end{equation}
(b) For $n=2,3,\dots $,
\begin{equation}
\|(v_n-v_{n-1},w_n-w_{n-1})\|_{L^\infty(\Omega_{T_0})} \leq \frac{1}{2}
 \|(v_{n-1}-v_{n-2},w_{n-1}-w_{n-2})\|_{L^\infty(\Omega_{T_0})}
\label{e3.18}
\end{equation}
\end{lemma}

\begin{proof}
First we prove the estimate \eqref{e3.17} for $r_n$, the estimate
for $s_n$ is similar. For a fixed $t>0$, let $y_0(t) = 0$ and
$y_i(t)$, $i=1,2,\dots $ are the points where $\partial_x r_n(x,t)$
changes sign
and let $k=0$ if $\partial_x r_n(x,t) \geq 0$ and $k=1$ if
$\partial_x r_n(x,t) \leq 0$. Following Oleinik \cite{o1}, we write,
\begin{equation}
\int_0^\infty|\partial_x r_n(x,t)| \, dx
= \sum_{i=0}^\infty (-1)^{i+k}
 \int_{y_i(t)}^{y_{i+1}(t)}\partial_x r_n(x,t) \, dx
\label{e3.19}
\end{equation}
Let us take the case $k=0$, the other case is similar. Differentiating
\eqref{e3.19}, we get
\begin{equation}
\frac{d}{dt}\int_0^\infty|\partial_x r_n(x,t)| \, dx = \sum_{i=0}^\infty
 (-1)^i \int_{y_i(t)}^{y_{i+1}(t)} \partial_t(\partial_x r_n(x,t)) \, dx
\label{e3.20}
\end{equation}
where we have used $\frac{d}{dt}(y_0(t)) =0$ and
$\partial_x r_n(y_i(t),t)=0$ if $i=1,2,\dots $.
 Now differentiating the first equation of \eqref{e3.6n}
with respect to $x$, multiplying the resulting equation by $(-1)^i$ and
then integrating
from $y_i(t)$ to $y_{i+1}(t)$, we get for $i=1,2,\dots $
\begin{equation}
\begin{aligned}
&(-1)^i \int_{y_i(t)}^{y_{i+1}(t)} \partial_t[\partial_x r_n](x,t)
\, dx \\
&= \epsilon[ (-1)^i {\partial_x(\partial_x
r_n)}(y_{i+1}(t),t)+ (-1)^{i+1} {\partial_x(\partial_x
r_n)}(y_i(t),t).
\end{aligned} \label{e3.21}
\end{equation}
 For $i=0$,
\begin{equation}
\int_{y_0(t)}^{y_{1}(t)} \partial_t[\partial_x r_n](x,t) \, dx =
\epsilon [{\partial_x(\partial_x v_n)}(y_1(t),t) - {\partial_x(\partial_x
r_n)}(0,t)]
 +(\lambda_{1,n-1} \partial_x r_n)(0,t),
 \label{e3.22}
\end{equation}
where we have used $(\partial_x r_n)(y_i(t),t) = 0$, for $i = 1,2,\dots $.
 From \eqref{e3.6n} and the boundary condition \eqref{e3.8n}, we have
\begin{equation}
\epsilon \partial_{xx} r_n(0,t) -
\lambda_{1,n-1}(0,t)\partial_{x}(0,t) = r'_B(t) \label{e3.23}
\end{equation}
Also in the present case $\partial_x r_n(x,t)$ changes from positive
to negative at $x = y_i(t)$ when $i$ is odd and negative to positive
when $i$ is even and hence $\partial_{xx}v_n(y_i(t),t) \leq 0$ when $i$
is odd and $\partial_{xx}v_n(y_i(t),t)\geq 0$ when $i$ is even. Using
these facts in \eqref{e3.20}--\eqref{e3.23} we get,
\[
\frac{d}{dt}\int_0^\infty |\partial_x r_n(x,t)| \, dx \leq | r'_B(t)|
\]
Integrating this from $0$ to $t$ and using initial
conditions \eqref{e3.7n}, we get,
\[
\int_0^\infty |\partial_x r_n(x,t)| \,dx \leq \int_0^\infty |r'_0(x)| \,
dx + \int_0^t |r'_B(t)| \, dt
\]
Thus for any $T>0$ fixed,we have
\begin{equation}
\int_0^\infty |\partial_x r_n(x,t)| \,dx \leq \int_0^\infty |r'_0(x)| \,
dx
 +\int_0^T |r'_B(s)| \, ds, \quad \text{if } 0 \leq t \leq T
\label{e3.24)}
\end{equation}
The estimate for $s_n$ is similar. To prove the second part we use the
integral representation \eqref{e3.15} to get
\begin{align*}
r_n(x,t)-r_{n-1}(x,t)
=& - \int_0^t \int_0^\infty p_\epsilon(x,y,t-s)
\big[\lambda_{1n-1}(y,s)\partial_y r_n(y,s)\\
&- \lambda_{1,n-2}(y,s)\partial_y
r_{n-1}(y,s)\big] \, dy \, ds
\end{align*}
This can be written as
\begin{equation}
r_n(x,t)-r_{n-1}(x,t) = a_n(x,t) + b_n(x,t) \label{e3.25}
\end{equation}
where
\begin{equation}
a_n(x,t) = -\int_0^t \int_0^\infty
p_\epsilon(x,y,t-s)(\frac{r_{n-1}-s_{n-1}} {2k} - k)
\partial_y(r_n -r_{n-1}) \, dy \, ds \label{e3.26}
\end{equation}
and
\begin{equation}
b_n(x,t) = -\int_0^t \int_0^\infty
p_\epsilon(x,y,t-s)(\frac{(s_{n-1}-s_{n-2})} {2k} -
\frac{(r_{n-1}-r_{n-2})}{2k}) \partial_y r_{n-1} \, dy \, ds
\label{e3.27}
\end{equation}
Integrating by parts and changing variables we get
\begin{equation}
\begin{aligned}
&a_n(x,t) \\
& = \frac{1}{(\pi \epsilon)^{1/2}} \int_0^t \frac{1}
 {(t-s)^{1/2}}
\int_{- \infty}^{x/(4(t-s)\epsilon)^{1/2}} z e^{-z^2}
(\frac{(s_{n-1}-r_{n-1})}{2k} + k)(r_n - r_{n-1}) \, dz \\
& \quad - \frac{1}{(\pi \epsilon)^{1/2}} \int_0^t \frac{1}{(t-s)}^{1/2}
\int_{x/(4(t-s)\epsilon)^{1/2}}^\infty z e^{-z^2}
(\frac{(s_{n-1}-r_{n-1})}{2k} + k)(r_n - r_{n-1}) \, dz \\
& \quad + \int_0^t\int_0^\infty p_\epsilon(x,y,t-s) \partial_y \frac{(s_{n-1}
-r_{n-1})}{2k}
(r_n - r_{n-1}) \, dy \, ds.
\end{aligned}
\label{e3.28)}
\end{equation}
So we get for $0 \leq t \leq t_0 \leq T$,
\begin{equation}
\begin{aligned}
|a_n(x,t)| &\leq \frac{t_0^{1/2}}{(\pi \epsilon)^{1/2}}
\|r_n - r_{n-1}\|_{L^\infty(\Omega_{t_0})}\Big[2 \lambda \\
&\quad + \frac{1}{2k}(
\int_0^\infty (|r'_0(x)|+|s'_0(x)|) \, dx
+ \int_0^T (|r'_B(t)|+|s'_B(t)|)\, dt)\Big].
\end{aligned}
\label{e3.29}
\end{equation}
Similarly, for $0 \leq t \leq t_0 \leq T$,
\begin{equation}
\begin{aligned}
|b_n(x,t)|
&\leq \frac{t_0^{1/2}}{(\pi \epsilon)^{1/2}}
\frac{(\|r_{n-1} - r_{n-2}\|_{L^\infty(\Omega_{t_0})}+
 \|s_{n-1} - s_{n-2}\|_{L^\infty(\Omega_{t_0})})}{2k}\\
&\quad \times\Big[\int_0^\infty |r'_0(x)| \, dx + \int_0^T |r'_B(t)| \, dt\Big]
\end{aligned} \label{e3.30}
\end{equation}
 From \eqref{e3.25}--\eqref{e3.30},
 we get for $0 \leq t \leq t_0 \leq T$,
\begin{equation}
\begin{aligned}
&|r_n(x,t) -r_{n-1}(x,t)|\\
 & \leq \frac{t_0^{1/2}}{ (\pi \epsilon)^{1/2}}\Big[2 \lambda +
\frac{1}{2k}\Big(\int_0^\infty (|r'_0(x)|+|s'_0(x)|) \, dx
+ \int_0^T (|r'_B(t)|+|s'_B(t)|) \, dt\Big)\Big]\\
&\quad \times
{\|r_n - s_{n-1}\|_{L^\infty(\Omega_{t_0})}}
 + \frac{t_0^{1/2}}{(\pi \epsilon)^{1/2}} \frac{1}{2k}
\Big( \int_0^\infty |r'_0(x)| \, dx +
\int_0^T |r'_B(t)| \, dt\Big)  \\
&\quad \times (\|r_{n-1} -
r_{n-2}\|_{L^\infty(\Omega_{t_0})}
+ \|s_{n-1} - s_{n-2}\|_{L^\infty(\Omega_{t_0})}) \\
\end{aligned}
\label{e3.31}
\end{equation}
and
\begin{equation}
\begin{aligned}
|s_n(x,t) - s_{n-1}(x,t)|
 & \leq \frac{t_0^{1/2}}{(\pi \epsilon)^{1/2}} [2 \lambda +
\frac{1}{2k}(\int_0^\infty (|r'_0(x)|+|s'_0(x)|) \, dx\\
&\quad + \int_0^T
(|r'_B(t)|+|s'_B(t)|) \, dt)]\times
 \|s_n - s_{n-1}\|_{L^\infty(\Omega_{t_0})}\\
&\quad  + \frac{(t_0^{1/2}}{(\pi \epsilon)^{1/2}} \frac{1}{2k}
 \Big(\int_0^\infty |s'_0(x)| \, dx +
\int_0^T |s'_B(t)| \, dt\Big)\\
&\quad\times (\|r_{n-1} -
r_{n-2}\|_{L^\infty(\Omega_{t_0})}
+ \|s_{n-1} - s_{n-2}\|_{L^\infty(\Omega_{t_0})}). \\
\end{aligned}
\label{e3.32}
\end{equation}
 From \eqref{e3.31} and \eqref{e3.32}, we get
\begin{equation}
\begin{aligned}
&\|r_n -r_{n-1}\|_{L^\infty(\Omega_{t_0})} + \|s_n -
s_{n-1}\|_{L^\infty(\Omega_{t_0})}\\
&\leq C (t_0)^{1/2} [\|r_n - r_{n-1}\|_{L^\infty(\Omega_{t_0})} +
\|s_n - s_{n-1}\|_{L^\infty(\Omega_{t_0})}] \\
&\quad + C (t_0)^{1/2}[\|r_{n-1} - r_{n-2}\|_{L^\infty(\Omega_{t_0})} +
\|s_{n-1} - s_{n-2}\|_{L^\infty(\Omega_{t_0})}]
\end{aligned}
\label{e3.33}
\end{equation}
where $C_0$ is given by \eqref{e3.16}. Now take
$t_0 = T_0 = \frac{1}{9 C_0^2}$ in \eqref{e3.16} and the estimate
\eqref{e3.18} follows. The proof of Lemma is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3.1}]
First we shall prove that there exists a
continuous function $(r,s)$ such that the sequence $(r_n,s_n)$ converges
uniformly to to $(r,s)$ on $\Omega_T$. Estimate \eqref{e3.18} shows that
$(r_n,s_n)$ converges uniformly to a continuous function
$(r_{T_0},s_{T_0})$ on $\Omega_{T_0}$. Now we consider the region
\[
\Omega_{T_0,2T_0} = [{(x,t):x \geq 0,T_0 \leq t \leq 2T_0}].
\]
Consider problem \eqref{e3.6n} in $\Omega_{T_0,2T_0}$ with initial
data at $T_0$ as $(r_n(x,T_0), s_n(x,T_0))$. Now use the estimates
\eqref{e3.10} and \eqref{e3.17} and using
the same argument to get the estimate \eqref{e3.18} to get
\begin{align*}
&\|(r_n-r_{n-1},s_n - s_{n-1})\|_{L^\infty(\Omega_{T_0,2T_0})}\\
& \leq \frac{1}{2}
 \|(r_{n-1}-r_{n-2},s_{n-1}-s_{n-2})\|_{L^\infty(\Omega_{T_0,2T_0})} \\
&\quad  + \frac{3}{2}\|(r_n(x,T_0)-r_{n-1}(x,T_0),s_n(x,T_0) -
          s_{n-1}(x,T_0)\|_{L^\infty [0,\infty)}.
\end{align*}
Iterating this inequality  leads to
\begin{align*}
&\|(r_n -r_{n-1},s_n - s_{n-1})\|_{L^\infty(\Omega_{T_0,2T_0})}\\
&\leq {(\frac{1}{2})^{(n-2)}}
 \|(r_2-r_1,s_2-s_1)\|_{L^\infty(\Omega_{T_0,2T_0})}\\
&\quad+ 3(n-1){(\frac{1}{2})^{(n-2)}}\|(r_n(x,T_0)-r_{n-1}(x,T_0),s_n(x,T_0 -
      s_{n-1}(x,T_0)\|_{L^\infty [0,\infty)} \\
\end{align*} %3.34
Using the estimate \eqref{e3.10} in the above equation, we get
\begin{equation}
\|(v_n-v_{n-1},w_n-w_{n-1})\|_{L^\infty(\Omega_{T_0,2T_0})}
  \leq C_T.6n (1/2)^{(n-2)}
\label{e3.35}
\end{equation}
where $C_T = \max[\|(r_0,s_0)\|_{L^\infty},\|(r_B,s_B)\|_{L^\infty
[0,T]}]$.
Estimate \eqref{e3.35} shows that $(r_n,s_n)$ is Cauchy sequence in
$\Omega_{T_0,2T_0}$ in the uniform norm and hence converges to a
continuous
function $(r,s)$. Repeating this for a finite number of time intervals
we get
$(r_n,s_n)$ converge uniformly to a continuous function $(r,s)$ in
$\Omega_T$. Now passing to the limit in \eqref{e3.15} we get $(r,s)$
satisfies the integral equation
\begin{align*}
r(x,t) & = \int_0^\infty r_0(y) p_\epsilon(x,y,t) \, dy + \int_0^t r_B(s)
 q_\epsilon(x,t,s) \, ds \\
 &\quad - \int_0^t \int_0^\infty p_\epsilon (x,y,t-s)
\lambda_{1}(r,s)(y,s)\partial_y r(y,s) \, dy \, ds ,\\
s(x,t) & = \int_0^\infty s_0(y) p_\epsilon(x,y,t) \, dy + \int_0^t s_B(s)
 q_\epsilon(x,t,s) \, ds \\
&\quad  - \int_0^t \int_0^\infty p_\epsilon (x,y,t-s)
\lambda_{2}(r,s)(y,s)\partial_y s(y,s) \, dy \, ds.
\end{align*}
 From this integral representation it follows that $(r,s)$ is once
continuously differentiable in $t$ and twice continuously differentiable
in $x$ and solves the problem \eqref{e3.3}--\eqref{e3.4}.
Further  the estimate \eqref{e3.1} and \eqref{e3.2}
 follows from \eqref{e3.10} and \eqref{e3.17}. The proof of the theorem is
complete.
\end{proof}

\subsection*{Acknowledgements}
This work is supported by a grant 2601-2 from the Indo-French
Centre for the promotion of advanced Research, IFCPAR
(Centre Franco-Indien pour la promotion de la Recherche Avancee, CEFIPRA),
New Delhi.

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\end{document}