\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, $$\begin{gathered} -\xi \frac {du}{d\xi} + u \frac {du}{d\xi} -\frac {d\sigma}{d\xi} = \epsilon \frac {d^2u}{d\xi^2}, \\ -\xi \frac {d\sigma}{d \xi} +u \frac {d\sigma}{d \xi} - k^2 \frac {du}{d \xi} = \epsilon \frac {d^2\sigma}{d \xi^2} \end{gathered}\label{e1.1}$$ for $\xi \in [0,\infty)$ with boundary conditions $$\begin{gathered} u(0) = u_B, u(\infty)=u_R, \\ \sigma(0) = \sigma_B, \sigma(\infty)=\sigma_R. \end{gathered} \label{e1.2}$$ Next we consider the initial boundary value problem, for a system of parabolic equations in $x>0$ $t>0$, $$\begin{gathered} u_t + u u_x - \sigma_x = \epsilon u_{xx}, \\ \sigma_t + u \sigma_x - k^2 u_x = \epsilon \sigma_{xx} \end{gathered} \label{e1.3}$$ in $\Omega ={(x,t) : x>0, t>0}$, with the initial condition at $t=0$ $$u(x,0) = u_0(x) , \sigma(x,0) = \sigma_0(x) \quad x>0, \label{e1.4}$$ and boundary condition, at $x =0$, $$u(0,t) = u_B(t) , \sigma(0,t) = \sigma_B(t) \quad t>0. \label{e1.5}$$ 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: $$\begin{gathered} u_t + u u_x - \sigma_x = 0,\\ \sigma_t + u \sigma_x - k^2 u_x = 0, \end{gathered} \label{e1.6}$$ 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 $$\lambda_1(u,\sigma) = u - k, \lambda_2(u,\sigma) = u + k \label{e1.7}$$ with Riemann invariants $$r(u,\sigma)=\sigma + k u, s(u,\sigma)= \sigma - k u \label{e1.8}$$ 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 $$r_B = \sigma_B + k u_B, r_R = \sigma_R + k u_R, s_B = \sigma_B - k u_B, s_R = \sigma_R - k u_R \label{e2.1}$$ 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 $$|u^\epsilon(\xi)| + |\sigma^\epsilon(\xi)| \leq C, \int_{0}^{\infty}| \frac {du^\epsilon}{d \xi}|d \xi + \int_{0}^{\infty}| \frac {d\sigma^\epsilon}{d \xi}(\xi)|d \xi \leq C, \label{e2.3}$$ If $\lambda_1^m >0$, then $$|u^\epsilon(\xi) - u_B| + |\sigma^\epsilon(\xi) -\sigma_B| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_1^m)^2}{2\epsilon} , \quad 0\leq \xi\leq\lambda_1^m-\delta \label{e2.4)}$$ If $\lambda_2^M>0$, then $$|u^\epsilon(\xi) - u_R| + |\sigma^\epsilon(\xi) -\sigma_R| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_2^M)^2}{2\epsilon} , \quad \xi\geq\lambda_2^M+\delta, \label{e2.5}$$ 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 $$-\xi \frac {dr}{d \xi} + \lambda_1(r,s) \frac {dr}{d \xi} = \epsilon \frac {d^2 r}{d \xi^2}, \quad -\xi \frac {ds}{d \xi} + \lambda_2(r,s) \frac {ds}{d \xi} = \epsilon \frac {d^2 s}{d \xi^2} \label{e2.6}$$ on $[0,\infty)$ with boundary conditions $$r(0) = r_{B} ,\quad r(\infty) = r_{R} ,\quad s(0) = s_{B},\quad s(\infty) = s_{R} \label{e2.7}$$ 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 $$\begin{gathered} r^\epsilon(\xi) \in [\min(r_{B},\quad r_{R}),\max(r_{B},r_{R})], \quad \xi \in [0,\infty),\\ s^\epsilon(\xi) \in [\min(s_{B}, \quad s_{R}),\max(s_{B},s_{R})], \quad \xi \in [0,\infty); \end{gathered} \label{e2.8}$$ $$\begin{gathered} |r^\epsilon(\xi) - r_{B}| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_1^m)^2}{2\epsilon} ,\quad \xi \leq \lambda_1^m - \delta,\\ |s^\epsilon(\xi) - s_{B}| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_2^m)^2}{2\epsilon} ,\quad \xi \leq \lambda_2^m - \delta; \end{gathered} \label{e2.9}$$ $$\begin{gathered} |r^\epsilon(\xi) - r_{R}| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_1^M)^2}{2\epsilon} , \quad \xi \geq \lambda_1^M + \delta,\\ |s^\epsilon(\xi) - s_{R}| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_2^M)^2}{2\epsilon} , \quad \xi \geq \lambda_2^M + \delta; \end{gathered} \label{e2.10}$$ $$\int_{0}^{\infty}| \frac {dr^\epsilon}{d \xi}|d \xi \leq |r_{R}-r_{B}|,\quad \int_{0}^{\infty}| \frac {ds^\epsilon}{d \xi}|d \xi \leq |s_{R}-s_{B}|. \label{e2.11}$$ 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 $$\begin{gathered} \frac{d^2r}{d\xi^2} = (\frac{\lambda_1(r,s) - \xi}{\epsilon}) \frac{dr}{d\xi},\\ \frac{d^2s}{d\xi^2} = (\frac{\lambda_2(r,s) - \xi}{\epsilon}) \frac{ds}{d\xi}. \end{gathered} \label{e2.12}$$ For $j=1,2$, let $$g^{\epsilon}{_j}(\xi)= \int_{\alpha_j}^\xi (y - \lambda_j(r,s)(y)) dy \label{e2.13}$$ Integrating the equation \eqref{e2.12} once leads to $$\begin{gathered} \frac {dr^\epsilon}{d\xi} = (r_{R} - r_{B})\frac{e^\frac{-g_1(\xi)}{\epsilon}} {\int_{0}^{\infty} e^\frac{-g_1(y)}{\epsilon} dy},\\ \frac {ds^\epsilon}{d\xi} = (s_{R} - s_{B})\frac{e^{-g_2(\xi)}{\epsilon}} {\int_{0}^{\infty} e^\frac{-g_2(y)}{\epsilon} dy}. \end{gathered} \label{e2.14}$$ On integrating \eqref{e2.14} using the boundary condition \eqref{e2.7} we get, $$\begin{gathered} r^\epsilon(\xi) = r_{B} + (r_{R} - r_{B})\frac{\int_{0}^{\xi}e^\frac{-g_1(y)}{\epsilon} dy} {\int_{0}^{\infty} e^\frac{-g_1(y)}{\epsilon} dy},\\ s^\epsilon(\xi) = s_{B} + (s_{R} - s_{B})\frac{\int_{0}^{\xi}e^\frac{-g_2(y)}{\epsilon} dy} {\int_{0}^{\infty} e^\frac{-g_2(y)}{\epsilon} dy}. \end{gathered} \label{e2.15}$$ 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 $$\begin{gathered} F_1(r,s)(\xi) = r_{B} + (r_{R} - r_{B})\frac{\int_{0}^{\xi}e^\frac{-g_1(y)}{\epsilon} dy} {\int_{0}^{\infty} e^\frac{-g_1(y)}{\epsilon} dy},\\ F_2(r,s)(\xi) = s_{B} + (s_{R} - s_{B})\frac{\int_{0}^{\xi}e^\frac{-g_2(y)}{\epsilon} dy} {\int_{0}^{\infty} e^\frac{-g_2(y)}{\epsilon} dy} \end{gathered} \label{e2.16}$$ 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 $$\begin{gathered} F_1(r,s)(\xi) \in [\min(r_{B},r_{R}),\max(r_{B},r_{R})],\\ F_2(r,s)(\xi) \in [\min(s_{B},s_{R}),\max(s_{B},s_{R})] \end{gathered} \label{e2.17}$$ 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 $$g_j(\xi) \geq 0 , \xi \in [0,\infty). \label{e2.18}$$ 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 $\lambda_j^M>0$, $$g_j(\xi)\geq \frac{(\xi - \lambda_j^M)^2}{2} , \text{ if } \xi \geq \lambda_j^M. \label{e2.19}$$ Similarly, for $\lambda_j^m>0$, we have $$g_j(\xi) \geq \frac{(\xi - \lambda_j^m)^2}{2} , if \xi \leq \lambda_j^m. \label{e2.20}$$ Further, $$\int_{0}^{\infty} e^\frac{-{g_j(\xi)}{\epsilon}} d\xi \geq \epsilon^{1/2}\int_{0}^{\infty} e^\frac{-{g_j(\rho_j+ {\epsilon}^{1/2}\xi)} {\epsilon}} d\xi \,.\label{e2.21}$$ Now \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} From \eqref{e2.21} and \eqref{e2.22}. we get for $j=1,2$ \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} From \eqref{e2.16} and \eqref{e2.23} we get for $j=1,2$ $$|\frac {d F_j(r,s)}{d \xi}(\xi)| \leq \frac{C}{\epsilon}. \label{e2.24}$$ 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 $$\begin{gathered} |F_1(r,s)(\xi) - r_{B}| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_1^m)^2}{2\epsilon} ,\quad \xi \leq \lambda_1^m - \delta,\\ |F_2(r,s)(\xi) - s_{B}| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_2^m)^2}{2\epsilon} , \quad \xi \leq \lambda_2^m - \delta; \end{gathered} \label{e2.25a}$$ $$\begin{gathered} |F_1(r,s)(\xi) - r_{R}| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_1^M)^2}{2\epsilon} , \quad\xi \geq \lambda_1^M + \delta,\\ |F_2(r,s)(\xi) - s_{R}| \leq \frac{C}{\delta} e^\frac{-(\xi -\lambda_2^M)^2}{2\epsilon} ,\quad \xi \geq \lambda_2^M + \delta. \end{gathered} \label{e2.25b)}$$ 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: $$\begin{gathered} \|u^\epsilon\|_{L^\infty(\Omega_T)} \leq \frac{1}{k} \max\big[\|\sigma^{\epsilon}_0\|_{L^\infty}+k\|u^{\epsilon}_0\|_{L^\infty}, \|\sigma^{\epsilon}_B\|_{L^\infty(0,T)}+k\|u^{\epsilon}_B\|_{L^\infty(0,T)}\big]\\ \|\sigma^\epsilon\|_{L^\infty(\Omega_T)} \leq \max\big[\|\sigma^{\epsilon}_0\|_{L^\infty}+k\|u^{\epsilon}_0\|_{L^\infty}, \|\sigma^{\epsilon}_B\|_{L^\infty(0,T)}+k\|u^{\epsilon}_B\|_{L^\infty(0,T)} \big]\\ \end{gathered} \label{e3.1}$$ \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{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 $$\begin{gathered} r_t + (\frac{{s-r}}{{2k}} - k) r_x = \epsilon r_{xx}, \\ s_t + (\frac{{s-r}}{{2k}} + k) s_x = \epsilon s_{xx}. \\ \end{gathered} \label{e3.3}$$ with initial conditions $$r(x,0) = r_0(x) = \sigma_0(x) + k u_0(x), s(x,0) = s_0(x) = \sigma_0(x) - k u_0(x) \label{e3.4}$$ and the boundary conditions $$r(0,t) = r_B(t) = \sigma_B(t) + k u_B(t), s(0,t) = s_B(t) = \sigma_B(t) - k u_B(t). \label{e3.5}$$ 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 $$\begin{gathered} (r_n)_t + (\frac{s_{n-1}-r_{n-1}}{2k} - k) (r_n)_x = \epsilon (r_n)_{xx}, \\ (s_n)_t + (\frac{s_{n-1}-r_{n-1}}{2k} + k) (s_n)_x = \epsilon (s_n)_{xx}. \\ \end{gathered} \label{e3.6n}$$ with initial conditions $$r_n(x,0) = r_0(x) , s_n(x,0) = s_0(x) \label{e3.7n}$$ and the boundary conditions $$r_n(0,t) = r_B(t) , s_n(0,t) = s_B(t). \label{e3.8n}$$ 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 $$\begin{gathered} \|r_1(x,t)\|_{L^\infty(\Omega_T)} = \max\big[\|r_0\|_{L^\infty[0,\infty)},\|r_B\|_{L^\infty[0,T]}\big], \\ \|s_1(x,t)\|_{L^\infty(\Omega_T)} = \max\big[\|s_0\|_{L^\infty[0,\infty)},\|s_B\|_{L^\infty[0,T]}\big]. \\ \end{gathered} \label{e3.9}$$ Iteratively we get unique solution $(r_n,s_n)$ of the problem \eqref{e3.6n}--\eqref{e3.8n} in $C^\infty(\Omega_T)$ and $$\begin{gathered} \|r_n(x,t)\|_{L^\infty(\Omega_T)} = \max\big[\|r_0\|_{L^\infty[0,\infty)},\|r_B\|_{L^\infty[0,T]}\big], \\ \|s_n(x,t)\|_{L^\infty(\Omega_T)} = \max\big[\|s_0\|_{L^\infty[0,\infty)},\|s_B\|_{L^\infty[0,T]}\big]. \\ \end{gathered} \label{e3.10}$$ Note that $$\begin{gathered} \lambda_{1n}(x,t) = \frac{s_n(x,t) - r_n(x,t)}{2k} - k, \\ \lambda_{2n}(x,t) = \frac{s_n(x,t) - r_n(x,t)}{2k} + k. \\ \end{gathered} \label{e3.11}$$ By \eqref{e3.9} and \eqref{e3.10}, we have there exists a constant $\lambda \geq 1$ such that $$\sup_{\Omega_T}|\lambda_{in}(x,t)| \leq \lambda ,\quad \text{ for } i=1,2,\; n=0,1,2,\dots \label{e3.12}$$ 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{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} 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 $$\label{e3.16} T_0 = \frac{1}{{9 C_{0}^2}}$$ 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 $0 \leq t \leq T$, $$\begin{gathered} \int_0^\infty|\partial_x r_n(x,t)| \, dx \leq \int_0^\infty|r'_0| \, dx + \int_0^T|r'_B(t)| \, dt , \\ \int_0^\infty|\partial_x s_n(x,t)| \, dx \leq \int_0^\infty|s'_0| \, dx + \int_0^T|s'_B(t)| \, dt . \\ \end{gathered} \label{e3.17}$$ (b) For $n=2,3,\dots$, $$\|(v_n-v_{n-1},w_n-w_{n-1})\|_{L^\infty(\Omega_{T_0})} \leq \frac{1}{2} \|(v_{n-1}-v_{n-2},w_{n-1}-w_{n-2})\|_{L^\infty(\Omega_{T_0})} \label{e3.18}$$ \end{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, $$\int_0^\infty|\partial_x r_n(x,t)| \, dx = \sum_{i=0}^\infty (-1)^{i+k} \int_{y_i(t)}^{y_{i+1}(t)}\partial_x r_n(x,t) \, dx \label{e3.19}$$ Let us take the case $k=0$, the other case is similar. Differentiating \eqref{e3.19}, we get $$\frac{d}{dt}\int_0^\infty|\partial_x r_n(x,t)| \, dx = \sum_{i=0}^\infty (-1)^i \int_{y_i(t)}^{y_{i+1}(t)} \partial_t(\partial_x r_n(x,t)) \, dx \label{e3.20}$$ 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{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} For $i=0$, $$\int_{y_0(t)}^{y_{1}(t)} \partial_t[\partial_x r_n](x,t) \, dx = \epsilon [{\partial_x(\partial_x v_n)}(y_1(t),t) - {\partial_x(\partial_x r_n)}(0,t)] +(\lambda_{1,n-1} \partial_x r_n)(0,t), \label{e3.22}$$ 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 $$\epsilon \partial_{xx} r_n(0,t) - \lambda_{1,n-1}(0,t)\partial_{x}(0,t) = r'_B(t) \label{e3.23}$$ 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 $$\int_0^\infty |\partial_x r_n(x,t)| \,dx \leq \int_0^\infty |r'_0(x)| \, dx +\int_0^T |r'_B(s)| \, ds, \quad \text{if } 0 \leq t \leq T \label{e3.24)}$$ 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 $$r_n(x,t)-r_{n-1}(x,t) = a_n(x,t) + b_n(x,t) \label{e3.25}$$ where $$a_n(x,t) = -\int_0^t \int_0^\infty p_\epsilon(x,y,t-s)(\frac{r_{n-1}-s_{n-1}} {2k} - k) \partial_y(r_n -r_{n-1}) \, dy \, ds \label{e3.26}$$ and $$b_n(x,t) = -\int_0^t \int_0^\infty p_\epsilon(x,y,t-s)(\frac{(s_{n-1}-s_{n-2})} {2k} - \frac{(r_{n-1}-r_{n-2})}{2k}) \partial_y r_{n-1} \, dy \, ds \label{e3.27}$$ Integrating by parts and changing variables we get \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)} So we get for $0 \leq t \leq t_0 \leq T$, \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} Similarly, for $0 \leq t \leq t_0 \leq T$, \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} From \eqref{e3.25}--\eqref{e3.30}, we get for $0 \leq t \leq t_0 \leq T$, \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} and \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} From \eqref{e3.31} and \eqref{e3.32}, we get \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} 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 $$\|(v_n-v_{n-1},w_n-w_{n-1})\|_{L^\infty(\Omega_{T_0,2T_0})} \leq C_T.6n (1/2)^{(n-2)} \label{e3.35}$$ 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. \begin{thebibliography}{0} \bibitem{c1} J. J. Cauret, J. F. Colombeau and A.-Y. LeRoux, \emph{Discontinous generalized solutions of nonlinear nonconservative hyperbolic equation}, J. Math. Anal. Appl. {\bf139} (1989), 552--573. \bibitem{f1}A . Friedman, \emph{Partial differential equations of parabolic type}, Printice-Hall, Englewood Cliffs, N. J., 1964. \bibitem{j1} K. T. Joseph, \emph{A Riemann problem for a model in Elastodynamics}, {Southeast Asian Bull. Math}, {\bf26}, (2003), 765--771. \bibitem{j2} K. T. Joseph, \emph{Generalized Solutions to a Cauchy Problem for a Nonconservative Hyperbolic System}, J. Math. Anal. Appl. 207 (1997), 361--387. \bibitem{j3} K. T. Joseph and P. L. 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