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%\issuemonth{February}                                             %
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%\issueyear{1999}                                                  %
\shortauthor{Yong Jung Kim}                                       %
\shorttitle{The Riemann Problem in Isentropic Gas Dynamics}       %
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% Macros specific to this paper   %
\overfullrule0pt
\def\supp{\text{supp}}
\def\loc{\text{ loc}}
\def\s{ }
\def\eone{$(2.24)$}
\def\etwo{$(2.25)$}
\def\ethree{$(2.26)$}

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\def\bbox{\vrule height6pt width4pt}
\def\blackbox{\bbox}
\def\eps{\varepsilon}
\def\bdelta{\bar \delta}
\def\bA{\bar A}
\def\bB{\bar B}
\def\bM{\bar M}
\def\grad{\nabla}
\def\rhoe{{\rho_\varepsilon}}
\def\rhoe{{\rho_\varepsilon}}
\def\u{u}
\def\P{P}
\def\U{V}
\def\ue{{\u_\varepsilon}}
\def\Ee{{\Cal E}_\varepsilon}
\def\E{{\Cal E}}
\def\F{{\Cal F}}
\def\T{{\Cal T}}
\def\C{{\Cal C}}
\def\S{{\Cal S}}
\def\equationE{\eqalign{\rho_t+(\rho \u)_x = 0 \cr
        (\rho \u)_t + (\rho \u^2 +p(\rho))_x = 0 \cr}}
\def\Pe{{\Cal P}_\varepsilon}
\def\eP{{\Cal P}}
\def\Pme{{\Cal P}_\varepsilon^\mu}
\define\loner{{L^1(\Bbb R)}}      %
\define\linfr{{L^\infty(\Bbb R)}} %
\define\bvr{{\roman{BV}(\Bbb R)}} %
\define\TV{{\roman {TV}}}         %
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\topmatter
\title
A SELF-SIMILAR VISCOSITY APPROACH FOR THE 
RIEMANN PROBLEM IN ISENTROPIC GAS DYNAMICS
AND THE STRUCTURE OF THE SOLUTIONS
\endtitle
\author
Yong Jung Kim\footnote{Department of Mathematics,
University of Wisconsin-Madison,
Madison, Wisconsin 53706. (jkim\@math.wisc.edu).}
\endauthor
\abstract
We study the Riemann problem for the system of conservation laws of
one dimensional isentropic gas dynamics in Eulerian coordinates.
We construct solutions of the Riemann problem by the method of self-similar
zero-viscosity limits, where the self-similar viscosity only appears
in the equation for the conservation of momentum.
No size restrictions on the data are imposed.
The structure of the obtained solutions is also analyzed.
\endabstract
\keywords
conservation law, weak solution, self-similar viscosity, self-similar solutions,
isentropic gas dynamics, zero-viscosity limit, singularity
\endkeywords
\subjclass
76N10(35L67,35L65)
\endsubjclass
\endtopmatter
\document
\subhead  1. Introduction\endsubhead
We consider the equations describing one dimensional isentropic motions of
inviscid gases,
$$
\eqalign{\rho_t+(\rho \u)_x = 0,\cr
        (\rho \u)_t + (\rho \u^2 +p(\rho))_x = 0,\cr}
\qquad x\in\Bbb R, \quad t>0,
                                                                  \tag 1.1 $$
in Eulerian coordinates. The functions $\rho=\rho(x,t)$, $\u=\u(x,t)$
and $p=p(\rho)$ represent density, velocity and pressure in that order.
The density $\rho$ takes nonnegative values and
the pressure function $p(\rho)$ is smooth and defined for $\rho\ge0$.
We assume that the pressure function $p(\rho)$ satisfies the hypothesis
$$  p'(\rho)>0\quad\text{for}\quad \rho>0.
                                                                     \tag H1 $$
Then (1.1) forms a strictly hyperbolic system with
characteristic speeds $\lambda_\pm(\rho,\u)=\u\pm\sqrt{p'(\rho)}$
for $\rho>0$. We do not assume the strict hyperbolicity for $\rho=0$
since it does not include the usual $\gamma$-laws,                  
$p(\rho)=k\rho^\gamma,\quad \gamma\ge1,\quad k>0$.
We see that it causes significant difficulties in the analysis
when vacuum is considered.
We also adopt a hypothesis
$$p(\rho)\to\infty\quad\text{as}\quad \rho\to\infty\quad \text{and}\quad
  p(\rho)\to 0\quad\text{as}\quad \rho\to 0,
                                                                    \tag H2 $$
which is natural for the pressure functions of gas dynamics.

We are interested in the Riemann problem: finding a weak solution for (1.1)
with initial data
$$(\rho(x,0),\u(x,0))=\left \{
		\eqalign{(\rho_-,\u_-)&,\quad x<0 \cr
			 (\rho_+,\u_+)&,\quad 0<x\cr }\right.       \tag 1.2 $$
with $\rho_\pm>0$.
Since homogeneous conservation laws and Riemann initial data are
invariant under the rescaling $(x,t)\to(\alpha x,\alpha t),\alpha>0$,
it is natural to expect that the solution of the Riemann problem should be
a function of the scaling invariance variable $\xi=x/t$ which is called
the self-similar variable of the Riemann problem. 
A simple computation shows that $(\u,\rho)(x,t)=(\u,\rho)(x/t)$ is a solution
of (1.1),(1.2) if $(\u,\rho)(\xi)$ is a solution of the boundary value problem
$(\eP)$ :
$$ \eqalign{-\xi\rho'+ (\rho &\u)' = 0 \cr
	-\xi(\rho\u)'+(\rho \u^2+ &p(\rho))' = 0 \cr}
                                                                   \tag 1.3 $$
$$\rho(\pm\infty)=\rho_\pm,\quad\u(\pm\infty)=\u_\pm,
                                                                   \tag 1.4 $$
where the ordinary differentiation is with respect to $\xi$.

It is will known that weak solutions are not unique and hence the problem
(1.3),(1.4) should be studied by introducing
an admissibility criterion to single out the physically admissible solution.
We refer to [4], [5] for the solution of the Riemann problem
for general strictly hyperbolic systems and to [2] for a discussion of
the issue of admissibility for hyperbolic systems of conservation laws.

In this article we study the solution of the Riemann problem $(\eP)$
as $\eps\to 0$ limit of the solutions of its perturbed problem $(\Pe)$ :
$$\eqalign {-\xi\rho'&+(\rho\u)'=0 \cr
  -\xi(\rho\u)'+(\rho\u^2&+p(\rho))'=\eps\u'' \cr}
                                                                    \tag 1.5 $$
$$\rho(\pm\infty)=\rho_\pm,\quad\u(\pm\infty)=\u_\pm,
                                                                   \tag 1.4 $$
which depends on the self-similarity of the original problem.
The method of self-similar viscous limits has been studied in [1], [3], [10],
and the problem with full viscosity matrices of $(\eP)$
has been studied in [8]. Here, we are interested on the effect of singular
diffusion matrices. In models of mechanics viscosity appears in the equation
of balance of momentum, but not in the equation of balance of mass.

This kind of approach is followed by Tzavaras in [9] for
the system of one-dimensional isothermal elastic response in Lagrangian
coordinates. Many of the ideas we use here are based on this work.
In both cases the interplay of hyperbolic and parabolic aspects of the problem
must be analyzed.  There are two differences in the present work.
While in [9] the singularity lies at a fixed point $\xi=0$(in Lagrangian
coordinates), the locations of the singularity in (1.5) are non-fixed points
$\xi$ with $u(\xi)=\xi$ and depends on the solution, what leads to solving
a free boundary problem.  We can easily see this if we consider
$$\eqalign{(\u-\xi)\rho'&+ \rho \u' = 0 \cr                        
(\u-\xi)\rho \u'&+ p(\rho)' = 0, \cr}              
                                                                   \tag 1.6 $$ 
$$\eqalign{(\u-\xi)\rho'&+ \rho \u' = 0 \cr                        
(\u-\xi)\rho \u'&+ p(\rho)' = \eps \u'', \cr}              
                                                                   \tag 1.7 $$ 
which are equivalent to (1,3),(1.5) for smooth solutions.
The second difference is that  zero pressure is associated with zero density
in gas dynamics models, what leads to loss of strict hyperbolicity at vacuum.
Therefore, strict hyperbolicity can be lost as $\eps\to 0$
(when vacuum appears), what complicates the analysis considerably.
In fact, a lot of the analytical effort in this work is directed towards
resolving the complications arising in dealing with vacuum.
The objectives of the article are {\rm (i)} to show the existence of
solutions to
the problem $(\Pe)$, {\rm (ii)} to solve the problem $(\eP)$ as the  $\eps\to 0$
limit of the solutions to $(\Pe) $
and {\rm (iii)} to study the structure of the emerging limit.

In Section 2 an analysis of regularity properties of weak solutions of $(\Pe)$
shows that solutions are smooth except for a unique singular point located at
$\u(\xi)=\xi$.  This singular point is induced by the singular diffusion
matrix in (1.7). The uniqueness of the singularity provide monotonicity of
at least one of $\rho,\u$ and unique critical point of the other.

For the existence of solutions of $(\Pe)$ we consider one-parameter family of
boundary value problems $(\Pme)$:
$$ \quad\eqalign{(\u-\xi)\rho'+ \rho \u' &= 0 \cr
	(\u-\xi)\rho \u'+ p(\rho)' &= \eps \u'',\cr}\qquad -\infty<\xi<\infty,
                                                                  \tag 1.8  $$
$$ \eqalign{\rho(\pm\infty)=\rho_\pm^\mu:=&\rho_-+\mu(\rho_\pm-\rho_-)\cr
	\u(\pm\infty)=\u_\pm^\mu:=&\u_-+\mu(\u_\pm-\u_-),\cr}\qquad0\le\mu\le1,
                                                                  \tag 1.9  $$
which connect the solutions of $(\Pe)$ to the trivial solution.
In lemmas 2.4 and 2.5 we establish a-priori estimates that are used in Section 3
to construct solutions of $(\Pe)$. In Lemma 2.4 an a-priori estimate
$$0<\delta_\eps<\rho(\xi)
                                                                     \tag A $$
is missing for the case that $\u$ is increasing on $\Bbb R$, which is the case
vacuum may appear. We have been unable, under the sole Hypotheses (H1) and (H2),
to obtain the estimate for general pressure laws. In Section 5 we establish
the missing estimate for special pressure functions or under a restriction on
the Riemann data which prevent vacuum.

In Section 3 we apply the Leray-Schauder degree theory to a construction
scheme suggested by the a-priori estimates of Section 2. The obtained
result on existence of the viscous problem $(\Pe)$ is stated in Theorem 3.4.

In Section 4 we consider a family of solutions $(\rhoe,\ue)$ to $(\Pe)$
and study the limit $\eps\to 0$.
We show that the total variation of $(\rhoe,\ue)$ is uniformly
bounded, and hence, by virtue of Helly's theorem, along a subsequence 
$\rho_{\eps_n}\to\rho,\u_{\eps_n}\to\u$ for some function $\rho\ge 0$
and $\u$ of bounded variation. In Theorem 4.1 the emerging limit
$(\rho,\u)$ turns out to be a solution of $(\eP)$ and a solution of the Riemann
problem $(\eP)$  is established through the method of self-similar
zero-viscosity limits.

Then we study the structure of $(\rho,\u)$.
For the case of convex pressure laws
$$ p''(\rho)\ge 0\quad\text{for}\quad \rho> 0,
                                                                     \tag H3 $$
the structure of $(\rho,\u)$ is established in Theorem 4.7 in terms of
eigenvalues. The solution consists of two waves separated by
a constant state. The waves are either a rarefaction or a shock.
From this study we get a better understanding of
the structure of the solutions near vacuum which was not clear in Lagrangian
coordinates. As an example to apply the structure, we consider the strictly
hyperbolic case in Corollary 4.8 and show that vacuum is not acceptable in the
case.  Note that Hypothesis (H3) includes both cases of genuine nonlinearity
and linear degeneracy. The usual $\gamma$-laws,
$p(\rho)=k\rho^\gamma,\gamma\ge1,k>0$, satisfy all the hypotheses (H1-3).


In the last section we complete the a-priori estimates for the systems
with convex pressure laws. First, if the system is strictly hyperbolic,
i.e. there exists a constant $c>0$ such that
$$ p'(\rho)\ge c^2,\qquad \rho>0,
                                                                    \tag 1.10 $$
then we obtain a lower bound for $\rho$ and have a complete theory:
\medskip\proclaim{Theorem} (Strictly Hyperbolic Convex Laws)
Suppose $p(\rho)$ satisfies (H1), (H2) and (H3).
If the system (1.1) is strictly hyperbolic,
then the boundary value problem $(\eP)$ has a solution $(\rho,\u)$
which is a $\eps\to0$ of solutions of $(\Pe)$.
The function $(\rho,\u)$ has the structure stated in Theorem 4.7 and 
does not contain vacuum.
\endproclaim\medskip
Second, for convex pressure laws (not necessarily strictly convex), we exhibit
a sufficient condition on the data $(\rho_\pm,\u_\pm)$
that prevents the appearance of vacuum and completes the theory.
The the construction of solutions of the Riemann problem
(1.1) with (1.2) via self-similar viscous limits 
and the structure of the emerging solution of these two cases
are stated in Theorem 5.3.

\subhead 2. Weak Solutions and Regularity Properties\endsubhead
We consider the nonlinear boundary value problem $(\Pe)$ 
$$\eqalign{-\xi\rho'&+ (\rho \u)' = 0 \cr
        -\xi(\rho \u)'+(\rho &\u^2+ p(\rho))' = \eps\u'' \cr}\quad
	-\infty < \xi < \infty
                                                                   \tag 2.1 $$
$$\rho(\pm\infty)=\rho_\pm,\quad\u(\pm\infty)
    =\u_\pm,\qquad\qquad\qquad\quad
                                                                     \tag 2.2 $$
with fixed boundary data $\rho_\pm>0$, $\u_\pm$ and $0<\eps<1$.
In this section we study regularity for solutions of this problem 
and establish a-priori estimates which are used to show the existence
of solutions to the problem $(\Pe)$.

\subhead 2.1 Regularity Properties\endsubhead
First, we give a definition of the solution of $(\Pe)$ in the weak sense.
\definition{Definition 2.1}
A pair of functions $\rho>0$ and $\u$ with
$p(\rho),\rho\in L_{\loc}^\infty(\Bbb R) $
and $\u\in W_{1 \loc}^1(\Bbb R)$ is a solution of $(\Pe)$ if $(\rho,\u) $
satisfies
$$
\int (\zeta -\u)\rho \varphi' d\zeta + \int \rho\varphi d\zeta=0
                                                                 \tag 2.3 $$
$$
\int [(\zeta-\u) \rho\u- p(\rho) + \eps\u']\varphi' d\zeta 
+\int \rho\u\varphi d\zeta=0
                                                                  \tag 2.4 $$ 
for all $\varphi\in C^1_c(\Bbb R)$, continuously differentiable functions
with compact support, and the essential limits $\rho(\pm\infty)$ and
$\u(\pm\infty)$ exist and satisfy (2.2).
\enddefinition
It is clear that the integrals in (2.3) and (2.4) are well defined.
%Since $\rho \in L_{\loc}^\infty(\Bbb R)$ and $\rho\u \in L_{\loc}^1(\Bbb R)$,
%$(\xi-\u)\rho$, $(\zeta-\u) \rho\u- p(\rho) + \eps\u'$
%are in $W^1_{1 \loc}(\Bbb R)$, and hence absolutely continuous.
The equations (2.3) and (2.4) imply that $\rho \in L_{\loc}^\infty(\Bbb R)$
and $\rho\u \in L_{\loc}^1(\Bbb R)$ are the weak derivatives of $(\xi-\u)\rho$
and $(\zeta-\u) \rho\u- p(\rho) + \eps\u'$ respectively.
So $(\xi-\u)\rho$, $(\zeta-\u) \rho\u- p(\rho) + \eps\u'$ are in
$W^1_{1 \loc}(\Bbb R)$, and hence absolutely continuous.
So $-p(\rho) + \eps\u'$ is also continuous.
 
\proclaim{Lemma 2.1} Let $(\rho,\u)$ be a solution of $(\Pe)$.
(i) For $a,b\in\Bbb R$,
$$
\Big [(\xi-\u(\xi))\rho(\xi)\Big ]^b_a - \int^b_a \rho(\zeta)d\zeta = 0
                                                                   \tag 2.5 $$
$$\Big [(\xi-\u(\xi))\rho(\xi)\u(\xi)-p(\rho(\xi))+\eps\u'(\xi)\Big ]^b_a-
   \int^b_a\rho(\zeta)\u(\zeta) d\zeta = 0.
                                                                     \tag 2.6 $$
(ii) $\u,(\xi-\u)\rho$ and $-p(\rho)+\eps\u'$ are continuous on $\Bbb R$.
   If $p\in C^n(\Bbb R^+)$ for $n\ge 0$, then $\rho$ and $\u$ are
$C^{n+1}$ for all $\xi$ such that $\xi\ne\u(\xi)$. 
\endproclaim\demo{Proof}
We already saw that $\u,(\xi-\u)\rho,- p(\rho) + \eps\u'$ and
$(\zeta-\u) \rho\u- p(\rho) + \eps\u'$ are continuous on $\Bbb R$.
Fix $a,b\in\Bbb R$ with $a<b$ and consider
$$
\psi_n(\xi) = \left\{\eqalign{
0, \qquad\qquad\qquad &-\infty < \xi \leq a - 1/n \cr
n(\xi-a)+1,\qquad &a - {1/n} \leq \xi \leq a \cr
1, \qquad\qquad\qquad &a \leq \xi \leq b\cr
-n(\xi-b)+1, \qquad &b \leq \xi \leq b + {1/n} \cr
0, \qquad\qquad\qquad &b + {1/n} \leq \xi < +\infty \; . \cr} \right.
$$
As $\psi_n \not\in C^1_c(\Bbb R)$, it cannot be directly used as a test 
function. However, since $\psi_n$ is Lipchitz continuous, it can be
approximated by $C^1_c$ functions. Let the sequence $\psi^k_n\in C^1_c(\Bbb R) $
converge to $\psi_n$ as $k\to\infty$.
If we put $\psi^k_n$ in the place of $\varphi$ in (2.4), then we get
$$ \int[(\zeta-\u)\rho\u-p(\rho)+\eps\u'](\psi^k_n)'d\zeta
+\int\rho\u\psi^k_n d\zeta =0.$$
Taking the limit $k\to \infty$, we obtain
$$\eqalign {
n\int^a_{a-{1/n}}[(\zeta-\u)\rho\u&-p(\rho)+\eps\u'] d\zeta
-n\int^{b+{1/n}}_b[(\zeta-\u)\rho\u-p(\rho)+\eps\u']d\zeta\cr
&+\int_{a-{1/n}}^{b+{1/n}}\rho\u\psi_n d\zeta =0.\cr}
$$
Since $(\zeta-\u)\rho\u-p(\rho)+\eps\u'$ is continuous and
$|\rho\u\psi_n|\le|\rho\u|\in L^1_{\loc}$, 
Lebesgue Differentiation Theorem and Lebesgue Dominated Convergence Theorem
imply that (2.6) holds in the limit $n\to\infty$.
Similar statements show (2.5).

Let us consider {\rm (ii)}.
Since $(\zeta-\u)\rho$ is continuous, $\rho$ is continuous at $\xi$
if $\xi\ne\u(\xi)$.  From (2.6),
$$ \eqalign {\eps\u'(\xi)=\int^\xi_a\rho(\zeta)\u(\zeta) d\zeta
-(\xi-\u(\xi)) &\rho(\xi) \u(\xi)+p(\rho(\xi))\cr
&+(a-\u(a))\rho(a) \u(a)-p(\rho(a))+\eps\u'(a).\cr}
                                                                     \tag 2.7 $$
If $p$ is $C^0(\Bbb R^+)$, $\u$ is $C^1$ at $\xi\ne\u(\xi)$. From (2.5),
$$ (\xi-\u(\xi))\rho(\xi)=\int^\xi_a\rho(\zeta)d\zeta+(a-\u(a))\rho(a),
                                                                    \tag 2.8 $$
and $\rho$ is $C^1$ at those points.
If $p$ is $C^n(\Bbb R)$,  we can consider (2.7) and (2.8) $n$ more times to get 
$C^{n+1}$ smoothness of $\rho$ and $\u$ at $\xi\ne\u(\xi)$. 
\qed\enddemo\medskip
Lemma 2.1 indicates that singularities may arise at $s$
when $\u(s)=s$, i.e. at the fixed points of $\u$.
Let $s$ be a singular point.
Since $\rho\in L^\infty_{\loc}(\Bbb R)$, (2.8) yields
$$\int^{\xi}_s\rho(\zeta) d\zeta=\rho(\xi)(\xi-\u(\xi)).
                                                                  \tag 2.9 $$
Now we prove the uniqueness of the singularity.
Note that the proof is closely related with the fact that
Definition 2.1 does not accept zero density.

\proclaim{Lemma 2.2}
The singular point of a solution $(\rho,\u)$ of $(\Pe)$ is unique. 
\endproclaim\demo{Proof}
Let $s$ be a singular point and suppose there are no singular points in the
interval $(s,s+\tau)$ for some $\tau>0$ small. From (2.9), we have
$$\xi -\u(\xi)=\int^\xi_s {\rho(\zeta)\over \rho(\xi)}d\zeta>0,
   \qquad \xi\in (s,s+\tau).
                                                                 \tag 2.10 $$
So $\u(\xi)<\xi$ on $(s,s+\tau)$.
Let us suppose there are no singular points in the
interval $(s-\tau,s)$ for some $\tau>0$ small. Then
$$\xi -\u(\xi)=\int^\xi_s {\rho(\zeta)\over \rho(\xi)}d\zeta<0,
   \qquad \xi\in (s-\tau,s).
$$
So $\xi<\u(\xi)$ on $(s-\tau,s)$. From these facts,
it is impossible that the graph of $y=\u(\xi) $
meets the diagonal $y=\xi$ twice. Therefore, either the singular point is
unique or the set of all singular points is a closed interval.

If the set of the singular points is a closed interval $[a,b]$ with $a\ne b$,
then (2.5) implies $\int^d_c \rho(\zeta)d\zeta = 0$ for any $[c,d]\subset[a,b]$,
and thus $\rho$ vanishes on $[a,b]$. This contradicts Definition 2.1.
\qed\enddemo\medskip
\subhead 2.2 Monotonicity Properties of Solutions\endsubhead
Monotonicity of solutions plays a key role in our problem.
It can be easily verified that, at a point of smoothness, the solution
$(\rho,\u)$ of $(\Pe)$ satisfies
$$\eqalign{(\u-\xi)\rho'&+ \rho \u' = 0 \cr
(\u-\xi)\rho \u'&+ p(\rho)' = \eps \u''. \cr}   
                                                                    \tag 2.11 $$

Next we analyze the behavior of $(\rho,\u)$ in a neighborhood of the
singular point $\xi=s$ and $\xi=\pm\infty$. From $(2.11)$ we obtain
$$\rho'={\rho\u'\over (\xi-\u)}               
                                                                    \tag 2.12 $$
$$ \eps\u''+{\{(\xi-\u)^2-p'(\rho)\}\rho\over \xi-\u}\u'=0. 
                                                                    \tag 2.13 $$
(2.13) can be written in a differential form
$$ {d\over d\xi} \Big [ \u'(\xi)\exp\Big \{ {1\over\eps}\int^\xi 
{ \{(\zeta-\u)^2-p'(\rho)\} \rho \over \zeta-\u} d\zeta \Big \} \Big ]=0,
                                                                   \tag 2.14 $$
and upon integrating (2.14) we get
$$ \u'(\xi)= \left\{\eqalign{
  \u'(\alpha_+)\exp\Big \{ -{1\over\eps}\int^\xi_{\alpha_+} {\{(\zeta-\u)^2
  -p'(\rho)\}\rho \over \zeta-\u} d\zeta\Big\},\qquad s<\xi \cr
  \u'(\alpha_-)\exp\Big \{ -{1\over\eps}\int^\xi_{\alpha_-} {\{(\zeta-\u)^2
  -p'(\rho)\}\rho \over \zeta-\u} d\zeta\Big\},\qquad \xi<s \cr} \right.
                                                                   \tag 2.15 $$
for any $\alpha_\pm$ such that $s<\alpha_+$ and $\alpha_-<s$,
where $s$ is the unique singular point.

Since ${\displaystyle \exp\Big \{ -{1\over\eps}\int^\xi_{\alpha_\pm}
{\{(\zeta-\u)^2 -p'(\rho)\}\rho \over \zeta-\u} d\zeta\Big\}}$
is always positive, (2.15) implies that 
either $\u$ is strictly monotone on $(s,\infty)$( or $(-\infty,s)$)
or identically constant.
From (2.12), $\rho$ has the same monotonicity as $\u$ on $(s,\infty) $
and the opposite one on $(-\infty,s)$.

Since $\rho\in L^\infty,\rho>0$ and $\rho$ is monotone on $(-\infty,s)$ and
$(s,\infty)$, the solution $\rho$ satisfies
$$ 0<k\le\rho(\xi)\le K<\infty,\qquad \xi\in\Bbb R,
                                                                   \tag 2.16 $$ 
where $k$ and $K$ depend only on $\rho_\pm$ and
$\rho(s\pm)=$ $\displaystyle{\lim_{\xi\to s\pm}\rho(\xi)}$.
Under Hypothesis (H1),
$p'(\rho)$ is bounded by
$$0< a_0\le p'(\rho(\xi)) \le A_0,\qquad\xi\in\Bbb R,
                                                                   \tag 2.17 $$
where $a_0$ and $A_0$  depend on $k$ and $K$ of (2.16).
\proclaim{Lemma 2.3}
Let $(\rho,\u)$ be a solution of $(\Pe)$ with the singular point $s$.
(i) There exist constants $\alpha_-<s<\alpha_+$, depending on $a_0$,
and $\alpha>0$, depending on $a_0$ and $k$, such that
$$\eqalign{
|\u'(\xi)|\le|\u'(\alpha_+)|\Big |{\xi-s\over\alpha_+-s}\Big
  |^{\alpha\over\eps}, \quad  \qquad &s<\xi<\alpha_+,\cr
|\u'(\xi)|\le|\u'(\alpha_-)|\Big |{\xi-s\over\alpha_- -s}\Big
  |^{\alpha\over\eps}, \quad \qquad &\alpha_-<\xi<s.\cr}
                                                                    \tag 2.18 $$
(ii) There exist constants $\beta_-<s<\beta_+$, depending on $A_0$,
and $\beta>0$, depending on $A_0$ and $k$, such that
$$ \eqalign{
|\u'(\xi)|\le|\u'(\beta_+)|\exp\Big\{-{\beta\over\eps}\Big(\Big({\xi-s\over
 \beta_+-s}\Big)^2-1\Big) \Big\},\quad \qquad &\beta_+<\xi,\cr 
|\u'(\xi)|\le|\u'(\beta_-)|\exp\Big\{-{\beta\over\eps}\Big(\Big({\xi-s\over
 \beta_- -s}\Big)^2-1\Big) \Big\},\quad \qquad &\xi<\beta_-.\cr}
                                                                  \tag 2.19 $$
\endproclaim\demo{Proof}
$\u(\xi)\to\u_+$ as $\xi\to\infty$ and
$\u'(s+)$ is finite from (2.7).
We have $\u(\xi)<\xi$ on $(s,\infty)$. Thus there is a positive constant
$b$ such that $-b(\xi-s)+s<\u(\xi)<\xi$ on $(s,\infty)$ (see Figure 1).
Let $\alpha_+$ be a constant such that
$s<\alpha_+<s+{\theta\over b+1}$ with $\theta=\sqrt{a_0}$.
Then for all $\zeta\in(s,\alpha_+)$,
$$
{\{(\zeta-\u)^2-p'(\rho)\}\rho \over \zeta-\u}\le
\{(1+b)(\alpha_+-s)^2-{\theta^2\over(1+b)}\}{\rho \over \zeta-s}\le
-\alpha{1\over\zeta-s}<0
$$
with
$$\alpha={(\theta^2-(1+b)^2(\alpha_+-s)^2)k\over (1+b)}.
                                                                   \tag 2.20 $$
Then $\alpha$ is positive and, from (2.14),
$$
|\u'(\xi)|\le
|\u'(\alpha_+)|\exp\Big \{ {\alpha\over\eps}\int^\xi_{\alpha_+}
{1\over\zeta-s}d\zeta\Big \} =
|\u'(\alpha_+)|\Big ({\xi-s\over\alpha_+-s}\Big )^{\alpha\over\eps}
$$
for all $\xi \in (s,\alpha_+)$.
The second statement of {\rm (i)} can be proved similarly.
Part {\rm (i)} implies regularity for $u'$ and especially that $\u'(s)=0$.
 
Now we prove {\rm (ii)}.
Fix $\beta_+>s+\max\{2(\u_+-\u_-),2\sqrt{2}\Theta\}$ with $\Theta=\sqrt{A_0}$.
Then, for any $\xi\in(\beta_+,\infty)$, $\xi-\u(\xi)\ge {1\over2}(\xi-s)$ and
$${\{(\zeta-\u)^2-p'(\rho)\}\rho \over \zeta-\u}
  \ge \{{1\over 2}-{2\Theta^2 \over(\zeta-s)^2}\}\rho(\zeta-s)
  \ge {k\over 4}(\zeta-s)> 0.
$$
Set
$$\beta={(\beta_+-s)^2\over2}{k\over 4}.
                                                                   \tag 2.21 $$
Then $\beta$ is positive and
$$
\eqalign {
|\u'(\xi)|
&\le|\u'(\beta_+)|\exp\Big \{ -{2\beta\over\eps(\beta_+-s)^2}
\int^\xi_{\beta_+}\zeta-s d\zeta \Big \} \cr
&=|\u'(\beta_+)|\exp\Big \{-{\beta\over\eps} \Big (({\xi-s\over\beta_+-s})^2-1
\Big )\Big \}. \cr }
$$ 
The proof of the second statement is similar.
\qed\enddemo\medskip
Lemma 2.3 implies that $\u'$ is continuous at the singular point $\xi=s$ and
thus $\u$ is $C^1(\Bbb R)$. Since $-p(\rho)+\eps\u'$ is continuous,
$\rho$ is also continuous (due to (H1)).
In summary, a solution $(\rho,\u)$ of $(\Pe)$ has the regularity
$$\rho\in C(\Bbb R)\cap C^{n+1} (\Bbb R-\{s\})\quad;\quad
  \u\in C^1(\Bbb R)\cap C^{n+1} (\Bbb R-\{s\}$$
for $p\in C^n(\Bbb R^+)$ with $n\ge 1$.

Note that $\alpha$ in Lemma 2.3 depends on the choice of $\alpha_+$.
For example, if we take $\alpha_+=s+{\theta\over \sqrt{2}(b+1)}$, we  get
$\alpha={\theta^2\over 2(1+b)}$.  Since $\u'(s)=0$, we will
get $\alpha\to{p'(\rho(s))\rho(s)}$ if we take $\alpha_+\to s$, 
So $\u$ has $C^2$ regularity if ${p'(\rho(s))\rho(s)}>\eps$.
\subhead 2.3 A-priori Estimates\endsubhead
Throughout this section we consider a solution $(\rho,\u)$ of the family
of boundary value problems $(\Pme)$
$$ \quad\eqalign{(\u-\xi)\rho'+ \rho \u' &= 0 \cr
	(\u-\xi)\rho \u'+ p(\rho)' &= \eps \u'',\cr}\qquad -\infty<\xi<\infty,
                                                                  \tag 2.22 $$
$$ \eqalign{\rho(\pm\infty)=\rho_\pm^\mu:=&\rho_-+\mu(\rho_\pm-\rho_-)\cr
	\u(\pm\infty)=\u_\pm^\mu:=&\u_-+\mu(\u_\pm-\u_-),\cr}\qquad0\le\mu\le1,
                                                                  \tag 2.23 $$
which connect the solutions of $(\Pe)$ to the trivial solution
associated with $\mu=0$.

Since $\rho_\pm>0$, the boundary
values $\rho_\pm^\mu$ in (2.23) are positive and the solutions
of $(\Pme)$ have the regularity derived from the previous sections.
In this section we derive a-priori estimates of solutions
$(\rho,\u)$ of $(\Pme)$, which are used to establish the existence
\noindent
of solutions to $(\Pe)$ in the next section. The a-priori estimates are:
$$\qquad\quad  0<\delta<\rho(\xi)<M,\qquad\xi\in(-\infty,\infty)
                                                                   \tag 2.24 $$
$$\qquad\qquad\qquad   |\u(\xi)|<M, \qquad\xi\in(-\infty,\infty)
                                                                   \tag 2.25 $$
$$ \eqalign{ -b(\xi-s)+s<\u(\xi)<a(\xi-s)+s,\qquad &\xi\in(s,\infty) \cr 
a(\xi-s)+s<\u(\xi)<-b(\xi-s)+s,\qquad &\xi\in(-\infty,s),\cr} 
                                                                   \tag 2.26 $$
\vfil\eject \qquad\quad \special{psfile="../figures/isen1.ps" hscale=80 vscale=80 hoffset=-130 voffset=-580} \vskip 3 truein 

\noindent
where $s$ is the singular point, the positive constants
$\delta$ and $M$ are independent of $\mu$, while the constants $a,b$ satisfy
$0<a<1$ and $0<b$ and depend only on $\delta,M,\rho_\pm$ and $\u_\pm$.

The main difference from the Lagrangian case is the estimate (2.26).
It implies that the graph of the velocity $\u(\xi)$ lies between two
straight lines which have
slopes less than 1 and pass through the point $(s,s)$(see Figure 1).
We can easily check that (2.26) is equivalent to
$$ A|\xi-s|<|u(\xi)-\xi|<B|\xi-s|,\qquad \xi\ne s,
                                                                   \tag 2.27 $$
with $0<A<1<B$.

The proof of the a-priori estimates strongly depends on the shape of
solutions $(\rho,\u)$. From the monotonicity property of the previous
section, we can classify solutions into 4 categories:

\noindent $C_1$: $\rho$ is increasing on $(-\infty,\infty), \; \u$ is
  decreasing on $(-\infty,s)$ and increasing on $(s,\infty)$.

\noindent $C_2$: $\rho$ is decreasing on $(-\infty,\infty), \; \u$ is
  increasing on $(-\infty,s)$ and decreasing on $(s,\infty)$.

\noindent $C_3$: $\rho$ is increasing on $(-\infty,s)$ and decreasing
  on $(s,\infty), \; \u$ is decreasing on $(-\infty,\infty)$.

\noindent $C_4$: $\rho$ is decreasing on $(-\infty,s)$ and increasing 
  on $(s,\infty), \; \u$ is increasing on $(-\infty,\infty)$.

\noindent
Furthermore, we may assume that the monotonicities are all strict.
If not, the solution is constant
on $(-\infty,s)$ or $(s,\infty)$ and the above estimates are trivial.


To establish those a-priori estimates we accept the second hypothesis on the
pressure function:
$$p(\rho)\to\infty\quad\text{as}\quad \rho\to\infty\quad \text{and}\quad
  p(\rho)\to 0\quad\text{as}\quad \rho\to 0.
                                                                      \tag H2 $$
\proclaim{Lemma 2.4}
Let $(\rho,\u)$ be a solution of $(\Pme)$.
If $(\rho,\u)$ belongs to the classes $C_1, C_2$ or $C_3$, there exist
positive constants $M$ and $\delta$ which are independent of $\mu$ and
$\eps$ so that $(\rho,\u)$ satisfies \eone\s and \etwo\.
If $(\rho,\u)$ belongs to the class of $C_4$, there exists a constant $M$
which is independent of $\mu$ and
$\eps$ so that $(\rho,\u)$ satisfies \eone\s and \etwo\.
\endproclaim\demo{Proof}
Let us consider (2.24) first.
If $(\rho,\u)$ is of class $C_1$ or $C_2$, we can take
$\delta=\min\{\rho_-,\rho_+\}$ and $M=\max\{\rho_-,\rho_+\}$ for the
estimations in \eone.
We can also take $\delta= \min\{\rho_-,\rho_+\}$ for the case of $C_3$
and $M=\max\{\rho_-,\rho_+\}$ for the case of $C_4$.
So \eone\s is completed,
if we prove the existence of an upper bound of $\rho$ for class $C_3$.

Let $\xi>s$. Then,
$$ \eqalign {\rho(\xi)&=\rho_+^\mu-\int^\infty_\xi \rho'd\zeta
	\le\rho_+^\mu-{1\over\xi-s}\int^\infty_\xi (\zeta-s)\rho'd\zeta
	=\rho_+^\mu-{1\over\xi-s}\int^\infty_\xi (\rho\u)'-s\rho'd\zeta  \cr
	&=\rho_+^\mu+{1\over\xi-s}(\rho_\xi\u_\xi-\rho_+\u_++s\rho_+
          -s\rho_\xi)
	\le\max\{\rho_-,\rho_+\}+{1\over\xi-s}\rho_+(\u_- -\u_+). \cr } $$

So $\rho(\xi)$ is bounded by a constant which is independent of $\mu$ and
$\eps$ for any fixed $\xi\ne0$.
Let $\tau\in[s+1,s+2]$ satisfy $\u'(\tau)=\u(s+2)-\u(s+1) > \u_+-\u_-$.
If we intergrate $(2.1)_2$ from $\xi>s$ to $\tau$, we get 
$$ \eqalign { \rho(&\xi)\u^2(\xi)
  +p(\rho(\xi))-\eps\u'(\xi) 
    -\rho(\tau)\u^2(\tau)-p(\rho(\tau))+\eps\u'(\tau)
    =-\int^\tau_\xi\zeta(\rho\u)'d\zeta  \cr
  &=\int^\tau_\xi\zeta(\rho(s-\u))'d\zeta -s\int^\tau_\xi\zeta\rho'd\zeta 
    =\Big [\zeta\rho(s-\u) \Big ]_\xi^\tau -\int^\tau_\xi\rho(s-\u)d\zeta
    -s\int^\tau_\xi(\rho\u)'d\zeta \cr
  &=\tau(\rho(\tau)(s-\u(\tau))-\xi(\rho(\xi)(s-\u(\xi)) -\int^\tau_\xi\rho
    (s-\u)d\zeta -s\rho(\tau)\u(\tau) +s\rho(\xi)\u(\xi) \cr
  &\le \tau(\rho(\tau)(s-\u(\tau))-s\rho(\tau)\u(\tau) +s\rho(\xi)\u(\xi)
\cr } $$
If we take the limit $\xi\to s$, then
$$p(\rho(s))\le{\displaystyle\max_{\rho_+<q<\rho(s+1)}}
	\{3q{\bar \u}^2+p(q)\}+(\u_- -\u_+)=:{\Cal A},$$
where $\bar \u=\max\{|\u_-|,|\u_+|\}$. ${\Cal A}$ is independent of $\mu$
and $\eps$ and, from (H2), $p(\rho)\to\infty$ as $\rho\to\infty$.
Hence $\rho(s)$ is bounded by a constant
which depends only on $\rho_\pm,\u_\pm$ and function $p$.

Now we consider (2.25). If the solution $(\rho,\u)$ is of class $C_3$ or
$C_4$, we can take $M=\max\{\u_-,\u_+\}$
for the estimation \etwo.
The proof of \etwo\s for solutions of class $C_1$ and $C_2$ are similar.
We prove (2.25) only for the case $C_2$.
Because of the shape of the solution we know that $|\u(\xi)|\le \max\{
|\u_\pm|,|\u(s)|\}$. So it is enough to prove that the singular point
$s=\u(s)$ is bounded by a constant $M$ which is independent of $\mu$ and $\eps$.
If $s\le 0$, we take $M=\max\{|\u_-|,|\u_+|\}$.
Assume that $s>0$.
Since $(\xi-\u)'=1-\u'>1$ on $(s,\infty)$ and $s-\u(s)=0$,
there is a constant $\alpha\in(s,s+1)$ such that $\alpha-\u(\alpha)=1$. 
Because of the monotonicity of $(\xi-\u)$, we can say
$(\xi-\u)<1$ on $(s,\alpha)$ and $(\xi-\u)>1$ on $(\alpha,\infty)$.
$$ \eqalign { \rho_+\u(\alpha)&=\rho_+\u_+-\int^\infty_\alpha\rho_+\u' d\zeta
	\le\rho_+\u_+-\int^\infty_\alpha\rho\u' d\zeta
	\le\rho_+\u_+-\int^\infty_\alpha(\zeta-\u)\rho\u' d\zeta \cr
	&\le\rho_+\u_+-\int^\infty_s(\zeta-\u)\rho\u' d\zeta \
        =  \rho_+\u_+-\int^\infty_s p(\rho)'d\zeta
           +\int^\infty_s \eps\u'' d\zeta   \cr
	&=  \rho_+\u_+-p(\rho_+)+p(\rho(s))
	\le\rho_+\u_+-p(\rho_+)+p(\rho_-). \cr } $$
(Note that the last inequality is from (H1).)
Hence $\u(\alpha)\le\u_+-{1\over \rho_+}\{p(\rho_+)-p(\rho_-)\}$.
Thus $\u(\alpha)$ is bounded by a constant which is independent of $\eps$ and
$\mu$.
$$  \eqalign { \rho_+\u(s)&=\rho_+\u(\alpha)-\int^\alpha_s\rho_+\u' d\zeta
	\le\rho_+\u(\alpha)-\int^\alpha_s\rho\u' d\zeta  \cr
	&=\rho_+\u(\alpha)-\int^\alpha_s(\zeta-\u)\rho' d\zeta
	\le\rho_+\u(\alpha)-\int^\alpha_s\rho' d\zeta \cr
	&=\rho_+\u(\alpha)-\rho(\alpha)+\rho(s)
	\le\rho_+\u(\alpha)-\{\rho_+-\rho_-\}, \cr }$$
and $\u(s)\le\u(\alpha)+{\rho_-\over\rho_+}-1$.
This means $\u(s)=s$ is bounded by a constant which is independent 
of $\eps$ and $\mu$.
\qed\enddemo\medskip

The a-priori estimates (2.24) and (2.25), established in Lemma 2.4,
are all independent of both of $\mu$ and $\eps$. It is well known that if
the boundary conditions $(\rho_-,\u_-)$ and $(\rho_+,\u_+)$ are not close
enough, the solution of the Riemann problem may have a vacuum state.
Therefore, the lower bound $\delta$ of \eone, for the case of $C_4$,
may depend on $\eps$.
We have been unable, under the sole Hypotheses (H1) and (H2),
to establish the bound: There exists $\delta_\eps>0$
independent of $\mu$ such that
$$0<\delta_\eps<\rho(\xi)
                                                                     \tag A $$
for any solution of $(\rho,\u)$ of Class $C_4$. From here on, in order to
simplify the exposition, we admit (A) as an assumption and present
the rest of the analysis in Section 2 and 3 under Hypothesis (A). In Section
5 we validate (A) under additional hypotheses on the pressure law $p(\rho)$:
for strictly hyperbolic systems with convex laws (cf. Lemma 5.1), or
for convex pressure laws under restrictions on $(\rho_\pm,\u_\pm)$ that exclude
vacuum (cf. Lemma 5.2).

\proclaim{Lemma 2.5}
Under Hypothesis (A), there exist constants $0\le a<1$ and $b\ge0$,
depending on $\rho_\pm,\u_\pm$ and $\delta$ and $M$ in $(2.24)$ and $(2.25)$,
such that $(\rho,\u)$ satisfies $(2.26)$.
\endproclaim\demo{Proof}
The proof of $\ethree_1$ and $\ethree_2$ are similar and we prove
$\ethree_1$ only.
Suppose $\u$ is increasing on $(s,\infty)$. Then we can take $b=0$.
Since $\u(\xi)<\xi,\u'(s)=0$ and $\u(\xi)\to\u_+<\infty$ 
as $\xi\to\infty$, there is a constant $a\in(0,1)$ such that the line
$y=a(\xi-s)+s$ is tangent to the graph of $\u(\xi)$ and
$\u(\xi)\le a(\xi-s)+s$ on $(s,\infty)$.
Now we need to show that the slope
$a$ is bounded above by a constant which is less than 1 and depends only on
$\rho_\pm,\u_\pm,\delta$ and $M$.
Let $\xi_1$ be the tangential point.
If the tangential point $\xi_1$ is far from the fixed point $s$, for example,
$\xi_1-s>2(\u_+-s)$, then $a<{1\over 2}$. Now we assume $\xi_1-s\le2(\u_+-s)$.
From $(2.11)$ we get
$$ \{p'(\rho(\xi_1))-(\u(\xi_1)-\xi_1)^2\}\rho'(\xi_1)=
   \eps\u''(\xi_1).
                                                               \tag 2.28 $$
Since $\rho'(\xi_1)>0$ and $\u''(\xi_1)<0$,
$$ \xi_1-\u(\xi_1)>\sqrt{p'(\rho(\xi_1))}\ge\theta,
									 $$
and 
$$a={\u(\xi_1)-s\over \xi_1-s} <{\xi_1-\theta-s\over\xi_1-s}
  \le {\xi_1-\theta\over\xi_1} \le {2(\u_+-s)-\theta \over 2(\u_+-s)}.
									 $$
So we get $a\le \max\{{1\over2},{2(\u_+-s)-\theta \over 2(\u_+-s)}\}$.

Now suppose $\u$ is decreasing on $(s,\infty)$. Then we can take $a=0$.
Since $\u'(s)=0$ and $\u(\xi)\searrow\u_+>-\infty$ as $\xi\to\infty$,
there exists a positive constant $b$ such that $y=-b(\xi-s)+s$ is tangent
to the graph of $y=\u(\xi)$ and $\u(\xi)\ge -b(\xi-s)+s$ on $(s,\infty)$.
Now we need to show that the slope $-b$ is bounded below by a constant which
depends only on $\rho_\pm,\u_\pm,\delta$ and $M$.
Let $\xi_1$ be the tangential point.
Since $|\u(\xi_1)-s|<|\u_+-s|$, it is enough to show that $\xi_1-s>M_0$ for
some constant $M_0$ which depends only on $\rho_\pm,\u_\pm,\delta$ and $M$.
In that case $b\le{|\u_+-s|\over M_0}$.

Since $\u''(\xi_1)>0$ and $\rho'(\xi_1)<0$, (2.28) yields
$$ \sqrt{a_0}\le \sqrt{p'(\rho(\xi_1))}\le (\xi_1-\u(\xi_1)).$$
Multiplying by $\rho(\xi_1)$, we get
$$\rho(\xi_1)\theta\le\rho(\xi_1)(\xi_1-\u(\xi_1))
   =\int^{\xi_1}_s\rho(\zeta)d\zeta
                                                                 \tag 2.29 $$
$$\delta\theta\le M\int^{\xi_1}_sd\zeta=M(\xi_1-s).
                                                                 \tag 2.30 $$
Hence $\xi_1-s\ge{\delta\theta\over M}$. The proof is complete.
\qed\enddemo\medskip
This Corollary is easily derived from Lemma 2.4 and the monotonicity
of solutions.

\proclaim{Corollary 2.6}
Let $\{(\rhoe,\ue)\}_{\eps>0}$ be a family of
solutions to the boundary-value problem $({\Cal P}_\varepsilon)$ corresponding
to fixed data $(\rho_\pm , \u_\pm)$. Then there exist constants $M$
and $\delta$ depending on the data  such that
$$
0 < \delta \leq \rhoe(\xi) \leq M\quad;\quad |\ue(\xi)| \leq M
                                                                 \tag 2.31 $$
$$
TV_{(-\infty,\infty)}\rhoe\leq M\quad;\quad TV_{(-\infty,\infty)}\ue\leq M,
                                                                   \tag 2.32 $$
where $\delta$ may depend on $\eps$ only if the solution belongs
to the category $C_4$.
\endproclaim

\subhead 3. Existence of Solutions of $(\Pe)$\endsubhead
To construct solutions of $(\Pe)$, we apply  the Leray-Schauder degree theory
(Rabinowitz [6, Ch V]) to a deformation of maps.
Degree theory has been successful to establish connecting trajectories
in problems of self-similar viscous limits (Dafermos [1],
Slemrod and Tzavaras [8] in parabolic problems and Tzavaras [9],
Slemrod [7] in hyperbolic-parabolic problems).
In this work we adapt the method in [9]
which captures the interplay between the hyperbolic and parabolic effects
in system (2.1).

We consider a Banach space $X=\{(\P,\U)\in C^0(\Bbb R)\times
C^1(\Bbb R):\|(\P,\U)\|_X <\infty \}$
which is equipped with the $C^0\times C^1$ norm,
$$\|(\P,\U)\|_X=   {\displaystyle \sup_{-\infty<\xi<\infty}}|\P(\xi)|
		+{\displaystyle \sup_{-\infty<\xi<\infty}}|\U(\xi)|
		+{\displaystyle \sup_{-\infty<\xi<\infty}}|\U'(\xi)|.
$$
To apply the degree theory we chose a bounded open subset of $X$ which is
based on the a-priori estimates established in the previous section.
The choice is proceeded in two steps. The reason is that we don't have any
a-priori estimate for the derivative $\u'$ yet. We establish it in the way
to show the compactness of an operator and see that the image of the operator
is bounded in $X$ even if the domain is not.

Let $Y$ be the set consisting of all $(P,V)\in X$ which are bounded by
$$ 0<\bar\delta<P(\xi)<\bar M,\qquad\xi\in\Bbb R
                                                                   \tag 3.1 $$
$$ |V(\xi)|<\bar M,\qquad\xi\in\Bbb R
                                                                   \tag 3.2 $$
and that satisfy
$$  \bA<1-V'(s)<\bB,
                                                                   \tag 3.3 $$
$$ \bA|\xi-s|<|\xi - V(\xi)|<\bB |\xi-s|,\qquad\xi\ne s
                                                                   \tag 3.4 $$
for some $s\in\Bbb R$.
Here, $\bar\delta,\bar M, \bA$ and $\bB$ are fixed constants which
satisfy $0<\bar\delta<\delta, 0<M<\bar M$ and $0<\bA<A<1<\bB<B$,
where $M,\delta,A$ and $B$ are the constants in the estimates
(2.24),(2.25) and $(2.27)$. Note that the point $s$ in (3.3) and (3.4) can be
different for each $(P,V)\in Y$ and $\bar\delta$ is the only constant which may
depend on $\eps$. If we consider the geometric meaning of (3.4),
{\bf Figure 1}, it is clear that the point $s$ is the fixed point of $V$
and unique for each $V$.
So these fixed points are bounded by (3.2), i.e., $|s|<\bar M$.

We transform the boundary value problem $(\Pme)$ into a fixed point
problem as follows.
For $(\P,\U)\in Y$ fixed we consider the linear problem:
$$ \qquad \eqalign { (\U-\xi)\rho'+\P\u'&=0 \cr
                (\U-\xi)\P\u'+p'(\P)\rho'&=\eps\u'' \cr } 
   \qquad -\infty<\xi<\infty             
                                                                   \tag 3.5 $$
$$ \eqalign { \rho(\pm\infty)=\rho_\pm^\mu&:=\rho_-+\mu(\rho_\pm-\rho_-) \cr
        \u(\pm\infty)=\u_\pm^\mu&:=\u_-+\mu(\u_\pm-\u_-). \cr }
   \hskip 50pt
                                                                   \tag 3.6 $$
This defines an operator $\F:[0,1]\times Y\to X$ that carries
$(\mu,(\P,\U))\in [0,1]\times Y$ to the solution of (3.5) and (3.6).

From the a-priori estimates of Section 2, any solution $(\rho,\u)$ of $(\Pme)$
belongs to $Y$ and, furthermore, is the fixed point of the operator
$\F(\mu,\cdot)$.

In our case there is a difference in the linearized problem (3.5) and (3.6) from
that of the Lagrangian case, [9].
In (3.5) we use two input functions
$\P$ and $\U$ to linearize the equation $(2.11)$. So the function space to
which we apply the degree theory  should be a space of two functions.
Furthermore, to complete the estimates of the operator
$\F$, we should estimate not only (2.24) and (2.25)
but also (2.27). Because of (2.27)
we need more regularity for the function $\U$ which corresponds to the 
solution $\u$. For this reason the space $X$ consists of
$C^0(\Bbb R)\times C^1(\Bbb R)$ functions,
what complicates the topology of the function space and requires
more analysis for the solutions.

To apply the Leray-Schauder degree theory we need a bounded open subset
of the Banach space $X$ which includes all possible solutions of $(\Pme)$.
The subset $Y\subset X$ defined by (3.1--4)
contains all solutions of $(\Pme)$ but is not bounded in
the $C^0\times C^1$ norm.
%Actually we do not have any a-priori estimate for the
%derivative $\u'$ of solutions of $(\Pme)$ yet. So we should find a generic
%constant $K$ so that $|\u'|<K$ for any solution $\u$ of $(\Pme)$.
%It is accomplished in the way to prove the compactness of the operator $\F$.
%After we find the generic constant $K$
We shall apply the degree theory to $\Omega=\{(\P,\U)\in Y:|V'(\xi)|<K\}$ with
a constant $K$ which shall be decided later.
It is clear that $\Omega$ is bounded.
Next we prove that $\Omega$ is open.

\proclaim{Lemma 3.1}
Let $Y\subset X$ be given by (3.1--4).
Then $\Omega=\{(\P,\U)\in Y:|V'(\xi)|<K\}$ is a bounded open subset of
the Banach space $X$ for all $K>0$.
\endproclaim\demo{Proof}
To show that $\Omega$ is open, we fix $(\P_o,\U_o)\in\Omega$ and find
a small positive number $\nu$ such that $\|(\P,\U)-(\P_o,\U_o)\|_X<\nu$
implies $(\P,\U)\in\Omega$.
The inequalities (3.1) and (3.2) for $(\P,\U)$ are easy to verify,
and we just show (3.3) and (3.4).

Since $\U_o(\xi)$ is continuous and bounded,
there exist positive constants $A_o$ and $B_o$ such that 
$$\bA|\xi-s_o|<A_o|\xi-s_o|<|\xi - V_o(\xi)|
<B_o|\xi-s_o|<\bB |\xi-s_o|,
\qquad \xi\ne s_o, $$
where $s_o$ is the fixed point of $\U_o(\xi)$ (i.e. $\U_o(s_o)=s_o$).
Let $\nu_o:=\min\{A_o-\bA,\bB-B_o\}<1$.
Since $A_o\le1-{V_o}'(s_o) \le B_o$, there exists a positive
constant $\kappa<1$ such that
$$
\bA +{\nu_o\over 2}<1-{V_o}'(\xi)<\bB-{\nu_o\over2}
,\qquad s_o-\kappa<\xi<s_o+\kappa.
                                                                    \tag 3.7 $$
If we take $\nu={1\over 2}\nu_o\kappa{\bA\over\bB}$, we can show that (3.3)
and (3.4) are satisfied for any $(\P,\U)$ such that
$\|(\P_o,\U_o)-(\P,\U)\|_X<\nu$.  So $\Omega$ is open.
%
% which is less than
%${1\over 2}\nu_o\kappa$ and ${1\over 2}\nu_o$.
%
%\noindent
%Suppose  $\|(\P_o,\U_o)-(\P,\U)\|_X<\nu$. Since $\U$ is bounded and continuous,
%$\U$ has a fixed point.  Let $\U(s)=s$. Then,
%$$ \nu\ge |\U(s)-\U_o(s)|=|s-\U_o(s)|\ge \bA|s-s_o|$$
%$$|s-s_o|\le{\nu\over\bA}.$$
%So $ |s-s_o|\le{\nu_o\kappa\over 2\bB}$.
%In particular $s\in [s_o-\kappa,s_o+\kappa]$.
%
%\noindent
%Since $1-\U'(\xi)=1-{\U_o}'(\xi) +({\U_o}'(\xi)-\U'(\xi))$ and
%$|{\U_o}'(\xi)-\U'(\xi)|<{\nu_o\over 2}$, $(3.7)$ implies
%$$ \bA<1-\U'(\xi)<\bB\qquad \text{ on } (s_o-\kappa,s_o+\kappa).$$
%Since $\U(s)=s$ and $s\in[s_o-\kappa,s_o+\kappa]$,
%$$ \bA|\xi-s|<|\xi - \U(\xi)|<\bB|\xi-s|\qquad \text{ on } 
%(s_o-\kappa,s_o+\kappa)\setminus \{s\}.$$
%Hence $(3.3)$ is satisfied at least on $(s_o-\kappa,s_o+\kappa)$.
%
%\noindent
%Now let $s_o+\kappa<\xi$ or $\xi<s_o-\kappa$. Then,
%from $\xi-\U(\xi)=\xi-\U_o(\xi)+(\U_o(\xi)-\U(\xi))$,
%$$|\xi-\U(\xi)|<|\xi-\U_o(\xi)|+\nu<B_o|\xi-s_o| + \nu\le\bB|\xi-s|,$$
%$$|\xi-\U(\xi)|>|\xi-\U_o(\xi)|-\nu>A_o|\xi-s_o|-\nu\ge\bA|\xi-s|.$$
%Hence $(\P,\U)$ satisfies $(3.3)$ and thus lies in $\Omega$.
\qed\enddemo\medskip



\subhead 3.1 Estimates of the Operator\endsubhead
In this section we check that the map $\F$ is well defined and establish
uniform estimates of $(\rho,\u)=\F(\mu,(\P,\U))$ and their derivatives.
The derived estimates may depend on $\rho_\pm,\u_\pm$ and
$\eps$ but are independent of the choice of $(\mu,(\P,\U))\in [0,1]\times Y$.
Consider a mapping
$\T$ which carries $(\P,\U)\in Y$ to a solution $\T(\P,\U):=(\rho,\u) $
of (3.5) with boundary conditions
$$ \rho(\pm\infty)=\rho_\pm-\rho_-\quad;\quad \u(\pm\infty)=\u_\pm-\u_-.
                                                                  \tag 3.8 $$
It can be easily verified that $(\rho_-,\u_-)+\mu \T(\P,\U)$ is a solution
of (3.5) and (3.6), and hence
$\F(\mu,(\P,\U))=(\rho_-,\u_-)+\mu \T(\P,\U)$.
%Now we derive the
%estimates of $(\rho,\u)\in \T(Y)$ to complete the estimates of $(\rho,\u)\in
%\F([0,1]\times Y)$ by analyzing the linear problem (3.5) with (3.8).

The bounds in (3.1) and Hypothesis (H1) imply the existence of
positive constants $a_0$ and $A_0$ which satisfy $0<a_0<p'(\P(\xi))<A_0<\infty$
for all $\xi\in\Bbb R$ and depend on $\bar\delta$ and $\bar M$.
From $(3.5)$, we get
$$\eps\u''+{\{p'(\P)-(\U-\xi)^2\}\P\over \U-\xi}\u'=0,\qquad \U(\xi)\ne\xi.
                                                                    \tag 3.9 $$
Since (3.9) has a unique singularity at the fixed point $s$ of $\U$,
$\u'$ is obtained by
$$ \u'(\xi)=\left\{\eqalign{
c_+\exp\Big \{ -{1\over\eps}\int^\xi_{\alpha_+} {\{(\zeta-\U)^2
  -p'(\P)\}\P \over \zeta-\U} d\zeta\Big\}=:c_+I_+,\qquad s<\xi \cr
c_-\exp\Big \{ -{1\over\eps}\int^\xi_{\alpha_-} {\{(\zeta-\U)^2
  -p'(\P)\}\P \over \zeta-\U} d\zeta\Big\}=:c_-I_-,\qquad \xi<s\cr}\right.
                                                                    \tag 3.10 $$
for any $\alpha_-<s<\alpha_+$. In turn, $\rho'$ is obtained by $(3.5)_1$
$$ \rho'(\xi)={\P(\xi)\over\xi-\U}\u'.
                                                                    \tag 3.11 $$

\proclaim{Lemma 3.2}
Let $(\P,\U)\in Y$ and $\U(s)=s$. Then there exist positive constants
$\alpha,\alpha',\beta, \beta'$ and $C_\eps$ which depend only on 
$a_0,A_0,\bar A,\bar B,\bar \delta$ and $\bar M$ ($C_\eps$
may  depend on $\eps$) and satisfy
$$ {1\over C_\eps}|\xi-s|^{\alpha'\over\eps}\le I_\pm(\xi)
   \le C_\eps|\xi-s|^{\alpha\over\eps},\qquad |\xi-s|<1,
                                                                    \tag 3.12 $$
$${1\over C_\eps}e^{-{\beta'\over\eps}(\xi-s)^2}\le I_\pm(\xi)
   \le C_\eps e^{-{\beta\over\eps}(\xi-s)^2},\qquad |\xi-s|>1.
                                                                   \tag 3.13 $$
\endproclaim\demo{Proof}
Let $s<\xi<s+1$. Then
$$\eqalign {&\int^{s+1}_\xi{\{(\zeta-\U)^2-p'(\P)\}\P \over \zeta-\U} d\zeta
  =\int^{s+1}_\xi(\zeta-\U)\P d\zeta-\int^{s+1}_\xi{p'(\P)\P\over \zeta-\U}d\zeta\cr
  &\le\bar M\bB\int^{1}_0\zeta d\zeta -{a_0\bar\delta\over
      \bB}\int^{1}_{\xi-s} {1\over \zeta}d\zeta 
  ={\Cal A}+\alpha\log|\xi-s|,\cr }
$$
where $\alpha:={a_0\bar \delta\over \bB}>0$ and 
${\Cal A}:={\bar M\bB\over 2}$. 
Also
$$
I_+(\xi)=\exp\Big \{{1\over\eps}\int^{s+1}_\xi
       		{\{(\zeta-\U)^2-p'(\P)\}\P \over \zeta-\U} d\zeta\Big \}
\le e^{{\Cal A}\over\eps}e^{{\alpha\over\eps}\log|\xi-s|}=
C_\eps|\xi-s|^{\alpha\over\eps}.
$$
Let $s+1<\xi$, then
$$\eqalign {&-\int^\xi_{s+1}{\{(\zeta-\U)^2-p'(\P)\}\P \over \zeta-\U} d\zeta
 =-\int^\xi_{s+1}(\zeta-\U)\P d\zeta+\int^\xi_{s+1}{p'(\P)\P\over \zeta-\U}
 d\zeta\cr
 &\le-\bar\delta\bA\int^{\xi-s}_{1}\zeta d\zeta
	+{A_0 \bar M\over \bA}\int^{\xi-s}_{1}{1\over\zeta} d\zeta 
  \le -\beta(\xi-s)^2+{\Cal A},\cr}
$$
where $\beta={\bar\delta\bA\over 2}+1$ and ${\Cal A}$ is a positive constant
which depends on $\beta$ and ${A_0 \bar M\over \bA}$. 
Also
$$
I_+(\xi)=\exp\Big \{-{1\over\eps}\int_{s+1}^\xi
		{\{(\zeta-\U)^2-p'(\P)\}\P \over \zeta-\U} d\zeta\Big \}
  \le e^{{\Cal A}\over\eps}e^{-{\beta\over\eps}(\xi-s)^2}
  = C_\eps e^{-{\beta\over\eps}(\xi-s)^2.}
$$
The rest follows by similar arguments.
\qed\enddemo\medskip
By Lemma 3.2, $\u'$ and $\rho'$ are integrable on $(-\infty,\infty)$ and thus
$(\rho,\u)$ can be calculated by the formulas
$$
\rho(\xi) = \left \{\eqalign{(\rho_+-\rho_-) - c_+ \int^\infty_\xi 
{\P(\zeta)I_+(\zeta)\over \zeta-\U(\zeta)}d\zeta,\qquad &s<\xi\cr
c_- \int^\xi_{-\infty} {\P(\zeta)I_-(\zeta)\over \zeta-\U(\zeta)}d\zeta, 
\qquad &\xi < s\cr} \right.
                                                                    \tag 3.14 $$
$$
\u(\xi) = \left \{\eqalign{(\u_+ - \u_-) - c_+ \int^\infty_\xi I_+(\zeta)
d\zeta,\qquad\qquad &s<\xi\cr
c_- \int^\xi_{-\infty} I_-(\zeta)d\zeta,\qquad\qquad &\xi < s \; . \cr} \right.
                                                                   \tag 3.15 $$
Expressing the continuity of $(\rho,\u)$ at $\xi=s$ gives
$$\eqalign {c_+\int^\infty_s{\P(\zeta)I_+(\zeta)\over \zeta-\U(\zeta)}d\zeta
  +c_-\int^s_{-\infty}{\P(\zeta)I_-(\zeta)\over \zeta-\U(\zeta)}d\zeta
  &=\rho_+-\rho_-\cr
  c_+\quad\int^\infty_s I_+(\zeta)d\zeta\quad\hskip 4pt
  +c_-\quad\int^s_{-\infty}I_-(\zeta)d\zeta
\quad\hskip 4pt  &=\u_+-\u_-.\cr}
                                                                   \tag 3.16 $$
The determinant $\Delta$ of the linear system (3.16) is
$$\int^\infty_s{\P(\zeta)I_+(\zeta)\over \zeta-\U(\zeta)}d\zeta
  \int^s_{-\infty}I_-(\zeta)d\zeta
 -\int^s_{-\infty}{\P(\zeta)I_-(\zeta)\over \zeta-\U(\zeta)}d\zeta
  \int^\infty_s I_+(\zeta)d\zeta>0.
                                                                    \tag 3.17 $$
So there exists a unique solution $(c_+,c_-)$ to (3.16) and the operators
$\T$ and $\F$ are well defined.

We now estimate $(\rho,\u)=\T(\P,\U)$, defined by (3.14) and (3.15).
Since our objective in this section is to get uniform bounds
which are independent of the choice of $(\P,\U)\in Y$, we consider a generic
constant $K_\eps$ which may depend on $a_0,A_0,\bar A,\bar B,\bar \delta$,
$\bar M$ and $\eps$ but does not depend on $(\P,\U)\in Y$.

Considering the lower bounds for $I_\pm$ in Lemma 3.2,
the determinant $\Delta$ of the linear system (3.16) is
bounded from below by a positive constant which depends
on $a_0,A_0,\bar A,\bar B,\bar \delta$ and $\bar M$.
So we get
$$ |c_+|+|c_-|<K_\eps.
                                                                   \tag 3.18 $$
Now we estimate $\rho,\u$ and their derivatives to see the regularity properties
of the operator. Since $\u',\rho'$ are given by (3.10) and (3.11) and $I_\pm$
are bounded by (3.12) and (3.13), we have
$$|\u'(\xi)|<K_\eps|\xi-s|^{\alpha\over\eps},\qquad |\xi-s|<1,  
                                                                   \tag 3.19 $$
$$ |\u'(\xi)|<K_\eps e^{-{\beta\over\eps}(\xi-s)^2},\qquad |\xi-s|>1,
                                                                   \tag 3.20 $$
$$  |\rho'(\xi)|={|\P(\xi)|\over|\xi-\U(\xi)|}|\u'(\xi)|
   <K_\eps|\xi-s|^{{\alpha\over\eps}-1},\qquad |\xi-s|<1,
                                                                   \tag 3.21 $$
$$  |\rho'(\xi)|={|\P(\xi)|\over|\xi-\U(\xi)|}|\u'(\xi)|
   <K_\eps e^{-{\beta\over\eps}(\xi-s)^2},\qquad |\xi-s|>1.
                                                                   \tag 3.22 $$
We also have
$$
\u''(\xi)={1\over\eps}{\{p'(\P)-(\xi-\U)^2\}\P\over\xi-\U}c_\pm I_\pm
    ,\qquad \xi\ne s,
                                                                   \tag 3.23 $$
and
$$
|\u''(\xi)| \le{1\over\eps}\Big ({A_0\over(1-a)|\xi-s|}+(1+b)|\xi-s|\Big)
\bar M c_\pm I_\pm \le K_\eps |\xi-s|^{{\alpha\over\eps}-1}, \qquad |\xi-s|<1,
                                                                   \tag 3.24 $$
$$
|\u''(\xi)|\le{1\over\eps}\Big ({A_0\over(1-a)|\xi-s|}+(1+b)|\xi-s|\Big )
  \bar Mc_\pm I_\pm
\le K_\eps e^{-{\beta\over\eps} (\xi-s)^2} ,\qquad |\xi-s|>1.
                                                                   \tag 3.25 $$
From these estimates we get equicontinuity of $\rho,\u$ and $\u'$ on any closed
set which does not contain the singular point $s$.
(3.19) and (3.20) imply
$$ |\u'(\xi)|< K_\eps,\qquad -\infty<\xi<\infty,
                                                                   \tag 3.26 $$
and hence $\u$ is equicontinuous.
From (3.12) and (3.13), we get
$$\int^\xi_s I_+(\zeta)d\zeta < K_\eps(\xi-s)^{{\alpha\over\eps}+1};\quad
\int^\xi_s {\P I_+\over|\zeta-\U|}d\zeta<K_\eps (\xi-s)^{\alpha\over\eps},\qquad
                                                                    s<\xi<s+1,$$
$$\int^s_\xi I_-(\zeta)d\zeta < K_\eps(s-\xi)^{{\alpha\over\eps}+1};\quad
\int^s_\xi {\P I_-\over|\zeta-\U|}d\zeta<K_\eps (s-\xi)^{\alpha\over\eps},\qquad
                                                                    s-1<\xi<s,$$
$$\int^\xi_{s+1}I_+(\zeta)d\zeta < K_\eps e^{-{\beta\over\eps}(\xi-s)^2};\quad
\int^\xi_{s+1}{\P I_+\over|\zeta-\U|}d\zeta<K_\eps e^{-{\beta\over\eps}(\xi-s)^2},\qquad
                                                                      s+1<\xi,$$
$$\int_\xi^{s-1}I_-(\zeta)d\zeta < K_\eps e^{-{\beta\over\eps}(\xi-s)^2};\quad
\int_\xi^{s-1}{\P I_-\over|\zeta-\U|}d\zeta<K_\eps e^{-{\beta\over\eps}(\xi-s)^2},\qquad
                                                                      s+1<\xi.$$
These estimates imply the boundedness of $\rho$ and $\u$
$$|\u(\xi)|<K_\eps;\qquad 
|\rho(\xi)|<K_\eps,\quad\qquad -\infty<\xi<\infty,
                                                                   \tag 3.27 $$
and establish estimates for $\rho(\xi)$ and $\u(\xi)$ from (3.14) and (3.15);
$$ |\u(\xi)-\u(s)|<K_\eps|\xi-s|^{{\alpha\over\eps}+1},\qquad |\xi-s|<1,
                                                                   \tag 3.28 $$
$$ |\u(\xi)-\u(s)|<K_\eps e^{-{\beta\over\eps}(\xi-s)^2},\qquad |\xi-s|>1,
                                                                   \tag 3.29 $$
$$ |\rho(\xi)-\rho(s)|<K_\eps|\xi-s|^{\alpha\over\eps},\quad\qquad |\xi-s|<1,
                                                                   \tag 3.30 $$
$$ |\rho(\xi)-\rho(s)|<K_\eps e^{-{\beta\over\eps}(\xi-s)^2},\qquad |\xi-s|>1.
                                                                   \tag 3.31 $$

The estimates (3.26) and (3.27) imply that the image $\T(Y)$ of $Y$
under the mapping $\T$ is bounded under the $C^0\times C^1$ norm.

\subhead 3.2 Existence of Solutions of $(\Pe)$\endsubhead
If $(\rho,\u)$ is a solution of $(\Pme)$, then $(\rho,\u)$ is a fixed point
under $\F(\mu,\cdot)$. So $(\rho,\u)$ is a image under the mapping
$\F(\mu,\cdot)$ and the previous estimate (3.26) can be considered as an
a-priori estimate of solutions of $(\Pme)$.
Now we fix $K$ of Lemma 3.1 with ${\bar K}:= K_\eps+1$ and consider
$$\Omega=\{(\P,\U)\in Y:|V'(\xi)|<{\bar K}\}.
                                                                   \tag 3.32 $$

\proclaim{Lemma 3.3}
The mapping $\T:\bar \Omega\to X$ is a compact operator.
\endproclaim\demo{Proof}
First, we show that $\T(\bar\Omega)$ is precompact in $X$.
Let $(\rho_n,\u_n)$ be a sequence in $\T(\bar\Omega)$. 
Since $\u_n'$ is uniformly bounded by (3.26), $\u_n$ is equicontinuous.
The equicontinuity of $\rho_n$ follows from (3.21), (3.22) and (3.30)
and the one of $\u_n'$ from (3.19),(3.24) and (3.25).

For example we consider $\rho_n$. Let $\eta>0$ be given. 
From (3.30) there exists $\delta_1>0$ such that $|\rho_n(\xi_1)-\rho_n(\xi_2)|
<\eta$ for all $\xi_1,\xi_2\in I= [-\delta_1,\delta_1]$. From (3.21) and (3.22)
$\rho_n'$ is uniformly bounded on $I^c=(-\infty,-\delta_1)
\cup (\delta_1,\infty)$ and there exists $\delta_2>0$ such that
$|\rho_n(\xi_1)-\rho_n(\xi_2)|<\eta$ for all $\xi_1,\xi_2\in I^c$
with $|\xi_1-\xi_2|<\delta_2$. If we take $\delta=\min(\delta_1,\delta_2)$,
we see that $|\rho_n(\xi_1)-\rho_n(\xi_2)|<2\eta$ for $|\xi_1-\xi_2|<\delta$.
So $\rho_n$ is equicontinuous.

From (2.25) and (2.26) $\rho_n,\u_n,\u_n'$ are also uniformly bounded and
the Ascoli-Arzela theorem implies the existence of a subsequence, rename it
$\rho_n, \u_n$, which converges uniformly on every compact set.
By taking a subsequence again, we
can assume that the singular points $s_n$ converge to $s$.

Let $\rho,\u$ and $\u_1$ be the limit of $\rho_n,\u_n$ and ${\u_n}'$.
We can easily verify that $\u'=\u_1$ and $(\rho,\u)\in X$.
Since the singular points $s_n$ are bounded by $\bar M$,
we can choose $L>0$ which satisfies
$$\eqalign {
|{\u_n}'(\xi)|<\eta    ,   \hskip 100pt                &L<|\xi|\cr 
|\u_n(\xi)|<\eta;\quad |\rho_n(\xi)|<\eta    ,\hskip 66pt   &\xi<-L\cr
|\u_n(\xi)-(\u_+-\u_-)|<\eta    ;\quad|\rho_n(\xi)-(\rho_+-\rho_-)|<\eta    ,
	                     \quad    &L<\xi\cr} 
$$
from (3.20), (3.31) and (3.32).
The limit $(\rho,\u)$ also satisfies these estimates.
From these estimates together with the fact that $\rho_n,\u_n,{\u_n}'$
converge uniformly on $[-L,L]$, we can take $N$ such that
$$ \|(\rho_n,\u_n)-(\rho,\u)\|_X
  =\sup_{-\infty<\xi<\infty}|\rho_n-\rho|
  +\sup_{-\infty<\xi<\infty}|\u_n-\u|
  +\sup_{-\infty<\xi<\infty}|{\u_n}'-\u'|
	       <6\eta    ,
$$
whenever $n>N$.
So $(\rho_n,\u_n)\to(\rho,\u)$ in $X$, and $\T(\bar\Omega)$ is precompact.

Now we show that the mapping $\T:\bar\Omega\to X$ is continuous.
Let $(\P_n,\U_n)\in\bar \Omega$ and $(\P_n,\U_n)\to(\P,\U)$ in $X$. 
Let $(\rho_n,\u_n)=\T(\P_n,\U_n)$ and $(\rho,\u)=\T(\P,\U)$.
The sequence $\{(\rho_n,\u_n)\}$ has at least
one limit point $(\rho^o,\u^o)$. Let $(\rho_{n_k},\u_{n_k})$ be a subsequence
converges to $(\rho^o,\u^o)$. Then,
$$
\rho_{n_k}(\xi) = \left \{\eqalign{(\rho_+-\rho_-) - c^{n_k}_+ \int^\infty_\xi {\P_{n_k}
  (\zeta)I^{n_k}_+(\zeta)\over\zeta-\U_{n_k}(\zeta)}d\zeta,\quad\quad&s_{n_k}<\xi\cr
c^{n_k}_- \int^\xi_{-\infty} {\P_{n_k}(\zeta)I^{n_k}_-(\zeta)\over
   \zeta-\U_{n_k}(\zeta)}d\zeta, \quad  \quad &\xi < s_{n_k} \cr} \right.
                                                                    \tag 3.33 $$
$$
\u_{n_k}(\xi)=\left \{\eqalign{
(\u_+-\u_-)-c^{n_k}_+\int^\infty_\xi I^{n_k}_+(\zeta) d\zeta,\qquad\qquad\quad &s_{n_k}<\xi\cr
c^{n_k}_-\int^\xi_{-\infty}I^{n_k}_-(\zeta)d\zeta,\qquad\qquad\quad&\xi<s_{n_k}\;,\cr} \right.
                                                                   \tag 3.34 $$
where $c^{n_k}_+$ and $c^{n_k}_-$ are solutions of (3.16) with $I^{n_k}_\pm$ given by
$$
I^{n_k}_\pm(\xi)=\exp\Big \{ -{1\over\eps}\int^\xi_{\pm 1}{\{(\zeta-\U_{n_k})^2
          -p'(\P_{n_k})\}\P_{n_k}\over \zeta-\U_{n_k}} d\zeta\Big\}.
$$
Because of the bounds in (3.12) and (3.13) we can take the limit as $k\to\infty$
inside the integrals in (3.33) and (3.34).
Since $(\rho_{n_k},\u_{n_k})\to (\rho^o,\u^o)$ in $X$, we get
$$ (\rho^o,\u^o)=\T(\P,\U)=(\rho,\u).$$
This says that $(\rho_n,\u_n)=\T(\P_n,\U_n)$ has exactly one limit point
$(\rho,\u)$, i.e. $(\rho_n,\u_n)\to(\rho,\u)$ in $X$.
\qed\enddemo\medskip
We conclude with a theorem guaranteeing existence for solutions of $(\Pe)$
under extra assumption (A). In Section 5 we justify (A) under additional
assumptions on the pressure function $p(\rho)$ or on the Riemann data
$(\rho_\pm,\u_\pm)$.

\proclaim{Theorem 3.4}
Suppose a pressure function $p(\rho)$ satisfies (H1), (H2). If solutions
$(\rho,\u)$ of $(\Pme)$ satisfy the a-priori lower bound
(A), then the boundary value problem $(\Pe)$ has a solution
$(\rho(\xi),\u(\xi))$ for any $\eps>0$.

\endproclaim\demo{Proof}
We define the map
${\Cal F} : [0,1] \times {\bar\Omega} \rightarrow X$ by
${\Cal F} (\mu,\P,\U) = (\rho_-,\u_-)+\mu \T(\P,\U)$.
If $(\rho,\u)$ is a solution of $(\rho,\u)= (\rho_-,\u_-)+\mu \T(\rho,\u)$ in 
$\Omega$, then $(\rho,\u)$ is a solution of $(\Pme)$ with $(\rho,\u)\in\Omega$.
We apply the Leray-Schauder degree theory (Rabinowitz [6, Ch V]) to solve
$$
(\rho,\u)-\mu \T(\rho,\u)= (\rho_-,\u_-)\; ,\quad\mu\in[0,1] \; .
                                                                    \tag 3.35 $$
We have already shown that $\T : {\bar\Omega} \rightarrow X$ is compact.
The map $\mu \T : [0,1] \times {\bar\Omega} \rightarrow X$ is also compact,
thus the Leray-Schauder degree of $I - \mu \T$ is well defined.
For any solution $(\rho,\u)$ of $(3.35)$, 
${1\over\mu}\{(\rho,\u)-(\rho_-,\u_-)\} \in \T(\bar\Omega)$. So $\u$ satisfies
(3.26). 
Hence by Lemma 2.4 and 2.5 and the definition of $\Omega$, any solution 
$(\rho,\u)$ of $(3.35)$ lies in the interior of $\Omega$.
Therefore
$$
d(I-\mu \T,\Omega,(\rho_-,\u_-))=d(I,\Omega,(\rho_-,\u_-))=1\; ,\quad\mu\in[0,1] 
\;
$$
 $(3.35)$ admits at least one solution for each $\mu \in [0,1]$.
\qed\enddemo\medskip

\subhead 4. The Structure of the Solution of the Riemann Problem\endsubhead
In this section we consider a sequence $(\rhoe,\ue)$ of solutions of $(\Pe)$
obeying the estimates
(2.24), (2.25) and (2.26). Since the bounds $M$ are independent of
$\eps$, $TV_{(-\infty,\infty)}\ue<2M$ and $TV_{(-\infty,\infty)}\rhoe<2M$.
On account of the Helly's theorem, there exists a
subsequence, which we call $(\rhoe,\ue)$ again, such that
$(\rhoe,\ue)$ converges pointwise to a function $(\rho,\u)$ of bounded
variation as $\eps\to 0$.
By taking further subsequences, if necessary, we assume that $(\rhoe,\ue) $
belongs to one of the four categories in Section 2.3.
Since the singular points of $(\rhoe,\ue)$ are uniformly bounded,
we may also assume that $s_\eps\to s$ as $\eps\to 0$, for some $s$.
%From the fact that $\u_\eps(s_\eps)=s_\eps$, we can also verify that $\u(s)=s$.
The limit $\rho$ and $\u$ inherit the monotonicity properties of
$\rhoe$ and $\ue$, but the monotonicities are no longer strict.

\subhead 4.1 Solution of the Riemann Problem\endsubhead
We construct solutions of $(\eP)$ as limits of solutions
$(\rhoe,\ue)$ of $(\Pe)$,
and study the structure of the emerging limit under the condition 
of $\rho>0$. In that case, we can assume that the lower
bound $\delta$ for $\rhoe$ is independent of $\eps$ and the constants in
(2.24)--(2.26) are independent of $\eps$.

\proclaim{Theorem 4.1}
Let $(\rhoe,\ue),\eps>0$ be the solution of $(\Pe)$. Then there exists
a subsequence $(\rho_{\eps_n},\u_{\eps_n}),\eps_n\to 0$ such that
the sequence of singular points $s_{\eps_n}\to s$
and $(\rho_{\eps_n},\u_{\eps_n})$ converges pointwise
to a weak solution $(\rho,\u)$ of $(\eP)$.
Furthermore, if $\rho> 0$, then there exist constants
$\beta_-<\alpha_-<s<\alpha_+<\beta_+$ such that
$$(\rho(\xi),\u(\xi))=\left \{ \eqalign
{
 (\rho_-,\u_-)&,\hskip 51pt \xi<\beta_-,\cr
 (\rho(s),\u(s))&,\hskip 26pt \alpha_-<\xi<\alpha_+,\cr
 (\rho_+,\u_+)&,\hskip 26pt \beta_+<\xi.\cr}\right.
                                                                   \tag 4.1 $$
\endproclaim\demo{Proof}
We omit the sub-index of $\eps_n$ and simply use $\eps$ for the subsequence.
The first part of the theorem has been established already and we consider the other parts.
Integrate $(2.1)_2$ on $(s_\eps,\xi)$ to get
$$ \eps\ue'=  \rhoe\ue^2+p(\rhoe)-\xi\rhoe\ue+\int^\xi_{s_\eps}\rhoe\ue
                                 d\zeta-p(\rhoe({s_\eps})).
                                                                     \tag 4.2 $$
Since $\rhoe,\ue$ and $p(\rhoe)$ are bounded by a constant $M$,
$$ |\ue'(\xi)|\le {M\over\eps}(|\xi|+1),\qquad -\infty<\xi<\infty.
                                                                \tag 4.3 $$
If we multiply $\ue$ to $(2.1)_2$ and integrate it on any bounded interval
$(a_0,b_0)$, then integration by parts gives
$$
\eps\int^{b_0}_{a_0}(\ue')^2 d\zeta
\le M(|a_0|+|b_0|+1)+\eps M(\ue'(b_0)-\ue'(a_0)).
                                                                   \tag 4.4 $$
From (4.3), we have $\eps(\ue'(b_0)-\ue'(a_0))\le M(|a_0|+|b_0|+1) $. So
$$\eps\int^{b_0}_{a_0}(\ue')^2 d\zeta\le M(|a_0|+|b_0|+1).
                                                                   \tag 4.5 $$
Let $\varphi$ be a test function with $\supp \varphi\subset[a_0,b_0]$. Then,
$$\eqalign {
\Big |\int^\infty_{-\infty}\eps\ue''\varphi d\zeta\Big | 
  &=\Big |\eps\int^\infty_{-\infty}\ue'\varphi'd\zeta\Big |
  \le \eps\Big (\int^{b_0}_{a_0}(\ue')^2 d\zeta\Big)^{1\over 2}
	  \Big (\int^{b_0}_{a_0}(\varphi')^2 d\zeta\Big)^{1\over 2}\cr
  &\le \eps^{1\over2}M(|a_0|+|b_0|+1)\Big (\int^{b_0}_{a_0}(\varphi')^2 d\zeta\Big)^{1\over2}.\cr}
                                                                     \tag 4.6 $$
Hence
$$ \lim_{\eps\to 0}\eps\int^\infty_{-\infty}\ue''\varphi d\zeta=0,
                                                                   \tag 4.7 $$
and 
$$ \eqalign {
    \int^\infty_{-\infty}&\rho\u(\zeta\varphi)'d\zeta
   -\int^\infty_{-\infty}(\rho\u^2+p(\rho))\varphi'd\zeta\cr
   &=\lim_{\eps\to 0}\int^\infty_{-\infty}\rhoe\ue(\zeta\varphi)'d\zeta
   -\lim_{\eps\to 0}\int^\infty_{-\infty}(\rhoe\ue^2+p(\rhoe))\varphi'd\zeta\cr
   &=\lim_{\eps\to 0}\eps\int^\infty_{-\infty}\ue''\varphi d\zeta=0.\cr} $$
That is $(\rho,\u)$ is a weak solution of $(\eP)$.

Now we consider the structure of the limit solution under the condition
$\rho>0$. In that case there exist constants $\delta>0$ and $M>0$ such that
$$ \delta<\rho(\xi)<M,\qquad\xi\in \Bbb R
                                                                    \tag 4.8 $$
for all small $\eps>0$. From (H1) there exist constants $a_0>0$ and $A_0>0$
such that
$$ a_0\le p'(\rho) \le A_0, \qquad \xi\in\Bbb R.
                                                                    \tag 4.9 $$
So the constants $\beta_-<\alpha_-<s<\alpha_+<\beta_+$ and $\alpha,\beta$ in
Lemma 2.3 are independent of $\eps$. Taking the limit $\eps\to 0$ in Lemma 2.3,
we obtain (4.1). 
\qed\enddemo\medskip
If $(\rho,\u)$ is a solution of $(\eP)$, then it can be easily verified that
$(\rho(x,t),\u(x,t))=(\rho({x\over t}),\u({x\over t}))$ is a weak
solution of the Riemann problem $(1.1)$ and $(1.2)$. So the solution of the
Riemann problem has been established through the vanishing viscosity process.
Theorem 4.1 provides information on the structure of solutions that
do not have a vacuum state. We conclude this section with establishing
the Rankine-Hugoniot jump conditions.
Since $(\rho,\u)$ is of bounded variation, it has right and left limits
at each $\xi\in\Bbb R$. Let $\S$ be the points of jump discontinuity of either
$\rho$ or $\u$.

\proclaim{Lemma 4.2}
The solution $(\rho,\u)$ satisfies the Rankine-Hugoniot jump condition
at all $\xi \in \S$:
$$
\xi[\rho(\xi+)-\rho(\xi-)]=\rho(\xi+)\u(\xi+)-\rho(\xi-)\u(\xi-)
                                                                  \tag 4.10 $$
$$\xi[\rho(\xi+)\u(\xi+)-\rho(\xi-)\u(\xi-)]
  =\rho(\xi+)\u^2(\xi+)+p(\rho(\xi+))-\rho(\xi-)\u^2(\xi-)-p(\rho(\xi-)).
                                                                 \tag 4.11 $$ 
\endproclaim\demo{Proof}
We integrate (2.5) on $(\xi-\delta,\xi +\delta)$, along 
$(\rhoe,\ue)$, and take the limit as $\eps\to 0$, to get
$$(\xi +\delta-\u(\xi +\delta))\rho(\xi +\delta)
  -(\xi-\delta-\u(\xi-\delta))\rho(\xi-\delta)=\int^{\xi +\delta}_{\xi-\delta}
  \rho(\zeta) d\zeta.
                                                		\tag 4.12 $$
The limit of (4.12) as $\delta\to 0$ gives
$$ 
\xi[\rho(\xi+)-\rho(\xi-)]=\rho(\xi+)\u(\xi+)-\rho(\xi-)\u(\xi-).
$$
If we integrate of (2.6) on $(\xi-\delta,\xi +\delta)$, along 
$(\rhoe,\ue)$, and perform the same process, then we get the other
equality.
\qed\enddemo\medskip

\subhead 4.2 Structure of the Limit Solutions\endsubhead
In this section we study the structure of the limit $(\rho,\u)$ under the
convexity hypothesis 
$$ p''(\rho)\ge 0\quad\text{ for}\quad \rho> 0.
                                                                   \tag H3 $$
Hypothesis (H3) includes both cases of genuine nonlinearity
and linear degeneracy.
Lemma 4.3 takes two important properties of solutions to $(\Pme) $
under Hypothesis (H3).
From $(2.11)$ we obtain 
$$(p'(\rhoe)-(\ue-\xi)^2)\rhoe'=\eps\ue''.
                                                                 \tag 4.13 $$ 

\proclaim{Lemma 4.3}
Let $(\rhoe,\ue)$ be a solution of $(\Pme)$ and $s_\eps$ be the singular point.
Then

\noindent{\rm (i)}  If $\ue $  is increasing on $(s_\eps,\infty)$, then
 $\ue' \leq 1$  on $(s_\eps,\infty)$.

\noindent{\rm (ii)}  If $\ue $ is increasing on $(-\infty,s_\eps)$, then
 $\ue' \leq 1$ on $(-\infty,s_\eps)$.

\noindent{\rm (iii)} If $\ue $ is decreasing on $(s_\eps,\infty)$ ,then
there exists exactly one $\xi \in (s_\eps,\infty)$  such that $\ue''(\xi)=0$.

\noindent{\rm (iv)} If $\ue $ is decreasing on $(-\infty,s_\eps)$, then 
there exists exactly one $\xi \in (-\infty,s_\eps)$  such that $\ue''(\xi)=0$.

\endproclaim\demo{Proof}
Let $\ue $  be increasing on $(s_\eps,\infty)$ and suppose there exist
$\xi \in (s_\eps,\infty)$ such that $\ue'(\xi) > 1$.
Since $\ue'(s_\eps)=0$ and $\ue'(\xi)\to 0$ as $\xi\to\infty$,
there exist $\xi_1,\xi_2 \in (s_\eps,\infty)$ such that $\ue'(\xi_1)=\ue'(\xi_2)=1$,
$\ue''(\xi_1)>0$, $\ue''(\xi_2)<0$ and $\ue'>1$ on $(\xi_1,\xi_2)$.
In that case, $(\xi-\ue)^2$ is decreasing on $(\xi_1,\xi_2)$ and
$$ \eqalign{p'(\rhoe(\xi_1))-(\ue(\xi_1)-\xi_1)^2
       =\eps{\ue''(\xi_1)\over \rho'(\xi_1)} > 0,\cr
   p'(\rhoe(\xi_2))-(\ue(\xi_2)-\xi_2)^2
       =\eps{\ue''(\xi_2)\over \rho'(\xi_2)} < 0 .\cr}
                                                                    \tag 4.14 $$
Hence 
$$p'(\rhoe(\xi_2))<(\ue(\xi_2)-\xi_2)^2<(\ue(\xi_1)-\xi_1)^2
<p'(\rhoe(\xi_1)).
                                                                    \tag 4.15 $$
From (H3) $p'$ is increasing for $\rho>0$. So $\rhoe(\xi_2)<\rhoe(\xi_1) $
which contradicts the fact that $\rhoe$ is strictly increasing on $(s_\eps,\infty)$
when $\ue$ is strictly increasing on $(s_\eps,\infty)$.
A similar argument gives {\rm (ii)}.

Let $\ue$ be decreasing on $(s_\eps,\infty)$.
Since $\ue$ is decreasing on $(s_\eps,\infty)$, $\rhoe$ is decreasing on
$(s_\eps,\infty)$, too.
Since $\ue'(\xi)\to 0$ as $\xi\to\infty$ and $\ue'(s_\eps)=0$,
there exists $\xi_1\in(s_\eps,\infty)$ such that $\ue''(\xi_1)=0$.
Now suppose that there is $\xi_2>\xi_1$ such that 
$\ue''(\xi_2)=0$, too.
If we put $\xi_1$ and $\xi_2$ into $(4.13)$, we get
$$ \eqalign{p'(\rhoe(\xi_1))-(\ue(\xi_1)-\xi_1)^2 = 0,\cr
   p'(\rhoe(\xi_2))-(\ue(\xi_2)-\xi_2)^2 = 0.\cr}
                                                                  \tag 4.16 $$
So we have
$$ p'(\rhoe(\xi_1))=(\ue(\xi_1)-\xi_1)^2<(\ue(\xi_2)-\xi_2)^2
=p'(\rhoe(\xi_2)).
                                                                    \tag 4.17 $$
Since $p'$ is increasing, $\rhoe(\xi_1)\leq\rhoe(\xi_2)$. 
But $\rhoe$ is an decreasing function.
So there exists exactly one point $\xi\in(s_\eps,\infty)$ 
such that $\ue''(\xi)=0$. Property {\rm (iv)} is proved similarly.
\qed\enddemo\medskip
Properties {\rm (i)} and {\rm (ii)} in Lemma 4.3 provide the structure of
rarefaction waves and {\rm (iii)} and {\rm (iv)} provide the structure of
shock waves.
Since (2.18) implies the convergence of the double index sequence
$\u_{\eps_1}(s_{\eps_2})$ for viscous solutions $\u_\eps$ which belong to
the category of $C_1,C_2,C_3$, we have $\u(s)=s$. 
For the case of $C_4$, the convergence is from (i) and (ii) of Lemma 4.3.
So we get $\u(s)=s$.
In the following two lemmas we study the continuity of the limit solution.

\proclaim{Lemma 4.4}
Let a solution $(\rho,\u)$ of $(\eP)$ be a limit of
viscous solutions $(\rhoe,\ue)$ of $(\Pe)$ and $s$ be
the limit of singular point $s_\eps$.
Then $\u(s)=s$ and $\rho$ and $\u$ are continuous at $\xi=s$.
\endproclaim\demo{Proof}
First, we suppose $\rho(s)>0$. 
Then, from Theorem 4.1, $\rho(\xi)$ and $\u(\xi) $
are constant on a neighborhood of $\xi = s$. So $\rho$ and $\u$ are
continuous at $\xi=s$.

Now suppose $\rho(s)=0$. Since $\rho_\pm$ are positive, this is possible
only when $(\rhoe,\ue)$ belongs to Category $C_4$ in Section 2.3.
So $\ue$ are increasing on $(-\infty,\infty)$ and $|\ue'(\xi)|\le1$.
We already know that $\{\ue\}$ is uniformly bounded.
The Ascoli-Arzela theorem implies the limit $\u$ is continuous on
$(-\infty,\infty)$.
From the Rankine-Hugoniot jump condition at $\xi=s$
we have $p(\rho(s+))=p(\rho(s-))$. 
Then (H1) implies $\rho(s+)=\rho(s-)$.
\qed\enddemo\medskip
Lemma 4.4 implies that the limit of viscous solutions is continuous at a
singular point $s=\u(s)$ in both cases of shock or rarefaction waves. 
Now we consider the continuity of the rarefaction waves on $\Bbb R$.

\proclaim{Lemma 4.5}
Let a solution $(\rho,\u)$ of $(\P)$ be the limit of
viscous solutions of $(\Pe)$ and $s$ be a singular point, i.e. $\u(s)=s$.
If $\ue(\xi)$ are increasing on $(s,\infty)$, 
then the limit $\u$ and $\rho$ are continuous on $(s,\infty)$.
If $\ue(\xi)$ are increasing on $(-\infty,s)$,
then the limit $\u$ and $\rho$ are continuous on $(-\infty,s)$.
\endproclaim\demo{Proof}
From Lemma 2.4, $|\ue(\xi)|$ is uniformly bounded by a constant $M$
which is independent of $\eps$.
We also know from Lemma 4.3 that $|\ue'(\xi)|<1$.
So $\{\ue\}$ is uniformly bounded and equicontinuous.
So, from the Ascoli-Arzela Theorem, $\u$ is continuous.
From $(2.11)_1$
$$ |\rhoe'|={|\rhoe \ue'|\over |\ue-\xi|}<{\rhoe \over |\ue-\xi|}.
                                                                  \tag 4.18 $$
Suppose $\rho(s) >0 $. Then, $\rho$ is constant on an open set $(s,s+\delta)$ 
for some $\delta$ which is independent of $\eps$. The constant $a$ in (2.26)
is independent of $\eps$. So $ |\rhoe'|$ is bounded above uniformly 
on the open interval $(s+c,\infty)$ for any $c>0$.
The Ascoli-Arzela theorem implies that $|\rho|$ is continuous on $(s+c,\infty)$
for any $c>0$, that is $\rho$ is continuous on $(s,\infty)$.  

Now we consider the case of $\rho(s)=0$.
This case is divided into two cases.

\noindent
Case 1 :\quad Suppose $\rho(\xi)>0$ on $(s,\infty)$. 
It is enough to prove that $\rho$ is continuous on
$(s+\delta,\infty)$ for any $\delta>0$.
Suppose $\u(s+\delta)=s+\delta$.
Then, since $\ue'(\xi)\le 1$
for all $\eps$, $\u(\xi)=\xi$ on $[s,s+\delta]$. From $(2.11)_1$, we have
$\rho=0$ on $[s,s+\delta]$. So $\u(s+\delta)<s+\delta$. Since $|\u(\xi)-\xi|$ is
increasing, we have $|\u(\xi)-\xi|>s+\delta-\u(s+\delta)$ on
$(s+\delta,\infty)$.
So, from (4.18), $\rhoe'$ is uniformly bounded on $(s+\delta,\infty)$. 
Hence $\rho$ is continuous on$(s+\delta,\infty)$ by the Ascoli-Arzela theorem.

\noindent
Case 2 :\quad Suppose $\rho(\xi)=0$ on $(s,s+\tau]$ and 
$\rho(\xi)>0$ on $(s+\tau,\infty)$ for some $\tau>0$.
The proof of the continuity on $(s+\tau,\infty)$ is the same as the Case 1.
We just prove the continuity at $s+\tau$.
The Rankine-Hugoniot jump
condition should be satisfied at $s+\tau$. If we write the condition with 
$\rho((s+\tau)-)=0$ we get
$$ \eqalign {(s+\tau)\rho((s+\tau)+)=\rho((s+\tau)+)\u((s+\tau)+) \cr
   (s+\tau)\rho((s+\tau)+) \u((s+\tau)+)=\rho((s+\tau)+) \u((s+\tau)+)^2+p(\rho((s+\tau)+)).\cr}
						                \tag 4.19 $$
So we get $\u((s+\tau)+)=s+\tau$ and $p(\rho((s+\tau)+))=0$.
So $\rho((s+\tau)+)=0$ and $\rho$ is continuous.
\qed\enddemo\medskip
The previous lemmas provide regularity properties for the limit solution
$(\rho,\u)$.
Let $\Cal S$ be the set of points of discontinuity of $(\rho,\u)$ and
$\Cal C$ be the set of points of continuity.
If $\u$ is increasing, $(\rho,\u)$ is continuous from Lemma 4.5. If $\u$ is
decreasing, we can easily verify that there exists at most one point of
discontinuity in $(-\infty,s)$ and $(s,\infty)$ from Lemma 4.3 (iii),(iv).

Now we consider the relationship between the characteristic speeds of the
problem $(\eP)$ and the weak derivative of the limit solution $(\rho,\u)$. Let
$$m=\rho\u,\quad U={\rho\choose m},\quad
F(U)={m \choose {m^2\over \rho}+p(\rho)}.                          \tag 4.20 $$
We know that the eigenvalues $\lambda_\pm$ of $\grad F$ are given by
$$\lambda_\pm(\rho,\u)=\u\pm\sqrt{p'(\rho)}.
                                                                  \tag 4.21 $$
We use the notation 
$\lambda_\pm(\xi)=\lambda_\pm(\rho(\xi),\u(\xi))$ and
$\lambda_\pm^\eps(\xi)=\lambda_\pm(\rhoe(\xi),\ue(\xi))$.

Let $d\mu=(d\mu_1,d\mu_2)={dU\over d\xi}$ be the vector valued measure which
corresponds to the weak derivative of $U$, i.e. corresponding to the linear
functional:
$$\phi\to-\int\phi'(\xi)U(\xi)d\xi,\qquad \phi\in C^1_c(\Bbb R).
                                                                   \tag 4.22 $$
We apply the Volpert product([11]) of $\grad F(U)$ and $d\mu$ to the equation
$(\eP)$ to get
$$\big(\Hat{\grad F}(U)-\xi I\big)d\mu=0
                                                                   \tag 4.23 $$
in the sense of measures, where the averaged superposition $\Hat{\grad F}(U)$
of $U$ by $\grad F$ is given by
$$ \Hat{\grad F}(U)(\xi)=\int^1_0 \grad F(U(\xi-)+s(U(\xi+)-U(\xi-)))ds.$$
Let $\xi\in\Cal C\cap\supp\mu$. Since there is at most one point of
discontinuity, $\Hat{\grad F}(U)=\grad F(U)$ in a
neighborhood of $\xi$. Suppose the determinant of $(\grad F(U)-\xi I)$
is not zero, for example $\det(\grad F(U)-\xi I)>0$.  Then there exists
a neighborhood $N$ of $\xi$ such that $\det(\grad F(U)-\zeta I)>\delta>0$
for all $\zeta\in N$.
But from (4.23), the measures $\det(\grad F(U)-\zeta I)d\mu_{1,2}=0$ on $N$,
which contradicts the fact that $\xi\in\supp\mu$.
So $\xi$ is an eigenvalue of $\grad F(U(\xi))$.
We summarize these facts in a lemma :

\proclaim{Lemma 4.6}
Let a solution $(\rho,\u)$ of $(\eP)$ be a limit
of viscosity solutions $(\rhoe,\ue)$ of $(\Pe)$ with a singular point $s$. Let $d\mu$
be the measure of $(4.22)$. Then we have:
(i) If $\u$ is increasing, then $(\rho,\u)$ is continuous.
If $\u$ is decreasing, then there exists at most one point of discontinuity
in $(-\infty,s)$ and $(s,\infty)$.
(ii)
If $(\rho,\u)$ is continuous at $\xi\in \supp \mu$,
$\xi=\lambda_+(\xi)$ on $(s,\infty)$ and $\xi=\lambda_-(\xi)$ on $(-\infty,s)$.
\endproclaim\medskip
We conclude the section with a theorem which provides the structure of
the limit solution $(\rho,\u)$ of the viscosity solutions $(\rhoe,\ue) $
which obey the a-priori estimates (2.24)--(2.26).

\proclaim{Theorem 4.7}
Let $(\rho,\u)$ be a solution of the Riemann
problem $(\eP)$  through the method of self-similar zero-viscosity limits
and $s$ be the limit of singular points.
(i)  If $\u$  is increasing on $(s,\infty)$,  then 
$\lambda_+(\xi)$  is continuous on $(s,\infty)$  and
$$ 
\lambda_+(\xi)=\left\{\eqalign{
\lambda_+(s)&,\hskip 40pt s<\xi<\lambda_+(s), \cr
\xi&,\hskip 40pt\lambda_+(s)<\xi <\lambda_+(\rho_+,\u_+), \cr
\lambda_+(\rho_+,\u_+)&,\hskip 40pt\lambda_+(\rho_+,\u_+)<\xi. \cr} \right.
                                                   		\tag 4.24 $$
(ii)  If $\u$  is increasing on $(-\infty,s)$,  then 
$\lambda_-(\xi)$  is continuous on $(-\infty,s)$  and
$$
\lambda_-(\xi)=\left\{\eqalign{
\lambda_-(s)&,\hskip 40pt \lambda_-(s)<\xi<s,\cr
\xi&,\hskip 40pt \lambda_-(\rho_-,\u_-)<\xi <\lambda_-(s),\cr
\lambda_-(\rho_-,\u_-)&,\hskip 40pt \xi<\lambda_-(\rho_-,\u_-).\cr}\right.
                                                                   \tag 4.25 $$
(iii)  If $\u$  is decreasing on $(s,\infty)$,  then
$\lambda_+(\xi)$  has an unique discontinuity on $(s,\infty)$  and
$$
\lambda_+(\xi)=\left\{\eqalign{
\lambda_+(s)&,\hskip 40pt s<\xi<{\rho_+\u_+\over (\rho_+-\rho(s))},\cr
\lambda_+(\rho_+,\u_+)&,\hskip 40pt{\rho_+\u_+\over (\rho_+-\rho(s))}<\xi.\cr}\right.
                                                                   \tag 4.26 $$
(iv)  If $\u$  is decreasing on $(-\infty,s)$,  then
$\u$  has an unique discontinuity on $(-\infty,s)$  and
$$
\lambda_-(\xi)=\left\{\eqalign{
\lambda_-(s)&,\hskip 40pt{\rho_-\u_-\over (\rho_- -\rho(s))}<\xi<s,\cr
\lambda_-(\rho_-,\u_-)&,\hskip 40pt\xi<{\rho_-\u_-\over (\rho_- -\rho(s))}.\cr}\right.
			                                          \tag 4.27 $$

\endproclaim\demo{Proof}
We always consider the eigenvalue $\lambda_-$ on the interval $(-\infty,s)$ and
$\lambda_+$ on the other side $(s,\infty)$.
If $\u$ is increasing, then $\u$ and $\rho$ are continuous.
So they should be constant out of $\supp\mu$ and $\lambda_\pm$
are continuous and increasing, too. With these facts
we can easily check that $\supp\mu$ is a
connected sub-intervals of $(-\infty,s]$ or $[s,\infty)$. If not,
$\lambda_\pm$ is discontinuous.
So the structure of $\lambda_\pm$ should follow {\rm (i)} and {\rm (ii)}.

If $\u$ is decreasing, then $\lambda_\pm$ are also decreasing
and there exists at most one point of discontinuity on 
$(-\infty,s)$ and $(s,\infty)$.
Since $\lambda_\pm=\xi$ on $\supp\mu$ and $\lambda_\pm$ are decreasing,
$\supp\mu$ should be the point of discontinuity and $\lambda_\pm$ be constant
before and after the discontinuity. We can find the point of the discontinuity
from the Rankine-Hugoniot jump condition, and
$\lambda_\pm$ should follow (iii) and (iv). 
\qed\enddemo\medskip
In summary, the limit of viscosity
solutions has an intermediate state which is connected to the boundary states
by rarefactions ({\rm (i)} and {\rm (ii)}) or shocks ({\rm (iii)}
and {\rm (iv)}).

\proclaim{Corollary 4.8}
If the system $(1.1)$ is strictly hyperbolic, i.e.
there exists $c>0$ such that
$$ p'(\rho)\ge c^2>0,\qquad \rho>0,
                                                                   \tag 4.28 $$
then the emerging limit does not have a vacuum state.

\endproclaim\demo{Proof}
Theorem 4.7 implies that $\lambda_+(\xi)$ is constant on the interval
$(s,\lambda_+(s))\ne\phi$
and $(\lambda_+(\rho_+,\u_+),\infty)$.
Since $\sqrt{p'(\rho)}$ is increasing on $(s,\infty)$, $\u(\xi)$ is also
constant on those intervals. From $(1.6)_1$ $\rho$ is also constant.
Now we consider the interval $(\lambda_+(s),\lambda_+(\rho_+,\u_+))$.
From $(1.6)_2$ we get
$ c^2\rho'\le(\xi-\u)\rho$,
and hence there exists a constant $C$ such that
$$\rho'\le C\rho.$$
So we have 
$$\rho(\xi)\le\rho(\lambda_+(s))e^{C(\xi-\lambda_+(s))}.$$
Suppose the solution has a vacuum state, i.e.
$\rho(s)=0$. Then, since $\rho$ is constant on $(s,\lambda_+(s))$,
$\rho(\lambda_+(s))=0$, and hence
$\rho$ is zero on $(\lambda_+(s),\lambda_+(\rho_+,\u_+))$.
So $\rho(\infty)=0$ which contradicts the boundary condition
$\rho(\infty)=\rho_+>0$.
\qed\enddemo\medskip
\subhead 5. CONVEX PRESSURE LAWS\endsubhead
In Lemma 2.4 the a-priori estimates (2.24), (2.25) are established except for
the lower bound for $\rho$ of the case $C_4$.
In this section we complete the a-priori estimates in two cases under the
convex pressure laws (H3).
First, we consider the case of strictly hyperbolic systems.

The equation $(2.22)_1$ can be written as a first order linear equation
for $\rho$ :
$$\rho'+{\u'\over u-\xi}\rho =0,\qquad \xi\ne s,
                                                                   \tag 5.1 $$
where s is the singular point.
Then the solution is given by
$$
\rho(\xi)= \left\{\eqalign{
\rho^\mu_+e^{-\int^\infty_\xi{\u'\over\zeta-\u}d\zeta}
,\qquad s<\xi \cr
\rho^\mu_-e^{-\int^\xi_{-\infty}{\u'\over\u-\zeta}d\zeta}
,\qquad \xi<s,\cr} \right.
                                                                   \tag 5.2 $$
where $\rho^\mu_- = \rho_-$ and $\rho^\mu_+=\rho_-+\mu(\rho_+-\rho_-)$ are the
boundary conditions of (2.23).

\proclaim{Lemma 5.1}
Let a solution $(\rho,\u)$ of $(\Pme)$ belong to
the class $C_4$. If the system $(1.1)$ is strictly hyperbolic, i.e.
$$ p'(\rho)\ge c^2>0,\qquad \rho>0,
                                                                    \tag 5.3 $$
then there exists a constant $\delta>0$ which satisfies $(2.24)$ and is
independent of $\eps$ and $\mu$.
\endproclaim\demo{Proof}
Since $\u$ is increasing on $\Bbb R$, $\u'\le 1$ by Lemma 4.3 and 
$\xi-\u(\xi)$ is also increasing.
So there exists $\xi_1>s$ such that $0<\xi-\u(\xi)\le{c\over 2}$ on $(s,\xi_1) $
and ${c\over 2}\le\xi-\u(\xi)$ on $(\xi_1,\infty)$.
Then
$$\rho(\xi_1)=\rho_+^\mu e^{-\int_{\xi_1}^{\infty}{\u'\over\zeta-\u}d\zeta}
   \ge \rho_+^\mu e^{-{2\over c}(\u_+-\u_-)}.
                                                                    \tag 5.4 $$
Integrating (2.28) on $(s,\xi_1)$ we get
$$\eps\int_s^{\xi_1} \u''(\zeta) d\zeta=\eps\u'(\xi_1)\le \eps
                                                                    \tag 5.5 $$
and
$$\int_s^{\xi_1}(p'(\rho)-(\zeta-\u)^2)\rho'd\zeta
\ge{3c^2\over 4}\int_s^{\xi_1}\rho'd\zeta
\ge{3c^2\over 4}(\rho(\xi_1)-\rho(s)).
                                                                    \tag 5.6 $$
From the above estimations $\rho$ is bounded below by
$$\rho(s)\ge \rho_+^\mu e^{-{2\over c}(\u_+-\u_-)}-{4\eps\over 3c^2}
\ge \min\{\rho_-,\rho_+\} e^{-{2\over c}(\u_+-\u_-)}-{4\eps\over 3c^2}.
                                                                    \tag 5.7 $$
The positive lower bound for $\rho$ is obtained for small $\eps$.
\qed\enddemo\medskip
We return to general convex pressure laws (H3) and consider the function
$$g(\rho)={p(\rho)\over\rho},\qquad \rho>0.
                                                                    \tag 5.8 $$
Either the function $g:\Bbb R^+\to\Bbb R^+$ is
invertible or the system is strictly hyperbolic.
Consider the case when $g$ has an inverse $g^{-1}$.

\proclaim{Lemma 5.2}
Let a solution $(\rho,\u)$ of $(\Pme)$ belong to
the class $C_4$. If the boundary conditions $(\rho_\pm,\u_\pm)$ satisfy
$$\u_+-\u_-<\max_{m>0}(m\ln({\rho_-^\mu\over g^{-1}(m^2)}))
           +\max_{m>0}(m\ln({\rho_+^\mu\over g^{-1}(m^2)})),
                                                                   \tag 5.9 $$
then there exists a constant $\delta>0$ which satisfies $(2.24)$ and
is independent of $\eps$ and $\mu$.
\endproclaim\demo{Proof}
Let $s$ be the singular point of the solution $(\rho,\u)$. Since $\u$ is
increasing in $\Bbb R$, we have $\u_-<\u(s)<\u_+$.
If (5.9) holds,
$$ \u_+-\u(s)<\max_{m>0}(m\ln({\rho_+^\mu\over g^{-1}(m^2)}))
                                                                   \tag 5.10 $$
or
$$ \u(s)-\u_-<\max_{m>0}(m\ln({\rho_-^\mu\over g^{-1}(m^2)})).
                                                                   \tag 5.11 $$
Let us assume that (5.10) holds. Then there exists $m>0$ such that
$\u_+-\u(s)<m\ln({\rho_+^\mu\over g^{-1}(m^2)})$ or equivalently
$$ g(\rho_+^\mu e^{-{(\u_+-\u(s))\over m}})-m^2>0.
                                                                   \tag 5.12 $$
Since $\xi-\u(\xi)$ is increasing, there exists $\xi_1>s$ such that
$0<\xi-\u(\xi)\le m$ on $(s,\xi_1)$ and $m\le\xi-\u(\xi)$ on $(\xi_1,\infty)$.
Then
$$\rho(\xi_1)=\rho_+^\mu e^{-\int_{\xi_1}^{\infty}{\u'\over\zeta-u}d\zeta}
    \ge \rho_+^\mu e^{-{(\u_+-\u(s))\over m}}.
                                                                   \tag 5.13 $$
We can easily check that $g(\rho)$ is increasing for $\rho>0$ and
$g(\rho(\xi_1))-m^2>0$.
Integrating (2.28) on $(s,\xi_1)$ we get
$$\eps\int_s^{\xi_1} \u''(\zeta) d\zeta=\eps\u'(\xi_1)\le \eps
                                                                   \tag 5.14 $$
and
$$\int_s^{\xi_1}(p'(\rho)-(\zeta-\u)^2)\rho'd\xi \ge({p(\rho(\xi_1))
   -p(\rho(s))\over\rho(\xi_1)-\rho(s)}-m^2)(\rho(\xi_1)-\rho(s)).
                                                                   \tag 5.15 $$
The convexity Hypothesis (H3) implies
$${p(\rho(\xi_1))-p(\rho(s))\over\rho(\xi_1)-\rho(s)}-m^2
           >{p(\rho(\xi_1))\over\rho(\xi_1)}-m^2=g(\rho(\xi_1))-m^2>0.
                                                                   \tag 5.16 $$
So the density $\rho$ is bounded below by
$$\rho(s)\ge \min\{\rho_+,\rho_-\}e^{-{1\over m}(\u_+-\u(s))}-
    {\eps\over g(\xi_1)-m^2}>0
                                                                   \tag 5.17 $$
for a sufficiently small $\eps>0$. The situation is similar if (5.11) holds.
\qed\enddemo\medskip
One can check that, if 
$$\u_+-\u_-<\max_{m>0}(m\ln({\rho_-\over g^{-1}(m^2)}))
           +\max_{m>0}(m\ln({\rho_+\over g^{-1}(m^2)}))
                                                                   \tag 5.18 $$
holds and $\rho_-\le\rho_+$, then (5.9) holds. If $\rho_->\rho_+$ and
instead of using the continuation of the boundary data (2.23) one uses
$$
\eqalign{\rho(\pm\infty)=\rho_\pm^\mu:=&\rho_++\mu(\rho_\pm-\rho_+)\cr
        \u(\pm\infty)=\u_\pm^\mu:=&\u_++\mu(\u_\pm-\u_+)\cr }
                                                                   \tag 5.19 $$
then again (5.9) holds. Thus (5.18) provides a sufficient condition which
prevents vacuum from appearing. It is known that admissible solutions of
(1.1) and (1.2) do not have a vacuum state if and only if
$$\u_+-\u_-<\int^{\rho_-}_0{\sqrt{p'(\rho)}\over\rho}d\rho
           +\int^{\rho_+}_0{\sqrt{p'(\rho)}\over\rho}d\rho.
                                                                   \tag 5.20 $$
While (5.18) is a sufficient condition to avoid vacuum, simple numerical
computations show that it is not a necessary.
In the case of $\gamma$-laws,
$p(\rho)=\rho^\gamma$ for $\gamma>1$, the condition (5.18) corresponds to
$$ \u_+-\u_-<\ln\big({2\over\gamma-1}\big)
    \Big[\big({2\over\rho_-(\gamma-1)}\big)^{1-\gamma\over 2}
        +\big({2\over\rho_+(\gamma-1)}\big)^{1-\gamma\over 2} \Big].
                                                                   \tag 5.21 $$

Now we summarize the previous lemmas and the results of Section 3 in the
theorem :

\proclaim{Theorem 5.3}
Suppose $p(\rho)$ satisfies (H1), (H2) and (H3).
If the system (1.1) is strictly hyperbolic or the initial data
$(\rho_\pm,\u_\pm)$ satisfy (5.18),
then the boundary value problem $(\eP)$ has a solution $(\rho,\u)$
which is a $\eps\to0$ of solutions of $(\Pe)$.
The function $(\rho,\u)$ has the structure stated in Theorem 4.7 and 
does not contain vacuum. $(\rho(x/t),\u(x/t))$ is a solution of Riemann
problem (1.1),(1.2).
\endproclaim

\medskip\demo
{\bf Acknowledgement} I would like to thank Professor A. E. Tzavaras, my
thesis advisor. He introduced me this problem and gave me valuable remarks
on this work.  Research partially supported by the National Science
Foundation under Grant DMS-9505342
\enddemo

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\enddocument