math0111281.tex
\documentstyle[12pt]{article}
%\renewcommand{\baselinestretch}{1.3}

\begin{document}
%\pagestyle{empty}

%\vspace*{.5cm}
%\begin{center}
\noindent Title: On the uniphase solutions of 
the nonlinear damped wave equation
%\footnote {AMS clssification: 39A13, 35B10, 35L70}\\
%\end{center}

%\begin{center}

\noindent Authors: {\bf S. Bir\u{a}ua\c{s}, D. Opri\c{s}} Departament of Mathematics 
The West University of Timi\c{s}oara Timi\c{s}oara 1900, Romania\\
%e-mail: birauas@tim1.uvt.ro\\
%~~~~~~~~opris@tim1.uvt.ro
\noindent Comments: 19 pages, Tex\\
\noindent Subj-class: Partial Differential Equations\\
\noindent MSC-class: 39A13 (Primary) 35B10 (Secondary)\\
%\end{center}
In this paper we study the steady uniphase and multiphase 
solutiuons of the discretized nonlinear damped wave equation. Conditions for the
stability and instability of the steady solutions are given; in the instability
case the linear stable and unstable manifolds are described.
\vspace*{1cm}

\begin{center}
\Large \bf 1. Introduction
\end{center}

\vspace*{.5cm}
%\footnote {AMS clssification: 39A13, 35B10, 35L70}
We consider the following mixed problem:
\begin{equation}
u_{tt}=(\sigma(u_{x}))_{x}+\varepsilon u_{xxt}, \; x\in (0,1),\; t>0,\;
\varepsilon \in {\bf R}
\end{equation}
\begin{equation}
u(0,t)=0,\; u(1,t)=P,\; t>0
\end{equation}
\begin{equation}
u(x,0)=u_{0}(x),\; u_{t}(x,0)=u_{1}(x),\; x\in (0,1)
\end{equation}
where $P>0$ and $\sigma :{\bf R} \to {\bf R}$ is an application with the following
properties: $\sigma (0)=0$, $\sigma$ is continously differentiable on ${\bf R}$,
$\sigma (\xi) >0$, $\forall \xi >0$, $\sigma (\xi )<0$, $\forall \xi <0$; there
exist $\overline{\alpha}$, $\underline{\beta}$ such that $0<\overline{\alpha}<
\underline{\beta}$ and $\sigma^{\prime}(\xi )>0$, $\forall \xi \in (-\infty ,
\overline{\alpha}) \cup (\underline{\beta} ,\infty )$, $\sigma^{\prime}(\xi)<0$,
$\forall \xi \in (\overline{\alpha}, \underline{\beta})$, and $\underline{\alpha}, \overline{\beta},$
such that $0<\underline{\alpha}<\overline{\beta}<\underline{\beta}<\overline{\beta},$
such that $\sigma(\underline{\alpha})=\sigma(\underline{\beta})$ and
$\sigma(\overline{\alpha})=\sigma(\overline{\beta}).$

In the paper [1] it is shown that for $\varepsilon=1$ the mixed problem 1-3 has a
unique solution on the appropiate function space. The uniphase and multiphase steady
solutions are defined and it is studied the stability of these steady solutions.
It is presented also a discretization in the spatial variable $x$ of the equation
({\bf 1}.1) and some numerical simulations are presented. For $\varepsilon =0$ the problem
was studied in [2].

For $\sigma (\xi )=a^{2}\xi$, $a\in {\bf R}$ we obtain the wave equation studied
in a lot of papers.

The aim of this paper is the study of uniphase steady solutions and partially
the study of multiphase steady solutions of the system obtained via a discretization
of ({\bf 1}.1) in spatial variable $x$ and then a discretization in temporal variable $t$.

In the section {\bf 2} it is shown that the system obtained from the discretization of
({\bf 1}.1) in the spatial variable $x$ represents an equation Euler-Lagrange by rapport
with a Lagrange function discrete continuous with damped term.

The system has a finite number of steady solutions (uniphase and multiphase).

In the section {\bf 3}, using the linearized system of the system used in the section
{\bf 2}, we study the stability of uniphase steady solutions. We proove that such
a solution is stable if $P\in (\underline{\alpha}, \overline{\alpha})$ or $P\in
(\underline{\beta}, \overline{\beta})$ and $\varepsilon >0$. If $P\in (\underline{\alpha},
\underline{\beta})$ the uniphase steady solution is hiperbolic so it is unstable.
We have determined a polycycle curve for this solution.

For $P\in (\underline{\alpha},\overline{\alpha})$ or $P\in (\underline{\beta},
\overline{\beta})$, $\varepsilon =0$, we find a curve $u^{c}(t)$, $t\in {\bf R}$
such that every of this components is periodic with the period
$$T_{k}=\frac{\pi}{n\sqrt{\tau}} cosec\; \frac{k\pi}{2n}$$

For the steady solution 2-phase we present stability conditions depending on $\varepsilon$,
$\tau_{1}=\sigma^{\prime}(\alpha)$, $\rho=\sigma^{\prime}(\beta)$, $\alpha \in
(\underline{\alpha},\overline{\alpha})$, $\beta \in (\underline{\beta},\overline{\beta})$.

In the section {\bf 4} it is shown that the discretization of the equation ({\bf 1}.1)
leads us to a difference system of equations which represents the equations Euler-Lagrange
with damping for an associated discrete Lagrange function. The steady solutions,
uniphase and multiphase, are presented and we proove that they are in finite
number.

Also, in the section {\bf 4} we are studing the stability of the uniphase solutions,
the conditions for a uniphase steady solution to be hiperbolic and the linear stable
and unstable associated manifolds.

A similar study for the multiphase steady solutions will be done in a future paper.

\setcounter{equation}{0}
\begin{center}
\Large \bf 2. Semidiscretization of the equation $u_{tt}=(\sigma (u_{x}))_{x}+\varepsilon
u_{xxt}$
\end{center}

\vspace*{.5cm}

For $\varepsilon =0$, the equation ({\bf 1}.1) represents the equation Euler-Lagrange
for the Lagrange function $L:J^{2}({\bf R}^{2}, {\bf R}) \to {\bf R}$,
$L(x,t,u,u_{x},u_{tx})=\frac{1}{2} u^{2}_{x}-\sigma (u_{x})$, where $J^{2}({\bf R}^{2},
{\bf R})$ is the bundle of the jets of order 2 of the fibrate $\pi :{\bf R}^{3}
\to{\bf R}^{2}$.

Be $x_{k}=kh_{1}$, $k=1,2,...,n$ the division points of the interval $[0,1]$,
$h_{1}=1/n$ and $u_{k}(t)=u(kh_{1},t)$, $k=1,2,...,n$. For the boundary conditions
we consider: $u_{0}=u(0,t)=0$, $u_{n}=u(1,t)=P$.

The partial derivatives $u_{x}$, $u_{xx}$, $(\sigma (u_{x}))_{x}$ will be approximate by:
$$u_{x}(kh_{1},t)\sim \frac{1}{h_{1}} (u_{k}(t)-u_{k-1}(t))$$
\begin{equation}
u_{xx}(kh_{1},t)\sim \frac{1}{h_{1}^{2}} (u_{k+1}(t)-2u_{k}(t)+u_{k-1}(t))
\end{equation}
$$(\sigma(u_{x}))_{x}(kh_{1},t)\sim \frac{1}{h_{1}} \left[ \sigma \left( \frac{1}{h_{1}}
(u_{k+1}(t)-u_{k}(t)\right) -\sigma \left( \frac{1}{h_{1}} (u_{k}(t)-u_{k-1}(t)
\right) \right] $$

Let be the sequence $(u_{k},\dot{u}_{k})\in T{\bf R}$, $k=1,...,n$ on the tangent bundle
at $\bf R$ and $L:T{\bf R}\to {\bf R}$ the Lagrange function defined by:
\begin{equation}
L(u_{k},\dot{u}_{k})=\frac{1}{2}\dot{u}^{2}_{k}+w \left( \frac{1}{h_{1}}
(u_{k}-u_{k-1})\right) \; ,\; k=1,...,n-1
\end{equation}
where $w :{\bf R}\to {\bf R}$ with $w^{\prime} (\xi)=\sigma (\xi)$. The
action of $L$ is defined by:
\begin{equation}
{\cal{A}}(u,\dot{u})=\sum_{k=1}^{n} L(u_{k},\dot{u}_{k})
\end{equation}

In order to obtain the first variation of $\cal{A}$ we consider the sequence
$(u_{k}(\eta),\dot{u}_{k}(\eta))\in T{\bf R}$, $k=1,...,n$ with $\eta \in (-a,a)$ and
$u_{k}(0)=u_{k}$, $\dot{u}_{k}(0)=\dot{u}_{k}$. The action ({\bf 2}.3) on this
sequence is:
$${\cal{A}} (\eta)=\sum_{k=1}^{n-1} L(u_{k}(\eta),\dot{u}_{k}(\eta)) $$

The first variation of ({\bf 2}.3) is given by
$$\left. \frac{\partial {\cal{A}} (\eta)}{\partial \eta} \right|_{\eta=0} =0 $$

The first variation of ({\bf 2}.3) is:
\begin{equation}
\ddot{u}_{k}-\frac{1}{h_{1}} \sigma (\Delta u_{k})+\frac{1}{h_{1}} \sigma (\Delta
u_{k-1})=0\; ,\; k=1,...,n-1
\end{equation}
where
$$\Delta u_{k+1}=\frac{1}{h_{1}} (u_{k+1}-u_{k})\; ,\; k=1,...,n-1$$
and
$$\Delta u_{n}=\frac{1}{h_{1}} (P-u_{n-1})$$

The system ({\bf 2}.4) represent the semidiscretized system of the equation ({\bf 1}.1)
for $\varepsilon =0$.

The semidiscretized system of the equation ({\bf 1}.1) for $\varepsilon \neq 0$ is:
\begin{equation}
\ddot{u}_{k}-\frac{1}{h_{1}} \sigma (\Delta u_{k+1})+\frac{1}{h_{1}} \sigma
(\Delta u_{k})=\frac{\varepsilon}{h_{1}^{2}} (\dot{u}_{k+1}-2\dot{u}_{k}+\dot{u}_{k-1})\;
, \; k=1,...,n-1
\end{equation}
and represents the equation Euler-Lagrange for the function $L$ defined in ({\bf 2}.2)
with the disperssion:
$$\frac{\varepsilon}{h_{1}^{2}} (\dot{u}_{k+1}-2\dot{u}_{k}+\dot{u}_{k-1}) $$

To the system ({\bf 2}.5) we associate the equivalent system on $T^{*}R$:
$$\dot{u}_{k}=v_{k}\; ,\; \; k=1,...,n-1 $$
\begin{equation}
\dot{v}_{k}=\frac{1}{h_{1}} \left[ \sigma (\Delta u_{k+1})-\sigma (\Delta u_{k}) 
\right] +\frac{\varepsilon}{h_{1}^{2}} (v_{k+1}-2v_{k}+v_{k-1})
\end{equation}
with the action:
\begin{equation}
V(u,v)=\sum_{k=1}^{n-1} \left[ \frac{1}{2}v^{2}_{k}+w(\Delta u_{k})\right] +w
(\Delta u_{n})
\end{equation}

From ({\bf 2}.3) and ({\bf 2}.5) it follows that:
\begin{equation}
\frac{d{\cal{A}}(u,\dot{u}}{dt}) =-\sum_{k=1}^{n} \left[ \frac{1}{h_{1}} (\dot{u}_{k+1}
-\dot{u}_{k}) \right]^{2}
\end{equation}

From ({\bf 2}.5) and ({\bf 2}.6) it follows that any steady solution of the system
({\bf 2}.5) satisfies the conditions:
\begin{equation}
\sigma(\Delta u_{k})=C,\; k=1,...,n,\; \; \; \sum_{k=1}^{n} \frac{1}{h_{1}} \Delta u_{k}
=nP, \; C\in {\bf R}
\end{equation}

Because the function $\sigma$ is not monotone we obtain two types of steady solutions:

\underline{\bf Definition 2.1.} A uniphase steady solution $\bar{u}$ for the
system ({\bf 2}.5) is a steady solution with the property: $\bar{u}_{k}=kh_{1}P$,
$k=1,...,n-1$.

\underline{\bf Definition 2.2.} A multiphase steady solution for the system ({\bf 2}.5)
is a steady solution with the property: $u_{k+1}-u_{k}=h_{1}f(k)$, $k=1,...,n-1$.

We denote by $E$ the set of multiphase steady solutions and:
$$E^{+}=\{ \bar{u}\in E : \sigma^{\prime} (\Delta u_{k})>0,\; k=1,...,n \} $$
$$E^{-}=\{ \bar{u}\in E : \exists k\in \{ 1,...,n\} :\sigma^{\prime} (\Delta u_{k})<0 \} $$

In [1] it is shown that:

\underline{\bf Lemma 2.1.} For any steady solution $\bar{u}\in E^{+}$ there is
$\alpha \in (\underline{\alpha},\bar{\alpha})$ and $\beta \in (\underline{\beta},
\bar{\beta})$, $C\in (\underline{\sigma}, \bar{\sigma})$, such that:

1) $\sigma (\Delta \bar{u}_{k})=C,\; \; k=1,...,n$

2) $\Delta (\bar{u}_{k})=\alpha$ or $\Delta (\bar{u}_{k})=\beta , \; k=1,...,n$

3) $k\alpha +(n-k)\beta =nP$

\underline{\bf Lemma 2.2.} If $P<\underline{\alpha}$ or $P>\bar{\beta}$ then
$E^{+}=\phi$ 

\underline{\bf Lemma 2.3.} The system ({\bf 2}.5) has a finite number of steady solutions.

\newpage
\setcounter{equation}{0}
\begin{center}
\Large \bf 3. Properties of the steady solutions of the system (2.5)
\end{center}

\vspace*{.5cm}

The general form of the system ({\bf 2}.6) is:
\begin{equation}
F_{k}(u_{k},u_{k+1},\dot{u}_{k-1},\dot{u}_{k},\dot{u}_{k+1},\ddot{u}_{k})=0,\; 
k=1,2,...,n
\end{equation}

Let $\bar{u}$ be a steady solution of the system ({\bf 2}.6). The linearized system
in a neighborhood of $\bar{u}$ is given by:
\begin{equation}
\left. 
\frac{\partial F_{k} \left( \bar{u} +\displaystyle \sum_{j=1}^{n} \eta^{j}w_{j}, \dot{\bar{u}}+\sum_{j=1}^{n}
\eta^{j}\dot{w}_{j}, \ddot{\bar{u}}+\sum_{j=1}^{n}\eta^{j}\ddot{w}_{j} \right) }{\partial
\eta^{j}} \right|_{\eta^{j}} =0
\end{equation}
$\hspace*{10cm} j=1,...,n$ \\
where $\eta_{j}\in (-a,a),\; j=1,...,n$.

The linearized system associated to ({\bf 2}.5) is:
\begin{equation}
\ddot{w}_{j}-\frac{1}{h_{1}^{2}} \sigma^{\prime} (\Delta \bar{u}_{j+1})(w_{j+1}-w_{j})+
\frac{1}{h_{1}^{2}} \sigma^{\prime} (\Delta \bar{u}_{j})(w_{j}-w_{j-1})=
\end{equation}
$$=\frac{\varepsilon}{h_{1}^{2}} (\dot{w}_{j+1}-2\dot{w}_{j}+\dot{w}_{j-1}),\; \;
j=1,...,n-1$$

A solution for the system ({\bf 3}.3) is:
\begin{equation}
w_{j}=exp\; (\lambda j)exp\; (ia_{k}j),\; a_{k}=\frac{\pi k}{n},\; k=1,...,n-1,
\; \lambda \in {\bf C}
\end{equation}

From ({\bf 3}.3) and ({\bf 3}.5) yields:
$$\lambda^{2}exp\; ia_{k}-\frac{1}{h_{1}^{2}} \sigma^{\prime} (\Delta \bar{u}_{k+1})
exp\; ia_{k}(exp\; ia_{k-1})+\frac{1}{h_{1}^{2}} \sigma^{\prime} (\Delta \bar{u}_{k})
(exp\; ia_{k-1}) =$$
\begin{equation}
=\frac{\varepsilon}{h_{1}^{2}} \lambda (exp\; ia_{k-1})^{2}
\end{equation}

\underline{\bf Theorem 3.1.} If $P\in (\underline{\alpha},\bar{\alpha})$ or $P\in
(\underline{\beta},\bar{\beta})$ the uniphase steady solution $\bar{u}_{k}=kh_{1}P$,
$k=1,...,n-1$ of the system ({\bf 2}.5) is asymptotically stable if $\varepsilon >0$
and unstable if $\varepsilon <0$.

\underline{\it Proof: } For the uniphase steady solution $\bar{u}$, the equations
({\bf 3}.5) become:
\begin{equation}
\lambda^{2} exp\; ia_{k}-\frac{1}{h_{1}^{2}} \tau (exp\; ia_{k-1})^{2}-\frac{\varepsilon}
{h_{1}^{2}} \lambda (exp\; ia_{k-1})^{2}=0,\; k=1,...,n-1
\end{equation}
where $\tau =\sigma^{\prime}(P)$. We denote:
$$\mu_{k}=\frac{(exp\; ia_{k-1})^{2}}{exp\; ia_{k}}=-4sin^{2}\; \frac{\pi k}{2n}$$

The equations ({\bf 3}.6) become:
\begin{equation}
\lambda^{2}-\varepsilon \mu_{k}n^{2}\lambda -\mu_{k}n^{2}\tau=0,\; k=1,...,n-1
\end{equation}
with the roots:
\begin{equation}
\lambda^{2}=\frac{1}{2} \mu_{k}n^{2} \left[ \varepsilon \pm \left( \varepsilon^{2}
+\frac{4\tau}{\mu_{k}n^{2}}\right)^{\frac{1}{2}} \right],\; k=1,...,n-1
\end{equation}

If $\varepsilon >0$, because $\mu_{k}<0$ yields that $Re\; \lambda_{k} <0$ so it
follows that the steady uniphase solution $\bar{u}$ is asymptotically stable.

If $\varepsilon <0$, it follows that $Re\; \lambda_{k} <0$ so the steady uniphase
solution is unstable.

\underline{\bf Theorem 3.2.} If $P\in (\bar{\alpha}, \underline{\beta})$ the uniphase
steady solution $\bar{u}$ of the system ({\bf 2}.5) is hyperbolic so it is unstable.

\underline{\it Proof: } For the uniphase steady solution $\bar{u}$ in the equations
({\bf 3}.7) become:
\begin{equation}
\lambda^{2}-\varepsilon \mu_{k}n^{2}\lambda +\mu_{k}n^{2}\rho =0,\; k=1,...,n-1
\end{equation}
where $\rho=-\sigma^{\prime}(P)$; the solutions of these equations are:
\begin{equation}
\lambda^{+}_{k}=\frac{1}{2}\mu_{k}n^{2}\left[ \varepsilon +\left( \varepsilon^{2}
-\frac{4\rho}{\mu_{k}n^{2}} \right)^{\frac{1}{2}} \right] 
\end{equation}
$$\lambda^{-}_{k}=\frac{1}{2}\mu_{k}n^{2}\left[ \varepsilon -\left( \varepsilon^{2}
-\frac{4\rho}{\mu_{k}n^{2}} \right)^{\frac{1}{2}} \right] ,\; k=1,...,n-1 $$

The real solutions $\lambda^{+}$, $\lambda^{-}$ have the property: $\lambda^{+}
\cdot \lambda^{-}=\mu_{k}n^{2}\rho <0$. It follows that they have opposite signs.
The steady solution is unstable, it is hyperbolic.

The linear manifolds associated to the steady solution of hyperbolic type are:
\begin{equation}
\tilde{u}_{k}^{u}(t)=h_{1}kP+exp\; (t\lambda_{k}^{+}) \sum_{j=1}^{n-1} \eta_{j}
exp\; (ia_{k}j),\; t\in{\bf R}, \; k=1,...,n-1
\end{equation}
\begin{equation}
\tilde{u}_{k}^{s}(t)=h_{1}kP+exp\; (t\lambda_{k}^{-}) \sum_{j=1}^{n-1} \eta_{j}
exp\; (ia_{k}j),\; t\in{\bf R}, \; k=1,...,n-1
\end{equation}

$\tilde{u}_{k}^{u}$ and $\tilde{u}_{k}^{s}$ have the properties:
$$\lim_{t\to -\infty} \tilde{u}_{k}^{u}(t)=\bar{u}_{k},\; \lim_{t\to \infty}
\tilde{u}_{k}^{s}(t)=\bar{u}_{k},\; k=1,...,n-1 $$

\underline{\bf Theorem 3.3.} The curves $u_{kk+1}^{h}$, $k=1,2,...,n$, $t\in {\bf R}$
given by:
\begin{equation}
u_{kk+1}^{h}=\frac{\tilde{u}_{k}^{s} (t)}{\tilde{u}_{k+1}^{s} (t)}\bar{u}_{k+1}+
\frac{\tilde{u}_{k+1}^{u} (t)}{\tilde{u}_{k}^{u} (t)}\bar{u}_{k},\; t\in {\bf R}
\end{equation}
are policycles for the uniphase steady solution, that is
$$\lim_{t\to \infty} u^{h}_{kk+1} (t)=\bar{u}_{k}$$

\underline{\it Proof: } From ({\bf 3}.12) and ({\bf 3}.13) we have:
$$\lim_{t\to \infty} u^{h}_{kk+1} (t)=\frac{\bar{u}_{k}}{\bar{u}_{k+1}}
\bar{u}_{k+1}+\bar{u}_{k}\lim_{t\to \infty} \frac{\tilde{u}_{k+1}^{u}(t)}
{\tilde{u}_{k}^{u}(t)}$$

From ({\bf 3}.11) yields:
$$\lim_{t\to \infty} \frac{\tilde{u}_{k+1}^{u}(t)}{\tilde{u}_{k}^{u}(t)}=
\lim_{t\to \infty} \frac{h_{1}(k+1)P+\displaystyle exp(t\lambda_{k+1}^{+})\sum_{j=1}^{n}
\eta_{j}exp(ia_{k+1}j)}{h_{1}(k)P+\displaystyle exp(t\lambda_{k}^{+})\sum_{j=1}^{n} \eta_{j}exp(ia_{k}j)}=$$
$$=\frac{\lambda^{+}_{k+1}}{\lambda^{+}_{k}} \lim_{t\to \infty} exp(t(\lambda_{k+1}^{+}-
\lambda^{+}_{k})) \frac{\displaystyle \sum_{j=1}^{n} \eta_{j}exp(ia_{k+1}j)}
{\displaystyle \sum_{j=1}^{n} \eta_{j}exp(ia_{k}j)}$$

From ({\bf 3}.10) we have $\lambda^{+}_{k+1}<\lambda^{+}_{k}$ and $\lambda^{-}_{k+1}
<\lambda^{-}_{k}$. It follows that
$$\lim_{t\to \infty} \frac{\tilde{u}_{k+1}^{h}(t)}{\tilde{u}_{k}^{h}(t)}=0$$

Thus we obtain:
$$\lim_{t\to \infty} u^{h}_{k+1} (t)=\bar{u}_{k}$$

Using ({\bf 3}.11) and ({\bf 3}.13) it follows:
$$\lim_{t\to -\infty} u^{h}_{kk+1} (t)=\bar{u}_{k+1}+\bar{u}_{k+1}
\lim_{t\to -\infty} \frac{\tilde{u}_{k}^{s}(t)}{\tilde{u}_{k+1}^{s}(t)}$$

From ({\bf 3}.12) yields:
$$\lim_{t\to -\infty} \frac{\tilde{u}_{k}^{s}(t)}{\tilde{u}_{k+1}^{s}(t)}=
\frac{\lambda^{-}_{k}}{\lambda^{-}_{k+1}} \lim_{t\to -\infty} exp\; (t(\lambda^{-}_{k}
-\lambda^{-}_{k+1})\cdot $$
$$\cdot \frac{\displaystyle \sum_{j=1}^{n} \eta_{j}exp(ia_{k}j)}{\displaystyle 
\sum_{j=1}^{n} \eta_{j}exp(ia_{k+1}j)}=0$$

Thus we obtain:
$$\lim_{t\to -\infty} u_{kk+1}^{h}=\bar{u}_{k+1}$$

The curves ({\bf 3}.13) are a polycycle considering:
$$u_{n+1}=u_{1}$$

\underline{\bf Theorem 3.4.} If $P\in(\underline{\alpha},\bar{\alpha})$ or
$P\in(\underline{\beta},\bar{\beta})$ and $\varepsilon =0$ the curves:
\begin{equation}
u_{k}^{c}(t)=\bar{u}_{k}+exp\; (\lambda^{c}t)\sum_{j=1}^{n} \eta_{j}exp\; (ia_{k}j),
\; k=1,...,n,\; t\in {\bf R}
\end{equation}
are periodic, with the period $\displaystyle T_{k}=\frac{\pi}{n\sqrt{\tau}} sec\;
\frac{k\pi}{2n}$, $k=1,...,n$, where $\lambda^{c}=\pm 2in\sqrt{\tau} sin\; \displaystyle
\frac{k\pi}{2n}$.

\underline{\it Proof: } For $\varepsilon =0$ the equation ({\bf 3}.7) is:
$$\lambda^{2}-\mu_{k}n^{2}\tau =0,\; k=1,...,n-1$$
with the roots:
$$\lambda^{c}=\pm 2in\sqrt{\tau} sin \; \frac{k\pi}{2n}$$

The uniphase steady solution $\bar{u}_{k}=kh_{1}P$ is center for the system ({\bf 2}.5)
with $\varepsilon=0$. From the condition $u_{k}^{c}(t)=u_{k+1}^{c}(t+T_{k})$,
$k=1,...,n-1$, it follows that $exp\; (\lambda^{c}(t+T_{k}))=exp\; (\lambda^{c}(t)$.
Hence: $2n\sqrt{\tau} sin \displaystyle \frac{k\pi}{2n} T_{k}=2\pi$ or
$$T_{k}=\frac{\pi}{n\sqrt{\tau}} cosec \frac{k\pi}{2n}$$

We continue our study with a multiphase steady solution $\bar{u}\in E^{+}$. To
be more specific we will consider only 2-phase steady solutions. From {\it Lemma
2.1.} it follows that for $\alpha \in (\underline{\alpha},\bar{\alpha})$, $\beta
\in (\underline{\beta},\bar{\beta})$, with $I(\alpha)=\{ i\in \{ 1,2,...,n\} $,
$\Delta \bar{u}_{i}=\alpha \} $, $I(\beta)=\{ i\in \{ 1,...,n\} ,\Delta \bar{u}_{i}
=\beta \} $ we have
$$I(\alpha) \cup I(\beta)=\{ 1,...,n\}$$
$$card\; I(\alpha)\cdot \alpha +(n-card\; I(\beta))\beta =nP$$

Let consider the following sets:
$$J(\alpha)=\{ i\in I(\alpha):\; i+1\in I(\alpha) \} $$
$$J(\beta)=\{ i\in I(\beta):\; i+1\in I(\beta) \} $$
and
$$K(\alpha ,\beta)=\{ i\in I(\alpha), \; i+1\in I(\beta) \} $$

From ({\bf 3}.5) we obtain:
\begin{equation}
\lambda^{2}+4\varepsilon n^{2}sin^{2}\frac{a_{p}}{2}\lambda +4n^{2}sin^{2}\frac{a_{p}}
{2}\tau =0, \; p\in J(\alpha)
\end{equation}
\begin{equation}
\lambda^{2}+4\varepsilon n^{2}\lambda sin^{2}\frac{a_{p}}{2}+4\rho n^{2}sin^{2}
\frac{a_{p}}{2}\tau =0, \; p\in J(\beta)
\end{equation}
\begin{equation}
\lambda^{2}+4\varepsilon n^{2}\lambda sin^{2}\frac{a_{p}}{2}-n^{2}[\rho (exp\; ia_{p-1})+
\tau (exp\; (-ia_{p})-1 ]=0,\; p\in K(\alpha ,\beta)
\end{equation}
where $\tau=\sigma^{\prime}(\alpha)$, $\rho=\sigma^{\prime}(\beta)$, $\rho >\tau$.

\underline{\bf Lemma 3.1.}

a). The equations ({\bf 3}.15) have complex roots with negative real part if and only if:
\begin{equation}
\varepsilon >0 \; \mbox{and} \; \varepsilon <\frac{\sqrt{\tau}}{\displaystyle 
nsin\frac{a_{r}}{2}},\; r=max\; J(\alpha)
\end{equation}

The equations ({\bf 3}.15) have real negative roots if and only if:
\begin{equation}
\varepsilon >0 \; \mbox{and} \; \varepsilon >\frac{\sqrt{\tau}}{\displaystyle
nsin\frac{a_{q}}{2}},\; q=min\; J(\alpha)
\end{equation}

b). The equations ({\bf 3}.16) have complex roots with negative real part if and only if:
\begin{equation}
\varepsilon >0 \; \mbox{and} \; \varepsilon <\frac{\sqrt{\rho}}{\displaystyle
nsin\frac{a_{r}}{2}},\; r=max\; J(\beta)
\end{equation}

The equations ({\bf 3}.16) have real negative roots if and only if:
\begin{equation}
\varepsilon >0 \; \mbox{and} \; \varepsilon >\frac{\sqrt{\rho}}{\displaystyle
nsin\frac{a_{q}}{2}},\; q=min\; J(\beta)
\end{equation}

c). The equations ({\bf 3}.17) have complex roots with negative real part if and only if:
\begin{equation}
\varepsilon >0 \;\frac{\sqrt{\rho+\tau-\sqrt{\Delta}}}{\displaystyle nsin\frac{a_{q}}
{2}}< \varepsilon <\frac{\sqrt{\rho+\tau+\sqrt{\Delta}}}{\displaystyle nsin\frac{a_{r}}{2}}
\end{equation}
$$q=min\; K(\alpha,\beta)),\; r=max\; K(\alpha,\beta))$$
where
$$\Delta =(\rho +\tau)^{2}-(\rho-\tau)^{2}cos^{2}\frac{a_{p}}{2}$$

\underline{\it Proof: } From the equations ({\bf 3}.15) and ({\bf 3}.16) it follows
directly a). and b).

The equations ({\bf 3}.17) may be written in the following form:
\begin{equation}
\lambda^{2}+4\varepsilon n^{2} sin^{2} \frac{a_{p}}{2}\lambda +2n^{2}(\rho +\tau)
sin^{2}\frac{a_{p}}{2}-in^{2}(\rho -\tau)sin\; a_{p}=0, p\in K(\alpha,\beta)
\end{equation}

The necessary and sufficient condition for these equations to admit complex roots
with negative real part is:
\begin{equation}
2n^{2}(\rho+\tau)sin^{2}\frac{a_{p}}{2} >\frac{16\varepsilon n^{4}sin^{4} \displaystyle
\frac{a_{p}}{2}}{8}+2\frac{n^{4}(\rho-\tau)^{2}sin^{2}a_{p}}{16\varepsilon^{2}n^{4}
\displaystyle sin^{4}\frac{a_{p}}{2}}
\end{equation}
and
$$4\varepsilon n^{2}sin^{2}\frac{a_{p}}{2}>0,\; 2n^{2}(\rho+\tau)sin^{2}\frac{a_{p}}{2}>0$$

\underline{\bf Theorem 3.5.} The 2-phase steady solution is stable if one of the
following conditions holds:

a).
$$\frac{\sqrt{\rho+\tau-\sqrt{\Delta}}}{\displaystyle nsin\frac{a_{q}}
{2}}< \varepsilon <\frac{\sqrt{\tau}}{n}, \; q=min\; K(\alpha,\beta)) $$

b).
$$\frac{\sqrt{\rho+\tau-\sqrt{\Delta}}}{\displaystyle nsin\frac{\pi}
{2n}}< \varepsilon <\frac{\sqrt{\rho+\tau+\sqrt{\Delta}}}{\displaystyle nsin\frac{a_{r}}{2}},
\; r=max\; K(\alpha,\beta)) $$

c).
$$max\left( \frac{\rho}{\displaystyle nsin\frac{a_{q_{2}}}{2}},\frac{\sqrt{\rho+\tau-
\sqrt{\Delta}}}{\displaystyle nsin\frac{a_{q_{3}}}{2}}\right) <\varepsilon <
min\left( \frac{\sqrt{\tau}}{\displaystyle nsin\frac{a_{r_{1}}}{2}},\frac{\sqrt{\rho+
\tau+\sqrt{\Delta}}}{\displaystyle nsin\frac{a_{r_{3}}}{2}}\right) $$
$$q_{2}=min\; J(\beta),\; q_{3}=min\; K(\alpha,\beta))$$
$$r_{1}=max\; J(\alpha),\; r_{3}=max\; K(\alpha,\beta))$$

d).
$$max\left( \frac{\sqrt{\tau}}{\displaystyle nsin\frac{a_{q_{1}}}{2}},\frac{\sqrt{\rho+\tau-
\sqrt{\Delta}}}{\displaystyle nsin\frac{a_{q_{3}}}{2}}\right) <\varepsilon <
min\left( \frac{\rho}{\displaystyle nsin\frac{a_{r_{2}}}{2}},\frac{\sqrt{\rho+
\tau+\sqrt{\Delta}}}{\displaystyle nsin\frac{a_{r_{3}}}{2}}\right) $$
$$q_{1}=min\; J(\alpha),\; q_{3}=min\; K(\alpha,\beta))$$
$$r_{2}=max\; J(\beta),\; r_{3}=max\; K(\alpha,\beta))$$

The proof is a direct consequence of the {\it Lemma 3.1.}

\underline{\bf Theorem 3.6.} If $\varepsilon =0$ the curves
$$u^{c}_{p}(t)=\bar{u}_{p}+exp\; (\lambda^{c}_{p}t) \sum_{k=1}^{n} \eta_{k}exp\;
(ia_{p}k), \; p\in J(\alpha)$$
$$u^{c}_{p}(t)=\bar{u}_{p}+exp\; (\lambda^{c}_{p}t) \sum_{k=1}^{n} \eta_{k}exp\;
(ia_{p}k), \; p\in J(\beta)$$
where $\lambda^{c}_{p},\; p\in J(\alpha)$ respectively $\lambda^{c}_{p},\; p\in J(\beta)$
are roots of the equations:
$$\lambda^{2}+4n^{2}sin^{2}\frac{a_{p}}{2}\tau =0,\; p\in J(\alpha) $$
$$\lambda^{2}+4n^{2}sin^{2}\frac{a_{p}}{2}\rho =0,\; p\in J(\beta) $$
are periodic with the period
$$T_{p}=\frac{\pi}{n\sqrt{\tau}} cosec \frac{p\pi}{2n}, \; p\in J(\alpha)$$
respectively
$$T_{p}=\frac{\pi}{n\sqrt{\rho}} cosec \frac{p\pi}{2n}, \; p\in J(\beta)$$

The proof follows from the above theorem.

\underline{\bf Remark.} For $\varepsilon<0$ the equations ({\bf 3}.15) and ({\bf 3}.16)
have positive roots and it follows that the 2-phase steady solution is unstable.

For $\varepsilon<0$ or $\varepsilon$ not satisfying one of the above conditions a).,
b)., c)., the roots of the equations ({\bf 4}.10), ({\bf 4}.11), ({\bf 4}.12)
may have positive or negative real part in the case they are complex, or they
are positive or negative real numbers. In this case the steady 2-phase solution
is unstable; it is hyperbolic saddle. The associated linear manifolds may be described
completely in the same manner we have done for the uniphase steady solutions.

\setcounter{equation}{0}
\begin{center}
\Large \bf 4. The discretization of the equation $u_{tt}=(\sigma (u_{x}))_{x}+
\varepsilon u_{xxt}$
\end{center}

\vspace*{.5cm}

Let $x_{k}=kh_{1}$, $k=1,...,n$ the division points of the interval $[0,1]$ with
$h_{1}=1/n$ and $t_{p}=ph_{2}$, $p=1,...,m$ the division points of the interval
$[0,1]$, $h_{2}=1/n$.

We$^{\prime}$ll approximate the derivates $\dot{u}_{k}(t)$, $\ddot{u}_{k}(t)$ by
$\displaystyle \frac{1}{h_{2}}[u_{k}^{p}-u_{k}^{p-1}]$ and $\displaystyle \frac{1}
{h_{2}^{2}}[u_{k}^{p+1}-2u_{k}^{p}+u_{k}^{p-1}]$, where $u_{k}^{p}=u(kh_{1},ph_{2})$.

We call the discrete system associated to the equation ({\bf 1}.1) the following
system:
$$\frac{1}{h_{2}^{2}} [u_{k}^{p+1}-2u_{k}^{p}+u_{k}^{p-1}]-\frac{1}{h_{1}} \sigma
\left( \frac{1}{h_{1}} (u_{k+1}^{p}-u_{k}^{p})-\frac{1}{h_{1}} \sigma (\frac{1}{h_{1}}
(u_{k}^{p}-u_{k-1}^{p})\right) =$$
\begin{equation}
=\frac{\varepsilon}{h_{1}^{2}h_{2}} [u_{k+1}^{p+1}-u_{k+1}^{p}-2(u_{k}^{p+1}-
u_{k}^{p})+u_{k-1}^{p+1}-u_{k-1}^{p}]
\end{equation}
$\hspace*{5cm} k=1,...,n-1,\; p=1,...,m-1$

The corresponding boundary conditions are:
\begin{equation}
u_{0}^{p}=0,\; u_{n}^{p}=P>0,\; p=1,...,m
\end{equation}

The system ({\bf 4}.1) represent the discrete Euler-Lagrange equations for the discrete
Lagrange function:
\begin{equation}
L(u_{k-1}^{p},u_{k}^{p},u_{k}^{p+1})=\frac{1}{2h_{2}^{2}} (u_{k}^{p+1}-u_{k}^{p})^{2}+
w(\frac{1}{h_{1}} (u_{k}^{p}-u_{k-1}^{p}))
\end{equation}
where $w:{\bf R}\to {\bf R}$, $w^{\prime}(\xi)=\sigma(\xi)$, and the dispersive term is:
$$\frac{\varepsilon}{h_{1}^{2}h_{2}} [u_{k+1}^{p+1}-u_{k+1}^{p}-2(u_{k}^{p+1}-
u_{k}^{p})+u_{k-1}^{p+1}-u_{k-1}^{p}]$$

For ({\bf 4}.1) it follows that the steady solutions $\bar{u}=\bar{u}_{k}$ satisfy:
\begin{equation}
\sigma (\frac{1}{h_{1}} (\bar{u}_{k+1}^{p}-\bar{u}_{k}^{p}))=\sigma (\frac{1}{h_{1}}
(\bar{u}_{k}^{p}-\bar{u}_{k-1}^{p})),\; k=1,...,n-1
\end{equation}

From ({\bf 4}.2) we obtain:
$$\sum_{k=1}^{n} \frac{1}{h_{1}} (\bar{u}_{k}^{p}-\bar{u}_{k-1}^{p})=\frac{1}{h_{1}}P $$

Thus a steady solution $\bar{u}$ satisfies the following:
\begin{equation}
\bar{u}_{k}^{p}-\bar{u}_{k-1}^{p}=P-\bar{u}_{n-1}^{p}, \; \sigma (\frac{1}{h_{1}}
(\bar{u}_{k}^{p}-\bar{u}_{k-1}^{p}))=C,\; k=1,...,n,\; C\in {\bf R}
\end{equation}

\underline{\bf Definition 4.1.} A steady solution $\bar{u}=(\bar{u}_{k}^{p})$ with
$\bar{u}_{k}^{p}=kh_{1}P$, $k=1,...,n-1$, $p=1,...,m$ is called uniphase steady solution.

A steady solution $\bar{u}=(\bar{u}_{k}^{p})$ with the property $\bar{u}_{k}^{p}-
\bar{u}_{k-1}^{p}=h_{1}f(k)$, $k=1,...,n-1$ is called multiphase steady solution.

Let $E=\{ \bar{u}=(\bar{u}_{k}^{p}) \} $ the set of multiphase steady solution. We
consider the following sets:
$$E^{+}=\left\{ \bar{u}=(\bar{u}_{k}^{p}):\sigma^{\prime} (\frac{1}{h_{1}}(\bar{u}_{k}^{p}-
\bar{u}_{k-1}^{p}))>0,\; k=1,...,n-1 \right\}$$
$$E^{-}=\left\{ \bar{u}=(\bar{u}_{k}^{p}):\sigma^{\prime} (\frac{1}{h_{1}}(\bar{u}_{k}^{p}-
\bar{u}_{k-1}^{p}))>0,\; k=1,...,n-1 \right\}$$

\underline{\bf Theorem 4.1.} The system ({\bf 4}.1) has a finite number of multiphase
steady solutions.

Let $\bar{u}=(\bar{u}_{k}^{p})$ a steady solution of ({\bf 4}.1) and $\tilde{u}_{k}^{p}=
\bar{u}_{k}^{p}+\displaystyle \sum_{l=1}^{n} \eta_{l}w_{l}^{p}(k)$, $k=1,...,n-1$
the components of a vector with $\eta_{l} \in (-a,a)$.

The linearized system associated to the system ({\bf 4}.1) in a neighborghood
of the steady solution $\bar{u}$ is:
$$h_{1}^{2}(w_{l}^{p+1}-2w_{l}^{p}+w_{l}^{p-1})-h_{2}^{2}h_{1}^{2}\left[ 
\sigma^{\prime}(\frac{1}{h_{1}} (\bar{u}_{k+1}^{p}-\bar{u}_{k}^{p})(w_{l+1}^{p}
-w_{l}^{p})-\right. $$
$$\left. \sigma^{\prime}(\frac{1}{h_{1}} (\bar{u}_{k}^{p}-\bar{u}_{k-1}^{p})
(w_{l}^{p}-w_{l-1}^{p})\right] =$$
\begin{equation}
=\varepsilon h_{2} [w_{l+1}^{p+1}+w_{l+1}^{p}-2(w_{l}^{p+1}-w_{l}^{p})+w_{l-1}^{p+1}-
w_{l-1}^{p}]
\end{equation}
$\hspace*{5cm} p=1,...,m,\; l=1,...,n-1$

A solution of ({\bf 4}.6) has the form:
\begin{equation}
w_{l}^{p}=\lambda^{p} exp\; (ia_{k}l),\; a_{k}=\frac{4\pi}{n},\; k=1,...,n,\; \lambda \in {\bf C}
\end{equation}

Replaing in ({\bf 4}.6) we obtain:

$$h_{1}^{2}(\lambda^{2}-2\lambda+1)exp\; (ia_{k})-h_{2}^{2}h_{1}^{2}(exp\; 2ia_{k}-
exp\; ia_{k})-$$
\begin{equation}
-s_{k}\lambda(exp\; ia_{k}-1)-\varepsilon h_{2}(\lambda^{2}-\lambda)
(exp\; ia_{k}-1)^{2}=0
\end{equation}

$\hspace*{5cm} k=1,...,n$ \\
where
$$s_{k}=\sigma^{\prime} \left( \frac{1}{h_{1}} (\bar{u}_{k}^{p}-\bar{u}_{k-1}^{p})\right) $$

From ({\bf 4}.8) it follows:
$$\lambda^{2}[h_{1}^{2}-\varepsilon h_{2}(exp\; ia_{k}-1)^{2}]-\lambda[2h_{1}^{2}
exp\; ia_{k}+h_{2}^{2}h_{1}(s_{k+1} exp\; ia_{k})(exp\; ia_{k}-1) -$$
\begin{equation}
-s_{k}(exp\; ia_{k}-1)-h_{2}\varepsilon (exp\; ia_{k}-1)^{2}]+h_{1}^{2}exp\; ia_{k}=0
\end{equation}

$\hspace*{5cm} k=1,...,n$

For a uniphase steady solution $\bar{u}=(\bar{u}_{k}^{p})$, $\bar{u}_{k}^{p}=
kh_{1}P$ and $\tau=\sigma^{\prime} (P)>0$, the equations ({\bf 4}.9) become:
\begin{equation}
\lambda^{2}(h_{1}^{2}-\varepsilon h_{2}\mu_{k})-\lambda(2h_{1}^{2}-h_{2}^{2}
h_{1}\tau \mu_{k}-\varepsilon h_{2}\mu_{k})+h_{1}^{2}=0
\end{equation}
$\hspace*{5cm} k=1,...,n$\\
where
$$\mu_{k}=\frac{(exp\; (ia_{k}-1))^{2}}{exp\; (ia_{k}-1)} =-4sin^{2}\frac{k\pi}{2n}$$

We suppose for the rest of this section that $h_{1},h_{2}\in (0,1)$ and $\varepsilon
\neq \displaystyle \frac{h_{1}^{2}}{h_{2}\mu_{k}}$.

\underline{\bf Theorem 4.1.} The uniphase steady solution for the system ({\bf 4}.1)
is assimptotically stable if and only if the following condition hold:
\begin{equation}
\tau >\frac{h_{1}}{h_{2}^{2}sin^{2}\frac{\pi}{2n}},\; \varepsilon \in \left(
-\frac{h_{1}^{2}}{2h_{2}sin^{2}\frac{\pi}{2n}}+\frac{h_{1}h_{2}}{2}\tau,\infty \right)
\end{equation}

\underline{\it Proof: } The equations ({\bf 4}.10) have the roots in modulus less
than 1 if and only if ({\bf 4}.11) hold.

A necessary and sufficient condition for the $k$-th equation in ({\bf 4}.10) to
have the modulus of its roots less than one is:
$$\frac{h_{1}^{2}}{h_{1}^{2}-\varepsilon h_{2}\mu_{k}}-1<0$$
\begin{equation}
-1-\frac{h_{1}^{2}-\varepsilon h_{2}\mu_{k}}{h_{2}^{2}}<\frac{|2h_{1}^{2}+h_{1}^{2}h_{1}
\tau \mu_{k}-\varepsilon h_{2}\mu_{k}|}{h_{1}^{2}}<1+\frac{h_{1}^{2}-\varepsilon
h_{2}\mu_{k}}{h_{1}^{2}}
\end{equation}

The inequatities ({\bf 4}.12) hold if and only if
\begin{equation}
\tau > -\frac{4h_{1}}{h_{2}^{2}\mu_{k}},\; \varepsilon \in \left( \frac{2h_{1}^{2}}
{\mu_{k}h_{2}}+\frac{h_{1}h_{2}}{2} \tau, \infty \right)
\end{equation}

But $\mu_{k}>\mu_{k+1}$ so it follows that ({\bf 4}.13) hold for any $k=1,...,n$
if and only if ({\bf 4}.11) hold.

\underline{\bf Theorem 4.2.} The uniphase steady solution $\bar{u}=(\bar{u}_{k}^{p})$
of the system ({\bf 4}.1) is unstable if one of the following conditions hold:

a).
$$\tau < -\frac{h_{1}}{h_{1}^{2}sin^{2}\frac{\pi}{2n}},\; \varepsilon \in \left( 
-\infty ,-\frac{h_{1}^{2}}{4h_{2}sin^{2}\frac{\pi}{2n}}+\frac{h_{1}h_{2}}{2} \tau \right) $$

b).
$$\varepsilon \in \left( -\infty ,-\frac{h_{1}^{2}}{2h_{2}sin^{2}\frac{\pi}{2n}}-h_{1}h_{2} \tau \right) $$

c).
$$\tau > \frac{h_{1}}{h_{1}^{2}sin^{2}\frac{\pi}{2n}},\; \varepsilon \in \left( 
-\frac{h_{1}^{2}}{4h_{2}sin^{2}\frac{\pi}{2n}},-\frac{h_{1}^{2}}{4h_{2}sin^{2}
\frac{\pi}{2n}}+h_{1}h_{2}\tau \right) $$

\underline{\it Proof :} a). The equations ({\bf 4}.10) have the roots in modulus
greater than one only if a). hold. A necessary and sufficient condition for the
$k$-th equation from ({\bf 4}.9) to have roots in modulus greater than one is:
$$\frac{h_{1}^{2}-\varepsilon h_{2}\mu_{k}}{h_{1}^{2}}-1 <0$$

\begin{equation}
-1-\frac{h_{1}^{2}-\varepsilon h_{2}\mu_{k}}{h_{1}^{2}}<\frac{|2h_{1}^{2}+h_{2}^{2}h_{1}
\tau \mu_{k}-\varepsilon h_{2}\mu_{k}|}{h_{1}^{2}}<\frac{h_{1}^{2}-\varepsilon
h_{2}\mu_{k}}{h_{1}^{2}}+1
\end{equation}

From ({\bf 4}.14) we obtain
\begin{equation}
\tau < -\frac{4h_{1}}{h_{2}^{2}\mu_{k}},\; \varepsilon \in \left( -\infty ,\frac{2h_{1}^{2}}
{h_{2}\mu_{k}}+\frac{h_{1}h_{2}}{2} \tau \right)
\end{equation}

b)., c). can be prooved in an analogous manner.

\underline{\bf Theorem 4.3.} The uniphase steady solution $\bar{u}=(\bar{u}_{k}^{p})$
for the system ({\bf 4}.1) with $\varepsilon =0$ is unstable if
$$\tau <\frac{h_{1}}{h_{2}^{2}sin^{2}\frac{\pi}{2n}} $$

\underline{\it Proof :} For $\varepsilon=0$, the equation ({\bf 4}.10) become:
\begin{equation}
\lambda^{2}-\lambda (2+\frac{h_{2}^{2}}{h_{1}}\tau \mu_{k})+1=0 ,\; k=1,...,n
\end{equation}

From ({\bf 4}.16) we have $\lambda_{1}\lambda_{2}=1$ and $\displaystyle 
\lambda_{1}+\lambda_{2}=2+\frac{h_{2}^{2}}{h_{1}}\tau \mu_{k}$. A necessary and
sufficient condition for $k$-th equation from ({\bf 4}.16) to have one root greater
than one and the other less than one is:
$$\tau <\frac{h_{1}}{h_{2}^{2}sin^{2}\frac{\pi}{2n}}$$

\underline{\bf Proposition 4.1.} Let $u=\bar{u}_{k}^{p}$ a uniphase steady solution
for the system ({\bf 4}.1) which satisfies one of the condition b). or c). from the
theorem 4.2. and $\lambda_{k}^{+}$, $\lambda_{k}^{-}$ the roots of the $k$-th
equation ({\bf 4}.9) with $|\lambda_{k}^{+}|>1$ and $|\lambda_{k}^{-}|<1$.

The linear stable manifold is given by:
$$u_{k}^{sp}=kh_{1}P+(\lambda^{-}_{k})^{p} \sum_{l=1}^{n} \eta_{l}exp\; (ia_{k}l)$$
$\hspace*{5cm} p=1,...,m,\; k=1,...,n-1$

The linear unstable manifold is given by:
$$u_{k}^{up}=kh_{1}P+(\lambda^{+}_{k})^{p} \sum_{l=1}^{n} \eta_{l}exp\; (ia_{k}l)$$
$\hspace*{5cm} p=1,...,m,\; k=1,...,n-1$

In these conditions it follows that:
$$\lim_{p\to \infty} u_{k}^{sp}=kh_{1}P \; \mbox{and} \; \lim_{p\to -\infty} u_{k}^{up}=kh_{1}P$$


\begin{center}
\bf References
\end{center}

[1]. S. Bir\u{a}ua\c{s}, S. Balint, A. M. Balint - {\it On an infinite system
of differential equations related to the hydrodynamical shock problems},
Bulletins for Appl. Math. (BAM), 1260/96 (LXXX), pp. 247-259\\

[2]. J. M. Greenberg - {\it Continuum limite of discret gases}, Arch. Rat. Mech.
Anal. 105 (1989), pp. 367-376\\

[3]. P. D. Lax - {\it On dispersive difference schemes}, Phisica D, 1986, pp. 250-254\\

[4]. J. von Neumann - {\it Proposal and analysis of a numerical method for the
treatment of hydrodynamical shock problems}, Collected Work, vol. VI, pp. 361-379,
Pergamon, Oxford, 1963






\end{document}