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\markboth{\hfil Singularity Formation in Systems \hfil  
EJDE--1995/09}%
{EJDE--1995/09\hfil R. Saxton \& V. Vinod \hfil}
\newtheorem{theorem}{Theorem}[section] %(If you want theorem numbered
\newtheorem{lemma}{Lemma}[section]
\begin{document}
\ifx\Box\undefined \newcommand{\Box}{\diamondsuit}\fi
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations, 
Vol. {\bf 1995}(1995), No. 09, pp. 1--15.} \newline
ISSN 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113 }
 \vspace{\bigskipamount} \\
 Singularity Formation in Systems of Non-strictly Hyperbolic Equations
\thanks{ {\em 1991 Mathematics Subject Classifications:}
35L45, 35L65, 35L67, 35L80.\newline\indent
{\em Key words and phrases:} Finite time breakdown, non-strict 
hyperbolicity, linear degeneracy.
\newline\indent
\copyright 1995 Southwest Texas State University  and University of
North Texas.\newline\indent
Submitted June 12, 1995. Published June 28, 1995. } }
\date{}
\author{ R. Saxton \& V. Vinod}
\maketitle

\begin{abstract}
We analyze finite time singularity formation for two systems of
hyperbolic equations. Our results extend previous proofs of breakdown
concerning $2\times 2$ non-strictly hyperbolic systems to $n \times  n$
systems, and to a situation where, additionally, the condition of
genuine nonlinearity is violated throughout phase space. The systems
we consider include as special cases those examined by Keyfitz and
Kranzer and by Serre. They take the form
        $$ u_{t} + (\phi(u)u)_{x} = 0, $$
where $\phi$ is a scalar-valued function of the $n$-dimensional
vector $u$, and
        $$ u_{t}+\Lambda(u)u_{x} = 0, $$
under the assumption $\Lambda = \mbox{diag}\,
\{\lambda^{1},\ldots,\lambda^{n}\}$ with
 $\lambda^{i}=\lambda^{i}(u-u^{i})$, \newline where
$u-u^{i}\equiv\{u^{1},\ldots,u^{i-1},u^{i+1},\ldots,u^{n}\}$.
\end{abstract}

\newcommand{\nablu}{\nabla_{u}}
\newcommand{\nablv}{\nabla_{v}}

\newcommand{\eqn}[1]{\label{eq:{#1}}}
\newcommand{\re}[1]{(\ref{eq:{#1}})}

\newcommand{\ct}[1]{\cite{kn:{#1}}}
\newcommand{\by}[1]{\bibitem{kn:{#1}}}
\newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}}

\newcommand{\uu}{{\bf u}}
\newcommand{\vv}{{\bf v}}
\newcommand{\rr}{{\bf r}}
\newcommand{\FF}{{\cal F}}
\newcommand{\GG}{{\cal G}}
\newcommand{\HH}{{\cal H}}
\newcommand{\QQ}{{\cal Q}}
\newcommand{\ao}{{\alpha}_{0}(a)}

\section{Introduction}
In this paper we examine the formation of singularities in solutions  
to two  $n\times n$ systems. The first of these is a conservation law 
\be {\uu}_{t} + {\FF}_{x}({\uu})={\bf 0}  \eqn a  \ee
which has $\FF=\phi (\uu)\uu $, and so the two vector fields $\FF$  
and $\uu$ are parallel. We call this situation {\it radial}. The  
second system takes the form
\be \uu_{t}+\Lambda(\uu)\uu_{x}={\bf 0}. \eqn b \ee
Here $\Lambda$ is a matrix-valued function of $\uu$, $\Lambda =  
\mbox{diag}\,\{\lambda^{1}(\uu),\ldots ,\lambda^{n}(\uu)\}$. The fact 
that $\Lambda$ is diagonal leads to the consideration of $n$ weakly  
coupled equations, coupled through the dependence of the  
$\lambda_{i}\ ^{\rq}s.$ These dependencies will be given the more  
explicit form 
$\lambda^{i}=\lambda^{i}(\uu-u^{i})$, where  
$\uu-u^{i}\equiv\{u^{1},\ldots,u^{i-1},u^{i+1},\ldots,u^{n}\}$ which  
we term {\it quasi-orthogonal}. We examine two special cases of this.  

Each system has $n$ eigenvalues some of which become equal on a  
submanifold $\Sigma$ in phase space. They are therefore non-strictly  
hyperbolic.  The principal distinguishing feature of the two systems  
turns out to be that while in \re{a} finite time breakdown can never  
take place on $\Sigma$, in {\re b} this can only take place there.  
The $2\times 2$ counterpart of \re a has been studied from a related  
perspective to ours in \ct{KK}, while \re b has been considered via  
compensated compactness in \ct{Se}.
Our approach to system \re a in Section~2 is first to examine the  
structure of simple waves in the case of general $\phi$, then to  
construct an invariant in the case that $\phi (\uu)$ has the simple  
dependence $\phi = \chi({1\over 2}|\uu|^2)$, and exploit its  
properties for general initial data. This leads to an approach for  
general initial data and with a larger class of functions, $\phi$.
%\ct{KK}
In Section~3, we find a necessary condition for finite time breakdown  
of solutions to \re{b}, while in Section~4 we demonstrate that this  
does indeed take place in the $2\times 2$ case. The proof of this  
last result is somewhat different from previous $2\times 2$ breakdown  
results  (\ct{KM}, \ct{KT}, \ct{L}, \ct{M}).
Finally, in Section~5, we present some numerical results showing  
the singularity formation  in the equation of Section~4.

\section{Radial Flux  $n\times n$ Systems}
In this section we briefly examine the system of equations
\be {\uu}_{t} + {\FF}_{x}({\uu})={\bf 0}, \eqn 1 \ee
where the flux function $\FF (\uu)$ takes the particular form $\FF  
(\uu)= \phi (\uu) \uu.  $ Here $\phi :\Bbb R^{n}\rightarrow\Bbb R$,  
and the  flux lies parallel to the vector field $\uu$, so for convenience we  
call this a {\it radial} flux function. Setting ${\cal  
A}(\uu)=\nabla_{u}(\phi (\uu)\uu)$ gives
\be {\cal A}(\uu)=\uu\otimes\nabla_{u}\phi(\uu) + \phi(\uu){\bf  
I}. \eqn 2 \ee
The first term in \re 2 has rank one, which reduces the  
characteristic polynomial for ${\cal A}(\uu)$ to
\begin{eqnarray} |\lambda{\bf I}-{\cal A}(\uu)|&=&|(\lambda  
-\phi(\uu)){\bf I}- \uu\otimes\nabla_{u}\phi(\uu)|\nonumber\\
&=&(\lambda -\phi(\uu))^{n}-(\lambda  
-\phi(\uu))^{n-1}tr((\uu\otimes\nabla_{u}\phi(\uu))\nonumber\\
&=&(\lambda -\phi(\uu))^{n-1}(\lambda - \phi (\uu)-\uu  
.\nabla_{u}\phi (\uu)).  \eqn 3\end{eqnarray}     
Labeling the characteristic speeds by
\be
\lambda_{i}=\left\{\begin{array}{ll}
\phi(\uu),& 1\leq i\leq n-1,\\
\phi(\uu)+\uu .\nabla_{u}\phi(\uu),&i=n, \eqn 4\end{array}
\right. \ee
implies the corresponding right eigenvectors, $\rr_{i}$,  
satisfy\newline $(\phi(\uu){\bf I}-{\cal A}(\uu))\rr_{i} =  
-\uu\otimes\nabla_{u}\phi(\uu)\rr_{i} =-\uu(\nabla_{u}\phi.\rr_{i}),  
1\leq i\leq n-1,$  and\newline $((\phi(\uu)+\uu .\nabla_{u}\phi  
(\uu)){\bf I}-{\cal A}(\uu))\rr _{n} =(\uu .\nabla_{u}\phi (\uu){\bf  
I} -\uu\otimes \nabla_{u}\phi (\uu))\rr_{n} =(\uu .\nabla  
_{u}\phi)\rr_{n}-\uu (\nabla_{u}\phi .\rr_{n}).$ 

Consequently, for $1\leq i\leq n-1,$ the $\rr_{i}$'s can be chosen  
proportional to a set of mutually orthogonal vectors  
$\{(\nablu^{\perp}\phi)_{i}, 1\leq i\leq n-1\}\equiv  
\nablu\phi^{\perp}$ perpendicular to $\nablu\phi$. $\rr_{n}$ is  
proportional to $\uu$
unless $\uu .\nabla _{u}\phi=0$, in which case  
$\rr_{n}\in\nablu\phi^{\perp}$. 

 Similarly, one finds that the first $n-1$ left eigenvectors, ${\bf  
l}_{i}$, belong to the set $\uu^{\perp}$ and that ${\bf l}_{n}$ is  
proportional to $\nablu\phi$ or ${\bf l}_{n}\in\uu^{\perp}$ if $\uu  
.\nabla _{u}\phi=0$. The first $n-1$ characteristic fields satisfy  
$\rr_{i}.\nablu\lambda_{i}\propto({\nablu^{\perp}\phi})_{i}.\nablu
\phi=0$ and are linearly degenerate, (\ct {KK}), while the $n$th  
characteristic field satisfies  
$\rr_{n}.\nablu\lambda_{n}\linebreak\propto\uu.\nablu(\phi+\uu.\nablu 
\phi).$ Set $\Upsilon = \{{\bf u}\in \Bbb R^n,  
\uu.\nablu(\phi+\uu.\nablu\phi)=0\}.$ Transforming to polar  
coordinates in $\Bbb R^{n},$ with $ u_{1}=r{\rm cos}\theta_{1},  
u_{j}=r\prod_{k=1}^{j-1}{\rm sin}\theta_{k}{\rm cos}\theta_{j},  
\linebreak u_{n}=r\prod_{k=1}^{n-1}{\rm sin}\theta_{k},$ implies that  
$\rr_{n}.\nablu\lambda_{n}\propto  
r\partial_{r}(\phi+r\partial_{r}\phi)=r\partial_{r}^{2}(r\phi), $ and  
so the $n$th characteristic field is genuinely nonlinear only when  
this term is nonzero. 

By equation \re{4}, all eigenvalues of $\cal A$ are equal where   
$\uu.\nablu\phi(\uu)\equiv\linebreak r\phi_{r} =0.$ Following the  
terminology and notation of \ct {KK}, we set \linebreak $\Sigma =  
\{{\bf u}\in \Bbb R^{n}, \uu.\nablu\phi=0\}$ and observe that for  
$n=2$  the system loses strict hyperbolicity on $\Sigma$, strict  
hyperbolicity being defined through the presence of real, distinct  
eigenvalues (\ct {L2}). For $n>2$ the system becomes non-strictly  
hyperbolic everywhere since by \re {3} there are $n-1$ identical,  
real, eigenvalues for any $\uu$. Some details of the behavior of  
solutions lying in $\Sigma\cap\Upsilon$ and $\Sigma\cap{\bf  
C}\Upsilon$ can be found in \ct{S2}.  

In the following Lemma, we consider the behavior of simple wave  
solutions, (\ct J), to \re{1}.
 
\begin{lemma} Let $\uu\in C^{1}([0, T]; C^{1}(\Bbb R))$ be a solution  
to  \re{1} of the form $\uu (t,x)=\vv (\psi (t,x))$ where $\psi (x,t)$ is  
a scalar function of $t$ and $x$. Then given data $\psi_{0}(x)=\psi  
(0,x)$,  $||\psi_{x}||_{\infty}(t)\rightarrow\infty$ can occur in  
finite time only if there is a point $x$ where $\vv  
(\psi_{0}(x))\notin \Sigma\cup\Upsilon$.   
\end{lemma}
\paragraph{Proof} For such solutions, \re{1} reduces to
\be \vv_{\psi}\psi_{t}+\phi  
(\vv)\vv_{\psi}\psi_{x}+(\nablv\phi(\vv).\vv_{\psi})\psi_{x}\vv={\bf  
0} \eqn 5 \ee 
or
\be (\psi_{t}+\phi  
(\vv)\psi_{x})\vv_{\psi}+\psi_{x}\vv\otimes\nablv\phi(\vv)\vv_{\psi}= 
{\bf 0}. \eqn 6 \ee
Consequently $\vv_{\psi}$ is a right eigenvector of the matrix
\be {\cal A}(\vv)=\vv\otimes\nablv\phi(\vv)+\phi(\vv){\bf I} \eqn 7  
\ee
having eigenvalue $\lambda$ such that $\psi_{t}+\lambda\psi_{x}=0.$
Now using \re{4}, $\lambda$ takes on either the value $\phi(\vv)$  
with corresponding right eigenvectors  
$\vv_{\psi}\in\nablv\phi^{\perp}(\vv)$, or the value  
$\phi(\vv)+\vv.\nablv\phi(\vv)$ with eigenvector  
$\vv_{\psi}\propto\vv$. 

In the first case, because of the linear  
degeneracy, linear waves maintain $\phi(\vv)$ constant on the  
hypersurface $\nablv\phi^{\perp}(\vv)$ while preventing singularity  
formation. 

In the second case, $\phi(\vv)+\vv.\nablv\phi(\vv)$ remains constant  
in the radial, $\vv$, direction, however singularities may form in  
finite time provided both $\vv.\nablv\phi$ and  
$\vv.\nablv(\phi+\vv.\nablv\phi) $ are nonzero. 
This can be seen as follows. Suppose first that $\lambda = \phi$, and  
so $\vv_{\psi}\linebreak\in\nablv\phi^{\perp}(\vv)$. Then $\psi$, and  
consequently $\phi$, remains constant along the ({\it straight})  
characteristic ${dx(t)\over dt}=\phi (\vv (\psi (t, x(t)))).$  
Differentiating $\psi_{t}+\phi\psi_{x}=0$ with respect to $x$, gives  
$\psi_{tx}+\phi\psi_{xx}+\phi_{x}\psi_{x}=0.$ However  
$\phi_{x}=\nablv\phi.\vv_{\psi}\psi_{x}=0$ since  
$\vv_{\psi}\in\nablv\phi^{\perp}(\vv)$, and so $\psi_{x}$ can only  
evolve linearly along the characteristic. It is simple to show (eg.  
\ct{S2}) that no other derivatives can blow up either in this case.
Next suppose that $\lambda = \phi+\vv.\nablv\phi$. Then  
$\psi_{t}+(\phi+\vv.\nablv\phi)\psi_{x}=0$, and $\psi$, therefore  
$\phi+\vv.\nablv\phi$, remains constant along the (again, {\it  
straight}) characteristic ${dx\over dt}=\phi+\vv.\nablv\phi.$  
Differentiating with respect to $x$ gives  
$\psi_{tx}+(\phi+\vv.\nablv\phi)\psi_{xx}\linebreak+\vv_{\psi}.\nablv 
(\phi+\vv.\nablv\phi)\psi_{x}^{2}=0.$ Since all the terms in brackets  
depend only on $\psi$, these are constant on the characteristic, and  
finite time blow up of $\psi$ will depend (together with the sign of  
the derivative of the initial data, $\psi_{0x}$) on the last term  
being nonzero. However by equation \re{5} it follows that for this  
value of $\lambda$,  
$\vv_{\psi}(\vv.\nablv\phi)=\vv(\nablv\phi.\vv_{\psi})$. So  
$\vv_{\psi}$ is parallel to $\vv$  unless $\vv.\nablv\phi=0$, in  
which case $\vv$ lies in $\Sigma$ and then $\nablv\phi.\vv_{\psi}=0$,  
{\it ie.} either
$\nablv\phi={\bf 0}$ or 
$\vv_{\psi}\in\nablv\phi^{\perp}(\vv)$. If  
$\vv_{\psi}\in\nablv\phi^{\perp}(\vv)$ and 
$\vv\in\Sigma$, we can argue as in the previous paragraph to show no  
blow up occurs, and if $\nablv\phi={\bf 0}$, it is straightforward to  
show the same thing directly.
We now assume $\vv\notin\Sigma$. In this case, for nontrivial  
solutions, the coefficient of $\psi_{x}^{2}$ above will be nonzero  
whenever the term $\vv.\nablv(\phi+\vv.\nablv\phi)$ is nonzero, {\it  
ie.} $\vv\notin\Upsilon$. This is simply the condition for genuine  
nonlinearity of the {\it n}th characteristic field above. Blow up is  
therefore possible only in this case, details of which can be  
supplied using standard techniques, (\ct M).         \quad~$\Box$

%Simple ends here.

\paragraph{Remark.} It can be seen from the above that in the case  
when  
$\vv\in\Sigma$, then all {\it n} eigenvectors $\vv_{\psi}$ must lie 
in the $n-1$  dimensional hyperplane   
$\nablv\phi^{\perp}(\vv)$. However it remains possible to construct  
a basis of %solutions in $\Bbb R^{n}$ by using {\it generalized}  
eigenvectors and appropriate %modifications to the simple solution  
definition.
Now we consider the possibility of introducing more general data than  
that in the above Lemma. We will assume that here  
$\phi(\uu)=\chi({1\over 2}|\uu|^{2})$. Our approach will be to  
extract a scalar conservation law from \re 1. This provides an  
invariant which we use to examine breakdown of solutions. In fact  
since the term $\FF (\uu)= \chi({1\over 2}|\uu|^{2}) \uu  $ in \re{1}  
is now a gradient, $  \chi({1\over 2}|\uu|^{2})  
\uu=\nablu\Psi({1\over 2}|\uu|^{2})$ where $\Psi^{\rq}\equiv \chi$,  
there exists an entropy, $\eta = {1\over 2}|\uu|^{2}$, for \re 1  
together with an entropy flux, $\nu = \Psi - |\uu|^{2}\chi$, such  
that $\eta_{t}+\nu_{x}=0, (\ct {L2}).$ Instead we choose another pair  
$\eta$, $\nu$,  with a more convenient functional relation to deduce  
breakdown.
\newpage

\begin{lemma}Let $\uu\in C^{1}([0, T]; C^{1}(\Bbb R))$ be a solution  
to  \re{1}, with $\phi(\uu)=\chi({1\over 2}|\uu|^{2}).$ Then given data  
${\uu}_{0}(x)=\uu(0,x)$, $||\uu_{x}||_{\infty}(t)\rightarrow\infty$  
can occur in finite time if there is a point $x$ where  
$\uu_{0x}\notin\uu_{0}^{\perp}$ and  
$\uu_{0}\notin\Sigma\cup\Upsilon.$ In particular, this will occur if  
$(3 \chi ' +\chi ''|\uu_{0}|^{2})\uu_{0}.\uu_{0x}<0.$ 
\end{lemma}

\paragraph{Proof} We attempt to extract a scalar conservation law  
from \re{1}  having the form 
\be \eta_{t}+f_{x}(\eta)=0. \eqn8 \ee
In other words, we require that $\nu = f(\eta).$ Once this is done,  
establishing breakdown becomes straightforward. Assuming it is  
possible to derive \re{8} from \re{1}, then $\eta = \eta (\uu)$ and  
so \re{8} implies
\be \nablu\eta .\uu_{t} + f\rq(\eta)\nablu\eta.\uu_{x}= 0 \eqn9 \ee
or
\be \nablu\eta .(\uu_{t}+ f\rq(\eta)\uu_{x})= 0. \eqn{10} \ee
But \re{1} implies
\be \nablu\eta .(\uu_{t}+\nablu\FF\uu_{x})= 0, \eqn{11} \ee
and so \re{10} and \re{11} show
\be \nablu\eta\nablu\FF = \nablu\eta f\rq(\eta) \eqn{12} \ee
which means that $f\rq (\eta)$ is an eigenvalue $\lambda(\uu)$ of  
$\nablu\FF$ having left eigenvector $\nablu\eta$.
Now, since $f\rq(\eta) = \lambda(\uu)$, then
\be f''(\eta)\nablu\eta=\nablu\lambda \eqn{13} \ee
which implies that $\nablu\lambda$ is also a left eigenvector of  
$\nablu\FF$ unless $f''(\eta) =0,$ and then  
$\eta~=~f'^{-1}(\lambda(\uu)).$ 

By \re{13}, if $\rr$ and ${\bf l}$ are right and left eigenvectors  
corresponding to $\lambda$, then 
$$\rr .\nablu\lambda  =f''(\eta)\rr.\nablu\eta \propto 
f''(\eta)\ \rr.{\bf l}\,.$$
So $f''(\eta) =0 \Rightarrow\uu\in\Upsilon.$
(Note also that $\rr.{\bf l}=0\Rightarrow\uu\in\Sigma$.)
Setting $g=f'^{-1}$, \re8 together with $\eta = g(\lambda(\uu))$  
gives
\be g'(\lambda)\lambda_{t}+f'(\eta)g'(\lambda)\lambda_{x}=0 \eqn{14}  
\ee
or, since $f'(\eta)=\lambda$,
\be \lambda_{t}+\lambda\lambda_{x}=0. \eqn{15} \ee 

Now in the case ${\cal F}=\chi ({1\over 2}|\uu|^{2})\uu$, we have  
from \re{4} that
\be
\lambda_{i}=\left\{\begin{array}{ll}
\chi ({1\over 2}|\uu|^{2}),& 1\leq i\leq n-1,\\
\chi ({1\over 2}|\uu|^{2})+\chi '({1\over 2}|\uu|^{2})|\uu|^{2},  
&i=n, \eqn {16}\end{array}
\right. \ee
and corresponding left eigenvectors ${\bf l}_{i}$ lie in the set $\uu  
^{\perp}, 1\leq i \leq n-1$, or are proportional to $\nablu\phi =  
\chi '\uu$ for $i=n$, unless $\uu\in\Sigma,\ ie.\ \chi '\neq 0.$
For the above procedure to be possible for some $\lambda  
=\lambda_{i}$, we recall that $\nablu\lambda$ must be a left  
eigenvector corresponding to some eigenvector $\lambda$. Since by  
\re{16}, all the $\lambda_{i}$ have $\nablu\lambda_{i}$ proportional  
to $\uu$, then it becomes possible to proceed only using  
$\lambda_{n}$. This leads to the result
\be \lambda_{nt}+\lambda_{n}\lambda_{nx}=0  \eqn{17} \ee
with $\lambda_{n}$ given by \re{16}, which then implies (\ct{L2})  
that on the characteristic ${dx\over dt}=\lambda_{n}$,
\be \lambda_{nx}={\lambda_{n0x}\over 1+\lambda_{n0x}t} \eqn{18} \ee
where $\lambda_{n0}=\chi ({1\over 2}|{\uu}_{0}|^{2})+\chi '({1\over  
2}|{\uu}_{0}|^{2})|{\uu}_{0}|^{2}$ and ${\uu}_{0}(x)=\uu(0,x)$. So  
\newline $\lambda_{n0x}=(3 \chi ' +\chi  
''|\uu_{0}|^{2})\uu_{0}.\uu_{0x}$. However, recalling the definition  
of $\Upsilon$, genuine nonlinearity requires the expression  
$\uu.\nablu(\phi+\uu.\nablu\phi)$ to be nonzero. With $\phi(\uu)=\chi  
({1\over 2}|\uu|^{2})$ this implies $(3 \chi ' +\chi  
''|\uu|^{2})|\uu|^2\neq 0$. So, for $\uu_{0}\notin\Sigma\cup\Upsilon  
,\uu_{0x}\notin\uu_{0}^{\perp}$ then $\lambda_{n0x}\neq 0$, and for  
$(3 \chi ' +\chi ''|\uu_{0}|^{2})\uu_{0}.\uu_{0x}<0,$ then  
$\lambda_{n0x}<0$ and finite time breakdown follows from \re{18}.    
\quad~$\Box$  

With the previous Lemma as motivation, we turn to the final result of  
this section. This is to obtain more general conditions on $\phi$  
under which breakdown can take place for arbitrary data. $\uu$ will  
be represented in terms of polar coordinates, $\uu = (r,  
\theta_{1},\dots,\theta_{n-1}),\ r=|\uu|.$

\begin{theorem}Let $\uu\in C^{1}([0, T]; C^{1}(\Bbb R))$ be a  
solution to  
\re{1}, with $\phi(\uu)={\cal J}(r{\cal  
K}(\theta_{1},\dots,\theta_{n-1}))$, ${\cal J}\in C^{2}(\Bbb R),  
{\cal  K}\in C^{1}(\Bbb R^{n-1}).$ Then  
$||\uu_{x}||_{\infty}(t)\rightarrow\infty$ in finite time if there is  
a point $x$ where $(2{\cal J}'+{\cal J}''r{\cal K})(r{\cal K})_{x}<0$  
at $t=0.$
\end{theorem}
\paragraph{Proof}
As before, we attempt to construct a convenient scalar conservation  
law. Rather than working with \re{12} and general $\phi$, it turns  
out to be convenient to proceed as follows. Observe that the general  
form of equation \re{15} could, by \re{16}, have been replaced by an  
equation of the form
\be \phi_{t}+h(\phi)\phi_{x}=0 \eqn{20} \ee
for an appropriate function $h$, depending on the choice of  
$\lambda$. With this as a starting point, we attempt to find the most  
general conditions on $\phi(\uu)$ for which \re{20} can be derived  
for some function $h$. 

Now by \re{1},
\be \uu_{t}+\phi_{x}\uu +\phi\uu_{x}={\bf 0}.  \eqn{21} \ee
Taking the scalar product of \re{21} with $\nablu\phi$ gives
\be \phi_{t}+\phi_{x}(\uu.\nablu)\phi  +\phi\phi_{x}={\bf  
0}, \eqn{22} \ee
and for this to be of the form \re{20} requires that
\be (\uu.\nablu)\phi + \phi = h(\phi). \eqn{23} \ee
We therefore solve the equation
\be (\uu.\nablu)\phi = h(\phi)-\phi\equiv \GG(\phi). \eqn{24} \ee
Define a curve $\Gamma$ by $x=x(s), {d\uu\over d s}=\uu,  
x(0)=\gamma.$ Then on $\Gamma$ (consider $t$ here as a parameter),  
${d\phi\over d s}(\uu(t, x(s)))=({d\uu\over d s}.\nablu)\phi =  
(\uu.\nablu)\phi =\GG(\phi).$ Solving for $\uu$ on $\Gamma$ gives 
\be \uu (t, x(s)) = \uu (t, \gamma )e^s \eqn{25} \ee
where 
\be {d\phi\over d s}=\GG(\phi). \eqn{26} \ee
Integrating \re{26} gives
\be \HH(\phi(\uu(t, x(s))))=\HH(\phi(\uu(t, \gamma)))+s \eqn{27} \ee
where $\HH'\equiv 1/ \GG $. Combining \re{25} with \re{27},
\be \HH(\phi(\uu(t, x(s))))=\HH(\phi(\uu(t, x(s))e^{-s}))+s \eqn{28}  
\ee
implies, together with the result from \re{25} with $\uu$ expressed  
in polar coordinates that $r(t, x(s))=r(t, \gamma)e^{s},  
\theta_{i}(t, x(s))=\theta_{i}(t, \gamma), 1\leq i\leq {n-1}$,
\be  \HH(\phi(\uu(t, x(s))))=\HH(\phi(\uu(t, x(s)))r(t, \gamma)/r(t,  
x(s)))+\ln(r(t, x(s))/r(t, \gamma)), \eqn{29} \ee
or
\begin{eqnarray}  
\phi(r,\theta_{1},\dots,\theta_{n-1})&=&\HH^{-1}\circ(\HH\circ
\phi(r_{0},\theta_{1},\dots,\theta_{n-1})+\ln(r/r_{0}))\nonumber\\
&\equiv&{\cal J}(r{\cal K}(\theta_{1},\dots,\theta_{n-1})),
 \eqn{30}
\end{eqnarray}
where we have set $r(t, \gamma)=r_{0},\ {\cal J}={\cal  
H}^{-1}\circ\ln$, and ${\cal K}=1/r_{0}\exp{\cal H}\circ\phi.$
Taking ${\cal J}$ from \re{30} and using \re{24} gives
\begin{eqnarray} r{\partial\over\partial  
r}\phi=\GG(\phi)&&\nonumber\\
\Rightarrow {\cal J}'(r{\cal K}(\theta_{1},\dots,\theta_{n-1}))r{\cal  
K}(\theta_{1},\dots,\theta_{n-1})=&{\cal G}\circ{\cal J}(r{\cal  
K}(\theta_{1},\dots,\theta_{n-1}))& \eqn{31}
\end{eqnarray}
or
\be{\cal J}'(z)z= h({\cal J}(z))-{\cal J}(z) \eqn{32} \ee
which gives a functional relation between ${\cal J}$ and $h$. ${\cal  
K}$ is unconstrained. Thus we obtain a single conservation law of the  
form \re{20} provided $\phi$ has the structure given by \re{30}, and  
then from \re{32}, \re{20} becomes
\be {\cal J}_{t}+({\cal J}+{\cal J}'z){\cal J}_{x}=0,\ \ z=r{\cal  
K}. \eqn{33} \ee
Alternatively, multiplying by $({\cal J}+{\cal J}'z)'$ and dividing  
by ${\cal J}'$ ( $\neq 0$ if $\uu\notin\Sigma$)  gives
\be ({\cal J}+{\cal J}'z)_{t}+ ({\cal J}+{\cal J}'z)({\cal J}+{\cal  
J}'z)_{x}=0 \eqn{34} \ee
which implies (cp. \re{18})
$ ({\cal J}+{\cal J}'z)_{x}\rightarrow\infty$ in finite time provided  
\newline $({\cal J}+{\cal J}'z)_{x}<0$ at $t=0.$ 
The result follows. \quad~$\Box$

\section{Quasi-orthogonal  $n\times n$ Systems}
Here we consider systems of the form 
\be \uu_{t}+\Lambda(\uu)\uu_{x}={\bf 0}, \eqn{c} \ee
with 
\be \Lambda = \mbox{diag}\,\{\lambda^{1}(\uu -u^{1}),\ldots  
,\lambda^{n}(\uu  
-u^{n})\}, \eqn{cc} \ee
where
\be  \uu-u^{i}=\{u^{1},\ldots,u^{i-1},u^{i+1},\ldots,u^{n}\},\ 1\leq  
i\leq n. \eqn{35} \ee
For simplicity, we make the additional hypothesis that the  
$\lambda^{i}$ admit either the following additive structure
\be \lambda^{i}(\uu-u^{i})=\sigma(\uu)-\nu^{i}(u^{i}) \eqn{h} \ee
where
\be \sigma(\uu)=\sum_{j=1}^{n}\nu^{j}(u^{j}), \eqn{hh} \ee
or the multiplicative structure
\be \lambda^{i}(\uu-u^{i})=\prod^{n}_{j\neq  
i}\mu^{j}(u^{j}). \eqn{hhh} \ee
Since the eigenvalues of $\Lambda$ are  
$\lambda^{1},\dots,\lambda^{n}$, equality of any pair defines a  
(possibly {\it empty}) set $\Sigma$ where \re{c} becomes non-strictly  
hyperbolic. The component $u^{i}$ of $\uu$ remains constant on the  
$i$-th characteristic, $dx^{i}/dt = \lambda^{i}(\uu-u^{i}),\ 1\leq  
i\leq n$, and so there exist at least $n$ Riemann invariants for  
\re{c}. The $i$-th right eigenvector, $\rr_{i}$, satisfies  
$\rr_{i}\propto{\bf e}_{i}$ where the set $\{{\bf e}_{i},\ 1\leq  
i\leq n\}$ makes up the standard Cartesian basis for $\Bbb R^{n}$,  
therefore by \re{35} $\rr_{i}.\nablu\lambda^{i}=0,\ 1\leq i\leq n.$  
So the set $\Upsilon$ where the problem becomes linearly degenerate  
comprises the full phase space $\Bbb R^{n}$. 

\begin{lemma} Let $\Lambda$ be a $C^{1}$ function, $\Lambda  
:\Bbb R^{n}\rightarrow\Bbb R^{n^{2}}$, and let ${\bf u}(t,  
x)\linebreak
\in C^{1}([0, t^{*}); C^{1}(\Bbb R^{n}))$ be a solution to \re{c},  
with  
${\bf u}(t, 0)={\bf u}_{0}(x),\ x\in\Bbb R$, for some maximal  
$t^{*}$.  
Then, under either \re{h} with \re{hh}, or \re{hhh}, $t^{*}<\infty$  
if and only if ${\bf u}:\Bbb R^{n}-\Sigma\rightarrow\Sigma$, as a map  
from ${\bf u}_{0}\rightarrow{\bf u}(t, .)$.  In addition, ${\bf  
u}:\Sigma\rightarrow\Sigma$ on any interval of existence.
\end{lemma}
\paragraph{Proof}
Define the characteristic $\Gamma_{i}$ by $x^{i}=x^{i}(t),\  
{dx^{i}\over dt}=\lambda^{i}, x^{i}(0)=\alpha^{i}, \linebreak 1\leq  
i\leq n.$ Differentiation along $\Gamma_{i}$ will be written as  
$D_{i}\equiv\partial/\partial t+\lambda^{i}\partial/\partial x$, from  
which it is immediate by \re{c} that $D_{i}u^{i}=0,\ 1\leq i\leq n$,  
{\it ie.} $u^{i}(t, x^{i}(t))=u^{i}_{0}(\alpha^{i}),$ where  
$u^{i}_{0}(x)\equiv u^{i}(0, x).$ 

Differentiating \re{c} with respect to $x$ implies
\be D_{i}u^{i}_{x}+u^{i}_{x}\sum^{n}_{j\neq  
i}{\partial\lambda^{i}\over\partial u^{j}}u^{j}_{x} =0. \eqn{36} \ee
Also, for $i\neq j$,
\be  
D_{i}u^{j}=D_{j}u^{j}+(\lambda^{i}-\lambda^{j})u^{j}_{x}=(\lambda^{i} 
-\lambda^{j})u^{j}_{x}. \eqn{37} \ee
Consequently, unless $\lambda^{i}=\lambda^{j}$,
\be D_{i}u^{i}_{x}+u^{i}_{x}\sum^{n}_{j\neq  
i}{\partial\lambda^{i}\over\partial u^{j}}{D_{i}u^{j}\over  
\lambda^{i}-\lambda^{j}}=0. \eqn{38} \ee
Adopting the additive assumptions \re{h}, \re{hh} reduces equation  
\re{38} to
\be D_{i}u^{i}_{x}+u^{i}_{x}\sum^{n}_{j\neq  
i}\nu^{j}(u^{j})'{D_{i}u^{j}\over  
\nu^{j}(u^{j})-\nu^{i}(u^{i})}=0 \eqn{39} \ee
implying
\be D_{i}u^{i}_{x}+u^{i}_{x}D_{i}\sum^{n}_{j\neq i}\ln  
|\nu^{j}(u^{j})-\nu^{i}(u^{i})|=0 \eqn{40} \ee
or
\be D_{i}(u^{i}_{x}\prod^{n}_{j\neq  
i}|\nu^{j}(u^{j})-\nu^{i}(u^{i})|)=0. \eqn{41} \ee
The multiplicative condition \re{hhh} instead reduces \re{38} to
\be D_{i}u^{i}_{x}+u^{i}_{x}\sum^{n}_{j\neq i}({\partial\over\partial  
u^{j}}\prod^{n}_{k\neq  
i}\mu_{k}(u^{k})){D_{i}u^{j}\over\prod^{n}_{l\neq  
i}\mu_{l}(u^{l})-\prod^{n}_{m\neq j}\mu_{m}(u^{m})}=0 \eqn{42}  \ee
and so, on simplifying,
\be D_{i}u^{i}_{x}+u^{i}_{x}\sum^{n}_{j\neq  
i}\mu_{j}(u^{j})'{D_{i}u^{j}\over  
\mu_{j}(u^{j})-\mu_{i}(u^{i})} \eqn{43} \ee
which takes the same form as \re{39}. We therefore have, as with  
\re{41},
\be D_{i}(u^{i}_{x}\prod^{n}_{j\neq  
i}|\mu^{j}(u^{j})-\mu^{i}(u^{i})|)=0. \eqn{44} \ee
Thus, both sets of hypotheses stated lead to analogous results,  
namely
that on any characteristic, $\Gamma_{i}$, one obtains a relation of  
the form 
\be u^{i}_{x}\prod^{n}_{j\neq  
i}|\kappa^{j}(u^{j})-\kappa^{i}(u^{i})|=
 u^{i}_{0x}\prod^{n}_{j\neq  
i}|\kappa^{j}(u^{j}_{0})-\kappa^{i}(u^{i}_{0})|,\ (t, x)\in  
\Gamma_{i}, \eqn{45} \ee
where $\kappa^{i}$ represents either $\mu^{i}$ or $\nu^{i}$.  
Accordingly,  if  
$\kappa^{j}(u^{j}_{0}(\alpha^{i}))=\kappa^{i}(u^{i}_{0}(\alpha^{i}))$  
for some $j\neq i$, then $\kappa^{j}(u^{j}(t,  
x^{i}(t)))=\kappa^{i}(u^{i}(t, x^{i}(t)))$, $t\in (0, t^{*})$ for  
some $t^{*}>0$, by local continuity in time. On the other hand, if  
the right side of \re{45} is nonzero, then $u_{x}(t,  
x^{i}(t))\rightarrow\infty$ if ever $\kappa^{j}(u^{j}(t,  
x^{i}(t)))\rightarrow\kappa^{i}(u^{i}(t, x^{i}(t)))$ for some $j\neq  
i.$ Both sets of hypotheses allow this form of behavior only in  
$\Sigma$. If \re{h}, \re{hh} hold, then $\nu^{i}(u^{i})  
=\nu^{j}(u^{j}),\ j\neq i$, implies $\sigma({\bf u})-\lambda^{i}({\bf  
u}-u^{i})\linebreak=\sigma({\bf u})-\lambda^{j}({\bf u}-u^{j})$ , so  
$\lambda^{i}({\bf u}-u^{i})=\lambda^{j}({\bf u}-u^{j})$. If however  
\re{hhh} holds, then $\mu^{i}(u^{i})=\mu^{j}(u^{j})$. But  
$\lambda^{j}({\bf u}-u^{j})/\lambda^{i}({\bf  
u}-u^{i})=\prod^{n}_{l\neq j}\mu^{l}(u^{l})/ \prod^{n}_{k\neq  
i}\mu^{k}(u^{k})\linebreak=\mu^{i}(u^{i})/\mu^{j}(u^{j})$, and so  
again $\lambda^{i}({\bf u}-u^{i})=\lambda^{j}({\bf u}-u^{j})$. 
\quad~$\Box$
\paragraph{Remark.} It is possible to obtain analogous results to the  
above  
under other conditions than \re{h}-\re{hhh}. Either condition can  
however apply to the system considered in the next section, and so we  
do not generalize further here. 

\section{Quasi-orthogonal  $2\times 2$ Systems}
Next, we consider the system of equations (\ct{Se}),
\begin{eqnarray}
u_{t}+vu_{x}&=&0, \eqn{101}\\
v_{t}+uv_{x}&=&0. \eqn{102}
\end{eqnarray}
In the following, we let $\Gamma$ denote the $v$-characteristic,  
defined by
\be {dx\over dt}(t, \alpha)=v(t, x(t, \alpha)), \eqn{103} \ee
where $\alpha$ is a Lagrangian coordinate, and
\be x(0, \alpha)=\alpha . \eqn{104} \ee
\begin{theorem} Let $(u, v)(t, x)\in C^{1}([0, t^{*}); C^{1}(\Bbb  
R))$ be a  
solution to \re{101}, \re{102}, for some maximal $t^{*}$. Then $(u,  
v)(t, .):\Bbb R^{2}-\Sigma\rightarrow\Sigma$ as $t\rightarrow  
t^{*}<\infty$ whenever $u_{0}'<0$ or $v_{0}'<0.$   
\end{theorem}
\paragraph{Proof}
Equation \re{101} implies that on $\Gamma$,
\be u(t, x(t, \alpha))=u_{0}(\alpha). \eqn{107} \ee
Now, from \re{102},
\be v_{t}+vv_{x}=(v-u)v_{x}, \eqn{111} \ee
and differentiating \re{101},
\be u_{tx}+vu_{xx}=-u_{x}v_{x}. \eqn{112} \ee
So \re{111} and \re{112} together give
\be (v-u)(u_{tx}+vu_{xx})+(v_{t}+vv_{x})u_{x}=0, \eqn{113} \ee
which reduces to
\be {d\over dt}((v-u)u_{x})=0, \eqn{114} \ee
where
\be {d\over dt}\equiv D_{1}={\partial\over\partial t}  
+v{\partial\over\partial x} \eqn{115} \ee
and we have used \re{107}.
As a result of \re{114}, then
\be (v(t, x(t, \alpha))-u_{0}(\alpha))u_{x}(t, x(t, \alpha))
=(v_{0}(\alpha)-u_{0}(\alpha))u_{0}'(\alpha). \eqn{116} \ee 


Now by \re{103}
\be {dx\over dt}=v\Rightarrow {dx_{\alpha}\over  
dt}=v_{x}x_{\alpha}\Rightarrow {d\ln|x_{\alpha}|\over  
dt}=v_{x} \eqn{117} \ee
and by \re{112},
\be u_{tx}+vu_{xx}=-u_{x}v_{x}\Rightarrow {d\ln|u_{x}|\over  
dt}=-v_{x}. \eqn{118} \ee
\re{117}, \re{118} therefore show
\be {d\ln|u_{x}|\over d\ln|x_{\alpha}|}=-1,\ (t, x)\in\Gamma,  
\eqn{119} \ee
from which it follows easily that
\be |u_{x}|\rightarrow\infty\ {\rm as}\ |x_{\alpha}|\rightarrow  
0 \eqn{120} \ee
since \re{119} implies
\be \int_{u_{0}'(\alpha)}^{u_{x}(t, x(t, \alpha))}d\ln |u_{x}| =  
-\int_{\alpha}^{x(t, \alpha)}d\ln|x_{\alpha}|, \eqn{121} \ee
and so
\be u_{x}(t, x(t, \alpha))=u_{0}'(\alpha)x_{\alpha}^{-1}(t,  
\alpha). \eqn{123} \ee
Here we have used continuity in time of the {\it local} initial value  
problem and \re{104} to remove the absolute value signs.
Together with \re{116}, \re{123} also gives
\be v(t, x(t,  
\alpha))-u_{0}(\alpha)=(v_{0}(\alpha)-u_{0}(\alpha))x_{\alpha}(t,  
\alpha). \eqn{124} \ee
Next, using \re{103}, \re{107} and \re{124}, we obtain
\be x_{t}+(u_{0}-v_{0})x_{\alpha}=u_{0}, \eqn{125} \ee
a linear, non-constant coefficient equation for $x(t, \alpha)$.
Introducing a second coordinate, $a$, for $(t, \alpha)$ space, such  
that
\be {d\alpha\over dt}(t, a)=u_{0}(\alpha(t, a))-v_{0}(\alpha(t, a))  
\equiv  
w_{0}(\alpha(t, a)),           \eqn{128} \ee
with $\alpha(0, a)\equiv\ao$, and denoting
\be {\cal D} ={\partial\over\partial t}  
+w_{0}{\partial\over\partial\alpha}, \eqn{129} \ee \re{125} then implies that
\be {\cal D}x(t, \alpha(t, a))=u_{0}(\alpha(t, a)), \eqn{130} \ee
where $x(0, \alpha(0, a))=\ao$.
Since initial data lie in $\Bbb R^{2}-\Sigma$, therefore \linebreak  
$w_{0}(\ao)\neq 0$ and \re{128} gives
\be \QQ(\alpha(t, a))-\QQ(\ao)\equiv\int_{\ao}^{\alpha (t,  
a)}{d\alpha\over  
w_{0}(\alpha)}=t \eqn{131} \ee
where $\QQ'(\alpha)\equiv1/w_{0}(\alpha)$. So provided $w_{0}(\alpha  
(t,  a))\neq 0$, 
\be \alpha(t, a)=\QQ^{-1}(\QQ(\ao)+t). \eqn{134} \ee
By \re{130}, then
\be {\cal D}x(t, \alpha(t, a))=u_{0}(\alpha(t,  
a))=u_{0}(\QQ^{-1}(\QQ(\ao)+t)). \eqn{135} \ee
If we now define a Lagrangian variable $X(t, a)$ by
\be X(t, a)=x(t, \alpha(t, a)),\ X(0, a)=\ao , \eqn{135.5} \ee
then $X_{t}={\cal D}x$ by \re{129}, and
\be X_{t}(t, a)=u_{0}(\alpha(t,  
a))=u_{0}(\QQ^{-1}(\QQ(\ao)+t)) \eqn{135.7} \ee
implies
\be X(t, a)=\ao + {\cal S}_{0}(\QQ(\ao)+t)-{\cal  
S}_{0}(\QQ(\ao)) \eqn{136} \ee
where ${\cal S}_{0}'=u_{0}\circ\QQ^{-1}.$ As a result, using  
\re{134}, \re{135.5} and \re{136},
\be x(t, \QQ^{-1}(\QQ(\ao)+t))=\ao + {\cal S}_{0}(\QQ(\ao)+t)-{\cal  
S}_{0}(\QQ(\ao)), \eqn{138} \ee 
or, since \re{134} implies $\QQ(\ao)=\QQ(\alpha(t, a))-t$, then  
\re{138}  reads
\be x(t, \alpha)=\QQ^{-1}(\QQ(\alpha)-t)+{\cal  
S}_{0}(\QQ(\alpha))-{\cal  
S}_{0}(\QQ(\alpha)-t). \eqn{140} \ee
In particular, on differentiating \re{140},
\be x_{\alpha}(t,  
\alpha)={\QQ'(\alpha)\over\QQ'(\QQ^{-1}(\QQ(\alpha)-t))}+{\cal  
S}_{0}'(\QQ(\alpha))\QQ'(\alpha)-{\cal  
S}_{0}'(\QQ(\alpha)-t)\QQ'(\alpha), \eqn{141} \ee
and so, since ${\cal S}_{0}'=u_{0}\circ\QQ^{-1},\ \QQ'=1/w_{0},$  
by means of \re{128}
\begin{eqnarray} x_{\alpha}(t, \alpha)&=&{1\over  
w_{0}(\alpha)}(w_{0}(\QQ^{-1}(\QQ(\alpha)-t))+u_{0}(\alpha)-u_{0}(\QQ 
^{-1}(\QQ 
(\alpha)-t)))\nonumber\\
&=&{u_{0}(\alpha)-v_{0}(\QQ^{-1}(\QQ(\alpha)-t))\over  
u_{0}(\alpha)-v_{0}(\alpha)}.
 \eqn{142}\end{eqnarray}
 This then implies breakdown, by \re{123}, provided there exists some  
positive time, $t$, at which  
$u_{0}(\alpha)=v_{0}(\QQ^{-1}(\QQ(\alpha)-t))$, {\it ie.} provided  
$t=  \QQ(\alpha)-\QQ(v_{0}^{-1}(u_{0}(\alpha)))>0$, if $v_{0}$ possesses a  
local inverse. Since $\QQ'=1/w_{0}$, then $\QQ(\alpha)$ is locally  
increasing if $u_{0}(\alpha)>v_{0}(\alpha)$ and locally decreasing if  
$u_{0}(\alpha)\linebreak <v_{0}(\alpha)$. It is an elementary  
exercise to show that this is consistent with $t>0$ only if  
$v_{0}'(\alpha)<0$.
 Then $t^{*}=inf_{\alpha}\ t.$
Interchanging $u$ and $v$ in the above proof gives the result stated  
in the Theorem, with $t^{*}$ the infimum, over $\alpha$, of all $t>0$  
constructed as above.
\quad~$\Box$

\paragraph{Remark.} Recalling \re{124}, which can be written
\be x_{\alpha}(t, \alpha)={u_{0}(\alpha)-v(t, x(t, \alpha))\over  
u_{0}(\alpha)-v_{0}(\alpha)}, \eqn{143} \ee
and comparing \re{142} with \re{143} shows that $v$ evolves along  
$\Gamma$ as
\be v(t, x(t, \alpha))=v_{0}(\QQ^{-1}(\QQ(\alpha)-t)). \eqn{144} \ee

\section{Numerical Results}
In order to examine the onset of singularity formation for  the system
$$
\left \{ \begin{array}{lcr}
u_{t} + v u_{x} & = & 0 \\
v_{t} + u v_{x} & = & 0
\end{array}
\right. 
$$
numerically, the graphics shown in Figure 1 were obtained using  
a simple finite difference scheme 
\begin{eqnarray}
\label{num}
u_{i}^{n+1}&=&u_{i}^{n}-0.02v^{n}_{i}(u^{n}_{i+1}-u^{n}_{i-1}),\\
v_{i}^{n+1}&=&v_{i}^{n}-0.02u^{n}_{i}(v^{n}_{i+1}-v^{n}_{i-1}).
\end{eqnarray}
Step sizes are $\Delta t = 0.01$ and $\Delta x = 1$, and initial data 
takes the form
 $$u_{0}=0.0095 j (150-j) \sin(0.06(j-37.5)),\ 0\leq j\leq 150 \,,$$
 and
$$v_{0}=.01 k (150-k),\ 0\leq k\leq 150\,.$$             
The singularity forms immediately the $u$ and $v$ curves  
touch, which takes place at  $t=0.11$.
\input epsf.tex
\begin{figure} 
\epsfxsize=13cm \epsfysize=17cm
\epsfbox{graph.ps}
\caption{Singularity formation for smooth initial data.}
\end{figure}



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\end{thebibliography}


\bigskip\noindent
{\sc 
 Department of Mathematics\newline
 University of New Orleans\newline
 New Orleans, LA 70148}\newline
E-mail address: rsaxton@math.uno.edu  \newline
E-mail address: vvinod@math.uno.edu

\end{document}




