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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 53, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/53\hfil Almost entire solutions]
{Almost entire solutions of the Burgers equation}

\author[N. D. Alikakos, D. Gazoulis \hfil EJDE-2018/53\hfilneg]
{Nicholas D. Alikakos, Dimitrios Gazoulis}

\address{Nicholas D. Alikakos \newline
Department of Mathematics, 
University of Athens,
panepistimioupolis 15784 Athens, Greece}
\email{nalikako@math.uoa.gr}

\address{Dimitrios Gazoulis \newline
Department of Mathematics, 
University of Athens,
panepistimioupolis 15784 Athens, Greece}
\email{dgazoulis@math.uoa.gr}

\dedicatory{Communicated by Robert V. Kohn}

\thanks{Submitted November 21, 2016. Published February 20, 2018.}
\subjclass[2010]{35B08, 35L65}
\keywords{Burgers equation; entropy solution; rigidity}

\begin{abstract}
 We consider Burgers equation on the whole $x-t$ plane.
 We require the solution to be classical everywhere, except possibly
 over a closed set $S$ of potential singularities, which is
 \begin{itemize}
 \item[(a)] a subset of a countable union of ordered graphs of differentiable
 functions,

 \item[(b)] has one dimensional Hausdorff measure, $H^1(S)$, equal to zero.
 \end{itemize}
 We establish that under these conditions the solution is identically equal
 to a constant.

\end{abstract}

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

\section{Introduction}\label{sec1}

In this note we establish a sort of rigidity theorem for solutions of the
Burgers equation
\begin{equation}
h_t(x,t)+h(x,t)h_x(x,t)=0 \label{eq1}
\end{equation}
in the plane $\mathbb{R}_x\times\mathbb{R}_t$. We consider functions $h(x,t)$ 
that solve
\eqref{eq1} classically, pointwise, except perhaps on a closed set $S$
of the $x-t$ plane as in the Abstract, and we show that $h$ must be
identically constant. We note that such a statement is false in the
half plane $\mathbb{R}_x\times\mathbb{R}^+_t$ because of rarefaction waves.
We also note that the conclusion of the theorem is relatively simple to
recover for entropy solutions. Indeed if $u(x,t)$ is an $L^\infty(\mathbb{R}_x\times\mathbb{R}_t)$
entropy solution to
\begin{equation}
\begin{gathered}
 u_t+\frac{1}{2}(u^2)_x=0 \\
 u(x,0)=u_0(x)
 \end{gathered}  \label{eq2}
\end{equation}
then we have the (well known) estimate
\begin{equation}
\frac{u(x+a,t)-u(x,t)}{a}<\frac{E}{t} \label{eq3}
\end{equation}
for every $a>0$, $t>0$ with $E$ depending only on $\|u_0\|_{L^\infty}=M$ 
 (see \cite[Theorem 16-4]{s1} or  \cite[Lemma in 3.4.3]{e1}).
 By shifting the origin of time all the way to $t=-\infty$, and by 
uniqueness in the entropy class, we conclude via \eqref{eq3} that 
$x\to u(x,t)$ is non-\linebreak increasing for every $t$.
Thus in particular $u_0$ is a nonincreasing $L^\infty$ function, and if 
$u_0$ is not identically constant (a.e.) then the solution of \eqref{eq2} 
will have a shock. Thus, the hypothesis $H^1(S)=0$ will force $u_0$ to be 
identically constant, and so also $u$.

There is a similar result for the eikonal equation
\begin{equation}
\Big(\frac{\partial u}{\partial t}\Big)^2
+\Big(\frac{\partial u}{\partial x}\Big)^2=1 \label{eq4}
\end{equation}
by Caffarelli and Crandall \cite{c1} which states that if $u$ solves
 \eqref{eq4} pointwise on $\mathbb{R}^2\setminus\widehat{S}$, with 
$H^1(\widehat{S})=0$, then necessarily $u$ is either affine or a double 
``cone function'', $u(y)=a\pm|y-z|$, $y=(x,t)$, $z=(x_0,t_0)$. 
The point in \cite{c1} again is that $u$ is not assumed a viscosity solution.

The proof of our result is based on a simple and explicit change of variables 
(see \eqref{eq6} below) that transforms \eqref{eq2} into \eqref{eq4}, 
and actually establishes almost the equivalence of the two problems in 
$\mathbb{R}^2$.
Note that for the set $\widehat{S}$ in \cite{c1} there 
is no extra hypothesis besides that $H^1(\widehat{S})=0$. 
Our only excuse for writing it down is that it concerns the Burgers equation, 
which in spite of its simplicity pervades the theory of hyperbolic
 conservation laws \cite{d1,e1}. 


\section{Main result}

\begin{theorem} \label{thm1}
Let $h(x,t)$ be a measurable function on $\mathbb{R}^2$ and suppose that $S$ 
is closed and on $\mathbb{R}^2\setminus S$ the following hold: $h(x,t)$ is continuous,
$\frac{\partial h}{\partial t}$, $\frac{\partial h}{\partial x}$ exist,
 $x\to\frac{\partial h}{\partial x}(x,t)$ is $L^1_{\rm loc}$ and moreover
\begin{equation}
h_t+hh_x=0, \quad \text{on } \mathbb{R}^2\setminus S. \label{eq5}
\end{equation}
If $H^1(S)=0$ and  $S\subset\cup_{i\in Z}\Gamma_i$, where 
$\Gamma_i:=\{(x,t): t=p_i(x)$, $p_i$ differentiable, 
$x\in \mathbb{R}\}$, 
\[
\ldots<p_{-n}(x)<\cdots<p_{-1}(x)<p_1(x)<p_2(x)<\dots<p_n(x)<\cdots,
\]
then $h\equiv$ constant on $\mathbb{R}^2$, and $S=\emptyset$.
\end{theorem}

\subsection*{Notes}
(1) The change of variables $h=c(v)$ converts $v_t+c(v)v_x=0$ into Burgers'
 equation $h_t+hh_x=0$, hence this more general equation is covered for 
differentiable $c$ provided that $c'\neq0$. Note that if we write the 
equation for $v$ in divergence form $v_t+(C(v))_x=0$, where $C'=c$, then the condition $c'\neq0$ corresponds to $C''\neq0$ which is naturally weaker than the usual condition of genuine nonlinearity $C''>0$, since we do not require any orientation of the $x-t$ plane.

(2) The change of variables relating \eqref{eq2} to \eqref{eq4} is basically
\begin{equation}
u(x,t)=\int^t_0\frac{ds}{\sqrt{h^2(x,s)+1}}+g(x) \label{eq6}
\end{equation}
where $g(x)=\int^x_0\frac{h(u,0)du}{\sqrt{h^2(u,0)+1}}$.
Note that the projected characteristics of the corresponding equations coincide,
\begin{gather*}
 \frac{dx}{d\tau}=h  \frac{dx}{d\tau}=u_x=\frac{h}{\sqrt{h^2+1}} \\ 
 \frac{dt}{d\tau}=1  \frac{dt}{d\tau}=u_t=\frac{1}{\sqrt{h^2+1}}.
\end{gather*}
The need for differentiating under the integral sign in \eqref{eq6} 
for obtaining \eqref{eq4} forces us to introduce the perhaps unnecessary 
hypothesis that $S$ lies on a set of graphs.

(3) The hypotheses on the singular set a priori do not exclude $S$ to be a 
countable union of Cantor sets arranged on a family of parallel lines in the 
$x-t$ plane.

\begin{proof}[Proof of Theorem \ref{thm1}]
 For the convenience of the reader we begin by giving the proof in the simple 
case where $S$ lies on a single differentiable graph contained inside a strip, 
$S\subset\Gamma:=\{(x,t)\mid t=p(x)$, $p$ differentiable, $0<p(x)<1$, $x\in\mathbb{R}\}$.
Set 
\[
\Omega^+=\{(x,t)\in\mathbb{R}^2\mid t\le p(x)\}, \quad
\Omega^-=\{(x,t)\in \mathbb{R}^2\mid t\ge p(x)\}.
\]
For $(x,t)\in\Omega^+$, we define
\begin{equation}
u^+(x,t)=\int^t_0\frac{ds}{\sqrt{h^2(x,s)+1}}+g^+(x), \label{eq7}
\end{equation}
where 
\[
g^+(x)=\int^x_0\frac{h(u,0)du}{\sqrt{h^2(u,0)+1}}
\]
and for $(x,t)\in\Omega^-$, we define
\begin{equation}
u^-(x,t)=\int^t_1\frac{ds}{\sqrt{h^2(x,s)+1}}+g^-(x), \label{eq8}
\end{equation}
where 
\[
g^-(x)=\int^x_0\frac{h(u,1)du}{\sqrt{h^2(u,1)+1}}\,.
\]
We begin with $u^+(x,t)$ for $t\le p(x)$, $(x,t)\in U:=\mathbb{R}^2\setminus S$, open.
By our hypothesis
\begin{equation}
\begin{aligned}
u^+_x(x,t)
&=\int^t_0\frac{-h(x,s)h_x(x,s)}{\big(\sqrt{h^2(x,s)+1}\big)^3}ds
 +\frac{h(x,0)}{\sqrt{h^2(x,0)+1}} \\
&=\int^t_0\frac{h_s(x,s)}{\big(\sqrt{h^2(x,s)+1}\big)^3}ds
 +\frac{h(x,0)}{\sqrt{h^2(x,0)+1}}  \\
&=\frac{h(x,t)}{\sqrt{h^2(x,t)+1}}.
\end{aligned} \label{eq9}
\end{equation}
On the graph we have
\begin{equation}
u^+_x(x,p(x))=\frac{h(x,p(x))}{\sqrt{h^2(x,p(x))+1}}, \quad
 (x,p(x))\notin S. \label{eq10}
\end{equation}
Differentiating in $t$ is straightforward, and holds quite generally that
\begin{equation}
u^+_t(x,t)=\frac{1}{\sqrt{h^2(x,t)+1}}, \quad
u^+_t(x,p(x))=\frac{1}{\sqrt{h^2(x,p(x))+1}}. \label{eq11}
\end{equation}
Thus from \eqref{eq9} and \eqref{eq11} we have
\begin{equation}
(u^+_x(x,t))^2+(u^+_t(x,t))^2=1 \quad \text{in }  \Omega^+\setminus S. \label{eq12}
\end{equation}
Analogously we argue for $u^-(x,t)$ and we obtain
\begin{gather}
u^-_x(x,t)=\frac{h(x,t)}{\sqrt{h^2(x,t)+1}} \quad \text{in }
 \Omega^-\setminus S, \label{eq13} \\
u^-_x(x,p(x))=\frac{h(x,p(x))}{\sqrt{h^2(x,p(x))+1}}, \quad
 (x,p(x))\notin S, \label{eq14} \\
u^-_t(x,t)=\frac{1}{\sqrt{h^2(x,t)+1}}, \quad
u^-_t(x,p(x))=\frac{1}{\sqrt{h^2(x,p(x))+1}} \label{eq15}
\end{gather}
and so once more
\begin{equation}
(u^-_x(x,t))^2+(u^-_t(x,t))^2=1 \quad \text{in }  \Omega^-\setminus S. \label{eq16}
\end{equation}
Also from \eqref{eq10} and \eqref{eq14} we obtain
\begin{equation}
u^+_x(x,p(x))=u^-_x(x,p(x)), \quad u^+_t(x,p(x))=u^-_t(x,p(x)), \quad
 (x,p(x))\notin S. \label{eq17}
\end{equation}
We now set
\begin{equation}
u(x,t)=\begin{cases}
 u^+(x,t), & (x,t)\in\Omega^+ \\
 u^-(x,t)+{\Delta}(x), & (x,t)\in\Omega^-
 \end{cases}  \label{eq18}
\end{equation}
where
\begin{equation}
{\Delta}(x):=u^+(x,p(x))-u^-(x,p(x)), \quad x\in\mathbb{R}. \label{eq19}
\end{equation}
Note that $\Gamma\setminus S$ is open in $\Gamma$ and so is its projection
 $\pi_x(\Gamma\setminus S)=\cup^\infty_{i=1}(a_i,b_i)=:O$, and for $x\in O$
\begin{equation}
\begin{aligned}
\frac{d{\Delta}(x)}{dx}
&=u^+_x(x,p(x))+u^+_t(x,p(x))p'(x)-(u^-_x(x,p(x)) \\
&\quad +u^-_t(x,p(x))p'(x))=0 
\end{aligned} \label{eq20}
\end{equation}
(by \eqref{eq17}). Therefore, by the continuity of $h$ and $p$, $u(x,t)$
is differentiable on $\mathbb{R}^2\setminus S$, and by
\eqref{eq12}, \eqref{eq16}, \eqref{eq18} and \eqref{eq20},
\begin{equation}
(u_x(x,t))^2+(u_t(x,t))^2=1 \quad \text{on }  \mathbb{R}^2\setminus S. \label{eq21}
\end{equation}
Hence, by the result in \cite{c1}, $u$ is of the form
\begin{equation}
u(x,t)=ax+bt+\gamma \quad (a^2+b^2=1), \label{eq22}
\end{equation}
or
\begin{equation}
u(x,t)=c\pm\sqrt{(x-x_0)^2+(t-t_0)^2}. \label{eq23}
\end{equation}
In the first case $u_t=b$ and so $h(x,t)\equiv$ constant.

On the other hand \eqref{eq23} gives
\begin{equation}
\begin{aligned}
u_t(x,t)
&=\pm\frac{t-t_0}{\sqrt{(x-x_0)^2+(t-t_0)^2}}  \\
&\Rightarrow h(x,t)=\frac{x-x_0}{t-t_0}
\end{aligned} \label{eq24}
\end{equation}
which is singular on $\{t=t_0\}$, and thus is excluded by the hypothesis
$H^1(S)=0$. Therefore $h(x,t)\equiv$ constant is the only option.


Note that ${\Delta}(x)$ is continuous for $x\in\mathbb{R};\mathcal{L}(\pi_x(S))=0$.

For the proof of the general case, we indicate the necessary modifications.
Suppose $p_\ell(x)<p_{\ell+1}(x)$, $a_\ell(x)\in C^1$, 
$p_\ell(x)<a_\ell(x)<p_{\ell+1}(x)$, $\ell=1,2,\dots$, $\ell=-2,-3,\dots$ 
(and $p_{-1}(x)<a_0(x)<p_1(x)$) where we have inserted the $C^1$ graphs 
$a_\ell(x)$ that will play the role of the horizontal lines $t=0$ and $t=1$ 
in the simple case treated above.
Let
\begin{gather}
\Omega^+_1=\{p_{-1}(x)\le t\le p_1(x)\}, \quad
\Omega^-_1=\{p_1(x)\le t\le a_1(x)\},  \label{eq25} \\
\begin{gathered}
u^+_1(x,t):=\int^t_{a_0(x)}\frac{ds}{\sqrt{h^2(x,s)+1}}+g^+_1(x), \\
g^+_1(x)=\int^x_0\frac{h(s,a_0(s))+a'_0(s)}{\sqrt{h^2(s,a_0(s))+1}}ds,
\end{gathered} \quad
 \text{on } \Omega^+_1;  \label{eq26} \\
\begin{gathered}
u^-_1(x,t):=\int^t_{a_1(x)}\frac{ds}{\sqrt{h^2(x,s)+1}}+g^-_1(x),\\
g^-_1(x)=\int^x_0 \frac{h(s,a_1(s))+a'_1(s)}{\sqrt{h^2(s,a_1(s))+1}}ds,
\end{gathered}\quad
\text{on }\Omega^-_1;  \label{eq27}  \\
{\Delta}_1(x):=u^+_1(x,p_1(x))-u^-_1(x,p_1(x)) ;\nonumber \\
u_1(x,t)=\begin{cases}
 u^+_1(x,t), & \text{on }  \Omega^+_1 \\
 u^-_1(x,t)+{\Delta}_1(x), & \text{on }  \Omega^-_1.
 \end{cases} \label{eq28}
\end{gather}
For $i=2,3,\dots$, set
\begin{gather}
\Omega^+_i=\{a_{i-1}(x)\le t\le p_i(x)\}, \quad
\Omega^-_i=\{p_i(x)\le t\le a_i(x)\}, \label{eq29} \\
u^+_i(x,t):=u^-_{i-1}(x,t)+{\Delta}_{i-1}(x), \quad \text{on }  \Omega^+_i, \label{eq30}\\
{\Delta}_j(x):=(u^+_j-u^-_j)(x,p_j(x)), \quad j=1,2,\dots \,. \label{eq31}
\end{gather}
Set
\begin{gather}
\begin{gathered}
u^-_i(x,t):=\int^t_{a_i(x)}\frac{ds}{\sqrt{h^2(x,s)+1}}+g^-_i(x),\\
g^-_i(x)=\int^x_0\frac{h(s,a_i(s))+a'_i(s)}{\sqrt{h^2(s,a_i(s))+1}}ds,
\end{gathered} \quad \text{in }\Omega^-_i ,  \label{eq32} \\
u_k(x,t)=\begin{cases}
 u^+_k(x,t), & \text{on }  \Omega^+_k \\
 u^-_k(x,t)+{\Delta}_k(x), & \text{on }  \Omega^-_k
 \end{cases}
\quad k=1,2,\dots \,. \label{eq33}
\end{gather}
Next we define $u$ below $a_0(x)$.
\begin{equation}
u^+_{-1}(x,t)=u^+_1(x,t) \quad \text{on } 
 \Omega^+_{-1}=\{p_{-1}(x)\le t\le a_0(x)\}, \label{eq34}
\end{equation}
with
\begin{equation}
u^-_{-1}(x,t):=\int^t_{a_{-1}(x)}\frac{ds}{\sqrt{h^2(x,s)+1}}+g^-_{-1}(x), 
 \label{eq35}
\end{equation}
on $\Omega^-_{-1}=\{a_{-1}(x)\le t\le p_{-1}(x)\}$, where  
\begin{gather}
g^-_{-1}(x)=\int^x_0\frac{h(s,a_{-1}(s))
+a'_{-1}(s)}{\sqrt{h^2(s,a_{-1}(s))+1}}ds, \\
{\Delta}_{-1}(x):=u^+_{-1}(x,p_{-1}(x))-u^-_{-1}(x,p_{-1}(x)), \label{eq36} \\
u_{-1}(x,t)=\begin{cases}
 u^+_{-1}(x,t), & \text{in } \Omega^+_{-1}, \\
 u^-_{-1}(x,t)+{\Delta}_{-1}(x), & \text{in }  \Omega^-_{-1}.
 \end{cases} \label{eq37}
\end{gather}
And further down $i=2,3,\dots$, we  set
\begin{gather}
\Omega^+_{-i}=\{p_{-i}(x)\le t\le a_{-i+1}(x)\}, \quad
\Omega^-_{-i}=\{a_{-i}(x)\le t\le p_{-i}(x)\}, \label{eq38} \\
u^+_{-i}(x,t):=u^-_{-i+1}(x,t)+{\Delta}_{-i+1}(x), \quad \text{on } 
\Omega^+_{-i}, \label{eq39} \\
{\Delta}_{-i}(x):=(u^+_{-i}-u^-_{-i})(x,p_{-i}(x)), \label{eq40}
\end{gather}
with
\begin{equation}
u^-_{-i}(x,t):=\int^t_{a_{-i}(x)}\frac{ds}{\sqrt{h^2(x,s)+1}}+g^-_{-i}(x), 
\quad \text{on } \Omega^-_{-i} \label{eq41}
\end{equation}
where 
\begin{gather}
g^-_{-i}(x)=\int^x_0\frac{h(s,a_{-i}(s))+a'_{-i}(s)}{\sqrt{h^2(s,a_{-i}(s))+1}}ds,\\
u_{-k}(x,t)=\begin{cases}
 u^+_{-k}(x,t), & \text{in }  \Omega^+_{-k}, \\
 u^-_{-k}(x,t)+{\Delta}_{-k}(x), & \text{in }  \Omega^-_{-k},
 \end{cases}\quad k=2,3,\dots \,. \label{eq42}
\end{gather}
Finally we set
\begin{equation}
u(x,t)=u_k(x,t) \quad \text{on } \Omega^+_k\cup\Omega^-_k, \; k\in Z\setminus\{0\}. \label{eq43}
\end{equation}
With this definition we note that $u(x,t)$ is differentiable on 
$\mathbb{R}^2\setminus S$, and
\begin{equation}
\Big(\frac{\partial u}{\partial t}\Big)^2
+\Big(\frac{\partial u}{\partial x}\Big)^2=1 \quad \text{on }  \mathbb{R}^2\setminus S.
 \label{eq44}
\end{equation}
and thus we conclude as before that $h(x,t)\equiv$\,constant and 
$S=\emptyset$. The proof is complete. 
\end{proof}

\subsection*{Acknowledgments}
We would like to thank the anonymous referees for their useful suggestions
 that improved significantly the presentation of our result.

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\end{thebibliography}




\end{document}