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\begin{document}
\noindent{\small {\sc Electronic Journal of Differential Equations}\newline
Vol. {\bf 1994}(1994), No. 05, pp. 1-4. Published July 19, 1994. \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{1.5cm}
\thanks{\copyright 1994 Southwest Texas State University  and 
University of North Texas
\newline\indent Submitted: June 12, 1994.} 
\title[\hfilneg EJDE--1994/05\hfil Uniqueness of Entropy Solutions \hfil ]
{A Note on the Uniqueness of Entropy Solutions to First Order Quasilinear Equations}
\author[\hfil D.J. Diller\hfil EJDE--1994/05\hfilneg]{David J. Diller}
\address{Lunt Hall, Northwestern University, 2033 Sheridan Rd., Evanston,
IL 60208}
\email{diller@@math.nwu.edu}
\keywords{ Burger's Equation, Entropy Solution, Scalar Conservation Law}
\subjclass{35L60,35L65}
\begin{abstract}
In this note, we consider entropy solutions to scalar conservation laws with
unbounded initial data.
In particular, we offer an extension of Kru$\check{\mbox{z}}$khov's
uniqueness proof (see [1]).
\end{abstract}
\maketitle
\newcommand{\absval}[1]{\mid\!\! {#1}\!\!\mid}
\newcommand{\av}[1]{\mid {#1} \mid}

\section{Introduction}
We are concerned with the following Cauchy problem:
\begin{equation}
\left\{
\begin{split}
u_t +\mbox{div}F(u) =0 & \;\;\;\mbox{in} \;\;S_T =\Bbb{R}^N\times(0,T)
\\ 
u(x,0)=u_0(x) & \;\;\; x\in\Bbb{R}^N .
\end{split}
\right.
\end{equation}
Here $F=(F_1,\cdots,F_N)\in [C^{0,1}(\Bbb{R})]^N$, 
and $u_0\in L^1_{loc}(\Bbb{R}^N)$.
In particular, we are interested in the entropy solutions to
(1).  We say that $u\in L^{\infty}_{loc}(S_T)$ is an
entropy solution to (1) if 
\begin{equation}
\iint_{S_T} \mbox{sign}(u-k)\left[ (u-k)\phi_t + (F(u)-F(k))\cdot
D\phi\right] \,dx\,dt\, \geq 0,
\end{equation}
for all $\phi\in C_0^{\infty}(S_T)$, $\phi\geq 0$, and all $k\in\Bbb{R}$,
and there exists a set $\Gamma_0 \subseteq [0,T]$ of measure zero, such
that for all compact sets $K\subseteq\Bbb{R}^N$
\begin{equation}
\lim_{\overset{t\rightarrow 0^+}{t\notin\Gamma_0}}\|u(\cdot,t)-u_0\|_{1,K} =0.
\end{equation}

In [1], Kru$\check{\mbox{z}}$khov proves existence and uniqueness of an entropy 
solution to (1) when $u_0$ is bounded and F is sufficiently smooth.  
If $u_0,v_0\in L^1(\Bbb{R}^N)\cap L^{\infty}(\Bbb{R}^N)$ with corresponding 
entropy solutions $u,v$ respectively then
\begin{displaymath}
\int_{\Bbb{R}^N} \absval{u(x,t)-v(x,t)} \,dx\,\leq
\int_{\Bbb{R}^N}\absval{u_0(x)-v_0(x)}\,dx\,
\end{displaymath}
for a.e. $t\in[0,T]$ (see [1] equation 3.1).  
If $u_0\in L^1(\Bbb{R}^N)$ (but not bounded) then there
is a natural candidate for an entropy solution with this initial data.  
This note is motivated by the following two questions:

(i) Is this candidate an entropy solution?

(ii) If it is an entropy solution then is it the unique entropy solution?

\noindent This note is a partial answer to the second of these two questions.

\section{Main Result}
In proving uniqueness Kru$\check{\mbox{z}}$khov proves the 
following Proposition:
\begin{Proposition} If $u$ and $v$ are entropy solutions to (1) satisfying 
\begin{displaymath}
\left\| \frac{F(u)-F(v)}{u-v} \right\|_{\infty,S_T}\leq M
\end{displaymath}
then $u=v$ almost everywhere in $S_T$.
\end{Proposition}
The primary result of this note is the following improvement of Proposition
2.1.
\begin{Proposition}
If $u$ and $v$ are entropy solutions to (1) satisfying
\begin{equation}
\left\| \frac{F(u(\cdot,t))-F(v(\cdot,t))}{u(\cdot,t)-v(\cdot,t)} 
\right\|_{\infty,B_{\rho}}\leq M(t,\rho)
\end{equation}
where M satisfies
\begin{equation}
\lim_{\rho\rightarrow\infty}\left(\rho-\int_0^T M(t,\rho) \,dt\,\right) =\infty
\end{equation}
then $u=v$ almost everywhere in $S_T$.
\end{Proposition}
The advantage of Proposition 2.2 over  Proposition 2.1 is that Proposition 
2.2 allows for $u_0$ to become unbounded.
Set $A(u)=(F_1'(u),\cdots,F_N'(u))$.
Then one can easily verify that Proposition 2.2 implies the following. 
\begin{Corollary}
There exists at most one entropy solution to (1) satisfying
\begin{displaymath}
\|A(u(\cdot,t))\|_{\infty,\Bbb{R}^N}\leq M(t)
\end{displaymath}
where M satisfies
\begin{displaymath}
\int_0^T M(t)\,dt\,<\infty.
\end{displaymath}
\end{Corollary}

As an example we apply Corollary 2.3 to the Burger's equation, i.e., $N=1$
and  $F(u)=\frac{1}{2} u^2$. 

\begin{Lemma}
If $u_0\in L^p(\Bbb{R}^N)$ with $1\leq p<\infty$ then there exists an entropy
solution u of (1) satisfying
\begin{equation}
\|u(\cdot,t)\|_{\infty,\Bbb{R}^N}\leq \frac{(2\|u_0\|_p)^{\frac{p}{p+1}}}
{t^{\frac{1}{p+1}}} \in L^1([0,1]).
\end{equation}
\end{Lemma}
\begin{pf}
Let $u$ be the almost everywhere unique minimizer of 
\begin{displaymath}
\psi(x,t,v)=\int_x^{x-vt} u_0(s)\,ds\, +\frac{tv^2}{2}.
\end{displaymath}
For details see [2].  
Then $u$ is an entropy solution to (1).  
Now $\psi(x,t,0)=0$ and  by Holder's inequality
\begin{align*}
\psi(x,t,v)\geq & -\|u_0\|_p\absval{vt}^{\frac{p-1}{p}} +\frac{tv^2}{2} \\
=&\absval{vt}^{\frac{p-1}{p}}(-\|u_0\|_p
+\frac{t^{\frac{1}{p}}\absval{v}^{\frac{p+1}{p}}}{2}) \\
>& 0
\end{align*}
when
\begin{displaymath}
v>\frac{(2\|u_0\|_p)^{\frac{p}{p+1}}}{t^{\frac{1}{p+1}}}.
\end{displaymath}
This implies (6).
\end{pf}
Thus from Corollary 2.3 the entropy solution to Burger's equation
satisfying (6) is unique.
One can prove a similar estimate for solutions when
$F(u)=\absval{u}^{\alpha}$ for any $\alpha>1$.
\begin{pf} (of Proposition 2.2).
Let $u$ and $v$ be any two entropy solutions to (1).  Let $J\in
C^{\infty}_{0}(-1,1)$ satisfy:
\begin{equation*}
\left\{
\begin{split}
\int_{-1}^1 J(x) \,dx\, =1 \\
J\geq 0. \;\;\;\;\;\;
\end{split}
\right.
\end{equation*}
Any two entropy solutions $u,v$ satisfy
\begin{equation}
\iint_{S_T} \mbox{sign}(u-v) [(u-v)\phi_t +(F(u)-F(v))\cdot D\phi]
\,dx\,dt\,\geq 0
\end{equation}
for all $\phi\in C_0^{\infty} (S_T)$ with $\phi\geq 0$ (see [1] equation
3.7).     
Set
\begin{displaymath}
r(t,\rho)=\rho -\int_0^t M(\tau,\rho)d\tau.
\end{displaymath}
Since $t\rightarrow r(t,\rho)$ is decreasing, (5) implies
\begin{displaymath}
\lim_{\rho\rightarrow\infty} r(t,\rho) = \infty \;\;\mbox{ for all }\;\;
0\leq t\leq T.
\end{displaymath}
Select $R>0$ such that if $\rho> R$ then $r(T,\rho)>0$.
Fix $\rho>R.$  Set $r(t)=r(t,\rho)$.
Let $0<t_1<t_2<T$.  
Let $0<\epsilon<\min\{t_1,T-t_2\}$. 
Consider the following test function:
\begin{displaymath}
\phi_{\epsilon}(x,t)=\epsilon^{-N-1} \int_{t-t_2}^{t-t_1}
J\left(\frac{s}{\epsilon}\right) \,ds\,
\int_{\av{x}-r(t)+\epsilon}^{\infty}
J\left(\frac{s}{\epsilon}\right) \,ds\,.
\end{displaymath}
Notice that
\begin{displaymath}
\mbox{supp} \phi_{\epsilon} \subseteq \{ (x,t) : t_1-\epsilon<t
<t_2+\epsilon \;\;\mbox{and}\;\; \absval{x}<r(0)+\epsilon\},
\end{displaymath}
so that $\phi_{\epsilon}$ is an admissible test function for (7).
Now we compute $\phi_{\epsilon,t}$ and $D\phi_{\epsilon}$.
\begin{equation*}
\begin{split}
\phi_{\epsilon,t}=&\epsilon^{-1}\left[J\left(\frac{t-t_1}{\epsilon}\right) 
-J\left(\frac{t-t_2}{\epsilon}\right)\right]
\epsilon^{-N}\int_{\av{x}-r(t)+\epsilon}^{\infty}
J\left(\frac{s}{\epsilon}\right) \,ds\, \\
&+ \left[M(t)\epsilon^{-N}J
\left(\frac{\absval{x}-r(t)+\epsilon}{\epsilon}\right)\right]
\epsilon^{-1}\int^{t-t_1}_{t-t_2} J\left(\frac{s}{\epsilon}\right)\,ds\,.
\end{split}
\end{equation*}
\begin{equation*}
D\phi_{\epsilon} =\frac{x}{\absval{x}}
\left[\epsilon^{-N}
J\left(\frac{\absval{x}-r(t)+\epsilon}{\epsilon}\right)\right]
\epsilon^{-1}\int^{t-t_1}_{t-t_2} J\left(\frac{s}{\epsilon}\right) \,ds\,.
\end{equation*}
Then using this test function in (7)  yields
\begin{equation*}
\begin{split}
\iint_{S_T} & \absval{u-v}\epsilon^{-1}
\left[J\left(\frac{t-t_1}{\epsilon}\right)
-J\left(\frac{t-t_2}{\epsilon}\right)\right]
\epsilon^{-N}\int^{\infty}_{\av{x}-r(t)+\epsilon} 
J\left(\frac{s}{\epsilon}\right) \,ds \\
\geq & \iint_{S_T} 
\epsilon^{-N}J\left(\frac{\absval{x}-r(t)+\epsilon}{\epsilon}\right)
\epsilon^{-1}\int_{t-t_2}^{t-t_1} J\left(\frac{s}{\epsilon}\right) \,ds \\
 & \times \left[\mbox{sign}(u-v)(F(u)-F(v))\cdot\frac{x}{\absval{x}}
+M(t,\rho)\absval{u-v}\right]\,dx\,dt\,\\
\geq & 0 \;\;\mbox{because}\;\; M(t,\rho)\absval{u-v}\geq \absval{F(u)-F(v)}.
\end{split}
\end{equation*}
So by letting $\epsilon\rightarrow 0$ we obtain
\begin{equation*}
\int_{B_{r(t_2)}} \absval{u-v}(x,t_2) \,dx\, \leq
\int_{B_{r(t_1)}} \absval{u-v}(x,t_1) \,dx\,.
\end{equation*}
Since $r(t_1)<\rho$
\begin{equation*}
\int_{B_{r(t_1)}} \absval{u-v}(x,t_2) \,dx\, \leq
\int_{B_{\rho}} \absval{u-v}(x,t_1) \,dx\,.
\end{equation*}
Combining these with (3) yields
\begin{equation*}
\int_{B_{r(t_2)}} \absval{u-v}(x,t_2) \,dx\, \leq
\int_{B_{\rho}} \absval{u-v}(x,0) \,dx\, = 0.
\end{equation*}
Thus, for $0<t_2<T$, $u(x,t_2)=v(x,t_2)$ for $x\in B_{r(t_2)}$.  Letting
$\rho\rightarrow\infty$ and using condition (5),
$u(\cdot,t_2)=v(\cdot,t_2)$.
Since $t_2$ is arbitrary, $u=v$ in $S_T$.
\end{pf}

As another example we consider Burger's equation with initial data
$u_0(x)=-x$.  
We show that in this situation (1) has a unique entropy solution 
in $S_1$ satisfying
\begin{displaymath}
\|u(\cdot,t)\|_{\infty,B_{\rho}}\leq \frac{\rho}{1-t}.
\end{displaymath}
First note that
\begin{displaymath}
u(x,t)=\frac{-x}{1-t}
\end{displaymath}
is one such solution.
Suppose $v$ is another. For $0<t<1$
\begin{displaymath}
r(t,\rho) = \rho - \int^t_0 \frac{\rho}{1-\tau} d\tau = \rho(1+\ln(1-t)),
\end{displaymath}
so Proposition~2.2 guarantees that $u=v$ in $S_{1-e^{-1}}$.
By reapplying Proposition~2.2 this time beginning at $t=1-e^{-1}$ we find
that $u=v$ in $S_{1-e^{-2}}$.  
Finally, by repeating this argument $n$ times, $u=v$ in $S_{1-e^{-n}}$.
Thus $u=v$ in $S_1$.

\subsection*{Acknowledgements}
I am indebted to Gui-Qiang Chen for reading this note and making
some valuable suggestions.

\begin{thebibliography}{0} 
\bibitem{} S.N. Kru$\check{\mbox{z}}$kov, 
{\it First Order Quasilinear Equations in Several
Independent Variables}, Math. Sb. 123 (1970) pp. 217-243. 

\bibitem{} P.D. Lax, {\it Hyperbolic Systems of Conservation Laws and the
Mathematical Theory of Shock Waves},  Society for Industrial and Applied
Mathematics, Philadelphia 1973.

\end{thebibliography}
\end{document}