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\begin{document}

\title{Introduction to Partial Differential Equations}
\author{Homework 7}
\date{due April 25, 2013}
\maketitle

\begin{enumerate}
\item Evans, p.~163 problem 3
\item (This question refers to the remark in Evans, p.~114.)  Consider
the scalar conservation law in one spatial dimension
\begin{gather*}
  u_t + F'(u) \, u_x = 0 
  \qquad \text{in } \mathbb R \times (0,\infty) \,, \\
  u = g \qquad \text{in } \mathbb R \times \{ t = 0 \} \,.
\end{gather*}
\begin{enumerate}
\item Show that the function $u$ defined through the implicit equation
\[
  u = g(x-t \, F'(u))
\]
is a solution provided
\begin{equation*}
  1 + t \, g'(x-t \, F'(u)) \, F''(u) \neq 0 \,.
  \tag{*}
\end{equation*}
\item 
Suppose that $F(u) = \tfrac12 \, u^2$.  When does solvability
condition (*) fail?  Is this only a failure of the method of
characteristics, or does it also correspond to a failure of the
solution to the PDE?
\end{enumerate}
\item  Evans, p.~163 problem 5
\item Let $L$ be convex with superlinear growth and suppose that $g$
is uniformly Lipshitz as defined in class (and in Evans, p.~124) via
\[
  \operatorname{Lip}(g) = \sup_{\substack{x,y \in \mathbb R^n\\x\neq y}}
  \frac{\lvert g(x)-g(y) \rvert}{\lvert x-y \rvert} < \infty \,.
\]
Show that if, for fixed $x \in \mathbb R^n$ and $t>0$,
\[
  u(x,t) = \inf_{y \in \mathbb R^n} 
   \biggl\{ t \, L \biggl( \frac{x-y}t \biggr) + g(y) \biggr\} \,,
\]
then the infimum in this expression actually is really a minimum.


\end{enumerate}


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