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\begin{document}

\title{Partial Differential Equations}
\author{Midterm Exam}
\date{Solutions}
\maketitle

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\begin{enumerate}



\item
\begin{enumerate}
\item
Solve the partial differential equation
\begin{gather*}
  u_t + x^2 \, u_x = 0 \,,
\end{gather*}
where $u=u(x,t)$ on the open first quadrant of the $(x,t)$ plane.

\emph{Hint:} Show that $z(s)=u(x(s),s)$ is constant if
$x'(s)=x^2(s)$.


\item Draw the characteristic curves, then state a set of boundary
and/or initial conditions that specify the solution uniquely in the
first quadrant of the $(x,t)$ plane.

\end{enumerate}
\hfill (10+10)

\subsubsection*{Solution:}
$$
  z'(s)=u_x(x(s),s) \, x'(s)+u_t(x(s),s)=u_t + x^2 \, u_x=0 \,.
$$
Now the characteristic curves are the solutions of the ordinary
differential equation $x'(s)=x^2(s)$:
$$
\mathrm dt=\dfrac{\mathrm d x}{x^2}, 
\qquad s=x_0^{-1}-x^{-1},\qquad x_0=\dfrac{x}{1+xs}.
$$
The characteristic curve which passes through any point $(x,t) \equiv
(x(s),s)$ in the (open) first quadrant also passes through the
positive $x$-axis at $(x_0,0) \equiv (x(0),0)$:
\begin{center}
  \includegraphics[width=8cm]{Charact.eps}
\end{center}
Moreover, the $t$-axis is also a characteristic line.  Thus, in the
first quadrant, we need to specify initial data 
\[
  u = g \qquad \text{on } [0,\infty) \times \{ t = 0 \}
\]
and the solution is given by
\[
  u(x,t) = g \biggl( \frac{x}{1+xt} \biggr) \,.
\]


\item Let $U \subset \R^n$ open and bounded and $u \in C^2(\bar
U)$.  We say that $u$ is \emph{superharmonic} if
\[
  -\Delta u \geq 0 \quad \text{in } U \,.
\]

\begin{enumerate}
\item Prove that for superharmonic $u$,
\[
  u(x) \geq \PVint_{B(x,r)} u(y) \, dy
\]
for all $B(x,r) \subset U$.

\emph{Note:} Do the computation explicitly.  Simply referring to the
homework is not enough.




\item Show that if $u$ is superharmonic and $u\geq 0$ on $\partial U$,
then $u \geq 0$ in $\bar U$.



\end{enumerate}
\hfill (10+10)
\subsubsection*{Solution:}

\begin{enumerate}
\item Set
$$
\phi(r)=\bint_{\partial B(x,r)}u(y)\;\mathrm{d}S(y)=\bint_{\partial B(x,1)}u(ry)\;\mathrm{d}S(y).
$$
Then
\begin{align*}
  \phi'(r) 
  & = \bint_{\partial B(x,1)}y\cdot Du(ry)\,\mathrm{d}S(y)\\
  & = \bint_{\partial B(x,r)}\dfrac{z}{r}\cdot Du(z)\,\mathrm{d}S(z)\\
  & = \dfrac{1}{n\alpha(n) r^{n-1}}\int_{B(x,r)}\Delta
      u(z)\,\mathrm{d}z\leq0 \,
\end{align*}
The rest is to mention that $\lim\limits_{r\to0}\phi(r)=u(x)$.
\item Let us prove that $\min\limits_{\bar U}u=\min\limits_{\partial
    U}u$. Assume that $\min\limits_{\bar U}u<\min\limits_{\partial
    U}u$ (it can not be greater, since $\partial U\subset\bar U$). Then
  $\exists x_0\in U^0$ such that $u(x_0)=\min\limits_{\bar U}u$. For
  $B(x_0,r)\subset U$ we have (see above) $$
  u(x_0)\geq\bint_{B(x_0,r)}u\,\mathrm{d}y\geq
  \min\limits_{B(x_0,r)}u \,.$$ Hence $u$ is constant on
  $B(x_0,r)$. Assuming that $U$ is path-connected, for any $x \in U$
  there exists a continuous curve $\gamma \colon [0,1] \to U$ with
  $\gamma(0) = x_0$ and $\gamma(1)=x$.  Set
\[
  s^* = \sup \{ s \in [0,1] \colon u(\gamma(t)) = \min_{\bar U} u
  \text{ for } t \in [0,s] \}
\]
By continuity, $u(\gamma(s^*))= \min u$.  Moreover, $s^*=1$ because if
not, we apply the first part of the argument with $x_0 = \gamma(s^*)$
and arrive at a contradiction to the maximality of $s^*$.
\end{enumerate}

\item Recall that the solution to the heat equation
\begin{gather*}
  u_t - \Delta u = 0
  \quad \text{in } \mathbb{R} \times (0,\infty) \,, \\
  u = g
  \quad \text{on } \mathbb{R} \times \{t=0\}
\end{gather*}
is given by
\[
  u(x,t) = \int_{\mathbb{R}} \Phi(x-y,t) \, g(y) \, dy \,,
\]
where, for $t>0$,
\[
  \Phi(z,t) = \frac1{\sqrt{4\pi t}} \,
              e^{-\tfrac{\lvert z \rvert^2}{4t}} \,.
\]

\begin{enumerate}
\item Show that if $g$ is an even function, i.e.\ if $g(x)=g(-x)$, then
$u(\cdot,t)$ is even for every $t\geq 0$.


\item What does this imply for the solution of the heat equation on
the halfline with Neumann boundary conditions, i.e.
\begin{gather*}
  u_t - \Delta u = 0
  \quad \text{in } (0,\infty) \times (0,\infty) \,, \\
  u_x = 0
  \quad \text{on } \{x=0\} \times (0,\infty) \,, \\
  u = g
  \quad \text{on } (0,\infty) \times \{t=0\} \,?
\end{gather*}

\end{enumerate}
\hfill (10+10)
\subsubsection*{Solution:}

\begin{enumerate}
\item 

\begin{align*}
  u(x,t) 
  & = \int_{\mathbb{R}} \Phi(x-y,t) \, g(y) \, dy \\
  &=\int_{\mathbb{R}} \Phi(x-y,t) \, g(-y) \, dy \\
  & =\int_{\mathbb{R}} \Phi(x+z,t) \, g(z) \, dz \\
  & = \int_{\mathbb{R}} \Phi(-(-x-z),t) \, g(z) \, dz \\
  & =\int_{\mathbb{R}} \Phi(-x-z,t) \, g(z) \, dz=u(-x,t).
\end{align*}
\item The problem is equivalent to
\begin{gather*}
  u_t - \Delta u = 0
  \quad \text{in } \mathbb{R} \times (0,\infty) \,, \\
  u(x,0) = g(x)
  \quad \text{on } (0,\infty) \times \{t=0\}\,,\\
  u(x,0) = g(-x)
  \quad \text{on } (-\infty,0) \times \{t=0\} \,.
\end{gather*}
Theorem 1 [Evans, p.47] shows that $u\in C^{\infty}$ $\forall\, t>0$.
Hence $u_x(0,t)$ is well defined $\forall\, t>0$ and is zero because
$u$ is even. 
\end{enumerate}

\item Let $U \subset \R^n$ open and bounded.  Recall that $1 \leq p <
\infty$, the $\Space{L}{}{p}$-norm of a suitably integrable function
$u$ is defined
\[
  \lVert u \rVert_{\Space{L}{}{p}}^p
  = \int_U \lvert u(x) \rvert^p \, dx
\]
while the $\Space{L}{}{\infty}$-norm is given by
\[
  \lVert u \rVert_{\Space{L}{}{\infty}}
  = \operatorname*{ess\,sup}_{x\in U} \, \lvert u(x) \rvert \,.
\]
\begin{enumerate}
\item Show that if $u \in \Space{C}{1}{2}(\bar
U\times[0,\infty))$ satisfies the heat equation
\begin{gather*}
  u_t - \Delta u = 0
  \quad \text{in } U \times (0,\infty) \,, \\
  u = g
  \quad \text{on } U \times \{t=0\} \,, \\
  \nu \cdot D u = 0
  \quad \text{on } \partial U \times (0,\infty) \,, \\
\end{gather*}
and $p \geq 2$, then
\[
  \lVert u (\cdot, t) \rVert_{\Space{L}{}{p}} \leq
  \lVert g \rVert_{\Space{L}{}{p}} \,.
\]

\emph{Note:} For simplicity, you may assume that $p$ is an even
integer.



\item Prove that, provided $u \in C(\bar U)$,
\[
  \lim_{p \to \infty}
  \lVert u  \rVert_{\Space{L}{}{p}}
  = \lVert u  \rVert_{\Space{L}{}{\infty}} \,.
\]

\emph{Note:} The statement is also true if $u$ is only
$\Space{L}{}{\infty}$, but the proof is not so elementary.



\item Explain why (a) and (b) imply yet another proof of the maximum
principle for the heat equation.



\end{enumerate}
\hfill (10+10+10)

\subsubsection*{Solution:}

\begin{enumerate}
\item 

\begin{eqnarray*}
&\displaystyle\dfrac{\mathrm d}{\mathrm d t}\lVert u(x,t) \rVert_{\Space{L}{}{p}}^p=\dfrac{\mathrm d}{\mathrm d t}\int_U \lvert u(x) \rvert^p \, dx=p\int_U \lvert u(x,t) \rvert^{p-1} u_{t}(x,t) \, dx\ =&\\
&\displaystyle=\ p\int_U \lvert u(x,t) \rvert^{p-1} \Delta u(x,t) \, dx=-p(p-1)\int_U \lvert u(x,t) \rvert^{p-2} \lvert D u(x,t)\rvert^2 \, dx\ +&\\
&\displaystyle +\ \int_{\partial U} \lvert u(x,t) \rvert^{p-2} \nu \cdot D u(x,t) \, dx=-p(p-1)\int_U \lvert u(x,t) \rvert^{p-2} \lvert D u(x,t)\rvert^2 \, dx\leq 0.&
\end{eqnarray*}
$$
\lVert u(x,0) \rVert_{\Space{L}{}{p}}^p=\lVert g(x) \rVert_{\Space{L}{}{p}}^p.
$$
\item 
$$
\lVert u  \rVert_{\Space{L}{}{p}}\leq \left(\int_{U} \lVert u  \rVert_{\Space{L}{}{\infty}}^p \, dx\right)^{1/p} =\lVert u  \rVert_{\Space{L}{}{\infty}}\cdot \mu(U)^{1/p}\rightarrow \lVert u  \rVert_{\Space{L}{}{\infty}}\qquad \textrm{as}\ p\to\infty.
$$
Then $\forall\; 0<M<\lVert u  \rVert_{\Space{L}{}{\infty}}$ consider $U_M:=\{|u|>M\}$. The set is nonempty, since $u \in C(\bar U)$. Then
$$
\lVert u  \rVert_{\Space{L}{}{p}}\geq \left(\int_{U_M} \lvert u(x) \rvert^p \, dx\right)^{1/p}\geq \left(\int_{U_M} \lvert M \rvert^p \, dx\right)^{1/p}=M\cdot \mu(U_M)^{1/p}\rightarrow M\qquad \textrm{as}\ p\to\infty.
$$
Since $M$ can be chosen arbitrarily close to $\lVert u
\rVert_{\Space{L}{}{\infty}}$, the statement follows.

\item 
Let $p\to\infty$.
\end{enumerate}


\item Show that for solutions of the wave equation
\[
  u_{tt} - \Delta u = 0
\]
on $\mathbb{R}^n \times (0,\infty)$, where $u(\,\cdot\,,0)$ has
compact support, the energy
\[
  E(t) = \int_{\R^n} \bigl( u_t^2 + \lvert Du \rvert ^2 \bigr) \, dx
\]
is constant in time.

\hfill (10)

\subsubsection*{Solution:}
\begin{align*}
  \dfrac{1}{2}  \dot{E}(t)
  & = \int_{\R^n} \bigl( u_tu_{tt} + Du \cdot Du_t \bigr) \,\mathrm{d} x
  \\
  & \lim\limits_{R\to \infty} \int\limits_{\partial B(0,R)}
  \nu\cdot Du \cdot u_t \,\mathrm dS + \int_{\R^n} \bigl( u_t\Delta u
  - \Delta u u_t \bigr) \, dx \\
  & = 0.
\end{align*}
From the first tot he second line, we used integration by parts.  The
first term in the second line vanishes due to the assumption on the
compact support of the initial data which, due to the finite speed of
propagation of the wave equation, remains compactly supported for all
times $t>0$, the second term vanishes when plugging in the wave
equation.

\end{enumerate}

\end{document}
