\documentclass[12pt]{article}
\usepackage{ccfonts,euler}
\usepackage{fullpage,amsmath,amsfonts}

\newcommand*{\Space}[3]{{#1}_{{\vphantom 1}#2}^{{\vphantom +}#3}}
\newcommand\PVint{\mathop{\setbox0\hbox{$\displaystyle\intop$}%
    \hskip0.2\wd0%
    \vcenter{\hrule width0.6\wd0height0.5pt depth0.5pt}%
    \hskip-0.8\wd0%
    }\mskip-\thinmuskip\intop\nolimits}

\def\R{\mathbb{R}}
\def\T{\mathbb{T}}
\def\e{\mathrm{e}}

\def\bint{{-\kern-0.6em{\int}}}

\begin{document}

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

\begin{enumerate}
\item Consider the homogeneous Helmholtz equation on $\R^3$,
\[
  (1-\Delta) u = 0 \,.
\]
\begin{enumerate}
\item Show that if $u$ is a radial solution, i.e.\ $u(x)=v(r)$ with $r
= \lvert x \rvert$, then
\[
  v - \frac2r \, v' - v'' = 0 \,.
\]

\verb"Solution".
$$
Du(x)=v'(r)Dr=v'(r)\dfrac{x}{r},
$$
$$
\Delta u(x)=D\cdot D u(x)=D\cdot\left[v'(r)\dfrac{x}{r}\right]=v''(r)+v'(r)\dfrac{3}{r}-v'(r)\dfrac{x\cdot x}{r^3}=v''(r)+v'(r)\dfrac{2}{r},
$$
$$
0=(1-\Delta) u(x) =v(r)-v''(r)-v'(r)\dfrac{2}{r}.
$$


\item Show that
\[
  v_\pm(r) = \frac{\e^{\pm r}}r
\]
are two (independent) radial solutions.  Which one would you consider
to use as a fundamental solution for the Helmholtz equation?

\verb"Solution".
$$
v_\pm'(r)=-\dfrac{\e^{\pm r}}{r^2}\pm \dfrac{\e^{\pm r}}{r},\qquad v''(r)=2\dfrac{\e^{\pm r}}{r^3}\mp \dfrac{\e^{\pm r}}{r^2}  \mp \dfrac{\e^{\pm r}}{r^2} + \dfrac{\e^{\pm r}}{r},
$$
$$
 \dfrac r {\e^{\pm r}}\left[v - \frac2r \, v' - v''\right]= 1 - \frac2r \left[ -\dfrac{1}{r}\pm 1 \right] - \dfrac{2}{r^2}\pm \dfrac{2}{r} - 1\ =1 +\dfrac{2}{r^2}\mp \dfrac{2}{r} - \dfrac{2}{r^2}\pm  \dfrac{2}{r} -1 =0.
$$
$v_{-}(r)$ vanishes at infinity and hence can be considered as a fundamental solution for the Helmholtz equation.
\end{enumerate}
\hfill(10+10)


\item Let $U \subset \R^n$ be open and assume that $u \colon U \to
\R$ is harmonic.
\begin{enumerate}
\item \label{1} Show that, for any $i=1,\dots,n$,
\[
  \lvert \partial_i u (x) \rvert
  \leq \frac{n}r \, \lVert u \rVert_{\Space{L}{}{\infty}(\partial
  B(x,r))}
\]
so long as $B(x,r) \subset U$.

\emph{Hint:} Mean value formula.

\verb"Solution". The mean value formula reads
$$
u(x)=\bint_{ B(x,r)} u(y) \mathrm dy=\bint_{\partial B(x,r)} u(y) \mathrm dS(y) = \bint_{\partial B(0,1)} u(x+ rz) \mathrm dS(z)= \bint_{ B(0,1)} u(x+ rz) \mathrm dz.
$$
Then, since $\partial u_i(x)$ is also harmonic,
$$
\partial_i u(x)=  \bint_{B(x,r)} \partial_i u(y) \mathrm dy = \dfrac{1}{\alpha(n)r^n} \int\limits_{B(x,r)} \partial_i u(y) \mathrm dy =
\dfrac{1}{\alpha(n)r^n} \int\limits_{\partial B(x,r)}  u(y) \mathrm dS(y). 
$$
Here we applied the Gauss formula. Then
$$
|\partial_i u(x)| \leq
\dfrac{\|u\|_{L^\infty(\partial B(x,r))}}{\alpha(n)r^n} \int\limits_{\partial B(x,r)}  \mathrm dS(y)=\|u\|_{L^\infty(\partial B(x,r))}\dfrac{n\alpha(n)r^{n-1}}{\alpha(n)r^n}=\dfrac{n}{r}\|u\|_{L^\infty(\partial B(x,r))}.
$$


\item Then conclude that
\[
  \lvert \partial_i u (x) \rvert
  \leq \frac{n}{\alpha(n)\,r^{n+1}} \,
       \lVert u \rVert_{\Space{L}{}{1}(B(x,2r))}
\]
provided $B(x,2r) \subset U$.

\emph{Hint:} Mean value formula.

\verb"Solution". First we note that if $y\in B(x,r)$, then $\forall\, z\in B(y,r) \,\Rightarrow\, z\in B(x,2r)$ due to the triangle inequality. Then
$$
|u(y)|\leq \dfrac{1}{\alpha(n)r^n}\int\limits_{B(y,r)}|u(z)|\;\mathrm dz \leq \dfrac{1}{\alpha(n)r^n}\|u\|_{L^1(B(x,2r))}, \quad \forall y\in B(x,r).
$$
In particular, it holds for all $y\in\partial B(x,r)$ Therefore
$$
\|u\|_{L^{\infty} (\partial B(x,r))}=\max\limits_{y\in \partial B(x,r)} |u(y)| \leq \dfrac{1}{\alpha(n)r^n}\|u\|_{L^1(B(x,2r))}.
$$
Result of \ref{1} gives
$$
|\partial_i u(x)| \leq  \dfrac{n}{r} \|u\|_{\Space{L}{}{\infty}(\partial B(x,r))}\leq \frac{n}{\alpha(n)\,r^{n+1}} \,\lVert u \rVert_{\Space{L}{}{1}(B(x,2r))}.
$$


\end{enumerate}
\hfill(10+5)


\item Let $U\subset \R^n$ be open and bounded.  Consider the Poisson
equation with so-called Neumann boundary conditions,
\begin{gather*}
  -\Delta u = f \quad \text{in } U \,, \\
  \nu \cdot D u = 0 \quad \text{on } \partial U \,.
\end{gather*}
Show that this equation cannot have a solution unless
\[
  \int_U f \, dx = 0 \,.
\]

\verb"Solution".
$$
\int\limits_U f\;\mathrm dx = -\int\limits_U \Delta u \;\mathrm dx= -\int\limits_{\partial U} \nu \cdot D u\;\mathrm dS=0.
$$

\hfill (10)


\item A function $u \in L^1_{\operatorname{loc}}(\R \times \R)$ is
called a \emph{weak solution} of the wave equation on the line if
\[
  \int_\R \int_\R u(x,t) \,
    \bigl( v_{tt} (x,t) - v_{xx}(x,t) \bigr) \, dx \, dt = 0
\]
for every $v \in C_0^\infty(\R \times \R)$.
\begin{enumerate}
\item Show that if $u \in C^2 (\R \times \R)$ is a classical solution
of the wave equation, it is also a weak solution.

\verb"Solution".
$$
\int\limits_\R \int\limits_\R u(x,t) \, \bigl( v_{tt} (x,t) - v_{xx}(x,t) \bigr) \, dx \, dt =
\int\limits_\R \int\limits_\R u_t(x,t) v_{t} (x,t) - u_x(x,t) v_{x}(x,t) \, dx \, dt\ =
$$
$$
=\ \int\limits_\R \int\limits_\R u_{tt}(x,t) v (x,t) - u_{xx}(x,t) v(x,t) \, dx \, dt=
\int\limits_\R \int\limits_\R (u_{tt}(x,t)  - u_{xx}(x,t)) v(x,t) \, dx \, dt =0.
$$
There are no boundary terms, since $v \in C_0^\infty(\R \times \R)$.

\item Verify, by explicit computation, that
\[
  u(x,t) =
  \begin{cases}
    1 & \text{for } x > t \\
    0 & \text{for } x \leq t
  \end{cases}
\]
is a weak solution of the wave equation.

\verb"Solution".
$$
\int\limits_\R \int\limits_\R u(x,t) \, \bigl( v_{tt} (x,t) - v_{xx}(x,t) \bigr) \, dx \, dt=
\int\limits_\R \int\limits_{-\infty}^{x} v_{tt} (x,t) \, dt \, dx - \int\limits_\R \int\limits_t^{+\infty} v_{xx}(x,t) \, dx \, dt =
$$
$$
=\ \int\limits_\R  v_{t} (x,x)  \, dx + \int\limits_\R  v_{x}(t,t) \, dt = \int\limits_\R  \bigl( v_{t} (y,y) + v_x(y,y)\bigr)  \, dy =0 .
$$

\end{enumerate}
\hfill (5+10)


\item Consider the Korteweg--de Vries equation
\[
  u_t - 6 u u_x + u_{xxx} = 0
\]
on $\T \times (0,\infty)$.  Show that
\begin{gather*}
  M = \int_\T u \, dx \,, \\
  E = \int_\T u^2 \, dx \,,
\intertext{and}
  H = \int_\T \bigl( \tfrac12 \, u_x^2 + u^3 \bigr) \, dx
\end{gather*}
are all constants of the motion.

\verb"Solution".
$$
\dfrac{\mathrm d}{\mathrm dt} M = \int\limits_\T u_t \,\mathrm dx= \int\limits_\T \left(6 u u_x - u_{xxx}\right) \,\mathrm dx\ = (3u^2+u_{xx})\Big|_\T=0,
$$
$$
\dfrac 1 2 \dfrac{\mathrm d}{\mathrm dt} E = \int\limits_\T uu_t \,\mathrm dx=\int\limits_\T \left(6 u^2 u_x - uu_{xxx}\right) \,\mathrm dx = 2u^3\Big|_\T+
\int\limits_\T  u_xu_{xx} \,\mathrm dx =u_x^2\Big|_\T=0,
$$
$$
\dfrac{\mathrm d}{\mathrm dt} H = \int\limits_\T \bigl(  u_xu_{xt} + 3u^2 u_t\bigr) \,\mathrm dx= \int\limits_\T  u_t \bigl( u_{xx} + 3u^2\bigr) \,\mathrm dx=
\int\limits_\T  \left(6 u u_x - u_{xxx}\right) \bigl( u_{xx} + 3u^2\bigr) \,\mathrm dx\ =
$$
$$
=\ \int\limits_\T  \bigl( 6 u u_x u_{xx} - u_{xxx}u_{xx}+ 18 u_x u^3  - 3u^2 u_{xxx} \bigr) \,\mathrm dx\ =
$$
$$
=\ \int\limits_\T  \bigl( 6 u u_x u_{xx} - [u^2_{xx}]_x+ \dfrac 9 2 [u^4]_x  - 6u u_x u_{xx} \bigr) \,\mathrm dx =
\left[\dfrac 9 2 u^4- u^2_{xx}\right]\Bigg|_\T= 0.
$$
There is no boundary terms, since $\T$ has no boundary.

\hfill(5+5+10)

\item Let $H$ be a Hilbert space and $u_n$ a sequence in $H$.  Show
that the following are equivalent.
\begin{itemize}
\item[(i)] $u_n \to u$ strongly;
\item[(ii)] $u_n \rightharpoonup u$ weakly and $\lVert u_n \rVert \to
\lVert u \rVert$.
\end{itemize}

\verb"Solution".
"$\Rightarrow$" By definition,
$$
u_n \to u \qquad \Leftrightarrow \qquad \|u_n-u\|_{H}\to 0.
$$
Using the triangle inequality we conclude
$$
\lVert u_n \rVert_{H} - \lVert u \rVert_{H} =  \lVert u_n-u+u \rVert_{H} - \lVert u \rVert_{H}  \leq \lVert u_n-u \rVert_{H}+\lVert u \rVert_{H} - \lVert u \rVert_{H} = \lVert u_n-u \rVert_{H},
$$
$$
\lVert u_n \rVert_{H} - \lVert u \rVert_{H} =  \lVert u_n \rVert_{H} - \lVert u-u_n+u_n \rVert_{H}  \geq \lVert u_n \rVert_{H}-\lVert u-u_n \rVert_{H} - \lVert u_n \rVert_{H} = - \lVert u-u_n \rVert_{H}.
$$
Hence,
$$
\big|\lVert u_n \rVert_{H} - \lVert u \rVert_{H} \big|\leq \lVert u_n-u \rVert_{H} \to 0.
$$
The Cauchy-Schwarz inequality gives
$$
\big|\langle u_n-u,\, w \rangle\big| \leq \lVert u_n-u \rVert_{H} \; \lVert w \rVert_{H} \to 0 \qquad \forall w\in H.
$$
Consequently,
$$
\langle u_n,\, w \rangle \to \langle u,\, w \rangle \quad \forall w\in H \qquad \Leftrightarrow \qquad u_n \rightharpoonup u.
$$

"$\Leftarrow$" By definition,
$$
\langle u_n-u,\, w \rangle \to 0 \qquad \forall w\in H.
$$
Take $w:=u_n-u$. Then
$$
\lVert u_n-u \rVert^2_{H}=\langle u_n-u,\, u_n-u \rangle = \lVert u_n \rVert^2_{H} -2\langle u_n,\,u \rangle + \lVert u \rVert^2_{H}\to \lVert u_n \rVert^2_{H} - \lVert u \rVert^2_{H}.
$$
Since the right hand side vanishes, we have $u_n \to u$ strongly.


\hfill (10+10)

\end{enumerate}

\end{document}
