\documentclass[minion]{homework}
\usepackage{cmacros}
\def\calB{\mathcal{B}}
\DeclareMathOperator{\Ker}{{\rm Ker}}
\DeclareMathOperator{\Image}{{\rm Image}}
\def\Fbb{\mathbb{F}}
\begin{document}
\doclabel{Math 617: Homework 11}
\docdate{Due: April 20, 2012}
\begin{aproblems}
\aproblem  Suppose $\alpha$, $\beta$, and $\gamma$ are continuous
on $[a,b]$ and $\alpha\neq 0$.  
Given $x_0,v_0\in\Reals$, and $f\in C[a,b]$,
show that there exists a unique $u\in C^2[a,b]$
such that
$$
\alpha u'' + \beta u' +\gamma u = f
$$
satisfying $u(b)=x_0$ and $u'(b)=v_0$.

\aproblem  Suppose $\{u_k\}$ is an orthonormal basis
for a Hilbert space $X$, let $\{b_k\}\in\ell_2$,
and let $\{c_k\}$ be a bounded sequence in 
$C^{2}[a,b]$.  Show that
\[
v(t) = \sum_{k=1}^\infty b_k c_k(t) u_k
\]
is a well defined function from $\Reals$ to $X$
Moreover, it is differentiable and
\[
v'(t) = \sum_{k=1}^\infty b_k c_k'(t) u_k
\]
You should decide for yourself what 
differentiability means in this context.

\aproblem D \& M 6.15

\aproblem Let $T:L^2_\omega\ra L^2_\omega $ be the compact operator we 
constructed in class for solving the equation
\begin{equation}\label{eq1}
-\frac{1}{\omega} L u = f
\end{equation}
together with separated boundary conditions.  We showed that if
$f$ is continuous then $Tf$ is a $C^2$ solution of this problem.
Now show that if $f\in L^2_\omega$, then $u$ solves equation
\eqref{eq1}
in the sense of distributions, and that $u\in H^2([a,b])$.
Hint: approximate $f$ with continuous functions.

\aproblem Suppose $\Omega\subseteq\Reals^n$ is open and 
$f\in L^1_{\rm loc}(\Omega)$.  Show that the map $T_f:\mathcal{D}(\Omega)\ra\Fbb$ given by 
\[
T_f(\phi) = \int_\Omega f\phi
\]
is an element of $\mathcal{D}'(\Omega)$.


\hrule

The following problem is part of the take-home final.  You can
start thinking about it now.

\aproblem Suppose $p\in C^1[a,b]$.
Show that given $f\in L^2[a,b]$,
and $T>0$, there exists a function $u:[0,T]\ra L^2[a,b]$
such that
\begin{enumerate}
\item $u(0)=f$.
\item $u$ is continuous on $[0,T]$.
\item $u$ is differentiable on $(0,T]$.
\item $\frac{d}{dt} u=(p u')'$; the derivative on the right is in the 
sense of distributions.
\item For all $t>0$, $u(t,a)=0=u(t,b)$.
\end{enumerate}



\end{aproblems}
\end{document}