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\def\Rl{\mathbb{R}}
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\centerline{{\sc Problem Set 5}}
\centerline{Math 207B, Winter 2012}
\centerline{Due: Fri., Feb. 17}


\bigskip\noindent
{\bf 1.} Consider the non-homogeneous regular Sturm-Liouville problem
\begin{align}
\begin{split}
&-[p(x)u']' + q(x) u = \lambda u + f(x),\qquad a<x<b,
\\
&u(a) = 0,\qquad u(b) = 0,
\end{split}
\label{sleq}
\end{align}
with eigenvalues $\lambda_1<\lambda_2<\dots<\lambda_n < \dots$, and (real-valued) orthonormal eigenfunctions
$\{\phi_1,\phi_2,\dots,\phi_n,\dots\}$.
Let $P_n : L^2(a,b) \to L^2(a,b)$ denote the projection onto the orthogonal complement of the $n$th eigenspace, defined by
\[
P_n f(x) = f(x) - \left(f,\phi_n\right) \phi_n(x).
\] 
(a) Suppose that $\lambda = \lambda_n$ in (\ref{sleq}). If $\left(f,\phi_n\right)=0$, show that a solution of (\ref{sleq})
is given by
\begin{align}
\begin{split}
&u(x) = \int_a^b G(x,\xi;\lambda_n) f(\xi)\,d\xi,
\\
&G(x,\xi;\lambda_n) = \sum_{k\ne n} \frac{\phi_k(x) \phi_k(\xi)}{\lambda_k-\lambda_n}.
\end{split}
\label{modgr}
\end{align}
Is this solution unique?

\smallskip\noindent
(b) For arbitrary $f\in L^2(a,b)$ show that $u(x)$ defined in (\ref{modgr}) is the unique solution of the following problem:
\begin{align*}
&-[p(x)u']' + q(x) u = \lambda_n u + P_n f(x),\quad a<x<b,
\\
&u(a) = 0,\quad u(b) = 0,\qquad P_n u = u.
\end{align*}
(The function $G$ is called the modified Green's function.)


\smallskip\noindent
(c) Compute the modified Green's function $G(x,\xi;\lambda_n)$ for the problem
\begin{align*}
&- u'' = \lambda_n u + f(x),\quad 0<x<1,
\\
&u'(0) = 0,\qquad u'(1) = 0
\end{align*}
in terms of solutions of the homogeneous equation, where $\lambda_n = n^2\pi^2$ for $n=0,1,2,\dots$.



\bigskip\noindent
{\bf 2.} Consider the functional $J : X \to \Rl$ defined by
\[
J(u) = \int_a^b \left[\frac{1}{2} (u'')^2 + \frac{1}{2}q u^2 - fu\right]\, dx
\]
where $f,q : [a,b]\to \Rl$ are continuous functions and $X= H_0^2(a,b)$ is the Sobolev space
\[
X = \left\{u : u, u', u''\in L^2(a,b), u(a)=u'(a) = u(b)=u'(b) = 0\right\}
\]
Derive the Euler-Lagrange equation satisfied by a smooth minimizer $u\in C^4[a,b]$ of $J$. What 
is the variational derivative ${\delta J}/{\delta u}$ of $J$?



\bigskip\noindent
{\bf 3.} Consider the regular weighted Sturm-Liouville eigenvalue problem
\begin{align*}
&-(pu')' + q u = \lambda r u \qquad a<x<b,
\\
&u(a) = 0,\qquad u(b) =0
\end{align*}
where $p(x)$, $q(x)$, $r(x)$ are smooth real-valued
functions and $p, r >0$.

\smallskip\noindent(a) If the eigenvalues are
$\lambda_1 < \lambda_2 < \dots, \lambda_n < \dots$, show that
the minimum eigenvalue $\lambda_1$ is given by
\[
\lambda_1 = \min_{u\in H_0^1(a,b)} R(u),\qquad R(u) = \frac{\int_a^b \left[p(u')^2 + q u^2\right]\, dx}{\int_a^b r u^2\, dx}.
\]
(b) Consider the (singular) Sturm-Liouville problem
\begin{align*}
-(xu')' = \lambda x u\qquad 0<x<1
\\
\mbox{$xu'\to 0$ as $x\to 0$},\qquad u(1) = 0 
\end{align*}
with Rayleigh quotient
\[
R(u) = \frac{\int_0^1 x(u')^2 \, dx}{\int_0^1 x u^2\, dx}.
\]
Assuming the result in (a) holds for this singular problem, use a trial function of the form
\[
u(x) = (1-x) (ax + b)
\]
to obtain an upper bound for the minimum eigenvalue $\lambda_1$. Compare your answer with the numerical result
$\lambda_1 = 5.783\dots$.

\smallskip\noindent
\textbf{Remark}. After a rescaling, this ODE is the zeroth-order Bessel equation. The eigenfunctions
are $J_0(\sqrt{\lambda_n}x)$ where $J_0(z)$ is the Bessel function of order zero
and $\sqrt{\lambda_n}=j_{0,n}$ is the $n$th zero of $J_0(z)$. The first zero is $j_{0,1} \approx 2.4048$.



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