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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
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\begin{document}


\begin{center}
Rayleigh Quotient
\end{center}

\bigskip 
\begin{align*}
-\nabla \cdot (p\nabla u)+qu& =\lambda \widehat{m}u\qquad \text{in }D \\
u& =0\qquad \quad \text{on }\partial D
\end{align*}%
We know that there are an infinite number of (positive) eigenvalues tending
to $\infty $ . \ \ \ \ \ \ \ \ Denote them by%
\begin{equation*}
0\,<\lambda _{1}\leq \lambda _{2}\leq \,...
\end{equation*}

\noindent Define%
\begin{equation*}
(f,g)=\tiiint \widehat{m}(x)f(x)\overline{g(x)}dx
\end{equation*}

\noindent Let%
\begin{equation*}
Q(w)=\frac{\tiiint \left[ p(x)|\nabla w|^{2}+q(x)|w|^{2}\right] dx}{\tiiint 
\widehat{m}(x)|w|^{2}dx}
\end{equation*}

\begin{theorem}
Define%
\begin{equation*}
m=\min Q(w)\qquad w=0\text{ on }\partial D\text{\quad }w\text{ smooth}
\end{equation*}
Then%
\begin{equation*}
\lambda _{1}=m
\end{equation*}

\noindent and the minimizing function $u(x)$is an eigenfunction, i.e.%
\begin{equation*}
-\nabla \cdot (p\nabla u)+qu=\lambda _{1}\widehat{m}u\text{ \ \ \ in }D\text{
\ }w=0\text{ on }\partial D
\end{equation*}
\end{theorem}

\begin{proof}
Consider the case $p=1$, $q=0$ and $\widehat{m}=1$ (we shall now use $m$ as
minimum of $Q$).

So for an eigenfunction $u$ with eigenvalue $\lambda $ 
\begin{equation*}
-\Delta u=\lambda u
\end{equation*}

Multiply by $u$ and integrate we get%
\begin{eqnarray*}
-\tiiint u\Delta udx &=&\lambda \tiiint u^{2}dx \\
\text{by Green's identity \ \ \ \ }\tiiint |\nabla u|^{2}dx &=&\lambda
\tiiint u^{2}dx \\
\lambda  &=&\frac{\tiiint |\nabla u|^{2}dx}{\tiiint u^{2}dx}
\end{eqnarray*}%
So we define%
\begin{equation*}
Q(w)=\frac{\tiiint |\nabla w|^{2}dx}{\tiiint |w|^{2}dx}
\end{equation*}

Then

\begin{equation*}
Q(w)=\frac{\tiiint |\nabla w|^{2}dx}{\tiiint |w|^{2}dx}
\end{equation*}

\noindent Since $m$ is a minimum and $u(x)$ is the minimizer function%
\begin{equation*}
m\leq f(\varepsilon )=\frac{\tiiint |\nabla \left( u+\varepsilon v\right)
|^{2}dx}{\tiiint \left( u+\varepsilon v\right) ^{2}dx}
\end{equation*}

\noindent Expanding%
\begin{equation*}
f(\varepsilon )=\frac{\tiiint |\nabla u|^{2}+2\varepsilon \nabla u\cdot
\nabla v+\varepsilon ^{2}|\nabla v|^{2}dx}{\tiiint \left( u^{2}+2\varepsilon
uv+v^{2}\right) dx}
\end{equation*}

\noindent Differentiating with respect to $\varepsilon $ and setting $%
\varepsilon =0$ we get%
\begin{equation*}
f^{^{\prime }}(0)=\frac{\tiiint u^{2}dx\cdot \tiiint 2\nabla u\cdot \nabla
vdx-\tiiint |\nabla u|^{2}dx\cdot 2\tiiint uvdx}{\left( \tiiint
u^{2}dx\right) ^{2}}
\end{equation*}

\noindent So%
\begin{equation*}
\tiiint \nabla u\cdot \nabla vdx=\frac{\tiiint |\nabla u|^{2}dx}{\tiiint
u^{2}dx}\tiiint uvdx=m\tiiint uvdx
\end{equation*}

\noindent By the first Green lemma%
\begin{equation*}
\tiiint \left( \Delta u+mu\right) vdx=0
\end{equation*}

\noindent Since $v$ is arbitarary 
\begin{equation*}
\Delta u+mu=0
\end{equation*}

\noindent i.e. $m$ is an eigenvalue of $-\Delta u$.

However,%
\begin{align*}
m &\leq \frac{\tiiint |\nabla v_{j}|^{2}dx}{\tiiint v_{j}^{2}dx}= \frac{%
\tiiint v_{j}\Delta v_{j}dx}{\tiiint v_{j}^{2}dx} \\
&=\frac{\tiiint \left( \lambda _{j}v_{j}\right) v_{j}dx}{\tiiint v_{j}^{2}dx}
=\lambda _{j} \qquad \text{ for all j}
\end{align*}

\noindent Hence, $m$ is the smallest eigenvalue of $-\Delta u$
\end{proof}

\noindent Using these ideas (integration by parts) we can prove that

\begin{theorem}
The eigenvalues are all non-negative if for every eigenfunction

\begin{itemize}
\item $-p u \frac{d u}{dx} {\large \vert_a^b \ge 0}$

\item $q \le 0$
\end{itemize}

\newpage

Other eigenvalues

\bigskip

\begin{theorem}
Minimum principle for n-th eigenvalue:

Suppose $\lambda _{1}$,...$\lambda _{n-1}$ are known with their
eigenfunctions $v_{1}(x)$,...$v_{n-1}(x)$. Then

\begin{align*}
\lambda _{n}& =\left\{ \min \frac{\tiiint |\nabla w|^{2}dx}{\tiiint |w|^{2}dx%
}\quad w\neq 0\quad w=0\text{ on }\partial D\quad w\text{ in }C^{2}\right. \\
& \left. \text{and }0=(w,v_{1})=...=(w,v_{n-1})\right\}
\end{align*}
\end{theorem}

\noindent We wish to find a formulation where $\lambda_n$ doesn't explicitly
depend on the previous eigenvalues and eigenfunctions.

\begin{theorem}
Minimax principle

Given n-1 linearly independent piecewise continuous functions $v_1,...,v_n$

we define%
\begin{equation*}
\Lambda _{n}(v_{1},...,v_{n})=\inf \tiiint |\nabla w|^{2}dx\quad
(w,v_{j})=0\quad (w,w)=1\quad w=0\text{ on }\partial D
\end{equation*}%
Then%
\begin{align*}
\lambda_n &= \sup \Lambda_n (v_{1},...,v_{n})\quad \text{all possible } v_{j}
\\
&= \max_{v_{1},,,v_{n}}\left\{ \min_{w}\tiiint |\nabla w|^{2}dx\quad
(w,v_{j})=0\quad (w,w)=1\quad w=0\text{ on }\partial D\right\}
\end{align*}
\end{theorem}

\newpage

example

\bigskip

\begin{align*}
u^{\prime \prime} + \lambda u &= 0 \qquad 0 \le x \le l \\
u(0)&=u(l)=0
\end{align*}

\noindent exact: $\ \ \lambda _{n}=\left( \frac{n\pi }{l}\right) ^{2}$ so $%
\lambda _{1}=\left( \frac{\pi }{l}\right) ^{2}$

\bigskip

\begin{equation*}
\lambda _{1}=\min_{w}\frac{\tint\limits_{0}^{l}\left( \frac{\partial w}{%
\partial x}\right) ^{2}dx}{\tint\limits_{0}^{l}w^{2}dx}\leq \frac{%
\tint\limits_{0}^{l}\left( \frac{\partial \widehat{w}}{\partial x}\right)
^{2}dx}{\tint\limits_{0}^{l}\widehat{w}^{2}dx}
\end{equation*}

\noindent choose%
\begin{align*}
\widehat{w}(x) &= x(l-x)\quad \text{so }\widehat{w}(0)=\widehat{w}(l)=0 \\
\widehat{w}^{\prime}(x) &= l-2x \\
\frac{\tint\limits_{0}^{l}\left( \frac{\partial \widehat{w}}{\partial x}%
\right)^2 dx}{\tint\limits_{0}^{l}\widehat{w}^{2}dx} &= \frac{%
\tint\limits_{0}^{l}\left( l-2x\right)^2 dx}{\tint%
\limits_{0}^{l}x^{2}(l-x)^{2}dx}=\frac{10}{l^{2}}
\end{align*}

\noindent So we have $\lambda _{1}\leq \frac{10}{l^{2}}$ while the exact
answer is $\lambda _{1}=\frac{\pi ^{2}}{l^{2}}\sim \frac{9.8}{l^{2}}$
\end{theorem}

\end{document}
