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\title{MAT 280: Laplacian Eigenfunctions: Theory, Applications, and Computations\\
Lectures 9+10: Laplacian Eigenvalue Problems for General Domains --
II. Computation of Eigenvalues}
\author{Lecturer: Naoki Saito\\
        Scribe: Patrick Satarzadeh/Allen Xue}
\date{May 1, 2007}

\begin{document}
\maketitle

The basic references for this lecture are the texts by
Strauss~\cite[Sec.~11.3]{STRAUSS}, and Courant and
Hilbert~\cite[Sec.~VI.1, VI.2]{COURANT-HILBERT}. For more advanced
treatments, see \cite[Chap. 4]{DAVIES} and \cite[Chap. 1]{HENROT}.

\ \\
The Dirichlet-Laplacian eigenvalues, in particular, $\lambda_1$ (the
smallest), is quite important in many applications. To compute them
we will use the {\em Rayleigh quotient}.

\ \\
For any trial function $w \in C_0^2(\Omega)$, with $w \not\equiv 0$,
we have
\[\lambda_1 \leq \frac{\rVert \nabla w
\rVert^2}{\rVert w \rVert^2}.\]

If $w=\varphi_1$, we get exactly $\lambda_1$, but we do not know
$\varphi_1$ at this point. So, we should be satisfied with a
moderately clever choice of $w$ that might provide a relatively good
approximation.

\ \\
\begin{ex}
In Lecture 8, we already discussed the approximate minimum of (MP):
\[m = \displaystyle\min_{{w\in C_{0}^2(\Omega)\atop{w\not\equiv
0}}} \dfrac{ \Int{0}{1} \left( w^{'} \right)^2 \, \dd{x} }{
\Int{0}{1} w^2 \, \dd{x}},\quad w\in C^2_0(0,1).\]

We choose the trial function $w=\sqrt{30}x(1-x)$, which gives an
estimate $\hat \lambda_1 = 10$ where the true eigenvalue is
$\lambda_1 = \pi^2\approx 9.8696$.

\ \\
In this case we have done well, but in general, it is too difficult
to come up with a good trial function to produce a good
approximation of $\lambda_1$ or $\lambda_j$.
\end{ex}

\section{Rayleigh-Ritz Approximation (RRA)}
Let $w_1, \cdots, w_n \in C_0^2(\Omega)$, $w_j\not\equiv 0$ be
arbitrary trial functions. Let $A=(a_{jk})$, $B=(b_{jk})$, where
$a_{jk}$, $b_{jk}$ are defined as:
\begin{equation}\label{ABentry}
\begin{array}{lcl}
a_{jk} & \define & \inner{\nabla w_j}{\nabla
w_k}=\Int{\Omega}{}\nabla
w_j\cdot\nabla w_k\,\dd{\bx},\\[8pt]
b_{jk} & \define & \inner{w_j}{w_k}=\Int{\Omega}{}w_jw_k\,\dd{\bx},
\end{array}
\end{equation}

for $j,k=1,\ldots,n$. Thus, $A$ and $B$ are $n\times n$ symmetric
matrices. Then

\begin{center}
\framebox[0.8\textwidth]{\parbox{0.8\textwidth}{The roots of the
polynomial equation $\text{det}(A-\lambda B)$ are approximation to
the first $n$ eigenvalues $\lambda_1,\cdots,\lambda_n$.}}
\end{center}

\ \\
Before presenting an informal proof of this, let us consider the
following example.

\begin{ex}
Consider the {\emph radial} vibrations of a circular membrane
$\Omega=\{(x,y)\cond x^2 + y^2 < 1\}$. Then the Dirichlet-Laplacian
(DL) problem is $-\varDelta u=\lambda u$, which can be written in
the polar coordinates $(r,\theta)$ as

\[(-r u_r)_r = \lambda
r u\quad\ \ (0<r<1),\quad u=0\ \text{at } r=1.\]

Here we apply the fact that radial function depends only on the
parameter $r$, i.e., $u(x,y)=u(r)$.

\ \\
Then the Rayleigh quotient is given by

\beq\label{ray} Q =\fr{\displaystyle\iint\rvert \nabla w
\rvert^2\dd{\bx}}{\displaystyle\iint w^2 \dd{\bx}} = \frac{2\pi
\Int{0}{1} r w_r^2 \dd{r}}{2\pi \Int{0}{1} r w_r^2 \dd{r}}. \eeq

Now we ask ``what trial functions should we use?'' The trial
functions are required to satisfy the boundary conditions
\[w(0):\ \text{finite},\qquad w_r(0)=0=w(1).\]

A simple choice of a pair of them is $1-r^2$ and $(1-r^2)^2$. In
this case, we can compute the entries for matrices $A$ and $B$ in
\eqref{ABentry} as follows
\[A=\left(
      \begin{array}{cc}
        2\pi & 4\pi/3 \\
        4\pi/3 & 4\pi/3 \\
      \end{array}
    \right),\qquad \quad B=\left(
                             \begin{array}{cc}
                               \pi/3 & \pi/4 \\
                               \pi/4 & \pi/5 \\
                             \end{array}
                           \right).\]

Hence
\begin{eqnarray*}
\text{det}(A-\lambda B) & = & \text{det}\left[\left(
                                            \begin{array}{cc}
                                              2\pi-\lambda\pi/3 & 4\pi/3-\lambda\pi/4 \\
                                              4\pi/3-\lambda\pi/4 & 4\pi/3-\lambda\pi/5 \\
                                            \end{array}
                                          \right)\right]\\[8pt]
                        & = &
                        \pi^2\left\{\left(2-\fr{\lambda}{3}\right)\left(\fr{4}{3}-\fr{\lambda}{5}\right)-\left(\fr{4}{3}-\fr{\lambda}{4}\right)^2\right\}.
\end{eqnarray*}

Solving the characteristic equation $\text{det}(A-\lambda B)=0$
gives us the estimates

\beq \hat \lambda_1 = \frac{64-8\sqrt{34}}{3} \simeq 5.7841, \qquad
\quad \hat \lambda_2 = \frac{64+8\sqrt{34}}{3} \simeq 36.8825 \eeq

On the otherhand, the true eigenvalues are given by \beq \lambda_1 =
j_{0,1}^2 = 5.783, \qquad \quad \lambda_2 = j_{0,2}^2 = 30.4705 \eeq
where $j_{0,k}$ denote the $k$th zero of the Bessel function of the
first kind of order $0$. It becomes clear that our estimate for the
first eigenvalue is good, but the second has significant error. To
improve, we need to use either better trial functions or possibly
use three trial functions.
\end{ex}

\subsection{Informal derivation of the Rayleigh-Ritz Approximation
(RRA)}

Let $\{w_j\}_{j=1}^n \subset C_0^2(\Omega)$ be arbitrary trial
functions and linearly independent. As an approximation to the true
minimum problem $\text{(MP)}_n$ (see Lecture 8), let us impose an
additional condition for $w$:

\beq\label{one} w(\bx) = \sum_{k=1}^n c_k w_k(\bx),\qquad c_k:\
\text{some constant}\eeq

So we only seek $w$ of the form \eqref{one}. Therefore, in general
$\hat{\lambda}_1\geq\lambda_1$, because we impose more constraints
in $\text{(MP)}_n$. If we are incredibly smart, so that $w(\bx)$
would be an eigenfunction, then we have $-\varDelta w = \lambda w$,
and
\begin{equation}
\begin{array}{lcl}
\inner{\nabla w}{\nabla w_j} & = & \Int{\Omega}{}\nabla w\cdot\nabla
w_j\,\dd{\bx}\\[8pt]
& \overset{(*)}{=} & -\Int{\Omega}{}\varDelta w\cdot
w_j\,\dd{\bx}+\Int{\partial\Omega}{}\normalderiv{w}w_j\,\dd{\bx}\\[8pt]
& = & \lambda\inner{w}{w_j},
\end{array}
\end{equation}

where we used Green's first identity in (*). Therefore, recalling
the definitions of $a_{jk}$ and $b_{jk}$, we get
\begin{equation}
\inner{\Sum{k}c_k\nabla w_k}{\nabla
w_j}=\lambda\inner{\Sum{k}c_kw_k}{w_j}\ \Rightarrow
\Sum{k}c_ka_{jk}=\lambda\Sum{k}c_kb_{jk}.
\end{equation}

So, we can write them in a matrix form
\[A\bc=\lambda B\bc,\qquad \bc\neq \bzero.\]
which leads to $\text{det}(A-\lambda B)=0$.

\ \\
If we are not so smart, which is usually the case, we can still use
this determinant as our approximation method. This leads to the {\em
Minimax Principle}. See \cite[Sec. 1.3]{HENROT} for more details.

\section{Minimax Principle}

In reality, what we really want is an exact formula to compute the
eigenvalues instead of an approximation.

\ \\
Let $\lambda_1^* \le \lambda_2^* \le \cdots \le \lambda_n^* $ be the
roots of $\text{det}(A-\lambda B)=0$. From numerical linear algebra
(see Sec.~\ref{sec:intermission}), it is easy to see that

\[\lambda_n^* = \max_{\bc \neq \bzero} \frac{\bc^T A \bc}{\bc^T B \bc} =
\max_{\bc \neq \bzero} \frac{\inner{A \bc}{\bc}}{\inner{B
\bc}{\bc}}\]

Thus, we have
\begin{equation}\label{lambdan}
\begin{array}{lcl}
\lambda_n^* & = & \displaystyle\max_{\bc \neq \bzero} \fr{\sum_{j,k}
a_{jk}
c_j c_k}{\sum_{j,k} b_{jk} c_j c_k}\\[14pt]
& = & \displaystyle\max_{\bc\neq\bzero}\fr{\inner{\nabla(\sum_j
c_jw_j)}{\nabla(\sum_k c_kw_k)}}{\inner{\sum_j c_jw_j}{\sum_k
c_kw_k}}\\[14pt]
& = &
\displaystyle\max_{\bc\neq\bzero\atop{w\in\text{span}\{w_1,\ldots,w_n\}}}\fr{||\nabla
w||^2}{||w||^2}
\end{array}
\end{equation}

with $w \in \text{span}\{w_1, \cdots w_n\}$. Hence we have from the
$\text{(MP)}_n$

\[\lambda_n \le \lambda_n^*\].

Thus this allows us to take the minimum on the RHS to get

\beq \lambda_n = \min\lambda_n^*, \eeq

where the minimization is taken over all possible choices of
$\{w_1,\ldots,w_n\}\subset C^2_0(\Omega)$.

\ \\
\begin{thm}[Minimax Principle]\label{thm:minimax}
Let $\{ w_j\}_{j=1}^n \subset C_0^2(\Omega)$ be an arbitrary set of
$n$ trial functions. Define $\lambda_n^*$ by \eqref{lambdan}, Then
the $n$th eigenvalue is

\[\lambda_n=\min\lambda^*_n\]

where $\min$ is taken over all possible choices of the $n$ trial
functions $\{ w_j\}_{j=1}^n \subset C_0^2(\Omega)$.
\end{thm}

\ \\
Before presenting the proof, consider the following intermission
which plays a role in the proof.

\subsection{Intermission: Rayleigh Quotient $\&$ Linear Algebra}\label{sec:intermission}

This intermission is taken from \cite[Lect. 27]{Trefethen}.

\ \\
Let $A \in \Rf^{m\times m}$ be symmetric and $x \in \Rf^m$. Then we
know $A$ has real eigenvalues $\lambda_1,\ldots,\lambda_m$ and real
orthogonal eigenvectors $\bvphi_1,\ldots,\bvphi_m$.

\ \\
Define

\beq r(\bx) \triangleq \frac{\bx^T A \bx}{\bx^T \bx} \eeq

Clearly if $\bx =\bvphi_j$, then $r(\bvphi_j) = \lambda_j$. Of
course, we do not know $\bvphi_j$ a priori. So a motivating question
is: ``Given $\bx$, what scalar $\alpha$ acts most like an eigenvalue
for $\bx$?''

\ \\
The answer to this question leads to the following least square
problem
\begin{equation}\label{LS}
\displaystyle\min_{\alpha\in\Rf} \rVert A\bx-\alpha \bx\rVert_2.
\end{equation}

The answer to this least square problem is $\hat \alpha = r(\bx)$.
The reason is as follows:
\begin{eqnarray*}
\rVert A\bx - \alpha\bx \rVert_2^2 & = & \inner{(A-\alpha
I)\bx}{(A-\alpha I)\bx}\\
& = & \bx^T(A-\alpha I)^T(A-\alpha I)\bx \\
& = & \bx^TA^2\bx -2\alpha \bx^TA\bx + \alpha^2\bx^T\bx
\end{eqnarray*}

By $\pdif{}{\alpha}\rVert A\bx-\alpha\bx\rVert^2_2=0$, we get
$2\alpha \bx^T\bx-2\bx^TA\bx=0$. So

\begin{equation}\label{alpha}
  \alpha=\frac{\bx^TA\bx}{\bx^T\bx}
\end{equation}

is the solution of the least square problem \eqref{LS}.

\ \\
Thus, if $\bx$ is close to one of the eigenvectors, then the
$\alpha$ in \eqref{alpha} should be close to the corresponding
eigenvalue. Also notice that

\beq \nabla r(\bx) = \frac{2}{\bx^T\bx}(A\bx- r(\bx)\bx). \eeq

So if $\bx = \bvphi_j$, then $\nabla r (\bvphi_j) = 0$. Conversely
if $\nabla r(\bx) = 0$ for $\bx \neq \bzero$, then $\bx$ is an
eigenvector of $A$ and $r(\bx)$ is the corresponding eigenvalue.
Thus the key point is that $\{\bvphi_j\}$ are the stationary points
of $r(\bx)$. Further we can show that \beq r(\bx) - r(\bvphi_j) =
O(\rVert \bx-\bvphi_j \rVert^2)\quad \text{as}\  \bx\rightarrow
\bvphi_j \eeq This property is the basis for the method of power
iterations. See \cite[Lecture 27]{Trefethen} and \cite[Sec.
8.2]{GOLUB-VANLOAN}.

\subsection{Proof of Theorem~\ref{thm:minimax}}
\begin{proof}
First let's fix $\{w_1, \cdots w_n\} \subset C_0^2(\Omega)$. Then
choose $c_1, c_2, \cdots c_n$ (not all zeros) such that

\[w(x)\define \Sm{k=1}{n} c_k w_k(x)\ \ \perp\ \ \{\varphi_j,\cdots,\varphi_{n-1}\},\]

i.e., $\inner{w}{\varphi_k}=\Sm{j=1}{n}c_j\inner{w_j}{\varphi_k}=0,\
\ k=1,\ldots,n-1.$

\ \\
It is clear now that we have $n$ unknowns and $n-1$ equations.
Therefore it is possible to find $\{ c_j \}_{j=1}^n$ not all zero.
By the minimum principle, we have

\beq \lambda_n \le \frac{\rVert \nabla w \rVert^2}{\rVert w
\rVert^2}. \eeq

On the other hand, the maximum in \eqref{lambdan} is taken over all
possible $\{c_j\}$. So

\beq \frac{\rVert \nabla w \rVert^2}{\rVert w \rVert^2} \le
\lambda_n^*. \eeq

Therefore, we have

\beq \lambda_n \le \frac{\rVert \nabla w \rVert^2}{\rVert w
\rVert^2} \le \lambda_n^* \label{three}. \eeq

Now \eqref{three} must be true for all possible $\{ w_1, \cdots
w_n\} \subset C_0^2 (\Omega)$, which allows us to take minimum on
the RHS, i.e., \beq \lambda_n \le \min \lambda_n^* \eeq where the
minimum is taken over all possible $w_j$. Yet we still need to show
that $\lambda_n = \min \lambda_n^*$ is attainable.

\ \\
Choose $w_j = \varphi_j$ for $j = 1,\cdots n$ where $\rVert
\varphi_j \rVert = 1$. With this choice we have

\beq \lambda_n^* = \max_{\bc \neq \bzero} \frac{\rVert \nabla
(\sum_{j} c_j \varphi_j) \rVert^2}{\rVert \sum_{j} c_j \varphi_j
\rVert^2}. \eeq

Using Green's first identity (G1) results in

\[\inner{\nabla \varphi_j}{\nabla
\varphi_k}=\inner{-\varDelta\varphi_j}{\varphi_k}=\lambda_j\delta_{jk}.\]

Therefore, we have

\beq \lambda_n^* = \max_{\bc \neq \bzero} \frac{\sum_{j} \lambda_j
c_j^2}{\sum_j c_j^2}\le \frac{\lambda_n \sum_j c_j^2}{\sum_j c_j^2}
= \lambda_n \eeq

so that $\lambda_n^* \le \lambda_n$. Therefore $\lambda_n = \min
\lambda_n^*$ is attainable.
\end{proof}

\begin{thebibliography}{10}

\bibitem{Trefethen}
{\sc L. N. Trefethen and D. Bau III}: ``Numerical Linear Algebra'',
SIAM, 1997

\bibitem{STRAUSS}
{\sc W.~A. Strauss}, {\em Partial Differential Equations: An
Introduction}, John Wiley \& Sons, 1992.

\bibitem{COURANT-HILBERT}
{\sc R. Courant and D. Hilbert}, {\em Methods of Mathematical
Physics}, Wiley-Interscience, 1953.

\bibitem{DAVIES}
{\sc E. B. Davies},{ \em Spectral Theory and Differential
Operators}, Cambridge Univ. Press, 1995.

\bibitem{HENROT}
{\sc A. Henrot},{ \em Extremum Problems for Eigenvalues of Elliptic
Operators}, Birkh$\ddot{a}$user, 2006.

\bibitem{GOLUB-VANLOAN}
{\sc G. H. Golub and C. F. Van Loan}, {\em Matrix Computations, 3rd
Ed.}, Johns Hopkins Univ. Press, 1996.

\end{thebibliography}


\end{document}
