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\title{MAT 280: Laplacian Eigenfunctions: Theory, Applications, and Computations\\
Lecture 4: Diffusions and Vibrations in 2D and 3D --- I. Basics}
\author{Lecturer: Naoki Saito\\
        Scribe: Brendan Farrell/Allen Xue}
\date{April 10, 2007}
\begin{document}
\maketitle

The basic reference for this lecture is \cite[Sec.10.1]{STRAUSS}.
\begin{section}{Wave Equation and Heat Equation}
%\begin{section}{Basics} Consider a finite domain
Consider $\Omega\subset\Rf^d,\quad d=2,3,\cdots.$
\begin{equation}
\label{waveheat}
\begin{array}{lcl}
\text{ \underline{wave equation}  } && \text{  \underline{heat equation}  }\\
 & &\\
u_{tt} =c^{2}\varDelta u \ \ \ \text{in }\Omega && u_{t} =k\varDelta
u\ \ \ \text{in }\Omega
\end{array}
\end{equation}
with one of the three boundary conditions (BC) on $\partial\Omega$:
\begin{equation}
\label{BCs}
\begin{array}{clccl}
u=0 & \text{(D)}&& \ u=0 & \text{(D)}\\[6pt]
\normalderiv{u}=0 & \text{(N)} &&\
\normalderiv{u}=0 & \text{(N)}\\[6pt]
\normalderiv{u}+au=0 & \text{(R)}&&\ \normalderiv{u}+au=0 &
\text{(R)}
\end{array}
\end{equation}
with initial conditions (IC):
\begin{equation}
\label{ICs}
\begin{array}{lcl}
u(\bx,0)=f(\bx) && u(\bx,0)=f(\bx)\\[6pt]
u_{t}(\bx,0)=g(\bx)&&u_{t}(\bx,0)=g(\bx).
\end{array}
\end{equation}

The abbreviations for the boundary conditions used here are:
Dirichlet (D), Neumann (N), Robin (R). For the Robin BC, $a$ is a
constant number.

\ \\
We use the separation of variables technique and set
$u(\bx,t)=T(t)v(\bx)$, which leads to the following equations
\begin{equation}
\begin{array}{ll} \text{From wave equation:} &
\fr{T''}{c^{2}T}=\fr{\triangle
v}{v}=-\lambda\\[8pt]
\text{From heat equation:} & \fr{T'}{kT}=\fr{\triangle
v}{v}=-\lambda
\end{array}
\end{equation}

Later in this lecture we will show that $\lambda\geq 0$, for at
least either (D), (N) or (R) from \eqref{BCs} is satisfied.

\ \\
Regardless of whether we consider the heat or the wave equation, we
reach
\begin{equation}
\label{star}
\begin{array}{c}
-\triangle v=\lambda v\qquad \text{in}\ \ \Omega\\
\text{where $v$ satisfies either (D), (N) or (R)}
\end{array}
\end{equation}

Lots of mathematics are involved to prove that the set of $\lambda$
satisfying \eqref{star} is discrete, i.e.,
$\lambda_{1},\lambda_{2},\cdots$, and there exist the corresponding
eigenfunctions $\phi_1,\phi_2,\cdots$ that are mutually orthogonal.
We'll cover those math later, but at this point, we assume the
existence of $\lambda_1,\lambda_2,\cdots$ and
$\phi_{1},\phi_{2},\cdots$. Once we have the eigenpairs
$\{(\lambda_{n},\phi_{n})\}_{n=1}^{\infty}$, we can write the
solutions for \eqref{waveheat} as
\begin{equation}
\label{solution}
\begin{array}{ll}
\text{wave equation:} &
u(\bx,t)=\sum_{n=1}^{\infty}\big[A_{n}\cos(\sqrt{\lambda_{n}}ct)+B_{n}\sin(\sqrt{\lambda_{n}}ct)\big]\phi_{n}(\bx)\\[6pt]
\text{heat equation:} &
u(\bx,t)=\sum_{n=1}^{\infty}A_{n}e^{-\lambda_{n}kt}\phi_{n}(\bx)
\end{array}
\end{equation}

\ \\
\textbf{Preliminary:} some important formulas used in the
following sections:
\begin{itemize}
\item Divergence Theorem
\[\int_{\Omega}\nabla\cdot f\dd{\bx}=\int_{\partial\Omega}\nu\cdot f\dd{S},\ \ \nu\text{ is normal vector and $\dd{S}$ is surface measure.}\]
\item Green's first identity
\[\int_{\Omega}u\varDelta v\dd{\bx}+\int_{\Omega}\nabla u\cdot \nabla v\dd{\bx}=\int_{\partial\Omega}u\normalderiv{v}\dd{S} \hspace{.4in}\textnormal{(G1)}\]
\item Green's second identity:
\[\int_{\Omega}(u\varDelta v-v\varDelta u)\dd{\bx}=\int_{\partial\Omega}(u\normalderiv{v}-v\normalderiv{u})\dd{S} \hspace{.4in}\textnormal{(G2)}\]
\item $\nu\cdot\nabla=\normalderiv{}$
\end{itemize}
\end{section}

\begin{section}{Orthogonality of the Eigenfunctions}
\noindent Define the inner-product
\[ \langle f,g\rangle \define \int_{\Omega}f(\bx)\overline{g(\bx)}\dd{\bx},\ \ \text{where }\Omega\in\Rf^{d},\ \
\dd{\bx}=\dd{x_1}\dd{x_2}\ldots\dd{x_d}.\]

Consider two functions $u,v\in C^{2}(\overline{\Omega})$, with
$\overline{\Omega}=\Omega\bigcup\partial\Omega$, ($C^{2}$ condition
can be weakened), we have
\[ u\varDelta v-(\varDelta u)v=\nabla\cdot[u\nabla v-(\nabla u)v]\]
Then integrate both sides in $\Omega$:

\begin{equation}
\begin{array}{rcl}
\Int{\Omega}{}(u\varDelta v-(\varDelta u)v)\dd{\bx} & = &
\Int{\Omega}{}\nabla\cdot[u\nabla v-(\nabla u)v]\dd{\bx}\\[8pt]
\text{(by divergence theorem)} & = &
\Int{\partial\Omega}{}\nu\cdot[u\nabla v-(\nabla u)v]\dd{S}\\[8pt]
\text{(by $\nu\cdot\nabla=\normalderiv{}$)} & = &
\Int{\partial\Omega}{}(u\normalderiv{v}-v\normalderiv{u})\dd{S}
\end{array}
\end{equation}

\ \\
Now we can show that any $u,v\in C^{2}(\overline{\Omega})$
satisfying either (D), (N) or (R) also satisfy
\[ \langle u,\varDelta v\rangle =\langle \varDelta u,v\rangle\]
Proof: We can have
\[\langle u,\varDelta v\rangle -\langle \varDelta
u,v\rangle=\Int{\partial\Omega}{}(u\normalderiv{v}-v\normalderiv{u})\dd{S}\]
If $u$ and $v$ satisfy (D) or (N), it is obvious that the above is equal to $0$.

If $u$ and $v$ satisfy (R), we get
\[u\normalderiv{v}-v\normalderiv{u}=u(-av)-v(-au)=0.\qquad\square\]

Therefore each of these three classical BC's is symmetric. Suppose
both $u,v$ are real eigenfunctions satisfying
\[-\varDelta u=\lambda_1 u,\qquad -\varDelta v=\lambda_2 v\]
and satisfying either (D), (N) or (R). Then
$\lambda_{1},\lambda_{2}$ are reals, and if
$\lambda_{1}\neq\lambda_{2}$, then $\langle u,v\rangle=0$.

\noindent Proof:
\begin{eqnarray*}
\lambda_{1}\langle u,u\rangle&=&\langle \lambda_{1} u,u\rangle\\
&=&\langle -\varDelta u,u\rangle\\
&=&\langle u, -\varDelta u\rangle\\
&=&\langle u, \lambda_{1} u\rangle\\
&=&\overline{\lambda_{1}}\langle  u,u\rangle,
\end{eqnarray*}
which implies $(\lambda_{1}-\overline{\lambda_{1}})\|u\|^2_{2} =0$.
Since $\|u\|_{2}\neq 0$,
$\lambda_{1}=\overline{\lambda_{1}}\Leftrightarrow
\lambda_{1}\in\Rf$. Similarly,
\begin{eqnarray*}
\lambda_{1}\langle u,v\rangle - \lambda_{2}\langle u,v\rangle&=&\langle \lambda_{1} u,v\rangle-\langle u,\lambda_{2}v\rangle\\
&=&\langle -\varDelta u,v\rangle-\langle u ,-\varDelta v\rangle\\
&=&\langle u, -\varDelta v\rangle-\langle u ,-\varDelta v\rangle\\
&=&0
\end{eqnarray*}
which implies $(\lambda_{1}-\lambda_{2})\langle u,v \rangle =0$.
Since $\lambda_{1}\neq\lambda_{2}$, $\langle u,v\rangle =0$.
$\square$

\ \\
We summarize with the following theorem.

\begin{thm}
\label{eigthm} In the eigenvalue problem \eqref{star}, the
followings hold:
\begin{itemize}
\item all the eigenvalues are real
\item the eigenfunctions can be chosen to be real-valued
\item the eigenfunctions corresponding to distinct eigenvalues are necessarily orthogonal
\item all the eigenfunctions can be chosen to be orthogonal, i.e., orthonormal.
\end{itemize}
\end{thm}
\end{section}

\begin{section}{Multiplicity of the Eigenvalues}
\begin{defn} An eigenvalue $\lambda$ has multiplicity $m$ if it has $m$
linearly independent eigenfunctions. The eigenspace $E_{\lambda}$ is
a linear space spanned by the set of eigenfunctions corresponding to
$\lambda$. So, in this case $\text{dim}(E_{\lambda})=m$.
\end{defn}

\textbf{Notice:} if $\text{dim}(E_{\lambda})=m$, and
$E_{\lambda}=\text{span}\{w_{1},...,w_{m}\}$, but $\langle
w_{i},w_{j}\rangle\neq \delta_{ij}$, then we use the Gram-Schmidt
orthogonalization method to get
\[E_{\lambda}=\text{span}\{\phi_{1},...,\phi_{m}\},\quad \text{with}\ \ \langle\phi_{i},\phi_{j}\rangle
=\delta_{ij}.\]
\end{section}

\begin{section}{Generalized Fourier Series}
\noindent Because of the Theorem \ref{eigthm}, we have for $f\in
L^2(\Omega)$
\[f(\bx)=\sum_{n=1}^{\infty}f_{n}\phi_{n}(\bx),\quad f_n=\langle f,\phi_{n}\rangle\]

This is a generalization of the Fourier series, and we can discuss
the decay of $\{f_{n}\}$, etc.

\begin{thm}
As in \eqref{star}, let $\lambda_{k}$ be the Dirichlet-Laplacian
eigenvalues, let $\nu_{k}$ be the Neumann-Laplacian eigenvalues, and
let $\rho_{k}$ be the Robin-Laplacian eigenvalues, where $k\in\N$.
Then
\[\lambda_{k}>0,\quad \nu_{k}\geq 0,\quad \text{and}\ \ \rho_{k}\geq 0,\ \text{if $a\geq
0$}\]
\end{thm}

\noindent Proof: Let $u$ and $v$ are corresponding eigenfunctions.
Use Green's first identity (G1):
\[ \int_{\Omega}u\varDelta v\dd{\bx}+\int_{\Omega}\nabla u\cdot \nabla v\dd{\bx}=\int_{\partial\Omega}u\normalderiv{v}\dd{S}\]
Set $v=u$,
\[\int_{\Omega}u\varDelta u\dd{\bx}+\int_{\Omega}|\nabla u|^{2}\dd{\bx}=\int_{\partial\Omega}u\normalderiv{u}\dd{S}\]
For boundary condition (D),
\[\int_{\Omega}u(-\lambda u)\dd{\bx}+\int_{\Omega}|\nabla u|^{2}\dd{\bx}=0\ \ \Rightarrow\ \lambda=\frac{\int_{\Omega}|\nabla
u|^{2}\dd{\bx}}{\int_{\Omega} u^{2}\dd{\bx}}\geq 0\]

But $|\nabla u|^{2}\neq 0$. Since if so, $u=\text{const}$, then
$u\equiv 0$, which conflicts with the fact that $u$ is
eigenfunction. Therefore, $\lambda>0$.\\

For boundary condition (N),
\[ \nu = \frac{\int_{\Omega}|\nabla u|^{2}\dd{\bx}}{\int_{\Omega} u^{2}\dd{\bx}}\geq 0\]
Here $|\nabla u|^{2}=0$ is acceptable, i.e.
$u\equiv\textnormal{const}\neq 0$. Then $\nu\geq 0$, where $\nu=0$
corresponds to eigenfunction $\phi_{0}(x)\equiv
\textnormal{const}\neq 0$.\\

For boundary condition (R), we can get
\[-\rho\int_{\Omega}|u|^{2}\dd{\bx}+\int_{\Omega}|\nabla u|^{2}\dd{\bx}=\int_{\partial\Omega}u(-au)\dd{S}\]
\[\Rightarrow \rho=\fr{a\int_{\partial\Omega}|u|^{2}\dd{S}+\int_{\Omega}|\nabla u|^{2}\dd{\bx}}{\int_{\Omega}|u|^{2}\dd{\bx}}\geq 0,\qquad\text{if $a\geq 0$}.\quad \square\]

\end{section}

\begin{section}{Completeness of $\{\phi_{n}\}_{n\in\N}$ in $L^{2}$-sense}
See \cite[Sec.3.3-3.4]{FOLLAND} and \cite[Chapter 4]{YOUNG} for the
elementary discussion on the completeness of a set of basis in
$L^2(\Omega)$.

\ \\
For all $f\in L^{2}(\Omega)$, we have
\[\|f-\sum_{n=1}^{N}f_{n}\phi_{n}\|_{L^{2}}\rightarrow 0\quad \text{as $N\rightarrow \infty$}.\]

Equivalently,
\[f=\sum_{n=1}^{\infty}f_{n}\phi_{n}\qquad \text{in the $L^{2}$
sense}.\]

This is important because if it were not the case, we cannot express
any $L^{2}(\Omega)$ function in terms of $\{\phi_{n}\}_{n\in\N}$. We
will discuss more about the completeness later.

\ \\
\textbf{Example:} Diffusion in a 3D cube.

Let $\Omega=Q=\{(x,y,z)\cond 0<x<\pi,\ 0<y<\pi,\ 0<z<\pi\}$.
\[\left\{\begin{array}{lll}
\textnormal{DE}:&u_{t}=k\varDelta u& \textnormal{in } \Omega\\
\textnormal{BC}:&u=0&\textnormal{on }\partial\Omega\\
\textnormal{IC}:&u=f(\bx)& \bx\in\Omega,\ \ t=0
\end{array}\right.\]

Then by separation of variables, let $u(\bx,t)=T(t)\cdot v(\bx)$ as
before, we can have
\[-\varDelta v=\lambda v\quad
\text{in }Q,\qquad\text{with}\ \ v=0\quad\text{on }\partial Q.\]

Because the sides of $Q$ are parallel to the axes, one can do the
separation of variables again.
\[ v(x,y,z)=X(x)Y(y)Z(z)\]
\[ \Rightarrow\;\;\; \frac{X''}{X}+\frac{Y''}{Y}+\frac{Y''}{Y}=-\lambda.\]
BC's are also separated as
\[ X(0)=X(\pi)=Y(0)=Y(\pi)=Z(0)=Z(\pi)=0.\]
Therefore, we can solve $v(x,y,z)$ for $(l,m,n)\in\N^{3}$,
\begin{eqnarray*}
 v(x,y,z)&=&\sin(lx)\sin(my)\sin(nz)\\
&=&v_{l,m,n}(\bx)
\end{eqnarray*}
whose orthonormal version is
$(\frac{2}{\pi})^{3}\sin(lx)\sin(my)\sin(nz)$.

Then
\[\lambda=\lambda_{l,m,n}=l^{2}+m^{2}+n^{2}\]

Finally we get the solution
\[u(\bx,t)=\sum_{l,m,n}A_{lmn}e^{-(l^{2}+m^{2}+n^{2})kt}\sin(lx)\sin(my)\sin(nz)\]
where $A_{l,m,n}=(\frac{2}{\pi})^{3}\langle
f,v_{l,m,n}\rangle$.\\

Here, different values for $l,m,n$ can result in the same
eigenvalue. For example, $\lambda=27$. The valid values for
$(l,m,n)$ are $(5,1,1)$, $(1,5,1)$, $(1,1,5)$, $(3,3,3)$.
\end{section}

\begin{thebibliography}{10}

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

\bibitem{FOLLAND}
{\sc G.~B. Folland}, {\em Fourier Analysis and Its Applications},
Brooks/Cole, 1992.

\bibitem{YOUNG}
{\sc N. Young}, {\em An Introduction to Hilbert Space}, Cambridge
Univ. Press, 1988.

\end{thebibliography}
\end{document}
