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\section{Property of eigenfunctions}\label{sect-13.3}




\subsection{Base state}\label{sect-13.3.1}
We assume that $\Omega$ is connected domain. Consider the lowest eigenvalue $\lambda_1$ and a corresponding eigenfunction $u_1$.

\begin{theorem}\label{thm-13.3.1}
Let $\Omega $ be a connected domain. Then

a. Eigenfunction $u_1$ does not change its sign;

b. $\lambda_1$ is a simple eigenvalue.
\end{theorem}

\begin{proof}a. Let $v = |u_1|$. Observe that $\|v\|=\|u_1\|$ and $Q(v)=Q(u_1)$. Then $|u_1|=v$ is also an eigenfunction corresponding to eigenvalue $\lambda_1$ and then $u_{1,\pm}=\frac{1}{2}(|u_1|\pm u_1)=\max (\pm u_1, 0)$ are also eigenfunctions corresponding to $\lambda_1$. At least one of these two eigenfunctions, say $u_{1,+}$ is positive on some open set
$\Omega'\subset \Omega$ and then $u_{1,-}=0$ on $\Omega'$. However
$(\Delta +\lambda_1)u_{1,-}=0$  and there are many different theorems implying that since $u_{1,-}=0$ on $\Omega'$ it must be $0$ on $\Omega$.
(For example: solutions of $(\Delta +\lambda_1)v=0$ are analytic functions and there for analytic functions there is a unique continuation theorem.) Then $u_1\ge 0$ in $\Omega$.

b. If $u_1$ and $v_1$ are two linearly independent eigenfunctions corresponding to the same lowest eigenvalue $\lambda_1$, then one can find $w_1=u_1+ \alpha u_1$ which also is an eigenfunction and $(u_1,w_1)=0$ which is impossible since both of them have constant signs in $\Omega$ and therefore $u_1w_1$ has a constant sign in $\Omega$ and does not vanish identically in virtue of argument of (a).
\end{proof}

\begin{remark}

a. Let $\lambda_1\ge 0$. Assuming that $u_1\ge 0$ we see that $u_1$ is subharmonic function ($\Delta u_1\le 0$) but such functions cannot reach minimum inside domain unless constant. So $u_1>0$ in $\Omega$.

b. Let $\lambda_1<0$. Assuming that $u_1\ge 0$ we see that $u_1$ is superharmonic function ($\Delta u_1\ge 0$) and such functions cannot reach maximum inside domain unless constant.

c. While $\lambda_1$ is always simple $\lambda_n$ with $n\ge 2$ may be multiple.
\end{remark}


\begin{corollary}
$u_n$ with  $n\ge 2$ changes sign.
\end{corollary}
Indeed, it is orthogonal to $u_1$ which does not change a sign.

\subsection{Nodal sets}

\begin{definition}

Let $u_n$ be an eigenfunction. Then $\{x:\, u_n(x)=0\}$ is called a *nodal set* (*nodal line* as $d=2$) and connected components of $\{x\in \Omega, u_n(x)\ne 0\}$ are called *nodal domains*.
\end{definition}-

We know that for $n=1$ is just one nodal domain and for $n\ge 2$ there are at least 2 nodal domains. We need the following theorem from *Ordinary Differential Equations*:

\begin{theorem}
For $d=1$   there are exactly $n-1$ nodal points and $n$ nodal intervals for $u_n$.
\end{theorem}Then we can use it in the case when variables are separated (we consider only $d=2$.

\begin{example}

a. Let $\Omega$ be a rectangular box and $u_{pq}=X_p(x_1) Y_q(x_2)$. Then there are $pq$ nodal domains:

b. Let $\Omega$ be a disk and $u_{pq}=R_p(r) \Theta_q(\theta)$. Similarly one can consider sectors, rings etc

c. Let $\Omega$ be an ellipse (or elliptical annuli, sector etc). 

d. 
\begin{figure}[h]
\centering
\subfloat[]{\begin{tikzpicture}
\draw (0,0) grid (4,3);
\draw[very thick] rectangle (4,3);
\end{tikzpicture}}\quad
\subfloat[]{\begin{tikzpicture}
\draw[very thick] (0,0) circle (1.5);
\draw (0,0) circle (0.65346642270974);
\draw (-1.5,0)--(1.5,0);
\draw (0,-1.5)--(0,1.5);
\end{tikzpicture}}\quad

\end{figure}
\end{example}

\begin{example}
a. Let $\Omega$ be a rectangular box and $\lambda$ be a multiple eigenvalue. Then there are several eigenfunctions considered above but also their linear combinations which could have rather complicated sets
\end{example}

\begin{example}
If $\Omega$ is a generic domain then there are 

\end{example}

But the argument proceeds as follows. Assume the number of nodal domains of the kth eigenfunction uk is larger than k, then you can find a function vanishing on a nodal domain, orthogonal to the first k−1 eigenfuncions and attaining the kth eigenvalue. Thus by the min-max principle it's an eigenfunction. But it vanishes on an open set, thus must be 0 by the unique continuation principle. This argument does not involve any regularity of the nodal sets. –  Kelei Wang Jan 7 '14 at 4:26

Chladni figures

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