# Ngô Quốc Anh

## August 8, 2010

### The nodal set of eigenfunctions

Filed under: PDEs — Ngô Quốc Anh @ 17:21

An eigenfunction $\varphi$ is meant to be a solution of Dirichlet’s problem $\begin{cases}-\Delta \varphi = \lambda\varphi,&\text{ in } \Omega,\\\varphi=0,&\text{ on } \partial\Omega,\end{cases}$

where $\displaystyle\Delta = \sum\limits_{k = 1}^n {\frac{{{\partial ^2}}}{{\partial x_k^2}}}$

is the Laplacian, $\Omega$ is a bounded smooth domain in $\mathbb R^n$, and $\lambda$ is a constant (i.e. the corresponding eigenvalue).

It is well known that the first eigenfunction is positive in $\Omega$, and all higher eigenfunctions must change sign.

Definition. The nodal set of an eigenfunction $\varphi$ is defined to be $\displaystyle N = \overline {\left\{ {x \in \Omega :\varphi (x) = 0} \right\}}$.

To gain a better understanding of nodal sets let us consider a few examples.

Example 1. Let us consider the case when $\Omega=(0,l)$ where the first four eigenfunctions had been shown. Red points are nodal sets. In this case, we simply call nodal nodes. This comes from the fact that all eigenvalues $\lambda_k$ and eigenfunctions $\varphi_k$ are already known $\displaystyle{\varphi _k}(x) = \sqrt {\frac{2}{l}} \sin \left( {\frac{{n\pi }}{l}x} \right), \quad k \in \mathbb{N}$.

Example 2. Let us consider the case when $\Omega=(0,\pi)\times (0,\pi)$. It is known that the eigenfunctions
and eigenvalues are $\lambda_{n,m}=n^2+m^2, \quad \varphi_{n,m}(x,y)=A\sin(nx)\sin(my), \quad n,m \in \mathbb N$.

Functions $\sin(nx)\sin(my)$ are usually denoted by $\varphi_{nm}$. We then write $\varphi_{n,m}=A\varphi_{nm}+B\varphi_{mn}$.

Since we know what the eigenvalues and functions are, we can tabulate them in order of increasing eigenvalues. For example, the first 4 pairs of eigenvalue and eigenfunction are of the following form

• $\lambda=2$, $\varphi_2(x,y)=A_2\sin x\sin y$,
• $\lambda=5$, $\varphi_5(x,y)=A_5\sin x\sin (2y)+B_5 \sin(2x)\sin y$,
• $\lambda=8$, $\varphi_8(x,y)=A_8\sin (2x)\sin (2y)$,
• $\lambda=10$, $\varphi_{10}(x,y)=A_{10}\sin x\sin (3y)+B_{10} \sin(3x)\sin y$,

where $A_k, B_k$ are appropriate coefficients. Let take all coefficients to be 1.

• For $\lambda=2$ we have the following picture Again, it is clear to see that the first eigenfunction is positive in the domain.

• For $\lambda=5$ we have In this case, the nodal set is a segment connected points $(3,0)$ and $(0,3)$.

• For $\lambda=8$, we have In this case, the nodal set is just a cross symbol.

• For $\lambda=10$, we have In this case, the nodal set is just a small closed curve which is almost round.

Remark. In the pictures above, the way to recognize nodal sets is to look at part of surface which is white. For the reader’s convenience, we summarize all above nodal sets If $A \ne 1$ and $B \ne 1$ then the nodal set changes. We have several nodal sets for the fourth eigenfunction as shown below. So, how many regions can the nodal set divide a general domain ­ into (assuming ­ is connected)? The following theorem limits the possibilities.

Theorem (Courant Nodal Domain Theorem).

1. The first eigenfunction $\varphi_1$ corresponding to the smallest eigenvalue $\lambda_1$ cannot have any nodes.
2. For $n \geqslant 2$, $\varphi_n$ corresponding to the $n$th eigenvalue counting multiplicity, divides the domain ­ into at least 2 and at most $n$ pieces.

Concerning part 2 of the theorem, the bounds 2 and $n$ are best possible as the following example shows. The nodal set for $\sin(x) \sin(12y)+\sin(12x)\sin(y)$

divides the square $(0,\pi)^2$ into 12 subregions. However, nodal set for $\sin(x) \sin(12y)+\mu\sin(12x)\sin(y)$

divides the square $(0,\pi)^2$ into 2 subregions where $\mu$ is near 1.

We do not know the topology of the nodal set in general, even for the simplest case $n = 2$. A conjecture about the nodal line (i.e. when $n = 2$) of a second eigenfunction states that

Conjecture. The nodal line of a second eigenfunction divides the domain $\Omega$ by intersecting its boundary at exactly two points if $\Omega$ is convex.

This conjecture was partially answered by C.S. Lin [here] in 1987 when when the domain is symmetric under a rotation with angle $\frac{2\pi p}{q}$, where $p, q$ are positive integers.

It is worth noticing that the nodal line has been recently used in a work of N. Ghoussoub and C.S. Lin [here] to further improve the Moser-Onofri-Aubin inequality.

Most of text in this entry comes from a book entitled Partial Differential Equations [Chapter 10] by Strauss.

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