13.3. Properties of eigenfunctions


## 13.3. Properties of eigenfunctions

### Base state

We assume that $\Omega$ is connected domain. Consider the lowest eigenvalue $\lambda_1$ and a corresponding eigenfunction $u_1$.

Theorem 1. Let $\Omega$ be a connected domain. Then

1. Eigenfunction $u_1$ does not change its sign;
2. $\lambda_1$ is a simple eigenvalue.

Proof.

1. 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 $\Omega$.

2. 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).

Remark 1.

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

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

3. While $\lambda_1$ is always simple $\lambda_n$ with $n\ge 2$ may be multiple.

Corollary 1. $u_n$ with $n\ge 2$ changes sign.

Indeed, it is orthogonal to $u_1$ which does not change a sign.

### Nodal sets

Definition 1. 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.

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:

Theorem 2. For $d=1$ there are exactly $n-1$ nodal points and $n$ nodal intervals for $u_n$.

Theorem 3. For $d\ge 2$ if $u$ is an eigenfunction with an eigenvalue $\lambda_n$ then $u_n$ has no more than $n$ nodal domains.

Proof. (i) Let $u$ have $m\ge n$ nodal domains. Consider $w_k$ coinciding with $u$ in $k$-th nodal domain $\Omega_k$ of $u$. Then $w_1,\ldots, w_m$ are linearly independent and \begin{gather*} \| c_1w_1+\ldots +c_m w_m\|^2 = \sum{1\le j\le m} c_j^2 \|w_j\|^2, \\ Q( c_1w_1+\ldots +c_m w_m) = \sum{1\le j\le m} c_j^2 Q(w_j). \end{gather*}

Consider the space $L$ of linear combinations of $w_1,\ldots, w_m$ which are orthogonal to $u_1,\ldots, u_{n-1}$; then $L\ne \{0\}$. By definition $Q(v)\ge \lambda_n \|v\|^2$ on $L$. Since $u\in L$ we conclude that $Q(u)\ge \lambda_n \|u\|^2$; however $Q(u)=\lambda \|u\|^2$ and then $\lambda\ge \lambda_n$. It proves theorem if $\lambda_{n+1}>\lambda_n$.

Observe that since $(\Delta + \lambda_n) w_k=0$ in $\Omega_k$ and $w_k=0$ in $\Omega\setminus \Omega_k$ we integrating by parts see that $Q(w_k)=\lambda_n\|w_k\|^2$ for all $k$. Then $Q(v)=\lambda_n\|v\|^2$ for all $v\in L$.

Assume now that $m>n$. Then there exists $0\ne v\in L$ which is a linear combination of $w_1,\ldots, w_n$. Then $v$ is an eigenfunction but it is $0$ in $\Omega_{n+1}$ and therefore it must be $0$. Contradiction.

Therefore if $\lambda_{n-1}<\lambda_n=\ldots =\lambda_m$ then each eigenfunction corresponding to multiple eigenvalue $\lambda_n$ has no more than $n$ nodal domains.

Then we can use it in the case when variables are separated (we consider only $d=2$ and only Dirichlet boundary condition:

Example 1. Let $\Omega=\{0< x< a, 0 < y < b\}$ be a rectangular box. Let us separate variables; then $$u_{pq}=\sin \Bigl(\frac{p\pi x}{a}\Bigr) \sin \Bigl(\frac{q\pi y}{b}\Bigr),\qquad \mu_{pq}= \pi^2 \Bigl(\frac{p^2}{a^2}+\frac{q^2}{b^2}\Bigr).$$ Then nodal lines form a rectangular grid (see below). Let $a=b=\pi$.

a. Then $\lambda_1=2$ and $\lambda_2=\lambda_3=10$ (where $\lambda_n$ are $\mu_{pq}$ ordered). First figure shows nodal lines for $u_{21}$ (and nodal lines for $u_{12}$ are exactly like this but flipped over $x=y$). Consider now linear combinations of $u_{21}$ and $u_{12}$:

 $u_{21}$ $u_{21}+\frac{1}{2} u_{12}$ $u_{21}+u_{12}$

b. Next $\lambda_4=8$:

 $u_{22}$

c. Further $\lambda_5=\lambda_6=10$. First figure shows nodal lines for $u_{31}$ (and nodal lines for $u_{13}$ are exactly like this but flipped over $x=y$). Consider now linear combinations of $u_{31}$ and $u_{13}$:

 $u_{31}$ $u_{31}+u_{13}$ $u_{31}+\frac{1}{3}u_{13}$ $u_{31}-\frac{1}{3}u_{13}$ $u_{31}-u_{13}$

Comparing two last pictures we see that crossing open under small perturbations.

d. Further $\lambda_7=\lambda_8=13$, First figure shows nodal lines for $u_{32}$ (and nodal lines for $u_{23}$ are exactly like this but flipped over $x=y$). Consider now linear combinations of $u_{32}$ and $u_{23}$:

 $u_{32}$ $u_{32}+\frac{1}{2}u_{23}$ $u_{32}+u_{23}$

e. Further $\lambda_9=\lambda_{10}=17$, First figure shows nodal lines for $u_{41}$ (and nodal lines for $u_{14}$ are exactly like this but flipped over $x=y$). Consider now linear combinations of $u_{412}$ and $u_{14}$:

 $u_{41}$ $u_{41}+\sqrt{\frac{2}{27}}u_{14}$ $u_{41}+{\frac{1}{2}}u_{14}$ $u_{41}+u_{14}$

f. Further $\lambda_{11}=18$ is simple $p=q=3$ and thus trivial; furthermore $\lambda_{12}=\lambda_{13}=20$ with $p=4, q=2$ is also trivial: we need just to take any picture for $p=2,q=1$ and make its double mirror reflection arriving to

 $u_{42}+u_{24}$ $u_{42}+\frac{1}{2}u_{24}$ $u_{42}-u_{24}$

and similar pictures (we do not draw $u_{pq}$ anymore).

f. Further $\lambda_{14}=\lambda_{15}=25$ does not produce anything much different from (d) but simply more nodal domains:

 $u_{43}+u_{34}$ $u_{43}+\frac{1}{2}u_{34}$

g. Further $\lambda_{16}=\lambda_{17}=26$:

 $u_{51}+u_{15}$ $u_{51}+\frac{4}{5}u_{15}$ $u_{51}+\frac{1}{2}u_{15}$ $u_{51}+\frac{1}{4}u_{15}$ $u_{51}-\frac{1}{5}u_{15}$ $u_{51}-\frac{1}{2}u_{15}$ $u_{51}-u_{15}$

h. Skipping $\lambda_{18}=\lambda_{19}=29$, $\lambda_{20}=32$, $\lambda_{21}=\lambda_{22}=34$, consider $\lambda_{23}=\lambda_{24}=37$

 $u_{61}+u_{16}$ $u_{61}+\frac{4}{5}u_{16}$ $u_{61}+\frac{1}{5}u_{16}$ $u_{61}+\frac{1}{10}u_{16}$

i. Starting from $\lambda=50=7^2+1^2=5^2+5^2$ multiplicities could be larger than $2$ and the following gallery is just a tiny sample

Example 2. Consider $\Omega=\{ 0< y < x <\pi\}$ which is a triangle.

1. If on the diagonal $x=y$ a Dirichlet condition is required, then eigenfunctions are $u_{pq}(x,y)-u_{pq}(y,x)$ with $p\ne q$ (or their linear combination like $u_{83}(x,y)-u_{83}(y,x)$ and $u_{74}(x,y)-u_{74}(y,x)$).

2. If on the diagonal $x=y$ a Neumann condition is required, then eigenfunctions are $u_{pq}(x,y)+u_{pq}(y,x)$ (or their linear combination like $u_{71}(x,y)+u_{71}(y,x)$ and $u_{55}(x,y)$).

Example 3.

a. Let $\Omega=\{r\le b \}$ be a disk. Then nodal lines form a circular grid: $u_{pq}(r,\theta)= \cos (p\theta) J_{p}(k_{pq}r)$ where $J_p(z)$ are Bessel functions and $k_{pq} b$ is $q$-th root of $J_p(z)$. The similar statement is true for circular sectors, rings etc.

b. Let $\Omega$ be an ellipse. Then (see Subsection 6.3.3) in the elliptical coordinates Laplacian is $$\Delta = \frac{1}{c^2\bigl(\sinh^2(\sigma)+\sin^2(\tau) \bigr)} (\partial_\sigma^2 +\partial_\tau^2 ) \label{eq-13.3.A}$$ and separating variables $u= S(\sigma)T(\tau)$ we get \begin{align} & S'' +\bigl( \lambda c^2 \sinh^2(\sigma) -k\bigr)S=0,\\ & T'' + \bigl( \lambda c^2 \sin^2(\tau) + k\bigr)T=0. \end{align} For $T$ we have either $\pi$-periodic or $\pi$-antiperiodic boundary condition: $T(\tau +\pi)=\pm T(\tau)$ and for $S$ we have Dirichlet boundary condition as $\cosh (\sigma)= a/c$ and respectively Neumann and Dirichlet boundary condition as $\sigma=0$ arising from $S(\pm \sigma)= \pm S(\sigma)$. So we have a grid consisting from confocal ellipses and hyperbolas. The similar statement is true for elliptical "sectors", rings etc.

c. Let $\Omega$ be an parabolic lense. Then (see Subsection 6.3.3) in the parabolic coordinates $$\Delta = \frac{1}{\sigma^2+\tau^2} (\partial_\sigma^2 +\partial_\tau^2 ) \label{eq-13.3.G}$$ and separating variables $u= S(\sigma)T(\tau)$ we get \begin{align} & S'' +\bigl( \lambda \sigma^2 -k\bigr)S=0,\\ & T'' + \bigl( \lambda c^2 \tau^2 + k\bigr)T=0. \end{align} So we have a grid consisting from confocal parabolas.

Remark . For generic domais (any sufficiently small perturbation of generic domain is a generic domain again but an arbitrarily small perturbation of non-generic domain may result in generic domain)

1. All eigenvalues are simple;
2. Nodal lines do not have self-intersections.

Remark . Historically interest to nodal lines appeared from Chladni plates but those are nodal lines for biharmonic operator $$(\Delta ^2-\lambda )u=0 \label{eq-13.3.H}$$ with free boundary conditions (appearing from variational problem $$\delta \Bigl (\iint \bigl(u_{xx}^2+ 2u_{xy}^2 + u_{yy}^2 -\lambda u^2\bigr)\,dxdy\Bigr)=0 \label{eq-13.3.I}$$ (without boundary conditions). This is much more complicated question which was (partially) solved by Marie-Sophie Germain.

### Hearing the shape of the drum

In 1965(?) Marc Kac asked the question"Can one hear the shape of the drum" which meant: if we know all the eigenvalues of the Dirichlet Laplacian $\Delta$: $0<\lambda_1 <\lambda_2 \le \lambda_3\le \ldots$ in the connected domain $\Omega$, can we restore $\Omega$ (up to isometric movements—shift and rotations). The extensive study was using the method of spectral invariants, which are numbers which have some geometric meaning and which can be calculated from $0<\lambda_1 <\lambda_2 \le \lambda_3\le \ldots$.

The main source of such invariants was the method of heat equation: Namely let $G(x,y,t)$ with $x,y\in \Omega$ and $t>0$ be a Green function: \begin{align} &G_t -\Delta_x G=0,\\ &G|_{t=0}=\delta (x-y),\\ &G|_{x\in \partial \Omega}=0 \end{align} and let $$\sigma(t)= \iint u(x,x,t)\, dx= \sum_{n\ge 1} e^{-\lambda_n t}$$ be a heat trace; then $$\sigma(t)\sim \sum _{k\ge -d} c_k t^{k}\qquad \text{as }\ t\to +0$$ Check! where $c_k$ are heat invariants. It was proven that (as $d=2$, for $d\ge 3$ similarly) area, perimeter, number of holes and many other geometric characteristic are spectral invariants but the final answer was negative: there are isospectral domains $\Omega$ and $\Omega'$ (so that eigenvalues of Laplacians in those are equal) which are not isometric (have different shapes).