13.2. Asymptotic distribution of eigenvalues

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\supp}{\operatorname{supp}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\mes}{\operatorname{mes}}$

13.2. Asymptotic distribution of eigenvalues


  1. Introduction
  2. Rectangular box
  3. Weyl formula
  4. Remarks

Introduction

We are interested how eignevalues are distributed. We introduce eigenvalue distribution function $N(\lambda)$ which is the number of eigenvalues (of operator $-\Delta$) which are less than $\lambda$. In other words, \begin{equation} N(\lambda)=\max _{\lambda_n< b} n \label{eq-13.2.1} \end{equation} where $\lambda_n$ are eigenvalues, or, using Corollary 13.1.1 we reformulate it as

Definition 1. Eignevalue distribution function is \begin{equation} N(\lambda)=\max_{L\subset \mathsf{H}:\ Q(v)<\lambda \|v\|^2\ \forall 0\ne v\in L }\ \dim L \label{eq-13.2.2} \end{equation}

Remark 1. While these two definitions coincide in our particular case there is a huge difference: formula (\ref{eq-13.2.1}) counts only eigenvalues. However formula (\ref{eq-13.2.2}) takes into account also a continuous spectrum: this way defined $N(\lambda)$ is $+\infty$ if $(-\infty,\lambda)$ contains (some) points of continuous spectrum. This definition is one widely used.

In the same way as Theorem 13.1.5 and Theorem 13.1.6 one can prove

Theorem 1.

  1. $N(\lambda;\alpha)$ does not increase as $\alpha$ increases;
  2. $N(\lambda;\Sigma^-)$ does not increase as $\Sigma^-$ increases;
  3. As $\alpha\to +\infty$ \ $N(\lambda;\alpha)\to N_D(\lambda)$ which is an eigenvalue counting function for Dirichlet problem (i.e. when $\Sigma^-=\Sigma$).
  4. $N_D(\lambda;\Omega)$ does not decrease as $\Omega$ increases.

Rectangular box

Consider rectangular box in $\mathbb{R}^d$: $\Omega =\{x:\, 0< x_1< a_1, 0< x_2< a_2, \ldots, 0< x_d< a_d\}$ with Dirichlet or Neumann boundary conditions. Then separation of variables brings us eigenfunctions and eigenvalues \begin{align} &X_m=\sin \Bigl(\frac{\pi m_1x_1}{a_1}\Bigr) \sin \Bigl(\frac{\pi m_2x_2}{a_2}\Bigr) \cdots \sin \Bigl(\frac{\pi m_dx_d}{a_d}\Bigr),\label{eq-13.2.3}\\ &\lambda_m= \pi^2\Bigl(\frac{m_1^2}{a_1^2}+\frac{m_2^2}{a_2^2}+\ldots +\frac{m_d^2}{a_d^2}\Bigr) \qquad m_1\ge 1, \ldots, m_d\ge 1 \label{eq-13.2.4} \end{align} for Dirichlet problem and \begin{align} &X_m=\cos \Bigl(\frac{\pi m_1x_1}{a_1}\Bigr) \cos \Bigl(\frac{\pi m_2x_2}{a_2}\Bigr)\cdots \cos \Bigl(\frac{\pi m_dx_d}{a_d}\Bigr),\label{eq-13.2.5}\\ &\lambda_m= \pi^2\Bigl(\frac{m_1^2}{a_1^2}+\frac{m_2^2}{a_2^2}+\ldots +\frac{m_d^2}{a_d^2}\Bigr) \qquad m_1\ge 0, \ldots, m_d\ge 0 \label{eq-13.2.6} \end{align} for Neumann problem where $m=(m_1,\ldots,m_d)\in \mathbb{Z}^d$ in both cases.

Exercise 1. Prove it by separation of variables.

Therefore

Theorem 2. For rectangular box $\Omega =\{x:\, 0< x_1< a_1, 0< x_2< a_2, \ldots, 0< x_d< a_d\}$

  1. $N_D(\lambda)$ equals to the number of integer points (points with all integer coordinates) in the domain \begin{multline*} E_-(\lambda)=\Bigl\{m=(m_1,\ldots,m_d):\, m_1>0, m_2>0,\ldots , m_d>0,\\ \frac{m_1^2}{a_1^2}+\frac{m_2^2}{a_2^2}+\ldots +\frac{m_d^2}{a_d^2}<\pi^{-2}\lambda\Bigr\}; \end{multline*}
  2. $N_N(\lambda)$ equals to the number of integer points (points with all integer coordinates) in the domain \begin{multline*} E_+(\lambda)=\Bigl\{m=(m_1,\ldots,m_d):\,m_1\ge 0, m_2>0,\ldots, m_d\ge 0,\\ \frac{m_1^2}{a_1^2}+\frac{m_2^2}{a_2^2}+\ldots +\frac{m_d^2}{a_d^2}<\pi^{-2}\lambda\Bigr\}. \end{multline*}

Calculation of the number of integer points in the "large" domains is an important problem of the Number Theory. For us the answer "the number of integer points inside of the large domain approximately equals to its volume" is completely sufficient but let us use the following

Theorem 3. As $d\ge 2$ the number of integer points inside ellipsoid \begin{equation*} \mathcal{E}(\lambda)=\Bigl\{m=(m_1,\ldots,m_d):\, \frac{m_1^2}{a_1^2}+\frac{m_2^2}{a_2^2}+\ldots +\frac{m_d^2}{a_d^2}<\pi^{-2}\lambda\Bigr\} \end{equation*} equals to the volume of $\mathcal{E}(\lambda)$ plus $o(\lambda^{(d-1)/2})$.

Remark 2. Actually much more precise results are known.

Observe that the volume of this ellipsoid $\mathcal{E}(\lambda)$ is $\pi^{-d}\omega_d a_1a_2\cdots a_d\lambda^{d/2}$ where $\omega_d$ is a volume of the unit ball in $\mathbb{R}^d$.

Exercise 2. Prove it observing that ellipsoid $\mathcal{E}(\lambda)$ is obtained by stretching of the unit ball.

Observe also that both $E_+(\lambda)$ and $E_-(\lambda)$ constitute $1/2^d$ part of $\mathcal{E}(\lambda)$ and therefore their volumes are $(2\pi)^{-d}\omega_da_1\cdots a_d \lambda^{d/2}$. However if we consider integer points we observe that

  1. the number of integer points in $E_-(\lambda)$ multiplied by $2^d$ equals to the number of integer points in $\mathcal{E}(\lambda)$ minus the number of integer points belonging to the intersection of $\mathcal{E}(\lambda)$ with one of the planes $\Pi_j=\{m=(m_1,\ldots,m_d):\, m_j=0\}$ while
  2. the number of integer points in $E_+(\lambda)$ multiplied by $2^d$ equals to the number of integer points in $\mathcal{E}(\lambda)$ plus the number of integer points belonging to the intersection of $\mathcal{E}(\lambda)$ with one of the planes $\Pi_j=\{m=(m_1,\ldots,m_d):\, m_j=0\}$.

Actually, it is not completely correct as one needs to take into account intersections of $\mathcal{E}(\lambda)$ with two different planes $\Pi_j$ and $\Pi_k$ but the number of integer points there is $O(\lambda^{(d-2)/2})$.

Since intersections of $\mathcal{E}(\lambda)$ with $\Pi_j$ is an ellipsoid we conclude that the number of integer points there is $\pi^{1-d}\omega_{d-1}a_1a_2 \cdots a_d/a_j\lambda^{(d-1)/2}$.

Therefore the number of integer points in $E_\pm (\lambda)$ is equal to \begin{multline*} (2\pi)^{-d}\omega_d a_1a_2\cdots a_d \lambda^{d/2}\pm\\ \frac{1}{2}(2\pi)^{1-d}\omega_{d-1} \underbracket{a_1a_2\cdots a_d \bigl(a_1^{-1}+a_2^{-1}+\ldots a_d^{-1}\bigr) } \lambda^{(d-1)/2} + o\bigl(\lambda^{(d-1)/2}\bigr). \end{multline*} Observe that $a_1a_2\cdots a_d$ is a volume of $\Omega$ and $a_1a_2\cdots a_d \bigl(a_1^{-1}+a_2^{-1}+\ldots a_d^{-1}\bigr)$ is a half of area of its boundary $\partial\Omega$.

So we arrive to

Theorem 4. For rectangular box $\Omega =\{x:\, 0< x_1< a_1, 0< x_2< a_2, \ldots, 0< x_d< a_d\}$, $d\ge 2$ \begin{equation} N(\lambda)=(2\pi)^{-d} \mes_d(\Omega)\lambda^{d/2}+ O\bigl(\lambda^{(d-1)/2}\bigr) \label{eq-13.2.7} \end{equation} and more precisely \begin{multline} N (\lambda)=(2\pi)^{-d} \mes_d(\Omega)\lambda^{d/2} \pm \\ \frac{1}{4} (2\pi)^{1-d} \mes_{d-1}(\partial\Omega)\lambda^{(d-1)/2} +o\bigl(\lambda^{(d-1)/2}\bigr) \label{eq-13.2.8} \end{multline} where "$+$" corresponds to Neumann and "$-$" to Dirichlet boundary conditions and $\mes_k$ means $k$-dimensional volume.

R. box

For Dirichlet boundary condition we do not count points with
$m_i=0$ (blue) and for Neumann problem we count them.

Remark 3.

  1. Asymptotics (\ref{eq-13.2.7}) holds for $d=1$ as well;
  2. Asymptotics (\ref{eq-13.2.7}) and (\ref{eq-13.2.8}) with the sign "$+$" hold for Robin boundary condition.

Weyl formula

Now we want to generalize these results for an arbitrary domain $\Omega$. To do so let us consider domain $\Omega^+_\varepsilon$ which includes $\Omega$ and all points on the distance $\le C\varepsilon$ from $\partial \Omega$ and partition it into cubic boxes such that

  1. The side of any box is at least $\varepsilon$;
  2. If the box is not inside of $\Omega$ completely its side is exactly $\varepsilon$.

The boxes which are completely inside of $\Omega$ are called inner boxes and the boxes which intersect $\partial \Omega$ are called boundary boxes.


Inner and boundary boxes

We consider now only Dirichlet conditions. Let us use Definition 1 for $N(\lambda)$; for Dirichlet boundary conditions we can consider $\mathsf{H}$ as the space of all functions which vanish outside of $\Omega$.

Let us tighten constrains to these functions: we assume that they are $0$ in the boundary boxes as well and that they vanish on the walls of all boxes. Then $N(\lambda)$ can only decrease: $N(\lambda)\ge N_*((\lambda)$ where $N_*(\lambda)$ is an eigenvalue counting function under these new constrains. But under these constrains calculation in each box becomes independent and we arrive to \begin{equation*} N_*(\lambda)= \Bigl[(2\pi)^{-d}\omega_d \lambda^{d/2}- c \lambda^{(d-1)/2} \varepsilon^{-1}\Bigr] \sum_\iota \mes_d (B_\iota) \end{equation*} where we sum over all inner boxes. As $\varepsilon \ge c_1 \lambda ^{-1/2}$ this expression is greater than $\Bigl[(2\pi)^{-d}\omega_d \lambda^{d/2}- c \lambda^{(d-1)/2} \varepsilon^{-1}\Bigr] \mes_d (\Omega^-_\varepsilon) $ where $\Omega^-_\varepsilon$ is the set of all points of $\Omega$ on the distance $> \varepsilon$ from $\partial\Omega$. Thus \begin{equation} N_D(\lambda)\ge \Bigl[(2\pi)^{-d}\omega_d\lambda^{d/2}- c\lambda^{(d-1)/2} \varepsilon^{-1}\Bigr] \bigl(\mes_d (\Omega) - \mes_d (\Omega \setminus \Omega^-_\varepsilon)\bigr). \label{eq-13.2.9} \end{equation}

Let us loosen constrains to these functions: we assume that they are arbitrary in the boundary boxes as well and that they can be discontinuous on the walls of all boxes (then $Q(u)$ is calculated as a sum of $Q_\iota (u)$ all over boxes). Then $N(\lambda)$ can only increase: $N(\lambda)\le N^*((\lambda)$ where $N^*(\lambda)$ is an eigenvalue counting function under these new constrains. But under these constrains calculation in each box becomes independent and we arrive to \begin{equation*} N_*(\lambda)= \Bigl[(2\pi)^{-d}\omega_d \lambda^{d/2}+ c \lambda^{(d-1)/2} \varepsilon^{-1}\Bigr] \sum_\iota \mes_d (B_\iota) \end{equation*} where we sum over all inner and boundary boxes. Thus \begin{equation} N_D(\lambda)\le \Bigl[(2\pi)^{-d}\omega_d \lambda^{d/2}+ c \lambda^{(d-1)/2} \varepsilon^{-1}\Bigr] \bigl(\mes_d(\Omega) + \mes_d(\Omega ^+_\varepsilon \setminus \Omega )\bigr). \label{eq-13.2.10} \end{equation} But what is the gap between the right-hand expressions of (\ref{eq-13.2.9}) and (\ref{eq-13.2.10})? It does not exceed \begin{equation*} C\lambda^{(d-1)/2}\mes_d \varepsilon^{-1} (\Omega ^+_\varepsilon ) + C\lambda^{d/2} \mes_d (\Sigma_\varepsilon ) \end{equation*} where $\Sigma_\varepsilon$ is the set of all points on the distance $\le C_0\varepsilon$ from $\Sigma=\partial \Omega$.

We definitely want $\mes_d (\Sigma_\varepsilon )\to 0$ as $\varepsilon\to +0$. According to Real Analysis we say that it means exactly that $\Sigma$ is a set of measure $0$.

Thus we arrive to

Theorem 5. If $\Omega$ is a bounded domain and $\partial \Omega$ is a set of measure $0$ then \begin{equation} N_D(\lambda)= (2\pi)^{-d}\omega_d \lambda^{d/2}+ o\bigl(\lambda^{d/2}\bigr) \label{eq-13.2.11} \end{equation} as $\lambda\to +\infty$.

In particular, it holds if $\Omega$ is a bounded domain and $\partial \Omega$ is smooth.

Remarks

Remark 4.

  1. This Theorem 5 (in a bit less general form) was proven by H.Weyl in 1911;
  2. In fact we need neither of assumptions: even if $\partial \Omega$ is not a set of measure $0$ asymptotics (\ref{eq-13.2.11}) holds but in this case $\mes_d (\Omega)$ means a Lebesgue measure of $\Omega$ (see Real Analysis).
  3. Further, as $d\ge 3$ we can assume only that $\mes_d (\Omega)<\infty$.

Remark 5.

  1. We can always make a more subtle partition into cubic boxes: we can assume additionally that the size of the inner box $B_\iota$ is of the same magnitude as a distance from $B_\iota$ to $\partial \Omega$; then we can get a remainder estimate \begin{equation} C\lambda^{(d-1)/2} \iiint_{{x:\,\gamma(x) \ge c\lambda^{-1/2}}} \gamma(x)^{-1}\,dx + C\lambda ^{d/2} \mes_d (\Sigma_{c\lambda^{-1/2}}) \label{eq-13.2.12} \end{equation} where $\gamma(x)$ is a distance from $x$ to $\Sigma=\partial\Omega$.
  2. If $\Omega$ is a bounded set and $\partial\Omega$ is smooth then for Dirichlet and Robin boundary condition \begin{equation} N_D(\lambda)= (2\pi)^{-d}\omega_d \lambda^{d/2}+ O\bigl(\lambda^{(d-1)/2} \ln \lambda \bigr). \label{eq-13.2.13} \end{equation}
  3. Even if the boundary is highly irregular one can define Neumann Laplacian. However in this case even if domain is bounded it can have non-empty continuous spectrum an then $N_N(\lambda)=\infty$ for $\lambda\ge C$.

Remark 6. Let $\Omega$ be bounded domain and $\partial \Omega$ be smooth.

  1. Much more delicate proof based on completely different ideas shows that for Dirichlet and Robin boundary condition \begin{equation} N (\lambda)= (2\pi)^{-d}\omega_d \lambda^{d/2}+ O\bigl(\lambda^{(d-1)/2} \bigr). \label{eq-13.2.14} \end{equation}
  2. Even more delicate proof shows that under some billiard condition asymptotics (\ref{eq-13.2.8}) holds with sign "$+$" for Robin boundary condition and "$-$" for Dirichlet boundary conditions.
  3. This billialrd condition is believed to be fulfilled for any domain in $\mathbb{R}^d$. However it is not necessarily true on manifolds; for Laplacian on the sphere only (\ref{eq-13.2.14}) holds but (\ref{eq-13.2.14}) fails.
  4. Actually, one can allow Dirichlet boundary condition on some part of the boundary ($\Sigma^-$) and Robin boundary condition on the remaining part of the boundary ($\Sigma^+=\Sigma\setminus\Sigma^-$) provided transition between those parts is regular enough. Then the formula becomes \begin{multline} N (\lambda)=(2\pi)^{-d} \mes_d(\Omega)\lambda^{d/2} +\\ \frac{1}{4} (2\pi)^{1-d} \bigl(\mes_{d-1}(\Sigma^+)-\mes_{d-1}(\Sigma^-)\bigr) \lambda^{(d-1)/2} +o\bigl(\lambda^{(d-1)/2}\bigr)\qquad \label{eq-13.2.15} \end{multline}
  5. There are generalizations allowing some kinds or "regular singularities" of $\partial\Omega$.

Remark 7. There are many generalizations of these asymptotics; let us mention only few. We consider Schrödinger operator \begin{equation} H=-\hbar^2\Delta + V(x). \label{eq-13.2.16} \end{equation} Then Weyl formula for this operator is \begin{equation} N(\lambda,\hbar)\approx (2\pi\hbar)^{-d}\omega_d \iiint (\lambda-V(x))_+^{d/2}\,dx \label{eq-13.2.17} \end{equation} where $z_+=\max(z,0)$. This formula is justified

  1. As $\hbar\to +0$ and $V(x)\ge \lambda +\epsilon $ as $|x|\ge C$;
  2. As $\hbar$ is either fixed or tends to $+0$ and $\lambda\to +\infty$ provided $V(x)\to +\infty$ as $|x|\to \infty$;
  3. As $\hbar$ is either fixed or tends to $+0$ and $\lambda\to -0$ provided $V(x)\to 0$ and $|x|^2(-V(x))_+\to \infty$ as $|x|\to \infty$.
  4. In some other cases

$\Leftarrow$  $\Uparrow$  $\Rightarrow$