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

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.**

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

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
$\Omega =\{x:\, 0< x_1< a_1, 0< x_2< a_2, \ldots, 0< x_d< a_d\}$

- $N_D(\lambda)$ 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*}
- $N_N(\lambda)$ 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

- 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
- 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.

For Dirichlet boundary condition we do not count points with

$m_i=0$ (blue) and for Neumann problem we count them.

**Remark 3.**

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

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

- The side of any box is at least $\varepsilon$;
- 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.

**Remark 4.**

- This Theorem 5 (in a bit less general form) was proven by H.Weyl in 1911;
- 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*). - Further, as $d\ge 3$ we can assume only that $\mes_d (\Omega)<\infty$.

**Remark 5.**

- 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 $\Sigma=\partial\Omega$.
- 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}
- 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.

- 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}
- 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. - 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.
- 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}
- 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 SchrÃ¶dinger 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

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