# Weyl's asympotic law for eigenvalue : $\lim_{\lambda \to \infty} \frac{M(\lambda)}{\lambda}=\sum_p \frac{A(D_p)}{4 \pi}$

In the book Strauss W.A. Partial differential equations - An introduction (Wiley, $2008$, $1$nd Ed.) page $310 - 311$, it is probably a silly question, but is there anyone could give me a hint how to obtain $\lim_{\lambda \to \infty} \frac{M(\lambda)}{\lambda}=\sum_p \frac{A(D_p)}{4 \pi}$? I am stocked on the problem for a while and I don't know how to solve it.

Edit : I am able to know why $\lim_{\lambda \to \infty} \frac{M(\lambda)}{\lambda}= \frac{A(\text{rectangle})}{4 \pi}$ (example on the rectangle page $307$), but when we let $\lambda$ in an arbitrary planar domain, I dont know why $\lim_{\lambda \to \infty} \frac{M(\lambda)}{\lambda}=\sum_p \frac{A(D_p)}{4 \pi}$ become true.

Proof : Let $D$ be the union of a finite number of rectangle $D = D_1 \cup D_2 \cup \cdots$ in the plane. Each particular $\mu_n$ corresponds to one of these rectangles, say $D_p$ (where $p$ depends on $n$). Let $A(D_p)$ denote the area of $D_p$. Let $M(\lambda)$ be the enumeration function for the sequence $\mu_1, \mu_2, \cdots$ define above $$M(\lambda) \equiv \text{the number of } \mu_1, \mu_2, \cdots \text{that do not exceed } \lambda.$$

Then, adding up the integer lattice points which are located within $D$, we get $$\text{(1)} \lim_{\lambda \to \infty} \frac{M(\lambda)}{\lambda}=\sum_p \frac{A(D_p)}{4 \pi}$$ as for the case of a single rectangle. Since $M(\mu_n)=n$, the reciprocal of $(1)$ take the form $$\lim_{n \to \infty} \frac{\mu_n}{n}=\sum_p \frac{4 \pi}{A(D)}.$$ By the theorem $5$ it follows that $$\lim_{n \to \infty} \frac{\lambda_n}{n}=\sum_p \frac{4 \pi}{A(D)}.$$

Clarification : For the whole domain $D$, we regard the eigenvalues $\lambda_n$. When we regard the domain as $D = D_1 \cup D_2 \cup \cdots$, we know that each $D_i$ has its own set of eigenvalues. Therefore, we combine all the Dirichlet eigenvalues of all of the subdomains $D_1, D_2, \cdots$ into a single increasing sequence $\mu_1 \leq \mu_2 \leq \cdots$.

• come on ... I'm not insulting you... and why don't you try understanding the exercice so that you can write it here directly without copying the book ? Jun 7 '16 at 3:01
• What have you tried and where are you stuck? It looks like it is explained in the text. Could you clarify? Jun 7 '16 at 6:46
• Hmmm... You are "J. Doe", the user who got kicked out of MSE a few days ago. Be careful this time, stop violating MSE rules. In particular, stop linking to pirated material. Jun 7 '16 at 12:29
• @SpinningAtInfinity: Don't introduce links to illegal copies of copyrighted books in your posts, anymore; if the copyright owners find out, they could sue MSE and nobody wants this, right? Jun 7 '16 at 13:15
• Check out section 2.2.3 of this: math.harvard.edu/~canzani/math253/Lecture3.pdf Jun 7 '16 at 19:04

Denote by $N_{D_p}(\lambda)$ the number of eigenvalues for the Dirichlet Laplacian on the rectangle $D_p$ that are less than $\lambda$.
Then, and this is what you seem to be missing, since the sequence $\mu_1\leq\mu_2\leq\cdots$ enumerate all eigenvalues on the different rectangles, the quantity $M(\lambda)$ that you define in the question is just the sum of the numbers of eigenvalues less than $\lambda$ for each rectangle, i.e $$M(\lambda)=\sum_p N_{D_p}(\lambda).$$
Once you see this, then the rest follows easily, $$\lim_{\lambda\to+\infty}\frac{M(\lambda)}{\lambda}=\lim_{\lambda\to+\infty}\sum_p\frac{N_{D_p}(\lambda)}{\lambda}=\sum_p\lim_{\lambda\to+\infty}\frac{N_{D_p}(\lambda)}{\lambda}=\sum_p\frac{A(D_p)}{4\pi}=\frac{A(D)}{4\pi}.$$