Weyl's asymptotic law for eigenvalues - Rectangle $D = \{0 < x < a, 0 < y < b \}$

Question

Let the domain $D = \{0 < x < a, 0 < y < b \}$ in the plane. We now that $$\lambda_{n,m} = \frac{n^2 \pi^2}{a^2}+\frac{m^2 \pi^2}{b^2}$$ with the eigenfunction $$u_{n,m}= \sin(\frac{nπ}{a}x) \sin(\frac{mπ}{b}y).$$ I would like to show the Weyl's asymptotic law for eigenvalues, i.e., $\lim_{l \to \infty} \frac{\lambda_l}{l}=\frac{4 \pi}{A}$ ($2$-dimensional), where $A$ is the area of the rectangle. I am blocked on this problem for a while. Is there anyone could give the principle steps how could realize this problem?

Answer 1

The notation can also be interpreted as follows: order the eigenvalues $\lambda_1\le\lambda_2\le \ldots$. Consider $\lambda_N$ for a large $N$, and now follow Willie's outline to approximately determine $N$: The number of smaller eigenvalues is one fourth the number of lattice points in the ellipse $$ x^2/a^2 + y^2/b^2 = \lambda_N/\pi^2 , $$ and this is approximately one fourth times the area $= (1/4)\pi ab\lambda_N/\pi^2=A\lambda_N/(4\pi)$. So $N\simeq \lambda_N A/(4\pi)$, as claimed.