# some question about weyl's law

Let $\Omega\subset\mathbb{R}^n$ be a bounded domain with smooth boundary, consider the following Dirichlet eigenvalue problem. $$\label{1} \left\{ \begin{array}{ll} -\Delta u=\lambda u, & \hbox{in \Omega;} \\[3mm] u=0, & \hbox{on \partial\Omega.} \end{array} \right.$$ The Weyl's law shows that $$N(\lambda)=\#\{k\mid 0<\lambda_k \leq \lambda\}\sim \frac{\omega_n}{(2\pi)^n} |\Omega| \lambda^{\frac{n}{2}},\qquad \lambda\to +\infty.$$ and it means that $$\lambda_k\sim \frac{(2\pi)^2}{(\omega_n|\Omega|)^{\frac{2}{n}}} k^{\frac{2}{n}},\qquad k\to +\infty.$$

I feel confused about how can we prove the second conclusion from the first conclusion. It seems we only know that $N(\lambda_k)\geq k$. I found in some books that use the fact $N(\lambda_k)=k$, I don't know how to get this since $\lambda_{k}$ may has multiplicty. For example, if $0<\lambda_{1}<\lambda_2=\lambda_3=\lambda_4<\lambda_5<\cdots$, then we have $N(\lambda_2)=N(\lambda_4)=4$. Can some one help me with this question? Thanks a lot!

What you have is a non-decreasing function on the natural numbers $\lambda(k)$. So in the first line you are really just saying, given $\bar{\lambda}$, let's solve $$\lambda(k) = \bar{\lambda}.$$ (this tells you how many numbers $k$ are such that $\lambda(k) \leq \bar{\lambda}$). The answer is $$k \approx \frac{\omega_n}{(2\pi)^n}|\Omega|\bar{\lambda}^{n/2}$$ But this means you essentially you have $$\frac{\omega_n}{(2\pi)^n}|\Omega|\lambda(k)^{n/2} = k$$ Then in the second line you are asking what is an explicit expression for $\lambda(k)$, so you just manipulate this expression and it clearly becomes $$\lambda(k) = \frac{(2\pi)^2}{(\omega_n|\Omega|)^{2/n}}k^{2/n}$$