some question about weyl's law

Question

Let $\Omega\subset\mathbb{R}^n$ be a bounded domain with smooth boundary, consider the following Dirichlet eigenvalue problem. \begin{equation}\label{1} \left\{ \begin{array}{ll} -\Delta u=\lambda u, & \hbox{in $\Omega$;} \\[3mm] u=0, & \hbox{on $\partial\Omega$.} \end{array} \right. \end{equation} 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!

Answer 1

I think you are overthinking it.

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 $$ \lambda(k) = \frac{(2\pi)^2}{(\omega_n|\Omega|)^{2/n}}k^{2/n} $$