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The following comes from Springer Online Reference Works:
Consider a bounded domain $\Omega\subset\mathbb{R}^n$ with a piecewise smooth boundary $\partial\Omega$. $\lambda$ is a Dirichlet eigenvalue of $\Omega$ if there exists a function $u\in C^2(\Omega)\cap C^0(\bar{\Omega})$ (a Dirichlet eigenfunction) satisfying the following Dirichlet boundary value problem $$ -\Delta u=\lambda u \qquad \text{in } \Omega $$ $$ u=0\qquad \text{in } \partial\Omega $$

Provided $\Omega$ is bounded and the boundary $\partial \Omega$ is sufficiently regular, the Dirichlet Laplacian has a discrete spectrum of infinitely many positive eigenvalues with no finite accumulation point: $$ 0<\lambda_1\le\lambda_2\le\cdots $$

The Weyl’s asymptotic law says that:
For large values of $k$ , if $\Omega \subset \mathbb{R}^n$ ,then $$ \lambda_k\approx\frac{4\pi^2k^{2/n}}{(C_n\vert\Omega\vert)^{2/n}} $$

where $\vert\Omega\vert$ and $C_n$ are the volumes of $\Omega$ and of the unit ball in $\mathbb{R}^n$.

I've found Weyl's original work (Das asymptotische Verteilungsgesetz der Eigenwerte linearer partieller Differentialgleichungen) but it is in German.

So is there an English translation or can anyone help? Thank you~

EDIT: Or, should this be a mathoverflow question?

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3 Answers 3

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Walter Strauss' book has a nice exposition of the proof. It uses comparison principles based on a variational characterization of the eigenvalues.

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  • $\begingroup$ Walter Strauss's book? This one? $\endgroup$
    – t.b.
    Dec 6, 2011 at 5:55
  • $\begingroup$ Close but not quite. I included a link. $\endgroup$
    – timur
    Dec 6, 2011 at 5:59
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I've found a related question here, and it was answered. Its answer gives a book where an English proof can be found: Elliptic operators, topology, and asymptotic method by J. Roe.

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  • $\begingroup$ Does this proof require the smoothness of the boundary? What if the boundary is nowhere differentiable like a fractal curve? $\endgroup$
    – Hans
    Jan 6, 2019 at 21:04
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A version of the theorem can also be found in Courant & Hilbert, Methods of Mathematical Physics Vol I, Chapter VI, Theorem 16.

The gist of the proof is to approximate the domain by triangles and rectangles, then obtain Weyl's law in the domain by using eigenvalue monotonicity results and Weyl's law for rectangles (which is established by explicit computation).

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  • $\begingroup$ It seems this proof requires the piecewise smoothness of the boundary. Is this true? If so, what is the proof for a boundary that is nowhere differentiable like a fractal curve? $\endgroup$
    – Hans
    Feb 1, 2019 at 0:06
  • $\begingroup$ No- the proof for Dirichlet boundary conditions works even for nowhere differentiable boundaries. You don't even have to use triangles and rectangles; just cover the domain by small squares. $\endgroup$
    – Neal
    Feb 1, 2019 at 14:24
  • $\begingroup$ For Neumann boundary conditions, however, if the boundary is too irregular, the Neumann Laplacian will have continuous spectrum. The "segment condition" is sufficient for the Neumann Laplacian to have discrete spectrum and obey Weyl's law. $\endgroup$
    – Neal
    Feb 1, 2019 at 14:26

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