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A planar region will have associated to it a spectrum consisting of Dirichlet eigenvalues, or parameters $\lambda$ for which it is possible to solve the Dirichlet problem for the Laplacian operator,

$$ \begin{cases} \Delta u + \lambda u = 0 \\ u|_{\partial R} = 0 \end{cases}$$

I'm wondering, if we have a sequence of boundaries $\partial R_n$ converging pointwise towards $\partial R$, then will the spectrums also converge? (I make the notion of convergence formal in the following manner: $\cap_{N=1}^\infty l(\cup_{n=N}^\infty\partial R_n)=\partial R$; $\cap_{N=1}^\infty l(\cup_{n=N}^\infty\mathrm{spec}(R_n))=\mathrm{spec}( R)$, where $ l(\cdot)$ denotes the set of accumulation points of a set and $\mathrm{spec}(\cdot)$ denotes the spectrum of a region.)

One motivating pathological example is the sequence of boundaries, indexed by $n$, defined by the polar equations $r=1+\frac{1}{n}\sin(n^2\theta)$. The boundaries converge to the unit circle. However, since the gradient of any eigenfunction must be orthogonal to the region boundary (as it is a level set), the eigenfunctions can't possibly converge to anything (under any meaningful notion) and so it makes me question if it's even possible for the eigenvalues to do so.

If the answer is "no, the spectrum doesn't necessarily converge," a much broader question arises: what are necessary and sufficient conditions for it to converge? Intuitively, I imagine a necessary condition is that the curvature of the boundaries also converge appropriately, but I have no idea if that's sufficient. EDIT: Another interesting question is if the principal eigenvalue (the smallest nonzero one) can grow arbitrarily large.

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Non-convergence of eigenvectors is probably not an obstruction. You can easily construct smooth family of matrices whose eigenvalues converge but whose eigenvectors won't. –  Willie Wong Jul 11 '11 at 10:18

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up vote 3 down vote accepted

There is a domain monotonicity of Dirichlet eigenvalues: if domains $\Omega^1\supset\Omega^2\supset\ldots\supset\Omega^n\supset\ldots\ $ then the corresponding eigenvalues $\lambda_k^1\ge\lambda_k^2\ge\ldots\ge\lambda_k^n\ge...\ $ so convergence of curvatures are not necessary in this case. There are also lots of more general results on spectral stability problems for elliptic differential operators.

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