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Let $D\subset\mathbb{R}^d$ be a bounded domain which is regular for the Dirichlet problem. There is then a complete set of orthonormal eigenfunctions $\phi_j$ with corresponding eigenvalues $0<\lambda_1<\lambda_2\leq\lambda_3\ldots$ satisfying $$\Delta \phi_j=-\lambda_j\phi_j\mbox{ on }D\mbox{ and }\phi_j=0\mbox{ on }\partial D.$$ I know from a paper of Morrey and Nirenberg (http://onlinelibrary.wiley.com/doi/10.1002/cpa.3160100204/abstract) that if the boundary $\partial D$ is real analytic, then each eigenfunction $\phi_j$ is analitycally continuable across the boundary. My naive question is the following:

Does it hold that there exists an open $\epsilon$-neighbourhood of $D$ on which all $\phi_j$ are real analytic ?

I understand that the proof of Morrey and Nirenberg's Theorem does not give any hint on the ''size'' of the set where a particular $\phi_j$ can be extended, but I'm wondering whether it is possible to exploit some property of the family of eigenfunctions $\{\phi_j\}$ to obtain a simultaneous continuation.

I'm not at all a specialist in this field, so I apologize for having no idea on how to proceed.

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  • $\begingroup$ What do you mean by "analytic"? The $\phi_j$ are certainly not holomorphic as functions of $z=x+iy$ because a holomorphic $u$ satisfies $\Delta u=0$. $\endgroup$
    – Christian Remling
    Commented May 17, 2014 at 17:46
  • $\begingroup$ I mean real analytic. I have edited the question. $\endgroup$
    – MassiveJack
    Commented May 17, 2014 at 22:00

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I doubt you can do what you want for the following reason. The eigenfunctions $\{\phi_j\}$ should form a basis of $L^2(D)$. Suppose that you could extend each $\phi_j$ across the boundary in the same way as you want. Then you would be able to extend any $L^2(D)$ function defined on $D$ to an $\epsilon$-neighbourhood of $D$ using this basis and I really doubt you can do this.

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  • $\begingroup$ Well, of course, it wouldn't quite follow that every $L^2$ function would continue across, even merely in a distributional sense, without further information on the sizes of the continuations. But, yes, this potential mechanism itself suggests that the sizes of the continuations of the eigenfunctions must behave badly. $\endgroup$
    – paul garrett
    Commented May 17, 2014 at 15:49
  • $\begingroup$ @paul garret: I take it you both read "analytic" as "real analytic" (in $x$ and $y$). Then we can of course have bases consisting of functions that are analytic everywhere. (Or how about the Dirichlet problem for $-u''$ on $[0,1]$?) $\endgroup$
    – Christian Remling
    Commented May 17, 2014 at 18:05
  • $\begingroup$ @ChristianRemling, yes, I did presume that in the context of the question "analytic" means "real-analytic". And your "analytic everywhere" presumably means "on opens containing $D$ but depending on which eigenfunction"? $\endgroup$
    – paul garrett
    Commented May 17, 2014 at 18:18
  • $\begingroup$ @paul garret: So if we just take the unit square, say, we'll get Dirichlet eigenfunctions that look something like $u_{mn}=\sin \pi mx\sin \pi ny$, so there's definitely nothing wrong (in general) with having a very nice analytic basis. $\endgroup$
    – Christian Remling
    Commented May 17, 2014 at 18:24
  • $\begingroup$ @ChristianRemling, no, I didn't mean to insinuate any problem with analytic eigenfunctions, but only with the complications in "continuing" more general functions outside the domain. The interval and/or a rectangle are anomalous in that they can tile the plane "too nicely". $\endgroup$
    – paul garrett
    Commented May 17, 2014 at 18:55
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[Not really an answer but, hopefully, may be of help.]

My feeling is that eigenfunctions can be continued, although I would not bet on this. At least, it is so in cases when they can be computed explicitely, like circle and ellipse http://arxiv.org/abs/1206.1278. My advice is to ask personally someone working on quantum unique ergodicity, these guys may know the right literature. (Do not assume that every mathematician read MO.)

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  • $\begingroup$ This seems to be a good advice. I will do that. Thank you ! $\endgroup$
    – MassiveJack
    Commented May 21, 2014 at 21:42

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