7
$\begingroup$

Let $f$ be a positive real-analytic function on the closed unit disk. Consider the eigenvalue problem $\Delta \phi = \lambda f \phi$, with $\phi = 0$ on the boundary. There exists a sequence of eigenvalues $\lambda_n$. Now suppose $f$ depends real-analytically on a parameter $t$ for $t$ in some interval containing $0$. Let $n$ be given and let $k$ be the dimension of the eigenspace of $\lambda_n$. Do there exist $k$ functions of $t$, defined on some interval about $t=0$, that are at least $C^1$ in $t$ and give $k$ eigenvalues of $t$, all of which equal $\lambda_n$ when $t=0$?

Remarks: (1) Courant-Hilbert, Methods of Mathematical Physics, vol. 1, page 419 proves continuity (i.e. $C^0$ dependence). You would think if $C^1$ dependence were known (in 1953) they would mention it.

(2) We do not require that the functions preserve the order of the eigenvalues. For example it might happen that $\lambda_2 < \lambda_3$ for $t < 0$ and $\lambda_2 > \lambda_3$ for $t > 0$, and maybe they cross transversally, though I do not know if this can really happen.

(3) Perhaps one can even get real-analytic dependence on $t$.

(4) There is a thick book by Kato on perturbation theory with a lot of theorems in it, one or more of which possibly contains the answer.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Kato has a lot of results on this. The basic picture is: as long as the eigenvalues are distinct, they are smooth (analytic). When the cross each other, singularities may arise, but typically (in the selfadjoint case) they remain Lipschitz. In general crossings do occur. $\endgroup$
    – Piero D'Ancona
    Commented Feb 14, 2011 at 22:11
  • $\begingroup$ So are you saying that the answer is "no"? Or can it be that this Lipschitz behavior is because of insisting that the eigenvalues remain in order, i.e. monotonic in the subscript? $\endgroup$
    – Michael Beeson
    Commented Feb 14, 2011 at 22:21

1 Answer 1

Reset to default
7
$\begingroup$

In your case I think you can apply Rellich's theorem, that is Theorem VII.3.9 in Kato's book (p.392 in my edition). The result states that, whenever you have a family of selfadjoint operators with compact resolvent, depending analytically on a real parameter on some open interval of the reals, with a common domain independent of the parameter (this is what Kato calls a family of type (A)), then you can enumerate both eigenvalues and eigenfunctions in such a way that they are analytic functions of the parameter in the same interval.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ +1. I should have written analytically instead of holomorphically, since the parameter is reel. Besides, it should be emphasized that the Theorem, true for one-parameter families, becaomes false for two parameters or more. The standard counter-example is $$\begin{pmatrix} x & y \\\\ y & -x \end{pmatrix}.$$ $\endgroup$
    – Denis Serre
    Commented Feb 15, 2011 at 8:02
  • $\begingroup$ Right, I meant analytically of course. Edited $\endgroup$
    – Piero D'Ancona
    Commented Feb 15, 2011 at 17:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .