Is it possible for a (non-symmetric) elliptic differential operator to have finite spectrum. If so, is there an explicit example?
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$\begingroup$ What is the domain and codomain for your operator? $\endgroup$– Bombyx moriCommented Dec 23, 2019 at 4:37
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$\begingroup$ @Bombxy mori. Smooth functions on some open subset of Euclidean space (or manifold) should be a core for the differential operator. Otherwise, I am open to any construction. $\endgroup$– Chris JudgeCommented Dec 24, 2019 at 8:52
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$\begingroup$ At first the question strikes me as odd, but now I think about it is actually not trivial. It seems the infinite spectrum only holds for positive-definite operators. I think you are essentially asking how to come up with a non-trivial example. $\endgroup$– Bombyx moriCommented Dec 24, 2019 at 20:50
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$\begingroup$ So here is the set up: Let $M$ be a $C^{\infty}$ manifold of dimension $n$, and let $A$ be an elliptic operator of order $m>0$ whose top order symbol $a_m$ satisfy that all the eigenvalues $\lambda$ of $a_m$ is in the region $|\arg(\lambda)-\theta|<\delta,\forall x\in U, |\xi|=1$. Then the pole of $A^{s}$ only happens at $s=\frac{k-n}{m}$ and they are simple. So with uniform ellipticity we can exclude the case of finite spectrum. However I still do not know how to construct a counter-example. $\endgroup$– Bombyx moriCommented Dec 24, 2019 at 20:57
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$\begingroup$ @Bombyx mori. I am unsure of what you have written. Symmetry is enough to obtain infinitely many eigenvalues. One doesn't need positivity. In the literature there are results that state that, for example, Weyl's law holds if the elliptic operator is not far from symmetric e.g. principal symbol symmetric but lower order terms not. The resolvent being compact does not immediately imply that the spectrum is infinite. A non-symmetric compact operator can have finite spectrum. $\endgroup$– Chris JudgeCommented Dec 25, 2019 at 21:39
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1 Answer
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This is more a long comment than an answer since some details should be checked. The answer shoould be not for (say) a second order differential operators with continuous top order coefficients and bounded first and zero order ones in bounded domain (with some reularity on the boundary), under say Dirichlet bc. A very nice theorem in the second volume of Dunford Schartz (Theorem 29 pag 1115 and subsequent corollaries) implies that if the resolvent is of trace class and decays like $1/|\lambda|$ on two lines from the origin such that the smallest angle between them can be arbitrarily small, then the linear span of the generalized eigenfunctions is dense in $L^2$. In our situation the domain of the operator is $H^{2}\cap H^1_0$ and the resolvent to the power $k$ is trace class if $k>d/2$ ($d$ is the dimension). The rays above from the origin exist since the operator generates an analytic semigroup of angle $\pi/2$. I admit that many details are left...but it should work.
answered Jan 11, 2020 at 10:53
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1$\begingroup$ I checked better. The completeness of the generalized eigenfuctions can be found in the last chapter of the second volume of DS or in the papers by S. Agmon, in more generality. $\endgroup$ Commented Jan 11, 2020 at 18:32
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$\begingroup$ In Dunford-Schwartz the most relevant results seem to be Coro XI.6.31 and Thm XIV.6.28, and the most relevant paper of Agmon seems to be `On the eigenfunctions and...' doi.org/10.1002/cpa.3160150203. There are additional hypotheses to these theorems. Are they necessary? $\endgroup$ Commented Jan 12, 2020 at 17:30
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1$\begingroup$ I checked on Agmon: On eigenvalues eigenfuctions and resolvents of general elliptic problems (this a course of lectures he gave in Italy in 1962 and is published by Springer, G. Fichera editor, not easy to find but more readable). For real coefficients and for a class of elliptic problems he calls "absolutely elliptic" (containing Dirichlet and oblique derivative) the result is true. In the general setting I see also some extra hypotheses. DS has smooth coefficients but Agmon only continuous. See also DS XIV.6.27 for the properties of the rays when the coefficeints are real. $\endgroup$ Commented Jan 12, 2020 at 18:02
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$\begingroup$ Thanks for the reference to the lectures of Agmon. Before asking the question I had looked at the CMP paper by Agmon and some other things, but they all have hypotheses in addition to elliptic. I just wonder whether these are really necessary...I don't want to assume real coefficients etc. $\endgroup$ Commented Jan 12, 2020 at 19:21