Skip to main content
13 events
when toggle format what by license comment
Oct 11, 2021 at 7:45 history edited Student CC BY-SA 4.0
added 293 characters in body
Oct 11, 2021 at 7:44 history edited YCor CC BY-SA 4.0
removed capitals from title
Oct 11, 2021 at 7:42 history edited Student CC BY-SA 4.0
added 293 characters in body
Oct 11, 2021 at 7:38 comment added Student I should just add that I am working on the space half-space $R^n_{+}$
Oct 11, 2021 at 7:37 comment added username To complement the comment above, assuming that you have some natural boundary conditions on $\partial\Omega$, Neumann or Dirichlet, and that you have setup your problem in the right space, such a Beppo-Levi space (see e.g. [Deny and Lions] (eudml.org/doc/73718) in French), you have a good chance of showing compactness.
Oct 11, 2021 at 2:22 comment added user378654 The most common reason eigenspaces are finite is that your operator's inverse is compact. This does not follow from the information you provided, but the information provided is kind of vague: you do not give a boundary condition along $\partial \Omega$, or the assumptions on $w$ and $a$. The point is that you explicitly state that the domain is unbounded, which is potentially dangerous for having eigenfunctions/values at all.
Oct 10, 2021 at 22:42 comment added Student Thanks for your comment, just a quick question how do you know that the dimension of the second eigenspace is finite?
Oct 10, 2021 at 17:30 comment added username I would be suprised if there was such a result. If you simply consider the Laplacian on a bounded domain, it is difficult to predict the multiplicity of the eigenvalues, without precise informations on the geometry. Naturally, the multiplicity of the first one is known, by Krein-Rutman. But after that you know that that the dimension of the eigenspace is finite and that's about it.
Oct 10, 2021 at 14:53 history edited Student CC BY-SA 4.0
edited title
Oct 10, 2021 at 14:21 history edited Student CC BY-SA 4.0
added 16 characters in body
Oct 10, 2021 at 14:21 comment added Student @username It's positive actually, let me add that to the question. Thanks!
Oct 10, 2021 at 14:06 comment added username Does w have an arbitrary sign?
Oct 10, 2021 at 13:48 history asked Student CC BY-SA 4.0