The set of $\lambda$’s such that this equation has a solution, denoted $\mathcal L(\Omega)$, is called the Laplace spectrum of $\Omega$. Does the Laplace spectrum determine $\Omega$? In general, the answer is negative. Consider the billiard problem inside $\Omega$. Call the length spectrum the closure of the set of perimeters of all periodic orbits of the billiard. Due to deep properties of the wave trace function, generically, the Laplace spectrum determines the length spectrum. We show that any generic axis symmetric planar domain with sufficiently smooth boundary is dynamically spectrally rigid, i.e. can’t be deformed without changing the length spectrum. This partially answers a question of P. Sarnak. This is a joint work with J. De Simoi, A. Figalli and Q. Wei.
Freitag, den 27. Oktober 2017 um 17.00-18.00 Uhr, in Mathematikon, INF 205, Konferenzraum, 5. OG Freitag, den 27. Oktober 2017 at 17.00-18.00 , in Mathematikon, INF 205, Konferenzraum, 5. OG
Der Vortrag folgt der Einladung von The lecture takes place at invitation by Prof. P. Albers