most recent 30 from http://mathoverflow.net 2013-06-20T06:18:18Z http://mathoverflow.net/feeds/question/104558 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104558/what-do-we-know-about-the-semigroup-eit-sqrt-delta Shanlin Huang 2012-08-12T13:40:44Z 2012-08-13T03:14:59Z

<p>I'm very interested in the properties of the semigroup $e^{it\sqrt{-\Delta}}$, it may has some fundamental differences(such as the kernel) with the well-known schrodinger semigroup $e^{it\Delta}$.</p> <p>Any properties (or references or books) that related this semigroup are appreciated.</p> <p>Thanks!</p>

http://mathoverflow.net/questions/104558/what-do-we-know-about-the-semigroup-eit-sqrt-delta/104581#104581 timur 2012-08-12T22:11:55Z 2012-08-13T00:54:13Z

<p>The wave operator decomposes as $$ \partial_t^2-\Delta = (\partial_t-i\sqrt{-\Delta})(\partial_t+i\sqrt{-\Delta}), $$ so you can think of $e^{it\sqrt{-\Delta}}$ as solving a "half of" the wave equation. In particular, it has a finite propagation speed. This can also be seen from the dispersion relation $\omega = |\xi|$, where $\omega$ and $\xi$ are the Fourier variables for $t$ and $x$, respectively. On the other hand, the Schrödinger propagator $e^{it\Delta}$ has the dispersions relation $\omega=|\xi|^2$, which makes it genuinely dispersive, i.e., the propagation speed depends on the frequency.</p> <p>Note that $e^{it\Delta}$ is <em>not</em> the heat semigroup, which the other answers and comments seem to suggest.</p>