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I am very interested in the properties of the semigroup $e^{it\sqrt{-\Delta}}$, which may has some fundamental differences (such as the kernel) with the well-known Schrödinger semigroup $e^{it\Delta}$.

Any properties (or references or books) that related this semigroup are appreciated.

Thanks!

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    $\begingroup$ Try putting "heat semigroup" in google and you'll get lots relevant references. $\endgroup$ Aug 12, 2012 at 13:51
  • $\begingroup$ I think no-one has mentioned yet that $e^{it\sqrt{-\Delta}}$ has an explicit kernel $c_n (-it) (|x|^2 - t^2 - 0i)^{-(n+1)/2}$, where $n$ is the dimension and $c_n$ is an appropriate constant. The kernel is singular around $|x|=t$. In particular, in dimension $n = 1$, the kernel is simply $$\operatorname{p.v.} \biggl(\frac{1}{\pi} \, \frac{-it}{x^2 - t^2}\biggr) + \frac{1}{2} (\delta_t(x) + \delta_{-t}(x)).$$ $\endgroup$ Feb 5, 2020 at 19:52
  • $\begingroup$ Also, the operator $\sqrt{-\Delta}$ is sometimes called the (quasi-)relativistic Hamiltonian for a massless particle, and has been studied heavily by, for example, Lieb and co-authors. $\endgroup$ Feb 5, 2020 at 19:56

2 Answers 2

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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. The dispersion relation is $\omega = |\xi|$, 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.

Note that $e^{it\Delta}$ is not the heat semigroup, which the other answers and comments seem to suggest.

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    $\begingroup$ The downvoter care to comment? $\endgroup$
    – timur
    Aug 13, 2012 at 0:23
  • $\begingroup$ You're right, $e^{it\Delta}$ isn't the heat semigroup. Editing my answer. $\endgroup$
    – Nik Weaver
    Aug 13, 2012 at 3:14
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    $\begingroup$ It is not actually true that this operator has a finite speed of propagation. Finite speed = compactly supported kernel = entire analytic symbol (by Paley-Wiener). The symbol $e^{it|\xi|}$ is not entire analytic $\endgroup$ Feb 5, 2020 at 15:51
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    $\begingroup$ Note that the wave equation has entire symbol, indeed it is $\cos(t|\xi|)$ plus $\sin(t|\xi|)|\xi|^{-1}$ $\endgroup$ Feb 5, 2020 at 15:52
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You can say a lot about the function $u(t,x)=e^{it\sqrt{-\Delta}}f$ by noticing, as remarked by others, that $u$ is also solution of the Cauchy problem $$ u_{tt}-\Delta u=0,\quad u(0,x)=f,\quad u_t(0,x)=i\sqrt{-\Delta}f. $$ Thus most estimates for solutions of the wave equation carry over to $u$. However there are some subtle differences, mostly due to the non--locality of the operator $\sqrt{-\Delta}$.

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