Timeline for Evolution operator for a linear parabolic equation

7 events

when toggle format	what		by	license	comment
Oct 26, 2015 at 23:46	answer	added	Delio Mugnolo		timeline score: 2
Oct 26, 2015 at 20:08	comment	added	Denis Serre		In the case $A$ is independent of $t$ and is elliptic, you need more to ensure that $e^{-tA}$ is of trace class. This is false if $Af=af$ for some $a\in(0,+\infty)$, but it is true if $A$ has order $m>0$. Presumably, you had in mind the case where $A$ is differential of second order.
May 29, 2015 at 19:18	comment	added	Maxim Braverman		I don't think so. If $A$ is independent of $t$, then $U(t)$ is the heat operator and is of trace class.
May 29, 2015 at 15:01	answer	added	Peter Michor		timeline score: 0
May 29, 2015 at 14:53	comment	added	Peter Michor		Don't you want $U(t)$ is identity plus trace class?
May 29, 2015 at 13:52	review	First posts
May 29, 2015 at 14:37
May 29, 2015 at 13:48	history	asked	Maxim Braverman	CC BY-SA 3.0