TITLE: Evolution operator for a linear parabolic equation
QUESTION [10 upvotes]: Let $A(t)$ be a smooth family of positive definite operators on a Hilbert space $H$. Consider the operator
$$D:= \frac{d}{dt}+A(t)$$
and let $U(t):H\to H$ be the evolution operator, i.e., $U(0)=I$ and $U'(t)+A(t)U(t)=0$.
Question: What condition  on $A(t)$ guarantees that $U(t)$ is of trace class for all $t>0$?
More specifically, I am interested in situation when the following situation: Let $M$ be a compact manifold, $H=L^2(M)$, and let $A(t)$ be a smooth family of positive definite elliptic operators on $M$. If $A(t)$ is independent of $t$, then $U(t)=e^{-tA}$ is of trace class. So I hope that if $A'(t)$ is not very large, then $U(t)$ is still of trace class. What is the precise condition on $A'(t)$ which guarantees this?

REPLY [2 votes]: You may want to take a look at this Ph.D. thesis, especially at Thm. 5.5.2, which as far as I know is the sharpest currently available result concerning regularity of nonautonomous problems. The theorem states that the solution to your problem is at most in $V$ for each $t$, under suitable assumptions. Now, if you are thinking of a time-dependent diffusion equation, $V=H^1(\Omega)$, which is never trace-class-embedded in $L^2(\Omega)$, even in 1-dimension. In the autonomous case this is not a problem, since the semigroup law still allows us to conclude that $e^{-tA}$ is of trace class for all $t>0$ iff $V\hookrightarrow H$ is of $p$-Schatten class for some $p<\infty$.
But in the autonomous case this smoothing property seems to fail, so in general you won't get any more than $p$-Schatten class for all $p>1$ in the case of a nonautonomoum problem associated with a differential operator of order 1.
As suggested by @DenisSerre, things change if you consider higher order operators: if $A(t)$ is for all $t$ a 2k-th order differential operator with domain $H^k(\Omega)$, then you do have trace class evolution operators as soon as $k>d$.