A question on the existence of an equation about non-autonomous semigroup

Question

In functional analysis, let $A$ be the infinitesimal generator of a $C_{0}$-semigroup $S(t)$ on a real Hilbert space $\mathbb{H}$, then the following properties are satisfied:

1. $\forall h \in \mathbb{H}$, $\int_{s}^{t} S(u)hdu \in \mathcal{D}(A)$, the domain of $A$; and $A\left(\int_{s}^{t} S(u)hdu\right) = S(t)h - S(s)h.$

2. $\forall h \in \mathbb{H}$, $S(t)h - S(s)h = \int_{s}^{t} AS(u) h du$.

In above, $A$ is time-independent, now I would like to consider a non-autonomous situation. Consider the family of operators $\{A(t): 0\leq t \leq T\}$ of linear operators such that

I. The domain $\mathcal{D}(A)$ of $A(t)$ is dense in $\mathbb{H}$ and independent of time $t$;

II. $\forall t \in [0, T]$, the resolvent $R(\lambda, A(t))$ exists for all $\lambda$ with $Re\lambda \geq 0$ and there exists $K > 0$ so that $\Vert R(\lambda, A(t)) \Vert \leq \frac{K}{(1+\vert \lambda \vert)}$;

III. There exists $0 < \delta \leq 1$ and $K > 0$ such that $\Vert (A(t) - A(s))A^{-1}(\tau) \Vert \leq K \vert t - s \vert^{\delta}$ for all $t,s,\tau \in [0, T]$;

IV. $\forall t \in [0, T]$ and some $\lambda \in \rho(A(t))$, the resolvent set of $A(t)$, the resolvent $R(\lambda, A(t))$, is a compact operator.

then $A(t)$ generates a unique linear evolution system $U(t,s), 0 \leq s \leq t \leq T$ which has the following properties:

a. $U(t,s)$ is compact if $t>s$;

b. $\frac{\partial}{\partial t}U(t,s) = A(t)U(t,s)$ if $t > s$.

My problems are as follows:

1). In the autonomous case, is Statement 2 a consequence of Statement 1?

2). For the non-autonomous case, for any $h \in \mathbb{H}$, is the following equation valid:

$U(t, s)h - h = \int_{s}^{t} (A(u) U(u,s) h) du$.

Or one can just do the integral on both sides of $\frac{\partial}{\partial t}U(t,s) = A(t)U(t,s)$ to get $U(t, s)h - h = \int_{s}^{t} (A(u) U(u,s) h) du$.

Thank you so much!! I would really appreciate it if any comments, answers or reference could be provided.

Answer 1

(1) In the autonomous case, the statement $$2.\; \forall h \in \mathbb{H},\; S(t)h - S(s)h = \int_{s}^{t} AS(u) h\,du \quad (*),$$ is not true in general. You need that $S(u)h\in \mathcal{D}(A)$.

If we assume that $S(u)$ map $\mathbb{H}$ into $\mathcal{D}(A)$, then $(*)$ follows by integrating, from $s$ to $t$, the following equality $$\frac{d}{du} S(u)h=AS(u)h.$$ You can conclude Statement 2 from Statement 1, but you need the following lemma:

If $f$ solves the Cauchy problem (classically) and $A$ is a closed operator, then $$\int_{s}^{t} f(u)\,du \in \mathcal{D}(A) \qquad \text{and} \qquad A\left(\int_{s}^{t} f(u)\,du\right)=\int_{s}^{t} Af(u)\,du.$$

(2) In the non-autonomous case, the same argument applies.