Let $X$ Banach and $A \colon D(A) \subset X \to X$ be a generator of a strongly continuous semigroup $e^{At}$. Let $B \in L(X)$ (linear bounded opearator on $X$) we then know that $B$ generates a uniform continuous semigroup $e^{Bt}$ given by the exponential series of powers of $B$.
Now consider $C=A+B$; under which conditions does it generate a strongly continuous semigroup $e^{Ct}$ given by $e^{Ct}=e^{At}e^{Bt}$? Can you provide some references in case?
Best Answer
Assume $G(t):=e^{At}e^{Bt}$ forms a strongly continuous semigroup. Then $$e^{At}e^{Bt}e^{As}e^{Bs}=G(t)G(s)=G(t+s)=e^{A(t+s)}e^{B(t+s)}=e^{At}e^{As}e^{Bt}e^{Bs}$$ Hence $$e^{Bt}e^{As}=e^{As}e^{Bt}$$ and $$Be^{As}=\lim_{t\to 0^+}{e^{Bt}-I\over t}e^{As}= \lim_{t\to 0^+}e^{As}{e^{Bt}-I\over t}=e^{As}B$$ Assume $x\in D(A).$ Then $$\lim_{s\to 0^+}B{e^{As}x-x\over s}=BAx$$ On the other hand, as $B$ and $e^{As},$ commute we get $$BAx=\lim_{s\to 0^+}B{e^{As}x-x\over s}=\lim_{s\to 0^+}{e^{As}Bx-Bx\over s}$$ Since the limit exists, then $Bx\in D(A)$ and the limit is equal $ABx.$
Summarizing we showed that $B(D(A))\subset D(A)$ and $BAx=ABx$ for $x\in D(A).$ These conditions are therefore necessary.