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I am considering the following abstract Cauchy problem on Banach space $X$: \begin{cases} u'(t)=Au(t)+\big(f(t)+Bu(t)\big),&t\in[0,T],\\ u(0)=x_0, \end{cases} Suppose $A$ generates an analyitc semi-group $S(t)$ on $X$, then under certain assumptions, we can study the wellposeness of the following integral equation for the original Cauchy problem: \begin{equation} u(t)=S(t)x_0+\int_0^t S(t-s)\big(f(s)+Bu(s)\big)\, ds.\tag{1}\label{eq:noper} \end{equation}

However, we can also view $C=A+B$ as a perturbation of $A$, hence suppose $A+B$ also generates an analytic semi-group $T(t)$ on $X$, then we can also study \begin{equation} u(t)=T(t)x_0+\int_0^t T(t-s)f(s)\, ds.\tag{2}\label{eq:per} \end{equation}

I hope to show the wellposedness of a mild solutions $u\in C([0,T];(D(A),\|\cdot\|_A)$ to \eqref{eq:noper} and \eqref{eq:per} should be equivalent, in the sense that a solution to \eqref{eq:noper} should also be a solution to \eqref{eq:per}, also conversely. Here $\|\cdot\|_A$ refers to the graph norm. I think it may be correct(at least with good assumptions for $x_0$ and $B$), since intuitively, they come from two different ways to interpret the same Cauchy problem.

For your information, the specific problem I am considering is $A$ be an elliptic 2nd order operator and $B$ be a first order operator.

I have read Engel-Nagel' book, which presents the following form of a variation of parameters formula for perturbed semigroups, for some operator $B$, $$T(t)x=S(t)x+\int_0^t S(t-s)BT(s)x\,ds,$$ but I failed to use it to show the desired result.

Is there any reference with rigorous arguments you can point out?

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You can derive that the mild solution $u$ as given in (2) satisfies (1) from the connection between the semigroups $S$ and $T$ as you cited from the Engel-Nagel book (as a minor remark, there is an $x$ missing right before the $ds$): Just insert the cited formula into (2), re-arrange and interchange order of integration. A function satisfying (1) must be unique (use Gronwall's Lemma), so the mild solution as given in (2) is the unique function satisfying (1).

Please also recall that $B$ must be a bounded operator for the cited formula to be true!

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  • $\begingroup$ Thanks for your hint. Following it, I have verified a mild solution given in (2) satisfied (1) using the cited formula. So I think conversely, just reversing the process, it should follow that a mild solution given in (1) should also satisfied (2). Hence if I should the cited formula holds and I can show the uniqueness of mild solution to one of them (in particular, I have show the uniqueness of solution to (2)), then the uniqueness of mild solution to the other should follow. Am I correct? $\endgroup$
    – John
    Commented May 4, 2016 at 10:41
  • $\begingroup$ Yes, basically. I think it is enough to observe that you need the same assumptions on your data to make both expressions (1) and (2) well-defined, so as soon as (1) is meaningful, the function defined in (2) is already the (unique, due to Gronwall) function satisfying (1). $\endgroup$
    – Hannes
    Commented May 4, 2016 at 11:26
  • $\begingroup$ > Please also recall that BB must be a bounded operator for the cited formula to be true! This is certainly not true. All you need is that the integrand is a continuous function (of the variable $s$) from $\mathbb R_+$ to the space of bounded linear operators on $X$. Since the semigroup generated by $A$ is analytic, the formula could make sense even for unbounded $B$ if $B$ is relatively $A$-bounded. $\endgroup$
    – Delio Mugnolo
    Commented May 5, 2016 at 14:07
  • $\begingroup$ @DelioMugnolo Thanks for your remark. I found from Engel-Nagel book page 168 that the formula also holds for bounded operator $B$ on $D(A)$ with $x\in D(A)$. $\endgroup$
    – John
    Commented May 6, 2016 at 18:00
  • $\begingroup$ @DelioMugnolo So let me make sure I understand your comments. What you are suggesting is that in my case, if $B$ is relatively $A$-bounded, then $A+B$ generates an analytic semigroup $T$ on $X$, hence $T(t)x\in D(A)$ for all $x\in X$. So by repeating Engel-Nagel's proof for the cited formula, in my case, the formula holds for $B\in L(D(A), X)$ with $x\in X$ even though $B$ is unbounded on $X$. Am I correct? $\endgroup$
    – John
    Commented May 6, 2016 at 18:09

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