Timeline for Equivalent ways to study a semilinear parabolic equation as a perturbed abstract Cauchy problem
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9 events
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| May 6, 2016 at 21:12 | comment | added | Delio Mugnolo | @JohnZHANG One has to be careful. What I am saying is that if $B$ is bounded from $D(A)$ to $X$ and $T(\cdot)$ is analytic, then $BT(\cdot)$ is a bounded operator on $X$, and of course so is $S(\cdot)$. Now, you may have enough information to conclude that the mapping $T(t-\cdot)BS(\cdot)$ (from $\mathbb R_+$ to the space of bounded linear operators on $X$) is integrable. | |
| May 6, 2016 at 18:09 | comment | added | John | @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? | |
| May 6, 2016 at 18:00 | comment | added | John | @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 $x\in D(A)$. | |
| May 5, 2016 at 14:07 | comment | added | Delio Mugnolo | > 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. | |
| May 4, 2016 at 13:52 | vote | accept | John | ||
| May 4, 2016 at 11:26 | comment | added | Hannes | 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). | |
| May 4, 2016 at 10:41 | comment | added | John | 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? | |
| May 4, 2016 at 10:05 | review | First posts | |||
| May 4, 2016 at 10:21 | |||||
| May 4, 2016 at 10:01 | history | answered | Hannes | CC BY-SA 3.0 |