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In a nutshell, here is my question. I read and know about the relation between the spectra of $L$ and $A$ if $A$ is a relatively compact perturbation of $L$. However, for my purpose, I am interested in the relation between the solutions of the equations $v_t=Lv$ and $v_t=Av$. For example, if the first has an unbounded solution, can we say the same about the second? Or, maybe, if all the solutions of the first are bounded, can we say the same for the second? Given the relation between the spectra of $A$ and $L$, one might think it is possible to make a connection, but I did not find one. Note that in my case, the operators $A$ and $L$ have the same spectrum in the sense that their resolvent sets coincide. Really, I am looking for references that would treat this kind of questions, not a proof.

The precise version of my question is below.

Let $L$ be a compactly defined differential operator on $L^2(\mathbb{R})$ such that it is possible to prove that the solutions to the IVP $$ v_t=Lv,\;\;v(0)=v_0\in {\rm Dom}(L)\;\;\;\;\;\;(1) $$ admits for any $v_0\in {\rm Dom}(L)$ the unique solution $v \in C(\mathbb{R},{\rm Dom}(L))$ such that the $L^2$ norm $v(t)$ is bounded by the $L^2$ norm of $v_0$. Then I would like the same to be true for $$ v_t=Av,\;\;v(0)=v_0\in {\rm Dom}(L)\;\;\;\;\;\;\;(2) $$ where $A$ is a relatively compact perturbation for $L$. More specifically, I am thinking of $A-L$ as a Hilbert-Schmidt integral operator.

Also, I would like the result that if there is a $v_0$ such that the solution to (1) has a norm that is not bounded, then there is also a solution to (2) that is not bounded.

If that helps, in my case, I can solve (1) by the method of characteristic and thus establish the boundedness or unboundedness of the solutions and I would like to extend that property to (2). Intuitively, this might be possible for a relatively compact perturbation. One important fact though is that the operators $A$ and $L$ have the same spectrum in the sense that their resolvent sets coincide.

Again, I am not interested in a proof but rather a reference where this type of problems could possibly be discussed.

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    $\begingroup$ In general, relatively compact perturbations do not preserve well-posedness except when L generates and analytic semigroup. Bounded perturbations do preserve well-posedness. You can find this is any text on operator semigroups. The condition that L and A have the same spectrum is a very strong one. From which property do you deduce that? $\endgroup$ May 27, 2021 at 17:29
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    $\begingroup$ There are two things in the question that are confusing me: (i) The sentence "I am thinking of $A-L$ as a Hilbert-Schmidt integral operator": On $L^2$? But then $A$ would even be a compact perturbation of $L$, not only a relatively compact one. (ii) The phrase "have the same spectrum in the sense that their resolvent sets coincide": The fact that the spectra of two operators coincide if and only if their resolvent sets coincide is an (admittedly, obvious) consequence of the definition of the spectrum and the resolvent set - so I don't understand the meaning of "in the sense" here. $\endgroup$ May 27, 2021 at 17:48
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    $\begingroup$ @MichaelRenardy: Another case where a relatively compact perturbation preserves well-posedness is as follows: if $L$ generates a $C_0$-semigroup on $X$, the operator $B: D(L) \to X$ is compact and $A+B$ is dissipative, then $A+B$ also generates a (contractive) $C_0$-semigroup. I'm not aware of a paper or book where this is stated (one of my students needed this for his Master's thesis, so we were looking around a bit), but it can be proved by the same argument as in the analytic case (i.e., as in the Desch-Schappacher theorem). $\endgroup$ May 27, 2021 at 17:59
  • $\begingroup$ @Jochen Glueck I used "in the sense" because when reading "they have the same spectrum", some people might interpret this as "they have the same point spectrum, the same continuous spectrum, and so on". So I erred on the on the side of caution. Concerning "relatively compact", yes, I have a compact perturbation. I used the term "relatively compact" because that's the condition used in all the theorems of the Weyl type. Maybe you are right and "compact perturbation" is the wording I should have used from the start. $\endgroup$ May 27, 2021 at 18:43
  • $\begingroup$ @Gateauaufromage: Thanks for your reply! Compactness (rather than relative compactness) of the perturbation makes the well-posedness question much easier: it's a classical result in $C_0$-semigroup theory that every bounded perturbation of a semigroup generator is again a semigroup generator. $\endgroup$ May 27, 2021 at 18:53

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General references.

The references that you're probably looking for are books on the theory of $C_0$-semigroups. Some classics are:

[1] Amnon Pazy: Semigroups of Linear Operators and Applications to Partial Differential Equations, 1983

[2] Klaus-Jochen Engel and Rainer Nagel: One-Parameter Semigroups of Linear Evolution Equations, 2000 (my personal favourite)

[3] Klaus-Jochen Engel and Rainer Nagel: A Short Course on Operator Semigroups, 2006 (the short version, naturally, of the previous reference)

Well-posedness.

Bounded perturbations of semigroup generators are again semigroup generators (see for instance [2, Section III.1].

Boundedness of the solutions.

That's essentially the reason why I wrote this answer: all solutions to $v_t = Lv$ are bounded if and only if the semigroup generated by $L$ is bounded. However, this property is not preserved by compact perturbations, even if they don't change the spectrum:

Counterexample. Let $L$ be the zero operator on $L^2(0,1)$ (which is everyhwere defined), and let $V: L^2(0,1) \to L^2(0,1)$ be the Volterra operator given by $$ (Vf)(x) = \int_0^x f(y) \, dy. $$ Then $V$ is compact and both $L=0$ and $L+V=V$ have spectrum $\{0\}$. But the semigroup generated by $L$ is bounded, while the semigroup generated by $L+V$ is not.

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  • $\begingroup$ I will certainly look at those references. Actually, I do have Pazy's book (in my office, which I have not visited for ages of course). However, if I can pick your brain once more, what about if $v_t=Lv$ has an unbounded solution, do you think (or know) if that would automatically mean that $v_t=Av$ has an unbounded solution? $\endgroup$ May 27, 2021 at 19:36
  • $\begingroup$ @Gateauaufromage: For this question you can use the same counterexample as in my answer, but with interchanged roles of "unperturbed" and "perturbed" operator. More precisely, let $L = V$ (the volterra operator from the answer). Then $v_t = Lv$ has an unbounded solution (since the semigroup generated by $L = V$ is unbounded). However, the operator $A = 0$ (which is a perturbation of $L$ by $-V$) generates a bounded semigroup, so all solutions to $v_t = Av$ are bounded. $\endgroup$ May 27, 2021 at 20:47

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