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I am dealing with the problem

\begin{align}u_t &= \nabla \cdot (a(x,t) \nabla u) + f(x,t) &\text{ on } \Omega \times (0,T)\\ \partial_{\nu} u &= 0 &\text{ on } \partial \Omega \times (0,T)\\ u(x,0) &= u_0(x) &\text{ in } \Omega\end{align}

on some bounded domain $\Omega$ with $a_1 \leq a(x,t) \leq a_2$ for constants $a_1,a_2 >0$.

We set $A(t)u = \nabla \cdot (a(\cdot,t) \nabla u)$ on some $D(A(t))$ that is independent from $t$ and includes the condition $\partial_{\nu} u = 0$ on $\partial \Omega$. For example $D(A(t)) = \{\varphi \in W^{2,q} (\Omega): \,\partial_{\nu} \varphi = 0 \}$?

If $a$ would not depend on $t$ the solution would (for appropriate $a$ and $f$) be given by \begin{align} u(t) = e^{tA}u_0 + \int_0^te^{(t-s)A} f(s) \,ds. \end{align}

Is there some similar result for this particular $A$ depending on $t$?

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  • $\begingroup$ "Nonautonomous semigroup" theory is treated in the standard textbooks, for example in Pazy "Semigroups of Linear Operators and Applications to Partial Differential Equations" Chapter 5 (or 5.6 for parabolic), or Engel/Nagel "One-Parameter Semigroups for Linear Evolution Equations" Chapter IV.9. That being said, it will in general be nontrivial to know that the domains of $A(t)$ are constant in time. (This is essentially elliptic regularity theory.) It will depend on the choice of the ambient Banach space $X$ and the properties/regularity of the coefficient function $a$. Thus your question ... $\endgroup$
    – Hannes
    Commented Apr 6, 2023 at 7:42
  • $\begingroup$ ... (cont'd) is very hard to answer in this generality. Let me also point out that there is the concept of nonautonomous maximal parabolic regularity which is a particular case of the semigroup approach and which is often convenient to use (I personally find it easier to handle than the evolution system), see e.g. Prüss/Schnaubelt "Solvability and Maximal Regularity of Parabolic Evolution Equations with Coefficients Continuous in Time" for an important contribution in this area which might help. $\endgroup$
    – Hannes
    Commented Apr 6, 2023 at 7:46

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