text:
\section{Central Limit Theorem for The Constant Step Size SGD}\label{sec:cltheory}

In this section, we establish an asymptotic central limit theorem (CLT) for the Polyak-Ruppert averaging of
the constant step size SGD iterates given in \eqref{ref:sgd_org}
when the objective function is potentially non-convex, non-smooth, and has quadratically growing tails.
More specifically, we first prove that there exists a unique stationary distribution~$\sdist\in\mathcal{P}_2(\rd)$ for the Markov chain defined by the SGD algorithm
when the objective function is dissipative (see Assumption~\ref{asm:dissipativity}) with gradient exhibiting at most linear growth (see Assumption~\ref{asm:growth}).
Furthermore, under the same conditions, we prove that a CLT holds for
the Polyak-Ruppert averaging, and it is independent of the initialization. In what follows, we list and discuss the main assumptions required to establish a CLT for the SGD iterates, and compare them to those existing in the literature.
\begin{assumption}[Linear growth]\label{asm:growth}
	The gradient of the objective function $f$
	has at most linear growth. That is, for some $L\ge 0$, we have
	\eq{
	\norm{\grad f(\theta)}\le L\big(1+\norm{\theta}\big)\ \ \text{ for all }\ \ \theta\in\rd.
	}
\end{assumption}
Majority of the results on SGD focus on smooth functions with  gradients satisfying
$\| \nabla f(\theta) - \nabla f(\theta') \| \leq \|\theta - \theta'\|$
for all $\theta, \theta' \in \mathbb{R}^d$; see e.g.~\cite{polyak1992acceleration,dieuleveut2017}.  The above condition allows for non-smooth objectives, and is a significant relaxation of the standard Lipschitz gradient condition.
\begin{assumption}[Dissipativity]\label{asm:dissipativity}
	The objective function~$f$ is $(\alpha,\beta)$-dissipative. That is,
	there exists positive constants~$\alpha,\beta$ such that
\eq{
	\inprod{\theta}{\grad f(\theta)}\ge \alpha\,\norm{\theta}^2-\beta\ \ \text{ for all} \ \ \theta\in\rd.
	}
\end{assumption}
The dissipativity assumption has its origins in the analysis of dynamical systems, and is used widely in the analysis of optimization and learning algorithms~\cite{mattingly2002,raginsky2017non, erdogdu2018global, xu2018global}. It could be viewed as a relaxation of strong convexity since it restricts the quadratic growth assumption to the tails of the function $f$, enforcing no local growth around the first-order critical points.
A canonical example for this condition is the sum of a quadratic and any non-convex function with bounded gradient.
For example, consider the function $x\to x^2+10\sin(x)$ which is clearly non-convex and $(1,25)$-dissipative.
It is worth mentioning that many statistical learning problems
such as phase retrieval~\cite{tan2019online} satisfy Assumption~\ref{asm:dissipativity}.

\begin{assumption}[Noise sequence]\label{asm:noise}
Gradient noise sequence~$\{\noise_k\}_{k\ge 1}$ is a collection of i.i.d. random fields satisfying
\eq{
\E[\noise_{1}(\theta)]=0\, \ \text{and } \  \E^{1/2} [\norm{\noise_{1}(\theta)}^2]\le L_{\noise}(1+\norm{\theta})\,,
}
for any $\theta\in\rd$ and a positive constant $L_{\noise}$. Moreover, for each $\theta \in \rd$ the distribution of the random variable $\noise_1(\theta)$ can be decomposed as $\mu_{1,\theta} + \mu_{2,\theta}$ where $\mu_{1,\theta}$ has a density, say $p_\theta$, with respect to Lebesgue measure which satisfies $\inf_{\theta \in C} p_{\theta}(t) > 0$ for any bounded set $C$ and any $t \in \rp$.
\end{assumption}
Assumption~\ref{asm:noise} as formulated above is stronger than what is used in the proofs. It can easily be seen that the lower bound on the density $p_\theta$ is only required to hold for a specific set whose form depends on $\eta$ and various constants from Assumptions~\ref{asm:growth}--\ref{asm:noise}. The form of this set is complicated, and an exact expression is given in the Appendix -- see equation~\eqref{ref:smallset}. We also emphasize that Assumption~\ref{asm:noise} also does not specify any explicit parametric form for the distribution of the noise sequence contrary to recent works in non-convex settings where dissipitavity condition has been heavily utilized~\cite{raginsky2017non,xu2018global,erdogdu2018global}.

We now establish the existence and uniqueness of the stationary distribution of the SGD iterates~\eqref{ref:sgd_org}.
\begin{proposition}[Ergodicity of SGD]
	\label{ergodicity}
	Let the Assumptions~\ref{asm:growth}-\ref{asm:noise} hold.
	For a step size satisfying
	$$
	\eta<\frac{\alpha-\sqrt{\bigl(\alpha^2-{(3L^2+L_\noise)} \bigr)\vee 0 }}{{3L^2+L_\noise}}\,,
	$$
	the following statements hold for the SGD~\eqref{ref:sgd_org}.

	\begin{enumerate}[label=(\alph*)]
		\item \label{exist_sdist} SGD iterates admit a unique stationary distribution~$\sdist\in\mathcal{P}_2(\rd)$, depending on the choice of step size $\eta$.
		\item \label{conv_to_sdist}
		For a test function $\phi:\rd\to \R$ satisfying
		$|\phi(\theta)| \le L_{\phi}(1+\norm{\theta})$ $\forall \theta\in\rd $ and some $L_\phi>0$,
		and for any initialization $\theta_0^{(\eta)}=\theta_0\in\rd$ of the SGD algorithm, there exists $\rho\in(0,1)$
		and $\kappa$ (both depending on $\eta$) such that we have
		$$\big|\E\big[\phi\big(\theta_k^{(\eta)}\big)\big]-\sdist(\phi)\big| \le \kappa\, \rho^{k} (1+\norm{\theta_0}^2),$$  where $\sdist(\phi) \coloneqq \int \phi(x) d\sdist(x)$.
	\end{enumerate}
\end{proposition}
The uniqueness of the stationary distribution of the constant step size SGD has been established in~\cite{dieuleveut2017} for strongly convex and smooth objectives.
In Proposition~\ref{ergodicity}, we relax both of these assumptions allowing for non-convex and non-smooth objectives.
Our proof relies on $V$-uniform ergodicity~\cite{meyn2012}, which is fundamentally different from the ergodicity analysis in~\cite{dieuleveut2017}.
Under the dissipativity condition (quadratic growth of $f$), geometric ergodicity in Proposition~\ref{ergodicity} is not surprising;
yet, it is worth highlighting that the function $f$ as well as the noise sequence require significantly less
structure than what was assumed in the literature.
The above step size assumption is almost standard
and it is required to obtain a uniform bound on the moments of SGD iterates.
We highlight that similar to the gradient descent algorithm, the step size depends on a quantity that serves as a \emph{surrogate} condition number in our setting, namely, $L/\alpha$.
For the purposes of establishing a CLT, it is sufficient to consider moments of order 4 (in fact any order larger than 2 suffices), but it is also worth
noting that any order moments of SGD can be controlled under Assumption~\ref{asm:dissipativity}
as long as the noise has the same order finite moment.

Next, we state our first principal contribution, a central limit theorem for the averaged SGD iterates starting from any initial distribution for a non-convex objective. For a test function~$\phi:\rd\to\R$, we denote the centered partial sums of $\phi$ evaluated at the SGD iterates with $S_n(\phi)$, i.e., we define
\eq{S_n(\phi)\coloneqq \sum_{k=0}^{n-1}\Big[\phi\big(\iter_{k}\big) - \pi_\eta(\phi)\Big]\ \ \text{ where }\ \ \sdist(\phi) \coloneqq \int \phi(x) d\sdist(x).}

\begin{theorem}[CLT]
	\label{general_clt_arb_sdist}
	Let the Assumptions~\ref{asm:growth}-\ref{asm:noise} hold.
	For a step size $\eta$ and a test function $\phi$ satisfying the conditions in Proposition~\ref{ergodicity}, we define $\sigma^2_{\sdist}(\phi)\coloneqq \lim_{n\to\infty} \frac{1}{n}\E_{\sdist} \bigl[S_n^2(\phi)\bigr]$.
	Then,
	\eq{
		n^{-1/2}S_n(\phi) \overset{\text{d}}{\longrightarrow} \mathcal{N}\big(0,\sigma^2_{\sdist}(\phi)\big)\,.
	}
\end{theorem}

The above result characterizes the fluctuations of a test function $\phi$
averaged across SGD iterates, even when the objective function is both non-convex and non-smooth.
The asymptotic variance in the above CLT can be equivalently stated in another compact form.
	If we define the centered test function as
	$h( \theta) = \phi(\theta)- \sdist(\phi)$,
	the asymptotic variance can be written as
        \eq{
        \sigma^2_{\sdist}(\phi)= 2\sdist(h\hat h)-\sdist(h^2)\ \ \text{ where }\ \ \hat h = \sum_{k=0}^\infty  \E\Big[h\big(\theta_k^{(\eta)}\big)\Big].
        }
	Indeed, this is the variance we compute at the end of our proof in Section~\ref{sec:cltproofs}.
	However, the expression in Theorem~\ref{general_clt_arb_sdist} is obtained by
	simply applying \cite[Thm 21.2.6]{douc2018}.
	For the case of strongly convex functions with decreasing step size schedule,
	it is well-known from the works of~\cite{polyak1992acceleration,ruppert1988efficient}
	that the limiting variance of the averaged SGD iterates
	achieves the Cramer-Rao lower bound for parameter estimation;
see also~\cite{moulines2011non, anastasiou2019normal} for non-asymptotic rates in various metrics.
The question of providing lower bounds for the limiting variance of the critical points in the non-convex setting is extremely subtle, and is often handled on a case-by-case basis.
We refer the interested reader to~\cite{geyer1994asymptotics,shapiro2000asymptotics,loh2017statistical}.

There are several important implications of the above CLT result for constructing confidence intervals in practice.
First note that, following the standard construction in inference, one can write the distribution of the sample mean approximately as
$n^{-1}S_n(\phi) \approx \mathcal{N}\bigl(0,n^{-1}\sigma^2_{\sdist}(\phi) \bigr)$. Here, one needs to estimate the population quantity,
the asymptotic variance  $\sigma^2_{\sdist}(\phi)$, for the purpose of obtaining confidence intervals. In Section~\ref{sec:discussion}, we discuss three strategies for estimating this quantity,
which could be eventually used for inference in practice. A theoretical analysis of the proposed approaches in Section~\ref{sec:discussion} is beyond the scope of this work.


TITLE: Euler's method proof of the limit formula for $e$
QUESTION [3 upvotes]: While using Euler's method for $\frac{dy}{dx}=y$ given $y(0)=1$ I noticed that the approximate solution is $(1+1/s)^{sx}$ where s is the interval size. Plug in one for $x$ and you get:
$$e^{1}=\lim_{s\to\infty}\,(1+1/s)^{s}$$
you get $e$. This has to be $e$ because that is that is the solution to the differential equation. Is this a new proof that
$$e=\lim_{s\to\infty}\,(1+1/s)^{s}$$

List item

REPLY [2 votes]: It depends on how you define $e$. If you define $e$ as
\begin{align}
e := \lim_{s \to \infty} \left(1 + \frac{1}{s} \right)^s
\end{align}
Then (of course) it trivially follows that it is true ($e = e$ end proof).
Although if you define $e$ as the solution to the Initial Value Problem
$$\left\{
\begin{aligned}
\frac{dy}{dx} &= y \\
y(0) &= 1
\end{aligned}
\right.
$$
at the value $x = 1$, or if you define $e$ some other way and prove that $e^x$ is the solution to this IVP, and then in addition you prove (or are allowed to assume) that Euler's Method will in fact converge upon the unique solution to the IVP, then yes: what you have written would suffice as another proof that
$$ e = \lim_{s \to \infty} \left(1 + \frac{1}{s} \right)^s $$


TITLE: A thought experiment to understand (literal) bootstrapping
QUESTION [1 upvotes]: The common explanation of why you can't lift yourself off the ground by pulling your feet up with your hands, or in more cliched terms "pull yourself up by your own bootstraps", is that if you consider your whole body as a system, the pull of your hands on your feet is just an internal force, so it can't pull your body off the ground.
But what about this thought experiment: let's say you say cross-legged with only your feet on the ground.  Then you place your hands under your feet, so that only your hands are on the ground.  Then you rapidly slide your hands way from under your feet and bearing your hands up.  What would happen?  Well, if your hands were 1 cm thick, then, the moment you removed your hands from under your feet, your feet would be 1 cm off the ground.  So wouldn't you have managed to successfully lift yourself off the ground, if only for a moment before you crash back down?

REPLY [1 votes]: Let's consider this thought experiment. Let's go to a very light planet where $g$ is only a 1000th of what it is on earth. Lets get a very light ladder. This makes the thought experiment easier to imagine. The kind of ladder that makes an "A" shape so you don't have to lean it against anything. The ladder is so light that its mass is negligible compared to yours. Now you climb up the ladder and put your feet on one of the top rungs.
Now remember we feel almost weightless and the ladder is almost massless. Therefore we are able to pull the ladder out from under our feet to above our heads. Now we are certainly off the ground, and we will float for a while. The reason we are off the ground is because we climbed up a ladder. I fail to see how this is the same as pulling ourselves up by our bootstraps. Even if the ladder was considered part of your body, we still just climbed up the ladder.
Now lets think about your experiment. It's the exact same thing except the ladder is your hands. But since its the same thing as climbing up a ladder, it's really more like climbing up a ladder than pulling yourself up by your bootstraps. Does this make sense? Tell me if some part of the argument doesn't make sense.
Edit 1
You could say that once I have raised the ladder over my head, I have lifted myself, since before the ladder-body system was touching that ground and now it is not.
However, this is not a fair definition of lifting yourself. For example, you could lift yourself in this sense by simply pulling your knees toward you chest so quickly that your feet leave the ground, but no one would call this lifting yourself because it would really seem like you are going down.
Also, if you tried to extend your feet again, you would find that your feet would hit the ground before becoming fully extended. Thus if your body could return to the same configuration it would actually be lower than it was originally. Comparing the minimum distance of the body off the ground between two configurations is misleading because some configuration can more easily have a large minimum distance from the ground.
A more sensible definition of pulling yourself up by your bootstraps would be if you could keep your body rigid while pulling on your bootstraps, but still cause an upward motion. We could even loosen this and say that you can move your body but you have to return to the same configuration.
It would be convenient if there were a way to compare the height of a body in two different configurations. So if I am a certain distance form the ground when my knees are pulled up to my chest, how far off the ground will I be when I extend my feet? It turns out there is a way to compare positions of two objects in different configurations. It is called the center of mass. If you have an object in any configuration you can calculate its center of mass. This center of mass is independent of configuration: the object can rearrange itself all it wants, but the center of mass you calculate won't change (if it isn't interacting with anything else).
Now it is sufficient for me to show that it is impossible to raise my center of mass by pulling on my bootstraps, for if I could raise my center of mass somehow, then I could return to my initial configuration and I would be higher off the ground, thus legitimately raising myself according to our more stringent definition.
However newton's laws say it is impossible for me to raise my center of mass by applying a force to myself, since only external forces can cause me to accelerate.
So it is impossible to pull yourself up by your bootstraps in the sense I described. You might say, "But it is possible to pull yourself up by your bootstraps in the sense I said." You would be right, but the statement is really as powerful. You are giving a way for someone to get there feet of the ground. The other statement is that the center of mass of any isolated system is always constant.


TITLE: How could it be possible if we have (a,b) then it is an open interval?
QUESTION [0 upvotes]: I am really wondered why do we consider these kind of intervals open, how come they are not considered closed? is there any mathematical proof or i just have to accept the way it is (as i did in high-school!).

REPLY [1 votes]: There is in fact a definition of what it means for sets to be open or closed, though it usually comes up in topology (so is maybe a little past what you would cover in a precalculus class).
The definition is based on the boundary points of a set.  A set is called open if it doesn't contain any of its boundary points. So for an interval $(a,b)$, the boundary points of this interval are $a$ and $b$, neither of which is included in the set. So this interval is called open. A set is closed if all of the boundary points are in the set.  So the interval $[a,b]$ would be closed, and $[a,b)$ would be neither open nor closed (since it contains some but not all of the boundary points).


TITLE: limit of floor function, ceiling function and fractions
QUESTION [0 upvotes]: We know that
$\lfloor x\rfloor$ : greatest integer $\leq x$
$\lceil x\rceil$ : least integer $\geq x$
$\mathrm{frac}(x)=x-\lfloor x\rfloor$
I can solve the questions limit of function like
$$
\lim\limits_{x\to n^\pm}\frac{\lfloor x-1\rfloor}{x-1}\\
\lim\limits_{x\to n^\pm}\frac{\lfloor x\rfloor}{x-1}
$$
As $x$ approaches $n$ from above, $\lfloor x-1\rfloor=n-1$; therefore,
$$
\lim_{\large x\to n^+}\frac{\lfloor x-1\rfloor}{x-1}=\frac{n-1}{n-1}
$$
but I can't solve the questions like
$$
\lim\limits_{x\to\infty}\frac{\lfloor x-3\rfloor}{x-1},\\
\lim\limits_{x\to\infty}\frac{\lceil x-3\rceil}{x-1},\\
\lim\limits_{x\to a} \lfloor x\rfloor,\\
\lim\limits_{x\to a} \lceil x\rceil,\\
 \lim\limits_{x\to a} \mathrm{frac}( x)
$$ for $a\in \mathbf{R}$
All questions are similar type, so I have given many problems in my questions.
Please help me.

REPLY [3 votes]: For the limit at infinity:
$$L=\lim_{x\to \infty}\frac{\lfloor x-3\rfloor}{x-1}=\lim_{x\to \infty}\frac{x-3-\text{frac}(x-3)}{x-1}=\lim_{x\to \infty}\frac{x-3}{x-1}-\frac{\text{frac}(x-3)}{x-1}$$
since
$$0\leq\frac{\text{frac}(x-3)}{x-1}<\frac{1}{x-1}$$
we have
$$L=\lim_{x\to \infty}\frac{x-3}{x-1}=1$$
For the limit at a:
Let $\epsilon = \frac{\text{frac}(a)}{2}$, we have that $\lfloor x\rfloor=\lfloor a\rfloor$ for $x\in(a-\epsilon,a+\epsilon).$ So $\lim_{x\to a} \lfloor x\rfloor=\lfloor a\rfloor.$
NOTE THIS ASSUMES $\text{frac}(a) > 0$, if $a\in \mathbb{Z}$ then the limit does not exist.


TITLE: Proving $|S/R^2|=\aleph$ , $(x_1,x_2)S(y_1,y_2)\iff x_1^2-x_2=y^2_1-y_2$
QUESTION [0 upvotes]: Let $S$ be an equivalence relation over $\mathbb R^2$ such that: $(x_1,x_2)S(y_1,y_2)\iff x_1^2-x_2=y^2_1-y_2$
Prove that $|S/R^2|=\aleph$

One side is pretty simple: $|S/R^2|\le |\mathbb R^2|=\aleph$.
We're left with showing $|S/R^2|\ge \aleph$, I thought of defining an injective function from $S$ to $\mathbb R$ like so:
$f((x_1,x_2),(y_1,y_2))=2^{x_1}3^{x_2}5^{y_1}7^{y_2}$
Is this the right way?
How does one of the equivalence classes look like here?

REPLY [1 votes]: First let me point out two notational issues:

It is important to be consistent. Don't write $R^2$ when you mean $\Bbb R^2$, or vice versa, and have used the same notation before in the same sentence.
The usual notation is $A/E$ for the quotient set, not $E/A$. So you would in fact want $\Bbb R^2/S$ rather than $S/R^2$.

While the trick of taking powers of prime numbers is a very good trick for showing that $\Bbb N^4$ and $\Bbb N$ have the same cardinality, it doesn't work for $\Bbb R$ since the function is continuous and there are no continuous injections from $\Bbb R^4$ into $\Bbb R$.
Additionally this is the wrong approach, since it establishes that $S$ and $\Bbb R$ have the same cardinality, not that $\Bbb R^2/S$ and $\Bbb R$ have the same cardinality.
What you should do, is understand what the equivalence relation mean, and find a way to pick a canonical representative from each equivalence class. For example, what happens if $x_2=0$? How many values could $x_1$ assume within the same equivalence class?
Now use that in order to define an injection from a subset of $\Bbb R$ into $\Bbb R^2/S$, namely find some $A\subseteq\Bbb R$ such that $|A|=|\Bbb R|$, and the map $a\mapsto a/S$ is injective (so no two elements of $A$ have the same equivalence class).


TITLE: Special derivative operator for complex functions
QUESTION [2 upvotes]: I have recently encountered a rather strange definition of a 'derivative' on functions with complex arguments. It was defined as
$$d^* f(z) = \frac{\partial f}{\partial z} dz - \frac{\partial f}{\partial z^*} dz^*$$
It does follow the Leibniz rule so that $d^* (fg) = (d^*f)g + f(d^*g)$, so it seems to be a valid derivation. However, I have no idea about the context where this could be useful. Has anybody seen this derivation and can point me to relevant literature? Any mathematical insights are very welcome too.
Edit: Please specifically note the negative sign in the definition. I've found references to a similar expression, but with a + sign. How are they related?
Edit2: Ok, so with user44874's hint I've performed the calculation explicitly in complex cartesian coordinates (x,y) and I get $$ d^* f(x+iy) = i \left( \frac{\partial f}{\partial x} dy - \frac{\partial f}{\partial y} dx\right)$$
which is a symplectic "gradient" like the one used in Hamiltonian mechanics, up to a factor of $i$. Can anyone confirm this?

REPLY [0 votes]: At "Notes on complex function theory" by Donald Sarason you can find this form of derivative under "III.16 COMPLEX PARTIAL DIFFERENTIAL OPERATORS".
Note that:
$$ \frac{\partial }{\partial z}  = \frac{1}{2} (\frac{\partial }{\partial x}-i\frac{\partial }{\partial y}  )$$
$$ \frac{\partial }{\partial z^*}  = \frac{1}{2} (\frac{\partial }{\partial x}+i\frac{\partial }{\partial y}  )$$
and that $f$ satisfies that Cauchy Reiman equation iff $\frac{\partial f }{\partial z^*}=0$ and thus holomorphic functions can be interpreted as those in the kernal of the operator $\frac{\partial f }{\partial z^*}$.


TITLE: Elliptic regularity with mixed boundary conditions
QUESTION [2 upvotes]: I'm looking for some results about elliptic regularity with mixed boundary conditions. I know they exist with non mixed boundary conditions but where can I find some results for the mixed case?
Thanks

REPLY [1 votes]: It's a fairly specialized area. I think most of the treatments of mixed boundary conditions would be in journal articles. As far as textbooks go, all I can find is these relatively brief discussions.
Carlo Miranda, "Partial differential equations of elliptic type", Springer 1970, pages 233–234 and 261.
Gilbarg and Trudinger, "Elliptic partial differential equations of second order", 3rd edition, Springer 1983, 1998, 2001, pages 47 and 215.
However, these just summarize some results and refer the reader to numerous journal articles.


text:
\begin{document}

\maketitle

\begin{abstract}
Consider the Navier-Stokes flow past a rotating obstacle with
a general time-dependent angular velocity and a time-dependent
outflow condition at infinity -- sometimes called an Oseen condition. By a suitable change
of coordinates the problem is transformed to an non-autonomous problem with unbounded drift terms
on a fixed exterior domain $\Omega\subset \R^d$. It is shown that the solution to the linearized problem is governed by a strongly
continuous evolution system $\{T_\Omega(t,s)\}_{t\geq s\geq0}$ on $L^p_\sigma(\Omega)$ for $1<p<\infty$. Moreover, $L^p$-$L^q$
smoothing properties and gradient estimates of $T_\Omega(t,s)$, $0\leq s \leq t$, are obtained. These results are the key ingredients to
show local in time existence of mild solutions to the full nonlinear problem for $p\geq d$ and initial value in $L^p_\sigma(\Omega)$.
\end{abstract}

\section{Introduction}
In this paper we consider the flow of an incompressible, viscous fluid
in the exterior of a rotating obstacle subject to an additional time-dependent outflow condition at
infinity. Here the angular velocity of the obstacle and the outflow condition at infinity may depend on time
and also the axis of rotation may change.
The equations describing this problem are the Navier-Stokes
equations in a time-dependent exterior domain with a prescribed velocity field at infinity.

After rewriting the problem on a fixed exterior domain $\Omega \subset \R^d$, we obtain an non-autonomous system of equations involving a family of time-dependent operators of the form
\begin{equation}\label{eq:Op}
A(t)u= \P_\Omega\left( \Delta u + (M(t)x + c(t))\cdot \nabla u - M(t) u\right), \qquad t\ge 0.
\end{equation}
where $\P_\Omega$ denotes the Helmholtz projection from $L^p(\Omega)^d$ onto the solenoidal space $L^p_\sigma(\Omega)$ and $M\in C^1([0,\infty);\R^{d\times d})$, $c\in C^1([0,\infty);\R^d)$.  The  main difficulty in dealing with these operators arises since the
term $M(t)x \cdot \nabla$ has unbounded coefficients in the exterior domain $\Omega$. In particular, the
lower order terms cannot be treated by classical perturbation theory for the
Stokes operator. In the autonomous case
$M(t)\equiv M$ and $c(t)\equiv 0$ such an operator was first considered by Hishida \cite{Hishida:1999} and then later by Geissert, Heck, Hieber \cite{Geissert/Heck/Hieber:2006a}.

It is the aim of this paper to extend their result to the non-autonomous case.

In the following let us briefly motivate our problem and let us show why it is interesting to study the non-autonomous case. For this purpose let $\mathcal O \subset \R^d$ be a compact obstacle with
smooth boundary and let $\Omega:=\R^d \setminus \mathcal O$ be the
exterior of the obstacle. Furthermore, let $m \in C^1([0,\infty); \R^{d\times d})$ be a
matrix-valued function that describes the velocity of the obstacle.
Then, the exterior of the rotated obstacle at time $t\ge 0$ is represented by $\Omega(t):=
Q(t)\Omega$ where $Q\in C^1([0,\infty),\R^{d\times d})$ solves the ordinary differential equation
\begin{equation}
\left\{\begin{array}{rcll}
\partial_t Q(t) &=& m(t)Q(t),& t\ge0,\\[0.2cm]
Q(0)&=&\mathrm{Id}.
\end{array}\right.
\end{equation}\normalsize
With a prescribed velocity field $v_\infty\in C^1([0,\infty);\R^d)$
at infinity, the equations for the fluid on the time-dependent
domain $\Omega(t)$ with no-slip boundary condition
take the form
\begin{align}\label{eq:NS_2}
v_t-\Delta v +v\cdot \nabla v+\nabla \q&=0&\quad\quad\quad\mbox{in $(0,\infty)\times \Omega(t)$,}\notag\\
\div v&=0&\quad\quad\quad\mbox{in $(0,\infty)\times \Omega(t)$,}\notag\\
v(t,y)&=m(t)y&\quad\quad\quad\mbox{ on $(0,\infty)\times \partial\Omega(t)$,}\\
\lim_{|y|\to \infty} v(t,y)&=v_\infty(t)&\quad\quad\quad\mbox{ \mbox{for} $t\in(0,\infty) $,}\notag\\
v(0,y)&=v_0(y)&\quad\quad\quad\mbox{in $\Omega$}\notag.
\end{align}
Here $v=v(y,t)$ and $\q=\q(y,t)$ are the unknown velocity field and the pressure of
the fluid, respectively.

As usual, it is reasonable to reduce \eqref{eq:NS_2} to a new problem on a fixed exterior domain by some suitable coordinate transformation. Since $m(\cdot)$ is the velocity of the rotated obstacle, it is natural to assume that $m(t)$ is skew
symmetric for all $t\ge0$. This implies that for all $t\ge 0$ the matrix $Q(t)$ is orthogonal. Thus, we can set\vspace{-0.2cm}
\begin{equation}
x=Q(t)^{\mathrm T}y, \quad u(t,x)=Q(t)^{\mathrm T}
(v(t,y)-v_\infty(t)), \quad \p(t,x)=\q(t,y).
\end{equation}
Then we obtain the following new system of equations on the reference domain $\Omega$:

\begin{align}\label{eq:NS_3}
	\left.
	\begin{array}{l}
u_t- \Delta u - M(t)x \cdot \nabla u + M(t)u  \\
\quad+ Q(t)^{\mathrm T}v_{\infty}(t)\cdot \nabla u
+Q(t)^{\mathrm T}\partial_t v_\infty(t)\\
\quad +u\cdot \nabla u
+\nabla \p
\end{array}\right\}&=0&\;\mbox{in   $(0,\infty)\times \Omega $,}\notag\\
\div u&=0&\quad\mbox{in $ (0,\infty)\times \Omega $,}\notag\\
u(t,x)&= M(t)x-Q(t)^{\mathrm T}v_\infty(t)&\;\mbox{on $(0,\infty)\times \partial\Omega $},\\
\lim_{|x|\to \infty} u(t,x)&= 0&\;\mbox{\mbox{for} $t\in(0,\infty) $,}\notag\\
u(0,x)&=u_0(x)&\;\mbox{in $\Omega$}.\notag
\end{align}
\normalsize
Here $M(t):=Q(t)^{\mathrm T}m(t)Q(t)$ is the transformed velocity of the obstacle. The coordinate transformation also ensures that the new velocity field $u$ vanishes at infinity, which is a natural condition in the $L^p$-setting.

Note that a problem of this type also arises in the analysis of a rotating body with a translational velocity $-v_\infty(t)$ by a similar coordinate transformation, see e.g. the explanations in \cite{Farwig:2005}.

Problem \eqref{eq:NS_2} was studied intensively for the special case of \emph{time-independent} matrices $M(t)\equiv M$ and
{\em without} an outflow condition, i.e.  $v_\infty\equiv0$. Hishida \cite{Hishida:1999} showed that the solution to the linearized problem is governed by a strongly continuous semigroup on $L^2$, which is however \emph{not analytic}. Moreover, he constructed local mild solutions
in $L^2$ by using the Fujita-Kato approach (cf. \cite{Fujita/Kato:1964}). Later this generation and existence result was extended to the general $L^p$-theory by Geissert, Heck,
Hieber \cite{Geissert/Heck/Hieber:2006a}. Hishida and Shibata \cite{Hishida/Shibata:2009} were even able to show global extistence for small data. The model problem in $\R^d$ was studied by Hieber and Sawada \cite{Hieber/
Sawada:2005} in the $L^p$-setting.

The case of \emph{time-dependent} angular velocities was considered by Hishida \cite{Hishida:2001} in the $L^2$-context, however he assumes that the axis of rotation is fixed and that the sign of his angular velocity does not change.

For the problem including an additional outflow condition at infinity,
there are only a few results.
The case, where $M(t)x=\omega(t) \times x$ and
$\omega:[0,\infty) \rightarrow \R^3$ is the angular velocity of the
obstacle and $v_\infty:[0,\infty)\to\R^3$ a time-dependent outflow velocity was considered by
Borchers \cite{Borcher:1992} in the framework of weak solutions. This work was somehow the starting point in the analysis of viscous fluid flow past a rotating obstacle. More recently,
Shibata \cite{Shibata:2008} studied
the special case where $M(t)\equiv M$, $v_\infty(t)\equiv v_\infty$ and
$Mv_\infty=0$. The additional condition $Mv_\infty=0$,  i.e.
$Q(t)^{\mathrm T}v_\infty= kv_\infty$ for $k\in \{+1,-1\}$, ensures that
\eqref{eq:NS_3} is still an autonomous equation.
The physical meaning of the additional condition is that the outflow
direction of the fluid is parallel to the axis of rotation of the
obstacle. The stationary problem of this latter situation was
analysed by Farwig \cite{Farwig:2005} for the whole space case $\R^3$.

In order to relax the assumption $Mv_\infty=0$, i.e. in order to allow a general  $v_\infty$, it is necessary to study a non-autonomous problem. Thus, in this context it is natural even to allow  time-dependent outflow velocities $v_\infty(\cdot)$ and time-dependent angular velocities $M(\cdot)$. The non-autonomous model problem of \eqref{eq:NS_3} in the case $\Omega=\R^d$ was recently studied by the first author \cite{Hansel:2009} and by Geissert and the first author \cite{Geissert/Hansel:2010}. Indeed, they were able to show that the family of operators in \eqref{eq:Op}, equipped with suitable domains, generate a strongly continuous evolution system on $L^p_\sigma(\R^d)$, $1<p<\infty$, which enjoys nice regularity properties. Their approach is based on an explicit solution formula for the linearized problem. By a version of Kato's iteration scheme (cf. \cite{Kato:1984,Giga:1986}) one obtains a (local) mild solution to the nonlinear problem on $\R^d$ for initial value $v_0 \in L^p_\sigma (\R^d)$, $d\leq p < \infty$. In this paper we use their results for the linearized problem to cover the physically more realistic situation of exterior domains by some cut-off techniques.

\subsection*{Notations}
The euclidian norm of $x\in \R^d$ will be denoted by $|x|$. By $B(R)$ we denote the open ball in $\R^d$ with centre
at the origin and radius $R$. For $T>0$ we use the notations:
\begin{eqnarray*}
\Lambda_T &:=& \{(t,s): 0\le s\le t\le T\},\\
\widetilde{\Lambda}_T &:=&  \{(t,s): 0\le s< t\le T\},\\
\Lambda &:=& \{(t,s): 0\le s\le t \},\\
\widetilde{\Lambda} &:=&  \{(t,s): 0\le s< t \}.
\end{eqnarray*}
Let us come to notation for function spaces. For a $C^{1,1}$ domain $\Omega\subset \R^d$ and $1\le p< \infty,\,j\in \N$, $W^{j,p}(\Omega)$ denotes the
classical Sobolev space of all $L^p(\Omega)$--functions having weak derivatives in
$L^p(\Omega)$ up to the order $j$. Its usual norm is denoted by $\|\cdot
\|_{j,p,\Omega}$ and by $\|\cdot \|_{p,\Omega}$ when $j=0$. If $\Omega =\R^d$ we drop $\Omega$ in the notations of the above norms.
We will use also the notation
$$\langle f,g\rangle_\Omega :=\int_\Omega fg\,dx,\quad f\in L^p(\Omega),\,g\in L^{p'}(\Omega)$$ with $\frac{1}{p}+\frac{1}{p'}=1$.
By $W_0^{s,p}(\Omega)$, $s\geq 0$, we denote the closure of the space of test functions $C_c^\infty(\Omega)$ with respect to the norm of $W^{s,p}(\Omega)$. For $s<0$ we set
\begin{equation*}
W^{s,p}_0(\Omega) := \left( W^{-s,p'}(\Omega) \right)',
\end{equation*}
where $\frac{1}{p}+\frac{1}{p'}=1$. We denote by $H^{s,p}(\Omega)$ with $s\in(0,2)$ the Bessel potential spaces, which are defined by complex interpolation
$$
H^{s,p}(\Omega):=[L^p(\Omega), W^{2,p}(\Omega)]_{\frac s 2}.
$$
Its norm will be denoted by $\| \cdot \|_{s,p,\Omega}$.
Moreover, we set
\begin{align*}
C_{c,\sigma}^{\infty}(\Omega)&:= \{f\in C_c^{\infty}(\Omega)^d: \; \div f = 0\},\\
L^p(\Omega)^d&:= \{f=(f_1,\ldots,f_d) : f_i \in L^p(\Omega), \, i=1,\ldots,d\}, \quad L_\sigma^p(\Omega):= \overline{C_{c,\sigma}^{\infty}(\Omega)}^{L^p(\Omega)^d},\\
G^p(\Omega)&:= \{\nabla \p : \,\p \in \hat{W}^{1,p}(\Omega)\}, \quad \hat{W}^{1,p}(\Omega):=\{\p \in L^p_{loc}(\overline{\Omega}) : \, \nabla \p \in L^p(\Omega)^d\}.
\end{align*}
It is well-known that for a Lipschitz domain $\Omega\subset \R^d$ with compact boundary the Helmholtz decomposition holds (see e.g. \cite{Galdi:1994} for more information):
\begin{equation*}
L^p(\Omega)^d=L^p_\sigma(\Omega) \,\oplus\,G^p(\Omega).
\end{equation*}
The projection from $L^p(\Omega)^d$ onto $L^p_\sigma(\Omega)$ is denoted by $\P_\Omega$.

\section{Main results and strategy of proofs}

In this section we present the main results and sketch the basic strategy of the proofs. In the following $\mathcal O \subset \R^d$ is always a compact obstacle with $C^{1,1}$-boundary and $\Omega:=\R^d\setminus \mathcal O$ is an exterior domain. Moreover, $M \in C^1([0,\infty),\R^{d\times d})$ and $v_\infty \in C^1([0,\infty),\R^d)$ are as described in the Introduction. In particular recall that $\mathrm{tr} \;M(t)=0$ for all $t\geq0$. To simplify our notation, we set $c(t):= - Q(t)^{\mathrm T}v_{\infty}(t)$. Since the term $Q(t)\partial_t v_\infty(t)$ in equation \eqref{eq:NS_3} is constant in space, we may put this term in the pressure $\p$. Thus, in the following we consider the system
\begin{align}\label{eq:NS_4}
\left.\begin{array}{l}
u_t- \Delta u - (M(t)x+c(t)) \cdot \nabla u + M(t)u \\
\quad +u\cdot \nabla u +\nabla \p
\end{array}\right\}&=0&\;\mbox{in   $(0,\infty)\times \Omega $,}\notag\\
\div u&=0&\quad\mbox{in $ (0,\infty)\times \Omega $,}\notag\\
u(t,x)&= M(t)x+c(t)&\;\mbox{on $(0,\infty)\times \partial\Omega $},\\
\lim_{|x|\to \infty} u(t,x)&= 0&\;\mbox{\mbox{for} $t\in(0,\infty) $,}\notag\\
u(0,x)&=u_0(x)&\;\mbox{in $\Omega$},\notag
\end{align}
\normalsize
where $\div u_0 = 0$. Moreover, we assume that the initial value $u_0$ satisfies the
compatibility assumption $u_0 \cdot \nu = (M(0)x+c(0)) \cdot \nu$ on $\partial \Omega$, where $\nu$ denotes the outer normal vector.

As a first step we construct a solenoidal extension in $\Omega$ of the boundary velocity $u(t,x)\vert_{\partial \Omega}$. For this
purpose we introduce the Bogovskii operator, which
concerns the solution of the equation $\div u = f$ in appropriate function spaces. This operator will also be needed later in Section
\ref{sect:exterior} to keep the solenoidal condition in our cut-off procedure. For proofs and more information on the Bogovskii operator we refer to \cite{Bogovskii:1979, Geissert/Heck/Hieber:2006b} and to the monograph \cite{Galdi:1994}.

\begin{lemma}\label{prop_Bog1}
Let $D \subset \R^d$, $d\geq 2$, be a bounded Lipschitz domain, $1<p<\infty$ and $k\in \N_0$.
\begin{itemize}
\item[(a)] There exists a continuous operator
\begin{equation*}
\BB_D : W^{k,p}_0(D)\to (W^{k+1,p}_0(D))^d
\end{equation*}
such that
\begin{equation*}
\div \BB_D f = f
\end{equation*}
for all $f\in W^{k,p}_0(D)$ satisfying $\int_D f \,\d x= 0$.
\item[(b)] For $k>-2+\frac 1 p$, the above operator $\BB_D$ can be continuously extended to a bounded operator from $W^{k,p}_0(D)$ to $(W^{k+1,p}_0(D))^d$.
\end{itemize}
\end{lemma}

Let $\zeta \in C_c^\infty(\R^d)$ be a cut-off function with $0\leq \zeta \leq 1$ and $\zeta = 1$ near $\partial \Omega$. Moreover we set $K:= \mathrm{supp} \, \nabla \zeta$ and define $b:[0,\infty)\times \R^d\rightarrow \R^d$ by
\begin{equation*}
b(t,x):=\zeta(x)(M(t)x + c(t)) - \BB_K((\nabla \zeta)\cdot (M(t)x+c(t))),
\end{equation*}
where $\BB_K$ is the operator from Lemma \ref{prop_Bog1} associated to the bounded domain $K$. Then $\div b(t,x)=0$ and
$b(t,x)=M(t)x + c(t)$ on $\partial \Omega$ for every $t\ge0$.

If we set $\tilde u = u - b$, then $u$ satisfies \eqref{eq:NS_4} if and only if $\tilde u$ satisfies
\begin{align}\label{eq:NS_5}
	\left.
	\begin{array}{l}
u_t- \Delta u - (M(t)x+c(t)) \cdot \nabla u + M(t)u \\
\quad + b(t,x)\cdot \nabla u + u\cdot \nabla b(t,x) +u\cdot \nabla u
+\nabla \p
\end{array}\right\}&=F_1(t,x)&\;\mbox{in   $(0,\infty)\times \Omega $,}\notag\\
\div u&=0&\quad\mbox{in $ (0,\infty)\times \Omega $,}\notag\\
u(t,x)&= 0&\;\mbox{on $(0,\infty)\times \partial\Omega $},\\
\lim_{|x|\to \infty} u(t,x)&= 0&\;\mbox{\mbox{for} $t\in(0,\infty) $,}\notag\\
u(0,x)&= f&\;\mbox{in $\Omega$},\notag
\end{align}

\normalsize
where $f(x):= u_0(x)-b(0,x)$ and
\begin{equation}\label{eq:inhom}
F_1(t,x):= \Delta b(t,x) + (M(t)x + c(t))\cdot \nabla b(t,x) - M(t) b(t,x) - b(t,x) \cdot \nabla b(t,x)-b_t(t,x).
\end{equation}
Note that $\div f = 0$ and that the compatibility assumption ensures that even $f\in L^p_\sigma (\Omega)$.

Our approach to system \eqref{eq:NS_5} is based on linear operators of the form
\begin{equation}
\OU (t)u(x) = \Big( \Delta u_i(x) + \langle M(t)x + c(t) , \nabla u_i(x) \rangle \Big)_{i=1}^{d} - M(t)u(x), \quad t\ge0,
\end{equation}
and perturbations of the form
\begin{equation*}
\mathcal B(t)u(x) = -b(t,x)\cdot \nabla u - u\cdot \nabla b(t,x), \quad t\ge0,
\end{equation*}
where $u=(u_1,\ldots,u_d)$ and $x$ is an element from $\Omega$, a bounded domain $D\subset \R^d$ or $\R^d$.  Note that the operators $\OU(\cdot)$ are of Ornstein-Uhlenbeck type (cf. \cite{Hansel/Rhandi:2010}). In the case of exterior domains $\Omega$ we define the $L^p$-realizations of $\OU(\cdot)$ as
\begin{equation}
\begin{array}{rcl}
\D(L_{\Omega}(t))&:=&\{u \in W^{2,p}(\Omega)^d \cap W^{1,p}_0(\Omega)^d: M(t)x \cdot \nabla u \in L^p(\Omega)^d\},\\[0.15cm]
L_{\Omega}(t)u & := & \OU(t)u,
\end{array}
\end{equation}
and the perturbed operators are defined by
\begin{equation}
\begin{array}{rcl}
\D(L_{\Omega,b}(t))&:=&\D(L_\Omega(t)),\\[0.15cm]
L_{\Omega,b}(t)u & := & \OU(t)u + \mathcal B(t) u.
\end{array}
\end{equation}
In the following, to simplify our notation, we do not distinguish between $L^p(\Omega)^d$ and $L^p(\Omega)$ and sometimes write
\begin{equation*}
L_\Omega(t)u(x)  := \Delta u(x)+\left(M(t)x + c(t)\right)\cdot\nabla u(x)
	-M(t)u(x),\quad t>0,\,x\in \Omega.
\end{equation*}

With these linear operators the linearization of the system \eqref{eq:NS_5} for some initial time $s\geq 0$ is now given by
\begin{equation}\label{eq:S_exterior}
\left\{\begin{array}{rcll}
u_t - L_{\Omega,b}(t)u + \nabla \p &=& 0, & \quad \mbox{in}\; (s,\infty)\times \Omega ,\\
\div u &=& 0,&  \quad \mbox{in}\; (s,\infty)\times \Omega ,\\
u &=& 0, & \quad \mbox{on}\; (s,\infty)\times \partial\Omega ,\\
u(s,\cdot) &=&f, &  \quad \mbox{in}\; \Omega.
\end{array}\right.
\end{equation}

As usual in the theory of the Navier-Stokes equations we shall later work in the space $L^p_\sigma(\Omega)$ of all solenoidal vector fields in $L^p$. Therefore we set
\begin{equation}
\begin{array}{rcl}
\D(A_{\Omega}(t))&:=&\{u \in W^{2,p}(\Omega)^d \cap W^{1,p}_0(\Omega)^d \cap L_\sigma^p(\Omega) : M(t)x \cdot \nabla u \in
L^p(\Omega)^d\},\\[0.15cm]
A_{\Omega}(t)u & := &\P_\Omega L_{\Omega}(t)u,
\end{array}
\end{equation}
and
\begin{equation}
\begin{array}{rcl}
\D(A_{\Omega,b}(t))&:=&\D(A_{\Omega}(t)),\\[0.15cm]
A_{\Omega,b}(t)u & := &\P_\Omega L_{\Omega,b}(t)u.
\end{array}
\end{equation}

By applying the Helmholtz projection $\P_\Omega$ to \eqref{eq:S_exterior} the pressure $\p$ can be eliminated and we may rewrite the equations as
an non-autonomous abstract Cauchy problem
\begin{equation}\label{eq:nACP_exterior}
\left\{\begin{array}{lcll}
u'(t) & = & A_{\Omega,b}(t)u(t),& 0\leq s < t ,\\[0.15cm]
u(s) & = & f.
\end{array}\right.
\end{equation}

It directly follows from \cite{Geissert/Heck/Hieber:2006a} that for fixed $s\ge 0$ the operator $A_{\Omega,b}(s)$ generates a $C_0$-semigroup
on $L^p_\sigma(\Omega)$, $1<p<\infty$, which is however \emph{not analytic}. Therefore we cannot apply standard generation results for evolution
systems of parabolic type (we refer to the monographs \cite{Lunardi:1995} and \cite{Tanabe:1997} for more information on this matter). Moreover, we note that the domain of $A_{\Omega,b}(t)$ depends on time $t$. Therefore, to
overcome this difficulty and in order to discuss well-posedness of \eqref{eq:nACP_exterior} we introduce the regularity space
\begin{equation*}
Y_{\Omega}:=\{u \in W^{2,p}(\Omega)^d \cap W^{1,p}_0(\Omega)^d \cap L^p_\sigma(\Omega) : |x| \nabla u_i(x) \in
L^p(\Omega)^d \;
\mathrm{for} \; i=1,\ldots, d\}
\end{equation*}
which is contained in $\D(A_{\Omega,b}(t))$ for every $t\ge0$.

Our first main result is the existence of a strongly
continuous evolution system on $L^p_\sigma(\Omega)$, $1<p<\infty$, that solves the
Cauchy problem \eqref{eq:nACP_exterior} on the regularity space $Y_\Omega$. This directly implies well-posedness of
\eqref{eq:nACP_exterior}. Moreover, we obtain
$L^p$-$L^q$ smoothing properties and gradient estimates for the evolution system. This is \emph{a priori} not obvious, since the
evolution system is not of parabolic type.

\begin{theorem}\label{thm:evolution_system_exterior}
Let $\Omega \subset \R^d$ be an exterior domain with $C^{1,1}$-boundary and $1<p<\infty$. Then there exists a unique evolution system $\{T_{\Omega,b}(t,s)\}_{(t,s)\in \Lambda}$ on $L^p_\sigma(\Omega)$ with the following properties.
\begin{itemize}
\item[(a)] For $(t,s)\in \Lambda$, the operator $T_{\Omega,b}(t,s)$ maps $Y_{\Omega}$ into $Y_{\Omega}$.
\item[(b)] For every $f\in Y_{\Omega}$ and $s\ge 0$, the map $t\mapsto T_{\Omega,b}(t,s)f$ is differentiable in $(s,\infty)$ and
\begin{equation}\label{eq:derivative_evolution_exterior}
\frac{\partial}{\partial t} T_{\Omega,b}(t,s)f = A_{\Omega,b}(t)T_{\Omega,b}(t,s)f .
\end{equation}
\item[(c)] For every $f\in Y_{\Omega}$ and $t>0$, the map $s\mapsto T_{\Omega,b}(t,s)f$ is differentiable in $[0,t)$ and
\begin{equation}\label{eq:derivative_evolution_exterior_2}
\frac{\partial}{\partial s} T_{\Omega,b}(t,s)f = - T_{\Omega,b}(t,s)A_{\Omega,b}(s)f.
\end{equation}
\item[(d)] For $T>0$ and $1<p<\infty$ there is a constant $C:=C(T)>0$ such that for every $f\in L^p_
\sigma(\Omega)$ $$\|\nabla T_{\Omega,b}(t,s)f\|_{p,\Omega} \leq C(t-s)^{-\frac {1}{2}}\|f\|_{p,\Omega}, \qquad (t,s)\in\tilde\Lambda_T.$$
\item[(e)] Let $T>0$ and $1< p\leq q < \infty$. Then there exists a constant $C:=C(T)>0$ such that for every $f\in L^p_
\sigma(\Omega)$
\begin{equation}\label{Lpq}
\|T_{\Omega,b}(t,s)f\|_{q,\Omega} \leq C(t-s)^{-\frac{d}{2}\left(\frac 1 p - \frac 1 q\right)}\|f\|_{p,\Omega}, \qquad
(t,s)\in\tilde\Lambda_T,
\end{equation}
\begin{equation}\label{nabla-Lpq}
\|\nabla T_{\Omega,b}(t,s)f\|_{q,\Omega} \leq C(t-s)^{-\frac{d}{2}\left(\frac 1 p - \frac 1 q\right)-\frac{1}{2}}\|f\|_{p,\Omega}, \qquad (t,s)\in\tilde\Lambda_T.
\end{equation}
\end{itemize}
\end{theorem}

\begin{remark}
\begin{itemize}
\item[(a)] The analogous result holds for the evolution system $\{T_\Omega(t,s)\}_{(t,s)\in\Lambda}$ associated to the operators $A_\Omega(\cdot)$.
\item[(b)] If we denote the evolution system on $L^p_\sigma(\Omega)$ by $\{T^p_{\Omega,b}(t,s)\}_{(t,s)\in \Lambda}$, then the family of evolution systems is consistent in the sense that
\begin{equation*}
T^p_{\Omega,b}(t,s) f = T^q_{\Omega,b}(t,s) f, \qquad \quad (t,s)\in \Lambda, \qquad \quad f \in L^p_\sigma(\Omega)\cap L^q_\sigma (\Omega),
\end{equation*}
holds for $1<p,q<\infty$.
\end{itemize}
\end{remark}
The basic idea to prove Theorem \ref{thm:evolution_system_exterior} is to study first the whole space case $\R^d$ and the case of a
bounded domain $D$, which is
done in Sections \ref{sect_whole} and \ref{sect_bounded} respectively. Then in Section \ref{sect:exterior} we use some cut-off techniques to construct the evolution system for the exterior domain $\Omega$. Our method, which was already presented in \cite{Hansel/Rhandi:2010} for treating non-autonomous Ornstein-Uhlenbeck equations, then also allows to obtain the $L^p$-$L^q$ smoothing properties and the gradient estimates from the corresponding estimates in $\R^d$ and $D$.

Next we come back to the nonlinear problem \eqref{eq:NS_5}. Again by applying the Helmholtz projection $\P_\Omega$ we can rewrite this system in abstract form as
\begin{equation}\label{eq:NS_abstract}
\left\{
\begin{array}{rclll}
u'(t)- A_{\Omega,b}(t)u(t) + \P_\Omega( u \cdot \nabla u)(t) &=&\P_\Omega F_1(t), & t>0,  \\[0.15cm]
u(0)&=&f,&
\end{array}\right.
\end{equation}
where $F_1$ is given in \eqref{eq:inhom}. By the Duhamel principle (variation of constant formula) this problem can be reduced to the
integral equation
\begin{align}\label{eq:Duhamel}
&u(t) = T_{\Omega,b}(t,0)f - \int_0^t T_{\Omega,b}(t,s) \mathbb P_\Omega(u\cdot \nabla u)(s) \d s\\
&\qquad\qquad\qquad +  \int_0^t T_{\Omega,b}(t,s) \mathbb P_\Omega F_1(s)  \d s,\quad \qquad t\geq0,\notag
\end{align}
in $L^p_{\sigma}(\R^d)$.
For given $0<T_0\leq \infty$,
we call a function $u\in C([0,T_0);L^p_{\sigma}(\R^d))$ a \textit{mild solution} of (\ref{eq:NS_abstract})  if $u$ satisfies
the integral equation (\ref{eq:Duhamel}).
By adjusting Kato's iteration scheme
to our situation the existence of a unique local mild solution follows.

\begin{theorem}\label{thm:Kato}
Let $2\leq d\leq p \le q < \infty$ and $f\in
L^p_{\sigma}(\Omega)$. Then there exists $T>0$ and a unique mild
solution $u\in C([0,T];L^p_{\sigma}(\Omega))$ of
(\ref{eq:NS_abstract}), which has the properties
\begin{equation}\label{eq:mild_solution_prop1}
t^{\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}u(t) \in
C([0,T];L^q_{\sigma}(\Omega)),
\end{equation}
\begin{equation}\label{eq:mild_solution_prop2}
t^{\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)+\frac{1}{2}}\nabla
u(t) \in C([0,T];L^p(\Omega)^{d\times d}).
\end{equation}
If $p<q$, then in addition
\begin{equation}\label{eq:mild_solution_prop3}
t^{\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}\|u(t)\|_{q,\Omega} +
t^{\frac{1}{2}}\|\nabla u(t)\|_{p,\Omega} \rightarrow 0 \qquad \mbox{as\;} t
\rightarrow 0^+.
\end{equation}
\end{theorem}

\section{The linearized problem in $\R^d$}\label{sect_whole}

In this section we study the linearized problem in the whole space $\R^d$. This situation was already studied in detail by
the first author in \cite{Hansel:2009} and by Geissert and the first author in \cite{Geissert/Hansel:2010}. Here we recall the main results. For this purpose we set
\begin{equation}\label{eq:operator_Rd}
\begin{array}{rcl}
\D(L_{\R^d}(t))&:=& \{ u\in W^{2,p}(\R^d)^d: M(t)x\cdot
	\nabla u\in L^p(\R^d)^d\},\\[0.2cm]
L_{\R^d}(t)u & := & \OU(t)u.
\end{array}
\end{equation}
and
\begin{equation}
\begin{array}{rcl}
\D(A_{\R^d}(t))&:=& \D(L_{\R^d}(t)) \cap L^p_\sigma (\R^d),\\[0.2cm]
A_{\R^d}(t)u & := & \OU(t)u.
\end{array}
\end{equation}
Note that $A_{\R^d}(t)$ is indeed an operator on $L^p_\sigma(\R^d)$, since an easy computation yields that $\div A_{\R^d}(t)u = 0$ for all $u\in C_{c,\sigma}^\infty(\R^d)$. In particular, the operator $L_{\R^d}(t)$ commutes with the Helmholtz projection $\P_{\R^d}$ in $\R^d$. Moreover we introduce the regularity space
\begin{equation*}
Y_{\R^d}:=\{u \in W^{2,p}(\R^d)^d \cap L^p_\sigma(\R^d) : |x| \nabla u_i(x) \in L^p(\R^d)^d \; \mathrm{for} \; i=1,\ldots, d\}.
\end{equation*}
For every $t\ge0$, the space $Y_{\R^d}$ is contained in $\D(A_{\R^d}(t))$.

In the following we denote by $\{U(t,s)\}_{t,s\ge 0}$ the evolution system in $\R^d$ that satisfies
\begin{equation*}
\left\{\begin{array}{lcll}
\frac{\partial}{\partial t} U(t,s) & = & -M(t)U(t,s),& t\ge s,\\[0.2cm]
U(s,s) & = & \mathrm{Id}.
\end{array}\right.
\end{equation*}
Now, for $f\in L^p_\sigma(\R^d)$ and $s\geq 0$, we set $T_{\R^d}(s,s)=\mathrm{Id}$ and for $(t,s)\in \widetilde{\Lambda}$ we define
\begin{equation}\label{eq:evol_sys_whole}
T_{\R^d}(t,s)f(x) = (k(t,s,\cdot)\ast f)(U(s,t)x+g(t,s)) \qquad x\in \R^d,
\end{equation}
where
\begin{equation}
k(t,s,x):= \frac{1}{(4\pi)^{\frac d 2} (\det Q_{t,s})^{\frac 1 2}} U(t,s) \mathrm{e}^{-\frac 1 4 \langle Q_{t,s}^{-1}x,x\rangle}, \qquad x\in \R^d,
\end{equation}
\begin{equation}
g(t,s) = \int_s^t U(s,r)c(r) \d r \qquad \mbox{and} \qquad Q_{t,s}=\int_s^t U(s,r)U^*(s,r)\d r.
\end{equation}

For the derivation of this solution formula we refer to \cite[Section 3]{Geissert/Hansel:2010}. The explicit formula now allows to prove the following result (see \cite{Geissert/Hansel:2010, Hansel:2009} and \cite[Section 2]{Hansel/Rhandi:2010} for details).

\begin{prop}\label{prop:evolution_whole}
Let $1<p<\infty$. Then the family of operators $\{T_{\R^d}(t,s)\}_{(t,s)\in \Lambda}$ defined in \eqref{eq:evol_sys_whole} is a strongly continuous evolution system on $L^p_\sigma(\R^d)$ with the following properties.
\begin{itemize}
\item[(a)] For $(t,s)\in \Lambda$, the operator $T_{\R^d}(t,s)$ maps $Y_{\R^d}$ into $Y_{\R^d}$.
\item[(b)] For every $f\in Y_{\R^d}$ and every $s\in[0,\infty)$, the map $t\mapsto T_{\R^d}(t,s)f$ is differentiable in $(s,\infty)$ and
\begin{equation}\label{eq:derivative_t}
\frac{\partial}{\partial t} T_{\R^d}(t,s)f = A_{\R^d}(t)T_{\R^d}(t,s)f .
\end{equation}
\item[(c)] For every $f\in Y_{\R^d}$ and $t\in (0,\infty)$, the map $s\mapsto T_{\R^d}(t,s)f$ is differentiable in $[0,t)$
and
\begin{equation}\label{eq:derivative_s}
\frac{\partial}{\partial s} T_{\R^d}(t,s)f = -T_{\R^d}(t,s)A_{\R^d}(s)f .
\end{equation}
\item[(d)] Let $1<p\leq q \leq \infty$. Then there exists a constant
	$C>0$ such that for every $f\in L^p_\sigma(\R^d)$
\begin{align}
	\|T_{\R^d}(t,s)f\|_{q}
	&\leq C
(t-s)^{-\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}\|f\|_{p},&
 (t,s)\in \widetilde{\Lambda},\label{eq:LpLqsmoothing}\\
\|\nabla T_{\R^d}(t,s)f\|_{q}
&\leq C
(t-s)^{-\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)-\frac{1}{2}}\|f\|_{p},
&  (t,s)\in \widetilde{\Lambda}. \label{eq:GradientEstimate}
\end{align}
\end{itemize}
\end{prop}

By Proposition \ref{prop:evolution_whole} and the fact that $L_{\R^d}(t)=A_{\R^d}(t)$ on $L^p_\sigma (\R^d)$ it follows that the solution to the problem
\begin{equation}\label{eq:problem_bounded}
\left\{\begin{array}{rcll}
u_t - L_{\R^d}(t)u &=& 0, & \quad \mbox{in}\; (s,\infty)\times \R^d ,\\
\div u &=& 0,&  \quad \mbox{in}\; (s,\infty)\times \R^d ,\\
u(s,\cdot) &=&f, &  \quad \mbox{in}\; \R^d,
\end{array}\right.
\end{equation}
for $s\geq 0$, is given by $u(t,x)=T_{\R^d}(t,s)f(x)$. So in the whole space case the pressure is constant, thus we may assume that the pressure is even zero.

\section{The linearized problem in bounded domains}\label{sect_bounded}

In this section let $D\subset \R^d$ be a bounded domain with $C^{1,1}$-boundary. We set
\begin{equation}\label{eq:operator_bounded}
\begin{array}{rcl}
\D(L_{D,b}(t))&:=& \D(L_{D,b}) := W^{2,p}(D)^d\cap W^{1,p}_0(D)^d,\\[0.2cm]
L_{D,b}(t)u & := & \OU_{D,b}(t)u.
\end{array}
\end{equation}
Moreover, we define the operators
\begin{equation}\label{eq:operator_bounded}
\begin{array}{rcl}
\D(A_{D,b}(t))&:=& \D(A_{D,b}) := W^{2,p}(D)^d\cap W^{1,p}_0(D)^d \cap L^p_\sigma(D),\\[0.2cm]
A_{D,b}(t)u & := & \P_D L_{D,b}(t)u.
\end{array}
\end{equation}

In a bounded domain also the coefficients of the term $M(t)x\cdot \nabla$ are bounded. Thus, it follows directly from the classical perturbation theory for the Stokes operator that for fixed $s\ge 0$ the operator $(A_{D,b}(s),\D(A_{D,b}))$ generates an analytic semigroup (cf. \cite[Proposition 3.1]{Geissert/Heck/Hieber:2006a}). Moreover, by our assumptions on the coefficients we obtain that the map $t\mapsto A_{D,b}(t)$ belongs to $C^1(\R_+, \L(\D(A_{D,b}),L^p_\sigma(D)))$. Thus, the following result follows from the theory of parabolic evolution systems (see \cite[Chapter 6]{Lunardi:1995} and \cite[Section 2.3]{Grisvard}).

\begin{prop}\label{prop:evolution_bounded}
Let $D\subset \R^d$ be a bounded domain with $C^{1,1}$-boundary and $1<p<\infty$. Then there is a unique evolution system $\{T_{D,b}(t,s)\}_{(t,s)\in \Lambda}$ on $L^p_\sigma(D)$ with the following properties.
\begin{enumerate}
\item[(a)] For $(t,s)\in \widetilde{\Lambda}$, the operator $T_{D,b}(t,s)$ maps $L^p_\sigma(D)$ into $\D(A_{D,b})$.
\item[(b)] The map $t \mapsto T_{D,b}(t,s)$ is differentiable in $(s,\infty)$ with values in $\L(L^p_\sigma(D))$ and
\begin{equation}
\frac{\partial}{\partial t} T_{D,b}(t,s) = A_{D,b}(t)T_{D,b}(t,s).
\end{equation}
\item[(c)] For every $f \in \D(A_{D,b})$ and $t>0$, the map $s\mapsto T_{D,b}(t,s)f$ is differentiable in $[0,t)$ and
\begin{equation}
\frac{\partial}{\partial s} T_{D,b}(t,s)f = - T_{D,b}(t,s)A_{D,b}(s)f.
\end{equation}
\item[(d)] Let $T>0$. Then there exists a constant $C:=C(T)>0$ such that
\begin{equation}\label{eq45}
\|T_{D,b}(t,s)f\|_{p,D}\leq C\|f\|_{p,D},
\end{equation}
and
\begin{equation}\label{eq46}
\|T_{D,b}(t,s)f\|_{2,p,D}\leq C(t-s)^{-1}\|f\|_{p,D}.
\end{equation}
for all $f\in L^p_\sigma(D)$ and all $(t,s)\in \widetilde{\Lambda}_T$.
\end{enumerate}
\end{prop}

For the following first estimate let us recall the Gagliardo-Nierenberg inequality
\begin{equation}\label{G-N}
\|f\|_{k,q,D}\le C \|f\|_{m,p,D}^\theta \|f\|_{r,D}^{(1-\theta)},\quad \hbox{\ for all }f\in W^{m,p}(D),
\end{equation}
where $p\ge 1,\,q\ge 1,\,r\ge 1,\,0<\theta \le 1$ and $k-\frac{d}{q}\le \theta\left(m-\frac{d}{p}\right)-(1-\theta)\frac{d}{r}.$

\begin{corollary}\label{prop:Lp_Lq_estimates_bounded}
Let $T>0$ and $1< p\leq q < \infty$. Then there exists a constant $C:=C(T)>0$ such that for every $f\in L^p_\sigma(D)$
and $(t,s)\in \widetilde{\Lambda}_T$
\begin{enumerate}
\item[(a)] $\|T_{D,b}(t,s)f\|_{q,D} \leq C(t-s)^{-\frac{d}{2}\left(\frac 1 p - \frac 1 q\right)}\|f\|_{p,D},$
\item[(b)] $\|\nabla T_{D,b}(t,s)f\|_{p,D} \leq C(t-s)^{-\frac{1}{2}}\|f\|_{p,D},$

\item[(c)]  $\|\nabla T_{D,b}(t,s)f\|_{q,D} \leq C(t-s)^{-\frac{d}{2}\left(\frac 1 p - \frac 1 q\right)-\frac 1 2}\|f\|_{p,D}$.
\end{enumerate}
Moreover, $$\|T_{D,b}(t,s)f\|_{k,p}\le C\|f\|_{k,p},\quad (t,s)\in \Lambda_T,$$
for all $f\in W^{k,p}(D)^d\cap L^p_\sigma(D),\,k=1,\,2,$ and
$$\|T_{D,b}(t,s)f\|_{2,p}\le C(t-s)^{-\frac{1}{2}}\|f\|_{1,p},\quad (t,s)\in \widetilde{\Lambda}_T,$$
for all $f\in W^{1,p}(D)^d\cap L^p_\sigma (D)$.
\end{corollary}

\begin{proof}
Let us assume first that $0<\frac{1}{p}-\frac{1}{q}\le \frac{2}{d}$. Then $0<\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)\le 1$.
Hence (a) follows by applying (\ref{G-N}) with $\theta=\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right),\,k=0,\,m=2,\,r=p$ and
(\ref{eq46}).\\
Assume now that $\frac{2}{d}<\frac{1}{p}-\frac{1}{q}\le \frac{4}{d}$. Set $\frac{1}{r}=\frac{1}{q}+\frac{2}{d}$. Then
$0<\frac{1}{p}-\frac{1}{r}\le \frac{2}{d}$. So, by the first step and (\ref{G-N}), we obtain
\begin{eqnarray*}
\|T_{D,b}(t,s)f\|_{q,D} &=& \left\|T_{D,b}\left(t,s+\tfrac{t-s}{2}\right)T_{D,b}\left(s+\tfrac{t-s}{2},s\right)f\right\|_{q,D}\\
&\le & C\left(\frac{t-s}{2}\right)^{-\frac{d}{2}\left(\frac 1 r - \frac 1 q\right)}\left\|T_{D,b}\left(s+\tfrac{t-s}{2},s\right)f\right\|_{r,D}\\
&\le & C\left(\frac{t-s}{2}\right)^{-\frac{d}{2}\left(\frac 1 r - \frac 1 q\right)}\left(\frac{t-s}{2}\right)^{-\frac{d}{2}\left(\frac 1 p - \frac 1 r\right)}\|f\|_{p,D}\\
&\le & C(t-s)^{-\frac{d}{2}\left(\frac 1 p - \frac 1 q\right)}\|f\|_{p,D},\quad (t,s)\in \widetilde{\Lambda}_T.
\end{eqnarray*}
By iterating this argument we obtain (a) also for $\frac{1}{p}-\frac{1}{q}>\frac{2}{d}$.
\\
Assertion (b) follows from (\ref{G-N}) with $\theta=\frac{1}{2},\,p=q=r,\,k=1,\,m=2$ and (\ref{eq46}).
\\
It follows from (a) and (b) that
\begin{eqnarray*}
\|\nabla T_{D,b}(t,s)f\|_{q,D} &=& \left\| \nabla T_{D,b}\left(t,s+\tfrac{t-s}{2}\right)T_{D,b}\left(s+\tfrac{t-s}{2},s\right)f\right\|_{q,D}\\
&\le & C\left(\frac{t-s}{2}\right)^{-\frac{1}{2}}\left\|T_{D,b}\left(s+\tfrac{t-s}{2},s\right)f\right\|_{q,D}\\
&\le & C(t-s)^{-\frac{d}{2}\left(\frac 1 p - \frac 1 q\right)-\frac 1 2}\|f\|_{p,D}, \quad (t,s)\in \widetilde{\Lambda}_T,
\end{eqnarray*}
and this proves (c).

For the last assertions we refer, for example, to \cite[Corollary 6.1.8]{Lunardi:1995}.
\end{proof}

By Proposition \ref{prop:evolution_bounded} the solution to the problem
\begin{equation}\label{eq:problem_bounded}
\left\{\begin{array}{rcll}
u_t - L_{D,b}(t)u + \nabla \p &=& 0, & \quad \mbox{in}\; (s,\infty)\times D ,\\
\div u &=& 0,&  \quad \mbox{in}\; (s,\infty)\times D ,\\
u &=& 0, & \quad \mbox{on}\; (s,\infty)\times \partial D ,\\
u(s,\cdot) &=&f, &  \quad \mbox{in}\; D,
\end{array}\right.
\end{equation}
for $s\geq 0$, is given by $u(t,x)=T_{D,b}(t,s)f(x)$. From \eqref{eq:problem_bounded} and the fact that $u_t \in L^p_\sigma(D)$ it follows that $\nabla \p(t) = (\mathrm{Id}-\P_D)L_{D,b}(t)u(t)$.  Since the pressure is only unique up to an additive constant, in the case of a bounded domain we can always assume that
$$
\p \in L^p_0(D) :=\left\{u \in L^p(D): \int_D u \,\d x = 0\right\}.
$$
By the abstract theory of parabolic evolution systems it is clear that $u\in C^1((s,\infty); L^p_\sigma(D)) \cap C((s,\infty); \D(A_{D,b}))$. So in particular, we can conclude that $\nabla \p \in C((s,\infty),L^p(D)^d)$. By Poincar\'{e}'s inequality we can conclude that also $\p \in C((s,\infty),L^p(D))$.

\begin{lemma}\label{lem:pressure}
Let $1<p<\infty$ and $D\subset \R^d$ be a bounded domain with $C^{1,1}$-boundary. Let $(u,\p)$ be the unique solution of
\eqref{eq:problem_bounded} with $\p\in L^p_0(D)$ and let $\gamma \in (1+\frac 1 p,2)$. Then for $T>s$ there exists a
constant $C:=C(T,\gamma)>0$ with
\begin{equation*}
\|\p(t)\|_{p,D}\leq C(t-s)^{-\frac \gamma 2}\|f\|_{p,D}
\end{equation*}
for all $t\in (s,T)$.
\end{lemma}

A similar estimate was proved in \cite[Lemma 3.5]{Geissert/Heck/Hieber:2006a} for the solution to the corresponding resolvent problem (see also \cite{Noll/Saal:2003} for the case of the Stokes operator). However, our proof follows the ideas in \cite{Shibata/Shimada:2007} (see also \cite[Section 4]{Shibata:2008}).

\begin{proof}[Proof of Lemma \ref{lem:pressure}]
Let $\varphi \in C_c^\infty(D)$ and set $\tilde \varphi = \varphi - |D|^{-1}\int_D \varphi\,\d x$. Then, in $L^{p'}(D)$ with
 $\frac{1}{p}+\frac{1}{p'}=1$, there exits a unique solution $\psi \in W^{2,p'}(D)$ of the Neumann problem
\begin{equation}\label{eq:Neumann}
\left\{\begin{array}{rcll}
\Delta \psi &=& \tilde\varphi &\mbox{in}\;D,\\
\nabla \psi \cdot\nu &=&0&\mbox{on}\; \partial D,
\end{array}\right.
\end{equation}
which satisfies the estimate
\begin{equation*}
\|\psi\|_{2,p',D}\leq C \|\tilde\varphi\|_{p',D}\le 2C\|\varphi \|_{p',D},
\end{equation*}
for some constant $C>0$ (cf. \cite[Proposition 5.5]{Shibata/Shimada:2007}). We obtain now
\begin{align*}
\langle \p, \varphi \rangle_D &= \langle \p, \tilde\varphi \rangle_D = \langle \p, \Delta\psi \rangle_D = -  \langle \nabla\p, \nabla\psi \rangle_D=-\langle (\mathrm{Id}-\P_D)L_{D,b}(t)u(t), \nabla\psi \rangle_D\\
&= -\langle \Delta u(t), \nabla\psi \rangle_D-\langle (M(t)x-c(t))\cdot \nabla u(t), \nabla\psi \rangle_D + \langle M(t)u(t), \nabla\psi \rangle_D\\
& \quad +\langle b\cdot \nabla u(t), \nabla \psi \rangle_D +\langle u(t)\cdot \nabla b, \nabla \psi \rangle_D \\
&= -\langle \nabla u(t)\cdot \nu,\nabla \psi \rangle_{\partial D}+ \langle \nabla u(t), \nabla^2\psi \rangle_D
-\langle (M(t)x-c(t))\cdot \nabla u(t), \nabla\psi \rangle_D \\
& \quad + \langle M(t)u(t), \nabla\psi \rangle_D
+\langle b\cdot \nabla u(t), \nabla \psi \rangle_D +\langle u(t)\cdot \nabla b, \nabla \psi \rangle_D.
\end{align*}
By using the embedding $H^{\frac 1 p + \eps, p}(D) \hookrightarrow L^p(\partial D)$ for $0<\eps \leq 1 - \frac 1 p$
(cf \cite[Theorem 1.5.1.2]{Grisvard}) we obtain
\begin{align*}
|\langle \p, \varphi \rangle_D| &\leq C \left( \|\nabla u \cdot \nu\|_{p,\partial D}  +\|\nabla u\|_{p,D}+\| u\|_{p,D}\right)\|\psi\|_{2,p',D}\\
&\leq C\|u \|_{1+\frac 1 p +\eps,p,D} \|\varphi\|_{p',D}.
\end{align*}
By taking now $\eps <1-\frac 1 p$ and $\gamma =1+\frac 1 p +\eps$, the assertion follows from Proposition \ref{prop:evolution_bounded}, Corollary \ref{prop:Lp_Lq_estimates_bounded} and simple interpolation .
\end{proof}

\section{The linearized problem in exterior domains}\label{sect:exterior}

In this section we prove Theorem \ref{thm:evolution_system_exterior}. The general idea is to derive the result for exterior domains from the corresponding results in the
case of $\R^d$ and bounded domains by
some cut-off techniques. For this purpose let $R>0$ be such that $\mathcal O \cup \mathrm{supp}\,b(\cdot,t) \subset B(R)$ for every $t>0$. We then set
\begin{align*}
D&=\Omega \cap B(R+8),\\
K_1 & = \Omega \cap B(R+2),\\
K_2 & = \{ x \in \Omega : R+2 < |x|<R+5\},\\
K_3 & = \{ x \in \Omega : R+5 < |x|<R+8\}.
\end{align*}
We denote by $\BB_i$ for $i\in\{1,2,3\}$ the operator defined in Lemma \ref{prop_Bog1} associated to the domain $K_i$. Moreover, by $\{T_{\R^d}(t,s)\}_{(t,s)\in\Lambda}$ we denote the evolution system in $L^p_\sigma(\R^d)$ from Proposition \ref{prop:evolution_whole} and by $\{T_{D,b}(t,s)\}_{(t,s)\in\Lambda}$
the evolution system in $L^p_\sigma(D)$ for the bounded domain $D$ from Proposition \ref{prop:evolution_bounded}.

Next, we choose cut-off functions $\varphi,\xi,\eta \in C^{\infty}(\R^d)$ such that $0\leq \varphi, \xi, \eta \leq 1$ and
$$
\varphi(x) :=
\left\{
\begin{array}{cc}
  1,&|x| \geq R+4,  \\
  0, & |x| \leq R+3,
\end{array}
\right.
$$
$$
\xi(x) :=
\left\{
\begin{array}{ll}
  1,&|x| \geq R+1,  \\
  0, & |x| \leq R,
\end{array}
\right.
$$
$$
\eta(x) :=
\left\{
\begin{array}{ll}
  1,&|x| \leq R+6,  \\
  0, & |x| \geq R+7.
\end{array}
\right.
$$

For a given $f\in L^p_{\sigma}(\Omega)$ we now define functions $f_R\in L^p_\sigma(\R^d)$ and $f_D\in L^p_\sigma(D)$, respectively, by
$$
f_R(x) :=
\left\{
\begin{array}{ll}
  \xi(x) f(x) - \BB_1((\nabla \xi)\cdot f)(x),&x\in\Omega,   \\
  0, & x\not\in \Omega,
\end{array}
\right.
$$
and
$$
f_D(x) := \eta(x)f(x) - \BB_3((\nabla \eta)\cdot f)(x), \qquad\quad x\in D.
$$
With
partial integration we see that for every $f\in \D(A_{\Omega,b}(t))$ we have $f_R\in \D(A_{\R^d}(t))$ and $f_D\in \D(A_{D,b}(t))$.

Now for $(t,s)\in\Lambda$, we define the operator $W(t,s)$ by  setting
\begin{equation*}
W(t,s)f := \varphi T_{\R^d}(t,s)f_R + (1-\varphi) T_{D,b}(t,s)f_D - \BB_2((\nabla\varphi)\cdot ( T_{\R^d}(t,s)f_R  -T_{D,b}(t,s)f_D))
\end{equation*}
for every $f\in L^p_\sigma(\Omega)$. By Lemma \ref{prop_Bog1} it is clear that $W(t,s)f \in L^p_\sigma(\Omega)$. Moreover, for $f
\in \D(A_{\Omega,b}(t))$ it follows from the properties of $\{T_{\R^d}(t,s)\}_{(t,s)\in\Lambda}$, $\{T_{D,b}(t,s)\}_{(t,s)\in\Lambda}$ and the operator $\BB_2$ that $W(t,s)f \in W^{2,p}(\Omega)\cap W^{1,p}_0(\Omega)\cap L^p_\sigma(\Omega)$
holds. Moreover, a short calculation yields
\begin{align*}
\nabla W(t,s)f & = \varphi \nabla T_{\R^d}(t,s)f_R + (1-\varphi) \nabla T_{D,b}(t,s)f_D \\
&\qquad + \nabla \varphi \left(T_{\R^d}(t,s)f_R - T_{D,b}(t,s)f_D\right)\\
&\qquad - \nabla \left[\BB_2((\nabla \varphi)\cdot (T_{\R^d}(t,s)f_R - T_{D,b}(t,s)f_D) )\right].
\end{align*}
Thus, it follows that $W(t,s)f \in \D(A_{\Omega,b}(t))$ if $f\in \D(A_{\Omega,b}(t))$.

Let us set $u_R(t):=T_{\R^d}(t,s)f_R$ and $u_D(t):=T_{D,b}(t,s)f_D$ and let $\p_D$ be the pressure that is associated
to $u_D$. We may assume that $\int_D \p_D \,\d x= 0$.

A short calculation yields that for some initial value $f\in Y_\Omega$, the function

$u(t):=W(t,s) f$ solves the inhomogeneous equation
\begin{equation}\label{eq:inhom_error_terms}
\left\{\begin{array}{rcll}
u_t - L_{\Omega,b} (t)u +\nabla \p_\varphi &=&- \tilde f, & \quad \mbox{in}\; (s,\infty)\times \Omega ,\\
\div u &=& 0,&  \quad \mbox{in}\; (s,\infty)\times \Omega ,\\
u &=& 0, & \quad \mbox{on}\; (s,\infty)\times \partial\Omega ,\\
u(s,\cdot) &=&f, &  \quad \mbox{in}\; \Omega,
\end{array}\right.
\end{equation}
where $\nabla\p_\varphi := \nabla\left((1-\varphi)\p_D\right)$ and
\begin{align}\label{eq:error_terms}
\tilde f := F(t,s)f & := +2\left(  \nabla  T_{\R^d}(t,s)f_R - \nabla T_{D,b}(t,s)f_D \right)\cdot  (\nabla \varphi)^* \notag \\
& \qquad + (\Delta \varphi + (M(t)x + c(t))\cdot (\nabla \varphi))\left(T_{\R^d}(t,s)f_R - T_{D,b}(t,s)f_D\right)\notag\\
&\qquad +\BB_2((\nabla \varphi)\cdot (\partial_t T_{\R^d}(t,s)f_R  -\partial_t T_{D,b}(t,s)f_D)) \\
&\qquad -  L_{\Omega,b}(t)\BB_2((\nabla \varphi)\cdot (T_{\R^d}(t,s)f_R - T_{D,b}(t,s)f_D) ) \notag\\
&\qquad + (\nabla \varphi) \p_D \notag\\
&=: I_1 + I_2 + I_3 + I_4 + I_5.\notag
\end{align}
Here we use the fact that $\mathrm{supp}\,b(t,\cdot)\subset B(R)$ for every $t>0$ and the expression of $f_R$.

Certainly, the function $F(t,s)f$ in (\ref{eq:error_terms}) is well-defined for every $f\in L^p_\sigma (\Omega)$ and $(t,s)\in \widetilde
\Lambda$. Later, we need a certain decay of $F(t,s)$ in $t$, stated in the following lemma.

\begin{lemma}\label{lemma:error_terms}
Let $1<p<\infty$. For every $f\in L^p_\sigma (\Omega)$ we have
\begin{equation*}
F(\cdot,\cdot)f \in C(\widetilde \Lambda; L^p(\Omega)^d).
\end{equation*}
Moreover, let $T>0$ be fixed and let $\gamma\in(1+\frac 1 p, 2)$. Then
\begin{equation}\label{eq:norm_estimate_error}
\|F(t,s)f\|_{p,\Omega}\leq C  (t-s)^{-\frac \gamma 2} \|f\|_{p,\Omega}, \qquad (t,s)\in \widetilde\Lambda_T,

\end{equation}
for  some constant $C:=C(T,\gamma)>0$.
\end{lemma}

\begin{proof}
Let us start with the norm estimates for $I_1$ and $I_2$.
By using Proposition \ref{prop:evolution_whole} and Corollary \ref{prop:Lp_Lq_estimates_bounded} we obtain
\begin{equation*}
\|I_1\|_{p,\Omega} \leq C (t-s)^{-\frac 1 2} \|f\|_{p,\Omega} \qquad \mbox{and} \qquad \|I_2\|_{p,\Omega} \leq C \|f\|_{p,\Omega}.
\end{equation*}
In order to estimate the norm of $I_4$, let us first note that Lemma \ref{prop_Bog1} implies that $L_{\Omega,b}(t)\BB_2 \in \L(W^{1,p}_0 (K_2), L^p(K_2))$. Thus again by using Proposition \ref{prop:evolution_whole} and Corollary \ref{prop:Lp_Lq_estimates_bounded} we obtain
\begin{equation*}
\|I_4\|_{p,\Omega} \leq C \|f\|_{p,\Omega}.
\end{equation*}
We come now to the term $I_3$. At first we note that we can write
\begin{align*}
&\BB_2((\nabla \varphi)\cdot (\partial_t u_R(t)  -\partial_t u_D(t))) \\
& \qquad = \BB_2((\nabla \varphi)\cdot (L_{\R^d}(t) u_R(t)))  -\BB_2((\nabla\varphi)\cdot (L_{D,b}(t) u_D(t) - \nabla \p_D)).
\end{align*}
Now, for a test function $\psi \in C_c^\infty (\R^d)$, we have
\begin{align*}
&|\langle(\nabla \varphi)\cdot L_{\R^d}(t) u_R(t), \psi\rangle| = |\langle L_{\R^d}(t) u_R(t), \psi (\nabla \varphi) \rangle|\\
&\qquad \quad \leq |\langle \nabla u_R(t), \nabla \left(\psi (\nabla \varphi)\right)  \rangle| +  |\langle (M(t)x+c(t)) \cdot \nabla u_R(t) - M(t)u_R(t), \psi (\nabla \varphi) \rangle|\\
&\qquad \quad \leq C \|u_R(t)\|_{1,p}\|\psi\|_{1,p'} + C \|u_R(t)\|_{1,p,K_2}\|\psi\|_{p',K_2}\\
&\qquad \quad \leq  C   \|u_R(t)\|_{1,p}\|\psi\|_{1,p'},
\end{align*}
where $\frac 1 p + \frac{1}{p'}=1$.  This shows that
\begin{equation*}
\|(\nabla \varphi)\cdot L_{\R^d}(t) u_R(t)\|_{W^{-1,p}(\R^d)}\leq  C   \|u_R(t)\|_{1,p}
\end{equation*}
holds. Analogously, we obtain
\begin{equation*}
\|(\nabla \varphi)\cdot L_{D,b}(t) u_D(t)\|_{W^{-1,p}(D)}\leq  C   \left(\|u_D(t)\|_{1,p,D} + \|\p_D\|_{p,D}\right).
\end{equation*}
Now, Lemma \ref{prop_Bog1} together with Proposition \ref{prop:evolution_whole}, Corollary \ref{prop:Lp_Lq_estimates_bounded} and Lemma \ref{lem:pressure} yield
\begin{equation*}
\|I_3\|_{p,\Omega} \leq C (t-s)^{-{\frac \gamma 2}} \|f\|_{p,\Omega}.
\end{equation*}
From Lemma \ref{lem:pressure} we can conclude that
\begin{equation*}
\|I_5\|_{p,\Omega} \leq C(t-s)^{-\frac \gamma 2}\|f\|_{p,\Omega}.
\end{equation*}
This proves \eqref{eq:norm_estimate_error}.

The continuity of the map $\widetilde{\Lambda}\ni (t,s)\mapsto F(t,s)f$ follows from the strong continuity of $T_{\R^d}(\cdot,\cdot)$ and $T_{D,b}(\cdot,\cdot)$, the continuity of the pressure $\p_D(\cdot)$ together with the properties of the operator $\BB_1$ stated in Lemma \ref{prop_Bog1}.
\end{proof}
Applying $\P_\Omega$ to (\ref{eq:inhom_error_terms}) we have
\begin{equation}\label{eq:inhom_error_terms+}
\left\{\begin{array}{rcll}
u_t &=&  A_{\Omega,b} (t)u - \P_\Omega \tilde f, & \quad \mbox{in}\; (s,\infty)\times \Omega ,\\
\div u &=& 0,&  \quad \mbox{in}\; (s,\infty)\times \Omega ,\\
u &=& 0, & \quad \mbox{on}\; (s,\infty)\times \partial\Omega ,\\
u(s,\cdot) &=&f, &  \quad \mbox{in}\; \Omega .
\end{array}\right.
\end{equation}
It is clear, that if an evolution system $\{T_{\Omega,b}(t,s)\}_{(t,s)\in\Lambda}$ exists on $L^p_\sigma(\Omega)$, then the solution $u(t)$ to the inhomogeneous equation (\ref{eq:inhom_error_terms+}) is given by the variation of constant formula
\begin{equation}\label{eq:integral}
u(t) = T_{\Omega,b}(t,s)f - \int_s^t T_{\Omega,b}(t,r) \P_\Omega F(r,s)f \d r.
\end{equation}
The integral in \eqref{eq:integral} exists because of Lemma \ref{lemma:error_terms}. This consideration suggests to consider the integral equation
\begin{equation}\label{eq:integral_eq}
T_{\Omega,b}(t,s)f = W(t,s)f + \int_s^t T_{\Omega,b}(t,r) \P_\Omega F(r,s)f \d r \qquad (t,s)\in\Lambda, f\in L^p_\sigma(\Omega).
\end{equation}

Let us now state a lemma which will be very useful in the proof of Theorem \ref{thm:evolution_system_exterior}. For the proof we refer to \cite{Hansel/Rhandi:2010}.

\begin{lemma}\label{lemma:iterated}
Let $X_1$ and $X_2$ be two Banach spaces, $T>0$ and let $R:\widetilde{\Lambda}_T \to \L(X_2,X_1)$ and $S:\widetilde{\Lambda}_T \to \L(X_2)$ be strongly continuous functions. Assume that
\begin{equation*}
\|R(t,s)\|_{\L(X_2,X_1)} \leq C_0(t-s)^{\alpha}, \quad \|S(t,s)\|_{\L(X_2)} \leq C_0(t-s)^{\beta}, \quad (t,s)\in \widetilde{\Lambda}_T,
\end{equation*}
holds for some $C_0=C_0(T)>0$ and $\alpha, \beta >-1$. For $f\in X_2$ and $(t,s)\in \widetilde{\Lambda}_T$, set
$T_0(t,s)f:=R(t,s)f$ and
\begin{equation*}
T_n(t,s)f:= \int_s^t T_{n-1}(t,r)S(r,s)f\d s, \qquad n\in \N, \; (t,s)\in \widetilde{\Lambda}_T.
\end{equation*}
Then there exists a constant $C > 0$ such that
\begin{equation}\label{eq:convergence_series}
\sum_{n=0}^\infty \|T_n(t,s)f\|_{X_1} \leq C (t-s)^{\alpha}\|f\|_{X_2}, \qquad (t,s)\in \widetilde{\Lambda}_T.
\end{equation}
Moreover, if $\alpha \geq 0$, the convergence of the series in (\ref{eq:convergence_series}) is uniform on $\Lambda_T$.
\end{lemma}

\begin{proof}[Proof of Theorem  \ref{thm:evolution_system_exterior}.]
Let $T>0$. By using Proposition \ref{prop:evolution_whole} and Corollary \ref{prop:Lp_Lq_estimates_bounded} we have
\begin{equation*}
\|W(t,s)f\|_{p,\Omega} \leq C\|f\|_{p,\Omega} \qquad \quad \mbox{for}\; f \in L^p_\sigma(\Omega), \; (t,s)\in \Lambda_T.
\end{equation*}
So, by (\ref{eq:norm_estimate_error}), we can apply Lemma \ref{lemma:iterated} with $R=W$, $S=\P_\Omega F,\,\alpha =0,\,\beta =-\frac{\gamma}{2}$ and $X_1=X_2=L^p_\sigma(\Omega)$. Thus, for any $f\in L^p_\sigma(\Omega)$,
the series $\sum_{k=0}^{\infty}T_k(t,s)f$ converges uniformly in $\Lambda_T$,
where $T_0(t,s)f=W(t,s)f$ and
\begin{equation}
T_{k+1}(t,s)f= \int_s^t T_k(t,r)\P_\Omega F(r,s)f \d r,\quad (t,s)\in \Lambda_T,\,f\in L^p_\sigma(\R^d).
\end{equation}
Since $T>0$ is arbitrary,
\begin{equation}\label{eq:representation_evolution_exterior}
T_{\Omega,b} (t,s):=\sum_{k=0}^{\infty}T_k(t,s),\quad(t,s)\in \Lambda
\end{equation}
is well-defined. It is easy to check that $T_{\Omega,b}(t,s)$ satisfies the integral equation \eqref{eq:integral_eq}. Moreover, from the strong continuity of $W(\cdot ,\cdot)$ and (\ref{eq:norm_estimate_error}) we deduce inductively that $T_k(\cdot ,\cdot)$ is strongly continuous
and hence, by the uniform convergence of the series we get the strong continuity of $T_{\Omega,b}(\cdot ,\cdot)$.

In order to show that $\{T_{\Omega ,b}(t,s)\}_{(t,s)\in \Lambda}$ leaves $Y_\Omega$ invariant, we proceed as in the proof of Theorem 3.1 in \cite{Hansel/Rhandi:2010} and consider the Banach space
$$Z :=\{f\in W_0^{1,p}(\Omega)^d\cap L^p_\sigma(\Omega) : |x|\cdot \nabla f_i(x) \in L^p(\Omega)^d\; \mbox{for} \; i=1,\ldots,d\}$$ endowed with the norm
$$\|f\|_{Z}:=\|f\|_{1,p,\Omega}+\||x|\cdot \nabla f\|_{p,\Omega},\quad f\in Z.$$
Proposition \ref{prop:evolution_whole}, Corollary \ref{prop:Lp_Lq_estimates_bounded} and (\ref{eq:norm_estimate_error}) permit us to apply
Lemma \ref{lemma:iterated} with $X_1=X_2=Z,\,R=W,\,S=\P_\Omega F,\alpha =0$ and $\beta =-\frac{\gamma}{2}$.
So, we obtain that $T_{\Omega ,b}(t,s)f \in Z$ for all $f\in Z$ and $(t,s)\in \Lambda$. Moreover, by taking
$X_1=W^{2,p}(\Omega)^d,\,X_2=W^{1,p}(\Omega)^d\cap L^p_\sigma(\Omega),\,R=W,\,S=\P_\Omega F,\,\alpha=-\frac{1}{2},\,\beta =-\frac{\gamma}{2}$ and applying Proposition \ref{prop:evolution_whole}, Corollary \ref{prop:Lp_Lq_estimates_bounded}
and (\ref{eq:norm_estimate_error}), it follows, by Lemma \ref{lemma:iterated}, that
  $T_{\Omega ,b}(t,s)f \in W^{2,p}(\Omega)$ for all $f\in W^{1,p}(\Omega)^d\cap L^p_\sigma(\Omega)$ and $(t,s)\in \widetilde{\Lambda}$. This yields that $\{T_{\Omega ,b}(t,s)\}_{(t,s)\in \Lambda}$ leaves $Y_\Omega$ invariant and
  \begin{align}\label{conv-in-Y_Omega}
  &\sum_{n=0}^\infty \left[\|T_k(t,s)f\|_{2,p,\Omega}+\||x|\cdot \nabla T_k(t,s)f\|_{p,\Omega}\right]\notag\\
  &\qquad\quad\quad <C_T(1+(t-s)^{-\frac{1}{2}})
   (\|f\|_{1,p,\Omega}+\||x|\cdot \nabla f\|_{p,\Omega}),\quad (t,s)\in \widetilde{\Lambda}_T ,\,f\in Y_\Omega.
  \end{align}

Let us now prove that for every $f\in Y_\Omega$ and for every $s\geq 0$ fixed, the map $t\mapsto T_{\Omega,b}(t,s)f$ is differentiable on $(s,\infty)$ and that \eqref{eq:derivative_evolution_exterior} holds. For $f\in Y_\Omega$ we compute
\begin{align*}
\partial_t T_0(t,s)f & = A_{\Omega,b}(t) T_0(t,s) f - \P_\Omega F(t,s)f\\
\partial_t T_1(t,s)f & = A_{\Omega,b}(t) T_1(t,s) f + \P_\Omega F(t,s)f - \int_s^t \P_\Omega F(t,r)\P_\Omega F(r,s)f \d r\\
\partial_t T_2(t,s)f & = A_{\Omega,b}(t) T_2(t,s) f +  \int_s^t \P_\Omega F(t,r)\P_\Omega F(r,s)f \d r \\
& \qquad - \int_s^t \int_{r_1}^t \P_\Omega F(t,r_2)\P_\Omega F(r_2,r_1)\P_\Omega F(r_1,s)f \d r_2 \d r_1.
\end{align*}
Inductively we see that
\begin{equation}\label{eq:derivative}
\partial_t \sum_{k=0}^n T_k(t,s)f  = A_{\Omega,b}(t) \sum_{k=0}^n T_k(t,s)f - R_n(t,s)f
\end{equation}
holds for $n\in \N$, where
\begin{equation*}
R_n(t,s)f := \int_s^t \int_{r_1}^t \ldots \int_{r_{n-1}}^t \P_\Omega F(t,r_n)\P_\Omega F(r_n,r_{n-1})\ldots \P_\Omega F(r_1,s)f\,\d r_n \ldots \d r_2 \d r_1.
\end{equation*}

We estimate now the norm of $R_n(t,s)f$. By Lemma \ref{lemma:error_terms} we obtain
\begin{align*}
\|R_1(t,s)f\|_{p,\Omega} &\leq C^2 \int_s^t (t-r)^{-\frac \gamma 2}(r -s)^{-\frac \gamma 2} \d r \|f\|_{p,\Omega}= C^2 \B(1-\gamma/2,1-\gamma/2)(t-s)^{1-\gamma}\|f\|_{p,\Omega},\\
\|R_2(t,s)f\|_{p,\Omega} & \leq C^3 \B(1-\gamma/2,1-\gamma/2)\int_s^t (t-r)^{1-\gamma}(r-s)^{-\frac{\gamma}{2}}\d r \|f\|_{p,\Omega}\\
& = C^3\B(1-\gamma/2,1-\gamma/2)\B(1-\gamma/2,2-\gamma)(t-s)^{2-\frac{3\gamma}{2}}\|f\|_{p,\Omega}.
\end{align*}
Inductively we see that
\begin{align}\label{eq:correction_terms}
\|R_n(t,s)f\|_{p,\Omega} &\leq C^{n+1}\B(1-\gamma/2,1-\gamma/2)\B(1-\gamma/2,2-\gamma)\ldots\notag\\
&\qquad\qquad  \ldots \B(1-\gamma/2, n-(n\gamma)/2)(t-s)^{n-\frac{(n+1)\gamma}{2}}\|f\|_{p,\Omega}\nonumber\\
& \leq \frac{C^{n+1}\Gamma(1-\gamma/2)^n}{\big[n-\frac{(n+1)\gamma}{2}\big]!}(t-s)^{n-\frac{(n+1)\gamma}{2}}\|f\|_{p,\Omega}
\end{align}
holds for $n\in \N$. Here the constant $C$ may change from line to line. From estimate (\ref{eq:correction_terms}) it follows that $\|R_n(t,s)\|_p$ tends to zero as $n\rightarrow \infty$. Since we used Lemma \ref{lemma:iterated}, we know that
 the convergence of $\sum_{k=0}^\infty T_k(t,s)f$ is uniform in $\Lambda_T$ for any $f\in Y_\Omega$ and any $T>0$ and so, by using
 (\ref{conv-in-Y_Omega}) and the closedness of $A_{\Omega,b}(t)$ we can conclude that
\begin{equation*}
\frac{\partial}{\partial t} \sum_{k=0}^{\infty} T_k(t,s)f  = A_{\Omega,b}(t) \sum_{k=0}^{\infty} T_k(t,s)f
\end{equation*}
 holds and this proves (\ref{eq:derivative_evolution_exterior}).

Let us show now the differentiability of the map $s\mapsto T_{\Omega,b}(t,s)f$ on $[0,t)$ for $t>0$ and $f\in Y_\Omega$. First we
note by a short calculation that for $f\in \D(L_{\Omega,b}(s))$
\begin{align*}
L_{\R^d}(s)f_R &= (L_{\Omega,b} (s)f)_R + 2 \nabla f \cdot (\nabla \xi)^* + (\Delta \xi + (M(s)x + c(s))\cdot \nabla \xi) f \\
&\qquad\quad - L_{\R^d}(s)\BB_1((\nabla\xi) \cdot f) + \BB_1((\nabla\xi)\cdot L_{\Omega,b}(s)f)
\end{align*}
 and
 \begin{align*}
 L_{D,b}(s)f_D & = (L_{\Omega,b}(s) f)_D + 2 \nabla f \cdot (\nabla \eta)^* + (\Delta \eta + (M(s)x + c(s))\cdot \nabla \eta) f \\
 &\qquad\quad - L_{D,b}(s)\BB_3((\nabla\eta) \cdot f) + \BB_3((\nabla\eta)\cdot L_{\Omega,b}(s)f).
 \end{align*}
 In the following we set $\nabla \tilde\p := (\Id-\P_\Omega)L_{\Omega,b}(s)f$. Then we can conclude that for any
 $f\in\D(A_{\Omega,b}(s))$ we have
 \begin{align*}
 \P_{\R^d}(L_{\Omega,b} (s)f)_R &= \P_{\R^d}\left[\xi A_{\Omega,b}(s)f - \BB_1((\nabla \xi)A_{\Omega,b}(s)f)+\xi\nabla\tilde \p - \BB_1((\nabla\xi)\nabla \tilde \p) \right]\\
 & = (A_{\Omega,b}f)_R - \P_{\R^d}[\tilde \p \nabla \xi  + \BB_1((\nabla \xi)\nabla \tilde \p)],
 \end{align*}
 and analogously
 \begin{align*}
 \P_{D}(L_{\Omega,b} (s)f)_D &= (A_{\Omega,b}f)_D - \P_{D}[\tilde \p \nabla \eta  + \BB_3((\nabla \eta)\nabla \tilde \p)].
 \end{align*}

 Thus, for $f\in Y_\Omega$, we obtain
 \begin{align*}
 \frac{\partial}{\partial s} W(t,s)f & = -\varphi T_{\R^d}(t,s)A_{\R^d}(s) f_R - (1-\varphi)T_{D,b}(t,s)A_{D,b}(s)f_D\\
 &\qquad \quad + \BB_2((\nabla \varphi)\cdot (T_{\R^d}(t,s)A_{\R^d}(s)f_R-T_{D,b}(t,s)A_{D,b}(s)f_D))\\[0.15cm]
 & = - W(t,s)A_{\Omega,b}(s)f - G(t,s)f
 \end{align*}
 where
 \begin{align*}
 G(t,s)f := &\; \varphi T_{\R^d}(t,s)\P_{\R^d}\left[2 \nabla f \cdot (\nabla \xi)^* + (\Delta \xi + (M(s)x + c(s))\cdot \nabla \xi) f \right.\\
&\left. \qquad\qquad - L_{\R^d}(s)\BB_1((\nabla\xi) \cdot f) + \BB_1((\nabla\xi)\cdot L_{\Omega,b}(s)f) - \tilde \p \nabla \xi  - \BB_1((\nabla \xi)\nabla \tilde \p)\right]\\
& \;+ (1-\varphi) T_{D,b}(t,s)\P_D\left[2\nabla f \cdot (\nabla \eta)^* + (\Delta \eta + (M(s)x + c(s))\cdot \nabla \eta)f \right.\\
 &\left. \qquad\qquad - L_{D,b}(s)\BB_3 ((\nabla\eta) \cdot f) + \BB_3((\nabla\eta)\cdot L_{\Omega,b}(s)f)-\tilde \p \nabla \eta  - \BB_3((\nabla \eta)\nabla \tilde \p)\right]\\
 &\; - \BB_2\left\{ (\nabla \varphi) \cdot T_{\R^d}(t,s)\P_{\R^d}\left[2 \nabla f \cdot (\nabla \xi)^* + (\Delta \xi + (M(s)x + c(s))\cdot \nabla \xi) f\right. \right. \\
&\left. \left. \qquad\qquad - L_{\R^d}(s)\BB_1((\nabla\xi) \cdot f) + \BB_1((\nabla\xi)\cdot L_{\Omega,b}(s)f) - \tilde \p \nabla \xi  - \BB_1((\nabla \xi)\nabla \tilde \p)\right]  \right\}\\
 & \; + \BB_2\left\{(\nabla \varphi) \cdot T_{D,b}(t,s)\P_D\left[2\nabla f \cdot (\nabla \eta)^* + (\Delta \eta + (M(s)x + c(s))\cdot \nabla \eta)f \right. \right. \\
 &\left. \left. \qquad\qquad - L_{D,b}(s)\BB_3 ((\nabla\eta) \cdot f) + \BB_3((\nabla\eta)\cdot L_{\Omega,b}(s)f) - \tilde \p \nabla \eta  - \BB_3((\nabla \eta)\nabla \tilde \p)\right]\right\}

 \end{align*}
 We estimate now the norm of $\tilde p$ in $L^p(D)$ since in the expression of $G(t,s)f$ the supports of the functions
 $\tilde p\nabla \xi$ and $\tilde p\nabla \eta$ are subsets of $D$.

Analogously to the proof of Lemma \ref{lem:pressure} we can show that for any $\varphi \in C_c^\infty(D)$
\begin{eqnarray*}
\left|\langle \tilde p ,\varphi \rangle_D\right| &\le &C
\|f \|_{\gamma,p,D}\|\varphi \|_{p',D}\\
&\le &C
\|f \|_{\gamma,p,\Omega}\|\varphi \|_{p',D}
\end{eqnarray*} and hence
$$\|\tilde p\|_{p,D}\le C\|f\|_{\gamma,p,\Omega}$$
holds, where $\gamma \in (1+\frac 1 p , 2)$. Now, using the same arguments as in the proof of Lemma \ref{lemma:error_terms} we obtain that
 $$
 \|G(t,s)f\|_{\gamma,p,\Omega}\leq C (t-s)^{-\frac {\gamma}{2}}\|f\|_{\gamma,p,\Omega}
 $$
 for some constant $C>0$ and $f\in H^{\gamma,p}(\Omega)^d\cap L^p_\sigma(\Omega)$. Now we apply Lemma \ref{lemma:iterated}  with $X_1=X_2=H^{\gamma,p}(\Omega)^d\cap L^p_\sigma(\Omega)$, $R=S=G$ and $\alpha=\beta=-\tfrac{\gamma}{2}$. So, the series
 \begin{equation*}
V(t,s)f := \sum_{k=0}^{\infty} V_k (t,s)f, \qquad (t,s)\in \widetilde{\Lambda},
\end{equation*}
is well-defined and
\begin{equation}\label{eq-22}
\|V(t,s)f\|_{\gamma,p,\Omega}\le C(t-s)^{-\frac{\gamma}{2}}\|f\|_{\gamma,p,\Omega},\quad (t,s)\in \widetilde{\Lambda}_T,
\end{equation}
 for $f\in H^{\gamma,p}(\Omega)^d\cap L^p_\sigma(\Omega)$. On the other hand, $V(\cdot,\cdot)$ satisfies the integral equation
\begin{equation}\label{eq:integral}
V(t,s)f = G(t,s)f + \int_s^t V(t,r)G(r,s)f \d r, \qquad (t,s)\in \widetilde{\Lambda},\,f\in H^{\gamma,p}(\Omega).
\end{equation}
In particular $V(t,\cdot)f$ is continuous on $[0,t)$ with respect to the $L^p$-norm for any $f\in H^{\gamma,p}(\Omega)^d\cap L^p_\sigma(\Omega)$ and
$t>0$. Now, for $f\in L^p_\sigma(\Omega)$ and $(t,s)\in \Lambda$ we set
\begin{equation*}
\tilde T(t,s)f := W(t,s)f+\int_s^t V(t,r)W(r,s)f \d r.
\end{equation*}
It follows from the continuity of $V(t,\cdot)W(\cdot ,s)f$ on $[s,t)$, Proposition \ref{prop:evolution_whole}, Proposition \ref{prop:evolution_bounded} and (\ref{eq-22}) that the above integral is well-defined for any $f\in L^p_\sigma(\Omega)$.
Computing the derivative with respect to $s\in [0,t)$ yields
\begin{align*}
\frac{\partial}{\partial s} \tilde T(t,s)f &= - W(t,s)L_\Omega(s)f - G(t,s)f+ V(t,s)f - \int_s^t V(t,r) W(r,s)L_\Omega(s) f \d r \\
& \qquad \quad - \int_s^t V(t,r)G(r,s)f \d r  \\[0.2cm]
& = - \tilde T(t,s)L_\Omega(s)f ,
\end{align*}
for any $f\in Y_\Omega$, due to (\ref{eq:integral}). From this equality together with (\ref{eq:derivative_evolution_exterior}) and since
$T_{\Omega,b}(t,s)Y_\Omega \subset Y_\Omega,\,(t,s)\in \Lambda$ we can conclude that
\begin{align*}
\frac{\partial}{\partial r} (\tilde T(t,r)T_{\Omega,b}(r,s)f)= 0
\end{align*}
holds for all $f\in Y_\Omega$. This yields that for $f\in Y_\Omega$, the function $\tilde T(t,r)T_{\Omega,b}(r,s)f$ is constant on $\Lambda_T $ and thus, by the density of $Y_\Omega$ in $L^p_\sigma(\Omega)$ and by the fact that $T>0$ was arbitrary, it follows that $\tilde T(t,s)f=T_{\Omega,b}(t,s)f$ holds for all $f\in L^p_\sigma(\Omega)$, $(t,s)\in \Lambda$. This proves  (\ref{eq:derivative_evolution_exterior_2}).

The uniqueness of the evolution system $\{T_{\Omega,b}(t,s)\}_{(t,s)\in \Lambda}$ follows by a similar argument from \eqref{eq:derivative_evolution_exterior} and \eqref{eq:derivative_evolution_exterior_2} (see the proof of Theorem 3.1 in \cite{Hansel/Rhandi:2010} for details).

The estimate in (d) can be obtained by applying Lemma \ref{lemma:iterated} with $X_1=W^{1,p}(\Omega),\,X_2=L^p_\sigma(\Omega),\,R=W,\,S=\P_\Omega F,\,\alpha =-\frac{1}{2},\,\beta=-\frac{\gamma}{2}$, Proposition \ref{prop:evolution_whole}  and Corollary  \ref{prop:Lp_Lq_estimates_bounded}.

Finally, the first estimate in (e) follows by applying Lemma \ref{lemma:iterated} with $X_1=L^q_\sigma(\Omega),\,X_2=L^p_\sigma(\Omega),\,R=W,\,S=\P_\Omega F,\,\alpha=
-\frac{d}{2}\left(\frac 1 p - \frac 1 q\right),\,\beta =-\frac{\gamma}{2}$,
Proposition \ref{prop:evolution_whole} and Corollary \ref{prop:Lp_Lq_estimates_bounded} if $\frac{1}{p}-\frac{1}{q}\le \frac{2}{d}$.
The case $\frac{1}{p}-\frac{1}{q}> \frac{2}{d}$ can be obtained by iteration as in the proof of Corollary \ref{prop:Lp_Lq_estimates_bounded}.
The second estimate follows now from the first one and (d) (see the proof of Corollary  \ref{prop:Lp_Lq_estimates_bounded}).
 \end{proof}

To conclude this section let us state the following lemma about the behavior of $T_{\Omega,b}(t,s)$ near $t=s$, which will be needed in the proof of Theorem \ref{thm:Kato}.

\begin{lemma}\label{lem5.3}
Let $s\geq0$ and $f\in L^p_{\sigma}(\Omega)$. Then we have
\begin{enumerate}
\item[(a)] for $1< p<q<\infty$
$$
	\lim\limits_{t\to s,\ t> s}(t-s)^{\frac d 2 \left(\frac 1 p - \frac 1 q\right)} \|T_{\Omega,b}(t,s)f\|_{q,\Omega}
	=0,$$
\item[(b)] for $1<p<\infty$
$$	\lim\limits_{t\to s,\ t> s}
(t-s)^{\frac 1 2 } \|\nabla T_{\Omega,b}(t,s)f\|_{p,\Omega}=0.$$
\end{enumerate}
\end{lemma}

\begin{proof}
Let $f\in L^p_\sigma(\Omega)$, $0\le t-s \leq 1$ and choose $(f_n)_{n\in\N}
\subset C_{c,\sigma}^{\infty}(\Omega) \subset L_{\sigma}^p(\Omega)$,
such that $\lim_{n\to\infty}\|f-f_n\|_{p,\Omega}=0$. The triangle inequality
together with the $L^p$-$L^q$ estimates stated in Theorem \ref{thm:evolution_system_exterior}
imply that there exist constants $C_1, C_2>0$ such that
\begin{align*}
&(t-s)^{\frac d 2 \left(\frac 1 p - \frac 1 q \right)}
\|T_{\Omega ,b}(t,s)f\|_{q,\Omega}\\
&\qquad \leq (t-s)^{\frac d 2 \left(\frac 1 p - \frac 1 q \right)}
\|T_{\Omega ,b}(t,s)(f-f_n)\|_{q,\Omega} + (t-s)^{\frac d 2 \left(\frac 1 p - \frac 1
q
\right)} \|T_{\Omega ,b}(t,s)f_n\|_{q,\Omega}\\
&\qquad \leq  C_1 \|f-f_n\|_{p,\Omega} + C_2 (t-s)^{\frac d 2 \left(\frac 1 p -
\frac 1 q \right)} \|f_n\|_{q,\Omega},\quad 0\leq t-s\leq 1,\
n\in\N.
\end{align*}
Hence,
$
	\lim\limits_{t\to s}(t-s)^{\frac d 2 \left(\frac 1 p - \frac 1 q
	\right)} \|T_{\Omega ,b}(t,s)f\|_{q,\Omega}=0
$
by letting first $t\rightarrow s$ and then $n\rightarrow \infty$.
The second assertion is proved in a similar
way (cf. \cite[Proposition 3.4]{Hansel:2009}).
\end{proof}

\section{Mild solutions to the nonlinear problem}

In this section we finally come back to the nonlinear problem \eqref{eq:NS_5} and its abstract formulation \eqref{eq:NS_abstract}. Based on the results proved in Section \ref{sect:exterior} and an adaptation of the Kato iteration procedure (see \cite{Kato:1984}) we now prove Theorem \ref{thm:Kato}.

\begin{proof}[Proof of Theorem  \ref{thm:Kato}.]
Let $d\le p<q$, $f\in L^p_\sigma(\Omega)$ and $T>0$. For $0<t\le T$ and $k\in \N$ we define
$u_1(t):=T_{\Omega ,b}(t,0)f$ and
$$u_{k+1}(t)=u_1(t)-\int_0^tT_{\Omega ,b}(t,s)\P_\Omega (u_k\cdot \nabla u_k)(s)\,\d s+
\int_0^tT_{\Omega ,b}(t,s)\P_\Omega F_1(s)\,\d s.$$
Let us prove that, for some $T_0>0$, the sequence $(u_k)_k$ converges in $C([0,T_0); L^p_\sigma(\Omega))$ to a mild solution $u$ of (\ref{eq:NS_abstract}) for $T\in (0,T_0)$.

We set $\beta :=\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)$ and define
$$L_k:=L_k(T):=\sup_{t\in (0,T]}t^\beta \|u_k(t)\|_{q,\Omega},\,\,L'_k:=L'_k(T):=\sup_{t\in (0,T]}t^{\frac{1}{2}}\|\nabla u_k\|_{p,\Omega}$$ and
$$M_k:=M_k(T):=\sup_{t\in (0,T]}t^\beta \|u_{k+1}(t)-u_k(t)\|_{q,\Omega},\,M'_k:=M'_k(T):=
\sup_{t\in (0,T]}t^{\frac{1}{2}}\|\nabla u_{k+1}(t)-\nabla u_k(t)\|_{p,\Omega}$$ for $k\in \N$.

It follows from the $L^p$-$L^q$ estimates (\ref{Lpq}) and the boundedness of $\P_\Omega$ from $L^r(\Omega)^d$ into $L^r_\sigma(\Omega)$ that
\begin{eqnarray*}
\|u_{k+1}(t)\|_{q,\Omega} &\le & \|u_1(t)\|_{q,\Omega}+\int_0^t\left\|T_{\Omega ,b}(t,s)\P_\Omega (u_k\cdot \nabla u_k)(s)\right\|_{q,\Omega}\,\d s\\
& & \quad +\int_0^t\left\|T_{\Omega ,b}(t,s)\P_\Omega F_1(s)\right\|_{q,\Omega}\,\d s\\
&\le & t^{-\beta}L_1+C\int_0^t(t-s)^{-\frac{d}{2}\left(\frac{1}{r}-\frac{1}{q}\right)}\|u_k(s)\cdot \nabla u_k(s)\|_{r,\Omega}\,\d s\\
& & \quad +C\int_0^t\|F_1(s)\|_{q,\Omega}\,\d s\\
&\le & t^{-\beta}L_1+C\int_0^t(t-s)^{-\frac{d}{2}\left(\frac{1}{r}-\frac{1}{q}\right)}\|u_k(s)\|_{q,\Omega}\|\nabla u_k(s)\|_{p,\Omega}\,\d s\\
& & \quad +C\int_0^t\|F_1(s)\|_{q,\Omega}\,\d s\\
&\le & t^{-\beta}L_1+Ct^{-\beta}L_kL'_k\int_0^t(t-s)^{-\frac{d}{2p}}t^{\beta}s^{-\beta -\frac{1}{2}}\,\d s+
Ct^{-\beta}\int_0^t t^{\beta}\|F_1(s)\|_{q,\Omega}\,\d s,
\end{eqnarray*}
where $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$.
Since $\beta <\frac{1}{2}$ we obtain, by multiplying with $t^\beta$,
$$t^\beta \|u_{k+1}(t)\|_{q,\Omega} \le  L_1+C_1L_kL'_k+C_2T,$$
and taking the supremum over $t\in (0,T]$ yields
\begin{equation}\label{suite L}
L_{k+1}\le L_1+C_1L_kL'_k+C_2T,
\end{equation}
where $C_1,\,C_2>0$ are constants independent of $k\in \N$ but depend on $T$. Similarly, with the gradient $L^p$-$L^q$ estimates (\ref{nabla-Lpq}) we have
\begin{eqnarray}\label{eq6.2}
\|\nabla u_{k+1}(t)\|_{p,\Omega} &\le & t^{-\frac{1}{2}}L'_1+C\int_0^t(t-s)^{-\frac{d}{2}\left(\frac{1}{r}-\frac{1}{p}\right)-\frac{1}{2}}\|u_k(s)\cdot \nabla u_k(s)\|_{r,\Omega}\,\d s \nonumber \\
& & \quad +C\int_0^t(t-s)^{-\frac{1}{2}}\|F_1(s)\|_{p,\Omega}\,\d s \nonumber \\
&\le & t^{-\frac{1}{2}}L'_1+C\int_0^t(t-s)^{-\frac{d}{2q}-\frac{1}{2}}\|u_k(s)\|_{q,\Omega}\|\nabla u_k(s)\|_{p,\Omega}\,\d s\\
& & \quad +C\int_0^t(t-s)^{-\frac{1}{2}}\|F_1(s)\|_{p,\Omega}\,\d s \nonumber
\end{eqnarray}
and hence, as above, we obtain
\begin{equation}\label{suite L'}
L'_{k+1}\le L'_1+C_3L_kL'_k+C_4T,
\end{equation}
where $C_3,\,C_4>0$ are constants independent of $k\in \N$ but depend on $T$.
By setting $R_k:=R_k(T):=\max\{L_k,L'_k\}$, it follows from (\ref{suite L}) and (\ref{suite L'}) that $R_{k+1}\le R_1+c_1R_k^2+c_2T$ for some constants $c_1,\,c_2\ge 1$. By Lemma \ref{lem5.3} for any $\varepsilon >0$, there is $\widetilde{T}_0>0$ such that $R_1<\varepsilon$ for $T<\widetilde{T}_0$.
So, for $\varepsilon \le \frac{1}{8c_1}$, it follows by induction that $$R_k\le 4\varepsilon ,\quad k\in \N,\,T<\min\left(\widetilde{T}_0,\frac{\varepsilon}{c_2}\right)=:T_0.$$ Thus, the sequences
\begin{equation}\label{suites}
(t\mapsto t^\beta u_k(t))_k\quad \hbox{\ and}\quad (t\mapsto t^{\frac{1}{2}}\nabla u_k(t))_k
\end{equation}
 are uniformly bounded in
$L^q_\sigma(\Omega)$ and $L^p(\Omega)^{d\times d}$ respectively for $t\le T$ and all $k\in \N$. The continuity of the maps
$t\mapsto t^\beta u_1(t)$ and $t\mapsto t^{\frac{1}{2}}\nabla u_1(t)$ at $t=0$ follows from Lemma \ref{lem5.3}. Hence the continuity
of $t\mapsto t^\beta u_k(t)$ and $t\mapsto t^{\frac{1}{2}}\nabla u_k(t)$ follows from similar calculations.

We prove now that the sequences in (\ref{suites}) are Cauchy sequences. Since
$$u_k\cdot \nabla u_k-u_{k-1}\cdot \nabla u_{k-1}=u_k\cdot \nabla (u_k-u_{k-1})+(u_k-u_{k-1})\cdot \nabla u_{k-1}$$ and by the same computations as above we obtain
\begin{eqnarray*}
\|u_{k+1}(t)-u_k(t)\|_{q,\Omega} &\le & \int_0^t \left\|T_{\Omega ,b}(t,s)\P_\Omega\left((u_k\cdot \nabla u_k)(s)-(u_{k-1}\cdot \nabla u_{k-1})(s)\right)\right\|_{q,\Omega}\,\d s\\
&\le & C\int_0^t (t-s)^{-\frac{d}{2p}}\left(\|u_k(s)\|_{q,\Omega}\|\nabla (u_k(s)-u_{k-1}(s))\|_{p,\Omega}\right. \\
& & \quad \quad \left. +\|u_k(s)-u_{k-1}(s)\|_{q,\Omega}\|\nabla u_{k-1}(s)\|_{p,\Omega}\right)\,\d s
\end{eqnarray*}
and
\begin{eqnarray*}
\|\nabla u_{k+1}(t)-\nabla u_k(t)\|_{p,\Omega} &\le & \int_0^t \left\|\nabla T_{\Omega ,b}(t,s)\P_\Omega\left((u_k\cdot \nabla u_k)(s)-(u_{k-1}\cdot \nabla u_{k-1})(s)\right)\right\|_{p,\Omega}\,\d s\\
&\le & C\int_0^t (t-s)^{-\frac{d}{2q}-\frac{1}{2}}\left(\|u_k(s)\|_{q,\Omega}\|\nabla (u_k(s)-u_{k-1}(s))\|_{p,\Omega}\right. \\
& & \quad \quad \left. +\|u_k(s)-u_{k-1}(s)\|_{q,\Omega}\|\nabla u_{k-1}(s)\|_{p,\Omega}\right)\,\d s.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
M_k &\le & C_5(M'_{k-1}L_k+M_{k-1}L'_{k-1})\le 2C_5R_1(M'_{k-1}+M_{k-1}),\\
M'_k &\le & C_6(M'_{k-1}L_k+M_{k-1}L'_{k-1})\le 2C_6R_1(M'_{k-1}+M_{k-1}),
\end{eqnarray*}
and hence,
\begin{eqnarray*}
(M_k+M'_k) &\le & 4\varepsilon (C_5+C_6)(M_{k-1}+M'_{k-1})\\
&\le & \frac{1}{2}(M_{k-1}+M'_{k-1})
\end{eqnarray*}
for $\varepsilon \le \min\left(\frac{1}{8(C_5+C_6)},\frac{1}{8c_1}\right)$ and $T<T_0$,
where $C_5,\,C_6>0$ are constants independent of $k\in \N$ but depend on $T$.
Thus,
$(t\mapsto t^\beta u_k(t))_k$ converges to some $t\mapsto t^\beta u(t)\in C([0,T], L^q_\sigma(\Omega))$ and
$(t\mapsto t^{\frac{1}{2}}\nabla u_k(t))_k$ converges to some $t\mapsto t^{\frac{1}{2}}v\in C([0,T], L^p(\Omega)^{d\times d})$. It is now clear that $v(t)=\nabla u(t)$ and $u$ is a mild solution of \eqref{eq:NS_abstract} on $[0,T]$.

The case $d<p=q$ follows by taking $\beta =\frac{1-\delta}{2}$ for $\delta \in (0,1)$ in the above computations. Here we need
$\frac{d}{2q}+\frac{1}{2}<1$ to make the integral in (\ref{eq6.2}) convergent.

Moreover, take any $p\ge d$ and any $r>p$ and using the same estimates as for $\|u_{k+1}(t)\|_{q,\Omega}$ above, we obtain
\begin{eqnarray*}
\|u(t)\|_{p,\Omega} &\le & \|T_{\Omega ,b}(t,0)f\|_{p,\Omega}+\int_0^t \left\|T_{\Omega ,b}(t,s)\P_\Omega (u\cdot \nabla u)(s)\right\|_{p,\Omega}\,\d s+CT\\
&\le & C\|f\|_{p,\Omega}+C(\sup_{t\in [0,T]}t^\beta \|u(t)\|_{r,\Omega})(\sup_{t\in [0,T]}t^{\frac{1}{2}} \|\nabla u(t)\|_{p,\Omega})\int_0^t (t-s)^{-\frac{d}{2r}}s^{-\beta -\frac{1}{2}}\,\d s\\
& & \quad \quad \quad \quad +CT
\end{eqnarray*}
and hence $\sup_{t\in [0,T]} \|u(t)\|_{p,\Omega}<\infty$. The continuity can be obtained similarly. Thus, $u\in C([0,T],L^p_\sigma(\Omega))$.

The case $d=p=q$ can be deduced as in the case $d<p=q$ by taking $\beta=\frac{1-\delta}{2}$ for $\delta \in (0,1)$ and making the some computations but with a $q'>d$ instead of $q=d$, since if $q=d$ then the integral in (\ref{eq6.2}) could diverge.

The property (\ref{eq:mild_solution_prop3}) follows from the construction of the solution $u$ and
Lemma \ref{lem5.3}.
Finally the uniqueness follows as in \cite{Hansel:2009}.
\end{proof}

\section*{Acknowledgments}

Parts of the paper were written while the first author stayed at Waseda University in Tokyo. He would like to express his thank to Professor Y. Shibata for the kind hospitality and for fruitful discussions.

The first author acknowledges the financial support of the DFG International Research Training Group 1529 on \emph{Mathematical Fluid
Dynamics} at TU Darmstadt and of the University of Salerno where the paper originated.

The second author acknowledges the financial support of the M.I.U.R. research project
Prin 2008 ``Metodi deterministici e stocastici nello studio di problemi di evoluzione''.

config: arxiv


TITLE: Evaluate $\int_0^{\infty}\frac{\ln x}{x^a(x+1)}dx$ where $0<a<1$
QUESTION [2 upvotes]: I'm trying to compute this integral, $$\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx \hbox{ where } 0<a<1$$
I drew a typical Pacman contour with branch cut at positive real axis. Then, we have $$\int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz=\left(\int_{L_{1}}+\int_{L_{2}}+\int_{C_{R}}+\int_{C_{\epsilon}}\right)\frac{\ln z}{z^{a}(z+1)}dz$$
The function has a pole at $z=-1$. Then, by residue theorem, the whole integral becomes
\begin{align*}
\int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz&=2\pi i\text{res}(\frac{\ln z}{z^{a}(z+1)};-1)\\&=2\pi i\lim_{z\rightarrow-1}(z+1)\frac{\ln z}{z^{a}(z+1)}\\&=2\pi i\lim_{z\rightarrow-1}\frac{\ln z}{z^{a}}\\&=2\pi i\frac{\ln e^{i\pi}}{e^{i\pi a}}\\&=2\pi i\frac{\ln1+i\pi}{e^{i\pi a}}\\&=-\frac{2\pi^{2}}{(e^{i\pi})^{a}}
\end{align*}
On $L_1$, $z=xe^{i\epsilon}$ where $\epsilon\leq x\leq R$
On $L_{2}$, $z=xe^{i(2\pi-\epsilon)}$ where $\epsilon\leq x\leq R$
On $C_{R}$, $z=Re^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$
Lastly on On $C_{\epsilon}$, $z=\epsilon e^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$
So I tried to compute them separately and I anticipate the curves will both go to $0$ by ML inequality, and two linear integrals will consist of the term $\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx$ and hence simply solve for it. But I'm having trouble physically solving it out. I think the log is throwing me off. Can anybody please help me?

REPLY [1 votes]: We seek to compute using contour integration the integral
$$J = \int_0^\infty \frac{\ln x}{x^a (1+x)} \; dx$$
where $0\lt a\lt 1.$ We work with
$$f(z) = \frac{\mathrm{Log}(z)}
{\exp(a\mathrm{Log}(z)) (1+z)}$$
where $\mathrm{Log}(z)$ is the branch  with argument in $[0,2\pi).$ We
use a keyhole contour wih radius $R$ and the slot on the positive real
axis.  Let $\Gamma_0$  be the  segment  on the  real axis  up to  $R$,
$\Gamma_1$ the big circle of  radius $R$, $\Gamma_2$ the segment below
the real  axis coming  in from  $R$ and  finally $\Gamma_3$  the small
circle of radius $\epsilon$ enclosing the origin. We then have
$$\left(\int_{\Gamma_0}
+ \int_{\Gamma_1}+ \int_{\Gamma_2}+ \int_{\Gamma_3}\right) f(z) \; dz
= 2\pi i \times \mathrm{Res}_{z=-1} f(z)
= 2\pi i \times \frac{\pi i}{\exp(a\pi i)}$$
Observe that in the limit
$$\int_{\Gamma_0}  f(z) \; dz =
\int_0^\infty \frac{\ln x}{x^a (1+x)} \; dx = J.$$
For $\Gamma_1$ we have by  ML estimate $\lim_{R\to\infty} 2\pi R \times
(\log R + 2\pi) / R^a / (R-1) $ or $\lim_{R\to\infty} 2\pi \times
(\log R + 2\pi) / R^a + \lim_{R\to\infty} 2\pi \times
(\log R + 2\pi) / R^a /(R-1) = 0,$ so this vanishes.
Furthermore for $\Gamma_2$ in the  limit (we get $\mathrm{Log}(z) =
\log |z| + 2\pi i$)
$$\int_{\Gamma_2}  f(z) \; dz =
\int_\infty^0 \frac{\ln x + 2\pi i}
{\exp(2a\pi i) x^a(1+x)} \; dx
\\ = - \exp(-2a\pi i) J
- \exp(-2a\pi i) 2\pi i \int_0^\infty \frac{1}{x^a(1+x)} \; dx
\\ = - \exp(-2a\pi i) J
- \exp(-2a\pi i) 2\pi i K.$$
For $\Gamma_3$  we again use  ML and  get $\lim_{\epsilon\to 0}  2 \pi
\epsilon \times (|\log\epsilon| +2\pi) / \epsilon^a / (1-\epsilon)$ which is
$\lim_{\epsilon\to  0} 3/2 \times  2  \pi  \epsilon^{1-a}   \times  (|\log\epsilon|
+2\pi)  = 0$, so this  too vanishes. We have  shown that in
the limit
$$2\pi i \times \frac{\pi i}{\exp(a\pi i)} =
J - \exp(-2a\pi i) J
- \exp(-2a\pi i) 2\pi i K.$$
This is
$$-2\pi^2 = 2 i J \sin(a\pi) - \exp(-a\pi i) 2\pi i K.$$
It seems we require $K.$ We use the same estimates and the same contour
to get
$$K (1-\exp(-2a\pi i)) = 2\pi i \times \exp(-a\pi i)$$
which is
$$K = 2\pi i \frac{\exp(-a\pi i)}{1-\exp(-2a\pi i)}
= 2\pi i \frac{1}{\exp(a\pi i)-\exp(-a\pi i)}
= \frac{\pi}{\sin(a\pi)}.$$
We then get
$$2i J\sin(a\pi) =
-2\pi^2 + \exp(-a\pi i)2\pi i \frac{\pi}{\sin(a\pi)}
\\ = 2\pi^2 \frac{-\sin(a\pi)+i\exp(-a\pi i)}{\sin(a\pi)}
\\ = 2\pi^2
\frac{-\exp(a\pi i)/2/i+\exp(-a\pi i)/2/i-\exp(-a\pi i)/i}
{\sin(a\pi)}
\\ = - 2\pi^2 \frac{\cos(a\pi)/i}{\sin(a\pi)}.$$
We thus have
$$J = - 2\pi^2 \frac{\cos(a\pi)/i}{\sin^2(a\pi)} \frac{1}{2i}$$
This is
$$\bbox[5px,border:2px solid #00A000]{
J = \pi^2 \frac{\cos(a\pi)}{\sin^2(a\pi)}.}$$