TITLE: Uniqueness in spectral theorem for commutative (non-unital) Banach $*$-algebras
QUESTION [1 upvotes]: $\require{AMScd}$I'm struggling to understand the uniqueness part of Theorem 1.54 in Folland's book on Harmonic analysis.
Theorem 1.54: Let $\mathcal{A}$ be a commutative Banach $*$-algebra (non necessarily unital) and let $\phi$ be a nondegenerate $*$-representation $\phi:\mathcal{A}\to \mathcal{L}(H)$ (the codomain is the set of bounded operators on a Hilbert space). There is a unique projection-valued measure $P$ on $\sigma(\mathcal{A})$ such that $\phi(x) = \int G_{\mathcal{A}}(x)dP
$ for all $x\in\mathcal{A}$. Here $G_{\mathcal{A}}(x)$ is the continuous function on the spectrum $\sigma(\mathcal{A})$ which is given by evaluation on $x$.
Attempt: (Uniqueness) Given two $P, Q$ satisfying the condition, we would have
$$
\int G_{\mathcal{A}}(x) dP_{u,v} = \int G_{\mathcal{A}}(x) dQ_{u,v}
$$
for all $u, v \in H$. In the case of (non-unital) $C^*$-algebras, one could use the fact that the $G_A(x)$ exhaust $C_0(\sigma(\mathcal{A}))$ and conclude equality of $P, Q$ by Riesz representation. However, this case is different since the set $\{G_A(x) : x\in \mathcal{A}\}$ might not be big. So how do I go about proving uniqueness?
I include the existence part of the theorem here, in case it helps. Let $\mathcal{B}$ denote the norm closure of $\phi(\mathcal{A})$, and let $\mathcal{\tilde{A}},\ \mathcal{\tilde{B}}$ denote the algebras with units attached. $\mathcal{\tilde{B}}$ is then $\mathcal{B}\oplus \mathbb{C}I$ and is a commutative unital $C^*$-algebra.
Moreover, the $*$-representatiom $\phi$ extends to give the diagram
$$
\begin{CD}
\mathcal{A} @>{\phi}>> \mathcal{B}  \\
@V{}VV @V{}VV \\
 \mathcal{\tilde{A}} @>{\phi}>> \mathcal{\tilde{B}}
\end{CD}  \qquad\quad (1)
$$
On identifying the spectrum of $\mathcal{\tilde{A}}$ with a subset of bounded linear functionals on $\mathcal{A}$, we can write $\sigma(\mathcal{\tilde{A}}) = \sigma(\mathcal{A})\cup \{0\}$ which,  if $\sigma(\mathcal{A})$ is noncompact, is the one-point compactification of $\sigma(\mathcal{A})$. Similarly, this holds for $\mathcal{B}$. Thus $(1)$ gives a dual diagram (the vertical arrows are inclusions and not the dual)
$$
\begin{CD}
\sigma(\mathcal{A}) @<{\phi^*}<< \sigma(\mathcal{B})  \\
@V{}VV @V{}VV \\
 \sigma(\mathcal{A})\cup \{0\} @<{\phi^*}<< \sigma(\mathcal{B})\cup \{0\}
\end{CD}  \qquad\quad (2)
$$
It follows from the definitions that the arrows are continuous injections. The bottom row is a homeomorphism onto the image. Taking duals again, we have the following commutative diagram of Gelfand transforms
$$
\begin{CD}
\mathcal{A} @>{\phi}>> \mathcal{B}  \\
@V{}VV @V{}VV \\
 \mathcal{\tilde{A}} @>{\phi}>> \mathcal{\tilde{B}}
\end{CD}  \qquad \overset{G_\cdot}{\longrightarrow}
\begin{CD}
C_0(\sigma(\mathcal{A})) @>{(\phi^*)\check{}}>> C_0(\sigma(\mathcal{B}))  \\
@V{}VV @V{}VV \\
 C(\sigma(\mathcal{A})\cup \{0\}) @>{(\phi^*)\check{}}>> C(\sigma(\mathcal{B})\cup \{0\})
\end{CD}
$$
Now the spectral theorem applies to $\mathcal{\tilde{B}}$ to give a unique projection-valued measure on Borel sets $Q:\text{Borel}(\sigma(\mathcal{B})\cup\{0\}) \to \mathcal{L}(H)$ such that for every $T\in \mathcal{\tilde{B}}$,
$$
T = \int G_\mathcal{\tilde{B}}(T) dQ.
$$
Now, we claim $Q(\{0\})=0$. Put $E=\{0\}$. If $u \in H$ is in the image of $Q(E)$, then we have, for any $S \in \mathcal{B}$ and $v\in H$, by the spectral theorem
$$(Su,v) = (SQ(E)u,v) = \int_{h\in E} G_{\mathcal{\tilde{B}}}(S)(h) dQ_{u,v}(h)=0.
$$
Since $\phi$ is non-degenerate, $Q(E)$ must be the zero operator.
Thus we have a projection-valued measure $R$ on $\sigma(\mathcal{B})$ uniquely defined by the property that, for every $T\in \mathcal{B}$ we have $$
T = \int G_{\mathcal{B}}(T) dR. \qquad (3)
$$
I guess the uniqueness here follows by the fact that $\{G_\mathcal{B}(T) : T\in \mathcal{B}\}=C_0(\sigma(\mathcal{B}))$.
Finally, we define a projection-valued measure $P$ on $\sigma(\mathcal{A})$ using the diagram
$$
\text{Borel}(\sigma(\mathcal{A})) \overset{(\phi^*)\check{}}{\longrightarrow} \text{Borel}(\sigma(\mathcal{B})) \overset{R}{\longrightarrow} \mathcal{L}(H)
$$
With this definition, it is easy to see that $P$ is the pushforward measure from $R$ by $\phi^*$. That is, $P_{u,v} = (\phi^*)_*R_{u,v}$.
To see that $P$ has the required property, take any $x\in\mathcal{A}, u, v \in H$ and observe that, by $(3)$, we have
\begin{equation}\begin{split}
(\phi(x)u,v) &= \int_{\sigma(\mathcal{B})}G_\mathcal{B}(\phi(x))(h)dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{B})} h(\phi(x)) dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{B})} \phi^*(h)(x) dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{B})} G_\mathcal{A}(x)(\phi^*(h)) dR_{u,v}(h)\\
&= \int_{\sigma(\mathcal{A})} G_\mathcal{A}(x)(l) d\left((\phi^*)_*R_{u,v}\right)(l)\\
&= \int_{\sigma(\mathcal{A})} G_\mathcal{A}(x)(l) dP_{u,v}(l)
\end{split}\end{equation}
Again, I don't see why such a measure should be unique. Please send help.

REPLY [1 votes]: The set of functions of the form $G_{\mathcal{A}}(x)$ is a $*$-subalgebra of $C_0(\sigma(\mathcal{A})$ that separates points from each other and from $0$.  So by Stone-Weierstrass, it is dense in $C_0(\sigma(\mathcal{A}))$.  It follows that the bounded functionals of the form $G_{\mathcal{A}}(x)\mapsto \int G_{\mathcal{A}}(x) dP_{u,v}$ extend uniquely to bounded functionals on all of $C_0(\sigma(\mathcal{A}))$, and you can apply the Riesz representation theorem.


TITLE: Is it equivalent to say that a formula is logically valid and that a formula follows from the axioms?
QUESTION [1 upvotes]: Lets say I have some predicate logic formula in Boolean algebra and I have to show whether the formula is logically valid. Can I show it by showing that the formula follows from the axioms of the Boolean algebra? Or can I only prove it for example by creating a semantic tree for negation of that formula and checking whether the formula is a contradiction and therefore not negated formula a logically valid formula?
Also, is it always the case that every axiom is logically valid? Why?

REPLY [1 votes]: No, the axioms of Boolean Algebra are not logically valid.
Logically valid means:

"true in every interpretation".

If they were, also the theorems (laws) of BA would be, and there are obvious laws of BA that are not true in every interpretation, like e.g.: $x+x=x$.
And yes, in order to prove that the above formula is a theorem of Boolean Algebra (i.e. that it follows form the axioms), you can apply the semantic tree proof procedure to the set $\mathsf {BA}$ of axioms and the negation of the law to be proved.


TITLE: Riemann Hypothesis: Is $1/2$ of critical line same as the $1/2$ of square-root accurately of error term of prime number theorem?
QUESTION [7 upvotes]: Here is a question about Riemann Hypothesis:
Is $1/2$ of critical line same as the $1/2$ of square-root accurately of error term of prime number theorem $?$
In other words, (just for some brain exercise), let's assume that critical line is at $x=3/4$, does this mean that the error term in prime number theorem will be proportional to the power of $3/4$ $?$ So $|\pi(x) - Li(x) |$ would be bounded by $x^{(3/4)}$ $?$
Which book has a simple proof of this $?$
Thank you.

REPLY [3 votes]: Yes.
Ingham, The Distribution of Prime Numbers, contains a proof that the lower bound $\Theta_1$ of numbers $\alpha$ for which it is true that
$$\psi(x)-x = O(x^{\alpha}) $$
is equal to the upper bound $\Theta_2$ of the real parts of the zeros of $\zeta(s).$
This is his Theorem 31 (with a proof) on page 84.


text:
\begin{document}

\title[Thermoelastic plate equation with free boundary conditions]
{Generation of semigroups for the thermoelastic plate equation with free boundary conditions}
\author{Robert Denk}
\address{Universit\"at Konstanz, Fachbereich f\"ur Mathematik und Statistik,
         78457 Konstanz, Germany}
\email{robert.denk@uni-konstanz.de}
\author{Yoshihiro Shibata}
\address{Department of Mathematical Sciences, School of Science and
Engineering, Waseda University, Ohkobu 3-4-1, Shinjuku-ku, Tokyo 169-8555,
Japan} \email{yshibata@waseda.jp}
\thanks{}
\date{June 26, 2018}

\begin{abstract}
We consider the linear thermoelastic plate equations with free boundary conditions in uniform $C^4$-domains, which includes the half-space, bounded and exterior domains. We show that the corresponding operator generates an analytic semigroup  in $L^p$-spaces for all $p\in(1,\infty)$ and has maximal $L^q$-$L^p$-regularity on finite time intervals. On bounded $C^4$-domains, we obtain exponential stability.
\end{abstract}

\subjclass[2010]{35K35; 35J40; 42B15}

\keywords{Thermoelastic plate equations; operator-valued Fourier multipliers; generation of analytic semigroups}

\maketitle

\section{Introduction}

Let $\Omega\subset\R^N$ be a domain with boundary $\Gamma$. We consider the linear thermoelastic plate equations
\begin{equation}
  \label{1-1}
  \begin{aligned}
    u_{tt} + \Delta^2 u + \Delta \theta & = f_1\quad \text{ in } (0,\infty)\times \Omega,\\
    \theta_t -\Delta \theta -\Delta u_t & = f_2 \quad \text{ in } (0,\infty)\times \Omega
  \end{aligned}
\end{equation}
with initial conditions
\begin{equation}\label{1-2}
\begin{aligned}
    u|_{t=0} & = u_0\quad\text{ in } \Omega,\\
    u_t|_{t=0} & = u_1\quad\text{ in } \Omega,\\
    \theta|_{t=0} & = \theta_0\quad\text{ in } \Omega.
  \end{aligned}
\end{equation}
System \eqref{1-1} serves as a standard simplified model for thin elastic plates with thermoelastic effects, see \cite{lagnese89}, Chapter~2, or \cite{chuesov-lasiecka10}, for a discussion of this and similar models. In \eqref{1-1},  $u(t,x)$ stands for the vertical displacement at time $t\ge 0$ and at position $x=(x_1,\dots,x_N)\in\Omega$, while $\theta(t,x)$ denotes the temperature (relative to some reference temperature) at time $t$ and position $x$. Note that we omitted all physical constants for simplicity.

Among the physically relevant boundary conditions, the maybe most complicated are the so-called free boundary conditions
\begin{equation}
  \label{1-3}
  \begin{aligned}
    \Delta u - (1-\beta) \Delta' u +\theta & = g_1 \quad \text{ on }(0,\infty)\times\Gamma,\\
    \partial_\nu \big( \Delta u + (1-\beta) \Delta' u + \theta\big) & = g_2 \quad\text{ on }(0,\infty)\times\Gamma,\\
    \partial_\nu\theta & = g_3 \quad\text{ on }(0,\infty)\times \Gamma.
  \end{aligned}
\end{equation}
which will be considered in the present paper. In \eqref{1-3}, $\Delta$ and $\Delta'$ stand for the Laplace operator in $\Omega$ and the Laplace-Beltrami operator on the boundary $\Gamma$, respectively, and $\partial_\nu$ denotes the derivative in outer normal direction. For a survey on other types of boundary conditions and generation of semigroups for them, we refer, e.g., to \cite{lasiecka-triggiani97}. The physically relevant situation is the two-dimensional case $N=2$, but we can consider \eqref{1-1}--\eqref{1-3} in any dimension.

One of the standard approaches to \eqref{1-1}-\eqref{1-3} is to set $v:=\partial_t u$ and obtain the first-order system  acting on $U:=(u,u_t,\theta)^\top$ and being of the form
\begin{equation}
  \label{1-1aa}
U_t - A(D) U = 0 \quad \text{ in } (0,\infty)\times \Omega
\end{equation}
with an operator-matrix $A(D)$ of mixed order. This equation is augmented by boundary conditions of the form
\begin{equation}
\label{1-2aa}
 B(D) U = 0 \quad \text{ on } (0,\infty)\times\Gamma.
\end{equation}
Here $A(D)$ and $B(D)$  are given by
\[ A(D) := \begin{pmatrix}
  0 & 1 & 0 \\
  -\Delta^2 & 0 & -\Delta\\
  0 &  \Delta & \Delta
\end{pmatrix},\quad
B(D) := \begin{pmatrix}
  \Delta - (1-\beta)\Delta' & 0 & 1\\
  \partial_\nu (\Delta + (1-\beta)\Delta') & 0 & 0\\
  0 & 0& \partial_\nu
\end{pmatrix}.\]
The natural space for the  $L^p$-realization of the mixed-order boundary value problem $(A(D),$ $B(D))$ is given by $\E_p^{(0)}(\Omega)$ and its solution space by $\E_p^{(2)}(\Omega)$, where
for $j\in\{0,1,2\}$ we set
\[ \E_p^{(j)}(\Omega) := H_p^{2+j}(\Omega) \times H_p^{j}(\Omega)\times H_p^j(\Omega).\]
More precisely, we define $A_{p,\Omega}$ as an unbounded operator in $\E_p^{(0)}(\Omega)$ with domain
 \[ D(A_{p,\Omega}) := \{ U\in \E_p^{(2)}(\Omega): B(D)U=0\}\]
acting as $A_{p,\Omega} U := A(D)U\;(U\in D(A_{p,\Omega}))$.
We consider uniform $C^4$-domains, see Definition~\ref{1.1} below. The main result of the present paper shows that for all $p\in (1,\infty)$, the operator $A_{p,\Omega}$ generates an analytic $C_0$-semigroup. This is a consequence of the stronger result that $A_{p,\Omega}$ has maximal $L^q$-$L^p$-regularity (Theorem~\ref{3.2}). On bounded $C^4$-domains, we obtain exponential stability (Theorem~\ref{3.6}).

\medskip

The thermoelastic plate equations has been studied by many authors, mostly in an $L^2$-setting. Many results deal with exponential stability of the associated semigroup, e.g.,  \cite{kim92},
 \cite{munos_rivera-racke95},
 \cite{liu-zheng97},
\cite{lasiecka-triggiani97},   \cite{shibata94}.
For the  analyticity of the semigroup, we refer to  \cite{liu-renardy95},
 \cite{liu-liu97}, and   \cite{liu-yong98}
in the $L_2$-setting. For the treatment of nonlinear problems, corresponding results in $L_p$ are of relevance.
In the whole-space case, analyticity of the
generated semigroup in $L_p$ was shown in  \cite{denk-racke06}.
In the case of the half-space and of bounded
domains, equations \eqref{1-1} with Dirichlet (clamped) boundary conditions
\[ u = \partial_\nu u = \theta = 0 \quad\text{ on }
(0,\infty)\times\Gamma\]
were studied in  \cite{naito-shibata09} and \cite{naito09}. In the paper \cite{lasiecka-wilke13}, a rather complete analysis in the $L^p$-setting can be found for hinged boundary conditions $u=\Delta u = \theta =0$.

System \eqref{1-1}--\eqref{1-3}, i.e. the thermoelastic plate equations with free boundary conditions in the $L_p$-setting, has been studied recently by the authors in \cite{Denk-Shibata17}. It was shown that the second-order (in time) system \eqref{1-1}--\eqref{1-3} has maximal $L_q$-$L_p$-regularity. However, this does not imply that the first-order system \eqref{1-1aa}--\eqref{1-2aa} generates an analytic $C_0$-semigroup. This was also observed in the case of the structurally damped plate equation with clamped boundary conditions in \cite{denk-schnaubelt15}. In fact, in the situation of \cite{denk-schnaubelt15}, we have maximal regularity, but no generation of semigroup unless additional conditions are included in the basic space. Roughly speaking, this is due to the fact that the standard resolvent estimates hold only for right-hand sides with vanishing first component, and the reformulation of \eqref{1-1} as a first-order system in fact leads to such a right-hand side.

In the present paper, however, we show that the operator related to the first-order system \eqref{1-1a}  generates an analytic $C_0$-semigroup without additional conditions on the basic space $\E_p^{(0)}(\Omega)$. The proofs are based on Fourier multiplier methods on one hand and  on the results from \cite{Denk-Shibata17} on the other hand. If the domain $\Omega$ is bounded, we obtain exponential stability apart from the kernel of the operator. In particular, we obtain generation of an analytic semigroup and exponential stability for the two-dimensional system which was studied in \cite{lasiecka-triggiani98c}, in this way generalizing the results in \cite{lasiecka-triggiani98c} from the $L^2$-case to the $L^p$-case.

\section{The whole space case}

In this section, we consider the whole-space case, i.e. system \eqref{1-1}--\eqref{1-2} with $\Omega=\R^N$. Our approach is based on the Fourier transform and results on vector-valued Fourier multipliers. In particular, the proof of maximal regularity in the sense of well-posedness in $L^q$-$L^p$-Sobolev spaces make use of the concept of $\mathcal R$-boundedness and variants of Michlin's theorem. As standard references, we mention \cite{denk-hieber-pruess03} and \cite{kunstmann-weis04}.

The Fourier transform $\mathscr F$ in $\R^N$ is given by
\[ (\mathscr F\varphi)(\xi) := (2\pi)^{-N/2} \int_{\R^N} \varphi(x)e^{-ix\xi} dx\quad (\xi\in\R^N)\]
for Schwartz functions $\varphi$ and extended by duality to tempered distributions. A symbol $m\in L^\infty(\R^N)$ is called a Fourier multiplier if $\mathscr F^{-1} m\mathscr F$ defines a bounded linear operator in $L^p(\R^N)$.
One of the key ingredients to show $\mathcal R$-sectoriality will be the vector-valued version of Michlin's theorem on Fourier multipliers due to Weis \cite{weis01} and Girardi and Weis \cite{girardi-weis03}.

The following definition is a variant of \cite{Denk-Shibata17}, Definition~3.2.

\begin{definition}
  \label{2.1}
  Let $\Sigma\subset \C$ be a set, let $m\colon (\R^{N}\setminus\{0\})\times \Sigma\to \C, \, (\xi,\lambda)\mapsto m(\xi,\lambda)$, be $C^\infty$ with respect to $\xi$. Let $s\in\R$.  Then $m$ is called a multiplier of order $s$ in $\Sigma$ if the estimates
  \[ |\partial_{\xi}^\alpha m(\xi,\lambda)| \le C_\alpha (|\lambda|^{1/2} + |\xi|)^{s} |\xi|^{-|\alpha|}\]
  hold for any multi-index $\alpha\in\N_0^N$ and $(\xi,\lambda)\in (\R^{N}\setminus\{0\})\times \Sigma$ with some constant $C_\alpha$ depending only on $\alpha$ and $\Sigma$. The set of all multipliers of order $s$  in $\Sigma$ will be denoted by $\mathbb M_s(\Sigma)$.
\end{definition}

It is easily seen that $\mathbb M_s(\Sigma)$ is a complex vector space and that  for $m_1\in \mathbb M_{s_1}(\Sigma)$ and $m_2\in \mathbb M_{s_2}(\Sigma)$ we have $m_1m_2\in \mathbb M_{s_1+s_2}(\Sigma)$ (see \cite{Denk-Shibata17}, Lemma~3.3).

\begin{example}
  \label{2.2} We mention some examples which will be useful below. Let $\theta\in (0,\pi)$, and let
\begin{equation}\label{2-1a}
 \Sigma_\theta := \{ \lambda\in\C\setminus\{0\}: |\arg\lambda|<\theta\}
\end{equation}
be the open sector in the complex plane.

a) Directly from the definition it can be seen that $ \lambda\in \mathbb M_2(\Sigma_\theta)$ (where $\lambda$ stands for the constant mapping $(\xi,\lambda)\mapsto \lambda$) and $|\xi|^{2k}\in M_{2k}(\Sigma_\theta)$ for all $k\in\N$.

b) Let $s\in\R\setminus\{0\}$, and let $m(\xi,\lambda):= (\lambda+|\xi|^2)^{s/2}$. Then $m\in \mathbb M_s(\Sigma_\theta)$. This can be seen by homogeneity: As $m$ is quasi-homogeneous of order $s$ in the sense that
\[ m(\rho\xi,\rho^2 \lambda) = \rho^s m(\xi,\lambda)\quad (\rho>0,\, \xi\in\R^N\setminus\{0\},\, \lambda\in\Sigma_\theta),\]
the derivative $\partial_\xi^\alpha m$ is quasi-homogeneous of degree $s-|\alpha|$. By a compactness argument,
\[ |\partial_\xi^\alpha m(\xi,\lambda)| \le c_\alpha (|\lambda|^{1/2} + |\xi|)^{s-|\alpha|} \le c_\alpha (|\lambda|^{1/2} + |\xi|)^s |\xi|^{-|\alpha|}\]
which shows $m\in\mathbb M_s(\Sigma_\theta)$.

c) By a similar homogeneity argument, we see that $(\xi,\lambda)\mapsto \frac{|\xi|}{(1+|\xi|^2)^{1/2}} \in \mathbb M_0(\Sigma_\theta)$.

d) Let $\lambda_0>0$. Then $(\xi,\lambda)\mapsto 1\in \mathbb M_2(\lambda_0+\Sigma_0)$ due to
\[ 1\le c_{\lambda_0} |\lambda| \le c_{\lambda_0} (|\lambda|^{1/2}+|\xi|)^2.\]
Therefore, $1+|\xi|^2\in \mathbb M_2(\lambda_0+\Sigma_\theta)$. Note that $1\not\in \mathbb M_2(\Sigma_\theta)$.
\end{example}

The following result is one main tool for the results below and was shown in \cite{enomoto-shibata13}, Theorem~3.3.

\begin{lemma}
  \label{2.3}
  In the situation of Definition~\ref{2.1}, let $m\in \mathbb M_0(\Sigma)$. For $\lambda\in\Sigma$, define the operator $m(D,\lambda)$ by $m(D,\lambda) f= \mathscr F_\xi^{-1}[m(\lambda,\xi)\mathscr F_\xi f(\xi)]$. Then, the family $\{m(D,\lambda):\lambda\in\Sigma\}\subset L(L^p(\R^N))$ is $\mathcal R$-bounded and
  \[ \mathcal R_{L(L^p(\R^N))}(\{m(D,\lambda):\lambda\in\Sigma\}) \le C_{p,N} \max_{|\alpha|\le N+1} C_\alpha\]
  with $C_{p,N}$ depending only on $p$ and $N$.
\end{lemma}

The analysis of the operator $A_{p,\R^N}$ in the whole space was essentially done in \cite{naito-shibata09} and \cite{naito09}. We summarize some results from these papers. Define $\gamma_1,\gamma_2,\gamma_3$ by the equality
\begin{equation}
\label{1-1a}
 p(t) := t^3+t^2+2t+1 = (t+\gamma_1)(t+\gamma_2)(t+\gamma_3)
\end{equation}
with $\gamma_1\in\R$, $\gamma_2=\bar\gamma_3$ and $ \Im \gamma_2>0$. Then $\gamma_1\in (0,1)$, $\Re\gamma_2=\Re\gamma_3\in (0,\frac12)$, and $\det(\lambda-A(\xi)) = \prod_{j=1}^3 (\lambda+\gamma_j|\xi|^2)$ (see \cite{naito-shibata09}, Lemma~2.3). We define $\vartheta_0:= \arg(-\gamma_3)\in (\frac\pi 2,\pi)$ (note that $-\gamma_3$ is the root of the polynomial $p$ with positive imaginary part).

We consider the whole space resolvent
\begin{equation}\label{1-4}
 R(\lambda) := (\lambda-A(D))^{-1} := \mathscr F_{\xi}^{-1} (\lambda-A(\xi))^{-1}\mathscr F
\end{equation}
with
\[ A(\xi) := \begin{pmatrix}
  0 & 1 & 0 \\
  -|\xi|^4 & 0 & |\xi|^2\\
  0 & -|\xi|^2 & -|\xi|^2
\end{pmatrix}.\]
For $j\in\{0,1,2\}$, define
\[ S_j(\xi) := (1+|\xi|^2)^{j/2}\diag((1+|\xi|^2), 1,1),\]
 where $\diag(\ldots)$ stands for the diagonal matrix with the corresponding elements on the diagonal.
For the next result, we use the fact that the induced operator
$S_j(D)$ defines an  isometric isomorphism
\begin{equation}
\label{2-1}
  S_j(D)\in L_{\text{Isom}}( \E_p^{(j)} , L^p(\R^N;\C^3)) \quad (j\in\{0,1,2\}).
\end{equation}

\begin{lemma}
  \label{2.4}
  For every $\vartheta<\vartheta_0$, $\lambda_0>0$ and $j\in\{0,1,2\}$ we have
  \[ \mathcal R_{L(\E_p^{(0)}(\R^N), \E_p^{(2-j)}(\R^N))}\big( \big\{ \lambda^{j/2} R(\lambda): \lambda\in\lambda_0+\Sigma_{\vartheta}\big\}\big) <\infty.\]
\end{lemma}

\begin{proof}
  Let $j\in\{0,1,2\}$. In view of \eqref{2-1} and Lemma~\ref{2.2}, we have to show that every entry of the matrix
  \[ M^{(j)}(\xi,\lambda) = \big( m^{(j)}_{kl}(\xi,\lambda)\big)_{k,l=1,2,3} := \lambda^{j/2} S_{2-j}(\xi)(\lambda-A(\xi))^{-1} S_0(\xi)^{-1}\]
  belongs to $\mathbb M_0(\lambda_0+\Sigma_{\vartheta})$.
  It was shown in \cite{naito-shibata09}, Section~2, that for all $\lambda\in\lambda_0+\Sigma_{\vartheta}$ we have
  \[ (\lambda-A(\xi))^{-1} = \frac1{\det(\lambda-A(\xi))}
  \begin{pmatrix}
    \lambda (\lambda+|\xi|^2) + |\xi|^4 & \lambda+|\xi|^2 & |\xi|^2\\
    -(\lambda+|\xi|^2)|\xi|^4 & \lambda(\lambda+|\xi|^2) & \lambda|\xi|^2\\
    |\xi|^6 & -\lambda|\xi|^2 & \lambda^2+|\xi|^4
  \end{pmatrix}\,.\]
Moreover, $\det(\lambda-A(\xi)) = \prod_{i=1}^3 (\frac\lambda{\gamma_i}+|\xi|^2)$ (\cite{naito-shibata09}, Lemma~2.3). As in Example~\ref{2.2} b), we see that
\[ (\xi,\lambda)\mapsto (\tfrac\lambda{\gamma_i}+|\xi|^2)^{-1}\in \mathbb M_{-2}(\Sigma_\vartheta)\]
and therefore $(\det(\lambda-A(\xi)))^{-1}\in\mathbb M_{-6}(\Sigma_\vartheta)$. For the left upper corner of $M^{(j)}$, we have
\[ m^{(j)}_{11}(\xi,\lambda) = (\det(\lambda-A(\xi)))^{-1} \lambda^{j/2} (1+|\xi|^2)^{(2-j)/2} (\lambda^2+\lambda|\xi|^2+|\xi|^4).\]
With Example~\ref{2.2} we see that
\begin{align*}
  \lambda^{j/2}& \in\mathbb M_j(\Sigma_\vartheta),\\
  (1+|\xi|^2)^{(2-j)/2} & \in \mathbb M_{2-j}(\lambda_0+\Sigma_\vartheta),\\
  \lambda^2+\lambda|\xi|^2+|\xi|^4 & \in \mathbb M_4(\Sigma_\vartheta),
\end{align*}
which yields $m_{11}^{(j)}\in \mathbb M_0(\lambda_0+\Sigma_\vartheta)$. Similarly,
\[ m_{21}^{(j)}(\xi,\lambda) = - (\det(\lambda-A(\xi)))^{-1} \lambda^{j/2} (1+|\xi|^2)^{(2-j)/2} (\lambda+|\xi|^2) \tfrac{|\xi|^2}{1+|\xi|^2}.\]
Using
\begin{align*}
  \lambda^{j/2} & \in\mathbb M_j(\Sigma_\vartheta),\\
  (1+|\xi|^2)^{(2-j)/2} & \in \mathbb M_{2-j}(\lambda_0+\Sigma_\vartheta),\\
  (\lambda+|\xi|^2) & \in \mathbb M_2(\Sigma_\vartheta),\\
  \tfrac{|\xi|^2}{1+|\xi|^2} & \in \mathbb M_0(\Sigma_\vartheta),
\end{align*}
we obtain $m_{21}\in\mathbb M_0(\lambda_0+\Sigma_\vartheta)$. All other entries of the matrix $M^{(j)}$ can be estimated similarly. Therefore, $M^{(j)}\in \mathbb M_0(\lambda_0+\Sigma_\vartheta)$ which finishes the proof.
 \end{proof}

\begin{corollary}
  \label{2.5} a) For all $\lambda\in\Sigma_{\vartheta_0}$, the operator $\lambda - A_{p,\R^N}\colon \E_p\to \F_p$ is invertible.

  b) The operator $A_{p,\R^N}$ is not sectorial for any angle and therefore does not generate a bounded $C_0$-semigroup on $\F_p$.

  c) For any $\lambda_0>0$, the operator $A_{p,\R^N}-\lambda_0$ is $\mathcal R$-sectorial with $\mathcal R$-angle $\vartheta_0$. Therefore, $A_{p,\R^N}-\lambda_0$ has maximal $L^q$-$L^p$-regularity in $(0,\infty)$, and $A_{p,\R^N}$ has maximal $L^q$-$L^p$-regularity in $(0,T)$ with $T<\infty$. In particular, $A_{p,\R^N}$ generates an analytic $C_0$-semigroup.
\end{corollary}

\begin{proof}
a) Let $\lambda\in\Sigma_{\vartheta_0}$ and choose $\vartheta<\vartheta_0$ and $\lambda_0>0$ such that $\lambda\in \lambda_0+\Sigma_\vartheta$. By Lemma~\ref{2.4} with $j=0$, we have $R(\lambda)\in L(\E_p^{(0)}, \E_p^{(2)})$. Obviously, $R(\lambda)$ is the inverse of $\lambda-A_{p,\R^N}$, and therefore $\lambda$ is in the resolvent set of $A_{p,\R^N}$.

b) Assume that $\|\lambda(\lambda-A_{p,\R^N})^{-1} \|_{L(\F_p)} \le C\;(\lambda\in (0,\infty))$ holds. Then the operator $M_0(D,\lambda)\in L(L^p(\R^N;\C^3))$ is uniformly (with respect to $\lambda$) bounded, where $M_0(\xi,\lambda) := \lambda S_1(\xi) (\lambda-A(\xi))^{-1} S_1(\xi)^{-1}$. In particular, every entry of $M_0(\xi,\lambda)$ is an $L^\infty$-function (see Prop.~3.17 in \cite{denk-hieber-pruess03}). For the last entry in the first row of $M_0(\xi,\lambda)$, we obtain
\[ \left| \frac{\lambda(1+|\xi|^2)|\xi|^2}{\prod_{j=1}^3 (\lambda+\gamma_j|\xi|^2)}\right| \le C<\infty \quad (\lambda\in (0,\infty),\, \xi\in\R^N).\]
However, setting $\lambda=k^{-2}$ and $|\xi|=k^{-1}$, we see that the left-hand side is unbounded for $k\to\infty$.

c) The $\mathcal R$-sectoriality  follows from Lemma~\ref{2.4} with $j=2$, and the other statements are consequences of the general theory on $\mathcal R$-sectorial operators.
\end{proof}

\section{The case of a uniform $C^4$-domain}

\begin{definition}
  \label{1.1}
A domain $\Omega$ is called a uniform $C^4$-domain if
there exist positive constants $\alpha$, $\beta$ and $K$
such that for any $x_0 \in \Gamma$ there exist a coordinate
number $j$ and a $C^4$-function
$h(x')$
defined on $B'_\alpha(x'_0)$ such that
$\|h\|_{H^4_\infty(B'_\alpha(x_0'))} \leq K$ and
\begin{align*}
\Omega\cap B_\beta(x_0) & = \{x \in \R^N \mid
x_j > h(x') \enskip(x' \in B'_\alpha(x'_0))\}
\cap B_\beta(x_0), \\
\Gamma\cap B_\beta(x_0) & = \{x \in \R^N \mid
x_j = h(x') \enskip(x' \in B'_\alpha(x'_0))\}
\cap B_\beta(x_0).
\end{align*}
Here, $x'$ has been defined by
$x' = (x_1, \ldots, x_{j-1}, x_{j+1}, \ldots, x_{_N})$
for $x=(x_1,\ldots, x_{_N})$,
$$B'_\alpha(x'_0)
= \{x' \in \R^{N-1} \mid |x'-x'_0| < \alpha\},
\quad B_\beta(x_0) = \{x \in \R^N \mid |x-x_0| < \beta\}.
$$
\end{definition}

Let $\Omega\subset\R^N$ be a uniform $C^4$-domain with boundary $\Gamma$, and let $p\in(1,\infty)$. To show that the operator $A_p\colon \E_p^{(0)}(\Omega)\supset D(A_p)\to \E_p^{(0)}(\Omega)$ generates an analytic $C_0$-semigroup, we first consider the boundary value problem
\begin{equation}
  \label{3-1}
  \begin{aligned}
    (\lambda-A(D))U & = 0\quad\text{ in }\Omega,\\
    B(D)U & = G\quad\text{ on }\Gamma.
  \end{aligned}
\end{equation}
Here, $G=(g_1,g_2,g_3)^\top$ is defined in the whole of $\Omega$. Similarly to \cite{Denk-Shibata17}, (1.8), we define the spaces
\begin{align*}
  \mathbb G_p(\Omega) & := H_p^2(\Omega)\times H_p^1(\Omega)\times H_p^1(\Omega),\\
  \mathbb X_p(\Omega) & := \mathbb G_p(\Omega)\times\big( H_p^1(\Omega)\times L^p(\Omega)\times L^p(\Omega)\big)\times L^p(\Omega).
\end{align*}
The $\lambda$-dependent map $H(\lambda)\colon \mathbb G_p(\Omega)\to \mathbb X_p(\Omega)$ is defined by
\[ H(\lambda)((g_1,g_2,g_3)^\top) := \big((g_1,g_2,g_3),\lambda^{1/2} (g_1,g_2,g_3), \lambda g_1\big)^\top.\]
The following result on the existence of $\mathcal R$-bounded solution operators was shown in \cite{Denk-Shibata17}, Theorem~1.4.

\begin{theorem}
  \label{3.1}
  There exist a number $\vartheta_1\in (\frac\pi2,\pi)$, a positive number $\lambda_0$ and an operator family
  \[ L(\lambda) = \big(L_1(\lambda),\lambda L_1(\lambda), L_2(\lambda)\big)^\top \in L(\mathbb X_p(\Omega), \E_p^{(2)}(\Omega))\]
  such that for every $\lambda\in\lambda_0+\Sigma_{\vartheta_1}$ and every $G\in\mathbb G_p(\Omega)$, problem \eqref{3-1} admits a unique solution $U\in \E_p^{(2)}(\Omega)$ given by $U = L(\lambda) H(\lambda)G$. Moreover,
  \begin{align*}
  \mathcal R_{L(\mathbb X_p(\Omega), H_p^{4-j}(\Omega))} \big(\big\{
  \lambda^{j/2} L_1(\lambda)\colon \lambda\in \lambda_0+\Sigma_{\vartheta_1}\big\} \big) \le C\quad (j=0,\dots,4),\\
  \mathcal R_{L(\mathbb X_p(\Omega), H_p^{2-j}(\Omega))} \big(\big\{
  \lambda^{j/2} L_2(\lambda)\colon \lambda\in \lambda_0+\Sigma_{\vartheta_1}\big\} \big) \le C\quad (j=0,1,2).
  \end{align*}
\end{theorem}

Let $R_\Omega\colon f\mapsto f|_\Omega$ denote the restriction of a function defined on $\R^N$ to $\Omega$. Obviously, we have $r_\Omega\in L(H_p^i(\R^N), H_p^i(\Omega))$ with norm 1 for any $i\in\N_0$. In fact, $r_\Omega$ is a retraction as a corresponding co-retraction (extension operator) exists for uniform $C^4$-domains. In the following, we fix an extension operator $e_\Omega\colon L^1_{\text{loc}}(\Omega)\to L^1_{\text{loc}}(\R^N)$ with the property that for any $p\in (1,\infty)$ and $f\in H_p^i(\Omega)$, we have $e_\Omega f\in H_p^i(\R^N)$, $r_\Omega e_\Omega f=f$, and $ \|e_\Omega\|_{L(H_p^i(\Omega), H_p^i(\R^N))} \le C_p$ for $i=0,\dots,4$. For the existence of such an extension operator, we refer to \cite{schade-shibata15}, Appendix~A.

The following theorem is the main result of the present paper.

\begin{theorem}
  \label{3.2}
  There exist $\lambda_0>0$ and $\vartheta>\frac\pi 2$ such that the operator $A_{p,\Omega}-\lambda_0$ is $\mathcal R$-sectorial with $\mathcal R$-angle $\vartheta$. Therefore, $A_{p,\Omega}$ has maximal $L^q$-$L^p$-regularity in every finite time interval. In particular, $A_{p,\Omega}$ generates an analytic $C_0$-semigroup in $\E_p^{(0)}(\Omega)$.
\end{theorem}

\begin{proof}
  We first obtain a description of the resolvent $(\lambda-A_{p,\Omega})^{-1}$. For this, let $F\in \E_p^{(0)}(\Omega)$ be given. We apply the extension operator $e_\Omega$ from above to every component of $F$ and obtain $e_\Omega F\in \E_p^{(0)}(\R^N)$. We set $U_1 := r_\Omega R(\lambda)e_\Omega F$ for $\lambda\in\Sigma_{\vartheta_0}$ with $R(\lambda)$ being the whole space resolvent defined in \eqref{1-4}.

  To solve
  \begin{equation}
  \label{3-2}
  \begin{aligned}
    (\lambda - A(D)) U & = F \quad\text{ in } \Omega,\\
    B(D) U & = 0 \quad\text{ on }\Gamma,
  \end{aligned}
  \end{equation}
  we set $U=U_1+U_2$ and obtain the boundary value problem
  \begin{align*}
    (\lambda - A(D)) U_2 & = 0 \quad \text{ in }\Omega,\\
    B(D) U_2 & = - B(D) U_1 \quad\text{ on }\Gamma
  \end{align*}
  for $U_2$. Due to Theorem~\ref{3.1}, there exist $\lambda_0$ and $\vartheta_1$ such this equation is uniquely solvable for $\lambda\in \lambda_0+\Sigma_{\vartheta_1}$, and its solution is given by
  \[ U_2 = - L(\lambda) H(\lambda) B(D) U_1.\]
  Therefore, for $\vartheta\in (\frac\pi 2, \min\{\vartheta_0,\vartheta_1\})$ and $\lambda\in \lambda_0+\Sigma_\vartheta$, the boundary value problem \eqref{3-2} is uniquely solvable with solution
  \[ U = U_1+U_2 = r_\Omega R(\lambda) e_\Omega F - L(\lambda)H(\lambda)B(D)r_\Omega R(\lambda)e_\Omega F.\]
  Consequently, we have to show the $\mathcal R$-boundedness of the operator family
  \begin{equation}
    \label{3-3}
    \lambda(\lambda-A_{p,\Omega})^{-1} = r_\Omega \lambda R(\lambda) e_\Omega - \lambda L(\lambda) H(\lambda) B(D) r_\Omega R(\lambda) e_\Omega\quad (\lambda\in\lambda_0+\Sigma_\vartheta).
  \end{equation}
  By Corollary~\ref{2.5} c), $\lambda R(\lambda)$ is $\mathcal R$-bounded. As $e_\Omega$ and $r_\Omega$ are continuous and $\lambda$-independent, we obtain
  \begin{equation}
    \label{3-4}
    \mathcal R_{L(\E_p^{(0)}(\Omega))} \big(\big\{ \lambda r_\Omega R(\lambda) e_\Omega\colon \lambda\in \lambda_0+\Sigma_\vartheta\big\}\big) <\infty.
  \end{equation}
  Similarly, by Theorem~\ref{3.1} with $j=2$ and $j=4$ we see
  \begin{equation}
    \label{3-5}
    \mathcal R_{L(\mathbb X_p(\Omega), \E_p^{(0)}(\Omega))} \big(\big\{ \lambda L(\lambda)\colon \lambda\in \lambda_0+\Sigma_\vartheta\big\}\big) <\infty.
  \end{equation}
  It remains to show that the family $H(\lambda)B(D)r_\Omega R(\Lambda) e_\Omega$ is $\mathcal R$-bounded. By the definition of the matrix $B(D)$ and the spaces, we see that the operators
  \begin{align*}
    B(D)&\colon \E_p^{(2)}(\Omega)\to \mathbb G_p(\Omega),\\
    B(D)&\colon \E_p^{(1)}(\Omega) \to H_p^1(\Omega)\times L^p(\Omega)\times L^p(\Omega),\\
    B_1(D)&\colon \E_p^{(0)}(\Omega)\to L^p(\Omega)
  \end{align*}
  are continuous, where $B_1(D)$ stands for the first row of $B(D)$, i.e. $B_1(D)(u,v,\theta)^\top = (\Delta-(1-\beta)\Delta') u + \theta$. Thus,
  \begin{equation}
    \label{3-6}
    \begin{pmatrix}
      B(D) \\ B(D)\\ B_1(D)
    \end{pmatrix}
    \colon \E_p^{(2)}(\Omega) \times \E_p^{(1)}(\Omega)\times \E_p^{(0)}(\Omega) \to \mathbb X_p(\Omega)
  \end{equation}
  is continuous (and independent of $\lambda$). By Corollary~\ref{2.5} c), the family
  \[ \left\{ \begin{pmatrix}
    R(\lambda)\\ \lambda^{1/2} R(\lambda)\\ \lambda R(\lambda)
  \end{pmatrix}
  \colon \lambda\in\lambda_0+\Sigma_\vartheta\right\}\subset L\big( \E_p^{(0)}(\Omega), \E_p^{(2)}(\Omega) \times \E_p^{(1)}(\Omega)\times \E_p^{(0)}(\Omega)\big)\]
  is $\mathcal R$-bounded. In combination with
  \[ H(\lambda)B(D)r_\Omega R(\lambda) e_\Omega F =\begin{pmatrix}
    B(D) r_\Omega R(\lambda)e_\Omega F\\
   B(D) r_\Omega \lambda^{1/2} R(\lambda)e_\Omega F\\
    B_1(D) r_\Omega \lambda R(\lambda)e_\Omega F
  \end{pmatrix}\]
  and \eqref{3-6}, this yields
  \begin{equation}
    \label{3-7}
    \mathcal R_{L(\E_p^{(0)}(\Omega), \mathbb X_p(\Omega))} \big\{ H(\lambda) B(D) r_\Omega R(\lambda) e_\Omega: \lambda\in \lambda_0+\Sigma_\vartheta\big\} <\infty.
  \end{equation}
  From \eqref{3-4}, \eqref{3-5}, and \eqref{3-7}, the first statement of the theorem follows by the description of the resolvent in \eqref{3-3}. As before, the other statements follow by the general theory of $\mathcal R$-boundedness.
\end{proof}

The results of Theorem~\ref{3.2} are preserved under lower-order perturbations of the operators $A(D)$ and $B(D)$. More precisely, we consider perturbation matrices of the form
\begin{align*}
  A'(D) & = \begin{pmatrix}
    0 & 0 & 0 \\
    a_{21}(x,D) & 0 & a_{23}(x,D)\\
    0 & a_{32}(x,D) & a_{33}(x,D)
  \end{pmatrix},\\
  B'(D) & = \begin{pmatrix}
    b_{11}(x,D) & 0 & 0 \\
    b_{21}(x,D) & 0 & 0 \\
    0 & 0 & b_{33}(x,D)
  \end{pmatrix}.
  \end{align*}
Here $a_{ij}(x,D)$ and $b_{ij}(x,D)$ are linear differential operators. With respect to the orders of the operators, we assume $\ord a_{ij}(x,D)\le s_{ij}$ and $\ord b_{ij}(x,D)\le t_{ij}$ with $s_{21}=3,\, s_{23}=s_{32}=s_{33}=1$ and $t_{11}=1,\, t_{21}=2,\, t_{33}=0$. The coefficients of $a_{ij}(x,D)$ are assumed to belong to $H_\infty^{s_{ij}-1}(\Omega)$, while the coefficients of $b_{11}(x,D),\, b_{21}(x,D)$, and $b_{33}(x,D)$ are assumed to belong to $H_\infty^2(\Omega)$, $H_\infty^1(\Omega)$, and $H_\infty^1(\Omega)$, respectively.

\begin{lemma}
  \label{3.3} Let $(A'(D),B'(D))$ be a lower-order perturbation as described above. Define the perturbed operator $\tilde A_{p,\Omega}\colon \E_p^{(0)}\supset D(\tilde A_{p,\Omega})\to \E_p^{(0)}(\Omega)$ by
  \[ D(\tilde A_{p,\Omega}) := \{ U\in \E_p^{(2)}(\Omega): \tilde B(D) U =0 \},\; \tilde A_{p,\Omega}U := \tilde A(D) U,\]
  where $\tilde A(D) := A(D) + A'(D)$ and $\tilde B(D) := B(D) + B'(D)$.

  Then there exist $\lambda_0>0$ and $\vartheta>\frac\pi 2$ such that the operator $\tilde A_{p,\Omega}-\lambda_0$ is $\mathcal R$-sectorial with $\mathcal R$-angle $\vartheta$. In particular, $\tilde A_{p,\Omega}$ has maximal $L^q$-$L^p$-regularity in every finite time interval and generates an analytic $C_0$-semigroup in $\E_p^{(0)}(\Omega)$.
\end{lemma}

\begin{proof}
  \textbf{(i)} First, we consider boundary perturbations, i.e. $A'(D)=0$. As in the proof of Theorem~\ref{3.2}, we have to find a solution $\tilde U_2$ of the boundary value problem
  \begin{equation}
    \label{3-8}
    \begin{aligned}
      (\lambda - A(D)) \tilde U_2 & = 0 \quad \text{ in }\Omega,\\
      \tilde B(D) \tilde U_2 & = G\quad\text{ on }\Gamma,
    \end{aligned}
  \end{equation}
  where $G:=-\tilde B(D) r_\Omega R(\lambda) e_\Omega F$. We set $U:=L(\lambda) H(\lambda) G$. Then $U$ solves
  \begin{align*}
    (\lambda-A(D)) U & = 0\quad\text{ in }\Omega,\\
    \tilde B(D) U & = G-B'(D) U = (1-B'(D)L(\lambda)H(\lambda)) G =: \tilde G \quad\text{ on }\Gamma.
  \end{align*}
  Let $\lambda_0$ and $\vartheta$ be as in the proof of Theorem~3.2. We show that the operator family
  \[ \big\{ \lambda^{1/2} H(\lambda) B'(D) L(\lambda):\lambda\in \lambda_0+\Sigma_\vartheta\big\}\subset L(\mathbb X_p(\Omega))\]
  is $\mathcal R$-bounded. In fact, due to the assumptions on $B'(D)$, we have
  \begin{align*}
    b_{11}(x,D) & \in L(H_p^{4-j}(\Omega),H_p^{3-j}(\Omega))\quad (j=1,2,3),\\
    b_{21}(x,D) & \in L(H_p^{4-j}(\Omega),H_p^{2-j}(\Omega))\quad(j=1,2),\\
    b_{33}(x,D) & \in L(H_p^{2-j}(\Omega), H_p^{2-j}(\Omega))\quad(j=1,2).
  \end{align*}
  By Theorem~\ref{3.1}, the families
  \begin{align*}
    \big\{ \lambda^{j/2} L_1(\lambda):\lambda\in\lambda_0+\Sigma_\vartheta\big\} & \subset L(\mathbb X_p(\Omega), H_p^{4-j}(\Omega))\quad (j=0,\ldots,4),\\
    \big\{ \lambda^{j/2} L_2(\lambda):\lambda\in\lambda_0+\Sigma_\vartheta\big\} & \subset L(\mathbb X_p(\Omega), H_p^{2-j}(\Omega))\quad (j=0,1,2)
  \end{align*}
  are $\mathcal R$-bounded. By composition, we see that the family
  \[ \left\{ \lambda^{1/2} H(\lambda) B'(D)L(\lambda) = \begin{pmatrix}
    \lambda^{1/2} B'(D) L(\lambda)\\
    \lambda B'(D) L(\lambda)\\
    \lambda^{3/2} b_{11}(x,D)L_1(\lambda)
  \end{pmatrix}: \lambda\in \lambda_0+\Sigma_\vartheta\right\}\subset L(\mathbb X_p(\Omega))\]
  is $\mathcal R$-bounded. Choosing $\lambda_1>\lambda_0$ sufficiently large, we obtain
  \begin{equation}
    \label{3-9}
    \mathcal R_{L(\mathbb X_p(\Omega))} \Big( \big\{H(\lambda)B'(D)L(\lambda)\colon \lambda\in\lambda_1+\Sigma_\vartheta\big\}\Big) \le \frac 12.
  \end{equation}
  Therefore, $1-H(\lambda)B'(D)L(\lambda)\in L(\mathbb X_p(\Omega))$ is invertible for all $\lambda\in \lambda_1+\Sigma_\vartheta$. A simple algebraic calculation shows that this implies that also $1-B'(D)L(\lambda)H(\lambda)\in L(\mathbb G_p(\Omega))$ is invertible, and that we have
  \begin{equation}
    \label{3-10}
    H(\lambda) (1-B'(D)L(\lambda)H(\lambda))^{-1} = (1-H(\lambda)B'(D)L(\lambda))^{-1} H(\lambda).
  \end{equation}
  Setting $\tilde U_2 := L(\lambda)H(\lambda)(1-B'(D)L(\lambda)H(\lambda))^{-1} G$, we obtain $(\lambda-A(D))\tilde U_2=0$ and
  \[ \tilde B(D)\tilde U_2 = (1-B'(D)L(\lambda)H(\lambda)) (1-B'(D)L(\lambda)H(\lambda))^{-1} G = G,\]
  i.e., $\tilde U_2$ is a solution of \eqref{3-8}. As in the proof of Theorem~\ref{3.2}, the solution $\tilde U$ of the resolvent equation is now given by $\tilde U = U_1 + \tilde U_2$ with $U_1 := r_\Omega R(\lambda) e_\Omega$ as in the proof of Theorem~\ref{3.2}. Therefore, the resolvent of $\tilde A_{p,\Omega}$ is given by
  \begin{align*}
    (\lambda-\tilde A_{p,\Omega})^{-1} & = r_\Omega R(\lambda) e_\Omega - L(\lambda) H(\lambda) (1-B'(D)L(\lambda)H(\lambda))^{-1}\tilde B(D) r_\Omega R(\lambda) e_\Omega\\
    & = r_\Omega R(\lambda) e_\Omega - L(\lambda) (1-H(\lambda) B'(D)L(\lambda))^{-1} H(\lambda) \tilde B(D) r_\Omega R(\lambda)e_\Omega,
  \end{align*}
  where we used \eqref{3-10} for the last equality. We have already seen in the proof of Theorem~\ref{3.2} that the operator families $\lambda L(\lambda)$ and $H(\lambda)\tilde B(D) r_\Omega R(\lambda) e_\Omega$ are $\mathcal R$-bounded. Using \eqref{3-9} and a Neumann series argument, we see that
  \[ \mathcal R_{L(\mathbb X_p)} \big( \big\{ (1-H(\lambda)B'(D)L(\lambda))^{-1}: \lambda\in\lambda_1+\Sigma_\vartheta\big\}\big) \le 2.\]
  Now the statements of the lemma follow in the same way as in the proof of Theorem~\ref{3.2}.

  \medskip

  \textbf{(ii)} In the case $A'(D)\not=0$, we consider $(\tilde A(D), \tilde B(D))$ as a perturbation of $(A(D),\tilde B(D))$. Let $\tilde A_{\tilde B}$ and $A_{\tilde B}$ denote the corresponding operators, respectively. Note that we have $D(\tilde A_{\tilde B})=D(A_{\tilde B})$. By the interpolation inequality, for every $\epsilon>0$ there exists $C_\epsilon>0$ such that
  \[ \| \tilde A_{\tilde B} u \|_{\E_p^{(0)}(\Omega)} \le \epsilon \| A_{\tilde B} u\|_{\E_p^{(0)}(\Omega)} + C_\epsilon \|u\|_{\E_p^{(0)}(\Omega)} \quad (u\in D(A_{\tilde B})).\]
  Due to part (i) of the proof, $A_{\tilde B}$ is $\mathcal R$-sectorial, and by an abstract perturbation result on $\mathcal R$-sectorial operators (\cite{kunstmann-weis04}, Corollary~6.7), the same holds for $\tilde A_{\tilde B}$.
\end{proof}

\begin{remark}
  \label{3.4}
  Whereas the lower-order perturbation of the operator $A(D)$ could be handled by an abstract perturbation result on $\mathcal R$-boundedness, to our knowledge there is no such theorem on boundary perturbation which could be applied to our situation. Therefore, the proof of Lemma~\ref{3.3} directly uses the structure of the solution operators.
\end{remark}

The results above were formulated in a general setting in $\R^N$ with $N\ge 2$. In the physically relevant case $N=2$, the modelling can be found in \cite{lagnese89}, Chapter~2. Apart from physical constants, the equation in a uniform $C^4$-domain $\Omega\subset\R^2$ is given by
\begin{equation}
  \label{3-11}
  \begin{aligned}
    u_{tt}+\Delta^2 u + \Delta \theta & = 0\quad\text{ in } (0,\infty)\times\Omega,\\
    \theta_t - \Delta \theta - \Delta u_t & = 0\quad\text{ in }(0,\infty)\times \Omega
  \end{aligned}
\end{equation}
with boundary conditions
\begin{equation}
  \label{3-12}
  \begin{aligned}
    \Delta u + (1-\mu) B_1 u + \theta & = 0 \quad\text{ on }(0,\infty)\times \Gamma,\\
    \partial_\nu \Delta u + (1-\mu)B_2 u + \partial_\nu\theta & = 0 \quad\text{ on } (0,\infty)\times \Gamma,\\
    \partial_\nu\theta & = 0\quad\text{ on }(0,\infty)\times\Gamma.
  \end{aligned}
\end{equation}
Here, the operators $B_1$ and $B_2$ are given by
\begin{align*}
  B_1 u & := 2\nu_1\nu_2 u_{xy} -\nu_1^2 u_{yy} - \nu_2^2 u_{xx},\\
  B_2 u &:= \partial_\tau\big[ (\nu_1^2-\nu_2^2) u_{xy} + \nu_1\nu_2 (u_{yy}-u_{xx})\big],
\end{align*}
where $\nu = \binom{\nu_1}{\nu_2}$ denotes the outer normal vector and $\tau:=\binom{-\nu_2}{\nu_1}$. In \cite{lasiecka-triggiani98c}, the following variant of boundary conditions was considered:
\begin{equation}
  \label{3-13}
  \begin{aligned}
    \Delta u + (1-\mu) B_1 u + \theta & = 0 \quad\text{ on }(0,\infty)\times \Gamma,\\
    \partial_\nu \Delta u + (1-\mu)B_2 u -u + \partial_\nu\theta & = 0 \quad\text{ on } (0,\infty)\times \Gamma,\\
    \partial_\nu\theta + b\theta & = 0\quad\text{ on }(0,\infty)\times\Gamma
  \end{aligned}
\end{equation}
with $b>0$.

\begin{corollary}
  \label{3.5} Let $N=2$, and let $\Omega\subset\R^N$ be a uniform $C^4$-domain. Then the statements of Theorem~\ref{3.2} hold for the operators related to the boundary value problems \eqref{3-11}, \eqref{3-12} and  \eqref{3-11}, \eqref{3-13}.
\end{corollary}

\begin{proof}
  A straight-forward calculation shows that $B_1u = -\Delta'u $ and $B_2 u = \partial_\nu\Delta'u$ holds up to lower-order terms. Therefore, we can apply Lemma~\ref{3.3} to both boundary value problems.
\end{proof}

Finally, we study exponential stability in the case of a bounded domain.

\begin{theorem}
  \label{3.6}
  Let $\Omega\subset\R^N$, $N\ge 2$, be a bounded $C^4$-domain, and let $(T(t))_{t\ge 0}\subset L(\E_{p}^{(0)}(\Omega))$ be the $C_0$-semigroup generated by $A_{p,\Omega}$, see Theorem~\ref{3.2}. Let $P_{p,\Omega}\in L(\E_{p}^{(0)}(\Omega))$ denote the spectral projection corresponding to the eigenvalue $0$ of $A_{p,\Omega}$, and let $(T_0(t))_{t\ge 0}\subset L(\ker P_{p,\Omega})$ be the part of $T(t)$ in $\ker P_{p,\Omega}$, i.e., $T_0(t):= T(t)|_{\ker P_{p,\Omega}}$.

  Then $(T_0(t))_{t\ge 0}$ is exponentially stable, i.e., there exist $C>0$ and $\epsilon>0$ such that $\|T(t)\|_{L(\ker P_{p,\Omega})}\le C e^{-\epsilon t}\;(t\ge 0)$. The same holds for the perturbed problem $\tilde A_{p,\Omega}$ as in Lemma~\ref{3.3}.
\end{theorem}

\begin{proof}
  As $\Omega$ is bounded, the operator $A_{p,\Omega}$ has compact resolvent and discrete spectrum. Moreover, the spectrum is independent of $p\in (1,\infty)$. It was shown in \cite{lasiecka-triggiani98c} that $A_{2,\Omega}$ is dissipative which implies $\sigma(A_{2,\Omega})\subset \{\lambda\in\C: \Re\lambda\le 0\}$. Moreover, 0 is the only eigenvalue on the imaginary axis. Now the statements of the theorem follow from general semigroup theory.
\end{proof}

\begin{corollary}
  \label{3.7}
  Let $N=2$, and let $\Omega\subset\R^2$ be a bounded $C^4$-domain. Then the analytic semigroup related to the boundary value problem \eqref{3-11}, \eqref{3-12} is exponentially stable in the space $\ker P_{p,\Omega}$, and the analytic semigroup related to \eqref{3-11}, \eqref{3-13} is exponentially stable in the whole space $\E_p^{(0)}(\Omega)$.
\end{corollary}

\begin{proof}
  This is a particular case of Theorem~\ref{3.6} where we note that in the case of \eqref{3-11}, \eqref{3-13} there is no eigenvalue on the imaginary axis due to \cite{lasiecka-triggiani98c}.
\end{proof}

config: arxiv


TITLE: Light entering in a 0° angle, is there any reflection?
QUESTION [4 upvotes]: Is there any reflection of light that enters a new medium at a 0° angle, if the electric field is such, that it is completely in the plane of the double layer?

REPLY [4 votes]: Everyday life experience says that light can be reflected in normal incidence (this condition).
Now, according to Fresnel equations, we have:
$$\tilde E_{0R}=\left(\frac{\alpha -\beta }{\alpha +\beta}\right)\tilde E_{0I} \tag{1}$$
$$\tilde E_{0T}=\left(\frac{2 }{\alpha +\beta}\right)\tilde E_{0I}\tag{2}$$
where
$\alpha=\dfrac{\cos \theta_T}{\cos \theta_I} $ and $\beta=\dfrac{\mu_1 n_2}{\mu_2 n_1}$.

Now, apply Snell's law to find the relation between $\theta_T$ and $\theta_I$:
$$\frac{\sin \theta_T}{\sin \theta_I}=\frac{n_1}{n_2}\tag{3}$$
In the case of zero incidence angle ($\theta_I=0\text{ or }\cos\theta_I=1$) , $(3)$ implies that $\cos \theta_T=1$ and according to $(1)$, it is evident that $\tilde E_{0R}\neq 0$:
$$\alpha=1\to \cases{\tilde E_{0R}=\left(\frac{1-\beta }{1 +\beta}\right)\tilde E_{0I}\\\tilde E_{0T}=\left(\frac{2 }{1 +\beta}\right)\tilde E_{0I}}$$


TITLE: Fundamental theorem of calculus corollary
QUESTION [0 upvotes]: May someone supply a proof to the following lemma:
If $ f $ is continuous on an open interval I and $ b $  is any point in I, then at each point in I $$\displaystyle \frac{d}{dx} \int _x ^b f(t) dt =-f(x)$$

REPLY [1 votes]: $$(\int_x^a f)'=\big((\int_p^af)- (\int_p^xf)\big)'=-f.$$


TITLE: What program/language can I use to handle large numbers (over 10^300)
QUESTION [0 upvotes]: I'm trying to multiply together a set of numbers.
However, the programs I have tried (Excel, R, FreeMat) do not allow numbers above around 300 digits. This limits me to sets of a few hundred numbers at most.
What program can I use that isn't limited by the number of digits? And what size numbers can I realistically manipulate on a (high spec) home PC?
I'm not a programmer, however I can learn if this is the only way around it!

REPLY [2 votes]: Every computer algebra system for symbolic manipulations, for example Sage or Maxima
Some programming languages have built-in support, for example Python, Haskell and Ruby
GMP is a well-established library for C and C++

The size of numbers is limited by the memory of your PC (few GiB typically, thus a few billion digits are possible). Of course you should think about the question which kind of numbers you want to handle (integers, floating point numbers, fractions?).


TITLE: What does the interaction between two charges depend on?
QUESTION [2 upvotes]: In our intro to electrodynamics course, we were told that the interaction between two charges depends on:
➢Magnitude of the charges
➢Separation distance between the charges
➢Velocities of the charges
➢Acceleration of the source charge
➢At a past instant of time!
(Electromagnetic radiation travels at the speed of light-Relativistic effects)
The first two are obvious, but how do the next few influence the interactions? Especially, the fifth one (That was the exact thing he wrote.)

REPLY [1 votes]: You get the magnitude of the charges and the separation distance. Consider one charge the source and the other the target. Practically we're talking about the source at the prior time and the target at the later time. Usually they talk like it's source at later time and target at later time, and then they have to go back and figure out where the source was at the prior time. There's no value in bothering with that.
We are computing the electric and magnetic fields at the later time at the target position. For that we don't need the target velocity. We only need target velocity to decide how the force affects the target. Which is of course the point of the exercise....
For your convenience, you might as well make the frame be the one where the target is stationary. Then you don't have to consider its velocity and you don't have to consider magnetic fields, which will not affect it. Unless for some reason that frame is not convenient.
You can divide up the source velocity into two parts. There's the velocity inline with the straight-line vector between the source and target. And there's the velocity normal to that vector. Those behave completely differently.
Imagine the source is moving directly toward the target at velocity v.
Integrate the force over a time interval of 1 unit, ignoring inverse square effects. In 1 time unit, the source will get v distance closer. So the force that leaves the target after 1 time unit will take only 1-v time units to arrive. That force is squeezed into a smaller distance, it isn't 1 distance unit but less. If you calculate the density of force along that distance it comes out to $\frac{1}{1-v}$. And if all that force arrives in $1-v$ time units, then the amount of force that arrives in 1 time unit is $1+v$ times as much.
So we get a factor of $\frac{1+v}{1-v}$ compared to a source that isn't moving at all. This is easy and obvious and there's nothing mysterious about it.
When the source moves sideways instead, the factor is $(1-v^2)\sqrt{1+v^2}$.
I can explain the $\sqrt{1+v^2}$. The original equation includes the sum of two vectors, n and -v. n is a unit vector, and this time -v is perpendicular to n. So their sum has length $\sqrt{1+v^2}$.
The other part is harder. I can imagine part of it.
We imagine that the force travels straight from the source to the target. But because the source is moving sideways, the force acts at an angle, not along the direction of movement. I can imagine that a part of the force is lost. If v is like the sine, then the part that isn't lost would be the cosine, $\sqrt{1-v^2}$. (This doesn't explain why v would be the sine when we started out with a ratio of 1:v.)
Something like that happens with light polarization. After polarizing to an angle, the remaining irradiance is $\cos^2$. But this is force, not irradiance, and even if the logic works (for force whose direction is rotated instead of polarization around that direction), still I only get $\sqrt{1-v^2}$ instead of $1-v^2$.
Magnetic fields are a property of the frame. In a frame where the source is stationary there is no magnetic field. In a frame where the target is stationary, the magnetic field has no effect on it. It's only in frames where both target and source move, and only in particular relative directions, that the magnetic field is relevant.
I'm not ready to talk much about acceleration, but notice that the magnitude factors that depend only on velocity also affect the acceleration terms.
We think of radiation as waves with a wavelength. Notice that the equation gives a radiation term whenever the source accelerates, whether there is any periodicity to it or not. Waves are just something that happens, that can give big effects.
If electrons move in a sine wave, in a radio tower, then you get radio waves that are strongest perpendicular to the tower and fade out to nothing directly above and below the tower. Linear polarization.
If electrons move in a circle, they create linear-polarized waves in the plane of the circle, circularly-polarized waves perpendicular to the circle, and elliptically polarized waves of radiation in between.
Anything that happens with radiation can be split up into a sum of sine waves, because Fourier transforms let you divide anything that happens into a sum of sine waves.


TITLE: Find subgroup $\langle a,b\rangle$ of $\Bbb Z_{20}^*$ which is not cyclic.
QUESTION [0 upvotes]: Find subgroup $\langle a,b\rangle$ of $\Bbb Z_{20}^*$ which is not cyclic.
I know that $\mathbb{Z}_{20}^{*}$ is not cyclic
But how can I find subgroup which is not cyclic?
e.g why is $\langle 3,11\rangle$ not cyclic?
Thanks and sorry if I have English mistakes :)

REPLY [1 votes]: Since you already know $G=\Bbb Z_{20}^*$ is not cyclic, and every group is a subgroup of itself, you could take the subgroup to be $\Bbb Z_{20}^*$.

Less glib, follow @MeanX's comment.
Since $3$ and $11$ in $G$, $3\times 11=11\times 3$, $3\notin\langle 11\rangle=\{1,11\}$, and $11\notin\langle 3\rangle =\{1,3,9,7\}$, we have
$$\begin{align}
\langle 3,11\rangle &\cong \langle x,y\mid x^2, y^4, xy=yx\rangle \\
&\cong \Bbb Z_2\times\Bbb Z_4,
\end{align}$$
which is not cyclic.
But $|\Bbb Z_2\times\Bbb Z_4|=8=\varphi(20)=|G|$.

A proper subgroup can be found similarly by considering $\langle 9, 11\rangle$.

 That subgroup is isomorphic to $\Bbb Z_2\times\Bbb Z_2$, the Klein four group.


TITLE: is intersection of a curve and a family of curves generically constant as a scheme?
QUESTION [3 upvotes]: (everything below is defined over an algebraically closed field)
Let $D$ be a (smooth) surface, and let $X \subset T \times D$ be a flat family of curves on $D$, where $T$ is irreducible. Let $E$ be a curve on $D$. Fix some $t_0 \in T$ such that $X_{t_0}$ does not contain $E$, the intersection  $X_{t_0} \times_D E$ is some not necessarily reduced closed subscheme of $D$ of dimension 0.
Is it true then that there exists an open dense subset $U \subset T$, such that $X_t \times_D E$ are all isomorphic for different $t \in U$? Does one actually need to assume $D$ to be projective? smooth? Is the previous statement true for an affine $D$?

REPLY [1 votes]: Let me just flesh out Laurent's answer. $X_t$ is the vanishing set of $f_t=xy(x-y)(x-ty)$, an equation of degree $4$. Let $g$ be any homogeneous polynomial of degree $5$, and let $E$ be the vanishing set of $G$. Then the intersection of $E$ and $X_t$ has a unique singular point, the origin, whose maximal ideal is $(x,y)$. Thus, the local ring at the origin, modulo the fifth power of the maximal ideal, is determined by the isomorphism class of the intersection.
This local ring is $k[x,y]/(f_t,x^5,x^4y,x^3y^2,x^2y^3,xy^4,y^5)$. There is a unique-up-to-scaling degree $4$ homogeneous polynomial on the Zariski tangent space that vanishes in this ring, and that polynomial is $f_i$. Degree-$4$ homogenous determine $4$-point subsets of $\mathbb P^1$ determine elliptic curves. The $j$-invariant of that curve is a nonconstant function of $t$ and is determined solely by the isomorphism type of the intersection.


TITLE: Maximum value of an entire function in a closed set.
QUESTION [0 upvotes]: Consider the entire function $f(z)=z(z-i)$. Put $$S=\bigg\{\frac{1}{|f(z)|}\ |\ |z|\geq 2\bigg\}.$$ At what value(s) of $z$ is the maximum of the set $S$ attained?
My Idea: As $f(z)$ is an entire function, we need to find the minimum value of $f(z)$ in $|z|\geq 2$. This must be attained on the boundary. But the boundary is whole $2e^{i\theta}$ for $0\leq \theta \leq 2\pi$. How to reach to the exact answer?

REPLY [2 votes]: Since $f$ is analytic and $\lim_{z\to\infty}f(z)=0$, the maximum of $|f|$ must be attained at the boundary of your region (by the maximum principle), that is, when $|z|=2$. But then$$\max\frac1{|f(z)|}=\max\frac1{2|z-i|}=\frac12\frac1{\min|z-i|},$$and the minimum of $|z-i|$ (when $|z|=2$) is attained at $2i$.


TITLE: Find a bound for approximating Taylor Series
QUESTION [1 upvotes]: I'm struggling to figure out how to find a bound on my error for this problem:
Let T_{6}(x) be the Taylor polynomial of degree 6 based at a = 0 for the function f(x)=\cos(x). Suppose you approximate f(x) by T_{6}(x). If |x|\leq 1, find a bound on the error in your approximation by using the alternating series estimate.
So far, I have
f(x) = cos(x), f(0) = 1
f^(1)(x) = -sin(x), f^(1)(0) = 0
f^(2)(x) = -cos(x), f^(2)(0) = -1
f^(3)(x) = sin(x), f^(3)(0) = 0
f^(4)(x) = cos(x), f^(4)(0) = 1
and
T6(x) = 1 - x^2/4! + x^4/8! -x^6/12!
and that [b_n] = x^8/16!
I'll be honest, this section is way over my head and I'm struggling to make any sense out of it, so I'm not sure that I'm taking the correct approach here in general.
Any help is appreciated!

REPLY [4 votes]: the mcclaurin series $\cos x$ is $$\cos x = 1 - \frac1{2!} x^2 + \frac 1{4!}x^4 - \frac 1{6!}x^6 + \frac 1{8!} x^8 + \cdots  $$ now, if you truncate this alternating series by $T_6,$  then the error committed  is of the same sign as the next term and smaller in absolute value. therefore $$0 \le  \cos x - T_6(x)\le \frac1{8!} x^8 \le \frac 18 \text{ for } |x| \le 1.$$


TITLE: Expected time to get x units away when only able to move 1 unit either way
QUESTION [2 upvotes]: I know this is a common problem, but this problem has been bugging me after someone asked me it, and I can't find the answer anywhere on the Internet.
Say we have a number line, and we start at the point 0. On each turn, we can move left with the probability of 1/2, or we can move right with the probability of 1/2. What is the expected number of turns until we get x units away on either side? (i.e., I don't care which side I get to, I'll stop when I get x units away)

REPLY [0 votes]: If $E_d$ is the expected number of steps to reach $\pm x$ when an absolute distance of $d$ from $0$ then $$E_d=1+\frac12 E_{d-1}+\frac12E_{d+1}$$ for $0 \lt d \lt x$,  but $$E_x=0$$ $$E_0=1+E_1$$ which leads to  $E_d = 2d+1 +E_{d+1}$ and so $E_d = x^2 - d^2$ and in particular $E_0=x^2$.