# On a Mathematical Model of the Rotating Atmosphere of the Earth

Tetu Makino 111Professor Emeritus at Yamaguchi University, Japan; E-mail:
<EMAIL_ADDRESS>
###### Abstract

In meteorology the analysis of motions of the atmosphere on the Earth has been
done using various mathematical models and using various approximations. In
this article as the simplest model the compressible Euler equations with
barotropic equation of state of the ideal gas is analyzed under the co-
ordinate system which rotates with constant angular velocity. Mathematically
rigorous inquiry is tried. Although problems remain to be open, some
fundamental results are exhibited.

Key Words and Phrases. Compressible Euler equations, Atmospheric moton,
Rotations, Hille-Yosida theory, Variational principle,

2020 Mathematical Subject Classification Numbers. 35L53, 76N15, 76U05, 86A10

##  1 Introduction

In meteorology the analysis of motions of the atmosphere on the Earth has been
done using various mathematical models and using various approximations.
Actually models used for numerical weather prediction or climate simulation
must take into account various factors besides the fundamental state variables
of the gas and must meet practical computational efficiency with moderate
accuracy in the process of the numerical simulations. Discussions of choice of
models have been accumulated as an enormous collection in the meteorological
literatures. However fundamental mathematically rigorous inquiries of the
structure of solutions of the equations adopted as models in atmospheric
dynamics have not yet been thoroughly done even for the simplest model. In
this article we try to investigate the fundamental mathematical properties of
the simplest model with the compressible Euler equations and the barotropic,
or, isentropic motion of the ideal gas described in the co-ordinate system
which rotates with a constant angular velocity. Although there remains open
problems to be solved in mathematically rigorous way, some elementary aspects
of the inquiry are exhibited in this article.

In the following in this section the equations and the boundary condition to
be considered are going to be described, and the definition of some concepts
of solutions are going to be given. In Section 2 stationary solutions with
comactly supported density and zero relative velocity will be discussed. The
limit of allowable magnitude of the angular velocity of the rotation should be
noted. In Section 3 Lagrangian co-ordinate description of the equations for
the perturbations near the stationary solutions will be discussed, and in
Section 4 we shall show that the linearized wave equation for the
perturbations allows an application of the Hille-Yosida theory of existence of
solutions. Higher order regularities of the solutions remains in an open
problem. In Section 5 the so called ’variational principle’ will be formurated
and its efficiency will be discussed. Although the concept of eigenfrequency
and eigenvectors to the wave equations for the perturbations is remarkable,
the existence and completeness of eigenvectors is in an open problem when the
rotation is present.

Let us describe the situation to be considred precisely.

We consider motions of an atmosphere governed by the compressible Euler
equations described by the uniformly rotating coordinate system
$(t,\mbox{\boldmath$x$})$:

$\displaystyle\frac{D\rho}{Dt}+\rho(\nabla|\mbox{\boldmath$v$})=0,$ (1.1a)
$\displaystyle\rho\Big{[}\frac{D\mbox{\boldmath$v$}}{Dt}+2\mbox{\boldmath$\Omega$}\times\mbox{\boldmath$v$}\Big{]}+\nabla
P+\rho\nabla\Phi^{\langle\Omega\rangle}=0$ (1.1b)

on
$t\in\mathbb{R},\mbox{\boldmath$x$}=(x^{1},x^{2},x^{3})\in\mathbb{R}^{3}\setminus{\mathfrak{B}}_{0}=\\{\mbox{\boldmath$x$}\in\mathbb{R}^{3}\
|\ r:=\|\mbox{\boldmath$x$}\|>R_{0}\\}$, where

${\mathfrak{B}}_{0}=\\{\mbox{\boldmath$x$}\in\mathbb{R}^{3}\ |\
r:=\|\mbox{\boldmath$x$}\|\leq R_{0}\\},$ (1.2)

$R_{0}$ being a positive number. The variables $\rho,P$ are the density, the
pressure and $\mbox{\boldmath$v$}=(v^{1},v^{2},v^{3})^{\top}$ is the velocity
field. We are denoting

$\frac{D}{Dt}:=\frac{\partial}{\partial
t}+(\mbox{\boldmath$v$}|\nabla)=\frac{\partial}{\partial
t}+\sum_{k=1}^{3}v^{k}\frac{\partial}{\partial x^{k}}$ (1.3)

and $\displaystyle(\nabla|\bm{v})=\sum_{k}\frac{\partial v^{k}}{\partial
x^{k}}$. On the other hand

$\mbox{\boldmath$\Omega$}=\Omega\frac{\partial}{\partial x^{3}},$ (1.4)

$\Omega$ being a constant, and $\Phi^{\langle\Omega\rangle}$ is the
geopotential given as

$\Phi^{\langle\Omega\rangle}(\mbox{\boldmath$x$})=-\frac{{\mathsf{G}M_{0}}}{r}-\frac{\Omega^{2}}{2}\varpi^{2}$
(1.5)

where

$r=\sqrt{(x^{1})^{2}+(x^{2})^{2}+(x^{3})^{2}},\quad\varpi=\sqrt{(x^{1})^{2}+(x^{2})^{2}},$

${\mathsf{G}M_{0}}$ being a positive constant. Since we are concerned with the
value of the potential $\Phi^{\langle\Omega\rangle}$ only on the domain
$\mathbb{R}^{3}\setminus\mathfrak{B}_{0}$, we assume that
$\Phi^{\langle\Omega\rangle}\in C^{\infty}(\mathbb{R}^{3})$ and (1.5) holds
for $r=\|\bm{x}\|>R_{0}$ by removing the singularity
$\displaystyle\frac{1}{r}$ at $r=0$ by a smooth function near $r=0$, namely,
we consider, instead of (1.5),

$\Phi^{\langle\Omega\rangle}(\bm{x})=\chi\Big{(}\frac{r}{R_{0}}\Big{)}\Big{[}-\frac{\mathsf{G}M_{0}}{r}-\frac{\Omega^{2}}{2}\varpi^{2}\Big{]},$
(1.6)

where $\chi\in C^{\infty}(\mathbb{R})$ such that
$1_{[\frac{1}{2},+\infty[}\leq\chi\leq 1_{[\frac{1}{4},+\infty[}$.

The atmosphere is surrounding the Earth of radius $R_{0}$, mass $M_{0}$ and
$\mathsf{G}$ is the gravitational constant. $\Omega$ is the angular velocity
of the rotation.

We assume

(A) : $P$ is the function of $\rho$ defined by

$P=\mathsf{A}\rho^{\gamma}\quad\mbox{for}\quad\rho>0,$ (1.7)

$\mathsf{A},\gamma$ being positive constants such that $1<\gamma<2$.

Under this assumption we introduce the variable $\varUpsilon$ by

$\varUpsilon:=\int_{0}^{\rho}\frac{dP}{\rho}=\frac{\mathsf{A}\gamma}{\gamma-1}\rho^{\gamma-1}\quad\mbox{for}\quad\rho>0.$
(1.8)

Since we are interested in solutions with $\rho$ which has a vacuum region on
which $\rho=0$, the concept of classical solutions needs a precise definition.
The following are the definitions adopted in this article.

###### Definition 1

An open connected subset $\mathfrak{D}$ of $\mathbb{R}^{3}$ is said to be an
admissible domain with cover $\tilde{\mathfrak{D}}$ if
$\tilde{\mathfrak{D}}=\mathfrak{D}\cup\mathfrak{B}_{0}$ is an open connected
subset of $\mathbb{R}^{3}$ such that
$\mathfrak{D}=\tilde{\mathfrak{D}}\setminus\mathfrak{B}_{0}$ and
$\partial\mathfrak{D}$ is smooth.

Here and hereafter we use the following

###### Notation 1

Let $A,B$ be subsets of $\mathbb{R}^{3}$. $\complement A$ denotes the
complement of $A$ and $A\setminus B$ stands for $A\cap\complement B$.
$\mathbf{Cl}A$ stands for the closure of $A$, $\mathbf{Int}A$ stands for the
interior of $A$, and $\partial A$ stands for the boundary of $A$, namely,
$\partial A=\mathbf{Cl}A\setminus\mathbf{Int}A$.

Note that, if $\mathfrak{D}$ is an admissible domain, then, for a sufficiently
small positive $\delta$, we have

$\\{R_{0}<r\leq(1+\delta)R_{0}\\}\subset\mathfrak{D}.$

###### Definition 2

Let $\mathfrak{D}$ be an admissible domain with cover $\tilde{\mathfrak{D}}$.
A vector field $\bm{v}$ defined on $[0,T[\times\mathfrak{D}$ is said to be a
classical admissible velocity field on $[0,T[\times\mathfrak{D}$ if 1)
$\bm{v}$ has an extension onto $[0,T[\times\tilde{\mathfrak{D}}$ of class
$C^{1}([0,T[\times\tilde{\mathfrak{D}})$ and 2) the boundary condition

$(\bm{n}|\bm{v})=0\quad\mbox{on}\quad[0,T[\times\partial\mathfrak{B}_{0}$
(1.9)

holds, where
$\mbox{\boldmath$n$}=\mbox{\boldmath$n$}(\mbox{\boldmath$x$})=\mbox{\boldmath$x$}/R_{0}$
is the inward normal vector at the boundary point
$\mbox{\boldmath$x$}\in\partial\mathfrak{B}_{0}$.

###### Definition 3

Let $\mathfrak{D}$ be an admissible doman with cover $\tilde{\mathfrak{D}}$. A
function $(\rho,\mbox{\boldmath$v$})$ defined on $[0,T[\times\mathfrak{D}$ is
said to be a classical $(\rho,\mbox{\boldmath$v$})$-solution on
$[0,T[\times\mathfrak{D}$ if 1) $(\rho,\mbox{\boldmath$v$})$ has an extension
onto $[0,T[\times\tilde{\mathfrak{D}}$ of class
$C^{1}([0,T[\times\tilde{\mathfrak{D}})$, 2) $\rho\geq 0$ on
$[0,T[\times\mathfrak{D}$, 3) $(\rho,\mbox{\boldmath$v$})$ satsfies
(1.1a)(1.1b) on $[0,T[\times\mathfrak{D}$, and 4) $\rho>0$ on
$[0,T[\times\partial\mathfrak{B}_{0}$ and the boundary condition

$(\bm{n}|\bm{v})=0\quad\mbox{on}\quad[0,T[\times\partial\mathfrak{B}_{0},$
(1.10)

holds, where
$\mbox{\boldmath$n$}=\mbox{\boldmath$n$}(\mbox{\boldmath$x$})=\mbox{\boldmath$x$}/R_{0}$
is the inward normal vector at the boundary point
$\mbox{\boldmath$x$}\in\partial\mathfrak{B}_{0}$.

We want to permit the variable $\varUpsilon$ to take negative values
somewhere. In order to do this we extend the equations devided by $\rho$ where
$\rho>0$ to the equations:

$\displaystyle\frac{D\varUpsilon}{Dt}+(\gamma-1)\varUpsilon(\nabla|\mbox{\boldmath$v$})=0,$
(1.11a)
$\displaystyle\frac{D\mbox{\boldmath$v$}}{Dt}+2\mbox{\boldmath$\Omega$}\times\mbox{\boldmath$v$}+\nabla(\varUpsilon+\Phi^{\langle\Omega\rangle})=0$
(1.11b)

###### Definition 4

Let $\mathfrak{D}$ be an admissible domain with cover $\tilde{\mathfrak{D}}$.
A function $(\varUpsilon,\mbox{\boldmath$v$})$ defined on
$[0,T[\times\mathfrak{D}$ is said to be a classical
$(\varUpsilon,\mbox{\boldmath$v$})$-solution on $[0,T[\times\mathfrak{D}$, if
1) $(\varUpsilon,\mbox{\boldmath$v$})$ has an extension onto
$[0,T[\times\tilde{\mathfrak{D}}$ of class
$C^{1}([0,T[\times\tilde{\mathfrak{D}})$, 2)
$(\varUpsilon,\mbox{\boldmath$v$})$ satisfies (1.11a)(1.11b) on
$[0,T[\times\mathfrak{D}$, and 4) $\varUpsilon>0$ on
$[0,T[\times\partial\mathfrak{B}_{0}$ and the boundary condition

$(\bm{n}|\bm{v})=0\quad\mbox{on}\quad[0,T[\times\partial\mathfrak{B}_{0}$
(1.12)

holds, where
$\mbox{\boldmath$n$}=\mbox{\boldmath$n$}(\mbox{\boldmath$x$})=\mbox{\boldmath$x$}/R_{0}$
is the inward normal vector at the boundary point
$\mbox{\boldmath$x$}\in\partial\mathfrak{B}_{0}$.

As for the variable $\varUpsilon$, $\varUpsilon\geq 0$ is not assumed. When
$(\varUpsilon,\mbox{\boldmath$v$})$ is a classical
$(\varUpsilon,\mbox{\boldmath$v$})$-solution on the admmisible domain
$[0,T[\times\mathfrak{D}$, then $(\rho,\mbox{\boldmath$v$})$ defined by

$\rho=\Big{(}\frac{\gamma-1}{\mathsf{A}\gamma}\varUpsilon\vee
0\Big{)}^{\frac{1}{\gamma-1}}$ (1.13)

turns out to be a classical $(\rho,\mbox{\boldmath$v$})$-solution on
$[0,T[\times\mathfrak{D}$. Here and hereafter we use the

###### Notation 2

We denote

$Q\vee Q^{\prime}=\max\\{Q,Q^{\prime}\\},\quad Q\wedge
Q^{\prime}=\min\\{Q,Q^{\prime}\\}.$ (1.14)

But the inverse may be impossible, namely, we cannot expect that any classical
$(\rho,\mbox{\boldmath$v$})$-solution comes from classical
$(\varUpsilon,\mbox{\boldmath$v$})$-solution by the above procedure, (1.13),
since $1<\frac{1}{\gamma-1}<+\infty$ but $0<\gamma-1<1$ so that $\rho\in
C^{1}$ does not imply $\rho^{\gamma-1}\in C^{1}$ at the vacumm boundary. In
this sense the concept of classical $(\rho,\mbox{\boldmath$v$})$-solutions is
weaker than that of classical $(\varUpsilon,\mbox{\boldmath$v$})$-solutions.

Moreover we note that the uniqueness of solution cannot be expected under the
concept of classical $(\rho,\mbox{\boldmath$v$})$-solutions in the following
sense:

Let $(\rho,\mbox{\boldmath$v$})$ be a classical
$(\rho,\mbox{\boldmath$v$})$-solution on $[0,T[\times\mathfrak{D}$,
$\mathfrak{D}$ being an admissible domain with cover $\tilde{\mathfrak{D}}$.
Suppose that
$(]0,T[\times\mathfrak{D})\setminus\mathbf{Cl}\\{\rho>0\\}\not=\emptyset$, and
consider a velocity field $\mbox{\boldmath$v$}^{\prime}\in
C^{1}([0,T[\times\tilde{\mathfrak{D}})$ such that
$\mbox{\boldmath$v$}^{\prime}(t,\mbox{\boldmath$x$})=\mbox{\boldmath$v$}(t,\mbox{\boldmath$x$})$
on $([0,T[\times\mathfrak{D})\setminus\mathcal{V}$ and
$\mbox{\boldmath$v$}^{\prime}(t,\mbox{\boldmath$x$})\not=\mbox{\boldmath$v$}(t,\mbox{\boldmath$x$})$
on $\mathcal{V}$, where $\mathcal{V}$ is a non-empty open subset of
$(]0,T[\times\mathfrak{D})\setminus\mathbf{Cl}\\{\rho>0\\}$. Clearly
$(\rho,\mbox{\boldmath$v$}^{\prime})$ is another classical
$(\rho,\mbox{\boldmath$v$})$-solution such that it coincides with the original
$(\rho,\mbox{\boldmath$v$})$ on $[0,t_{1}]\times\mathfrak{D}$ but it is
different from the original $(\rho,\mbox{\boldmath$v$})$ after $t=t_{1}$ with
$0<\exists t_{1}<T$.

We are interested in motions with compactly supported $\rho$. Namely we use

###### Definition 5

Let $\mathfrak{D}$ be an admisible domain. A classical
$(\rho,\bm{v})$-solution $(\rho,\bm{v})$ on $[0,T[\times\mathfrak{D}$ is said
to be compactly supported if, for any fixed $t\in[0,T[$,
$\mathbf{Cl}\\{\bm{x}|\rho(t,\bm{x})>0\\}$ is a comoact subset of
$\partial\mathfrak{B}_{0}\cup\mathfrak{D}$.

##  2 Stationary solutions

We are lokking for stationary solutions
$\varUpsilon=\varUpsilon(\bm{x}),\bm{v}=\bm{v}(\bm{x})$. The equation to be
satisfied are

$\displaystyle(\bm{v}|\nabla)\varUpsilon+(\gamma-1)\varUpsilon(\nabla|\bm{v})=0,$
(2.1a)
$\displaystyle(\bm{v}|\nabla)\bm{v}+2\bm{\Omega}\times\bm{v}+\nabla(\varUpsilon+\Phi^{\langle\Omega\rangle})=0.$
(2.1b)

Let us use the cylindrical co-ordinate system $(\varpi,\phi,z)$ defined by

$x^{1}=\varpi\cos\phi,\quad x^{2}=\varpi\sin\phi,\quad x^{3}=z.$

The basis of the co-ordinates consists of the unit vectors

$\displaystyle\bm{e}_{\varpi}$
$\displaystyle=\frac{1}{\varpi}\frac{\partial}{\partial\varpi}=\frac{1}{\varpi}\Big{(}\cos\phi\frac{\partial}{\partial
x^{1}}+\sin\phi\frac{\partial}{\partial x^{2}}\Big{)},$
$\displaystyle\bm{e}_{\phi}$
$\displaystyle=\frac{1}{\varpi}\frac{\partial}{\partial\phi}=-\sin\phi\frac{\partial}{\partial
x^{1}}+\cos\phi\frac{\partial}{\partial x^{2}},$ $\displaystyle\bm{e}_{z}$
$\displaystyle=\frac{\partial}{\partial x^{3}}.$

Suppose that the velocity field $\bm{v}$ is of th form

$\bm{v}=V^{\phi}\bm{e}_{\phi}=\frac{V^{\phi}}{\varpi}\frac{\partial}{\partial\phi}.$
(2.2)

Since

$(\bm{v}|\nabla)=\frac{V^{\phi}}{\varpi}\frac{\partial}{\partial\phi},\quad(\nabla|\bm{v})=\frac{1}{\varpi}\frac{\partial
V^{\phi}}{\partial\phi},$

the equation of continuity (2.1a) reduces to

$\frac{V^{\phi}}{\varpi}\frac{\partial\varUpsilon}{\partial\phi}+(\gamma-1)\frac{\varUpsilon}{\varpi}\frac{\partial
V^{\phi}}{\partial\phi}=0.$

This equation holds if

$\frac{\partial
V^{\phi}}{\partial\phi}=0,\quad\frac{\partial\varUpsilon}{\partial\phi}=0.$
(2.3)

So, supposing (2.3), we solve te equation of motion (2.1b). Since

$\displaystyle(\bm{v}|\nabla)\bm{v}$
$\displaystyle=-(V^{\phi})^{2}\bm{e}_{\varpi}+\frac{V^{\phi}}{\varpi}\frac{\partial
V^{\phi}}{\partial\phi}\bm{e}_{\phi}=-(V^{\phi})^{2}\bm{e}_{\varpi},$
$\displaystyle 2\bm{\Omega}\times\bm{v}$ $\displaystyle=-2\Omega\varpi
V^{\phi}\bm{e}_{\varpi},$
$\displaystyle\nabla(\varUpsilon+\Phi^{\langle\Omega\rangle})$
$\displaystyle=\frac{\partial}{\partial\varpi}(\varUpsilon+\Phi^{\langle\Omega\rangle})\bm{e}_{\varpi}+\frac{1}{\varpi}\frac{\partial}{\partial\phi}(\varUpsilon+\Phi^{\langle\Omega\rangle})\bm{e}_{\phi}+$
$\displaystyle+\frac{\partial}{\partial
z}(\varUpsilon+\Phi^{\langle\Omega\rangle})\bm{e}_{z},$

the equation (2.1b) reduces to

$\displaystyle-(V^{\phi})^{2}-2\Omega\varpi
V^{\phi}+\frac{\partial}{\partial\varpi}(\varUpsilon+\Phi^{\langle\Omega\rangle})$
$\displaystyle=0,$ (2.4a) $\displaystyle\frac{\partial}{\partial
z}(\varUpsilon+\Phi^{\langle\Omega\rangle})=0.$ (2.4b)

Here recall that $\frac{\partial}{\partial\phi}\Phi^{\langle\Omega\rangle}=0$
and that we are supposing $\frac{\partial}{\partial\phi}\varUpsilon=0$.

Taking

$\frac{\partial}{\partial
z}\mbox{\eqref{605a}}-\frac{\partial}{\partial\varpi}\mbox{\eqref{605b}}=0,$

we see $\partial V^{\phi}/\partial z=0$. Therefore there should exist a
function $\omega$ such that ${V}^{\phi}=\varpi\omega(\varpi)$, namely, we
consider

${\mbox{\boldmath$v$}}=\omega(\varpi)\frac{\partial}{\partial\phi}=\varpi\omega(\varpi)\mbox{\boldmath$e$}_{\phi}.$
(2.5)

Integration of (2.4a), (2.4b) gives

$\varUpsilon+\Phi^{\langle\Omega\rangle}+\frac{\Omega^{2}}{2}\varpi^{2}-\mathsf{B}(\varpi)=\mbox{Const.},$
(2.6)

where

$\mathsf{B}(\varpi):=\int_{0}^{\varpi}(\omega(\acute{\varpi})+\Omega)^{2}\acute{\varpi}d\acute{\varpi}.$
(2.7)

Recall

$\Phi^{\langle\Omega\rangle}+\frac{\Omega^{2}}{2}\varpi^{2}=-\frac{{\mathsf{G}M_{0}}}{r}.$

Therefore (2.6) reads

$\Upsilon+\Phi^{\langle\Omega,\omega\rangle}=\mbox{Const.},$ (2.8)

where

$\displaystyle\Phi^{\langle\Omega,\omega\rangle}$
$\displaystyle:=-\frac{\mathsf{G}M_{0}}{r}-\mathsf{B}(\varpi)$
$\displaystyle=-\frac{\mathsf{G}M_{0}}{r}-\int_{0}^{\varpi}(\omega(\acute{\varpi})+\Omega)^{2}\acute{\varpi}d\acute{\varpi}.$
(2.9)

As in Section 2, let us specify the constant so that

${\varUpsilon}={\mathsf{G}M_{0}}\Big{(}\frac{1}{r}-\frac{1}{R}\Big{)}+\mathsf{B}(\varpi).$
(2.10)

Thus we have stationary soltion

$\varUpsilon=\mathsf{G}M_{0}\Big{(}\frac{1}{r}-\frac{1}{R}\Big{)}+\mathsf{B}(\varpi),\qquad\bm{v}=\varpi\omega(\varpi)\bm{e}_{\phi}.$
(2.11)

Here, in order to fix the idea, let us suppose the following assumption:

(B): The function $\omega$ belongs to the class $C^{1}([0,+\infty[)\cap
L^{\infty}(0,+\infty)$.

Now let us consider the static stationary solution
$(\rho,\bm{v})=(\rho(\bm{x}),\mathbf{0})$: that is, $\omega(\varpi)=0$, and

$\displaystyle\varUpsilon:$
$\displaystyle=-\Phi^{\langle\Omega\rangle}-\frac{{\mathsf{G}M_{0}}}{R}$
(2.12) so that $\displaystyle\varUpsilon$
$\displaystyle={\mathsf{G}M_{0}}\Big{(}\frac{1}{r}-\frac{1}{R}\Big{)}+\frac{\Omega^{2}}{2}\varpi^{2}\quad\mbox{for}\quad
r>R_{0},$

where $R$ is a positive constant at the present. If $R\geq R_{0}$, then
$R-R_{0}$ is the height of the stratosphere at the North and South Poles, and

$\varUpsilon_{\mathsf{P}}:={\mathsf{G}M_{0}}\Big{(}\frac{1}{R_{0}}-\frac{1}{R}\Big{)}\geq
0$ (2.13)

gives the density of the atmosphere at the Poles

$\rho_{\mathsf{P}}:=\Big{(}\frac{\gamma-1}{\mathsf{A}\gamma}\varUpsilon_{\mathsf{P}}\Big{)}^{\frac{1}{\gamma-1}}=\Big{(}\frac{(\gamma-1){\mathsf{G}M_{0}}}{\mathsf{A}\gamma}\Big{(}\frac{1}{R_{0}}-\frac{1}{R}\Big{)}\Big{)}^{\frac{1}{\gamma-1}}.$
(2.14)

Now we are going to observe the shape of the set

$\mathfrak{U}:=\\{(\varpi,z)\ |\
0\leq\varpi,|z|<\infty,R_{0}<r,-\Phi^{\langle\Omega\rangle}-\frac{{\mathsf{G}M_{0}}}{R}>0\\}.$
(2.15)

Let us introduce the non-dimensional variables $X=\varpi/R_{0},Z=z/R_{0}$ and
put

$\displaystyle F(X^{2},Z^{2};\kappa):=\frac{1}{\sqrt{X^{2}+Z^{2}}}+\kappa
X^{2},$ (2.16)
$\displaystyle\kappa=\frac{\Omega^{2}R_{0}^{3}}{2{\mathsf{G}M_{0}}}.$ (2.17)

Then

$\Phi^{\langle\Omega\rangle}=-\frac{{\mathsf{G}M_{0}}}{R_{0}}F(X^{2},Z^{2};\kappa)$
(2.18)

and
$\\{\displaystyle-\Phi^{\langle\Omega\rangle}-\frac{{\mathsf{G}M_{0}}}{R}>0\\}=\\{\displaystyle
F>\frac{R_{0}}{R}\\}$. Let us observe te shape of the level set
$\\{F(X^{2},Z^{2};\kappa)=\lambda\\}$ of $F$, $\lambda$ being a positive
number. To do so, solving

$F(X^{2},Z^{2};\kappa)=\lambda,$ (2.19)

we consider

$Z^{2}=\frac{g(X^{2})}{(\lambda-\kappa X^{2})^{2}},$ (2.20)

for $\displaystyle X^{2}<{\frac{\lambda}{\kappa}}$, where

$g(Q)=g(Q;\kappa,\lambda):=1-\lambda^{2}Q+2\lambda\kappa
Q^{2}-\kappa^{2}Q^{3}.$ (2.21)

We see

$g(0)=1,\qquad g\Big{(}\frac{\lambda}{\kappa}\Big{)}=1,$

and

$Dg(Q)=(\lambda-\kappa Q)(3\kappa Q-\lambda).$

Note that

$g\Big{(}\frac{\lambda}{3\kappa}\Big{)}>0\quad\Leftrightarrow\lambda^{3}<\frac{27}{4}\kappa.$

We see the shape of the graph of the function $g$ according to the following
three cases:

Case(L): $\lambda^{3}<\frac{27}{4}\kappa$.

Then $g(Q)>0$ for $0\leq Q<Q_{\infty}$, $g(Q_{\infty})=0$, $g(Q)<0$ for
$Q_{\infty}<Q<+\infty$. Here
$Q_{\infty}=Q_{\infty}(\frac{\lambda^{3}}{\kappa})$ is a number such that
$Q_{\infty}>\frac{\lambda}{\kappa}$.

Case (M): $\lambda^{3}=\frac{27}{4}\kappa$.

Then $g(Q)>0$ for $0\leq Q<\frac{\lambda}{3\kappa}(=\frac{9}{4\lambda^{2}})$,
$g(\frac{\lambda}{3\kappa})=0$, $g(Q)>0$ for
$\frac{\lambda}{3\kappa}<Q<Q_{\infty}$, $g(Q_{\infty})=0$, $g(Q)<0$ for
$Q_{\infty}<Q<+\infty$.

Case (H): $\lambda^{3}>\frac{27}{4}\kappa$.

Then $g(Q)>0$ for $0\leq Q<Q_{-}$, $g(Q_{-})=0$, $g(Q)<0$ for $Q_{-}<Q<Q_{+}$,
$g(Q_{+})=0$, $g(Q)>0$ for $Q_{+}<Q<Q_{\infty}$, $g(Q_{\infty})=0$, $g(Q)<0$
for $Q_{\infty}<Q<+\infty$. Here $Q_{\pm}=Q_{\pm}(\frac{\lambda^{3}}{\kappa})$
are numbers such that
$0<Q_{-}<\frac{\lambda}{3\kappa}<Q_{+}<\frac{\lambda}{\kappa}$ and
$Q_{\pm}\rightarrow\frac{\lambda}{3\kappa}$ as
$\frac{\lambda^{3}}{\kappa}\rightarrow\frac{27}{4}$.

Recall that we need the function $g(Q)$ only for $0\leq
Q\leq\frac{\lambda}{\kappa}$, and we do not take care of the behavior of
$g(Q)$ beyond $Q=\frac{\lambda}{\kappa}$, a fortiori, neither near nor beyond
$Q=Q_{\infty}$. Anyway, correspondingly we see the shape of the set
$\\{F>\lambda\\}$ as follows:

In Case (L), the set $\\{F>\lambda\\}$ is unbounded.

In Case (M), the set $\\{F>\lambda\\}$ consistes of two connected components,
say, $\mathfrak{U}_{0}$ and $\mathfrak{U}_{\infty}$, where $\mathfrak{U}_{0}$
is bounded and included in $\\{0\leq
X<\sqrt{\frac{\lambda}{3\kappa}}(=\frac{3}{2\lambda})\\}$ but
$\mathfrak{U}_{\infty}$ is an unbounded subset of
$\\{\sqrt{\frac{\lambda}{3\kappa}}<X<\sqrt{\frac{\lambda}{\kappa}}\\}$. Note
that $Dg(\frac{\lambda}{3\kappa})=0$ so that
$\partial\mathfrak{U}_{0}=\\{F=\lambda,X\leq\frac{3}{2\lambda}\\}$ has a
corner at the point $(X,Z)=(\sqrt{\frac{\lambda}{3\kappa}},0)$. Namely
$\partial\mathfrak{U}_{0}$ near this point can be described as

$Z=\pm
C\Big{(}\frac{3}{2\lambda}-X\Big{)}\Big{(}1+\Big{[}X-\frac{3}{2\lambda}\Big{]}_{1}\Big{)}\quad\mbox{as}\quad
X\rightarrow\frac{3}{2\lambda}-0,$

where $C=C(\kappa)$ is a positive constant and $[Y]_{1}$ stads for a
convergent power series of the form $\sum_{k\geq 1}a_{k}Y^{k}$.

In Case (H), the set $\\{F>\lambda\\}$ consistes of two connected components,
say $\mathfrak{U}_{0}\subset\\{0\leq X<\sqrt{Q_{-}}\\}$ and
$\mathfrak{U}_{\infty}\subset\\{\sqrt{Q_{+}}<X<\sqrt{\frac{\lambda}{\kappa}}\\}$.
$\mathfrak{U}_{0}$ is bounded but $\mathfrak{U}_{\infty}$ is unbounded. Note
that $X$ along $\partial\mathfrak{U}_{0}$ can be solved as a smooth ( real
analytic) function of $Z,|Z|\ll 1$, near the point $(X,Z)=(\sqrt{Q_{-}},0)$.
Namely the shape of
$\partial\mathfrak{U}_{0}=\\{F=\lambda,X<\sqrt{\frac{\lambda}{3\kappa}}\\}$
near this point is

$Z=\pm C\sqrt{Q_{-}-X}(1+[X-Q_{-}]_{1})\quad\mbox{as}\quad X\rightarrow
Q_{-}-0,$

$C=C(\kappa,\lambda)$ beng a positive constant. The point is not a corner, and
$\partial\mathfrak{U}_{0}$ is smooth. However it should be noticed that
$\mathfrak{U}_{0}$ is not an ellipse.

Therefore, applying the above observations to
$\displaystyle\lambda=\frac{R_{0}}{R}$, we claim

###### Theorem 1

Compactly supported axially and equatorially symmetric static stationary
solutions with the height of the stratosphere at the North Pole $R>R_{0}$ and
constant angular velocity $\Omega$ exist if and only if
$\displaystyle\kappa=\frac{\Omega^{2}R_{0}^{3}}{2{\mathsf{G}M_{0}}}\leq\frac{4}{27}\Big{(}\frac{R_{0}}{R}\Big{)}^{3}$.

In this sense we should restrict ourselves to the case
$0\leq\kappa<\frac{4}{27}\lambda^{3}$, that is, we require the following
assumption

(K):  $R>R_{0}$ and it holds

$0\leq\Big{(}\frac{R}{R_{0}}\Big{)}^{3}\kappa=\frac{\Omega^{2}R^{3}}{2{\mathsf{G}M_{0}}}<\frac{4}{27}.$
(2.22)

We put

$\mathcal{R}_{1}(\Omega^{2}):=\begin{cases}+\infty\quad\mbox{for}\quad\Omega^{2}=0\\\
\\\
\displaystyle\frac{2}{3}\Big{(}\frac{{\mathsf{G}M_{0}}}{\Omega^{2}}\Big{)}^{\frac{1}{3}}\quad\mbox{for}\quad\Omega^{2}>0\end{cases}.$
(2.23)

Then the condition (K) reads

$R_{0}<R<\mathcal{R}_{1}(\Omega^{2}).$ (2.24)

In order that $R_{0}<\mathcal{R}_{1}(\Omega^{2})$ it is necessary that

$\Omega^{2}<\Omega^{2}_{\mathrm{max}}:=\frac{4}{9}\frac{{\mathsf{G}M_{0}}}{(R_{0})^{3}}.$
(2.25)

Under the assumption (K), or, in Case (H) with $\lambda=\frac{R_{0}}{R}$, we
consider the number $\frac{3R}{2R_{0}}=\frac{3}{2\lambda}$. Since
$\frac{3}{2\lambda}<\sqrt{\frac{\lambda}{3\kappa}}$ and since $\displaystyle
g\Big{(}\frac{9}{4\lambda^{2}}\Big{)}<0$, which is not obvious but can be
shown, we see that $\sqrt{Q_{-}}<\frac{3}{2\lambda}$. On the other hand, we
see obviosly that
$\frac{3}{2\lambda}<\sqrt{\frac{\lambda}{3\kappa}}<\sqrt{Q_{+}}$. Therefore we
have the estimate

$\sqrt{Q_{-}}<\frac{3}{2\lambda}=\frac{3R}{2R_{0}}<\sqrt{Q_{+}}.$

Thus the closure of $\mathfrak{U}_{0}$ is included in $\\{R_{0}\leq
r,\varpi<{3R}/{2}\\}$, and $\varUpsilon$ is a classical solution on
$[0,+\infty[\times\mathfrak{D}$, where
$\mathfrak{D}=\\{R_{0}<r,\varpi<{3R}/{2}\\}$. So

$\bar{\rho}=\Big{(}\frac{\gamma-1}{\mathsf{A}\gamma}(\varUpsilon\restriction\mathfrak{D})\vee
0\Big{)}^{\frac{1}{\gamma-1}}$

turns out to be a classical solution on $[0,+\infty[\times\mathfrak{D}$. In
other words, if we put

$\varUpsilon_{\blacktriangle}=\begin{cases}\varUpsilon\quad\mbox{on}\quad\mathfrak{U}_{0}\\\
0\quad\mbox{on}\quad\\{R_{0}<r\\}\setminus\mathfrak{U}_{0},\end{cases}$ (2.26)

that is,

$\displaystyle\varUpsilon_{\blacktriangle}$ $\displaystyle=(\varUpsilon\vee
0)\cdot 1_{\varpi<3R/2}=(\varUpsilon\cdot 1_{\varpi<3R/2})\vee 0$
$\displaystyle=\begin{cases}\varUpsilon\quad\mbox{if}\quad\varUpsilon>0\quad\mbox{and}\quad\varpi<\frac{3R}{2}\\\
0\quad\mbox{otherwise}\end{cases},$ (2.27)

the density distribution $\bar{\rho}$ given by

$\bar{\rho}=\Big{(}\frac{\gamma-1}{\mathsf{A}\gamma}\varUpsilon_{\blacktriangle}\Big{)}^{\frac{1}{\gamma-1}}$
(2.28)

gives a classical solution $(\bar{\rho},\mathbf{0})$ on
$[0,\infty[\times(\mathbb{R}^{3}\setminus{\mathfrak{B}}_{0})$. Note that
$(\varUpsilon\cdot 1_{\varpi<3R/2},\mathbf{0})$ is not a classical
$(\varUpsilon,\mbox{\boldmath$v$})$-solution on
$[0,+\infty[\times(\mathbb{R}^{3}\setminus{\mathfrak{B}}_{0})$ in the sense of
Definition 4.

We are considering the uniformly rotating atmosphere which occupies

$\mathfrak{R}:=\\{\mbox{\boldmath$x$}\in\mathbb{R}^{3}\ |\
R_{0}<r,\rho^{\flat}(\mbox{\boldmath$x$})=\rho(\varpi,z)>0\\},$ (2.29)

provided (K). Then the boundary $\partial\mathfrak{R}$ of he atmosphere
consists of the two connected components
$\Sigma_{0}=\partial\mathfrak{B}_{0}=\\{r=R_{0}\\}$, the surface of the Earth,
and

$\Sigma_{1}=\\{\frac{1}{\sqrt{X^{2}+Z^{2}}}+\kappa
X^{2}=\frac{R_{0}}{R},\quad\mbox{with}\quad\varpi=R_{0}X,z=R_{0}Z\\},$

the stratosphere of the atmosphere. We are going to show that the boundary
$\Sigma_{1}$ is a physical vacuum boundary, that is, at each boundary point
$\mathrm{P}\in\Sigma_{1}$ we have
($\nabla\varUpsilon^{\flat}|\mbox{\boldmath$n$})<0$, where $n$ is the outer
normal vector at $\mathrm{P}$ and
$\varUpsilon^{\flat}(\mbox{\boldmath$x$})=\varUpsilon(\varpi,z)$. Recall that
$\displaystyle\varUpsilon=\frac{1}{\gamma-1}\frac{dP}{d\rho}$, while
$\displaystyle\frac{dP}{d\rho}$ is the square of the sound speed.

In fact if we consider the situation in the $(X,Z)$-plane at
$\mathrm{P}:X=X_{1},Z=Z_{1}$ with $0<X_{1}<\sqrt{Q_{-}},Z_{1}=f(X_{1})$ where

$f(X):=\frac{\sqrt{g(X^{2})}}{\lambda-\kappa
X^{2}}\quad\mbox{with}\quad\lambda=\frac{R_{0}}{R}.$

By a straight calculation, it can be shown that $Df(X)<0$ for
$0<X<\sqrt{Q_{-}}$. We have

$\mbox{\boldmath$n$}=\frac{1}{1+Df(X_{1})^{2}}\Big{(}-Df(X_{1})\frac{\partial}{\partial
X}+\frac{\partial}{\partial Z}\Big{)}$

so that

$\displaystyle(\nabla\varUpsilon^{\flat}|\mbox{\boldmath$n$})$
$\displaystyle=\frac{1}{1+Df(X_{1})^{2}}\Big{(}-\frac{\partial\varUpsilon}{\partial
X}Df(X_{1})+\frac{\partial\varUpsilon}{\partial Z}\Big{)}$
$\displaystyle=\frac{1}{1+Df(X_{1})^{2}}\frac{{\mathsf{G}M_{0}}}{R_{0}}\Big{(}\frac{XDf(X)-f(X)}{(X^{2}+f(X)^{2})^{3/2}}-2\kappa
XDf(X)\Big{)}_{X=X_{1}}.$

Since $0<X_{1},0<f(X_{1}),Df(X_{1})<0,0<\kappa$, we see
$(\nabla\varUpsilon^{\flat}|\mbox{\boldmath$n$})<0$ at $\mathrm{P}$. The
exceptional cases, the North Pole ($X=0$) and the Equator ($X=\sqrt{Q_{-}}$ )
, can be checked easily.

Now we consider a not static stationary solution

$\varUpsilon=\mathsf{G}M_{0}\Big{(}\frac{1}{r}-\frac{1}{R}\Big{)}+\mathsf{B}(\varpi),\qquad\bm{v}=\varpi\omega(\varpi)\bm{e}_{\phi}$
(2.30)

under the assumption:

(B): The function $\omega$ belongs to the class $C^{1}([0,+\infty[)\cap
L^{\infty}(0,+\infty)$.

Under this assumption, we put

$\tilde{\kappa}(X^{2})=\frac{R_{0}}{\mathsf{G}M_{0}}\frac{\mathsf{B}(R_{0}X)}{X^{2}}=\frac{R_{0}}{\mathsf{G}M_{0}}\frac{1}{X^{2}}\int_{0}^{R_{0}X}(\omega(\acute{\varpi})+\Omega)^{2}\acute{\varpi}d\acute{\varpi}.$
(2.31)

Note $X^{2}\mapsto\tilde{\kappa}(X^{2})X^{2}$ is continuos and montone
nondecreasing.

Put

$\kappa:=\sup_{X^{2}>0}\tilde{\kappa}=\frac{R_{0}^{3}}{2\mathsf{G}M_{0}}\|\omega+\Omega\|^{2},$
(2.32)

where

$\|\omega+\Omega\|=\sup_{\varpi>0}|\omega(\varpi)+\Omega|.$ (2.33)

Let us consider the stationary solution with compactly supported $\rho$ under
the assumption

($\tilde{\mathbf{K}}$): It holds that

$R_{0}<R,\qquad\Big{(}\frac{R}{R_{0}}\Big{)}^{3}\kappa=\frac{R^{3}}{2\mathsf{G}M_{0}}\|\omega+\Omega\|^{2}<\frac{4}{27}.$
(2.34)

The density is given by

${\rho}=\Big{(}\frac{\gamma-1}{\mathsf{A}\gamma}\Big{)}^{\frac{1}{\gamma-1}}({\varUpsilon}_{\blacktriangle})^{\frac{1}{\gamma
1-1}}\quad\mbox{with}\quad{\varUpsilon}_{\blacktriangle}=({\varUpsilon}\vee
0)\cdot 1_{\varpi<3R/2}.$ (2.35)

In fact, $\\{\varUpsilon>0\\}$ is
$\\{F(X^{2},Z^{2};\tilde{\kappa})>\lambda=\frac{R_{0}}{R}\\}$, where

$F(X^{2},Z^{2};\tilde{\kappa})=\frac{1}{\sqrt{X^{2}+Z^{2}}}+\tilde{\kappa}(X^{2})X^{2}.$
(2.36)

Here we are using the change of varialble $\varpi=R_{0}X,z=R_{0}Z$

Then $F(X^{2},Z^{2};\tilde{\kappa})>\lambda$ if and only if either 1)
$\tilde{\kappa}(X^{2})X^{2}\geq\lambda$, or 2)

$\tilde{\kappa}(X^{2})X^{2}<\lambda\quad\mbox{and}\quad
Z^{2}<f(X^{2};\tilde{\kappa})$

where

$f(X^{2};\tilde{\kappa})=\frac{1}{(\lambda-\tilde{\kappa}(X^{2})X^{2})^{2}}-X^{2}.$

Note that $f(X^{2};\tilde{\kappa})\leq f(X^{2};\kappa)$. Now we are supposing

$\tilde{\kappa}(X^{2})\leq\kappa<\frac{4}{27}\Big{(}\frac{R_{0}}{R}\Big{)}^{3}.$

So, $\\{\varUpsilon>0\\}$ has a bounded connected component of the form
$\\{0<X^{2}<Q_{-},Z^{2}<f(X^{2},\tilde{\kappa})\\}$. Here $Q_{-}$ is a
positive number $\leq Q_{-}(\frac{\lambda^{3}}{\kappa})<\frac{3}{2}\lambda$
such that $f(X^{2};\tilde{\kappa})>0$ for $0<X^{2}<Q_{-}$, and
$f(Q_{-};\tilde{\kappa})=0$.

###### Remark 1

In this situation with $R(>R_{0})$ being fixed, it is sufficient that the
function $\omega$ is given on the finite interval $[0,{3R}/{2}]$ as a function
of $C^{1}$-class on this interval, under a weaker assumption than (B), and we
can put

$\|\omega+\Omega\|_{\infty}=\sup_{0\leq\varpi\leq
3R/2}|\omega(\varpi)+\Omega|$

instead of (2.33).

##  3 Description of the flow by the Lagrangian co-ordinate

Let $\mathfrak{D}_{0}$ be an admissible domain with cover
$\tilde{\mathfrak{D}_{0}}$, and let $v$ be a classical admissible velocity
field on $[0,T_{0}[\times\mathfrak{D}_{0}$. Let $\mathfrak{D}$ with a cover
$\tilde{\mathfrak{D}}$ be an admissible subdomain of $\mathfrak{D}$, which we
shall call a proper asdmissible subdomain of $\mathfrak{D}$, such that
$\tilde{\mathfrak{D}}\Subset\tilde{\mathfrak{D}_{0}}$, namely, there is a
compact set $K$ such that
$\mathbf{Cl}\tilde{\mathfrak{D}}\subset\mathbf{Int}K\subset
K\subset\tilde{\mathfrak{D}_{0}}$.

Then there is a positive number $T(<T_{0})$ such that for any
$(\tau,\bar{\bm{x}})\in[0,T]\times\tilde{\mathfrak{D}}$ the initial value
problem of the ordinary differential equation

$\displaystyle\frac{d\bm{x}}{dt}=\mbox{\boldmath$v$}(t,\bm{x}),$ (3.1a)
$\displaystyle\bm{x}|_{t=\tau}=\bar{{\bm{x}}}$ (3.1b)

admits the unique solution $\bm{x}=\bm{\varphi}(t,\tau,\bar{{\bm{x}}})$ which
exists on $t\in[0,T]$ while $\bm{x}\in\tilde{\mathfrak{D}_{0}}$. We shall call
$\varphi$ the flow associated with the velocity field $\bm{v}$. The flow
$\varphi$, as a function, belongs to the class
$C^{1}([0,T]\times[0,T]\times\mathbf{Cl}\tilde{\mathfrak{D}}\ ;\
\tilde{\mathfrak{D}_{0}})$.

In fact the fundamental theorems of ordinary differential equations (see e.g.,
[6] ) applied to the vector field $v$ considered as a $C^{1}$-field on
$[0,T_{0}[\times\tilde{\mathfrak{D}_{0}}$ guarantee the existence and
uniqueness of the flow on $[0,T]$ valued in $\tilde{\mathfrak{D}_{0}}$ for
$\mbox{\boldmath$x$}\in\mathbf{Int}K$. On the other hand the boundary
condition (1.9) of $v$ on $\partial\mathfrak{B}_{0}$ guarantees that
$\partial\mathfrak{B}_{0}$ is an invarinat set of the equation so that the
flow starting with $\bar{\bm{x}}\in\mathfrak{D}$ cannot touch
$\partial\mathfrak{B}_{0}$ and must remain inside of
$\complement\mathfrak{B}_{0}$, namely,
$\bm{x}\in\tilde{\mathfrak{D}}\setminus\mathfrak{B}_{0}=\mathfrak{D}$.

Of course, abstractly speaking, we can consider the existence domain for
$\bm{\varphi}$ of the form
$\mathcal{O}=\bigcup_{(\tau,\bar{\bm{x}})\in[0,T_{0}[\times\tilde{\mathfrak{D}_{0}}}I_{\tau,\bar{\bm{x}}}\times\\{\tau\\}\times\\{\bar{\bm{x}}\\}$,
where $I_{\tau,\bar{\bm{x}}}\subset[0,T_{0}[$ is the maximal interval of
existence of the solution of (3.1a)(3.1b). The above observation says that
$[0,T]\subset I_{\tau,\bar{\bm{x}}}$ uniformly for
$\forall(\tau,\bar{\bm{x}})\in[0,T]\times\tilde{\mathfrak{D}}$.

Hereafter, we consider $\bm{\varphi}(t,0,\bar{\bm{x}})$ and denote, for the
sake of brevity,

$\bm{\varphi}(t,\bar{\bm{x}}):=\bm{\varphi}(t,0,\bar{\bm{x}}).$

Thus

$\displaystyle\frac{\partial}{\partial
t}\bm{\varphi}(t,\bar{\bm{x}})=\bm{v}(t,\bm{\varphi}(t,\bar{\bm{x}})),$ (3.2a)
$\displaystyle\bm{\varphi}(0,\bar{\bm{x}})=\bar{\bm{x}},$ (3.2b)

for $(t,\bar{\bm{x}})\in[0,T]\times\tilde{\mathfrak{D}}$.

Now let us suppose that the velocity field $\bm{v}$ is that of a classical
$(\varUpsilon,\bm{v})$-solution $(\varUpsilon,\bm{v})$ on
$[0,T_{0}[\times\mathfrak{D}_{0}$. We consider the associated flow
$\bm{\varphi}\in C^{1}([0,T]\times\mathbf{Cl}\tilde{\mathfrak{D}})$ which
satisfies (3.2a)(3.2b) for
$\forall(t,\bar{\bm{x}})\in[0,T]\times\tilde{\mathfrak{D}}$.

Hereafter we denote

$(D_{\bar{\bm{x}}}{\bm{x}})(t,\bar{\bm{x}})=D\bm{\varphi}(t,\bar{\bm{x}})=\Big{(}\frac{\partial}{\partial\bar{\bm{x}}^{\alpha}}\Big{(}\bm{\varphi}(t,\bar{\bm{x}})\Big{)}^{j}\Big{)}_{j,\alpha}.$
(3.3)

We shall use the inverse matrix of $D_{\bar{\bm{x}}}\bm{x}(t,\bar{\bm{x}})$.
Actually thereis a positive number $T_{1}(\leq T)$ such that
$D_{\bar{\bm{x}}}\bm{x}(t,\bar{\bm{x}})=D\bm{\varphi}(t,\bar{\bm{x}})$ is
invertible when $(t,\bar{\bm{x}})\in[0,T_{1}]\times\tilde{\mathfrak{D}}$. (
Proof. Let

$\Delta=D\bm{\varphi}(t,\bar{\bm{x}})-I=D\bm{\varphi}(t,\bar{\bm{x}})-D\bm{\varphi}(t,\bar{\bm{x}}).$

Since $\bm{\varphi}\in C^{1}([0,T]\times\mathbf{Cl}\tilde{\mathfrak{D}})$, for
any $\Theta,0<\Theta<1,$ there is $T_{1}\leq T$ such that $\|\Delta\|<\Theta$
for $(t,\bar{\bm{x}})\in[0,T_{1}]\times\tilde{\mathfrak{D}}$. then
$(D\bm{\varphi}(t,\bar{\bm{x}}))^{-1}$ is given by the Neumann series
$\sum_{k=0}^{\infty}(-\Delta)^{k}$. $\square$. )

If we denote by $\bm{\psi}_{t}$ the inverse mapping of
$\bm{\varphi}(t,\cdot)\restriction_{\mathfrak{V}}$, $\mathfrak{V}$ being a
neighborhood of $\bar{\bm{x}}_{0}\in\tilde{\mathfrak{D}},t\in[0,T_{1}]$, then
for $\bar{\bm{x}}\in\mathfrak{V}$,
$(D_{\bar{\bm{x}}}\bm{x}(t,\bar{\bm{x}}))^{-1}$ is nothing but the matrix
$\displaystyle\Big{(}\frac{\partial}{\partial
x^{j}}\Big{(}\psi_{t}(\bm{x})\Big{)}^{\alpha}\Big{)}_{\alpha,j}$ at
$\bm{x}=\bm{\varphi}(t,\bar{\bm{x}})$.

We note that it holds

$D_{\bar{\bm{x}}}\bm{x}(t,\bar{\bm{x}})=D\bm{\varphi}(t,\bar{\bm{x}})=\exp\Big{[}\int_{0}^{t}(D_{\bm{x}}\bm{v})(s,\bm{\varphi}(s,\bar{\bm{x}}))ds\Big{]}$
(3.4)

for $(t,\bar{\bm{x}})\in[0,T]\times\tilde{\mathfrak{D}}$. Here, of course,
$D_{\bm{x}}\bm{v}$ stands for the matrix $\displaystyle\Big{(}\frac{\partial
v^{k}}{\partial x^{j}}\Big{)}_{k,j}$.

In fact, we see

$\displaystyle\frac{\partial}{\partial t}D_{\bar{\bm{x}}}x(t,\bar{\bm{x}})$
$\displaystyle=\frac{\partial}{\partial t}D\bm{\varphi}(t,\bar{\bm{x}})=$
$\displaystyle=D_{\bar{\bm{x}}}\frac{\partial}{\partial
t}\bm{\varphi}(t,\bar{\bm{x}})=D_{\bar{\bm{x}}}\bm{v}(t,\bm{\varphi}(t,\bm{x}))=$
$\displaystyle=D_{\bm{x}}\bm{v}(t,\bm{x})\Big{|}_{\bm{x}=\bm{\varphi}(t,\bar{\bm{x}})}.D_{\bar{\bm{x}}}\bm{x}(t,\bar{\bm{x}}).$

and

$D_{\bar{\bm{x}}}(0,\bar{\bm{x}})=I.$

.

Let us put

$\overset{\circ}{\varUpsilon}(\bm{x})=\varUpsilon(0,\bm{x}).$ (3.5)

and

$\varUpsilon^{L}(t,\bar{\bm{x}})=\varUpsilon(t,\bm{\varphi}(t,\bar{\bm{x}}))$
(3.6)

for $(t,\bar{\bm{x}})\in[0,T]\times\tilde{\mathfrak{D}}$.

Integrating the equation of continuity (1.11a), we claim

###### Proposition 1

It holds that

$\varUpsilon^{L}(t,\bar{\bm{x}})=\varUpsilon(t,\bm{\varphi}(t,\bar{\bm{x}}))=\overset{\circ}{\varUpsilon}(\bar{\bm{x}})\mathrm{det}D\bm{\varphi}(t,\bar{\bm{x}})^{-(\gamma-1)}$
(3.7)

for $t\in[0,T],\bar{\bm{x}}\in\mathfrak{D}$.

Proof. The equation (1.11a) reads

$\frac{D}{Dt}\log\varUpsilon=-(\gamma-1)(\nabla_{\bar{x}}|\mbox{\boldmath$v$})(t,{\mbox{\boldmath$x$}})$

where $\varUpsilon\not=0$. Therefore

$\varUpsilon(t,\bm{x})=\overset{\circ}{\varUpsilon}(\bar{\bm{x}})\exp\Big{[}-(\gamma-1)\int_{0}^{t}(\nabla_{\bar{x}}|\mbox{\boldmath$v$})(t^{\prime},\bm{\varphi}(t^{\prime},\bar{\mbox{\boldmath$x$}}))dt^{\prime}\Big{]}$

We note

$(\nabla_{\bar{x}}|\mbox{\boldmath$v$})(t,{\mbox{\boldmath$x$}})=\mathrm{tr}D_{\bar{x}}\mbox{\boldmath$v$}(t,{\mbox{\boldmath$x$}}).$

But, since

$\displaystyle\Big{(}\frac{\partial}{\partial
t}\Big{)}_{\bar{\mbox{\boldmath$x$}}}D\bm{\varphi}(t,\bar{\mbox{\boldmath$x$}})$
$\displaystyle=D_{\bar{\mbox{\boldmath$x$}}}\Big{(}\frac{\partial}{\partial
t}\Big{)}_{\bar{\mbox{\boldmath$x$}}}{\mbox{\boldmath$x$}}(t,0,\bar{\mbox{\boldmath$x$}})=D_{\bar{\mbox{\boldmath$x$}}}\mbox{\boldmath$v$}(t,{\mbox{\boldmath$x$}})$
$\displaystyle=D_{{\mbox{\boldmath$x$}}}\mbox{\boldmath$v$}(t,{\mbox{\boldmath$x$}}).D_{\bar{\mbox{\boldmath$x$}}}{\mbox{\boldmath$x$}}=D_{{\mbox{\boldmath$x$}}}\mbox{\boldmath$v$}(t,{\mbox{\boldmath$x$}}).D\bm{\varphi}(t,\bar{\mbox{\boldmath$x$}}),$

we have

$D_{\bar{x}}\mbox{\boldmath$v$}(t,{\mbox{\boldmath$x$}})=\Big{(}\frac{\partial}{\partial
t}\Big{)}_{\bar{\mbox{\boldmath$x$}}}D\bm{\varphi}(t,\bar{\mbox{\boldmath$x$}}).D\bm{\varphi}(t,\bar{\mbox{\boldmath$x$}})^{-1}.$

Thus

$(\nabla_{\bar{x}}|\mbox{\boldmath$v$})(t,{\mbox{\boldmath$x$}})=\mathrm{tr}\Big{(}\Big{(}\frac{\partial}{\partial
t}\Big{)}_{\bar{\mbox{\boldmath$x$}}}D\bm{\varphi}(t,\bar{\mbox{\boldmath$x$}}).D\bm{\varphi}(t,\bar{\mbox{\boldmath$x$}})^{-1}\Big{)}=\Big{(}\frac{\partial}{\partial
t}\Big{)}_{\bar{\bm{x}}}\log\mathrm{det}D\bm{\varphi}(t,\bar{\mbox{\boldmath$x$}}).$

Since $D\bm{\varphi}(0,\bar{\mbox{\boldmath$x$}})=I$, it follows that

$\int_{0}^{t}(\nabla_{\bar{x}}|\bm{v})(t^{\prime},\varphi(t^{\prime},\bar{\bm{x}}))dt^{\prime}=\log\mathrm{det}D\bm{\varphi}(t,\bar{\mbox{\boldmath$x$}}).$

$\square$

Let us consider the equation of motion (1.11b), namely

$\frac{D\bm{v}}{Dt}+B\bm{v}+\mathrm{glad}_{\bm{x}}(\varUpsilon+\Phi^{\langle\Omega\rangle})=0,$
(3.8)

where we denote

$B\bm{v}=2\Omega\begin{bmatrix}0\\\ \\\ 0\\\ \\\
1\end{bmatrix}\times\bm{v}=2\Omega\begin{bmatrix}-v^{2}\\\ \\\ v^{1}\\\ \\\
0\end{bmatrix}=2\Omega\begin{bmatrix}0&-1&0\\\ \\\ 1&0&0\\\ \\\
0&0&0\end{bmatrix}\bm{v}.$ (3.9)

We put

$\displaystyle\bm{v}^{L}$
$\displaystyle:=\Big{(}D\bm{\varphi}(t,\bar{\bm{x}})\Big{)}^{-1}\bm{v}(t,\bm{\varphi}(t,\bar{\bm{x}}))$
$\displaystyle=\Big{(}D\bm{\varphi}(t,\bar{\bm{x}})\Big{)}^{-1}\frac{\partial}{\partial
t}\bm{\varphi}(t,\bar{\bm{x}}).$ (3.10)

Then (3.8) reads

$\displaystyle\frac{\partial\bm{v}^{l}}{\partial
t}+(D_{\bar{\bm{x}}}\bm{x})^{-1}\frac{\partial}{\partial
t}(D_{\bar{\bm{x}}}\bm{x})\bm{v}^{L}+$
$\displaystyle+(D_{\bar{\bm{x}}}\bm{x})^{-1}B(D_{\bar{\bm{x}}}\bm{x})\bm{v}^{L}+(D_{\bar{\bm{x}}}\bm{x})^{-1}((D_{\bar{\bm{x}}}\bm{x})^{-1})^{\top}\mathrm{grad}_{\bar{\bm{x}}}(\Upsilon^{L}+\Phi^{\langle\Omega\rangle
L})=0$ (3.11)

on $(t,\bar{\bm{x}})\in[0,T]\times\tilde{\mathfrak{D}}$. Here

$\Phi^{\langle\Omega\rangle
L}(t,\bar{\bm{x}}):=\Phi^{\langle\Omega\rangle}(\bm{\varphi}(t,\bar{\bm{x}})).$
(3.12)

Let us consider tge boundary condition

$(\bm{n}(\bm{x})|\bm{v}(t,\bm{x}))=0\quad\forall(t,\bm{x})\in[0,T]\times\partial\mathfrak{B}_{0},$
(3.13)

where

$\bm{n}(\bm{x})=\frac{1}{R_{0}}\bm{x}\quad\mbox{for}\quad\bm{x}\in\partial\mathfrak{B}_{0}=\\{\|\bm{x}\|=R_{0}\\}.$
(3.14)

Put

$\bm{n}^{L}(t,\bar{\bm{x}})=\frac{1}{R_{0}}(D_{\bar{\bm{x}}}\bm{x}(t,\bar{\bm{x}}))^{\top}\bm{\varphi}(t,\bar{\bm{x}})$
(3.15)

for $(t,\bar{\bm{x}})\in[0,T]\times\partial\mathfrak{B}_{0}$. Then (3.13)
reads

$(\bm{n}^{L}(t,\bar{\bm{x}})|\bm{v}^{L}(t,\bar{\bm{x}}))=0\quad\mbox{for}\quad(t,\bar{\bm{x}})\in[0,T]\times\partial\mathfrak{B}_{0}.$
(3.16)

Note that (3.16) means

$\frac{1}{2}\frac{\partial}{\partial
t}\|\bm{\varphi}(t,\bar{\bm{x}})\|^{2}=\Big{(}\bm{\varphi}\Big{|}\frac{\partial}{\partial
t}\bm{\varphi}\Big{)}=0\quad\mbox{if}\quad\bar{\bm{x}}\in\partial\mathfrak{B}_{0},$
(3.17)

so that , if $\bar{\bm{x}}\in\partial\mathfrak{B}_{0}$, then
$\bm{\varphi}(t,\bar{\bm{x}})\in\partial\mathfrak{B}_{0}$ for $\forall t$.

Now let us suppose that there is a vector field
${\bm{\varphi}}(t,\bar{\bm{x}}),(t,\bar{\bm{x}})\in[0,T]\times\tilde{\mathfrak{D}},$
such that
$\bm{\varphi}\in\bigcap_{\ell=1,2}C^{2-\ell}([0,T];C^{\ell}(\mathbf{Cl}\tilde{\mathfrak{D}}))$
and

$\bm{\varphi}(0,\bar{\bm{x}})=\bar{\bm{x}}$

which satisfies the equation (3.11), where we read

$\displaystyle D_{\bar{\bm{x}}}\bm{x}=D\bm{\varphi}(t,\bar{\bm{x}}),$ (3.18a)
$\displaystyle\bm{v}^{L}=\Big{(}D\bm{\varphi}(t,\bar{\bm{x}})\Big{)}^{-1}\frac{\partial}{\partial
t}\bm{\varphi}(t,\bar{\bm{x}}),$ (3.18b)
$\displaystyle\varUpsilon^{L}=\overset{\circ}{\varUpsilon}(\bar{\bm{x}})\det\Big{(}D\bm{\varphi}(t,\bar{\bm{x}})\Big{)}^{-(\gamma-1)},$
(3.18c) $\displaystyle\Phi^{\langle\Omega\rangle
L}=\Phi^{\langle\Omega\rangle}(\bm{\varphi}(t,\bar{\bm{x}})),$ (3.18d)

and the boundary condition (3.16), where we read

$\bm{n}^{L}(t,\bar{\bm{x}})=\frac{1}{R_{0}}(D\bm{\varphi}(t,\bar{\bm{x}}))^{\top}\bm{\varphi}(t,\bar{\bm{x}}),$
(3.19)

forgetting that the vector field $\bm{\varphi}$ comes from the flow generated
by the velocity field $\bm{v}$ of a classical solution $(\varUpsilon,\bm{v})$.
Namely, we use the following definitions:

###### Definition 6

Let $\mathfrak{D}$ be an admissible domain with cover $\tilde{\mathfrak{D}}$.
A scalar field $\overset{\circ}{\varUpsilon}$ defined on
$\tilde{\mathfrak{D}}$ is called an admissible $\varUpsilon$-data, if
$\overset{\circ}{\varUpsilon}\in C^{1}(\mathbf{Cl}\tilde{\mathfrak{D}})$ and
$\overset{\circ}{\varUpsilon}>0$ on $\partial\mathfrak{B}_{0}$.

###### Definition 7

Let $\mathfrak{D}$ be an admissible domain with cover $\tilde{\mathfrak{D}}$,
and let $\overset{\circ}{\varUpsilon}$ be an admissible $\varUpsilon$-data on
$\mathfrak{D}$. A vector field $\varphi$ is said to be an admissible flow on
$[0,T]\times\mathfrak{D}$ associated with $\overset{\circ}{\varUpsilon}$, if
1)
$\bm{\varphi}\in\bigcap_{\ell=1,2}C^{2-\ell}([0,T];C^{\ell}(\mathbf{Cl}\tilde{\mathfrak{D}}))$,
2) the equation (3.11) and the boundary condition (3.16) are satisfied on
$[0,T]\times\mathfrak{D}$ and on $[0,T]\times\partial\mathfrak{B}_{0}$, and 3)
the initial condition

$\bm{\varphi}(0,\bar{\bm{x}})=\bar{\bm{x}}\quad\mbox{for}\quad\forall\bar{\bm{x}}\in\tilde{\mathfrak{D}}$

holds.

As for the existence of the inverse mapping of $\bm{\varphi}(t,\cdot)$, we
have

###### Proposition 2

Let
$\bm{\varphi}\in\bigcap_{\ell=1,2}C^{2-\ell}([0,T];C^{\ell}(\mathbf{Cl}\tilde{\mathfrak{D}})$
be a vector field such that

$\bm{\varphi}(0,\bar{\bm{x}})=\bar{\bm{x}}\quad\mbox{for}\quad\forall\bar{\bm{x}}\in\tilde{\mathfrak{D}},$

Let ${\mathfrak{O}},\mathfrak{O}_{0},\mathfrak{O}_{1}$ be connected open
subsets of $\mathbb{R}^{3}$ such that

$\mathfrak{B}_{0}\subset\mathfrak{O}_{0}\Subset{\mathfrak{O}}\Subset\mathfrak{O}_{1}\Subset\tilde{\mathfrak{D}}$
(3.20)

and $\mathfrak{O}_{1}$ is convex. Then for a sufficiently small ${T_{1}}(<T)$
there is a mapping
$\bm{\psi}(t,\cdot):{\mathfrak{O}}\rightarrow\mathfrak{O}_{1}$, $t$ being
$\in[0,{T_{1}}]$, such that

$\bm{x}=\bm{\varphi}(t,\bm{\psi}(t,\bm{x}))$ (3.21)

for $\forall(t,\bm{x})\in[0,{T_{1}}]\times{\mathfrak{O}}$. Moreover

$\mathfrak{O}_{0}\subset\bm{\psi}(t,{\mathfrak{O}})=\\{\bm{\psi}(t,\bm{x})\ |\
\bm{x}\in{\mathfrak{O}}\\}$ (3.22)

for $\forall t\in[0,{T_{1}}]$.

Proof. Fixing $t$, we are going to solve the equation for unknown
$\bar{\bm{x}}$

$\bm{x}=\bm{\varphi}(t,\bar{\bm{x}})$

for gven $\bm{x}$. Writing

$F(\bar{\bm{x}})=\bm{x}+\bar{\bm{x}}-\bm{\varphi}(t,\bar{\bm{x}}),$

we convert the equation to the fixed point problem

$\bar{\bm{x}}=F(\bar{\bm{x}}).$

Note

$\displaystyle\|F(\bar{\bm{x}})-\bm{x}\|=\|\bar{\bm{x}}-\bm{\varphi}(t,\bar{\bm{x}})\|=o(1),$
$\displaystyle\|DF(\bar{\bm{x}})\|=\|I-D\bm{\varphi}(t,\bar{\bm{x}})\|=o(1)$

uniformly for $\bar{\bm{x}}\in\mathfrak{O}_{1}$ as $t\rightarrow 0$, since
$\bm{\varphi}(0,\bar{\bm{x}})=\bar{\bm{x}}$. Therefore, if
$\bm{x}\in{\mathfrak{O}}$, we see $F(\bar{\bm{x}})\in\mathfrak{O}_{1}$ for
$\bar{\bm{x}}\in\mathfrak{O}_{1}$ and
$\|F(\bar{\bm{x}}^{\prime})-F(\bar{\bm{x}})\|\leq\Theta\|\bar{\bm{x}}^{\prime}-\bar{\bm{x}}\|$
for $\bar{\bm{x}}^{\prime},\bar{\bm{x}}\in\mathfrak{O}_{1}$, $\Theta$ being
$\in]0,1[$, provided that $0\leq t\leq{T_{1}}\ll 1$. Here recall that
$\mathfrak{O}_{1}$ is supposed to be convex. Therefore the fixed point problem
admits the unique soluton
$\bar{\bm{x}}=\bm{\psi}(t,\bm{x})\in\mathfrak{O}_{1}$ for
$(t,\bm{x})\in[0,{T_{1}}]\times{\mathfrak{O}}$. Then

$\bm{x}=\bm{\varphi}(t,\bm{\psi}(t,\bm{x}))$

for $(t,\bm{x})\in[0,{T_{1}}]\times{\mathfrak{O}}$.

Since $\mathbf{Cl}\mathfrak{O}_{0}\subset{\mathfrak{O}}$, taking ${T_{1}}$
smaller if necessary, we can assume that
$\bm{\varphi}(t,\bar{\bm{x}})\in\tilde{\mathfrak{D}}$ if
$(t,\bar{\bm{x}})\in[0,{T_{1}}]\times\mathfrak{O}_{0}$. Thus, if
$\bar{\bm{x}}\in\mathfrak{O}_{0}$, there is $\bm{x}\in{\mathfrak{O}}$ such
that $\bar{\bm{x}}=\bm{\psi}(t,\bm{x})$ provided that $t\in[0,{T_{1}}]$.
$\square$

In this situation, we can claim

###### Theorem 2

Let $\mathfrak{D}$ be an admissible domain with cover $\tilde{\mathfrak{D}}$,
$\overset{\circ}{\varUpsilon}$ be an admissible $\varUpsilon$-data on
$\mathfrak{D}$, and $\varphi$ be an admissible flow associated with
$\overset{\circ}{\varUpsilon}$. Let ${T_{1}},{\mathfrak{O}}$ be those of
Proposition 2. Put

$\displaystyle\varUpsilon(t,\bm{x})=\overset{\circ}{\varUpsilon}(\bm{\psi}(t,\bm{x}))\Big{(}\mathrm{det}D\bm{\varphi}(t,\bar{\bm{x}})\Big{|}_{\bar{\bm{x}}=\bm{\psi}(t,\bm{x})}\Big{)}^{-(\gamma-1)}$
(3.23) $\displaystyle\bm{v}(t,\bm{x})=\frac{\partial}{\partial
t}\bm{\varphi}(t,\bar{\bm{x}})\Big{|}_{\bar{\bm{x}}=\bm{\psi}(t,\bm{x})}$
(3.24)

for $(t,\bm{x})\in[0,{T_{1}}]\times{\mathfrak{O}}$. Then
$(\varUpsilon,\bm{v})$ is a classical $(\varUpsilon,\bm{v})$-solution on
$[0,{T_{1}}]\times\underline{\mathfrak{D}}$, where
$\underline{\mathfrak{D}}={\mathfrak{O}}\setminus\mathfrak{B}_{0}$.

• Now let us derive the linearized approximation of (3.7), (3.11), (3.16).
Namely, we fix a stationary solution $(\bar{\Upsilon},\bm{0})$, say, we take

$\bar{\varUpsilon}(\bm{x})=-\Phi^{\langle\Omega\rangle}(\bm{x})-\frac{\mathsf{G}M_{0}}{R}$
(3.25)

under the assumption (K). Let us denote

$\mathfrak{R}:=\\{\bar{\Upsilon}>0,\varpi<3R/2,r>R_{0}\\}=\\{\bar{\rho}>0,r>R_{0}\\}.$
(3.26)

Considering small $\varepsilon$, a quantity $Q$ will denoted by
$\bm{O}(\varepsilon)$ if $Q$ and its derivatives are of order $O(\varepsilon)$
uniformly on each bounded interval of $t$. We assume that
$\Upsilon-\bar{\Upsilon},\bm{v}$ are of $\bm{O}(\varepsilon)$.

Then (3.4) shows

$\displaystyle D_{\bar{\bm{x}}}\bm{x}(t,\bar{\bm{x}})$
$\displaystyle=I+\int_{0}^{t}(D_{\bm{x}}\bm{v}(s,\bm{\varphi}(s,\bar{\bm{x}}))ds+\bm{O}(\varepsilon^{2})$
(3.27) $\displaystyle=I+\bm{O}(\varepsilon),$ (3.28) $\displaystyle
D_{\bm{x}}\bar{\bm{x}}(t,\bm{x})$ $\displaystyle=I+\bm{O}(\varepsilon),$
(3.29)

and so on.

The equation (3.7) gives

$\Upsilon^{L}(t,\bar{\bm{x}})=\overset{\circ}{\Upsilon}(\bar{\bm{x}})\Big{(}1-(\gamma-1)\mathrm{div}_{\bar{\bm{x}}}(\bm{\varphi}(t,\bar{\bm{x}})-\bar{\bm{x}})\Big{)}+\bm{O}(\varepsilon^{2}).$
(3.30)

Recalling (3.25), we have

$\Upsilon^{L}(t,\bar{\bm{x}})+\Phi^{\langle\Omega\rangle}(\bm{\varphi}(t,\bar{\bm{x}}))=\Upsilon^{L}(t,\bar{\bm{x}})-\bar{\Upsilon}_{L}(\bar{\bm{x}})-\frac{\mathsf{G}M_{0}}{R}.$
(3.31)

Here we define $\bar{\Upsilon}_{L}$ by

$\bar{\Upsilon}_{L}(\bar{\bm{x}})=\bar{\Upsilon}(\bm{\varphi}(t,\bar{\bm{x}})).$
(3.32)

Then

$\bar{\Upsilon}_{L}(\bar{\bm{x}})=\bar{\Upsilon}(\bar{\bm{x}})+(\mathrm{grad}_{\bar{\bm{x}}}\bar{\Upsilon}_{L}(\bar{\bm{x}})|\bm{\varphi}(t,\bar{\bm{x}})-\bar{\bm{x}})+\bm{O}(\varepsilon^{2}).$
(3.33)

Summing up, putting

$\displaystyle G(t,\bar{\bm{x}})$
$\displaystyle:=\Upsilon(t,\bm{\varphi}(t,\bar{\bm{x}})+\Phi^{\langle\Omega\rangle}(\bm{\varphi}(t,\bar{\bm{x}}))+\frac{\mathsf{G}M_{0}}{R},$
$\displaystyle=\Upsilon^{L}(t,\bar{\bm{x}})-\bar{\Upsilon}_{L}(\bar{\bm{x}}),$
(3.34)

we see

$\displaystyle G(t,\bar{\bm{x}})$
$\displaystyle=\overset{\circ}{\Upsilon}(\bar{\bm{x}})-\bar{\Upsilon}(\bar{\bm{x}})-(\gamma-1)\overset{\circ}{\Upsilon}(\bar{\bm{x}})\mathrm{div}_{\bar{\bm{x}}}(\bm{\varphi}(t,\bar{\bm{x}})-\bar{\bm{x}})+$
$\displaystyle-(\mathrm{grad}_{\bar{\bm{x}}}\bar{\Upsilon}_{L}(\bar{\bm{x}})|\bm{\varphi}(t,\bar{\bm{x}})-\bar{\bm{x}})+\bm{O}(\varepsilon^{2}).$
(3.35)

Let a constant vector field
$\overset{\circ}{\bm{\xi}}(\bar{\bm{x}})=\bm{O}(\varepsilon)$ on
$\mathfrak{R}$ be given arbitrarily. We take $\overset{\circ}{\varUpsilon}$
given by

$\overset{\circ}{\varUpsilon}(\bar{\bm{x}})=\bar{\varUpsilon}(\bar{\bm{x}})-(\gamma-1)\bar{\varUpsilon}(\bar{\bm{x}})\mathrm{div}_{\bar{\bm{x}}}\overset{\circ}{\bm{\xi}}-(\mathrm{grad}_{\bar{\bm{x}}}\bar{\varUpsilon}_{L}(\bar{\bm{x}})|\overset{\circ}{\bm{\xi}}).$
(3.36)

Then it turns out to be that

$\displaystyle
G(t,\bar{\bm{x}})=-(\gamma-1)\bar{\Upsilon}(\bar{\bm{x}})\mathrm{div}_{\bar{\bm{x}}}\hat{\bm{\xi}}-(\mathrm{grad}_{\bar{\bm{x}}}\bar{\Upsilon}_{L}(\bar{\bm{x}})\Big{|}\hat{\bm{\xi}})+\bm{O}(\varepsilon^{2}),$
(3.37)
$\displaystyle=-(\gamma-1)\bar{\Upsilon}(\bar{\bm{x}})\mathrm{div}_{\bar{\bm{x}}}\hat{\bm{\xi}}-(\mathrm{grad}_{\bar{\bm{x}}}\bar{\Upsilon}(\bar{\bm{x}})\Big{|}\hat{\bm{\xi}})+\bm{O}(\varepsilon^{2}),$
(3.38)

where

$\hat{\bm{\xi}}(t,\bar{\bm{x}})=\bm{\varphi}(t,\bar{\bm{x}})-\bar{\bm{x}}+\overset{\circ}{\bm{\xi}}(\bar{\bm{x}})$
(3.39)

and $\mathrm{grad}_{\bar{\bm{x}}}\bar{\varUpsilon}(\bar{\bm{x}})$ means
$\Big{(}\mathrm{grad}_{\bm{x}}\bar{\varUpsilon}(\bm{x})\Big{)}\Big{|}_{\bm{x}=\bar{\bm{x}}}$
so that

$\displaystyle\mathrm{grad}_{\bar{\bm{x}}}\bar{\varUpsilon}_{L}(\bar{\bm{x}})$
$\displaystyle=(D_{\bar{\bm{x}}}\bm{x}(t,\bar{\bm{x}}))^{\top}\mathrm{grad}_{\bar{\bm{x}}}\bar{\varUpsilon}(\bar{\bm{x}})$
$\displaystyle=\mathrm{grad}_{\bar{\bm{x}}}\bar{\varUpsilon}(\bar{\bm{x}})+\bm{O}(\varepsilon).$

Then we see

$G(t,\bar{\bm{x}})=-\bar{\sigma}(\bar{\bm{x}})\mathrm{div}_{\bar{\bm{x}}}(\bar{\rho}(\bar{\bm{x}})\hat{\bm{\xi}}(t,\bar{\bm{x}}))+\bm{O}(\varepsilon^{2})\quad\mbox{on}\quad\mathfrak{R},$
(3.40)

where

$\sigma=\frac{d\Upsilon}{d\rho}=\mathsf{A}\gamma\rho^{\gamma-2}.$ (3.41)

Moreover we introduce

$\bm{\xi}(t,\bar{\bm{x}}):=(D\bm{\varphi}(t,\bar{\bm{x}}))^{-1}\hat{\bm{\xi}}(t,\bar{\bm{x}})-\int_{0}^{t}\frac{\partial}{\partial
s}\Big{[}(D\bm{\varphi}(s,\bar{\bm{x}}))^{-1}\Big{]}\hat{\bm{\xi}}(s,\bar{\bm{x}})ds.$
(3.42)

Then we have

$\bm{\xi}=\hat{\bm{\xi}}+\bm{O}(\varepsilon^{2}),$ (3.43)

$\frac{\partial\bm{\xi}}{\partial t}=\bm{v}^{L},$ (3.44)
$\bm{\xi}(0,\bar{\bm{x}})=\overset{\circ}{\bm{\xi}}(\bar{\bm{x}})$ (3.45)

and

$G(t,\bar{\bm{x}})=-\bar{\sigma}(\bar{\bm{x}})\mathrm{div}_{\bar{\bm{x}}}(\bar{\rho}(\bar{\bm{x}})\bm{\xi}(t,\bar{\bm{x}}))+\bm{O}(\varepsilon^{2})$
(3.46)

The approximation of the equation of motion (3.11) is clearly given by

$\frac{\partial\bm{v}^{L}}{\partial
t}+B\bm{v}^{L}+\mathrm{grad}_{\bar{\bm{x}}}G=\bm{O}(\varepsilon^{2}).$ (3.47)

Inserting (3.44), (3.46) into (3.47), we have

$\frac{\partial^{2}\bm{\xi}}{\partial t^{2}}+B\frac{\partial\bm{\xi}}{\partial
t}+\mathrm{grad}_{\bar{\bm{x}}}\Big{(}-\bar{\sigma}(\bar{\bm{x}})\mathrm{div}_{\bar{\bm{x}}}(\bar{\rho}(\bar{\bm{x}})\bm{\xi})\Big{)}=\bm{O}(\varepsilon^{2}).$
(3.48)

The approximation of the boundary condition (3.16) is given by

$\Big{(}\bm{n}(\bar{\bm{x}})\Big{|}\frac{\partial\bm{\xi}}{\partial
t}(t,\bar{\bm{x}})\Big{)}=0\quad\mbox{for}\quad(t,\bar{\bm{x}})\in[0,T]\times\partial\mathfrak{B}_{0}.$
(3.49)

In fact we have

$\bm{n}^{L}=\frac{1}{R_{0}}(I+\bm{O}(\varepsilon))(\bar{\bm{x}}+\bm{O}(\varepsilon))=\bm{n}(\bar{\bm{x}})+\bm{O}(\varepsilon)$

and

$\bm{v}^{L}=\frac{\partial\bm{\xi}}{\partial t}=\bm{O}(\varepsilon).$

Of course (3.49) is equivalent to

$(\bm{n}(\bar{\bm{x}})|\bm{\xi}(t,\bar{\bm{x}}))=0\quad\mbox{for}\quad(t,\bar{\bm{x}})\in[0,T]\times\partial\mathfrak{B}_{0},$
(3.50)

provided that

$(\bm{n}(\bar{\bm{x}})|\overset{\circ}{\bm{\xi}}(\bar{\bm{x}}))=0\quad\mbox{on}\quad\partial\mathfrak{B}_{0}.$
(3.51)

====

Now the domain
$\mathfrak{R}=\\{\bar{\rho}>0\\}=\\{\bar{\varUpsilon}>0\\}\cap\\{\varpi<3R/2\\}$
has the form

$\mathfrak{R}=\\{R_{0}<r<R\cdot
H(\zeta^{2};\kappa,\lambda),\quad-1\leq\zeta:=\frac{z}{r}\leq 1\\},$

where $\zeta^{2}\mapsto H(\zeta^{2};\kappa,\lambda)$ is a smooth monotone
function on $[0,1]$ such that $1=H(1;\kappa,\lambda)\leq H(0;\kappa,\lambda)$.
(Note $H(1)=H(0)\Leftrightarrow\kappa=0,\mbox{that is,}\Omega=0$.) Given a
small $\overset{\circ}{\bm{\xi}}$, we consider $\overset{\circ}{\varUpsilon}$
determined by (3.36). Of course
$\overset{\circ}{\varUpsilon}-\bar{\varUpsilon}$ is small, but the topology of
$\\{\overset{\circ}{\varUpsilon}>0\\}\cap\\{\varpi<3R/2\\}$ is not clear,
generaly speaking. In fact, as for
$\\{\overset{\circ}{\varUpsilon}>0\\}=\Big{\\{}\Big{(}1-(\gamma-1)(\nabla|\overset{\circ}{\bm{\xi}})\Big{)}\bar{\varUpsilon}>(\nabla\bar{\varUpsilon}|\overset{\circ}{\bm{\xi}})\Big{\\}}$,
we have that $\displaystyle\frac{\nabla\bar{\varUpsilon}}{\bar{\varUpsilon}}$
may diverge along the vacuum boundary
$\Sigma_{1}=\\{\bar{\varUpsilon}=0,\varpi<\frac{3R}{2}\\}=\\{\bar{\rho}=0\\}$
of $\bar{\rho}$. At least, if, for example,
$\overset{\circ}{\mbox{\boldmath$\xi$}}_{1}\in
C_{0}^{\infty}(\Sigma_{0}\cup\mathfrak{R})$, then for
$\overset{\circ}{\mbox{\boldmath$\xi$}}=\varepsilon\overset{\circ}{\mbox{\boldmath$\xi$}}_{1}$
with $\varepsilon\ll 1$, it is guaranteed that
$\\{\overset{\circ}{\varUpsilon}>0\\}\cap\\{\varpi<3R/2\\}=\mathfrak{R}(=\\{\bar{\varUpsilon}>0\\}\cap\\{\varpi<3R/2\\})$.
More generally, if
$\displaystyle\frac{1}{\bar{\rho}}(\nabla|\bar{\rho}\overset{\circ}{\bm{\xi}}_{1})=(\nabla\log\bar{\rho}|\overset{\circ}{\bm{\xi}}_{1})+(\nabla|\overset{\circ}{\bm{\xi}}_{1})\subset\frac{1}{\gamma-1}(\nabla\log\bar{\varUpsilon}|\overset{\circ}{\bm{\xi}}_{1})+(\nabla|\overset{\circ}{\bm{\xi}}_{1})$
is bounded on a neighborhood of the vacuum boundary $\Sigma_{1}$ of
$\bar{\rho}$, then $\bar{\rho}>0\Leftrightarrow\overset{\circ}{\varUpsilon}>0$
there for $\overset{\circ}{\bm{\xi}}=\varepsilon\overset{\circ}{\bm{\xi}}_{1}$
with $\varepsilon\ll 1$.

Historical Remark: The derivation of the linearized approximation of the
equations in Lagrangian co-ordinate system can be found [14, Sect. 56], [1,
pp. 139-140.], [2, p.11, (A)], [16], [13, p.500, (1)] and so on. But there was
considered only the case of $\overset{\circ}{\rho}=\bar{\rho}$ and
$\overset{\circ}{\bm{\xi}}=0$.

##  4 Linearized equations for perturbations from a static stationary solution

Let us fix a static stationary solution

$\bar{\varUpsilon}=-\Phi^{\langle\Omega\rangle}-\frac{{\mathsf{G}M_{0}}}{R},\quad\bar{\bm{v}}=\mathbf{0}$
(4.1)

under the assumption (K). We are concerned with the domain

$\mathfrak{R}=\\{\bar{\rho}>0\\}=\\{\bar{\varUpsilon}>0,\varpi<\frac{3R}{2},R_{0}<r\\}.$

We consider the initial boundary problem

$\displaystyle\frac{\partial^{2}\bm{\xi}}{\partial
t^{2}}+\bm{B}\frac{\partial\bm{\xi}}{\partial t}+\bm{L}\bm{\xi}=0$
$\displaystyle\quad\mbox{on}\quad[0,+\infty[\times\mathfrak{R},$ (4.2a)
$\displaystyle(\bm{n}|\bm{\xi})=0$
$\displaystyle\quad\mbox{on}\quad[0,+\infty[\times\Sigma_{0},$ (4.2b)
$\displaystyle\bm{\xi}\Big{|}_{t=0}=\overset{\circ}{\bm{\xi}}(\bm{x}),\quad\frac{\partial\bm{\xi}}{\partial
t}\Big{|}_{t=0}=\overset{\circ}{\bm{v}}(\bm{x})$
$\displaystyle\quad\mbox{on}\quad\mathfrak{R}.$ (4.2c)

Here

$\displaystyle\bm{B}\bm{v}=2\Omega\bm{J}\bm{v},\quad\bm{J}=\begin{bmatrix}0&-1&0\\\
1&0&0\\\ 0&0&0\end{bmatrix},$ (4.3)
$\displaystyle\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}=\nabla G,\quad
G=-\overline{\frac{d\varUpsilon}{d\rho}}g,\quad
g=(\nabla|\bar{\rho}\mbox{\boldmath$\xi$}),$ (4.4)
$\displaystyle\bm{n}=\frac{\bm{x}}{R_{0}}$ (4.5)

and $\overset{\circ}{\bm{\xi}},\overset{\circ}{\bm{v}}$ are given vector
fields on $\mathfrak{R}$ such that
$(\bm{n}|\overset{\circ}{\bm{\xi}})=(\bm{n}|\overset{\circ}{\bm{v}})=0$ on
$\Sigma_{0}$, while

$\Sigma_{0}=\partial\mathfrak{B}_{0}=\\{r=R_{0}\\}.$

Here and hereafter the Lagrangian coordinate, denoted by $\bar{\bm{x}}$ in thr
previous section, is denoted by $\bm{x}$, since we do not refer the Eulerian
coordinate so for.

We consider the differential operator $L$ in the Hilbert space $\mathfrak{H}$
of all measurable functions $\xi$ defined on $\mathfrak{R}$ such that
$\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}}<\infty$, where

$\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}}^{2}=\int_{\mathfrak{R}}\|\mbox{\boldmath$\xi$}(\mbox{\boldmath$x$})\|^{2}\bar{\rho}(\mbox{\boldmath$x$})d\mbox{\boldmath$x$},$
(4.6)

that is, the inner product of $\mathfrak{H}$ is

$(\mbox{\boldmath$\xi$}_{1}|\mbox{\boldmath$\xi$}_{2})_{\mathfrak{H}}=\int_{\mathfrak{R}}(\mbox{\boldmath$\xi$}_{1}(\mbox{\boldmath$x$})|\mbox{\boldmath$\xi$}_{2}(\mbox{\boldmath$x$}))\bar{\rho}(\mbox{\boldmath$x$})d\mbox{\boldmath$x$}.$
(4.7)

Of course

$(\mbox{\boldmath$\xi$}_{1}(\mbox{\boldmath$x$})|\mbox{\boldmath$\xi$}_{2}(\mbox{\boldmath$x$})):=\sum_{k}\xi_{1}^{k}(\mbox{\boldmath$x$})(\xi_{2}^{k}(\mbox{\boldmath$x$}))^{*}\quad\mbox{for}\quad\mbox{\boldmath$\xi$}_{\mu}(\mbox{\boldmath$x$})=\begin{bmatrix}\xi_{\mu}^{1}(\mbox{\boldmath$x$})\\\
\\\ \xi_{\mu}^{2}(\mbox{\boldmath$x$})\\\ \\\
\xi_{\mu}^{3}(\mbox{\boldmath$x$})\end{bmatrix},\mu=1,2.$

Here and hereafter $Z^{*}$ denotes the complex conjugate $X-\mathrm{i}Y$of
$Z=X+\mathrm{i}Y$, while $\mathrm{i}$ stands for the imaginary unit,
$\sqrt{-1}$.

Briefly speaking, we consider
$\mathfrak{H}=L^{2}(\mathfrak{R},\bar{\rho}dx;\mathbb{C}^{3})$.

For $\mbox{\boldmath$\xi$}_{\mu}\in C_{0}^{\infty}(\mathfrak{R}),\mu=1,2$, we
have

$(\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}_{1}|\mbox{\boldmath$\xi$}_{2})_{\mathfrak{H}}=\int_{\mathfrak{R}}\overline{\frac{d\varUpsilon}{d\rho}}g_{1}g_{2}^{*}d\mbox{\boldmath$x$},\quad\mbox{with}\quad
g_{\mu}=(\nabla|\bar{\rho}\mbox{\boldmath$\xi$}_{\mu}).$ (4.8)

So we put

$Q(\mbox{\boldmath$\xi$}_{1},\mbox{\boldmath$\xi$}_{2})=\int_{\mathfrak{R}}\overline{\frac{d\varUpsilon}{d\rho}}g_{1}g_{2}^{*}d\mbox{\boldmath$x$}\quad\mbox{with}\quad
g_{\mu}=(\nabla|\bar{\rho}\mbox{\boldmath$\xi$}_{\mu}),$ (4.9)

and

$Q[\mbox{\boldmath$\xi$}]=Q(\mbox{\boldmath$\xi$},\mbox{\boldmath$\xi$})=\int_{\mathfrak{R}}\overline{\frac{d\varUpsilon}{d\rho}}|g|^{2}d\mbox{\boldmath$x$}\quad\mbox{with}\quad
g=(\nabla|\bar{\rho}\mbox{\boldmath$\xi$}).$ (4.10)

We start from the operator $\mathfrak{T_{1}}_{c}$ in $\mathfrak{H}$ defined by
$\mathfrak{T_{1}}_{c}:\mbox{\boldmath$\xi$}\mapsto\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}$
on the domain $\mathsf{D}(\mathfrak{T_{1}}_{c})=C_{0}^{\infty}(\mathfrak{R})$.
Then $\mathfrak{T_{1}}_{c}$ is densely defined, symmetric and bounded from
below as

$(\mathfrak{T_{1}}_{c}\mbox{\boldmath$\xi$}|\mbox{\boldmath$\xi$})_{\mathfrak{H}}=Q[\mbox{\boldmath$\xi$}]\geq
0.$

Therefore, seeing, e.g., [12, Chapter VI, Section 2.3], we have that
$\mathfrak{T_{1}}_{c}$ admits the Friedrichs extension $\mathfrak{T_{1}}$
which is a self-adjoint operator in $\mathfrak{H}$. The domain of
$\mathfrak{T_{1}}$ is

$\mathsf{D}(\mathfrak{T_{1}})=\\{\mbox{\boldmath$\xi$}\in\mathfrak{H}_{0}^{\mathrm{div}}\
|\ \mbox{\boldmath$L$}\mbox{\boldmath$\xi$}\in\mathfrak{H}\quad\mbox{in the
sense of distribution}\\}.$ (4.11)

Here $\mathfrak{H}_{0}^{\mathrm{div}}$ is the Hilbert space of all
$\mbox{\boldmath$\xi$}\in\mathfrak{H}$ such that there is a sequence
$\mbox{\boldmath$\varphi$}_{n}\in C_{0}^{\infty}(\mathfrak{R})$ such that

$\displaystyle\mbox{\boldmath$\varphi$}_{n}\rightarrow\mbox{\boldmath$\xi$}\quad\mbox{in}\quad\mathfrak{H}\quad\mbox{as}\quad
n\rightarrow\infty,$ $\displaystyle
Q[\mbox{\boldmath$\varphi$}_{m}-\mbox{\boldmath$\varphi$}_{n}]\rightarrow
0\quad\mbox{as}\quad n,m\rightarrow\infty.$

In order to fix the idea we use

###### Definition 8

We put

$\mathfrak{H}^{\mathrm{div}}=\\{\bm{\xi}\in\mathfrak{H}\ |\
g=(\nabla|\bar{\rho}\bm{\xi})\in
L^{2}(\mathfrak{R};\overline{\frac{d\varUpsilon}{d\rho}}d\bm{x})\\},$ (4.12)

and regard it as a Hilbert space endowed with the inner product

$(\mbox{\boldmath$\xi$}_{1}|\mbox{\boldmath$\xi$}_{2})_{\mathfrak{H}^{\mathrm{div}}}=Q(\mbox{\boldmath$\xi$}_{1},\mbox{\boldmath$\xi$}_{2})+(\mbox{\boldmath$\xi$}_{1}|\mbox{\boldmath$\xi$}_{2})_{\mathfrak{H}}.$
(4.13)

Thus we are saying

###### Definition 9

$\mathfrak{H}_{0}^{\mathrm{div}}$ is the closure of
$C_{0}^{\infty}(\mathfrak{R})$ in $\mathfrak{H}^{\mathrm{div}}$.

Hereafter we shall denote by $L$ the Friedrichs extension $\mathfrak{T_{1}}$,
diverting the letter. So we have

###### Theorem 3

The operator $L$ is a self-adjoint operator bounded from below by $0$ in the
Hilbert space $\mathfrak{H}$, whose domain is

$\mathsf{D}(\mbox{\boldmath$L$})=\\{\mbox{\boldmath$\xi$}\in\mathfrak{H}_{0}^{\mathrm{div}}\
|\ \mbox{\boldmath$L$}\mbox{\boldmath$\xi$}\in\mathfrak{H}\\}.$ (4.14)

Note that
$\mbox{\boldmath$\xi$}\in\mathsf{D}(\mbox{\boldmath$L$})\subset\mathfrak{H}_{0}^{\mathrm{div}}$
enjoys the boundary condition

$(\bm{n}|\bm{\xi})=0\quad\mbox{on}\quad\Sigma_{0}=\\{r=R_{0}\\}$ (4.15)

in the following sense: There is known to uniquely exist the ‘normal trace
operator’ $\gamma_{\mathrm{n}}$ which maps $\mathfrak{H}^{\mathrm{div}}$ into
$H^{-1/2}(\Sigma_{0})$ continuously such that
$\gamma_{\mathrm{n}}\mbox{\boldmath$\xi$}=-(\bm{n}|\mbox{\boldmath$\xi$})\in
C^{1}(\mathfrak{R}\cup\partial\mathfrak{R})$, and
$\gamma_{\mathrm{n}}(\mbox{\boldmath$\xi$})=0$ when
$\mbox{\boldmath$\xi$}\in\mathfrak{H}_{0}^{\mathrm{div}}$. (See e.g., [9,
Chapter I].) Here we have used the fact that
$\displaystyle\bar{\rho},\overline{\frac{d\varUpsilon}{d\rho}}\in
C^{\infty}(\mathfrak{R}\cup\Sigma_{0})$ and

$\displaystyle\bar{\rho}\geq\Big{(}\frac{\gamma-1}{\mathsf{A}\gamma}\Big{)}^{\frac{1}{\gamma-1}}\Big{(}{\mathsf{G}M_{0}}\Big{(}\frac{1}{R_{0}}-\frac{1}{R}\Big{)}\Big{)}^{\frac{1}{\gamma-1}}>0\quad\mbox{on}\
\Sigma_{0},$
$\displaystyle\overline{\frac{d\varUpsilon}{d\rho}}\geq(\gamma-1)\Big{(}\frac{\mathsf{A}\gamma}{\gamma-1}\Big{)}^{\frac{1}{\gamma-1}}\Big{(}{\mathsf{G}M_{0}}\Big{(}\frac{1}{R_{0}}-\frac{1}{R}\Big{)}\Big{)}^{-\frac{2-\gamma}{\gamma-1}}>0\quad\mbox{on}\
\Sigma_{0},$

thanks to the assumption $R>R_{0}$, so that

$\displaystyle\|\mbox{\boldmath$\xi$}\|^{2}\bar{\rho}d\mbox{\boldmath$x$}=\frac{1}{\bar{\rho}}\|\hat{\mbox{\boldmath$\xi$}}\|^{2}d\mbox{\boldmath$x$}\geq\frac{1}{C}\|\hat{\mbox{\boldmath$\xi$}}\|^{2}d\mbox{\boldmath$x$}\quad\mbox{on}\
\mathfrak{R},$
$\displaystyle\overline{\frac{d\varUpsilon}{d\rho}}|(\nabla|\bar{\rho}\mbox{\boldmath$\xi$})|^{2}d\mbox{\boldmath$x$}={\mathsf{A}\gamma}{\bar{\rho}^{\gamma-2}}|(\nabla|\hat{\mbox{\boldmath$\xi$}})|^{2}d\mbox{\boldmath$x$}\geq\frac{1}{C}|(\nabla|\hat{\mbox{\boldmath$\xi$}})|^{2}d\mbox{\boldmath$x$}\quad\mbox{on}\
\mathfrak{R},$

where $\hat{\mbox{\boldmath$\xi$}}:=\bar{\rho}{\mbox{\boldmath$\xi$}}$ and $C$
is a sufficiently large finite positive number, and the theory on the
functional spaces
$H({\mathrm{div}};\mathfrak{R}),H_{0}({\mathrm{div}};\mathfrak{R}),H^{1/2}(\partial\mathfrak{R}),H^{-1/2}(\partial\mathfrak{R})$
and the normal trace operator $\gamma_{\mathrm{n}}$ seen in [9] can be applied
to the vector filed
$\hat{\mbox{\boldmath$\xi$}}=\bar{\rho}\mbox{\boldmath$\xi$}$, since the above
estimates guarantee that the space
$\hat{\mathfrak{H}}^{\mathrm{div}}:=\\{\hat{\mbox{\boldmath$\xi$}}|\mbox{\boldmath$\xi$}\in\mathfrak{H}^{\mathrm{div}}\\},\hat{\mathfrak{H}}_{0}^{\mathrm{div}}:=\\{\hat{\mbox{\boldmath$\xi$}}|\mbox{\boldmath$\xi$}\in\mathfrak{H}_{0}^{\mathrm{div}}\\}$
are continuously imbedded into
$H(\mathrm{div};\mathfrak{R}),H_{0}(\mathrm{div};\mathfrak{R})$.

Therefore, if $\bm{\xi}(t,\cdot)\in\mathsf{D}(\bm{L})\quad\forall t$, then the
boundary condition (4.2b) is satisfied in this sense.

As for the operator
$\mbox{\boldmath$J$}\cdot:\bm{v}\mapsto\mbox{\boldmath$J$}\bm{v}$, it is clear
that it is a bounded linear operator on $\mathfrak{H}$. But, for the sake of
generality, we replace $2\Omega\bm{J}$ by the operator $\bm{B}:\bm{v}\mapsto
B\bm{v}$ defined by an arbitrary smooth matrix-valued function
$B:\bm{x}\mapsto(B_{j}^{i}(\bm{x}))_{i,j}\in
C^{\infty}(\mathfrak{R}\cup\partial\mathfrak{R};\mathbb{R}^{3\times 3})$. Of
course $\bm{B}$ is a bounded operator on $\mathfrak{H}$, and

$|(\bm{B}\bm{v}|\bm{v})_{\mathfrak{H}}|\leq\beta\|\bm{v}\|_{\mathfrak{H}},$
(4.16)

where $\beta=\|B(\cdot)\|_{L^{\infty}}$. The adjoint operator $\bm{B}^{*}$ is
defined by the transposed matrix $B(\bm{x})^{\top}$. Later we shall use the
following densely defined closed operator

$\bm{L}^{c\bm{B}}:=\bm{L}+c\bm{B},\quad\mathsf{D}(\bm{L}^{c\bm{B}})=\mathsf{D}(\bm{L}),$
(4.17)

$c$ being a real number. When $c\not=0$, we cannot say that it is self-
adjoint, in genaral, but we can claim the folowing

###### Proposition 3

For any $c\in\mathbb{R}$ and $\lambda>|c\beta|$ the operator
$\mbox{\boldmath$L$}+c\bm{B}+\lambda$ has the bounded linear inverse operator
$(\mbox{\boldmath$L$}+c\bm{B}+\lambda)^{-1}$ defined on the whole space
$\mathfrak{H}$ such that

$|\|(\mbox{\boldmath$L$}+c\bm{B}+\lambda)^{-1}\||_{\mathcal{B}(\mathfrak{H})}\leq\frac{1}{\lambda-|c\beta|}.$
(4.18)

Proof. First we see $\mbox{\boldmath$L$}+c\bm{B}+\lambda$ is invertible. In
fact, if

$(\mbox{\boldmath$L$}+c\bm{B}+\lambda)\mbox{\boldmath$\xi$}=\mbox{\boldmath$f$},\quad\mbox{\boldmath$\xi$}\in\mathsf{D}(\mbox{\boldmath$L$}),\quad\mbox{\boldmath$f$}\in\mathfrak{H},$

then we see

$(\lambda-|c\beta|)\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}}^{2}\leq
Q[\mbox{\boldmath$\xi$}]+(c\bm{B}\mbox{\boldmath$\xi$}|\mbox{\boldmath$\xi$})_{\mathfrak{H}}+\lambda\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}}^{2}=(\mbox{\boldmath$\xi$}|\mbox{\boldmath$f$})_{\mathfrak{H}}\leq\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}}\|\mbox{\boldmath$f$}\|_{\mathfrak{H}},$

therefore we have

$\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}}\leq\frac{1}{\lambda-|c\beta|}\|\mbox{\boldmath$f$}\|_{\mathfrak{H}}.$

We claim that the range $\mathsf{R}(\mbox{\boldmath$L$}+c\bm{B}+\lambda)$ is
dense in $\mathfrak{H}$. In fact, suppose

$((\mbox{\boldmath$L$}+c\bm{B}+\lambda)\mbox{\boldmath$\xi$}|\mbox{\boldmath$f$})=0\quad\forall\mbox{\boldmath$\xi$}\in\mathsf{D}(\mbox{\boldmath$L$}).$

Then

$\displaystyle(\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}|\mbox{\boldmath$f$})$
$\displaystyle=-((c\bm{B}+\lambda)\bm{\xi}|\bm{f})$
$\displaystyle=(\bm{\xi}|(-c\bm{B}^{*}-\lambda)\bm{f})$

for $\forall\bm{\xi}\in\mathsf{D}(\bm{L})$. Hence
$\mbox{\boldmath$f$}\in\mathsf{D}(\mbox{\boldmath$L$}^{*})$ and

$\mbox{\boldmath$L$}^{*}\mbox{\boldmath$f$}=-c\bm{B}^{*}-\lambda\mbox{\boldmath$f$}.$

Since $\mbox{\boldmath$L$}=\mbox{\boldmath$L$}^{*}$, this means that
$\mbox{\boldmath$f$}\in\mathsf{D}(\mbox{\boldmath$L$})$ and

$(\mbox{\boldmath$L$}+c\bm{B}^{*}+\lambda)\mbox{\boldmath$f$}=0.$

Since $\mbox{\boldmath$L$}+c\bm{B}^{*}+\lambda$ is invertible, for
$\|B^{\top}\|=\|B\|=\beta$, it follows that $\mbox{\boldmath$f$}=0$. Summing
up, we have the assertion. $\square$

We are going to apply the Hille-Yosida theory to the initial-boundary value
problem

$\displaystyle\frac{\partial^{2}\mbox{\boldmath$\xi$}}{\partial
t^{2}}+\bm{B}\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t}+\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}=\bf{0},$
$\displaystyle\mbox{\boldmath$\xi$}=\overset{\circ}{\bm{\xi}},\quad\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t}=\overset{\circ}{\mbox{\boldmath$v$}}=\begin{bmatrix}v^{1}(0,\mbox{\boldmath$x$})\\\
\\\ v^{2}(0,\mbox{\boldmath$x$})\\\ \\\
v^{3}(0,\mbox{\boldmath$x$})\end{bmatrix}\quad\mbox{at}\quad t=0,$
$\displaystyle\mbox{\boldmath$\xi$}(t,\cdot)\in\mathsf{D}(\mbox{\boldmath$L$})\quad\mbox{for}\quad\forall
t\geq 0.$ (4.19)

We put

$\displaystyle U=\begin{bmatrix}\mbox{\boldmath$\xi$}\\\ \\\
\dot{\mbox{\boldmath$\xi$}}\end{bmatrix},\quad\dot{\mbox{\boldmath$\xi$}}=\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t},$ (4.20) $\displaystyle\mathbf{A}U=\begin{bmatrix}O&-I\\\ \\\
\mbox{\boldmath$L$}&\bm{B}\end{bmatrix}U=\begin{bmatrix}-\dot{\mbox{\boldmath$\xi$}}\\\
\\\
\bm{B}\dot{\mbox{\boldmath$\xi$}}+\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}\end{bmatrix},$
(4.21)
$\displaystyle\mathfrak{E}=\mathfrak{H}_{0}^{\mathrm{div}}\times\mathfrak{H}$
with
$\displaystyle(U_{1}|U_{2})_{\mathfrak{E}}=(\mbox{\boldmath$\xi$}_{1}|\mbox{\boldmath$\xi$}_{2})_{\mathfrak{H}^{\mathrm{div}}}+(\dot{\mbox{\boldmath$\xi$}}_{1}|\dot{\mbox{\boldmath$\xi$}}_{2})_{\mathfrak{H}}=$
$\displaystyle=Q(\mbox{\boldmath$\xi$}_{1},\mbox{\boldmath$\xi$}_{2})+(\mbox{\boldmath$\xi$}_{1}|\mbox{\boldmath$\xi$}_{2})_{\mathfrak{H}}+(\dot{\mbox{\boldmath$\xi$}}_{1}|\dot{\mbox{\boldmath$\xi$}}_{2})_{\mathfrak{H}},$
(4.22)
$\displaystyle\mathsf{D}(\mathbf{A})=\mathsf{D}(\mbox{\boldmath$L$})\times\mathfrak{H}_{0}^{\mathrm{div}}.$
(4.23)

Then the initial-boundary value problem (4.19) can be written as

$\frac{dU}{dt}+\mathbf{A}U=\mathbf{0},\quad U|_{t=0}=U_{0},$ (4.24)

where

$U_{0}=\begin{bmatrix}\overset{\circ}{\bm{\xi}}\\\ \\\
\overset{\circ}{\mbox{\boldmath$v$}}\end{bmatrix}.$ (4.25)

Applying [3, Theorem 7.4], we can claim

###### Proposition 4

If $U_{0}\in\mathsf{D}(\mathbf{A})$, say, if
$\overset{\circ}{\bm{\xi}}\in\mathsf{D}(\bf{L})$ and
$\overset{\circ}{\mbox{\boldmath$v$}}\in\mathfrak{H}_{0}^{\mathrm{div}}$, then
there exists a unique solution $U\in C^{1}([0,+\infty[,\mathfrak{E})\cap
C([0,+\infty[,\mathsf{D}(\mathbf{A}))$ to the problem (4.24). Moreover
$E(t)=\|U(t)\|_{\mathfrak{E}}^{2}$ enjoys

$E(t)\leq e^{2\Lambda t}E(0),$ (4.26)

where $\Lambda=1\vee\beta$.

Here we consider that $\mathsf{D}(\mathbf{A})$ is equipped with the operator
norm $(\|U\|_{\mathfrak{E}}^{2}+\|\mathbf{A}U\|_{\mathfrak{E}}^{2})^{1/2}$.

Proof of Proposition 4. Firstly $\mathbf{A}+1\vee\beta$ is monotone, that is,
for $\forall U\in\mathsf{D}(\mathbf{A})$ we have

$\displaystyle\mathfrak{Re}[(\mathbf{A}U|U)_{\mathfrak{E}}]+(1\vee\beta)\|U\|_{\mathfrak{E}}^{2}$
$\displaystyle=\mathfrak{Re}\Big{[}-Q(\dot{\mbox{\boldmath$\xi$}},\mbox{\boldmath$\xi$})-(\dot{\mbox{\boldmath$\xi$}}|\mbox{\boldmath$\xi$})_{\mathfrak{H}}+(\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}|\dot{\mbox{\boldmath$\xi$}})_{\mathfrak{H}}+(\mbox{\boldmath$B$}\dot{\mbox{\boldmath$\xi$}}|\dot{\mbox{\boldmath$\xi$}})_{\mathfrak{H}}\Big{]}+$
$\displaystyle+(1\vee\beta)(Q[\mbox{\boldmath$\xi$}]+\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}}^{2}+\|\dot{\mbox{\boldmath$\xi$}}\|_{\mathfrak{H}}^{2})=$
$\displaystyle\geq-\mathfrak{Re}[(\dot{\bm{\xi}}|\bm{\xi})_{\mathfrak{H}}]-\beta\|\dot{\bm{\xi}}\|_{\mathfrak{H}}^{2}+(1\vee\beta)\|\bm{\xi}\|_{\mathfrak{H}}^{2}+(1\vee\beta)\|\bm{\xi}\|_{\mathfrak{H}}^{2}$
$\displaystyle\geq\|\dot{\bm{\xi}}\|_{\mathfrak{H}}^{2}-\mathfrak{Re}[(\dot{\bm{\xi}}|\bm{\xi})_{\mathfrak{H}}]+\|\bm{\xi}\|_{\mathfrak{H}}^{2}$
$\displaystyle\geq 0,$

since
$(\bm{L}\bm{\xi}|\dot{\bm{\xi}})=Q(\bm{\xi},\dot{\bm{\xi}})=Q(\dot{\bm{\xi}},\bm{\xi})^{*}$
and
$|\mathfrak{Re}[(\bm{B}\dot{\bm{\xi}}|\dot{\bm{\xi}})_{\mathfrak{H}}|\leq\beta\|\dot{\bm{\xi}}\|_{\mathfrak{H}}^{2}$.

If $\Lambda>\beta$, then the operator $\mathbf{A}+\Lambda$ has the bounded
inverse defined on $\mathfrak{E}$. Actually the equation

$\mathbf{A}U+\Lambda U=F=\begin{bmatrix}\mbox{\boldmath$f$}\\\ \\\
\mbox{\boldmath$g$}\end{bmatrix}\in\mathfrak{E}$

means

$\begin{cases}-\dot{\mbox{\boldmath$\xi$}}+\Lambda\mbox{\boldmath$\xi$}=\mbox{\boldmath$f$}\in\mathfrak{H}_{0}^{\mathrm{div}}\\\
\\\
\bm{B}\dot{\mbox{\boldmath$\xi$}}+\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}+\Lambda\dot{\mbox{\boldmath$\xi$}}=\mbox{\boldmath$g$}\in\mathfrak{H},\end{cases}$

which can be solved as

$\begin{cases}\mbox{\boldmath$\xi$}=(\mbox{\boldmath$L$}+\Lambda\bm{B}+\Lambda^{2})^{-1}(\bm{B}\mbox{\boldmath$f$}+\Lambda\mbox{\boldmath$f$}+\mbox{\boldmath$g$})\in\mathsf{D}(\mbox{\boldmath$L$}),\\\
\\\
\dot{\mbox{\boldmath$\xi$}}=(\mbox{\boldmath$L$}+\Lambda\bm{B}+\Lambda^{2})^{-1}(\bm{B}\mbox{\boldmath$f$}+\Lambda\mbox{\boldmath$f$}+\mbox{\boldmath$g$})-\mbox{\boldmath$f$}\in\mathfrak{H}_{0}^{\mathrm{div}},\end{cases}$

thanks to Proposition 3, since $\Lambda^{2}>|\Lambda\beta|$ for
$\Lambda>\beta$. $\square$

Therefore, considering the problem

$\displaystyle\frac{\partial^{2}\mbox{\boldmath$\xi$}}{\partial
t^{2}}+\bm{B}\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t}+\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}=\bf{0},$
$\displaystyle\mbox{\boldmath$\xi$}(t,\cdot)\in\mathsf{D}(\mbox{\boldmath$L$})\quad\mbox{for}\quad\forall
t\geq 0,$
$\displaystyle\mbox{\boldmath$\xi$}=\overset{\circ}{\bm{\xi}},\quad\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t}=\overset{\circ}{\mbox{\boldmath$v$}}\quad\mbox{at}\quad t=0.,$ (4.27)

we can claim

###### Theorem 4

Suppose $\overset{\circ}{\bm{\xi}}\in\mathsf{D}(\bm{L})$ and
$\overset{\circ}{\mbox{\boldmath$v$}}\in\mathfrak{H}_{0}^{\mathrm{div}}$. Then
the initial-boundary value problem (4.27) admits a unique solution

$\mbox{\boldmath$\xi$}\in C^{2}([0,+\infty[,\mathfrak{H})\cap
C^{1}([0,+\infty[,\mathfrak{H}_{0}^{\mathrm{div}})\cap
C([0,+\infty[,\mathsf{D}(\mbox{\boldmath$L$}))$

and the energy

$\displaystyle E(t)$
$\displaystyle=\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}^{\mathrm{div}}}^{2}+\|\dot{\mbox{\boldmath$\xi$}}\|_{\mathfrak{H}}^{2}$
$\displaystyle=\|\mbox{\boldmath$\xi$}||_{\mathfrak{H}}^{2}+Q[\mbox{\boldmath$\xi$}]+\Big{\|}\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t}\Big{\|}_{\mathfrak{H}}^{2}$ (4.28)

enjoys the estimate

$\sqrt{E(t)}\leq e^{\Lambda t}\cdot\sqrt{E(0)},$ (4.29)

where $\Lambda=1\vee\beta$.

Here $\mathsf{D}(\mbox{\boldmath$L$})$ is equipped with the norm
$(\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}^{\mathrm{div}}}^{2}+\|\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}\|_{\mathfrak{H}}^{2})^{1/2}$.

Correspondingly we may consider the inhomogeneous initial-boundary value
problem

$\frac{dU}{dt}+\mathbf{A}U=F(t),\quad U|_{t=0}=U_{0}.$ (4.30)

We can claim

###### Proposition 5

If $U_{0}\in\mathsf{D}(\mathbf{A})$ and $F\in C([0,+\infty[;\mathfrak{E})$,
then there exists a unique solution

$U\in C^{1}([0,+\infty[;\mathfrak{E})\cap
C([0,+\infty[;\mathsf{D}(\mathbf{A}))$

to the problem (4.30), and it enjoys the estimate

$\|U(t)\|_{\mathfrak{E}}\leq e^{\Lambda
t}\Big{(}\|U_{0}\|_{\mathfrak{E}}+\int_{0}^{t}e^{-\Lambda
s}\|F(s)\|_{\mathfrak{E}}ds\Big{)},$ (4.31)

where $\Lambda=1\vee 2|\Omega|$.

Therefore, considering the problem

$\displaystyle\frac{\partial^{2}\mbox{\boldmath$\xi$}}{\partial
t^{2}}+\bm{B}\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t}+\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}=\mbox{\boldmath$f$}(t,\mbox{\boldmath$x$}),$
$\displaystyle\mbox{\boldmath$\xi$}=\overset{\circ}{\bm{\xi}},\quad\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t}=\overset{\circ}{\mbox{\boldmath$v$}}\quad\mbox{at}\quad t=0,$
$\displaystyle\mbox{\boldmath$\xi$}(t,\cdot)\in\mathsf{D}(\mbox{\boldmath$L$})\quad\mbox{for}\quad\forall
t\geq 0.,$ (4.32)

we can claim

###### Theorem 5

Suppose $\overset{\circ}{\bm{\xi}}\in\mathsf{D}(\bm{L})$,
$\overset{\circ}{\mbox{\boldmath$v$}}\in\mathfrak{H}_{0}^{\mathrm{div}}$ and
$\mbox{\boldmath$f$}\in C([0,+\infty[;\mathfrak{H})$. Then the initial-
boundary value problem (4.32) admits a unique solution

$\mbox{\boldmath$\xi$}\in C^{2}([0,+\infty[,\mathfrak{H})\cap
C^{1}([0,+\infty[,\mathfrak{H}_{0}^{\mathrm{div}})\cap
C([0,+\infty[,\mathsf{D}(\mbox{\boldmath$L$}))$

and the energy

$\displaystyle E(t)=E(t,\mbox{\boldmath$\xi$})$
$\displaystyle:=\|\mbox{\boldmath$\xi$}\|_{\mathfrak{H}^{\mathrm{div}}}^{2}+\|\dot{\mbox{\boldmath$\xi$}}\|_{\mathfrak{H}}^{2}$
$\displaystyle=\|\mbox{\boldmath$\xi$}||_{\mathfrak{H}}^{2}+Q[\mbox{\boldmath$\xi$}]+\Big{\|}\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t}\Big{\|}_{\mathfrak{H}}^{2}$ (4.33)

enjoys the estimate

$\sqrt{E(t)}\leq e^{\Lambda t}\Big{(}\sqrt{E(0)}+\int_{0}^{t}e^{-\Lambda
s}\|\mbox{\boldmath$f$}(s)\|_{\mathfrak{H}}ds\Big{)}$ (4.34)

for $\Lambda=1\vee\beta$.

##  5 Eigenfrequency, eigenvector, the variational principle

Astrophysicists used to discuss on the so called ‘variational principle’. See
[4], [5], [16], and so on. Although they discuss about self-gravitating
gaseous masses, we would like to follow their discussions by applying them to
the present case of the model of rotating atmosphere on the Earth, namely, we
consider the linearized wave equation (4.2a).

Since the operator $\displaystyle\bm{J}:\bm{v}\mapsto\begin{bmatrix}0&-1&0\\\
1&0&0\\\ 0&0&0\end{bmatrix}\bm{v}$ is skew symmetric, we introduce the
operator $\mbox{\boldmath$J$}_{\star}$ defined by

$\mbox{\boldmath$J$}_{\star}=\mathrm{i}\mbox{\boldmath$J$}.$ (5.1)

Then the operator $\mbox{\boldmath$J$}_{\star}$ is a bounded self-adjoint
operator on $\mathfrak{H}$. Recall

$(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\xi$}|\mbox{\boldmath$\xi$})_{\mathfrak{H}}=-2\int_{\mathfrak{R}}\mathfrak{Im}[\xi^{1}(\xi^{2})^{*}]\bar{\rho}(\mbox{\boldmath$x$})d\mbox{\boldmath$x$}.$
(5.2)

The equation (4.2a) reads

$\frac{\partial^{2}\mbox{\boldmath$\xi$}}{\partial
t^{2}}-2\Omega\mathrm{i}\mbox{\boldmath$J$}_{\star}\frac{\partial\mbox{\boldmath$\xi$}}{\partial
t}+\mbox{\boldmath$L$}\mbox{\boldmath$\xi$}=0$ (5.3)

Let us suppose that there exists a solution $\xi$ to (5.3) of the form

${\mbox{\boldmath$\xi$}}(t,\mbox{\boldmath$x$})=e^{\mathrm{i}\sigma
t}\mbox{\boldmath$\Xi$}(\mbox{\boldmath$x$})$ (5.4)

where $\sigma\in\mathbb{C}$ and
$\mbox{\boldmath$\Xi$}\in\mathsf{D}({\mbox{\boldmath$L$}})$. Then the equation
(5.3) reduces to

$-\sigma^{2}\mbox{\boldmath$\Xi$}+\sigma
2\Omega\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}+\mbox{\boldmath$L$}\mbox{\boldmath$\Xi$}=\mathbf{0},$
(5.5)

or the equation (4.2a)reduces to

$-\sigma^{2}\mbox{\boldmath$\Xi$}+\mathrm{i}\sigma
2\Omega\mbox{\boldmath$J$}\mbox{\boldmath$\Xi$}+\mbox{\boldmath$L$}\mbox{\boldmath$\Xi$}=\mathbf{0},.$
(5.6)

So, we use

###### Definition 10

When the equation (5.5) [( (5.6) )] is satisfied for $\sigma\in\mathbb{C}$
and $\bm{\Xi}\in\mathsf{D}(\bm{L}),\not=\mathbf{0}$, then $\sigma$ is called
an eigenfrequency of the wave equation (5.3) [( (4.2a))] and $\bm{\Xi}$ is
called an eigenvector associated with the eigenfrequency $\sigma$.

Note that $0$ is an eigenfrequency. In fact the vector

$\bm{\Xi}=\frac{1}{\bar{\rho}}\nabla\times\bm{a}$

belongs to $\mathrm{Ker}\bm{L}$ for any $\bm{a}\in
C_{0}^{\infty}(\mathfrak{R};\mathbb{C}^{3})$, and turns out to be an
eigenvector associated with the eingenfrequency $0$ provided that
$\bm{\Xi}\not=\mathbf{0}$.

Here let us recall the operator $\mathbf{A}$ defined as

$\mathbf{A}U=\begin{bmatrix}O&-I\\\ \\\
\bm{L}&2\Omega\bm{J}\end{bmatrix}U=\begin{bmatrix}-\dot{\bm{\Xi}}\\\ \\\
2\Omega\bm{J}\dot{\bm{\Xi}}+\bm{L}\bm{\Xi}\end{bmatrix}$

for

$U=\begin{bmatrix}\bm{\Xi}\\\ \\\
\dot{\bm{\Xi}}\end{bmatrix}\in\mathsf{D}(\mathbf{A})=\mathsf{D}(\bm{L})\times\mathfrak{H}_{0}^{\mathrm{div}},$

which was introduced in order to alpply the Hille-Yosida theory to the linear
evolution equation in the preceeding section. Obviously we can claim

###### Proposition 6

If and only if $\sigma\in\mathbb{C}$ is an eigenfrequency of the equation
(5.3) [( (4.2a))] , $\Lambda=\mathrm{i}\sigma$ is an eigenvalue of the
operator $-\mathbf{A}$.

Let us introduce the following

###### Definition 11

We denote by $\mathfrak{L}$ the one parameter family of operators
$(-\sigma^{2}+\sigma 2\Omega\bm{J}_{\star}+\bm{L})_{\sigma\in\mathbb{C}}$,
which is called ’quadratic pencil’. We denote

$\mathfrak{L}(\sigma):=-\sigma^{2}+\sigma 2\Omega\bm{J}_{\star}+\bm{L}.$ (5.7)

If the operator $\mathfrak{L}(\sigma)=-\sigma^{2}+\sigma
2\Omega\bm{J}_{\star}+\bm{L}$ admits the bounded inverse defined on
$\mathfrak{H}$, $\sigma$ is said to belong to the resolvent set
$\varrho(\mathfrak{L})$. We denote
$\sigma(\mathfrak{L})=\mathbb{C}\setminus\varrho(\mathfrak{L})$, and call it
the spectrum of the quadratic pencil $\mathfrak{L}$.

If $\sigma$ is an eigenfrequency, then it belongs to the spectrum
$\sigma(\mathfrak{L})$, so, $0\in\sigma(\mathfrak{L})$, but belonging to
$\sigma(\mathfrak{L})$ does not mean being an eigenfrequency a priori, of
course.

###### Proposition 7

$\varrho(\mathfrak{L})$ is an open subset of $\mathbb{C}$, and
$\sigma(\mathfrak{L})$ is closed.

Proof. Let us consider $\sigma\in\varrho(\mathfrak{L})$. Then

$\mathfrak{L}(\sigma+\Delta\sigma)=\mathfrak{L}(\sigma)\Big{[}I+\mathfrak{L}(\sigma)^{-1}(\Delta\sigma(-2\sigma+2\Omega\bm{J}_{\star}-\Delta\sigma)\Big{]}$

admits the bounded inverse and $\sigma+\Delta\sigma\in\varrho(\mathfrak{L})$,
if

$|\|\mathfrak{L}(\sigma)^{-1}\Delta\sigma(-2\sigma+2\Omega\bm{J}_{\star}-\Delta\sigma)\||_{\mathcal{B}(\mathfrak{H})}<1.$

For this inequality, it is sufficient that

$|\|\mathfrak{L}(\sigma)^{-1}\||_{\mathcal{B}(\mathfrak{H})}\cdot|\Delta\sigma|\cdot|(2(|\sigma|+|\Omega|)+|\Delta\sigma|)<1,$

or

$|\Delta\sigma|<-(|\sigma|+|\Omega|)+\sqrt{\frac{1}{|\|\mathfrak{L}(\sigma)^{-1}\||_{\mathcal{B}(\mathfrak{H})}}+(|\sigma|+|\Omega|)^{2}}.$

This means $\varrho(\mathfrak{L})$ is open. $\square$

We claim

###### Proposition 8

It holds that

$\mathrm{i}\varrho(\mathfrak{L})=\varrho(-\mathbf{A}),\quad\mathrm{i}\sigma(\mathfrak{L})=\sigma(-\mathbf{A}).$
(5.8)

Here $\varrho(-\mathbf{A}),\sigma(-\mathbf{A})$ stand for the usual resolvent
set, the spectrum of the operator $-\mathbf{A}$ in the Hilbert space
$\mathfrak{E}=\mathfrak{H}_{0}^{\mathrm{div}}\times\mathfrak{H}$.

Proof. Let $\sigma\in\varrho(\mathfrak{L})$ and $\Lambda=\mathrm{i}\sigma$.
Consider the equation

$\mathbf{A}U+\Lambda U=F=\begin{bmatrix}\bm{f}\\\ \\\
\bm{g}\end{bmatrix}\in\mathfrak{E},$

or,

$\begin{cases}&-\dot{\bm{\Xi}}+\mathrm{i}\sigma\bm{\Xi}=\bm{f}\in\mathfrak{H}_{0}^{\mathrm{div}},\\\
&\\\ &2\Omega
J\dot{\bm{\Xi}}+\bm{L}\bm{\Xi}+\mathrm{i}\sigma\dot{\bm{\Xi}}=\bm{g}\in\mathfrak{H}.\end{cases}$

This system of equations can be solved as

$\begin{cases}\bm{\Xi}&=(-\sigma^{2}-\mathrm{i}\sigma
2\Omega\bm{J}+\bm{L})^{-1}(2\Omega
J\bm{f}-\mathrm{i}\sigma\bm{f}+\bm{g})\in\mathsf{D}(\bm{L}),\\\ &\\\
\dot{\bm{\Xi}}&=(-\sigma^{2}-\mathrm{i}\sigma
2\Omega\bm{J}+\bm{L})^{-1}(2\Omega
J\bm{f}-\mathrm{i}\sigma\bm{f}+\bm{g})-\bm{f}\in\mathfrak{H}_{0}^{\mathrm{div}},\end{cases}$

since $\sigma\in\varrho(\mathfrak{L})$, while $F\mapsto U$ is continuous.
Therefore $\Lambda=\mathrm{i}\sigma\in\varrho(-\mathbf{A})$, or,
$\mathrm{i}\varrho(\mathfrak{L})\subset\varrho(-\mathbf{A})$.

Inversely let $\Lambda\in\varrho(-\mathbf{A})$ and
$\sigma=-\mathrm{i}\Lambda$. Consider the equation

$(-\sigma^{2}+\mathrm{i}\sigma
2\Omega\bm{J}+\bm{L})\bm{\Xi}=\bm{f}\in\mathfrak{H},$

or

$(\Lambda^{2}+\Lambda 2\Omega\bm{J}+\bm{L})\bm{\Xi}=\bm{f},$

which is equivalent to the system of equations

$\begin{cases}&\Lambda\dot{\bm{\Xi}}+2\Omega\bm{J}\dot{\bm{\Xi}}+\bm{L}\bm{\Xi}=\bm{f},\\\
&\\\ &\dot{\bm{\Xi}}=\Lambda\bm{\Xi}.\end{cases}$

But this is nothing but

$\mathbf{A}U+\Lambda U=\begin{bmatrix}\mathbf{0}\\\ \\\ \bm{f}\end{bmatrix}.$

Since $\Lambda\in\varrho(-\mathbf{A})$ is supposed, this admits the solution

$U=\begin{bmatrix}\bm{\Xi}\\\ \\\
\dot{\bm{\Xi}}\end{bmatrix}=(\mathbf{A}+\Lambda)^{-1}\begin{bmatrix}\mathbf{0}\\\
\\\ \bm{f}\end{bmatrix},$

and $\bm{f}\mapsto\bm{\Xi}$ is continuous, that is,
$\sigma=-\mathrm{i}\Lambda\in\varrho(\mathfrak{L})$. $\square$

When $\Omega=0$, we have

$\sigma(\mathfrak{L})=\\{\sigma\in\mathbb{C}\ |\
\lambda=\sigma^{2}\in\sigma(\bm{L})\\},$

where $\sigma(\bm{L})$ is the spectrum of the self-adjoint operator $\bm{L}$
in the Hilbert space $\mathfrak{H}$. Since $\bf{L}$ is self-adjoint and
$\bm{L}\geq 0$, we have $\sigma(\bm{L})\subset\\{\lambda\in\mathbb{R}\ |\
\lambda\geq 0\\}$. Thus it holds that

$\sigma(\mathfrak{L})\subset\mathbb{R}\quad\mbox{and}\quad\\{\Lambda\in\mathbb{C}\
|\ \mathfrak{Re}[\Lambda]\not=0\\}\subset\varrho(-\mathbf{A}),$ (5.9)

when $\Omega=0$. But, when $\Omega\not=0$, the situation is not so evident. At
least, we can claim

$\sigma(\mathfrak{L})\subset\mathbb{C}\setminus]-\infty,-2|\Omega|[\mathrm{i},$

since

$\\{\Lambda\in\mathbb{R}\ |\ \Lambda>2|\Omega|\\}\subset\varrho(-\mathbf{A}),$

as shown in the proof of Proposition 4. Moreover, it is clear that
$\mathfrak{L}(\sigma^{*})=(\mathfrak{L}(\sigma))^{*}$, since
$2\Omega\bm{J}_{\star}$ and $\bm{L}$ are self-adjoint. Therefore we have

$\sigma\in\varrho(\mathfrak{L})\quad\Leftrightarrow\quad\sigma^{*}\in\varrho(\mathfrak{L}).$
(5.10)

This means that $\varrho(\mathfrak{L})$ and $\sigma(\mathfrak{L})$ are
symmetric re the real axis in the complex number plane. Correspondingly,
$\sigma(-\mathbf{A})$ and $\varrho(-\mathbf{A})$ are symmetric re the
imaginary axis. Thus we can claim

$\displaystyle\sigma(\mathfrak{L})\subset\mathbb{C}\setminus(]-\infty,-2|\Omega|[\cup]2|\Omega|,+\infty[)\mathrm{i}\quad\mbox{and}$
$\displaystyle\\{\Lambda\in\mathbb{R}\ |\
|\Lambda|>2|\Omega|\\}\subset\varrho(-\mathbf{A}).$ (5.11)

However the gap between the information (5.9) for $\Omega=0$ and that (5.11)
for $\Omega\not=0$ is too much. So, we are want to strengthen (5.11) when
$\Omega\not=0$. In order to do it, we use the following

###### Proposition 9

If $\sigma\in\varrho(\mathfrak{L})$, then it holds that

$|\|\mathfrak{L}(\sigma)^{-1}\||_{\mathcal{B}(\mathfrak{H})}\geq\frac{1}{d(2(|\sigma|+|\Omega|)+d)},$
(5.12)

where $d:=\mathrm{dist}(\sigma,\sigma(\mathfrak{L}))$.

Proof. Let $\sigma\in\varrho(\mathfrak{L})$. Then, for
$\Delta\sigma\in\mathbb{C}$, the operator

$\mathfrak{L}(\sigma+\Delta\sigma)=\mathfrak{L}(\sigma)\Big{[}I+\mathfrak{L}(\sigma)^{-1}\Delta\sigma(-2\sigma+2\Omega\bm{J}_{\star}-\Delta\sigma)\Big{]}$

admits the bounded inverse in $\mathcal{B}(\mathfrak{H})$ and
$\sigma+\Delta\sigma\in\varrho(\mathfrak{L})$, if

$|\|\mathfrak{L}(\sigma)^{-1}\Delta\sigma(-2\sigma+2\Omega\bm{J}_{\star}-\Delta\sigma)\||_{\mathcal{B}(\mathfrak{H})}<1.$

For this inequality, it is sufficient that

$|\|\mathfrak{L}(\sigma)^{-1}\||_{\mathcal{B}(\mathfrak{H})}\cdot|\Delta\sigma|\cdot|(-2\sigma+2\Omega\bm{J}_{\star}-\Delta\sigma|<1.$

In other words, if $\sigma+\Delta\sigma\in\sigma(\mathfrak{L})$, then it
should hold

$|\|\mathfrak{L}(\sigma)^{-1}\||_{\mathcal{B}(\mathfrak{H})}\cdot|\Delta\sigma|\cdot|(-2\sigma+2\Omega\bm{J}_{\star}-\Delta\sigma|\geq
1,$

and necessarily

$|\|\mathfrak{L}(\sigma)^{-1}\||_{\mathcal{B}(\mathfrak{H})}\cdot|\Delta\sigma|\cdot|(2(|\sigma|+|\Omega|)+|\Delta\sigma|)\geq
1.$

If $d<+\infty$, then there is a sequence
$\sigma+(\Delta\sigma)_{n}\in\sigma(\mathfrak{L})$ such that
$|(\Delta\sigma)_{n}|\rightarrow d$, and the assertion follows. $\square$

Let us fix $\sigma_{\infty}\in\partial\sigma(\mathfrak{L})$. We are going to
show $\sigma_{\infty}\in\mathbb{R}$.

Let us consider a sequance $(\sigma_{n})_{n}$ such that
$\sigma_{n}\in\varrho(\mathfrak{L})$ and
$\sigma_{n}\rightarrow\sigma_{\infty}$ as $n\rightarrow\infty$. By Proposition
9 we have
$|\|(\mathfrak{L}(\sigma_{n})^{-1}\||_{\mathcal{B}(\mathfrak{H})}\rightarrow+\infty$,
therefore there are $\bm{f}_{n}\in\mathfrak{H}$ such that
$\|\bm{f}_{n}\|_{\mathfrak{H}}=1$ and
$\|\mathfrak{L}(\sigma_{n}))^{-1}\bm{f}_{n}\|_{\mathfrak{H}}\rightarrow+\infty$
as $n\rightarrow\infty$. Put
$\bm{\xi}_{n}=\mathfrak{L}(\sigma_{n})^{-1}\bm{f}_{n}(\in\mathsf{D}(\bm{L}))$
and $\bm{\eta}_{n}=\bm{\xi}_{n}/\|\bm{\xi}_{n}\|_{\mathfrak{H}}$. Then
$\|\bm{\eta}_{n}\|_{\mathfrak{H}}=1$ and

$\Big{|}(\mathfrak{L}(\sigma_{n})\bm{\eta}_{n}|\bm{\eta}_{n})_{\mathfrak{H}}\Big{|}=\Big{|}\frac{1}{\|\bm{\xi}_{n}\|_{\mathfrak{H}}^{2}}(\bm{f}_{n}|\bm{\xi}_{n})_{\mathfrak{H}}\Big{|}\leq\frac{1}{\|\bm{\xi}_{n}\|_{\mathfrak{H}}}\rightarrow
0.$

But we see

$(\mathfrak{L}(\sigma_{n})\bm{\eta}_{n}|\bm{\eta}_{n})_{\mathfrak{H}}=-(\sigma_{n})^{2}+\sigma_{n}b_{n}+c_{n},$

where

$b_{n}:=2\Omega(\bm{J}_{\star}\bm{\eta}_{n}|\bm{\eta}_{n})_{\mathfrak{H}},\quad
c_{n}=(\bm{L}\bm{\eta}_{n}|\bm{\eta}_{n})_{\mathfrak{H}}=Q[\bm{\eta}_{n}].$

Therefore $b_{n},c_{n}$ are real and

$|b_{n}|\leq 2|\Omega|,\qquad c_{n}\geq 0.$

Hence, by taking a subsequence if necessary, we can suppose that $b_{n}$ tends
to a limit $b_{\infty}$ such that $|b_{\infty}|\leq 2|\Omega|$. Put
$c_{\infty}:=(\sigma_{\infty})^{2}-\sigma_{\infty}b_{\infty}$. Then we see
$c_{n}\rightarrow c_{\infty}$. Hence $c_{\infty}$ is real and $\geq 0$, and
$\sigma_{\infty}$ turns out to enjoy the quadratic equation

$-(\sigma_{\infty})^{2}+b_{\infty}\sigma_{\infty}+c_{\infty}=0.$

Consequently,

$\sigma_{\infty}=\frac{b_{\infty}}{2}+\sqrt{\frac{b_{\infty}^{2}}{4}+c_{\infty}}\quad\mbox{or}\quad\sigma_{\infty}=\frac{b_{\infty}}{2}-\sqrt{\frac{b_{\infty}^{2}}{4}+c_{\infty}},$

so, anyway, $\sigma_{\infty}\in\mathbb{R}$. This was to be prooved.

Summing up, we can claim

###### Proposition 10

It holds that

$\partial\sigma(\mathfrak{L})\subset\mathbb{R}.$ (5.13)

This conclusion owes to [8, Theorem 1]. But their original proof is little bit
logically weak, and we have needed to edit it as above.

Let us consider
$\sigma_{0}=\alpha_{0}+\beta_{0}\mathrm{i}\in\sigma(\mathfrak{L})$, where
$\alpha_{0},\beta_{0}\in\mathbb{R}$. We are going to show that
$|\beta_{0}|\not=0$ implies a contradiction. By the symmetricity of
$\sigma(\mathfrak{L})$ we can suppose $\beta_{0}>0$ without loss of
generality. Choosing $K>2|\Omega|\vee\beta_{0}$, we consider the segment

$\displaystyle I$ $\displaystyle=[\sigma,K\mathrm{i}]$
$\displaystyle=\\{\sigma(t)=(1-t)\alpha_{0}+(\beta_{0}+(K-\beta_{0})t)\mathrm{i}\
|\ 0\leq t\leq 1\\}.$

Note that $\sigma(0)=\sigma_{0}\in\sigma(\mathfrak{L})$ and
$\sigma(1)=K\mathrm{i}\in\varrho(\mathfrak{L})$ by (5.11), since
$K>2|\Omega|$. Put

$\bar{t}:=\sup\\{t\in[0,1]\ |\ \sigma(t)\in\sigma(\mathfrak{L})\\}.$

Then $0\leq\bar{t}<1,\sigma(\bar{t})\in\sigma(\mathfrak{L})$ and
$\sigma(t)\in\varrho(\mathfrak{L})$ for $t>\bar{t}$. Hence
$\sigma(\bar{t})\in\partial\sigma(\mathfrak{L})$. But
$\mathfrak{Im}[\sigma(\bar{t})]\geq\beta_{0}>0$, a contradiction to
$\partial\sigma(\mathfrak{L})\in\mathbb{R}$. $\square$

Therefore we can claim

###### Theorem 6

It holds that

$\sigma(\mathfrak{L})\subset\mathbb{R}$ (5.14)

even when $\Omega\not=0$.

Now we note that it is known that there is a sequence of eigenfrequencies
$\sigma_{n,\pm}=\pm\sqrt{\lambda_{n}^{\mathcal{N}}},n\in\mathbb{N}$, when
$\Omega=0$. ( See the discussion given later.) However, up to now, we have no
knowledge on the existence of eigenfrequencies when $\Omega\not=0$. When
$\Omega=0$, then (5.6) reads

$-\sigma^{2}\bm{\Xi}+\bm{L}\bm{\Xi}=\mathbf{0},$

so that the eigenvctor $\bm{\Xi}$ associated with the eigenfrequency
$\sigma\not=0$ can be supposed to be real, since $\sigma\in\mathbb{R}$.
However, when $\Omega\not=0$, the situation is different. Namely, we claim

###### Proposition 11

Suppose $\Omega\not=0$. Let $\sigma\not=0$ be an eigenfrequency of the
equation (5.3) [( (4.2a))] and $\bm{\Xi}$ be an associated eigenvector. Then
the eigenvector $\bm{\Xi}$ is impossible to be real, that is,
$\mathfrak{Im}[\bm{\Xi}(\bm{x})]$ cannot vanish everywhere.

Proof. By Theorem 6 we see $\sigma\in\mathbb{R}$. Let us look at (5.6):

$-\sigma^{2}\bm{\Xi}+\mathrm{i}\sigma
2\bm{\Omega}\times\bm{\Xi}+\bm{L}\bm{\Xi}=\mathbf{0}..$ (5.15)

Let us denote
$\bm{X}(\bm{x})=\mathfrak{Re}[\bm{\Xi}(\bm{x})],\bm{Y}(\bm{x})=\mathfrak{Im}[\bm{\Xi}(\bm{x})]$
so that
$\bm{X}(\bm{x}),\bm{Y}(\bm{x})\in\mathbb{R},\bm{\Xi}=\bm{X}+\mathrm{i}\bm{Y}$,
Suppose that $\bm{Y}=0$ and deduce a contradiction. Now (5.15) means

$\displaystyle-\sigma^{2}\bm{X}-\sigma
2\bm{\Omega}\times\bm{Y}+\bm{L}\bm{X}=\mathbf{0},$ (5.16a)
$\displaystyle-\sigma\bm{Y}+\sigma
2\bm{\Omega}\times\bm{X}+\bm{L}\bm{Y}=\mathbf{0}.$ (5.16b)

Since $\bm{Y}=\mathbf{0}$ is supposed, this reads

$\displaystyle-\sigma^{2}\bm{X}+\bm{L}\bm{X}=\mathbf{0},$ (5.17a)
$\displaystyle\sigma 2\bm{\Omega}\times\bm{X}=\mathbf{0}$ (5.17b)

Since $\sigma\not=0,\Omega\not=0$, (5.17b) implies

$X^{1}=X^{2}=0,$ (5.18)

where $\bm{X}=(X^{1},X^{2},X^{3})^{\top}$. Then (5.17a) reads

$\displaystyle\frac{\partial}{\partial
x^{1}}\Big{(}-\overline{\frac{d\varUpsilon}{d\rho}}\frac{\partial}{\partial
x^{3}}(\bar{\rho}X^{3})\Big{)}$ $\displaystyle=0,$ (5.19a)
$\displaystyle\frac{\partial}{\partial
x^{2}}\Big{(}-\overline{\frac{d\varUpsilon}{d\rho}}\frac{\partial}{\partial
x^{3}}(\bar{\rho}X^{3})\Big{)}$ $\displaystyle=0,$ (5.19b)
$\displaystyle-\sigma^{2}X^{3}+\frac{\partial}{\partial
x^{3}}\Big{(}-\overline{\frac{d\varUpsilon}{d\rho}}\frac{\partial}{\partial
x^{3}}(\bar{\rho}X^{3})\Big{)}$ $\displaystyle=0.$ (5.19c)

Consequently, (5.19a) and (5.19b) imply that
$\displaystyle-\overline{\frac{d\varUpsilon}{d\rho}}\frac{\partial}{\partial
x^{3}}(\bar{\rho}X^{3})$ is a function of $x^{3}$ independent of
$x^{1},x^{2}$, and, since $\sigma\not=0$, (5.19c) implies that $X^{3}$ is so,
too. However $\bm{\Xi}=\bm{X}\in\mathsf{D}(\bm{L})$ suppose the boundary
condition

$\Big{(}\bm{\Xi}\Big{|}\frac{\partial}{\partial r}\Big{)}=0\quad\mbox{on}\quad
r=R_{0}.$

Namely,

$X^{3}(x^{3})x^{3}=0\quad\mbox{on}\quad\Sigma_{0}=\\{\ r=R_{0}\ \\},$

therefore $X^{3}=0$, and $\bm{\Xi}=(X^{1},X^{2}.X^{3})^{\top}=\mathbf{0}$, a
contradiction. $\square$

Let $\sigma$ be an eigenfrequency of the equation (4.2a)and $\bm{\Xi}$ be an
associated eigenvector.

Then $\bm{\xi}(t,\bm{x})=e^{\mathrm{i}\sigma t}\bm{\Xi}(\bm{x})$ is a solution
of the equation (LABEL:3.12). Since the coefficients of the equation (4.2a)are
real and the equation is linear, we can claim that
$\bm{\xi}(t,\bm{x})^{*},\mathfrak{Re}[\bm{\xi}(t,\bm{x})],\mathfrak{Im}[\bm{\xi}(t,\bm{x})]$
are solutions of (LABEL:3.12), too. But, since $\sigma$ is real by Theorem 6,
we see

$\displaystyle\bm{\xi}_{\mathfrak{R}}(t,\bm{x}):=\mathfrak{Re}[\bm{\xi}(t,\bm{x})]$
$\displaystyle=\mathfrak{Re}[e^{\mathrm{i}\sigma t}\bm{\Xi}(\bm{x})]=$
$\displaystyle=\cos(\sigma t)\bm{\Xi}_{\mathfrak{R}}(\bm{x})-\sin(\sigma
t)\bm{\Xi}_{\mathfrak{I}}(\bm{x})=$
$\displaystyle=\begin{bmatrix}|\Xi(\bm{x})^{1}|\cos(\sigma
t+\alpha_{1}(\bm{x}))\\\ \\\ |\Xi(\bm{x})^{2}|\cos(\sigma
t+\alpha_{2}(\bm{x}))\\\ \\\ |\Xi(\bm{x})^{3}|\cos(\sigma
t+\alpha_{3}(\bm{x}))\end{bmatrix},$

where

$\displaystyle\bm{\Xi}_{\mathfrak{R}}(\bm{x}):=\mathfrak{Re}[\bm{\Xi}(\bm{x})],\quad\bm{\Xi}_{\mathfrak{I}}(\bm{x}):=\mathfrak{Im}[\bm{\Xi}(\bm{x})]$
$\displaystyle\tan\alpha_{j}(\bm{x})=\frac{{\Xi}_{\mathfrak{I}}(\bm{x})^{j}}{{\Xi}_{\mathfrak{R}}(\bm{x})^{j}},\quad\mbox{so
that}\quad{\Xi}(\bm{x})^{j}=|{\Xi}(\bm{x})^{j}|e^{\mathrm{i}\alpha_{j}(\bm{x})}.$

Note that the field $\bm{\xi}_{\mathfrak{R}}(t,\bm{x})$ is a real-valued
solution of (4.2a) such that
$\bm{\xi}_{\mathfrak{R}}(0,\bm{x})=\bm{\Xi}_{\mathfrak{R}}(\bm{x})$.

Let $\sigma,\mbox{\boldmath$\Xi$}(\not=\mathbf{0})$ be an eigenfrequency and
an associated eigenvector. Multiplying (5.5) by
$\mbox{\boldmath$\Xi$}^{*}\bar{\rho}(\mbox{\boldmath$x$})$ and integrating it,
we have

$-\sigma^{2}\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}^{2}+\sigma
2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\Xi$})_{\mathfrak{H}}+Q[\mbox{\boldmath$\Xi$}]=0.$
(5.20)

Recall the quadratic form $Q$ is defined by (4.10). If we write

$a=\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}^{2},\quad
b=2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\Xi$})_{\mathfrak{H}},\quad
c=Q[\mbox{\boldmath$\Xi$}],$ (5.21)

then $a,b,c$ are real numbers, and $\sigma$ satisfies the quadratic equation

$-a\sigma^{2}+b\sigma+c=0,$ (5.22)

whoe roots are

$\sigma=\frac{b}{2a}\pm\sqrt{\frac{b^{2}}{4a^{2}}+\frac{c}{a}}.$ (5.23)

Here D. Lynden-Bell and J. P. Ostriker, [16, p.301, line 18], say:

> Equation (36) [ read (5.23) ] shows that the system is stable if $c$ is
> positive for each eigen $\xi$ [ read $\Xi$ ] . This assured if $\mathbf{C}$
> [ read $L$] is positive definite. Thus:
>
> A sufficient condition for stability is that $\mathbf{C}$ [read $L$ ] is
> positive definite. This is the  condition for secular stability.

This saying sounds strange. In fact, we may suppose the meaning of the words
‘stability’ and ‘secular stability’ as C. Hunter [10] defines:

> A general system is said to be ordinarily or dynamically unstable if the
> amplitude of some mode grows exponentially in time, but ordinarily stable if
> every mode is oscillatory in time. An ordinarily stable system can be said
> to be secularly unstable if small additional dissipative forces can cause
> some perturbation to grow. Otherwise, the system is sad to be secularly
> stable.

As C. Hunter says in the same article, the definition of secular instability
does not always confirm to that given above, so, we consider the (ordinary)
stability. If there is an eigenfrequancy $\sigma$ which is not real, then the
system described by (5.3) is unstable. It is true. But this means that there
is an eigenfrequency $\sigma$ and an associated eigenvector $\Xi$ such that

$\frac{b^{2}}{4a^{2}}+\frac{c}{a}<0.$

Normalizing $a=\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}^{2}=1$, this means

$\frac{b^{2}}{4}+c<0.$

Therefore we can claim that the system is unstable if there is an eigenvector
$\Xi$ such that

$\frac{b^{2}}{4}+c=\Omega^{2}((\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\Xi$})_{\mathfrak{H}})^{2}+Q[\mbox{\boldmath$\Xi$}]<0.$

Of course, in the situation under consideration, this cannot happen, since
$Q[\bm{\Xi}]\geq 0$ for $\forall\bm{\Xi}\in\mathsf{D}(\bm{L})$ and
$b\in\mathbb{R}$.

However, logically speaking, the condition $c=Q[\mbox{\boldmath$\Xi$}]>0$ for
each eigenvector $\Xi$ is far from the condition of the stability, contrary to
the saying of D. Lynden-Bell and J. P. Ostriker.

Moreover let us note the following fact: $Q[\mbox{\boldmath$\Xi$}]\geq 0$ for
any $\mbox{\boldmath$\Xi$}\in\mathfrak{H}^{\mathrm{div}}$, but
$Q[\mbox{\boldmath$\Xi$}]=0$ does not imply
$\mbox{\boldmath$\Xi$}=\mathbf{0}$; In fact

$\mbox{\boldmath$\Xi$}(\mbox{\boldmath$x$})=\frac{1}{\bar{\rho}(\mbox{\boldmath$x$})}\nabla\times\mbox{\boldmath$a$},$

$a$ being an arbitrary vector field belonging to
$C_{0}^{\infty}(\mathfrak{R})$, belongs to the kernel of $L$, that is,
$Q[\mbox{\boldmath$\Xi$}]=0$.

Anyway, we are going to describe the ‘variational principle’.

Let us suppose that there exist $\sigma_{0}\in\mathbb{R}$ and
$\mbox{\boldmath$\Xi$}_{0}\in\mathfrak{H}^{\mathrm{div}}\ (\not=\mathbf{0})$
such that

$-\sigma_{0}^{2}\|\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}}^{2}+\sigma_{0}2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}_{0}|\mbox{\boldmath$\Xi$}_{0})_{\mathfrak{H}}+Q[\mbox{\boldmath$\Xi$}_{0}]=0.$
(5.24)

Of course if $\sigma_{0},\mbox{\boldmath$\Xi$}_{0}$ are real eigenfrequency
and an associated eigenvector then (5.24) is satisfied. Now we assume that

$a_{0}=\|\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}}^{2},\quad
b_{0}=2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}_{0}|\mbox{\boldmath$\Xi$}_{0})_{\mathfrak{H}},\quad
c_{0}=Q[\mbox{\boldmath$\Xi$}_{0}]$ (5.25)

satisfies

$\frac{b_{0}}{4a_{0}^{2}}+\frac{c_{0}}{a_{0}}>0.$ (5.26)

Then

$\displaystyle\sigma_{0}=\frac{b_{0}}{2a_{0}}+\sqrt{\frac{b_{0}^{2}}{4a_{0}^{2}}+\frac{c_{0}}{a_{0}}}$
(5.27a) or
$\displaystyle\sigma_{0}=\frac{b_{0}}{2a_{0}}-\sqrt{\frac{b_{0}^{2}}{4a_{0}^{2}}+\frac{c_{0}}{a_{0}}}$
(5.27b)

In order fix the idea, suppose that (5.27a) is the case. Then we can consider
$\sigma$ as a functional of
$\mbox{\boldmath$\Xi$}\in\mathfrak{H}^{\mathrm{div}}$ near
$\mbox{\boldmath$\Xi$}_{0}$, say,
$\|\mbox{\boldmath$\Xi$}-\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}^{\mathrm{div}}}\leq\delta(\ll
1)$, defined by

$\sigma(\mbox{\boldmath$\Xi$})=\frac{b(\mbox{\boldmath$\Xi$})}{2a(\mbox{\boldmath$\Xi$})}+\sqrt{\frac{b(\mbox{\boldmath$\Xi$})^{2}}{4a(\mbox{\boldmath$\Xi$})^{2}}+\frac{c(\mbox{\boldmath$\Xi$})}{a(\mbox{\boldmath$\Xi$})}}$
(5.28)

with

$a(\mbox{\boldmath$\Xi$})=\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}^{2},\quad
b(\mbox{\boldmath$\Xi$})=2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\Xi$})_{\mathfrak{H}},\quad
c(\mbox{\boldmath$\Xi$})=Q[\mbox{\boldmath$\Xi$}].$ (5.29)

Here we take $\delta$ so small that

$\frac{b(\mbox{\boldmath$\Xi$})^{2}}{4a(\mbox{\boldmath$\Xi$})^{2}}+\frac{c(\mbox{\boldmath$\Xi$})}{a(\mbox{\boldmath$\Xi$})}>0\quad\mbox{for}\quad\|\mbox{\boldmath$\Xi$}-\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}^{\mathrm{div}}}\leq\delta.$
(5.30)

The variation
$\mbox{\boldmath$\delta$}\sigma=\mbox{\boldmath$\delta$}\sigma(\mbox{\boldmath$\Xi$})$
of $\sigma$ at
$\mbox{\boldmath$\Xi$},\|\mbox{\boldmath$\Xi$}-\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}^{\mathrm{div}}}\leq\delta$,
is the linear functional on $\mathfrak{H}^{\mathrm{div}}$ defined by

$\langle\mbox{\boldmath$\delta$}\sigma(\mbox{\boldmath$\Xi$})|\mbox{\boldmath$h$}\rangle=\lim_{\tau\rightarrow
0}\frac{1}{\tau}(\sigma(\mbox{\boldmath$\Xi$}+\tau\mbox{\boldmath$h$})-\sigma(\mbox{\boldmath$\Xi$})).$
(5.31)

It follows from (5.28) the equation

$-\sigma^{2}\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}^{2}+\sigma
2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\Xi$})_{\mathfrak{H}}+Q[\mbox{\boldmath$\Xi$}]=0.$
(5.32)

holds for
$\sigma=\sigma(\mbox{\boldmath$\Xi$}),\mbox{\boldmath$\Xi$}\in\mathfrak{H}^{\mathrm{div}},\|\mbox{\boldmath$\Xi$}-\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}^{\mathrm{div}}}\leq\delta$.
Therefore we have

$\displaystyle(-2\sigma\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}^{2}+2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\Xi$})_{\mathfrak{H}})\mbox{\boldmath$\delta$}\sigma+$
$\displaystyle-\sigma^{2}\mbox{\boldmath$\delta$}\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}^{2}+\sigma\mbox{\boldmath$\delta$}(2\Omega
J_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\Xi$})_{\mathfrak{H}}+\mbox{\boldmath$\delta$}Q[\mbox{\boldmath$\Xi$}]=0,$

or, precisely writing,

$\displaystyle(-2\sigma\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}^{2}+2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\Xi$})_{\mathfrak{H}})\langle\mbox{\boldmath$\delta$}\sigma(\mbox{\boldmath$\Xi$})|\mbox{\boldmath$\delta$}\mbox{\boldmath$\Xi$}\rangle+$
$\displaystyle+2\mathfrak{Re}\Big{[}-\sigma^{2}(\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\delta$}\mbox{\boldmath$\Xi$})_{\mathfrak{H}}+\sigma
2\Omega(J_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\delta$}\mbox{\boldmath$\Xi$})_{\mathfrak{H}}+Q(\mbox{\boldmath$\Xi$},\mbox{\boldmath$\delta$}\mbox{\boldmath$\Xi$})\Big{]}=0$

for
$\forall\mbox{\boldmath$\delta$}\mbox{\boldmath$\Xi$}\in\mathfrak{H}^{\mathrm{div}}$.
Here we note that

$\displaystyle-2\sigma\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}^{2}+2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$\Xi$})_{\mathfrak{H}}$
$\displaystyle=-2\sigma a(\mbox{\boldmath$\Xi$})+b(\mbox{\boldmath$\Xi$})=$
$\displaystyle=-\sqrt{\frac{b(\mbox{\boldmath$\Xi$})^{2}}{4a(\mbox{\boldmath$\Xi$})^{2}}+\frac{c(\mbox{\boldmath$\Xi$})}{a(\mbox{\boldmath$\Xi$})}}\not=0$

for
$\|\mbox{\boldmath$\Xi$}-\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}^{\mathrm{div}}}\leq\delta$.
Therefore $\mbox{\boldmath$\delta$}\sigma(\mbox{\boldmath$\Xi$})=0$ if and
only if

$-\sigma^{2}(\mbox{\boldmath$\Xi$}|\mbox{\boldmath$h$})_{\mathfrak{H}}+\sigma
2\Omega(J_{\star}\mbox{\boldmath$\Xi$}|\mbox{\boldmath$h$})_{\mathfrak{H}}+Q(\mbox{\boldmath$\Xi$},\mbox{\boldmath$h$})=0$

for $\forall\mbox{\boldmath$h$}\in\mathfrak{H}^{\mathrm{div}}$. Thus we can
claim the following ‘variational principle’:

###### Theorem 7

Let $\mbox{\boldmath$\Xi$}_{0}\in\mathfrak{H}^{\mathrm{div}}$ satisfy

$\|\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}}^{2}>0,\quad\frac{(2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}_{0}|\mbox{\boldmath$\Xi$}_{0})_{\mathfrak{H}})^{2}}{4\|\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}}^{4}}+\frac{Q[\mbox{\boldmath$\Xi$}_{0}]}{\|\mbox{\boldmath$\Xi$}_{0}\|_{\mathfrak{H}}^{2}}>0.$

The variation $\mbox{\boldmath$\delta$}\sigma$ of $\sigma$ (specified by
(5.28)) vanishes at $\mbox{\boldmath$\Xi$}_{0}$ if and only if
$\mbox{\boldmath$\Xi$}_{0}\in\mathsf{D}(\mbox{\boldmath$L$})$ and enjoys the
equation (5.5):

$-\sigma_{0}^{2}\mbox{\boldmath$\Xi$}_{0}+\sigma_{0}2\Omega\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}_{0}+\mbox{\boldmath$L$}\mbox{\boldmath$\Xi$}_{0}=\mathbf{0}.$

Then $e^{\mathrm{i}\sigma_{0}t}\mbox{\boldmath$\Xi$}_{0}(\mbox{\boldmath$x$})$
is a solution of the equation (5.3). Here
$\sigma_{0}=\sigma(\mbox{\boldmath$\Xi$}_{0})$.

This principle tells us that, if we want to find an eigenfrequency, we may try
to find a stationary point of the functional

$\sigma(\mbox{\boldmath$\Xi$})=\frac{b(\mbox{\boldmath$\Xi$})}{2}\pm\sqrt{\frac{b(\mbox{\boldmath$\Xi$})^{2}}{4}+c(\mbox{\boldmath$\Xi$})}$

under the constraint $\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}=1$. But it
seems that this principle is far from the solution of the problem to establish
the existence and completeness of the system of eigenvectors.

For example, as a ‘practical use of the variational principle’, D. Lynden-Bell
and J. P. Ostriker, [16, Section 2.6], proposed the following scheme:

> Take sufficiently many functions $\mbox{\boldmath$\Xi$}_{(i)},i=1,\cdots,N$
> as those who consist a base, and consider the trial function
>
>
> $\mbox{\boldmath$\Xi$}=\sum_{i=1}^{N}a^{i}\mbox{\boldmath$\Xi$}_{(i)},\quad\mbox{\boldmath$a$}=\begin{bmatrix}a^{1}\\\
> \cdot\\\ \cdot\\\ a^{N}\end{bmatrix}.$
>
> Put
>
> $\displaystyle\mbox{\boldmath$A$}=(A_{ij})_{i,j},\quad
> A_{ij}=(\mbox{\boldmath$\Xi$}_{(i)}|\mbox{\boldmath$\Xi$}_{(j)})_{\mathfrak{H}},$
> $\displaystyle\mbox{\boldmath$B$}=(B_{ij})_{i,j},\quad
> B_{ij}=2\Omega(\mbox{\boldmath$J$}_{\star}\mbox{\boldmath$\Xi$}_{(i)}|\mbox{\boldmath$\Xi$}_{(j)})_{\mathfrak{H}},$
> $\displaystyle\mbox{\boldmath$C$}=(C_{ij})_{i,j},\quad
> C_{ij}=Q(\mbox{\boldmath$\Xi$}_{(i)},\mbox{\boldmath$\Xi$}_{(j)}).$
>
> Then
>
>
> $\sigma(\mbox{\boldmath$\Xi$})=\Big{(}(-\sigma^{2}\mbox{\boldmath$A$}+\sigma\mbox{\boldmath$B$}+\mbox{\boldmath$C$})\mbox{\boldmath$a$}\Big{|}\mbox{\boldmath$a$}\Big{)}.$
>
> Thus, D. Lynden-Bell and J. P. Ostriker say, the variational principle reads
>
>
> $\mbox{\boldmath$\delta$}\Big{(}(-\sigma^{2}\mbox{\boldmath$A$}+\sigma\mbox{\boldmath$B$}+\mbox{\boldmath$C$})\mbox{\boldmath$a$}\Big{|}\mbox{\boldmath$a$}\Big{)}=0$
>
> and, varying $a$, we obtain the ‘secular determinant’
>
>
> $\mathrm{det}(-\sigma^{2}\mbox{\boldmath$A$}+\sigma\mbox{\boldmath$B$}+\mbox{\boldmath$C$})=0$
> (5.33)
>
> for the determination of the variationally best eigenfrequencies $\sigma$.

But it is doubtful that this scheme is so practtical, since we have no
confidence that the equation (5.33), which is an algebraic equation for the
unknown $\sigma$ of degree is $2N$, can be numerically solved to determine an
approximating eigenfrequency $\sigma_{N}$ so that they converge to a true
eigenfrequency as $N\rightarrow\infty$.

On the other hand, since

$\sigma(\mbox{\boldmath$\Xi$})=\frac{b(\mbox{\boldmath$\Xi$})}{2}+\sqrt{\frac{b(\mbox{\boldmath$\Xi$})^{2}}{4}+c(\mbox{\boldmath$\Xi$})}\geq
0$

for $\forall\mbox{\boldmath$\Xi$}\in\mathfrak{H}^{\mathrm{div}}$, we can
consider

$\sigma_{*}:=\inf\\{\sigma(\mbox{\boldmath$\Xi$})|\mbox{\boldmath$\Xi$}\in\mathfrak{H}^{\mathrm{div}},\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}=1\\}$

and we may expect that the minimum might give an eigenfrequency. But, when
$\Omega\not=0$, we are not sure about the existence of a
$\mbox{\boldmath$\Xi$}_{*}$ which attains the infimum, namely
$\sigma_{*}=\sigma(\mbox{\boldmath$\Xi$}_{*})$, in general, since the
imbedding $\mathfrak{H}^{\mathrm{div}}\hookrightarrow\mathfrak{H}$ is not
compact and we may be unable to extract convergent subsequences from a
minimizing sequence, say, a sequence $(\bm{\Xi}_{n})_{n\in\mathbb{N}}$ such
that
$\bm{\Xi}_{n}\in\mathfrak{H}^{\mathrm{div}},\|\bm{\Xi}_{n}\|_{\mathfrak{H}}=1,\sigma(\bm{\Xi}_{n})\rightarrow\sigma_{*}$
as $n\rightarrow\infty$.

When $\Omega=0$, then $b(\mbox{\boldmath$\Xi$})=0$, and the variational
problem

$\lambda_{*}(=\sigma_{*}^{2})=\inf\\{c(\mbox{\boldmath$\Xi$})(=Q[\mbox{\boldmath$\Xi$}])|\mbox{\boldmath$\Xi$}\in\mathfrak{H}^{\mathrm{div}},\|\mbox{\boldmath$\Xi$}\|_{\mathfrak{H}}=1\\}$

actually admits the trivial solution $\lambda_{*}=0$ with any eigenvector
$\in\mathrm{Ker}(\mbox{\boldmath$L$})$. But the Mini-Max principle does not
work, for, since $\mathrm{dim.Ker}(\mbox{\boldmath$L$})=\infty$, we cannot go
ahead across the $0$ eigenvalue eternally, although there actually remain
infinitely many positive eigenvalues. Thus also in this case the variational
principle seems to be not so useful.

However, when $\Omega=0$, the equation (5.5) reduces to

$-\lambda\bm{\Xi}+\bm{L}\bm{\Xi}=\mathbf{0},$ (5.34)

where $\lambda=\sigma^{2}$, and this eigenvalue problem is completely solved
as follows:

Suppose $\Omega=0$. We note that the background stationary solution
$\bar{\rho}$ is a spherically symmetric equilibrium, say,

$\bar{\rho}(\bm{x})=\Big{(}\frac{(\gamma-1)\mathsf{G}M_{0}}{\mathsf{A}\gamma}\Big{(}\frac{1}{r}-\frac{1}{R}\Big{)}\Big{)}^{\frac{1}{\gamma-1}}$
(5.35)

and $\mathfrak{R}=\\{\bar{\rho}>0\\}=\\{R_{0}<r<R\\}$ is an annulus. Taking
the divergence of $\bar{\rho}$ times (5.34), the problem reduces to

$-\lambda
g-\mathrm{div}\bar{\rho}\Big{(}\mathrm{grad}\overline{\frac{d\varUpsilon}{d\rho}}g\Big{)}=0,$
(5.36)

where

$g=\mathrm{div}(\bar{\rho}\bm{\Xi}).$ (5.37)

Note that we can treat the operator
$\mathcal{N}:g\mapsto\mathrm{div}\bar{\rho}\Big{(}\mathrm{grad}\overline{\frac{d\varUpsilon}{d\rho}}g\Big{)}$
as that with the similar property to the Laplacian operator
$\triangle=\mathrm{div.grad}$, taking into account the singular behaviors of
$\bar{\rho}$ and $\displaystyle\overline{\frac{d\varUpsilon}{d\rho}}$ at the
physical vacuum boundary, say,

$\displaystyle\bar{\rho}\sim\Big{(}\frac{(\gamma-1)\mathsf{G}M_{0}}{\mathsf{A}\gamma}\Big{)}^{\frac{1}{\gamma-1}}(R-r)^{\frac{1}{\gamma-1}},$
$\displaystyle\overline{\frac{d\varUpsilon}{d\rho}}\sim\mathsf{A}\gamma\Big{(}\frac{(\gamma-1)\mathsf{G}M_{0}}{\mathsf{A}\gamma}\Big{)}^{\frac{\gamma-2}{\gamma-1}}(R-r)^{-\frac{2-\gamma}{\gamma-1}}$

near $\Sigma_{1}=\\{r=R\\}$. Therefore we can claim:

The operator $\mathcal{N}$ can be considered as a self-adjoint operator in the
Hilbert space $\mathfrak{G}$ and its spectrum $\sigma(\mathcal{N})$ is of the
Sturm-Liouville type, that is,
$\sigma(\mathcal{N})=\\{\lambda_{n}^{\mathcal{N}}|n\in\mathbb{N}\\}$, where
$\lambda_{n}^{\mathcal{N}}$ is an eigenvalue with finite multiplicity,
$0<\lambda_{0}^{\mathcal{N}}<\cdots<\lambda_{n}^{\mathcal{N}}<\lambda_{n+1}^{\mathcal{N}}$,
and $\lambda_{n}^{\mathcal{N}}\rightarrow+\infty$ as $n\rightarrow\infty$.
Here

$\mathfrak{G}=L^{2}\Big{(}\mathfrak{R};\overline{\frac{d\varUpsilon}{d\rho}}d\bm{x}\Big{)}\cap\Big{\\{}g\
\Big{|}\ \int_{\mathfrak{R}}gd\bm{x}=0\Big{\\}}.$

Note that the imbedding of
$L^{2}(\mathfrak{R};\overline{\frac{d\varUpsilon}{d\rho}}d\bm{x})$ into
$L^{2}(\mathfrak{R};d\bm{x})(\hookrightarrow L^{1}(\mathfrak{R};d\bm{x}))$ is
continuous. For a proof see the proof of [11, Theorem 2]. Hence the argument
of [11] leads us to the following conclusion:

When $\Omega=0$, $\bm{L}$ can be considered as a self-adjoint operator in the
Hilbert space $\mathfrak{F}$ and its spectrum $\sigma(\bm{L})$ coincides with
$\sigma(\mathcal{N})\cup\\{0\\}$, while $\mathrm{dim.Ker}(\bm{L})=\infty$ and
$\lambda=\lambda_{n}^{\mathcal{N}}\not=0$ is an eigenvalue with finite
multilicity. Here

$\mathfrak{F}=\\{\bm{\Xi}\in\mathfrak{H}\ |\
\mathrm{div}(\bar{\rho}\bm{\Xi})\in\mathfrak{G}\\}=\Big{\\{}\bm{\Xi}\in\mathfrak{H}^{\mathrm{div}}\
\Big{|}\
\int_{\mathfrak{R}}\mathrm{div}(\bar{\rho}\bm{\Xi})d\bm{x}=0\Big{\\}}.$

Note that $\mathfrak{F}$ is a closed subspace of
$\mathfrak{H}^{\mathrm{div}}$, since $\mathfrak{G}\ni
g\mapsto\int_{\mathfrak{R}}gd\bm{x}$ is constinuous. In fact, we have
$\displaystyle\int_{\mathfrak{R}}\overline{\frac{d\rho}{d\Upsilon}}d\bm{x}<+\infty$
thanks to $\gamma>1$ so that

$\Big{|}\int
g\Big{|}\leq\Big{[}\int\overline{\frac{d\Upsilon}{d\rho}}|g|^{2}\Big{]}^{1/2}\Big{[}\int_{\mathfrak{R}}\overline{\frac{d\rho}{d\Upsilon}}\Big{]}^{1/2}\lesssim\|g\|_{L^{2}(\frac{d\Upsilon}{d\rho}d\bm{x})}.$

We have $\mathfrak{H}_{0}^{\mathrm{div}}\subset\mathfrak{F}$, since

$\int\mathrm{div}(\bar{\rho}\bm{\varphi})d\bm{x}=0\quad\mbox{for}\quad\forall\bm{\phi}\in
C_{0}^{\infty}(\mathfrak{R}).$

Therefore by the well-known theorem ( [7, p.905, X.3.4.Theorem] complemented
by [12, p. 177, Chapter III, Theorem 6.15]), we can say that the eigenvectors
of a CONS of $\mathrm{Ker}(\bm{L})$ and all

$\psi_{n}=\frac{1}{\lambda_{n}^{\mathcal{N}}}\bar{\rho}\mathrm{grad}\Big{(}-\overline{\frac{d\varUpsilon}{d\rho}}\varphi_{n}^{\mathcal{N}}\Big{)}\Big{\|}\frac{1}{\lambda_{n}^{\mathcal{N}}}\bar{\rho}\mathrm{grad}\Big{(}-\overline{\frac{d\varUpsilon}{d\rho}}\varphi_{n}^{\mathcal{N}}\Big{)}\Big{\|}^{-1},$

$\varphi_{n}^{\mathcal{N}}$ being an eigenvalue of $\mathcal{N}$ associated
with the eigenvalue $\lambda_{n}^{\mathcal{N}}\not=0$ form a complete
ortogonal system of the Hilbert space $\mathfrak{F}$.

In this sense, when $\Omega=0$, the eigenfrequency problem is completely
solved.

###### Remark 2

Let $\Omega=0$. If we consider the operator $\bm{L}$ in the space
$\mathfrak{H}$, we can claim that
$\\{0\\}\cap\sigma(\mathcal{N})=\\{0,\lambda_{0}^{\mathcal{N}},\lambda_{1}^{\mathcal{N}},\cdots\\}\subset\sigma(\bm{L})$,
but we do not know whether they coincide or not, say, we do not know whether
there are real continuous spectrum between the eigenvalues or not.

However, when $\Omega\not=0$, the above discussion seems not to work. In fact
the term $2\Omega(\nabla|\bar{\rho}J_{\star}\bm{\Xi})$ may cause trouble,
since it cannot be reduced to a quantity determined by
$g=(\nabla|\bar{\rho}\bm{\Xi})$.

Acknowledgment

The author expresses his sincere thanks to Professor Akitaka Matsumura (Osaka
University) for kindly having helpful discussions on this research and
especially enlghting the author on the boundary trace of the divergence spaces
$H(\mathrm{div}),H_{0}(\mathrm{div})$. This work is supported by JSPS KAKENHI
Grant Number JP21K03311.

The data availability statement

No new data were created or analyzed in this study.

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