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\documentstyle[12pt]{article}
\textwidth = 6.5 in \hoffset = -.5 in \textheight = 8 in
\title{Sojourn times of trapping rays and the behavior of the modified
resolvent of the Laplacian}
\author{by\\Vesselin Petkov and Latchezar Stoyanov\\}
\date{}
\begin{document}
\maketitle
\def\sn{S^{n-1}}
\def\ft{{\cal F}_{t\rightarrow \lambda}}
\def\xl{(x,\lambda)}
\def\lto{(\lambda,\theta,\omega)}
\def\ssu{{\rm supp}}
\def\hs{\hat{s}}
\def\hu{\hat{u}}
\def\do{\partial \Omega}
\def\ued{U_{\epsilon,d}}
\def\tto{(t,\theta,\omega)}
\def\rl{R(\lambda)}
\def\rrr{R_{\varphi,\psi}(\lambda)}
\section{Introduction}
Let $\Omega\subset {\bf R}^n, n\geq 3, n$ odd, be an open and connected
domain with $C^{\infty}$ boundary $\do$ and bounded complement
$$K = \overline{{\bf R}^n\setminus \Omega} \subset \{ x\in {\bf R}^n :
\mid x\mid \leq \rho_0 \}.$$
Consider the problem
\begin{eqnarray}
\cases{ (\partial_t^2 - \Delta_x)u = 0 \:\: {\rm in} \:\: {\bf R}\times
\Omega,\cr u = 0 \:\:\: {\rm on} \:\: {\bf R}\times\do,\cr
u(0,x) = f_1(x), \partial_t u(0,x) = f_2(x).\cr }
\end{eqnarray}
We can associate to (1) a {\it scattering operator}
$$S(\lambda) : L^2({\bf R}^{n-1}) \longrightarrow L^2({\bf R}^{n-1}),
\:\: \lambda \in {\bf R},$$
whose kernel $a\lto$, called {\it scattering amplitude}, depends
analytically on $\omega, \theta \in \sn$ (see \cite{kn:LP}).
For fixed $(\theta,\omega)\in \sn\times\sn$, $a\lto$ is a tempered
distribution in $\lambda$ and
$$a\lto = (\frac{2\pi}{i\lambda})^{(n-1)/2}\ft s\tto.$$
Here $\ft$ denotes the {\it Fourier transform} and the distribution
$s\tto$ is called the {\it scattering kernel} (see \cite{kn:Ma},
\cite{kn:P1}).
\def\va{\varphi_a}
\def\ei{e^{i\lambda\langle x,\omega\rangle}}
\def\rab{R_{a,b}}
\def\cc{C_0^{\infty}}
The operator $S(\lambda)$ and the distribution $a\lto$ admit
meromorphic continuation in ${\bf C}$ with poles $\lambda_j$,
Im $\lambda_j < 0$, which are independent of $\theta$ and $\omega$.
Notice that in \cite{kn:LP} the scattering poles are related to
$\overline{a\lto}$.
One can characterize the poles $\lambda_j$ using the modified
resolvent of the Laplacian in $\Omega$ given by
$$\rrr = \varphi(x)\rl\psi(x).$$
Here the operator
$$\rl : \cc(\overline{\Omega}) \ni f \mapsto u(x,\lambda)\in C^{\infty}
(\overline{\Omega}), \: {\rm Im} \lambda \geq 0,$$
is determined by the $(-i\lambda)$-outgoing solution of the Dirichlet
problem for the reduced wave equation (see Section 2). The functions
$\varphi(x), \psi(x) \in \cc ({\bf R}^n)$ are chosen to be equal to 1
in some neighbourhood of $K$. Then $\rrr$ admits a meromorphic
continuation in ${\bf C}$ the poles of which and their multiplicities
coincide with these of the $\lambda_j$'s. Moreover, the poles
$\lambda_j$ do not depend on the choice of $\varphi$ and $\psi$
(see \cite{kn:LP}, \cite{kn:V}).
\def\ot{(\omega,\theta)}
It is well known that if $K$ is non-trapping, there exist
$\epsilon > 0$ and $d > 0$ such that $S(\lambda)$ has no poles in the
domain
\begin{equation}
\ued = \{ \lambda\in {\bf C} : d -\epsilon {\rm Log}(1+\mid\lambda\mid)
\leq \: {\rm Im}\:\lambda \leq 0 \: \}.
\end{equation}
On the other hand, Vainberg \cite{kn:V} proved that for non-trapping obstacles
the estimate
\begin{equation}
\| \rrr f\|_{H^1(\Omega)} \leq C e^{\alpha \mid {\rm Im}\lambda\mid}
\| f\|_{L^2(\Omega)}
\end{equation}
holds for each $\lambda\in\ued$. Here the constants $C > 0, \alpha > 0$
depend on supp $\varphi \: \cup$ supp $\psi$.
The situation changes when the obstacle is trapping. In this case one
expects to have an infinite number of scattering poles in $\ued$ for
all $\epsilon > 0$ and all $d > 0$. Such a result is proved in
\cite {kn:BGR} provided the obstacle is a union of two disjoint strictly
convex bodies. In the same case more precise results concerning the
distribution of poles are obtained by Ikawa \cite{kn:I1} and G\'{e}rard
\cite{kn:G}.
The case
$K = \bigcup_{j=1}^N K_j, K_i \cap\ K_j = \emptyset$,
$i\neq j$, $K_j$ strictly convex, has been examined by Ikawa \cite{kn:I2},
\cite{kn:I3} under the condition
\begin{eqnarray*}
{\rm (H)} \:\:\:\:\:\:\:\:\:\:\:\:\:\: \cases{ K_{\nu}\cap \:{\rm convex} \:
{\rm hull}\: (K_i\cup K_j) = \emptyset \:\: {\rm for} \: {\rm each} \:\cr
{\rm triple} \:\: (i,j,\nu)\in \{ 1,\ldots,N \}^3, i\neq j,
i\neq \nu, j\neq\nu. \cr}
\end{eqnarray*}
This condition implies the non-existence of periodic rays in $\Omega$ with
segments tangent to $\do$. The case when the obstacles $K_j$ are in
generic position has been studied in \cite{kn:PS1}.
To describe the results in this paper we need several definitions.
Given an obstacle $K$, fix an open ball $B_0$ of radius $\rho_0$
containing $K$. For
each $\xi$ we shall denote by $Z_{\xi}$ the hyperplane tangent to
$B_0$ and orthogonal to $\xi$ such that the halfspace $H_{\xi}$, determined by
$Z_{\xi}$ and having $\xi$ as an inner normal, contains $K$. Let
$\omega \in\sn, \theta\in\sn$ and let $\gamma$ be an $\ot$-ray
in $\Omega$, that is an infinite ray in $\Omega$ with incoming direction
$\omega$ and outgoing direction $\theta$ (see \cite{kn:PS3} for the
precise definition). The {\it sojourn time} $T_{\gamma}$ of $\gamma$
is defined by $T_{\gamma} = T'_{\gamma} - 2\rho_0$, where $T'_{\gamma}$
is the length of this part of $\gamma$ which is contained in
$H_{\omega}\cap H_{-\theta}$. If $\gamma$ has no gliding segments on
$\do$ and only finitely many reflection points, then $\gamma$ is called
a {\it reflecting $\ot$-ray} in $\Omega$. If moreover $\gamma$ has no
segments tangent to $\partial K$, then it is called an {\it ordinary}
reflecting $\ot$-ray.
Let $\gamma$ be an ordinary reflecting $\ot$-ray in $\Omega$ with
successive reflection points $x_1,\ldots,x_k$ on $\partial K$. Note that
in this case
$$T_{\gamma} = \langle \omega,x_1\rangle + \sum_{i=1}^{k-1} \|x_i -
x_{i+1} \| - \langle \theta,x_k\rangle$$
where $<\: , \: >$ denotes the standard inner product in ${\bf R}^n$.
(see Section 2.4 in \cite{kn:PS3}). Denote by $u_{\gamma}$ the
orthogonal projection
of $x_1$ on $Z = Z_{\omega}$. Then there exists a neighbourhood
$W = W_{\gamma}$ of $u_{\gamma}$ in $Z$ such that for every $u\in W$
there are unique
$\theta(u)\in \sn$ and points $x_1(u),\ldots,x_k(u) \in \partial K$
which are the successive reflection points of a reflecting
$(\omega,\theta(u))$-ray in $\Omega$ passing through $u$. Setting
$J_{\gamma}(u) = \theta(u)$, one obtains a smooth map
$$J_{\gamma} : W_{\gamma} \longrightarrow \sn.$$
The ray $\gamma$ is called {\it non-degenerate} if
$\det dJ_{\gamma}(u_{\gamma})\neq 0$.
The aim of this paper is to establish a link between the existence of
a sequence $\{ -T_m\}$ of singularities of $s(t,\theta_m,\omega_m)$
and the estimate
\begin{equation}
\|\rrr f\|_{H^1(\Omega)} \leq C e^{\alpha\mid {\rm Im}\lambda\mid}
(1+\mid\lambda\mid)^p \| f\|_{H^k(\Omega)},
\end{equation}
for $\lambda\in\ued$ and some constants $C > 0, \alpha \geq 0$,
$p\in {\bf N}, k \in {\bf N}$. It is proved in Theorem 2..3 below that
if there exist ordinary reflecting $(\omega_m,\theta_m)$-rays $\gamma_m$
with sojourn times $T_m \rightarrow \infty$ such that
\begin{equation}
-T_m \in {\rm sing} \: {\rm supp} \: s(t,\theta_m,\omega_m),
\end{equation}
then either there exist poles in $\ued$ for all $\epsilon > 0, d > 0$,
or there exist $\epsilon > 0, d > 0$ for which the domain $\ued$ is free
of poles but (4) fails for every choice of $C, \alpha, p$ and $k$.
If $K$ is a trapping obstacle, the existence of a sequence $\{\gamma_m\}$
of reflecting $(\omega_m,\theta_m)$-rays with
$T_{\gamma_m} \rightarrow \infty$ follows from the continuity of the
generalized Hamiltonian flow related to the wave operator
$\Box = \partial_t^2 - \Delta_x$ (see Section 5). The main difficulty is
to prove that these rays produce singularities leading to (5).
According to the results in \cite{kn:CPS} (see also Chapters 8 and 9 in
\cite{kn:PS3}), to obtain (5)
it is sufficient that for all $m$ the pair
$(\omega_m,\theta_m)$ has the following three properties:
\def\ot{(\omega,\theta)}
(a) if $\delta$ and $\gamma$ are different ordinary reflecting
$(\omega_m,\theta_m)$-rays, then $T_{\delta}\neq T_{\gamma}$;
(b) for every $m$, $\gamma_m$ is non-degenerate;
(c) there are no $(\omega_m,\theta_m)$-rays in $\Omega$
containing gliding segments on $\do$ or tangent segments to $\do$.
In Section 4 we show that given an arbitrary obstacle $K$,
for almost all $\ot \in \sn\times\sn$ (with respect to the Lebesgue
measure in $\sn\times\sn$) every two different ordinary reflecting
$\ot$-rays have distinct sojourn times. Using this, the property (a)
can be arranged by an apropriate approximation.
To obtain an analogue of this for (c), one
must study all $\ot$-rays in $\Omega$, i.e. all projections of
generalized $\ot$-bicharacteristics of the wave operator. In general
this is a difficult problem, especially when the set
$G^{\infty}(K)$ of points $z\in \partial K$, where the curvature of $K$
along some direction $\xi \in T_z(\partial K)$ vanishes of infinite
order, is not empty. In Section 3 we prove that for each obstacle $K$
for almost all $\ot\in \sn\times\sn$ all reflecting $\ot$-rays in
$\Omega$ are ordinary. This result can be applied in the case when
the obstacle has the form
\begin{equation}
K = \bigcup_{j=1}^N K_j \: , K_i \cap K_j = \emptyset \:\: {\rm for}\:
i\neq j,\\
\:\:\:\:\:\: K_j \:\: {\rm convex} \: {\rm for} \: {\rm all}\:\:
j = 1,\ldots,N,
\end{equation}
to show that (c) is satisfied for almost
all $\ot\in \sn\times\sn$ even if $G^{\infty}(K)\neq \emptyset$.
To arrange (b) for obstacles of the form (6) we use the
representation of the (linear)
Poincar\'{e} map (see \cite{kn:PS3}, Proposition 2.4.2). On the other hand,
using Sard's theorem, it is possible to arrange (b) for a dense set of
directions $\ot \in \sn\times\sn$. The results in Sections 2-5 are
established for obstacles with arbitrary geometry. For several convex
obstacles we make two applications.\\
{\bf 1.1. Theorem.} {\it Let $K$ have the form {\rm (6)}. Assume that
there are no points $z\in \partial K$ such that the Gauss curvature
${\cal K}(u)$ of $\partial K$ at $u$ vanishes for each $u$ in some
neighbourhood $U_z$ of $z$ in $\partial K$. Then there exists a
sequence of ordinary reflecting $(\omega_m,\theta_m)$-rays in $\Omega$
with sojourn times $T_m \rightarrow \infty$ such that for $t$ near $-T_m$
we have
$$s(t,\theta_m,\omega_m) = A_m \delta^{(n-1)/2} (t+T_m) +
\: {\rm lower} \: {\rm order} \: {\rm singularities},$$
with $A_m \neq 0$. Moreover,
either the assertion {\rm (i)} or the assertion {\rm (ii)} in Theorem 2.3
in Section 2 holds.}\\
{\bf 1.2. Theorem.} {\it Let $K$ have the form {\rm (6)}. Then there
exists a set ${\cal R}\subset \sn\times\sn$, the complement of which has
zero Lebesgue measure in $\sn\times\sn$, such that for all
$\ot \in {\cal R}$ we have
$${\rm sing}\:{\rm supp} s(t,\theta,\omega) = \{ -T_{\gamma} :
\gamma \in {\cal L}_{\omega,\theta} \: \},$$
where ${\cal L}_{\omega,\theta}$ is the set of all $\ot$-rays in $\Omega$.
Moreover, for $t$ near $-T_{\gamma}$ we have
$$s(t,\theta,\omega) = c_{\gamma} \delta^{(n-1)/2} (t+T_{\gamma}) +
\: {\rm lower} \: {\rm order} \: {\rm singularities}$$
with $c_{\gamma} \neq 0$.}\\
For $n = 3$ the assumption of Theorem 1.1 means that there are no points
$z\in \partial K$ such that the standard metric on $\partial K$ is
locally flat around $z$. For strictly convex obstacles the assertion of
Theorem 1.2 was proved in \cite{kn:PS2}.
In the works of Ikawa \cite{kn:I2}, \cite{kn:I3} and Sj\"{o}strand and
Zworski \cite{kn:SjZ} the singularities of the distribution
\begin{eqnarray*}
u(t) = \cases{\sum_j e^{i\lambda_j t} \: , \:\: t > 0, \cr
\sum_j e^{-i\lambda_j t} \: , \:\: t < 0,\cr}
\end{eqnarray*}
have been exploited. Here the sumation is over all poles $\lambda_j$
including their multiplicities.
It is well-known that
${\rm sing}\: {\rm supp} u(t)$ is contained in the set of the
periods (lengths) of all
periodic rays in $\Omega$ (see \cite{kn:PS3}, Chapter 5). Thus, to prove that a period $T_{\gamma}$
belongs to sing supp $u(t)$, it is sufficient to establish some
properties $(a')-(c')$ similar to $(a)-(c)$. The reader may consult
\cite{kn:PS3} for results in this direction. Notice that in general,
even for several strictly convex obstacles, there might be different
periodic rays with rationally dependent periods or periodic reflecting
rays with segments tangent to the boundary. Thus, the singularities produced
by periodic rays could be canceled and it seems difficult to prove that there
exists a sequence $d_j \rightarrow \infty$ of singularities of $u(t)$.The condition (H) in \cite{kn:I2}, \cite{kn:I3} has been used to avoid the existence of
periodic rays with tangent segments. Notice also that the results in \cite{kn:SjZ} can be applied for all logarithmic domains if a sequence $d_j$ of singularities going to infinity exists.
In this paper we study trapping $(\omega,\theta)$-rays with suitable
directions $\omega, \theta$ instead
of periodic rays. This makes it possible to examine the general case of
several convex obstacles and to find a sequence $T_m \rightarrow \infty$ of
singularities of the scattering kernel. One could expect that such sequence of
singularities exists for every trapping obstacle $K$ with
$G^{\infty}(K) = \emptyset$ and the conclusion of Theorem 2.3 holds for
trapping obstacles of this type.
It is an open problem if the existence of a sequence
$T_m \rightarrow \infty$ of singularities always implies the assertion (i)
of Theorem 2.3. From general point of view there might exist some very
degenerate examples of trapping obstacles for which there are logarithmic
domains free of poles and the assertion (ii) of Theorem 2.3 holds.
\section{The behavior of the resolvent of the Laplacian}
Consider the problem
\begin{eqnarray}
\cases{(\Delta +\lambda^2)u\xl = f \:\: {\rm in} \:\: \Omega \cr
u\xl = 0 \:\: {\rm on} \:\: \do \cr
u\xl \:\: {\rm is} \:\: (-i\lambda){\rm -outgoing}. \cr }
\end{eqnarray}
The last condition means that for $x = \mid x\mid\omega, \omega\in\sn,$
and $\mid x\mid \rightarrow\infty$ we have
$$u(\mid x\mid\omega,\lambda) = \frac{e^{i\lambda\mid x\mid}}
{\mid x\mid^{(n-1)/2}} (b(\omega,\lambda) + {\rm O}(\frac{1}{\mid x\mid})).$$
For Im $\lambda \geq 0$ the problem (7) has an unique solution
for $f\in L^2(\Omega)$ and the operator
$$R(\lambda) : C_0^{\infty}(\overline{\Omega}) \ni f \mapsto u\xl
\in C^{\infty}(\overline{\Omega}) \:\: , \: {\rm Im} \lambda \geq 0,$$
admits a meromorphic extension in ${\bf C}$ with poles $\lambda_j$,
Im $\lambda_j < 0$ (see \cite{kn:LP}).
Let $\hs\lto = \ft s(t,\theta,\omega)$ be the {\it Fourier transform}
of the scattering kernel $s(t,\theta,\omega)$. It is well known (
\cite{kn:LP}, \cite{kn:P1}, \cite{kn:Ma}) that $\hs\lto$ has the
representation
$$\hs\lto = c_n \lambda^{n-2} \int_{\do} e^{-i\lambda\langle x,\theta\rangle}
(\frac{\partial v}{\partial \nu} + i\lambda\langle \nu,\theta\rangle v)
\xl dS_x,$$
where $c_n =$ const, $\nu(x)$ is the {\it unit normal} to $\do$ at
$x\in \do$ pointing into $\Omega$, and $v\xl$ is the solution of the problem
\begin{eqnarray*}
\cases{(\Delta +\lambda^2)v\xl = 0 \:\: {\rm in} \:\: \Omega \cr
v + e^{i\lambda\langle x,\omega\rangle} = 0 \:\: {\rm on} \:\: \do \cr
v \:\: {\rm is} \:\: (-i\lambda){\rm -outgoing}. \cr }
\end{eqnarray*}
\def\va{\varphi_a}
\def\ei{e^{i\lambda\langle x,\omega\rangle}}
\def\rab{R_{a,b}}
\def\cc{C_0^{\infty}}
We shall express $\hs\lto$ by using the operator $R(\lambda)$.
Let $a > \rho_0$. Consider a function
$\va \in C_0^{\infty}({\bf R}^n)$ such that
$$\va(x) = 1 \:\: {\rm for} \:\: \mid x\mid \leq a.$$
Setting
$$F_a(\lambda) = {[ \Delta\va + 2i\lambda\langle\nabla\va,\omega\rangle
]} \ei,$$
we get
$$v\xl + \va(x)\ei = R(\lambda){[(\Delta +\lambda^2)(\va\ei)]} =
R(\lambda)F_a(\lambda).$$
Next, choose a function $\chi_b(x)\in C_0^{\infty}({\bf R}^n)$
such that $\chi_b(x) = 1$ on a neighbourhood of $K$ and
$\va(x) = 1$ on supp $\chi_b.$
Then the normal derivative becomes
$$\frac{\partial v}{\partial \nu}_{\mid \do} = -i\lambda
e^{i\lambda\langle x,\omega\rangle}
\langle\nu,\omega\rangle_{\mid \do} + \frac{\partial}{\partial \nu} (\chi_b
R(\lambda)F_a(\lambda))_{\mid \do}.$$
On the other hand, for $\theta\neq -\omega$ by Green's formula we have
$$\int_{\do} e^{-i\lambda\langle x,\theta - \omega\rangle}
\langle \nu,\theta+\omega\rangle dS_x = \int_{K}
\frac{\partial}{\partial(\theta +\omega)}(e^{-i\lambda \langle x,\theta
- \omega\rangle} ) dx = 0.$$
Thus, using Green's formula once more, we obtain
\begin{eqnarray}
\hs\lto & = & - c_n \lambda^{n-2}\int_{\Omega} e^{-i\lambda\langle x,
\theta\rangle} (\Delta + \lambda^2)(\chi_b R(\lambda)F_a) dx \nonumber\\
& = & - c_n \lambda^{n-2} \int_{\Omega} e^{-i\lambda\langle x,
\theta\rangle}{[(\Delta \chi_b) R(\lambda)F_a(\lambda) +2\langle
\nabla_x \chi_b,\nabla_x (R(\lambda)F_a(\lambda)\rangle]}dx.
\end{eqnarray}
Let $\psi_c(x)\in C_0^{\infty}({\bf R}^n), c = a,b,$ be cut-off
functions such that
$$\psi_a(x) = 1 \:\: {\rm on} \:\: \ssu \va \: , \: \psi_b(x) = 1 \:\:
\:\: {\rm on} \:\: \ssu \chi_b.$$
Then in the representation of $\hs\lto$ we can replace the resolvent
$R(\lambda)$ by the {\it modified resolvent}
$$\rab (\lambda) = \psi_b(x) R(\lambda) \psi_a(x).$$
Below we assume that $\psi_a$ and $\psi_b$ are fixed.
Next, consider the domain $\ued$ defined in the introduction and assume
that $\rab(\lambda)$ satisfies the condition
\begin{eqnarray*}
{\rm (B)} \:\:\:\:\:\:\:\: \cases{\rab (\lambda) \:\: {\rm is} \:
{\rm analytic}\: {\rm in}\:\:
\ued \:\: {\rm and}\: {\rm there} \:
{\rm exist}\:\: C > 0, \alpha\geq 0, p\in {\bf N}, k\in {\bf N} \cr
{\rm such} \: {\rm that}\:\: \|\rab(\lambda)\varphi\|_{H^1(\Omega)}
\leq C(1 +\mid
\lambda\mid)^p e^{\alpha\mid {\rm Im} \lambda\mid}\|\varphi\|_{H^k
(\Omega)} \cr
{\rm for}\: {\rm each} \:\: \varphi \in \cc(\Omega) \:\: {\rm and}\:
{\rm each}\:\: \lambda \in \ued.\cr }
\end{eqnarray*}
Since the integration in (8) is over a compact set, we conclude that
the condition (B) implies the estimate
\begin{equation}
\mid\hs\lto\mid \leq C_1 (1+\mid\lambda\mid)^m e^{\beta\mid {\rm Im}
\lambda\mid} \:\: , \:\: \lambda\in \ued,
\end{equation}
with $\beta\geq 0, m\in {\bf N},$ uniformly with respect to
$(\theta,\omega)\in \sn\times\sn$.\\
{\bf 2.1. Remark}. The analysis in Chapter III of \cite{kn:LP} shows
that the norm of the {\it scattering operator}
$$S(\lambda) : L^2(\sn) \longrightarrow L^2(\sn) \:\: , \:\:
\lambda\in \ued,$$
is less than the norm of the modified resolvent
$\varphi(x) R(\lambda)\psi(x)$, where
$\varphi(x),\psi(x)\in\cc(\overline{\Omega})$ and $\varphi(x) = 1$,
$\psi(x) = 1$ in a neighbourhood of $K$. The inequality (9) is
sharper since it gives an estimate for the kernel of the scattering
operator $S(\lambda)$.\\
Now we shall deduce from (9) some regularity of $s(t,\theta,\omega)$
for $t \rightarrow -\infty$. To do this we need the following lemma,
which is similar to Theorem 7.3.8 in \cite{kn:H1}.\\
{\bf 2.2. Lemma.} {\it Let $u\in {\cal S}'({\bf R})$ be a distribution
with \ssu $\: u\subset \{t : t\leq \tau\}$. Assume that the Fourier
transform $\hu (\xi)$ admits an analytic continuation in the domain
$\ued$ such that for all $\zeta\in\ued$ we have
\begin{equation}
\mid\hu(\zeta)\mid \leq C(1+\mid\zeta\mid)^Ne^{\alpha\mid {\rm Im} \zeta
\mid} \:\: , \:\: \alpha \geq 0.
\end{equation}
Then for each $q\in {\bf N}$ there exist $t_q < \tau$ and
$v_q \in C^q({\bf R})$ such that $u = v_q$ for $t\leq t_q$}.\\
{\it Proof.} Choose a function $\varphi\in \cc({\bf R})$ such that
$\ssu \: \varphi\subset (-1,1)$ and $\int_{\bf R}\varphi(t) dt = 1$.
Set $\varphi_{\delta}(t) = \frac{1}{\delta}\varphi(\frac{t}{\delta}),$
$0 < \delta\leq 1$, and consider $u*\varphi_{\delta}$. Consider the
path
$$\Gamma_{\epsilon} : {\bf R}\setminus {[-\gamma,\gamma]} \ni
\xi\mapsto\zeta = \xi +
i(d- \epsilon \: {\rm Log}\: (1+\mid\xi\mid)\in \ued,$$
where $\gamma = \exp(\frac{d}{\epsilon})-1$ and $d$ is given in the
definition of $\ued$. Clearly, (10) implies
$$\mid\hu(\zeta)\hat{\varphi}(\delta \zeta)\mid \leq C_M
(1+\mid\zeta\mid)^N (1+\mid \delta \zeta\mid
)^{-M} e^{(a+\delta)\mid {\rm Im}\zeta\mid} \:\: , \:\: \zeta\in\ued.$$
Using the analyticity of $\hu(\zeta)$ in $\ued$ and the above estimate
for fixed $\delta > 0$, we can write the integral
$$(u*\varphi_{\delta})(t) = (2\pi)^{-1} \int_{\bf R} e^{it\xi}\hu(\xi)
\hat{\varphi}(\delta\xi)d\xi$$
as a sum of two integrals
$$(2\pi)^{-1}\int_{\Gamma_{\epsilon}} e^{it\zeta}\hu (\zeta)
\hat{\varphi}(\delta\zeta) d\zeta + (2\pi)^{-1}\int_{\mid\xi\mid\leq
\gamma} e^{it\zeta} \hu(\xi)\hat{\varphi}(\delta\xi)d\xi.$$
The second integral is over a compact interval, therefore passing to
a limit as $\delta \rightarrow 0$, we obtain a $C^{\infty}$ function.
Next, for $\zeta \in \Gamma_{\epsilon}, 0 <\delta\leq 1,$ we have the
estimate
$$\mid\hu(\zeta)\hat{\varphi}(\delta\zeta)e^{it\zeta}\mid\leq C(1+
\mid\zeta\mid)^N e^{-(a+1+t){\rm Im}\zeta} \leq C'(1+ \mid {\rm Re}
\zeta\mid)^{N+\epsilon(a+1+t)},$$
provided $a + 1+ t < 0$. Given $q\in {\bf N}$, take
$$t_q = \frac{1}{\epsilon} (-N-n-1-q-\epsilon(a+1)).$$
Then $t\leq t_q$ implies $\epsilon(a+1+t)\leq -N-n-1-q$ and, since
$d\zeta = F(\xi)d\xi \rightarrow d\xi$ as
$\mid\xi\mid \rightarrow \infty$, for $t\leq t_q$, the integral on
$\Gamma_{\epsilon}$ is uniformly convergent for $0\leq \delta\leq 1$.
The same is true if we take the derivatives with respect to $t$ up to
order $q$. Letting $\delta \rightarrow 0$ and using the fact that
$\hat{\varphi}(\delta\zeta) \rightarrow 1$, by Lebesgue theorem we
conclude that for $t\leq t_q$ we have
$$u*\varphi_{\delta} \longrightarrow_{\delta \rightarrow 0} f,$$
$f$ being a $C^{q}$ function. This completes the proof of the lemma.\\
Combining the estimate (9) and Lemma 2.2, we obtain the following.\\
{\bf 2.3. Theorem.} {\it Assume that there exists a sequence of
ordinary reflecting $(\omega_m,\theta_m)$-rays in $\Omega$ with
sojourn times $T_m$ such that
\begin{equation}
T_m \longrightarrow_{m \rightarrow\infty} \infty.
\end{equation}
Let $\Phi\in \cc({\bf R})$ be such that $\ssu \: \Phi\subset (-1,1)$,
$\Phi(t) = 1$ for $\mid t\mid\leq \frac{1}{2}$. Assume that there
exists a sequence $\gamma_m \rightarrow 0$ of non-zero real numbers
and an integer $k$ independent of $m$ such that
\begin{equation}
\mid \ft {[ \Phi(\frac{t+T_m}{\gamma_m})s(t,\theta_m,\omega_m)]}\mid
\geq (c_m - o_m(1))\mid\lambda\mid^k \:\: , \: \mid\lambda\mid
\rightarrow \infty,
\end{equation}
where $c_m > 0$. Then there are two possibilities:
{\rm (i)} For each $\epsilon > 0$ and each $d > 0$, $\rab(\lambda)$
and $S(\lambda)$ have poles in the domain $\ued$;
{\rm (ii)} For some $\epsilon > 0, d > 0$, $\rab(\lambda)$ is analytic
in $\ued$ but for all $\alpha\geq 0, p\in {\bf N}, k\in {\bf N}$ we have}
$$\sup_{\lambda\in\ued,\|\varphi\|_{H^k(\Omega)} = 1} (1+\mid\lambda\mid
)^{-p} e^{-\alpha\mid {\rm Im}\lambda\mid} \|\rab(\lambda)\varphi
\|_{H^1(\Omega)} = +\infty.$$
{\bf 2.4. Remark.} Notice that if $\rab(\lambda)$ is
analytic in $\ued$, the same is true for $\hs\lto$ for all
$\omega,\theta \in \sn$, and this implies the analyticity of the
operator $S(\lambda)$ in $\ued$. This shows that the cases (i) and (ii)
do not depend on the choice of $\psi_a(x)$ and $\psi_b(x)$.\\
Now we are going to find a link between (ii) and the behavior of the
resolvent of the generator $B_c$ of the Lax-Phillips semigroup
$Z_c(t)$. Let $H_D(\Omega)$ be the closure of $\cc (\Omega)$ with respect
to the norm
$$\|\varphi\|_{H_D} = (\int_{\Omega} \mid\nabla\varphi\mid^2 dx)^{1/2}.$$
In the energy space $H = H_D(\Omega)\oplus L^2(\Omega)$ consider the
unitary group $U(t) = e^{tG}$ with generator
$$G = \left\lgroup\matrix{0&1\cr \Delta&0\cr }\right\rgroup,$$
whose domain is
$$D(G) = \{ (f_1,f_2)\in H : \: \Delta f_1 \in L^2(\Omega),
\nabla f_2 \in H_D(\Omega) \: \}.$$
Let $H_0 = H_D({\bf R}^n)\oplus L^2({\bf R}^n)$ be the energy space
for $\Omega$ replaced by ${\bf R}^n$, and let $U_0(t) = e^{tG_0}$ be
the unitary group related to the Cauchy problem for
$\partial_t^2 - \Delta_x$ (see \cite{kn:LP}). Consider the spaces
$$D_{\pm}^c = \{ f\in H_0 : U_0(t)f = 0 \:\: {\rm for}\:\: \mid x\mid
\leq c\pm t, \pm t \geq 0 \: \},$$
where $c \geq \rho_0$. Denote by $P_{\pm}^c$ the orthogonal projections
on $(D_{\pm}^c)^{\bot}$ and by $B^c$ the generator of the semigroup
$Z_c(t) = P_+^c U(t)P_-^c $.
Let $\chi\in\cc({\bf R}^n)$ be a function such that
$\chi \equiv 1$ in a neighbourhood of $K$ and $\chi(x) = 1$
on supp $\psi_a(x) \cup {\rm supp} \: \psi_b(x)$. Therefore
$$\rab(\lambda)f = \psi_b \chi R(\lambda)\chi \psi_a f,$$
and an estimate for $\rab(\lambda)$ can be deduced from that for
$\|\chi R(\lambda)\chi \|_{H^1(\Omega)}$, which in turn can be
estimated by $\| \chi(G + i\lambda)^{-1}\chi\|_{H}$.
Moreover, if supp $\chi\subset \{ x : \mid x\mid \leq c \}$, we have
$$ (\chi(G + i\lambda)^{-1}\chi f,g)_H = ((B^c + i\lambda)^{-1}\chi f,
\chi g)_H ,$$
where $( \: , \: )$ denotes the inner product in $H$. Thus an estimate
\begin{equation}
\| (B^c - \mu)^{-1}\|_H \leq C e^{\alpha\mid {\rm Re}\mu\mid} (1+
\mid\mu\mid)^p
\end{equation}
in the domain $W_{\epsilon,d} =
\{\mu\in {\bf C} : d-\epsilon {\rm Log}(1+\mid\mu\mid ) < {\rm Re}\mu \leq 0\}$
leads to an estimate for $\rab(\lambda)$ in $\ued$, which is a
contradiction with (ii).
The existence of ordinary reflecting trapping rays implies that for some
$t_0 > 0$ the spectral radius of the semigroup $Z^c(t_0)$ is 1 (see
\cite{kn:R}). Therefore, there exists $\alpha \in {\bf R}$ such that
$$e^{it_0\alpha} \in \sigma (Z^c(t_0)),$$
$\sigma (Z^c(t_0))$ being the spectrum of $Z^c(t_0)$. Consequently,
the result of I.Herbst \cite{kn:H} shows that we must have
$$\sup_{{\rm Re} \mu = 0} \| (B^c - \mu)^{-1}\|_H = +\infty.$$
The estimate (13) concerns the case when $\mu\in {\bf C}$ runs over
a logarithmic domain, and the assumptions in
Theorem 2.3 imply that even a polynomial bound of
$\sup_{\mu\in W_{\epsilon,d}} \| (B^c - \mu)^{-1}\|_H$ is impossible.
\def\ot{(\omega,\theta)}
\def\sn{S^{n-1}}
\def\ssn{S^{n-1}\times S^{n-1}}
\section{Directions of tangency}
Let $X$ be a compact smooth $(n-1)$-dimensional submanifold of
${\bf R}^n , n \geq 2$. In this section we consider pairs
$(\omega,\theta) \in S^{n-1}\times S^{n-1}$ for which there exists
at least one $\ot$-ray for $X$ which is not ordinary, i.e. it
is tangent to $X$ at some of its points. It is shown that the set of
these pairs has Lebesgue measure zero in $\ssn$.
Before proceeding with the statement of the main result in this section,
we introduce a notion that is slightly different from the notion of an
$\ot$-ray but it is rather convenient for our next considerations. It was
also used in our paper \cite{kn:CPS}.
A curve $\gamma$ in ${\bf R}^n$ is called an $\ot${\it-trajectory for} $X$
if it has the form
$$\gamma = \bigcup_{i=0}^{s} l_i,$$
where $l_i = {[x_i,x_{i+1}]}, i = 1,\ldots,s-1, x_i \in X$ for all
$i = 1,\ldots,s,$ while $l_0$ (resp. $l_s$) is the infinite ray starting
at $x_1$ (resp. $x_s$) with direction $-\omega$ (resp. $\theta$), and for
every $i = 0,1,\ldots,s-1,$ $l_i$ and $l_{i+1}$ satisfy the law of
reflection at $x_i$ with respect to $X$.
Clearly, every reflecting $\ot$-ray is a $\ot$-trajectory, but the
converse is not true in general, since an $\ot$-trajectory may intersect
transversally $X$.\\
{\bf 3.1. Theorem.} {\it There exists ${\cal R} \subset \ssn$ the
complement of which is a countable union of compact subsets of
measure zero in $\ssn$ such that for every pair $\ot \in {\cal R}$
all $\ot$-trajectories for $X$ are ordinary.}\\
Fix a hyperplane $Z$ in ${\bf R}^n$ such that $X$ is contained in one of
the open halfspaces determined by $Z$.
As a simple corollary of our argument in this section we get also the
following theorem, which is in fact a consequence of a result of
Melrose and Sj\"{o}strand \cite{kn:MS}, see also Chapter 24 in \cite{kn:H2}.\\
{\bf 3.2. Theorem.} {\it There exists ${\cal T} \subset Z\times \sn$, the
complement of which is a countable union of compact subsets of measure
zero in $Z\times\sn$, such that for every $(x,\omega)\in {\cal T}$ the
trajectory of the generalized geodesic flow (in the exterior domain
determined by $X$) starting at $x$ in direction $\omega$ has no tangencies
to $X$}.\\
\def\gr{{\rm grad}}
We now turn to the proofs of the above statements.
Fix two integers $k$ and $s$ with $s \geq 1$ and $0\leq k \leq s$.
Denote by $M(s,k)$ the set of those
$$\tilde{\xi} = (\omega;x;y;\theta) \in M_s = \sn\times X^{(s)}\times
X\times\sn,$$
with $x = (x_1,\ldots,x_s)$, such that there exists an $\ot$-trajectory
for $X$ with successive (transversal) reflection points
$x_1,\ldots,x_s$, the segment ${[x_k,x_{k+1}]}$ of which is tangent
to $X$ at the point $y \in (x_k,x_{k+1})$. Here by $x_0$ (resp. $x_{s+1}$)
we denote the orthogonal projection of $x_1$ on $Z_{\omega}$ (resp. of
$x_s$ on $Z_{-\theta}$), and by definition
$$X^{(s)} = \{ (x_1,\ldots,x_s)\in X^s \: : \: x_i\neq x_j \: ,
i\neq j \: \}.$$
It is convenient to consider also the following
subsets of $M(s,k)$:
$$M_r(s,k) = \{ (\omega;x;y;\theta)\in M(s,k) : x_k^{(r)}
\neq x_{k+1}^{(r)} \} \:\:\: , \: r = 1,\ldots,n.$$
Clearly, these sets are open in $M(s,k)$ and
$$M(s,k) = \bigcup_{r=1}^{n} M_r(s,k).$$
{\bf 3.3. Lemma.} {\it Each $M_r(s,k)$ is a smooth submanifold of
$M_s$ of codimension} $2n-3$.\\
{\it Proof.} We consider only the case $r = n$, the other cases are the same.
Fix an arbitrary
$\eta = (\tilde{\omega};\tilde{x};\tilde{y};\tilde{\theta})\in M_n(s,k)$.
Choose smooth charts
$$\varphi_i : U_i \longrightarrow X$$
of $X$ around $\tilde{x}_i$ and
$$\psi : V \longrightarrow X$$
of $X$ around $\tilde{y}$ such that
$\varphi_i(U_i)\cap \varphi_{i+1}(U_{i+1}) = \emptyset$,
$i = 1,\ldots,s-1$, $\varphi_k(U_k)\cap \psi(V) = \emptyset$,
$\varphi_{k+1}(U_{k+1})\cap \psi(V) = \emptyset$ (for $k = 0$ or $s$
the corresponding condition is to be deleted). Assuming
$\tilde{\omega}_n > 0$, we may parametrize $\sn$ around $\tilde{\omega}$ by
$$D_0\ni \omega' = (\omega_1,\ldots,\omega_{n-1}) \mapsto \omega =
(\omega';\omega_n)\in \sn,$$
where $\omega_n = (1 - {\omega_1}^2 -\ldots -{\omega_{n-1}}^2)^{1/2}$ and
$D_0$ is the unit open ball in ${\bf R}^{n-1}$. Similarly, we may assume
that $\sn$ is parametrized around $\tilde{\theta}$ by
$\theta'' = (\theta_2,\ldots,\theta_n) \in D_0$. In this way we get a chart
$$\chi : U = D_0\times U_1\times \ldots \times U_s\times V\times D_0
\longrightarrow D \subset M_s,$$
defined by
$$\chi(\xi) = (\omega;\varphi_1(u_1),\ldots,\varphi_s(u_s);\psi(v);\theta)$$
for
$$\xi = (\omega';u;v;\theta'')\in U.$$
Here $\omega = (\omega';\omega_n), \theta = (\theta_1;\theta'')$,
$u_i = (u_i^{(1)},\ldots,u_i^{(n-1)}) \in U_i$.
Next, define
$$F : U \longrightarrow {\bf R}$$
by
$$F(\xi) = \sum_{i=1}^{s-1} \| \varphi_i(u_i) - \varphi_{i+1}
(u_{i+1})\|.$$
{\bf case 1.} $0 < k < s$. Let
$$\xi = (\omega';u;v;\theta'') \in U$$
be such that $\chi(\xi) \in M_n(s,k)$. Then we have
\begin{equation}
\gr_{u_i} F(\xi) = 0 \:\: , \: i = 2,\ldots,s-1,
\end{equation}
\begin{equation}
\langle \frac{\varphi_2(u_2) - \varphi_1(u_1)}{\|\varphi_2(u_2) -
\varphi_1(u_1)\|} - \omega , \frac{\partial \varphi_1}{\partial u_1^{(j)}}
(u_1) \rangle = 0 \:\: , \: j = 1,\ldots,n-1
\end{equation}
\begin{equation}
\langle \frac{\varphi_s(u_s) - \varphi_{s-1}(u_{s-1})}{\|\varphi_s(u_s) -
\varphi_{s-1}(u_{s-1})\|} - \theta , \frac{\partial \varphi_s}{\partial
u_s^{(j)}}(u_s) \rangle = 0 \:\: , \: j = 1,\ldots,n-1,
\end{equation}
\begin{equation}
\frac{\psi(v) - \varphi_k(u_k)}{\|\psi(v) - \varphi_k(u_k)\|} +
\frac{\psi(v) - \varphi_{k+1}(u_{k+1})}{\|\psi(v) - \varphi_{k+1}(u_{k+1})\|}
= 0,
\end{equation}
\begin{equation}
\langle \varphi_{k+1}(u_{k+1}) - \varphi_k(u_k) , N(\xi)\rangle = 0,
\end{equation}
where
$$N(\xi) = \det \left\lgroup\matrix{f_1 & f_2 &\ldots & f_n \cr
\frac{\partial \psi^{(1)}}{\partial v^{(1)}}(v) & \frac{\partial \psi^{(2)}}
{\partial v^{(1)}}(v) & \ldots & \frac{\partial \psi^{(n)}}{\partial v^{(1)}}
(v) \cr \ldots & \ldots & \ldots & \ldots \cr
\frac{\partial \psi^{(1)}}{\partial v^{(n-1)}}(v) & \frac{\partial \psi^{(2)}}
{\partial v^{(n-1)}}(v) & \ldots & \frac{\partial \psi^{(n)}}{\partial
v^{(n-1)}}\cr }\right\rgroup $$
is a normal vector to $X$ at $\psi(v)$. Here $f_1,\ldots,f_n$ are the
standard basis vectors in ${\bf R}^n$. In correspondence with these
conditions we introduce the functions
$$K_i^{(j)}(\xi) = \frac{\partial F}{\partial u_i^{(j)}}(\xi) \:\:\: ,
\: i = 2,\ldots,s-1, j = 1,\ldots,n-1,$$
$$L_j(\xi) =
\langle \frac{\varphi_2(u_2) - \varphi_1(u_1)}{\|\varphi_2(u_2) -
\varphi_1(u_1)\|} - \omega , \frac{\partial \varphi_1}{\partial u_1^{(j)}}
(u_1) \rangle \:\: , \: j = 1,\ldots,n-1,$$
$$M_j(\xi) =
\langle \frac{\varphi_s(u_s) - \varphi_{s-1}(u_{s-1})}{\|\varphi_s(u_s) -
\varphi_{s-1}(u_{s-1})\|} - \theta , \frac{\partial \varphi_s}{\partial
u_s^{(j)}}(u_s) \rangle \:\: , \: j = 1,\ldots,n-1,$$
$$P_j (\xi) =
\frac{\psi^{(j)}(v) - \varphi_k^{(j)}(u_k)}{\|\psi(v) - \varphi_k(u_k)\|} +
\frac{\psi^{(j)}(v) - \varphi_{k+1}^{(j)}(u_{k+1})}{\|\psi(v) -
\varphi_{k+1}(u_{k+1})\|} \:\: , \: j = 1,\ldots,n-1,$$
$$Q(\xi) = \langle \varphi_{k+1}(u_{k+1}) - \varphi_k(u_k), N(\xi)
\rangle.$$
Finally, define the map
\def\rn{{\bf R}^{n-1}}
$$G : U \longrightarrow (\rn)^{s-2}\times \rn\times\rn\times\rn\times
{\bf R}$$
by
$$G(\xi) = \big( (K_i^{(j)}(\xi))_{2\leq i\leq s-1}^{1\leq j\leq n-1};
(L_j(\xi))_{1\leq j\leq n-1};(M_j(\xi))_{1\leq j\leq n-1};
(P_j(\xi))_{1\leq j\leq n-1}; Q(\xi) \big).$$
Then $G$ is smooth and, according to the above, we have
$$\chi^{-1}(D\cap M_n(s,k)) = G^{-1}(0).$$
Thus, to prove the lemma we have to establish that $G^{-1}(0)$ is a
smooth submanifold of $U$ of dimension $2n-3$. To do this it is
sufficient to show that $G$ is submersion at any point of $G^{-1}(0)$.
Indeed, if this is true, then $G^{-1}(0)$ is a smooth submanifold with
\def\gm{G^{-1}(0)}
$$\dim \gm = (s+3)(n-1) - {[(s+1)(n-1)+1]} = 2n-3.$$
Fix $\xi \in \gm$ and assume that
\begin{equation}
\sum_{i=2}^{s-1} \sum_{j=1}^{n-1} A_i^{(j)} \gr K_i^{(j)}(\xi) +
\sum_{j=1}^{n-1} B_{j} \gr L_j(\xi) + \sum_{j=1}^{n-1} C_j
\gr M_j(\xi)\\
+ \sum_{j=1}^{n-1} p_j \gr P_j(\xi) + q \gr Q(\xi) = 0
\end{equation}
for some real coefficients $A_i^{(j)}, B_j, C_j, p_j, q$. We have to show
that these constants are zero.
First, consider in (19) the derivatives with respect to
$\omega_1,\ldots,\omega_{n-1}$. According to
$\omega_n = (1- \omega_1^2 -\ldots -\omega_{n-1}^2)^{1/2}$, we get
\begin{equation}
\sum_{j=1}^{n-1} B_j (- \frac{\partial \varphi_1^{(r)}}{\partial
u_1^{(j)}}(u_1) + \frac{\omega_r}{\omega_n}\cdot \frac{\partial
\varphi_1^{(n)}}{\partial u_1^{(j)}}(u_1)) = 0
\end{equation}
for $r = 1,\ldots,n-1$. Note that (20) holds also for $r = n$. Setting
$$c = \frac{1}{\omega_n} \sum_{j=1}^{n-1} B_j \frac{\partial
\varphi_1^{(n)}}{\partial u_1^{(j)}}(u_1),$$
(20) implies
$$c\omega_r = \sum_{j=1}^{n-1} B_j \frac{\partial \varphi_1^{(r)}}
{\partial u_1^{(j)}}(u_1) \:\:\: , \: r = 1,\ldots,n,$$
that is
\begin{equation}
c\omega = \sum_{j=1}^{n-1} B_j \frac{\partial \varphi_1}
{\partial u_1^{(j)}}(u_1).
\end{equation}
Consequently, $c\omega$ is a tangent vector to $X$ at $\varphi_1(u_1)$.
However, $\xi = (\omega;u;v;\theta)\in \gm$ implies
$\chi(\xi) \in M_n(s,k) \subset M(s,k)$, and so $\varphi_1(u_1)$ is the
first (transversal) reflection point of an $\ot$-trajectory for $X$.
In particular, $\omega$ is not tangent to $X$ at $\varphi_1(u_1)$. Hence
$c = 0$ and now (21) yields $B_1 = \ldots = B_{n-1} = 0$.
In a similar way, considering in (19) the derivatives with respect to
$\theta_2,\ldots,\theta_n$, we find $C_1 = \ldots = C_{n-1} = 0$.
Next, we are going to show that $A_m^{(1)} = \ldots = A_m^{(n-1)} = 0$
for each $m = 2,\ldots,k$. Here we assume $k \geq 2$, for $k =1$ there
is nothing to be proved in this step. Since $k \geq 2$, the functions
$P_j, Q$ and $K_i^{(j)}$ for $i \geq 3$ do not depend on the variables
$u_1^{(r)}$. On the other hand,
$$K_2^{(j)}(\xi) = \frac{\partial F}{\partial u_2^{(j)}}(\xi) =
\langle \frac{\varphi_2(u_2) - \varphi_1(u_1)}{\|\varphi_2(u_2)-
\varphi_1(u_1)\|} + \frac{\varphi_2(u_2) - \varphi_3(u_3)}{\|\varphi_2(u_2)
- \varphi_3(u_3)\|} , \frac{\partial \varphi_2}{\partial u_2^{(j)}}
(u_2)\rangle ,$$
and therefore
$$\frac{\partial K_2^{(j)}}{\partial u_1^{(r)}}(\xi) = - a_1 {[\langle
\frac{\partial \varphi_1}{\partial u_1^{(r)}}(u_1) , \frac{\partial
\varphi_2}{\partial u_2^{(j)}}(u_2)\rangle - \langle e_1, \frac{\partial
\varphi_1}{\partial u_1^{(r)}}(u_1)\rangle \langle e_1 , \frac{\partial
\varphi_2}{\partial u_2^{(j)}}(u_2)\rangle ]},$$
where
$$a_i = \frac{1}{\|\varphi_i(u_i) - \varphi_{i+1}(u_{i+1})\|} \:\: ,
\:\: e_i = \frac{\varphi_i(u_i) - \varphi_{i+1}(u_{i+1})}
{\|\varphi_i(u_i) - \varphi_{i+1}(u_{i+1})\|}.$$
Now we consider in (19) the derivatives with respect to $u_1^{(r)}$ and find
\begin{equation}
\sum_{j=1}^{n-1} A_2^{j)} \frac{\partial K_2^{(j)}}{\partial
u_1^{(r)}}(\xi) = 0.
\end{equation}
Setting
\begin{equation}
w = \sum_{j=1}^{n-1} A_2^{(j)}\frac{\partial\varphi_2}{\partial
u_2^{(j)}}(u_2) \in T_{\varphi_2(u_2)} X,
\end{equation}
and using the expression for
$\frac{\partial K_2^{(j)}}{\partial u_1^{(r)}}$ found above, we can rewrite
(22) in the form
$$\langle \frac{\partial\varphi_1}{\partial u_1^{(r)}}(u_1),w\rangle -
\langle e_1 , w\rangle \langle e_1 , \frac{\partial \varphi_1}{\partial
u_1^{(r)}}(u_1)\rangle = 0.$$
That is,
$$\langle \frac{\partial \varphi_1}{\partial u_1^{(r)}}(u_1) , w -
\langle e_1 , w\rangle e_1 \rangle = 0,$$
and this is true for all $r = 1,\ldots,n-1$. Consequently,
\begin{equation}
w - \langle e_1 , w\rangle e_1 = \lambda N_1
\end{equation}
for some $\lambda \in {\bf R}$, $N_1$ being an unit normal vector to $X$
at $\varphi_1(u_1)$. Note that $\langle e_1,N_1\rangle\neq 0$. Taking
inner product of (24) with $e_1$ gives $0 = \lambda\langle N_1,e_1 \rangle$,
and so $\lambda = 0$. Thus, by (24), $w = \langle e_1,w\rangle e_1$. If
$\langle e_1,w\rangle \neq 0$, the latter would imply
$e_1 \in T_{\varphi_2(u_2)} X$, which would be a contradiction with
$\xi \in \gm$. Hence $\langle e_1,w\rangle = 0$ and so $ w = 0$. Now
(23) yields $A_2^{(1)} = \ldots = A_2^{(n-1)} = 0$.
Using the above procedure, by induction, we get $A_m^{(j)} = 0$ for all
$m = 2,\ldots,k$ and $j = 1,\ldots,n-1$.
Next, if $k < s-1$,
we repeat the same argument, considering the derivatives with respect
to $u_s^{(r)}$, to show that $A_{s-1}^{(j)} = 0$ for $j = 1,\ldots,n-1$.
In a similar way, by induction, one gets $A_m^{(j)} = 0$ for all
$m = s-1, s-2,\ldots,k+1$ and $j = 1,\ldots,n-1$. Therefore all coeffiecients
$A_m^{(j)}$ in (19) are zero.
\def\vk{\varphi_k(u_k)}
\def\vkk{\varphi_{k+1}(u_{k+1})}
\def\pv{\psi(v)}
Now (19) has the form
\begin{equation}
\sum_{j=1}^{n-1} p_j \: \gr \: P_j(\xi) + q\: \gr\: Q(\xi) = 0.
\end{equation}
Set for convenience
$$b_1 = \frac{1}{\|\psi(v) - \varphi_k(u_k)\|} \:\: , \:\:
b_2 = \frac{1}{\|\psi(v) - \varphi_{k+1}(u_{k+1})\|},$$
and note that
$$\frac{\psi(v) - \varphi_k(u_k)}{\|\psi(v) - \varphi_k(u_k)\|} =
- \frac{\pv - \vkk}{\|\pv - \vkk \|} = e_k .$$
\def\dvk{\partial \varphi_k}
\def\dvkk{\partial \varphi_{k+1}}
\def\dvkj{\partial \varphi_k^{(j)}}
\def\ukr{\partial u_k^{(r)}}
\def\ukkr{\partial u_{k+1}^{(r)}}
\def\dvkkj{\partial \varphi_{k+1}^{(j)}}
We have
$$\frac{\partial P_j}{\ukr}(\xi) = - b_1 {[ \frac{\dvkj}{\ukr}(u_k) -
\langle e_k, \frac{\dvk}{\ukr}(u_k)\rangle e_k^{(j)} ]}$$
and
$$\frac{\partial P_j}{\ukkr}(\xi) = -b_2 {[ \frac{\dvkkj}{\ukkr}(u_{k+1})
- \langle e_k, \frac{\dvkk}{\ukkr}(u_{k+1})\rangle e_{k+1}^{(j)} ]}.$$
Further, set $p_n = 0$ and $p = (p_1,\ldots,p_n) \in {\bf R}^n$, and
consider in (25) the derivatives with respect to $u_k^{(r)}$. We get
$$-b_1\sum_{j=1}^{n-1} p_j {[ \frac{\dvkj}{\ukr}(u_k) - \langle e_k,
\frac{\dvk}{\ukr}(u_k)\rangle e_k^{(j)} ]} - q\langle \frac{\dvk}{\ukr}
(u_k), N(\xi)\rangle = 0,$$
which is equivalent to
$$-b_1\langle p,\frac{\dvk}{\ukr}(u_k)\rangle + b_1 \langle e_k,p\rangle
\langle e_k,\frac{\dvk}{\ukr}(u_k)\rangle - q\langle N(\xi),\frac{\dvk}
{\ukr}(u_k)\rangle = 0.$$
The latter can be rewritten as
$$\langle b_1 p - \langle e_k,b_1 p\rangle e_k + q N(\xi) ,
\frac{\dvk}{\ukr}(u_k)\rangle = 0.$$
Since this is true for all $r = 1,\ldots,n-1$, it implies
\begin{equation}
b_1 p - \langle e_k, b_1 p\rangle e_k + q N(\xi) = \mu N_k
\end{equation}
for some $\mu \in {\bf R}$,
$N_k$ being a unit normal vector to $X$ at $\vk$. Taking inner product
of (26) with $e_k$, one finds
$$\mu \langle N_k,e_k\rangle = q\langle N(\xi),e_k\rangle = 0,$$
since the segment ${[\vk,\vkk]}$ is tangent to $X$ at $\pv$ and $N(\xi)$
is a normal vector to $X$ at $\pv$. On the other hand,
$\langle N_k,e_k\rangle \neq 0$, and so $\mu = 0$. Now (26) implies
\begin{equation}
b_1 p - \langle e_k,b_1 p\rangle e_k + q N(\xi) = 0.
\end{equation}
In a similar way, considering in (25) the derivatives with respect to
$u_{k+1}^{(r)}$, one obtains
\begin{equation}
b_2 p - \langle e_k,b_2 p\rangle e_k - q N(\xi) = 0.
\end{equation}
Since $b = b_1 + b_2 > 0$, combining (27) and (28) gives
\begin{equation}
p = \langle e_k,p\rangle e_k.
\end{equation}
First, note that the latter and (27) imply $q = 0$. Next, $\xi \in \gm$
yields $\chi(\xi) \in M_n(s,k)$ and so
$\varphi_k^{(n)}(u_k)\neq \varphi_{k+1}^{(n)}(u_{k+1})$. Hence
$e_k^{(n)} \neq 0$. On the other hand, $p_n = 0$ by definition, therefore
(29) implies $0 = \langle e_k,p\rangle e_k^{(n)}$, i.e.
$\langle e_k,p\rangle = 0$. Again by (29) we get $p = 0$. Thus, all
coefficients in (19) are zero.\\
{\bf case 2.} $k = s$. In this case for $\chi(\xi) \in M_n(s,k)$ the
conditions (14), (15) and (16) remain the same, while (17) and (18) are
replaced by
$$\frac{\pv - \varphi_s(u_s)}{\|\pv - \varphi_s(u_s)\|} = \theta$$
and
$$\langle \theta, N(\xi)\rangle = 0,$$
respectively. Correspondingly, we define the functions $K_i^{(j)}$,
$L_j$ and $M_j$ in the same way and change the definitions of $P_j$
and $Q$ as follows:
$$P_j(\xi) = \frac{\psi^{(j)}(v) - \varphi_s^{(j)}(u_s)}{\|\pv -
\varphi_s(u_s)\|} - \theta_j \:\: , \:\: Q(\xi) = \langle \theta,
N(\xi)\rangle.$$
Then the map $G$ is defined in the same way.
Let $\xi \in \gm$ and assume (19) holds. Using the argument from case 1,
we first show that $B_j = 0$ for all $j$. Then, by induction on
$i = 2,\ldots,s-1$, considering the derivatives with respect to
$u_{i-1}^{(r)}$, respectively, we show that $A_i^{(j)} = 0$ for all
$i = 2,\ldots,s-1, j = 1,\ldots,n-1$. Here we use the fact that $P_j, Q$
and $M_j$ do not depend on $u_i$ for $i\leq s-2$. Next, considering
in (19) the derivatives with respect to $u_{s-1}^{(r)}$ and $u_s^{(r)}$,
we find $C_j = 0$ and $p_j = 0$ for all $j = 1,\ldots,n-1$. Finally,
$q = 0$ follows easily considering the derivatives with respect to
$\theta_r$. Hence in this case $G$ is again submersion at any point of
$\gm$, and so $M_n(s,k)$ is a smooth submanifold of $M_s$ of dimension
$2n-3$.
{\bf case 3.} $k = 0$. The same as case 2.
This concludes the proof of the lemma.\\
{\it Proof of Theorem 3.1}. For given $s$ and $k$ consider the projection
$$\pi_s : M_s = \sn\times X^{(s)}\times X\times \sn \longrightarrow
\sn \times \sn,$$
defined by
$$\pi_s(\omega;x;y;\theta) = \ot.$$
Since $\pi_s$ is smooth and $M_r(s,k)$ is a smooth submanifold of
dimension
$$2n-3 < \dim (\sn\times\sn),$$
the set
$$L_r(s,k) = \pi_s(M_r(s,k)) \subset \ssn$$
has measure zero. Let $M_r(s,k) = \cup_{j=1}^{\infty} K_j$ with $K_j$
compact. Then $L_r(s,k) = \cup_{j=1}^{\infty} \pi_s(K_j)$ is a
countable union of compact subsets of $\ssn$ with measure zero.
Finally, set
$$L = \bigcup_{0\leq k\leq s} \bigcup_{r=1}^n L_r(s,k)$$
and ${\cal R} = (\ssn)\setminus L$. Then ${\cal R}$ has the desired
properties.\\
{\it Proof of Theorem 3.2}. Consider the projections
$$p_s : M_s \longrightarrow \sn\times X,$$
defined by
$$p_s(\omega;x_1,\ldots,x_s;y;\theta) = (\omega,x_1),$$
and set
$${\cal T} = (\sn\times X)\setminus \bigcup_{0\leq k\leq s}
\bigcup_{r=1}^n p_s(M_r(s,k)).$$
The same argument as that in the proof of Theorem 1 shows that
${\cal T}$ has the desired properties.\\
{\bf 3.3. Remark.} In the case when $K$ has the form (6) with
$G^{\infty}(K) = \emptyset$ it is possible to prove a slightly stronger
result than Theorem 3.1. Namely, for each fixed $\omega\in \sn$
there exists a set ${\cal R}(\omega)\subset\sn$, the complement of which
has Lebesgue measure zero, such that for each
$\theta\in {\cal R}(\omega)$ all $\ot$-rays are ordinary reflecting ones
(see \cite{kn:P2}).
\section{Sojourn times}
Let $\Omega$ be a domain in
${\bf R}^n, n \geq 2,$ with bounded complement and smooth boundary
$X = \partial \Omega$, and let $\omega$ be a fixed unit vector in
${\bf R}^n$. Our aim in this section is to prove the following.\\
{\bf 4.1. Proposition.} {\it There exists ${\cal S}(\omega)\subset\sn$
the complement of which is a countable union of compact subsets of
measure zero of $\sn$ such that if $\theta \in {\cal S}(\omega)$, then
any two different ordinary reflecting $\ot$-rays in $\Omega$ have
distinct sojourn times}.\\
The rest of the section is devoted to the proof of this proposition.
Let $B = B_R$ be the open ball in ${\bf R}^n$ with center 0 and radius $R$
and fix $R$ so big that $B$ contains $X$. Let
$Z = Z_{\omega}$ be the hyperplane tangent to $B$ and orthogonal to
$\omega$ and such that the halfspace determined by $Z$ and having
$\omega$ as an inner normal contains $X$.
For a given integer $k \geq 1$ denote by $U_k$ the set of those
$u \in Z$ such that the trajectory $\gamma(u)$ of the generalized
deodesic flow in $\Omega$ is an ordinary reflecting ray with exactly
$k$ reflection points. Denote by $J_{k}(u) \in\sn$ the direction of
$\gamma(u)$ after the last reflection.
Clearly, $U_k$ is open in $Z$ and
$$J_k : U_k \longrightarrow \sn$$
is smooth.
Next, we fix two arbitrary integers $k\geq 1, s\geq 1$. For $u\in U_k$
denote by $f(u)$ the sojourn time of the scattering ray $\gamma(u)$
(more precisely, the scattering ray determined by $\gamma(u)$). Thus,
we get a smooth function
$f: U_k \longrightarrow {\bf R}$. For convenience the same function on
$U_s$ will be denoted by $g$, so $g : U_s \longrightarrow {\bf R}$.
\def\ra{{\rm rank} \: }
\def\dd{{\rm d}}
For $u\in U_k$ let $x_1(u)\in X,\ldots,x_k(u)\in X$ be the successive
reflection points of $\gamma(u)$. Then $x_i : U_k \longrightarrow X$
are smooth maps. Let for $y\in X$, $N(y)$ denotes the {\it unit normal to}
$X$ at $y$ pointing into $\Omega$. Then for $u\in U_k$ we have
$$J_k(u) = \frac{x_k(u) - x_{k-1}(u)}{\|x_k(u) - x_{k-1}(u)\|} - 2\langle
\frac{x_k(u) - x_{k-1}(u)}{\|x_k(u) -
x_{k-1}(u)\|},N(x_k(u))\rangle N(x_k(u))$$
and
$$f(u) = \sum_{i=0}^{k-1} \|x_{i+1}(u) - x_i(u)\| + t -2R,$$
where $x_0(u)$ (resp. $x_{k+1}(u)$) denotes the orthogonal projection of
$x_1(u)$ (resp. $x_k(u)$) on $Z$ (resp. $Z_{-\theta}$), $\theta = J_k(u)$,
and $t = \|x_k(u) - x_{k+1}(u)\|$. It is easy to see that
$$\langle \theta, x_k + t\theta -R\theta\rangle = 0.$$
Therefore $t = R - \langle \theta,x_k\rangle$, and so
$$f(u) = \sum_{i=0}^{k-1}\|x_{i+1}(u) - x_i(u)\| - \langle x_k(u),
J_k(u)\rangle - R.$$
For $v\in U_s$ the successive reflection points of $\gamma(v)$ will be
denoted by $y_1(v),\ldots,y_s(v)$. We set $y_0(v) = v$ and define
$y_{s+1}(v)$ in the same way as $x_{k+1}(u)$.
Further, denote by $W(k,s)$ the set of those $(u,v)\in U_k\times U_s$ such that
$J_k(u) = J_s(v), f(u) = g(v)$ and $\ra\dd J_k(u) = \ra\dd J_s(v) = n-1.$\\
{\bf 4.2. Lemma.} {\it $W(k,s)$ is a smooth $(n-2)$-dimensional submanifold
of $U_k\times U_s$.}\\
{\it Proof.} Consider an arbitrary point $w_0 = (u_0,v_0)$ in $W(k,s)$.
Then $\ra\dd J_k(u_0) = \ra\dd J_s(v_0) = n-1$. Clearly, there exists
a neighbourhood $U$ of $w_0$ in $U_k\times U_s$ such that for every
$(u,v)\in W$ we have $\ra\dd J_k(u) = \ra\dd J_s(v) = n-1$.
Define the map
$$L : U \longrightarrow {\bf R}^n$$
by
$$L(u,v) = (\lambda(u,v);(\chi^{(j)}(u,v))_{1\leq j\leq n-1}),$$
where
$$\lambda(u,v) = f(u) - g(v) \:\: , \:\: \chi(u,v) = J_k(u) - J_s(v).$$
Clearly,
$$W(k,s)\cap U \subset L^{-1}(0),$$
and so it is sufficient to show that $L$ is submersion at any point of
$L^{-1}(0)$.
We shall show that $L$ is submersion at $w_0$. For the other points of
$L^{-1}(0)$ the argument is the same. Set $\theta = J_k(u_0)$.
Without loss of generality we may assume that
\begin{equation}
\theta^{(n)} \neq 0.
\end{equation}
Suppose that
\begin{equation}
\sum_{j=1}^{n-1} A_j \gr \: \chi^{(j)}(w_0) + C \gr\: \lambda (w_0) = 0
\end{equation}
for some constants $A_j, C$. Set $A_n = 0$ and
$A = (A_1,\ldots,A_n) \in {\bf R}^n$.
\def\xiu{\frac{\partial x_i}{\partial u_p}(u_0)}
\def\xiiu{\frac{\partial x_{i+1}}{\partial u_p}(u_0)}
Before going on, we have to compute several derivatives. Setting
$$e_i = \frac{x_{i+1}(u_0) - x_i(u_0)}{\| x_{i+1}(u_0) - x_i(u_0)\|},$$
for $p = 1,\ldots,n-1$ and $i = 1,\ldots,k-1$ we have
$$\frac{\partial}{\partial u_p} \| x_{i+1} - x_i\|(u_0) = \frac{1}{\| x_{i+1} -
x_i \|} \langle x_{i+1} - x_i,\xiiu - \xiu\rangle = \langle e_i,
\xiiu - \xiu\rangle.$$
Note that
$$\langle e_{i-1},\xiu\rangle = \langle e_i,\xiu\rangle,$$
since $\xiu$ is tangent to $X$ at $x_i(u_0)$. Consequently,
$$\frac{\partial f}{\partial u_p}(u_0) = \sum_{i=0}^{k-1}
\langle e_i, \xiiu - \xiu\rangle -
\langle \frac{\partial x_k}{\partial u_p}(u_0), J_k(u_0)\rangle -
\langle x_k,\frac{\partial J_k}{\partial u_p}(u_0)\rangle,$$
and using the fact that
$$\langle \frac{\partial x_k}{\partial u_p}(u_0), J_k(u_0)\rangle =
\langle e_k,\frac{\partial x_k}{\partial u_p}(u_0)\rangle,$$
we find
$$\frac{\partial f}{\partial u_p}(u_0) =
\sum_{i=1}^{k} \langle e_{i-1},\xiu\rangle - \sum_{i=0}^{k-1}
\langle e_i,\xiu\rangle - \langle e_k, \frac{\partial x_k}{\partial u_p}(u_0)
\rangle - \langle x_k(u_0), \frac{\partial J_k}{\partial u_p}(u_0)\rangle.$$
Since $e_0 = \omega$, this yields
$$\frac{\partial f}{\partial u_p}(u_0) =
- \langle e_0,\frac{\partial x_0}{\partial u_p}(u_0)\rangle -
\langle x_k(u_0),\frac{\partial J_k}{\partial u_p}(u_0)\rangle = -
\langle x_k(u_0),\frac{\partial J_k}{\partial u_p}(u_0)\rangle.$$
According to $\langle x_k(u_0),\theta\rangle = 0$, we deduce
$$\frac{\partial \lambda}{\partial u_p}(u_0) =
\frac{\partial f}{\partial u_p}(u_0) = -
\langle x_k(u_0),\frac{\partial J_k}{\partial u_p}(u_0)\rangle = 0.$$
Moreover, we have
$$\frac{\partial \chi^{j)}}{\partial u_p}(u_0) =
\frac{\partial J_k^{(j)}}{\partial u_p}(u_0).$$
Hence, considering in (31) the derivatives with respect to $u_p$,
we obtain
\begin{eqnarray*}
0 = \sum_{j=1}^{n-1} A_j \frac{\partial J_k^{(j)}}{\partial u_p}(u_0)
- C\langle x_k(u_0), \frac{\partial J_k^{(j)}}{\partial u_p}(u_0)\rangle \cr
= \langle A,\frac{\partial J_k}{\partial u_p}(u_0)\rangle
- C\langle x_k(u_0), \frac{\partial J_k^{(j)}}{\partial u_p}(u_0)\rangle
= \langle A-Cx_k(u_0),\frac{\partial J_k^{(j)}}{\partial u_p}(u_0)\rangle
\end{eqnarray*}
for all $p = 1,\ldots,n-1$. Since $\ra\dd J_k(u_0) = n-1$, this yields
\begin{equation}
A - Cx_k(u_0) = a\theta
\end{equation}
for some $a \in {\bf R}$. In a similar way,
considering in (31) the derivatives with respect to $v_p$, one gets
\begin{equation}
A - Cy_s(v_0) = b\theta
\end{equation}
for some $b \in {\bf R}$. Now combining (32) and (33) gives
$C(x_k(u_0) - y_s(v_0)) = 0$. Since $x_k(u_0)\neq y_s(v_0)$ and the
vector $x_k(u_0) - y_s(v_0)$ is orthogonal to $\theta$, this implies
$C = 0$. Using (32) again and taking into account that $A_n = 0$ by
definition, we find $a\theta^{(n)} = 0.$ Hence, by (30), $a = 0$.
Applying again (32), we find $A = 0$, i.e. $A_1 = \ldots = A_{n-1} = 0$.
Thus, $L$ is submersion at $w_0$ which concludes the proof of the lemma.\\
{\it Proof of Proposition} 4.1.
Define the map
$$\varphi : U_k\times U_s \longrightarrow \ssn$$
by $\varphi (u,v) = J_k(u)$. Then $\varphi$ is smooth and since
$\dim W(s,k) = n-2$,
$\varphi (W(s,k))$ is a countable union of compact subsets of $\sn$ of measure
zero.
Set
$$F_k = \{ u \in U_k : \ra\dd J_k(u) \leq n-2 \}.$$
Then $F_k$ is closed in $U_k$ and so it can be represented as a countable
union of compact subsets
$$F_k = \cup_i F_{k,i}.$$
Then by Sard's theorem $J_k(F_{k,i})$ has measure zero in $\sn$
for all $k$ and $i$. Therefore
$$F = \cap_k \cap_i J_k(F_{k,i})$$
has measure zero in $\sn$. Finally, setting
$${\cal S}(\omega) = \sn\setminus (F\cup \cup_k \cup_s J_k(W(k,s))),$$
we get a subset of $\sn$ the complement of which is a countable union
of compact subsets of $\sn$ with measure zero.
To show that ${\cal S}(\omega)$ has the desired property, consider and
arbitrary $\theta \in {\cal S}(\omega)$. Assume that there exists $u\neq v$
in $Z$ which determine ordinary reflecting $\ot$-rays in $\Omega$
with coinciding sojourn times. Then we have $u \in U_k$ and
$v \in U_s$ for some $k$ and $s$. If $\ra\dd J_k(u)\leq n-2$, then
$u\in F_k$ and so $u\in F_{k,i}$ for some $i$. However, this implies
$\theta = J_k(u) \in J_k(F_{k,i})$ which is a contradiction with
$\theta \in {\cal S}(\omega)$. Hence $\ra\dd J_k(u) = n-1$. In the same way
we find $\ra\dd J_s(v) = n-1$ and therefore $(u,v) \in W(k,s)$.
This implies $\theta = J_k(u) = \varphi(u,v)$ which is again a
contradiction with $\theta\in {\cal S}(\omega)$. In this way we have seen that
any two different ordinary reflecting $\ot$-rays in $\Omega$ have
distinct sojourn times which proves the proposition.\\
Introduce the set
$${\cal S} = \{ \ot \in \ssn : \theta \in {\cal S}(\omega) \}.$$
It follows by the properties of the sets
${\cal S}(\omega)$ that the complement of ${\cal S}$ in $\ssn$ has
measure zero and for each $\ot \in {\cal S}$ any two different ordinary
reflecting $\ot$-rays in $\Omega$ have distinct sojourn times.
\section{Existence of scattering rays with sojourn times tending
to infinity}
Let $\Omega$ be a closed domain in ${\bf R}^n, n \geq 2,$ with bounded
complement and smooth boundary $\partial \Omega$. In this
section we show that if the obstacle
$$K = {\bf R}^n\setminus \Omega^{\circ}$$
is trapping, then there exists a sequence
of reflecting $(\omega_m,\theta_m)$-rays $\gamma_m$ in $\Omega$
with sojourn times $T_{\gamma_m} \rightarrow \infty$.
\def\sii{\Sigma_{\infty}}
\def\do{\partial\Omega}
\def\tq{T^* ({\bf Q})}
\def\tdq{T^* (\partial {\bf Q})}
\def\tzdq{T^*_z (\partial {\bf Q})}
\def\tt{\tilde{T}^*}
\def\qq{\overline{{\bf Q}}}
\def\dq{\partial {\bf Q}}
\def\sn{S^{n-1}}
\def\qc{{\bf Q}^{\circ}}
Set ${\bf Q} = {\bf R}\times \Omega$ and denote by $(\tau,\xi)$ the
variables dual to $(t,x)$ in $\tq$. The {\it characteristic set} of
the wave operator $\Box$ is defined by
$$\Sigma = \{ (t,x,\tau,\xi)\in T^*(\qq)\setminus\{0\} : \tau^2 =
\mid\xi\mid^2 \}.$$
For $z = (t,x) \in {\bf R}\times\do$ consider the {\it compressed
cotangent bundle}
$$\tt_z (\qq) = T_z^*(\qq)/N_z(\dq) \cong \tzdq,$$
$N_z(\dq)$ being the fiber of one-forms vanishing on $T_z(\do)$. Setting
$$\tt(\qq) = T^*(\qc)\cup\tdq,$$
introduce the map
$$\sim : T^* (\qq) \longrightarrow \tt (\qq)$$
which coincides with the identity on $\qc$, while for
$(z,\zeta)\in T^*(\qq), z\in \dq,$ we set
$$\sim (z,\zeta) = (z,\eta)\in T_z^*(\dq) \:\: , \:\: \eta =
\zeta\mid_{T_z(\dq)}.$$
The image $\Sigma_b = \tilde{\Sigma} = \sim(\Sigma)$ is called
the {\it compressed characteristic set}, and if $\gamma$ is a generalized
bicharacteristic of $\Box$, its image $\tilde{\gamma} = \sim(\gamma)$
is called a {\it compressed generalized bicharacteristic (ray)}.
Let $\rho_0 > 0$ be fixed so that
$$K \subset B_{0} = \{ x\in {\bf R}^n : \:\:
\mid x \mid \leq \rho_0 \: \}.$$
Given a point $\nu = (0,x,1,\xi)\in \Sigma_b, (x,\xi)\in T^*(\do)$,
consider the compressed generalized bicharacteristic
$$\gamma_{\nu}(t) = (t,x(t),1,\xi(t))\in \tt(\qq)$$
of $\Box$, parametrized by the time $t$ and passing through $\nu$ for
$t = 0$. Denote by $T(\nu) \in {\bf R}^+\cup{\infty}$ the maximal
$T > 0$ such that
$$x(t) \in B_{0} \:\:\: {\rm for} \:\:\: 0\leq t\leq T(\nu).$$
Denote by $\sii$ the set of those
$\nu = (0,x,1,\xi)\in \Sigma_b, (x,\xi)\in T^*(\do),$ such that
$T(\nu) = \infty$. Using the continuity of the generalized Hamiltonian
flow of $\Box$ (cf. \cite{kn:MS}), it is easy to see that $\sii$ is
closed in $\Sigma_b$. On the other hand, $\sii \neq \Sigma_b$. Indeed,
take a hyperplane $\Pi$ tangent to $\do$ such that $K$ is contained in
a half-space determined by $\Pi$. Consider an arbitrary
$\nu_0 = (0,x_0,1,\xi_0)\in \Sigma_b$ with $(x_0,\xi_0)\in T^*(\do),$
$x_0 \in \do\cap \Pi$. Then we have $T(\nu_0) < \infty$, since
$\gamma_{\nu_0}(t)$ leaves $B_0$ for $t > 2\rho_0$.
\def\hn{\hat{\nu}}
By definition the obstacle $K$ is {\it trapping} if $\sii\neq\emptyset$.
Therefore the boundary $\partial \sii$ in $\Sigma_b$ is not empty.
Take an arbitrary $\hn \in \partial \sii$. Since
$\Sigma_b\setminus \sii \neq \emptyset$, there exists a sequence
of elements $\nu_m = (0,x_m,1,\xi_m)$ of $\Sigma_b$ with
$(x_m,\xi_m)\in T^*(\dq)$ such that $\nu_m \notin \sii$ for all $m$
and $\nu_m \rightarrow \hn$. Consider the compressed generalized
bicharacteristics
$$\gamma_{\nu_m}(t) = (t,x_m(t),1,\xi_m(t))$$
passing through $\nu_m$ for $t = 0$ and such that $T(\nu_m) < \infty$.
If the sequence $\{ T(\nu_m)\}$ is bounded, this would imply
$T(\hn) < \infty$ in contradiction with $\hn \in \sii$. Therefore
$\{T(\nu_m)\}$ is unbounded and we may assume
$$T(\nu_m) \longrightarrow_{m\rightarrow\infty} + \infty.$$
Set
$$y_m = x_m(T(\nu_m)) \in \partial B_0 \:\: , \:\:
\omega_m = \xi_m(T(\nu_m)) \in \sn.$$
Passing to a subsequence, we may assume that
$y_m \rightarrow z \in \partial B_0$ and
$\omega_m \rightarrow \omega \in \sn$. Consider the generalized
bicharacteristic $\gamma_{\mu}(t) = (t,y(t),1,\xi(t))$ of $\Box$
issued from $\mu = (0,z,1,\omega)$. By continuity we have
$T(\mu) = \infty$ and $y(t) \in B_0$ for $t\geq 0$.
Let $Z_{\omega}$ be the hyperplane passing through $z$ and
orthogonal to $\omega$. Denote by $Z_{\infty}$ the set of those
points $y \in Z_{\omega}$ such that the generalized bicharacteristic
$\gamma_{\mu_y}$ passing through $\mu_y = (0,y,1,\omega)$ has the
property $T(\mu_y) = \infty$. A simple argument shows that $Z_{\infty}$
is closed in $Z_{\omega}$. Clearly $Z_{\infty}\neq Z_{\omega}$.
Consequently, there exists a sequence
$$z_m \rightarrow y_0 \:\: , \:\: z_m \in Z_{\omega}\setminus
Z_{\infty}$$
such that
$$T(\mu_{z_m}) < \infty \:\:\: , \:\: T(\mu_{z_m}) \rightarrow \infty.$$
In general the bicharacteristic $\gamma_{\mu_{z_m}}$ could contain
gliding or glancing segments. However, if $G^{\infty}(K) = \emptyset$,
then $\gamma_{\mu_z}$ can be approximated by multiple reflecting rays
(cf. \cite{kn:MS}). Thus, taking $(z'_m,\omega'_m)$ sufficiently close
to $(z_m,\omega)$, we obtain the following result.\\
{\bf 5.1. Proposition.} {\it Let the obstacle $K$ be such that
$G^{\infty}(K) = \emptyset$ and $\sii \neq \emptyset$. Then there exists
a sequence of ordinary reflecting $(\omega'_m,\theta'_m)$-rays
$\gamma_m$ such that}
$$T_{\gamma_m} \longrightarrow_{m \rightarrow \infty} \infty.$$
{\bf 5.2. Corollary.} {\it Let the obstacle $K$ have the form {\rm (6)}
and let $\Sigma_{\infty} \neq \emptyset$. Then there exists a sequence
of ordinary reflecting $(\omega'_m,\theta'_m)$-rays
$\gamma_m$ such that}
$$T_{\gamma_m} \longrightarrow_{m \rightarrow \infty} \infty.$$
{\it Proof.} In this case every generalized bicharacteristic $\gamma$
of $\Box$ in $\Omega$ is uniquelly determined by each of its points.
Moreover, every such bicharacteristic can be approximated by ordinary
reflecting ones (see \cite{kn:MS}). Thus, the argument of the proof
of Proposition 5.1 works even if $G^{\infty} (K) \neq \emptyset$.\\
Next, we consider a fixed ordinary reflecting $(\omega'_m,\theta'_m)$-ray
$\gamma_m$ which is non-degenerate (see Section 1).
We wish to replace $(\omega'_m,\theta'_m)$ by a pair
$(\omega''_m,\theta''_m)$ sufficiently close to $(\omega'_m,\theta'_m)$
for which there exist ordinary reflecting $(\omega''_m,\theta''_m)$-rays
$\delta_m$ such that $\{ T_{\delta_m} \}$ is an infinite sequence of
isolated points in sing supp $s(-t,\theta''_m,\omega''_m)$.
Let ${\cal S}$ be the subset of $\sn\times\sn$ introduced at the end of
Section 4.
Assume that $K$ satisfies the following condition:
\begin{eqnarray*}
{\bf (G)}\:\:\:\:\:\:\:\:
\cases{{\rm There}\: {\rm exits}\: {\rm a}\: {\rm subset}\:
{\cal G} \subset \sn\times\sn \: {\rm the}\: {\rm complement}\:{\rm of}\cr
{\rm which} \: {\rm has}\: {\rm Lebesgue}\: {\rm measure}\:
{\rm zero}\: {\rm such}\: {\rm that} \: {\rm for}\: {\rm each}\:
\ot\in {\cal G} \cr
{\rm all}\: \ot-{\rm rays} \: {\rm in}\: \Omega \: {\rm are}\: {\rm ordinary}\:
{\rm reflecting}\: {\rm ones}.\cr }
\end{eqnarray*}
Setting
$\Xi = {\cal G}\cap {\cal S}$, it suffices to take an approximation by
$(\omega''_m,\theta''_m)\in \Xi$.
On the other hand, to
guarantee the existence of such $(\omega''_m,\theta''_m)$-rays,
we shall use a corollary of the inverse mapping theorem (cf.
\cite{kn:H1}, Theorem 1.1.7). Let $U$ and $V$ be open subsets of
${\bf R}^m$ and let
$$F : U\ni x\mapsto f(x)\in V$$
be a $C^{\infty}$ map. Suppose that $x_0 \in U$ is such that
$\det df(x_0) \neq 0$. Then
$$\alpha = 1/\| df(x_0)^{-1}\| > 0,$$
$\| . \|$ being the standard norm in ${\cal L}({\bf R}^m,{\bf R}^m)$ --
the space of linear maps.
Set $y_0 = f(x_0)$ and choose $\delta > 0$ so small that
$$U_{\delta} = \{ x\in {\bf R}^m : \| x-x_0\| < \delta \}\subset U,$$
$$\| df(x) - df(x_0)\| \leq \frac{\alpha}{2} \:\: {\rm for} \:\:
x\in U_{\delta},$$
$$V_{\delta} = \{ y\in {\bf R}^m : \|y - y_0\| < \frac{\delta\alpha}
{2} \} \subset V.$$
Then it follows by the inverse mapping theorem that
the map $f$ is injective on $U_{\delta}$ and surjective on
$V_{\delta}$.
\def\ot{(\omega,\theta)}
In what follows we are going to construct suitable approximations for
$x_0 = z'_m$ and $\omega_0 = \omega'_m$. Consider the hyperplane
$Z = Z_{\omega_0}$. For $\omega$ sufficiently close to $\omega_0$,
the $\ot$-rays issued from $y\in Z_{\omega}$ in direction $\omega$
can be considered as suitable $\ot$-rays issued from a point
$x\in Z$, provided $y$ is close enough to $x_0$. Thus we obtain
a $C^{\infty}$ map
$$ U = {\cal O}\times \Gamma \ni (x,\omega) \mapsto f(x,\omega)\in\sn.$$
Here ${\cal O}\subset Z$ is a small neighbourhood of $x_0$,
$\Gamma \subset\sn$ is a small neighbourhood of $\omega_0$, and
$f(x,\omega)$ is the outgoing direction of the ray issued from $x$
in direction $\omega$. Since
$\gamma_m$ is non-degenerate by assumption, we have
$$\det f'_x (x_0,\omega_0) \neq 0.$$
We may assume that $U$ is chosen so small that
$\det f'_x (x,\omega) \neq 0$ holds for all
$(x,\omega)\in \overline{U}$. Set
$$\max_{(x,\omega)\in U} \|(f'_x(x,\omega))^{-1}\| = \frac{1}{\alpha}.$$
Then there exists $\delta > 0$ such that for $(x,\omega)\in U$ with
$\|x-x_0\| < \delta, \|\omega - \omega_0\| < \delta$ we have
$$\|f'_x(x,\omega) - f'_x(x_0,\omega_0)\| \leq \frac{1}{4}\alpha.$$
We may assume that $\delta $ is so small that
$${\cal O}_{\delta} = \{ x\in Z : \|x-x_0\| < \delta \} \subset
{\cal O} \:\: ,
\:\: \Gamma_{\delta} = \{ \omega\in \sn : \|\omega - \omega_0\| <
\delta \} \subset \Gamma.$$
Clearly, for $\omega \in \Gamma_{\delta}$ fixed the map
$${\cal O}_{\delta} \ni x \mapsto f(x,\omega) \in \sn$$
is injective. Denote $\theta_0 = f(x_0,\omega_0)$ and consider the set
$$W_{\delta} = \{ \theta\in\sn : \|\theta - \theta_0\| < \frac{\delta
\alpha}{4} \}.$$
Choose $\delta' \in (0,\delta)$ so small that
$$\| f(x_0,\omega) - \theta_0\| < \frac{\delta\alpha}{4} \:\:\:
{\rm for} \:\:\: \omega\in \Gamma_{\delta'}.$$
Then for $\omega\in \Gamma_{\delta'}$ and $\theta\in W_{\delta}$
we deduce
$$\|\theta - f(x_0,\omega)\| < \frac{\delta\alpha}{2},$$
therefore, according to the above remark, for each fixed
$\omega \in \Gamma_{\delta'}$ and each fixed $\theta\in W_{\delta}$
we can find $x_{\ot}\in {\cal O}_{\delta}$ with
$$f(x_{\ot},\omega) = \theta.$$
This shows that, exploiting the density of $\Xi$ in $\sn\times\sn$,
we can approximate $(\omega_0,\theta_0) = (\omega'_m,\theta'_m)$ by pairs
$(\omega_j,\theta_j)\in \Xi$ for which there exists $(\omega_j,\theta_j)$-
rays $\nu_j$ with sojourn times $T_j$ converging to $T_{\gamma_m}$
as $j \rightarrow\infty$. Applying the results of \cite{kn:CPS},
\cite{kn:PS3}, we obtain a sequence of ordinary reflecting
$(\omega''_m,\theta''_m)$-rays $\delta_m$ such that
$$T_{\delta_m} \longrightarrow_{m \rightarrow\infty} \infty,$$
and
$$- T_{\delta_m} \in {\rm sing}\: {\rm supp}\: s(t,\theta''_m,\omega''_m)
\:\:\: , \:\: m = 1,2, \ldots.$$
Moreover, near $-T_{\delta_m}$ the scattering kernel
$s(t,\theta''_m,\omega''_m)$ has a singularity with leading term
$A_m \delta^{(n-1)} (t+T_{\delta_m})$ with $A_m \neq 0$. Thus, if
the reflecting rays $\gamma_m$ in Proposition 5.1 are non-degenerate,
we conclude that the assertion of Theorem 2.3 is true.
Finally, notice that by Theorem 3.1 the condition {\bf (G)} holds for
obstacles $K$ having the form (6). It seems natural to make the
following\\
{\bf Conjecture.} {\it For each obstacle $K$ the condition
{\bf (G)} is satisfied}.
\def\kk{{\cal K}}
\def\dk{\partial K}
\section{Proofs of Theorems 1.1 and 1.2}
Throughout this section we assume that $K$ has the form (6). For
$z\in \partial K$ we denote by $\kk (z)$ the {\it Gauss curvature}
of $\partial K$ at $z$. Following the argument at the end of the
previous section, we need to construct a sequence of ordinary
reflecting non-degenerate $(\omega_m,\theta_m)$-rays $\gamma_m$ in
$\Omega$ with $T_m = T_{\gamma_m} \rightarrow \infty$. To do this
we use the following.\\
{\bf 6.1. Proposition.} {\it Let $K$ have the form {\rm (6)} and let
$\gamma$ be an ordinary reflecting $\ot$-ray with reflection points
$x_1,\ldots,x_k$. Assume that there exists $j$ such that
$\kk(x_j) > 0$. Then $\gamma$ is non-degenerate.}\\
{\it Proof.} Consider the map $J_{\gamma}$ defined in Section 1. Then
the map d$\: J_{\gamma}(u_{\gamma})$ has the following representation
(see \cite{kn:PS3}, Proposition 2.4.2)
$$\dd J_{\gamma}(u_{\gamma})u = M_k \sigma_k (I+\lambda_kM_{k-1})
\sigma_{k-1}(I+\lambda_{k-1}M_{k-2})\cdots \sigma_2 (I+\lambda_2M_1)
\sigma_1 u.$$
Here
$$\lambda_i = \|x_{i-1} - x_i\| \:\: , \: i = 1,\ldots,k, \:
x_0 = u_{\gamma},$$
$\sigma_i$ is a linear map related to the symmetry with respect to the
tangent plane to $\dk$ at $x_i$, and $M_i$ are symmetric linear maps
defined by
$$M_1 = \tilde{\psi}_1 \: , \: M_i = \sigma_i M_{i-1}(I+ \lambda_i
M_{i-1})^{-1} \sigma_i + \tilde{\psi}_i \:\: , \: i = 2,\ldots,k,$$
$\tilde{\psi}_i$ being linear symmetric maps depending on the second
fundamental form of $\partial K$ at $x_i$. Since $K_j$ are convex, we have
$\tilde{\psi}_i \geq 0$ for all $i = 1,\ldots,k$. Hence $M_i \geq 0$
for each $i$. By assumption, there exists $j$ with $\kk(x_j) > 0$.
Then $\tilde{\psi}_j > 0$, and so $M_i > 0$ for $i = j,j+1,\ldots,k$.
Consequently, $\dd J_{\gamma}(u_{\gamma})u = 0$ implies $u = 0$
which proves the assertion.\\
{\it Proof of Theorem 1.1.} By Corollary 5.2, there exists a sequence
of ordinary reflecting $(\omega'_m,\theta'_m)$-rays $\delta_m$ with
sojourn times $T_{\delta_m} \rightarrow \infty$. Each point $z\in \dk$ can be
approximated by points $z' \in \dk$ such that $\kk(z') > 0$. Using
this and Proposition 6.1, we find a sequence of ordinary reflecting
non-degenerate $(\omega_m,\theta_m)$-rays $\gamma_m$ with sojourn times
$T_{\gamma_m} \rightarrow \infty$. To complete the proof we use the
argument at the end of Section 5.\\
{\it Proof of Theorem 1.2.} Let ${\cal R}$ and ${\cal S}$ be the sets
from Theorem 3.1 and the end of Section 4, respectively, and let
$$\Xi = \{ \ot\in {\cal R}\cap {\cal S} : \omega\neq \theta \} \subset
\sn\times\sn.$$
Given $\ot\in \Xi$, there are two possibilities:
1) there are no $\ot$-rays in $\Omega$;
2) each $\ot$-ray is ordinary reflecting and different $\ot$-rays have
different sojourn times.
Let ${\cal O}$ be a sphere in ${\bf R}^n$ which contains the obstacle
$K$ in its interior.
Consider the set $\Gamma$ of those
$(x,\omega,y,\theta)\in {\cal O}\times\sn\times {\cal O}\times\sn$ such that
$\ot\in \Xi$ and there exists an ordinary reflecting $\ot$-ray passing
through $x$ and $y$. Clearly, $\Gamma$ is an open submanifold of
${\cal O}\times\sn\times {\cal O}\times\sn$. Since the projection
$$\pi : \Gamma \ni (x,\omega,y,\theta) \mapsto (\omega,\theta) \in
\sn\times\sn$$
is smooth, it follows by Sard's theorem that there exists a set
$\Sigma(\Gamma)\subset \sn\times\sn$ of measure zero such that
d$\pi (x,\omega,y,\theta) \neq 0$ whenever $\ot\notin \Sigma(\Gamma)$.
As one can easily check, the last condition means that all
$\ot$-rays with $\ot \notin \Sigma(\Gamma)$ are non-degenerate.
Consequently, for $\ot \in \Xi \setminus \Sigma(\Gamma)$, all
$\ot$-rays have the properties (a)-(c). Applying the results of Chapter
9 in \cite{kn:PS3}, we complete the proof.
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{V.P. D\'epartement de Math\'ematiques, Universit\'e Bordeaux I,
351, Cours de la Lib\'eration, 33405 Talence, FRANCE}
{L.S. Institute of Mathematics of Bulgarian Acad. of Sci., P.O.Box
373, 1090 Sofia, BULGARIA}
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