information: LaTex file, 13 pages BODY \documentstyle[12pt]{article} \textwidth = 5.8 in \hoffset = -0.5 in \textheight = 8.2 in \title{An inverse scattering result for several\\ convex bodies} \author{Latchezar Stoyanov} \date{} \begin{document} \maketitle \def\sn{S^{n-1}} \def\ssn{S^{n-1}\times S^{n-1}} \def\do{\partial \Omega} \section{Introduction} Let $K$ be a compact subset of ${\bf R}^n, n\geq 3, n$ odd, with $C^{\infty}$ boundary $\partial K$ such that $$\Omega_K = \overline{{\bf R}^n\setminus K}$$ is connected. The {\it scattering operator} related to the wave equation in ${\bf R}\times \Omega$ with Dirichlet boundary condition on ${\bf R}\times \do$ can be represented as a unitary operator $$S : L^2({\bf R}\times \sn ) \longrightarrow L^2({\bf R}\times \sn)$$ (see \cite{kn:LP1}). The kernel of $S - $Id, which can be considered as a distribution $$s_K (t,\theta,\omega) \in {\cal D}'({\bf R}\times \ssn ),$$ is called the {\it scattering kernel}. \def\cc{C_0^{\infty}} \def\ot{(\omega,\theta)} It was shown by A.Majda \cite{kn:Ma} that for each obstacle $K$ the convex hull of $K$ can be recovered if one knows sing supp $s_K (t,-\omega,\omega)$ for a dense set of $\omega$'s in $\sn$ (see also \cite{kn:LP2} for a similar result). Consequently, an obstacle $K$ is completely determined by the singularities of the scattering kernel, provided we know in advance that $K$ is convex. Moreover, one can distinguish between convex and strictly convex obstacles by the same data (see \cite{kn:So} and \cite{kn:Y}). A general problem arises: is an obstacle $K$ completely determined if we know sing supp $s_K(t,\theta,\omega)$ for all $\theta, \omega \in \sn$? In this paper we consider obstacles $K$ of the form \begin{equation} K = \bigcup_{j=1}^s K_j \:\:\: , \:\:\:K_i \cap K_j = \emptyset \:\:\:\:\: {\rm for}\:\:\: i\neq j, \:\:\:\:\:\: K_j \:\: {\rm convex} \: {\rm for} \: {\rm all}\:\: j = 1,\ldots,s, \end{equation} which satisfy the following condition introduced by M.Ikawa \cite{kn:I}: \begin{eqnarray*} {\rm (H)} \:\:\:\:\:\:\:\:\:\:\:\:\: \:\:\:\: (K\setminus (K_i \cup K_j)) \cap \:{\rm convex} \: {\rm hull}\: (K_i\cup K_j) = \emptyset \:\:\:\: {\rm for} \: {\rm all} \:\:\: i,j = 1,\ldots,s. \end{eqnarray*} We show that among the obstacles with these properties, $K$ can be completely recovered if one knows sing supp $s_K (t,\theta,\omega)$ for almost all $\ot \in \ssn$. To state precisely the corresponding result, consider another obstacle $L$ of the same type, namely \begin{equation} L = \bigcup_{j=1}^r L_j \:\:\:\: ,\:\:\: L_i \cap L_j = \emptyset \:\:\:\: {\rm for}\:\:\: i\neq j, \:\:\:\:\:\: L_j \:\: {\rm convex} \: {\rm for} \: {\rm all}\:\: j = 1,\ldots,r. \end{equation} {\bf 1.1. Theorem.} {\it Let $K$ and $L$ have the form {\rm (1)} and {\rm (2)}, respectively, and let $K$ and $L$ satisfy the condition {\rm (H)}. Assume that there exists a subset ${\cal R}$ of full Lebesgue measure in $\sn\times\sn$, such that for all $\ot \in {\cal R}$ we have \begin{equation} {\rm sing}\:{\rm supp} s_K(t,\theta,\omega) \subset {\rm sing}\:{\rm supp} s_L(t,\theta,\omega). \end{equation} Then for every $i = 1,\ldots,s$ there exists $j_i = 1,\ldots,r$ with $K_i = L_{j_i}$. In particular, $K \subset L$.}\\ As an immediate consequence one gets the following.\\ {\bf 1.2. Corollary.} {\it Let $K$ and $L$ have the form {\rm (1)} and {\rm (2)}, respectively, and let both $K$ and $L$ satisfy the condition {\rm (H)}. If there exists a subset ${\cal R}$ of full Lebesgue measure in $\sn\times\sn$ such that $${\rm sing}\:{\rm supp} \; s_K(t,\theta,\omega) = {\rm sing}\:{\rm supp} \; s_L(t,\theta,\omega)$$ holds for all $\ot \in {\cal R}$, then $K = L$.}\\ It is seen from the proof that the assertions of both Theorem 1.1 and Corollary 1.2 remain true if $R$ is {\it residual } in $\ssn$ (instead of having full Lebesgue measure), i.e. if $R$ contains a countable inersection of open dense subsets of $\ssn$.\\ {\footnotesize Acknowledgement. This paper has been written while the author was visiting the University of Bordeaux I as chercheur associ\'{e} of C.N.R.S.} \section{Proof of Theorem 1.1} Throughout we assume that $K$ and $L$ satisfy the assumptions of Theorem 1.1. Let ${\cal R}$ be a subset of full Lebesgue measure of $\ssn$ such that (3) holds for all $(\omega,\theta)\in {\cal R}$. We are going to show that each convex component of $K$ coincides with some convex component of $L$. To do this we shall use the connection between singularities of the scattering kernel and sojourn times of reflecting $\ot$-rays. Let $\omega \in\sn, \theta\in\sn$. By a {\it reflecting $\ot$-ray} in $\Omega_K$, we mean an infinite ray in $\Omega_K$ with incoming direction $\omega$ and outgoing direction $\theta$ making finitely many reflections at $\do$ according to the law of the geometrical optics (see \cite{kn:PS1} or \cite{kn:PS2} for the precise definition). The {\it sojourn time} $T_{\gamma}$ of $\gamma$ is defined by $$T_{\gamma} = \langle \omega , x_1\rangle + \sum_{i=1}^{s-1} \| x_i - x_{i+1} \| - \langle \theta , x_s \rangle,$$ where $x_1,\ldots,x_s$ are the successive reflection points of $\gamma$ (\cite{kn:G}). If $\gamma$ has no segments tangent to $\partial K$, then it is called an {\it ordinary} reflecting $\ot$-ray. By ${\cal L}_{\omega,\theta}(K)$ we denote the {\it set of all reflecting $\ot$-rays in} $\Omega_K$. It follows from Theorem 1.2 in \cite{kn:PS3} that there exists ${\cal R}' \subset \ssn$ having full Lebesgue measure in $\ssn$ (and residual at the same time) such that for $\ot \in {\cal R}'$ we have $${\rm sing} \: {\rm supp} \: s_K (t,\theta,\omega) = \{ -T_{\gamma} : \gamma \in {\cal L}_{\omega,\theta}(K) \}$$ and $${\rm sing} \: {\rm supp} \: s_L (t,\theta,\omega) = \{ -T_{\gamma} : \gamma \in {\cal L}_{\omega,\theta}(L) \}.$$ Then $${\cal R}'' = {\cal R} \cap {\cal R}'$$ is a subset of full Lebesgue measure in $\ssn$, and according to (3) and the above inclusions, one gets \begin{equation} \{ -T_{\gamma} : \gamma \in {\cal L}_{\omega,\theta}(K) \} \subset \{ -T_{\gamma} : \gamma \in {\cal L}_{\omega,\theta}(L) \} \:\:\:\: , \:\:\:\: \ot \in {\cal R}''. \end{equation} \def\dk{\partial K} \def\dl{\partial L} Given $x \in \dk$, we denote by $N_K(x)$ the {\it unit normal vector} to $\dk$ at $x$ pointing into $\Omega_K$. Thus $$N_K : \dk \longrightarrow \sn$$ is the {\it Gauss map} of $\dk$. Theorem 1.1 will be easily derived by the following lemma.\\ {\bf 2.1. Lemma.} {\it For every $x \in \partial K$ there exists $y \in \partial L$ such that} \begin{equation} N_L(y) = \pm N_K(x) \:\:\:\: , \:\:\:\: \langle x-y, N_K(x)\rangle = 0. \end{equation} That is, the tangential hyperplane to $K$ at $x$ coincides with the tangential hyperplane to $L$ at some $y \in \dl$. The proof of Lemma 2.1 is rather long and will be devided into several steps.\\ {\bf 2.2. Lemma.} {\it Let $Y$ and $Z$ be smooth convex (concave) hypersurfaces in ${\bf R}^n$ with $Y \cap Z = \emptyset$. Let $\Sigma(Y,Z)$ be the set of those $(y,z)\in Y\times Z$ such that the tangent hyperplanes to $Y$ at $y$ and $Z$ at $z$ coincide and the Gauss maps of $Y$ at $y$ and of $Z$ at $z$ are regular. Then $\Sigma (Y,Z)$ is a smooth $(n-2)$-dimensional submanifold of $Y\times Z$}.\\ {\it Proof.} Take smooth charts $$\varphi : V \longrightarrow Y \:\:\: , \:\:\: \psi : W \longrightarrow Z,$$ $V, W \subset {\bf R}^{n-1}$, around some points $y_0\in Y$ and $z_0\in Z$ with $(y_0,z_0)\in \Sigma(Y,Z)$. For $v\in V$, denote by $N_Y(v)$ one of the unit normal vectors to $Y$ at $\varphi(y)$ so that $N_Y$ is continuos on $V$. In the same way we define $N_Z$ on $W$. Since $(y_0,z_0)\in \Sigma(Y,Z)$, the normals to $Y$ at $y_0$ and to $Z$ at $z_0$ are collinear, and (choosing appropriately the normal fields $N_Y$ and $N_Z$) we may assume that these normals coincide. We may also assume that $N_Y$ is regular on $V$, $N_Z$ is regular on $W$ and $$N_Y^{(n)}(v) \neq 0 \:\:\: , \:\: v\in V \: \:\: , \:\:\: N_Z^{(n)}(w)\neq 0 \:\:\: , \:\:\: w\in W.$$ Then $N_Y^{(n)}(v)$ and $N_Z^{(n)}(w)$ have one and the same sign for all $v\in V$ and $w\in W$. Define the functions $$f_i, g : V\times W \longrightarrow {\bf R}$$ by $f_i(v,w) = N_Y^{(i)}(v) - N_Z^{(i)}(w), i = 1,\ldots,n-1$ and $g(v,w) = \langle \varphi(v) - \psi(w), N_Y(v)\rangle$, and consider $F: V\times W \longrightarrow {\bf R}^n$ given by $$F(v,w) = (f_1(v,w),\ldots,f_{n-1}(v,w),g(v,w)).$$ Then $F$ is smooth and we have $$ \{ (v,w)\in V\times W : (\varphi(v),\psi(w))\in \Sigma(Y,Z) \} = F^{-1}(0).$$ Therefore the assertion will be proved if we show that $F^{-1}(0)$ is a smooth $(n-2)$-dimensional submanifold of $V\times W$. To do this it is sufficient to establish that $F$ is submersion on $F^{-1}(0)$ (cf. \cite{kn:GG}). \def\dnw{\frac{\partial N_Z^{(i)}}{\partial w_j}(w)} \def\dpw{\frac{\partial \psi}{\partial w_j}(w)} \def\dnv{\frac{\partial N_Y^{(i)}}{\partial v_j}(v)} \def\dvv{\frac{\partial \varphi}{\partial v_j}(v)} Fix $(v,w)\in F^{-1}(0)$ and let \begin{equation} \sum_{i=1}^{n-1} A_i {\rm grad}_{v,w} f_i(v,w) + B {\rm grad}_{v,w} g(v,w) = 0 \end{equation} for some reals $A_i$, $B$. Considering in (6) the derivatives with respect to $w_j$ ($1\leq j\leq n-1$), we get \begin{equation} - \sum_{i=1}^{n-1} A_i \dnw - B \langle \dpw , N_Y(v)\rangle = 0. \end{equation} Set $A_n = 0$ and $A = (A_1,\ldots,A_n) \in {\bf R}^n$. Since $F(v,w) = 0$, we have $f_i(v,w) = 0$ for all $i = 1,\ldots,n-1$, and therefore $N_Y(v) = N_Z(w)$. Therefore (7) implies $$0 = \langle A, \frac{\partial N_Z}{\partial w_j}(w) \rangle + B\langle \dpw, N_Z(w)\rangle = \langle A, \frac{\partial N_Z}{\partial w_j}(w) \rangle.$$ Since $N_Z$ is regular at $w$, we have that $\{ \frac{\partial N_Z}{\partial w_j}(w) \}_{j=1}^{n-1}$ is a linear basis in the tangent hyperplane $T_{\psi(w)}Z$. Hence the latter equality implies $A = \lambda N_Z(w)$ for some $\lambda \in {\bf R}$. Consequently, $0 = A_n = \lambda N_Z^{(n)}(w)$, which shows that $\lambda = 0$. Thus, $A = 0$. Next, consider in (6) the derivatives with respect to $v_j$ ($ j = 1,\ldots,n-1$). We get $$B(\langle \dvv , N_Y(v)\rangle + \langle \varphi(v) - \psi (w), \frac{\partial N_Y}{\partial v_j}(v) \rangle ) = 0,$$ and consequently $$B \langle \varphi(v) - \psi (w), \frac{\partial N_Y}{\partial v_j}(v) \rangle = 0.$$ Since $F(v,w) = 0$, we have that $N_Y(v) = N_Z(w)$ and $\varphi(v) - \psi(w)$ is orthogonal to $N_Y(v)$. On the other hand, $\{ \frac{\partial N_Y}{\partial v_j}(v) \}_{j=1}^{n-1}$ is a linear basis of $T_{\varphi(v)}Y$, so there exists $j = 1,\ldots,n-1$ with $$\langle \varphi (v) - \psi(w), \frac{\partial N_Y}{\partial v_j}(v) \rangle \neq 0.$$ This yields $B = 0$ which proves the assertion. $\spadesuit$\\ Now we begin with the {\it proof of Lemma} 2.1. For $i = 1,\ldots, s$ denote by ${\cal N}_{K,i}$ the set of those $\omega \in \sn$ such that $\omega$ is a regular value of the restriction of $N_K$ to $\dk_i$. In the same way we indtroduce the sets ${\cal N}_{L,j}$ for $j = 1,\ldots,r$. It follows by the Sard theorem that the sets ${\cal N}_{K,i}, {\cal N}_{L,j}$ have full Lebesgue measure in $\sn$. Moreover, they are open and dense in $\sn$, so $${\cal N} = (\bigcap_{i=1}^s {\cal N}_{K,i}) \bigcap (\bigcap_{j=1}^r {\cal N}_{L,j})$$ is an open and dense subset of $\sn$ having full Lebesgue measure. Note that to prove Lemma 2.1 it is sufficient to consider the case when $N_K(x)\in {\cal N}$. Indeed, assume that for every $x'\in \dk$ with $N_K(x')\in {\cal N}$ there exists $y'\in \dl$ with $N_L(y') = \pm N_K(x')$, $\langle x' - y',N_K(x')\rangle = 0$. Given an arbitrary $x\in \dk$, there exists a sequence $\{ x_m \} \subset \dk$ with $N_K(x_m) \rightarrow N_K(x)$ and $N_K(x_m)\in {\cal N}$ for each $m$. Taking a subsequence, we may assume that all $x_m$ belong to one and the same convex component $K_i$ of $K$ and $x_m \rightarrow x'' \in \dk$. Then $N_K(x'') = N_K(x)$ and the convexity of $K_i$ implies $x\in T_{x''} K_i$. Since $N_K(x_m) \in {\cal N}$, by our assumption there exists $y_m \in \dl$ with $N_L(y_m) = \pm N_K(x_m)$, $\langle x_m - y_m,N_K(x_m)\rangle = 0$. Passing again to an appropriate subsequence, we may assume $y_m \rightarrow y \in \dl$. Then we have $N_L(y) = \pm N_K(x'')$, $\langle x'' - y,N_K(x'')\rangle = 0$. Combining this with $N_K(x) = N_K(x'')$ and $x\in T_{x''}\dk$, we see that (5) hold. Fix an arbitrary $x^{(0)} \in \dk$ with $N_K(x^{(0)})\in {\cal N}$. We want to find $y^{(0)}\in \dl$ such that $T_{x^{(0)}} \dk = T_{y^{(0)}} \dl$. We may assume that $x^{(0)} \in \dk_1$. It is sufficient to show that for each sufficiently small neighbourhood $U$ of $x^{(0)}$ in $\dk_1$ there exist $x \in U$ and $y \in \dl$ such that $T_x \dk = T_y \dl$. If this is so, then the compactness of $\dl$ implies the existence of a point $y^{(0)} \in \dl$ with $T_{x^{(0)}} \dk = T_{y^{(0)}} \dl$. Let $U_0$ be a small neighbourhood of $x^{(0)}$ in $\dk_1$ such that $N_K(x)\in {\cal N}$ for all $x\in U_0$. The first step in our proof will be to choose in a special way a vector $\omega_0 \in \sn$ tangent to $\dk_1$ at some $x\in U_0$. Given $i\neq j, i,j = 1,\ldots,r,$ denote by $\Sigma_{i,j}$ the set of those $\omega \in \sn$ such that $\omega = \frac{z-y}{\|z-y\|}$ for some $(y,z)\in \Sigma(\dl_i,\dl_j)$ (see the notation in Lemma 2.2). That is, $$\Sigma_{i,j} = h_{i,j} (\Sigma(\dl_i,\dl_j)),$$ where $$h_{i,j} : \dl_i \times \dl_j \longrightarrow \sn$$ is defined by $$h_{i,j}(y,z) = \frac{z-y}{\|z-y\|}.$$ Since $h_{i,j}$ is smooth, it follows by the Sard theorem and Lemma 2.2 that $\Sigma_{i,j}$ is a finite union of compact subsets of measure zero in $\sn$. Therefore $${\cal S} = \sn \setminus ( \bigcup_{i\neq j} \Sigma_{i,j})$$ is a subset of full Lebesque measure in $\sn$. In particular, ${\cal S}$ is dense in $\sn$. \def\tx{T_{x^{(0)}} \dk} Next, using the fact that $K$ satisfies the condition (H), one can find a line $l$ in ${\bf R}^n$ with $l\cap K = \{ x^{(0)}\}$. Indeed, for each $i = 2,\ldots,s,$ $K'_i = K_i \cap T_{x^{(0)}} \dk$ is a convex domain in $T_{x^{(0)}} \dk$ with $x^{(0)} \notin K'_i$. Let $$C_i = \{ x^{(0)} + tz : t \geq 0, z \in K'_i \}.$$ Then $C_i$ is a cone in $\tx$ with vertex $x^{(0)}$. The condition (H) implies $C_i \cap (\pm C_j) = \{ x^{(0)}\}$ for all $i\neq j, i,j = 2,\ldots,s$. It is then clear that there exists a line $l \subset \tx$ through $x^{(0)}$ with $l\cap C_i = \{x^{(0)}\}$ for all $i = 2,\ldots,s$. Consider an arbitrary line $l$ with this property and denote by $\xi$ one of the directions of $l$, $\xi \in \sn$. The compactness of $K\setminus K_1$ and the regularity of $N_K$ at $x^{(0)}$ show that for all $\omega\in\sn$ sufficiently close to $\xi$ there exists $x\in U_0$ (not unique) so that $\omega$ is tangent to $\dk_1$ at $x$ and the line $l(x,\omega)$, determined by $x$ and $\omega$ has no common points with $K\setminus K_1$. Since ${\cal S}$ is dense in $\sn$, there exist $\omega_0 \in {\cal S}$ and $x\in U_0$ with $$l(x,\omega_0)\cap (K\setminus K_1) = \emptyset,$$ $l(x,\omega_0)$ being tangent to $\dk_1$ at $x$. {\bf Fix $\omega_0$ and $x$ with these properties}. Since $x$ will be fixed in the next arguments, without loss of generality we may {\bf assume that $x = 0$}, i.e. $x$ coincides with the initial point of the coordinate system in ${\bf R}^n$. We shall also assume that $$N_K (0) = e = (0,\ldots,0,1).$$ \def\to{\theta(\omega)} \def\ott{(\omega,\theta(\omega))} For each \begin{equation} \omega \in \sn \:\:\: , \:\:\: \langle \omega,e\rangle < 0, \end{equation} define $$\theta(\omega) = \omega - 2\langle \omega,e\rangle e \in \sn.$$ Then $\theta(\omega)$ is symmetric to $\omega$ with respect to $T_0 \dk$. Consequently, for each $\omega$ with (8) there exists a reflecting $\ott$-ray $\gamma(\omega)$ in $\Omega_{K_1}$ with exactly one reflection point 0. It then follows by the choice of $\omega_0$ that if $\delta > 0$ is small enough and \begin{equation} \mid \omega - \omega_0 \mid < \delta, \end{equation} then $\gamma(\omega)$ is contained in $\Omega_K$ and moreover $\gamma(\omega)$ is an ordinary reflecting $\ott$-ray in $\Omega_K$. Fix $\delta > 0$ with the latter property. Note that $$T_{\gamma(\omega)} = \langle \omega, x\rangle - \langle \theta, x\rangle = 0,$$ since $x = 0$ by our assumption.\\ {\bf 2.3. Lemma.} {\it For each $\omega$ with {\rm (8)} and {\rm (9)} there exists a reflecting $\ott$-ray $\gamma'(\omega)$ in $\Omega_L$ with $T_{\gamma'(\omega)} = 0$.}\\ \def\omtm{(\omega_m,\theta_m)} {\it Proof of Lemma 2.3.} Fix $\omega$ with (8) and (9). Since ${\cal R}''$ is dense in $\ssn$, $\ott$ can be approximated by elements of ${\cal R}''$. Let $\{(\omega_m,\theta_m)\}\subset {\cal R}''$ be a sequence with $\omega_m \rightarrow \omega$, $\theta_m \rightarrow \to$. It is easy to see that for sufficiently large $m$, there exists a (unique) reflecting $\omtm$-ray $\gamma_m$ in $\Omega_K$ with one reflection point $x_m \in \dk_1$. We may assume that there exists $\lim x_m = y \in \dk_1$. Since $$N_K(y) = \lim N_K(x_m) = \lim \frac{\theta_m - \omega_m}{\| \theta_m - \omega_m \|} = \frac{\to - \omega}{\| \to - \omega \|} = N_K(0),$$ and $N_K$ is regular at $0 = x \in U_0$, we find that $y = 0$, i.e. $x_m \rightarrow 0$. So we may also assume that $\| x_m \| \leq 1/2$ for all $m$. It follows from $\omtm \in {\cal R}''$ and (4) that there exists a reflecting $\omtm$-ray $\gamma'_m$ in $\Omega_L$ with \begin{equation} T_{\gamma'_m} = T_{\gamma_m} = \langle \omega_m, x_m\rangle - \langle \theta_m, x_m \rangle = \langle \omega_m - \theta_m, x_m \rangle. \end{equation} In particular, $T_{\gamma'_m} \leq 1$ for all $m$. This implies that the reflecting rays $\gamma'_m$ have bounded number of reflection points. Indeed, let $d$ be the {\it minimal distance} between convex components of $L$ and let $D$ be the {\it radius of a ball with center $0$ containing} $L$. If $k_m$ is the number of reflection points of $\gamma'_m$, then we clearly have $$1 \geq T_{\gamma'_m} \geq (k_m - 1)d - 2D,$$ and so $k_m \leq \frac{1+2D}{d} + 1$. Now passing to an appropriate subsequence, we may assume that each $\gamma'_m$ has exactly $k$ reflection points ($k$ does not depend on $m$). Moreover, we may assume that for each $m$ and each $i = 1,\ldots,k$ the $i$-th reflection point $x_i^{(m)}$ of $\gamma'_m$ belongs to one and the same convex component $L_{j_i}$ of $L$ and there exists $\lim_{m \rightarrow \infty} x_i^{(m)} = x_i \in \dl_{j_i}$. It is then easy to see that $x_1,\ldots,x_k$ are the successive reflection points of a reflecting (not necessarily ordinary) $\ott$-ray $\gamma'(\omega)$ in $\Omega_L$ with (cf (10)) $$T_{\gamma'(\omega)} = \lim_{m \rightarrow \infty} T_{\gamma'_m} = 0.$$ This proves the assertion. $\spadesuit$\\ Denote by $k(\omega)$ the number of reflection points of the reflecting $\ott$-ray $\gamma'(\omega)$ from Lemma 2.3. The next step will be to show that $k(\omega) \leq 2$ for all $\omega$ with (8) and (9), provided $\delta$ is chosen small enough. To do this we use the fact that $L$ satisfies the condition (H).\\ {\bf 2.4. Lemma.} {\it There exists $\varphi_0 \in (0,\frac{\pi}{2})$ such that if $X, Y, Z$ are three convex components of $L$ with $X \neq Y, Y \neq Z$ and $\xi \in X, \eta \in Y, \zeta \in Z$, then the angle between the vectors $\eta - \xi$ and $\zeta - \eta$ is not less than $\varphi_0$.}\\ The proof of this assertion is an easy consequence of the condition (H) and the compactness of $L$. We omit the details. $\spadesuit$. We continue with the proof of Lemma 2.1. Assume that \begin{equation} \delta < \frac{d}{D} \sin^2 \frac{\varphi_0}{4}, \end{equation} and fix $\omega$ with (8) and (9). Let $y_1,\ldots, y_k \:\: (k = k(\omega))$ be the successive reflection points of the reflecting $\ott$-ray $\gamma'(\omega)$ in $\Omega_L$ (see Lemma 2.3). Then we have \begin{eqnarray*} 0 = T_{\gamma'(\omega)} & = & \langle \omega, y_1\rangle + \sum_{i=1}^{k-1} \| y_i - y_{i+1}\| - \langle \to , y_k\rangle \cr & = & \langle \omega - \to, y_k\rangle + \sum_{i=1}^{k-1} {[\| y_{i+1} - y_i \| - \langle \omega, y_{i+1} - y_i\rangle ]}.\cr \end{eqnarray*} Since $\mid \langle \omega - \to, y_k \rangle \mid \leq 2D\delta$ and $$\| y_{i+1} - y_i\| - \langle \omega, y_{i+1} - y_i\rangle = (1- \cos \varphi_i ) \| y_{i+1} - y_i\|,$$ $\varphi_i \in {[0,\frac{\pi}{2}]}$ being the angle between the vectors $\omega$ and $y_{i+1} - y_i$, we find $$2D\delta \geq 2\sum_{i=1}^{k-1} \|y_{i+1} - y_i\| \sin^2\frac{\varphi_i} {2} \geq 2d \sum_{i=1}^{k-1} \sin^2\frac{\varphi_i}{2}.$$ Combining this with (11), one obtains $$\sin^2\frac{\varphi_i}{2} \leq \frac{D}{d}\delta < \sin^2\frac{\varphi_0}{4}$$ and therefore $\varphi_i < \frac{\varphi_0}{2}$ for each $i = 1,\ldots,k-1$. Assume that $k > 2$ and denote by $\varphi$ the angle between the vectors $y_2 - y_1$ and $y_3 - y_2$. Since $\varphi \leq \varphi_1 + \varphi_2$, we get $\varphi < \varphi_0$, which is a contradiction with the choice of $\varphi_0$ (see Lemma 2.4). Thus, $k = k(\omega) \leq 2$. In this way we have shown that, if $\delta$ satisfies (11), then $k(\omega) \leq 2$ for each $\omega$ with (8) and (9). Next, we assume that $\delta$ satisfies (11). Given $\omega$ with (8) and (9), we have $\omega \neq \to$, and so $\gamma'(\omega)$ has at least one {\it transversal reflection point} $y$, that is $\gamma'(\omega)$ is not tangent to $\dl$ at $y$. The final step in the proof of Lemma 2.1 will be to show that there exists $\omega$ with (8) and (9) so that $\gamma'(\omega)$ has exactly one transversal reflection point. This will be derived by the following lemma.\\ {\bf 2.5. Lemma.} {\it Let $P\neq Q$ be two convex components of $L$ and let ${\cal T}$ be the set of those $\omega$ with {\rm (8)} and {\rm (9)} such that $\gamma'(\omega)$ has two transversal reflection points, the first of which belongs to $P$ and the second to $Q$. Assume that $\omega_0$ is a point of Lebesgue measure density of ${\cal T}$ in $\sn$. Then there exist $y\in P, z\in Q$ such that} $$N_L(y) = \pm N_L(z) = \pm e \:\:\: , \:\:\: \frac{z-y}{\| z-y\|} = \omega_0 .$$ {\it Proof of Lemma 2.5.} Recall that by our assumptions we have $x = 0$ and $N_K (0) = e = (0,\ldots,0,1)$. Consequently, $\omega_0 = (u_0;0)$ for some $u_0 \in {\bf R}^{n-1}$ with $\| u_0\| = 1$. Set $$B_0 = \{ u\in {\bf R}^{n-1} : \| u\| < 1 \}.$$ For $u = (u_1,\ldots,u_{n-1})\in B_0$ define $$\omega(u) = (u; -\sqrt{1-\| u\|^2}) \:\: , \:\: \theta(u) = \theta(\omega(u)) = (u;\sqrt{1-\| u\|^2}).$$ Clearly, if $u$ is sufficiently close to $u_0$, then $\omega(u)$ satisfies (8) and (9). Therefore $u_0$ is a point of positive Lebesgue measure density of $$U' = \{ u \in B_0 : \omega(u) \in {\cal T} \: \}$$ in ${\bf R}^{n-1}$. \def\ou{\omega(u)} \def\tu{\theta(u)} \def\du{\partial u_i} \def\dtu{\frac{\partial T}{\partial u_i}} \def\dou{\frac{\partial \omega}{\partial u_i}} \def\dttu{\frac{\partial \theta}{\partial u_i}} \def\dyu{\frac{\partial y}{\partial u_i}} \def\dzu{\frac{\partial z}{\partial u_i}} \def\zy{\frac{z(u) - y(u)}{\|z(u)- y(u)\|}} Let $u' \in U'$ be sufficiently close to $u_0$. Then there exist $y(u') \in P$ and $z(u') \in Q$ which are the successive reflection points of a reflecting $(\ou,\tu)$-ray in $\Omega_L$. This is in fact the ray $\gamma'(\ou)$, and so $T_{\gamma'(\ou)} = 0$. Set $L' = P\cup Q$. It is clear that there exists an open $U$ with $U' \subset U\subset B_0$ such that for each $u\in U$ there exist unique $y(u) \in \partial P, z(u)\in \partial Q$ which are the successive reflection points of a reflecting $(\ou,\tu)$-ray $\gamma'(u)$ in $\Omega_{L'}$. However, the sojourn time $T(u) = T_{\gamma'(u)}$ might be non-zero. It is easy to see that $y(u)$ and $z(u)$ are smooth maps of $u\in U$ (cf. for example Appendix b in \cite{kn:Sj} or Chapter 10 in \cite{kn:PS2}). Hence $$T(u) = \langle \ou, y(u)\rangle + \|y(u) - z(u)\| - \langle \tu, z(u)\rangle $$ is a smooth function of $u$. Let us calculate $\dtu$ for $u\in U$ and $i = 1,\ldots,n-1$. We have \begin{eqnarray*} \dtu (u) & = & \langle \dou (u), y(u)\rangle + \langle \ou, \dyu (u)\rangle + \langle \zy, \dzu (u) - \dyu (u)\rangle \cr & & - \langle \dttu (u), z(u)\rangle - \langle \tu ,\dzu (u)\rangle.\cr \end{eqnarray*} Since $\dyu (u)$ is tangent to $\partial P$ at $y(u)$, it follows by the reflection law at $y(u)$ that $$\langle \zy , \dyu (u) \rangle = \langle \ou ,\dyu (u)\rangle.$$ Similarly, one finds $$\langle \zy , \dzu (u) \rangle = \langle \tu ,\dzu (u)\rangle.$$ Using the latter two equalities, we deduce from above that \begin{equation} \dtu (u) = \langle \dou (u), y(u) \rangle - \langle \dttu (u), z(u) \rangle. \end{equation} \def\tiu{\tilde{u}} Fix $\epsilon \in (0,1)$ such that $u\in U, \|u- u_0\| < \epsilon$ imply $\|\ou - \omega_0\| < \delta$. Clearly, there exist points $u \in U$ with $\|u - u_0\| < \epsilon$ which are points of positive Lebesgue measure density of $U'$ (otherwise $U'$ would have measure zero). Fix a point $\tilde{u}$ with these properties. We claim that \begin{equation} \dtu (\tiu) = 0 \:\:\:\: , \:\:\: i = 1,\ldots,n-1. \end{equation} Consider an arbitrary $i =1,\ldots,n-1$. It is sufficient to show that each neighbourhood $V$ of $\tiu$ in ${\bf R}^{n-1}$ contains a point $v$ with $\dtu (v) = 0$. Fix a cubic neighbourhood $V$ of $\tiu$ and let $e_i = (0,\ldots,0,1,0,\ldots,0)$ be the $i$-th basis vector in ${\bf R}^{n-1}$. Since $V\cap U'$ has positive measure in ${\bf R}^{n-1}$, there exists $v = (v_1,\ldots,v_{n-1}) \in V$ such that $v+se_i \in V\cap U'$ for at least two different values of $s \in {\bf R}$. For such values of $s$ the definition of $U'$ implies $T(v+ s e_i) = 0$. Therefore there exists $v' = v+ s' e_i \in V$ with $\dtu (v') = 0$. This proves (13). Now it follows from (12) and (13) that $$\langle \dou (\tiu), y(\tiu)\rangle - \langle \dttu (\tiu), z(\tiu) \rangle = 0 \:\:\:\: , \:\: i =1,\ldots,n-1.$$ On the other hand, $$\dou (\tiu) = e_i + \frac{\tiu_i}{\sqrt{1-\| \tiu \|^2}} e \:\:\: , \dttu (\tiu) = e_i - \frac{\tiu_i}{\sqrt{1-\| \tiu \|^2}} e,$$ therefore $$y_i(\tiu ) + \frac{\tiu_i}{\sqrt{1-\| \tiu \|^2}} y_n(\tiu) - z_i(\tiu) + \frac{\tiu_i}{\sqrt{1-\| \tiu \|^2}} z_n(\tiu) = 0.$$ That is $$z_i(\tiu) - y_i (\tiu) = \lambda \tiu_i \:\:\: , \:\: i =1,\ldots,n-1$$ with $\lambda = \frac{z_n(\tiu) + y_n(\tiu)}{\sqrt{1 - \| \tiu \|^2}}$. Consequently, there exists $\mu \in {\bf R}$ with $$z(\tiu) - y(\tiu) = \lambda \tiu + \mu e.$$ In particular, we see that the vector $z(\tiu) - y(\tiu)$ is collinear with the two-dimensional plane $\Pi(\tiu)$, determined by the point $y(\tiu)$ and the vectors $\tiu$ and $e$. Since the vector $\omega(\tiu)$ is collinear with this plane, the law of reflection at $y(\tiu)$ yields that $N_L(y(\tiu))$ is also collinear with $\Pi(\tiu)$. Using the same argument for $\theta(\tiu)$ and the law of reflection at $z(\tiu)$, we find that $N_L(z(\tiu))$ is collinear with $\Pi(\tiu)$, too. In this way we have shown that the vectors $\tiu, \omega(\tiu), \theta(\tiu), N_L(y(\tiu)), N_L(z(\tiu))$ are collinear. Since $u_0$ can be approximated by vectors $\tiu \in U$ having the latter property, using the compactness of $\partial P$ and $\partial Q$, we deduce that there exist $y\in \partial P$ and $z\in \partial Q$ such that $\zy = \omega_0$, $\langle z-y, N_L(y)\rangle = 0$, $\langle z-y,N_L(z)\rangle = 0$ and the vectors $N_L(y)$ and $N_L(z)$ are collinear with the two-dimensional plane, determined by $\omega_0$ and $e$. This clearly implies that both $N_L(y)$ and $N_L(z)$ are collinear with $e$, i.e. $N_L(y) = \pm N_L(z) = \pm e $. $\spadesuit$\\ Now assume that for each $\omega$ with (8) and (9), $\gamma'(\omega)$ has two transversal reflection points. For $i,j = 1,\ldots,r$ denote by ${\cal T}_{i,j}$ the set of those $\omega$ with (8) and (9) such that the first reflection point of $\gamma'(\omega)$ belongs to $L_i$ and the second one to $L_j$. Since $\cup_{i,j=1}^r {\cal T}_{i,j}$ is an open subset of $\sn$ (so it has positive measure) and $\omega_0$ belongs to its closure, there exist $i \neq j$ such that $\omega_0$ is a point of positive Lebesgue measure density of ${\cal T}_{i,j}$. Fix $i$ and $j$ with this property and apply Lemma 2.5 to $P = L_i$, $Q = L_j$ and ${\cal T} = {\cal T}_{i,j}$. We get that there exist $y\in\dl_i$ and $z\in\dl_j$ such that $N_L(y) = \pm N_L(z) = \pm e$ and $\omega_0 = \frac{z-y}{\|z-y\|}$. Since $e = N_K(0)\in {\cal N}$, it follows by the definition of ${\cal N}$ that $N_L$ is regular at both $y$ and $z$. Consequently, $(y,z) \in \Sigma(\dl_i , \dl_j)$ and $\omega_0 = h_{i,j}(y,z) \in \Sigma_{i,j}$ in contradiction with $\omega_0 \in {\cal S}$. Therefore there exists $\omega$ with (8) and (9) such that $\gamma'(\omega)$ has exactly one transversal reflection point $y(\omega)$. Fix $\omega$ with this property and set $y = y(\omega)$. Then $\omega - \theta(\omega)$ is a non-zero vector collinear with both $N_K(0) = N_K(x)$ (recall that $x = 0$ by our assumption) and $N_L(y)$. Hence $N_K(x) = \pm N_L(y)$. On the other hand, $$0 = T_{\gamma'(\omega)} = \langle \omega - \theta(\omega), y \rangle$$ shows that $\langle x - y, N_K(x)\rangle = 0$ which implies that $T_x \dk = T_y \dl$. This completes the proof of Lemma 2.1. $\spadesuit$\\ {\it Proof of Theorem 1.1}. It is sufficient to show that $K_1$ coincides with some convex component of $L$. Fix an element $x$ of $\dk_1$. It follows from Lemma 2.1 that there exists $y\in \dl$ with $T_x \dk = T_y \dl$. Fix $y$ with this property and let $y \in L_i$ for some $i = 1,\ldots,r$. We will show that $K_1 = L_i$. According to the condition (H) and Lemma 2.1, one gets easily that the set $W$ of those $x'\in \dk_1$ for which there exists $y'\in \dl_i$ with $T_{x'} \dk = T_{y'} \dl$ is open in $\dk_1$. On the other hand, $W$ is also closed in $\dk_1$, and since $W \neq \emptyset$ and $\dk_1$ is connected, we find $W = \dk_1$. Consequently, any supporting hyperplane to $K_1$ is a supporting hyperplane to $L_i$ as well. Now using the convexity of $K_1$ and $L_i$ we conclude that $K_1 = L_i$. \begin{thebibliography}{4} \bibitem[GG]{kn:GG} M. Golubitskii and V. Guillemin, {\em Stable Mappings and their Singularities}, Berlin, Springer, 1973. \bibitem[G]{kn:G} V. Guillemin, {\em Sojourn time and asymptotic properties of the scattering matrix}, Publ. 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Equations {\bf 64} (1986), 283-293.\\ \end{thebibliography} $\:\:\:\:$ Department of Mathematics, University of Western Australia, Nedlands, Perth 6009, Western Australia; stoyanov@maths.uwa.edu.au \end{document}