\magnification=1200
\noindent {\bf Inverse Scattering at Fixed Energy for Layered Media}
\medskip
\noindent {\bf Jean-Claude Guillot}
\noindent D\'epartement de Math\'ematiques, UMR 7539, Universit\'e de
Paris 13, 93430 Villetaneuse, France
\medskip
\noindent{\bf James Ralston}
\noindent Department of Mathematics, UCLA, Los Angeles, CA 90095-1555,
USA
\medskip
\medskip
In this article we prove a uniqueness theorem in
inverse scattering for the wave equation in a layered
medium. We consider the wave equation in $R\times R^n, n \geq 3$,
with a variable sound speed, $c(x)$,
$$\partial^2_t u = c^2(x)\Delta u$$
as a perturbation of the wave equation with a sound speed, $c_0(x_n)$,
which is a function of one variable,
$$\partial^2_t u = c^2_0(x_n)\Delta u.$$
Thus the unperturbed wave equation could
be used to model wave propagation in a medium composed of uniform layers
with different physical properties.
When one takes the scattering amplitude at fixed energy as the observed data,
simple examples, e.g. infinitesimal perturbations of
a homogeneous medium, show that it is not reasonable to expect to recover
more than the Fourier transform of the perturbation restricted to a ball
from this data. Hence one needs to assume that the
perturbation will be determined by this restricted Fourier transform, and a
natural way to do this is to assume exponential decay of the perturbation.
\medskip
The precise formulation of the problem we study here is as follows.
We consider perturbations of the operator
$L_0 =-c_0^2(x_n)\Delta$ in $R^n$, where $c_0\in L^\infty(R)$ satisfies
$c_0(s)\geq c_{min}>0$ for all $s$ and
$$c_0(s)=\left\{ \matrix c_+,\hbox{ for } s>s_+>0,\\
c_- < c_+, \hbox{ for } s< s_-<0.\endmatrix \right.$$
The perturbed operators have the form $L=-c^2(x)\Delta$, where
$c\in L^\infty(R^n)$ approaches $c_0$ at an exponential rate, i.e.
$|c(x)-c_0(x_n)| < C\hbox{exp}(-\delta |x|)$. We assume
that $c$ is bounded
away from zero, and without loss of generality can take $c(x)\geq
c_{min}$ for $x\in R^n$. With these hypotheses we have the following
\medskip
\noindent {\bf Theorem}. The coefficient $c_0(x_n)$ and the scattering
amplitude at energy $k^2>0$ determine $c(x)$.
\medskip
In this setting the scattering amplitude is
more complicated than in two-body potential
scattering because of the presence of
critical angles of reflection and guided waves. In Section 1 of this
paper we define the scattering amplitude, but the proof that this
scattering amplitude can be determined from the asymptotic behavior of
distorted plane waves is postponed until the Appendix. The organization
of the rest of the paper is as follows. In Section 2 we outline
the proof, based on [ER], of the theorem above.
The proof involves a sequence of integral equations
which must be solved to connect to scattering amplitude with the
perturbation. In Section 3 we derive the estimates on integral operators
needed for this argument, and in Section 4 we discuss the analytic
continuation which connects the analogue of Faddeev's scattering amplitude
with
the Fourier transform of the perturbation.
\medskip
That the method of [ER] could be applied in layered media was first recognized
by Isozaki in [I]. He proved the result that we prove here in the case
$$c_0(s)=\left\{ \matrix c_+,\hbox{ for } s>0,\\
c_- < c_+, \hbox{ for } s< 0.\endmatrix \right.$$
The omission of the layer $s_-< s< s_+$ precludes the existence of guided
waves, though the other features of the problem remain the same. This
was rectified in [W] and a preliminary version of this article, [GR].
In [W] Weder introduced the layer $s_- < s k_+$.
In all cases
$$-{d^2\phi_\alpha\over ds^2} -{k^2\over c_0^2(s)}\phi_\alpha = \lambda
\phi_\alpha.$$
\medskip
\noindent For $\alpha =1$
$$\phi_1(s,\lambda ) = a_+(\lambda)\hbox{exp}(is(\lambda+k_+^2)^{1/2})
,$$
when $s>s_+$. For $\alpha =2$
$$\phi_2(s,\lambda) = a_-(\lambda)\hbox{exp}(-is(\lambda +k_-^2)^{1/2})
,$$
when $s-k_+^2$,
and we define them to be
zero for $\lambda \leq -k_+^2$. For $\alpha =3$
$$\phi_3(s,\lambda) = a_0(\lambda)\hbox{exp}(-s(-\lambda-k_+^2)^{1/2})
,$$
when $s>s_+$.
The function $\phi_3$ is defined with a positive square root for $-k^2_- <
\lambda
< -k^2_+$, and we define it to be zero outside this interval.
We define $\phi_4(s,\lambda)$ to be a normalized
$\lambda$-eigenfunction of
$G$ for $\lambda$ in the point spectrum
$\{\lambda_1,\dots,\lambda_N\}$ of $G$,
and zero elsewhere. Then, following [Wi], we may choose the coefficients
$a_*(\lambda)$ so
that the following Parseval formula holds (the actual coefficients are
given in the Appendix)
$$f(s) = \sum^2_{\alpha =
1}\int^\infty_{-k^2_+}\phi_\alpha(s,\lambda)\int_R\overline{\phi_
\alpha (t,\lambda)}f(t)dtd\lambda +\int_{-k^2_-}^{-k^2_+}\phi_3(s,\lambda)
\int_R\overline{\phi_3(t,\lambda)}f(t)dtd\lambda
$$
$$ +\sum ^N_{j=1}\phi_4(s,\lambda_j)\int_R
\overline{\phi_4(t,\lambda_j)}f(t)dt$$
for $f\in C_0^\infty(R)$. We let $\Psi f$ denote the associated spectral
representation of $f$, i.e.
$$\Psi f(\lambda) =(\Psi_1f(\lambda),
\dots,\Psi_4f(\lambda)) =
\int_R f(t)(\overline{\phi_1(t,\lambda)},
\overline{\phi_2
(t,\lambda)},\overline{\phi_3(t,\lambda)},\overline{\phi_4(t,\lambda)})dt.$$
\medskip
\noindent {\bf Definition.} The scattering amplitude is defined to be
$[\Psi h(\eta,\cdot)](-|\eta |^2)$, where $h$ is defined in (2).
In the Appendix we will show that $[\Psi h(\eta,\cdot)](-|\eta |^2)$ can be
recovered from the asymptotics of $v$.
\medskip
\medskip
\noindent{\bf 2. Formal Solution of the Problem}
\medskip
The methods in this section are taken largely from [ER] with the
adaptations for layered media introduced in [I]. We will assume that
the reader has some familiarity with this approach; [R]
would be an adequate introduction. For vectors in $R^n$ and
$C^n$ we will now use
the notation $\zeta^2$ for $\zeta_1^2 +\dots +\zeta_n^2$.
\medskip
>From (1) we have the analog of (10) in [ER],
$$h(\xi, x_n) + (2\pi)^{1-n}\int_{R^{n-1}}\hat q(\xi -\eta,x_n)[(G
+\eta^2-i0)^{-1}h(\eta,
\cdot)](x_n)d\eta$$
$$=-\hat q(\xi -\zeta,x_n)\phi_\alpha(x_n,\lambda),\eqno{(3)}$$
where $\hat q(\xi-\zeta,x_n)$ is the
Fourier transform of $q$ in the first $(n-1)$-variables. Note that, if one is
completely explicit $h=h(\xi,x_n;\zeta,\lambda,\alpha)$, but for simplicity
of notation
we will often suppress the $(\zeta,\lambda,\alpha)$-dependence, and write
$h(\xi,x_n)$.
Since $(L_0- k^2)\phi_\alpha(x_n,\lambda)e^{ix^\prime\cdot\zeta}=0$
only for
$\lambda =-\zeta^2$, the scattering amplitude is
given more explicitly by
$$\{[\Psi_\beta h(\eta,\cdot;\zeta,-\zeta^2,\alpha)](-\eta^2) :
\alpha,\beta=1,2,3,4,\
(\eta,\zeta)
\in R^{n-1}\times R^{n-1}\}.$$
However, it will be convenient to continue to
allow $\lambda$ to take arbitrary
real values.
\medskip
In analogy with (16) in [ER] we will define the Faddeev extension, $h^*$,
of the
scattering data by means of the equation
$$h^*(\xi,x_n,\sigma) +(2\pi)^{1-n}\int_{R^{n-1}}\hat q(\xi-\eta,x_n)
[(G +
\eta^2
+0i\ \hbox{sgn}(\eta\cdot\nu -\sigma))^{-1}h^*(\eta,\cdot,
\sigma)](x_n)d\eta$$
$$=-\hat q (\xi-\zeta,x_n)\phi_\alpha(x_n). \eqno{(4)}$$
Here $\sigma \in R$ and $\nu \in S^{n-2}$, and from here on we will
suppress
the $\nu$-dependence in $h^*$. To see the relation between $h$ and $h^*$
recall
that
$$\lim_{\epsilon\to 0_+} \int_{R^{n-1}}\hat q(\xi-\eta,x_n)[((G+\eta^2
+i\epsilon)^{-1}
-(G+\eta^2 -i\epsilon)^{-1})f(\eta, \cdot)](x_n)d\eta$$
$$=\lim_{\epsilon\to 0_+}\int_{R^{n-1}}\hat
q(\xi-\eta,x_n)[\Psi^*({-2i\epsilon\over
(\lambda +\eta^2)^2 +\epsilon^2}\Psi f(\eta,\cdot))](x_n)d\eta$$
$$=-\sum^3_{\alpha =1} 2\pi i\int_{R^{n-1}}\hat
q(\xi-\eta,x_n)\phi_\alpha(x_n,-\eta^2)
[\Psi_\alpha f(\eta,\cdot)](-\eta^2)d\eta$$
$$-\sum_{j=1}^N{\pi i\over \sqrt{-\lambda_j}}
\int_{\eta^2=-\lambda_j}\hat
q(\xi-\eta,x_n)
\phi_4(x_n,-\eta^2)[\Psi_4f(\eta,\cdot)](-\eta^2)dm_j,$$
where $m_j$ is the volume measure on the sphere $\eta^2=-\lambda_j$.
Thus, substituting
$$h(\xi,x_n;\delta,\lambda,\alpha)+(2\pi)^{1-n}\int_{R^{n-1}}\hat
q(\xi-\eta,x_n)
[(G+\eta^2-0i)^{-1}h(\eta,\cdot;\delta,\lambda,\alpha)](x_n)d\eta \eqno{
(5)}$$
for $-\hat q(\xi-\delta,x_n)\phi_\alpha(x_n,\lambda)$ in all places where it
appears in
(4), and then applying the inverse of the operator acting on $h$ in (5) we
arrive at the
analog of (24) in [ER],
$$h^*(\xi,x_n;\zeta,\lambda,\alpha;\sigma) +2\pi i\sum^3_{\beta =
1}\int_{\eta\cdot\nu
>\sigma}h(\xi,x_n;\eta,-\eta^2,\beta)[\Psi_\beta
h^*(\eta,\cdot;\zeta,\lambda,\alpha;\sigma)]
(-\eta^2)d\eta$$
$$+\sum^N_{j=1}{\pi i\over
\sqrt{-\lambda_j}}\int_{\{\eta^2=-\lambda_j,\eta\cdot\nu>\sigma\}}
h(\xi,x_n;\eta,-\eta^2,4)[\Psi_4h^*(\eta,\cdot;\zeta,\lambda,\alpha;\sigma)]
(-\eta^2)dm_j$$
$$=h(\xi,x_n;\zeta,\lambda,\alpha). \eqno{(6)}$$
Hence, applying $\Psi$ to both sides of (6) and evaluating it at $
-\xi^2$, we
see, assuming that the resulting system of integral equations
for $[\Psi (h^*(\xi,\cdot))](-\xi^2)$ is uniquely
solvable, that
$$\{[\Psi h(\xi,\cdot;\zeta,-\zeta^2,\alpha;\sigma)](-\xi^2):\alpha
=1,2,3,4,\ (\xi,\zeta)
\in R^{n-1}\times R^{n-1}\}$$
determines
$$\{[\Psi h^*(\xi,\cdot;\zeta,-\zeta^2,\alpha;\sigma)](-\xi^2):\alpha
=1,2,3,4,\ (\xi,\zeta)
\in R^{n-1}\times R^{n-1}\}$$
\medskip
We define $h_*(\xi,x_n;\zeta,\lambda,\alpha;\sigma)=h^*(\xi +\sigma
\nu,x_n;\zeta+\sigma\nu,
\lambda,\alpha;\sigma)$, and then (4) becomes
$$h_*(\xi,x_n;\sigma)+(2\pi)^{1-n}\int_{R^{n-1}}\hat
q(\xi-\eta,x_n)[(G+(\eta +\sigma\nu)^2
+0i\ \hbox{sgn}(\eta\cdot\nu))^{-1}h_*(\eta,\cdot;\sigma)](x_n)d\eta$$
$$=-\hat q(\xi-\zeta,x_n)\phi_\alpha(x_n,\lambda).\eqno{(7)}$$
We wish to extend $h_*(\sigma)$ to $h_*(z)$ for $z$ in a set ${\Cal
D}_\epsilon$ of the form
${\Cal D}_\epsilon =\{z: |\hbox{Re}\{z\}|
<\epsilon,\ \hbox{Im}\{z\}>0\}$. For $z=i\tau$ this extension
can be done directly. We define $h_*(\xi,x_n;i\tau)$ as the solution of
$$h_*(\xi,x_n;i\tau)+(2\pi)^{1-n}\int_{R^{n-1}}\hat
q(\xi-\eta,x_n)[(G+(\eta +i\tau \nu)^2)
^{-1}h_*(\eta,\cdot;i\tau)](x_n)d\eta$$
$$=-\hat q(\xi-\zeta,x_n)\phi_\alpha(x_n,\lambda),\eqno{(8)}$$
or, more compactly,
$$h_*(i\tau)+A(i\tau)h_*(i\tau)=\tilde q.$$
\medskip
We will show that on a suitable Banach space ${\Cal A}$, containing $\tilde
q$,
$A(i\tau)$ has an
extension to $A(z)$, a compact operator-valued analytic function on ${\Cal
D}_\epsilon$
for $\epsilon$ sufficiently small. Moreover, $A(z)$ extends continuously to
$A(\sigma)$,
the operator in (7), as $z$ goes to the real axis. Since we will also show
that the norm
of $A(i\tau)$ goes to zero as $\tau$ goes to infinity, the inverse of
$I+A(z)$ is
meromorphic on ${\Cal D}_\epsilon$ with a continuous extension to
$-\epsilon <\sigma
<\epsilon$ outside a closed set of measure zero. The functions in ${\Cal
A}$ will be
holomorphic in a neighborhood of Im$\{\xi\}=0$, and, since $h_*$ will
inherit the analyticity
of $\hat q(\xi-\zeta,x_n)\phi_\alpha(x_n,\lambda)$ in
$\zeta$ and $\lambda$,
$h_*(\xi,x_n;\zeta,
\lambda,\alpha;z)$ will by analytic in $(\xi,\zeta,\lambda,z)$ on
$${\Cal S}_\epsilon = \{|\hbox{Im}\{\xi\}|<\epsilon\}
\times \{|\hbox{Im}\{\zeta\}|<\epsilon\}\times \{\hbox{Re}\{\lambda\}>
-k^2_+,\ |\hbox
{Im}\{\lambda\}|<\epsilon\}\times\{ {\Cal D}_\epsilon\cap F^c\}$$
for a discrete set $F$,
when $\epsilon$ is sufficiently small.
\medskip
Now consider an analytic curve $(\xi(s), \zeta(s), z(s))$, defined in
$s_0<\hbox{Re}\{s\}<\infty$, $0< \hbox{Im}\{s\}<\epsilon^\prime$
with a continuous extension to Im$\{s\}=0$,
such that $(\xi(s),\zeta(s),-(\zeta(s) +
z(s)\nu)^2,z(s))$
lies in ${\Cal S}_\epsilon$. Since the scattering amplitude
determines $[\Psi h_*(\xi,\cdot;\zeta,-(\zeta+\sigma\nu)^2,\alpha;\sigma
)](-(\xi+\sigma\nu)^2)$, as long as $(\xi(s),\zeta(s),z(s))$ is
real-valued for $s$ in an interval on the real axis,
the scattering amplitude will determine
$$[\Psi h_*(\xi(s),\cdot;\zeta(s),-(\zeta(s)+z(s)\nu)^2,\alpha;z(s)
)](-(\xi(s)+z(s)\nu)^2).$$
If $z(s)\to i\infty$ as $s\to \infty$, then the integral term in (8) goes
to zero, and we
may conclude that the scattering amplitude determines the asymptotic
behavior of
$$[\Psi(\hat
q(\xi(s)-\zeta(s),\cdot)\phi_\alpha(\cdot,-(\zeta(s)+z(s)\nu)^2,\alpha;z(s))]
(-(\xi(s) +z(s)\nu)^2)$$
as $s\to\infty$. The choice of $(\xi(s), \zeta(s),z(s))$ that we make
here is precisely
the one used in [I] and [ER]. Given $p\in R^n$ with $p_n\neq 0$ and $|p| <
2k_+$,
we choose $\nu\in S^{n-2}$ with $\nu\cdot p^\prime =0$ and $\mu\in S^{n-1}$
with
$\mu^\prime \cdot \nu =0$, $\mu\cdot p =0$ and $\mu_n >0$. Then we set
$$\xi(s)=p^\prime/2 + s\mu^\prime,\ \zeta(s) =-p^\prime/2 +s\mu
^\prime
,\hbox{ and
}z(s)=i\sqrt{s^2
+|p|^2/4-k_+^2}.$$
With this choice of $z(s)$ for $s>>0$
$$\sqrt{k_+^2 -(\xi(s) +z(s)\nu)^2} =\sqrt{(p_n/2 +s\mu_n)^2} =p_n/2
+s\mu_n,$$
and
$$\sqrt{k_+^2 -(\zeta(s) +z(s)\nu)^2} =\sqrt{(p_n/2 -s\mu_n)^2} =-p_n/2
+s\mu_n.$$
>From the definition of $\phi_1(x_n,\lambda)$, setting $\lambda(s)
=-(\zeta(s) +z(s)\nu)^2$
and $\Lambda(s) =-(\xi(s) +z(s)\nu)^2$, we have
$$\lim_{s\to
\infty}(\overline{a_+(\Lambda(s))}a_+(\lambda(s)))^{-1}[\Psi_1(\hat q(\xi(s)-
\zeta(s),\cdot)\phi_1(\cdot,\lambda(s))](\Lambda(s)) = {\Cal F}q(p),
\eqno{(9)}
$$
where ${\Cal F}$ denotes the Fourier transform on $R^n$. In deriving (9)
one only needs to
use $|a_+(\lambda)^{-1}\phi_1(x_n,\lambda)-\hbox{exp}(ix_n\sqrt\lambda)|\to
0$ as
$\lambda\to\infty$. Thus we have shown that the scattering amplitude
determines the
Fourier transform of the perturbation $q(x)$ on an open set. Since
exponential decay of
$q(x)$ implies that its Fourier tranform is analytic, this implies that the
scattering
amplitude determines $q$.
\medskip
\medskip
\noindent {\bf 3. Integral Equations}
\medskip
This section is devoted to the integral equations in {\bf 2.} The
ingredient in all these equations which requires further study
is the resolvent $(G-\lambda)^{-1}$. While we could represent
$(G-\lambda)^{-1}$ in terms of generalized eigenfunctions, as we
already did in the derivation of (6), it is simpler to use the
standard Green's function construction here.
When Im$\{\lambda\}\neq 0$, the operator $(G-\lambda)
^{-1}$ is an integral operator on $L^2(R)$ with the kernel
$$g(s,t.\lambda)= {\phi_+(s,\lambda)\phi_-
(t,\lambda)\over W(\lambda)}\hbox{ for }s>t,\hbox{ and }g(s,t,\lambda)=
{\phi_-(s,\lambda)\phi_+(t,\lambda)\over W(\lambda)}\hbox{ for }ss_+,\hbox{ and }
\phi_-(s,\lambda)= e^{-is(\lambda +k_-^2)^{1/2}},
\hbox{ for } s
0\}$ and its restriction to $R\times R\times \{|\lambda|C|\lambda|^{1/2}$ on $\{|\lambda| > R_0\}\cap\{\hbox{Im}\{\lambda\}
\neq 0\}$
for $R_0$ sufficiently large.
\medskip
\noindent {\bf Proof.} Since we assume that $c_0(s)$ is constant outside
the interval $s_-~~s_+$.
\medskip
\noindent {\bf Lemma 2}. The Wronskian $W(\lambda)$ does not vanish
for Im$\{\lambda\}\neq 0$. $W(\lambda)$ has continuous extensions
to Im$\{\lambda\} = 0$ from the upper and lower half-planes. These
extensions agree for $\lambda <-k^2_-$ and hence $W(\lambda)$
is analytic on Re$\{\lambda\} <-k^2_-$, vanishing only at $\lambda =
\lambda_j,\ j=1,\dots,N$, where it has simple zeros. The extensions
of $W(\lambda)$ to Im$\{\lambda\}=0$ do not vanish for $\lambda >
-k_-^2$. At $\lambda = -k^2_-$ one either has $W(-k_-^2)\neq 0$
or $\lim_{\lambda \to -k^2_-}
(\lambda +k_-^2)^{-1/2}W(\lambda)=i\gamma\neq 0$, $\gamma
\in R$.
\medskip
\noindent {\bf Proof.} The statements in the first two sentences follow
from the observation that for $\lambda \in \{\hbox{Im}\lambda \neq 0\}
\cup\{\hbox{Re}\lambda <-k^2_-\}$ $W(\lambda) = 0$ is equivalent to
$\lambda$ being an eigenvalue of $G$. The continuity
of the extensions of $W(\lambda)$ to the real axis follows from
(10). For the statements in the last two sentences we
appeal to [CK]. In the notation of [CK] $\phi_+(s,\lambda) =
f_+(s, (\lambda +k_-^2)^{1/2})$ and $\phi_-(s,\lambda) = f_-(s,
(\lambda +k_-^2)^{1/2})$, where $f_\pm$ are the Jost functions
for the potential $v(s)= -k^2/c_0^2(s)+k_-^2$. With these
identifications Lemma 1.2 and Prop. 2.4 of [CK] contain the
desired results. Note that (10) implies that $W(\lambda)=
R(\lambda,\sqrt{\lambda +k^2_-})$, where $R(\lambda,z)$ is analytic
near $(\lambda,z)=(-k^2_-,0)$.
\medskip
To solve the integral equations (3), (4) and (8) in {\bf
3.} we will use the weighted $L^2$-spaces, $L^2_r(R^n)$, with
norms
$$||f||_{2,r}=\left(\int_{R^n}|f(x)|^2(1 + x^2)^rdx\right)^{1/2}.$$
In each case we will invert the partial Fourier transform in (2). Thus,
(3)
becomes
$$g(x)+q(x)[E_0g](x)=q(x)e^{i\zeta\cdot
x^\prime}\phi_\alpha(x_n,\lambda),
\eqno{(11)}$$
where $E_0$ is the operator given by
$$E_0f=(-\Delta -{k^2\over c^2_0}-i0)^{-1}f,$$
and for $f\in C^\infty_0(R^n)$ we have
$$[E_0f](x)=(2\pi)^{1-n}\int_{R^{n-1}}\int_{R^{n-1}}
e^{i\eta\cdot(x^\prime -y^\prime)}
[(G+\eta^2-i0)^{-1}f(y^\prime,\cdot)](x_n)dy^\prime d\eta.$$
We claim that (11) is a Fredholm equation in $L^2_r(R^n)$ for $r >r_0$.
To prove this it will suffice to show that $qE_0$ is compact. We begin
by showing that $E_0$ is a bounded operator from $L^2_r(R^n)$ to
$L^2_{-r}(R^n)$ for $r>r_0$. Since $q$ decays exponentially, any value
of
$r_0$ will suffice for us, and the value which we will use is far from
optimal.
We will bound the norm of $E_0$ as an
operator from $L^2_r$ to $L^2_{-r}$ by duality.
Thus it suffices to bound
$$\int_{R^n}\bar f E_0gdx =(2\pi)^{1-n}\int_{R^{n-1}\times
R}\overline{\hat
f(\eta,x_n)}[(G+\eta^2-i0)^{-1}\hat g(\eta,\cdot)](x_n)d\eta dx_n
\eqno{(12)}$$
by $C||f||_{2,r}||g||_{2,r}$. Since $G$ is a self-adjoint operator on
$L^2(R^n)$ with spectrum $$\{\lambda_1,\dots,\lambda_N\}\cup
[-k^2_-,\infty),$$
we have
$$(2\pi)^{1-n}\int_R dx_n\int_{\{|\eta|> 1
+\sqrt{-\lambda_1}\}}\overline{\hat
f(\eta,x_n)}[(G+\eta^2-i0)^{-1}\hat g(\eta,\cdot)](x_n)d\eta \leq
||f||_{2,0}||g||_{2,0}.$$
Thus we may assume that the integration in $\eta$ in (12) is restricted
to
the ball $|\eta| < R_0 = 1 +\sqrt{-\lambda_1}$.
By Lemma 2 $W(-\eta^2 )$ vanishes to first order on the spheres
$\{\eta^2=-\lambda_j\}$, and we may have
$${1\over W(-\eta^2 +i0)} = {r(\eta^2)\over (-\eta^2 + k_-^2
+i0)^{1/2}},
$$
where $r(k^2_-)\neq 0$, but elsewhere $1/W$ is continuous. Using the
notation of Lemma 1,
$$[(G+\eta^2-i\epsilon)^{-1}\hat g](x_n)={1\over W(-\eta^2
+i\epsilon)}\int_R f(x_n,t,
-\eta^2 +i\epsilon)\hat g(\eta,t)dt = {h(\eta,x_n,\epsilon)\over
W(-\eta^2 +i\epsilon)}.$$
>From Lemma 1(ii)
it follows that the H\"older norm
$||h(\cdot,x_n,\epsilon)||_\alpha,\ \alpha <1/2$,
over $|\eta| 1 +n/2$, and the distribution
$${\Cal L}(\rho) =\lim_{\epsilon\to 0_+}\int_{-1}^1 {\rho(s)\over
s+i\epsilon}ds$$
is bounded by the $\alpha$-H\"older norm of $\rho$ for any $\alpha >0$,
it follows that the quantity in (12) is bounded by
$C||f||_{2,r}||g||_{2,r}$
for $r> 3+n/2$. Thus by duality $E_0$ is bounded from $L^2_r$ to
$L^2_{-r}$
in this range, and by exponential decay $qE_0$ is bounded from $L^2_r$
to $L^2_{r^\prime}$ for all $r^\prime$.
\medskip
To complete the proof that $qE_0$ is compact on $L^2_r(R^n)$ for $r>r_0$
we note that the Sobolev estimate
$$||u||_{H^2(|x-x_0|<1)}r_0$.
If $(I + qE_0)h=0$, for $h\in L^2_r(R^n)$, then $u=E_0h$ is an outgoing
solution to $(L-k^2)u=0$. The limiting absorption theorem of
[dBP] shows
that $u=0$ for any $k>0$. Thus we have
\medskip
\noindent {\bf Prop. 1} $I +qE_0$ is invertible on $L^2_r(R^n)$ for
$r>3+n/2$, and
equation (3) is solvable in the partial Fourier transform (as in (2)) of
this space.
\medskip
The preceding arguments apply equally well to show that (4) is a
Fredholm
equation. Changing $(G+\eta^2 -i0)^{-1}$ to $(G+\eta^2 +0i \hbox{ sgn}(
\eta\cdot\nu-\sigma))^{-1}$ introduces a jump discontinuities across
$\eta\cdot\nu
=\sigma$ in $\phi_-(s,-\eta^2-0i \hbox{ sgn}( \eta\cdot\nu-\sigma))$ when
$\eta^2 0$, we have
\medskip
\noindent {\bf Prop. 2} The equation (4) is Fredholm in the partial
Fourier transform (as in (2)) of $L^2_r(R^n)$ for $r>3+n/2.$
\medskip
Next we need to show that the Faddeev equation, derived by evaluating
(6) at $\lambda = -\zeta^2$, applying
$\Psi$ to both sides, and then evaluating $\Psi h$ and $\Psi h^*$
at $-\xi^2$, will be uniquely solvable when (4) is uniquely solvable.
$[\Psi f](\lambda)=0$ when $\lambda <\lambda_1$, and $[\Psi
h(\xi,\cdot;
\zeta,-\zeta^2;\alpha](-\xi^2)$ is continuous in all variables,
except in the case that $W(-k_-^2)=0$. In that case the normalizing
coefficient $a_0(\lambda)$ behaves like $|\lambda +k_-^2|^{-1/4}$
near $\lambda =-k_-^2$. This follows from the explicit formulas for
the normalizing coefficients in the Appendix. In all cases the
Faddeev
equation is Fredholm, if we pose it in the space of
functions
$$S=C(\{\xi^2\leq k^2_+\})^2\times C(\{k_+^2\leq\xi^2\leq
k_-^2\},a_0(-\xi^2))
\times \prod_{j=1}^N C(\{\xi^2=-\lambda_j\})
,$$
where $C({\Cal X},a(x))$ denotes the space of functions continuous
on the interior of ${\Cal X}$ with the norm
$$||f||=\sup_{\Cal X}|f(x)|/a(x).$$
If $f=(f_1,f_2,f_3,f_{4,1},\dots,f_{4,N})\in S$ is a solution
of the corresponding homogeneous equation, we have
$$ f_\alpha(\xi) +2\pi i\sum^3_{\beta =1}\int_{\eta\cdot\nu>\sigma}
[\Psi_\alpha h(\xi,\cdot;\eta,-\eta^2,\beta)]
(-\xi^2)f_\beta(\eta)d\eta$$
$$+\sum^N_{j=1}{\pi i\over\sqrt{-\lambda_j}}\int_{\{\eta^2=
-\lambda_j,\eta
\cdot\nu>\sigma\}}[\Psi_\alpha h(\xi,\cdot;\eta,-\eta^2,4)]
(-\xi^2)f_{4,j}(\eta)dm_j
=0 \eqno{(13)}$$
for $\alpha = 1,2,3,(4,1),\dots,(4,N)$. If we evaluate
$\Psi h$ at $\lambda$
instead of $-\xi^2$, (13) gives an extension of $f$ to $f(\xi,\lambda)$.
Then, applying $\Psi^{(-1)}$ to the resulting equation gives a
solution to the homogeneous version of (6), and we can reverse the
derivation of (6) from (4) to get a solution -- nontrivial since we
have only applied invertible operators -- to the homogeneous version
of (4). Thus the Faddeev equation is uniquely solvable whenever
(4) is uniquely solvable. This argument is analogous to the one given
in [ER] in the paragraph following formula (26) in that article.
\medskip
The critical step in the argument outlined in {\bf 2} is the proof that
the norm of $A(i\tau)$ goes to zero as $\tau\to \infty$. This not
only implies that (8) is solvable for $\tau$ large, but also
by the analytic continuation argument
it implies that (4) will be uniquely
solvable for $\sigma$ in
an open interval on the real axis.
\medskip
To bound the norm $A(i\tau)$ as an operator on
$L^2_r(R^n)$ we will use the method used to prove Prop. 1.
As in the proof of Prop.1, we first need to bound
$$I(f,g)= \int_{R^{n-1}\times R}\overline{ \hat f(\eta,x_n)}
[(G +\eta^2-\tau^2
+2i\tau \eta\cdot\nu)^{-1}\hat g(\eta,\cdot)](x_n)d\eta dx_n $$
in terms of the norms of $f$ and $g$ in $L^2_r$.
We have
$$|W(-(\xi +i\tau\nu)^2)|>C|(\xi +i\tau\nu)^2|^{1/2}$$
for $|(\xi +i\tau\nu)^2|>R_0$ from Lemma 1(iii), and from Lemma 2
$$|W(-(\xi +i\tau\nu)^2|>C|(\xi +i\tau\nu)^2 +\lambda_j|$$
for $-(\xi+i\tau\nu)^2$ in a neighborhood of $\lambda_j,\ j=1,\dots,N$,
and
$$|W(-(\xi+i\tau\nu)^2)|>C|(\xi +i\tau\nu)^2-k_-^2|^{1/2}$$ for
$-(\xi+i\tau\nu)^2$ in a neighborhood of $-k_-^2$.
Elsewhere W is nonzero
and
analytic on $C-[-k_-^2,\infty)$ with continuous limits on
$[-k_-^2,\infty)$.
Since
$$\eqalign{|(\xi+i\tau\nu)^2-b^2|&=((\xi^2-\tau^2+b^2)^2 +
(2\tau\xi\cdot \nu)^2)^{1/2}\cr
&\geq {1\over 2}(||\xi|^2-\tau^2+b^2| +2\tau|\xi\cdot\nu|)\geq{\tau\over
2}
(||\xi|-\sqrt{\tau^2+b^2}| +|\xi\cdot\nu|),\cr}$$
it follows that $I(f,g)$ is bounded by a finite sum of terms of the form
$${C\over \tau^{1/2}}\int_{R^{n-1}\times R}{|\hat f(\eta,x_n)|(\int_R(1
+s^2)
|\hat g(\eta,s)|^2ds)^{1/2}\over
(||\eta|-\sqrt{\tau^2+b^2}|+|\eta\cdot\nu|)^\alpha}d\eta
dx_n, \eqno{(14)}$$
where $b^2 =0,k^2_-$, or $-\lambda_j,\ j=1,\dots,N$, and $\alpha = 1/2$
or 1.
The only complication in estimating (14) is that the sphere of
codimension
two where the denominator vanishes has radius $(\tau^2+b^2)^{1/2}$.
Thus to bound the
integral we need to have the numerator of the integrand bounded in the
variables normal to the sphere and integrable in the variables tangent
to
the sphere. For $\tau >2$ we introduce coordinates
$(t_1,t_2)=(\xi\cdot\nu,
|\xi|-(\tau^2+b^2)^{1/2})$ near $|\xi|-(\tau^2+b^2)^{1/2}=\xi\cdot\nu=0$.
Then the Jacobian of the
transformation from $\xi$ to $(t_1,t_2,\tau\omega^\prime)$, where
$\omega^\prime$ is the polar angular coordinate on the sphere
$|\xi|-(\tau^2+b^2)^{1/2}=\xi\cdot\nu=0$, will be bounded above and
below independently of $\tau$. Thus, letting $S_{t_1,t_2}$ be the
sphere of codimension two
obtained by fixing $(t_1,t_1)$ and $dA_{t_1,t_2}$ be the volume on it
induced from Euclidean measure, (14) is bounded by
$${C_1\over \tau^{1/2}}\int_{t_1^2+t_2^2<1}dt_1dt_2
\int_{S_{t_1,t_2}}{(\int_R(1+s^2)|\hat f(\eta,s)|^2ds)^{1/2}(\int_R(1
+s^2)
|\hat g(\eta,s)|^2ds)^{1/2}\over (|t_1| +|t_2|)^\alpha}dA_{t_1,t_2}$$
$$+{C_2\over \tau^{1/2}}\int_{t_1^2+t_2^2>1}(\int_R(1+s^2)|\hat
f(\eta,s)|^2ds)^{1/2}(\int_R(1 +s^2)
|\hat g(\eta,s)|^2ds)^{1/2}d\eta .\eqno{(15)}$$
The second integral in (15) is bounded by
$\tau^{-1/2}||f||_{2,1}||g||_{2,1}$. To bound the first integral we
observe that
$$\int_{S_{t_1,t_2}}(\int_R(1+s^2)|\hat
f(\eta,s)|^2ds)^{1/2}(\int_R(1 +s^2)
|\hat g(\eta,s)|^2ds)^{1/2}dA_{t_1,t_2}$$
$$\leq(\int_{R\times S_{t_1,t_2}}
(1+s^2)|\hat f(\eta,s)|^2dA_{t_1,t_2}ds)^{1/2}(\int_{R\times
S_{t_1,t_2}}
(1+s^2)|\hat g(\eta,s)|^2dA_{t_1,t_2}ds)^{1/2}.\eqno{(16)}
$$
The $L^2$-norm of a function $h$ over the sphere $|\xi|=r_0,\
\xi\cdot\nu
=s_0$ with $2<2s_0 0,$$
with the norms
$$|| f||_\mu= \left(\int_{R^n}e^{2\mu|x|}|f(x)|^2dx \right)^{1/2}.$$
Note that $f\in {\Cal A}_\mu$ implies that the partial Fourier
transform of $f$ satisfies
$$\int_R|\hat f(\xi,x_n)|e^{\mu|x_n|/3}dx_nR_T\}} \hat q(\xi -
\eta,x_n)
[(G+(\eta +z\nu)^2)^{-1}\hat f(\eta,\cdot)](x_n)d\eta,
$$
where $R_T$ is chosen large enough that $|\eta|>R_T-1$ implies
$-(\eta +\sigma\nu)^2+\tau^2 <\lambda_1-1$ on ${\Cal D}_{\epsilon
,T}$. Since this implies $(G+(\eta +z\nu)^{-1}$ is analytic on
${\Cal D}_{\epsilon,T}$ as an operator on $L^2(R)$ for $|\eta|>R_T$
with norm bounded by $C(1 +|\eta|^2)^{-1}$, it follows that the operator
$$[E_1(z)f](x) = \int_{\{|\eta|>R_T\}}e^{ix^\prime\cdot\eta}[(G+(\eta
+z\nu)^2)^{-1}\hat f(\eta,\cdot)](x_n)d\eta$$
is analytic on ${\Cal D}_{\epsilon,T}$ as an operator on $L^2(R^n)$.
Thus, since $|q(x)|
2\epsilon\}} \hat q(\xi-\eta)[(G+(\lambda +z\nu)^2)^{-1}\hat f(\eta,
\cdot)](x_n)d\eta,$$
then the analyticity of $(G-\lambda)^{-1}$ for Im$\{\lambda\}\neq 0$
allows us to re-use the preceding argument, and conclude that Prop.
4 holds for $A_{2,1}$.
\medskip
The analytic continuation of the remainder of $A(i\tau)$ is a little
more complicated, and makes essential use of the exponential decay of
$q$. To continue $A_{2,2}(i\tau)=A(i\tau)-A_1(i\tau)-A_{2,1}(i\tau)$
we will deform the integration in the following way. We introduce
cylindrical coordinates on $R^{n-1}$ by defining
$$\eta_\nu =\eta\cdot\nu,\ r=|\eta -\eta_\nu\nu|,\hbox{ and }
\omega = r^{-1}(\eta -\eta_\nu\nu).$$
Writing the integral defining $A_{2,1}(i\tau)$ as an iterated
integral over
$$\{\omega \in S^{n-2}\}\times\{|\eta_\nu|<2\epsilon\}\times\{
00$ the contour $\Gamma(\eta_\nu)$ is given by
$$\{s:0\leq s\leq a\}\cup\{a+it:0\leq t\leq b\}\cup\{s+ib:a\leq s\leq c
\}$$
$$\cup\{c+i(b-t):0\leq t\leq b\}\cup
\{s:c\leq s\leq(R^2_T-\eta^2_\nu)^{1/2}\}.$$
We now need to choose $a$, $b$ and $c$.
\medskip
The choice of $a$ and $c$ is
determined by the requirement that for $r$ on the segments $[0,a]$ and
and $[c, R_T]$ and $|\eta_\nu|\leq 2\epsilon$
$$[(G+(\eta +z\nu)^2)^{-1}\hat f(\eta,\cdot)](x_n)
=[(G + r^2 +(\eta_\nu +\sigma +i\tau)^2)^{-1}\hat f(\eta,\cdot)](x_n)$$
must be analytic in $z$ on ${\Cal D}_{\epsilon,T}$.
Our choice for $a$ is $a=k^2_+/2$. This insures that, taking
$\epsilon$ sufficiently small ($\epsilon \leq k_+/8$ will suffice), for $r\in
[0,a]$, $|\eta_\nu|\leq 2\epsilon$ and $z\in {\Cal D}_{\epsilon,T}$
we can define $[(G +(\eta +(\sigma +i\tau)\nu)^2)^{-1}(\hat
f(\eta,\cdot)](x_n)
$ as the analytic continuation
of $([G +(\eta +i\tau))^2)^{-1}\hat f(\eta,\cdot)](x_n)$,
across $[-k^2_+,\infty)$.
The estimate (18) implies that we may do this for
$f\in {\Cal A}_{\delta/3}$ as long as
$$|\hbox{Im}\sqrt{-(\eta + (\sigma +i\tau)\nu)^2+k^2_\pm}|\leq
\delta/4.\eqno{(19)}$$
Note that (19) will hold for $\epsilon$ sufficiently small for
$r\in [0,a]$, $|\eta_\nu|\leq 2\epsilon$ and
$z\in {\Cal D}_{\epsilon, T}$,
and this choice of $\epsilon$ can be made independent of $T$. For the segment
$[c,R_T]$ we will have this analyticity as long as $R_T$ is chosen large
enough that
$-r^2 +\tau^2 +16\epsilon^2 <\lambda_1$ for $|\eta|>R_T-4\epsilon^2$.
Our earlier choice of $R_T$ suffices for this, if $\epsilon $ is sufficiently
small.
We choose the constant $b$ so that the domain of integration stays
within the domain of analyticity of $\hat q(\xi - \eta,x_n)$ in $\eta$ for
$|\hbox{Im}\{\xi\}|\leq \delta/2$ and within the domain of analyticity
of $\hat f(\eta,x_n)$ for $f\in {\Cal A}_{\delta/3}$. In view of (17)
it suffices to take $b<\delta/9$ for this.
\medskip
Now we are ready to continue $A_{2,2}(i\tau)$ to ${\Cal D}_{\epsilon,T}$ as an
operator on ${\Cal A}_{\delta/3}$. The procedure now is completely
analogous to
the one used in Section 2 of [ER]. We begin with the integral representation
$$[A_{2,2}(i\tau)\hat f](\xi,x_n)= (2\pi)^{1-n}
\int_{\{|\eta|-k_+^2$ and
$|\hbox{Im} \{ (-r^2
-(\eta_\nu+\sigma +i\tau)^2+k^2_\pm)^{1/2}\}|<\delta/4$, when $0
\leq\tau\leq T$,
$|\eta_\nu|\leq 2\epsilon$, $|\sigma|\leq \epsilon$, and $r\in [0,k_+^2/2]$
or
$r=\hbox{sgn}(\eta_\nu)(
k_+^2/2 +i(b-t)$, where $0\leq t\leq b$,
\medskip
\noindent ii) Im$\{r^2 +(\eta_\nu +\sigma +i\tau)^2\}\neq 0$, when $\tau\geq 0$,
$|\eta_\nu|\leq 2\epsilon$, $|\sigma|\leq \epsilon$, and
$r=\hbox{sgn}(\eta_\nu)(
s +ib)$, where $k_+^2/2\leq s\leq c$, and
\medskip
\noindent iii) Re$\{-r^2 -(\eta_\nu +\sigma +i\tau)^2\}<\lambda_1$, when $0
\leq\tau\leq T$,
$|\eta_\nu|\leq 2\epsilon$, $|\sigma|\leq \epsilon$, and
$r=\hbox{sgn}(\eta_\nu)(
c +i(b-t)$, where $0\leq t\leq b$ or $r\in [c,R_T]$.
\medskip
\noindent Our choices of $a$, $b$ and $c$ guarantee that all of these
properties
hold except for i) and iii) on the short segments $r=k_+^2/2+i\hbox{sgn}(\eta_\nu)
t$, where $0\leq t\leq b$, and $r= c+i\hbox{sgn}(\eta_\nu)
(b-t)$, where $0\leq t\leq b$, respectively. To correct this we
just choose $b$ smaller and then take $\epsilon$ small enough that
ii) holds again.
\medskip
This completes the analytic continuation of $A_{2,2}$, and, since the
proof from [ER] applies to show that $A_{2,2}(z)$ is compact, we conclude that
Prop. 4 holds for $A_{2,2}(z)$, and hence for $A(z)$.
If $I + A(i\tau)$
had a nontrivial null space in ${\Cal A}_{\delta/3}$, it would have a
nontrivial
null space in $L^2_r(R^n)$, contradicting Prop. 3 for $\tau$ sufficiently
large.
Thus $(I + A(z))^{-1}$ is meromorphic on ${\Cal D}_{\epsilon,T}$. It is
also clear, as in
[ER], from the contruction of the analytic continuation that $A(z)$ extends
continuously to the operator $A(\sigma)$ in (7) as $\tau\to 0_+$. Thus we
have justified the analytic continuation argument in {\bf 2}.
\medskip
\medskip
\noindent {\bf Appendix}
\medskip
In this appendix we show that the scattering amplitude defined in
{\bf 1.} can be recovered from the asymptotics of the scattered wave $v$.
Because of the form of the leading terms in these asymptotic expansions,
it will always suffice to compute asymptotics modulo functions in $L^2(R^n)$.
Our starting point is the representation of the scattered wave $v$ in
terms of the "free" Green's function. This is given in (1), but in
view of (3) and (11) we also have
$v = E_0\tilde h$, where
$\tilde h$ is the inverse Fourier transform in $\eta$ of $h(\eta,x_n)$.
Prop. 1 shows that $\tilde h$ is in $L^2_r(R^n)$, and from the equation
$$\tilde h +qE_0\tilde h= -q\Phi,$$
one sees that $\tilde h$ belongs to ${\Cal A}_\mu$
for any $\mu <\delta$. Let $\chi\in C_0^\infty(R^{n-1})$ satisfy
$\chi(\eta)=1$ for $|\eta|<1 +\sqrt{-\lambda_1}$, and also let $\chi$
denote the operator multiplying the Fourier transform in $x^\prime$
by $\chi$. Since the norm of $(1-\chi(\eta))(G+\eta^2)^{-1}$ as
an operator from $L^2(R)$ to $L^2(R)$ is bounded uniformly in $\eta$,
one sees from (11) that $E_0(I-\chi)$ is bounded from $L^2(R^n)$ to
$L^2(R^n)$. Thus $E_0(I-\chi)\tilde h\in L^2(R^n)$. Likewise, let
$\beta\in C^\infty(R)$ satisfy $\beta(\lambda)=1$ for $\lambda <1$
and $\beta(\lambda)=0$ for $\lambda >2$, and also let $\beta$
denote the operator on $L^2(R)$ obtained by multiplying the
spectral representation, $\Psi f$ by $\beta$. One sees directly
that $E_0\chi(1-\beta)$ is bounded from $L^2(R^n)$ to $L^2(R^n)$,
and thus $E_0\chi(1-\beta)\tilde h \in L^2(R^n)$.
\medskip
For the remainder of $v$, i.e. $E_0\chi\beta\tilde h$, we have the
representation
$$[E_0\chi\beta\tilde h](x) = (2\pi)^{1-n}\int_{R^{n-1}}e^{ix^\prime
\cdot\eta}\chi(\eta)[(G+\eta^2 -i0)^{-1}\beta h(\eta,\cdot)](x_n)d\eta.
\eqno{(A.1)}$$
We can express $\chi(\eta)(G+\eta^2 -i0)^{-1}\beta h(\eta,\cdot)
$ in terms of the spectral representation $\Psi$:
$$(2\pi)^{n-1}[E_0\chi\beta\tilde h](x)=\sum_{\alpha
=1}^2\int_{R^{n-1}}\int^\infty_{-k^2_+}
e^{ix^\prime\cdot\eta}
\chi(\eta)\beta(\lambda){\phi_\alpha(x_n,\lambda)
[\Psi_\alpha h(\eta,\cdot)](\lambda)\over
\lambda +\eta^2-i0}d\lambda d\eta\ +$$
$$\int_{R^{n-1}}\int^{-k^2_+}_{-k^2_-}
e^{ix^\prime\cdot\eta}\chi(\eta){ \phi_3(x_n,\lambda)
[\Psi_3 h(\eta,\cdot)](\lambda)\over
\lambda +\eta^2-i0}d\lambda d\eta\ +$$
$$\sum_{j=1}^N\phi_4(x_n,\lambda_j)\int_{R^{n-1}}
e^{ix^\prime\cdot\eta}{\chi(\eta)
[\Psi_4h(\eta,\cdot)](\lambda_j)\over
\lambda_j +\eta^2-i0}d\eta.\eqno{(A.2)}$$
We index the integrals in (A.2) by the components of $\Psi$ that they
contain and write
$$[E_0\chi\tilde h](x)=I_1(x)+I_2(x)+I_3(x)+I_{4,1}(x)+\dots+
I_{4,N}(x).$$
In what follows we will usually suppress the cutoffs $\chi$ and
$\beta$ to simplify the notation, and ask the reader to remember
that all integrands can be assumed to have compact support in
$\eta$ and $\lambda$.
\medskip
To compute the asymptotics of the integrals $I_j(x)$ as $|x|\to\infty$ we will
use the following lemma.
\medskip
\noindent{\bf Lemma A.1} Assume that $w$ and $g$ are real-valued smooth
functions
with the properties:
\medskip
i) $\nabla g\neq 0$ on the level set $\{g=0\}$,
ii) the critical points of $w_0$,
the restriction
of $w$ to $\{g=0\}$, are nondegenerate, and
iii) $\nabla g\cdot \nabla w\neq 0$ at
each
of these critical points.
\medskip
\noindent Then for $f\in C^\infty_0(R^m)$ the function
$$I(r)=(2\pi)^{-m}\int_{R^m}{e^{irw(\xi)}f(\xi)\over g(\xi)-i0}d\xi$$
satisfies
$$I(r\theta)=\sum_{p\in S_+}C_p{e^{iw(p)r}\over r^{(m-1)/2}}(f(p) +
O(r^{-1}))$$
as $r\to \infty$. The set $S_+ =\{p\in \{g=0\}: p \hbox{ is a critical
point for } w_0,
\hbox{ and } \nabla w(p)\cdot \nabla g(p)>0\}$. The coefficient $C_p$ is
determined by
the gradients of $w$ and $g$ at $p$ and the Hessian of $w_0$. In the case
that
interests us most $g(\xi) = \xi^2-k^2$, $w(\xi)=\xi\cdot\theta,\ \theta\in
S^{m-1}$,
$S_+$ is the point $p=k\theta$ and $C_p=
C_m(k)
=(4\pi)^{-1}k^{(m-3)/2}(2\pi)^{(3-m)/2}\hbox
{exp}(-i\pi(m-3)/4)$.
\medskip
This lemma is well-known; it is contained in Theorem 8.3 of [AH], if
one takes $w$ as a coordinate function near $\{g=0\}$.
\medskip
\noindent This lemma applies directly with $m=n-1$ to the integrals
$I_{4,j}$, $j=1,\dots,N$,
yielding ($r=|x^\prime|,\theta =x^\prime/|x^\prime|$)
$$I_{4,j}(r\theta,x_n)=C_{n-1}((-\lambda_j)^{1/2})
{e^{ir(-\lambda_j)^{1/2}}\over
r^{(n-2)/2}}
(\phi_4(x_n,
\lambda_j)[\Psi_4h((-\lambda_j)^{1/2}\theta,\cdot)](\lambda_j)
+O(r^{-1})).\eqno{(A.3)}$$
Since the $\lambda_j$ and the eigenfunctions $\phi_j(x_n,\lambda_j)$
are assumed to be known, and the $\lambda_j$ are distinct,
one can determine
$[\Psi_4h((-\lambda_j)^{1/2}\theta,\cdot)](\lambda_j),\ j=1,\dots,N,\
\theta \in S^{n-2}$
from the asymptotics of $I_{4,1}(r\theta,x_n)+\dots +I_{4,N}(r\theta,x_n)$
as $r\to\infty$.
\medskip
We cannot apply Lemma A.1 directly to $I_\alpha (x),\ \alpha =1,2,3$,
because the functions $\phi(x_n,\lambda)[\Psi h(\eta,\cdot)](\lambda)$
have singularities at $\lambda=-k_\pm^2$. To make these singularities
explicit we need to know the normalizing coefficients $a_+$, $a_-$ and $a_0$.
These may be computed as in
[Wi], beginning with the identity
$$||P(E)f||^2 =\lim_{\epsilon\to 0_+}{1\over 2\pi i}\int_E(f,(G-(\lambda
+i\epsilon)I)^{-1}
f)-(f,G-(\lambda +i\epsilon)I)^{-1}f)d\lambda,$$
where $P(E)$ is the spectral family for $G$,
and then using the explicit formula for the kernel of $(G-\lambda)^{-1}$. The
results are
$$a_+(\lambda)={(\lambda +k_-^2)^{1/4}\over \sqrt{\pi}|W(\lambda)|},\
\lambda\in
[-k_+^2,\infty),$$
$$a_-(\lambda)={(\lambda +k_+^2)^{1/4}\over \sqrt{\pi}|W(\lambda)|},\
\lambda\in
[-k_+^2,\infty),\hbox{ and}
\eqno{(A.4)}$$
$$a_0(\lambda)={(\lambda +k_-^2){1/4}\over \sqrt{\pi}|W(\lambda)|},\
\lambda\in
[-k_-^2,-k_+^2].$$
Note that, since $W(\overline{\lambda})=\overline {W(\lambda)}$,
$|W(\lambda)|$
is well-defined for $\lambda$ real. Now, since the analogue of (10) holds for
$\phi_\alpha(s,\lambda)$, we see that for $\alpha =1,2$, near $\lambda =
-k_+^2$
$$\phi_\alpha(x_n,\lambda)[\Psi_\alpha h(\eta,\cdot)](\lambda)=
R_\alpha(x_n,\eta,\lambda,(\lambda+k_+^2)^{1/2}),\eqno{(A.5)}$$
where $R_1(x_n,\eta,\lambda, \beta)$ and $R_2(x_n,\eta,\lambda, \beta)$
are restrictions of smooth functions to $[-k_+^2,\infty)$. This follows
from Lemma 2, since only the squares of the coefficients $a_\pm$ enter in the
formulas, and from the rapid decay of $h(\eta,x_n)$ in $x_n$.
For $\alpha =3$ the situation is a little more complicated, since
now
both $\lambda =-k_+^2$ and $\lambda=-k_-^2$ are in the support. Lemma 2
implies that near $\lambda =-k_-^2$ we either have
$$(a_0(\lambda))^2 = f(\lambda,(\lambda+k_-^2)^{1/2}),$$
where $f(\lambda,\tilde\beta)$ is a smooth function with $f(-k_-^2,0)=0$ and
$\partial f/\partial\tilde\beta(-k^2_-,0)\neq 0$, or $(a_0(\lambda))^2$ is the
reciprocal of a function of this form. Thus we either have
$$\phi_3(x_n,\lambda)[\Psi_3 h(\eta,\cdot)](\lambda)=
R(x_n,\eta,\lambda,(\lambda+k_-^2)^{1/2})\eqno{(A.6)}$$
near $\lambda =-k^2_-$ with $R$ smooth, or
$$\phi_3(x_n,\lambda)[\Psi_3 h(\eta,\cdot)](\lambda)= (\lambda +k_-^2)^{-1/2}
R(x_n,\eta,\lambda,(\lambda+k_-^2)^{1/2}).\eqno{(A.6^\prime)}$$
However, $\phi_3(x_n,\lambda)[\Psi_3 h(\eta,\cdot)](\lambda)$ simply has the
form given in (A.5) near $\lambda =k_+^2$. This is the information on the singularities
that we will need.
\medskip
To study $I_\alpha (x),\ \alpha =1,2$ and $I_3$ when $x_n>0$, we first make
the
change of
variables $\lambda = -k^2_+ +\beta^2$ in $I_1(x)$ and $I_2(x)$, and $\lambda =
-k^2_+-\tilde \beta^2$ in $I_3(x)$. Then for $\alpha =1,2,$
$$I_\alpha(x)=\int_{R^{n-1}}\int^\infty_0
e^{ix^\prime\cdot\eta}
{\phi_\alpha(x_n,-k^2_++\beta^2)[\Psi_\alpha
h(\eta,\cdot)](-k^2_++\beta^2)\over
\eta^2+\beta^2-k^2_+-i0}2\beta d\beta d\eta$$
and
$$I_3(x)=\int_{R^{n-1}}\int^{(k_-^2-k_+^2)^{1/2}}_0
e^{ix^\prime\cdot\eta}
{\phi_3(x_n,-k^2_+-\tilde \beta^2)[\Psi_3
h(\eta,\cdot)](-k^2_+-\tilde \beta^2)\over
\eta^2-\tilde\beta^2-k^2_+-i0}2\tilde\beta d\tilde\beta d\eta.$$
>From (A.5) we see that the numerators of the integrands of $I_1(x)$ and
$I_2(x)$ are smooth, and the numerator of $I_3(x)$ is smooth near
$\tilde\beta=0$.
\medskip
We will only need the asymptotics of $I_\alpha
(r\theta),\ \alpha =1,2,3,$
as $r\to
\infty$ for $\theta\in S^{n-1},\ \theta_n\neq 0, -(1-(k_+/k_-)^2)^{1/2}$. The
results will
be uniform for $\theta$ in compact subsets of this set. Getting uniform
asymptotics on
all of $S^{n-1}$ would be much more difficult, cf. [C]. For $\theta_n>0$
the functions
$\phi_1(r\theta_n,\beta^2-k_+^2)$ and $\phi_2(r\theta_n,\beta^2-k_+^2)$
become linear
combinations of exp$(ir\theta_n\beta)$ and exp$(-ir\theta_n\beta)$ for $r$
sufficiently
large. The function $\phi_3(r\theta_n,-k_-^2 +\tilde\beta^2)$ becomes a
multiple of
exp$(-r\theta_n\tilde\beta)$. The exponential decrease of this function
implies that the only significant contributions to $I_3(r\theta)$ when
$\theta_n>0$ come from a neighborhood of $\tilde \beta=0$,
and we will see that
$I_3(r\theta)$
will not contribute to
the leading asymptotics of $v(r\theta)$ for $\theta_n>0$.
For $\theta_n<0$ and $r$ sufficiently large
the functions $\phi_1(r\theta_n,\beta^2-k_+^2)$ and
$\phi_2(r\theta_n,\beta^2-k_+^2)$ are linear combinations of
exp$(ir\theta_n(\beta^2
+k_-^2-k_+^2)^{1/2})$ and exp$(-ir\theta_n(\beta^2 +k_-^2-k_+^2)^{1/2})$.
Thus for $r$ sufficiently large $I_1(r\theta)$ and $I_2(r\theta)$ are
sums of integrals of the form
$$I(r\theta)=\int_{R^{n-1}}\int^\infty_0
e^{irw(\theta,\eta,\beta)}
{f(-k^2_++\beta^2)[\Psi_\alpha
h(\eta,\cdot)](-k^2_++\beta^2)\over
\eta^2+\beta^2-k^2_+-i0}2\beta d\beta d\eta,\eqno{(A.7)}$$
where $w(\theta,\eta,\beta)=\theta^\prime\cdot\eta \pm \theta_n\beta$ when
$\theta_n
>0$, and $w(\theta,\eta,\beta)=\theta^\prime\cdot\eta \pm
\theta_n(\beta^2+k_-^2-k_+^2
)^{1/2}$ when $\theta_n<0$.
\medskip
To study $I_3(x)$ when $x_n<0$ we need to remove the singularities in the
integrand. In
order to do this we split $I_3(x)$ into two integrals using a cutoff
$\rho(\lambda)$ which vanishes
near $\lambda=-k_-^2$, and is chosen so that $1-\rho(\lambda)$ vanishes near
$\lambda=-k_+^2$. In the integral containing $\rho$
we make the change of variables $\lambda
=-k_+^2-\tilde\beta^2$ and
integral containing $1-\rho$ we make the change of variables $\lambda
=-k_-^2+\hat\beta^2$. Then,
for $r$ sufficently large and $\theta_n<0$, $I_3(r\theta)$ is a sum of
integrals
either of the form
$$I(r\theta)=\int_{R^{n-1}}\int^\infty_0
e^{irw(\theta,\eta,\tilde\beta)}
{f(-k^2_+-\tilde\beta^2)[\Psi_\alpha
h(\eta,\cdot)](-k^2_+-\tilde\beta^2)\over
\eta^2-\tilde\beta^2-k^2_+-i0}2\tilde\beta d\tilde\beta d\eta\eqno{(A.8)}$$
with $w(\theta,\eta,\tilde\beta)=\theta^\prime\cdot\eta \pm
\theta_n(-\beta^2+k_-^2-k_+^2
)^{1/2}$, or of the form
$$I(r\theta)=\int_{R^{n-1}}\int^\infty_0
e^{irw(\theta,\eta,\hat\beta)}
{f(-k^2_-+\hat\beta^2)[\Psi_\alpha
h(\eta,\cdot)](-k^2_-+\hat\beta^2)\over
\eta^2+\hat\beta^2-k^2_--i0}2\hat\beta d\hat\beta d\eta\eqno{(A.9)}$$
with $w(\theta,\eta,\beta)=\theta^\prime\cdot\eta \pm \theta_n\hat\beta$.
Note that the factor of $\hat\beta$ in (A.9) cancels the singularity from
$(A.6^\prime)$.
\medskip
Now all the integrands that we have to consider (except for $I_3(r\theta)
$ when $\theta_n>0$) have the form required for Lemma A.1. However,
we still need to show that the endpoints at $\beta =0,\ \tilde\beta=0$ and
$\hat\beta =0$ do not contribute to the leading term in the asymptotics.
In (A.7) and (A.9) the level sets of the denominators of the integrands
are spheres. Since we are assuming that $\theta_n\neq
0,-(1-(k_+/k_-)^2)^{1/2}$,
the critical point of $w$ on the sphere where the denominator vanishes is
not at $\beta =0$ in (A.7) or $\hat \beta=0$ in (A.9). Introducing spherical
coordinates $(s,\omega)$ in (A.7) and (A.9) with
$\beta=\hat\beta=s\omega_n$,
one sees that $\partial w/\partial \omega_n\neq 0$ on
$(s,\omega)=(k_+,\omega^\prime,0)$
in (A.7) and on $(s,\omega)=(k_-,\omega^\prime,0)$ in (A.9). We can cut
the integrands off to small neighborhoods of $s =k_+$ in (A.7) and $s=k_-$ in
(A.9), and the discarded portions will be Fourier transforms of functions in
$L^2(R^n)$, and hence will not contribute to the leading asymptotics. Next we
write the remaining integrals as sums of integrals with integrands supported
in the interior of $\{\omega_n>0\}$ so that Lemma A.1 applies to them, and integrals with
integrands
with supports so close to $s=k_\pm,\ \omega_n=0$ that $\partial w/\partial
\omega_n\neq 0$
on them. In these integrals we can write exp$(irw)=(ir\partial w/\partial
\omega_n)^{-1}
\partial/\partial \omega_n$exp$(irw)$ and integrate by parts in $\omega_n$
. Each integration by parts gives an additional factor of $r^{-1}$
and a boundary term which is an integral over $\omega_n=0$. Lemma A.1 with
$m=n-1$ can be applied to the boundary terms -- and we can repeat the
integration by
parts as often as needed -- because integration by parts in $\omega_n$ does
not change the denominators. Thus the contributions from the the boundaries
$\beta =0$
and $\hat \beta =0$ will be $O(r^{-1+(2-n)/2})=O(r^{-n/2})$, and will not
contribute
to the leading term in the asymptotics.
\medskip
The argument of the preceding paragraph applies to the integrals (A.8) as
well. The
only change is that the level sets of the denominators are not spheres.
However, again
since we assume that $\theta_n\neq -(1-(k_+/k_-)^2)^{1/2}$, we can choose a
coordinate
$\tilde \omega_n$ near $\eta^2 =k_+^2,\ \beta=0$ transverse to
$\tilde\beta=0$
by parts such that $\eta^2 -\beta^2$ is independent of
$\tilde \omega_n$. Then we can repeat the argument of the preceding paragraph
using integration by parts in $\tilde\omega$. This shows that the contribution from the
boundary $\tilde \beta
=0$ will not contribute to the leading term in the asymptotics. The same
reasoning
shows that the
contribution of $I_3(r\theta)$ to the asymptotics when $\theta_n>0$ is
negligible as well.
\medskip
Applying Lemma A.1 to (A.7), and absorbing the terms from the
boundary $\beta =0$ in the $O(r^{-n/2})$ correction, we have from (A.7)
with $\theta_n>0$ modulo terms in $L^2(R^n)$
$$ I(r\theta)=C_n(k_+){e^{ik_+r}\over r^{(n-1)/2}}
(f(-k^2_+(1-\theta_n^2)) [\Psi_\alpha
h(k_+\theta^\prime,\cdot)](-k^2_+(1-\theta_n^2))
+ O(r^{-1/2})),\eqno{(A.10)}$$
when $w(\theta,\eta,\beta)=\theta^\prime\cdot\eta + \theta_n\beta$, and
$I(r\theta)=
O(r^{-n/2})$, when $w(\theta,\eta,\beta)=\theta^\prime\cdot\eta -
\theta_n\beta$, because
there are no critical points in the interior in this case.
When $\theta_n <0$, the contribution from the integral in (A.7) modulo
terms in $L^2(R^n)$ is
$$I(r\theta)=C_n(k_-){e^{ik_-r}\over r^{(n-1)/2}}(f(-k_-^2(1-\theta_n^2))
[\Psi_\alpha(k_-\theta^\prime,\cdot)](-k^2_-(1-\theta_n^2))
+ O(r^{-1/2})),\eqno{(A.11)}$$
when $w(\theta,\eta,\beta)=\theta^\prime\cdot\eta -
\theta_n(\beta^2+k_-^2-k_+^2
)^{1/2}$, and $I(r\theta)=
O(r^{-n/2})$, when $w(\theta,\eta,\beta)=\theta^\prime\cdot\eta +
\theta_n(\beta^2+k_-^2-k_+^2
)^{1/2}$.
Likewise applying Lemma A.1) to (A.9) and absorbing the contributions
from the
boundary $\hat\beta=0$ in $O(r^{n/2})$, we have modulo terms in $L^2(R^n)$
$$I(r\theta)=C_n(k_-){e^{irk_-}\over
r^{(n-1)/2}}(f(-k_-^2(1-\theta_n^2))
[\Psi_3h(k_-\theta^\prime,\cdot)](-k^2_-(1-\theta_n^2))
+ O(r^{-1/2})),\eqno{(A.12)}$$
when $w(\theta,\eta,\beta)=\theta^\prime\cdot\eta -
\theta_n(\beta^2+k_-^2-k_+^2
)^{1/2}$, and $I(r\theta)=O(r^{-n/2})$ when
$w(\theta,\eta,\beta)=\theta^\prime\cdot\eta +\theta_n(\beta^2+k_-^2-k_+^2
)^{1/2}$. Finally, since $\theta_n\neq -(1-(k_+/k_-)^2)^{1/2}$, we can
assume that the
support of $1-\rho$ was chosen sufficiently small that there are no
critical points
of $w_0$ in the support of the integrand in (A.8), and thus the contribution
from (A.8)
is $O(r^{-n/2})$.
\medskip
Using (A.10-12) we can now give the leading asymptotics of
$[E_0\chi\beta\tilde h](r\theta)$ for $\theta_n \neq 0,-(1-(k_+/k_-)^2)^{1/2}$. Note
that
the terms $I_{4,j}(r\theta)$ are exponentially decreasing in this case, and do
not contribute.
To specify the functions $"f(-k^2_\pm(1-\theta_n^2)^{1/2})"$ which appear, we
introduce
the following notation. For $ss_+$
$$\phi_2(s,\lambda) =a_2(\lambda)\hbox{exp}(is(\lambda+k_+^2)^{1/2})
+b_2(\lambda)\hbox{exp}(-is(\lambda+k_+^2)^{1/2}).$$
and for $s0$ we have
$$[E_0\chi\beta\tilde h](r\theta) =C_n(k_+){e^{ik_+r}\over
r^{(n-1)/2}}(a_+(-k^2_+(1-\theta_n^2))
[\Psi_1h(k_+\theta^\prime,\cdot)](-k^2_+(1-\theta_n^2))+$$
$$a_2(-k^2_+(1-\theta_n^2))
[\Psi_2h(k_+\theta^\prime,\cdot)](-k^2_+(1-\theta_n^2))+O(r^{-1/2})),\eqno{(
A.13)}$$
and for $\theta_n<-(1-(k_+/k_)^2)^{1/2}$
$$[E_0\chi\beta\tilde h](r\theta)=C_n(k_-){e^{ik_-r}\over
r^{(n-1)/2}}(b_1(-k^2_-(1-\theta_n^2))
[\Psi_1h(k_-\theta^\prime,\cdot)](-k^2_-(1-\theta_n^2))+$$
$$a_-(-k^2_-(1-\theta_
n^2))
[\Psi_2h(k_-\theta^\prime,\cdot)](-k^2_-(1-\theta_n^2))+O(r^{-1/2})),\eqno{(
A.14)}$$
For the remaining interval, $-(1-(k_+/k_-)^2)^{1/2} <\theta_n<0$, we have
$$[E_0\chi\beta\tilde h](r\theta)=
C_n(k_-){e^{ik_-r}\over
r^{(n-1)/2}}(\overline {a_3(-k_-^2(1-\theta_n^2))}
[\Psi_3h(k_-\theta^\prime,\cdot)](-k^2_-(1-\theta_n^2))
+ O(r^{-1/2})),\eqno{(A.15)}$$
Since $a_3(\lambda) \neq 0$ for $-k_-^2<\lambda<-k_+^2$, we conclude from
(A.15) that $\Psi_3h(\xi,\cdot)](-\xi^2)$ is determined by the
asymptotics of $v$. To reach this conclusion for
$[\Psi_\alpha h(\xi,\cdot)](-\xi^2),\ \alpha =1,2$, using (A.13) and (A.14),
it now suffices to know that the determinant of the matrix
$$\left(\matrix a_+(\lambda) & a_2(\lambda)\\
b_1(\lambda) & a_-(\lambda)\endmatrix\right)$$
does not vanish for $\lambda >-k_+^2$. Computing the Wronskian of
$\phi_1(s,\lambda)$
and $\overline{\phi_2(s,\lambda)}$ for $ss_+$, and
equating the results
(as in Lemma 1.1(iii) of [CK]), we have for $\lambda >-k_+^2 $
$$-a_+(\lambda)\overline {a_2(\lambda)}(\lambda
+k_+^2)^{1/2}=a_-(\lambda)b_1(\lambda)
(\lambda +k_-^2)^{1/2}.$$
Thus the determinant is strictly positive for $\lambda >-k_+^2$.
\medskip
All that we still need to do to complete this argument is show that
$I_1(x)+I_2(x)+I_3(x)$ decays sufficiently rapidly as $|x^\prime|\to
\infty$ that
$\Psi_4 h(\xi,\cdot)](-\xi^2)$ will be determined by the asymptotics given
in (A.3).
For this it suffices to show that $I_1(x)+I_2(x)+I_3(x)$ is a sum of terms
which are
square-integrable over the slab $\{0- 2\epsilon$. Applying Lemma A.1 with
$m=n-1$ to the
integration in $\eta$ in the integrals over $\lambda \leq-\epsilon<0$, for
$\alpha =1,2,3$,
we get a sum of terms of the form
$$r^{(2-n)/2}\int^{-\epsilon}_{-k^2_\pm} e^{i(-\lambda)^{1/2}r}
C_{n-1}((-\lambda)^{1/2}\phi_\alpha(x_n,\lambda)
[\Psi_\alpha h((-\lambda)^{1/2})\theta^\prime,\cdot)](\lambda)d\lambda,$$
plus lower order terms. These are $o(r^{(2-n)/2})$ by the Riemann-Lebesgue
lemma.
Finally, in the integrals over
$\lambda>-2\epsilon$ we may integrate by parts
in $\lambda$ reduce the singularity from $(\eta^2+\lambda-i0)^{-1}$ to
log$(\eta^2+\lambda +i0)$. Thus these integrals also give terms which
are
square-integrable over $0~~