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Inverse scattering on  the line
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\begin{document}

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\title{Inverse Scattering with Partial Information on the Potential
\thanks{Mathematics Subject Classification (2000): 34L25 34A55 81U40;
PACS Numbers (1999): 02.60.Lj 03.65.Nk;
Research partially supported by the National Science Foundation under grant 
DMS-9803219 and by Universidad Nacional Aut\'onoma de
M\'exico under Proyecto PAPIIT-DGAPA IN 105799.}}

\author{Tuncay Aktosun \\ Department of Mathematics\\ North Dakota State University
\\ Fargo, ND 58105, USA\\ Tuncay\_Aktosun@ndsu.nodak.edu
\\
\\
\and
Ricardo Weder\thanks{Fellow Sistema Nacional de 
Investigadores}\\
Instituto de Investigaciones en Matem\'aticas Aplicadas y en Sistemas\\
Universidad Nacional Aut\'onoma de M\'exico
\\Apartado Postal 20-726\\ M\'exico DF 01000\\
weder@servidor.unam.mx}
\date{}
\maketitle


\begin{center}
\begin{minipage}{5.75in}
\centerline{{\bf Abstract}}\bigskip

The one-dimensional Schr\"{o}dinger equation is considered
when the potential is real valued and integrable and has a finite
first moment. The recovery of such a potential is analyzed
in terms of the scattering data consisting of a 
reflection coefficient, all the
bound-state energies, knowledge of the potential
on a finite interval, and all of the
bound-state norming constants except one.
It is shown that a missing norming constant in the data can cause at most
a double nonuniqueness in the recovery.
In the particular case when the missing norming constant
in the data corresponds to the lowest-energy bound state,
the necessary and sufficient conditions are obtained for the
nonuniqueness, and
the two norming constants and the corresponding 
potentials are determined.
Some explicit examples are provided to illustrate
the nonuniqueness.

\end{minipage}
\end{center}

\newpage
\section{Introduction}\sss

Consider the one-dimensional Schr\"odinger equation
\beq
-\psi''(k,x)+ V(x)\,\psi(k,x) = k^2\psi(k,x),
\qquad x\in\bR,
\label{1.1}
\ene
where the potential $V$ is real valued and belongs to
$L^1_1(\bR),$
the set of measurable potentials
such that
$\int_{-\infty}^\infty dx\, (1+|x|)\, |V(x)|$ is finite.
In our notation, the prime denotes the derivative with respect to
the spatial variable $x,$
$\bCp$ is the upper-half
complex plane, 
${\bR}^-:=(-\infty,0),$
and ${\bR}^+:=(0,+\infty).$
For a given subset $J$ of
$\bR,$ we will use $V|_J$
to denote the fragment of $V$ supported on $J,$ i.e.
$$V|_J(x):=\cases {V(x),& $x\in  J,$\cr\cr
0,& $x\notin J.$}$$

Recall \cite{1}-\cite{6} that the scattering solutions of (1.1)
are asymptotic to linear combinations
of $e^{\pm ikx}$ as $x\to-\infty$ and 
$x\to+\infty,$ and they occur for
all $k\in{\bR}\setminus\{0\}.$
Among such solutions are the
Jost solution from the left, $f_{\rm l}(k,x),$ and the
Jost solution from the right, $f_{\rm r}(k,x),$ satisfying
the boundary conditions
$$e^{-ikx}\,f_{\rm l}(k,x)=1+o(1),\quad
e^{-ikx}\,f'_{\rm l}(k,x)=ik+o(1),\qquad
x\to+\infty,$$
$$e^{ikx}\,f_{\rm r}(k,x)=1+o(1),\quad
e^{ikx}\,f'_{\rm r}(k,x)=-ik+o(1),\qquad
x\to-\infty.$$
 From the spatial asymptotics
\beq
f_{\rm l}(k,x)=\ds\frac{e^{ikx}}{T(k)}+\ds\frac{L(k)\,
e^{-ikx}}{T(k)}+o(1),\qquad x\to-\infty,
\label{1.2}
\ene
\beq
f_{\rm r}(k,x)=\ds\frac{e^{-ikx}}{T(k)}+\ds\frac{R(k)\,e^{ikx}}{T(k)}+o(1),
\qquad x\to+\infty,
\label{1.3}
\ene
we obtain the scattering coefficients,
namely, the transmission coefficient $T,$
and the reflection coefficients
$L$ and $R,$ from the left and right, respectively.
We have
\beq
R(k)\, T(k)^* = -L(k)^* \, T(k), \qquad k \in \bR ,
\label{1.4}
\ene
where the asterisk denotes complex conjugation.

The bound-state solutions
decay exponentially as $x\to\pm\infty,$
and they can occur only at certain $k$ values
on the positive imaginary axis in $\bCp.$
The number of bound states is finite, and we use
$N$ to denote that number.
We let the bound states occur at $k=i\kappa_j$ with
$0<\kappa_1<\dots<\kappa_N.$
Each bound state at $k=i\kappa_j$
is simple, i.e.
there is only one linearly independent bound-state
solution of
(1.1) at $k=i\kappa_j.$ 
The bound-state norming constants 
$c_{{\rm l}j}$ from the left and $c_{{\rm r}j}$ from the right, respectively,
are defined as
$$c_{{\rm l}j}:=\left[\ds\int_{-\infty}
^\infty dx\, f_{\rm l}(i\kappa_j,x)^2\right]^{-1/2},\qquad
c_{{\rm r}j}:=\left[\ds\int_{-\infty}
^\infty dx\, f_{\rm r}(i\kappa_j,x)^2\right]^{-1/2} ,$$
and they are
related to each other via the residues of $T$ as
\beq
\hbox{Res}\,(T,i\kappa_j)=i\,c_{{\rm l}j}^2\,\gamma_j
=i\,\ds\frac{c_{{\rm r}j}^2}{\gamma_j},
\label{1.5}
\ene
where the $\gamma_j$ are the dependency constants
defined as
\beq
\gamma_j:=\ds\frac{f_{\rm l}(i\kappa_j,x)}{f_{\rm r}(i\kappa_j,x)},
\qquad x\in\bR.
\label{1.6}
\ene
Recall \cite{2} that when $\beta\in(0,+\infty),$ the quantity
$1/T(i\beta)$ is real and continuous,
has simple zeros at $\beta=\kappa_j,$ and behaves like
$1+O(1/\beta)$
as $\beta\to+\infty.$ Thus, $(-1)^NT(i\beta)>0$ when
$\beta\in(0,\kappa_1),$ $T(i\beta)>0$ when $\beta>\kappa_N,$ and
$(-1)^{N-j+1}T(i\beta)>0$ when
$\beta\in(\kappa_{j-1},\kappa_j)$ for $j=2,\dots,N.$ Hence, with the help
of (\ref{1.5}), it is seen that $\gamma_j=(-1)^{N-j}c_{{\rm r}j}/c_{{\rm l}j}.$ 

It is already known \cite{1}-\cite{6}
that $V$ is uniquely determined 
by either the left scattering data \linebreak
$\{R,\{\kappa_j\},\{c_{{\rm l}j}\}\}$ or by
the right scattering data
$\{L,\{\kappa_j\},\{c_{{\rm r}j}\}\}.$
Given $\{|R|,\{\kappa_j\}\},$ we can construct
$T$ explicitly (e.g. see Lemma~2.5 of \cite{2}).
Then, when $\{R,\{\kappa_j\}\}$ is known, from (\ref{1.5})
it follows that there is a one-to-one
correspondence between each left norming constant
$c_{{\rm l}j}$ and the dependency constant
$\gamma_j;$ similarly, given
$\{L,\{\kappa_j\}\},$ each $c_{{\rm r}j}$ determines
$\gamma_j$ uniquely and vice versa.
Hence, $V$ is uniquely determined
by either $\{R,\{\kappa_j\},\{\gamma_j\}\}$ or
$\{L,\{\kappa_j\},\{\gamma_j\}\}.$

We are interested in the following inverse problem.
Suppose we know one reflection
coefficient, the bound-state energies $-\kappa_j^2,$
and the potential $V$ on a finite interval $I$ of positive length.
We may also know some but not all of
the bound-state norming constants.
In other words, we are missing
the full information on
the norming constants for a unique determination
of the potential, and we would like to know if
knowledge of $V$ on $I$ compensates for the
missing information on the
norming constants.
We wish to find out
if our data determines the potential uniquely on the entire
line or if there are two or more potentials corresponding to our data.
Clearly, if there are
no bound states, $V$ is uniquely
determined by $R$ or $L$ even without 
knowing a fragment of the potential.
Hence, our inverse problem
has relevance only in the
presence of bound states. 

If the interval
in which the potential is known is a 
half line, then we already know 
\cite{7}-\cite{13}
the answer to our question;
namely, if $V|_{\bR^+}$
and $R$ are known, then
$V$ is uniquely determined without
the bound-state information. In fact,
the bound-state energies and norming
constants are uniquely determined
by $\{R,V|_{\bR^+}\}.$
On the other hand, $V$
is not uniquely determined by
$\{L,V|_{\bR^+}\}$ alone.
However, if we know $L$ and all the bound-state
energies, then we can construct
the corresponding $R$ (e.g. see Lemma~2.5 of \cite{2}). Thus,
$V$ is uniquely determined by
$\{L,V|_{\bR^+},\{\kappa_j\}\}.$
In a similar way, $V$ is
uniquely determined by $\{L,V|_{\bR^-}\}$ without
the bound-state information,
and in fact, the bound-state energies and norming
constants are
uniquely determined
by $\{L,V|_{\bR^-}\};$
however, $V$
is not uniquely determined by
$\{R,V|_{\bR^-}\}$ alone,
but is uniquely determined
by $\{R,V|_{\bR^-},\{\kappa_j\}\}.$
Therefore, our inverse problem needs to be studied
only when our interval is finite.

In this paper we analyze and solve 
our inverse problem when only one norming constant
is missing from the scattering data. If the data
lacks two or more norming constants, the
problem is still open; in that case, it would be
desirable either to produce an example where
infinitely many distinct potentials
correspond to the same scattering data or to prove that
there can only be a finite number of such potentials
and to determine an upper bound for that number.

Let us say a few words about the 
existence aspect of our inverse problem.
Clearly, we must expect a severe restriction on the 
fragment of the potential contained in our data.
We cannot specify that fragment arbitrarily,
and in general a solution to our inverse problem does not
exist. In our paper, we are only interested
in the uniqueness aspect of our inverse problem;
namely, we assume that there exists at least one potential
corresponding to our data and we would like to find out if 
there are more than one.

Our paper is organized as follows.
In Section~2 we consider the scattering
data consisting of one reflection coefficient,
knowledge of the potential on a finite interval $I,$
all the bound-state energies, and all the norming
constants except one; we show that
a missing norming constant in our data can cause at most a double
nonuniqueness in the determination of the potential.
In Section 3 we analyze in detail the uniqueness and nonuniqueness when
our data lacks only 
the norming constant
corresponding to the lowest-energy bound state;
we give the necessary and sufficient
conditions on the scattering data so that it corresponds to
two distinct potentials,
and we also determine the
corresponding dependency constants and
those two potentials.
In Section 4 we illustrate, with explicit examples, various cases of
uniqueness and nonuniqueness.

\section{Nonuniqueness in the General Case}\sss

When exactly one norming constant is missing from our data,
as the next theorem shows,
there can be at most
two distinct potentials corresponding to that data.

\begin{theorem} Let $V$ be a real-valued
potential belonging to $L^1_1(\bR),$ and consider
the scattering data
consisting of one reflection coefficient,
all the $N$ bound-state energies, knowledge of
$V$ on a finite interval $I$ of positive length, and $N-1$ 
of the bound-state norming constants.
Then, besides $V,$ there can be at most
one other potential corresponding to this scattering data.
\end{theorem}

\noindent PROOF: Without loss of generality
it is enough to give the proof
when our data consists of $R(k)$ for $k\in\bR,$
the $N$ constants $\kappa_j,$ knowledge of $V|_I,$
and $N-1$ norming constants.
Let us use the superscript $[0]$ to indicate the 
quantities associated with the potential
$V^{[0]}$ that is obtained from $V$ by removing all
the $N$ bound states. Note \cite{2} that $V^{[0]}\in
L^1_1(\bR)$ and it can uniquely be constructed
 from the corresponding right reflection
coefficient $R^{[0]}$ that is given by
$$R^{[0]}(k)=(-1)^N\left(\prod_{j=1}^{N} \frac{k-i\kappa_j}{k+i\kappa_j}\right)
R(k).$$
The transmission coefficient $T^{[0]}$ 
corresponding to $V^{[0]}$ can be uniquely
constructed \cite{2} from $|R(k)|$ alone, and we have
\beq
T^{[0]}(k)= \left(\prod_{j=1}^{N} \frac{k-i\kappa_j}{k+i\kappa_j}\right)
T(k).\label{2.1}
\ene
Define
\beq
\alpha_j:=\frac{2 (-1)^{j-1}\kappa_j}{c_{{\rm l}j}^2}
\left(\prod_{m\neq j}^N\, \frac{\kappa_m +\kappa_j}{\kappa_m-\kappa_j}\right) 
T^{[0]}(i\kappa_j),\qquad j=1,2, \dots ,N.
\label{2.2}
\ene
 From (\ref{1.5}), (\ref{2.1}), and (\ref{2.2}), it follows that
\beq
\alpha_j=|\gamma_j|=(-1)^{N-j}
\gamma_j=
\frac{(-1)^{N-j}\,\hbox{Res}\,(T,i\kappa_j)}{i\,c_{{\rm l}j}^2}
=\frac{(-1)^{N-j}\,i\,c_{{\rm r}j}^2}{\hbox{Res}\,(T,i\kappa_j)}
,\qquad j=1,\dots,N.\label{2.3}
\ene
Thus, in our data, for each fixed $j,$
we can use any one of
$c_{{\rm l}j},$ $c_{{\rm r}j},$
$\alpha_j,$ and
$\gamma_j$ interchangeably. So, instead of the $j$th norming constant,
we can simply use $\alpha_j$ in our analysis.
In fact, each $\alpha_j$ acts as a dependency constant 
(cf. (\ref{1.6})) in the sense that
\beq
\alpha_j=\ds\frac{f^{[j]}_{\rm l}(i\kappa_j,x)}{f^{[j]}_{\rm r}(i\kappa_j,x)},
\qquad x\in\bR,
\label{2.4}
\ene
where $f^{[j]}_{\rm l}$ and $f^{[j]}_{\rm r}$ are the Jost solutions
 from the left and from the right, respectively, for the
potential $V^{[j]};$
here $\{V^{[j]}\}_{j=0}^N$ is
the sequence of potentials given in Theorem~3.6
of \cite{2}. Note that $V^{[N]}=V$ and that $V^{[j]}$
is obtained from $V^{[j-1]}$ by
adding the bound state at $k=i\kappa_j.$

Define
\beq
\omega_j(x):= (-1)^{j+1} f^{[0]}_{\rm l}(i\kappa_j,x) +\alpha_j\,
f^{[0]}_{\rm r}(i\kappa_j,x)
,\qquad j=1,\dots,N,
\label{2.5}
\ene
and let $\Gamma(k,x)$ be
the $N\times N$ matrix whose entries are defined as
$$\Gamma_{2j-1,s}(k,x):= \kappa_s^{2j-2}\, \omega_s(x),
\qquad \Gamma_{2j,s}(k,x):=\kappa_s^{2j-2}\, \omega'_s(x).$$
Let us use the absolute value bars
to denote the matrix determinant.
We have \cite{2}
\beq
V(x)=V^{[0]}(x)- 2\frac{d}{dx} \left[
\frac{|\Gamma(x)|'}{|\Gamma(x)|}\right],\qquad
x\in\bR.
\label{2.6}
\ene  

Now let us assume that our data contains all the dependency
constants except one, namely $\alpha_m.$
In view of (\ref{2.5}) and (\ref{2.6}),
we need to determine how many distinct positive
$\alpha_m$ values can be found in such a way that
\beq
V(x)=V^{[0]}(x)-2\frac{d}{dx} \left[
\frac{|\Gamma(x)|'}{|\Gamma(x)|}\right],\qquad x \in I.
\label{2.7}
\ene

When we replace $\omega_m(x)$ by $f^{[0]}_{\rm r}(i\kappa_m,x)$
in $\Gamma(x),$ let us denote the resulting matrix by $A(x).$ Similarly, let
$B(x)$ denote the matrix obtained from $\Gamma(x)$
by replacing $\omega_m(x)$ by $(-1)^{m+1}f^{[0]}_{\rm l}(i\kappa_m,x).$
As seen from (\ref{2.5}) we have
\beq
\Gamma(x)=\alpha_m\, A(x)+B(x).
\label{2.8}
\ene
The elements in the first row of $A(x)$ are
eigenfunctions of the Schr\"{o}dinger equation with distinct eigenvalues, and hence
they are linearly independent on $\bR.$
The same property holds also for the elements in the first row of
$B(x)$ and in the first row of $\Gamma(x).$
Consequently, none of
$|A(x)|,$ $|B(x)|,$ and $|\Gamma(x)|$ can vanish
identically on the interval $I.$ In fact, because
$V^{[0]}$ has no bound states, for any $1\le j\le N$ we have
$f^{[0]}_{\rm l}(i\kappa_j,x)>0$ and
$f^{[0]}_{\rm r}(i\kappa_j,x)>0$ on $\bR;$
by using induction (cf. pp.~179--180 of \cite{2})
one can prove that 
these three determinants are strictly positive for $x\in\bR.$

By letting
$$G(x):= - \frac{1}{2} \left[V(x)- V^{[0]}(x)\right],$$
we write (\ref{2.7}) as
\beq
G(x)\, |\Gamma(x)|^2 = |\Gamma(x)|''-(|\Gamma(x)|')^2,\qquad x \in I.
\label{2.9}
\ene
Let us suppress the $x$-dependence in the rest of the proof.
 From (\ref{2.5}) and (\ref{2.6}), we see that $|A|'$ is equal to
the determinant of the matrix that is obtained
by taking the derivative of only the last row of $A;$
similar remarks also apply to $|A|''$ and to
$|B|',$ $|\Gamma|',$ and their derivatives.
With the help of (\ref{2.8}),
we can write (\ref{2.9}) as
\begin{eqnarray}
\left(G\,|A|^2- |A|''\,| A| +(|A|')^2 \right)\alpha_m^2
+
\! \! \! \!
&\left(2 G\,|A|\,|B|+ 2|A|'\,|B|'-|A|''\,|B|-
|A|\,|B|''\right) \alpha_m \nonumber \\
&
+ \left(G|B|^2 -|B|''\,|B|
+(|B|')^2 \right)=0,\qquad  x \in I.
\label{2.10}
\end{eqnarray}
Let us analyze
(\ref{2.10}) as a quadratic equation
in $\alpha_m$ where the coefficients
depend on $x\in I.$
It has at most two distinct positive solutions 
(that are independent of $x\in I$) unless
the coefficients are identically zero for all $x\in I,$
in which case (\ref{2.7}) would hold for all 
$\alpha_m >0$ and there would be infinitely many
distinct potentials corresponding to our data.
We prove below that this is not possible. 

If the coefficients in (\ref{2.10}) vanished
for all $x \in I,$ since $|A(x)|>0,$ $|B(x)|>0,$ and
$|\Gamma(x)|>0,$ we would have for $x\in I$
$$G=\frac{|A|''}{|A|}- \left( \frac{|A|'}{|A|}\right)^2=
\left( \frac{|A|'}{|A|} \right)^{\prime},$$
$$2G=\frac{|A|''}{|A|}+ \frac{|B|''}{|B|}-2
\frac{|A|'\, |B|'}{|A|\, |B|}= \left( \frac{|A|'}{|A|}\right)^{\prime}+
\left( \frac{|B|'}{|B|}\right)^{\prime} 
+\left( \frac{|A|'}{|A|}- \frac{|B|'}{|B|}\right)^2,$$
$$G=\frac{|B|''}{|B|}- \left(\frac{|B|'}{|B|}\right)^2=
\left( \frac{|B|'}{|B|}\right)^{\prime},$$
which, with the help of (\ref{2.9}), would imply that
\beq
G=\left(\frac{|\Gamma|'}{|\Gamma|}\right)'= \left( \frac{|A|'}{|A|} 
\right)'=\left( \frac{|B|'}{|B|} \right)', \quad \frac{|A|'}{|A|}=
\frac{|B|'}{|B|},\qquad x\in I.
\label{2.11}
\ene
The second
equation in
(\ref{2.11}) would imply that $|A|$ and $|B|$ are linearly dependent 
on $I,$ and this would mean that, for some constant $c,$ the matrix
$A-cB$ would have zero determinant for all $x\in I.$ However,
this is impossible because the entries in the
first row of $A-cB$ are eigenvectors of
the Schr\"odinger operator with distinct eigenvalues and
hence are linearly independent on
$I.$ Thus, (\ref{2.10})
can at most have two distinct positive roots. \qed

\section {Characterization of the Nonuniqueness}\sss

Let us use the notation introduced below
(\ref{2.4}); namely, $f_{\rm l}^{[j]}$ and $f_{\rm r}^{[j]}$
denote the Jost solutions for the potential
$V^{[j]}$ obtained from $V^{[j+1]}$ by removing
the bound state at $k=i\kappa_j,$ where
$V^{[N]}:=V$ In terms of the $\alpha_j$ appearing in
(\ref{2.2})-(\ref{2.4}), let us define
\beq
g^{[j]}(x):=f_{\rm l}^{[j-1]}(i\kappa_j,x)+\alpha_j\, f_{\rm r}^{[j-1]}(i\kappa_j,x),
\qquad j=1,\dots,N.
\label{3.1}
\ene
Note that $g^{[j]}(x)>0$ on $\bR$ for any $\alpha_j\ge 0$
because it is known \cite{2} that $f_{\rm l}^{[j-1]}(i\kappa_j,x)>0$ and
$f_{\rm r}^{[j-1]}(i\kappa_j,x)>0$ on $\bR.$
According to the
Darboux transformation formulas \cite{2}, \cite{5} we have
\beq
V^{[j]}(x)=-V^{[j-1]}(x)-2\kappa_j^2
+2\left(\frac{g^{[j]\prime}(x)}{g^{[j]}(x)}\right)^2,
\qquad j=1,\dots,N,\quad x\in\bR.
\label{3.2}
\ene
Thus, we can unambiguously define the (nonnegative and real-valued)
quantities
\beq
\Lambda^{[j]}(x):=\ds\frac{1}{\sqrt{2}}
\ds\sqrt{V^{[j]}(x)+V^{[j-1]}(x)+2\kappa_j^2},
\qquad j=1,\dots,N.
\label{3.3}
\ene

In Theorem~2.1 we have seen that a missing norming constant
in our data can cause at most a double nonuniqueness. We will now analyze
how this happens
when the norming constant missing in our data
corresponds to the lowest-energy bound state.

\begin{theorem}
Assume $V$ is real valued, belongs to
$L^1_1(\bR),$ and has $N$ bound states
at $k=i\kappa_j$ with
$0<\kappa_1<\dots<\kappa_N.$
Consider the scattering data consisting of
one reflection coefficient,
knowledge of $V$ on a finite interval
$I$ of positive length, all the $N$ bound-state energies,
and all the $N$ norming
constants except for the one corresponding
to the $N$th bound state. 
Then, either $V$ is the only potential on
$\bR$ corresponding to this data, or there is
exactly one other such potential;
the latter happens if and only if all the following four conditions are satisfied.
\item{{\rm (i)}} $V|_I\equiv V^{[N-1]}|_I.$
\item{{\rm (ii)}} $V|_I(x)$ is constant.
\item{{\rm (iii)}} $V|_I(x)> -\kappa_N^2.$
\item{{\rm (iv)}} The quantities $\alpha_{N;\pm}$ given in
{\rm (\ref{3.10})} are finite and strictly positive
at any one particular $x$ value in $I.$

When {\rm (i)-(iv)} are satisfied, the two potentials
corresponding to the aforementioned scattering data are
given by 
\beq 
V(x;\alpha_{N;\pm})=-V^{[N-1]}(x)-2\kappa_N^2
+2\left(\frac{f_{\rm l}^{[N-1]\prime
}(i\kappa_N,x)+\alpha_{N;\pm}\,f_{\rm r}^{[N-1]\prime
}(i\kappa_N,x)}
{f_{\rm l}^{[N-1]}(i\kappa_N,x)+\alpha_{N;\pm}\,f_{\rm r}^{[N-1]}(i\kappa_N,x)}
\right)^2,\qquad
x\in\bR.\label{3.4}
\ene
\end{theorem}

\noindent PROOF:
Our data uniquely determines (see e.g. Theorems~3.3 and 3.6
of \cite{2}) the potentials
$V^{[j]}$ on $\bR$
for $0\le j\le N-1.$ Because of
(\ref{2.3}), we can use $\alpha_N$ as the missing
dependency constant in our data. There can be at most two distinct
potentials corresponding to the two possible distinct
positive values of $\alpha_N,$ say
$\alpha_{N;\pm},$ and let us denote the
corresponding potentials by $V^{[N;\pm]}.$
Thus, $V^{[N;\pm]}\equiv V_I$ on $I$ even though
$V^{[N;+]}\not \equiv V^{[N;-]}$ on $\bR.$
Then, as seen from (\ref{3.3}), our data uniquely determines $\Lambda^{[N]}$
on $I,$ even though there can be two distinct
$\Lambda^{[N]}$ on $\bR.$

Using (\ref{3.1}) and (\ref{3.3}), from (\ref{3.2}) we get
\beq
[\Lambda^{[N]}(x)]^2=
\left(\ds\frac{f^{[N-1]\prime}_{\rm l}(i\kappa_N,x)
+\alpha_N\,f^{[N-1]\prime}_{\rm r}(i\kappa_N,x)}
{f^{[N-1]}_{\rm l}(i\kappa_N,x)+\alpha_N\,f^{[N-1]}_{\rm r}(i\kappa_N,x)}\right)^2,
\qquad x\in I.
\label{3.5}
\ene
Note that (\ref{3.5}) is a quadratic equation in $\alpha_N$ with
the $x$-dependent coefficients.
We can write the two solutions $\alpha_{N;\pm}$ 
of (\ref{3.5}) as
\beq
\alpha_{N;\pm}=\ds\frac{f^{[N-1]\prime}_{\rm l}(i\kappa_N,x)
\mp \Lambda^{[N]}(x)\,f^{[N-1]}_{\rm l}(i\kappa_N,x)}
{-f^{[N-1]\prime}_{\rm r}(i\kappa_N,x)\pm \Lambda^{[N]}(x)\,
f^{[N-1]}_{\rm r}(i\kappa_N,x)},
\qquad x\in I.
\label{3.6}
\ene
For the nonuniqueness we must have $\alpha_{N;\pm}$ finite, positive,
distinct, and independent of
$x.$ Differentiating the right-hand side of (\ref{3.6})
with respect to $x$ and setting the result to zero,
after some simplifications, we obtain
\beq
\left[V^{[N-1]}(x)+\kappa_N^2-\Lambda^{[N]}(x)^2\mp \Lambda^{[N]\prime}(x)
\right]\,[f^{[N-1]}_{\rm r}(i\kappa_N,x);f^{[N-1]}_{\rm l}(i\kappa_N,x)]=0,
\qquad x\in I,\label{3.7}
\ene
where $[F;G]:=FG'-F'G$ denotes the Wronskian.
Note that the Wronskian in (\ref{3.7}) is equal (see e.g.
\cite{2}, \cite{4}, \cite{5}) to
$-2\kappa_N/T^{[N-1]}(i\kappa_N),$ which
is a negative constant. Thus, from (\ref{3.7}) we get
\beq
\Lambda^{[N]}(x)^2-V^{[N-1]}(x)-\kappa_N^2=
\mp \Lambda^{[N]\prime}(x),
\qquad x\in I.
\label{3.8}
\ene
Using (\ref{3.3}) we can write
(\ref{3.8}) as
\beq
\ds\frac{1}{2}[V^{[N]}(x)-V^{[N-1]}(x)]=\mp \Lambda^{[N]\prime}(x),
\qquad x\in I. 
\label{3.9}
\ene
Note that $\kappa_N,$ $\Lambda^{[N]}$ on $I,$ and
$V^{[N-1]}(x)$ on $\bR$ are
all uniquely determined by our data, and hence
the left-hand side in (\ref{3.9}) is uniquely specified.
Thus, the two equations in (\ref{3.8}) specified by
the sign $\mp$ cannot simultaneously
be satisfied
unless $\Lambda^{[N]\prime}(x)\equiv 0$ on $I.$
However, in that case (\ref{3.9}) implies that
$V^{[N]}\equiv V^{[N-1]}$ on $I.$
Thus, (i) is proved.
There are now two possibilities: either
$\Lambda^{[N]}_I(x)$ is identically zero or
a positive constant.
In either case, we see from (\ref{3.8}) and (i) that
(ii) is also proved.
 From (\ref{3.6}) we get
$\alpha_{N;+}=\alpha_{N;-}$ if and only if
$\Lambda^{[N]}_I(x)\,
[f^{[N-1]}_{\rm r}(i\kappa_N,x);f^{[N-1]}_{\rm l}(i\kappa_N,x)]=0;$
however, the Wronskian on the left-hand side
is a negative constant, and hence
$\alpha_{N;+}=\alpha_{N;-}$ if and only if $\Lambda^{[N]}_I(x)\equiv 0.$
Thus, for the nonuniqueness to occur,
$\Lambda^{[N]}_I$ cannot be zero
and instead it must be a positive constant.
Then, however, with the help of
(i), (ii), and (\ref{3.8}), we see that
(iii) must hold for the nonuniqueness.
Thus, the nonuniqueness occurs when
$V|_I+\kappa_N^2$ is a positive constant
on $I,$ in which case
 from (\ref{3.6}) we get
\beq
\alpha_{N;\pm}=
\ds\frac{f^{[N-1]\prime}_{\rm l}(i\kappa_N,x)
\mp \sqrt{V|_I+k_N^2}\,f^{[N-1]}_{\rm l}(i\kappa_N,x)}
{-f^{[N-1]\prime}_{\rm r}(i\kappa_N,x)\pm \sqrt{V|_I+k_N^2}\,
f^{[N-1]}_{\rm r}(i\kappa_N,x)},
\qquad x\in I,
\label{3.10}
\ene
where the right-hand side is independent of
$x$ and can be evaluated at any $x\in I.$ Note that
we do not have to require $\alpha_{N;+}\neq \alpha_{N;-}$
because this is automatically satisfied when (i)-(iii) 
are satisfied. Finally, (\ref{3.4})
is obtained from (\ref{3.1}) and (\ref{3.2}) with $j=N$
by replacing $\alpha_N$ with $\alpha_{N;\pm},$
which are the constants in (\ref{3.10}). \qed

\begin{prop}
Under the assumptions of Theorem~{\rm 3.1}, suppose
{\rm (i)-(iv)} are satisfied so that there
are two distinct potentials
corresponding to the scattering data.
Assume further that $V^{[N-1]}$ is evenly symmetric
with respect to the midpoint $a$ of the interval $I.$ Then
we have the following:
\item{{\rm (i)}} The $\alpha_{N;\pm}$ given in
{\rm (\ref{3.10})} satisfy
$\alpha_{N;+}\,\alpha_{N;-}=e^{-4\kappa_N\,a}.$
\item{{\rm (ii)}} The two potentials $V(x;\alpha_{N;\pm})$
corresponding to the scattering data stated in
Theorem~{\rm 3.1} satisfy
$V(x;\alpha_{N;+})=V(2a-x;\alpha_{N;-})$ for
all $x\in\bR.$
\end{prop}

\noindent PROOF: From (\ref{3.10}) we get
\beq
\alpha_{N;+}\,\alpha_{N;-}=\ds\frac{[f^{[N-1]\prime}_{\rm l}(i\kappa_N,x)]^2
-(V|_I+k_N^2)\,[f^{[N-1]}_{\rm l}(i\kappa_N,x)]^2}
{[f^{[N-1]\prime}_{\rm r}(i\kappa_N,x)]^2-(V|_I+k_N^2)\,[f^{[N-1]}_{\rm r}(i\kappa_N,x)]^2},
\qquad x\in I.
\label{3.11}
\ene
When $V^{[N-1]}(x)=V^{[N-1]}(2a-x)$ for $x\in\bR,$ we have
\beq
f^{[N-1]}_{\rm l}(k,x)=e^{2ika}\,f^{[N-1]}_{\rm r}(k,2a-x),\quad
f^{[N-1]\prime}_{\rm l}(k,x)=-e^{2ika}\,f^{[N-1]\prime}_{\rm r}(k,2a-x),\qquad
x\in \bR,
\label{3.12}
\ene
and hence, in particular
\beq
f^{[N-1]}_{\rm l}(k,a)=e^{2ika}\,f^{[N-1]}_{\rm r}(k,a),\quad
f^{[N-1]\prime}_{\rm l}(k,a)=-e^{2ika}\,f^{[N-1]\prime}_{\rm r}(k,a),\qquad
x\in \bR.
\label{3.13}
\ene
Since the right-hand side of (\ref{3.11})
is independent of $x$ on $I,$ we can evaluate it at
$x=a$ and use (\ref{3.13}) to see that (i) holds.
Next, using $\alpha_{N;-}=e^{-4\kappa_N a}/\alpha_{N;+}$ and
the first equality in (\ref{3.12}),
for $x\in\bR$ we get
\beq
f^{[N-1]}_{\rm l}(k,2a-x)+\alpha_{N;-}\,
f^{[N-1]}_{\rm r}(k,2a-x)=\alpha_{N;-}\,e^{-2ika}\,[f^{[N-1]}_{\rm l}(k,x)+\alpha_{N;+}\,
e^{4(\kappa_N+ik)a}\,
f^{[N-1]}_{\rm r}(k,x)].
\label{3.14}
\ene
Using (\ref{3.14}) in (\ref{3.1}) and (\ref{3.2})
with $j=N,$ we get (ii). \qed

In the rest of this section, we will express
$\alpha_{N;\pm}$ in (\ref{3.10}) in other equivalent forms,
which will be useful in the analysis of our inverse problem.

Without any loss of generality, we may choose our interval $I$ as $(0,1),$
and in the rest of this section we will do so.
Let us fragment the potential $V^{[j]}$ (see below
(\ref{2.4}) and above (\ref{3.1}))
as $V^{[j]}=V^{[j]}_1+V^{[j]}_2+V^{[j]}_3,$ where we have defined
\beq
V_1^{[j]}:=V^{[j]}|_{(-\infty,0)},\qquad
V_0^{[j]}:=V^{[j]}|_{(0,1)},\qquad
V_2^{[j]}:=V^{[j]}|_{(1,+\infty)},
\label{3.15}
\ene
and let $f^{[j]}_{ls}$ and $f^{[j]}_{rs}$
denote the Jost solutions from the left and from the right,
respectively,
corresponding to the fragment $V_s^{[j]}$ for $s=0,1,2.$
Similarly, let us use $T^{[j]}_s,$ $R^{[j]}_s,$ and $L^{[j]}_s$ for
the scattering coefficients corresponding to
the fragment $V_s^{[j]};$ in the same way,
let $T^{[j]},$ $R^{[j]},$ and $L^{[j]}$ denote
the scattering coefficients corresponding to $V^{[j]}.$
Since we have $V^{[N]}:=V,$ we will use
$V_0:=V_0^{[N]}.$ In case of the nonuniqueness, as seen
 from Theorem~3.1, $V_0$ is a constant strictly
greater than $-\kappa_N^2,$ and
$V_0^{[N-1]}\equiv V_0.$
In that case, we have
\beq
f^{[N-1]}_{\rm l}(k,x)=\cases{
f^{[N-1]}_{{\rm l}2}(k,x),& $x\ge 1,$\cr\cr
E_+(k)\,e^{i\Delta x}+E_-(k)\,e^{-i\Delta x},
& $0\le x\le 1,$\cr\cr
\ds\frac{f^{[N-1]}_{{\rm r}1}(-k,x)}{T^{[N-1]}(k)}
+\ds\frac{L^{[N-1]}(k)\,f_{{\rm r}1}(k,x)}{T^{[N-1]}(k)},& $x\le 0,$}
\label{3.16}
\ene
\beq
f^{[N-1]}_{\rm r}(k,x)=\cases{
\ds\frac{f^{[N-1]}_{{\rm l}2}(-k,x)}{T^{[N-1]}(k)}
+\ds\frac{R^{[N-1]}(k)\,f^{[N-1]}_{{\rm l}2}(k,x)}{T^{[N-1]}(k)},
& $x\ge 1,$\cr\cr
D_+(k)\,e^{i\Delta x}+D_-(k)\,e^{-i\Delta x},
& $0\le x\le 1,$\cr\cr
f^{[N-1]}_{{\rm r}1}(k,x),& $x\le 0,$}
\label{3.17}
\ene
where we have defined
$\Delta:=\sqrt{k^2-V_0}.$
Using the boundary conditions at $x=0$ and $x=1$
resulting from the continuity of $f^{[N-1]}_{\rm l},$ $f^{[N-1]}_{\rm r},$
$f^{[N-1]\prime}_{\rm l},$ and $f^{[N-1]\prime}_{\rm r},$ we get
\beq
E_+(k)=\ds\frac{e^{-i\Delta}}{2}\left[
f^{[N-1]}_{{\rm l}2}(k,1)-i\,f^{[N-1]\prime}_{{\rm l}2}(k,1)/\Delta\right]
=\ds\frac{1}{2}\left[
f^{[N-1]}_{\rm l}(k,0)-i\,f^{[N-1]\prime}_{\rm l}(k,0)/\Delta\right],
\label{3.18}
\ene
\beq
E_-(k)=\ds\frac{e^{i\Delta}}{2}\,
\left[
f^{[N-1]}_{{\rm l}2}(k,1)+i\,f^{[N-1]\prime}_{{\rm l}2}(k,1)/\Delta\right]
=\ds\frac{1}{2}\left[
f^{[N-1]}_{\rm l}(k,0)+i\,f^{[N-1]\prime}_{\rm l}(k,0)/\Delta\right],
\label{3.19}
\ene
\beq
D_+(k)=\ds\frac{1}{2}\left[
f^{[N-1]}_{{\rm r}1}(k,0)-i\,f^{[N-1]\prime}_{{\rm r}1}(k,0)/\Delta\right]
=\ds\frac{e^{-i\Delta}}{2}
\left[
f^{[N-1]}_{\rm r}(k,1)-i\,f^{[N-1]\prime}_{\rm r}(k,1)/\Delta\right],
\label{3.20}
\ene
\beq
D_-(k)=\ds\frac{1}{2}\,\left[
f^{[N-1]}_{{\rm r}1}(k,0)+i\,f^{[N-1]\prime}_{{\rm r}1}(k,0)/\Delta\right]
=\ds\frac{e^{i\Delta}}{2}
\left[
f^{[N-1]}_{\rm r}(k,1)+i\,f^{[N-1]\prime}_{\rm r}(k,1)/\Delta\right].
\label{3.21}
\ene
In this case, using (\ref{3.18})-(\ref{3.21})
we can write $\alpha_{N;\pm}$ given in
(\ref{3.10}) in various forms such as
\beq
\alpha_{N;\pm}=-\ds\frac{E_\pm(i\kappa_N)}{D_\pm(i\kappa_N)},
\label{3.22}
\ene
\beq
\alpha_{N;\pm}=\ds\frac{
e^{\pm\sqrt{\kappa_N^2+V_0}}\left[
f^{[N-1]}_{{\rm l}2}(i\kappa_N,1)\mp 
f^{[N-1]\prime}_{{\rm l}2}(i\kappa_N,1)/\sqrt{\kappa_N^2+V_0}\right]}
{-f^{[N-1]}_{{\rm r}1}(i\kappa_N,0)\pm 
f^{[N-1]\prime}_{{\rm r}1}(i\kappa_N,0)/\sqrt{\kappa_N^2+V_0}},
\label{3.23}
\ene
$$\alpha_{N;\pm}=\ds\frac{
e^{\mp\sqrt{\kappa_N^2+V_0}}\left[
f^{[N-1]}_{\rm l}(i\kappa_N,0)\mp f^{[N-1]\prime}_{\rm l}
(i\kappa_N,0)/\sqrt{\kappa_N^2+V_0}\right]}
{-f^{[N-1]}_{\rm r}(i\kappa_N,1)\pm f^{[N-1]\prime}_{\rm r}
(i\kappa_N,1)/\sqrt{\kappa_N^2+V_0}}.$$

When (i)-(iv) of Theorem~3.1 are satisfied,
 from (\ref{3.16}) and (\ref{3.17}), for $x\in [0,1]$ we get
\begin{eqnarray*}
f^{[N-1]}_{\rm l}(i\kappa_N,x)+\alpha_{N;\pm}\,f^{[N-1]}_{\rm r}(i\kappa_N,x)=
\! \! \! \! \! \! \! \! \! \! \! \! \! \! \!
&
[E_+(i\kappa_N)+\alpha_{N;\pm}\,D_+(i\kappa_N)]
e^{-\sqrt{\kappa_N^2+V_0}}\\
&
\; \; \; \; \; \; \; \; \; \; \; \; 
+[E_-(i\kappa_N)+\alpha_{N;\pm}\,D_-(i\kappa_N)]
e^{\sqrt{\kappa_N^2+V_0}},\end{eqnarray*}
which , with the help of (\ref{3.22}), simplifies to
$$f^{[N-1]}_{\rm l}(i\kappa_N,x)+\alpha_{N;\pm}\,f^{[N-1]}_{\rm r}(i\kappa_N,x)
=[E_\mp(i\kappa_N)+\alpha_{N;\pm}\,D_\mp(i\kappa_N)]
e^{\pm\sqrt{\kappa_N^2+V_0}},
\qquad x\in[0,1].$$
Thus, in (\ref{3.4}) we get
$$\frac{f^{[N-1]\prime}_{\rm l}(i\kappa_N,x)+\alpha_{N;\pm}\,
f^{[N-1]\prime}_{\rm r}(i\kappa_N,x)}
{f^{[N-1]}_{\rm l}(i\kappa_N,x)+\alpha_{N;\pm}\,
f^{[N-1]}_{\rm r}(i\kappa_N,x)}=\pm \sqrt{\kappa_N^2+V_0},$$
and we see why the two distinct $\alpha_{N;\pm}$ lead to the
same potential on $[0,1].$

At the end of the proof of Theorem~3.1 we have indicated that
$\alpha_{N;+}\ne\alpha_{N;-}$ when (i)-(iii) in Theorem~3.1
are satisfied. Next we elaborate on this point by using
the representations of $\alpha_{N;\pm}$ given in (\ref{3.22}).

\begin{prop}
Assume $V$ is real valued, belongs to
$L^1_1(\bR),$ and has $N$ bound states
at $k=i\kappa_j$ with
$0<\kappa_1<\dots<\kappa_N.$
Consider the scattering data consisting of
one reflection coefficient,
knowledge of $V|_{(0,1)},$ all the $N$ bound-state energies,
and all the $N$ norming
constants except for the one corresponding
to the $N$th bound state. Moreover, suppose that
$V|_{(0,1)}$ and $V^{[N-1]}|_{(0,1)}$ are both equal
to the same constant $V_0$ that is greater than $-\kappa_N^2.$ 
Then, we have the following.
\item{{\rm (i)}} The quantity
$E_+(i\kappa_N)\,D_-(i\kappa_N)-E_-(i\kappa_N)\,D_+(i\kappa_N)$
is nonzero, and hence $D_+(i\kappa_N)$ and $D_-(i\kappa_N)$ cannot simultaneously
be zero.
\item{{\rm (ii)}} If either of $D_+(i\kappa_N)$ and $D_-(i\kappa_N)$ is zero,
then the scattering data uniquely determines $V$ on $\bR.$
\item{{\rm (iii)}} If both $D_+(i\kappa_N)$ and $D_-(i\kappa_N)$ are
nonzero, then $\alpha_{N;+}$ and
$\alpha_{N;-}$ given in {\rm (\ref{3.22})}
are necessarily distinct.
\end{prop}

\noindent PROOF: Using (\ref{3.18})-(\ref{3.21}), we see that
\beq
E_+(k)\,D_-(k)-E_-(k)\,D_+(k)=\ds\frac{i}{2\sqrt{k^2-V_0}}
\left[f^{[N-1]}_{\rm l}(k,0)\,f^{[N-1]\prime}_{{\rm r}1}(k,0)-
f^{[N-1]\prime}_{\rm l}(k,0)\,f^{[N-1]}_{{\rm r}1}(k,0)\right].
\label{3.24}
\ene
Note that the quantity in the brackets
on the right-hand side in (\ref{3.24})
is nothing but the Wronskian $[f_{\rm l}(k,x);f_{{\rm r}1}(k,x)]$
at $x=0.$ This Wronskian is independent of $x$ when
$x\in\bR^-$ because both $f_{\rm l}(k,x)$ and
$f_{{\rm r}1}(k,x)$ are solutions of (\ref{1.1}) for $x\in\bR^-.$
In fact, by using
$$f^{[N-1]}_{{\rm r}1}(k,0)=f^{[N-1]}_{\rm r}(k,0),\qquad
f^{[N-1]\prime}_{{\rm r}1}(k,0)=f^{[N-1]\prime}_{\rm r}(k,0),$$
$$\ds\frac{2ik}{T^{[N-1]}(k)}=[f^{[N-1]}_{\rm r}(k,x);f^{[N-1]}_{\rm l}(k,x)],$$
we can write (\ref{3.24}) in the form
$$E_+(k)\,D_-(k)-E_-(k)\,D_+(k)=\ds\frac{k}{T^{[N-1]}(k)\,
\sqrt{k^2-V_0}},$$
which implies that
$$E_+(i\kappa_N)\,D_-(i\kappa_N)-E_-(i\kappa_N)
\,D_+(i\kappa_N)=\ds\frac{\kappa_N}{T^{[N-1]}(i\kappa_N)\,
\sqrt{\kappa_N^2+V_0}}.$$
By using an argument similar to
that given below (\ref{1.6}) we get $T^{[N-1]}(i\kappa_N)>0$
and hence (i) holds. From (i) it follows that
$E_+(i\kappa_N)$ and $D_+(i\kappa_N)$ cannot simultaneously vanish;
similarly, $E_-(i\kappa_N)$ and $D_-(i\kappa_N)$ cannot simultaneously vanish.
Hence, $|\alpha_{N;\pm}|=+\infty$ if and only if
$D_\pm(i\kappa_N)=0.$ However, there cannot be a nonuniqueness
unless $\alpha_{N;\pm}$ are both finite and positive, and
thus, we have proved (ii). Finally, from (\ref{3.22}) we see that
$\alpha_{N;+}=\alpha_{N;-}$ would be possible
either when 
$E_+(i\kappa_N)\,D_-(i\kappa_N)-E_-(i\kappa_N)
\,D_+(i\kappa_N)=0$ or 
$D_-(i\kappa_N)=D_+(i\kappa_N)=0,$
but this is not possible because of (i). \qed

Next, when the assumptions in Proposition~3.3 are
satisfied, we will write the numerators and the denominators
in (\ref{3.23}) by using the scattering
coefficients of $V^{[N-1]}_1$ and $V^{[N-1]}_2,$ and
this will enable us to study the signs of
$\alpha_{N;\pm}$ when those
scattering coefficients are known.
Then, in view of Theorem~3.1(iv), we already know that
our scattering data uniquely determines
$V$ unless $\alpha_{N;\pm}$ are both finite and positive.

Since $V^{[N-1]}_1$ is supported
in $\bR^-,$ $R^{[N-1]}_1$ has (see e.g.
\cite{7}-\cite{13}) a meromorphic extension
 from $k\in\bR$ to $k\in\bCp$ having its poles exactly coinciding with the
poles of $T^{[N-1]}_1$ there. Similarly, since $V^{[N-1]}_2$ is supported
in $(1,+\infty),$
$L^{[N-1]}_2$ has a meromorphic extension
to $\bCp$ having its poles coinciding with the poles of
$T^{[N-1]}_2$ there. Thus, we have
$$f^{[N-1]}_{{\rm r}1}(i\kappa_N,0)=\ds\frac{1+R^{[N-1]}_1
(i\kappa_N)}{T^{[N-1]}_1(i\kappa_N)},
\qquad
f^{[N-1]\prime}_{{\rm r}1}(i\kappa_N,0)=\kappa_N\,
\ds\frac{1-R^{[N-1]}_1(i\kappa_N)}{T^{[N-1]}_1(i\kappa_N)},$$
$$f^{[N-1]}_{{\rm l}2}(i\kappa_N,1)=\ds\frac{e^{-\kappa_N}+L^{[N-1]}_2(i\kappa_N)
\,e^{\kappa_N}}{T^{[N-1]}_2(i\kappa_N)},\qquad
f^{[N-1]\prime}_{{\rm l}2}(i\kappa_N,1)=-\kappa_N\,\ds\frac{e^{-\kappa_N}-
L^{[N-1]}_2(i\kappa_N)\,e^{\kappa_N}}{T^{[N-1]}_2(i\kappa_N)}.$$
Let us define
\beq
\epsilon_\pm:=\sqrt{\kappa_N^2+V_0}\pm\kappa_N,
\label{3.25}
\ene
\beq
q_{1;\pm}:=\epsilon_\pm\,R^{[N-1]}_1(i\kappa_N)+\epsilon_\mp,
\qquad
q_{2;\pm}:=\epsilon_\mp\,
L^{[N-1]}_2(i\kappa_N)+\epsilon_\pm\,e^{-2\kappa_N}.
\label{3.26}
\ene
Then, from (\ref{3.23}) we get
\beq
\alpha_{N;\pm}=-\ds{e^{\pm\epsilon_\pm}}\,
\ds\frac{T^{[N-1]}_1(i\kappa_N)}{T^{[N-1]}_2(i\kappa_N)}\,
\ds\frac{q_{2;\pm}}{q_{1;\pm}}.
\label{3.27}
\ene

\section {Examples}\sss

In this section we illustrate the
uniqueness and nonuniqueness
stated in Theorem~3.1 with some examples. Without 
any loss of any generality
we use the interval $(0,1)$ as $I.$
In a typical example
we use the fragmentation given in (\ref{3.15}). As in (i)-(iii)
of Theorem~3.1, we assume that $V_0$ and
$V_0^{[N-1]}$ are equal to each other and equal to
a constant greater than $-\kappa_N^2.$
Since the scattering data
specified in Theorem~3.1 is equivalent to
$\{V_0,N,V^{[N-1]}_1,V^{[N-1]}_2,\kappa_N\},$
which is also equivalent to
$\{V_0,N,R^{[N-1]}_1,L^{[N-1]}_2,\kappa_N\},$ we use
either of these two sets as our scattering data in our examples.
We then use (\ref{3.27}) to determine the signs
of $\alpha_{N;\pm}.$ As in (iv) of Theorem~3.1,
the nonuniqueness occurs when both $\alpha_{N;\pm}$ are
finite and positive. Otherwise, our scattering data uniquely determines $V.$

For the sake of exact analytical calculations,
our $V^{[N-1]}_1$ and $V^{[N-1]}_2$ consist
of constant steps
or pieces of Bargmann potentials
(recall \cite{5} that a potential is called a Bargmann
potential if the corresponding scattering matrix
is a rational function of $k$).
Such choices allow
us to explicitly and uniquely evaluate $V^{[N-1]}_1$ and $T^{[N-1]}_1$
when $R^{[N-1]}_1$ is known, and conversely evaluate
$R^{[N-1]}_1$ and $T^{[N-1]}_1$ when $V^{[N-1]}_1$ is known.
Similarly, we are able to explicitly and uniquely evaluate
$V^{[N-1]}_2$ and $T^{[N-1]}_2$ when $L^{[N-1]}_2$ is known, and evaluate
$L^{[N-1]}_2$ and $T^{[N-1]}_2$ when $V^{[N-1]}_2$ is known.
In our examples, we choose
$R^{[N-1]}_1$ and $L^{[N-1]}_2$ in such a way that
the existence in $L^1_1$ of $V^{[N-1]}_1$ and $V^{[N-1]}_2,$
respectively, is assured by the characterization results
\cite{1}-\cite{6} or by the results
in Section~XVII.3.2 of \cite{5}. 

When $V_0$ is supported on $(0,1)$ and equal to a
constant, the corresponding transmission coefficient
$T_0$ and the left reflection coefficient
$L_0$ are obtained by
using (\ref{1.2}), (\ref{1.3}), (\ref{3.16}), and (\ref{3.17}),
and we get
\beq
\ds\frac{1}{T_0(k)}=e^{ik}\left[\cos\left(\sqrt{k^2-V_0}\right)+
\ds\frac{2k^2-V_0}{2ik\sqrt{k^2-V_0}}\sin\left(\sqrt{k^2-V_0}\right)\right],
\label{4.1}
\ene
\beq
\ds \frac{L_0(k)}{T_0(k)}=
\ds\frac{V_0\,e^{ik}}{2ik\sqrt{k^2-V_0}}\sin\left(\sqrt{k^2-V_0}\right).
\label{4.2}
\ene
When $V_0\ge 0$ we have $N_0=0,$ where
$N_0$ denotes the number of bound states of $V_0.$
When $V_0<0$ we have $N_0\ge 1,$ and
recall \cite{14} that $N_0$
is evaluated as the
unique integer satisfying
\beq
(N_0-1)\pi<\sqrt{-V_0}\le N_0\pi.
\label{4.3}
\ene
Recall also that under the shift $V(x)\mapsto
\tilde V(x):=V(x-b),$ the corresponding scattering
coefficients undergo the changes
$\tilde T(k)=T(k),$ $\tilde L(k)=e^{2ikb}L(k),$
and $\tilde R(k)=e^{-2ikb}R(k).$ This observation
will help us to evaluate
$R_1^{[N-1]}$ and $L_2^{[N-1]}$ in some of our examples
when $V^{[N-1]}_1$ and $V^{[N-1]}_2$
are square-well potentials.

In order to know $N$ in our examples, we count
the number of zeros of $1/T^{[N-1]}(i\beta)$ for $\beta\in(0,+\infty),$
which is equal to $N-1.$
For this purpose, we use the following formula, see e.g. \cite{15}, 
in order to express $T^{[N-1]}(k)$ in terms of the appropriate
scattering coefficients for
the pieces $V^{[N-1]}_1,$ $V_0,$ and $V^{[N-1]}_2:$
\begin{eqnarray}
\ds\frac{1}{T^{[N-1]}(k)}=
\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!
&\ds\frac{1}{T^{[N-1]}_1(k)}\ds\frac{1}
{T_0(k)}\ds\frac{1}{T^{[N-1]}_2(k)}
-\ds\frac{R^{[N-1]}_1(k)}{T^{[N-1]}_1(k)}\ds\frac{L_0(k)}{T_0(k)}
\ds\frac{1}{T^{[N-1]}_2(k)}\nonumber \\
&
\; \; \; \; \; \; \; \; \; 
+\ds\frac{1}{T^{[N-1]}_1(k)}\ds\frac{L_0(-k)}{T_0(-k)}
\ds\frac{L^{[N-1]}_2(k)}{T^{[N-1]}_2(k)}
-\ds\frac{R^{[N-1]}_1(k)}{T^{[N-1]}_1(k)}\ds\frac{1}{T_0(-k)}
\ds\frac{L^{[N-1]}_2(k)}{T^{[N-1]}_2(k)}.
\label{4.4}
\end{eqnarray}
Recall that
we know the set $\{T^{[N-1]}_1,R^{[N-1]}_1,L_0,T_0,L^{[N-1]}_2,T^{[N-1]}_2\}$
when $\{V^{[N-1]}_1,V_0,V^{[N-1]}_2\}$ is known.

For our examples, it is also helpful to remember,
see e.g. \cite{16}, that
if one fragment has $n_1$ bound states and the other
has $n_2$ bound states, then the combination
of these two fragments has either $n_1+n_2$ or
$n_1+n_2-1$ bound states. This helps us to
estimate the number of bound states of the whole potential
in terms of the number of
bound states of its fragments.
In particular, if none of the fragments
have any bound states, then the total potential
does not have any bound states either.

\begin{example}
{\rm 
Note that in the special case $V_0\equiv 0,$ from (\ref{3.25}) we get
$\epsilon_+=2\kappa_N$ and $\epsilon_-=0,$
and hence (\ref{3.27}) leads to}
\beq
\alpha_{N;+}=-\ds\frac{T^{[N-1]}_1(i\kappa_N)}{R^{[N-1]}_1(i\kappa_N)}
\ds\frac{1}{T^{[N-1]}_2(i\kappa_N)},\qquad
\alpha_{N;-}=-T^{[N-1]}_1(i\kappa_N)\,\ds\frac{L^{[N-1]}_2(i\kappa_N)}
{T^{[N-1]}_2(i\kappa_N)}.
\label{4.5}
\ene
{\rm Let us choose}
$$V^{[N-1]}_1(x)=\cases{v_1,& $x\in(-1,0),$\cr\cr
0,& $x\not\in (-1,0),$}
\qquad V^{[N-1]}_2(x)=\cases{v_2,& $x\in(1,2),$\cr\cr
0,& $x\not\in(1,2),$}$$
{\rm where $v_1$ and $v_2$ are
adjustable constant parameters.
When $v_1>0$ and $v_2>0,$ we have
$N=1,$ and with the help of (\ref{4.1}),
(\ref{4.2}), and the observation below
(\ref{4.3}), for any $\kappa_1>0$
we see from (\ref{4.5}) that $\alpha_{1;\pm}>0$ and
hence the nonuniqueness follows.
When at least one of
$v_1$ and $v_2$ is zero, we see from (\ref{4.5}) that
one of $\alpha_{N;\pm}$ is either zero or infinite,
leading to the uniqueness.
On the other hand, when at least one of
$v_1$ and $v_2$ is negative, we mostly
get the uniqueness.}
\end{example}

\begin{example}
{\rm 
In the special case $V^{[N-1]}_1\equiv 0$ and
$V^{[N-1]}_2\equiv 0,$ we have
$R^{[N-1]}_1=L^{[N-1]}_2\equiv 0$ and
$T^{[N-1]}_1=T^{[N-1]}_2\equiv 1;$ hence, from 
(\ref{3.25})-(\ref{3.27})
we see that 
$\alpha_{N;\pm}=-\epsilon_{\pm}e^{\pm\epsilon_{\mp}}/\epsilon_{\mp}.$
Thus, $\alpha_{N;\pm}>0$ if and only if $-\kappa_N^2<V_0<0.$
As indicated in Proposition~3.2(ii),
the two potentials corresponding to our data satisfy
the symmetry $V(x;\alpha_{N;+})=
V(1-x;\alpha_{N;-})$ for $x\in\bR.$
For example, with $V_0=-1,$ with the help of (\ref{4.3}) we get
$N=2.$ Using (\ref{4.1}), we obtain $\kappa_1=0.43513\overline{1},$
where we use the overline on the last digit to indicate
roundoff. Then, as long as $\kappa_2>1,$ there are two distinct
potentials corresponding to our data. 
With $\kappa_2=\sqrt{2}$ so that
$\alpha_{2;\pm}=(\sqrt{2}\pm1)^2 e^{-\sqrt{2}\pm 1},$
the graphs of the two resulting potentials
are given in Figures~1 and 2, from which the symmetry
mentioned above is observed.}
\end{example}

\begin{example}
{\rm Let}
\beq
R^{[N-1]}_1(k)=e^{-2ik}
L^{[N-1]}_2(k)=\ds\frac{2}{(k+i)(k+2i)},\qquad
T^{[N-1]}_1(k)=T^{[N-1]}_2(k)=\ds\frac{k(k+\sqrt{5}i)}{(k+i)(k+2i)},
\label{4.6}
\ene
{\rm so that $V^{[N-1]}_1$ is supported on $(-\infty,0)$ and
$V^{[N-1]}_2$ on $(1,+\infty).$ Neither $V^{[N-1]}_1$ nor $V^{[N-1]}_2$
has any bound states, and $V^{[N-1]}_1(x)=V^{[N-1]}_2(1-x)$
for $x\in{\bR}.$
Thus, when $V_0$ is constant, the potential $V^{[N-1]}$ has even symmetry
with respect to the point $x=1/2.$
Due to that symmetry, when there is a nonuniqueness,
Proposition~3.2 applies and we have
$\alpha_{N;+}=e^{-2\kappa_N}/\alpha_{N;-},$ and
$V(x;\alpha_{N;+})=V(1-x;\alpha_{N;-})$
for $x\in\bR.$
Using (\ref{4.6}) in (\ref{3.27}), we get
\beq
\alpha_{N;\pm}=e^{\pm\epsilon_{\mp}}\frac{V_0 (\kappa_N+1)(\kappa_N+2)-
2(\sqrt{\kappa_N^2+V_0}\mp\kappa_N)^2}
{2V_0-(\kappa_N+1)(\kappa_N+2)(\sqrt{\kappa_N^2+V_0}\mp\kappa_N)^2},
\label{4.7}
\ene
where $\epsilon_{\pm}$ are the constants defined in (\ref{3.25}).
With the help of (\ref{4.4}), we can get $N-1$ by counting the
number of zeros of $1/T^{[N-1]}$ on the positive imaginary axis.
We find that $N=1$ when $V_0\ge -2.16681\overline{5},$
$N=2$ when $-15.54\overline{8}\le V_0<-2.16681\overline{5},$
$N=3$ when $-44.5877\overline{5}\le V_0<-15.54\overline{8},$ etc.
As indicated in Theorem~3.1(iii), for the nonuniqueness 
we need $V_0+\kappa_N^2>0.$ From (\ref{4.7}) we see 
that $\alpha_{N;\pm}>0$ when $V_0<0.$ When $V_0=0,$ we see that
$\alpha_{1;-}=0$ while $\alpha_{1;+}$ is not finite, and hence
we have the uniqueness. When $V_0>0,$ we may or may not
have the nonuniqueness depending on whether $\alpha_{1;\pm}>0$ or not.
For example, for $V_0=1$ we get the nonuniqueness for any
$\kappa_1>0.$
When $V_0=5,$ we get
$\alpha_{1;\pm}<0$ for
$\kappa_1\in(0,\beta_0)$ and
$\alpha_{1;\pm}>0$ for
$\kappa_1>\beta_0,$
where $\beta_0=4.2659\overline{7};$
for $\kappa_1=\beta_0$ we get $\alpha_{1;-}=0$ while $\alpha_{1;+}$ is not finite.
On the other hand, for $V_0=10,$ we get
$\alpha_\pm<0$ for any $\kappa_1>0.$
When $V_0=1$ and $\kappa_1=1,$
we show the potentials
$V^{[0]}(x),$ $V(x;\alpha_{1;+}),$ and $V(x;\alpha_{1;+})$
in Figures~3, 4, 5,
respectively.}
\end{example}

\begin{example}
{\rm Let $V_2^{[N-1]}\equiv 0,$ $V_0=(t^2-1)\,\kappa_N^2,$
where $t$ is a positive parameter
and $N$ and $\kappa_N$ are to be specified.
Note that (iii) of Theorem~3.1 is assured because $t>0.$
Let us choose}
$$R^{[N-1]}_1(k)=\ds\frac{-2(k-it\kappa_N)}{(k+i)(k+2i)(k+it\kappa_N)},\qquad
T^{[N-1]}_1(k)=\ds\frac{k(k+\sqrt{5}i)}{(k+i)(k+2i)},$$
{\rm so that $V^{[N-1]}_1$ is supported on $\bR^-$ and has no bound states.
Thus, $T^{[N-1]}_1(i\kappa_N)>0.$ Using (\ref{3.25})-(\ref{3.27}) we get}
\beq
\alpha_{N;+}=\frac{(\kappa_N+\sqrt{5})e^{(\sqrt{1+t}-1)\kappa_N}}
{(1-t)(\kappa_N+3)},\qquad
\alpha_{N;-}=\frac{(1-t)(1+t)\kappa_N(\kappa_N+\sqrt{5})e^{-(\sqrt{1+t}+1)\kappa_N}}
{(1+t)^2(\kappa_N+1)(\kappa_N+2)-2(1-t)^2},
\label{4.8}
\ene
{\rm and hence $\alpha_{N;\pm}>0$ only if $t<1.$
Therefore, in order to have the
nonuniqueness, we must impose the restriction $t\in (0,1).$
By analyzing the zeros (on the 
positive imaginary axis) of $1/T^{[N-1]}$ given in (\ref{4.4}), 
we are able to get $N$ for any specific value of $\kappa_N$ in
our data. For example, if $\kappa_N=1,$ we find that
$N=1$ for $t\in[t_1,1)$ and $N=2$ for $t\in (0,t_1),$ where
$t_1=0.7639\overline{8}.$ For the particular
values $\kappa_2=1$ and $t=0.5,$
 from (\ref{4.8}) we obtain $\alpha_{2;+}=2.0257\overline{8}$ and
$\alpha_{2;-}=0.020180\overline{9}.$}
\end{example}

\noindent {\bf Acknowledgment.} The research leading to this paper started
as a collaboration between B. Gr\'ebert and R. Weder, who proved
a particular case of Theorem~3.1, namely that the unique
recovery with $N=1$
holds when $V|_I$ is not constant, and who also obtained the results 
in Example~4.2. R. Weder thanks B. Gr\'ebert for that
collaboration.

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\begin{figure}[pos]
\begin{center}
\leavevmode
\epsffile{graph1.ps}
\end{center}
\caption{The potential $V(x;\alpha_{2;+})$ in Example~4.2
with $\kappa_2=\sqrt{2}$ and $\alpha_{2;+}=(\sqrt{2}+1)^2 e^{-\sqrt{2}+1}$}
\label{xx}
\end{figure}

\begin{figure}[pos]
\begin{center}
\leavevmode
\epsffile{graph2.ps}
\end{center}
\caption{The potential $V(x;\alpha_{2;-})$ in Example~4.2
with $\kappa_2=\sqrt{2}$ and $\alpha_{2;-}=(\sqrt{2}-1)^2 e^{-\sqrt{2}-1}$}
\label{xxx}
\end{figure}

\begin{figure}[pos]
\begin{center}
\leavevmode
\epsffile{graph3.ps}
\end{center}
\caption{The potential $V^{[0]}(x)$ in Example~4.3
with $V_0=1.$}
\label{xxxx}
\end{figure}

\begin{figure}[pos]
\begin{center}
\leavevmode
\epsffile{graph4.ps}
\end{center}
\caption{The potential $V(x;\alpha_{1;+})$ in Example~4.3
with $V_0=1,$ $\kappa_1=1,$ and $\alpha_{1;+}=8.8194\overline{6}.$}
\label{xxxxx}
\end{figure}

\begin{figure}[pos]
\begin{center}
\leavevmode
\epsffile{graph5.ps}
\end{center}
\caption{The potential $V(x;\alpha_{1;-})$ in Example~4.3
with $V_0=1,$ $\kappa_1=1,$ and $\alpha_{1;-}=0.015345\overline{1}.$}
\label{xxxxxx}
\end{figure}

\end{document}
---------------0103141035714--
