\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2019 (2019), No. 23, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2019 Texas State University.}
\vspace{8mm}}
\begin{document}
\title[\hfilneg EJDE-2019/23\hfil
Bound states of the discrete Schr\"odinger equation]
{Bound states of the discrete Schr\"odinger equation with
compactly supported potentials}
\author[T. Aktosun, A. E. Choque-Rivero, V. G. Papanicolaou \hfil EJDE-2019/23\hfilneg]
{Tuncay Aktosun, Abdon E. Choque-Rivero, Vassilis G. Papanicolaou}
\address{Tuncay Aktosun \newline
Department of Mathematics,
University of Texas at Arlington,
Arlington, TX 76019-0408, USA}
\email{aktosun@uta.edu}
\address{Abdon E. Choque-Rivero \newline
Instituto de F\'{\i}sica y Matem\'aticas,
Universidad Michoacana de San Nicol\'as de Hidalgo,
Ciudad Universitaria, C.P. 58048,
Morelia, Michoac\'an, M\'exico}
\email{abdon@ifm.umich.mx}
\address{Vassilis G. Papanicolaou \newline
Department of Mathematics,
National Technical University of Athens,
Zografou Campus, 157 80,
Athens, Greece}
\email{papanico@math.ntua.gr}
\thanks{Submitted December 16, 2018. Published February 11, 2019.}
\subjclass[2010]{39A70, 47B39, 81U15, 34A33}
\keywords{Discrete Schr\"odinger operator; half-line lattice; bound states;
\hfill\break\indent resonances; compactly-supported potential;
number of bound states}
\begin{abstract}
The discrete Schr\"odinger operator is considered on the half-line
lattice $n\in \{1,2,3,\dots\}$ with the Dirichlet boundary condition at
$n=0$. It is assumed that the potential
belongs to class $\mathcal{A}_b$, i.e.\ it is real valued, vanishes
when $n>b$ with $b$ being a fixed positive integer, and is nonzero
at $n=b$. The proof is provided to show that the corresponding
number of bound states, $N$, must satisfy the inequalities
$0\le N\le b$. It is shown that for each fixed nonnegative integer $k$ in the
set $\{0,1,2,\ldots,b\}$, there exist infinitely many potentials in
class $\mathcal{A}_b$ for which the corresponding Schr\"odinger
operator has exactly $k$ bound states. Some auxiliary results are
presented to relate the number of bound states to the number of real
resonances associated with the corresponding Schr\"odinger operator.
The theory presented is illustrated with some explicit examples.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks
\section{Introduction}\label{sec1}
We consider the discrete Schr\"odinger equation on the half-line
lattice, i.e.\, the difference equation
\begin{equation}
-\psi_{n+1}+2\psi_n-\psi_{n-1}+V_n\psi_n=\lambda \psi_n, \quad
n\ge 1, \label{1.1}
\end{equation}
with the Dirichlet boundary condition
\begin{equation}
\psi_0=0. \label{1.2}
\end{equation} Here, the discrete independent variable $n$
takes positive integer values, the boundary of the half-line lattice
corresponds to $n=0$,
$V_n$ denotes the value of the potential $V$ at the lattice
location $n$, $\lambda$ is the spectral parameter, and $\psi_n$
denotes the value of the wave function at the location $n$. We
assume that the potential is real valued, i.e.\
$$
V_n=V_n^\ast, \quad n\ge 1,
$$
where the asterisk is used for complex conjugation.
There are various physical models \cite{6} governed by the
discrete Schr\"odinger equation on a half-line lattice,
and such models describe the
quantum mechanical behavior of a particle of energy $\lambda$
in a semi-infinite crystal where the particle experiences at each lattice point
the force associated with the potential value $V_n$.
In this article we assume that the potential belongs to class
$\mathcal{A}_b$, which is defined below as in \cite{2}.
\begin{definition}\label{def1.1} \rm
The potential $V$ appearing in \eqref{1.1} belongs to the class
$\mathcal{A}_b$ if the $V_n$-values are real and the support of the potential
$V$ is confined to the finite set $\{1,2,\dots,b\}$ for some
positive integer $b$, i.e.\ $V_n=0$ for $n>b$ and $V_b\ne 0$.
\end{definition}
We refer to a potential $V$ in class $\mathcal{A}_b$ as a
compactly-supported potential, and we see that $b$ in the definition of
$\mathcal{A}_b$ refers to the smallest positive integer beyond which the potential
vanishes.
In class $\mathcal{A}_b$, it is possible to have
$V_n=0$ for some or all
$n$-values with $1\le n**0,\\
-1,& x<0 .\end{cases}
$$
In the following proposition we consider the sign of the product
appearing in the numerator of \eqref{2.13}.
\begin{proposition}\label{prop3.1}
Assume that the potential $V$ appearing in
\eqref{1.1} belongs to class $\mathcal{A}_b$, and further assume
that there exists at least one bound state. Let
$\{\alpha_j\}_{j=1}^{2b-1}$ denote the set of zeros of $f_0(z)$
ordered as indicated in \eqref{3.5}-\eqref{3.9}. Then, for the bound
state occurring at $z=\alpha_k$ we have
\begin{equation}
\operatorname{sgn}\Big(\prod_{j=1}^{2b-1}(1-\alpha_k
\alpha_j)\Big)=\begin{cases}
\operatorname{sgn}\big(\prod_{j=1}^{p}(1-\alpha_k
\alpha_j)\big), & \alpha_k\in(-1,0), \\[4pt]
\operatorname{sgn}\big(\prod_{j=r+1}^{s}(1-\alpha_k
\alpha_j)\big), &\alpha_k\in(0,1),
\end{cases}\label{3.10}
\end{equation}
where $p$, $r$, and $s$ are the integers defined in \eqref{3.1},
\eqref{3.3}, and \eqref{3.4}, respectively.
\end{proposition}
\begin{proof}
Let us omit $(1-\alpha_k \alpha_j)$ from the argument of the
product and e.g.\ write $\prod_1^p$ to denote
$\prod_{j=1}^p (1-\alpha_k \alpha_j)$. We have
\begin{equation}
\prod_1^{2b-1}=\Big(\prod_1^p\big) \Big(\prod_{p+1}^q\Big)
\Big(\prod_{q+1}^r \Big)\Big(\prod_{r+1}^s\Big)
\Big( \prod_{s+1}^{2b-1}\Big). \label{3.11}
\end{equation}
With the help of \eqref{3.6} and \eqref{3.7} we see that each
of $\prod_{p+1}^q$ and $\prod_{q+1}^r$ is positive
because $|\alpha_j|<1$ when $p+1\le j\le r$.
By Theorem~\ref{t2.1}(g), the zeros $\alpha_j$ for $s+1\le j\le 2b-1$
occur in complex-conjugate pairs and hence
the quantity $\prod_{s+1}^{2b-1}$ appearing in \eqref{3.11}
consists of the products of the form $|1-\alpha_k \alpha_j|^2$ when the
$\alpha_j$-values consist of all nonreal zeros of $f_0(z)$ in the
upper-half complex-$z$ plane. Since $\alpha_k$ is real and such
$\alpha_j$-values are nonreal, we conclude that $\prod_{s+1}^{2b-1}$
is positive. Let us now consider the sign of $\prod_1^p$
appearing in \eqref{3.11}. As seen from \eqref{3.5} and \eqref{3.7},
for $1\le j\le p$ we have $1-\alpha_k \alpha_j>0$
if $\alpha_k\in(0,1)$. Let us also consider the sign of $\prod_{r+1}^s$
appearing in \eqref{3.11}. As seen from \eqref{3.6} and \eqref{3.8},
for $r+1\leq j\leq s$ we have $1-\alpha_k \alpha_j>0$
if $\alpha_k\in (-1,0)$. Thus, from \eqref{3.11} we directly conclude
\eqref{3.10}.
\end{proof}
In the following proposition, we consider the sign of the product
appearing in the denominator of \eqref{2.13}.
\begin{proposition}\label{prop3.2}
Assume that the potential $V$ appearing in \eqref{1.1} belongs to class
$\mathcal{A}_b$, and further assume that there exists at least one
bound state. Let $\{\alpha_j\}_{j=1}^{2b-1}$ denote the set of zeros
of $f_0(z)$ ordered as indicated in \eqref{3.5}-\eqref{3.9}. Then,
for the bound state occurring at $z=\alpha_k$ we have
\begin{equation}
\operatorname{sgn}\Big(\prod_{j\ne k}(\alpha_k-\alpha_j)
\Big)=(-1)^{k-1}, \quad k=p+1,\dots, r, \label{3.12}
\end{equation}
where $p$ and $r$ are the respective integers appearing in \eqref{3.1}
and \eqref{3.3}, the product $\prod_{j\ne k}$ is over
all $j$-values with $1\le j\le 2b-1$ except $j=k$, and
we recall that $\operatorname{sgn}$ denotes the signum function.
\end{proposition}
\begin{proof}
Let us drop $(\alpha_k-\alpha_j)$ from the argument of the
product and use
$\prod_{j\ne k}$ to denote the product on the left-hand
side of \eqref{3.12}. We have
\begin{equation}
\prod_{j\ne k}=\Big(\prod_1^{k-1}\Big)
\Big( \prod_{k+1}^s\Big)\Big(\prod_{s+1}^{2b-1}\Big),\label{3.13}
\end{equation}
where $\prod_1^{k-1}$ denotes
$\prod_{j=1}^{k-1}(\alpha_k-\alpha_j)$, etc. From
\eqref{3.5}-\eqref{3.7} we see that $\prod_1^{k-1}$
is positive because $\alpha_k>\alpha_j$ when $1\le j0$ and hence from \eqref{3.20}
we conclude that
\begin{equation}
\operatorname{sgn} \Big(\prod_{j=1}^p
(1-\alpha_k \alpha_j)\Big)=(-1)^{k-1}, \quad k=p+1,\dots,q,
\label{3.21}
\end{equation}
where $q$ is the integer appearing in \eqref{3.2} and
\eqref{3.6}. Defining
\begin{equation}
P_-(z):=\prod_{j=1}^p (1-\alpha_j\,z),\label{3.22}
\end{equation}
let us investigate the sign of $P_-(z)$ when $z$ takes values in
the interval $z\in(-1,0)$.
From \eqref{3.21} and \eqref{3.22} we know that
\begin{equation}
\operatorname{sgn}( P_{-}(\alpha_k)) =(-1)^{k-1},\quad
k=p+1,\dots, q. \label{3.23}
\end{equation}
Thus, from \eqref{3.23} we conclude that in the interval
$z\in(-1,0)$, the polynomial $P_-(z)$ defined in \eqref{3.22}
changes sign at least $(q-p-1)$ times.
This indicates that the polynomial $P_{-}(z)$ must have degree no
less than $q-p-1$.
From \eqref{3.1} and \eqref{3.2} we have $q-p=Z(-1,0)$, and
from \eqref{3.22} we know that the degree of $P_-(z)$ is the
same as $p$, which is equal to $Z(-\infty,-1]$ as seen from
\eqref{3.1}.
Thus, we have proved that
\begin{equation}
Z(-\infty,-1)\ge Z(-1,0)-1,\label{3.24}
\end{equation}
when $Z(-1,0)\ge 1$. We can then write \eqref{3.24} as \eqref{3.15},
and hence the proof of (b) is complete. In a similar way, let us
prove (d). If $Z(0,1)\ge 1$ then there exists at least one
$\alpha_k$-value in the interval $z\in(0,1)$, which corresponds to a
bound state. Let $c_k$ be the corresponding Marchenko norming
constant. From \eqref{2.13}, \eqref{3.12}, and the second line of
\eqref{3.10}, we obtain
\begin{equation}
\operatorname{sgn}(c_k^2)
=(-1)^{k-1}\,\operatorname{sgn}\Big(\prod_{j=r+1}^s
(1-\alpha_k \alpha_j)\Big),\label{3.25}
\end{equation}
where $r$ and $s$ are the integers appearing in \eqref{3.3},
\eqref{3.4}, \eqref{3.7}, and \eqref{3.8}. The norming constant
$c_k$ is positive and hence from \eqref{3.25} we conclude that
\begin{equation}
\operatorname{sgn} \Big(\prod_{j=r+1}^s
(1-\alpha_k\alpha_j)\Big)=(-1)^{k-1},
\quad k=q+1,\dots, r,\label{3.26}
\end{equation}
where we recall that $q$ is the integer appearing in \eqref{3.2}, \eqref{3.6},
and \eqref{3.7}.
Letting
\begin{equation}
P_+(z):=\prod_{j=r+1}^s (1-\alpha_j\,z),\label{3.27}
\end{equation}
we notice that the degree of $P_+(z)$ is $s-r$, which is equal
to $Z[1,+\infty)$ as seen from \eqref{3.3} and \eqref{3.4}.
Let us investigate the sign of $P_+(z)$ when $z$ takes values
in the interval $z\in(0,1)$. From \eqref{3.26} and \eqref{3.27}
we see that
\begin{equation}
\operatorname{sgn} \left(P_{+}(\alpha_k)\right)=(-1)^{k-1},\quad
k=q+1,\dots, r. \label{3.28}
\end{equation}
Thus, from \eqref{3.28} we conclude that in the interval $z\in(0,1)$
the polynomial $P_+(z)$ changes sign at least $(r-q-1)$ times.
Hence, the degree of $P_+(z)$ cannot be less than $(r-q-1)$. From
\eqref{3.2} and \eqref{3.3} we see that $r-q=Z(0,1)$, and we have
already seen that the degree of $P_+(z)$ is equal to
$Z[1,+\infty)$. Thus, we have proved that
\begin{equation}
Z[1,+\infty)\geq Z(0,1)-1, \label{3.29}
\end{equation}
when $Z(0,1)\ge 1$. We can
write \eqref{3.29} in the form
given in \eqref{3.17}, and hence the proof of (d) is complete.
The proof of (e) is obtained as follows.
Using \eqref{3.14}-\eqref{3.17} in
\eqref{2.10} we obtain
$$2b-1=2\left[Z(-1,0)+Z(0,1)-1+Z_c\right]+\varepsilon_{-}+\varepsilon_{+},$$
where the left-hand side is an odd integer
and we have $\varepsilon_{-}+\varepsilon_{+}\ge 0$
on the right-hand side. Thus, (e) must hold.
The result in (f) directly follows from (a)-(d).
\end{proof}
To prepare the proof of one of our two main theorems,
namely that for
any potential in class $\mathcal{A}_b$ having $N$ bound states we must
have $0\le N\le b$, we first prove that for any $b\ge 1$
there exist infinitely many nontrivial potentials in class $\mathcal{A}_b$ for which $N=0$ and also
there exist infinitely many
potentials in class $\mathcal{A}_b$ for which $N=b$. We then prove our other main theorem,
namely, that for each
$k$ in the set $\{0,1,\dots,b\}$ there are infinitely many potentials
in class $\mathcal{A}_b$ having exactly $k$ bound states. We remark
that in the limiting case where $b=0$ the potential class $\mathcal{
A}_b$ contains only the trivial potential where $V_n\equiv 0$, for
which we already know
that $N=0$. Thus, our main result $0\le N\le b$
automatically and trivially holds
also when $b=0$.
In the next theorem we prove that for any fixed $b\ge 1$ the class
$\mathcal{A}_b$ contains infinitely many potentials with
$N=0$. We recall that for each potential in class $\mathcal{A}_b$
we have $V_b\ne 0$.
\begin{theorem} \label{t3.4}
For any fixed $b$ with $b\ge 1$, the class
$\mathcal{A}_b$ specified in Definition \ref{def1.1} contains
infinitely many potentials having no bound states.
\end{theorem}
\begin{proof}
From Theorem~\ref{t2.1}(b) we already know that
the Jost function $f_0(z)$ is a polynomial in the multivariable
$(V_1,V_2,\dots,V_b)$. In fact, $f_0(z)$ considered as such
a polynomial consists of terms where the degree of each term is
between $0$ and $b$. In Theorem~\ref{t2.1}(b) we have seen that the term
with the largest degree is the unique monomial
given by $(V_1\cdots V_b)z^b$.
In fact, $f_0(z)$ as a polynomial in $(V_1,\dots, V_b)$ contains
a single monomial term with zero degree and that term is the
term given by $1$ in $f_0(z)$. We already know that the zero potential
$V_n\equiv 0$ corresponds to $N=0$ with the corresponding Jost function
being $f_0\equiv 1$.
By using a small perturbation on the trivial potential
appropriately, we can get a nontrivial potential with $N=0$.
This can be seen as follows. In \eqref{2.6} every term in
$f_0(z)-1$, viewed as
a polynomial in $(V_1,\dots, V_b)$, has degree at least one.
Thus, we can choose each
$|V_j|$ small enough and with $V_b\ne 0$ so that the corresponding $K_{0j}$
appearing in \eqref{2.6} satisfies
\begin{equation}
|K_{0j}|<\frac{1}{2b}, \quad 1\le j\le 2b-1.\label{3.30}
\end{equation}
By Theorem~\ref{t2.1}(a) we know that each $K_{0j}$ is real and hence
\eqref{3.30} implies that when $z\in (-1,1)$ we obtain
\begin{equation}
-\frac{1}{2b}1$ and
\begin{equation}
|V_j|=|V_b|, \quad j=1,\dots, b-1.\label{3.33}
\end{equation} Let us estimate
the corresponding $|F(z)|$
and $|G(z)|$ on the unit circle $|z|=1$.
Using \eqref{3.33} in \eqref{2.8} we obtain
\begin{equation}
|F(z)|\big|_{|z|=1}=|V_b|^b.\label{3.34}
\end{equation} On the other hand, using
\eqref{3.33} in \eqref{2.9} we obtain
\begin{equation}
|G(z)|\big|_{|z|=1}\le c |V_b|^{b-1}, \label{3.35}
\end{equation}
for some positive constant $c$, and this can be seen as follows.
Note that $G(z)$ can be viewed as a finite
sum of monomials in $(V_1,\dots, V_b)$ of degree $b-1$ or less
multiplied with some nonnegative integer power of $|z|$.
Thus, \eqref{2.9} implies that each such monomial is bounded
by a constant multiple of $|V_b|^{b-1}$, and since there are only
a finite number of such monomials, there exists a positive constant
$c$ depending on $b$ for which \eqref{3.35} holds.
We can choose $|V_b|$ large enough so that $|V_b|>c$, and
hence from \eqref{3.34} and \eqref{3.35} we obtain
\begin{equation}
|G(z)|\big|_{|z|=1}<|F(z)|\big|_{|z|=1}. \label{3.36}
\end{equation}
Using \eqref{3.36} in the decomposition \eqref{2.7}, we can
apply Rouch\'e's theorem of complex variables and conclude that the
number of zeros of $f_0(z)$ inside the unit circle $|z|=1$
coincides with the number of zeros of $F(z)$ inside that unit
circle. By Theorem~\ref{t2.1}(d), the number of zeros of $f_0(z)$
inside the unit circle is equal to $N$, the number of bound states.
On the other hand, using \eqref{3.33} in \eqref{2.8} we have
$$
|F(z)|=|V_b|^b\, |z|^b,$$ and hence $F(z)$ has exactly $b$ zeros
inside the unit circle, and in fact all such zeros occur at $z=0$.
Thus, we have proved that there exists at least one potential with
$V_b\ne 0$ in class $\mathcal{A}_b$ for which $N=b$. In fact, since
we can choose $V_b$ in an infinite number of ways such that
\eqref{3.33} and $|V_b|>c$ are satisfied, it follows that there
are indeed an infinite number of potentials in class $\mathcal{A}_b$ for
which we have $N=b$.
\end{proof}
The next theorem shows that for each integer $k$ in the set
$\{0,1,\dots,b\}$ there are an infinite number of potentials in
class $\mathcal{A}_b$ having exactly $k$ bound states. We recall
that $V_b\ne 0$ for each potential in $\mathcal{A}_b$.
\begin{theorem} \label{t3.6}
For any fixed $b$ with $b\ge
1$ and for each nonnegative integer $k$ in the set
$\{0,1,\dots,b\}$, there correspond infinitely many potentials in
class $\mathcal{A}_b$ having exactly $k$ bound states.
\end{theorem}
\begin{proof}
The proof for $k=0$ is given in Theorem~\ref{t3.4} and the proof for
$k=b$ is given in Theorem~\ref{t3.5}. So, it is enough to provide
the proof for $k$ in the set $\{1,\dots,b-1\}$. In the proof of
Theorem~\ref{t3.5}, the potentials explicitly constructed in class
$\mathcal{A}_b$ with $b$ bound states and with $V_b\ne 0$ are all
generic, i.e.\ the corresponding Jost functions $f_0(z)$ do not
vanish at $z=\pm 1$. This can be seen
from \eqref{2.7} and \eqref{3.36} by noting that on the unit circle
$|z|=1$ we have
$$
|f_0(z)|\big|_{|z|=1}=|F(z)+G(z)|\big| _{|z|=1}
\ge |F(z)|\big| _{|z|=1}-|G(z)|\big| _{|z|=1}>0,
$$
and hence $f_0(z)$ cannot vanish on the unit circle and in
particular it cannot vanish at $z=\pm 1$. Thus, for each integer $k$
in the set $\{1,\dots,b-1\}$ we conclude that there is at least one
generic potential $V$ with exactly $k$ bound states in class
$\mathcal{A}_k$ with $V_k\ne 0$ and $V_n=0$ for $n> k$. Let us
assume that the corresponding bound-state zeros of the Jost function
$f_0(z)$ occur at $z=z_l$ with $l=1,\dots,k$. We know from
Theorem~\ref{t2.1}(d) that each such $z_l$ is simple and belongs to the set
$z\in(-1,0)\cup(0,1)$. Let us continuously perturb the potential $V$
to $\tilde V$ in such a way that $\tilde V_n=V_n$ for $n\le k$,
$\tilde V_n=\epsilon$ for $kb$,
where $\epsilon$ is a nonzero real parameter. The perturbed
potential $\tilde V$ belongs to class $\mathcal{A}_b$, and let us
use $\tilde f_0(z)$ for the corresponding Jost function. By
Theorem~\ref{t2.1}(b) we know that as we perturb $V$ to $\tilde V$
continuously, the coefficients of $f_0(z)$ change continuously, and
hence the zeros of $f_0(z)$ also change continuously. We claim that
when $\epsilon$ is small the number of zeros of $\tilde f_0(z)$ in
$z\in(-1,0)\cup(0,1)$ must be equal to $k$. This can be seen as
follows. Because of Theorem~\ref{t2.1}(g), as we introduce the
perturbation none of $z_l$-values can abruptly change to nonreal
values, and because of Theorem~\ref{t2.1}(c) none of those
$z_l$-values can change to zero. Thus, the only way to change the
number of bound states during the continuous perturbation would be
for a zero of $f_0(z)$ to move into or out of $z\in(-1,0)\cup(0,1)$
through $z=-1$ or $z=1$. By choosing $\epsilon$ small enough, we
know that we must have $\tilde f_0(\pm 1)\ne 0$ because we have
$f_0(\pm 1)\ne 0$. Thus, we have shown that for small enough
$\epsilon$ we have $\tilde V$ belonging to class $\mathcal{A}_b$
with $\tilde V_b\ne 0$ and the corresponding perturbed operator has
exactly $k$ bound states. Let us remark that, as we continuously
perturb the potential $V$ to $\tilde V$, the degree of $f_0(z)$
abruptly changes from $2\,k-1$ to $2\,b-1$, which is the degree of
$\tilde f_0(z)$. However, the additional zeros of the Jost function
enter the complex-$z$ plane
from $z=\infty$ and hence
they do not increase the number of bound states.
Thus, our proof is complete.
\end{proof}
Having proved one of our main results in Theorem~\ref{t3.6}, in the
next theorem we prove our other main result.
\begin{theorem} \label{t3.7}
Assume that the potential $V$ appearing in \eqref{1.1}
belongs to class $\mathcal{A}_b$ for some positive integer $b$.
Then, the number of bound states, denoted by $N$,
for the corresponding discrete Schr\"odinger operator
associated with \eqref{1.1} and \eqref{1.2}
satisfies the inequalities
\begin{equation}
0\le N\le b.\label{3.37}
\end{equation} We remark that the result trivially
holds also when $b=0$.
\end{theorem}
\begin{proof}
Let us first indicate that \eqref{3.37} holds when $b=0$ because in
that case $\mathcal{A}_b$ consists of the trivial potential
$V_n\equiv 0$ for which we already know \cite{2} that $N=0$. By
Theorem~\ref{t2.1}(d) we have
\begin{equation}
Z(-1,0)+Z(0,1)=N,\label{3.38}
\end{equation}
where we recall that $Z(-1,0)$ and $Z(0,1)$ denote the number
of zeros of the Jost function $f_0(z)$ in the respective intervals
$z\in (-1,0)$ and $z\in(0,1)$.
With the help of \eqref{3.38}, from \eqref{3.18}
and \eqref{3.19} we conclude that
\begin{equation}
Z(-\infty,-1]+Z[1,+\infty)\geq \begin{cases}
0, & N=0, \\
N-1,& N=1, \\
N-2, & N\geq 2.
\end{cases} \label{3.39}
\end{equation}
Using \eqref{3.38} and \eqref{3.39} in \eqref{2.10}, we obtain
\begin{equation}
2b-1\ge N+2\,Z_c+
\begin{cases}
0,& N=0, \\
N-1,& N=1, \\
N-2,& N\geq 2.
\end{cases} \label{3.40}
\end{equation}
Since $Z_c\ge 0$, from \eqref{3.40} we obtain
$$2b-1\ge
\begin{cases}
N, & N=0, \\
2 N-1, & N=1, \\
2 N-2, & N\geq 2,\end{cases}
$$
or equivalently
\begin{equation}
2b-1\geq \begin{cases}
0,& N=0, \\
1,& N=1, \\
2 N-2,& N\geq 2.
\end{cases} \label{3.41}
\end{equation}
From \eqref{3.41} we see that
\begin{equation}
b\geq \begin{cases}
1/2, & N=0, \\
1,& N=1, \\
N-\frac{1}{2},& N\geq 2.
\end{cases} \label{3.42}
\end{equation}
From \eqref{3.42} we infer that
\begin{equation}
N\leq \begin{cases}
b, & N=0, \\
b, & N=1, \\
b+\frac{1}{2}, & N\geq 2.
\end{cases} \label{3.43}
\end{equation}
From the third line of \eqref{3.43} we conclude that any
potential in class $\mathcal{A}_b$ satisfies $N****1$ and hence $\alpha_1\not\in (-1,0)\cup(0,1)$,
indicating that $N=0$. On the other hand, when $|V_1|>1$,
we have $0<|\alpha_1|<1$ and hence $\alpha_1\in (-1,0)\cup(0,1)$,
indicating that $N=1$. Let us also remark that the inequalities given
is \eqref{3.18} and \eqref{3.19} hold when $N=0$ and also when $N=1$.
\end{example}
In the next example, we explore all the possibilities for the bound
states in class $\mathcal{A}_b$ with $b=2$.
\begin{example}\label{exa4.2}\rm
Consider the potential class $\mathcal{A}_b$ with $b=2$, and hence
we have $V_n=0$ for $n>2$ and $V_2\ne 0$. From \eqref{2.11} we
obtain the corresponding Jost function as
\begin{equation}
f_0(z)=1+\left(V_1+V_2\right)z+V_1V_2z^2+V_2z^3.\label{4.2}
\end{equation}
In this case $f_0(z)$ has three zeros $\alpha_1$, $\alpha_2$,
$\alpha_3$ and we have
\begin{equation}
f_0(z)=\Big(1-\frac{z}{\alpha_1}\Big)
\Big(1-\frac{z}{\alpha_2}\Big)
\Big(1-\frac{z}{\alpha_3}\Big). \label{4.3}
\end{equation}
Equating the corresponding coefficients in \eqref{4.2} and
\eqref{4.3} we obtain
\begin{equation}
\left\{\begin{aligned}
&V_1+V_2 =-\frac{1}{\alpha_1}-\frac{1}{\alpha_2}-\frac{1}{\alpha_3},\\
&V_1V_2 =\frac{1}{\alpha_1\alpha_2}
+\frac{1}{\alpha_1\alpha_3}+\frac{1}{\alpha_2\alpha_3},\\
&V_2= -\frac{1}{\alpha_1\alpha_2\alpha_3}.
\end{aligned}\right. \label{4.4}
\end{equation}
The nonlinear algebraic equations given in \eqref{4.4}
impose certain restrictions on
the locations of $\alpha_1$, $\alpha_2$, $\alpha_3$ in the
complex-$z$ plane when $V_1$ and $V_2$ take real values. We can
view $V_1$ and $V_2$ as the two solutions to the quadratic equation
\begin{equation}
x^2-\left(V_1+V_2\right) x+V_1V_2=0,\label{4.5}
\end{equation}
where the solutions are given by
\begin{equation}
x=\frac{V_1+V_2\pm \sqrt{(V_1+V_2)^2-4V_1V_2}}{2}.\label{4.6}
\end{equation}
Thus, $V_2$ must be equal to one of the two quantities on the
right-hand side of \eqref{4.6}.
Using the first two lines of \eqref{4.4} in \eqref{4.6} we obtain
\begin{equation}
x=
\frac{-\big(\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+\frac{1}{\alpha_3}
\big)\pm
\sqrt{\big(\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+\frac{1}{\alpha_3}
\big)^2-4\big(\frac{1}{\alpha_1\alpha_2}
+\frac{1}{\alpha_1\alpha_3}+\frac{1}{\alpha_2\alpha_3}
\big)}}{2}.
\label{4.7}
\end{equation}
Comparing \eqref{4.7} with the third line of \eqref{4.4} we
see that we must have
\begin{equation}
\begin{split}
&-\frac{2}{\alpha_1\alpha_2\alpha_3}+
\Big(\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+\frac{1}{\alpha_3} \Big)\\
& =
\sqrt{\big(\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+\frac{1}{\alpha_3}
\big)^2-4\big(\frac{1}{\alpha_1\alpha_2}
+\frac{1}{\alpha_1\alpha_3}+\frac{1}{\alpha_2\alpha_3}
\big)},\label{4.8}
\end{split}
\end{equation}
or
\begin{equation}
\begin{split}
&-\frac{2}{\alpha_1\alpha_2\alpha_3}+
\Big(\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+\frac{1}{\alpha_3}
\Big)\\
&=-
\sqrt{\big(\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+\frac{1}{\alpha_3}
\big)^2-4\big(\frac{1}{\alpha_1\alpha_2}
+\frac{1}{\alpha_1\alpha_3}+\frac{1}{\alpha_2\alpha_3}
\big)}\,.\label{4.9}
\end{split}
\end{equation}
In order for \eqref{4.8} or \eqref{4.9} to hold, we must have
\begin{equation}
-\left(\alpha_1\alpha_2\alpha_3\right) \left(\alpha_1+\alpha_2+\alpha_3\right)+\left(\alpha_1\alpha_2
+\alpha_1\alpha_3+\alpha_2\alpha_3\right)=1,
\label{4.10}
\end{equation} which
is obtained by squaring both sides of \eqref{4.8} and \eqref{4.9}
and then simplifying the resulting expressions. We can write
\eqref{4.10} in the equivalent form as
\begin{equation}
\alpha_1\alpha_2\left(1-\alpha_3^2\right)+
\alpha_1\alpha_3\left(1-\alpha_2^2\right)+
\alpha_2\alpha_3\left(1-\alpha_1^2\right)=1. \label{4.11}
\end{equation}
The conditions \eqref{4.8} and \eqref{4.9} impose various
restrictions on $\alpha_1$, $\alpha_2$, $\alpha_3$. For example,
we cannot have $\alpha_1$, $\alpha_2$, $\alpha_3$ all located in the
interval $[1,+\infty)$. Otherwise, the left-hand side of
\eqref{4.11} would be negative. Similarly, we cannot have $\alpha_1$, $\alpha_2$, $\alpha_3$ all
located in the interval $(-\infty,-1]$. Otherwise, the left-hand
side of \eqref{4.10} would again be negative.
On the other hand, for example, we have a double real resonance and a bound
state with
\begin{equation}
\alpha_1=\alpha_2=2,\quad
\alpha_3=-\frac{3}{2}+\sqrt{3}=0.23205\overline{1},\label{4.12}
\end{equation}
which
correspond to
\begin{equation}
V_1=-\frac{5}{2}-\sqrt{3}=-4.2320\overline{5}, \quad
V_2=-\frac{1}{2}-\frac{1}{\sqrt{3}}=-1.0773\overline{15}.
\label{4.13}
\end{equation}
Here, we use an overline on a digit to indicate a round off.
For example, we have a double real resonance and another real resonance with
$$
\alpha_1=\alpha_2=2, \quad
\alpha_3=-\frac{3}{2}-\sqrt{3}=-3.23205\overline{1},
$$
with the corresponding potential values given by
$$
V_1=-\frac{5}{2}+\sqrt{3}=-0.7679\overline{5},\quad
V_2=-\frac{1}{2}+\frac{1}{\sqrt{3}}=0.0773\overline{5}.
$$
We also get a double real resonance and another real resonance with
\begin{equation}
\alpha_1=\alpha_2=-2,\quad
\alpha_3=\frac{3}{2}+\sqrt{3}=3.2320\overline{5},\label{4.14}
\end{equation}
which correspond to
\begin{equation}
V_1=\frac{5}{2}-\sqrt{3}=0.7679\overline{5}, \quad
V_2=\frac{1}{2}-\frac{1}{\sqrt{3}}=-0.0773\overline{5}. \label{4.15}
\end{equation}
The restriction \eqref{3.18} indicates that if we have two bound
states with $\alpha_1$ and $\alpha_2$
both being in the interval $(-1,0)$, then we must have
$\alpha_3\le -1$. However, in that case the restriction
\eqref{4.11} rules out the possibility $\alpha_3=-1$ and dictates that
we must have $\alpha_3\le -1.5610\overline{5}$.
Similarly, the restriction
\eqref{3.19} indicates that if we have two bound states with
$\alpha_1$ and $\alpha_2$ both being in
the interval $(0,1)$, then we must have a real resonance
with $\alpha_3\ge 1.5610\overline{5}$.
Since nonreal resonances must occur in complex-conjugate pairs
it is impossible to have a complex resonance
and two bound states.
We can have two bound
states and one real resonance by choosing $\alpha_1,\alpha_2,\alpha_3$
appropriately so that the corresponding $V_1$ and $V_2$ are
real valued. For example, for
$$
V_1=-\sqrt{5}=-2.2360\overline{7},\quad V_2=\frac{4}{\sqrt{5}}
=1.7888\overline{5},
$$
we obtain
$$
\alpha_1=\frac{1}{2},\quad \alpha_2=-\frac{1}{2},\quad \alpha_3
=\sqrt{5}.
$$
Choosing
$$
V_1=\sqrt{5},\quad
V_2=-\frac{4}{\sqrt{5}}=-1.7888\overline{5},
$$
we obtain
$$
\alpha_1=\frac{1}{2},\quad \alpha_2=-\frac{1}{2},\quad \alpha_3=-\sqrt{5}.
$$
Choosing
$$
V_1=\frac{-13+\sqrt{22}}{3}=-2.7698\overline{6},\quad
V_2=-4-4\sqrt{\frac{2}{11}}=-5.7056\overline{1},$$
we obtain
$$
\alpha_1=\frac{1}{6},\quad \alpha_2=\frac{1}{2},\quad
\alpha_3= \frac{11-\sqrt{22}}{3}=2.1031\overline{9},
$$
and choosing
$$
V_1=\frac{13-\sqrt{22}}{3}=2.7698\overline{6},\quad
V_2=4+4\sqrt{\frac{2}{11}}=5.7056\overline{1},
$$
we obtain
$$
\alpha_1=-\frac{1}{6},\quad
\alpha_2=-\frac{1}{2},\quad \alpha_3= \frac{-11+\sqrt{22}}{3}=-2.1031\overline{9}.
$$
\end{example}
In the following example, we illustrate some of the possibilities
for the number of bound states and resonances for potentials in
class $\mathcal{A}_b$ with $b=3$.
\begin{example}\label{exa4.3} \rm
Consider the potential class $\mathcal{A}_b$ with $b=3$, and hence
$V_n=0$ for $n>3$ and $V_3\ne 0$. From \eqref{2.11} we see that the
corresponding Jost function $f_0(z)$
is expressed in terms of $V_1$, $V_2$, $V_3$ as
\begin{equation}
\begin{aligned}
f_0(z)&=1+\left(V_1+V_2+V_3\right)z+\left[V_1V_2+(V_1+V_2)V_3\right]z^2\\
&\quad + \left[V_2+V_3(1+V_1V_2)\right]z^3 +V_3\left(V_1+V_2\right)z^4+V_3\,z^5.
\end{aligned}\label{4.16}
\end{equation}
In terms of the zeros $\alpha_1$, $\alpha_2$, $\alpha_3$,
$\alpha_4$, $\alpha_5$ of $f_0(z)$ we have the representation
\begin{equation}
f_0(z)=\Big(1-\frac{z}{\alpha_1}\Big)\Big(1-\frac{z}{\alpha_2}\Big)
\Big(1-\frac{z}{\alpha_3}\Big) \Big(1-\frac{z}{\alpha_4}\Big)
\Big(1-\frac{z}{\alpha_5}\Big). \label{4.17}
\end{equation}
By equating the corresponding coefficients in \eqref{4.16}
and \eqref{4.17} we express
$\alpha_1$,
$\alpha_2$, $\alpha_3$, $\alpha_4$, $\alpha_5$
in terms of $V_1$, $V_2$, $V_3$ as a nonlinear system of five equations
given by
\begin{equation}
\left\{\begin{aligned}
& V_1+V_2+V_3 =-\Big(\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+
\frac{1}{\alpha_3}+\frac{1}{\alpha_4}+\frac{1}{\alpha_5} \Big),\\
&V_1V_2+\left(V_1+V_2\right)V_3
= \frac{1}{\alpha_1\alpha_2}+\frac{1}{\alpha_1\alpha_3}+\dots+
\frac{1}{\alpha_4\alpha_5},\\
&V_2+V_3\left(1+V_1V_2\right)
=-\Big(\frac{1}{\alpha_1\alpha_2\alpha_3}+
\frac{1}{\alpha_1\alpha_2\alpha_4}+\dots+
\frac{1}{\alpha_3\alpha_4\alpha_5} \Big), \\
&\left(V_1+V_2\right)V_3=\frac{1}{\alpha_1\alpha_2\alpha_3\alpha_4}+
\frac{1}{\alpha_1\alpha_2\alpha_3\alpha_5}+\dots
+\frac{1}{\alpha_2\alpha_3\alpha_4\alpha_5},\\
&V_3=-\frac{1}{\alpha_1\alpha_2\alpha_3\alpha_4\alpha_5}.
\end{aligned} \right.\label{4.18}
\end{equation}
Notice that if \eqref{4.18} has a solution, then we can change
the sign of each of $V_1$, $V_2$, $V_3$ and $\alpha_1$, $\alpha_2$, $\alpha_3$,
$\alpha_4$, $\alpha_5$ and get another solution. The nonlinear relations given
in \eqref{4.18} put certain restrictions on the locations of $\alpha_1$,
$\alpha_2$, $\alpha_3$, $\alpha_4$,
$\alpha_5$ on the complex-$z$ plane in order to have $V_1$, $V_2$, $V_3$
as real-valued constants. The system in \eqref{4.18} can be solved
to express $V_1$, $V_2$, $V_3$, $\alpha_5$
in terms of $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$ by solving the
fifth line in \eqref{4.18} for $V_3$, then solving the fourth line
for $V_2$, then solving the first line for $\alpha_5$, and
solving the second line for $V_1$. Then, we can use the resulting
expressions for $V_1$, $V_2$, $V_3$, $\alpha_5$ in the third line of (4.18)
to get the consistency. We then obtain a consistency equation involving
$\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$. By assigning various
allowable values for $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$,
we can then produce some explicit examples. For example, by choosing
$$
\alpha_2=\alpha_1,\quad \alpha_3=\alpha_4=\alpha_1^\ast,
$$
we can obtain the existence of a double nonreal resonance with
$\alpha_1=-0.3196\overline{8}+2i$ corresponding to
$$
\alpha_5=-0.60017\overline{2},\quad
V_1=1.1327\overline{9},\quad V_2=0.74610\overline{6},
\quad V_3=0.099012\overline{9},
$$
and we observe that this case has exactly one bound state at $z=\alpha_5$.
We obtain another example with
\begin{gather}
\alpha_1=1.161\overline{3}+i,
\quad \alpha_2=\alpha_1,\quad
\alpha_3=\alpha_4=\alpha_1^\ast,\quad
\alpha_5=0.2779\overline{7},\label{4.19} \\
V_1=-1.8911\overline{4},\quad
V_2=-3.0320\overline{2},\quad
V_3=-0.652\overline{2},\label{4.20}
\end{gather}
which indicates that we have one bound state at $z=\alpha_5$,
a double complex resonance at $z=\alpha_1$, and a double complex resonance at
$z=\alpha_1^\ast$.
\end{example}
In the final example below we present a specific example where $N=b$
is attained in class $\mathcal{A}_b$.
\begin{example}\label{exa4.4}\rm
For a fixed positive value of $b$, let us choose the potential appearing
in \eqref{1.1} as
\begin{equation}
V_n=\begin{cases}
(-1)^n 2, & 1\le n\le b,\\
0, & n>b,
\end{cases} \label{4.21}
\end{equation}
so that it belongs to class $\mathcal{A}_b$.
Using \eqref{2.11} we then get the Jost function $f_0(z)$
explicitly expressed as a polynomial in $z$ of degree $2b-1$.
Using the symbolic computing system Mathematica,
we evaluate the zeros of $f_0(z)$ numerically and observe
that e.g.\ for each $b=1,2,\dots, 110$ the resulting $f_0(z)$
has exactly $b$ real zeros in $z\in(-1,0)\cup(0,1)$.
This numerically confirms the result presented in Theorem~\ref{t3.5}.
We remark that as $b$ increases some of the zeros of $f_0(z)$
start getting closer to $z=\pm 1$. In that case, one needs to increase
the accuracy of the numerical program used in Mathematica to evaluate
the zeros of a polynomial function to avoid any discrepancies. If one replaces
the value of $2$ in the first line of \eqref{4.21} with a larger value,
then the zeros of $f_0(z)$ in the set
$(-1,0)\cup(0,1)$ move away from $z=\pm 1$ and hence
it becomes easier to confirm
$N=b$ for large $b$-values during the
numerical evaluation of the zeros of $f_0(z)$.
\end{example}
\subsection*{Acknowledgments}
T. Aktosun wants to express his gratitude to the Institute of Physics
and Mathematics of the Universidad Michoacana de San Nicol\'as de
Hidalgo, M\'exico for its hospitality.
A. E. Choque-Rivero was partially supported by SNI-CONACYT and CIC-UMNSH, Mexico.
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\end{document}
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