#  Interior inverse problem for global conservative multipeakon solutions of
the Camassa-Holm equation

Tao Liu  School of Mathematics and Statistics, Shaanxi Normal University,
Xi’an 710062, PR China<EMAIL_ADDRESS>and  Kang Lyu  School of
Mathematics and Statistics, Nanjing University of Science and Technology,
Nanjing 210094, PR China<EMAIL_ADDRESS>
###### Abstract.

We consider the interior inverse problem associated with the global
conservative multipeakon solution of the Camassa-Holm equation. Based on the
inverse spectral theory on the half-line and the oscillation property of
eigenfunctions, some (non)uniqueness results of the interior inverse problem
are obtained. In addition, we give the trace formula, which connects the
global conservative multipeakon solution with the corresponding eigenvalues
and normalized eigenfunctions.

###### Key words and phrases:

Interior inverse problem; The Camassa-Holm equation; Weyl-Titchmarsh function

###### 2020 Mathematics Subject Classification:

Primary: 37K10, 34B07; Secondary: 34B09, 34A55

##  1\. Introduction

The Camassa-Holm (CH) equation

$u_{t}-u_{xxt}=2u_{x}u_{xx}-3uu_{x}+uu_{xxx}$ (1.1)

has received a lot of attention during the last two decades (cf [6, 7, 33, 13,
14, 12, 26, 34, 35, 36, 37, 38, 44, 45]). The isospectral problem [1, 2, 42]
of the CH equation is the weighted Sturm-Liouville equation

$-f^{\prime\prime}(x)+\frac{1}{4}f(x)=z\omega(x,t)f(x),\ x\in\mathbb{R}.$
(1.2)

Here $\omega=u-u_{xx}$ is called the weight function, $z\in\mathbb{C}$ is the
complex parameter. Inverse spectral theory for (1.2) is useful for solving the
Cauchy problem of the CH equation.

The multipeakon solution of the CH equation is the solution of the form

$u(x,t)=\frac{1}{2}\sum_{i=1}^{N}\omega_{i}(t)e^{-|x-x_{i}(t)|},$ (1.3)

which represents the interaction of $N$ peaks at position
$x_{i},i=1,\cdots,N$. In this case,

$\omega(x,t)=u(x,t)-u_{xx}(x,t)=\sum_{i=1}^{N}\omega_{i}(t)\delta_{x_{i}(t)},$
(1.4)

where $\delta_{x_{i}(t)}$ is the Dirac measure at $x_{i}(t)$. Beals, Sattinger
and Szmigielski [2] studied the multipeakon solution by solving the inverse
problem of (1.2). They observed that if $\omega$ is definite, then the inverse
problem is solvable for any $t\in\mathbb{R}$. Otherwise, the inverse problem
is not always solvable. This causes the collision between some peaks, which
leads to finite-time blow-up resembling wave breaking. Bressan and Constantin
[5] introduced the pair $(u,\mu)$ to describe the overall picture of the
collision, where $\mu$ is a non-negative Borel measure with the absolutely
continuous part determined by $u$ via

$\mu_{ac}(B,t)=\int_{B}|u(x,t)|^{2}+|u_{x}(x,t)|^{2}dx,\ t\in\mathbb{R},$

for each Borel set $B\in\mathcal{B}(\mathbb{R})$. The pair $(u,\mu)$ is called
the global conservative multipeakon solution ([5, 19, 30]). In [19], Eckhardt
and Kostenko introduced a generalized isospectral problem

$-f^{\prime\prime}(x)+\frac{1}{4}f(x)=z\omega(x,t)f(x)+z^{2}v(x,t),\
x\in\mathbb{R}.$ (1.5)

In the indefinite case, the inverse problem is always solvable within the
class of spectral problem (1.5). $v(\cdot,t)$ is the singular part of
$\mu(\cdot,t)$ and get switched on exact the time of collisions.

Previous literature on spectral analysis and inverse spectral analysis for
(1.5) always considered the case $v=0$ (cf [2, 1, 3, 12, 23, 25, 8, 11, 10,
9]). For example, In [8], Chu and Meng gave the maximization of eigenvalue
ratios. If $v=0$, $\omega$ is definite and smooth enough, this problem can be
transformed into a Sturm-Liouville problem [1, 38] and hence direct and
inverse spectral conclusions may be drawn from this observation. We refer to
[4, 46, 48, 49, 29, 51, 52, 54] for some spectral and inverse spectral results
of Sturm-Liouville operators. Inverse spectral problems for (1.5) with $v\neq
0$ were widely studied by Eckhardt and his collaborators (cf[18, 19, 21, 17]
and the literature cited therein). In [19], Eckhardt and Kostenko showed that
$\omega$ and $v$ are uniquely determined by the spectral measure in the case
that $\omega$ and $v$ are discrete and supported on a finite set. In [21],
Eckhardt and Kostenko obtained some uniqueness results, if $\omega$ and $v$
are periodic and discrete. In the case that $\omega$ and $v$ are general Borel
measures, Eckhardt and Kostenko [18] showed that $\omega$ and $v$ are uniquely
determined by the spectral measure under additional conditions on the growth
property of solutions. We mention that the problem (1.5) with $f(x)/4$
vanished has been studied in [31, 32].

We want to mention a result about inverse Jacobi matrix problem. In [28],
Gesztesy and Simon reconstructed the Jacobi matrix from a diagonal Green’s
function, which motivates us to consider the following interior inverse
problem: For given $a\in\mathbb{R}$, reconstruct $\omega,v$ from
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ (applying Lemma 3.1), where
$\varphi_{i}(a)$ is the normalized eigenfunction defined by (3.1).

There are two main difficulties in solving this interior inverse problem.
Firstly, we study the diagonal Green’s function
$z\phi_{+}(z,a)\phi_{-}(z,a)/W(z)$ (see the definitions of
$\phi_{\pm}(z,a),W(z)$ in Section 2) so that we can determine the Weyl-
Titchmarsh function from the interior spectral data. The diagonal Green’s
function in [28] is a rational Herglotz-Nevanlinna function with

$\displaystyle c:=\deg(den)-\deg(num)=1,$

where $\deg(den)$ and $\deg(num)$ are the degrees of the denominator and
numerator, respectively. However, there are three cases in our problem:
$c=0,1,-1$. This will yield more complicated results. Moreover, we also obtain
various interior inverse spectral conclusions in terms of the signs of
eigenvlalues (see Theorem 3.5 for example). Secondly, to determine the Weyl-
Titchmarsh functions ($M_{\pm}(z,a)$) from the interior spectral data, we need
to divide the poles of $M_{+}(z,a)+M_{-}(z,a)$ into two parts: the poles of
$M_{+}(z,a)$, the poles of $M_{-}(z,a)$. However, the change of the signs of
$\varphi_{i}(a)$ induces that it is hard to study the poles, especially when
$\varphi_{i}(a)=0$ for some $i$. This is different from the previous papers
(cf[16, 20, 25, 48, 29, 47, 46, 50, 51, 53, 52, 24, 39, 40, 41]).

The second main result of this paper is the trace formula, which connects the
global conservative multipeakon solution $(u,\mu)$ with the corresponding
eigenvalues and normalized eigenfunctions of (1.5). Let
$\\{\lambda_{i}\\}_{i=1}^{N}$ be the eigenvalues of (1.5). For a fixed $t$,
let $\\{\varphi_{i}(x,t)\\}_{i=1}^{N}$ be the corresponding normalized
eigenfunctions. Then we have (see Theorem 5.1)

$u(x,t)=\frac{1}{2}\sum_{i=1}^{N}\frac{|\varphi_{i}(x,t)|^{2}}{\lambda_{i}}.$
(1.6)

It is well-known that $u$ consists of $N$ widely separated peakons in the long
time limit. By (1.6) we know that there is one
$|\varphi_{i}|^{2}/2\lambda_{i}$ associated with each peakon. This type of
trace formula first appeared in [27, Theorem 3.4] for the KdV equation.

The structure of this paper is as follows. Necessary facts on the generalized
spectral problem are collected in Section 2. In Section 3, we are concerned
with the inverse interior problem under the condition that $\varphi_{i}(a)\neq
0$ for any $i=1,\cdots,N$. Section 4 considers the inverse interior problem
with $\varphi_{i}(a)=0$ for some $i$. The complete characterization of the
interior spectral data corresponding to discrete measures $\omega$ and $v$ is
given. In Section 5, we give the trace formula (1.6). By virtue of (1.6), we
give the supremum of $|u(x,t)|$, which depends only on the height(depth) of
the highest(deepest) peakon. In Section 6, we provide a proof for oscillation
of eigenfunctions, which is needed for proving some interior inverse spectral
results.

##  2\. The Generalized Spectral Problem

In this section, we give some preliminary assertions about the generalized
spectral problem. We will closely follow the notation employed in [19].

Let $n\in\mathbb{N}$ be fixed. Suppose that $x_{1}<x_{2}<\cdots<x_{n}$. Define
the measures $\omega$ and $v$ by

$\omega=\sum_{i=1}^{n}\omega_{i}\delta_{x_{i}},\
v=\sum_{i=1}^{n}v_{i}\delta_{x_{i}},$ (2.1)

where $\omega_{i}\in\mathbb{R}$, $v_{i}\geq 0$, $\delta_{x_{i}}$ is the Dirac
measure at $x_{i},\ i=1,\cdots,n$. We always assume that

$|\omega_{i}|+v_{i}>0,\ i=1,\cdots,n.$ (2.2)

Consider the generalized spectral problem

$Hf=-f^{\prime\prime}(x)+\frac{1}{4}f(x)=z\omega(x)f(x)+z^{2}v(x)f(x),\
x\in\mathbb{R},$ (2.3)

where $\omega$ and $v$ are defined by (2.1). Note that equation (2.3) has to
be interpreted in the sense of distributions, namely, a function $f$ satisfies
(2.3) if and only if

$-f^{\prime\prime}(x)+\frac{1}{4}f(x)=0,x\in\mathbb{R}\backslash\\{x_{1},\cdots,x_{n}\\},$
(2.4)

and for any $i=1,\cdots,n,$

$\displaystyle f(x_{i}-)=f(x_{i}+),$ (2.5) $\displaystyle
f^{\prime}(x_{i}-)=f^{\prime}(x_{i}$
$\displaystyle+)+z\omega_{i}f(x_{i}+)+z^{2}v_{i}f(x_{i}+).$ (2.6)

We mention that $f$ is differentiable on
$\mathbb{R}\setminus\\{x_{1},\cdots,x_{n}\\}$. We define

$\displaystyle f^{\prime}(x_{i}):=f^{\prime}(x_{i}-),\ i=1,\cdots,n,$ (2.7)

so that $f^{\prime}$ is left-continuous on $\mathbb{R}$.

The values of the parameter $z$ for which (2.3) has nontrivial bounded
solutions are called eigenvalues of $H$, and the corresponding nontrivial
bounded solutions are called eigenfunctions. All the eigenvalues are real (see
for example Theorem 6.8 or [19]). The set of eigenvalues is called the
spectrum of $H$ (denoted by $\sigma(H)$).

For any $z\in\mathbb{C}$, there exist solutions $\phi_{\pm}(z,\cdot)$ of (2.3)
satisfying

$\phi_{\pm}(z,x)=e^{\mp\frac{x}{2}}$ (2.8)

for all $x$ near $\pm\infty$. The Wronskian of $\phi_{+}(z,x),\phi_{-}(z,x)$,
which is independent of $x$, is given by

$W(z)=\phi_{+}(z,x)\phi_{-}^{\prime}(z,x)-\phi_{+}^{\prime}(z,x)\phi_{-}(z,x),z\in\mathbb{C}.$
(2.9)

Note that by (2.3), (2.5) and (2.6), $W(z)$ is a polynomial with degree $\leq
2n$. For $\lambda\in\mathbb{R}$, $W(\lambda)=0$ if and only if
$\phi_{+}(\lambda,x)$ and $\phi_{-}(\lambda,x)$ are linear dependent. Namely,
there exists a nonzero constant $c_{\lambda}$ such that

$\phi_{-}(\lambda,x)=c_{\lambda}\phi_{+}(\lambda,x).$ (2.10)

By (2.8), we know that the zeros of $W(z)$ coincide with $\sigma(H)$.
Therefore, $\sigma(H)$ consists of finitely many eigenvalues. Notice that
$0\notin\sigma(H)$ because of

$\displaystyle{W(0)=1}.$ (2.11)

For any eigenvalue $\lambda$, denote by

$\gamma_{\lambda}^{2}:=\int_{\mathbb{R}}|\phi_{+}(\lambda,x)|^{2}d\omega(x)+2\lambda\int_{\mathbb{R}}|\phi_{+}(\lambda,x)|^{2}dv(x).$
(2.12)

Applying (2.3), by some calculations one can obtain

$\lambda\gamma_{\lambda}^{2}=\frac{1}{4}\int_{\mathbb{R}}|\phi_{+}(\lambda,x)|^{2}dx+\int_{\mathbb{R}}|\phi_{+}^{\prime}(\lambda,x)|^{2}dx+\lambda^{2}\int_{\mathbb{R}}|\phi_{+}(\lambda,x)|^{2}dv(x)>0.$

We call $\lambda\gamma_{\lambda}^{2}$ the (modified) norming constant
associated with the eigenvalue $\lambda$.

######  Lemma 2.1.

[19, Proposition 3.1] For any $\lambda\in\sigma(H)$, $\lambda$ is real and

$\dot{W}(\lambda)=-c_{\lambda}\gamma_{\lambda}^{2}\neq 0,$ (2.13)

where $c_{\lambda}$ is defined by (2.10).

We need to introduce more concepts for studying the interior spectral problem
discussed in the following. For a fixed $a\in\mathbb{R}$, consider
$H_{\pm}(\omega,v,a)$ defined by restricting (2.3) to the intervals
$I_{\pm}(a)$ with Dirichlet boundary condition at $a$, where
$I_{+}(a)=[a,+\infty)$ and $I_{-}(a)=(-\infty,a)$. The Weyl-Titchmarsh
functions of $H_{+}(\omega,v,a)$ and $H_{-}(\omega,v,a)$ are defined by

$M_{+}(z,a)=\frac{\phi_{+}^{\prime}(z,a)}{{z}\phi_{+}(z,a)}$ (2.14)

and

$M_{-}(z,a)=-\frac{\phi_{-}^{\prime}(z,a)}{{z}\phi_{-}(z,a)},$ (2.15)

respectively. Clearly, ${M_{+}(z,a)}$, ${M_{-}(z,a)}$ are Herglotz-Nevanlinna
functions (see [18, Lemma 4.9]). Note that

$\displaystyle\phi_{\pm}(0,a)=e^{\mp\frac{a}{2}}.$ (2.16)

Then by (2.14) and (2.15), we have that

$\displaystyle{\rm{Res}}_{z=0}M_{+}(z,a)={\rm{Res}}_{z=0}M_{-}(z,a)=-\frac{1}{2}.$
(2.17)

Denote by

$l_{i}:=2\left(\tanh\left(\frac{x_{i+1}-a}{2}\right)-\tanh\left(\frac{x_{i}-a}{2}\right)\right),i=0,\cdots,n,$
(2.18)

and

$m_{i}(z):=(zv_{i}+\omega_{i})\cosh^{2}\left(\frac{x_{i}-a}{2}\right),z\in\mathbb{C},i=1,\cdots,n,$
(2.19)

where $x_{0}:=-\infty$ and $x_{n+1}:=+\infty$.

The proofs of the following two lemmas are similar to that of [21, Lemma 3.4].
For readers’ convenience, we include a proof for Lemma 2.4.

######  Lemma 2.2.

Suppose that $a\in(x_{n_{0}},x_{n_{0}+1}]$ for some $n_{0}$, then the function
$M_{+}(z,a)$ has the following finite continued fraction expansion

$M_{+}(z,a)=\cfrac{1}{-l(n_{0})z+\cfrac{1}{m_{n_{0}+1}(z)+\cfrac{1}{-l_{n_{0}+1}z+\cfrac{1}{\ddots+\cfrac{1}{-l_{n-1}z+\cfrac{1}{m_{n}(z)-\cfrac{1}{l_{n}z}}}}}}},$
(2.20)

where

$l(n_{0})=2\tanh\left(\frac{x_{n_{0}+1}-a}{2}\right).$ (2.21)

######  Remark 2.3.

By (2.20), we have

$M_{+}(z,x_{n_{0}+1})=m_{n_{0}+1}(z)+\cfrac{1}{-l_{n_{0}+1}z+\cfrac{1}{m_{n_{0}+2}(z)+\cfrac{1}{\ddots+\cfrac{1}{-l_{n-1}z+\cfrac{1}{m_{n}(z)-\cfrac{1}{l_{n}z}}}}}}.$
(2.22)

######  Lemma 2.4.

Suppose that $a\in(x_{n_{0}},x_{n_{0}+1}]$ for some $n_{0}$, then the function
$M_{-}(z,a)$ has the following finite continued fraction expansion

$M_{-}(z,a)=\cfrac{1}{-\tilde{l}(n_{0})z+\cfrac{1}{m_{n_{0}}(z)+\cfrac{1}{-l_{n_{0}-1}z+\cfrac{1}{\ddots+\cfrac{1}{-l_{1}z+\cfrac{1}{m_{1}(z)-\cfrac{1}{l_{0}z}}}}}}},$
(2.23)

where

$\tilde{l}(n_{0})=-2\tanh\Big{(}\frac{{x_{n_{0}}}-a}{2}\Big{)}.$ (2.24)

###### Proof.

If $a\in(-\infty,x_{1}]$, it is obvious that (2.23) holds. We next show that
(2.23) holds in the case that $a\in(x_{n_{0}},x_{n_{0}+1}]$ with $n_{0}\geq
1$. Define

$\Lambda_{-}(z,x)=-\frac{\phi_{-}^{\prime}(z,x)}{\phi_{-}(z,x)}\cosh^{2}\left(\frac{x-a}{2}\right)+\frac{1}{2}\sinh\left(\frac{x-a}{2}\right)\cosh\left(\frac{x-a}{2}\right).$
(2.25)

From (2.7), (2.8) and (2.25), one can obtain

$\Lambda_{-}(z,x_{1})=-\frac{e^{-(x_{1}-a)}+1}{4}=-\frac{1}{l_{0}}.$

Applying (2.5), (2.6), (2.7) and (2.19), for $i=1,\cdots,n$, one has

$\Lambda_{-}(z,x_{i}+)-\Lambda_{-}(z,x_{i})=(z\omega_{i}+z^{2}v_{i})\cosh^{2}\left(\frac{x_{i}-a}{2}\right)=zm_{i}(z).$
(2.26)

Differentiating (2.25) with respect to $x$, after some calculations, we have
that for any $x\in\mathbb{R}\setminus\\{x_{1},\cdots,x_{n}\\}$,

$\frac{\Lambda_{-}^{\prime}(z,x)}{\Lambda_{-}(z,x)^{2}}=\frac{1}{\cosh^{2}\left(\frac{x-a}{2}\right)}.$
(2.27)

Integrating (2.27) from $x_{i}+$ to $x_{i+1}$, and from $x_{n_{0}}+$ to $a$,
we conclude that

$\displaystyle\frac{1}{\Lambda_{-}(z,x_{i+1})}-\frac{1}{\Lambda_{-}(z,x_{i}+)}$
$\displaystyle=-2\tanh\left(\frac{x-a}{2}\right)\Big{|}_{x_{i}}^{x_{i+1}}=-l_{i},$
(2.28)
$\displaystyle\frac{1}{\Lambda_{-}(z,a)}-\frac{1}{\Lambda_{-}(z,x_{n_{0}}+)}$
$\displaystyle=-2\tanh\left(\frac{x-a}{2}\right)\Big{|}_{x_{n_{0}}}^{a}=-\tilde{l}(n_{0}).$
(2.29)

Thus from (2.26), (2.28) and (2.29), we have

$\frac{1}{\Lambda_{-}(z,x_{i+1})}=-l_{i}+\frac{1}{zm_{i}(z)+\Lambda_{-}(z,x_{i})},$

and

$\frac{1}{\Lambda_{-}(z,a)}=-\tilde{l}(n_{0})+\frac{1}{zm_{n_{0}}(z)+\Lambda_{-}(z,x_{n_{0}})}.$

Note that

$M_{-}(z,a)=\frac{\Lambda_{-}(z,a)}{z}.$

Therefore, one can obtain (2.20). ∎

######  Remark 2.5.

By Lemma 2.4, we have that

$\displaystyle\lim_{\left\lvert z\right\rvert\to+\infty}M_{-}(z,a)=0.$ (2.30)

The proofs of the following two lemmas are similar to that of [21, Theorem
5.1]. For reader’s convenience, we give a proof of Lemma 2.7.

######  Lemma 2.6.

Let $M_{+}(z)$ be a rational Herglotz-Nevanlinna function with

$\displaystyle{\rm{Res}}_{z=0}M_{+}(z)=-\frac{1}{2}.$ (2.31)

Then for any $a\in\mathbb{R}$, there exists unique $\omega|_{[a,+\infty)}$,
$v|_{[a,+\infty)}$ of the form $\eqref{eq2.111}$, so that the Weyl-Titchmarsh
function of $H_{+}(\omega,v,a)$ coincides with $M_{+}(z)$.

######  Lemma 2.7.

Let $M_{-}(z)$ be a rational Herglotz-Nevanlinna function with

$\displaystyle\lim_{\left\lvert z\right\rvert\to+\infty}M_{-}(z)=0$ (2.32)

and

$\displaystyle{\rm{Res}}_{z=0}M_{-}(z)=-\frac{1}{2}.$ (2.33)

Then for any $a\in\mathbb{R}$, there exists unique $\omega|_{(-\infty,a)}$,
$v|_{(-\infty,a)}$ of the form $\eqref{eq2.111}$, so that the Weyl-Titchmarsh
function of $H_{-}(\omega,v,a)$ coincides with $M_{-}(z)$.

###### Proof.

By (2.32) and [22, Corollarys 4.3 and 4.4], $M_{-}(z)$ has a finite continued
fraction expansion of the form (2.23) for some positive
$\tilde{l}(n_{0}),l_{n_{0}-1},\cdots,l_{0},$ and some real, nonzero
polynomials $m_{i}(z)$ of degree at most one with $\dot{m}_{i}(0)\geq 0$ for
all $i=1,\cdots,n_{0}$. Note that

$\tilde{l}(n_{0})+\sum_{j=0}^{n_{0}-1}l_{j}=-\lim_{z\rightarrow
0}\frac{1}{zm(z)}=2.$ (2.34)

Then for any $a\in\mathbb{R}$, we could define $x_{n_{0}},\cdots,x_{1},$ by

$\displaystyle-2\tanh\left(\frac{{x_{n_{0}}}-a}{2}\right)$
$\displaystyle=\tilde{l}(n_{0}),$ (2.35)
$\displaystyle-2\tanh\left(\frac{{x_{i}}-a}{2}\right)$
$\displaystyle=\tilde{l}(n_{0})+\sum_{j=i}^{n_{0}-1}l_{j},i={n_{0}-1},\cdots,1.$
(2.36)

By (2.34), (2.35) and (2.36), we can conclude that (2.18) holds for
$i=0,\cdots,n_{0}-1$. For $i=1,\cdots,n_{0}$, define $\omega_{i},v_{i}$ such
that (2.19) holds. Define

$\omega=\sum_{i=1}^{n_{0}}\omega_{i}\delta_{x_{i}},\
v=\sum_{i=1}^{n_{0}}v_{i}\delta_{x_{i}}.$ (2.37)

By the above construction, we know that the Weyl-Titchmarsh function of
$H_{-}(\omega,v,a)$ coincides with $M_{-}(z)$. Since the fraction expansion of
the form (2.23) for $M_{-}(z)$ is unique, then $\omega|_{(-\infty,a)}$ and
$v|_{(-\infty,a)}$ are uniquely determined. ∎

##  3\. Interior Inverse Problem I

In this section we consider the interior inverse problem with the case that
$\varphi_{i}(a)\neq 0$ (see (3.1) for the definition of $\varphi_{i}(a)$) for
any $i=1,\cdots,N$. Since $\sigma(H)$ consists of finitely many eigenvalues,
we suppose that

$\displaystyle\sigma(H)=\\{\lambda_{i}\\}_{i=1}^{N},$

where $N\leq 2n$, $\lambda_{1}<\lambda_{2}<\cdots<\lambda_{N}$. We mention
that $0\notin\sigma(H).$ Define

$\varphi_{i}(x)=\frac{\phi_{+}(\lambda_{i},x)}{\sqrt{\lambda_{i}\gamma_{\lambda_{i}}^{2}}}$
(3.1)

as the normalized eigenfunction associated with the eigenvalue $\lambda_{i}$.
We consider the following interior inverse problem: For given
$a\in\mathbb{R}$, reconstruct $\omega$ and $v$ from
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. The set
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is called the interior spectral
data of $H=H(\omega,v)$. We mention that $\omega$ and $v$ are uniquely
determined by the spectral data
$\\{\lambda_{i},\lambda_{i}\gamma_{\lambda_{i}}^{2}\\}_{i=1}^{N}$ ([19,
Theorem 4.3]).

Define

$G(z,a)=\sum_{i=1}^{N}\frac{\lambda_{i}^{2}|\varphi_{i}(a)|^{2}}{\lambda_{i}-z}.$
(3.2)

Obviously, $G(z,a)$ is a Herglotz-Nevanlinna function. Since
$0\notin\sigma(H)$, $G(z,a)$ has the following Taylor series at $z=0$,

$G(z,a)=\sum_{i=1}^{N}\lambda_{i}|\varphi_{i}(a)|^{2}+z\sum_{i=1}^{N}|\varphi_{i}(a)|^{2}+z^{2}\sum_{i=1}^{N}\frac{|\varphi_{i}(a)|^{2}}{\lambda_{i}}+O(z^{3}).$
(3.3)

Let us introduce some basic lemmas. Denote by

$\displaystyle\alpha:=1-\sum_{i=1}^{N}|\varphi_{i}(a)|^{2},\
\beta:=-\sum_{i=1}^{N}\lambda_{i}|\varphi_{i}(a)|^{2}.$

######  Lemma 3.1.

We have

$-\frac{1}{M_{+}(z,a)+M_{-}(z,a)}=\alpha z+\beta+G(z,a)$ (3.4)

with $\alpha\geq 0.$

###### Proof.

By the definitions of $M_{+}(z,a)$ and $M_{-}(z,a)$, we have

$\displaystyle-\frac{1}{M_{+}(z,a)+M_{-}(z,a)}=\frac{z\phi_{+}(z,a)\phi_{-}(z,a)}{W(z)}.$
(3.5)

Combining Lemma 2.1 with (2.10) and (3.1), we can obtain

$\displaystyle{\rm{Res}}_{z=\lambda_{i}}\frac{z\phi_{+}(z,a)\phi_{-}(z,a)}{W(z)}=\frac{\lambda_{i}\phi_{+}(\lambda_{i},a)\phi_{-}(\lambda_{i},a)}{\dot{W}(\lambda_{i})}=-\lambda_{i}^{2}|\varphi_{i}(a)|^{2}.$
(3.6)

Since $M_{+}(z,a),M_{-}(z,a)$ are Herglotz-Nevanlinna functions,

$-\frac{1}{M_{+}(z,a)+M_{-}(z,a)}$

is also a Herglotz-Nevanlinna function. Combining with (3.6), there exist
$t\geq 0$, $s\in\mathbb{R}$ such that

$-\frac{1}{M_{+}(z,a)+M_{-}(z,a)}=tz+s+G(z,a).$ (3.7)

Let $z\to 0$ in (3.7), we obtain $s=\beta$. From (2.11) and (2.16), letting
$z\rightarrow 0$, one can obtain

$\frac{z\phi_{+}(z,a)\phi_{-}(z,a)}{W(z)}=z+O(z^{2}).$ (3.8)

By (3.3), (3.5), (3.7) and (3.8), we conclude that

$t=\alpha\geq 0.$

∎

######  Lemma 3.2.

We have

$\alpha=0$

if and only if $a=x_{i_{0}}$ for some $i_{0}\in\\{1,\cdots,n\\}$. Furthermore,
we have

$\displaystyle\beta=0$ (3.9)

if and only if $v_{i_{0}}>0.$

###### Proof.

From (2.19), (2.20) and (2.22), one has that $a=x_{i_{0}}$ for some
$i_{0}\in\\{1,\cdots,n\\}$ if and only if there exists $C\in\mathbb{R}$ with

$\displaystyle\lim_{\left\lvert
z\right\rvert\to+\infty}\frac{1}{M_{+}(z,a)}=C.$ (3.10)

By (2.30), we have that (3.10) holds if and only if

$\lim_{|z|\rightarrow+\infty}-\frac{1}{M_{+}(z,a)+M_{-}(z,a)}=-C,$

which is equivalent to $\alpha=0$ by $\eqref{eq3.2}$. Furthermore, by (2.19)
and (2.22), $v_{i_{0}}>0$ if and only if $C=0$, which is equivalent to
$\alpha=\beta=0$ by $\eqref{eq3.2}$. ∎

Suppose that $A_{1}(z),A_{2}(z),B_{1}(z),B_{2}(z)$ are real polynomials such
that

$\displaystyle F_{1}(z)=\frac{A_{1}(z)}{B_{1}(z)},\
F_{2}(z)=\frac{A_{2}(z)}{B_{2}(z)}$ (3.11)

are Herglotz-Nevanlinna functions. Assume that $A_{i}(z)$ and $B_{i}(z)$ have
no common zeros, $i=1,2$. Define

$F(z):=F_{1}(z)+F_{2}(z).$ (3.12)

Clearly, for any adjacent zeros of $F(z)$ (denoted by $\nu_{j},\nu_{j+1}$),
there exists exactly one pole $\mu$ of $F(z)$ with $\nu_{j}<\mu<\nu_{j+1}$.

We need the following lemma.

######  Lemma 3.3.

Assume that $\mu$ is not a common pole of $F_{1}(z)$ and $F_{2}(z)$. If

$\displaystyle B_{2}(\nu_{j})B_{2}(\nu_{j+1})>0,$ (3.13)

then $\mu$ is a pole of $F_{1}(z)$. Otherwise, $\mu$ is a pole of $F_{2}(z)$.

###### Proof.

Note that for any $\lambda\in[\nu_{j},\nu_{j+1}]\setminus\\{\mu\\}$,

$\displaystyle\ B_{i}(\lambda)\neq 0,\quad i=1,2.$ (3.14)

Otherwise, for example, assume $B_{1}(\lambda)=0$. Since polynomials
$A_{1}(z)$ and $B_{1}(z)$ have no common zeros, one has $A_{1}(\lambda)\neq
0$. Because $F_{1}(z)$ is a Herglotz-Nevanlinna function, one has

$\displaystyle{\rm{Res}}_{z=\lambda}F_{1}(z)<0.$

Then by the fact that $F(\lambda)\in\mathbb{R}$ we obtain that
$B_{2}(\lambda)=0$ and

$\displaystyle{\rm{Res}}_{z=\lambda}F_{2}(z)>0,$

this is in contradiction with that $F_{2}(z)$ is a Herglotz-Nevanlinna
function.

From (3.12), we have that

$F(z)=\frac{f(z)}{B_{1}(z)B_{2}(z)},$ (3.15)

where

$f(z)=A_{1}(z)B_{2}(z)+A_{2}(z)B_{1}(z).$

Since $F(z)$ is a Herglotz-Nevanlinna function, its zeros are simple. By
(3.14) and (3.15), we know that $\nu_{j}$, $\nu_{j+1}$ are simple zeros of
$f(z)$. For any $\lambda\in(\nu_{j},\nu_{j+1})\setminus\\{\mu\\}$, combining
(3.14) with the fact that $F(\lambda)\neq 0$, we can conclude that
$f(\lambda)\neq 0$. By the assumption, $\mu$ is not a common pole of
$F_{1}(z)$ and $F_{2}(z)$. Then by (3.11), $\mu$ is a simple zero of
$B_{1}(z)B_{2}(z)$. Since $\mu$ is a pole of $F(z)$, by (3.15), one obtains
that $f(\mu)\neq 0$. Hence one can conclude that $\nu_{j}$ and $\nu_{j+1}$ are
adjacent simple zeros of $f(z)$. Then one has that

$\dot{f}(\nu_{j})\dot{f}(\nu_{j+1})<0.$ (3.16)

Combining (3.15) with the fact that $F(z)$ is a Herglotz-Nevanlinna function,
we obtain

$\displaystyle\dot{F}(\nu_{j})$
$\displaystyle=\frac{\dot{f}(\nu_{j})}{B_{1}(\nu_{j})B_{2}(\nu_{j})}>0,$
(3.17) $\displaystyle\dot{F}(\nu_{j+1})$
$\displaystyle=\frac{\dot{f}(\nu_{j+1})}{B_{1}(\nu_{j+1})B_{2}(\nu_{j+1})}>0.$
(3.18)

Then from (3.16), (3.17) and (3.18), one has

$\displaystyle
B_{1}(\nu_{j})B_{2}(\nu_{j})B_{1}(\nu_{j+1})B_{2}(\nu_{j+1})<0.$

Applying (3.13), we have that $B_{1}(\nu_{j})B_{1}(\nu_{j+1})<0$. Therefore,
$B_{1}(z)$ has a zero in interval $(\nu_{j},\nu_{j+1})$, which implies that
$\mu$ is a pole of $F_{1}(z)$.

Otherwise, if $B_{2}(\nu_{j})B_{2}(\nu_{j+1})\leq 0$, by (3.14), one has

$\displaystyle B_{2}(\nu_{j})B_{2}(\nu_{j+1})<0.$

Then $B_{2}(z)$ has a zero in the interval $(\nu_{j},\nu_{j+1})$. This implies
that $\mu$ is a pole of $F_{2}(z)$. ∎

Now we are going to introduce some inverse interior spectral results under the
condition that $\varphi_{i}(a)\neq 0$ for any $i=1,\cdots,N$.

######  Theorem 3.4.

For given $a$, suppose that $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is
the interior spectral data for some $\omega,v$. If

$\alpha=\beta=0$

and

$\varphi_{i}(a)\neq 0,\ i=1,\cdots,N,$ (3.19)

then $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ uniquely determines
$\omega,v$.

###### Proof.

Since $\alpha=\beta=0,$ by (3.2) and Lemma 3.1, we have

$-\frac{1}{M_{+}(z,a)+M_{-}(z,a)}=\sum_{i=1}^{N}\frac{\lambda_{i}^{2}|\varphi_{i}(a)|^{2}}{\lambda_{i}-z}.$
(3.20)

Namely, $-\frac{1}{M_{+}(z,a)+M_{-}(z,a)}$ is completely determined by the
interior spectral data $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. Since
$M_{+}(z,a)+M_{-}(z,a)$ is a Herglotz-Nevanlinna function, by (3.19) and
(3.20), there exist $\gamma>0,\zeta,\beta_{i},\mu_{i},i=1,\cdots,N-1$, (
determined by $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ ) so that

$M_{+}(z,a)+M_{-}(z,a)=\gamma
z+\zeta+\sum_{i=1}^{N-1}\frac{\beta_{i}}{\mu_{i}-z},\ {\beta_{i}>0},$ (3.21)

where $\\{\mu_{i}\\}_{i=1}^{N-1}$ are the poles of
${M_{+}(z,a)}+{M_{-}(z,a)}$. We emphasize that ${M_{+}(z,a)}+{M_{-}(z,a)}$ has
one more zero than poles.

By (2.17), we have $\mu_{j_{0}}=0$ for some $j_{0}\in\\{1,\cdots,N-1\\}$.
Clearly, $\\{\mu_{i}\\}_{i=1}^{N-1}$ and $\\{\lambda_{i}\\}_{i=1}^{N}$ are
interlaced:

$\lambda_{1}<\mu_{1}<\lambda_{2}<\cdots<\lambda_{j_{0}}<\mu_{j_{0}}=0<\lambda_{j_{0}+1}<\cdots<\lambda_{N}.$
(3.22)

Now, we are going to determine the poles of ${M_{+}(z,a)}$ and $M_{-}(z,a)$,
respectively. Then by (3.21) we can completely determine $M_{+}(z,a)$ and
$M_{-}(z,a).$ Note that the only common pole of $M_{+}(z,a)$ and $M_{-}(z,a)$
is $z=0$. Indeed, if $\mu_{j}\neq 0$ is a pole of $M_{+}(z,a)$ and
$M_{-}(z,a)$. Namely,

$\phi_{+}(\mu_{j},a)=\phi_{-}(\mu_{j},a)=0.$

Then by (2.9) we have that

$W(\mu_{j})=\phi_{+}(\mu_{j},a)\phi_{-}^{\prime}(\mu_{j},a)-\phi_{+}^{\prime}(\mu_{j},a)\phi_{-}(\mu_{j},a)=0.$

This implies that $\mu_{j}$ is an eigenvalue, which leads to a contradiction.

Let

$\displaystyle A_{1}(z)$ $\displaystyle=-\phi_{-}^{\prime}(z,a),\
B_{1}(z)=z\phi_{-}(z,a),$ $\displaystyle A_{2}(z)$
$\displaystyle=\phi_{+}^{\prime}(z,a),\ B_{2}(z)=z\phi_{+}(z,a),$

in Lemma 3.3. Then we have

$F_{1}(z)=M_{-}(z,a),F_{2}(z)=M_{+}(z,a).$ (3.23)

For any $j\neq j_{0}$ with

$\displaystyle\varphi_{j}(a)\varphi_{j+1}(a)<0,$

from (3.1) and (3.22), one has

$\displaystyle
B_{2}(\lambda_{j})B_{2}(\lambda_{j+1})=\lambda_{j}\lambda_{j+1}\phi_{+}(\lambda_{j},a)\phi_{+}(\lambda_{j+1},a)<0.$

By Lemma 3.3, we obtain that $\mu_{j}$ is a pole of $M_{+}(z,a)$. Similarly,
for any $j\neq j_{0}$ with $\varphi_{j}(a)\varphi_{j+1}(a)>0,$ $\mu_{j}$ is a
pole of $M_{-}(z,a)$. Hence, we can completely determine, by the signs of
$\varphi_{i}(a),i=1,\cdots,N$, all poles of $M_{+}(z,a)$ and $M_{-}(z,a)$,
respectively. Denote the poles of $M_{+}(z,a)$ and $M_{-}(z,a)$ by
$\\{0\\}\cup\\{u_{i}\\}_{i\in I_{1}}$ and $\\{0\\}\cup\\{u_{i}\\}_{i\in
I_{2}}$, respectively. Firstly, by (2.17) and (2.30), one has

$\displaystyle M_{-}(z,a)=-\frac{1}{2z}+\sum_{i\in
I_{2}}\frac{\beta_{i}}{\mu_{i}-z}.$

Using (2.17) and (3.21) we obtain

$\displaystyle M_{+}(z,a)=\gamma
z+\zeta-\frac{1}{2z}+\sum_{i\in{I_{1}}}\frac{\beta_{i}}{\mu_{i}-z}.$

By Lemmas 2.6 and 2.7, $M_{+}(z,a)$ and $M_{-}(z,a)$ uniquely determine
$\omega|_{[a,+\infty)}$, $v|_{[a,+\infty)}$ and
$\omega|_{(-\infty,a)},v|_{(-\infty,a)}$, respectively. Hence $\omega$ and $v$
are uniquely determined.

Now we show that the $\omega,v$ determined above has the interior spectral
data $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. Assume that
$\\{\hat{\lambda}_{i},\hat{\varphi}_{i}(a)\\}_{i=1}^{N}$ is the interior
spectral data for the determined $\omega,v$. Firstly, $M_{\pm}(z,a)$ are the
Weyl-Titchmarsh functions of $H_{\pm}(\omega,v,a)$, then by Lemma 3.1 we have

$-\frac{1}{M_{+}(z,a)+M_{-}(z,a)}=\sum_{i=1}^{N}\frac{\hat{\lambda}_{i}^{2}|\hat{\varphi}_{i}(a)|^{2}}{\hat{\lambda}_{i}-z}.$

Combining with (3.20), for any $i=1,\cdots,N$, one has

$\hat{\lambda}_{i}=\lambda_{i},|\hat{\varphi}_{i}(a)|=|\varphi_{i}(a)|.$
(3.24)

Therefore, it suffices to show that for any $i=1,\cdots,N$, we have

$\displaystyle{\rm{sgn}}(\hat{\varphi}_{i}(a))={\rm{sgn}}(\varphi_{i}(a)).$
(3.25)

By (3.22) and Remark 6.9, one has (3.25) holds for $i=j_{0},j_{0}+1$. For any
$j\neq j_{0}$, we know that $\mu_{j}$ is a pole of $M_{\pm}(z,a)$ if and only
if

$\pm\hat{\varphi}_{j}(a)\hat{\varphi}_{j+1}(a)<0.$

Hence, the signs of $\hat{\varphi}_{j}(a)$ for $j\neq j_{0},j_{0}+1,$ are
uniquely determined by ${\rm{sgn}}(\hat{\varphi}_{j_{0}}(a))$ and
${\rm{sgn}}(\hat{\varphi}_{j_{0}+1}(a))$. Therefore, we can obtain that (3.25)
holds for any ${i}=1,\cdots,N$. Then
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is the interior spectral data of
the determined $\omega,v$. ∎

######  Theorem 3.5.

For given $a$, suppose that $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is
the interior spectral data for some $\omega,v$. Assume that

$\alpha=0,\ \beta\neq 0$

and

$\varphi_{i}(a)\neq 0,i=1,\cdots,N.$ (3.26)

${\rm{(i)}}$ If we have $\lambda_{1}>0$ or $\lambda_{N}<0$, then
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ uniquely determines $\omega,v$.
${\rm{(ii)}}$ If for some $j_{0}\in\\{1,\cdots,N-1\\}$ we have

$\lambda_{j_{0}}<0<\lambda_{j_{0}+1},$

then two $\omega,v$’s yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$.

###### Proof.

Arguing as in Theorem 3.4, there exist $\zeta,\beta_{i},\mu_{i},i=1,\cdots,N$,
( determined by $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ ) so that

$M_{+}(z,a)+M_{-}(z,a)=\zeta+\sum_{i=1}^{N}\frac{\beta_{i}}{\mu_{i}-z},\
{\beta_{i}>0},$ (3.27)

where $\\{\mu_{i}\\}_{i=1}^{N}$ are the poles of ${M_{+}(z,a)}+{M_{-}(z,a)}$.
We emphasize that ${M_{+}(z,a)}+{M_{-}(z,a)}$ has the same number of zeros and
poles.

Let us show ${\rm{(i)}}$ first. We only consider the case that
$\lambda_{1}>0$, the proof is similar for the case that $\lambda_{N}<0$.
Clearly, $\\{\mu_{i}\\}_{i=1}^{N}$ and $\\{\lambda_{i}\\}_{i=1}^{N}$ are
interlaced:

$\mu_{1}=0<\lambda_{1}<\mu_{2}<\lambda_{2}<\cdots<\mu_{N-1}<\lambda_{N-1}<\mu_{N}<\lambda_{N}.$
(3.28)

Following the similar steps in the proof of Theorem 3.4, we can completely
determine all poles of $M_{+}(z,a)$ and $M_{-}(z,a)$, respectively, by the
signs of $\varphi_{i}(a),i=1,\cdots,N$. Then by (2.17), (2.30) and (3.27),
$M_{+}(z,a)$ and $M_{-}(z,a)$ are uniquely determined. Hence, we can obtain
exactly one acceptable $\omega,v$.

Now we are going to prove ${\rm(ii)}$. Since $\\{\mu_{i}\\}_{i=1}^{N}$ and
$\\{\lambda_{i}\\}_{i=1}^{N}$ are interlaced, we have either

$\lambda_{1}<\mu_{1}<\cdots<\lambda_{j_{0}}<\mu_{j_{0}}=0<\lambda_{j_{0}+1}<\cdots<\lambda_{N}<\mu_{N},$
(3.29)

or

$\mu_{1}<\lambda_{1}<\cdots<\lambda_{j_{0}-1}<\mu_{j_{0}}=0<\lambda_{j_{0}}<\cdots<\mu_{N}<\lambda_{N}.$
(3.30)

We only consider the case that (3.29) holds. Following the similar steps in
the proof of Theorem 3.4, we can completely determine all poles of
$M_{+}(z,a)$ and $M_{-}(z,a)$, respectively, except for $\mu_{N}$. Since
$\mu_{N}$ can be a pole of $M_{+}(z,a)$ or $M_{-}(z,a)$, we have two
acceptable $\omega,v$’s. Therefore, we can conclude that two $\omega,v$’s
yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. ∎

######  Theorem 3.6.

For given $a$, suppose that $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is
the interior spectral data for some $\omega,v$. Assume that $\alpha\neq 0$ and

$\varphi_{i}(a)\neq 0,i=1,\cdots,N.$

$\rm{(i)}$ If we have $\lambda_{1}>0$ or $\lambda_{N}<0$, then two
$\omega,v$’s yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$.
$\rm{(ii)}$ If for some $j_{0}\in\\{1,\cdots,N-1\\}$ we have

$\lambda_{j_{0}}<0<\lambda_{j_{0}+1},$

then four $\omega,v$’s yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$.

###### Proof.

Since $M_{+}(z,a)+M_{-}(z,a)$ is a Herglotz-Nevanlinna function, by
$\alpha\neq 0$ and Lemma 3.1, there exist
${\beta_{i}>0},\mu_{i},i=0,\cdots,N$, ( determined by
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ ) so that

$M_{+}(z,a)+M_{-}(z,a)=\sum_{i=0}^{N}\frac{\beta_{i}}{\mu_{i}-z},$ (3.31)

where $\\{\mu_{i}\\}_{i=0}^{N}$ are the poles of ${M_{+}(z,a)}+{M_{-}(z,a)}$.
We emphasize that ${M_{+}(z,a)}+{M_{-}(z,a)}$ has one more pole than zeros.

Let us show ${\rm{(i)}}$ first. We only consider the case that
$\lambda_{1}>0$. Clearly, $\\{\mu_{i}\\}_{i=0}^{N}$ and
$\\{\lambda_{i}\\}_{i=1}^{N}$ are interlaced:

$\mu_{0}=0<\lambda_{1}<\mu_{1}<\lambda_{1}<\cdots<\lambda_{N-1}<\mu_{N-1}<\lambda_{N}<\mu_{N}.$

Then we can completely determine all poles of $M_{+}(z,a)$ and $M_{-}(z,a)$,
respectively, except for $\mu_{N}$. Since $\mu_{N}$ can be a pole of
$M_{+}(z,a)$ or $M_{-}(z,a)$, we have two acceptable $\omega,v$’s. Therefore,
we can conclude that two $\omega,v$’s yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$.

Now we are going to prove ${\rm(ii)}$. Clearly, $\\{\mu_{i}\\}_{i=0}^{N}$ and
$\\{\lambda_{i}\\}_{i=1}^{N}$ are interlaced:

$\mu_{0}<\lambda_{1}<\cdots<\lambda_{j_{0}}<\mu_{j_{0}}=0<\lambda_{j_{0}+1}<\cdots<\mu_{N-1}<\lambda_{N}<\mu_{N}.$

Hence, we can completely determine all poles of $M_{+}(z,a)$ and $M_{-}(z,a)$,
respectively, except for $\mu_{0},\mu_{N}$. Since $\mu_{0},\mu_{N}$ can be the
poles of $M_{+}(z,a)$ or $M_{-}(z,a)$, we have four acceptable $\omega,v$’s.
Therefore, we can conclude that four $\omega,v$’s yield the given interior
spectral data $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. ∎

######  Example 3.7.

Suppose that $\sigma(H)=\\{\lambda_{1}\\}$, thus there exist
$c,d\in\mathbb{R}$, so that $W(z)=cz+d$. Then by
$\eqref{lianxu},\eqref{tiaoyue}$ and $\eqref{Wzdefine}$, one obtains that

$\omega=\omega_{1}\delta_{x_{1}},\ v=0$

for some $\omega_{1},x_{1}$. After some calculations, we have

$\displaystyle\phi_{+}(z,x)=\begin{cases}z\omega_{1}e^{\frac{x}{2}-x_{1}}+(1-z\omega_{1})e^{-\frac{x}{2}},&x<x_{1},\\\
e^{-\frac{x}{2}},&x\geq x_{1}.\\\ \end{cases}$ (3.32)

Then one has $1-\lambda_{1}\omega_{1}=0$. Namely,

$\omega_{1}=\frac{1}{\lambda_{1}}.$ (3.33)

By (2.12), (3.1), (3.32) and (3.33), we can obtain

$\displaystyle\varphi_{1}(x)=\begin{cases}e^{\frac{x-x_{1}}{2}},&x<x_{1},\\\
e^{\frac{x_{1}-x}{2}},&x\geq x_{1}.\\\ \end{cases}$ (3.34)

Then we have the following two cases:
$\rm{(i)}$ $\alpha=0$, namely, $\varphi_{1}(a)=1$, then
$\\{\lambda_{1},\varphi_{1}(a)\\}$ uniquely determines $\omega,v$. By (3.34),
one has $x_{1}=a$. Hence

$\omega=\frac{1}{\lambda_{1}}\delta_{a},\ v=0.$

$\rm{(ii)}$ $\alpha\neq 0$, namely, $\varphi_{1}(a)=|\varphi_{1}(a)|<1$, then
two $\omega,v$’s yield the given interior spectral data
$\\{\lambda_{1},\varphi_{1}(a)\\}$. From (3.34), one obtains

$x_{1}=a+2\ln\varphi_{1}(a)\ {\rm{or}}\ x_{1}=a-2\ln\varphi_{1}(a).$

Hence we can conclude that

$\omega=\frac{1}{\lambda_{1}}\delta_{a+2\ln\varphi_{1}(a)},\ v=0,$

or

$\omega=\frac{1}{\lambda_{1}}\delta_{a-2\ln\varphi_{1}(a)},\ v=0.$

For any $N\in\mathbb{N}$, let $\mathcal{T}$ be the set of real numbers
$\\{\lambda_{i},\alpha_{i}\\}_{i=1}^{N}$ with
$\lambda_{1}<\lambda_{2}<\cdots<\lambda_{N}$ satisfying

$\sum_{i=1}^{N}\alpha^{2}_{i}\leq 1$ (3.35)

and

$\displaystyle\lambda_{j_{0}}<0<\lambda_{j_{0}+1},\ \alpha_{j_{0}}>0,\
\alpha_{j_{0}+1}>0,$

for some $j_{0}\in\\{0,\cdots,N\\}$, where $j_{0}=0$ means that
$\lambda_{1}>0,\alpha_{1}>0$ and $j_{0}=N$ means that
$\lambda_{N}<0,\alpha_{N}>0$.

We next discuss the existence of the interior spectral problem.

######  Lemma 3.8.

Assume that

$\alpha_{i}\neq 0,i=1,\cdots,N.$

Then for given $a$, $\\{\lambda_{i},\alpha_{i}\\}_{i=1}^{N}$ is the interior
spectral data for some $\omega,v$ at $a$ if and only if
$\\{\lambda_{i},\alpha_{i}\\}_{i=1}^{N}\in\mathcal{T}$.

###### Proof.

The necessity is obtained from Lemma 3.1 and Remark 6.9. Following the similar
steps in the proof of Theorem 3.4, we can obtain the sufficiency. ∎

In the following, we require that if a certain symbol $\gamma$ denotes an
object related to $H(\omega,v)$, then the corresponding symbol
$\tilde{\gamma}$ with tilde denotes the analogous object related to
$H(\tilde{\omega}$, $\tilde{v})$.

######  Theorem 3.9.

For given $a$, suppose that $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is
the interior spectral data for some $\omega,v$ with

$\varphi_{i}(a)\neq 0,i=1,\cdots,N.\\\ $

$\rm{(i)}$ If

$\alpha=\beta=0,$

then there are precisely $2^{N-2}$ $\tilde{\omega},\tilde{v}$’s, so that

$\tilde{\lambda}_{i}=\lambda_{i},|\tilde{\varphi}_{i}(a)|=|\varphi_{i}(a)|,i=1,\cdots,N.$

$\rm{(ii)}$ If

$\alpha=0,\ \beta\neq 0,$

then there are precisely $2^{N-1}$ $\tilde{\omega},\tilde{v}$’s, so that

$\tilde{\lambda}_{i}=\lambda_{i},|\tilde{\varphi}_{i}(a)|=|\varphi_{i}(a)|,i=1,\cdots,N.$

$\rm{(iii)}$ If

$\alpha\neq 0,$

then there are precisely $2^{N}$ $\tilde{\omega},\tilde{v}$’s, so that

$\tilde{\lambda}_{i}=\lambda_{i},|\tilde{\varphi}_{i}(a)|=|\varphi_{i}(a)|,i=1,\cdots,N.$

###### Proof.

We first show that ${\rm{(iii)}}$ holds. We split the proof into two cases.
Case 1: $\lambda_{1}>0$ or $\lambda_{N}<0$. We only consider the case that
$\lambda_{1}>0$. By Remark 6.9, we have that $\tilde{\varphi}_{1}(a)>0$. From
Lemma 3.8, for any $i\in\\{2,\cdots,N\\}$, the sign of
$\tilde{\varphi_{i}}(a)$ is not determined. Then we can choose the signs of
$\tilde{\varphi}_{1}(a),\cdots,\tilde{\varphi}_{N}(a)$ in $2^{N-1}$ ways.
Every such choice yields two $\tilde{\omega},\tilde{v}$’s by (i) of Theorem
3.6. Hence, we determine $2^{N}$ $\tilde{\omega},\tilde{v}$’s.
Case 2: For some $j_{0}\in\\{1,\cdots,N-1\\}$, we have

$\lambda_{j_{0}}<0<\lambda_{j_{0}+1}.$

By Remark 6.9, we have
$\tilde{\varphi}_{j_{0}}(a),\tilde{\varphi}_{j_{0}+1}(a)>0$. From Lemma 3.8,
for any $i\in\\{1,\cdots,j_{0}-1,j_{0}+2,\cdots,N\\}$, the sign of
$\tilde{\varphi}_{i}(a)$ is not determined. Then we can choose the signs of
$\tilde{\varphi}_{1}(a),\cdots,\tilde{\varphi}_{N}(a)$ in $2^{N-2}$ ways.
Every such choice yields four $\tilde{\omega},\tilde{v}$’s by (ii) of Theorem
3.6. Hence, we determine $2^{N}$ $\tilde{\omega},\tilde{v}$’s.

Note that in both cases, we have (iii) holds. Applying the above method with
slight modification, we know that (i) and (ii) hold. ∎

##  4\. Interior Inverse Problem II

In this section, we consider the interior inverse problem with the case that
$\varphi_{i}(a)=0$ for some $i$. We need the following lemmas.

######  Lemma 4.1.

For any $a\in\mathbb{R}$, if $\varphi_{i_{0}}(a)=0$ for some
$i_{0}\in\\{2,\cdots,N-1\\}$, then

$\varphi_{i_{0}-1}(a)\varphi_{i_{0}+1}(a)<0.$

###### Proof.

If $\varphi_{i_{0}}(a)=0$, by (2.10) and (3.1), we have

$\phi_{+}(\lambda_{i_{0}},a)=\phi_{-}(\lambda_{i_{0}},a)=0.$ (4.1)

Hence, by (3.5), one has that

$\displaystyle-\frac{1}{M_{+}(\lambda_{i_{0}},a)+M_{-}(\lambda_{i_{0}},a)}=0.$
(4.2)

We first show that

$\varphi_{i_{0}-1}(a)\neq 0,\ \varphi_{i_{0}+1}(a)\neq 0.$

Otherwise, for example, assume that $\varphi_{i_{0}-1}(a)=0$. Then

$\displaystyle-\frac{1}{M_{+}(\lambda_{i_{0}-1},a)+M_{-}(\lambda_{i_{0}-1},a)}=0.$
(4.3)

Since $-1/(M_{+}(z,a)+M_{-}(z,a))$ is a Herglotz-Nevanlinna function, by (4.2)
and (4.3), the function $-1/(M_{+}(z,a)+M_{-}(z,a))$ has a pole $\nu$ in
$(\lambda_{i_{0}-1},\lambda_{i_{0}})$. This contradicts (3.4).

Since $\varphi_{i_{0}-1}(a)\neq 0$ and $\varphi_{i_{0}+1}(a)\neq 0$, applying
(3.4) and (4.2), $M_{+}(z,a)+M_{-}(z,a)$ has exactly one pole
$\lambda_{i_{0}}$ in the interval $[\lambda_{i_{0}-1},\lambda_{i_{0}+1}]$.
Then by (4.1), we know that $M_{+}(z,a)$ has exactly one pole
$\lambda_{i_{0}}$ in $[\lambda_{i_{0}-1},\lambda_{i_{0}+1}]$. From (2.14),
$z\phi_{+}(z,a)$ has exactly one zero $\lambda_{i_{0}}$ in
$[\lambda_{i_{0}-1},\lambda_{i_{0}+1}]$. Therefore, we have

$\lambda_{i_{0}-1}\phi_{+}(\lambda_{i_{0}-1},a)\lambda_{i_{0}+1}\phi_{+}(\lambda_{i_{0}+1},a)<0.$
(4.4)

Note that Theorem 6.8 implies $\lambda_{i_{0}-1}\lambda_{i_{0}+1}>0$. Then one
obtains

$\phi_{+}(\lambda_{i_{0}-1},a)\phi_{+}(\lambda_{i_{0}+1},a)<0.$

By (3.1), one has that $\varphi_{i_{0}-1}(a)\varphi_{i_{0}+1}(a)<0$. ∎

Suppose that there are $k$ many $\varphi_{i}(a)$ such that $\varphi_{i}(a)=0$,
and $\varphi_{1}(a)\neq 0$, $\varphi_{N}(a)\neq 0$. For example, assume that
$\varphi_{m_{1}}(a)=\cdots=\varphi_{m_{k}}(a)=0$. By Lemma 4.1, for any
$j=1,\cdots,k$, we obtain that $M_{+}(z,a)+M_{-}(z,a)$ has exactly one pole
$\lambda_{m_{j}}$ in $(\lambda_{m_{j}-1},\lambda_{m_{j}+1})$. For any
$i\notin\\{m_{1}-1,m_{1},\cdots,m_{k}-1,m_{k}\\}$, $M_{+}(z,a)+M_{-}(z,a)$ has
exactly one pole in $(\lambda_{i},\lambda_{i+1})$.

######  Lemma 4.2.

$\rm{(i)}$ Suppose that

$\alpha=\beta=0.$

Then $\varphi_{1}(a)\neq 0$ and $\varphi_{N}(a)\neq 0$.
$\rm{(ii)}$ Suppose that

$\alpha=0,\ \beta\neq 0.$

Then $\varphi_{1}(a),\varphi_{N}(a)$ cannot both be zero. If moreover,
$\lambda_{1}>0$ or $\lambda_{N}<0$, then we have that $\varphi_{1}(a)\neq 0$
and $\varphi_{N}(a)\neq 0$.

###### Proof.

Assume that there are $k$ many $\varphi_{i}(a)$ such that $\varphi_{i}(a)=0$.
Let us show (i) first. Otherwise, for example, assume that $\varphi_{1}(a)=0$.
From the proof of Lemma 4.1, we know that

$\displaystyle-\frac{1}{M_{+}(\lambda_{1},a)+M_{-}(\lambda_{1},a)}=0.$ (4.5)

Since $\alpha=\beta=0$, by Lemma 3.1, $-1/\left(M_{+}(z,a)+M_{-}(z,a)\right)$
has $N-k$ poles and $N-k-1$ zeros. We emphasize that
$-1/({M_{+}(z,a)}+{M_{-}(z,a)})$ has one more pole than zeros. Combining with
(4.5), one concludes that $-1/(M_{+}(z,a)+M_{-}(z,a))$ has one pole in
$(-\infty,\lambda_{1})$. This contradicts (3.4).

Now we are going to prove (ii). Otherwise, assume that
$\varphi_{1}(a)=\varphi_{N}(a)=0$. Then one has that

$\displaystyle-\frac{1}{M_{+}(\lambda_{1},a)+M_{-}(\lambda_{1},a)}=-\frac{1}{M_{+}(\lambda_{N},a)+M_{-}(\lambda_{N},a)}=0.$
(4.6)

Since $\alpha=0,\beta\neq 0$, by Lemma 3.1, we know that
$-1/({M_{+}(z,a)}+{M_{-}(z,a)})$ has the same number of zeros and poles.
Combining with (4.6), $-1/(M_{+}(z,a)+M_{-}(z,a))$ has one pole in
$(-\infty,\lambda_{1})\cup(\lambda_{N},+\infty)$. This contradicts (3.4).

It remains to establish that if moreover, $\lambda_{1}>0$ or $\lambda_{N}<0$,
then $\varphi_{1}(a)\neq 0,\varphi_{N}(a)\neq 0$. We only consider the case
that $\lambda_{1}>0$. By Theorem 6.8, we know that $\varphi_{1}(a)\neq 0$. We
show that $\varphi_{N}(a)\neq 0$. Otherwise, we obtain that

$\displaystyle-\frac{1}{M_{+}(\lambda_{N},a)+M_{-}(\lambda_{N},a)}=0.$ (4.7)

Since $\varphi_{1}(a)\neq 0$, by (3.4), $\lambda_{1}$ is a pole of
$-1/({M_{+}(z,a)}+{M_{-}(z,a)})$. Using (3.4) and (3.5), the smallest zero and
pole of $-1/({M_{+}(z,a)}+{M_{-}(z,a)})$ are $\mu_{1}=0$ and $\lambda_{1}$,
respectively, which satisfy

$\mu_{1}=0<\lambda_{1}.$ (4.8)

Note that $-1/({M_{+}(z,a)}+{M_{-}(z,a)})$ has the same number of zeros and
poles. Then by (4.7) and (4.8), $-1/(M_{+}(z,a)+M_{-}(z,a))$ has one pole in
$(\lambda_{N},+\infty)$. This contradicts (3.4). ∎

Let $\mathcal{F}$ be the collection of spectral data of inverse spectral
family, which yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. The following theorems tell us
some properties of $\mathcal{F}$ including connectedness. Compared with the
results in Section 3, the connectedness of $\mathcal{F}$ also depends on
whether $\varphi_{1}(a)$ and $\varphi_{N}(a)$ are zero.

######  Theorem 4.3.

For given $a$, suppose that $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is
the interior spectral data for some $\omega,v$, and

$\alpha=\beta=0.$

If there are $k$ many $\varphi_{i}(a)$ such that $\varphi_{i}(a)=0$, then
infinitely many $\omega,v$’s yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. $\mathcal{F}$ is a connected
manifold of dimension $k$.

###### Proof.

Since $\alpha=\beta=0$, by (3.2) and Lemma 3.1, we know that

$M_{+}(z,a)+M_{-}(z,a)=\gamma
z+\zeta+\sum_{i=1}^{N-k-1}\frac{\beta_{i}}{\mu_{i}-z},$ (4.9)

where $\\{\mu_{i}\\}_{i=1}^{N-k-1}$ are the poles of
${M_{+}(z,a)}+{M_{-}(z,a)}$. Denote the zeros of ${M_{+}(z,a)}+{M_{-}(z,a)}$
by $\\{\lambda_{l_{i}}\\}_{i=1}^{N-k}$. Clearly, $\\{\mu_{i}\\}_{i=1}^{N-k-1}$
and $\\{\lambda_{l_{i}}\\}_{i=1}^{N-k}$ are interlaced:

$\lambda_{l_{1}}<\mu_{1}<\lambda_{l_{2}}<\cdots<\lambda_{l_{j_{1}}}<\mu_{j_{1}}=0<\lambda_{l_{j_{1}+1}}<\cdots<\lambda_{l_{N-k}}.$
(4.10)

We divide the $N-k-1$ poles of $M_{+}(z,a)+M_{-}(z,a)$ into four parts:
(1) 0;
(2) $A$ is the set of zeros of both functions $\phi_{+}(z,a)$ and
$\phi_{-}(z,a)$. We mention that $A$ coincides with the set of $\lambda_{i}$
with $\varphi_{i}(a)=0$;
(3) $B$ is the set of zeros of $\phi_{+}(z,a)$ alone;
(4) $C$ is the set of zeros of $\phi_{-}(z,a)$ alone.

Following the similar steps in the proof of Theorem 3.4, by (4.10), the sets
$B$ and $C$ are uniquely determined by the signs of
$\varphi_{i}(a),i=1,\cdots,N$. Then we make the following choice: for any
$\mu_{i}\in A$, let $\beta_{i}^{(1)},\beta_{i}^{(2)}>0$ be arbitrary with

$\beta_{i}=\beta_{i}^{(1)}+\beta_{i}^{(2)},$ (4.11)

and take

$\displaystyle M_{+}(z,a)$ $\displaystyle=\gamma
z+\zeta-\frac{1}{2z}+\sum_{\mu_{i}\in
B}\frac{\beta_{i}}{\mu_{i}-z}+\sum_{\mu_{i}\in
A}\frac{\beta_{i}^{(1)}}{\mu_{i}-z},$ (4.12) $\displaystyle M_{-}(z,a)$
$\displaystyle=-\frac{1}{2z}+\sum_{\mu_{i}\in
C}\frac{\beta_{i}}{\mu_{i}-z}+\sum_{\mu_{i}\in
A}\frac{\beta_{i}^{(2)}}{\mu_{i}-z}.$ (4.13)

Every such choice yields an acceptable $\omega$, $v$ (see Theorem 4.8 for the
acceptance of $\omega,v$).

Notice that by (2.11) and (2.16), one has

$\displaystyle W(z)=\prod_{\lambda_{i}\in\sigma(H)}(1-\frac{z}{\lambda_{i}}),$
(4.14) $\phi_{+}(z,a)=e^{-\frac{a}{2}}\prod_{\mu_{i}\in A\cup
B}(1-\frac{z}{\mu_{i}}),\phi_{-}(z,a)=e^{\frac{a}{2}}\prod_{v_{i}\in A\cup
C}(1-\frac{z}{v_{i}}).$ (4.15)

Therefore, $W(z)$ and $\phi_{\pm}(z,a)$ are uniquely determined by
$\sigma(H)=\\{\lambda_{i}\\}_{i=1}^{N}$ and the sets $A,B$ and $C$. For any
$j$ with $\varphi_{j}(a)\neq 0$, by (2.10), (2.13) and (4.15), we have

${\lambda_{j}}\gamma_{\lambda_{j}}^{2}=e^{-a}\prod_{\lambda_{i}\in\sigma(H)\setminus\\{\lambda_{j}\\}}(1-\frac{\lambda_{j}}{\lambda_{i}})\prod_{\mu_{i}\in
B}(1-\frac{\lambda_{j}}{\mu_{i}})\prod_{v_{i}\in
C}(1-\frac{\lambda_{j}}{v_{i}})^{-1},$ (4.16)

which is uniquely determined by $\sigma(H)$ and the sets $B,C$.

For any $j$ with $\varphi_{j}(a)=0$, by (3.5), one has that

${\beta_{j}}={\rm{Res}}_{z=\lambda_{j}}\frac{W(z)}{z\phi_{+}(z,a)\phi_{-}(z,a)}=\frac{\dot{W}(\lambda_{j})}{\lambda_{j}\dot{\phi}_{+}(\lambda_{j},a)\dot{\phi}_{-}(\lambda_{j},a)}>0.$
(4.17)

Hence, one obtains that

$\displaystyle{\rm
sgn}\Big{(}{\frac{\dot{W}(\lambda_{j})}{\dot{\phi}_{+}(\lambda_{j},a)}}\Big{)}={\rm
sgn}\big{(}\lambda_{j}\dot{\phi}_{-}(\lambda_{j},a)\big{)}.$ (4.18)

Letting $x=a$ in (2.9), and then differentiating with respect to $z$ at
$z=\lambda_{j}$, one has

$\displaystyle\dot{\phi}_{+}(\lambda_{j},a)\phi_{-}^{\prime}(\lambda_{j},a)-\phi_{+}^{\prime}(\lambda_{j},a)\dot{\phi}_{-}(\lambda_{j},a)=\dot{W}(\lambda_{j}).$
(4.19)

By (2.10), we know

$\displaystyle\phi_{-}^{\prime}(\lambda_{j},a)=c_{\lambda_{j}}\phi_{+}^{\prime}(\lambda_{j},a).$
(4.20)

Substituting (4.20) into (4.19), one can obtain

$\displaystyle\phi_{+}^{\prime}(\lambda_{j},a)=\frac{\dot{W}(\lambda_{j})}{c_{\lambda_{j}}\dot{\phi}_{+}(\lambda_{j},a)-\dot{\phi}_{-}(\lambda_{j},a)}.$
(4.21)

From (2.14), (4.12) and (4.21), we have

$\displaystyle-\beta_{j}^{(1)}=\frac{\phi_{+}^{\prime}(\lambda_{j},a)}{\lambda_{j}\dot{\phi}_{+}(\lambda_{j},a)}=-\frac{\dot{W}(\lambda_{j})}{\dot{\phi}_{+}(\lambda_{j},a)}\frac{1}{\frac{\lambda_{j}^{2}\dot{W}(\lambda_{j})\dot{\phi}_{+}(\lambda_{j},a)}{\lambda_{j}\gamma_{\lambda_{j}}^{2}}+\lambda_{j}\dot{\phi}_{-}(\lambda_{j},a)},$

which implies

$\displaystyle\lambda_{j}\gamma_{\lambda_{j}}^{2}=\frac{\lambda_{j}^{2}\dot{W}(\lambda_{j})\dot{\phi}_{+}(\lambda_{j},a)}{\frac{\dot{W}(\lambda_{j})}{\beta_{j}^{(1)}\dot{\phi}_{+}(\lambda_{j},a)}-\lambda_{j}\dot{\phi}_{-}(\lambda_{j},a)}.$
(4.22)

Since we can choose any $\beta_{j}^{(1)}\in(0,\beta_{j})$, by (4.17), (4.18)
and (4.22), we conclude that
$\lambda_{j}\gamma_{\lambda_{j}}^{2}\in(0,+\infty)$ can be arbitrary.

Hence, we have that

$\displaystyle\mathcal{F}$
$\displaystyle=\\{(\lambda_{1},\cdots,\lambda_{N},\lambda_{1}\gamma_{1}^{2},\cdots,\lambda_{N}\gamma_{N}^{2})\in\mathbb{R}^{2N}|\lambda_{j}\
{\rm{is\ given\ for\ any}}\ j,$ $\displaystyle\lambda_{j}\gamma_{j}^{2}\
{\rm{is\ given\ by\ \eqref{lambdajgammalambdaj}}}\ {\rm{for}}\
\varphi_{j}(a)\neq 0,\lambda_{j}\gamma_{j}^{2}>0\ {\rm{for}}\
\varphi_{j}(a)=0\\},$

which is a connected manifold of dimension $k$. ∎

######  Remark 4.4.

Note that there are $k$ many eigenvalues
$\lambda_{j}\in\\{\mu_{i}\\}_{i=1}^{N-k-1}\setminus\\{0\\}$ with
$\varphi_{j}(a)=0$. Hence we have that $N-k-2\geq k$, which means that

$\displaystyle N-2k-2\geq 0.$

From the proof of Thoerem 4.3, we can compute $\mathcal{F}$ explicitly, which
is a manifold of dimension $k$. Then by the reconstruction method in [19,
Theorem 4.3], we can obtain all the $w,v$ which yield the given interior
spectral data $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$.

######  Theorem 4.5.

For given $a$, suppose that $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is
the interior spectral data for some $\omega,v$, and

$\alpha=0,\ \beta\neq 0.$

If there are $k$ many $\varphi_{i}(a)$ such that $\varphi_{i}(a)=0$, then
infinitely many $\omega,v$’s yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. We have
$\rm{(i)}$ if $\lambda_{1}>0$ or $\lambda_{N}<0$, then $\mathcal{F}$ is a
connected manifold of dimension $k$;
$\rm{(ii)}$ if for some $j_{0}\in\\{1,\cdots,N-1\\}$,

$\lambda_{j_{0}}<0<\lambda_{j_{0}+1},$

and $\varphi_{1}(a)=0$ or $\varphi_{N}(a)=0$, then $\mathcal{F}$ is a
connected manifold of dimension $k$;
$\rm{(iii)}$ if for some $j_{0}\in\\{1,\cdots,N-1\\}$,

$\lambda_{j_{0}}<0<\lambda_{j_{0}+1},$

and

$\varphi_{1}(a)\neq 0,\ \varphi_{N}(a)\neq 0,$ (4.23)

then $\mathcal{F}$ is a collection of 2 disjoint manifolds, each of dimension
$k$.

###### Proof.

Since $\alpha=0,\beta\neq 0$, by (3.2) and Lemma 3.1, we know that

$M_{+}(z,a)+M_{-}(z,a)=\zeta+\sum_{i=1}^{N-k}\frac{\beta_{i}}{\mu_{i}-z},$
(4.24)

where $\\{\mu_{i}\\}_{i=1}^{N-k}$ are the poles of
${M_{+}(z,a)}+{M_{-}(z,a)}$. Note that there are $k$ many eigenvalues
$\lambda_{j}\in\\{\mu_{i}\\}_{i=1}^{N-k}\setminus\\{0\\}$ with
$\varphi_{j}(a)=0$. Hence we have that $N-2k-1\geq 0$.

Let us show ${\rm{(i)}}$ first. We only consider the case that
$\lambda_{1}>0$. Denote the zeros of ${M_{+}(z,a)}+{M_{-}(z,a)}$ by
$\\{\lambda_{l_{i}}\\}_{i=1}^{N-k}$. Then one has

$\mu_{1}=0<\lambda_{l_{1}}<\mu_{2}<\lambda_{l_{2}}<\cdots<\lambda_{l_{N-k-1}}<\mu_{N-k}<\lambda_{l_{N-k}}.$
(4.25)

We have that $\lambda_{l_{N-k}}=\lambda_{N}$. Then the sets $B$ and $C$, which
are defined in the proof of Theorem 4.3, are uniquely determined by the signs
of $\varphi_{i}(a),i=1,\cdots,N$.

Now we are going to prove (ii) and (iii). Without loss of generality, assume
that

$\lambda_{l_{1}}<\mu_{1}<\lambda_{l_{2}}<\cdots<\lambda_{l_{j_{1}}}<\mu_{j_{1}}=0<\lambda_{l_{j_{1}+1}}<\cdots<\lambda_{l_{N-k}}<\mu_{N-k}.$
(4.26)

If $\varphi_{N}(a)=0$, then $\mu_{N-k}=\lambda_{N}$. Namely, $\mu_{N-k}\in A$.
Therefore, the sets $B$ and $C$ are uniquely determined by the signs of
$\varphi_{i}(a),i=1,\cdots,N$. This yields (ii).

If $\varphi_{N}(a)\neq 0$, then $\mu_{N-k}>\lambda_{N}=\lambda_{l_{N-k}}$.
Namely, $\mu_{N-k}\notin A$. Since $\mu_{N-k}\in B\ {\rm{or}}\ C$, then we can
obtain two disjoint manifolds of dimension $k$. This yields (iii). ∎

######  Theorem 4.6.

For given $a$, suppose that $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is
the interior spectral data for some $\omega,v$, and

$\alpha\neq 0.$

Assume that $\lambda_{1}>0$ (or $\lambda_{N}<0$). If there are $k$ many
$\varphi_{i}(a)$ such that $\varphi_{i}(a)=0$, then infinitely many
$\omega,v$’s yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. We have
$\rm{(i)}$ if $\varphi_{N}(a)=0$ (or $\varphi_{1}(a)=0$), then $\mathcal{F}$
is a connected manifold of dimension $k$;
$\rm{(ii)}$ if $\varphi_{N}(a)\neq 0$ (or $\varphi_{1}(a)\neq 0$), then
$\mathcal{F}$ is a collection of 2 disjoint manifolds, each of dimension $k$.

###### Proof.

Since $\alpha\neq 0$, by (3.2) and Lemma 3.1, there exist
$\beta_{i},\mu_{i},i=0,\cdots,N$, ( determined by
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N-k}$ ) so that

$M_{+}(z,a)+M_{-}(z,a)=\sum_{i=0}^{N-k}\frac{\beta_{i}}{\mu_{i}-z},$ (4.27)

where $\\{\mu_{i}\\}_{i=0}^{N}$ are the poles of ${M_{+}(z,a)}+{M_{-}(z,a)}$.
Note that there are $k$ many eigenvalues
$\lambda_{j}\in\\{\mu_{i}\\}_{i=1}^{N-k-1}\setminus\\{0\\}$ with
$\varphi_{j}(a)=0$. Hence we have $N-2k\geq 0$.

Denote the zeros of ${M_{+}(z,a)}+{M_{-}(z,a)}$ by
$\\{\lambda_{l_{i}}\\}_{i=1}^{N-k}$. Then $\\{\mu_{i}\\}_{i=0}^{N-k}$ and
$\\{\lambda_{l_{i}}\\}_{i=1}^{N-k}$ are interlaced:

$\mu_{0}=0<\lambda_{l_{1}}<\mu_{1}<\lambda_{l_{2}}<\cdots<\lambda_{l_{N-k-1}}<\mu_{N-k-1}<\lambda_{l_{N-k}}<\mu_{N-k}.$
(4.28)

If $\varphi_{N}(a)=0$, then $\mu_{N-k}=\lambda_{N}$. Namely, $\mu_{N-k}\in A$.
Here the set $A$ is defined in the proof of Theorem 4.3. Then by (4.28), the
sets $B$ and $C$ are uniquely determined by the signs of
$\varphi_{i}(a),i=1,\cdots,N$. This yields (i).

If $\varphi_{N}(a)\neq 0$, then $\mu_{N-k}>\lambda_{N}=\lambda_{l_{N-k}}$.
Namely, $\mu_{N-k}\notin A$. Since $\mu_{N-k}\in B\ {\rm{or}}\ C$, we can
obtain two disjoint manifolds of dimension $k$. This yields (ii). ∎

######  Theorem 4.7.

For given $a$, suppose that $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is
the interior spectral data for some $\omega,v$, and

$\alpha\neq 0.$

Assume that for some $j_{0}\in\\{1,\cdots,N-1\\}$, we have

$\lambda_{j_{0}}<0<\lambda_{j_{0}+1}.$ (4.29)

If there are $k$ many $\varphi_{i}(a)$ such that $\varphi_{i}(a)=0$, then
infinitely many $\omega,v$’s yield the given interior spectral data
$\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$. We have
$\rm{(i)}$ if $\varphi_{1}(a)=0$ and $\varphi_{N}(a)=0$, then $\mathcal{F}$ is
a connected manifold of dimension $k$;
$\rm{(ii)}$ if $\varphi_{1}(a)=0$ or $\varphi_{N}(a)=0$ (cannot both be zero),
then $\mathcal{F}$ a collection of 2 disjoint manifolds, each of dimension
$k$;
$\rm{(iii)}$ if $\varphi_{1}(a)\neq 0$ and $\varphi_{N}(a)\neq 0$, then
$\mathcal{F}$ is a collection of 4 disjoint manifolds, each of dimension $k$.

###### Proof.

By (4.27) and (4.29), we have that

$\mu_{0}<\lambda_{l_{1}}<\cdots<\lambda_{l_{j_{1}}}<\mu_{j_{1}}=0<\lambda_{l_{j_{1}+1}}<\cdots<\lambda_{l_{N-k}}<\mu_{N-k}.$
(4.30)

If $\varphi_{1}(a)=0$ and $\varphi_{N}(a)=0$, then

$\mu_{0}=\lambda_{1},\ \mu_{N-k}=\lambda_{N}.$

Namely, $\mu_{0},\mu_{N-k}\in A$. Therefore, the sets $B$ and $C$ are uniquely
determined by the signs of $\varphi_{i}(a),i=1,\cdots,N$. This yields (i).

Let us show (ii). We only consider the case that
$\varphi_{1}(a)=0,\varphi_{N}(a)\neq 0$. Then

$\mu_{0}=\lambda_{1},\ \mu_{N-k}>\lambda_{N}=\lambda_{l_{N-k}}.$

Namely, $\mu_{0}\in A$ and $\mu_{N-k}\notin A$. Since $\mu_{N-k}\in B\
{\rm{or}}\ C$, we can obtain two disjoint manifolds. This yields (ii).

If $\varphi_{1}(a)\neq 0$ and $\varphi_{N}(a)\neq 0$, then

$\mu_{0}<\lambda_{1}=\lambda_{l_{1}},\
\mu_{N-k}>\lambda_{N}=\lambda_{l_{N-k}}.$

Namely, $\mu_{0},\mu_{N-k}\notin A$. Since $\mu_{0},\mu_{N-k}\in B\ {\rm{or}}\
C$, we can obtain four disjoint manifolds. This yields (iii). ∎

Now we discuss the existence of the interior inverse problem.

######  Theorem 4.8.

For given $a$, $\\{\lambda_{i},\alpha_{i}\\}_{i=1}^{N}\subset\mathbb{R}$ is
the interior spectral data for some $\omega,v$ at $a$, if and only if
$\rm{(i)}$ for any $j$ with $\alpha_{j}=0$, $\lambda_{j}$ is the zero of the
function

$\displaystyle
G(z):=(1-\sum_{i=1}^{N}\alpha_{i}^{2})z-\sum_{i=1}^{N}\lambda_{i}\alpha_{i}^{2}+\sum_{i=1}^{N}\frac{\lambda_{i}^{2}\alpha_{i}^{2}}{\lambda_{i}-z};$
(4.31)

$\rm{(ii)}$ if $\alpha_{j}=0$ for some $j\neq 1,N$, then we have
$\alpha_{j-1}\alpha_{j+1}<0$;
$\rm{(iii)}$ $\\{\lambda_{i},\alpha_{i}\\}_{i=1}^{N}\in\mathcal{T}$, where
$\mathcal{T}$ is defined in Section 3.

###### Proof.

The necessity is obtained from Lemma 3.1, Lemma 4.1 and Remark 6.8.

Let us show the sufficiency. Assume that there are $k$ many $\alpha_{i}$ such
that $\alpha_{i}=0$. If $k=0$, the sufficiency holds by Lemma 3.8. We consider
the case that $k>0$. From (3.35), we know that $G(z)$ is a Herglotz-Nevanlinna
function. Let

$M(z):=-\frac{1}{G(z)}=M_{+}(z)+M_{-}(z),$ (4.32)

where $M_{+}(z)$ and $M_{-}(z)$ will be determined later. Then there exist
$\gamma\geq 0$, $\zeta$, $\beta_{i},\mu_{i},i=1,\cdots,N(k)$, so that

$M(z)=\gamma z+\zeta+\sum_{i=1}^{N(k)}\frac{\beta_{i}}{\mu_{i}-z}.$ (4.33)

Let $A$ be the set of $\lambda_{j}$ with $\alpha_{j}=0$. By (i), we know that
$A\subseteq\\{\mu_{i}\\}_{i=1}^{N(k)}$.

We make the following choice. For $\mu_{i}\neq
0\in(\lambda_{j},\lambda_{j+1})$ for some $j$, let $\mu_{i}$ be the pole of
$M_{\pm}(z)$ if

$\pm\alpha_{j}\alpha_{j+1}<0.$ (4.34)

For $\mu_{i}\in(-\infty,\lambda_{1})\cup(\lambda_{N},+\infty)$ with
$\mu_{i}\neq 0$ (if exist), we can choose $\mu_{i}$ to be the pole of
$M_{+}(z)$. Denote the above poles of $M_{+}(z)$ and $M_{-}(z)$ by $B$ and
$C$, respectively. For any $\mu_{i}\in A$, let
$\beta_{i}^{(1)},\beta_{i}^{(2)}>0$ satisfy

$\beta_{i}=\beta_{i}^{(1)}+\beta_{i}^{(2)},$ (4.35)

and take

$\displaystyle M_{+}(z)$ $\displaystyle=\gamma
z+\zeta-\frac{1}{2z}+\sum_{\mu_{i}\in
B}\frac{\beta_{i}}{\mu_{i}-z}+\sum_{\mu_{i}\in
A}\frac{\beta_{i}^{(1)}}{\mu_{i}-z},$ (4.36) $\displaystyle M_{-}(z)$
$\displaystyle=-\frac{1}{2z}+\sum_{\mu_{i}\in
C}\frac{\beta_{i}}{\mu_{i}-z}+\sum_{\mu_{i}\in
A}\frac{\beta_{i}^{(2)}}{\mu_{i}-z}.$ (4.37)

Applying Lemma 2.6 and Lemma 2.7, $M_{+}(z)$ and $M_{-}(z)$ uniquely determine
$\omega|_{[a,+\infty)},v|_{[a,+\infty)}$ and
$\omega|_{(-\infty,a)},v|_{(-\infty,a)}$, respectively. Hence $\omega$ and $v$
are uniquely determined.

It remains to establish that the $\omega,v$ determined above has the interior
spectral data $\\{\lambda_{i},\alpha_{i}\\}_{i=1}^{N}$. Note that the Weyl-
Titchmarsh functions $M_{\pm}(z,a)$ of $H_{\pm}(\omega,v,a)$ satisfy

$M_{\pm}(z,a)=M_{\pm}(z).$ (4.38)

If $\lambda_{j}\in\\{\lambda_{i}\\}_{i=1}^{N}\setminus A$, by (4.32), (4.36),
(4.37) and (4.38), $\lambda_{j}$ is a pole of $-1/(M_{+}(z,a)+M_{-}(z,a))$.
Then by Lemma 3.1, $\lambda_{j}$ is an eigenvalue of $H(\omega,v)$. If
$\lambda_{j}\in A$, by (2.14) and (2.15), we obtain

$\phi_{+}(\lambda_{j},a)=\phi_{-}(\lambda_{j},a)=0.$

Then from (2.9), we know that $W(\lambda_{j})=0$, which implies that
$\lambda_{j}$ is also an eigenvalue of $H(\omega,v)$. Since $H(\omega,v)$ has
no other eigenvalues, one concludes that $\\{\lambda_{i}\\}_{i=1}^{N}$ is the
spectrum of $H(\omega,v)$.

Now we show that for any $i=1,\cdots,N$, one has $\varphi_{i}(a)=\alpha_{i}$.
We consider the case that $\lambda_{1}>0$. By (4.31), (4.32), (4.38) and Lemma
3.1, one has

$\left\lvert\varphi_{i}(a)\right\rvert=|\alpha_{i}|,i=1,\cdots,N.$

Therefore, it suffices to show that for any $i=1,\cdots,N$, we have

$\displaystyle{\rm{sgn}}(\varphi_{i}(a))={\rm{sgn}}(\alpha_{i}).$ (4.39)

Applying Remark 6.9, one has that (4.39) holds for $i=1$. If
$\mu_{i}\in(\lambda_{j},\lambda_{j+1})$ for some $j$, we know that $\mu_{i}$
is a pole of $M_{\pm}(z,a)$ if and only if
$\pm\varphi_{j}(a)\varphi_{j+1}(a)<0$. Combining with Lemma 4.1, the signs of
${\varphi}_{i}(a)$ for $i\neq 1$ are uniquely determined by
${\rm{sgn}}({\varphi}_{1}(a))$. Therefore, by (ii) and the definition of
$\mathcal{T}$, we can obtain that (4.39) holds for any $i=1,\cdots,N$.

Following the similar steps, we can obtain that $\varphi_{i}(a)=\alpha_{i}$
for any $i=1,\cdots,N,$ under the condition that $\lambda_{N}<0$ or

$\lambda_{j_{0}}<0<\lambda_{j_{0}+1}$

for some $j_{0}$. Therefore, we conclude that
$\\{\lambda_{i},\alpha_{i}\\}_{i=1}^{N}$ is the interior spectral data for
some $\omega,v$ at $a$. ∎

######  Theorem 4.9.

For given $a$, suppose that $\\{\lambda_{i},\varphi_{i}(a)\\}_{i=1}^{N}$ is
the interior spectral data for some $\omega,v$, and there are $k$ many
$\varphi_{i}(a)$ such that $\varphi_{i}(a)=0$. Then there are infinitely many
$\tilde{\omega},\tilde{v}$, such that

$\tilde{\lambda}_{i}=\lambda_{i},|\tilde{\varphi}_{i}(a)|=|\varphi_{i}(a)|,i=1,\cdots,N.$

We have
$\rm{(i)}$ if

$\alpha=\beta=0,$

then the spectral data of the inverse spectral family is a collection of
$2^{N-2k-2}$ disjoint manifolds, each of dimension $k$;
$\rm{(ii)}$ if

$\alpha=0,\ \beta\neq 0,$

then the spectral data of the inverse spectral family is a collection of
$2^{N-2k-1}$ disjoint manifolds, each of dimension $k$;
$\rm{(iii)}$ if

$\alpha\neq 0,$

then the spectral data of the inverse spectral family is a collection of
$2^{N-2k}$ disjoint manifolds, each of dimension $k$.

###### Proof.

For example, assume that $\varphi_{m_{1}}(a)=\cdots=\varphi_{m_{k}}(a)=0$. We
first show that ${\rm{(iii)}}$ holds.
Case 1: $\lambda_{1}>0$ or $\lambda_{N}<0$. We only consider the case that
$\lambda_{1}>0$. By Remark 6.9, we have that $\tilde{\varphi}_{1}(a)>0$. If
$\varphi_{N}(a)\neq 0$, namely $m_{k}<N$, by Theorem 4.8, for any

$i\in\\{2,\cdots,N\\}\setminus\\{{m_{1}},{m_{1}+1},\cdots,{m_{k}},{m_{k}+1}\\},$

the sign of $\tilde{\varphi}_{i}(a)$ is not determined. We can choose the
signs of $\tilde{\varphi}_{1}(a),\cdots,\tilde{\varphi}_{N}(a)$ in
$2^{N-2k-1}$ ways. Every such choice yields 2 disjoint manifolds by (ii) of
Theorem 4.6. Hence, we can determine $2^{N-2k}$ disjoint manifolds.

If $\varphi_{N}(a)=0$, namely $m_{k}=N$, by Theorem 4.8, for any

$i\in\\{2,\cdots,N\\}\setminus\\{{m_{1}},{m_{1}+1},\cdots,{m_{k-1}},{m_{k-1}+1},{m_{k}}\\},$

the sign of $\tilde{\varphi}_{i}(a)$ is not determined. We can choose the
signs of $\tilde{\varphi}_{1}(a),\cdots,\tilde{\varphi}_{N}(a)$ in $2^{N-2k}$
ways. Every such choice yields a connected manifold by (i) of Theorem 4.6.
Hence, we can determine $2^{N-2k}$ disjoint manifolds.
Case 2: For some $j_{0}\in\\{1,\cdots,N-1\\}$, we have

$\lambda_{j_{0}}<0<\lambda_{j_{0}+1}.$

By Remark 6.9, we know that
$\tilde{\varphi}_{j_{0}}(a),\tilde{\varphi}_{j_{0}+1}(a)>0$.

If $\varphi_{1}(a)=\varphi_{N}(a)=0$, by Theorem 4.8, for any

$i\in\\{1,\cdots,N\\}\setminus\\{j_{0},j_{0}+1,{m_{1}},{m_{2}},{m_{2}+1},\cdots,{m_{k-1}},{m_{k-1}+1},{m_{k}}\\},$

the sign of $\tilde{\varphi}_{i}(a)$ is not determined. We can choose the
signs of $\tilde{\varphi}_{1}(a),\cdots,\tilde{\varphi}_{N}(a)$ in $2^{N-2k}$
ways. Every such choice yields a connected manifold by (i) of Theorem 4.6.
Hence, we can determine $2^{N-2k}$ disjoint manifolds.

Similarly, if $\varphi_{1}(a)\neq 0,\varphi_{N}(a)=0$ or $\varphi_{1}(a)\neq
0,\varphi_{N}(a)\neq 0$, we can also determine $2^{N-2k}$ disjoint manifolds.

Note that in both cases, the spectral data of the inverse spectral family is a
collection of $2^{N-2k}$ disjoint manifolds, each of dimension $k$. Applying
the above method with slight modification, we can show (i) and (ii). ∎

##  5\. The Camassa-Holm Equation

Let $(u,\mu)$ be a multipeakon global conservative solution of the CH equation
(see [19]). Here $\mu$ is a non-negative Borel measure with the absolutely
continuous part determined by $u$ via

$\mu_{ac}(B,t)=\int_{B}|u(x,t)|^{2}+|u_{x}(x,t)|^{2}dx,\ t\in\mathbb{R},$
(5.1)

for each Borel set $B\in\mathcal{B}(\mathbb{R})$. Define

$\displaystyle\omega(x,t)$
$\displaystyle=u(x,t)-u_{xx}(x,t)=\sum_{i=1}^{n(t)}\omega_{i}(t)\delta_{x_{i}(t)},$
(5.2) $\displaystyle v(B,t)$
$\displaystyle=\mu(B,t)-\int_{B}|u(x,t)|^{2}+|u_{x}(x,t)|^{2}dx,\
B\in\mathcal{B}(\mathbb{R}),$ (5.3)

for each $t\in\mathbb{R}$.

Consider the family (parametrized by time $t\in\mathbb{R})$ of spectral
problems

$-f^{\prime\prime}(x)+\frac{1}{4}f(x)=z\omega(x,t)f(x)+z^{2}v(x,t)f(x),\
x\in\mathbb{R},$ (5.4)

with $\omega(x,t)$ and $v(x,t)$ defined by (5.2) and $\eqref{eq4.4}$,
respectively. We will denote all spectral quantities associated with this
spectral problem as in the preceding sections but with an additional time
parameter. In particular, the spectrum of (5.4) will be denoted by
$\sigma(H(t))$. By[19, Theorem 5.1], the pair $(u,\mu)$ is a global
conservative multipeakon solution of the CH equation if and only if the
problems (5.4) are isospectral with

$\gamma_{\lambda_{i}}^{2}(t)=e^{-\frac{t-t_{0}}{2\lambda_{i}}}\gamma_{\lambda_{i}}^{2}(t_{0}),\
t\in\mathbb{R},\ \lambda_{i}\in\sigma(H(t_{0})).$ (5.5)

Here $\gamma_{\lambda_{i}}^{2}(t)$ is defined by (2.12) with an additional
time parameter.

Now in a position to state the trace formula.

######  Theorem 5.1.

Suppose that $(u,\mu)$ is a global conservative multipeakon solution of the CH
equation and let $\\{\lambda_{i}\\}_{i=1}^{N}$ be eigenvalues of (5.4) with
$\omega(x,t)$ and $v(x,t)$ defined by (5.2) and (5.3), respectively. For a
fixed $t\in\mathbb{R}$, let $\\{\varphi_{i}(x,t)\\}_{i=1}^{N}$ be the
corresponding normalized eigenfunctions given by (3.1). Then one has

$u(x,t)=\frac{1}{2}\sum_{i=1}^{N}\frac{|\varphi_{i}(x,t)|^{2}}{\lambda_{i}}.$
(5.6)

###### Proof.

For a fixed $t\in{\mathbb{R}}$, by Lemma 3.1, for any $x\in\mathbb{R}$, one
has

$-\frac{1}{M_{+}(z,x)+M_{-}(z,x)}=\alpha z+\beta+G(z,x).$ (5.7)

Here

$\displaystyle\alpha=1-\sum_{i=1}^{N}|\varphi_{i}(x)|^{2},\beta=-\sum_{i=1}^{N}\lambda_{i}|\varphi_{i}(x)|^{2}.$

Here we omit the $t$-dependance in functions $G(z,x),M(z,x),\varphi_{i}(x)$
for simplicity. From [19, p. 901], we know that

$\displaystyle\dot{\phi}_{\pm}(0,x)$
$\displaystyle=e^{\mp\frac{x}{2}}\int_{I_{\pm(x)}}(e^{-|x-s|}-1)d\omega(s),$
$\displaystyle\dot{W}(0)$ $\displaystyle=-\int_{\mathbb{R}}\omega(s)ds,$

where $I_{+}(x)=[x,+\infty)$ and $I_{-}(x)=(-\infty,x)$. Combining with (2.11)
and (2.16), one can obtain that

$\displaystyle\phi_{\pm}(z,x)$
$\displaystyle=e^{\mp\frac{x}{2}}+ze^{\mp\frac{x}{2}}\int_{I_{\pm}(x)}(e^{-|x-s|}-1)d\omega(s)+O(z^{2}),\
z\rightarrow 0,$ $\displaystyle W(z)$
$\displaystyle=1-z\int_{\mathbb{R}}\omega(s)ds+O(z^{2}),\ z\rightarrow 0.$

Then from (3.5), we have

$-\frac{1}{M_{+}(z,x)+M_{-}(z,x)}=z+z^{2}\int_{\mathbb{R}}e^{-|x-s|}d\omega(s)+O(z^{3}),\
z\rightarrow 0.$ (5.8)

By (3.3) and (5.8), one obtains

$\int_{\mathbb{R}}e^{-|x-s|}d\omega(s)=\sum_{i=1}^{N}\frac{|\varphi_{i}(x)|^{2}}{\lambda_{i}}.$
(5.9)

Then from (5.2), we obtain that

$u(x)=\frac{1}{2}\int_{\mathbb{R}}e^{-|x-s|}d\omega(s)=\frac{1}{2}\sum_{i=1}^{N}\frac{|\varphi_{i}(x)|^{2}}{\lambda_{i}}.$

∎

The Cauchy problem for the CH equation is solving $u(x,t)$ from $u(x,0)$.
Namely, solve $u(x,t)$ by $x_{i}(0),\omega_{i}(0),i=1,\cdots,N$ in (1.3).
Similarly, (5.6) enlightens us to consider the following question:
For given $x_{0}\in\mathbb{R}$, can we solve $u(x,t)$ from
$\lambda_{i},|\varphi_{i}(x_{0},0)|,i=1,\cdots,N$? We have the following
theorem.

######  Theorem 5.2.

Suppose that $(u,\mu)$ is a global conservative multipeakon solution of the CH
equation.
$\rm{(i)}$ For a fixed $x_{0}\in\mathbb{R}$, if

$\varphi_{i}(x_{0},0)\neq 0,i=1,\cdots,N,$

then there exist finite global conservative multipeakon solutions
$(\tilde{u},\tilde{\mu})$, so that

$\tilde{\lambda}_{i}=\lambda_{i},|\tilde{\varphi}_{i}(x_{0},0)|=|\varphi_{i}(x_{0},0)|,\
\ i=1,\cdots,N.$

Namely,

$\tilde{u}(x_{0},0)=\frac{1}{2}\sum_{i=1}^{N}\frac{|\varphi_{i}(x_{0},0)|^{2}}{\lambda_{i}}.$

$\rm{(ii)}$ For a fixed $x_{0}$, if $\varphi_{i}(x_{0},0)=0$ for some $i$,
then there exist infinite many global conservative multipeakon solutions
$(\tilde{u},\tilde{\mu})$, so that

$\tilde{\lambda}_{i}=\lambda_{i},|\tilde{\varphi}_{i}(x_{0},0)|=|\varphi_{i}(x_{0},0)|,\
\ i=1,\cdots,N.$

Namely,

$\tilde{u}(x_{0},0)=\frac{1}{2}\sum_{i=1}^{N}\frac{|\varphi_{i}(x_{0},0)|^{2}}{\lambda_{i}}.$

###### Proof.

By Theorem 3.9 and Theorem 5.1, we have (i). Applying Theorem 4.9 and Theorem
5.1, we obtain (ii). ∎

Denote by

$\lambda^{\star}=\min\\{|\lambda_{i}|:i=1,\cdots,N\\}.$ (5.10)

Combining Lemma 3.1, Lemma 3.2 and Theorem 5.1, we obtain the supremum of
$|u(x,t)|$. This supremum only depends on $\lambda^{\star}$.

######  Theorem 5.3.

Assume that $(u,\mu)$ is a global conservative multipeakon solution of the CH
equation. Let $\\{\lambda_{i}\\}_{i=1}^{N}$ be eigenvalues of (5.4) with
$\omega(x,t)$ and $v(x,t)$ defined by (5.2) and (5.3), respectively. Then we
have

$\sup\\{|u(x,t)|:x,t\in\mathbb{R}\\}=\frac{1}{2\lambda^{\star}}.$ (5.11)

In particular, $|u(x,t)|$ obtains the supremum at some
$(x,t)\in\mathbb{R}^{2}$ if and only if $N=1$.

###### Proof.

By (5.6), (5.10) and Lemma 3.1, we have

$\displaystyle|u(x,t)|$
$\displaystyle\leq\frac{1}{2}\sum_{i=1}^{N}\frac{|\varphi_{i}(x,t)|^{2}}{|\lambda_{i}|}$
(5.12)
$\displaystyle\leq\frac{1}{2\lambda^{\star}}\sum_{i=1}^{N}|\varphi_{i}(x,t)|^{2}\leq\frac{1}{2\lambda^{\star}}.$

Hence, $1/2\lambda^{\star}$ is a upper bound for $|u(x,t)|$. Combining with
$(6.5)$ in [2], one obtains that (5.11) holds (in the normalization used in
this paper).

If $N=1$, for any $t\in\mathbb{R}$, $|\varphi_{1}(x_{1}(t)),t)|=1$ by Lemma
3.2. Then by (5.6), one has

$|u(x_{1}(t),t)|=\frac{1}{2\lambda^{\star}}.$

Therefore for any $t\in\mathbb{R}$, $|u(x,t)|$ obtains the supremum at the
point $(x_{1}(t),t)$.

If $N=2$, for any $t\in\mathbb{R}$, if $x\neq x_{j}(t),j=1,2,$ one has

$\sum_{i=1}^{2}|\varphi_{i}(x,t)|^{2}\ <1$

by Lemma 3.2. Then we obtain that for $x\neq x_{j}(t),j=1,2,$

$\displaystyle|u(x,t)|\leq\frac{1}{2\lambda^{\star}}\sum_{i=1}^{2}|\varphi_{i}(x,t)|^{2}<\frac{1}{2\lambda^{\star}}.$
(5.13)

Note that

$\varphi_{i}(x_{j}(t),t)\neq 0,\quad i,j=1,2.$ (5.14)

Otherwise, for example, assume that

$\varphi_{1}(x_{1}(t),t)=0.$ (5.15)

Since for $x\leq x_{1}(t)$, $\varphi_{1}(x,t)=Ae^{\frac{x}{2}}$ for some
constant $A$. By (5.15), one has $A=0$. Then from (2.5), (2.6), for any
$x\in\mathbb{R}$, one obtains $\varphi_{1}(x,t)=0$. This is in contradiction
with the fact that $\varphi_{1}(x,t)$ is an eigenfunction. By (5.14), for
$j=1,2$,

$\displaystyle|u(x_{j}(t),t)|<\frac{1}{2\lambda^{\star}}\sum_{i=1}^{2}|\varphi_{i}(x_{j}(t),t)|^{2}=\frac{1}{2\lambda^{\star}}.$
(5.16)

Hence from (5.13) and (5.16), we conclude that if $N=2$, $|u(x,t)|$ cannot
obtain the supremum at some $(x,t)\in\mathbb{R}^{2}$.

If $N\geq 3$, we need to consider the following two cases.
Case 1. $\lambda_{1}>0$ or $\lambda_{N}<0$. For any $t\in\mathbb{R}$, by
Corollary 6.11, we have

$\displaystyle|u(x,t)|<\frac{1}{2\lambda^{\star}}\sum_{i=1}^{N}|\varphi_{i}(x,t)|^{2}\leq\frac{1}{2\lambda^{\star}}.$
(5.17)

Case 2. For some $j_{0}\in\\{1,\cdots,N-1\\}$, we have

$\lambda_{j_{0}}<0<\lambda_{j_{0}+1}.$ (5.18)

Then by Theorem 6.8 one has that

$\Big{|}\frac{|\varphi_{j_{0}}(x,t)|^{2}}{2\lambda_{j_{0}}}+\frac{|\varphi_{j_{0}+1}(x,t)|^{2}}{2\lambda_{j_{0}+1}}\Big{|}<\frac{|\varphi_{j_{0}}(x,t)|^{2}+|\varphi_{j_{0}+1}(x,t)|^{2}}{2\lambda^{\star}}.$

This implies that (5.17) holds. Therefore, if $N\geq 3$, then $|u(x,t)|$
cannot obtain the supremum at some $(x,t)\in\mathbb{R}^{2}$.

From all the above, we conclude that $|u(x,t)|$ obtains the supremum at some
$(x,t)\in\mathbb{R}^{2}$ if and only if $N=1$. ∎

If $N=1$, corresponding to the single peakon, $|u(x,t)|$ obtains the supremum.
If $u$ is a multipeakon solution with $N\geq 2$, $|u(x,t)|\leq
1/2\lambda^{\star}$ means other peakons have a negative effect on this highest
(deepest) peakon which has a height (depth) of $1/2\lambda^{\star}$.
$|u(x,t)|$ cannot obtain the supremum implies that as $t$ becomes large, the
negative effect becomes smaller but never disappear.

##  6\. Oscillation of eigenfunctions

In this section, we give a proof of the oscillation theorem for problem (2.3).
The approach we are taking in this section follows in part the one in [43].
But we take advantage of the property of Herglotz-Nevanlinna function and
greatly simplify the proof.

In this part, if an eigenvalue $\lambda$ is positive (or negative), we denote
it by $\lambda^{+}$ (or $\lambda^{-}$).

For $x_{i}=1,\cdots,n-1$, one has

$\displaystyle\phi_{+}^{\prime}(z,x_{i}+)$
$\displaystyle=\frac{1}{2\sinh\left(\frac{x_{i+1}-x_{i}}{2}\right)}\phi_{+}(z,x_{i+1})-\frac{1}{2}\coth\left(\frac{x_{i+1}-x_{i}}{2}\right)\phi_{+}(z,x_{i}).$
(6.1)

Indeed, for any $i=1,\cdots,n-1$, there exists $A,B\in\mathbb{R}$, so that

$\phi_{+}(z,x)=Ae^{\frac{x}{2}}+Be^{-\frac{x}{2}},\ x_{i}\leq x\leq x_{i+1}.$
(6.2)

Then we have

$\left\\{\begin{aligned}
Ae^{\frac{x_{i}}{2}}+Be^{-\frac{x_{i}}{2}}&=\phi_{+}(z,x_{i}),\\\
\frac{1}{2}Ae^{\frac{x_{i}}{2}}-\frac{1}{2}Be^{-\frac{x_{i}}{2}}&=\phi^{\prime}_{+}(z,x_{i}+).\\\
\end{aligned}\right.$

As a consequence, one obtains

$A=\frac{1}{2}e^{-\frac{x_{i}}{2}}(\phi_{+}(z,x_{i})+2\phi^{\prime}_{+}(z,x_{i}+)),B=\frac{1}{2}e^{\frac{x_{i}}{2}}(\phi_{+}(z,x_{i})-2\phi^{\prime}_{+}(z,x_{i}+)).$
(6.3)

Substituting (6.3) into (6.2) and let $x=x_{i+1},$ we obtain

$\displaystyle\phi_{+}(z,x_{i+1})=$
$\displaystyle\frac{1}{2}e^{\frac{x_{i+1}-x_{i}}{2}}(\phi_{+}(z,x_{i})+2\phi^{\prime}_{+}(z,x_{i}+))$
$\displaystyle+\frac{1}{2}e^{-\frac{x_{i+1}-x_{i}}{2}}(\phi_{+}(z,x_{i})-2\phi^{\prime}_{+}(z,x_{i}+)).$
(6.4)

Therefore, (6.1) holds. Similarly, for $i=2,\cdots,n$,

$\displaystyle\phi_{+}^{\prime}(z,x_{i}-)$
$\displaystyle=-\frac{1}{2\sinh\left(\frac{x_{i}-x_{i-1}}{2}\right)}\phi_{+}(z,x_{i-1})+\frac{1}{2}\coth\left(\frac{x_{i}-x_{i-1}}{2}\right)\phi_{+}(z,x_{i}).$
(6.5)

Denote by

$\phi_{i}(z)=\phi_{+}(z,x_{n-i}),i=1,\cdots,n,$ (6.6)

and let

$\displaystyle a_{i}$
$\displaystyle=\frac{1}{2\sinh\left(\frac{x_{n-i+1}-x_{n-i}}{2}\right)},i=1,\cdots,n-1,$
(6.7) $\displaystyle b_{i}$
$\displaystyle=\frac{1}{2}\left(\coth\left(\frac{x_{n-i+1}-x_{n-i}}{2}\right)+\coth\left(\frac{x_{n-i}-x_{n-i-1}}{2}\right)\right),i=0,\cdots,n-1.$
(6.8)

Here $x_{0}:=-\infty$ and $x_{n+1}:=+\infty$. By (2.8) and (6.6), we have

$\displaystyle\phi_{0}(z)$ $\displaystyle=e^{-\frac{x_{n}}{2}}.$ (6.9)

Letting $i=n$ in (6.5), then substituting equations (6.5) and

$\phi_{+}^{\prime}(z,\lambda_{n}+)=-\frac{1}{2}e^{-\frac{x_{n}}{2}}$

into (2.6), we can obtain

$\displaystyle-a_{1}\phi_{1}(z)$
$\displaystyle+b_{0}\phi_{0}(z)=(z^{2}v_{n}+z\omega_{n})\phi_{0}(z).$ (6.10)

Substituting (6.1) and (6.5) into (2.6) with $i=n-j+1$, one has that for
$j=2,\cdots,n-1$,

$\displaystyle-
a_{j}\phi_{j}(z)+b_{j-1}\phi_{j-1}(z)-a_{j-1}\phi_{j-2}(z)=(z^{2}v_{n-j+1}+z\omega_{n-j+1})\phi_{j-1}(z).$
(6.11)

Therefore, one can obtain $\phi_{j}(z),j=0,\cdots,n-1,$ from (6.9), (6.10) and
(6.11).

Recall that $\lambda$ is an eigenvalue of $H$ if and only if there exists
$A\in\mathbb{R}$, so that $\phi_{+}(\lambda,x)=Ae^{\frac{x}{2}}$ for $x\leq
x_{1}$. Namely,

$\phi_{+}^{\prime}(\lambda,x_{1}-)=\frac{1}{2}\phi_{+}(\lambda,x_{1}).$ (6.12)

Substituting (6.12) and (6.1) into (2.6) with $i=1$, we can obtain

$b_{n-1}\phi_{n-1}(\lambda)-a_{n-1}\phi_{n-2}(\lambda)=(\lambda^{2}v_{1}+\lambda\omega_{1})\phi_{n-1}(\lambda).$

Therefore, $\lambda$ is an eigenvalue of $H$ if and only if $\lambda$ is a
root of

$b_{n-1}\phi_{n-1}(z)-a_{n-1}\phi_{n-2}(z)=(z^{2}v_{1}+z\omega_{1})\phi_{n-1}(z).$
(6.13)

For any $i\in\\{1,\cdots,n\\}$, define

$\displaystyle i_{v}$ $\displaystyle=\\#\\{j\in\\{1,\cdots,i\\}|v_{n-j+1}\neq
0\\},$ $\displaystyle i_{+}$
$\displaystyle=\\#\\{j\in\\{1,\cdots,i\\}|v_{n-j+1}=0,\omega_{n-j+1}>0\\},$
(6.14) $\displaystyle i_{-}$
$\displaystyle=\\#\\{j\in\\{1,\cdots,i\\}|v_{n-j+1}=0,\omega_{n-j+1}<0\\}.$
(6.15)

For any $j$ with $v_{n-j+1}\neq 0$ and $j_{v}=k$, define

$\varphi_{k}(z):=z\sqrt{v_{n-j+1}}\phi_{j-1}(z).$ (6.16)

Denote by

$y=(\phi_{0},\cdots,\phi_{n-1},\varphi_{1},\cdots,\varphi_{n_{v}})^{T}.$

Then by (6.10), (6.11), (6.13) and (6.16), we can rewrite the spectral problem
$H$ in the matrix form

$Jy=zDy.$ (6.17)

Here

$\displaystyle J=\left(\begin{array}[]{cc}J_{11}&J_{12}\\\ J_{21}&J_{22}\\\
\end{array}\right),$ (6.20)

$J_{11}$ is an $n\times n$ matrix with

$\displaystyle
J_{11}=\left(\begin{array}[]{ccccccccc}b_{0}&-a_{1}&0&\cdots&0\\\
-a_{1}&b_{1}&-a_{2}&\cdots&0\\\ 0&-a_{2}&b_{2}&\cdots&0\\\
\cdots&\cdots&\cdots&\cdots&\cdots\\\
0&0&0&-a_{n-1}&b_{n-1}\end{array}\right),$ (6.26)

$J_{22}$ is an $n_{v}\times n_{v}$ identity matrix, $J_{12}$ is an $n\times
n_{v}$ zero matrix, $J_{21}$ is an $n_{v}\times n$ zero matrix. $D$ is a
symmetric matrix with

$\displaystyle D=\left(\begin{array}[]{cc}D_{11}&D_{12}\\\ D_{21}&D_{22}\\\
\end{array}\right),$ (6.29)

$D_{11}$ is an $n\times n$ matrix with

$\displaystyle
D_{11}=\left(\begin{array}[]{ccccccccc}\omega_{n}&0&0&\cdots&0\\\
0&\omega_{n-1}&0&\cdots&0\\\ 0&0&\omega_{n-2}&\cdots&0\\\
\cdots&\cdots&\cdots&\cdots&\cdots\\\ 0&0&0&0&\omega_{1}\end{array}\right),$
(6.35)

$D_{22}$ is an $n_{v}\times n_{v}$ zero matrix, $D_{12}$ is an $n\times n_{v}$
matrix with entries

$d_{jk}(D_{12})=\left\\{\begin{aligned} \sqrt{v_{n-j+1}},\quad j_{v}=k,\\\
0,\quad\rm{otherwise}.\\\ \end{aligned}\right.$

Define

$\displaystyle J_{i}=\left(\begin{array}[]{cc}J_{11,i}&J_{12,i}\\\
J_{21,i}&J_{22,i}\\\ \end{array}\right),$ (6.38)

where $J_{11,i}$ is an $i$-$th$ principal submatrix of $J_{11}$, $J_{22,i}$ is
an $i_{v}\times i_{v}$ identity matrix, $J_{12,i}$ is an $i\times i_{v}$ zero
matrix, $J_{21,i}$ is an $i_{v}\times i$ zero matrix. Define the symmetric
matrix

$\displaystyle D_{i}=\left(\begin{array}[]{cc}D_{11,i}&D_{12,i}\\\
D_{21,i}&D_{22,i}\\\ \end{array}\right),$ (6.41)

where $D_{11,i}$ is an $i$-$th$ principal submatrix of $D_{11}$, $D_{22,i}$ is
an $i_{v}\times i_{v}$ zero matrix, $D_{12,i}$ is an $i\times i_{v}$ matrix
with entries

$d_{jk}(D_{12,i})=\left\\{\begin{aligned} \sqrt{v_{n-j+1}},\quad j_{v}=k,\\\
0,\quad\rm{otherwise}.\\\ \end{aligned}\right.$

Let $Q_{i}(z)=\det(J_{i}-zD_{i})$ and suppose that

$Q_{0}(z)=1.$ (6.42)

Then from (6.17), after some calculations, one has

$\displaystyle Q_{1}(z)$ $\displaystyle=b_{0}-\omega_{n}z-v_{n}z^{2};$ (6.43)
$\displaystyle Q_{i}(z)$
$\displaystyle=(b_{i-1}-\omega_{n-i+1}z-v_{n-i+1}z^{2})Q_{i-1}(z)-a_{i-1}^{2}Q_{i-2}(z),\
i=2,\cdots,n.$ (6.44)

######  Lemma 6.1.

For any $i=1,\cdots,n$, $Q_{i-1}(z)$ and $Q_{i}(z)$ have no common zeros.

###### Proof.

Otherwise, if there exists $z=\lambda_{0}$ so that $Q_{i-1}(\lambda_{0})=0$
and $Q_{i}(\lambda_{0})=0$. By (6.44), one has $Q_{i-2}(\lambda_{0})=0$. Then
we can obtain that $Q_{i-3}(\lambda_{0})=\cdots=Q_{0}(\lambda_{0})=0,$ which
contradicts $\eqref{q0}$. ∎

######  Lemma 6.2.

For any $i=1,\cdots,n$, the zeros of $Q_{i}(z)$ are real and simple. Moreover,
$Q_{i}(z)$ has $i_{v}+i_{+}$ positive zeros and $i_{v}+i_{-}$ negative zeros.

###### Proof.

For any $\xi=(\xi_{1},\cdots,\xi_{i+i_{v}})\in\mathbb{R}^{i+i_{v}}$, we have
that

$\displaystyle\xi J_{i}\xi^{T}=$
$\displaystyle(b_{0}-a_{1})\xi_{1}^{2}+(b_{i-1}-a_{i-1})\xi_{i}^{2}+\sum_{j=2}^{i-1}(b_{j-1}-a_{j-1}-a_{j})\xi_{j}^{2}$
$\displaystyle+\sum_{j=1}^{i-1}a_{j}(\xi_{j+1}-\xi_{j})^{2}+\sum_{j=i+1}^{i+i_{v}}\xi_{j}^{2}.$

By (6.7) and (6.8), one obtains $b_{0}-a_{1}>0,\ b_{i-1}-a_{i-1}>0$ and

$b_{j-1}-a_{j-1}-a_{j}>0,\ j=2,\cdots,i-1.$

Then one has that

$\xi J_{i}\xi^{T}\geq 0,$ (6.45)

and $\xi J_{i}\xi^{T}=0$ if and only if $\xi=0$. Therefore, $J_{i}$ is
positive definite.

For any $j\in\\{1,\cdots,i\\}$ with $v_{n-j+1}=0$, by (6.41), $\omega_{n-j+1}$
is an eigenvalue of $D_{i}$. Obviously, for any $j\in\\{1,\cdots,i\\}$ with
$v_{n-j+1}\neq 0$, all the eigenvalues of the matrix

$\displaystyle\left(\begin{array}[]{cc}\omega_{n-j+1}&\sqrt{v_{n-j+1}}\\\
\sqrt{v_{n-j+1}}&0\\\ \end{array}\right)$ (6.48)

are also eigenvalues of the matrix $D_{i}$. By direct calculations,

$\lambda_{\pm}=\frac{\omega_{n-j+1}\pm\sqrt{\omega_{n-j+1}^{2}+4v_{n-j+1}}}{2}$

are eigevalues of the matrix (6.48). Hence, $\lambda_{\pm}$ are also
eigevalues of the matrix $D_{i}$ with $\lambda_{\pm}\gtrless 0$. By the
definitions of $i_{+},i_{-}$ and $i_{v}$, we conclude that the matrix $D_{i}$
has $i_{v}+i_{+}$ positive zeros and $i_{v}+i_{-}$ negative zeros.

Since $J_{i}$ is positive definite, $J_{i}$ has the form $B^{2}$, where the
matrix $B$ is symmetric and positive definite. Then the spectral problem

$J_{i}\xi=zD_{i}\xi$ (6.49)

is equivalent to $B^{-1}D_{i}B^{-1}w=z^{-1}w,w=B\xi$. Note that the geometric
multiplicity of each eigenvalue for problem (6.49) is one, this also holds for
the matrix $B^{-1}D_{i}B^{-1}$. Since $B^{-1}D_{i}B^{-1}$ is a symmetric
matrix, the eigenvalues of $B^{-1}D_{i}B^{-1}$ are real and the algebraic
multiplicity of each eigenvalue equals geometric multiplicity, which equals
$1$. Therefore, we conclude that the zeros of $Q_{i}(z)$ are real and simple.

Applying Sylvester’s Law of Inertia, the matrix $B^{-1}D_{i}B^{-1}$ has the
same number of positive and negative eigenvalues as $D_{i}$. Then problem
(6.49) has $i_{v}+i_{+}$ positive zeros and $i_{v}+i_{-}$ negative zeros. ∎

######  Remark 6.3.

(i) If $v_{n-i}=0$, $\omega_{n-i}>0$, then $Q_{i+1}(z)$ and $Q_{i}(z)$ has the
same number of negative zeros, $Q_{i+1}(z)$ has one more positive zero than
$Q_{i}(z)$;
(ii) if $v_{n-i}=0$, $\omega_{n-i}<0$, then $Q_{i+1}(z)$ and $Q_{i}(z)$ has
the same number of positive zeros, and $Q_{i+1}(z)$ has one more negative zero
than $Q_{i}(z)$;
(iii) if $v_{n-i}>0$, then $Q_{i+1}(z)$ has one more positive zero and
negative zero than $Q_{i}(z)$.

######  Lemma 6.4.

For any $i=1,\cdots,n-1,$ one has that

$Q_{i}(z)=\left(e^{\frac{x_{n}}{2}}\prod_{j=1}^{i}a_{j}\right)\phi_{i}(z),$
(6.50)

and

$Q_{n}(z)=\left(e^{\frac{x_{n}}{2}-\frac{x_{1}}{2}}\prod_{j=1}^{n-1}a_{j}\right)W(z),$
(6.51)

where $W(z)$ is defined by (2.9).

###### Proof.

We first show that (6.50) holds. Let

$u_{0}=e^{\frac{x_{n}}{2}}\phi_{0}(z),u_{i}(z)=\left(e^{\frac{x_{n}}{2}}\prod_{j=1}^{i}a_{j}\right)\phi_{i}(z),\
{i=1,\cdots,n-1}.$

Then

$\displaystyle u_{0}(z)$ $\displaystyle=1;$ $\displaystyle u_{1}(z)$
$\displaystyle=b_{0}-\omega_{n}z-v_{n}z^{2};$ $\displaystyle u_{i}(z)$
$\displaystyle=(b_{i-1}-\omega_{n-i+1}z-v_{n-i+1}z^{2})u_{i-1}(z)-a_{i-1}^{2}u_{i-2}(z),\
i=2,\cdots,n-1.$

By (6.42), (6.43) and (6.44), for any $i=1,\cdots,n-1,$ we have
$u_{i}(z)=Q_{i}(z)$.

Now we are going to prove that (6.51) holds. By (2.13) and Lemma 6.2, $W(z)$
and $Q_{n}(z)$ have the same zeros and multiplicities. Then there exists
$\gamma\in\mathbb{R}$, so that

$Q_{n}(z)=\gamma W(z).$ (6.52)

From (6.44) and (6.50),

$\displaystyle Q_{n}(0)$
$\displaystyle=b_{n-1}Q_{n-1}(0)-a_{n-1}^{2}Q_{n-2}(0)$
$\displaystyle=e^{\frac{x_{n}}{2}}(b_{n-1}\phi_{n-1}(0)-a_{n-1}\phi_{n-2}(0))\prod_{j=1}^{n-1}a_{j}.$
(6.53)

Substituting (2.16), (6.7) and (6.8) into (6.53), by some calculations, we
have that

$Q_{n}(0)=e^{\frac{x_{n}}{2}-\frac{x_{1}}{2}}\prod_{j=1}^{n-1}a_{j}.$ (6.54)

Combining (6.54) with (2.11), we can obtain (6.51). ∎

Let $\mathcal{A}$ be the set of pairs $(A,B)$ of real polynomials satisfying
the following conditions:
$\rm{(i)}$ $\rm{deg}(\emph{A}(\emph{z}))+1\leq
deg(\emph{B}(\emph{z}))\leq\rm{deg}(\emph{A}(\emph{z}))+2$;
$\rm{(ii)}$ the zeros of $A(z)$ and $B(z)$ are real and simple, and
$A(0)>0,B(0)>0$;
$\rm{(iii)}$ the largest negative zero $\lambda_{1}^{-}(A)$ and the smallest
positive zero $\lambda_{1}^{+}(A)$ of $A(z)$, the largest negative zero
$\lambda_{1}^{-}(B)$ and the smallest positive zero $\lambda_{1}^{+}(B)$ of
$B(z)$ satisfy

$\lambda_{1}^{-}(A)<\lambda_{1}^{-}(B)<\lambda_{1}^{+}(B)<\lambda_{1}^{+}(A);$

$\rm{(iv)}$ the positive zeros of $A(z)$ and $B(z)$ separate one another; the
negative zeros of $A(z)$ and $B(z)$ separate one another.

######  Lemma 6.5.

For any $i=1,\cdots,n$, we have that

$(Q_{i-1}(z),Q_{i}(z))\in\mathcal{A}.$ (6.55)

###### Proof.

Let us show that for any $i=1,\cdots,n$, the function
$-{Q_{i}(z)}/{zQ_{i-1}(z)}$ is a Herglotz-Nevanlinna function by induction. By
(6.42) and (6.43), we obtain that

$-\frac{Q_{1}(z)}{zQ_{0}(z)}=v_{n}z+\omega_{n}-\frac{b_{0}}{z}.$

Hence, $-Q_{1}(z)/zQ_{0}(z)$ is a Herglotz-Nevanlinna function.

Assume that $-Q_{p-1}(z)/zQ_{p-2}(z)$ is a Herglotz-Nevanlinna function. Then
$zQ_{p-2}(z)/Q_{p-1}(z)$ is also a Herglotz-Nevanlinna function. Let
$\lambda_{0}$ be the zero of $Q_{p-1}(z)$. Then $\lambda_{0}\neq 0$ and

${\rm{Res}}_{z=\lambda_{0}}\frac{zQ_{p-2}(z)}{Q_{p-1}(z)}=\frac{\lambda_{0}Q_{p-2}(\lambda_{0})}{\dot{Q}_{p-1}(\lambda_{0})}<0.$
(6.56)

By (6.44), one obtains that

$-\frac{Q_{p}(z)}{zQ_{p-1}(z)}=v_{n-p+1}z+\omega_{n-p+1}-\frac{b_{p-1}}{z}+\frac{a_{p-1}^{2}Q_{p-2}(z)}{zQ_{p-1}(z)}.$

Combining with (6.56), we have that

${\rm{Res}}_{z=\lambda_{0}}-\frac{Q_{p}(z)}{zQ_{p-1}(z)}=\frac{a_{p-1}^{2}Q_{p-2}(\lambda_{0})}{\lambda_{0}\dot{Q}_{p-1}(\lambda_{0})}<0.$
(6.57)

Note that $z=0$ is also a pole of $-Q_{p}(z)/zQ_{p-1}(z)$. From (2.11), (2.16)
and Lemma 6.4, one has

$\displaystyle Q_{i}(0)>0,i=0,\cdots,n.$ (6.58)

Therefore,

${\rm{Res}}_{z=0}-\frac{Q_{p}(z)}{zQ_{p-1}(z)}=-\frac{Q_{p}(0)}{Q_{p-1}(0)}<0.$
(6.59)

By (6.57), (6.59) and the fact that $v_{n-p+1}\geq 0$, we obtain that
$-Q_{p}(z)/zQ_{p-1}(z)$ is a Herglotz-Nevanlinna function. Hence, for any
$i=1,\cdots,n$, the function $-Q_{i}(z)/zQ_{i-1}(z)$ is a Herglotz-Nevanlinna
function.

By Lemma 6.1, for any $i=1,\cdots,n$, $zQ_{i-1}(z)$ and $Q_{i}(z)$ have no
common zeros. Then the zeros of $zQ_{i-1}(z)$ and $Q_{i}(z)$ are simple and
interlaced. Combining with (6.58), we have
$(Q_{i-1}(z),Q_{i}(z))\in\mathcal{A}$. ∎

By Lemma 6.2, we can denote the zeros of $Q_{i}(z)$ by

$\lambda_{i^{v}+i^{-}}^{-}(Q_{i})<\cdots<\lambda_{1}^{-}(Q_{i})<0<\lambda_{1}^{+}(Q_{i})<\cdots<\lambda_{i^{v}+i^{+}}^{+}(Q_{i}).$

From Lemma 6.5, we know
(i) if
$\lambda_{p_{v}+p_{+}}^{+}(Q_{p})<\lambda_{(p-1)_{v}+(p-1)_{+}}^{+}(Q_{p-1})$
for some $p$, then $Q_{p}(z)$ and $Q_{p-1}(z)$ has the same number of positive
zeros. Namely,

$p_{v}+p_{+}=(p-1)_{v}+(p-1)_{+};\\\ $ (6.60)

(ii) if
$\lambda_{(p-1)_{v}+(p-1)_{+}}^{+}(Q_{p-1})<\lambda_{p_{v}+p_{+}}^{+}(Q_{p})$
for some $p$, then $Q_{p}(z)$ has one more positive zero than $Q_{p-1}(z)$.
Namely,

$p_{v}+p_{+}=(p-1)_{v}+(p-1)_{+}+1.\\\ $ (6.61)

Let $S_{i}(z)$ be the sign-changing number of the sequence $Q_{0}(z),\cdots$,
$Q_{i}(z)$. Then we have the following lemma.

######  Lemma 6.6.

We have that

$S_{i}(z)=\begin{cases}0,\quad&z\in[0,\lambda_{1}^{+}(Q_{i})),\\\
j,\quad&z\in(\lambda_{j}^{+}(Q_{i}),\lambda_{j+1}^{+}(Q_{i})),1\leq j\leq
i_{v}+i_{+}-1,\\\
i_{v}+i_{+},\quad&z\in(\lambda_{i_{v}+i_{+}}^{+}(Q_{i}),+\infty).\end{cases}$
(6.62)

Here if $i_{v}+i_{+}=0$, (6.62) means that $S_{i}(z)=0$ for $z\in[0,\infty)$.

###### Proof.

We can proceed by induction. Obviously, $S_{0}(z),S_{1}(z)$ has the
representation (6.62).

Assume that for $i\leq p-1$, (6.62) holds. We show that for $i=p$, (6.62) also
holds.

For $z\in[0,\lambda_{1}^{+}(Q_{p}))$, by Lemma 6.5, one has

$Q_{i}(z)>0,i=1,\cdots,p.$

Namely, $S_{p}(z)=0$.

For $z\in(\lambda_{j}^{+}(Q_{p}),\lambda_{j+1}^{+}(Q_{p}))$ with $1\leq j\leq
p_{v}+p_{+}-1,$ using (6.58), one has

$\displaystyle{\rm sgn}\ Q_{p}(z)$ $\displaystyle=(-1)^{j}.$ (6.63)

Notice that by Lemma 6.5, we have

$\lambda_{j}^{+}(Q_{p})<\lambda_{j}^{+}(Q_{p-1})<\lambda_{j+1}^{+}(Q_{p}).$

Then we consider the following three cases:
Case 1. $z\in(\lambda_{j}^{+}(Q_{p}),\lambda_{j}^{+}(Q_{p-1}))$. From the
induction hypothesis, we have

$S_{p-1}(z)=j-1$ (6.64)

and hence

$\displaystyle{\rm sgn}\ Q_{p-1}(z)$ $\displaystyle=(-1)^{j-1}.$ (6.65)

By (6.63), (6.64) and (6.65), one obtains that $S_{p}(z)=j$.
Case 2. $z\in(\lambda_{j}^{+}(Q_{p-1}),\lambda_{j+1}^{+}(Q_{p}))$. By the
induction hypothesis, we obtain that $S_{p-1}(z)=j$ and hence

${\rm sgn}\ Q_{p-1}(z)=(-1)^{j}.$

Combining with (6.63), we have that $S_{p}(z)=j$.
Case 3. $z=\lambda_{j}^{+}(Q_{p-1})$. by (6.63), we obtain

${\rm sgn}\ Q_{p}(\lambda_{j}^{+}(Q_{p-1}))=(-1)^{j}.$ (6.66)

Then from (6.44) and the fact that

$Q_{p-1}(\lambda_{j}^{+}(Q_{p-1}))=0,$ (6.67)

one can conclude

${\rm sgn}\ Q_{p-2}(\lambda_{j}^{+}(Q_{p-1}))=(-1)^{j-1}.$ (6.68)

By the continuity of $Q_{p-2}(z)$, there exists $\varepsilon>0$, for any
$z\in(\lambda_{j}^{+}(Q_{p-1})-\varepsilon,\lambda_{j}^{+}(Q_{p-1})+\varepsilon)$,
we have that

${\rm sgn}\ Q_{p-2}(z)=(-1)^{j-1}.$ (6.69)

By the induction hypothesis, for any
$z\in(\lambda_{j}^{+}(Q_{p-1})-\varepsilon,\lambda_{j}^{+}(Q_{p-1})+\varepsilon)$,
one has

$S_{p-2}(z)=C,$

where $C=1$ or $-1$. Note that for any
$z\in(\lambda_{j}^{+}(Q_{p-1})-\varepsilon,\lambda_{j}^{+}(Q_{p-1}))$, by
(6.64), (6.65) and (6.69), we deduce

$S_{p-2}(z)=(-1)^{j-1}$

and hence $C=(-1)^{j-1}.$ Then we can obtain that

$S_{p-2}(\lambda_{j}^{+}(Q_{p-1}))=(-1)^{j-1}.$

Combining with (6.66) and (6.67), we conclude that

$S_{p}(\lambda_{j}^{+}(Q_{p-1}))=(-1)^{j}.$

Hence, for $z\in(\lambda_{j}^{+}(Q_{p}),\lambda_{j+1}^{+}(Q_{p}))$ with $1\leq
j\leq p_{v}+p_{+}-1,$ one has that

$S_{p}(z)=(-1)^{j}.$

For $z\in(\lambda_{p_{v}+p_{+}}^{+}(Q_{p}),+\infty)$, using (6.58), one can
obtain

${\rm sgn}\ Q_{p}(z)=(-1)^{p_{v}+p_{+}}.$ (6.70)

If
$\lambda_{(p-1)_{v}+(p-1)_{+}}^{+}(Q_{p-1})<\lambda_{p_{v}+p_{+}}^{+}(Q_{p})$,
by the induction hypothesis and (6.61), we have that

$S_{p-1}(z)=(p-1)_{v}+(p-1)_{+}=p_{v}+p_{+}-1.$ (6.71)

and hence

${\rm sgn}\ Q_{p-1}(z)=(-1)^{p_{v}+p_{+}-1}.$ (6.72)

By (6.70), (6.71) and (6.72), one can obtain that $S_{p}(z)=p_{v}+p_{+}.$

If
$\lambda_{p_{v}+p_{+}}^{+}(Q_{p})<\lambda_{(p-1)_{v}+(p-1)_{+}}^{+}(Q_{p-1})$,
we split the proof into three cases:
Case 1.
$z\in(\lambda_{{p}_{v}+{p}_{+}}^{+}(Q_{p}),\lambda_{(p-1)_{v}+(p-1)_{+}}^{+}(Q_{p-1}))$.
Due to our induction hypothesis and (6.60), one can obtain that

$S_{p-1}(z)=(p-1)_{v}+(p-1)_{+}-1=p_{v}+p_{+}-1.$ (6.73)

and hence

${\rm sgn}\ Q_{p-1}(z)=(-1)^{p_{v}+p_{+}-1}.$ (6.74)

By (6.70), (6.73) and (6.74) , we have that $S_{p}(z)=p_{v}+p_{+}$.
Case 2. $z\in(\lambda_{(p-1)_{v}+(p-1)_{+}}^{+}(Q_{p-1}),+\infty).$ By our
induction hypothesis and (6.60), we can obtain

$S_{p-1}(z)=(p-1)_{v}+(p-1)_{+}=p_{v}+p_{+}$ (6.75)

and hence

${\rm sgn}\ Q_{p-1}(z)=(-1)^{p_{v}+p_{+}}.$ (6.76)

By (6.70), (6.75) and (6.76), we have $S_{p}(z)=p_{v}+p_{+}$.
Case 3. $z=\lambda_{(p-1)_{v}+(p-1)_{+}}^{+}(Q_{p-1})$. Following the similar
steps in the proof of Case 3 above, we obtain that $S_{p}(z)=p_{v}+p_{+}$.

Therefore, for $z\in(\lambda_{p_{v}+p_{+}}^{+}(Q_{p}),+\infty)$, if
$\lambda_{p_{v}+p_{+}}^{+}(Q_{p})<\lambda_{(p-1)_{v}+(p-1)_{+}}^{+}(Q_{p-1})$,
we can conclude that

$S_{p}(z)=p_{v}+p_{+}.$

As a consequence, for $z\in(\lambda_{p_{v}+p_{+}}^{+}(Q_{p}),+\infty)$, one
has $S_{p}(z)=p_{v}+p_{+}$. ∎

######  Lemma 6.7.

For any $z\in\mathbb{R}$, if $\phi_{+}(z,x_{j})\phi_{+}(z,x_{j+1})>0$, then
$\phi_{+}(z,x)$ has no zero in the interval $[x_{j},x_{j+1}]$. If
$\phi_{+}(z,x_{j})\phi_{+}(z,x_{j+1})\leq 0$, then $\phi_{+}(z,x)$ has exactly
one zero in the interval $[x_{j},x_{j+i}]$.

###### Proof.

We first show that if $\phi_{+}(z,x_{j})\phi_{+}(z,x_{j+1})>0$, then
$\phi_{+}(z,x)$ has no zero in the interval $[x_{j},x_{j+1}]$. Otherwise, let
$y_{1}\in[x_{j},x_{j+i}]$ be the zero of $\phi_{+}(z,x)$. Note that for
$x\in[x_{j},x_{j+1}]$, there exists $A,B\in\mathbb{R}$, so that

$\phi_{+}(z,x)=Ae^{\frac{x}{2}}+Be^{-\frac{x}{2}}.$ (6.77)

Then we have $Ae^{\frac{y_{1}}{2}}+Be^{-\frac{y_{1}}{2}}=0$. That is,
$B=-Ae^{y_{1}}$. Substituting this into (6.77) and letting $x=x_{j},x_{j+1}$,
we have that

$\displaystyle\phi_{+}(z,x_{j})=Ae^{\frac{x_{j}}{2}}-Ae^{y_{1}}e^{-\frac{x_{j}}{2}},\
\phi_{+}(z,x_{j+1})=Ae^{\frac{x_{j+1}}{2}}-Ae^{y_{1}}e^{-\frac{x_{j+1}}{2}}.$

Therefore,

$\displaystyle\phi_{+}(z,x_{j})\phi_{+}(z,x_{j+1})=A^{2}(e^{\frac{x_{j}}{2}}-e^{y_{1}-\frac{x_{j}}{2}})(e^{\frac{x_{j+1}}{2}}-e^{y_{1}-\frac{x_{j+1}}{2}})\leq
0.$

This is in contradiction with the assumption that
$\phi_{+}(z,x_{j})\phi_{+}(z,x_{j+1})>0$.

If $\phi_{+}(z,x_{j})\phi_{+}(z,x_{j+1})\leq 0$, then $\phi_{+}(z,x)$ has at
least one zero in the interval $[x_{j},x_{j+1}]$. We show $\phi_{+}(z,x)$ has
exactly one zero in the interval $[x_{j},x_{j+1}]$. Otherwise, let
$y_{1},y_{2}\in[x_{j},x_{j+1}]$ be zeros of $\phi_{+}(z,x)$. Hence by (6.77),
one has

$\displaystyle Ae^{\frac{y_{1}}{2}}+Be^{-\frac{y_{1}}{2}}=0,\
Ae^{\frac{y_{2}}{2}}+Be^{-\frac{y_{2}}{2}}=0.$

Then one obtains that $A=B=0$. By (2.5) and (2.6), we have $\phi_{+}(z,x)=0$
for all $x\in\mathbb{R}$. This contradicts (2.8). ∎

######  Theorem 6.8.

$H$ has $n+n_{v}$ real and simple eigenvalues:

$\lambda_{n_{v}+n_{-}}^{-}<\cdots<\lambda_{1}^{-}<0<\lambda_{1}^{+}<\cdots<\lambda_{n_{v}+n_{+}}^{+}.$

The eigenfunction of the eigenvalue $\lambda_{i}^{\pm}$ has exactly $i-1$
zeros.

###### Proof.

From Lemma 6.2 and Lemma 6.5, we obtain that $H$ has $n+n_{v}$ real and simple
eigenvalues.

If $n_{v}+n_{+}=(n-1)_{v}+(n-1)_{+}$. By (6.51) and Lemma 6.5, one has

$0<\lambda_{1}<\lambda_{1}^{+}(Q_{n-1})<\cdots<\lambda_{n_{v}+n_{+}-1}^{+}(Q_{n-1})<\lambda_{n_{v}+n_{+}}^{+}<\lambda_{n_{v}+n_{+}}^{+}(Q_{n-1}).$

If $n_{v}+n_{+}=(n-1)_{v}+(n-1)_{+}+1$. Then we have

$0<\lambda_{1}<\lambda_{1}^{+}(Q_{n-1})<\cdots<\lambda_{n_{v}+n_{+}-1}^{+}<\lambda_{n_{v}+n_{+}-1}^{+}(Q_{n-1})<\lambda_{n_{v}+n_{+}}^{+}<+\infty.$

Applying Lemma 6.6, for both cases, we have that
$S_{n-1}(\lambda_{i}^{+})=i-1$. Then by Lemma 6.4 and (6.6), we know that
$\phi_{+}(\lambda_{i}^{+},x_{1}),\cdots,\phi_{+}(\lambda_{i}^{+},x_{n})$
change signs for $i-1$ times. By Lemma 6.7, $\phi_{+}(\lambda_{i}^{+},x)$ has
$i-1$ zeros in the interval $(x_{1},x_{n})$. Note that
$\phi_{+}(\lambda_{i}^{+},x)$ has no zero in the interval $(-\infty,x_{1}]$
and $[x_{n},+\infty)$. Therefore, the eigenfunction of the eigenvalue
$\lambda_{i}^{+}$ has exactly $i-1$ zeros.

By a similar argument, the eigenfunction of the eigenvalue $\lambda_{i}^{-}$
also has exactly $i-1$ zeros. ∎

######  Remark 6.9.

Combining the fact that $S_{n-1}(\lambda_{1}^{+})=0$ with (6.42) and (6.50),
for any $x\in\mathbb{R}$, we have that

$\phi_{+}(\lambda_{1}^{-},x),\phi_{+}(\lambda_{1}^{+},x)>0.$

######  Remark 6.10.

Let us mention that if $n_{v}+n_{-}=0$, then $n_{v}+n_{+}=n$. $H$ has $n$
positive and simple eigenvalues:

$0<\lambda_{1}^{+}<\cdots<\lambda_{n}^{+}.$

The eigenfunction of the eigenvalues $\lambda_{i}^{+}$ has exactly $i-1$
zeros.

By Lemma 4.1 and Theorem 6.8, we have the following corollary.

######  Corollary 6.11.

Assume that $n+n_{v}\geq 3$. Then for any $x\in\mathbb{R}$, there exist
$\lambda_{i},\lambda_{j}\in\sigma(H)$, so that the corresponding
eigenfunctions $\varphi_{i}$ and $\varphi_{j}$ satisfy

$\varphi_{i}(x)\neq 0,\varphi_{j}(x)\neq 0.$

Acknowledgement. The authors are indebted to Guangsheng Wei for bringing this
problem to their attention and for stimulating discussions. The authors would
like to thank Dmitry Pelinovsky and Yiteng Hu for their valuable comments and
additional references.

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