#  On spectral and orbital stability for the Klein–Gordon equation coupled to
an anharmonic oscillator

Andrew Comech
Texas A&M University, College Station, TX
Institute for Information Transmission Problems, Moscow, Russia

Elena A. Kopylova
Institute for Information Transmission Problems, Moscow, Russia
Vienna University, Vienna, Austria

(December 8, 2020)

###### Abstract

We obtain explicit characterization of spectral and orbital stability of
solitary wave solutions to the $\mathbf{U}(1)$-invariant Klein–Gordon equation
in one spatial dimension coupled to an anharmonic oscillator. We also give the
complete analysis of the spectrum of the linearization at a solitary wave.

##  1 Introduction

In the present article we study the spectral stability of the
$\mathbf{U}(1)$-invariant Klein–Gordon equation on a line with a concentrated
nonlinearity:

$\ddot{\psi}(x,t)=\partial_{x}^{2}\psi(x,t)-m^{2}\psi(x,t)+\delta(x)a(|\psi(0,t)|^{2})\psi(0,t),\qquad\psi(x,t)\in{\mathbb{C}},\quad
x\in{\mathbb{R}},$ (1.1)

where $m>0$. Above, $a(\cdot)$ is a real-valued differentiable function, so
that the model is $\mathbf{U}(1)$-invariant. The equation is understood in the
sense of distributions. Physically, equation (1.1) describes the Klein–Gordon
field coupled to a nonlinear oscillator located at $x=0$, with
$a(|\psi|^{2})\psi$ being the oscillator force. Equation (1.1) admits finite
energy solutions of the form $\phi(x,\omega)e^{-\mathrm{i}\omega t}$,
$\omega\in{\mathbb{R}}$, called _solitary waves_. The solitary waves form a
two-dimensional _solitary manifold_ in the Hilbert space of finite energy
states of the system.

We will determine conditions for orbital and spectral stability of solitary
waves, formulating the results in terms of values of $a$ and $a^{\prime}$
evaluated at $|\phi(0,\omega)|^{2}$; see Theorem 3.6 below. Then, in Theorem
4.5, we give the complete description of the spectrum of the linearization at
a solitary wave.

Equation (1.1) was proposed to model electron’s transitions between Bohr’s
_quantum orbits_ , which is one of the fundamental problems of Quantum
Mechanics. More precisely, (1.1) models the interaction of the electron
(represented by the anharmonic oscillator) with the electromagnetic radiation
(represented by the Klein–Gordon field). The global attraction of any finite
energy solution to the set of all solitary waves in local energy norms, which
was established in [KK07], can be interpreted as the relaxation of a perturbed
electron to the Bohr orbit, where it no longer loses the energy via the
radiation, in the complete agreement with Bohr’s postulate on quantum jumps.

We pursue further properties of the model (1.1) since it is a convenient
playground for establishing asymptotic stability results, similarly to
[BKKS08, KKS12], allowing one to explicitly check all the spectral properties
of the linearized equation. In the present article, we will obtain the
spectral and orbital stability results. Moreover, we obtain the complete
description of the spectrum of the linearized equation: these results will be
needed for the subsequent proof of asymptotic stability. We point out that
many pieces of the spectral analysis which we develop in the present article
can be carried over essentially verbatim to models similar to (1.1), such as
the models where the concentrated nonlinearity is substituted by its
regularized versions, such as the self-interaction based on the mean field
[KK09].

We mention that the local and global well-posedness of (1.1) has already been
proved in [KK07]. In [Kop09, Kop10], asymptotic stability of solitary waves
was obtained for discrete Schrödinger and Klein-Gordon equations. Let us also
mention that related results on local well-posedness, orbital stability, and
linear instability of solitary waves in the nonlinear Klein–Gordon equation in
the external $\delta$-function potential were obtained in [CGOR19].

The paper is organized as follows. The model and its solitary wave solutions
are described in Section 2. The linearization at a solitary wave is carried
out in Section 3, where the standard properties of the linearized operator are
obtained. The detailed structure of the spectrum of the linearized operator is
derived in Section 4 (see Theorem 4.5).

##  2 The model

We define $\varPsi(x,t)=\begin{bmatrix}\psi(x,t)\\\
\pi(x,t)\end{bmatrix}\in{\mathbb{C}}^{2}$ and rewrite (1.1) in the vector
form:

$\dot{\varPsi}(t)=\begin{bmatrix}0&1\\\
\partial_{x}^{2}-m^{2}&0\end{bmatrix}\varPsi(t)+\delta(x)\begin{bmatrix}0\\\
a(|\psi|^{2})\psi\end{bmatrix},\qquad\varPsi=\begin{bmatrix}\psi\\\
\pi\end{bmatrix}\in{\mathbb{C}}^{2},\qquad\varPsi\big{|}_{t=0}=\varPsi_{0}:=\begin{bmatrix}\psi_{0}\\\
\pi_{0}\end{bmatrix}.$ (2.1)

The oscillator force $a(|\psi|^{2})\psi$ admits a real-valued potential,

$a(|\psi|^{2})\psi=-\nabla_{\,\mathrm{Re}\,\psi,\,\mathrm{Im}\,\psi}U(\psi),\qquad\psi\in{\mathbb{R}}^{2},\qquad
U\in C^{2}({\mathbb{R}}^{2}),$ (2.2)

where $U(\psi)=u(|\psi|^{2})$, with
$u(\tau)=\frac{1}{2}\int_{0}^{\tau}a(s)\,ds$. Then (2.1) is formally a
Hamiltonian system with the Hamiltonian functional

${{\cal
H}}(\varPsi)=\frac{1}{2}\int_{{\mathbb{R}}}\Big{(}|\pi|^{2}+|\partial_{x}\psi|^{2}+m^{2}|\psi|^{2}\Big{)}\,dx+U(\psi(0)),\qquad\varPsi=\begin{bmatrix}\psi(x)\\\
\pi(x)\end{bmatrix}\in H^{1}({\mathbb{R}})\times L^{2}({\mathbb{R}}),$ (2.3)

which is conserved for sufficiently regular finite energy solutions. Equation
(1.1) is $\mathbf{U}(1)$-invariant: if $\psi(x,t)$ is a solution, then so is
$e^{\mathrm{i}\theta}\psi(x,t)$ for any $\theta\in{\mathbb{R}}$. The Nöther
theorem implies the _charge_ conservation: the value of the functional

$\mathcal{Q}(\varPsi)=-\frac{1}{2}\,\mathrm{Im}\,\int_{{\mathbb{R}}}\big{(}\overline{\psi}(x)\pi(x)-\overline{\pi}(x)\psi(x)\big{)}\,dx,\qquad\varPsi=\begin{bmatrix}\psi(x)\\\
\pi(x)\end{bmatrix},$ (2.4)

is conserved for solutions to (1.1).

The local and global existence result for the Cauchy problem (2.1) proved in
[KK07, Theorem 2.1]:

######  Theorem 2.1.

Assume that the potential is represented by $U(\psi)=u(|\psi|^{2})$ with $u\in
C^{2}({\mathbb{R}})$ and that $U$ satisfies the inequality

$U(\psi)\geq
A-B|\psi|^{2}\quad\mbox{for}\quad\psi\in{\mathbb{C}},\qquad\mbox{where}\quad
A\in{\mathbb{R}},\quad 0\leq B<m.$

Then:

  1. 1.

For every $\varPsi_{0}\in{\mathcal{E}}:=H^{1}({\mathbb{R}})\oplus
L^{2}({\mathbb{R}})$ the Cauchy problem (2.1) has a unique solution
$\varPsi\in C_{b}({\mathbb{R}},{\mathcal{E}})$.

  2. 2.

The energy of $\varPsi$ is conserved: for all $t\in{\mathbb{R}}$, ${\cal
H}(\varPsi(t))={\cal H}(\varPsi(0))$.

  3. 3.

There exists $\varLambda(\varPsi_{0})>0$ such that
$\sup\limits_{t\in{\mathbb{R}}}\|{\varPsi(t)}\|_{\mathcal{E}}\leq\varLambda(\varPsi_{0})<\infty$.

The main subject of this paper is the analysis of the spectral and orbital
stability of “quantum stationary states”, or _solitary waves_ [GSS87], which
are finite energy solutions of the form

$\varPsi(x,t)=\varPhi(x)e^{-\mathrm{i}\omega
t},\qquad\omega\in{\mathbb{R}},\qquad\varPhi(x)=\begin{bmatrix}\phi(x)\\\
-\mathrm{i}\omega\phi(x)\end{bmatrix},\qquad\phi\in H^{1}({\mathbb{R}}).$
(2.5)

Above, $H^{1}({\mathbb{R}})$ is the Sobolev space of complex-valued measurable
functions satisfying
$\int_{{\mathbb{R}}}(|\partial_{x}\psi|^{2}+|\psi|^{2}\bigr{)}\,dx<\infty$. By
(2.1), the frequency $\omega\in{\mathbb{R}}$ and the amplitude $\phi(x)$ solve
the following _nonlinear eigenvalue problem_ :

$\omega^{2}\phi(x)=-\partial_{x}^{2}\phi(x)+m^{2}\phi(x)-\delta(x)a(|\phi(0)|^{2})\phi(0),\qquad
x\in{\mathbb{R}}.$ (2.6)

######  Definition 2.2.

The solitary manifold ${\cal S}$ is the set of all amplitudes
$\varPhi(\cdot,\omega)=\begin{bmatrix}\phi(\cdot,\omega)\\\
-\mathrm{i}\omega\phi(\cdot,\omega)\end{bmatrix}$, where
$\phi(x,\omega)e^{-\mathrm{i}\omega t}$ is a nonzero solitary wave solution to
(1.1), with all possible $\omega\in{\mathbb{R}}$.

######  Lemma 2.3.

The set ${\cal S}$ is given by

${\cal
S}=\Bigl{\\{}\varPhi(x,\omega)e^{\mathrm{i}\theta}=\begin{bmatrix}\phi(x,\omega)\\\
-\mathrm{i}\omega\phi(x,\omega)\end{bmatrix}e^{\mathrm{i}\theta},\qquad\phi(x,\omega)=Ce^{-\varkappa|x|},\qquad\theta\in[0,2\pi]\Bigr{\\}},$

where $|\omega|<m$, $\varkappa=\sqrt{m^{2}-\omega^{2}}>0$, and $C>0$ satisfies
the relation

$\displaystyle a(C^{2})=2\varkappa.$ (2.7)

If $a^{\prime}(C^{2})\neq 0$, then $C$ is locally a $C^{1}$-function of
$\omega$.

###### Proof.

Equation (2.6) implies that
$\partial_{x}^{2}\phi(x,\omega)=(m^{2}-\omega^{2})\phi(x,\omega)$, $x\neq 0$,
and hence $\phi(x,\omega)=C_{\pm}e^{-\varkappa_{\pm}|x|}$ for $\pm x>0$, where
$\varkappa_{\pm}$ satisfy $\varkappa_{\pm}^{2}=m^{2}-\omega^{2}$. Since
$\phi(\cdot,\omega)\in H^{1}({\mathbb{R}})$, we conclude that
$\varkappa_{\pm}>0$, $|\omega|<m$, and that
$\varkappa_{\pm}=\sqrt{m^{2}-\omega^{2}}>0$. Since the function
$\phi(x,\omega)$ is continuous at $x=0$, one has $C_{-}=C_{+}=C$. Thus, the
solitary waves are solutions of the form

$\phi(x,\omega)=Ce^{-\varkappa|x|},\qquad\varkappa=\sqrt{m^{2}-\omega^{2}}>0,\qquad
C\not=0.$ (2.8)

For our convenience, we assume that $C>0$, and write solitary wave solutions
in the form

$\phi(x,\omega)=e^{\mathrm{i}\theta}Ce^{-\varkappa|x|},\qquad\varkappa=\sqrt{m^{2}-\omega^{2}}>0,\qquad
C>0,\quad\theta\in{\mathbb{R}}\mod 2\pi.$ (2.9)

The algebraic equation satisfied by the constant $C>0$ is obtained by
collecting coefficients at $\delta(x)$ in (2.6):

$0=\partial_{x}\phi(0+,\omega)-\partial_{x}\phi(0-,\omega)+a(|\phi(0,\omega)|^{2})\phi(0,\omega).$
(2.10)

This implies that $0=-2\varkappa C+a(C^{2})C$, hence $2\varkappa=a(C^{2})$.

We note that there is the following relation:

$\displaystyle-2\omega=\frac{d}{d\omega}\varkappa^{2}=\frac{d}{d\omega}\frac{a(C^{2})^{2}}{4}=a(C^{2})a^{\prime}(C^{2})C\frac{dC}{d\omega},$
(2.11)

which shows that $C$ is locally a $C^{1}$-function of $\omega$ as long as
$a^{\prime}(C^{2})\neq 0$. ∎

##  3 Linearization at a solitary wave

Let us linearize the nonlinear Klein–Gordon (2.1) at a solitary wave

$e^{-\mathrm{i}\omega
t+\mathrm{i}\theta}\varPhi(x,\omega)=e^{-\mathrm{i}\omega
t+\mathrm{i}\theta}\begin{bmatrix}\phi(x,\omega)\\\
-\mathrm{i}\omega\phi(x,\omega)\end{bmatrix},$

where $\omega\in(-m,m)$, $\theta\in{\mathbb{R}}\mod 2\pi$, and
$\phi(x,\omega)=C(\omega)e^{-\varkappa|x|}$, with
$\varkappa=\sqrt{m^{2}-\omega^{2}}>0$ and $C(\omega)>0$ (see Lemma 2.3).
Substituting

$\varPsi(x,t)=e^{-\mathrm{i}\omega
t+\mathrm{i}\theta}(\varPhi(x,\omega)+\mathcal{X}(x,t)),\qquad\mathcal{X}(x,t)=\begin{bmatrix}\mathcal{X}_{1}(x,t)\\\
\mathcal{X}_{2}(x,t)\end{bmatrix}\in{\mathbb{C}}^{2},$ (3.1)

into (2.1), we obtain:

$-\mathrm{i}\omega(\varPhi+\mathcal{X})+\dot{\mathcal{X}}=\begin{bmatrix}0&1\\\
\partial_{x}^{2}-m^{2}&0\end{bmatrix}(\varPhi+\mathcal{X})+\delta(x)\begin{bmatrix}0\\\
a(|C+\mathcal{X}_{1}|^{2})(C+\mathcal{X}_{1})\end{bmatrix}.$

Equality $\varPhi_{2}=-\mathrm{i}\omega\varPhi_{1}$ and equation (2.6) lead to

$\dot{\mathcal{X}}=\begin{bmatrix}\mathrm{i}\omega&1\\\
\partial_{x}^{2}-m^{2}&\mathrm{i}\omega\end{bmatrix}\mathcal{X}(x,t)+\delta(x)\begin{bmatrix}0\\\
a(|C+\mathcal{X}_{1}|^{2})(C+\mathcal{X}_{1})-a(|C|^{2})C\end{bmatrix}.$ (3.2)

The first order part of (3.2) is given by

$\dot{\mathcal{X}}(x,t)=\begin{bmatrix}\mathrm{i}\omega&1\\\
\partial_{x}^{2}-m^{2}&\mathrm{i}\omega\end{bmatrix}\mathcal{X}(x,t)+\delta(x)\begin{bmatrix}0\\\
a(C^{2})\mathcal{X}_{1}(0,t)+C^{2}a^{\prime}(C^{2})2\,\mathrm{Re}\,\mathcal{X}_{1}(0,t)\end{bmatrix}.$
(3.3)

In the case $a^{\prime}(C^{2})\not=0$, the operator in the right-hand side is
${\mathbb{R}}$-linear but not ${\mathbb{C}}$-linear; to study its spectrum, it
is convenient to rewrite (3.3) in the real form. For
$\varPsi=\begin{bmatrix}\psi\\\ \dot{\psi}\end{bmatrix}\in{\mathbb{C}}^{2}$,
we denote $\Psi=\begin{bmatrix}\,\mathrm{Re}\,\psi\\\ \,\mathrm{Im}\,\psi\\\
\,\mathrm{Re}\,\partial_{t}\psi\\\
\,\mathrm{Im}\,\partial_{t}\psi\end{bmatrix}\in{\mathbb{R}}^{4}$. The solitary
wave $\varPhi(x,\omega)e^{-\mathrm{i}\omega t}\in{\mathbb{C}}^{2}$ is then
represented by $e^{\mathbf{J}\omega t}\Phi(x,\omega)\in{\mathbb{R}}^{4}$,
where

$\Phi(x,\omega)=\begin{bmatrix}\phi(x,\omega)\\\ 0\\\ 0\\\
-\omega\phi(x,\omega)\end{bmatrix},\qquad\mathbf{J}=\begin{bmatrix}0&1&0&0\\\
-1&0&0&0\\\ 0&0&0&1\\\ 0&0&-1&0\end{bmatrix}.$

We identify the perturbation $\mathcal{X}\in{\mathbb{C}}^{2}$ with the real
vector $X=\begin{bmatrix}\,\mathrm{Re}\,\mathcal{X}_{1}\\\
\,\mathrm{Im}\,\mathcal{X}_{1}\\\ \,\mathrm{Re}\,\mathcal{X}_{2}\\\
\,\mathrm{Im}\,\mathcal{X}_{2}\end{bmatrix}\in{\mathbb{R}}^{4}$. Then (3.3)
becomes

$\dot{X}(x,t)=\begin{bmatrix}0&-\omega&1&0\\\ \omega&0&0&1\\\
\partial_{x}^{2}-m^{2}&0&0&-\omega\\\
0&\partial_{x}^{2}-m^{2}&\omega&0\end{bmatrix}X(x,t)+\delta(x)\begin{bmatrix}0\\\
0\\\ \alpha(1+2\kappa)X_{1}(0,t)\\\ \alpha X_{2}(0,t)\end{bmatrix},$ (3.4)

where (cf. (2.7))

$\alpha:=a(C^{2})=2\varkappa>0,\qquad\kappa:={C^{2}a^{\prime}(C^{2})}/{a(C^{2})}.$
(3.5)

This gives a system which is ${\mathbb{C}}$-linear.

######  Remark 3.1.

We note that the above definition of $\kappa$ is compatible with the pure
power case $a(\tau)=\tau^{\kappa}$, $\kappa>0$, $\tau\geq 0$, when
$a^{\prime}(|\psi|^{2})|\psi|^{2}=\kappa a(|\psi|^{2})$.

Let us denote

$\begin{array}[]{l}L_{\kappa}(\omega)=-\partial_{x}^{2}+m^{2}-\alpha(1+2\kappa)\delta(x)-\omega^{2}.\end{array}$
(3.6)

######  Remark 3.2.

The domains of all operators which we consider require a careful definition.
For example, $A=-\partial_{x}^{2}+c\delta(x)$, $c\in{\mathbb{R}}$, is defined
as $A:\,L^{2}({\mathbb{R}})\to L^{2}({\mathbb{R}})$ with the domain

$\mathfrak{D}(A)=\\{\psi\in H^{2}({\mathbb{R}}_{-},{\mathbb{C}})\cup
H^{2}({\mathbb{R}}_{+},{\mathbb{C}}),\qquad\partial_{x}\psi(0+)-\partial_{x}\psi(0-)=c\psi(0)\\},$

where it is selfadjoint.

######  Lemma 3.3 (Spectrum of $L_{\kappa}$).

Let $\omega\in(-m,m)$, $\kappa\in{\mathbb{R}}$. The operator
$L_{\kappa}(\omega)$ is selfadjoint and satisfies

$\displaystyle\sigma_{\mathrm{ess}}(L_{\kappa}(\omega))=[\varkappa^{2},+\infty),$
$\displaystyle\sigma_{\mathrm{p}}(L_{\kappa}(\omega))=\begin{cases}\emptyset,&\kappa\leq-1/2,\\\
\Lambda_{\kappa}(\omega):=-4\varkappa^{2}(\kappa+\kappa^{2}),&\kappa>-1/2,\end{cases}$

where $\varkappa=\sqrt{m^{2}-\omega^{2}}$. The eigenvalue
$\Lambda_{\kappa}(\omega)$ is simple.

###### Proof.

Solving the equation

$(-\partial_{x}^{2}-\alpha(1+2\kappa)\delta(x)+\varkappa^{2})\psi=\Lambda\psi,\qquad\psi\in
L^{2}({\mathbb{R}}),$

we find that $\Lambda$ has to satisfy $\Lambda<\varkappa^{2}$ and that

$\displaystyle\psi(x)=Ce^{-|x|\sqrt{\varkappa^{2}-\Lambda}},\qquad
x\in{\mathbb{R}},\quad C\neq 0,$ (3.7)

while $\Lambda$ is obtained from the jump condition at $x=0$:

$\alpha(1+2\kappa)=2\sqrt{\varkappa^{2}-\Lambda}.$

This shows that there could only be an eigenvalue if $\kappa>-1/2$. Taking
into account that $\alpha=2\varkappa$, one arrives at
$\varkappa^{2}(1+2\kappa)^{2}=\varkappa^{2}-\Lambda$, so
$\Lambda=-4\varkappa^{2}(\kappa+\kappa^{2})$. In particular, setting
$\kappa=0$, one obtains the point spectrum $\\{0\\}$ of $L_{0}$; by (2.6), the
corresponding eigenvector is $\phi$. ∎

In terms of operators (3.6), the system (3.4) reads as

$\dot{X}(x,t)=\mathbf{A}(\omega,\kappa)X(x,t),\qquad\mathbf{A}(\omega,\kappa):=\begin{bmatrix}0&-\omega&1&0\\\
\omega&0&0&1\\\ -L_{\kappa}(\omega)-\omega^{2}&0&0&-\omega\\\
0&-L_{0}(\omega)-\omega^{2}&\omega&0\end{bmatrix}.$ (3.8)

Theorem 2.1 generalizes to equation (3.8): for every initial function
$X(x,0)=X_{0}\in\mathcal{E}$, the equation admits a unique solution $X(x,t)\in
C_{b}({\mathbb{R}},\mathcal{E})$. Denote

$\bm{\Sigma}=\begin{bmatrix}0&I_{2}\\\
-I_{2}&0\end{bmatrix},\qquad\mathbf{H}(\omega,\kappa)=\begin{bmatrix}L_{\kappa}+\omega^{2}&0&0&\omega\\\
0&L_{0}+\omega^{2}&-\omega&0\\\ 0&-\omega&1&0\\\ \omega&0&0&1\end{bmatrix};$

then the operator $\mathbf{A}(\omega,\kappa)$ from (3.8) corresponding to a
linearization at the solitary wave is factored into

$\mathbf{A}(\omega,\kappa)=\bm{\Sigma}\mathbf{H}(\omega,\kappa).$

Denote

$\displaystyle\mathbf{G}_{1}=\mathbf{G}_{1}^{-1}=\begin{bmatrix}1&0&0&0\\\
0&0&0&1\\\ 0&0&1&0\\\
0&1&0&0\end{bmatrix},\qquad\mathbf{G}_{2}=\mathbf{G}_{2}^{-1}=\begin{bmatrix}I_{2}&0\\\
0&\sigma_{2}\end{bmatrix}$ (3.9)

and consider the conjugated version of $\mathbf{H}(\omega,\kappa)$:

$\displaystyle\tilde{\mathbf{H}}(\omega,\kappa)=\mathbf{G}_{2}\mathbf{G}_{1}\mathbf{H}(\omega,\kappa)\mathbf{G}_{1}^{-1}\mathbf{G}_{2}^{-1}=\begin{bmatrix}L_{\kappa}+\omega^{2}&\omega&0&0\\\
\omega&1&0&0\\\ 0&0&L_{0}+\omega^{2}&\omega\\\ 0&0&\omega&1\\\
\end{bmatrix}=:\begin{bmatrix}H_{\kappa}(\omega)&0\\\
0&H_{0}(\omega)\end{bmatrix}.$ (3.10)

We start with the spectrum of

$\displaystyle
H_{\kappa}(\omega)=\begin{bmatrix}L_{\kappa}+\omega^{2}&\omega\\\
\omega&1\end{bmatrix}.$ (3.11)

######  Lemma 3.4.

For any $\kappa\in{\mathbb{R}}$, the essential spectrum of
$H_{\kappa}(\omega)$ is given by

  1. 1.

$\sigma_{\mathrm{ess}}(H_{\kappa}(\omega))=\begin{cases}[m^{2},+\infty),&\omega=0,\\\
[c^{-}(\omega),1]\cup[c^{+}(\omega),+\infty),&\omega\in(-m,m)\setminus\\{0\\},\end{cases}$

where

$\displaystyle
c^{\pm}(\omega)=\frac{m^{2}+1\pm\sqrt{(m^{2}-1)^{2}+4\omega^{2}}}{2}.$

We note that $0<c^{-}(\omega)\leq\min\\{1,m^{2}\\}$,
$c^{+}(\omega)\geq\max\\{1,m^{2}\\}$.

  2. 2.

$\sigma_{\mathrm{p}}(H_{\kappa}(\omega))=\begin{cases}\emptyset,&\kappa\leq-1/2,\quad\omega\in(-m,m)\setminus\\{0\\},\\\
\\{1\\},&\kappa\leq-1/2,\quad\omega=0,\\\
\\{\lambda^{\pm}_{\kappa}(\omega)\\},&\kappa>-1/2,\quad\omega\in(-m,m),\end{cases}$

with the eigenvalues $\lambda^{\pm}_{\kappa}(\omega)$ given by

$\displaystyle\lambda^{\pm}_{\kappa}(\omega)={\Big{(}\Lambda_{\kappa}+\omega^{2}+1\pm\sqrt{(\Lambda_{\kappa}+\omega^{2}+1)^{2}-4\Lambda_{\kappa}}\Big{)}}/{2},$
(3.12)

with $\Lambda_{\kappa}(\omega)=-4\varkappa^{2}(\kappa+\kappa^{2})$ (when
$\kappa>-1/2$) from Lemma 3.3. In particular,
$0\in\sigma_{\mathrm{p}}(H_{\kappa}(\omega))$ if and only if $\kappa=0$, and

$\sigma_{\mathrm{p}}(H_{0}(\omega))=\\{0\\}\cup\\{\omega^{2}+1\\},$

with the corresponding eigenvectors

$\psi_{0}(x)=\begin{bmatrix}\phi(x)\\\
-\omega\phi(x)\end{bmatrix},\qquad\psi_{\omega^{2}+1}(x)=\begin{bmatrix}\omega\phi(x)\\\
\phi(x)\end{bmatrix}.$

###### Proof.

Let us compute the essential spectrum of $H_{\kappa}(\omega)$. Since
$\delta(x)$ is a relatively compact perturbation of the Laplacian in one
dimension, by the Weyl theorem, when computing the essential spectrum of
$H_{\kappa}(\omega)$, we can replace $L_{\kappa}$ in (3.11) by
$L_{\kappa}^{(0)}=-\partial_{x}^{2}+m^{2}-\omega^{2}$. Then the values of the
essential spectrum are those $\lambda\in{\mathbb{C}}$ such that

$\displaystyle\det\begin{bmatrix}\xi^{2}+\omega^{2}-\lambda&\omega\\\
\omega&1-\lambda\end{bmatrix}=(1-\lambda)\Big{(}\xi^{2}+m^{2}-\lambda-\frac{\omega^{2}}{1-\lambda}\Big{)}=0\qquad\mbox{for
some $\xi\in{\mathbb{R}}$.}$ (3.13)

Thus, either $\lambda=1$, or, expressing $\xi^{2}$ in terms of $\lambda$, we
arrive at the inequality

$\displaystyle-m^{2}+\lambda+\frac{\omega^{2}}{1-\lambda}=\frac{(\lambda-m^{2})(1-\lambda)+\omega^{2}}{1-\lambda}=\frac{\lambda^{2}-\lambda(m^{2}+1)+m^{2}-\omega^{2}}{\lambda-1}\geq
0.$ (3.14)

There are two roots $\lambda=c^{\pm}(\omega)$ of the numerator and root
$\lambda=1$ of the denominator; these roots satisfy the inequalities

$0<c^{-}(\omega)<1<c^{+}(\omega),\qquad\omega\in(-m,m)\setminus\\{0\\}.$

(We note that when $\omega=0$, one has $c^{-}(0)=1$ if $m^{2}\geq 1$ and
$c^{+}(0)=1$ if $m^{2}\leq 1$.) We conclude that the essential spectrum (the
set where (3.13) is satisfied) is given by the intervals

$\lambda\in[c^{-}(\omega),1]\cup[c^{+}(\omega),+\infty).$

Let us now study the point spectrum of $H_{\kappa}(\omega)$. First let us
consider when $\lambda=1$ is an eigenvalue of $H_{\kappa}(\omega)$. One can
see from (3.11) that this is only possible when $\omega=0$ (with
eigenfunctions of the form $\begin{bmatrix}0\\\ u_{2}(x)\end{bmatrix}$,
$u_{2}\in L^{2}({\mathbb{R}})$); in this case, the rest of the point spectrum
comes from Lemma 3.3 (($\Lambda_{\kappa}(0)=-4m^{2}(\kappa+\kappa^{2})$ in the
case $\kappa>-1/2$), agreeing with $\lambda^{-}_{\kappa}(0)=1$ and
$\lambda^{+}_{\kappa}(0)=\Lambda_{\kappa}(0)$ from (3.12).

Now we consider the case $\lambda\neq 1$. The operator

$\displaystyle H_{\kappa}(\omega)-\lambda
I_{2}=\begin{bmatrix}L_{\kappa}+\omega^{2}-\lambda&\omega\\\
\omega&1-\lambda\end{bmatrix}$ (3.15)

has zero eigenvalue if and only if so does the Schur complement of
$1-\lambda$, which is given by

$L_{\kappa}(\omega)+\omega^{2}-\lambda-\frac{\omega^{2}}{1-\lambda}=L_{\kappa}(\omega)-\lambda-\frac{\lambda\omega^{2}}{1-\lambda}.$

Thus, $\lambda\neq 1$ is an eigenvalue of $H_{\kappa}(\omega)$ if and only if

$\displaystyle\Lambda=\lambda+\frac{\lambda\omega^{2}}{1-\lambda}$ (3.16)

is an eigenvalue of $L_{\kappa}(\omega)$; see Figure 1. Now the proof follows
from Lemma 3.3, leading to the expressions (3.12). Let us point out that the
eigenfunctions corresponding to $\lambda\neq 1$ are given by (cf. (3.7))

$\displaystyle u^{\pm}(x)=\begin{bmatrix}1-\lambda^{\pm}_{\kappa}\\\
-\omega\end{bmatrix}e^{-|x|\sqrt{\varkappa^{2}-\Lambda}}.$ (3.17)

∎

$\Lambda_{\kappa}$$\lambda^{\\!-}_{\\!\kappa}$$0$$1$$\lambda^{\\!+}_{\\!\kappa}$$\lambda^{\\!+}_{\\!0}$$\sigma(L_{\kappa})$$\sigma(H_{\kappa}(\omega))$$\varkappa^{2}$$c^{-}$$c^{+}$$\Lambda=\lambda+{\lambda\omega^{2}}/{(1-\lambda)}$
Figure 1:  The relation between $\sigma(L_{\kappa})$ and
$\sigma(H_{\kappa}(\omega))$ for $\omega\neq 0$, $\kappa>0$.

Lemma 3.4 gives the spectrum of the selfadjoint operator
$\mathbf{H}(\omega,\kappa):\,L^{2}({\mathbb{R}},{\mathbb{C}}^{4})\to
L^{2}({\mathbb{R}},{\mathbb{C}}^{4})$:

######  Corollary 3.5.

For any $\kappa\in{\mathbb{R}}$, the essential spectrum of
$\mathbf{H}(\omega,\kappa)$is given by

$\sigma_{\mathrm{ess}}(\mathbf{H}(\omega,\kappa))=[c^{-}(\omega),1]\cup[c^{+}(\omega),\infty).$

For $\kappa<0$, one has $\mathbf{H}(\omega,\kappa)\geq 0$, with eigenvalue
$\lambda=0$ of multiplicity one. For $\kappa=0$, one has
$\mathbf{H}(\omega,0)\geq 0$, with eigenvalue $\lambda=0$ of multiplicity two.
For $\kappa>0$ (when $\Lambda_{\kappa}(\omega)$ from Lemma 3.3 is negative),
the operator $\mathbf{H}(\omega,\kappa)$ has exactly one simple negative
eigenvalue $\lambda_{\kappa}^{-}$ given by (3.12).

The following theorem gives the orbital stability result for solitary wave
solutions to (1.1).

######  Theorem 3.6.

Assume that there are solitary wave solutions for
$\omega\in(\omega_{1},\omega_{2})\subset(-m,m)$. The solitary wave
$\phi(x,\omega)e^{-\mathrm{i}\omega t}$ is orbitally stable if and only if

$\kappa<\frac{\omega^{2}}{m^{2}}.$

Above,
$\kappa:={|\phi(0,\omega)|^{2}a^{\prime}(|\phi(0,\omega)|^{2})}/{a(|\phi(0,\omega)|^{2})}$
could be negative (see (3.5)).

Let us recall that for nonzero solitary waves $a(|\phi(0,\omega)|^{2})\neq 0$.

###### Proof.

The Grillakis–Shatah–Strauss theory [GSS87] applies when either
$\mathbf{H}\geq 0$, with $\lambda=0$ a simple eigenvalue with the
corresponding eigenvector $\mathbf{J}\Phi$, satisfying the assumptions of
[GSS87, Theorem 1]. (when $\kappa<0$), or $\mathbf{H}$ has a negative spectrum
consisting of one simple eigenvalue, has its kernel spanned by
$\mathbf{J}\Phi$, while the rest of its spectrum is positive and separated
away from $z=0$, satisfying the assumptions of [GSS87, Theorem 2].

In the case $\kappa=0$, the operator
$A_{0}(\omega)=\begin{bmatrix}\mathrm{i}\omega&1\\\
-L_{0}(\omega)-\omega^{2}&\mathrm{i}\omega\end{bmatrix}$ in the right-hand
side of (3.3) is ${\mathbb{C}}$-linear. Moreover, it can be represented as
$A_{0}(\omega)=J\tilde{H}_{0}(\omega)$, where

$J=\begin{bmatrix}0&1\\\
-1&0\end{bmatrix},\quad\tilde{H}_{0}(\omega)=\begin{bmatrix}L_{0}(\omega)+\omega^{2}&-\mathrm{i}\omega\\\
\mathrm{i}\omega&1\end{bmatrix}.$

The operator $\tilde{H}_{0}(\omega)$ has the same spectral properties as the
operator $H_{0}(\omega)$. Namely,

$\sigma_{\mathrm{ess}}\big{(}\tilde{H}_{0}(\omega)\big{)}=[c^{-}(\omega),1]\cup[c^{+}(\omega),+\infty),\quad\sigma_{\mathrm{p}}\big{(}\tilde{H}_{0}(\omega)\big{)}=\\{0\\}\cup\\{\omega^{2}+1\\},$

with eigenvalue $\lambda=0$ being simple. Hence, [GSS87, Theorems 1] applies,
showing that the solitary wave $\phi(x,\omega)e^{-\mathrm{i}\omega t}$ is
orbitally stable.

The instability in the case $\kappa>\omega^{2}/m^{2}$ follows from [GSS87,
Theorem 3]; in the critical case, the instability follows from the Jordan
block structure of the zero eigenvalue; see e.g. [CP03]. ∎

The spectrum of the linearization at a solitary wave also follows from the
arguments in [Kol73] and [GSS87]; we consider it next.

######  Lemma 3.7 (Spectrum of $\mathbf{A}(\omega,\kappa)$: basic properties).

  1. 1.

If $\lambda$ belongs to $\sigma_{\mathrm{p}}(\mathbf{A}(\omega,\kappa))$, then
so do $\bar{\lambda}$, $-\lambda$, and $-\bar{\lambda}$.

  2. 2.

$\sigma_{\mathrm{ess}}(\mathbf{A}(\omega,\kappa))=\mathrm{i}\big{(}{\mathbb{R}}\setminus(-m+|\omega|,m-|\omega|)\big{)}$.

  3. 3.

$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,\kappa))\subset{\mathbb{R}}\cup\mathrm{i}{\mathbb{R}}$.
There is a pair of a positive and a negative eigenvalues of
$\mathbf{A}(\omega,\kappa)$ if and only if $\kappa>0$ and
$|\omega|<\min\\{m,\Omega_{\kappa}\\}$, with

$\displaystyle\Omega_{\kappa}:=m\sqrt{\kappa}.$ (3.18)

  4. 4.

If $\omega\in(-m,m)$ and
$\kappa\not\in\big{\\{}0,\,\omega^{2}/m^{2}\big{\\}}$, then the zero
eigenvalue of $\mathbf{A}(\omega,\kappa)$ is of geometric multiplicity $1$ and
of algebraic multiplicity $2$.

  5. 5.

If $\omega\in(-m,m)\setminus\\{0\\}$ and $\kappa=\omega^{2}/m^{2}$, then the
zero eigenvalue of $\mathbf{A}(\omega,\kappa)$ is of geometric multiplicity
$1$ and of algebraic multiplicity $4$.

  6. 6.

If $\omega\in(-m,m)\setminus\\{0\\}$ and $\kappa=0$, then the zero eigenvalue
of $\mathbf{A}(\omega,\kappa)$ is of geometric multiplicity $2$ and of
algebraic multiplicity $2$.

  7. 7.

If $\omega=0$ and $\kappa=0$, then eigenvalue $z=0$ of
$\mathbf{A}(\omega,\kappa)$ is of geometric multiplicity $2$ and of algebraic
multiplicity $4$.

###### Proof.

We follow the arguments of Kolokolov [Kol73] and Grillakis–Shatah–Strauss
[GSS87]. For Part (1), one notices that $\mathbf{A}(\omega,\kappa)$ has real
coefficients (hence $\bar{\lambda}$ is also an eigenvalue) and then that
$\mathbf{A}(\omega,\kappa)^{*}=-\mathbf{H}(\omega,\kappa)\bm{\Sigma}$, which
is conjugate to $-\bm{\Sigma}\mathbf{H}(\omega,\kappa)$ (hence
$-\bar{\lambda}$ is also an eigenvalue). Part (2) follows by fixing
$\lambda\in\mathrm{i}{\mathbb{R}}$ with $|\lambda|\geq m-|\omega|$ and
considering the Weyl sequences supported away from $x=0$.

To prove Part (3), it is convenient to write $\mathbf{A}(\omega,\kappa)$ from
(3.8) in the form similar to the nonlinear Schrödinger theory [Kol73]. We
consider the conjugated versions of $\bm{\Sigma}$ and
$\mathbf{H}(\omega,\kappa)$, with

$\tilde{\mathbf{H}}(\omega,\kappa)=\mathbf{G}_{2}\mathbf{G}_{1}\mathbf{H}(\omega,\kappa)\mathbf{G}_{1}^{-1}\mathbf{G}_{2}^{-1}$

from (3.10) and with

$\displaystyle\tilde{\bm{\Sigma}}=\mathbf{G}_{2}\mathbf{G}_{1}\bm{\Sigma}\mathbf{G}_{1}^{-1}\mathbf{G}_{2}^{-1}=-\mathrm{i}\begin{bmatrix}0&\sigma_{1}\\\
\sigma_{1}&0\end{bmatrix}.$ (3.19)

Now we are in the framework of A. Kolokolov [Kol73]. Consider the eigenvalue
problem for $\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(\omega,\kappa)$:

$\displaystyle\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(\omega,\kappa)\psi=-\mathrm{i}\begin{bmatrix}0&\sigma_{1}\\\
\sigma_{1}&0\end{bmatrix}\begin{bmatrix}H_{\kappa}(\omega)&0\\\
0&H_{0}(\omega)\end{bmatrix}\begin{bmatrix}u\\\
v\end{bmatrix}=\lambda\begin{bmatrix}u\\\
v\end{bmatrix},\qquad\psi=\begin{bmatrix}u\\\ v\end{bmatrix}\in
L^{2}({\mathbb{R}},{\mathbb{C}}^{4}).$ (3.20)

First, we notice that if $\kappa\leq 0$, then
$\tilde{\mathbf{H}}(\omega,\kappa)$ is nonnegative and selfadjoint, hence one
can extract the square root (also nonnegative and selfadjoint); therefore,

$\sigma_{\mathrm{d}}(\tilde{\bm{\Sigma}}\tilde{\mathbf{H}})\setminus\\{0\\}=\sigma_{\mathrm{d}}\big{(}\tilde{\mathbf{H}}^{1/2}\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}^{1/2}\big{)}\setminus\\{0\\}\subset\mathrm{i}{\mathbb{R}}$

since $\tilde{\mathbf{H}}^{1/2}\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}^{1/2}$ is
antiselfadjoint.

From now on, we assume that $\kappa>0$. Eliminating $v$, we get

$-\sigma_{1}H_{0}(\omega)\sigma_{1}H_{\kappa}(\omega)u=\lambda^{2}u,\qquad
u\in L^{2}({\mathbb{R}},{\mathbb{C}}^{2}).$

For $\lambda\neq 0$, one can see that $u$ is orthogonal to

$\displaystyle\ker(\sigma_{1}H_{0}(\omega)\sigma_{1})=\ker\left(\begin{bmatrix}1&\omega\\\
\omega&L_{0}+\omega^{2}\end{bmatrix}\right)=\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix}$ (3.21)

(cf. (2.6)), hence we can write

$H_{\kappa}(\omega)u=-\lambda^{2}(\sigma_{1}H_{0}(\omega)\sigma_{1})^{-1}u=-\lambda^{2}\sigma_{1}H_{0}(\omega)^{-1}\sigma_{1}u.$

Coupling this relation with $u$ and taking into account that $\langle
u,(\sigma_{1}H_{0}(\omega)\sigma_{1})^{-1}u\rangle>0$, we see that
$\lambda^{2}\in{\mathbb{R}}$. To find whether $-\lambda^{2}$ can be negative,
one considers the minimization problem

$\displaystyle\mu=\inf\left\\{\langle u,H_{\kappa}(\omega)u\rangle:\ \langle
u,u\rangle=1,\quad
u\in\big{(}\ker(\sigma_{1}H_{0}(\omega)\sigma_{1})\big{)}^{\perp}=\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix}^{\perp}\right\\},$ (3.22)

which implies that $u$ satisfies

$\displaystyle H_{\kappa}(\omega)u=\mu u+\nu\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix},$ (3.23)

with $\mu,\,\nu\in{\mathbb{R}}$ the Lagrange multipliers. We note that if
$\mu\leq 0$, then $\nu\neq 0$, or else one would have
$\mu=\lambda^{-}_{\kappa}$ (the only negative eigenvalue of
$H_{\kappa}(\omega)$), which is not possible since $u^{-}(x)$ from (3.17)
corresponding to eigenvalue $\lambda^{-}_{\kappa}$ is not orthogonal to
$\ker(\sigma_{1}H_{0}(\omega)\sigma_{1})$: one has

$(u^{-})^{*}\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix}=\begin{bmatrix}1-\lambda^{-}_{\kappa}\\\
-\omega\end{bmatrix}^{*}\begin{bmatrix}-\omega\\\
1\end{bmatrix}e^{-|x|\sqrt{\varkappa^{2}-\Lambda}}e^{-|x|\varkappa}=-(2-\lambda^{-}_{\kappa})\omega
e^{-|x|\sqrt{\varkappa^{2}-\Lambda}}e^{-|x|\varkappa},$

where $2-\lambda^{-}_{\kappa}$ is strictly positive. So, $\nu\neq 0$ (or else
$\mu$ would have to be positive); then one concludes from (3.22) that
$\mu>\lambda^{-}_{\kappa}$. We rewrite (3.23) as

$(H_{\kappa}(\omega)-\mu)u=\nu\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix},\qquad
u=\nu(H_{\kappa}(\omega)-\mu)^{-1}\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix}.$

The sign of $\mu$ could be found from the condition that $u$ is orthogonal to
$\ker(\sigma_{1}H_{0}(\omega)\sigma_{1})$:

$\left\langle\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix},(H_{\kappa}(\omega)-z)^{-1}\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix}\right\rangle=0.$

We consider

$h(z)=\left\langle\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix},(H_{\kappa}(\omega)-z)^{-1}\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix}\right\rangle,\qquad
z\in(\lambda^{-}_{\kappa},c^{-})\subset\rho(H_{\kappa}(\omega)).$

Since $h(z)$ is monotonically increasing on $\rho(H_{\kappa}(\omega))$, the
sign of $\mu$ is opposite to the sign of $h(0)$, which is given by

$\displaystyle h(0)=\left\langle\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix},H_{\kappa}(\omega)^{-1}\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix}\right\rangle=\left\langle\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix},\begin{bmatrix}-\partial_{\omega}\phi\\\
\omega\partial_{\omega}\phi+\phi\end{bmatrix}\right\rangle=\partial_{\omega}\left(\omega\|\phi\|^{2}\right).$
(3.24)

Above, we used the relation

$\displaystyle H_{\kappa}(\omega)\begin{bmatrix}-\partial_{\omega}\phi\\\
\omega\partial_{\omega}\phi+\phi\end{bmatrix}=\begin{bmatrix}L_{\kappa}+\omega^{2}&\omega\\\
\omega&1\end{bmatrix}\begin{bmatrix}-\partial_{\omega}\phi\\\
\omega\partial_{\omega}\phi+\phi\end{bmatrix}=\begin{bmatrix}-\omega\phi\\\
\phi\end{bmatrix}$ (3.25)

which in turn follows from taking the $\omega$-derivative of (2.6) (after we
substitute $\phi(x,\omega)$), which yields

$\displaystyle L_{\kappa}\partial_{\omega}\phi=2\omega\phi.$ (3.26)

We conclude from (3.24) that there is $\lambda^{2}>0$ (hence, there is a pair
of a positive and a negative eigenvalues) if and only if
$h(0)=\partial_{\omega}(\omega\|\phi\|^{2})>0$.

Let us prove Parts (4) and (5). Given $\omega\in(-m,m)$, let us compute the
geometric multiplicity of $\lambda=0$. We consider
$\tilde{\mathbf{A}}=\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(\omega,\kappa)$,
with $\tilde{\bm{\Sigma}}$ from (3.19) and $\tilde{\mathbf{H}}(\omega,\kappa)$
from (3.10). Since $\kappa\neq 0$, by Lemma 3.4 the geometric multiplicity of
$\lambda=0$ equals $1$, with

$\ker(\tilde{\mathbf{H}}(\omega,\kappa))=\mathop{\mathrm{Span}}\left\\{\phi(x)\bm{e}_{3}-\omega\phi(x)\bm{e}_{4}\right\\},$

with $\\{\bm{e}_{i}\\}_{1\leq\mathrm{i}\leq 4}$ the standard basis in
${\mathbb{C}}^{4}$.

Now let us compute the algebraic multiplicity of $\lambda=0$. By (3.21) and
(3.25) we have

$\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(\omega,\kappa)\begin{bmatrix}0\\\ 0\\\
\phi\\\ -\omega\phi\end{bmatrix}=-\mathrm{i}\begin{bmatrix}0&\sigma_{1}\\\
\sigma_{1}&0\end{bmatrix}\begin{bmatrix}H_{\kappa}(\omega)&0\\\
0&H_{0}(\omega)\end{bmatrix}\begin{bmatrix}0\\\ 0\\\ \phi\\\
-\omega\phi\end{bmatrix}=0,$
$\displaystyle\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(\omega,\kappa)\begin{bmatrix}-\partial_{\omega}\phi\\\
\omega\partial_{\omega}\phi+\phi\\\ 0\\\
0\end{bmatrix}=-\mathrm{i}\begin{bmatrix}0&\sigma_{1}\\\
\sigma_{1}&0\end{bmatrix}\begin{bmatrix}H_{\kappa}(\omega)&0\\\
0&H_{0}(\omega)\end{bmatrix}\begin{bmatrix}-\partial_{\omega}\phi\\\
\omega\partial_{\omega}\phi+\phi\\\ 0\\\
0\end{bmatrix}=-\mathrm{i}\begin{bmatrix}0\\\ 0\\\ \phi\\\
-\omega\phi\end{bmatrix};$ (3.27)

thus $\lambda=0$ is an eigenvalue of
$\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(\omega,\kappa)$ of multiplicity at
least two. To be able to extend this Jordan chain, solving

$\displaystyle\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(\omega,\kappa)\Xi=\begin{bmatrix}-\partial_{\omega}\phi\\\
\omega\partial_{\omega}\phi+\phi\\\ 0\\\
0\end{bmatrix},\qquad\Xi\in{\mathbb{C}}^{4},$ (3.28)

we need to make sure that the right-hand side is orthogonal to the kernel of
the adjoint of the operator in the left-hand side,

$\ker\big{(}\tilde{\mathbf{H}}(\omega,\kappa)\tilde{\bm{\Sigma}}\big{)}=\mathop{\mathrm{Span}}\left\\{\begin{bmatrix}-\omega\phi\\\
\phi\\\ 0\\\ 0\end{bmatrix}\right\\};$

thus, the condition to have a Jordan block of a larger size is

$\displaystyle 0=\left\langle\begin{bmatrix}-\omega\phi\\\ \phi\\\ 0\\\
0\end{bmatrix},\ \begin{bmatrix}-\partial_{\omega}\phi\\\
\omega\partial_{\omega}\phi+\phi\\\ 0\\\
0\end{bmatrix}\right\rangle=\partial_{\omega}\left(\omega\|\phi\|^{2}\right).$
(3.29)

Let us compute explicitly the right-hand side in (3.29). Since
$\|\phi(\omega)\|^{2}=\frac{C(\omega)^{2}}{\varkappa}$ (see Lemma 2.3), one
computes:

$\displaystyle\frac{d}{d\omega}(\omega\|\phi(\omega)\|^{2})=\frac{C^{2}}{\varkappa}+\omega\Big{(}\frac{2C}{\varkappa}\frac{dC}{d\omega}+\frac{\omega
C^{2}}{\varkappa^{3}}\Big{)}=\frac{1}{\varkappa^{3}}\Big{(}m^{2}C^{2}-\frac{\omega^{2}a(C^{2})}{a^{\prime}(C^{2})}\Big{)}=\frac{C^{2}}{\varkappa^{3}}\Big{(}m^{2}-\frac{\omega^{2}}{\kappa}\Big{)},$
(3.30)

where $C=C(\omega)$; to express $dC/d\omega$, we used (2.11). Thus, if
$\kappa\neq 0$, $\partial\big{(}\omega\|\phi_{\omega}\|^{2}\big{)}=0$ if and
only if $\kappa=\omega^{2}/m^{2}$ (we will consider the case $\kappa=0$
separately).

If $\kappa\neq\omega^{2}/m^{2}$, $\kappa\neq 0$, since the condition (3.29) is
not satisfied, equation (3.28) has no $L^{2}$-solutions hence the algebraic
multiplicity is exactly two. If $\kappa=\omega^{2}/m^{2}>0$, then equation
(3.28) has an $L^{2}$-solution; the algebraic multiplicity jumps by at least
one. This means that an eigenvalue arrives at $z=0$; by the symmetry of
$\sigma(\mathbf{A}(\omega,\kappa))$ with respect to ${\mathbb{R}}$ and
$\mathrm{i}{\mathbb{R}}$, there is also the opposite sign eigenvalue arriving
at $z=0$, thus the algebraic multiplicity of $z=0$ jumps by two. Standard
considerations (see e.g. [CP03]) show that for $\omega\neq\pm\Omega_{\kappa}$
the algebraic multiplicity of zero eigenvalue can not be more than four.
Indeed, let us show that the equation

$(\tilde{\bm{\Sigma}}\tilde{\mathbf{H}})^{4}\Gamma=\phi(x)\bm{e}_{3}-\omega\phi(x)\bm{e}_{4},\qquad\Gamma\in
L^{2}({\mathbb{R}},{\mathbb{C}}^{4}),$

has no solutions. If the algebraic multiplicity of $\lambda=0$ is at least
four (hence $\kappa=\omega^{2}/m^{2}$) so that there are $\Xi,\,\Theta\in
L^{2}({\mathbb{R}},{\mathbb{C}}^{4})$ such that

$\displaystyle(\tilde{\bm{\Sigma}}\tilde{\mathbf{H}})\Xi=-\partial_{\omega}\phi\bm{e}_{1}+(\omega\partial_{\omega}\phi+\phi)\bm{e}_{2},\qquad(\tilde{\bm{\Sigma}}\tilde{\mathbf{H}})\Theta=\Xi,$
(3.31)

where, as one can readily see from the block form of
$\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}$ (see (3.20)), one can decompose
$\Xi(x)=\xi_{3}(x)\bm{e}_{3}+\xi_{4}(x)\bm{e}_{4}$, then, trying to solve

$(\tilde{\bm{\Sigma}}\tilde{\mathbf{H}})^{2}\Gamma=\Xi,\qquad\Gamma\in
L^{2}({\mathbb{R}},{\mathbb{C}}^{4}),$

we need to make sure that $\Xi$ is orthogonal to the kernel of
$((\tilde{\bm{\Sigma}}\tilde{\mathbf{H}})^{2})^{*}=(\tilde{\mathbf{H}}\tilde{\bm{\Sigma}})^{2}$
which contains in particular
$\tilde{\bm{\Sigma}}((-\partial_{\omega}\phi\bm{e}_{1}+\omega\partial_{\omega}\phi+\phi)\bm{e}_{2})=\tilde{\bm{\Sigma}}(\tilde{\bm{\Sigma}}\tilde{\mathbf{H}})\Xi=-\tilde{\mathbf{H}}\Xi$.
We arrive at the following necessary condition:

$\displaystyle\langle\Xi,\tilde{\mathbf{H}}(\omega,\kappa)\Xi\rangle=0.$
(3.32)

Taking into account that
$\tilde{\mathbf{H}}(\omega,\kappa)=\begin{bmatrix}H_{\kappa}(\omega)&0\\\
0&H_{0}(\omega)\end{bmatrix}$ is semi-positive-definite on vectors of the form
$\xi_{3}(x)\bm{e}_{3}+\xi_{4}(x)\bm{e}_{4}$, with the one-dimensional kernel
(this follows from Lemma 3.4), while by (3.31) the function $\Xi$ is not in
this kernel, we conclude that the left-hand side is strictly positive, hence
the condition (3.32) can not be satisfied.

For Part (6), we consider the case $\omega\neq 0$, $\kappa=0$. By Lemma 3.4,
the geometric multiplicity of $\lambda=0$ equals $2$, with

$\ker\big{(}\tilde{\mathbf{H}}(\omega,0)\big{)}=\mathop{\mathrm{Span}}\left\\{\phi(x)\bm{e}_{1}-\omega\phi(x)\bm{e}_{2},\
\phi(x)\bm{e}_{3}-\omega\phi(x)\bm{e}_{4}\right\\}.$

To have a Jordan block, we would need to solve

$\displaystyle\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(\omega,0)\phi(x)\bm{e}_{4}=-\mathrm{i}\begin{bmatrix}0&\sigma_{1}\\\
\sigma_{1}&0\end{bmatrix}\begin{bmatrix}H_{0}(\omega)&0\\\
0&H_{0}(\omega)\end{bmatrix}\Xi=-\mathrm{i}\phi(x)\bm{e}_{1}.$ (3.33)

Since $\omega\neq 0$, the right-hand side of (3.33) is not orthogonal to

$\ker\big{(}(\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(\omega,0))^{*}\big{)}=\ker\big{(}\tilde{\mathbf{H}}(\omega,0)\tilde{\bm{\Sigma}}\big{)}\ni\begin{bmatrix}-\omega\phi(x)\\\
\phi(x)\\\ 0\\\ 0\end{bmatrix},$

hence (3.33) has no solutions.

Finally, for Part (7), we consider the case $\omega=0$, $\kappa=0$. In this
case, the geometric multiplicity of $\lambda=0$ equals $2$, with

$\ker\big{(}\tilde{\mathbf{H}}(0,0)\big{)}=\mathop{\mathrm{Span}}\left\\{\phi(x)\bm{e}_{1},\
\phi(x)\bm{e}_{3}\right\\}.$

The Jordan block corresponding to $\phi(x)\bm{e}_{1}$ is of size at least two
since

$\tilde{\bm{\Sigma}}\tilde{\mathbf{H}}(0,0)\phi(x)\bm{e}_{4}=-\mathrm{i}\begin{bmatrix}0&\sigma_{1}\\\
\sigma_{1}&0\end{bmatrix}\begin{bmatrix}H_{0}(0)&0\\\
0&H_{0}(0)\end{bmatrix}\phi(x)\bm{e}_{4}=-\mathrm{i}\phi(x)\bm{e}_{1}.$

Now we notice that $\phi(x)\bm{e}_{4}$ is not orthogonal to
$\ker\big{(}\tilde{\mathbf{H}}(0,0)\tilde{\bm{\Sigma}}\big{)}$ which contains
$\phi(x)\bm{e}_{4}$; therefore, the corresponding Jordan block is of size
exactly two. Similarly, there is a Jordan block of size exactly two
corresponding to $\phi(x)\bm{e}_{3}$.

This completes the proof of Lemma 3.7. ∎

######  Remark 3.8.

By Lemma 3.7 (3), there is a positive eigenvalue of
$\mathbf{A}(\omega,\kappa)$ if and only if
$\omega^{2}<\min\\{m^{2},m^{2}\kappa\\}$. We see from (3.30) that in this case
one has $\partial_{\omega}(\omega\|\phi(\omega)\|^{2})>0$. Since the charge of
the Klein–Gordon field is given by
$\mathcal{Q}(\phi(\omega))=\omega\|\phi(\omega)\|^{2}$ (see (2.4)), the above
is in agreement with the Kolokolov stability condition

$\displaystyle\partial_{\omega}\mathcal{Q}(\phi(\omega))<0$ (3.34)

derived in [Kol73] in the context of the nonlinear Schrödinger equation.

##  4 The spectrum of the linearization operator

Now we are going to perform a complete analysis of the spectrum of the
linearization operator. For convenience, we consider the linearization
operator $\mathbf{A}(\omega,\kappa)$ from (3.8) using $\kappa>0$ as a
parameter. Recall that

$\displaystyle\mathbf{A}(\omega,\kappa)=\begin{bmatrix}0&-\omega&1&0\\\
\omega&0&0&1\\\
\partial_{x}^{2}-m^{2}+(1+2\kappa)\alpha\delta(x)&0&0&-\omega\\\
0&\partial_{x}^{2}-m^{2}+\alpha\delta(x)&\omega&0\end{bmatrix},\qquad\alpha=2\sqrt{m^{2}-\omega^{2}}.$
(4.1)

For $x\neq 0$, substituting $\Psi(x)=\bm{w}e^{-\nu|x|}$ with some
$\nu\in{\mathbb{C}}$, $\,\mathrm{Re}\,\nu\geq 0$, into the equation
$(\mathbf{A}(\omega,\kappa)-\lambda)\Psi=0$, we get

$\begin{bmatrix}-\lambda&-\omega&1&0\\\ \omega&-\lambda&0&1\\\
\nu^{2}-m^{2}&0&-\lambda&-\omega\\\
0&\nu^{2}-m^{2}&\omega&-\lambda\end{bmatrix}\bm{w}=0,\qquad\nu\in{\mathbb{C}},\quad\bm{w}\in{\mathbb{C}}^{4},\quad\bm{w}\neq
0,$ (4.2)

which, via the Schur complement idea, is equivalent to

$\begin{bmatrix}\begin{array}[]{cc}-\lambda&-\omega\\\
\omega&-\lambda\end{array}&\begin{array}[]{cc}1&0\\\ 0&1\end{array}\\\
S(\omega,\lambda,\nu)&\begin{array}[]{cc}0&0\\\
0&0\end{array}\end{bmatrix}\bm{w}=0,\qquad\bm{w}\in{\mathbb{C}}^{4},\quad\bm{w}\neq
0,$ (4.3)

with $S(\omega,\lambda,\nu)\in\mathop{\mathrm{End}}({\mathbb{C}}^{2})$ the
Schur complement of the top right block $I_{2}$:

$\displaystyle S(\omega,\lambda,\nu)=\begin{bmatrix}\nu^{2}-m^{2}&0\\\
0&\nu^{2}-m^{2}\end{bmatrix}-\begin{bmatrix}-\lambda&-\omega\\\
\omega&-\lambda\end{bmatrix}^{2}=\begin{bmatrix}\nu^{2}-m^{2}+\omega^{2}-\lambda^{2}&-2\lambda\omega\\\
2\lambda\omega&\nu^{2}-m^{2}+\omega^{2}-\lambda^{2}\end{bmatrix}.$ (4.4)

The condition to have nonzero solution $\bm{w}\in{\mathbb{C}}^{4}$ to (4.3) is
equivalent to

$\det
S(\omega,\lambda,\nu)=(m^{2}-\omega^{2}+\lambda^{2}-\nu^{2})^{2}+4\lambda^{2}\omega^{2}=0.$

This gives

$m^{2}-\omega^{2}+\lambda^{2}-\nu^{2}=-2\mathrm{i}\lambda\omega,\qquad
m^{2}-\omega^{2}+\lambda^{2}-\nu^{2}=2\mathrm{i}\lambda\omega,$
$m^{2}-(\omega-\mathrm{i}\lambda)^{2}=\nu^{2},\qquad
m^{2}-(\omega+\mathrm{i}\lambda)^{2}=\nu^{2},$

allowing one to express $\nu\in{\mathbb{C}}$ in terms of $\omega$ and
$\lambda$:

$\nu=\sqrt{m^{2}-(\omega\pm\mathrm{i}\lambda)^{2}}.$

We choose the cuts in the complex plane $\lambda$ from the branching points to
infinity:

$\displaystyle\mathcal{C}_{+}:=(-\mathrm{i}\infty,-\mathrm{i}(m-\omega)]\cup[\mathrm{i}(m+\omega,\mathrm{i}\infty),$
$\displaystyle\mathcal{C}_{-}:=(-\mathrm{i}\infty,-\mathrm{i}(m+\omega)]\cup[\mathrm{i}(m-\omega,\mathrm{i}\infty),$

defining

$\displaystyle\nu_{\pm}(\omega,\lambda)=\sqrt{m^{2}-(\omega\pm\mathrm{i}\lambda)^{2}},$
(4.5)

with

$\,\mathrm{Re}\,\nu_{\pm}(\omega,\lambda)>0,\qquad\lambda\in{\mathbb{C}}\setminus\mathcal{C}_{\pm}.$
(4.6)

The zero eigenvalue of the Schur complement $S$ from (4.4) corresponds to two
eigenvectors $u_{\pm}\in{\mathbb{C}}^{2}$, depending on the choice
$\nu=\nu_{\pm}$; these eigenvectors are given by

$\begin{bmatrix}2\lambda\omega\\\
-m^{2}+\omega^{2}+\nu_{\pm}^{2}-\lambda^{2}\end{bmatrix}=\begin{bmatrix}2\lambda\omega\\\
-m^{2}+\omega^{2}+(m^{2}-\omega^{2}+\lambda^{2}\mp
2\mathrm{i}\lambda\omega)-\lambda^{2}\end{bmatrix}=2\lambda\omega\begin{bmatrix}1\\\
\mp\mathrm{i}\end{bmatrix},$

so we can use $u_{\pm}=\begin{bmatrix}1\\\ \mp\mathrm{i}\end{bmatrix}$. By
(4.3), the corresponding vector from the null space of
$\mathbf{A}(\omega,\kappa)-\lambda I$ is then
$\bm{w}_{\pm}=\begin{bmatrix}u_{\pm}\\\
v_{\pm}\end{bmatrix}\in{\mathbb{C}}^{4}$, with
$v_{\pm}=-\begin{bmatrix}-\lambda&-\omega\\\
\omega&-\lambda\end{bmatrix}u_{\pm}=\begin{bmatrix}\lambda\mp\mathrm{i}\omega\\\
-\omega\mp\mathrm{i}\lambda\end{bmatrix}\in{\mathbb{C}}^{2}.$ Thus, one has

$\bm{w}_{\pm}=\begin{bmatrix}1\\\ \pm\mathrm{i}\\\
\lambda\mp\mathrm{i}\omega\\\
\omega\pm\mathrm{i}\lambda\end{bmatrix}\in{\mathbb{C}}^{4}.$

Therefore, an eigenfunction corresponding to the eigenvalue
$\lambda\in{\mathbb{C}}$ is of the form

$\displaystyle\Psi(x,\omega,\lambda)=A\bm{w}_{+}e^{-\nu_{+}|x|}+B\bm{w}_{-}e^{-\nu_{-}|x|}=A\begin{bmatrix}1\\\
\mathrm{i}\\\ \lambda-\mathrm{i}\omega\\\
\omega+\mathrm{i}\lambda\end{bmatrix}e^{-\nu_{+}|x|}+B\begin{bmatrix}1\\\
-\mathrm{i}\\\ \lambda+\mathrm{i}\omega\\\
\omega-\mathrm{i}\lambda\end{bmatrix}e^{-\nu_{-}|x|},$ (4.7)

with $A,\,B\in{\mathbb{C}}$ not simultaneously zeros and
$\,\mathrm{Re}\,\nu_{\pm}>0$ (when the corresponding coefficient is nonzero).
The values of $A$ and $B$ are obtained from the jump conditions at $x=0$:
substituting $\Psi$ into $(\mathbf{A}(\omega,\kappa)-\lambda)\Psi=0$ and
collecting the terms with $\delta$-function gives:

$\displaystyle\begin{cases}(-2\nu_{+}+\alpha(1+2\kappa))A+(-2\nu_{-}+\alpha(1+2\kappa))B=0,\\\
(-2\nu_{+}+\alpha)\mathrm{i}A+(-2\nu_{-}+\alpha)(-\mathrm{i}B)=0.\end{cases}$
(4.8)

The condition to have $A,\,B\in{\mathbb{C}}$ not simultaneously zeros,

$\det\begin{bmatrix}-2\nu_{+}+\alpha(1+2\kappa)&-2\nu_{-}+\alpha(1+2\kappa)\\\
-2\nu_{+}+\alpha&2\nu_{-}-\alpha\end{bmatrix}=0,$

takes the form
$-8\nu_{+}\nu_{-}+4(\nu_{+}+\nu_{-})\alpha(1+\kappa)-2\alpha^{2}(1+2\kappa)=0,$
which we rewrite as

$\displaystyle D_{\omega,\kappa}(\lambda)=0$ (4.9)

with

$\displaystyle
D_{\omega,\kappa}(\lambda)=\alpha^{2}(1+\kappa)^{2}-2(\nu_{+}+\nu_{-})\alpha(1+\kappa)+4\nu_{+}\nu_{-}-\alpha^{2}\kappa^{2},$
(4.10)

with $\alpha=2\sqrt{m^{2}-\omega^{2}}$ from (3.5) and
$\nu_{\pm}(\omega,\lambda)=\sqrt{m^{2}-(\omega\pm\mathrm{i}\lambda)^{2}}$ from
(4.5). The function $D_{\omega,\kappa}(\lambda)$ is analytic in $\lambda$ in
${\mathbb{C}}\setminus(\mathcal{C}_{-}\cup\mathcal{C}_{+})$. Since there are
two possible values for each of the square roots in the definition (4.5) of
$\nu_{\pm}$, $D_{\omega,\kappa}(\lambda)$ can be continuied analytically
through the cuts $\mathcal{C}_{-}$ and $\mathcal{C}_{+}$ to an analytic
function on the four-sheet cover of ${\mathbb{C}}$, which we also denote by
$D_{\omega,\kappa}(\lambda)$. We call the sheet defined by conditions (4.6)
the _physical sheet_ of $D_{\omega,\kappa}(\lambda)$.

###  4.1 Embedded eigenvalues and virtual levels

We start by studying embedded eigenvalues and virtual levels of the operator
$\mathbf{A}(\omega,\kappa)$. Before formulating our results, let us mention
that a _virtual level_ (also known as a _threshold resonance_) can be defined
as a limit point of an eigenvalue family which corresponds to values of a
perturbation parameter in an interval when this limit point no longer
corresponds to a square-integrable eigenfunction. The virtual levels usually
occur at thresholds of the essential spectrum (the endpoints of the essential
spectrum or the points where the continuous spectrum changes its
multiplicity). For more on the phenomenon of virtual levels, see e.g. [JK79,
JN01, Yaf10, GN20, EGT19].

######  Lemma 4.1 (Embedded eigenvalues and virtual levels of
$\mathbf{A}(\omega,\kappa)$).

  1. 1.

There are embedded eigenvalues $\lambda=\pm 2\omega\mathrm{i}$ if and only if
$\kappa=0$ and $|\omega|\geq m/3$.

  2. 2.

For any $\kappa\in{\mathbb{R}}$ and $\omega\in(-m,m)\setminus 0$ there are no
virtual levels at the embedded thresholds $\lambda=\pm\mathrm{i}(m+|\omega|)$.

  3. 3.

There are virtual levels at $\lambda=\pm\mathrm{i}(m-|\omega|)$ if and only if
$\kappa\in\big{[}-\frac{1}{2},\frac{1}{\sqrt{2}}\big{)}$ and
$\omega=\pm\mathcal{T}_{\kappa}$, where

$\displaystyle\mathcal{T}_{\kappa}=m\frac{(1+2\kappa)^{2}}{3+4\kappa};$ (4.11)

in particular, $\lambda=\pm\mathrm{i}m$ are virtual levels if and only if
$\kappa=-1/2$ and $\omega=0$. By Part 1, these are genuine virtual levels
(with non-$L^{2}$ eigenfunction) if and only if $\kappa\neq 0$.

###### Proof.

Because of the symmetry with respect to the sign of $\omega$, we will assume
that

$0\leq\omega<m.$

For Part (1), it suffices to consider $\lambda=\mathrm{i}\Lambda$ with
$\Lambda\geq m-\omega$. In this case

$\nu_{-}=\sqrt{m^{2}-(\omega+\Lambda)^{2}}\in\mathrm{i}{\mathbb{R}};$

therefore, in the expression for the corresponding eigenfunction (4.7) one
would need to take $B=0$. There are two cases to consider:

  * •

If $\kappa\neq 0$, the system (4.8) shows that one would also have $A=0$;
hence, $\lambda$ can not be an eigenvalue;

  * •

If $\kappa=0$ and $\omega\in[m/3,m)$, there is an embedded eigenvalue
$\lambda=2\omega\mathrm{i}$ since the system (4.8) is satisfied for $B=0$ and
any $A\in{\mathbb{C}}$. (Note that in this case $\Lambda=2\omega\geq m-\omega$
and $\nu_{+}=\varkappa=\alpha/2$.)

For Part (2), it is enough to consider $\omega\in(0,m)$,
$\lambda=\mathrm{i}(m+\omega)$. At this value of $\lambda$, one has
$\nu_{+}=0$,
$\nu_{-}=\sqrt{m^{2}-(m+2\omega)^{2}}\in\mathrm{i}{\mathbb{R}}\setminus\\{0\\}$.
Thus,

$D_{\omega,\kappa}(\mathrm{i}(m+\omega))=\alpha^{2}(1+2\kappa)-2\nu_{-}\alpha(1+\kappa)$

can not be zero since $\alpha>0$. In the case $\omega=0$, one has
$\nu_{\pm}=0$, and then $D_{\omega,\kappa}(\mathrm{i}m)=0$ if and only if
$\kappa=-1/2$.

For Part (3), we need to consider $\omega\in(0,m)$,
$\lambda=\mathrm{i}(m-\omega)$. In this case

$\nu_{-}=0,\quad\nu_{+}=\sqrt{m^{2}-(2\omega-m)^{2}}=2\sqrt{\omega(m-\omega)}>0$

by (4.6). We need to solve:

$0=D_{\omega,\kappa}(\mathrm{i}(m-\omega))=\alpha^{2}(1+2\kappa)-2\nu_{+}\alpha(1+\kappa).$

The equation has no solution for $\kappa=-1$. In the case $\kappa\not=-1$, it
is equivalent to

$\sqrt{m^{2}-\omega^{2}}\frac{1+2\kappa}{1+\kappa}=2\sqrt{\omega(m-\omega)},$

which leads to

$\displaystyle\omega=\mathcal{T}_{\kappa}=m\frac{(1+2\kappa)^{2}}{3+4\kappa}=m\Big{(}\kappa+\frac{1}{4}+\frac{1}{4(3+4\kappa)}\Big{)}.$
(4.12)

We note that $0\leq\mathcal{T}_{\kappa}<m$ as long as
$\kappa\in[-1/2,1/\sqrt{2})$. ∎

###  4.2 Discrete spectrum

Now we will study the discrete spectrum of the operator
$(\mathbf{A}(\omega,\kappa))$. Recall that for any $\kappa\in{\mathbb{R}}$ and
$\omega\in(-m,m)$ the operator has zero eigenvalue of of multiplicity $2$. We
start with the special cases $\kappa=0$ and $\omega=0$.

######  Lemma 4.2 (Eigenvalues of $\mathbf{A}(\omega,0)$, $\omega\neq 0$).

$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,0))=\\{0,\pm 2\omega\mathrm{i}\\}\quad
0<|\omega|<m,$

with eigenvalues $\lambda=\pm 2\omega\mathrm{i}$ being simple. The only
eigenvalue of $\mathbf{A}(0,0)$ is $\lambda=0$ of algebraic multiplicity $4$
and geometric multiplicity $2$.

###### Proof.

Considering for which values $\lambda$ the matrix

$\displaystyle\mathbf{A}(\omega,0)-\lambda
I_{4}=\begin{bmatrix}-\lambda&-\omega&1&0\\\ \omega&0-\lambda&0&1\\\
-L_{0}(\omega)-\omega^{2}&0&-\lambda&-\omega\\\
0&-L_{0}(\omega)-\omega^{2}&\omega&-\lambda\end{bmatrix}$ (4.13)

has eigenvalue zero reduces to studying this question for the Schur complement
of the top right block $I_{2}$,

$\displaystyle S=\begin{bmatrix}-L_{0}(\omega)-\omega^{2}&0\\\
0&-L_{0}(\omega)-\omega^{2}\end{bmatrix}-\begin{bmatrix}-\lambda&-\omega\\\
\omega&-\lambda\end{bmatrix}^{2}=\begin{bmatrix}-L_{0}(\omega)-\lambda^{2}&-2\lambda\omega\\\
2\lambda\omega&-L_{0}(\omega)-\lambda^{2}\end{bmatrix}.$ (4.14)

By Lemma 3.3, $L_{0}(\omega)$ has the only eigenvalue $0$; we conclude that
$\lambda$ can be obtained from the equation

$\det\begin{bmatrix}-\lambda^{2}&-2\lambda\omega\\\
2\lambda\omega&-\lambda^{2}\end{bmatrix}=0,$

which gives $\lambda=0$ and $\lambda=\pm 2\omega\mathrm{i}$. Recall that the
eigenvalues $\pm 2\omega\mathrm{i}$ are embedded into the essential spectrum
if $|\omega|\geq m/3$. In the case $\omega=0$, the geometric multiplicity
could be obtained from substituting $\lambda=0$, $\omega=0$, and $\kappa=0$
into (4.13) and taking into account that $\Lambda_{0}(\omega)=0$ is a simple
eigenvalue of $L_{0}(\omega)$ (Lemma 3.3). ∎

######  Lemma 4.3 (Eigenvalues of $\mathbf{A}(0,\kappa)$).

$\sigma_{\mathrm{p}}(\mathbf{A}(0,\kappa))=\left\\{\begin{array}[]{ll}0,\quad\quad\kappa\leq-1/2,\\\
\pm 2m\sqrt{\kappa(1+\kappa)},\quad\kappa>-1/2.\end{array}\right.$

The eigenvalues $\lambda=\pm 2m\sqrt{\kappa(1+\kappa)}$ are simple (real if
$\kappa>0$ and purely imaginary if $-1/2<\kappa<0$).

###### Proof.

In the case $\omega=0$, one has $\nu_{\pm}=\sqrt{m^{2}+\lambda^{2}}$ and
$\alpha=2m$. Then

$\displaystyle D_{0,\kappa}(\lambda)$ $\displaystyle=$ $\displaystyle
4m^{2}(1+2\kappa)-8m(1+\kappa)\sqrt{m^{2}+\lambda^{2}}+4(m^{2}+\lambda^{2})$
$\displaystyle=$ $\displaystyle
8m^{2}(1+\kappa)+4\lambda^{2}-8m(1+\kappa)\sqrt{m^{2}+\lambda^{2}}.$

In the case $\kappa=-1$, $D_{0,\kappa}(\lambda)$ has the only root $\lambda=0$
of multiplicity $2$. For $\kappa\not=-1$, $D_{0,\kappa}(\lambda)=0$ if and
only if

$2m^{2}+\frac{\lambda^{2}}{1+\kappa}>0~{}~{}{\rm
and}~{}~{}\Big{(}2m^{2}+\frac{\lambda^{2}}{1+\kappa}\Big{)}^{2}=4m^{2}(m^{2}+\lambda^{2}).$
(4.15)

The second condition gives $\lambda^{2}=0$ and
$\frac{\lambda^{2}}{1+\kappa}=4m^{2}\kappa$. Using the first condition of
(4.15), we conclude that $\kappa>-1/2$, and then
$\lambda=2m\sqrt{\kappa(1+\kappa)}$. ∎

Denote

$\displaystyle
K_{\omega}=\frac{\sqrt{m^{2}-(2|\omega|-m)^{2}}-\varkappa}{2\varkappa-\sqrt{m^{2}-(2|\omega|-m)^{2}}}=\frac{2\sqrt{\frac{|\omega|}{m+|\omega|}}-1}{2-2\sqrt{\frac{|\omega|}{m+|\omega|}}},\qquad\omega\in(-m,m).$
(4.16)

We note that

$K_{\omega}\leq 0~{}~{}{\rm for}~{}~{}\omega\leq m/3,\quad{\rm
and}~{}~{}K_{\omega}\geq 0~{}~{}{\rm for}~{}~{}\omega\geq m/3.$ (4.17)

We note also that for $\omega\in[0,m)$ the function $\omega\mapsto K_{\omega}$
is the inverse to the function $\kappa\mapsto\mathcal{T}_{\kappa}$ defined in
(4.11).

######  Lemma 4.4.

For $\omega\in(-m,m)$ and $\kappa\in(-\infty,0)\cup(0,K_{\omega})$,
$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,\kappa))=\\{0\\}$.

###### Proof.

For $\omega=0$, the statement follows from Lemma 4.3; because of the symmetry
with respect to the sign of $\omega$, it suffices to consider
$\omega\in(0,m)$. Due to Lemma 3.7 (3), it suffices to prove that

$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,\kappa))\cap\mathrm{i}{\mathbb{R}}=\\{0\\},\qquad\kappa\in(-\infty,0)\cup(0,K_{\omega}),\quad\omega\in(0,m).$

It suffices to prove that

$D_{\omega,\kappa}(\lambda)\neq
0,\quad\lambda\in\mathrm{i}{\mathbb{R}}\setminus\\{0\\},\qquad\kappa\in(-\infty,0)\cup(0,K_{\omega}),\quad\omega\in(0,m),$
(4.18)

with $D_{\omega,\kappa}(\lambda)$ defined in (4.10). We rewrite
$D_{\omega,\kappa}(\lambda)$ as

$D_{\omega,\kappa}(\lambda)=(\alpha-2\nu_{+})(\alpha-2\nu_{-})+2\kappa\alpha(\alpha-\nu_{+}-\nu_{-}).$
(4.19)

Note that $\lambda=\pm 2\omega\mathrm{i}$ are not roots of
$D_{\omega,\kappa}(\lambda)\neq 0$ for $\omega>0$ and $\kappa\neq 0$. Indeed,
for $\lambda=2\omega\mathrm{i}$ one has $\alpha-2\nu_{+}=0$, and then

$D_{\omega,\kappa}(2\omega\mathrm{i})=\alpha\kappa(\alpha-2\nu_{-})=\alpha\kappa\Big{(}2\sqrt{m^{2}-\omega^{2}}-2\sqrt{m^{2}-9\omega^{2}}\Big{)}\neq
0;$

the case $\lambda=-2\omega\mathrm{i}$ is treated similarly.

For $\lambda\neq\\{0,\pm 2\omega\mathrm{i}\\}$, we may rewrite (4.19) as

$D_{\omega,\kappa}(\lambda)=(\alpha-2\nu_{+})(\alpha-2\nu_{-})\Big{(}1+\kappa\alpha\Big{(}\frac{1}{\alpha-2\nu_{+}}+\frac{1}{\alpha-2\nu_{-}}\Big{)}\Big{)}.$

It remains to prove that for $\omega\in(0,m)$ and
$\kappa\in(-\infty,0)\cup(0,K_{\omega})$, the equation

$1+\kappa\alpha\Big{(}\frac{1}{\alpha-2\nu_{+}}+\frac{1}{\alpha-2\nu_{-}}\Big{)}=0$

has no solutions $\lambda\in\mathrm{i}{\mathbb{R}}\setminus\\{0,\pm
2\omega\mathrm{i}\\}$. Denoting $\Lambda=\mathrm{i}\lambda$, we rewrite the
above equation as

$1+\kappa\varkappa Q(\Lambda)=0,$ (4.20)

where

$\displaystyle Q(\Lambda)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{m^{2}-\omega^{2}}-\sqrt{m^{2}-(\omega+\Lambda)^{2}}}+\frac{1}{\sqrt{m^{2}-\omega^{2}}-\sqrt{m^{2}-(\omega-\Lambda)^{2}}}$
(4.21) $\displaystyle=$
$\displaystyle\frac{1}{\Lambda}\biggr{[}\frac{\varkappa+\sqrt{\varkappa^{2}-\lambda^{2}-2\omega\Lambda}}{\Lambda+2\omega}+\frac{\varkappa+\sqrt{\varkappa^{2}-\Lambda^{2}+2\omega\Lambda}}{\Lambda-2\omega}\biggr{]}.$

Due to (4.17) it suffices to prove that there are no solution to (4.20) in the
following domains:

(1) $0<\omega<m/3$, $\ \kappa<T_{\kappa}(\omega)<0$, $\
2\omega<\Lambda<m-\omega$;

(2) $0<\omega<m/3$, $\ \kappa<T_{\kappa}(\omega)<0$, $\ 0<\Lambda<2\omega$;

(3) $m/3<\omega<m$, $\ 0<\kappa<T_{\kappa}(\omega)$, $\ 0<\Lambda<m-\omega$.

(1) Note that $Q(\Lambda)>0$ for $2\omega<\Lambda<m-\omega$, and

$Q^{\prime}(\Lambda)=-\sum\limits_{\pm}\frac{\Lambda\pm\omega}{\sqrt{m^{2}-(\Lambda\pm\omega)^{2}}\,\Big{(}\varkappa-\sqrt{m^{2}-(\Lambda\pm\omega)^{2}}\Big{)}^{2}}<0.$

Hence,

$\displaystyle\inf\limits_{(2\omega,m-\omega)}Q(\Lambda)$ $\displaystyle=$
$\displaystyle
Q(m-\omega)=\frac{1}{\varkappa}+\frac{1}{\varkappa-\sqrt{m^{2}-(2\omega-m)^{2}}}=\frac{1}{\varkappa}\frac{2\sqrt{m^{2}-\omega^{2}}-2\sqrt{\omega(m-\omega)}}{\sqrt{m^{2}-\omega^{2}}-2\sqrt{\omega(m-\omega)}}$
(4.22) $\displaystyle=$
$\displaystyle\frac{1}{\varkappa}\frac{2\sqrt{m+\omega}-2\sqrt{\omega}}{\sqrt{m+\omega}-2\sqrt{\omega}}=\frac{1}{\varkappa}\frac{2-2\sqrt{\frac{\omega}{m+\omega}}}{1-2\sqrt{\frac{\omega}{m+\omega}}}=\frac{1}{\varkappa}|K_{\omega}|^{-1}>0.$

and

$|\kappa|\varkappa Q(\Lambda)>|K_{\omega}|\varkappa Q(m-\omega)=1.$

Therefore (4.20) has no solutions in the first domain.

(2) Note that $0<\Lambda<\min\\{2\omega,m-\omega\\}$ in the second and the
third domains. Hence $|\Lambda(\Lambda\pm 2\omega)|<\varkappa^{2}$, and the
following expansion holds:

$\displaystyle\sqrt{m^{2}-(\omega\pm\Lambda)^{2}}$ $\displaystyle=$
$\displaystyle\sqrt{\varkappa^{2}-\Lambda(\Lambda\pm
2\omega)}=\varkappa\bigg{(}1-\frac{\Lambda(\Lambda\pm
2\omega)}{2\varkappa^{2}}-\sum_{n=2}^{\infty}\frac{(2n-3)!!\Lambda^{n}(\Lambda\pm
2\omega)^{n}}{n!2^{n}\varkappa^{2n}}\bigg{)}.$

The above leads to

$\frac{\varkappa+\sqrt{m^{2}-(\omega\pm\Lambda)^{2}}}{\Lambda(\Lambda\pm
2\omega)}=\frac{2\varkappa}{\Lambda(\Lambda\pm
2\omega)}-\frac{1}{2\varkappa}-\sum_{n=2}^{\infty}\frac{(2n-3)!!\Lambda^{n-1}(\Lambda\pm
2\omega)^{n-1}}{n!2^{n}\varkappa^{2n-1}},$

and then we derive from (4.21):

$\displaystyle Q(\Lambda)$ $\displaystyle=$
$\displaystyle\frac{\varkappa+\sqrt{m^{2}-(\omega+\Lambda)^{2}}}{\Lambda(\Lambda+2\omega)}+\frac{\varkappa+\sqrt{m^{2}-(\omega-\Lambda)^{2}}}{\Lambda(\Lambda-2\omega)}$
(4.23) $\displaystyle=$
$\displaystyle-\frac{1}{\varkappa}-\frac{4\varkappa}{4\omega^{2}-\Lambda^{2}}-\sum_{n=1}^{\infty}\frac{(2n-1)!!\Lambda^{n}\big{(}(\Lambda+2\omega)^{n}+(\Lambda-2\omega)^{n}\big{)}}{(n+1)!2^{n+1}\varkappa^{2n+1}}<0$

since each summand in $\sum\limits_{n=1}^{\infty}$ is positive for
$0<\Lambda<2\omega$. This immediately implies that (4.20) has no solution in
the second domain.

(3) In the third domain $m-\omega<2\omega$. Let us show that $Q(\Lambda)$
decreases monotonically on $(0,m-\omega)$. Indeed, (4.23) implies

$\displaystyle Q^{\prime}(\Lambda)$ $\displaystyle=$
$\displaystyle-\frac{8\Lambda\varkappa}{(4\omega^{2}-\Lambda^{2})^{2}}-\sum_{n=1}^{\infty}\frac{(2n-1)!!n\Lambda^{n-1}\big{(}(\Lambda+2\omega)^{n}+(\Lambda-2\omega)^{n}\big{)}}{(n+1)!2^{n+1}\varkappa^{2n+1}}$
$\displaystyle-$
$\displaystyle\sum_{n=1}^{\infty}\frac{(2n-1)!!n\Lambda^{n}\big{(}(\Lambda+2\omega)^{n-1}+(\Lambda-2\omega)^{n-1}\big{)}}{(n+1)!2^{n+1}\varkappa^{2n+1}}<0.$

Hence, similarly to (4.22),

$\sup\limits_{(0,m-\omega)}|Q(\Lambda)|=|Q(m-\omega)|=\frac{1}{\sqrt{m^{2}-(2\omega-m)^{2}}-\varkappa}-\frac{1}{\varkappa}=\frac{1}{\varkappa}K_{\omega}^{-1},$

and

$\kappa\varkappa|Q(\Lambda)|<K_{\omega}\varkappa|Q(m-\omega)|=1.$

Hence (4.20) has no solutions in this domain. ∎

Now we collect all the facts about the eigenvalues obtained above.

######  Theorem 4.5 (Eigenvalues of $\mathbf{A}(\omega,\kappa)$).

  1. 1.

For $\kappa=\omega^{2}/m^{2}$, one has
$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,\kappa))=\\{0\\}$ of algebraic
multiplicity $4$; its geometric multiplicity is $1$ if $\omega\neq 0$ and $2$
if $\omega=0$;

  2. 2.

For $\omega\neq 0$, $\kappa=0$, one has
$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,0))=\\{0,\,\pm 2\omega\mathrm{i}\\}$
($\pm 2\omega\mathrm{i}\in\sigma_{\mathrm{ess}}(\mathbf{A}(\omega,\kappa))$
when $|\omega|\geq m/3$);

  3. 3.

For $\kappa>\omega^{2}/m^{2}$, one has
$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,\kappa))=\\{0,\,\pm\lambda\\}$ with
some $\lambda>0$;

  4. 4.

For $\kappa\in(K_{\omega},\omega^{2}/m^{2})$, one has
$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,\kappa))=\\{0,\pm\mathrm{i}\Lambda\\}$,
$\Lambda>0$;

  5. 5.

$\kappa\leq K_{\omega}$, $\kappa\neq 0$, one has
$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,\kappa))=\\{0\\}$.

Figure 2:  Location of simple eigenvalues $\pm\lambda$ for different values of
parameters $\kappa\in{\mathbb{R}}$ and $\omega\in(-m,m)$. The values of
$\omega$ on the Kolokolov curve $\Omega_{\kappa}$ ($\kappa=\omega^{2}/m^{2}$)
correspond to collision of two simple eigenvalues at $\lambda=0$ as indicated
by the Kolokolov condition. The values of $\omega$ on the virtual level curve
$\mathcal{T}_{\kappa}$ (blue curves for $\kappa\in(-1/2,1/\sqrt{2}$)
correspond to virtual levels at thresholds $\pm\mathrm{i}(m-|\omega|)$. The
regions between the Kolokolov and virtual level lines (dots on the plot)
correspond to two simple purely imaginary eigenvalues $\pm\mathrm{i}\Lambda$
in the spectral gap. The values $|\omega|\geq m/3$ at $\kappa=0$ correspond to
embedded eigenvalues $\pm 2\omega\mathrm{i}$. Besides eigenvalues mentioned on
this plot, there is always eigenvalue $\lambda=0$.

###### Proof.

Part (1) in the case $\omega\neq 0$ follows from Lemma 3.7 (5) and from Lemma
4.2 in the case $\omega=0$. Moreover, in the case $\omega\neq 0$, one can
arrive at the same conclusion from considering the function
$D_{\omega,\kappa}(\lambda)$. We have:

$\displaystyle
D_{\omega,\kappa}(\lambda)=(\alpha-2\nu_{+})(\alpha-2\nu_{-})(1+\kappa\varkappa
Q(\mathrm{i}\lambda)).$ (4.24)

We notice that

$\alpha-2\nu_{\pm}=2(\varkappa-\sqrt{\varkappa^{2}+\lambda\pm
2\mathrm{i}\lambda\omega})=O(\lambda),\quad\lambda\to 0.$

Moreover, by (4.23),

$Q(\mathrm{i}\lambda)=-\frac{1}{\varkappa}-\frac{\varkappa}{\omega^{2}(1+\frac{\lambda^{2}}{4\omega^{2}})}+O(\lambda^{2})=-\frac{1}{\varkappa}-\frac{\varkappa}{\omega^{2}}+O(\lambda^{2})=-\frac{m^{2}}{\varkappa\omega^{2}}+O(\lambda^{2}),\quad|\lambda|<2|\omega|,$

therefore,

$1+\kappa\varkappa Q(\mathrm{i}\lambda)=1-\frac{\kappa
m^{2}}{\omega^{2}}+O(\lambda^{2})=\left\\{\begin{array}[]{cc}O(1),\quad\omega\neq\Omega_{\kappa},\\\
O(\lambda^{2}),\quad\omega=\Omega_{\kappa},\end{array}\right.\qquad|\lambda|\leq
2|\omega|.$

It follows that if $\omega=\Omega_{\kappa}$, then $\lambda=0$ is a root of
(4.24) of order four.

Part (2) follows from Lemma 4.2, and Part (3) follows from Lemma 3.7 (3). Let
us prove Part (4). By Part (1), for $\omega=\Omega_{\kappa}$,
$\sigma_{\mathrm{p}}(\mathbf{A}(\omega,\kappa))=\\{0\\}$ of algebraic
multiplicity $4$. For $\omega\gtrless\Omega_{\kappa}$, the two nonzero
eigenvalues start moving away from the origin, becoming purely real for
$\omega<\Omega_{\kappa}$ and purely imaginary for $\omega>\Omega_{\kappa}$.
These two purely imaginary eigenvalues hit the threshold points
$\pm\mathrm{i}(m-|\omega|)$ at some values $\pm\mathcal{T}_{\kappa}$, with
$\Omega_{\kappa}<\mathcal{T}_{\kappa}<m$, becoming virtual levels (see Lemma
4.1), since by Part (1), there are no eigenvalues at this threshold. It
remains to note that $\omega=\mathcal{T}_{\kappa}$ corresponds to
$\kappa=K_{\omega}$. Indeed, solving (4.12), we get

$\kappa=K_{\omega}=\frac{1}{2}\Big{(}\sqrt{\frac{\omega^{2}}{m^{2}}+\frac{\omega}{m}}+\frac{\omega}{m}-1\Big{)}=\frac{2\Big{(}\frac{|\omega|}{m+|\omega|}\Big{)}^{1/2}-1}{2-2\Big{(}\frac{|\omega|}{m+|\omega|}\Big{)}^{1/2}}.$

The condition for $\mathcal{T}_{\kappa}$ to be smaller than $m$ (so that the
root of $D_{\omega,\kappa}(\lambda)$ indeed arrives at
$\mathrm{i}(m-|\omega|)$ when $\omega=\pm\mathcal{T}_{\kappa}\in(-m,m)$) is
$(1+2\kappa)^{2}<3+4\kappa$, which gives the requirement
$\kappa\in(-\frac{1}{2},\frac{1}{\sqrt{2}})$.

Part (5) follows from Lemma 4.4. The absence of nonzero roots for
$\kappa<K_{\omega}$, $\kappa\neq 0$, is explained by the fact that that when
we passing the thresholds $\kappa=K_{\omega}$ the nonzero root
$\lambda(\omega,\kappa)$ of $D_{\omega,\kappa}(\lambda)$ moves onto the
unphysical sheet of the Riemann surface of $D_{\omega,\kappa}(\lambda)$.
Recall that in the case $\kappa=0$, $|\omega|\geq m/3$ the point spectrum also
contains embedded eigenvalues $\pm 2\omega\mathrm{i}$ by Lemma 4.1 1).

This completes the proof of Theorem 4.5. ∎

##  Appendix A Appendix: Reducing $D_{\omega,\kappa}(\lambda)=0$ to a cubic
with explicit solution

For the completeness, we provide an explicit solution to the equation
$D_{\omega,\kappa}(\lambda)=0$, with $D_{\omega,\kappa}(\lambda)$ from (4.10),
which we can cast as a cubic equation. Albeit explicit, this solution does not
yield the spectral properties of $\mathbf{A}(\omega,\kappa)$ as readily as the
methods employed in Section 4; yet this solution enables, for example, the
analysis of large $\kappa$ asymptotics of roots of
$D_{\omega,\kappa}(\lambda)$ on physical and unphysical sheets of the Riemann
surface of $D_{\omega,\kappa}(\lambda)$. Denote $\varSigma=\nu_{+}+\nu_{-}$,
with $\nu_{\pm}=\nu_{\pm}(\omega,\lambda)$ from (4.5); one has

$-2\nu_{+}\nu_{-}=-\varSigma^{2}+(\nu_{+}^{2}+\nu_{-}^{2})=-\varSigma^{2}+(2m^{2}-2\omega^{2}+2\lambda^{2})=-\varSigma^{2}+2\lambda^{2}+\alpha^{2}/2.$

Then (4.9) takes the form

$\alpha^{2}(1+\kappa)^{2}-2\alpha(1+\kappa)\varSigma-2(-\varSigma^{2}+2\lambda^{2}+\alpha^{2}/2)-\alpha^{2}\kappa^{2}=0$
(A.1)

and then
$\varSigma^{2}-\alpha(1+\kappa)\varSigma+(\alpha^{2}\kappa-2\lambda^{2})=0.$
Equation (A.1) implies

$\varSigma={\big{(}\alpha(1+\kappa)\pm\sqrt{\alpha^{2}(1-\kappa)^{2}+8\lambda^{2}}\,\big{)}}/{2}.$
(A.2)

Further, from (A.1) and (A.2),

$\displaystyle-2\nu_{+}\nu_{-}$ $\displaystyle=$
$\displaystyle-\varSigma^{2}+2\lambda^{2}+\alpha^{2}/2=\alpha^{2}(1+\kappa)^{2}/2-\alpha^{2}\kappa^{2}/2-\alpha(1+\kappa)\varSigma$
$\displaystyle=$
$\displaystyle\frac{\alpha^{2}(1+\kappa)^{2}}{2}-\frac{\alpha^{2}\kappa^{2}}{2}-\frac{\alpha(1+\kappa)}{2}\Big{(}\alpha(1+\kappa)\pm\sqrt{\alpha^{2}(1-\kappa)^{2}+8\lambda^{2}}\Big{)}$
$\displaystyle=$
$\displaystyle-\frac{\alpha^{2}\kappa^{2}}{2}\mp\frac{\alpha(1+\kappa)}{2}\sqrt{\alpha^{2}(1-\kappa)^{2}+8\lambda^{2}},$

so

$\displaystyle\nu_{+}^{2}\nu_{-}^{2}=\frac{\alpha^{4}\kappa^{4}}{16}+\frac{\alpha^{2}(1+\kappa)^{2}}{16}\big{(}\alpha^{2}(1-\kappa)^{2}+8\lambda^{2}\big{)}\pm\frac{\alpha^{3}(1+\kappa)\kappa^{2}}{8}\sqrt{\alpha^{2}(1-\kappa)^{2}+8\lambda^{2}}.$
(A.3)

On the other side, the definition of $\nu_{\pm}$ (4.5) implies that

$\displaystyle\nu_{+}^{2}\nu_{-}^{2}$ $\displaystyle=$
$\displaystyle((\omega+\mathrm{i}\lambda)^{2}-m^{2})((\omega-\mathrm{i}\lambda)^{2}-m^{2})=((\omega-m)^{2}+\lambda^{2})((\omega+m)^{2}+\lambda^{2})$
(A.4) $\displaystyle=$
$\displaystyle(\omega^{2}-m^{2})^{2}+\lambda^{4}+2\lambda^{2}(\omega^{2}+m^{2})=\frac{\alpha^{4}}{16}+\lambda^{4}+4\lambda^{2}m^{2}-\frac{\lambda^{2}\alpha^{2}}{2}.$

From (A.3) and (A.4), simplifying and denoting $x=\lambda^{2}$, we obtain:

$\displaystyle
x^{2}+4xm^{2}-x\alpha^{2}-\frac{\alpha^{4}\kappa^{4}}{8}+\frac{\alpha^{4}\kappa^{2}}{8}-\frac{x\alpha^{2}\kappa^{2}}{2}-x\alpha^{2}\kappa=\pm\frac{\alpha^{3}(1+\kappa)\kappa^{2}}{8}\sqrt{\alpha^{2}(1-\kappa)^{2}+8x}.$
(A.5)

Defining

$\displaystyle
c(\omega,\kappa)=4m^{2}-\alpha^{2}-\alpha^{2}\kappa-{\alpha^{2}\kappa^{2}}/{2}$
(A.6)

and squaring (A.5) yields

$\Big{(}x^{2}+cx+\frac{\alpha^{4}\kappa^{2}(1-\kappa^{2})}{8}\Big{)}^{2}=\frac{\alpha^{8}(1-\kappa^{2})^{2}\kappa^{4}}{64}+\frac{\alpha^{6}(1+\kappa)^{2}\kappa^{4}x}{8}.$

There is a root $x=0$. To find nonzero roots, we simplify the above and cancel
$x$, arriving at

$\displaystyle
x^{3}+2cx^{2}+x\Big{[}c^{2}+\frac{\alpha^{4}\kappa^{2}(1-\kappa^{2})}{4}\Big{]}+c\frac{\alpha^{4}\kappa^{2}(1-\kappa^{2})}{4}-\frac{\alpha^{6}(1+\kappa)^{2}\kappa^{4}}{8}=0.$
(A.7)

Writing

$\displaystyle x=y-{2c}/{3},$ (A.8)

we reduce the cubic equation (A.7) to the form

$\displaystyle y^{3}+py+q=0$ (A.9)

with

$\displaystyle
p=-\frac{c^{2}}{3}+\frac{\alpha^{4}\kappa^{2}(1-\kappa^{2})}{4},\qquad
q=-\frac{2c^{3}}{27}+\frac{c\alpha^{4}\kappa^{2}(1-\kappa^{2})}{12}-\frac{\alpha^{6}(1+\kappa)^{2}\kappa^{4}}{8}.$
(A.10)

If the discriminant

$\Delta(\omega,\kappa)=-4p^{3}-27q^{2}$ (A.11)

with $c=c(\omega,\kappa)$ from (A.6), is negative, then there is exactly one
real root $y\in{\mathbb{R}}$ of equation (A.9) (and two complex conjugate
roots with nonzero imaginary part). The real root is given by
$y_{1}=\big{(}-\frac{q}{2}+\frac{(-\Delta)^{1/2}}{108}\big{)}^{1/3}+\big{(}-\frac{q}{2}-\frac{(-\Delta)^{1/2}}{108}\big{)}^{1/3},$
and then (A.8) yields

$\displaystyle
x_{1}=\Big{(}-\frac{q}{2}+\frac{(-\Delta)^{1/2}}{108}\Big{)}^{1/3}+\Big{(}-\frac{q}{2}-\frac{(-\Delta)^{1/2}}{108}\Big{)}^{1/3}-\frac{2c}{3}.$
(A.12)

Let us mention that roots of (A.9) with nonzero imaginary part correspond to
roots $\lambda=\pm\sqrt{y-2c/3}$ of $D_{\omega,\kappa}(\lambda)$ with nonzero
real and imaginary parts. By Lemma 3.7 (3), these roots do not correspond to
eigenvalues of $\mathbf{A}(\omega,\kappa)$.

We are going to show that the discriminant $\Delta(\omega,\kappa)$ is negative
either if $|\omega|<m$ is sufficiently close to $m$ or if $|\kappa|$ is
sufficiently large, so that there is exactly one real solution to (A.9) and
therefore exactly one pair of roots to $D_{\omega,\kappa}(\lambda)=0$ which
are located on the Riemann sheet corresponding to eigenvalues of
$\mathbf{A}(\omega,\kappa)$ (that is, when the real parts of $\nu_{+}$ and
$\nu_{-}$ from (4.5) are positive).

######  Lemma A.1.

  1. 1.

There is $\kappa_{0}>0$ such that $\Delta(\omega,\kappa)<0$ for
$|\kappa|>\kappa_{0}$.

  2. 2.

For each $\kappa\in{\mathbb{R}}\setminus\\{-1,0\\}$, there is
$\omega_{\kappa}\in(0,m)$ such that $\Delta(\omega,\kappa)<0$ for
$|\omega|\in(\omega_{\kappa},m)$.

###### Proof.

Using (A.10), one computes:

$\displaystyle\hskip-20.0pt\Delta(\omega,\kappa)=-4p^{3}-27q^{2}$
$\displaystyle\hskip-20.0pt=-\frac{\alpha^{6}\kappa^{4}(1+\kappa)^{2}}{2}\left(c^{3}-\frac{c^{2}\alpha^{2}(1-\kappa)^{2}}{8}-\frac{9c\alpha^{4}\kappa^{2}(1-\kappa^{2})}{8}+\frac{\alpha^{6}\kappa^{2}(1-\kappa^{2})(1-\kappa)^{2}}{8}+\frac{27\alpha^{6}(1+\kappa)^{2}\kappa^{4}}{32}\right).$

Since $c=4m^{2}-\alpha^{2}(1+\kappa+\kappa^{2}/2)$, one concludes that, for
each fixed $\kappa\in{\mathbb{R}}$ different from $-1$ and $0$,
$\Delta(\omega,\kappa)<0$ if $\alpha=2\sqrt{m^{2}-\omega^{2}}$ is sufficiently
small (that is, if $|\omega|$ is sufficiently close to $m$).

Alternatively, one can keep the highest order powers of $\kappa$, substituting
$c\sim-\alpha^{2}\kappa^{2}/2$ and getting

$\Delta(\omega,\kappa)\sim-\alpha^{6}\kappa^{6}\Big{(}-\frac{1}{2}\frac{\alpha^{6}\kappa^{6}}{8}-\frac{\alpha^{4}\kappa^{4}}{4}\frac{\alpha^{2}\kappa^{2}}{16}+\frac{\alpha^{2}\kappa^{2}}{2}\frac{9\alpha^{4}\kappa^{4}}{16}+\frac{\alpha^{6}\kappa^{6}}{16}+\frac{27\alpha^{6}\kappa^{6}}{64}\Big{)}=-\frac{\alpha^{12}\kappa^{12}}{8},$

showing that $\Delta(\omega,\kappa)$ is strictly negative for $|\kappa|$ large
enough. ∎

######  Remark A.2.

Let us note that for $|\kappa|\to\infty$, one has:

$c\sim-\frac{\alpha^{2}\kappa^{2}}{2},\qquad(-\Delta)^{1/2}\sim\frac{\alpha^{6}\kappa^{6}}{2\sqrt{2}},\qquad
q\sim-\frac{2}{27}c^{3}-\frac{\alpha^{6}\kappa^{6}}{8}\sim-\frac{29}{27}\frac{\alpha^{6}\kappa^{6}}{8},$

and (A.12) yields

$x_{1}\sim\Big{(}\frac{\alpha^{6}\kappa^{6}}{8}+\frac{1}{108}\frac{\alpha^{6}\kappa^{6}}{2\sqrt{2}}\Big{)}^{1/3}+\Big{(}\frac{\alpha^{6}\kappa^{6}}{8}-\frac{1}{108}\frac{\alpha^{6}\kappa^{6}}{2\sqrt{2}}\Big{)}^{1/3}+\frac{2}{3}\frac{\alpha^{2}\kappa^{2}}{2}\sim
c\alpha^{2}\kappa^{2},$

with some $c>1$, showing that $\lambda_{1,2}=\pm\sqrt{x_{1}}$ are real. For
$\kappa>0$ these values $\lambda_{1,2}$ are exactly the two real nonzero
eigenvalues of $\mathbf{A}(\omega,\kappa)$ which appear in Theorem 4.5 (3).
For $\kappa<0$, since $\mathbf{A}(\omega,\kappa)$ has no real eigenvalues
(Lemma 3.7 (3)), $\lambda_{1,2}$ correspond to resonances of
$\mathbf{A}(\omega,\kappa)$ (that is, to zeros of (4.10) on one of the
unphysical Riemann sheets of $D_{\omega,\kappa}(\lambda)$).

The expression (A.12) can be used to analyze zero eigenvalues of
$\mathbf{A}(\omega,\kappa)$.

######  Lemma A.3.

$x_{1}=0$ in the following cases: $\kappa=-1$, $\kappa=0$, and
$\kappa=\omega^{2}/m^{2}$.

###### Proof.

Since $x_{1}=0$ corresponds to $y_{1}=2c/3$ (see (A.8)), we need to solve

$(2c/3)^{3}+(2c/3)p+q=0.$

Substituting $c$ from (A.6), $p$, $q$ from (A.10), and simplifying, one
arrives at

$\Big{(}4m^{2}-(1+\kappa+\kappa^{2}/2)\alpha^{2}\Big{)}\frac{\alpha^{4}\kappa^{2}(1-\kappa^{2})}{4}-\frac{\alpha^{6}(1+\kappa)^{2}\kappa^{4}}{8}=0.$

Canceling common factors (corresponding to solutions $\kappa=-1$ and
$\kappa=0$), one obtains $\omega^{2}=\kappa m^{2}$. ∎

######  Remark A.4.

The value $\kappa=\omega^{2}/m^{2}$ is in agreement with the value of
$\Omega_{\kappa}$ from (3.18). By Lemma 3.7, we know that only the cases
$\kappa=0$ and $\kappa=\omega^{2}/m^{2}$ correspond to higher algebraic
multiplicity of eigenvalue $\lambda=0$, while the case $\kappa=-1$ has to
correspond to resonances: one can see from (4.10) that in this case the values
$\lambda=0$ are located on the Riemann sheet of $D_{\omega,\kappa}(\lambda)$
characterized by $\nu_{\pm}<0$.

## References

  * [BKKS08] V. Buslaev, A. Komech, E. Kopylova, and D. Stuart, On asymptotic stability of solitary waves in Schrödinger equation coupled to nonlinear oscillator, Communications in partial differential equations 33 (2008), pp. 669–705.
  * [CGOR19] E. Csobo, F. Genoud, M. Ohta, and J. Royer, Stability of standing waves for a nonlinear Klein–Gordon equation with delta potentials, Journal of Differential Equations 268 (2019), pp. 353–388.
  * [CP03] A. Comech and D. Pelinovsky, Purely nonlinear instability of standing waves with minimal energy, Comm. Pure Appl. Math. 56 (2003), pp. 1565–1607.
  * [EGT19] M. B. Erdoğan, W. R. Green, and E. Toprak, Dispersive estimates for Dirac operators in dimension three with obstructions at threshold energies, American Journal of Mathematics 141 (2019), pp. 1217–1258.
  * [GN20] F. Gesztesy and R. Nichols, On absence of threshold resonances for Schrödinger and Dirac operators, Discrete & Continuous Dynamical Systems-S (2020).
  * [GSS87] M. Grillakis, J. Shatah, and W. Strauss, Stability theory of solitary waves in the presence of symmetry. I, J. Funct. Anal. 74 (1987), pp. 160–197.
  * [JK79] A. Jensen and T. Kato, Spectral properties of Schrödinger operators and time-decay of the wave functions, Duke Math. J. 46 (1979), pp. 583–611.
  * [JN01] A. Jensen and G. Nenciu, A unified approach to resolvent expansions at thresholds, Rev. Math. Phys. 13 (2001), pp. 717–754.
  * [KK07] A. Komech and A. Komech, Global attractor for a nonlinear oscillator coupled to the Klein–Gordon field, Arch. Ration. Mech. Anal. 185 (2007), pp. 105–142.
  * [KK09] A. Komech and A. Komech, Global attraction to solitary waves for Klein–Gordon equation with mean field interaction, Ann. Inst. H. Poincaré Anal. Non Linéaire 26 (2009), pp. 855–868.
  * [KKS12] A. Komech, E. Kopylova, and D. Stuart, On asymptotic stability of solitons in a nonlinear Schrödinger equation, Communications on Pure and Applied Analysis 11 (2012), pp. 1063–1079.
  * [Kol73] A. A. Kolokolov, Stability of the dominant mode of the nonlinear wave equation in a cubic medium, J. Appl. Mech. Tech. Phys. 14 (1973), pp. 426–428.
  * [Kop09] E. Kopylova, On the asymptotic stability of solitary waves in the discrete Schrödinger equation coupled to a nonlinear oscillator, Nonlinear Analysis: Theory, Methods & Applications 71 (2009), pp. 3031–3046.
  * [Kop10] E. Kopylova, On asymptotic stability of solitary waves in discrete Klein–Gordon equation coupled to a nonlinear oscillator, Applicable Analysis 89 (2010), pp. 1467–1492.
  * [Yaf10] D. R. Yafaev, Mathematical scattering theory, vol. 158 of Mathematical Surveys and Monographs, American Mathematical Society, Providence, RI, 2010.