text:
\begin{document}

\title{Spectral stability and instability of solitary waves
of the Dirac equation with concentrated nonlinearity}

\author{
{\sc Nabile Boussa{\"\i}d}
\\
{\it\small Universit\'e Bourgogne Franche-Comt\'e, 25030 Besan\c{c}on CEDEX, France}
\\~\\
{\sc Claudio Cacciapuoti}
\\
{\it\small DiSAT, Sezione di Matematica, Universit\`a dell'Insubria, via Valleggio 11, I-22100 Como, Italy}
\\~\\
{\sc Raffaele Carlone}
\\
{\it\small Dipartimento di Matematica e Applicazioni R. Caccioppoli,  Universit\`a Federico II di Napoli,}
\\
{\it\small MSA, via Cinthia, I-80126, Napoli, Italy}

\\~\\
{\sc Andrew Comech}
\\
{\it\small Texas A\&M University, College Station, TX, USA}

\\~\\
{\sc Diego Noja}
\\
{\it\small Dipartimento di Matematica e Applicazioni, Universit\`a di Milano Bicocca,}
\\
{\it\small via R. Cozzi 55, I-20126 Milano, Italy}

\\~\\
{\sc Andrea Posilicano}
\\
{\it\small DiSAT, Sezione di Matematica, Universit\`a dell'Insubria, via Valleggio 11, I-22100 Como, Italy}

}

\date{\version}

\maketitle

\begin{abstract}

We consider the nonlinear Dirac equation with Soler-type nonlinearity
concentrated at one point
and present a detailed study of the spectrum of linearization at solitary waves.
We then consider two different perturbations of the nonlinearity which break the $\mathbf{SU}(1,1)$-symmetry:
the first preserving and the second breaking the parity symmetry.
We show that a perturbation
which breaks the $\mathbf{SU}(1,1)$-symmetry
but not the parity symmetry also preserves the spectral stability
of solitary waves.
Then we consider a perturbation
which breaks both the $\mathbf{SU}(1,1)$-symmetry
and the parity symmetry and show that this perturbation
destroys the stability of weakly relativistic solitary waves.
The developing instability is due to the bifurcations of positive-real-part eigenvalues
from the embedded eigenvalues $\pm 2\omega\jj$.
\end{abstract}

\section{Introduction}

In this article, we consider a nonlinear Dirac equation (NLD)
in one dimension
with a nonlinearity concentrated at a point,
\begin{eqnarray}\label{nld-point-0}
\jj \p_t\psi=D_m\psi
-\delta(x)f(\psi\sp\ast\sigma_3\psi)\sigma_3\psi,
\qquad
\psi(t,x)\in\C^2,
\quad
x\in\R,
\quad
t\in\R,
\end{eqnarray}
and study stability of its solitary wave solutions.
We also consider
how this stability is affected by certain perturbations.
Above, the free Dirac operator
in one spatial dimension
is taken in the form
\[
D_m=\jj\sigma_2\p_x+m\sigma_3
=\begin{bmatrix}m&\p_x\\-\p_x&-m\end{bmatrix},
\qquad
m>0\textup;
\]
the standard Pauli matrices are given by
\[
\sigma_{1}=\begin {bmatrix}0&1 \\1&0\end {bmatrix},
\qquad
\sigma_{2}=\begin {bmatrix}0&-\jj \\\jj&0\end {bmatrix},
\qquad
\sigma_{3}=\begin {bmatrix}1&0 \\0&-1\end {bmatrix},
\]
and $f$
is a differentiable real-valued function.
The presence of the delta distribution means that the nonlinearity
is supported at the origin only,
from which the name \emph{concentrated nonlinearity} comes.

A rigorous definition of the model is given in Section~\ref{sect-model}, while for a complete and general treatment we refer to \cite{cacciapuoti2017one}. In the usual setting (where the Dirac distribution is missing and the nonlinearity is everywhere distributed), the nonlinearity $f(\psi\sp\ast\sigma_3\psi)\sigma_3\psi$ appearing in \eqref{nld-point-0} defines what is known as Soler model (also called Gross--Neveau model in one spatial dimension, $(1+1)$D); by analogy, the above equation describes a Soler-type concentrated nonlinearity.
We mention that the analysis of various PDEs with concentrated nonlinearities is now a well-developed subject.
Rigorous studies have been performed especially, but not only,
in the Schr\"odinger case (see \cite{adami2001class, adami2003cauchy, noja2005wave, carlone2019well} and references therein).
The well-posedness of the nonlinear Dirac equation (NLD) with concentrated nonlinearity is given
in the already cited \cite{cacciapuoti2017one}
and the extension to quantum graphs has also been considered
in \cite{borrelli2019nonlinear,Carlone2019extended}.

The local and global well-posedness for the NLS with concentrated nonlinearity
is given in \cite{komech2007global,cacciapuoti2014nls},
starting from extended nonlinearities and taking the point limit on solutions.

A similar derivation and well-posedness of the one-dimensional NLD
with concentrated nonlinearity
could be treated along similar lines, although up to now this problem is open.
Our interest in this kind of nonlinearity is raised by the possibility of characterizing explicitly the solitary waves of the model and of giving a fairly complete spectral theory of the linearization around solitary waves. While in the usual Soler model it seems difficult to have complete and definite results on the spectral stability of solitary waves, in the present example, simplified yet nontrivial, spectral stability and instability of some classes of solitary waves can be established.
(We recall that a solitary wave solution $\phi_\omega(x) e^{-\jj\omega t}$ of the NLD is \emph{spectrally stable} if the spectrum of the linearization operator around the solitary wave has no points
in the right half of the complex plane; in the opposite case we say that the solitary wave is linearly unstable.)
Knowledge of the linearization spectrum and in particular the spectral stability of solitary waves is important because it is a fundamental step towards the analysis of their asymptotic stability.
In previous works on asymptotic stability of solitary waves of NLD
\cite{boussaid2006stable,boussaid2008asymptotic,boussaid2012stability,MR2985264,MR3592683}, their spectral stability was either taken as an assumption,
or checked numerically.
For analytical approaches to the spectral stability in NLD, see \cite{berkolaiko2012spectral,boussaid2016spectral,boussaid2017nonrelativistic,boussaid2018spectral,linear-b,aldunate2021results},
and also the monograph \cite{opus}.
Let us mention a recent related article
on the orbital stability of solitary waves
in the Klein--Gordon equation with concentrated
nonlinearity \cite{comech2021orbital},
with the complete analysis of the spectrum
of the linearized equation.

In Section~\ref{sect-model},
after the study of the solitary waves (Lemma~\ref{lemma-sw}),
we treat the spectrum of the linearized system.
Let us give the essence of
our Theorem~\ref{theorem-sect2}
on a particular case of a pure power nonlinearity,
with $f(\tau)=\abs{\tau}^\kappa$, $\kappa>0$.
Considering solitary waves with frequencies in the gap $\omega\in(-m,m)$,
the spectrum of the linearization is as follows:
There are always eigenvalues $\pm 2\omega\jj$
(embedded into the continuous spectrum when $\abs{\omega}>m/3$);
When
$\kappa\in (0,1]$,
the entire spectrum is located on the imaginary axis;
There are two nontrivial eigenvalues when $\kappa\in \big(2^{-1/2},1\big]$
and the frequency satisfies $\omega>\mathcal{T}_\kappa^{+}$, with $\mathcal{T}_\kappa^{+}\in(0,m)$
a certain threshold frequency.
For $\kappa>1$, these two imaginary eigenvalues collide at zero
when $\omega=\Omega_\kappa$,
with $\Omega_\kappa\in(\mathcal{T}_\kappa^{+},m)$ the second threshold value,
and a couple of real eigenvalues appear from this collision
when $\omega\in(\Omega_\kappa,m)$.
This second threshold value is the one corresponding to algebraic multiplicity
of the null space of the linearization jumping from two to four.
This value satisfies the Kolokolov condition \cite{kolokolov-1973}:
$\p_\omega \norm{\phi_\omega}_{L^2}^2$ vanishes at $\omega=\Omega_\kappa$.
The statement and proof of these results are in Section~\ref{sect-spectrum}.
A relevant part of the analysis relies on the parity symmetry of the Soler model,
which allows one to split the Hilbert space into two invariant subspaces:
odd-even-odd-even and even-odd-even-odd subspaces.
In the former subspace live the "trivial" eigenvalues $\pm 2\omega\jj$
and in the latter subspace live the possibly further "nontrivial" eigenvalues.

The presence of real eigenvalues for $\kappa>1$ rules out spectral stability of the corresponding solitary waves. As explained above, for any positive power $\kappa$ and any $\omega\in (-m,m)$, besides eigenvalue $0$ and possible nontrivial eigenvalues referred above, the point spectrum contains purely imaginary eigenvalues $\pm 2\omega\jj$. These eigenvalues are related to the $\mathbf{SU}(1,1)$ of the Soler model (see \cite{galindo1977remarkable})
and to the existence of bi-frequency solitary waves
(see \cite{berkolaiko2012spectral,boussaid2018spectral} and Remark~\ref{remark-bi-frequency}
in the present paper).
It turns out that the spectral stability
of small amplitude solitary waves of the Soler model heavily relies
on the presence of $\pm 2\omega\jj$ eigenvalues in the spectrum
of the linearized equation \cite{linear-b}.
On the other hand, when the symmetry responsible of the $\pm 2\omega\jj$ eigenvalues is broken,
then in principle one could expect that the eigenvalues $\pm 2\omega\jj$
bifurcate off the imaginary axis,
either becoming eigenvalues with nonzero real part,
or turning into resonances,
that is, poles of the resolvent on the unphysical sheet of its Riemann surface.
The second part of the paper is dedicated to the analysis of this issue.
We will consider examples of perturbations of the Soler concentrated nonlinearity
which destroy the $\mathbf{SU}(1,1)$ symmetry;
we are interested in the fate of the eigenvalues $\pm 2\omega \jj$
associated to the $\mathbf{SU}(1,1)$ symmetry.
In Section~\ref{sect-nld-perturbation},
we consider a perturbation which preserves the parity
(the self-interaction is based on the quantity
$\psi\sp\ast(\sigma_3+\epsilon I_2)\psi$, $\epsilon\ne 0$,
instead of $\psi\sp\ast\sigma_3\psi$)
and show that solitary waves remain spectrally stable (if they were stable in the Soler model).
We say that this class of perturbations preserves the parity in the sense that the
operator corresponding to the linearization at a solitary wave is invariant
in odd-even-odd-even and even-odd-even-odd subspaces.
In Section~\ref{sect-nld-perturbation-broken-parity}
we consider a perturbation
when
the self-interaction is based on the quantity $\psi\sp\ast(\sigma_3+\epsilon\sigma_1)\psi$, $\epsilon\ne 0$,
instead of $\psi\sp\ast\sigma_3\psi$.
This perturbation breaks not only $\mathbf{SU}(1,1)$ symmetry,
but also the parity symmetry (in the above sense).
We show that such a perturbation leads to linear instability
of weakly relativistic solitary waves
with $\omega<m$ close enough to $m$.
We point out that in the model under consideration
the $\pm 2\omega\jj$ eigenvalues of the linearized operator,
the ones which are due to the $\mathbf{SU}(1,1)$ symmetry of the model, are simple (in the sense
that they correspond to a one-dimensional eigenspace). Due to the symmetries of the spectrum with respect to the real and imaginary axes,
these two eigenvalues could not bifurcate off the imaginary axis if they were isolated
(this is the case when $\abs{\omega}<m/3$).
The linear instability that we prove in the nonrelativistic regime ($\omega$ is close to $m$)
is only possible since these two eigenvalues are embedded into the essential spectrum:
in this case, an eigenvalue corresponding to one-dimensional eigenspace
can bifurcate to both sides of the imaginary axis.

Let us make a further perspective comment. The $\mathbf{SU}(1,1)$ symmetry is absent
for the physically interesting Dirac--Maxwell system,
of which the nonlinear Dirac equation is an effective reduction
in a suitable approximation.
We presently do not know
whether solitary waves in the Dirac--Maxwell system are spectrally stable.
In this respect,
a further motivation for studying the present problem is to
understand what
could happen to the spectral stability
when the symmetry $\mathbf{SU}(1,1)$ is absent.

\section{The Soler model with concentrated nonlinearity}
\label{sect-model}

We are looking for solitary wave solutions
$\psi(t,x)=\phi(x)e^{-\jj\omega t}$
to the nonlinear Dirac equation with nonlinear point interaction at the origin of Soler type. This reads formally as
\begin{eqnarray}\label{nld-point}
\jj \p_t\psi=D_m\psi
-\delta(x)f(\psi\sp\ast\sigma_3\psi)\sigma_3\psi,
\qquad
\psi(t,x)\in\C^2,
\quad
x\in\R,
\quad
t\in\R,
\end{eqnarray}
with
\begin{eqnarray}\label{def-dm}
D_m=\jj\sigma_2\p_x+\sigma_3 m
=
\begin{bmatrix}m&\p_x
\\-\p_x&-m
\end{bmatrix},

\end{eqnarray}
and with the nonlinearity represented by
\[
f\in C(\R)\cap C^1(\R\setminus\{0\}).
\]

Let us give a formalized version of \eqref{nld-point}.

Denote
$\R_{-}=(-\infty,0)$,
$\R_{+}=(0,\infty)$,
and let
$H_{-}$ and $H_{+}$ be the free Dirac  operators on
$L^{2}(\R_{-},0)\otimes\C^2$ and $L^{2}(\R_{+})\otimes\C^2$,
formally given by $D_m$,
with domains
\[
\dom(H_{-})=H^{1}(\R_{-})\otimes\C^{2},
\qquad
\dom(H_{+})=H^{1}(\R_{+})\otimes\C^{2}.
\]
Denoting by $H_{\circ}$ the restriction of $D_m$ onto the domain $\dom(H_{\circ}):=\{\psi\in H^{1}(\R,\C):\psi(0)=0\}$,  one has that $H_{\circ}$ is closed, symmetric, has defect indices $(2,2)$, and  adjoint $H_{\circ}^{*}=H_{-}\oplus H_{+}$. We define a Dirac operator $H^{\mathrm{nl}}_{f}$ with concentrated nonlinearity
so that the coupling between the jump and the mean value of the spinor function
is given by a nonlinear relation (self-interaction); see~\cite{cacciapuoti2017one}.
To this aim we define the nonlinear domain
\begin{equation}\label{domH}
\dom(H^{\mathrm{nl}}_{f})  =
\left\{\psi \in L^2(\R,\C)\otimes \C^2:\psi \in H^1(\R\setminus\{0\}) \otimes \C^{2},\,
\jj\sigma_{2}[\psi ]_{0}-f(\hat\psi\sp\ast\sigma_3\hat\psi)\sigma_3\hat\psi=0  \right\},
\end{equation}
where the two-component vector
\begin{eqnarray}\label{psi-hat}
\hat\psi:=\frac12(\psi(0^+)+\psi(0^-))
\end{eqnarray}
is the ``mean value'' of the spinor $\psi$ at $x=0$,
and
\begin{eqnarray}\label{psi-zero}
[\psi]_{0}:=\psi(0^+)-\psi(0^-)
\end{eqnarray}
is the jump of the spinor $\psi$ at $x=0$.
The operator
$H^{\mathrm{nl}}_{f}$ is then defined as the restriction
of $H^{*}_{\circ}=H_{-}\oplus H_{+}$ to
the domain $\dom(H^{\mathrm{nl}}_{f})$.
Thus, the Hamiltonian system
\[
\jj\p_t\psi=H^{\mathrm{nl}}_{f}\psi,
\qquad
\psi(t)\in \dom(H^{\mathrm{nl}}_{f}),
\]
is a formalized version of the Soler model with point interaction \eqref{nld-point}.

We will refer to the boundary condition defining the operator domain
$\dom(H^{\mathrm{nl}}_{f})$ from \eqref{domH}
as to the \emph{jump condition},
rewriting it in the form
\begin{equation}\label{jumpeq}

[\psi ]_{0}=f(\hat\psi\sp\ast\sigma_3\hat\psi)\sigma_1\hat\psi.
\end{equation}
\nn{See if we need to add a word on the $\delta$ notation.}
\ak{Not necessary... let us leave it for distribution theory course.}
\subsection{Solitary waves}

Below, we will use the following notations:
\begin{eqnarray}\label{def-def}
\varkappa(\omega)=\sqrt{m^2-\omega^2},
\qquad
\mu(\omega)=\sqrt{\frac{m-\omega}{m+\omega}}.
\end{eqnarray}

First
let us describe all
solitary waves to \eqref{nld-point},
which are defined as solutions of the form
\begin{eqnarray}\label{s-w}
\psi(t,x)=\phi_\omega(x)e^{-\jj\omega t},
\qquad
\phi_\omega\in H^1(\R,\C^2),
\quad
\omega\in\R.
\end{eqnarray}

\begin{lemma}\label{lemma-sw}
\begin{enumerate}
\item
\label{lemma-sw-1}
There are no nonzero solitary waves
with $\omega\in\R\setminus(-m,m)$.
\item
\label{lemma-sw-2}
For
$\omega\in(-m,m)\setminus\{0\}$,
there are two types of solitary waves:
\begin{eqnarray}\label{s-w-1}
\psi(t,x)
=
\begin{bmatrix}a\\a\mu(\omega) \sgn x\end{bmatrix}
e^{-\varkappa(\omega)\abs{x}}e^{-\jj\omega t},
\qquad
\mu(\omega)=\sqrt{\frac{m-\omega}{m+\omega}},
\end{eqnarray}
where $a\in\C$ satisfies the relation
\begin{equation}\label{Eq:ConditionAlpha}
f(\abs{a}^2)=2\mu(\omega),
\end{equation}
and
\begin{eqnarray}\label{s-w-2}
\psi(t,x)
=
\begin{bmatrix}b\mu(\omega)\sgn x\\b\end{bmatrix}
e^{-\varkappa(\omega)\abs{x}}e^{-\jj\omega t},
\end{eqnarray}
where $b\in\C$
satisfies the relation
\[
f(-\abs{b}^2)=\frac{2}{\mu(\omega)}.
\]
\item
\label{lemma-sw-3}
For $\omega=0$, there are solitary waves of the form
\begin{eqnarray}\label{s-w-3}
\psi(x)
=
\begin{bmatrix}
a+b\sgn x\\
b+a\sgn x
\end{bmatrix}e^{-m\abs{x}},
\end{eqnarray}
with $a,\,b\in\C$ satisfying the relation
\[
f(\abs{a}^2-\abs{b}^2)=2.
\]
\end{enumerate}
\end{lemma}

\begin{proof}
The amplitude $\phi(x)$
of a solitary wave $\phi(x)e^{-\jj\omega t}$ is to satisfy,
formally,
\[
(D_m-\omega I_2-\delta(x)f\sigma_3)\phi=0,
\]
where $f=f(\hat\phi^*\sigma_3\hat\phi)$.
On $\R\setminus\{0\}$, one has:
\begin{eqnarray}\label{m-o}
\omega\phi=D_m\phi,
\qquad
\mbox{hence}\quad
\begin{bmatrix}
m-\omega&\p_x\\-\p_x&-m-\omega
\end{bmatrix}\phi(x)=0,
\qquad
x\in\R\setminus\{0\};
\end{eqnarray}
we conclude that for $x\in\R_\pm$
the amplitude $\phi(x)$
is given by $\phi\sb\pm(x)=\bm{v}\sb\pm e^{-\varkappa\sb\pm x }$,
$\bm{v}\sb\pm\in\C^2$ and $\varkappa\sb\pm\in \R^\pm$, which we write as
\begin{eqnarray}\label{phi-p-m}
\phi\sb\pm(x)=\begin{bmatrix}a+b\sgn x\\c+d\sgn x\end{bmatrix}
e^{-\varkappa\sb\pm x},
\qquad
x\in\mathbb{R},
\qquad
a,\,b,\,c,\,d\in\C.
\end{eqnarray}
\nn{Shall we add that for $\omega=\pm m$
then the solution is either constant or linearly growing ?}
\ak{Adding:}
\begin{remark}
If $\omega=m$ and $\omega=-m$,
then, besides constant solutions,
equation \eqref{m-o}
has solutions
$\phi(x)=\begin{bmatrix}2m x\\-1\end{bmatrix}$
and
$\phi(x)=\begin{bmatrix}-1\\2m x\end{bmatrix}$,
respectively;
we do not need to consider them
since they do not belong to $L^2(\R)$.
\end{remark}

Substituting \eqref{phi-p-m} into \eqref{m-o}
leads to the relations
\begin{eqnarray}\label{two-two}
\begin{bmatrix}
m-\omega&-\varkappa_{+}\\\varkappa_{+}&-m-\omega
\end{bmatrix}
\begin{bmatrix}a+b\\c+d\end{bmatrix}=0,
\qquad
\begin{bmatrix}
m-\omega&-\varkappa_{-}\\\varkappa_{-}&-m-\omega
\end{bmatrix}
\begin{bmatrix}a-b\\c-d\end{bmatrix}=0,
\end{eqnarray}
hence $\varkappa\sb\pm^2=m^2-\omega^2$;
we see that one needs to take
$\varkappa_{+}=\varkappa(\omega)=\sqrt{m^2-\omega^2}$,
$\varkappa_{-}=-\varkappa(\omega)$,
and that one also needs to assume that
$\omega\in(-m,m)$ (or else the $L^2$-norm of $\phi$ is infinite
unless $\phi=0$).
Substituting $\varkappa_\pm=\pm\varkappa(\omega)$,
we derive from \eqref{two-two} the relations
\[
c=\frac{\varkappa(\omega)}{m+\omega}b=\mu(\omega) b,
\qquad d=\frac{\varkappa(\omega)}{m+\omega}a=\mu(\omega) a,
\]
with $\mu(\omega)$ from \eqref{def-def}.
Thus,
\begin{eqnarray}\label{phi-a-b}
\phi(x)=\begin{bmatrix}
a+b\sgn x
\\
\mu(\omega)b+\mu(\omega)a\sgn x
\end{bmatrix}e^{-\varkappa(\omega)\abs{x}},
\qquad
x\in\R.
\end{eqnarray}
The jump condition at $x=0$
(that is, \eqref{jumpeq}
with $[\phi]_0=2\begin{bmatrix}b\\\mu(\omega) a\end{bmatrix}$
and $\hat\phi=\begin{bmatrix}a\\\mu(\omega) b\end{bmatrix}$
coming from \eqref{phi-a-b})
takes the form
\begin{eqnarray}\label{str}
2\begin{bmatrix}
b\\\mu(\omega) a
\end{bmatrix}
=f
\begin{bmatrix}
\mu(\omega) b\\a
\end{bmatrix},
\end{eqnarray}
with
$f=f(\tau)$ evaluated at
\begin{eqnarray}\label{at-tau}
\tau:=\hat\phi\sp\ast\sigma_3\hat\phi=\abs{a}^2-\abs{b}^2.
\end{eqnarray}
We conclude from \eqref{str} that
if $\mu\ne 1$ (that is, $\omega\ne 0$),
then either
$b=0$ and $2\mu(\omega)=f(\abs{a}^2)$,
or $a=0$ and $2=f(-\abs{b}^2)\mu(\omega)$.
These two cases correspond to solutions
\eqref{s-w-1} and \eqref{s-w-2},
respectively.

If $\mu(\omega)=1$ (that is, $\omega=0$), then \eqref{str}
will be satisfied if and only if
$a,\,b\in\C$ satisfy
$2=f(\abs{a}^2-\abs{b}^2)$;
this corresponds to the solution \eqref{s-w-3}.
\end{proof}

\begin{remark}\label{remark-bi-frequency}

Just like the standard Soler model
\cite{PhysRevD.1.2766},
equation \eqref{nld-point} has the
$\mathbf{SU}(1,1)$-symmetry:
if $\psi(t,x)$ is a solution, then so is
\[
(A+B\sigma_1\bm{K})\psi(t,x),
\]
where
$A,\,B\in\C$ satisfy $\abs{A}^2-\abs{B}^2=1$
and $\bm{K}:\,\C^2\to\C^2$ is the complex conjugation.
In particular,
if $\phi\sb\omega(x) e^{-\jj\omega t}$
is a solitary wave solution to \eqref{nld-point}, then
there is also a bi-frequency solution
\begin{eqnarray}\label{bi-frequency-waves}
A\phi\sb\omega(x) e^{-\jj\omega t}
+B\phi\sb\omega\sp{C}(x) e^{\jj\omega t}
\qquad
A,\,B\in\C,
\qquad
\abs{A}^2-\abs{B}^2=1,
\end{eqnarray}
with
$

\phi\sb\omega\sp{C}(x):=\sigma_1\bm{K}\phi\sb\omega(x)
$.

For more details on bi-frequency solutions,
see \cite{opus}.

We also note that the $\omega=0$ solitary waves
of the form \eqref{s-w-3},
with $a,\,b\in\C$ satisfying $f(\abs{a}^2-\abs{b}^2)=2$,

with $\abs{a}>\abs{b}$,
can be written as
$(A+B\sigma_1\bm{K})
\begin{bmatrix}a\\a\sgn x\end{bmatrix}e^{-m\abs{x}}$,
with
$A=a/\sqrt{\abs{a}^2-\abs{b}^2}$,
$B=b/\sqrt{\abs{a}^2-\abs{b}^2}$,
and $a_0=\sqrt{\abs{a}^2-\abs{b}^2}$
satisfying
$\abs{A}^2-\abs{B}^2=1$
and $f(\abs{a_0}^2)=2$,
hence their stability properties
(both linear and nonlinear)
follow from the corresponding stability properties of
the solitary wave
$\begin{bmatrix}a_0\\a_0\sgn x\end{bmatrix}e^{-m\abs{x}}$.
Similarly, the stability of solitary waves
of the form \eqref{s-w-3}
in the case $\abs{a}<\abs{b}$
can be reduced to the stability properties
of solitary waves \eqref{s-w-2}.
\end{remark}

\begin{remark}\label{remark-even}

One can see from  \eqref{nld-point}
that if
$\psi(t,x)$ is its solution,
then
$\psi\sp{C}(t,x)
=\sigma_1\bm{K}\psi(t,x)$
is a solution to \eqref{nld-point}
with the nonlinearity represented by
the function $\tilde f(\tau)=f(-\tau)$, $\tau\in\R$,
instead of $f$.
Consequently, if
\begin{eqnarray}\label{psi-1}
\psi(t,x)=\beta\begin{bmatrix}
\mu(\omega)\sgn x\\ 1
\end{bmatrix}
e^{-\jj\omega t},
\qquad
\beta\in\C,
\qquad
f(-\abs{\beta}^2)=2/\mu(\omega),
\end{eqnarray}
is a solitary wave solution to
\eqref{nld-point},
then
\begin{eqnarray}\label{psi-1-c}
\psi\sp{C}(t,x)
=\sigma_1\bm{K}\psi(t,x)
=\beta\begin{bmatrix}
1\\\mu(\omega)\sgn x
\end{bmatrix}
e^{\jj\omega t},
\qquad
\beta\in\C,
\qquad
\tilde f(\abs{\beta}^2)=2/\mu(\omega)=2\mu(-\omega),
\end{eqnarray}
is a solitary wave solution to
\eqref{nld-point}
with $\tilde f(\tau)=f(-\tau)$, $\tau\in\R$,
and
the solitary waves \eqref{psi-1} and \eqref{psi-1-c}
share all their stability properties.
In particular, in the case when
the nonlinearity is represented by $f$ which is even,

if $\phi_\omega(x)e^{-\jj\omega t}$
from \eqref{s-w}
is a solitary wave solution, then so is
$\phi_\omega\sp{C}(x)e^{\jj\omega t}$,
with the same stability properties.

\end{remark}

\subsection{Spectrum of the linearization operator}
\label{sect-spectrum}

\ak{\sout{
The stability of solitary waves \eqref{s-w-1} and \eqref{s-w-2}
is considered in the same way
(we will not consider the stationary solitary wave
\eqref{s-w-3});
}}

In the present work,
we focus on stability of solitary waves
of the form \eqref{s-w-1}:
\ak{\sout{that is, when $\omega\neq0$} [we no longer need this
restriction :):)]}

\begin{eqnarray}\label{solitary-wave}
\psi(t,x)=\phi_\omega(x)e^{-\jj\omega t},
\quad
\mbox{with}
\quad
\phi_\omega(x)=
\alpha
\begin{bmatrix}1\\\mu(\omega)\sgn x\end{bmatrix}e^{-\varkappa(\omega)\abs{x}},
\end{eqnarray}
with
\[
\varkappa(\omega)=\sqrt{m^2-\omega^2},
\qquad
\mu(\omega)=\sqrt{\frac{m-\omega}{m+\omega}},
\]
and with $\alpha>0$ satisfying the
relation
\begin{eqnarray}\label{a-b-f}
f(\alpha^2)=2\mu(\omega).
\end{eqnarray}

We assume that
\[
\alpha>0;
\]
due to
$\mathbf{U}(1)$-invariance of equation \eqref{nld-point},
there is no loss of generality in this assumption.

The spectral stability of solitary waves of the form
\eqref{s-w-2}
is obtained in the same way
(one can use Remark~\ref{remark-even});
for definiteness, we restrict our attention to the solitary
waves of the form \eqref{s-w-1}.

The spectral stability of solitary waves of the form
\eqref{s-w-3}

with $\abs{a}\ne\abs{b}$
follows from Remark~\ref{remark-bi-frequency}.

We will not consider the stability of solitary waves
of the form \eqref{s-w-3} with $\abs{a}=\abs{b}$;
such solitary waves are only present in the case
when $f(0)=2$.

For the future use, let us mention that
for a family of solitary waves with $\omega$ from an interval
of $(-m,m)$, the relation
\eqref{a-b-f}
allows us to compute $\p_\omega\alpha(\omega)$ when $f$ is $C^1$ and its derivative does not vanish except possibly at $\alpha^2=0$:
\begin{eqnarray}\label{p-alpha}
f'(\alpha^2)\alpha\p_\omega\alpha
=\p_\omega\mu(\omega)
=-\frac{m}{(m+\omega)\varkappa}.
\end{eqnarray}

Let us consider the spectrum of the operator
corresponding to the linearization at
the solitary wave $\phi\sb\omega e^{-\jj\omega t}$
from \eqref{solitary-wave}.
We use the Ansatz
\begin{eqnarray}\label{Ansatz-br-0}
\psi(t,x)=(\phi(x)+r(t,x)+\jj s(t,x))e^{-\jj\omega t},
\end{eqnarray}
where
\[
\big(r(t,x),\,s(t,x)\big)\in \R^2\times\R^2.
\]
A substitution of the Ansatz
\eqref{Ansatz-br-0}
into equation \eqref{nld-point} shows that the perturbation
$(r(t,x),\,s(t,x))$
satisfies in the first order the following system:
\begin{eqnarray}\label{l-p-l-m}
\begin{cases}
-\dot s=(D_m-\omega) r
-\delta(x)f\sigma_3\, r
-\delta(x)(\phi\sp\ast\sigma_3\, r)2g\sigma_3\,\phi

,
\\[1ex]

\dot r=(D_m -\omega) s
-\delta(x)f\sigma_3\, s

.
\end{cases}
\end{eqnarray}
Above,
$D_m$ is from \eqref{def-dm}
and $f,\,g\in\R$ are given by
\begin{equation}\label{def-f-g-ch2}
f=f(\alpha^2),
\qquad
g=f'(\alpha^2).
\end{equation}
In \eqref{l-p-l-m},
we abuse notation considering $\delta$
as an operator acting on $L^2(\R)$.
We refer to Remark~\ref{Rem:RigorousNotation} below
for a more rigorous formulation.

We define
\begin{eqnarray}\label{def-kappa}
\kappa=\frac{\alpha^2 f'(\alpha^2)}{f(\alpha^2)}\in\R.
\end{eqnarray}
By \eqref{Eq:ConditionAlpha}, $f(\alpha^2)=2\mu>0$
since $\omega\in(-m,m)$.
Let us point out that the definition \eqref{def-kappa} is compatible
with the pure power case,
\begin{eqnarray}\label{pure-power}
f(\tau)=\abs{\tau}^\kappa,
\qquad
\tau\in\R,
\qquad
\kappa>0.
\end{eqnarray}
\begin{remark}
Nonlinearities giving rise to $\kappa<0$ are not covered from the known well-posedness results (\cite{cacciapuoti2017one}); in this section for completeness we give the analysis of the linearization operator for any value of $\kappa$.
\end{remark}

Using the relation \eqref{a-b-f} and the definition \eqref{def-kappa},
we simplify the system \eqref{l-p-l-m} to
\begin{eqnarray}\label{l-p-l-m-1}
\begin{cases}
-\dot s=\big(D_m-\omega I_2
-2\mu\delta(x)\sigma_3
-4\mu\kappa\delta(x)\Pi_1\big) r
=:L_{+}r,
\\[1ex]
\dot r=\big(D_m -\omega I_2
-2\mu\delta(x)\sigma_3\big) s
=:L_{-}s,
\end{cases}
\end{eqnarray}
with $I_2$ the identity matrix in $\C^2$
and with
\begin{equation}\label{Pi-one}
\Pi_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}.
\end{equation}

In the matrix form,
the linearized system
\eqref{l-p-l-m-1}
can be written as
\begin{eqnarray}\label{nld-point-linearization}
\p_t
\begin{bmatrix}r(t,x)\\s(t,x)\end{bmatrix}
=\bfA
\begin{bmatrix}r(t,x)\\s(t,x)\end{bmatrix},
\qquad
\bfA=\begin{bmatrix}
0&L_{-}\\-L_{+}&0
\end{bmatrix},
\end{eqnarray}
where the operator $\bfA$
is given explicitly by
\begin{eqnarray}\label{def-a-ch2}
\bfA
=
\begin{bmatrix}
0&D_m-\omega I_2-2\mu(\omega)\delta(x)\sigma_3
\\-D_m+\omega I_2
+2\mu(\omega)\delta(x)\sigma_3
+4\mu(\omega)\kappa\delta(x)\Pi_1
&0
\end{bmatrix}.
\end{eqnarray}
We consider $\bfA$ as an operator-valued function
\[
\bfA=\bfA(\omega,\kappa),
\qquad
\omega\in(-m,m),
\quad
\kappa\in\R.
\]

\begin{remark}\label{Rem:RigorousNotation}
Formally, one should consider $\bfA$ as an operator
\[
 \begin{bmatrix}
0&D_m-\omega I_2
\\-D_m+\omega I_2
&0
\end{bmatrix}
\]
acting on $H^1(\R\setminus\{0\},\C^2\times\C^2)$,
with the domain
\begin{equation}
\label{Eq:DomainOfA}
\dom(\bfA)=
\left\{
(r,s)\in
H^1(\R\setminus\{0\},\C^2\times\C^2):
\,
\jj\sigma_2[r]_0
=2\mu(\sigma_3+2\kappa\Pi_1)\hat r,
\quad
\jj\sigma_2 [s]_0=2\mu\sigma_3\hat s  \right\},
\end{equation}
with
\[
\hat r=\frac 1 2\big(r(0^{+})+r(0^{-})\big),
\qquad
[r]_0=r(0^{+})- r(0^{-}),
\]
and similarly for $s$.
\end{remark}

Before we formulate the results,
let us mention that
a \emph{virtual level}
\ac{\sout{(also known as a \emph{threshold resonance})}}
can be defined as a limit point of an eigenvalue family (dependent on a parameter)
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),
\ac{Dec 6, 2021: added:}
when they are referred to as \emph{threshold resonances}.
For more on the phenomenon of virtual levels,
see e.g. \cite{MR544248,MR1841744,MR2598115,erdogan2019dispersive}.
The general theory is developed in \cite{virtual-levels-review,virtual-levels}.

\medskip

We start with the spectra of $L_\pm$.
We consider the closed densely defined operator
\begin{eqnarray}\label{l-kappa}
L(\omega,\kappa)
=
D_m-\omega I_2-2\mu\delta(x)\sigma_3-4\mu\kappa\delta(x)\Pi_1,
\end{eqnarray}
so that
\[
L_{-}=L(\omega,0),
\qquad
L_{+}=L(\omega,\kappa).
\]
Denote
\begin{eqnarray}\label{def-x-e-o-o-e}
\begin{array}{l}
\bfX_{\mbox{\rm\footnotesize even-odd}}
=L^2_{\mbox{\rm\footnotesize even}}(\R,\C)
\times L^2_{\mbox{\rm\footnotesize odd}}(\R,\C)\subset L^2(\R,\C^2),
\\[2ex]
\bfX_{\mbox{\rm\footnotesize odd-even}}
=L^2_{\mbox{\rm\footnotesize odd}}(\R,\C)
\times L^2_{\mbox{\rm\footnotesize even}}(\R,\C)\subset L^2(\R,\C^2),
\end{array}
\end{eqnarray}
with
$L^2_{\mbox{\rm\footnotesize odd}}$
and
$L^2_{\mbox{\rm\footnotesize even}}$
the subspaces of $L^2$ consisting of odd and even functions of $x\in\R$,
respectively;
note that these are invariant subspaces for the operator $L(\omega,\kappa)$
while $L^2(\R,\C^2)
=
\bfX_{\mbox{\rm\footnotesize odd-even}}\oplus \bfX_{\mbox{\rm\footnotesize even-odd}}$,
hence it suffices to study the spectra of the restrictions of $L$ onto these subspaces.

\begin{lemma}\label{lemma-l}

Let $\omega\in(-m,m)$, $\kappa\in\R$.
\begin{enumerate}
\item
\label{lemma-l-1}
$\sigma_{\mathrm{ess}}\big(L(\omega,\kappa)\big)
=\R\setminus(-m-\omega,m-\omega)$,
$
\ \sigma_{\mathrm{ess}}\big(L(\omega,\kappa)\big)
\cap
\sigma_{\mathrm{p}}\big(L(\omega,\kappa)\big)=\emptyset$\textup;
\item
\label{lemma-l-2}
$
\sigma_{\mathrm{p}}\big(
L(\omega,\kappa)\at{\bfX_{\mbox{\rm\tiny{odd-even}}}}
\big)
=\{-2\omega\},
$
and eigenvalue $\lambda=-2\omega$ is simple\textup;
\item
\label{lemma-l-3}
$
\sigma_{\mathrm{p}}\big(
L(\omega,\kappa)\at{\bfX_{\mbox{\rm\tiny{even-odd}}}}
\big)
=
\big\{Z(\omega,\kappa)\big\},
$
where
\[
Z(\omega,\kappa)
=
-4(m-\omega)
\frac{\kappa(\kappa+1)}{1+(1+2\kappa)^2\mu^2}\in(-m-\omega,m-\omega),

\]
and is a simple eigenvalue.
\end{enumerate}
\end{lemma}

\begin{remark}\label{Rem:OnZ}
We note that for each $\omega\in(-m,m)$,
$Z(\omega,\kappa)$ is symmetric with respect to $\kappa=-\frac12$,
\[
Z\Big(\omega,-\frac12\Big)=m-\omega,
\qquad
Z(\omega,0)=0,
\qquad
Z(\omega,-1)=0,
\qquad
Z\Big(\omega,-\frac12\pm\frac1{2\mu^2}\Big)=-2\omega,
\]
and
$Z(\omega,\kappa)\to -m-\omega+0$ when
$\kappa\to \pm\infty$.

\end{remark}

\begin{proof}
The conclusion about the essential spectrum is standard:
the Weyl sequences for $L(\omega,\kappa)$, which is selfadjoint, do not depend on the jump condition in~\eqref{Eq:DomainOfA}, hence they are Weyl sequences for the standard Dirac operator and are associated with values in $\R\setminus(-m-\omega,m-\omega)$.
The conclusion on embedded eigenvalues will follow from the proof of Parts~\itref{lemma-l-2} and~\itref{lemma-l-3}.

Let us now prove Part~\itref{lemma-l-2}.
To find the point spectrum of the restriction of $L(\omega,\kappa)$
onto
$\bfX_{\mbox{\rm\footnotesize odd-even}}$,
we need to consider the spectral problem
\[
\left(
\begin{bmatrix}
m-\omega-\lambda&\p_x\\-\p_x&-m-\omega-\lambda
\end{bmatrix}
-2\mu\delta(x)\sigma_3
-4\mu\kappa\delta(x)\Pi_1
\right)
\psi=0,
\qquad
x\in\R,
\]
with
\[
\psi(x)=
\begin{bmatrix}
c_1\sgn x\\c_2
\end{bmatrix}e^{-\gamma\abs{x}},
\qquad x\in\R
\]
\begin{eqnarray}\label{here-nu-a-b}
\gamma=\sqrt{(m-\omega-\lambda)(m+\omega+\lambda)},
\qquad
c_1=m+\omega+\lambda,
\qquad
c_2=\gamma
\end{eqnarray}
(the  values of $c_1$ and $c_2$ are defined up to a nonzero coefficient)
and the jump condition
\begin{eqnarray}\label{ttc}
-2c_1+2\mu c_2=0.
\end{eqnarray}
We note that
since $\psi$ is square-integrable, we need $\Re\gamma>0$
and so
\begin{eqnarray}\label{spectral-gap}
-m-\omega<\lambda<m-\omega.
\end{eqnarray}

The relation \eqref{ttc} takes the form
\[
-(m+\omega+\lambda)+\mu
\sqrt{(m-\omega-\lambda)
(m+\omega+\lambda)}
=0;
\]
it is satisfied only for $\lambda=-2\omega$.

Let us prove Part~\itref{lemma-l-3}.
For the spectrum of the restriction of $L(\omega,\kappa)$
onto
$\bfX_{\mbox{\rm\footnotesize even-odd}}$,
we consider the spectral problem
\[
\left(
\begin{bmatrix}
m-\omega-\lambda&\p_x\\-\p_x&-m-\omega-\lambda
\end{bmatrix}
-2\mu\delta(x)\sigma_3
-4\mu\kappa\delta(x)\Pi_1
\right)\psi=0,
\qquad
x\in\R,
\]
with
\[
\psi(x)=
\begin{bmatrix}
c_1\\c_2\sgn x
\end{bmatrix}e^{-\gamma\abs{x}},
\qquad x\in\R.
\]
This again leads to the values \eqref{here-nu-a-b}.
Substituting these values
into the jump condition
\[
2c_2-2(1+2\kappa)\mu c_1=0
\]
results in
\[
\sqrt{(m-\omega-\lambda)(m+\omega+\lambda)}
=(1+2\kappa)\mu(m+\omega+\lambda).
\]
By \eqref{spectral-gap}, $\lambda>-m-\omega$,
hence
the above relation can hold only for $1+2\kappa>0$;
squaring it,
we arrive at
\[
m-\omega-\lambda
=(1+2\kappa)^2\mu^2(m+\omega+\lambda),
\]
which leads to
\[
\lambda
=m\frac{1-(1+2\kappa)^2\mu^2}{1+(1+2\kappa)^2\mu^2}-\omega

=-4(m-\omega)\frac{\kappa(1+\kappa)}{1+(1+2\kappa)^2\mu^2}.

\]
Note that $\lambda$ belongs to $(-m-\omega,m-\omega)$,
it is symmetric with respect to $\kappa=-\frac12$,
it has value $m-\omega$ at $\kappa=-\frac12$,
and tends to $-m-\omega$ at $\kappa\to\pm\infty$.
This finishes the proof of the lemma.
\end{proof}

We now consider the operator $\bfA$ from \eqref{def-a-ch2}
as an operator-valued function of $\omega\in(-m,m)$ and $\kappa\in\R$. In the following theorem the point spectrum and virtual levels of $\bfA(\omega,\kappa)$ are fully and explicitly described.
Similarly to
\eqref{def-x-e-o-o-e},
we introduce the subspaces
\begin{eqnarray}\label{def-oeoe}
\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}
=
L^2\sb{\mbox{\rm\footnotesize odd}}(\R,\C)
\times
L^2\sb{\mbox{\rm\footnotesize even}}(\R,\C)
\times
L^2\sb{\mbox{\rm\footnotesize odd}}(\R,\C)
\times
L^2\sb{\mbox{\rm\footnotesize even}}(\R,\C)
\subset L^2(\R,\C^4),
\end{eqnarray}
\begin{eqnarray}\label{def-eoeo}
\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}
=
L^2\sb{\mbox{\rm\footnotesize even}}(\R,\C)
\times
L^2\sb{\mbox{\rm\footnotesize odd}}(\R,\C)
\times
L^2\sb{\mbox{\rm\footnotesize even}}(\R,\C)
\times
L^2\sb{\mbox{\rm\footnotesize odd}}(\R,\C)
\subset L^2(\R,\C^4).
\end{eqnarray}
These are invariant subspaces for $\bfA$,
and there is a decomposition of
$
\bfX:=L^2(\R,\C^4)
$
into a direct sum
\begin{eqnarray}\label{def-x-x}
\bfX
=\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}
\oplus
\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}.
\end{eqnarray}

\begin{theorem}\label{theorem-sect2}
Let $\omega\in(-m,m)$ and $\kappa\in\R$.
\nn{We have to be more careful on $\kappa=0$.}
\begin{enumerate}
\item
\label{theorem-sect2-1}
There is the symmetry
of the spectrum
with respect to $\R$ and $\jj\R$:
\begin{eqnarray}\label{symmetry}
\lambda\in\sigma\sb{\mathrm{p}}(\bfA(\omega,\kappa))
\qquad
\Leftrightarrow
\qquad
-\lambda\in\sigma\sb{\mathrm{p}}(\bfA(\omega,\kappa))
\qquad
\Leftrightarrow
\qquad
\bar\lambda\in\sigma\sb{\mathrm{p}}(\bfA(\omega,\kappa)).
\end{eqnarray}
\item
\label{theorem-sect2-2}
For all $\kappa\in\R$ and $\omega\in(-m,m)$,
one has
\[
\pm 2\omega\jj\in\sigma\sb{\mathrm{p}}(\bfA(\omega,\kappa)\at{\bfX_{\mbox{\rm\tiny odd-even-odd-even}}})\subset\sigma\sb{\mathrm{p}}(\bfA(\omega,\kappa)).
\]
Moreover, $\pm 2\omega\jj$ are simple eigenvalues $\bfA(\omega,\kappa)\at{\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}}$ when $\omega\neq 0$\textup;
they are of geometric multiplicity two when $\omega=0$.
\item
\label{theorem-sect2-3}
For $\omega\ne 0$,
the virtual levels
at the thresholds
$\lambda=\pm(m-\abs{\omega})\jj$ exist
and they occur when
\[
\omega=
\begin{cases}
\displaystyle
\mathcal{T}_\kappa^{-}:=\frac{(\kappa+1)^2}{(\kappa+2)\kappa}m,
&\qquad
-1<\kappa<2^{-1/2}-1\textup;

\\[1ex]
\displaystyle
\mathcal{T}_\kappa^{+}:=\frac{(\kappa+1)^2}{(3\kappa+2)\kappa}m,
&\qquad
\kappa<-1
\ \ \mbox{and}\ \ \kappa>2^{-1/2}.

\end{cases}
\]
We note that $\mathcal{T}_\kappa^{-}\in(-m,0)$
and $\mathcal{T}_\kappa^{+}\in(0,m)$
in the relevant ranges of $\kappa$.
\item
\label{theorem-sect2-4}
Denote
\begin{eqnarray}\label{def-Omega-kappa}
\Omega_\kappa=\frac{\kappa+1}{2\kappa}m,
\qquad
\kappa\in\R\setminus\{0\}\textup;
\qquad
\mbox{
$\abs{\Omega_\kappa}<m$ as long as
$\kappa\in\R\setminus[-1/3,1]$.}

\end{eqnarray}
Besides $\lambda=0$ and $\lambda=\pm 2\omega\jj$,
the point spectrum of the linearization operator $\bfA(\omega,\kappa)$
contains the following additional eigenvalues:

\begin{enumerate}
\item
\label{theorem-sect2-4-a}
$\kappa<-1$\textup:
\begin{itemize}
\item
Two purely imaginary eigenvalues in the spectral gap for
$\omega\in(\mathcal{T}_\kappa^{+},\Omega_\kappa)$\textup;
\item
Two real eigenvalues
(hence linear instability) for $\omega\in(\Omega_\kappa,2\Omega_\kappa)$,
with eigenvalues going to $\pm\infty$ as
$\omega\to 2\Omega_\kappa-0$\textup;
\item
No additional eigenvalues for other values of $\omega$.
\end{itemize}

\item
\label{theorem-sect2-4-b}
$\kappa=-1$\textup:
\begin{itemize}
\item
For $\omega=0$,
one has

$\sigma\sb{\mathrm{p}}(\bfA)
=\C\setminus
\big(
\jj(-\infty,-m]\cap\jj[m,+\infty)
\big)$\textup;
\item
No additional eigenvalues for other values of $\omega$.
\end{itemize}

\item
\label{theorem-sect2-4-c}
$-1<\kappa<2^{-1/2}-1$\textup:
\begin{itemize}

\item
Two real eigenvalues
(hence linear instability) for
$\omega\in(\max(2\Omega_\kappa,-m),
\max(\Omega_\kappa,-m))$
(this case is vacuous for $-1/3\le\kappa<2^{-1/2}-1$).
When
$-1<\kappa\le-2/3$
(so that $2\Omega_k\in[-m,0)$),
these eigenvalues go to $\pm\infty$ as
$\omega\to 2\Omega_\kappa+0$\textup;
\item
Two purely imaginary eigenvalues in the spectral gap for
$\omega\in(\max(\Omega_\kappa,-m),\mathcal{T}_\kappa^{-})$\textup;
\item
No additional eigenvalues for other values of $\omega$.
\end{itemize}

\item
\label{theorem-sect2-4-d}
$2^{-1/2}-1\le\kappa\le 2^{-1/2}$\textup:
No additional eigenvalues for any $\omega\in(-m,m)$.

\item
\label{theorem-sect2-4-e}
$2^{-1/2}<\kappa\le 1$\textup:

\begin{itemize}
\item
Two purely imaginary eigenvalues
in the spectral gap
for
$\omega\in(\mathcal{T}_\kappa^{+},m)$\textup;
\item
No additional eigenvalues for other values of $\omega$.
\end{itemize}

\item
\label{theorem-sect2-4-f}
$\kappa>1$\textup:
\begin{itemize}
\item
Two purely imaginary eigenvalues in the spectral gap for
$\omega\in(\mathcal{T}_\kappa^{+},\min(\Omega_\kappa,m))$\textup;
\item
Two real eigenvalues
(hence linear instability) for
$\omega\in(\min(\Omega_\kappa,m),m)$
(this case is vacuous
for $\ 0<\kappa\le 1$)\textup;
\nn{$\omega=0$ is not considered !}
\ak{Now it is :):)}
\item
No additional eigenvalues
for other values of $\omega$.
\end{itemize}
\end{enumerate}
\item
\label{theorem-sect2-5}
For all $\kappa\in\R$ and $\omega\in(-m,m)$, one has
\[
0\in\sigma\sb{\mathrm{p}}(\bfA(\omega,\kappa)).
\]

Moreover,
\[
\dim\ker(\bfA(\omega,\kappa))
=
\begin{cases}
1,&
\omega\ne 0,\ \kappa\not\in\{-1,0\},
\\
2,&
\omega\ne 0,\ \kappa\in\{-1,0\},
\\
3,&
\omega=0,\ \kappa\not\in\{-1,0\},
\\
4,&
\omega=0,\ \kappa\in\{-1,0\}.
\end{cases}
\]
\item
\label{theorem-sect2-6}
The algebraic multiplicity of $\lambda=0$ is given by
\begin{eqnarray}
\dim
\mathfrak{L}(\bfA(\omega,\kappa))
=\begin{cases}
2,&\omega\in(-m,m)\setminus\{0,\Omega_\kappa\},
\quad
\kappa\in\R\setminus[-1/3,1]\textup;
\\
4, &\omega
=\Omega_\kappa,
\quad
\kappa\in\R\setminus(\{-1\}\cup[-1/3,1])\textup;
\\
4,&\omega\in(-m,m),\quad\kappa=0\textup;
\\
6,&\omega=0,\quad\kappa=-1.
\end{cases}
\end{eqnarray}
Above, $\mathfrak{L}(\bfA(\omega,\kappa))$
is the notation for the generalized eigenspace
of $\bfA(\omega,\kappa)$ corresponding to $\lambda=0$.

\item
\label{theorem-sect2-7}

\[
\sigma\sb{\mathrm{ess}}(\bfA(\omega,\kappa))
=
\begin{cases}
\jj\big(\R\setminus(-m+\abs{\omega},m-\abs{\omega})\big),
\quad
&(\omega,\kappa)\ne (0,-1)\textup;
\\
\C,
&(\omega,\kappa)=(0,-1).
\end{cases}
\]
\end{enumerate}
\end{theorem}

\begin{figure}
\begin{center}

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        xmin=-2,xmax=2,
        ymin=-1,ymax=1,
        axis x line=middle,
        axis y line=middle,

        ylabel={$\omega$},
        xlabel={$\kappa$},
x tick label style={yshift={(\tick==-1)*2em}},
y tick label style={xshift={(\tick==-1)*2em}},
        ]

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    \end{axis}
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\begin{comment}
\else
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\textcolor{red}{*** TO SEE THE RIGHT PICTURE, PLEASE COMPILE WITH PDFLATEX ***}}
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\end{comment}
\caption{
Location of simple eigenvalues
of the linearized operator $\bfA(\omega,\kappa)$
for different values of parameters
$\omega\in(-m,m)$ and $\kappa\in\R$.
In the unshaded regions, there are no simple eigenvalues
besides $\lambda=0$ and $\lambda=\pm 2\omega\jj$.

The Kolokolov curve
$\omega=\Omega_\kappa$

(\textcolor{red}{\bf thick solid lines} for $\kappa<-0.5$ and $\kappa>1$)
correspond to
collision of two simple eigenvalues at $\lambda=0$
as indicated by the Kolokolov condition.
The virtual level curve
$\omega=\mathcal{T}_\kappa^{-}$

(\textcolor{blue}{\bf dotted line} for $\kappa\in(-1,2^{-1/2}-1)$)
corresponds to
virtual levels at thresholds
$\pm\jj(m-\abs{\omega})$ when $\omega<0$.
The virtual level curves
$\omega=\mathcal{T}_\kappa^{+}\in(0,m)$

(\textcolor{green}{\bf dash-dot lines} for $\kappa<-1$ and $\kappa>2^{-1/2}$)
correspond to virtual levels at thresholds
$\pm\jj(m-\abs{\omega})$  when $\omega>0$.
The \textcolor{blue}{dotted regions}
between the Kolokolov and virtual level lines
correspond to two simple purely imaginary eigenvalues
in the spectral gap.
On the other side of the Kolokolov line
there is a pair of real eigenvalues
(linear instability; regions on the graph
filled with \textcolor{red}{thin lines}).
In the region where there are two real eigenvalues $\pm\lambda\in\R$,
one has $\lambda\to +\infty$ as $\omega$ and $\kappa$ approach
the curve $\omega=2\Omega_\kappa$
(\textcolor{red}{\bf dashed line} for $\kappa<-2/3$).

At $\kappa=-1$, $\omega=0$ ({\bf bullet} on the plot),
the spectrum consists of the whole complex plane.
}
\label{fig-omegak}
\end{center}
\end{figure}

We depict these cases on Figure~\ref{fig-omegak}.

\begin{remark}
The spectrum of the linearization operator for a one-dimensional Schr\"odinger model with concentrated nonlinearity is studied in \cite{buslaev2008asymptotic} and more thoroughly in \cite{komech2012asymptotic}.
\end{remark}

\begin{remark}
For $\kappa<-1$,
there are real eigenvalues
for $\omega\in(\Omega_\kappa,2\Omega_\kappa)$
which bifurcate from zero (when $\omega=\Omega_\kappa$)
and slip off to $\pm\infty$
as $\omega\ge 2\Omega_\kappa$,
as one can see from explicit
expressions for eigenvalues;
see Lemma~\ref{lemma-w-kappa}
below.

As a result,
for $\omega$ between $\Omega_\kappa$ and $2\Omega_\kappa$,
there are
eigenvalue families $\pm\lambda(\omega,\kappa)\in\sigma\sb{\mathrm{p}}(\bfA(\omega,\kappa))$
such that
\[

\begin{cases}
\lim\limits\sb{\omega\to 2\Omega_\kappa-0}\lambda(\omega,\kappa)\to+\infty,
&\kappa<-1;
\\[2ex]
\lim\limits\sb{\omega\to 2\Omega_\kappa+0}\lambda(\omega,\kappa)\to+\infty,
&-1<\kappa<-1/2.
\end{cases}
\]
\end{remark}

We give the proof
of Theorem~\ref{theorem-sect2}
In the next section.

\section{Proof of Theorem~\ref{theorem-sect2}}
\label{sect-theorem-sect2-proof}

For Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-1},
we notice that the symmetry
\[
\lambda\in\sigma\sb{\mathrm{p}}(\bfA)
\qquad
\Leftrightarrow
\qquad
\bar\lambda\in\sigma\sb{\mathrm{p}}(\bfA)
\]
follows from $\bfA$ having real coefficients,
while the symmetry
\[
\bar\lambda\in\sigma\sb{\mathrm{p}}(\bfA)
\qquad
\Leftrightarrow
\qquad
-\lambda\in\sigma\sb{\mathrm{p}}(\bfA)
\]
follows from
\begin{equation}\label{Eq:Symmetries}
\bfA
=
\mathbf{J}\mathbf{L},\quad\bfA^*
=
(\mathbf{J}\mathbf{L})^*=\mathbf{L}^* \mathbf{J}^*=-\mathbf{L}\mathbf{J},
\quad
\mbox{where}
\quad
\mathbf{L}=
\begin{bmatrix}L_{+}&0\\0&L_{-}\end{bmatrix},
\quad
\mathbf{J}=\begin{bmatrix}0&I_2\\-I_2&0\end{bmatrix},
\end{equation}
with $L_\pm$ defined in \eqref{l-p-l-m-1},
while $\sigma\sb{\mathrm{p}}(\mathbf{L}\mathbf{J})
=\sigma\sb{\mathrm{p}}(\mathbf{J}\mathbf{L})$
due to $\mathbf{J}$ being bounded and invertible.
This proves
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-1}.

We start with the case $\kappa=0$.
In this case, one has
\[
\bfA(\omega,0)
=
\begin{bmatrix}0&I_2\\-I_2&0\end{bmatrix}
\otimes
L(\omega,0),
\]
and the statement of Theorem~\ref{theorem-sect2}
immediately follows from Lemma~\ref{lemma-l}.
\nn{This should be emphasized and recalled in Section~\ref{Sec:Kolokolov}}
In the rest of the argument, we make the assumption
that $\kappa$ is nonzero:
\begin{eqnarray}\label{kappa-nonzero}
\kappa\in\R\setminus\{0\}.
\end{eqnarray}

Below, we will use the fact that the operator $\bfA$
from \eqref{def-a-ch2}
is invariant in the subspaces

$\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$
and
$\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$
(see \eqref{def-oeoe} and \eqref{def-eoeo}) of $L^2(\R,\C^4)$,

so the search for eigenvalues and eigenvectors can be restricted to the analysis
of the spectrum of $\bfA$ in these two subspaces.

\subsection{
The spectrum of $\bfA$ in odd-even-odd-even subspace
and eigenvalue $2\omega\jj$}
\label{sect-spectrum-oe}

\nn{All this part assumes that $\nu_+\neq\nu_-$ which is equivalent to $\omega\neq 0$ or $\lambda\neq 0$. So $\omega =0$ has to be treated apart and as soon as we deal with the kernel, we have the same issue.}

To prove Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-2},
we restrict $\bfA$ to the subspace
$\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$.

For $x\ne 0$,
a representation for a solution in $L^2$ of the equation $\bfA\Psi=\lambda\Psi $,
$\lambda\in\C$
belonging to the subspace
$\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$
(see \eqref{def-x-x})
is given by
\begin{eqnarray}\label{psi-o-e}
\Psi(x)
=
c_1
\begin{bmatrix}
\nu_{+}(\omega,\Lambda)\sgn x\\S_{+}(\omega,\Lambda)
\\-\jj \nu_{+}(\omega,\Lambda)\sgn x\\-\jj S_{+}(\omega,\Lambda)
\end{bmatrix}e^{-\nu_{+}(\omega,\Lambda)\abs{x}}
+
c_2
\begin{bmatrix}
\nu_{-}(\omega,\Lambda)\sgn x\\S_{-}(\omega,\Lambda)
\\\jj \nu_{-}(\omega,\Lambda)\sgn x\\\jj S_{-}(\omega,\Lambda)
\end{bmatrix}e^{-\nu_{-}(\omega,\Lambda)\abs{x}},
\qquad
c_1,\,c_2\in\C,
\end{eqnarray}
where we used the notations
\begin{eqnarray}\label{def-m-n}
\begin{array}{l}
\nu_{+}(\omega,\Lambda):=\sqrt{m^2-(\omega+\jj\lambda)^2}=\sqrt{m^2-(\omega-\Lambda)^2},
\qquad
\Re\nu_{+}(\omega,\lambda)\ge 0,
\\[2ex]
\nu_{-}(\omega,\Lambda):=\sqrt{m^2-(\omega-\jj\lambda)^2}=\sqrt{m^2-(\omega+\Lambda)^2},
\qquad
\Re\nu_{-}(\omega,\lambda)\ge 0
\end{array}
\end{eqnarray}

(these expressions
come from considering the characteristic equation of the homogeneous system
with constant coefficients,
$(\bfA-\jj\Lambda I_4)\Psi=0$,
with $\Psi\in C^{\infty}(\R\setminus\{0\})$)
and
\begin{eqnarray}\label{def-r-s}
S_{+}(\omega,\Lambda)=m-\omega+\Lambda,
\qquad
S_{-}(\omega,\Lambda)=m-\omega-\Lambda;
\end{eqnarray}
above, the value $\Lambda\in\C$
is defined by
\begin{eqnarray}\label{lambda-lambda}
\lambda=\jj\Lambda.
\end{eqnarray}
\ac{\sout{In \eqref{def-m-n}
,
we take the branch
of the square root
with $\Re\sqrt{z}\geq 0$
for $z\in\C\setminus\R_{-}$.}
[Added $\Re\nu_\pm\ge 0$
directly in the definition; more convenient for referencing.]}

We assume that
$\nu_{-}$ and $\nu_{+}$ are non-vanishing
and with positive real part
as long as the corresponding coefficient
in \eqref{psi-o-e} is nonzero,
so that $\Psi\in L^2(\R,\C^4)$.

\begin{remark}\label{remark-sometimes}

We note that
the vectors in \eqref{psi-o-e}
are linearly independent
unless either $S_{+}=\nu_{+}=0$
or $S_{-}=\nu_{-}=0$.
This happens at the threshold point
$\Lambda=m-\omega$
(where $S_{-}=\nu_{-}=0$)
and at the threshold point $\Lambda=-(m-\omega)$
(where $S_{+}=\nu_{+}=0$);
note that for $\omega<0$ these threshold points
are embedded into the essential spectrum.
At
$\Lambda=m-\omega$,
one needs to consider
\begin{eqnarray}\label{psi-o-e-1}
\Psi(x)
=
c_1
\begin{bmatrix}
\nu_{+}\sgn x\\S_{+}\\-\jj \nu_{+}\sgn x\\-\jj S_{+}
\end{bmatrix}e^{-\nu_{+}\abs{x}}
+
c_2
\begin{bmatrix}
\nu_{-}^{\mathrm{reg}}\sgn x\\S_{-}^{\mathrm{reg}}
\\\jj \nu_{-}^{\mathrm{reg}}\sgn x\\\jj S_{-}^{\mathrm{reg}}
\end{bmatrix},

\qquad
c_1,\,c_2\in\C,
\end{eqnarray}

with
$\nu_{-}^{\mathrm{reg}}:=
\lim\sb{\Lambda\to m-\omega+0}
\frac{\nu_{-}(\omega,\Lambda)}{\sqrt{m-\omega-\Lambda}}

=\sqrt{2m+2\omega}$
and
$S_{-}^{\mathrm{reg}}:=
\lim\sb{\Lambda\to m-\omega+0}
\frac{S_{-}(\omega,\Lambda)}{\sqrt{m-\omega-\Lambda}}
=0$.
The case $\Lambda=-(m-\omega)$
is treated similarly.
\end{remark}

We note that
if $\lambda\in\jj\R$, $\abs{\lambda}\ge m+\abs{\omega}$
(so that $\lambda$ is beyond the embedded
thresholds at $\pm \jj(m+\abs{\omega})$),
then
both
$m^2-(\omega-\jj\lambda)^2\le 0$ and $m^2-(\omega+\jj\lambda)^2\le 0$,
hence
there is no corresponding square-integrable
function $\Psi\ne 0$ of the form \eqref{psi-o-e}.

An eigenvector is an element of the domain of  $\dom(\bfA)$ given in \eqref{Eq:DomainOfA}; it has

to satisfy the jump condition at the origin,
which is given by
\begin{eqnarray}\label{nu-a-b}
\begin{cases}
2\jj \nu_{+} c_1-2\jj \nu_{-} c_2
+2\mu(-\jj S_{+} c_1+\jj S_{-} c_2)=0,
\\2\nu_{+} c_1+2\nu_{-} c_2
-2\mu(S_{+} c_1+S_{-} c_2)=0.
\end{cases}
\end{eqnarray}

Since $c_1,\,c_2\in\C$ are not simultaneously zeros,
the compatibility condition leads to
\begin{eqnarray}\label{det-zero}
\det\begin{bmatrix}
\jj\nu_{+}-\jj S_{+} \mu&-\jj \nu_{-}+\jj S_{-}\mu
\\\nu_{+}-S_{+}\mu&\nu_{-}-S_{-}\mu
\end{bmatrix}
=
2\jj (\nu_{-}-S_{-} \mu)(\nu_{+}-S_{+} \mu)
=0.
\end{eqnarray}
The relation $\nu_{-}-S_{-} \mu=0$
results in
\[
m^2-(\omega+\Lambda)^2
=
\frac{m-\omega}{m+\omega}(m-\omega-\Lambda)^2,
\]
which takes the form
\[
(m+\omega)^2-(m-\omega)^2
=-2m\Lambda,
\]
leading to $\Lambda=-2\omega$.
Similarly,
the relation
$\nu_{+}-S_{+}\mu=0$
leads to $\Lambda=2\omega$.

Thus, since $\omega\neq 0$,
the factors $\nu_{+}-S_{+}\mu$ and $\nu_{-}-S_{-}\mu$
in \eqref{det-zero}
can not vanish simultaneously.
We conclude from \eqref{nu-a-b}
that either $c_1$ or $c_2$ vanishes,
hence the geometric multiplicity of $\pm 2\jj \omega$ is equal to one.

\nn{Added a remark on the multiplicity.}
As to the algebraic multiplicity,
we recall that $\bfA^\ast=-\mathbf{J}^\ast \bfA \mathbf{J}$
(see~\eqref{Eq:Symmetries})
and hence if $\Psi\in \range(\bfA-\lambda)$ then $\Psi\in \bf J \ker(\bfA-\lambda)^\bot$. So if the the algebraic multiplicity is larger than one
(while the geometric multiplicity equals one),
then necessarily $\Psi$ and $\mathbf{J} \Psi$ are orthogonal.
For $\Lambda=2\omega$,
this orthogonality is equivalent to $\nu_+(\omega,2\omega)=S_+(\omega,2\omega)=0$;

for $\Lambda=-2\omega$,
it is equivalent to $\nu_-(\omega,-2\omega)=S_-(\omega,-2\omega)=0$.
This is impossible as $\omega\in (-m,m)$.

We thus conclude that $\pm 2 \omega$ are simple eigenvalues
for $\omega\in(-m,m)\setminus\{0\}$.

We point out that in the case $\omega=0$,
the jump condition \eqref{nu-a-b} becomes trivial,
and $c_1$ and $c_2$ in \eqref{psi-o-e}
could take arbitrary values.
Then the two terms in the right-hand side of
\eqref{psi-o-e}
are linearly independent eigenvectors
corresponding to eigenvalue $\lambda=2\omega\jj=0$,
proving that the geometric multiplicity
of eigenvalue zero
of $\bfA$ restricted to
$\bfX\sb{\mbox{\rm\footnotesize odd-even-odd-even}}$
is equal to two.

This proves
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-2}.

\begin{remark}

Let us point out that for $m/3\le \abs{\omega}<m$
the eigenvalue $\lambda=2\jj\omega$
is
embedded in the essential spectrum of $\bfA$.
For example, if $m/3\le\omega<m$,
then
$\nu_{+}=\varkappa$,
$\nu_{-}=\sqrt{m^2-9\omega^2}$ is purely imaginary,
$S_{+}=m+\omega$,
$S_{-}=m-3\omega$,
$\nu_{+}=S_{+}\mu$,
and the system \eqref{nu-a-b}
takes the form
\[
\nu_{-} c_2
-\mu S_{-} c_2=0,
\]
which results in $c_1\in\C$ arbitrary and $c_2=0$;
due to $\nu_{+}>0$, one can see that
$\Psi$ from \eqref{psi-o-e}
belongs to $L^2$.

\end{remark}

\begin{remark}
The eigenvalues
$\lambda=\pm 2\omega\jj$
are present in the spectrum of $\bfA$
due to the $\mathbf{SU}(1,1)$-invariance of the Soler model
\cite{boussaid2018spectral}.
\end{remark}

\subsection{The spectrum of $\bfA$
in even-odd-even-odd subspace
and virtual levels at thresholds}
\label{sect-spectrum-eo}

\nn{We can make a remark later on the multiplicity. The proof of the previous case may work in the case $\lambda\neq 0$.}

In this Section we prove
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-3}
and
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4}.
Similarly to our approach in Section~\ref{sect-spectrum-oe},
any square-integrable solution of $\bfA\Psi=\lambda\Psi $  with
$\lambda=\jj\Lambda$ in the subspace
$\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$
of $L^2$ (see \eqref{def-x-x}) can be represented as
(cf. \eqref{psi-o-e})
\begin{eqnarray}\label{psi-x-e-o-e-o}
\Psi(x)
=
c_1
\begin{bmatrix}
\nu_{+}(\omega,\Lambda)\\S_{+}(\omega,\Lambda)\sgn x\\-\jj \nu_{+}(\omega,\Lambda)\\-\jj S_{+}(\omega,\Lambda)\sgn x
\end{bmatrix}e^{-\nu_{+}(\omega,\Lambda)\abs{x}}
+
c_2
\begin{bmatrix}
\nu_{-}(\omega,\Lambda)\\S_{-}(\omega,\Lambda)\sgn x\\\jj \nu_{-}(\omega,\Lambda)\\\jj S_{-}(\omega,\Lambda)\sgn x
\end{bmatrix}e^{-\nu_{-}(\omega,\Lambda)\abs{x}},
\quad
c_1,\,c_2\in\C,
\end{eqnarray}
with
$\nu_\pm(\omega,\Lambda)$, $S_\pm(\omega,\Lambda)$
from \eqref{def-m-n} and \eqref{def-r-s},
where we will assume

that both $\nu_{-}$ and $\nu_{+}$ are non-vanishing
and with positive real part, so that $\Psi\in L^2(\R,\C^4)$.
Again, by Remark~\ref{remark-sometimes},
at $\Lambda=m-\omega$,
the vectors in \eqref{psi-x-e-o-e-o}
become linearly dependent
(the second vector vanishes)
and one can use
the following decomposition (cf. \eqref{psi-o-e-1}):
\begin{eqnarray}\label{psi-e-o-1}
\Psi(x)
=
c_1
\begin{bmatrix}
\nu_{+}\\S_{+}\sgn x\\-\jj \nu_{+}\\-\jj S_{+}\sgn x
\end{bmatrix}e^{-\nu_{+}\abs{x}}
+
c_2
\begin{bmatrix}
\nu_{-}^{\mathrm{reg}}\\S_{-}^{\mathrm{reg}}\sgn x
\\\jj \nu_{-}^{\mathrm{reg}}\\\jj S_{-}^{\mathrm{reg}}\sgn x
\end{bmatrix},

\qquad
c_1,\,c_2\in\C.
\end{eqnarray}

The jump condition for $\Psi$ at the origin
takes the form
\begin{eqnarray}\label{Eq:SystemSandR-e-o-e-o}
\begin{cases}
-2\jj S_{+} c_1+2\jj S_{-} c_2
-2(
-\jj \nu_{+} c_1
+\jj \nu_{-} c_2)\mu
=0,
\\-(2 S_{+} c_1+2 S_{-} c_2)
+2(\nu_{+} c_1+\nu_{-} c_2)(1+2\kappa)\mu
=0.
\end{cases}
\end{eqnarray}

To find eigenvalues,
we need to consider the compatibility condition
for the system \eqref{Eq:SystemSandR-e-o-e-o},

\begin{eqnarray}\label{c-cond}
\det
\begin{bmatrix}
\mu\nu_{+}-S_{+}&-\mu\nu_{-}+S_{-}
\\(1+2\kappa)\mu\nu_{+}-S_{+}&(1+2\kappa)\mu\nu_{-}-S_{-}
\end{bmatrix}
=0.
\end{eqnarray}

\begin{lemma}\label{lemma-no}

For
$\omega\in(-m,m)$ and $\kappa\in\R\setminus\{0\}$,
the operator $\bfA(\omega,\kappa)$
restricted onto
$\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$
has no embedded eigenvalues.
\end{lemma}

\begin{proof}

In order to have square-integrable solutions,
the coefficients $c_1,\,c_2$ in \eqref{psi-x-e-o-e-o}
corresponding to
purely imaginary or zero values of $\nu_{+}$, $\nu_{-}$
have to vanish.
If $\Re\nu_{-}=0$, then $c_2=0$ in \eqref{psi-x-e-o-e-o}
(if $\Lambda=m-\omega$, then, rather,
$c_2=0$ in the expression similar to \eqref{psi-o-e-1}
but of different parity,
so that $\Psi\in\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$).
Then the jump condition \eqref{Eq:SystemSandR-e-o-e-o} yields
\[
S_{+}-\nu_{+} \mu=0,
\qquad
S_{+}-(1+2\kappa)\mu\nu_{+}=0.
\]
Since $\mu(\omega)>0$,
the assumption $\kappa\ne 0$ leads to
$\nu_{+}=0$, but then
\eqref{psi-x-e-o-e-o} would not be in $L^2$ unless $c_1=0$.
The case when $\Re\nu_{+}=0$,
so that $c_1=0$

is treated similarly.

One concludes that there are no embedded eigenvalues
corresponding to eigenfunctions from $\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$.
\end{proof}

Let us now study isolated eigenvalues.
We rewrite
the compatibility condition \eqref{c-cond}
as
\begin{eqnarray}
\label{condition-e-o-e-o}

\Gamma(\Lambda):&=&-\nu_{-} \nu_{+}\mu^2
(2\kappa+1)
+\mu \nu_{-}(\kappa+1)(m-\omega+\Lambda)
\nonumber
\\&&
\quad
+
(m-\omega-\Lambda)(\kappa+1)\mu \nu_{+}
-(m-\omega-\Lambda)(m-\omega+\Lambda)
=0,
\end{eqnarray}

where
$\mu=\mu(\omega)=\sqrt{(m-\omega)/(m+\omega)}$
and
$\nu_\pm=\nu_\pm(\omega,\Lambda)$ were introduced in \eqref{def-m-n}.
One can see
that $\nu_{-}(\omega,\Lambda)$
vanishes at
$\Lambda=m-\omega$ and $\Lambda=-m-\omega$
while
$\nu_{+}(\omega,\Lambda)$ vanishes at $\Lambda=m+\omega$ and at $\Lambda=-m+\omega$;
it follows that
$\Gamma(\Lambda)$ vanishes at $\Lambda=m-\omega$ and $\Lambda=-m+\omega$.
We will define the ``first", or ``physical", sheet of the  Riemann surface
of the function $\Gamma(\Lambda)$ to be the one where $\Re \nu_{-}\ge 0$
and $\Re\nu_{+}\ge 0$ (in the following referred to as the $(+,+)$ Riemann sheet).

We now consider $\Lambda$ outside of thresholds
$$\Lambda\not\in\left\{ -m-\omega, -m+\omega, m-\omega, m+\omega\right\}$$
so that $\nu_{-}\nu_{+}$ does not vanish. Let us find the solutions of $\Gamma(\Lambda)=0$
on the first Riemann sheet.
We divide
\eqref{condition-e-o-e-o}
by $\nu_{-} \nu_{+}$
(this corresponds to ``normalizing'' the vectors from
\eqref{psi-x-e-o-e-o}
near $\nu_\pm\to 0$; now the resulting function will not vanish identically
near $\Lambda=m-\omega$ and $\Lambda=m+\omega$).
Taking into account the fact that $z = (\sqrt z)^2$ for all $z\in \C\backslash \R_-$,
and that $\sqrt{c z} = \sqrt c \sqrt z$ for all $c>0$ and $z\in\C\backslash \R_-$,
after some manipulations (dividing by $\mu$ and factorizing), we end up with the equation
\begin{equation}\label{EQ1prov}
\kappa^2
=
\left(\kappa+1
-
\frac{\sqrt{1-\frac{\Lambda}{m-\omega}}}{\sqrt{1+\frac{\Lambda}{m+\omega}}}
\right)
\left(\kappa+1
-
\frac{\sqrt{1+\frac{\Lambda}{m-\omega}}}{\sqrt{1-\frac{\Lambda}{m+\omega}}}
\right).
\end{equation}

In this formula,
we choose the branch of $\sqrt z$,
$z\in \C\backslash \R_-$,
such that
$\Re \sqrt z\ge 0$.

\begin{remark}\label{remark-root}
One has $\sqrt{z w} = \sqrt z \sqrt w$ by analytical extension from $\R_{+}$ to $z\in\C\setminus\R_{-}$ and $w\in\C\setminus\R_{-}$,
with $\arg(z)\neq \arg(w) +\pi \mod 2\pi$ for instance.
\end{remark}

Let us first consider the case
$\kappa=-1$.
In this case, \eqref{EQ1prov}
leads to
\[
1-\frac{\Lambda^2}{(m-\omega)^2}
=
1-\frac{\Lambda^2}{(m+\omega)^2},
\]
and thus
$\kappa=-1$
corresponds to the following two cases:
\begin{enumerate}
\item
$\Lambda=0$ and $\omega\in(-m,m)$,
so that
\[
\sigma\sb{\mathrm{p}}\big(\bfA(\omega,-1)
\at{\bfX
\sb{\mbox{\rm\tiny even-odd-even-odd}}}\big)
=\{0\};
\]
\item
$\omega=0$ and $\Lambda\in\C$ is arbitrary;
the values corresponding to the point spectrum
are $\Lambda\in\C\setminus((-\infty,m]\cup[m,+\infty))$,
corresponding to $\Re\nu_\pm(0,\Lambda)=\sqrt{m^2-\Lambda^2}>0$.
We conclude that
\[
\sigma\sb{\mathrm{p}}(\bfA(0,-1))
=\C\setminus\big(\jj(-\infty,-m]\cap\jj[m,+\infty)\big),
\qquad
\sigma(\bfA(0,-1))
=\sigma\sb{\mathrm{ess}}(\bfA(0,-1))
=\C.
\]
\end{enumerate}
This proves Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4-b}.

The above consideration also shows that
$\lambda=0$ is an eigenvalue
of the restriction of $\bfA(\omega,\kappa)$
with $\kappa=-1$
onto $\bfX\sb{\mbox{\rm\footnotesize even-odd-even-odd}}$
for any value of $\omega\in(-m,m)$,
which is needed
for
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-5}
and
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-6}
(in the case $\kappa=-1$).

\medskip

In what follows, we only consider
the spectrum of the restriction of $\bfA$
onto $\bfX\sb{\mbox{\footnotesize even-odd-even-odd}}$
in the case
$\kappa\ne -1$;
together with \eqref{kappa-nonzero},
this reduces our consideration to the situation
\begin{eqnarray}\label{kappa-nonzero-1}
\kappa\in\R\setminus\{-1,\,0\}.
\end{eqnarray}

Due to Remark~\ref{remark-root}, we claim that
on the first Riemann sheet of $\Gamma(\Lambda)$ one has
\begin{equation}\label{claim}
\frac{\sqrt{1-\frac{\Lambda}{m-\omega}}}{\sqrt{1+\frac{\Lambda}{m+\omega}}} =
\sqrt{\frac{1-\frac{\Lambda}{m-\omega}}{1+\frac{\Lambda}{m+\omega}}}
\qquad \textrm{and} \qquad
\frac{\sqrt{1+\frac{\Lambda}{m-\omega}}}{\sqrt{1-\frac{\Lambda}{m+\omega}}} =
\sqrt{\frac{1+\frac{\Lambda}{m-\omega}}{1-\frac{\Lambda}{m+\omega}}}.
\end{equation}

Let us prove the first relation in \eqref{claim}. Note that
\[
\frac{\sqrt{1-\frac{\Lambda}{m-\omega}}}{\sqrt{1+\frac{\Lambda}{m+\omega}}}=
\frac{\sqrt{1-\frac{\Lambda}{m-\omega}}\sqrt{1+\frac{\overline\Lambda}{m+\omega}}}{\left|\sqrt{1+\frac{\Lambda}{m+\omega}}\right|^2},
\]
where we used $\overline{\sqrt{z}} = \sqrt{\overline z}$,
which holds true for all $z\in \C\backslash\R_-$.
It is enough to prove that
\[
\sqrt{1-\frac{
\phantom{\int}
\Lambda
\phantom{\int}
}{m-\omega}}
\,
\sqrt{1+\frac{\overline\Lambda}{m+\omega}}
=
\sqrt{\left(1-\frac{\Lambda}{m-\omega}\right)\left(1+\frac{\overline\Lambda}{m+\omega}\right)}.
\]
Since
\[
\Im \left(1-\frac{\Lambda}{m-\omega} \right) \left(1+\frac{\overline\Lambda}{m+\omega}\right)
= - \frac{2 m}{m^2-\omega^2} \Im \Lambda,
\]
from Remark~\ref{remark-root}
we have the first relation \eqref{claim}.

Similarly, to prove the second relation in \eqref{claim}, it is enough to note that
\[
\Im\left(1+\frac{\Lambda}{m-\omega}\right) \left(1-\frac{\overline \Lambda}{m+\omega}\right)  =  \frac{2 m}{m^2-\omega^2} \Im \Lambda
\]
and use again  Remark~\ref{remark-root}.

The conclusion is that on the first Riemann sheet of $\Gamma(\Lambda)$,

equation \eqref{EQ1prov}
can be rewritten equivalently as
\begin{equation}\label{EQ1}
\kappa^2
=
\left(\kappa+1
-
\sqrt{\frac{1-\frac{\Lambda}{m-\omega}}{1+\frac{\Lambda}{m+\omega}}}
\right)
\left(\kappa+1
-
\sqrt{\frac{1+\frac{\Lambda}{m-\omega}}{1-\frac{\Lambda}{m+\omega}}}
\right),
\qquad \Lambda \in \C.
\end{equation}
To solve this equation, we set
\begin{equation}
\label{X-vs-Lambda}
X =
\sqrt{\frac{1-\frac{\Lambda}{m-\omega}}{1+\frac{\Lambda}{m+\omega}}},
\qquad \Re X \geq 0,
\end{equation}
or equivalently
$X^2 = \frac{1-\frac{\Lambda}{m-\omega}}{1+\frac{\Lambda}{m+\omega}}$;
the right hand side defines a proper homography since $\omega\in (-m,m)$.
This is a bijection from $\C\setminus\{-m-\omega\}$ to $\C\setminus\{-\frac{m+\omega}{m-\omega}\}$.

Notice that $X=1$ if and only if $\Lambda=0$.
Then
\begin{eqnarray}\label{Lambda-vs-X}
\Lambda = \frac{1-X^2}{\frac{1}{m-\omega} +\frac{X^2}{m+\omega}}.
\end{eqnarray}
For $\Lambda\neq m+\omega$ (equivalently, for $X^2\neq -\frac{\omega}{m-\omega}$),
we also have
\[
\frac{1+\frac{\Lambda}{m-\omega}}{1-\frac{\Lambda}{m+\omega}} = \frac{m+\omega - \omega X^2}{\omega+(m-\omega)X^2},
\]
so that we rewrite equation \eqref{EQ1} as
\begin{equation}\label{across}
\kappa^2
=
\left(\kappa+1 - X
\right)
\left(\kappa+1
-
\sqrt{\frac{m+\omega - \omega X^2}{\omega+(m-\omega)X^2}}
\right).
\end{equation}

If $\kappa+1 - X = 0$,
then $\kappa =0$ and $X=1$, leading to $\Lambda=0$.
Now we assume that $\kappa+1 - X \ne 0$;
then \eqref{across} is equivalent to
\begin{equation}\label{EQ2}
\kappa+1 - \frac{\kappa^2}{\kappa+1 - X}
=
\sqrt{\frac{m+\omega - \omega X^2}{\omega+(m-\omega)X^2}}\,.
\end{equation}

Squaring the above relation, we arrive at
\begin{equation}\label{EQ4}
\left(\kappa+1 - \frac{\kappa^2}{\kappa+1 - X}\right)^2
=
\frac{m+\omega - \omega X^2}{\omega+(m-\omega)X^2},
\qquad \Re\left( \kappa+1 - \frac{\kappa^2}{\kappa+1 - X} \right)  \geq 0.
\end{equation}

Recall that we consider the situation when the following conditions are satisfied:
$X^2\neq -\frac{\omega}{m-\omega}$ (which is equivalent to $\Lambda\neq m+\omega$),
$X^2\neq -\frac{m+\omega}{m-\omega}$ (which is equivalent to $\Lambda\neq -m-\omega$),
and $ X\neq \kappa +1$.
\ac{\sout{(which leads to $\Lambda=0$)}
[``or else'' would be more appropriate
than ``which leads'', but not needed anyways]}
We note the condition on the real part of $\kappa+1 - \frac{\kappa^2}{\kappa+1 - X}$
must be satisfied by the solution $X$;

this is due to our choice of the square root,
for the square-integrability of \eqref{psi-x-e-o-e-o}.

Equation \eqref{EQ4} can be rewritten as
\begin{equation}\label{Eq:equation0}
\frac{((\kappa+1)^2 - \kappa^2  -(\kappa+1)X)^2}{(\kappa+1 - X)^2}
=
\frac{m+\omega - \omega X^2}{\omega+(m-\omega)X^2}.
\end{equation}
With a straightforward calculation,

the above relation can be rewritten
equivalently
due to the conditions $X^2\neq -\frac{\omega}{m-\omega}$ and $ X\neq \kappa +1$,

as
\begin{equation}\label{Eq:equation1}
((\kappa+1)^2 - \kappa^2  -(\kappa+1)X)^2(\omega+(m-\omega)X^2)
=
(m+\omega - \omega X^2)(\kappa+1 - X)^2,
\end{equation}
or
\begin{equation}\label{equation}
(X-1)^2
\big(
a(\omega,\kappa) X^2 - 2b(\omega,\kappa) X - c(\omega,\kappa)
\big) = 0,
\end{equation}
with \nn{$a$ and $b$ are already used ....}
\ak{changed the coefficients in
\eqref{psi-x-e-o-e-o}
to $c_1$ and $c_2$...}
\begin{eqnarray}\label{a-b-c}
\begin{array}{l}
a(\omega,\kappa) = m (\kappa+1)^2 -\omega  \kappa(\kappa+ 2),
\\[1ex]
b(\omega,\kappa)= \kappa\big( m (\kappa+ 1) -  \omega \kappa\big),
\\[1ex]
c(\omega,\kappa) = m (\kappa+1)^2 - \omega \kappa (3\kappa +2).
\end{array}
\end{eqnarray}

We have:

\begin{itemize}
 \item $a(\omega,\kappa) = 0$
if and only if
$\kappa\ne -2$,
$\omega=\mathcal{T}_\kappa^{-}:=\frac{(\kappa+1)^2}{(\kappa+2)\kappa}m$
(we recall that we already assumed that $\kappa\ne 0$,
see \eqref{kappa-nonzero-1})\textup;
\item $b(\omega,\kappa) = 0$
if and only if
$\omega= 2\Omega_\kappa:=\frac{\kappa+1}{\kappa}m$\textup;
 \item $c(\omega,\kappa) = 0$
if and only if
$\kappa\neq -2/3$
and moreover
$\omega=\mathcal{T}_\kappa^{+}:=\frac{(\kappa+1)^2}{\kappa(3\kappa+2)}m$.
\end{itemize}
  Note that:
 \begin{itemize}
  \item $\mathcal{T}_\kappa^{-}\in(-m,m)$ if and only if $ \kappa \in(-2^{-\frac12}-1, 2^{-\frac12}-1)$\textup;
  \item $2\Omega_k\in(-m,m)$ if and only if $\kappa<-2^{-1}$\textup;
  \item $\mathcal{T}_\kappa^{+}\in(-m,m)$ if and only if $|\kappa|>2^{-\frac12}$.
 \end{itemize}

Equation \eqref{equation} has
the root $X_0=1$ of multiplicity two.

Let us consider the case $a(\omega,\kappa)=0$.
In this case, as we pointed out above,
$\kappa\neq-2$ and $\omega=\mathcal{T}_\kappa^{-}$;
one has:
$$b(\mathcal{T}_\kappa^{-},\kappa) = m\frac{\kappa(\kappa+1)}{\kappa+2},
\qquad
c(\mathcal{T}_\kappa^{-},\kappa) = -2m\frac{\kappa(\kappa+1)^2}{\kappa+2}.
$$

We note that
$b(\mathcal{T}_\kappa^{-},\kappa)\ne 0$
since
$\kappa\ne -1$, $\kappa\ne 0$
(see \eqref{kappa-nonzero-1})
and $\omega=\mathcal{T}_\kappa^{-}\ne 2\Omega_\kappa$,
and hence
equation~\eqref{equation} has the root
\[
X_-(\mathcal{T}_\kappa^{-},\kappa) = -\frac{c(\mathcal{T}_\kappa^{-},\kappa) }{2b(\mathcal{T}_\kappa^{-},\kappa)}=\kappa+1.
\]
In this case, we have
$X_-(\mathcal{T}_\kappa^{-},\kappa)^2
=(\kappa+1)^2
=-\frac{\mathcal{T}_\kappa^{-}}{m-\mathcal{T}_\kappa^{-}}$ (this corresponds to $\Lambda=m+\omega$).
\nn{In the limit $\omega\to \mathcal{T}_{-1}^{-}$, $X_+(\omega,\kappa)\to \infty$ if $\kappa \in (-2,-1)$ or $i\infty$ if $\kappa <-2$ or $\kappa \in (-1,0)$.}

   From now on, we can assume that
$a(\omega,\kappa) \neq 0$;
that is, either

$\kappa=-2$, or $\omega\ne \mathcal{T}_\kappa^{-}$

and $\kappa\not\in\{-2,\,-1,\,0\}$.
Hence, besides $X=1$, equation~\eqref{equation} has the roots
\begin{equation}\label{EQ3}
X_{\pm}(\omega,\kappa) = \frac{b(\omega,\kappa) \pm \sqrt{b^2(\omega,\kappa) + a(\omega,\kappa) c(\omega,\kappa) }}{ a(\omega,\kappa)},
\qquad \Re \sqrt{b^2(\omega,\kappa) + a(\omega,\kappa) c(\omega,\kappa) }\geq 0
\end{equation}

under the
conditions

$\omega+(m-\omega)X_{\pm}(\omega,\kappa)^2\neq 0$ and $ \kappa +1-X_{\pm}(\omega,\kappa)\neq 0$.
\nn{Added} These are

thus roots of \eqref{Eq:equation1},
and $\omega+(m-\omega)X_{\pm}(\omega,\kappa)^2= 0$ implies:

\begin{itemize}
 \item either $m+\omega-\omega X_{\pm}(\omega,\kappa)^2= 0$,

which contradicts $m>0$;
 \item or $ \kappa +1-X_{\pm}(\omega,\kappa)= 0$,
which leads to $\omega=\mathcal{T}_\kappa^{-}$.
\end{itemize}
Hence the conditions $\omega+(m-\omega)X_{\pm}(\omega,\kappa)^2\neq 0$ and $ \kappa +1-X_{\pm}(\omega,\kappa)\neq 0$ are here satisfied.

The solutions corresponding to bound states
or virtual levels
must satisfy the
following two necessary conditions
(coming from \eqref{X-vs-Lambda} and \eqref{EQ4}):
\begin{equation}\label{conditions}
\Re X \geq 0,
\qquad
\Re\left( \kappa+1 - \frac{\kappa^2}{\kappa+1 - X} \right)  \geq 0.
\end{equation}
The root
$X_0=1$ satisfies both conditions \eqref{conditions}
and corresponds to $\Lambda=0$.

To understand for what values of the parameters the functions $X_+(\omega,\kappa) $ and $X_-(\omega,\kappa) $ are
\emph{admissible solutions},
in the sense that they correspond to either eigenvalues
or virtual levels of the linearized operator, one could check directly conditions \eqref{conditions} on $X_\pm$.

However, the analysis of the second condition in \eqref{conditions}

requires rather lengthy calculations. For this reason, we choose a different approach which exploits a symmetry of equation \eqref{EQ1}.

We note that if $\Lambda$ is a solution of the original equation \eqref{EQ1},
then so is $-\Lambda$. This symmetry has a counterpart on the equation
in the variable $X$:
if $X$, with $\Re X\geq 0$,
is a solution of equation \eqref{across}, then so is
\begin{equation}\label{mountains}
Y = \sqrt{\frac{m+\omega - \omega X^2}{\omega+(m-\omega)X^2}},
\qquad
\Re Y\ge 0.
\end{equation}

\[
X =\sqrt{ \frac{m+\omega - \omega Y^2}{\omega+(m-\omega)Y^2} }.
\]

The functions $X_+(\omega,\kappa)$ and $X_-(\omega,\kappa)$ are conjugated
with respect to this symmetry of the equation,
in the sense that there are the following relations:
\begin{equation}\label{for}
X_+^2 = \frac{m+\omega - \omega X_-^2}{\omega+(m-\omega)X_-^2},
\qquad
X_-^2 = \frac{m+\omega - \omega X_+^2}{\omega+(m-\omega)X_+^2}.
\end{equation}
These relations hold true for all $\kappa\in\R$ and $-m<\omega<m$

and can be deduced by noting that the identities in equation \eqref{for} are equivalent to
\begin{eqnarray}\label{x2x2}
\omega(X_+^2+X_-^2) = (m+\omega) - (m-\omega)X_+^2X_-^2
\end{eqnarray}
and by using relations
$X_+^2+X_-^2 = 2(2b^2 + ac)/a^2$ and $X_+^2X_-^2 = c^2/a^2$.\nn{These last three identities, including the display, are useful but I do not see easily where they come from. We need anyway to rephrase it.}

Hence,
for all $\kappa\in \R$ and $-m<\omega<m$,
 $X_+(\omega,\kappa)$ and $X_-(\omega,\kappa)$ satisfy the
relations
\begin{equation}\label{dune}
 \kappa^2=(\kappa+1 - X_+)(\kappa+1 - X_-)
\end{equation}
and
\begin{equation}\label{Eq:Dune2}
X_++X_- =2b/a,
\qquad X_+X_- = - c/a.
\end{equation}
Identity \eqref{dune} implies
\[
\Re X_+ =\Re\left(\kappa+1 - \frac{\kappa^2}{\kappa+1 - X_-} \right)\quad \textrm{and} \quad  \Re X_- =\Re\left(\kappa+1 - \frac{\kappa^2}{\kappa+1 - X_+} \right) .
\]
\nn{Changes:} We also have
\[
\Re X_++\Re X_- =2b/a,
\qquad \Im X_+=-\Im X_-
\]
and
\[
\Re X_+\Re X_--\Im X_+\Im X_- = - c/a,
\qquad \Re X_+ \Im X_- +\Im X_+\Re X_-= 0.
\]

Hence either $X_+$ and $X_-$ are real,
or they are conjugate to each other, with real part $b/a$.

With \eqref{dune}, the double root is either $1$ or $2\kappa+1$ which is possible (and distinct from $1$). The latter holds if and only if
\begin{equation}\label{def-W-kappa}
\omega=W_\kappa:=m\frac{2\kappa^2+2\kappa+1}{2\kappa(\kappa+1)}.
\end{equation}
We note that the denominator is nonzero
in view of \eqref{kappa-nonzero-1}
and that $\abs{W_\kappa}>m$ for
$\kappa\in\R\setminus\{-1,-1/2,0\}$,
$\abs{W_\kappa}=m$ for $\kappa=-1/2$.
In the other cases, the roots
$X_\pm$
are conjugate with positive real part or real positive if and only if $b$ and $a$ have same sign and $c$ has opposite sign (due to \eqref{Eq:Dune2}). For instance, since
\begin{itemize}
 \item $a(\omega,\kappa) > 0$ is equivalent
to $\omega>\mathcal{T}_\kappa^{-}$
for $\kappa\in (-2,0)$,
or
to $\omega<\mathcal{T}_\kappa^{-}$
for $\kappa\not\in [-2,0]$\textup;
\item $b(\omega,\kappa) > 0$ is equivalent
to $\omega< 2\Omega_\kappa$ for $\kappa\neq 0$\textup;
\item $c(\omega,\kappa) > 0$ is equivalent
to $\omega>\mathcal{T}_\kappa^{+}$ for $\kappa\in(-\frac23,0)$,
or to $\omega<\mathcal{T}_\kappa^{+}$
for $\kappa\not \in[-\frac23,0]$\textup;
\end{itemize}
the roots are positive reals or conjugate with positive real part if and only if
\begin{itemize}
 \item  $\kappa\in (-\infty,-2)\cup (-2,-1)$ and  $\mathcal{T}_\kappa^{+}<\omega<2\Omega_\kappa$\textup;

 \item $\kappa\in (-1, -\frac23)$ and

$2\Omega_k<\omega<\mathcal{T}_\kappa^{-}$\textup;
 \item $\kappa\in (-\frac23,2^{-\frac12}-1)$ and

$-m<\omega<\mathcal{T}_\kappa^{-}$\textup;

\item  $\kappa\in (2^{-\frac12},+\infty)$ and $\mathcal{T}_\kappa^{+}<\omega<m$.

\end{itemize}

Outside these values of $\kappa$ and $\omega$,
equation \eqref{across} has only the trivial solution $X_0 =1$ corresponding to the double eigenvalue $\lambda_0 =\jj\Lambda_0 = 0$. Otherwise,
two additional  solutions $X_+(\omega,\kappa)$ and  $X_-(\omega,\kappa)$ appear. These correspond to two simple eigenvalues $\lambda_+ =\jj\Lambda_+ $ and $\lambda_- =\jj\Lambda_-=-\lambda_+$ with
\[
\Lambda_+ = \frac{1-X_+^2}{\frac{1}{m-\omega} +\frac{X_+^2}{m+\omega}},  \qquad
\Lambda_- = \frac{1-X_-^2}{\frac{1}{m-\omega} +\frac{X_-^2}{m+\omega}}=-\Lambda_+.
\]
Hence,
$\Lambda_+$ and $\Lambda_- = - \Lambda_+$
-- the roots of $\Gamma(\Lambda)$, when they exist --

are either real or purely imaginary.
We give now an explicit expression for the eigenvalues. One has:
\[\begin{aligned}
\Lambda_+ =&\frac12 \left( \frac{1-X_+^2}{\frac{1}{m-\omega} +\frac{X_+^2}{m+\omega}} - \frac{1-X_-^2}{\frac{1}{m-\omega} +\frac{X_-^2}{m+\omega}} \right) =
- \frac12
\frac{ \frac{X_+^2-X_-^2}{m+\omega}  +\frac{X_+^2-X_-^2}{m-\omega}}{\frac{1}{(m-\omega)^2} + \frac{X_+^2+X_-^2}{m^2-\omega^2} + \frac{X_+^2X_-^2}{(m+\omega)^2} }
\\=&- \frac{m}{m^2-\omega^2}
\frac{ X_+^2-X_-^2}{\frac{1}{(m-\omega)^2} + \frac{1}{m^2-\omega^2}  \frac{1}{\omega} (m+\omega - (m-\omega) X_+^2X_-^2)  + \frac{X_+^2X_-^2}{(m+\omega)^2} },
\end{aligned}
\]
where we used the identity
\eqref{x2x2}.

After straightforward calculations one arrives
at the following expression:
\begin{eqnarray}\label{lambda-plus}
\Lambda_+ = - \frac{4 b\, \omega  \sqrt{b^2 + ac}}{\frac{m+\omega}{m-\omega} a^2 - \frac{m-\omega}{m+\omega} c^2},
\end{eqnarray}
with $a=a(\omega,\kappa)$, $b=b(\omega,\kappa)$, and $c=c(\omega,\kappa)$.

The expression for the eigenvalues given in \eqref{lambda-plus} can be further elaborated and made explicit. Taking into account that
\begin{equation}\label{b2+ac}
b^2(\omega,\kappa) + a(\omega,\kappa) c(\omega,\kappa)
=(\kappa+1)\big(m(\kappa+1)-2\kappa\omega\big)
\big(m(2\kappa^2+2\kappa+1)-2\kappa(\kappa+1)\omega\big)
\end{equation}
and noticing that for $\omega\in[-m,m]$ one has  $\R\ni m(2\kappa^2+2\kappa +1)-2\kappa(\kappa+1)\omega>0\ $,
after a straightforward calculation one gets
\begin{align}\label{lambda-plus-minus}
\Lambda_\pm &= \pm\frac{m^2-\omega^2}{2\Omega_\kappa-\omega}\frac{\sqrt{2\kappa(\kappa+1)(\Omega_\kappa -\omega)(m(2\kappa^2+2\kappa +1)-2\kappa(\kappa+1)\omega)}}
{m(2\kappa^2+2\kappa +1)-2\kappa(\kappa+1)\omega}
\nonumber
\\
&=\pm\frac{m^2-\omega^2}{2\Omega_\kappa-\omega}\sqrt{\frac{2\kappa(\kappa+1)(\Omega_\kappa -\omega)}
{m(2\kappa^2+2\kappa +1)-2\kappa(\kappa+1)\omega}}
=\pm\frac{m^2-\omega^2}{2\Omega_\kappa-\omega}\sqrt{\frac{\Omega_\kappa -\omega}{W_\kappa-\omega}},
\end{align}
\nn{Maybe the last identity is valid only if $\kappa\neq-1$ and $\kappa\neq0$ due to the division by $2\kappa(\kappa+1)$. But $\kappa=0$ is an issue for defining $\Omega_\kappa$.}
\ak{Ok, now we seem safe, see the next sentence:}
where $\Omega_\kappa$ is from \eqref{def-Omega-kappa}
and $W_\kappa$ is from \eqref{def-W-kappa}.
Above, we factored out $\kappa(\kappa+1)$,
which is nonzero due to \eqref{kappa-nonzero-1}.

It then follows that for $\mathcal{T}_\kappa^{+} \leq \omega \leq \Omega_\kappa $,
where $\mathcal{T}_\kappa^{+} \leq W_\kappa$
\ac{Dec 6, 2021: The previous looks strange and the comma is missing}

the values $\Lambda_\pm$ are real
(hence the corresponding
eigenvalues $\lambda_\pm=\jj\Lambda_\pm$ are purely imaginary),
while for $\omega >\Omega_\kappa$ they are purely imaginary.
For $\omega=\Omega_\kappa$ the two eigenvalues coincide and are both equal to zero, and $\lambda =0$ is an eigenvalue with total algebraic  multiplicity four. Notice also that $\Lambda_\pm$ are going to infinity as $\omega$ approaches $2\Omega_\kappa$ if $\kappa<-1$.
The analysis just given covers the cases of Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4-a},
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4-e}
and Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4-f}. The other cases can be treated exactly the same way. The explicit expression of eigenvalues is the same, and the ranges of $\kappa$ and $\omega$ are determined as before from the requirement that eigenvalues belong to the first Riemann sheet of $\Gamma(\Lambda)$, and that they are in the gap $(-\omega,\omega)$.\medskip
\nn{Correction}
Since $W_\kappa\not\in(-m,m)$
(see our comment after \eqref{def-W-kappa}),
\ac{Dec 6, 2021: ``it''... this value??}
it is negative for $\kappa\in(-1,0)$ and positive for $\kappa\in(-\infty,-1)\cup(0,+\infty)$ and $\Omega_\kappa\in(-m,m)$ if and only if $\kappa\not\in (-1/3,1)$ (it is decreasing in $\kappa$),
it then follows that
\begin{itemize}
  \item for $\kappa\in(-\infty,-1)$\footnote{for $\kappa=-1/2$, $W_\kappa=-m$} or $\kappa\in (1,\infty)$\footnote{for $\kappa=1$, $\Omega_\kappa=m$}
and
  \begin{itemize}
   \item any $\omega\in(-m,\Omega_k)$ the values $\Lambda_\pm$ are real\textup;
   \item any $\omega\in(\Omega_k,m)$ the values $\Lambda_\pm$ are purely imaginary\textup;
   \item $\omega=\Omega_\kappa$ then $\Lambda_\pm=0$\textup;
  \end{itemize}
  \item for $\kappa\in\{-1,0\}$ and any $\omega\in(-m,m)$ then $\Lambda_\pm=0$\textup;

  \item for $\kappa\in(-1,-1/3)$\footnote{for $\kappa=-1/3$, $\Omega_\kappa=-m$}
  \begin{itemize}
   \item any $\omega\in(-m,\Omega_k)$ the values $\Lambda_\pm$ are purely imaginary\textup;
   \item any $\omega\in(\Omega_k,m)$ the values $\Lambda_\pm$ are real\textup;
   \item $\omega=\Omega_\kappa$ then $\Lambda_\pm=0$\textup;
  \end{itemize}

 \item for $\kappa\in[-1/3,0)$ and any $\omega\in(-m,m)$ the values $\Lambda_\pm$ are purely imaginary\textup;
 \item for $\kappa\in(0,1]$ and any $\omega\in(-m,m)$ the values $\Lambda_\pm$ are real.
 \end{itemize}

The analysis just given covers the cases of Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4-a},
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4-e}
and Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4-f}. The other cases can be treated exactly the same way. The explicit expression of eigenvalues is the same, and the ranges of $\kappa$ and $\omega$ are determined as before from the requirement that eigenvalues belong to the first Riemann sheet of $\Gamma(\Lambda)$,

and that they are in $\jj(-m+\abs{\omega},m-\abs{\omega})$.

\medskip

Once we know that there are exactly two zeros of the function $\Gamma(\Lambda)$
which could correspond to eigenvalues,
and therefore located on the real or on the imaginary axis
(let us mention that
the spectrum of $\bfA$
remains symmetric with respect to $\R$ and $\jj\R$
even after its restriction
onto $\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$),
we can use
a simpler argument to locate the eigenvalues and threshold resonances.
\nn{Added:}
Due to spectral stability,
we know that for each $\kappa\in\R\setminus[-1/3,1]$,
at $\omega=\Omega_\kappa\in(-m,m)$,
\nn{added $...\setminus[-1/3,1]$ and $...\in(-m,m)$}
there is a collision of two eigenvalues
since the dimension of the generalized null space of $\bfA$
jumps at this value of $\omega$.
\nn{due to spectral stability ?}
\nn{Perhaps; added (above).}
All we need to do is to find when these eigenvalues disappear
from the spectral gap
\nn{?? real spectral gap ?}
\nn{OK, adding:}
$\jj(-m+\abs{\omega},m-\abs{\omega})$;
that is, when the points
$\lambda=\pm\jj(m-\abs{\omega})$
become \nn{embedded ? threshold ?}\nn{OK, adding ``threshold'':}
threshold eigenvalues or virtual levels.
Moreover, if these points become virtual levels, this means that
the corresponding $\nu_\pm$ changes the sign at this point,
so the zeros of $\Gamma(\Lambda)$ move onto the unphysical sheet of its
Riemann surface, becoming resonances (corresponding to antibound states).

\begin{lemma}\label{lemma-t-kappa}
The restriction of $\bfA(\omega,\kappa)$
onto $\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$
has virtual levels at the thresholds of the essential spectrum
$\lambda=\pm\jj(m-\abs{\omega})$
at the following values of
$\omega\in(-m,m)\setminus\{0\}$
and $\kappa\in\R$:
\begin{enumerate}
\item
For $\kappa<-1$
or $\kappa>2^{-1/2}$,
there are virtual levels at $\lambda=\pm\jj(m-\omega)$
when $\omega=\mathcal{T}_\kappa^{+}:=\frac{(k+1)^2 m}{(3\kappa+2)\kappa}>0$.
\item
For $-1<\kappa<2^{-1/2}-1$,
there are virtual levels at $\lambda=\pm\jj(m+\omega)$
when $\omega=\mathcal{T}_\kappa^{-}:=\frac{(k+1)^2 m}{(\kappa+2)\kappa}<0$.

\end{enumerate}
\end{lemma}

\begin{proof}
We first consider the case $\omega>0$.
Let us find when an imaginary eigenvalue touches the
essential spectrum at the endpoint
$\lambda=\jj(m-\omega)$.
(By Lemma~\ref{lemma-no},
the endpoints never correspond to eigenvalues.)
\nn{Has to be shown}
\nn{trying to show; see Lemma~\ref{lemma-no} above.}
One needs
\[
\kappa^2
=
(\kappa+1)
\left(\kappa+1
-
\sqrt{\frac{m+\omega}{m-\omega}}
\sqrt{2\frac{m-\omega}{2\omega}}
\right)
=
(\kappa+1)
\left(\kappa+1
-
\sqrt{\frac{m+\omega}{\omega}}
\right).
\]
That is,
\begin{eqnarray}\label{bad-kappa}
2\kappa+1
=
(\kappa+1)
\sqrt{\frac{m+\omega}{\omega}}
,
\qquad
\frac{2\kappa+1}{\kappa+1}
=
\sqrt{1+\frac{m}{\omega}}.
\end{eqnarray}
We point out that if $\kappa\in(-1,-1/2)$,
the fraction on the left is strictly negative
while the square root is nonnegative; these values of $\kappa$
can not correspond to virtual levels at the threshold
$\jj(m-\omega)$ with $\omega>0$.
The condition to have a virtual level or an eigenvalue
at some value $0<\omega<m$
takes the form
\[
\frac{2\kappa+1}{\kappa+1}>\sqrt{2},
\]
which leads to $\kappa>2^{-1/2}$.
Let us compute the value of $\omega$
corresponding to a virtual level:
\[
\frac{m}{\omega}=
\frac{(2\kappa+1)^2}{(\kappa+1)^2}-1
=\frac{(3\kappa+2)\kappa}{(\kappa+1)^2},
\]
hence the critical value which corresponds to virtual levels
at the thresholds $\lambda=\pm \jj(m-\omega)$ is given by
\begin{eqnarray}\label{def-omega-kappa}
\mathcal{T}_\kappa^{+}:=m\frac{(\kappa+1)^2}{(3\kappa+2)\kappa},
\end{eqnarray}
which is non-negative and satisfies $0\le\omega<m$ for
$\kappa>2^{-1/2}$ and for $\kappa\le -1$.

\nn{Minor changes in the next two sentences.}
Now we consider the case $\omega<0$.
To find the value of $\omega$ which corresponds
to a bifurcation of an eigenvalue from the threshold
of the essential spectrum (that is, when there is
a virtual level or an eigenvalue at the threshold
$\lambda=\jj(m-\abs{\omega})=\jj(m+\omega)$),
we consider the equation $\Gamma(\Lambda)=0$,
with $\Lambda=m+\omega$
and with $\Gamma(\Lambda)$ defined in \eqref{condition-e-o-e-o}.
Taking into account that $\nu_{+}(\omega,\Lambda)=0$
at $\Lambda=m+\omega$,
we are to solve:
\[
0=\Gamma(m+\omega)=2(\kappa+1)m\mu\nu_{-}+4m\omega.
\]
This leads to
\[
(\kappa+1)\mu\nu_{-}(\omega,\Lambda)=-2\omega.
\]
Since $\omega<0$,
we conclude that
the eigenvalue touches the threshold
$\lambda=\jj(m+\omega)$
(that is, there is a virtual level at threshold)
if $\kappa>-1$.
Squaring the above equation and substituting $\nu_{-}$, we arrive at

\[
(\kappa+1)^2\frac{m-\omega}{m+\omega}
(-4\omega m-4\omega^2)=4\omega^2,
\]
hence
$
(\kappa+1)^2(m-\omega)=-\omega,

$
which leads to
\[
\omega=\frac{(\kappa+1)^2 m}{(\kappa+2)\kappa}.
\]
This value satisfies
$-m<\omega<0$ when $\kappa<2^{-1/2}-1$
and we pointed out that this case is relevant if $\kappa>-1$.
\end{proof}

\ac{Rewritten,
both lemma and its proof; July 10, 2021}
\nn{The case $\omega\to-m+$ for $\kappa=-1/2$.}
\ac{Indeed; added $\kappa=-1/2$}
\begin{lemma}\label{lemma-w-kappa}
\ac{
\begin{enumerate}
\item
\sout{Let $\kappa\in\R\setminus\{-1,0\}$.
One has
$2\Omega_\kappa\in(-m,m)$ as long as $\kappa<-1/2$\textup;
}
\nn{already stated}
\item
\sout{
As $\omega$ approaches
$2\Omega_\kappa+0$ for $-1<\kappa<-1/2$
or
$2\Omega_\kappa-0$ for $\kappa<-1$,
\nn{This case does not seem to allow eigenvalues ??}
the two real eigenvalues
go to $\pm\infty$.
}
\end{enumerate}
}
\ac{REFORMULATED:}
As
\[
\omega\to
\begin{cases}
2\Omega_\kappa-0,&\kappa<-1,
\\
2\Omega_\kappa+0,&-1<\kappa\le -1/2,
\end{cases}
\]
one has
$\lambda=\jj\Lambda_\pm\to\pm\infty$.
\end{lemma}
\ac{\sout{\nn{Added:} Recall that for $\kappa=-1$, $\Lambda_\pm=0$.}}

\begin{proof}
\ac{TO REMOVE:
\sout{The first part follows from \eqref{def-Omega-kappa}.
we have:
$2\Omega_\kappa=\frac{\kappa+1}{\kappa}m$.
It follows that $\abs{2\Omega_\kappa}<m$
when $\kappa<-1/2$.}
}

\nn{The proof of the second part is strange !
I would write that as long as $\Lambda_\pm$ is real, $\Omega_\kappa\neq 2\Omega_\kappa$
(that is, $\kappa\neq -1$),
and $2\Omega\neq \pm m$ (that is, $\kappa\neq -1/2$),
then the limits of $\Lambda_\pm$ for $\omega\to 2\Omega_\kappa$
are infinite (one being positive and the other negative).}
\ac{Shortened:}
The statement of the lemma
follows from \eqref{lambda-plus-minus},
\[
\Lambda_\pm=\pm\jj\frac{m^2-\omega^2}{2\Omega_\kappa-\omega}
\sqrt{\frac{\Omega_\kappa-\omega}{W_\kappa-\omega}}.
\]
We note that
$\Omega_\kappa$ from \eqref{def-Omega-kappa}
and $W_\kappa$ from \eqref{def-W-kappa}
satisfy
\[
\begin{cases}
0<\Omega_\kappa<2\Omega_\kappa<m,\qquad W_\kappa>m, &\kappa<-1,
\\
-m\le 2\Omega_\kappa<\Omega_\kappa<0,\qquad W_\kappa\le -m,& -1<\kappa\le -1/2,
\end{cases}
\]
so that
$\lambda=\jj\Lambda_{\pm}$
are real and approach $\pm\infty$
as
$\omega\to 2\Omega_\kappa$.
\end{proof}

\ac{OLDER PROOF; TO REMOVE:
\sout{
For the second part,
to know when two real eigenvalues
go to $\pm\infty$ along the real axis,
it is enough to analyze when $\Lambda$
takes infinite values.
By \eqref{Lambda-vs-X},
this happens if and only if
}
\begin{eqnarray}\label{a-c}
X^2=-\frac{m+\omega}{m-\omega}.
\end{eqnarray}
\sout{If $a=0$, then
\eqref{equation}
can have at most one root different from $X=1$;
this root can not correspond to a nonzero eigenvalue
since eigenvalues come in pairs.
Thus, we may assume that $a\ne 0$.
Since $X=(b\pm\sqrt{b^2+ac})/a$ (see \eqref{EQ3})
with $a,\,b,\,c\in\R$ from \eqref{a-b-c},
one can see that $b=0$
(or else $X$ would not be purely imaginary).
By \eqref{a-b-c},
the relation $b(\omega,\kappa)=0$ is satisfied
if $\omega=(1+\kappa^{-1})m$,
and then
}
\[
X^2
=-\frac{c}{a}
=
-\frac{m(\kappa+1)^2-\omega\kappa(3\kappa+2)}
{m(\kappa+1)^2-\omega\kappa(\kappa+2)}
=
-\frac{(\kappa+1)^2-(\kappa+1)(3\kappa+2)}
{(\kappa+1)^2-(\kappa+1)(\kappa+2)}
=1+2\kappa
=1-\frac{2m}{m-\omega},
\]
\sout{
 hence \eqref{a-c} is satisfied.
 We notice that for the proof
 one can also make use of the explicit formula \eqref{lambda-plus-minus}.
 }
}

Lemmata~\ref{lemma-t-kappa} and~\ref{lemma-w-kappa}
complete the proof of
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4}.

\subsection{Zero eigenvalue and the Kolokolov condition}\label{Sec:Kolokolov}

Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-5}
immediately follows from Lemma~\ref{lemma-l} (see also Remark~\ref{Rem:OnZ}).

\noindent
Now let us prove
Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-6}.

\nn{Small changes here;
moved
$\bfX\sb{\mbox{\rm\footnotesize odd-even-odd-even}}$
to the end as suggested by Nabile.
}
Let us consider
$\bfA(\omega,\kappa)$
in the invariant subspace
$\bfX\sb{\mbox{\rm\footnotesize even-odd-even-odd}}$
(see \eqref{def-eoeo}).
We notice that for the restriction of $\bfA(\omega,\kappa)$
onto this subspace
one has:
\begin{eqnarray}\label{a-jordan}
\bfA \begin{bmatrix}0\\\phi_\omega\end{bmatrix}= \begin{bmatrix}0\\0\end{bmatrix},
\qquad
\bfA \begin{bmatrix}\p_\omega\phi_\omega\\0\end{bmatrix}=\begin{bmatrix}0\\\phi_\omega \end{bmatrix}.
\end{eqnarray}
By \eqref{a-jordan},
we already know that
the generalized null space of $\bfA$ is at least
two-dimensional.
Whether there are more elements in the generalized null space of $\bfA$,
depends on whether there is a solution
$\theta\in L^2(\R,\C^2)$ to
\begin{eqnarray}\label{ltp}
L_{-}\theta=\p_\omega\phi,
\end{eqnarray}
so that
\[
\bfA\begin{bmatrix}0\\\theta
\end{bmatrix}=\begin{bmatrix}\p_\omega\phi\\0\end{bmatrix}.
\]
Since the range of $L_{-}$
is closed,
there is a solution to \eqref{ltp}
if and only if
its right-hand side,
$\p_\omega\phi$, is orthogonal to $\ker(L_{-}^*)=\ker(L_{-})$.
By Lemma~\ref{lemma-l},
the kernel of $L_{-}(\omega)=L(\omega,0)$
on
$\bfX\sb{\mbox{\rm\footnotesize odd-even}}$
is zero (since $\omega\ne 0$),
while its kernel on
$\bfX\sb{\mbox{\rm\footnotesize even-odd}}$
is spanned
by $\phi_\omega$.
Thus, the condition to have a solution to \eqref{ltp}
is given by
\[
\langle\phi_\omega,\p_\omega\phi_\omega\rangle
=\frac{1}{2}\p_\omega Q(\phi_\omega).
\]
We conclude that whether there are more elements
in the generalized null space of $\bfA$,
depends on the Kolokolov condition
$\p_\omega Q(\phi_\omega)=0$ \cite{kolokolov-1973};
this condition gives the value of the threshold
$\omega=\Omega_\kappa$ at which the dimension
of the generalized null space
$\mathfrak{L}(\bfA(\omega,\kappa))$
changes.
Let us compute $\p_\omega Q(\phi_\omega)$.
For the
$L^2$-norm of a solitary wave profile
$\phi_\omega(x)=\begin{bmatrix}v(x)\\u(x)\end{bmatrix}$
from \eqref{solitary-wave},
we have:
\[
Q(\phi_\omega)
=
\int_{\R}
(v^2+u^2)\,dx
=\alpha^2(1+\mu^2)
\int_{\R}
e^{-2\varkappa\abs{x}}\,dx
=\frac{\alpha^2(1+\mu^2)}{\varkappa}.
\]
\ak{Moved this part here [from the beginning of
Section~\ref{sect-theorem-sect2-proof}
where it was not needed]
}
Combining \eqref{a-b-f}, \eqref{p-alpha},
and the definition \eqref{def-kappa},
we can express
\begin{eqnarray}\label{p-alpha-kappa}
2\kappa\mu\p_\omega\alpha=\alpha\p_\omega\mu.
\end{eqnarray}
Using the relations
\begin{eqnarray}\label{p-of}
\p_\omega\alpha=\frac{\alpha\p_\omega\mu}{2\kappa\mu},
\qquad
\p_\omega\varkappa=-\frac{\omega}{\varkappa},
\qquad
\p_\omega\mu=-\frac{m}{
(m+\omega)\varkappa
}
\end{eqnarray}
(see \eqref{def-def} and \eqref{p-alpha-kappa}),
we derive:
\begin{eqnarray}\label{q-prime}
&&\hskip -54pt
\p_\omega Q(\phi(\omega))
=
\frac{
2\alpha(1+\mu^2)\varkappa\p_\omega\alpha
+
2\alpha^2\varkappa\mu\p_\omega\mu
-
\alpha^2(1+\mu^2)\p_\omega\varkappa
}{\varkappa^2}
\nonumber

\\
&&
=
\frac{2m\alpha^2}{(m+\omega)\varkappa^2}
\Big(
-
\frac{m}{\varkappa \kappa}
-
\mu
+
\frac{\omega}{\varkappa}
\Big)
=
\frac{2m\alpha^2}{(m+\omega)\varkappa^3}
\Big(
-
\frac{m}{\kappa}
-
m+2\omega
\Big).
\end{eqnarray}
Thus, we reduce the Kolokolov condition
$\p_\omega Q=0$
to the form

\begin{eqnarray}\label{omega-m-kappa}

\frac{\omega}{m}=\frac{1+\kappa}{2\kappa}.
\end{eqnarray}
We use this relation
to define the critical value $\Omega_\kappa$ in \eqref{def-Omega-kappa}
which corresponds to the critical point of $Q(\phi_\omega)$.
We point out that there is no critical value
$\omega\in(-m,m)$ of $Q(\phi(\omega))$
for
$-1/3\le\kappa\le 1$
since in this case \eqref{def-Omega-kappa}
yields $\abs{\Omega_\kappa}\ge m$.

\begin{remark}
Let us point out that,
in the context of the nonlinear Dirac equation,
the sign of $\p_\omega Q(\phi_\omega)$
is not directly related to the spectral stability:
by Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4},
the spectral regions correspond to
$\omega<\Omega_\kappa$
for $\kappa<-1$ and $\kappa>2^{-1/2}-1$
(hence $\p_\omega Q(\phi_\omega)<0$ by \eqref{q-prime})
and to
$\omega>\Omega_\kappa$ for $-1<\kappa<2^{-1/2}-1$
(hence $\p_\omega Q(\phi_\omega)>0$).
\end{remark}

\begin{remark}
One may check explicitly that indeed
for $\kappa\ne 0$
the equation $L_{-}\theta=\p_\omega\phi$ has a solution $\theta\in L^2(\R,\C^2)$
if and only if $\p_\omega Q=0$.
We have:
\[
\p_\omega\phi_\omega
=\p_\omega
\left(
\alpha\begin{bmatrix}1\\\mu\sgn x\end{bmatrix}e^{-\varkappa\abs{x}}
\right)
=
\left(
\begin{bmatrix}\p_\omega\alpha\\\p_\omega(\alpha\mu)\sgn x\end{bmatrix}
-\alpha\begin{bmatrix}1\\\mu\sgn x\end{bmatrix}\abs{x}\p_\omega\varkappa
\right)
e^{-\varkappa\abs{x}}.
\]
The solution to
\begin{eqnarray}\label{l-theta}
\begin{bmatrix}
m-\omega&\p_x
\\-\p_x&-m-\omega
\end{bmatrix}
\theta(x)
=\p_\omega\phi(x)
\end{eqnarray}
has the form
\begin{eqnarray}\label{theta}
\theta(x)=
\begin{bmatrix}
B\abs{x}+C\abs{x}^2
\\
(E\abs{x}+F\abs{x}^2)\sgn x
\end{bmatrix}
e^{-\varkappa\abs{x}},
\qquad
x\in\R;
\qquad
B,\,C,\,E,\,F\in\C;
\end{eqnarray}
when writing this expression,
we took into account that
the operator $D_m-\omega I_2$ is invariant in
the subspace of functions with even first component
and odd second component.
There is no jump condition to worry about since
$\theta(x)$ vanishes at the origin.
Because of the spatial symmetry,
it suffices to consider $x>0$.
Substitution of \eqref{theta}
into \eqref{l-theta} leads to the system
\[
\begin{cases}
(m-\omega)(B x+C x^2)+E+2F x
-\varkappa(E x+F x^2)=\p_\omega\alpha-\alpha\p_\omega\varkappa x,
\\
-(B+2C x)+\varkappa(Bx+C x^2)-(m+\omega)(E x+F x^2)
=\p_\omega(\alpha\mu)
-\alpha\p_\omega\varkappa \mu x,
\end{cases}
\qquad
x>0.
\]
The above system allows us to express
$E=\p_\omega\alpha$
and
$F=\mu C$
(from the first equation)
and also
$B=-\p_\omega(\alpha\mu)$
(from the second equation),
and then we derive an overdetermined system
\[
\begin{cases}
-(m-\omega)\p_\omega(\alpha\mu)+2\mu C-\varkappa\p_\omega\alpha
=-\alpha\p_\omega\varkappa,
\\
-2C-\varkappa\p_\omega(\alpha\mu)-(m+\omega)\p_\omega\alpha
=-\alpha\mu\p_\omega\varkappa
\end{cases}
\]
with the only unkonwn $C\in\C$.
This system yields the compatibility condition
\[
2\varkappa\p_\omega\alpha
+(m-\omega)\alpha\p_\omega\mu-\frac{\alpha m}{m+\omega}\p_\omega\varkappa=0.
\]
Substituting the expression for
$\p_\omega \alpha$ from \eqref{p-alpha-kappa}
(note that
it is for the finiteness of $\p_\omega\alpha$
that we needed the condition $\kappa\ne 0$)
and using the relations \eqref{p-of},
one again arrives at \eqref{omega-m-kappa}.
\end{remark}
\nn{We should use this to show that the Jordan block cannot be continued.}
\ac{I think this is done in the next paragraph;
is there anything else we need to do?}
Notice that
\[
\begin{bmatrix}0\\\theta
\end{bmatrix}
\mbox{
is orthogonal to
 }
 \begin{bmatrix}\phi_\omega\\0\end{bmatrix}
\]
and hence the Jordan chain can be continued. Hence the algebraic multiplicity of $0$ jumps at least by $2$ when \eqref{omega-m-kappa} is satisfied.
As the matter of fact, it is exactly two,
since, as we have seen
in the proof of
of Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4},
when \eqref{omega-m-kappa} is not satisfied,
there are at most two simple nonzero eigenvalues
of the restriction
$\bfA(\omega,\kappa)$
onto
$\bfX\sb{\mbox{\rm\footnotesize even-odd-even-odd}}$.
As long as $a(\omega,\kappa)\ne 0$
(see \eqref{a-b-c}),
these eigenvalues
are locally continuous functions of parameters,
moving to $\pm\infty$ along the real axis
as
$X$ approaches
$\pm \jj\sqrt{\frac{m+\omega}{m-\omega}}$
(cf. \eqref{Lambda-vs-X})
or equivalently
as $\omega$ approaches $2\Omega_\kappa$
(see Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-4}).

We note that if
\eqref{omega-m-kappa} is satisfied,
then, taking into account that $\kappa\ne 0$,
\[
a(\omega,\kappa) = m (\kappa+1)^2 -\omega  \kappa(\kappa+ 2)
=m(1+\kappa)^2
-\frac{1}{2}m(1+\kappa)(2+\kappa)

=\frac 1 2 m(\kappa+\kappa^2).
\]
Since $\kappa\ne 0$,
the function
$a(\omega,\kappa)$
vanishes only when
$\kappa=-1$
(then $\omega=0$ by \eqref{omega-m-kappa});
so, outside of the point
$(\omega,\kappa)=(0,-1)$,
the value of
$X$ is a continuous function of $\omega$ and $\kappa$
in an open neighborhood of the curve $\omega=\Omega_\kappa$.
If we consider $\Lambda\in\mathbb{D}_\delta$
in the disc of some fixed radius $\delta>0$,
then one can see

from \eqref{Lambda-vs-X}
that $\Lambda$
is also a continuous function
of $\omega$ and $\kappa$.
It follows that
there could be
at most two eigenvalues $\pm\jj\Lambda$
colliding at $\lambda=0$,
hence the algebraic multiplicity of
eigenvalue $\lambda=0$ can not jump by more than two.

\medskip

\nn{Minor changes below}
Now we consider
$\bfA(\omega,\kappa)$
in the invariant subspace
$\bfX\sb{\mbox{\rm\footnotesize odd-even-odd-even}}$
of $L^2(\R,\C^4)$
(see~\eqref{def-oeoe}).
By Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-2},
the restriction of $\bfA(\omega,\kappa)$
to this subspace contains eigenvalue
$\lambda=0$
only when $\omega=0$,
with both the geometric and algebraic multiplicities
being equal to two.
This completes the proof
of Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-6}.

\medskip

\ak{Needed to move this part to be the last one
since we refer from here to other parts of
Theorem~\ref{theorem-sect2}}
Finally, let us consider the essential spectrum
(Theorem~\ref{theorem-sect2}~\itref{theorem-sect2-7}).
For $\kappa=0$,
the essential spectrum
could be obtained from Lemma~\ref{lemma-l}:
\begin{eqnarray}\label{mm}
\sigma\sb{\mathrm{ess}}(\bfA\at{\kappa=0})
=
\sigma\sb{\mathrm{ess}}
\left(
\begin{bmatrix}0&1\\-1&0\end{bmatrix}
\otimes
(D_m-\omega I_2-2\mu\delta(x))
\right)
=
\jj\big(\R\setminus(-m+\abs{\omega},m-\abs{\omega})\big).
\end{eqnarray}
Then, for the general case,
the Weyl sequences do not depend on the value of
$\kappa\in\R$;
this proves that the Weyl spectrum
$\sigma\sb{\mathrm{ess},4}(\bfA)$
(in the terminology of \cite[\S I.4]{edmunds2018spectral})
coincides with \eqref{mm}.
\nn{the text before Section~\ref{sect-spectrum-oe} is rewritten:}
Its complement in the complex plane,
consists of a single connected component,
\begin{eqnarray}
\label{component}
\C\setminus\big(
\jj(-\infty,-(m-\abs{\omega})]
\cup
\jj[(m-\abs{\omega}),+\infty)
\big).
\end{eqnarray}
If $(\omega,\kappa)=(0,-1)$, then,
by Part~\itref{theorem-sect2-4-b},
this connected component consists of the point spectrum;
it follows that in this case one has
$
\sigma(\bfA(0,-1))
=\sigma\sb{\mathrm{ess}}(\bfA(0,-1))=\C$.

If $(\omega,\kappa)\ne(0,-1)$,
then, as we know from
Part~\itref{theorem-sect2-4},
the component \eqref{component}
contains at most discrete spectrum,
and we deduce that
the essential spectrum
$\sigma\sb{\mathrm{ess}}(\bfA)
:=\sigma\sb{\mathrm{ess},5}(\bfA)$
also coincides with \eqref{mm}
(for more details, see
\cite{edmunds2018spectral,opus}).

This proves
Part~\itref{theorem-sect2-7},
completing the proof of
Theorem~\ref{theorem-sect2}.

\section{Parity-preserving
perturbation of the Soler model}
\label{sect-nld-perturbation}

In this section we address by perturbative analysis
the effect of changing the Soler nonlinearity
by the term which breaks the $\mathbf{SU}(1,1)$-invariance
while preserving the parity:
the equation is invariant in subspaces
$\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$
and
$\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$
consisting of odd-even and in even-odd wave functions.

\subsubsection*{Model}

We consider the perturbation of the Soler model
changing the Lagrangian density
so that the self-interaction is based on the quantity
$\psi\sp\ast(\sigma_3+\epsilon I_2)\psi$,
$\epsilon\ne 0$, and formally
the dynamics is governed by the equation

\begin{eqnarray}\label{nld-point-perturbed-1}
\jj \p_t\psi=(\jj\sigma_2\p_x+\sigma_3 m)\psi
-\delta(x)f(\psi\sp\ast(\sigma_3+\epsilon I_2)\psi)
(\sigma_3+\epsilon I_2)\psi,
\quad
x\in\R,
\quad
t\in\R.
\end{eqnarray}
Above,
$f\in C^1(\R)$, $f(0)=0$,
and
the following jump condition on $\psi$ is understood
(cf. \eqref{jumpeq}):
\begin{equation}\label{jumpeq_paritypres}
\jj\sigma_2[\psi]_{0}
=f(\hat\psi\sp\ast(\sigma_3+\epsilon I_2)\hat\psi)(\sigma_3+\epsilon I_2)\hat\psi.
\end{equation}
Just like \eqref{nld-point},
this is a Hamiltonian $\mathbf{U}(1)$-invariant system,
but for $\epsilon\ne 0$
it is no longer $\mathbf{SU}(1,1)$-invariant.

\subsubsection*{Solitary waves}

Just like in \eqref{solitary-wave},
there are solitary wave solutions $\phi(x)e^{-\jj\omega t}$
to \eqref{nld-point-perturbed-1}
with
\[
\phi(x)
=\alpha
\begin{bmatrix}1\\\mu\sgn x\end{bmatrix}
e^{-\varkappa\abs{x}},
\qquad
\varkappa=\sqrt{m^2-\omega^2},
\qquad
\mu=\sqrt{\frac{m-\omega}{m+\omega}},
\qquad
\alpha>0.
\]
The value of $\alpha=\alpha(\epsilon)$
is to satisfy the jump condition
\eqref{jumpeq_paritypres}
with
\[
[\phi]_0=
2\begin{bmatrix}0\\\mu(\omega)\alpha\end{bmatrix},
\qquad
\hat\phi=
\begin{bmatrix}\alpha\\0\end{bmatrix},
\]
which leads to
$

2\jj\sigma_2\begin{bmatrix}0\\\mu(\omega)\alpha\end{bmatrix}
=
(\sigma_3+\epsilon I_2)
f
\begin{bmatrix}\alpha\\0\end{bmatrix},

$
resulting in
\begin{eqnarray}\label{alpha-mu}
2\mu(\omega)=(1+\epsilon)f;
\end{eqnarray}
above, $f=f(\tau)$ is evaluated at
\begin{eqnarray}\label{def-tau-ch3}
\tau:=\phi\sp\ast(\sigma_3+\epsilon I_2)\phi\at{x=0}
=(1+\epsilon)\alpha^2.
\end{eqnarray}

\subsubsection*{Linearization}

Let us consider the linearization at a solitary wave.
Using the Ansatz
\[
\psi(t,x)=(\phi(x)+\,r(t,x)+\jj s(t,x))e^{-\jj\omega t},
\qquad
r(t,x),\,s(t,x)\in\R^2;
\]
we derive that the perturbation
$(r(t,x),\,s(t,x))$
satisfies the following system (where we omit explicit and repetitive domain definition):
\[
\begin{cases}
-\dot s=D_m r-\omega r
-f\delta(x)(\sigma_3+\epsilon I_2)r
-2g\delta(x)(\phi\sp\ast(\sigma_3+\epsilon I_2)r)(\sigma_3+\epsilon I_2)\phi
=:L_{+}(\epsilon)r,
\\[1ex]
\dot r=D_m s-\omega s
-f\delta(x)(\sigma_3+\epsilon I_2)s
=:L_{-}(\epsilon)s,
\end{cases}
\]
where
\begin{eqnarray}\label{def-f-g-ch3}
f=f(\tau),
\qquad
g=f'(\tau)
\end{eqnarray}
evaluated at $\tau$ from \eqref{def-tau-ch3}.
Explicitly,
\[
L_{-}(\epsilon)s
=
D_m s-\omega s
-f\delta(x)(\sigma_3+\epsilon I_2)s
\]
and
\[
L_{+}(\epsilon)r
=
(D_m-\omega)r
-f\delta(x)(\sigma_3+\epsilon I_2)r
-2g\delta(x)\phi^*(\sigma_3+\epsilon I_2)r(\sigma_3+\epsilon I_2)\phi
\]
\[
=
(D_m-\omega)r
-f\delta(x)(\sigma_3+\epsilon I_2)r
-2\alpha g\delta(x)(1+\epsilon)r_1
\begin{bmatrix}
(1+\epsilon)\alpha\\0
\end{bmatrix}
\]
\[
=
D_m r-\omega r
-f\delta(x)(\sigma_3+\epsilon I_2)r
-2(1+\epsilon)^2 g\alpha^2\delta(x)\Pi_1 r,
\]
with $\Pi_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
and with
with $f$, $g$ from \eqref{def-f-g-ch3}.
Thus, the linearization operator is given by
\begin{eqnarray}\label{def-a-not-broken}
\bfA(\epsilon)
=
\begin{bmatrix}
0&D_m-\omega I_2-f\delta(x)(\sigma_3+\epsilon I_2)
\\-D_m+\omega I_2
+
\delta(x)(f\sigma_3+f\epsilon I_2
+2(1+\epsilon)^2 g\alpha^2\Pi_1)
)
&0
\end{bmatrix}.
\end{eqnarray}

We are going to prove that
there are no unstable eigenvalues
bifurcating from $\pm 2m\jj$ for $\epsilon\ne 0$.
Since these eigenvalues correspond to the invariant subspace
$\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$ of $\bfA$
(see \eqref{def-x-x}),
which is also an invariant subspace for $\bfA(\epsilon)$,
it is enough to consider this operator in this subspace.
(As in the even-odd-even-odd subspace analysis
in Section~\ref{sect-spectrum-eo},
the spectrum of the restriction of $\bfA(\epsilon)$
on the invariant subspace
$\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$
contains no eigenvalues in the vicinity of the essential spectrum
except possibly near the thresholds $\jj(\pm m\pm\omega)$.)

Both
$L\sb\pm$ are invariant in the subspace
of $L^2(\R,\C^2)$
consisting of odd-even (and, similarly, even-odd) functions.
Moreover, the restrictions of
$L_{-}(\epsilon)$ and $L_{+}(\epsilon)$
onto odd-even spaces are equal,
therefore
\[
\bfA(\epsilon)\big\vert\sb{\bfX\sb{\mbox{\rm\footnotesize odd-even-odd-even}}}
=
\begin{bmatrix}0&L_{-}(\epsilon)\\-L_{-}(\epsilon)&0\end{bmatrix}
=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\otimes L_{-}(\epsilon)
\]
has purely imaginary spectrum.
Let us give a more accurate argument.

\begin{theorem}\label{theorem-instability}
\ac{Added the requirement that $\omega$ is close to $m$.
Not sure that we actually need it...
but at least this was used in the display
after \eqref{def-kappa-xi}...
}
There is
$\omega_0\in(0,m)$ and an open neighborhood $U\subset\R$,
$U\ni 0$,
such that for
$\omega\in(\omega_0,m)$ and $\epsilon\in U$
the operator $\bfA(\epsilon)$ has two eigenvalues
\begin{eqnarray*}
\lambda(\epsilon)=\pm\jj(2\omega+\zeta(\epsilon)),
\qquad
\zeta(\epsilon)\in\R\quad \forall\epsilon\in U,
\qquad
\lim\sb{\epsilon\to 0}\zeta(\epsilon)=0.
\end{eqnarray*}

\end{theorem}
\nn{It seems to me that it is just a spectral stability result on $\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$ and $\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$ separately ? Do we really need a theorem ?}
\ak{We need to formulate this result somehow. For uniformity
[see Theorem~\ref{theorem-sect2} and Theorem~\ref{theorem-instability-broken}],
let us keep it as a theorem.}
\begin{proof}
To study
whether $\lambda(\epsilon)=\jj\Lambda(\epsilon)$
is an eigenvalue of the operator $\bfA(\epsilon)$
from \eqref{def-a-not-broken},
we consider the action of $\bfA(\epsilon)-\jj\Lambda(\epsilon) I_4$
onto the superposition

\[
\Psi
=
a
\begin{bmatrix}
\nu_{+}\sgn x\\S_{+}\\-\jj\nu_{+}\sgn x\\-\jj S_{+}
\end{bmatrix}e^{-\nu_{+}\abs{x}}
+
b
\begin{bmatrix}
-\jj\xi\sgn x\\S_{-}\\\xi\sgn x\\\jj S_{-}
\end{bmatrix}e^{\jj\xi\abs{x}}
+
c
\begin{bmatrix}
\nu_{+}\\S_{+}\sgn x\\-\jj\nu_{+}\\-\jj S_{+}\sgn x
\end{bmatrix}e^{-\nu_{+}\abs{x}}
+
d
\begin{bmatrix}
-\jj\xi\\S_{-}\sgn x\\\xi\\\jj S_{-}\sgn x
\end{bmatrix}e^{\jj\xi\abs{x}},
\]
with $S_{+}=S_{+}(\omega,\Lambda)$ and $S_{-}=S_{-}(\omega,\Lambda)$ from \eqref{def-r-s}
and with $\nu_{+}$ and $\xi$
defined by
\begin{equation}\label{def-kappa-xi}
\nu_{+}(\omega,\Lambda)=\sqrt{m^{2}-(\Lambda-\omega)^{2}},
\qquad
\xi(\omega,\Lambda)=-\sqrt{(\omega+\Lambda)^{2}-m^{2}}
\end{equation}
(cf. \eqref{def-m-n});
we consider $\lambda=\jj\Lambda$ in the first quadrant,
so that $\Lambda$ (and thus $\zeta$)
has non-positive imaginary part;
then, for $\omega$ sufficiently close to $m$,
\[
\Re\xi
=-\Re((3\omega+\zeta)^2-m^2)^{1/2}
=-\Re(9\omega^2-m^2+6\omega\zeta+\zeta^2)^{1/2}
=-\sqrt{9\omega^2-m^2}
+\mathcal{O}(\zeta)
<0,
\]
\[
\Im\xi

=-(2\sqrt{9\omega^2-m^2})^{-1}6\omega\Im\zeta
+\mathcal{O}(\zeta)\Im\zeta
\ge 0.
\]
 \nn{?? What does $\approx$ mean ? Where does it come from ?}
 \ac{above, added explicit expressions.}
We note that
the first two terms in the above Ansatz for $\Psi$
are obtained from \eqref{psi-o-e} by substituting
$\nu\sb{-}(\omega,\Lambda)$ with $-\jj\xi(\omega,\Lambda)$
(we note that both expressions have
positive real part)
\ac{corrected:}
and correspond to perturbations from
the invariant subspace $\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$;
the second two terms
  \ak{\sout{come from changing the parity
  since $\bfA(\epsilon)$ for $\epsilon\ne 0$
  is no longer invariant
  in the subspaces
  \eqref{def-oeoe} and \eqref{def-eoeo}}}
  \nn{??? This is not what is claimed before the statement of the theorem ?
  Moreover \eqref{j-c-0} does not mix $(a,b)$ and $(c,d)$ !}
  \ak{Indeed! Correcting.}
correspond to perturbations
from the invariant subspace $\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$.
The relation $(\bfA-\jj\Lambda I_4)\Psi=0$
leads to the following jump condition:
\begin{eqnarray}\label{j-c-0}
\begin{cases}
2(-\jj S_{+} c
+\jj S_{-} d)
-(1+\epsilon)(-\jj\nu_{+} c+\xi d)f
=0
\\-2(-\jj\nu_{+} a+\xi b)
+(1-\epsilon)(-\jj S_{+} a+\jj S_{-}b)f
=0
\\-2(S_{+} c+S_{-} d)
+\big((1+\epsilon)f+2 g\alpha^2(1+\epsilon)^2\big)(\nu_{+} c-\jj\xi d)=0
\\2(\nu_{+} a-\jj\xi b)-(1-\epsilon)(S_{+} a+S_{-} b)f=0.
\end{cases}
\end{eqnarray}
As in the case of the unperturbed operator $\bfA$ (see \eqref{def-a-ch2}),
there are two invariant subspaces
of $\bfA(\epsilon)$
defined in \eqref{def-x-x}:
$\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$

corresponding to $a=b=0$
and
$\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$

corresponding to $c=d=0$
\nn{added:}
(note that the system \eqref{j-c-0} does not mix $a,\,b$ and $c,\,d$).
We are interested in the deformation of eigenvalues $\pm 2\omega\jj$
corresponding to $\bfX\sb{\mbox{\rm\footnotesize odd-even-odd-even}}$.

\bigskip

\noindent
$\bullet$
The spectrum of
$\bfA(\epsilon)$ restricted onto
$\bfX\sb{\mbox{\rm\footnotesize even-odd-even-odd}}$.
We do not need to consider this case
since $\bfA(0)$ restricted onto
$\bfX\sb{\mbox{\rm\footnotesize even-odd-even-odd}}$
only has simple isolated purely imaginary eigenvalues,
which have to stay on imaginary axes because of the symmetries
\eqref{symmetry}.
  \nn{It is surprising as $\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$
  there is only simple eigenvalues for $\epsilon=0$ while on
  $\bfX\sb{\mbox{\rm\footnotesize even-odd-even-odd}}$ the kernel
  provides some extra degeneracy .... so on
  $\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$
  one has to consider embedded $\pm \jj \omega$,
  $|\omega| \ge m/3$}
\ac{
$\pm 2\jj\omega$ is in $\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$;
when it is embedded ($\omega>m/3$),
then perturbations could produce bifurcations even if
it were simple.
Since we have only simple eigenvalues and no embedded eigenvalues
in $\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$,
here we have nothing to worry about.
}
For completeness, we mention that
in this case
the jump condition \eqref{j-c-0} takes the form
\[
\begin{cases}
2(-\jj S_{+} c+\jj S_{-} d)
-(1+\epsilon)(-\jj\nu_{+} c+\xi d)f
=0
\\-2(S_{+} c+S_{-} d)
+\big((1+\epsilon)f+2g\alpha^2(1+\epsilon)^2\big)(\nu_{+} c-\jj\xi d)=0,
\end{cases}
\]
and the compatibility condition
for having a nontrivial solution $c,\,d\in\C$
is given by
\[
\det
\begin{bmatrix}
-2\jj S_{+}+\jj(1+\epsilon)f\nu_{+}
&
2\jj S_{-}
-(1+\epsilon)f\xi
\\
-2 S_{+}
+\big((1+\epsilon)f+2 g\alpha^2(1+\epsilon)^2\big)\nu_{+}
&
-2 S_{-}
-\big((1+\epsilon)f+2 g\alpha^2(1+\epsilon)^2\big)\jj\xi
\end{bmatrix}
=0.
\]

\noindent
$\bullet$
The spectrum of
$\bfA(\epsilon)$ restricted onto
$\bfX\sb{\mbox{\rm\footnotesize odd-even-odd-even}}$.
The jump condition \eqref{j-c-0} takes the form
\[
\begin{cases}
-2(-\jj\nu_{+} a+\xi b)
+(1-\epsilon)(-\jj S_{+} a+\jj S_{-}b)f
=0
\\2(\nu_{+} a-\jj\xi b)-(1-\epsilon)(S_{+} a+S_{-} b)f=0.
\end{cases}
\]
The compatibility condition is:
\[
\det
\begin{bmatrix}
2\jj\nu_{+}-\jj(1-\epsilon)S_{+} f
&
-2\xi
+\jj(1-\epsilon)S_{-} f
\\2\nu_{+}-(1-\epsilon)S_{+} f
&
-2\jj\xi-(1-\epsilon)S_{-} f
\end{bmatrix}
=
-2\jj (2\nu_{+}-(1-\epsilon)S_{+} f)
(2\jj\xi+(1-\epsilon)S_{-} f)=0.
\]
The deformation of the eigenvalue $2\omega\jj$
corresponds to vanishing of the first factor;
thus,
$
\nu_{+}=\frac{1}{2}(1-\epsilon)S_{+} f;
$
squaring this relation, we arrive at
\[
m^2-(\omega+\zeta)^2
=\frac{1}{4}(1-\epsilon)^2
(m+\omega+\zeta)^2 f^2.
\]
This allows us to write
\[
-
\Big(
2\omega-\frac{1}{2}(1-\epsilon)^2(m+\omega)f^2+\frac14(1-\epsilon)^2\zeta^2
\Big)\zeta
=\frac{1}{4}(1-\epsilon)^2
(m+\omega)^2 f^2
-m^2+\omega^2.
\]
Using \eqref{alpha-mu},
we arrive at
\[
-
\Big(
2\omega-(1-\epsilon)^2(m+\omega)
\frac{2\mu^2}{(1+\epsilon)^2}
\Big)\zeta+\frac{(1-\epsilon)^2\zeta^2}{4}
=\frac{(1-\epsilon)^2(m+\omega)^2 \mu^2}{(1+\epsilon)^2}
-m^2+\omega^2
=-\frac{4(m^2-\omega^2)\epsilon}{(1+\epsilon)^2}.
\]

This relation shows that,
for $\abs{\epsilon}$ small enough,
there is a real-valued solution
$\zeta$ which satisfies
\[
\zeta=2\epsilon\frac{m^2-\omega^2}{m}
\big(1+\mathcal{O}(\epsilon)\big).
\]
This completes the proof
of Theorem~\ref{theorem-instability}.
\end{proof}

\section{Broken parity perturbation of the Soler model}
\label{sect-nld-perturbation-broken-parity}

Now
we consider the perturbation
that breaks not only the $\mathbf{SU}(1,1)$-symmetry of the
Soler model,
but also the parity symmetry:
the linearized equation is no longer invariant
in the subspaces
$\bfX_{\mbox{\rm\footnotesize even-odd-even-odd}}$
and
$\bfX_{\mbox{\rm\footnotesize odd-even-odd-even}}$
of $L^2(\R,\C^4)$,
consisting of even-odd-even-odd and odd-even-odd-even components.
We show that under this perturbation
the weakly relativistic solitary waves
become linearly unstable, because
the spectrum of the corresponding linearization
contains the eigenvalues with positive real part;
these eigenvalues bifurcate from $\pm 2\jj\omega$
(see Theorem \ref{theorem-instability-broken}).

\subsubsection*{Model}

We consider the perturbed
self-interaction based on the term
\begin{eqnarray}\label{perturbation-sigma-1}
\psi\sp\ast(\sigma_3+\epsilon \sigma_1)\psi,
\end{eqnarray}
$\epsilon\ne 0$,
so that the dynamics is described formally by the equation
\begin{eqnarray}\label{nld-point-perturbed-2}
\jj \p_t\psi=(\jj\sigma_2\p_x+\sigma_3 m)\psi
-\delta(x)f(\psi\sp\ast(\sigma_3+\epsilon\sigma_1)\psi)
(\sigma_3+\epsilon\sigma_1)\psi,
\qquad
x\in\R,
\quad
t\in\R,
\end{eqnarray}
with the pure power nonlinearity
\begin{eqnarray}\label{pp}
f(\tau)=|\tau|^\kappa,
\qquad
\tau\in\R,
\qquad
\kappa>0,
\end{eqnarray}
and where the following boundary condition for domain elements is everywhere understood in this section
(see \eqref{psi-hat}, \eqref{psi-zero}, and \eqref{jumpeq}):
\begin{equation}\label{jumpeq_paritymotpres}
\jj\sigma_2[\psi ]_{0}
-f(\hat\psi\sp\ast(\sigma_3+\epsilon \sigma_1)\hat\psi)(\sigma_3+\epsilon\sigma_1)\hat\psi=0.
\end{equation}
Equation \eqref{nld-point-perturbed-2}
is a Hamiltonian $\mathbf{U}(1)$-invariant system
which is no longer $\mathbf{SU}(1,1)$-invariant.
We will show that the perturbation
\eqref{perturbation-sigma-1}
breaks the parity symmetry:
components of the solitary waves are no longer
even or odd,
and the linearization operator at a solitary wave
is no longer invariant
in $\bfX\sb{\mbox{\rm\footnotesize even-odd-even-odd}}$
or $\bfX\sb{\mbox{\rm\footnotesize odd-even-odd-even}}$.

\subsubsection*{Solitary waves}

The first step of the analysis is to construct solitary waves
$\phi(x)e^{-\jj\omega t}$.
Instead of \eqref{solitary-wave},
$\phi(x)$ is now to be of the form
\begin{equation}\label{phibroken}
\phi(x)
=
\left(
\alpha(\epsilon)\begin{bmatrix}1\\\mu\sgn x\end{bmatrix}
+
\beta(\epsilon)\begin{bmatrix}\sgn x\\\mu\end{bmatrix}
\right)
e^{-\varkappa\abs{x}},
\end{equation}
where
\begin{equation}
\varkappa=\sqrt{m^2-\omega^2},
\qquad
\mu=\sqrt{\frac{m-\omega}{m+\omega}}.
\end{equation}
The conditions on $\alpha=\alpha(\epsilon)$ and $\beta=\beta(\epsilon)$
come from the jump condition
(cf. \eqref{jumpeq})
\begin{eqnarray}\label{j-c}
\jj\sigma_2
\begin{bmatrix}2\beta\\2\alpha\mu
\end{bmatrix}
-(\sigma_3+\epsilon\sigma_1)f
\begin{bmatrix}\alpha\\\beta\mu\end{bmatrix}=0,
\end{eqnarray}
where
$f=f(\tau)$
with
\[
\tau:=\hat\phi\sp\ast(\sigma_3+\epsilon\sigma_1)\hat\phi.
\]
The jump condition \eqref{j-c} takes the shape of the following system:
\begin{eqnarray}\label{f-a-b}
\begin{cases}
(f-2\mu)\alpha+f\epsilon\mu \beta=0,
\\f\epsilon \alpha+(2-f\mu)\beta=0.
\end{cases}
\end{eqnarray}
The compatibility condition leads to
$
{(f-2\mu)(2-f\mu)-f^{2} \epsilon^2\mu=0},
$
which we rewrite as
\[
{f^2\mu(1+\epsilon^{2})-2(1+\mu^2)f+4\mu=0},
\]
hence
\[
f=\frac{1+\mu ^2\pm\sqrt{1-2 \mu ^2+\mu ^4-4 \mu ^2 \epsilon^2}}{\mu +\mu  \epsilon^2}.
\]
We need to choose the negative sign at the square root,
so that
$f

=
2\mu+\mathcal{O}(\epsilon^2)$;
then we are consistent with the case
$\epsilon=0$,
$\omega\in(0,m)$
(see \eqref{a-b-f}).
\nn{This comment is not clear as the other case provides $f=
2/\mu+\mathcal{O}(\epsilon^2)$ which correspond to the $-\omega$ wave ?}
\ac{added $\omega\in(0,m)$ which implies that $\mu$ is small.}
Therefore, one has:
\begin{eqnarray}\label{f-is-f}
&&
f
=
\frac{
1+\mu^2-\sqrt{1-2\mu^2+\mu^4-4\mu^2\epsilon^2}
}{(1+\epsilon^2)\mu}
=\frac{1}{(1+\epsilon^2)\mu}
\left(
1+\mu^2-(1-\mu^2)\sqrt{1-\frac{4\mu^2\epsilon^2}{(1-\mu^2)^2}}
\right)
\nonumber

\\
&&\quad =\frac{2}{1+\epsilon^2}
\Big(
\mu
+\frac{\mu\epsilon^2}{1-\mu^2}
+\mathcal{O}(\epsilon^4\mu^3)
\Big)
=\frac{2}{(1+\epsilon^2)(1-\mu^2)}
\left(
\mu-\mu^3
+\mu\epsilon^2
+\mathcal{O}(\epsilon^4\mu^3)
\right)
\nonumber
\\
&&\quad =
\frac{2}{(1+\epsilon^2)(1-\mu^2)}
\left(
\mu-\mu^3
+\mu\epsilon^2
-\mu^3\epsilon^2
+\mu^3\epsilon^2
+\mathcal{O}(\epsilon^4\mu^3)
\right)
=
2\mu
\left(1+\mathcal{O}(\epsilon^2\mu^2)\right).
\end{eqnarray}
The second equation from \eqref{f-a-b} yields:
\begin{eqnarray}\label{bbroken}
\beta=-\frac{f\epsilon\alpha}{2-f\mu}
=
-
\frac{
2\mu(1+\mathcal{O}(\epsilon^2\mu^2)
)\epsilon\alpha}{
2-
2\mu^2(1+\mathcal{O}(\epsilon^2\mu^2))
}
=-\frac{\epsilon\mu}{1-\mu^2}(1
+\mathcal{O}(\epsilon^2\mu^2)
)\alpha.
\end{eqnarray}
For the future use, we compute
\begin{eqnarray}\label{alpha-beta}
\alpha-\frac{\beta\mu}{\epsilon}
=
\Big(1+
\frac{\mu^2}{1-\mu^2}
(1+\mathcal{O}(\epsilon^2\mu^2))
\Big)\alpha,
\end{eqnarray}
and by \eqref{phibroken} one has
\begin{eqnarray}\label{tau-ch4}
\tau
&=&\phi\sp\ast\at{x=0}(\sigma_3+\epsilon\sigma_1)\phi\at{x=0}
=
\begin{bmatrix}\alpha&\beta\mu\end{bmatrix}
\begin{bmatrix}1&\epsilon\\\epsilon&-1\end{bmatrix}
\begin{bmatrix}\alpha\\\beta\mu\end{bmatrix}
=\alpha^2+2\epsilon\alpha\beta\mu-\beta^2\mu^2
\\
\nonumber
&=&
\Big(1
-2\epsilon^2\frac{\mu^2}{1-\mu^2}
(1+\mathcal{O}(\epsilon^2\mu^2))
-\epsilon^2\frac{\mu^2}{(1-\mu^2)^2}
(1+\mathcal{O}(\epsilon^2\mu^2))
\mu^2
\Big)\alpha^2
=
\big(1+\mathcal{O}(\epsilon^2\mu^2)
\big)\alpha^2.
\end{eqnarray}

Combining the above expression for $\tau$
with the relation \eqref{f-is-f} satisfied by $f$,
we derive:
\begin{eqnarray}\label{abroken}
2\mu\big(1+\mathcal{O}(\epsilon^2\mu^2)\big)
=f=\abs{\tau}^\kappa
=
\alpha^{2\kappa}
(1+\mathcal{O}(\epsilon^2\mu^2)).
\end{eqnarray}
The solitary wave
is given by the expression \eqref{phibroken} with $\alpha$ and $\beta$
from \eqref{abroken} and \eqref{bbroken}.

\subsubsection*{Linearization}

Let us consider the linearization at the solitary wave
\eqref{phibroken}.
We use the Ansatz
\begin{eqnarray}\label{Ansatz-br}
\psi(t,x)=(\phi(x)+r(t,x)+\jj s(t,x))e^{-\jj\omega t},
\end{eqnarray}
where
$\big(r(t,x),\,s(t,x)\big)\in \R^2\times\R^2$.
A substitution of the Ansatz
\eqref{Ansatz-br}
into equation \eqref{nld-point-perturbed-2}
shows that the perturbation
$(r(t,x),\,s(t,x))$
satisfies the following linearized system:
\[
\begin{cases}
-\dot s=(D_m-\omega) r
-f\delta(x)(\sigma_3+\epsilon \sigma_1)r
-2g\delta(x)(\phi\sp\ast(\sigma_3+\epsilon \sigma_1)r)(\sigma_3+\epsilon \sigma_1)\phi
=:L_{+}(\epsilon)r,
\\[1ex]
\dot r=\left(D_m -\omega\right) s
-f\delta(x)(\sigma_3+\epsilon \sigma_1)s
=:L_{-}(\epsilon)s.
\end{cases}
\]
Above,
\begin{eqnarray}\label{def-f-g-ch4}
f=f(\tau),
\qquad
g=f'(\tau)
\end{eqnarray}
are evaluated at
\[
\tau:=\phi\sp\ast(\sigma_3+\epsilon \sigma_1)\phi\at{x=0}
=\alpha(\alpha+\epsilon\beta\mu)
+\beta\mu(\epsilon\alpha-\beta\mu)
=(\alpha+\epsilon\beta\mu)^2
-\beta^2\mu^2
+\mathcal{O}(\epsilon^4).
\]
Thus, we have:
\begin{eqnarray}
&&
\hskip -25pt
L_{+}r
=
(D_m-\omega)r
-f\delta(x)(\sigma_3+\epsilon\sigma_1)r
-2g\delta(x)(\phi\sp\ast(\sigma_3+\epsilon\sigma_1)r)
(\sigma_3+\epsilon\sigma_1)\phi
\nonumber
\\&&
=(D_m-\omega)r
-f\delta(x)\cdot(\sigma_3+\epsilon\sigma_1)r
-2g\delta(x)(\alpha (r_1+\epsilon r_2)+\beta\mu(-r_2+\epsilon r_1))
\begin{bmatrix}
\alpha+\epsilon\beta\mu
\\\epsilon\alpha-\beta\mu
\end{bmatrix}
\nonumber
\\&&
=(D_m-\omega)r
-f\delta(x)(\sigma_3+\epsilon\sigma_1)r
-2g\delta(x)((\alpha+\epsilon\beta\mu)r_1+(\alpha\epsilon-\beta\mu)r_2)
\begin{bmatrix}
\alpha+\epsilon\beta\mu
\\\epsilon\alpha-\beta\mu
\end{bmatrix}

\nonumber

\\&&
=
\Big(
D_m-\omega
-f\delta(x)(\sigma_3+\epsilon\sigma_1)
-\delta(x)
\big[
X\Pi_1
+\epsilon Y\sigma_1
+\epsilon^2 Z\Pi_2
\big]
\Big)r,
\end{eqnarray}
where we used the projectors
\begin{eqnarray}\label{def-pi1-pi2}
\Pi_1=\begin{bmatrix}1&0\\0&0\end{bmatrix},
\qquad
\Pi_2=\begin{bmatrix}0&0\\0&1\end{bmatrix}
\end{eqnarray}
and the constants
$X,\,Y,\,Z\in\R$
defined by
\begin{eqnarray}\label{def-x-y-z}
X=2(\alpha+\epsilon\beta\mu)^2g,
\qquad
Y=2
(\alpha+\epsilon\beta\mu)\Big(\alpha-\frac{\beta\mu}{\epsilon}\Big)g
,
\qquad
Z=2\Big(\alpha-\frac{\beta\mu}{\epsilon}\Big)^2g.
\end{eqnarray}
In the pure power case, by \eqref{f-is-f},
one has

\[
\tau g
=\tau f'(\tau)
=\kappa f(\tau)
=2\kappa\mu(1+\mathcal{O}(\epsilon^2\mu^2)),
\]
with $\tau$ from \eqref{tau-ch4};
hence,
using
\eqref{alpha-beta} in \eqref{def-x-y-z},
we have the following estimates:
\begin{eqnarray}\label{x-y-z-such}

X=4\kappa\mu(1+\mathcal{O}(\epsilon^2\mu^2)),
\ \quad
Y=4\kappa\mu(1+\mathcal{O}(\mu)),
\ \quad
Z=4\kappa\mu(1+\mathcal{O}(\mu)).
\end{eqnarray}
We denote
\begin{eqnarray}\label{def-f-f}
F=f+Y
=
f+4\kappa\mu(1+\mathcal{O}(\mu))
=2(1+2\kappa)\mu(1+\mathcal{O}(\mu)),
\end{eqnarray}
so that
\begin{eqnarray}\label{def-dp-lm}
\begin{array}{l}
L_{+}
=
D_m-\omega
-\delta(x)(f\sigma_3
+F\epsilon\sigma_1
+X\Pi_1
+\epsilon^2 Z\Pi_2),
\\[1ex]
L_{-}
=
D_m-\omega
-f\delta(x)(\sigma_3+\epsilon\sigma_1).
\end{array}
\end{eqnarray}
Now we can write
$\bfA(\epsilon)=\begin{bmatrix}
0&L_{-}\\-L_{+}&0\end{bmatrix}$
in the explicit form
as
\begin{eqnarray}\label{def-a-broken}
\bfA(\epsilon)
=
\begin{bmatrix}
0&\!\!\!\!\!\!
D_m-\omega I_2-f\delta(x)(\sigma_3+\epsilon\sigma_1)
\\-D_m+\omega I_2
+\delta(x)
\big(
f\sigma_3+\epsilon F\sigma_1
+X\Pi_1
+Z\epsilon^2
\Pi_2
\big)
&0
\end{bmatrix},
\end{eqnarray}
with constants
$f$, $g$ from \eqref{def-f-g-ch4},
$X$, $Z$ from \eqref{def-x-y-z},
$F$ from \eqref{def-f-f},
and with projectors
$\Pi_1$, $\Pi_2$ from \eqref{def-pi1-pi2}.

\subsection*{Bifurcations
of eigenvalues from the essential spectrum}

Let $\lambda(\epsilon)$
be the deformation of the eigenvalue $2\omega\jj$ of $\bfA(\epsilon)$ from \eqref{def-a-ch2}
under the perturbation \eqref{nld-point-perturbed-2}.
As before
(see \eqref{lambda-lambda} and Theorem~\ref{theorem-instability}),
let $\Lambda\in\C$ and $\zeta\in\C$
be defined by relations
\begin{eqnarray}\label{lambda-lambda-zeta}
\lambda(\epsilon)=\jj\Lambda(\epsilon),
\qquad
\Lambda(\epsilon)=2\omega+\zeta(\epsilon).
\end{eqnarray}
The condition for the eigenvalue $\lambda(\epsilon)$
bifurcating from $2\omega\jj$
to be inside the first quadrant
(that is, the linear instability condition, $\Re\lambda>0$)
is now $\Im\zeta<0$.

\begin{theorem}\label{theorem-instability-broken}

Let $f(\tau)=\abs{\tau}^\kappa$, $\tau\in\R$\textup; $\kappa>0$.
There is
$\omega_0<m$ and an open neighborhood
$U\subset\R$, $U\ni 0$,
such that for
$\omega\in(\omega_0,m)$
and $\epsilon\in U\setminus\{0\}$

the spectrum
$\sigma\sb{\mathrm{p}}(\bfA(\epsilon))$
contains an eigenvalue
$\lambda(\epsilon)$
with positive real part:
\[
\lambda(\epsilon)=\jj(2\omega+\zeta(\epsilon)),
\qquad
\Im\zeta(\epsilon)<0\quad \forall\epsilon\in U\setminus\{0\},
\qquad
\lim\sb{\epsilon\to 0}\zeta(\epsilon)=0.
\]

\end{theorem}

\begin{proof}
As in the proof of Theorem~\ref{theorem-instability},
to study
whether
$
\lambda(\epsilon)=\jj\Lambda(\epsilon)
$
is an eigenvalue
of the operator $\bfA(\epsilon)$
from \eqref{def-a-broken},
we consider the action of $\bfA(\epsilon)-\jj\Lambda(\epsilon) I_4$
onto the superposition
\[
\Psi
=
a
\begin{bmatrix}
\nu_{+}\sgn x\\S_{+}\\-\jj\nu_{+}\sgn x\\-\jj S_{+}
\end{bmatrix}e^{-\nu_{+}\abs{x}}
+
b
\begin{bmatrix}
-\jj\xi\sgn x\\S_{-}\\\xi\sgn x\\\jj S_{-}
\end{bmatrix}e^{\jj\xi\abs{x}}
+
c
\begin{bmatrix}
\nu_{+}\\S_{+}\sgn x\\-\jj\nu_{+}\\-\jj S_{+}\sgn x
\end{bmatrix}e^{-\nu_{+}\abs{x}}
+
d
\begin{bmatrix}
-\jj\xi\\S_{-}\sgn x\\\xi\\\jj S_{-}\sgn x
\end{bmatrix}e^{\jj\xi\abs{x}}.
\]
Above,
$S_{+}=S_{+}(\omega,\Lambda)$ and $S_{-}=S_{-}(\omega,\Lambda)$ are from \eqref{def-r-s}
and
$\nu_{+}=\nu_{+}(\omega,\Lambda)$ and $\xi=\xi(\omega,\Lambda)$ are given by \eqref{def-kappa-xi}.

The jump condition at $x=0$
leads to the relations
\[
\begin{cases}
2(-\jj S_{+} c+\jj S_{-} d)
-(-\jj\nu_{+} c+\xi d)f
-\epsilon (-\jj S_{+} a+\jj S_{-} b)f
=0,
\\-2(-\jj\nu_{+} a+\xi b)
+(-\jj S_{+} a+\jj S_{-} b)f
-\epsilon (-\jj\nu_{+} c+\xi d)f
=0,
\\-2(S_{+} c+S_{-} d)
+(f+X)(\nu_{+} c-\jj\xi d)
+\epsilon (S_{+} a+S_{-} b)F
=0,
\\2(\nu_{+} a-\jj\xi b)
-(f-\epsilon^2 Z)(S_{+} a+S_{-} b)
+\epsilon (\nu_{+} c-\jj\xi d)F
=0,
\end{cases}
\]
with
$f$ from \eqref{def-f-g-ch4},
$F=f+Y$ from \eqref{def-f-f},
and $X$, $Y$, $Z$ from \eqref{def-x-y-z}.
Above, the first terms in the left-hand side
correspond to the contributions from the derivative.
The assumption that
$a,\,b,\,c,\,d\in\C$ are not simultaneously zeros
leads to the condition
\[
\det
\begin{bmatrix}
\jj\epsilon S_{+} f&-\jj\epsilon S_{-} f&-2\jj S_{+}+\jj f\nu_{+}&2\jj S_{-}-f\xi
\\
2\jj\nu_{+}-\jj S_{+} f&-2\xi+\jj S_{-} f&\jj\epsilon f \nu_{+}&-\epsilon f \xi
\\
\epsilon S_{+} F&\epsilon S_{-} F&-2 S_{+}+(f+X)\nu_{+}&-2 S_{-}-\jj(f+X)\xi
\\
2\nu_{+}-(f-\epsilon^2 Z)S_{+}&-2\jj\xi-(f-\epsilon^2 Z)S_{-}&\epsilon F\nu_{+}&-\jj\epsilon F\xi
\end{bmatrix}
=0,
\]
which we rewrite as

\begin{eqnarray}\label{det-det}
\hskip -10pt
\det
\begin{bmatrix}
2\nu_{+}-S_{+} f&-2\jj\xi-S_{-} f&\epsilon f \nu_{+}&-\jj\epsilon f \xi
\\2\nu_{+}-(f-\epsilon^2 Z)S_{+}&2\jj\xi+(f-\epsilon^2 Z)S_{-}&\epsilon F\nu_{+}&\jj\epsilon F\xi
\\\epsilon S_{+} f&\epsilon S_{-} f&-2 S_{+}+f\nu_{+}&-2 S_{-}-\jj f\xi
\\\epsilon S_{+} F&-\epsilon S_{-} F&-2 S_{+}+(f+X)\nu_{+}&2 S_{-}+\jj(f+X)\xi
\end{bmatrix}
=0.
\end{eqnarray}
Let $A$, $B$, $C$, and $D$
be the $2\times  2$ matrices
so that the above matrix
is written in the block form as
$
\begin{bmatrix}
A&\epsilon B\\\epsilon C&D
\end{bmatrix}
$;
that is,
\begin{eqnarray}
\label{matrix-a}
A=\begin{bmatrix}
2\nu_{+}-S_{+} f&-S_{-} f-2\jj\xi
\\2\nu_{+}-S_{+} f+\epsilon^2 Z S_{+}&2\jj\xi+S_{-} f-\epsilon^2 Z S_{-}
\end{bmatrix},
\qquad
B=
\begin{bmatrix}
f \nu_{+}&-\jj f \xi
\\F\nu_{+}&\jj F\xi
\end{bmatrix},
\end{eqnarray}
\begin{eqnarray}
\label{matrix-c-d}
C=
\begin{bmatrix}
S_{+} f&S_{-} f
\\S_{+} F&-S_{-} F
\end{bmatrix},
\qquad
D=
\begin{bmatrix}
-2 S_{+}+f\nu_{+}&-2 S_{-}-\jj f\xi
\\-2 S_{+}+(f+X)\nu_{+}&2 S_{-}+\jj(f+X)\xi
\end{bmatrix}.
\end{eqnarray}
Since (see \eqref{det-d} below) one has
$\lim\sb{\omega\to m,\,\Lambda\to 2m}\det D=32 m^2$
(we recall that $S_{+}$ and $S_{-}$ are defined in \eqref{def-r-s}
and $\nu_{+}$ and $\xi$ are defined in \eqref{def-kappa-xi}),
we can use the Schur complement of $D$
to write
the condition \eqref{det-det}
as
\[
\det(A-\epsilon^2 M)=0,
\qquad
\mbox{with}
\qquad
M=B D^{-1}C.
\]

We have:
\begin{eqnarray}\label{def-m}
M=\frac{1}{\det D}
\begin{bmatrix}
f\nu_{+}&-\jj f\xi
\\F\nu_{+}&\jj F\xi
\end{bmatrix}
\begin{bmatrix}
2 S_{-}+\jj (f+X) \xi&2 S_{-}+\jj f \xi
\\2 S_{+}-(f+X)\nu_{+}&-2 S_{+}+f\nu_{+}
\end{bmatrix}
\begin{bmatrix}
S_{+} f&S_{-} f
\\S_{+} F&-S_{-}F
\end{bmatrix}.
\end{eqnarray}
Taking
into account that $f=\mathcal{O}(\mu)$ (see \eqref{f-is-f}),
$F=\mathcal{O}(\mu)$ (see \eqref{def-f-f}),
$S_{+}+S_{-}=2(m-\omega)=\mathcal{O}(\mu^2)$,
\nn{I think that below the remainder is $\mathcal{O}(\mu^2)$ and hence it is the same for \eqref{n-d} and \eqref{finalformula} and everything in the middle ....}
\ak{Nabile, thank you for pointing out this trouble!
Please note that the matrices $B$ and $C$ (the first and the last
matrices in the next line)
have all entries $\mathcal{O}(\mu)$.
The leading contribution
into $M$ is only $\mathcal{O}(\mu^2)$,
so we certainly need the error to be smaller,
$\mathcal{O}(\mu^3)$,
but there was a typo: terms with $\nu_{+}$ were missing!
Correcting....
Fortunately, this will not affect the conclusions since
we show later that $\nu_{+}=O(\mu)$;
see \eqref{r-s-m} below.
}

\begin{eqnarray*}
M&=&\frac{1}{\det D}
\begin{bmatrix}
f\nu_{+}&-\jj\xi f
\\
F\nu_{+}&\jj \xi F
\end{bmatrix}
\begin{bmatrix}
2 S_{-}&2 S_{-}
\\2 S_{+}&-2 S_{+}
\end{bmatrix}
\begin{bmatrix}
S_{+} f&S_{-} f
\\S_{+} F&-S_{-} F
\end{bmatrix}
+\mathcal{O}(\mu^3)
\\
&=&\frac{2 S_{+}^2}{\det D}
\begin{bmatrix}
f\nu_{+}&-\jj\xi f\\F\nu_{+}& \jj\xi F
\end{bmatrix}
\begin{bmatrix}
-1&-1\\1&-1
\end{bmatrix}
\begin{bmatrix}
f&-f\\F&F
\end{bmatrix}
+\mathcal{O}(\mu^3)
\\
&=&\frac{2 S_{+}^2}{\det D}
\begin{bmatrix}
f\nu_{+}&-\jj\xi f\\F\nu_{+}& \jj\xi F
\end{bmatrix}
\begin{bmatrix}-f-F&f-F\\f-F&-f-F\end{bmatrix}
+\mathcal{O}(\mu^3)
\\
&=&\frac{2 S_{+}^2}{\det D}
\begin{bmatrix}
(-f-F)f\nu_{+}-\jj(f-F)f\xi
&
(f-F)f\nu_{+}-\jj(-f-F)f\xi
\\
(-f-F)F\nu_{+}+\jj(f-F)F\xi
&
(f-F)F\nu_{+}
+\jj(-f-F)F\xi
\end{bmatrix}
+\mathcal{O}(\mu^3).
\end{eqnarray*}
In the second line, we substituted $S_{-}$ by $-S_{+}$,
with the error counted in the $\mathcal{O}(\mu^3)$ term.
It follows that
\begin{eqnarray}\label{m11-m21}
M_{11}+M_{21}
=
\frac{2 S_{+}^2}{\det D}
\big(
-(f+F)^2\nu_{+}
-\jj(F-f)^2\xi
\big)
+\mathcal{O}(\mu^3)
\nonumber
\\
=
-\frac{2 S_{+}^2}{\det D}
\big(
(2f+Y)^2\nu_{+}
+\jj Y^2\xi
\big)
+\mathcal{O}(\mu^3),
\end{eqnarray}
with $Y$ from \eqref{def-x-y-z}.
Taking into account that
$A_{21}=A_{11}+\epsilon^2 Z S_{+}$
and
$A_{22}=-A_{12}-\epsilon^2 Z S_{-}$,
we derive:
\begin{eqnarray*}
&&\det(A-\epsilon^2 B D^{-1}C)
=
(A_{11}-\epsilon^2 M_{11})(A_{22}-\epsilon^2 M_{22})
-
(A_{12}-\epsilon^2 M_{12})(A_{21}-\epsilon^2 M_{21})
\\&&
=
(A_{11}-\epsilon^2M_{11})(-A_{12}-\epsilon^2 Z S_{-} -\epsilon^2 M_{22})
-
(A_{12}-\epsilon^2 M_{12})(A_{11}+\epsilon^2 Z S_{+}-\epsilon^2 M_{21})=0,
\end{eqnarray*}
\[
-2A_{11}A_{12}
-A_{11}
(\epsilon^2 Z S_{-} +\epsilon^2 M_{22}-\epsilon^2 M_{12})
+A_{12}(\epsilon^2 M_{11}-\epsilon^2 Z S_{+}+\epsilon^2 M_{21})
\]
\[
+\epsilon^4 M_{11}(Z S_{-}+M_{22})
+\epsilon^4 M_{12}(Z S_{+}-M_{21})
=0,
\]

\[
A_{11}
=
\epsilon^2
\frac{A_{12}(M_{11}+M_{21}-Z S_{+})
+\epsilon^2
(M_{11}(Z S_{-}+M_{22})+M_{12}(Z S_{+}-M_{21}))}
{2A_{12}+\epsilon^2(Z S_{-}+M_{22}-M_{12})},
\]
\[
2\nu_{+}=S_{+} f
+
\epsilon^2
\frac{A_{12}(M_{11}+ M_{21}-Z S_{+})
+\epsilon^2
(M_{11}(Z S_{-}+M_{22})+M_{12}(Z S_{+}-M_{21}))}
{2A_{12}+\epsilon^2(Z S_{-}+M_{22}-M_{12})}.
\]
Substituting
$\nu_{+}=\sqrt{m^2-(\omega-\Lambda)^2}=\sqrt{m^2-(\omega-(2\omega+\zeta))^2}$
(see \eqref{lambda-lambda-zeta}),

we arrive at
\begin{eqnarray}\label{zeta-is-0}
&&
\hskip -10pt
\zeta^2+2\omega\zeta
\\\nonumber
&&
\hskip -10pt
=
m^2-\omega^2
-\left(
\frac{S_{+} f}{2}
+
\epsilon^2
\frac{M_{11}+M_{21}-Z S_{+}
+\frac{\epsilon^2}{A_{12}}
(M_{11}(Z S_{-}+M_{22})+M_{12}(Z S_{+}-M_{21}))}
{4+2\epsilon^2(Z S_{-}+M_{22}-M_{12})/A_{12}}
\right)^2.
\end{eqnarray}
Taking into account \eqref{f-is-f} and \eqref{x-y-z-such},
the entries of the matrix $M$ from \eqref{def-m}
are estimated by
\[
M_{i j}=\mathcal{O}(\mu^2),
\qquad
1\le i,\,j\le 2;
\]
since
$A_{12}=-S_{-} f-2\jj\xi\longrightarrow  -4\jj m\sqrt{2}$
in the limit $\epsilon\to 0$, $\omega\to m$,
$\Lambda\to 2m$,
\eqref{zeta-is-0} yields the relation
\begin{eqnarray}\label{zeta-is}
\zeta^2+2\omega\zeta
=
m^2-\omega^2
-\left(
\frac{S_{+} f}{2}
+
\epsilon^2
\frac{M_{11}+M_{21}-S_{+}Z
+
\mathcal{O}(\epsilon^2\mu^3)
}
{4-
\mathcal{O}(\epsilon^2\mu)}
\right)^2.
\end{eqnarray}
Writing
\begin{eqnarray}\label{large-error}
\zeta^2+2\omega\zeta
&=&
(m+\omega)^2\mu^2
-
\Big(
\frac{S_{+} f}{2}
+\mathcal{O}(\epsilon^2\mu)
\Big)^2

\nonumber
\\&=&(m+\omega)^2\mu^2
-\frac{(m+\omega+\zeta)^2f^2}{4}
-S_{+} f
\epsilon^2 \mathcal{O}(\mu)
+\mathcal{O}(\epsilon^4\mu^2)
\end{eqnarray}
(let us point out that
the largest error term,
$\mathcal{O}(\epsilon^4\mu^2)$,
is contributed by squaring
$\epsilon^2 S_{+}Z$
in the right-hand side of \eqref{zeta-is}),
we have
\begin{eqnarray*}
\zeta^2+2\omega\zeta
&=&
(m+\omega)^2
\Big(\mu^2-\frac{f^2}{4}\Big)
-
\frac{2(m+\omega)\zeta+\zeta^2}{4}f^2
+\mathcal{O}(\epsilon^2\mu^2)
\\&=&
(m+\omega)^2\mathcal{O}(\epsilon^2\mu^2)
-4\mu^2
\frac{2(m+\omega)\zeta+\zeta^2}{4}
+\mathcal{O}(\epsilon^2\mu^2),
\end{eqnarray*}
hence
\begin{eqnarray}\label{zeta-small}
\zeta=\mathcal{O}(\epsilon^2\mu^2).
\end{eqnarray}
In view of \eqref{zeta-small},
\begin{eqnarray}\label{r-s-m}
\begin{array}{l}
S_{+}(\omega,\Lambda)=m+\Lambda-\omega
=m+\omega+\zeta=2m+\mathcal{O}(\mu^2),
\\[1ex]
S_{-}(\omega,\Lambda)=m-\omega-\Lambda=-2m+\mathcal{O}(\mu^2),
\\[1ex]
\nu_{+}(\omega,\Lambda)
=\sqrt{m^2-(\Lambda-\omega)^2}
=\sqrt{m^2-(\omega+\zeta)^2}
=\mathcal{O}(\mu).
\end{array}
\end{eqnarray}
Now we can compute the determinant of the matrix
$D$ from \eqref{matrix-c-d}:
\begin{eqnarray}\label{det-d}
&&
\det D
=
(-2 S_{+}+f\nu_{+})(2 S_{-}+\jj(f+X)\xi)
+(2 S_{-}+\jj f\xi)(-2 S_{+}+(f+X)\nu_{+})
\nonumber
\\
&&
=
-8S_{+}S_{-}
+2 (2f+X)S_{-}\nu_{+}
+\jj
\big(
2(f+X)f\nu_{+}\xi
-2 (2f+X)S_{+}\xi
\big)
=32m^2+\mathcal{O}(\mu),
\qquad
\end{eqnarray}
with the error term being complex-valued.
Taking the imaginary part of
\eqref{zeta-is},
we obtain:
\begin{eqnarray}
\label{n-d}
2(\omega+\Re\zeta)
\Im\zeta
=
-
\epsilon^2
S_{+} f\Im
\frac{M_{11}+M_{21}-S_{+}Z
+
\mathcal{O}(\epsilon^2\mu^3)}
{4-\mathcal{O}(\epsilon^2\mu)}
+\mathcal{O}(\epsilon^4\mu^3).
\end{eqnarray}
\begin{remark}
Note that in the right-hand side the error term is
$\mathcal{O}(\epsilon^4\mu^3)$
(instead of $\mathcal{O}(\epsilon^4\mu^2)$ as in
\eqref{large-error});
indeed, by \eqref{r-s-m},
\begin{eqnarray}\label{z-r-small}
S_{+}Z
=(m+\omega+\mathcal{O}(\epsilon^2\mu^2))Z,
\end{eqnarray}
and since $Z$ from \eqref{def-x-y-z} is real-valued,
$(\epsilon^2 S_{+}Z)^2$
can not contribute
$\mathcal{O}(\epsilon^4\mu^2)$
to the imaginary part of the right-hand side.
\end{remark}

Since the numerator in \eqref{n-d}
is $\mathcal{O}(\mu)$,
and so is the factor $S_{+} f$,
we conclude that neglecting
$\mathcal{O}(\epsilon^2\mu)$ terms from
the denominator contributes the error absorbed into
$\mathcal{O}(\epsilon^4\mu^3)$, so
\[
2(\omega+\Re\zeta)
\Im\zeta
=
-
\frac{\epsilon^2}{4}
S_{+} f\Im
\left(M_{11}+M_{21}-S_{+}Z
\right)
+\mathcal{O}(\epsilon^4\mu^3).
\]
Using \eqref{m11-m21}
(where in view of \eqref{r-s-m}
one has $(2f+Y)^2\nu_{+}=\mathcal{O}(\mu^3)$), \eqref{det-d},
and taking into account
the fact that
$Z=\mathcal{O}(\mu)$ is real-valued
while $S_{+}=m+\omega+\zeta$,
we continue:
\begin{eqnarray}\label{finalformula}
&&
2(\omega+\Re\zeta)\Im\zeta
=
-\frac{\epsilon^2}{4}
S_{+} f
\Im
\big(M_{11}+M_{21}-S_{+}Z\big)
+\mathcal{O}(\epsilon^4\mu^3)
\\&&
\nonumber
=
\frac{\epsilon^2}{4}
S_{+} f
\Im
\Big(
\frac{2\jj\xi S_{+}^2}{\det D}Y^2+\mathcal{O}(\mu^3)
+Z\zeta
\Big)
+\mathcal{O}(\epsilon^4\mu^3)
=
\frac{\epsilon^2}{4}
f
\frac{\xi S_{+}^3}{16 m^2}Y^2
+\mathcal{O}(\epsilon^2\mu^4)
+\mathcal{O}(\epsilon^4\mu^3).
\end{eqnarray}
Taking into account the relations

\[
\lim\sb{\omega\to m,\,\Lambda\to 2m}S_{+}(\omega,\Lambda)=2m,
\qquad
\lim\sb{\omega\to m,\,\Lambda\to 2m}\xi(\omega,\Lambda)=-2m\sqrt{2},
\qquad
Y=4\kappa\mu(1+\mathcal{O}(\mu))
\]
(see \eqref{def-r-s},
\eqref{def-kappa-xi},
\eqref{x-y-z-such}),
we conclude from \eqref{finalformula} that
there is $c>0$ such that
$
\Im\zeta
<-c\epsilon^2\mu^3$,
as long as
$\abs{\epsilon}$ and $\mu>0$ are sufficiently small.
It follows that the eigenvalue
$\lambda=(2\omega+\zeta)\jj$ moves to the right of the imaginary axis,
becoming an eigenvalue with positive real part and indicating
the linear instability of the corresponding solitary wave.
\end{proof}

\vskip10pt
\noindent
{\bf Acknowledgments.}\ D. Noja
acknowledges for funding the EC grant IPaDEGAN (MSCA-RISE-778010).
This work was supported by a grant from the Simons Foundation (851052, A.C.).

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