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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 56, pp. 1--26.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/56\hfil  Breather solutions]
{On the variational structure of breather solutions II:
periodic mKdV equation}

\author[M. A. Alejo, C. Mu\~noz, J. M. Palacios \hfil EJDE-2017/56\hfilneg]
{Miguel A. Alejo, Claudio Mu\~noz, Jos\'e M. Palacios}

\address{Miguel A. Alejo \newline
Departamento de Matem\'atica,
Universidade Federal de Santa Catarina, Brasil}
\email{miguel.alejo@ufsc.br}

\address{Claudio Mu\~noz \newline
CNRS and Departamento de Ingenier\'ia Matem\'atica DIM,
 Universidad de Chile, Chile}
\email{cmunoz@dim.uchile.cl}

\address{Jos\'e M. Palacios \newline
Departamento de Ingenier\'ia Matem\'atica DIM,
Universidad de Chile, Chile}
\email{jpalacios@dim.uchile.cl}

\dedicatory{Communicated by Jerry Bona}

\thanks{Submitted January 21, 2017. Published February 22, 2017.}
\subjclass[2010]{35Q51, 35Q53, 37K10, 37K40}
\keywords{Modified KdV; sine-Gordon equation; periodic mKdV; integrability;
\hfill\break\indent  breather; stability}

\begin{abstract}
 We study the periodic modified KdV equation, where a periodic in space
 and time breather solution is known from the work of Kevrekidis et al.\
 \cite{KKS2}. We show that these breathers satisfy a suitable elliptic
 equation, and we also discuss via numerics its spectral stability.
 We also identify a source of nonlinear instability for the case described
 in \cite{KKS2}, and we conjecture that, even if spectral stability is satisfied,
 nonlinear stability/instability depends only
 on the sign of a suitable discriminant function, a condition that is trivially
 satisfied in the case of non-periodic (in space) mKdV breathers.
 Finally, we present a new class of  breather solution for mKdV,
 believed to exist from geometric considerations, and which is periodic
 in time and space, but has nonzero mean, unlike standard breathers.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

\subsection{Setting of the problem}
In this article, we consider the breather solution of the  periodic
modified Korteweg-de Vries (mKdV) equation
\begin{equation}\label{mKdV}
u_{t} +(u_{xx} + u^3)_x=0, \quad u(t,x) \in \mathbb{R}, \; t\in\mathbb{R},\;
 x \in  \mathbb{T}_x:= \mathbb{R}/L\mathbb{Z};
\end{equation}
Here $L>0$ is a fixed length to be determined later on. The above equation
is a  well-known \emph{completely integrable} model
\cite{AC,Ga,La}, with infinitely many conserved quantities, and a
suitable Lax-pair formulation.

Solutions of \eqref{mKdV} are an invariant under space and time translations.
Indeed, for any $t_0, x_0\in \mathbb{R}$, $u(t-t_0, x-x_0)$
is also a solution. Additionally, if $c>0$ is any number, then
$\sqrt{c}\, u(c^{3/2}t ,\sqrt{c} x)$ is also solution.


In addition to standard solitons, mKdV \eqref{mKdV} do have \emph{periodic in space breather solutions} \cite{KKS2}.

\begin{definition}[Periodic breather] \label{DEF_B_per} \rm
A KKSH periodic breather is a solution of \eqref{mKdV} of the form
\begin{gather*}
B  = B(t,x; \alpha, \beta, k, m) :=2\sqrt{2} \partial_x
\Big[ \arctan \Big( \frac{\beta}{\alpha}\frac{\operatorname{sn}(\alpha (x+ \delta t),k)}
{\operatorname{nd}(\beta (x+\gamma t),m)}\Big) \Big], \\
 \delta  := \alpha^2(1+k) +3\beta^2(m-2), \quad
\gamma := 3\alpha^2(1+k) +\beta^2(m-2),
\end{gather*}
where $\operatorname{sn}(\cdot)$ and $\operatorname{nd}(\cdot)$ are the standard Jacobi elliptic functions.
\end{definition}

(See \eqref{Bper} for a more detailed description and general definition of
periodic breather.) These breathers can be written using only two parametric
variables, say $\beta$ and $k$, and have a characteristic period $L=L(\beta,k)$,
with $L\to +\infty$ as $k\to 0$.

\subsection{Main results}

In \cite{AM0,AM,AM1}, we showed that aperiodic (in space) mKdV breath\-ers are
stable in their energy space. In \cite{AMP1},
we also showed evidence that sine-Gordon breathers are stable.
In this paper, we will make use of a combined theoretical-numerical
approach to conclude why periodic breathers may and may not be stable in some
particular regimes. We will need to introduce the following assumptions:
\begin{itemize}
\item[(A1)]
 The kernel of a linearized operator around a breather is nondegenerate
and it satisfies the gap condition; and

\item[(A2)] There is a unique simple negative eigenvalue associated
to this linear operator. (See p. \pageref{A14} for a precise description.)

\item[(A3)] The following sign condition is satisfied:
if $M_\#[B]$ is the mass of the breather solution in terms of $\beta$ and $k$,
and $a_1, a_2$ are ``variational'' parameters given in \eqref{a1}-\eqref{a2}, then
\begin{equation} \label{Weinstein22}
\frac{\partial_k a_1 \partial_\beta M_\#[B]
-\partial_\beta a_1 \partial_k M_\#[B]}{ \partial_{k}a_1
\partial_\beta a_2 -\partial_{k}a_2\partial_{\beta}a_1} >0.
\end{equation}
(See  \eqref{Cond3} for more details.)
\end{itemize}


\begin{theorem}\label{T1p8}
Under spectral assumptions {\rm (A1)--(A3)}, KKSH breathers are stable for
$L$-periodic $H^2$-perturbations.
\end{theorem}

We also present numerical evidence (see Subsection \ref{Subsection_NUM})
that assumptions (A1) and (A2) are valid in generality.
Assumption (A3) is only verified in a certain regime of parameters, therefore
it is expected that KKHS breathers may and may not be stable,
according to some particular condition. In Figure \ref{KKHS_d} we  describe
numerically the meaning of assumption (A3). In particular,
it is inferred that for $k$ small enough (depending only on $\beta$),
assumption (A3) is satisfied.
Additionally, note that the condition $k\to 0^+$ is equivalent to take the
spatial period $L\to +\infty$,
and formally recovering the standard mKdV aperiodic breather.
(See \eqref{Largo} for this fact.)


Moreover, we believe that the lack or invalidity of assumption (A3) is precisely
the source of instability in KKHS breathers, in the sense that if this
assumption is not satisfied, then they should be unstable,
as it happens in the soliton case in the regime of supercritical nonlinearities
(see \cite[Lemma 1.6]{Mu3} for example),
and just in agreement with the numerical computations performed by Kevrekidis
et al.\ \cite{KKS2}, for which \eqref{Weinstein22} is not satisfied.

\subsection{Discussion}
Assumption (A3) above is a sort of generalization of the Weinstein's sign condition
for soliton solutions, but it is different from the former because it also
considers a certain variation of energy (and not only mass),
which has been written above in terms of the mass only. Additionally,
this sign condition can be evaluated for non-periodic breathers such
as mKdV and SG, and it is trivially satisfied, see e.g. \cite[eq. (3.23)]{AMP1}
for the corresponding computation in the SG case.

Additionally, assumption (A3) is needed for ensuring that one can ``replace''
the first eigenfunction appearing from assumption (A2), which is an
instability direction, by the breather itself. The advantage of this
replacement comes from the fact that a perturbative dynamics lying in the $L^2$
orthogonal space to the breather allows to control variations of the scalings
of the breather, and moreover, their dynamics has formally
\emph{quadratic variation} in time, meaning that if the error in our initial
data is of order $\eta$, then any bad variation of the scalings will be
of order at most $\eta^2$, and therefore negligible for the stability result.
At the rigorous level, the conservation of mass allows to control the orthogonal
direction  to the breather itself in terms of quadratic terms only.
The procedure to replace the first eigenfunction by the breather is standard,
but needs to ensure that a certain denominator is never zero, see e.g.
\cite[eq. (3.30)]{AMP1}. Under assumption (A3), such a denominator is never zero.
Recall that a similar result has been proved in \cite[eq. (4.24)]{AM},
see \cite{Dyuck,Raphael,Mu3} for more examples in the case of soliton solutions.
Adding this new assumption, our method of proof works as in the SG case.

\subsection{New breathers}
 Finally, Section \ref{Sect:7} deals with a new class of mKdV periodic breathers,
which have the nonstandard property of being of nonzero mean.

\begin{theorem}\label{T1p9}
Given any parameter $\mu>0$, there exists a periodic in space breather solution
$B_\mu$, which is solution of \eqref{mKdV} and
satisfies the decomposition $B_\mu =\mu + \widetilde B_{\mu}$, where
$\widetilde B_\mu$ has zero mean.
\end{theorem}

For an explicit formula for $B_\mu$, the reader can consult Definition
\ref{NEW_BREATHER}. As it is explained in the last comments of this paper,
it also satisfies a proper fourth order elliptic equation.

This new breather has been conjectured to exist by several works on curvature
motion of closed curves on the plane, see e.g. \cite{Ale,Khare} and
references therein. However, its description does not follow the ideas
from \cite{KKS2}, because KKHS breathers are zero mean solutions. Instead we use,
in a slightly different way,  the method of proof employed in our
work \cite{AM1}, which links zero mean breathers with the corresponding zero
solution of the equation. In order to find a breather with a nonzero mean,
we use as starting point the nonzero constant solution $\mu$, and then apply twice a
suitable B\"acklund transformation, as is done in \cite{AM1}.
Since the mean of a modified KdV solution is a conserved quantity,
this property will be also preserved by the B\"acklund transformation,
leading to the desired solution with the  property sought. Concerning this
new breather, in this work we only study its simplest properties,
and describe its main differences with KKSH breathers, leaving its deep
understanding, by length reasons, to a forthcoming publication.
 For the moment, we only advance that these breathers satisfy a suitable
elliptic equation, as any other breather in this article.
 Additionally, we conjecture that this breather should be as stable as
the constant solution $u=\mu$ is for the mKdV periodic dynamics.


\subsection{Previous results}
If one studies  perturbations of solitons in \eqref{mKdV} and more general
equations,  the concepts of \emph{orbital}, and \emph{asymptotic stability}
emerge naturally. In particular, since energy and mass are conserved quantities,
it is natural to expect that solitons are stable in a suitable energy space.
Indeed, $H^1$-stability of mKdV and more general solitons and multi-solitons
has been considered e.g. in Benjamin \cite{Benj}, Bona-Souganidis-Strauss \cite{BSS},
Weinstein \cite{We2}, Maddocks-Sachs \cite{MS},
Martel-Merle-Tsai \cite{MMT}, Martel-Merle \cite{MMcol2} and M. \cite{Mu}.
$L^2$-stability of KdV solitons has been proved by Merle-Vega \cite{MV}.
Moreover, asymptotic stability properties for gKdV equations have been studied
 by Pego-Weinstein \cite{PW} and Martel-Merle \cite{MMarma,MMnon},
among many other authors.



We also mention that, in addition to Theorems \cite[Thm. 1.3]{AMP1}, \eqref{T1p8}
and \eqref{T1p9},
the more involved problem of asymptotic stability for breathers
could been also considered, as far as a good and rigorous understanding of the
associated spectral problem is at hand.
Usually, these spectral properties are harder to establish than the ones
involved in the stability problem (because the convergence problem requires
the use of weighted functions, which destroy most of breather's algebraic
properties). In addition, breathers can have zero, positive or negative velocity,
which means that they do not necessarily decouple from radiation. However, it
is worth to mention that if the velocity of a periodic
mKdV breather is positive, then there is local strong asymptotic stability
in the energy space, see \cite{AM}.

This article is organized as follows.
Section \ref{Sect:5}, and in particular, Theorem \ref{GBp} are devoted to the
proof that periodic KKHS breathers satisfy suitable elliptic equations,
and we find its variational structure.  In Section \ref{Sect:6}, after some
numerical tests, we sketch the proof of Theorem \ref{T1p8}
(see Theorem \ref{Thm000}) and we conjecture (see p. \pageref{Conj1})
that KKHS breathers have a dual stability/instability regime.
Moreover, assuming the validity of some numerical computations, we show
that KKHS are stable in a particular set of parameters.
Finally, in Section \ref{Sect:7} we present a new kind of periodic mKdV
breather whose main property is the fact that it has nonzero mean.


\section{Periodic mKdV breathers}\label{Sect:5}

\subsection{Definitions} We consider now the case of the periodic
(in space) mKdV equation. Periodic mKdV breathers (or KKSH breathers),
in the sense of Definition \ref{DEF_B_per}, were found by Kevrekidis, Khare,
Saxena and Herring  \cite{KKS1,KKS2} by using elliptic functions and a
matching of free parameters. More precisely, we consider the equation
\eqref{mKdV} where
\[
u:  \mathbb{R}_t\times \mathbb{T}_x \mapsto \mathbb{R}_x,
\]
is periodic in space, and $\mathbb{T}_x =\mathbb{T} = \mathbb{R}/ L\mathbb{Z}  =(0,L)$ denotes a torus
with period $L$, to be fixed later. Given $\alpha,\beta>0$, $x_1,x_2\in \mathbb{R}$ and
$k,m\in [0,1]$, KKSH breathers are given by the explicit formula \cite{KKS1}
\begin{equation} \label{Bper}
B= B(t,x; \alpha, \beta, k, m, x_1,x_2)
:=  \partial_x \tilde B := 2\sqrt{2} \partial_x \Big[ \arctan
\Big( \frac{\beta}{\alpha}\frac{\operatorname{sn}(\alpha y_1,k)}
{\operatorname{nd}(\beta y_2,m)}\Big) \Big],
\end{equation}
with $\operatorname{sn}(\cdot ,k)$ and $\operatorname{nd} (\cdot , m)$ the standard Jacobi elliptic
functions of elliptic modulus $k$ and $m$, respectively, but now
\begin{gather} \label{Y1_Y2}
y_1:=x+\delta t + x_1,  \quad y_2:=x+\gamma t +x_2, \\
\label{DG}
 \delta := \alpha^2(1+k) +3\beta^2(m-2), \quad
 \gamma := 3\alpha^2(1+k) +\beta^2(m-2).
\end{gather}

See Figure \ref{One_case} for a description of a KKHS breather solution and
 \cite{AS,By} for a more detailed account on
the Jacobi elliptic functions $sn$ and $nd$ presented in \eqref{Bper}.
Additionally, in order to be a periodic solution of mKdV,
the parameters $m,k,\alpha$ and $\beta$ must satisfy the commensurability conditions on
the spatial periods
\begin{equation} \label{Cond1}
\frac{\beta^4}{\alpha^4}=\frac{k}{1-m}, \quad K(k)=\frac{\alpha}{2\beta} K(m),
\end{equation}
where $K$ denotes the complete elliptic integral of the first kind,
defined as \cite{By}
\begin{equation} \label{Kint}
K(r):=\int_0^{\pi/2}(1-r\sin^2(s))^{-1/2}ds
= \int_0^1((1-t^2)(1-rt^2))^{-1/2}dt,
\end{equation}
and which satisfies
\[
K(0) =\frac \pi2, \quad \lim_{k\to 1^-}K(k)=\infty.
\]
Under these assumptions, the spatial period is given by
\begin{equation} \label{Largo}
L:=\frac{4}{\alpha}K(k) =\frac{2}{\beta} K(m).
\end{equation}
Note that conditions \eqref{Cond1}  formally imply that $B$ has only
four independent parameters (e.g. $\beta,k$ and translations $x_1,x_2$).
Additionally, if we assume that
the ratio $\beta/\alpha$ stays bounded, we have that $k$ approaches $0$ as $m$
is close to $1$. Using this information, the standard non
periodic mKdV breather \cite[eq. (1.8)]{AM} can be formally recovered as
the limit of very large spatial period $L\to +\infty$, obtained e.g.
if $k\to 0$. In that sense, we can think of \eqref{Bper} as a nontrivial
periodic bifurcation at infinity of the aperiodic mKdV breather.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=8cm, height=4cm]{fig1} %Breather_periodico_small.pdf}
\end{center}
\caption{Graph of the KKSH breather at time $t=0$, for $\beta=1$, $k=0.001$, $m=0.999$
 and $L=12.398$.}\label{One_case}
\end{figure}

It is important to mention that conditions \eqref{Cond1} impose a particular
set of restrictions on $k$ and $m$.
In terms of $k$, one has $k\in [0,k^*)$, with $k^*\sim 0.058$, while $m$
is decreasing with respect to $k$, with $m(k=0)=1$ and $m(k\sim k^*) \sim 0$.
Below, Figure \ref{km} describes the behavior of $k$ and $m$ more clearly.

In what follows, we will use the convention that
\begin{equation} \label{dependence}
m=m(k), \quad \alpha=\alpha(\beta,k),
\end{equation}
obtained by solving $m=m(k)$ in \eqref{Cond1} numerically, and then solving
 for $\alpha$ algebraically.


\begin{figure}[ht]
\begin{center}
\includegraphics[width=6cm, height=3cm]{fig2} % comm11.pdf
\end{center}
\caption{Graph of $m$ as a function of
$k$ obtained solving conditions \eqref{Cond1}. Although $m$ runs from 0 to 1,
not all values of $k \in [0,1]$ are allowed, being the limiting value of
$k\sim0.05883626$.}\label{km}
\end{figure}

\subsection{Variational characterization}
Our first result is the following theorem.

\begin{theorem}\label{GBp}
Assume \eqref{Cond1}. Let $B$ be any KKSH breather. Define
\begin{gather} \label{a1}
a_1 :=2(\beta^2(2-m)-\alpha^2(1+k)) = -\frac12(\delta+\gamma), \\
\label{a2}
a_2:=   \alpha^4(1+k^2-26k) + 2\alpha^2\beta^2(2-m)(1+k) + \beta^4m^2 .
\end{gather}
Then $B$ satisfies the generalized nonlinear elliptic equation
\begin{equation} \label{EcBp}
B_{(4x)} + 5 BB_x^2 + 5B^2 B_{xx} + \frac 32 B^5 - a_1(B_{xx}+B^3)  + a_2 B  =0.
\end{equation}
\end{theorem}

See Section \ref{Id522} for the proof of this result.
We emphasize that the first condition in \eqref{Cond1} is essential for
the proof of \eqref{EcBp}. However, we do not need the
second one. Note additionally that $a_1$ and $a_2$ converge to the
corresponding constants for the aperiodic mKdV case when $k\to 0$ and $m\to 1$,
see \cite[eq. (1.8)]{AM}.

Now we introduce the conserved quantities
\begin{gather} \label{Mass_per}
M_\#[u](t)  :=  \frac 12 \int_\mathbb{T} u^2(t,x)dx, \\
E_\#[u](t)  :=  \frac 12 \int_\mathbb{T} u_x^2(t,x)dx -\frac 14 \int_\mathbb{T} u^4(t,x)dx, \\
F_\#[u](t)  :=    \frac 12 \int_\mathbb{T} u_{xx}^2(t,x)dx -\frac 52 \int_\mathbb{T} u^2u_x^2(t,x)dx
  + \frac 14 \int_\mathbb{T} u^6(t,x)dx,
\end{gather}
which are preserved in the space
\begin{equation} \label{H2T}
H^2(\mathbb{T}) := \{ v \in H^2(0,L)  :  v(0)=v(L), \; v_x(0) =v_x(L)\},
\end{equation}
if $u(t,\cdot) \in H^2(\mathbb{T})$ is a solution of \eqref{mKdV}. The proof of this
result is straightforward if we work by density
in a space of smooth functions and we note that
$u(t,0) = u(t,L)$ and $u_x(t,0) = u_x(t,L) $  for all time imply
\[
u_t(t,0) = u_t(t,0), \quad  u_{xt}(t,0) = u_{xt}(t,L) ,
\]
and therefore, using \eqref{mKdV}, $u_{xxx}(t,0)=u_{xxx}(t,L)$.

\begin{corollary}\label{timeD}
KKSH breathers are critical points of the functional
\[
\mathcal H_\#[u] := F_\#[u](t) +a_1E_\#[u](t) + a_2M_\#[u](t),
\]
defined in the space $H^2(\mathbb{T})$ and preserved along the mKdV periodic flow.
\end{corollary}

As in the previous sections, the next step is the study of the nonlinear
stability of this solution. However, we must emphasize that the existence of a
suitable elliptic equation does not imply stability. Even worse, breathers
with standard spectrum may be nonlinearly unstable. Indeed,
we present below compelling evidence that
periodic mKdV breathers are \emph{unstable} under a suitable type of periodic
perturbations, at least for $L$ not so large.
(If $L$ is large it seems that the periodic breather is close in a certain
topology to the aperiodic breather,
which satisfies stability properties in a very general open neighborhood.)
Our results do agree with the numerical ones obtained by
Kevrekidis et al.\ \cite{KKS1,KKS2}, and our numerical computations below.
In this case, the periodic character of the solution leads to
nontrivial interactions between adjacent  breathers,
which probably play an important role in the instability character of this solution.

\section{Spectral analysis of the periodic mKdV breather}\label{Sect:6}

\subsection{Mathematical description}
In this section we give further evidence of the stable-unstable character
of the KKSH breather solution depending on the parameters phase space.
First of all, let us notice that, thanks to \eqref{EcBp},
the linearized operator for a KKSH breather is given by the expression
\begin{equation} \label{L_per}
\begin{aligned}
\mathcal L_\#[z]
& :=z_{(4x)} +(5 B^2 - a_1) z_{xx} + 10 B B_x z_x  \\
&\quad + \Big(a_2 + 5B_x^2 +  10BB_{xx} + \frac{15}2 B^4 - 3a_1 B^2 \Big)z.
\end{aligned}
\end{equation}
This operator is defined acting on functions in $H^2(\mathbb{T})$, $\mathbb{T}=(0,L)$,
 see \eqref{H2T}, and it reduces to the standard aperiodic mKdV breather operator
(see \cite[(4.1)]{AM}) as the length of the interval tends to infinity
(or $k\to 0$).
The constants $a_1$ and $a_2$ were introduced in \eqref{a1} and \eqref{a2},
and we assume the convention \eqref{dependence}.

In the following lines, we analyze the spectral stability of the KKSH breather,
namely the understanding of the spectrum of $\mathcal L_\#$ in \eqref{L_per}.
First, we prove some useful expression for the mass of a breather.
As a second step, we compute numerically the spectrum of $\mathcal L_\#$,
and conclude that it has the desired spectral properties.
Then, we analyze which property is the main responsible of the KKSH stability.

\subsection{Mass calculations}

The purpose of this paragraph is to compute the mass of the KKSH breather
as a function of $k$ and $\beta$.
This explicit function will be essential for the study of the nonlinear
stability of the solution.

First of all, recall the breather profile in \eqref{Bper}. Since the mass $M_\#$
in \eqref{Mass_per} is conserved,
we can simply assume $t=x_1=x_2=0$ in \eqref{Bper}-\eqref{Y1_Y2}.
Consider the functions $F$ and $G$ defined in \eqref{F_G},
and the length of the interval $L$ defined in \eqref{Largo}.
We have from \eqref{Mass_x},
\begin{equation}
\frac 12\int_0^L B^2 =  4\alpha^2\beta^2 \frac{F}{G}(x=L)
-4\alpha^2\beta^2 \frac{F}{G}(x=0)  -2\alpha^2\frac{k}{\beta^2}
\int_0^L (\beta^2\operatorname{sn}_1^2 - \alpha^2\operatorname{nd}_2^2).
\end{equation}
Now, note that from \eqref{Cond1} and the periodic character of the involved
functions, we have
\begin{equation} \label{B23}
\begin{aligned}
 \frac12\int_0^L B^2
& = 4\beta m \frac{ \operatorname{cd}(\frac{\beta}{2}L,m)\operatorname{sd}(\frac{\beta}{2}L,m) }{\operatorname{nd}(\frac{\beta}{2}L,m)}
 - 4\beta m \frac{\operatorname{cn}(\frac{\beta}{2}L,m)
 \operatorname{sn}(\frac{\beta}{2}L,m)}{\operatorname{dn}(\frac{\beta}{2}L,m)} \\
 & \quad -8\alpha K(k) + 4\alpha E( A(2 K(k),k),k)
 + 4\beta E\big(A\big(\frac{\beta}{2}L,m\big),m \big),
\end{aligned}
\end{equation}
where $E$ denotes the {\it complete elliptic integral of the second kind} \cite{By},
defined as
\begin{equation} \label{Eint}
E(r):=\int_0^{\pi/2}(1-r\sin^2(s))^{1/2}ds
= \int_0^1(1-t^2)^{-1/2}(1-rt^2)^{1/2}dt,
\end{equation}
and $A$ is the {\it Jacobi amplitude}, that can be defined by
\begin{equation} \label{Aint}
A(x,r):=\int_0^{x}dn(s,r)ds.% = \int_0^1(1-t^2)^{-1/2}(1-rt^2)^{1/2}dt.
\end{equation}
Some simple exact values for $E$ are $E(0)=\pi/2$ and $E(1)=1$, and for
$A$ are $A(0,r)=0$, $A(K(r),r)=\frac{\pi}{2}$. Note also that
$E(A(2 K(k),k),k) = E(A(0, k) + \pi,k)=2E(k)$.
Using \eqref{Cond1}, \eqref{B23} simplifies as follows:
\begin{equation} \label{B24}
\frac12 \int_0^L B_0^2 = 4\beta E(m) + 8\alpha E(k) - 8\alpha K(k).
\end{equation}

We are able to go one step further in the simplification of \eqref{B24},
 using \eqref{Cond1} again. We have
\begin{equation}\label{relper}
\frac k{1-m}=\frac{1}{16}\frac{K(m)^4}{K(k)^4}.
\end{equation}
Hence, with these relations, \eqref{B24} simplifies to the compact expression:
\begin{equation} \label{B25}
M_\#[B]= \frac{1}{2}\int_0^L B_0^2 = 4\beta\Big(E(m) + 4\frac{ K(k)}{ K(m)}( E(k)
-  K(k))\Big).
\end{equation}

When $k\to 0$, we have $m\to 1$, and $ M_\#[B] \to 4\beta$, which is the value
 of the mass of the aperiodic mKdV breather solution in the
real line (see \cite[p.6, Lemma 2.1]{AM}).

\begin{figure}[htb]
\begin{center}
 \includegraphics[width=10cm,height=9cm]{fig3} % mass2a.pdf}
\end{center}
\caption{Mass of the periodic breather $M_\#[B]/4\beta$ as a function of $m$.
Note that the resulting function is  decreasing except for $m\gtrsim 0.98$
(see zoom figure below),
corresponding to $k\lesssim 0.025$,  a parameter region very close
to the stable, aperiodic mKdV breather.} \label{MassBper}
\end{figure}

For $k\neq0,m\neq1$, the dependence of the {periodic mass} \eqref{B25} with
 respect to the parameter $m$ is computed in the following way:
for each value of $m$, we solve numerically the implicit equation \eqref{relper}
 in $k$. We then substitute these two pairs
of values $(m,k)$ verifying \eqref{relper} inside the expression \eqref{B25}.
The resulting plot of \eqref{B25} versus $m$ is given in Figure \ref{MassBper}.


\subsection{Numerical analysis}\label{Subsection_NUM}
Given the complicated functions that define the KKSH breather and its
 related linearized operator, a rigorous description of the spectra of
$\mathcal L_\#$ \eqref{L_per} has escaped to us. 
We perform then some numerical simulations
to get a good understanding of these spectral properties. We could expect,
given the instability results in \cite{KKS2}, that the KKSH breather
should give rise to a very different spectrum, with respect to the previous
aperiodic breathers. However, as we will show later, this is not the case.

In the following lines we explain the main ideas of the numerical method used
to compute the eigenvalues of $\mathcal L_\#$.
The core of the algorithm is the same as in the previous paper \cite{AMP1},
the main difference being the required test functions,
which now must be periodic on $[0,L]$. We have used the classical orthonormal
 basis of $L^2(0,L)$:
\[
\Big\{ \frac 1{\sqrt{L}}, ~\sqrt{\frac{2}L}
\cos \Big( \frac{2\pi n x}{L}\Big),~\sqrt{\frac{2}L}
\sin \Big( \frac{2\pi n x}{L}\Big) \Big\},
\]
where $n \in \{1, \ldots, N\}$. For the standard numerical computations,
 it is sufficient to take $N=40$, although when approaching the critical
values $k\to k^*$ ($m\to 0$) or $k\to 0$ ($m\to 1$), more and more test
 functions are naturally required.


\begin{figure}[ht]
\begin{center}
 \includegraphics[width=8cm,height=4cm]{fig4} % KKHS_breather.pdf}
\end{center}
\caption{Periodic KKSH breather \cite{KKS2} at time $t=0$ for
$\beta=1$, $m=0.5$, $k=0.057$, here $L\sim 3.71$.
Although numerically unstable \cite{KKS2}, this breather leads to a
linearized operator $\mathcal L_\#$ that possesses
a ``standard'' one negative eigenvalue and a two dimensional kernel
\eqref{results1}.}\label{KKHS_b}
\end{figure}

First of all, we run a specific computation for the case described in \cite{KKS2},
which the authors present as numerically unstable. In this case, $\beta=1$,
$m=0.5$, $k=0.057$, and $L\sim 3.71$ (see Figure \ref{KKHS_b}).
Numerically, we have found that this breather possesses the same spectral
structure of all our previous breather solutions.
To be more precise, running our numerical algorithm with $N= 40$ test functions,
we have found the following approximations of the first four eigenvalues of
$\mathcal L_\#:$
\begin{equation} \label{results1}
\{-4.86, ~ -1.23\times 10^{-8}, ~ 3.22\times 10^{-10}, ~ 35.35\}.
\end{equation}
Clearly the two components of the kernel are recovered with high precision (recall that a second negative eigenvalue but
very close to zero is unlikely just by continuity arguments on the coefficients of the original breather),
and a distinctive negative eigenvalue appears, far from the kernel itself. This behavior repeats for all cases we have studied.

\begin{figure}[ht]
\begin{center}
 \includegraphics[width=11cm, height=6cm]{fig5} % v261.pdf}
\end{center}
\caption{The  double zero kernel (purple box and brown diamond) and the first
 positive eigenvalue (green triangle) of the linearized operator $\mathcal L_\#$
 around a periodic KKSH breather for $N=40$, $\beta=1$ and $k$ increasing.
Note that the numerical method returns only two eigenvalues very
close to zero, as expected from the conjectured linear spectral stability.
Inside, the representation of the negative (blue circle) and double zero kernel.
We used the notation $.x242\equiv 0.058836242$, $x=058836$.}\label{v26}
\end{figure}

As a second test, we perform several eigenvalue computations for the same
parameter $\beta=1$, and $k$ moving.
For the most difficult case,  the one where $k$ approaches the critical value
$\sim 0.0588$, we obtain the results described in Figure   \ref{v26}.
Also, in Figure \ref{TablePer1}, we exactly describe those eigenvalues,
that we obtain for different values of $k$.
It is important to mention that we always get one negative eigenvalue and a
two dimensional kernel, as well as a clearly defined spectral gap.


{\small
\begin{table}[!h]
\begin{center}
 \begin{tabular}{|c c c c c|}
 \hline
 $k$ & 1st. eig & 2nd & 3rd & 4th \\ [0.5ex]
 \hline
 $.x240$ & -0.00008  & $-7.750\cdot10^{-10}$&  $-1.565\cdot10^{-9}$ & 75.490 \\
 \hline
 $.x242$  & -0.00006 & $-6.449\cdot10^{-10}$ &  $-1.561\cdot10^{-9}$ & 75.502 \\
 \hline
 $.x244$  & -0.00005 & $-5.952\cdot10^{-10}$ & $-1.558\cdot10^{-9}$ & 75.514\\
 \hline
 $.x246$  & -0.00004 & $-6.333\cdot10^{-10}$ & $-1.554\cdot10^{-9}$ & 75.528 \\
 \hline
 $.x248$  & -0.00003 & $-6.830\cdot10^{-10}$ & $-1.552\cdot10^{-9}$ & 75.544 \\
 \hline
 $.x250$  & -0.00002 & $-6.160\cdot10^{-10}$ & $-1.546\cdot10^{-9}$ &  75.563\\
 \hline
 $.x252$  & $-9.189\cdot10^{-6}$ & $-6.774\cdot10^{-10}$ & $-1.541\cdot10^{-9}$ & 75.589 \\
 \hline
 $.x254$  & $-3.463\cdot10^{-7}$ & $-1.135\cdot10^{-6}$ & $-1.869\cdot10^{-6}$ & 75.640 \\
 \hline
\end{tabular}
\end{center}
\caption{The first four eigenvalues of $\mathcal L_\#$ for
$\beta=1,~ x_1 = 0.1,~ x_2 = 0$, and
$k$ varying from $.x240\equiv0.058836240,~~x=058836$
to $.x254$, as in Figure \ref{v26}. All computations were made with
$N=40$ test functions.
The third and fourth columns represent  approximate kernel of
$\mathcal L_\#$.}\label{TablePer1}
\end{table}
}

In the intermediate case, for  pairs $(k,m)$, with $k=0.01,\dots,0.04$,
 we obtain the results described in Figure \ref{kinterm}. Also, in the
Table \ref{TablePer2}, we describe the eigenvalues  of $\mathcal L_\#$
 that we obtain for different values of $k$.  Once again, we obtain one negative
 eigenvalue and a two dimensional kernel.

\begin{figure}[ht]
\begin{center}
 \includegraphics[width=9cm, height=4.5cm]{fig6} % kinterm31.pdf}
\end{center}
\caption{For intermediate values of $k$: we plot the negative eigenvalue,
the double zero kernel and the fourth eigenvalue of the
linearized operator $\mathcal L_\#$ around a periodic KKSH breather,
for $\beta=1$ and $k$ increasing from $0.01$ to $0.05$, as expressed before
in Table \ref{TablePer2}.
Note that the numerical method returns only two eigenvalues very close to zero,
as expected from the conjectured linear spectral stability.
Computations were made with $N=50$ test functions.}\label{kinterm}
\end{figure}


{\small
\begin{table}[ht]
\begin{center}
 \begin{tabular}{|c c c c c|}
 \hline
 $k$ & 1st eig. & 2nd & 3rd  & 4th  \\ [0.5ex]
 \hline
 %0.09 & -4.226  & -0.028 &  0.028 & 1.776 \\
 0.01 & -5.343  & $1.023\cdot10^{-6}$ &  0.0001 & 2.540 \\
 \hline
 0.02 & -6.751  & $5.905\cdot10^{-10}$ &  $2.683\cdot10^{-6}$ & 3.561 \\
 \hline
 0.03 & -8.216 & $-2.651\cdot10^{-9}$ & $1.163\cdot10^{-6}$ & 5.067\\
 \hline
 0.04 & -9.623 & $3.997\cdot10^{-9} $ & $4.896\cdot10^{-8}$ & 7.756 \\
 \hline
 0.05 & -9.922 & $-2.173\cdot10^{-7}$ & $-1.143\cdot10^{-8}$ & 14.329 \\
 \hline
\end{tabular}
\end{center}
\caption{The first four eigenvalues of  $\mathcal L_\#$ for
$\beta=1$, $x_1 = 0.1$, $x_2 = 0$, and $k$ varying, as corresponding to Figure
 \ref{kinterm}.
All computations were made with $N=50$ test functions. The third and fourth
columns represent the approximate kernel  of $\mathcal L_\#$.}\label{TablePer2}
\end{table}
}



 Finally, for the  case,  with $k$ small, i.e. $k\approx10^{-3}$,
 we obtain in Figure \ref{kpeq} the description of the discrete spectra for
the first four eigenvalues for the linearized operator $\mathcal L_\#$.
 Also, in the Table \ref{TablePer3}, we show the explicit numerical eigenvalues
 which correspond to these small values of $k$.  Once again, we obtain one negative
eigenvalue and a two dimensional kernel.


\begin{figure}[ht]
\begin{center}
\includegraphics[width=10cm, height=5cm]{fig7} % kpeq.pdf
\end{center}
\caption{For $k$ (\emph{small}): the negative eigenvalue, the double zero
kernel and the fourth eigenvalue for the
linearized operator $\mathcal L_\#$ around
a periodic KKSH breather, for $\beta=1$ and $k$ increasing from $0.0005$
to $0.0095$, as expressed before in Table \ref{TablePer3}.
Note that the numerical method returns only two eigenvalues very close to zero,
as expected from the conjectured linear spectral stability. Computations were
made with $N=50$ test functions.}\label{kpeq}
\end{figure}


{\small
\begin{table}[htb]
\begin{center}
\begin{tabular}{|c c c c c |}
 \hline
 $k$ &   1st eig. &  2nd$\cdot10^{-6}$&  3rd & 4th\\
 \hline
0.0095 & -5.271  & 1.265 &  0.0001 & 2.495 \\
 \hline
 0.0085 & -5.127  & 1.962 & 0.0002 & 2.405 \\
 \hline
 0.0075 & -4.971 & 1.248 & 0.0142 & 2.316 \\
 \hline
 0.0065 & -4.828 & 5.092 & 0.0005 & 2.225 \\
 \hline
 0.0055 & -4.671 & 8.609 & 0.0009 & 2.134 \\
 \hline
 $k$ &   1st eig. &  2nd$\cdot10^{-5}$ &  3rd  & 4th \\
 \hline
 0.0045 & -4.505  & 1.531 & 0.0017 & 2.040 \\
 \hline
  0.0035 & -4.327  & 2.919 & 0.0034 & 1.941\\
 \hline
  0.0025 & -4.127  & 6.234 & 0.0073 & 1.833\\
 \hline
 0.0015 & -3.883 & 16.47 & 0.0198 & 1.708  \\
 \hline
 0.0005 & -3.497& 79.70 & 0.0983 & 1.530 \\
 \hline
\end{tabular}
\end{center}
\caption{The first four eigenvalues for the linearized operator
$\mathcal L_\#$ around a periodic KKSH breather for $\beta=1$, $x_1 = 0.1$,
$x_2 = 0$, and $k$ varying in a sample of points of Figure \ref{kpeq}.
All computations were made with $N=50$ test functions. The third and fourth
columns  represent respectively the approximate kernel of $\mathcal L_\#$.}
\label{TablePer3}
\end{table}
}

Therefore, we can certainly claim that there is strong evidence that,
according to numerical simulations, KKSH breathers have standard spectrum,
in the sense that they have only one negative eigenvalue, and a
nondegenerate, two-dimensional kernel. For clarity reasons, we present
a rigorous statement of both properties for the case of the periodic KKHS breather.
\begin{itemize}  \label{A14}
\item[(A4)]  (Nondegeneracy of the kernel)
 For each $k\in (0,k^*)$, $x_1,x_2\in \mathbb{R}$ and $\beta>0$,
$\ker \mathcal L_\#$ is spanned by the two elements $\partial_{x_1}B$ and
$\partial_{x_2} B$;
 and there is a uniform gap between the kernel and the bottom of the positive
spectrum;

\item[(A5)] (Unique, simple negative eigenvalue)
For each $k\in (0,k^*)$, $x_1,x_2\in \mathbb{R}$ and $\beta>0$,
the operator  $\mathcal L_\#$ has a unique simple, negative eigenvalue
$\lambda_{1}=\lambda_{1}(\beta,k,x_1,x_2)<0$ associated to the unit $L^2$-norm
eigenfunction $B_{-1}$. Moreover, there is $\lambda_{1}^0<0$ depending on
$\beta$ and $k$ only, such that  $\lambda_{1} \leq \lambda_{1}^0$ for all $x_1,x_2$.
\end{itemize}

\subsection{Duality stability/instability}
 Now the main problem is to figure out
where our nonlinear stability proof does/does not work. For this purpose,
the \emph{discriminant}
\begin{equation} \label{Discr}
D= D(\beta,k):= \partial_{k}a_1 \partial_\beta a_2 -\partial_{k}a_2 \partial_\beta a_1
\quad (a_1,a_2 \text{ from } \eqref{a1}-\eqref{a2}),
\end{equation}
is the key element to check.

\begin{figure}[ht]
\begin{center}
 \includegraphics[width=10cm,height=5cm]{fig8} % D1.pdf}
\end{center}
\caption{Discriminant function $D(1,k)$ in terms of $k$, as described
in \eqref{Discr}. Note that $D(1,k)$ changes sign (see zoom figure inside)
for $k \geq k_* \sim 0.0545$.
}\label{KKHS_c}
\end{figure}

To explain why this element is important, let us notice that from
\eqref{EcBp} and \eqref{L_per},
we readily have (compare with \cite[Corollary (3.7)]{AMP1})
\begin{gather*}
\mathcal L_\# (\partial_k B)  = \partial_k a_1 (B_{xx}+B^3) -\partial_k a_2 B, \\
\mathcal L_\# (\partial_\beta B)  = \partial_\beta a_1 (B_{xx}+B^3) -\partial_\beta a_2 B.
\end{gather*}
Therefore, as soon as $D\neq 0$,
\begin{equation} \label{B_0per}
B_{0,\#} := \frac1D (\partial_k a_1 \partial_\beta B - \partial_\beta a_1 \partial_k B ),
\end{equation}
satisfies the equation
\[
\mathcal L_\# (B_{0,\#}) =-B,
\]
see also \cite[Corollary 4.5]{AM}. Using this fact, we can easily prove,
as in \cite[ Proposition (3.11)]{AMP1},
that the eigenfunction associated to the negative eigenvalue of
$\mathcal L_\#$ can be replaced by the breather itself,
which has better behavior in terms of error controlling, unlike the first
eigenfunction. This simple fact allows us to prove the nonlinear stability
result as in the standard approach, without using scaling modulations.
Recall that using the first eigenfunction as orthogonality condition does
not guaranty a suitable control on the scaling modulation
parameter, because the control given by this direction might be not good
enough to close the stability estimates. However,
the breather can be used as an alternative direction, and all these previous
arguments remain valid, exactly as in \cite{AM},
provided the Weinstein's sign condition
\begin{equation} \label{Cond3}
\int_0^L B_{0,\#} B >0  \quad \text{\big(or equivalently $\int_0^L B_{0,\#
 \mathcal L_\#[B_{0,\#}] <0}$\big),}
\end{equation}
do hold.  Using \eqref{B_0per}, we are lead to the understanding of the quantity
\begin{equation} \label{HG}
\begin{aligned}
\int_0^L B_{0,\#} B
& =  \frac1D \int_0^L (\partial_k a_1 \partial_\beta B
 - \partial_\beta a_1 \partial_k B ) B \\
& = \frac1D (\partial_k a_1 \partial_\beta M_\#[B]
 -\partial_\beta a_1 \partial_k M_\#[B]) =: HG(\beta,k),
\end{aligned}
\end{equation}
where $ M_\#[B]$ was computed in \eqref{B25}. Recall that $a_1$ and $a_2$
are almost explicit from \eqref{a1}-\eqref{a2}. An exact expression for
$HG(\beta,k)$ has escaped to us, however, we can graph this new function
in some interesting cases. In particular, for the case considered in \cite{KKS2},
we assume $\beta=1$ and we graph $D=D(1,k)$ and $HG(1,k)$, to obtain the results
in Figure \ref{KKHS_c} and Figure \ref{KKHS_d}. We note that condition \eqref{Cond3}
holds provided $k$ is small enough.  However, the values for
which $HG(\beta,k)>0$ do not coincide with the values for which the standard
 Weinstein's condition
(positive derivative with respect to the scaling), deduced from Figure \ref{MassBper},
 holds true, and this is totally natural for the case of breathers,
as it was explained in \cite[Corollary 2.2]{AM}.

\begin{figure}[ht]
\begin{center}
 \includegraphics[width=11cm, height=7.5cm]{fig9} % HG4.pdf
\end{center}
\caption{Weinstein's condition $HG(\beta,k)$ \eqref{HG}, for the periodic  KKSH
breather, in the case $\beta=1$, for $k\in [0.0045,0.01]$ (above left),
for $k\in [0.01,0.054]$ (above right), $k\in [0.054,0.056]$ (below left),
and $k\in [0.056, 0.058]$ (below right). In order to run our argument for a
stability proof \cite{AM}, we require $HG(1,k)>0$,
which is only satisfied for $k< k_*$, where $k_*\sim 0.0545$ is the
approximate point where $D(1,k)$ vanishes (see Figure \ref{KKHS_c}).
Note also that the case $k\sim 0.057$ assumed in \cite{KKS2},
that leads to instability, is not included in the stability region
described in Theorem \ref{Thm000}, but in the region where $HG(1,k)<0$.}
\label{KKHS_d}
\end{figure}

Now we are ready to fully state assumption (A3) described in the introduction
of this paper.
\begin{itemize}
\item[(A6)]    (Positive generalized Weinstein's condition)
 The following generalized Weinstein's type sign condition is satisfied:
 if $M_\#[B]$ is the mass of the breather solution  \eqref{B25}  in terms of
$\beta$ and $k$, and $a_1, a_2$ are the variational parameters given
in \eqref{a1}-\eqref{a2}, then
\begin{equation} \label{Weinstein23}
 HG(\beta,k) =  \frac{1}{D}(\partial_k a_1 \partial_\beta M_\#[B]
-\partial_\beta a_1 \partial_k M_\#[B]) >0.
\end{equation}
\end{itemize}
Now we can present a rigorous statement for Theorem \ref{T1p8}.

\begin{theorem}\label{Thm000}
Under assumptions (A4)--(A6), KKHS breathers are orbitally stable under small,
 $L$-periodic $H^2$ perturbations. More precisely, there are $\eta_0>0$ and
$K_0>0$, only depending on $\beta$ and $k$, such that if $0<\eta<\eta_0$ and if
 $u_0$ is $L$-periodic with
\[
\| u_0 - B(t=0,\cdot ; 0, 0)    \|_{H^2(\mathbb{T})} <\eta,
\]
then there are real-valued parameters $x_1(t)$ and $x_2(t)$ for which the
global $H^2$, $L$-periodic solution $u(t)$ of \eqref{mKdV} with initial data
$u_0$ satisfies
\[
\sup_{t\in \mathbb{R}} \| u(t) - B(t,\cdot ; x_1(t), x_2(t))    \|_{H^2(\mathbb{T})} <K_0\eta,
\]
with similar estimates for the derivatives of the shift parameters $x_1,x_2$.
\end{theorem}

\begin{proof}
The proof of this result is completely similar to the proof of
\cite[Theorem (2.5)]{AMP1}, after following the same steps (see also \cite{AM}
for a proof in the scalar mKdV case).
Assumption (A3) is used to ensure that an expression like  \cite[eq. (3.30)]{AMP1}
 has a nonzero denominator.
\end{proof}

Additionally, we conjecture the following alternative stability theory
for KKSH breathers:


\begin{conjecture} \label{Conj1}
 Assume that $HG(\beta,k)<0$. Then $B$ is  unstable under small $H^2(\mathbb{T})$
perturbations.
\end{conjecture}

Examples of unstable structures that have a
nondegenerate kernel and only one negative eigenvalue are solitons for
the nonlinear Klein-Gordon equation in $\mathbb{R}_t \times \mathbb{R}_x^d$:
\[
u_{tt} - \Delta u + u + u^p =0, \quad p>1.
\]
Note that this result cannot be deduced from the Grillakis-Shatah-Strauss
method, since breathers are not simple solitary waves. Here,
unlike our case, the lack of stability is related to the absence of a scaling
symmetry controlling the negative direction (appearing
because of the negative eigenvalue). Another type of instability result
suggested very recently is motivated by the existence of an stable
``internal mode'' which triggers a nonlinear unstable dynamics in NLS
and discrete models. See the works \cite{Cue,KPS} for further
reading and more references.

\section{Periodic mKdV breathers with nonzero mean}\label{Sect:7}

\subsection{Introduction}
Although KKSH breathers are periodic in space, the are still zero-mean solutions.
We would like to see if nonzero mean periodic breather solutions may exist.
(Non periodic and nonzero mean breather solutions of mKdV
were already known, see \cite{Ale0, Ale}.)
 By periodic breather we refer to the object in Definition \ref{DEF_B_per},
that is, any solution that is periodic in time and space, having two
independent and different space variables (i.e. not being a one profile solution
being translated over time, as solitons are),
and finally, having oscillatory behavior, unlike standard 2-soliton solutions.
Numerical evidence of the existence of these solutions
was given by the first author in \cite{Ale}, since these solutions are
connected with the (nonzero) curvature of the planar curve evolving
according to a particular law. For more details about this connection,
see e.g. \cite{Ale}.

For the case of the mKdV equation, we have been able to obtain a new set
of periodic breather solutions of \eqref{mKdV} \emph{with nonzero mean}.
More explicitly,

\begin{definition}\label{NEW_BREATHER} \rm
Given $c_1,c_2,\mu>0$, $p,q$ nonzero integers, with $p,q$ coprime, such that
the following \emph{commensurability} condition is
satisfied\footnote{Note that each $c_i$ is less than $2\mu^2$.}
\begin{equation} \label{Cons2}
 \frac{2\mu^2-c_1}{2\mu^2-c_2}=\frac{p^2}{q^2},
\end{equation}
we define the breather $B=B(t,x;c_1,c_2,\mu,p,q)$ by the formula
\begin{equation} \label{BP1}
 B:= \mu + 2\sqrt{2}\partial_x\arctan\Big(\frac{f(t,x)}{g(t,x)} \Big),
\end{equation}
where
\begin{gather*}
\rho:=\frac{\sqrt{c_1}+\sqrt{c_2}}{\sqrt{c_1}-\sqrt{c_2}}, \\
 f(t,x) := -\sqrt{2}\mu \rho\Big(\sqrt{c_1}-\sqrt{c_2}
+ \sqrt{2\mu^2-c_2}\tan y_2-\sqrt{2\mu^2-c_1}\tan y_1\Big), \\
g(t,x) := 2\mu^2 + \Big( \sqrt{2\mu^2-c_1}\tan y_1 -\sqrt{c_1}
\Big)\Big( \sqrt{2\mu^2-c_2}\tan y_2 -\sqrt{c_2}  \Big).
\end{gather*}
Here
\begin{equation} \label{y1y2_fin}
y_1=\frac12\sqrt{2\mu^2-c_1}(x-\delta t),\quad
y_2=\frac12\sqrt{2\mu^2-c_2}(x-\gamma t),
\end{equation}
and the speeds are
\[
 \delta:=\mu^2+ c_1,\quad \gamma:=\mu^2+ c_2.
\]
\end{definition}
Note that condition \eqref{Cons2} is imposed in order to obtain
a truly periodic solution, see the formulae for $f$  and $g$.
The spatial period of this breather is given by
$L = \frac{2\pi q}{\sqrt{2\mu^2 -c_1}}$. See also Figure \ref{fig1}-\ref{fig2}
for some drawings of different breather solutions, depending on the parameters
$c_1,c_2,\mu, p$ and $q$.


\begin{figure}[htb]
\begin{center}
 \includegraphics[width=12cm,height=6cm]{fig10} % l1a.pdf
\end{center}
\caption{Periodic breather of \eqref{mKdV} with non zero mean.
The parameters here are $c_1=1.65$, $c_2=2.95$, $p=23$, $q=22$, and the period
$L$ is $\sim 35.7$}\label{fig1}
\end{figure}


\begin{figure}[htb]
\begin{center}
 \includegraphics[width=11cm,height=5cm]{fig11} % l2a.pdf
\end{center}
\caption{Periodic breather of \eqref{mKdV} with non zero mean,
for parameters $c_1=20.65$, $c_2=21.95$, $p=17$, $q=16.$}\label{fig2}
\end{figure}

Recall that KKSH breathers cannot have any admissible set of parameters $k$ and $m$;
they are constrained by conditions in \eqref{Cond1}. Here, the only condition
that we need to satisfy is \eqref{Cons2}. Therefore, given nonnegative integers
$p$ and $q$, with $p$ and $q$ coprime, and given $c_1<2\mu^2$, then there is a
unique $c_2$ solution to \eqref{Cons2}. In that sense,
$\mu$ being fixed, the breather in \eqref{BP1} has three different degrees of
freedom (in addition to shifts), two of them being discrete.
Finally, in terms of $p$ and $q$, the larger these integers are, the more
oscillatory the breather solution is. Note additionally that this breather
has not been constructed by using Jacobi functions, but only standard
periodic functions, a fact that simplifies many computations.

It is also relevant to mention that the construction of a breather solution
with different patterns as the usually required can be very involved.
For example, a different class
of periodic breather was discovered by Blank et al. in \cite{BCLS}.
This particular solution is constructed for the so called $\phi^4$ model,
by assuming that the equation is no longer autonomous, but it has suitable,
well-chosen periodic coefficients.


\subsection{Sketch of proof of Theorem \ref{T1p9}}
We will use the B\"acklund Transformation for mKdV (see \cite{AM1} for a
rigorous setting of the computations below) to construct this solution.
Given a solution $u_0$ of the form $u_0 =\partial_x \widetilde u_0$ of mKdV
and fixed constants $a_1,a_2\in \mathbb{R}$, $a_1\neq a_2$, we can construct a
second $u_1 =\partial_x \widetilde u_1$ and third solution
$u_2 =\partial_x \widetilde u_2$ by setting
\[
u_1 - u_0 = a_1 \sin\Big(\frac{\widetilde u_1
+ \widetilde u_0}{\sqrt{2}} \Big), \quad \text{and} \quad
u_2 - u_0 = a_2 \sin\Big(\frac{\widetilde u_2 + \widetilde u_0}{\sqrt{2}} \Big).
\]
These two solutions can be combined to construct a fourth solution $u$ through
the \emph{permutability} condition \cite{AM1}:
\begin{equation} \label{permu4}
\tan\Big(\frac{\widetilde u -\widetilde u_0}{2\sqrt{2}}\Big)
= -\Big(\frac{a_1+a_2}{a_1-a_2}\Big) \tan\Big(\frac{\widetilde u_2
 - \widetilde u_1}{2\sqrt{2}}\Big).
\end{equation}
Fix $\mu>0$. Starting with the constant solution $u_0 =\mu$,
$\widetilde u_0= \mu x$ and two constants $a_1=\sqrt{2c_1}$,
$a_2=\sqrt{2c_2}$, $c_1,c_2>0$, we have
\begin{equation} \label{bt1}
u_1 - \mu =  \sqrt{2c_1}\sin \Big( \frac{\tilde u_1 + \mu x}{\sqrt{2}} \Big), \quad
u_2 - \mu =  \sqrt{2c_2}\sin \Big( \frac{\tilde u_2 + \mu x}{\sqrt{2}} \Big).
\end{equation}
Looking for a solution of the form $u_i= \partial_x \tilde u_i$, $i=1,2$, we obtain
\begin{equation} \label{bt12}
\begin{aligned}
\widetilde u_i(t,x)
& = -\mu x + 2\sqrt{2}\arctan\Big(\frac{1}{2\mu}
\Big(- \sqrt{2c_i}\\
&\quad +\sqrt{4\mu^2-2c_i} 
 \tan\Big(\frac{\sqrt{4\mu^2-2c_i}}{2\sqrt{2}}(x-(\mu^2+ c_i)t)
 \Big) \Big)\Big).
\end{aligned}
\end{equation}
The factor $(\mu^2+ c_i)t$ in the above solution appears as a constant of
integration, and it is chosen in such a form that $u_i$ is actually solution
to mKdV. Calling $y_1 =x+\delta t$, $y_2= x+ \gamma t$ as in \eqref{y1y2_fin},
and $\rho=\frac{\sqrt{c_1}+\sqrt{c_2}}{\sqrt{c_1}-\sqrt{c_2}}$,
the desired breather \eqref{BP1} is obtained by using \eqref{permu4} with
$u_0=\mu$, $u_1,u_2$ from \eqref{bt12}:
\[
B =\mu + 2\sqrt{2}\partial_x\arctan\Big(\frac{f(t,x)}{g(t,x)} \Big),
\]
where,
\begin{gather*}
 f(t,x) := -\sqrt{2}\mu \rho\Big(\sqrt{c_1}-\sqrt{c_2} 
+ \sqrt{2\mu^2-c_2}\tan y_2-\sqrt{2\mu^2-c_1}\tan y_1\Big), \\
g(t,x) := 2\mu^2 + \Big( \sqrt{2\mu^2-c_1}\tan y_1 -\sqrt{c_1}  \Big)
\Big( \sqrt{2\mu^2-c_2}\tan y_2 -\sqrt{c_2}  \Big).
\end{gather*}
The fact that this is a solution of mKdV is a tedious, lengthy but
straightforward computation.


\subsection{Final comments}
A detailed study of this new breather solution will be done elsewhere.
For the moment, we advance that $B$ \eqref{BP1} satisfies the elliptic ODE
\begin{align*}
& B_{4x}  -   (c_1+c_2-4\mu^2) (B_{xx} +3\mu (B-\mu)^2 + (B-\mu)^3) \\
&+(c_1-2\mu^2)(c_2-2\mu^2)(B-\mu)
  + 5 (B-\mu)B_{x}^2 + 5(B-\mu)^2 B_{xx} \\
&+ \frac 32 (B-\mu)^5  + 5\mu B_{x}^2 
 + \frac{15}{2} \mu (B-\mu)^4  + 10\mu  (B-\mu) B_{xx}  + 10\mu^2 (B-\mu)^3=0.
\end{align*}
In particular, $B$ is a critical point of the functional
\[
\mathcal H_{\mu p}[w](t) := F_{\mu p}[w](t)
+  (c_1+c_2-4\mu^2)E_{\mu p}[w](t)
+(c_1-2\mu^2)(c_2-2\mu^2) M_{\mu p}[w](t),
\]
where $M_{\mu p}$ and $E_{\mu p}$ are defined as follows:
\begin{gather} \label{M1p}
M_{\mu p}[w](t)  :=  \frac 12 \int_\mathbb{R} (w-\mu)^2(t,x)dx = M_{\mu p}[w](0), \\
 \label{E2p}
E_{\mu p}[w](t)  :=  \frac 12 \int_\mathbb{R} w_x^2 -\mu\int_\mathbb{R} (w-\mu)^3
- \frac 14 \int_\mathbb{R} (w-\mu)^4= E_{\mu p}[w](0),\\
\label{F1p}
\begin{aligned}
 F_{\mu p}[w](t)
& :=   \frac 12 \int_\mathbb{R} w_{xx}^2dx - 5\mu\int_\mathbb{R} (w-\mu)w^2_x\,dx
+ \frac{5}{2}\mu^2\int_\mathbb{R} (w-\mu)^4dx \\
& \quad   - \frac{5}{2} \int_\mathbb{R} (w-\mu)^2w_x^2dx
+ \frac{3\mu}{2} \int_\mathbb{R} (w-\mu)^5+ \frac{1}{4} \int_\mathbb{R} (w-\mu)^6dx
\end{aligned}
\end{gather}
is a third conserved quantity for the mKdV equation. Finally, it is
 interesting to note that we can recover the aperiodic
breather with nonzero mean (see \cite{Ale0,Ale1}) choosing in \eqref{BP1}
complex conjugate scalings $\sqrt{c_1}=\beta+i\alpha,~\sqrt{c_2}=\beta-i\alpha$.
For the sake of simplicity, we only show a picture of it in Figure \ref{mKdV_per1}.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=9cm, height=5cm]{fig12} %perBN0mean.pdf
\end{center}
\caption{Aperiodic mKdV breather with non zero mean at $t=0.2$ with
$\beta=1$, $\alpha=4$, $\mu=0.5$, obtained from the periodic breather \eqref{BP1}.}
\label{mKdV_per1}
\end{figure}


\section{Appendix: Proof of Theorem \ref{GBp}}\label{Id522}

Let $B$ be a periodic KKSH breather. Without loss of generality, we can
assume $x_1 =x_2 =0$, and after taking time derivative we
assume $t=0$, since \eqref{mKdV} is invariant under space and time translations,
as well as \eqref{GBp}. Recall that from \eqref{Bper}
\[
\tilde B_t := \delta \tilde B_1 + \gamma \tilde B_2,
\]
where $\tilde B_j := \partial_{x_j}\tilde B$. We also have \cite{AS,By}:
\begin{gather*}
\operatorname{sn}'(s,k) = \operatorname{cn}(s,k) \operatorname{dn}(s,k), \quad \Big(\operatorname{dn}(s,k) := \frac1{\operatorname{nd}(s,k)} \Big), \\
\operatorname{cn}' (s,k)=- \operatorname{sn}(s,k)\operatorname{dn}(s,k), \quad
\operatorname{dn}' (s,k)= -k \operatorname{sn} (s,k)\operatorname{cn}(s,k),
\end{gather*}
and $\operatorname{cn}(0,k)=\operatorname{dn}(0,k) =1$, $\operatorname{sn}(0,k)=0$.

We start with some notation. Let
\begin{gather*}
\operatorname{sn}_1 := \operatorname{sn}(\alpha y_1,k), \quad \operatorname{cn}_1 := \operatorname{cn}(\alpha y_1,k) , \quad
 \operatorname{dn}_1 := \operatorname{dn}(\alpha y_1,k); \\
\operatorname{sn}_2 := \operatorname{sn}(\beta y_2,m), \quad \operatorname{cn}_2 := \operatorname{cn}(\beta y_2,m) , \quad
\operatorname{dn}_2 := \operatorname{dn}(\beta y_2,m).
\end{gather*}
We have
\[
\tilde B = 2\sqrt{2} \arctan \Big(\frac\beta\alpha \operatorname{sn}_1\operatorname{dn}_2 \Big).
\]
Define
\begin{equation} \label{hth}
h := \alpha \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2 - \beta m \operatorname{sn}_1\operatorname{sn}_2\operatorname{cn}_2, \quad
\tilde h:=\alpha \delta \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2 - \beta m\gamma \operatorname{sn}_1\operatorname{sn}_2\operatorname{cn}_2,
\end{equation}
so that
\begin{equation} \label{hx}
\begin{aligned}
h_x &= - \Big[ \alpha^2\operatorname{sn}_1\operatorname{dn}_2(k\operatorname{cn}_1^2+\operatorname{dn}_1^2)  + 2\alpha\beta m\operatorname{cn}_1\operatorname{dn}_1\operatorname{sn}_2\operatorname{cn}_2 \\
&\quad + \beta^2 m \operatorname{sn}_1\operatorname{dn}_2(\operatorname{cn}_2^2-\operatorname{sn}_2^2) \Big],
\end{aligned}
\end{equation}
and
\begin{equation} \label{ht}
\begin{aligned}
h_t &= - \Big[\alpha^2 \delta \operatorname{sn}_1 \operatorname{dn}_2 (k \operatorname{cn}_1^2 + \operatorname{dn}_1^2)
+ \alpha\beta m(\delta+\gamma) \operatorname{cn}_1 \operatorname{dn}_1 \operatorname{sn}_2\operatorname{cn}_2  \\
&\quad + \beta^2 m\gamma \operatorname{sn}_1\operatorname{dn}_2 (\operatorname{cn}_2^2 -\operatorname{sn}_2^2) \Big].
\end{aligned}
\end{equation}
Similarly,
\begin{gather} \label{gg}
g := \alpha^2 + \beta^2 \operatorname{sn}_1^2\operatorname{dn}_2^2, \quad
g_x = 2\beta^2\operatorname{sn}_1 \operatorname{dn}_2 ( \alpha\operatorname{cn}_1\operatorname{dn}_1 \operatorname{dn}_2 - \beta m\operatorname{sn}_1 \operatorname{sn}_2\operatorname{cn}_2), \\
\label{gt}
g_t =2\beta^2 \operatorname{sn}_1 \operatorname{dn}_2 (\alpha \delta \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2
 - \beta m \gamma \operatorname{sn}_1\operatorname{sn}_2\operatorname{cn}_2) =2\beta^2 \operatorname{sn}_1 \operatorname{dn}_2 \tilde h.
\end{gather}
Then
\begin{gather*}
B= \frac{2\sqrt{2}\alpha\beta h}{g}, \quad B(x=0) =2\sqrt{2} \beta, \\
\tilde B_t = \delta \tilde B_1 + \gamma \tilde B_2
 =  \frac{2\sqrt{2}\alpha\beta \tilde h}{g}, \quad
\tilde B_t(x=0) = 2\sqrt{2}\beta \delta.
\end{gather*}
Similarly,
\begin{equation} \label{Bxper}
B_{x} = \frac{2\sqrt{2}\alpha\beta  \hat h}{g^2}, \quad \hat h:= h_x g- g_xh.
\end{equation}
It is not difficult to check that $h_x(x=0) =g_x(x=0) =0$.
In particular $B_x(x=0)=0$. From this identity we see that
\[
B_{xx}(x=0) =\frac{2\sqrt{2}\alpha\beta \hat h_x}{g^2}(x=0)
= -2\sqrt{2} \beta [ (2+3m)\beta^2 +(1+k)\alpha^2 ].
\]
First of all,  from \eqref{mKdV} and \eqref{DG} we have
\[
\tilde B_t +B_{xx} + B^3 = (\tilde B_t +B_{xx} + B^3)(x=0) = 0.
\]
This identity can be proved to hold \emph{for any} $t,x_1,x_2\in \mathbb{R}$.
 On the other hand, if
\[
\mathcal M:=  \frac 12\int_0^x B^2, \quad \mathcal M_t =  \int_0^x B B_t,
\]
we have
\[
B \tilde B_t - \mathcal M_t + \frac 12 B_{x}^2 + \frac 14B^4
=(B \tilde B_t +  \frac 12 B_{x}^2 + \frac 14B^4)(x=0) =:  \frac 12c_0,
\]
where for $t=x_1=x_2=0$ we have that  $c_0$ is explicitly given by
\begin{equation} \label{c0}
c_0 := 16\beta^2 [\alpha^2 (1+k) + \beta^2(3m-4)].
\end{equation}
However, in the general case,  $c_0$ may depend on time.

Replacing in \eqref{EcBp} we obtain
\begin{equation}
\begin{aligned}
&B_{(4x)} + 5 BB_x^2 + 5B^2 B_{xx} + \frac 32 B^5 - a_1(B_{xx}+B^3)  + a_2 B
  \\
& = -(B_t + 3B^2B_x)_{x}+ 5 BB_x^2 + 5B^2 B_{xx} + \frac 32 B^5 - a_1(B_{xx}+B^3)
 + a_2 B   \\
& = -B_{tx}  - BB_x^2 + 2 B^2 B_{xx} + \frac 32 B^5 - a_1(B_{xx}+B^3)  + a_2 B   \\
&= -B_{tx}  - B(c_0 - 2 B\tilde B_t +2\mathcal M_t -\frac 12 B^4) -  2 B^2 ( B^3
  + \tilde B_t) \\
&\quad + \frac 32 B^5 + a_1\tilde B_t  + a_2 B   \\
&= -B_{tx}  - 2 B \mathcal M_t   + a_1\tilde B_t  + (a_2-c_0) B .
\end{aligned} \label{Ecuacion}
\end{equation}
Now we prove that this last quantity is identically zero.
We compute $\mathcal M_t$. Denote
\begin{equation} \label{F_G}
F:= \frac{1}{2\alpha} \operatorname{sn}_1 \operatorname{sn}_1' + \frac 1{2\beta } \operatorname{nd}_2 \operatorname{nd}_2' , \quad
G:= \beta^2 \operatorname{sn}_1^2 + \alpha^2 \operatorname{nd}_2^2,
\end{equation}
where, with a slight abuse of notation we denote
\[
\operatorname{sn}_1' := \operatorname{sn}'(\alpha y_1,k), \quad \operatorname{cn}_1' := \operatorname{cn}'(\alpha y_1,k) , \quad
\operatorname{dn}_1' := \operatorname{dn}'(\alpha y_1,k);
\]
and so on.  We claim that
\begin{equation} \label{B22}
B^2 = 8\alpha^2\beta^2(\frac{F}{G})_x-4\alpha^2\frac{k}{\beta^2}(\beta^2 \operatorname{sn}_1^2
- \alpha^2 \operatorname{nd}_2^2).
\end{equation}
Indeed, from \eqref{Bper} we have
\begin{equation} \label{Bquad}
B^2 =8\alpha^2\beta^2\frac{\alpha^2 \operatorname{nd}_2^2\operatorname{sn}_1'^2 + \beta^2\operatorname{sn}_1^2\operatorname{nd}_2'^2
- 2\alpha\beta \operatorname{sn}_1 \operatorname{nd}_2 \operatorname{sn}_1 ' \operatorname{nd}_2'}{(\beta^2 \operatorname{sn}_1^2 + \alpha^2 \operatorname{nd}_2^2)^2}.
\end{equation}
Note that
\begin{align*}
F_xG - FG_x
&=  \frac12  (  \operatorname{sn}_1'^2 +  \operatorname{sn}_1\operatorname{sn}_1''  +  \operatorname{nd}_2'^2
  +  \operatorname{nd}_2\operatorname{nd}_2''  )(\beta^2 \operatorname{sn}_1^2 + \alpha^2 \operatorname{nd}_2^2) \\
&\quad -( \beta \operatorname{sn}_1 \operatorname{sn}_1' +\alpha  \operatorname{nd}_2 \operatorname{nd}_2')
(\beta \operatorname{sn}_1\operatorname{sn}_1' + \alpha \operatorname{nd}_2 \operatorname{nd}_2') .
\end{align*}
Now using the well-known JEF identities (see \cite{AS,By})
and replacing above we obtain
\begin{align*}
F_xG - FG_x
&=  \alpha^2 \operatorname{nd}_2^2  \operatorname{sn}_1'^2 + \beta^2 \operatorname{sn}_1^2  \operatorname{nd}_2'^2
  - 2\alpha\beta \operatorname{sn}_1 \operatorname{sn}_1'\operatorname{nd}_2 \operatorname{nd}_2' \\
&\quad +  \frac12 (\beta^2k \operatorname{sn}_1^6 + \alpha^2  k  \operatorname{nd}_2^2 \operatorname{sn}_1^4
 + \beta^2 (m-1) \operatorname{sn}_1^2 \operatorname{dn}_2^4 + \alpha^2 (m-1) \operatorname{dn}_2^6) \\
& =  \alpha^2 \operatorname{nd}_2^2  \operatorname{sn}_1'^2 + \beta^2 \operatorname{sn}_1^2  \operatorname{nd}_2'^2
 - 2\alpha\beta \operatorname{sn}_1 \operatorname{nd}_2 \operatorname{sn}_1' \operatorname{nd}_2'\\
&\quad +  \frac{k}2 \operatorname{sn}_1^4(\beta^2 \operatorname{sn}_1^2+\alpha^2 \operatorname{nd}_2^2)
  + \frac{(m-1)}2 \operatorname{nd}_2^4(\beta^2 \operatorname{sn}_1^2+\alpha^2\operatorname{nd}_2^2).
\end{align*}
Therefore, from \eqref{Bquad},
\begin{equation} \label{B2per}
B^2  =  8\alpha^2 \beta^2 \frac{(F_xG-FG_x)}{G^2}
- 4\alpha^2\beta^2 \frac{k\operatorname{sn}_1^4
+ (m-1)\operatorname{nd}_2^4}{(\beta^2 \operatorname{sn}_1^2 + \alpha^2 \operatorname{nd}_2^2)}.
\end{equation}
Since $\frac{\beta^4}{\alpha^4}=\frac{k}{1-m}$, \eqref{B2per} simplifies as follows:
\[
B^2 = 8\alpha^2\beta^2(\frac{F}{G})_x-4\alpha^2 \frac{k}{\beta^2}(\beta^2\operatorname{sn}_1^2 - \alpha^2\operatorname{nd}_2^2),
\]
as desired. From \eqref{B22} we have
\begin{equation} \label{Mass_x}
\frac 12\int_0^x B^2 =  4\alpha^2\beta^2 \frac{F}{G} -4\alpha^2\beta^2 \frac{F}{G}(0)  -2\alpha^2\frac{k}{\beta^2} \int_0^x (\beta^2\operatorname{sn}_1^2 - \alpha^2\operatorname{nd}_2^2).
\end{equation}
Since
\[
\frac{F}{G} = \frac{\beta \operatorname{sn}_1 \operatorname{cn}_1 \operatorname{dn}_1\operatorname{dn}_2^2
+\alpha m \operatorname{sn}_2\operatorname{cn}_2 \operatorname{nd}_2}{2\alpha\beta  g} =: \frac{\bar h}{2\alpha\beta  g} ,
\]
we obtain
\begin{align*}
\mathcal M_t &= 2\alpha\beta \frac{(g \bar h_t - g_t \bar h)}{g^2} 
 -2\alpha\beta \Big(\frac{(g \bar h_t - g_t \bar h)}{g^2} (x=0)\Big) \Big|_{t=0}\\
&\quad -2\alpha^2\frac{k}{\beta^2}  (\delta \beta^2 \operatorname{sn}_1^2 -  \gamma \alpha^2 \operatorname{nd}_2^2  +\gamma \alpha^2).
\end{align*}
The second term above can be computed explicitly. We have
\[
\mathcal M_t = 2\alpha\beta \frac{(g \bar h_t - g_t \bar h)}{g^2}
 -2\beta^2 (\delta+m\gamma)  -2\alpha^2\frac{k}{\beta^2}  (\delta \beta^2 \operatorname{sn}_1^2 -
\gamma \alpha^2 \operatorname{nd}_2^2  +\gamma \alpha^2).
\]
Let us calculate $B_{tx}$. Using \eqref{Bxper}, we have
\[
B_{tx} =  \frac{2\sqrt{2} \alpha\beta}{g^3} ( g \hat h_t -2g_t \hat h).
\]
Now we compute the term $-B_{tx} - 2B\mathcal M_t$. We have
\begin{equation} \label{Final0}
\begin{aligned}
-B_{tx} - 2B\mathcal M_t 
& =  - \frac{2\sqrt{2} \alpha\beta}{g^3} \Big[ \hat h_t g -2  g_t \hat h
  + 4 \alpha\beta (g \bar h_t - g_t \bar h) h \\
&\quad  - 4 \alpha^2\frac{k}{\beta^2} g^2 (\delta \beta^2 \operatorname{sn}_1^2 -  \gamma \alpha^2 \operatorname{nd}_2^2
 +\gamma \alpha^2)  h  \Big] 
 + 4\beta^2 (\delta+m\gamma) B.
\end{aligned}
\end{equation}
We consider the term $\hat h + 2\alpha\beta h \bar h$. We have
\begin{equation} \label{Final1}
\hat h + 2\alpha\beta h \bar h = h_x g +  h( 2\alpha\beta  \bar h -g_x).
\end{equation}
The term $2\alpha\beta  \bar h -g_x$ reads now
\begin{align*}
2\alpha\beta  \bar h -g_x
& =  2\alpha\beta (\beta \operatorname{sn}_1 \operatorname{cn}_1 \operatorname{dn}_1\operatorname{dn}_2^2 +\alpha m \operatorname{sn}_2\operatorname{cn}_2 \operatorname{nd}_2)  \\
&\quad - 2\beta^2 ( \alpha \operatorname{sn}_1 \operatorname{cn}_1\operatorname{dn}_1 \operatorname{dn}_2^2 -m \beta \operatorname{sn}_1^2 \operatorname{sn}_2\operatorname{cn}_2 \operatorname{dn}_2) \\
& =  2\beta m\operatorname{sn}_2\operatorname{cn}_2 (\alpha^2 \operatorname{nd}_2 + \beta^2 \operatorname{sn}_1^2 \operatorname{dn}_2 ) \\
& =  2\beta m\operatorname{sn}_2\operatorname{cn}_2 \operatorname{nd}_2 g.
\end{align*}
Consequently,  $\eqref{Final1} = g(h_x  + 2\beta m\operatorname{sn}_2\operatorname{cn}_2 \operatorname{nd}_2  h)$,
and replacing in \eqref{Final0},
\begin{equation} \label{Final3}
\begin{aligned}
&-B_{tx} - 2B\mathcal M_t  \\
& =  - \frac{2\sqrt{2} \alpha\beta}{g^2} \Big[ \hat h_t
 -2  g_t (h_x  + 2\beta m\operatorname{sn}_2\operatorname{cn}_2 \operatorname{nd}_2  h) + 4 \alpha\beta  \bar h_t  h \\
&\quad - 4 \alpha^2\frac{k}{\beta^2} g (\delta \beta^2 \operatorname{sn}_1^2
 -  \gamma \alpha^2 \operatorname{nd}_2^2  +\gamma \alpha^2)  h  \Big] + 4\beta^2 (\delta+m\gamma) B.
\end{aligned}
\end{equation}
Since $ \hat h_t = h_{tx} g + h_x g_t - g_{xt}h - g_x h_t$, we are left
to compute the term
\begin{equation} \label{Final2}
4\beta h (\alpha \bar h_t  -\frac1{4\beta} g_{tx} - m \operatorname{sn}_2\operatorname{cn}_2 \operatorname{nd}_2 g_t ) -h_x g_t   -g_x h_t.
\end{equation}
Note that
\[
\alpha \bar h_t  -\frac1{4\beta} g_{tx} = ( \alpha \bar h -\frac1{4\beta} g_x)_t.
\]
From \eqref{gg} we have
\[
 \alpha \bar h -\frac1{4\beta} g_x = m \operatorname{sn}_2 \operatorname{cn}_2 \operatorname{nd}_2 g
+  \frac12 \beta  \Big[ \alpha \operatorname{sn}_1 \operatorname{cn}_1 \operatorname{dn}_1\operatorname{dn}_2^2
-  \beta m \operatorname{sn}_1^2 \operatorname{sn}_2 \operatorname{cn}_2\operatorname{dn}_2   \Big],
\]
and
\begin{align*}
\eqref{Final2}
&=  4\beta m  ( \operatorname{sn}_2 \operatorname{cn}_2 \operatorname{nd}_2)_t  g h
 +  2\beta^2 h \Big[  \alpha\operatorname{sn}_1 \operatorname{cn}_1 \operatorname{dn}_1\operatorname{dn}_2^2
 -\beta m \operatorname{sn}_1^2 \operatorname{sn}_2 \operatorname{cn}_2\operatorname{dn}_2 \Big]_t \\
&\quad   -(h_x g_t   + g_x h_t) \\
& = 4\beta^2 m \gamma   ( \operatorname{sn}_2 \operatorname{cn}_2 \operatorname{nd}_2)' g h  + 2\beta^2 h
 \Big[ \alpha \operatorname{sn}_1 \operatorname{cn}_1 \operatorname{dn}_1\operatorname{dn}_2^2  -\beta m \operatorname{sn}_1^2 \operatorname{sn}_2 \operatorname{cn}_2\operatorname{dn}_2 \Big]_t\\
&\quad  -(h_x g_t   + g_x h_t) .
\end{align*}
Now we have
\begin{align*}
& (\alpha \operatorname{sn}_1 \operatorname{cn}_1 \operatorname{dn}_1\operatorname{dn}_2^2  -\beta m \operatorname{sn}_1^2 \operatorname{sn}_2 \operatorname{cn}_2\operatorname{dn}_2 )_t \\
& =  \alpha^2 \delta  \operatorname{dn}_2^2 [\operatorname{dn}_1^2(\operatorname{cn}_1^2 -\operatorname{sn}_1^2) -k \operatorname{sn}_1^2\operatorname{cn}_1^2] 
  -2 \alpha\beta m (\delta+\gamma) \operatorname{sn}_1\operatorname{cn}_1\operatorname{dn}_1\operatorname{sn}_2\operatorname{cn}_2\operatorname{dn}_2 \\
&\quad -\beta^2 m \gamma \operatorname{sn}_1^2 [  \operatorname{dn}_2^2 (\operatorname{cn}_2^2-\operatorname{sn}_2^2) -m \operatorname{sn}_2^2\operatorname{cn}_2^2 ],
\end{align*}
and since $h = \alpha \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2 - \beta m \operatorname{sn}_1\operatorname{sn}_2\operatorname{cn}_2$,
\begin{align*}
&   2 \beta^2 h (  \alpha \operatorname{sn}_1 \operatorname{cn}_1 \operatorname{dn}_1\operatorname{dn}_2^2
 -\beta m \operatorname{sn}_1^2 \operatorname{sn}_2 \operatorname{cn}_2\operatorname{dn}_2 )_t   \\
& =  2\beta^2 \Big[  \alpha^3 \delta \operatorname{cn}_1\operatorname{dn}_1 \operatorname{dn}_2^3[\operatorname{dn}_1^2(\operatorname{cn}_1^2
 -\operatorname{sn}_1^2) -k\operatorname{sn}_1^2\operatorname{cn}_1^2 ]  \\
&\quad - 2 \alpha^2 \beta m(\delta+\gamma) \operatorname{sn}_1\operatorname{cn}_1^2 \operatorname{dn}_1^2 \operatorname{sn}_2\operatorname{cn}_2 \operatorname{dn}_2^2 \\
&\quad - \alpha\beta^2 m \gamma \operatorname{sn}_1^2\operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2  [\operatorname{dn}_2^2(\operatorname{cn}_2^2-\operatorname{sn}_2^2)
 -m\operatorname{sn}_2^2\operatorname{cn}_2^2]  \\
&\quad -  \alpha^2\beta m \delta \operatorname{sn}_1\operatorname{sn}_2\operatorname{cn}_2 \operatorname{dn}_2^2  [\operatorname{dn}_1^2(\operatorname{cn}_1^2 -\operatorname{sn}_1^2)
  -k\operatorname{sn}_1^2\operatorname{cn}_1^2 ]   \\
&\quad+  2\alpha\beta^2 m^2(\delta+\gamma) \operatorname{sn}_1^2 \operatorname{cn}_1\operatorname{dn}_1\operatorname{sn}_2^2 \operatorname{cn}_2^2 \operatorname{dn}_2 \\
&\quad + \beta^3 m^2\gamma  \operatorname{sn}_1^3 \operatorname{sn}_2 \operatorname{cn}_2  [\operatorname{dn}_2^2(\operatorname{cn}_2^2-\operatorname{sn}_2^2)
 -m\operatorname{sn}_2^2\operatorname{cn}_2^2] \Big] .
\end{align*}
On the other hand, using \eqref{ht}, \eqref{hx}, \eqref{gg} and \eqref{gt},
\begin{align*}
&-(g_t h_x   + g_x h_t) \\
&= 2\beta^2 \operatorname{sn}_1\operatorname{dn}_2 \Big[  ( \alpha \delta\operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2
 -  \beta m \gamma \operatorname{sn}_1\operatorname{sn}_2\operatorname{cn}_2)   (\alpha^2 \operatorname{sn}_1 \operatorname{dn}_2 (k\operatorname{cn}_1^2 +\operatorname{dn}_1^2)\\
&\quad + 2 \alpha\beta m \operatorname{cn}_1\operatorname{dn}_1\operatorname{sn}_2 \operatorname{cn}_2 +\beta^2m \operatorname{sn}_1\operatorname{dn}_2(\operatorname{cn}_2^2-\operatorname{sn}_2^2) )\\
&\quad +  (\alpha \operatorname{cn}_1\operatorname{dn}_1 \operatorname{dn}_2 -\beta m\operatorname{sn}_1\operatorname{sn}_2\operatorname{cn}_2) 
 ( \alpha^2\delta \operatorname{sn}_1 \operatorname{dn}_2 (k\operatorname{cn}_1^2 +\operatorname{dn}_1^2)\\
&\quad +  \alpha\beta m (\delta+\gamma) \operatorname{cn}_1\operatorname{dn}_1\operatorname{sn}_2 \operatorname{cn}_2
 +\beta^2m \gamma \operatorname{sn}_1\operatorname{dn}_2(\operatorname{cn}_2^2-\operatorname{sn}_2^2)) \\
&= 2\beta^2 \Big[  2\alpha^3\delta \operatorname{sn}_1^2 \operatorname{cn}_1 \operatorname{dn}_1\operatorname{dn}_2^3 (k\operatorname{cn}_1^2 +\operatorname{dn}_1^2)\\
&\quad +    \alpha^2\beta m (3\delta+\gamma)\operatorname{sn}_1 \operatorname{cn}_1^2\operatorname{dn}_1^2\operatorname{sn}_2 \operatorname{cn}_2\operatorname{dn}_2^2 \\
&\quad + \alpha\beta^2m (\delta+\gamma) \operatorname{sn}_1^2 \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2^3 (\operatorname{cn}_2^2-\operatorname{sn}_2^2)\\
&\quad -\alpha^2\beta m (\delta+\gamma) \operatorname{sn}_1^3 \operatorname{sn}_2\operatorname{cn}_2\operatorname{dn}_2^2(k\operatorname{cn}_1^2+\operatorname{dn}_1^2)\\
&\quad -\alpha \beta^2 m^2(\delta +3\gamma) \operatorname{sn}_1^2  \operatorname{cn}_1 \operatorname{dn}_1 \operatorname{sn}_2^2 \operatorname{cn}_2^2 \operatorname{dn}_2
 - 2\beta^3 m^2\gamma \operatorname{sn}_1^3 \operatorname{sn}_2\operatorname{cn}_2\operatorname{dn}_2^2 (\operatorname{cn}_2^2-\operatorname{sn}_2^2)  \Big].
\end{align*}
Rearranging similar terms, and using the identities \cite{By}
\[
\operatorname{sn}_1^2 + \operatorname{cn}_1^2 =1, \quad k \operatorname{sn}_1^2 + \operatorname{dn}_1^2 =1, \quad
m \operatorname{cn}_2^2 + 1-m =\operatorname{dn}_2^2, \ldots,
\]
and the fact that $k\alpha^4 = (1-m)\beta^4$, we obtain
\begin{align*}
& 2\beta^2 h ( \alpha \operatorname{sn}_1 \operatorname{cn}_1 \operatorname{dn}_1\operatorname{dn}_2^2
 -\beta m \operatorname{sn}_1^2 \operatorname{sn}_2 \operatorname{cn}_2\operatorname{dn}_2 )_t -(h_x g_t   + g_x h_t) \\
& = 2\beta^2 \Big[  \alpha^3 \delta \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2^3
  ( \operatorname{dn}_1^2 + k\operatorname{sn}_1^2 \operatorname{cn}_1^2) + \beta^3m^2 \gamma \operatorname{sn}_1^3
 \operatorname{sn}_2 \operatorname{cn}_2(\operatorname{sn}_2^2 \operatorname{dn}_2^2-\operatorname{cn}_2^2 )  \\
&\quad  -\alpha^2\beta m \gamma \operatorname{sn}_1\operatorname{sn}_2\operatorname{cn}_2\operatorname{dn}_2^2
 ( \operatorname{dn}_1^2 + k\operatorname{sn}_1^2 \operatorname{cn}_1^2  ) \\
&\quad  - \alpha\beta^2 m \delta \operatorname{sn}_1^2 \operatorname{cn}_1 \operatorname{dn}_1 \operatorname{dn}_2  (  \operatorname{sn}_2^2 \operatorname{dn}_2^2 - \operatorname{cn}_2^2   )
 \Big] \\
&= 2\beta^2 (\alpha\delta \operatorname{cn}_1\operatorname{dn}_1 \operatorname{dn}_2 -\beta m \gamma \operatorname{sn}_1 \operatorname{sn}_2 \operatorname{cn}_2 )
 ( \alpha^2 \operatorname{dn}_2^2 (\operatorname{dn}_1^2 + k \operatorname{sn}_1^2 \operatorname{cn}_1^2) \\
&\quad  + \beta^2 m \operatorname{sn}_1^2 (\operatorname{cn}_2^2 -\operatorname{sn}_2^2 \operatorname{dn}_2^2) )\\
&= 2\beta^2  \tilde h [    \alpha^2 \operatorname{dn}_2^2 (1-k \operatorname{sn}_1^4 )
 + \beta^2 \operatorname{sn}_1^2 \operatorname{dn}_2^2 -(1-m) \beta^2 \operatorname{sn}_1^2 -\beta^2 m \operatorname{sn}_1^2\operatorname{sn}_2^2 \operatorname{dn}_2^2   ] \\
&= 2\beta^2  \tilde h [  \operatorname{dn}_2^2 \{ \alpha^2  (1-k \operatorname{sn}_1^4 ) + \beta^2 \operatorname{sn}_1^2
 -\beta^2 \operatorname{sn}_1^2(1-\operatorname{dn}_2^2) \}  -   \alpha^4 \beta^{-2} k \operatorname{sn}_1^2   ] \\
&= 2\beta^2  \tilde h g ( \operatorname{dn}_2^2   -  \alpha^2 \beta^{-2}k \operatorname{sn}_1^2  ).
\end{align*}
From \eqref{Final3} we conclude that
\begin{equation}
\begin{aligned}
&-B_{tx} - 2B\mathcal M_t\\
& =  - \frac{2\sqrt{2} \alpha\beta}{g} \Big[ h_{tx} + 2  \tilde h  (  \beta^2 \operatorname{dn}_2^2
 -  \alpha^2 k \operatorname{sn}_1^2  )  + 4\beta^2 m \gamma   ( \operatorname{sn}_2 \operatorname{cn}_2 \operatorname{nd}_2)'  h   \\
 & \quad - 4 \alpha^2\frac{k}{\beta^2}  (\delta \beta^2 \operatorname{sn}_1^2
 -  \gamma \alpha^2 \operatorname{nd}_2^2  +\gamma \alpha^2)  h  \Big] + 4\beta^2 (\delta+m\gamma) B.
\end{aligned}  \label{Final4}
\end{equation}
Note that \cite{By} $ ( \operatorname{sn}_2 \operatorname{cn}_2 \operatorname{nd}_2)'  =  \operatorname{cn}_2^2-\operatorname{sn}_2^2
 + m \operatorname{sn}_2^2\operatorname{cn}_2^2 \operatorname{nd}_2^2$. Consequently,
\begin{align*}
&  4\beta^2 m \gamma   ( \operatorname{sn}_2 \operatorname{cn}_2 \operatorname{nd}_2)'
- 4 \alpha^2\frac{k}{\beta^2}  (\delta \beta^2 \operatorname{sn}_1^2
-  \gamma \alpha^2 \operatorname{nd}_2^2  +\gamma \alpha^2)  \\
&=    4\beta^2 m \gamma (\operatorname{cn}_2^2-\operatorname{sn}_2^2)  -4\alpha^2 k \delta \operatorname{sn}_1^2
 +  4 \beta^2  \gamma \operatorname{nd}_2^2 (   m^2 \operatorname{sn}_2^2\operatorname{cn}_2^2
 +  k \frac{\alpha^4}{\beta^4}(1-\operatorname{dn}_2^2) ) \\
&=    4\beta^2 m \gamma (\operatorname{cn}_2^2-\operatorname{sn}_2^2)  -4\alpha^2 k \delta \operatorname{sn}_1^2
 +  4  \beta^2 m  \gamma  \operatorname{sn}_2^2 \operatorname{nd}_2^2 ( m \operatorname{cn}_2^2 + k\frac{\alpha^4}{\beta^4} )\\
&=    4\beta^2 m \gamma (\operatorname{cn}_2^2-\operatorname{sn}_2^2)  -4\alpha^2 k \delta \operatorname{sn}_1^2
 +  4  \beta^2 m \gamma   \operatorname{sn}_2^2  =   4\beta^2 m \gamma \operatorname{cn}_2^2   -4\alpha^2 k \delta \operatorname{sn}_1^2 .
\end{align*}
Finally we compute $h_{tx}$. From \eqref{ht} we have
\begin{align*}
h_{tx}
& =  -\Big[  \alpha^2(k\operatorname{cn}_1^2 +\operatorname{dn}_1^2) ( \alpha\delta \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2
-\beta m(2\delta+\gamma) \operatorname{sn}_1\operatorname{sn}_2 \operatorname{cn}_2 ) \\
&\quad + \beta^2 m(\operatorname{cn}_2^2 -\operatorname{sn}_2^2) (\alpha(\delta+2\gamma) \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2
  -\beta m\gamma \operatorname{sn}_1\operatorname{sn}_2 \operatorname{cn}_2 ) \\
&\quad - 4\alpha^2 k  \delta  \operatorname{sn}_1^2 (\alpha \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2)
 - 4 \beta^2  \gamma  ( \beta m \operatorname{sn}_1\operatorname{sn}_2 \operatorname{cn}_2) \operatorname{dn}_2^2  \Big].
\end{align*}
Using the last two identities and some standard simplifications, \eqref{Final4}
 becomes
\begin{align*}
&-B_{tx} - 2B\mathcal M_t \\
& = - \frac{2\sqrt{2} \alpha\beta}{g} \Big[  [-\alpha^2 (1+k) \delta
 + \beta^2((2-m)\delta + 2m\gamma)] (\alpha \operatorname{cn}_1\operatorname{dn}_1\operatorname{dn}_2)  \\
&\quad + [ \alpha^2 (1+k) (2\delta+\gamma) +\beta^2 (2-3m)\gamma ]
  ( \beta m \operatorname{sn}_1\operatorname{sn}_2 \operatorname{cn}_2)\Big]  + 4\beta^2 (\delta+m\gamma) B\\
& = -a_1 \tilde B_t - \tilde a_2 B,
\end{align*}
where $a_1$ is defined in \eqref{a1}, and
\[
\tilde a_2 =\alpha^4 (1+k)^2 - 2\alpha^2\beta^2(1+k)(m+6) + \beta^4(2-m)(18-m).
\]
Comparing this with  \eqref{Ecuacion} we have ($c_0$ is given by \eqref{c0})
\[
a_2 = \tilde a_2 + c_0 = \alpha^4(1+k)^2 + 2\alpha^2\beta^2 (1+k)(2-m) +\beta^4(m^2+28m-28).
\]
Finally we use that $k\alpha^4 = \beta^4(1-m)$ to obtain
\[
a_2 =\alpha^4(1+k^2 -26 k) + 2\alpha^2\beta^2 (1+k)(2-m) +\beta^4m^2,
\]
as in \eqref{a2}. The proof is complete.

\subsection*{Acknowledgments}
 M. A. Alejo likes to express his gratitude
with professors L. Vega, P. Kevrekidis and A. Khare
for several enlightening  discussions and  valuable comments.
He also would like to thank to the Departamento de Ingenier\'ia
Matem\'atica (U. Chile)
for the support where part of this work was done.

C. Mu\~noz was partially funded by ERC Blowdisol (France), Fondecyt no.
1150202 Chile, Fondo Basal CMM (U. Chile),
and Millennium Nucleus Center for Analysis of PDE NC130017.
He also would like to thank to the Laboratoire de Math\'ematiques d'{}Orsay
for his kind hospitality during past years, and where part of this
work was completed.

J. M. Palacios was partially funded by Fondecyt no. 1150202 Chile.

This work is part of our paper at arxiv.org/pdf/1309.0625v2.pdf.


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