\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 43, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/43\hfil Entire functions related to stationary solutions]
{Entire functions related to stationary solutions of
 the Kawahara equation}

\author[A. L. C. dos Santos, P. N. da Silva,C. F. Vasconcellos \hfil EJDE-2016/43\hfilneg]
{Andr\'e Luiz C. dos Santos, Patr\'icia N. da Silva, \\
Carlos Frederico Vasconcellos}

\address{Andr\'e Luiz C. dos Santos \newline
DEMAT/CEFET/Maracan\~{a}; PPG-EM/UERJ
- Rua Fonseca Teles, 121, 1o. andar, S\~ao Crist\'ov\~ao,
 Rio de Janeiro, RJ. CEP: 20940-903, Brazil}
\email{andreluiz.cordeiro@gmail.com}

\address{Patr\'icia N. da Silva \newline
IME/UERJ, Rua S\~ao Francisco Xavier, 524, 6o.
andar, Rio de Janeiro, RJ,  CEP 20550-900, Brazil}
\email{nunes@ime.uerj.br}

\address{Carlos Frederico Vasconcellos \newline
IME/UERJ, Rua S\~ao Francisco Xavier, 524, 6o.~andar,
 Rio de Janeiro, RJ, CEP 20550-900,  Brazil}
\email{cfredvasc@ime.uerj.br}

\thanks{Submitted November 20, 2014. Published January 29, 2016.}
\subjclass[2010]{30D20, 35Q53}
\keywords{Entire functions; M\"obius transformations; stationary solutions;
\hfill\break\indent  Kawahara equation}

\begin{abstract}
 In this study, we characterize  the lengths of intervals for
 which the linear Kawahara equation has a non-trivial solution,
 whose energy is stationary. This gives rise to a family of complex
 functions. Characterizing the lengths amounts to deciding which members
 of this family are entire functions.  Our approach is essentially
 based on determining the existence of certain M\"obius transformation.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

In the Kawahara equation
\begin{equation} \label{berloffe13}
u_t +u_x+\kappa u_{xxx}+\eta  u_{xxxxx}+uu_x=0,
\end{equation}
the conservative dispersive effect is represented by the
term $(\kappa u_{xxx}+\eta u_{xxxxx})$. This equation is a  model for plasma
wave, capilarity-gravity water waves and other
dispersive phenomena when the cubic KdV-type equation is
weak. Kawahara~\cite{ka} pointed out that it happens
 when the coefficient of the third
order derivative in the KdV equation becomes very small
or even zero. It is then necessary to take into account  the
higher order effect of
dispersion in order to balance the nonlinear effect.  Kakutani and
Ono~\cite{kakutani} showed that for a critical value of angle between
the magneto-acoustic wave in a cold collision-free plasma and the
external magnetic field, the third order derivative term in the KdV
equation vanishes and may be replaced by the fifth order derivative
term. Following this idea, Kawahara~\cite{ka}
studied a generalized nonlinear
dispersive equation which has a form of the KdV
equation with an additional fifth order derivative
term. This equation has also been obtainded by
Hasimoto~\cite{hasimoto} for the shallow wave near critical values of
surface tension. More precisely, in this work  Hasimoto found
these critical values  when the Bond number is near one third.

While analyzing the evolution of solutions of the water
wave-problem, Schneider and  Wayne~\cite{sw}
also showed that the coefficient of the third
order dispersive term in nondimensionalized statements of
the KdV equation vanishes when the Bond number is
equal to one third.
The Bond number is proportional to the strength of the
surface tension and in the KdV equation it is related
to the leading order dispersive effects in the
water-waves problem. With its
disappearance, the resulting equation is just
Burger's equation whose solutions typically form shocks
in finite time. Thus, if we wish to model interesting
behavior in the water-wave problem it is necessary to
include higher order terms. That is, it is necessary to
consider the Kawahara equation.
In any case, the inclusion of the fifth order derivative term takes
into account the comparative magnitude of the coefficients of the third and
fifth power terms in the linearized dispersion relation.

 Berloff and Howard~\cite{berloff} presented the
Kawahara equation as the purely dispersive form of the
nonlinear partial differential equation
\[
u_t+u^ru_x+a u_{xx}+bu_{xxx}+cu_{xxxx}+du_{xxxxx}=0.
\]
The above equation describes the evolution of long
waves in various problems in fluid dynamics. The  Kawahara equation
corresponds to the choice $a=c=0$ and $r=1$ and
describes water waves with surface tension. Bridges and
Derks~\cite{bd}  presented the Kawahara equation -- or
fifth-order KdV-type equation -- as a particular case of the
general form
\begin{equation} \label{mbridgese11}
u_t+\kappa u_{xxx}+\eta  u_{xxxxx}=
\frac{\partial}{\partial x}f(u,u_x,u_{xx})
\end{equation}
where $u(x,t)$ is a scalar real valued function,
$\kappa $ and $\eta \not=0$ are real parameters and
$f(u,u_x,u_{xx})$ is some smooth function. The
form \eqref{berloffe13} occurs most often in
applications and corresponds to the choice of $f$ in
\eqref{mbridgese11} with the form $f(u,u_x,u_{xx})=-u^2/2$.

As noted by Kawahara \cite{ka}, we may assume without loss of
generality that $\eta <0$ in \eqref{berloffe13}. In fact, if we
introduce the following simple transformations
\[
u\to -u,\quad x\to -x, \quad t\to t,
\]
we can obtain an equation of the form of
\eqref{berloffe13} in which $\kappa $
and $\eta $ are replaced, respectively, by $-\kappa $ and $-\eta $.

Hagarus, Lombardi and Scheel \cite{hagarus} pointed out that
the Kawahara equation
\begin{equation}  \label{eq:h11}
 u_t=u_{xxxxx}-\varepsilon u_{xxx}+uu_x,
\end{equation}
in which $\varepsilon$ is a real parameter,  models water waves in the
long-wave regime for moderate values of surface tension (Weber numbers
close to $1/3$). For such
Weber numbers, the usual description of long water waves via the Korteweg-de
Vries (KdV)
equation fails since the cubic term in the linear dispersion relation
vanishes and fifth order
dispersion becomes relevant at leading order,
$\omega(k) = k^5 + \varepsilon k^3$. Positive (resp.~negative) values
of the parameter $\varepsilon$ in \eqref{eq:h11} correspond to
Weber numbers larger
(resp.~smaller) than $1/3$.

Dispersive problems have been object of
 intensive research (see, for instance, the classical paper of Benjamin,
Bona and Mahoni \cite{bbm},   Biagioni and Linares \cite{lin},
Bona and Chen \cite{bona}, Menzala {\it et al.} \cite{mvz}, Rosier
  \cite{rosier}, and references therein).
 Recently global stabilization of the generalized KdV system have
been obtained by Rosier and Zhang \cite{rz} and Linares and Pazoto
\cite{linpa} studied the stabilization of the generalized KdV system
with critical exponents. For the stabilization of global solutions of
the Kawahara under the effect of a localized damping mechanism,
see Vasconcellos and Silva~\cite{vs,vs1,vs2}.

We consider the linear Kawahara equation
\begin{equation} \label{eq:kawa}
u_t +\beta u_x+ \kappa u_{xxx} +\eta u_{xxxxx} = 0 \quad \text{with }
(x,t) \in (0,L_0)\times (0,\infty),
\end{equation}
where the coefficients $\beta,\kappa$ and $\eta$ are real numbers such that
$\eta < 0$, $\kappa\not=0$, $\beta\in\{0,1\}$.
Sometimes, while discussing the existence of solutions of certain partial
differential equations, it is necessary to establish when a certain
quotient of entire functions still turns out to be an entire function
(see, for instance, Rosier~\cite{rosier}, Vasconcellos and Silva~\cite{vs}).
The problem of factoring an entire function is solved by the famous
Weierstrass factorization theorem and its corollaries.
However, applying this result may not be very practical or even viable
in some cases.

Now we proceed to a general description of such kind of factoring problems.
 We have a polynomial $p:\mathbb{C} \to \mathbb{C}$ and a family
of functions
\[
N_{\mathbf{a}}: \mathbb{C} \times (0,\infty) \to \mathbb{C},
\]
$\mathbf{a} \in \mathbb{C}^{4} \thicksim \{\mathbf{0}\}$,
whose restriction
$N_{\mathbf{a}}(\cdot,L)$
is entire for each $L > 0$. We consider a family of functions
$f_{\mathbf{a}}(\cdot,L)$
defined by

\begin{equation}
N_{\mathbf{a}}(\xi,L) = f_{\mathbf{a}}(\xi,L)\, p(\xi)\label{first}
\end{equation}
in its maximal domain.
For a given polynomial $p(\cdot)$, the problem of characterizing the set
of values $L_0 > 0$, for which it is possible to find a
 non null $\mathbf{a}_0 \in \mathbb{C}^4$ such that the function
$f_{\mathbf{a}_0}(\cdot,L_{0})$
is entire, is a challenging problem and of particular interest of
the academic community.

Vasconcellos and Silva \cite[Lemma 2.1]{vs} discussed the existence
of non-zero  solutions for \eqref{eq:kawa} whose energy is constant over time.
Their results show that the existence of such solutions
is equivalent to determining the lengths of interval
$(0,L_0)$ for which it is possible to verify that the condition
\begin{equation} \label{eq:lambda}
(\exists\, \lambda \in \mathbb{C}, \; u_0 \in (H^{3}_{0}(0,L_0)
\cap H^5(0,L_0),\mathbb{C}) \;\Rightarrow \;
 \lambda u_0+\beta u_0' +\kappa u_0'''+ \eta u_0'''''=0)
\end{equation}
is valid.
Such condition in turn reduces to the problem of characterizing the set
$\mathcal{X}$ of $L_0 > 0$ values, for which exist $r$ and
$\mathbf{a}_0$ providing that function  $f_{\mathbf{a}}(\cdot,L)$
is entire for  $L = L_0$ and $\mathbf{a} = \mathbf{a}_0$.
In this case, using \eqref{first},
$f_{\mathbf{a}}(\cdot,L)$ is defined by
\begin{equation}
\begin{gathered}
N_{\mathbf{a}}(\xi,L) = a_1 i\xi - a_2 i\xi e^{-i\xi L}+ a_3
- a_4 e^{-i\xi L} \\
p(\xi) = r+\beta\xi-\kappa\xi^{3}+\eta\xi^{5}
\end{gathered} \label{N(xi)}
\end{equation}
where $\mathbf{a} = (a_1,a_2,a_3,a_4)$ and $r\in\mathbb{R}$.
It follows from \eqref{eq:lambda} that $\lambda$ is a pure imaginary number.
Thus, we only have to consider polynomials $p(\cdot)$ with $r\in\mathbb{R}$.

For each $r\in\mathbb{R}$ and $\mathbf{a}_0\in \mathbb{C}^{4}
\thicksim \{\mathbf{0}\}$, let $\mathcal{X}_{\mathbf{a}_0r}$ be the set
of $L_0 > 0$ values, for which the function $f_{\mathbf{a}}(\cdot,L)$
is entire for $L = L_0$ and $\mathbf{a} = \mathbf{a}_0$.
The set $\mathcal{X}$ is the union of $\mathcal{X}_{\mathbf{a}_0r}$
for $r\in\mathbb{R}$ and $\mathbf{a}_0\in \mathbb{C}^{4}
\thicksim \{\mathbf{0}\}$.


Here, we place emphasis on the following statements:
\begin{itemize}
\item[(S1)] $f_{\mathbf{a}_0}(\cdot,L_0)$ is entire;

\item[(S2)] all the zeros, taking the respective multiplicities
into account, of the polynomial $p$ are zeros of $N_{\mathbf{a}_0}(\cdot,L)$;

\item[(S3)] the maximal domain of $f_{\mathbf{a}_0}(\cdot,L_0)$ is $\mathbb{C}$;
\end{itemize}
which are, clearly, equivalent and will be widely used throughout this article.
A closer look shows that determining the solution to the problem guarantees
the existence of a M\"{o}bius transformation in some circumstances.
Further, for the function $f_\mathbf{a}(\cdot,L)$,
defined by \eqref{first} and \eqref{N(xi)}, to be entire, given the equivalence
between statements (S1) and (S2); informally, we must have
\begin{equation}
\frac{a_1 i \xi_0 + a_3}{a_2 i \xi_0 + a_4} = e^{-i L\xi_0} \label{M-sem}
\end{equation}
for each root $\xi_0$ of the polynomial $p$.
We note that for $\mathbf{a}$ such that
$a_1 a_3 - a_2 a_4 \neq 0$, the left side of \eqref{M-sem} suggests that a
M\"{o}bius transformation is defined.
Note that we already have an indication that for
a polynomial $p$ with at least two roots differing by an integer multiple
of $2\pi/L$, we obtain $L\notin\mathcal{X}$.
With this, a method for solving the problem is revealed:
we must verify for which structures
of the roots of the polynomial $p$ is it possible to define a M\"{o}bius
transformation $M$ that satisfies
$M(\xi_0) = e^{-iL\xi_0}$ for each zero $\xi_0$ of polynomial $p$
(See Lemma~\ref{lem:raizesbeta01}).

Taking \eqref{M-sem}, it is essential to define, for each non null
$\mathbf{a} \in \mathbb{C}^4$, the discriminant of $\mathbf{a}$,
specifically, the complex number $d(\mathbf{a}) = a_1 a_4 - a_2 a_3$.
It is natural, however, to consider:
\begin{itemize}
\item[(i)] $d(\mathbf{a}) = 0$ or
\item[(ii)] $d(\mathbf{a}) \neq 0$.
\end{itemize}
The main result shown in this article guarantees that the existence
of pairs $(\mathbf{a}_0, L_0)$  that make $f_{\mathbf{a}}(\cdot,L)$
entire is intimately linked to whether or not the discriminant is zero.
In fact, when the discriminant of $\mathbf{a}$ is zero, such pairs
do not exist for any $r\in\mathbb{R}$.
On the other hand, if the discriminant of $\mathbf{a}$ is non-zero,
we identify situations where the pairs $(\mathbf{a}_0, L_0)$  can exist or not.
Whereas case (i) has been completely solved here, in case (ii)
there are situations where the problem remains to be solved, i.e.,
in some cases, we do not know whether or not it is possible to satisfy
 \eqref{M-sem}.
As far as we know, Rosier~\cite{rosier} was the first to analyze these
kinds of problems. In fact, he showed that the existence of non-trivial
solutions for the Kortweg de Vries equation, whose energies do not decay
over time, is equivalent to determining the set $\mathcal{U}$ of values
$l_0>0$, for which  there exists a non null
$\mathbf{k}_0 \in \mathbb{C}^{2}$ and $s\in\mathbb{C}$, so that the function
$g_{\mathbf{k}}(\cdot,L)$ with $\mathbf{k} = (k_1,k_2)$, defined by
\begin{equation} \label{eq:frosier}
M_{\mathbf{k}}(\xi,l) = g_{\mathbf{k}}(\xi,l)\, q(\xi),
\end{equation}
is entire for $\mathbf{k} = \mathbf{k}_0$ and $l = l_0$.
Here, in particular,
$M_{\mathbf{k}}(\xi,l) = k_1- k_2 e^{-iL\xi}$ and
$q(\xi) = \xi^3-\xi+ s$. Then  Rosier~\cite{rosier} proves that
$$
\mathcal{U}=\Big\{2\pi\sqrt{\frac{m^2+mn+n^2}{3}}:\,\, n,m\in\mathbb{N}\Big\}.
$$
Let us take case (i) from the same starting point as
Rosier~\cite{rosier}, i.e., the analysis of zeros of
$N_{\mathbf{a}}(\cdot,L)$. Here, it makes no sense to argue about
the existence of a M\"{o}bius transformation.
Case (ii) is completely based on equation \eqref{M-sem}. Our strategy
is quite efficient. It proved to be efficient in this situation,
where using previously established results, such as the Weierstrass
factorization theorem, is not possible.

Notice that, for each choice of the coefficients 
$\beta,\kappa$ and $\eta$, condition \eqref{eq:lambda} associates
the Kawahara equation
\[
u_t +\beta u_x+ \kappa u_{xxx} +\eta u_{xxxxx} = 0
\]
to a family of polynomials
\[
p(\xi) = r+\beta\xi-\kappa\xi^{3}+\eta\xi^{5},\quad r\in\mathbb{R}.
\]
Let $\mathcal{X}$ be the set of the lengths of interval
$(0,L_0)$ for which exist non-zero solutions for \eqref{eq:kawa}
whose energy is constant over time.
Consider for each $r\in \mathbb{R}$ and
$\mathbf{a} \in \mathbb{C}^4 \thicksim \{\mathbf{0}\}$,
the set $\mathcal{X}_{\mathbf{a}r}$ of values $L_0>0$ for which the
function $f_{\mathbf{a}}(\cdot,L)$ defined by \eqref{first} and
\eqref{N(xi)} is entire for $L = L_0$. We can decompose $\mathcal{X}$
as the union of the sets $\mathcal{X}_{\mathbf{a}r}$ for $r\in \mathbb{R}$
and non null $\mathbf{a} \in \mathbb{C}^4$

We extend the results obtained by Vasconcellos and Silva~\cite{vs,vs1}
for characterizing the set $\mathcal{X}$ for the Kawahara equation
\eqref{eq:kawa}. They have partially analyzed the case $\kappa=0$ in
\eqref{eq:kawa} and did not deal with the case $\kappa=1$ in
\eqref{eq:kawa}.
In our proof, we argue by exhaustion characterizing the sets
$ \mathcal{X}_{\mathbf{a}r}$.
In Part (I) of Theorem~\ref{teo:andre}, we see that if $d(\mathbf{a}) = 0$,
then $\mathcal{X}_{\mathbf{a}r}=\emptyset$ for all $r\in\mathbb{R}$.
As a consequence of this result, it follows that for any Kawahara equation
\eqref{eq:kawa}, the set $\mathcal{X}$ is given by the union of the
set $\mathcal{X}_{\mathbf{a}r}$ for $r\in \mathbb{R}$ and
$d(\mathbf{a}) \neq 0$.
Our results for $d(\mathbf{a}) \neq 0$ allow to partially describe the
set $\mathcal{X}$ for Kawahara equations \eqref{eq:kawa} with $\beta = 1$
and $\kappa\neq 0$ or $\beta = 0$ and $\kappa < 0$.
For Kawahara equations \eqref{eq:kawa} with $\beta = 0$, $\kappa > 0$,
as a consequence of Theorem~\ref{teo:andre}, we obtain that $\mathcal{X}$
is empty.

Now we summarize the results obtained in this article in the following theorem
guided by the roots of polynomial $p$, as we will shortly see.

\begin{theorem} \label{teo:andre}
Let $r\in\mathbb{R}$,
a non null $\mathbf{a} \in \mathbb{C}^4$ and $L > 0$,
and consider the function $f_{\mathbf{a}}(\cdot,L)$
defined by the product
\begin{equation} \label{eq:ffinal}
N_{\mathbf{a}}(\xi,L) = f_{\mathbf{a}}(\xi,L)\, p(\xi)
\end{equation}
in its maximal domain. Let us suppose that $N_{\mathbf{a}}(\xi,L)$ and $p(\xi)$
are as in  \eqref{N(xi)}.
Let $\mathcal{X}_{\mathbf{a}r}$ be the set
of values $L_0>0$ for which the function
$f_{\mathbf{a}}(\cdot,L)$ defined by \eqref{eq:ffinal}
is entire for $L = L_0$.

(I) %\label{item:teodeq0}
If $L_0 > 0$ is such that $f_{\mathbf{a}_0}(\cdot,L_0)$ is entire for some
 non null $\mathbf{a}_0 \in \mathbb{C}^4$, then
$d(\mathbf{a}_0) \neq 0$. In other words, for any non null $\mathbf{a}$,
if $d(\mathbf{a}) = 0$,
 we obtain $\mathcal{X}_{\mathbf{a}r}=\emptyset$,
for any $r\in\mathbb{R}$. The reciprocal, however, is false.

(II) % \label{item:teodneq0vazio}
If $\mathbf{a}$ is such that $d(\mathbf{a}) \neq 0$ and one of
following three items occurs:
\begin{itemize}
\item[(a)] %\label{item:teodneq0beta12pares}
$\beta = 1$ and $|r| > z-\kappa z^{3}+\eta z^{5}$, where
$z=\sqrt{\frac{3\kappa-\sqrt{9\kappa^2-20\eta}}{10\eta}}$;

\item[(b)] % \label{item:teodneq0beta0kapapositivo}
$\beta = 0$, $\kappa > 0$ and $r\in\mathbb{R}$;

\item[(c)] %\label{item:teodneq0beta0kapanegativo2pares}
$\beta = 0$, $\kappa < 0$ and $|r| > -\kappa z^{3}+\eta z^{5}$,
where $z=\sqrt{\frac{3\kappa}{5\eta}}$.
\end{itemize}
Then there is no  $L > 0$  that renders the function
$f_{\mathbf{a}}(\cdot,L)$ entire. Therefore,
 $\mathcal{X}_{\mathbf{a}r}=\emptyset$.

(III) % \label{item:teodneq0enumeravel}
\begin{itemize}
\item[(a)] %\label{item:teodneq0beta1r0}
If $\beta = 1$ and $r=0$, then there exist $L_0 > 0$ and non null
$\mathbf{a}_0$ such that $f_{\mathbf{a}_0}(\cdot,L_0)$ is entire if and only if
\[
L_0 \in \Big\{L\in\mathbb{R},\; k\operatorname{cotanh}\Big(\frac{Lk}{2}\Big)
=-\rho\cot \Big(\frac{L\rho}{2}\Big)\Big\}
\]
where
\[
\rho=\sqrt{\frac{\kappa-\sqrt{\kappa^2-4\eta}}{2\eta}}
\quad\text{and}\quad
k=\sqrt{\left|\frac{\kappa+\sqrt{\kappa^2-4\eta}}{2\eta}\right|}.
\]
\item[(b)] %\label{item:teodneq0beta0r0kapanegativo}
If $\beta = 0$, $\kappa < 0$ and $r=0$, then there exist $L_0 > 0$ and non null
$\mathbf{a}_0$ such that $f_{\mathbf{a}_0}(\cdot,L_0)$ is entire if and only if
\[
L_0 \in \Big\{L>0,\;\tan \frac{\rho L}{2} =\frac{\rho L}{2}\Big\},
\]
where $\rho=\sqrt{\kappa/\eta}$.
\end{itemize}
The sets in (a) and (b) are enumerable.
\end{theorem}

The knowledge of the zeros of $g_{\mathbf{k}}(\xi,l)$ in \eqref{eq:frosier}
plays a key role in the Rosier's analysis of the existence of
non-trivial solutions for the Kortweg de Vries
equation, whose energies do not decay over time.
The function $f_{\mathbf{a}}(\xi,L)$ related to Kawahara equation does
not resemble this fact and its structure together with the order of
the polynomial turn the analysis of the Kawahara case into a hard problem.
Many other authors have made efforts to tackle this problem
(see for instance,  Glass and Guerrero~\cite{glass}, for a particular
case of (III)(b); Araruna, Capistrano-Filho and Doronin \cite{araruna}, 
for an example of a critical set). 
Our results take their contributions into account.
We show they can be presented and obtained in a systematic way and we
go a step further.

\section{Proof of main results} \label{sec:resaux}

First we establish some lemmas need for proving the three parts of
Theorem~\ref{teo:andre}.

\subsection*{Part (I)}
The main idea behind Part (I) of Theorem~\ref{teo:andre} is to find out
whether there is at least one zero of polynomial $p$ that is not a zero of
$N_\mathbf{a}(\cdot,L)$. The following lemma is a decisive factor in
obtaining this result.

\begin{lemma} \label{lem:zerosN}
Let non null $\mathbf{a}\in\mathbb{C}^4$ with
$d(\mathbf{a}) =0$ and $L > 0$.
 Then the set of the imaginary parts of the zeros of $N_\mathbf{a}(\cdot,L)$
has at most two elements.
\end{lemma}

\begin{proof}
Fix an arbitrarily $L > 0$ and  a non null $\mathbf{a}$
such that $d(\mathbf{a}) =0$.
The nullity of the discriminant of
$\mathbf{a}$ guarantees that the vectors $(a_1,a_3)$ and $(a_2,a_4)$
are linearly dependent. We can then assume that there exists some complex
number $\lambda$ such that $(a_1,a_3) = \lambda (a_2,a_4)$. Thus
$(a_2,a_4)$ cannot be zero and, if  $(a_1,a_3)$ is zero, then
$\lambda = 0$. Therefore, we can write
\begin{equation}
N_\mathbf{a}(\xi,L)=\left(a_2 i\xi+a_4\right)(\lambda-e^{-iL\xi}) \label{num-max}
\end{equation}
Finally, we see that in one of the factors of \eqref{num-max}
there is at most one zero and, in the other, an infinite number of
zeros all with the same imaginary part. Thus, the set of the imaginary
parts of the zeros of  $N_\mathbf{a}(\cdot,L)$ has no more than two elements.
If we assume that  $(a_2,a_4) = \lambda (a_1,a_3)$, the same conclusion
about the zeros of  $N_\mathbf{a}(\cdot,L)$  is valid, proving the result.
\end{proof}

By quickly verifying the result shown in Lemma~\ref{lem:raizesbeta01},
we conclude the set of the imaginary parts of the polynomial $p$ has
at least three elements, except for
$r=\beta=0$ and $\kappa<0$, when all the roots are real.
With the aim of proving part (I) of Theorem~\ref{teo:andre},
we fix an arbitrary  $L > 0$ and a non null $\mathbf{a}$ such that
$d(\mathbf{a}) = 0$. We initially consider any complementar case to $r=\beta=0$
and $\kappa<0$. The Lemmas~\ref{lem:raizesbeta01} and \ref{lem:zerosN}
combined guarantee that there will always be a zero of the polynomial $p$
that is not a zero of $N_\mathbf{a}(\cdot,L)$.
Consequently, the function $f_\mathbf{a}(\cdot,L)$ is not entire.
Finally, we suppose that  $r=\beta=0$ and $\kappa<0$ and assume that
$f_\mathbf{a}(\cdot,L)$ is entire. Since $0$ is a root of multiplicity three,
given the equivalence between (S1) and (S2), we have
$N_\mathbf{a}(0,L) = N'_\mathbf{a}(0,L) = N''_\mathbf{a}(0,L) = 0$
(differentiating with respect to $\xi$).
These three equations imply that
$\mathbf{a}=\mu (-\frac{L}{2},\frac{L}{2},1,1)$ for a complex number
$\mu \neq 0$. Here, we obtain
$d(\mathbf{a}) = -\mu L \neq 0$, which contradicts the hypothesis
of the nullity of the discriminant of $\mathbf{a}$.
Therefore, $f_\mathbf{a}(\cdot,L)$ is not entire for any value of $L > 0$
and non null $\mathbf{a}$ such that $d(\mathbf{a}) = 0$.
This proves part (I) of Theorem~\ref{teo:andre}.

\subsection*{Part (II)}

The following lemma essentially states that if the polynomial
 $p$ has ``too many'' complex roots, equation \eqref{M-sem}
cannot be satisfied.

\begin{lemma} \label{lem:2paresconjugadas}
For any $L > 0$, there is no M\"{o}bius transformation $M$ such that
\[
M(\xi)=e^{-iL\xi}, \quad \xi\in\{\xi_1,\xi_2,\overline{\xi_1},\overline{\xi_2}\}
\]
with $\xi_1,\xi_2,\overline{\xi_1},\overline{\xi_2}$ all distinct
in $\mathbb{C}$.
\end{lemma}

\begin{proof}
Write  $\xi_j=x_j+iy_j$ and $ \omega_j=M(\xi_j)$ for $j = 1,2$
 and note that  $M(\overline{\xi_j}) = \frac{1}{\overline{\omega_j}}$.
The M\"{o}bius transformation $M$ that satisfies
\[
M(\xi)=e^{-iL\xi}, \quad
\xi\in\{\xi_1,\xi_2,\overline{\xi_1},\overline{\xi_2}\}
\]
exists if, and only if, the equality
$$
(\xi_1,\xi_2;\overline{\xi_1},\overline{\xi_2})=
\left(\omega_1,\omega_2;\frac{1}{\overline{\omega_1}},
\frac{1}{\overline{\omega_2}}\right)
$$
is valid  \cite[Theorem VI,p.~178]{town}.
(Note that $(\xi_1,\xi_2;\xi_3,\xi_4)$
 stands for the cross ratio of pairs $(\xi_1,\xi_2)$ and $(\xi_3,\xi_4)$)
However, this does not occur, as
\begin{gather*}
\Big(\omega_1,\omega_2;\frac{1}{\overline{\omega_1}},
\frac{1}{\overline{\omega_2}}\Big)=
\frac{4\sinh Ly_1\sinh Ly_2}{
2\cosh L(y_1+y_2)-2\cos L(x_2-x_1)}
>\frac{4y_1 y_2}{K}
=(\xi_1,\xi_2;\overline{\xi_1},\overline{\xi_2}),
\end{gather*}
where $K = |\xi_1 - \overline{\xi_2}|$.
 To prove this statement, let us assume, without loss of generality,
that $y_1, y_2 >0$ and define the function:
 \[
F(t)= K \sinh (y_1 t) \sinh (y_2 t)
-2 y_1 y_2[\cosh ((y_1+y_2)t)-\cos ((x_2-x_1)t)]
\]
for $t \in \mathbb{R}$. A direct calculation (see the Appendix)
shows that $F(0)= F'(0) = 0$ and $F''(t)>0$ for all $t>0$.
Using a second order Taylor expansion for the function $F$,
we conclude that $F(t) > 0$  for all $t > 0$. In particular,
this means that for $L > 0$, the inequality
$$
4K\sinh (Ly_1)\sinh (Ly_2)
>4y_1y_2(2\cosh (L(y_1+y_2))-2\cos (L(x_2-x_1)))
%\label{eq:2pares}
$$
is valid. This completes the proof.
\end{proof}

Thus, now we can prove part (II) of Theorem~\ref{teo:andre}.
 Let us suppose there exists $\mathbf{a}$ such that $d(\mathbf{a}) \neq 0$
and that one of (a), (b) with $r\neq 0$, or (c) of part (II) of
the theorem occurs  (case (b) with $r = 0$
is proven in Lemma~\ref{lem:r0semux}).

Here, Lemma~\ref{lem:raizesbeta01} guarantees that
polynomial $p$ has a single real root, whose multiplicity is equal to $1$.
This means that this polynomial has two pairs of complex conjugate roots.
Let us assume, in contradiction, that there exists $L>0$ such that the
function  $f_\mathbf{a}(\cdot,L)$ is entire. Then, all roots of polynomial
$p$ must satisfy \eqref{M-sem}; i.e., there exists a M\"{o}bius transformation
that takes each root $\xi_0$  of $p$ into $e^{-iL\xi_0}$.
However, this contradicts Lemma~\ref{lem:2paresconjugadas} and proves
part (II) of the theorem except for the case where
$\beta=0$, $\kappa>0$ and $r=0$
that will be shown in Lemma~\ref{lem:r0semux}.


\subsection*{Part (III)}
Lemma~\ref{lemazeroeimaginarias} below, unlike
Lemma~\ref{lem:2paresconjugadas}, guarantees the existence of
a M\"{o}bius transformation in a case when the polynomial $p$ has
exactly three real roots whose multiplicities are equal to $1$.
Lemma~\ref{lem:r0semux} below guarantees the existence of a M\"{o}bius
transformation when all roots of polynomial $p$ are real.

\begin{lemma} \label{lemazeroeimaginarias}
Let $L, k$ and $\rho$ be real numbers with $L>0$ and $k \neq 0$.
There is a unique  M\"{o}bius transformation $M$ which satisfies
$M(0) = 1$, $M(\pm \rho) = e^{\mp i L \rho}$
and $M(\pm i k) = e^{\pm L k}$
if and only if the following equality occurs
\[
k\operatorname{cotanh}\Big(\frac{Lk}{2}\Big)
=-\rho\cot \Big(\frac{L\rho}{2}\Big).
\]
\end{lemma}

\begin{proof}
First we prove the necessity part. Assume that exists a unique
M\"{o}bius transformation $M$ that satisfies
$M(0) = 1$, $M(\pm \rho) = e^{\mp i L \rho}$
and $M(\pm i k) = e^{\pm L k}$. We know
\cite[Theorem V, p. 176]{town} that the pairs $(0,1)$, $(ik,e^{Lk})$ and
$(-ik,e^{- Lk})$ unambiguously determine the M\"{o}bius transformation
of $M_1$, whose expression is
\begin{equation} \label{eq:zeroeimaginarias}
M_1(z)
=-\frac{z+ik\operatorname{cotanh}\big(\frac{Lk}{2}\big)}{z-ik\operatorname{cotanh}\big(\frac{Lk}{2}\big)}
\end{equation}
and can be obtained by \eqref{eq:Mcoefficients}.
Thus, we must obtain $M = M_1$.
(For the coefficients  $M_1(z)$ and $M_2(z)$, see the Appendix)

Similarly, the injectivity of $M$ implies that $L$ is not an integer
multiple of $\frac{\pi}{\rho}$.
Thus the pairs $(0,1)$, $(\rho,e^{-iL\rho})$ and
$(-\rho,e^{iL\rho})$ determine a unique M\"{o}bius transformation $M_2$.
The formula  \eqref{eq:Mcoefficients} guarantees that $M_2$ is expressed
by
\begin{equation} \label{eq:zeroereaisopostas}
M_2(z)
=-\frac{z-i\rho\cot \big(\frac{L\rho}{2}\big)}
{z+i\rho\cot \big(\frac{L\rho}{2}\big)}.
\end{equation}
Consequently, we obtain $M = M_2$. Thus, $M_1 = M_2$.
We know \cite[Theorem 3, p. 26]{kova} that if
$M_1$ and $M_2$ coincide, their coefficients must be such that
\begin{gather*}
\begin{bmatrix}
-1&-ik\operatorname{cotanh}\big(\frac{Lk}{2}\big)\\
1&-ik\operatorname{cotanh}\big(\frac{Lk}{2}\big)
\end{bmatrix}
=
\begin{bmatrix}
-1&i\rho\cot \big(\frac{L\rho}{2}\big)\\
1&i\rho\cot \big(\frac{L\rho}{2}\big)
\end{bmatrix}.
\end{gather*}
The same result guarantees the sufficiency part.
\end{proof}

\begin{lemma}
\label{lemaenumeravel}
Let $k$ and $\rho$ be real numbers with  $k \neq 0$.
The positive solutions of the equation
\[
k\operatorname{cotanh}\Big(\frac{Lk}{2}\Big)
=-\rho\cot \Big(\frac{L\rho}{2}\Big)
\]
 form a countable set.
\end{lemma}

\begin{proof}
If $\rho = 0$, there is nothing to prove. Thus, suppose that $\rho \neq 0$.
For every $n \in \mathbb{N}$, let $J_n = ((n-1)\pi,n\pi)$ and
$J = \cup_{n\in \mathbb{N}} J_n$.
The continuous function $g:J \to \mathbb{R}$,
defined by
$g(t)=k\operatorname{cotanh}(\frac{kt}{2})+\rho\cot (\frac{\rho t}{2})$,
is strictly decreasing on each $J_n$
and such that $g[J_n]= \mathbb{R}$. Therefore, there exists a unique $L_n$
in each $J_n$ such that $g(L_n)=0$, proving that the solutions of the
 equation form a countable set. This completes the proof.
\end{proof}

We have all the ingredients to prove item (a) of part (III) of
 Theorem~\ref{teo:andre}. If $\beta = 1$ and $r = 0$,
the polynomial $p$ has exactly three real roots, whose multiplicities
are equal to $1$. Now, it is sufficient to observe that
Lemma~\ref{lemazeroeimaginarias} together with equation \eqref{M-sem},
as well as Lemma~\ref{lemaenumeravel}, conclude this. Finally,
we prove the final lemma that supports item (b) of Theorem~\ref{teo:andre}.

\begin{lemma} \label{lem:r0semux}
Consider the family of functions $f_\mathbf{a}(\cdot,L)$ defined by
\eqref{N(xi)} and let us suppose that $r = \beta = 0$.
Let $\mathcal{X}_{\mathbf{a}r}$ be the set
of values $L_0>0$ for which the function $f_{\mathbf{a}}(\cdot,L)$
is entire for $L = L_0$.
\begin{enumerate}
\item If $\kappa<0$,
there exist $L_0 > 0$ and non null
$\mathbf{a}_0$ such that $f_{\mathbf{a}_0}(\cdot,L_0)$ is entire if and only if
\[
L_0\in \Big\{L>0,\tan \frac{\rho L}{2} =\frac{\rho L}{2}\Big\}
\]
where $\rho=\sqrt{\kappa/\eta}$.

\item If $\kappa>0$, then $f_\mathbf{a}(\cdot,L)$
is not entire for any non null $\mathbf{a}$ and $L>0$.
That is, $\mathcal{X}_{\mathbf{a}0} = \emptyset$.
\end{enumerate}
\end{lemma}

\begin{proof} In the case $\kappa < 0$, the polynomial $p$ allows
the decomposition
$p(\xi) = \eta\xi^3 (\xi - \rho)(\xi + \rho)$ where
$\rho = \sqrt{\kappa/\eta}$. Assume $L_0$ is in
\[
\mathcal{P}=\Big\{L>0,\;\tan \frac{\rho L}{2} =\frac{\rho L}{2}\Big\}
\]
and set  $\mathbf{a}_0 = (-\frac{L_0}{2},\frac{L_0}{2},1,1)$. We obtain
$$
N_{\mathbf{a}_0}(0,L_0) = N'_{\mathbf{a}_0}(0,L_0)
 = N''_{\mathbf{a}_0}(0,L_0) = 0,
$$
i.e., $0$ is the root of $N_{\mathbf{a}_0}(0,L_0)$
with multiplicity  at least equal to $3$.
Now, if we prove that $N_{\mathbf{a}_0}(\pm\rho,L_0) = 0$,
then we will be done. To this end,  observe that
$N_{\mathbf{a}_0}(\rho,L_0) = 0$
if and only if
\begin{equation}
\Big(\frac{i\rho L_0}{2}+1\Big) e^{\left(\frac{i\rho L_0}{2}+1\right)}
= \overline{\Big(\frac{i\rho L_0}{2}+1\Big)
e^{(\frac{i\rho L_0}{2}+1)}} \label{argumento}
\end{equation}
This equality, in turn, is equivalent to
$\sin\frac{\rho L_0}{2}
= \frac{\rho L_0}{2}\cos\frac{\rho L_0}{2}$,
which is valid, since we chose $L_0$ in $\mathcal{P}$.
The same argument proves that $N_{\mathbf{a}_0}(-\rho,L_0) = 0$.
Then, the function $f_{\mathbf{a}_0}(\cdot,L_0)$
is entire, proving that the condition is sufficient.
Now, we will prove that it is necessary.
Suppose that $f_{\mathbf{a}_0}(\cdot,L_0)$ is entire for a
certain pair $(\mathbf{a}_0,L_0)$. This tells us that
\begin{equation}
N_{\mathbf{a}_0}(0,L_0) = N'_{\mathbf{a}_0}(0,L_0)
= N''_{\mathbf{a}_0}(0,L_0) = N_{\mathbf{a}_0}(\pm\rho,L_0) = 0.
\label{igualdades}
\end{equation}
The three first equalities in \eqref{igualdades}
can be used to explicitly determine
$\mathbf{a}_0$, i.e.,
$\mathbf{a}_0 = \mathbf{a}(L_0) = \gamma (-\frac{L_0}{2},\frac{L_0}{2},1,1)$
for $\gamma \neq 0$. As $N_{\mathbf{a}(L_0)}(\pm\rho,L_0) = 0$,
we conclude from argument \eqref{argumento}, that $L_0$  must be in
$\mathcal{P}$, proving that the condition is also necessary.

Finally, we assume $\kappa > 0$.
Here, the polynomial $p$ admits the factorization
$p(\xi) =\eta \xi^3 (\xi - \widetilde{\rho})(\xi + \widetilde{\rho})$
where $\widetilde{\rho} = i\rho$ and
$\rho =\sqrt{-\kappa/\eta}$.
Suppose that function $f_\mathbf{a}(\cdot,L)$
is entire for the pair  $(\mathbf{a},L)$.
This implies
\begin{equation}
N_\mathbf{a}(0,L) = N'_\mathbf{a}(0,L) = N''_\mathbf{a}(0,L)
= N_\mathbf{a}(\pm\widetilde{\rho},L) = 0.\label{igualdades1}
\end{equation}
As in the first part, we can obtain
$\mathbf{a}(L) = \gamma (-\frac{iL}{2},\frac{iL}{2},1,1)$
for $\gamma \neq 0$. Thus
\[
N_{\mathbf{a}(L)}(\xi,L)=\gamma\Big[-\frac{iL}{2}\xi+1
-\Big(\frac{iL}{2}\xi+1\Big)e^{-i\xi L}
\Big].
\]
On the other hand, as $N_{\mathbf{a}(L)}(\pm\widetilde{\rho},L) = 0$,
the following equalities must be valid
\begin{equation}
\frac{\pm\rho L}{2}+1-\Big(\frac{\mp\rho L}{2}+1\Big)e^{\pm\rho L} = 0
\label{semnome}
\end{equation}
 However, this is absurd. To prove this, note that
$h:\mathbb{R} \to \mathbb{R}$
defined by $h(t) = e^{t}(2 - t) - (2 + t)$
is such that $h(0) = 0$ and $h'(t) < 0$ for all
$t \in \mathbb{R}$. This completes the proof.
\end{proof}

Note that the sets obtained from the final lemma are clearly countable.
Thus Theorem~\ref{teo:andre} is proved.

\subsection{Describing the roots}

The following lemma separates the roots into groups according to
their algebraic structure. To characterize a group, we consider
the quantity of real roots and their respective multiplicities.
It is worth noting that for each polynomial $p$, the relation of
its roots with these groups determines whether or not a solution exists
for the problem of determining an entire member of family
$f_\mathbf{a}(\cdot,L)$.

\begin{lemma} \label{lem:raizesbeta01}
Consider the polynomial $p(\xi)=r + \beta \xi -\kappa\xi^{3}+\eta\xi^{5}$
where $r, \beta, \kappa$ and $\eta$ are real such that $\beta \in \{0,1\}$,
$\kappa \neq 0$ and $\eta < 0$.

(1) If $\beta = 0$,
\begin{itemize}
\item[(a)] $\kappa>0$ and
\begin{itemize}
\item[(i)] $r=0$, then $0$ is the only real root of  $p$ with multiplicity $3$.
\item[(ii)] $r\neq 0$, then $p$ has a single real root, with multiplicity 1.
\end{itemize}

\item[(b)] $\kappa<0$ and
\begin{itemize}
\item[(i)] $r=0$,  then $0$ and $\pm\rho$ are roots of $p$, where
$\rho=\sqrt{\frac{\kappa}{\eta}}$,
being $0$ the root of multiplicity $3$.

\item[(ii)] $0< |r|<-\kappa z^{3}+\eta z^{5}$, with
$z=\sqrt{\frac{3\kappa}{5\eta}}$, then $p$
has exactly three real roots, all of which are of multiplicity $1$.

\item[(iii)] $|r|=-\kappa z^{3}+\eta z^{5}$, with
$z=\sqrt{\frac{3\kappa}{5\eta}}$, then $p$ has exactly three real roots,
one of which has multiplicity $2$.

\item[(iv)] $|r|>-\kappa z^{3}+\eta z^{5}$, with
$z=\sqrt{\frac{3\kappa}{5\eta}}$, then $p$ has exactly one real root,
 with multiplicity $1$.
\end{itemize}
\end{itemize}

(2) If $\beta = 1$ and
\begin{itemize}
\item[(a)] $r=0$, then the roots of $p$ are $0,\pm\rho,\pm ik$ where
\[
\rho=\sqrt{\frac{\kappa-\sqrt{\kappa^2-4\eta}}{2\eta}}
\quad\text{and}\quad
k=\sqrt{\left|\frac{\kappa+\sqrt{\kappa^2-4\eta}}{2\eta}\right|}.
\]

\item[(b)] $0<|r|<z-\kappa z^{3}+\eta z^{5}$, where
$z=\sqrt{\frac{3\kappa-\sqrt{9\kappa^2-20\eta}}{10\eta}}$, then $p$
has exactly three non null real roots, with multiplicity $1$.

\item[(c)] $|r|=z-\kappa z^{3}+\eta z^{5}$, where
$z=\sqrt{\frac{3\kappa-\sqrt{9\kappa^2-20\eta}}{10\eta}}$, then $p$
has exactly three real roots, one of which has multiplicity $2$.

\item[(d)] $|r|>z-\kappa z^{3}+\eta z^{5}$, where
$z=\sqrt{\frac{3\kappa-\sqrt{9\kappa^2-20\eta}}{10\eta}}$, then $p$
 has exactly one real root with multiplicity $1$.
\end{itemize}
\end{lemma}

\section{Final Remarks} \label{sec:main}

It is worth noting that when $d(\mathbf{a})=0$, the sets
$\mathcal{X}_{\mathbf{a}r}$ were completely characterized
for any $r\in\mathbb{R}$. The same happens when we consider
$d(\mathbf{a})\neq 0$, $\beta=0$, $\kappa>0$ and $r\in\mathbb{R}$;
i.e., when $p(\xi)=r-\kappa\xi^3+\eta\xi^5$.
In particular, in this case, Theorem~\ref{teo:andre} tells us that
the sets $\mathcal{X}_{\mathbf{a}r}$ are empty for all non null
$\mathbf{a}\in \mathbb{C}^4$ and $r\in \mathbb{R}$.
Thus the set $\mathcal{X}$ is empty and
the problem of the initial and boundary value, analyzed by Vasconcellos
and Silva \cite{vs} and  associated with the linear Kawahara equation
$u_t+\kappa u_{xxx}+\eta u_{xxxxx}=0$, does not admit non-trivial
solutions whose energies do not decay over time.
 Note that for $p(\xi)=r+\beta\xi-\kappa\xi^3+\eta\xi^5$, the case
$d(\mathbf{a})\neq 0$ remains to be solved in two situations:
(a) when $r\neq 0$ and $p$ has exactly three real roots, with all
the multiplicities being equal to $1$ and (b) $p$ has exactly three
real roots with one of them having a multiplicity of $2$.


\section{Appendix: Derivatives of $F(t)$}

Consider the function
\[
F(t)= K \sinh y_1t \sinh y_2t
-2 y_1 y_2(\cosh ((y_1+y_2)t)-\cos ((x_2-x_1)t))
\]
for  $t \in \mathbb{R}$.
Note that $F(0)=0$ and that
\begin{gather*}
\begin{aligned}
F'(t) &=K(y_1\cosh y_1t\sinh y_2t +y_2\sinh y_1t\cosh y_2t)\\
 &\quad -2y_1y_2((y_1+y_2)\sinh ((y_1+y_2)t)
 +(x_2-x_1)\sin ((x_2-x_1)t))
\end{aligned} \\
\begin{aligned}
F''(t) &=(x_2-x_1)^2
(y^2_1+y^2_2)\sinh y_1t\sinh y_2t\\
&\quad +(y_1+y_2)^2
\left(y_1-y_2\right)^2\sinh y_1t\sinh y_2t
+2(x_2-x_1)^2y_1y_2(\cosh y_1t\cosh y_2t\\
&\quad -\cos ((x_2-x_1)t))
\end{aligned}
\end{gather*}
Note that $F'(0)=0$ and that, for $y_1,y_2>0$,  $F''(t)>0$ for all $t>0$.

\section{Appendix: Coefficients of $M_1$ and $M_2$}

Let $\xi_1, \xi_2$ and $\xi_3$ be three distinct points that are mapped by $M$ 
into three distinct points $w_1, w_2$ and $w_3$. Since there is always one and 
only one linear fractional transformation that transforms any three distinct
points into three given distinct points (see, for instance, 
\cite[Theorem V p.~176]{town}, if we take
\begin{equation}\label{eq:Mcoefficients}
\begin{gathered}
a= \begin{vmatrix}
w_1\xi_1&w_1&1\\
w_2\xi_2&w_2&1\\
w_3\xi_3&w_3&1
\end{vmatrix},
\quad 
b=\begin{vmatrix}
w_1\xi_1&\xi_1&w_1\\
w_2\xi_2&\xi_2&w_2\\
w_3\xi_3&\xi_3&w_3
\end{vmatrix},
\\
c=\begin{vmatrix}
\xi_1&w_1&1\\
\xi_2&w_2&1\\
\xi_3&w_3&1
\end{vmatrix},
\quad
d=\begin{vmatrix}
w_1\xi_1&\xi_1&1\\
w_2\xi_2&\xi_2&1\\
w_3\xi_3&\xi_3&1
\end{vmatrix},
\end{gathered}
\end{equation}
then $M(\xi)=\frac{a\xi+b}{c\xi+d}$ is the M\"{o}bius transformation 
$M(\xi)$ such that $M(\xi_j)=w_j$ ($j=1,2,3$).

\subsection{Coefficients of  $M_1$}
Let $M_1(\xi)$ be the linear fractional transformation such that
$M_1(0)=1$, $ M_1(ik)=e^{Lk}$, $M_1(-ik)=e^{-Lk}$.
By \eqref{eq:Mcoefficients}, the coefficients of $M_1$ may be determined by
\begin{gather*}
a_I =-ik\big(w_I+\frac{1}{w_I}-2\big)
=-2ik(\cosh(Lk)-1)\\
b_I =k^2\big(w_I-\frac{1}{w_I}\big)
=d_I
=2k^2\sinh(Lk)\\
c_I =ik\big(w_I+\frac{1}{w_I}-2\big)
=2ik(\cosh(Lk)-1)
\end{gather*}
Therefore,
\[
M_1(\xi)=\frac{-2ik(\cosh(Lk)-1)\xi+2k^2\sinh(Lk)}{2ik(\cosh(Lk)-1)\xi+2k^2\sinh(Lk)}
=-\frac{\xi+ik\operatorname{cotanh}\big(\frac{Lk}{2}\big)}{\xi-ik\operatorname{cotanh}\big(\frac{Lk}{2}\big)}.
\]

\subsection{Coefficients of $M_2$}

Let $M_2(\xi)$ be the linear fractional transformation
such that
$M_2(0)=1$, $w_\rho=M_2(\rho)=e^{-iL\rho}$, 
$M_2(-\rho)=e^{iL\rho}$.
By \eqref{eq:Mcoefficients},
the coefficients of $M_2$ may be determined by
\begin{gather*}
a_\rho =-\rho(w_\rho+\overline{w_\rho}-2)
=-2\rho(\cos(L\rho)-1)\\
b_\rho =-\rho^2\left(w_\rho-\overline{w_\rho}\right)
=d_\rho
=-2i\rho^2\sin(L\rho)\\
c_\rho =\rho\left(w_\rho+\overline{w_\rho}-2\right)
=2\rho(\cos(L\rho)-1)\,.
\end{gather*}
Therefore,
\[
M_2(\xi)=\frac{-2\rho(\cos(L\rho)-1)\xi
-2i\rho^2\sin(L\rho)}{2\rho(\cos(L\rho)-1)\xi-2i\rho^2\sin(L\rho)}
=-\frac{\xi-i\rho\cot \big(\frac{L\rho}{2}\big)}{\xi+i\rho\cot 
\big(\frac{L\rho}{2}\big)}.
\]

\subsection*{Acknowledgments}
 This work was partially supported by FAPERJ under grant E26/010.002646/2014.


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