# Critical Graph of a Polynomial Quadratic Differential related to a
Schrödinger Equation with Cubic Oscillator : WKB Approach

Gliia Braek, Mondher Chouikhi, and Faouzi Thabet

###### Abstract

In this paper, we give a full description of the critical graph of the
quadratic differential $\varpi_{a,\theta}$ defined on the Riemann sphere
$\widehat{\mathbb{C}}$ by
$\varpi_{a,\theta}=-e^{2i\theta}\left(z-a\right)\left(z^{2}-1\right)dz^{2},$
where $\theta\in\mathbb{R},$ and $a\in\mathbb{C}.$. We prove that the
existence and the number of short trajectories of $\varpi_{a,\theta}$ depend
on the location of $a$ in certain curves defined on the complex plane as the
level sets of some harmonic functions. More focus will be to the cases
$\theta\in\left\\{0,\pi/4\right\\}.$ We investigate these classifications to
study an inverse spectral problem related to the complex cubic oscillator for
Schrödinger equation.

2020 Mathematics subject classification: Primary 30E25, 34A26; Secondary
30F15, 34A26

Keywords and phrases: Schrödinger equation. PT symmetry. WKB method. Quadratic
differentials. Horizontal and vertical trajectories. Stokes line. Critical
graph. Half-plane domain. Strip domain. Teichmüller Lemma.

###### Contents

  1. 1 Introduction
  2. 2 Level set of harmonic functions
  3. 3 A quadratic differential
    1. 3.1  The case $\theta=0$
    2. 3.2 The case $\theta=\pi/4$
  4. 4  Eigenvalues problems from cubic oscillator
    1. 4.1 Preliminary results from complex WKB
      1. 4.1.1 Topology of Stokes complex
      2. 4.1.2 WKB solutions
      3. 4.1.3 Elementary basis
      4. 4.1.4 Transition matrices
    2. 4.2 Eingenvalue problems related to the cubic oscillator
      1. 4.2.1 Admissible half planes
      2. 4.2.2 Classification of Eingen-value problem

##  1 Introduction

This paper is motivated by the enumeration of short trajectories (or saddle
connections) related to a polynomial quadratic differential in the Riemann
sphere $\widehat{\mathbb{C}}=\mathbb{C}\cup\\{\infty\\}$. Quadratic
differentials in a Riemann surface $\mathcal{R}$ equipped with a conformal
structure $\left\\{U_{\alpha},\varphi_{\alpha}\right\\}$ is given locally by
an expression $\varpi(z)dz^{2}$, with $\varpi$ is holomorphic (or meromorphic)
function. These differentials define two important tools on $\mathcal{R}$: a
flat metric with singularities at the critical points of $\varpi$ (zeros and
poles); and a foliation (or trajectories). The trajectories play an important
role in different area of mathematics and mathematical physics, in particular
certain geodesics with respect to the flat metric are called short
trajectories (for definition see 2). For the general theory of quadratic
differentials the reader can see ([1], [2], [3], [15], [16],…). In this paper
we consider the case

$\varpi(z)dz^{2}=A\left(z-a\right)\left(z^{2}-1\right)dz^{2},$ (1)

where $A$ is a non vanishing complex number, $a\in\mathbb{C}$ and
$\mathcal{R}=\widehat{\mathbb{C}}$ . Taking to account that the critical graph
of a quadratic differential is invariant under multiplication by a non
vanishing positive real number, we consider the set:

$\mathcal{S}_{\theta}=\left\\{a\in\mathbb{C};\text{
}-\exp(2i\theta)\left(z-a\right)\left(z^{2}-1\right)dz^{2}\text{ admit at
least a short trajectories}\right\\}$ (2)

where $\theta\in[0,\pi[$. We give a full description of $\Lambda_{\theta}$ and
so the critical graph of $\varpi$ (see 2). We prove that the number of short
trajectories depend on the location of $a$ in a curve $\Sigma$ defined in the
complex plane as the level sets of some harmonic functions (see again 2). As
an application, we treat an inverse spectral problem related to cubic
oscillator (see 4). But before going in details let give some motivations and
backgrounds to this work.

Quadratic differentials was originally initiated by O.Teichmuller in his work
on extremal quasi-conformal mapping, Teichmuller gives a solution to the
moduli problem for Riemann surfaces. Since there, quadratic differentials
become a fundamental tool used to solve many problems in geometric function
theory (see again [1], [2], [4], [5] ) and moduli spaces of Riemann surfaces
([6], [7],)

Recently, in potential theory, short trajectories are candidate to define the
support of some equilibrium measure $\mu$ ([12], [13]). The measure $\mu$ used
in some cases to describe the weak limit of zeros distribution related to some
family of orthogonal polynomials that satisfies non-Hermitian orthogonality
([8], [10], [14]). For the strong asymptotic of zeros and by the use of
Riemann-Hilbert approach, short trajectories of quadratic differentials are
curves where the "jump function" is defined, so they plays a crucial role in
the construction of the solutions to such problem (see again [8], [14]). As
the theory of orthogonal polynomials is one of the keys used to solve some
spectral problems in the random matrix theory, so trajectories of quadratic
differentials define some spectral curves useful in the asymptotic study of
spectrum in some model ([11], [19])

In physical level, Gaiotto, Moore, and Neitzke ([18]), introduced spectral
networks of trajectories on a Riemann surface $\mathcal{R}$ to characterize
possible spectra of BPS state. When $\mathcal{R}$ is equipped with a quadratic
differential $\varpi$, these spectral networks are the trajectories of
$\varpi$. Short trajectories of $\varpi$ play a crucial role in counting BPS
states (for more details see [16], [18]). Gaiotto and al pay more attention to
some particular quadratic differentials, called sometimes GMN differentials.
These differentials obeys some local conditions:

  * •

GMN differential has no zeros of order $>1.$

  * •

GMN differential has at least one pole

  * •

GMN differential has at least one finite critical point

GMN is said to be complete if it has no simple pole. Polynomials quadratic
differential in the Riemann sphere are typical example a complete GMN
differential, in particular the example studied in this paper.

In ([33]) B.Shapiro, introduced very flat quadratic differentials to be
polynomial differentials with only simple roots and maximal number of strip
domains (2). The author asked if it’s true that for any polynomials $P$ there
exist $t\in[0,2\pi)$ such that the differential $\exp(it)P(z)dz^{2}$ is very
flat ([33, Problem 3]). In our work give a positive answer to this problem in
the case $\deg P=3$.

Finally, as an application to our results we consider an inverse spectral
problem related to a second order ordinary differential equation (ODE) with
polynomial coefficients in complex domain of the form:

$-y^{\prime\prime}(z)+\lambda^{2}P(z)y(z)=0$ (3)

where the potential $P(z)=\underset{j=1}{\overset{d}{\prod}}(z-a_{j})$, $d\geq
2$ is an integer and $(a_{1},...,a_{d})$ are complex numbers. A global study
of this ODE can be found in the book of Sibuya ([20]). The equation (3) is an
example of ODE with irregular singularity at infinity so an eigenvalue problem
is given with boundary conditions at this point. The Stokes sector related to
this equation are defined by:

$S_{j}=\left\\{z\in\mathbb{C}\text{; }\left|\arg
z-\frac{2\arg\lambda}{d+2}-\frac{2j\pi}{d+2}\right|<\frac{\pi}{d+2}\right\\}$

where $j=0,...,d+1$. Observe that $S_{j}$ are cyclically ordered on
$\mathbb{C}$ and that $S_{j}=S_{k}$ if and only if $k\equiv j[d+2]$. A
classical theorem of Sibuya ( [20]) stat that in each Stokes sector, equation
(3) had a unique up to a scalar factor non trivial subdominant solution
$\varphi_{j}$ which decays exponentially along any ray in this sector. Two
“consecutive” subdominant solution
$\left\\{\varphi_{j},\varphi_{j+1}\right\\}$ constitute a fundamental system
of solution (F.S.S) of equation (3). To avoid repetition a solution to an
eigenvalue problem means a non trivial one.

Fedoryuk ([25], [26]) used the complex WKB, reformulated from different point
of view by Voros ([29]), to study an eigenvalue problem with spectral
parameter $\lambda$ and boundary conditions imposed on the real line:

$\begin{cases}{-y^{\prime\prime}(z)+\lambda^{2}P(z)y(z)=0}\\\
{y(-\infty,\lambda)=y(+\infty,\lambda)=0}\end{cases}$ (4)

where $\lambda$ is a non-vanishing complex number and $P$ is an entire
function in the complex plane $\mathbb{C}.$
Fedoryuk established an algorithm that enables one to continue an asymptotic
solution of (4), known in a domain or along a line (the real line in the case
of (4)), over the whole $z-$complex plane. The boundary conditions are
equivalent to $y\in L^{2}(\mathbb{R})$. The solutions used by Fedoryuk are
subdominant in some canonical domains (4.1.1). The spectrum of (4), namely the
set of all non-vanishing complex numbers $\lambda$ such that there exist a non
trivial $L^{2}(\mathbb{R})-$solution, is discreet. In the case when $P$ is a
polynomial with real coefficients with at least two simple real zeros, a full
asymptotic study of the spectrum is given in ([25, chapter 3 §5]). In the case
when $P$ is a complex polynomial, some additional conditions was established
on the Stokes complexes related to the ODE. In fact, Fedoyruk claimed that the
spectrum of (4), for the case of complex polynomial, tend to accumulate near
only finite rays in the complex plane with slope $\tan\theta_{1}$,
$\tan\theta_{2}...\tan\theta_{k}$ (see again ([25, Chapter 3 §6]), ([33])).
It’s difficult, in general, to calculate $\tan\theta_{i}$, even in the case
where $\deg P$ is small. Conversely, we think that describing the conditions
to a given angle $\theta_{0}$ to be an "accumulation direction" (for
definition see 18) will be very useful especially for the exact WKB method
based on Borel re-summation method ([27]). To be more precise, inspired by the
works of Sibuya ([20]) and Fedoryuk ([25], [26]), we treat an inverse problem
related to ODE with polynomial potential:

######  Problem 1.

For a given $\theta_{0}\in[0,\pi[$ and $d$ integer $\geq 3$, can we construct
a polynomials $P$ with at least two distinct zeros such that the differential
equation (3) admit a subdominant solution in at least two different Stokes
sectors. If this set is non empty, can we compute the number of eigenvalue
problems related to this differential equation, what are the conditions such
that $\theta_{0}$ is an "accumulation direction" (see again 18) of the
spectrum?

We will give a short resume of Fedoyruk’s work in (see 4.1). He was, probably,
the first to point out the importance of quadratic differential in the
asymptotic studies of such eigenvalue problem. In paragraph (4), we try to
give answers to these questions for $d=3$. More precisely, we construct
$P(z)=(z-a)(z^{2}-1)$ using $\mathcal{S}_{\theta_{0}}$ (2), and we prove that
the number of eigenvalue problems such that $\theta_{0}$ is an accumulation
direction of the spectrum, depend on the location of $a$ on
$\mathcal{S}_{\theta_{0}}$ (see theorem$!!!!)$. In fact, for $\theta_{0}$
fixed, $\mathcal{S}_{\theta_{0}}$ divide the complex plane $\mathbb{C}$ into
$n_{0}$ simply connected domains $\left(\Omega_{i}\right)_{1\leq i\leq
n_{0}}$. The critical graph of the quadratic differential
$-e^{2i\theta_{0}}P(z)dz^{2}$ change as $a$ goes $\Omega_{i}$ to $\Omega_{j}$
with $i\neq j$ crossing $\mathcal{S}_{\theta_{0}}$. Theses mutations of the
critical graph affect the number of the eigenvalue problems. These problems
are not-self adjoint in general, since $P$ take complex values. The spectrum
is always discrete. We investigate the asymptotic formula established by
Fedoruyk ([25],[26]) to prove that for a fixed
$\theta_{0}\in[0,\frac{\pi}{2}]$, the series of eigenvalues
$\left\\{\lambda_{n}\text{, }n\in\mathbb{N}\right\\}$ tend to accumulate near
$\theta_{0}$ in the $z-$plane if and only if $a$ describe some levels of the
curve $\mathcal{S}_{\theta_{0}}$. For the cubic oscillator with two distinct
real zeros, the eigenvalue problem with boundary conditions imposed on the
real line (4) is one of the eigenvalue problems given in our work.

K.C. Shin in ([28]) had studied an inverse problem related to an eigenvalue
problem of type ([20, chapter 6]). The methods used by the author is analytic
and in ([28, theorem 1.6]) he investigate the asymptotic distribution of the
eigenvalue to reconstruct "some" coefficients of the polynomial potential. Our
methods are different for the case $d=3$. We use geometric method to construct
a "full" polynomial potential.

The paper is organised as follow:………………………………………………………………………….

##  2 Level set of harmonic functions

For $\theta\in\left[0,\pi/2\right[,$ we consider the sets

$\displaystyle\Sigma_{1,\theta}$
$\displaystyle=\left\\{a\in\mathbb{C}\setminus\left]-\infty,-1\right]:\Re\left(\int_{\left[1,a\right]}e^{i\theta}\sqrt{p_{a}\left(z\right)}dz\right)=0\right\\};$
$\displaystyle\Sigma_{-1,\theta}$
$\displaystyle=\left\\{a\in\mathbb{C}\setminus\left[1,+\infty\right[:\Re\left(\int_{\left[-1,a\right]}e^{i\theta}\sqrt{p_{a}\left(z\right)}dz\right)=0\right\\};$
$\displaystyle\Sigma_{\theta}$
$\displaystyle=\left\\{a\in\mathbb{C}\setminus\left[-1,1\right]:\Re\left(\int_{\left[-1,1\right]}e^{i\theta}\sqrt{p_{a}\left(z\right)}dz\right)=0\right\\};$

where $p_{a}\left(z\right)$ is the complex polynomial defined by

$p_{a}\left(z\right)=\left(z-a\right)\left(z^{2}-1\right).$

######  Lemma 2.

Let $\theta\in\left[0,\pi/2\right[.$ Then

Each of the sets $\Sigma_{1,\theta}$ and $\Sigma_{-1,\theta}$ is formed by two
smooth curves that are locally orthogonal respectively at $z=1$ and $z=-1;$
more precisely :

$\displaystyle\lim\limits_{\begin{subarray}{c}a\longrightarrow-1\\\
a\in\Sigma_{-1,\theta}\end{subarray}}\arg\left(a+1\right)$
$\displaystyle=\frac{-2\theta+\left(2k+1\right)\pi}{4},k=0,1,2,3;$
$\displaystyle\lim\limits_{\begin{subarray}{c}a\longrightarrow+1\\\
a\in\Sigma_{1,\theta}\end{subarray}}\arg\left(a-1\right)$
$\displaystyle=\frac{-\theta+k\pi}{2},k=0,1,2,3.$

The two curves defining $\Sigma_{1,\theta}$ (resp. $\Sigma_{-1,\theta}$ ) meet
only in $z=1$ (resp. $z=-1$ ). Moreover, for
$\theta\notin\left\\{0,\frac{\pi}{2}\right\\},$ they diverge differently to
$\infty$ following one of the directions

$\lim_{\begin{subarray}{c}\left|a\right|\longrightarrow+\infty\\\
a\in\Sigma_{\pm 1,\theta}\end{subarray}}\arg
a=\frac{-2\theta+2k\pi}{5},k=0,1,2,3,4.$

For $\theta=0,$ (resp. $\theta=\frac{\pi}{2}$ ), one ray of
$\Sigma_{1,\theta}$ (resp. $\Sigma_{-1,\theta}$ ) diverges to $z=-1$ (resp.
$z=1$)

###### Proof.

Given a non-constant harmonic function $u$ defined in some domain
$\mathcal{D}$ of $\mathbb{C}.$ The critical points of $u$ are precisely those
where

$\frac{\partial u}{\partial z}=\frac{1}{2}\left(\frac{\partial u}{\partial
x}-i\frac{\partial u}{\partial y}\right)=0.$

They are isolated. If $v$ is the harmonic conjugate of $u$ in $\mathcal{D},$
say, $f\left(z\right)=u\left(z\right)+iv\left(z\right)$ analytic in
$\mathcal{D},$ then, by Cauchy-Riemann,

$f^{\prime}\left(z\right)=0\Longleftrightarrow u^{\prime}\left(z\right)=0.$

The level-set

$\Sigma_{z_{0}}=\left\\{z\in\mathcal{D}:u\left(z\right)=u\left(z_{0}\right)\right\\}$

of $u$ through a point $z_{0}\in\mathcal{D}$ depends on the behavior of $f$
near $z_{0}.$ More precisely, if $z_{0}$ is a critical point of $u,$ (
$u^{\prime}\left(z_{0}\right)=0$ ), then, there exists a neighborhood
$\mathcal{U}$ of $z_{0},$ an a holomorphic function $g\left(z\right)$ defined
in $\mathcal{U},$ such that

$\forall
z\in\mathcal{U},f\left(z\right)=\left(z-z_{0}\right)^{m}g\left(z\right);g\left(z\right)\neq
0.$

Taking a branch of the $m$-th root of $g\left(z\right),$ $f$ has the local
structure

$f\left(z\right)=\left(h\left(z\right)\right)^{m},\forall z\in\mathcal{U}.$

It follows that $\Sigma_{z_{0}}$ is locally formed by $m$ analytic arcs
passing through $z_{0}$ and intersecting there with equal angles $\pi/m.$
Through a regular point $z_{0}\in\mathcal{D},$ (
$u^{\prime}\left(z_{0}\right)\neq 0$ ), the implicit function theorem asserts
that $\Sigma_{z_{0}}$ is locally a single analytic arc. Notice that the level-
set of a harmonic function cannot terminate at a regular point; (see
[34],[35]…).

Let consider the multi-valued function

$f_{1,\theta}\left(a\right)=\int_{1}^{a}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt,a\in\mathbb{C}.$

Integrating along the segment $\left[1,a\right],$ we may assume without loss
of generality that

$f_{1,\theta}\left(a\right)=ie^{i\theta}\left(a-1\right)^{2}\int_{0}^{1}\sqrt{t\left(1-t\right)}\sqrt{t\left(a-1\right)+2}dt=\left(a-1\right)^{2}g\left(a\right);g\left(1\right)\neq
0.$ (5)

It is obvious that :

$\forall
a\in\mathbb{C}\setminus\left]-\infty,-1\right],\left\\{t\left(a-1\right)+2;t\in\left[0,1\right]\right\\}=\left[2,a+1\right]\subset\mathbb{C}\setminus\left]-\infty,0\right].$

Therefore, with a fixed choice of the argument and square root inside the
integral, $f_{1,\theta}$ and $g$ are single-valued analytic functions in
$\mathbb{C}\setminus\left]-\infty,-1\right].$

Suppose that for some $a\in\mathbb{C}\setminus\left]-\infty,-1\right],a\neq
1,$

$u\left(a\right)=\Re
f_{1,\theta}\left(a\right)=0;f_{1,\theta}^{\prime}\left(a\right)=0.$

Then,

$\left(a-1\right)^{3}g^{\prime}\left(a\right)+2f_{1,\theta}\left(a\right)=0.$

Taking the real parts, we get

$\displaystyle 0$
$\displaystyle=\int_{0}^{1}\sqrt{t\left(1-t\right)}\Im\left(e^{i\theta}\left(a-1\right)^{2}\sqrt{t\left(a-1\right)+2}\right)dt;$
$\displaystyle 0$
$\displaystyle=\Re\left(\left(a-1\right)^{3}g^{\prime}\left(a\right)\right)=\int_{0}^{1}t\sqrt{t\left(1-t\right)}\Im\left(\frac{e^{i\theta}\left(a-1\right)^{3}}{2\sqrt{t\left(a-1\right)+2}}\right)dt.$

By continuity of the functions inside these integrals along the segment
$\left[0,1\right],$ there exist $t_{1},t_{2}\in\left[0,1\right]$ such that

$\Im\left(e^{i\theta}\left(a-1\right)^{2}\sqrt{t_{1}\left(a-1\right)+2}\right)=\Im\left(\frac{e^{i\theta}\left(a-1\right)^{3}}{2\sqrt{t_{2}\left(a-1\right)+2}}\right)=0;$

and then

$e^{2i\theta}\left(a-1\right)^{4}\left(t_{1}\left(a-1\right)+2\right)>0,\left(\frac{e^{2i\theta}\left(a-1\right)^{6}}{t_{2}\left(a-1\right)+2}\right)>0.$

Taking their ratio, we get

$\frac{\left(t_{1}\left(a-1\right)+2\right)\left(t_{2}\left(a-1\right)+2\right)}{\left(a-1\right)^{2}}>0.$

which cannot hold, since, if $\Im a>0,$ then

$0<\arg\left(t_{1}\left(a-1\right)+2\right)+\arg\left(\left(t_{2}\left(a-1\right)+2\right)\right)<2\arg\left(a+1\right)<\arg\left(\left(a-1\right)^{2}\right)<2\pi.$

The case $\Im a<0$ is in the same vein, while the case $a\in\mathbb{R}$ can be
easily discarded. Thus, $a=1$ is the unique critical point of $\Re
f_{1,\theta}.$ Since
$f^{\prime}{}^{\prime}{}_{1,\theta}(1)=2g\left(1\right)\neq 0,$ we deduce the
local behavior of $\Sigma_{1,\theta}$ near $a=1.$

Suppose now that for some $\theta\in\left]0,\frac{\pi}{2}\right[,$ a ray of
$\Sigma_{\pm 1,\theta}$ diverges to a certain point in
$\left]-\infty,-1\right[;$ for example,

$\left(\overline{\Sigma_{1,\theta}}\setminus\Sigma_{1,\theta}\right)\cap\left\\{z\in\overline{\mathbb{C}}:\Im
z\geq 0\right\\}=\left\\{x_{\theta}\right\\}.$

Let $\epsilon>0$ such that $0<\theta-2\epsilon.$ For $a\in\mathbb{C}$
satisfying $\pi-\epsilon<\arg a<\pi,$

$0<\theta-2\epsilon<\theta+2\arg
a+\arg\int_{0}^{1}\sqrt{t\left(1-t\right)}\sqrt{t\left(a-1\right)+2}dt<\frac{\pi}{2}+\theta-\frac{\epsilon}{2}<\pi,$

which contradicts (5). The other cases are similar. Thus, any ray of
$\Sigma_{\pm 1,\theta}$ should diverge to $\infty.$ The case $\theta=0$ is
quiet easier.

If $a\longrightarrow\infty,$ then
$\left|f_{1,\theta}\left(a\right)\right|\longrightarrow+\infty;$ since $\Re
f_{1,\theta}\left(a\right)=0,$ we have $\left|\Im
f_{1,\theta}\left(a\right)\right|\longrightarrow+\infty.$ It follows that

$\arg\left(f\left(a\right)\right)\sim\arg\left(\frac{4}{15}e^{i\theta}a^{5/2}\right)\longrightarrow\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\text{
as }a\longrightarrow\infty.$

We get the behavior of any arc of $\Sigma_{1,\theta}$ that diverges to
$\infty.$ In particular, from the maximum modulus principle, two rays of
$\Sigma_{1,\theta}$ cannot diverge to $\infty$ in the same direction.

If $\Sigma_{1,\theta}$ contains a regular point $z_{0}$ (for example, $\Im
z_{0}>0$) not belonging to the arcs of $\Sigma_{1,\theta}$ emerging from
$a=1.$ The two rays of the level set curve $\gamma$ going through $z_{0}$
diverge to $\infty$ in two different directions. It follows that $\gamma$
should go through $z_{1}=1+iy,$ for some $y>0,$ or $z_{1}=y,$ for some $y>1.$
It is easy to check that in the two cases, for any choice of the argument,

$\Re\int_{1}^{z_{1}}\left(e^{i\theta}\sqrt{p_{z_{1}}\left(t\right)}dt\right)\neq
0;$

and we get a contradiction. Thus, $\Sigma_{1,\theta}$ is formed only by the
two curves going through $a=1.$ The same idea gives the structure of
$\Sigma_{-1,\theta};$ even more, from the relation

$\Re f_{\pm 1,\theta}\left(a\right)=0\Longleftrightarrow\Re f_{\pm
1,\frac{\pi}{2}-\theta}\left(-\overline{a}\right)=0,$ (6)

available for any $\theta\in\left[\pi/4,\pi/2\right[,$ one can see easily that
$\Sigma_{-1,\frac{\pi}{2}-\theta}$ and $\Sigma_{1,\theta}$ are symmetric with
respect to the imaginary axis; (6). This, leads us to restrict our
investigation to the case . ∎

######  Proposition 3.

For $\theta\in\left[0,\pi/4\right],$ we denote by $\Sigma_{\pm 1,\theta}^{l}$
and $\Sigma_{\pm 1,\theta}^{r}$ respectively the left and right curves
defining $\Sigma_{\pm 1,\theta}$ in the upper half plane, see Fig(). Then

| $\Sigma_{1,\theta}^{l}\cap\Sigma_{-1,\theta}^{r}$ | $\Sigma_{1,\theta}^{l}\cap\Sigma_{-1,\theta}^{l}$ | $\Sigma_{1,\theta}^{r}\cap\Sigma_{-1,\theta}^{r}$ | $\Sigma_{1,\theta}^{r}\cap\Sigma_{-1,\theta}^{l}$
---|---|---|---|---
$\theta\in\left[0,\pi/8\right[$ | $\left\\{t_{\theta}\right\\}$ | $\left\\{e_{\theta}\right\\}$ | $\emptyset$ | $\emptyset$
$\theta\in\left[\pi/8,\pi/4\right]$ | $\left\\{t_{\theta}\right\\}$ | $\emptyset$ | $\emptyset$ | $\emptyset$

where $t_{\theta}$ and $e_{\theta}$ are two complex number varying
respectively in two smooth curves

$\mathcal{A\cap}\left\\{a\in\mathbb{C}:-1\leq\Re a\leq 0,\Im a\geq
0\right\\},$

and

$\mathcal{E\cap}\left\\{a\in\mathbb{C}:\Re a\leq-1,\Im a\geq 0\right\\};$

where $\mathcal{A}$ and $\mathcal{E}$ are defined by

$\displaystyle\mathcal{A}$
$\displaystyle\mathcal{=}\left\\{a\in\mathbb{C}:-1\leq\Re a\leq 1,\Re
f_{1,\theta}\left(a\right)=\Re f_{-1,\theta}\left(a\right)=0\right\\};$
$\displaystyle\mathcal{E}$
$\displaystyle\mathcal{=}\left\\{a\in\mathbb{C}:\left|\Re a\right|\geq 1,\Re
f_{1,\theta}\left(a\right)=\Re f_{-1,\theta}\left(a\right)=0\right\\}.$

Moreover, the set $\Sigma_{\theta}$ is a smooth curve included in the part of
the upper half plane of the strip bounded by the segment $\left[-1,1\right]$
and the lines $y=-\tan\left(2\theta\right)\left(x\pm 1\right).$ It goes
through $t_{\theta}$ and $e_{\theta}$ (if exists); its two rays diverge
differently to $\infty$ following the direction $\arg a=\pi-2\theta,$ and to
the unique $s_{\theta}\in\left[-1,0\right],$ such that

$\Re\int_{-1}^{1}e^{i\theta}\left(\sqrt{p_{s_{\theta}}\left(t\right)}\right)_{+}dt=0.$

In particular

$\Sigma_{1,\theta}\cap\Sigma_{-1,\theta}\cap\left\\{a\in\mathbb{C}:\Im
a<0\right\\}=\emptyset.$

###### Proof.

Let consider the functions defined as in the proof of lemma.2 :

$f_{\theta}\left(a\right)=e^{i\theta}\int_{-1}^{1}\sqrt{p_{a}\left(t\right)}dt;f_{\pm
1,\theta}\left(a\right)=e^{i\theta}\int_{\pm
1}^{a}\sqrt{p_{a}\left(t\right)}dt.$

Then, looking to the real parts of these functions as harmonic ones in two
variables $a=\left(x,y\right)\in\mathbb{R}\times\mathbb{R}^{\ast+},$ and
taking the arguments of the square roots in
$\left]-\frac{\pi}{2},\frac{\pi}{2}\right[,$ we may assume that their
gradients are :

If we denote by

$\displaystyle\Sigma_{\pm 1,\theta}^{l}\cap\left\\{z\in\mathbb{C}:\Im z\geq
0\right\\}$ $\displaystyle=\left\\{a_{\pm
1}^{l}\left(s\right),s\in\left[0,+\infty\right[,a_{\pm
1}^{l}\left(0\right)=\pm 1\right\\};$ $\displaystyle\Sigma_{\pm
1,\theta}^{r}\cap\left\\{z\in\mathbb{C}:\Im z\geq 0\right\\}$
$\displaystyle=\left\\{a_{\pm
1}^{r}\left(s\right),s\in\left[0,+\infty\right[,a_{\pm
1}^{r}\left(0\right)=\pm 1\right\\},$

then, with (5),

$\displaystyle\Im\left(e^{i\theta}\left(a_{1}^{l}\left(s\right)-1\right)^{2}\int_{0}^{1}\sqrt{t\left(1-t\right)}b\left(t\right)dt\right)$
$\displaystyle=0;$ (7)
$\displaystyle\Im\left(e^{i\theta}\left(a_{1}^{l}\right)^{\prime}\left(s\right)\left(a_{1}^{l}\left(s\right)-1\right)\int_{0}^{1}\sqrt{\frac{t}{1-t}}b\left(t\right)dt\right)$
$\displaystyle=0;$ (8)

where $b\left(t\right)=\sqrt{t\left(a_{1}^{l}\left(s\right)-1\right)+2}.$
Taking the arguments in $\left[-\pi,\pi\right[,$ we get

$\displaystyle\theta+2\arg\left(a_{1}^{l}\left(s\right)-1\right)+\arg\int_{0}^{1}\sqrt{t\left(1-t\right)}b\left(t\right)dt$
$\displaystyle\equiv 0\operatorname{mod}\left(\pi\right);$ (9)
$\displaystyle\theta+\arg\left(a_{1}^{l}\right)^{\prime}\left(s\right)+\arg\left(a_{1}^{l}\left(s\right)-1\right)+\arg\int_{0}^{1}\sqrt{\frac{t}{1-t}}b\left(t\right)dt$
$\displaystyle\equiv 0\operatorname{mod}\left(\pi\right),$ (10)

and then

$\arg\frac{\left(a_{1}^{l}\right)^{\prime}\left(s\right)}{a_{1}^{l}\left(s\right)-1}\equiv\arg\int_{0}^{1}\sqrt{\frac{t}{1-t}}b\left(t\right)dt-\arg\int_{0}^{1}\sqrt{t\left(1-t\right)}b\left(t\right)dt\operatorname{mod}\left(\pi\right).$

Since for $s>0,$

$0<\arg\int_{0}^{1}\sqrt{\frac{t}{1-t}}b\left(t\right)dt-\arg\int_{0}^{1}\sqrt{t\left(1-t\right)}b\left(t\right)dt<\frac{\arg\left(a\left(s\right)+1\right)}{2},$

and considering the continuity of the function
$s\longmapsto\arg\frac{\left(a_{1}^{l}\right)^{\prime}\left(s\right)}{a_{1}^{l}\left(s\right)-1}$
in $\left]0,+\infty\right[,$ we deduce the existence of
$k\in\left\\{0,1\right\\},$ such that :

$\forall s\geq
0,-\frac{\pi}{2}+k\pi<-\frac{\arg\left(a_{1}^{l}\left(s\right)+1\right)}{2}+k\pi<\arg\frac{\left(a_{1}^{l}\right)^{\prime}\left(s\right)}{a_{1}^{l}\left(s\right)-1}<k\pi.$

In other word,
$\Im\left(\frac{\left(a_{1}^{l}\right)^{\prime}\left(s\right)}{a_{1}^{l}\left(s\right)-1}\right)$
and
$\Re\left(\frac{\left(a_{1}^{l}\right)^{\prime}\left(s\right)}{a_{1}^{l}\left(s\right)-1}\right)$
preserve the same signs in $\left[0,+\infty\right[,$ and we get the monotony
of the functions $s\longmapsto\arg\left(a_{1}^{l}\left(s\right)-1\right),$ and
$s\longmapsto\left|a_{1}^{l}\left(s\right)-1\right|$ in
$\left[0,+\infty\right[.$ Taking into account their boundary values at $s=0$
and $\infty,$ we get

$\forall s\geq
0,-\frac{\arg\left(a_{1}^{l}\left(s\right)+1\right)}{2}<\arg\left(\frac{\left(a_{1}^{l}\right)^{\prime}\left(s\right)}{a_{1}^{l}\left(s\right)-1}\right)<0.$
(11)

In the same vein, one can show that

$\forall t\geq
0,\frac{\pi}{2}<\arg\left(\frac{\left(a_{-1}^{l}\right)^{\prime}\left(t\right)}{a_{-1}^{l}\left(t\right)+1}\right)<\frac{\pi}{2}+\frac{\arg\left(a_{-1}^{l}\left(t\right)-1\right)}{2}.$
(12)

Since $\Re a_{-1}^{l}\left(t\right)<-1$ for any $t>0,$ it follows that, if
$a_{-1}^{l}\left(t\right)=a_{1}^{l}\left(s\right),$ then

$-\frac{\pi}{2}<\arg\left(a_{-1}^{l}\left(t\right)+1\right)-\arg\left(a_{1}^{l}\left(s\right)-1\right)<0;$

in the later case, after making the difference between (12) and (11), we
obtain

$0<\arg\left(\frac{\left(a_{-1}^{l}\right)^{\prime}\left(t\right)}{\left(a_{1}^{l}\right)^{\prime}\left(s\right)}\right)<\pi.$

Thus, $\Sigma_{1,\theta}^{l}$ and $\Sigma_{-1,\theta}^{l}$ cannot meet twice
in the upper half-plane. Repeating the same reasoning for the three other
cases, we get all the points of the lemma about the uniqueness of intersection
when exists.

For $\theta\in\left[0,\pi/4\right],$ there exist two directions that should
follow any curves of $\Sigma_{\pm 1,\theta}$ diverging to $\infty$ in the
upper half plane, $D_{1}:\arg a=\frac{-2\theta+2\pi}{5}$ and $D_{2}:\arg
a=\frac{-2\theta+4\pi}{5}.$ More precisely, since for $a=\pm 1,$
$\Sigma_{a,\theta}^{r}$ and $\Sigma_{a,\theta}^{l}$ cannot diverge to $\infty$
in the same direction, we deduce that, $\Sigma_{a,\theta}^{r}$ and
$\Sigma_{a,\theta}^{l}$ follow respectively the direction $D_{1}$ and $D_{2}.$
Therefore, $\Sigma_{1,\theta}^{l}$ and $\Sigma_{-1,\theta}^{r}$ should meet at
some point, say $t_{\theta}.$ We claim that for
$\theta\in\left[0,\frac{\pi}{4}\right],$

$-1\leq\Re t_{\theta}\leq 0.$

The first inequality is obvious since
$\Sigma_{-1,\theta}^{r}\cap\left\\{z\in\mathbb{C}:\Re
z<-1\right\\}=\emptyset.$ If $\Re t_{\theta}>0,$ then,

$\left|\sqrt{t_{\theta}-t}\right|\leq\left|\sqrt{t_{\theta}+t}\right|,\forall
t\in\left[0,1\right],$

and

$0<\frac{\arg\left(t_{\theta}+1\right)}{2}\leq\arg\int_{-1}^{1}\sqrt{p_{t_{\theta}}\left(t\right)}=\arg\int_{0}^{1}\sqrt{1-t^{2}}\left(\sqrt{t_{\theta}-t}+\sqrt{t_{\theta}+t}\right)dt\leq\frac{\arg
t_{\theta}}{2}<\frac{\pi}{4}.$

Therefore,

$0<\arg\int_{-1}^{1}e^{i\theta}\sqrt{p_{t_{\theta}}\left(t\right)}<\frac{\pi}{2},$
(13)

and $t_{\theta}\notin\Sigma_{\theta};$ a contradiction. Moreover, relation (6)
shows that the sets $\Sigma_{-1,\pi/4}$ and $\Sigma_{1,\pi/4}$ are symmetric
with respect to the imaginary axis; it follows that $a_{\frac{\pi}{4}}\subset
i\mathbb{R}^{+\ast}.$ The monotony of $\arg a_{1}^{l}\left(s\right)$ and $\arg
a_{-1}^{r}\left(s\right)$ prove that for
$\theta\in\left[0,\frac{\pi}{4}\right],$

$\Im t_{\theta}\in\left[0,a_{\frac{\pi}{4}}/i\right].$ (14)

Implicit function theorem, and relation (14), prove that the set
$\left\\{t_{\theta};\theta\in\left[0,\pi\right]\right\\}$ defined by $\Re
f_{1,\theta}\left(t_{\theta}\right)=\Re f_{1,\theta}\left(t_{\theta}\right)=0$
is a closed curve encircling $\left[-1,1\right],$ that is smooth in
$\mathcal{A}\setminus\left\\{\pm 1\right\\}.$ The relation
$t_{\theta}\in\Sigma_{1,\theta}^{l}\cap\Sigma_{-1,\theta}^{l}$ implies that

$a_{0}=-1;\lim\limits_{\theta\longrightarrow
0^{\pm}}\arg\left(t_{\theta}+1\right)=\pm\frac{\pi}{4}.$

See Fig.1.

For $\Delta=\left[-1,+\infty\right[$ or $\Delta=\left]-\infty,1\right],$ the
function $\Re f_{\theta}$ is harmonic in $\mathbb{C}\setminus\Delta,$ which
implies the smoothness of any curve in its zero-level set. If $\gamma$ is a
ray of $\Sigma_{\theta}$ going through $t_{\theta}$ and diverging to $\infty,$
then, from the relation

$\left(\int_{-1}^{1}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt\right)^{2}\sim\frac{\pi^{2}}{4}ae^{2i\theta},a\longrightarrow\infty,$

we deduce that

$\arg a\sim\pi-2\theta,a\in\gamma,a\longrightarrow\infty.$

The relation for $\theta\in\left[0,\pi/4\right],$

$\frac{-2\theta+4\pi}{5}<\pi-2\theta\Longleftrightarrow\theta\in\left[0,\pi/8\right[,$

implies the existence of $e_{\theta}\in\mathbb{C}$ ($\Re e_{\theta}<-1,\Im
e_{\theta}>0$) as a unique intersection between $\Sigma_{-1,\theta}^{l}$ and
$\Sigma_{1,\theta}^{l}.$ If for some $\theta\in\left[\pi/8,\pi/4\right],$
$\Sigma_{-1,\theta}^{l}$ and $\Sigma_{1,\theta}^{l}$ meet, then they should do
it at least twice because $\Sigma_{\theta}$ diverges to $\infty$ in the
direction $\pi-2\theta\in\left[\pi/2,\frac{-2\theta+4\pi}{5}\right];$ which
violates the uniqueness of the intersection if exists. Moreover,
straightforward calculations show that the set $\mathcal{E}$ is a smooth curve
satisfying,

$\displaystyle\lim\limits_{\theta\longrightarrow 0^{+}}e_{\theta}$
$\displaystyle=-1,\lim\limits_{\theta\longrightarrow
0^{+}}\arg\left(e_{\theta}+1\right)=\left(\frac{3\pi}{4}\right)^{-};$
$\displaystyle\lim\limits_{\theta\longrightarrow\left(\frac{\pi}{8}\right)^{-}}e_{\theta}$
$\displaystyle=\infty,\lim\limits_{\theta\longrightarrow\left(\frac{\pi}{8}\right)^{-}}\arg\left(e_{\theta}+1\right)=\left(\frac{3\pi}{4}\right)^{-}.$

See Fig.1.

The zero-level sets $\Sigma_{1,\theta}$ (resp.$\Sigma_{-1,\theta}$) of the
harmonic functions $\Re f_{1,\theta}$ (resp.$\Re f_{-1,\theta}$) split
$\mathbb{C}\setminus\left]-\infty,-1\right]$
(resp.$\mathbb{C}\setminus\left[1,+\infty\right[$) into four disjoint
connected domains. In each domain, $\Re f_{1,\theta}$ (resp.$\Re
f_{-1,\theta}$) preserves a fixed sign, and in two adjacent domains, it has
opposite signs. It follows that $\Sigma_{1,\theta}\cup\Sigma_{-1,\theta}$
split the complex plane into a finite number $n_{\theta}$ of disjoint
connected domains $\Omega_{k},k=1,...,n_{\theta}.$ In particular, the set
$\Sigma_{\theta}$ is included in those on which $\Re f_{1,\theta}$ and $\Re
f_{-1,\theta}$ have the same sign. See Fig…

Taking the arguments in $\left]-\pi,\pi\right]$ and the square roots such that
$\sqrt{1}=1,$ we have for $\theta\in\left]0,\pi/4\right]$ and
$a\in\mathbb{C}\setminus\mathbb{R}$ $:$

$\frac{\arg\left(a+1\right)}{2}+\theta<\arg\left(e^{i\theta}\int_{\left[-1,1\right]}\sqrt{p_{a}\left(t\right)}dt\right)<\frac{\arg\left(a-1\right)}{2}+\theta.$
(15)

It follows that

$\frac{\Im a}{\Re\left(a+1\right)}<-\tan 2\theta<\frac{\Im
a}{\Re\left(a-1\right)}.$

If $\Im a<0,$ then (15) gives

$-\frac{\pi}{2}<-\frac{\pi}{2}+\theta<\arg\left(e^{i\theta}\int_{\left[-1,1\right]}\sqrt{p_{a}\left(t\right)}dt\right)<\theta<\frac{\pi}{2};$

and then

$\Sigma_{\theta}\cap\left\\{a\in\mathbb{C}:\Im a<0\right\\}=\emptyset.$

Thus,

$\left(\Sigma_{1,\theta}\cup\Sigma_{-1,\theta}\right)\cap\left\\{a\in\mathbb{C}:\Im
a<0\right\\}=\emptyset.$

Finally, standard arguments of complex analysis show that, the functions

$\displaystyle t$
$\displaystyle\longmapsto\left(\sqrt{p_{a}\left(t\right)}\right)_{+}=\lim\limits_{s\longrightarrow
t,\Im s>0}\sqrt{p_{a}\left(s\right)},$ $\displaystyle t$
$\displaystyle\longmapsto\left(\sqrt{p_{a}\left(t\right)}\right)_{-}=\lim\limits_{s\longrightarrow
t,\Im s<0}\sqrt{p_{a}\left(s\right)}$

are well defined and continuous in $\left[-1,1\right].$ An elementary study
shows that in $\left[-1,1\right],$

$\displaystyle a$
$\displaystyle\longmapsto\Re\int_{\left[-1,1\right]}\left(\sqrt{p_{a}\left(t\right)}\right)_{+}dt,$
$\displaystyle a$
$\displaystyle\longmapsto\Im{\displaystyle\int_{\left[-1,1\right]}}\left(\sqrt{p_{a}\left(t\right)}\right)_{-}dt$

respectively, increases from $0$ to $\frac{16\sqrt{2}}{15}$ and decreases from
$\frac{16\sqrt{2}}{15}$ to $0.$ It follows that the function

$\begin{array}[c]{cc}\left[-1,1\right]\longrightarrow\left]-\pi,\pi\right[&a\longmapsto\arg{\displaystyle\int_{\left[-1,1\right]}}\left(\sqrt{p_{a}\left(t\right)}\right)_{-}dt\end{array}$

decreases from $\frac{\pi}{2}$ to $0;$ which proves the existence and
uniqueness of $s_{\theta}$ in $\left[-1,1\right].$ Relation (6) shows that

$s_{\theta}\in\left\\{\begin{array}[c]{c}\left[-1,0\right[\text{ if
}\theta\in\left[0,\frac{\pi}{4}\right[;\\\ \left\\{0\right\\}\text{ if
}\theta=\frac{\pi}{4};\\\ \left]0,1\right]>0\text{ if
}\theta\in\left]\frac{\pi}{4},\frac{\pi}{2}\right].\end{array}\right.$

The observation that for
$a\in\mathbb{C}\setminus\left[-1,1\right],\int_{-1}^{1}\sqrt{\frac{1-t^{2}}{a-t}}dt\neq
0,$ (as a period of complete elliptic integral) shows that the harmonic
function $f_{\theta}^{\prime}\left(a\right)$ defining $\Sigma_{\theta}$ has no
critical points. If $\gamma$ is another arc in $\Sigma_{\theta},$ not going
through $t_{\theta},$ its two rays diverge differently to $s_{\theta}$ and to
$\infty$ in direction $\arg a=\pi-2\theta,$ which violates the maximum modulus
principle. Thus, $\Sigma_{\theta}=\gamma.$ Notice that, if we parametrize
$\Sigma_{\theta}$ by $\left\\{a\left(s\right),s\geq 0\right\\},$ then it
follows from (15), and (13) that

$0\leq\arg\left(\frac{a^{\prime}\left(s\right)}{a\left(s\right)+1}\right)\leq\pi;\frac{-\pi}{2}\leq\arg\left(\frac{a^{\prime}\left(s\right)}{a\left(s\right)-1}\right)\leq
0.$ (16)

Therefore, the functions $s\longmapsto\arg$
$\left(a\left(s\right)+1\right),s\longmapsto\arg$
$\left(a\left(s\right)-1\right),$ and $s\longmapsto\arg$
$\left|a\left(s\right)-1\right|,$ respectively increases from $0$ to
$\pi-2\theta,$ decreases from $\pi$ to $\pi-2\theta,$ and increases from $0$
to $+\infty.$ ∎

Figure 1: Sets of $e_{\theta}$ (left) and $a_{\theta}$ (right).

######  Examples 4.

$\diagdown\theta$ | $0$ | $\arctan\left(1/2\right)$ | $\pi/8$ | $\pi/4$
---|---|---|---|---
$t_{\theta}$ | $-1$ | $-0.69..+i0.25..$ | $-0.49..+i0.36..$ | $i0.462..$
$s_{\theta}$ | $-1$ | $-0,37..$ |  | $0$
$e_{\theta}$ | $-1$ | $-1.67..+i0.79$ | none | none

See Fig.2.

Figure 2: Approximate plots of the sets $\Sigma_{1,\theta}$ (blue),
$\Sigma_{-1,\theta}$ (red), and $\Sigma_{\theta}$ (green) for $\theta=0$
(left), $\theta=\arctan(0.5)/2$ (middle) and $\theta=\pi/4$ (right)

##  3 A quadratic differential

In this section, we consider the quadratic differential on the Riemann sphere
$\widehat{\mathbb{C}},$

$\varpi=\lambda p\left(z\right)dz^{2},$

where $p\left(z\right)$ is a 3-degree complex polynomial with simple zeros,
and $\lambda$ is a non-vanishing complex number. A full description of its
critical graph can be found in ([23]), but independentely of the location of
its zeros, which is the key of the eigenvalue problem’s classification in the
upcoming sections.

By a linear change of variable, and taking into account that the critical
graph is invariant under multiplication of $\varpi$ by a positive real, we may
assume, without loss of the generality that

$\varpi=\varpi_{a,\theta}=-e^{2i\theta}p_{a}\left(z\right)dz^{2}=-e^{2i\theta}\left(z-a\right)\left(z^{2}-1\right)dz^{2},$

for some $a\in\mathbb{C}\setminus\left\\{\pm 1\right\\},$ and
$\theta\in\left[0,\pi\right[.$ The relations

$\displaystyle\varpi_{a,\theta}$
$\displaystyle>0\Longleftrightarrow\varpi_{-a,\theta+\frac{\pi}{2}}>0,$
$\displaystyle\varpi_{a,\theta}$
$\displaystyle>0\Longleftrightarrow\varpi_{-\overline{a},\frac{\pi}{2}-\theta}>0,$

let us take the restriction $\theta\in\left[0,\pi/4\right].$ More focus will
be on the two cases $\theta=0,$ and $\theta=\pi/4.$

In a first plan, we give some immediate and brief observations from the theory
of quadratic differentials:

Critical points of $\varpi_{a,\theta}$ are its zero’s and poles in
$\widehat{\mathbb{C}}$. Zeros are called _finite critical points_ , while
poles of order 2 or greater are called _infinite critical points_. All other
points of $\widehat{\mathbb{C}}$ are called _regular points_. Finite critical
points of the quadratic differential $\varpi_{a,\theta}$ are $\pm 1$ and $a,$
as simple zeros; while, by the change of variable $y=1/z,$ $\varpi_{a,\theta}$
has a unique infinite critical point that is located at $\infty$ as a pole of
order $7.$

_Horizontal trajectories,_ or _Stokes lines_(or just trajectories, in this
section) of the quadratic differential $\varpi_{a,\theta}$ are the level
curves defined by

$\Re\int^{z}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt={const;}$ (17)

or equivalently

$-e^{2i\theta}p_{a}\left(z\right)dz^{2}>0.$

If $z\left(t\right),t\in\mathbb{R}$ is a horizontal trajectory, then the
function

$t\longmapsto\Im\int^{t}e^{i\theta}\sqrt{p\left(z\left(s\right)\right)}z^{\prime}\left(s\right)ds$

is monotonic.

The _vertical_(or, _orthogonal_) _trajectories_ are obtained by replacing
$\Im$ by $\Re$ in equation (17). Horizontal and vertical trajectories of the
quadratic differential $\varpi_{a,\theta}$ produce two pairwise orthogonal
foliations of the Riemann sphere $\widehat{\mathbb{C}}.$ A trajectory passing
through a critical point of $\varpi_{a,\theta}$ is called _critical
trajectory_. In particular, if it starts and ends at a finite critical point,
it is called _finite critical trajectory_ or _short trajectory_ , otherwise,
we call it an _infinite critical trajectory_. The closure of the set of finite
and infinite critical trajectories, that we denote by $\Gamma_{a,\theta}$ is
called the _critical_(or _Stokes_ ) _graph_. The local and global structure of
the trajectories are studied in [1],[2],…etc (see Fig.3). At any regular
point, horizontal (resp. vertical) trajectories look locally as simple
analytic arcs passing through this point, and through every regular point of
$\varpi_{a,\theta}$ passes a uniquely determined horizontal (resp. vertical)
trajectory of $\varpi_{a,\theta};$ the are locally orthogonal at this point.
From each zero with multiplicity $r$ of $\varpi_{a,\theta},$ there emanate
$r+2$ critical trajectories spacing under equal angles $2\pi/(r+2).$ Since
$\infty$ is a pole of order $7,$ there are $5$ asymptotic directions (called
_critical directions_) spacing under equal angle $2\pi/5,$ and a neighborhood
$\mathcal{V}$ of $\infty,$ such that each trajectory entering $\mathcal{V}$
stays in $\mathcal{V}$ and tends to $\infty$ following one of the critical
directions:

$D_{\theta,k}=\left\\{z\in\mathbb{C}:\arg\left(z\right)=\frac{-2\theta+\left(2k+1\right)\pi}{5}\right\\};k=0,...,4.$
(18)

Analogues behavior happens to the orthogonal trajectories at $\infty$, but
their directions are :

$D_{\theta,k}^{\perp}=\left\\{z\in\mathbb{C}:\arg\left(z\right)=\frac{-2\theta+2k\pi}{5}\right\\};k=0,...,4.$

Observe that for $\theta\in\left[0,\frac{\pi}{2}\right[,$ exactly three
critical directions, $D_{\theta,0},D_{\theta,1},$ and $D_{\theta,2}$ are in
the upper half-plane.

Figure 3: Local structure of the trajectories near a simple zero (left) and a
pole of order $7$ (right).

In the large, any trajectory of $\varpi_{a,\theta}$ is either an arc
connecting two critical points, or, is an arc that diverges to $\infty$ at
least in one of its rays. Thanks to Jenkins Three pole Theorem, the quadratic
differential $\varpi_{a,\theta}$ has no recurrent trajectory ( trajectory that
is dense in some domain of $\mathbb{C}$). The structure of the set
$\widehat{\mathbb{C}}\setminus\Gamma_{a,\theta}$ depends on the local and
global behaviors of trajectories. It consists of a finite number of domains
called _domain configurations_ of $\varpi_{a,\theta}.$ Jenkins Theorem (see
[2, Theorem3.5],[1]) asserts that there are two kinds of domain configurations
of $\varpi_{a,\theta}$ :

  * •

Half-plane domain : It is swept by trajectories diverging to $\infty$ in its
two ends, and along consecutive critical directions. Its boundary consists of
a path with two unbounded critical trajectories, and possibly a finite number
of short ones. It is conformally mapped to a vertical half plane
$\left\\{w\in\mathbb{C}:\Re w>c\right\\}$ for some real $c$ by the function
$\int_{z_{0}}^{z}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt$ with suitable
choices of $z_{0}\in\mathbb{C}$ and the branch of the square root;

  * •

strip domain : It is swept by trajectories which both ends tend $\infty.$ Its
boundary consists of a disjoint union of two paths, each of them consisting of
two unbounded critical trajectories diverging to $\infty,$ and possibly a
finite number of short trajectories. It is conformally mapped to a vertical
strip $\left\\{w\in\mathbb{C}:c_{1}<\Re w<c_{2}\right\\}$ for some reals
$c_{1}$ and $c_{2}$ by the function
$\int_{z_{0}}^{z}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt$ with suitable
choices of $z_{0}\in\mathbb{C}$ and the branch of the square root.

A necessary condition for the existence of a short trajectory connecting two
finite critical points of $\varpi_{a,\theta}$ is the existence of a Jordan arc
$\gamma$ connecting them, such that

$\Re\int_{\gamma}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt=0.$ (19)

However, this condition is not sufficient in general as we will see in
theorem.9. A powerful tool which will be used in our investigation is the
following lemma :

######  Lemma 5 (Teichmüller ).

Let $\mathcal{C}$ be a closed curve formed by a finite number of parts of
horizontal and/or vertical trajectories of $\varpi_{a,\theta}$ (with their
endpoints). Let $z_{j}$ be the critical points on $\mathcal{C},$ $\theta_{j}$
be the corresponding interior angles at vertices $z_{j},$ and $n_{j}$ be the
multiplicities of $z_{j}.$Then

$\sum_{j}\left(1-\frac{\left(n_{j}+2\right)\theta_{j}}{2\pi}\right)=2+\sum_{i}m_{i}$
(20)

where $m_{i}$ are the multiplicities of critical points inside $\mathcal{C}.$

######  Remark 6.

As an immediate observation from lemma 5 is that $\varpi_{a,\theta}$ has at
most two short trajectories connecting two different couples of two different
finite critical points.

######  Lemma 7.

For any $a\in\mathbb{C}\setminus\mathbb{R},$ a short trajectory of
$\varpi_{a,\theta}$ if exists, it is entirely included in the half plane
bordered by $\mathbb{R}$ and containing $a.$

###### Proof.

It is straightforward that for $\Im a>0$ and
$\alpha,\beta,\gamma\in\mathbb{R},\alpha<\beta\leq\gamma<\delta:$

$0<\arg\int_{\gamma}^{\delta}\sqrt{p_{a}\left(t\right)}dt<\arg\int_{\alpha}^{\beta}\sqrt{p_{a}\left(t\right)}dt<\frac{\pi}{2},$

where the arguments are taken in $\left[-\pi,\pi\right[,$ and the square roots
such that $\sqrt{1}=1.$ It follows that the two properties

$\displaystyle\Re\left(e^{i\theta}\int_{\alpha}^{\beta}\sqrt{p_{a}\left(t\right)}dt\right)$
$\displaystyle=0,$
$\displaystyle\Re\left(e^{i\theta}\int_{\gamma}^{\delta}\sqrt{p_{a}\left(t\right)}dt\right)$
$\displaystyle=0$

cannot hold together.

Suppose that a short trajectory of $\varpi_{a,\theta}$ connects $z=1$ to
$z=a,\Im a>0,$ going through some point $b$ in the real line, for example
$b<-1.$ If a critical trajectory of $\varpi_{a,\theta}$ cuts the real line in
some point $c,$ then $\alpha=c,\beta=b,$ and $\gamma=1$ give the previous
contradiction. If not, the two remaining critical trajectories should diverge
to $\infty$ in the directions $D_{\theta,1}$ and $D_{\theta,2}.$ Lemma.5
insures that two remaining critical trajectories emerging from $z=1$ should
diverge to $\infty$ in the lower half plane, one of them, cuts the real line
in some $\gamma>1.$ Again, $\alpha=b,\beta=1,$ and $\gamma$ give the same
contradiction. ∎

######  Notation 8.

For $\theta\in\left[0,\pi/4\right],$ we denote by

  * •

$S_{\pm 1,\theta}:$ the sets $\Sigma_{\pm 1,\theta}$ minus the arc starting at
$e_{\theta}$ and diverging to $\infty$ in the upper half plane;

  * •

$S_{\theta}:$ the part of $\Sigma_{\theta}$ starting at $t_{\theta}$ and
diverging to $\infty.$

  * •

$n_{\theta}:$ the finite number ($n_{0}=8,n_{arct\left(0.5\right)/2}=10,$and
$n_{\pi/4}=9$) of the connected components
$\Omega_{1},...,\Omega_{n_{\theta}}$ of $\mathbb{C}\setminus\left(S_{\pm
1,\theta}\cup S_{\theta}\right).$

The main result of this section is the following :

######  Theorem 9.

Let $\theta\in\left[0,\pi/2\right[.$ In each of the domains
$\Omega_{1},...,\Omega_{n_{\theta}},$ the critical graph $\Gamma_{a,\theta}$
of the quadratic differential $\varpi_{a,\theta}$ has the same structure; it
splits the Riemann sphere into five half-plane and two strip domains.
Moreover, $\varpi_{a,\theta}$

  * •

has a short trajectory connecting $z=\pm 1$ if and only if $a\in S_{\theta};$
it has a short trajectory connecting $z=\pm 1$ to $z=a$ if and only if $a\in
S_{\pm 1,\theta}.$ In all these cases, $\Gamma_{a,\theta}$ splits the Riemann
sphere into five half-plane domains and exactly one strip domain;

  * •

It has a tree (juxtaposition of two short trajectories) with summit
$t_{\theta};$ it has a tree with summit $z=-1$ or $z=+1$ respectively for
$\theta\in$ $\left[0,\pi/8\right[$ and $\theta\in$
$\left[3\pi/8,\pi/2\right[.$ In these cases, $\Gamma_{a,\theta}$ splits the
Riemann sphere into five half-plane domains.

In particular, any change of a critical graph structure should pass by a
critical graph with at least one short trajectory.

###### Proof.

We begin by the observation that if $\gamma$ is a short trajectory of
$\varpi_{a,\theta}$ connecting $\pm 1,$ then $z=a$ belongs to the part of
$\Sigma_{\theta}$ included in the exterior of the open bounded domain bordered
by $\gamma\cup\left[-1,1\right];$ otherwise, all critical trajectories
emerging from $a$ should cut the segment $\left[-1,1\right],$ and we get the
same contradiction as in the proof of lemma.7. It follows that $\gamma$ is
homotopic to $\left[-1,1\right]$ in $\mathbb{C}\setminus\left\\{a\right\\},$
and then $a\in\Sigma_{\theta}.$ Let parametrize $\Sigma_{\theta},$ for
$\theta\in\left]0,\frac{\pi}{4}\right],$ by :

$\Sigma_{\theta}=\left\\{a=a\left(s\right);s\geq
0\right\\}\subset\left\\{z\in\mathbb{C}:\Re z\leq 0,\Im z\geq
0\right\\},a\left(0\right)=1.$

Since $\left]-1,1\right[\cap\Sigma_{\pm 1,\theta}=\emptyset,$ the quadratic
differential $\varpi_{s_{\theta},\theta}$ has no short trajectory connecting
$\pm 1;$ more precisely, if $\varpi_{s_{\theta},\theta}$ has only one critical
trajectory emerging from $z=s_{\theta}$ that diverges to $\infty$ in lower
half-plane in the direction $D_{\theta,k},k\in\left\\{3,4\right\\},$ then,
lemma.5 asserts that two critical trajectories $\gamma_{1}$ and $\gamma_{-1}$
emerging respectively from $z=1$ and $z=-1,$ should diverge to $\infty$ in the
same direction $D_{\theta,k}.$ From the local behavior of the trajectories at
$\infty,$ there exists a neighborhood $\mathcal{V}$ of $\infty,$ such that any
vertical trajectory in $\mathcal{V}$ have its two rays diverging to $\infty$
following the two adjacent critical directions $D_{\theta,k}^{\perp}$ and
$D_{\theta,\left(k+1\right)\operatorname{mod}(5)}^{\perp}.$ Let be $\sigma$
such a vertical trajectory. It is obvious that $\sigma$ intersects
$\gamma_{1}$ and $\gamma_{-1}$ respectively in two points $z=m_{1}$ and
$z=m_{-1}.$ Consider the path $\gamma$ connecting $z=1$ and $z=-1$ formed by,
the piece of $\gamma_{1}$ from $z=1$ to $z=m_{1},$ the piece of $\sigma$ from
$z=m_{1}$ to $z=m_{-1},$ and the piece of $\gamma_{-1}$ from $z=m_{-1}$ to
$z=-1.$ Integrating along $\gamma,$ we get

$\displaystyle\Re\int_{\gamma}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt$
$\displaystyle=\Re\int_{1}^{m_{1}}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt+\Re\int_{m_{1}}^{m_{-1}}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt+\Re\int_{m_{-1}}^{-1}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt$
(21)
$\displaystyle=\Re\int_{m_{1}}^{m_{-1}}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt\neq
0;$

which violates (19), more precisely

$\Re\int_{\gamma}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt=\Re\int_{\left[-1,1\right]}e^{i\theta}\left(\sqrt{p_{a}\left(t\right)}\right)_{-}dt=0.$

Thus, critical trajectories of $\varpi_{s_{\theta},\theta},$ emerging from
$s_{\theta},$ say
$\gamma_{\theta,1}\left(0\right),\gamma_{\theta,3}\left(0\right)$ and
$\gamma_{\theta,4}\left(0\right),$ diverge to $\infty$ respectively in the
directions $D_{\theta,1},D_{\theta,3}$ and $D_{\theta,4}.$ Moreover, by
continuity (in the Haussdörff metric) of the critical graph
$\Gamma_{s_{\theta},\theta}$ with respect to $s\geq 0,$ there exists
$s_{0}>0,$ ($s_{0}$ is taken maximum) such that, for $0<s<s_{0},$
$\gamma_{\theta,3}\left(s\right)$ and $\gamma_{\theta,4}\left(s\right)$
intercept the real line respectively in
$t_{3}\left(s\right)\in\left]-1,s_{\theta}\right[$ and
$t_{4}\left(s\right)\in\left]s_{\theta},1\right[;$ see Fig.theo.Clearly, if
$\gamma_{\theta,3}\left(s\right)$ (resp. $\gamma_{\theta,4}\left(s\right)$)
changes its behavior for some $s>0,$ then it should go through $z=-1$ for some
$s_{-1}>0,$ (resp. $s_{1}>0$) and necessary,
$a\left(s_{-1}\right)\in\Sigma_{-1,\theta}$ (resp.
$a\left(s_{1}\right)\in\Sigma_{1,\theta}$). These facts, besides the same idea
of the choice of paths $\gamma$ joining $z=a$ to $\pm 1,$ leads us to prove
that, $a\left(s_{0}\right)=t_{\theta},$ $\varpi_{t_{\theta},\theta}$ has a
tree with summit $t_{\theta},$ and, $\varpi_{a,\theta}$ has no short
trajectory for any $a\in\Omega_{1}\cap\Sigma_{\theta}.$

Conversely, let $s>s_{0}$ and suppose that $\varpi_{a\left(s\right),\theta}$
has no short trajectory connecting $\pm 1.$ The critical graph of
$\varpi_{a\left(s\right),\theta}$ cannot be the same as in Fig.theo, because,
since $a\in\Omega\cup\Omega,$
$\Re\int_{-1}^{a}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt$ and
$\Re\int_{1}^{a}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt$ change their signs
both, and necessarily we get a strip domain containing $\left]-1,x\right[,$
for some $x\in$ $\left]-1,1\right],$ formed by two critical trajectories
emerging from $z=\pm 1.$ Integrating along a path as in (21), we get a similar
contradiction.

Now, for $\theta\in\left]0,\pi/8\right[,$ we consider three reals $u,v_{1},$
and $v_{2}$ such that

$x<\Re
e_{\theta}<-1;a_{1}=u+iv_{1}\in\Sigma_{\theta};a_{2}=x+iv_{2}\in\Sigma_{1,\theta}^{l}.$

Let $a=u+iv,$ $v\in\left]v_{1},v_{2}\right[.$ Taking the principal branch of
the square roots, we get :

$\lim_{x\longrightarrow-1^{+}}\int_{1}^{x}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt=\int_{1}^{-1}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt<0;$
$\displaystyle\lim_{x\longrightarrow
1^{-}}\arg\int_{1}^{x}e^{i\theta}\sqrt{p_{a}\left(t\right)}dt$
$\displaystyle=\lim_{x\longrightarrow
1^{-}}\arg\left(\frac{3}{\sqrt{2}}\left(1-x\right)^{3/2}e^{i\left(\theta+\pi\right)}\sqrt{a-1}\right)$
$\displaystyle=.$

Since $\arg\left(a\left(s\right)-1\right)$have with notations of the proof of
lemma.3, to show that, for , $\varpi_{a_{1}^{l}\left(s\right),\theta}$ has no
short trajectory for $s>s_{1},$ $a_{1}^{l}\left(s_{1}\right)=e_{\theta},$ the
segment $\Re a<\Re e_{\theta}$let

From lemma.5, the knowledge of the behavior of critical trajectories of
$\varpi_{a,\theta}$ emerging from two zeros, gives a full idea about its
critical graph $\Gamma_{a,\theta}.$ For
$i\in\left\\{1,...,n_{\theta}\right\\},$ we consider the equivalence relation
$\mathcal{R}_{i}$ in $\Omega_{i}$ : $a\mathcal{R}_{i}a^{\prime}$ if and only
if, the critical graphs $\Gamma_{a,\theta}$ and $\Gamma_{a^{\prime},\theta}$
have the same structure. In other words, for any critical trajectory of
$\varpi_{a,\theta}$ emerging from $z=-1,$ (resp. $z=+1$), there exists a
correspondent critical trajectory of $\varpi_{a^{\prime},\theta}$ that emerges
from $z=-1$ (resp. $z=+1$), and diverges to $\infty$ following the same
critical direction. Let be $\Lambda_{1},...,\Lambda_{m},m\geq 1,$ the
equivalence classes of $\mathcal{R}_{i},$ and $b_{n}$ a sequence of points in
$\Lambda_{j},1\leq j\leq m,$ converging to some $b\in\Omega_{i}.$ It is
obvious, with an adequate choice of the square roots, that the sequence of
functions

$f_{n}\left(z\right)=e^{i\theta}\sqrt{\left(b_{n}-z\right)\left(1-z^{2}\right)}$

converges uniformly on compact sub-sets of $\Omega_{i}$ to the function

$f\left(z\right)=e^{i\theta}\sqrt{\left(b-z\right)\left(1-z^{2}\right)}.$

Therefore, if, for example, $\gamma_{-1,b_{n}}$ is a critical trajectory of
$\varpi_{b_{n},\theta}$ emerging from $z=-1,$ and diverging to $\infty$ in a
direction $D_{\theta,k},$ then, it should converge (in the Haussdörf metric)
to a correspondent critical trajectory $\gamma_{-1,b}$ of $\varpi_{b,\theta}$
diverging to $\infty$ in the same direction. Indeed, consider a sequence
$\left(y_{n}\right)\subset\gamma_{-1,b_{n}},$ and converging to some
$y\in\mathbb{C}.$ It is obvious then that :

$\Re\left(e^{i\theta}\int_{-1}^{y_{n}}\sqrt{\left(b_{n}-z\right)\left(1-z^{2}\right)}dt\right)=\Re\left(e^{i\theta}\int_{-1}^{y}\sqrt{\left(b-z\right)\left(1-z^{2}\right)}dt\right)=0,$

and $y\in\gamma_{-1,b}.$ Moreover, if $\gamma_{-1,b}$ diverges to $\infty,$
then it cannot follow a direction different to $D_{\theta,k},$ unless he meets
another critical trajectory, and we get a short one, which cannot hold since
$b\in\Omega_{i}.$ Hence, $b_{n}\mathcal{R}_{i}b,$ and, $\Lambda_{j}$ is a
closed sub-set of $\Omega_{i}.$ Since $\Omega_{i}$ is a non-empty connected
set, it cannot be a disjoint finite union of non-empty closed sub-sets. Thus,
$m=1,$ and all critical graphs
$\left(\Gamma_{a,\theta}\right)_{a\in\Omega_{i}}$ have the same structure. ∎

###  3.1  The case $\theta=0$

######  Lemma 10.

Let $a\in\mathbb{C}^{+}.$ Then,

  * •

no short trajectory of $\varpi_{a,0}$ connects $z=1$ and $z=-1;$

  * •

no critical trajectory of the quadratic differential $\varpi_{a,0}$ emerging
from $z=\pm 1$ intersects $\mathbb{R}\setminus\left\\{\pm 1\right\\};$

  * •

there exist at most one critical trajectory of $\varpi_{a,0}$ emerging from
$z=a,$ intersecting the real line; if it exists, the intersection is in a
unique point;

  * •

for $\Re a\leq 1,$ there is no critical trajectory of $\varpi_{a,0}$ emerging
from $z=1$ and intersecting $i\mathbb{R}^{+}.$

###### Proof.

Let be $\gamma$ a critical trajectory of $\varpi_{a,0}$ emerging for example
from $z=1$ (the case $z=-1$ is similar ) and diverging to $\infty.$ Suppose
for example that $x\in\gamma\cap\left]1,+\infty\right[.$ Then

$\Re\left(\int_{1}^{x}\sqrt{p_{a}\left(t\right)}dt\right)=\int_{1}^{x}\Re\left(\sqrt{t-a}\right)\sqrt{t^{2}-1}dt=0.$
(22)

As in the proof of lemma 4, if $\Re\left(\sqrt{t-a}\right)=0$ for some
$t_{0}>1,$ then $\left(t_{0}-a\right)<0,$ which cannot hold. By an adequate
choice of the square roots, and continuity of the function
$t\longmapsto\Re\left(\sqrt{t-a}\right),$ each of the quantities
$\Re\left(\sqrt{t-a}\right)$ and $\sqrt{t^{2}-1}$ preserve the same sign for
every $t>1,$ and the integral
$\int_{1}^{x}\Re\left(\sqrt{t-a}\right)\sqrt{t^{2}-1}dt$ cannot vanish, which
contradicts (17). If a critical trajectory emerges from $z=a,$ and intersects
$\mathbb{R}\setminus\left\\{\pm 1\right\\}$ in two different points, then,
obviously from lemma 5, they should be both on a same side with $\pm 1.$ Again
we get a contradiction by considering the integral in (22) between these two
points.

For the last point, let us begin by the case $\Re a\leq 0,$ and suppose that
such a critical trajectory $\gamma$ of $\varpi_{a,0}$ does exist. Let be
$bi,b>0$ the first intersection of $\gamma$ with $i\mathbb{R}^{+}.$ Taking the
paths of integration in $\left[0,1\right]$ and $\left[0,bi\right]$
respectively in the first and in the second integral, the arguments in
$\left[0,2\pi\right[$ and the square roots in $\left[0,\pi\right[,$ we get

$\displaystyle 0$
$\displaystyle=\Re\left(\int_{1}^{bi}\sqrt{p_{a}\left(z\right)}dz\right)=-\Re\left(\int_{0}^{1}\sqrt{p_{a}\left(z\right)}dz\right)+\Re\left(\int_{0}^{bi}\sqrt{p_{a}\left(z\right)}dz\right)$
$\displaystyle=-\Re\left(\int_{0}^{1}\sqrt{\left(1-t^{2}\right)\left(a-t\right)}dt\right)-b\Im\left(\int_{0}^{1}\sqrt{\left(1+b^{2}t^{2}\right)\left(a-ibt\right)}dt\right).$
(23)

We have for any $t\in\left[0,1\right],$

$\displaystyle\arg\left(\sqrt{\left(1-t^{2}\right)\left(a-t\right)}\right)$
$\displaystyle\in\left[\frac{\pi}{4},\frac{\pi}{2}\right],$
$\displaystyle\arg\left(\sqrt{\left(1+b^{2}t^{2}\right)\left(a-ibt\right)}\right)$
$\displaystyle\in\left[\frac{\pi}{4},\frac{3\pi}{4}\right],$

and then

$\Re\left(\int_{0}^{1}\sqrt{\left(1-t^{2}\right)\left(a-t\right)}dt\right)>0,\Im\left(\int_{0}^{1}\sqrt{\left(1+b^{2}t^{2}\right)\left(a-ibt\right)}dt\right)>0$

which violates (23).

The case $\Re a\in\left]0,1\right]$ can be treated by the same idea, with the
choice of the arguments in $\left]-\pi,\pi\right[$ and the square roots in
$\left]-\frac{\pi}{2},\frac{\pi}{2}\right[.$ ∎

In the following proposition we give a detailed description of the critical
graph $\Gamma_{a,0}$ of $\varpi_{a,0}$ byby summarizing the preceding results.
The general case of $\theta$ can be treated in the same vein but more
fastidious discussions and calculations.

######  Proposition 11.

For any $a\in\mathbb{C}^{+},$ the quadratic differential $\varpi_{a,0}$ has
five half-plane domains, and one or two strip domains, as follows :

  1. 1.

If $a$ $\in\Omega_{1},$ a strip domain, bordered by two pairs of critical
trajectories emerging from $z=-1$ and $z=a,$ and diverging to $\infty$
following the critical directions $D_{0,1}$ and $D_{0,3};$ it contains
$\left]\Delta,-1\right[,\Delta<-1.$ A second strip domain, bordered by two
pairs of critical trajectories emerging from $z=-1$ and $z=+1,$ and diverging
to $\infty$ following the critical directions $D_{0,0}$ and $D_{0,3};$ it
contains $\left]-1,1\right[.$

  2. 2.

If $a$ $\in\Sigma_{-1,0}^{l},$ a unique strip domain, bordered by the short
trajectory, two pairs of critical trajectories emerging from $z=-1$ and $z=a,$
and diverging to $\infty$ following the critical directions $D_{0,0}$ and
$D_{0,3};$ it contains $\left]-1,1\right[.$

  3. 3.

If $a$ $\in\Omega_{2},$ a strip domain in the upper half plane, bordered by
two pairs of critical trajectories emerging from $z=-1$ and $z=a,$ and
diverging to $\infty$ following the critical directions $D_{0,0}$ and
$D_{0,2}.$ A second strip domain, bordered by two pairs of critical
trajectories emerging from $z=-1$ and $z=+1,$ and diverging to $\infty$
following the critical directions $D_{0,0}$ and $D_{0,3};$ it contains
$\left]-1,1\right[.$

  4. 4.

If $a$ $\in\Sigma_{-1,0}^{r},$ a unique strip domain, bordered by two pairs of
critical trajectories emerging from $z=-1$ and $z=+1,$ and diverging to
$\infty$ following the critical directions $D_{0,0}$ and $D_{0,3};$ it
contains $\left]-1,1\right[.$

  5. 5.

If $a$ $\in\Omega_{3},$ a strip domain, bordered by two pairs of critical
trajectories emerging from $z=-1$ and $z=a,$ and diverging to $\infty$
following the critical directions $D_{0,1}$ and $D_{0,3};$ it contains
$\left]-1,\Delta\right[,$ $\Delta\in\left]-1,1\right[.$ A second strip domain,
bordered by two pairs of critical trajectories emerging from $z=+1$ and $z=a,$
and diverging to $\infty$ following the critical directions $D_{0,0}$ and
$D_{0,3};$ it contains $\left]\Delta,1.\right[$

  6. 6.

If $a$ $\in\Sigma_{1,0},$ a unique strip domain, bordered by the short
trajectory, two pairs of critical trajectories emerging from $z=-1$ and
$z=+1,$ and diverging to $\infty$ following the critical directions $D_{0,1}$
and $D_{0,3};$ it contains $\left]-1,1\right[.$

  7. 7.

if $a$ $\in\Omega_{4},$ a strip domain, bordered by two pairs of critical
trajectories emerging from $z=-1$ and $z=+1,$ and diverging to $\infty$
following the critical directions $D_{0,1}$ and $D_{0,3};$ it contains
$\left]-1,1\right[.$ A second strip domain; bordered by two pairs of critical
trajectories emerging from $z=+1$ and $z=a,$ and diverging to $\infty$
following the critical directions $D_{0,1}$ and $D_{0,4};$ it contains
$\left]1,\Delta\right[,$ $\Delta>1.$

###### Proof.

Let be $\gamma_{1}\left(t\right),\gamma_{2}\left(t\right),$ and
$\gamma_{3}\left(t\right),$ $t\geq 0,$ the three critical trajectories of
$\varpi_{a,0}$ emerging from $z=-1,$ with
$\gamma_{1}\left(0\right)=\gamma_{2}\left(0\right)=\gamma_{3}\left(0\right)=-1.$
From the local behavior of the trajectories near a simple zero and lemma.10,
we may assume that there exists
$\alpha_{k}\left(t\right)\in\left]0,\frac{2\pi}{3}\right[$ such that :

$\arg\left(\gamma_{k}\left(t\right)+1\right)=\alpha_{k}\left(t\right)+\frac{2\left(k-1\right)\pi}{3},t\longrightarrow
0^{+};k=1,2,3.$

Setting

$\displaystyle\gamma_{1}\left(t\right)+1$
$\displaystyle=r_{1}\left(t\right)e^{i\alpha_{1}\left(t\right)},r_{1}\left(t\right)>0,t>0,$
$\displaystyle a+1$
$\displaystyle=\left|a+1\right|e^{i\beta_{a}},\beta_{a}\in\left]0,\pi\right[;$

we have :

$\displaystyle\int_{0}^{t}\sqrt{p_{a}\left(\gamma_{1}\left(s\right)\right)}\gamma_{1}^{\prime}\left(s\right)ds$
$\displaystyle\sim\frac{2\sqrt{2}}{3}\left(\sqrt{a+1}\left(\gamma_{1}\left(t\right)+1\right)^{\frac{3}{2}}\right),t\longrightarrow
0^{+};$
$\displaystyle\arg\left(\int_{0}^{t}\sqrt{p_{a}\left(\gamma_{1}\left(s\right)\right)}\gamma_{1}^{\prime}\left(s\right)ds\right)$
$\displaystyle\equiv\frac{\pi}{2}\operatorname{mod}\left(\pi\right).$

It follows that

$\arg\left(\sqrt{a+1}\left(\gamma_{1}\left(t\right)+1\right)^{\frac{3}{2}}\right)\equiv\frac{\pi}{2}\operatorname{mod}\left(\pi\right),t\longrightarrow
0^{+},$

and then

$\alpha_{k}\left(0^{+}\right)=\frac{-\beta_{a}+\left(2k-1\right)\pi}{3};k\in\left\\{1,2,3\right\\}.$

Since $\beta_{a}\in\left]0,\pi\right[,$ we get

$\alpha_{1}\left(0^{+}\right)=\frac{-\beta_{a}+\pi}{3}\in\left]0,\frac{\pi}{3}\right[,$

which shows with lemma 10 that $\gamma_{1}$ and $\gamma_{2}$ belong the upper
half plane, while $\gamma_{3}$ belongs the lower half plane, and then, it
diverges to $\infty.$ The same reasoning shows that from $z=1$ emerges a
trajectory, say $\beta_{1},$ staying in the upper half plane; while the two
remaining trajectories, taken in the counter clockwise orientation, say
$\beta_{2},\beta_{3},$ they belong to the lower half plane and diverge to
$\infty.$ We deduce that $\gamma_{3}$ diverges to $\infty$ following the
critical direction $D_{0,2}$ or $D_{0,3}.$

Let us assume first that $\gamma_{2}$ diverges to $\infty.$ In the large,
$\gamma_{2}$ forms with $\gamma_{3}$ an angle equal to $\frac{2k\pi}{5}$ for a
certain $k=0,...,4.$ Consider the curve formed by $\gamma_{2}$ and
$\gamma_{3}$ as the border of domain $\Omega$ encircling
$\left]-\infty,-1\right[.$ Applying lemma 5, equality (20) becomes

$1+k=2+\sum m_{i}.$ (24)

Lemma 10 asserts that the only possible critical point in $\Omega$ is $z=a,$
in this case, $k=2;$ otherwise, $k=1.$

  1. 1.

If $k=2$ and $\gamma_{3}$ diverges to $\infty$ following $D_{0,2},$ then
$\gamma_{1}$ and $\gamma_{2}$ should diverge to $\infty$ following the same
critical direction $D_{0,0},$ which is impossible by lemma 5. Then,
$\gamma_{3}$ and $\gamma_{2}$ diverge to $\infty$ respectively following
$D_{0,3}$ and $D_{0,1},\gamma_{1}$ and $\beta_{1}$ diverge towards to $\infty$
following $D_{0,0},$ while $\beta_{2}$ and $\beta_{3}$ diverge to $\infty$
respectively following $D_{0,3}$ and $D_{0,4}.$ Critical trajectories that
emerge from $z=a,$ say, $\delta_{1},\delta_{2},$ and $\delta_{3}$ diverge to
$\infty$ respectively following $D_{0,1},D_{0,2},$ and $D_{0,3}.$ The last one
intersects the real line in some $\Delta<-1.$ In this case, $\varpi_{a,0}$ has
two strip domains, the first one contains $\left]\Delta,-1\right[,$ and it is
bordered by $\gamma_{2},\gamma_{3},\delta_{1}\ $and $\delta_{2}.$ The second
one contains $\left]-1,1\right[$, and it is bordered by
$\gamma_{1},\gamma_{3},\beta_{1}$ and $\beta_{2};$ see Fig.11.

  2. 2.

If $k=1,$ then $\Omega$ is a half plane domain of $\varpi_{a,0}$ containing
$\left]-\infty,-1\right[.$ Suppose that $\gamma_{3}$ diverges to $\infty$
following the critical direction $D_{0,2}.$ Lemma 5 imposes on $\gamma_{2}$ to
follow the critical direction $D_{0,1},$ which is impossible; indeed :

If $\gamma_{1}$ is not a short trajectory, then, again we get equality (24)
for the domain bordered by $\gamma_{1}$ and $\gamma_{2}$ in the upper half
plane. It implies that $\gamma_{1}$ diverges to $\infty$ following $D_{0,0}.$
By the other hand, $\beta_{1}$ cannot be a short trajectory meeting $z=a,$
because if the two critical trajectories emerging from $z=a$ diverge to
$\infty$ in the upper half plane, then their only possible critical direction
is $D_{0,0},$ which violates lemma 5; moreover, lemma 10 guarantees that none
of these critical trajectories can intersect the real line. Thus,
$\beta_{1}$diverges to $\infty$ following $D_{0,0}.$ Again, lemma 5 applied on
the domain containing $z=0,$ bordered by $\beta_{1},\beta_{1},\gamma_{1}$ and
$\gamma_{3},$ with lemma 10 require that $z=a$ should belong to the domain
containing $\left]1,+\infty\right[$, bordered by $\beta_{1}$ and $\beta_{2};$
and we get the same contradiction.

If $\gamma_{1}$ is a short trajectory meeting $z=a,$ then, considering the two
remaining critical trajectories emerging from $z=a,$ we get the same previous
contradiction.

We just proved that $\gamma_{3}$ diverges to $\infty$ following $D_{0,3},$
$\gamma_{2}$ diverges to $\infty$ following $D_{0,2},$ and then, $\beta_{2}$
and $\beta_{3}$ diverge to $\infty$ following respectively $D_{0,3}$ and
$D_{0,4}.$ There are three possibilities for $\gamma_{1},$ either it is a
short trajectory, or it diverges to $\infty$ following $D_{0,0}$ or $D_{0,1}.$

    1. (a)

If $\gamma_{1}$ is a short trajectory (meeting $z=a$), then the two remaining
critical trajectories emerging from $z=a$ diverge to $\infty$ following
respectively $D_{0,0}$ and $D_{0,1};$ $\beta_{1}$ diverge to $\infty$
following $D_{0,0}.$ In this case, $\left]-1,1\right[$ is included in a strip
domain of $\varpi_{a,0}$ defined by
$\beta_{1},\beta_{2},\gamma_{3},\gamma_{1},$ and the critical trajectory
emerging from $z=a$ and diverging to $\infty$ following $D_{0,0};$ see Fig.11.

    2. (b)

If $\gamma_{1}$ diverges to $\infty$ following $D_{0,0},$ then from lemma 5,
$z=a$ belongs to the domain in the upper half plane bordered by $\gamma_{1}$
and $\gamma_{2},$ critical trajectories emerging from $z=a$ diverge to
$\infty$ respectively following $D_{0,0},D_{0,1},$ and $D_{0,2};$ $\beta_{1}$
diverges to $\infty$ following $D_{0,0}.$ In this case, $\varpi_{a,0}$ has two
strip domains, the first, bordered by $\gamma_{1},\gamma_{2},$ and critical
trajectories emerging from $z=a$ that follow $D_{0,0}$ and $D_{0,2}.$ The
second, bordered by $\beta_{1},\beta_{2},\gamma_{3},$ and $\gamma_{1};$ it
contains $\left]-1,1\right[;$ see Fig.11.

    3. (c)

If $\gamma_{1}$ diverges to $\infty$ following $D_{0,1},$ then, either
$\beta_{1}$ is a short trajectory, or it diverges to $\infty$ following
$D_{0,0}$ or $D_{0,1};$

      1. i.

if $\beta_{1}$ is a short trajectory, the two remaining critical trajectories
emerging from $z=a$ diverge to $\infty$ respectively following
$D_{0,0},D_{0,1}.$ In this case, $\left]-1,1\right[$ is included in a strip
domain of $\varpi_{a,0}$ defined by
$\gamma_{2},\gamma_{3},\beta_{1},\beta_{2},$ and the critical trajectory
emerging from $z=a$ and diverging to $\infty$ following $D_{0,1};$ see Fig.11.

      2. ii.

if $\beta_{1}$ diverges to $\infty$ following $D_{0,0},$ then two critical
trajectories emerging from $z=a,$ say $\delta_{1},\delta_{2},$ diverge to
$\infty$ following respectively $D_{0,0},D_{0,1};$ the remaining one, say
$\delta_{3},$ diverges to $\infty$ following $D_{0,3},$ it intersects the real
line in some $\Delta\in$ $\left]-1,1\right[.$ In this case, the sets
$\left]-1,\Delta\right[$ and $\left]\Delta,1\right[$ are included in two strip
domains, respectively bordered by
$\gamma_{1},\gamma_{3},\delta_{2},\delta_{3},$ and
$\beta_{1},\beta_{2},\delta_{1},\delta_{2};$ see Fig.11.

      3. iii.

if $\beta_{1}$ diverges to $\infty$ following $D_{0,1},$ then from lemma 5,
$z=a$ belongs the upper half plane bordered by $\beta_{1}$ and $\beta_{3},$
two critical trajectories emerge from $z=a,$ say $\delta_{1},\delta_{2},$
diverge to $\infty$ following respectively $D_{0,0}$ and $D_{0,1};$ the third
one diverges to $\infty$ following $D_{0,4};$ it intersects the real line in
some $\Delta>1;$ it forms with $\beta_{1},\beta_{3},$ and $\delta_{2},$ a
strip domain containing $\left]1,\Delta\right[.$ The set $\left]-1,1\right[$
is included in a second strip domain bordered by
$\gamma_{1},\gamma_{3},\beta_{1},$ and $\beta_{2};$ see Fig.11.

If $\gamma_{2}$ is a short trajectory meeting $z=a,$ and $\gamma_{3}$ diverges
to $\infty$ following $D_{0,2},$ then the two critical trajectories emerging
from $z=a$ diverge to $\infty$ following $D_{0,0}$ and $D_{0,1};$ one of them
diverges together with $\gamma_{1}$ following $D_{0,0,}$ which violates lemma
5. Thus, $\gamma_{3}$ diverges to $\infty$ following $D_{0,3},$ two critical
trajectories emerging from $z=a$ diverge to $\infty$ following $D_{0,1}$ and
$D_{0,2},$ $\gamma_{1}$ and $\beta_{1}$ diverge to $\infty$ following
$D_{0,0}.$ The set $\left]-1,1\right[$ is included in the unique strip domain
of $\varpi_{a,0}$ defined by $\beta_{1},\beta_{2},\gamma_{3},$and
$\gamma_{1}$; see Fig.11.

To finish the proof, the previous reasoning shows is that the knowledge of the
behavior of the critical trajectories emerging from $z=a,$ suffices to
determine the whole structure of the critical graph $\Gamma_{a,0}$ of
$\varpi_{a,0}.$ It follows then, in order to see in each case the belonging of
$a$ in which of the sets defined in notation.4, it is enough to see the
structure of $\Gamma_{a,0}$ when $\Im a\longrightarrow 0^{+}.$ For example, it
is straightforward to check that for $a<-1,$ the critical trajectories
emerging from $z=a,$ diverge to $\infty$ following the critical directions
$D_{0,1},D_{0,2}$ and $D_{0,3};$ by continuity of the structure of the
trajectories, we deduce the first case $a\in\Omega_{1}$; see Fig.11. ∎

Figure 4: $\Gamma_{-2.5+0.7i,0}$

Figure 5: $\Gamma_{-1.22i,0}$

Figure 6: $\Gamma_{-0.8+2i,0}$

Figure 7: $\Gamma_{1.3+2.4i,0}$

Figure 8: $\Gamma_{0.5+i,0}$

Figure 9: $\Gamma_{2+i,0}$

Figure 10: $\Gamma_{-1.95+0.87i,0}$

Figure 11: $\Gamma_{-2,0}$

###  3.2 The case $\theta=\pi/4$

In the following proposition, we describe the critical graph of
$\varpi_{a,\pi/4}$ when it has at least one short trajectory. The symmetry of
$\Gamma_{a,\pi/4}$ and $\Gamma_{-\overline{a},\pi/4}$ with respect to the
imaginary axis let us restrict the investigation for $a\in\Sigma_{1,\pi/4}\cup
S_{\pi/4}.$ The proof is a same as in proposition.11.

######  Proposition 12.

Let $a\in\Sigma_{1,\pi/4}\cup S_{\pi/4}.$ Then, besides five half-plane
domains, the configuration domain of

$\varpi_{a,\pi/4}$ is as follow :

  1. 1.

For $a\in\Sigma_{1,\pi/4}^{r},$ exactly one strip domain; it includes
$\left]-1,1\right[$ and it is bounded by, the short trajectory of
$\varpi_{a,\pi/4},$ two critical trajectories diverging to $\infty$ following
the direction $D_{\pi/4,1},$ and two critical trajectories diverging to
$\infty$ following the direction $D_{\pi/4,3};$ see Fig.17;

  2. 2.

for $a\in\Sigma_{1,\pi/4}^{l}\cap\left\\{z:\Re z>0\right\\},$ exactly one
strip domain, it includes $\left]-1,\delta\right[$ for some
$\delta\in\left]-1,1\right],$and it is bounded by two critical trajectories
diverging to $\infty$ following the direction $D_{\pi/4,1},$ and two critical
trajectories diverging to $\infty$ following the direction $D_{\pi/4,3};$ see
Fig.17;

  3. 3.

$\varpi_{a,\pi/4}$ has a tree with summit $a_{\pi/4},$ and no strip domain;
see Fig.17;

  4. 4.

for $a\in\Sigma_{1,\pi/4}^{l}\cap\left\\{z:\Re z<0\right\\},$ exactly one
strip domain that includes $\left]\delta,1\right[$ for some
$\delta\in\left[-1,1\right[,$ and it is bounded by, the short trajectory of
$\varpi_{a,\pi/4},$ two critical trajectories diverging to $\infty$ following
the direction $D_{\pi/4,2},$ and two critical trajectories diverging to
$\infty$ following the direction $D_{\pi/4,4};$ see Fig.17,Fig.17;

  5. 5.

for $a=xi,x>\left|a_{\pi/4}\right|,$ $\varpi_{a,\pi/4}$ has exactly one short
trajectory connecting $\pm 1;$ it is included in $\left\\{z:-1<\Re z<1,0<\Im
z<\left|a_{\pi/4}\right|\right\\};$ exactly one strip domain that is bounded
by the short trajectory of $\varpi_{a,\pi/4},$ two critical trajectories
diverging to $\infty$ following the direction $D_{\pi/4,0},$ and two critical
trajectories diverging to $\infty$ following the direction $D_{\pi/4,2};$ see
Fig.17.

The short trajectory of $\varpi_{a,\pi/4}$ is a path between two half plane
domains in all cases excepted the case.2, where it joins the strip domain and
a half plane domain.

Figure 12: $\Gamma_{1.4+0.85i,\frac{\pi}{4}}$

Figure 13: $\Gamma_{0.3+0.31i,\pi/4}$

Figure 14: $\Gamma_{0.462i,\pi/4}$

Figure 15: $\Gamma_{-0.5+0.77i,\pi/4}$

Figure 16: $\Gamma_{-1+1.19i,\pi/4}$

Figure 17: $\Gamma_{0.7i,\pi/4}$

######  Remark 13.

The case of real cubic polynomial quadratic differential $\varpi$ with three
real zeros is trivial; when the polynomial has two conjugate complex zeros :

$\varpi_{x_{0},b}=\left(z-x_{0}\right)\left(z-b\right)\left(z-\overline{b}\right)dz^{2},x_{0}\in\mathbb{R},b\in\mathbb{C}\setminus\mathbb{R},$

by the linear change of variable

: $z=i\Im\left(b\right)y+\Re b,$ the quadratic differential
$\lambda\varpi,\lambda\in\mathbb{C}^{\ast},$ has the same critical graph then

$-e^{2i\left(\alpha+\pi/4\right)}\left(y^{2}-1\right)\left(y-i\frac{\Re
b-x_{0}}{\Im b}\right)dy^{2},\alpha\in\left[0,\pi\right[.$

In particular, $\varpi_{x_{0},b}$ has a tree if, and only if,

$i\frac{\Re b-x_{0}}{\Im b}=t_{\pi/4}.$

##  4  Eigenvalues problems from cubic oscillator

###  4.1 Preliminary results from complex WKB

In this section, we give a short review of some ingredients and notations in
complex WKB, for more details one can see ([25, §3 ], [24]).

####  4.1.1 Topology of Stokes complex

Let $\Lambda=\left\\{-1,1,a\right\\}$ the set of zeros of
$P(z)=(z^{2}-1)(z-a)$ called turning points. The Stokes lines associated to
the ODE (3) are the horizontal trajectories of the quadratic differential

$\varpi(z)dz^{2}=-\lambda^{2}P(z)dz^{2}.$

The Stokes graph (the union of the Stokes lines) is then the critical graph of
the quadratic differential $\varpi(z)dz^{2}$. Notice that the critical graph
of a quadratic differential is invariant under multiplication by a positive
scalar, we deduce from (3) that the Stokes lines are exactly the horizontal
trajectories of

$\varpi(z)dz^{2}=\varpi_{a,2\arg\lambda}(z)dz^{2}.$

The components of the complement of the union of turning points and Stokes
lines will be called the Stokes regions. These regions are unbounded and
simply connected. The closure of each Stokes regions (in $\mathbb{C}$) is
either of half-plane type $(H_{k})_{0\leq k\leq 4}$ (ordered always
anticlockwise) or strip (=band) type $(B_{k})_{0\leq k\leq 1}$. In the general
case if $\deg P=d$, then the Stokes lines divide up the $z$-plane into $d+2$
domains of half plane type and $N$ strip domains such that $0\leq N\leq d-1$.
The connected components of the Stokes graph are called the Stokes complexes.

####  4.1.2 WKB solutions

Let

$h(\lambda,z_{0},z)=\lambda\int_{z_{0}}^{z}\sqrt{(t-a)(t^{2}-1)}dt,z_{0}\in\Lambda.$

A domain $D$ in the complex $z$-plane is called canonical if there is a
holomorphic branch of $h(\lambda,z_{0},z)$ in $D$ that maps $D$ onto the whole
complex plane with a finite number of vertical cuts. The domain $D$ is simply
connected and contains no turning points and $\partial D$ consists of Stokes
lines (the pre-images of the sides of the cuts). A canonical domain is the
union of two domains of half-plane type and up to two strip-domains.

Let $\varepsilon>0$ be arbitrary. We denote $D_{\varepsilon}$ for the preimage
of $h(D)$ with $\varepsilon$-neighborhoods of the cuts and
$\varepsilon$-neighborhoods of the turning points removed. A canonical path in
$D$ is a path such that $Re(h(\lambda,z_{0},z))$ is monotone along the path.
For example the anti-Stokes lines are canonical paths. For every point $z$ in
$D_{\varepsilon}$, there are always canonical paths
$\gamma_{\varepsilon}^{+}(z)$ and $\gamma_{\varepsilon}^{-}(z)$ from $z$ to
$\infty$, such that $\Re(h(\lambda,z_{0},z))\rightarrow+\infty$ and
$\Re(h(\lambda,z_{0},z))\rightarrow-\infty$, respectively.

With $D_{\varepsilon}$, $\gamma_{\varepsilon}^{-}$ and
$\gamma_{\varepsilon}^{+}$ as above and the sector
$\Omega_{\alpha}^{\delta}=\left\\{\lambda\in\mathbb{C}^{\ast};\text{
}\left|\arg\lambda-\alpha\right|\leq\delta;\text{
}\left|\lambda\right|\geq\lambda_{0}>0\right\\}$, up to constant multiple,
equation (3) has a unique solutions $y_{1}(z,\lambda)$ and $y_{2}(z,\lambda)$,
such that

$\begin{cases}y_{1}(z,\lambda)=(P(z))^{-\frac{1}{4}}\exp(h(\lambda,z_{0},z))[1+\frac{\phi_{1}(z,\lambda)}{\lambda}]\\\
{y_{2}(z,\lambda)=(P(z))^{-\frac{1}{4}}\exp(-h(\lambda,z_{0},z))[1+\frac{\phi_{2}(z,\lambda)}{\lambda}]}\par\end{cases}$
(25)

with dual asymptotic behaviour, that means this is true if:

  * •

$z\rightarrow\infty$, $z\in\gamma_{\varepsilon}^{\pm}$ so that
$\Re(h(\lambda,z_{0},z))\rightarrow\pm\infty$ for $\lambda$ fixed in
$\lambda\in\Omega_{\alpha}^{\delta}$.

  * •

$\lambda\rightarrow+\infty$ and $\lambda\in\Omega_{\alpha}^{\delta}$,
uniformly for $z\in D_{\varepsilon}$.

######  Remark 14.

  1. 1.

We mention only the principal part of the asymptotic.

  2. 2.

If $y_{l}(z,\lambda)$ $(l=1,2)$ is decaying (or blowing-up) exponentially
along a canonical path $\gamma$ in $D_{\epsilon}$, so it decays (blows-up)
exponentially along any path homotopic to $\gamma$ in $D_{\epsilon}.$ We call
so this solution subdominant (dominant) in $D_{\epsilon}.$

  3. 3.

If $y_{l}(z,\lambda)$ $(l=1,2)$ is subdominant in $H_{k}$ so it will be
dominant in $H_{j}$ for all $j\neq k.$

  4. 4.

We will not deal with the proof of this classical result. We mention
([30],[22],[21])for a proof based on Liouville transformation of the ODE; and
([25],[26]) using a matrix gauge transformation to a differential system.

####  4.1.3 Elementary basis

We note the Stokes data $(D,l,z_{0})$ for: $D$ a canonical domain, $l$ a
Stokes line in $D$ and $z_{0}$ a turning point. We select the branch of
$h(\lambda,z_{0},z)$ in $D$ such that $\Im[h(\lambda,z_{0},z)]>0$ for $z\in
l$. The elementary basis $\left\\{u(z),v(z)\right\\}$ associated to
$(D,l,z_{0})$ is uniquely determined by

$\begin{cases}{u(z)=cy_{1}(z,\lambda)\text{, and \ }v(z)=cy_{2}(z,\lambda)}\\\
{\left|c\right|=1\text{, \ and }\arg(c)=\underset{z\rightarrow z_{0},z\in
l}{\lim}\arg((q(z))^{\frac{1}{4}})}\end{cases}$ (26)

where $y_{1}(z,\lambda)$ and $y_{2}(z,\lambda)$ are given in (25).

####  4.1.4 Transition matrices

Let’s $(D,l,z)_{k}$ and $(D,l,z)_{j}$ be two Stokes data and
$\beta_{k,j}=\left\\{u_{k,j}\text{, }v_{k,j}\right\\}$ their corresponding
elementary basis. The matrix $\Omega_{kj}$ which changes the basis $\beta_{k}$
to $\beta_{j}$ is called transition matrix from $\beta_{k}$ to
$\beta_{j}.$It’s clear that

$\Omega_{kj}=\Omega_{rj}\Omega_{kr}\text{ , }\Omega_{kj}^{-1}=\Omega_{jk}$

The classification made by Fedyoruk ([25, page 98-100]) gives three type of
"elementary"transition matrix. Any transition matrix from one elementary
(F.S.S) to another is the product of a finite number of this elementary
transition matrix.

  1. 1.

The transition matrix for $(D,l,z_{0})\rightarrow(D,l,z_{1})$. This transition
matrix exist only for a finite Stokes line remaining in the same canonical
domain. It’s given by

$\Omega(\lambda)=\exp(i\sigma)\left(\begin{array}[]{cc}0&\exp(-i\left|\lambda\right|\alpha)\\\
\exp(i\left|\lambda\right|\alpha)&0\end{array}\right)\text{,
}\alpha=\left|\varphi(z_{0},z_{1})\right|\text{,
}\exp(i\sigma)=\frac{c_{2}}{c_{1}}$

  2. 2.

The transition matrix for $(D,l_{0},z_{0})\rightarrow(D,l_{1},z_{1})$. Here
the rays $h(l_{0})$ and $h(l_{1})$ are directed to one side and $l_{0}$ lies
on the left of $l_{1}$. This is the transition from one turning point to
another along an anti-Stokes line, remaining in the same domain $D.$

$\Omega(\lambda)=\exp(i\sigma)\left(\begin{array}[]{cc}\exp(-\lambda\xi)&0\\\
0&\exp(\lambda\xi)\end{array}\right)\text{, }\xi=\varphi(z_{0},z_{1})\text{,
}\Re(\lambda\xi)>0\text{, }\exp(i\sigma)=\frac{c_{2}}{c_{1}}$

  3. 3.

The transition matrix $(D_{0},l_{0},z_{0})\rightarrow(D_{1},l_{1},z_{0}).$ Let
$\left\\{l_{0}\text{, }l_{1}\text{, }l_{2}\right\\}$ be Stokes lines starting
at $z_{0}$, and let $l_{j+1}$ lie to the left to $l_{j}$ (the order is
counter-clockwise and indexed thus $4=1,...$). We choose the canonical domain
$D_{j}$ on the left of $l_{j}$ coincides with the part of $D_{j+1}$ on the
right of $l_{j+1}$. Then

$\left\\{\begin{array}[]{c}\Omega_{j,j+1}(\lambda)=\exp(-\frac{i\pi}{6})\left(\begin{array}[]{cc}0&\alpha_{j,j+1}^{-1}(\lambda)\\\
1&i\alpha_{j+1,j+2}(\lambda)\end{array}\right)\\\
\alpha_{j,j+1}(\lambda)=1+O(\lambda^{-1})\text{, \ \ }1\leq j\leq 3\\\
\alpha_{1,2}(\lambda)\alpha_{2,3}(\lambda)\alpha_{3,1}(\lambda)=1\text{, and
}\alpha_{j,j+1}(\lambda)\alpha_{j+1,j}(\lambda)=1\end{array}\right.$ (27)

######  Remark 15.

  1. 1.

The total classification of transition matrices gives four type of such
matrices (see [25]). We need to compute the transitions matrices only to
within $O(\lambda^{-1})$ and so it’s sufficient to work with this three type.

  2. 2.

The relations in 27 are direct conclusions from the fact that
$\Omega_{3,1}\Omega_{2,3}\Omega_{1,2}=I$ and
$\Omega_{j,j+1}^{-1}=\Omega_{j+1,j}$ , respectively.

Let now $H^{\prime}$ and $H^{\prime\prime}$ be non admissible half plane. From
(25) there exist a unique solution $y_{H^{\prime}}$ subdominant in
$H^{\prime}$: $y_{H^{\prime}}(z,\lambda)\rightarrow 0$ as $z\rightarrow\infty$
along any canonical path in $H^{\prime}$. Similarly, by (25) there exist a
unique solution $y_{H^{\prime\prime}}$ subdominant in $H^{\prime\prime}$:
$y_{H^{{}^{\prime\prime}}}(z,\lambda)\rightarrow 0$ as $z\rightarrow\infty$
along any canonical path in $H^{\prime\prime}$. If $y$ is a general solution
of an eigenvalue problem, then for some constants
$c^{\prime},c^{\prime\prime}$ we have
$y(z,\lambda)=c^{\prime}y_{H^{\prime}}(z,\lambda)=c^{\prime\prime}y_{H^{\prime\prime}}(z,\lambda)$.
Let $\Omega^{\prime\prime\prime}(\lambda)$ be the transition matrix connecting
two canonical domains containing $H^{\prime}$ and $H^{\prime\prime}$ and let

$\binom{a(\lambda)}{b(\lambda)}=\Omega^{\prime\prime\prime}(\lambda)\binom{0}{1}$

then we have $b(\lambda)=0$ which is the equation that determines the
eigenvalues $\lambda_{n}$. To calculate $\Omega^{\prime\prime\prime}(\lambda)$
and hence $b(\lambda)$ we have to write this matrix as a product of finitely
many elementary transition matrices connecting $H^{\prime}$ to
$H^{\prime\prime}.$

###  4.2 Eingenvalue problems related to the cubic oscillator

####  4.2.1 Admissible half planes

For $\theta_{0}\in[0,\frac{\pi}{2}[$ we denote
$L(\theta_{0})=\left\\{\lambda\in\mathbb{C}^{\ast};\text{
}\lambda=\rho\exp(i\theta_{0});\text{ }\rho>0\right\\}$. We introduce the
admissibility of half plane domains in order to classify eigenvalue problems:

######  Definition 16.

A differentiable curve $\gamma:[0,1]\rightarrow\mathbb{C}$ is said to be
admissible if:

  * •

$\gamma$ is injective on $[0,1[$

  * •

$z_{i}\notin\gamma([0,1])$ for all $z_{i}\in\Lambda$

  * •

$\Re h(\lambda,\gamma(0),\gamma(t))$ is a monotone function of $t\in[0,1]$

We say that two half plane $H_{i}$ and $H_{k}$ are admissible if for every
$x_{i}\in H_{i}$, $x_{k}\in H_{k}$ there exist an admissible path $\gamma$
such that $\gamma(0)=x_{i}$ and $\gamma(1)=x_{k}.$

######  Remark 17.

  1. 1.

This definition is adopted from ([23]). The author give a global
classification of critical graph in the case $\deg P=3$, but he did not
studied the dependence of critical graph on a complex parameter which is the
key of the eigenvalue problem’s classification.

  2. 2.

For example two adjacent (share a stokes line) half plane are admissible.

  3. 3.

Notice that, from the classification of the Stokes graph, two non admissible
half plane exist if and only if there exist a finite Stokes line. Recall that
a canonical domain is swept by exactly two half plane (left and right) and up
to two band domains. We easily deduce that this two half plane are admissible
and reciprocally if two half plane are admissible then there exist a canonical
domain that it contains.

####  4.2.2 Classification of Eingen-value problem

Recall the notations from the section (2):

$\begin{array}[]{c}\Sigma_{\theta_{0}}=\Sigma^{\prime}_{\theta_{0}}\cup\Sigma_{-1,\theta_{0}}\cup\Sigma_{1,\theta_{0}}\\\
\Sigma_{1,\theta_{0}}=\Sigma_{1,\theta_{0}}^{l}\cup\Sigma_{1,\theta_{0}}^{r}\\\
\Sigma_{-1,\theta_{0}}=\Sigma_{-1,\theta_{0}}^{l}\cup\Sigma_{-1,\theta_{0}}^{r}\\\
\Sigma_{-1,\theta_{0}}\cap\Sigma_{1,\theta_{0}}\subseteq\left\\{a_{\theta_{0}},e_{\theta_{0}}\right\\}\end{array}$

######  Definition 18.

We say that $\theta_{0}$ is an accumulation direction to the ODE:

$-y^{\prime\prime}(z)+\lambda^{2}(z-a)(z^{2}-1)y(z)=0$ (28)

for $a\in\mathbb{C}$ and $\lambda\in L(\theta_{0})$, if it exists a pair of
distinct half planes $(H^{\prime},H^{\prime\prime})$ such that the (28) admit
a subdominant solution (non-trivial) in $H^{\prime}$ and $H^{\prime\prime}$.

We will give a necessary and sufficient conditions on $a$, such that
$\theta_{0}$ will be an accumulation direction to (28). We begin our study
with this useful lemma which describe the topology of
$\left(\Sigma_{\theta_{0}}\right)_{\theta_{0}\in[0,\pi[}$:

######  Lemma 19.

If $\alpha\neq\theta_{0}$, $\alpha\in[0,\frac{\pi}{2}[$ then:

$\begin{array}[]{c}\Sigma_{-1,\theta_{0}}\cap\Sigma_{-1,\alpha}=\left\\{-1\right\\}\\\
\Sigma_{1,\theta_{0}}\cap\Sigma_{1,\alpha}=\left\\{1\right\\}\\\
\Sigma^{\prime}_{\theta_{0}}\cap\Sigma^{\prime}_{\alpha}=\varnothing\end{array}$

###### Proof.

To prove the first part of the lemma it is sufficient to see, from the
definition of $\Sigma_{1,\theta_{0}}$ for example, that there exist a unique
value of $\theta_{0}$ in $[0,\pi[$ such that the two complex number
$\exp(i\theta_{0})$ and $Z=\int_{1}^{a}\sqrt{P\left(z\right)}dz$ have real
part zeros for their product and so
$\Sigma_{1,\theta_{0}}\cap\Sigma_{1,\alpha}=\left\\{1\right\\}$. By the same
way $\Sigma_{-1,\theta}\cap\Sigma_{-1,\alpha}=\left\\{-1\right\\}$. Finally,
$\Sigma^{\prime}_{\theta_{0}}$ originate from the point $s_{\theta_{0}}$ in
the real axis and passes throw
$\Sigma_{1,\theta_{0}}\cap\Sigma_{-1,\theta_{0}}\subseteq\left\\{a_{\theta_{0}},e_{\theta_{0}}\right\\}$,
so by the previous argument, it’s easy to deduce that
$\Sigma^{\prime}_{\alpha}$ and $\Sigma^{\prime}_{\theta_{0}}$ had no
intersection. ∎

######  Proposition 20.

$\theta_{0}$ is not an accumulation direction to
$-y^{\prime\prime}(z)+\lambda^{2}(z-a)(z^{2}-1)y(z)=0$ for all $\lambda\in
L(\theta_{0})$ and $a\in K$, where $K$ is a compact subset
$\mathbb{C}\diagdown\Sigma_{\theta_{0}}$.

###### Proof.

Let $K$ a compact subset $K\subset\mathbb{C}\diagdown\Sigma_{\theta_{0}}$. For
$\lambda\in L(\theta_{0})$ and $a\in K$ all the five half planes are
admissible. Let $H_{z_{0}}$ be a half plane domain such that $\partial
H_{z_{0}}$ contain the turning point $z_{0}$. From $z_{0}$ rise three Stokes
lines $l_{z_{0}}^{0},l_{z_{0}}^{1}$ and $l_{z_{0}}^{2}$. Let us make a cut
along $l_{z_{0}}^{2}$ and remove an $\epsilon$ neighbourhood of it, such that
$Re(h(\lambda,z_{0},z))>0$. The formula (25) is then applicable as $\lambda$
fixed in $L(\theta_{0})$ and $z\rightarrow\infty$ in $H_{z_{0}}^{\epsilon}$
(or for $z$ fixed in a compact set $\Omega$ and
$\left|\lambda\right|\rightarrow\infty$, $\lambda$ $\in$ $L(\theta_{0})$ ) .
$H_{z_{0}}$ border a domain of band type $B_{0}$ such that $\partial B_{0}$
contains $l_{z_{0}}^{0},l_{z_{0}}^{2}$and another turning point $z_{1}$. Any
point in $B_{0}$ can be joined with any point of $H_{z_{0}}^{\epsilon}$ by an
admissible curve and so (25) is applicable in $H_{z_{0}}^{\epsilon}\cup
B_{0}$. From $z_{1}$ rise three Stokes lines $l_{z_{1}}^{0},l_{z_{1}}^{1}$ and
$l_{z_{1}}^{2}$ such that $l_{z_{1}}^{0},l_{z_{1}}^{1}\in\partial B_{0}$. Let
us make a cut along $l_{z_{1}}^{2}$ and remove an $\epsilon$ neighborhood of
it. The half plane $H_{z_{1}}$ ,delimited for example by $l_{z_{1}}^{0}$ and
$l_{z_{1}}^{2},$ is admissible to $H_{z_{0}}$ and so (25) is then applicable
in $H_{z_{0}}^{\epsilon}\cup B_{0}\cup H_{z_{1}}^{\epsilon}$. By the same way
we extant the applicability of (25) to the whole complex plane from which
neighbourhoods of some Stokes lines have been removed. This achieve the proof.
∎

Figure 18: Stokes graph structure when $a\notin\Sigma_{\theta_{0}}$

######  Lemma 21.

Let $a,b\in\mathbb{C}$, $a\neq b$, $C$ be a simple closed curve encircling the
line segment $l_{a,b}$ and oriented as in figure 19 then

$\oint_{C}{\sqrt{(z-a)(z-b)}dz}=2\int_{a}^{b}{\sqrt{(z-a)(z-b)}dz}$

Figure 19:

###### Proof.

Let $\gamma_{a}$ and $\gamma_{b}$ be two Jordan curve respectively from $a$
two infinity and from $b$ to infinity as shown in figure19. The function
$\sqrt{(z-a)(z-b)}$ has two holomorphic branches in
$\mathbb{C}\setminus(\gamma_{a}\cup\gamma_{b})$, denoted by $f_{+}$ and
$f_{-}$, where $f_{+}=-f_{-}$.
Now suppose $f$ is holomorphic branch of $\sqrt{(z-a)(z-b)}$ in
$\mathbb{C}\setminus l_{a,b}$, if $f$ coincides with $f_{+}$ in one connected
component $\mathbb{C}\setminus(\gamma_{a}\cup\gamma_{b}\cup l_{a,b})$ then $f$
should coincide $f_{-}$ in the other component, Let $x_{a}$ and $x_{b}$
represent the intersection points of the contour $C$ respectively with the
curves $\gamma_{a}$ and $\gamma_{b}$. Then

$\oint_{C}{fdz}=\int_{x_{a}}^{x_{b}}f_{+}dz+\int_{x_{b}}^{x_{a}}{f_{-}}dz=2\int_{x_{a}}^{x_{b}}f_{+}dz=2\int_{a}^{b}{f}dz$

∎

The main result of this section is the following:

######  Proposition 22.

  1. 1.

Let $a\in\Sigma_{\theta_{0}}\setminus\\{a_{\theta_{0}},e_{\theta_{0}}\\}$
$\theta_{0}$ is an accumulation direction to (28) if and only if
$a\in(\Sigma_{-1,\theta_{0}}^{l}\cap\\{z,\Re(z)<\Re(e_{\theta_{0}})\\})\cup(\Sigma_{-1,\theta_{0}}^{r}\cap\\{z,\Re(z)<\Re(e_{\theta_{0}})\\})\cup(\Sigma_{1,\theta_{0}}^{l}\cap\\{z,\Re(z)>\Re(a_{\theta_{0}})\\}).$
In addition, the spectrum is a discrete sequence
$\left\\{\lambda_{n},\leavevmode\nobreak\ n\in\mathbb{N}\right\\}$. Infinitely
many eigenvalue accumulated near$L(\theta_{0})$ with the asymptotic formulae :

$\left|\lambda_{n}\right|=(2n+1)\pi|{\oint\limits_{C}\sqrt{P(z)}dz}|^{-1}+O(n^{-1})$
(29)

Here $C$ is a simple closed contour encircling the short trajectory.

  2. 2.

Let $\theta_{0}\in\\{a_{\theta},e_{\theta}\\}$. $\theta$ is an accumulation
direction to (28) if and only if
$\frac{|\oint_{C_{1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|}{|\oint_{C_{2}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|}\in\mathbb{Q}$,
where $C_{1}$ and $C_{2}$ are two simple closed contours encircling
respectively the two unbroken short trajectories.

Before going to the proof, let us give some remarks:

######  Remark 23.

  1. 1.

The table 20 gives all possible cases for
$\theta\in]0,\frac{\pi}{8}[\cup]\frac{3\pi}{8},\frac{\pi}{2}[$, where

$S_{1,\theta_{0}}=(\Sigma_{-1,\theta_{0}}^{l}\cap\\{z,\Re(z)<\Re(e_{\theta_{0}})\\})\cup(\Sigma_{-1,\theta_{0}}^{r}\cap\\{z,\Re(z)<\Re(e_{\theta_{0}})\\})$
$\cup(\Sigma_{1,\theta_{0}}^{l}\cap\\{z,\Re(z)>\Re(a_{\theta_{0}})\\})\cup(\Sigma^{\prime}_{\theta_{0}}\cap\\{z,\Re(z)>\Re(e_{\theta_{0}}\\}))$
$S_{2,\theta_{0}}=(\Sigma_{-1,\theta}^{l}\cap\\{z,\Re(z)>\Re(e_{\theta})\\})\cup(\Sigma_{-1,\theta}^{r}\cap\\{z,\Re(z)<\Re(a_{\theta})\\})\cup(\Sigma_{1,\theta}^{l}$
$\cap\\{z,\Re(e_{\theta})<\Re(z)<\Re(a_{\theta})\\})\cup\Sigma_{1,\theta}^{r}\cup(\Sigma^{\prime}_{\theta_{0}}\cap\\{z,\Re(e_{\theta_{0}})>\Re(z)>\Re(a_{\theta_{0}})\\})$

Figure 20: Tab 2

  2. 2.

For simplicity we will deal with the proof for $\theta_{0}=\frac{\pi}{4}$. As
it was shown in figure 20 the other cases are similar.

  3. 3.

For $\theta_{0}\in\left\\{0,\frac{\pi}{2}\right\\}$ we have
$a_{\theta_{0}}\in\left\\{-1,1\right\\}$ and in this case we may have a double
turning point at $\pm 1$. The study of this case will be a project of a
forthcoming paper.

###### Proof.

For $\lambda=re^{i\frac{\pi}{4}}$ fixed, from 4.1.2 the equation (28) has
unique (to within constant multiple) solutions $y_{1}(z,\lambda)$ and
$y_{2}(z,\lambda)$ which are subdominant respectively in $H^{\prime}$ and
$H^{\prime\prime}$. If $\lambda$ is an eigenvalue then
$y_{1}(z,\lambda)=c\leavevmode\nobreak\ y_{2}(z,\lambda)$.

  1. 1.

Let $a\in S_{1,\frac{\pi}{4}}$, without loss of generality we can take $a$ in
$\Sigma_{1,\frac{\pi}{4}}^{l}\cap\\{z,\Re(z)>\Re(a_{\frac{\pi}{4}})\\}$, there
exist a unique pair of non admissible domains $(H_{1},H_{2})$ (see the stokes
graph in figure 22 ) . The eigenvalue problem is given by:

$\begin{cases}-y^{\prime\prime}(z)+\lambda^{2}(z-a)(z^{2}-1)y(z)=0\\\
\lim_{|z|\to+\infty}y(z,\lambda)=0,\leavevmode\nobreak\ \leavevmode\nobreak\
z\in H_{1}\cup H_{2}\end{cases}$ (30)

where $H^{\prime}$=$H_{1}$ and $H^{\prime\prime}=H_{2}$.

    1. (a)

Let $\varepsilon>0$, the branch of $(\sqrt{P(z)})_{1}$ in ${H_{1}}$ be chosen
such that $\Re{h(\lambda,a,z)}>0$ and $\gamma^{+}$ be an antistokes line in
$H_{1,\varepsilon}$, then by 4.1.2 the solution $y_{1}$ has the asymptotic
expansion:

$y_{1}(z,\lambda)=(P(z))_{1}^{-\frac{1}{4}}(e^{-h(\lambda,a,z)}(1+O({\lambda}^{-1}))$
(31)

as $|\lambda|\to+\infty$, $z$ uniformly in $\gamma^{+}$, The domain $H^{+}$ of
applicability of 31 is obtained by removing an $\varepsilon-$neighborhood of
the stokes lines $l_{0},l_{1}\text{ and }l_{1}^{*}$ and the domain $H_{2}$
lying to the left of $l_{1}\text{ and }l_{1}^{*}$ (see Fig 22), indeed, for
all $z\in H^{+}$ there exists an admissible curve $\alpha$ such that
$\alpha(0)=z$ and $\gamma^{+}\subset\alpha$.

    2. (b)

The solution $y_{2}$ is defined with the asymptotic expansion:

$y_{2}(z,\lambda)=(P(z))_{2}^{\frac{-1}{4}}(e^{h(\lambda,1,z)}(1+O({\lambda}^{-1}))$
(32)

$|\lambda|\to+\infty$, $z$ uniformly in $H_{2,\varepsilon}$, the branch of
square root is chosen such that $\Re{h(\lambda,1,z)}<0$, with the same reason
in 1a the maximal domain $H^{-}$ applicability of the asymptotic expansion 32
is given by removing $\varepsilon-$ neighborhood of the cuts
$l_{0},l_{a},l_{a}^{*}$ and the half plane domain $H_{1}$ (see fig 22).

Figure 21: $H^{+}$

Figure 22: $H^{-}$

Let $D=H^{+}\cap H^{-}\cap V$ and $z^{+},z^{-}\in D$, where $V$ is a
$2\varepsilon-$neighborhood of $l_{0}$, such that $z^{+},z^{-}$ do not belong
to the same connected component of D. then 31 and 32 are applicable in $z^{+}$
and $z^{-}$, we obtain the equation for the eigenvalue

$\frac{y_{1}(z^{+},a,\lambda)y_{2}(z^{-},1,\lambda)}{y_{1}(z^{-},a,\lambda)y_{2}(z^{+},1,\lambda)}=1.$
(33)

While the branches $(\sqrt{P(z)})_{1}\text{ and }(\sqrt{P(z)})_{2}$ are chosen
such that $(\Re{h(\lambda,a,z)})_{1}(\Re{h(\lambda,a,z)})_{2}<0$ in $H^{+}\cap
H^{-}$ then$(\sqrt{P(z)})_{1}=-(\sqrt{P(z)})_{2}$. Additionally, the branches
$(P(z)^{\frac{1}{4}})_{1}$ and $(P(z)^{\frac{1}{4}})_{2}$ are chosen such that
$(P(z^{+})^{\frac{1}{4}})_{1}=i(P(z^{+})^{\frac{1}{4}})_{2}$, then it follows
that $(P(z^{-})^{\frac{1}{4}})_{1}=-i(P(z^{-})^{\frac{1}{4}})_{2}$, from 33 we
obtain an equation of eigenvalue :

$exp(i\pi+2|\lambda|(h(\lambda,1,a))_{2})(1+O(\frac{1}{\lambda}))=1.$

from lemma21 $h(\lambda,1,a))$ is purely imaginary equals
$\frac{1}{2}\oint_{C}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt$ , where $C$ is simple
closed curve go around the short trajectory $l_{0}$.
Consequently, the set of eigenvalues $\\{\lambda_{n},\leavevmode\nobreak\
n\in\mathbb{N}\\}$ is discrete, and verifie the asymptotic formula:

$|\lambda_{n}|\underset{n\to\infty}{\sim}(2n+1)\pi(|\oint_{C}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|)^{-1}$

then this set can be arranged as
$|\lambda_{n_{0}}|<|\lambda_{n_{0}+1}|<...<|\lambda_{n}|....\rightarrow\infty$.

  2. 2.

Let $a\in S_{2,\theta}$, without loss of generality, we suppose a in
$\Sigma_{1,\frac{\pi}{4}}^{l}\cap\\{z,\Re(e_{\frac{\pi}{4}})<\Re(z)<\Re(a_{\frac{\pi}{4}}\\}$,
there are precisely two pairs of non admissible half-plane domains, as
depicted in figure24. It is sufficient to concentrate on the eigenvalue
defined within one of these pairs :

$\begin{cases}-y^{\prime\prime}(z)+\lambda^{2}(z-a)(z^{2}-1)y(z)=0\\\
\lim_{|z|\to+\infty}y(z,\lambda)=0,\leavevmode\nobreak\ \leavevmode\nobreak\
z\in H_{1}\cup H_{2}\end{cases}$ (34)

By following the same steps as in the proof 31 we obtain:

     * •

The subdominant solutions $y_{1}(z,\lambda)$ and $y_{2}(z,\lambda)$ in $H_{1}$
and $H_{2}$ respectively are given by:

$\begin{cases}y_{1}(z,\lambda)=(P(z))_{2}^{-\frac{1}{4}}(e^{h(\lambda,a,z)}(1+O({\lambda}^{-1}))\\\
y_{2}(z,\lambda)=(P(z))_{2}^{-\frac{1}{4}}(e^{-h(\lambda,-1,z)}(1+O({\lambda}^{-1}))\end{cases}$
(35)

Here, the maximal domains of applicability of these expansions are denoted by
$H_{1}^{+}$ and $H_{2}^{-}$, which are the non-shaded domains illustrated in
figures 24 and 24.

     * •

Substituting (35) in 33, we obtain the equation for the eigenvalues:

$exp(i\pi+2|\lambda|(h(\lambda,-1,a))_{2})(1+O(\frac{1}{\lambda}))=1.$

We have the real part of the term $h(\lambda,-1,a)$ equal
$\Re(h(\lambda,-1,1))$ which is not vanishing (there is no trajectory joins 1
and -1), then $\frac{\pi}{4}$ can not be an accumulation direction of 28 in
this case.

Figure 23: $H_{1}^{+}$

Figure 24: $H_{2}^{-}$

  3. 3.

Let $a=a_{\frac{\pi}{4}}$. As seen in fig 25 there exists a unique stokes
complex that splits the complex plane into five half plane domains
$H_{0},H_{1},H_{2},H_{3},H_{4}$. There three pairs of non admissible domains,
so we have three possible eigenvalue problem:

$\begin{cases}-y^{\prime\prime}(z)+\lambda^{2}(z-a)(z^{2}-1)y(z)=0\\\
\lim_{|z|\to+\infty}y(z,\lambda)=0,\leavevmode\nobreak\ \leavevmode\nobreak\
z\in H_{0}\cup H_{2}\end{cases}$ (36)
$\begin{cases}-y^{\prime\prime}(z)+\lambda^{2}(z-a)(z^{2}-1)y(z)=0\\\
\lim_{|z|\to+\infty}y(z,\lambda)=0,\leavevmode\nobreak\ \leavevmode\nobreak\
z\in H_{1}\cup H_{3}\end{cases}$ (37)
$\begin{cases}-y^{\prime\prime}(z)+\lambda^{2}(z-a)(z^{2}-1)y(z)=0\\\
\lim_{|z|\to+\infty}y(z,\lambda)=0,\leavevmode\nobreak\ \leavevmode\nobreak\
z\in H_{0}\cup H_{3}\end{cases}$ (38)

36 and 37 are given with boundary conditions respectively in $(H_{0},H_{2})$
and $(H_{1},H_{3})$, where are connected with unbroken short trajectory. This
is similar to the case when $a\in S_{1,\theta_{0}}$ in 1. We deduce that the
spectrum is discrete, accumulating near $L(\frac{\pi}{4})$ and verify,
respectively for $k=1$ and $k=-1$ the asymptotic formula

$|\lambda_{n}|\underset{n\to\infty}{\sim}(2n+1)\pi(|\oint_{C_{k}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|)^{-1}.$

Where $C_{k}$ is a simple closed contour go around the short trajectory
$l_{0}^{k}$ .

Figure 25:

For the remaining case 38, we use transition matrices to obtain asymptotic
formula for the eigenvalue . We define the canonical domains
$D_{1},D_{2},D_{3}$ and $D_{4}$ by
$l_{1}\subset D_{1},\leavevmode\nobreak\ l_{1}^{*}\cup l_{0}^{1}\cup
l_{a}\subset\partial D_{1},\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ D_{1}=H_{0}\cup H_{1}$
$l_{0}^{1}\subset D_{1},\leavevmode\nobreak\ l_{1}^{*}\cup l_{1}\cup
l_{a}\subset\partial D_{1},\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ D_{2}=H_{1}\cup H_{4}$
$l_{0}^{-1}\subset D_{1},\leavevmode\nobreak\ l_{1}^{*}\cup l_{a}\cup
l_{-1}\subset\partial D_{1},\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ D_{3}=H_{2}\cup H_{4}$
$l_{-1}\subset D_{1},\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ D_{4}=H_{3}\cup H_{4}.$ Then following this sequence of
triples :
$\left(D_{1},1,l_{1}\right)_{1}\rightarrow\left(D_{2},1,l_{0}^{1}\right)_{2}\rightarrow\left(D_{2},a,l_{0}^{1}\right)_{3}\rightarrow\left(D_{3},a,l_{0}^{-1}\right)_{4}\rightarrow\left(D_{3},-1,l_{0}^{-1}\right)_{5}\rightarrow(D_{4},-1,l_{1}^{*})_{4}.$
From the paragraph 4.1.3, let $\\{(u_{i},v_{i})\\}_{1\leq i\leq 5}$ the
elementary F.S.S corresponding respectively to the triples above. Suppose that
$y(z,\lambda)$ is a solution of 28 with asymptotic expansion 31 in
$H_{1,\varepsilon}$. Then $y(z,\lambda)=cv_{1}$, we will express $y$ in terms
of $(u_{4},v_{4})$: $y=au_{4}+bv_{4}$
$\left(\begin{array}[]{c}a\left(\lambda\right)\\\
b\left(\lambda\right)\end{array}\right)=\begin{pmatrix}0&\alpha_{0,1}^{-1}(\lambda)\\\
1&i\alpha_{2,1}(\lambda)\end{pmatrix}\begin{pmatrix}0&e^{-i|h(a,-1,\lambda)|}\\\
e^{i|h(a,-1,\lambda)|}&0\end{pmatrix}\begin{pmatrix}0&\beta_{0,1}^{-1}(\lambda)\\\
1&i\beta_{1,2}(\lambda)\end{pmatrix}\begin{pmatrix}0&e^{-i|h(1,a,\lambda)|}\\\
e^{i|h(1,a,\lambda)|}&0\end{pmatrix}$

$\begin{pmatrix}0&\gamma_{0,1}^{-1}(\lambda)\\\
1&i\gamma_{1,2}(\lambda)\end{pmatrix}\begin{pmatrix}0\\\
1\end{pmatrix}=\frac{1}{\beta_{1,2}\gamma_{1,2}}\allowbreak\begin{pmatrix}\Delta_{1}\\\
\Delta_{2}\end{pmatrix}\allowbreak$

Where

$\begin{cases}\Delta_{1}=\alpha_{1,2}^{-1}e^{i|h(-1,a,\lambda)|+i|h(1,a\lambda)|}\\\
\Delta_{2}=ie^{-i|h(-1,a,\lambda)-i|h(1,a,\lambda)|}(\alpha_{2,3}e^{2i(|ih(-1,a,\lambda)|+|h(1,a,\lambda)|)}+\beta^{-1}_{1,3}e^{2i|h(1,a,\lambda)}+\beta_{1,2}\gamma^{-1}_{1,3})\end{cases}$
(39)

$\lambda$ is eigenvalue of 38 if and only if $b(\lambda)=0$.
$b(\lambda)=0$
$\iff\alpha_{2,3}e^{2i(|h(-1,a,\lambda)|+|h(1,a,\lambda)|)}+\beta_{1,2}\beta_{2,3}e^{2i|h(1,a,\lambda)|}=-1+O(\frac{1}{\lambda})$

$\iff
e^{i(|\lambda\oint_{C_{1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|+\frac{1}{2}|\lambda\oint_{C_{-1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|)}2\cos(\frac{1}{2}\leavevmode\nobreak\
|\lambda\oint_{C_{-1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|)=-1+O(\frac{1}{\lambda})$

$\iff\begin{cases}|\lambda_{n}|\underset{|\lambda_{n}|\to+\infty}{\sim}(2n+1)\pi(|\oint_{C_{1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|+\frac{1}{2}|\oint_{C_{-1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|)^{-1}\\\
|\lambda_{n}|\underset{|\lambda_{n}|\to+\infty}{\sim}(2\varphi(n)+\frac{2\pi}{3})\pi(|\oint_{C_{-1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|)^{-1}\end{cases}$
Then $\lambda_{n}$ is an eigenvalue of 38 if and only if
$\frac{|\oint_{C_{-1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|}{|\oint_{C_{1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|}=\frac{n}{\varphi(n)}-\frac{1}{2}\in\mathbb{Q}$
Here the symmetry observed in the Stokes graph of
$e^{\frac{\pi}{2}}(z^{2}-1)(z-a_{\frac{\pi}{4}})$ implies that
$|\oint_{C_{1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|=|\oint_{C_{-1}}e^{i\frac{\pi}{4}}\sqrt{P(t)}dt|$,
so the condition of the existence of a solution to the eigenvalue problem 38
is verified. However, it is important to note that this condition does not
hold universally for $a\in\\{a_{\theta},e_{\theta}\\}$ with
$\theta\notin\\{0,\frac{\pi}{4}\\}$.

∎

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