# Completeness theorem for the system of eigenfunctions of the complex
Schrödinger operator $\mathscr{L}_{c,\alpha}=-d^{2}/dx^{2}+cx^{\alpha}$

Sergey Tumanov<EMAIL_ADDRESS>Moscow Center of Fundamental and Applied
Mathematics at Lomonosov Moscow State University, Russian Federation

###### Abstract

The completeness of the system of eigenfunctions of the complex Schrödinger
operator $\mathscr{L}_{c,\alpha}=-d^{2}/dx^{2}+cx^{\alpha}$ on the semi-axis
with Dirichlet boundary conditions is proved for an arbitrary $\alpha\in(0,2)$
and $|\arg c|<2\pi\alpha/(\alpha+2)+\Delta t(\alpha)$ with some $\Delta
t(\alpha)>0$.

##  1 Introduction

We consider the operator

$\mathscr{L}_{c,\alpha}=-\frac{d^{2}}{dx^{2}}+cx^{\alpha}$

in $L_{2}(\mathbb{R}_{+})$ with Dirichlet boundary conditions with
$c\in\mathbb{C}$, $|\arg c|<\pi$, $\alpha>0$.

It’s known, $\mathscr{L}_{c,\alpha}$ has a compact inverse, the spectrum is
discrete, root subspaces are one-dimensional [1].

For $0<|\arg c|<\pi$ it is not self-adjoint, moreover, it has bad spectral
properties: the norm of the resolvent [2, 3] and the norms of spectral
projectors [4] grow exponentially. Under these conditions, the operator cannot
be similar to self-adjoint, its eigenfunctions do not form a Riesz basis in
$L_{2}(\mathbb{R}_{+})$. Nevertheless, the completeness of the system of
eigenfunctions (S.E.) in $L_{2}(\mathbb{R}_{+})$ is an open problem, which is
what our work is devoted to.

For $\alpha\geq 2$ the problem of completeness of S.E. of
$\mathscr{L}_{c,\alpha}$ is fully explored [2, 5, 6]: the system is complete
for all $c\in\mathbb{C}$: $|\arg c|<\pi$.

For $\alpha\in(0,2)$ completeness is proved in case $|\arg
c|<t_{0}(\alpha)=2\pi\alpha/(\alpha+2)$ [6]. At the same time, the case
$t_{0}(\alpha)\leq|\arg c|<\pi$ is almost not studied, since it is a much more
difficult one. The corresponding arguments are given in [1, 6].

Perhaps the first result for the case $t_{0}(\alpha)\leq|\arg c|<\pi$ was the
study of Savchuk and Shkalikov [6] of the complex Airy operator ($\alpha=1$).
The authors proved the completeness of S.E. of $\mathscr{L}_{c,1}$ in case
$|\arg c|<t_{0}(1)+\pi/6=5\pi/6$.

In our recent paper [1] a positive answer was given regarding the completeness
of S.E. of $\mathscr{L}_{c,2/3}$ in case $|\arg c|<t_{0}(2/3)+\Delta
t=\pi/2+\Delta t$, where $\Delta t>\pi/10$ is the only zero of some
transcendental equation.

It turns out that these results generalize for all $\alpha\in(0,2)$: there
exists $\Delta t=\Delta t(\alpha)>0$ continuously depending on $\alpha$, such
that the S.E. of $\mathscr{L}_{c,\alpha}$ is complete in case $|\arg
c|<t_{0}(\alpha)+\Delta t(\alpha)$. This fact is the subject of our study.

The interest to the operators of the form $\mathscr{L}_{c,\alpha}$ has
increased especially in recent decades due to the popularity of the ideas of
non-Hermitian quantum mechanics [7, 8], as well as the general interest in
spectral properties of non-self-adjoint operators.

Let us note in this regard the problem posed by Almog in 2015 [9]: is the S.E.
of $\mathscr{L}_{i,\alpha}$ complete for $0<\alpha\leq 2/3$? The answer is
affirmative if $\theta_{0}(\alpha)=t_{0}(\alpha)+\Delta t(\alpha)>\pi/2$, in
particular, for $\alpha=2/3$. Taking into account the continuity of
$\theta_{0}(\alpha)$, there exists $\alpha_{0}<2/3$ such that the S.E. of
$\mathscr{L}_{i,\alpha}$ is complete for all $\alpha>\alpha_{0}$. This
hypothesis was expressed by Savchuk and Shkalikov [6]. In our paper it is a
simple Corollary to the main Theorem.

For complex $\zeta=|\zeta|e^{i\arg\zeta}$, $-\pi<\arg\zeta\leq\pi$ and real
$\beta$ we use the term $\zeta^{\beta}$ to denote the main branch:
$\zeta^{\beta}=|\zeta|^{\beta}e^{i\beta\arg\zeta}$.

Let $\theta\in[t_{0}(\alpha),\pi)\cap[t_{0}(\alpha),\pi\alpha)$, and

$\zeta_{0}(\theta)=e^{i(t_{0}(\alpha)-\theta)/\alpha},\quad
Z_{0}(\theta)=(\sin t_{0}(\alpha)/\sin\theta)^{1/\alpha},$

we determine

$\displaystyle\rho(\theta)=\operatorname{Re}\,\Bigl{\\{}\int\limits_{0}^{Z_{0}(\theta)}\sqrt{e^{i\theta}\zeta^{\alpha}-e^{it_{0}(\alpha)}}\,d\zeta-2\int\limits_{0}^{\zeta_{0}(\theta)}\sqrt{e^{i\theta}\zeta^{\alpha}-e^{it_{0}(\alpha)}}\,d\zeta\Bigr{\\}}=$
$\displaystyle=\operatorname{Re}\,\Bigl{\\{}\int\limits_{\zeta_{0}(\theta)}^{Z_{0}(\theta)}\sqrt{e^{i\theta}\zeta^{\alpha}-e^{it_{0}(\alpha)}}\,d\zeta-\int\limits_{0}^{\zeta_{0}(\theta)}\sqrt{e^{i\theta}\zeta^{\alpha}-e^{it_{0}(\alpha)}}\,d\zeta\Bigr{\\}},$

where the integration is performed over segments and the branch of the square
root is chosen so that

$\operatorname{Re}\int\limits_{0}^{Z_{0}(\theta)}\sqrt{e^{i\theta}\zeta^{\alpha}-e^{it_{0}(\alpha)}}\,d\zeta>0,\quad\operatorname{Re}\int\limits_{0}^{\zeta_{0}(\theta)}\sqrt{e^{i\theta}\zeta^{\alpha}-e^{it_{0}(\alpha)}}\,d\zeta>0.$

######  Theorem 1.

Given $\alpha\in(0,2)$, the function $\rho(\theta)$ has the only zero
$\theta_{0}(\alpha)$ within the interval:
$\theta_{0}(\alpha)\in(t_{0}(\alpha),\pi)\cap(t_{0}(\alpha),\pi\alpha)$;
$\theta_{0}(\alpha)=t_{0}(\alpha)+\Delta t(\alpha)$, $\Delta t(\alpha)>0$.

The function $\theta_{0}(\alpha)$ is continuous for $\alpha\in(0,2)$.

For $|\arg c|<\theta_{0}(\alpha)$ the S.E. of $\mathscr{L}_{c,\alpha}$ is
complete in $L_{2}(\mathbb{R}_{+})$.

Corollary.  There exists $\alpha_{0}<2/3$ such that the S.E. of
$\mathscr{L}_{i,\alpha}$ is complete in $L_{2}(\mathbb{R}_{+})$ for all
$\alpha>\alpha_{0}$.

The previously known completeness boundary $t_{0}(\alpha)$ and the one
obtained in the present work $\theta_{0}(\alpha)$ are shown in the figure 1.

Figure 1: Schematic chart $t_{0}(\alpha)$ and $\theta_{0}(\alpha)$.

##  2 Auxiliary results

The operator $\mathscr{L}_{c,\alpha}$ in $L_{2}(\mathbb{R}_{+})$ with
Dirichlet boundary conditions is defined by differential expression

$l_{c,\alpha}(y)=-\frac{d^{2}y}{dx^{2}}+cx^{\alpha}y,\quad x\in[0,+\infty),$

and is considered on the domainn

$\mathfrak{D}(\mathscr{L}_{c,\alpha})=\bigl{\\{}y\in
L_{2}(\mathbb{R}_{+})\,\bigl{|}\bigr{.}\,y\in
W_{2,\operatorname{loc}}^{2}(\mathbb{R}_{+}),\ l_{c,\alpha}(y)\in
L_{2}(\mathbb{R}_{+}),\ y(0)=0\bigr{\\}}.$

The following statements underlie the proof of the Theorem 1. They correspond
to Proposition A.1 and Lemmas 1 and 2 of [1].

######  Lemma 1.

Given $|\arg c|<\pi$, the operator $\mathscr{L}_{c,\alpha}$ is closed, has
compact inverse. The eigenvalues $\\{\lambda_{n}\\}$, $n\in\mathbb{N}$ are
simple (the root subspaces are one-dimensional), and have the form of
$\lambda_{n}=c^{2/(\alpha+2)}\tau_{n}$, $n\in\mathbb{N}$, where $\tau_{n}>0$
don’t depend on $c$.

Consider the equation

$y^{\prime\prime}=(cx^{\alpha}-\lambda)y.$ (1)

The existence of the so-called Weyl solution — the solution in
$L_{2}(\mathbb{R}_{+})$ is known [10, Ch II] for any homogeneous
Sturm–Liouville equation with real locally integrable potential. This theory
was generalized by Lidskii to the case of some complex potentials [11]. The
following Lemma states for the equation (1):

######  Lemma 2 (On the Weyl solution).

Given $0<|\arg c|<\pi$, there exists $\mathcal{Y}_{0}(x,\lambda)$
($x\in\mathbb{R}_{+}$, $\lambda\in\mathbb{C}$) — non-trivial solution to (1)
with properties:

  * •

the function $\mathcal{Y}_{0}(x,\lambda)$ is continuous of two variables
$(x,\lambda)\in\mathbb{R}_{+}\times\mathbb{C}$;

  * •

for any $\lambda\in\mathbb{C}$, $\mathcal{Y}_{0}(x,\lambda)\in
L_{2}(\mathbb{R}_{+})$ as a function of $x$;

  * •

for any $x\geq 0$, $\mathcal{Y}_{0}(x,\lambda)$ is an entire function of
$\lambda$ of the order of growth of $\leq(\alpha+2)/(2\alpha)$;

  * •

the zeros of $\mathcal{Y}_{0}(0,\lambda)$ coincide with the eigenvalues of
$\mathscr{L}_{c,\alpha}$;

  * •

for an arbitrary $f\in L_{2}(\mathbb{R}_{+})$ the integral

$\mathcal{G}(\lambda)=\int\limits_{0}^{+\infty}\mathcal{Y}_{0}(x,\lambda)f(x)\,dx$
(2)

is an entire function of the order of growth of $\leq(2+\alpha)/(2\alpha)$;

  * •

the function
$\mathcal{F}(\lambda)=\mathcal{G}(\lambda)/\mathcal{Y}_{0}(0,\lambda)$ is
bounded in any closed sector for which
$\lambda\not\in\Lambda=\\{\lambda\in\mathbb{C}\,\Bigl{|}\Bigr{.}\,\arg\lambda\in[0,\arg
c]\\}$.

Referring to Levin [12], the system of eigenfunctions $\\{y_{n}\\}$,
$n\in\mathbb{N}$ of the operator $\mathscr{L}_{c,\alpha}$ is generated by
$\mathcal{Y}_{0}(x,\lambda)$ and the sequence of eigenvalues
$\\{\lambda_{n}\\}$: $y_{n}(x)=\mathcal{Y}_{0}(x,\lambda_{n})$. It turns out
that $\mathcal{Y}_{0}(x,\lambda)$ is a closed kernel in terms of Levin: if for
some $f\in L_{2}(\mathbb{R}_{+})$ holds $\mathcal{G}(\lambda)\equiv 0$, then
$f\equiv 0$. Moreover,

######  Lemma 3 (On the closed kernel).

Given $0<|\arg c|<\pi$ and some $f\in L_{2}(\mathbb{R}_{+})$. If the function

$\mathcal{F}(\lambda)=\frac{1}{\mathcal{Y}_{0}(0,\lambda)}\int\limits_{0}^{+\infty}\mathcal{Y}_{0}(x,\lambda)f(x)\,dx$
(3)

is a constant, then $f\equiv 0$.

##  3 The proof of the Theorem 1

In our proof we use some notation.

In estimates, $C$ denotes arbitrary positive constants, possibly different on
both sides of the inequalities.

If for some $a>0$ we say that the statement is true for $a\ll 1$ ($a\gg 1$),
we mean the validity of this statement asymptotically — for all $0<a<a_{0}$
($a>a_{0}$) for some $a_{0}>0$.

Sometimes we use short notation: $\phi\equiv\arg\lambda$, $\theta\equiv\arg
c$.

Let fix some $\alpha\in(0,2)$. Without loss of generality,
$\operatorname{Im}c>0$, $|c|=1$ — with complex conjugation and scaling in $x$
one can cover any $c\in\mathbb{C}$: $0<|\arg c|<\pi$.

Restrict ourselves to the case $\arg
c\in[t_{0}(\alpha),\pi)\cap[t_{0}(\alpha),\pi\alpha)$.

Let fix some $\delta>0$: $0<\delta<(\pi\min\\{1,\alpha\\}-\theta)/2$.
Additional restrictions on $\delta>0$ will be introduced if necessary.

Let
$\mathfrak{l}_{\delta}=\\{\lambda\in\mathbb{C}\,\bigl{|}\bigr{.}\,t_{0}(\alpha)-\delta<\arg\lambda<t_{0}(\alpha)\\}$.
With our restrictions, $\arg c>\arg\lambda>0$ for
$\lambda\in\mathfrak{l}_{\delta}$.

Let’s walk through the main idea of the proof and its stages.

Assuming that for some $c$ the S.E. $\\{y_{n}\\}$, $n\in\mathbb{N}$ is
incomplete, one can find $f\not\equiv 0$, $f\in L_{2}(\mathbb{R}_{+})$, such
that $\mathcal{F}(\lambda)$ (3) turns out to be an entire function. If we show
that $\mathcal{F}(\lambda)$ is constant, then in view of the Lemma 3 we come
to a contradiction with $f\not\equiv 0$.

It is enough to prove that $\mathcal{F}(\lambda)$ grows not faster than a
polynomial just in the small sector $\mathfrak{l}_{\delta}$. Indeed, on the
one hand, the order of growth of $\mathcal{F}(\lambda)$ is not higher than
$\pi/t_{0}(\alpha)$, and on the other, — $\mathcal{F}(\lambda)$ is bounded in
closed sectors outside $\Lambda$ (Lemma 2). The central angles of the sectors
complementing $\mathfrak{l}_{\delta}$ to $\Lambda$ are strictly less than
$t_{0}(\alpha)$, so referring to Phragmén-Lindelöf Principle, we obtain that
$\mathcal{F}(\lambda)$ is a polynomial. Taking into account the boundedness of
$\mathcal{F}(\lambda)$ outside $\Lambda$, we obtain that
$\mathcal{F}(\lambda)$ is a constant.

The condition $0<\theta<\theta_{0}(\alpha)$ turns out to be sufficient for
such estimate of $\mathcal{F}(\lambda)$ in $\mathfrak{l}_{\delta}$.

We split the proof into several steps:

  * •

Step 1. The independent variable and parameter substitution in (1).

  * •

Step 2. Construction of the Weyl solution in terms of new variable and
parameter, investigation of its uniform in $x\in\mathbb{R}_{+}$ asymptotics
for $\mathfrak{l}_{\delta}\ni\lambda\to\infty$.

  * •

Step 3. Proving that the condition $\rho(\arg c)<0$ is sufficient for the
completeness of the S.E. of $\mathscr{L}_{c,\alpha}$.

  * •

Step 4. Verification of the sufficient condition. The proof of the existence
of a critical $\Delta t(\alpha)>0$ continuously depending on $\alpha$ such
that the condition $\rho(\theta)<0$ is satisfied for all $\theta$:
$t_{0}(\alpha)\leq\theta<t_{0}(\alpha)+\Delta t(\alpha)$. This completes the
proof of the Theorem 1.

###  3.1 Step 1. The independent variable and parameter substitution

Let $\lambda\in\mathfrak{l}_{\delta}$. We fix some $\arg\lambda$, at the same
time $|\lambda|>0$ can be arbitrarily large.

For any solution $y(x)=y(x,\lambda)$ to the equation (1), let

$w(t)=y(|\lambda|^{1/\alpha}t),\quad k=|\lambda|^{1/2+1/\alpha}.$

The equation is converted to the form:

$w^{\prime\prime}=k^{2}(ct^{\alpha}-\mu)w,$ (4)

with $\mu=e^{i\phi}$, $c=e^{i\theta}$, $0<\phi<\theta$.

######  Proposition 1.

The following inequalities hold:

$0<\arg c/\mu<\frac{\pi}{3},\quad 0<\arg(c/\mu)^{1/\alpha}<\frac{\pi}{3}.$ (5)

Proof. Only the upper bounds are of interest.

For $0<\alpha<1$ holds $\delta+\arg c<\pi\alpha$. Thus

$\begin{split}\arg c/\mu&<\pi\alpha-
t_{0}(\alpha)=\frac{\pi\alpha^{2}}{\alpha+2}<\pi/3,\\\
\arg(c/\mu)^{1/\alpha}&<\frac{\pi\alpha}{\alpha+2}<\pi/3,\end{split}$

For $1\leq\alpha<2$ holds $\delta+\arg c<\pi$. Thus

$\begin{split}\arg c/\mu&<\pi-
t_{0}(\alpha)=\frac{\pi(2-\alpha)}{\alpha+2}\leq\pi/3,\\\
\arg(c/\mu)^{1/\alpha}&<\frac{\pi(2-\alpha)}{\alpha(\alpha+2)}\leq\pi/3.\end{split}$

In each case, the estimates reach their maximum at $\alpha=1$. $\Box$

###  3.2 Step 2. Construction of the Weyl solution

Further, let $\mu\in\mathfrak{l}_{\delta}$, $\mu=e^{i\phi}$.

Determine a multivalued analytic function in
$\mathbb{C}_{r}=\\{z\in\mathbb{C}\,|\,\operatorname{Re}z>0\\}$:

$S(z)=\int\limits_{0}^{z}\sqrt{c\zeta^{\alpha}-\mu}\,d\zeta,$

integrating over rectifiable paths in the right half-plane that do not pass
through the only singularity $\zeta_{0}=(\mu/c)^{1/\alpha}$ in the IVth
quarter (5).

Determine the horizontal cut $\Delta=\\{z=\zeta_{0}+t,\ t\geq 0\\}$ (see
fig.2).

Figure 2: The paths of monotonicity of $\operatorname{Re}S(z)$. The wavy line
corresponds to the cut.

For $z\in\mathbb{C}_{r}$ we define $q(z)=cz^{\alpha}-\mu$.

We also define the main branches of $q^{1/2}(z)$ and $S(z)$ in the simply
connected domain $\mathbb{C}_{r}\setminus\Delta$ by the conditions: $S(0)=0$
(by continuity); $\operatorname{Re}q^{1/2}(z)>0$ and $\operatorname{Re}S(z)>0$
for $z>0$, $z\ll 1$. In what follows, $\sqrt{cz^{\alpha}-\mu}$ and $S(z)$ will
denote the main branches of $q^{1/2}(z)$ and $S(z)$. Cases of the continuation
of these branches through the cut $\Delta$ will be specially noted.

We also fix in $\mathbb{C}_{r}\setminus\Delta$ any branch of $q^{1/4}(z)$,
which we call the main one.

Denote $Z_{0}=(\operatorname{Im}\mu/\operatorname{Im}c)^{1/\alpha}>0$.

######  Proposition 2.

There exist $\varepsilon>0$, $R_{0}>0$, the value $\operatorname{Re}S(z)$ is
monotone along each of the following paths for $R>R_{0}$:

$\displaystyle l_{1}=\\{\zeta_{0}e^{-i\varepsilon}t,\ 0\leq t<+\infty\\},$
$\displaystyle l_{2}=\\{Z_{0}+it,\ t\geq 0\\}\cup[Z_{0},+\infty),$
$\displaystyle l_{2,R}=\\{Z_{0}+it,\ t\geq
0\\}\cup[Z_{0},R]\cup\Gamma_{R}\cup\\{z\in l_{1},\ |z|\geq R\\},$
$\displaystyle l_{3}=[0,Z_{0}]\cup\\{Z_{0}+it,\ t\geq 0\\},$

where $\Gamma_{R}=\\{Re^{it},\ t\in[\arg\zeta_{0}-\varepsilon,0]\\}$ (see
fig.2). The analytic continuation of the main branch of $S(z)$ is considered
along $l_{2,R}$, since $l_{2,R}$ is crossing $\Delta$.

When $z$ moves from the origin to an infinite point along each of the paths
$l_{1}$ and $l_{3}$, the value of $\operatorname{Re}S(z)$ increases
monotonically from $0$ to $+\infty$.

When $z$ moves from $+\infty$ to $Z_{0}+i\infty$ along $l_{2}$, the value of
$\operatorname{Re}S(z)$ increases monotonically from $-\infty$ to $+\infty$.

When $z$ moves from $\zeta_{0}e^{-i\varepsilon}\infty$ to $Z_{0}+i\infty$
along $l_{2,R}$, the value of $\operatorname{Re}S(z)$ decreases monotonically
from $+\infty$ to $-\infty$. The continuation of $S(z)$ through $\Delta$ is
considered here.

Along the real axis the value $\operatorname{Re}S(z)$: is increasing on
$[0,Z_{0}]$, has a single maximum at $z=Z_{0}$, is decreasing to $-\infty$ on
$(Z_{0},+\infty)$ with asymptotics:

$S(z)\sim-\ \frac{2\operatorname{Re}c^{1/2}}{\alpha+2}z^{\alpha/2+1},\mbox{ as
}z\to+\infty.$ (6)

For all $z>0$ the value $\operatorname{Im}q^{1/2}(z)<0$ for the main branch of
$q^{1/2}$.

The inequality holds: $\operatorname{Re}S(\zeta_{0})>0$.

Proof. We use the asymptotic formula following from the integral
representation of $S(z)$ in neighborhoods of the infinite parts of the curves
$l_{j}$:

$S(z)\sim\int\limits_{0}^{z}\sqrt{c\zeta^{\alpha}}\,d\zeta\sim\pm
c^{1/2}\frac{z^{\alpha/2+1}}{\alpha/2+1},\quad z\to\infty,$ (7)

the choice of the sign will be further specified for each curve $l_{j}$.

It also follows from the integral representation of $S(z)$:

$S(z)\sim\int\limits_{0}^{z}\sqrt{-\mu}\,d\zeta\sim-i\mu^{1/2}z\mbox{, as
}z\to 0,\ \operatorname{Re}z>0,$ (8)

the choice of the sign is determined by the condition
$\operatorname{Re}S(z)>0$ for $z>0$, $z\ll 1$.

It follows from (8) that $\operatorname{Im}S(z)<0$ for $z\ll 1$.

Further we explore the extrema of $\operatorname{Re}S(z)$ along the real axis
for $z=t\geq 0$:

$\frac{d}{dt}\operatorname{Re}S(t)=\operatorname{Re}\sqrt{ct^{\alpha}-\mu}=0\
\Leftrightarrow\ \sqrt{ct^{\alpha}-\mu}=i\beta_{1}$

for some $t\geq 0$, $\beta_{1}\in\mathbb{R}\setminus\\{0\\}$, that is
equivalent to $ct^{\alpha}-\mu=-\beta_{1}^{2}<0$. Considering the imaginary
part, $t=Z_{0}=(\operatorname{Im}\mu/\operatorname{Im}c)^{1/\alpha}$. For the
real part,

$\operatorname{Re}cZ_{0}^{\alpha}-\operatorname{Re}\mu=\frac{\operatorname{Im}\\{\mu\overline{c}\\}}{\operatorname{Im}c}=\frac{|c|^{2}}{\operatorname{Im}c}\operatorname{Im}\frac{\mu}{c}<0,$

taking (5) into account. Thus there is a unique extremum of
$\operatorname{Re}S(z)$ at $Z_{0}>0$ — the global maximum for $z\geq 0$.

Exploring the extrema of $\operatorname{Im}S(z)$ for $z=t\geq 0$:

$\frac{d}{dt}\operatorname{Im}S(t)=0\ \Leftrightarrow\
\sqrt{ct^{\alpha}-\mu}=\beta_{2}\in\mathbb{R}\setminus\\{0\\},$

that is equivalent to $ct^{\alpha}-\mu=\beta_{2}^{2}>0$. Considering the
imaginary part, again we get $t=Z_{0}$, but
$cZ_{0}^{\alpha}-\mu=-\beta_{1}^{2}<0$, thus, $\operatorname{Im}S(z)$ has no
extrema for $z\geq 0$.

To summarize: for $z\geq 0$ the function $\operatorname{Re}S(z)$ increases,
reaches its maximum at $z=Z_{0}$, then decreases. The function
$\operatorname{Im}S(z)$ decreases for all $z\geq 0$. In particular for the
main branch of $q^{1/2}(z)$ we obtain
$\operatorname{Im}q^{1/2}(z)=\operatorname{Im}S^{\prime}(z)<0$ for $z>0$.

Further we explore the extrema of $\operatorname{Re}S(z)$ along the vertical
rays $\\{a+it,\ t\geq 0\\}$ for $a>0$:

$\frac{d}{dt}\operatorname{Re}S(a+it)=0\ \Leftrightarrow\
c(a+it)^{\alpha}-\mu=\beta_{3}^{2}>0\ \Leftrightarrow\
(a+it)^{\alpha}=\frac{\mu}{c}+\frac{\beta_{3}^{2}}{c}.$

Both terms on the right-hand side lie in the lower half-plane (5). At the same
time, since $\alpha\in(0,2)$, the value $(a+it)^{\alpha}$ lies in the upper
half-plane, or on the real axis. In other words, there are no extrema,
$\operatorname{Re}S(z)$ is strictly monotone. In particular, along the
vertical ray $l_{3}\cap l_{2}$.

Let’s move along $l_{3}$ from the origin and observe the curve in the image of
$S(z)$. The value $\operatorname{Re}S(z)$ is strictly increasing, and the
value $\operatorname{Im}S(z)$ is strictly decreasing on the segment
$[0,Z_{0}]$. The tangent to the image $S([0,Z_{0}])$ at $S(Z_{0})$ is strictly
vertical and directed downward. Since $S(z)$ is univalent in $Z_{0}$, the
angles are preserved: when moving along $l_{3}$ from $Z_{0}$ to
$Z_{0}+i\infty$, the value $\operatorname{Re}S(z)$ will increase.

The same reasoning shows that $\operatorname{Re}S(z)$ is monotone along
$l_{2}$.

We take $\varepsilon>0$ so that $l_{1}$ lies in the IVth quarter. We reduce
$\varepsilon>0$ so that $S(z)$ lies in the IVth quarter for arbitrarily small
points of $l_{1}$. This is possible, as $\delta<(\pi\alpha-\theta)/2$. Indeed,

$\displaystyle\lim_{t\to+0}\arg
S(\zeta_{0}e^{-i\varepsilon}t)=\arg\bigl{\\{}-i\zeta_{0}e^{-i\varepsilon}\mu^{1/2}\bigr{\\}}=$
$\displaystyle=-\frac{\pi}{2}+\frac{t_{0}(\alpha)}{2}-\frac{\delta_{1}}{2}+\frac{t_{0}(\alpha)}{\alpha}-\frac{\delta_{1}}{\alpha}-\frac{\theta}{\alpha}-\varepsilon=\frac{\pi}{2}-\frac{\theta}{\alpha}-\delta_{1}\frac{\alpha+2}{2\alpha}-\varepsilon>$
$\displaystyle>\frac{\pi}{2}-\frac{\theta}{\alpha}-\delta\frac{\alpha+2}{2\alpha}-\varepsilon$
(9)

where $\delta_{1}=t_{0}(\alpha)-\phi\in(0,\delta)$. Since $\alpha\in(0,2)$,

$\frac{\pi}{2}-\frac{\theta}{\alpha}-\delta\frac{\alpha+2}{2\alpha}>\frac{\pi}{2}-\frac{\theta}{\alpha}-\frac{1}{2}(\pi\alpha-\theta)\frac{\alpha+2}{2\alpha}>\frac{\pi}{2}-\frac{\theta}{\alpha}-\frac{2}{\alpha+2}(\pi\alpha-\theta)\frac{\alpha+2}{2\alpha}=-\frac{\pi}{2},$

therefore, we reduce $\varepsilon>0$ to make the final estimate (9) greater
than $-\pi/2$. On the other hand, as
$\pi/2<t_{0}(\alpha)/\alpha\leq\theta/\alpha$, we have

$\lim_{t\to+0}\arg
S(\zeta_{0}e^{-i\varepsilon}t)=\frac{\pi}{2}-\frac{\theta}{\alpha}-\delta_{1}\frac{\alpha+2}{2\alpha}-\varepsilon<0.$

Further we explore the extrema of $\operatorname{Re}S(z)$ along $l_{1}$:

$\frac{d}{dt}\operatorname{Re}S(\zeta_{0}e^{-i\varepsilon}t)=0\
\Leftrightarrow\
\operatorname{Re}\bigl{\\{}\zeta_{0}e^{-i\varepsilon}\mu^{1/2}\sqrt{e^{-i\varepsilon\alpha}t^{\alpha}-1}\bigr{\\}}=0.$
(10)

The value $\zeta_{0}e^{-i\varepsilon}\mu^{1/2}$ lies in the Ist quarter (see
the first equality in (9)); for $t>0$, the expression under the root in (10)
lies is in the lower half-plane, that is, the root itself is in the IVth
quarter (up to a sign). The whole expression in curly braces in (10) lies in
the right half-plane (up to a sign), i.e. equality (10) is impossible, and
$\operatorname{Re}S(z)$ has no extrema along $l_{1}$.

With $\varepsilon=0$, the same reasoning proves the monotonic growth of
$\operatorname{Re}S(\zeta_{0}t)$, along $0\leq t<1$. Whence
$\operatorname{Re}S(\zeta_{0})>0$.

Now we show that $\operatorname{Re}S(z)$ is monotonic along
$\Gamma_{R}=\\{Re^{it},\ t\in[\arg\zeta_{0}-\varepsilon,0]\\}$:

$\frac{d}{dt}\operatorname{Re}S(Re^{it})=0\ \Leftrightarrow\
e^{2it}(cR^{\alpha}e^{it\alpha}-\mu)=\beta_{4}^{2}>0.$ (11)

For $R\gg 1$, the argument of the left part of the last equality in (11) is
determined by the argument of $ce^{it(\alpha+2)}$, but

$\displaystyle\arg\\{ce^{it(\alpha+2)}\\}=\theta+t(\alpha+2)\geq\theta+\bigl{(}\frac{\phi-\theta}{\alpha}-\varepsilon\bigr{)}(\alpha+2)=$
$\displaystyle=2\frac{\pi\alpha-\theta}{\alpha}-\delta_{1}\frac{\alpha+2}{\alpha}-\varepsilon(\alpha+2),$
(12)

where $\delta_{1}=t_{0}(\alpha)-\phi\in(0,\delta)$. As
$\delta<(\pi\alpha-\theta)/2$, we have:

$\displaystyle
2\frac{\pi\alpha-\theta}{\alpha}-\delta_{1}\frac{\alpha+2}{\alpha}>2\frac{\pi\alpha-\theta}{\alpha}-\delta\frac{\alpha+2}{\alpha}=\frac{\alpha+2}{\alpha}\Bigl{(}\frac{2}{\alpha+2}(\pi\alpha-\theta)-\delta\Bigr{)}>$
$\displaystyle>\frac{\alpha+2}{\alpha}\Bigl{(}\frac{1}{2}(\pi\alpha-\theta)-\delta\Bigr{)}>0,$

reducing $\varepsilon>0$ if needed, we achieve $\arg\\{ce^{it(\alpha+2)}\\}>0$
(12).

On the other hand, $\arg\\{ce^{it(\alpha+2)}\\}\leq\arg c<\pi$ for $t\leq 0$,
i.e. $ce^{it(\alpha+2)}$ lies in the upper half-plane, therefore, for
sufficient large $R>R_{0}>0$ the left part of the last equality in (11) also
lies strictly in the upper half-plane, i.e. cannot be real.

For $z=Re^{it}\in\Gamma_{R}$ the value $ce^{it(\alpha+2)}$ lies in the upper
half-plane, $c^{1/2}e^{it(\alpha/2+1)}$ — in the Ist quarter, in particular
for $t=\arg\zeta_{0}-\varepsilon$. Thus, for $z\in l_{1}$, $z=|z|e^{it}$,
$|z|\to\infty$:

$S(z)\sim c^{1/2}e^{it(\alpha/2+1)}\frac{|z|^{\alpha/2+1}}{\alpha/2+1},$

the value $S(z)$ has to lie in the right half-plane due to the monotonic
growth of $\operatorname{Re}S(z)$ along $l_{1}$, this determines the choice of
the sign in the last formula, and $\operatorname{Re}S(z)\to+\infty$ as
$l_{1}\ni z\to\infty$.

Let $z\in\Gamma_{R}$, consider the analytic continuation of $S(z)$ through the
cut $\Delta$ from below, (7) takes the form:

$S(z)\sim c^{1/2}\frac{z^{\alpha/2+1}}{\alpha/2+1},\mbox{ uniformly in
}z\in\Gamma_{R},\ R\to+\infty.$ (13)

For $R\gg 1$, $z\in\Gamma_{R}$ the value $\operatorname{Re}S(z)$ lies in the
Ist quarter. Taking into account the monotonicity of $\operatorname{Re}S(z)$
along $\Gamma_{R}$, it follows from (13) that $\operatorname{Re}S(z)$
decreases with $z$ moving along $\Gamma_{R}$ counterclockwise for $R\gg 1$. In
view of the previous considerations along the rest parts of $l_{2,R}$, this
justifies the monotonicity of $\operatorname{Re}S(z)$ along the entire
$l_{2,R}$ for $R\gg 1$, without loss of generality, for $R>R_{0}>0$.

The formula (6) follows from (7) up to a sign. We conclude that
$\operatorname{Re}S(z)$ is unbounded for $z\to+\infty$. Since
$\operatorname{Re}S(z)$ decreases on $z\in[Z_{0},+\infty)$, we obtain
$\operatorname{Re}S(z)\to-\infty$, and the sign “minus” is selected in (6).

Let $z\in l_{3}$, $z\to\infty$. Then $z=|z|e^{i\pi/2+o(1)}\sim|z|e^{i\pi/2}$.
Using the asymptotics of the integral $S(z)$, we obtain:

$S(z)\sim-c^{1/2}e^{i\pi(\alpha/2+1)/2}\frac{|z|^{\alpha/2+1}}{\alpha/2+1},$

the term $e^{i\pi(\alpha/2+1)/2}$ lies in the IInd quarter, as
$\alpha\in(0,2)$; the term $c^{1/2}$ lies in the Ist quarter. The product
$c^{1/2}e^{i\pi(\alpha/2+1)/2}$ lies in the left half-plane, thus
$\operatorname{Re}S(z)$ is unbounded, $\operatorname{Re}S(z)\to+\infty$, and
the sign “minus” is used in the last asymptotic formula. $\Box$

The next Proposition is the main one in the current section.

######  Proposition 3.

For $k\gg 1$ and $t\geq 0$ there is the solution $\mathcal{W}_{0}(t,k)$ to
(4), for which the uniform asymptotic formulas are valid:

$\begin{split}\mathcal{W}_{0}(t,k)&=(ct^{\alpha}-\mu)^{-1/4}\bigl{\\{}-ie^{k(S(t)-2S(\zeta_{0}))}[1]+e^{-kS(t)}[1]\bigr{\\}},\quad
t\in[0,Z_{0}],\\\
\mathcal{W}_{0}(t,k)&=(ct^{\alpha}-\mu)^{-1/4}\bigl{\\{}-ie^{k(S(t)-2S(\zeta_{0}))}[1]\bigr{\\}},\quad
t>Z_{0},\end{split}$ (14)

where $[1]=1+O(k^{-1})$ is uniform in $t$ on the corresponding interval.

For $k\gg 1$, $\mathcal{W}_{0}(t,k)\in L_{2}(\mathbb{R}_{+})$ as a function of
$t$.

Proof. Further, let $z\in\mathbb{C}_{r}\cup\\{0\\}$.

The function $\operatorname{Re}S(z)$ increases along $l_{1}$. For $k\gg 1$, we
construct the subordinate solution to (4) at the infinite point of $l_{1}$,
according to [13, Ch.6 §12], of the form

$\mathcal{W}_{0}(z,k)=(cz^{\alpha}-\mu)^{-1/4}e^{-kS(z)}(1+\epsilon(z,k)),$
(15)

where

$|\epsilon(z,k)|\leq\exp\Bigl{\\{}\frac{\mathscr{V}_{z,\infty}(F)}{2k}\Bigr{\\}}-1,\
F=\int\frac{1}{q^{1/4}}\frac{d^{2}}{dz^{2}}\Bigl{(}\frac{1}{q^{1/4}}\Bigr{)}\,dz.$

The symbol $\mathscr{V}_{z,\infty}(F)$ denotes to the variation of the error-
control function $F(z)$ along the unbounded part of $l_{1}$, starting from
$z\in l_{1}$.

We estimate:

$\mathscr{V}_{z,\infty}(F)=\int\limits_{[z,\infty)\subset
l_{1}}\Bigl{|}\frac{1}{q^{1/4}}\frac{d^{2}}{dz^{2}}\Bigl{(}\frac{1}{q^{1/4}}\Bigr{)}\Bigr{|}\,|d\zeta|\leq
C\int\limits_{[z,\infty)\subset
l_{1}}\frac{|d\zeta|}{1+|\zeta|^{2+\alpha/2}}\leq\frac{C}{1+|z|^{1+\alpha/2}},$

thus $\epsilon(z,k)=O(k^{-1})$ uniformly in $z\in l_{1}$.

According to [13, Ch.5 §3], the solution $\mathcal{W}_{0}(z,k)$ can be
continued analytically to $z\in\mathbb{C}_{r}$.

Consider $l_{2,R}$ for $R>R_{0}>0$ (Proposition 2) and construct the solution
to (4) along $l_{2,R}$ subordinate at the infinite point of the ray $l_{1}\cap
l_{2,R}$. By construction, it coincides with $\mathcal{W}_{0}(z,k)$ along
$l_{1}\cap l_{2,R}$, thus it coincides with $\mathcal{W}_{0}(z,k)$ in
$\mathbb{C}_{r}$ due to the uniqueness of the analytic continuation.

The formula (15) remains along the entire $l_{2,R}$ with the remark, that when
passing through the cut $\Delta$ from below, one should consider the analytic
continuation of the main branches $q^{1/4}(z)$ and $S(z)$.

The new branches of $q^{1/4}(z)$ and $S(z)$ obtained as a result of this
continuation will be also considered in the domain
$\mathbb{C}_{r}\setminus\Delta$ (on the second sheet). Eventually, for
$z\in\mathbb{C}_{r}\setminus\Delta$, the new branch of $q^{1/4}(z)$ takes the
form of $iq^{1/4}(z)$. The new branch of $S(z)$ (denoted as $\tilde{S}(z)$)
takes the form of:

$\tilde{S}(z)=2S(\zeta_{0})-S(z).$ (16)

In terms of new branches, for $z\in(l_{3}\cap l_{2,R})\cup[Z_{0},R]$, $k\gg
1$:

$\displaystyle\mathcal{W}_{0}(z,k)=-i(cz^{\alpha}-\mu)^{-1/4}e^{-k\tilde{S}(z)}(1+\epsilon(z,k)),$
(17)
$\displaystyle|\epsilon(z,k)|\leq\exp\Bigl{\\{}\frac{\mathscr{V}_{z,\infty}(F)}{2k}\Bigr{\\}}-1,$

where the variation is considered along the unbounded part of $l_{2,R}$ with
the end at the infinite point of the ray $l_{1}\cap l_{2,R}$. Obviously:

$\displaystyle\mathscr{V}_{z,\infty}(F)<C,\mbox{ for }z\in l_{2,R},$
$\displaystyle\mathscr{V}_{z,\infty}(F)<C\Bigl{(}\int\limits_{z}^{R}+\int\limits_{\Gamma_{R}}+\int\limits_{[z,\infty)\subset
l_{1}}\Bigr{)}\frac{|d\zeta|}{|\zeta|^{2+\alpha/2}}\leq\frac{C}{z^{1+\alpha/2}}+\frac{C}{R^{1+\alpha/2}},\mbox{
for }z\in[Z_{0},R],$

where the constant $C>0$ does not depend on $R>R_{0}$.

Since the choice of $R>R_{0}$ is arbitrary, the formula (17) defines a
representation for $\mathcal{W}_{0}(z,k)$ along $l_{2}$ for $k\gg 1$ with the
uniform in $z\in l_{2}$ estimate $\epsilon(z,k)=O(k^{-1})$. This proves the
second formula in (14).

We turn to the path $l_{3}$, along which we construct two solutions to (4):
$u(z,k)$ — subordinate at $z=0$ and $v(z,k)$ — subordinate at the infinite
point of $l_{3}$. Like $\mathcal{W}_{0}(z,k)$, both $u(z,k)$ and $v(z,k)$ can
be analytically continued to $\mathbb{C}_{r}$. The following asymptotics are
valid for $k\gg 1$ uniformly in $z\in l_{3}$:

$\begin{split}u(z,k)&=(cz^{\alpha}-\mu)^{-1/4}e^{kS(z)}(1+O(k^{-1})),\\\
v(z,k)&=(cz^{\alpha}-\mu)^{-1/4}e^{-kS(z)}(1+O(k^{-1})).\end{split}$ (18)

Moreover, (18) guarantees the linear independence of $u(z,k)$ and $v(z,k)$ for
$k\gg 1$.

The constructed functions are solutions to a homogeneous second-order
equation. For each $k\gg 1$ there exist two constants $A(k)$ and $B(k)$:

$\mathcal{W}_{0}(z,k)=A(k)u(z,k)+B(k)v(z,k).$ (19)

To find $A(k)$ and $B(k)$, we use (15), (17), (18) at $z=0$ and at some
arbitrary large in absolute value point $z_{*}\in\\{Z_{0}+it,\ t\geq
0\\}\subset l_{3}$:

$\left\\{\begin{aligned} A(k)[1]+B(k)[1]&=1\\\
A(k)e^{kS(z_{*})}[1]+B(k)e^{-kS(z_{*})}[1]&=-ie^{-k\tilde{S}(z_{*})}\end{aligned}\right.,$

where $[1]=1+O(k^{-1})$ for $k\gg 1$ uniformly in $z\in\\{Z_{0}+it,\ t\geq
0\\}$.

Taking into account (16), the limit $\operatorname{Re}S(z)\to+\infty$ for
$l_{3}\ni z\to\infty$, the inequality $\operatorname{Re}S(\zeta_{0})>0$
(Proposition 2), finally:

$A(k)=-ie^{-2kS(\zeta_{0})}[1],\quad B(k)=[1].$

Combining this result with the formula (19) and the asymptotics (18), we get
the first formula in (14).

For $k\gg 1$, taking into account (6) and (14), we obtain that
$\mathcal{W}_{0}(t,k)$ is bounded and exponentially decaying as $t\to+\infty$,
thus $\mathcal{W}_{0}(t,k)\in L_{2}(\mathbb{R}_{+})$ as a function of $t$.
$\Box$

###  3.3 Step 3. The sufficient condition for the completeness

Note that the function $S(z)$, the critical points
$Z_{0}=(\operatorname{Im}\mu/\operatorname{Im}c)^{1/\alpha}>0$ and
$\zeta_{0}=(\mu/c)^{1/\alpha}$ depend on $\theta=\arg
c\in[t_{0}(\alpha),\pi)\cap[t_{0}(\alpha),\pi\alpha)$, and on
$\phi=\arg\mu\in(t_{0}(\alpha)-\delta,t_{0}(\alpha))$.

We determine:

$\rho(\theta,\phi)=\operatorname{Re}\bigl{\\{}S(Z_{0})-2S(\zeta_{0})\bigr{\\}}.$

For each $\theta\in[t_{0}(\alpha),\pi)\cap[t_{0}(\alpha),\pi\alpha)$ the
function $\rho(\theta,\phi)$ can be continuously extender to the point
$\phi=t_{0}(\alpha)$. Determine

$\rho(\theta)=\rho(\theta,t_{0}(\alpha))=\operatorname{Re}\bigl{\\{}S(Z_{0})-2S(\zeta_{0})\bigr{\\}}\Bigl{|}_{\phi=t_{0}(\alpha)}\Bigr{.}.$

######  Proposition 4.

The condition $\rho(\theta)<0$, $\theta=\arg
c\in[t_{0}(\alpha),\pi)\cap[t_{0}(\alpha),\pi\alpha)$ is sufficient for the
completeness of S.E. of the operator $\mathscr{L}_{c,\alpha}$.

Proof. Let $\rho(\theta)<0$. We take $\delta>0$,
$\delta<(\pi\min\\{1,\alpha\\}-\theta)/2$ to satisfy $\rho(\theta,\phi)<0$ for
all $\mu\in\mathfrak{l}_{\delta}$.

Recently with the substitution of the variable and the parameter, we
constructed the solution $\mathcal{W}_{0}(t,k)\in L_{2}(\mathbb{R}_{+})$ to
(4) for $k\gg 1$. The function
$\mathcal{Y}_{0}(|\lambda|^{1/\alpha}t,\lambda)\in L_{2}(\mathbb{R}_{+})$ is
also the solution to (4). Hereinafter $k=|\lambda|^{1/2+1/\alpha}$.

The eigenvalues of $\mathscr{L}_{c,\alpha}$ lie on the ray
$\arg\lambda=2\theta/(\alpha+2)$ (Lemma 1). As $\delta<(\pi\alpha-\theta)/2$,
we have: $2\theta/(\alpha+2)<t_{0}(\alpha)-\delta$, thus, there are no
eigenvalues inside $\mathfrak{l}_{\delta}$. Therefore for
$\lambda\in\mathfrak{l}_{\delta}$, the solution $y(x)$ to (1) with the initial
conditions $y(0)=0$, $y^{\prime}(0)=1$ is not in $L_{2}(\mathbb{R}_{+})$; the
subspace of solutions in $L_{2}(\mathbb{R}_{+})$ is one-dimensional. The same
is true for the equation (4). Thus for $k\gg 1$, there exists the function
$A(k)$ such that

$\mathcal{Y}_{0}(|\lambda|^{1/\alpha}t,\lambda)=A(k)\mathcal{W}_{0}(t,k).$
(20)

Let us assume the contrary, that the S.E. of $\mathscr{L}_{c,\alpha}$ is
incomplete. Then there exists $f\in L_{2}(\mathbb{R}_{+})$, $f\not\equiv 0$,
and for all eigenvalues $\\{\lambda_{n}\\}$ of the operator
$\mathscr{L}_{c,\alpha}$, $\mathcal{G}(\lambda_{n})=0$ (2). In this case
$\mathcal{F}(\lambda)$ (3) is an entire function. We estimate it for
$|\lambda|\gg 1$ on an arbitrary ray
$\mathfrak{l}\subset\mathfrak{l}_{\delta}$ with fixed $\phi=\arg\lambda$.

$\displaystyle|\mathcal{F}(\lambda)|^{2}\leq
C\frac{1}{|\mathcal{Y}_{0}(0,\lambda)|^{2}}\int\limits_{0}^{+\infty}|\mathcal{Y}_{0}(x,\lambda)|^{2}\,dx=C\frac{1}{|\mathcal{Y}_{0}(0,\lambda)|^{2}}|\lambda|^{1/\alpha}\int\limits_{0}^{+\infty}|\mathcal{Y}_{0}(|\lambda|^{1/\alpha}t,\lambda)|^{2}\,dt=$
$\displaystyle=C\frac{1}{|\mathcal{Y}_{0}(0,\lambda)|^{2}}|\lambda|^{1/\alpha}\int\limits_{0}^{+\infty}|A(k)\mathcal{W}_{0}(t,k)|^{2}\,dt=C\frac{1}{|\mathcal{W}_{0}(0,k)|^{2}}|\lambda|^{1/\alpha}\int\limits_{0}^{+\infty}|\mathcal{W}_{0}(t,k)|^{2}\,dt\leq$
$\displaystyle\leq
C|\lambda|^{1/\alpha}\Bigl{\\{}\int\limits_{0}^{Z_{0}}(e^{2k\operatorname{Re}(S(t)-2S(\zeta_{0}))}+e^{-2k\operatorname{Re}S(t)})\,dt+\int\limits_{Z_{0}}^{+\infty}e^{2k\operatorname{Re}(S(t)-2S(\zeta_{0}))}\,dt\Bigr{\\}}\leq$
$\displaystyle\leq
C|\lambda|^{1/\alpha}\bigl{\\{}e^{2k\operatorname{Re}(S(Z_{0})-2S(\zeta_{0}))}+1\bigr{\\}}=C|\lambda|^{1/\alpha}\bigl{\\{}e^{2k\rho(\theta,\phi)}+1\bigr{\\}}<C|\lambda|^{1/\alpha}.$

The first estimation follows from the Cauchy–Bunyakovsky inequality and the
condition $f\in L_{2}(\mathbb{R}_{+})$; then we make the substitution
$x=|\lambda|^{\alpha}t$; then we use (20) and boundedness of
$|ct^{\alpha}-\mu|^{-1/4}$ along $\mathfrak{l}$. We take into account that
$\operatorname{Re}S(t)\geq 0$ for $t\in[0,Z_{0}]$. The final estimation
follows from Laplace method. Everywhere $C$ does not depend on $k$.

By construction $0<t_{0}(\alpha)-\delta<\phi<t_{0}(\alpha)\leq\theta$. As
$\delta<\pi\alpha-\theta$, it follows: $\delta<4\pi\alpha/(\alpha+2)-\theta$,
thus $\theta-t_{0}(\alpha)+\delta<t_{0}(\alpha)$.

In view of these estimates, we can form two adjacent sectors $\Lambda_{1}$,
$\Lambda_{2}$ with a common boundary passing along the ray $\mathfrak{l}$, so
that 1) the central angle of each sector is less than $t_{0}(\alpha)$, 2) the
second boundary of each sector lies outside
$\Lambda=\\{\arg\lambda\in[0,\theta]\\}$.

The order of growth of $\mathcal{F}(\lambda)$ is at most $\pi/t_{0}(\alpha)$
(Lemma 2). The function $\mathcal{F}(\lambda)$ is bounded on the rays
$\arg\lambda\not\in\Lambda$ and grows no faster than a polynomial on the ray
$\mathfrak{l}$. It follows from the Phragmén-Lindelöf Principle that in each
of the sectors $\Lambda_{j}$ ($j=1,2$) the function $\mathcal{F}(\lambda)$
grows no faster than a polynomial, and therefore in the whole plane
$\mathbb{C}=\Lambda\cup\Lambda_{1}\cup\Lambda_{2}$. In this way, the entire
function $\mathcal{F}(\lambda)$ is a polynomial itself. Because of its
boundedness on the rays $\arg\lambda\not\in\Lambda$, it follows that
$\mathcal{F}(\lambda)\equiv\mbox{const}$.

Applying Lemma 3, we conclude that $f\equiv 0$. We arrived at a contradiction
by assuming that there is no completeness with $\rho(\theta)<0$. $\Box$

###  3.4 Step 4. Verification of the sufficient condition

######  Proposition 5.

The function $\rho(\theta)$ increases for
$\theta\in[t_{0}(\alpha),\pi)\cap[t_{0}(\alpha),\pi\alpha)$, takes values of
different signs in the neighborhoods of the boundaries of this interval.

The only zero of $\rho(\theta)$, $\theta_{0}(\alpha)=t_{0}(\alpha)+\Delta
t(\alpha)$ lies strictly inside the interval, in particular, $\Delta
t(\alpha)>0$, and is a continuous function for $\alpha\in(0,2)$.

Thus, the sufficient conditions for the completeness of S.E. of the operator
$\mathscr{L}_{c,\alpha}$ are satisfied for all $\arg
c\in[t_{0}(\alpha),\theta_{0}(\alpha))$.

Proof. Denote, $\mu_{0}=e^{it_{0}(\alpha)}$.

The function $\rho(\theta)$ is continuous at $\theta=t_{0}(\alpha)$ and
$\rho(t_{0}(\alpha))=-\operatorname{Re}S(\zeta_{0})<0$, since in this case
$Z_{0}=\zeta_{0}$.

Exploring the second boundary value, consider two cases: $0<\alpha<1$ and
$1\leq\alpha<2$.

Let $0<\alpha<1$ and $\theta\to\pi\alpha-0$. We substitute the variable in the
integral:

$S(\zeta_{0})=-i\frac{\mu_{0}^{1/\alpha+1/2}}{c^{1/\alpha}}\int\limits_{0}^{1}\sqrt{1-\xi^{\alpha}}\,d\xi=\frac{i}{c^{1/\alpha}}\int\limits_{0}^{1}\sqrt{1-\xi^{\alpha}}\,d\xi,$

Obviously, $\operatorname{Re}S(\zeta_{0})\to 0$, therefore
$\rho(\theta)\sim\operatorname{Re}S(Z_{0})\Bigl{|}_{\theta=\pi\alpha}\Bigr{.}>0$.

Let $1\leq\alpha<2$ and $\theta\to\pi-0$. Then $\operatorname{Im}c\to 0$,
$\operatorname{Re}c\to-1$, $\operatorname{Re}c^{1/2}\sim\operatorname{Im}c/2$,
$Z_{0}\to+\infty$. In this case $\operatorname{Re}S(\zeta_{0})$ is bounded.

Let us estimate $\operatorname{Re}S(Z_{0})$:

$\displaystyle\operatorname{Re}S(Z_{0})>\int\limits_{Z_{0}/2}^{Z_{0}}\operatorname{Re}\sqrt{c\zeta^{\alpha}-\mu_{0}}\,d\zeta=Z_{0}\int\limits_{1/2}^{1}\operatorname{Re}\sqrt{c\xi^{\alpha}\frac{\operatorname{Im}\mu_{0}}{\operatorname{Im}c}-\mu_{0}}\,d\xi>$
$\displaystyle>CZ_{0}\frac{(\operatorname{Im}\mu_{0})^{1/2}}{(\operatorname{Im}c)^{1/2}}\operatorname{Re}c^{1/2}\int\limits_{1/2}^{1}\xi^{\alpha/2}(1+O(\operatorname{Im}c))\,d\xi>\frac{C}{(\operatorname{Im}c)^{1/\alpha-1/2}};$

the first estimate is legal, since $\operatorname{Re}S(z)$ is positive and
increasing in the interval $[0,Z_{0}]$. Then we substitute the variable
$\zeta=Z_{0}\xi$; the explicit expression is used for
$Z_{0}=(\operatorname{Im}\mu_{0}/\operatorname{Im}c)^{1/\alpha}$. The two
final inequalities are valid for $\operatorname{Im}c\ll 1$. The estimation
$O(\operatorname{Im}c)$ is uniform in $\xi\in[1/2,1]$.

As well as $\alpha<2$, we conclude $\operatorname{Re}S(Z_{0})\to+\infty$, thus
$\rho(\theta)\to+\infty$.

Now we prove the strict increase of $\rho(\theta)$ on the interval
$\theta\in(t_{0}(\alpha),\pi)\cap(t_{0}(\alpha),\pi\alpha)$. Note,

$\displaystyle\rho(\theta)=\operatorname{Re}\Bigl{\\{}\int\limits_{\zeta_{0}}^{Z_{0}}\sqrt{c\zeta^{\alpha}-\mu_{0}}\,d\zeta-\int\limits_{0}^{\zeta_{0}}\sqrt{c\zeta^{\alpha}-\mu_{0}}\,d\zeta\Bigr{\\}}=$
$\displaystyle=\operatorname{Re}\int\limits_{\zeta_{0}}^{Z_{0}}\sqrt{c\zeta^{\alpha}-\mu_{0}}\,d\zeta-\sin\frac{\theta}{\alpha}\int\limits_{0}^{1}\sqrt{1-\xi^{\alpha}}\,d\xi.$
(21)

We turn to the first term in (21), denoting

$I(\theta)=\operatorname{Re}\int\limits_{\zeta_{0}}^{Z_{0}}\sqrt{c\zeta^{\alpha}-\mu_{0}}\,d\zeta.$

For $I(\theta)$ we will carry out the integration along the path
$\gamma=\\{\zeta=\zeta(t),\ t\in[0,\tau]\\}$,
$\zeta(t)=((\mu_{0}-t)/c)^{1/\alpha}$,
$\tau=\operatorname{Im}(c/\mu_{0})/\operatorname{Im}c>0$.

The $(\mu_{0}-t)/c$ – image of the interval $t\in[0,\tau]$ is a segment in the
IVth quarter (due to the location of endpoints — Proposition 1). The IVth
quarter also contains the path $\gamma$ — this is true for endpoints by virtue
of the same Proposition 1, and taking into account the continuity of the
$\arg$ function — for the whole path $\gamma$.

For the main branch
$q^{1/2}(\zeta)=\sqrt{c\zeta^{\alpha}-\mu_{0}}=-it^{1/2}(\zeta)$ along
$\gamma$. Indeed, according to Proposition 2,
$\operatorname{Im}q^{1/2}(Z_{0})<0$, i.e $q^{1/2}(Z_{0})=-i\tau^{1/2}$; for
other points of $\gamma$ the sign is preserved due to the continuity of the
square root.

To shorten the notation, we denote the partial derivatives by indices with
respect to the corresponding variables. For example, $\zeta_{t}$ is the
derivative of the parameterization function for $\gamma$.

We have:

$I(\theta)=\int\limits_{0}^{\tau}t^{1/2}\operatorname{Im}\zeta_{t}\,dt,\quad
I_{\theta}(\theta)=\tau^{1/2}\operatorname{Im}\zeta_{t}(\tau)\tau_{\theta}+\int\limits_{0}^{\tau}t^{1/2}\operatorname{Im}\zeta_{\theta
t}\,dt.$ (22)

Obviously, $c_{\theta}=ic$,
$\tau_{\theta}=-(\operatorname{Re}c/\operatorname{Im}c)_{\theta}\operatorname{Im}\mu_{0}=-(\operatorname{Re}c_{\theta}\operatorname{Im}c-\operatorname{Im}c_{\theta}\operatorname{Re}c)\operatorname{Im}\mu_{0}/(\operatorname{Im}c)^{2}$,
thus $\tau_{\theta}=\operatorname{Im}\mu_{0}/(\operatorname{Im}c)^{2}$, as
well as $|c|=1$.

Further, $\zeta_{t}(\tau)=-(1/\alpha)\zeta(\tau)/(\mu_{0}-\tau)$. As
$\zeta(\tau)=Z_{0}$ and
$\mu_{0}-\tau=cZ_{0}^{\alpha}=c\operatorname{Im}\mu_{0}/\operatorname{Im}c$,
we obtain $\operatorname{Im}\zeta_{t}(\tau)\tau_{\theta}=Z_{0}/\alpha$. Also
note, that $\zeta_{\theta}=\zeta_{c}c_{\theta}=ic\zeta_{c}=-i\zeta/\alpha$.

Combining these results with (22), we get:

$I_{\theta}(\theta)=\frac{1}{\alpha}\tau^{1/2}Z_{0}-\frac{1}{\alpha}\int\limits_{0}^{\tau}t^{1/2}\operatorname{Re}\zeta_{t}\,dt.$

Applying the second mean value theorem for the integral term, we find some
intermediate point $\vartheta\in[0,\tau]$ for which

$I_{\theta}(\theta)=\frac{1}{\alpha}\tau^{1/2}Z_{0}-\frac{1}{\alpha}\tau^{1/2}(\operatorname{Re}\zeta(\tau)-\operatorname{Re}\zeta(\vartheta))=\frac{1}{\alpha}\tau^{1/2}\operatorname{Re}\zeta(\vartheta)>0,$

we have used the facts that all points of $\gamma$ lie strictly in the right
half-plane, and $\tau>0$.

Consider the second term in (21):

$J(\theta)=\sin\frac{\theta}{\alpha}\int\limits_{0}^{1}\sqrt{1-\xi^{\alpha}}\,d\xi.$

Since $t_{0}(\alpha)<\theta<\pi\alpha$, then
$\pi/2<2\pi/(\alpha+2)<\theta/\alpha<\pi$, therefore:

$J_{\theta}(\theta)=\frac{1}{\alpha}\cos\frac{\theta}{\alpha}\int\limits_{0}^{1}\sqrt{1-\xi^{\alpha}}\,d\xi<0,$

after all we obtain that
$\rho_{\theta}(\theta)=I_{\theta}(\theta)-J_{\theta}(\theta)>0$. The function
$\rho(\theta)$ is strictly increasing, has the only zero
$\theta_{0}=\theta_{0}(\alpha)=t_{0}(\alpha)+\Delta
t(\alpha)\in(t_{0}(\alpha),\pi)\cap(t_{0}(\alpha),\pi\alpha)$. For all
$\theta\in[t_{0}(\alpha),\theta_{0}(\alpha))$, the value $\rho(\theta)<0$.

For any $\alpha_{0}\in(0,2)$, $\theta_{0}=\theta_{0}(\alpha_{0})$, consider
$\rho(\theta)=R(\alpha,\theta)$ as a function of two variables: $\alpha$ and
$\theta$ in the small neighborhood $\Upsilon$ of $(\alpha_{0},\theta_{0})$.

From the representation (21) and the first formula of (22) we obtain the
existence of a small neighborhood $\Upsilon=\Upsilon(\alpha_{0},\theta_{0})$,
in which $R(\alpha,\theta)$ is continuous as a function of two variables.

Since $R(\alpha_{0},\theta_{0})=0$ and $R(\alpha_{0},\theta)$ is strictly
monotone as a function of $\theta$ for $(\alpha_{0},\theta)\in\Upsilon$, the
Implicit Function Theorem implies the continuity of $\theta_{0}(\alpha)$ in
some small neighborhood of the point $\alpha_{0}$. Since $\alpha_{0}$ is
arbitrary, it’s valid for all $\alpha\in(0,2)$. $\Box$

We have proved the completeness of the of S.E. of the operator
$\mathscr{L}_{c,\alpha}$ in case $t_{0}(\alpha)\leq|\arg
c|<\theta_{0}(\alpha)$. As already noted, the completeness in case $|\arg
c|<t_{0}(\alpha)$ is a known fact [6]. This completes the proof of the Theorem
1.

##  4 Annex

We turn to the complex Airy operator to evaluate $\theta_{0}(1)$ as an
example. Let $\alpha=1$; taking into account (21), (22) and the evaluation
$\zeta_{t}(t)=-1/c$, we obtain:

$\rho(\theta)=\int\limits_{0}^{\tau(\theta)}t^{1/2}\sin\theta\,dt-\frac{2}{3}\sin\theta=\frac{2}{3}(\tau^{3/2}(\theta)-1)\sin\theta.$

We solve the equation $\rho(\theta_{0})=0$ for
$\theta_{0}=\theta_{0}(1)\in(t_{0}(1),\pi)=(2\pi/3,\pi)$. The only possible
case — $\tau(\theta_{0})=1$. As well as
$\tau(\theta)=\operatorname{Im}(c/\mu_{0})/\operatorname{Im}c=\sin(\theta-2\pi/3)/\sin\theta$,
the equation on $\theta_{0}$ takes the form of:

$\sin(\theta_{0}-2\pi/3)=\sin\theta_{0},$

and it has the only solution $\theta_{0}=5\pi/6$, which exactly corresponds to
the result of Savchuk and Shkalikov [6].

The author expresses his deep appreciation to Andrei Andreyevich Shkalikov for
his attention to the work and valuable advice, as well as to the team of the
scientific seminar ”Operator Models in Mathematical Physics“ for their
support.

This research was supported by RFBR grant No 19-01-00240.

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