\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 95, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/95\hfil Two-point boundary-value problems]
{Two-point boundary-value problems with
nonclassical asymptotics on the spectrum}

\author[A. Makin \hfil EJDE-2018/95\hfilneg]
{Alexander Makin}

\address{Alexander Makin \newline
Moscow Technological University,
Moscow, Russia}
\email{alexmakin@yandex.ru}

\dedicatory{Communicated by Ludmila S. Pulkina}

\thanks{Submitted February 25, 2018. Published April 19, 2018.}
\subjclass[2010]{34L20}
\keywords{Ordinary differential operator; degenerate boundary conditions; 
\hfill\break\indent spectrum}

\begin{abstract}
 In this article, we consider the spectral problem for an nth-order 
 ordinary differential operator with degenerate boundary conditions.
 For even $n$, we construct nontrivial examples of boundary-value
 problems which have nonclassical asymptotics on the spectrum.
\end{abstract}

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

\section{Introduction}

 Let us consider the boundary-value problem generated by the $n$-th order 
differential equation
\begin{equation}
u^{(n)}(x)+\sum_{m=1}^{n}p_m(x)u^{(n-m)}(x)+\lambda u(x)=0,\label{e1.1}
\end{equation}
where the complex-valued coefficients $p_m(x)$ are functions in 
$L_1(0,\pi)$, with the linearly independent boundary conditions
\begin{equation}
\sum_{k=0}^{n-1}\alpha_{i,k}u^{(k)}(0)+\beta_{i,k}u^{(k)}(\pi)=0,\quad
i=1,\ldots,n,\label{e1.2}
\end{equation}
where $\alpha_{i,k}, \beta_{i,k}$ are complex numbers.
It is well known that the characteristic determinant of \eqref{e1.1}, \eqref{e1.2}
is an entire analytical function of spectral parameter $\lambda$.
Consequently, for operator \eqref{e1.1}, \eqref{e1.2} we have only the
following possibilities:
\begin{itemize}
\item [(a)] the spectrum is absent;
\item [(b)] the spectrum is a finite nonempty set;
\item [(c)] the spectrum is a countable set without finite limit points;
\item [(d)] the spectrum fills the entire complex plane.
\end{itemize}
We say that  problem \eqref{e1.1}, \eqref{e1.2} has the classical asymptotics
on the spectrum if the case (c) is realized,  moreover the multiplicities
of the eigenvalues are bounded by a single constant.
For the Sturm-Liouville equation
\begin{equation}
Lu+\lambda u=0, \label{e1.3}
\end{equation}
where $Lu=u''-q(x)u$, with nondegenerate boundary conditions the spectrum
always has the classical asymptotics \cite{Mar}. For equation \eqref{e1.3} with
degenerate boundary conditions
\begin{equation}
u'(0)+du'(\pi)=0,\quad u(0)-du(\pi)=0, \label{e1.4}
\end{equation}
another situation takes place.
In particular,  under  the condition that $d\ne0$ it follows from \cite{Mak}
that for any natural $m$ there exist potentials $q(x)$ in the class
$W_2^m(0,\pi)$ such that the root function system of problem
 \eqref{e1.3}, \eqref{e1.4} contains associated functions of arbitrary high order.
If $d=0$, problem \eqref{e1.3}, \eqref{e1.4} is the Cauchy problem which
has no spectrum. Note, that for the Sturm-Liouville operator
any two-point conditions  are nondegenerate except \eqref{e1.4}.
There is an enormous literature related to the spectral theory for operators
with nondegenerate boundary conditions.The case of  degenerate boundary
conditions has been investigated much less. However,
 it is known \cite{Akh,Loc,Sad} that there exist operators of high order,
where any complex number is an eigenvalue. The main goal of present
paper is to construct nontrivial examples of boundary value problems
for high order operators such that the spectrum is absent or the spectrum
is a countable set but the multiplicities of eigenvalues infinitely grow.

\section{Unbounded growth of order for associated functions}

For any even $n=2\nu$ with $\nu>1$, let us build  an example of boundary-value 
problem \eqref{e1.1}, \eqref{e1.2},  for which the multiplicities of 
eigenvalues grow infinitely. Consider problem \eqref{e1.3}, \eqref{e1.4} 
$(d\ne0)$ with a potential $q(x)\in W_2^m(0,\pi)$, where  $m=2\nu+2$, 
providing infinite growth  of the  multiplicities of eigenvalues. 
Then by the embedding theorem  $q(x)\in C^{(2\nu+1)}[0,\pi]$. 
Let $\{u_n(x)\}$ be the root function system of problem \eqref{e1.3}, \eqref{e1.4} 
with the above-mentioned potential.  Obviously, $u_n(x)\in C^{(2\nu+1)}[0,\pi]$. 
Let us prove that for any $j=0,1,\ldots,2\nu$,
\begin{equation}
q(0)=(-1)^jq(\pi). \label{e2.1}
\end{equation}
Denote by $c(x,\mu)$, $s(x,\mu)$ $(\lambda=\mu^2)$ the fundamental system
of solutions to \eqref{e1.3} with the initial conditions
 $c(0,\mu)=s'(0,\mu)=1$, $c'(0,\mu)=s(0,\mu)=0$.
In \cite{Mar} simple computations show  that  the characteristic equation of problem
\eqref{e1.3}, \eqref{e1.4} can be reduced to the form $\Delta(\mu)=0$,
where
\begin{equation}
\Delta(\mu)=\frac{d^2-1}{d}+c(\pi,\mu)-s'(\pi,\mu)=
\frac{d^2-1}{d}+\int_0^\pi r(t)\frac{\sin\mu t}{\mu}dt, \label{e2.2}
\end{equation}
where $r(t)\in C^{(2\nu+1)}[0,\pi]$. Let $k$ be the least whole number
$(0\le k\le2\nu)$, provided that equality \eqref{e2.1} does not hold.
Integrating  by parts $k+1$ times the last addend on the right-hand side
of  equality \eqref{e2.2}, from  \cite{Mal}, we obtain
$$
\Delta(\mu)=\sum_{j=1}^{k+1}\frac{\alpha_j}{\mu^{j+1}}
+\frac{B_{k+1}\sin\pi\mu}{\mu^{k+2}}
-\frac{1}{\mu^{k+2}}\int_0^\pi r^{(k+1)}(t)\sin\mu t dt
$$
for odd $k$ and
$$
\Delta(\mu)=\sum_{j=1}^{k+1}\frac{\alpha_j}{\mu^{j+1}}
+\frac{B_{k+1}\cos\pi\mu}{\mu^{k+2}}
+\frac{1}{\mu^{k+2}}\int_0^\pi r^{(k+1)}(t)\cos\mu t dt
$$
for even $k$. In both cases coefficients $\alpha_j$ are some numbers, and
$$
B_{k+1}=(-1)^{k+1}r^{(k)}(\pi)=(-1)^{k+1}(q^{(k)}(\pi)-(-1)^kq^{(k)}(0))/2^{k+1}
\ne0.
$$
Hence, it follows that problem \eqref{e1.3}, \eqref{e1.4} is almost-regular
in sense of \cite{Shk}, therefore, the multiplicities of eigenvalues are bounded
by a single constant, i.e. we receive a contradiction, hence,
equality \eqref{e2.1} is valid.

Further, consider the problem
\begin{gather}
L^\nu u+(-1)^{\nu-1}\lambda^\nu u=0, \label{e2.3} \\
u^{(2\nu-j)}(0)+d(-1)^{j+1}u^{(2\nu-j)}(\pi)=0,\label{e2.4}
\end{gather}
$j=1,\ldots,2\nu$, where $d$ is an arbitrary complex number $(d\ne0)$.

\begin{lemma} \label{lem2.1}
  The functions $u_n(x)$ satisfy boundary conditions \eqref{e2.4}.
  \end{lemma}

 \begin{proof}
Let us prove the lemma by induction.
 Obviously, equalities \eqref{e2.4} hold
if $j=2\nu, 2\nu-1$. Suppose, that the functions $u_n(x)$ satisfy 
equalities \eqref{e2.4} if $j=2\nu,2\nu-1,\ldots,2\nu-l$, where 
$1\le l\le2\nu-1$. Consider equality
\begin{equation}
u_n''(x)-q(x)u_n(x)+\lambda_n u_n(x)=u_{n-1}(x), \label{e2.5}
\end{equation}
 where $u_{n-1}(x)$  is an associated function per unit of lower order
corresponding to a function $u_n(x)$. If $u_n(x)$ is an eigenfunction
then the right-hand side of  equality \eqref{e2.5} equals zero identically.
Differentiating equality \eqref{e2.5} $2\nu-l-1$ times we obtain
\begin{equation}
\begin{aligned}
& u_n^{(2\nu-l+1)}(x)-\sum_{m=0}^{2\nu-l-1}C_{2\nu-l-1}^mq^{(m)}(x)
 u_n^{(2\nu-l-1-m)}(x)  +\lambda_n u_n^{(2\nu-l-1)}(x) \\
&=u_{n-1}^{(2\nu-l-1)}(x).
\end{aligned} \label{e2.6}
\end{equation}
It follows by the inductive hypothesis, equalities \eqref{e2.1} and \eqref{e2.6}
that
\begin{equation}
\begin{aligned}
&u_n^{(2\nu-l+1)}(0)+d(-1)^{l}u_n^{(2\nu-l+1)}(\pi) \\
&=\sum_{m=0}^{2\nu-l-1}C_{2\nu-l-1}^mq^{(m)}(0)u_n^{(2\nu-l-3-m)}(0)
 -\lambda_nu_n^{(2\nu-l-1)}(0) \\
&\quad +u_{n-1}^{(2\nu-l-1)}(0)
 +d(-1)^{l}[\sum_{m=0}^{2\nu-l-1}C_{2\nu-l-1}^mq^{(m)}(\pi)u_n^{(2\nu-l-1-m)}(\pi) \\
&\quad -\lambda_nu_n^{(2\nu-l-1)}(\pi)+u_{n-1}^{(2\nu-l-1)}(\pi)] \\
&=\sum_{m=0}^{2\nu-l-1}C_{2\nu-l-1}^m(q^{(m)}(0)u_n^{(2\nu-l-1-m)}(0) \\
&\quad +d(-1)^{l}q^{(m)}(\pi)u_n^{(2\nu-l-1-m)}(\pi)) \\
&\quad -\lambda_n(u_n^{(2\nu-l-1)}(0)+d(-1)^{l}u_n^{(2\nu-l-1)}(\pi)) \\
&\quad +(u_{n-1}^{(2\nu-l-1)}(0)+d(-1)^{l}u_{n-1}^{(2\nu-l-1)}(\pi)) \\
&=\sum_{m=0}^{2\nu-l-1}q^{(m)}(0)(C_{2\nu-l-1}^m(u_n^{(2\nu-l-1-m)}(0)
+d(-1)^{m+l}u_n^{(2\nu-l-1-m)}(\pi)) \\
&=0.
\end{aligned} \label{e2.7}
\end{equation}
\end{proof}

Let a function $\stackrel{0}{u}(x)$ be an arbitrary solution of the equation
\begin{equation}
L\stackrel{0}{u}+\stackrel{0}{u}=0,\label{e2.8}
\end{equation}
and a function $\stackrel{i}{u}(x)$ be an arbitrary solution of the equation
\begin{equation}
L\stackrel{i}{u}+\lambda\stackrel{i}{u}=\stackrel{i-1}{u},\label{e2.9}
\end{equation}
$i=1,2,\ldots$. Formally set $\stackrel{i}{u}\equiv0$ if $i=-1,-2,\ldots$.


\begin{lemma} \label{lem2.2}
For any $p=1,2,\ldots$ we have
\begin{equation}
L^p\stackrel{i}{u}=\sum_{k=0}^p(-1)^{p-k}C_p^k\lambda^{p-k}\stackrel{i-k}{u}.
\label{e2.10}
\end{equation}
 \end{lemma}

\begin{proof}
Let us prove the lemma by induction with respect to $p$. If $p=1$ then 
relations \eqref{e2.8}, \eqref{e2.9} imply \eqref{e2.10}. 
Let the lemma be valid for a natural  $p$.
It follows by the inductive hypothesis and the properties of the binomial 
coefficients that
\begin{align*}
L^{p+1}\stackrel{i}{u}
&=L(\sum_{k=0}^p(-1)^{p-k}C_p^k\lambda^{p-k}\stackrel{i-k}{u}) \\
&=\sum_{k=0}^p(-1)^{p-k}C_p^k\lambda^{p-k}L(\stackrel{i-k}{u}) \\
&=\sum_{k=0}^p(-1)^{p-k}C_p^k\lambda^{p-k}(\stackrel{i-k-1}{u}
 -\lambda\stackrel{i-k}{u}) \\
&=\sum_{k=0}^p(-1)^{p-k}C_p^k\lambda^{p-k}\stackrel{i-(k+1)}{u}
 -\sum_{k=0}^p(-1)^{p-k}C_p^k\lambda^{p+1-k}\stackrel{i-k}{u} \\
&=\sum_{m=1}^{p+1}(-1)^{p-m+1}C_p^{m-1}\lambda^{p-m+1}\stackrel{i-m)}{u}
 -\sum_{m=0}^p(-1)^{p-m}C_p^m\lambda^{p+1-m}\stackrel{i-m}{u} \\
&=\stackrel{i-(p+1)}{u}+\sum_{m=1}^{p}[(-1)^{p-m+1}C_p^{m-1}\lambda^{p-m+1}
 -(-1)^{p-m}C_p^m\lambda^{p+1-m}]\stackrel{i-m}{u} \\
&\quad -(-1)^{p}\lambda^{p+1}\stackrel{i}{u}=\stackrel{i-(p+1)}{u}
 +\sum_{m=1}^{p}[(-1)^{p-m+1}\lambda^{p-m+1}(C_p^{m-1}+C_p^m]\stackrel{i-m}{u} \\
&\quad -(-1)^{p}\lambda^{p+1}\stackrel{i}{u} \\
&=\sum_{m=0}^{p+1}(-1)^{p-m+1}\lambda^{p-m+1}C_{p+1}^m\stackrel{i-m}{u}.
\end{align*}
\end{proof}

Denote $\Lambda=(-1)^{p-1}\lambda^p$ $(p=1,2,\ldots)$.

\begin{lemma} \label{lem2.3}
Let $\lambda\ne0$.  If $\stackrel{i}{v}=\sum_{j=0}^i a_j\stackrel{j}{u}$, 
where  $a_j$ are some numbers, and $a_i\ne0$, then  there exists a function
$\stackrel{i+1}{v}=\sum_{j=0}^{i+1}b_j\stackrel{j}{u}$, where  $b_j$ 
are some numbers, and $b_{i+1}\ne0$ such that 
$L^p\stackrel{i+1}{v}+\Lambda\stackrel{i+1}{v}=\stackrel{i}{v}$.
\end{lemma}

\begin{proof}
It follows by Lemma \ref{lem2.2} that
\begin{align*}
& L^p\stackrel{i+1}{v}+\Lambda\stackrel{i+1}{v} \\
& =L^p(\sum_{j=0}^{i+1}b_j\stackrel{j}{u})+(-1)^{p-1}\lambda^p\stackrel{i+1}{v} \\
& =\sum_{j=0}^{i+1}b_jL^p\stackrel{j}{u}+(-1)^{p-1}\lambda^p\stackrel{i+1}{v} \\
& =\sum_{j=0}^{i+1}b_j\sum_{k=0}^p(-1)^{p-k}C_p^k\lambda^{p-k}\stackrel{j-k}{u} \\
&\quad +(-1)^{p-1}\lambda^p\stackrel{i+1}{v}=(-1)^{p}\lambda^p
 \sum_{j=0}^{i+1}b_j\stackrel{j}{u}+\sum_{j=0}^{i+1}b_j\sum_{k=1}^p(-1)^{p-k}
 C_p^k\lambda^{p-k}\stackrel{j-k}{u} \\
&\quad +(-1)^{p-1}\lambda^p\stackrel{i+1}{v} \\
&=\sum_{j=0}^{i+1}b_j\sum_{k=1}^p(-1)^{p-k}C_p^k\lambda^{p-k}\stackrel{j-k}{u}\\
&=\sum_{j=0}^{i+1}b_j\sum_{k=1}^p\gamma_k\stackrel{j-k}{u},
\end{align*}
where $\gamma_k=(-1)^{p-k}C_p^k\lambda^{p-k}$. 
Equating the coefficients  at the functions $\stackrel{i-m}{u}$
in the relation
$$
\sum_{j=0}^{i+1}b_j\sum_{k=1}^p\gamma_k\stackrel{j-k}{u}
=\sum_{l=0}^i a_l\stackrel{l}{u},
$$
we obtain the system of linear equations
\begin{equation}
\sum_{l=0}^{m}\gamma_{l+1}b_{i+1-l}=a_{i-m},\label{e2.11}
\end{equation}
$m=0,\ldots,i$. The matrix of system \eqref{e2.11} is lower triangular,
and all the elements of the principal diagonal are equal to
 $\gamma_1=(-1)^{p-1}p\lambda^{p-1}\ne0$.
Therefore, system \eqref{e2.11} has the unique solution.
Since $a_i\ne0$, we have  $b_{i+1}=a_i/\gamma_1\ne0$.
\end{proof}

Let $u_n(x)$ be an associated  function of order  $k$ corresponding to 
an eigenvalue  $\lambda_n\ne0$, and functions  
$\{u_{n-j}(x)\}$  $(j=0,\ldots,k)$ form the corresponding Jordan chain, i.e.
\begin{gather*}
Lu_{n-j}(x)+\lambda_n u_{n-j}(x)=u_{n-j-1}(x)\quad 
(j=0,\ldots,k-1), \\
Lu_{n-k}(x)+\lambda_n u_{n-k}(x)=0.
\end{gather*}
 By lemma \ref{lem2.1}, the function $u_{n-k}(x)$ is an eigenfunction  of 
problem \eqref{e2.3}, \eqref{e2.4} corresponding to the eigenvalue  
$\Lambda_n=(-1)^{\nu-1}\lambda_n^\nu$.
Set $v_{n-k}(x)=u_{n-k}(x)$. Then, by lemma \ref{lem2.3}, it follows that there 
exist functions
$$
v_{n-k+i}(x)=\sum_{j=1}^ib_{ij}u_{n-k+i}(x)
$$
$(i=0,\ldots,k)$ such that
$$
L^\nu v_{n-k+i}(x)+\Lambda_nv_{n-k+i}(x)=v_{n-k+i-1}(x).
$$
 By lemma \ref{lem2.1}, all the functions $v_{n-k+i}(x)$ satisfy boundary 
conditions \eqref{e2.4}, then the functions $v_{n-k+i}(x)$ form the 
Jordan chain corresponding to the eigenvalue $\Lambda_n$ of problem 
\eqref{e2.3}, \eqref{e2.4}. Thus we have that the function $v_{n}(x)$ 
is an associated  function of order  $k$ of problem \eqref{e2.3}, \eqref{e2.4}.
Whence, the following assertion is valid.

\begin{theorem} \label{thm2.1}
  The root function system of problem \eqref{e2.3}, \eqref{e2.4} contains
 associated functions of arbitrary high order.
\end{theorem}

\section{Empty spectrum}

Consider  boundary-value problem \eqref{e1.1}, \eqref{e2.4}, where 
$n=2\nu$ $(\nu>1)$. Suppose that $p_m(x)=(-1)^mp_m(\pi-x)$  
almost everywhere on the segment $[0,\pi]$,
$m=1,\ldots, n$. We will study the spectrum of problem \eqref{e1.1}, \eqref{e2.4}.

\begin{theorem} \label{thm3.1}
If $d\ne\pm1$ the spectrum of problem \eqref{e1.1}, \eqref{e2.4} is empty.
\end{theorem}

\begin{proof}
 Let a function $\hat u_k(x)$ be the solution of equation \eqref{e1.1} 
with initial conditions
\begin{equation}
u_k^{(j)}(\pi/2)=\delta_{k,j},\label{e3.1}
\end{equation}
where $k=0,\ldots,n-1$, $j=0,\ldots,n-1$.
Denote
$$
\hat u_-(x)=\sum_{k=0}^{\nu-1}c_{2k+1}\hat u_{2k+1}(x),\quad
\hat u_+(x)=\sum_{k=0}^{\nu-2}c_{2k}\hat u_{2k}(x),
$$
where $c_i$ are arbitrary constants $(i=0,\ldots,n-1)$.
Then
$$
\hat u_-^{(2k)}(\pi/2)=0,\quad \hat u_+^{(2k+1)}(\pi/2)=0,\quad k=0,\ldots,\nu-1.
$$

Obviously, that the functions $w_-(x)=-\hat u_-(\pi-x)$  and 
$w_+(x)=\hat u_+(\pi-x)$ are the solutions of equation \eqref{e1.1} and satisfy 
the same initial conditions at the point $\pi/2$ as well as the functions   
$\hat u_-(x)$ and $\hat u_+(x)$, correspondingly. This, together with the 
uniqueness of the solution of Cauchy  problem \eqref{e1.1}, \eqref{e3.1} implies
that
$\hat u_-(x)=-\hat u_-(\pi-x)$ and $\hat u_+(x)=\hat u_+(\pi-x)$,
if $0\le x\le1$.
It follows that
\begin{equation}
\hat u_-^{(n-j)}(0)+(-1)^{j+1}\hat u_-^{(n-j)}(\pi)=0,\quad
\hat u_+^{(n-j)}(0)+(-1)^{j}\hat u_+^{(n-j)}(\pi)=0\label{e3.2}
\end{equation}
$(j=1,\ldots, n)$. It follows from \eqref{e3.2} that for any complex
 number $\lambda$ the function $\hat u_-(x)$ is a solution of
problem \eqref{e1.1}, \eqref{e2.4} if $d=1$, and for any  complex number
$\lambda$ the function $\hat u_+(x) $ is a solution of problem
\eqref{e1.1}, \eqref{e2.4} if $d=-1$. Thus, we establish that if $d=\pm1$
the spectrum of problem \eqref{e1.1}, \eqref{e2.4} fills all complex plane.
If $p_m(x)\in C^m(0,1)$, $m=1,\ldots, n$, this assertion was proved in \cite{Sad}.

Assume, for a number $\lambda$ a function $\tilde u(x)$ is a solution of 
problem \eqref{e1.1}, \eqref{e2.4} if $d\ne\pm1$. 
Then $\tilde u(x)=\hat u_+(x)+\hat u_-(x)$. We see that
\begin{equation}
\hat u_-^{(n-j)}(0)+\hat u_+^{(n-j)}(0)+
(-1)^{j+1}d(\hat u_-^{(n-j)}(\pi)+\hat u_+^{(n-j)}(\pi))=0\label{e3.3}
\end{equation}
$(j=1,\ldots, n)$.
It follows from \eqref{e3.2}, \eqref{e3.3} that
$$
\hat u_-^{(n-j)}(0)(1-d)+\hat u_+^{(n-j)}(0)(1+d)=0
$$
$(j=1,\ldots, n)$. From this and the definition of the functions
$\hat u_+(x)$ è $\hat u_-(x)$, we have
$$
(1+d)\sum_{k=0}^{\nu-2}c_{2k}\hat u_{2k}^{(n-j)}(0)
+(1-d)\sum_{k=0}^{\nu-1}c_{2k+1}\hat u_{2k+1}^{(n-j)}(0)=0
$$
$(j=1,\ldots, n)$, hence, the constants $c_i$ $(i=0,\ldots, n-1)$
satisfy the system of linear equations
\begin{equation}
\sum_{i=0}^{n-1}c_i(1+d(-1)^i)\hat u_{i}^{(n-j)}(0)=0\label{e3.4}
\end{equation}
$(j=1,\ldots, n)$. The determinant of linear system \eqref{e3.4} is
$$
\Delta=(1-d^2)^\nu \det||\hat u_{i}^{(n-j)}(0)||.
$$
Since the last determinant  is the Wronskian of the fundamental
system of the solutions of equation \eqref{e1.1}, it is nonzero.
Therefore, system \eqref{e3.4} has only trivial solution, i.e.\
 the function $\tilde u(x)\equiv0$. Hence, if $d\ne\pm1$ problem
\eqref{e1.1}, \eqref{e2.4} has no eigenvalues.
\end{proof}

Problem \eqref{e1.3}, \eqref{e1.4} was first  investigated in \cite{Sto}. 
In particular,  it was shown  that under  the conditions $d=\pm1$, $q(x)\equiv0$
any complex number is eigenvalue of the considered problem.

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\end{document}