\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 15, pp. 1--18.\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-2018/15\hfil Characterization of domains]
{Characterization of domains of symmetric and
self-adjoint ordinary differential operators}

\author[A. Wang, A. Zettl \hfil EJDE-2018/15\hfilneg]
{Aiping Wang, Anton Zettl}

\address{Aiping Wang \newline
 School of Mathematics and Physics,
North China Electric Power University,
 Beijing 102206, China}
\email{wapxf@163.com}

\address{Anton Zettl \newline
Math. Dept., Northern Illinois University, 
DeKalb, IL 60115, USA}
\email{zettl@msn.com}

\thanks{Submitted September 14, 2017. Published January 10, 2018.}
\subjclass[2010]{34B20, 34B24, 47B25}
\keywords{Symmetric domains; differential operators; LC solutions}

\begin{abstract}
 We characterize the two point boundary conditions which determine symmetric
 ordinary differential operators of any order, even or odd, with complex
 coefficients and arbitrary deficiency index, in a Hilbert space. The
 self-adjoint characterizations are a special case.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}

We consider the equation
\begin{equation}
My=\lambda w y\quad\text{on }J=(a,b),\quad -\infty \leq a<b\leq \infty,  \label{1.1}
\end{equation}
where $M$ is a general symmetric ordinary quasi-differential expression of
any order, even or odd.

For the case when $M$ is regular M\"oller and Zettl \cite{moze95}
characterized the two-point boundary conditions which generate symmetric
operator realizations of equation \eqref{1.1} in the Hilbert space
 $H=L^{2}(J,w)$. Here we extend this result to singular $M$ of even or odd
order with complex coefficients and arbitrary deficiency index. Self-adjoint
operators have recently been characterized by Wang et al in \cite{wasz09}
when one endpoint is regular and by Hao et al in \cite{hswz12,hzsz16}
when both endpoints are singular. The symmetric characterizations
in \cite{moze95}, and the self-adjoint characterizations in
\cite{hswz12,hzsz16} are a special case of our main result.

Our proof is in the spirit of the proofs in \cite{hswz12,moze95,wasz09},
but there are some significant differences between even and
odd order differential operators and real and complex coefficients. In
particular, although our construction of the symmetric operators uses LC
solutions for real values of the spectral parameter $\lambda $, these
solutions cannot be chosen to be real valued in contrast to the even order
case with real coefficients. Also the extension of the heavy dose of linear
algebra analysis using nonsquare matrices introduced in \cite{moze95} for
regular problems is extended to singular problems. In particular, this
involves an extension of the Naimark Patching Lemma and the use of Lagrange
brackets in place of quasi-derivatives.

The organization of the paper is as follows: This Introduction is followed
by a brief discussion of the basic theory of first order systems of
differential equations and their relationship to very general $n$-th order
scalar equations in Section 2. Section 3 discusses the minimal and maximal
operators, Section 4 the Lagrange Identity, Section 5 the construction of LC
solutions and the decomposition of the maximal domain. The characterization
of symmetric operators is given in Section 6 and illustrated with examples
in Section 7.

\section{Preliminaries}

In this section we summarize some basic facts about general symmetric
quasi-differential equations of even and odd order with real or complex
coefficients for the convenience of the reader. For a comprehensive
discussion of these equations and their relationship to the classical
symmetric (formally self-adjoint) case discussed in the well known books by
Coddington and Levinson \cite{cole55}, Dunford and Schwartz \cite{dusc63}
and \cite{evze79,fren82,moze95a,zett75} and for the
`special' symmetric quasi-differential expressions studied in Naimark
\cite{naim68}, as well as additional references, historical remarks and other
comments, notation, definitions, etc., the reader is referred to the recent
survey article by Sun and Zettl \cite{zesu15}.

These expressions generate symmetric differential operators in the Hilbert
space $L^{2}(J,w)$ and it is these operators which are studied here. Let $
J=(a,b)$ be an interval with $-\infty \leq a<b\leq \infty $ and let $n>1$ be
a positive integer (even or odd).

\subsection*{Notation}
 Let $\mathbb{R}$ denote the real numbers, $\mathbb{N}_2=\mathbb{\{}2,3,4,\dots \}$,
$\mathbb{C}$ the complex numbers, 
$M_{n,k}(X)$ the $n\times k$ matrices with entries from $X$,
$M_{n}(X)=M_{n,k}(X)$ when $n=k$, $M_{n,1}(X)$ be also denoted
by $X^n$, $M_{n,k}(X)$ be abbreviated by $M_{n,k}$ when
$X=\mathbb{C}$; $L(J,\mathbb{R})$ and $L(J,\mathbb{C})$
the Lebesgue integrable real and
complex valued functions on $J$, respectively, $L_{\rm loc}(J,\mathbb{R})$ and
$L_{\rm loc}(J,\mathbb{C})$ the real and complex valued functions which are
Lebesgue integrable on all compact subintervals of $J$, respectively. We
also use $L_{\rm loc}(J)=L_{\rm loc}(J,\mathbb{C})$ and $L(J)=L(J,\mathbb{C})$.
$AC_{\rm loc}(J)$ denotes the complex valued functions which are absolutely
continuous on compact subintervals of $J$ and $AC(J)$ denotes the absolutely
continuous functions on $J$. $D(S)$ denotes the domain of the operator $S$.


\begin{definition} \label{d01} \rm
For $w\in L_{\rm loc}(J,\mathbb{R})$, $w>0$ a.e. in $J$, $L^{2}(J,w)$
denotes the Hilbert space of functions $f:J\to \mathbb{C}$
satisfying $\int_{J}|f|^{2}w<\infty $ with inner product
$(f,g)_{w}=\int_{J}f\overline{g}w$. Such a $w$ is called a `weight
function'.
\end{definition}

Let
\begin{equation}  \label{2.1}
\begin{aligned}
Z_{n}(J):=\Big\{&Q=(q_{rs})_{r,s=1}^n:   q_{r,r+1} \neq 0\text{ a.e. on }J,\;
q_{r,r+1}^{-1}\in L_{\rm loc}(J),\\
& 1\leq r\leq n-1, \;
 q_{rs}=0\text{ a.e. on }J,\;
 2\leq r+1<s\leq n; \\
& q_{rs}\in L_{\rm loc}(J),\;s\neq r+1,\; 1\leq r\leq n-1\Big\}.
\end{aligned}
\end{equation}
For $Q\in {Z}_{n}(J)$, define
\begin{gather*}
V_0:=\{y:J\to \mathbb{C}, y\text{ is measurable}\}, \\
y^{[0]}:=y\quad (y\in V_0).
\end{gather*}
Inductively, for $r=1,\dots ,n$, we define
\begin{gather*}
V_{r}=\{y\in V_{r-1}:y^{[r-1]}\in AC_{\rm loc}(J)\}, \\
y^{[r]}=q_{r,r+1}^{-1}\{y^{[r-1]'}
-\sum_{s=1}^{r}q_{rs}y^{[s-1]}\}\quad (y\in V_{r}),
\end{gather*}
where $q_{n,n+1}:=1$. Then we set
\begin{equation}
My=M_{Q}y:=i^ny^{[n]}\quad\text{on }J \quad (y\in V_{n},\;i=\sqrt{-1}).
\label{2.6}
\end{equation}
The expression $M=M_{Q}$ is called the quasi-differential expression
associated with $Q$. For $V_{n}$ we also use the notations $D(M_{Q})$ and
$D(Q)$. The function $y^{[r]}\ (0\leq r\leq n)$ is called the $r$-th
quasi-derivative of $y$. Since the quasi-derivative depends on $Q$, we
sometimes write $y_{Q}^{[r]}$ instead of $y^{[r]}$.

\begin{remark} \rm
Note that the operator $M:D(Q)\to L_{\rm loc}(J)$ is linear.
Also note that the differential expression $M_{Q}$ in equation \eqref{2.6}
requires only local integrability assumptions on the coefficients \eqref{2.1}.
\end{remark}

The initial value problem associated with $Y'=Q Y+F$ has a unique
solution.

\begin{proposition} \label{P2-2}
Suppose $Q\in Z_{n}(J)$. For each $F\in (L_{\rm loc}(J))^n$, each $
\alpha $ in $J$ and each $C\in \mathbb{C}^n$ there is a unique $Y\in
(AC_{\rm loc}(J))^n$ such that
\begin{equation*}
Y'=Q Y+F\quad and\quad Y(\alpha )=C.
\end{equation*}
\end{proposition}

For a proof of the above proposition, see \cite[Chapter 1]{zett05}.
From Proposition \ref{P2-2}, we immediately infer the following result.

\begin{corollary} \label{C2-2}
For each $f\in L_{\rm loc}(J)$, each $\alpha \in J$ and $
c_0,\dots ,c_{n-1}\in \mathbb{C}$ there is a unique $y\in D(Q)$ such that
\begin{equation*}
y^{[n]}=f\quad and\quad y^{[r]}(\alpha )=c_{r}\quad (r=0,\dots ,n-1).
\end{equation*}
If $f\in L(J)$, \ $J$ is bounded and all components of $Q$ are in $L(J)$,
then $y\in AC(J)$.
\end{corollary}

\begin{definition}[Regular endpoints]\label{D3-3} \rm
Let $Q\in Z_{n}(J)$, $J=(a,b)$. The expression $M=M_{Q}$ is said
to be regular at $a$ if for some $c$, $a<c<b$, we have
\begin{gather*}
q_{r,r+1}^{-1}\in L(a,c),\quad r=1,\dots ,n-1; \\
q_{rs}\in L(a,c),\quad 1\leq r,s\leq n,\;s\neq r+1.
\end{gather*}
Similarly the endpoint $b$ is regular if for some $c$, $a<c<b$, we have
\begin{gather*}
q_{r,r+1}^{-1}\in L(c,b),\;r=1,\dots ,n-1; \\
q_{rs}\in L(c,b),\;1\leq r,\,s\leq n,\;s\neq r+1.
\end{gather*}
Note that, from \eqref{2.1} it follows that if the above hold for some 
$c\in J$ then they hold for any $c\in J$. We say that $M$ is regular on $J$, or
just\ $M$ is regular, if $M$ is regular at both endpoints.

An endpoint is called singular if it is not regular.
\end{definition}

\begin{remark} \rm
In much of the literature when an endpoint of the underlying interval is
infinite the problem is automatically classified as singular; note that in
Definition \ref{D3-3} $a=-\infty $ or $b=\infty $ is allowed. For any $J$
observe that $M$ is regular on any compact subinterval of $J$. Although we
focus on the singular case because there the results are new but the results
hold when each endpoint is either regular or singular.
\end{remark}

Next we give the definition of symmetric quasi-differential expressions. For
examples and illustrations see \cite{zesu15}.

\begin{remark} \rm
The symplectic matrix
\begin{equation}
E_{k}=((-1)^{r}\delta _{r,k+1-s})_{r,s=1}^{k},\;\,k\in \mathbb{N}_2
\label{2.15}
\end{equation}
plays an important role in the construction of symmetric quasi-differential
expressions as well as in the characterization of symmetric differential
operators.
\end{remark}

\begin{definition} \rm
Let $Q\in Z_{n}(J)$ and let $M=M_{Q}$ be defined as in \eqref{2.6}. Assume
that
\begin{equation}
Q=-E_{n}^{-1}Q^{\ast }E_{n}.  \label{2.14}
\end{equation}
Then we call $Q$ a Lagrange symmetric matrix and $M=M_{Q}$ is called a
symmetric differential expression.
\end{definition}

\section{Minimal and maximal operators}

In this section we recall the minimal and maximal operators and their basic
properties.

\begin{definition} \rm
Let $Q\in Z_{n}(J)$ satisfy \eqref{2.14} and let $M=M_{Q}$ be the
corresponding symmetric differential expression. The maximal operator 
$S_{\rm max}$ generated by $M$ is defined by
\begin{gather*}
\begin{aligned}
D_{\rm max}=\Big\{ & y\in L^{2}(J,w):  y^{[0]},y^{[1]},\dots
,y^{[n-1]}\text{ are absolutely continuous}\\
&\text{in $J$, and } w^{-1}My\in L^{2}(J,w) \Big\}, 
\end{aligned}\\
S_{\rm max}y=w^{-1}My, \quad  y\in D_{\rm max}.
\end{gather*}
The minimal operator $S_{\rm min}$ is defined by
\begin{equation*}
S_{\rm min}=S_{\rm max}^{\ast }.
\end{equation*}
\end{definition}

\begin{lemma} \label{lem1}
Suppose $M$ is regular at $c$. Then for any $y\in D_{\rm max}$ the limits
\begin{equation*}
y^{[r]}(c)=\lim_{t\to c}y^{[r]}(c)
\end{equation*}
exist and are finite, $r=0,\dots ,n-1$. In particular this holds
at any regular endpoint and at each interior point of $J$. At an endpoint
the limit is the appropriate one sided limit.
\end{lemma}

For a proof of the above lemma see \cite[Lemma 2, p.63]{naim68}.

Let $a<c<b$. Below we will also consider \eqref{2.6} and the operators
generated by it on the intervals $(a,c)$ and $(c,b)$. Note that if $Q\in
Z_{n}(J)$, then it follows that $Q\in Z_{n}(a,c)$, $Q\in Z_{n}(c,b)$ and we
can study equation \eqref{2.6} on $(a,c)$ and $(c,b)$ as well as on $
J=(a,b)$. Also \eqref{2.14} holds on $(a,c)$ and on $(c,b)$. In particular,
the minimal and maximal operators are defined on these two subintervals and
we can also study the operator theory generated by \eqref{2.6} in the
Hilbert spaces $L^{2}((a,c),w)$ and $L^{2}((c,b),w)$. Below we will use the
notation $S_{\rm min}(I)$, $S_{\rm max}(I)$ for the minimal and maximal
operators on the interval $I$ for $I=(a,c)$, $I=(c,b)$, $I=(a,b)=J$. The
interval $J=(a,b)$ may be omitted when it is clear from the context. So we
make the following definition.

\begin{definition} \label{def2} \rm
Let $a<c<b$. Let $d_a^{+}$, $d_b^{+}$ denote the dimension
of the solution space of $My=i\,wy$ lying in $L^{2}((a,c),w)$ and 
$L^{2}((c,b),w)$, respectively, and let $d_a^{-},\,d_b^{-}$ denote the
dimension of the solution space of $My=-i\,wy$ lying in $L^{2}((a,c),w)$ and
$L^{2}((c,b),w)$, respectively. Then $d_a^{+}$ and $d_a^{-}$ are called
the positive deficiency index and the negative deficiency index of 
$S_{\rm min}(a,c)$, respectively. Similarly for $d_b^{+}$ and $d_b^{-}$.
Also $d^{+},\,d^{-}$ denote the deficiency indices of $S_{\rm min}(a,b)$;
these are the dimensions of the solution spaces of $My=iwy,\;My=-iwy$ lying in 
$L^{2}((a,b),w)$.  If $d_a^{+}=d_a^{-}$, then the common value is
denoted by $d_a$ and is called the deficiency index of $S_{\rm min}(a,c)$,
or the deficiency index at $a$. Similarly for $d_b$. \ Note that $d_a$,
$d_b$ are independent of $c$. If $d^{+}=d^{-}$, then we denote the common
value by $d$ and call it the deficiency index of $S_{\rm min}(a,b)$ or just of
$S_{\rm min}$.
\end{definition}

The relationships between $d_a$, $d_b$ and $d$ are well known and given
in the next lemma which is well known, see for example the book \cite{weid87}.

\begin{lemma}\label{lemma2} 
For $d^+_a,d^+_b,d^-_a,d^-_b,d^+,d^-,d_a,d_b$ defined as
Definition \ref{def2}, we have
\begin{enumerate}
\item $d^{+}=d_a^{+}+d_b^{+}-n,\;d^{-}=d_a^{-}+d_b^{-}-n;$

\item if $d^+_a=d^-_a=d_a,d^+_b=d^-_b=d_b$, then $[\frac{n+1}{2}]\leq
d_a,d_b\leq n$;

\item the minimal operator $S_{\rm min}$ has self-adjoint extensions in $H$ if
and only if\ $d=d^{+}=d^{-}$. If $d=0$ then $S_{\rm min}$ is self-adjoint with
no proper self-adjoint extension. In all other cases $S_{\rm min}$ has an
uncountable number of self-adjoint extensions, i.e. there are an uncountable
number of operators $S$ in $H$ satisfying
\begin{equation*}
S_{\rm min}\subset S=S^{\ast }\subset S_{\rm max}.
\end{equation*}
\end{enumerate}
\end{lemma}

\section{Lagrange identity}

In the study of boundary value problems the Lagrange identity is fundamental.

\begin{lemma}[Lagrange identity \cite{moze95a}] \label{L3-3} 
Let $Q\in Z_{n}(J)$ satisfy \eqref{2.14} and let $M=M_{Q}$ be
the corresponding differential expression. Let the quasi-derivatives $y$, $
y^{[1]}$, $\dots $, $y^{[n-1]}$ be defined as above. Then for any $y,z\in
D(Q)$, we have
\begin{equation}
\overline{z}My-(\overline{Mz})y=[y,z]',  \label{2.16}
\end{equation}
where
\begin{equation*}
[ y,z]= i^n\sum_{r=0}^{n-1}(-1)^{n+1-r}\bar{z}^{[n-r-1]}y^{[r]}.
\end{equation*}
Here $[y,z]$ or just $[\cdot ,\cdot ]$ is called a Lagrange bracket.
\end{lemma}

\begin{lemma} \label{l41} 
For any $y,z$ in $D_{\rm max}$ we have
\begin{equation*}
\int_a^{b}\{\overline{z}My-y\overline{Mz}\}=[y,z](b)-[y,z](a),
\end{equation*}
where $[y,z](b)=\lim_{t\to b^{-}}[y,z](t)$, and $[y,z](a)=\lim_{t
\to a^{+}}[y,z](t)$, $t\in (a,b)$.
\end{lemma}

The above lemma follows by integrating \eqref{2.16}.
The finite limits guaranteed by Lemma \ref{l41} play a fundamental role in
the characterization of the symmetric and self-adjoint domains.

\begin{corollary}\label{cor3.2} 
If $M\,y=\lambda w y$ and $M\,z=\overline{\lambda }w z$ on
some interval $(a,b)$, then $[y,z]$ is constant on $(a,b)$. In particular,
if $\lambda $ is real and $M\,y=\lambda w\,y,~Mz=\lambda wz$ on some
interval $(a,b)$, then $[y,z]$ is constant on $(a,b)$.
\end{corollary}

The above corollary follows directly from \eqref{2.16}.
For real $\lambda $, the solutions of \eqref{1.1} are not, in general,
real-valued. However, the Lagrange bracket of two linearly independent
solutions of \eqref{1.1} for real $\lambda $ is a constant. For $n$ even and
real coefficients, if there are $d$ linearly independent solutions of 
\eqref{1.1} in $H$, then there are $d$ linearly independent real-valued solutions
in $H$. This is one of the important differences between the equation 
\eqref{1.1} studied here and the equations studied in \cite{wasz09,hswz12}.

Following Everitt and Zettl \cite{evze79} we call the next lemma, the
Naimark Patching Lemma or just the Patching Lemma.

\begin{lemma} \label{lem2.4}
Let $Q\in Z_{n}(J)$ and assume that $ M $ is regular on $J$. 
Let 
\[
\alpha _0,\dots ,\alpha _{n-1},\beta_0,\dots ,\beta _{n-1}\in \mathbb{C}.
\]
 Then there is a function $y\in D_{\rm max}$ such that
\begin{equation*}
y^{[r]}(a)=\alpha _{r},\quad y^{[r]}(b)=\beta _{r}\quad (r=0,\dots ,n-1).
\end{equation*}
\end{lemma}

\begin{corollary}\label{C3-1}
Let $a<c<h<b$ and $\alpha _0,\dots ,\alpha _{n-1},\beta
_0,\dots ,\beta _{n-1}\in \mathbb{C}$. Then there is a $y\in D_{\rm max}$
such that $y$ has compact support in $J$ and satisfies:
\begin{equation*}
y^{[r]}(c)=\alpha _{r},\quad y^{[r]}(h)=\beta _{r}\quad (r=0,\dots ,n-1).
\end{equation*}
\end{corollary}

\begin{proof}
The proof in Naimark \cite{naim68} can easily be adapted to prove the above
corollary.
\end{proof}

\begin{corollary}
Let $a_1<\dots <a_{k}\in J$, where $a_1$ and $a_{k}$ can also be
regular endpoints. Let $\alpha _{jr}\in \mathbb{C}$  
$(j=1,\dots ,k; r=0,\dots ,n-1)$. Then there is a $y\in D_{\rm max}$ such that
\begin{equation*}
y^{[r]}(a_j)=\alpha _{jr}\quad (j=1,\dots ,k;\ r=0,\dots ,n-1).
\end{equation*}
\end{corollary}

The above corollary follows from repeated applications of  Corollary \ref{C3-1}.

\begin{lemma}
For $d_a,d_b$ given in Definition \ref{def2}, we have
\begin{enumerate}
\item If $r_a(\lambda )$ denotes the number of linearly independent
solutions of \eqref{1.1} lying in $L^{2}((a,c),w)$ for 
$\lambda \in \mathbb{R}$, then $r_a(\lambda )\leq d_a$.
Similarly $r_b(\lambda )\leq d_b$.

\item If $r_a(\lambda )<d_a$ or $r_b(\lambda )<d_b$ for some $
\lambda \in \mathbb{R}$, then $\lambda $ is in the essential spectrum of
every self-adjoint extension of $S_{\rm min}$.
\end{enumerate}
\end{lemma}

For a proof of the above lemma see \cite{hzsz16,hasz12}.

\section{LC solutions and the decomposition of the maximal domain}

In this section we recall some properties of the maximal and minimal
operators, construct limit-circle (LC) solutions and discuss the
decomposition of the maximal domain used below in Section 6 to prove our
main theorem. The next theorem is well known.

\begin{theorem} \label{t22}
Let $M=M_{Q}$, $Q\in Z_{n}(J)$, $n>1$, satisfy \eqref{2.14} and
let $w$ be a weight function. Then $D_{\rm max}(Q)$ is dense in $H$. 
Let $ S_{\rm min}=S_{\rm min}(Q)=S_{\rm max}^{\ast }(Q)=S_{\rm max}^{\ast }$. 
Then $S_{\rm min}$ is a closed symmetric operator in $H$ with dense domain and 
$S_{\rm min}^{\ast }=S_{\rm max}$.
\end{theorem}

\begin{proof}
The method of Naimark \cite[Chapter V]{naim68} can be adapted to prove this
theorem with minor modifications. See also \cite{evze79}.
\end{proof}

For the rest of this article we assume that the hypothesis  holds.
\begin{itemize}
\item[(H1)]  Let $a<c<b$ and assume that the equation \eqref{1.1} on $(a,c)$
has $d_a$ linearly independent solutions, denoted by $u_1$, $u_2,
\dots,u_{d_a}$, in $L^{2}((a,c),w)$ for some real $\lambda =$ $\lambda _a$
and that \eqref{1.1} has $d_b$ linearly independent solutions, denoted by $
v_1$, $v_2, \dots,v_{d_b}$, in $L^{2}((c,b),w)$ for some real 
$\lambda =\lambda _b$. Note that $d_a$ and $d_b$ are independent of $c$.
\end{itemize}

Regarding hypothesis (H1), note that 
 $d_a^{+}=d_a^{-}=d_a$, $d_b^{+}=d_b^{-}=d_b$ and $d^{+}=d^{-}=d$.
Recall that $r_a(\lambda )$ denotes the number of linearly independent
solutions of \eqref{1.1} on $(a,c)$ which lie in $L^{2}((a,c),w)$
for real $\lambda $. For any real $\lambda $ it is known 
\cite{hzsz16,hasz12} that $r_a(\lambda )\leq d_a$ and if
$r_a(\lambda )<d_a$ then $\lambda $ is in the essential spectrum of
every self-adjoint extension of $S_{\rm min}(a,c)$ and of $S_{\rm min}(a,b)$. 
Thus if there does not exist a real $\lambda _a$ such that \eqref{1.1} on $(a,c)$
has $d_a$ linearly independent solutions in $L^{2}((a,c),w)$ then the essential
spectrum of all self-adjoint extensions $S_{\rm min}(a,c)$ and of 
$S_{\rm min}(a,b)$ covers the whole real line. Similarly for the endpoint $b$. 
If the essential spectrum of every self-adjoint realization of \eqref{1.1} in
$L^{2}((a,b),w)$ covers the whole real line then any eigenvalue, if there is
one, is embedded in the essential spectrum. In this case the dependence of
such eigenvalues on the boundary condition seems to be `coincidental' and
nothing seems to be known, aside from examples, about this dependence.

The next theorem constructs LC solutions at each endpoint.

\begin{theorem} \label{t41}
Suppose that $Q\in Z_{n}(J,\mathbb{C})$, $J=(a,b)$,
 $-\infty \leq a<b\leq \infty $, is Lagrange symmetric, 
$M=M_{Q }$ and $w$ is a weight function. Let $a<c<b$ and assume {\rm (H1)}
holds. Consider the equation
\begin{equation*}
My=\lambda wy\quad\text{on }J.
\end{equation*}
Then
\begin{enumerate}
\item For $m_a=2d_a-n$ the solutions $u_1,\dots ,u_{d_a}$
can be ordered such that the $m_a\times m_a$ matrix $\widehat{U}
=([u_{i},u_j](a))_{1\leq i,j\leq m_a}$ is given by
\begin{equation*}
\widehat{U}=\begin{bmatrix}
[ u_1,u_1](a) & \dots & [u_{m_a},u_1](a) \\
\dots & \dots & \dots \\
[u_1,u_{m_a}](a) & \dots & [u_{m_a},u_{m_a}](a)
\end{bmatrix}
 =-i^nE_{m_a}
\end{equation*}
and is therefore nonsingular.

\item For $m_b=2d_b-n$ the solutions $v_1,\dots ,v_{d_b}$
on $(c,b)$ can be ordered such that the $m_b\times m_b$ matrix $\widehat{
V}=([v_{i},v_j](b))_{1\leq i,j\leq m_b}$ is given by
\begin{equation*}
\widehat{V}=\begin{bmatrix}
[ v_1,v_1](b) & \dots & [v_{m_b},v_1](b) \\
\dots & \dots & \dots \\
[v_1,v_{m_b}](b) & \dots & [v_{m_b},v_{m_b}](b)
\end{bmatrix}
=-i^nE_{m_b}
\end{equation*}
and is therefore nonsingular.

\item For every $y\in D_{\rm max}(a,b)$ we have
$[ y,u_j](a)=0$ for $j=m_a+1,\dots ,d_a$.  %\label{4.4}

\item For every $y\in D_{\rm max}(a,b)$ we have
$[ y,v_j](b)=0$ for $j=m_b+1,\dots ,d_b$. % \label{4.5}

\item For $1\leq i,j\leq d_a$, we have
$[ u_{i,}u_j](a)=[u_{i,}u_j](c)$. % \label{4.6}

\item For $1\leq i,j\leq d_b$, we have
$[ v_{i,}v_j](b)=[v_{i,}v_j](c)$.  %\label{4.7}

\item The solutions $u_1,\dots ,u_{d_a}$ can be extended to
 $(a,b)$ such that the extended functions, also denoted by 
$u_1,\dots ,u_{d_a}$, satisfy $u_j\in D_{\rm max}(a,b)$ and $u_j$ is
identically zero in a left neighborhood of $b$, $j=1,\dots,d_a $.

\item The solutions $v_1,\dots ,v_{d_b}$ can be extended to $
(a,b)$ such that the extended functions, also denoted by 
$v_1,\dots ,v_{d_b}$, satisfy $v_j\in D_{\rm max}(a,b)$ and $v_j$ is
identically zero in a right neighborhood of $a$, $j=1,\dots,d_b$.
\end{enumerate}
\end{theorem}

A proof of the above theorem can be found in \cite[Theorem 1]{hzsz16}.

\begin{definition}\label{d14.1} \rm
The solutions $u_1,\dots ,u_{m_a}$ and $v_1,\dots ,v_{m_b}$ are called 
LC solutions at $a$ and $b$, respectively. 
The solutions $u_{m_{a+1}},\dots ,u_{d_a}$ and
 $v_{m_b+1},\dots ,v_{d_b}$ are called LP solutions at $a$ and
$b$, respectively. The definitions of LC solutions and LP solutions were
proposed by Wang et al in \cite{wasz09}.
\end{definition}

\begin{remark} \rm
Only the LC solutions are used in the construction of the boundary
conditions which characterize the self-adjoint and symmetric operators in
the Hilbert space $L^{2}(J,w)$. The LP solutions and the solutions not in
this space make no contribution to the construction of the self-adjoint and
symmetric boundary conditions.
\end{remark}

Our proof of the symmetric operator characterization uses the
decomposition of the maximal domain in terms of LC solutions given by the
next theorem.

\begin{theorem}[\cite{hzsz16}] \label{t42}
Let the notation and hypotheses of Theorem \ref{t41} hold. Then
\begin{equation}
D_{\rm max}(a,b)=D_{\rm min}(a,b)\dotplus span\{u_1,\dots
,u_{m_a}\}\dotplus \operatorname{span}\{v_1,\dots ,v_{m_b}\}.  \label{5.1}
\end{equation}
\end{theorem}

\section{Symmetric operators}

In this section we state and prove our main result: the characterization of
two-point boundary conditions which determine symmetric operators in the
Hilbert space $L^{2}(J,w)$. The proof depends on several lemmas; some of
these are stated as Theorems because we believe they are of independent
interest.

\begin{definition} \label{d51} \rm
Let the hypothesis and notation of Theorem \ref{t41} hold. For
any $y\in D_{\rm max}$ define
\begin{equation}
Y_{a,b}=\begin{bmatrix}
Y(a) \\
Y(b)
\end{bmatrix}, \quad
Y(a)=\begin{pmatrix}
[ y,u_1](a) \\
\dots \\
[ y,u_{m_a}](a)
\end{pmatrix},\quad
Y(b)=\begin{pmatrix}
[ y,v_1](b) \\
\dots \\
[ y,v_{m_b}](b)
\end{pmatrix}  \label{5.4}
\end{equation}
and recall that the Lagrange brackets $[y,u_j](a)$ and $[y,v_j](b)$
exist as finite limits by Lemma \ref{l41}.
\end{definition}

\begin{definition} \rm
A matrix $U\in M_{l,2d}$ with rank $l$, $0\leq l\leq 2d$, $2d=m_a+m_b$
is called a boundary condition matrix. And for $y\in D_{\rm max}$ and $Y_{a,b}$
given by \eqref{5.4} the equation
\begin{equation}
U Y_{a,b}=0  \label{5.5}
\end{equation}
is called a boundary condition. The null space of $U$ is denoted by $
\mathcal{N(}U)$ and $\mathcal{R}(U)$ denotes its range, $U^{\ast }$ is its
adjoint.
\end{definition}

Note that any boundary condition \eqref{5.5} can be reduced by elementary
matrix operations to the case that the rank of $U$ is the number of its rows.

\begin{definition} \label{d53} \rm
Suppose $U\in M_{l,2d}$ is a boundary condition matrix. Define an
operator $S(U)$ in $L^{2}(J,w)$ by
\begin{equation}
\begin{gathered}
D(S(U)) =\big\{y\in D_{\rm max}:U Y_{a,b}=0\big\},   \\
S(U)y =My\quad \text{for }y\in D(S(U)). 
\end{gathered}  \label{5.6}
\end{equation}
\end{definition}

\begin{remark} \label{r51} \rm
If $l=0$, then $U=0$ and $S(U)=S_{\rm max}$. If $l=2d$, and $
I_{2d} $ denotes the $2d\times 2d$ identity matrix, 
then $S(I_{2d})=S_{\rm min} $\ by Theorem \ref{t50} below and for any 
nonsingular boundary condition matrix $U$ we have
\begin{equation*}
S(U)=S(I_{2d})=S_{\rm min}.
\end{equation*}
Hence for any boundary condition matrix $U$, $D(S(U))$ is a linear
submanifold of $D_{\rm max}$ and we have
\begin{equation*}
S_{\rm min}\subset S(U)\subset S_{\rm max}
\end{equation*}
and consequently, since $S_{\rm max}$ is a closed finite dimensional extension
of $S_{\rm min}$, \ it follows that every operator $S(U)$ is a closed finite
dimensional extension of $S_{\rm min}$. For which matrices $U$ is $S(U)$ a
symmetric operator in $L^{2}(J,w)?$ This is the question we answer below.
\end{remark}

We start by recalling the well known abstract von Neumann charaterization of
the domain of the adjoint of a densely defined closed symmetric operator in
Hilbert space.

\begin{lemma}\label{l51} 
Let $T$ be a closed densely defined symmetric operator on a
complex Hilbert space $H$, and let $N_{+}$ and $N_{-}$ be the deficiency
spaces of $T$. Then we have
\begin{equation*}
D{(}T^{\ast }{)}=D(T)\dotplus N_{+}\dotplus N_{-}\;
\end{equation*}

An operator $S$ is a closed symmetric extension of $T$ if and only if there
exist closed subspaces $F_{+}$ of $N_{+}$ and $F_{-}$ of $N_{-}$ and an
isometric mapping $V$ of $F_{+}$ onto $F_{-}$ such that
\begin{equation*}
D(S)=D(T)+\{g+Vg:\ g\in F_{+}\}.
\end{equation*}
Furthermore, $S$ is self-adjoint if and only if $F_{+}=N_{+}$ and $
F_{-}=N_{-}$.
\end{lemma}

\begin{proof}
For the definition of deficiency spaces and a proof of the lemma see any
classical book on operator theory, e.g. \cite{dusc63,naim68,weid80}.
\end{proof}

When applied to the minimal operator $S_{\rm min}=S_{\rm min}(Q)$, where $Q\in
Z_{n}(J)$ is Lagrange symmetric, the von Neumann formula yields the following result.

\begin{lemma} \label{l52}
\begin{equation*}
D(S_{\rm max})=D(S_{\rm min})\dotplus N_{\lambda }\dotplus N_{\overline{\lambda }
},\quad \operatorname{Im}(\lambda )\neq 0,
\end{equation*}
where
\begin{equation*}
N_{\lambda }=\{y\in D(S_{\rm max}):M_{Q}y
=\lambda w\,y, \operatorname{Im}(\lambda)\neq 0\}.
\end{equation*}

Since the solution bases of $M_{Q}\,y=\lambda w\,y$ have dimension $d$ when $
\operatorname{Im}(\lambda )\neq 0$, where $d$ is the deficiency index, it is clear
that $D_{\rm max}$ is a $2d$ dimensional extension of $D_{\rm min}$. Therefore $
S_{\rm min}$ has self-adjoint extensions and every self-adjoint extension is a
$d$ dimensional extension. Furthermore, every $d$ dimensional symmetric
extension of $S_{\rm min}$ is self-adjoint. Moreover, every symmetric
extension of $S_{\rm min}$ is an $m$ dimensional extension with
\begin{equation*}
0\leq m\leq d
\end{equation*}
and an $l=2d-m$ dimensional restriction of $S_{\rm max}$ with
\begin{equation*}
d\leq l\leq 2d.
\end{equation*}
\end{lemma}

The decomposition of $D_{\rm max}$ given by Lemma \ref{l52} is well known 
\cite{naim68,weid80},  and the furthermore and moreover statements follow
from Lemma \ref{l51}.


By Lemma \ref{l52} the operator $S(U)$ is not symmetric if $l<d$. But its
adjoint operator ($S(U))^{\ast }$ may be symmetric. For example, when 
$d\neq 0$, $S_{\rm max}$ is not symmetric but its adjoint
 $S_{\rm min}=S_{\rm max}^{\ast }$ is symmetric. When $d=0$, then
 $S_{\rm min}=S_{\rm max}$ and $S_{\rm max}$ is symmetric and self-adjoint. 
So we will continue to study $S(U)$ for $
U\in M_{l,2d}$ with rank $l$ for $0\leq l\leq 2d$.

The next theorem extends the well known characterization of the domain of
the minimal operator
\begin{equation*}
D_{\rm min}=\{y\in D_{\rm max}:y^{[i]}(a)=0=y^{[i]}(b),\;i=0,1,2,\dots ,n-1\}
\end{equation*}
for regular problems to singular ones.

\begin{theorem} \label{t50}
Let the notation and hypotheses of Theorem \ref{t41} hold. Then
\begin{align*}
D_{\rm min}
=\Big\{&y\in D_{\rm max}: [y,u_j](a)=0,\;for\;j=1,\dots
,m_a; \\
&[y,v_j](b)=0,\text{ for }j=1,\dots ,m_b\Big\}.
\end{align*}
\end{theorem}

\begin{proof}
Recall that $S_{\rm min}=S_{\rm max}^{\ast }$ and $S_{\rm min}^{\ast }=S_{\rm max}$
and that in the decomposition of  $D_{\rm max}$ given by Theorem \ref{t42}
the $u_j$ are identically $0$ in a neighborhood of $b$ and the $v_j$ are
identically zero in a neighborhood of $a$. From the definitions of the
maximal and minimal domains and the Lagrange Identity we get
\begin{gather*}
[ y,z](b)-[y,z](a)=0\quad\text{for $z\in D_{\rm max}$ and all }y\in D_{\rm min},\\
[ y,z](b)-[y,z](a)=0\quad\text{for $y\in D_{\rm min}$ and all }z\in D_{\rm max}.
\end{gather*}

Suppose that $y\in D_{\rm max}$ with $[y,u_j](a)=0$, for
$j=1,\dots ,m_a$ and $[y,v_j](b)=0$, for $j=1,\dots ,m_b$. 
Let $ z=z_0+c_1u_1+\dots +c_{m_a}u_{m_a}+h_1v_1+\dots +h_{m_b}v_{m_b}$
 where $z_0\in D_{\rm min}$. Then
\begin{equation*}
[ y,z](b)-[y,z](a)=\sum_{j=1}^{m_b}\bar{h}_j[y,v_j](b)-
\sum_{j=1}^{m_a}\bar{c}_j[y,u_j](a)=0
\end{equation*}
and hence $y\in D_{\rm min}$.

For the converse we assume that $y\in D_{\rm min}$, then for all 
$z\in D_{\rm max} $, $[y,z](b)-[y,z](a)=0$. Therefore for the functions 
$u_j$, $j=1,2,\dots,m_a $, $[y,u_j](b)-[y,u_j](a)=0$, i.e. $[y,u_j](a)=0$.
Similarly, $[y,v_j](b)=0$, for $j=1,\dots ,m_b$.
\end{proof}

The next lemma extends the `Naimark Patching Lemma' \ref{lem2.4} from
regular to singular problems. It says that in our search for solutions of
the algebraic equation $U Y_{a,b}=0$ the whole space $\mathbb{C}^{2d}$ is
available, i.e. the range of $Y_{a,b}$ as $y$ runs through $D_{\rm max}$ is
the whole space $\mathbb{C}^{2d}$.

\begin{lemma}[Singular patching lemma] \label{l53}
For any complex numbers $\alpha_1 $, $\alpha _2,\dots ,\alpha _{m_a}$,
$\beta _1,\beta _2,\dots ,\beta _{m_b}$, there exists
$y\in D_{\rm max}$ such that
\begin{equation}
\begin{gathered}
[ y,u_1](a)=\alpha _1,\quad [y,u_2](a)=\alpha _2,\quad \dots ,\quad
[y,u_{m_a}](a)=\alpha _{m_a}, \\
[ y,v_1](b)=\beta _1,\quad [y,v_2](b)=\beta _2,\quad \dots ,\quad
[y,v_{m_b}](b)=\beta _{m_b}.  \label{eq1}
\end{gathered}
\end{equation}
\end{lemma}

\begin{proof}
Consider the equation
\begin{equation*}
\begin{pmatrix}
[ u_1,u_1](a) & \dots & [u_{m_a},u_1](a) \\
\dots & \dots & \dots \\
[u_1,u_{m_a}](a) & \dots & [u_{m_a},u_{m_a}](a)
\end{pmatrix}
\begin{pmatrix}
c_1 \\
\dots \\
c_{m_a}
\end{pmatrix}
 =\begin{pmatrix}
\alpha _1 \\
\dots \\
\alpha _{m_a}
\end{pmatrix},
\end{equation*}
namely
\begin{equation}
\widehat{U}
\begin{pmatrix}
c_1 \\
\dots \\
c_{m_a}
\end{pmatrix}
=\begin{pmatrix}
\alpha _1 \\
\dots \\
\alpha _{m_a}
\end{pmatrix}.  \label{3.12}
\end{equation}
Since the $\widehat{U}$ defined in Theorem \ref{t41} is nonsingular, 
\ref{3.12} has a unique solution $c_1,\dots ,c_{m_a}$. Similarly,
since $\widehat{V}$ is nonsingular, the following equation
\begin{equation}
\begin{pmatrix}
[ v_1,v_1](b) & \dots & [v_{m_b},v_1](b) \\
\dots & \dots & \dots \\
[v_1,v_{m_b}](b) & \dots & [v_{m_b},v_{m_b}](b)
\end{pmatrix}
\begin{pmatrix}
h_1 \\
\dots \\
h_{m_b}
\end{pmatrix}
=\begin{pmatrix}
\beta _1 \\
\dots \\
\beta _{m_b}
\end{pmatrix},  \label{3.13}
\end{equation}
i.e. 
\[
\widehat{V} \begin{pmatrix}
h_1 \\
\dots \\
h_{m_b}
\end{pmatrix}
=\begin{pmatrix}
\beta _1 \\
\dots \\
\beta _{m_b}
\end{pmatrix}
\]
 has a unique solution $h_1,\dots ,h_{m_b}$.
Set
\begin{equation*}
y=y_0+c_1u_1+\dots+c_{m_a}u_{m_a}+h_1v_1+
\dots+h_{m_b}v_{m_b},
\end{equation*}
where $y_0\in D_{\min}$. Obviously $y\in D_{\rm max}(a,b) $ and then
\begin{gather*}
[ y,u_1](a)=c_1[u_1,u_1](a)+c_2[u_2,u_1](a)+\dots
+c_{m_a}[u_{m_a},u_1](a)=\alpha_1, \\
[ y,u_2](a)=c_1[u_1,u_2](a)+c_2[u_2,u_2](a)+\dots
+c_{m_a}[u_{m_a},u_2](a)=\alpha_2, \\
\dots \\
[y,u_{m_a}](a)=c_1[u_1,u_{m_a}](a)+c_2[u_2,u_{m_a}](a)+
\dots+c_{m_a}[u_{m_a},u_{m_a}](a)=\alpha_{m_a}.
\end{gather*}
Similarly,
\begin{equation*}
[ y,v_1](b)=\beta_1,\quad [y,v_2](b)=\beta_2,\quad \dots ,\quad
[y,v_{m_b}](b)=\beta_{m_b}.
\end{equation*}
This completes the proof.
\end{proof}

For the benefit of the reader, we include the next two lemmas that
show some basic results from linear algebra which are used below. 
We do not have specific references,  but the discussions on pages 7-17 
of Horn and Johnson  \cite{hojo85} are helpful, and so is
Kato \cite[Chapter 1]{kato66}.

\begin{lemma} \label{l54} 
If $S$ is a subset of $\mathbb{C}^n$, $n\in \mathbb{N}_2$, then
\begin{enumerate}
\item $S^{\bot }$ is a subspace of $\mathbb{C}^n$.

\item $(S^{\bot })^{\bot }=$ span of $S$.

\item $(S^{\bot })^{\bot }=S$, if $S$ is a subspace.

\item $n=\dim S^{\bot }+\dim (S^{\bot })^{\bot }$.

\item Suppose $A\in M_{l,m}$. Then $\mathcal{R}(A)=(\mathcal{N}(A^{\ast
}))^{\bot }$ i.e. $Ax=y$ has a solution (not necessarily unique) if and only
if $y^{\ast }z=0$ for all $z\in\mathbb{\ C}^{l}$ such that $A^{\ast }z=0$.
\end{enumerate}
\end{lemma}

\begin{lemma} \label{l55}
Let $G$ be any invertible $p\times p$ matrix and $F$ an $l\times
p $ matrix with $\operatorname{rank}F=l$. Then the following assertions are
equivalent:
\begin{enumerate}
\item[(i)] $\mathcal{N}(F)\subset \mathcal{R}(GF^{\ast })$;

\item[(ii)] $\operatorname{rank}(FGF^{\ast })\leq 2l-p$;

\item[(iii)] $\operatorname{rank}(FGF^{\ast })=2l-p$;

\item[(iv)] $\mathcal{N}(F)=GF^{\ast }\big(\mathcal{N}(FGF^{\ast })\big)$.
\end{enumerate}
\end{lemma}

The next lemma `connects' the Lagrange identity with the boundary condition 
\ref{5.5}).

\begin{lemma} \label{l56}
Assume that $U\in M_{l,2d}$, $\operatorname{rank}U=l$, $d\leq l\leq 2d$. 
Let $y,z\in D_{\rm max}$ and define $Y_{a,b}$, $Z_{a,b}$ by \eqref{5.4}. 
Let \begin{equation}
P=i^n\begin{pmatrix}
E_{m_a} & 0 \\
0 & -E_{m_b}
\end{pmatrix}  \label{weq8}
\end{equation}
and note that $P^{-1}=-P=P^{\ast }$. Then $S(U)$ is symmetric if and only if
\begin{equation}
Z_{a,b}^{\ast }PY_{a,b}=0,\quad\text{for all }y,z\in D(S(U)).  \label{5.9}
\end{equation}
\end{lemma}

\begin{proof}
By Lemma \ref{l41} for any $y,z\in D_{\rm max}$, we have
\begin{equation*}
\int_a^{b}\{\overline{z}My-y\overline{Mz}\}=[y,z](b)-[y,z](a).
\end{equation*}
Therefore, it follows from the definition of $S(U)$ given in \eqref{5.6}
that $S(U)$ is symmetric if and only if for all $y,z\in D(S(U))$,
\begin{equation*}
\int_a^{b}\{\overline{z}S(U)y-y\overline{S(U)z}\}
=\int_a^{b}\{\overline{z}My-y\overline{Mz}\}=[y,z](b)-[y,z](a)=0.
\end{equation*}
By \eqref{5.1}, functions $y,z\in D_{\rm max}$ can be represented as
\begin{gather*}
y=y_0+c_1u_1+c_2u_2+\dots
+c_{m_a}u_{m_a}+h_1v_1+h_2v_2+\dots +h_{m_b}v_{m_b}, \\
z=z_0+\widehat{c}_1u_1+\widehat{c}_2u_2+\dots +\widehat{c}
_{m_a}u_{m_a}+\widehat{h}_1v_1+\widehat{h}_2v_2+\dots +\widehat{
h}_{m_b}v_{m_b},
\end{gather*}
where $y_0,z_0\in D_{\rm min}$ and $c_j$, 
$\widehat{c}_j\in \mathbb{C}$,
$j=1,\dots ,m_a;$ $h_j$, $\widehat{h}_j\in \mathbb{C}$,
$j=1,\dots ,m_b$. From \eqref{3.13}, Lemma \ref{l53} and the
definition of $\widehat{V}$ it follows that
\begin{align*}
[ y,z](b)
&= (\overline{\widehat{h}}_1, \overline{\widehat{h}}_2,
\dots ,\overline{\widehat{h}}_{m_b})\widehat{V}
\begin{pmatrix}
h_1 \\
\dots \\
h_{m_b}
\end{pmatrix} \\
&= \big(\overline{[z,v_1](b)},\dots ,\overline{[z,v_{m_b}](b)}\big)
({\widehat{V}}^{-1})^*\widehat{V}\widehat{V}^{-1}
\begin{pmatrix}
[y,v_1](b) \\
\dots \\
 [y,v_{m_b}](b)
\end{pmatrix} \\
&=-i^n\big(\overline{[z,v_1](b)},\dots ,\overline{[z,v_{m_b}](b)}
\big)E_{m_b}
\begin{pmatrix}
[y,v_1](b) \\
\dots \\
[y,v_{m_b}](b)
\end{pmatrix}.
\end{align*}
Similarly, \eqref{3.13}, Lemma \ref{l53} and the definition of $\widehat{U}$
lead to
\begin{align*}
[ y,z](a)
&= (\overline{\widehat{c}}_1,\ \overline{\widehat{c}}_2,
\dots ,\overline{\widehat{c}}_{m_a})\widehat{U}
\begin{pmatrix}
c_1 \\
\dots \\
c_{m_a}
\end{pmatrix} \\
&= -i^n(\overline{[z,u_1](a)},\dots ,\overline{[z,u_{m_a}](a)}
)E_{m_a}\begin{pmatrix}
[y,u_1](a) \\
\dots \\
[y,u_{m_a}](a)
\end{pmatrix}.
\end{align*}
Therefore,
\begin{align*}
&[ y, z](b)-[y,z](a)=\\
&= \big(\overline{[z,u_1](a)},\dots ,\overline{[z,u_{m_a}](a)},
\overline{[z,v_1](b)},\dots ,\overline{[z,v_{m_b}](b)}\,\big)P
\begin{pmatrix}
[ y,u_1](a) \\
\dots \\
[y,u_{m_a}](a) \\
[y,v_1](b) \\
\dots \\
[y,v_{m_b}](b)
\end{pmatrix}.
\end{align*}
Hence, the operator $S(U)$ is symmetric if and only if
\begin{equation*}
[ y,z](b)-[y,z](a)=0\quad\text{for all } y,z\in D(S(U)),
\end{equation*}
i.e.
\begin{equation*}
Z_{a,b}^{\ast }PY_{a,b}=0\quad\text{for all } y,z\in D(S(U)).
\end{equation*}
\end{proof}

\begin{lemma}\label{l57}
Each of the following statements is equivalent to \eqref{5.9}:
\begin{enumerate}
\item For all $Y,Z\in \mathcal{N}(U),\ Z^{\ast }PY=0$;

\item $\mathcal{N}(U)\bot P(\mathcal{N}(U))$;

\item $P(\mathcal{N}(U))\subset \mathcal{N}(U)^{\bot }=\mathcal{R}(U^{\ast
}) $;

\item $\mathcal{N}(U)\subset \mathcal{R}(P^{-1}U^{\ast })=\mathcal{R}
(PU^{\ast })$.
\end{enumerate}
\end{lemma}

\begin{proof}
Statements (1) and (2) are the same statements, just written differently.
The equivalence of (2) and (3) follows from Lemma \ref{l54}. Whereas the
equivalence of (3) and (4) immediately follows from the fact that $P$ is an
invertible matrix and $P^{-1}=-P$.
\end{proof}

\begin{theorem} \label{t53}
Let $U$ be an $l\times 2d$ matrix with $\operatorname{rank}U=l$, where 
$d\leq l\leq 2d$, $d=d_a+d_b-n$. Then the operator $S(U)$ is symmetric if and
only if
\begin{equation*}
\mathcal{N}(U)\subset \mathcal{R}(PU^{\ast }),
\end{equation*}
where $P$ is defined by \eqref{weq8}.
\end{theorem}

\begin{proof}
This follows from the Singular patching lemma \ref{l53}, 
Lemma \ref{l56} and Lemma \ref{l57}.
\end{proof}

The result given by the next lemma is not new, it is 
\cite[Theorem 3]{hzsz16}. 
The decomposition \eqref{5.1} of the maximal domain plays an
important role in our proof of Theorem \ref{t52}. It is based on the
construction of LC solutions and the decomposition of the maximal domain due
to Wang et al  \cite{wasz09}, which, in turn, was influenced by a method
of Sun \cite{sun86}. We give this lemma here because of its relationship to
Theorem \ref{t52} and because our proof is different.

\begin{lemma} \label{l58}
Suppose $U\in M_{l,2d}$. Let $U=(A:B)$ where $A\in M_{l,m_a}$ 
consists of the first $m_a$ columns of $U$ in the same order as they are
in $U$ and $B\in M_{l,m_b}$ \ consists of the other $m_b$ columns in the
same order as in $U$ (recall that $m_a+m_b=2d$) and assume that $
\operatorname{rank}U=l$. Then the operator $S(U)$ is self-adjoint if and only if
\begin{equation*}
l=d\quad\text{and}\quad  AE_{m_a}A^{\ast }-BE_{m_b}B^{\ast }=0.
\end{equation*}
\end{lemma}

\begin{proof}
It follows from Lemma \ref{l52} and Theorem \ref{t53} that $S(U)$ is
self-adjoint if and only if $S(U)$ is a $d$ dimensional symmetric extension
of the minimal operator $S_{\rm min}$, i.e. if and only if $l=d$ and 
$\mathcal{N}(U)\subset \mathcal{R}(PU^{\ast })$. 
When $l=d$, one has $\dim (\mathcal{N}(U))=d$ and 
$\dim (\mathcal{R}(PU^{\ast }))=d$. Hence 
$\mathcal{N}(U)\subset \mathcal{R}(PU^{\ast })$ is equivalent to 
$\mathcal{R}(PU^{\ast })\subset \mathcal{N}(U)$, and this is equivalent to
 $UPU^{\ast }=0$, i.e. $AE_{m_a}A^{\ast }-BE_{m_b}B^{\ast }=0$.
\end{proof}

Next we study matrices $U$ such that $(S(U))^{\ast }$ is symmetric.

\begin{theorem}\label{t54}
Let $U\in M_{l,2d}$, $0\leq l\leq 2d$ and assume that $\operatorname{rank}U=l$. 
Then
\begin{equation*}
D((S(U))^{\ast })=\{z\in D_{\rm max}:Z_{a,b}
=\begin{pmatrix}
[ z,u_1](a) \\
\dots \\
[z,u_{m_a}](a) \\
[z,v_1](b) \\
\dots \\
[z,v_{m_b}](b)
\end{pmatrix} \in \mathcal{R}(PU^{\ast })\}.
\end{equation*}
\end{theorem}

\begin{proof}
Let $z\in D_{\rm max}$. Then $z\in D((S(U))^{\ast })$ if and only if
\begin{equation*}
(S_{\rm max}y,z)=(y,S_{\rm max}z),\quad\text{for all } y\in D(S(U)).
\end{equation*}
This is equivalent to $Z_{a,b}^{\ast }PY_{a,b}=0$ for all $y\in D(S(U))$.
Therefore $z\in D((S(U))^{\ast })$ if and only if 
$Y_{a,b}^{\ast }P^{\ast}Z_{a,b}=0$, i.e. 
$P^{\ast }Z_{a,b}\in \mathcal{N}(U)^{\perp }=\mathcal{R}
(U^{\ast })$. This completes the proof.
\end{proof}

\begin{lemma} \label{l511}
Let $U\in M_{l,2d}$ and assume $\operatorname{rank}U=l$ and $0 \leq l\leq d$.
Then the following assertions are equivalent:
\begin{enumerate}
\item $(S(U))^{\ast }$ is symmetric;

\item $\mathcal{N}(U)\supset \mathcal{R}(PU^{\ast })$;

\item $UPU^{\ast }=0$.
\end{enumerate}
\end{lemma}

\begin{proof}
From Lemma \ref{l56} and Theorem \ref{t54}, it follows that $(S(U))^{\ast }$
is symmetric if and only if
\begin{equation}
Z_{a,b}^{\ast }PY_{a,b}=0,\quad\text{for all } y,z\in D((S(U))^{\ast }),
\label{1.23}
\end{equation}
where $Y_{a,b},\,Z_{a,b}\in \mathcal{R}(PU^{\ast })$ are defined as in 
\eqref{5.4}. By Lemma \ref{l53} and Theorem \ref{t54}, \eqref{1.23} is equivalent
to $Z^{\ast }PY=0$ for all $Z,Y\in \mathcal{R}(PU^{\ast })$. 
Since $P^{2}=-I$, this is equivalent to 
$\mathcal{R}(PU^{\ast })\perp \mathcal{R}(U^{\ast })$. 
From Lemma \ref{l54} we know that 
$\mathcal{R}(U^{\ast })=(\mathcal{N}(U))^{\perp }$, so that 
$\mathcal{R}(PU^{\ast })\perp \mathcal{R}(U^{\ast })$
is equivalent to (2), which proves $(1)\Longleftrightarrow (2)$. The
equivalence of (2) and (3) can be obtained immediately.
\end{proof}

\begin{lemma} \label{l512}
Let $U\in M_{l,2d}$ and assume that $\operatorname{rank}U=l$ and 
$d\leq l\leq 2d=m_a+m_b$. Then the following statements are equivalent:
\begin{enumerate}
\item $S(U)$ is a symmetric extension of the minimal operator $S_{\rm min}$;

\item $\mathcal{N}(U)\subset \mathcal{R}(PU^{\ast })$;

\item There exists a $d\times 2d$ matrix $\tilde{{U}}$ satisfying $\text{rank
}\, \tilde{U}=d$, $\mathcal{N}(U)\subset \mathcal{N}(\tilde{U})$ and $\tilde{
U}P\tilde{U}^{\ast }=0$;

\item There exists a $d\times l$ matrix $\tilde{V}$ satisfying $\operatorname{rank}
\tilde{V}=d$ and $\tilde{{V}}UPU^{\ast }\tilde{V}^{\ast }=0$;

\item $\operatorname{rank}(UPU^{\ast })=2l-(m_a+m_b)=2(l-d)$;

\item $\operatorname{rank}(UPU^{\ast })\leq 2l-(m_a+m_b)=2(l-d)$;

\item $\mathcal{N}(U)=PU^{\ast }(\mathcal{N}(UPU^{\ast }))$.
\end{enumerate}
\end{lemma}

\begin{proof}
The equivalence of (1) and (2) is given in Theorem \ref{t53}.

(1) $\Rightarrow $ (3): Note that every symmetric extension of $S_{\rm min}$
is a restriction of a self-adjoint extension of $S_{\rm min}$. By (1), $S(U)$
is a symmetric extension of $S_{\rm min}$, and by Lemma \ref{l58}, $S(\tilde{U}
)$ is self-adjoint. Therefore (3) holds.

(3) $\Rightarrow $ (2): By matrix algebra and condition (3), we obtain that $
\mathcal{N}(U)\subset \mathcal{N}(\tilde{U})=\mathcal{R}(P\tilde{U}^{\ast })$
. It follows from $\mathcal{N}(U)\subset \mathcal{N}(\tilde{U})$ that
\begin{equation*}
\mathcal{R}(\tilde{U}^{\ast })=\mathcal{N}(\tilde{U})^{\perp }\subset
\mathcal{N}(U)^{\perp }=\mathcal{R}(U^{\ast }).
\end{equation*}
Thus $\mathcal{R}(P\tilde{U}^{\ast })\subset \mathcal{R}(PU^{\ast })$, and
then it follows that $\mathcal{N}(U)\subset \mathcal{R}(PU^{\ast })$. This
shows that (2) holds.

(3) $\Rightarrow $ (4): Since $\mathcal{N}(U)\subset \mathcal{N}(\tilde{U}) $
, we have $\mathcal{R}(U^{\ast }) \supset \mathcal{R}(\tilde{U}^{\ast })$.
Therefore there exists a $d\times l$ matrix $\tilde{V}$ such that $\tilde{U}
^{\ast }=U^{\ast }\tilde{V}^{\ast }$, i.e. $\tilde{U}=\tilde{V}U$. From $
\tilde{U}P\tilde{U}^{\ast }=0$, it follows that $\tilde{V}UPU^{\ast }\tilde{V
}^{\ast }=\tilde{U}P\tilde{U}^{\ast }=0$. By $\operatorname{rank}U =l$, one has $
\operatorname{rank}\tilde{V }
=\operatorname{rank}(\tilde{V}U)=\operatorname{rank} \tilde{U} =d$.

(4) $\Rightarrow $ (3): Set $\tilde{U}=\tilde{V}U$. Then $\tilde{U}P\tilde{U}
^{\ast }=\tilde{V}UPU^{\ast }\tilde{V}^{\ast }=0$. It follows from 
$\operatorname{rank}U=l$ that 
$\operatorname{rank}\tilde{U}=\operatorname{rank}(\tilde{V}U)=\operatorname{rank}
\tilde{V}=d$. For any $Y\in N(U)$, $\tilde{U}Y=\tilde{V}UY=0$ which shows
that $\mathcal{N}(U)\subset \mathcal{N}(\tilde{U})$.

The equivalence of (2), (5), (6) and (7) can be obtained by from the Linear
Algebra Lemma \ref{l55}.
\end{proof}

Based on the above lemmas and theorems we now obtain our main result: the
characterization of symmetric operators in the Hilbert space $L^{2}(J,w)$
determined by two-point boundary conditions.

\begin{theorem} \label{t52}
Suppose $M$ is a symmetric differential expression on the
interval $(a,b)$, $-\infty \leq a<b\leq \infty $, of order 
$n\in \mathbb{N}_2$. Let $a<c<b$. Assume that the deficiency indices of $M$ 
on $(a,c)$, $(c,b)$ are $d_a$, $d_b$, respectively, and hypothesis 
(H1) holds.
Let $u_1,u_2,\dots ,u_{m_a}$ , $m_a=2d_a-n$, and 
$v_1,v_2,\dots ,v_{m_b}$, $m_b=2d_b-n$, be LC solutions
on $(a,c)$, $(c,b)$ as constructed by Theorem \ref{t41}, respectively, and
extended to maximal domain functions in $D_{\rm max}=D_{\rm max}(a,b)$ as in
Theorem \ref{t41}. Define $Y_{a,b}$ by \eqref{5.4}. Assume $U\in M_{l,2d}$
has rank $l$, $0\leq l\leq m_a+m_b=2d$ and let $U=(A:B)$ with 
$A\in M_{l,m_a}$ consisting of the first $m_a$ columns of $U$ in the same
order as they are in $U$ and $B\in M_{l,m_b}$ consisting of the next 
$m_b $ columns of $U$ in the same order as they are in $U$. Define the
operator $S(U)$ in $L^{2}(J,w)$ by \eqref{5.6} and let
\begin{equation*}
C=C(A,B)=AE_{m_a}A^{\ast }-BE_{m_b}B^{\ast },\;and\;let\;r=\operatorname{rank}C.
\end{equation*}
Then we have
\begin{enumerate}
\item If $l<d_a+d_b-n=d$, then $S(U)$ is not symmetric.

\item If $l=d_a+d_b-n=d$, then $S(U)$ is self-adjoint (and hence also
symmetric) if and only if $r=0$.

\item Let $l=d+s$, $0<s\leq d$. Then $S(U)$ is symmetric if and only if $
r=2s$.
\end{enumerate}
\end{theorem}

\begin{proof}
Part (1) follows from the abstract von Neumann formula stated by Lemma 
\ref{l51} and Lemma \ref{l52}.

Part (2) is given by Lemma \ref{l58}.

Part (3): $d<l\leq 2d$. From Lemma \ref{l512} it follows that $S(U)$ is
symmetric if and only if $rankC=rankUPU^{\ast }=2(l-d)=2s$.
\end{proof}

\section{Examples of symmetric operators}

In this section, based on Theorem \ref{t52}, we construct examples of
symmetric operators for the\ symmetric expressions $M$ of order $5$ based on
Section 2 above: Let $Q\in Z_{5}(J)$ satisfy \eqref{2.14} and let $M=M_{Q}$.

Let $l=\operatorname{rank}U$. By (1) of Theorem \ref{t52} $S(U)$ is not 
symmetric when $l<d$. When $l=d$ Lemma \ref{l58} characterizes the self-adjoint (and
therefore also symmetric) operators $S(U)$.

\begin{example} \rm
Let the hypotheses and notation of Theorem \ref{t52} hold. It follows from
Lemma \ref{lemma2} that the deficiency indices $d_a$ and $d_b$ satisfy $
3\leq d_a,d_b\leq 5$.

Assume that $d_a=4$, $d_b=5$, then $d=4$, $m_a=3$ and $m_b=5$. In
this case, the endpoint $a$ is singular and the endpoint $b$ is regular or
limit-circle (LC). The LC solutions at $a$ are $u_1,u_2,u_{3}$ and the
LC solutions at $b$ are $v_1,v_2,\dots ,v_{5}$. If $b$ is a regular
endpoint for this $M$ then, in the discussion below, simply replace $
[y,v_1](b)$, $[y,v_2](b)$, $[y,v_{3}](b)$, $[y,v_{4}](b)$, $[y,v_{5}](b)$
with $y(b)$, $y^{[1]}(b),\,y^{[2]}(b),\,\,y^{[3]}(b),\ \ y^{[4]}(b)$.

It follows from Theorem \ref{t52} that if $l=d+s=4+s$, $0<s<4$, then $S(U)$
is symmetric if and only if $r=rank(AE_{m_a}A^{\ast }-BE_{m_b}B^{\ast })
=2s$. We construct examples for each $s=1,2,3$.

(1) If $s=1$, then $l=5$ and $r=2$.

(i) Let
\begin{equation*}
A=\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},\quad 
B=\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0
\end{pmatrix}.
\end{equation*}
Then $l=\operatorname{rank}U=\operatorname{rank}(A:B)=5$ and 
$r=\operatorname{rank}(AE_{3}A^{\ast }-BE_{5}B^{\ast})=2$. 
Therefore, by Theorem \ref{t52}, the operator $S(U)$ determined by
the following boundary condition is symmetric:
\begin{gather*}
[y,u_1](a)=0,\quad [y,u_2](a)=0, \\
[y,v_1](b)=0,\quad [y,v_2](b)=0,\quad [y,v_{3}](b)=0.
\end{gather*}
Note that the boundary conditions are strictly separated.

(ii) Let
\begin{equation*}
A=\begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},\quad 
B=\begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0
\end{pmatrix}.
\end{equation*}
Then $l=\operatorname{rank}U=\operatorname{rank}(A:B)=5$ and 
$r=\operatorname{rank}(AE_{3}A^{\ast }-BE_{5}B^{\ast})=2$. 
By Theorem \ref{t52}, the operator $S(U)$ is symmetric with mixed
boundary condition:
\begin{gather*}
[y,u_2](a)=0,\quad [y,v_{3}](b)=0,\quad [y,v_{4}](b)=0, \\
[y,u_1](a)+[y,v_{5}](b)=0, \quad
[y,u_{3}](a)+[y,v_1](b)=0.
\end{gather*}

(2) If $s=2$, then $l=6$ and $r=4$.

(i) Let
\begin{equation*}
A=\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},\quad 
B=\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0
\end{pmatrix}.
\end{equation*}
A direct computation shows that 
$l=\operatorname{rank}U=\operatorname{rank}(A:B)=6$ and 
$r=\operatorname{rank}(AE_{3}A^{\ast }-BE_{5}B^{\ast })=4$. 
Therefore, the following
boundary conditions determine a symmetric operator $S(U)$:
\begin{gather*}
[y,u_1](a)=0,\quad [y,u_2](a)=0, \quad [y,v_1](b)=0,\\
[y,v_2](b)=0,\quad [y,v_{3}](b)=0,\quad [y,v_{4}](b)=0.
\end{gather*}

(ii) Choose
\begin{equation*}
A=\begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},\quad 
B=\begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0
\end{pmatrix}.
\end{equation*}
Then $l=\operatorname{rank}U=\operatorname{rank}(A:B)=6$ and 
$r=\operatorname{rank}(AE_{3}A^{\ast }-BE_{5}B^{\ast})=4$. 
Therefore $S(U)$ determined by the following mixed boundary condition
is symmetric:
\begin{gather*}
[y,u_2](a)=0,\quad [y,v_2](b)=0, \\
[y,v_{3}](b)=0,\quad [y,v_{4}](b)=0, \\
[y,u_1](a)+[y,v_{5}](b)=0, \quad
[y,u_{3}](a)+[y,v_1](b)=0.
\end{gather*}

(3) If $s=3$, then $l=7$ and $r=6$.

(i) Choose
\begin{equation*}
A=\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},\quad 
B=\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1
\end{pmatrix}.
\end{equation*}
By a direct computation we have: $l=\operatorname{rank}U=7$ and 
$r=\operatorname{rank}(AE_{3}A^{\ast}-BE_{5}B^{\ast })=6$.
 Therefore the operator $S(U)$ determined by the
following boundary condition is symmetric:
\begin{gather*}
[y,u_1](a)=0,\quad [y,u_2](a)=0, \\
[y,v_{i}](b)=0,\quad i=1,2,3,4,5.
\end{gather*}
Note that this is a symmetric operator with strictly separated boundary
conditions: there are 2 conditions at the endpoint $a$, 5 at $b$ and no
coupled condition.
\end{example}

\begin{example} \rm
(ii) Let
\begin{equation*}
A=\begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},\quad 
B=\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -i \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0
\end{pmatrix}.
\end{equation*}
Then one has $l=\operatorname{rank}U=7$ and 
$r=\operatorname{rank}(AE_{3}A^{\ast }-BE_{5}B^{\ast })=6$.
By Theorem \ref{t52}, the following mixed boundary condition determines a
symmetric operator:
\begin{gather*}
[y,u_2](a)=0,\quad [y,u_{3}](a)=0, \\
[y,v_1](b)=0,\quad  [y,v_2](b)=0, \\
[y,v_{3}](b)=0,\quad  [y,v_{4}](b)=0, \\
[y,u_1](a)=i[y,v_{5}](b).
\end{gather*}
Note that here there are 2 separated conditions at $a$; 4 separated
conditions at $b$ and 1 nonreal coupled condition.
\end{example}

\subsection*{Acknowledgements}
The first author was supported by the China Postdoctoral Science Foundation
(project 2014M561336).

We thank the anonymous referees for their careful reading of the manuscript and
for their specific suggestions. These have significantly improved the
presentation of this article and eliminated two errors.


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\end{document}
