On symmetry of the product of two higher-order quasi-differential operators

1.   Introduction

Consider the equation

where $M$ is a symmetric differential expression of order $n$, and $w$ is a positive weight function.

In 1995, M$\ddot{o}$ller-Zettl ^[12] first characterized the symmetric realizations of (1.1) for regular problems. Motivated by the method used by Sun ^[10] for self-adjoint operators and by the method of M$\ddot{o}$ller-Zettl ^[12], Wang-Zettl ^[3,4] characterized the domains of the symmetric realizations of (1.1) for singular problems with any deficiency index.

Based on the self-adjoint GKN theorem and von Neumann's formula for the adjoint of a symmetric operator in Hilbert space, for classical expression $M$, Sun ^[10] gave a decomposition of the maximal domain using certain solutions for non-real value of the spectral parameter $\lambda$, and then characterized all the self-adjoint realizations. Sun's work is an important contribution to the study of self-adjoint domains. Wang et al. ^[6] established a new representation in terms of certain solutions for real $\lambda$. This leads to a classification of solutions as limit-point(LP) or limit-circle(LC). The LC solutions contribute to the singular boundary conditions, but the LP solutions do not. In 2012, Hao et al. ^[18] extended this result to the case when both endpoints are singular. This real $\lambda$ decomposition of the maximal domain and the construction of LC and LP solutions also play a critical role in the investigation of the spectrum of self-adjoint operators, the classification of self-adjoint boundary conditions, and the characterization of symmetric domains.

In ^[1,4,5], Wang-Zettl characterized the symmetric operators in $H = L^2(J, w)$ and proved a symmetric GKN-Type theorem. The result contains the self-adjoint GKN theorem as a special case. It is well known that the self-adjoin GKN theorem is widely used to study self-adjoint operators, difference operators, Hamiltonian systems, multi-interval operators, etc. The symmetric GKN-Type theorem maybe will have similar extensions for symmetric problems.

For products and powers of differential expressions, in ^[2,15,16], the deficiency indices of powers of classical expressions and of quasi-differential expressions were discussed. In ^[20], the self-adjointness of the product of two second-order differential operators was obtained. Based on ^[10,20], An-Sun ^[8] characterized the self-adjointness of product of two $n$th-order real classical differential expressions with two regular endpoints and extended to problems with one regular endpoint and one singular endpoint ^[9]. In recent years, there are some works on the self-adjointness of products of differential operators, see ^{[7,11,13,14,17,19]}.

In this paper, we study the symmetric domain characterization for product of two quasi-differential expressions of order $n$, even or odd, with complex coefficients. We consider the cases when each endpoint is either regular or LC singular. The self-adjoint characterization for product of two differential expressions is a special case.

The organization of this paper is as follows. Following this introduction, the quasi-differential expressions, Lagrange identity, maximal and minimal operators, and powers of differential expressions are given in Section 2. Section 3 is devoted to the symmetric domain characterization of product of two differential operators. In subsection 3.1, we consider the case when one endpoint is regular and the other LC singular. In subsection 3.2, we consider the case when both endpoints are singular. Some examples are given in Section 4.

2.   Preliminaries

We first repeat some definitions and basic properties of quasi-differential expressions. See the book ^[1] for more details.

Definition 1. For $n > 1$, let

For $Q \in Z_n(J)$ we define

and $y^{[0]} = y \left(y \in V_0\right)$. Inductively, for $r = 1, \ldots, n$, we define

where $q_{n, n+1}: = 1$. Finally we set

The expression $M = M_Q$ is called the quasi-differential expression generated by $Q$. For $V_{n}$ we also use the notations $D(Q)$ and $D(M)$.

Definition 2. Let $Q \in Z_n(J), J = (a, b)$. The expression $M = M_Q$ is said to be regular at $a$ or we say $a$ is a regular endpoint, if for some $c, a < c < b$, we have

Similarly the endpoint $b$ is regular if for some $c, a < c < b$, we have

Note that from the definition of $Q \in Z_n(J)$ 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 $J$, if $M$ is regular at both endpoints. An endpoint is singular if it is not regular.

Definition 3. Let $\mathbb{N}_{2} = \{2, 3, 4, \cdots, \}$. For $k \in \mathbb{N}_{2}$, we define the matrix $E_k$ as follows:

where $\delta_{i, j}$ is the Kronecker $\delta$. Note that

Lemma 1 (Lagrange Identity). Let $Q \in Z_{n}(J), P = -E^{-1} Q^{*} E$ where $E = E_{n}$ is defined in Definition 3. Then $P \in Z_{n}(J)$ and for any $y \in D\left(M_{Q}\right), z \in D\left(M_{P}\right)$, we have

where

Here we call $[y, z]$ or just $[\cdot, \cdot]$ a Lagrange bracket.

Corollary 1. If $My = \lambda wy$ and $Mz = \bar{\lambda}wz$, then $[y, z]$ is constant on $J$. In particular, if $\lambda$ is real and $My = \lambda wy$, $Mz = \lambda w z$, then $[y, z]$ is constant on $J$.

The above symplectic matrix $E_k$ and the Lagrange Identity play an important role in the study of general symmetric differential expressions and the characterization of domains of symmetric and self-adjoint boundary conditions.

Definition 4. Let $Q \in Z_n(J)$ and suppose that $Q$ satisfies

Then $Q$ is called a Lagrange symmetric matrix and the expression $M = M_Q$ is called a Lagrange symmetric, or just a symmetric, differential expression.

Definition 5. Let $Q \in Z_n(J)$, $Q = -E^{-1} Q^* E$, let $M = M_Q$, $H =$ $L^2(J, w)$, and let $w$ be a weight function. The maximal operator $S_{\max } = S_{\max }(Q, J)$ with domain $D_{\max } = D_{\max }(Q, J)$ is defined by:

$D_{\max }(Q, J)$ is dense in $H$. The minimal operator $S_{\min } = S_{\min }(Q, J)$ with domain $D_{\min } = D_{\min }(Q, J)$ is defined as $S_{\min } = S_{\max }^*$. $S_{\min }$ is a closed symmetric operator in $H$ with dense domain and $S_{\min }^* = S_{\max }$.

The following result is immediate from the Lagrange Identity and integration.

Corollary 2. For any $y, z \in D(M), J = (a, b)$, the limits $\lim _{t \rightarrow b^{-}}[y, z](t), \lim _{t \rightarrow a^{+}}[y, z](t)$ exist and are finite. And then we have:

Lemma 2. Let $a_1 < \cdots < a_k \in J$, where $a_1$ and $a_k$ can also be regular endpoints. Let $\alpha_{j r} \in \mathbb{C}(j = 1, \ldots, k; r = 0, \ldots, n-1)$. Then there is a $y \in D_{\max }$ such that

Lemma 3. Let $Q \in Z_n(J)$, $Q = -E^{-1} Q^* E$. Then

Given a Lagrange symmetric matrix $Q\in Z_n(J)$ and its associated symmetric expression $M = M_Q$ the construction and properties of powers $M^s$ ($s \in \mathbb{N}_{2}$) of quasi-differential expression $M$ was given in ^[1,2].

Lemma 4. Assume that $Q \in Z_{n}(J)$ is a Lagrange symmetric matrix. Let $M = M_{Q}$ and define $M^{2}$ by $M^{2} y = M(M y), \cdots, M^{s} y = M (M^{s-1} y).$ Let $Q^{[1]} = Q$ and for $s \in \mathbb{N}_{2}$, let $Q^{[s]}$ denote the block diagonal matrix

where there are $s$ matrices $Q$ on the diagonal and all other entries in this $s n \times s n$ matrix are zero except for the entries in positions $(n, n+1), (2 n, 2 n+1), \cdots, ((s-1) n, (s-1) n+1)$, these are all equal to 1. Then, for any positive integer $s$, the matrices $Q^{[s]}$ are in $Z_{s n}(J)$, and are Lagrange symmetric and the symmetric differential expression $M^{S}$ is given by

Lemma 5. Let $Q\in Z_n(J)$ be Lagrange symmetric and let $M = M_Q$ be the associated symmetric expression. If all solutions of $My = \lambda w y$ are in $L^2(J, w)$ for some $\lambda\in \mathbb{C}$, then this is true for all solutions of $M^sy = \lambda wy$ for every $\lambda\in \mathbb{C}$ and every $s\in \mathbb{N}$.

3.   Symmetric domain characterization of product of two differential operators

Wang-Zettl ^[1] characterized the domains of symmetric realizations $S$ of the equation

in the Hilbert space $H = L^2(J, w)$, where $M = M_Q$ is Lagrange symmetric, $w$ is a positive weighted function. Based on the work of ^[1], we now study the symmetric realizations of product of two quasi-differential expressions in the case that each endpoint is either regular or limit-circle(LC) singular.

In the following we always let $Q\in Z_n(J), J = (a, b), n\in \mathbb{N}_{2},$ be a Lagrange symmetric matrix and $M = M_Q$ the corresponding symmetric differential expression of order $n$, even or odd, with real or complex coefficients.

3.1.   The case when $a$ is regular and $b$ is LC singular

In this subsection, we consider the case when one endpoint of $J = (a, b)$ is regular and the other LC singular. By the Patching Lemma 2 and the decomposition of the maximal domain given by Theorem 4.4.3 in ^[1], we have the following result.

Lemma 6. Let the endpoint $a$ be regular, $b$ LC singular, and let $a < c < b$. Then the deficiency index of $M$ is $n$ and there exist $n$ linearly independent solutions $u_1(t), \ldots, u_n(t)$ of $M y = 0$ in $L^2(J)$ such that

And then we have the decomposition of $D_{\max }$:

where $z_{i} \in D_{\max }, i = 1, \cdots n$ such that $z_{i}(t) = 0$ for $t \geq c$ and $z_{i}^{[j-1]}(a) =$ $\delta_{i j}, i, j = 1, \cdots, n$, and $\delta_{i j}$ is the Kronecker $\delta$.

By the Lagrange Identity and Corollary 1, we have

where $[\cdot, \cdot]_n$ denotes the Lagrange bracket of differential expression $M$.

The following theorem can be found in Chapter 6 of ^[1].

Theorem 1. Let $M = M_{Q}, Q \in Z_{n}(J), J = (a, b), -\infty \leq a < b \leq+\infty$ be Lagrange symmetric, $w$ a weight function, and let $a$ be regular and $b$ LC singular. Let the composed matrix $U = (A: B)$ be a boundary condition matrix with $\operatorname{rank}(U) = l, 0 \leq l \leq 2 n$, where $A_{l, n}, B_{l, n}$ are complex matrices. Define the operator $S(U)$ in $L^{2}(J, w)$ by

where

functions $u_1, \cdots, u_n$ are given by (3.2). Let $C = A E_n A^{*}- BE_{m_b}(b)B^{*}$ and $r = \mathit{\text{rank}}(C)$, where $E_{n} = \left((-1)^{r} \delta_{r, n+1-s}\right)_{r, s = 1}^{n}$. Then we have:

(i) If $l < n$, then $S(U)$ is not symmetric;

(ii) If $l = n$, then $S(U)$ is self-adjoint (and hence symmetric) if and only if $r = 0$.

(iii) Let $l = n+s, 0 < s\leq n$. Then $S(U)$ is symmetric and not self-adjoint if and only if $r = 2s$.

Proof. This follows directly from Theorem 6.3.3 of ^[1]. Note that here we choose the solutions $u_1(t), \ldots, u_n(t)$ of $M y = 0$. In terms of (3.4) and (3.2), for the case when $a$ is regular and $b$ is LC singular(then $m_b = n$), the matrix $E_{m_b}(b)$ given by Theorem 6.3.3 ^[1] can be written as

where $i = \sqrt{-1}$. This concludes the proof.

For $M^2y = M(My)$, we obviously have $D_{\max}(M^2)\subset D_{\max}(M)$ and $D_{\min}(M^2)\subset D_{\min}(M)$.

Lemma 7. For any $y, z \in D_{\max}(M^{2})$, we have

where $[\cdot, \cdot]_{2n}$ denotes the Lagrange bracket of differential expression $M^2$.

Proof. By the Lagrange Identity (2.6), we have

Therefore

Now we study the symmetric domain characterizations of product of two $n$th-order differential operators $L_1$ and $L_2$ which are generated by the same symmetric differential expression $M$ (may be with same or different boundary conditions). Let

where $U_i = \left(A_i: B_i\right)$ is a composed matrix with $\operatorname{rank}\left(U_i\right) = n+s \, (0\leq s\leq n)$, $A_{i}, B_{i}$ are $(n+s) \times n$ complex matrices, and $Y_{a, b}$ is given by (3.5).

Set $L(y) = (L_{2}{ }^{\circ} L_{1})(y) = L_2(L_1(y))$, $y \in D_{1}, My \in D_{2}$. By (3.8), we have

where

and the functions $u_1, u_2, \cdots, u_n$ are defined in Lemma 6.

Theorem 2. The quasi-derivative $(M y)^{[m]} = i^n y^{[n+m]}, \, 0 \leq m \leq n.$

Proof. From Lemma 4, we have

When $n = 2$, we have

From $M y = i^2 y^{[2]}$, it follows that

Therefore $(M y)^{[1]} = i^2 y^{[3]}$.

When $n > 2$, we have

where $z = 1, 2, \cdots, 2n$. Since

it follows that $(M y)^{[1]} = i^n y^{[n+1]}$. Assume that for any given $m = k-1 (2\leq k < n)$, $(M y)^{[k-1]} = i^n y^{[n+k-1]}$ holds. Consider the case when $m = k$, then we have

and

Hence $(M y)^{[k]} = i^n y^{[n+k]}$ holds. This proof is completed by mathematical induction.

Corollary 3.

where $O$ is the $n\times n$ zero matrix, $I_n$ is the $n\times n$ identity matrix, and $\left(O: I_n\right)$ is a composed matrix.

Proof. This result follows directly from Theorem 2.

In the following, we will rewrite the operator defined in (3.9) in a clear form. We first give the maximal domain decomposition of the differential expression $M^2$.

It is obvious that $u_{1}, \cdots, u_{n}$ defined in Lemma 6 are linearly independent solutions of $M^{2}(y) = 0$. Let $\psi_{1}, \psi_2, \cdots, \psi_n$ be $n$ solutions of $M^{2}(y) = 0$ which satisfy

Combining the conditions of (3.2), we can obtain that $u_1, \cdots, u_n, \psi_1, \cdots, \psi_n$ are $2n$ linearly independent solutions of $M^2y = 0$. Similar to Lemma 6, we have the following decomposition.

Corollary 4. The maximal domain $D_{\max}(M^2)$ can has the representation:

where $z_{i} \in D_{\max }(M^2), i = 1, \cdots 2n$ such that $z_{i}(t) = 0$ for $t \geq c$ and $z_{i}^{[j-1]}(a) =$ $\delta_{i j}, i, j = 1, \cdots, 2n$.

Proof. This proof can be completed by using the same method as the proof of Theorem 4.4.3 in ^[1].

Since $\psi_{i} \in D_{\max }(M^2)$ and $D_{\max }(M^2) \subset D_{\max }(M)$, by Lemma 6, each $\psi_{i}$ has a unique representation:

where $y_{i 0}\in D_{\min }(M)$, $d_{i j}, a_{i j}\in \mathbb{C}$. Set

where the entries $a_{ij}\, (i, j = 1, 2, \cdots, n)$ of $N$ are given in (3.14).

Theorem 3. The operator $L$ defined in (3.9) can be rewritten as

where $U = (A : B)$ is a composed matrix of matrices

and matrices $A_1, A_2, B_1, B_2$ are given in (3.8).

Proof. By (3.14) and Corollary 4, every $y \in D_{\max }(M^2)$ can be uniquely written as

where $y_{0} \in D_{\min }(M^2),$ $d_i, c_i, c_i^*\in \mathbb{C}$. It follows from Lemma 6 and Lemma 7 that

Let

From (2.7), we have

Moreover

Therefore $(-1)^n D^* E_n+E_n D = 0$. Note that $u_j, \psi_j, \, j = 1, 2, \cdots, n$ are solutions of $M^2y$. By the Lagrange identity, we know $W(t)$ is constant on $J = (a, b)$. By (2.9), (3.2) and (3.19), we have

Hence

From the decomposition (3.19), we have

For any $y\in D_{\max }(M^{2})$, it follows from Lemma 7 and $M u_{i} = 0$ that

Namely

The proof of Theorem 3 is now immediate from the above discussion and (3.9). Furthermore, we have the following result.

Theorem 4. The relationship $\overline{N} E_n+(-1)^nE_n N^T = 0$ holds.

Proof. By Lemmas 7 and Lemma 3, we have

Let $\overline{N} E_n+(-1)^nE_n N^T = (b_{i j})_{1\leq i, j \leq n}$. Then by (3.23), we have

This completes the proof.

Based on the above lemmas and theorems, we now obtain our main result: the symmetric characterization of product of two differential operators.

Theorem 5. Let the hypothesis and notations of Theorem 3 hold. Then the product operator $L = L_{2}{ }^{\circ} L_{1}$ is symmetric if and only if

where $0\leq s\leq n$.

Proof. Since $\operatorname{rank}(A_i:B_i) = n+s \, (i = 1, 2)$ and

we can obtain $\operatorname{rank}(U) = 2 n+2 s$. By computation, we have

Now we will use the basic Theorem 1 to prove our result. Note that here the matrix $E_{m_b}(b)$ given in Theorem 1 is $W(b)$. Then, combining with Theorem 4, we have

Therefore, by Theorem 1, the operator $L$ is symmetric if and only if

which is equal to

Thus the proof is completed.

Remark 1. For theorem 5, if $s = 0$, then $L$ is self-adjoint if and only if $A_1 E_n A_2^* = B_1 E_n B_2^*$.

3.2.   The case when both endpoints are LC singular

We will consider the symmetric characterization of product of two differential expressions for the case when both endpoints are LC singular.

In the following we always let $Q \in Z_n(J), J = (a, b), n \in \mathbb{N}_2$, be Lagrange symmetric, $w$ a weight function, the two endpoints $a$ and $b$ LC singular, and let $M y = M_Q y = \lambda w y$ be the corresponding symmetric differential equation.

By Lemma 5, we have the next lemma.

Lemma 8. If the expression $M = M_Q$ is LC singular at both endpoints of $J = (a, b)$, then $M^2$ is LC singular at both endpoints.

We first reduce the decomposition Theorem 4.4.4 of ^[1] to the case when both endpoints are LC singular.

Lemma 9. Let $M$ be Lagrange symmetric, $a$ and $b$ LC singular, and let $c \in(a, b)$. Then $Q \in Z_n((a, c))$, $Q \in Z_n((c, b))$, and the deficiency indices of My $= \lambda w y$ on $(a, c), (c, b)$ and $(a, b)$ are all $n$. Then

(1) There exist $n$ linearly independent solutions $p_1, \cdots, p_n$ of $M y = 0$ on $(a, c)$ such that $p_i^{[j-1]}(c) = \delta_{ij}$ $(i, j = 1, 2, \ldots, n)$. For $1 \leq i, j \leq n$, we have

The solutions $p_1, \cdots, p_n$ can be extended to $(a, b)$ such that the extended functions, also denoted by $p_1, \cdots, p_n$, are in $D_{\max }(a, b)$ and are identically 0 near $b$.

(2) There exist $n$ linearly independent solutions $v_1, \cdots, v_n$ of $M y = 0$ on $(c, b)$ such that $v_i^{[j-1]}(c) = \delta_{i j}$ $(i, j = 1, 2, \ldots, n)$. For $1 \leq i, j \leq n$, we have

The solutions $v_1, \cdots, v_n$ can be extended to $(a, b)$ such that the extended functions, also denoted by $v_1, \cdots, v_n$, are in $D_{\max }(a, b)$ and are identically 0 near $a$.

(3) By the extended functions $p_i$ and $v_i$, the maximal domain has the following decomposition:

Proof. This lemma can be directly obtained from Theorem 4.4.4 of ^[1]. Since $p_i$ and $v_i$ are respectively the solutions of $M y = 0$ on $(a, c)$ and $(c, b)$, together with the Lagrange identity, we can obtain (3.30) and (3.31) hold.

Let

By (1) and (2) of Lemma 9, we have $\widehat{E}(a) = \widehat{E}(b) = -i^n E_n$.

Let

The following is a minor modification of Theorem 6.3.3 of ^[1].

Lemma 10. Let the hypothesis and notations of Lemma 9 hold. Suppose $U$ is a boundary matrix with $\operatorname{rank}(U) = l, 0 \leq l \leq 2 n$. Let $U = (A: B)$, where $A_{l \times n}$ and $B_{l \times n}$ are complex matrices. Define the operator $S(U)$ in $L^2(J, w)$ by

Let $\operatorname{rank}(A \widehat{E}(a) A^*-B \widehat{E}(b) B^*) = r$, where $\widehat{E}(a)$ and $\widehat{E}(b)$ are given in (3.33). Then we have:

(i) If $l < n$, then $S(U)$ is not symmetric;

(ii) Let $l = n+s, 0 \leq s \leq n$. Then $S(U)$ is symmetric if and only if $r = 2 s$.

Let the operators $\widehat{L}_1$ and $\widehat{L}_2$ be generated by the same LC symmetric differential expression $M$ :

where $U_i = \left(A_i: B_i\right), \operatorname{rank}\left(U_i\right) = n+s,$ $0 \leq s \leq n,$ $A_i, B_i$ are $(n+s) \times n$ matrices, and $\widehat{Y}_{a, b}$ is given in (3.34).

Let $\widehat{L} = \widehat{L}_2^{\circ} \widehat{L}_1,$ $y \in \widehat{D}_1,$ $M y \in \widehat{D}_2$, then we have:

where

It is obvious that $p_1, \cdots, p_n$ are solutions of $M^2(y) = 0$ on $(a, c)$, and $v_1, \cdots, v_n$ are solutions of $M^2(y) = 0$ on $(c, b)$. Let $\theta_1, \theta_2, \cdots, \theta_n$ be solutions of $M^2(y) = 0$ on $(a, c)$, and let $\beta_1, \beta_2, \cdots, \beta_n$ be solutions of $M^2(y) = 0$ on $(c, b)$, which satisfy

It is clear that $p_1, \cdots, p_n, \theta_1, \cdots, \theta_n$ are $2n$ linearly independent solutions of $M^2 y = 0$ on $(a, c)$ and $v_1, \cdots, v_n, \beta_1, \cdots, \beta_n$ are $2n$ linearly independent solutions of $M^2 y = 0$ on $(c, b)$. By Naimark Patching Lemma 2, $\theta_1, \cdots, \theta_n$ can be extended to $(a, b)$ such that the extended functions, also denoted by $\theta_1, \cdots, \theta_n$, are in $D_{\max }(a, b)$ and are identically 0 near $b$. Similarly, $\beta_1, \cdots, \beta_n$ can be extended to $(a, b)$ such that the extended functions, also denoted by $\beta_1, \cdots, \beta_n$, are in $D_{\max }(a, b)$ and are identically 0 near $a$. Similar to Lemma 9, we have the following decomposition.

Corollary 5. The maximal domain $D_{\max }(M^2) = D_{\max }(M^2, (a, b))$ has the representation:

Since $\theta_i, \beta_i \in D_{\max }(M^2)$ and $D_{\max }(M^2) \subset D_{\max }(M)$, by (3.32), each $\theta_i$ and $\beta_i$ has a unique representation:

where $y_{i 0}, y_{0 i} \in D_{\min }(M), \, a_{i j}, b_{i j}, c_{i j}, d_{i j} \in \mathbb{C}$.

Set

where the entries $a_{i j}, d_{ij}$ are given in (3.41) and (3.42).

Theorem 6. The operator $\widehat{L}$ defined in (3.36) can be rewritten as

where $U = (A: B)$,

and matrices $A_1, A_2, B_1, B_2$ are given in (3.35).

Proof. By Corollary 5, for every $y\in D_{\max}(M^2)$, we have

where $y_0 \in D_{\min }(M^2), a_i, b_i, c_i, d_i \in \mathbb{C}$.

Let

From (2.7), we have

By Lemma 7, we have

Therefore

At the same time, we obtain that $(-1)^n D_i^* E_n+E_n D_i = 0$, $i = 1, 2$. Since $p_j, \theta_j$ $(j = 1, 2, \cdots, n)$ are solutions of $M^2 y = 0$ on $(a, c)$, $v_j, \beta_j$ $(j = 1, 2, \cdots, n)$ are solutions of $M^2 y = 0$ on $(c, b)$. It follows from the Lagrange identity that $\widehat{W}(t)$, $\widehat{Z}(t)$ are constant on $(a, c)$ and $(c, b)$, respectively. Note that $p_i, \theta_i$ are identically 0 near $b$ and $v_i, \beta_i$ are identically 0 near $a$. Then, for $i, j = 1, 2\cdots, n$, we have

By (2.9), (3.47), (3.54) and (3.55), we have

Hence

By (3.41), (3.42), (3.47), (3.52) and (3.53), we have

and

For any $y \in D_{\max }(M^2)$, it follows from Lemma 7, together with $M p_i = 0, M v_i = 0$, that

Thus, in terms of above discussion, the proof for Theorem 6 is completed.

Theorem 7. The relationships $\overline{N_1} E_n+(-1)^n E_n N_1^T = 0,$ $\overline{N_2} E_n+(-1)^n E_n N_2^T = 0$ hold.

Proof. Similar to the proof of Theorem 4, we have

and

Then we can easily obtain $\overline{N}_1 E_n+(-1)^n E_n N_1^T = 0$ and $\overline{N}_2 E_n+(-1)^n E_n N_2^T = 0$.

Based on above discussion, we present the main result:

Theorem 8. Let the hypothesis and notations of Theorem 6 hold. Then the product operator $\widehat{L} = \widehat{L}_{2}{ }^{\circ} \widehat{L}_{1}$ is symmetric if and only if

where $0\leq s\leq n$.

Proof. Since

and $\operatorname{rank}(A_i:B_i) = n+s (i = 1, 2)$, we have $\operatorname{rank}(U) = 2 n+2 s$.

By computation, we get

and

To prove the furthermore part we note that here the matrices $\widehat{E}(a)$, $\widehat{E}(b)$ given in Lemma 10 are $\widehat{W}(a)$, $\widehat{Z}(b)$, respectively. Thus, by Lemma 10 and Theorem 7, we have

It follows from Lemma 10 that $L$ is symmetric if and only if

which is equal to

The next corollary is the self-adjoint special case of Theorem 8.

Corollary 6. Let the hypothesis and notations of Theorem 6 hold. Then the product operator $\widehat{L} = \widehat{L}_{2}{ }^{\circ} \widehat{L}_{1}$ is self-adjoint if and only if

Proof. This is the special case $s = 0$ of Theorem 8.

Remark 2. Recall that the Lagrange brackets $[y, z]$ are well defined at each singular endpoint. These brackets can be used to replace the quasi-derivatives. Our symmetric domain characterization Theorem 8 can be adapted to the maximal deficiency case which occurs when each endpoint is either regular or LC singular. Namely, our result can be used for the four cases for the endpoints: $R/R$, $R/LC$, $LC/R$, $LC/LC$.

4.   Examples

Example 1. Consider $My = -(py')'+qy$ on $J = (a, b), -\infty \leq a < b \leq+\infty$, where $p^{-1}, q\in L_{loc}(J, \mathbb{R})$. Let $a$ and $b$ be LC singular. Set

The operator $L_1y = My$ is determined by the boundary conditions:

The operator $L_2y = My$ is determined by the boundary conditions:

Since here $s = 1$ and $\operatorname{rank}(A_1 E_2 A_2^*-B_1 E_2 B_2^*) = 2s = 2$, by Theorem 8, the product operator $L = L_2{ }^{\circ} L_1$ is symmetric.

Example 2. Consider $My = -(py')'+qy$ on $J = (a, b), -\infty \leq a < b \leq+\infty$, where $p^{-1}, q\in L_{loc}(J, \mathbb{R})$. Let $a$ and $b$ be LC singular. Set

The operator $L_1y = My$ is determined by the boundary conditions:

The operator $L_2y = My$ is determined by the boundary conditions:

Since here $s = 1$ and $\operatorname{rank}(A_1 E_2 A_2^*-B_1 E_2 B_2^*) = 3$, by Theorem 8, the product operator $L = L_2{ }^{\circ} L_1$ is not symmetric.

Remark 3. For the above examples, when the endpoint $a$ is regular for this $M$, we can simply replace $[y, p_1](a), [y, p_2](a)$ with $y(a)$, $y^{[1]}(a)$, respectively. Similarly for a regular endpoint $b$, we can replace $[y, v_1](b), [y, v_2](b)$ with $y(b)$, $y^{[1]}(b)$, respectively.

Example 3. Let $Q \in Z_n(J), n \in \mathbb{N}_2,$ $J = (a, b), -\infty \leq a < b \leq+\infty$, $M = M_Q$ be the symmetric expression, and let $N_{s\times n}$ $(0\leq s\leq n)$ be of full row rank, $I_n$ the $n\times n$ identity matrix. Assume that $a$ and $b$ are LC singular. Choose

Let

We define the operators $L_1$, $L_2$ as (3.35). By computation, we have

Obviously, $\operatorname{rank}(U_1) = \operatorname{rank}(U_2) = n+s$, $\operatorname{rank}(A_1 E_n A_2^*-B_1 E_n B_2^*) = 2 s$. Then, by Theorem 8, we know $L = L_2{ }^{\circ} L_1$ is a symmetric operator.

Conflict of interest

The authors declare that there is no conflicts of interest in this paper.