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Abstract and Applied Analysis

Volume 2013 (2013), Article ID 202851, 6 pages

http://dx.doi.org/10.1155/2013/202851

## Simplicity and Spectrum of Singular Hamiltonian Systems of Arbitrary Order

Department of Mathematics, Shandong University at Weihai, Weihai, Shandong 264209, China

Received 22 August 2013; Accepted 5 December 2013

Academic Editor: Douglas Anderson

Copyright © 2013 Huaqing Sun. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The paper is concerned with singular Hamiltonian systems of arbitrary order with arbitrary equal defect indices. It is proved that the minimal operator generated by the Hamiltonian system is simple. As a consequence, a sufficient condition is obtained for the continuous spectrum of every self-adjoint extension of the minimal operator to be empty in some interval and for the spectrum to be nowhere dense in this interval in terms of the numbers of linearly independent square integrable solutions.

#### 1. Introduction

Consider the following Hamiltonian differential system of arbitrary order : where is a regular point, while is singular, that is, , or or is not integrable near ; and are Hermitian matrices and locally integrable on ; is an constant nonsingular matrix satisfying ( denotes the complex conjugate transpose of ); is a weight function; and is a complex parameter. System (1) includes higher-order symmetric differential expressions of both even-order and odd-order (cf. [1, 2]).

Introduce the following space: with inner product for . Set for . We remark that if is singular, then is a quotient space in the sense that if . In this case, is a Hilbert space.

Let denote the set of functions which are locally and absolutely continuous on an interval . In order to ensure the Hamiltonian operators generated by (1) to be single-valued, it is always assumed that

Clearly, assumption implies the Atkinson definiteness condition (see [3, Page 253]): for all and for all nontrivial solutions of system (1), the following inequality always holds:

The maximal and preminimal operators and corresponding to system (1) are defined by It can be easily verified that is single-valued under assumption . The operator is called the minimal operator corresponding to system (1).

Spectra of operators have been studied by many authors using various theories such as the oscillation theory, asymptotic analysis, singular sequences, and square-integrable solutions for real values of the spectral parameter (cf., e.g., [4–13]). Among these methods, the last one has attracted lots of attention because it takes advantage of using numerous tools available in the fundamental theory of differential equations. This method was first explored by Hartman and Wintner for the Sturm-Liouville differential expression and a series of results were given (cf. [14–16]). The following is one of them (cf. [15]).

Theorem 1. *If has exactly one linearly independent square integrable solution for all in an interval , then for every self-adjoint realization, one has the following:*(1)*the intersection of continuous spectrum and is empty;*(2)*the point spectrum is nowhere dense in .*

It is well-known that can be classified into the limit point case and the limit circle case at the singular endpoint , and it is only needed to consider the former case since every self-adjoint realization corresponding to has only a pure discrete spectrum in the limit circle case at . However, for higher-order differential expressions, there are some intermediate limit cases at besides the above two cases (cf. [17]). Weidmann’s monograph [13] proved that Theorem 1 holds for higher-order differential equations in the limit point case. This result of [13] was further extended to the case of arbitrary equal defect indices in [8, 12]. Simple operators in Hilbert spaces have their special properties and have been studied (cf., e.g., [18]). It has been shown that the minimal operators generated by higher-order symmetric differential equations with a regular endpoint are simple (cf., [10, 19]). Recently, Mogilevskii [20] considered simple symmetric operators in Hilbert spaces with equal defect indices and derived that the continuous spectrum of every self-adjoint extension of such an operator is empty in some interval under the assumption that the number of linearly independent solutions of an abstract equation is equal to the defect index of the corresponding operator for all except an at most countable set and that the spectrum of every self-adjoint extension of such an operator is nowhere dense in this interval under the same assumption (see Lemma 4). In addition, these abstract results were applied to higher-order differential operators over arbitrary interval in [20].

In this paper, we will prove that with arbitrary equal defect indices is simple inspired by the method in [19] and then apply the above results in [20] to . As a direct consequence, similar spectral properties of every self-adjoint extension of are obtained. Note that we use Mogilevskii's results of [20] to get spectral properties in this paper. Then the method in this paper for the proof of the results on spectral properties of all self-adjoint extensions of is different from those used in [8, 12].

The rest of this present paper is organized as follows. In Section 2, some basic concepts and results about linear operators in Hilbert spaces are introduced. In Section 3, the simplicity of is proved, the results on spectral properties of every self-adjoint extension of is given, and two examples are presented.

#### 2. Basic Concepts and Results about Linear Operators in Hilbert Spaces

Let be a Hilbert space over with inner product . The norm is for . Let be a (linear) operator in . We denote by , , and the domain, the range, and the kernel of , respectively, and by the adjoint of a densely defined operator . For a densely defined operator , it is called to be symmetric if and self-adjoint if . An operator is called a self-adjoint extension of a symmetric operator if is self-adjoint, , and for .

Let be an operator in . If there exists some nontrivial such that , then is called an eigenvalue of , while is called an eigenvector of associated with , and is called the multiplicity of . We denote by the set of all eigenvalues of , where is called the point spectrum of . Let . Then, the set is called the resolvent set of , and the set is called the spectrum of . When is a self-adjoint operator, it is known in [21, Page 209] that the spectrum of admits the representation where is called the continuous spectrum of .

For an operator , the subspace is called the defect space of and , and the number is called the defect index of and . By [21, Theorem 8.1], for a symmetric operator , the defect index is constant in the upper and lower half-planes, respectively, and hence are called the positive and negative defect indices of . From [21, Theorem 4.13], Hence, . Furthermore, by [21, Theorem 8.6], a symmetric operator has a self-adjoint extension if and only if

For a self-adjoint operator, the following is the well-known spectral decomposition theorem.

Lemma 2 (see [21, Page 191]). *For every self-adjoint operator in , there exists exactly one spectral family such that . Conversely, the spectral family is given by
**
for all and .*

Finally, we recall two results for simple symmetric operators in . A symmetric operator in is called simple if there is not an orthogonal decomposition with a self-adjoint operator acting in a nontrivial subspace (cf., [18]). For a simple symmetric operator, the result below holds.

Lemma 3 (see [19]). *A symmetric operator is simple if and only if for all implies that .*

Let be symmetric and (12) holds. Set

Then, the following result was given.

Lemma 4 (see [20]). *Assume that is a simple symmetric operator in and (12) holds with . Let , , be an interval such that is an at most countable set. Then, for every self-adjoint extension of , is empty and is nowhere dense in .*

#### 3. Main Results

With a similar argument to that of [13, Theorem 3.9], the minimal operator is symmetric and . Then, it must have self-adjoint extensions if

In this section, we will prove that is simple under condition (15) and then get results about the spectrum of every self-adjoint extension of using Mogilevskii’s results in [20]. First, we present the following lemma.

Lemma 5. *Assume that holds and . Then, there exists an matrix-valued function satisfying*(1)* is continuous and has first derivative with respect to on and ;*(2)*;
*(3)* satisfies that (as a function of ) for ;*(4)* for ;*(5)*, and for , the function
is in and satisfies that .*

*Proof. *This lemma holds by [22, Formulae (3.23) and (3.24), Lemma 3.8, and Proposition 3.11].

Theorem 6. *Assume that (15) holds. Then is simple.*

*Proof. *Since (15) holds, must have a self-adjoint extension . Let . Then, is a bounded operator on and for ,
Hence, for each , , and by of Lemma 5,
which, together with , implies that . Let be an orthonormal basis for . Then
where is a constant dependent on and only. It is evident that by the self-adjointness of . Furthermore, is an integral operator defined on satisfying by [22, (iii) of Proposition 3.11]. Then, (19) implies that for ,
while
Hence,
Let be an orthonormal basis for . Then, there exist constants dependent on such that
Inserting (23) into (20), we get , and then (19) yields that

Now, let and be fundamental solution matrices for (1) and (1) with being replaced by satisfying , respectively, where is the unit matrix. Then,
which implies that . We define
Then (as a function of ) for , and
Clearly,
by (2) of Lemma 5. Therefore, for any fixed . On the other hand, for by (27), which, together with and of Lemma 5, implies that for some . Furthermore,
for , which, together with by the regularity of the endpoint , implies that . Thus, for every . In addition, it can be verified that, as a function of ,
for any fixed . Consequently, . Therefore, from (24), (27), and , we get that if for all , then
This implies that for , is a continuous function in for and all since is regular. Let be a set of which has a compact support in . Then, for , the function
admits an extension to the continuous function in **C**. This, together with Lemma 2, gives that

Then, letting and , we have , and hence since is dense in . From Lemma 3, is simple. This completes the proof.

Let be the number of linearly independent solutions of (1) in and (15) holds. Then, we can get the following from , Lemma 4, and Theorem 6.

Theorem 7. *Assume that (15) holds. Moreover, let , , be an interval such that for all except an at most countable set. Then, for any self-adjoint extension of , is empty and is nowhere dense in .*

*Example 8. *Consider (1) on with , and
where and are real numbers. Then, (1) satisfies assumption since for . Let . Then, the system has two linearly independent solutions
where are roots of the algebraic equation , that is,
Let be the roots of , that is,
Then, and if . It can be verified that yields that and . Thus, if , then . In addition, (15) holds since the coefficients of (1) are real-valued in this case. Furthermore, by [17, Theorem 3.3]. If , then for all , all the solutions of (1) are in by [3, Theorem ] in this case. This is a contradiction. Therefore, , and then by Theorem 6, for every self-adjoint extension of and , is empty and is nowhere dense in .

Consider the Hamiltonian system
where with given in Example 8, in which , is a constant matrix such that and ,
on any subinterval with some constant , and is a matrix-valued integrable function on satisfying that
Then, by [23, Theorem 3.1], for every , system (1) has two linearly independent solutions
where for some constant , , and exist and are finite for any given .

*Example 9. *Consider (1) on with , and
It can be easily verified that (1) satisfies assumption under the conditions (42). Furthermore, in this case system (1) can be rewritten as (38) with , , , and . It is clear that (39) and (40) hold and . Therefore, the system has two linearly independent solutions given by (41) with . Clearly, and for all . Thus, if , then . Here, let be the minimal operator corresponding to this case. With a similar argument to Example 8, it can be obtained that the positive and negative indices of are equal to . Note that the corresponding results to Theorem 7 hold for the case that is singular and is regular. Then, for every self-adjoint extension of , is empty and is nowhere dense in .

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The author is indebted to the referees for their helpful suggestions and corrections. This research was supported by the NNSFs of China (Grants 11101241 and 11071143), the NNSF of Shandong Province (Grants ZR2011AQ002 and ZR2012AM002), the special fund for postdoctoral innovative programs of Shandong Province (Grant 201301010), and the independent innovation fund of Shandong University (Grant 2011ZRYQ003).

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