Simplicity and Spectrum of Singular Hamiltonian Systems of Arbitrary Order
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
Theorem 1. If τx = λx has exactly one linearly independent square integrable solution for all λ in an interval I, then for every self-adjoint realization, one has the following:
- (1)
the intersection of continuous spectrum and I is empty;
- (2)
the point spectrum is nowhere dense in I.
It is well-known that τ can be classified into the limit point case and the limit circle case at the singular endpoint t = ∞, and it is only needed to consider the former case since every self-adjoint realization corresponding to τ has only a pure discrete spectrum in the limit circle case at t = ∞. However, for higher-order differential expressions, there are some intermediate limit cases at t = ∞ 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 I under the assumption that the number r(λ) of linearly independent solutions of an abstract equation is equal to the defect index of the corresponding operator for all λ ∈ I except an at most countable set X ⊂ I and that the spectrum of every self-adjoint extension of such an operator is nowhere dense in this interval I 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 H0 with arbitrary equal defect indices is simple inspired by the method in [19] and then apply the above results in [20] to H0. As a direct consequence, similar spectral properties of every self-adjoint extension of H0 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 H0 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 H0 is proved, the results on spectral properties of every self-adjoint extension of H0 is given, and two examples are presented.
2. Basic Concepts and Results about Linear Operators in Hilbert Spaces
Let ℋ be a Hilbert space over C with inner product 〈·, ·〉. The norm ∥·∥ is ∥f∥ = 〈f, f〉 1/2 for f ∈ ℋ. Let T be a (linear) operator in ℋ. We denote by D(T), R(T), and N(T) the domain, the range, and the kernel of T, respectively, and by T* the adjoint of a densely defined operator T. For a densely defined operator T, it is called to be symmetric if T ⊂ T* and self-adjoint if T = T*. An operator S is called a self-adjoint extension of a symmetric operator T if S is self-adjoint, D(T) ⊂ D(S), and Tf = Sf for f ∈ D(T).
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 T in ℋ, there exists exactly one spectral family {E(t)} t∈R such that T = ∫R t dE(t). Conversely, the spectral family {E(t)} t∈R is given by
Finally, we recall two results for simple symmetric operators in ℋ. A symmetric operator T in ℋ is called simple if there is not an orthogonal decomposition T = T1 ⊕ T2 with a self-adjoint operator T1 acting in a nontrivial subspace ℋ1 ⊂ ℋ (cf., [18]). For a simple symmetric operator, the result below holds.
Lemma 3 (see [19].)A symmetric operator T is simple if and only if f⊥N(λ − T*) for all λ ∈ C∖R implies that f = 0.
Then, the following result was given.
3. Main Results
In this section, we will prove that H0 is simple under condition (15) and then get results about the spectrum of every self-adjoint extension of H0 using Mogilevskii’s results in [20]. First, we present the following lemma.
Lemma 5. Assume that (A) holds and λ ∈ C∖R. Then, there exists an n × n matrix-valued function G(t, s, λ) satisfying
- (1)
G(t, s, λ) is continuous and has first derivative with respect to t on a < t ≤ s < b and a < s ≤ t < b;
- (2)
lim t→s−0G(t, s, λ) − lim t→s+0G(t, s, λ) = J;
- (3)
G satisfies that ℒG = λWG (as a function of t) for t ≠ s;
- (4)
for t ≠ s;
- (5)
, and for , the function
()is in D(H) and satisfies that (ℒ − λW)𝒢(λ)f = Wf.
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 H0 is simple.
Proof. Since (15) holds, H0 must have a self-adjoint extension H1. Let λ ∈ C∖R. Then, R(λ, H1) is a bounded operator on and for ,
Now, let Φ(t, λ) and be fundamental solution matrices for (1) and (1) with λ being replaced by satisfying , respectively, where In is the n × n unit matrix. Then,
Then, letting t2 → +∞ and t1 → −∞, we have 〈g, f〉 W = 0, and hence f = 0 since 𝒞0 is dense in . From Lemma 3, H0 is simple. This completes the proof.
Let r(λ) 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 I = (μ1, μ2), −∞ ≤ μ1 < μ2 ≤ ∞, be an interval such that r(λ) = d for all λ ∈ Iexcept an at most countable setX ⊂ I. Then, for any self-adjoint extension H1 of H0, σc(H1)∩I is empty and σ(H1)∩I is nowhere dense in I.
Example 8. Consider (1) on [1, ∞) with m = 2, and
Consider the 2 × 2 Hamiltonian system
Example 9. Consider (1) on (t0, b] with m = 2, and
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).