Renormalized Oscillation Theory for Linear Hamiltonian Systems on [0, 1] Via the Maslov Index

Abstract

Working with a general class of regular linear Hamiltonian systems, we show that renormalized oscillation results can be obtained in a natural way through consideration of the Maslov index associated with appropriately chosen paths of Lagrangian subspaces of ${\mathbb {C}}^{2n}$. We verify that our applicability class includes Dirac and Sturm–Liouville systems, as well as a system arising from differential-algebraic equations for which the spectral parameter appears nonlinearly.

Appendix

In this short appendix, we briefly discuss the view of our operator

$$\begin{aligned} {\mathcal {L}} (\lambda ) := J \frac{d}{dx} - {\mathbb {B}} (\cdot ; \lambda ) \end{aligned}$$

(5.14)

as an operator pencil. In order to keep the discussion brief, we focus on the case of boundary conditions (BC1), for which the domain of ${\mathcal {L}} (\lambda )$ can be taken to be

$$\begin{aligned} \begin{aligned} {\mathcal {D}} ({\mathcal {L}} (\lambda ))&:= \{y \in L^2 ((0,1), {\mathbb {C}}^{2n}): y \in {\text {AC}}\,([0,1], {\mathbb {C}}^{2n}), \\&\quad \quad {\mathcal {L}} y \in L^2 ((0,1), {\mathbb {C}}^{2n}), \, \alpha y(0) = 0, \, \beta y(1) = 0 \}. \end{aligned} \end{aligned}$$

We will confine the discussion in this appendix to the case in which ${\mathbb {B}} (\cdot ; \lambda ) \in L^2 ((0,1), {\mathbb {C}}^{2n \times 2n})$ for all $\lambda \in I$. Under this additional assumption, ${\mathcal {D}} ({\mathcal {L}} (\lambda ))$ is independent of $\lambda$, and in order to emphasize this independence we will express ${\mathcal {D}} ({\mathcal {L}} (\lambda ))$ as ${\mathcal {D}}$. Here, $I \subset {\mathbb {R}}$ continues to be the interval specified in the introduction containing all values $\lambda$ under consideration.

Following the development of [6], we specify the resolvent set of ${\mathcal {L}}$ as

$$\begin{aligned} \rho ({\mathcal {L}}) := \{\lambda \in I: {\mathcal {L}} (\lambda )^{-1} \in {\mathcal {B}} (L^2 ((0,1), {\mathbb {C}}^{2n}))\}, \end{aligned}$$

(5.15)

where ${\mathcal {B}} (L^2 ((0,1), {\mathbb {C}}^{2n}))$ denotes the linear space of all bounded linear operators mapping $L^2 ((0,1), {\mathbb {C}}^{2n})$ to itself, and we specify the spectrum of ${\mathcal {L}}$ as $\sigma ({\mathcal {L}}) = I \backslash \rho ({\mathcal {L}})$. More generally, operator pencils are often defined on open sets of the complex plane $\Omega \subset {\mathbb {C}}$, but such a specification is not necessary for this brief discussion. In order to be precise about terminology, we define what we mean by the essential spectrum and the point spectrum (adapted from [20]). For this, we assume, as in the current setting, that ${\mathcal {D}} := {\text {dom}}\,({\mathcal {L}} (\lambda ))$ is independent of $\lambda$, and we denote by ${\mathbb {L}} (L^2 ((0,1),{\mathbb {C}}^{2n}))$ the space of all closed linear operators mapping ${\mathcal {D}} \subset L^2 ((0,1),{\mathbb {C}}^{2n})$ to $L^2 ((0,1),{\mathbb {C}}^{2n})$.

Definition 5.1

We define the essential spectrum $\sigma _{{\text {ess}}\,} ({\mathcal {L}})$ of an operator pencil ${\mathcal {L}}: I \rightarrow {\mathbb {L}} (L^2 ((0,1),{\mathbb {C}}^{2n}))$ as the set of $\lambda \in I$ for which either ${\mathcal {L}} (\lambda )$ is not Fredholm or ${\mathcal {L}} (\lambda )$ is Fredholm with Fredholm index ${\text {ind}}\,({\mathcal {L}} (\lambda )) \ne 0$. We define the point spectrum $\sigma _{{\text {pt}}\,} ({\mathcal {L}})$ as the set of $\lambda \in I$ so that ${\text {ind}}\,({\mathcal {L}} (\lambda )) = 0$, but ${\mathcal {L}} (\lambda )$ is not invertible.

With these definitions, we see that the sets $\rho ({\mathcal {L}})$, $\sigma _{{\text {ess}}\,} ({\mathcal {L}})$, and $\sigma _{{\text {pt}}\,} ({\mathcal {L}})$ are mutually exclusive, and

$$\begin{aligned} I = \rho ({\mathcal {L}}) \cup \sigma _{{\text {ess}}\,} ({\mathcal {L}}) \cup \sigma _{{\text {pt}}\,} ({\mathcal {L}}); \quad \sigma ({\mathcal {L}}) = \sigma _{{\text {ess}}\,} ({\mathcal {L}}) \cup \sigma _{{\text {pt}}\,} ({\mathcal {L}}). \end{aligned}$$

Another way to view the definitions is as follows. A value $\lambda _0 \in I$ is categorized as an element of $\rho ({\mathcal {L}})$, $\sigma _{{\text {ess}}\,} ({\mathcal {L}})$, or $\sigma _{{\text {pt}}\,} ({\mathcal {L}})$ according precisely to the categorization of 0 relative to the operator ${\mathcal {L}} (\lambda _0)$.

Returning to our particular operator pencil from (5.14), it’s a straightforward application of the methods of [32] to verify that under our Assumptions (A), we have the following: for each $\lambda \in I$, ${\mathcal {L}} (\lambda )$ is Fredholm with index zero, and indeed is self-adjoint. We can conclude that $\sigma ({\mathcal {L}})$ is comprised entirely of point spectrum, and in particular that for each $\lambda \in \sigma ({\mathcal {L}})$ there exist a finite number of linearly independent eigenfunctions $\{y_i (x; \lambda )\}_{i=1}^m \subset {\mathcal {D}}$ so that ${\mathcal {L}} (\lambda ) y_i (\cdot ; \lambda ) = 0$ for all $i \in \{1, 2, \dots , m\}$. In addition, our Assumption (B1) ensures that the eigenvalues are all discrete (i.e., isolated). Our Theorems 1.1 and 1.2 count the number of such discrete eigenvalues, including geometric multiplicity, and it’s natural to consider how this relates to the same count using algebraic multiplicity. First, proceeding as in [19], we can define the algebraic multiplicity of an eigenvalue $\lambda _0$ of ${\mathcal {L}}$ in terms of the nature of the Jordan chains associated with it. Readers interested in a complete definition along these lines can find it in Definition 6 of [19], but for our purposes, we only require the following.

Definition 5.2

Let $\lambda _0 \in I$ be an eigenvalue of an operator pencil ${\mathcal {L}}: I \rightarrow {\mathbb {L}} (L^2 ((0,1),{\mathbb {C}}^{2n}))$ with geometric multiplicity m, and assume ${\mathcal {L}}'(\lambda _0) \in {\mathbb {L}} (L^2 ((0,1),{\mathbb {C}}^{2n}))$ exists, with additionally ${\text {dom}}\,({\mathcal {L}}'(\lambda _0)) = {\mathcal {D}}$. Suppose that for any pair $(y, \zeta )$ with $y \in \ker {\mathcal {L}} (\lambda _0)$, and $\zeta \in {\text {dom}}\,({\mathcal {L}} (\lambda _0))$ satisfying

$$\begin{aligned} {\mathcal {L}} (\lambda _0) \zeta = {\mathcal {L}}' (\lambda _0) y, \end{aligned}$$

(5.16)

we must have $y \equiv 0$. Then $\lambda _0$ has algebraic multiplicity m.

We are now in a position to verify that under slightly stronger conditions on ${\mathbb {B}} (x; \lambda )$ than assumed for Theorems 1.1 and 1.2, the geometric and algebraic multiplicies of eigenvalues of the operator pencil ${\mathcal {L}}: I \rightarrow {\mathbb {L}} (L^2 ((0,1),{\mathbb {C}}^{2n}))$ coincide.

Claim 5.1

Let Assumptions (A) and (B1) hold, and additionally assume that for all $\lambda \in I$, we have ${\mathbb {B}} (\cdot ; \lambda ), {\mathbb {B}}_{\lambda } (\cdot ; \lambda ) \in L^2 ((0,1),{\mathbb {C}}^{2n \times 2n})$. Then for any eigenvalue $\lambda _0$ of the operator pencil ${\mathcal {L}}$, geometric and algebraic multiplicities agree.

Proof

In our setting, ${\mathcal {L}}' (\lambda ) = {\mathbb {B}}_{\lambda } (x; \lambda )$. Suppose $\lambda _0$ is an eigenvalue of ${\mathcal {L}}$, and that for some $y (\cdot ; \lambda _0) \in \ker {\mathcal {L}} (\lambda _0)$, there is a corresponding $\zeta (\cdot ; \lambda _0) \in {\mathcal {D}} ({\mathcal {L}})$ so that

$$\begin{aligned} {\mathcal {L}} (\lambda _0) \zeta (x; \lambda _0) = {\mathbb {B}}_{\lambda } (x; \lambda _0) y (x; \lambda _0), \quad \text {a.e. } x \in (0,1). \end{aligned}$$

(Our additional assumption ${\mathbb {B}}_{\lambda } (\cdot ; \lambda ) \in L^2 ((0,1),{\mathbb {C}}^{2n \times 2n})$ ensures that ${\mathcal {L}}' (\lambda _0)$ maps ${\mathcal {D}}$ to $L^2 ((0,1),{\mathbb {C}}^{2n})$, and in particular that ${\mathbb {B}}_{\lambda } (x; \lambda _0) y (x; \lambda _0)$ is in the range of ${\mathcal {L}} (\lambda _0)$.) If we take an $L^2$ inner product of this equation with y, we obtain the relation

$$\begin{aligned} \langle {\mathcal {L}} (\lambda _0) \zeta , y \rangle = \langle {\mathbb {B}}_{\lambda } (x; \lambda _0) y, y\rangle . \end{aligned}$$

Since ${\mathcal {L}} (\lambda _0)$ is self-adjoint, the left-hand side can be computed as

$$\begin{aligned} \langle {\mathcal {L}} (\lambda _0) \zeta , y \rangle = \langle \zeta , {\mathcal {L}} (\lambda _0) y \rangle = 0. \end{aligned}$$

We see that the right-hand side satisfies $\langle {\mathbb {B}}_{\lambda } (\cdot ; \lambda _0) y, y \rangle = 0$, and by our positivity condition (B1) this means $y = 0$ for a.e. $x \in (0,1)$. According to Definition 5.2, we can conclude that the algebraic multiplicity of $\lambda _0$ agrees with the geometric multiplicity of $\lambda _0$. $\square$