On the $j$-th Eigenvalue of Sturm-Liouville Problem and the Maslov Index

TL;DR

This paper establishes a precise relationship between the jump phenomena of the $j$-th eigenvalue in Sturm-Liouville problems and the Maslov index, providing new insights into eigenvalue behavior and boundary condition limits.

Contribution

It demonstrates that the jump number of eigenvalue branches equals the Maslov index and determines the eigenvalue range for boundary condition layers.

Findings

Jump number equals Maslov index for boundary conditions

Sharp eigenvalue range on boundary condition layers

Monodromy matrix approaches Dirichlet condition as spectral parameter decreases

Abstract

In the previous papers \cite{HLWZ,KWZ}, the jump phenomena of the $j$-th eigenvalue were completely characterized for Sturm-Liouville problems. In this paper, we show that the jump number of these eigenvalue branches is exactly the Maslov index for the path of corresponding boundary conditions. Then we determine the sharp range of the $j$-th eigenvalue on each layer of the space of boundary conditions. Finally, we prove that the graph of monodromy matrix tends to the Dirichlet boundary condition as the spectral parameter goes to $-\infty$.

Equations

$$-\frac{d}{dt}\left(P(t)\frac{d}{dt}x(t)+Q(t)x(t)\right)+Q(t{)}^T\frac{d}{dt}x(t)+R(t)x(t)=\lambda D(t)x(t),$$

$$(\mathcal{V},\Omega )=({\mathbf{C}}^{2n}\oplus {\mathbf{C}}^{2n},-{\omega }_n\oplus {\omega }_n),$$

$$\left[\begin{array}{c}z(0)\\ z(T)\end{array}\right]\in {\Lambda }_0,$$

$${\Lambda }_N=\left\{\left[\begin{array}{c}z(0)\\ z(T)\end{array}\right]\in {\mathbf{C}}^{4n}:y(0)=y(T)=0\right\},{\Lambda }_D=\left\{\left[\begin{array}{c}z(0)\\ z(T)\end{array}\right]\in {\mathbf{C}}^{4n}:x(0)=x(T)=0\right\},$$

$$\mathcal{A}:=-\frac{d}{dt}\left(P\frac{d}{dt}+Q\right)+Q^T\frac{d}{dt}+R.$$

$$E_{{\Lambda }_0}(0,T):=\left\{x\in H^2([0,T],{\mathbf{C}}^n):\left[\begin{array}{c}z(0)\\ z(T)\end{array}\right]\in {\Lambda }_0\right\}.$$

$${\mathcal{A}}_{{\Lambda }_0}x=\lambda Dx,$$

$$D^{-\frac{1}{2}}{\mathcal{A}}_{{\Lambda }_0}{D}^{-\frac{1}{2}}y=\lambda y,$$

$${\lambda }_1({\Lambda }_0)\le \cdots \le {\lambda }_j({\Lambda }_0)\le \cdots$$

$$\underset{s\to 0^{\pm }}{\lim }\dim ({\Lambda }_s\cap {\Lambda }_D)\ne \dim ({\Lambda }_0\cap {\Lambda }_D),$$

$$\dim {\Lambda }_s\cap {\Lambda }_D=c_{-}\text{for}s\in [-ϵ,0),\dim {\Lambda }_0\cap {\Lambda }_D=c_0,\dim {\Lambda }_s\cap {\Lambda }_D=c_{+}\text{for}s\in (0,+ϵ],$$

$$\mu ({\Lambda }_D,{\Lambda }_s,s\in [-ϵ,0])=-k_{-},\mu ({\Lambda }_D,{\Lambda }_s,s\in [0,+ϵ])=k_{+},$$

$$\underset{s\to 0^{-}}{\lim }{\lambda }_j({\Lambda }_s)={\lambda }_{j-k_{-}}({\Lambda }_0),\underset{s\to 0^{+}}{\lim }{\lambda }_j({\Lambda }_s)={\lambda }_{j-k_{+}}({\Lambda }_0).$$

$${\Sigma }_r=\{\Lambda \in \text{Lag}(\mathcal{V},\Omega ):\dim \left(\Lambda \cap {\Lambda }_D\right)=r\},$$

$${\lambda }_{j-(2n-r)-b_1}({\Lambda }_D)=\cdots ={\lambda }_{j-(2n-r)}({\Lambda }_D)=\cdots ={\lambda }_{j-(2n-r)+b_2}({\Lambda }_D)$$

$${\lambda }_{j-c_1}({\Lambda }_D)=\cdots ={\lambda }_j({\Lambda }_D)=\cdots ={\lambda }_{j+c_2}({\Lambda }_D)$$

$${\lambda }_j({\Sigma }_r)=(-\infty ,{\lambda }_j({\Lambda }_D)).$$

$${\lambda }_j({\Sigma }_r)=(-\infty ,{\lambda }_j({\Lambda }_D)].$$

$${\lambda }_j({\Sigma }_r)=({\lambda }_{j-(2n-r)}({\Lambda }_D),{\lambda }_j({\Lambda }_D)).$$

$${\lambda }_j({\Sigma }_r)=({\lambda }_{j-(2n-r)}({\Lambda }_D),{\lambda }_j({\Lambda }_D)].$$

$${\lambda }_j({\Sigma }_r)=[{\lambda }_{j-(2n-r)}({\Lambda }_D),{\lambda }_j({\Lambda }_D)).$$

$${\lambda }_j({\Sigma }_r)=[{\lambda }_{j-(2n-r)}({\Lambda }_D),{\lambda }_j({\Lambda }_D)].$$

$$\dot{z}(t)=J_n{\mathcal{B}}_{\lambda }(t)z(t),$$

$${\mathcal{B}}_{\lambda }(t)=\left[\begin{array}{cc}{P}^{-1}(t)& -P^{-1}(t)Q(t)\\ -Q^T(t)P^{-1}(t)& Q^T(t)P^{-1}(t)Q(t)-R(t)+\lambda I_n\end{array}\right].$$

$${\dot{\gamma }}_{\lambda }(t)=J_n{\mathcal{B}}_{\lambda }(t){\gamma }_{\lambda }(t),{\gamma }_{\lambda }(0)=I_{2n}.$$

$${\gamma }_{\lambda }(t)\in \mathrm{Sp}(2n):=\{M\in GL({\mathbf{R}}^{2n}):M^T{J}_{n}M=J_n\}.$$

$$Gr({\gamma }_{\lambda }(T)):=\{(x,{\gamma }_{\lambda }(T)x),x\in {\mathbf{C}}^{2n}\}$$

$$\underset{\lambda \to -\infty }{\lim }Gr({\gamma }_{\lambda }(T))={\Lambda }_D.$$

$$\mu ({\Lambda }_D,Gr({\gamma }_{\lambda (s)}(T)),s\in [-\pi /2,-\pi /2+ϵ])=2n$$

$$\mathcal{Z}v=\left[\begin{array}{c}(X-iY)v/2\\ (Y+iX)v/2\end{array}\right]+\left[\begin{array}{c}(X+iY)v/2\\ (Y-iX)v/2\end{array}\right]=P_{+}((X-iY)v/2)+P_{-}((X+iY)v/2).$$

Full text

AMS Subject Classification: 34B24, 34L15, 53D12, 34B09.

Key Words. Sturm-Liouville problem, eigenvalue, boundary condition, Maslov index, Lagrangian subspace.

1 Introduction

In this paper, we consider the Sturm-Liouville problem

[TABLE]

where $P,Q\in H^{1}([0,T],L(n))$, $R,D\in C([0,T],L(n))$, $P(t),D(t)$ are positive definite, and $P(t),R(t),D(t)$ are symmetric for all $t\in[0,T]$. Here $L(n)$ is the set of $n\times n$ real-valued matrices and $Q(t)^{T}$ is the transpose of $Q(t)$. We describe a self-adjoint boundary condition of (1.1) by a Lagrangian subspace of $\mathbf{C}^{2n}\oplus\mathbf{C}^{2n}$ as follows. Consider $(\mathbf{C}^{2n},\omega_{n})$ as a complex symplectic vector space with the symplectic form $\omega_{n}(x,y)=\langle J_{n}x,y\rangle$ for any $x,y\in\mathbf{C}^{2n},$ where $\langle\cdot,\cdot\rangle$ is the standard Hermitian inner product in $\mathbf{C}^{2n}$ and $J_{n}=\begin{bmatrix}0&-I_{n}\\ I_{n}&0\end{bmatrix}$. Denote

[TABLE]

which is a $4n$-dimensional symplectic space. A subspace $\Lambda\subset\mathcal{V}$ is called Lagrangian if $\Omega\mid_{\Lambda}=0$ and \hbox{\rm dim,}_{\mathbf{C}}\Lambda=\frac{1}{2}\hbox{\rm dim,}_{\mathbf{C}}\mathcal{V}=2n. Denote the set of Lagrangian subspaces by $\text{Lag}(\mathcal{V},\Omega)$. Then $\text{Lag}(\mathcal{V},\Omega)$ is a compact metric space [2]. Let $\dot{x}=\frac{d}{dt}x$, $y(t)=P(t)\dot{x}(t)+Q(t)x(t),$ and $z(t)=(y(t)^{T},x(t)^{T})^{T}$. Then any self-adjoint boundary condition can be written as

[TABLE]

where $\Lambda_{0}\in\text{Lag}(\mathcal{V},\Omega)$. In particular, the Neumann and Dirichlet boundary conditions are given by

[TABLE]

respectively. Moreover, $\text{Lag}(\mathcal{V},\Omega)$ is exactly the space of self-adjoint boundary conditions.

The formal differential operator corresponding to (1.1) is

[TABLE]

Define an operator $\mathcal{A}_{\Lambda_{0}}x:=\mathcal{A}x$ on $L^{2}([0,T],\mathbf{C}^{n})$ with the domain

[TABLE]

Then $\mathcal{A}_{\Lambda_{0}}$ is a self-adjoint operator. Note that (1.1)–(1.2) are equivalent to

[TABLE]

which can be written as

[TABLE]

where $y=D^{\frac{1}{2}}x$. Therefore, without loss of generality, we always assume that $D=I_{n}$.

The spectrum of $\mathcal{A}_{\Lambda_{0}}$ is bounded from below and consists of discrete eigenvalues, which are listed as follows:

[TABLE]

counting multiplicities, with $\lambda_{j}(\Lambda_{0})\to\infty$ as $j\to\infty$. Thus the $j$-th eigenvalue $\lambda_{j}$ can be regarded as a function $\lambda_{j}:\text{Lag}(\mathcal{V},\Omega)\to\mathbf{R}$ in the sequel. $\lambda_{j}$ is not always continuously dependent on $\Lambda_{0}$. Recently, Kong, Wu and Zettl completely characterized the discontinuity of $\lambda_{j}$ for $1$-dimensional case in [8], while we characterized it for $n$-dimensional case with $n\geq 2$ in [5]. In fact, discontinuity may occur only at such boundary condition $\Lambda_{0}$ that

[TABLE]

where $\Lambda_{s}$, $s\in[-\epsilon,+\epsilon]$, is a continuous path in $\text{Lag}(\mathcal{V},\Omega)$. Near such boundary condition $\Lambda_{0}$, the $j$-th eigenvalue always jumps in certain directions. In this paper, as a continuous work of [5, 8], we use the Maslov index to count the jump number of $\lambda_{j}$. For readers’ convenience, we give a brief introduction of the Maslov index in Section 2.

To describe the discontinuity in the framework of Lagrangian subspaces, we consider a continuous path $\Lambda_{s}\in\text{Lag}(\mathcal{V},\Omega)$, $s\in[-\epsilon,+\epsilon]$, with the isolated singularity at $s=0$. More precisely,

[TABLE]

and

[TABLE]

where $\mu(\cdot,\cdot,\cdot)$ is the Maslov index, see Definition 2.1.

For convenience, we set $\lambda_{j}=-\infty$ for $j\leq 0$. Our first main result is stated as follows.

Theorem 1.1**.**

Let $\Lambda_{s}\in\text{Lag}(\mathcal{V},\Omega)$, $s\in[-\epsilon,\epsilon]$, be a continuous path, and satisfy (1.3)–(1.4). Then $\lambda_{j}(\Lambda_{\cdot})$ is continuous on $s\in[-\epsilon,0)\cup(0,+\epsilon]$ and

[TABLE]

By Corollary 2.4, $k_{\pm}$ are non-negative. Our new contribution is that the jump number $n^{+}-n_{0}^{+}$ in Theorem 7.1 of [5] is exactly the Maslov index $k_{\pm}$ in Theorem 1.1. The idea of the proof of Theorem 1.1 is to express the $j$-th eigenvalue by the index form. We refer the readers to [6] for the introduction of index form. Then we study the monotone property of index form and give proper estimates for the eigenvalues.

The range of the $j$-th eigenvalue $\lambda_{j}$ on the whole space of boundary conditions was given in Theorem 4.1 of [8] for $1$-dimensional Sturm-Liouville problems. To the best of our knowledge, there are no results for high dimensional case. Define the $r$-th layer on $\text{Lag}(\mathcal{V},\Omega)$ to be

[TABLE]

where $0\leq r\leq 2n$. Our second result is to determine the sharp range of $\lambda_{j}$ on each layer of $\text{Lag}(\mathcal{V},\Omega)$ for $n$-dimensional Sturm-Liouville problems.

Theorem 1.2**.**

Fix any $j\geq 1$ and $0\leq r\leq 2n$. Let

[TABLE]

with multiplicity to be $b_{1}+b_{2}+1$, and

[TABLE]

with multiplicity to be $c_{1}+c_{2}+1$, where $b_{i},c_{i}\geq 0$, $i=1,2$. If $j\leq 2n-r$, then we have two cases.

Case 1: $r\leq c_{2}$.

[TABLE]

Case 2: $r>c_{2}$.

[TABLE]

If $j>2n-r$, then we have four cases.

Case 1: $r\leq\min\{b_{1},c_{2}\}.$

[TABLE]

Case 2: $c_{2}<r\leq b_{1}.$

[TABLE]

Case 3: $b_{1}<r\leq c_{2}$.

[TABLE]

Case 4: $r>\max\{b_{1},c_{2}\}$.

[TABLE]

Theorem 1.2 indicates that the “left-multiplicity” $b_{1}$ of $\lambda_{j-(2n-r)}(\Lambda_{D})$, the “right-multiplicity” $c_{2}$ of $\lambda_{j}(\Lambda_{D})$ and the layer’s number $r$ determine whether the endpoints $\lambda_{j-(2n-r)}(\Lambda_{D}),\lambda_{j}(\Lambda_{D})\in\lambda_{j}(\Sigma_{r})$ or not. Then the range of $\lambda_{j}$ on the whole space $\text{Lag}(\mathcal{V},\Omega)$ and on the [math]-th layer $\Sigma_{0}$ is a direct consequence.

Corollary 1.3**.**

For any $j\geq 1$, we have

(1) $\lambda_{j}(\text{Lag}(\mathcal{V},\Omega))=(\lambda_{j-2n}(\Lambda_{D}),\lambda_{j}(\Lambda_{D})]$;

(2) $\lambda_{j}(\Sigma_{0})=(\lambda_{j-2n}(\Lambda_{D}),\lambda_{j}(\Lambda_{D}))$.

Corollary 1.3 (1) generalizes Theorem 4.1 in [8] for $1$-dimensional result to any dimension. As an example of Theorem 1.2, we provide the sharp range of $\lambda_{j}$ on each layer for $1$-dimensional case, which is more accurate than the conclusions in [8]:

Corollary 1.4**.**

For any given $1$-dimensional Sturm-Liouville equation, we have for any $j\geq 1$,

(1) $\lambda_{j}(\Sigma_{0})=(\lambda_{j-2}(\Lambda_{D}),$ $\lambda_{j}(\Lambda_{D}))$;

(2) $\lambda_{1}(\Sigma_{1})=(-\infty,\lambda_{1}(\Lambda_{D})]$, and $\lambda_{j}(\Sigma_{1})=[\lambda_{j-1}(\Lambda_{D}),\lambda_{j}(\Lambda_{D})]$ for $j\geq 2$;

(3) $\lambda_{j}(\Sigma_{2})=\{\lambda_{j}(\Lambda_{D})\}$.

Since the multiplicity of an eigenvalue of $\mathcal{A}_{\Lambda_{D}}$ is at most $n$, we get the following result.

Corollary 1.5**.**

Let $n\leq r\leq 2n$. Then $\lambda_{j}(\Sigma_{r})=(-\infty,\lambda_{j}(\Lambda_{D})]$ for any $1\leq j\leq 2n-r$, and $\lambda_{j}(\Sigma_{r})=[\lambda_{j-(2n-r)}(\Lambda_{D}),$ $\lambda_{j}(\Lambda_{D})]$ for any $j>2n-r$.

By Theorem 1.2, we also get the following interesting fact.

Corollary 1.6**.**

Let $\lambda_{{j_{0}}-(r_{0}-1)}({\Lambda_{D}})=\cdots=\lambda_{j_{0}}({\Lambda_{D}})$ with multiplicity to be $r_{0}$, where $1\leq r_{0}\leq n$ and $j_{0}\geq r_{0}$. Then $\lambda_{j_{0}}(\Lambda)\equiv\lambda_{j_{0}}(\Lambda_{D})$ for all $\Lambda\in\Sigma_{r}$, where $2n-r_{0}+1\leq r\leq 2n$.

By the standard Legendre transformation, equation (1.1) with $D=I_{n}$ becomes

[TABLE]

where

[TABLE]

Let $\gamma_{\lambda}(t)$, $t\in[0,T]$, be the fundamental solution of (1.6), that is, for $\lambda\in\mathbf{C}$,

[TABLE]

It is well-known that for any $t\in[0,T]$,

[TABLE]

Since $\gamma_{\lambda}(T)\in{\mathrm{Sp}}(2n)$, it is obvious that $Gr(\gamma_{\lambda}(T))\in\text{Lag}(\mathcal{V},\Omega)$, where

[TABLE]

is the graph of $\gamma_{\lambda}(T)$.

Theorem 1.7**.**

Under the above notation, we have

[TABLE]

Furthermore, let $\lambda(s)=\tan(s)$ for $s\in[-\pi/2,+\pi/2]$, then

[TABLE]

for $\epsilon>0$ small enough.

From (1.8), we define $Gr({\gamma}_{-\infty}(T))=Gr({\gamma}_{\lambda(-\pi/2)}(T))=\Lambda_{D}$ and thus $Gr({\gamma}_{\lambda(s)}(T)),s\in[-\pi/2,-\pi/2+\epsilon]$ is a continuous path in $\text{Lag}(\mathcal{V},\Omega)$. (1.9) implies that $Gr(\gamma_{\lambda(s)}(T)),s\in[-\pi/2,-\pi/2+\epsilon]$ can be considered as a positive path in the sense that all the eigenvalues of the corresponding path of unitary matrices are rotated counterclockwise. Here we use the definition of Maslov index in Remark 2.2.

The rest of this paper is organized as follows. In Section $2$, we briefly review the Maslov index theory for Lagrangian subspaces. In Section 3, we give the proof of Theorem 1.1. Theorem 1.2 is shown in Section 4. In Section 5, we provide the proof of Theorem 1.7.

2 Maslov Index for the Lagrangian subspaces

In this section, we briefly introduce the general Maslov index theory for the Lagrangian subspaces. Then we apply it to our framework for Sturm-Liouville problems. For this theory, we refer the readers to [1, 2, 3, 9] and references therein.

The Lagrangian frame of a given $\Lambda\in\text{Lag}(\mathbf{C}^{2m},\omega_{m})$ is defined by an injective linear map $\mathcal{Z}:\mathbf{C}^{m}\rightarrow\Lambda$ with the form $\mathcal{Z}=\begin{bmatrix}X\\ Y\end{bmatrix}$, where $X$ and $Y$ are $m\times m$ complex matrices such that $X^{*}Y=Y^{*}X$ and $rank(\mathcal{Z})=m$. Here $X^{*}$ is the conjugate transpose of $X$. The Lagrangian subspace is represented by a corresponding Lagrangian frame in the sequel. Clearly, $P_{+}=\begin{bmatrix}I_{m}\\ iI_{m}\end{bmatrix}$ is a bijection from $\mathbf{C}^{m}$ to $\Lambda^{+}=\ker(iJ_{m}-I_{2m})$ and $P_{-}=\begin{bmatrix}I_{m}\\ -iI_{m}\end{bmatrix}$ is a bijection from $\mathbf{C}^{m}$ to $\Lambda^{-}=\ker(iJ_{m}+I_{2m})$. For any $v\in\mathbf{C}^{m}$, we decompose ${\mathcal{Z}}v$ to be

[TABLE]

Then we get a unitary operator

[TABLE]

from $\Lambda^{+}$ to $\Lambda^{-}$. Correspondingly, $P_{-}^{-1}WP_{+}=(X+iY)(X-iY)^{-1}$ is an $m\times m$ unitary matrix. So we define a map $\mathcal{U}:\text{Lag}(\mathbf{C}^{2m},\omega_{m})\rightarrow\mathbf{U}(m)$ as follows:

[TABLE]

Note that $\mathcal{U}$ is a homeomorphic (isomorphic) map [6] and $\mathcal{U}(\Lambda)$ is independent of the choice of frame. Let $\Lambda_{k}\in\text{Lag}(\mathbf{C}^{2m},\omega_{m})$ with Lagrangian frame to be $\begin{bmatrix}X_{k}\\ Y_{k}\end{bmatrix}$, and $W_{k}=P_{-}(X_{k}+iY_{k})(X_{k}-iY_{k})^{-1}P_{+}^{-1}$, $k=1,2$. Then it follows that $\mathcal{U}(\Lambda_{2})^{-1}\mathcal{U}(\Lambda_{1})=P_{+}^{-1}W_{2}^{-1}W_{1}P_{+}$ and the spectrum of $\mathcal{U}(\Lambda_{2})^{-1}\mathcal{U}(\Lambda_{1})$ is the same with that of $W_{2}^{-1}W_{1}$. The metric of Lagrangian subspaces is defined as the metric of corresponding unitary matrices on $\mathbf{U}(m)$, that is,

[TABLE]

where $\|\cdot\|$ is the operator norm. Moreover, \hbox{\rm dim,}\left(\Lambda_{1}\cap\Lambda_{2}\right)=\hbox{\rm dim,}\left(\ker(\mathcal{U}(\Lambda_{2})^{-1}\mathcal{U}(\Lambda_{1})-I_{m})\right). For any fixed $U_{0}\in\mathbf{U}(m)$, the singular cycle $\Sigma_{U_{0}}$ of $U_{0}$ is defined by

[TABLE]

Now we introduce the Maslov index. Consider a continuous path $U_{t},t\in[a,b]$, in $\mathbf{U}(m)$ and the small perturbation $e^{is}U_{t}$, $s\in[-\varepsilon,\varepsilon]$. For any fixed $t_{0}\in[a,b]$, the path $e^{is}U_{t_{0}}$, $s\in[-\varepsilon,\varepsilon]$, is transversal to $\Sigma_{U_{0}}$. Furthermore, $e^{-is_{0}}U_{a},e^{-is_{0}}U_{b}\notin\Sigma_{U_{0}}$ for any $s_{0}>0$ sufficiently small. Thus the intersection number $[e^{-is_{0}}U_{t}:\Sigma_{U_{0}}]$ can be well-defined. Then we give the concept of Maslov index:

Definition 2.1**.**

Let $\Lambda(t),t\in[a,b]$, be a continuous path in $\text{Lag}(\mathbf{C}^{2m},\omega_{m})$ and $\Lambda_{0}\in\text{Lag}(\mathbf{C}^{2m},\omega_{m})$. Then the Maslov index is defined by

[TABLE]

where $s_{0}>0$ is sufficiently small.

Remark 2.2**.**

Definition 2.1 is equivalent to the definition of Maslov index in Section 2.2 of [2], which is defined as follows. There are $m$ continuous functions $\theta_{j}\in C([a,b],\mathbf{R})$ such that $e^{i\theta_{j}(t)},1\leq j\leq m,$ are all the eigenvalues of $\mathcal{U}(\Lambda_{0})^{-1}\mathcal{U}(\Lambda(t))$ (counting algebraic multiplicities). Denote by $[a]$ the integer part of $a\in\mathbf{R}$. Define $E(a)=-[-a]$. Then the Maslov index can be defined as

[TABLE]

Then we provide some properties of Maslov index and we refer to [2] for the details.

Property I (Reparametrization invariance) Let $\phi:[c,d]\rightarrow[a,b]$ be a continuous and piecewise smooth function with $\phi(c)=a$, $\phi(d)=b$. Then

[TABLE]

Property II (Homotopy invariant with endpoints) For two continuous families of Lagrangian paths $\Lambda_{1}(s,t)$, $\Lambda_{2}(s,t)$, $0\leq s\leq 1$, $a\leq t\leq b$, such that both \hbox{\rm dim,}\left(\Lambda_{1}(s,a)\cap\Lambda_{2}(s,a)\right) and \hbox{\rm dim,}\left(\Lambda_{1}(s,b)\cap\Lambda_{2}(s,b)\right) are constants, we have

[TABLE]

Property III (Path additivity) If $a<c<b$, then

[TABLE]

Property IV (Symplectic invariance) Let ${\gamma}(t)$, $t\in[a,b],$ be a continuous path in ${\mathrm{Sp}}(2m)$. Then

[TABLE]

Property V (Symplectic additivity) Let $W_{i}$, $i=1,2$, be symplectic spaces, $\Lambda_{1}(\cdot),\Lambda_{2}(\cdot)\in C([a,b],$ $\text{Lag}(W_{1}))$ and ${\Lambda}_{3}(\cdot),{\Lambda}_{4}(\cdot)\in C([a,b],\text{Lag}(W_{2}))$. Then we have

[TABLE]

Now we turn back to the framework for boundary conditions of Sturm-Liouville problems. Firstly, we change the basis of $(\mathcal{V},\Omega)$ such that the symplectic structure becomes the standard form $\omega_{2n}$. Recall that $\Omega=-\omega_{n}\oplus\omega_{n}$ corresponds to the matrix

[TABLE]

Direct computation implies

[TABLE]

where

[TABLE]

Under the new basis $\omega_{2n}$, the boundary condition (1.2) becomes

[TABLE]

Next, we provide a Lagrangian frame of any given $\Lambda_{0}\in\text{Lag}(\mathcal{V},\omega_{2n})$. Let $V(\Lambda_{0})$ be the subspace of $\Lambda_{N}$ defined by

[TABLE]

Then $(x(0)^{T},x(T)^{T})^{T}\in V(\Lambda_{0})$. Thanks to the splitting $\mathbf{C}^{2n}\cong\Lambda_{N}=V(\Lambda_{0})\oplus V(\Lambda_{0})^{\perp}$, we get $(-y(0)^{T},y(T)^{T})^{T}$ $=(-y_{1}(0)^{T},y_{1}(T)^{T})^{T}+(-y_{2}(0)^{T},y_{2}(T)^{T})^{T}$, where $(-y_{1}(0)^{T},y_{1}(T)^{T})^{T}\in J_{2n}V(\Lambda_{0})^{\perp}$ and $(-y_{2}(0)^{T},y_{2}(T)^{T})^{T}\in J_{2n}V(\Lambda_{0})$. Then we have a linear map $A$ from $V(\Lambda_{0})$ to $J_{2n}V(\Lambda_{0})$ such that $(-y_{2}(0)^{T},y_{2}(T)^{T})^{T}=A(x(0)^{T},$ $x(T)^{T})^{T}$. $A$ is Hermitian since $\Lambda_{0}$ is Lagrangian. By assuming \hbox{\rm dim,}V(\Lambda_{0})=k_{0}, we can choose a suitable basis of $\mathcal{V}$ such that

[TABLE]

is a Lagrangian frame of $\Lambda_{0}$ under the symplectic form $J_{2n}$. The left column corresponds to $\Lambda_{0}\cap\Lambda_{D}$ and the right column corresponds to $\left\{\begin{bmatrix}Au\\ u\end{bmatrix}\ :\ u\in V(\Lambda_{0})\right\}$. In addition, $\begin{bmatrix}-I_{n}&0\\ D_{1}&D_{2}\\ 0&I_{n}\\ D_{3}&D_{4}\end{bmatrix}$ is a frame of $Gr(\mathcal{D})$ for $\mathcal{D}=\begin{bmatrix}D_{1}&D_{2}\\ D_{3}&D_{4}\end{bmatrix}\in{\mathrm{Sp}}(2n)$. By (2.1) we have

[TABLE]

where $U_{A}=(A+iI_{k_{0}})(A-iI_{k_{0}})^{-1}$. Especially,

[TABLE]

Let $\Lambda_{s}$, $s\in[\tau_{0},\tau_{1}],$ be a continuous path with $\tau_{1}-\tau_{0}$ small enough. From the definition of Maslov index, we will show that $\mu(\Lambda_{D},\Lambda_{s},s\in[\tau_{0},\tau_{1}])$ is to count the number of eigenvalues of $\mathcal{U}(\Lambda_{s})$ passing $1$. More precisely, we choose $\theta_{0}\in(0,2\pi)$ such that $e^{i\theta_{0}}\notin\sigma(\mathcal{U}(\Lambda_{s}))$ for any $s\in[\tau_{0},\tau_{1}]$. Denote $\nu^{+}(\Lambda_{s})$ to be the number of total eigenvalues of $\mathcal{U}(\Lambda_{s})$ in the region $\{e^{i\theta},\theta\in(0,\theta_{0})\}$. Then we have the following lemma.

Lemma 2.3**.**

Under the above notation, we have

[TABLE]

Proof.

We use the definition of Maslov index in Remark 2.2. Since $e^{i\theta_{0}}\notin\sigma(\mathcal{U}(\Lambda_{s}))$, we see that there exist $2n$ continuous functions $\theta_{j}\in C([\tau_{0},\tau_{1}],(\theta_{0},\theta_{0}+2\pi))$ such that $e^{i\theta_{j}(s)},1\leq j\leq 2n,$ are the spectrum of $\mathcal{U}(\Lambda_{s})$ and

[TABLE]

Since $\theta_{i}(t)\in(\theta_{0},\theta_{0}+2\pi)$ for all $t\in[\tau_{0},\tau_{1}]$, we obtain

[TABLE]

The lemma then follows. ∎

Corollary 2.4**.**

Under the assumptions (1.3)–(1.4), we have

[TABLE]

Proof.

It suffices to prove $0\leq k_{+}\leq c_{0}-c_{+}$. Note that c_{+}=\hbox{\rm dim,}\left(\Lambda_{s}\cap\Lambda_{D}\right)=\hbox{\rm dim,}\left(\ker(\mathcal{U}(\Lambda_{s})-I_{2n})\right),s\in(0,+\epsilon]. Choose $\theta_{0}>0$ such that $e^{it}\notin\sigma(\mathcal{U}(\Lambda_{0}))$ for all $t\in[-\theta_{0},0)\cup(0,\theta_{0}]$. Then there exists $\epsilon_{0}\in(0,\epsilon)$ such that $e^{\pm i\theta_{0}}\notin\sigma(\mathcal{U}(\Lambda_{s}))$ for all $s\in[0,\epsilon_{0}]$. Therefore, the number of eigenvalues of $\mathcal{U}(\Lambda_{s})$ in the region $\{e^{i\theta},\theta\in(-\theta_{0},\theta_{0})\}$is a constant for all $s\in[0,\epsilon_{0}]$ and it is exactly $c_{0}$. Hence we get $\nu^{+}(\Lambda_{\epsilon_{0}})\leq c_{0}-c_{+}$. By Lemma 2.3 and Property III of Maslov index, we have $k_{+}=\mu(\Lambda_{D},\Lambda_{s},s\in[0,\epsilon_{0}])+\mu(\Lambda_{D},\Lambda_{s},s\in[\epsilon_{0},\epsilon])=\nu^{+}(\Lambda_{\epsilon_{0}})-\nu^{+}(\Lambda_{0})=\nu^{+}(\Lambda_{\epsilon_{0}})$. Since $0\leq\nu^{+}(\Lambda_{\epsilon_{0}})\leq c_{0}-c_{+}$, the conclusion then follows. ∎

We have the following result for small perturbation of $\Lambda_{s_{0}}$.

Lemma 2.5**.**

Let $\alpha\in\text{Lag}(\mathcal{V},\omega_{2n})$ and

[TABLE]

with $A_{s}=A_{0}+\tan(s)I_{2n-r}$. Then for any $s_{0}\in(-{\pi\over 2},{\pi\over 2})$, \hbox{\rm dim,}(\alpha\cap\Lambda_{s})\leq r for $|s-s_{0}|\neq 0$ small enough.

Proof.

Without loss of generality, we assume that $s_{0}=0$. Since $\mathcal{U}(\Lambda_{s})=I_{r}\oplus(A_{s}+iI_{2n-r})(A_{s}-iI_{2n-r})^{-1}$, we have

[TABLE]

Define $B=\frac{d}{ds}\mathcal{U}(\Lambda_{s})|_{s=0}\mathcal{U}(\Lambda_{0})^{-1}$ and $C=\frac{d}{ds}(\mathcal{U}(\Lambda_{s})\mathcal{U}(\alpha)^{-1}-I_{2n})|_{s=0}$. Let $x\in\ker(\mathcal{U}(\Lambda_{0})\mathcal{U}(\alpha)^{-1}-I_{2n})$, then

[TABLE]

It follows that $C|_{\ker(\mathcal{U}(\Lambda_{0})\mathcal{U}(\alpha)^{-1}-I_{2n})}=B|_{\ker(\mathcal{U}(\Lambda_{0})\mathcal{U}(\alpha)^{-1}-I_{2n})}$.

Note that $\mathbf{C}^{2n}=\ker(\mathcal{U}(\Lambda_{0})\mathcal{U}(\alpha)^{-1}-I_{2n})\,\oplus\,\text{Ran}(\mathcal{U}(\Lambda_{0})\mathcal{U}(\alpha)^{-1}-I_{2n})$ and it is an orthogonal decomposition. Let $P_{1}$ and $P_{2}$ be the orthogonal projections from $\mathbf{C}^{2n}$ to $\text{Ran}(\mathcal{U}(\Lambda_{0})\mathcal{U}(\alpha)^{-1}-I_{2n})$ and $\ker(\mathcal{U}(\Lambda_{0})\mathcal{U}(\alpha)^{-1}-I_{2n})$, respectively. Then we have $\mathcal{U}(\Lambda_{s})\mathcal{U}(\alpha)^{-1}-I_{2n}=\begin{bmatrix}A_{11}(s)&A_{12}(s)\\ A_{21}(s)&A_{22}(s)\end{bmatrix}$ with $A_{ij}(s)=P_{i}(\mathcal{U}(\Lambda_{s})\mathcal{U}(\alpha)^{-1}-I_{2n})|_{\text{Ran}P_{j}}$, $A_{12}(0)=A_{21}(0)=A_{22}(0)=0$, and $\frac{d}{ds}A_{22}(s)|_{s=0}=P_{2}B|_{\text{Ran}P_{2}}$.

Since $iB=0_{r}\,\oplus\,(2(A_{0}^{2}+I_{2n-r})^{-1})$ is a positive semi-definite matrix, it follows that \text{rank}\,(P_{2}BP_{2})\geq\hbox{\rm dim,}\text{Ran}P_{2}-r. For $|s|\neq 0$ small enough, we have

[TABLE]

Since $\lim_{s\to 0}(A_{22}(s)-A_{21}(s)A_{11}(s)^{-1}A_{12}(s))/s=P_{2}B|_{\text{Ran}P_{2}}$, we have

[TABLE]

The lemma then follows. ∎

3 Proof of Theorem 1.1

In this section, we provide the proof of Theorem 1.1. To this end, we use the index form associated with $\mathcal{A}_{\Lambda_{0}}$ to study the properties of eigenvalues of Sturm-Liouville problems. Let $A$ be the Hermitian matrix determined by (2.2). The index form $I_{\Lambda_{0}}$ is given by

[TABLE]

on

[TABLE]

Obviously, $H_{\Lambda_{D}}=W^{1,2}_{0}([0,T],\mathbf{C}^{n})$. For any $\xi\in E_{\Lambda_{0}}(0,T)$ and $\eta\in H_{\Lambda_{0}}$, we get by the definition of $\mathcal{A}_{\Lambda_{0}}$ and integration by parts that

[TABLE]

where $y(t)=P(t)\dot{\xi}(t)+Q(t)\xi(t)$. Here the third equality holds due to the boundary condition $\Lambda_{0}$ in the frame (2.2).

Let $\lambda_{j}(\Lambda_{0})$ be the $j$-th eigenvalue of $\mathcal{A}_{\Lambda_{0}}$. From the minmax property of eigenvalues (see [4, 10]), we have

[TABLE]

where $E_{j-1}$ is any $j-1$ dimensional closed subspace of $H_{\Lambda_{0}}$.

Next, we will decompose $H_{\Lambda_{0}}$ into two subspaces. Let \hbox{\rm dim,}V(\Lambda_{0})=k_{0} and ${\tilde{\lambda}}_{k_{0}}(\Lambda_{0})\leq\cdots\leq\tilde{\lambda}_{1}(\Lambda_{0})$ be all the eigenvalues of $A$ with $e_{i}\in V(\Lambda_{0})$, $1\leq i\leq k_{0}$, to be the correspondingly normalized eigenvectors. For any $1\leq i\leq k_{0}$, we can construct a linear function $\xi_{i}$ with $\xi_{i}(t)=\xi_{i}(0)+\frac{t}{T}(\xi_{i}(T)-\xi_{i}(0))$, $t\in[0,T]$, such that $(\xi_{i}(0)^{T},\xi_{i}(T)^{T})^{T}=e_{i}$. For any $1\leq l\leq k_{0}$, we set

[TABLE]

which is a $l$ dimensional subspace of $H_{\Lambda_{0}}$. We define a new norm of $\xi\in\mathcal{X}_{k_{0}}(\Lambda_{0})$ by

[TABLE]

It is well-defined since $\mathcal{X}_{k_{0}}(\Lambda_{0})$ consists of linear functions. Since $\mathcal{X}_{l}(\Lambda_{0})$ is finite dimensional, we shall show that this norm is equivalent to the $L^{2}$ norm and the $W^{1,2}$ norm. More precisely, we have the following lemma.

Lemma 3.1**.**

There exist $c^{\pm}_{1},c^{\pm}_{2}>0$, which depend only on $T$, such that

[TABLE]

Proof.

Let $V=\{\xi:\xi(t)=a+{t\over T}(b-a),t\in[0,T],(0,0,a^{T},b^{T})^{T}\in\Lambda_{N}\}$. Then the dimension of $V\cong\Lambda_{N}$ is $2n$. So there exist $c^{\pm}_{1},c^{\pm}_{2}>0$ such that

[TABLE]

Now (3.3) follows from the fact that $\mathcal{X}_{k_{0}}(\Lambda_{0})\subset V$ for any $\Lambda_{0}$. On the other hand, we can also prove (3.3) by direct computation. More precisely,

[TABLE]

and

[TABLE]

which gives the exact values of $c^{\pm}_{1},c^{\pm}_{2}$. ∎

Denote $H_{0}=H_{\Lambda_{D}}$ for convenience. Then we get the decomposition of $H_{\Lambda_{0}}$.

Lemma 3.2**.**

[TABLE]

Proof.

For any $x\in H_{\Lambda_{0}}$, we have $x\in W^{1,2}([0,T],\mathbf{C}^{n})$ and $(x(0)^{T},x(T)^{T})^{T}\in V(\Lambda_{0})$. We choose $\tilde{x}\in\mathcal{X}_{k_{0}}(\Lambda_{0})$ such that $(\tilde{x}(0),\tilde{x}(T))=(x(0),x(T))$. Then $x-\tilde{x}\in H_{0}$. Since $\mathcal{X}_{k_{0}}(\Lambda_{0})$ consists of linear functions, we get $\mathcal{X}_{k_{0}}(\Lambda_{0})\cap H_{0}=\{0\}$. ∎

Now we are ready to give the relationship of $\lambda_{j}$ and $\tilde{\lambda}_{j}$.

Proposition 3.3**.**

Let $\mathcal{S}\subset\text{Lag}(\mathcal{V},\omega_{2n})$ and $c\in\mathbf{R}$ such that $\tilde{\lambda}_{j}(\Lambda)\leq c$ and \hbox{\rm dim,}V(\Lambda)=k_{0} for all $\Lambda\in\mathcal{S}$. Then $\lambda_{j}$ has a uniformly lower bound on $\mathcal{S}$.

Proof.

For any $\xi\in H_{0}\oplus(\mathcal{X}_{k_{0}}(\Lambda)\ominus\mathcal{X}_{j-1}(\Lambda))$ and any $\varepsilon>0$, there exists $C_{\varepsilon}>0$, independent of $\Lambda\in\mathcal{S}$, such that

[TABLE]

and

[TABLE]

Then we get by (3.1) that there exists $c_{1}>0$, $c_{2}<0$ and $\tilde{c}_{1}\in\mathbf{R}$, independent of $\Lambda\in\mathcal{S}$, such that

[TABLE]

for $\xi\in H_{0}\oplus(\mathcal{X}_{k_{0}}(\Lambda)\ominus\mathcal{X}_{j-1}(\Lambda))$, where $\varepsilon>0$ is small enough such that $c_{1}+c_{2}\varepsilon-2c\varepsilon>0$. By Lemma 3.2, $H_{0}\oplus(\mathcal{X}_{k_{0}}(\Lambda)\ominus\mathcal{X}_{j-1}(\Lambda))$ is a closed subspace of $H_{\Lambda}$ with codimension $j-1$. Then we get by (3.2) that $\lambda_{j}(\Lambda)\geq{c_{2}\over\varepsilon}+\tilde{c}_{1}-2cC_{\varepsilon}$ for all $\Lambda\in\mathcal{S}$. ∎

Proposition 3.4**.**

Let $\mathcal{S}\subset\text{Lag}(\mathcal{V},\omega_{2n})$ and \hbox{\rm dim,}V(\Lambda)=k_{0} for all $\Lambda\in\mathcal{S}$. Then for any $1\leq j\leq k_{0}$, there exist $c_{3}>0$ and $c_{4}\in\mathbf{R}$, which are independent of $\Lambda$, such that

[TABLE]

Proof.

For any $\xi\in\mathcal{X}_{j}(\Lambda)$, we have by (3.1) and Lemma 3.1 that

[TABLE]

where $c$ and $c_{4}$ depend only on $P,Q,R$ and $T$, $c_{3}=(c_{1}^{+})^{-2}>0$ if $\tilde{\lambda}_{j}(\Lambda)>0$, and $c_{3}=(c_{1}^{-})^{-2}>0$ if $\tilde{\lambda}_{j}(\Lambda)<0$. Since \hbox{\rm dim,}\mathcal{X}_{j}(\Lambda)=j, we have $\mathcal{X}_{j}(\Lambda)\cap E^{\perp}_{j-1}\neq\{0\}$ for any fixed $j-1$ dimensional subspace $E_{j-1}$ of $H_{\Lambda}$, and thus

[TABLE]

The proof is complete by (3.2). ∎

Then we give some criteria for the continuity of $\lambda_{j}$.

Lemma 3.5**.**

(1) Let $\mathcal{S}\subset\text{Lag}(\mathcal{V},\omega_{2n})$ and $\lambda_{1}$ be uniformly bounded from below on $\mathcal{S}$. Then $\lambda_{j}$ is continuous on $\mathcal{S}$ for all $j\geq 1$.

(2) Let $\Lambda_{s},s\in[0,\epsilon],$ be a continuous path in $\text{Lag}(\mathcal{V},\omega_{2n})$. If $\lim_{s\to 0^{+}}\lambda_{j}(\Lambda_{s})=-\infty$ for all $1\leq j\leq j_{0}$, and $\lambda_{j_{0}+1}(\Lambda_{s})$, $s\in(0,\epsilon]$, have a uniformly lower bound, then we have

[TABLE]

for all $j>j_{0}$.

Proof.

We first prove (1). Let $r_{1}<\inf_{\Lambda\in\mathcal{S}}\lambda_{1}(\Lambda)$, $\Lambda_{0}\in\mathcal{S}$, and $j_{0}\geq 1$ such that $\lambda_{j_{0}+1}(\Lambda_{0})>\lambda_{j_{0}}(\Lambda_{0})$. Choose $r_{2}\in(\lambda_{j_{0}}(\Lambda_{0}),\lambda_{j_{0}+1}(\Lambda_{0}))$. It follows from Theorem 3.16 in [7] that there exists a neighborhood $\mathcal{S}_{0}\subset\mathcal{S}$ of $\Lambda_{0}$ such that there are exactly $j_{0}$ eigenvalues (counting multiplicity) of $\mathcal{A}_{\Lambda}$ with $\Lambda\in\mathcal{S}_{0}$ in $(r_{1},r_{2})$. Since $\lambda_{1}(\Lambda)>r_{1}$ for $\Lambda\in\mathcal{S}_{0}$, the above $j_{0}$ eigenvalues are exactly $\lambda_{j}(\Lambda),1\leq j\leq j_{0}$. Let $\epsilon>0$ be small enough such that the intervals with radius $\epsilon>0$ centred at the non-equal ones of $\lambda_{j}(\Lambda_{0}),1\leq j\leq j_{0}$, are contained in $(r_{1},r_{2})$. By Theorem 3.16 in [7] again, there exists $\mathcal{S}_{1}\subset\mathcal{S}_{0}$ such that $|\lambda_{j}(\Lambda)-\lambda_{j}(\Lambda_{0})|<\epsilon$ for any $\Lambda\in\mathcal{S}_{1}$. Therefore, (1) holds.

(2) can be shown by a similar method, and thus we omit the details. ∎

Next, we study the asymptotic behavior of $\tilde{\lambda}_{j}$.

Lemma 3.6**.**

Assume that $\Lambda_{s},s\in[-\epsilon,\epsilon],$ satisfy (1.3) and (1.4). Then we have

[TABLE]

and there exists $M^{-}>0$ such that $\tilde{\lambda}_{j}(\Lambda_{s})\leq M^{-}$ on $s\in[-\epsilon,0)$ for $j>k_{-}$. Similarly,

[TABLE]

and there exists $M^{+}>0$ such that $\tilde{\lambda}_{j}(\Lambda_{s})\leq M^{+}$ on $s\in(0,\epsilon]$ for $j>k_{+}$.

Proof.

We only prove the first conclusion, since others can be shown similarly. For any $\beta\in(0,\pi),$ there exists $\alpha\in(0,\beta)$ such that $S_{\alpha}\cap\sigma(\mathcal{U}(\Lambda_{0}))=\emptyset$, where $S_{\alpha}=\{e^{i\theta}|\theta\in(0,\alpha]\}$. So there exists $r\in(0,\epsilon)$ such that $e^{i\alpha}\notin\sigma(\mathcal{U}(\Lambda_{s})),-r<s<0$. It follows from Lemma 2.3 that $\#(S_{\alpha}\cap\sigma(\mathcal{U}(\Lambda_{s})))=k_{-}$. Note that $\mathcal{U}(\Lambda_{s})=\begin{pmatrix}I_{c_{-}}&0\\ 0&(A_{s}+iI_{2n-c_{-}})(A_{s}-iI_{2n-c_{-}})^{-1}\end{pmatrix}$, and thus there are exactly $k_{-}$ eigenvalues, denoted by $\tilde{\lambda}_{j}(\Lambda_{s}),1\leq j\leq k_{-}$, of $A_{s}$ such that $(\tilde{\lambda}_{j}(\Lambda_{s})+i)(\tilde{\lambda}_{j}(\Lambda_{s})-i)^{-1}\in S_{\alpha}$ with $s\in(-r,0)$. This implies $\tilde{\lambda}_{j}(\Lambda_{s})>i(e^{i\alpha}+1)/(e^{i\alpha}-1)=\cot(\alpha/2)>\cot(\beta/2)$. By the arbitrary choice of $\beta$, we have

[TABLE]

Fix any $\beta_{0}\in(0,\pi).$ Since $(\tilde{\lambda}_{j}(\Lambda_{s})+i)(\tilde{\lambda}_{j}(\Lambda_{s})-i)^{-1}\notin S_{\alpha_{0}}$ for all $-r_{0}<s<0$ and all $j>k_{-}$, we infer that $\tilde{\lambda}_{j}(\Lambda_{s})<\cot(\alpha_{0}/2)$. ∎

Then we study the asymptotic behavior of $\lambda_{j}$ using that of $\tilde{\lambda}_{j}$.

Proposition 3.7**.**

Assume that $\Lambda_{s},s\in[-\epsilon,\epsilon],$ satisfy (1.3) and (1.4). Then for any $j\geq 1$,

[TABLE]

and

[TABLE]

Proof.

We only prove (3.5), and (3.6) can be shown in a similar way.

Let $1\leq j\leq k_{-}$. Then by Proposition 3.4, $\lambda_{j}(\Lambda_{s})\leq-c_{3}\tilde{\lambda}_{j}(\Lambda_{s})+c_{4},$ where $c_{3}>0$, $s\in[-\epsilon,0)$. Thanks to (3.4), we have $\lim_{s\to 0^{-}}\lambda_{j}(\Lambda_{s})=-\infty$.

Let $j>k_{-}$. By Lemma 3.6, there exists $M^{-}>0$ such that $\tilde{\lambda}_{j}(\Lambda_{s})\leq M^{-}$ on $s\in[-\epsilon,0)$ for $j>k_{-}$. In view of Proposition 3.3, we have $\lambda_{j}(\Lambda_{s})$, $s\in[-\epsilon,0)$, have a uniformly lower bound for any $j>k_{-}$. Then it follows from (2) of Lemma 3.5 that $\lim_{s\to 0^{-}}\lambda_{j}(\Lambda_{s})=\lambda_{j-k_{-}}(\Lambda_{0})$. ∎

Now we are in a position to show Theorem 1.1.

Proof of Theorem 1.1.

We first prove that $\lambda_{j}$, $j\geq 1$, are all continuous on $\{\Lambda_{s}:s\in[-\epsilon,0)\}$. Since $\mu(\Lambda_{D},\Lambda_{s},s\in[-\epsilon,s_{0}])=0$ for any $s_{0}\in(-\epsilon,0)$, we have by Lemma 3.6 that $\tilde{\lambda}_{1}(\Lambda_{s})<M^{-}$ for all $s\in[-\epsilon,s_{0})$. Thanks to Proposition 3.3, we get that $\lambda_{1}(\Lambda_{s})$, $s\in[-\epsilon,s_{0})$, have a uniformly lower bound. Then by (1) of Lemma 3.5 and the arbitrary choice of $s_{0}\in(-\epsilon,0)$, we obtain the result. The continuity of $\lambda_{j}$ on $\{\Lambda_{s}:s\in(0,\epsilon]\}$ can be shown similarly.

Please note that (1.5) is obtained by Proposition 3.7. ∎

4 Proof of Theorem 1.2

In this section, we give the proof of Theorem 1.2.

Proof of Theorem 1.2.

The proof is complete by Propositions 4.1, 4.2 and 4.3. ∎

Proposition 4.1**.**

Fix any $j\geq 1$ and $0\leq r\leq 2n$. Then

[TABLE]

Proof.

Let $A_{0}$ be the Hermitian matrix in the Lagrangian frame of $\Lambda_{0}\in\Sigma_{r}$. We define $\Lambda_{s}\in\text{Lag}(\mathcal{V},\omega_{2n})$ by (2.3), where $A_{s}=A_{0}+\tan(s)I_{2n-r}$ for $s\in(-\pi/2,\pi/2)$. Noting that $\Lambda_{\pm\pi/2}:=\lim_{s\to\pm{\pi\over 2}}\Lambda_{s}=\Lambda_{D}$, $\{\Lambda_{s},s\in[-\pi/2,\pi/2]\}$ is a continuous loop. It is obvious that

[TABLE]

Direct computation gives

[TABLE]

Recall that $I_{\Lambda_{s}}$ is the corresponding index form, we have

[TABLE]

By (3.2) we get

[TABLE]

Letting $s_{2}=0$ and $s_{1}\to(-\pi/2)^{+}$ in (4.2), we get by (4.1) and Theorem 1.1 that

[TABLE]

On the other hand, letting $s_{1}=0$ and $s_{2}\to(\pi/2)^{-}$ in (4.2), we infer again from (4.1) and Theorem 1.1 that

[TABLE]

Then the conclusion is proved by (4.3) and (4.4). ∎

Next, we study the left endpoint of the range $\lambda_{j}(\Sigma_{r})$.

Proposition 4.2**.**

Fix any $0\leq r\leq 2n$ and $j>2n-r$. Let

[TABLE]

with multiplicity to be $b_{1}+b_{2}+1$, where $b_{i}\geq 0$, $i=1,2$. Then we have two cases.

(1) If $r\leq b_{1}$, then for any $\Lambda\in\Sigma_{r}$,

[TABLE]

(2) If $r>b_{1}$, then

[TABLE]

Proof.

Firstly, we prove (1). Suppose that there exists $\Lambda_{0}\in\Sigma_{r}$ such that $\lambda_{j}(\Lambda_{0})=\lambda_{j-(2n-r)}(\Lambda_{D})$. Since $0\leq r\leq b_{1}$, $\lambda_{j-(2n-r)-b_{1}}(\Lambda_{D})=\lambda_{j-2n}(\Lambda_{D})=\lambda_{j-(2n-r)}(\Lambda_{D})$. By Proposition 4.1, we have $\lambda_{j-2n}(\Lambda_{D})\leq\lambda_{j-r}(\Lambda_{0})$. Thus

[TABLE]

Let $\Lambda_{s}$ be defined by (2.3), where $A_{s}=A_{0}+\tan(s)I_{2n-r}$ for $s\geq 0$. Thanks to Proposition 4.1 and the fact that $I_{\Lambda_{s}}\leq I_{\Lambda_{0}}$ for $s>0$, we have $\lambda_{i-(2n-r)}(\Lambda_{D})\leq\lambda_{i}(\Lambda_{s})\leq\lambda_{i}({\Lambda_{0}})$ for all $i\geq 1$. By (4.5) we get that for $s\geq 0$,

[TABLE]

Then $\lambda$ is an eigenvalue of $\mathcal{A}_{\Lambda_{s}}$ with multiplicity to be at least $r+1$ and thus

[TABLE]

On the other hand, we get by Lemma 2.5 that for $s>0$ small enough,

[TABLE]

which is a contradiction to (4.6).

Next, we show that (2) holds. Let $l_{1}=b_{1}+b_{2}+1$ and $\alpha_{0}=Gr(\gamma_{\lambda}(T))\cap\Lambda_{D}$ with $\lambda:=\lambda_{j-(2n-r)}(\Lambda_{D})$ for convenience. Then \hbox{\rm dim,}\alpha_{0}=l_{1}. We divide the proof into two cases.

Case 1: $r\geq l_{1}$. Let $\tilde{\Lambda}_{1}=\alpha_{0}\oplus V\oplus W_{s_{0}}$, where $V\subset\Lambda_{D}\ominus\alpha_{0}$, \hbox{\rm dim,}V=r-l_{1} and

[TABLE]

for ${\pi\over 2}-s_{0}>0$ small enough. Then $\tilde{\Lambda}_{1}\in\Sigma_{r}$. By Lemma 2.5 and the construction of $\tilde{\Lambda}_{1}$, \hbox{\rm dim,}(\tilde{\Lambda}_{1}\cap Gr(\gamma_{\lambda}(T)))=l_{1}. Let $\epsilon>0$ be small enough such that $\lambda$ is the only eigenvalue of $\mathcal{A}_{\Lambda_{D}}$ in $[\lambda-\epsilon,\lambda+\epsilon]$. By Theorem 3.16 in [7], there are exactly $l_{1}$ eigenvalues (counting multiplicity) of $\mathcal{A}_{\tilde{\Lambda}_{1}}$ in $[\lambda-\epsilon,\lambda+\epsilon]$. They are $\lambda_{j-b_{1}}(\tilde{\Lambda}_{1})\leq\cdots\leq\lambda_{j}(\tilde{\Lambda}_{1})\leq\cdots\leq\lambda_{j+b_{2}}(\tilde{\Lambda}_{1})$ by Theorem 1.1. Since \hbox{\rm dim,}(\tilde{\Lambda}_{1}\cap Gr(\gamma_{\lambda}(T)))=l_{1}, we have $\lambda_{j-b_{1}}(\tilde{\Lambda}_{1})=\cdots=\lambda_{j}(\tilde{\Lambda}_{1})=\cdots=\lambda_{j+b_{2}}(\tilde{\Lambda}_{1})=\lambda$. Therefore, $\lambda_{j}(\tilde{\Lambda}_{1})=\lambda_{j-(2n-r)}(\Lambda_{D})$.

Case 2: $b_{1}<r<l_{1}$. Let $\tilde{\Lambda}_{2}=U\oplus W_{s_{0}}$, where $U\subset\alpha_{0}$, \hbox{\rm dim,}U=r and $W_{s_{0}}$ is given in (4.7) for ${\pi\over 2}-s_{0}>0$ small enough. Then $\tilde{\Lambda}_{2}\in\Sigma_{r}$. By Lemma 2.5, $s_{0}$ can be chosen such that \hbox{\rm dim,}(\tilde{\Lambda}_{2}\cap Gr(\gamma_{\lambda}(T)))=r. Similar to Case 1, $\lambda_{j-b_{1}}(\tilde{\Lambda}_{2})\leq\cdots\leq\lambda_{j}(\tilde{\Lambda}_{2})\leq\cdots\leq\lambda_{j+b_{2}}(\tilde{\Lambda}_{2})$ are all the eigenvalues of $\mathcal{A}_{\tilde{\Lambda}_{2}}$ in $[\lambda-\epsilon,\lambda+\epsilon]$. Since \hbox{\rm dim,}(\tilde{\Lambda}_{2}\cap Gr(\gamma_{\lambda}(T)))=r and $\lambda_{i-(2n-r)}(\Lambda_{D})\leq\lambda_{i}(\tilde{\Lambda}_{2})$ for $j-b_{1}\leq i\leq j+b_{2}$, we have $\lambda_{j-b_{1}}(\tilde{\Lambda}_{2})=\cdots=\lambda_{j}(\tilde{\Lambda}_{2})=\cdots=\lambda_{j+(r-b_{1}-1)}(\tilde{\Lambda}_{2})=\lambda<\lambda_{j+(r-b_{1})}(\tilde{\Lambda}_{2})$. Therefore, $\lambda_{j}(\tilde{\Lambda}_{2})=\lambda_{j-(2n-r)}(\Lambda_{D})$. ∎

Finally, we study the right endpoint of the range $\lambda_{j}(\Sigma_{r})$.

Proposition 4.3**.**

Fix any $j\geq 1$ and $0\leq r\leq 2n$. Let

[TABLE]

with multiplicity to be $c_{1}+c_{2}+1$, where $c_{i}\geq 0$, $i=1,2$. Then we have two cases.

(1) If $r\leq c_{2}$, then for any $\Lambda\in\Sigma_{r}$,

[TABLE]

(2) If $r>c_{2}$, then

[TABLE]

Proof.

The method is similar as Proposition 4.2 and we give the proof here for completeness. We first prove (1). Suppose that there exists $\Lambda_{0}\in\Sigma_{r}$ such that $\lambda_{j}(\Lambda_{0})=\lambda_{j}(\Lambda_{D})$. Since $0\leq r\leq c_{2}$, $\lambda_{j}(\Lambda_{D})=\lambda_{j+r}(\Lambda_{D})=\lambda_{j+c_{2}}(\Lambda_{D})$. By Proposition 4.1, we have $\lambda_{j+r}(\Lambda_{0})\leq\lambda_{j+r}(\Lambda_{D})$. Thus

[TABLE]

Let $\Lambda_{s}$ be given by (2.3), where $A_{s}=A_{0}+\tan(s)I_{2n-r}$ for $s\leq 0$. Thanks to Proposition 4.1 and the fact that $I_{\Lambda_{s}}\geq I_{\Lambda_{0}}$ for $s<0$, we have $\lambda_{i}({\Lambda_{0}})\leq\lambda_{i}(\Lambda_{s})\leq\lambda_{i}(\Lambda_{D})$ for all $i\geq 1$. By (4.8) we get that for $s\leq 0$,

[TABLE]

Then

[TABLE]

However, we get by Lemma 2.5 that for $s<0$ small enough,

[TABLE]

which contradicts (4.9).

Next, we prove (2). Let $l_{2}=c_{1}+c_{2}+1$ and $\beta_{0}=Gr(\gamma_{\lambda}(T))\cap\Lambda_{D}$ with $\lambda:=\lambda_{j}(\Lambda_{D})$. Then \hbox{\rm dim,}\beta_{0}=l_{2}. We divide the proof into two cases.

Case 1: $r\geq l_{2}$. Let $\hat{\Lambda}_{1}=\beta_{0}\oplus V\oplus W_{s_{0}}$, where $V\subset\Lambda_{D}\ominus\beta_{0}$, \hbox{\rm dim,}V=r-l_{2} and $W_{s_{0}}$ is given by (4.7) for $s_{0}+{\pi\over 2}>0$ small enough. Then $\hat{\Lambda}_{1}\in\Sigma_{r}$. By Lemma 2.5 and the construction of $\hat{\Lambda}_{1}$, \hbox{\rm dim,}(\hat{\Lambda}_{1}\cap Gr(\gamma_{\lambda}(T)))=l_{2}. Let $\epsilon>0$ be small enough such that $\lambda$ is the only eigenvalue of $\mathcal{A}_{\Lambda_{D}}$ in $[\lambda-\epsilon,\lambda+\epsilon]$. By Theorem 3.16 in [7], there are exactly $l_{2}$ eigenvalues of $\mathcal{A}_{\hat{\Lambda}_{1}}$ in $[\lambda-\epsilon,\lambda+\epsilon]$. They are $\lambda_{j-c_{1}}(\hat{\Lambda}_{1})=\cdots=\lambda_{j}(\hat{\Lambda}_{1})=\cdots=\lambda_{j+c_{2}}(\hat{\Lambda}_{1})=\lambda$ by Theorem 1.1 and the fact that \hbox{\rm dim,}(\hat{\Lambda}_{1}\cap Gr(\gamma_{\lambda}(T)))=l_{2}. Therefore, $\lambda_{j}(\hat{\Lambda}_{1})=\lambda_{j}(\Lambda_{D})$.

Case 2: $c_{2}<r<l_{2}$. Let $\hat{\Lambda}_{2}=U\oplus W_{s_{0}}$, where $U\subset\beta_{0}$ and \hbox{\rm dim,}U=r and $W_{s_{0}}$ is given in (4.7) for $s_{0}+{\pi\over 2}>0$ small enough. Then $\hat{\Lambda}_{2}\in\Sigma_{r}$. By Lemma 2.5, $s_{0}$ can be chosen such that \hbox{\rm dim,}(\hat{\Lambda}_{2}\cap Gr(\gamma_{\lambda}(T)))=r. Similar to Case 1, $\lambda_{j-c_{1}}(\hat{\Lambda}_{2})\leq\cdots\leq\lambda_{j}(\hat{\Lambda}_{2})\leq\cdots\leq\lambda_{j+c_{2}}(\hat{\Lambda}_{2})$ are all the eigenvalues of $\mathcal{A}_{\hat{\Lambda}_{2}}$ in $[\lambda-\epsilon,\lambda+\epsilon]$. Since \hbox{\rm dim,}(\hat{\Lambda}_{2}\cap Gr(\gamma_{\lambda}(T)))=r and $\lambda_{i}(\hat{\Lambda}_{2})\leq\lambda_{i}(\Lambda_{D})$ for $j-c_{1}\leq i\leq j+c_{2}$, we have $\lambda_{j-(r-c_{2})}(\hat{\Lambda}_{2})<\lambda_{j-(r-c_{2}-1)}(\hat{\Lambda}_{2})=\cdots=\lambda_{j}(\hat{\Lambda}_{2})=\cdots=\lambda_{j+c_{2}}(\hat{\Lambda}_{2})=\lambda$. Therefore, $\lambda_{j}(\hat{\Lambda}_{2})=\lambda_{j}(\Lambda_{D})$. ∎

5 Proof of Theorem 1.7

Recall that ${\gamma}_{\lambda}$ defined in (1.7) is the fundamental solution of (1.6). In this section, we give the asymptotic behavior of $Gr(\gamma_{\lambda}(T))$ as $\lambda\to-\infty$.

Proof of Theorem 1.7.

We first prove (1.8). It is equivalent to show that $\lim\limits_{\lambda\to-\infty}\mathcal{U}(Gr({\gamma}_{\lambda}(T)))=\mathcal{U}(\Lambda_{D})$. Suppose otherwise, there exist $\Lambda_{0}\neq\Lambda_{D}\in\text{Lag}(\mathcal{V},\omega_{2n})$ and a sequence $\{v_{m}\}_{m=1}^{\infty}$ such that $\lim\limits_{m\to\infty}v_{m}=-\infty$ and $\lim\limits_{m\to\infty}\mathcal{U}(Gr({\gamma}_{v_{m}}(T)))=\mathcal{U}(\Lambda_{0})$. Let $\hat{\Lambda}_{m}:=Gr(\gamma_{v_{m}}(T))$ for convenience.

Firstly, we claim that

[TABLE]

In fact, since $\mathcal{U}(\Lambda_{0})\neq I_{2n}$, there exists $\kappa\in\sigma(\mathcal{U}(\Lambda_{0}))$ such that $\kappa\neq 1$. Obviously, $\kappa I_{2n}\notin\Sigma_{\mathcal{U}(\Lambda_{D})}$ and $\kappa I_{2n}\in\Sigma_{\mathcal{U}(\Lambda_{0})}$. Direct computation shows that $\inf_{U\in\Sigma_{\mathcal{U}(\Lambda_{D})}}\|\kappa I_{2n}-U\|>0$, and thus (5.1) holds.

Then we claim that

[TABLE]

as $m\to\infty.$

Since $\mathcal{U}(\hat{\Lambda}_{m})\to\mathcal{U}(\Lambda_{0})$, we infer that for any $\epsilon>0$, there exists $N>0$ such that $\|\mathcal{U}(\Lambda_{0})^{-1}\mathcal{U}(\hat{\Lambda}_{m})-I_{2n}\|<\epsilon$ for each $m>N$. Let $U\in\Sigma_{\mathcal{U}(\Lambda_{0})}$. Then $U\mathcal{U}(\Lambda_{0})^{-1}\mathcal{U}(\hat{\Lambda}_{m})\in\Sigma_{\mathcal{U}(\hat{\Lambda}_{m})}$. It follows that

[TABLE]

where $m>N$. Therefore, we get (5.2).

By (5.1), there exists $\mathcal{U}(\Lambda_{3})\in\Sigma_{\mathcal{U}(\Lambda_{0})}$ such that $\mathcal{U}(\Lambda_{3})\notin\Sigma_{\mathcal{U}(\Lambda_{D})}$, and there exists a compact neighborhood $V_{\mathcal{U}(\Lambda_{3})}$ of $\mathcal{U}(\Lambda_{3})$ such that $V_{\mathcal{U}(\Lambda_{3})}\cap\Sigma_{\mathcal{U}(\Lambda_{D})}=\emptyset$. This deduces that $\lambda_{1}(V_{\Lambda_{3}})$ is bounded from below by Proposition 3.3, where $V_{\Lambda_{3}}=\mathcal{U}^{-1}(V_{\mathcal{U}(\Lambda_{3})})$. On the other hand, we get by (5.2) that $\lim\limits_{m\to\infty}\inf_{U\in\Sigma_{\mathcal{U}(\hat{\Lambda}_{m})}}\|\mathcal{U}$ $(\Lambda_{3})-U\|=0$. So there exists $U_{m}\in\Sigma_{\mathcal{U}(\hat{\Lambda}_{m})}$ such that $\lim\limits_{m\to\infty}\|\mathcal{U}(\Lambda_{3})-U_{m}\|=0$. It follows that $\Sigma_{\mathcal{U}(\hat{\Lambda}_{m})}\cap V_{\mathcal{U}(\Lambda_{3})}\neq\emptyset$ when $m$ is sufficiently large. Choose $\mathcal{U}(\tilde{\Lambda}_{m})\in\Sigma_{\mathcal{U}(\hat{\Lambda}_{m})}\cap V_{\mathcal{U}(\Lambda_{3})}$. Then $v_{m}$ is an eigenvalue of $\mathcal{A}_{\tilde{\Lambda}_{m}}$. However, $\lim\limits_{m\to\infty}v_{m}=-\infty$ contradicts that $\lambda_{1}(V_{\Lambda_{3}})$ is bounded from below.

Next, we prove (1.9). Using the fact that ${\mathcal{A}}_{Gr(\gamma_{\lambda(s)}(T))}$ has an eigenvalue $\lambda(s)$ with the multiplicity to be $2n$, we obtain that ${\mathcal{A}}_{Gr(\gamma_{\lambda(s)}(T))}$ has at least $2n$ eigenvalues such that they tend to $-\infty$ as $s\to(-\pi/2)^{+}$. Then we have

[TABLE]

by Theorem 1.1. Since $Gr(\gamma_{\lambda(s)}(T))\cap{\Lambda_{D}}=\{0\}$ for $s\in(-\pi/2,-\pi/2+\epsilon]$ with $\epsilon>0$ small enough, we have $\mu(\Lambda_{D},Gr(\gamma_{\lambda(s)}(T)),s\in[-\pi/2,-\pi/2+\epsilon])\leq 2n$. It then follows that

[TABLE]

∎

As a consequence of Theorem 1.7, we get the following result.

Proposition 5.1**.**

Let $\lambda(s)=\tan(s)$ for $s\in[-\pi/2,+\pi/2]$. Then there exists $\epsilon_{0}>0$ such that $Gr(\gamma_{\lambda(s)}(T))\cap Gr(\gamma_{\lambda(t)}(T))=\{0\}$ for any $-\pi/2<s<t<-\pi/2+\epsilon_{0}$.

Proof.

Denote $Gr(\gamma_{\lambda(r)}(T))$ by $\Lambda_{r}$. Suppose that for any $\epsilon>0$, there exist $-\pi/2<s_{\epsilon}<t_{\epsilon}<-\pi/2+\epsilon$ such that $\Lambda_{s_{\epsilon}}\cap\Lambda_{t_{\epsilon}}\neq\{0\}$. Then ${\mathcal{A}}_{\Lambda_{t_{\epsilon}}}$ has an eigenvalue $\lambda(t_{\epsilon})$ with its multiplicity to be $2n$ and another eigenvalue $\lambda(s_{\epsilon})$. It follows that at least $2n+1$ eigenvalues of ${\mathcal{A}}_{\Lambda_{t_{\epsilon}}}$ tend to $-\infty$ as $\epsilon\to 0^{+}$. By Theorem 1.7, $\mu(\Lambda_{D},\Lambda_{t},t\in[-\pi/2,-\pi/2+\epsilon])=2n$ with $\epsilon>0$ small enough. So by Theorem 1.1, the $(2n+1)$-th eigenvalue of ${\mathcal{A}}_{\Lambda_{t}}$ is bounded from below as $t\to(-\pi/2)^{+}$, which is a contradiction. Then the proposition follows. ∎

Acknowledgements. The authors sincerely thank Prof. Yiming Long for his interest. X. Hu is partially supported by NSFC (Nos.11425105, 11790271). H. Zhu is partially supported by PITSP (No. BX20180151) and CPSF (No. 2018M630266).