Index theory and multiple solutions for asymptotically linear second-order delay differential equations

Abstract

This paper is concerned with the existence of periodic solutions for asymptotically linear second-order delay differential equations. We will establish an index theory for the linear system directly in the sense that we do not need to change the problem of the original linear system into the problem of an associated Hamiltonian system. By using the critical point theory and the index theory, some new existence results are obtained.

1 Introduction

In the past few decades, many results on the existence of periodic solutions for delay differential equations were obtained by several different approaches, including various fixed point theorems, Hopf bifurcation theorems, coincidence degree theory, coupled system methods, Poincaré–Bendixson theorem, and so on. One can refer to [9, 12, 26, 27] for detailed discussions.

In [28], Kaplan and Yorke introduced a technique studying the existence of periodic solutions of the first-order delay differential equation

$$\begin{aligned} \dot{x}(t)=-f\bigl(x(t-\tau )\bigr) \end{aligned}$$

(1)

that may reduce the existence problem of periodic solutions of (1) to a problem of finding periodic solutions of an associated plane ordinary differential system. Afterwards, Kaplan and York’s original idea was used by many authors to study some general differential delay equations by transforming them into Hamiltonian systems, for example, [7, 8, 10, 20–22, 29–31, 33]. Especially, some known results involving various index theories in Hamiltonian systems were generalized to study delay differential equations. For example, in [29–31], by using Morse–Ekeland index theory for the associated Hamiltonian system, Li and He firstly studied the periodic solutions of delay differential equations with multiple delays via critical point theory. Some multiplicity results on periodic solutions in delay differential equations were given by Fei [20, 21] using $S^{1}$-pseudo index theory. In [32, 42], by relative Fredholm index and spectral flow, Liu and Wang constructed an index theory for the coupled Hamiltonian system and obtained some interesting existence and multiplicity results for delay differential equations.

In [24], Guo and Yu firstly applied the critical point theory to (1) directly in the sense that one does not need to transform the original existence problem of (1) to the existence problem for an associated Hamiltonian system while allowing a delay in the variational functional. By using pseudo-index theory introduced by Benci and Rabinowitz [4], they obtained multiple periodic solutions of (1) with odd nonlinearity that grows asymptotically linear both at the origin and at infinity. By establishing suitable variational frameworks for the second-order delay differential equation and high-dimensional case, Guo and Guo [23], Guo and Yu [25] obtained some sufficient conditions on the existence of periodic solutions.

Motivated by [23, 32, 42], we consider the existence of 2τ-periodic solutions for the following nonautonomous second-order delay system:

$$\begin{aligned} \ddot{x}(t)=-\nabla F\bigl(t, x(t-\tau )\bigr), \end{aligned}$$

(2)

where $\tau >0$ is a given constant.

Throughout this paper, we make use of the following hypotheses:

$(F_{1})$:

$F\in C^{1}({\mathbb{R}}\times {\mathbb{R}}^{n}, {\mathbb{R}})$, $F(t, 0)=0$.

Furthermore, we assume there exist continuous 2τ periodic symmetric matrix functions $A_{0}(t)$ and $A_{\infty}(t)$ such that

$(F_{0})$:

$f(t, x)=A_{0}(t)x+o(x)\text{ as }|x|\rightarrow 0$,

$(F_{\infty})$:

$A_{\infty}>A_{0}$ and $f(t, x)=A_{\infty}(t)x+o(x)\text{ as }|x|\rightarrow \infty$,

where we denote $\nabla F=f$, and for two continuous $n\times n$ matrix-valued functions $B_{1}(t)$, $B_{2}(t)$, we say that $B_{1}(t)\leq B_{2}(t)$ if and only if $\max_{\xi \in {\mathbb{R}}^{n}, |\xi |=1} (B_{1}(t)- B_{2}(t))\xi \xi \leq 0$, and $B_{1}(t)> B_{2}(t)$ if and only if $B_{1}(t)\leq B_{2}(t)$ does not hold. Furthermore, we write $B_{1}\leq B_{2}$ if $B_{1}(t)\leq B_{2}(t)$ for a.e. $t\in S^{1}$ and denote by $B_{1}< B_{2}$ if $B_{1}\leq B_{2}$ and $B_{1}(t)< B_{2}(t)$ on a subset of $S^{1}$ with nonzero measure.

Conditions ($F_{0}$) and ($F_{\infty}$) are referred to in the literature as asymptotically linear quadratic for the nonlinearity F. This type of conditions was widely used in the problems of periodic solutions of Hamiltonian systems. It is well known that the existence and multiplicity of the periodic solutions are related to the difference between $A_{0}$ and $A_{\infty}$, and a quantitative way to measure the difference is the index theory. Ekeland established an index theory for convex linear Hamiltonian systems in [17–19]. In [11, 34, 35, 37, 39], Conley et al. introduced an index theory for symplectic paths. In [38, 44], Long and Zhu defined spectral flows for paths of linear operators, the relative Morse index between two linear operators, and redefined Maslov index for symplectic paths. Liu [32] defined the $(\mathcal{J}, \mathcal{M})$-index and studied the delay Hamiltonian systems. Dong [15] developed an index theory for abstract operator equations with compact resolvent. We also mention that the authors in [6, 40, 43] established various index theories when essential spectrum emerges. We refer to the excellent books of Abbondandolo [1], Ekeland [18], and Long [36] for a more detailed account of the concept.

In this paper, we establish the existence and multiplicity of periodic solutions for (2) by combining the index theory with a generalized linking theorem developed by Ding and Liu [14]. More precisely, for any 2τ-periodic continuous matrix-valued function $B(t)$, we consider the following linear delay differential equation:

$$\begin{aligned} \ddot{x}(t+\tau )-\lambda ^{2}B(t)x(t)=0,\quad x(t)=x(t+2\tau ). \end{aligned}$$

(3)

Note that the spectrum of the operator $A: Ax=\ddot{x}(t+\tau )$ consists of a sequence of eigenvalues with finite multiplicity, which is unbounded from above and below. This implies that the Morse index of the energy functional corresponding to (3) is infinite. Thus, the main difficulty is to find the finite representation of the indefinite Morse index. By taking advantage of the spectral properties of A, we will modify the linear system (3) so that the Morse index of the modified linear system is finite. Then we will define a relative Morse index. Our approach follows closely that in [16]. Moreover, to combine the relative Morse index and the generalized linking theorem developed by Ding and Liu [14], we need to make a suitable choice of the reduction very carefully and overcome some involved issues.

To state our main results, we first introduce some notations. We denote by $\nu (B)$ the dimension of the kernel of (3) and call $\nu (B)$ the nullity of B. For any 2τ-periodic continuous matrix-valued functions $B_{1}$, $B_{2}$ with $B_{2}>B_{1}$, we define

$$\begin{aligned} I(B_{1}, B_{2})=\sum _{s=0}^{1}\nu \bigl((1-s)B_{1}+sB_{2}\bigr). \end{aligned}$$

We call $I(B_{1}, B_{2})$ the relative Morse index between $B_{1}$ and $B_{2}$ (see Definition 2). For the definitions of the relative Morse index, we refer to [15, 18, 43].

Our main results are stated as follows.

Theorem 1

Let $(F_{1})$, $(F_{0})$, $(F_{\infty})$ be satisfied. Moreover, we assume that $(F_{0})$ holds with $\nu (A_{0})=0$ and $(F_{\infty})$ holds with $\nu (A_{\infty})=0$. Then equation (2) has at least one nonconstant 2τ periodic solution.

Theorem 2

Under the assumption of Theorem 1, if furthermore f is odd, then (2) has at least $I(A_{0}, A_{\infty})$ pairs of 2τ periodic solutions.

Remark 1

The condition $\nu (A_{\infty})= 0$ implies that the nonlinearity f is nonresonant at infinity. This assumption is crucial for the proof of the Palais–Smale condition (PS condition for short). We refer to [4, 41] and the references therein for results that allow resonance at infinity under some additional technical conditions such as Landsmann–Lazer condition, Rabinowitz resonant condition, strong resonant conditions, etc.

The paper is organized as follows. In Sect. 2, we formulate the variational setting and develop an index theory to classify the associated linear second-order delay differential equations and define the relative Morse index. After collection in Sect. 2.3 of the abstract critical point theorems, which we needed, we prove our theorems in Sect. 3.

2 Variational structure and classification theory

2.1 Variational setting

In this paper, we establish a variational structure that enables us to reduce the existence of 2τ-periodic solutions of (2) to the existence of critical points of corresponding functional defined on some appropriate function space.

Assume that $\lambda =\frac{\tau}{\pi}$, $s=\frac{\pi}{\tau}t$. Then equation (2) is transformed to

$$\begin{aligned} \ddot{x}(t)=-\lambda ^{2}f\bigl(x(t-\pi )\bigr), \end{aligned}$$

(4)

and we seek 2π periodic solutions of (4), which, of course, correspond to the 2τ periodic solutions of (2). Define $Ax=-\ddot{x}(t+\pi )$. Then A is self-adjoint on $H^{1}(S^{1}, {\mathbb{R}}^{n})$ with domain $D(A)=H^{2}(S^{1}, {\mathbb{R}}^{n})$ (see Lemma 2.2 of [23]). Let $\sigma (A)$, $\sigma _{d}(A)$, $\sigma _{e}(A)$ denote, respectively, the spectrum, the discrete spectrum, and the essential spectrum of A. It is easy to calculate that

$$\begin{aligned} \sigma (A)=\sigma _{d}(A). \end{aligned}$$

(5)

For any 2π-period continuous matrix-valued function $B_{0}(t)$, we denote $B_{0}$ as the multiplication operator by $B_{0}(t)$ in $L^{2}(S^{1}, {\mathbb{R}}^{n})$. Let $A_{B_{0}}=A-B_{0}$. By (5), we also have

$$\begin{aligned} \sigma (A_{B_{0}})=\sigma _{d}(A_{B_{0}}). \end{aligned}$$

(6)

Throughout this paper, we require $0\notin \sigma (A_{B_{0}})$, which means Ker$(A-B_{0})=\emptyset$. In fact, if $0\in \sigma (A_{B_{0}})$, by property (6), there exists $\epsilon _{0}$ such that, for any $\epsilon \in (0, \epsilon _{0})$, we have $0\notin \sigma (A_{B_{0}+\epsilon})$, and then we can replace $B_{0}$ by $B_{0}+\epsilon$.

Moreover, (4) can be transformed to

$$\begin{aligned} -\ddot{x}(t+\pi )-\lambda ^{2} B_{0}(t)x(t)= \lambda ^{2}f\bigl(x(t-\pi )\bigr)- \lambda ^{2} B_{0}(t)x(t). \end{aligned}$$

(7)

This modification is crucial for the establishment of the linking structure for the problem. More precisely, in Sect. 3, we may assume that $B_{0}=A_{0}-\epsilon$ for some ϵ small, where $A_{0}$ is defined as in condition ($f_{0}$).

Let $|A_{B_{0}}|$ be the absolute value of $A_{B_{0}}$ and $E_{B_{0}}=D(|A_{B_{0}}|^{\frac{1}{2}})$ be the domain of the self-adjoint operator $|A_{B_{0}}|^{\frac{1}{2}}$, which is a Hilbert space equipped with the inner product

$$\begin{aligned} (z,w)_{B_{0}}=\bigl( \vert A_{B_{0}} \vert ^{\frac{1}{2}}z, \vert A_{B_{0}} \vert ^{\frac{1}{2}}w\bigr)_{2} \end{aligned}$$

and the induced norm $\|z\|_{B_{0}}=(z,z)_{B_{0}}^{\frac{1}{2}}$. By the spectral properties (6), we have $|A_{B_{0}}|^{-\frac{1}{2}}$ is a compact operator and it follows that

Lemma 1

$E_{B_{0}}$ embeds continuously into $H^{1}(S^{1}, {\mathbb{R}}^{n})$, and $E_{B_{0}}$ embeds compactly into $L^{p}(S^{1}, {\mathbb{R}}^{n})$ for all $p\geq 2$.

Let $\{G_{\lambda}\}_{\lambda \in {\mathbb{R}}}$ denote the spectral family of $A_{B_{0}}$. We define the following projections:

$$\begin{aligned} P_{\beta,B_{0}}^{+}= \int _{0}^{+\infty}\,dG_{\lambda},\qquad P_{\beta,B_{0}}^{0}= \int _{-\beta}^{0}\,dG_{\lambda},\qquad P_{\beta,B_{0}}^{-}= \int _{- \infty}^{-\beta}\,dG_{\lambda}. \end{aligned}$$

Here $\beta >0$ is large enough and $\pm \beta \notin \sigma (A_{B_{0}})$. This induces an orthogonal decomposition on $E_{B_{0}}$:

$$\begin{aligned} E_{B_{0}}=E_{\beta,B_{0}}^{+}\oplus E_{\beta,B_{0}}^{0} \oplus E_{ \beta,B_{0}}^{-}, \quad\text{where } E_{\beta,B_{0}}^{\pm}=P_{\beta,B_{0}}^{ \pm }E_{B_{0}} \text{ and } E_{\beta,B_{0}}^{0}=P_{\beta,B_{0}}^{0} E_{B_{0}}. \end{aligned}$$

For any $x\in E_{B_{0}}$, we have $x=x^{+} + x^{0} + x^{-}\in E_{\beta,B_{0}}^{+}\oplus E_{\beta,B_{0}}^{0} \oplus E_{\beta,B_{0}}^{-}$ and $|A_{B_{0}}|x=A_{B_{0}}(x^{+} - x^{0} - x^{-})$.

On $E_{B_{0}}$ we define the functional

$$\begin{aligned} I_{\beta, B_{0}}(x)={}&\frac{1}{2} \bigl\Vert x^{+} \bigr\Vert _{B_{0}}^{2}-\frac{1}{2} \bigl\Vert x^{0} \bigr\Vert _{B_{0}}^{2}- \frac{1}{2} \bigl\Vert x^{-} \bigr\Vert _{B_{0}}^{2} \\ &{}-\lambda ^{2} \int _{0}^{2\pi}F\bigl(x(t)\bigr)+\frac{1}{2} \lambda ^{2}(B_{0}x,x)_{2}. \end{aligned}$$

(8)

By a standard argument as in [4, 23], the functional $I_{\beta, B_{0}}$ is continuously differentiable on $E_{B_{0}}$ and the existence of 2π-periodic solutions $x(t)$ for (2) is equivalent to the existence of critical points of functional $I_{\beta, B_{0}}(x)$.

2.2 Index theory

In this subsection, we investigate the linear second-order delay differential equations

$$\begin{aligned} \begin{aligned}&\ddot{x}(t+\pi )-\lambda ^{2}B_{0}(t)x(t)= \lambda ^{2}\bigl(B(t)-B_{0}(t)\bigr)x(t), \\ &x(t+\pi )=x(t-\pi ), \end{aligned} \end{aligned}$$

(9)

where $B_{0}(t)$, $B(t)$ are 2π-period continuous matrix-valued functions. Recall that we define $(A_{B_{0}}x)(t)=\ddot{x}(t+\pi )-\lambda ^{2} B_{0}(t)x(t)$ and $(Bx)(t)=B(t)x(t)$. Then (9) can be rewritten as

$$\begin{aligned} A_{B_{0}}x(t)-\lambda ^{2}(B-B_{0})x(t)=0. \end{aligned}$$

(10)

Let $x\in E_{B_{0}}$ be a solution of (10). Set $u=|A_{B_{0}}|^{\frac{1}{2}}x$. Then $u\in L^{2}(S^{1}, {\mathbb{R}}^{n})$. The projections $P_{\beta, B_{0}}^{+}$, $P_{\beta, B_{0}}^{0}$, $P_{\beta, B_{0}}^{-}$ defined as in Sect. 2.1 also induce a decomposition on $L^{2}(S^{1}, {\mathbb{R}}^{N})$:

$$\begin{aligned} L^{2}\bigl(S^{1}, {\mathbb{R}}^{n} \bigr)=L_{\beta, B_{0}}^{+}\oplus L_{\beta,B_{0}}^{0} \oplus L_{\beta,B_{0}}^{-}, \end{aligned}$$

where $L_{\beta,B_{0}}^{\pm}=P_{\beta,B_{0}}^{\pm }L^{2}(S^{1}, { \mathbb{R}}^{n})$ and $L_{\beta,B_{0}}^{0}=P_{\beta,B_{0}}^{0} L^{2}(S^{1}, {\mathbb{R}}^{n})$. For any $u\in L^{2}(S^{1}, {\mathbb{R}}^{n})$, we have $u=u^{+} + u^{0} + u^{-}\in L_{\beta,B_{0}}^{+}\oplus L_{\beta,B_{0}}^{0} \oplus L_{\beta,B_{0}}^{-}$. Then (10) is equivalent to

$$\begin{aligned} u^{+}-u^{0}-u^{-}- \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}u=0,\quad u\in L^{2} \bigl(S^{1}, {\mathbb{R}}^{n}\bigr). \end{aligned}$$

(11)

Define the associated bilinear form

$$\begin{aligned} q_{\beta, B_{0},B}(u,v)={}&\frac{1}{2}\bigl(u^{+}, v^{+} \bigr)_{2}-\frac{1}{2}\bigl(u^{0}, v^{0} \bigr)_{2}-\frac{1}{2}\bigl(u^{-}, v^{-} \bigr)_{2} \\ &{}-\frac{1}{2}\bigl( \vert A_{B_{0}} \vert ^{-\frac{1}{2}} \lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}u,v\bigr)_{2} \end{aligned}$$

for any $u, v\in L^{2}(S^{1}, {\mathbb{R}}^{n})$. From (11), it is easy to get

$$\begin{aligned} -u^{-}-P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}u=0 \end{aligned}$$

and

$$\begin{aligned} &{-}\bigl(P_{\beta,B_{0}}^{-}+P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}P_{\beta,B_{0}}^{-} \bigr)u^{-} \\ &\quad=P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\bigl(u^{+}+u^{0}\bigr). \end{aligned}$$

Since $P_{\beta,B_{0}}^{-}|A_{B_{0}}|^{-\frac{1}{2}}=\int _{-\infty}^{- \beta}|\lambda |^{-\frac{1}{2}}\,d F_{\lambda}$, we have

$$\begin{aligned} \bigl\Vert P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{-\frac{1}{2}} \bigr\Vert \leq \frac{1}{\sqrt{\beta}}. \end{aligned}$$

(12)

Thus, by choosing $\beta >\lambda ^{2}(\|B\|+ \|B_{0}\|)$, we have

$$\begin{aligned} \bigl\Vert P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{-\frac{1}{2}} \lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}P_{\beta,B_{0}}^{-} \bigr\Vert \leq \frac{\lambda ^{2}( \Vert B \Vert + \Vert B_{0} \Vert ) }{\beta}< 1. \end{aligned}$$

It follows that $P_{\beta,B_{0}}^{-}+P_{\beta,B_{0}}^{-}|A_{B_{0}}|^{-\frac{1}{2}} \lambda ^{2}(B-B_{0})|A_{B_{0}}|^{-\frac{1}{2}}P_{\beta,B_{0}}^{-}$ is invertible. Let $L_{\beta, B_{0}}^{\ast}=L_{\beta, B_{0}}^{+}\oplus L_{\beta, B_{0}}^{0}$ and define $L_{\beta, B_{0}, B }:L_{\beta, B_{0}}^{\ast}\rightarrow L_{\beta, B_{0}}^{-}$:

$$\begin{aligned} L_{\beta, B_{0}, B }={}&{-}\bigl(P_{\beta,B_{0}}^{-}+P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{- \frac{1}{2}} \lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}P_{\beta,B_{0}}^{-} \bigr)^{-1} \\ &{}\times P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\bigl(u^{+}+u^{0}\bigr). \end{aligned}$$

Then it is easy to calculate that, for fixed $u^{\ast}\in L_{\beta, B_{0}}^{\ast}$, $L_{\beta, B_{0}, B }u^{\ast}$ is the unique solution for

$$\begin{aligned} -y-P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\bigl(u^{\ast}+y\bigr)=0,\quad y\in L_{\beta, B_{0}}^{-} . \end{aligned}$$

Moreover, for any $u^{\ast }\in L_{\beta, B_{0}}^{\ast}$, we define

$$\begin{aligned} \psi _{u^{\ast}}(y) = q_{\beta, B_{0}, B}\bigl(u^{\ast}+y,u^{\ast}+y \bigr),\quad \forall y\in L_{\beta, B_{0}}^{-}. \end{aligned}$$

Then $\psi _{u^{\ast}}(y)$ is of class $C^{2}$ on $L_{\beta, B_{0}}^{-}$, and for any $w\in L_{\beta, B_{0}}^{-}$,

$$\begin{aligned} D^{2} \psi _{u^{\ast}}(y) (w, w)=- \Vert w \Vert ^{2}_{2}-\bigl( \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}w, w\bigr)_{2}. \end{aligned}$$

And it follows from (12) that $D^{2} \psi _{u^{\ast}}(y)(w, w)<0$. Thus $\psi _{u^{\ast}}(y)$ has unique maximum at the point $L_{\beta, B_{0}, B }u^{\ast}$. This yields

$$\begin{aligned} q_{\beta, B_{0}, B}\bigl(u^{\ast}+y,u^{\ast}+y\bigr) \leq q_{\beta, B_{0}, B}\bigl(u^{ \ast}+L_{\beta, B_{0}, B }u^{\ast},u^{\ast}+L_{\beta, B_{0}, B }u^{ \ast} \bigr). \end{aligned}$$

(13)

Now we define a quadratic form $\tilde{q}_{\beta, B_{0}, B}$ on $L^{\ast}_{\beta, B_{0}}$ as

$$\begin{aligned} \tilde{q}_{\beta, B_{0}, B}\bigl(u^{\ast},v^{\ast} \bigr)={}&q_{\beta, B_{0}, B}\bigl(u^{ \ast}+L_{\beta, B_{0}, B }u^{\ast},v^{\ast}+L_{\beta, B_{0}, B }v^{ \ast} \bigr) \\ ={}&\frac{1}{2}\bigl(u^{+}, v^{+}\bigr)_{2}- \frac{1}{2}\bigl(u^{0}, v^{0}\bigr)_{2}- \frac{1}{2}\bigl(L_{\beta, B_{0}, B }u^{\ast}, L_{\beta, B_{0}, B }v^{ \ast} \bigr)_{2} \\ &{}-\bigl( \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\bigl(u^{\ast}+L_{\beta, B_{0}, B }u^{\ast} \bigr),v^{\ast}+L_{ \beta, B_{0}, B }v^{\ast}\bigr)_{2}. \end{aligned}$$

By the definition of $L_{\beta, B_{0}, B }$, we have

$$\begin{aligned} L_{\beta, B_{0}, B }u^{\ast}=-P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\bigl(u^{\ast}+L_{ \beta, B_{0}, B }u^{\ast} \bigr),\quad \forall u^{\ast}\in L^{\ast}_{\beta, B_{0}}, \end{aligned}$$

and for any $u^{\ast}, v^{\ast}\in L^{\ast}_{\beta, B_{0}}$,

$$\begin{aligned} &\bigl(L_{\beta, B_{0}, B }u^{\ast}, L_{\beta, B_{0}, B }v^{\ast} \bigr)_{2} \\ &\quad=-\bigl( \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\bigl(u^{\ast}+L_{\beta, B_{0}, B }u^{\ast}\bigr), L_{\beta, B_{0}, B }v^{\ast}\bigr)_{2}. \end{aligned}$$

This yields

$$\begin{aligned} \tilde{q}_{\beta, B_{0}, B}\bigl(u^{\ast},v^{\ast} \bigr)={}&\frac{1}{2}\bigl(u^{+}, v^{+} \bigr)_{2}- \frac{1}{2}\bigl(u^{0}, v^{0} \bigr)_{2} \\ &{}-\frac{1}{2}\bigl( \vert A_{B_{0}} \vert ^{-\frac{1}{2}} \lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}u^{\ast}, v^{\ast} \bigr)_{2} \\ &{}-\frac{1}{2}\bigl( \vert A_{B_{0}} \vert ^{-\frac{1}{2}} \lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}} L_{\beta, B_{0}, B }u^{\ast}, v^{\ast} \bigr)_{2}. \end{aligned}$$

(14)

Remark 2

In the spirit, the saddle-point reduction process was developed by Amann and Zehnder [2], Chang [5], and Long [36], although technically our constructions of reduction are different.

Lemma 2

For any 2π-periodic continuous matrix-valued function B, there is a splitting

$$\begin{aligned} L_{\beta, B_{0}}^{\ast}=L_{\beta, B_{0}}^{+}(B)\oplus L_{\beta, B_{0}}^{0}(B) \oplus L_{\beta, B_{0}}^{-}(B) \end{aligned}$$

such that

(1) $L_{\beta, B_{0}}^{+}(B)$, $L_{\beta, B_{0}}^{0}(B)$, $L_{\beta, B_{0}}^{-}(B)$ are $\tilde{q}_{\beta, B_{0}, B}$-orthogonal, and $\tilde{q}_{\beta, B_{0}, B}$ is positive definite, negative definite on $L_{\beta, B_{0}}^{+}(B)$ and $L_{\beta, B_{0}}^{-}(B)$, respectively. Moreover, $\tilde{q}_{\beta, B_{0}, B}(u^{\ast}, u^{\ast})=0$, $\forall u^{\ast }\in L_{\beta, B_{0}}^{0}(B)$.

(2) $L_{\beta, B_{0}}^{0}(B)$, $L_{\beta, B_{0}}^{-}(B)$ are two finite dimensional subspaces.

Proof

Define the self-adjoint operator $\Lambda _{\beta, B_{0}, B }$ on $L^{\ast}_{\beta, B_{0}}$ as follows:

$$\begin{aligned} \Lambda _{\beta, B_{0}, B }u^{\ast}={}&2 P_{\beta, B_{0}}^{0}u^{\ast}+ \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}u^{\ast} \\ &{}+ \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}L_{\beta, B_{0}, B }u^{\ast}. \end{aligned}$$

Then, by (14), we have

$$\begin{aligned} \tilde{q}_{\beta, B_{0}, B}\bigl(u^{\ast},u^{\ast}\bigr)= \frac{1}{2}\bigl( (\mathrm{Id}- \Lambda _{\beta, B_{0}, B }) u^{\ast}, v^{\ast}\bigr)_{2},\quad \forall u^{ \ast}, v^{\ast} \in L^{\ast}_{\beta, B_{0}}, \end{aligned}$$

where Id denotes the identity map on $L_{\beta, B_{0}}^{\ast}$. Recalling that $|A_{B_{0}}|^{-\frac{1}{2}}$ is a compact operator, we have $\Lambda _{\beta, B_{0}, B }: L_{\beta, B_{0}}^{\ast}\rightarrow L_{ \beta, B_{0}}^{\ast}$ is self-adjoint and compact. Then there is a basis $\{ e_{j}\}\in L_{\beta, B_{0}}^{\ast}$ and a sequence $\mu _{j}\rightarrow 0$ in ${\mathbb{R}}$ such that

$$\begin{aligned} \Lambda _{\beta, B_{0}, B } e_{j}=\mu _{j} e_{j};\quad (e_{j}, e_{i})_{2}= \delta _{i, j }. \end{aligned}$$

(15)

Thus, for any $u^{\ast}\in L_{\beta, B_{0}}^{\ast}$, which can be expressed as $u^{\ast}=\sum _{j=1}^{\infty }c_{j} e_{j}$, we have

$$\begin{aligned} \tilde{q}_{\beta, B_{0}, B}\bigl(u^{\ast},u^{\ast}\bigr)&= \frac{1}{2} \bigl\Vert u^{ \ast} \bigr\Vert _{2}^{2}- \frac{1}{2}\bigl(\Lambda _{\beta, B_{0}, B } u^{\ast}, u^{ \ast} \bigr)_{2} \\ &=\frac{1}{2}\sum _{j=1}^{\infty }(1-\mu _{j})c_{j}^{2}. \end{aligned}$$

Since $\mu _{j}\rightarrow 0$, all the coefficients $(1-\mu _{j})$ are positive except a finite number. Thus our lemma follows by

$$\begin{aligned} &L_{\beta, B_{0}}^{+}(B)=\Biggl\{ \sum _{j=1}^{\infty }c_{j} e_{j} | c_{j}=0, \text{ if } 1-\mu _{j}\leq 0 \Biggr\} , \\ &L_{\beta, B_{0}}^{0}(B)=\Biggl\{ \sum _{j=1}^{\infty }c_{j} e_{j} | c_{j}=0, \text{ if } 1-\mu _{j}\neq 0 \Biggr\} , \\ &L_{\beta, B_{0}}^{-}(B)=\Biggl\{ \sum _{j=1}^{\infty }c_{j} e_{j} | c_{j}=0, \text{ if } 1-\mu _{j}\geq 0 \Biggr\} . \end{aligned}$$

 □

Definition 1

For any 2τ-periodic continuous matrix-valued function B, we define

$$\begin{aligned} \nu _{\beta, B_{0}}(B)=\dim L_{\beta, B_{0}}^{0}(B);\qquad i_{ \beta, B_{0}}(B)=\dim L_{\beta, B_{0}}^{-}(B). \end{aligned}$$

We call $i_{\beta, B_{0}}(B)$ and $\nu _{\beta, B_{0}}(B)$ the index and nullity of B, respectively.

Lemma 3

$\nu _{\beta, B_{0}}(B)=\dim \ker (A-B)$.

Proof

For any $u^{\ast}\in L_{\beta, B_{0}}^{0}(B)$, we have

$$\begin{aligned} \tilde{q}_{\beta, B_{0}, B}\bigl(u^{\ast}, v^{\ast}\bigr)=0,\quad \forall v^{ \ast}\in L_{\beta, B_{0}}^{\ast}. \end{aligned}$$

This implies

$$\begin{aligned} &\bigl(P_{\beta,B_{0}}^{+}-P_{\beta,B_{0}}^{0} \bigr)u^{\ast} \\ &\quad{}-\bigl(P_{\beta,B_{0}}^{+}+P_{\beta,B_{0}}^{0}\bigr) \vert A_{B_{0}} \vert ^{- \frac{1}{2}} \lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\bigl(u^{\ast}+L_{ \beta,B_{0},B}u^{\ast} \bigr)=0. \end{aligned}$$

(16)

By the definition of $L_{\beta,B_{0},B}$, we have

$$\begin{aligned} -L_{\beta,B_{0},B}u^{\ast}-P_{\beta,B_{0}}^{-} \vert A_{B_{0}} \vert ^{- \frac{1}{2}} \lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\bigl(u^{\ast}+L_{ \beta,B_{0},B}u^{\ast} \bigr)=0. \end{aligned}$$

(17)

Combining (16) and (17), we obtain

$$\begin{aligned} &\bigl(P_{\beta,B_{0}}^{+}-P_{\beta,B_{0}}^{0} \bigr)u^{\ast}-L_{\beta,B_{0},B}u^{ \ast} \\ &\quad{} - \vert A_{B_{0}} \vert ^{-\frac{1}{2}} \lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\bigl(u^{\ast}+L_{\beta,B_{0},B}u^{\ast} \bigr)=0. \end{aligned}$$

Let $x=|A_{B_{0}}|^{-\frac{1}{2}}(u^{+}- u^{0}- L_{\beta,B_{0},B}u^{\ast})$, where $u^{+}=P_{\beta, B_{0}}^{+}u^{\ast}$, $u^{0}=P_{\beta, B_{0}}^{0}u^{\ast}$. Then $x\in E_{B_{0}}$ and

$$\begin{aligned} Ax-Bx=0. \end{aligned}$$

Hence, $\dim L_{\beta, B_{0}}^{0}(B)=\dim \ker (A-B)$. □

Remark 3

From Lemma 3, we observe that $\nu _{\beta, B_{0}}(B)$ is independent of β and $B_{0}$. Thus, we will write $\nu (B)$ for convenience.

Recall that for a symmetric bilinear form φ defined on a Hilbert space X, its Morse index is defined as $m^{-}(\phi )=\max \{ \dim X_{1} | X_{1}\text{ is a subspace of }X \text{ such that } \phi (x,x)<0\text{ for any }x\in X_{1}\backslash \{0\} \}$.

Lemma 4

$i_{\beta, B_{0}}(B)$ is the Morse index of $\tilde{q}_{\beta, B_{0},B}$.

Proof

Let $E_{1}\subset L^{\ast}_{\beta, B_{0}}$ with $\dim E_{1}=k$ such that

$$\begin{aligned} \tilde{q}_{\beta, B_{0},B}\bigl(u^{\ast},u^{\ast} \bigr)< 0,\quad \forall u^{ \ast}\in E_{1}\backslash \{0\}. \end{aligned}$$

Let $\{e_{j}\}_{j=1}^{k}$ be linear independent in $E_{1}$. We have the following decomposition:

$$\begin{aligned} &e_{j}= e_{j}^{+} + e_{j}^{0} +e_{j}^{-},\quad j=1,\ldots, k. \\ &e_{j}^{+}\in L_{\beta, B_{0}}^{+}(B),\qquad e_{j}^{0}\in L_{\beta, B_{0}}^{0}(B),\qquad e_{j}^{-}\in L_{\beta, B_{0}}^{-}(B). \end{aligned}$$

We claim that $e_{1}^{-}, \ldots, e_{k}^{-}$ are linear independent. Arguing indirectly, we assume that there exist not all zero numbers $\alpha _{j}\in {\mathbb{R}}$ such that

$$\begin{aligned} \sum _{j=1}^{k}\alpha _{j}e_{j}^{-}=0. \end{aligned}$$

Denote $e=\sum _{j=1}^{k}\alpha _{j}e_{j}$. On the one hand, we have $e\in E_{1}$ and

$$\begin{aligned} \tilde{q}_{\beta, B_{0},B}(e,e)< 0. \end{aligned}$$

On the other hand, $e=\sum _{j=1}^{k}\alpha _{j}(e_{j}^{0}+e_{j}^{+})\in L_{\beta, B_{0}}^{0}(B) \oplus L_{\beta, B_{0}}^{+}(B)$, we have

$$\begin{aligned} \tilde{q}_{\beta, B_{0},B}(e,e)\geq 0. \end{aligned}$$

This is a contradiction. Thus, $e_{1}^{-}, \ldots, e_{k}^{-}$ are linear independent, which implies dim$L_{\beta, B_{0}}^{-}\geq k$ and $i_{\beta, B_{0}}(B)\geq m^{-}(\tilde{q}_{\beta, B_{0},B})$.

In addition, by the definition of $L_{\beta, B_{0}}^{-}$, we have

$$\begin{aligned} \tilde{q}_{\beta, B_{0},B}\bigl(u^{\ast}, u^{\ast}\bigr)< 0,\quad \forall u^{ \ast }\in L_{\beta, B_{0}}^{-} \backslash \{0\}. \end{aligned}$$

Thus, $m^{-}(\tilde{q}_{\beta, B_{0},B})\geq$ dim$L_{\beta, B_{0}}^{-}=i_{ \beta, B_{0}}(B)$. This completes the proof. □

For any $B_{1}< B_{2}$, let $B_{s}=(1-s)B_{1}+sB_{2}$, $s\in (0,1]$, and let $i_{\beta, B_{0}}(s)=i_{\beta, B_{0}}(B_{s})$, $\nu (s)=\nu (B_{s})$.

Lemma 5

For any $s_{0}\in [0,1)$, there exists $\delta >0$ such that

$$\begin{aligned} i_{\beta, B_{0}}(s_{0})+\nu (s_{0})\leq i_{\beta, B_{0}}(s), \quad\forall s\in (s_{0}, s_{0}+\delta ). \end{aligned}$$

Proof

Let $C_{B_{s}}=(P_{\beta, B_{0}}^{-} + P_{\beta, B_{0}}^{-} | A_{B_{0}}|^{- \frac{1}{2}} \lambda ^{2}(B_{s}-B_{0})| A_{B_{0}}|^{-\frac{1}{2}} P_{ \beta, B_{0}}^{-} )^{-1}$. By direct calculation, we have

$$\begin{aligned} C_{B_{s}}=C_{B_{s_{0}}}-(s-s_{0})C_{B_{s_{0}}}P_{\beta, B_{0}}^{-} \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\lambda ^{2}(B_{2}-B_{1}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}P_{ \beta, B_{0}}^{-}C_{B_{s_{0}}}+o(s-s_{0}) \end{aligned}$$

and

$$\begin{aligned} &\Lambda _{\beta, B_{0}, B_{s} }-\Lambda _{\beta, B_{0}, B_{s_{0}} } \\ &\quad=- (s-s_{0}) \bigl(P_{\beta, B_{0}}^{\ast} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}-P_{ \beta, B_{0}}^{\ast} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}B_{s_{0}} \vert A_{B_{0}} \vert ^{- \frac{1}{2}}C_{B_{s_{0}}}P_{\beta, B_{0}}^{-} \vert A_{B_{0}} \vert ^{- \frac{1}{2}} \bigr) \\ &\qquad{}\times \lambda ^{2}(B_{2}-B_{1}) \bigl(P_{\beta, B_{0}}^{\ast} \vert A_{B_{0}} \vert ^{- \frac{1}{2}}- \vert A_{B_{0}} \vert ^{-\frac{1}{2}}C_{B_{s_{0}}} \vert A_{B_{0}} \vert ^{- \frac{1}{2}}B_{s_{0}} P_{\beta, B_{0}}^{\ast} \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\bigr)+o(s-s_{0}), \end{aligned}$$

where $P_{\beta, B_{0}}^{\ast}=P_{\beta, B_{0}}^{+}+ P_{\beta, B_{0}}^{0}$. Thus, for any $u^{\ast}\in L_{\beta, B_{0}}^{\ast}(B_{s_{0}})$, there exists $\delta >0$ such that if $0< s-s_{0}<\delta$, we have

$$\begin{aligned} \tilde{q}_{\beta, B_{0},B_{s}}\bigl(u^{\ast},u^{\ast}\bigr)< 0. \end{aligned}$$

Note that $\dim L_{\beta, B_{0}}^{\ast}(B_{s_{0}})<+\infty$, $L_{\beta, B_{0}}^{\ast}(B_{s_{0}})$ is compact. It is easy to deduce that there exists $\delta _{0}$ independent of $u^{\ast}$ such that

$$\begin{aligned} \tilde{q}_{\beta, B_{0},B_{s}}\bigl(u^{\ast},u^{\ast}\bigr)< 0,\quad u^{\ast} \in L_{\beta, B_{0}}^{\ast}(B_{s_{0}}) \end{aligned}$$

for all $s\in (s_{0}, s_{0}+\delta )$. Thus, by Lemma 4, we have $i_{\beta, B_{0}}(s_{0})+\nu (s_{0})\leq i_{\beta, B_{0}}(s)$. □

Let $i_{\beta, B_{0}}(s+0)=\lim_{t\rightarrow s^{+}}i_{\beta, B_{0}}(t)$. We have

Lemma 6

$i_{\beta, B_{0}}(s+0)=i_{\beta, B_{0}}(s)+\nu (s)$.

Proof

From Lemma 5, we have $i_{\beta, B_{0}}(s+0)\geq i_{\beta, B_{0}}(s)+\nu (s)$. Thus, it suffices to prove $i_{\beta, B_{0}}(s+0)\leq i_{\beta, B_{0}}(s)+\nu (s)$. Let $i_{\beta, B_{0}}(s+0)=k$. Since $i_{\beta, B_{0}}$ is a finite integer, there exists $s'>s$ such that

$$\begin{aligned} i_{\beta, B_{0}}(p)=i_{\beta, B_{0}}(s+0),\qquad \nu (p)=0,\quad \forall p\in \bigl(s, s'\bigr). \end{aligned}$$

Similar to (15), for each p, there is a basis $\{ e_{p,j}\}_{j=1}^{k}\subset L_{\beta, B_{0}}^{-}(B_{p})$ such that

$$\begin{aligned} \Lambda _{\beta, B_{0}, B_{p} }e_{p,j}=\mu _{p,j} e_{p,j};\quad (e_{p,j}, e_{p,i})_{2}=\delta _{i,j}. \end{aligned}$$

(18)

Here, $1-\mu _{p,j}<0$. Since $\Lambda _{\beta, B_{0}, B_{p} }$ is bounded, we obtain that $\mu _{p,j}=(\Lambda _{\beta, B_{0}, B_{p} }e_{p,j},e_{p,j})_{2}$ is bounded. Then, for any j, there exists $\{p_{l}\}_{l}\subseteq (s,s')$ with $p_{l}\rightarrow s+0$ such that

$$\begin{aligned} \mu _{p_{l},j}\rightarrow \mu _{j}, \qquad e_{p_{l},j} \rightharpoonup e_{j} \quad\text{in }L^{2}\bigl(S^{1}, {\mathbb{R}}^{n}\bigr) \text{ as }l\rightarrow \infty. \end{aligned}$$

Recall that $1-\mu _{p_{l},j}<0$. We have $\{\frac{1}{\mu _{p_{l},j}}\}$ is bounded. Taking the limit in (18), we have

$$\begin{aligned} 1-\mu _{j}\leq 0, \quad\Lambda _{\beta, B_{0}, B_{s} }e_{j}=\mu _{j}e_{j}, \quad\forall j=1,2,\ldots,k. \end{aligned}$$

Moreover, for all $i, j=1,2,\ldots,k$,

$$\begin{aligned} e_{p_{l}, j}=\frac{1}{\mu _{p_{l},j}}\Lambda _{\beta, B_{0}, B_{p_{l}} }e_{p_{l}, j} \rightarrow \frac{1}{\mu _{j}}\Lambda _{\beta, B_{0}, B_{s} }e_{ j} = e_{ j} \quad\text{and}\quad (e_{j}, e_{i})_{2}= \delta _{i,j}. \end{aligned}$$

This means, by Definition 1, that $i_{\beta, B_{0}}(s)+\nu (s)\geq k=i_{\beta, B_{0}}(s+0)$. This completes the proof. □

By Lemma 5 and Lemma 6, we conclude that the index function $i_{\beta, B_{0}}(s)$ is integer-valued and nondecreasing on $[0, 1)$. Its value at any point s must be equal to the sum of the jumps it incurred in $[0, 1)$. Hence,

Lemma 7

For any $B_{1}< B_{2}$, we have

$$\begin{aligned} i_{\beta, B_{0}}(B_{2})-i_{\beta,B_{0}}(B_{1})=\sum _{s\in [0,1)} \nu \bigl(B_{1}+s(B_{2}-B_{1}) \bigr). \end{aligned}$$

By Lemma 7, we observe that the difference between $i_{\beta, B_{0}}(B_{1})$ and $i_{\beta, B_{0}}(B_{2})$ is independent of β and $B_{0}$. We define

Definition 2

For any 2π periodic continuous matrix-valued functions $B_{1}$ and $B_{2}$ with $B_{1}< B_{2}$, we define

$$\begin{aligned} I(B_{1}, B_{2})=\sum_{s\in [0,1)}\nu \bigl(B_{1}+s(B_{2}-B_{1})\bigr), \end{aligned}$$

and for any $B_{1}$, $B_{2}$ we define

$$\begin{aligned} I(B_{1}, B_{2})=I(B_{1}, K\mathrm{id})-I(B_{2}, K\mathrm{id}), \end{aligned}$$

where K is a constant and id is the identity map on $L_{B_{0}}^{\ast}$, satisfying $K\mathrm{id}>B_{1}$ and $K\mathrm{id}>B_{2}$. We call $I(B_{1}, B_{2})$ the relative Morse index between $B_{1}$ and $B_{2}$.

Remark 4

If we choose $k_{1}$, $k_{2}$ such that $k_{1}\mathrm{id}$, $k_{2}\mathrm{id}>B_{1}$ and $B_{2}$, we have

$$\begin{aligned} I(B_{2}, k_{1}\mathrm{id})-I(B_{1}, k_{1}\mathrm{id})=I(B_{2}, k_{2}\mathrm{id})-I(B_{1}, k_{2}\mathrm{id}). \end{aligned}$$

Thus, the relative Morse index $I(B_{1}, B_{2})$ depends only on $B_{1}$, $B_{2}$ and the operator A and is well defined.

Lemma 8

If $\nu (B)=0$ for some β large enough, then $( \tilde{q}_{\beta, B_{0}, B}(u^{\ast},u^{\ast}))^{\frac{1}{2}}$ and $(-\tilde{q}_{\beta,B_{0},B}(u^{\ast},u^{\ast}))^{\frac{1}{2}}$ are equivalent norms on $L^{+}_{\beta, B_{0}}(B)$ and $L^{-}_{\beta, B_{0}}(B)$, respectively.

Proof

It is sufficient to prove that, for sufficiently large β, there exists $\delta >0$ independent of β such that

$$\begin{aligned} (1-\delta, 1+\delta ) \cap \sigma (\Lambda _{\beta,B_{0}, B})= \emptyset. \end{aligned}$$

Arguing indirectly, there exist $\beta _{k}\rightarrow \infty$, $\mu _{k}\rightarrow 1$ such that

$$\begin{aligned} \Lambda _{\beta _{k}, B_{0}, B}e_{k}^{\ast}=\mu _{k} e_{k}^{\ast}, \end{aligned}$$

where $e_{k}^{\ast}\in L^{\ast}_{\beta _{k}, B_{0}}$ with $\| e_{k}^{\ast}\|_{2}=1$. This yields

$$\begin{aligned} &\bigl(2P_{\beta _{k}, B_{0}}^{0}+ \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}} \\ &\quad{}+ \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}L_{\beta _{k}, B_{0}, B} \bigr)e_{k}^{\ast}=\mu _{k} e_{k}^{ \ast}. \end{aligned}$$

(19)

Assume that $e_{k}^{\ast}=e_{k}^{+}+e_{k}^{0} \in L^{+}_{\beta _{k}, B_{0}}(B) \oplus L^{0}_{\beta _{k}, B_{0}}(B)$. Then, up to a subsequence,

$$\begin{aligned} e_{k}^{\ast}\rightharpoonup e^{\ast},\qquad e_{k}^{+}\rightharpoonup e^{+},\qquad e_{k}^{-}\rightharpoonup e^{-}. \end{aligned}$$

From (19) we have

$$\begin{aligned} \mu _{k} e_{k}^{+}={}&\bigl(P_{\beta _{k}, B_{0}}^{+} \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}} \\ &{}+ P_{\beta _{k}, B_{0}}^{+} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}L_{\beta _{k}, B_{0}, B}\bigr)e_{k}^{\ast} \end{aligned}$$

and

$$\begin{aligned} (\mu _{k}-2) e_{k}^{0}={}&\bigl(P_{\beta _{k}, B_{0}}^{0} \vert A_{B_{0}} \vert ^{- \frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}} \\ &{}+ P_{\beta _{k}, B_{0}}^{0} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{-\frac{1}{2}}L_{\beta _{k}, B_{0}, B}\bigr)e_{k}^{\ast}. \end{aligned}$$

Recall that $|A_{B_{0}}|^{-\frac{1}{2}}$ is compact and $\|L_{\beta _{k}, B_{0}, B}\|\rightarrow 0$ as $\beta _{k}\rightarrow \infty$. Taking the limit as $k\rightarrow \infty$, we have

$$\begin{aligned} &e^{+}=P^{+}_{B_{0}} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}e^{\ast}, \\ &e^{0}=P^{-}_{B_{0}} \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}e^{\ast}, \end{aligned}$$

and

$$\begin{aligned} e^{\ast}= \vert A_{B_{0}} \vert ^{-\frac{1}{2}}\lambda ^{2}(B-B_{0}) \vert A_{B_{0}} \vert ^{- \frac{1}{2}}e^{\ast}. \end{aligned}$$

This yields

$$\begin{aligned} \bigl(A-\lambda ^{2} B\bigr)e^{\ast}=0. \end{aligned}$$

This is a contradiction to $\nu (B)=0$. □

Note that, for any $x\in E_{B_{0}}=D(|A_{B_{0}}|^{\frac{1}{2}})$, we have $|A_{B_{0}}|^{\frac{1}{2}}x\in L^{2}(S^{1}, {\mathbb{R}}^{n})$. Thus $L^{2}(S^{1}, {\mathbb{R}}^{n})$ and $E_{B_{0}}$ are isomorphic. Define $E_{\beta,B_{0}}^{\ast}=E_{\beta,B_{0}}^{+} \oplus E_{\beta,B_{0}}^{0}$ and $\tilde{L}_{\beta, B_{0}, B}:E_{\beta,B_{0}}^{\ast }\rightarrow E_{ \beta,B_{0}}^{-}$

$$\begin{aligned} \tilde{L}_{\beta, B_{0}, B}x^{\ast}= \vert A_{B_{0}} \vert ^{-\frac{1}{2}}L_{ \beta, B_{0}, B} \vert A_{B_{0}} \vert ^{\frac{1}{2}}x^{\ast}. \end{aligned}$$

Definition 3

For any $x, y\in E_{B_{0}}$ and $x^{\ast}=P^{\ast}_{\beta, B_{0}}x$, $y^{\ast}=P^{\ast}_{\beta, B_{0}}y$, we define

• $Q_{\beta, B_{0}, B}(x,y)=q_{\beta, B_{0}, B}(|A_{B_{0}}|^{ \frac{1}{2}}x,|A_{\epsilon}|^{\frac{1}{2}}x) = \frac{1}{2}(x^{+}, y^{+})_{B_{0}} -\frac{1}{2}(x^{0}, y^{0})_{B_{0}} - \frac{1}{2}(x^{-}, y^{-})_{B_{0}} - \frac{1}{2}(\lambda ^{2}(B-B_{0}) x,y)_{2}$.

• For the reduced functional, we define

  $$\begin{aligned} \tilde{Q}_{\beta, B_{0}, B}\bigl(x^{\ast}, y^{\ast}\bigr) ={}&\tilde{q}_{\beta, B_{0}, B}\bigl( \vert A_{B_{0}} \vert ^{\frac{1}{2}}x^{\ast}, \vert A_{B_{0}} \vert ^{\frac{1}{2}}x^{ \ast}\bigr) \\ ={}& \frac{1}{2}\bigl(x^{+}, y^{+} \bigr)_{B_{0}} - \frac{1}{2}\bigl(x^{0}, y^{0} \bigr)_{B_{0}} \\ &{}-\frac{1}{2}\bigl(\lambda ^{2}(B-B_{0}) x^{\ast},y^{\ast}\bigr)_{2} - \frac{1}{2}\bigl( \lambda ^{2}(B-B_{0})\tilde{L}_{\beta, B_{0}, B} x^{\ast},y^{ \ast}\bigr)_{2}. \end{aligned}$$

  (20)

Also, by (13), for any $x^{\ast}\in E_{\beta, B_{0}}^{\ast}$, we have

$$\begin{aligned} Q_{\beta, B_{0}, B}\bigl(x^{\ast }+y, x^{\ast }+y \bigr) \leq \tilde{Q}_{\beta, B_{0}, B}\bigl(x^{\ast}, x^{\ast}\bigr)\quad \text{for any } y\in E_{\beta, B_{0}}^{-}. \end{aligned}$$

(21)

Applying Lemma 2, Lemma 7, and Lemma 8 to $E_{B_{0}}^{\ast}$ and $\tilde{Q}_{\beta,B}$, we conclude

Lemma 9

(1) The $E_{B_{0}}^{\ast}$ has the following decomposition:

$$\begin{aligned} E_{B_{0}}^{\ast}=E^{+}_{\beta, B_{0}}(B) \oplus E^{0}_{\beta, B_{0}}(B) \oplus E^{-}_{\beta, B_{0}}(B) \end{aligned}$$

such that $\tilde{Q}_{\beta,B}$ is positive definite, zero, and negative definite on $E^{+}_{\beta, B_{0}}(B)$, $E^{0}_{\beta, B_{0}}(B)$, and $E^{-}_{\beta, B_{0}}(B)$, respectively. Furthermore, $E^{0}_{\beta, B_{0}}(B)$ and $E^{-}_{\beta, B_{0}}(B)$ are finitely dimensional with

$$\begin{aligned} \nu (B)=\operatorname{dim } E^{0}_{\beta, B_{0}}(B),\qquad i_{\beta, B_{0}}(B)= \operatorname{dim } E^{-}_{\beta, B_{0}}(B). \end{aligned}$$

(2) For any $x^{\ast}\in E_{\beta, B_{0}}^{\ast}$, $({\bar{Q}}_{\beta,B}(x^{\ast},x^{\ast}))^{\frac{1}{2}}$ and $(-{\bar{Q}}_{\beta,B}(x^{\ast},x^{\ast}))^{\frac{1}{2}}$ are equivalent norms on $E^{+}_{\beta, B_{0}}(B)$ and $E^{-}_{\beta, B_{0}}(B)$.

(3) There exists $\epsilon _{0}>0$ such that, for any $\epsilon \in (0, \epsilon _{0}]$, we have

$$\begin{aligned} &\nu (B+\epsilon )=0=\nu (B-\epsilon ), \\ &i_{\beta, B_{0}}(B-\epsilon )=i_{\beta, B_{0}}(B) \\ &i_{\beta, B_{0}}(B+\epsilon )=i_{\beta, B_{0}}(B)+\nu (B). \end{aligned}$$

Proof

(1) and (2) of Lemma 9 come from Lemma 2 and Lemma 8 directly. By Lemma 7, (3) follows by the fact that the index function and the relative Morse index are all integer-valued. □

2.3 Critical point theorem

To prove Theorem 1 and Theorem 2, we use the following critical point theorems.

Let E be a real Hilbert space with $E=X\oplus Y$. A sequence $(z_{n})\subset E$ is said to be a $(PS)_{c}$-sequence if $\Phi (z_{n})\rightarrow c$ and $\Phi '(z_{n})\rightarrow 0$. Φ is said to satisfy the $(PS)_{c}$-condition if any $(PS)_{c}$-sequence has a convergent subsequence.

Theorem 3

([14, Theorem 2.5])

Let $e\in Y \backslash \{0\}$ and $\Omega =\{u=se+v: \|u\|< R, s>0, v\in X \}$. Suppose that

$(I_{1})$ $\Phi \in C^{1}(E, {\mathbb{R}})$ satisfies the $(PS)_{c}$-condition for any $c\in {\mathbb{R}}$;

$(I_{2})$ there is $r\in (0, R)$ such that $\rho:=\inf \Phi (X\cap \partial B_{r})>\omega:=\sup \Phi ( \partial \Omega )$, where ∂Ω refers to the boundary of Ω relative to span$\{e\}\oplus X$, and $B_{r}=\{u\in E: \|u\|< r \}$.

Then Φ has a critical value $c\geq \rho$ with

$$\begin{aligned} c=\inf_{h\in \Gamma}\sup_{u\in \Omega} \Phi \bigl(h(u) \bigr), \end{aligned}$$

where

$$\begin{aligned} \Gamma =\bigl\{ h\in C(E,E): h|_{\partial \Omega}=\mathrm{id}, \Phi \bigl(h(u)\bigr)\leq \Phi (u) \textit{ for } u\in \bar{\Omega} \bigr\} . \end{aligned}$$

Theorem 4

([14, Theorem 2.8])

Assume that ϕ is even and satisfies $(\Phi _{1})$. If

$(I_{3})$ there exists $r>0$ with $\inf \Phi (S_{r}Y) >\Phi (0)=0$, where $S_{r}=\partial B_{r}$;

$(I_{4})$ there exists a finite dimensional subspace $Y_{0}\subset Y$ and $R>r$ such that, for $E_{\ast}=X\oplus Y_{0}$, $M_{\ast}=\sup \Phi (E_{\ast})<+\infty$ and $\sigma:=\sup \Phi (E_{\ast}\backslash B_{R})<\rho$,

then Φ possesses at least m distinct pairs of critical points, where $m=\dim Y_{0}$.

Remark 5

(1) By using the abstract critical point theorems, we do not need to do the saddle point reduction procedure for the variational function $I_{\beta, B_{0}}$. Correspondingly, the nonlinearity f does not need to be $C^{1}$. For this kind of critical point theorems, we also refer to [3, 13].

(2) Recall that, in this paper, we only consider the nonresonance case ($\nu (A_{\infty})=0$). The PS-condition is sufficient to prove our theorems. Thus we use the PS-condition instead of the Cerami condition in the theorems.

3 Linking structure and proof of the main results

In view of condition $(F_{0})$, we assume $B_{0}=A_{0}-\epsilon$ for some $\epsilon >0$ small enough. By (3) of Lemma 9, we have $\nu (A_{0}-\epsilon )=\nu (A_{0})=0$. Define $E_{A_{0}-\epsilon}=D(|A_{A_{0}-\epsilon}|^{\frac{1}{2}})$ with the inner product $(\cdot, \cdot )_{A_{0}-\epsilon}$ and the norm $\|\cdot \|_{A_{0}-\epsilon}$. In what follows, we will write $(\cdot, \cdot )$ and $\|\cdot \|$ for short. Define the following projections:

$$\begin{aligned} P_{\beta,A_{0}-\epsilon}^{+}= \int _{0}^{+\infty}\,dF_{\lambda},\qquad P_{ \beta,A_{0}-\epsilon}^{0}= \int _{-\beta}^{0}\,dF_{\lambda},\qquad P_{ \beta,A_{0}-\epsilon}^{-}= \int _{-\infty}^{-\beta}\,dF_{\lambda}. \end{aligned}$$

Here $\{F_{\lambda}\}_{\lambda \in {\mathbb{R}}}$ denotes the spectral family of $A_{A_{0}-\epsilon}$. This deduces the decomposition on $E_{A_{0}-\epsilon}$:

$$\begin{aligned} E_{A_{0}-\epsilon}=E_{\beta,A_{0}-\epsilon}^{+}\oplus E_{\beta,A_{0}- \epsilon}^{0} \oplus E_{\beta,A_{0}-\epsilon}^{-},\quad x=x^{+} + x^{0} + x^{-}. \end{aligned}$$

(22)

Substituting $A_{0}-\epsilon$ for $B_{0}$ in (8), we define

$$\begin{aligned} I_{\beta, A_{0}-\epsilon}(x)={}&\frac{1}{2} \bigl\Vert x^{+} \bigr\Vert ^{2}-\frac{1}{2} \bigl\Vert x^{0} \bigr\Vert ^{2} - \frac{1}{2} \bigl\Vert x^{-} \bigr\Vert ^{2} \\ &{}-\lambda ^{2} \int _{S^{1}}F(x)\,dt + \frac{1}{2} \lambda ^{2} \bigl((A_{0}- \epsilon ) x,x \bigr)_{2}, \quad\forall x\in E_{A_{0}-\epsilon}. \end{aligned}$$

(23)

Applying Lemma 9 to $E_{A_{0}-\epsilon}^{\ast}=E_{\beta,A_{0}-\epsilon}^{+} \oplus E_{ \beta,A_{0}-\epsilon}^{0}$ for any B, there is a splitting

$$\begin{aligned} E_{A_{0}-\epsilon}^{\ast}=E_{A_{0}-\epsilon}^{+}(B) \oplus E_{A_{0}- \epsilon}^{0}(B) \oplus E_{A_{0}-\epsilon}^{-}(B). \end{aligned}$$

To apply the abstract Theorems 3 and 4 to $I_{\beta, A_{0}-\epsilon}$, we choose

$$\begin{aligned} X=E^{-}_{A_{0}-\epsilon}\oplus E^{0}_{A_{0}-\epsilon},\qquad Y=E^{+}_{A_{0}- \epsilon},\qquad Y_{0}= E_{A_{0}-\epsilon}^{-}(A_{\infty}+2 \epsilon ) \cap E^{+}_{A_{0}-\epsilon}. \end{aligned}$$

Since $Y_{0}\subseteq E_{A_{0}-\epsilon}^{-}(A_{\infty}+2\epsilon )$, we have $Y_{0}$ is a finite dimensional subspace with $\dim Y_{0}\leq i_{\beta, A_{0}-\epsilon}(A_{\infty}+2\epsilon )$. Moreover

Lemma 10

$\dim Y_{0} \geq I(A_{0}, A_{\infty})$.

Proof

By (20), for any B, we have

$$\begin{aligned} &\tilde{Q}_{\beta, A_{0}-\epsilon, B}\bigl(x^{\ast}, x^{\ast}\bigr) \\ &\quad = \frac{1}{2} \bigl\Vert x^{+} \bigr\Vert ^{2}- \frac{1}{2} \bigl\Vert x^{0} \bigr\Vert ^{2} \\ &\qquad{}-\frac{1}{2}\bigl(\lambda ^{2}\bigl(B-(A_{0}- \epsilon )\bigr) x^{\ast},x^{\ast}\bigr)_{2} - \frac{1}{2}\bigl(\lambda ^{2}\bigl(B-(A_{0}-\epsilon ) \bigr)\tilde{L}_{\beta, (A_{0}- \epsilon ), B} x^{\ast},x^{\ast} \bigr)_{2}. \end{aligned}$$

Then

$$\begin{aligned} \tilde{Q}_{\beta, A_{0}-\epsilon,A_{0}-\epsilon}\bigl(x^{\ast},x^{\ast}\bigr)= \frac{1}{2} \bigl\Vert x^{+} \bigr\Vert ^{2}- \frac{1}{2} \bigl\Vert x^{0} \bigr\Vert ^{2},\quad \forall x^{ \ast}\in E_{A_{0}-\epsilon}^{\ast}. \end{aligned}$$

It is easy to get $E^{+}_{\beta, A_{0}-\epsilon}(A_{0}-\epsilon )= E_{\beta, A_{0}- \epsilon}^{+}$, $E^{-}_{\beta, A_{0}-\epsilon}(A_{0}-\epsilon )= E_{\beta, A_{0}- \epsilon}^{0}$. Recall that $Y_{0}\subseteq E_{A_{0}-\epsilon}^{\ast}=E^{0}_{\beta, A_{0}- \epsilon}\oplus E^{+}_{\beta, A_{0}-\epsilon}$, we have

$$\begin{aligned} Y_{0}&= E_{A_{0}-\epsilon}^{-}(A_{\infty}+2\epsilon )- E^{0}_{A_{0}- \epsilon} \\ &=E_{A_{0}-\epsilon}^{-}(A_{\infty}+2\epsilon )- E^{-}_{\beta, A_{0}- \epsilon}(A_{0}-\epsilon ) \end{aligned}$$

and

$$\begin{aligned} \dim Y_{0}&\geq \dim E^{-}_{\beta, A_{0}-\epsilon}(A_{\infty}+2 \epsilon ) - \dim E^{-}_{\beta, A_{0}-\epsilon}(A_{0}-\epsilon ) \\ &= i_{\beta, A_{0}-\epsilon}(A_{\infty}+2\epsilon )-i_{\beta, A_{0}- \epsilon}( A_{0}-\epsilon ) \\ &=i_{\beta, A_{0}-\epsilon}(A_{\infty})-i_{\beta, A_{0}-\epsilon}( A_{0}) \\ &=I( A_{0}, A_{\infty }). \end{aligned}$$

 □

Now we are ready to verify the conditions of Theorems 3 and 4 for the functional $I_{\beta, A_{0}-\epsilon}$.

Lemma 11

Let $(F_{0})$ and $(F_{\infty})$ be satisfied. Then there exist $r>0$ and $\rho >0$ such that

$$\begin{aligned} I_{\beta, A_{0}-\epsilon}(x)\geq \rho,\quad \forall x\in \partial B_{r} \cap Y. \end{aligned}$$

Proof

Assume that $f_{1}(t, x)=f(t, x)-A_{0}(t)x(t)$. By $(F_{0})$, we have $f_{1}(t, x)=o(x)$ as $|x|\rightarrow 0$ and $f_{1}(t, x)-(A_{\infty}-A_{0})x=o(x)$ as $|x|\rightarrow \infty$. Then, for any $\epsilon >0$, there exist δ, $M_{1}$, $M_{2}>0$ such that

$$\begin{aligned} \bigl\vert f_{1}(t, x) \bigr\vert \leq \epsilon \vert x \vert , \quad\forall \vert x \vert < \delta, \quad\text{and}\quad \bigl\vert f_{1}(t, x) \bigr\vert \leq M_{2} \vert x \vert ,\quad \forall \vert x \vert >M_{1}. \end{aligned}$$

In conclusion, under $(F_{0})$ and $(F_{\infty})$, given $p>2$, for $\epsilon >0$, there is $C_{\epsilon}>0$ such that

$$\begin{aligned} \bigl\vert f_{1}(t, x) \bigr\vert \leq \epsilon \vert x \vert + C_{\epsilon } \vert x \vert ^{p-1} \end{aligned}$$

and

$$\begin{aligned} \bigl\vert F_{1}(t, x) \bigr\vert \leq \frac{1}{2}\epsilon \vert x \vert ^{2}+ \frac{C_{\epsilon}}{p} \vert x \vert ^{p}, \end{aligned}$$

where $F_{1}(t, x)=\int _{S^{1}}f_{1}(t, \theta x)\,d\theta x$. This yields

$$\begin{aligned} I_{\beta,A_{0}-\epsilon}(x)&=\frac{1}{2} \bigl\Vert x^{+} \bigr\Vert ^{2}-\frac{1}{2} \bigl\Vert x^{0} \bigr\Vert ^{2} - \frac{1}{2} \bigl\Vert x^{-} \bigr\Vert ^{2} -\lambda ^{2} \int _{S^{1}}F(t, x)\,dt + \frac{1}{2} \lambda ^{2}\bigl((A_{0}-\epsilon ) x,x \bigr)_{2} \\ & \geq \frac{1}{2} \bigl\Vert x^{+} \bigr\Vert ^{2}-\frac{1}{2} \bigl\Vert x^{0} \bigr\Vert ^{2} - \frac{1}{2} \bigl\Vert x^{-} \bigr\Vert ^{2}-C \Vert x \Vert _{p}^{p}, \quad\forall x\in E_{A_{0}- \epsilon}. \end{aligned}$$

Thus, for $x\in E_{A_{0}-\epsilon}^{+}$ with $\|x\|=r$, we have $x=x^{+}$ and

$$\begin{aligned} I_{A_{0}-\epsilon}(x) &\geq \frac{1}{2} \Vert x \Vert ^{2}-C \Vert x \Vert _{p}^{p} \\ &\geq \frac{1}{2} \Vert x \Vert ^{2}-C \Vert x \Vert ^{p} \\ &=\biggl(\frac{1}{2}-Cr^{p-2}\biggr)r^{2}. \end{aligned}$$

Thus, if we assume $\rho =\frac{1}{4}r^{2}$, then the lemma follows by choosing r small enough. □

Lemma 12

Let ($F_{0}$) and ($F_{\infty}$) be satisfied and $\epsilon >0$ be given by Lemma 11. We have

$$\begin{aligned} I_{\beta, A_{0}-\epsilon}(x)\rightarrow -\infty \quad\textit{as }x\in X \oplus Y_{0} \textit{ and } \Vert x \Vert \rightarrow \infty. \end{aligned}$$

Proof

Arguing indirectly, we assume that for some sequence $\{x_{j}\}\in X\oplus Y_{0}$ with $\| x_{j}\|\rightarrow \infty$ there is $a>0$ such that $I_{A_{0}-\epsilon}(x_{j})>-a$ for all j. Setting $w_{j}=x_{j}/ \| x_{j}\|$, we have $w_{j}\in X\oplus Y_{0}$ and $\| w_{j}\|=1$. And according to (22), we have the decomposition

$$\begin{aligned} w_{j}=w_{j}^{-}+ w_{j}^{0} +w_{j}^{+},\qquad w_{j}^{-}\in E^{-}_{A_{0}- \epsilon},\qquad w_{j}^{0}\in E^{0}_{A_{0}-\epsilon}, w_{j}^{+}\in Y_{0}. \end{aligned}$$

Since $Y_{0}$ is a finite dimensional subspace, we assume that $w_{j}\rightharpoonup w \in E_{A_{0}-\epsilon}$ and

$$\begin{aligned} w_{j}^{-}\rightharpoonup w^{-} \in E^{-}_{ A_{0}-\epsilon},\qquad w_{j}^{0} \rightarrow w^{0}\in E^{0}_{ A_{0}-\epsilon},\qquad w_{j}^{+} \rightarrow w^{+} \in Y_{0}. \end{aligned}$$

Assume that $f_{2}(t,x)=f(t, x)-A_{\infty}(t)x(t)$. Under $(F_{0})$ and $(F_{\infty})$, there is $M>0$ such that

$$\begin{aligned} \bigl\vert f_{2}(t, x) \bigr\vert \leq \epsilon \vert x \vert + M \end{aligned}$$

and

$$\begin{aligned} \bigl\vert F_{2}(t, x) \bigr\vert \leq \frac{1}{2}\epsilon \vert x \vert ^{2}+ M \vert x \vert \leq \epsilon \vert x \vert ^{2}+M_{1}, \end{aligned}$$

where $F_{2}(t, x)=\int _{S^{1}}f_{2}(t, \theta x)\,d\theta x$. Thus,

$$\begin{aligned} \frac{-a}{ \Vert x_{j} \Vert }\leq {}&\frac{I_{A_{0}-\epsilon}(x_{j})}{ \Vert x_{j} \Vert } \\ ={}& \frac{1}{2} \bigl\Vert w_{j}^{+} \bigr\Vert ^{2}-\frac{1}{2} \bigl\Vert w_{j}^{0} \bigr\Vert ^{2}- \frac{1}{2} \bigl\Vert w_{j}^{-} \bigr\Vert ^{2} \\ &{}-\frac{1}{2} \lambda ^{2}\bigl(\bigl(A_{\infty}-(A_{0}- \epsilon )\bigr) w_{j},w_{j} \bigr)_{2}-\lambda ^{2} \int _{S^{1}}\frac{F_{2}(t, x)}{ \Vert x_{j} \Vert ^{2}}\,dt \\ \leq{}& \frac{1}{2} \bigl\Vert w_{j}^{+} \bigr\Vert ^{2} -\frac{1}{2} \bigl\Vert w_{j}^{0} \bigr\Vert ^{2} - \frac{1}{2} \bigl\Vert w_{j}^{-} \bigr\Vert ^{2} \\ &{} - \frac{1}{2} \lambda ^{2}\bigl(\bigl(A_{\infty}+2 \epsilon -(A_{0}-\epsilon )\bigr) w_{j},w_{j} \bigr)_{2}+\lambda ^{2} \int _{S^{1}} \frac{M_{1}}{ \Vert x_{j} \Vert ^{2}}\,dt \\ ={}& Q_{\beta, A_{0}-\epsilon, A_{\infty}+2\epsilon}(w_{j}, w_{j})+o(1). \end{aligned}$$

(24)

Firstly, by the definition of $\tilde{Q}_{\beta,A_{0}-\epsilon, A_{\infty}+2\epsilon}$ and taking the limit in (24), we have

$$\begin{aligned} o(1)&\leq Q_{\beta, A_{0}-\epsilon, A_{\infty}+2\epsilon}(w_{j}, w_{j}) \\ &\leq \tilde{Q}_{\beta,A_{0}-\epsilon, A_{\infty}+2\epsilon}\bigl(w_{j}^{ \ast}, w_{j}^{\ast}\bigr) \\ &\rightarrow \tilde{Q}_{\beta, A_{0}-\epsilon, A_{\infty}+2 \epsilon}\bigl(w^{\ast}, w^{\ast} \bigr). \end{aligned}$$

Here, $w^{\ast}_{j}=w_{j}^{+}+w_{j}^{0}$ and $w^{\ast}=w^{+}+w^{0}$. Besides, by the definition of $\tilde{Q}_{\beta, A_{0}-\epsilon, A_{\infty}+2\epsilon}$, we have

$$\begin{aligned} &\tilde{Q}_{\beta, A_{0}-\epsilon, A_{\infty}+2\epsilon}\bigl(w^{0}, w^{0}\bigr)\\ &\quad=- \frac{1}{2} \bigl\Vert w^{0} \bigr\Vert ^{2}- \frac{1}{2} \bigl\Vert \tilde{L}_{\beta, A_{0}- \epsilon, A_{\infty}+2\epsilon}w^{0} \bigr\Vert ^{2} \\ &\qquad{}-\frac{1}{2}\lambda ^{2}\bigl((A_{\infty}-A_{0}-3 \epsilon ) \bigl(w^{0}+ \tilde{L}_{\beta, A_{0}-\epsilon, A_{\infty}+2\epsilon}w^{0} \bigr), w^{0}+ \tilde{L}_{\beta, A_{0}-\epsilon, A_{\infty}+2\epsilon}w^{0}\bigr) \end{aligned}$$

Since $A_{\infty }>A_{0}$ and $w^{0}\neq 0$, we have $\tilde{Q}_{\beta, A_{0}-\epsilon, A_{\infty}+2\epsilon}(w^{0}, w^{0})<0$. Thus,

$$\begin{aligned} E^{0}_{\beta, A_{0}-\epsilon}\subset E^{-}_{\beta, A_{0}-\epsilon}(A_{ \infty}+2 \epsilon ), \end{aligned}$$

and $w^{\ast}\in E^{-}_{\beta, A_{0}-\epsilon}(A_{\infty}+2\epsilon )$. Moreover, since $\nu (A_{\infty})=0$, we have $\nu (A_{\infty}+2\epsilon )=0$ for ϵ small enough. By (2) of Lemma 9, we have $\sqrt{- \tilde{Q}_{A_{0}-\epsilon, A_{\infty}+2\epsilon}(w^{\ast}, w^{ \ast})}$ is an equivalent norm of $w^{\ast}$ on $E^{-}_{A_{0}-\epsilon}(A_{\infty}+2\epsilon )$. Thus, $w_{j}^{\ast }\rightarrow w^{\ast}=0$.

Secondly, define the following projections:

$$\begin{aligned} P_{1}= \int _{-\infty}^{-\beta}\,dF_{\lambda},\qquad P_{2}= \int _{-\beta}^{ \beta}\,dF_{\lambda},\qquad P_{3}= \int _{\beta}^{+\infty}\,dF_{\lambda}. \end{aligned}$$

We have $P_{1}E_{A_{0}-\epsilon}=E^{-}_{A_{0}-\epsilon}$, $(P_{2}+P_{3})E_{A_{0}-\epsilon}=E^{\ast}_{A_{0}-\epsilon}$, and $P_{2}E_{A_{0}-\epsilon}$ is a finite dimensional subspace. Recalling that $w_{j}^{0}\rightarrow 0$ and $w_{j}^{+}\rightarrow 0$, we obtain

$$\begin{aligned} P_{2}w_{j}\rightarrow 0,\qquad P_{3}w_{j} \rightarrow 0. \end{aligned}$$

Moreover, by the definition of $P_{1}$ and $P_{3}$, we have

$$\begin{aligned} \Vert P_{1}x \Vert _{2}\leq \frac{1}{\sqrt{\beta}} \Vert P_{1}x \Vert , \qquad \Vert P_{3}x \Vert _{2} \leq \frac{1}{\sqrt{\beta}} \Vert P_{3}x \Vert ,\quad \forall x\in E_{A_{0}- \epsilon}. \end{aligned}$$

Suppose that $\lambda \|A_{0}-A_{\infty}+\epsilon \|< M_{3}<\beta$. By (24), we obtain

$$\begin{aligned} o(1)&\leq \frac{1}{2} \bigl\Vert w_{j}^{+} \bigr\Vert ^{2} - \frac{1}{2} \bigl\Vert w_{j}^{-} \bigr\Vert ^{2} - \frac{1}{2} \lambda ^{2}\bigl( \bigl(A_{\infty}+2\epsilon -(A_{0}-\epsilon )\bigr) w_{j},w_{j} \bigr)_{2} \\ & =o(1) - \frac{1}{2} \bigl\Vert w_{j}^{-} \bigr\Vert ^{2}- \frac{M_{3}}{2} \Vert w_{j} \Vert _{2}^{2} \\ &\leq o(1)-\frac{1}{2}\biggl(1-\frac{M_{3}}{\beta} \biggr) \Vert P_{1}w_{j} \Vert ^{2}. \end{aligned}$$

This implies that $P_{1}w_{j}\rightarrow 0$ in $E_{A_{0}-\epsilon}$. Hence, $w_{j}=(P_{1}+P_{2}+P_{3})w_{j}\rightarrow 0$. This is a contradiction to $\|w_{j}\|=1$. □

Lemma 13

Let $(F_{0})$ and $(F_{\infty})$ be satisfied. Then any $(PS)_{c}$-sequence of $I_{\beta, A_{0}-\epsilon}$ is bounded.

Proof

Let $\{x_{j}\}\subset E$ be such that $I_{\beta, A_{0}-\epsilon}(x_{j})\rightarrow c$ and $I_{\beta, A_{0}-\epsilon}'(x_{j})\rightarrow 0$. To prove that $\{x_{j}\}$ is bounded, we develop a contradiction argument. We assume that, up to a subsequence, $\| x_{j}\|\rightarrow \infty$ and set $v_{j}=\frac{ x_{j}}{\| x_{j}\|}$. Then $\|v_{j}\|=1$. Without loss of generality, we assume that

$$\begin{aligned} v_{j} \rightharpoonup v \quad\text{in } E_{A_{0}-\epsilon}\quad \text{and}\quad v_{j} \rightarrow v \quad\text{in } L^{2}\bigl(S^{1}, {\mathbb{R}}^{n}\bigr). \end{aligned}$$

Then, up to a subsequence, $v_{j}(t)\rightarrow v(t)$ a.e. on $S^{1}$. Since, by ($F_{\infty}$), $|f_{2}(t, x)|=|f(t, x)-A_{\infty}(t) x|=o(|x|)$ as $|x|\rightarrow \infty$ and $x_{j}(t)\rightarrow \infty$ if $v(t)\neq 0$, it is easy to see that

$$\begin{aligned} \int _{S^{1}}\frac{f(t, x_{j})\varphi (t)}{ \Vert x_{j} \Vert }\rightarrow \int _{S^{1}}A_{\infty }(t)v \varphi \end{aligned}$$

for all $\varphi \in L^{2}(S^{1}, {\mathbb{R}}^{n})$. From this we deduce that

$$\begin{aligned} A_{A_{0}-\epsilon}v-A_{\infty }v+(A_{0}-\epsilon )v=0 \end{aligned}$$

and

$$\begin{aligned} \ddot{v}(t+\tau )- A_{\infty }v(t)=0. \end{aligned}$$

Since $i(A_{\infty})=0$, we have $v=0$. Thus, $v_{j}\rightharpoonup 0$ in $E_{A_{0}-\epsilon}$ and $v_{j}\rightarrow 0$ in $L^{2}(S^{1}, {\mathbb{R}}^{n})$. Note that

$$\begin{aligned} o(1)&=\frac{I_{\beta,A_{0}-\epsilon}'(x_{j})}{ \Vert x_{j} \Vert } \bigl(v_{j}^{+}-v_{j}^{0}-v_{j}^{-} \bigr) \\ &=1- \int _{S^{1}}\frac{f(t, x_{j})}{ \vert x_{j} \vert } \vert v_{j} \vert \bigl(v_{j}^{+}-v_{j}^{0}-v_{j}^{-} \bigr)+ \bigl((A_{0}-\epsilon )v_{j},v_{j}^{+}-v_{j}^{0}-v_{j}^{-} \bigr)_{2} \\ & \geq 1-C \int _{S^{1}} \vert v_{j} \vert \bigl\vert v_{j}^{+}-v_{j}^{0}-v_{j}^{-} \bigr\vert \\ & \geq 1-C \Vert v_{j} \Vert _{2}^{2}=1-o(1). \end{aligned}$$

This is a contradiction. □

Lemma 14

Any $(PS)_{c}$-sequence of $I_{A_{0}-\epsilon}$ has a convergence subsequence.

Proof

By Lemma 13, the $(PS)_{c}$-sequence is bounded. Assume that $v_{j}\rightharpoonup v$ and let $w_{j}=v_{j}-v$. Then we have $w_{j}^{+}\rightharpoonup 0$ in $E_{A_{0}-\epsilon}$ and $w_{j}^{+}\rightarrow 0$ in $L^{2}(S^{1}, {\mathbb{R}}^{n})$. To establish strong convergence, it suffices to prove that $\|w_{j}\|\rightarrow 0$. Similar to Lemma 13, the proof follows from the procedure in Lemma 13 to $\frac{I_{\beta, A_{0}-\epsilon}'(x_{j})}{\|x_{j}\|} (w_{j}^{+}-w_{j}^{0}-w_{j}^{-})$. □

Proof of Theorems 1 and 2

(Existence) With $X=E^{-}_{A_{0}-\epsilon}\oplus E^{0}_{A_{0}-\epsilon}, Y=E^{+}_{A_{0}- \epsilon}, Y_{0}= E_{A_{0}-\epsilon}^{-}(A_{\infty}+2\epsilon ) \cap E^{+}_{A_{0}-\epsilon}$, the condition $(I_{2})$ of Theorem 3 holds by Lemma 12. Lemma 14 shows that $I_{\beta, A_{0}-\epsilon}$ satisfies the $(PS)_{c}$-condition and $(I_{1})$ of Theorem 3 is verified. Therefore, $I_{\beta, A_{0}-\epsilon}$ has at least one critical point by Theorem 3.

(Multiplicity) Lemma 11 implies that $I_{\beta, A_{0}-\epsilon}$ satisfies $(I_{3})$ of Theorem 4. Lemma 10 and Lemma 12 say that $I_{\beta, A_{0}-\epsilon}$ satisfies $(I_{4})$ of Theorem 4 with dim$Y_{0}\geq I(A_{0}, A_{\infty})$. Moreover, $I_{\beta, A_{0}-\epsilon}$ is even if f is odd. Therefore, $I_{\beta, A_{0}-\epsilon}$ has at least $I(A_{0}, A_{\infty})$ pairs of nontrivial critical points by Theorem 4. □