Multiple periodic solutions of high order differential delay equations with $2k-1$ lags

Abstract

In this paper, we study the periodic solutions of high order differential delay equations with $2k-1$ lags. The 4k-periodic solutions are obtained by using the variational method and the method of Kaplan–Yorke coupling system. These are new types of differential delay equations compared with all previous research. And it provides a precise counting method for the number of periodic solutions. Two examples are given to demonstrate our main results.

1 Introduction

Given $f\in C^{0}(R,R)$ with $f(-x)=-f(x)$, $xf(x)>0$, $x\neq 0$. Kaplan and Yorke [13] studied the existence of 4-periodic and 6-periodic solutions to the differential delay equations

$$ x'(t)=-f\bigl(x(t-1)\bigr) $$

(1)

and

$$\begin{aligned} x'(t)=-f\bigl(x(t-1)\bigr)-f\bigl(x(t-2)\bigr), \end{aligned}$$

(2)

respectively. The method they applied is transforming the two equations into adequate ordinary differential equations by regarding the retarded functions $x(t-1)$ and $x(t-2)$ as independent variables. They guessed that the existence of $2(n+1)$-periodic solution to the equation

$$\begin{aligned} x'(t)=-\sum_{i=1}^{n}f \bigl(x(t-i)\bigr) \end{aligned}$$

(3)

could be studied under the restriction

$$ x\bigl(t-(n+1)\bigr)=-x(t), $$

which was proved by Nussbaum [20] in 1978 by the use of a fixed point theorem on cones.

After that a lot of papers [3,4,5,6,7,8,9,10,11,12, 14,15,16,17,18] discussed the existence and multiplicity of $2(n+1)$-periodic solutions to Eq. (3) and its extension

$$\begin{aligned} x'(t)=-\sum_{i=1}^{n} \nabla F\bigl(x(t-i)\bigr), \end{aligned}$$

(4)

where $F\in C^{1}(R^{N},R)$, $\nabla F(-x)=-\nabla F(x)$, $F(0)=0$. But they all studied first order equations.

In this paper, we study the periodic orbits to two types of high order differential delay equations with $2k-1$ lags in the form

$$\begin{aligned} x^{(2s+1)}(t)=-\sum_{i=1}^{2k-1}f \bigl(x(t-i)\bigr), \end{aligned}$$

(5)

and

$$\begin{aligned} x^{(2s+1)}(t)=-\sum_{i=1}^{2k-1}(-1)^{i+1}f \bigl(x(t-i)\bigr), \end{aligned}$$

(6)

which are different from (3) and can be regarded as a new extension of (3). The method applied in this paper is the variational approach in the critical point theory [1, 2, 19].

We suppose that

$$\begin{aligned} f\in C^{0}(R,R),\quad f(-x)=-f(x) \end{aligned}$$

(7)

and there are $\alpha ,\beta \in R$ such that

$$\begin{aligned} \lim_{x \to 0} \frac{f(x)}{x}= \alpha , \qquad \lim_{x \to \infty } \frac{f(x)}{x}= \beta . \end{aligned}$$

(8)

Let $F(x)=\int_{0}^{x}f(s)\,ds$. Then $F(-x)=F(x)$ and $F(0)=0$. For convenience, we make the following assumptions:

$(S_{1})$ :

f satisfies (7) and (8),

$(S_{2})$ :

there exist $M>0$ and a function $r\in C^{0}(R^{+},R^{+})$ satisfying $r(s)\rightarrow \infty$, $r(s)/s\rightarrow 0$ as $s\rightarrow \infty$ such that

$$ \biggl\vert F(x)-\frac{1}{2}\beta x^{2} \biggr\vert >r \bigl( \vert x \vert \bigr)-M, $$

$(S_{3}^{\pm })$ :

$\pm [F(x)-\frac{1}{2}\beta x^{2}]>0$, $|x|\rightarrow \infty$,

$(S_{4}^{\pm })$ :

$\pm [F(x)-\frac{1}{2}\alpha x^{2}]>0$, $0<|x|\ll 1$.

In this paper, we need the following lemma as the base of our discussion.

Let X be a Hilbert space, $L:X\rightarrow X$ be a linear operator, and $\varPhi :X\rightarrow R$ be a differentiable functional.

Lemma 1.1

([3], Lemma 2.4)

Assume that there are two closed $s^{1}$-invariant linear subspaces, $X^{+}$ and $X^{-}$, and $r>0$ such that

• (a) $X^{+}\cup X^{-}$ is closed and of finite codimensions in X,

• (b) $\widehat{L}(X^{-})\subset X^{-}$, $\widehat{L}=L+P^{-1}A_{0}$ or $\widehat{L}=L+P^{-1}A_{\infty }$,

• (c) there exists $c_{0}\in R$ such that

  $$ \inf_{x\in X^{+}}\varPhi (x)\geq c_{0}, $$

• (d) there is $c_{\infty }\in R$ such that

  $$\varPhi (x)\leq c_{\infty }< \varPhi (0)=0,\quad \forall x \in X^{-} \cap S_{r}=\bigl\{ x\in X^{-}: \Vert x \Vert =r\bigr\} , $$

• (e) Φ satisfies the $(P.S)_{c}$-condition, $c_{0}< c< c_{\infty }$, i.e., every sequence $\{x_{n}\}\subseteq X$ with $\varPhi (x_{n})\rightarrow c$ and $\varPhi '(x_{n})\rightarrow 0$ possesses a convergent subsequence.

Then Φ has at least $\frac{1}{2}[\dim (X^{+}\cap X^{-})- \operatorname{codim}_{X}(X^{+}\cup X^{-})]$ generally different critical orbits in $\varPhi^{-1}([c_{0},c_{\infty }])$ if $[\dim (X^{+}\cap X^{-})- \operatorname{codim}_{X}(X^{+}\cup X^{-})]>0$.

Definition 1.2

We say that Φ satisfies the $(P.S)$-condition if every sequence $\{x_{n}\}$ with $\varPhi (x_{n})$ is bounded and $\varPhi '(x_{n})\rightarrow 0$ possesses a convergent subsequence.

Remark 1.3

We may use the $(P.S)$-condition to replace condition (e) in Lemma 1.1 since the $(P.S)$-condition implies that the $(P.S)_{c}$-condition holds for each $c\in R$.

2 Space X, functional Φ, and its differential $\varPhi '$

We are concerned with the 4k-periodic solutions to (5) and (6) and suppose

$$\begin{aligned} x(t-2k)=-x(t),\quad k\geq 1. \end{aligned}$$

(9)

Let

$$\begin{aligned} &\widehat{X}=\bigl\{ x\in L^{2}:x(t-2k)=-x(t)\bigr\} \\ &\hphantom{\widehat{X}}= \Biggl\{ \sum_{i=0}^{\infty } \biggl(a_{i}\cos \frac{(2i+1)\pi t}{2k}+b_{i}\sin \frac{(2i+1)\pi t}{2k}\biggr):a _{i},b_{i}\in R \Biggr\} , \\ &X=\operatorname{cl} \Biggl\{ \sum_{i=0}^{\infty } \biggl(a_{i}\cos \frac{(2i+1) \pi t}{2k}+b _{i}\sin \frac{(2i+1)\pi t}{2k}\biggr):a_{i},b_{i}\in R,\\ &\hphantom{X=}\sum _{i=0}^{\infty }(2i+1) \bigl(a_{i}^{2}+b_{i}^{2} \bigr)< \infty \Biggr\} \subset \widehat{X}, \end{aligned}$$

and define $P:X\rightarrow L^{2}$ by

$$\begin{aligned} \mathrm{Px}(t)&= P\Biggl(\sum_{i=0}^{\infty } \biggl(a_{i}\cos \frac{(2i+1)\pi t}{2k}+b_{i} \sin \frac{(2i+1)\pi t}{2k}\biggr)\Biggr) \\ &= \sum_{i=0}^{\infty }(2i+1)^{2s+1} \biggl(a_{i}\cos \frac{(2i+1)\pi t}{2k}+b _{i}\sin \frac{(2i+1)\pi t}{2k}\biggr). \end{aligned}$$

(10)

Let

$$ P^{-1}x(t)=\sum_{i=0}^{\infty } \frac{1}{(2i+1)^{2s+1}}\biggl(a_{i}\cos \frac{(2i+1) \pi t}{2k}+b_{i} \sin \frac{(2i+1)\pi t}{2k}\biggr). $$

Then the inverse $P^{-1}$ of P exists. For $x\in X$, define

$$\begin{aligned} &\langle x,y\rangle = \int_{0}^{4k}\bigl(\mathrm{Px}(t),y(t)\bigr)\,dt, \quad \Vert x \Vert =\sqrt{ \langle x,x\rangle }, \\ &\langle x,y\rangle_{2}= \int_{0}^{4k}\bigl(x(t),y(t)\bigr)\,dt,\quad \Vert x \Vert _{2}=\sqrt{ \langle x,x\rangle_{2}}. \end{aligned}$$

Therefore $(X,\|\cdot \|)$ is an $H^{\frac{1}{2}}$ space.

For (5), define a functional $\varPhi :X\rightarrow R$ by

$$\begin{aligned} \varPhi (x)=\frac{1}{2}\langle \mathit{Lx},x\rangle + \int_{0}^{4k}F\bigl(x(t)\bigr)\,dt, \end{aligned}$$

(11)

where

$$\begin{aligned} \mathit{Lx}=-P^{-1}\sum_{i=1}^{2k-1}(-1)^{i+1}x^{(2s+1)}(t-i). \end{aligned}$$

(12)

Let $m=k-1$ and

$$ X(i)=\biggl\{ x(t)=a_{i}\cos \frac{(2i+1)\pi t}{2k}+b_{i}\sin \frac{(2i+1) \pi t}{2k}:a_{i},b_{i}\in R\biggr\} . $$

We have

$$\begin{aligned} X=\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m} \bigl(X(2lk+i)+X(2lk+2k-i-1) \bigr) \Biggr]. \end{aligned}$$

(13)

If $x_{i}(t)=a_{i}\cos \frac{(2i+1)\pi t}{2k}+b_{i}\sin \frac{(2i+1) \pi t}{2k}\in X(i)$, $i\in N$, we have

$$\begin{aligned} \mathit{Lx}=(-1)^{s+1} \biggl(\frac{\pi }{2k} \biggr)^{2s+1} \Biggl(\sum_{i=0} ^{\infty }x_{i}\tan \frac{(2i+1)\pi }{4k} \Biggr). \end{aligned}$$

(14)

Obviously, $L|_{X(i)}:X(i)\rightarrow X(i)$ is invertible.

Based on the theorem given by Mawhin and Willem [19, Theorem 1.4] the differential of functional Φ is differentiable, and its differential is

$$\begin{aligned} P^{-1}\varPhi '(x)=\mathit{Lx}+K(x), \end{aligned}$$

(15)

where $K(x)=P^{-1}f(x)$. It is easy to prove that $K:(X,\|x\|^{2}) \rightarrow (X,\|x\|_{2}^{2})$ is compact.

Therefore, from (14) we have that if

$$ x(t)=\sum_{i=0}^{\infty }\biggl(a_{i} \cos \frac{(2i+1)\pi t}{2k}+b_{i} \sin \frac{(2i+1)\pi t}{2k}\biggr), $$

then

$$\begin{aligned} \langle \mathit{Lx},x\rangle =&(-1)^{s+1}\sum _{i=0}^{\infty } \biggl(\frac{ \pi }{2k} \biggr)^{2s+1} 2k(2i+1)^{2s+1}\bigl(a_{i}^{2}+b_{i}^{2} \bigr)\tan \frac{(2i+1) \pi }{4k} \\ =&\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m}(-1)^{s+1} \biggl( \frac{ \pi }{2k} \biggr)^{2s+1} 2k(4lk+2i+1)^{2s+1} \bigl(a_{2lk+i}^{2} +b_{2lk+i} ^{2}\bigr)\tan \frac{(2i+1)\pi }{4k} \\ &{}-\sum_{i=0}^{m}(-1)^{s+1} \biggl(\frac{\pi }{2k} \biggr)^{2s+1} 2k(4lk+4k-2i-1)^{2s+1}\\ &{}\times \bigl(a _{2lk+2k-i-1}^{2} +b_{2lk+2k-i-1}^{2}\bigr)\tan \frac{(2i+1) \pi }{4k} \Biggr]. \end{aligned}$$

On the other hand,

$$\begin{aligned} &\begin{aligned} \bigl\langle P^{-1}\beta x,x\bigr\rangle &=\sum _{i=0}^{\infty }2k\beta \bigl(a_{i} ^{2}+b_{i}^{2}\bigr) \\ &=\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m}2k\beta \bigl(a_{2lk+i}^{2}+b _{2lk+i}^{2}\bigr)+\sum_{i=0}^{m}2k \beta \bigl(a_{2lk+2k-i-1}^{2}+b_{2lk+2k-i-1} ^{2}\bigr) \Biggr], \end{aligned}\\ &\begin{aligned} \bigl\langle P^{-1}\alpha x,x\bigr\rangle &=\sum _{i=0}^{\infty }2k\alpha \bigl(a _{i}^{2}+b_{i}^{2} \bigr) \\ &=\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m}2k\alpha \bigl(a_{2lk+i}^{2}+b _{2lk+i}^{2}\bigr)+\sum_{i=0}^{m}2k \alpha \bigl(a_{2lk+2k-i-1}^{2}+b_{2lk+2k-i-1} ^{2} \bigr) \Biggr]. \end{aligned} \end{aligned}$$

Therefore, we have

$$\begin{aligned} &\bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x\bigr\rangle \\ &\quad =2k\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m} \biggl((-1)^{s+1} \biggl(\frac{ \pi }{2k} \biggr)^{2s+1}(4lk+2i+1)^{2s+1} \tan \frac{(2i+1)\pi }{4k} + \beta \biggr)\bigl(a_{2lk+i}^{2}+b_{2lk+i}^{2} \bigr) \\ &\qquad {}+\sum_{i=0}^{m} \biggl(-(-1)^{s+1} \biggl(\frac{\pi }{2k} \biggr)^{2s+1}(4lk+4k-2i-1)^{2s+1} \tan \frac{(2i+1)\pi }{4k} +\beta \biggr)\\ &\qquad {}\times \bigl(a_{2lk+2k-i-1}^{2} +b_{2lk+2k-i-1} ^{2}\bigr) \Biggr], \\ &\bigl\langle \bigl(L+P^{-1}\alpha \bigr) x,x\bigr\rangle \\ &\quad =2k\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m} \biggl((-1)^{s+1} \biggl(\frac{ \pi }{2k} \biggr)^{2s+1}(4lk+2i+1)^{2s+1} \tan \frac{(2i+1)\pi }{4k} + \alpha \biggr) \bigl(a_{2lk+i}^{2}+b_{2lk+i} ^{2}\bigr) \\ &\qquad {}+\sum_{i=0}^{m} \biggl(-(-1)^{s+1} \biggl(\frac{\pi }{2k} \biggr)^{2s+1}(4lk+4k-2i-1)^{2s+1} \tan \frac{(2i+1)\pi }{4k} +\alpha \biggr)\\ &\qquad {}\times \bigl(a_{2lk+2k-i-1}^{2} +b_{2lk+2k-i-1} ^{2}\bigr) \Biggr]. \end{aligned}$$

For (6), define the functional $\varPsi :X\rightarrow R$ by

$$\begin{aligned} \varPsi (x)=\frac{1}{2}\langle \mathit{Lx},x\rangle + \int_{0}^{4k}F\bigl(x(t)\bigr)\,dt, \end{aligned}$$

(16)

where

$$\begin{aligned} \mathit{Lx}=-P^{-1}\sum_{i=1}^{2k-1}x^{(2s+1)}(t-i). \end{aligned}$$

(17)

Let $m=k-1$ and

$$ X(i)=\biggl\{ x(t)=a_{i}\cos \frac{(2i+1)\pi t}{2k}+b_{i}\sin \frac{(2i+1) \pi t}{2k}:a_{i},b_{i}\in R\biggr\} . $$

We have

$$\begin{aligned} X=\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m} \bigl(X(2lk+i)+X(2lk+2k-i-1) \bigr) \Biggr]. \end{aligned}$$

(18)

If $x_{i}(t)=a_{i}\cos \frac{(2i+1)\pi t}{2k}+b_{i}\sin \frac{(2i+1) \pi t}{2k}\in X(i)$, $i\in N$, we have

$$\begin{aligned} \mathit{Lx}=(-1)^{s+1} \biggl(\frac{\pi }{2k} \biggr)^{2s+1} \Biggl(\sum_{i=0} ^{\infty }x_{i}\cot \frac{(2i+1)\pi }{4k} \Biggr). \end{aligned}$$

(19)

Obviously, $L|_{X(i)}:X(i)\rightarrow X(i)$ is invertible.

Based on the theorem given by Mawhin and Willem [16, Theorem 1.4] the differential of functional Φ is differentiable, and its differential is

$$\begin{aligned} P^{-1}\varPsi '(x)=\mathit{Lx}+K(x), \end{aligned}$$

(20)

where $K(x)=P^{-1}f(x)$. It is easy to prove that $K:(X,\|x\|^{2}) \rightarrow (X,\|x\|_{2}^{2})$ is compact.

Therefore, from (14) we have that if

$$ x(t)=\sum_{i=0}^{\infty }\biggl(a_{i} \cos \frac{(2i+1)\pi t}{2k}+b_{i} \sin \frac{(2i+1)\pi t}{2k}\biggr), $$

then

$$\begin{aligned} \langle \mathit{Lx},x\rangle =&(-1)^{s+1}\sum _{i=0}^{\infty } \biggl(\frac{ \pi }{2k} \biggr)^{2s+1} 2k(2i+1)^{2s+1}\bigl(a_{i}^{2}+b_{i}^{2} \bigr)\cot \frac{(2i+1) \pi }{4k} \\ =&\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m}(-1)^{s+1} \biggl( \frac{ \pi }{2k} \biggr)^{2s+1} 2k(4lk+2i+1)^{2s+1} \bigl(a_{2lk+i}^{2} +b_{2lk+i} ^{2}\bigr)\cot \frac{(2i+1)\pi }{4k} \\ &{}-\sum_{i=0}^{m}(-1)^{s+1} \biggl(\frac{\pi }{2k} \biggr)^{2s+1} 2k(4lk+4k-2i-1)^{2s+1} \bigl(a _{2lk+2k-i-1}^{2} +b_{2lk+2k-i-1}^{2}\bigr)\\ & {}\times\cot \frac{(2i+1) \pi }{4k} \Biggr]. \end{aligned}$$

On the other hand,

$$\begin{aligned} &\begin{aligned} \bigl\langle P^{-1}\beta x,x\bigr\rangle &=\sum _{i=0}^{\infty }2k\beta \bigl(a_{i} ^{2}+b_{i}^{2}\bigr) \\ &=\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m}2k\beta \bigl(a_{2lk+i}^{2}+b _{2lk+i}^{2}\bigr)+\sum_{i=0}^{m}2k \beta \bigl(a_{2lk+2k-i-1}^{2}+b_{2lk+2k-i-1} ^{2}\bigr) \Biggr]. \end{aligned} \\ &\begin{aligned} \bigl\langle P^{-1}\alpha x,x \bigr\rangle &=\sum_{i=0}^{\infty }2k\alpha \bigl(a _{i}^{2}+b_{i}^{2}\bigr) \\ &=\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m}2k\alpha \bigl(a_{2lk+i}^{2}+b _{2lk+i}^{2}\bigr)+\sum_{i=0}^{m}2k \alpha \bigl(a_{2lk+2k-i-1}^{2}+b_{2lk+2k-i-1} ^{2} \bigr) \Biggr]. \end{aligned} \end{aligned}$$

Therefore, we have

$$\begin{aligned} &\bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x\bigr\rangle \\ &\quad =2k\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m} \biggl((-1)^{s+1} \biggl(\frac{ \pi }{2k} \biggr)^{2s+1}(4lk+2i+1)^{2s+1} \cot \frac{(2i+1)\pi }{4k} + \beta \biggr)\\ &\qquad {}\times \bigl(a_{2lk+i}^{2}+b_{2lk+i}^{2} \bigr) \\ &\qquad {}+\sum_{i=0}^{m} \biggl(-(-1)^{s+1} \biggl(\frac{\pi }{2k} \biggr)^{2s+1}(4lk+4k-2i-1)^{2s+1} \cot \frac{(2i+1)\pi }{4k} +\beta \biggr)\\ &\qquad {}\times \bigl(a_{2lk+2k-i-1}^{2} +b_{2lk+2k-i-1} ^{2}\bigr) \Biggr], \\ &\bigl\langle \bigl(L+P^{-1}\alpha \bigr) x,x\bigr\rangle \\ &\quad =2k\sum_{l=0}^{\infty } \Biggl[\sum _{i=0}^{m} \biggl((-1)^{s+1} \biggl(\frac{ \pi }{2k} \biggr)^{2s+1}(4lk+2i+1)^{2s+1} \cot \frac{(2i+1)\pi }{4k} + \alpha \biggr) \\ &\qquad {}\times \bigl(a_{2lk+i}^{2}+b_{2lk+i} ^{2}\bigr) \\ &\qquad {}+\sum_{i=0}^{m} \biggl(-(-1)^{s+1} \biggl(\frac{\pi }{2k} \biggr)^{2s+1}(4lk+4k-2i-1)^{2s+1} \cot \frac{(2i+1)\pi }{4k} +\alpha \biggr) \\ &\qquad {}\times\bigl(a_{2lk+2k-i-1}^{2} +b_{2lk+2k-i-1} ^{2}\bigr) \Biggr]. \end{aligned}$$

Lemma 2.1

Each critical point of the functional Φ is a 4k-periodic classical solution of Eq. (5) satisfying (9).

Proof

Let x be a critical point of the functional Φ. Then $x(t)$ satisfies

$$\begin{aligned} -\sum_{i=1}^{2k-1}{{{(-1)}^{i+1}}} {{x}^{(2s+1)}}(t-i)+f\bigl(x(t)\bigr)=0,\quad \mbox{a.e. } t\in [0,4k]. \end{aligned}$$

(21)

Consequently,

$$\begin{aligned} &{-}\sum_{i=1}^{2k-1}{{{(-1)}^{i+1}}} {{x}^{(2s+1)}}(t-i-1)+f\bigl(x(t-1)\bigr)=0, \end{aligned}$$

(21.1)

$$\begin{aligned} &{-}\sum_{i=1}^{2k-1}{{{(-1)}^{i+1}}} {{x}^{(2s+1)}}(t-i-2)+f\bigl(x(t-2)\bigr)=0, \end{aligned}$$

(21.2)

$$\begin{aligned} &{-}\sum_{i=1}^{2k-1}{{{(-1)}^{i+1}}} {{x}^{(2s+1)}}(t-i-3)+f\bigl(x(t-3)\bigr)=0, \end{aligned}$$

(21.3)

$$\begin{aligned} &\vdots \\ &{-}\sum_{i=0}^{2k-1}{{{(-1)}^{i+1}}} {{x}^{(2s+1)}}\bigl(t-i-(2k-1)\bigr)+f\bigl(x\bigl(t-(2k-1)\bigr)\bigr)=0. \end{aligned}$$

(21.(2k-1))

Calculating (21.1) + (21.2) + (21.3) + ⋯ + (21.(2k-1)), we can get

$$ x^{(2s+1)}(t)+\sum_{i=1}^{2k-1}f \bigl(x(t-i)\bigr)=0,\quad \mbox{a.e. }t\in [0,4k], $$

namely

$$ x^{(2s+1)}(t)=-\sum_{i=1}^{2k-1}f \bigl(x(t-i)\bigr),\quad \mbox{a.e. } t\in [0,4k], $$

which implies that x satisfies the above equation for all $t\in [0,4k]$ since the function on the right-hand side is continuous, then x is a classical solution to (5). □

Lemma 2.2

Each critical point of the functional Ψ is a 4k-periodic classical solution of Eq. (6) satisfying (9).

Proof

Let x be a critical point of the functional Ψ. Then $x(t)$ satisfies

$$\begin{aligned} -\sum_{i=1}^{2k-1}{x}^{(2s+1)}(t-i)+f \bigl(x(t)\bigr)=0,\quad \mbox{a.e. } t\in [0,4k]. \end{aligned}$$

(22)

Consequently,

$$\begin{aligned} &{-}\sum_{i=1}^{2k-1}{x}^{(2s+1)}(t-i-1)+f \bigl(x(t-1)\bigr)=0, \end{aligned}$$

(22.1)

$$\begin{aligned} &{-}\sum_{i=1}^{2k-1}{x}^{(2s+1)}(t-i-2)+f \bigl(x(t-2)\bigr)=0, \end{aligned}$$

(22.2)

$$\begin{aligned} &{-}\sum_{i=1}^{2k-1}{x}^{(2s+1)}(t-i-3)+f \bigl(x(t-3)\bigr)=0, \end{aligned}$$

(22.3)

$$\begin{aligned} &\vdots \\ &{-}\sum_{i=1}^{2k-1}{x}^{(2s+1)} \bigl(t-i-(2k-1)\bigr)+f\bigl(x\bigl(t-(2k-1)\bigr)\bigr)=0. \end{aligned}$$

(22.(2k-1))

Calculating (22.1) − (22.2) + (22.3) − ⋯ + (22.(2k-1)), we can get

$$ x^{(2s+1)}(t)+\sum_{i=1}^{2k-1}(-1)^{i+1}f \bigl(x(t-i)\bigr)=0,\quad \mbox{a.e. }t\in [0,4k], $$

namely

$$ x^{(2s+1)}(t)=-\sum_{i=1}^{2k-1}(-1)^{i+1}f \bigl(x(t-i)\bigr),\quad \mbox{a.e. }t\in [0,4k], $$

which implies that x satisfies the above equation for all $t\in [0,4k]$ since the function on the right-hand side is continuous, then x is a classical solution to (6). □

3 Partition of space X and symbols

For (5), let

$$\begin{aligned} &\begin{aligned} X_{\infty }^{+} ={}& \biggl\{ X(2lk+i):l\geq 0,0 \leq i\leq m, \\ &(-1)^{s+1} \biggl(\frac{(4lk+2i+1)\pi }{2k} \biggr)^{2s+1} \tan \frac{(2i+1)\pi }{4k} + \beta >0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ &{-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta >0 \biggr\} , \end{aligned} \\ &\begin{aligned} X_{\infty }^{-} ={}& \biggl\{ X(2lk+i):l\geq 0,0\leq i\leq m, \\ &(-1)^{s+1} \biggl( \frac{(4lk+2i+1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} + \beta < 0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ & {-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta < 0 \biggr\} , \end{aligned} \\ &\begin{aligned} X_{0}^{+} ={}& \biggl\{ X(2lk+i):l\geq 0,0 \leq i\leq m, \\ & (-1)^{s+1} \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1} \tan \frac{(2i+1) \pi }{4k} +\alpha >0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ & {-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\alpha >0 \biggr\} , \end{aligned} \\ &\begin{aligned} X_{0}^{-} ={}& \biggl\{ X(2lk+i):l\geq 0,0\leq i\leq m, \\ &(-1)^{s+1} \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1) \pi }{4k} +\alpha < 0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ & {-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\alpha < 0 \biggr\} . \end{aligned} \end{aligned}$$

On the other hand,

$$\begin{aligned} &\begin{aligned} X_{\infty }^{0} ={}& \biggl\{ X(2lk+i):l\geq 0,0 \leq i\leq m, \\ &(-1)^{s+1} \biggl(\frac{(4lk+2i+1)\pi }{2k} \biggr)^{2s+1} \tan \frac{(2i+1)\pi }{4k} + \beta =0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ &{-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta =0 \biggr\} , \end{aligned} \\ &\begin{aligned} X_{0}^{0} ={}& \biggl\{ X(2lk+i):l\geq 0,0\leq i\leq m, \\ &(-1)^{s+1} \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1) \pi }{4k} +\alpha =0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ &{-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\alpha =0 \biggr\} . \end{aligned} \end{aligned}$$

Obviously, $\dim X_{\infty }^{0}<\infty$ and $\dim X_{0}^{0}< \infty$.

Lemma 3.1

Under assumptions $(S_{1})$ and $(S_{2})$, there is $\sigma >0$ such that

$$\begin{aligned} \begin{aligned} &\bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle >\sigma \Vert x \Vert ^{2}, \quad x\in X_{\infty }^{+} \quad \textit{and}\quad \\ &\bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle < -\sigma \Vert x \Vert ^{2}, \quad x\in X_{\infty }^{-}. \end{aligned} \end{aligned}$$

(23)

Proof

First, we have that, for $\beta \geq 0$ and $s=\mathrm{even}$, $(-1)^{s+1}=-1$, $i \in \{0,1,\ldots ,m\}$,

$$\begin{aligned} &{-} \biggl(\frac{(4lk+2i+1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta\\ &\quad >- \biggl(\frac{(4l^{+}(i)k+2i+1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1) \pi }{4k} +\beta >0, \end{aligned}$$

where $l^{+}(i)=\max \{l\in N:- (\frac{(4lk+2i+1)\pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta >0 \}$ and

$$\begin{aligned} &{-} \biggl(\frac{(4lk+2i+1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta\\ &\quad < - \biggl(\frac{(4l^{-}(i)k+2i+1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1) \pi }{4k} +\beta < 0, \end{aligned}$$

where $l^{-}(i)=\min \{l\in N:- (\frac{(4lk+2i+1)\pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta <0 \}$.

In this case, we may choose

$$\begin{aligned} \sigma_{i} =&\min \biggl\{ - \biggl(\frac{\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1) \pi }{4k} +\frac{\beta }{(4l^{+}(i)k+2i+1)^{2s+1}}, \\ &{} \biggl(\frac{\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} - \frac{ \beta }{(4l^{-}(i)k+2i+1)^{2s+1}} \biggr\} >0, \end{aligned}$$

and let $\sigma =\min \{\sigma_{0},\sigma_{1},\ldots ,\sigma_{m}\}>0$.

Second, we have that, for $\beta \geq 0$ and $s=\mathrm{odd}$, $(-1)^{s+1}=1$, $i \in \{0,1,\ldots ,m\}$,

$$\begin{aligned} &{-} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta\\ &\quad >- \biggl( \frac{(4l^{+}(i)k+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1) \pi }{4k} + \beta >0, \end{aligned}$$

where $l^{+}(i)=\max \{l\in N:- (\frac{(4lk+4k-2i-1)\pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta >0 \}$ and

$$\begin{aligned} &{-} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta\\ &\quad < - \biggl( \frac{(4l^{-}(i)k+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1) \pi }{4k} + \beta < 0, \end{aligned}$$

where $l^{-}(i)=\min \{l\in N:- (\frac{(4lk+4k-2i-1)\pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\beta <0 \}$.

In this case, we may choose

$$\begin{aligned} \sigma_{i} =&\min \biggl\{ - \biggl(\frac{\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1) \pi }{4k} +\frac{\beta }{(4l^{+}(i)k+4k-2i-1)^{2s+1}}, \\ & {} \biggl(\frac{\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} - \frac{ \beta }{(4l^{-}(i)k+4k-2i-1)^{2s+1}} \biggr\} >0, \end{aligned}$$

and let $\sigma =\min \{\sigma_{0},\sigma_{1},\ldots ,\sigma_{m}\}>0$. The proof for the case $\beta <0$ is similar. We omit it. The inequalities in (23) are proved. □

Lemma 3.2

Under conditions $(S_{1})$ and $(S_{2})$, the functional Φ defined by (11) satisfies the $(P.S)$-condition.

Proof

Let Π, $\mathbb{N}$, $\mathbb{Z}$ be the orthogonal projections from X onto $X_{\infty }^{+}$, $X_{\infty }^{-}$, $X_{\infty }^{0}$, respectively. From the second condition in (8) it follows that

$$\begin{aligned} \bigl\vert \bigl\langle P^{-1}\bigl(f(x)-\beta x \bigr),x \bigr\rangle \bigr\vert < \frac{\sigma }{2} \Vert x \Vert ^{2}+M,\quad x\in X \end{aligned}$$

(24)

for some $M>0$.

Assume that $\{x_{n}\}\subset X$ is a subsequence such that $\varPhi '(x_{n})\rightarrow 0$ and $\varPhi (x_{n})$ is bounded. Let $w_{n}=\varPi x_{n}$, $y_{n}=\mathbb{N}x_{n}$, $z_{n}=\mathbb{Z}x_{n}$. Then we have

$$\begin{aligned} \varPi \bigl(L+P^{-1}\beta \bigr)=\bigl(L+P^{-1} \beta \bigr)\varPi ,\qquad \mathbb{N}\bigl(L+P^{-1}\beta \bigr)= \bigl(L+P^{-1}\beta \bigr)\mathbb{N}. \end{aligned}$$

(25)

From

$$ \bigl\langle \varPhi '(x_{n}),x_{n} \bigr\rangle = \bigl\langle Lx_{n}+P ^{-1}f(x_{n}),x_{n} \bigr\rangle = \bigl\langle \bigl(L+P^{-1}\beta \bigr)x_{n},x _{n} \bigr\rangle + \bigl\langle P^{-1}\bigl(f(x_{n})- \beta x_{n}\bigr),x_{n} \bigr\rangle $$

and (25), we have

$$\begin{aligned} \bigl\langle \varPi \varPhi '(x_{n}),x_{n} \bigr\rangle =& \bigl\langle \varPi \bigl(L+P^{-1}\beta \bigr)x_{n},x_{n} \bigr\rangle + \bigl\langle \varPi P^{-1}\bigl(f(x _{n})-\beta x_{n}\bigr),x_{n} \bigr\rangle \\ =& \bigl\langle \bigl(L+P^{-1}\beta \bigr)w_{n},w_{n} \bigr\rangle + \bigl\langle \varPi P^{-1}\bigl(f(x_{n})-\beta x_{n}\bigr),w_{n} \bigr\rangle , \end{aligned}$$

and then, by (23), we have

$$ \bigl\langle \bigl(L+P^{-1}\beta \bigr)w_{n},w_{n} \bigr\rangle + \bigl\langle \varPi P^{-1}\bigl(f(x_{n})-\beta x_{n}\bigr),w_{n} \bigr\rangle >\frac{\sigma }{2} \Vert w_{n} \Vert ^{2}-M \Vert w_{n} \Vert , $$

which, together with $\varPi \varPhi '(x_{n})\rightarrow 0$, implies the boundedness of $w_{n}$. Similarly we have the boundedness of $y_{n}$. At the same time, $(S_{2})$ yields

$$\begin{aligned} \varPhi (x_{n}) =&\frac{1}{2} \bigl\langle \bigl(L+P^{-1} \beta \bigr)x_{n},x_{n} \bigr\rangle + \int_{0}^{4k}F(x_{n})\,dt- \frac{\beta }{2} \Vert x_{n} \Vert _{2} ^{2} \\ =&\frac{1}{2} \bigl\langle \bigl(L+P^{-1}\beta \bigr)w_{n},w_{n} \bigr\rangle + \frac{1}{2} \bigl\langle \bigl(L+P^{-1}\beta \bigr)y_{n},y_{n} \bigr\rangle \\ & {} + \int_{0}^{4k}F(x_{n})\,dt - \frac{\beta }{2} \bigl( \Vert w_{n} \Vert _{2}^{2}+ \Vert y _{n} \Vert _{2}^{2}+ \Vert z_{n} \Vert _{2}^{2} \bigr). \end{aligned}$$

Then the boundedness of $\varPhi (x)$ implies that $\|z_{n}\|_{2}$ is bounded. Consequently $\|z_{n}\|$ is bounded since $X_{\infty }^{0}$ is finite-dimensional. Therefore, $\|x_{n}\|$ is bounded.

It follows from (15) that

$$\begin{aligned} (\varPi +N)\varPhi '(x_{n}) =&(\varPi +\mathbb{N})Lx_{n}+( \varPi +\mathbb{N})K(x _{n}) \\ =&L(w_{n}+y_{n})+(\varPi +\mathbb{N})K(x_{n}). \end{aligned}$$

From the compactness of operator K and the boundedness of $x_{n}$, we have that $K(x_{n})\rightarrow u$. Then

$$\begin{aligned} L|_{x_{\infty }^{+}+x_{\infty }^{-}}(w_{n}+y_{n})\rightarrow -( \varPi + \mathbb{N})u. \end{aligned}$$

(26)

The finite-dimensionality of $X_{\infty }^{0}$ and the boundedness of $z_{n}=\mathbb{Z}x_{n}$ imply $z_{n}\rightarrow \varphi \in X_{\infty }^{0}$. Therefore,

$$ x_{n}=z_{n}+w_{n}+y_{n}\rightarrow \varphi -(L|_{x_{\infty }^{+}+x _{\infty }^{-}})^{-1} (\varPi +\mathbb{N})u, $$

which implies the $(P.S)$-condition.

For (6), let

$$\begin{aligned} &\begin{aligned} X_{\infty }^{+} ={}& \biggl\{ X(2lk+i):l\geq 0,0 \leq i\leq m,\\ & (-1)^{s+1} \biggl(\frac{(4lk+2i+1)\pi }{2k} \biggr)^{2s+1} \cot \frac{(2i+1)\pi }{4k} + \beta >0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ &{-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} +\beta >0 \biggr\} , \end{aligned} \\ &\begin{aligned} X_{\infty }^{-} ={}& \biggl\{ X(2lk+i):l\geq 0,0\leq i\leq m, \\ &(-1)^{s+1} \biggl( \frac{(4lk+2i+1)\pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} + \beta < 0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ &{-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} +\beta < 0 \biggr\} , \end{aligned} \\ &\begin{aligned} X_{0}^{+} ={}& \biggl\{ X(2lk+i):l\geq 0,0\leq i\leq m,\\ & (-1)^{s+1} \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1) \pi }{4k} +\alpha >0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ &{-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} +\alpha >0 \biggr\} , \end{aligned} \\ &\begin{aligned} X_{0}^{-} ={}& \biggl\{ X(2lk+i):l\geq 0,0\leq i\leq m, \\ &(-1)^{s+1} \biggl( \frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1) \pi }{4k} +\alpha < 0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ &{-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} +\alpha < 0 \biggr\} . \end{aligned} \end{aligned}$$

On the other hand,

$$\begin{aligned} &\begin{aligned} X_{\infty }^{0} ={}& \biggl\{ X(2lk+i):l\geq 0,0 \leq i\leq m, \\ &(-1)^{s+1} \biggl(\frac{(4lk+2i+1)\pi }{2k} \biggr)^{2s+1} \cot \frac{(2i+1)\pi }{4k} + \beta =0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m,\\ & {-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} +\beta =0 \biggr\} , \end{aligned} \\ &\begin{aligned} X_{0}^{0} ={}& \biggl\{ X(2lk+i):l\geq 0,0\leq i\leq m,\\ & (-1)^{s+1} \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1) \pi }{4k} +\alpha =0 \biggr\} \\ &{}\cup \biggl\{ X(2lk+2k-i-1):l\geq 0,0\leq i\leq m, \\ &{-}(-1)^{s+1} \biggl(\frac{(4lk+4k-2i-1)\pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} +\alpha =0 \biggr\} . \end{aligned} \end{aligned}$$

Obviously, $\dim X_{\infty }^{0}<\infty$ and $\dim X_{0}^{0}< \infty$. □

Lemma 3.3

Under assumptions $(S_{1})$ and $(S_{2})$, there is $\sigma >0$ such that

$$\begin{aligned} \begin{aligned} &\bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle >\sigma \Vert x \Vert ^{2}, \quad x\in X_{\infty }^{+} \quad \textit{and}\\ & \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle < -\sigma \Vert x \Vert ^{2}, \quad x\in X_{\infty }^{-}. \end{aligned} \end{aligned}$$

(27)

Proof

The proof is similar to Lemma 3.1, we omit it. □

Lemma 3.4

Under conditions $(S_{1})$ and $(S_{2})$, the functional Ψ defined by (16) satisfies the $(P.S)$-condition.

Proof

The proof is similar to Lemma 3.2, we omit it. □

4 Notations and main results of this paper

We first give some notations.

For (5), if $(-1)^{s+1}=-1$, denote

$$\begin{aligned} &N_{1}(\alpha )= \textstyle\begin{cases} -\sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+4k-2i-1) \pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} < -\alpha \}, \quad \alpha < 0, \\ \sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< ( \frac{(4lk+2i+1) \pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} < \alpha \}, \quad \alpha \geq 0. \end{cases}\displaystyle \\ &N_{1}(\beta )= \textstyle\begin{cases} -\sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+4k-2i-1) \pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} < -\beta \}, \quad \beta < 0, \\ \sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< ( \frac{(4lk+2i+1) \pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} < \beta \}, \quad \beta \geq 0. \end{cases}\displaystyle \end{aligned}$$

And

$$\begin{aligned} &N_{1}^{0}(\alpha_{-}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+4k-2i-1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} =-\alpha \biggr\} , \quad \alpha < 0, \\ &N_{1}^{0}(\alpha_{+}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k}= \alpha \biggr\} , \quad \alpha \geq 0, \\ &N_{1}^{0}(\beta_{-}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+4k-2i-1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} =-\beta \biggr\} , \quad \beta < 0, \\ &N_{1}^{0}(\beta_{+}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k}= \beta \biggr\} , \quad \beta \geq 0. \end{aligned}$$

Alternatively, if $(-1)^{s+1}=1$, denote

$$\begin{aligned} &N_{1}(\alpha )= \textstyle\begin{cases} -\sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+2i+1) \pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} < -\alpha \}, \quad \alpha < 0, \\ \sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+4k-2i-1) \pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} < \alpha \}, \quad \alpha \geq 0. \end{cases}\displaystyle \\ &N_{1}(\beta )= \textstyle\begin{cases} -\sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+2i+1) \pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} < -\beta \}, \quad \beta < 0, \\ \sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+4k-2i-1) \pi }{2k} )^{2s+1}\tan \frac{(2i+1)\pi }{4k} < \beta \}, \quad \beta \geq 0. \end{cases}\displaystyle \end{aligned}$$

And

$$\begin{aligned} &N_{1}^{0}(\alpha_{-}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} =-\alpha \biggr\} , \quad \alpha < 0, \\ &N_{1}^{0}(\alpha_{+}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+4k-2i-1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k}=\alpha \biggr\} , \quad \alpha \geq 0, \\ &N_{1}^{0}(\beta_{-}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} =-\beta \biggr\} , \quad \beta < 0, \\ &N_{1}^{0}(\beta_{+}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+4k-2i-1) \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k}=\beta \biggr\} , \quad \beta \geq 0. \end{aligned}$$

For (6), if $(-1)^{s+1}=-1$, denote

$$\begin{aligned} &N_{2}(\alpha )= \textstyle\begin{cases} -\sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+4k-2i-1) \pi }{2k} )^{2s+1}\cot \frac{(2i+1)\pi }{4k} < -\alpha \}, \quad \alpha < 0, \\ \sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< ( \frac{(4lk+2i+1) \pi }{2k} )^{2s+1}\cot \frac{(2i+1)\pi }{4k} < \alpha \}, \quad \alpha \geq 0. \end{cases}\displaystyle \\ &N_{2}(\beta )= \textstyle\begin{cases} -\sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+4k-2i-1) \pi }{2k} )^{2s+1}\cot \frac{(2i+1)\pi }{4k} < -\beta \}, \quad \beta < 0, \\ \sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< ( \frac{(4lk+2i+1) \pi }{2k} )^{2s+1}\cot \frac{(2i+1)\pi }{4k} < \beta \}, \quad \beta \geq 0. \end{cases}\displaystyle \end{aligned}$$

And

$$\begin{aligned} &N_{2}^{0}(\alpha_{-}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+4k-2i-1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} =-\alpha \biggr\} , \quad \alpha < 0, \\ &N_{2}^{0}(\alpha_{+}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k}= \alpha \biggr\} , \quad \alpha \geq 0, \\ &N_{2}^{0}(\beta_{-}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+4k-2i-1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} =-\beta \biggr\} , \quad \beta < 0, \\ &N_{2}^{0}(\beta_{+}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k}= \beta \biggr\} , \quad \beta \geq 0. \end{aligned}$$

Alternatively, if $(-1)^{s+1}=1$, denote

$$\begin{aligned} &N_{2}(\alpha )= \textstyle\begin{cases} -\sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+2i+1) \pi }{2k} )^{2s+1}\cot \frac{(2i+1)\pi }{4k} < -\alpha \}, \quad \alpha < 0, \\ \sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+4k-2i-1) \pi }{2k} )^{2s+1}\cot \frac{(2i+1)\pi }{4k} < \alpha \}, \quad \alpha \geq 0. \end{cases}\displaystyle \\ &N_{2}(\beta )= \textstyle\begin{cases} -\sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+2i+1) \pi }{2k} )^{2s+1}\cot \frac{(2i+1)\pi }{4k} < -\beta \}, \quad \beta < 0, \\ \sum_{i=0}^{m}\operatorname{card} \{l\geq 0:0< (\frac{(4lk+4k-2i-1) \pi }{2k} )^{2s+1}\cot \frac{(2i+1)\pi }{4k} < \beta \}, \quad \beta \geq 0. \end{cases}\displaystyle \end{aligned}$$

And

$$\begin{aligned} &N_{2}^{0}(\alpha_{-}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} =-\alpha \biggr\} , \quad \alpha < 0, \\ &N_{2}^{0}(\alpha_{+}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+4k-2i-1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k}=\alpha \biggr\} , \quad \alpha \geq 0, \\ &N_{2}^{0}(\beta_{-}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+2i+1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k} =-\beta \biggr\} , \quad \beta < 0, \\ &N_{2}^{0}(\beta_{+}) =\sum _{i=0}^{m}\operatorname{card} \biggl\{ l\geq 0:0< \biggl(\frac{(4lk+4k-2i-1) \pi }{2k} \biggr)^{2s+1}\cot \frac{(2i+1)\pi }{4k}=\beta \biggr\} , \quad \beta \geq 0. \end{aligned}$$

Now we give the main results of this paper.

Theorem 4.1

Suppose that $(S_{1})$ and $(S_{2})$ hold. Then Eq. (5) possesses at least

$$ n=\max \bigl\{ N_{1}(\beta )-N_{1}(\alpha )-N_{1}^{0}(\beta_{-})-N_{1} ^{0}(\alpha_{+}), N_{1}(\alpha )-N_{1}( \beta )-N_{1}^{0}(\alpha_{-})-N _{1}^{0}(\beta_{+}) \bigr\} $$

4k-periodic solutions satisfying $x(t-2k)=-x(t)$ provided that $n > 0$.

Theorem 4.2

Suppose that $(S_{1})$, $(S_{2})$, $(S_{3}^{+})$, and $(S_{4}^{-})$ hold. Then Eq. (5) possesses at least

$$ n=N_{1}(\beta )-N_{1}(\alpha )+N_{1}^{0}( \beta_{+})+N_{1}^{0}(\alpha_{-}) $$

4k-periodic solutions satisfying $x(t-2k)=-x(t)$ provided that $n > 0$.

Theorem 4.3

Suppose that $(S_{1})$, $(S_{2})$, $(S_{3}^{-})$, and $(S_{4}^{+})$ hold. Then Eq. (5) possesses at least

$$ n=N_{1}(\alpha )-N_{1}(\beta )+N_{1}^{0}( \alpha_{+})+N_{1}^{0}(\beta_{-}) $$

4k-periodic solutions satisfying $x(t-2k)=-x(t)$ provided that $n > 0$.

Theorem 4.4

Suppose that $(S_{1})$ and $(S_{2})$ hold. Then Eq. (6) possesses at least

$$ n=\max \bigl\{ N_{2}(\beta )-N_{2}(\alpha )-N_{2}^{0}(\beta_{-})-N_{2} ^{0}(\alpha_{+}), N_{2}(\alpha )-N_{2}(\beta )-N_{2}^{0}(\alpha_{-})-N _{2}^{0}(\beta_{+}) \bigr\} $$

4k-periodic solutions satisfying $x(t-2k)=-x(t)$ provided that $n > 0$.

Theorem 4.5

Suppose that $(S_{1})$, $(S_{2})$, $(S_{3}^{+})$, and $(S_{4}^{-})$ hold. Then Eq. (6) possesses at least

$$ n=N_{2}(\beta )-N_{2}(\alpha )+N_{2}^{0}( \beta_{+})+N_{2}^{0}(\alpha_{-}) $$

4k-periodic solutions satisfying $x(t-2k)=-x(t)$ provided that $n > 0$.

Theorem 4.6

Suppose that $(S_{1})$, $(S_{2})$, $(S_{3}^{-})$, and $(S_{4}^{+})$ hold. Then Eq. (6) possesses at least

$$ n=N_{2}(\alpha )-N_{2}(\beta )+N_{2}^{0}( \alpha_{+})+N_{2}^{0}(\beta_{-}) $$

4k-periodic solutions satisfying $x(t-2k)=-x(t)$ provided that $n > 0$.

5 Proof of main results of this paper

Proof of Theorem 4.1

Suppose without loss of generality that

$$ n=N_{1}(\beta )-N_{1}(\alpha )-N_{1}^{0}( \beta_{-})-N_{1}^{0}(\alpha_{+}). $$

Let $X^{+}=X_{\infty }^{+}$ and $X^{-}=X_{0}^{-}$. Then

$$ X\setminus \bigl(X^{+}\cup X^{-}\bigr)=X\setminus \bigl(X_{\infty }^{+}\cup X_{0} ^{-}\bigr) \subseteq X_{\infty }^{0}\cup X_{0}^{0}\cup \bigl(X_{\infty }^{+} \cap X_{0}^{-}\bigr). $$

Obviously,

$$ \operatorname{codim}_{X}\bigl(X^{+}+X^{-}\bigr) \leq \dim X_{\infty }^{0}+\dim X_{0}^{0}+ \dim \bigl(X_{\infty }^{+}\cap X_{0}^{-}\bigr)< \infty , $$

which implies that condition (a) in Lemma 1.1 holds. Let $A_{\infty }= \beta$. Then condition (b) in Lemma 1.1 holds since, for each $j\in N$, we have that $x\in X(j)$ yields $(L+P^{-1}\beta )x\in X(j)$.

At the same time, Lemma 3.2 gives the $(P.S)$-condition.

Now it suffices to show that conditions (c) and (d) in Lemma 1.1 hold under assumptions $(S_{1})$ and $(S_{2})$.

In fact, we have shown in Lemma 3.1 that there is $\sigma >0$ such that $\langle (L+P^{-1}\beta )x,x \rangle >\sigma \|x\|^{2}$, $x \in X_{\infty }^{+}$. And the second condition in (8) implies that $|F(x)-\frac{1}{2}\beta x^{2}|<\frac{1}{4}\sigma |x|^{2}+M_{1}$, $x \in R$ for some $M_{1}>0$.

Then

$$\begin{aligned} \varPhi (x) =&\frac{1}{2} \langle \mathit{Lx},x \rangle + \int _{0}^{4k}F\bigl(x(t)\bigr)\,dt \\ =&\frac{1}{2} \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle + \int_{0} ^{4k} \biggl[F\bigl(x(t)\bigr)- \frac{1}{2}\beta \bigl\vert x(t) \bigr\vert ^{2} \biggr]\,dt \\ \geq& \frac{1}{2}\sigma \Vert x \Vert ^{2}- \frac{1}{4}\sigma \Vert x \Vert ^{2}-4kM _{1} \\ \geq& \frac{1}{4}\sigma \Vert x \Vert ^{2}-4kM_{1} \end{aligned}$$

if $x\in X^{+}$. Clearly, there is $c_{0}\in R$ such that

$$ \inf_{x\in X^{+}}\varPhi (x)\geq c_{0}. $$

On the other hand, we have shown in Lemma 3.1 that there is $\sigma >0$ such that $\langle (L+P^{-1}\beta )x,x \rangle <- \sigma \|x\|^{2}$, $x\in X_{\infty }^{-}$. And we can show that there are $r, \sigma >0$ such that $|F(x)-\frac{1}{2}\beta x^{2}|< \frac{1}{4} \sigma |x|^{2}$, $\|x\|=r$. So

$$\begin{aligned} \varPhi (x) =&\frac{1}{2} \langle \mathit{Lx},x \rangle + \int _{0}^{4k}F\bigl(x(t)\bigr)\,dt \\ =&\frac{1}{2} \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle + \int_{0} ^{4k} \biggl[F\bigl(x(t)\bigr)- \frac{1}{2}\beta \bigl\vert x(t) \bigr\vert ^{2} \biggr]\,dt \\ \leq& -\frac{1}{2}\sigma \Vert x \Vert ^{2}+ \frac{1}{4}\sigma \Vert x \Vert ^{2} \\ \leq& -\frac{1}{4}\sigma \Vert x \Vert ^{2}. \end{aligned}$$

That is, there are $r>0$ and $c_{\infty }<0$ such that

$$ \varPhi (x)\leq c_{\infty }< 0=\varPhi (0),\quad \forall x\in X^{-} \cap S_{r}= \bigl\{ x\in X: \Vert x \Vert =r\bigr\} . $$

Our last task is to compute the value of

$$\begin{aligned} n =&\frac{1}{2} \Bigl[\dim \Bigl(X^{+}\cap X^{-}\Bigr)-\operatorname{codim}_{X}\bigl(X^{+}+X ^{-}\bigr) \Bigr] \\ =&\frac{1}{2} \Bigl[\dim \Bigl(X_{\infty }^{+}\cap X_{0}^{-}\Bigr)- \operatorname{codim} _{X}\bigl(X_{\infty }^{+}+X_{0}^{-} \bigr) \Bigr] \\ =&\frac{1}{2}\sum_{j=0}^{\infty } \Bigl[ \dim \Bigl(X_{\infty }^{+}(j)\cap X_{0}^{-}(j) \Bigr)-\operatorname{codim}_{X(j)}\bigl(X_{\infty }^{+}(j)+X_{0}^{-}(j) \bigr) \Bigr]. \end{aligned}$$

By computation we get that, for each $i\in \{0,1,\ldots ,m\}$,

$$\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle = \biggl((-1)^{s+1} \biggl(\frac{ \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} + \frac{\beta }{(4lk+2i+1)^{2s+1}} \biggr) \Vert x \Vert ^{2}, \\ &\quad x\in X(2lk+i), \\ & \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle = \biggl(-(-1)^{s+1} \biggl(\frac{ \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} + \frac{\beta }{(4lk+4k-2i-1)^{2s+1}} \biggr) \Vert x \Vert ^{2}, \\ &\quad x\in X(2lk+2k-i-1), \end{aligned}$$

and

$$\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\alpha \bigr)x,x \bigr\rangle = \biggl((-1)^{s+1} \biggl(\frac{ \pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} + \frac{\alpha }{(4lk+2i+1)^{2s+1}} \biggr) \Vert x \Vert ^{2},\\ &\quad x\in X(2lk+i), \\ & \bigl\langle \bigl(L+P^{-1}\alpha \bigr)x,x \bigr\rangle = \biggl(-(-1)^{s+1} \biggl(\frac{\pi }{2k} \biggr)^{2s+1}\tan \frac{(2i+1)\pi }{4k} +\frac{ \alpha }{(4lk+4k-2i-1)^{2s+1}} \biggr) \Vert x \Vert ^{2},\\ &\quad x\in X(2lk+2k-i-1). \end{aligned}$$

Therefore,

• $X_{\infty }^{+}(2lk+i)=X_{\infty }^{+}\cap X(2lk+i)=\emptyset$,

• $X_{\infty }^{+}(2lk+2k-i-1)=X_{\infty }^{+}\cap X(2lk+2k-i-1)=X(2lk+2k-i-1)$,

• $X_{0}^{-}(2lk+i)=X_{0}^{-}\cap X(2lk+i)=X(2lk+i)$,

• $X_{0}^{-}(2lk+2k-i-1)=X_{0}^{-}\cap X(2lk+2k-i-1)=\emptyset$

if $i\in \{0,1,\ldots ,m\}$ and $(-1)^{s+1}=1$ and $l\geq 0$ is large enough, which means that there is $M>0$ such that $\dim (X_{\infty } ^{+}(j)\cap X_{0}^{-}(j))-\operatorname{codim}_{X}(X_{\infty }^{+}(j)+X_{0} ^{-}(j))=0$, $j>M$, from which it follows that

$$\begin{aligned} n =&\frac{1}{2}\sum_{j=0}^{M} \Bigl[\dim \Bigl(X_{\infty }^{+}(j)\cap X _{0}^{-}(j)\Bigr)-\operatorname{codim}_{X(j)} \bigl(X_{\infty }^{+}(j)+X_{0}^{-}(j)\bigr) \Bigr] \\ =&\frac{1}{2}\sum_{j=0}^{M} \bigl[ \dim X_{\infty }^{+}(j)+\dim X_{0} ^{-}(j)-2 \bigr] \\ =&\frac{1}{2}\sum_{j=0}^{M} \bigl[ \dim X_{\infty }^{+}(j)+\dim X_{0} ^{-}(j) \bigr]-(M+1). \end{aligned}$$

Then we have

$$\begin{aligned} &\sum_{j=0}^{M}\dim \bigl(X_{\infty }^{+}(j) \bigr) \\ &\quad =2 \textstyle\begin{cases} N_{1}(\beta )+\operatorname{card}\{2lk+2k-i-1:0\leq 2lk+2k-i-1\leq M\}, \quad \beta \geq 0, \\ N_{1}(\beta )-N_{1}^{0}(\beta_{-})+\operatorname{card}\{2lk+2k-i-1:0\leq 2lk+2k-i-1 \leq M\}, \quad \beta < 0, \end{cases}\displaystyle \\ &\sum_{j=0}^{M}\dim \bigl(X_{0}^{-}(j) \bigr) =2 \textstyle\begin{cases} -N_{1}(\alpha )-N_{1}^{0}(\alpha_{+})+\operatorname{card}\{2lk+i:0\leq 2lk+i\leq M\}, \quad \alpha \geq 0, \\ -N_{1}(\alpha )+\operatorname{card}\{2lk+i:0\leq 2lk+i\leq M\}, \quad \alpha < 0, \end{cases}\displaystyle \end{aligned}$$

and

$$\begin{aligned} \sum_{j=0}^{M} \bigl[\dim X_{\infty }^{+}(j)+\dim X_{0}^{-}(j) \bigr]= 2 \bigl[N_{1}(\beta )-N_{1}(\alpha )-N_{1}^{0}( \beta_{-})-N_{1}^{0}( \alpha_{+}) \bigr]+2(M+1). \end{aligned}$$

(28)

Therefore

$$ n=N_{1}(\beta )-N_{1}(\alpha )-N_{1}^{0}( \beta_{-})-N_{1}^{0}(\alpha_{+}). $$

The proof for the case $(-1)^{s+1}=-1$ is similar.

Theorem 4.1 is proved. □

Proof of Theorem 4.2 and Theorem 4.3

Since the proof for the two theorems is similar, we prove only Theorem 4.2.

Let $X^{+}=X_{\infty }^{+}+X_{\infty }^{0}$, $X^{-}=X_{-}^{0}+X_{0} ^{0}$. Then, as in the proof of Theorem 4.1, we check conditions (a), (b), (c), (d), and (e). In the present case, we may suppose that (28) still holds for some $M>0$. Let $X_{\infty }^{0}(i)=X_{\infty }^{0} \cap X(i)$, $X _{0}^{0}(i)=X_{0}^{0}\cap X(i)$. Then

$$\begin{aligned} n =&\frac{1}{2}\sum_{i=0}^{M} \Bigl[\dim \Bigl(X_{\infty }^{+}(i)\cap X _{0}^{-}(i)\Bigr)-\operatorname{codim}_{X(i)} \bigl(X_{\infty }^{+}(i)+X_{0}^{-}(i)\bigr) \Bigr]+ \bigl(\dim X_{\infty }^{0}+\dim X_{0}^{0} \bigr) \\ =&\frac{1}{2}\sum_{i=0}^{M} \bigl[ \dim X_{\infty }^{+}(i)+\dim X_{0} ^{-}(i)-2 \bigr]+ \bigl(\dim X_{\infty }^{0}+\dim X_{0}^{0} \bigr) \\ =&\frac{1}{2}\sum_{i=0}^{M} \bigl[ \dim X_{\infty }^{+}(i)+\dim X_{0} ^{-}(i) \bigr]-(M+1)+ \bigl(\dim X_{\infty }^{0}+\dim X_{0}^{0} \bigr) \\ =&N(\beta )-N(\alpha )-N^{0}(\beta_{-})-N^{0}( \alpha_{+})+ \bigl(N ^{0}(\beta_{+})+N^{0}( \beta_{-})+N^{0}(\alpha_{+})+N^{0}( \alpha_{-}) \bigr) \\ =&N(\beta )-N(\alpha )+N^{0}(\beta_{+})+N^{0}( \alpha_{-}). \end{aligned}$$

 □

Proof of Theorem 4.4, Theorem 4.5, and Theorem 4.6

Since the proof of Theorem 4.4 is similar to that of Theorem 4.1, and the proofs of Theorems 4.5 and 4.6 are similar to that of Theorem 4.2, we omit it. Our proof is completed. □

6 Examples

Example 6.1

Suppose that $f\in C^{0}(R,R)$ satisfies

$$f(x)= \textstyle\begin{cases} 5\pi^{3} x+x^{\frac{1}{5}}, \quad \vert x \vert \gg 1, \\ -\pi^{3} x-x^{5}, \quad \vert x \vert \ll 1. \end{cases} $$

We are to discuss the multiplicity of 8-periodic solutions of the equation

$$\begin{aligned} x^{(3)}(t)= -\sum_{i=1}^{3}f \bigl(x(t-i)\bigr). \end{aligned}$$

(29)

In this case, $s=1$, $(-1)^{s+1}=1$, $k=2$, $m=1$, $\alpha =-\pi^{3}$, $\beta =5\pi^{3}$. This yields that

$$\begin{aligned} &\begin{aligned} N_{1}(\alpha )={} &{-}\operatorname{card}\biggl\{ l \geq 0:0< \biggl( \frac{(8l+1)\pi }{4} \biggr)^{3} \tan \frac{\pi }{8}< \pi^{3}\biggr\} \\ &{}-\operatorname{card}\biggl\{ l\geq 0:0< \biggl(\frac{(8l+3)\pi }{4} \biggr)^{3} \tan \frac{3 \pi }{8}< \pi^{3}\biggr\} =-1, \end{aligned} \\ &\begin{aligned} N_{1}(\beta )= {}&\operatorname{card} \biggl\{ l\geq 0:0< \biggl( \frac{(8l+8-1)\pi }{4} \biggr)^{3} \tan \frac{\pi }{8}< 5\pi^{3}\biggr\} \\ &{}+\operatorname{card}\biggl\{ l\geq 0:0< \biggl(\frac{(8l+8-2-1)\pi }{4} \biggr)^{3} \tan \frac{3 \pi }{8}< 5\pi^{3}\biggr\} =2, \end{aligned} \end{aligned}$$

and $N_{1}^{0}(\alpha_{+})=N_{1}^{0}(\beta_{-})=N_{1}^{0}(\alpha_{-})=N _{1}^{0}(\beta_{+})=0$.

Applying Theorem 4.2, we conclude that Eq. (29) possesses at least three different 8-periodic orbits satisfying $x(t-4)=-x(t)$.

Example 6.2

Suppose that $f\in C^{0}(R,R)$ satisfies

$$f(x)= \textstyle\begin{cases} 3\pi x+x^{\frac{1}{3}}, & \vert x \vert \gg 1, \\ \pi x-x^{3}, & \vert x \vert \ll 1. \end{cases} $$

We are to discuss the multiplicity of 12-periodic solutions of the equation

$$\begin{aligned} x'(t)= -\sum_{i=1}^{5}(-1)^{i+1}f \bigl(x(t-i)\bigr). \end{aligned}$$

(30)

In this case, $s=0$, $(-1)^{s+1}=-1$, $k=3$, $m=2$, $\alpha =\pi$, $\beta =3\pi$. This yields that

$$\begin{aligned} &\begin{aligned} N_{2}(\alpha )= {}&\operatorname{card}\biggl\{ l\geq 0:0< \frac{(12l+1)\pi }{6} \cot \frac{ \pi }{12}< \pi \biggr\} \\ &{}+\operatorname{card}\biggl\{ l\geq 0:0< \frac{(12l+3)\pi }{6}\cot \frac{3\pi }{12}< \pi \biggr\} \\ &{}+\operatorname{card}\biggl\{ l\geq 0:0< \frac{(12l+5)\pi }{6}\cot \frac{5\pi }{12}< \pi \biggr\} =4, \end{aligned} \\ &\begin{aligned} N_{2}(\beta )= {}&\operatorname{card} \biggl\{ l\geq 0:0< \frac{(12l+1)\pi }{6}\cot \frac{ \pi }{12}< 3 \pi \biggr\} \\ &{}+\operatorname{card}\biggl\{ l\geq 0:0< \frac{(12l+3)\pi }{6}\cot \frac{3\pi }{12}< 3\pi \biggr\} \\ &{}+\operatorname{card}\biggl\{ l\geq 0:0< \frac{(12l+5)\pi }{6}\cot \frac{5\pi }{12}< 3\pi \biggr\} =9, \end{aligned} \end{aligned}$$

and $N_{2}^{0}(\alpha_{+})=N_{2}^{0}(\beta_{-})=N_{2}^{0}(\alpha_{-})=N _{2}^{0}(\beta_{+})=0$.

Applying Theorem 4.5, we conclude that Eq. (30) possesses at least five different 12-periodic orbits satisfying $x(t-6)=-x(t)$.