Periodic orbits to Kaplan-Yorke like differential delay equations with two lags of ratio $(2k-1)/2$

Abstract

In this paper, we study the periodic solutions to a type of differential delay equations with two lags of ratio $(2k-1)/2$ in the form

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(2k-1) \bigr) \bigr),\quad k\geq2. $$

The 4k-periodic solutions are obtained by using the variational method and the method of Kaplan-Yorke coupling system. This is a new type of differential-delay equations compared with all previous researches since the ratio of two lags is not an integer. Three functionals are constructed for a discussion on critical points. An example is given to demonstrate our main results.

1 Introduction

The differential delay equations have useful applications in various fields such as age-structured population growth, control theory, and any models involving responses with nonzero delays [1–5].

Given $f\in C^{0}(R^{+},R^{-})$ with $f(-x)=-f(x),xf(x)>0,x\neq0$. Kaplan and Yorke [6] studied the existence of 4-periodic and 6-periodic solutions to the differential delay equations

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

(1.1)

and

$$ x'(t)=-f \bigl(x(t-1) \bigr)-f \bigl(x(t-2) \bigr), $$

(1.2)

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

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

(1.3)

could be studied under the restriction

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

which was proved by Nussbaum [7] in 1978 by use of a fixed point theorem on cones.

After then, a lot of papers [8–21] discussed the existence and multiplicity of $2(n+1)$-periodic solutions to equation (1.3) and its extension

$$ x'(t)=-\sum_{i=1}^{n} \operatorname{grad} F \bigl(x(t-i) \bigr), $$

(1.4)

where $F\in C^{1}(R^{N},R),F(-x)=F(x),F(0)=0$.

Recently, Zhang and Ge [22] studied the multiplicity of 2n-periodic solutions to a type of differential delay equations of the form

$$ x'(t)=-f \bigl(x(t-1) \bigr)-f \bigl(x(t-n) \bigr),\quad n \geq2, $$

(1.5)

and obtained new results.

In this paper, we study the periodic orbits to a type of differential delay equations with two lags of ratio $(2k-1)/2$ in the form

$$ x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(2k-1) \bigr) \bigr),\quad k \geq2, $$

(1.6)

which is different from (1.3) and can be regarded as a new extension of (1.2). The method applied in this paper is the variational approach in the critical point theory [23, 24].

Since the equation

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x(t-2k) \bigr),\quad k\geq2, $$

can be changed into the form of equation (1.5) by the transformation

$$t=2s,\qquad x(t)=x(2s)=y(s),\qquad\widehat{f}(y)=2f(x), $$

this paper completes the research of the equations in the form

$$ x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x(t-n) \bigr),\quad n \geq3. $$

(1.7)

In fact, it follows from

$$y'(s)=2x'(t)=-2f \bigl(x(t-2) \bigr)-2f \bigl(x(t-2k) \bigr)=-2f \bigl(x \bigl(2(s-1) \bigr) \bigr)-2f \bigl(x \bigl(2(s-k) \bigr) \bigr) $$

that

$$y'(s)=-\widehat{f} \bigl(y(s-1) \bigr)-\widehat{f} \bigl(y(s-k) \bigr) $$

for $\widehat{f}=2f$, which is much the same as equation (1.5).

We suppose that

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

(1.8)

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

$$ \lim_{x \to0} \frac{f(x)}{x}= \alpha,\qquad\lim _{x \to\infty} \frac{f(x)}{x}= \beta. $$

(1.9)

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.

(S₁):

f satisfies (1.8) and (1.9),

(S₂):

there exist $M>0$ and a function $r\in C^{0}(R^{+},R^{+})$ satisfying $r(s)\rightarrow\infty$, $r(s)\rightarrow0$ as $s\rightarrow \infty$ such that

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

($\mathrm{S}_{3}^{\pm}$):

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

($\mathrm{S}_{4}^{\pm}$):

$\pm[F(x)-\frac{1}{2}\alpha x^{2}]>0, 0<|x|\ll1$.

In this paper, we need the following lemma.

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

Lemma 1.1

([24], Theorem 2.4; [8], 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

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

• (d) there is $c_{\infty}\in R$ such that $\Phi(x)\leq c_{\infty}<\Phi(0)=0, \forall x\in X^{-}\cap S_{r}=\{x\in X^{-}:\|x\|=r\}$,

• (e) Φ satisfies $(P.S)_{c}$-condition, $c_{0}< c< c_{\infty}$. Then Φ has at least $\frac{1}{2}[\dim(X^{+}\cap X^{-})-\operatorname{co\,dim} _{X}(X^{+}\cup X^{-})]$ generally different critical orbits in $\Phi ^{-1}([c_{0},c_{\infty}])$ if

  $$\bigl[\dim\bigl(X^{+}\cap X^{-}\bigr)-\operatorname{co \,dim}_{X}\bigl(X^{+}\cup X^{-}\bigr) \bigr]>0. $$

Remark 1.1

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

In order to construct adequate functional whose critical points are the solutions of equation (1.6), we need to distinguish our problem into three cases:

$$\begin{aligned}& k=3l+2, \qquad k=3l+3, \end{aligned}$$

and

$$k=3l+4. $$

Then we construct the corresponding three functionals.

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

2.1 4k-Periodic orbits to equation (1.6) when $k=3l+2$

We are concerned at the 4k-periodic solutions to (1.6) and suppose that

$$ x \bigl(t-(6l+4) \bigr)=-x(t),\quad l\geq0. $$

(2.1)

We transform (1.6) into

$$ x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+3) \bigr) \bigr). $$

(2.2)

Let

$$\begin{aligned}& \begin{aligned}[b] \widehat{X}&= \bigl\{ x\in C_{T}:x \bigl(t-(6l+4) \bigr)=-x(t) \bigr\} \\ &= \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr):a_{i},b_{i}\in R \Biggr\} , \end{aligned} \\& \begin{aligned}[b] X={}&cl \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin \frac{(2i+1)\pi t}{6l+4} \biggr):a_{i},b_{i}\in R, \\ &{}\sum_{i=0}^{\infty}(2i+1) \bigl(a_{i}^{2}+b_{i}^{2} \bigr)< \infty \Biggr\} , \end{aligned} \end{aligned}$$

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

$$\begin{aligned} Px(t) =&P \Biggl(\sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr) \Biggr) \\ =& \sum_{i=0}^{\infty}(2i+1) \biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr). \end{aligned}$$

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

$$\begin{aligned}& \langle x,y\rangle= \int_{0}^{12l+8} \bigl(Px(t),y(t) \bigr)\,dt,\quad\|x\|= \sqrt{ \langle x,x\rangle}, \\& \langle x,y\rangle_{2}= \int_{0}^{12l+8} \bigl(x(t),y(t) \bigr)\,dt,\quad\|x \|_{2}=\sqrt {\langle x,x\rangle_{2}}. \end{aligned}$$

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

Define the functional $\Phi:X\rightarrow R$ by

$$\begin{aligned} \Phi(x)&=\frac{1}{2}\langle Lx,x\rangle+ \int_{0}^{4k}F \bigl(x(t) \bigr)\,dt \\ &=\frac{1}{2}\langle Lx,x\rangle+ \int _{0}^{12l+8}F \bigl(x(t) \bigr)\,dt, \end{aligned}$$

(2.3)

where

$$ Lx=\frac{1}{2}P^{-1} \Biggl[\sum_{i=0}^{2l+1}x'(t-3i)- \sum_{i=0}^{2l}x'(t-3i-1)-\sum _{i=0}^{2l}x'(t-3i-2) \Biggr]. $$

(2.4)

Let $k_{1}= [\frac{2k-2}{3} ]= [\frac{6l+2}{3} ]=2l, k_{2}=k-1=3l+1$, and

$$X(i)= \biggl\{ x(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin \frac{(2i+1)\pi t}{6l+4}:a_{i},b_{i}\in R \biggr\} . $$

We have

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

(2.5)

For each

$$x(t)=\sum_{i=0}^{\infty} \biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr)\in X, $$

define $\Omega:X\rightarrow X$ by

$$(\Omega x) (t)=\sum_{i=0}^{\infty} \biggl(b_{i}\cos\frac{(2i+1)\pi t}{6l+4}-a_{i}\sin\frac{(2i+1)\pi t}{6l+4} \biggr)\frac{\sin\frac{(2i+1)\pi }{12l+8}}{\sin\frac{(2i+1)3\pi}{12l+8}}. $$

If $x_{i}(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin\frac{(2i+1)\pi t}{6l+4}\in X(i), i\in N$, then we have

$$ Lx=-\frac{\pi}{12l+8} \Biggl(\Omega x+\sum_{i=0}^{\infty}x_{i} \frac{\cos\frac {(2i+1)\pi}{12l+8}}{\sin\frac{(2i+1)3\pi}{12l+8}} \Biggr). $$

(2.6)

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

By the theorem of Mawhin and Willem [25], Theorem 1.4, the functional Φ is differentiable, and its differential is

$$ \Phi'(x)=Lx+K(x), $$

(2.7)

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

It is easy to see that $\langle\Omega x,x\rangle=0$. Therefore, from (2.6) we have that if

$$x(t)=\sum_{i=0}^{\infty} \biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+4}+b_{i}\sin \frac{(2i+1)\pi t}{6l+4} \biggr), $$

then

$$\begin{aligned} &\langle Lx,x\rangle \\ &\quad=-\sum_{i=0}^{\infty}\frac{(2i+1)\pi }{2} \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \\ &\quad=\sum_{m=0}^{\infty} \Biggl[-\sum _{i=0}^{2l}\frac{(4m(3l+2)+2i+1)\pi }{2} \bigl(a_{2m(3l+2)+i}^{2} +b_{2m(3l+2)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \\ &\qquad{}+\sum_{i=0}^{2l}\frac{(4(m+1)(3l+2)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+2)-i-1}^{2} +b_{2(m+1)(3l+2)-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \\ &\qquad{}-\sum_{i=2l+1}^{3l+1}\frac{(4m(3l+2)+2i+1)\pi}{2} \bigl(a_{2m(3l+2)+i}^{2} +b_{2m(3l+2)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \\ &\qquad{}+\sum_{i=2l+1}^{3l+1}\frac{(4(m+1)(3l+2)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+2)-i-1}^{2} +b_{2(m+1)(3l+2)-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}} \Biggr]. \end{aligned}$$

On the other hand,

$$\begin{aligned} \bigl\langle P^{-1}\beta x,x \bigr\rangle =&\sum _{i=0}^{\infty}(6l+4)\beta \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \\ =&\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l}(6l+4)\beta \bigl(a_{2m(3l+2)+i}^{2}+b_{2m(3l+2)+i}^{2} \bigr) \\ &{}+\sum_{i=0}^{2l}(6l+4)\beta \bigl(a_{2m(3l+2)+2(3l+2)-i-1}^{2}+b_{2m(3l+2)+2(3l+2)-i-1}^{2} \bigr) \\ &{}+\sum_{i=2l+1}^{3l+1}(6l+4)\beta \bigl(a_{2m(3l+2)+i}^{2}+b_{2m(3l+2)+i}^{2} \bigr) \\ &{}+\sum_{i=2l+1}^{3l+1}(6l+4)\beta \bigl(a_{2m(3l+2)+2(3l+2)-i-1}^{2}+b_{2m(3l+2)+2(3l+2)-i-1}^{2} \bigr) \Biggr]. \end{aligned}$$

Therefore, we have

$$\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x \bigr\rangle \\ &\quad=(6l+4)\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l} \biggl(-\frac{(4m(3l+2)+2i+1)\pi}{12l+8} \frac{\cos \frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2m(3l+2)+i}^{2} +b_{2m(3l+2)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=0}^{2l} \biggl( \frac{(4(m+1)(3l+2)-2i-1)\pi}{12l+8}\frac{\cos \frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+2)-i-1}^{2} +b_{2(m+1)(3l+2)-i-1}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+1}^{3l+1} \biggl(- \frac{(4m(3l+2)+2i+1)\pi}{12l+8}\frac{\cos \frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}}+\beta \biggr) \bigl(a_{2m(3l+2)+i}^{2} +b_{2m(3l+2)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+1}^{3l+1} \biggl( \frac{(4(m+1)(3l+2)-2i-1)\pi}{12l+8}\frac {\cos\frac{(2i+1)\pi}{12l+8}}{ \sin\frac{(2i+1)3\pi}{12l+8}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+2)-i-1}^{2} +b_{2(m+1)(3l+2)-i-1}^{2} \bigr) \Biggr]. \end{aligned}$$

(2.8)

Lemma 2.1

Each critical point of the functional Φ is a $(12l+8)$-periodic solution of equation (1.6) satisfying (2.1).

Proof

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

$$ \frac{1}{2} \Biggl[\sum_{i=0}^{2l+1}x'(t-3i)- \sum_{i=0}^{2l}x'(t-3i-1)-\sum _{i=0}^{2l}x'(t-3i-2) \Biggr]+f \bigl(x(t) \bigr)=0. $$

(2.9)

Consequently,

$$\begin{aligned}& \frac{1}{2} \Biggl[\sum_{i=0}^{2l+1}x'(t-3i-2)- \sum_{i=1}^{2l}x'(t-3i)-\sum _{i=1}^{2l}x'(t-3i-1) \Biggr]+f \bigl(x(t-2) \bigr)=0, \end{aligned}$$

(2.10)

$$\begin{aligned}& \frac{1}{2} \Biggl[\sum_{i=0}^{2l+1}x'(t-3i+1)- \sum_{i=0}^{2l}x'(t-3i)-\sum _{i=0}^{2l}x'(t-3i-1) \Biggr]+f \bigl(x(t+1) \bigr)=0. \end{aligned}$$

(2.11)

Subtracting (2.10) from (2.11), we have

$$-x'(t)+f \bigl(x(t+1) \bigr)-f \bigl(x(t-2) \bigr)=0, $$

that is,

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+3) \bigr) \bigr), $$

which implies that x is a solution to (1.6). □

2.2 4k-Periodic orbits to equation (1.6) when $k=3l+3$

We are concerned at the 4k-periodic solutions to (1.6) and suppose that

$$ x \bigl(t-(6l+6) \bigr)=-x(t),\quad l\geq0. $$

(2.12)

We transform (1.6) into

$$ x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+5) \bigr) \bigr). $$

(2.13)

Let

$$\begin{aligned}& \begin{aligned}[b] \widehat{X}&= \bigl\{ x\in C_{T}:x \bigl(t-(6l+6) \bigr)=-x(t) \bigr\} \\ &= \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr):a_{i},b_{i} \in R \Biggr\} , \end{aligned} \\& \begin{aligned}[b] X={}&cl \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr):a_{i},b_{i}\in R, \\ &{}\sum_{i=0}^{\infty}(2i+1) \bigl(a_{i}^{2}+b_{i}^{2} \bigr)< \infty \Biggr\} , \end{aligned} \end{aligned}$$

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

$$\begin{aligned} Px(t) =&P \Biggl(\sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr) \Biggr) \\ =& \sum_{i=0}^{\infty}(2i+1) \biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr). \end{aligned}$$

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

$$\begin{aligned}& \langle x,y\rangle= \int_{0}^{12l+12} \bigl(Px(t),y(t) \bigr)\,dt,\quad\|x\|= \sqrt{ \langle x,x\rangle} , \\& \langle x,y\rangle_{2}= \int_{0}^{12l+12} \bigl(x(t),y(t) \bigr)\,dt,\quad\|x \|_{2}=\sqrt {\langle x,x\rangle_{2}} . \end{aligned}$$

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

Define the functional $\Phi:X\rightarrow R$ by

$$ \Phi(x)=\frac{1}{2}\langle Lx,x\rangle+ \int _{0}^{12l+12}F \bigl(x(t) \bigr)\,dt, $$

(2.14)

where

$$ Lx=-\frac{1}{2}P^{-1}\sum_{i=0}^{2l+1}x'(t-3i-1). $$

(2.15)

Let $k_{1}= [\frac{2k-2}{3} ]= [\frac{6l+4}{3} ]=2l+1,k_{2}=k-1=3l+2$, and

$$X(i)= \biggl\{ x(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6}:a_{i},b_{i}\in R \biggr\} . $$

We have

$$ \begin{aligned}[b] X={}&\sum_{m=0}^{\infty} \Biggl[ \sum_{i=0}^{k_{1}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr)\\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr) \Biggr]. \end{aligned} $$

(2.16)

Define $\Omega:X\rightarrow X$ by

$$(\Omega x) (t)=\sum_{i=0}^{\infty} \biggl(b_{i}\cos\frac{(2i+1)\pi t}{6l+6}-a_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr)\frac{\sin\frac{(2i+1)\pi }{12l+12}}{\sin\frac{(2i+1)3\pi}{12l+12}}. $$

Then if

$$x_{i}(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6}\in X(i),\quad i\in N, $$

then we have

$$ Lx=-\frac{\pi}{12l+12} \Biggl(\Omega x+\sum_{i=0}^{\infty}x_{i} \frac{\cos \frac{(2i+1)\pi}{12l+12}}{\sin\frac{(2i+1)3\pi}{12l+12}} \Biggr). $$

(2.17)

The functional Φ is differentiable, and its differential is

$$ \Phi'(x)=Lx+K(x), $$

(2.18)

where $K(x)=P^{-1}f(x)$. The mapping $K:(X,\|x\|^{2})\rightarrow(X,\| x\|_{2}^{2})$ is compact.

Therefore, from (2.17) it follows that, for each

$$x(t)=\sum_{i=0}^{\infty} \biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+6}+b_{i}\sin \frac{(2i+1)\pi t}{6l+6} \biggr), $$

we have

$$\begin{aligned} \langle Lx,x\rangle ={}&-\sum_{i=0}^{\infty}\frac{(2i+1)\pi}{2} \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \frac {\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \\ ={}&\sum_{m=0}^{\infty} \Biggl[-\sum _{i=0}^{2l+1}\frac{(4m(3l+3)+2i+1)\pi }{2} \bigl(a_{2m(3l+3)+i}^{2} +b_{2m(3l+3)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \\ &{}+\sum_{i=0}^{2l+1}\frac{(4(m+1)(3l+3)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+3)-i-1}^{2} +b_{2(m+1)(3l+3)-i-1}^{2} \bigr)\\ &{}\times \frac{\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \\ &{}-\sum_{i=2l+2}^{3l+2}\frac{(4m(3l+3)+2i+1)\pi}{2} \bigl(a_{2m(3l+3)+i}^{2} +b_{2m(3l+3)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \\ &{}+\sum_{i=2l+2}^{3l+2}\frac{(4(m+1)(3l+3)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+3)-i-1}^{2} +b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \\ &{}\times \frac{\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}} \Biggr]. \end{aligned}$$

On the other hand,

$$\begin{aligned}[b] \bigl\langle P^{-1}\beta x,x \bigr\rangle ={}&\sum _{m=0}^{\infty}(6l+6)\beta \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \\ ={}&\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l+1}(6l+6)\beta \bigl(a_{2m(3l+3)+i}^{2}+b_{2m(3l+3)+i}^{2} \bigr) \\ &{}+\sum_{i=0}^{2l+1}(6l+6)\beta \bigl(a_{2(m+1)(3l+3)-i-1}^{2}+b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \\ &{}+\sum_{i=2l+2}^{3l+2}(6l+6)\beta \bigl(a_{2m(3l+3)+i}^{2}+b_{2m(3l+3)+i}^{2} \bigr) \\ &{}+\sum_{i=2l+2}^{3l+2}(6l+6)\beta \bigl(a_{2(m+1)(3l+3)-i-1}^{2}+b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \Biggr]. \end{aligned} $$

Therefore, we have

$$\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x \bigr\rangle \\ &\quad=(6l+6)\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l+1} \biggl(-\frac{(4m(3l+3)+2i+1)\pi}{12l+12} \frac {\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2m(3l+3)+i}^{2} +b_{2m(3l+3)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=0}^{2l+1} \biggl( \frac{(4(m+1)(3l+3)-i-1)\pi}{12l+12}\frac{\cos \frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+3)-i-1}^{2} +b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+2}^{3l+2} \biggl(- \frac{(4m(3l+3)+2i+1)\pi}{12l+12}\frac {\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}}+\beta \biggr) \bigl(a_{2m(3l+3)+i}^{2} +b_{2m(3l+3)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+2}^{3l+2} \biggl( \frac{(4(m+1)(3l+3)-2i-1)\pi}{12l+12}\frac {\cos\frac{(2i+1)\pi}{12l+12}}{ \sin\frac{(2i+1)3\pi}{12l+12}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+3)-i-1}^{2} +b_{2(m+1)(3l+3)-i-1}^{2} \bigr) \Biggr]. \end{aligned}$$

(2.19)

Lemma 2.2

Each critical point of the functional Φ is a $(12l+12)$-periodic solution of equation (1.6) satisfying (2.12).

Proof

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

$$ -\frac{1}{2}\sum_{i=0}^{2l+1}x'(t-3i-1)+f \bigl(x(t) \bigr)=0. $$

(2.20)

Consequently, we have

$$\begin{aligned}& -\frac{1}{2}\sum_{i=0}^{2l+1}x'(t-3i-3)+f \bigl(x(t-2) \bigr)=0, \end{aligned}$$

(2.21)

$$\begin{aligned}& -\frac{1}{2}\sum_{i=0}^{2l+1}x'(t-3i)+f \bigl(x(t+1) \bigr)=0. \end{aligned}$$

(2.22)

Subtracting (2.21) from (2.22), we have

$$-x'(t)+f \bigl(x(t+1) \bigr)-f \bigl(x(t-2) \bigr)=0, $$

that is,

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+5) \bigr) \bigr), $$

which implies that x is a solution to (1.6). □

2.3 4k-Periodic orbits to equation (1.6) when $k=3l+4$

We are concerned at the 4k-periodic solutions to (1.6) and suppose that

$$ x \bigl(t-(6l+8) \bigr)=-x(t),\quad l\geq0. $$

(2.23)

We transform (1.6) into

$$ x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+7) \bigr) \bigr). $$

(2.24)

Let

$$\begin{aligned}& \begin{aligned}[b] \widehat{X}&= \bigl\{ x\in C_{T}:x \bigl(t-(6l+8) \bigr)=-x(t) \bigr\} \\ &= \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr):a_{i},b_{i} \in R \Biggr\} , \end{aligned} \\& \begin{aligned}[b] X={}&cl \Biggl\{ \sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr):a_{i},b_{i}\in R, \\ &{}\sum_{i=0}^{\infty}(2i+1) \bigl(a_{i}^{2}+b_{i}^{2} \bigr)< \infty \Biggr\} , \end{aligned} \end{aligned}$$

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

$$\begin{aligned} Px(t) =&P \Biggl(\sum_{i=0}^{\infty}\biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr) \Biggr) \\ =& \sum_{i=0}^{\infty}(2i+1) \biggl(a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr). \end{aligned}$$

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

$$\begin{aligned}& \langle x,y\rangle= \int_{0}^{12l+16} \bigl(Px(t),y(t) \bigr)\,dt,\quad\|x\|= \sqrt{ \langle x,x\rangle} , \\& \langle x,y\rangle_{2}= \int_{0}^{12l+16} \bigl(x(t),y(t) \bigr)\,dt,\quad\|x \|_{2}=\sqrt {\langle x,x\rangle_{2}} . \end{aligned}$$

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

Define the functional $\Phi:X\rightarrow R$ by

$$ \Phi(x)=\frac{1}{2}\langle Lx,x\rangle+ \int _{0}^{12l+16}F \bigl(x(t) \bigr)\,dt, $$

(2.25)

where

$$ Lx=\frac{1}{2}P^{-1} \Biggl[-\sum_{i=0}^{2l+2}x'(t-3i)- \sum_{i=0}^{2l+2}x'(t-3i-1)+\sum _{i=0}^{2l+1}x'(t-3i-2) \Biggr]. $$

(2.26)

Let $k_{1}= [\frac{2k-2}{3} ]= [\frac{6l+6}{3} ]=2l+2,k_{2}=k-1=3l+3$, and

$$X(i)= \biggl\{ x(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8}:a_{i},b_{i}\in R \biggr\} . $$

We have

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

(2.27)

Define $\Omega:X\rightarrow X$ by

$$(\Omega x) (t)=\sum_{i=0}^{\infty} \biggl(b_{i}\cos\frac{(2i+1)\pi t}{6l+8}-a_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr)\frac{\sin\frac{(2i+1)\pi }{12l+16}}{\sin\frac{(2i+1)3\pi}{12l+16}}. $$

Then if

$$x_{i}(t)=a_{i}\cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8}\in X(i),\quad i\in N, $$

then we have

$$ Lx=-\frac{\pi}{12l+16} \Biggl(\Omega x+\sum_{i=0}^{\infty}x_{i} \frac{\cos \frac{(2i+1)\pi}{12l+16}}{\sin\frac{(2i+1)3\pi}{12l+16}} \Biggr). $$

(2.28)

The functional Φ is differentiable, and its differential is

$$ \Phi'(x)=Lx+K(x), $$

(2.29)

where $K(x)=P^{-1}f(x)$. The mapping $K:(X,\|x\|^{2})\rightarrow(X,\| x\|_{2}^{2})$ is compact.

For

$$x(t)=\sum_{i=0}^{\infty} \biggl(a_{i} \cos\frac{(2i+1)\pi t}{6l+8}+b_{i}\sin \frac{(2i+1)\pi t}{6l+8} \biggr), $$

we have

$$\begin{aligned} &\langle Lx,x\rangle \\ &\quad=-\sum_{i=0}^{\infty}\frac{(2i+1)\pi}{2} \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \frac {\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \\ &\quad=\sum_{m=0}^{\infty} \Biggl[-\sum _{i=0}^{2l+2}\frac{(4m(3l+4)+2i+1)\pi }{2} \bigl(a_{2m(3l+4)+i}^{2} +b_{2m(3l+4)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \\ &\qquad{}+\sum_{i=0}^{2l+2}\frac{(4(m+1)(3l+4)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+4)-i-1}^{2} +b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \\ &\qquad{}-\sum_{i=2l+3}^{3l+3}\frac{(4m(3l+4)+2i+1)\pi}{2} \bigl(a_{2m(3l+4)+i}^{2} +b_{2m(3l+4)+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \\ &\qquad{}+\sum_{i=2l+3}^{3l+3}\frac{(4(m+1)(3l+4)-2i-1)\pi}{2} \bigl(a_{2(m+1)(3l+4)-i-1}^{2} +b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \\ &\qquad{}\times \frac{\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}} \Biggr]. \end{aligned}$$

On the other hand,

$$\begin{aligned} \bigl\langle P^{-1}\beta x,x \bigr\rangle =&\sum _{m=0}^{\infty}(6l+8)\beta \bigl(a_{i}^{2}+b_{i}^{2} \bigr) \\ =&\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l+2}(6l+8)\beta \bigl(a_{2m(3l+4)+i}^{2}+b_{2m(3l+4)+i}^{2} \bigr) \\ &{}+\sum_{i=0}^{2l+2}(6l+8)\beta \bigl(a_{2(m+1)(3l+4)-i-1}^{2}+b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \\ &{}+\sum_{i=2l+3}^{3l+3}(6l+8)\beta \bigl(a_{2m(3l+4)+i}^{2}+b_{2m(3l+4)+i}^{2} \bigr) \\ &{}+\sum_{i=2l+3}^{3l+3}(6l+8)\beta \bigl(a_{2(m+1)(3l+4)-i-1}^{2}+b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \Biggr]. \end{aligned}$$

Therefore, we have

$$\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x \bigr\rangle \\ &\quad=(6l+8)\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{2l+2} \biggl(-\frac{(4m(3l+4)+2i+1)\pi}{12l+16} \frac {\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2m(3l+4)+i}^{2} +b_{2m(3l+4)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=0}^{2l+2} \biggl( \frac{(4(m+1)(3l+4)-i-1)\pi}{12l+16}\frac{\cos \frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+4)-i-1}^{2} +b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+3}^{3l+3} \biggl(- \frac{(4m(3l+4)+2i+1)\pi}{12l+16}\frac {\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}}+\beta \biggr) \bigl(a_{2m(3l+4)+i}^{2} +b_{2m(3l+4)+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=2l+3}^{3l+3} \biggl( \frac{(4(m+1)(3l+4)-2i-1)\pi}{12l+16}\frac {\cos\frac{(2i+1)\pi}{12l+16}}{ \sin\frac{(2i+1)3\pi}{12l+16}}+\beta \biggr) \\ &\qquad{}\times \bigl(a_{2(m+1)(3l+4)-i-1}^{2} +b_{2(m+1)(3l+4)-i-1}^{2} \bigr) \Biggr]. \end{aligned}$$

(2.30)

Lemma 2.3

Each critical point of the functional Φ is a $(12l+16)$-periodic solution of equation (1.6) satisfying (2.23).

Proof

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

$$ \frac{1}{2} \Biggl[-\sum_{i=0}^{2l+2}x'(t-3i)- \sum_{i=0}^{2l+2}x'(t-3i-1)+\sum _{i=0}^{2l+1}x'(t-3i-2) \Biggr]+f \bigl(x(t) \bigr)=0. $$

(2.31)

Consequently,

$$\begin{aligned}& \frac{1}{2} \Biggl[-\sum_{i=0}^{2l+2}x'(t-3i-2)- \sum_{i=0}^{2l+2}x'(t-3i-3)+\sum _{i=0}^{2l+1}x'(t-3i-4) \Biggr]+f \bigl(x(t-2) \bigr)=0, \end{aligned}$$

(2.32)

$$\begin{aligned}& \frac{1}{2} \Biggl[-\sum_{i=0}^{2l+2}x'(t-3i+1)- \sum_{i=0}^{2l+2}x'(t-3i)+\sum _{i=0}^{2l+1}x'(t-3i-1) \Biggr]+f \bigl(x(t+1) \bigr)=0. \end{aligned}$$

(2.33)

Subtracting (2.32) from (2.33), we have

$$-x'(t)+f \bigl(x(t+1) \bigr)-f \bigl(x(t-2) \bigr)=0, $$

that is,

$$x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x \bigl(t-(6l+7) \bigr) \bigr), $$

which implies that x is a solution to (1.6). □

3 Partition of space X and symbols

In fact, in Sections 2.1, 2.2, and 2.3, we could let

$$\begin{aligned}& k_{1}= \biggl[\frac{2k-2}{3} \biggr],\qquad k_{2}=k-1, \\& X(i)= \biggl\{ x(t)=a_{i}\cos\frac{(2i+1)\pi t}{2k}+b_{i}\sin \frac{(2i+1)\pi t}{2k}:a_{i},b_{i}\in R \biggr\} , \end{aligned}$$

and

$$X=\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{k_{1}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr)+\sum _{i=k_{1}+1}^{k_{2}} \bigl(X(2mk+i)+X(2mk+2k-i-1) \bigr) \Biggr] $$

and define $\Omega:X\rightarrow X$ by

$$(\Omega x) (t)=\sum_{i=0}^{\infty} \biggl(b_{i}\cos\frac{(2i+1)\pi t}{2k}-a_{i}\sin \frac{(2i+1)\pi t}{2k} \biggr)\frac{\sin\frac{(2i+1)\pi }{4k}}{\sin\frac{(2i+1)3\pi}{4k}}. $$

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

$$Lx=-\frac{\pi}{4k} \Biggl(\Omega x+\sum_{i=0}^{\infty}x_{i} \frac{\cos\frac {(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} \Biggr), $$

and, for each

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

we have

$$\begin{aligned} \langle Lx,x\rangle =&-\sum_{i=0}^{\infty} \frac{(2i+1)\pi }{2} \bigl(a_{i}^{2}+b_{i}^{2} \bigr)\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \\ =&\sum_{m=0}^{\infty} \Biggl[-\sum _{i=0}^{k_{1}}\frac{(4mk+2i+1)\pi }{2} \bigl(a_{2mk+i}^{2} +b_{2mk+i}^{2} \bigr)\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \\ &{}+\sum_{i=0}^{k_{1}}\frac{(4(m+1)k-2i-1)\pi}{2} \bigl(a_{2(m+1)k-i-1}^{2} +b_{2(m+1)k-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \\ &{}-\sum_{i=k_{1}+1}^{k_{2}}\frac{(4mk+2i+1)\pi}{2} \bigl(a_{2mk+i}^{2} +b_{2mk+i}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}}\frac{(4(m+1)k-2i-1)\pi}{2} \bigl(a_{2(m+1)k-i-1}^{2} +b_{2(m+1)k-i-1}^{2} \bigr) \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \Biggr]. \end{aligned}$$

On the other hand,

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

Therefore, we have

$$\begin{aligned} & \bigl\langle \bigl(L+P^{-1}\beta \bigr) x,x \bigr\rangle \\ &\quad=2k\sum_{m=0}^{\infty} \Biggl[\sum _{i=0}^{k_{1}} \biggl(-\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac {(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta \biggr) \bigl(a_{2mk+i}^{2} +b_{2mk+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=0}^{k_{1}} \biggl( \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac {(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta \biggr) \bigl(a_{2(m+1)k-i-1}^{2} +b_{2(m+1)k-i-1}^{2} \bigr) \\ &\qquad{}+\sum_{i=k_{1}+1}^{k_{2}} \biggl(- \frac{(4mk+2i+1)\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta \biggr) \bigl(a_{2mk+i}^{2} +b_{2mk+i}^{2} \bigr) \\ &\qquad{}+\sum_{i=k_{1}+1}^{k_{2}} \biggl( \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta \biggr) \bigl(a_{2(m+1)k-i-1}^{2} +b_{2(m+1)k-i-1}^{2} \bigr) \Biggr]. \end{aligned}$$

Let

$$\begin{aligned} X_{\infty}^{+} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{} \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \biggr\} , \\ X_{\infty}^{-} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0 \biggr\} , \\ X_{0}^{+} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha>0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{} \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha>0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha>0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha>0 \biggr\} , \\ X_{0}^{-} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha< 0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha< 0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha< 0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha< 0 \biggr\} . \end{aligned}$$

On the other hand,

$$\begin{aligned} X_{\infty}^{0} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta=0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{} \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta=0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta=0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta=0 \biggr\} , \\ X_{0}^{0} ={}& \biggl\{ X(2mk+i):m\geq0,0\leq i\leq k_{1}, -\frac{(4mk+2i+1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha=0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,0\leq i\leq k_{1}, \\ &{} \frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha=0 \biggr\} \\ &{}\cup \biggl\{ X(2mk+i):m\geq0,k_{1}+1\leq i\leq k_{2}, - \frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha=0 \biggr\} \\ &{}\cup \biggl\{ X \bigl(2(m+1)k-i-1 \bigr):m\geq0,k_{1}+1 \leq i\leq k_{2}, \\ &{}\frac{(4(m+1)k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\alpha=0 \biggr\} . \end{aligned}$$

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

Lemma 3.1

Under assumptions (S₁) and (S₂), there is $\sigma>0$ such that

$$\bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle >\sigma\|x \|^{2},\quad x\in X_{\infty}^{+}, $$

and

$$ \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle < -\sigma\|x \|^{2},\quad x\in X_{\infty}^{-} . $$

(3.1)

Proof

First, we have that, for $\beta\geq0$ and $i\in\{0,1,\ldots ,k_{1}\}$,

$$-\frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>-\frac{(4m_{0}^{+}(i)k+2i+1)\pi }{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0, $$

where $m_{0}^{+}(i)=\max \{m\in N:-\frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac {(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \}$, and

$$-\frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< -\frac{(4m_{0}^{-}(i)k+2i+1)\pi }{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< 0, $$

where $m_{0}^{-}(i)=\min \{m\in N:-\frac{(4mk+2i+1)\pi}{4k}\frac{\cos\frac {(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta<0 \}$.

In this case, we may choose

$$\begin{aligned} \sigma_{i} =&\min \biggl\{ -\frac{\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+ \frac{\beta}{4m_{0}^{+}(i)k+2i+1}, \\ &{}\frac{\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}-\frac{\beta}{4m_{0}^{-}(i)k+2i+1} \biggr\} >0. \end{aligned}$$

and let $\sigma^{0}=\min\{\sigma_{0},\sigma_{1},\ldots,\sigma_{k_{1}}\}>0$. Further, for $\beta\geq0$ and $i\in\{k_{1}+1,\ldots,k_{2}\}$, we have

$$\frac{(4mk+4k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>\frac{(4m_{1}^{+}(i)k+4k-2i-1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+ \beta>0, $$

where $m_{1}^{+}(i)=\max \{m\in N:\frac{(4mk+4k-2i-1)\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta>0 \}$, and

$$\frac{(4mk+4k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta< \frac{(4m_{1}^{-}(i)k+4k-2i-1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+ \beta< 0, $$

where $m_{1}^{-}(i)=\min \{m\in N:\frac{(4mk+4k-2i-1)\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\beta<0 \}$.

In this case, we may choose

$$\begin{aligned} \sigma_{i} =&\min \biggl\{ \frac{\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+ \frac{\beta}{4m_{1}^{+}(i)k+4k-2i-1}, \\ &{}-\frac{\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}-\frac{\beta}{4m_{1}^{-}(i)k+4k-2i-1} \biggr\} >0, \end{aligned}$$

and let $\sigma^{1}=\min\{\sigma_{k_{1}+1},\sigma_{k_{1}+2},\ldots ,\sigma_{k_{2}}\}>0$.

Let $\sigma=\min\{\sigma^{0},\sigma^{1}\}=\min\{\sigma_{1},\sigma _{2},\ldots,\sigma_{k_{2}}\}$. The proof for the case $\beta<0$ is similar. We omit it. The inequalities in (3.1) are proved. □

Lemma 3.2

Under conditions (S₁) and (S₂), the functional Φ defined by (2.3) satisfies $(P.S)$-condition if

$$\biggl|\Phi(x)-\frac{1}{2}\beta x^{2}\biggr|>r\bigl(|x|\bigr)-M_{0},\quad x \in R, $$

for some $M_{0}>0$ and some function $r\in C^{0}(R^{+},R^{+})$ that satisfies

$$ r(s)\rightarrow\infty,\qquad r(s)/s\rightarrow0 \quad\textit{as } s\rightarrow \infty. $$

(3.2)

Proof

Let $\Pi,N,Z$ be the orthogonal projections from X onto $X_{\infty}^{+},X_{\infty}^{-},X_{\infty}^{0}$, respectively. From the second condition in (1.9) it follows that

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

(3.3)

for some $M>0$.

Assume that $\{x_{n}\}\subset X$ is a subsequence such that $\Phi '(x_{n})\rightarrow0$ and $\Phi(x_{n})$ is bounded. Let $w_{n}=\Pi x_{n},y_{n}=Nx_{n},z_{n}=Zx_{n}$. Then we have

$$ \Pi \bigl(L+P^{-1}\beta \bigr)= \bigl(L+P^{-1}\beta \bigr) \Pi,\qquad N \bigl(L+P^{-1}\beta \bigr)= \bigl(L+P^{-1}\beta \bigr)N. $$

(3.4)

From

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

and (3.4) we have

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

and then, by (3.1),

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

which, together with $\Pi\Phi'(x_{n})\rightarrow0$, implies the boundedness of $w_{n}$. Similarly, we have the boundedness of $y_{n}$. At the same time, (S₂) yields

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

Then the boundedness of $\Phi(x)$ 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 (2.7) that

$$\begin{aligned} (\Pi+N)\Phi'(x_{n}) =&(\Pi+N)Lx_{n}+( \Pi+N)Kx_{n}\\ =&L(w_{n}+y_{n})+(\Pi+N)Kx_{n}. \end{aligned}$$

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

$$ L|_{x_{\infty}^{+}+x_{\infty}^{-}}(w_{n}+y_{n})\rightarrow-(\Pi +N)u. $$

(3.5)

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

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

which implies $(P.S)$-condition. □

Lemma 3.3

Under conditions (S₁) and (S₂), the functional Φ defined by (2.14) and (2.25) satisfies $(P.S)$-condition if

$$\biggl|\Phi(x)-\frac{1}{2}\beta x^{2}\biggr|>r\bigl(|x|\bigr)-M_{0},\quad x \in R, $$

for some $M_{0}>0$ and some function $r\in C^{0}(R^{+},R^{+})$, which satisfies

$$r(s)\rightarrow\infty,\qquad r(s)/s\rightarrow0 \quad\textit{as } s\rightarrow \infty. $$

The proof of Lemma 3.3 is the same as that of Lemma 3.2, and we omit it.

4 Notation and main results of this paper

We first give some notation.

Denote

$$\begin{aligned}& N(\alpha)=\left \{ \textstyle\begin{array}{@{}l} -\sum_{i=0}^{k_{1}}\operatorname{card} \{m\geq0:0< \frac{(4mk+4k-2i-1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}}< -\alpha \}\\ \quad{}-\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \{m\geq0:0< \frac{(4mk+2i+1)\pi }{4k} |\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} |< -\alpha \}, \quad\alpha< 0,\\ \sum_{i=0}^{k_{1}}\operatorname{card} \{m\geq0:0< \frac{(4mk+2i+1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}}< \alpha \}\\ \quad{}+\sum_{i=k_{1}+1}^{k_{2}}\operatorname{card} \{m\geq0:0< \frac{(4mk+4k-2i-1)\pi}{4k} |\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} |< \alpha \}, \quad\alpha\geq0, \end{array}\displaystyle \right . \\& N(\beta)=\left \{ \textstyle\begin{array}{ll} -\sum_{i=0}^{k_{1}}\operatorname{card} \{m\geq0:0< \frac{(4mk+4k-2i-1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}}< -\beta \}\\ \quad{}-\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \{m\geq0:0< \frac{(4mk+2i+1)\pi }{4k} |\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} |< -\beta \}, \quad\beta< 0,\\ \sum_{i=0}^{k_{1}}\operatorname{card} \{m\geq0:0< \frac{(4mk+2i+1)\pi }{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}}< \beta \}\\ \quad{}+\sum_{i=k_{1}+1}^{k_{2}}\operatorname{card} \{m\geq0:0< \frac{(4mk+4k-2i-1)\pi}{4k} |\frac{\cos\frac{(2i+1)\pi}{4k}}{\sin\frac{(2i+1)3\pi}{4k}} |< \beta \}, \quad \beta\geq0, \end{array}\displaystyle \right . \end{aligned}$$

and

$$\begin{aligned} N^{0}(\alpha_{-}) =&\sum_{i=0}^{k_{1}} \operatorname{\operatorname{card}} \biggl\{ m\geq0:0< \frac {(4mk+4k-2i-1)\pi}{4k} \frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}=-\alpha \biggr\} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \biggl\{ m\geq0:0< \frac{(4mk+2i+1)\pi }{4k} \biggl|\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \biggr|=-\alpha \biggr\} , \quad \alpha< 0, \\ N^{0}(\alpha_{+}) =&\sum_{i=0}^{k_{1}} \operatorname{card} \biggl\{ m\geq0:0< \frac {(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}=\alpha \biggr\} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \biggl\{ m\geq0:0< \frac{(4mk+4k-2i-1)\pi }{4k} \biggl|\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \biggr|=\alpha \biggr\} , \quad \alpha\geq0, \\ N^{0}(\beta_{-}) =&\sum_{i=0}^{k_{1}} \operatorname{card} \biggl\{ m\geq0:0< \frac {(4mk+4k-2i-1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}=-\beta \biggr\} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \biggl\{ m\geq0:0< \frac{(4mk+2i+1)\pi }{4k} \biggl|\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \biggr|=-\beta \biggr\} , \quad \beta< 0, \\ N^{0}(\beta_{+}) =&\sum_{i=0}^{k_{1}} \operatorname{card} \biggl\{ m\geq0:0< \frac {(4mk+2i+1)\pi}{4k}\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}=\beta \biggr\} \\ &{}+\sum_{i=k_{1}+1}^{k_{2}} \operatorname{card} \biggl\{ m\geq0:0< \frac{(4mk+4k-2i-1)\pi }{4k} \biggl|\frac{\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}} \biggr|=\beta \biggr\} ,\quad \beta\geq0. \end{aligned}$$

Now we give the main results of this paper.

Theorem 4.1

Suppose that (S₁) and (S₂) hold. Then equation (1.6) possesses at least

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

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

Theorem 4.2

Suppose that (S₁), (S₂), ($\mathrm{S}_{3}^{+}$), and ($\mathrm{S}_{4}^{-}$) hold. Then equation (1.6) possesses at least

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

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

Theorem 4.3

Suppose that (S₁), (S₂), ($\mathrm{S}_{3}^{-}$), and ($\mathrm{S}_{4}^{+}$) hold. Then equation (1.6) possesses at least

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

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

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

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,

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

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 (S₁) and (S₂).

In fact, condition (S₁) implies that on $X^{-}$ we have $\Phi (x)>0$ if $0<\|x\|\ll1$, that is, there are $r>0$ and $c_{\infty}<0$ such that

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

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

Then

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

if $x\in X^{+}$. Clearly, there is $c_{0}< c_{\infty}$ such that $\Phi (x)\geq c_{0},x\in X^{+}$.

Our last task is to compute the value of

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

By computation we get that, for each $i\in\{0,1,\ldots,k_{2}\}$,

$$\begin{aligned}& \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle = \biggl(- \frac{\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\frac{\beta}{4mk+2i+1} \biggr)\|x \|^{2},\quad x\in X(2mk+i), \\& \bigl\langle \bigl(L+P^{-1}\beta \bigr)x,x \bigr\rangle = \biggl( \frac{\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\frac{\beta}{4mk+4k-2i-1} \biggr)\|x\| ^{2},\\& \quad x\in X(2mk+2k-i-1), \end{aligned}$$

and

$$\begin{aligned}& \bigl\langle \bigl(L+P^{-1}\alpha \bigr)x,x \bigr\rangle = \biggl(- \frac{\pi}{4k}\frac {\cos\frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\frac{\alpha}{4mk+2i+1} \biggr)\|x\| ^{2},\quad x\in X(2mk+i), \\& \bigl\langle \bigl(L+P^{-1}\alpha \bigr)x,x \bigr\rangle = \biggl( \frac{\pi}{4k}\frac{\cos \frac{(2i+1)\pi}{4k}}{ \sin\frac{(2i+1)3\pi}{4k}}+\frac{\alpha}{4mk+4k-2i-1} \biggr)\|x\| ^{2}, \\& \quad x\in X(2mk+2k-i-1). \end{aligned}$$

Therefore,

$$\begin{aligned}& X_{\infty}^{+}(2mk+i)=X_{\infty}^{+}\cap X(2mk+i)=\emptyset, \\& X_{\infty}^{+}(2mk+2k-i-1)=X_{\infty}^{+}\cap X(2mk+2k-i-1)=X(2mk+2k-i-1), \\& X_{0}^{-}(2mk+i)=X_{0}^{-}\cap X(2mk+i)=X(2mk+i), \\& X_{0}^{-}(2mk+2k-i-1)=X_{0}^{-}\cap X(2mk+2k-i-1)=\emptyset \end{aligned}$$

if $i\in\{0,1,\ldots,k_{1}\}$ and $m\geq0$ is large enough,

$$\begin{aligned}& X_{\infty}^{+}(2mk+i)=X_{\infty}^{+}\cap X(2mk+i)=X(2mk+i), \\& X_{\infty}^{+}(2mk+2k-i-1)=X_{\infty}^{+}\cap X(2mk+2k-i-1)=\emptyset, \\& X_{0}^{-}(2mk+i)=X_{0}^{-}\cap X(2mk+i)= \emptyset, \\& X_{0}^{-}(2mk+2k-i-1)=X_{0}^{-}\cap X(2mk+2k-i-1)=X(2mk+2k-i-1) \end{aligned}$$

if $i\in\{k_{1}+1,\ldots,k_{2}\}$ and $m\geq0$ is large enough, which means that there is $M>0$ such that $\dim(X_{\infty}^{+}(j)\cap X_{0}^{-}(j))-\operatorname{co\,dim}_{X}(X_{\infty }^{+}(j)+X_{0}^{-}(j))=0, j>M$, from which it follows that

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

Then we have

$$\begin{aligned}& \sum_{j=0}^{M}\dim \bigl(X_{\infty}^{+}(j) \bigr) \\& \quad=2\left \{ \textstyle\begin{array}{@{}l} N(\beta)+\operatorname{card}\{2mk+2k-i-1:0\leq2mk+2k-i-1\leq M,0\leq i\leq k_{1}\}\\ \quad{}+\operatorname{card}\{2mk+i:0\leq2mk+i\leq M,k_{1}+1\leq i\leq k_{2}\}, \quad\beta>0,\\ N(\beta)-N^{0}(\beta_{-})+\operatorname{card}\{2mk+2k-i-1:0\leq2mk+2k-i-1\leq M,0\leq i\leq k_{1}\}\\ \quad{}+\operatorname{card}\{2mk+i:0\leq2mk+i\leq M,k_{1}+1\leq i\leq k_{2}\}, \quad\beta< 0, \end{array}\displaystyle \right . \\& \sum_{j=0}^{M}\dim \bigl(X_{0}^{-}(j) \bigr) \\& \quad=2\left \{ \textstyle\begin{array}{@{}l} -N(\alpha)-N^{0}(\alpha_{+})+\operatorname{card}\{2mk+i:0\leq2mk+i\leq M,0\leq i\leq k_{1}\}\\ \qquad{}+\operatorname{card}\{2mk+2k-i-1:0\leq2mk+2k-i-1\leq M,k_{1}+1\leq i\leq k_{2}\},\\ \quad\alpha>0,\\ -N(\alpha)+\operatorname{card}\{2mk+i:0\leq2mk+i\leq M,0\leq i\leq k_{1}\}\\ \qquad{}+\operatorname{card}\{2mk+2k-i-1:0\leq2mk+2k-i-1\leq M,k_{1}+1\leq i\leq k_{2}\},\\ \quad\alpha< 0, \end{array}\displaystyle \right . \end{aligned}$$

and

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

(5.1)

Therefore,

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

Theorem 4.1 is proved. □

Proof of Theorem 4.2 and Theorem 4.3

Since the proof for the two theorems is almost the same, 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 (5.1) 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

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

Our proof is completed. □

6 An example

Example

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

$$f(x)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{5\pi}{2}x+x^{\frac{1}{3}}, &|x|\gg1,\\ -\frac{7\pi}{2}x-x^{3}, & |x|\ll1. \end{array}\displaystyle \right . $$

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

$$ x'(t)=-f \bigl(x(t-2) \bigr)-f \bigl(x(t-5) \bigr). $$

(6.1)

In this case, $k=3,k_{1}=1,k_{2}=2,\alpha=-\frac{7\pi}{2},\beta=\frac {5\pi}{2}$. This yields that

$$\begin{aligned}& \begin{aligned}[b] N(\alpha)={}&{-}\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+12-1)\pi}{12} \frac{\cos\frac{\pi }{12}}{\sin\frac{3\pi}{12}}< \frac{7\pi}{2} \biggr\} \\ &{}-\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+12-3)\pi}{12}\frac{\cos\frac{3\pi }{12}}{\sin\frac{9\pi}{12}}< \frac{7\pi}{2} \biggr\} \\ &{}-\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+5)\pi}{12} \biggl|\frac{\cos\frac{5\pi }{12}}{\sin\frac{15\pi}{12}} \biggr|< \frac{7\pi}{2} \biggr\} =-15, \end{aligned}\\& \begin{aligned}[b] N(\beta)={}&\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+1)\pi}{12} \frac{\cos\frac{\pi }{12}}{\sin\frac{3\pi}{12}}< \frac{5\pi}{2} \biggr\} \\ &{}+\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+3)\pi}{12}\frac{\cos\frac{3\pi}{12}}{\sin \frac{9\pi}{12}}< \frac{5\pi}{2} \biggr\} \\ &{}+\operatorname{card} \biggl\{ m\geq0:0< \frac{(12m+12-5)\pi}{12} \biggl|\frac{\cos\frac{5\pi }{12}}{\sin\frac{15\pi}{12}} \biggr|< \frac{5\pi}{2} \biggr\} =12, \end{aligned}\\& N^{0}(\alpha_{+})=N^{0}(\beta_{-})=N^{0}( \alpha_{-})=N^{0}(\beta_{+})=0. \end{aligned}$$

Applying Theorem 4.2, we conclude that equation (6.1) possesses at least 27 different 12-periodic orbits satisfying $x(t-6)=-x(t)$ since f satisfies (S₁), (S₂), ($\mathrm{S}_{3}^{+}$), and ($\mathrm{S}_{4}^{-}$).