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# Multiple periodic solutions to a class of nonautonomous second-order delay differential equation

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## Abstract

The existence of the nontrivial periodic solutions for nonautonomous second-order delay differential equation

$x ″ (t)+λx(t)=−f ( t , x ( t ) , x ( t − τ ) )$

is investigated, where $λ> π 2 / τ 2$, $τ>0$, $f∈ C 1 (R× R 2 ,R)$. Multiple periodic solutions are obtained by some recent critical point theorems.

MSC:34K13, 34K18, 58E50.

## 1 Introduction

It is well known that the critical point theory is a powerful tool to deal with the multiplicity of periodic solutions to ordinary differential systems as well as partial differential equations (see ). In 1998, Li and He  first applied the critical point theory to study the multiplicity of periodic solutions for delay differential equations. Especially, in 2005, Guo and Yu  established a variational framework for delay differential autonomous systems. In the past several years, some results on the existence of periodic solutions for the functional differential equation have been obtained by the critical point theory (see ). However, most of these functional differential equations are autonomous, the results on the non-autonomous functional differential equations are relatively few (see [12, 13]).

Motivated by the work of [11, 12], we consider a class of nonautonomous second-order delay differential equation

$x ″ (t)+λx(t)=−f ( t , x ( t ) , x ( t − τ ) ) ,$
(1.1)

where $λ> π 2 / τ 2$, $τ>0$, $f∈ C 1 (R× R 2 ,R)$.

In this paper, we have the following conditions on f.

(f 1) $f(t+τ,x)=f(t,x)$, $f(t,−x,−y)=−f(t,x,y)$ and

$∂ f ( t , x , y ) ∂ y = ∂ f ( t , y , x ) ∂ x$

for all $x,y∈R$, $t∈[0,τ]$;

(f 2) there exist four τ-periodic and continuous functions $a(t)$, $b(t)$, $c(t)$ and $d(t)$ such that

and

uniformly for $t∈[0,τ]$;

(f 3±) $|∇F(t,z)−B(t)z|$ is bounded for all $z=(x,y)∈ R 2$, $t∈[0,τ]$ and

uniformly for $t∈[0,τ]$, where

$F ( t , x , y ) = ∫ 0 x f ( t , s , y ) d s + ∫ 0 y f ( t , s , 0 ) d s , A ( t ) = ( a ( t ) b ( t ) b ( t ) a ( t ) ) , B ( t ) = ( c ( t ) d ( t ) d ( t ) c ( t ) ) .$

Theorem 1.1 Assume that f satisfies (f 1)-(f 3) with $(A(t)z,z)≤0$, $(B(t)z,z)>(λ− π 2 / τ 2 ) | z | 2 >0$ for $z∈ R 2 ∖{0}$, $t∈[0,τ]$. Then (1.1) possesses at least 2m pairs 2τ-periodic solutions, where

$Z +$ is the set of all positive integers.

Theorem 1.2 Assume that f satisfies (f 1)-(f 3+) with $(A(t)z,z)≥0$, $(B(t)z,z)<(λ− π 2 m 0 2 / τ 2 ) | z | 2 <0$ for $z∈ R 2 ∖{0}$, $m 0 ∈ Z +$, $t∈[0,τ]$. Then (1.1) possesses at least $2[ m 1 − m 0 +1]$ pairs 2τ-periodic solutions, where

In this paper, the main purpose is to study the multiplicity of periodic solutions for systems (1.1) via some recent critical point theorems for strongly indefinite functionals. In order to achieve this, some preliminaries are necessary. Let X and Y be Banach spaces with X being separable and reflexive, and set $E=X⊕Y$. Let $S⊂ X ∗$ be a dense subset. For each $s∈S$, there is a semi-norm on E defined by

We denote by $T S$ the topology on E induced by the semi-norm family ${ p s }$, and let ω and $ω ∗$ denote the weak-topology and weak*-topology, respectively. Clearly, the $T S$ topology contains the product topology on $E=X⊕Y$ produced by the weak topology on X and the strong topology on Y.

For a functional $Φ∈ C 1 (E,R)$, we write $Φ a ={u∈E:Φ(u)≥a}$. Recall that $Φ ′$ is weakly sequentially continuous if $u k ⇀u$ in E, one has $lim k → ∞ Φ ′ ( u k )v→ Φ ′ ( u k )v$ for each $v∈E$, i.e., $Φ ′ :(E,ω)→( E ∗ , ω ∗ )$ is sequentially continuous. For $c∈R$, we say that Φ satisfies the $( C ) c$ condition if any sequence ${ u k }⊂E$, such that $Φ( u k )→c$ and $(1+∥ u k ∥) Φ ′ ( u k )→0$ as $k→∞$, contains a convergent subsequence.

Suppose that

($Φ 0$) $Φ∈ C 1 (E,R)$, $Φ c$ is $T S$-closed for every $c∈R$ and $Φ ′ :( Φ c , T S )→( E ∗ , ω ∗ )$ is continuous;

($Φ 1$) there exists $ρ>0$ such that $κ:=infΦ( B ρ ∩Y)>0=Φ(0)$, where $B ρ ={u∈E:∥u∥=ρ}$;

($Φ 2$) there exist a finite dimensional subspace $Y 0 ⊂Y$ and $R>ρ$ such that $c ¯ :=supΦ( E 0 )<∞$ and $supΦ( E 0 ∖ S 0 ), where $E 0 :=X⊕ Y 0$, and $S 0 ={u∈ E 0 :∥u∥≤R}$.

The following critical point theorem will be used later (see [6, 14]).

Theorem A Assume that Φ is even and ($Φ 0$)-($Φ 2$) are satisfied. Then Φ has at least $m=dim Y 0$ pairs of critical points with critical values less than or equal to $c ¯$ provided Φ satisfies the $( C ) c$ condition for all $c∈[κ, c ¯ ]$.

## 2 Preliminaries

In this section, we establish a variational structure which enables us to reduce the existence of 2τ periodic solutions of (1.1) to a classic Hamiltonian system. First, we have the following lemma by using similar arguments .

Lemma 2.1 Suppose that $f∈ C 1 (R× R 2 ,R)$ satisfies (f 1), then the function

$H(t,x,y)= ∫ 0 x f(t,s,y)ds+ ∫ 0 y f(t,s,0)ds$
(2.1)

satisfies

$∂ H ( t , x , y ) ∂ x =f(t,x,y), ∂ H ( t , x , y ) ∂ y =f(t,y,x)$

and $H(t,x,y)∈ C 2 (R× R 2 ,R)$.

Proof It is obvious that $∂ H ∂ x =f(t,x,y)$. By (f 1), $∂ f ( t , x , y ) ∂ y = ∂ f ( t , y , x ) ∂ x$. Then we have

$∂ H ( t , x , y ) ∂ y = ∫ 0 x ∂ f ( t , y , s ) ∂ s ds+f(t,y,0)=f(t,y,x).$

The proof of Lemma 2.1 is complete. □

Suppose that x is a periodic solution of (1.1) with 2τ, let $y(t)=x(t−τ)$, then

$y ″ (t)=−f ( t − τ , x ( t − τ ) , x ( t ) ) =−f ( t , y ( t ) , x ( t ) ) .$

We denote $z=(x,y)$. Then

$z ″ (t)+λz(t)=−∇H ( t , z ( t ) ) ,$
(2.2)

where $H(t,z)$ is defined in (2.1) and $∇H(t,z)$ denotes the gradient of $H(t,z)$ with respect to the z variable. It is easy to obtain the following lemma.

Lemma 2.2 For any 2τ periodic solution of (1.1), let $y(t)=x(t−τ)$, then $z=(x,y)$ is a 2τ periodic solution of (2.2). Conversely, for any 2τ periodic solution $z=(x,y)$ of (2.2) satisfying $y(t)=x(t−τ)$, x is a 2τ periodic solution of (1.1).

For $S 1 =R/(2τZ)$, let $C ∞ ( S 1 , R 2 )$ denote the space of 2τ periodic $C ∞$ functions on R with values in $R 2$. For any $z∈ C ∞ ( S 1 , R 2 )$, it has the following Fourier expansion in the sense that it is convergent in the space $L 2 ( S 1 , R 2 )$,

$z(t)= a 0 2 τ + 1 τ ∑ j = 1 ∞ [ a j cos ( π τ j t ) + b j sin ( π τ j t ) ] ,$

where $a 0 , a j , b j ∈ R 2$, $j=1,2,…$ .

Let $z∈ L 2 ( S 1 , R 2 )$. If there exists a function $y∈ L 2 ( S 1 , R 2 )$ such that, for every $x∈ C ∞ ( S 1 , R 2 )$,

$∫ 0 2 τ ( z ( t ) , x ′ ( t ) ) dt=− ∫ 0 2 τ ( y ( t ) , x ( t ) ) dt,$

then y is called a weak derivative of z denoted by $y= z ′ (t)$.

Let

$H 1 ( S 1 , R 2 ) = { z ∈ L 2 ( S 1 , R 2 ) | ∫ 0 2 τ ( | z ( t ) | 2 + | z ′ ( t ) | 2 ) d t < + ∞ } .$

For any $z,y∈ H 1 ( S 1 , R 2 )$, $〈⋅,⋅〉$ and $∥⋅∥$ can be explicitly expressed by

$〈z,y〉= ∫ 0 2 τ [ ( z ( t ) , y ( t ) ) + ( z ′ ( t ) , y ′ ( t ) ) ] dt$

and

$∥z∥= [ | a 0 | 2 + ∑ j = 1 ∞ ( 1 + j 2 ) ( | a j | 2 + | b j | 2 ) ] 1 / 2 .$

We define an operator $L: H 1 ( S 1 , R 2 )→ H 1 ( S 1 , R 2 )$ by the Riesz representation theorem

$〈Lz,y〉= ∫ 0 2 τ ( z ˙ ( t ) , y ˙ ( t ) ) dt−λ ∫ 0 2 τ ( z ( t ) , y ( t ) ) dt.$

By a direct computation, L is a bounded self-adjoint linear operator on $H 1 ( S 1 , R 2 )$.

By (f 2) and (3.6) in Section 3, there exist two continuous functions $α(t)>0$ and $β(t)≥0$ such that

$H(t,z)≤α(t)|z | 2 +β(t),∀z∈ R 2 .$

By using similar arguments as in , we have that

$φ(z)= 1 2 〈Lz,z〉+ψ(z),$

where

$ψ(z)= ∫ 0 2 τ H ( t , z ( t ) ) dt.$

Thus the critical points of $φ(z)$ in $H 1 ( S 1 , R 2 )$ are classical solutions of (2.2).

Denote $E={z=(x,y)∈ H 1 ( S 1 , R 2 ):y(t)=x(t−τ)}$. By a direct computation, we have

$E = { z ∈ H 1 ( S 1 , R 2 ) | z ( t ) = 1 2 τ a 0 ( 1 1 ) + 1 τ ∑ j = 1 ∞ ( a j cos ( π τ j t ) + b j sin ( π τ j t ) ) ( 1 ( − 1 ) j ) } ,$

where $a 0 , a j , b j ∈R$, $j=1,2,…$ .

Note that for any $z=(x,y)∈E$, $y(t)=x(t−τ)$ and $∇H(t,z)=(f(t,x,y),f(t,y,x))$. Due to the fact that $f(t,x(t−τ),y(t−τ))=f(t,y(t),x(t))$, we know that $∇H(t,z)∈E$ whatever $z∈E$. Therefore, we have the following lemma.

Lemma 2.3 If $z(t)$ is a critical point of φ in E, then $z(t)$ is a critical point of φ in $H 1 ( S 1 , R 2 )$.

Moreover, we also denote by $M + (⋅)$, $M − (⋅)$ and $M 0 (⋅)$ the positive definite, negative definite and null subspaces of the self-adjoint linear operator defining it, respectively.

Then E has an orthogonal decomposition

$E= M + (L)⊕ M − (L)⊕ M 0 (L).$

Let

$z j (t)= 1 τ ( a j cos ( π τ j t ) + b j sin ( π τ j t ) ) ( 1 ( − 1 ) j ) ,j=1,2,…,$

where $a j , b j ∈R$. We have

$〈L z j , z j 〉= 1 1 + j 2 ( π 2 τ 2 j 2 − λ ) ∥ z j ∥,j=1,2,….$

There exists $σ>0$ such that

where $J + ={j≥1: 1 1 + j 2 ( π 2 τ 2 j 2 −λ)>0}$, $J − ={j≥1: 1 1 + j 2 ( π 2 τ 2 j 2 −λ)<0}$, $J 0 ={j≥1: 1 1 + j 2 ( π 2 τ 2 j 2 −λ)=0}$. Therefore, we get

(2.3)
(2.4)

Remark 2.1 The condition $λ> π 2 / τ 2$ is to ensure $J − ≠∅$.

## 3 Proofs of theorems

In this section, $c i$ stand for different positive constants for $i∈ Z +$.

By a direct computation, (f 1) and (f 2) imply that $H(t,z)$ is even and satisfies

(3.1)
(3.2)

uniformly for $t∈[0,τ]$.

(f 3±) implies that (3.3±)

uniformly for $t∈[0,τ]$.

Lemma 3.1 Suppose that f satisfies (f 1)-(f 3±). Then the function φ satisfies the $( C ) c$ condition for any $c∈R$.

Proof First we define an operator

$〈Bz,y〉= ∫ 0 2 τ ( B ( t ) z ( t ) , y ( t ) ) dt$

for any $z,y∈E$. By a direct computation, B is a bounded self-adjoint linear operator on E. Thus $L+B$ is also a self-adjoint linear operator on E. Then E has an orthogonal decomposition

$E= M + (L+B)⊕ M − (L+B)⊕ M 0 (L+B).$

Also there exists $σ 1 >0$ such that

(3.4)
(3.5)

So, set

$Φ(z)=−φ(z)=− 1 2 〈 ( L + B ) z , z 〉 − ∫ 0 2 τ [ H ( t , z ) − 1 2 ( B ( t ) z , z ) ] dt$

and

$〈 Φ ′ ( z ) , y 〉 =− 〈 ( L + B ) z , y 〉 − ∫ 0 2 τ ( ∇ H ( t , z ) − B ( t ) z , y ) dt$

for any $z,y∈E$, $φ(z)$ is defined in Section 2.

Let ${ z k }⊂E$ be any sequence such that

We first prove that ${ z k }$ is bounded. For any $ε>0$, (3.2) implies that there exists a constant $C ε >0$ such that

$|∇H(t,z)−B(t)z|<ε|z|+ C ε$
(3.6)

for all $z∈ R 2$ and $t∈[0,2τ]$. Since $z k ∈E$, we have

$z k = z k + + z k − + z k 0 ,$

where $z k + ∈ M + (L+B)$, $z k − ∈ M − (L+B)$, $z k 0 ∈ M 0 (L+B)$. By (3.3±), there exists a constant $d>0$ such that

$|∇H(t,z)−B(t)z|≤d.$

Therefore we have

$( 1 + c ) ∥ z k + ∥ ≥ 〈 − Φ ′ ( z k ) , z k + 〉 = 〈 ( L + B ) z k + , z k + 〉 + ∫ 0 2 τ ( ∇ H ( t , z k ) − B ( t ) z k , z k + ) d t ≥ σ 1 ∥ z k + ∥ 2 − d ∥ z k + ∥ .$

Thus we have that ${∥ z k + ∥}$ is bounded. Using similar arguments, we can prove that ${∥ z k − ∥}$ is bounded. Consider ${∥ z k 0 ∥}$. Arguing indirectly, we suppose ${∥ z k 0 ∥}$ is unbounded, then we have $∥ z k 0 ∥→∞$. According to the definition of $M 0 (L+B)$, this implies that there are constants $d 1 , d 2 >0$ such that

$d 1 | z k 0 |≤ ∥ z k 0 ∥ ≤ d 2 | z k 0 |.$
(3.7)

By (3.7), we have

(3.8)

Then

$2 c = lim k → ∞ [ 2 Φ ( z k ) − Φ ′ ( z k ) z k ] = lim k → ∞ { ∫ 0 2 τ ( ∇ H ( t , z k ) − B ( t ) z k , z k ) d t − 2 ∫ 0 2 τ [ H ( t , z k ) − 1 2 ( B ( t ) z k , z k ) ] d t } .$
(3.9)

By (3.8) and (3.3±), (3.9) is a contradiction. Hence ${∥ z k 0 ∥}$ is bounded. Therefore there exists a constant $d 3 >0$ such that

$∥ z k ∥= ∥ z k + ∥ + ∥ z k − ∥ + ∥ z k 0 ∥ ≤ d 3 .$

So, we get ${ z k }$ is bounded, and going if necessary to a subsequence, we can assume that $z k ⇀z$ in E and $z k →z$ in $C( S 1 , R N )$. Write $z k = z k + + z k − + z k 0$ and $z= z + + z − + z 0$, then $z k ± ⇀ z ±$ in E, $z k 0 → z 0$ in E and $z k ± → z ±$ in $C( S 1 , R N )$.

In view of (3.6) and $z k − → z −$ in $C( S 1 , R 2 )$, it is easy to verify

$∫ 0 2 τ ( ∇ H ( t , z k ) − B ( t ) z k , z k − − z − ) dt→0$

and

$∫ 0 2 τ ( ∇ H ( t , z ) − B ( t ) z , z k − − z − ) dt→0.$

But then $〈 Φ ′ ( z k )− Φ ′ (z), z k − − z − 〉→0$ as $k→∞$, and

$〈 Φ ′ ( z k ) − Φ ′ ( z ) , z k − − z − 〉 = − 〈 ( L + B ) ( z k − − z − ) , z k − − z − 〉 − ∫ 0 2 τ ( ∇ H ( t , z k ) − B ( t ) z k , z k − − z − ) d t + ∫ 0 2 τ ( ∇ H ( t , z ) − B ( t ) z , z k − − z − ) d t ≥ σ 1 ∥ z k − − z − ∥ − ∫ 0 2 τ ( ∇ H ( t , z k ) − B ( t ) z k , z k − − z − ) d t + ∫ 0 2 τ ( ∇ H ( t , z ) − B ( t ) z , z k − − z − ) d t .$

This yields $z k − → z −$ in E. Similarly, $z k + → z +$ in E and hence $z k →z$ in E, that is, Φ satisfies the $( C ) c$ condition. Thus φ satisfies the $( C ) c$ condition. The proof of Lemma 3.1 is complete. □

Proof of Theorem 1.1 For $z∈E$, let $X= M + (L)⊕ M 0 (L)$, $Y= M − (L)$, $E=X⊕Y$, and

$Φ(z)=−φ(z),Ψ(z)=ψ(z),$
(3.10)

where $φ(z)$ and $ψ(z)$ are from Section 2.

In order to obtain this theorem, we apply Theorem A to the functional $Φ(z)$. The proof of this theorem is divided into the following three steps.

Step 1. Φ satisfies ($Φ 0$).

We first check that $Φ c$ is $T S$ -closed for any $c∈R$. Let ${ z k }$ be any sequence $T S$-converging to some $z∈E$. Write $z k = z k + + z k 0 + z k −$ and $z= z + + z 0 + z −$, then $z k − → u −$ in E and hence ${ z − }$ is bounded in the norm topology. Note that $(B(t)z,z)>λ− π 2 / τ 2 >0$ for $z∈ R 2 ∖{0}$ and for $t∈[0,2τ]$, then for any $z∈E$ and small $ε>0$, we have by (3.10) and (3.6)

$Ψ ( z ) = ∫ 0 2 τ H ( t , z ) d t = ∫ 0 2 τ ∫ 0 1 ( ∇ H ( t , s z ) , z ) d s d t ≥ ∫ 0 2 τ ∫ 0 1 ( B ( t ) s z , z ) d s d t − ∫ 0 2 τ ∫ 0 1 | ∇ H ( t , z ) − ( B ( t ) s z , z ) | | z | d s d t ≥ 1 2 ∫ 0 2 τ ( B ( t ) z , z ) d t − ∫ 0 2 τ ∫ 0 1 ( ε | s z | + C ε ) | z | d s d t ≥ 1 2 ( λ − π 2 τ 2 − ε ) ∥ z ∥ L 2 2 − C ε ∥ z ∥ L 2 ≥ c 1 ∥ z ∥ L 2 2 − c 2 ∥ z ∥ L 2 ,$

which implies that Ψ is bounded from below on E. Consequently, combining $z k ∈ Φ c$ and $1 2 〈L z k + , z k + 〉=− 1 2 〈L z k − , z k − 〉−Φ( z k )−Ψ( z k )$ shows that ${ z k + }$ is bounded in E by (2.3) and hence

$Ψ ( z k ) = − 1 2 〈 L z k + , z k + 〉 − 1 2 〈 L z k − , z k − 〉 − Φ ( z k ) ≤ − 1 2 〈 L z k + , z k + 〉 − 1 2 〈 L z k − , u k − 〉 − c ≤ c 3 .$
(3.11)

Moreover, since $∥ z k ∥ L 2 2 = ∥ z k + ∥ L 2 2 + ∥ z k 0 ∥ L 2 2 + ∥ z k − ∥ L 2 2$, we have

$Ψ( z k )≥ c 1 ∥ z k ∥ L 2 2 − c 2 ∥ z k ∥ L 2 ≥ ∥ z k ∥ L 2 − c 4 ≥ ∥ z k 0 ∥ L 2 − c 4 .$

It follows from (3.11) and the above inequality that ${ z k 0 }$ is also bounded in E since all norms are equivalent in a finite dimensional space. Then $∥ z k ∥ 2$ ($= ∥ z k + ∥ 2 + ∥ z k 0 ∥ 2 + ∥ z k − ∥ 2$) is bounded, and hence we can assume that ${ z k }$ converges weakly to $z= z + + z 0 + z −$ in E. Thus we have $Ψ( z k )→Ψ(z)$. Note that $〈 L y , y 〉 1 / 2$ is an equivalent norm on $H +$. By the lower semi-continuity of the norm, we get

$c ≤ lim sup k → ∞ Φ ( z k ) = lim sup k → ∞ ( − 1 2 〈 L z k + , z k + 〉 − 1 2 〈 L z k − , z k − 〉 − Ψ ( z k ) ) = − lim inf k → ∞ 1 2 〈 L z k + , z k + 〉 − 1 2 〈 L z − , z − 〉 − Ψ ( z ) ≤ Φ ( z ) ,$

that is, $z∈ Φ c$ and hence $Φ c$ is $T S$-closed.

Next, we prove that $Φ ′ :( Φ c , T S )→( E ∗ , ω ∗ )$ is continuous. To achieve this, it is sufficient to demonstrate that $Ψ ′$ has the same property. Suppose $z k ⇀u$ in E, then ${ z k }$ converges uniformly to z on $[0,2τ]$. Hence, for every given $y∈E$, we see that $(∇H(t, z k (t)),y(t))$ converges to $(∇H(t,z(t)),y(t))$ in measure on $[0,2τ]$. Moreover, by (3.6), one has

$|∇H ( t , z k ( t ) ) ,y(t))|≤ ( ε ∥ z k ∥ ∞ + b ¯ ∥ z k ∥ ∞ + C ε ) ∥ y ∥ ∞ ≤ c 5$

for all k and $t∈[0,2τ]$, where $b ¯ = max t ∈ [ 0 , τ ] {|B(t)|}$ and $∥ ⋅ ∥ ∞$ denotes the natural norm of $C( S 1 , R 2 )$. Thus, the Vitali theorem is applicable and

$〈 Ψ ′ ( z k ) , y 〉 = ∫ 0 2 τ ( ∇ H ( t , z k ) , y ) dt→ ∫ 0 2 τ ( ∇ H ( t , z ) , y ) dt= 〈 Ψ ′ ( z ) , y 〉$

for any $y∈E$. So, Φ satisfies ($Φ 0$).

Step 2. Φ satisfies ($Φ 1$).

By (3.1), there exist $0<ε<σ/2$, $ρ 0 >0$ such that

(3.12)

By Proposition 1.1 in , there is a positive constant ξ such that $∥ z ∥ ∞ ≤ξ∥z∥$. Set small $ρ< ρ 0 /ξ$, then for each $z∈Y$ with $∥z∥≤ρ$, one has $∥ z ∥ ∞ ≤ ρ 0$, and hence by (3.12)

$Φ ( z ) = − 1 2 〈 L z , z 〉 − ∫ 0 2 τ H ( t , z ) d t ≥ σ 2 ∥ z ∥ 2 − ∫ 0 2 τ [ ε | z | 2 + 1 2 ( A ( t ) z , z ) ] d t ≥ ( σ 2 − ε ) ∥ z ∥ 2 .$

Therefore,

$κ:=infΦ( B ρ ∩Y)≥ ( σ 2 − ε ) ρ 2 >0,$

and hence ($Φ 1$) holds.

Step 3. Φ satisfies ($Φ 2$).

Let

$Y 0 =span { cos ( π τ j t ) ( 1 ( − 1 ) j ) , sin ( π τ j t ) ( 1 ( − 1 ) j ) : j ∈ Z + , j ≤ m } .$

Obviously, $Y 0 ⊂Y$ and $dim Y 0 =2m$. In order to obtain the desired conclusion, it is sufficient to prove that $Φ(z)→−∞$ as $∥z∥→∞$ on $E 0 :=X⊕ Y 0$.

Let $Γ=λ− π 2 τ 2$. By the definition of m, there exists a constant $0<δ<σ$ such that

$B(t)≥Γ+δ$
(3.13)

for $t∈[0,τ]$. Clearly, for any $y∈ Y 0$,

$〈Ly,y〉≥−Γ ∥ y ∥ L 2 2 .$
(3.14)

Let $F ˜ (t,z)=H(t,z)− 1 2 (B(t)z,z)$. We claim that

(3.15)

Indeed, for $0≠z∈E$, by (3.6), one has

$∥ z ∥ − 2 | ∫ 0 2 τ F ˜ ( t , z ) d t | = ∥ z ∥ − 2 | ∫ 0 2 τ ∫ 0 1 ( ∇ H ( t , s z ) − B ( t ) s z , z ) d s d t | ≤ ∥ z ∥ − 2 ∫ 0 2 τ ( ε | z | + C ε ) | z | d t ≤ ∥ z ∥ − 2 ( ε ∥ z ∥ L 2 2 + C ε ∥ z ∥ L 2 ) ≤ ε + C ε ∥ z ∥ ,$

which implies that (3.15) is true by the arbitrariness of ε. Then, for $z= z + + z 0 + z − ∈ E 0$, by (2.3), (3.13) and (3.14), one has

$Φ ( z ) = − 1 2 〈 L z − , z − 〉 − 1 2 〈 L z + , z + 〉 − ∫ 0 2 τ F ( t , z ) d t ≤ Γ 2 ∥ z − ∥ L 2 2 − δ 2 ∥ z + ∥ 2 − 1 2 ∫ 0 2 τ ( B ( t ) z , z ) d t − ∫ 0 2 τ F ˜ ( t , z ) d t ≤ Γ 2 ∥ z − ∥ L 2 2 − δ 2 ∥ z + ∥ 2 − 1 2 ( Γ + δ ) ∥ z ∥ L 2 2 − ∫ 0 2 τ F ˜ ( t , z ) d t ≤ Γ 2 ∥ z − ∥ L 2 2 − δ 2 ∥ z + ∥ 2 − 1 2 ( Γ + δ ) ( ∥ z − ∥ L 2 2 + ∥ z 0 ∥ L 2 2 ) − ∫ 0 2 τ F ˜ ( t , z ) d t ≤ − δ 2 ( ∥ z − ∥ L 2 2 + ∥ z + ∥ 2 + ∥ z 0 ∥ L 2 2 ) − ∫ 0 2 τ F ˜ ( t , z ) d t .$

Since $M 0$ and $Y 0$ are finitely dimensional, (3.15) and the above estimate imply that $Φ(z)→−∞$ as $∥z∥→∞$, $z∈ E 0 :=X⊕ Y 0$. Hence ($Φ 2$) holds.

The proof of Theorem 1.1 is complete. □

Proof of Theorem 1.2 For $z∈E$, let $X= M − (L)⊕ M 0 (L)$, $Y= M + (L)$, $E=X⊕Y$, and

$Φ(z)=φ(z),Ψ(z)=−ψ(z)$

and

$Y 0 =span { cos ( π τ j t ) ( 1 ( − 1 ) j ) , sin ( π τ j t ) ( 1 ( − 1 ) j ) : j = m 0 , m 0 + 1 , … , m 1 } .$

Then the conclusion is obtained by the same argument as in the proof of Theorem 1.1. The proof of Theorem 1.2. is complete. □

Example 3.1 Consider the following equation:

$x ″ (t)+λx(t)=−α(t)x(t)−β(t)x(t−π)− x ( t ) x 2 ( t ) + x 2 ( t − π ) + γ 2 ( t ) ,$
(3.16)

where $x(t)≥0$. Let

$f(t,x,y)=α(t)x+β(t)y+ x x 2 + y 2 + γ 2 ( t ) .$

Then

$∂ f ( t , x , y ) ∂ y =β(t)− 2 x y ( x 2 + y 2 + γ 2 ( t ) ) 2 = ∂ f ( t , y , x ) ∂ x .$

Let

$H(t,x,y)= 1 2 α(t) x 2 +β(t)xy+ 1 2 α(t) y 2 + 1 2 ln x 2 + y 2 + γ 2 ( t ) γ 2 ( t ) .$

Then

$∂ H ∂ x =f(t,x,y), ∂ H ∂ y =f(t,y,x).$

By a straightforward computation, we have

and

uniformly for $t∈[0,π]$.

Take $λ=6$, $α(t)=−5+sinπt$, $β(t)=1−sinπt$, $γ 2 (t)= 1 6$. By Theorem 1.2, we have $a(t)= 1 γ 2 ( t ) +α(t)≥0$, $b(t)=d(t)=1−sinπt≥0$, $c(t)=−5+sinπt$ for $t∈[0,π]$, $z=(x,y)$, $x≥0$, $y≥0$,

$( A ( t ) z , z ) = ( ( a ( t ) b ( t ) b ( t ) a ( t ) ) ( x y ) , ( x y ) ) ≥ 0 , ( B ( t ) z , z ) = ( ( c ( t ) d ( t ) d ( t ) c ( t ) ) ( x y ) , ( x y ) ) ( B ( t ) z , z ) = c ( t ) ( x 2 + y 2 ) + 2 d ( t ) x y ≤ [ c ( t ) + d ( t ) ] ( x 2 + y 2 ) ( B ( t ) z , z ) = − 4 ( x 2 + y 2 ) < ( 6 − 3 2 ) ( x 2 + y 2 ) < 0 .$

Then we have $m 0 = m 1 =3$. By Theorem 1.2, (3.16) possesses at least two pairs 2π-periodic solutions.

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## Acknowledgements

The authors are grateful for the referees’ careful reviewing and their valuable suggestions. The work is partially supported by the Natural Science Foundation of Shanxi Province (No. 2012011004-1) of China.

## Author information

Correspondence to Qiong Meng.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors read and approved the final manuscript.

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Meng, Q., Zhang, X. Multiple periodic solutions to a class of nonautonomous second-order delay differential equation. Bound Value Probl 2013, 244 (2013). https://doi.org/10.1186/1687-2770-2013-244 