Multiple periodic solutions of high order differential delay equations with \(2k 1\) lags

(1)

R E S E A R C H

Open Access

Multiple periodic solutions of high order

differential delay equations with

2 k

– 1

lags

Lin Li

1

_{, Huafei Sun}

1*

_{and Weigao Ge}

1

*_{Correspondence:}

[email protected]

1_{School of Mathematics and} Statistics, Beijing Institute of Technology, Beijing, P.R. China

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.

Keywords: High order diﬀerential delay equation; Periodic solutions; Critical point theory; Variational method

1 Introduction

Givenf ∈C0₍_R_,_R_{) with}_f_(–_x_{) = –}_f₍_x_),_xf₍_x_{) > 0,}_x_{= 0. Kaplan and Yorke [}₁₃_{] studied the}

existence of 4-periodic and 6-periodic solutions to the diﬀerential delay equations

x(t) = –fx(t– 1) (1)

and

x(t) = –fx(t– 1)–fx(t– 2), (2)

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

x(t) = –

n

i=1

fx(t–i) (3)

could be studied under the restriction

xt– (n+ 1)= –x(t),

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

(2)

After that a lot of papers [3–12,14–18] discussed the existence and multiplicity of 2(n+ 1)-periodic solutions to Eq. (3) and its extension

x(t) = –

n

i=1

∇Fx(t–i), (4)

whereF∈C1₍_RN_,_R_),_∇_F_(–_x_{) = –}_∇_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

x(2s+1)(t) = –

2k–1

i=1

fx(t–i), (5)

and

x(2s+1)(t) = –

2k–1

i=1

(–1)i+1fx(t–i), (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

f ∈C0(R,R), f(–x) = –f(x) (7)

and there areα,β∈Rsuch that

lim x→0

f(x)

x =α, xlim→∞ f(x)

x =β. (8)

LetF(x) =₀xf(s)ds. ThenF(–x) =F(x) andF(0) = 0. For convenience, we make the fol-lowing assumptions:

(S1) f satisﬁes (7) and (8),

(S2) there existM> 0and a functionr∈C0(R+,R+)satisfyingr(s)→ ∞,r(s)/s→0as s→ ∞such that

F(x) –1

2βx

2_>_r_|_x_|_–_M_,

(S±₃) ±[F(x) –1₂βx2_{] > 0}_,_|_x_{| → ∞}_,

(S±₄) ±[F(x) –1₂αx2_{] > 0}_,_{0 <}_|_x_|₁_.

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

LetXbe a Hilbert space,L:X→Xbe a linear operator, andΦ:X→Rbe a diﬀerentiable functional.

Lemma 1.1([3], Lemma 2.4) Assume that there are two closed s1_{-invariant linear} sub-spaces,X+_{and X}–_,_{and r}_{> 0}_{such that}

(a) X+∪X–is closed and of ﬁnite codimensions inX, (b) L(X–₎_⊂_X–_,_L₌_L₊_P–1_A0_or_L₌_L₊_P–1_A

(3)

(c) there existsc0∈Rsuch that

inf

x∈X+Φ(x)≥c0,

(d) there isc∞∈Rsuch that

Φ(x)≤c∞<Φ(0) = 0, ∀x∈X–∩Sr=

x∈X–:x=r ,

(e) Φsatisﬁes the(P.S)c-condition,c0<c<c∞,i.e.,every sequence{xn} ⊆Xwith Φ(xn)→candΦ(xn)→0possesses a convergent subsequence.

ThenΦhas at least1₂[dim(X+_∩_X–_{) –}_codim

X(X+∪X–)]generally diﬀerent critical orbits inΦ–1([c0,c∞])if[dim(X+∩X–) –codimX(X+∪X–)] > 0.

Deﬁnition 1.2 We say thatΦ satisﬁes the (P.S)-condition if every sequence{xn} with Φ(xn) is bounded andΦ(xn)→0 possesses a convergent subsequence.

Remark1.3 We may use the (P.S)-condition to replace condition (e) in Lemma1.1since the (P.S)-condition implies that the (P.S)c-condition holds for eachc∈R.

2 SpaceX, functional

Φ

, and its differential

Φ

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

x(t– 2k) = –x(t), k≥1. (9)

Let

X=x∈L2:x(t– 2k) = –x(t)

=

_∞

i=0

aicos

(2i+ 1)πt

2k +bisin

(2i+ 1)πt

2k

:ai,bi∈R

,

X=cl

_∞

i=0

aicos

(2i+ 1)πt

2k +bisin

(2i+ 1)πt

2k

:ai,bi∈R,

∞

i=0

(2i+ 1)a2_i+b2_i<∞

⊂X,

and deﬁneP:X→L2_by

Px(t) =P

_∞

i=0

aicos

(2i+ 1)πt

2k +bisin

(2i+ 1)πt

2k

= ∞

i=0

(2i+ 1)2s+1

aicos

(2i+ 1)πt

2k +bisin

(2i+ 1)πt

2k

. (10)

Let

P–1x(t) = ∞

i=0

1 (2i+ 1)2s+1

aicos

(2i+ 1)πt

2k +bisin

(2i+ 1)πt

2k

(4)

Then the inverseP–1_of_P_{exists. For}_x_∈_X_{, deﬁne}

x,y= 4k

0

Px(t),y(t)dt, x=x,x,

x,y2=

4k

0

x(t),y(t)dt, x2=

x,x2.

Therefore (X, · ) is anH12 _space.

For (5), deﬁne a functionalΦ:X→Rby

Φ(x) = 1

2Lx,x+ 4k

0

Fx(t)dt, (11)

where

Lx= –P–1 2k–1

i=1

(–1)i+1x(2s+1)(t–i). (12)

Letm=k– 1 and

X(i) =

x(t) =aicos

(2i+ 1)πt

2k +bisin

(2i+ 1)πt

2k :ai,bi∈R

.

We have

X= ∞

l=0

_m

i=0

X(2lk+i) +X(2lk+ 2k–i– 1)

. (13)

Ifxi(t) =aicos(2i+1)π₂_k t+bisin(2i+1)π₂_k t∈X(i),i∈N, we have

Lx= (–1)s+1

π

2k

2s+1∞

i=0 xitan

(2i+ 1)π

4k

. (14)

Obviously,L|X(i):X(i)→X(i) is invertible.

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

P–1Φ(x) =Lx+K(x), (15)

whereK(x) =P–1f(x). It is easy to prove thatK: (X,x2)→(X,x2₂) is compact. Therefore, from (14) we have that if

x(t) = ∞

i=0

aicos

(2i+ 1)πt

2k +bisin

(2i+ 1)πt

2k

(5)

then

On the other hand,

(6)

For (6), deﬁne the functionalΨ :X→Rby

Ψ(x) =1

2Lx,x+ 4k

0

Fx(t)dt, (16)

where

Lx= –P–1 2k–1

i=1

x(2s+1)(t–i). (17)

Letm=k– 1 and

X(i) =

x(t) =aicos

(2i+ 1)πt

2k +bisin

(2i+ 1)πt

2k :ai,bi∈R

.

We have

X= ∞

l=0

_m

i=0

X(2lk+i) +X(2lk+ 2k–i– 1)

. (18)

Ifxi(t) =aicos(2i+1)π₂_k t+bisin(2i+1)π₂_k t∈X(i),i∈N, we have

Lx= (–1)s+1

π

2k

2s+1∞

i=0 xicot

(2i+ 1)π

4k

. (19)

Obviously,L|X(i):X(i)→X(i) is invertible.

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

P–1Ψ(x) =Lx+K(x), (20)

whereK(x) =P–1_f₍_x_{). It is easy to prove that}_K_{: (}_X_,_x2₎_→₍_X_,_x2

2) is compact.

Therefore, from (14) we have that if

x(t) = ∞

i=0

aicos

(2i+ 1)πt

2k +bisin

(2i+ 1)πt

2k

,

then

Lx,x= (–1)s+1 ∞

i=0

π

2k

2s+1

2k(2i+ 1)2s+1a_i2+b2_icot(2i+ 1)π

4k

= ∞

l=0

_m

i=0

(–1)s+1

π

2k

2s+1

2k(4lk+ 2i+ 1)2s+1a2₂_lk₊_i+b2₂_lk₊_icot(2i+ 1)π

4k

–

m

i=0

(–1)s+1

π

2k

2s+1

2k(4lk+ 4k– 2i– 1)2s+1a2₂_lk₊₂_k_–_i_–1+b2₂_lk₊₂_k_–_i_–1

×cot(2i+ 1)π

4k

(7)

On the other hand,

Lemma 2.1 Each critical point of the functionalΦ is a4k-periodic classical solution of

(8)

Consequently,

–

2k–1

i=1

(–1)i+1x(2s+1)(t–i– 1) +fx(t– 1)= 0, (21.1)

–

2k–1

i=1

(–1)i+1x(2s+1)(t–i– 2) +fx(t– 2)= 0, (21.2)

–

2k–1

i=1

(–1)i+1x(2s+1)(t–i– 3) +fx(t– 3)= 0, (21.3)

.. .

–

2k–1

i=0

(–1)i+1x(2s+1)t–i– (2k– 1)+fxt– (2k– 1)= 0. (21.(2k-1))

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

x(2s+1)(t) +

2k–1

i=1

fx(t–i)= 0, a.e.t∈[0, 4k],

namely

x(2s+1)(t) = –

2k–1

i=1

fx(t–i), a.e.t∈[0, 4k],

which implies thatxsatisﬁes the above equation for allt∈[0, 4k] since the function on the right-hand side is continuous, thenxis a classical solution to (5). Lemma 2.2 Each critical point of the functionalΨ is a4k-periodic classical solution of Eq. (6)satisfying(9).

Proof Letxbe a critical point of the functionalΨ. Thenx(t) satisﬁes

–

2k–1

i=1

x(2s+1)(t–i) +fx(t)= 0, a.e.t∈[0, 4k]. (22)

Consequently,

–

2k–1

i=1

x(2s+1)(t–i– 1) +fx(t– 1)= 0, (22.1)

–

2k–1

i=1

x(2s+1)(t–i– 2) +fx(t– 2)= 0, (22.2)

–

2k–1

i=1

x(2s+1)(t–i– 3) +fx(t– 3)= 0, (22.3)

(9)

–

2k–1

i=1

x(2s+1)t–i– (2k– 1)+fxt– (2k– 1)= 0. (22.(2k-1))

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

x(2s+1)(t) +

2k–1

i=1

(–1)i+1fx(t–i)= 0, a.e.t∈[0, 4k],

namely

x(2s+1)(t) = –

2k–1

i=1

(–1)i+1_f_x₍_t_–_i₎_, _a.e._t_∈_{[0, 4}_k_],

which implies thatxsatisﬁes the above equation for allt∈[0, 4k] since the function on the right-hand side is continuous, thenxis a classical solution to (6).

3 Partition of spaceXand symbols

For (5), let

X_∞+ =

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

4k +β> 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

tan(2i+ 1)π

4k +β> 0

,

X_∞– =

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

4k +β< 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

tan(2i+ 1)π

4k +β< 0

,

X₀+=

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

4k +α> 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

tan(2i+ 1)π

4k +α> 0

(10)

X₀–=

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

4k +α< 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

tan(2i+ 1)π

4k +α< 0

.

On the other hand,

X_∞0 =

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

4k +β= 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

tan(2i+ 1)π

4k +β= 0

,

X₀0=

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

4k +α= 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

tan(2i+ 1)π

4k +α= 0

.

Obviously,dimX0

∞<∞anddimX00<∞.

Lemma 3.1 Under assumptions(S1)and(S2),there isσ> 0such that

L+P–1βx,x>σx2, x∈X_∞+ and

L+P–1βx,x< –σx2, x∈X_∞–.

(23)

Proof First, we have that, forβ≥0 ands= even, (–1)s+1_{= –1,}_i_{∈ {}_{0, 1, . . . ,}_m_}_,

–

(4lk+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

4k +β

> –

(4l+(i)k+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

(11)

wherel+₍_i_{) =}_max_{_l_∈_N_{: –(}(4lk+2i+1)π 2k )2s+1tan

(2i+1)π

4k +β> 0}and

–

(4lk+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

4k +β

< –

(4l–(i)k+ 2i+ 1)π

2k

2s+1

tan(2i+ 1)π

4k +β< 0,

wherel–₍_i_{) =}_min_{_l_∈_N_{: –(}(4lk+2i+1)π 2k )2s+1tan

(2i+1)π

4k +β< 0}.

In this case, we may choose

σi=min

–

π

2k

2s+1

tan(2i+ 1)π

4k +

β

(4l+₍_i₎_k_{+ 2}_i_{+ 1)}2s+1,

π

2k

2s+1

tan(2i+ 1)π

4k –

β

(4l–₍_i₎_k_{+ 2}_i_{+ 1)}2s+1

> 0,

and letσ=min{σ0,σ1, . . . ,σm}> 0.

Second, we have that, forβ≥0 ands= odd, (–1)s+1_{= 1,}_i_{∈ {}_{0, 1, . . . ,}_m_}_,

–

(4lk+ 4k– 2i– 1)π

2k

2s+1

tan(2i+ 1)π

4k +β

> –

(4l+(i)k+ 4k– 2i– 1)π

2k

2s+1

tan(2i+ 1)π

4k +β> 0,

wherel+₍_i_{) =}_max_{_l_∈_N_{: –(}(4lk+4k–2i–1)π 2k )2s+1tan

(2i+1)π

4k +β> 0}and

–

(4lk+ 4k– 2i– 1)π

2k

2s+1

tan(2i+ 1)π

4k +β

< –

(4l–₍_i₎_k_{+ 4}_k_{– 2}_i_{– 1)}_π

2k

2s+1

tan(2i+ 1)π

4k +β< 0,

wherel–(i) =min{l∈N: –((4lk+4k₂–2_ki–1)π)2s+1_tan(2i+1)π

4k +β< 0}.

In this case, we may choose

σi=min

–

π

2k

2s+1

tan(2i+ 1)π

4k +

β

(4l+₍_i₎_k_{+ 4}_k_{– 2}_i_{– 1)}2s+1,

π

2k

2s+1

tan(2i+ 1)π

4k –

β

(4l–₍_i₎_k_{+ 4}_k_{– 2}_i_{– 1)}2s+1

> 0,

and letσ=min{σ0,σ1, . . . ,σm}> 0. The proof for the caseβ< 0 is similar. We omit it. The

inequalities in (23) are proved.

Lemma 3.2 Under conditions(S1)and(S2),the functionalΦdeﬁned by(11)satisﬁes the

(12)

Proof LetΠ,N,Zbe the orthogonal projections fromXontoX+

∞,X–∞,X0∞, respectively. From the second condition in (8) it follows that

P–1f(x) –βx,x<σ 2x

2₊_M_, _x_∈_X ₍₂₄₎

for someM> 0.

Assume that{xn} ⊂Xis a subsequence such thatΦ(xn)→0 andΦ(xn) is bounded. Let wn=Πxn,yn=Nxn,zn=Zxn. Then we have

ΠL+P–1β=L+P–1βΠ, NL+P–1β=L+P–1βN. (25) From

Φ(xn),xn

=Lxn+P–1f(xn),xn

=L+P–1βxn,xn

+P–1f(xn) –βxn

,xn

and (25), we have

Π Φ(xn),xn

=ΠL+P–1βxn,xn

+ΠP–1f(xn) –βxn

,xn

=L+P–1βwn,wn

,wn

,

and then, by (23), we have

L+P–1βwn,wn

,wn

>σ

2wn

2_–_M_w n,

which, together withΠ Φ(xn)→0, implies the boundedness ofwn. Similarly we have the

boundedness ofyn. At the same time, (S2) yields

Φ(xn) =

1 2

L+P–1βxn,xn

+

4k

0

F(xn)dt– β

2xn

2 2

=1 2

L+P–1βwn,wn

+1

2

L+P–1βyn,yn

+ 4k

0

F(xn)dt– β

2

wn22+yn22+zn22

.

Then the boundedness of Φ(x) implies that zn2 is bounded. Consequently zn is

bounded sinceX0

∞is ﬁnite-dimensional. Therefore,xnis bounded.

It follows from (15) that

(Π+N)Φ(xn) = (Π+N)Lxn+ (Π+N)K(xn)

=L(wn+yn) + (Π+N)K(xn).

From the compactness of operatorKand the boundedness ofxn, we have thatK(xn)→u.

Then

(13)

The ﬁnite-dimensionality ofX0

∞ and the boundedness ofzn=Zxnimplyzn→ϕ∈X∞0. Therefore,

xn=zn+wn+yn→ϕ– (L|x+∞+x–∞)–1(Π+N)u,

which implies the (P.S)-condition.

For (6), let

X_∞+ =

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

cot(2i+ 1)π

4k +β> 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

cot(2i+ 1)π

4k +β> 0

,

X_∞– =

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

cot(2i+ 1)π

4k +β< 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

cot(2i+ 1)π

4k +β< 0

,

X₀+=

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

cot(2i+ 1)π

4k +α> 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

cot(2i+ 1)π

4k +α> 0

,

X₀–=

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

cot(2i+ 1)π

4k +α< 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

cot(2i+ 1)π

4k +α< 0

(14)

On the other hand,

X_∞0 =

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

cot(2i+ 1)π

4k +β= 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

cot(2i+ 1)π

4k +β= 0

,

X₀0=

X(2lk+i) :l≥0, 0≤i≤m,

(–1)s+1

(4lk+ 2i+ 1)π

2k

2s+1

cot(2i+ 1)π

4k +α= 0

∪

X(2lk+ 2k–i– 1) :l≥0, 0≤i≤m,

–(–1)s+1

(4lk+ 4k– 2i– 1)π

2k

2s+1

cot(2i+ 1)π

4k +α= 0

.

Obviously,dimX_∞0 <∞anddimX0₀<∞.

Lemma 3.3 Under assumptions(S1)and(S2),there isσ> 0such that

L+P–1βx,x>σx2, x∈X_∞+ and

L+P–1βx,x< –σx2, x∈X_∞–.

(27)

Proof The proof is similar to Lemma3.1, we omit it.

Lemma 3.4 Under conditions(S1)and(S2),the functionalΨ deﬁned by(16)satisﬁes the

(P.S)-condition.

Proof The proof is similar to Lemma3.2, we omit it.

4 Notations and main results of this paper

We ﬁrst give some notations. For (5), if (–1)s+1_{= –1, denote}

N1(α) = ⎧ ⎨ ⎩

–m_i₌₀card{l≥0 : 0 < ((4lk+4k₂–2_ki–1)π)2s+1tan(2i₄+1)π_k < –α}, α< 0,

m

i=0card{l≥0 : 0 < ((4lk+22ki+1)π)2s+1tan (2i+1)π

4k <α}, α≥0.

N1(β) = ⎧ ⎨ ⎩

–m_i₌₀card{l≥0 : 0 < ((4lk+4k₂–2_ki–1)π)2s+1_tan(2i+1)π

4k < –β}, β< 0,

m

i=0card{l≥0 : 0 < ((4lk+22ki+1)π)

2s+1_tan(2i+1)π

(15)

And

Alternatively, if (–1)s+1_{= 1, denote}

(16)

N₂0(α+) =

Alternatively, if (–1)s+1_{= 1, denote}

N2(α) =

Now we give the main results of this paper.

(17)

Theorem 4.4 Suppose that(S1)and(S2)hold.Then Eq. (6)possesses at least

n=maxN2(β) –N2(α) –N20(β–) –N20(α+),N2(α) –N2(β) –N20(α–) –N20(β+)

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

Theorem 4.5 Suppose that(S1), (S2), (S+

3),and(S–4)hold.Then Eq. (6)possesses at least n=N2(β) –N2(α) +N₂0(β+) +N20(α–)

Theorem 4.6 Suppose that(S1), (S2), (S3–),and(S+4)hold.Then Eq. (6)possesses at least

n=N2(α) –N2(β) +N₂0(α+) +N20(β–)

5 Proof of main results of this paper

Proof of Theorem4.1 Suppose without loss of generality that

n=N1(β) –N1(α) –N₁0(β–) –N10(α+).

LetX+=X_∞+ andX–=X0–. Then

X\X+∪X–=X\X_∞+ ∪X₀–⊆X_∞0 ∪X₀0∪X_∞+ ∩X₀–.

Obviously,

codimX

X++X–≤dimX_∞0 +dimX₀0+dimX+_∞∩X₀–<∞,

which implies that condition (a) in Lemma1.1holds. LetA∞=β. Then condition (b) in Lemma1.1holds since, for eachj∈N, we have thatx∈X(j) yields (L+P–1_β₎_x_∈_X₍_j_).

At the same time, Lemma3.2gives the (P.S)-condition.

Now it suﬃces to show that conditions (c) and (d) in Lemma1.1hold under assumptions (S1) and (S2).

In fact, we have shown in Lemma3.1that there isσ> 0 such that(L+P–1_β₎_x_,_x_>_σ_x2_, x∈X+

∞. And the second condition in (8) implies that|F(x) –12βx2|< 1

4σ|x|2+M1,x∈R

for someM1> 0. Then

Φ(x) = 1

2Lx,x+ 4k

0

Fx(t)dt

= 1 2

L+P–1βx,x+ 4k

0

Fx(t)–1 2βx(t)

(18)

≥1

Our last task is to compute the value of

(19)

and

L+P–1αx,x=

(–1)s+1

π

2k

2s+1

tan(2i+ 1)π

4k +

α

(4lk+ 2i+ 1)2s+1

x2,

x∈X(2lk+i),

L+P–1αx,x=

–(–1)s+1

π

2k

2s+1

tan(2i+ 1)π

4k +

α

(4lk+ 4k– 2i– 1)2s+1

x2,

x∈X(2lk+ 2k–i– 1).

Therefore,

X+

∞(2lk+i) =X∞+ ∩X(2lk+i) =∅,

X_∞+(2lk+ 2k–i– 1) =X_∞+ ∩X(2lk+ 2k–i– 1) =X(2lk+ 2k–i– 1),

X–

0(2lk+i) =X0–∩X(2lk+i) =X(2lk+i), X–

0(2lk+ 2k–i– 1) =X0–∩X(2lk+ 2k–i– 1) =∅

ifi∈ {0, 1, . . . ,m}and (–1)s+1_{= 1 and}_l_≥_{0 is large enough, which means that there is}_M_{> 0}

such thatdim(X+

∞(j)∩X0–(j)) –codimX(X∞+(j) +X0–(j)) = 0,j>M, from which it follows that

n=1 2

M

j=0

dim

X_∞+(j)∩X₀–(j) –codimX(j)

X_∞+(j) +X₀–(j)!

=1 2

M

j=0

"

dimX_∞+(j) +dimX₀–(j) – 2#

=1 2

M

j=0

"

dimX_∞+(j) +dimX₀–(j)#– (M+ 1).

Then we have

M

j=0

dimX_∞+(j)

= 2 ⎧ ⎨ ⎩

N1(β) +card{2lk+ 2k–i– 1 : 0≤2lk+ 2k–i– 1≤M}, β≥0,

N1(β) –N0

1(β–) +card{2lk+ 2k–i– 1 : 0≤2lk+ 2k–i– 1≤M}, β< 0, M

j=0

dimX₀–(j)= 2 ⎧ ⎨ ⎩

–N1(α) –N₁0(α+) +card{2lk+i: 0≤2lk+i≤M}, α≥0,

–N1(α) +card{2lk+i: 0≤2lk+i≤M}, α< 0,

and

M

j=0

"

dimX_∞+(j) +dimX–₀(j)#= 2"N1(β) –N1(α) –N₁0(β–) –N10(α+)

#

+ 2(M+ 1). (28)

Therefore

n=N1(β) –N1(α) –N₁0(β–) –N10(α+).

The proof for the case (–1)s+1_{= –1 is similar.}

(20)

Proof of Theorem4.2and Theorem4.3 Since the proof for the two theorems is similar, we prove only Theorem4.2.

LetX+₌_X+

∞+X∞0,X–=X–0+X00. Then, as in the proof of Theorem4.1, we check

condi-tions (a), (b), (c), (d), and (e). In the present case, we may suppose that (28) still holds for someM> 0. LetX_∞0(i) =X_∞0 ∩X(i),X₀0(i) =X0₀∩X(i). Then

n=1 2

M

i=0

dim

X_∞+(i)∩X₀–(i) –codimX(i)

X_∞+(i) +X₀–(i)!+dimX_∞0 +dimX₀0

=1 2

M

i=0

"

dimX_∞+(i) +dimX₀–(i) – 2#+dimX0_∞+dimX0₀

=1 2

M

i=0

"

dimX_∞+(i) +dimX₀–(i)#– (M+ 1) +dimX_∞0 +dimX₀0

=N(β) –N(α) –N0(β–) –N0(α+) +

N0(β+) +N0(β–) +N0(α+) +N0(α–)

=N(β) –N(α) +N0(β+) +N0(α–).

Proof of Theorem4.4,Theorem4.5,and Theorem4.6 Since the proof of Theorem4.4is similar to that of Theorem4.1, and the proofs of Theorems4.5and4.6are similar to that

of Theorem4.2, we omit it. Our proof is completed.

6 Examples

Example6.1 Suppose thatf∈C0₍_R_,_R_{) satisﬁes}

f(x) = ⎧ ⎨ ⎩

5π3x+x15, |x| 1,

–π3_x_–_x5_, _|_x_|_1.

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

x(3)(t) = –

3

i=1

fx(t–i). (29)

In this case,s= 1, (–1)s+1_{= 1,}_k_{= 2,}_m_{= 1,}_α_{= –}_π3_,_β_{= 5}_π3_{. This yields that}

N1(α) = –card

l≥0 : 0 <

(8l+ 1)π

4 3

tanπ

8 <π

3

–card

l≥0 : 0 <

(8l+ 3)π

4 3

tan3π

8 <π

3

= –1,

N1(β) =card

l≥0 : 0 <

(8l+ 8 – 1)π

4

3 tanπ

8 < 5π

3

+card

l≥0 : 0 <

(8l+ 8 – 2 – 1)π

4

3 tan3π

8 < 5π

3

= 2,

andN₁0(α+) =N10(β–) =N10(α–) =N10(β+) = 0.

(21)

Example6.2 Suppose thatf∈C0₍_R_,_R_{) satisﬁes}

f(x) = ⎧ ⎨ ⎩

3πx+x13_, |_x| _1,

πx–x3_, _|_x_|_1.

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

x(t) = –

5

i=1

(–1)i+1fx(t–i). (30)

In this case,s= 0, (–1)s+1_{= –1,}_k_{= 3,}_m_{= 2,}_α₌_π_,_β_{= 3}_π_{. This yields that}

N2(α) =card

l≥0 : 0 <(12l+ 1)π

6 cot

π

12<π

+card

l≥0 : 0 <(12l+ 3)π

6 cot

3π

12 <π

+card

l≥0 : 0 <(12l+ 5)π

6 cot

5π

12 <π

= 4,

N2(β) =card

l≥0 : 0 <(12l+ 1)π

6 cot

π

12< 3π

+card

l≥0 : 0 <(12l+ 3)π

6 cot

3π

12 < 3π

+card

l≥0 : 0 <(12l+ 5)π

6 cot

5π

12 < 3π

= 9,

andN₂0(α+) =N20(β–) =N20(α–) =N20(β+) = 0.

Applying Theorem4.5, we conclude that Eq. (30) possesses at least ﬁve diﬀerent 12-periodic orbits satisfyingx(t– 6) = –x(t).

Acknowledgements

The present research is supported by the National Natural Science Foundations of China (No. 61179031). The authors thank the referees for carefully reading the manuscript and for their valuable suggestions which have signiﬁcantly improved the paper.

Funding

Not applicable.

Availability of data and materials

Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally and signiﬁcantly in writing this article. All authors read and approved the ﬁnal manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

(22)

References

1. Benci, V.: On critical point theory for indeﬁnite functionals in the presence of symmetries. Transl. Am. Math. Soc. 274(2), 533–572 (1982)

2. Fannio, L.: Multiple periodic solutions of Hamiltonian systems with strong resonance at inﬁnity. Discrete Contin. Dyn. Syst.3, 251–264 (1997)

3. Fei, G.: Multiple periodic solutions of diﬀerential delay equations via Hamiltonian systems (I). Nonlinear Anal.65, 25–39 (2006)

4. Fei, G.: Multiple periodic solutions of diﬀerential delay equations via Hamiltonian systems (II). Nonlinear Anal.65, 40–58 (2006)

5. Ge, W.: On the existence of periodic solutions of diﬀerential delay equations with multiple lags. Acta Math. Appl. Sin. 17(2), 173–181 (1994) (in Chinese)

6. Ge, W.: Two existence theorems of periodic solutions for diﬀerential delay equations. Chin. Ann. Math.15(B2), 217–224 (1994)

7. Ge, W.: Periodic solutions of the diﬀerential delay equationx(t) = –f(x(t– 1)). Acta Math. Sin. New Ser.12, 113–121 (1996)

8. Ge, W.: Oscillatory periodic solutions of diﬀerential delay equations with multiple lags. Chin. Sci. Bull.42(6), 444–447 (1997)

9. Ge, W., Zhang, L.: Multiple periodic solutions of delay diﬀerential systems with 2k– 1 lags via variational approach. Discrete Contin. Dyn. Syst.36(9), 4925–4943 (2016)

10. Ge, W., Zhang, L.: Multiple periodic solutions of delay diﬀerential systems with 2klags via variational approach. Preprint

11. Guo, Z., Yu, J.: Multiple results for periodic solutions to delay diﬀerential equations via critical point theory. J. Diﬀer. Equ.218, 15–35 (2005)

12. Guo, Z., Yu, J.: Multiple results on periodic solutions to higher dimensional diﬀerential equations with multiple delays. J. Dyn. Diﬀer. Equ.23, 1029–1052 (2011)

13. Kaplan, J., Yorke, J.: Ordinary diﬀerential equations which yield periodic solution of delay equations. J. Math. Anal. Appl.48, 317–324 (1974)

14. Li, J., He, X.: Multiple periodic solutions of diﬀerential delay equations created by asymptotically linear Hamiltonian systems. Nonlinear Anal. TMA31, 45–54 (1998)

15. Li, J., He, X.: Proof and generalization of Kaplan–Yorke conjecture under the conditionf(0) > 0 on periodic solution of diﬀerential delay equations. Sci. China Ser. A42(9), 957–964 (1999)

16. Li, J., He, X., Liu, Z.: Hamiltonian symmetric group and multiple periodic solutions of diﬀerential delay equations. Nonlinear Anal. TMA35, 957–964 (1999)

17. Li, L., Xue, C., Ge, W.: Periodic orbits to Kaplan–Yorke like diﬀerential delay equations with two lags of ratio. Adv. Diﬀer. Equ.2016, 247 (2016)

18. Li, S., Liu, J.: Morse theory and asymptotically linear Hamiltonian systems. J. Diﬀer. Equ.78, 53–73 (1989) 19. Mawhen, J., Willem, M.: Critical Point Theory and Hamiltonian Systems. Springer, New York (1989)