R E S E A R C H
Open Access
Multiple periodic solutions to a class of
nonautonomous second-order delay
differential equation
Qiong Meng
1*and Xiaosheng Zhang
2*Correspondence:
1School of Mathematical Science,
Shanxi University, Wucheng Road, Taiyuan, 030006, P.R. China Full list of author information is available at the end of the article
Abstract
The existence of the nontrivial periodic solutions for nonautonomous second-order delay differential equation
x(t) +
λ
x(t) = –f(
t,x(t),x(t–τ
))is investigated, where
λ
>π
2/τ2,τ
> 0,f∈C1(R×R2,R). Multiple periodic solutionsare obtained by some recent critical point theorems.
MSC: 34K13; 34K18; 58E50
Keywords: second-order delay differential equation; nonautonomous; periodic solution; critical point theorems
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 equa-tions (see [–]). In , Li and He [] first applied the critical point theory to study the multiplicity of periodic solutions for delay differential equations. Especially, in , 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 func-tional 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 [, ]).
Motivated by the work of [, ], we consider a class of nonautonomous second-order delay differential equation
x(t) +λx(t) = –ft,x(t),x(t–τ), (.)
whereλ>π/τ,τ> ,f∈C(R×R,R).
In this paper, we have the following conditions onf.
(f) 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 allx,y∈R,t∈[,τ];
(f) there exist fourτ-periodic and continuous functionsa(t),b(t),c(t)andd(t)such that
f(t,x,y) =a(t)x+b(t)y+◦(r) asr=x+y→
and
f(t,x,y) =c(t)x+d(t)y+◦(r) asr=x+y→ ∞
uniformly fort∈[,τ];
(f±) |∇F(t,z) –B(t)z|is bounded for allz= (x,y)∈R,t∈[,τ]and
F(t,z) –
B(t)z,z→ ±∞ as|z| → ∞
uniformly fort∈[,τ], where
F(t,x,y) = x
f(t,s,y)ds+ y
f(t,s, )ds,
A(t) =
a(t) b(t) b(t) a(t)
, B(t) =
c(t) d(t) d(t) c(t)
.
Theorem . Assume that f satisfies (f)-(f–) with (A(t)z,z) ≤ , (B(t)z,z) > (λ –
π/τ)|z|> for z∈R\{},t∈[,τ].Then(.)possesses at leastm pairsτ-periodic solutions,where
m=max
j∈Z+:B(t)z,z> λ–π
τj
|z|> for z∈R\{},t∈[,τ]
,
Z+is the set of all positive integers.
Theorem . Assume that f satisfies (f)-(f+) with (A(t)z,z)≥ , (B(t)z,z) < (λ –
πm/τ)|z|< for z∈R\{},m∈Z+,t∈[,τ].Then(.)possesses at least[m– m+ ]pairsτ-periodic solutions,where
m=max
j∈Z+:B(t)z,z< λ–π
τj
|z|< for z∈R\{},t∈[,τ]
.
In this paper, the main purpose is to study the multiplicity of periodic solutions for sys-tems (.) via some recent critical point theorems for strongly indefinite functionals. In order to achieve this, some preliminaries are necessary. Let XandY be Banach spaces withXbeing separable and reflexive, and setE=X⊕Y. LetS⊂X*be a dense subset. For eachs∈S, there is a semi-norm onEdefined by
ps:E→R, ps(u) =s(x)+y foru=x+y∈X⊕Y.
We denote byTS the topology onEinduced by the semi-norm family{ps}, and letωand
ω*denote the weak-topology and weak*-topology, respectively. Clearly, theT
contains the product topology onE=X⊕Yproduced by the weak topology onXand the strong topology onY.
For a functional∈C(E,R), we write
a={u∈E:(u)≥a}. Recall thatis weakly
sequentially continuous ifukuinE, one haslimk→∞(uk)v→(uk)vfor eachv∈E,
i.e.,: (E,ω)→(E*,ω*) is sequentially continuous. Forc∈R, we say thatsatisfies the (C)ccondition if any sequence{uk} ⊂E, such that(uk)→cand ( +uk)(uk)→ as
k→ ∞, contains a convergent subsequence. Suppose that
() ∈C(E,R),cisTS-closed for everyc∈Rand: (c,TS)→(E*,ω*)is
continu-ous;
() there existsρ> such thatκ:=inf(Bρ∩Y) > =(), whereBρ={u∈E:u= ρ};
() there exist a finite dimensional subspaceY⊂YandR>ρsuch thatc:=sup(E) <
∞andsup(E\S) <inf(Bρ∩Y), whereE:=X⊕Y, andS={u∈E:u ≤R}.
The following critical point theorem will be used later (see [, ]).
Theorem A Assume thatis even and()-()are satisfied.Thenhas at least m= dimYpairs of critical points with critical values less than or equal to c providedsatisfies the(C)ccondition for all c∈[κ,c].
2 Preliminaries
In this section, we establish a variational structure which enables us to reduce the exis-tence of τ periodic solutions of (.) to a classic Hamiltonian system. First, we have the following lemma by using similar arguments [].
Lemma . Suppose that f ∈C(R×R,R)satisfies(f),then the function
H(t,x,y) = x
f(t,s,y)ds+ y
f(t,s, )ds (.)
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(R×R,R).
Proof It is obvious that ∂H
∂x =f(t,x,y). By (f),
∂f(t,x,y)
∂y =
∂f(t,y,x)
∂x . Then we have
∂H(t,x,y)
∂y = x
∂f(t,y,s)
∂s ds+f(t,y, ) =f(t,y,x).
The proof of Lemma . is complete.
Suppose thatxis a periodic solution of (.) with τ, lety(t) =x(t–τ), then
We denotez= (x,y). Then
z(t) +λz(t) = –∇Ht,z(t), (.)
whereH(t,z) is defined in (.) and∇H(t,z) denotes the gradient ofH(t,z) with respect to thezvariable. It is easy to obtain the following lemma.
Lemma . For anyτ periodic solution of(.),let y(t) =x(t–τ),then z= (x,y)is aτ
periodic solution of(.).Conversely,for anyτperiodic solution z= (x,y)of(.)satisfying y(t) =x(t–τ),x is aτ periodic solution of(.).
ForS=R/(τZ), letC∞(S,R) denote the space of τperiodicC∞functions onRwith
values inR. For anyz∈C∞(S,R), it has the following Fourier expansion in the sense that it is convergent in the spaceL(S,R),
z(t) =√a τ +
√ τ
∞
j=
ajcos
π τjt
+bjsin
π τjt
,
wherea,aj,bj∈R,j= , , . . . .
Let z∈L(S,R). If there exists a function y∈ L(S,R) such that, for every x ∈ C∞(S,R),
τ
z(t),x(t)dt= – τ
y(t),x(t)dt,
thenyis called a weak derivative ofzdenoted byy=z(t). Let
HS,R=
z∈LS,R τ
z(t)+z(t)dt< +∞
.
For anyz,y∈H(S,R),·,·and · can be explicitly expressed by
z,y= τ
z(t),y(t)+z(t),y(t)dt
and
z=
|a|+
∞
j=
+j|aj|+|bj|
/ .
We define an operatorL:H(S,R)→H(S,R) by the Riesz representation theorem
Lz,y= τ
˙
z(t),y˙(t)dt–λ
τ
z(t),y(t)dt.
By (f) and (.) in Section , there exist two continuous functionsα(t) > andβ(t)≥ such that
H(t,z)≤α(t)|z|+β(t), ∀z∈R.
By using similar arguments as in [], we have that
ϕ(z) =
Lz,z+ψ(z),
where
ψ(z) = τ
Ht,z(t)dt.
Thus the critical points ofϕ(z) inH(S,R) are classical solutions of (.).
DenoteE={z= (x,y)∈H(S,R) :y(t) =x(t–τ)}. By a direct computation, we have
E=
z∈HS,R|z(t) =√ τa
+√
τ ∞
j= ajcos
π τjt
+bjsin
π τjt
(–)j
⎫⎬
⎭,
wherea,aj,bj∈R,j= , , . . . .
Note that for anyz= (x,y)∈E,y(t) =x(t–τ) and∇H(t,z) = (f(t,x,y),f(t,y,x)). Due to the fact thatf(t,x(t–τ),y(t–τ)) =f(t,y(t),x(t)), we know that∇H(t,z)∈Ewhateverz∈E. Therefore, we have the following lemma.
Lemma . If z(t)is a critical point ofϕin E,then z(t)is a critical point ofϕin H(S,R).
Moreover, we also denote byM+(·),M–(·) andM(·) the positive definite, negative defi-nite and null subspaces of the self-adjoint linear operator defining it, respectively.
ThenEhas an orthogonal decomposition
E=M+(L)⊕M–(L)⊕M(L).
Let
zj(t) =
√
τ ajcos π τjt
+bjsin
π τjt
(–)j
, j= , , . . . ,
whereaj,bj∈R. We have
Lzj,zj=
+j
π τj
–λ
zj, j= , , . . . .
There existsσ> such that
Lzj,zj ≤–σzj forj∈J–,
whereJ+={j≥ : +j(π
τj–λ) > },J–={j≥ :+j(π
τj–λ) < },J={j≥ :+j(π
τj–
λ) = }. Therefore, we get
Lz,z ≥σz forz∈M+(L), (.)
Lz,z ≤–σz forz∈M–(L). (.)
Remark . The conditionλ>π/τis to ensureJ–=∅.
3 Proofs of theorems
In this section,cistand for different positive constants fori∈Z+.
By a direct computation, (f) and (f) imply thatH(t,z) is even and satisfies
∇H(t,z) =A(t)z+◦|z| as|z| →, (.)
∇H(t,z) =B(t)z+◦|z| as|z| → ∞ (.)
uniformly fort∈[,τ]. (f±) implies that
∇H(t,z) –B(t)z is bounded and
H(t,z) –
B(t)z,z→ ±∞ as|z| → ∞
(.±)
uniformly fort∈[,τ].
Lemma . Suppose that f satisfies(f)-(f±).Then the functionϕsatisfies the(C)c
con-dition for any c∈R.
Proof First we define an operator
Bz,y= τ
B(t)z(t),y(t)dt
for anyz,y∈E. By a direct computation,Bis a bounded self-adjoint linear operator onE. ThusL+Bis also a self-adjoint linear operator onE. ThenEhas an orthogonal decompo-sition
E=M+(L+B)⊕M–(L+B)⊕M(L+B).
Also there existsσ> such that
(L+B)z,z≥σz forz∈M+(L+B), (.)
So, set
(z) = –ϕ(z) = –
(L+B)z,z– τ
H(t,z) –
B(t)z,zdt
and
(z),y= –(L+B)z,y– τ
∇H(t,z) –B(t)z,ydt
for anyz,y∈E,ϕ(z) is defined in Section . Let{zk} ⊂Ebe any sequence such that
(zk)→c,
+zk
(zk)→ ask→ ∞.
We first prove that{zk}is bounded. For anyε> , (.) implies that there exists a constant
Cε> such that
∇H(t,z) –B(t)z<ε|z|+Cε (.)
for allz∈Randt∈[, τ]. Sincez
k∈E, we have
zk=zk++z–k+zk,
wherez+
k∈M+(L+B),z–k ∈M–(L+B),zk∈M(L+B). By (.±), there exists a constant
d> such that
∇H(t,z) –B(t)z≤d.
Therefore we have
( +c)z+k≥–(zk),z+k
=(L+B)z+k,z+k+ τ
∇H(t,zk) –B(t)zk,z+k
dt
≥σz+k
–dz+k.
Thus we have that{z+
k}is bounded. Using similar arguments, we can prove that{z–k}
is bounded. Consider{zk}. Arguing indirectly, we suppose{zk}is unbounded, then we havezk → ∞. According to the definition ofM(L+B), this implies that there are constantsd,d> such that
dzk≤zk≤dzk. (.)
By (.), we have
Then
c= lim
k→∞
(zk) –(zk)zk
= lim
k→∞
τ
∇H(t,zk) –B(t)zk,zk
dt
– τ
H(t,zk) –
B(t)zk,zk
dt
. (.)
By (.) and (.±), (.) is a contradiction. Hence{z
k}is bounded. Therefore there
exists a constantd> such that
zk=z+k+z–k+zk≤d.
So, we get{zk}is bounded, and going if necessary to a subsequence, we can assume that
zkzinE andzk→zinC(S,RN). Write zk=z+k +z–k +zk andz=z++z–+z, then
z±k z±inE,z
k→zinEandz±k →z±inC(S,RN).
In view of (.) andz–
k →z–inC(S,R), it is easy to verify
τ
∇H(t,zk) –B(t)zk,z–k–z–
dt→
and τ
∇H(t,z) –B(t)z,zk––z–dt→.
But then(zk) –(z),z–k–z– → ask→ ∞, and
(zk) –(z),z–k–z–
= –(L+B)z–k–z–,z–k–z–
– τ
∇H(t,zk) –B(t)zk,z–k–z–
dt
+ τ
∇H(t,z) –B(t)z,z–k–z–dt
≥σz–k–z––
τ
∇H(t,zk) –B(t)zk,z–k–z–
dt
+ τ
∇H(t,z) –B(t)z,z–k–z–dt.
This yieldsz–
k →z– inE. Similarly,z+k →z+inEand hencezk→zinE, that is,
sat-isfies the (C)ccondition. Thusϕ satisfies the (C)ccondition. The proof of Lemma . is
complete.
Proof of Theorem. Forz∈E, letX=M+(L)⊕M(L),Y=M–(L),E=X⊕Y, and
(z) = –ϕ(z), (z) =ψ(z), (.)
In order to obtain this theorem, we apply Theorem A to the functional(z). The proof of this theorem is divided into the following three steps.
Step .satisfies ().
We first check that c is TS -closed for any c∈R. Let {zk} be any sequence TS
-converging to somez∈E. Writezk=z+k+zk+z–k andz=z++z+z–, thenz–k →u– in
Eand hence{z–}is bounded in the norm topology. Note that (B(t)z,z) >λ–π/τ> for z∈R\{}and fort∈[, τ], then for anyz∈Eand smallε> , we have by (.) and (.)
(z) = τ
H(t,z)dt
= τ
∇H(t,sz),zds dt
≥ τ
B(t)sz,zds dt– τ
∇H(t,z) –B(t)sz,z|z|ds dt
≥
τ
B(t)z,zdt– τ
ε|sz|+Cε
|z|ds dt
≥
λ–
π τ –ε
zL–CεzL≥cz
L–czL,
which implies thatis bounded from below onE. Consequently, combiningzk∈cand
Lz
+
k,zk+= –Lz –
k,z–k–(zk) –(zk) shows that{zk+}is bounded inEby (.) and hence
(zk) = –
Lz+k,z+k–
Lz–k,zk––(zk)
≤–
Lz+k,z+k–
Lz–k,u–k–c≤c. (.)
Moreover, sincezkL =z+kL+zkL+z–kL, we have
(zk)≥czkL–czkL≥ zkL–c≥zk
L–c.
It follows from (.) and the above inequality that{zk}is also bounded inE since all norms are equivalent in a finite dimensional space. Thenzk(=z+k+zk+z–k)
is bounded, and hence we can assume that{zk}converges weakly toz=z++z+z–inE.
Thus we have(zk)→(z). Note thatLy,y/is an equivalent norm onH+. By the lower
semi-continuity of the norm, we get
c≤lim sup
k→∞
(zk) =lim sup k→∞
–
Lzk+,z+k–
Lz–k,z–k–(zk)
= –lim inf
k→∞
Lzk+,z+k–
Lz–,z––(z)≤(z),
that is,z∈cand hencecisTS-closed.
Next, we prove that: (c,TS)→(E*,ω*) is continuous. To achieve this, it is sufficient
to demonstrate thathas the same property. SupposezkuinE, then{zk}converges
con-verges to (∇H(t,z(t)),y(t)) in measure on [, τ]. Moreover, by (.), one has
∇Ht,zk(t)
,y(t))≤εzk∞+b¯zk∞+Cε
y∞≤c
for allkandt∈[, τ], whereb¯=maxt∈[,τ]{|B(t)|}and · ∞denotes the natural norm of C(S,R). Thus, the Vitali theorem is applicable and
(zk),y
=
τ
∇H(t,zk),y
dt→
τ
∇H(t,z),ydt=(z),y
for anyy∈E. So,satisfies (). Step .satisfies ().
By (.), there exist <ε<σ/,ρ> such that
H(t,z)≤ε|z|+
A(t)z,z for all|z|<ρ, t∈[, τ]. (.)
By Proposition . in [], there is a positive constantξ such thatz∞≤ξz. Set small
ρ<ρ/ξ, then for eachz∈Ywithz ≤ρ, one hasz∞≤ρ, and hence by (.)
(z) = –
Lz,z– τ
H(t,z)dt
≥ σ
z –
τ
ε|z|+
A(t)z,zdt≥ σ –ε
z.
Therefore,
κ:=inf(Bρ∩Y)≥ σ
–ε
ρ> ,
and hence () holds. Step .satisfies (). Let
Y=span
cos π
τjt
(–)j
,sin π
τjt
(–)j
:j∈Z+,j≤m
.
Obviously,Y⊂YanddimY= m. In order to obtain the desired conclusion, it is suffi-cient to prove that(z)→–∞asz → ∞onE:=X⊕Y.
Let=λ–πτ. By the definition ofm, there exists a constant <δ<σ such that
B(t)≥+δ (.)
fort∈[,τ]. Clearly, for anyy∈Y,
Ly,y ≥–yL. (.)
LetF(t,z) =H(t,z) –(B(t)z,z). We claim that
z– τ
Indeed, for =z∈E, by (.), one has
z– τ
F(t,z)dt
=z– τ
∇H(t,sz) –B(t)sz,zds dt
≤ z– τ
ε|z|+Cε
|z|dt
≤ z–εzL+CεzL
≤ε+ Cε
z,
which implies that (.) is true by the arbitrariness ofε. Then, forz=z++z+z–∈E , by (.), (.) and (.), one has
(z) = –
Lz–,z––
Lz+,z+– τ
F(t,z)dt
≤
z –
L–
δ
z +–
τ
B(t)z,zdt– τ
F(t,z)dt
≤
z –
L–
δ
z +
–
(+δ)z
L– τ
F(t,z)dt
≤
z –
L–
δ
z +
–
(+δ)z –
L+z L – τ
F(t,z)dt
≤–δ z
–
L+z+
+z
L
– τ
F(t,z)dt.
SinceMandYare finitely dimensional, (.) and the above estimate imply that(z)→ –∞asz → ∞,z∈E:=X⊕Y. Hence () holds.
The proof of Theorem . is complete.
Proof of Theorem. Forz∈E, letX=M–(L)⊕M(L),Y=M+(L),E=X⊕Y, and
(z) =ϕ(z), (z) = –ψ(z)
and
Y=span cos π τjt (–)j ,sin π
τjt
(–)j
:j=m,m+ , . . . ,m
.
Then the conclusion is obtained by the same argument as in the proof of Theorem .. The
proof of Theorem .. is complete.
Example . Consider the following equation:
x(t) +λx(t) = –α(t)x(t) –β(t)x(t–π) – x(t)
wherex(t)≥. Let
f(t,x,y) =α(t)x+β(t)y+ x x+y+γ(t).
Then
∂f(t,x,y)
∂y =β(t) –
xy
(x+y+γ(t)) =
∂f(t,y,x)
∂x .
Let
H(t,x,y) = α(t)x
+β(t)xy+ α(t)y
+ ln
x+y+γ(t)
γ(t) .
Then
∂H
∂x =f(t,x,y),
∂H
∂y =f(t,y,x).
By a straightforward computation, we have
f(t,x,y) =
γ(t)+α(t)
x+β(t)y+◦(r) asr=x+y→
and
f(t,x,y) =α(t)x+β(t)y+◦(r) asr=x+y→ ∞
uniformly fort∈[,π].
Takeλ= ,α(t) = – +sinπt,β(t) = –sinπt,γ(t) =
. By Theorem ., we havea(t) =
γ(t)+α(t)≥,b(t) =d(t) = –sinπt≥,c(t) = – +sinπtfort∈[,π],z= (x,y),x≥, y≥,
A(t)z,z=
a(t) b(t) b(t) a(t)
x y
,
x y
≥,
B(t)z,z=
c(t) d(t) d(t) c(t)
x y
,
x y
=c(t)x+y+ d(t)xy≤c(t) +d(t)x+y
= –x+y< – x+y< .
Then we havem=m= . By Theorem ., (.) possesses at least two pairs π-periodic solutions.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Author details
1School of Mathematical Science, Shanxi University, Wucheng Road, Taiyuan, 030006, P.R. China.2School of Mathematical
Sciences, Capital Normal University, Beijing, 100048, P.R. China.
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.
Received: 18 May 2013 Accepted: 12 September 2013 Published:19 Nov 2013 References
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