The quantity of 2(2k+1)-periodic solutions to a specific differential delay system with 2k lags is studied and resolved by variational methods. Several results are revealed and two examples are given to illustrate the application of the main results.
Citation: Li Zhang, Huihui Pang, Weigao Ge. Multiple periodic solutions of differential delay systems with 2k lags[J]. AIMS Mathematics, 2021, 6(7): 6815-6832. doi: 10.3934/math.2021399
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The quantity of 2(2k+1)-periodic solutions to a specific differential delay system with 2k lags is studied and resolved by variational methods. Several results are revealed and two examples are given to illustrate the application of the main results.
Delay differential equations naturally appear in many fields of science and engineering. Such equations have been proposed as models for a variety of physiological processes and conditions including production of blood cells, respiration and cardiac arrhythemias. In recent years, the existence of positive periodic solutions for delay differential equations has been received considerable attention. J. Kaplan and J. Yorke [10] gave conditions for the existence of 4− or 6− periodic solutions of equation
x′(t)=−n∑i=1f(x(t−i)),x∈R | (1.1) |
while n=1 and n=2, respectively. After then, lots of results are achieved for this equation in general cases [1,2,3,4,5,6,7,8,9,10,11,12]. W. Ge [4,5,6] proved several existence theorems of periodic solutions for (1.1) in the general case by use of the fixed-point theorem in cone. After transforming (1.1) into a Hamiltonian system and applying a theorem given by Mawhin and Willem [13], J.Li and X. He [11,12] proved a theorem for the existence of several periodic solutions under the condition xf(x)>0(xf(x)<0) for x≠0. G. Fei [1,2] studied the multiplicity of periodic solutions of (1.1) and proved theorems for the cases n=2k−1 and n=2k by using the index theory. In his work, the way to construct the functional for the second case is quite different from that for the first case.
Applying S1 index theory Z. Guo and J. Yu [8] gave a theorem for the multiple periodic solutions of the delay-differential system with one lag in the form
x′(t)=−∇F(x(t−l)),x∈RN,F∈C0(RN,RN). | (1.2) |
And by applying the same index theory, in [9], they obtained results on the multiplicity of 2(n+1)-periodic solutions to the delay differential system
x′(t)=−n∑t=1∇F(x(t−i)),x∈RN,F∈C0(RN,RN). | (1.3) |
when n=2k−1. After then B. Zheng and Z. Guo studied (1.3) while n=2k and gave two criteria for the multiplicity of 2(2k+1)-periodic solutions. The theorems given in both[9] and [14] contain a condition to calculate the following function
Ψ(A,B)=∞∑j=1Ψj(A,B),n=2k−1, |
Ψ(A,B)=∞∑j=1,j∉ΓΨj(A,B),Γ⊂N+,n=2k, |
for symmetric matrices A and B, respectively. This is a hard task to be fulfilled since it is not definite for the problem how large the j must be to ensure Ψj(A,B)=0.
In [7], we researched the same problem for system (1.3) when n=2k−1 and by constructing a new functional we gave criteria for the multiplicity of periodic solutions only based on the eigenvalues of symmetric matrices A and B.
In this paper we are to study the multiplicity of 2(2k+1)-periodic solutions for system (1.3) when n=2k, i.e.,
x′(t)=−2k∑i=1∇F(x(t−i)),x∈RN,F∈C0(RN,RN). | (1.4) |
with the conditions F∈C1(RN,R),∇F(−x)=−F(x),F(0)=0,
x(t−k−1)=−x(t). | (1.5) |
At the same time, assume there are real symmetric matrices A∞ and A0 such that
∇F(x)=A∞x+o(|x|)as|x|→∞,∇F(x)=A0x+o(|x|)as|x|→0. |
In section 2, some notations and variation structure for system (1.4) are introduced. Meanwhile, a lemma about the relation between the periodic solutions of system(1.4) and the critical points of the functional Φ are also proved in section 2. Section 3 presents and proves the main results of this paper. Before the proof of the main results some lemmas about the calculation of the differential of functional Φ are proven. Two examples are given in section 4 to illustrate the applications of the main results.
Firstly, we consider a linear space which consists of all the 2(2k+1)-periodic functions satisfying (1.5) in RN, i.e.
H12={x(t)=∞∑i=1(aicos(2i−1)π2k+1t+bisin(2i−1)π2k+1t):ai,bi∈RN,∞∑i=1(2i−1)(|ai|2+|bi|2)<∞}
Since our goal is to find the 2(2k+1)-periodic solutions for system (1.4) with condition (1.5), we discuss now the special requirement for a solution of system (1.4) in H12. After then, we are able to choose a suitable Hilbert space for our problem. Assume n∑i=mai=0 if n<m.
If x(t) is a 2(2k+1)-periodic solution of (1.4) satisfying (1.5), then, x′(t−l)=l−1∑i=0∇F(x(t−i))−2k∑i=l+1∇F(x(t−i)),l=0,1,⋯,2k. Hence,
2k∑l=0(−1)lx′(t−l)=0. | (2.1) |
From the equation above, one has 2k∑l=0(−1)lx(t−l)=c. However, c=2k∑l=0(−1)lx(t−l−2k−1)=−2k∑l=0(−1)lx(t−l)=−c, which means c=0. Then,
2k∑l=0(−1)lx(t−l)=0. | (2.2) |
By Fourier's expansion theory and (1.5), it holds that
x(t)=∞∑i=1(aicos(2i−1)π2k+1t+bisin(2i−1)π2k+1t)=∞∑l=02k+1∑i=1(al(2k+1)+icos(2l(2k+1)+2i−1)π2k+1t+bl(2k+1)+isin(2l(2k+1)+2i−1)π2k+1t). |
Substituting the above expression into (2.2) we have
al(2k+1)+k+1=bl(2k+1)+k+1=0. | (2.3) |
Then suppose
X={x∈H12:x(t−2k−1)=−x(t),2k∑i=0(−1)ix(t−i)=0}. |
For x∈X, define P:X→X:
x↦∞∑l=0(k∑i=1+2k+1∑i=k+2)(2l(2k+1)+2i−1)(al(2k+1)+icos(2l(2k+1)+2i−1)π2k+1t+bl(2k+1)+isin(2l(2k+1)+2i−1)π2k+1t), |
and denote
⟨x,y⟩X=∫2(2k+1)0(Px(t),y(t))dt,⟨x,y⟩=∫2(2k+1)0(x(t),y(t))dt,x,y∈X. |
Define
‖x‖X=√⟨x,x⟩X,‖x‖2=√⟨x,x⟩=√⟨P−1x,x⟩X. |
It is obvious that (X,‖⋅‖X) is a Hilbert space and P:X→X is an invertible and self-adjoint operator. Define
Z={z(t)}={(z1(t),z2(t),⋯,z2k(t))}={(x(t),x(t−1),⋯,x(t−2k+1)):x∈X}⊂X2k. |
Therefore z1(t)=x(t)∈X. Furthermore, let
X(i)={acos(2i−1)π2k+1t+bsin(2i−1)π2k+1t:a,b∈RN}, |
Z(i)={(x(t),x(t−1),⋯,x(t−2k+1)):x(t)∈X(i)}. |
Then,
Z=∞∑i=1i≠l(2k+1)+k+1Z(i)=∞∑l=0k∑i=1Z(l(2k+1)+i). |
Denote H(z)=2k−1∑i=0F(x(t−i))+F(2k−1∑i=0(−1)i+1x(t−i)), then,
∇H(z)=(∂H(z)∂z1,∂H(z)∂z2,⋯,∂H(z)∂z2k)=(∇F(x(t))−∇F(x(t−2k)),∇F(x(t−1))+∇F(x(t−2k)),⋯,∇F(x(t−1))−(−1)i∇F(x(t−2k)),⋯,∇F(x(t−2k+1))+∇F(x(t−2k))) |
where x(t−2k)=2k−1∑i=0(−1)i+1x(t−i), zl=x(t−l+1),l=1,2,⋯,2k, and from (2.2),
z2k+1=2k−1∑i=0(−1)i+1x(t−i)=2k∑i=1(−1)izi. |
Define Φ:Z→R as a functional of z:
Φ(z)=12⟨Lz,z⟩+G(z) | (2.4) |
where
Lz=P−1(2k−1∑i=1(−1)i+1x′(t−i),⋯,l−1∑i=0(−1)i+lx′(t−i)+2k−1∑i=l+1(−1)i+l+1x′(t−i),⋯,2k−2∑i=0(−1)i+2k−1x′(t−i)),G(z)=−∫2(2k+1)0(2k−1∑i=0F(x(t−i))+F(2k−1∑i=0(−1)i+1x(t−i)))dt,⟨z,y⟩=∫2(2k+1)0(ˆPz,y)dt,z,y∈Z, |
ˆP=(P,P,⋯,P):Z→Z. Then
‖z‖=⟨z,z⟩=2k⟨x,x⟩X=2k‖x‖2X, | (2.5) |
Φ(z)=12∫2(2k+1)02k−1∑l=0(l−1∑i=0(−1)i+lx′(t−i)+2k−1∑i=l+1(−1)i+l+1x′(t−i),x(t−l))dt−∫2(2k+1)0[2k−1∑i=0F(x(t−i))+F(2k−1∑i=0(−1)i+1x(t−i))]dt=12∑0≤i<l≤2k−1∫2(2k+1)0(−1)i+l+1[(x(t−i),x′(t−l))−(x′(t−i),x(t−l))]dt−∫2(2k+1)0[2k−1∑i=0F(x(t−i))+F(2k−1∑i=0(−1)i+1x(t−i))]dt. |
By Mawhin-Willem's theorem [13, Theorem 1.4], Φ is continuously differentiable and
⟨Φ′(z),v⟩=⟨Lu,v⟩Z−∫2(2k+1)0(2k−1∑l=0(∇F(x(t−l))−(−1)l∇F(x(t−2k)),y(t−l)))dt=2k−1∑l=0∫2(2k+1)0(l−1∑i=0(−1)i+lx′(t−i)+2k−1∑i=l+1(−1)i+l+1x′(t−i)−∇F(x(t−l))−(−1)l+1∇F(x(t−2k)),y(t−l)))dt |
where z(t)=(x(t),x(t−1),⋯,x(t−2k+1)),v(t)=(y(t),y(t−1),⋯,y(t−2k+1)).
Denote Φ′(z)=(Φ′1(z),Φ′2(z),⋯,Φ′2k(z)), where
Φ′l+1(z)=P−1(l−1∑i=0(−1)i+lx′(t−i)+2k−1∑i=l(−1)i+l+1x′(t−i)−∇F(x(t−l))+(−1)l∇F(x(t−2k))),l=0,1,⋯,2k−1. | (2.6) |
If a point z∈Z satisfies Φ′(z)=0, then z is called a critical point of Φ. The following lemma reveals the relation between the 2(2k+1)-periodic solutions of system (1.4) and the critical points of Φ. Based on this lemma we can discuss the multiplicity of critical points of Φ instead of discussing that of 2(2k+1)-periodic solutions of system (1.4). In the following, denote x(t) as the first N components of z(t) in Z and set −1∑i=0x′(t−i)=0.
Lemma 2.1 If x∈X, z=(x(t),x(t−1),⋯,x(t−2k+1))∈Z, then the following propositions are equivalent to each other.
1) x(t) is one of the solutions of system (1.4),
2) z(t) satisfies Φ′1(z)=0, i.e.
2k−1∑i=1(−1)i+1x′(t−i)−∇F(x(t))+∇F(x(t−2k))=0, | (2.7) |
3) z(t) is one of the critical points of Φ on Z, i.e., z(t) satisfies
l−1∑i=0(−1)i+lx′(t−i)+2k−1∑i=l+1(−1)i+l+1x′(t−i)−∇F(x(t−l))+(−1)l∇F(x(t−2k))=0,l=0,1,⋯,2k−1(2.8−l)
Proof. 1)⇒ 2).
Suppose x(t) is a solution of system (1.4).
By the periodicity of x(t) and (1.5), one has
x′(t−l)=l−1∑i=0∇F(x(t−i))−2k−1∑i=l+1∇F(x(t−i))−∇F(x(t−2k))(2.9−l) |
l=1,2,⋯,2k−1, x(t−2k)=2k−1∑i=0(−1)i+1x(t−i). Multiplying (−1)l+1 to the equalities (2.9−l) and summing them up, we have
2k−1∑l=1(−1)l+1x′(t−l)=∇F(x(t))−∇F(x(t−2k)), |
which means (2.7) holds.
2)⇒ 3).
System (2.7) is the same as (2.8−0).
From (2.7) one has
2k−1∑i=1(−1)i+1x′(t−i−1)−∇F(x(t−1))+∇F(x(t−2k−1))=0, |
i.e.,
0=2k−1∑i=2(−1)ix′(t−i)+x′(t−2k)−∇F(x(t−1))−∇F(x(t))=2k−1∑i=2(−1)ix′(t−i)+2k−1∑i=0(−1)i+1x′(t−i)−∇F(x(t−1))−∇F(x(t)). | (2.10) |
Then (2.7) and (2.10) yield
−x′(t)+2k−1∑i=2(−1)ix′(t−i)−∇F(x(t−1))−∇F(x(t−2k))=0, |
which means (2.8−1).
Suppose (2.8−l) holds. Then,
0=l∑i=1(−1)i+l+1x′(t−i)+2k−1∑i=l+2(−1)i+l+2x′(t−i)+(−1)lx′(t−2k)−∇F(x(t−l−1))+(−1)l+1∇F(x(t))=l∑i=1(−1)i+l+1x′(t−i)+2k−1∑i=l+2(−1)i+l+2x′(t−i)+(−1)l+12k−1∑i=0(−1)ix′(t−i)−∇F(x(t−l−1))+(−1)l+1∇F(x(t)). | (2.11) |
Multiplying (2.7)with (−1)l+1 and add it to (2.11), one can obtain (2.8−(l+1)).
3) ⇒ 1).
Summing the equalities from (2.8−1) to (2.8−(2k−1)), one has
−x′(t)=2k∑i=1∇F(x(t−i)), |
which implies x(t) is a solution of system (1.4).
Denote ^Pi:Z→Z(i), Z(i)={(x(t),x(t−1),⋯,x(t−2k+1)):x(t)=aicos(2i−1)π2k+1t+bisin(2i−1)π2k+1t,ai,bi∈RN},i≠(2l+1)(2k+1); Z((2l+1)(2k+1))=0,l∈N+∪{0}. Furthermore, let Pi=i∑j=1ˆPj:Z→i∑j=1Z(i),i=1,2,⋯. Then, one has LPi=PiL.
Let −G′(z)=(P−1(∇F(x(t))−∇F(x(t−2k))),P−1(∇F(x(t−1))+∇F(x(t−2k))),⋯,P−1(∇F(x(t−l))+(−1)l+1∇F(x(t−2k))),⋯,P−1(∇F(x(t−2k+1))−∇F(x(t−2k)))). Then Φ′(z)=Lz+G′(z). L is a bounded self-adjoint operator and G′(z):Z→X2k is compact since (G′z)(t) is differentiable with respect to t.
Denote B∞,B0∈R2kN×2kN as
B∞=(2k+12kA∞,⋯,2k+12kA∞),B0=(2k+12kA0,⋯,2k+12kA0) |
and for z(t)=(x(t),x(t−1),⋯,x(t−2k+1)),
B∞z=(2k+12kA∞x(t),2k+12kA∞x(t−1),⋯,2k+12kA∞x(t−2k+1)),B0z=(2k+12kA0x(t),2k+12kA0x(t−1),⋯,2k+12kA0x(t−2k+1)). | (2.12) |
Define δz(t)=T2kNz(t):Z→Z, where
T2kN=[I−I⋯I−IIO⋯OOOI⋯OO⋮⋮⋯⋮⋮OO⋯IO] | (2.13) |
is a 2kN×2kN matrix, I is a N×N unit matrix and O is a N×N zero matrix.
From (1.5) and (2.2), one has δz(t)=z(t+1)andδ2k+1z(t)=−z(t). Then, {1,δ,δ2,⋯,δ4k+1} is a Lie group with δ4k+2=1, where 1 stands for the identity transform in Z.
Now we have
Φ(δz)=Φ(z), |
Φ′(δz)=δΦ′(z), |
i.e., with respect to the Lie group {1,δ,δ2,⋯,δ4k+1}, Φ is invariant and Φ′ is δ−equivariant.
In order to prove our results we shall apply the following lemma.
Lemma 2.2 [2, Lemma 2.4]
Assume there are two S1-invariant linear subspaces, Z+ and Z−⊂Z, and r>0 such that
(a) Z+∪Z− is a closed and finite codimension in Z,
(b) ˆL(Z−)⊂Z− with ˆL=L−P−1B∞ or ˆL=L−P−1B0,
(c) there exists c∞∈R such that
Φ(z)≤c∞<Φ(0),∀z∈Z−∩Sr={z∈Z−:‖z‖=r}, |
(d) there exists c0∈R such that
infz∈Z+Φ(z)≥c0, |
(e) Φ satisfies (P.S)c-condition for c0≤c≤c∞.
Then Φ possesses at least m=12[dim(Z−∩Z+)−codimZ(Z−∪Z+)] different critical orbits in Φ−1([c0,c∞]) provided m>0.
Remark 2.1 (P.S)c-condition in (e) can be replaced by (P.S)-condition since (P.S)-condition implies (P.S)c-condition for each c∈R.
Remark 2.2 If (P.S)c-condition in (e) is replaced by (P.S)-condition, (c) and (d) can be changed into
(c′) there exists c∞∈R such that
Φ(z)≥c∞>Φ(0),∀z∈Z−∩Sr={z∈Z−:‖z‖=r}, |
(d′) there exists c0∈R such that
infz∈Z+Φ(z)≤c0 |
if Φ(z) is replaced by −Φ(z).
Remark 2.3 When Z=∞⨁i=1Z(i), Z+(i)=Z+∩Z(i), Z−(i)=Z−∩Z(i), we have dimZ(i)=2N and
m=12∞∑i=1[dim(Z+(i)∩Z−(i))−codimZ(i)(Z+(i)+Z−(i))]=12∞∑i=1[dim(Z+(i)∩Z−(i))−(dimZ(i)−dim(Z+(i)+Z−(i)))]=12∞∑i=1[dimZ+(i)+dimZ−(i)−2N]. | (2.14) |
Suppose {αj:j=1,2,⋯,N} are the eigenvalues of A∞ and {βj:j=1,2,⋯,N} are the eigenvalues of A0, dj,ej∈RN are the eigenvectors of A∞ and A0 with respect to αj and βj, respectively. Assume {d1,d2,⋯,dN} and {e1,e2,⋯,eN} are two orthogonal bases of RN.
Lemma 3.1 If z∈Z(i),i≠l(k+1), then there is γ=π2k+1 such that
⟨Lz,z⟩=2k+12k(2i−1)γtan(2i−1)γ2⟨z,z⟩. | (3.1) |
Proof. Denote |⋅| as the norm in RN, then
⟨z,z⟩=2k∫2(2k+1)0|x(t)|2dt=2k(2k+1)(|ai|2+|bi|2). | (3.2) |
It follows that
⟨Lz,z⟩=∑0≤m<l≤2k−1(−1)m+l+1∫2(2k+1)0[(x(t−m),x′(t−l))−(x′(t−m),x(t−l))]dt=2∑0≤m<l≤2k−1(−1)m+l+1∫2(2k+1)0(x(t−m),x′(t−l))dtj=l−m=22k−1∑j=12k−1∑l=j(−1)2l−j+1∫2(2k+1)0(x(t−l+j),x′(t−l))dt=22k−1∑j=1(−1)j+1(2k−j)∫2(2k+1)0(x(t+j),x′(t))dt=22k−1∑j=1(2k−j)(−1)j+1(2i−1)γ(2k+1)(|a|2+|b|2)sinj(2i−1)γ=2(2i−1)γ(2k+1)2k−1∑j=1(2k−j)(−1)j+1sinj(2i−1)γ(|a|2+|b|2).
Since
2k−1∑j=1(2k−j)(−1)j+1sinj(2i−1)γ=12cos(2i−1)γ22k−1∑j=1(−1)j+1(2k−j)[sin(2j−1)(2i−1)γ2+sin(2j+1)(2i−1)γ2]=12cos(2i−1)γ2[2ksin(2i−1)γ2+2k∑j=1(−1)jsin(2j−1)(2i−1)γ2]=ktan(2i−1)γ2+12cos(2i−1)γ22k∑j=1(−1)j[sinj(2i−1)γcos(2i−1)γ2−cosj(2i−1)γsin(2i−1)γ2]=(k+12)tan(2i−1)γ2, |
one has
⟨Lz,z⟩=(2k+1)2(2i−1)γtan(2i−1)γ2(|ai|2+|bi|2)=2k+12k(2i−1)γtan(2i−1)γ2⟨z,z⟩. |
Lemma 3.1 is proved.
Denote Ej={λej:λ∈R}, Dj={λdj:λ∈R}, j=1,2,⋯,N, l≥0 and
indαj={k∑i=1♯{l:0<[2l(2k+1)+(2i−1)]γtan(2i−1)γ2<αj},αj>0,0,αj=0,−k∑i=1♯{l:0<[(2l+1)(2k+1)+2(k+1−i)]γtan(2i−1)γ2<−αj},αj<0, |
indβj={k∑i=1♯{l:0<[2l(2k+1)+(2i−1)]γtan(2i−1)γ2<βj},βj>0,0,βj=0,−k∑i=1♯{l:0<[(2l+1)(2k+1)+2(k+1−i)]γtan(2i−1)γ2<−βj},βj<0, |
then RN=N∑j=1Dj=N∑j=1Ej. Suppose
indA∞=N∑j=1indαj,indA0=N∑j=1indβj, | (3.3) |
and
XD,j(i)={x(t)=aicos(2i−1)γ2t+bisin(2i−1)γ2t,ai,bi∈Dj}, |
XE,j(i)={x(t)=aicos(2i−1)γ2t+bisin(2i−1)γ2t,ai,bi∈Ej}, |
therefore,
X(i)=N∑j=1XD,j(i)=N∑j=1XE,j(i). | (3.4) |
Denote
x(t)=∞∑l=0k∑i=1[al(2k+1)+icos(2l(2k+1)+2i−1)γt+bl(2k+1)+isin(2l(2k+1)+2i−1)γt+al(2k+1)+k+1+icos((2l+1)(2k+1)+2i)γt+bl(2k+1)+k+1+isin((2l+1)(2k+1)+2i)γt]. |
It follows from (3.1) and (3.2) that for z1(t)=x(t), one has
⟨Lz,z⟩=γ(2k+1)2∞∑l=0k∑i=1[(2l(2k+1)+2i−1)tan(2l(2k+1)+2i−1)γ2(|al(2k+1)+i|2+|bl(2k+1)+i|2)+(2l(2k+1)+2(k+1+i)−1)tan(2l(2k+1)+2(k+1+i)−1)γ2(|al(2k+1)+k+1+i|2+|bl(2k+1)+k+1+i|2)]=γ(2k+1)2∞∑l=0k∑i=1[(2l(2k+1)+2i−1)tan(2i−1)γ2(|al(2k+1)+i|2+|bl(2k+1)+i|2)−((2l+1)(2k+1)+2(k+1−i))tan(2i−1)γ2(|al(2k+1)+2k+2−i|2+|bl(2k+1)+2k+2−i|2)]. | (3.5) |
and
⟨P−1A∞x,x⟩=∫2(k+1)0(A∞x,x)dt=(2k+1)αj(|ai|2+|bi|2),x∈XD,j(i), |
⟨P−1A0x,x⟩=∫2(k+1)0(A0x,x)dt=(2k+1)βj(|ai|2+|bi|2),x∈XE,j(i), |
and
⟨ˆP−1B∞z,z⟩=αj(2k+1)2(|ai|2+|bi|2),x∈XD,j,⟨ˆP−1B0z,z⟩=βj(2k+1)2(|ai|2+|bi|2),x∈XE,j. | (3.6) |
Denote
Z−∞={z∈Z:⟨Lz−ˆP−1B∞z,z⟩<0},Z−0={z∈Z:⟨Lz−ˆP−1B0z,z⟩<0}, |
Z0∞={z∈Z:⟨Lz−ˆP−1B∞z,z⟩=0},Z00={z∈Z:⟨Lz−ˆP−1B0z,z⟩=0}, |
Z+∞={z∈Z:⟨Lz−ˆP−1B∞z,z⟩>0},Z+0={z∈Z:⟨Lz−ˆP−1B0z,z⟩>0}. |
Then
Z=Z−∞⨁Z0∞⨁Z+∞=Z−0⨁Z00⨁Z+0. | (3.7) |
From (3.1) and Lemma 3.1 [7], we have the following lemmas.
Lemma 3.2 ˆL(Z−∞)⊂Z−∞, ˆL(Z0∞)⊂Z0∞, ˆL(Z+∞)⊂Z+∞, ˆL(Z−0)⊂Z−0, ˆL(Z00)⊂Z00, ˆL(Z+0)⊂Z+0, (ˆL=L−ˆP−1B∞ or ˆL=L−ˆP−1B0).
Lemma 3.3 All the subspaces of Z,
Z−0+Z+∞,Z+0+Z−∞,Z+0+Z−∞+Z00,Z−0+Z+∞+Z00,Z+0+Z−∞+Z0∞, |
Z−0+Z+∞+Z0∞,Z−0+Z+∞+Z00+Z0∞,Z+0+Z−∞+Z00+Z0∞, |
are of finite codimensions in Z.
Denote Γ+∞={αj>0:there arel≥0,i∈{1,2,⋯,k}such thatαj=γ(2l(2k+1)+2i−1)tan(2i−1)γ2},Γ−∞={αj<0:there arel≥0,i∈{1,2,⋯,k}such thatαj=−γ((2l+1)(2k+1)+2(k+1−i))tan(2i−1)γ2},Γ+0={βj>0:there arel≥0,i∈{1,2,⋯,k}such thatβj=γ(2l(2k+1)+2i−1)tan(2i−1)γ2},Γ−0={βj<0:there arel≥0,i∈{1,2,⋯,k}such thatβj=−γ((2l+1)(2k+1)+2(k+1−i))tan(2i−1)γ2},Γ∞=Γ+∞∪Γ−∞,Γ0=Γ+0∪Γ−0.
For i∈{1,2,⋯,k} and l≥0, denote
η+∞=♯{(l,i):there isαj∈Γ+∞such thatγ(2l(2k+1)+2i−1)tan(2i−1)γ2=αj},
η−∞=♯{(l,i):there isαj∈Γ−∞such that−γ((2l+1)(2k+1)+2(k+1−i))tan(2i−1)γ2=αj},
η+0=♯{(l,i):there isβj∈Γ+0such thatγ(2l(2k+1)+2i−1)tan(2i−1)γ2=βj},
η−0=♯{(l,i):there isβj∈Γ−0such that−γ((2l+1)(2k+1)+2(k+1−i))tan(2i−1)γ2=βj}.
Let D=⨁{Dj:αj∈Γ∞} and Π:RN→D be an orthogonal projection with ΠRN=D. Assume
(A1) F∈C1(RN,R) satisfies (1.4) and F(−x)=F(x),F(0)=0,
(A2) there exist M and r∈C0(R+,R+) with r(∞)=+∞, r(s)s→0 as s→∞, such that |F(x)−12(A∞x,x)|≥−M+r(∣Πx∣) whenever x∈∪{Dj:αj∈Γ∞},
(A±3) ±[F(x)−12(A∞x,x)]>0,as|x|→∞,
(A±4) ±[F(x)−12(A0x,x)]>0,0<|x|≪1.
By a standard argument as the proof of Lemma 2.1 [13] and Lemma 3.3 [8], we have
Lemma 3.4 Assume (A1) and (A2) hold. Then, Φ(x) defined by (2.4) satisfies (P.S)-condition.
Lemma 3.5 Suppose (A1) and (A2) hold. Then there is I>0 such that
m=∑1≤i<∞i≠l(2k+1)+k+1[dim(Z+(i)∩Z−(i))−codimZ(i)(Z+(i)+Z−(i))]=∑1≤i≤Ii≠l(2k+1)+k+1[dim(Z+(i)∩Z−(i))−codimZ(i)(Z+(i)+Z−(i))], |
if
(Z+,Z−)∈{(Z+∞,Z−0),(Z+0,Z−∞),(Z−∞,Z+0),(Z−0,Z+∞),(Z+∞+Z0∞,Z−0),(Z+∞,Z−0+Z00),(Z−0,Z+∞+Z0∞),(Z−0+Z00,Z+∞),(Z+0,Z−∞+Z0∞),(Z+0+Z00,Z−∞),(Z−∞+Z0∞,Z+0),(Z−∞,Z+0+Z00),(Z+∞+Z0∞,Z−0+Z00),(Z+0+Z00,Z−∞+Z0∞),(Z−∞+Z0∞,Z+0+Z00),(Z−0+Z00,Z+∞+Z0∞)}. |
Now we give the main results of this paper.
Theorem 3.1 Assume (A1) and (A2) hold. Then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)−η−∞−η+0,ind(A0)−ind(A∞)−η−0−η+∞} |
different 2(2k+1)-periodic orbits satisfying x(t−(2k+1))=−x(t) provided m>0.
Corollary 3.1 Suppose (A1) and (A2) hold.
ⅰ) If Γ+∞∪Γ−∞=∅, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)−η+0,ind(A0)−ind(A∞)−η−0} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
ⅱ) If Γ+0∪Γ−0=∅, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)−η−∞,ind(A0)−ind(A∞)−η+∞} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
ⅲ) If Γ+0∪Γ−0=Γ+∞∪Γ−∞=Γ+0∪Γ−0=∅, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0),ind(A0)−ind(A∞)} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
Corollary 3.1 can be directly obtained from Theorem 3.1.
Theorem 3.2 Suppose (A1) and (A2) hold.
ⅰ) If (A+3) holds, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)+η+∞−η+0,ind(A0)−ind(A∞)−η−0−η+∞} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
ⅱ) If (A−3) holds, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)−η−∞−η+0,ind(A0)−ind(A∞)−η−0+η−∞} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
ⅲ) If (A+4) holds, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)−η−∞−η+0,ind(A0)−ind(A∞)+η+0−η+∞} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
ⅳ) If (A−4) holds, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)−η−∞+η−0,ind(A0)−ind(A∞)−η−0−η+∞} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
ⅴ) If (A+3), (A−4) hold, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)+η+∞+η−0,ind(A0)−ind(A∞)−η−0−η+∞} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
ⅵ) If (A−3), (A+4) hold, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)−η−∞−η+0,ind(A0)−ind(A∞)+η−∞+η+0} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
ⅶ) If (A+3), (A+4) hold, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)+η+∞−η−0,ind(A0)−ind(A∞)+η+0−η+∞} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
ⅷ) If (A−3), (A−4) hold, then system (1.4) possesses at least
m=max{ind(A∞)−ind(A0)−η−∞+η−0,ind(A0)−ind(A∞)+η+∞−η−0} |
different 2(2k+1)-periodic orbits satisfying x(t−2k−1)=−x(t) provided that m>0.
Proof of Theorem 3.1.
Suppose m=ind(A0)−ind(A∞)−η−0−η+∞>0.
Let Z+=Z−∞ and Z−=Z+0,∀z∈Z+. There is σ>0 such that
12⟨Lz−ˆP−1B∞z,z⟩≥σ‖x‖22. |
Then,
|F(x)−12⟨P−1Ax,x⟩X|≤σ2(2k+1)‖x‖22+M. |
Therefore,
Φ(z)=12⟨Lz,z⟩+G(z)=12⟨Lz,z⟩−∫2(2k+1)02k∑i=0F(x(t−i))dt=12⟨Lz,z⟩−(2k+1)∫2(2k+1)0F(x(t))dt=12⟨Lz−12ˆP−1B∞z,z⟩−(2k+1)∫2(2k+1)0[F(x(t))−12⟨P−1Ax,x⟩X]dt≥12σ‖x‖22−2(2k+1)M→+∞,‖x‖22→∞, |
which implies that there is c0∈R such that
Φ(z)≥c0,z∈Z+. |
At the same time, there are c∞∈R− and γ>0 such that
Φ(z)≤c∞. |
Suppose c0<c∞. Then conditions (c) and (d) in Lemma 2.2 are satisfied.
Since Lemmas 2.1 and 3.2–3.4 imply (P.S)-condition as well as conditions (a) and (b) in Lemma 2.2 hold under the requirement of Theorem 3.1, we only need to compute m.
From Lemma 3.5 we have
m=12∑1≤i≤Ii≠l(2k+1)+k+1[dimZ+(i)∩Z−(i)−codimZ(i)(Z+(i)+Z−(i))]=12∑1≤i≤Ii≠l(2k+1)+k+1[dimZ+(i)+dimZ−(i)−dimZ(i)]=12∑1≤i≤Ii≠l(2k+1)+k+1[dimZ+(i)+dimZ−(i)−2N]. | (3.8) |
Then,
m=12∑1≤i≤Ii≠l(2k+1)+k+1[dim(Z+∞(i)+dimZ−0(i))−2N]=12∑l(2k+1)+i≤Ii∈{1,2,⋯,k},l≥0[dimZ+∞(l(2k+1)+i)+dimZ−0(l(2k+1)+i)−2N]+12∑l(2k+1)+k+1+i≤Ii∈{1,2,⋯,k},l≥0[dimZ+∞(l(2k+1)+k+1+i)+dimZ−0(l(2k+1)+k+1+i)−2N]=∑l(2k+1)+i≤Ii∈{1,2,⋯,k},l≥0[dimZ+∞(l(2k+1)+i)+dimZ−0(l(2k+1)+i)−2N]+12∑l(2k+1)+2(k+1)−i≤Ii∈{1,2,⋯,k},l≥0[dimZ+∞(l(2k+1)+2(k+1)−i)+dimZ−0(l(2k+1)+2(k+1)−i)−2N]. |
Denote I1=♯{l(2k+1)+i≤I:l≥0,i∈{1,2,⋯,k}},I2=♯{l(2k+1)+2(k+1)−i≤I:l≥0,i∈{1,2,⋯,k}}, and indA+∞=∑αj>0indαj,indA−∞=∑αj<0indαj,indA+0=∑βj>0indβj,indA−0=∑βj<0indβj, then,
indA∞=indA+∞+indA−∞,indA0=indA+0+indA−0. |
Obviously,
∑l(2k+1)+i≤Ii∈{1,2,⋯,k},l≥0dimZ−0(l(2k+1)+i)=2indA+0,
∑l(2k+1)+i≤Ii∈{1,2,⋯,k},l≥0dimZ+∞(l(2k+1)+i)=∑l(2k+1)+i≤Ii∈{1,2,⋯,k},l≥0[2N−dimZ−∞(l(2k+1)+i)−dimZ0∞(l(2k+1)+i)]=2NI1−2indA+∞−2η+∞,
∑l(2k+1)+2(k+1)−i≤Ii∈{1,2,⋯,k},l≥0dimZ−0(l(2k+1)+2(k+1)−i)=∑l(2k+1)+2(k+1)−i≤Ii∈{1,2,⋯,k},l≥0[2N−dimZ00−dimZ+0]=2NI2+2indA–02η−0.
Therefore,
m=indA+0+indA−0−indA+∞−indA−∞−η+∞−η−0=indA0−indA∞−η+∞−η−0. |
Theorem 3.1 is proved.
Proof of Theorem 3.2.
Since the proof of Theorem 3.2 for each case is similar, we prove the theorem only for case (ⅱ).
Without loss of generality, suppose
indA0−indA∞−η−0+η−∞>max{0,indA∞−indA0−η−∞−η+0}, |
and denote Z+=Z+∞+Z0∞,Z−=Z−0.
From (A−3), there is M>0 such that
−[F(x)−12(A∞x,x)]>−M,x∈RN. |
Then, for z∈Z+∞+Z0∞,
Φ(z)=12⟨Lz,z⟩+G(z)=12⟨Lz−12ˆP−1B∞z,z⟩+G(z)+12⟨ˆP−1B∞z,z⟩≥G(z)+12⟨ˆP−1B∞z,z⟩=−(2k+1)∫2(2k+1)0[F(x(t))−12(A∞x,x)]dt>(2k+1)∫2(2k+1)0(−M)dt=−2M(2k+1)2, |
which implies that condition (d) in Lemma 2.2 holds.
From (3.8), one has
m=12∑1≤i≤Ii≠l+k+1[dimZ+(i)+dimZ−(i)−2N]=12∑1≤i≤Ii≠l(2k+1)+k+1[dimZ+∞(i)+dimZ0∞(i)+dimZ−0(i)−2N]=12∑1≤i≤Ii≠l(2k+1)+k+1[dimZ+∞(i)+dimZ0∞(i)−2N]+12∑1≤i≤Ii≠l(2k+1)+k+1dimZ0∞(i)=indA0−indA∞−η+∞−η−0+12dimZ0∞=indA0−indA∞−η+∞−η−0+η+∞+η−∞=indA0−indA∞−η−0+η−∞.
Theorem 3.2 is proved.
Example 4.1 Consider the number of 6-periodic solutions of the following system
x′(t)=−∇F(x(t−1))−∇F(x(t−2)) | (4.1) |
where x=(x1,x2)∈R2,
F(x)=−π12√3(9x21+22x1x2+9x22)−(x431+2x432),|x|→∞, |
F(x)=π12√3(29x21+22√3x1x2+7x22)+(x21+x22)32,|x|→0. |
Obviously, F∈C1(R2,R),F(0)=0,F(−x)=F(x),
∇F(x)=[−3π2√3−11π6√3−11π6√3−3π2√3][x1x2]−[43x13183x132],|x|→∞ |
∇F(x)=[29π6√311√3π6√311√3π6√37π6√3][x1x2]+[3(x21+x22)2x13(x21+x22)12x2],|x|→0 |
and
A∞=[−3π2√3−11π6√3−11π6√3−3π2√3],A0=[−29π6√311π611π6−7π6√3]. |
The eigenvalues of A∞ and A0 are
α1=π3√3,α2=−10π3√3andβ1=−π3√3,β2=10π3√3, |
respectively. Since k=1, one has
indα1=♯{l≥0:0<13(6l+1)π1√3=π3√3(6l+1)<π3√3}=0,indα2=−♯{l≥0:0<π3√3(6l+5)<10π3√3}=−1,indβ1=−♯{l≥0:0<π3√3(6l+5)<π3√3}=0,indβ2=♯{l≥0:0<π3√3(6l+1)<10π3√3}=2. |
Therefore,
indA∞=−1,indA0=2. |
At the same time,
η+∞=1,η−∞=0,η+0=0,η−0=0, |
and all the conditions of Theorem 3.2 (vi) are satisfied. Then system (4.1) has at least
m=indA0−indA∞+η−∞+η+0=3 |
different 6-periodic orbits satisfying x(t−3)=−x(t).
Example 4.2 Let N=2. We discuss the multiplicity of 6-periodic solutions of system (4.1), where x=(x1,x2)∈R2,
F(x)=(−25π6√3x21+5π3x1x2−5π2√3x22−x41−x62)(1−φ(|x|))+(5π6√3x21−5π√3x1x2+5π6√3x22+x431+x852)φ(|x|). |
Then
∇F(x)=A0x−[4x316x52],|x|→0, |
∇F(x)=A∞x+[43x13185x352],|x|→∞, |
where
A0=[−25π3√35π35π3−5π√3],A∞=[5π3√3−5π√3−5π√35π3√3]. |
A∞ and A0 have their eigenvalues α1=−10π3√3,α2=20π3√3, and β1=−10π√3,β2=−10π3√3, respectively.
So
indα1=−♯{l≥0:0<π3√3(6l+5)<10π3√3}=−1,indα2=♯{l≥0:0<π3√3(6l+1)<20π3√3}=4,indβ1=−♯{l≥0:0<π3√3(6l+5)<10π√3}=−5,indβ2=−♯{l≥0:0<π3√3(6l+5)<10π3√3}=−1, |
and then
indA∞=3,indA0=−6. |
On the other hand, we have
η+∞=η−∞=η+0=η−0=0. |
Therefore,
m=indA∞−indA0=9, |
and system (4.1) has at least 9 different 6-periodic orbits satisfying x(t−3)=−x(t) by Theorem 3.1.
Supported by R & D Program of Beijing Municipal Education Commission (No. KM202011417010), Academic Research Projects of Beijing Union University (No. ZB10202004), Scientific Research Projects of Beijing Union University (No. WZ10201902).
The authors declare that there is no conflicts of interest regarding the publication of this paper.
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