For convenience, we introduce some notations, definitions, and lemmas. If g(t) is a continuous ω-periodic function defined on R, denote
(3)g_=mint∈[0,ω]|g(t)|, g¯=maxt∈[0,ω]|g(t)|,m(g)=1ω∫0ωg(t)dt.
We also denote the spectral radius of the matrix A by ρ(A). Denote
(4)X={x(t)=(x1(t),x2(t),…,xn(t))T ∈C1(R,Rn)∣x(t+ω)=x(t) ∀t∈R},Z={x(t)=(x1(t),x2(t),…,xn(t))T ∈C(R,Rn)∣x(t+ω)=x(t) ∀t∈R(x1(t),x2(t),…,xn(t))T}.
Proof.
Note that every solution y(t)=(y1(t),y2(t),...yn(t))T of system (1) with the initial value condition is positive. Make the change of variables
(7)yi(t)=exi(t), i=1,2,…,n.
Then system (1) is the same as
(8)xi(t)=ri(t)-aii(t)eαij(t)+∑j=1,j≠inaij(t)eαij(t) +∑j=1,j≠inbij(t)eβij(t-τij) +∑j=1,j≠incij(t)∫0∞Hij(s)eγij(t+s)ds,mmmmmmmmmmmmmmli=1,2,…n.
Obviously, if system (8) has at least one ω-periodic solution, then system (1) has at least one ω-periodic solution. To prove Theorem 4, we should find an appropriate open set Ω satisfying Lemma 1. We divide the proof into three steps.
Step
1. We verify that (i) of Lemma 1 is satisfied. For any x(t)∈X, by periodicity, it is easy to check that
(9)Δi(x,t)=ri(t)-aii(t)eαij(t)+∑j=1,j≠inaij(t)eαij(t) +∑j=1,j≠inbij(t)eβij(t-τij) +∑j=1,j≠incij(t)∫0∞Hij(s)eγij(t+s)ds.
And define L:DomL⊂X→Z and N:X→Z as follows:
(10) X∋x(t)⟶(Lx)(t)=dx(t)dt∈Z,X∋x(t)⟶ (Nx)(t)=((Nx)1(t),(Nx)2(t),…,(Nx)n(t))T∈Z,
where
(11)(Nx)i(t)=Δi(x,t), i=1,2,…,n.
The projectors are defined by P:X→X and Q:Z→Z by
(12)Px(t)=1ω∫0ωx(t)dt, Qz(t)=1ω∫0ωz(t)dt,mmmmmmmmmmmmmmmmnx∈X, z∈Z.
It is easy to follow that L is a Fredholm mapping of index zero. Furthermore, the generalized inverse (to L) KP:ImL→DomL∩KerP exists, which is given by
(13)KP(y)=∫0ty(s)ds-1ω∫0ω∫0ty(s)ds dt.
Then QN:X→Z and KP(I-Q)N:X→X are defined by
(14)QNx=(1ω∫0ωΔ1(x,t)dt,1ω∫0ωΔ2(x,t)dt,mmmmm…,1ω∫0ωΔn(x,t)dt)T,KP(I-Q)Nx=(Ψ1(x,t),Ψ2(x,t),…,Ψn(x,t))T,
where
(15)Ψk(x,t)=∫0tΔk(x,s)ds-1ω∫0ω∫0tΔk(x,s)ds dt -(tω-12)∫0ωΔk(x,s)ds, k=1,2,…,n.
Using similar arguments to Step 1 in [2], it is easy to show that (KP(I-Q)Nx)(Ω¯) is relatively compact in the space (X,∥·∥1).
Step
2. In this step, we are in a position to search for an appropriate open bounded subset Ω satisfying condition (i) of Lemma 1. Specifically, our aim is to search for an appropriate hi defined by Ω in Step 1 such that Ω satisfies condition (i) of Lemma 1. To this end, assume that x(t)∈X is a solution of the equation Lx=λNx for each λ∈(0,1); that is,
(16)x˙i(t)=λ[ri(t)-aii(t)eαij(t)+∑j=1,j≠inaij(t)eαij(t)mmll+∑j=1,j≠inbij(t)eβij(t-τij)mmll+∑j=1,j≠incij(t)∫0∞Hij(s)eγij(t+s)ds],mmmmmmmmmmmmmmmmi=1,2,…n.
Since x(t)∈X, each xi(t),i=1,2,…,n, as components of x(t), is continuously differentiable and ω-periodic. In view of continuity and periodicity, there exists ti∈[0,ω] such that xi(ti)=maxt∈[0,ω]|xi(t)|, i=1,2,…,n. Accordingly, x˙i(ti)=0 and we arrive at
(17)ri(ti)-aii(ti)eαiixi(ti)+∑j=1,j≠inaij(ti)eαijxj(ti) +∑j=1,j≠inbij(ti)eβijxj(ti-τij) +∑j=1,j≠incij(t)∫0∞Hij(s)eγijxj(ti+s)ds=0,mmmmmmmmmmmmmmi=1,2,…,n.
That is,
(18)aii(ti)eαiixi(ti)=ri(ti)+∑j=1,j≠inaij(ti)exj(ti)+∑j=1,j≠inbij(ti)eβijxj(ti-τij) +∑j=1,j≠incij(t)∫0∞Hij(s)eγijxj(ti+s)ds, i=1,2,…n.
Noticing that xj(tj)=maxt∈[0,ω]|xj(t)| implies
(19)xj(ti-τij)≤xj(tj).
It follows from (H2) that
(20)a_iieαiixi(ti)≤|aii(ti)eαiixi(ti)|=|ri(ti)+∑j=1,j≠inaij(ti)eαijxj(ti)+∑j=1,j≠inbij(ti)eβijxj(ti-τij) +∑j=1,j≠incij(t)∫0∞Hij(s)eγijxj(ti+s)ds|≤r¯i+∑j=1,j≠ina¯ijeαijxj(ti)+∑j=1,j≠inb¯ijeβijxj(ti-τij) +∑j=1,j≠inc¯ij(t)eγijxj(ti+s)∫0∞Hij(s)ds=r¯i+∑j=1,j≠ina¯ijeαijxj(ti)+∑j=1,j≠inb¯ijeβijxj(ti-τij) +∑j=1,j≠inc¯ij(t)eγijxj(ti+s)≤r¯i+∑j=1,j≠ina¯ijeαjjxj(tj)+∑j=1,j≠inb¯ijeαjjxj(tj) +∑j=1,j≠inc¯ij(t)eαjjxj(tj)=r¯i+∑j=1,j≠in(a¯ij+b¯ij+c¯ij)eαjjxj(tj).
Here we used (H2). Letting (a_ii+b_ii)eαiixi(ti)=zi(ti), it follows from (20) that
(21)zi(ti)≤r¯i+∑j=1,j≠in(a¯ij+b¯ij+c¯ij)a_jj-1zj(tj),
or
(22)zi(ti)-∑j=1,j≠ina¯ij+b¯ij+c¯ija_jjzj(tj)≤r¯i,
which implies
(23)(1-a¯12+b¯12+c¯12a_22⋯-a¯1n+b¯1n+c¯1na_nn-a¯21+b¯21+c¯21a_111⋯-a¯2n+b¯2n+c¯2na_nn⋯ ⋯⋯⋯-a¯n1+b¯n1+c¯n1a_11-a¯n2+b¯n2+c¯n2a_22⋯1) ×(z1(t1)z2(t2)⋯zn(tn))≤(r¯1r¯2⋯r¯n)
Set D=(D1,D2,…,Dn)T=(r¯1,r¯2,…,r¯n)T. It follows from (23) that
(24)(E-K)(z1(t1),z2(t2),…,zn(tn))T≤D.
In view of ρ(K)<1 and Lemma 3, (En-K)-1≥0. Let
(25)H=(h~1,h~2,…,h~n)T∶=(E-K)-1D≥0.
Then it follows from (24) and (25) that
(26)(z1(t1),z2(t2),…,zn(tn))T≤H, orzi(ti)≤h~i, i=1,2,…,n,
which implies
(27)|xi(t)|0=maxt∈[0,ω]|xi(t)|=xi(ti)≤1αiilnh~ia_ii+b_ii,mmmmmmmmmmmmlmmmmmi=1,2,…,n.
On the other hand, it follows from (25) that
(28)(E-K)H=D, or H=KH+D, that is,h~i=∑j=1nΓijh~j+Di, i=1,2,…,n.
Estimating (16), by using (26) and (28), we have
(29)|x˙i(t)|0=λ|ri(t)-aii(t)eαiixi(t)+∑j=1,j≠inaij(t)eαijxj(t)mmll+∑j=1,j≠inbij(t)eβijxj(t-τij)mmll+∑j=1,j≠incij(t)∫0∞Hij(s)eγijxj(ti+s)ds|0≤r¯i+|aii(t)eαiixi(ti)+∑j=1,j≠inaij(t)eαijxj(t)mmmm+∑j=1,j≠inbij(t)eβijxj(tj)mmmm+∑j=1,j≠incij(t)∫0∞Hij(s)eγijxj(ti+s)ds|0=r¯i+∑j=1,j≠ina¯ij+b¯ij+c¯ija_jjzj(tj)+a¯iia_iizi(ti)≤Di+∑j=1nΓijh~j+a¯iia_iizi(ti)≤h~i+a¯iia_iih~i=[1+a¯iia_ii]h~i.
We can choose a large enough real number (d>1) such that
(30)1αiilndh~ia_ii>1αiilnh~ia_ii+[1+a¯iia_ii]h~i.
Let hi=(1/αii)ln(dh~i/a_ii). Then, for any solution of Lx=λNx, we have |xi(t)|1=|xi(t)|0+|x˙i(t)|0≤ln(h~i/a_ii)+2h~i<hi for all i=1,2,…,n. Obviously, hi are independent of λ and the choice of x(t). Thus, taking hi=ln(dh~i/a_ii), the open subset Ω satisfies that Lx≠λNx for each λ∈(0,1), x∈∂Ω∩DomL; that is, the open subset Ω satisfies the assumption (i) of Lemma 1.
Using similar arguments to Step 3 in [2], it is not difficult to show that, for each x∈∂Ω∩KerL, QNx≠0 and deg{JQN,Ω∩KerL,0}≠0.
Hence, by Lemma 1, system (8) has at least one positive ω-periodic solution in DomL∩Ω¯. By (7), system (1) has at least one positive ω-periodic solution, denoted by y~(t). This completes the proof of Theorem 4.