We are ready to state and prove our main theorem.

Proof.
It is easy to see that if the system (2.3) has a ω-periodic solution (u1*(n),u2*(n)), then (x*(n),y*(n))=(exp(u1*(n)-u2*(n)),exp(u2*(n))) is a positive ω-periodic solution to the system (1.7). Therefore, to complete the proof, it suffices to show that the system (2.3) has at least two ω-periodic solutions.

We take
(3.1)X=Y={(u1(n),u2(n))∣ui(n+ω)=ui(n), i=1,2, n∈ℤ}
and define the norm of X and Y(3.2)∥u∥=maxn∈Iω|u1(n)|+maxn∈Iω|u2(n)|,
for u=(u1,u2)∈X or Y. Then X and Y are Banach spaces when they are endowed with the previous norm ∥·∥.

For any u=(u1,u2)∈X, because of its periodicity, it is easy to verify that
(3.3)Λ1(u,n)=a(n)+d(n)-b(n)∑l=0ω-1G(l)exp[u1(n-l)+u2(n-l)] -c(n)m2+exp[2u1(n)]-h(n)exp[u1(n-τ(n))]m2+exp[2u1(n-τ(n))],Λ2(u,n)=-d(n)+h(n)exp[u1(n-τ(n))]m2+exp[2u1(n-τ(n))]
are ω-periodic with respect to n.

Set
(3.4)L:DomL∩X→Y, (Lu)(n)=(L(u1,u2))(n)=(Δu1(n),Δu2(n)),N:X→Y, (Nu)(n)=(N(u1,u2))(n)=(Λ1(u,n),Λ2(u,n)).
Obviously, kerL=ℝ2, ImL={(u1,u2)∈Y:∑n=0ω-1ui(n)=0, i=1,2} is closed in Y, and dim kerL=codim ImL=2. Therefore, L is a Fredholm mapping of index zero.

Define two mappings P and Q as
(3.5)P:X→X, Pu=(1ω∑n=0ω-1u1(n),1ω∑n=0ω-1u2(n)), u=(u1,u2)∈X,Q:Y→Y, Qv=(1ω∑n=0ω-1v1(n),1ω∑n=0ω-1v2(n)), v=(v1,v2)∈Y.
It is easy to prove that P and Q are two projectors such that ImP=kerL and ImL=kerQ=Im(I-Q). Furthermore, by a simple computation, we find that the inverse Kp of Lp:ImL→DomL∩kerP has the form
(3.6)Kp(u1,u2)=(∑k=0n-1u1(k)-1ω∑k=0ω-1(ω-k)u1(k),∑k=0n-1u2(k)-1ω∑k=0ω-1(ω-k)u2(k)).
Evidently,
(3.7)QN(u1,u2)=(1ω∑n=0ω-1Λ1(u,n),1ω∑n=0ω-1Λ2(u,n))
and Kp(I-Q)N are continuous by the Lebesgues convergence theorem. Moreover, by Arzela Ascolis theorem, QN(Ω-) and Kp(I-Q)N(Ω-) are relatively compact for the open bounded set Ω⊂X. Therefore, N is L-compact on Ω- for the open bounded set Ω⊂X.

Corresponding to the operator equation (2.11), we get the following system:
(3.8)Δu1(n)=λΛ1(u,n),Δu2(n)=λΛ2(u,n),
where λ∈(0,1). Suppose that (u1(n),u2(n))∈X is an arbitrary solution of system (3.8) for a constant λ∈(0,1). Summing (3.8) over Iω, we obtain
(3.9)a-ω=∑n=0ω-1{b(n)∑l=0ω-1G(l)exp[u1(n-l)+u2(n-l)]+c(n)m2+exp[2u1(n)]},(3.10)d-ω=∑n=0ω-1h(n)exp[u1(n-τ(n))]m2+exp[2u1(n-τ(n))].
From system (3.8), we have
(3.11)∑n=0ω-1|Δu1(n)|<∑n=0ω-1(|a(n)|+|d(n)|) +∑n=0ω-1{b(n)∑l=0ω-1G(l)exp[u1(n-l)+u2(n-l)] +c(n)m2+exp[2u1(n)]+h(n)exp[u1(n-τ(n))]m2+exp[2u1(n-τ(n))]∑l=0ω-1},∑n=0ω-1|Δu2(n)|<∑n=0ω-1|d(n)|+∑n=0ω-1h(n)exp[u1(n-τ(n))]m2+exp[2u1(n-τ(n))].
By using (3.9) and (3.10), we obtain
(3.12)∑n=0ω-1|Δu1(n)|<(|a|¯+|d|¯+a-+d-)ω,(3.13)∑n=0ω-1|Δu2(n)|<(|d|¯+d-)ω.

Obviously, there exist ξi,ηi∈Iω, such that
(3.14)ui(ξi)=min n∈Iωui(n), ui(ηi)=max n∈Iωui(n), i=1,2.
From (3.10), it follows that
(3.15)d-ω≤h-ωexp[u1(η1)]m2+exp[2u1(ξ1)],
therefore
(3.16)u1(η1)≥ln[d-h-(m2+exp[2u1(ξ1)])].
By using Lemma 2.3 and (3.12), we obtain
(3.17)u1(n)≥u1(η1)-∑s=0ω-1|Δu1(s)|>ln[d-h-(m2+exp[2u1(ξ1)])]-(|a|¯+|d|¯+a-+d-)ω.
In particular, we have
(3.18)u1(ξ1)>ln[d-h-(m2+exp[2u1(ξ1)])]-(|a|¯+|d|¯+a-+d-)ω,
or
(3.19)d-exp[2u1(ξ1)]-h-exp[(|a|¯+|d|¯+a-+d-)ω]exp[u1(ξ1)]+m2d-<0.
The assumption (H1) implies that h-exp[(|a|¯+|d|¯+a-+d-)ω]>2md-. So we have
(3.20)lnl-<u1(ξ1)<lnl+.
From (3.10), we also have
(3.21)d-ω≥h-ωexp[u1(ξ1)]m2+exp[2u1(η1)],
it follows that
(3.22)u1(ξ1)≤ln[d-h-(m2+exp[2u1(η1)])].
By using Lemma 2.3 and (3.12) again, we have
(3.23)u1(n)≤u1(ξ1)+∑s=0ω-1|Δu1(s)|<ln[d-h-(m2+exp[2u1(η1)])]+(|a|¯+|d|¯+a-+d-)ω.
In particular, we have
(3.24)u1(η1)<ln[d-h-(m2+exp[2u1(η1)])]+(|a|¯+|d|¯+a-+d-)ω,
or
(3.25)d-exp[2u1(η1)]-h-exp[-(|a|¯+|d|¯+a-+d-)ω]exp[u1(η1)]+m2d->0.
Therefore,
(3.26)u1(η1)<lnv- or u1(η1)>lnv+.

From (3.12) and (3.20), we have
(3.27)u1(n)≤u1(ξ1)+∑s=0ω-1|Δu1(s)|<lnl++(|a|¯+|d|¯+a-+d-)ω∶=B11.
Similarly, from (3.12) and (3.26), we have
(3.28)u1(n)≥u1(η1)-∑s=0ω-1|Δu1(s)|>lnv+-(|a|¯+|d|¯+a-+d-)ω:=B12.

By using (3.14), (3.27), and (3.28), it follows from (3.9) that
(3.29)a-ω≥b-ωexp[u2(ξ2)+B12],(3.30)a-ω≤b-ωexp[u2(η2)+B11]+c-ωm2.
From (3.29), we have
(3.31)u2(ξ2)≤lna-b--B12.
In view of (3.12), we obtain
(3.32)u2(n)≤u2(ξ2)+∑s=0ω-1|Δu2(s)|<lna-b--B12+(|d|¯+d-)ω:=B21.
Under the assumption (H2), it follows from (3.30) that
(3.33)u2(η2)≥lna--(c-/m2)b--B11.
By using (3.12), we obtain again
(3.34)u2(n)≥u2(η2)-∑s=0ω-1|Δu2(s)|>lna--(c-/m2)b--B11-(|d|¯+d-)ω∶=B22.
It follows from (3.32) and (3.34) that
(3.35)maxn∈Iω|u2(n)|<max{|B21|,|B22|}∶=B2.

Notice that
(3.36)QN(u1,u2)=[a-+d--b-exp(u1+u2)-c-+h-exp(u1)m2+exp(2u1),-d-+h-exp(u1)m2+exp(2u1)]
for u=(u1,u2)∈ℝ2. Under the conditions (H1) and (H2), we can obtain two distinct solutions of QN(u1,u2)=0(3.37)u-=(u1-,u2-)=(lnu-,lna-(m2+u-2)-c-b-u-(m2+u-2)),u+=(u1+,u2+)=(lnu+,lna-(m2+u+2)-c-b-u+(m2+u+2)).
After choosing a constant C>0 such that
(3.38)C>max{|lna-(m2+u-2)-c-b-u-(m2+u-2)|,|lna-(m2+u+2)-c-b-u+(m2+u+2)|},
we can define two bounded open subsets of X as follows:
(3.39)Ω1={u=(u1,u2)∈X∣u1∈(lnl-,lnv-), maxn∈Iω|u2|<B2+C},Ω2={u=(u1,u2)∈X∣minn∈Iωu1∈(lnl-,lnl+), maxn∈Iωu1∈(lnv+,B11), maxn∈Iω|u2|<B2+C,}.
It follows from (2.10) and (3.38) that u-∈Ω1 and u+∈Ω2. Because of lnv-<lnv+, it is easy to see that Ω1∩Ω2 is empty, and Ωi satisfies the condition (a) in Lemma 2.2 for i=1,2. Moreover, QNu≠0 for u∈∂Ωi⋂kerL=∂Ωi⋂ℝ2. This shows that the condition (b) in Lemma 2.2 is satisfied.

Because ImQ=kerL, we can take the isomorphic J as the identity mapping, then we have
(3.40)deg(JQN(u1,u2),Ωi∩kerL,(0,0))=deg(QN(u1,u2),Ωi∩kerL,(0,0)).
From (3.37), QN(u1,u2)=0 has two solutions u-=(u1-,u2-)∈Ω1∩KerL and u+=(u1+,u2+)∈Ω2∩KerL. Therefore we have
(3.41)deg(QN(u1,u2),Ω1∩kerL,(0,0)) =sign|-b-exp(u1-+u2-)-h-exp(u1-)(m2-exp(2u1-))(m2+exp(2u1-))2-b-exp(u1-+u2-)h-exp(u1-)(m2-exp(2u1-))(m2+exp(2u1-))20| =sign(b- h-exp(2u1-+u2-)(m2-exp(2u1-))(m2+exp(2u1-))2)=sign(m-exp(u1-)) =sign(e--2md-(e-+2md--e--2md-)2d-) =1≠0.
Similarly, we can obtain that
(3.42)deg(QN(u1,u2),Ω2∩kerL,(0,0)) =sign(m-exp(u1+))=sign(-e--2md-(e-+2md-+e--2md-)2d-)=-1≠0.
So the condition (c) in Lemma 2.2 is also satisfied.

By now we know that Ωi (i=1,2) satisfies all the requirements of Lemma 2.2. Hence the system (2.3) has at least two ω-periodic solutions. This completes the proof.