In this section, we prove every solution of system (4) is persistent. In addition to this, we prove that there exists a bounded solution for (4).
Definition 3 System (4) is said to be persistent if there are positive constants , , and such that
and
for each positive solution of (4).
We assume the following condition:
(H.2) .
Set
Lemma 3 Let (H.1), (H.2) hold. Then, every solution of system (4) satisfies
Proof Let be a solution of (4). To prove that , we consider two cases:
Case I. There exists such that . By the first equation of (4), we have
which implies that
Thus,
We claim that for . Indeed, if there is an integer such that and is the least integer between and such that , then and which implies that . This is a contradiction. This proves the claim.
Case II. Let for . Then, exists and equals . Taking the limit of the first equation in (4), we have
Hence . This proves the claim.
Now, we prove that . For any , there exists a large enough integer such that for . By the second equation of (4), we get
Since , we can find a positive number d such that . Thus, by using Stolz’s theorem, we obtain
Hence
By the arbitrariness of ε, we obtain . The proof of Lemma 3 is complete. □
Set
Lemma 4 Let (H.1), (H.2) hold. Then every solution of system (4) satisfies
Proof Let be a solution of (4). By virtue of Lemma 3, one can figure out that for any which satisfies , there exists such that
To prove that , we consider two cases:
Case I. There exists such that . We observe that for , we have
For , we get
which implies that
It follows that
Let
We claim that
For the sake of contradiction, assume that there exists such that . Then . Let be the smallest integer such that . Then . The above arguments imply that which is a contradiction. This proves the claim.
Case II. Let for all . Then, exists and it is equal to . Taking the limit of the first equation of (4), we have
Hence and . This proves the claim.
By applying the same arguments followed in the proof of Lemma 3, one can easily show that . The proof of Lemma 4 is complete. □
The results of Lemma 3 and Lemma 4 can be concluded in the following theorem:
Theorem 1 Let (H.1), (H.2) hold. Then system (4) is persistent.
Let Ω be the set of all solutions of system (4) satisfying and for all . By virtue of Theorem 1, it should be noted that Ω is an invariant set of system (4).
In view of Lemma 2, we need to show that there exists a bounded solution of system (4). The following result proves the existence of such a solution.
Theorem 2 Let (H.1), (H.2) hold. Then .
Proof By the almost periodicity of , , , , and , there exists an integer valued sequence with as such that , , , , , as . Let ε be an arbitrary small positive number. It follows from Lemma 3 and Lemma 4 that there exists a positive integer such that
Let and for , . For any positive integer q, it is easy to see that there exist sequences and such that the sequences and have subsequences, denoted by and again, converging on any finite interval of as , respectively. Thus, we have sequences and such that
Therefore, the system
(9)
implies
(10)
We can easily see that is a solution of system (4) and , for . Since ε is arbitrary, it follows that , for . This completes the proof. □