In this section, by employing Kranoselskii’s fixed point theorem, we establish the existence criteria for each type of eventually positive solutions to (1).
Theorem 3.1 Equation (1) has an eventually positive solution in for some positive b if and only if there exists some constant such that
(11)
Proof Suppose that is an eventually positive solution of (1) in , i.e.,
(12)
Assume that (or ). By Lemma 2.3 we have (or ), which contradicts (12). Then we have
and there exists such that , , for . Integrating (1) from to , we obtain
Letting , we have
In view of (C4), it follows that
and
which means that (11) holds. The necessary condition is proved.
Conversely, suppose that there exists some constant such that (11) holds. There will be two cases to be considered: and .
Case 1: . Take such that , then .
When , since and (11) hold, we can choose a sufficiently large such that for , and
(13)
(14)
When , we can choose and the above such that
(15)
Furthermore, from (C3) there exists such that and for .
Define the Banach space as in (3) with , and let
(16)
It is easy to prove that is a bounded, convex, and closed subset of . By (C4), we have, for any ,
Now we define two operators and : as follows
(17)
Next, we will prove that and satisfy the conditions in Lemma 2.2.
(i) We prove that for any . Note that, for any , and . For any and , by (13), (14), and (16) we obtain
On the other hand, for and , we have
For , , and , we have and (15), and
Similarly, we can prove that for any and . Then we prove that for any and . In fact, for and , we have
For , , and , we have and (15), and
Therefore, we obtain for any .
(ii) We prove that is a contraction mapping. In fact, noting that and for , for we have
for , and
for . It follows that
for any . Therefore, is a contraction mapping.
(iii) We prove that is a completely continuous mapping.
Firstly, for , we have
and
That is, maps into .
Secondly, we prove the continuity of . For and , letting and as , we have
(18)
and
as . For , we have
For , we have . Then we obtain
Similar to Chen [7], by (18) and employing Lebesgue’s dominated convergence theorem [[5], Chapter 5], we conclude that
as . That is, is continuous.
Thirdly, we prove that is relatively compact. According to Lemma 2.1, it suffices to show that is bounded, uniformly Cauchy and equi-continuous. It is obvious that is bounded. Since and as , for any given there exists a sufficiently large such that and . Then, for any and , we have
Hence, is uniformly Cauchy.
Then, for , if with , we have
If with , we have
If , we always have
Therefore, there exists such that
whenever and . That is, is equi-continuous.
It follows from Lemma 2.1 that is relatively compact, and then is completely continuous.
By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Since
for and
we have
It is obvious that
The sufficiency holds when .
Case 2: . Take so that , then . Since and (11) hold, we can choose a sufficiently large such that
(19)
From (C3) there exists such that and for . Similarly, we introduce the Banach space and its subset as in (16). Define the operator as in (17) and the operator on as follows:
Next, we prove that for any . In fact, for any and , by (14) and (19) we obtain
and
That is, for any .
The remainder of the proof is similar to the case and we omit it here. By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Letting , we have
It is obvious that
The sufficiency holds when .
The proof is complete. □
Theorem 3.2 Equation (1) has an eventually positive solution in for some positive b if and only if there exists some constant such that
(20)
Proof Suppose that is an eventually positive solution of (1) in , i.e.,
Similarly, we have
and there exists such that , , for . Integrating (1) from to , we obtain
Letting , we have
or
(21)
Integrating (21) from to , we have
Letting , we have
In view of (C4), it follows that
and
which means that (20) holds. The necessary condition is proved.
Conversely, suppose that there exists some constant such that (20) holds. There will be two cases to be considered: and .
Case 1: . Take such that , then .
When , since and (20) hold, we can choose a sufficiently large such that for , and
When , we can choose and the above such that
Furthermore, from (C3) there exists such that and for .
Define the Banach space as in (3) with , and let
(22)
It is easy to prove that is a bounded, convex, and closed subset of . By (C4), we have, for any ,
Now we define two operators and as follows:
(23)
Next, we can prove that and satisfy the conditions in Lemma 2.2. The proof is similar to the case of Theorem 3.1 and omitted here.
By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Since
for and
we have
which implies that
The sufficiency holds when .
Case 2: . We introduce the Banach space and its subset as in (22). Define the operator as in (23) and the operator on as follows:
The following proof is similar to the case in Theorem 3.1 and we omit it here. By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Similarly, we have
which implies that
The sufficiency holds when .
The proof is complete. □
Theorem 3.3 Equation (1) has an eventually positive solution in for some positive b if and only if there exists some constant such that
(24)
Proof Suppose that is an eventually positive solution of (1) in , i.e.,
Then
and there exists such that , , for . Integrating (1) from to , we obtain
Letting , we have
or
(25)
Integrating (25) from to , we have
Letting , we have
or
(26)
Integrating (26) from to , we have
Letting , we have
In view of (C4), it follows that
and
which means that (24) holds. The necessary condition is proved.
Conversely, suppose that there exists some constant such that (24) holds. There will be two cases to be considered: and .
Case 1: . Take such that , then .
When , since and (24) hold, we can choose a sufficiently large such that for , and
When , we can choose and the above such that
Furthermore, from (C3) there exists such that and for .
Define the Banach space as in (3) with , and let
(27)
It is easy to prove that is a bounded, convex, and closed subset of . By (C4), we have, for any ,
Now we define two operators and as follows:
(28)
Next, we can prove that and satisfy the conditions in Lemma 2.2. The proof is similar to the case of Theorem 3.1 and omitted here.
By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Letting , we have
which implies that
The sufficiency holds when .
Case 2: . We introduce the Banach space and its subset as in (27). Define the operator as in (28) and the operator on as follows:
The following proof is similar to the case in Theorem 3.1 and we omit it here. By Lemma 2.2, there exists such that , which implies that is a solution of (1). In particular, for we have
Similarly, we have
which implies that
The sufficiency holds when .
The proof is complete. □
Theorem 3.4 Equation (1) has an eventually positive solution in , then
(29)
Conversely, if there exists a nonnegative constant M such that and
(30)
then (1) has an eventually positive solution in .
Proof Suppose that is an eventually positive solution of (1) in , i.e.,
Similarly, we have
and there exists such that for . From (C3) there exists such that and for . Integrating (1) from to , we obtain
Letting , we have
(31)
which implies that
by the monotonicity of f and for . Substituting s for in (31), we have
(32)
Integrating (32) from to , we have
Letting , we have
By the monotonicity of f and for , it follows that
and
which means that (29) holds. The necessary condition is proved.
Conversely, if there exists a positive constant M such that and (30) hold, then and we can choose a sufficiently large such that
From (C3) there exists such that and for .
Define the Banach space as in (3) with , and let
It is easy to prove that is a bounded, convex, and closed subset of . According to (C3) and (C4), we have, for any ,
Now we define two operators and as follows:
Next, we can prove that and satisfy the conditions in Lemma 2.2. The proof is similar to Theorem 3.1 and omitted here. By Lemma 2.2, there exists such that