Now we investigate sufficient conditions for the existence of uncountably many bounded nonoscillatory solutions of (1.12) under various ranges of the function p. The proofs of the results presented in this section are based on the Krasnoselskii and Schauder fixed point theorems and a few new and key techniques, one of which is to construct the mappings and satisfying the conditions in the cited fixed point theorems for each constant L, which belongs to certain interval. Let
Theorem 2.1 Assume that there exist and constants M, N, , and c satisfying
(2.1)
(2.2)
(2.3)
Then (1.12) has uncountably many bounded nonoscillatory solutions in .
Proof Let . It follows from (2.2) and (2.3) that there exist constants and sufficiently large satisfying
(2.4)
and
(2.5)
Define two mappings and by
(2.6)
for each .
First of all we show that
(2.7)
Let and . By (2.3)-(2.6), we get
and
which imply (2.7).
Second, we show that is continuous in and is relatively compact. Let be an arbitrary sequence in with
(2.8)
Since is a closed subset of , it follows that . Put
From (2.8) and the continuity of f and for we infer that
which together with (2.6) and the Lebesgue dominated convergence theorem yields for any
which gives
which implies that is continuous in .
Using (2.1), (2.5), and (2.6), we conclude that for any and
which yields
(2.9)
That is, is uniformly bounded in . In order to prove that is relatively compact, we have to prove that is equicontinuous in . Let be given. Equation (2.2) ensures that there exists satisfying
(2.10)
Put
(2.11)
Now we consider the following possible cases:
-
(i)
with . By (2.1), (2.6), and (2.10) we have
(2.12)
-
(ii)
with and . By means of (2.1), (2.6), and (2.11) we infer that
(2.13)
-
(iii)
with and . In light of (2.1), (2.5), and (2.6), we get
which together with the mean value theorem and (2.11) yields
(2.14)
-
(iv)
with . Clearly (2.6) means that
(2.15)
It follows from (2.12)-(2.15) that is equicontinuous in . Thus Lemma 1.1 means that there exists such that . That is,
which implies that
that is, is a bounded nonoscillatory solution of (1.12) in .
Finally, we show (1.12) has uncountably many bounded nonoscillatory solutions in . Let with . As in the above proof we can deduce that for each , there exist constants , , and mappings satisfying (2.4)-(2.6), where θ, T, L, and are replaced by , , , , , respectively, and has a fixed point . That is, and are also bounded nonoscillatory solutions of (1.12) in . We now need to show that . In view of (2.2) there exists satisfying
(2.16)
Note that (2.6) means that for ,
(2.17)
It follows from (2.1), (2.4), and (2.17) that for
which together with (2.16) implies that
which yields . This completes the proof. □
Theorem 2.2 Assume that there exist and constants M, N, and c satisfying (2.1), (2.2), and
(2.18)
Then (1.12) has uncountably many bounded nonoscillatory solutions in .
Proof Let . By (2.2) and (2.18), we choose constants and satisfying (2.4) and
(2.19)
Define two mappings and by (2.6).
Let and . In terms of (2.6), (2.18), and (2.19), we arrive at
which yields for any . The rest of the proof is similar to that of Theorem 2.1 and is omitted. This completes the proof. □
Theorem 2.3 Assume that there exist and constants M, N, , , and c satisfying (2.1), (2.2), and
(2.20)
Then (1.12) has uncountably many bounded nonoscillatory solutions in .
Proof Let . It follows from (2.2) and (2.20) that there exist constants and satisfying
(2.21)
and
(2.22)
Define two mappings and by
(2.23)
We show that (2.7) holds. In fact, for every and , by (2.1) and (2.20)-(2.23), we get
and
which means that we have (2.7).
Next we show that is equicontinuous in . For any given , (2.2) guarantees that (2.10) holds for some sufficiently large . Set
(2.24)
It follows from the uniform continuity of p in that there exists such that
(2.25)
whenever with . Put
(2.26)
We have to consider the following possible cases:
-
(i)
with . It follows from (2.1), (2.10), (2.20), (2.23), and (2.26) that
(2.27)
-
(ii)
with . For each , it follows from the mean value theorem that there exists satisfying
which together with (2.1), (2.20), (2.22), and (2.26) yields for each
which implies that for each
(2.28)
By means of (2.1), (2.20), (2.22)-(2.26), and (2.28), we get
(2.29)
-
(iii)
with . Obviously, (2.23) guarantees that
(2.30)
Using (2.27), (2.29), and (2.30), we conclude that is equicontinuous in . As in the proof of Theorem 2.1, we prove similarly that is continuous in and is uniformly bounded. It follows that is relatively compact. Consequently, Lemma 1.1 shows that there is such that . That is,
which yields
which implies that
that is, is a bounded nonoscillatory solution of (1.12) in .
Finally we show (1.12) has uncountably many bounded nonoscillatory solutions. Let with . As in the above proof, we infer that for each , there exist constants , , and mappings satisfying (2.21)-(2.23), where θ, T, L, , are replaced by , , , , , respectively, and (1.12) possesses a bounded nonoscillatory solution . In terms of (2.2), we select satisfying
(2.31)
It follows from (2.21), (2.23), and (2.31) that for
which yields
that is, . This completes the proof. □
Theorem 2.4 Assume that there exist and constants M, N, , , and c satisfying (2.1), (2.2), and
(2.32)
Then (1.12) has uncountably many bounded nonoscillatory solutions in .
Proof Let . It follows from (2.2) and (2.32) that there exist constants and satisfying (2.21) and
(2.33)
Let the mappings and be defined by (2.23).
Note that (2.1), (2.21), (2.23), and (2.33) imply that for each , and
and
which yields for any . The rest of the proof is similar to that of Theorem 2.3 and is omitted. This completes the proof. □
Theorem 2.5 Let . Assume that there exist and constants M, N, and c satisfying (2.1), (2.2), and
(2.34)
Then (1.12) has uncountably many bounded nonoscillatory solutions in .
Proof Let . It follows from (2.2) and (2.34) that there exists a constant satisfying
(2.35)
Define a mapping by
(2.36)
For every and , by (2.1), (2.35), and (2.36), we deduce that
which yield and hence is uniformly bounded in .
Let be a sequence in and satisfying (2.8) and let
(2.37)
Using (2.8), (2.37), and the continuity of f, , , and for and , we obtain for all . In light of (2.36) and the Lebesgue dominated convergence theorem, we conclude that for any
which means that
which implies that is continuous in .
Let ε be an arbitrary positive number. It follows from (2.2) that there exists large enough such that
(2.38)
Set
(2.39)
where satisfies
(2.40)
We consider the following possible cases:
-
(i)
with . From (2.1), (2.36), and (2.38), we conclude immediately that
(2.41)
-
(ii)
with . In terms of (2.1) and (2.36)-(2.40), we deduce that
(2.42)
-
(iii)
with . Equation (2.36) means that
(2.43)
It follows from (2.41)-(2.43) that is equicontinuous in . Thus Lemma 1.2 means that has a fixed point , that is, for any
and
which give for any
which implies that
that is, is a bounded nonoscillatory solution of (1.12). The rest of the proof is similar to that of Theorem 2.1 and is omitted. This completes the proof. □
Theorem 2.6 Let . Assume that there exist and constants M, N, and c satisfying (2.1), (2.2), and (2.34). Then (1.12) has uncountably many bounded nonoscillatory solutions in .
Proof Let . It follows from (2.2) and (2.34) that there exists a constant satisfying
(2.44)
Define a mapping by
(2.45)
for each .
Let and . By (2.44) and (2.45), we get
which gives that and is uniformly bounded in .
Let satisfy (2.8) for some and be defined by (2.37). Using (2.8), (2.37), (2.45), the continuity of f and for , and the Lebesgue dominated convergence theorem, we conclude that for any
which yields
that is, is continuous in .
Next we show that is equicontinuous in . Let . Equation (2.34) ensures that there exists large enough satisfying
(2.46)
Let B and K be defined by (2.39) and (2.40), respectively. Put
(2.47)
Now we have to consider the following possible cases:
-
(i)
with