Define an operator as
(3.1)
Observe that the problem (1.1) has a solution if and only if the operator T has a fixed point.
Lemma 3.1 The operator defined by (3.1) is completely continuous.
Proof It is obvious that is continuous in view of continuity of f, and .
Let be bounded. Then, there exist positive constants () such that , and , . Thus, , we have
(3.2)
which implies that .
On the other hand, for any , , we have
Hence, for , , , we have
This implies that is equicontinuous on all , . Thus, by the Arzela-Ascoli theorem, the operator is completely continuous. □
For the sake of convenience, we set the following notations:
Theorem 3.1
Assume that
(H1) there exist nonnegative functions and positive constants () such that
for , and .
Then the problem (1.1) has at least one solution.
Proof Define a ball , we just need to show that the operator , as it has already been proved that the operator is completely continuous in the previous lemma. Let us choose . For any , by the assumption , we have
(3.5)
which implies that
where τ and ν are given by (3.3) and (3.4). So, . Thus is completely continuous. Therefore, by the Schauder fixed point theorem, the operator has at least one fixed point. Consequently, the problem (1.1) has at least one solution in . □
Theorem 3.2
Assume that
() there exist nonnegative functions and positive constants () such that
for , and .
Then the problem (1.1) has at least one solution.
Proof The proof is similar to that of Theorem 3.1, so we omit it. □
Theorem 3.3 ([32])
Let E be a Banach space. Assume that is a completely continuous operator and the set is bounded. Then has a fixed point in E.
Theorem 3.4 If . In addition, assume that
() there exist nonnegative functions and positive constants () such that
for , and .
Then the problem (1.1) has at least one solution.
Proof Let us consider the set
where the operator is defined by (3.1). We just need to show that the set V is bounded as it has already been proved that the operator is completely continuous in the previous lemma. Let , then , . For any , we have
which implies that is bounded for any . So, the set V is bounded. Thus, by the conclusion of Theorem 3.3, the operator has at least one fixed point, which implies that (1.1) has at least one solution. □
Corollary 3.1 Assume that functions f, , () are bounded. Then the nonlinear problem (1.1) has at least one solution.
Theorem 3.5
Assume that
() there exist a nonnegative function and positive constants () such that
for , and .
Then the problem (1.1) has a unique solution if
(3.6)
Proof For , we can get
Consequently, we have , where is given by (3.6). As , the conclusion of the theorem follows by the contraction mapping principle. This completes the proof. □