Lemma 2.1 (comparison result).

Assume that satisfies

where , , , satisfy

,

;

here

Then , .

Proof.

Supposing that contrary (i.e., for some ), we consider the following two possible cases:

(1) for all ;

(2)there exist such that and .

Let ; we have

Case 1.

Equation (2.3) implies that for and , hence, is nondecreasing on . By (2.3), we can get

Integrating the above inequality from 0 to , we have

Thus,

Noting condition (i), we have . Besides, , that is,. Since is nondecreasing on , then we have , for all . That is, , for all .

Case 2.

Firstly, we consider (2.3). Let , then , and for some , there exists a , such that or . We only consider , for the case , and the proof is similar.

By (2.3), we have

Let in (2.7); we have

So,

On the other hand,

Let in (2.10), then

That is,

By (2.3), we have

Thus, by (2.9), (2.13), and , we obtain

So, , which contradicts condition (ii). Hence, on .

The proof of Lemma 2.1 is complete.

Corollary 2.2.

Assume that , , , , and condition (ii) in Lemma 2.1 hold. Let satisfy (2.1). Then , .

Proof.

The proof of Corollary 2.2 is easy, so we omit it.

Lemma 2.3 (comparison result).

Let satisfy (2.1). Assume that , , and condition (i) in Lemma 2.1 hold. In addition assume that

.

Then , .

Proof.

The proof is similar to the proof of Lemma 2.1 [28], so we omit it.

Corollary 2.4.

Assume that , , , , and condition (iii) in Lemma 2.3 hold. Let satisfy (2.1). Then , .

Proof.

The proof of Corollary 2.4 is easy, so we omit it.

Remark 2.5.

Corollary 2.4 holds for if we delete .

Remark 2.6.

In the special case where (2.1) does not contain the operators and , Lemmas 2.1 and 2.3 develop Lemma 2.1 [28], and Corollaries 2.2 and 2.4 develop Corollary 2.1 [28]. Moreover, the condition in Lemma 2.1 and Corollary 2.2 is more extensive than the corresponding condition in [28], and if we let in Lemma 2.3 and Corollary 2.4, we can obtain Lemma 2.1 and Corollary 2.1 in [28], respectively. Therefore, our comparison results in this paper develop and generalize the corresponding results in [28].

To study the nonlinear problem (1.1), we first consider the associated linear problem

where , .

Definition 2.7.

One says is a solution of (2.15) if it satisfies (2.15).

Definition 2.8.

One says that is called a lower solution of (2.15) if

and it is an upper solution of (2.15) if the above inequalities are reversed.

Lemma 2.9.

Let all assumptions of Lemma 2.1 hold. In addition assume that are lower and upper solutions of (2.15), respectively, and , for all . Then the problem (2.15) has a unique solution .

Proof.

The proof is similar to the proof of Lemma 2.2 [28], so we omit it.

Remark 2.10.

In Lemma 2.9, if we replace "Lemma 2.1" by any of "Corollary 2.2", "Lemma 2.3", or "Corollary 2.4", then the conclusion of Lemma 2.9 holds.