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.