For convenience, we denote
.

Lemma 2.1.

Let

. Then for any

, the BVP

Proof.

Let

be a solution of the BVP (2.1). Then, we may suppose that

By the boundary conditions in (2.1), we have

Therefore, the BVP (2.1) has a unique solution

Lemma 2.2 (see [12]).

For any

,

Lemma 2.3 (see [26]).

For any

,

In the remainder of this paper, we always assume that
,
and
.

Lemma 2.4.

If
and
for
, then the unique solution
of the BVP (2.1) satisfies

(1)
,
,

(2)
,
and
, where
.

Proof.

Since (1) is obvious, we only need to prove (2). By (2.2), we get

which indicates that
for
.

On the one hand, by (2.9) and Lemma 2.3, we have

On the other hand, in view of (2.2) and Lemma 2.2, we have

It follows from (2.10) and (2.11) that

which together with Lemma 2.2 implies that

Let

be equipped with the norm

. Then

is a Banach space. If we denote

then it is easy to see that

is a cone in

. Now, we define an operator

on

by

Obviously, if
is a fixed point of
, then
is a monotone nonnegative solution of the BVP (1.3).

Lemma 2.5.

is completely continuous.

Proof.

First, by Lemma 2.4, we know that
.

Next, we assume that

is a bounded set. Then there exists a constant

such that

for any

. Now, we will prove that

is relatively compact in

. Suppose that

. Then there exist

such that

. Let

Then for any

, by Lemma 2.2, we have

which implies that

is uniformly bounded. At the same time, for any

, in view of Lemma 2.3, we have

which shows that

is also uniformly bounded. This indicates that

is equicontinuous. It follows from Arzela-Ascoli theorem that

has a convergent subsequence in

. Without loss of generality, we may assume that

converges in

. On the other hand, by the uniform continuity of

, we know that for any

, there exists

such that for any

with

, we have

Let

. Then for any

,

with

, we have

which implies that
is equicontinuous. Again, by Arzela-Ascoli theorem, we know that
has a convergent subsequence in
. Therefore,
has a convergent subsequence in
. Thus, we have shown that
is a compact operator.

Finally, we prove that

is continuous. Suppose that

and

. Then there exists

such that for any

,

. Let

Then for any

and

, in view of Lemmas 2.2 and 2.3, we have

By applying Lebesgue Dominated Convergence theorem, we get

which indicates that
is continuous. Therefore,
is completely continuous.