Lemma 2.1 (see [23]).

If conditions

and

are satisfied, then the boundary value problem

has a unique solution for any

. Moreover, this unique solution can be expressed in the form

where

is defined by

Remark 2.2.

It is easy to prove that
has the following properties:

(1)
is continuous on

(2)
is continuous differentiable on
, except
,

(3)
,

(4)
,

(5)
,

(6) for all
,
,
, where

Obviously,
.

For the interval
, and the corresponding
in Remark 2.2, we define
=
:
,
and
exist,
.
=
exists
.
,
and
. It is easy to see that
is a Banach space with the norm
, and
is a positive cone in
. For details of the cone theory, see [1].
is called a positive solution of BVP (1.1) if
for all
and
satisfies (1.1).

As we know that the Ascoli-Arzela Theorem does not hold in infinite interval
, we need the following compactness criterion:

Lemma 2.3 (see [22]).

Let
. Then
is relatively compact in
if the following conditions hold.

(i)
is uniformly bounded in
.

(ii) The functions from
are equicontinuous on any compact interval of
.

(iii) The functions from
are equiconvergent, that is, for any given
, there exists a
such that
, for any
,
.

The main tool of this work is a fixed point theorem in cones.

Lemma 2.4 (see [4]).

Let X be a Banach space and
is a positive cone in
. Assume that
are open subsets of
with
,
. Let
be a completely continuous operator such that

(i)
for all
.

(ii) there exists a
such that
, for all
and
.

Then
has a fixed point in
.

Remark 2.5.

If (i) is satisfied for
and (ii) is satisfied for
, then Lemma 2.4 is still true.

Lemma 2.6 (see [3]).

The function

is a solution of the BVP (1.1) if and only if

satisfies the equation

The proof of this result is based on the properties of the Green function, so we omit it as elementary.

Define

Obviously, the BVP (1.1) has a solution
if and only if
is a fixed point of the operator
defined by (2.6).

Let us list some conditions as follows.

(*A*_{1}) There exist two nonnegative functions:
,
such that
.
,
may be singular at
.
,
, are continuous.

(*A*_{2})
,

Lemma 2.7.

If
are satisfied, then for any bounded open set
,
is a completely continuous operator.

Proof.

For any bounded open set
, there exists a constant
such that
for any
.

First, we show that
is well defined. Let
. From
, we have
, where
,
,
and

Hence,

is well defined. For any

, we have

Thus, by the Lebesgue dominated convergence theorem and the fact that

is continuous on

, we have, for any

,

,

Therefore,
. By the property (3) of
, it is easy to get
.

On the other hand, by (2.6) we have, for any
and
,

Then by

, the property (5) of Remark 2.2 and the Lebesgue dominated convergence theorem, we have

Thus
.

For any
, we get

On the other hand, for

we obtain

Thus
.

Next, we prove that
is continuous. Let
in
, then
We prove that
. For any
, by
, there exists a constant
such that

On the other hand, by the continuities of

on

and the continuities of

on

, for the above

, there exists a

such that, for any

,

,

From

, for the above

, there exists a sufficiently large number

such that, when

, we have

Therefore, by (2.15)–(2.17), we have, for

,

This implies that the operator
is continuous.

Finally we show that
is a compact operator. In fact for any bounded set
, there exists a constant
such that
for any
. Hence, we obtain

Therefore,
is uniformly bounded in
.

Given
, for any
, as the proof of (2.9), we can get that
are equicontinuous on
. Since
is arbitrary,
are locally equicontinuous on
. By (2.6),
,
, and the Lebesgue dominated convergence theorem, we have

Hence, the functions from
are equiconvergent. By Lemma 2.3, we have that
is relatively compact in
. Therefore,
is completely continuous. This completed the proof of Lemma 2.7.