Evidently,

. It is easy to see that

is a Banach space with norm

and

is also a Banach space with norm

Let

with norm

Then
is also a Banach space. The basic space using in this paper is
.

Let
be a normal cone in
with normal constant
which defines a partial ordering in
by
. If
and
, we write
. Let
. So,
if and only if
. For details on cone theory, see [4].

In what follows, we always assume that
. Let
. Obviously,
for any
. When
, we write
, that is,
. Let
and
. It is clear,
are cones in
and
, respectively. A map
is called a positive solution of BVP (1.2) if
and
satisfies (1.2).

Let

denote the Kuratowski measure of noncompactness in

and

, respectively. For details on the definition and properties of the measure of noncompactness, the reader is referred to [

1–

4]. Let

be all Lebesgue measurable functions from

to

. Denote

Let us list some conditions for convenience.

(H

_{1})

for any

and there exist

and

such that

uniformly for

, and

(H

_{2}) For any

and countable bounded set

, there exist

such that

where
.

(H

_{3})

imply

In what follows, we write
and
. Evidently,
, and
are closed convex sets in
and
, respectively.

We will reduce BVP (1.2) to a system of integral equations in

. To this end, we first consider operator

defined by

Lemma 2.1.

If condition
is satisfied, then operator
defined by (2.12) is a continuous operator from
into
.

Proof.

By virtue of condition

, there exists an

such that

Let

, we have, by (2.19)

which together with condition

implies the convergence of the infinite integral

which together with (2.13) and

implies that

Therefore, by (2.15) and (2.20), we get

Differentiating (2.13), we obtain

It follows from (2.24) and (2.25) that

So,

. On the other hand, it can be easily seen that

So,

. In the same way, we can easily get that

where

Thus,

maps

into

and we get

Finally, we show that

is continuous. Let

. Then

is a bounded subset of

. Thus, there exists

such that

for

and

. Similar to (2.24) and (2.26), it is easy to have

It follows from (2.33) and (2.34) and the dominated convergence theorem that

It follows from (2.32) and (2.35) that
as
. By the same method, we have
as
. Therefore, the continuity of
is proved.

Lemma 2.2.

If condition
is satisfied, then
is a solution of BVP (1.2) if and only if
is a fixed point of operator
.

Proof.

Suppose that

is a solution of BVP (1.2). For

integrating (1.2) from

to

, we have

Integrating (2.36) from 0 to

, we get

which together with the boundary value conditions imply that

Substituting (2.40) and (2.41) into (2.37) and (2.38), respectively, we have

It follows from Lemma 2.1 that the integral
and the integral
are convergent. Thus,
is a fixed point of operator
.

Conversely, if
is fixed point of operator
, then direct differentiation gives the proof.

Lemma 2.3.

Let

be satisfied,

is a bounded set. Then

and

are equicontinuous on any finite subinterval of

and for any

there exists

such that

uniformly with respect to
as

Proof.

We only give the proof for operator

, the proof for operator

can be given in a similar way. By (2.13), we have

For

we obtain by (2.44)

Then, it is easy to see by (2.45) and
that
is equicontinuous on any finite subinterval of
.

Since

is bounded, there exists

such that for any

. By (2.25), we get

It follows from (2.46) and
and the absolute continuity of Lebesgue integral that
is equicontinuous on any finite subinterval of
.

In the following, we are in position to show that for any

there exists

such that

uniformly with respect to
as

Combining with (2.45), we need only to show that for any

there exists sufficiently large

such that

for all
as
The rest part of the proof is very similar to Lemma
in [5], we omit the details.

Lemma 2.4.

Let

be a bounded set in

. Assume that

holds. Then

Proof.

The proof is similar to that of Lemma
in [5], we omit it.

Lemma 2.5 (see [1, 2]).

Mönch Fixed-Point Theorem. Let
be a closed convex set of
and
Assume that the continuous operator
has the following property:
countable,
is relatively compact. Then
has a fixed point in
.

Lemma 2.6.

If
is satisfied, then for
imply that

Proof.

It is easy to see that this lemma follows from (2.13), (2.25), and condition
. The proof is obvious.

Lemma 2.7 (see [16]).

Let

and

are bounded sets in

, then

where
and
denote the Kuratowski measure of noncompactness in
and
, respectively.

Lemma 2.8 (see [16]).

Let
be normal (fully regular) in
,
then
is normal (fully regular) in
.