Let
Evidently,
. It is easy to see that
is a Banach space with norm
and
is also a Banach space with norm
where
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.
(H1)
for any
and there exist
and
such that
uniformly for
, and
(H2) For any
and countable bounded set
, there exist
such that
with
where
.
(H3)
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
where
Lemma 2.1.
If condition
is satisfied, then operator
defined by (2.12) is a continuous operator from
into
.
Proof.
Let
By virtue of condition
, there exists an
such that
where
Hence
Let
, we have, by (2.19)
which together with condition
implies the convergence of the infinite integral
Thus, we have
which together with (2.13) and
implies that
Therefore, by (2.15) and (2.20), we get
Differentiating (2.13), we obtain
Hence,
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
where
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 is clear,
and by (2.20),
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
Thus, we obtain
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
.