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.

(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

with

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

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 .