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 .
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
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
If condition is satisfied, then operator defined by (2.12) is a continuous operator from into .
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
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
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 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.
If condition is satisfied, then is a solution of BVP (1.2) if and only if is a fixed point of operator .
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.
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
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 , we omit the details.
Let be a bounded set in . Assume that holds. Then
The proof is similar to that of Lemma in , 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 .
If is satisfied, then for imply that
It is easy to see that this lemma follows from (2.13), (2.25), and condition . The proof is obvious.
Lemma 2.7 (see ).
Let and are bounded sets in , then
where and denote the Kuratowski measure of noncompactness in and , respectively.
Lemma 2.8 (see ).
Let be normal (fully regular) in , then is normal (fully regular) in .