In this section, we apply Lemma 2.6 to establish the existence of at least two positive solutions for BVP (1.1).

The following assumptions will stand throughout this paper.

(*H*_{1})There exist such that

where and are defined in (1.4), respectively.

(*H*_{2})We have

for and given in (2.2) and (2.12), respectively.

If holds, then we can show that have the following properties.

Proposition 3.1.

If (1.2)–(1.4) and hold, then from (2.15), for , one has

where

Proof.

Let

Then from (1.2)–(1.4) and , we obtain Therefore,

On the other hand, since

we have So one has

This and imply (3.3) holds.

Proposition 3.2.

If (1.2)–(1.4) and hold, then from (2.16), , one has

Proof.

The proof is similar to that of Proposition 3.1. So we omit it.

For the sake of applying fixed point theorem on cone, we construct a cone in by

where is defined in (2.10).

Define by

By (2.14), it is well known that the problem (1.1) has a positive solution if and only if is a fixed point of .

Lemma 3.3.

Suppose that (1.2)–(1.4) and - hold. Then and is completely continuous.

Proof.

For by (2.14), we have and

On the other hand, for , by (3.9),(3.10) and (2.7), we obtain

Therefore , that is, .

Next by standard methods and the Ascoli-Arzela theorem one can prove that is completely continuous. So it is omitted.

Theorem 3.4.

Suppose that (1.2)–(1.4) and - hold. Then problem (1.1) has at least two positive solutions provided

where and are defined in (3.3), (3.7) and in Proposition 3.1, respectively.

Proof.

Let be the cone preserving, completely continuous operator that was defined by (3.9).

Let , where . Choosing and satisfy

Now we prove that

In fact, if there exists such that , then for , we have

where defined by (2.17).

Therefore , that is, , which is a contradiction. Hence (3.14) holds.

Next, turning to (3.15). If there exists such that , then for , we have

where are defined by (2.17).

Therefore , that is, , which is a contradiction. Hence (3.15) holds.

It remains to prove

In fact, if there exists such that , then for , we have

that is,

which is a contradiction, where are defined by (2.17). Hence (3.18) holds. From Lemma 2.6, (3.14), (3.15) and (3.18) yield that the problem (1.1) has at least two solutions and . The proof is complete.