In this section, we investigate the conditions for the existence of at least one positive solution of the BVP (1.3).

In the next theorem, we will also assume that the following condition on .

(H6):

where is large enough such that

and is small enough such that

where is the eigenfunction related to the smallest eigenvalue of the eigenvalue problem:

Theorem 3.1.

If (H1)–(H6) are satisfied, then the BVP (1.3) has at least one positive solution.

Proof.

Fix and let for . Then, satisfies (2.27). Define by

where

Then is a completely continuous operator. One has from Lemma 2.6 that there exist such that

Define by then is a completely continuous operator. By the first equality in (H6) and the definition of , there are and such that

We now prove that for all and . In fact, if there exists and such that , then satisfies the equation

and the boundary conditions

Multiplying both sides of (3.10) by , then integrating from to , and using integration by parts in the left-hand side two times, we obtain

Combining (3.9) and (3.12), we get

We also have

Equations (3.13) and (3.14) lead to

This is impossible. Thus for and . By (3.7) and the homotopy invariance of the fixed-point index (see [11]), we get that

On the other hand, according to the second inequality of (H6), there exist and such that

We define

then it follows that

Define by , then is a completely continuous operator. We claim that there exists such that

In fact, if for some and , then

where Combining (3.21) with (3.22), we have

Let Then we get

Consequently, by the homotopy invariance of the fixed-point index, we have

where is zero operator. Use (3.16) and (3.25) to conclude that

Hence, has a fixed point in .

Let . Since for and .

(H7)

Theorem 3.2.

If (H1)–(H5) and (H7) are satisfied, then the BVP (1.3) has at least one positive solution.

Proof.

Define by , then is a completely continuous operator. By the first inequality in (H7), there exist and such that

We claim that for and . In fact, if there exist and such that , then satisfies the boundary condition (3.11). Since , we have . Then we have

Multiplying the last equation by integrating from to , by (3.28), we obtain

then we have

Equations (3.30) and (3.31) lead to

This is impossible. By homotopy invariance of the fixed-point index, we get that

Define by , then is a completely continuous operator. By the second inequality in (H7), and definition of , there exist and such that

We define

then, it is obvious that

We claim that there exists such that

In fact, if for some and , then using (3.36), it is analogous to the argument of (3.13) and (3.14) that

Equation (3.38) leads to Let . Then we get

Consequently, by (3.8) and the homotopy invariance of the fixed-point index, we have

In view of (3.33) and (3.40), we obtain

Therefore, has a fixed point in . The proof is completed.

Corollary 3.3.

Using the following (H8) or (H9) instead of (H6) or (H7), the conclusions of Theorems 3.1 and 3.2 are true. For ,

(H8)

(H9)