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)