Lemma 2.1 (see [15, Proposition
]).

For any

, the linear problem

has a unique solution

and

, such that

Furthermore, if

, then

Let

be the Banach space with the norm

, and

Let

be an operator defined by

Then, from Lemma 2.1,
is well defined.

Lemma 2.2.

Let

and

be the first eigenfunction of (1.17). Then for all

, one has

Proof.

For any

, integrating by parts, we have

Since

and

, then

Therefore, we only need to prove that

Let us deal with the first equality, the second one can be treated by the same way. Note that

, then

which implies that

. Then

is bounded on

. Now, we claim that

Suppose on the contrary that

, then for

small enough, we have

which is a contradiction. Combining (1.19) with (2.13), we have

This completes the proof.

Remark 2.3.

Under the conditions of Lemma 2.2, for the later convenience, (2.8) is equivalent to

Lemma 2.4 (see [1, Lemma
]).

For every

, the subset

defined by

is precompact in
.

Let

be the closure of the set of positive solutions of the problem

We extend the function

to an

-Carathéodory function

defined on

by

Then

for

and a.e.

. For

, let

be an arbitrary solution of the problem

Since
for a.e.
, Lemma 2.2 yields
for
. Thus,
is a nonnegative solution of (2.19), and the closure of the set of nontrivial solutions
of (2.21) in
is exactly
.

Let

be an

-Carathéodory function. Let

be the Nemytskii operator associated with the function

as follows:

Lemma 2.5.

Let

on

. Let

be such that

in

,

. Then,

Moreover,
, whenever
.

Let

be the Nemytskii operator associated with the function

as follows:

Then (2.21), with

, is equivalent to the operator equation

Lemma 2.6.

Let (H1) and (H2) hold. Then the operator
is completely continuous.

Proof.

From (1.10) in (H1), there exists

, such that, for a.e.

and

,

Since

is an

-Carathéodory function, then there exists

, such that, for a.e.

and

,

. Therefore, for a.e.

and

, we have

For convenience, let

. We first show that

is continuous. Suppose that

in

as

. Clearly,

as

for a.e.

and there exists

such that

for every

. It is easy to see that

By the Lebesgue dominated convergence theorem, we have that
in
as
. Thus,
is continuous.

Let
be a bounded set in
. Lemma 2.4 together with (2.28) shows that
is precompact in
. Therefore,
is completely continuous.

In the following, we will apply the Leray-Schauder degree theory mainly to the mapping

,

For
, let
, let
denote the degree of
on
with respect to
.

Lemma 2.7.

Let

be a compact interval with

, then there exists a number

with the property

Proof.

Suppose to the contrary that there exist sequences
and
in
in
, such that
for all
, then,
in
.

Set

. Then

and

. Now, from condition (H1), we have the following:

Let

and

denote the nonnegative eigenfunctions corresponding to

and

, respectively, then we have from the first inequality in (2.33) that

From Lemma 2.2, we have that

Since

in

, from (1.12), we have that

By the fact that

, we conclude that

in

. Thus,

Combining this and (2.35) and letting

in (2.34), it follows that

Similarly, we deduce from second inequality in (2.33) that

Thus,
. This contradicts
.

Corollary 2.8.

For
and
,
.

Proof.

Lemma 2.7, applied to the interval

, guarantees the existence of

such that for

This together with Lemma 2.6 implies that for any

,

which ends the proof.

Lemma 2.9.

Suppose

, then there exists

such that

with

,

,

where
is the nonnegative eigenfunction corresponding to
.

Proof.

Suppose on the contrary that there exist

and a sequence

with

and

in

such that

for all

. As

and

in

, it concludes from Lemma 2.2 that

Notice that

has a unique decomposition

where
and
. Since
on
and
, we have from (2.46) that
.

Choose

such that

By (H1), there exists

, such that

Therefore, for a.e.

,

Since

, there exists

, such that

Applying (2.51), it follows that

This contradicts (2.47).

Corollary 2.10.

For
and
,
.

Proof.

Let

, where

is the number asserted in Lemma 2.9. As

is bounded in

, there exists

such that

,

. By Lemma 2.9, one has

This together with Lemma 2.6 implies that

Now, using Theorem A, we may prove the following.

Proposition 2.11.

is a bifurcation interval from the trivial solution for (2.30). There exists an unbounded component

of positive solutions of (2.30) which meets

. Moreover,

Proof.

For fixed
with
, let us take that
,
and
. It is easy to check that, for
, all of the conditions of Theorem A are satisfied. So there exists a connected component
of solutions of (2.30) containing
, and either

(i)
is unbounded, or

(ii)
.

By Lemma 2.7, the case (ii) can not occur. Thus,
is unbounded bifurcated from
in
. Furthermore, we have from Lemma 2.7 that for any closed interval
, if
, then
in
is impossible. So
must be bifurcated from
in
.