We first prove Theorem 2.2 via the following Lemmas.
Lemma 3.1.
If
,
, and
, then
is a Fredholm operator of index zero. Furthermore, the linear continuous projector operator
can be defined by
and the linear operator
can be written by
Furthermore,
Proof.
It is clear that
Obviously, the problem
has a solution
satisfying
,
, if and only if
which implies that
In fact, if (3.5) has solution
satisfying
,
, then from (3.5) we have
According to
,
, we obtain
then
On the other hand, if (3.6) holds, setting
where
is an arbitrary constant, then
is a solution of (3.5), and
, and from
and (3.6), we have
Then
. Hence (3.7) is valid.
For
, define
Let
, and we have
then
, thus
. Hence,
, where
, also
. So we have
, and
Thus,
is a Fredholm operator of index zero.
We define a projector
by
. Then we show that
defined in (3.2) is a generalized inverse of
.
In fact, for
, we have
and, for
, we know
In view of
,
, and
, thus
This shows that
. Also we have
then
. The proof of Lemma 3.1 is finished.
Lemma 3.2.
Under conditions (2.5) and (2.9), there are nonnegative functions
satisfying
Proof.
Without loss of generality, we suppose that
. Take
, then there exists
such that
Let
Obviously,
, and
Then
and from (2.5) and (3.21), we have
Hence we can take
,
, 0, and
to replace
,
,
, and
, respectively, in (2.5), and for the convenience omit the bar above
,
, and
, that is,
Lemma 3.3.
If assumptions (H1), (H2) and
,
, and
hold, then the set
for some
is a bounded subset of
.
Proof.
Suppose that
and
. Thus
and
, so that
thus by assumption (H2), there exists
, such that
. In view of
then, we have
Again for
,
, then
,
thus from Lemma 3.1, we know
From (3.29) and (3.30), we have
If (2.5) holds, from (3.31), and (3.26), we obtain
Thus, from
and (3.32), we have
From
, (3.32), and (3.33), one has
that is,
From (3.35) and (3.33), there exists
, such that
Thus
Again from (2.5), (3.35), and (3.36), we have
Then we show that
is bounded.
Lemma 3.4.
If assumption (H2) holds, then the set
is bounded.
Proof.
Let
, then
and
; therefore,
From assumption (H2),
, so
, clearly
is bounded.
Lemma 3.5.
If the first part of condition (H3) of Theorem 2.2 holds, then
for all
. Let
where
is the linear isomorphism given by
, for all
,
. Then
is bounded.
Proof.
Suppose that
, then we obtain
or equivalently
If
, then
. Otherwise, if
, in view of (3.40), one has
which contradicts
. Then
= 



and we obtain
; therefore, 


is bounded.
The proof of Theorem 2.2 is now an easy consequence of the above lemmas and Theorem 2.1.
Proof of Theorem 2.2.
Let
such that
. By the Ascoli-Arzela theorem, it can be shown that
is compact; thus
is
-compact on
. Then by the above Lemmas, we have the following.
(i)
for every
.
(ii)
for every
.
(iii)Let
, with
as in Lemma 3.5. We know 
, for 
. Thus, by the homotopy property of degree, we get
According to definition of degree on a space which is isomorphic to
,
, and
We have
and then
Then by Theorem 2.1,
has at least one solution in
, so that the BVP (1.1), (1.2) has at least one solution in
. The proof is completed.
Remark 3.6.
If the second part of condition (H3) of Theorem 2.2 holds, that is,
for all
, then in Lemma 3.5, we take
and exactly as there, we can prove that
is bounded. Then in the proof of Theorem 2.2, we have
since
. The remainder of the proof is the same.
By using the same method as in the proof of Theorem 2.2 and Lemmas 3.1–3.5, we can show Lemma 3.7 and Theorem 2.3.
Lemma 3.7.
If
,
, and
, then
is a Fredholm operator of index zero. Furthermore, the linear continuous projector operator
can be defined by
and the linear operator
can be written by
Furthermore,
Notice that
Proof of Theorem 2.3.
Let
Then, for
,
; thus
,
; hence
thus, from assumption (H4), there exists
, such that
and in view of
, we obtain
From
, there exists
, such that
. Thus, from
, one has
We let
; hence from (3.58) and (3.59), we have
thus, by using the same method as in the proof of Lemmas 3.2 and 3.3, we can prove that
is bounded too. Similar to the other proof of Lemmas 3.4–3.7 and Theorem 2.2, we can verify Theorem 2.3.
Finally, we give two examples to demonstrate our results.
Example 3.8.
Consider the following boundary value problem:
where
,
and
satisfying
,
, and
, then we can choose
,
, and
, for
; thus
Since
and
has the same sign as
when
, we may choose
, and then the conditions (H1)–(H3) of Theorem 2.2 are satisfied. Theorem 2.2 implies that BVP (3.61) has at least one solution,
.
Example 3.9.
Consider the following boundary value problem:
where
,
and
satisfying
, and
, then we can choose
,
, and
, for
; thus
Similar to Example 3.8, we have
and
has the same sign as
when
, we may choose
, and then all conditions of Theorem 2.3 are satisfied. Theorem 2.3 implies that BVP (3.65) has at least one solution
.