Denote
, where
. Let
For any
, define
then
is a Banach space with the norm
(see [17]).
The Arzela-Ascoli theorem fails to work in the Banach space
due to the fact that the infinite interval
is noncompact. The following compactness criterion will help us to resolve this problem.
Lemma 2.1 (see [17]).
Let
. Then,
is relatively compact in
if the following conditions hold:
(a)
is bounded in
;
(b) the functions belonging to
and
are locally equicontinuous on
;
(c) the functions from
and
are equiconvergent, at
.
Throughout the paper we assume the following.
Suppose that
, and there exist nonnegative functions
with
such that
, where
Denote
Lemma 2.2.
Supposing that
with
, then BVP
has a unique solution
where
in which
, and
for
.
Proof.
Integrating the differential equation from
to
, one has
Then, integrating the above integral equation from
to
, noticing that
and
, we have
Since
, it holds that
By using arguments similar to those used to prove Lemma 2.2 in [9], we conclude that (2.7) holds. This completes the proof.
Now, BVP (1.1) is equivalent to
Letting
, (2.12) becomes
For
, define operator
by
Then,
Set
Remark 2.3.
is the Green function for the following associated homogeneous BVP on the half-line:
It is not difficult to testify that
Let us first give the following result of completely continuous operator.
Lemma 2.4.
Supposing that
and
hold, then
is completely continuous.
Proof.
-
(1)
First, we show that
is well defined.
For any
, there exists
such that
. Then,
so
Similarly,
Further,
On the other hand, for any
and
, by Remark 2.3, we have
Hence, by
, the Lebesgue dominated convergence theorem, and the continuity of
, for any
, we have
So,
for any
.
We can show that
. In fact, by (2.23) and (2.24), we obtain
Hence,
is well defined.
-
(2)
We show that
is continuous.
Suppose
, and
. Then,
as
, and there exists
such that
. The continuity of
implies that
as
. Moreover, since
we have from the Lebesgue dominated convergence theorem that
Thus,
is continuous.
-
(3)
We show that
is relatively compact.
(a) Let
be a bounded subset. Then, there exists
such that
for all
. By the similar proof of (2.20) and (2.22), if
, one has
which implies that
is uniformly bounded.
(b) For any
, if
, we have
Thus, for any
there exists
such that if
, then
Since
is arbitrary, then
and
are locally equicontinuous on
.
(c) For
, from (2.27), we have
which means that
and
are equiconvergent at
. By Lemma 2.1,
is relatively compact.
Therefore,
is completely continuous. The proof is complete.
Lemma 2.5 (see [28, 29]).
Let
be Banach space,
be a bounded open subset of
, and
be a completely continuous operator. Then either there exist
such that
, or there exists a fixed point
.
Lemma 2.6 (see [28, 29]).
Let
be a bounded open set in real Banach space
, let
be a cone of
, and let
be completely continuous. Suppose that
Then,