Let
be a nonnegative integer and
, then we define a sequence of functions
as follows:
To obtain our main results, we need the following lemmas.
Lemma 2.1.
Let
be a positive integer, then there exists
, such that
Proof.
We will prove the above by induction. First, if
, then we take
. Thus,
Next, we assume that there exists
, such that
for
and
with
, then
from which it follows that there exists
, such that
for
and
. The proof is completed.
Lemma 2.2 (see [24]).
Let
, then
Lemma 2.3 (see [2]).
Let
and
, then
implies
Lemma 2.4 (see [2]).
Let
be a positive integer. Suppose that
is
times differentiable on T. Let
and
, then
Lemma 2.5 (see [2]).
Assume that
and
are differentiable on T with
. If there exists
, such that
then
Lemma 2.6 (see [23]).
Let
be defined on
, and
with
for
and not eventually zero. If
is bounded, then
-
(1)
for
,
-
(2)
for all
and
.
Now, one states and proves the main results.
Theorem 2.7.
Assume that there exists
, such that the function
satisfies
where
are nonnegative functions on
and
with
, then every solution
of (1.1) satisfies one of the following conditions:
-
(1)
,
-
(2)
there exist constants
with
, such that
Proof.
Let
be a solution of (1.1), then it follows from Lemma 2.4 that for
,
By (2.11) and Lemma 2.1, we see that there exists
, such that for
and
,
Then we obtain
where
with
Using (2.16) and (2.17), it follows that
By Lemma 2.3, we have
with
. Hence from (2.12), there exists a finite constant
, such that
for
. Thus, inequality (2.20) implies that
By (1.1), we see that if
, then
Since condition (2.12) and Lemma 2.2 implies that
we find from (2.11) and (2.21) that the sum in (2.22) converges as
. Therefore,
exists and is a finite number. Let
. If
, then it follows from Lemma 2.5 that
and
has the desired asymptotic property. The proof is completed.
Theorem 2.8.
Assume that there exist functions
, and nondecreasing continuous functions
, and
such that
with
then every solution
of (1.1) satisfies one of the following conditions:
-
(1)
,
-
(2)
there exist constants
with
such that
Proof.
Let
be a solution of (1.1), then it follows from Lemma 2.4 that for
,
By Lemma 2.1 and (2.25), we see that there exists
, such that for
and
,
Then, we obtain
where
with
Using (2.30) and (2.31), it follows that
Write
then
from which it follows that
Since
and
is strictly increasing, there exists a constant
, such that
for
. By (2.30), (2.33), and (2.34), we have
It follows from (1.1) that if
, then
Since (2.38) and condition (2.25) implies that
we see that the sum in (2.39) converges as
. Therefore,
exists and is a finite number. Let
. If
, then it follows from Lemma 2.5 that
and
has the desired asymptotic property. The proof is completed.
Theorem 2.9.
Assume that there exist positive functions
, and nondecreasing continuous functions
, and
, such that
with
then every solution
of (1.1) satisfies one of the following conditions:
-
(1)
,
-
(2)
there exist constants
with
, such that
Proof.
Arguing as in the proof of Theorem 2.8, we see that there exists
, such that for
and
,
from which we obtain
where
Using (2.46) and (2.47), it follows that
Write
then
from which it follows that
The rest of the proof is similar to that of Theorem 2.8, and the details are omitted. The proof is completed.
Theorem 2.10.
Assume that the function
satisfies
-
(1)
for all
,
-
(2)
for
and
,
-
(3)
for
and
is continuous at
with
,
then (1) if
is even, then every bounded solution of (1.1) is oscillatory; (2) if
is odd, then every bounded solution
of (1.1) is either oscillatory or tends monotonically to zero together with
.
Proof.
Assume that (1.1) has a nonoscillatory solution
on
, then, without loss of generality, there is a
, sufficiently large, such that
for
. It follows from (1.1) that
for
and not eventually zero. By Lemma 2.6, we have
and
is eventually monotone. Also
for
if
is even and
for
if
is odd. Since
is bounded, we find
. Furthermore, if
is even, then
.
We claim that
. If not, then there exists
, such that
since
is continuous at
by the condition (3). From (1.1) and (2.55), we have
Multiplying the above inequality by
, and integrating from
to
, we obtain
Since
we get
where
. Thus,
since
is bounded, which gives a contradiction to the condition (2). The proof is completed.