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.