Throughout this paper, we assume that:
-
(1)
, where the forward jump operator by .
-
(2)
and such that and are differentiable.
Write
(3.1)
(3.2)
Theorem 3.1 Suppose that (2.1) and (2.2) hold. If there exist differentiable functions and with being differentiable such that for all sufficiently large and for any positive constants , , there is a with such that
(3.3)
then every solution of (1.1) is either oscillatory or tends to 0.
Proof Assume that (1.1) has a nonoscillatory solution x on . Then, without loss of generality, there is a sufficiently large such that for . Therefore from Lemma 2.3, we know that there exists sufficiently large such that:
-
(1)
for .
-
(2)
Either or for and .
Let for and . Consider
(3.4)
Then and for .
By the product rule and the quotient rule
Since (), we get
(3.5)
Using the fact that x and τ are differentiable functions and , we see that is a differentiable function and . Note . From Lemma 2.5, we get
which implies
(3.6)
We choose such that for . Then, from (2.3) and the fact that for , we get
(3.7)
(3.8)
From (2.3), we have
Thus
which combines with (3.8) to imply
(3.9)
Combining (3.9) with (3.6) and from , we obtain that
Now we consider the following three cases.
Case (i). If , then for and . Thus
Case (ii). If , then
Case (iii). If , then from (2.4) we get that there exist and a constant such that
Thus
where .
We obtain from the above that
Thus
(3.10)
Since
(3.11)
from (3.10) and (3.11) and the definitions of and , we get
(3.12)
It is easy to check that
Integrating both sides of the above inequality from to t, we get
which leads to a contradiction to (3.3). The proof is completed. □
Theorem 3.2 Suppose that (2.1) and (2.2) hold. If there exist differentiable functions and with being differentiable such that for all sufficiently large and for any positive constants , , there is a with such that
(3.13)
then every solution of (1.1) is either oscillatory or tends to 0.
Proof Assume that (1.1) has a nonoscillatory solution x on . Then, without loss of generality, there is a sufficiently large such that for . Therefore from Lemma 2.3, we know that there exists sufficiently large such that:
-
(1)
for .
-
(2)
Either or for and .
Let for and . From (3.7), we get
(3.14)
Define as (3.4). Choosing such that for . Combining (3.14) with (3.6), we see that for ,
Note since , we obtain
Now we consider the following three cases.
Case (i). If , then
since for .
Case (ii). If , then
Case (iii). If , then we get from (2.4) that there exist and a constant such that
Thus
where .
We obtain from the above
Then
(3.15)
From Lemma 2.7, we have
(3.16)
Combining (3.16) with (3.15) and the definitions of and , we get
(3.17)
Let
(3.18)
(3.19)
We have from Lemma 2.6
Then
Integrating both sides of the above inequality from to t, we get
which leads to a contradiction to (3.13). The proof is completed. □