Theorem 3.1 If for any , there exist constants for such that , and
For each , let be defined as in Lemma 2.3. Assume further that there exist and such that
and is defined as (9). Then Eq. (1) is oscillatory.
Proof Assume, for the sake of contradiction, that there exists a solution of Eq. (1) which does not have zero in . Without loss of generality, we may assume that for . When for , the proof follows the same argument by using the interval instead of . Put
By Lemma 2.2, we have that for
Similar to the analysis in the proof of Theorem 2.1 in , we can get from Lemmas 2.3 and 2.4 that
By the definition of , multiplying both sides of (16) by , integrating over and using integration by parts, we get
Noting that the second term of the right-hand side of (17) is nonnegative, we get
This contradicts the assumption. □
Next, we will establish a Kamenev type interval oscillation criterion for Eq. (1). First, we introduce a class of functions ℋ which will be used in the sequel. Denote and . A function H is said to belong to the class ℋ if there exist satisfying the following conditions:
(A1) , for ;
(A2) , .
For two constants θ, λ (), we define two operators , by
where is defined as in Theorem 3.1.
Noticing that are all impulse moments in the interval for , we denote the number of impulse moments between and by for . We also mean if .
Theorem 3.2 Suppose that for any , there exist nontrivial subintervals and of , satisfying the conditions of Theorem 3.1. Further assume that for , there exist constants and a function such that
when () is an odd number,
when () is an even number,
Then Eq. (1) is oscillatory.
Proof Otherwise, we may assume that for . Proceeding as in the proof of Theorem 3.1, we have that (16) holds for and . Next, we consider the following two cases:
, the number of impulse moments in the interval , is odd;
is an even number.
For the case (i), we first consider the subinterval . Multiplying both sides of (16) by , then integrating it from to , and using integration by parts, we obtain
It implies that
On the other hand, multiplying both sides of (16) by , integrating it from to , and similar to the above analysis, we can get
Dividing (20) and (21) by and , respectively, and adding them, we have
For the remaining intervals, similar to the analysis in Theorem 2.2 in , we have that for ,
It can be concluded similarly for the case (ii) that
We see that (22) and (23) contradict (18) and (19), respectively. The proof is complete. □