Following [21, 22, 30], denote for
with
the admissible set
The main results of this paper are contained in the following three theorems. The arguments used in the proofs have common features with the ones developed in [22, 30, 34].
Theorem 3.1.
Suppose that for any given
there exist subintervals
and
of
such that
Let
be an
-tuple satisfying (2.2) in Lemma 2.1. If there exists a function
,
, such that
for
, where
then (1.1) is oscillatory.
Proof.
To arrive at a contradiction, let us suppose that
is a nonoscillatory solution of (1.1). First, we assume that
is positive for all
, for some
.
Let
, where
is sufficiently large. Define
It follows that
and hence
By our assumptions (3.2) we have
and
for
. Set
Then (3.7) becomes
In view of the arithmetic-geometric mean inequality, see [35],
and equality (3.9) we obtain
Multiplying both sides of inequality (3.11) by
and then using the identity
result in
where
As demonstrated in [7, 12], we know that
, and that
if and only if
where
stands for the inverse function. In our case, since
, dynamic equation (3.15) has a unique solution satisfying
. Clearly, the unique solution is
. Therefore,
on
.
For the benefit of the reader we sketch a proof of the fact that
. Note that if
is a right-dense point, then we may write
Applying Young's inequality (Lemma 2.3) with
we easily see that
holds. If
is a right-scattered point, then
can be written as a function of
and
as
Using differential calculus, see [7], the result follows.
Now integrating the inequality (3.13) from
to
and using
on
we obtain
which of course contradicts (3.3). This completes the proof when
is eventually positive. The proof when
is eventually negative is analogous by repeating the arguments on the interval
instead of
.
A close look at the proof of Theorem 3.1 reveals that one cannot take
. The following theorem is a substitute in that case.
Theorem 3.2.
Suppose that for any given
there exists a subinterval
of
such that
Let
be an
-tuple satisfying (2.5) in Lemma 2.2. If there exists a function
such that
where
then (1.1) with
is oscillatory.
Proof.
We proceed as in the proof of Theorem 3.1 to arrive at (3.7) with
, that is,
Setting
and using again the arithmetic-geometric mean inequality
we have
The remainder of the proof is the same as that of Theorem 3.1.
As it is shown in [34] for the sublinear terms case, we can also remove the sign condition imposed on the coefficients of the sub-half-linear terms to obtain interval criterion which is applicable for the case when some or all of the functions
,
, are nonpositive. We should note that the sign condition on the coefficients of super-half-linear terms cannot be removed alternatively by the same approach. Furthermore, the function
cannot take the value zero on intervals of interest in this case. We have the following theorem.
Theorem 3.3.
Suppose that for any given
there exist subintervals
and
of
such that
If there exist a function
,
, and positive numbers
and
with
such that
for
, where
with
then (1.1) is oscillatory.
Proof.
Suppose that (1.1) has a nonoscillatory solution. We may assume that
is eventually positive on
when
is sufficiently large. If
is eventually negative, then one can repeat the proof on the interval
. Rewrite (1.1) as follows:
with
Clearly,
where
Applying (2.8) and (2.9) to each summation on the right side with
we see that
where
From (3.32) and inequality (3.37) we obtain
where
Set
In view of inequality (3.39) it follows that
The remainder of the proof is the same as that of Theorem 3.1, hence it is omitted.