#### Abstract

We establish sufficient conditions for the oscillation of solutions of even order neutral type differential equations of the form .

#### 1. Introduction

This paper concerns the oscillatory behavior of solutions of higher order neutral type nonlinear differential equations of the following form: where is even and the following conditions are assumed to hold: (H1), , , where is constant;(H2), , ;(H3), , ,, , and ;(H4) and , for , and is a constant.

Further, we will consider the two cases By a solution of (1), we mean a real-valued function which satisfies (1) and for any . Such a solution is said to be oscillatory if it has arbitrarily large zeros and nonoscillatory otherwise.

Neutral differential equations arise in a number of important applications in natural science and technology. For instance, they are used in problems dealing with vibrating masses attached to an elastic bar and in the study of distributed networks containing lossless transmission lines which appears in high speed computers where the lossless transmission lines are used to interconnect switching circuits; see Hale [1].

During the last 20 years, significant efforts have been devoted to investigate the oscillatory behaviour of neutral differential equations; see [1–16] and the references cited therein. In particular, (1) and related forms have been considered by several authors; see [10, 11, 13, 14, 16]. Several recent results are surveyed in Sun et al. [13]. In addition, we refer to [3–5, 8], where the oscillatory behaviour of solutions of (1) with was studied.

In this paper, we establish oscillation theorems for solutions of (1). Our results generalize the results of Grammatikopoulos et al. [6] in some sense. Also, our results, in some sense, agree with the results of Sun et al. [13]. Here, we remove the restriction of [13] for and to be commute.

#### 2. Auxiliary Lemmas

The following lemmas will be needed in the proof of our main results.

Lemma 1 (see [7, page 193]). *Let be an times differentiable function on of constant sign, let be of constant sign and not identically equal to zero in any interval , , and let . Then,*(i)*there exists a such that , , is of constant sign on ;*(ii)*there exists an integer , , with odd, such that
*

Lemma 2 (see [12]). *Let be a function as in Lemma 1. If
**
then, for every , there exists a constant , such that
**
for sufficiently large .*

Lemma 3 (see [2, page 169]). *Let be a function as in Lemma 1. If
**
and , then for every **
for sufficiently large .*

#### 3. The Main Result

Theorem 4. *Assume that (2) holds. If
**
then every solution of (1) is oscillatory.*

*Proof. *Let be a nonoscillatory solution of (1). Without loss of generality, we may assume that is eventually positive (the proof is similar when is eventually negative). That is, let , , and let for .

Set
Since is nonnegative, for .

From (1) and (11), we have
Thus, is decreasing and is eventually of one sign. Hence, either
or
If (14) holds, then
Dividing this inequality by and integrating from to , then by using (2), we get
This result along with (14) leads to . But this contradicts the fact that. Thus, (13) holds. Then, from (12) and the fact that is a positive nondecreasing function, we conclude that , for . It follows that is strictly monotonic and of constant sign eventually.

By applying Lemma 1, satisfies (4) and (5). Since is even, the integer associated with is odd; that is, . Hence, is increasing for .

Then, from (11) and the fact that is increasing, we have
Let be such that for all . Combining (H4) and (17), we get
It is clear that we can apply Lemma 2. Then, from (7) and the decreasing character of , we have
where .

Define
and then for .

By differentiating and using (12), (18), and (19), we obtain
Since and , the term . Hence, (21) reduces to
Integrating this inequality from to , , and using assumption (10), we see that as . But this contradicts the positivity of . Hence, the theorem is proved.

In the above proof, being plays an important role. In fact, is possible only for odd orders. In this case, the solutions are bounded. For unbounded solutions with being odd, the integer must be greater than or equal to . Thus, it is easy to show that if is odd and the conditions of Theorem 4 are satisfied, then every unbounded solution of (1) is oscillatory.

Notice that if the solutions are assumed to be unbounded, then the restriction on in (H2) can be improved to be . Indeed, under the assumption of unboundedness, is increasing. This modifies (19) as where the rest of the proof stays as above.

*Remark 5. *The condition (10) can be rewritten as
Here, there is no need for abounded value for the function ; that is, . When we take , , , and , we recover the results of Grammatikopoulos et al. [6]. In this case, we consider unbounded solutions.

*Remark 6. *Theorem 4 remains true if the function satisfies the condition that and there exists a nondecreasing function with

Theorem 7. *Assume that (3) and (10) hold and is even. Further, suppose that
**
where , is a constant, and . Then, every solution of (1) either is oscillatory or tends to zero as .*

*Proof. *Assume that (1) has a nonoscillatory solution . Without loss of generality, we assume that there exists a such that , , and for all .

Proceeding as in the proof of Theorem 4, we conclude that is decreasing and is eventually of one sign. Hence, either (13) or (14) holds.

If (13) holds, we obtain a contradiction by proceeding as in the proof of Theorem 4.

Suppose that (14) holds; that is, , for . Now, we consider two assumptions: unbounded solutions and bounded solutions.*If the solution ** is unbounded*, it is obvious that is also unbounded. Since and is odd, we have by Lemma 1 that (if , then is bounded). Hence, from (4), we have that , and . Therefore, .

Since is increasing, we obtain
By Lemma 3 and the fact that is decreasing, we get
Combining (H4), (27), and (28), we obtain
where with .

Define
and then for .

Differentiating and using (12) and (29), we obtain
Following [8, 13], we can show that . Indeed, since is decreasing,
Dividing by , then integrating from to , and letting , we get
Thus, we obtain
Hence,
Multiplying inequality (31) by and integrating from to , we get
or
Thus,
Using assumption (26), we see that as . But this contradicts (35).*If the solution ** is bounded*, then is also bounded. Since and is odd, we have by Lemma 1 that (otherwise, is not bounded). Hence, from (4) and (5), we have
From (11) and the fact that , we obtain
or
From (39) , , and , we have . Now, we consider two cases.*Case I*. Consider that . Since is decreasing, there is an and such that, for ,
From this, we can conclude that
Choose such that, for , we have , for some . Thus,
Using this inequality in (41) and the fact that is decreasing, we obtain
By Lemma 3 and the fact that is decreasing, we get
Combining (H4), (45), and (46), we obtain
where with .

By using the transformation (30) and proceeding as in the previous assumption ( unbounded), again we obtain a contradiction with (26). *Case II.* Consider that , since , tends to zero as , and this completes the proof.

#### 4. Examples

In this section, we present some examples to illustrate the above results.

*Example 1. *Consider the following even order nonlinear neutral differential equation:
Here, , , , , , and . We can see that all conditions of Theorem 4 are satisfied. Thus, every solution of (48) is oscillatory.

The function in (48) lies in the interval ; that is, . Now, for the same equation, if is replaced by , then there is no such . In this case, by using condition (24), we conclude again that every solution of (48) with is oscillatory.

*Example 2. *Consider the following nonlinear neutral differential equation:
where is even, , , , and .

We can see that all conditions of Theorem 7 are satisfied. Thus, every solution of (49) either is oscillatory or tends to zero as .

*Example 3. *Consider the following even order nonlinear neutral differential equation:
where and . It is easy to check that all conditions of Theorem 7 are satisfied. Thus, every solution of (50) either is oscillatory or tends to zero as . Indeed, is a solution that tends to zero as .

Note that the results of Sun et al. [13] cannot be applied in the above examples, since the and are not commute.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.