We establish sufficient conditions for the oscillation of solutions of even order neutral type differential equations of the form
[r(t)[x(t)+p(t)x(τ(t))](n−1)]′+q(t)f(x(σ(t)))=0.

1. Introduction

This paper concerns the oscillatory behavior of solutions of higher order neutral type nonlinear differential equations of the following form:(1)[r(t)[x(t)+p(t)x(τ(t))](n-1)]′+q(t)f(x(σ(t)))=0,[r(t)[x(t)+p(t)x(τ(t))](n-1)]′+q(t)(x(σ(t)))t≥t0,
where n≥2 is even and the following conditions are assumed to hold:

p,q∈C([t0,∞)), q(t)>0, 0≤p(t)≤p0<1, where p0 is constant;

r∈C1([t0,∞)), r(t)>0, r′(t)≥0;

τ∈C([t0,∞),ℝ), σ∈C1([t0,∞),ℝ), τ(t)<t,σ(t)≤λt<t, σ′(t)>0, and limt→∞τ(t)=limt→∞σ(t)=∞;

f∈C(ℝ,ℝ) and f(x)/x≥K>0, for x≠0, and K is a constant.

Further, we will consider the two cases(2)∫∞1r(t)dt=∞,(3)∫∞1r(t)dt<∞.
By a solution of (1), we mean a real-valued function x which satisfies (1) and sup{|x(t)|:t≥tx}>0 for any tx≥t0. 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 n=2 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 τ and σ to be commute.

2. Auxiliary Lemmas

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

Lemma 1 (see [<xref ref-type="bibr" rid="B7">7</xref>, page 193]).

Let y(t) be an n times differentiable function on [0,∞) of constant sign, let y(n)(t) be of constant sign and not identically equal to zero in any interval [t0,∞), t0≥0, and let y(t)y(n)(t)≤0. Then,

there exists a t1≥t0 such that y(k)(t), k=1,…,n-1, is of constant sign on [t1,∞);

there exists an integer l, 0≤l≤n-1, with n-l odd, such that
(4)y(t)y(k)(t)>0,k=0,1,…,l,t≥t1,(5)(-1)n+k-1y(t)y(k)(t)>0,k=l+1,…,n-1,t≥t1.

Lemma 2 (see [<xref ref-type="bibr" rid="B12">12</xref>]).

Let y(t) be a function as in Lemma 1. If
(6)y(n-1)(t)y(n)(t)≤0,
then, for every θ∈(0,1), there exists a constant M>0, such that
(7)|y′(θt)|≥Mtn-2|y(n-1)(t)|,
for sufficiently large t.

Lemma 3 (see [<xref ref-type="bibr" rid="B1">2</xref>, page 169]).

Let y(t) be a function as in Lemma 1. If
(8)y(n-1)(t)y(n)(t)≤0,
and limt→∞y(t)≠0, then for every θ∈(0,1)(9)y(t)≥θ(n-1)!tn-1y(n-1)(t),
for sufficiently large t.

3. The Main ResultTheorem 4.

Assume that (2) holds. If
(10)∫∞q(t)dt=∞,
then every solution x(t) of (1) is oscillatory.

Proof.

Let x(t) be a nonoscillatory solution of (1). Without loss of generality, we may assume that x(t) is eventually positive (the proof is similar when x(t) is eventually negative). That is, let x(t)>0, x(τ(t))>0, and let x(σ(t))>0 for t≥t1≥t0.

Set
(11)z(t)=x(t)+p(t)x(τ(t)).
Since p(t) is nonnegative, z(t)>x(t)>0 for t≥t1.

From (1) and (11), we have
(12)(r(t)z(n-1)(t))′=r(t)z(n)(t)+r′(t)z(n-1)(t)=-q(t)f(x(σ(t)))<0.
Thus, r(t)z(n-1)(t) is decreasing and z(n-1)(t) is eventually of one sign. Hence, either
(13)z(n-1)(t)>0,fort≥t2≥t1
or
(14)z(n-1)(t)<0,fort≥t2≥t1.
If (14) holds, then
(15)r(t)z(n-1)(t)≤r(t2)z(n-1)(t2)<0,fort≥t2.
Dividing this inequality by r(t) and integrating from t2 to t, then by using (2), we get
(16)limt→∞z(n-2)(t)=-∞.
This result along with (14) leads to limt→∞z(t)=-∞. But this contradicts the fact thatz(t)>0. Thus, (13) holds. Then, from (12) and the fact that r(t) is a positive nondecreasing function, we conclude that z(n)(t)<0, for t≥t2. It follows that z(i)(t)(i=0,1,…,n-1) is strictly monotonic and of constant sign eventually.

By applying Lemma 1, z(t) satisfies (4) and (5). Since n is even, the integer l associated with z(t) is odd; that is, l≥1. Hence, z(t) is increasing for t≥t3≥t2.

Then, from (11) and the fact that z(t) is increasing, we have
(17)x(t)=z(t)-p(t)x(τ(t))≥z(t)-p(t)z(τ(t))≥(1-p(t))z(t)≥(1-p0)z(t),000000000000000000000fort≥t3.
Let t4≥t3 be such that σ(t)≥t3 for all t≥t4. Combining (H4) and (17), we get
(18)f(x(σ(t)))≥K(1-p0)z(σ(t)),fort≥t4.
It is clear that we can apply Lemma 2. Then, from (7) and the decreasing character of z(n-1)(t), we have
(19)z′(σ(t))≥Mλn-2σn-2(t)z(n-1)(σ(t)λ)≥M0σn-2(t)z(n-1)(t),0000000fort≥t5≥t3,
where M0=M/λn-2.

Define
(20)w(t)=r(t)z(n-1)(t)z(σ(t)),
and then w(t)>0 for t≥t6=max{t4,t5}.

By differentiating w and using (12), (18), and (19), we obtain
(21)w′(t)=(r(t)z(n-1)(t))′z(σ(t))-σ′(t)z′(σ(t))r(t)z(n-1)(t)z2(σ(t))≤-K(1-p0)q(t)-M0σ′(t)σn-2(t)r(t)w2(t).
Since r(t)>0 and σ′(t)>0, the term (M0σ′(t)σn-2(t)/r(t))w2(t)>0. Hence, (21) reduces to
(22)w′(t)≤-K(1-p0)q(t).
Integrating this inequality from t7 to t, t7>t6, and using assumption (10), we see that w(t)→-∞ as t→∞. But this contradicts the positivity of w(t). Hence, the theorem is proved.

In the above proof, being l≥1 plays an important role. In fact, l=0 is possible only for odd orders. In this case, the solutions are bounded. For unbounded solutions with n being odd, the integer l must be greater than or equal to 2. Thus, it is easy to show that if n 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 σ(t) in (H2) can be improved to be σ(t)≤t. Indeed, under the assumption of unboundedness, z′(t) is increasing. This modifies (19) as
(23)z′(σ(t))≥z′(θσ(t))≥Mσn-2(t)z(n-1)(σ(t))≥Mσn-2(t)z(n-1)(t),θ∈(0,1),
where the rest of the proof stays as above.

Remark 5.

The condition (10) can be rewritten as
(24)∫∞q(t)[1-p(σ(t))]dt=∞.
Here, there is no need for abounded value p0 for the function p(t); that is, 0≤p(t)<1. When we take n=2, r(t)=1, f(x)=x, and σ(t)=t-σ0, 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 f satisfies the condition that xf(x)>0 and there exists a nondecreasing function ϕ∈C([t0,∞),(0,∞)) with
(25)|f(x)|≥ϕ(|x|).

Theorem 7.

Assume that (3) and (10) hold and n≥4 is even. Further, suppose that
(26)∫∞[Cq(s)(σ(s))n-2δ(s)-14r(s)δ(s)]ds=∞,
where C=αK/(n-2)!, α∈(0,1) is a constant, and δ(t)=∫t∞(1/r(s))ds. Then, every solution x(t) of (1) either is oscillatory or tends to zero as t→∞.

Proof.

Assume that (1) has a nonoscillatory solution x(t). Without loss of generality, we assume that there exists a t1≥t0 such that x(t)>0, x(τ(t))>0, and x(σ(t))>0 for all t≥t1.

Proceeding as in the proof of Theorem 4, we conclude that r(t)z(n-1)(t) is decreasing and z(n-1)(t) 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, z(n-1)(t)<0, for t≥t2≥t1. Now, we consider two assumptions: unbounded solutions and bounded solutions.

If the solution x(t) is unbounded, it is obvious that z(t) is also unbounded. Since z(t)z(n-1)(t)<0 and n-1 is odd, we have by Lemma 1 that l≥2 (if l=0, then z(t) is bounded). Hence, from (4), we have that z′(t)>0, and z′′(t)>0. Therefore, limt→∞z(t)>0.

Since z(t) is increasing, we obtain
(27)x(t)≥(1-p0)z(t),fort≥t3≥t2.
By Lemma 3 and the fact that z(n-2)(t) is decreasing, we get
(28)z(σ(t))≥θ[σ(t)]n-2(n-2)!z(n-2)(t),t≥t4≥t2.
Combining (H4), (27), and (28), we obtain
(29)f(x(σ(t)))≥C(σ(t))n-2z(n-2)(t),iiiiiiiiiifort≥t5=max{t3,t4},
where C=αK/(n-2)! with α=(1-p0)θ∈(0,1).

Define
(30)w(t)=r(t)z(n-1)(t)z(n-2)(t),
and then w(t)<0 for t≥t5.

Differentiating w and using (12) and (29), we obtain
(31)w′(t)=(r(t)z(n-1)(t))′z(n-2)(t)-w2(t)r(t)≤-Cq(t)(σ(t))n-2-w2(t)r(t).
Following [8, 13], we can show that -1≤w(t)δ(t)<0. Indeed, since r(t)z(n-1)(t) is decreasing,
(32)r(s)z(n-1)(s)≤r(t)z(n-1)(t),s≥t≥t5.
Dividing by r(s), then integrating from t to l, and letting l→∞, we get
(33)0≤z(n-2)(t)+r(t)z(n-1)(t)δ(t).
Thus, we obtain
(34)-1≤r(t)z(n-1)(t)z(n-2)(t)δ(t).
Hence,
(35)-1≤w(t)δ(t)<0,t≥t5.
Multiplying inequality (31) by δ(t) and integrating from t5 to t, we get
(36)w(t)δ(t)-w(t5)δ(t5)+∫t5tw(s)r(s)ds+∫t5tw2(s)δ(s)r(s)ds≤-∫t5tCq(s)(σ(s))n-2δ(s)ds,
or
(37)w(t)δ(t)-w(t5)δ(t5)≤-∫t5t[Cq(s)(σ(s))n-2δ(s)-14r(s)δ(s)]ds-∫t5t[w(s)δ(s)+1/2]2r(s)δ(s)ds.
Thus,
(38)w(t)δ(t)≤-∫t5t[Cq(s)(σ(s))n-2δ(s)-14r(s)δ(s)]ds+w(t5)δ(t5).
Using assumption (26), we see that w(t)δ(t)→-∞ as t→∞. But this contradicts (35).

If the solution x(t) is bounded, then z(t) is also bounded. Since z(t)z(n-1)(t)<0 and n-1 is odd, we have by Lemma 1 that l=0 (otherwise, z(t) is not bounded). Hence, from (4) and (5), we have
(39)(-1)iz(i)(t)>0,i=0,1,…,n-2fort≥t3≥t2.
From (11) and the fact that z(t)>x(t), we obtain
(40)x(t)≥z(t)-p0x(τ(t))≥z(t)-p0z(τ(t)),
or
(41)x(t)≥z(τ(t))[z(t)z(τ(t))-p0].
From (39) z(t)>0, z′(t)<0, and z′′(t)>0, we have limt→∞z(t)=λ≥0. Now, we consider two cases.

Case I. Consider that λ>0. Since z(t) is decreasing, there is an ɛ>0 and t4≥t3 such that, for t≥t4,
(42)λ≤z(t)≤z(τ(t))≤λ+ɛ.
From this, we can conclude that
(43)z(t)z(τ(t))≥λλ+ɛ.
Choose t5≥t4 such that, for t≥t5, we have p0+ε1≤λ/(λ+ɛ), for some ε1>0. Thus,
(44)z(t)z(τ(t))≥p0+ε1,t≥t5.
Using this inequality in (41) and the fact that z(t) is decreasing, we obtain
(45)x(t)≥ε1z(t),t≥t5.
By Lemma 3 and the fact that z(n-2)(t) is decreasing, we get
(46)z(σ(t))≥θ[σ(t)]n-2(n-2)!z(n-2)(t),t≥t6≥t3.
Combining (H4), (45), and (46), we obtain
(47)f(x(σ(t)))≥C(σ(t))n-2z(n-2)(t),iiiiiiiiiifort≥t7=max{t5,t6},
where C=αK/(n-2)! with α=ε1θ∈(0,1).

By using the transformation (30) and proceeding as in the previous assumption (x(t) unbounded), again we obtain a contradiction with (26).

Case II. Consider that λ=0, since x(t)≤z(t), x(t) tends to zero as t→∞, 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:
(48)[t[x(t)+(12-12t+2)x(t-3)](n-1)]′+t2x(t-22)(1+1x2((t-2)/2)+1)=0,0000000000000000000000000000000000t>0.
Here, r(t)=t, p(t)=(1/2-1/(2t+2)), q(t)=t2, τ(t)=t-3, σ(t)=(t-2)/2, and f(x)/x≥K=1. We can see that all conditions of Theorem 4 are satisfied. Thus, every solution of (48) is oscillatory.

The function p(t)=(1/2-1/(2t+2)) in (48) lies in the interval (0,1/2); that is, p0=1/2<1. Now, for the same equation, if p(t) is replaced by p(t)=(1-1/(2t+2)), then there is no such p0<1. In this case, by using condition (24), we conclude again that every solution of (48) with p(t)=(1-1/(2t+2)) is oscillatory.

Example 2.

Consider the following nonlinear neutral differential equation:
(49)[eαt[x(t)+p0x(t-τ0)](n-1)]′+eβtx(λt)=0,t>0,
where n≥4 is even, τ0>0, 0<λ<1, 0<p0<1, and β≥α>0.

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

Example 3.

Consider the following even order nonlinear neutral differential equation:
(50)[et[x(t)+1ax(ta-nlna)](n-1)]′+(1-1a)e(1+1/b-1/a)tx(tb)=0,00000000000000000000000000t>0,
where n≥4 and a>b>1. 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 t→∞. Indeed, x(t)=e-t is a solution that tends to zero as t→∞.

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

Conflict of Interests

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

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