# Oscillation Criteria for Certain Even Order Neutral Delay Differential Equations with Mixed Nonlinearities

Authors: Zhen-Lai Han; Yi-Bing Sun; Yan Zhao; Dian-Wu Yang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We establish some oscillation criteria for the following certain even order neutral delay differential equations with mixed nonlinearities:rtzn-1tα-1zn-1t'+q0(t)(xτ0tα-1x(τ0(t))+q1t(x(τ1(t))β-1x(τ1(t))+q2t(x(τ2(t))γ-1x(τ2(t))=0,t≥t0, where z(t)=x(t)+p(t)x(σ(t)),n is even integer, and γ>α>β>0. Our results generalize and improve some known results for oscillation of certain even order neutral delay differential equations with mixed nonlinearities.


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## Body

## 1. Introduction

In this paper, we are concerned with oscillation behavior of the certain even order neutral delay differential equations with mixed nonlinearities:(1)(r(t)|z(n-1)(t)|α-1z(n-1)(t))′+q0(t)|x(τ0(t))|α-1x(τ0(t))+q1(t)|x(τ1(t))|β-1x(τ1(t))+q2(t)|x(τ2(t))|γ-1x(τ2(t))=0,t≥t0,
where z(t)=x(t)+p(t)x(σ(t)), n is even integer, and γ>α>β>0 are constants. r,qi∈C([t0,∞),R+), r′(t)≥0, p,τi∈C([t0,∞),R) satisfy that τi(t)≤t, i=0,1,2, and there exists a function σ∈C([t0,∞),R), such that σ(t)≤t, lim⁡t→∞σ(t)=∞. We assume that there exists a function τ∈C1([t0,∞),R), such that τ(t)≤τi(t), i=0,1,2, τ(t)≤t, τ′(t)>0, and lim⁡t→∞τ(t)=∞.We will consider the two cases(2)∫t0∞‍1r1/α(t)dt=∞,(3)∫t0∞‍1r1/α(t)dt<∞.Recently, there have been a large number of papers devoted to the oscillation of the delay differential equations; see [1–8]. Furthermore, there have been a large number of works on the oscillation of the neutral differential equations, and we refer the readers to the articles [9–25].Agarwal and Grace [3] studied the oscillation for functional differential equations of higher order,
(4)((x(n-1)(t))α)′+q(t)f(x(g(t)))=0,
and established some sufficient conditions for oscillation of (4).Sun and Meng [6] examined the oscillation of (1), where p(t)=0, n=2.Xu and Xia [7], by means of Riccati transformation technique, established some oscillation criteria for certain even order delay differential equations:
(5)(|(x(t))(n-1)|α-1(x(t))(n-1))′+F(t,x(g(t)))=0.In 2011, Zhang et al. [8] studied the oscillatory behavior of the following higher-order half-linear delay differential equation:
(6)(r(t)(x(n-1)(t))α)′+q(t)xβ(τ(t))=0,t≥t0,
where ∫t0∞‍r-(1/α)(t)dt<∞.In 2013, Zhang et al. [23] improved those reported in [8].Han et al. [9] studied the oscillation of second-order neutral differential equations:
(7)(r(t)ψ(x(t))|Z′(t)|α-1Z′(t))′+q(t)f(x(σ(t)))=0,
where Z(t)=x(t)+p(t)x(t-τ) and α>0, 0≤p(t)<1. Some new oscillation criteria are established for the second-order nonlinear neutral delay differential equations:
(8)[r(t)[x(t)+p(t)x(τ(t))]′]′+q(t)f(x(σ(t)))=0,
where ∫t0∞‍1/r(t)dt<∞, 0≤p(t)≤p0<+∞.Meng and Xu [19], by using the Riccati transformation technique and inequalities, considered the oscillation for even order quasilinear neutral differential equations:
(9)(r(t)|(x(t)+p(t)x(t-σ))(n-1)|α-1×(x(t)+p(t)x(t-σ))(n-1)|(x(t)+p(t)x(t-σ))(n-1)|α-1)′+q(t)f(x(σ(t)))=0,
where t≥t0, 0≤p(t)<1.In 2012, Sun et al. [22] considered the oscillation criteria for even order nonlinear neutral differential equations:
(10)(r(t)z(n-1)(t))′+q(t)f(x(σ(t)))=0,
where z(t)=x(t)+p(t)x(τ(t)), n≥2 is even integer, and 0≤p(t)≤p0<+∞. The results are obtained when ∫∞r-1(t)dt=∞ or ∫∞r-1(t)dt<∞. These criteria obtained in this paper extended and improved some known results in the literatures.In 2013, Agarwal et al. [24] considered the oscillation criteria for even order neutral differential equations:
(11)(x(t)+p(t)x(τ(t)))n+q(t)x(σ(t))=0.
Some new criteria are established that improve a number of related results reported in the literature and can be used in cases where known theorems fail to apply.In 2014, Zhang et al. [25] study oscillation and asymptotic behavior of solutions to two classes of higher-order delay damped differential equations with p-Laplacian like operators:
(12)(a(t)|x(n-1)(t)|α-1x(n-1)(t))′+r(t)|x(t)|α-1x(n-1)(t)+q(t)|x(g(t))|α-1x(g(t))=0,t≥t0,
where α>0. Some new criteria are presented that improve the related contributions to the subject.Clearly, the equations (4)–(12) are special cases of (1). The purpose of this paper is to extend and improve the abovementioned oscillation theorems for certain even order neutral delay differential equations with mixed nonlinearities (1).The paper is organized as follows. In the next section, we present some lemmas which will be used in the following results. In Sections3 and 4, by developing Riccati transformations technique and inequalities, some sufficient conditions for oscillation of all solutions of (1) are established. In Section 5, we give an example to illustrate Theorem 11.


## 2. Lemmas

In this section, in order to prove our main results, we need the following lemmas.Lemma 1 (see [5]).
Letu∈Cn([t0,∞),R+). If u(n)(t) is eventually of one sign for all large t, then there exist tx>t1, for some t1>t0, and an integer l, 0≤l≤n, with n+l even for u(n)(t)≥0 or n+l odd for u(n)(t)≤0 such that l>0 implies that u(k)(t)>0 for t>tx, k=0,1,…,l-1, and l≤n-1 implies that (-1)l+ku(k)(t)>0 for t>tx, k=l,l+1,…,n-1.Lemma 2 (see [1], Lemma 2.2.2).
If the functionu is as in Lemma 1 and u(n-1)(t)u(n)(t)≤0 for t>tx, then, for every λ, 0<λ<1, there exists a constant M>0 such that
(13)u(λt)≥Mtn-1|u(n-1)(t)|,
for all large t.Lemma 3 (see [1], Lemma 2.2.3).
If the functionu is as in Lemma 1 and u(n-1)(t)u(n)(t)≤0 for t>tx, lim⁡t→∞u(t)≠0, then, for every λ, 0<λ<1, such that
(14)u(t)≥λ(n-1)!tn-1|u(n-1)(t)|,
for all large t.Lemma 4 (see [2, 5]).
Consider the half-linear differential equations(15)(a(t)|x′(t)|α-1x′(t))′+q(t)|x(t)|α-1x(t)=0,
where α>0, a, q∈C([t0,∞),R+). Then every solution of (15) is nonoscillatory if and only if there exist a real number T≥t0 and a function ν∈C1([t0,∞),R), such that
(16)ν′(t)+αa-(1/α)(t)|ν(t)|(α+1)/α+q(t)≤0,t∈[T,∞).Lemma 5.
IfA and B are nonnegative constants, then
(17)Aγ-γABγ-1+(γ-1)Bγ≥0,γ>1,
and the equality holds if and only if A=B.In the next section, by developing Riccati transformations technique and inequalities, some sufficient conditions for oscillation of all solutions of (1) are established.


## 3. Oscillation Criteria for an Oscillating Functionp

In this section, we assume the following.(H)p is an oscillating function, and lim⁡t→∞p(t)=0.Lemma 6.
Assume that (2) holds. Furthermore, assume that x is an eventually positive solution of (1), which is bounded and does not converge to zero. Then there exists t1≥t0, such that
(18)z(t)>0,z′(t)>0,z(n-1)(t)>0,z(n)(t)≤0,∀t≥t1.Proof.
Sincex is an eventually positive solution of (1), there exists a constant t1≥t0, such that x(t)>0, x(σ(t))>0, and x(τi(t))>0, i=0,1,2, for all t≥t1. Then, by (1), we have (r(t)|z(n-1)(t)|α-1z(n-1)(t))′≤0, t≥t1.
Furthermore, sincex is a bounded solution and lim⁡t→∞x(t)≠0, by (H) we know that lim⁡t→∞p(t)x(σ(t))=0; then there exists t2≥t1, such that z(t)=x(t)+p(t)x(σ(t))>0, t≥t2. So z is eventually positive and bounded.
The rest of the proof is similar to that of Meng and Xu [19, Lemma 2.3], so it is omitted.Theorem 7.
Assume that (H) and (2) hold. Furthermore, assume that there exists a constant λ, 0<λ<1, and, for every constant M>0, assume that there exists a positive function ρ∈C1([t0,∞),R), for sufficiently large t1≥t0, such that
(19)limsup⁡t→∞∫t1t‍(-r(s)(ρ′(s))α+1(α+1)α+1(λMρ(s)τn-2(s)τ′(s))αρ(s)Q(s)-r(s)(ρ′(s))α+1(α+1)α+1(λMρ(s)τn-2(s)τ′(s))α)ds=∞,
where
(20)Q(t):=12α{q0(t)+[k1q1(t)]1/k1[k2q2(t)]1/k2},k1:=(γ-β)(γ-α),k2:=(γ-β)(α-β).
Then every bounded solution of (1) is oscillatory or converges to zero.Proof.
Suppose that (1) has a bounded nonoscillatory solution x. We may assume without loss of generality that there exists a number t1≥t0, such that x(t)>0, x(σ(t))>0, and x(τ(t))>0, for all t≥t1. Furthermore, we assume that lim⁡t→∞x(t)≠0. Using the definition of z and Lemma 6, we have z(t)>0, z′(t)>0, z(n-1)(t)>0, and z(n)(t)≤0, t≥t1. Hence there exists t2≥t1, such that
(21)x(t)=z(t)-p(t)x(σ(t))≥z(t)2,t≥t2.
From (1) and the above inequality, we obtain
(22)(r(t)(z(n-1)(t))α)′+12αq0(t)zα(τ(t))+12βq1(t)zβ(τ(t))+12γq2(t)zγ(τ(t))≤0,t≥t3≥t2.
Because of z′(t)>0, by Lemma 2, z(n-1)(t)>0, and z(n)(t)≤0, there exists t4≥t3, and, for every 0<λ<1, there exists a constant M>0, we have
(23)z′(λτ(t))≥Mτn-2(t)z(n-1)τ(t)≥Mτn-2(t)z(n-1)(t),
for t≥t4. We define the function ω by
(24)ω(t)=ρ(t)r(t)(z(n-1)(t)z(λτ(t)))α,t≥t4.
Then ω(t)>0, t≥t4. Next differentiating (24), we get
(25)ω′(t)=ρ′(t)r(t)(z(n-1)(t)z(λτ(t)))α+ρ(t)(r(t)(z(n-1)(t))α)′zα(λτ(t))-αλρ(t)r(t)(z(n-1)(t))ατ′(t)z′(λτ(t))zα+1(λτ(t)).
So by (22) and (23), we obtain
(26)ω′(t)≤ρ′(t)ρ(t)ω(t)-ρ(t)×(12αq0(t)+12βq1(t)zβ-α(τ(t))+12γq2(t)zγ-α(τ(t)))(z(τ(t))z(λτ(t)))α-αλMτn-2(t)τ′(t)(ρ(t)r(t))1/αω(α+1)/α(t).
Let
(27)a=k112βq1(t)zβ-α(τ(t)),b=k212γq2(t)zγ-α(τ(t)),
where k1 and k2 are defined as in Theorem 7. Using the inequality
(28)|a|p+|b|q≥|a|1/p|b|1/q,p>1,q>1,1p+1q=1,
we have
(29)12βq1(t)zβ-α(τ(t))+12γq2(t)zγ-α(τ(t))≥[k112βq1(t)zβ-α(τ(t))]1/k1[k212γq2(t)zγ-α(τ(t))]1/k2=12α[k1q1(t)]1/k1[k2q2(t)]1/k2,
so we get
(30)ω′(t)≤ρ′(t)ρ(t)ω(t)-ρ(t)Q(t)-αλMτn-2(t)τ′(t)(ρ(t)r(t))1/αω(α+1)/α(t).
Let
(31)A=(αλMτn-2(t)τ′(t)(ρ(t)r(t))1/α)1/γω(t),B=(ρ′(t)γρ(t)(αλMτn-2(t)τ′(t)(ρ(t)r(t))1/α)-(1/γ))1/(γ-1),
where γ=(α+1)/α>1. Applying the inequality in Lemma 5, we obtain
(32)ρ′(t)ρ(t)ω(t)-αλMτn-2(t)τ′(t)(ρ(t)r(t))1/αωγ(t)≤r(t)(ρ′(t))α+1(α+1)α+1(λMρ(t)τn-2(t)τ′(t))α.
Thus, by (30) and (32), we get
(33)ω′(t)≤-(r(t)(ρ′(t))α+1(α+1)α+1(λMρ(t)τn-2(t)τ′(t))αρ(t)Q(t)-r(t)(ρ′(t))α+1(α+1)α+1(λMρ(t)τn-2(t)τ′(t))α).
Integrating (33) from t1 to t, we have
(34)ω(t)≤ω(t1)-∫t1t‍(r(t)(ρ′(t))α+1(α+1)α+1(λMρ(t)τn-2(t)τ′(t))αρ(s)Q(s)-r(s)(ρ′(s))α+1(α+1)α+1(λMρ(s)τn-2(s)τ′(s))α)ds.
Let t→∞ in (34), which leads to a contradiction with (19). The proof is complete.Theorem 8.
Assume that (H) and (2) hold, and there exists a constant λ, 0<λ<1, and, for every constant M>0, such that
(35)(((r(t))1/αλMτn-2(t)τ′(t))α|x′(t)|α-1x′(t))′+Q(t)|x(t)|α-1x(t)=0,
is oscillatory, where Q is defined as in Theorem 7. Then every bounded solution of (1) is oscillatory or converges to zero.Proof.
Suppose that (1) has a bounded nonoscillatory solution x. We may assume without loss of generality that there exists t1≥t0, such that x(t)>0, x(σ(t))>0, and x(τ(t))>0, for all t≥t1. Furthermore, we assume that lim⁡t→∞x(t)≠0. We define ν by
(36)ν(t)=r(t)(z(n-1)(t)z(λτ(t)))α.
Proceeding as in the proof of Theorem 7, for every 0<λ<1, there exists M>0, and we have
(37)ν′(t)+Q(t)+αλMτn-2(t)τ′(t)(r(t))1/αν(α+1)/α(t)≤0.
That is,
(38)ν′(t)+Q(t)+α[((r(t))1/αλMτn-2(t)τ′(t))α]-(1/α)ν(α+1)/α(t)≤0.
Based on Lemma 4, we obtain that (35) is nonoscillatory, which leads to a contradiction. The proof is complete.Theorem 9.
Assume that (H) and (3) hold. Furthermore, assume that there exists a constant λ, 0<λ<1, and, for every constant M>0, assume that there exists a positive function ρ∈C1([t0,∞),R), such that (19) holds. If, for sufficiently large t1≥t0,
(39)limsup⁡t→∞∫t1t‍[(λ(n-2)!)αQ(s)τα(n-2)(s)δα(s)-(αα+1)α+11r1/α(s)δ(s)]ds=∞,
where Q and h are defined as in Theorem 7, and δ(t)=∫t∞‍(1/r1/α(s))ds, then every bounded solution of (1) is oscillatory or converges to zero.Proof.
Suppose that (1) has a bounded nonoscillatory solution x. We may assume without loss of generality that there exists t1≥t0, such that x(t)>0, x(σ(t))>0, and x(τ(t))>0, for all t≥t1. Then it follows from (1) that
(40)(r(t)|z(n-1)(t)|α-1z(n-1)(t))′≤0,t≥t1.
Therefore, r(t)|z(n-1)(t)|α-1z(n-1)(t) is a nonincreasing function on [t1,∞). Consequently, it is easy to conclude that there exist two possible cases of the sign of z(n-1)(t). Furthermore, we assume that lim⁡t→∞x(t)≠0.
Case I. If z(n-1)(t)>0, for t≥t1, then we go back to the proof of Theorem 7, and we get a contradiction to (19), so we omit the details.
Case II. z(n-1)(t)<0, for t≥t1. Applying Lemma 1, we get z(n-2)(t)>0. Define the function υ by
(41)υ(t)=r(t)(-z(n-1)(t))α-1z(n-1)(t)(z(n-2)(t))α,t≥t1.
Then υ(t)<0 for t≥t1. Noting that r(t)|z(n-1)(t)|α-1z(n-1)(t) is nonincreasing, we obtain
(42)r1/α(s)z(n-1)(s)≤r1/α(t)z(n-1)(t),s≥t.
Dividing (42) by r1/α(s) and integrating it from t to l(l≥t), we have
(43)z(n-2)(l)≤z(n-2)(t)+r1/α(t)z(n-1)(t)∫tl‍1r1/α(s)ds.
Letting l→∞ in the above inequality, we get
(44)0≤z(n-2)(t)+r1/α(t)z(n-1)(t)δ(t),
which implies that
(45)-1≤r1/α(t)z(n-1)(t)z(n-2)(t)δ(t),t≥t1,
where δ is defined as in Theorem 9. Hence, by (41), we obtain
(46)-1≤υ(t)δα(t)≤0,t≥t1.
Differentiating (41), we have
(47)υ′(t)=(r(t)(-z(n-1)(t))α-1z(n-1)(t))′(z(n-2)(t))α-αr(t)(-z(n-1)(t))α-1z(n-1)(t)z(n-1)(t)(z(n-2)(t))α+1.
From (22), we get
(48)υ′(t)≤[-12αq0(t)zα(τ(t))-12βq1(t)zβ(τ(t))-12γq2(t)zγ(τ(t))]1(z(n-2)(t))α-αr(t)(-z(n-1)(t))α-1z(n-1)(t)z(n-1)(t)(z(n-2)(t))α+1.
On the other hand, by lim⁡t→∞z(t)≠0 and Lemma 3, we obtain
(49)z(t)≥λ(n-2)!tn-2z(n-2)(t);
that is, because of z(n-1)<0,
(50)z(τ(t))≥λ(n-2)!τn-2(t)z(n-2)(τ(t))≥λ(n-2)!τn-2(t)z(n-2)(t),
for every 0<λ<1 and t≥t1. Then from (41), (48), and (50), we have
(51)υ′(t)≤[-12αq0(t)-12βq1(t)zβ-α(τ(t))-12γq2(t)zγ-α(τ(t))]zα(τ(t))(z(n-2)(t))α-αr(t)(-z(n-1)(t))α-1z(n-1)(t)z(n-1)(t)(z(n-2)(t))α+1≤-(λ(n-1)!)αQ(t)τα(n-2)(t)-αr1/α(t)(-υ(t))(α+1)/α,
where Q is defined as in Theorem 7. Multiplying (51) by δα(t) and integrating it from t1 to t, we get
(52)δα(t)υ(t)-δα(t1)υ(t1)+α∫t1t‍δα-1(s)r1/α(s)υ(s)ds+(λ(n-2)!)α∫t1t‍Q(s)τα(n-2)(s)δα(s)ds+α∫t1t‍δα(s)r1/α(s)(-υ(s))(α+1)/αds≤0.
Let
(53)A=(αδα(s)r1/α(s))1/γ(-υ(s)),B=(α2α+1δα-1(s)r1/α(s)(αδα(s)r1/α(s))-(1/γ))1/(γ-1),
where γ=(α+1)/α>1. Applying the inequality in Lemma 5, we obtain
(54)αδα-1(s)r1/α(s)(-υ(s))-αδα(s)r1/α(s)(-υ(s))(α+1)/α≤(αα+1)α+11r1/α(s)δ(s).
Therefore, it follows from (52) that
(55)δα(t)υ(t)+∫t1t‍[(λ(n-2)!)αQ(s)τα(n-2)(s)δα(s)-(αα+1)α+11r1/α(s)δ(s)]ds≤δα(t1)υ(t1).
From (39) and the above inequality, we get a contradiction to (46). The proof is complete.Theorem 10.
Assume that (H) and (3) hold. Furthermore, assume that there exists a constant λ, 0<λ<1, and for every constant M>0 such that (35) is oscillatory. If, for sufficiently large t1≥t0, one has (39), where Q and h are defined as in Theorem 7 and δ is defined as in Theorem 9, then every bounded solution of (1) is oscillatory or converges to zero.


## 4. Oscillation Criteria for0≤p(t)<1

In this section, we assume that0≤p(t)<1.Theorem 11.
Assume that (2) holds, there exists a constant λ, 0<λ<1, and, for every constant M>0, there exists a positive function ρ∈C1([t0,∞),R), such that, for sufficiently large t1≥t0,
(56)limsup⁡t→∞∫t1t‍(r(s)(ρ′(s))α+1(α+1)α+1(λMρ(s)τn-2(s)τ′(s))αρ(s)Q(s)¯-r(s)(ρ′(s))α+1(α+1)α+1(λMρ(s)τn-2(s)τ′(s))α)ds=∞,
where
(57)Q(t)¯=ξα(t){q0(t)+[k1q1(t)]1/k1[k2q2(t)]1/k2},ξ(t)=min⁡{1-p(τ0(t)),1-p(τ1(t)),1-p(τ2(t))},
and k1 and k2 are defined as in Theorem 7. Then every solution of (1) is oscillatory.Proof.
Suppose that (1) has a nonoscillatory solution x. Without loss of generality, we may assume that there exists t1≥t0, such that x(t)>0, x(σ(t))>0, and x(τ(t))>0, for all t≥t1. Similar to the proof of Lemma 2.3 in [19], there exists t2≥t1, such that
(58)z(t)>0,z′(t)>0,z(n-1)(t)>0,z(n)(t)≤0,t≥t2.
From the definition of z, we have
(59)x(t)=z(t)-p(t)x(σ(t))≥z(t)-p(t)z(σ(t))≥(1-p(t))z(t),t≥t3≥t2.
Since lim⁡t→∞τ(t)=∞, there exists t4≥t3, such that τ(t)≥t4, t≥t4, so
(60)x(τ(t))≥(1-p(τ(t)))z(τ(t)),t≥t4.
From (1), (59), and (60), we get
(61)(r(t)(z(n-1)(t))α)′+1-p(τ0(t))αq0(t)zα(τ(t))+1-p(τ1(t))βq1(t)zβ(τ(t))+1-p(τ2(t))γq2(t)zγ(τ(t))≤0,t≥t4.
For every 0<λ<1, we define the function
(62)ω(t)=ρ(t)r(t)(z(n-1)(t)z(λτ(t)))α,t≥t4.
Then ω(t)>0, t≥t4. Next differentiating (62), we obtain
(63)ω′(t)=ρ′(t)r(t)(z(n-1)(t)z(λτ(t)))α+ρ(t)(r(t)(z(n-1)(t))α)′zα(λτ(t))-αλρ(t)r(t)(z(n-1)(t))αz′(λτ(t))zα+1(λτ(t)).
From (23), (61), and (62), we have
(64)ω′(t)≤ρ′(t)ρ(t)ω(t)-ρ(t)×(ξα(t)q0(t)+ξβ(t)q1(t)zβ-α(τ(t))+ξγ(t)q2(t)zγ-α(τ(t))ξα(t)q0(t)+ξβ(t)q1(t)zβ-α(τ(t)))(z(τ(t))z(λτ(t)))α-αλMτn-2(t)τ′(t)(ρ(t)r(t))1/αω(α+1)/α(t),
where ξ is defined as in Theorem 11. Setting
(65)a=k1ξβ(t)q1(t)zβ-α(τ(t)),b=k2ξγ(t)q2(t)zγ-α(τ(t)),
by inequality (28), we get
(66)ξβ(t)q1(t)zβ-α(τ(t))+ξγ(t)q2(t)zγ-α(τ(t))≥[k1ξβ(t)q1(t)zβ-α(τ(t))]1/k1×[k2ξγ(t)q2(t)zγ-α(τ(t))]1/k2=ξα(t)[k1q1(t)]1/k1[k2q2(t)]1/k2;
hence,
(67)ω′(t)≤ρ′(t)ρ(t)ω(t)-ρ(t)Q(t)¯-αλMτn-2(t)τ′(t)(ρ(t)r(t))1/αω(α+1)/α(t).
Let
(68)A=(αλMτn-2(t)τ′(t)(ρ(t)r(t))1/α)1/γω(t),B=(ρ′(t)γρ(t)(αλMτn-2(t)τ′(t)(ρ(t)r(t))1/α)-(1/γ))1/(γ-1),
where γ=(α+1)/α>1. Applying the inequality in Lemma 5, we obtain
(69)ρ′(t)ρ(t)ω(t)-αλMτn-2(t)τ′(t)(ρ(t)r(t))1/αωγ(t)≤r(t)(ρ′(t))α+1(α+1)α+1(λMρ(t)τn-2(t)τ′(t))α.
Thus, by (67) and (69), we get
(70)ω′(t)≤-(r(t)(ρ′(t))α+1(α+1)α+1(λMρ(t)τn-2(t)τ′(t))αρ(t)Q(t)¯-r(t)(ρ′(t))α+1(α+1)α+1(λMρ(t)τn-2(t)τ′(t))α).
Integrating (70) from t1 to t, we have
(71)ω(t)≤ω(t1)-∫t1t‍(r(t)(ρ′(t))α+1(α+1)α+1(λMρ(t)τn-2(t)τ′(t))αρ(s)Q(s)¯-r(s)(ρ′(s))α+1(α+1)α+1(λMρ(s)τn-2(s)τ′(s))α)ds.
Letting t→∞ in (71), we get a contradiction with (56). This completes the proof of Theorem 11.Remark 12.
From Theorem11, we can obtain different conditions for oscillation of all solutions of (1) with different choices of ρ.Theorem 13.
Assume that (3) holds, assume that there exists a constant λ, 0<λ<1, and, for every constant M>0, there exists a positive function ρ∈C1([t0,∞),R), such that, for sufficiently large t1≥t0, (56) holds. If there exists a positive function η∈C1([t0,∞),R), η′(t)≥0, such that
(72)∫t1∞‍(1η(t)r(t)∫t1t‍(λM(n-2)!τn-2(s)δ(τ(s)))α×η(s)Q(s)¯ds1η(t)r(t)∫t1t‍(λM(n-2)!τn-2(s)δ(τ(s)))α)1/αdt=∞,(73)∫t1∞‍(1η(t)r(t)∫t1t‍(λM(n-2)!sn-2δ(s))α×η(s)Q(s)¯ds1η(t)r(t)∫t1t‍(λM(n-2)!τn-2(s)δ(τ(s)))α)1/αdt=∞,
where δ is defined as in Theorem 9 and Q¯ is defined as in Theorem 11, then every solution of (1) is oscillatory or converges to zero.Proof.
Suppose that (1) has a nonoscillatory solution x. We may assume without loss of generality that there exists t1≥t0, such that x(t)>0, x(σ(t))>0, x(τ(t))>0, and z(t)>0, for all t≥t1. Furthermore, we assume that lim⁡t→∞x(t)≠0. Similar to the proof of Theorem 9, we find that r(t)|z(n-1)(t)|α-1z(n-1)(t) is a nonincreasing function on [t1,∞) and there exist two possible cases of the sign of z(n-1)(t).
Case I. If z(n-1)(t)>0, for t≥t1, then we go back to the proof of Theorem 11, and we get a contradiction to (56), so we omit the details.
Case II. z(n-1)(t)<0, for t≥t1. Applying Lemma 1, we get z(n-2)(t)>0, z′(t)>0 or z(n-2)(t)>0, z′(t)<0.
Ifz(n-2)(t)>0, z′(t)>0. Since r(t)|z(n-1)(t)|α-1z(n-1)(t) is nonincreasing, we obtain
(74)r(s)|z(n-1)(s)|α-1z(n-1)(s)≤r(t)|z(n-1)(t)|α-1z(n-1)(t)≤r(t1)|z(n-1)(t1)|α-1z(n-1)(t1),
for s≥t≥t1; that is,
(75)-z(n-1)(s)≥(r(t)r(s))1/α(-z(n-1)(t)).
Integrating (75) from t to ∞, we get
(76)z(n-2)(t)≥r1/α(t)(-z(n-1)(t))∫t∞‍1r1/α(s)ds≥Mδ(t),t≥t1,
where M=r1/α(t1)(-z(n-1)(t1)). From (49) and the above inequality, we obtain
(77)z(t)≥Mλ(n-2)!tn-2δ(t).
Using (59) and (77) in (1) and noting that z′(t)>0, we have
(78)-(r(t)|z(n-1)(t)|α-1z(n-1)(t))′=q0(t)xα(τ0(t))+q1(t)xβ(τ1(t))+q2(t)xγ(τ2(t))≥q0(t)(1-p(τ0(t)))αzα(τ0(t))+q1(t)(1-p(τ1(t)))βzβ(τ1(t))+q2(t)(1-p(τ2(t)))γzγ(τ2(t))≥ξα(t)q0(t)zα(τ(t))+ξβ(t)q1(t)zβ(τ(t))+ξγ(t)q2(t)zγ(τ(t)).
Setting
(79)a=k1ξβ(t)q1(t)zβ(τ(t)),b=k2ξγ(t)q2(t)zγ(τ(t)),
by inequality (28), we get
(80)ξβ(t)q1(t)zβ(τ(t))+ξγ(t)q2(t)zγ(τ(t))≥[k1ξβq1(t)zβ(τ(t))]1/k1[k2ξγq2(t)zγ(τ(t))]1/k2≥ξα(t)[k1q1(t)]1/k1[k2q2(t)]1/k2zα(τ(t)).
Therefore, combining (77), (78), and (80), we obtain
(81)-(r(t)|z(n-1)(t)|α-1z(n-1)(t))′≥Q(t)¯zα(τ(t))≥(λM(n-2)!τn-2(t)δ(τ(t)))αQ(t)¯.
Define the function u by
(82)u(t)=η(t)r(t)|z(n-1)(t)|α-1z(n-1)(t)=-η(t)r(t)(-z(n-1)(t))α,t≥t1.
Then u(t)<0. Differentiating u(t) and from (81), we find that
(83)u′(t)=-η′(t)r(t)(-z(n-1)(t))α-η(t)(r(t)(-z(n-1)(t))α)′≤-η(t)(q0(t)xα(τ0(t))+q1(t)xβ(τ1(t))+q2(t)xγ(τ2(t))q0(t)xα(τ0(t))+q1(t)xβ(τ1(t)))≤-(λM(n-2)!τn-2(t)δ(τ(t)))αη(t)Q(t)¯.
Integrating (83) from t1 to t, we get
(84)u(t)-u(t1)≤-∫t1t‍(λM(n-2)!τn-2(s)δ(τ(s)))αη(s)Q(s)¯ds.
Therefore,
(85)-η(t)r(t)(-z(n-1)(t))α≤-∫t1t‍(λM(n-2)!τn-2(s)δ(τ(s)))αη(s)Q(s)¯ds;
that is,
(86)z(n-1)(t)≤-(1η(t)r(t)∫t1t‍(λM(n-2)!τn-2(s)δ(τ(s)))α×η(s)Q(s)¯ds1η(t)r(t)∫t1t‍(λM(n-2)!τn-2(s)δ(τ(s)))α)1/α.
Integrating the above inequality from t1 to l(l>t1), we obtain
(87)z(n-2)(l)-z(n-2)(t1)≤-∫t1l‍(1η(t)r(t)∫t1t‍(λM(n-2)!τn-2(s)δ(τ(s)))α×η(s)Q(s)¯ds1η(t)r(t)∫t1t‍(λM(n-2)!τn-2(s)δ(τ(s)))α)1/αdt.
Letting l→∞ and using (73) in (97), we have lim⁡l→∞z(n-2)(l)=-∞, which is a contradiction with the fact that z(n-2)(t)>0.
Ifz(n-2)(t)>0, z′(t)<0. Because of lim⁡t→∞x(t)≠0, lim⁡t→∞z(t)≠0. By Lemma 3, we obtain (49). Proceeding as in the proof of the above, (77) holds. Using (59) and (77) in (1) and noting that z′(t)<0, we have
(88)-(r(t)|z(n-1)(t)|α-1z(n-1)(t))′=q0(t)xα(τ0(t))+q1(t)xβ(τ1(t))+q2(t)xγ(τ2(t))≥q0(t)(1-p(τ0(t)))αzα(τ0(t))+q1(t)(1-p(τ1(t)))βzβ(τ1(t))+q2(t)(1-p(τ2(t)))γzγ(τ2(t))≥ξα(t)q0(t)zα(t)+ξβ(t)q1(t)zβ(t)+ξγ(t)q2(t)zγ(t).
Setting(89)a1=k1ξβ(t)q1(t)zβ(t),b1=k2ξγ(t)q2(t)zγ(t),
by inequality (28), we get
(90)ξβ(t)q1(t)zβ(t)+ξγ(t)q2(t)zγ(t)≥[k1ξβq1(t)zβ(t)]1/k1[k2ξγq2(t)zγ(t)]1/k2≥ξα(t)[k1q1(t)]1/k1[k2q2(t)]1/k2zα(t).
Therefore, combining (77), (88), and (90), we obtain
(91)-(r(t)|z(n-1)(t)|α-1z(n-1)(t))′≥Q(t)¯zα(t)≥(λM(n-2)!tn-2δ(t))αQ(t)¯.
Define the function u by
(92)u(t)=η(t)r(t)|z(n-1)(t)|α-1z(n-1)(t)=-η(t)r(t)(-z(n-1)(t))α,t≥t1.
Then
(93)u′(t)=-η′(t)r(t)(-z(n-1)(t))α-η(t)(r(t)(-z(n-1)(t))α)′≤-η(t)(q0(t)xα(τ0(t))+q1(t)xβ(τ1(t))+q2(t)xγ(τ2(t))q0(t)xα(τ0(t))+q1(t)xβ(τ1(t)))≤-(λM(n-2)!tn-2δ(t))αη(t)Q(t)¯.
Integrating from t1 to t, we get
(94)u(t)-u(t1)≤-∫t1t‍(λM(n-2)!sn-2δ(s))αη(s)Q(s)¯ds.
Therefore,
(95)-η(t)r(t)(-z(n-1)(t))α≤-∫t1t‍(λM(n-2)!sn-2δ(s))αη(s)Q(s)¯ds;
that is,
(96)z(n-1)(t)≤-(1η(t)r(t)∫t1t‍(λM(n-2)!sn-2δ(s))α×η(s)Q(s)¯ds1η(t)r(t)∫t1t‍(λM(n-2)!sn-2δ(s))α)1/α.
Integrating the above inequality from t1 to l(l>t1), we obtain
(97)z(n-2)(l)-z(n-2)(t1)≤-∫t1l‍(1η(t)r(t)∫t1t‍(λM(n-2)!sn-2δ(s))α×η(s)Q(s)¯ds1η(t)r(t)∫t1t‍(λM(n-2)!sn-2δ(s))α)1/αdt.
Letting l→∞ and using (73) in (97), we have lim⁡l→∞z(n-2)(l)=-∞, which is a contradiction with the fact that z(n-2)(t)>0. This completes the proof.


## 5. Example

In this section, we will give an example to illustrate Theorem11.Example 1.
Consider the even order neutral delay differential equations with mixed nonlinearities:(98)(tθ|z(n-1)(t)|1/2z(n-1)(t))′+1t|x(τ0(t))|1/2x(τ0(t))+1t3/2|x(τ1(t))|-(1/2)x(τ1(t))+1t3/4|x(τ2(t))|x(τ2(t))=0,t≥t0,
where z(t)=x(t)+1/2x(σ(t)) and n is even integer.
Setr(t)=tθ, p(t)=1/2, q0(t)=1/t, q1(t)=b/t3/2, q2(t)=c/t3/4, γ=2, α=2/3, β=1/2, and θ≤2/3. Then (2) holds, k1=3, and k2=3/2.
Takeρ(t)=1. It is easy to show that
(99)limsup⁡t→∞∫t1t‍(r(s)(ρ′(s))α+1(α+1)α+1(f(n,λ)ρ(s)τn-2(s)τ′(s))αρ(s)Q(s)¯-r(s)(ρ′(s))α+1(α+1)α+1(f(n,λ)ρ(s)τn-2(s)τ′(s))α)ds=limsup⁡t→∞∫t1t‍(12)3/2×{1t+(3·1t3/2)1/3(32·1t3/4)2/3}ds=∞.
Hence, by Theorem 11, every solution of (98) is oscillatory.



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*Source: 629074-2014-08-03.xml*