# Oscillation Criteria for Certain Second-Order Nonlinear Neutral Differential Equations of Mixed Type

Authors: Zhenlai Han; Tongxing Li; Chenghui  Zhang; Ying  Sun
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

Some oscillation criteria are established for the second-order nonlinear neutral differential equations of  mixed type[(x(t)+p1x(t−τ1)+p2x(t+τ2))γ]′′=q1(t)xγ(t−σ1)+q2(t)xγ(t+σ2), t≥t0,  where γ≥1 is a quotient of odd positive integers. Our results generalize the results given in the literature.


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## Body

## 1. Introduction

This paper is concerned with the oscillatory behavior of the second-order nonlinear neutral differential equation of mixed type(1.1)[(x(t)+p1x(t-τ1)+p2x(t+τ2))γ]′′=q1(t)xγ(t-σ1)+q2(t)xγ(t+σ2),t≥t0.Throughout this paper, we will assume the following conditions hold.(A1)
pi, τi, and σi, i=1,2, are positive constants;(A2)
qi∈C([t0,∞),[0,∞)),
i=1,2.By a solution of (1.1), we mean a function x∈C([Tx,∞),ℝ) for some Tx≥t0 which has the property that (x(t)+p1x(t-τ1)+p2x(t+τ2))γ∈C2([Tx,∞),ℝ) and satisfies (1.1) on [Tx,∞). As is customary, a solution of (1.1) is called oscillatory if it has arbitrarily large zeros on [t0,∞), otherwise, it is called nonoscillatory. Equation (1.1) is said to be oscillatory if all its solutions are oscillatory.Neutral functional differential equations have numerous applications in electric networks. For instance, they are frequently used for the study of distributed networks containing lossless transmission lines which rise in high speed computers where the lossless transmission lines are used to interconnect switching circuits; see [1].Recently, many results have been obtained on oscillation of nonneutral continuous and discrete equations and neutral functional differential equations, we refer the reader to the papers [2–35], and the references cited therein.Philos [2] established some Philos-type oscillation criteria for the second-order linear differential equation (1.2)(r(t)x′(t))′+q(t)x(t)=0,t≥t0.In [3–5], the authors gave some sufficient conditions for oscillation of all solutions of second-order half-linear differential equation (1.3)(r(t)|x′(t)|γ-1x′(t))′+q(t)|x(τ(t))|γ-1x(τ(t))=0,t≥t0
by employing a Riccati substitution technique.Zhang et al. [15] examined the oscillation of even-order neutral differential equation (1.4)[x(t)+p(t)x(τ(t))](n)+q(t)f(x(σ(t)))=0,t≥t0.Some oscillation criteria for the following second-order quasilinear neutral differential equation(1.5)(r(t)|z′(t)|γ-1z′(t))′+q(t)|x(σ(t))|γ-1x(σ(t))=0,forz(t)=x(t)+p(t)x(τ(t)),t≥t0
were obtained by [12–17].However, there are few results regarding the oscillatory properties of neutral differential equations with mixed arguments, see the papers [20–24]. In [25], the authors established some oscillation criteria for the following mixed neutral equation:(1.6)(x(t)+p1x(t-τ1)+p2x(t+τ2))′′=q1(t)x(t-σ1)+q2(t)x(t+σ2),t≥t0;
here q1 and q2 are nonnegative real-valued functions. Grace [26] obtained some oscillation theorems for the odd order neutral differential equation (1.7)(x(t)+p1x(t-τ1)+p2x(t+τ2))(n)=q1x(t-σ1)+q2x(t+σ2),t≥t0,
where n≥1 is odd. Grace [27] and Yan [28] obtained several sufficient conditions for the oscillation of solutions of higher-order neutral functional differential equation of the form (1.8)(x(t)+cx(t-h)+Cx(t+H))(n)+qx(t-g)+Qx(t+G)=0,t≥t0,
where q and Q are nonnegative real constants.Clearly, (1.6) is a special case of (1.1). The purpose of this paper is to study the oscillation behavior of (1.1).In the sequel, when we write a functional inequality without specifying its domain of validity we assume that it holds for all sufficiently larget.


## 2. Main Results

In the following, we give our results.Theorem 2.1.
Assume thatσi>τi, i=1,2. If
(2.1)limsup⁡t→∞∫tt+σ2-τ2(t+σ2-τ2-s)Q2(s)ds>(2γ-1)2(1+p1γ+p2γ2γ-1),(2.2)limsup⁡t→∞∫t-σ1+τ1t(s-t+σ1-τ1)Q1(s)ds>(2γ-1)2(1+p1γ+p2γ2γ-1),
where
(2.3)Qi(t)=min⁡{qi(t-τ1),qi(t),qi(t+τ2)},
for i=1,2, then every solution of (1.1) oscillates.Proof.
Letx be a nonoscillatory solution of (1.1). Without loss of generality, we assume that there exists t1≥t0 such that x(t)>0, x(t-τ1)>0,x(t+τ2)>0, x(t-σ1)>0, and x(t+σ2)>0 for all t≥t1. Setting
(2.4)z(t)=(x(t)+p1x(t-τ1)+p2x(t+τ2))γ,y(t)=z(t)+p1γz(t-τ1)+p2γ2γ-1z(t+τ2).
Thus z(t)>0, y(t)>0, and
(2.5)z′′(t)=q1(t)xγ(t-σ1)+q2(t)xγ(t+σ2)≥0.
Then, z′(t) is of constant sign, eventually. On the other hand,
(2.6)y′′(t)=q1(t)xγ(t-σ1)+q2(t)xγ(t+σ2)+p1γq1(t-τ1)xγ(t-τ1-σ1)+p1γq2(t-τ1)xγ(t-τ1+σ2)+p2γ2γ-1q1(t+τ2)xγ(t+τ2-σ1)+p2γ2γ-1q2(t+τ2)xγ(t+τ2+σ2).
Note that g(u)=uγ, γ≥1, u∈(0,∞) is a convex function. Hence, by the definition of convex function, we obtain
(2.7)aγ+bγ≥12γ-1(a+b)γ.
Using inequality (2.7), we get
(2.8)xγ(t-σ1)+p1γxγ(t-τ1-σ1)≥12γ-1(x(t-σ1)+p1x(t-τ1-σ1))γ,12γ-1(x(t-σ1)+p1x(t-τ1-σ1))γ+p2γ2γ-1xγ(t+τ2-σ1)≥1(2γ-1)2(x(t-σ1)+p1x(t-τ1-σ1)+p2x(t+τ2-σ1))γ=z(t-σ1)(2γ-1)2.
Similarly, we obtain
(2.9)xγ(t+σ2)+p1γxγ(t-τ1+σ2)+p2γ2γ-1xγ(t+τ2+σ2)≥z(t+σ2)(2γ-1)2.
Thus, from (2.6), we have
(2.10)y′′(t)≥1(2γ-1)2(Q1(t)z(t-σ1)+Q2(t)z(t+σ2)).
In the following, we consider two cases.Case 1.
Assume thatz′(t)>0. Then, y′(t)>0. In view of (2.10), we see that
(2.11)y′′(t+τ2)≥1(2γ-1)2Q2(t+τ2)z(t+τ2+σ2).
Applying the monotonicity of z, we find
(2.12)y(t+σ2)=z(t+σ2)+p1γz(t-τ1+σ2)+p2γ2γ-1z(t+τ2+σ2)≤(1+p1γ+p2γ2γ-1)z(t+τ2+σ2).
Combining the last two inequalities, we obtain the inequality
(2.13)y′′(t+τ2)≥Q2(t+τ2)(2γ-1)2(1+p1γ+p2γ/2γ-1)y(t+σ2).
Therefore, y is a positive increasing solution of the differential inequality
(2.14)y′′(t)≥Q2(t)(2γ-1)2(1+p1γ+p2γ/2γ-1)y(t-τ2+σ2).
However, by [11], condition (2.1) contradicts the existence of a positive increasing solution of inequality (2.14).
Case 2.
Assume thatz′(t)<0. Then, y′(t)<0. In view of (2.10), we see that
(2.15)y′′(t-τ1)≥1(2γ-1)2Q1(t-τ1)z(t-τ1-σ1).
Applying the monotonicity of z, we find
(2.16)y(t-σ1)=z(t-σ1)+p1γz(t-τ1-σ1)+p2γ12γ-1z(t+τ2-σ1)≤(1+p1γ+p2γ2γ-1)z(t-τ1-σ1).
Combining the last two inequalities, we obtain the inequality
(2.17)y′′(t-τ1)≥Q1(t-τ1)(2γ-1)2(1+p1γ+p2γ/2γ-1)y(t-σ1).
Therefore, y is a positive decreasing solution of the differential inequality
(2.18)y′′(t)≥Q1(t)(2γ-1)2(1+p1γ+p2γ/2γ-1)y(t+τ1-σ1).
However, by [11], condition (2.2) contradicts the existence of a positive decreasing solution of inequality (2.18).Remark 2.2.
Whenγ=1, Theorem 2.1 involves results of [25, Theorem 1].Theorem 2.3.
Letβi=(σi-τi)/2>0, i=1,2. Suppose that, for i=1,2, there exist functions
(2.19)ai∈C1[t0,∞),ai(t)>0,(-1)iai′(t)≤0,
such that
(2.20)Qi(t)≥(2γ-1)2(1+p1γ+p2γ2γ-1)ai(t)ai(t+(-1)iβi),
where Qi are as in (2.3) for i=1,2. If the first-order differential inequality
(2.21)v′(t)+(-1)i+1ai(t+(-1)iβi)v(t+(-1)iβi)≥0
has no eventually negative solution for i=1 and no eventually positive solution for i=2, then (1.1) is oscillatory.Proof.
Letx be a nonoscillatory solution of (1.1). Without loss of generality, we assume that there exists t1≥t0 such that x(t)>0,  x(t-τ1)>0, x(t+τ2)>0, x(t-σ1)>0, and x(t+σ2)>0 for all t≥t1. Define z and y as in Theorem 2.1. Proceeding as in the proof of Theorem 2.1, we get (2.10).
In the following, we consider two cases.Case 1.
Assume thatz′(t)>0. Clearly, y′(t)>0. Then, just as in Case 1 of Theorem 2.1, we find that y is a positive increasing solution of inequality (2.14). Let b2(t)=y′(t)+a2(t)y(t+β2). Then b2(t)>0. Using (2.19) and (2.20), we obtain
(2.22)b2′(t)-a2′(t)a2(t)b2(t)-a2(t)b2(t+β2)=y′′(t)-a2′(t)a2(t)y′(t)-a2(t)a2(t+β2)y(t+2β2)≥y′′(t)-a2(t)a2(t+β2)y(t+2β2)≥y′′(t)-Q2(t)(2γ-1)2(1+p1γ+(p2γ/2γ-1))y(t-τ2+σ2)≥0.
Define b2(t)=a2(t)v(t). Then, v is a positive solution of (2.21) for i=2, which is a contradiction.
Case 2.
Assume thatz′(t)<0. Clearly, y′(t)<0. Then, just as in Case 2 of Theorem 2.1, we find that y is a positive decreasing solution of inequality (2.18). Let b1(t)=y′(t)-a1(t)y(t-β1). Then b1(t)<0. Using (2.19) and (2.20), we obtain
(2.23)b1′(t)-a1′(t)a1(t)b1(t)+a1(t)b1(t-β1)=y′′(t)-a1′(t)a1(t)y′(t)-a1(t)a1(t-β1)y(t-2β1)≥y′′(t)-a1(t)a1(t-β1)y(t-2β1)≥y′′(t)-Q1(t)(2γ-1)2(1+p1γ+p2γ/2γ-1)y(t+τ1-σ1)≥0.
Define b1(t)=a1(t)v(t). Then, v is a negative solution of (2.21) for i=1. This contradiction completes the proof of the theorem.Remark 2.4.
Whenγ=1, Theorem 2.3 involves results of [25, Theorem 2].From Theorem2.3 and the results given in [12], we have the following oscillation criterion for (1.1).Corollary 2.5.
Letβi=(σi-τi)/2>0, i=1,2. Assume that (2.19) and (2.20) hold for i=1,2. If
(2.24)liminf⁡t→∞∫t-β1ta1(s-β1)ds>1e,(2.25)liminf⁡t→∞∫tt+β2a2(s+β2)ds>1e,
then (1.1) is oscillatory.Proof.
It is known (see [12]) that condition (2.24) is sufficient for inequality (2.21) (for i=1) to have no eventually negative solution. On the other hand, condition (2.25) is sufficient for inequality (2.21) (for i=2) to have no eventually positive solution.For an application of our results, we give the following example.Example 2.6.
Consider the second-order differential equation(2.26)[(x(t)+p1x(t-τ1)+p2x(t+τ2))γ]′′=q1xγ(t-σ1)+q2xγ(t+σ2),t≥t0,
where qi>0 are constants and σi>τi for i=1,2.
It is easy to see thatQi(t)=qi, i=1,2. Assume that ε>0. Let ai(t)=(2+ε)/(e(σi-τi)), i=1,2. Clearly, (2.19) holds. If
(2.27)qi>[2(e(σi-τi))]2(2γ-1)2(1+p1γ+p2γ2γ-1)
for i=1,2, then (2.20) holds. Moreover, we see that
(2.28)liminf⁡t→∞∫t-β1ta1(s-β1)ds=2+ε2e>1e,liminf⁡t→∞∫tt+β2a2(s+β2)ds=2+ε2e>1e.
Hence by applying Corollary 2.5, we find that (2.26) is oscillatory.



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