# Existence for Nonoscillatory Solutions of Higher-Order Nonlinear Differential Equations

Authors: Yazhou Tian; Fanwei Meng
Journal: ISRN Mathematical Analysis
 (2011)

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## Abstract

The existence of nonoscillatory solutions of the higher-order nonlinear differential equation[r(t)(x(t)+P(t)x(t-τ))(n-1)]′+∑i=1mQi(t)fi(x(t-σi))=0,t≥t0, where m≥1,n≥2 are integers, τ>0,σi≥0,r,P,Qi∈C([t0,∞),R),fi∈C(R,R)(i=1,2,…,m), is studied. Some new sufficient conditions for the existence of a nonoscillatory solution of above equation are obtained for general Qi(t)(i=1,2,…,m) which means that we allow oscillatory Qi(t)(i=1,2,…,m). In particular, our results improve essentially and extend some known results in the recent references.


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## Body

## 1. Introduction

Consider the higher-order nonlinear neutral differential equation(1.1)[r(t)(x(t)+P(t)x(t-τ))(n-1)]′+∑i=1mQi(t)fi(x(t-σi))=0,t≥t0.
With respect to (1.1), throughout, we shall assume the following:(i)
m≥1,n≥2 are integers, τ>0,σi≥0,(ii)
r,P,Qi∈C([t0,∞),R),r(t)>0,fi∈C(R,R),i=1,2,…,m.Letρ=max1≤i≤m{τ,σi}. By a solution of (1.1), we mean a function x(t)∈C([t1-ρ,∞),R) for some t1≥t0 which has the property that x(t)+P(t)x(t-τ)∈Cn-1([t1,∞),R) and r(t)(x(t)+P(t)x(t-τ))(n-1)∈C1([t1,∞),R) and satisfies (1.1) on [t1,∞).A nontrivial solution of (1.1) is called oscillatory if it has arbitrarily large zeros, and, otherwise, it is nonoscillatory.The existence of nonoscillatory solutions of higher-order nonlinear neutral differential equations received much less attention, which is due mainly to the technical difficulties arising in its analysis.In 1998, Kulenovic and Hadziomerspahic [1] investigated the existence of nonoscillatory solutions of second-order nonlinear neutral differential equation
(E0)(x(t)+cx(t-τ))′′+Q1(t)x(t-σ1)-Q2(t)x(t-σ2)=0,t≥t0,
where c is a constant.In 2006, Zhang and Wang [2] investigated the second neutral delay differential equation with positive and negative coefficients:
(E)[r(t)(x(t)+P(t)x(t-τ))′]′+Q1(t)f(x(t-σ1))-Q2(t)g(x(t-σ2))=0,t≥t0,
where τ>0,σi≥0,Q1,Q2∈C([t0,∞),R+),f,g∈C(R,R),xf(x)>0,xg(x)>0,(x≠0). By using Banach contraction mapping principle, they proved the following theorem which extends the results in [1].Theorem A. ([2, Theorem  2.3]).
Assume that(H1)
f and g satisfy local Lipschitz condition and xf(x)>0,xg(x)>0,forx≠0;(H2)
Qi(t)≥0,i=1,2,aQ1(t)-Q2(t) is eventually nonnegative for every a>0;(H3)
∫t0∞∫t0t(Qi(t)/r(s))dsdt<∞,i=1,2 holdif one of the following two conditions is satisfied:(H4)
P(t)>1 eventually, and 0<P2≤P1<P22<+∞,(H5)
P(t)<-1 eventually, and -∞<P2≤P1<-1,where P1=limsupt→∞P(t), P2=liminft→∞P(t), then (1.1) has a nonoscillatory solution.In 2007, Zhou [3] studies the existence of nonoscillatory solution of the following second-order nonlinear differential equation.
(E')[r(t)(x(t)+P(t)x(t-τ))′]′+∑i=1mQi(t)fi(x(t-σi))=0,t≥t0,
where fi∈C(R,R)(i=1,2,…,m). By using Krasnoselskii’s fixed point theorem, they proved the following theorem.Theorem B. ([3, Theorem  1]).
Assume that there exist nonnegative constantsc1 and c2 such that c1+c2<1, -c2≤P(t)≤c1. Further, assume that
(1.2)∫t0∞∫t0t|Qi(t)|r(s)dsdt<∞,i=1,2,…,m.
Then (1.1) has a bounded nonoscillatory solution.In this paper, by using Krasnoselskii’s fixed point theorem and some new techniques, we obtain some sufficient conditions for the existence of a nonoscillatory solution of (1.1) for general Qi(t)(i=1,2,…,m) which means that we allow oscillatory Qi(t)(i=1,2,…,m). Meanwhile, we extend the main results of [2, 3].


## 2. Main Result

The following fixed point theorem will be used to prove the main results in this section.Lemma 2.1 (see [3, Krasnoselskii’s fixed point theorem]).
LetX be a Banach space, let Ω be a bounded closed convex subset of X, and let S1, S2 be maps of Ω into X such that S1x+S2y∈Ω for every pair x,y∈Ω. If S1 is a contraction and S2 is completely continuous, then the equation
(2.1)S1x+S2x=x
has a solution in Ω.Theorem 2.2.
Assume that there exist nonnegative constantsc1 and c2 such that c1+c2<1, -1<-c2≤P(t)≤c1<1. Further, assume that
(2.2)∫t0∞∫t0tsn-2|Qi(t)|r(s)dsdt<∞,i=1,2,…,m.
Then (1.1) has a bounded nonoscillatory solution.Proof.
By interchanging the order of integral, we note that (2.2) is equivalent to
(2.3)∫t0∞sn-2∫s∞|Qi(t)|r(s)dsdt<∞,i=1,2,…,m.
By (2.3), we choose T>t0 sufficiently large such that
(2.4)1(n-2)!∫T∞sn-2∫s∞Mr(s)∑i=1m|Qi(u)|duds<1-c1-c24,
where M=max((1-c1-c2)/2)≤x≤1{|fi(x)|:1≤i≤m}.
LetC([t0,∞),R) be the set of all continuous functions with the norm ∥x∥=supt≥t0|x(t)|<∞. Then C([t0,∞),R) is a Banach space. We define a bounded, closed, and convex subset Ω of C([t0,∞),R) as follows:
(2.5)Ω={x=x(t)∈C([t0,∞),R):1-c1-c22≤x(t)≤1,t≥t0}.
Define two mapsS1 and S2:Ω→C([t0,∞),R) as follows:
(2.6)(S1x)(t)={3+c1-3c24-P(t)x(t-τ),t≥T,(S1x)(T),t0≤t≤T,(S2x)(t)={(-1)n-1(n-2)!∫t∞(s-t)n-2∫s∞1r(s)(∑i=1mQi(u)fi(x(u-σi)))duds,t≥T,(S2x)(T),t0≤t≤T.
(i) We shall show that for anyx,y∈Ω, S1x+S2y∈Ω.
In fact,x,y∈Ω, and t≥T, we get
(2.7)(S1x)(t)+(S2y)(t)≤3+c1-3c24-P(t)x(t-τ)+1(n-2)!∫t∞(s-t)n-2∫s∞1r(s)(∑i=1m|Qi(u)fi(y(u-σi))|)duds≤3+c1-3c24+c2+1(n-2)!∫T∞sn-2∫s∞Mr(s)∑i=1m|Qi(u)|duds≤3+c1-3c24+c2+1-c1-c24=1.
Furthermore, we have(2.8)(S1x)(t)+(S2y)(t)≥3+c1-3c24-P(t)x(t-τ)-1(n-2)!∫t∞sn-2∫s∞1r(s)(∑i=1m|Qi(u)fi(y(u-σi))|)duds≥3+c1-3c24-c1-1(n-2)!∫T∞sn-2∫s∞Mr(s)∑i=1m|Qi(u)|duds≥3+c1-3c24-c1-1-c1-c24=1-c1-c22.
Hence,
(2.9)1-c1-c22≤(S1x)(t)+(S2y)(t)≤1,fort≥t0.
Thus, we have proved that S1x+S2y∈Ω for any x,y∈Ω.
(ii) We shall show thatS1 is a contraction mapping on Ω.
In fact, forx,y∈Ω and t≥T, we have
(2.10)|(S1x)(t)-(S1y)(t)|≤|P(t)||x(t-τ)-y(t-τ)|≤c0‖x-y‖,
where c0=max{c1,c2}. This implies that
(2.11)‖S1x-S1y‖≤c0‖x-y‖.
Since 0<c0<1, we conclude that S1 is a contraction mapping on Ω.
(iii) We now show thatS2 is completely continuous.
First, we will show thatS2 is continuous. Let xk=xk(t)∈Ω be such that xk(t)→x(t) as k→∞. Because Ω is closed, x=x(t)∈Ω. For t≥T, we have
(2.12)|(S2xk)(t)-(S2x)(t)|≤1(n-2)!∫t∞sn-2∫s∞1r(s)(∑i=1m|Qi(u)||fi(xk(u-σi))-fi(x(u-σi))|)duds≤1(n-2)!∫T∞sn-2∫s∞1r(s)(∑i=1m|Qi(u)||fi(xk(u-σi))-fi(x(u-σi))|)duds.
Since |fi(xk(t-σi))-fi(x(t-σi))|→0 as k→∞ for i=1,2,…,m, by applying the Lebesgue dominated convergence theorem, we conclude that limk→∞∥(S2xk)(t)-(S2x)(t)∥=0. This means that S2 is continuous.
Next, we show thatS2Ω is relatively compact. It suffices to show that the family of functions {S2x:x∈Ω} is uniformly bounded and equicontinuous on [t0,∞). The uniform boundedness is obvious. For the equicontinuity, according to Levitan's result [4], we only need to show that, for any given ɛ>0, [T,∞) can be decomposed into finite subintervals in such a way that on each subinterval all functions of the family have change of amplitude less than ɛ. By (2.3), for any ɛ>0, take T*≥T large enough so that
(2.13)1(n-2)!∫T*∞sn-2∫s∞Mr(s)∑i=1m|Qi(u)|duds<ɛ2.
Then, for x∈Ω, t2≥t1≥T*,
(2.14)|(S2x)(t2)-(S2x)(t1)|≤1(n-2)!∫t2∞sn-2∫s∞1r(s)(∑i=1m|Qi(u)||fi(x(u-σi))|)duds+1(n-2)!∫t1∞sn-2∫s∞1r(s)(∑i=1m|Qi(u)||fi(x(u-σi))|)duds≤1(n-2)!∫t2∞sn-2∫s∞Mr(s)∑i=1m|Qi(u)|duds+1(n-2)!∫t1∞sn-2∫s∞Mr(s)∑i=1m|Qi(u)|duds<ɛ2+ɛ2=ɛ.
For x∈Ω, T≤t1<t2≤T*+1,
(2.15)|(S2x)(t2)-(S2x)(t1)|≤1(n-2)!|∫t1t2(s-t1)n-2∫s∞1r(s)(∑i=1mQi(u)fi(x(u-σi)))duds|+1(n-2)!|∫t2∞[(s-t2)n-2-(s-t1)n-2]∫s∞1r(s)(∑i=1mQi(u)fi(x(u-σi)))duds|≤1(n-2)!∫t1t2sn-2∫s∞Mr(s)∑i=1m|Qi(u)|duds+1(n-3)!(t2-t1)∫t2∞(s-ξ)n-3∫s∞Mr(s)∑i=1m|Qi(u)|duds≤1(n-2)!∫t1t2sn-2∫s∞Mr(s)∑i=1m|Qi(u)|duds+1(n-3)!(t2-t1)∫T∞sn-2∫s∞Mr(s)∑i=1m|Qi(u)|duds,
where t1<ξ<t2.
Then there existsδ>0 such that
(2.16)|(S2x)(t2)-(S2x)(t1)|<ɛ,if0<t2-t1<δ.
For any x∈Ω, t0≤t1<t2≤T, it is easy to see that
(2.17)|(S2x)(t2)-(S2x)(t1)|=0<ɛ.
Therefore, {S2x:x∈Ω} is uniformly bounded and equicontinuous on [t0,∞), and hence S2Ω is relatively compact. By Lemma 2.1, there is x0∈Ω such that S1x0+S2x0=x0. It is easy to see that x0(t) is a nonoscillatory solution of (1.1). The proof is complete.Theorem 2.3.
Assume that-∞<c1≤P(t)≤c2<-1 and (2.2) holds. Then (1.1) has a bounded nonoscillatory solution.Proof.
We choose positive constantsM1,M2,α such that -c1M1<α<(-c2-1)M2. c=min{(α+M1c1)c2/c1,((-c2-1)M2)-α}. Choosing T>t0 sufficiently large such that
(2.18)1(n-2)!∫T∞sn-2∫s∞M′r(s)∑i=1m|Qi(u)|duds<c,
where M′=maxM1≤x≤M2{|fi(x)|:1≤i≤m}.
LetC([t0,∞),R) be the set as in the proof of Theorem 2.2. We define a bounded, closed, and convex subset Ω of C([t0,∞),R) as follows:
(2.19)Ω={x=x(t)∈C([t0,∞),R):M1≤x(t)≤M2,t≥t0}.
Define two mapsS1 and S2:Ω→C([t0,∞),R) as follows:
(2.20)(S1x)(t)={-αP(t+τ)-x(t+τ)P(t+τ),t≥T,(S1x)(T),t0≤t≤T,(S2x)(t)={(-1)n-1(n-2)!1P(t+τ)∫t+τ∞(s-t-τ)n-2∫s∞1r(s)(∑i=1mQi(u)fi(x(u-σi)))duds,t≥T,(S2x)(T),t0≤t≤T.
(i) We shall show that for anyx,y∈Ω, S1x+S2y∈Ω.
In fact, for everyx,y∈Ω, and t≥T, we get
(2.21)(S1x)(t)+(S2y)(t)≥-αc1+cc2≥M1,(S1x)(t)+(S2y)(t)≤-αc2-M2c2-cc2≤M2.
Thus, we have proved thatS1x+S2y∈Ω. Since -∞<c1≤P(t)≤c2<-1, we get that S1 is a contraction mapping. We also can prove that {S2x:x∈Ω} is uniformly bounded and equicontinuous on [t0,∞), and hence S2Ω is relatively compact. So by Lemma 2.1, there is x0∈Ω such that S1x0+S2x0=x0. That is,
(2.22)x0(t)=-αP(t+τ)-x0(t+τ)P(t+τ)+(-1)n-1(n-2)!1P(t+τ)×∫t+τ∞(s-t-τ)n-2∫s∞1r(s)(∑i=1mQi(u)fi(x0(u-σi)))duds.
It is easy to see that x0(t) is a bounded nonoscillatory solution of (1.1).
The proof is complete.Theorem 2.4.
Assume that1<c1≤P(t)≤c2<+∞ and (2.2) holds. Then (1.1) has a bounded nonoscillatory solution.Proof.
We choose positive constantsM3,M4,α such that M4+c2M3<α<c1M4. c=min{α-M4-c2M3,c1M4-α}. Choosing T>t0 sufficiently large such that
(2.23)1(n-2)!∫T∞sn-2∫s∞M′′r(s)∑i=1m|Qi(u)|duds<c,
where M′′=maxM3≤x≤M4{|fi(x)|:1≤i≤m}.
LetC([t0,∞),R) be the set as in the proof of Theorem 2.2. We define a bounded, closed, and convex subset Ω of C([t0,∞),R) as follows:
(2.24)Ω={x=x(t)∈C([t0,∞),R):M3≤x(t)≤M4,t≥t0}.
Define two mapsS1 and S2:Ω→C([t0,∞),R) as follows:
(2.25)(S1x)(t)={αP(t+τ)-x(t+τ)P(t+τ),t≥T,(S1x)(T),t0≤t≤T,(S2x)(t)={(-1)n-1(n-2)!1P(t+τ)∫t+τ∞(s-t-τ)n-2∫s∞1r(s)(∑i=1mQi(u)fi(x(u-σi)))duds,t≥T,(S2x)(T),t0≤t≤T.
(i) We shall show that for anyx,y∈Ω, S1x+S2y∈Ω.
In fact, for everyx,y∈Ω and t≥T, we get
(2.26)(S1x)(t)+(S2y)(t)≥1c2(α-M4-c)≥M3,(S1x)(t)+(S2y)(t)≤αc1+cc1≤M4.
Thus, we have proved thatS1x+S2y∈Ω. Since 1<c1≤P(t)≤c2<+∞, we get S1 is a contraction mapping. We also can prove that {S2x:x∈Ω} is uniformly bounded and equicontinuous on [t0,∞), and, hence, S2Ω is relatively compact. So by Lemma 2.1, there is x0∈Ω such that S1x0+S2x0=x0. That is,
(2.27)x0(t)=αP(t+τ)-x0(t+τ)P(t+τ)+(-1)n-1(n-2)!1P(t+τ)×∫t+τ∞(s-t-τ)n-2∫s∞1r(s)(∑i=1mQi(u)fi(x0(u-σi)))duds.
It is easy to see that x0(t) is a bounded nonoscillatory solution of (1.1).
The proof is complete.Remark 2.5.
If we letn=2 in Theorem 2.2, we get the Theorem 1 in [3]. In the case where n=2, r(t)≡1, Theorem 2.2 improves essentially Theorem 2.2 in [5].Remark 2.6.
The conditions of Theorem2.4 relaxing the hypotheses (H4) of Theorem 3 in [2].Remark 2.7.
Theorems2.3 and 2.4 improve essentially Theorem 3 in [2], we allow that Qi(t)(i=1,2,…,m) are oscillatory.



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