# On the Nonoscillation of Second-Order Neutral Delay Differential Equation with Forcing Term

Authors: Jin-Zhu Zhang; Zhen Jin; Tie-Xiong Su; Jian-Jun Wang; Zhi-Yu Zhang; Ju-Rang Yan
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

This paper is concerned with nonoscillation of second-order neutral delay
differential equation with forcing term. By using contraction mapping principle,
some sufficient conditions for the existence of nonoscillatory solution are established.
The criteria obtained in this paper complement and extend several known results in
the literature. Some examples illustrating our main results are given.


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## Body

## 1. Introduction

During  the  last
two decades, there has been much research activity concerning the oscillation
and nonoscillation of solutions of neutral type delay differential equations
(see [1–9]). Investigation of such
equations or systems, besides of their theoretical interest, have some
importance in modelling of the networks containing lossless transmission lines,
in the study of vibrating masses attached to an elastic bar and also in
population dynamics, and so forth (see [1, 2, 8, 10]
and the references cited therein).In this paper, we consider the second-order neutral
delay differential equation with forcing term of the form(1.1)[x(t)+P(t)x(t−τ)]′′+Q1(t)x(t−σ1)−Q2(t)x(t−σ2)=h(t),t≥t0,where(1.2)τ>0,σ1≥0,σ2≥0,P,Q1,Q2,h∈C([t0,∞),ℝ).Letφ∈C([t0−σ,t0),ℝ),
where σ=max{τ,σ1,σ2},
be a given function and let x0 be a given constant. By the method of steps (see [2]), it
is easy to know that (1.1) has a unique solution x∈C([t0−σ,∞),ℝ) in the sense that x(t)+P(t)x(t−τ) is twice continuously differential for t≥t0, x(t) satisfies (1.1) and(1.3)x(s)=φ(s)fors∈[t0−σ,t0],[x(t)+P(t)x(t−τ))]t=t0′=x0.As  is customary, a solution of (1.1) is called
oscillatory if it has arbitrarily large zeros, and otherwise it is
nonoscillatory. Equation  (1.1) is oscillatory if all its solutions are oscillatory.WhenP(t)=p and the forcing term h(t)≡0, (1.1) reduces to(1.4)[x(t)+px(t−τ)]′′+Q1(t)x(t−σ1)−Q2(t)x(t−σ2)=0,t≥t0, where p∈ℝ(p≠±1) and ∫∞tQi(t)dt<∞,i=1,2.
The first global result of (1.4) (with respect to p),  which is a sufficient condition for the
existence of a nonoscillatory solution for all values of p≠±1,
have been examined by Kulenović and Hadžiomerspahić
[4].Recently, Parhi and Rath [7] studied oscillation
behaviors for forced first-order neutral differential equations as
follows (1.5)[x(t)−P(t)x(t−τ)]′+Q(t)G(x(t−σ))=h(t),t≥t0. Necessary and sufficient
conditions are obtained in various ranges for P(t)≠±1 so that every solution of (1.5) is
oscillatory or tends to zero or to ±∞ as t→∞.Motivated by the idea of [4, 7], in present paper we
establish sufficient conditions for existence of a nonoscillatory solution to (1.1) depending on various ranges of P(t)≠±1.
Hereinafter, we assume that the following conditions hold,(H1)
Qi≥0, and ∫∞tQi(t)dt<∞,i=1,2.(H2)
There exists a functionH(t)∈C2([t0,∞),ℝ) such that H′′(t)=h(t) and limt→∞H(t)=M∈ℝ.


## 2. Main results

Theorem 2.1.
Suppose
that conditions(H1) and (H2) hold. If P(t) is in one of the following ranges: (2.1)(i)0<P(t)≤p1<1,(ii)1<p2≤P(t)≤p1,(iii)−1<−p2≤P(t)<0,(iv)−p2≤P(t)≤−p1<−1, then (1.1) has a
nonoscillatory solution.Proof.
The proof of
this theorem will be divided into four cases in terms of the four different
ranges ofP(t).
Case (i) (0<P(t)≤p1<1).
Choose at1>t0+σ sufficiently large such that (2.2)∫t1∞s[Q1(s)+Q2(s)]ds<3(1−p1)4,(2.3)∫t1∞sQ1(s)ds≤p1+N1−1N1,(2.4)∫t1∞sQ2(s)ds≤1−p1(1+2N1)−2M12N1,(2.5)|H(t)−M|≤1−p14, where M1 and N1 are positive constants such that(2.6)1−N1<p1<1−2M11+2N1.
LetX be the set of all continuous and bounded
functions on [t0,∞) with the sup norm. Set(2.7)A={x∈X:M1≤x(t)≤N1,t≥t0}.Define a mapping T:A→X as follows:(2.8)(Tx)(t)={3(1−p1)4−P(t)x(t−τ)+t∫t∞[Q1(s)x(s−σ1)−Q2(s)x(s−σ2)]ds+∫t1ts[Q1(s)x(s−σ1)−Q2(s)x(s−σ2)]ds+H(t)−M,t≥t1(Tx)(t1),t0≤t≤t1.
Clearly,Tx is continuous. For every x∈A and t≥t1,
using (2.3) and (2.5),
we get (2.9)(Tx)(t)≤1−p1+N1∫t1∞sQ1(s)ds≤N1. Furthermore, from (2.4) and (2.5),
we have(2.10)(Tx)(t)≥1−p12−p1N1−N1∫t1∞sQ2(s)ds≥M1.Thus we prove that TA⊂A.
To apply the contraction principle, we need to prove that T is a contraction mapping on A since A is a bounded, closed, and convex subset of X.
Now, forx1,x2∈A and t≥t1 we have(2.11)|(Tx1)(t)−(Tx2)(t)|≤∥x1−x2∥{p1+∫t1∞s[Q1(s)+Q2(s)]ds}=q1∥x1−x2∥,where we used sup norm. From (2.2),
we obtain q1<1,
which completes the proof of Case (i).
Example 2.2.
Consider the second-order neutral delay
differential equation(2.12)(x(t)+e−tx(t−1))′′+e−t−1x(t−1)−4e−tx(t−1)=h(t),t≥1,where P(t)=e−t,Q1(t)=e−t−1,Q2(t)=4e−t,h(t)=e−t+e−2t such that 0<P(t)≤e−1<1. Since H(t)=e−t+1/4e−2t→0 as t→∞, ∫t1∞sQ1(s)ds=2e−t,
and ∫t1∞sQ2(s)ds=8e−t,
then the sufficient conditions—in Case (i) of Theorem 2.1—are satisfied.
Therefore, the equation has a positive solution. In fact y=e−t is a positive solution of this equation.
Case (ii) (1<p2≤P(t)≤p1).
Choose at1>t0+σ sufficiently large such that (2.13)∫t1∞s[Q1(s)+Q2(s)]ds<3(p2−1)4,(2.14)∫t1∞sQ1(s)ds≤1−p2(1−N2)N2,(2.15)∫t1∞sQ2(s)ds≤p2(p2−1)−2p1(N2+p2M2)2p1N2,(2.16)|H(t)−M|≤p2−14,where M2 and N2 are positive constants such that(2.17)1−1p2<N2<p2(p2−1−2p1M2)2p1.
LetX be the set as in Case (i). Set (2.18)A={x∈X:M2≤x(t)≤N2,t≥t0}.Define a mapping T:A→X as follows:(2.19)(Tx)(t)={p2−14P(t+τ)−x(t+τ)P(t+τ)+t+τP(t+τ)∫t+τ∞[Q1(s)x(s−σ1)−Q2(s)x(s−σ2)]ds+1P(t+τ)∫t1t+τs[Q1(s)x(s−σ1)−Q2(s)x(s−σ2)]ds+H(t+τ)−MP(t+τ),t≥t1(Tx)(t1),t0≤t≤t1.
Clearly,Tx is continuous. For every x∈A and t≥t1,
using (2.14) and (2.16),
we get(2.20)(Tx)(t)≤1−1p2+N2p2∫t1∞sQ1(s)ds≤N2.Furthermore, from (2.15) and (2.16) we have(2.21)(Tx)(t)≥p2−12p1−N2p2−N2p2∫t1∞sQ2(s)ds≥M2.Thus we prove that TA⊂A.
To apply the contraction principle, we need to prove that T is a contraction mapping on A since A is a bounded, closed, and convex subset of X.
Now, forx1,x2∈A and t≥t1,
we have(2.22)|(Tx1)(t)−(Tx2)(t)|≤1p2∥x1−x2∥{1+∫t1∞s[Q1(s)+Q2(s)]ds}=q2∥x1−x2∥,where we used sup norm. From (2.13),
we obtain q2<1,
which completes the proof of Case (ii).
Example 2.3.
Consider the second-order neutral delay
differential equation(2.23)(x(t)+(2+e−t)x(t−1))′′+e−t−1x(t−1)−4e−tx(t−1))=h(t),t≥1,where P(t)=2+e−t,Q1(t)=e−t−1,Q2(t)=4e−t,h(t)=e−t+2e−t+1+e−2t such that P(t)≥2+e−1>1. Since H(t)=e−t+1/4e−2t+2e−t+1→0 as t→∞, ∫t1∞sQ1(s)ds=2e−t,
and ∫t1∞sQ2(s)ds=8e−t,
then the sufficient conditions—in Case (ii) of Theorem 2.1—are satisfied.
Therefore, the equation has a positive solution. In fact y=e−t is a positive solution of this equation.
Case (iii) (−1<−p2≤P(t)<0).
Choose at1>t0+σ sufficiently large such that (2.24)∫t1∞s[Q1(s)+Q2(s)]ds<3(1−p2)4,(2.25)∫t1∞sQ1(s)ds≤(1−p2)(N3−1)N3,(2.26)∫t1∞sQ2(s)ds≤(1−p2)−2M32N3,(2.27)|H(t)−M|≤1−p24,where M3 and N3 are positive constants such that(2.28)2M3+p2<1<N3.
LetX be the set as in Case (i). Set (2.29)A={x∈X:M3≤x(t)≤N3,t≥t0}.Define a mapping T:A→X as follows:(2.30)(Tx)(t)={3(1−p2)4−P(t)x(t−τ)+t∫t1∞[Q1(s)x(s−σ1)−Q2(s)x(s−σ2)]ds+∫t1ts[Q1(s)x(s−σ1)−Q2(s)x(s−σ2)]ds+H(t)−M,t≥t1(Tx)(t1),t0≤t≤t1.
Clearly,Tx is continuous. For every x∈A and t≥t1,
using (2.25) and (2.27),
we get(2.31)(Tx)(t)≤1−p2+p2N3+N3∫t1∞sQ1(s)ds≤N3.Furthermore, from (2.26) and (2.27),
we have(2.32)(Tx)(t)≥1−p22−N3∫t1∞sQ2(s)ds≥M3.Thus we prove that TA⊂A.
To apply the contraction principle, we need to prove that T is a contraction mapping on A since A is a bounded, closed, and convex subset of X.
Now, forx1,x2∈A and t≥t1,
we have(2.33)|(Tx1)(t)−(Tx2)(t)|≤∥x1−x2∥{p2+∫t1∞s[Q1(s)+Q2(s)]ds}=q3∥x1−x2∥,where we used sup norm. From (2.24),
we obtain q3<1,
which completes the proof of Case (iii).
Example 2.4.
Consider the second-order neutral delay
differential equation(2.34)(x(t)−e−tx(t−1))′′+e−t−1x(t−1)−4e−tx(t−1)=h(t),t≥1,where h(t)=e−t+e−2t−8e−2t+1.
This equation has a nonoscillatory solution y=e−t since the sufficient conditions—in Case
(iii) of Theorem 2.1—are satisfied.
Case (iv) (−p2≤P(t)≤−p1<−1).
Choose at1>t0+σ sufficiently large such that(2.35)∫t1∞s[Q1(s)+Q2(s)]ds<3(p1−1)4,(2.36)∫t1∞sQ1(s)ds≤N4(p1−1)−p1N4,(2.37)∫t1∞sQ2(s)ds≤p2−p1(1+M4)(p2−1)p2N4,(2.38)|H(t)−M|≤p1−1,where M4 and N4 are positive constants such that(2.39)N4>p1p1−1,M4<p2−p1(p2−1)p1(p2−1).
LetX be the set as in Case (i). Set(2.40)A={x∈X:M4≤x(t)≤N4,t≥t0}.Define a mapping T:A→X as follows(2.41)(Tx)(t)={−1P(t+τ)−x(t+τ)P(t+τ)+t+τP(t+τ)∫t+τ∞[Q1(s)x(s−σ1)−Q2(s)x(s−σ2)]ds+1P(t+τ)∫t1t+τs[Q1(s)x(s−σ1)−Q2(s)x(s−σ2)]ds+H(t+τ)−MP(t+τ),t≥t1(Tx)(t1),t0≤t≤t1.
Clearly,Tx is continuous. For every x∈A and t≥t1,
using (2.37) and (2.38),
we get(2.42)(Tx)(t)≤−1+x(t+τ)p(t+τ)−1p(t+τ)∫t1∞sQ2(s)x(s−σ2)ds+H(t+τ)−Mp(t+τ)≤1+N4p1+N4p1∫t1∞sQ2(s)ds+p1−1p1≤N4.Furthermore, from
(2.36) and (2.38),
we have(2.43)(Tx)(t)≥−1+x(t+τ)p(t+τ)+1p(t+τ)∫t1∞sQ1(s)x(s−σ1)ds+H(t+τ)−Mp(t+τ)≥1+M4p2−N4p1∫t1∞sQ1(s)ds−p1−1p1≥M4.Thus we prove that TA⊂A.
To apply the contraction principle, we need to prove that T is a contraction mapping on A since A is a bounded, closed, and convex subset of X.
Now, forx1,x2∈A and t≥t1,
we have(2.44)|(Tx1)(t)−(Tx2)(t)|≤1p1∥x1−x2∥{1+∫t1∞s[Q1(s)+Q2(s)]ds}=q4∥x1−x2∥,where we used sup norm. From (2.35),
we obtain q4<1,
which completes the proof of Case (iv).
Example 2.5.
Consider the second-order neutral delay
differential equation(2.45)(x(t)−(1+e−t)x(t−1))′′+e−t−1x(t−1)−e−t−1x(t−1)=h(t),t≥1,where h(t)=e−t+4e−2t+1−e−t+1.
This equation has a nonoscillatory solution y=e−t since the sufficient conditions—in Case (iv)
of Theorem 2.1—are satisfied.



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