# Some Oscillation Results for Even Order Delay Difference Equations with a Sublinear Neutral Term

Authors: Govindasamy Ayyappan; Gunasekaran Nithyakala
Journal: Abstract and Applied Analysis
 (2018)

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## Abstract

In this paper, some new results are obtained for the even order neutral delay difference equationΔanΔm-1xn+pnxn-kα+qnxn-lβ=0, where m≥2 is an even integer, which ensure that all solutions of the studied equation are oscillatory. Our results extend, include, and correct some of the existing results. Examples are provided to illustrate the importance of the main results.


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## Body

## 1. Introduction

The aim of this paper is to investigate the oscillatory behavior of even order nonlinear neutral difference equation(1)ΔanΔm-1xn+pnxn-kα+qnxn-lβ=0,n∈Nn0where N(n0)={n0,n0+1,…},n0 is a positive integer, subject to the following conditions:(C1)
m≥2 is an even integer, and α and β are ratio of odd positive integers with 0<α≤1 and β∈(0,∞);(C2)
{an} is a positive increasing sequence of real number for all n∈N(n0);(C3)
{pn} and {qn} are positive real sequences for all n∈N(n0) with 0≤pn≤p<1;(C4)
l and k are positive integers.Letθ=max⁡{k,l}. Under a solution of (1), we mean a real sequence {xn} defined for n≥n0-θ and satisfying (1) for all n∈N(n0). As usual a solution of (1) is said to be oscillatory if it is neither eventually positive nor eventually negative; else it is nonoscillatory.In the past few years, there is a great interest in studying the oscillatory and asymptotic behavior of solutions of higher order neutral type difference equations, since such type of equations naturally arises in the applications including problems in population dynamics or in cobweb models in economics and so on. The problem of finding sufficient conditions which ensure that all solutions of the neutral type difference equations are oscillatory has been investigated by many authors; see, for example, [1–12] and the references cited therein. In all the results the neutral term is linear and few results are available when the neutral term is nonlinear; see [13–21].In [20], the authors considered (1) with α≥1 and an≡1 and established sufficient conditions for the oscillation of all solutions. In view of these facts, in this paper our purpose is to obtain sufficient conditions for the oscillation of solution of (1) when(2)∑n=n0∞1an=∞,or(3)∑n=n0∞1an<∞.Thus the results presented here extend and generalize some of the results in [13, 14, 16, 18, 19, 21], complement the results in [20], and correct some of the results in [8].


## 2. Some Preliminary Lemmas

In this section, we provide some lemmas which will be useful in proving our main results. We begin with the following lemma that can be found in [22, Theorem 41, page 39].Lemma 1.
If0<α≤1 and a>0, then aα≤αa+(1-α).Lemma 2 (Discrete Kneser’s Theorem).
Let{un} be a sequence of real number and un>0 with Δmun being of constant sign eventually and not identically zero eventually. Then there exists an integer j,0≤j≤m, with (m+j) odd for Δmun≤0 and (m+j) even for Δmun≥0 and N∈N(n0) such that(4)Δiun>0for  alli=1,2,…,jand(5)-1i+jΔiun>0for  alli=j+1,j+2,…,m-1for all n≥N.Lemma 3.
Let{un} be defined for n∈N(n0) and un>0 with Δmun≤0 for all n∈N(n0) and not identically zero. Then there exists a large n1∈N(n0) such that(6)un≥n-n1m-1m-1!Δm-1u2m-j-1n,n≥n1where j is defined in Lemma 2. Further, if {un} is increasing, then(7)un≥1m-1!n2m-1m-1Δm-1un,n≥2m-1n1.The proof of the last two lemmas can be found in [1].Next we define the sequence{zn} by(8)zn=xn+pnxn-kα.Lemma 4.
Assume condition (2) holds. Let {xn} be a positive solution of (1). Then there is an integer n1∈N(n0) such that(9)zn>0,Δzn>0,Δm-1zn>0,Δmzn≤0for all n≥n1.Proof.
The proof is similar to that of Lemma 3 of [8], and hence the details are omitted.


## 3. Oscillation Theorems

In this section, we present some sufficient conditions for the oscillation of all solutions of (1). To simplify our notation, for any positive real sequence {ρn} which is decreasing to zero, we set(10)Pn=1-αpn-1-αpnρn,Qn=qnPβn-l,and(11)An=∑s=n∞1as.Theorem 5.
Let condition (2) hold. Assume that there is a positive decreasing real sequence {ρn} tending to zero such that P(n) is positive for all n≥N∈N(n0). If(12)∑n=N∞Qn=∞,then every solution of (1) is oscillatory.Proof.
Let{xn} be a nonoscillatory solution of (1). Without loss of generality, we may assume that {xn} is a positive solution of (1). Then there exists an integer n1∈N(n0) such that xn>0,xn-k>0 and xn-l>0 for all n≥n1. From Lemma 4, we have zn>0,Δzn>0,Δm-1zn>0 and Δmzn≤0 for all n≥n1.
From the definition ofzn, we have(13)xn=zn-pnxn-kα≥zn-pnznα≥zn-αpnzn-1-αpn=1-αpnzn-1-αpn,where we have used Lemma 1. Since zn is positive and increasing and ρn is positive and decreasing to zero, there is an integer n2≥n1 such that(14)zn≥ρnfor  alln≥n2.Using (14) in (13), one obtains(15)xn≥1-αpn-1ρn1-αpnzn=Pnznand substituting this in (1) yields(16)ΔanΔm-1zn+qnPβn-lzn-lβ≤0,n≥n2.Now summing the last inequality from n2 to n-1, we obtain(17)anΔm-1zn-an2Δm-1zn2+∑s=n2n-1qsPβs-lzs-lβ≤0for all n≥n2. That is(18)∑s=n2n-1Qszs-lβ≤an2Δm-1zn2-anΔm-1zn,n≥n2.Since Δzn>0 and zn>0 eventually, there exists a positive constant M such that zn-l≥M for all n≥n2. Using this and the positivity of anΔm-1zn in (18) and letting n→∞, we obtain(19)∑n=n1∞Qn<∞which is a contradiction to (12). This completes the proof.Remark 6.
In the above theorem, we did not impose any condition onβ and hence our result is more general than some of the existing results in the literature.In the following, we present other oscillation criteria using Lemma3.Theorem 7.
Let condition (2) hold. Assume that there is a positive decreasing real sequence {ρn} tending to zero such that P(n) is positive for all n≥N∈N(n0). If(i)
(20)limn→∞inf⁡∑s=n-ln-1Qsas-ls-lm-1>1λll+1l+1forβ=1,(ii)
(21)∑n=N∞Qnan-lβn-lβm-1=∞for0<β<1,(iii)
there exists aδ>1/llog⁡β such that(22)limn→∞⁡inf⁡Qnan-lβn-lβm-1exp⁡e-δn>0forβ>1,
then every solution of (1) is oscillatory.Proof.
Let{xn} be a nonoscillatory solution of (1). Without loss of generality, we may assume that there is an integer n1∈N(n0) such that xn>0,xn-k>0 and xn-l>0 for all n≥n1. Now proceeding as in the proof of the previous theorem, we obtain (16). That is,(23)ΔanΔm-1zn+Qnzn-lβ≤0,n≥n2.Since Δm-1zn>0, Δmzn≤0 and using Lemma 3, we have from (23) that(24)ΔanΔm-1zn+Qn1m-1!n-l2m-1m-1βΔm-1zn-lβ≤0,n≥n2.Set wn=anΔm-1zn. Then wn>0 and the last inequality becomes(25)Δwn+λQnan-lβn-lβm-1wn-lβ≤0,n≥n2,where λ=1/(m-1)!1/2m-1m-1β>0. Now, using Lemma 1.1 of [20], we see that the equation(26)Δwn+λQnan-lβn-lβm-1wn-lβ=0,n≥n2has an eventually positive solution.
(i) If (20) holds, then by Theorem 7.6.1 of [23], (26) with β=1 has no positive solution, which is a contradiction.
(ii) If (21) holds, then by Theorem 1 of [24], (26) with 0<β<1 has no positive solution, which is a contradiction.
(iii) If (22) holds, then by Theorem 2 of [24], (26) with β>1 has no positive solution which is a contradiction. This completes the proof of the theorem.Theorem 8.
Assume that (3) and β=1 hold. Assume that there is a positive decreasing real sequence {ρn} tending to zero such that P(n) is positive for all n≥N∈N(n0). If (20) holds and(27)limn→∞⁡sup⁡∑s=NnMQsAs+1s-lm-2-14asAs+1=∞where M=1/(m-2)!1/2m-2m-2, then every solution of (1) either is oscillatory or tends to zero as n→∞.Proof.
Assume that (1) has a nonoscillatory solution {xn} which is eventually positive such that limn→∞xn≠0. From the definition of zn, we have zn>0 for all n≥n1∈N(n0). By virtue of (1) and Lemma 2 there are two possibilities, either(28)zn>0,Δzn>0,Δm-1zn>0,Δmzn≤0,ΔanΔm-1zn≤0or(29)zn>0,Δzn>0,Δm-2zn>0,Δm-1zn<0,ΔanΔm-1zn≤0for all n≥n1≥n0.
Case (i). Suppose conditions (28) hold for all n≥n1; then the proof for this case is similar to that of Case (i) of Theorem 7 and hence the details are omitted.
Case (ii). Assume now that conditions (29) hold for all n≥n1. Since anΔm-1zn is decreasing, then we have(30)ajΔm-1zj≤anΔm-1znforj≥n≥n1.Dividing the last inequality by aj and summing the resulting inequality from n to j-1, we obtain(31)Δm-2zj-Δm-2zn≤anΔm-1zn∑s=nj-11as.Letting j→∞, we obtain(32)0≤Δm-2zn+AnanΔm-1znforn≥n1.Define(33)wn=An1An+anΔm-1znΔm-2zn,n≥n1and then wn>0, and using (16), we have(34)Δwn=-1anAnwn+An+11anAnAn+1+ΔanΔm-1znΔm-2zn+1-anΔm-1znΔm-2znΔm-2zn+1Δm-1zn≤1-wnanAn-An+1Qnzn-lΔm-2zn+1-An+1ananΔm-1znΔm-2zn2where we have used Δm-2zn as positive and decreasing. Now using wn-1/An=anΔm-1zn/Δm-2zn in the above inequality, it follows that(35)Δwn≤1-wnanAn-An+1Qnzn-lΔm-2zn+1-An+1anAn21-wn2,n≥n1.Now from Lemma 3, we obtain(36)zn-l≥1m-2!n-l2m-2m-2Δm-2zn-l.Since Δm-1zn<0 and n-l<n+1, we have(37)Δm-2zn+1<Δm-2zn-l.Combining the inequalities (35) and (37), we have(38)Δwn≤-MAn+1Qnn-lm-2+1-wnanAn-An+1anAn21-wn2,n≥n1,where M=1/(m-2)!1/2m-2m-2. Completing the square in the above inequality, we have(39)Δwn≤-MAn+1Qnn-lm-2-An+1anAn21-wn-12AnAn+12+14anAn+1or(40)Δwn≤-MAn+1Qnn-lm-2+14anAn+1,n≥n1.By summing the last inequality from n1 to n, we obtain(41)∑s=n1nMAs+1Qss-lm-2-14asAs+1≤wn1.Taking lim sup as n→∞, in the above inequality we obtain a contradiction with (27). This completes the proof.Theorem 9.
Assume that (3) and 0<β<1 hold. Further assume that there is a positive decreasing real sequence {ρn} tending to zero such that P(n) is positive for all n≥N∈N(n0). If (21) holds and(42)limn→∞⁡sup⁡∑s=NnMβAs+1Qss-lβm-2-M11-β4asAs+1=∞for some constant M1>0, then every solution of (1) either is oscillatory or tends to zero as n→∞.Proof.
Assume that{xn} is an eventually positive solution of (1) such that limn→∞xn≠0. Proceeding as in the proof of Theorem 8, we see that {zn} satisfies two possible cases (28) and (29) for all n≥n1.
Case (i). Suppose conditions (28) hold for all n≥n1; then the proof for this case is similar to that of Case (ii) of Theorem 7 and hence the details are omitted.
Case (ii). Assume now that conditions (29) hold for all n≥n1; proceeding as in Case (ii) of Theorem 8 we have(43)Δwn≤1-wnanAn-An+1Qnzn-lβΔm-2zn+1-An+1anAn21-wn2,n≥n1.Now using (36) and (37) in (43), we obtain(44)Δwn≤-MβAn+1Qnn-lβm-2Δm-2zn-lβ-1+14anAn+1.Since {Δm-2zn} is positive and decreasing and β<1, there is a constant M1>0 such that Δm-2zn-lβ-1≥M1β-1 for all n≥n2≥n1. Using this in (44) and then summing the resulting inequality from n2 to n, we obtain(45)∑s=n2nMβAs+1Qss-lβm-2-M11-β4asAs+1≤M11-βwn2<∞.Taking lim sup as n→∞, in the above inequality, we obtain a contradiction with (42). This completes the proof.Theorem 10.
Assume that (3) and β>1 hold. Further assume that there is a positive decreasing and sequence {ρn} tending to zero such that P(n) is positive for all n≥N∈N(n0). If (22) holds and(46)limn→∞⁡sup⁡∑s=NnMβAs+1βQss-lβm-2-14M2β-1asAs+1=∞for some constant M2>0, then every solution of (1) either is oscillatory or tends to zero as n→∞.Proof.
Let us assume that{xn} is an eventually positive solution of (1) such that limn→∞xn≠0. Proceeding as in the proof of Theorem 8, we see that {zn} satisfies two possible cases (28) and (29) for all n≥n1.
Case (i). If conditions (28) hold for all n≥n1, then the proof is similar to that of Case (iii) of Theorem 7 and hence the details are omitted.
Case (ii). Assume now that conditions (29) hold for all n≥n1. Proceeding as in Case (ii) of Theorem 9, we have(47)Δwn≤-MβAn+1Qnn-lβm-2Δm-2zn-lβ-1+14anAn+1,n≥n1.Now from (32), one can see that Δm-2zn/An is nondecreasing and hence there is a constant M2>0 such that Δm-2zn/An≥M2 for all n≥n1. Using this in (47) and since β>1, we have(48)Δwn≤-MβM2β-1An+1βQnn-lβm-2+14anAn+1,n≥n1.Summing the last inequality from n1 to n, we obtain(49)∑s=n1nMβAs+1βQss-lβm-2-14M2β-1asAs+1≤wn1M2β-1.Taking lim sup as n→∞ in the above inequality, we get a contradiction with (46). This completes the proof.


## 4. Examples

In this section, we present two examples to illustrate the importance of the main results.Example 1.
Consider the neutral difference equation(50)ΔnΔm-1xn+1nxn-21/3+1nxn-13=0,n≥2,where m≥2 is an even integer. Here an=n,pn=1/n,qn=1/n,k=2,l=1,α=1/3, and β=3. By taking ρn=1/n, we see that P(n)=1/3n-1/n>0 for all n≥2. Now it is easy to see that the hypotheses (C1)-(C4) are satisfied. Also condition (12) holds and therefore, by Theorem 5, every solution of (50) is oscillatory.Example 2.
Consider the neutral difference equation(51)Δnn+1Δm-1xn+1nxn-21/3+nxn-11/3=0,n≥2,where m≥2 is an even integer. Here an=n(n+1),pn=1/n,qn=n,k=2,l=1,α=β=1/3. By taking ρn=1/n, we see that P(n)=1/3n-1/n>0 for all n≥2. Now condition (21) becomes(52)∑n=2∞nn-2/n-11/331/3n1/3n-11/3n-11/3m-1=∑n=2∞n2/3n-21/331/3n-11/3m-3=∞since m≥2. Also a simple calculation shows that An=1/n and, using this, condition (42) becomes(53)limn→∞⁡sup⁡∑s=2nM1/3ss+1s-21/3s-11/3m-3-M12/34s=∞.Thus all conditions of Theorem 9 are satisfied and hence every solution of (51) either is oscillatory or tends to zero as n→∞.


## 5. Conclusion

The results obtained in this paper extend and complement some of the results reported in the literature. Further, Theorem8, where α=1, corrects the conclusion of Theorem 4 established in [8]. The results reported in the papers [3, 4, 6–12, 17, 20] cannot be applicable to (50) and (51) to yield this conclusion since these equations have sublinear neutral terms. It would be interesting to improve Theorems 8, 9, and 10 so that all solutions are oscillatory instead of either being oscillatory or tending to zero.



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*Source: 2590158-2018-11-08.xml*