# New Oscillation Criteria for Third-Order Nonlinear Mixed Neutral Difference Equations

Authors: Elmetwally Mohammed Elabbasy; Magdy Yoseph Barsom; Faisal Saleh AL-dheleai
Journal: Chinese Journal of Mathematics
 (2014)

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## Abstract

Some new oscillation criteria are established for a third-order nonlinear mixed neutral difference equation. Our results improve and extend some known results in the literature. Several examples are given to illustrate the importance of the results.


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## Body

## 1. Introduction

By a Riccati transformation technique, we present some new oscillation criteria for the nonlinear difference equation of the form(1)Δ(a(n)Δ(b(n)Δ(x(n)+p1(n)x(n-σ1)+p2(n)x(n+σ2))))+q1(n)x(n-σ3)+q2(n)x(n+σ4)=0,n≥n0,
where n0∈N is a fixed integer and Δ denotes the forward difference operator defined by Δx(n)=x(n+1)-x(n) and Δix(n)=Δ(Δi-1x(n)). Throughout this paper, we will assume the following hypotheses: (A1)
a(n), b(n)>0 for n∈N(n0),(A2)
pi(n) is positive, 0≤pi(n)≤pi, where pi are constants for i=1,2,(A3)
qi(n)>0 on N(n0) for i=1,2,(A4)
σi≥0 are constants for i=1,2,3,4.We set z(n):=x(n)+p1(n)x(n-σ1)+p2(n)x(n+σ2). By a solution of (1), we mean a nontrivial sequence {x(n)} defined on N(n0), which satisfies (1) for all n≥n0. A solution {x(n)} of (1) is said to be oscillatory if it is neither eventually positive nor eventually negative and nonoscillatory otherwise. Equation (1) is called oscillatory if all its solutions are oscillatory. In recent years, there has been an increasing interest in the study of the oscillatory behavior of solutions of difference equations; we refer to the books [1–5] and the papers [6–8] and the references cited therein.In [9], via comparison with first-order oscillatory difference equations, Agarwal et al. proved that several theorems provided sufficient conditions for oscillation of all solutions of the third-order difference equation of the form
(2)Δ(1a2(n)(Δ(1a1(n))(Δx(n))α1)α2)±q(n)f(x(g(n)))=0,
depending on the condition
(3)∑∞(ai(n))1/αi=∞,i=1,2.In [10], by a Riccati transformation technique, Schmeidel studied the oscillatory and asymptotic behavior of solutions of the third-order difference equation
(4)Δ(a(n)Δ(b(n)(Δy(n))))+p(n)f(y(n+l))=0,
using the condition
(5)∑n=1∞a-1(n)=∞,∑n=1∞b-1(n)=∞.In [11], by the generalized Riccati transformation technique, Saker et al. established some new oscillation criteria for a certain class of third-order nonlinear delay difference equation of the form
(6)Δ(c(n)Δ(d(n))(Δx(n))γ)+q(n)f(x(n-σ))=0.In [12], by a Riccati transformation technique, Selvaraj et al. established some sufficient conditions for oscillation of all solutions of the third-order nonlinear difference equation of the form
(7)Δ(1a(n)Δ2y(n))+p(n)f(y(σ(n)))=0,
using the condition
(8)∑n=n0∞a(n)=∞.In [13], by means of a Riccati transformation technique, Thandapani et al. discussed the oscillatory behavior of the solutions of the difference equation of the form
(9)Δ(a(n)(Δ2(x(n)+p(n)x(n-δ)))α)+q(n)xα(n-τ)=0,
using the condition
(10)∑n=n0∞(a(n))-1/α=∞.In [14], by using condition (5) and a Riccati transformation technique, Aktas et al. obtained some new oscillation criteria of third-order nonlinear delay difference equation of the form
(11)Δ(a(n)Δ(b(n))(Δx(n)))+p(n)Δx(n+1)+q(n)f(x(n-σ))=0.In [15], using condition (8), Selvaraj et al. considered nonlinear third-order difference equations of the form
(12)Δ2(1a(n)Δx(n))+q(n)f(x(σ(n)))=0,
and they studied the oscillatory behavior of solutions of (12).In [16], Saker investigated the third-order difference equation
(13)Δ(a(n)(Δ(b(n))(Δx(n)))γ)+q(n)f(x(g(n)))=0,
using the condition
(14)∑n=n0∞a-γ(n)=∞,∑n=n0∞b-1(n)=∞.In [17], under condition (5), Selvaraj et al. discussed the oscillatory behavior of neutral delay difference equation of the form
(15)Δ(a(n)Δ(b(n)Δ(x(n)+c(n)x(n-k))))+q(n+1)f(x(n+1-l))=0.In [18], by comparison with some first difference equations whose oscillatory characters are known and by means of a Riccati transformation technique, Elabbasy et al. obtained several new sufficient conditions for the oscillation of all solutions of the nonlinear difference equation with Delay of the form
(16)Δ(a(n)(Δ(b(n)(Δx(n))α1))α2)+∑i=1nqi(n)f(x(gi(n)))=0.In this paper, the details of the proofs of results for nonoscillatory solutions will be carried out only for eventually positive solutions, since the arguments are similar for eventually negative solutions. The paper is organized as follows. In Section2, we will state and prove the main oscillation theorems and in Section 3 will we provide some examples to illustrate the main results.


## 2. Main Results

In this section, we establish some new oscillation criteria for (1) under the following condition
(17)∑n=n0∞a-1(n)=∑n=n0∞b-1(n)=∞.In the following results, we will use the following notations:(18)Q1(n)≔min⁡{q1(n),q1(n-σ1),q1(n+σ2)},Q2(n)≔min⁡{q2(n),q2(n-σ1),q2(n+σ2)},Q(n)=Q1(n)+Q2(n),δ(n)≔∑v=n∞1a(v),φ1(n)≔ρ(n)ρ2(n+1)b(n-σ3)∑s=n2n-σ3-1a-1(s),φ2(n):=ρ(n)ρ2(n+1)b(n-σ1)∑s=n2n-σ1-1a-1(s),ϑ(m,n)≔(Δρ(n)ρ(n+1)-h(m,n)H(m,n)).
We assume that there exists a double sequence {H(m,n)∣m≥n≥0} and h(m,n) such that(i)
H(m,m)=0 for m≥0,(ii)
H(m,n)>0 for m>n>0,(iii)
Δ2H(m,n)=H(m,n+1)-H(m,n)≤0 for m>n≥0,(iv)
h(m,n)=-Δ2H(m,n)/H(m,n).We begin with these lemmas, which will be used in obtaining our main results. The proof of the following lemmas are similar to that of Lemmas 2.1, 2.2, and 2.3, respectively, in [18], and hence the details are omitted.Lemma 1.
Let{x(n)} be an eventually positive solution of (1) and suppose that {z(n)} satisfies z(n)>0, Δ(b(n)Δz(n))>0, and Δ(a(n)Δ(b(n)Δz(n)))≤0, for all n≥N.Then there exists n≥n1≥n2 such that
(19)Δz(n)≥b-1(n)(a(n)Δ(b(n)Δz(n)))∑s=n2n-1a-1(s).Lemma 2.
Assume that (17) holds. Let {x(n)} be an eventually positive solution of (1). Then for sufficiently large n, there are only two possible cases: (I)
Δz(n)>0, Δ(b(n)(Δz(n)))>0,or(II)
Δz(n)<0,Δ(b(n)(Δz(n)))>0.Lemma 3.
Assume that (17) holds. Let {x(n)} be an eventually positive solution of (1) and suppose that (II) of Lemma 2 holds. If
(20)∑v=n0∞(b-1(v)(∑u=n0v-1a-1(u)×(∑s=n2u-1(q1(s)+q2(s)))))=∞,
then x(n)→0 as n→∞.Next, we state and prove the main theorems.Theorem 4.
Assume that (20) holds. Further, assume that σ1≤σ3 and there exists a positive nondecreasing sequence {ρ(n)}, such that
(21)∑s=n0n-1((Δρ(s))2b(s-σ3)ρ(s)∑u=n2s-σ3-1a-1(u)ρ(s)Q(s)-(1+p1+p2)4(Δρ(s))2b(s-σ3)ρ(s)∑u=n2s-σ3-1a-1(u))=∞.
Then every solution of (1) is oscillatory or lim⁡n→∞⁡x(n)=0.Proof.
Assume that (1) has a nonoscillatory solution, say x(n)>0, x(n-σ1)>0, x(n+σ2)>0, x(n-σ3)>0, and x(n+σ4)>0 for all n≥n0. From (1), we see that z(n)>x(n)>0 and
(22)Δ(a(n)Δ(b(n)Δz(n)))=-q1(n)x(n-σ3)-q2(n)x(n+σ4)≤0.
Then, a(n)Δ(b(n)Δz(n)) is nonincreasing sequence and thus Δz(n) and Δ(b(n)Δz(n)) are eventually of one sign. By Lemma 2, there exist two possible cases (I) and (II). Assume that (I) holds. From (1) and the definition of z(n), we have
(23)Δ(a(n)Δ(b(n)Δz(n)))+q1(n)x(n-σ3)+q2(n)x(n+σ4)+p1Δ(a(n-σ1)Δ(b(n-σ1)Δz(n-σ1)))+p1q1(n-σ1)x(n-σ1-σ3)+p1q2(n-σ1)x(n+σ4-σ1)+p2Δ(a(n+σ2)Δ(b(n+σ2)Δz(n+σ2)))+p2q1(n+σ2)x(n+σ2-σ3)+p2q2(n+σ2)x(n+σ2+σ4)=0.
Thus,
(24)Δ(a(n)Δ(b(n)Δz(n)))+p1Δ(a(n-σ1)Δ(b(n-σ1)Δz(n-σ1)))+p2Δ(a(n+σ2)Δ(b(n+σ2)Δz(n+σ2)))+Q1(n)z(n-σ3)+Q2(n)z(n+σ4)≤0.
It follows from Δz(n)>0 that z(n+σ4)≥z(n-σ3). Thus, by (24), we obtain
(25)Δ(a(n)Δ(b(n)Δz(n)))+p1Δ(a(n-σ1)Δ(b(n-σ1)Δz(n-σ1)))+p2Δ(a(n+σ2)Δ(b(n+σ2)Δz(n+σ2)))+Q(n)z(n-σ3)≤0.
Define a Riccati substitution
(26)ω1(n)≔ρ(n)a(n)Δ(b(n)Δz(n))z(n-σ3).
Then ω1(n)>0. From (26), we have
(27)Δω1(n)=Δρ(n)a(n+1)(Δ(b(n+1)Δz(n+1)))z(n+1-σ3)+ρ(n)Δ(a(n)(Δ(b(n)Δz(n))))z(n-σ3)-ρ(n)a(n+1)Δ(b(n+1)Δz(n+1))Δ(z(n-σ3))z(n-σ3)z(n+1-σ3).
From Lemma 1,Δ(a(n)(Δ(b(n)Δz(n))))≤0, and n-σ3<n, we get
(28)Δ(z(n-σ3))≥b-1(n-σ3)(a(n-σ3)(Δ(b(n-σ3)Δz(n-σ3))))×∑s=n2n-σ3-1a-1(s)≥b-1(n-σ3)(a(n)(Δ(b(n)Δz(n))))×∑s=n2n-σ3-1a-1(s)≥b-1(n-σ3)(a(n+1)(Δ(b(n+1)Δz(n+1))))×∑s=n2n-σ3-1a-1(s).
From (27) and (28), we obtain
(29)Δω1(n)≤Δρ(n)ρ(n+1)ω1(n+1)+ρ(n)Δ(a(n)Δ(b(n)Δz(n)))z(n-σ3)-ρ(n)ρ2(n+1)b(n-σ3)(ω1(n+1))2×∑s=n2n-σ3-1a-1(s).
Next, define another sequence ω2(n) by
(30)ω2(n)≔ρ(n)a(n-σ1)Δ(b(n-σ1)Δz(n-σ1))z(n-σ3).
Then ω2(n)>0. From (30), we have
(31)Δω2(n)=Δρ(n)ρ(n+1)ω2(n+1)+ρ(n)Δ(a(n-σ1)Δ(b(n-σ1)Δz(n-σ1)))z(n-σ3)-ρ(n)((z(n+1-σ1)z(n-σ3))-1a(n+1-σ1)(Δ(b(n+1-σ1)Δz(n+1-σ1)))×Δ(z(n-σ3))×(z(n+1-σ1)z(n-σ3))-1).
From Lemma 1, and σ3≥σ1, we get
(32)Δ(z(n-σ3))≥b-1(n-σ3)(a(n-σ3)(Δ(b(n-σ3)Δz(n-σ3))))×∑s=n2n-σ3-1a-1(s)≥(a(n-σ1)(Δ(b(n-σ1)Δz(n-σ1))))×b-1(n-σ3)∑s=n2n-σ3-1a-1(s)≥(a(n+1-σ1)(Δ(b(n+1-σ1)Δz(n+1-σ1))))×b-1(n-σ3)∑s=n2n-σ3-1a-1(s).
Then from (30), (31), and the above inequality, we have
(33)Δω2(n)≤ρ(n)Δ(a(n-σ1)Δ(b(n-σ1)Δz(n-σ1)))z(n-σ3)+Δρ(n)ρ(n+1)ω2(n+1)-ρ(n)ρ2(n+1)b(n-σ3)(ω2(n+1))2∑s=n2n-σ3-1a-1(s).
Similarly, we define another sequence ω3(n) by
(34)ω3(n)≔ρ(n)a(n+σ2)Δ(b(n+σ2)Δz(n+σ2))z(n-σ3).
Then ω3(n)>0. From (34), we have
(35)Δω3(n)=Δρ(n)ρ(n+1)ω3(n+1)+ρ(n)Δ(a(n+σ2)Δ(b(n+σ2)Δz(n+σ2)))z(n-σ3)-ρ(n)(Δ(z(n-σ3))×(z(n+1-σ3)z(n-σ3))-1a(n+1+σ2)×(Δ(b(n+1+σ2)Δz(n+1+σ2)))×Δ(z(n-σ3))×(z(n+1-σ3)z(n-σ3))-1).
From Lemma 1 and n-σ3<n+σ2, we get
(36)Δz(n-σ3)≥b-1(n-σ3)×(a(n-σ3)(Δ(b(n-σ3)Δz(n-σ3))))×∑s=n2n-σ3-1a-1(s)≥(a(n+σ2)(Δ(b(n+σ2)Δz(n+σ2))))×b-1(n-σ3)∑s=n2n-σ3-1a-1(s)≥(a(n+1+σ2)×(Δ(b(n+1+σ2)Δz(n+1+σ2))))×b-1(n-σ3)∑s=n2n-σ3-1a-1(s).
Then from (34), (35), and the above inequality, we have
(37)Δω3(n)≤Δρ(n)ρ(n+1)ω3(n+1)+ρ(n)Δ(a(n+σ2)Δ(b(n+σ2)Δz(n+σ2)))z(n-σ3)-ρ(n)ρ2(n+1)b(n-σ3)(ω3(n+1))2∑s=n2n-σ3-1a-1(s).
From (25), (29), (33), and (37), we obtain
(38)Δω1(n)+p1Δω2(n)+p2Δω3(n)≤-ρ(n)Q(n)+Δρ(n)ρ(n+1)ω1(n+1)-ρ(n)(ω1(n+1))2ρ2(n+1)b(n-σ3)∑s=n2n-σ3-1a-1(s)+p1Δρ(n)ρ(n+1)ω2(n+1)-p1ρ(n)(ω2(n+1))2ρ2(n+1)b(n-σ3)∑s=n2n-σ3-1a-1(s)+p2Δρ(n)ρ(n+1)ω3(n+1)-p2ρ(n)ρ2(n+1)b(n-σ3)×(ω3(n+1))2∑s=n2n-σ3-1a-1(s).
Using (38) and the inequality
(39)Bu-Au2≤B24A,A>0,
we have
(40)Δω1(n)+p1Δω2(n)+p2Δω3(n)≤-ρ(n)Q(n)+14(Δρ(n))2b(n-σ3)ρ(n)∑s=n2n-σ3-1a-1(s)+p14(Δρ(n))2b(n-σ3)ρ(n)∑s=n2n-σ3-1a-1(s)+p24(Δρ(n))2b(n-σ3)ρ(n)∑s=n2n-σ3-1a-1(s).
Summing the last inequality from n3 to n-1, we obtain
(41)∑s=n3n-1((Δρ(s))2b(s-σ3)ρ(s)∑u=n2s-σ3-1a-1(u)ρ(s)Q(s)-(1+p1+p2)4(Δρ(s))2b(s-σ3)ρ(s)∑u=n2s-σ3-1a-1(u))≤ω1(n3)+p1ω2(n3)+p2ω3(n3),
which yields
(42)∑s=n3n-1((1+p1+p2)4(Δρ(s))2b(s-σ3)ρ(s)∑u=n2s-σ3-1a-1(u)ρ(s)Q(s)-(1+p1+p2)4(Δρ(s))2b(s-σ3)ρ(s)∑u=n2s-σ3-1a-1(u))≤c1,
where c1>0 is a finite constant. But this contradicts (21). Next we assume that (II) holds. We are then back to the proof of Lemma 3 to show that lim⁡n→∞⁡x(n)=0. The proof is complete.Corollary 5.
Assume that all the assumptions of Theorem4 hold, except that condition (21) is replaced by
(43)lim⁡n→∞⁡sup⁡∑s=n0n-1ρ(s)Q(s)=∞,lim⁡n→∞⁡sup⁡∑s=n0n-1(Δρ(s))2b(s-σ3)ρ(s)∑u=n2s-σ3-1a-1(u)<∞.
Then every solution of (1) is oscillatory or lim⁡n→∞⁡x(n)=0.Remark 6.
Note that, from Theorem4, we can obtain different conditions for oscillation of all solutions of (1) by different choices of {ρ(n)}. Let ρ(n)=1,ρ(n)=n and ρ(n)=nλ, n≥n0, and λ>1 is a constant. By Theorem 4, we have the following results.Corollary 7.
Assume that all the assumptions of Theorem4 hold, except that condition (21) is replaced by
(44)lim⁡n→∞⁡sup⁡∑s=n0n-1Q(s)=∞.
Then every solution of (1) is oscillatory or lim⁡n→∞⁡x(n)=0.Corollary 8.
Assume that all the assumptions of Theorem4 hold, except that condition (21) is replaced by
(45)lim⁡n→∞⁡sup⁡∑s=n0n-1(b(s-σ3)s∑u=n2s-σ3-1a-1(u)sQ(s)-(1+p1+p2)4×b(s-σ3)s∑u=n2s-σ3-1a-1(u))=∞.
Then every solution of (1) is oscillatory or lim⁡n→∞⁡x(n)=0.Corollary 9.
Assume that all the assumptions of Theorem4 hold, except that condition (21) is replaced by
(46)lim⁡n→∞⁡sup⁡∑l=n0n(((l+1)λ-lλ)2b(l-σ3)lλ∑u=n2l-σ3-1a-1(u)lλQ(l)-(1+p1+p2)4×((l+1)λ-lλ)2b(l-σ3)lλ∑u=n2l-σ3-1a-1(u))=∞.
Then every solution of (1) is oscillatory or lim⁡n→∞⁡x(n)=0.Theorem 10.
Assume that (20) holds. Let σ1≤σ3 and {ρ(n)} be a positive sequence. Furthermore, one assumes that there exists a double sequence {H(m,n)∣m≥n≥0}. If
(47)lim⁡m→∞⁡sup⁡1H(m,0)×∑n=0m-1(-(1+p1+p2)ϑ2(m,n)H(m,n)4φ1(n)H(m,n)ρ(n)Q(n)-(1+p1+p2)ϑ2(m,n)H(m,n)4φ1(n))=∞,
then every solution of (1) is oscillatory or lim⁡n→∞⁡x(n)=0.Proof.
Proceeding as in the proof of Theorem4, we assume that (1) has a nonoscillatory solution, say x(n)>0, x(n-σ1)>0, x(n+σ2)>0, x(n-σ3)>0, and x(n+σ4)>0 for all n≥n0. By Lemma 2, there are two possible cases. If (I) holds, from the proof of Theorem 4, we find that (38) holds for all n≥n1. From (38), we have
(48)ρ(n)Q(n)≤-Δω1(n)-p1Δω2(n)-p2Δω3(n)+Δρ(n)ρ(n+1)ω1(n+1)-φ1(n)(ω1(n+1))2+p1Δρ(n)ρ(n+1)ω2(n+1)-p1φ1(n)(ω2(n+1))2+p2Δρ(n)ρ(n+1)ω3(n+1)-p2φ1(n)(ω3(n+1))2.
Therefore, we have
(49)∑n=km-1H(m,n)ρ(n)Q(n)≤-∑n=km-1H(m,n)Δω1(n)-p1∑n=km-1H(m,n)Δω2(n)-p2∑n=km-1H(m,n)Δω3(n)+∑n=km-1H(m,n)Δρ(n)ρ(n+1)ω1(n+1)-∑n=km-1H(m,n)φ1(n)(ω1(n+1))2+p1∑n=km-1H(m,n)Δρ(n)ρ(n+1)ω2(n+1)-p1∑n=km-1H(m,n)φ1(n)(ω2(n+1))2+p2∑n=km-1H(m,n)Δρ(n)ρ(n+1)ω3(n+1)-p2∑n=km-1H(m,n)φ1(n)(ω3(n+1))2,
which yields after summing by parts
(50)∑n=km-1H(m,n)ρ(n)Q(n)≤H(m,k)ω1(k)+∑n=km-1ϑ(m,n)H(m,n)ω1(n+1)-∑n=km-1H(m,n)φ1(n)(ω1(n+1))2+p1H(m,k)ω2(k)+p1∑n=km-1ϑ(m,n)H(m,n)ω2(n+1)-p1∑n=km-1H(m,n)φ1(n)(ω2(n+1))2+p2H(m,k)ω3(k)+p2∑n=km-1ϑ(m,n)H(m,n)ω3(n+1)-p2∑n=km-1H(m,n)φ1(n)(ω3(n+1))2.
By completing the square, we have
(51)∑n=km-1H(m,n)ρ(n)Q(n)≤H(m,k)ω1(k)+∑n=km-1ϑ2(m,n)H(m,n)4φ1(n)-∑n=km-1[H(m,n)φ1(n)H(m,n)φ1(n)ω1(n+1)-ϑ(m,n)2H(m,n)φ1(n)]2+p1H(m,k)ω2(k)+p1∑n=km-1ϑ2(m,n)H(m,n)4φ1(n)-p1∑n=km-1[H(m,n)φ1(n)H(m,n)φ1(n)ω2(n+1)-ϑ(m,n)2H(m,n)φ1(n)]2+p2H(m,k)ω3(k)+p2∑n=km-1ϑ2(m,n)H(m,n)4φ1(n)-p2∑n=km-1[H(m,n)φ1(n)H(m,n)φ1(n)ω3(n+1)-ϑ(m,n)2H(m,n)φ1(n)]2.
Then,
(52)∑n=km-1(ϑ2(m,n)H(m,n)4φ1(n)H(m,n)ρ(n)Q(n)-(1+p1+p2)ϑ2(m,n)H(m,n)4φ1(n))≤H(m,k)ω1(k)+p1H(m,k)ω2(k)+p2H(m,k)ω3(k),
which implies
(53)∑n=km-1(ϑ2(m,n)H(m,n)4φ1(n)H(m,n)ρ(n)Q(n)-(1+p1+p2)ϑ2(m,n)H(m,n)4φ1(n))≤H(m,0)|ω1(k)|+p1H(m,0)|ω2(k)|+p2H(m,0)|ω3(k)|.
Hence,
(54)∑n=0m-1(ϑ2(m,n)H(m,n)4φ1(n)H(m,n)ρ(n)Q(n)-(1+p1+p2)ϑ2(m,n)H(m,n)4φ1(n))≤H(m,0){∑n=0k-1|ρ(n)Q(n)|+|ω1(k)|+p1|ω2(k)|+p2|ω3(k)|∑n=0k-1|ρ(n)Q(n)|+|ω1(k)|}.
Hence,
(55)lim⁡m→∞⁡sup⁡1H(m,0)∑n=0m-1(ϑ2(m,n)H(m,n)4φ1(n)H(m,n)ρ(n)Q(n)-(1+p1+p2)ϑ2(m,n)H(m,n)4φ1(n))≤{∑n=0k-1|ρ(n)Q(n)|+|ω1(k)|+p1|ω2(k)|+p2|ω3(k)|∑n=0k-1|ρ(n)Q(n)|+|ω1(k)|}<∞,
which is contrary to (47). If (II) holds, then we are back to the proof of Lemma 3 to show that lim⁡n→∞⁡x(n)=0. This completes the proof of Theorem 10.Corollary 11.
Assume that all the assumptions of Theorem10 hold, except that condition (47) is replaced by
(56)lim⁡n→∞⁡sup⁡∑s=n0n-1H(m,s)ρ(s)Q(s)=∞,lim⁡n→∞⁡sup⁡∑s=n0n-1ϑ2(m,s)H(m,s)4φ1(s)<∞.
Then every solution of (1) is oscillatory or lim⁡n→∞⁡x(n)=0.Theorem 12.
Assume that (20) holds. Further, assume that σ3≤σ1 and there exists a positive nondecreasing sequence {ρ(n)}, such that
(57)lim⁡n→∞⁡sup⁡∑s=n0n-1((Δρ(s))2b(s-σ1)ρ(s)∑u=n2s-σ1-1a-1(u)ρ(s)Q(s)-(1+p1+p2)4(Δρ(s))2b(s-σ1)ρ(s)∑u=n2s-σ1-1a-1(u))=∞.
Then every solution of (1) is oscillatory or lim⁡n→∞⁡x(n)=0.Proof.
Assume that (1) has a nonoscillatory solution, say x(n)>0, x(n-σ1)>0, x(n+σ2)>0, x(n-σ3)>0 and x(n+σ4)>0, for all n≥n0. Proceeding as in the proof of Theorem 4, we get (24). By Lemma 2, there exist two possible cases (I) and (II). Assume that (I) holds. Then, we obtain (25).
FromΔz(n)>0, σ3≤σ1, we obtain
(58)Δ(a(n)Δ(b(n)Δz(n)))+p1Δ(a(n-σ1)Δ(b(n-σ1)Δz(n-σ1)))+p2Δ(a(n+σ2)Δ(b(n+σ2)Δz(n+σ2)))+Q(n)z(n-σ1)≤0.
Using the Riccati transformation
(59)ω1(n)≔ρ(n)a(n)Δ(b(n)Δz(n))z(n-σ1).
Then, ω1(n)>0. From (59), we have
(60)Δω1(n)=Δρ(n)a(n+1)(Δ(b(n+1)Δz(n+1)))z(n+1-σ1)+ρ(n)Δ(a(n)(Δ(b(n)Δz(n))))z(n-σ1)-ρ(n)a(n+1)(Δ(b(n+1)Δz(n+1)))Δ(z(n-σ1))z(n-σ1)z(n+1-σ1).
From Lemma 1,Δ(a(n)(Δ(b(n)Δz(n))))≤0, and n-σ1<n, we get
(61)Δz(n-σ1)≥b-1(n-σ1)(a(n+1)(Δ(b(n+1)Δz(n+1))))×∑s=n2n-σ1-1a-1(s).
From (60) and (61), we obtain
(62)Δω1(n)≤Δρ(n)ρ(n+1)ω1(n+1)+ρ(n)Δ(a(n)Δ(b(n)Δz(n)))z(n-σ1)-ρ(n)(ω1(n+1))2ρ2(n+1)b(n-σ1)∑s=n2n-σ1-1a-1(s).
Next, define another sequence ω2(n) by
(63)ω2(n)≔ρ(n)a(n-σ1)Δ(b(n-σ1)Δz(n-σ1))z(n-σ1).
Then, ω2(n)>0. From (63), we have
(64)Δω2(n)=Δρ(n)ρ(n+1)ω2(n+1)+ρ(n)Δ(a(n-σ1)Δ(b(n-σ1)Δz(n-σ1)))z(n-σ1)-ρ(n)((z(n+1-σ1)z(n-σ1))-1a(n+1-σ1)×(Δ(b(n+1-σ1)Δz(n+1-σ1)))×Δ(z(n-σ1))×(z(n+1-σ1)z(n-σ1))-1).
From Lemma 1, we get
(65)Δ(z(n-σ1))≥(a(n+1-σ1)×(Δ(b(n+1-σ1)Δz(n+1-σ1))))×b-1(n-σ1)∑s=n2n-σ1-1a-1(s).
Then from (63), (64), and the above inequality, we have
(66)Δω2(n)≤ρ(n)Δ(a(n-σ1)Δ(b(n-σ1)Δz(n-σ1)))z(n-σ1)+Δρ(n)ρ(n+1)ω2(n+1)-ρ(n)ρ2(n+1)b(n-σ1)×(ω2(n+1))2∑s=n2n-σ1-1a-1(s).
Similarly, we define another sequence ω3(n) by
(67)ω3(n)≔ρ(n)a(n+σ2)Δ(b(n+σ2)Δz(n+σ2))z(n-σ1).
Then, ω3(n)>0. From (67), we have
(68)Δω3(n)=Δρ(n)ρ(n+1)ω3(n+1)+ρ(n)Δ(a(n+σ2)Δ(b(n+σ2)Δz(n+σ2)))z(n-σ1)-ρ(n)((z(n+1-σ1)z(n-σ1))-1a(n+1+σ2)×(Δ(b(n+1+σ2)Δz(n+1+σ2)))×Δ(z(n-σ1))×(z(n+1-σ1)z(n-σ1))-1).
From Lemma 1 and n-σ1<n+σ2, we get
(69)Δz(n-σ1)≥(a(n+1+σ2)×(Δ(b(n+1+σ2)Δz(n+1+σ2))))×b-1(n-σ1)∑s=n2n-σ1-1a-1(s).
Then from (67), (68), and the above inequality, we have
(70)Δω3(n)≤Δρ(n)ρ(n+1)ω3(n+1)+ρ(n)Δ(a(n+σ2)Δ(b(n+σ2)Δz(n+σ2)))z(n-σ1)-ρ(n)ρ2(n+1)b(n-σ1)(ω3(n+1))2×∑s=n2n-σ1-1a-1(s).
From (58), (62), (66) and (70), we obtain
(71)Δω1(n)+p1Δω2(n)+p2Δω3(n)≤-ρ(n)Q(n)+Δρ(n)ρ(n+1)ω1(n+1)-ρ(n)ρ2(n+1)b(n-σ1)(ω1(n+1))2×∑s=n2n-σ1-1a-1(s)+p1Δρ(n)ρ(n+1)ω2(n+1)-p1ρ(n)ρ2(n+1)b(n-σ1)(ω2(n+1))2×∑s=n2n-σ1-1a-1(s)+p2Δρ(n)ρ(n+1)ω3(n+1)-p2ρ(n)ρ2(n+1)b(n-σ1)(ω3(n+1))2×∑s=n2n-σ1-1a-1(s).
Using (39) and (71), we have
(72)Δω1(n)+p1Δω2(n)+p2Δω3(n)≤-ρ(n)Q(n)+14(Δρ(n))2b(n-σ1)ρ(n)∑s=n2n-σ1-1a-1(s)+p14(Δρ(n))2b(n-σ1)ρ(n)∑s=n2n-σ1-1a-1(s)+p24(Δρ(n))2b(n-σ1)ρ(n)∑s=n2n-σ1-1a-1(s).
Summing the last inequality from n3 to n-1, we obtain
(73)∑s=n3n-1((Δρ(s))2b(s-σ1)ρ(s)∑u=n2s-σ1-1a-1(u)ρ(s)Q(s)-(1+p1+p2)4(Δρ(s))2b(s-σ1)ρ(s)∑u=n2s-σ1-1a-1(u))≤ω1(n3)+p1ω2(n3)+p2ω3(n3),
which yields
(74)∑s=n3n-1((Δρ(s))2b(s-σ1)ρ(s)∑u=n2s-σ1-1a-1(u)ρ(s)Q(s)-(1+p1+p2)4(Δρ(s))2b(s-σ1)ρ(s)∑u=n2s-σ1-1a-1(u))≤c1,
where c1>0 is a finite constant. But this contradicts (57). Next, we assume that (II) holds. We are then back to the proof of Lemma 3 to show that lim⁡n→∞⁡x(n)=0. The proof is complete.Theorem 13.
Assume that (20) holds. Let σ3≤σ1 and {ρ(n)} be a positive sequence. Furthermore, assume that there exists a double sequence {H(m,n)∣m≥n≥0}. If
(75)lim⁡m→∞⁡sup⁡1H(m,0)×∑n=0m-1((1+p1+p2)ϑ2(m,n)H(m,n)4φ2(n)H(m,n)ρ(n)Q(n)-(1+p1+p2)ϑ2(m,n)H(m,n)4φ2(n))=∞,
then every solution of (1) is oscillatory or lim⁡n→∞⁡x(n)=0.Proof.
Proceeding as in the proof of Theorem12, we assume that (1) has a nonoscillatory solution, say x(n)>0, x(n-σ1)>0, x(n+σ2)>0, x(n-σ3)>0, and x(n+σ4)>0 for all n≥n0. By Lemma 2, there are two possible cases. If (I) holds, from the proof of Theorem 12, we find that (71) holds for all n≥n1. From (71), we have
(76)ρ(n)Q(n)≤-Δω1(n)-p1Δω2(n)-p2Δω3(n)+Δρ(n)ρ(n+1)ω1(n+1)-φ2(n)(ω1(n+1))2+p1Δρ(n)ρ(n+1)ω2(n+1)-p1φ2(n)(ω2(n+1))2+p2Δρ(n)ρ(n+1)ω3(n+1)-p2φ2(n)(ω3(n+1))2.
The remainder of the proof is similar to that of the proof of Theorem 10 and hence the details are omitted.


## 3. Examples

In this section, we will show the applications of our oscillation criteria by three examples. We will see that the equations in the example are oscillatory or tend to zero based on the results in Section2.Example 1.
Consider the following nonlinear neutral equation:(77)Δ(a(n)Δ(b(n)Δ(x(n)+p1(n)x(n-1)+p2(n)x(n+σ2))))+1nx(n-σ3)+1nx(n+σ4)=0,
where σ1=1, qi(n)=1/n for i=1,2. Assume that (17) holds. If we take n0=1 and ρ(n)=1, thus Corollary 7, asserts that every solution of (77) is oscillatory or tends to zero.Example 2.
Consider the following nonlinear neutral equation(78)Δ(nΔ(nΔ(x(n)+p1(n)x(n-σ1)+p2(n)x(n+σ2))))+βnx(n-σ3)+γnx(n+σ4)=0,
where σ3≥σ1, p1+p2≤3, and β and γ are positive constants. We see that (17) holds. If we take n0=1, ρ(n)=1, then by Theorem 4, every solution of (78) is oscillatory or tend to zero provided that β+γ>1.Example 3.
Consider the following nonlinear neutral equation:(79)Δ(1nΔ(1nΔ(x(n)+p1(n)x(n-2)+p2(n)x(n+1))1n))+βn2x(n-σ3)+γn2x(n+σ4)=0,
where σ1≥σ3, p1+p2≤3, β and γ are positive constants. We see that (17) holds. If we take n0=1, ρ(n)=1 then by Theorem 12, every solution of (79) is oscillatory or tends to zero provided that β+γ>1.



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