# Qualitative Property of Third-Order Nonlinear Neutral Distributed-Delay Generalized Difference Equations

Authors: M. Sumathy; P. Venkata Mohan Reddy; M. Maria Susai Manuel
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

This paper investigates the qualitative property of third-order nonlinear neutral distributed-delay generalized difference equations. By utilizing Philos-type technique and Riccati transformation, some oscillation criteria are presented to ensure that every solution of this equation oscillates or converges to zero. To illustrate the significance of our main result, we provide a suitable example.


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## Body

## 1. Introduction

In several areas, such as electrical circuit analysis, finance insurance, dynamic systems, computing, and physical field, third-order difference equations appeared to scrutinize discrete models, naturally occurring in discrete models pertaining physical, biological, and chemical phenomena (see, for example, [1–8]). In many engineering problems, analyzing the existence of oscillatory solutions performs an essential role. Notably, numerous monographs concern with issues of the existence and multiplicity of solutions using different methods, such as critical point theory, topological degree theory, fixed-point index theory, and Lie theory. In recent years, there has been a continual interest in getting sufficient conditions for oscillatory behavior of different classes of third-order difference equations with or without deviating arguments (see [8–23] and the references cited therein).The third-order nonlinear neutral distributed-delay generalized differential equation is of the form(1)Δℓa1kΔℓa2kΔℓzkγ1γ2+∑s=cdqk,sfxk+sℓ−σℓ=0,where zk=xk+∑s=abpk,sxk+sℓ−τℓ and Δℓ is the forward generalized difference operator well defined by Δℓxk=xk+ℓ−xk, ℕℓk0=k0,k0+ℓ,k0+2ℓ,…,, k0∈0,∞, ℓ∈0,∞, and a, b, c, d∈ℕk0, which subject to the following conditions:c1: the sequence aik is positive real and ∑k=k0∞1/ai1/γik=∞, for i=1,2.c2: pk,s and qk,s are nonnegative real sequences along with 0≤pk≡∑s=abpk,s≤p<1.c3: γ1 and γ2 are a quotient of odd positive integers with γ=γ1γ2.c4: the function f:ℝ⟶ℝ is continuous with fx/xγ≥L>0, where x≠0 and L is a constant.c5: mik=k−ki−j−ℓ/ℓ, ki¯=ki+j, and j=k−k0−k−k0/ℓℓ.By a solution of equation (1), this means a real sequence xk, satisfying equation (1) for all k∈ℕℓk0: the solution xk of equation (1) which satisfies supxk:k≥K>0, for K∈ℕℓk0.Section2 provides some standard definitions and proves some lemmas necessary for obtaining the main results. Section 3 offers newest oscillation results for equation (1), and finally, in Section 4, we provide a suitable example to determine the major findings.


## 2. Preliminaries

We illustrate a few basic definitions and primary results in this section that will be included in the forthcoming discussions.Definition 1 (see [16]).
Ifxk, k∈0,∞ is a real or complex valued function and ℓ∈0,∞, at that point the generalized difference operators Δℓ is predefined as(2)Δℓxk=xk+ℓ−xk≡yk,and then, its inverse is defined by(3)xk=xk0+j+∑r=0m0kyk0+j+rℓ,k∈ℕℓj.Definition 2 (see [16]).
Forλ∈ℕ1, the generalized polynomial factorial is defined by(4)kℓλ=kk−ℓk−2ℓ,…,k−λ−1ℓ=ℓkΓ1+k/ℓΓk/ℓ−λ−1.Lemma 1 (see [16]).
Letℓ∈0,∞. Then, Δℓnℓλ=λℓnℓλ−1.Lemma 2 (see [16]).
Letun and vn be real-valued functions. In addition,(5)Δℓukvk=uk+ℓΔℓvk+vkΔℓuk=vk+ℓΔℓuk+ukΔℓvk.Lemma 3.
Letxk be a positive solution of equation (1). Then, function zk has any one of the given properties:P1: zk>0, Δℓzk>0, Δℓa2kΔℓzkγ1>0, Δℓa1kΔℓa2kΔℓzkγ1γ2<0P2: zk>0, Δℓzk<0, Δℓa2kΔℓzkγ1>0, Δℓa1kΔℓa2kΔℓzkγ1γ2<0
In the above properties,k≥k2 for sufficiently large k2∈Nℓk0.Proof.
Letxk be a positive solution of equation (1) for every k≥k0. By defining zk, with zk≥xk>0 for k≥k1∈Nℓk0, and in addition to equation (1), we have(6)Δℓa1kΔℓa2kΔℓzkγ1γ2=−∑s=cdqk,sfxk+sℓ−σℓ<0.
We can know thata1kΔℓa2kΔℓzkγ1γ2 is clearly a decreasing function on k1,∞ with positive or negative finally. Furthermore, we must have to prove that a1kΔℓa2kΔℓzkγ1γ2>0 for k≥k1≥k0. Otherwise, we have a constant M1γ2>0 such that(7)Δℓa2kΔℓzkγ1<−M1a11/γ2k<0,fork≥k1.
Hence, by Definition1,(8)a2kΔℓzkγ1≤a2k1¯Δℓzk1¯γ1−M1∑r=0m1k1a11/γ2k1¯+rℓ.
Lettingk⟶∞ and then using condition c1, we have limn⟶∞a2kΔℓzkγ1=−∞. Subsequently, there exists a k2≥k1 also a constant M2γ1>0 so that(9)a2kΔℓzkγ1<−M2γ1,fork≥k2.
Dividing the above inequality bya2k and by applying summation from k2 to k−ℓ, we obtain(10)zk<zk2¯−M2∑t=0m2k1a21/γ1k2¯+tℓ,
Lettingk⟶∞ and using condition c1, we have zn⟶−∞. Thus, zn<0 eventually which is contradictory with zn>0. Consequently, Δℓa2kΔℓzkγ1 is positive, that is, Δℓa2kΔℓzkγ1γ2>0 holds.
It can be known fromΔℓa2kΔℓzkγ1>0, that is, a monotonically increasing sign in the interval k2,∞. Therefore, Δℓzk is ultimately either positive or negative. We have only property P1 or P2 for zk. Hence, the proof is completed.Lemma 4.
Letxk be a positive solution of equation (1), and zk has P2 of Lemma 3. If(11)∑t2=0∞1a21/γ1k3¯+t2ℓ∑t1=t2m2k1a11/γ2k2¯+t1ℓ∑t=t1m1k∑s=cdqk1¯+tℓ,s1/γ21/γ1=∞,then xk of equation (1) converges to zero when k⟶∞.Proof.
Letxk be a positive solution of equation (1). Since zk satisfies the property P2 of Lemma 3, thereby there exist β≥0 such that(12)limk⟶∞zk=β≥0.
Now, we shall prove thatβ=0. Let β>0; then, we have β+ɛ>zk>β for every ϵ>0, and k is enough large. Choosing 0<ɛ<1−p/pβ, from zk, we have(13)xk=zk−∑s=abpk,sxk+sℓ−τℓ>β−∑s=abpk,sxk+sℓ−τℓ>β−pβ+ɛ=β−pβ+ɛβ+ɛβ+ɛ>M3zk,where M3=β−pβ+ɛ/β+ɛ>0. Thus, from equation (1) and c4, we have(14)Δℓa1kΔℓa2kΔℓzkγ1γ2=−∑s=cdqk,sfxk+sℓ−σℓ≤−∑s=cdqk,sLxγk+sℓ−σℓ.
Now, using (13), we obtain(15)Δℓa1kΔℓa2kΔℓzkγ1γ2≤−M3γL∑s=cdqk,szγk+sℓ−σℓ.
Summing the abovementioned inequality fromk1 to k−ℓ and from Definition 1, we obtain(16)−a1k1¯Δℓa2k1¯Δℓzk1¯γ1γ2≤−M3γL∑t=0m1k∑s=cdqk1¯+tℓ,szγk1¯+tℓ+sℓ−σℓ.
The above equation can also be rewritten as(17)Δℓa2k1¯Δℓzk1¯γ1≥M3βγ/γ2L1/γ2a11/γ2k1¯∑t=0m1k∑s=cdqk1¯+tℓ,s1/γ2.
Summing again fromk2≥k1 to k−ℓ, we obtain(18)−Δℓzk2¯≥M3βL1/γa21/γ1k2¯∑t1=0m2k1a11/γ2k2¯+t1ℓ∑t=t1m1k∑s=cdqk1¯+tℓ,s1/γ21/γ1.
Summing the last inequality with the limit fromk3 to ∞, we obtain(19)zk3¯≥M3βL1/γ∑t2=0∞1a21/γ1k3¯+t2ℓ∑t1=t2m2k1a11/γ2k2¯+t1ℓ∑t=t1m1k∑s=cdqk1¯+tℓ,s1/γ21/γ1.
This contradicts to condition (11). Thus, β=0. Furthermore, the inequality 0<xk<zk implies that limn⟶∞xn=0. Now, the proof is complete.Lemma 5.
Letxk be a positive solution of (1), and zk has the property P1. Then,(20)a1k+ℓΔℓa2k+ℓΔℓzk+ℓγ1γ21/γRk≤Δℓzk,where Rk=1/a21/γ1k∑r=0m1k1/a11/γ2k1+rℓ1/γ1.Proof.
Letxk be a positive solution of (1). Since zk has the property P1, we know(21)Δℓa1kΔℓa2kΔℓzkγ1γ2<0,from the Definition 1 and for all k1≥k0, we have(22)a2kΔℓzkγ1=a2k1¯Δℓzk1¯γ1+∑r=0m1kΔℓa2k1¯+rℓΔℓzk1¯+rℓγ1≥∑r=0m1ka1k1¯+rℓΔℓa2k1¯+rℓΔℓzk1¯+rℓγ1γ21/γ2a11/γ2k1¯+rℓ≥a1k+ℓΔℓa2k+ℓΔℓzk+ℓγ1γ21/γ2∑r=0m1k1a11/γ2k1¯+rℓ.
Hence, we obtain(23)Δℓzk≥a1k+ℓΔℓa2k+ℓΔℓzk+ℓγ1γ21/γRk.
This completes the proof.


## 3. Main Results

This section establishes criteria on oscillation and convergent solutions to (1) with the help of generalized Riccati transformation and Philos-type technique. Let us define functions h, H:ℕℓ×ℕℓ⟶ℝ such that(1
)Hk,k=0 with k≥k0≥0.(2)
Hk,s>0 with k>s≥k0.(3)
ΔℓsHk,s=Hk,s+ℓ−Hk,s≤0 for k>s≥k0, and there exists a positive real sequence ρk such that(24)ΔℓsHk,s+Δℓρsρs+ℓHk,s=−hk,sHk,s.Theorem 1.
Assume thatzk holds and there exists ρk a positive real-valued sequence in such a way that(25)limk⟶∞∑r=0m1kCk1¯+rℓ−γγB1+γk1¯+rℓ1+γ1+γAγk1¯+rℓ=∞,where(26)Ak=γρkRkργ+1/γk+ℓ,Bk=Δℓρkρk+ℓ,Ck=ρkq1k,(27)q1k=L1−pγ∑s=cdqk,s,and then, each solution of equation (1) is either xk⟶0 as k⟶∞ or oscillatory.Proof.
Suppose thatxk is a nonoscillatory solution of equation (1). By assuming xk>0 and xk+sℓ−τℓ>0 for k≥k1≥k0∈ℕℓ and zn satisfies two properties of Lemma 3, we have(28)xk≥zk−∑s=abpk,szk+sℓ−τℓ≥1−∑s=abpk,szk≥1−pzk.
Using conditionc3 in equation (1), we obtain(29)Δℓa1kΔℓa2kΔℓzkγ1γ2≤−∑s=cdqk,sLxγk+sℓ−σℓ.
By applying equation (28) in the aforementioned inequality, we obtain(30)Δℓa1kΔℓa2kΔℓzkγ1γ2≤−L1−pγ∑s=cdqk,szγk+sℓ−σℓ≤−q1kzγk+cℓ−σℓ.
Define(31)wk=ρka1kΔℓa2kΔℓzkγ1γ2zγk,k≥k1.
Then,wk>0, for every k≥k1, and from equations (20) and (30), we have(32)Δℓwk≤−ρkq1k+Δℓρkρk+ℓwk+ℓ−γρkRkργ+1/γk+ℓwγ+1/γk+ℓ.
The above equation is also expressed as(33)Δℓwk≤−Ck+Bkwk+ℓ−Akwγ+1/γk+ℓ,where(34)Ak=γρkRkργ+1/γk+ℓ,Bk=Δℓρkρk+ℓ,Ck=ρkq1k.
Using the inequality,(35)Au−Bu1+β/β≤ββ1+β1+β×A1+βBβ.
Now, using the above inequality, it is possible to write equation (33) as(36)Ck−γγ1+γ1+γ×B1+γAγ≤−Δℓwk.
By applying summation in the last inequality with the limits fromk1 to k−ℓ, we have(37)∑r=0m1kCk1¯+rℓ−γγB1+γk1¯+rℓ1+γ1+γAγk1¯+rℓ≤wk1¯−wk≤wk1¯,from wk>0, which contradicts (25) as k⟶∞, and then, the solution xk of (1) is oscillatory. When zk has property P2, from (11), we know limk⟶∞xk=0 by Lemma 4. The proof is complete.Theorem 2.
Suppose that (1) holds. If there exists ρk a positive real sequence such that(38)limsupk⟶∞1Hk,k2∑r=0m2kHk,k2¯+rℓCk2¯+rℓ−γγ1+γ1+γ−hk,k2¯+rℓ1+γAγk2¯+rℓHγ−1/2k,k2¯+rℓ=∞.
Subsequently, all solutions of equation (1) is either xk⟶0 as k⟶∞ or oscillatory.Proof.
Suppose thatxk is a nonoscillatory solution of equation (1). Proceeding as the proof of Theorem 1, we got equation (33). Now, multiplying inequality (33) by Hk,s and then summing the aforementioned inequality from k2 to k−ℓ, for all k≥k2≥k0, we have(39)∑r=0m2kHk,k2¯+rℓCk2¯+rℓ≤−∑r=0m2kHk,k2¯+rℓΔℓwk2¯+rℓ+∑r=0m2kBk2¯+rℓwk2¯+rℓ+ℓ−Ak2¯+rℓwγ+1/γk2¯+rℓ+ℓHk,k2¯+rℓ.
By summation by parts, we obtain(40)∑r=0m2kHk,k2¯+rℓCk2¯+rℓ≤Hk,k2¯wk2¯+∑r=0m2kΔℓk2¯Hk,k2¯+Bk2¯+rℓHk,k2¯+rℓwk2¯+rℓ+ℓ−∑r=0m2kAk2¯+rℓHk,k2¯+rℓwγ+1/γk2¯+rℓ+ℓ.
Using inequality (35), we have(41)∑r=0m2kHk,k2¯+rℓCk2¯+rℓ≤Hk,k2¯wk2¯+∑r=0m2kγγ1+γ1+γΔℓk2¯Hk,k2¯+Bk2¯+rℓHk,k2¯+rℓ1+γAk2¯+rℓHk,k2¯+rℓγ,∑r=0m2kHk,k2¯+rℓCk2¯+rℓ−γγ1+γ1+γ−hk,k2¯+rℓ1+γAγk2¯+rℓHγ−1/2k,k2¯+rℓ≤Hk,k2¯wk2¯,where(42)Δℓk2¯Hk,k2¯+Δℓρk2¯+rℓρk2¯+rℓ+ℓHk,k2¯+rℓ=−hk,k2¯+rℓHk,k2¯+rℓ,1Hk,k2¯∑r=0m2kHk,k2¯+rℓCk2¯+rℓ−γγ1+γ1+γ−hk,k2¯+rℓ1+γAγk2¯+rℓHγ−1/2k,k2¯+rℓ≤wk2¯,which is a contradiction to inequality (29). If zk has satisfied the property (ii) of Lemma 3, then, by condition (11), we have xk⟶0 as k⟶0. Hence, the theorem is proved.Corollary 1.
IfHk,s=k−sℓm for all k≥s≥0, ρk=1, and(43)limsupk⟶∞1kℓm∑r=0k−s−ℓ/ℓk−s¯−rℓℓmq1s¯+rℓ−1+γ−1+γ−mℓk−s¯−rℓℓm/2−11+γRγs¯+rℓk−s¯−rℓℓmγ−1/2=∞,for every m≥1, then each and every solution of equation (1) is oscillatory.Corollary 2.
IfHk,s=logk+ℓ/s+ℓm, for all k≥s≥0, ρk=1, and(44)limsupk⟶∞1logk+ℓ/ℓm∑r=0k−s−ℓ/ℓlogk+ℓs¯+rℓ+ℓmq1s¯+rℓ−1+γ−1+γlogk+ℓ/s¯+rℓ+ℓm−21+γRγs¯+rℓlogk+ℓ/s¯+rℓ+ℓmγ−1/2=∞,for every m≥1, then each and every solution of equation (1) is oscillatory.Theorem 3.
Suppose that condition (11) holds. Also, let(45)0<infs≥k0liminfk⟶∞Hk,sHk,k0≤∞(46)limsupk⟶∞1Hk,k0∑r=0m2kh2k,k2¯+rℓAk2¯+rℓw1−γ/γk2¯+rℓ<∞holds. If there is a sequence Φn such that(47)∑r=0m2kAk2¯+rℓΦ+2k2¯+rℓ+ℓ=∞(48)limsupk⟶∞1Hk,k2∑r=0m2kHk,k2¯+rℓCk2¯+rℓ−h2k,k2¯+rℓ1+γ4Ak2+rℓw1−γ/γk,k2¯+rℓ+ℓ≥Φs,where(49)Φ+k2+rℓ+ℓ=maxΦk2+rℓ+ℓ,0,where Ak, Bk, Ck, and hk,s are defined in (26) and (24), respectively, then each and every solution of equation (1) is either oscillatory or converges to zero.Proof.
Letxk be a nonoscillatory solution of (1), and proceeding as in the proof of Theorem 2, when zk has a property P1, from (40), and rearranging, we obtain(50)wk2¯≥limsupk⟶∞1Hk,k2¯∑r=0m2kHk,k2¯+rℓCk2¯+rℓ−h2k,k2¯+rℓ1+γ4Ak2+rℓw1−γ/γk,k2¯+rℓ+ℓ+liminfk⟶∞1Hk,k2¯∑r=0m2kHk,k2¯+rℓAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ+hk,k2¯+rℓ2Ak2¯+rℓw1−γ/γk,k2¯+rℓ+ℓ2,for k≥k2. It is derived from (48) that(51)wk2¯≥Φk2¯+liminfk⟶∞1Hk,k2¯∑r=0m2kHk,k2¯+rℓAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ+hk,k2¯+rℓAk2¯+rℓw1−γ/γk,k2¯+rℓ+ℓ2,which means that(52)wk2¯≥Φk2¯,and then,(53)liminfk⟶∞1Hk,k2¯∑r=0m2kHk,k2¯+rℓAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ+hk,k2¯+rℓAk2¯+rℓw1−γ/γk,k2¯+rℓ+ℓ2<∞.
Therefore,(54)liminfk⟶∞1Hk,k2¯∑r=0m2kHk,k2¯+rℓAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ+1Hk,k2¯∑r=0m2khk,k2¯+rℓHk,k2¯+rℓwk2¯+rℓ+ℓ+14Hk,k2¯∑r=0m2kh2k,k2¯+rℓAk2¯+rℓw1−γ/γk2¯+rℓ+ℓ<∞.
Then,(55)liminfk⟶∞1Hk,k2¯∑r=0m2kHk,k2¯+rℓAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ+1Hk,k2¯∑r=0m2khk,k2¯+rℓHk,k2¯+rℓwk2¯+rℓ+ℓ<∞.
The abovementioned inequality can also be expressed as(56)liminfk⟶∞Uk+Vk<∞,fork≥k2,where(57)Uk=1Hk,k2¯∑r=0m2kHk,k2¯+rℓAk2¯+rℓwγ+1/γk2¯+rℓ+ℓVk=1Hk,k2¯∑r=0m2khk,k2¯+rℓHk,k2¯+rℓwk2¯+rℓ+ℓ.
Here, we assert(58)∑r=0m2kAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ<∞.
Conversely, suppose that(59)∑r=0m2kAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ=∞.
From equation (45), we have(60)infs≥k0liminfk⟶∞Hk,sHk,k0>μ,for μ>0; then, Hk,s/Hk,k0>μ for k≥k2≥k1. There exists a positive constant M4>0 such that(61)∑r=0m2kAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ≥M4μ.
Thus, fork≥k3 and using equation (74), we have(62)Uk=1Hk,k2¯∑r=0m2kHk,k2¯+rℓΔℓ∑r1=0r−k2¯−ℓ/ℓAk2¯+r1ℓwγ+1/γk2¯+r1ℓ+ℓ+Ak2¯+rℓwγ+1/γk2¯+rℓ+ℓ=−1Hk,k2¯∑r=0k−k3¯−ℓ/ℓ∑r1=0r−k2¯−ℓ/ℓAk2¯+r1ℓwγ+1/γk2¯+r1ℓ+ℓΔℓk2¯Hk,k3¯+rℓ−Ak2¯wγ+1/γk2¯+ℓHk,k2¯∑r=0k−k3¯−ℓ/ℓHk,k3¯+rℓAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ≥1Hk,k2¯∑r=0k−k3¯−ℓ/ℓ∑r1=0r−k2¯−ℓ/ℓAk2¯+r1ℓwγ+1/γk2¯+r1ℓ+ℓ−Δℓk2¯Hk,k3¯+rℓ≥M4μHk,k2¯∑r=0k−k3¯−ℓ/ℓ−Δℓk2¯Hk,k3¯+rℓ≥M4Hk,k3¯μHk,k2¯≥M4.
SinceM4 is an arbitrary constant,(63)limk⟶∞Uk=∞.
Furthermore, consider a sequencekn⟶∞ such that(64)limn⟶∞Ukn+Vkn=liminfk⟶∞Uk+Vk.
From (56), there exists a number M5 such that(65)Ukn+Vkn≤M5,forn=0,1,2,.
Resulting from (64), we conclude that(66)limn⟶∞Vkn=−∞.
By (65), for a large value of n, we have(67)1+VknUkn≤M5Ukn<12.
From (66), we obtain(68)limn⟶∞V2knUkn=∞.
Nevertheless, by Schwarz’s inequality, we have(69)V2kn=1Hkn,k2¯∑r=0kn−k2¯−ℓ/ℓhkn,k2¯+rℓHkn,k2¯+rℓwk2¯+rℓ+ℓ2≤1Hkn,k2¯∑r=0kn−k2¯−ℓ/ℓHkn,k2¯+rℓAk2¯+rℓwγ+1/γk2¯+rℓ+ℓ1Hkn,k2¯∑r=0kn−k2¯−ℓ/ℓh2kn,k2¯+rℓAk2¯+rℓw1−γ/γk2¯+rℓ+ℓ≤Ukn1Hkn,k2¯∑r=0kn−k2¯−ℓ/ℓh2kn,k2¯+rℓAk2¯+rℓw1−γ/γk2¯+rℓ+ℓ.
Consequently,(70)V2knUkn≤1Hkn,k2¯∑r=0kn−k2¯−ℓ/ℓh2kn,k2¯+rℓAk2¯+rℓw1−γ/γk2¯+rℓ+ℓ≤1μHkn,k0∑r=0kn−k2¯−ℓ/ℓh2kn,k2¯+rℓAk2¯+rℓw1−γ/γk2¯+rℓ+ℓ.
It follows from (68) that(71)limn⟶∞1Hkn,k0∑r=0kn−k2−ℓ/ℓh2kn,k2+rℓAk2+rℓw1−γ/γk2¯+rℓ+ℓ=∞,which is a contradiction to inequality (46). Hence, by (52), equation (72) holds:(72)∑r=0m2kAk2¯+rℓΦ+2k2¯+rℓ+ℓ≤∑r=0m2kAk2¯+rℓw2k2¯+rℓ+ℓ<∞,which is a contradiction to inequality (47) and concludes the proof. If zk fulfills the property P2 of Lemma 3, then, by equation (11), we have obtained xk⟶0 as k⟶∞. Hence, the proof is complete.Theorem 4.
Suppose that all conditions of Theorem3 are satisfied excluding condition (46). Also, let(73)liminfk⟶∞1Hk,k0∑r=0k−k2−ℓ/ℓHk,s+rℓCs+rℓ<∞and(74)liminfk⟶∞1Hk,s∑r=0k−s−ℓ/ℓHk,s+rℓCs+rℓ−h2k,s+rℓ4As+rℓ≥Φs.
Furthermore, each and every solution of equation (1) is either xk⟶0 as k⟶∞ or oscillatory.Proof.
The proof is approximately the same as Theorem3, and in consequence, the details are excluded.Example 1.
The third-order neutral generalized difference equation with distributed delay is given as(75)ΔℓΔℓΔℓxk+∑s=231sxk+sℓ−ℓ31/3+∑s=1327/3sxk+sℓ−ℓ=0.
Here,a1k=a2k=1, γ1=3, γ2=1/3, pk,s=1/s, qk,s=27/3k/2+1/s, τ=1, σ=1, and ρk=1. Then, Rk=k−k1/ℓ and q1k=21/3L/96k+11 which imply(76)limk⟶∞∑r=0m1k21/3L96k1+6rℓ+11=∞.
Henceforth, by Theorem1, each and every solution of equation (75) is oscillatory. Moreover, xn=−1n/ℓ is one such oscillatory solution of equation (75).


## 4. Conclusion

In this paper, we present the qualitative properties such as convergence and oscillatory behaviors of third-order nonlinear neutral distributed-delay generalized difference equation. The results we obtained in this paper for difference equation involving generalized difference operatorΔℓ with distributed delay are rare and new in the literature. Also, the technique adopted is different from that which already exists. The significance of the results is also well established by an example presented in this paper.



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