Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 31 No. 2 (2024) 

 

 

119 
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Oscillatory Properties  of Second Order Half-Linear Delay 

Difference Equations 

 

K.  Masaniammal 1,  I. Mohammed Ali Jaffer 2 

1, 2 Department of Mathematics, Government Arts College, Udumalpet-642126, Tamilnadu, India. 

1  reka.maths@gmail.com   2 jaffermathsgac@gmail.com 

 

Article History: 

Received: 25-01-2024 

Revised: 02-04-2024 

Accepted: 25-04-2024 

Abstract:  

This study explores, some necessary and sufficient conditions that are established for 

oscillatory  properties of second order half-linear delay difference equations of the form 

   Δ(𝑝(𝜉)(Δ𝑥(𝜉))𝑟) + 𝑞(𝜉)𝑥𝑠(𝜎(𝜉)) = 0,  for 𝜉 ≥ 𝜉0 .  

Under the assumption  ∑t=𝜉0
𝜉−1

 
1

𝑝
1
𝑟(t)

= ∞. Two cases are considered for 𝑟 < 𝑠 and 𝑟 >

𝑠, where 𝑟 and 𝑠 are the quotients of two positive odd integers. The effectiveness and 

applicability of the result are illustrated through few examples. 

 

Keywords: Half-Linear, Delay Difference Equation, Oscillation. 

1. Introduction 

We consider the second order half-linear Delay difference equations of the form 

 Δ(𝑝(𝜉)(Δ𝑥(𝜉))𝑟) + 𝑞(𝜉)𝑥𝑠(𝜎(𝜉)) = 0,          for 𝜉 ≥ 𝜉0.                   (1.1) 

where 𝑟 and 𝑠 are the quotient of two positive odd integers, and Δ is the forward difference 

operator defined by Δ𝑥(𝜉) = 𝑥(𝜉 + 1) − 𝑥(𝜉). 

The following assumptions are used in this paper to obtain the result: 

H1) {𝑝(𝜉)} is sequence of positive real numbers, 0 < 𝑝 < 1, 𝜎(𝜉) < 𝜉, lim𝜉→∞  𝜎(𝜉) = ∞. 

H2) {𝑞(𝜉)} is a sequence of nonnegative real numbers and 𝑞(𝜉) is not identically zero for 

sufficiently large values of 𝜉. 

H3) 𝑣(𝜉) = ∑𝑡=𝜉1
𝜉−1

 𝑝−
1

𝑟(𝑡) with lim𝜉→∞  𝑣(𝜉) = ∞. 

H4) 0 < 𝜎0(𝜉) ≤ 𝜎(𝜉), for Δ𝜎0(𝜉) ≥ 𝜎0 > 0, for 𝜉 ≥ 𝜉0. 

2. Preliminary Results 

In this section, we provide useful lemma that will be essential in the analysis of the 

oscillation behavior of (1.1). 

Lemma 2.1.  Assuming (𝐻1) − (𝐻3) hold and that 𝑥(𝜉) is an eventually positive solution of 

(1.1). Then, there exists 𝜉1 ≥ 𝜉0 and 𝑑 > 0 such that 

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                                                                     0 < 𝑥(𝜉) ≤ 𝑑𝑣(𝜉),                                              (2.1) 

                                      𝑣(𝜉) [∑  

∞

𝜁=𝜉

 𝑞(𝜁)𝑥𝑠(𝜎(𝜁))]

1
𝑟

≤ 𝑥(𝜉),   for 𝜉 ≥ 𝜉1.
                          (2.2) 

Proof.  Assume that 𝑥(𝜉) be an eventually positive solution of (1.1). Then, by (H1), there 

exists a 𝜉∗ such that 𝑥(𝜉) > 0 and 𝑥(𝜎(𝜉)) > 0 for all 𝜉 ≥ 𝜉∗ 

It follows from (1.1) that 

                     Δ(𝑝(𝜉)(Δ𝑥(𝜉))𝑟) =  −𝑞(𝜉)𝑥𝑠(𝜎(𝜉)) ≤ 0.                                                     (2.3) 

Consequently, 𝑝(𝜉)(Δ𝑥(𝜉))𝑟 is nonincreasing for 𝜉 ≥ 𝜉∗. Next, we establish that 

𝑝(𝜉)(Δ𝑥(𝜉))𝑟 is positive. By contradiction, let 𝑝(𝜉)(Δ𝑥(𝜉))𝑟 ≤ 0 at a certain time 𝜉 ≥ 𝜉∗. 

In accordance to 𝑞 is not identically zero  and by (2.3),  

there exists 𝜉1 ≥ 𝜉∗ such that 

                            𝑝(𝜉)(Δ𝑥(𝜉))𝑟 ≤ 𝑝(𝜉1)(Δ𝑥(𝜉1))
𝑟
< 0,       𝜉 ≥ 𝜉1 .                              (2.4) 

Remember that 𝑟 is the quotient of two positive odd integers. Then, 

   Δ𝑥(𝜉) ≤ (
𝑝(𝜉1)

𝑝(𝜉)
)

1

𝑟
Δ𝑥(𝜉1),       for 𝜉 ≥ 𝜉1. 

                                  (2.5) 

Summing (2.5) from 𝜉1 to 𝜉 − 1, we arrive at the result 

                                            𝑥(𝜉) ≤ 𝑥(𝜉1) + (𝑝(𝜉1))
1

𝑟Δ𝑥(𝜉1)𝑣(𝜉).                                      (2.6) 

By (𝐻3), the approach of the right hand side is −∞ then, lim𝜉→∞  𝑣(𝜉) = −∞. 

This is a contradiction to the fact that 𝑥(𝜉) > 0. 

Thus,                       

     𝑝(𝜉)(Δ𝑥(𝜉))𝑟 > 0,             for all 𝜉 ≥ 𝜉∗. 

From 𝑝(𝜉)(Δ𝑥(𝜉))𝑟 being nonincreasing, we have 

Δ𝑥(𝜉) ≤ (
𝑝(𝜉1)

𝑝(𝜉)
)

1

𝑟
Δ𝑥(𝜉1),  for 𝜉 ≥ 𝜉1. 

                              (2.7) 

Summing (2.7) from 𝜉1 to 𝜉 − 1, we obtain 

𝑥(𝜉) ≤ 𝑥(𝜉1) + (𝑝(𝜉1))
1

𝑟Δ𝑥(𝜉1)𝑣(𝜉) 

.                          (2.8) 

Since lim𝜉→∞  𝑣(𝜉) = ∞, there exists a positive constant 𝑑 such that (2.1) holds. 



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Since 𝑝(𝜉)(Δ𝑥(𝜉))𝑟 is positive and nonincreasing, lim𝜉→∞  𝑝(𝜉)(Δ𝑥(𝜉))
𝑟 exists and is 

nonnegative. 

Summing (1.1) from 𝜉 to 𝑏 − 1, we get 

𝑝(𝑏)(Δ𝑥(𝑏))𝑟 − 𝑝(𝜉)(Δ𝑥(𝜉))𝑟 + ∑  𝑏−1
𝑡=𝜉 𝑞(𝑡)𝑥

𝑠(𝜎(𝑡)) =

0.                 (2.9) 

Letting limit as 𝑏 → ∞, we obtain 

                                                 𝑝(𝜉)(Δ𝑥(𝜉))𝑟 ≥

∑  ∞
𝑡=𝜉 𝑞(𝑡)𝑥

𝑠(𝜎(𝑡)).                                            (2.10) 

Then, 

            Δ𝑥(𝜉) ≥ [
1

𝑝(𝜉)
∑  ∞
𝑡=𝜉  𝑞(𝑡)𝑥

𝑠(𝜎(𝑡))]

1

𝑟
.      

                           (2.11) 

Since 𝑥(𝜉1) > 0, summing (2.11) from 𝜉1 to 𝜉 − 1, we have 

                                                     𝑥(𝜉) ≥ ∑  

𝑛−1

𝑡=𝜉1

[
1

𝑝(𝑡)
∑  

∞

𝜁=𝑡

 𝑞(𝜁)𝑥𝑠(𝜎(𝜁))]

1
𝑟

  .                         (2.12) 

Use the definition of 𝑣(𝜉) to obtain 

                                                       𝑥(𝜉)

≥ 𝑣(𝜉) [∑  

∞

𝜁=𝜉

 𝑞(𝜁)𝑥𝑠(𝜎(𝜁))]

1
𝑟

    .                                     (2.13) 

This yields (2.2). 

3. Main Results 

Theorem 3.1.  Assume that there exists a constant 𝛽1, the quotient of two positive odd 

integers, such that 0 < 𝑠 < 𝛽1 < 𝑟. If (𝐻1) − (𝐻3) hold, then each solution of (1.1) is 

oscillatory if and only if 

       ∑  ∞
𝜁=0 𝑞(𝜁)𝑣

𝑠(𝜎(𝜁)) = ∞ 

.                                                                (3.1) 

Proof. On the contrary, let 𝑥(𝜉)  be an eventually positive solution. So Lemma 2.1 holds, and 

then there exists 𝜉1 ≥ 𝜉0 such that 

𝑥(𝜉) ≥ 𝑣(𝜉)𝑤
1

𝑟(𝜉) ≥ 0,    for 𝜉 ≥ 𝜉1 ,                                               (3.2) 

 

 



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where 

                                   𝑤(𝜉) =∑  

∞

𝜁=𝜉

𝑞(𝜁)𝑥𝑠(𝜎(𝜁)) .                                                                     (3.3) 

Computing  we have , 

                                 Δ𝑤(𝜉) = −𝑞(𝜉)𝑥𝑠(𝜎(𝜉)) .                                                                        (3.4) 

Thus, 𝑤 is nonnegative and nonincreasing. Since 𝑥 > 0, by (𝐻2), in continuation 

𝑞(𝜉)𝑥𝑠(𝜎(𝜉)) cannot be identically zero. Thus, Δ𝑤 cannot be identically zero, and 𝑤 cannot 

be constant.  

Therefore, 𝑤(𝜉) > 0 for 𝜉 ≥ 𝜉1. 

Computing we get, 

Δ𝑤1−
𝛽1
𝑟 (𝜉) ≥ (1 −

𝛽1

𝑟
)𝑤

−𝛽1
𝑟 (𝜉)Δ𝑤(𝜉).                                                           

(3.5) 

Summing (3.5) from 𝜉2 to 𝜉 − 1 and using that 𝑤 > 0, we have 

 

       𝑤1−
𝛽1
𝑟 (𝜉2) ≥ (1 −

𝛽1
𝑟
) [− ∑  

𝜉−1

𝜁=𝜉2

 𝑤
−𝛽1
𝑟 (𝜁)Δ𝑤(𝜁)]

 

≥ (1 −
𝛽1

𝑟
) [∑  

𝜉−1
𝜁=𝜉2

 𝑤
−𝛽1
𝑟 (𝜁)(𝑞(𝜁)𝑥𝑠(𝜎(𝜁)))].              

(3.6) 

By (2.1) and (3.2), we obtain 

𝑥𝑠(𝜉) = 𝑥𝑠−𝛽1(𝜉)𝑥𝛽1(𝜉)

 ≥ (𝑑𝑣(𝜉))𝑠−𝛽1𝑥𝛽1(𝜉)

 ≥ (𝑑𝑣(𝜉))𝑠−𝛽1 (𝑣(𝜉)𝑤
1
𝑟(𝜉))

𝛽1

 = 𝑑𝑠−𝛽1𝑣𝑠(𝜉)𝑤
𝛽1
𝑟 (𝜉),   for 𝜉 ≥ 𝜉2.

 

Since 𝑤 is nonincreasing , 
𝛽1

𝑟
> 0, and 𝜎(𝑡) < 𝑡, it follows that 

                                                       𝑥𝑠(𝜎(t)) ≥ 𝑑𝑠−𝛽1𝑣𝑠(𝜎(𝑡))𝑤
𝛽1
𝑟 𝜎((𝑡))

                                                                   
 

                                  ≥ 𝑑𝑠−𝛽1𝑣𝑠(𝜎(𝑡))𝑤
𝛽1
𝑟 (𝑡).                                                     (3.7) 

 

 



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Going back to (3.6), we obtain 

                                               𝑤1−
𝛽1
𝑟 (𝜉2)

≥ (1 −
𝛽1
𝑟
) 𝑑𝑠−𝛽1 [∑  

𝜉−1

𝑡=𝜉2

 𝑞(𝑡)𝑥𝑠(𝜎(𝑡))]   .                     (3.8) 

Since (1 −
𝛽1

𝑟
) > 0, by (3.1) the right-hand side approaches +∞ as 𝜉 → ∞. 

In contradiction with (3.8), this completes the sufficiency proof for eventually positive 

solutions. 

Similar to this the eventually negative solution can be dealt by introducing the variables 𝜎 =

−𝑥. Then, the necessary part can be shown by the contrapositive argument. If (3.1) is not 

hold, then for each 𝛼 > 0 there exists 𝜉1 ≥ 𝜉0 such that 

∑  ∞
𝜁=𝑡 𝑞(𝜁)𝑣

𝑠(𝜎(𝜁)) ≤
𝛼 
(1 − 

𝑠
𝑟
)

2
,      for all 𝜉 ≥ 𝜉1 .                                                                    

 (3.9) 

 

We define 

𝑇 = {𝑥: (
𝛼

2
)

1

𝑟
𝑣(𝜉) ≤ 𝑥(𝜉) ≤ 𝛼

1

𝑟𝑣(𝜉),  𝜉 ≥ 𝜉1}.                                    

 (3.10) 

An operator 𝜙 is defined on T by 

(𝜙𝑥)(𝜉)

=

{
 
 

 
 
0,  if 𝜉 ≤ 𝜉1,

∑  

𝜉−1

𝑡=𝜉1

  [
1

𝑝(𝑡)
[
𝛼

2
+∑  

∞

𝜁=𝑡

 𝑞(𝜁)𝑥𝑠(𝜎(𝜁))]]

1
𝑟

,  if 𝜉 > 𝜉1.
                                               (3.11) 

If 𝑥 is a fixed point of 𝜙, i.e., 𝜙𝑥 = 𝑥, then 𝑥 is a solution of (1.1). First, we estimate 

(𝜙𝑥)(𝜉) . By (𝐻3), we have 

                      (𝜙𝑥)(𝜉) ≥ ∑  

𝜉−1

𝑡=𝜉1

[
1

𝑝(𝑡)
(
𝛼

2
+ 0)]

1
𝑟

= (
𝛼

2
)

1
𝑟
𝑣(𝜉)  .                                                       (3.12) 

Now, we establish (𝜙𝑥)(𝜉) from above. For 𝑥 in 𝑇, as we have 𝑥𝑠(𝜎(𝜁)) ≤ (𝛼
1

𝛼𝑣(𝜎(𝜁)))
𝑠

. 

 



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 Then, by (3.9), 

(𝜙𝑥)(𝜉) ≤ ∑  

𝜉−1

𝑡=𝜉1

  [
1

𝑝(𝑡)
[
𝛼

2
+∑  

∞

𝜁=𝑡

 𝑝(𝜁)𝑥𝑠(𝜎(𝜁))]]

1
𝑟

 

≤ 𝛼
1
𝑟𝑣(𝜉).                                                                                  (3.13) 

Therefore, 𝜙 maps 𝑇 to 𝑇, 

Next, we find a fixed point for 𝜙 in 𝑇. 

Let us define a sequence of functions in 𝑇 by the recurrence relation 

𝜎0(𝜉) = 0,   for 𝜉 ≥ 𝜉0,

            𝜎1(𝜉) = (𝜙𝜎0)(𝜉) = {
0,  if 𝜉 < 𝜉1,

𝛼
1

𝑟𝑣(𝜉),  if 𝜉 ≥ 𝜉1,

𝜎n+1(𝜉) = (𝜙𝜎n)(𝜉),   for  n ≥ 1,   𝜉 ≥ 𝜉1.

                                                                                  (3.14) 

Note that for each fixed 𝜉, we have 𝜎1(𝜉) ≥ 𝜎0(𝜉). 

Using Mathematical induction, we can show that 𝜎n+1(𝜉) ≥ 𝜎n(𝜉). Therefore, the sequence 

{𝜎n} converges pointwise to a sequence 𝜎. 

Using the Lebesgue dominated convergence theorem, we can show that 𝜎 is a fixed point of 

𝜙 in 𝑇. This shows under assumption (3.9), there is a nonoscillatory solution that dose not 

converge to zero. 

This concludes the proof. 

Theorem 3.2. Assume that there exists a constant 𝛽2, the quotient of two positive odd 

integers, such that 0 < 𝑟 < 𝛽2 < 𝑠. If (𝐻1) − (𝐻4) hold and 𝑝(𝜉) is nondecreasing, then 

each solution of (1.1) is oscillatory if and only if 

                                          ∑  ∞
𝑠=𝜉1

[
1

𝑝(𝑠)
∑  ∞
𝜁=𝑠  𝑞(𝜁)]

1

𝑟
= ∞.                                                     (3.15) 

Proof.  On the contrary, consider that 𝑥(𝜉) is an eventually positive solution that does not 

converge to zero. Using the same argument as in  Lemma 2.1, there exits 𝜉1 ≥ 𝜉0 such that 

𝑥(𝜎(𝜉)) > 0 and 𝑝(𝜉)(Δ𝑥(𝜉))𝑟 is positive and nonincreasing.  

Since 𝑝(𝜉) > 0, 𝑥(𝜉) is increasing for 𝜉 ≥ 𝜉1. Using 𝑥(𝜉) ≥ 𝑥(𝜉1), we have 

                                    𝑥𝑠(𝜉) ≥ 𝑥𝑠−𝛽2(𝜉)𝑥𝛽2(𝜉) ≥ 𝑥𝑠−𝛽2(𝜉1)𝑥
𝛽2(𝜉),                                    (3.16) 

and hence 

                               𝑥𝑠 (𝜎(𝜉)) ≥ 𝑥𝑠−𝛽2(𝜉1)𝑥
𝛽2(𝜎(𝜉)),   for  𝜉 ≥ 𝜉2                                         (3.17) 

 



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Using (3.17) and 𝜎(𝜉) ≥ 𝜎0(𝜉), from (2.10), we have 

                       𝑝(𝜉)(Δ𝑥(𝜉))𝑟 ≥ 𝑥𝑠−𝛽2(𝜉1)𝑥
𝛽2(𝜎0(𝜉))∑  ∞

𝑡=𝜉 𝑞(𝑡),   for  𝜉 ≥ 𝜉2.               (3.18) 

From 𝑝(𝜉)(Δ𝑥(𝜉))r being nonincreasing and 𝜎0(𝜉) ≤ 𝜉, we have 

       𝑝(𝜎0(𝜉))(Δ𝑥(𝜎0(𝜉)))
𝑟
≥ 𝑝(𝜉)(Δ𝑥(𝜉))𝑟.                                                                           (3.19) 

We apply this in the left-hand side of (3.18). Then, dividing by 𝑝(𝜎0(𝜉))𝑥
𝛽2(𝜎0(𝜉)) > 0 and 

raising both side to the  
1

𝑟
  power, we get 

                                 
Δ𝑥(𝜎0(𝜉))

𝑥
𝛽2
𝑟 (𝜎0(𝜉))

≥ [
𝑥𝑠−𝛽2(𝜉1)

𝑝(𝜎0(𝜉))
∑  ∞
𝑡=𝜉  𝑞(𝑡)]

1

𝑟
  ,      for 𝜉 ≥ 𝜉2.                               (3.20) 

 Multiplying the left - hand side by 
Δ𝜎0(𝜉)

𝜎0
≥ 1 and summing from 𝜉2 to 𝜉 − 1, we have 

1

  𝜎0
∑

Δ𝑥(𝜎0(t))Δ𝜎0(t)

𝑥
𝛽2
𝑟 (𝜎0(t))

𝜉−1

𝑡=𝜉2

≥ 𝑥𝑠−𝛽2(𝜉1) [∑
1

𝑝(𝜎0(t))

𝜉−1

𝑡=𝜉2

∑𝑞(𝜁)

∞

𝜁=𝑡

]

1
r

.                                      (3.21) 

On the left-hand side, since 𝑟 < 𝛽2, using summation by parts, we have 

                        

𝑥
−𝛽2
𝑟 𝜎0(𝜉)𝑥(𝜎0(𝜉)) − x

−𝛽2
𝑟 𝜎0(𝜉2)𝑥(𝜎0(𝜉2))

≤ ∑  
𝜉−1
𝑠=𝜉2

 𝑥(𝜎0(𝑠 + 1)) [
−(

𝛽2
𝑟
)𝑥
𝛽2
𝑟  −1 𝜎0(𝑠)

𝑥
𝛽2
𝑟 (𝜎0(𝑠))𝑥

𝛽2
𝑟 (𝜎0(𝑠+1))

] < ∞.
                                      

(3.22) 

On the right-hand side of (3.21), we use that 𝑝(𝜎0(𝑡)) ≤ 𝑝(𝑡) to conclude that (3.15) implies 

the right hand side approaching +∞ as 𝑦 ⟶ ∞, which is a contradiction. 

Hence, the solution 𝑥(𝜉) cannot be eventually positive. For eventually negative solutions, the 

same change of variables is used as in Theorem 3.1 and is proceed above. 

In order to prove the necessity part, we assume that (3.15) does not hold and obtain an 

eventually positive solution that does not converge to zero. 

If (3.15) does not hold, then for each 𝛼 > 0 there exists 𝜉1 ≥ 𝜉0 such that 

         ∑  ∞
𝑡=𝜉1

[
1

𝑝(𝑡)
∑  ∞
𝜁=𝑡  𝑞(𝜁)]

1

𝑟
≤

𝛼
(1−

𝑠
𝑟
)

2
,  for all 𝜉 ≥ 𝜉1.                                                  (3.23) 

We define 

                          𝑇 = {𝑥:
𝛼

2
≤ 𝑥(𝜉) ≤ 𝛼,   for  𝜉 ≥ 𝜉1} .                                                          (3.24) 

 

 

 



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we define an operator 𝜙 on 𝑇 by 

(𝜙𝑥)(𝜉) = {

0,    if 𝜉 ≤ 𝜉1,

𝛼

2
+ ∑  

𝜉−1
𝑡=𝜉1

1

𝑝(𝑡)
[∑  ∞

𝜁=𝑡  𝑞(𝜁)𝑥
𝑠(𝜎(𝜁))]

1

r   ,  if 𝜉 > 𝜉1.
                             (3.25) 

If 𝑥 is a fixed point of 𝜙, i.e., 𝜙𝑥 = 𝑥, then, 𝑥 is a solution of (1.1). First, we estimate 

(𝜙𝑥)(𝜉). Let 𝑥 ∈ 𝑀, we have 

(𝜙𝑥)(𝜉) ≥
𝛼

2
+ 0, 

Now, we estimate (𝜙𝑥)(𝜉) from above. Let 𝑥 ∈ 𝑀. Then 𝑥 ≤ 𝛼 and by (3.23), we have 

(𝜙𝑥)(𝜉) ≤
𝛼

2
+ 𝛼

𝑠
𝑟 ∑  

𝜉−1

𝑡=𝜉1

  [
1

𝑝(𝑡)
∑  

∞

𝜁=𝑡

 𝑞(𝜁)]

1
𝑟

 

                                                            ≤
𝛼

2
+
𝛼

2
= 𝛼.                                                                   (3.26) 

Therefore, 𝜙 maps 𝑇 to 𝑇, 

We find a fixed point for 𝜙 in 𝑇. 

Let us define a sequence of functions in T by the recurrence relation 

    𝜎0(𝜉) = 0,                                for 𝜉 ≥ 𝜉0, 

     𝜎1(𝜉) = (𝜙𝜎0)(𝜉) = 1,     for  𝜉 ≥ 𝜉0, 

                                                𝜎n+1(𝜉) = ( 𝜙𝜎n)(𝜉),         for n≥ 1, 𝜉 ≥ 𝜉1.                       (3.27) 

Note that for each fixed 𝜉, we have 𝜎1(𝜉) ≥ 𝜎0(𝜉). 

Using Mathematical induction, we can show that 𝜎n+1(𝜉) ≥ 𝜎n(𝜉). Therefore, the sequence 

{𝜎n} converges pointwise to a sequence 𝜎 in 𝑇. 

Then, 𝜎 is a fixed point of 𝜙 and a positive solution of (1.1). 

This complete the proof. 

4. Example 

Example 4.1. Consider the second order half-linear delay difference equation 

                               Δ [
1

𝜉
(Δ𝑥(𝜉))

7

3] + 2 
7

3 [
2𝜉+1

𝜉2+𝜉
] (𝑥(7𝜉 − 3))

1

3 = 0.                                     (4.1) 

where, 𝑝(𝜉) =
1

𝜉
,  𝑞(𝜉) = 2

 7

 3  [
2𝜉 + 1

𝜉 2 + 𝜉
] ,  𝜎(𝜉) = 7𝜉 − 3, 𝑠 =

1

3
,  𝑟 =

7

3
 , 

𝛽1 =
5

3
    we have 0 < 𝑠 < 𝛽1 < 𝑟. 

∑  ∞
𝜁=0 𝑞(𝜁)𝑣 

𝑠(𝜎(𝜁)) = ∑  ∞
𝜁=0 2 

7

3 [
2𝜁+1

𝜁 2+𝜁
]∑  

𝜉−1
𝑡=𝜉 𝑡 

3

7 = ∞. 



Communications on Applied Nonlinear Analysis 

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127 
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Hence all the conditions of Theorem 3.1 are satisfied. Hence every solution of (4.1) is 

oscillatory. One of such solution of equation (1.1) is 𝑥(𝜉) = (−1)𝜉+1. 

Example 4.2. Consider the second order half-linear delay difference equation 

                                Δ [
1

𝜉2
(Δ𝑥(𝜉))

1

3] + 2 
1

3 [
2𝜉2+2𝜉+1

𝜉4+2𝜉3+𝜉2
] (𝑥(𝜉 − 2))

7

3 = 0.                                                 

(4.2) 

where, 𝑝(𝜉) =
1

𝜉 2
,  𝑞(𝜉) = 2 

1

3  [
2 𝜉 2+ 2𝜉 + 1

𝜉 4+ 2𝜉 3+𝜉 2
] ,  𝜎(𝜉) = 𝜉 − 2, 𝑠 =

7

3
,   𝑟 =

1

3
 , 

𝛽1 =
5

3
 we have 𝑠 > 𝛽1 > 𝑟. 

∑  ∞
𝑠=𝜉1

[
1

𝑝(𝑠)
∑  ∞
𝜁=𝑠  𝑞(𝜁)]

1

𝑟
= ∑  ∞

𝑠=𝜉1
[𝑠2∑  ∞

𝜁=𝑠  2
1

3 [
2 𝜁 2+2𝜁+1

𝜁4+2𝜁3+𝜁2
]]

3

= ∞. 

Hence all the conditions of Theorem 3.2 are satisfied. Hence every solution of (4.2) is 

oscillatory. One of such solution of equation (1.1) is 𝑥(𝜉) = (−1)𝜉+1. 

5. Conclusion 

In this paper, we established necessary and sufficient conditions for the oscillation of solution 

to second order half-linear delay difference equation. The above discussed examples illustrate 

the significance and relevance of the proven results. 

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ISSN: 1074-133X 

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