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OSCILLATORY PROPERTIES OF THIRD-ORDER

NEUTRAL DELAY DIFFERENCE EQUATIONS.

S. Revathy 1 and R. Kodeeswaran 2

1 Department of Mathematics, Selvam College of Technology, Namakkal, Tamilnadu, India. 2 Department of Mathematics, Kandaswami Kandar’s College, P.Velur, Namakkal,

Tamilnadu, India.

1 E-Mail Address :[email protected], 2 E-mail Address: [email protected]

ABSTRACT: The aim of this paper is to investigate the oscillatory behavior of solutions of

third-order linear neutral delay difference equation of the term

,0,0)()()()()( 012 tttxtptytctc

where ).()()()( txtqtxty By using comparison principles with associated first and

second-order delay difference inequalities. Examples are given to illustrate the main results.

KEYWORDS: linear difference equation, delay, third -order

2010 Mathematics Subject Classification : 39A10.

INTRODUCTION

This research, we are considered with oscillation for the third order linear neutral

delay difference equation of the term

,0,0)()()()()( 012 tttxtptytctc (1)

where ).()()()( txtqtxty We make the below assumptions:

(LH1): )(1 tc and )(2 tc are the sequences of positive integers;

(LH2): )(tp and )(tq are the positive real sequences such that q(t) ≥ q0 > 1 𝑎𝑛𝑑 𝑝(t) ≠ 0;

(LH3): σ, are positive integers, such that 𝜎 < ;

(LH4): t + − σ ≤ t and t + − σ ≥ t − σ

For the sake of simplicity, we define the operators

E0y = y, E1y = c1∆y, E2y = c2∆(c1(∆y)), E3y = ∆(c2∆(c1(∆y))),

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Volume 10 Issue 9 - 2020

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mailto:[email protected]:[email protected]

and assume without further mention that E3y is of noncanonical type, (ie)

∑1

𝑐1(𝑠)< ∞

∞

𝑠=𝑡0

𝑎𝑛𝑑 ∑1

𝑐2(𝑠)< ∞.

∞

𝑠=𝑡0

(2)

By a solution of equation (1), we mean a real sequence {𝑥(𝑡)} defined for t ≥ t0 and

satisfies equation (1). We taken only those solutions {𝑥(𝑡)} for equation (1) which satisfy this

𝑠𝑢𝑝{|𝑥(𝑡)|: 𝑡 ≥ 𝑇} > 0 for all 𝑡 ≥ 𝑇 and assuming that (1) possesses such solutions. A

solution of equation (1) is called oscillatory if it is neither eventually positive nor eventually

negative; otherwise it is called non oscillatory.

We say that (1) have property V2 if any solution 𝑥(𝑡) of (1) is either oscillatory of

satisfy this lim𝑡→∞

𝑥(𝑡) = 0.

Oscillation problems for third-order difference equations have been investigated in

recent years, see for example, [2-10 , 12-14 ] and the references contained therein.

In [13 ] author consider the following equation

∆(𝑎𝑛∆(𝑏𝑛(∆𝑥𝑛)𝛼)) + 𝑝𝑛(∆𝑥𝑛+1)

𝛼 + 𝑞𝑛𝑓(𝑥𝜎(𝑛)) = 0, 𝑛 ≥ 𝑛0, (3)

and established some criteria for the oscillation of certain third-order difference equations

using comparison principles with a suitable couple of first order difference equations.

The above observation motivated us to study oscillation criteria for third order neutral

delay difference with noncanonical operators. In section 2, we present the oscillation of all

solutions of equation (1) and section 3, we provide some examples to illustrative the main

result.

2. Main Results

For the sake of convenience, we list the following notations to be used in the research

paper.

μ1(t) = ∑1

c1(s)

t−1

s=t1

, μ2(t) = ∑1

c2(s)

t−1

s=t1

, μ(n) = ∑μ2(s)

c1(s)

t−1

s=t1

,

ψ1(𝑡) = ∑1

𝑐1(𝑠)

∞

𝑠=𝑡

, ψ2(𝑡) = ∑1

𝑐2(𝑠)

∞

𝑠=𝑡

, ψ(𝑡) = ∑ψ1(𝑠)

𝑐2(𝑠)

∞

𝑠=𝑡

,

𝜇(𝑡, 𝑡1) = ∑1

𝑐1(𝑠)

𝑡−1

𝑠=𝑡1

∑1

𝑐2(𝑢)

𝑡−1

𝑢=𝑠

, 𝜇(𝑡, 𝑡1) = ∑1

𝑐1(𝑠)∑

1

𝑐2(𝑢)𝑢𝛽,

𝑡−1

𝑢=𝑠

𝑡−1

𝑠=𝑡1

where β is a constant satisfying



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0 ≤𝑞0𝛽

𝑞0 − 1≤

𝑡𝑝(𝑡)𝜇(𝑡, 𝑡 + − 𝜎)

𝑞(𝑡 + − 𝜎) (4)

Lemma 1: Suppose that (LH1) – (LH3) are satisfied and 𝑥(𝑡) is an eventually positive

solution of equation (1). Then

𝑦(𝑡) > 𝑥(𝑡) ≥1

𝑞(𝑡 + )[𝑦(𝑡 + ) −

𝑦(𝑡 + 2 )

𝑞(𝑡 + 2 )] (5)

and the corresponding sequence 𝑦(𝑡) belongs to one of following cases;

𝑦(𝑡) ∈ 𝐺1 ⇔ 𝑦 > 0, 𝐸1𝑦 < 0, 𝐸2𝑦 < 0,

𝑦(𝑡) ∈ 𝐺2 ⇔ 𝑦 > 0, 𝐸1𝑦 < 0, 𝐸2𝑦 > 0,

𝑦(𝑡) ∈ 𝐺3 ⇔ 𝑦 > 0, 𝐸1𝑦 > 0, 𝐸2𝑦 > 0,

𝑦(𝑡) ∈ 𝐺4 ⇔ 𝑦 > 0, 𝐸1𝑦 > 0, 𝐸2𝑦 < 0,

eventually.

Proof: Choose 𝑡1 > 𝑡0 such that 𝑥(𝑡 − 𝜎) > 0 𝑎𝑛𝑑 𝑥(𝑡 − ) > 0. From the definition of 𝑦,

we see that 𝑦(𝑡) > 𝑥(𝑡) > 0 and

𝑥(𝑡) =𝑦(𝑡 + ) − 𝑥(𝑡 + )

𝑞(𝑡 + )≥

1

𝑞(𝑡 + )(𝑦(𝑡 + ) −

𝑦(𝑡 + 2 )

𝑞(𝑡 + 2 ))

for 𝑡 ≥ 𝑡1. Obviously, 𝐸3𝑦(𝑡) is non-increasing, since 𝐸3𝑦(𝑡) = −𝑝(𝑡)𝑥(𝑡 − 𝜎) ≤ 0.

Hence 𝐸1𝑦(𝑡) and 𝐸2𝑦(𝑡) are of one sign eventually, which implies that four cases 𝐺1 − 𝐺4

are possible for 𝑦(𝑡).

Next we state an auxiliary criterion for the nonexistence of positive increasing

solutions of (1). As will be shown later, this condition is already included in those eliminating

solutions from the class 𝐺1. In the proof, we will take advantage of the useful fact

lim𝑡→∞

𝜇(𝑡 + )

𝜇(𝑡)= lim

𝑡→∞

𝜇1(𝑡 + )

𝜇1(𝑡)= 1, (6)

which follows from (2).

Lemma 2: Suppose that (LH1) – (LH3) are satisfied. If

∑ψ2(𝑠)𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

∞

𝑠=𝑡0

= ∞, (7)

then 𝐺3 = 𝐺4 = 𝜑.



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Proof: For the sake of contradiction, let (7) be satisfied but 𝑦 𝜖 𝐺3 ∪ 𝐺4. Choose 𝑡1 > 𝑡0

such that 𝑥(𝑡) > 0, 𝑥(𝑡 − 𝜎) > 0 𝑎𝑛𝑑 𝑥(𝑡 − ) > 0.

Assume that 𝑦 𝜖 𝐺3. Since 𝐸2𝑦 is decreasing, we have

𝐸1𝑦(𝑡) ≥ ∑1

𝑐2(𝑠)

𝑡−1

𝑠=𝑡1

𝐸2𝑦(𝑠) ≥ 𝐸2𝑦(𝑡)𝜇2(𝑡).

Thus,

∆ (𝐸1𝑦(𝑡)

𝜇2(𝑡)) =

𝐸2𝑦(𝑡)𝜇2(𝑡) − 𝐸1𝑦(𝑡)

𝑐2(𝑡)𝜇22(𝑡 + 1)

≤ 0.

Therefore,)1(

)(

2

1

t

tyE

is non-increasing and moreover, this fact yields

𝑦(𝑡) ≥ ∑𝜇2(𝑡)

𝑐1(𝑠)𝜇2(𝑡)

𝑡−1

𝑠=𝑡1

𝐸1𝑦(𝑠) ≥𝐸1𝑦(𝑡)𝜇(𝑡)

𝜇2(𝑡)𝑓𝑜𝑟 𝑡 ≥ 𝑡1.

Consequently,)(

)(

t

ty

is non-increasing, since

∆ (𝑦(𝑡)

𝜇(𝑡)) =

𝐸1𝑦(𝑡)𝜇(𝑡) − 𝑦(𝑡)𝜇2(𝑡)

𝑐1(𝑡)𝜇2(𝑡 + 1)≤ 0.

From 𝑡 + 2 ≥ 𝑡 + , we have

𝑦(𝑡 + 2 ) ≤𝜇(𝑡 + 2 )

𝜇(𝑡 + )𝑦(𝑡 + ) (8)

using this in (5), we find that

𝑥(𝑡) ≥𝑦(𝑡 + )

𝑞(𝑡 + )[1 −

𝜇(𝑡 + 2 )

𝜇(𝑡 + )𝑞(𝑡 + 2 )] , 𝑡 ≥ 𝑡1

By virtue of (LH2) and (6), there is 𝑡2 ≥ 𝑡1 such that for any constant 𝜀 ∈ (0, 𝑞0 − 1) and 𝑡 ≥

𝑡2,

𝜇(𝑡 + 2 )

𝜇(𝑡 + )𝑞(𝑡 + 2 )≤

1 + 𝜀

𝑞0,

which implies,

𝑥(𝑡) ≥𝑦(𝑡 + )

𝑞(𝑡 + )[1 −

1 + 𝜀

𝑞0] > 0. (9)



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Combining (9) with (1) we have

0 ≥ 𝐸3𝑦(𝑡) + (1 −1 + 𝜀

𝑞0)

𝑝(𝑡)

𝑞(𝑡 + − 𝜎)𝑦(𝑡 + − 𝜎),

≥ 𝐸3𝑦(𝑡) + 𝑘 (1 −1 + 𝜀

𝑞0)

𝑝(𝑡)

𝑞(𝑡 + − 𝜎), (10)

where we used the fact that 𝑦 is increasing, and set 𝑘 = 𝑦(𝑡2 + − 𝜎) < 𝑦(𝑡 + − 𝜎).

Summing (10) from 𝑡2 to 𝑡 − 1, we get

𝐸2𝑦(𝑡) ≤ 𝐸2𝑦(𝑡2) − 𝑘 (1 −1 + 𝜀

𝑞0) ∑

𝑝(𝑠)

𝑞(𝑠 + − 𝜎).

𝑡−1

𝑠=𝑡2

(11)

On the other hand, from (2) and (7), it follows that

∑𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

∞

𝑠=𝑡0

= ∞,

which in view of (11), contradicts the positivity of 𝐸2𝑦. Now, assume that 𝑦 ∈ 𝐺4 for 𝑡 ≥ 𝑡1.

Using the monotonicity of 𝐸1𝑦, we find

𝑦(𝑡) ≥ ∑1

𝑐1(𝑠)

𝑡−1

𝑠=𝑡1

𝐸1𝑦(𝑠) ≥ 𝐸1𝑦(𝑡)𝜇1(𝑡).

Thus, one can see that

∆ (𝑦(𝑡)

𝜇1(𝑡)) =

𝐸1𝑦(𝑡)𝜇1(𝑡) − 𝑦(𝑡)

𝑐1(𝑡)𝜇12(𝑡 + 1)

≤ 0,

which implies that )(

)(

1 t

ty

is non-increasing. Hence,

𝑦(𝑡 + 2 ) ≤𝜇1(𝑡 + 2 )

𝜇1(𝑡 + )𝑦(𝑡 + ).

we use (6) to arrive at (10), which holds for any 𝜀 > 0 𝑎𝑛𝑑 𝑡 ≥ 𝑡2 𝑓𝑜𝑟 𝑡2 ≥ 𝑡1 sufficiently

large. Summing (10) from 𝑡2 to 𝑡 − 1, we have

−∆(𝐸1𝑦(𝑡)) ≥ 𝑘 (1 −1 + 𝜀

𝑞0)

1

𝑐2(𝑡)∑

𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

𝑡−1

𝑠=𝑡2

.

Summing the above inequality again from 𝑡2 to 𝑡 − 1, we find that



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𝐸1𝑦(𝑡) ≤ 𝐸1𝑦(𝑡2) − 𝑘 (1 −1 + 𝜀

𝑞0) ∑

1

𝑐2(𝑢)

𝑡−1

𝑢=𝑡2

∑𝑝(𝑠)

𝑞(𝑠 + − 𝜎).

𝑢−1

𝑠=𝑡2

Letting 𝑡 𝑡𝑜 ∞ changing the summation and using (7) we obtain

0 ≤ 𝐸1𝑦(∞) ≤ 𝐸1𝑦(𝑡2) − 𝑘 (1 −1 + 𝜀

𝑞0) ∑

1

𝑐2(𝑢)

∞

𝑢=𝑡2

∑𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

𝑢−1

𝑠=𝑡2

= 𝐸1𝑦(𝑡2) − 𝑘 (1 −1 + 𝜀

𝑞0) ∑

𝑝(𝑠)ψ2(𝑠)

𝑞(𝑠 + − 𝜎)

∞

𝑢=𝑡2

= −∞,

a contradiction. The proof is complete.

Theorem 1: Suppose that (LH1) – (LH3) are satisfied. If

∑ψ(𝑠)𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

∞

𝑠=𝑡0

= ∞, (12)

then (1) has property V2.

Proof: Assume that 𝑥(𝑡) is a non-oscillatory solution of (1). Without loss of generality, we

make it eventually positive. We suppose that 𝑥(𝑡) > 0, 𝑥(𝑛 − 𝜎) > 0 𝑎𝑛𝑑 𝑥(𝑛 − ) > 0. By

conclusion of Lemma 1, 𝑦 ∈ 𝐺𝑖 , 𝑖 = 1,2,3, … . 𝑓𝑜𝑟 𝑡 ≥ 𝑡1. First it is easy to see that in view of

(2), condition (12) implies

∑ψ2(𝑠)𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

∞

𝑠=𝑡0

= ∑𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

∞

𝑠=𝑡0

= ∞.

Thus by Lemma 2, 𝐺3 = 𝐺4 = 𝜑 and so either 𝑦 𝜖 𝐺1 𝑜𝑟 𝑦 𝜖 𝐺2. Using (LH2) and the fact that

𝑦 is non-increasing in (5), we have,

𝑥(𝑡) ≥𝑦(𝑡 + )

𝑞(𝑡 + )[1 −

1

𝑞(𝑡 + 2 )] ≥ (1 −

1

𝑞0)

𝑦(𝑡 + )

𝑞(𝑡 + ) (13)

On the other hand, since ∆𝑦 < 0 there is a constant 𝑙 > 0 such that

lim𝑡→∞

𝑦(𝑡) = 𝑙 < ∞

If 𝑙 > 0, there exists a 𝑡2 ≥ 𝑡1 such that 𝑦(𝑡) ≥ 𝑙 𝑓𝑜𝑟 𝑡 ≥ 𝑡2. Hence, from (13), we see that

𝑥(𝑡) ≥𝑙(𝑞0 − 1)

𝑞0

1

𝑞(𝑡 + ), 𝑡 ≥ 𝑡2.

Using this in (1), we find



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𝐸3𝑦(𝑡) +𝑙(𝑞0 − 1)

𝑞0

𝑝(𝑡)

𝑞(𝑡 + − 𝜎)≤ 0, 𝑡 ≤ 𝑡2. (14)

If we assume that 𝑦 𝜖 𝐺1, then by summing (14) from 𝑡2 to 𝑡 − 1 we have

−∆(𝐸1𝑦(𝑡)) ≥𝑙(𝑞0 − 1)

𝑞0

1

𝑐2(𝑡)∑

𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

𝑡−1

𝑠=𝑡2

.

Summing the above inequality from𝑡2 to 𝑡 − 1, we obtain

−∆𝑦(𝑡) ≥𝑙(𝑞0 − 1)

𝑞0

1

𝑐1(𝑡)∑

1

𝑐2(𝑢)

𝑡−1

𝑢=𝑡2

∑𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

𝑢−1

𝑠=𝑡2

. (15)

Summing (15) from 𝑡2 to t-1, letting 𝑡 to ∞ and changing the summation in the resulting

inequality, and taking (12), we get

𝑙 = 𝑦(∞) ≤ 𝑦(𝑡2) −𝑙(𝑞0 − 1)

𝑞0∑

1

𝑐1(𝑣)

∞

𝑣=𝑡2

∑1

𝑐2(𝑢)

𝑣−1

𝑢=𝑡2

∑𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

𝑢−1

𝑠=𝑡2

(16)

= 𝑦(𝑡2) −𝑙(𝑞0 − 1)

𝑞0∑

ψ(𝑠)𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

∞

𝑠=𝑛2

= −∞,

is contradiction. Hence, 𝑙 = 0 in this case.

If we take 𝑦 ∈ 𝐺2, then by summing (14) from 𝑡2 to 𝑡 −1 and using (12) we arrive at

𝐸2𝑦(𝑡) ≤ 𝐸2𝑦(𝑡2) −𝑙(𝑞0 − 1)

𝑞0∑

𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

𝑡−1

𝑠=𝑡2

→ −∞ 𝑎𝑠 𝑡 → ∞, (17)

which contradicts the positivity of 𝐸2𝑦 and so 𝑙 = 0. Since 𝑦(𝑡) ≥ 𝑥(𝑡), we find lim𝑡→∞

𝑥(𝑡) =

0. This proof is complete.

In the following result, we present a criterion for nonexistence of 𝐺1type solutions, based on

comparison of the studied Equation (1) with an associated first - order delay difference

inequality. The given criterion also excludes solutions from classes 𝐺3and 𝐺4.


lim inf𝑡→∞

∑𝑝(𝑠)ψ(𝑠)

𝑞(𝑠 + − 𝜎)

𝑡−1

𝑠=𝑡+𝜉−𝜎

> ∞, (18)

then 𝐺1 = 𝐺3 = 𝐺4 = 𝜑.



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Proof: For the sake of contradiction, let (18) be satisfied but 𝑦 ∈ 𝐺1 ∪ 𝐺3 ∪ 𝐺4. Choose 𝑡1 >

𝑡0 such that 𝑥(𝑡) > 0, 𝑥(𝑛 − 𝜎) > 0 𝑎𝑛𝑑 𝑥(𝑛 − ) > 0. Assume first that𝑦 ∈ 𝐺1. As in the

proof of Theorem 1, we arrive at (13), which in view of (1) yields

𝐸3𝑦(𝑡) +𝑞0 − 1

𝑞0

𝑝(𝑡)

𝑞(𝑡 + − 𝜎)𝑦(𝑡 + − 𝜎) ≤ 0. (19)

Define the function

𝑤(𝑡) = ψ1(𝑡)𝐸1𝑦(𝑡) + 𝑦(𝑡) (20)

From

𝑦(𝑡) ≥ − ∑1

𝑐1(𝑠)𝐸1𝑦(𝑠) ≥ −𝐸1𝑦(𝑡)ψ1(𝑡)

∞

𝑠=𝑡

= −𝐸1𝑦(𝑡 + 1)ψ1(𝑡) (21)

and

∆𝑤(𝑡) = ψ1(𝑡)Δ(𝐸1𝑦(𝑡)) =ψ1(𝑡)

𝑐2(𝑡)𝐸2𝑦(𝑡) < 0,

we see that 𝑤(𝑡) is a strictly decreasing eventually positive function. Using the definition of

𝑤 in (19), we have

Δ (𝑐2(𝑡)

ψ1(𝑡)Δ𝑤(𝑡)) +

𝑞0 − 1

𝑞0

𝑝(𝑡)𝑦(𝑡 + − 𝜎)

𝑞(𝑡 + − 𝜎)≤ 0.

Hence 𝑤 is a solution of the second-order delay difference inequality

∆ (𝑐2(𝑡)∆𝑤(𝑡)

ψ1(𝑡)) +

𝑞0 − 1

𝑞0 𝑝(𝑡)𝑤(𝑡 + − 𝜎)

𝑞(𝑡 + − 𝜎)≤ 0. (22)

Similarly as before, we define the function u by

𝑢(𝑡) =ψ(𝑡)𝑐2(𝑡)

ψ1(𝑡)Δ𝑤(𝑡) + 𝑤(𝑡)

From

Δ𝑢(𝑡) = Δ (𝑐2(𝑡)Δ𝑤(𝑡)

ψ1(𝑡)) ψ(𝑡)

= 𝐸3𝑦(𝑡)ψ(𝑡) ≤ 0

and



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𝑤(𝑡) ≥ − ∑ψ1(𝑠)𝑐2(𝑠)

𝑐2(𝑠)ψ1(𝑠)

∞

𝑠=𝑡

Δ𝑤(𝑠) ≥ −𝑐2(𝑡)

ψ1(𝑡)Δ𝑤(𝑡)ψ(𝑡)

= −𝑐2(𝑡 + 1)

ψ1(𝑡 + 1)Δ𝑤(𝑡 + 1)ψ(𝑡), (23)

We conclude that 𝑢 is eventually positive and non-increasing. Using the definition of

𝑢 in (22), it is easy to see that 𝑢 satisfies the first-order delay difference inequality

Δ𝑢(𝑡) +𝑞0 − 1

𝑞0

𝑝(𝑡)ψ(𝑡)

𝑞(𝑡 + − 𝜎)𝑢(𝑡 + − 𝜎) ≤ 0. (24)

However, by [1 Theorem 6.20.5], condition (18) ensures that the above inequality does not

posses a positive solution, which is a contradiction.

To show that also 𝐺3 = 𝐺4 = 𝜑, it suffices to note that (12) is necessary for validity of

(18) since otherwise, the left-hand side of (18) would equal zero. The conclusion then

immediately follows from Theorem 1. The proof is complete.

Lemma 4: Suppose that (LH1) – (LH4) are satisfied and (7) holds. If for any 𝑡1 ≥ 𝑡0 large

enough,

lim𝑡→∞

𝑠𝑢𝑝 ∑ (ψ(𝑠)𝑝(𝑠)

𝑞(𝑠 + − 𝜎)− (

𝑞0𝑞0 − 1

)ψ1(𝑠)

4ψ(𝑠)𝑐2(𝑠))

𝑡−1

𝑠=𝑡1

>𝑞0

𝑞0 − 1 (25)

then 𝐺1 = 𝐺3 = 𝐺4 = 𝜑.

Proof: For the sake of contradiction, let (18) be satisfied but 𝑦 ∈ 𝐺1 ∪ 𝐺3 ∪ 𝐺4. Choose 𝑡1 >

𝑡0 such that 𝑥(𝑡) > 0, 𝑥(𝑡 − 𝜎) > 0 𝑎𝑛𝑑 𝑥(𝑡 − ) > 0. Assume that 𝑦 ∈ 𝐺1. We proceed as

in the proof of Lemma 3 to obtain (22), where 𝑤 is given by (20). Consider the function ρ

defined by

𝜌(𝑡) =𝑐2(𝑡)∆𝑤(𝑡)

ψ1(𝑡)𝑤(𝑡) , (26)

Clearly, 𝜌 < 0, from (23), it is easy to see that

−1 ≤ ψ(𝑡)𝜌(𝑡) < 0 (27)

Using (22) together with (26), we have

Δ𝜌(𝑡) = Δ (𝑐2(𝑡)Δ𝑤(𝑡)

ψ1(𝑡))

1

𝑤(𝑡)−

𝑐2(𝑡 + 1)[Δ𝑤(𝑡 + 1)]2

ψ1(𝑡 + 1)𝑤2(𝑡 + 1)

≤ − (𝑞0 − 1

𝑞0)

𝑝(𝑡)

𝑞(𝑡 + − 𝜎)

𝑤(𝑡 + − 𝜎)

𝑤(𝑡)−

ψ1(𝑡 + 1)𝜌2(𝑡 + 1)

𝑐2(𝑡 + 1) (28)



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≤ − (𝑞0 − 1

𝑞0)

𝑝(𝑡)

𝑞(𝑡 + − 𝜎)−

ψ1(𝑡 + 1)𝜌2(𝑡 + 1)

𝑐2(𝑡 + 1).

Multiplying both side of (28) by ψ(𝑡) and summing the resulting inequality from 𝑡1 to 𝑡 −

1, we have

ψ(𝑡)𝜌(𝑡) ≤ ψ(𝑡1)𝜌(𝑡1) + ∑𝜌(𝑠 + 1)ψ1(𝑠 + 1)

𝑐2(𝑠 + 1)

𝑡−1

𝑠=𝑡1

−𝑞0 − 1

𝑞0∑

ψ(𝑠)𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

𝑡−1

𝑠=𝑡1

− ∑ψ1(𝑠 + 1)𝜌

2(𝑠 + 1)ψ(𝑠)

𝑐2(𝑠 + 1)

𝑡−1

𝑠=𝑡1

= ψ(𝑡1)𝜌(𝑡1) −𝑞0 − 1

𝑞0∑

ψ(𝑠)𝑝(𝑠)

𝑞(𝑠 + − 𝜎)

𝑡−1

𝑠=𝑡1

+ ∑ψ1(𝑠 + 1)ψ(𝑠)

𝑐2(𝑠 + 1)

𝑡−1

𝑠=𝑡1

[𝜌(𝑠 + 1)

ψ(𝑠)− 𝜌2(𝑠 + 1)]

≤ − (𝑞0 − 1

𝑞0) ∑ [

ψ(𝑠)𝑝(𝑠)

𝑞(𝑠 + − 𝜎)− (

𝑞0𝑞0 − 1

)ψ1(𝑠 + 1)

4ψ(𝑠)𝑐2(𝑠 + 1)]

𝑡−1

𝑠=𝑡1

However, in view of (27), this inequality contradicts (25). Hence 𝐺1 = 𝜑. By Lemma 2, 𝐺3 =

𝐺4 = 𝜑 due to (7). The proof is complete.

Corollary 1: Suppose that (LH1) - (LH3) are satisfied and (7) holds. If there is a constant 𝐶𝓀

such that

ψ2(𝑡)𝑝(𝑡)𝑐2(𝑡)

𝑞(𝑡 + − 𝜎)ψ1(𝑡)≥ 𝐶𝓀 >

𝑞04(𝑞0 − 1)

, (29)

then 𝐺1 = 𝐺3 = 𝐺4 = 𝜑.

To attain the oscillation of all solutions, it remains to eliminate the solutions of 𝐺2

type.


lim𝑡→∞

𝑠𝑢𝑝 ∑𝑝(𝑠)𝜇(𝑡 + − 𝜎, 𝑠 + − 𝜎)

𝑞(𝑠 + − 𝜎)

𝑡−1


>𝑞0

𝑞0 − 1. (30)

then 𝐺2 = 𝜑.

Proof: For sake of the contradiction, let (30) be satisfied 𝑦 ∈ 𝐺2. Choose 𝑡1 > 𝑡0 such that

𝑥(𝑡) > 0, 𝑥(𝑡 − 𝜎) > 0 𝑎𝑛𝑑 𝑥(𝑡 − ) > 0. Using (13) in (1), we obtain

𝐸3𝑦(𝑡) +𝑞𝑜 − 1

𝑞0

𝑝(𝑡)

𝑞(𝑡 + − 𝜎)𝑦(𝑡 + − 𝜎) ≤ 0. (31)



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Using the monotonicity of 𝐸2𝑦, one can easily verify that

−𝐸1𝑦(𝑢) ≥ 𝐸1𝑦(𝑣) − 𝐸1𝑦(𝑢) = ∑𝐸2𝑦(𝑠)

𝑐2(𝑠)

𝑣−1

𝑠=𝑢

≥ 𝐸2𝑦(𝑣) ∑1

𝑐2(𝑠)

𝑣−1

𝑠=𝑢

(32)

for 𝑣 ≥ 𝑢 ≥ 𝑡1. Summing the latter inequality from 𝑢 𝑡𝑜 𝑣 − 1, we obtain,

𝑦(𝑢) ≥ 𝐸2𝑦(𝑣) ∑1

𝑐1(𝑠)

𝑣−1

𝑠=𝑢

∑1

𝑐2(𝑥)

𝑣−1

𝑥=𝑠

= 𝐸2𝑦(𝑣)𝜇(𝑣, 𝑢). (33)

setting 𝑢 = 𝑠 + − 𝜎 and 𝑣 = 𝑡 + − 𝜎 in (33), we find

𝑦(𝑠 + − 𝜎) ≥ 𝐸2𝑦(𝑡 + − 𝜎)𝜇(𝑡 + − 𝜎, 𝑠 + − 𝜎). (34)

On the other hand, summing (31) from 𝑡 + − 𝜎 to 𝑡 − 1 and using (34), we see that

𝐸2𝑦(𝑡 + − 𝜎) ≥ 𝐸2𝑦(𝑡 + − 𝜎) − 𝐸2𝑦(𝑡) ≥𝑞0 − 1

𝑞0∑

𝑝(𝑠)𝑦(𝑠 + − 𝜎)

𝑞(𝑠 + − 𝜎)

𝑡−1


≥𝑞0 − 1

𝑞0𝐸2𝑦(𝑡 + − 𝜎) ∑

𝑝(𝑠)𝜇(𝑡 + − 𝜎, 𝑠 + − 𝜎)

𝑞(𝑠 + − 𝜎).

𝑡−1


Dividing the above inequality by 𝐸2𝑦(𝑡 + − 𝜎) and taking the limsup on both sides of the

resulting inequality as 𝑡 → ∞, we obtain a contradiction with (30). The proof is complete.

Lemma 6: Suppose that (LH1) – (LH4) are satisfied and let β be a constant satisfying (4)

eventually. If

lim𝑡→∞

𝑠𝑢𝑝(𝑡 + − 𝜎)𝛽 ∑𝑝(𝑠)�̃�(𝑡 + − 𝜎, 𝑠 + − 𝜎)

𝑞(𝑠 + − 𝜎)

𝑡−1


>𝑞0

𝑞0 − 1, (35)

then 𝐺2 = 𝜑.

Proof: Setting 𝑢 = 𝑡 + − 𝜎 and 𝑣 = 𝑡 in (33), we obtain

𝑦(𝑡 + − 𝜎) ≥ 𝐸2𝑦(𝑡)𝜇(𝑡, 𝑡 + − 𝜎) = 𝐸2𝑦(𝑡 + 1)𝜇(𝑡, 𝑡 + − 𝜎) (36)

By (4), (31) and (36), we have



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∆ (𝑡𝛽𝐸2𝑦(𝑡)) = 𝛽𝑡𝛽−1𝐸2𝑦(𝑡 + 1) + 𝑡

𝛽𝐸3𝑦(𝑡)

≤ 𝛽𝑡𝛽−1𝐸2𝑦(𝑡 + 1) − (𝑞0 − 1

𝑞0)

𝑡𝛽𝑝(𝑡)𝑦(𝑡 + − 𝜎)

𝑞(𝑡 + − 𝜎)

≤ 𝛽𝑡𝛽−1𝐸2𝑦(𝑡 + 1) − (𝑞0 − 1

𝑞0)

𝑡𝛽𝑝(𝑡)𝐸2𝑦(𝑡 + 1)𝜇(𝑡, 𝑡 + − 𝜎)

𝑞(𝑡 + − 𝜎)

= 𝑡𝛽−1𝐸2𝑦(𝑡 + 1) [𝛽 − (𝑞0 − 1

𝑞0)

𝑡𝑝(𝑡)𝜇(𝑡, 𝑡 + − 𝜎)

𝑞(𝑡 + − 𝜎)] ≤ 0.

That is 𝑡𝛽 𝐸2𝑦(𝑡 + 1) is eventually non-increasing. From here, we obtain that

−𝐸1𝑦(𝑢) ≥ 𝐸1𝑦(𝑣) − 𝐸1𝑦(𝑢) = ∑𝐸2𝑦(𝑠)𝑠

𝛽

𝑠𝛽𝑐2(𝑠)

𝑣−1

𝑠=𝑢

≥ 𝐸2𝑦(𝑣)𝑣𝛽 ∑

1

𝑠𝛽𝑐2(𝑠)

𝑣−1

𝑠=𝑢

(37)

for 𝑣 ≥ 𝑢 ≥ 𝑡1. Proceeding as in the proof of Lemma 5 with (32) replaces by (37), one arrives

at a contradiction with (35). The proof is complete.

Theorem 2: Suppose that (LH1) – (LH4) are satisfied. If (18) (or (25)) and (30) (or (35))

hold, then (1) is oscillatory.

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[10] W.G., Kelley, A.C. Peterson, Difference Equations; An Introduction with

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