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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 Kandarβs 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) β₯ q0 > 1 πππ π(t) β 0;
(LH3): Ο, are positive integers, such that π < ;
(LH4): t + β Ο β€ t and t + β Ο β₯ t β Ο
For the sake of simplicity, we define the operators
E0y = y, E1y = c1βy, E2y = c2β(c1(βy)), E3y = β(c2β(c1(βy))),
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and assume without further mention that E3y is of noncanonical type, (ie)
β1
π1(π )< β
β
π =π‘0
πππ β1
π2(π )< β.
β
π =π‘0
(2)
By a solution of equation (1), we mean a real sequence {π₯(π‘)} defined for t β₯ t0 and
satisfies equation (1). We taken only those solutions {π₯(π‘)} for equation (1) which satisfy this
π π’π{|π₯(π‘)|: π‘ β₯ π} > 0 for all π‘ β₯ π 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 π₯(π‘) of (1) is either oscillatory of
satisfy this limπ‘ββ
π₯(π‘) = 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
β(ππβ(ππ(βπ₯π)πΌ)) + ππ(βπ₯π+1)
πΌ + πππ(π₯π(π)) = 0, π β₯ π0, (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.
ΞΌ1(t) = β1
c1(s)
tβ1
s=t1
, ΞΌ2(t) = β1
c2(s)
tβ1
s=t1
, ΞΌ(n) = βΞΌ2(s)
c1(s)
tβ1
s=t1
,
Ο1(π‘) = β1
π1(π )
β
π =π‘
, Ο2(π‘) = β1
π2(π )
β
π =π‘
, Ο(π‘) = βΟ1(π )
π2(π )
β
π =π‘
,
π(π‘, π‘1) = β1
π1(π )
π‘β1
π =π‘1
β1
π2(π’)
π‘β1
π’=π
, π(π‘, π‘1) = β1
π1(π )β
1
π2(π’)π’π½,
π‘β1
π’=π
π‘β1
π =π‘1
where Ξ² is a constant satisfying
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0 β€π0π½
π0 β 1β€
π‘π(π‘)π(π‘, π‘ + β π)
π(π‘ + β π) (4)
Lemma 1: Suppose that (LH1) β (LH3) are satisfied and π₯(π‘) is an eventually positive
solution of equation (1). Then
π¦(π‘) > π₯(π‘) β₯1
π(π‘ + )[π¦(π‘ + ) β
π¦(π‘ + 2 )
π(π‘ + 2 )] (5)
and the corresponding sequence π¦(π‘) belongs to one of following cases;
π¦(π‘) β πΊ1 β π¦ > 0, πΈ1π¦ < 0, πΈ2π¦ < 0,
π¦(π‘) β πΊ2 β π¦ > 0, πΈ1π¦ < 0, πΈ2π¦ > 0,
π¦(π‘) β πΊ3 β π¦ > 0, πΈ1π¦ > 0, πΈ2π¦ > 0,
π¦(π‘) β πΊ4 β π¦ > 0, πΈ1π¦ > 0, πΈ2π¦ < 0,
eventually.
Proof: Choose π‘1 > π‘0 such that π₯(π‘ β π) > 0 πππ π₯(π‘ β ) > 0. From the definition of π¦,
we see that π¦(π‘) > π₯(π‘) > 0 and
π₯(π‘) =π¦(π‘ + ) β π₯(π‘ + )
π(π‘ + )β₯
1
π(π‘ + )(π¦(π‘ + ) β
π¦(π‘ + 2 )
π(π‘ + 2 ))
for π‘ β₯ π‘1. Obviously, πΈ3π¦(π‘) is non-increasing, since πΈ3π¦(π‘) = βπ(π‘)π₯(π‘ β π) β€ 0.
Hence πΈ1π¦(π‘) and πΈ2π¦(π‘) are of one sign eventually, which implies that four cases πΊ1 β πΊ4
are possible for π¦(π‘).
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 πΊ1. In the proof, we will take advantage of the useful fact
limπ‘ββ
π(π‘ + )
π(π‘)= lim
π‘ββ
π1(π‘ + )
π1(π‘)= 1, (6)
which follows from (2).
Lemma 2: Suppose that (LH1) β (LH3) are satisfied. If
βΟ2(π )π(π )
π(π + β π)
β
π =π‘0
= β, (7)
then πΊ3 = πΊ4 = π.
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Proof: For the sake of contradiction, let (7) be satisfied but π¦ π πΊ3 βͺ πΊ4. Choose π‘1 > π‘0
such that π₯(π‘) > 0, π₯(π‘ β π) > 0 πππ π₯(π‘ β ) > 0.
Assume that π¦ π πΊ3. Since πΈ2π¦ is decreasing, we have
πΈ1π¦(π‘) β₯ β1
π2(π )
π‘β1
π =π‘1
πΈ2π¦(π ) β₯ πΈ2π¦(π‘)π2(π‘).
Thus,
β (πΈ1π¦(π‘)
π2(π‘)) =
πΈ2π¦(π‘)π2(π‘) β πΈ1π¦(π‘)
π2(π‘)π22(π‘ + 1)
β€ 0.
Therefore,)1(
)(
2
1
t
tyE
is non-increasing and moreover, this fact yields
π¦(π‘) β₯ βπ2(π‘)
π1(π )π2(π‘)
π‘β1
π =π‘1
πΈ1π¦(π ) β₯πΈ1π¦(π‘)π(π‘)
π2(π‘)πππ π‘ β₯ π‘1.
Consequently,)(
)(
t
ty
is non-increasing, since
β (π¦(π‘)
π(π‘)) =
πΈ1π¦(π‘)π(π‘) β π¦(π‘)π2(π‘)
π1(π‘)π2(π‘ + 1)β€ 0.
From π‘ + 2 β₯ π‘ + , we have
π¦(π‘ + 2 ) β€π(π‘ + 2 )
π(π‘ + )π¦(π‘ + ) (8)
using this in (5), we find that
π₯(π‘) β₯π¦(π‘ + )
π(π‘ + )[1 β
π(π‘ + 2 )
π(π‘ + )π(π‘ + 2 )] , π‘ β₯ π‘1
By virtue of (LH2) and (6), there is π‘2 β₯ π‘1 such that for any constant π β (0, π0 β 1) and π‘ β₯
π‘2,
π(π‘ + 2 )
π(π‘ + )π(π‘ + 2 )β€
1 + π
π0,
which implies,
π₯(π‘) β₯π¦(π‘ + )
π(π‘ + )[1 β
1 + π
π0] > 0. (9)
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Combining (9) with (1) we have
0 β₯ πΈ3π¦(π‘) + (1 β1 + π
π0)
π(π‘)
π(π‘ + β π)π¦(π‘ + β π),
β₯ πΈ3π¦(π‘) + π (1 β1 + π
π0)
π(π‘)
π(π‘ + β π), (10)
where we used the fact that π¦ is increasing, and set π = π¦(π‘2 + β π) < π¦(π‘ + β π).
Summing (10) from π‘2 to π‘ β 1, we get
πΈ2π¦(π‘) β€ πΈ2π¦(π‘2) β π (1 β1 + π
π0) β
π(π )
π(π + β π).
π‘β1
π =π‘2
(11)
On the other hand, from (2) and (7), it follows that
βπ(π )
π(π + β π)
β
π =π‘0
= β,
which in view of (11), contradicts the positivity of πΈ2π¦. Now, assume that π¦ β πΊ4 for π‘ β₯ π‘1.
Using the monotonicity of πΈ1π¦, we find
π¦(π‘) β₯ β1
π1(π )
π‘β1
π =π‘1
πΈ1π¦(π ) β₯ πΈ1π¦(π‘)π1(π‘).
Thus, one can see that
β (π¦(π‘)
π1(π‘)) =
πΈ1π¦(π‘)π1(π‘) β π¦(π‘)
π1(π‘)π12(π‘ + 1)
β€ 0,
which implies that )(
)(
1 t
ty
is non-increasing. Hence,
π¦(π‘ + 2 ) β€π1(π‘ + 2 )
π1(π‘ + )π¦(π‘ + ).
we use (6) to arrive at (10), which holds for any π > 0 πππ π‘ β₯ π‘2 πππ π‘2 β₯ π‘1 sufficiently
large. Summing (10) from π‘2 to π‘ β 1, we have
ββ(πΈ1π¦(π‘)) β₯ π (1 β1 + π
π0)
1
π2(π‘)β
π(π )
π(π + β π)
π‘β1
π =π‘2
.
Summing the above inequality again from π‘2 to π‘ β 1, we find that
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πΈ1π¦(π‘) β€ πΈ1π¦(π‘2) β π (1 β1 + π
π0) β
1
π2(π’)
π‘β1
π’=π‘2
βπ(π )
π(π + β π).
π’β1
π =π‘2
Letting π‘ π‘π β changing the summation and using (7) we obtain
0 β€ πΈ1π¦(β) β€ πΈ1π¦(π‘2) β π (1 β1 + π
π0) β
1
π2(π’)
β
π’=π‘2
βπ(π )
π(π + β π)
π’β1
π =π‘2
= πΈ1π¦(π‘2) β π (1 β1 + π
π0) β
π(π )Ο2(π )
π(π + β π)
β
π’=π‘2
= ββ,
a contradiction. The proof is complete.
Theorem 1: Suppose that (LH1) β (LH3) are satisfied. If
βΟ(π )π(π )
π(π + β π)
β
π =π‘0
= β, (12)
then (1) has property V2.
Proof: Assume that π₯(π‘) is a non-oscillatory solution of (1). Without loss of generality, we
make it eventually positive. We suppose that π₯(π‘) > 0, π₯(π β π) > 0 πππ π₯(π β ) > 0. By
conclusion of Lemma 1, π¦ β πΊπ , π = 1,2,3, β¦ . πππ π‘ β₯ π‘1. First it is easy to see that in view of
(2), condition (12) implies
βΟ2(π )π(π )
π(π + β π)
β
π =π‘0
= βπ(π )
π(π + β π)
β
π =π‘0
= β.
Thus by Lemma 2, πΊ3 = πΊ4 = π and so either π¦ π πΊ1 ππ π¦ π πΊ2. Using (LH2) and the fact that
π¦ is non-increasing in (5), we have,
π₯(π‘) β₯π¦(π‘ + )
π(π‘ + )[1 β
1
π(π‘ + 2 )] β₯ (1 β
1
π0)
π¦(π‘ + )
π(π‘ + ) (13)
On the other hand, since βπ¦ < 0 there is a constant π > 0 such that
limπ‘ββ
π¦(π‘) = π < β
If π > 0, there exists a π‘2 β₯ π‘1 such that π¦(π‘) β₯ π πππ π‘ β₯ π‘2. Hence, from (13), we see that
π₯(π‘) β₯π(π0 β 1)
π0
1
π(π‘ + ), π‘ β₯ π‘2.
Using this in (1), we find
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πΈ3π¦(π‘) +π(π0 β 1)
π0
π(π‘)
π(π‘ + β π)β€ 0, π‘ β€ π‘2. (14)
If we assume that π¦ π πΊ1, then by summing (14) from π‘2 to π‘ β 1 we have
ββ(πΈ1π¦(π‘)) β₯π(π0 β 1)
π0
1
π2(π‘)β
π(π )
π(π + β π)
π‘β1
π =π‘2
.
Summing the above inequality fromπ‘2 to π‘ β 1, we obtain
ββπ¦(π‘) β₯π(π0 β 1)
π0
1
π1(π‘)β
1
π2(π’)
π‘β1
π’=π‘2
βπ(π )
π(π + β π)
π’β1
π =π‘2
. (15)
Summing (15) from π‘2 to t-1, letting π‘ to β and changing the summation in the resulting
inequality, and taking (12), we get
π = π¦(β) β€ π¦(π‘2) βπ(π0 β 1)
π0β
1
π1(π£)
β
π£=π‘2
β1
π2(π’)
π£β1
π’=π‘2
βπ(π )
π(π + β π)
π’β1
π =π‘2
(16)
= π¦(π‘2) βπ(π0 β 1)
π0β
Ο(π )π(π )
π(π + β π)
β
π =π2
= ββ,
is contradiction. Hence, π = 0 in this case.
If we take π¦ β πΊ2, then by summing (14) from π‘2 to π‘ β1 and using (12) we arrive at
πΈ2π¦(π‘) β€ πΈ2π¦(π‘2) βπ(π0 β 1)
π0β
π(π )
π(π + β π)
π‘β1
π =π‘2
β ββ ππ π‘ β β, (17)
which contradicts the positivity of πΈ2π¦ and so π = 0. Since π¦(π‘) β₯ π₯(π‘), we find limπ‘ββ
π₯(π‘) =
0. This proof is complete.
In the following result, we present a criterion for nonexistence of πΊ1type 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 πΊ3and πΊ4.
Lemma 3: Suppose that (LH1) β (LH4) are satisfied. If
lim infπ‘ββ
βπ(π )Ο(π )
π(π + β π)
π‘β1
π =π‘+πβπ
> β, (18)
then πΊ1 = πΊ3 = πΊ4 = π.
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Proof: For the sake of contradiction, let (18) be satisfied but π¦ β πΊ1 βͺ πΊ3 βͺ πΊ4. Choose π‘1 >
π‘0 such that π₯(π‘) > 0, π₯(π β π) > 0 πππ π₯(π β ) > 0. Assume first thatπ¦ β πΊ1. As in the
proof of Theorem 1, we arrive at (13), which in view of (1) yields
πΈ3π¦(π‘) +π0 β 1
π0
π(π‘)
π(π‘ + β π)π¦(π‘ + β π) β€ 0. (19)
Define the function
π€(π‘) = Ο1(π‘)πΈ1π¦(π‘) + π¦(π‘) (20)
From
π¦(π‘) β₯ β β1
π1(π )πΈ1π¦(π ) β₯ βπΈ1π¦(π‘)Ο1(π‘)
β
π =π‘
= βπΈ1π¦(π‘ + 1)Ο1(π‘) (21)
and
βπ€(π‘) = Ο1(π‘)Ξ(πΈ1π¦(π‘)) =Ο1(π‘)
π2(π‘)πΈ2π¦(π‘) < 0,
we see that π€(π‘) is a strictly decreasing eventually positive function. Using the definition of
π€ in (19), we have
Ξ (π2(π‘)
Ο1(π‘)Ξπ€(π‘)) +
π0 β 1
π0
π(π‘)π¦(π‘ + β π)
π(π‘ + β π)β€ 0.
Hence π€ is a solution of the second-order delay difference inequality
β (π2(π‘)βπ€(π‘)
Ο1(π‘)) +
π0 β 1
π0 π(π‘)π€(π‘ + β π)
π(π‘ + β π)β€ 0. (22)
Similarly as before, we define the function u by
π’(π‘) =Ο(π‘)π2(π‘)
Ο1(π‘)Ξπ€(π‘) + π€(π‘)
From
Ξπ’(π‘) = Ξ (π2(π‘)Ξπ€(π‘)
Ο1(π‘)) Ο(π‘)
= πΈ3π¦(π‘)Ο(π‘) β€ 0
and
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π€(π‘) β₯ β βΟ1(π )π2(π )
π2(π )Ο1(π )
β
π =π‘
Ξπ€(π ) β₯ βπ2(π‘)
Ο1(π‘)Ξπ€(π‘)Ο(π‘)
= βπ2(π‘ + 1)
Ο1(π‘ + 1)Ξπ€(π‘ + 1)Ο(π‘), (23)
We conclude that π’ is eventually positive and non-increasing. Using the definition of
π’ in (22), it is easy to see that π’ satisfies the first-order delay difference inequality
Ξπ’(π‘) +π0 β 1
π0
π(π‘)Ο(π‘)
π(π‘ + β π)π’(π‘ + β π) β€ 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 πΊ3 = πΊ4 = π, 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) β (LH4) are satisfied and (7) holds. If for any π‘1 β₯ π‘0 large
enough,
limπ‘ββ
π π’π β (Ο(π )π(π )
π(π + β π)β (
π0π0 β 1
)Ο1(π )
4Ο(π )π2(π ))
π‘β1
π =π‘1
>π0
π0 β 1 (25)
then πΊ1 = πΊ3 = πΊ4 = π.
Proof: For the sake of contradiction, let (18) be satisfied but π¦ β πΊ1 βͺ πΊ3 βͺ πΊ4. Choose π‘1 >
π‘0 such that π₯(π‘) > 0, π₯(π‘ β π) > 0 πππ π₯(π‘ β ) > 0. Assume that π¦ β πΊ1. We proceed as
in the proof of Lemma 3 to obtain (22), where π€ is given by (20). Consider the function Ο
defined by
π(π‘) =π2(π‘)βπ€(π‘)
Ο1(π‘)π€(π‘) , (26)
Clearly, π < 0, from (23), it is easy to see that
β1 β€ Ο(π‘)π(π‘) < 0 (27)
Using (22) together with (26), we have
Ξπ(π‘) = Ξ (π2(π‘)Ξπ€(π‘)
Ο1(π‘))
1
π€(π‘)β
π2(π‘ + 1)[Ξπ€(π‘ + 1)]2
Ο1(π‘ + 1)π€2(π‘ + 1)
β€ β (π0 β 1
π0)
π(π‘)
π(π‘ + β π)
π€(π‘ + β π)
π€(π‘)β
Ο1(π‘ + 1)π2(π‘ + 1)
π2(π‘ + 1) (28)
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β€ β (π0 β 1
π0)
π(π‘)
π(π‘ + β π)β
Ο1(π‘ + 1)π2(π‘ + 1)
π2(π‘ + 1).
Multiplying both side of (28) by Ο(π‘) and summing the resulting inequality from π‘1 to π‘ β
1, we have
Ο(π‘)π(π‘) β€ Ο(π‘1)π(π‘1) + βπ(π + 1)Ο1(π + 1)
π2(π + 1)
π‘β1
π =π‘1
βπ0 β 1
π0β
Ο(π )π(π )
π(π + β π)
π‘β1
π =π‘1
β βΟ1(π + 1)π
2(π + 1)Ο(π )
π2(π + 1)
π‘β1
π =π‘1
= Ο(π‘1)π(π‘1) βπ0 β 1
π0β
Ο(π )π(π )
π(π + β π)
π‘β1
π =π‘1
+ βΟ1(π + 1)Ο(π )
π2(π + 1)
π‘β1
π =π‘1
[π(π + 1)
Ο(π )β π2(π + 1)]
β€ β (π0 β 1
π0) β [
Ο(π )π(π )
π(π + β π)β (
π0π0 β 1
)Ο1(π + 1)
4Ο(π )π2(π + 1)]
π‘β1
π =π‘1
However, in view of (27), this inequality contradicts (25). Hence πΊ1 = π. By Lemma 2, πΊ3 =
πΊ4 = π due to (7). The proof is complete.
Corollary 1: Suppose that (LH1) - (LH3) are satisfied and (7) holds. If there is a constant πΆπ
such that
Ο2(π‘)π(π‘)π2(π‘)
π(π‘ + β π)Ο1(π‘)β₯ πΆπ >
π04(π0 β 1)
, (29)
then πΊ1 = πΊ3 = πΊ4 = π.
To attain the oscillation of all solutions, it remains to eliminate the solutions of πΊ2
type.
Lemma 5: Suppose that (LH1) β (LH4) are satisfied. If
limπ‘ββ
π π’π βπ(π )π(π‘ + β π, π + β π)
π(π + β π)
π‘β1
π =π‘+πβπ
>π0
π0 β 1. (30)
then πΊ2 = π.
Proof: For sake of the contradiction, let (30) be satisfied π¦ β πΊ2. Choose π‘1 > π‘0 such that
π₯(π‘) > 0, π₯(π‘ β π) > 0 πππ π₯(π‘ β ) > 0. Using (13) in (1), we obtain
πΈ3π¦(π‘) +ππ β 1
π0
π(π‘)
π(π‘ + β π)π¦(π‘ + β π) β€ 0. (31)
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Using the monotonicity of πΈ2π¦, one can easily verify that
βπΈ1π¦(π’) β₯ πΈ1π¦(π£) β πΈ1π¦(π’) = βπΈ2π¦(π )
π2(π )
π£β1
π =π’
β₯ πΈ2π¦(π£) β1
π2(π )
π£β1
π =π’
(32)
for π£ β₯ π’ β₯ π‘1. Summing the latter inequality from π’ π‘π π£ β 1, we obtain,
π¦(π’) β₯ πΈ2π¦(π£) β1
π1(π )
π£β1
π =π’
β1
π2(π₯)
π£β1
π₯=π
= πΈ2π¦(π£)π(π£, π’). (33)
setting π’ = π + β π and π£ = π‘ + β π in (33), we find
π¦(π + β π) β₯ πΈ2π¦(π‘ + β π)π(π‘ + β π, π + β π). (34)
On the other hand, summing (31) from π‘ + β π to π‘ β 1 and using (34), we see that
πΈ2π¦(π‘ + β π) β₯ πΈ2π¦(π‘ + β π) β πΈ2π¦(π‘) β₯π0 β 1
π0β
π(π )π¦(π + β π)
π(π + β π)
π‘β1
π =π‘+πβπ
β₯π0 β 1
π0πΈ2π¦(π‘ + β π) β
π(π )π(π‘ + β π, π + β π)
π(π + β π).
π‘β1
π =π‘+πβπ
Dividing the above inequality by πΈ2π¦(π‘ + β π) and taking the limsup on both sides of the
resulting inequality as π‘ β β, we obtain a contradiction with (30). The proof is complete.
Lemma 6: Suppose that (LH1) β (LH4) are satisfied and let Ξ² be a constant satisfying (4)
eventually. If
limπ‘ββ
π π’π(π‘ + β π)π½ βπ(π )οΏ½ΜοΏ½(π‘ + β π, π + β π)
π(π + β π)
π‘β1
π =π‘+πβπ
>π0
π0 β 1, (35)
then πΊ2 = π.
Proof: Setting π’ = π‘ + β π and π£ = π‘ in (33), we obtain
π¦(π‘ + β π) β₯ πΈ2π¦(π‘)π(π‘, π‘ + β π) = πΈ2π¦(π‘ + 1)π(π‘, π‘ + β π) (36)
By (4), (31) and (36), we have
Journal of Information and Computational Science
Volume 10 Issue 9 - 2020
ISSN: 1548-7741
www.joics.org448
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β (π‘π½πΈ2π¦(π‘)) = π½π‘π½β1πΈ2π¦(π‘ + 1) + π‘
π½πΈ3π¦(π‘)
β€ π½π‘π½β1πΈ2π¦(π‘ + 1) β (π0 β 1
π0)
π‘π½π(π‘)π¦(π‘ + β π)
π(π‘ + β π)
β€ π½π‘π½β1πΈ2π¦(π‘ + 1) β (π0 β 1
π0)
π‘π½π(π‘)πΈ2π¦(π‘ + 1)π(π‘, π‘ + β π)
π(π‘ + β π)
= π‘π½β1πΈ2π¦(π‘ + 1) [π½ β (π0 β 1
π0)
π‘π(π‘)π(π‘, π‘ + β π)
π(π‘ + β π)] β€ 0.
That is π‘π½ πΈ2π¦(π‘ + 1) is eventually non-increasing. From here, we obtain that
βπΈ1π¦(π’) β₯ πΈ1π¦(π£) β πΈ1π¦(π’) = βπΈ2π¦(π )π
π½
π π½π2(π )
π£β1
π =π’
β₯ πΈ2π¦(π£)π£π½ β
1
π π½π2(π )
π£β1
π =π’
(37)
for π£ β₯ π’ β₯ π‘1. 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) β (LH4) are satisfied. If (18) (or (25)) and (30) (or (35))
hold, then (1) is oscillatory.
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ISSN: 1548-7741
www.joics.org450