Oscillation of Even Order Nonlinear Neutral Differential Equations of E

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@ IJTSRD | Available Online @ www ISSN No: 245 Inte R Oscillation of Even G. Pushpalatha Assistant Professor, Department of M Vivekanandha College of Arts and S Women, Tiruchengode, Namakkal, Tam ABSTRACT This paper presently exhibits about the even order nonlinear neutral differentia E of the form ቀ() (ଵ) ()ቁ + ()(ℎ൫()+ =0 Where () = () + ()൫(), integer. The output we considered and () < ∞ . This canon h enhanced and developed a few know literature. Some model are given to main results. INTRODUCTION We apprehensive with the oscillation the following half-linear even order n differential equation ቀ() (ଵ) ()ቁ + ()(൫()+( 0, ≥ , (1) Where() = () + ()൫(),≥ integer .Every part of this paper, we assu ( ) ∈ ([ , ), ), () > 0, () ( ) , ∈ ([ , ), ), 0 ≤ () ≤ < , () > 0, where w.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 56 - 6470 | www.ijtsrd.com | Volum ernational Journal of Trend in Sc Research and Development (IJT International Open Access Journ n Order Nonlinear Neutral Dif Equations of E Mathematics, Sciences for milnadu, India S. A. Vijaya L Research Scholar, Departm Vivekanandha College of A Women, Tiruchengode, Nama e oscillation of al equations of ()൫()൯ 2, is a even () = ∞, here extracted wn results in embellish our heorems for the neutral delay ()൫()൯ = 2, is a even ume that: 0; is a constant; ( ) 1 ([ , ), ), ∈ ([ , ), ), () ≥ () ≤ , () ≤ , ∘= ∘ = ∘, lim ௧→() = , lim ௧→( constant. ( ) ∈ (, ) and () , >0, for constant. Then the two cases are (௧) = (௧) < , By a solution of (1) a functio ([ , ), ) for some Where () = () + () 1 ([ , ), ) and s Then (1) satisfies {|) is called oscillatory. In certain case when =2 th the following equations r 2018 Page: 632 me - 2 | Issue 3 cientific TSRD) nal fferential Lakshmi ment of Mathematics, Arts and Sciences for akkal, Tamilnadu, India ([ , ), ), > 0, ∘ , () = , where is a ≠0 , where , is (2) (3) on be e , ()൯, has a property satisfies (1) on ( , ). ): ≥ |} > 0 for all he equation (1) lessen to

description

This paper presently exhibits about the oscillation of even order nonlinear neutral differential equations of E of the form e t z^ n 1 t ^ r t f h t v t f d t =0Where z t =x t p t x t ,n=2, is a even integer. The output we considered t o ^8¦ e^ 1 t dt=8 , and t o ^8¦ e^ 1 t dt 8 . This canon here extracted enhanced and developed a few known results in literature. Some model are given to embellish our main results. G. Pushpalatha | S. A. Vijaya Lakshmi "Oscillation of Even Order Nonlinear Neutral Differential Equations of E" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-2 | Issue-3 , April 2018, URL: https://www.ijtsrd.com/papers/ijtsrd11060.pdf Paper URL: http://www.ijtsrd.com/mathemetics/other/11060/oscillation-of-even-order-nonlinear-neutral-differential-equations-of-e/g-pushpalatha

Transcript of Oscillation of Even Order Nonlinear Neutral Differential Equations of E

Page 1: Oscillation of Even Order Nonlinear Neutral Differential Equations of E

@ IJTSRD | Available Online @ www.ijtsrd.com

ISSN No: 2456

InternationalResearch

Oscillation of Even Order

G. Pushpalatha Assistant Professor, Department of Mathematics, Vivekanandha College of Arts and Sciences for

Women, Tiruchengode, Namakkal, Tamilnadu, India

ABSTRACT This paper presently exhibits about the even order nonlinear neutral differential equations of E of the form

𝑒(𝑡)𝑧( )(𝑡) + 𝑟(𝑡)𝑓(ℎ 𝛾(𝑡) + 𝑣

= 0 Where 𝑧(𝑡) = 𝑥(𝑡) + 𝑝(𝑡)𝑥 𝜌(𝑡) , 𝑛 ≥

integer. The output we considered ∫ 𝑒

and ∫ 𝑒 (𝑡)𝑑𝑡 < ∞. This canon here extracted

enhanced and developed a few known results in literature. Some model are given to embellish our main results.

INTRODUCTION

We apprehensive with the oscillation theorems for the following half-linear even order neutral delay differential equation

𝑒(𝑡)𝑧( )(𝑡)′+ 𝑟(𝑡)𝑓(ℎ 𝛾(𝑡) + 𝑣(

0, 𝑡 ≥ 𝑡 , (1)

Where𝑧(𝑡) = 𝑥(𝑡) + 𝑝(𝑡)𝑥 𝜌(𝑡) , 𝑛 ≥integer .Every part of this paper, we assume that

(𝐸 ) 𝑒 ∈ 𝐶([𝑡 ,∞), 𝐸), 𝑒(𝑡) > 0, 𝑒′(𝑡) ≥

(𝐸 ) 𝑝, 𝑞 ∈ 𝐶([𝑡 ,∞), 𝐸),

0 ≤ 𝑝(𝑡) ≤ 𝑝 < ∞, 𝑞(𝑡) > 0, where 𝑝

@ IJTSRD | Available Online @ www.ijtsrd.com | Volume – 2 | Issue – 3 | Mar-Apr 2018

ISSN No: 2456 - 6470 | www.ijtsrd.com | Volume

International Journal of Trend in Scientific Research and Development (IJTSRD)

International Open Access Journal

f Even Order Nonlinear Neutral DifferentialEquations of E

Department of Mathematics, Vivekanandha College of Arts and Sciences for

Women, Tiruchengode, Namakkal, Tamilnadu, India

S. A. Vijaya LakshmiResearch Scholar, Department of Mathematics, Vivekanandha College of Arts and Science

Women, Tiruchengode, Namakkal, Tamilnadu, India

exhibits about the oscillation of even order nonlinear neutral differential equations of

𝑣(𝑡)𝑓 𝛿(𝑡)

≥ 2, is a even

𝑒 (𝑡)𝑑𝑡 = ∞,

here extracted

enhanced and developed a few known results in . Some model are given to embellish our

We apprehensive with the oscillation theorems for the neutral delay

(𝑡)𝑓 𝛿(𝑡) =

2, is a even we assume that:

≥ 0;

is a constant;

(𝐸 ) 𝜌 ∈ 𝐶1([𝑡 ,∞), 𝐸), 𝛾 ∈ 𝐶

𝛿 ∈ 𝐶([𝑡 ,∞), 𝐸), 𝜌′(𝑡) ≥ 𝜌

𝛾(𝑡) ≤ 𝑡, 𝛿(𝑡) ≤ 𝑡 , 𝜌 ∘ 𝛾 = 𝛾

𝜌 ∘ 𝛿 = 𝛿 ∘ 𝜌,

lim →∞ 𝛾(𝑡) = ∞ , lim →∞ 𝛿(constant.

(𝐸 ) 𝑓 ∈ 𝐶(𝐸, 𝐸) and

𝑓(𝑥) 𝑥⁄ ≥ 𝑀 , 𝑀 > 0, for 𝑥constant.

Then the two cases are

∫( )

𝑑𝑡 = ∞∞

∫( )

𝑑𝑡 < ∞∞

,

By a solution 𝑧 of (1) a function be

𝑒 ∈ 𝐶 ([𝑡 ,∞), 𝐸) for some

Where 𝑧(𝑡) = 𝑥(𝑡) + 𝑎(𝑡)𝑥

𝑒𝑧 ∈ 𝐶1([𝑡 ,∞), 𝐸) and satisfiesThen (1) satisfies 𝑠𝑢𝑝{|𝑥)𝑡𝑇 ≥ 𝑡 is called oscillatory.

In certain case when 𝑛 = 2 the equation (1) lessen to the following equations

Apr 2018 Page: 632

6470 | www.ijtsrd.com | Volume - 2 | Issue – 3

Scientific (IJTSRD)

International Open Access Journal

Nonlinear Neutral Differential

Vijaya Lakshmi Department of Mathematics,

Vivekanandha College of Arts and Sciences for Women, Tiruchengode, Namakkal, Tamilnadu, India

([𝑡 ,∞), 𝐸),

> 0,

𝛾 ∘ 𝜌 ,

(𝑡) = ∞ , where 𝜌 is a

𝑥 ≠ 0 , where 𝑀 , 𝑀 is

(2)

(3)

of (1) a function be

for some 𝑡 ≥ 𝑡 ,

( ) 𝜌(𝑡) , has a property and satisfies (1) on (𝑡 ,∞). { 𝑡): 𝑡 ≥ 𝑇|} > 0 for all

the equation (1) lessen to

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(𝑒(𝑡) 𝑥(𝑡) + 𝑝(𝑡)𝑥 𝜌(𝑡)′ ′

+ 𝑟(𝑡)𝑓(ℎ 𝛾(𝑡) +

𝑣(𝑡)𝑓 𝛿(𝑡) = 0, 𝑡 ≥ 𝑡 ,

(4)

Where ∫ 𝑒 (𝑡)𝑑𝑡 = ∞,

𝜌(𝑡) ≤ 𝑡 , 𝛾(𝑡) ≤ 𝑡, 𝛿(𝑡) ≤ 𝑡,

0 ≤ 𝑝(𝑡) ≤ 𝑝 < ∞.

Then the oscillatory behavior of the solutions of the neutral differential equations of the second order

(𝑒(𝑡) 𝑥(𝑡) + 𝑝(𝑡)𝑥 𝜌(𝑡)′ ′

+ 𝑟(𝑡)(ℎ 𝛾(𝑡) +

𝑣(𝑡) 𝛿(𝑡) = 0, 𝑡 ≥ 𝑡 (5)

Where ∫ 𝑒 (𝑠)𝑑𝑠 = ∞,

0 ≤ 𝑝(𝑡) ≤ 𝑝 < ∞.

The usual limitations on the coefficient of (5) be 𝜌(𝑡) ≤ 𝑡, 𝛾(𝑡) ≤ 𝜌(𝑡), 𝛿(𝑡) ≤ 𝜌(𝑡),

𝛾(𝑡) ≤ 𝑡, 𝛿(𝑡) ≤ 𝑡, 0 ≤ 𝑝(𝑡) < 1, are not assumed.

𝜌 Could be a advanced argument and 𝛾, 𝛿 could be a delay argument ,

Some known expand results are seen in [1,5]. Then the

Even-order nonlinear neutral functional differential equations

(𝑥(𝑡) + 𝑝(𝑡)𝑥 𝜌(𝑡) )′(n) + 𝑟(𝑡)𝑓(ℎ 𝛾(𝑡) +

𝑣(𝑡)𝑓 𝛿(𝑡) = 0, 𝑡 ≥ 𝑡 , (6)

Where n is even 0 ≤ 𝑝(𝑡) < 1 and 𝜌(𝑡) ≤ 𝑡.

(A) Lemma. 1

The oscillatory behavior of solutions of the following linear differential inequality

𝑤 ′(𝑡) + 𝑟(𝑡)ℎ(𝛾)) + 𝑣(𝑡)𝑔(𝛿(𝑡)) ≤ 0

Where 𝑟, 𝑣, 𝛾, 𝛿 ∈ 𝐶([𝑡 ,∞  )),

𝛾(𝑡)𝑡 , 𝛿(𝑡) ≤ 𝑡

lim→∞

𝛾(𝑡) = ∞ , lim→∞

𝛿(𝑡) = ∞

Now integrating from 𝛾(𝑡) to 𝑡 and 𝛿(𝑡) 𝑡𝑜 𝑡

If

lim→∞

𝑖𝑛𝑓 𝑟(𝑠)𝑑𝑠 +( )

lim→∞

𝑖𝑛𝑓 𝑣(𝑠)𝑑𝑠 >1

𝑒,

( )

Then it has no finally positive solutions.

Main results

The main results which covenant that every solution of (1) is oscillatory

(1) lim →∞ 𝑖𝑛𝑓 ∫ 𝐽(𝑠)𝑑𝑠( )

>

(2)lim→∞

𝑠𝑢𝑝 𝐽(𝑠)𝑑𝑠 > 1( )

B.Theorem. 2.1

Assume that ∫( )

𝑑𝑡 = ∞∞

holds. If

𝑆 (𝑡) +∞

𝑆 (𝑡) 𝑑𝑡 = ∞ ∞

Where 𝑆 (𝑡)= min {𝑟(𝑡), 𝑟(𝜌(𝑡))} 𝑆 (𝑡)= min{𝑣(𝑡), 𝑣(𝜌(𝑡))}, then every solutions of (1) is oscillatory.

Proof

Suppose, on the contradictory,𝑥 is a nonoscillatory solutions of (1). Without loss of generality, we may assume that there exists a constant 𝑡 ≥ 𝑡 , such that

𝑥(𝑡) > 0, 𝑥(𝜌(𝑡)) > 0 and (𝛾(𝑡)) > 0 ,

𝑥(𝛿(𝑡)) > 0 for all 𝑡 ≥ 𝑡 . Using the definitions of 𝑧 and x is a eventually positive solution of (1). Then there exists 𝑡 ≥ 𝑡 , such that

𝑧(𝑡) > 0, 𝑧 ′(𝑡) > 0, 𝑧( )(𝑡) > 0 and 𝑧 (𝑡) ≤ 0 for all 𝑡 ≥ 𝑡 .

.Hencelim →∞ 𝑧(𝑡) ≠ 0.

Applying ( 𝐸 ) and (1) we get

𝑒(𝑡)𝑧( )(𝑡)′

≤ −𝑀 𝑟(𝑡)ℎ 𝛾(𝑡) < 0, 𝑡 ≥ 𝑡

𝑒(𝑡)𝑧( )(𝑡)′

≤ −𝑀 𝑣(𝑡)𝑔 𝛿(𝑡) < 0, 𝑡 ≥ 𝑡 .

Therefore (𝑒(𝑡)𝑧( )(𝑡)) is a nonincreasing function. Besides, from the above inequality and the definition of 𝑧, we get

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Case (1)

𝑒(𝑡)𝑧( )(𝑡)′+ 𝑀 𝑟(𝑡)ℎ 𝛾(𝑡) +

′( )(𝑒 𝜌(𝑡) 𝑧( ) 𝜌(𝑡)

′+

𝑀 𝑝 𝑟((𝜌(𝑡))ℎ(𝛾((𝜌(𝑡))) ≤ 0

(𝑒(𝑡)𝑧( )(𝑡))′ + 𝑀 𝑅(𝑡)𝑧 𝛾(𝑡) +

(𝑒 𝜌(𝑡) 𝑧( )(𝜌(𝑡)))′ ≤ 0 (7)

Integrating (7) from 𝑡 𝑡𝑜 𝑡, we have

(𝑒(𝑠)𝑧( )(𝑠))′𝑑𝑠

+ 𝑀 𝑅(𝑠)𝑧 𝛾(𝑠) 𝑑𝑠

+𝑝

𝜌(𝑒 𝜌(𝑠) 𝑧( )(𝜌(𝑠)))′ 𝑑𝑠

≤ 0

Pointing that 𝜌′(𝑡) ≥ 𝜌 > 0

𝑀 𝑅(𝑠)𝑧 𝛾(𝑠) 𝑑𝑠 ≤ − 𝑒(𝑠)𝑧( )(𝑠))′𝑑𝑠

−𝑝

𝜌

1

𝜌′(𝑠)(𝑒 𝜌(𝑠) 𝑧( )(𝜌(𝑠)))′ 𝑑𝑠 𝑑(𝜌(𝑠)

𝑒(𝑡 )𝑧( )𝑡 −(𝑒(𝑡)𝑧( )(𝑡)) +

(𝑒 𝜌(𝑡 ) 𝑧( )(𝜌(𝑡 ) −

(𝑒(𝜌(𝑡))𝑧( )(𝑒(𝑡))) (8)

Since 𝑧 ′(𝑡) > 0 for 𝑡 ≥ 𝑡 . We can find a invariable 𝑐 > 0 such that

𝑧 𝛾(𝑡) ≥ 𝑐, 𝑡 ≥ 𝑡 .

Then from (8) and the fact that (𝑒(𝑡)𝑧( )(𝑡)) is non increasing, we obtain

∫ 𝑆 (𝑡) < ∞∞

(9)

Case (2)

(𝑒(𝑡)𝑧( )(𝑡))′ + 𝑀 𝑣(𝑡)𝑔 𝛿(𝑡)

+𝑝

𝜌′(𝑡)(𝑒 𝜌(𝑡) 𝑧( ) 𝜌(𝑡)

+ 𝑀 𝑝 𝑣((𝜌(𝑡))𝑔(𝛿((𝜌(𝑡))) ≤ 0

𝑒(𝑡)𝑧( )(𝑡)′+ 𝑀 𝑉(𝑡)𝑧 𝛿(𝑡) +

(𝑒 𝜌(𝑡) 𝑧( ) 𝜌(𝑡)′ (10)

Integrating (10) from 𝑡 𝑡𝑜 𝑡, we have

𝑒(𝑠)𝑧( )(𝑠))′𝑑𝑠

+ 𝑀 𝑉(𝑠)𝑧 𝛿(𝑠) 𝑑𝑠

+𝑝

𝜌(𝑒 𝜌(𝑠) 𝑧( ) 𝜌(𝑠)

′𝑑𝑠 ≤ 0

Pointing that 𝜌′(𝑡) ≥ 𝜌 > 0

𝑀 ∫ 𝑉(𝑠)𝑧 𝛿(𝑠) 𝑑𝑠 ≤ − ∫ 𝑒(𝑠)𝑧( )(𝑠))′𝑑𝑠 −

∫ ′( )(𝑒 𝜌(𝑠) 𝑧( )(𝜌(𝑠)))′ 𝑑𝑠 𝑑(𝜌(𝑠))

≤(𝑒(𝑡 )𝑧( )𝑡 −(𝑒(𝑡)𝑧( )(𝑡)) +

(𝑒 𝜌(𝑡 ) 𝑧( )(𝜌(𝑡 ) −

𝑒 𝜌(𝑡) 𝑧( ) 𝑒(𝑡) (11)

Since 𝑧 ′(𝑡) > 0 for 𝑡 ≥ 𝑡 . We can find a invariable 𝑐 > 0 such that

𝑧 𝛿(𝑡) ≥ 𝑐, 𝑡 ≥ 𝑡 .

Then from (8) and the fact that (𝑒(𝑡)𝑧( )(𝑡)) is nonincreasing, we obtain ∫ 𝑆 (𝑡) < ∞

(12)

We get inconsistency with (9),(12).

𝑆 (𝑡) +∞

𝑆 (𝑡) = ∞ ∞

Theorem. 2.2

Assume that ∫( )

𝑑𝑡 = ∞∞

holds and 𝜌(𝑡) ≥ 𝑡. if

either

lim →∞ 𝑖𝑛𝑓(∫( ) ( )

( ( ))𝑑𝑠 +

( )

∫( ) ( )

( ( ))( )𝑑𝑠) >

( )( )!, (13)

Or when 𝛾 , 𝛿 is increasing,

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lim →∞ 𝑠𝑢𝑝(∫( ) ( )

( ( ))𝑑𝑠 +

( )

∫( ) ( )

( ( ))( )𝑑𝑠) >

( )( )!, (14)

Where (𝑡) = min 𝑀 𝑗(𝑡), 𝑀 𝑗 𝜌(𝑡) ,

then every solution of (1) is oscillatory .

Proof

Suppose, on the contrary 𝑥 is a oscillatory solution of (1). Without loss of generality, we may assume that there exists a constant 𝑡 ≥ 𝑡 , such that t 𝑥(𝑡) > 0,

𝑥(𝜌(𝑡)) > 0 and 𝑥(𝛾(𝑡)) > 0 , 𝑥(𝛿(𝑡)) > 0 for all 𝑡 ≥ 𝑡 .

Assume that 𝑥( )(𝑡) is not consonantly zero on any interval [𝑡 ,∞), and there exists a 𝑡 > 𝑡 . Such that𝑥( )(𝑡)𝑢( )(𝑡) ≤ 0 for all 𝑡 ≥ 𝑡 . If lim →∞ 𝑥(𝑡) ≠ 0,then for every 𝜆, 0 < 𝜆 < 1, there exists𝑇 ≥ 𝑡 , such that for all 𝑡 ≥ 𝑇,

𝑥(𝑡) ≥( )!

𝑡 𝑢( )(t) forever

0 < 𝜆 < 1 , we obtain

(𝑒(𝑡)𝑧( )(𝑡))′ + (𝑒 𝜌(𝑡) 𝑧( ) 𝜌(𝑡)′+

( )!

𝛾 (𝑡)𝐽(𝑡)𝑧( )(𝛾((t)) +

( )!

𝛿 (𝑡)𝐾(𝑡)𝑧( )(𝛾((t)) ≤ 0,

For every 𝑡 sufficiently large.

Let 𝑥(𝑡) = 𝑒(𝑡)𝑧( )(𝑡) > 0. Then for all 𝑡 large

enough, we have

𝑥(𝑡) +𝑝

𝜌𝑥 𝜌(𝑡)

+𝜆

(𝑛 − 1)!

𝛾 (𝑡)𝐽(𝑡)

𝑒(𝛾(𝑡)) 𝑥 𝛾(𝑡)

+𝜆

(𝑛 − 1)!

𝛿 (𝑡)𝐾(𝑡)

𝑒(𝛿(𝑡)) 𝑥 𝛿(𝑡) ≤ 0

(15)

Next , let us denote

𝑦(𝑡) = 𝑥(𝑡) + 𝑥 𝜌(𝑡) . Since 𝑥 is non increasing,

it follows from 𝜌(𝑡) ≥ 𝑡 that

𝑦(𝑡) ≤ 1 + 𝑥(𝑡) . (16)

By combining (15) and (16), we get

𝑦 ′(𝑡) +( )!

( ) ( )

( )𝑦 𝛾(𝑡) +

( ) ( )

( )𝑦 𝛿(𝑡) ≤ 0 (17)

Therefore , 𝑦 is a non negative solutions of (17).

Then there will be two cases

Case (1)

If

lim →∞ 𝑖𝑛𝑓(∫( ) ( )

( ( ))𝑑𝑠 +

( )

∫( ) ( )

( ( ))( )𝑑𝑠) >

( )( )! holds then a

constant be 0 < 𝜆 < 1, such that

lim →∞ 𝑖𝑛𝑓(∫( )!

(( ) ( )

( )𝑑𝑠 +

( )

∫( ) ( )

( )( )𝑑𝑠)) > , (18)

By lemma (1), (18) holds that (17) has negative solutions, which is contradictory.

Case (2)

By the definition of 𝑦 and

( 𝑒(𝑡)𝑧( )(𝑡))′ + 𝑀 𝑅(𝑡)𝑧 𝛾(𝑡) +

(𝑒 𝜌(𝑡) 𝑧( )(𝜌(𝑡)))′ ≤ 0

we get ,

𝑦 ′(𝑡) = 𝑥 ′(𝑡) + 𝑥 𝜌(𝑡)′

≤ −𝐽(𝑡)𝑧 𝛾(𝑡) −

𝐾(𝑡)𝑧(𝛿(𝑡)) < 0 (19) pointing that 𝛾(𝑡) ≤ 𝑡, 𝛿(𝑡) ≤ 𝑡, there exists 𝑡 ≥ 𝑡 , such that

𝑦 𝛾(𝑡) ≥ 𝑦(𝑡), 𝑦 𝛿(𝑡) ≥ 𝑦(𝑡), 𝑡 ≥ 𝑡 . (20)

Integrating (17) from 𝛾(𝑡)𝑡𝑜 𝑡 and 𝛿(𝑡) to 𝑡 and applying 𝛾 , 𝛿 is increasing, we have

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𝑦(𝑡) − (𝑦 𝛾(𝑡) + 𝑦 𝛿(𝑡) ) +

( )!∫

( ) ( )

( )( 𝑦 𝛾(𝑠) +

( )

∫( ) ( )

( ) 𝑦 𝛿(𝑠)

( ))𝑑𝑠 ≤ 0, 𝑡 ≥ 𝑡 .

Thus

𝑦(𝑡) − (𝑦 𝛾(𝑡) + 𝑦(𝛿(𝑡))) +

( )!((𝑦 𝛾(𝑡) ∫

( ) ( )

( ) +

( )

𝑦 𝛿(𝑡) ) ∫( ) ( )

( ) )

( )𝑑𝑠 ≤ 0, 𝑡 ≥ 𝑡

From the above the inequality,we get

𝑦(𝑡)

(𝑦 𝛾(𝑡) + 𝑦(𝛿(𝑡)))− 1

+𝜌

𝑝 + 𝜌

𝜆

(𝑛 − 1)!

𝛾 (𝑠)𝐽(𝑠)

𝑒 𝛾(𝑠)( )

+𝛿 (𝑠)𝐾(𝑠)

𝑒 𝛿(𝑠) )

( )

𝑑𝑠 ≤ 0

From (20), we have

( )!∫

( ) ( )

( )+ ∫

( ) ( )

( ) )

( )𝑑𝑠

( )≤

1,𝑡 ≥ 𝑡 (21)

Taking upper limits as 𝑡 → ∞ in (21) we get

lim →∞ 𝑠𝑢𝑝(∫( ) ( )

( ( ))𝑑𝑠 +

( )

∫( ) ( )

( ( ))( )𝑑𝑠) ≤

( )( )!, (22)

If (14) holds, we choose a constant

0 < 𝜆 < 1 such that

lim →∞ 𝑠𝑢𝑝(∫( ) ( )

( )𝑑𝑠 +

( )

∫( ) ( )

( )( )𝑑𝑠) >

( )( )!,

Which is in contrary with (22).

Theorem 2.3

Assume that ∫( )

𝑑𝑡 = ∞∞

holds and𝛾(𝑡) ≤ 𝜌(𝑡) ≤

𝑡, 𝛿(𝑡) ≤ 𝜌(𝑡) ≤ 𝑡. If either

lim →∞ 𝑖𝑛𝑓(∫( ) ( )

( ( ))𝑑𝑠 +

( )

∫( ) ( )

( ( ))( )𝑑𝑠) >

( )( )!, (23)

Or when 𝜌 𝜊𝛾 , 𝜌 𝜊𝛿 is increasing,

lim →∞ 𝑠𝑢𝑝(∫( ) ( )

( ( ))𝑑𝑠 +

( )

∫( ) ( )

( ( ))( )𝑑𝑠) >

( )( )! (24)

Where 𝐽, 𝐾 is defined as in theorem (2.2), then every solution of (1) is oscillatory.

Proof

Suppose, on the contrary 𝑥 is a oscillatory solution of (1). Without loss of generality, we may assume that there exists a constant 𝑡 ≥ 𝑡 , such that

𝑥(𝑡) > 0, 𝑥(𝜌(𝑡)) > 0 and

(𝛾(𝑡)) > 0𝑥(𝛿(𝑡)) > 0 for all 𝑡 ≥ 𝑡 . Continuing as in the proof of the theorem (2.2), we have

𝑥(𝑡) + 𝑥 𝜌(𝑡)

+

( )!

( ) ( )

( ( )) 𝑥 𝛾(𝑡) +

( )!

( ) ( )

( ( )) 𝑥 𝛿(𝑡) ≤ 0. Let

𝑦(𝑡) = 𝑥(𝑡) + 𝑥 𝜌(𝑡) again. Since 𝑥 is non

increasing, it follows from𝜌(𝑡)) ≤ 0 that

𝑦(𝑡) ≤ (1 + )𝑥 𝜌(𝑡) (25)

By combining (15) and (24), we get

𝑦 ′(𝑡) +( )!

( ) ( )

( )𝑦 𝜌 𝛾(𝑡) +

( ) ( )

( ( )) 𝑦 𝜌 𝛿(𝑡) ≤ 0 (26)

Therefore, 𝑦 is a positive solution of (26). Now, we consider the following two cases , on (23) and (24) holds .

Case (1)

If

lim →∞ 𝑖𝑛𝑓(∫( ) ( )

( ( ))𝑑𝑠 +

( )

∫( ) ( )

( ( ))( )𝑑𝑠) >

( )( )! holds then a

constant be 0 < 𝜆 < 1, such that

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International Journal of Trend in Scientific Research and Development (IJTSRD) ISSN: 2456-6470

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lim →∞ 𝑖𝑛𝑓(∫( )!

(( ) ( )

( )𝑑𝑠 +

( )

𝛿𝑡𝑡𝛿𝑛−1𝑠𝐾𝑠𝑒𝛿𝑠𝑑𝑠))>1𝑒 (27)

Therefore (27) holds (26) has no positive solutions which is a contradiction.

Case (2)

From (19) and the condition

𝛾(𝑡) ≤ 𝜌(𝑡), 𝛿(𝑡) ≤ 𝜌(𝑡), there exists

𝑡 ≥ 𝑡 , such that 𝑦 𝜌 𝛾(𝑡) ≥ 𝑦(𝑡) ,

𝑦 𝜌 𝛿(𝑡) ≥ 𝑦(𝑡) 𝑡 ≥ 𝑡 . (28)

Integrating (26) from 𝜌 𝛾(𝑡) to𝑡, 𝜌 𝛿(𝑡) and

applying 𝜌 𝜊𝛾 is nondecreasing, then we get

𝑦(𝑡) − 𝑦 𝜌 𝛾(𝑡) +

( )!∫

( ) ( )

( )𝑦 𝜌 𝛾(𝑡) +

( )

𝜌−1𝛿(𝑡)𝑡𝛿𝑛−1𝑠𝐾(𝑠)𝑒(𝛿𝑠)𝑦𝜌−1𝛿𝑡𝑑𝑠≤ 0 , 𝑡≥𝑡2.

Thus

𝑦(𝑡) − 𝑦 𝜌 𝛾(𝑡) +

( )!𝑦 𝜌 𝛾(𝑡) ∫

( ) ( )

( ( ))+

( )

𝑦𝜌−1𝛿𝑡𝜌−1𝛿(𝑡)𝑡𝛿𝑛−1𝑠𝐾(𝑠)𝑒(𝛿𝑠)𝑑𝑠≤0, 𝑡≥𝑡2.

From the inequality , we obtain

𝑦(𝑡)

𝑦 𝜌 𝛾(𝑡)

− 1 𝜌

𝑝 + 𝜌

𝜆

(𝑛 − 1)!

𝛾 (𝑠)𝐽(𝑠)

𝑒(𝛾(𝑠))( )

+𝛿 (𝑠)𝐾(𝑠)

𝑒(𝛿(𝑠))( )

𝑑𝑠 ≤ 0 .

From (28) we get

( )!∫

( ) ( )

( ( ))+

( )

𝜌−1𝛿(𝑡)𝑡𝛿𝑛−1𝑠𝐾(𝑠)𝑒(𝛿𝑠)𝑑𝑠≤1, 𝑡≥𝑡2. (29)

Taking the upper limit as 𝑡 → ∞ in (29), we get

lim → 𝑠𝑢𝑝(∫( ) ( )

( ( ))𝑑𝑠 +

( )

𝛿(𝑡)𝑡𝛿𝑛−1𝑠𝐾(𝑠)𝑒(𝛿𝑠)𝑑𝑠)≤(𝑝0+𝜌0)(𝑛−1)!𝜆𝜌0, (30)

Then the proof is similar to that of the theorem (2.2) then it is contradiction to (24).

Reference

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3) Z. L. Han, T. X. Li, S. R. Sun, Y.B. Sun, Remarks on the paper [Appl. Math. Comput. 207(2009) 388-396], Appl. Math. Comput. 215 (2010) 3998-4007.

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