Oscillatory Behavior of Euler Type Delay Dynamic Equations with p

MATEC Web of Conferences 228, 01004 (2018) CAS 2018

https://doi.org/10.1051/matecconf/201822801004

Oscillatory Behavior of Euler Type Delay Dynamic Equations with p-Laplacian Like Operators Ying Sui, Yulong Shi, Yige Zhao and Zhenlai Han a College of Science, University of Jinan, Jinan, Shandong 250000, P R China

Abstract.

We

consider

the

Euler

type

delay

dynamic

equation

with

p-Laplacin

like

operators

( x (v)  x (v)  p2 )  a(v) x (v)  x (v)  p2 r(v) x( (v))  x( (v))  p2  0, where v [v0 )T . By using new inequality technique, we give some new criteria, which complement related contributions results.

1 Introduction We consider the Euler type dynamic equation with a pLaplacin like operator

( x  (v )  x  (v )  p  2 )   a ( v ) x  (v )  x  (v )  p  2  r (v) x( (v))  x( (v))  p  2  0 (1) where

v  V , V  [v0 , )T .

Assume that: (H1) p  2 is a real number, a, r

Crd (V, ℝ) satisfying

1  a ( v )  ( v )  0 , r (v )  0 . t 



p 2

In (1), a (v) x (v )  x (v )  also called the damping term of this equation, oscillation of equations with damping term has attracted more and more attention, we refer the reader to [1-3]. Zhan [1] studied a dynamic equation

(a (v) ( x  (v)))  q(v) f ( ( x( (v))))  p(v) ( x  (v))  0

(2)

( s )  s   2 s ,   1 , and a p q : T  R p  satisfying a  R . The author consider the two cases where

2 Main

results

Lemma 1. ([4]) Assume q Crd (V, ℝ), q (v) μ (v) + 1 > 0. Then eq ( v0 )  0 is a solution of the initial value

(H2)   Crd (V  T ) , v   ( v ) , lim  (v )   . 

by using new technique, some new properties of (2) are obtained. Our objective of this paper are establishing some new oscillation criteria for equation (1) by using new transformation technique, which complement some known results. We consider that all functional inequalities are eventually established when v is sufficiently large in this paper.

problem

 y  (v)  y (v)q (v)   y (v0 )  1 Lemma 2. ([4]) If

f is differentiable, then 1

( g n )  ng   [hg   (1  h) g ]n1 dh 0

where n is a constant. Further, letting 1









v0

1

1/( 1)

(a (v)e p  a (v v0 ))



C (v ) 

1

(e a (v v0 )) p1 ( e (( v(vv))) ) p1 p (v) a

0

 ( (v))

  (v )    ( (v))

and

v0

a

(a 1 (v)e p  a (v v0 ))1/( 1)  

Corresponding author: [email protected]

© The Authors, published by EDP Sciences. This is an open access article distributed under the terms of the Creative Commons Attribution License 4.0 (http://creativecommons.org/licenses/by/4.0/).

MATEC Web of Conferences 228, 01004 (2018) CAS 2018

https://doi.org/10.1051/matecconf/201822801004

p 1 1    (v) p 1  ( e ( s  v ))  s a 0  v r (v )   (v)   (v)   0 v 1     v (e a ( s v0 )) p1 s  e a (v v0 )   0



 (v )

(



e a (v v0 )

where

We claim that

that, for

satisfying

 (v )  0

v  [v3  )T ,

x  (v)  x  (v)  p  2 ea1 (v v0 )

p p 1

 x  (v3 )  x  (v3 )  p  2 ea1 (v3  v0 )   M  0



    Crd1 (V  R)

v3  [v2  )T and M  0 such

from (6) there exists

1  ) (e a (v v0 )) p1  , e a (v v0 ) 

 ( (v))

x  (v)  0 v  [v2  )T . If not, then

which yields

,

1

 (v )  0 .

1

x  (v)  (e a (v v0 )) p1 M p1 ,

Theorem 1. If (H1), (H2),





v0

then 1 p 1

(e a (v v0 )) v  

(3)

v

1

1

x(v)  x(v3 )  M p1  (e a ( s v0 )) p1 s v3

hold. If

lim sup v 

C p ( s ) p ( ( s )) v0 [ ( s )  p p  p 1 ( s ) e  a ( s  v 0 ) ] s    v

From (3),

(4) 

( x  (v)) p 1 a (v)  (( x  (v)) p 1 )  ( x ( (v ))) p 1 r (v)  0 (7) In addition, from (6), for

v  [v1  )T , v1  v0 , assume that x(v)  0 , x( (v))  0 . From (1),

then

x(v)  (e a (v v0 ))

e a ( v0 ) is a positive

solution of the problem

  x (v ) 

Then from (5), 

(v)  p2 ea1 (v v0 ) 



( x (v)  x (v) 



p 2 





p 2

 (v )

v2

  (e v

v2

a

1 p 1

( s v0 )) s



1



   0 

1

(e a ( s v0 )) p1 s



v

v2

1

(e a ( s v0 )) p1 s



(8)

Define a Ricati transformation (6) 

)  a(v) x (v)  x (v)  ea (v v0 )

r (v) x( (v))  x( (v))  ea (v v0 )

Hence,

1

v2

x( (v)) x 1 (v) 

x (v)  x (v)  p2 (ea (v v0 ))  ea (v v0 )e a (v v0 ) 

v

x  (v )  (e a ( s v0 )) p1 s

then

ea (v v0 )( x (v)  x (v)  p2 )  ea (v v0 )ea (v v0 )



 p11

Therefore,

 y  (v)   a(v ) y (v)   y (v0 )  1



s  v , we have

( x  ( s )) p 1 ea1 ( s v0 )  ( x  (v)) p 1 ea1 (v v0 )

a ( v ) x  (v )  x  (v )  p  2  ( x  (v )  x  ( v )  p  2 )    r (v) x( (v))  x( (v))  p  2  0 (5)

 x (v)  x

t 

rewrite as

then (1) is oscillatory. Proof. Let x is a non-oscillatory solution of (1). For

From Lemma 1 we know that

lim x (v )   . Therefore, (1) can be

 (v)  ( x  (v)) p 1 (v)ea1 (v v0 ) x1 p (v)   (v) (v)e1a (v v0 )

p 2

Obviously,

0

x  (v)  0 or x  (v)  0 , v  [v2  )T ,

v2  [v1  )T .

2

 (v)  0 . Then

(9)

MATEC Web of Conferences 228, 01004 (2018) CAS 2018

https://doi.org/10.1051/matecconf/201822801004

  (v )

  (v )

 x 1  p ( v )  ( v ) ( e  a1 ( v  v 0 ) ( x  (v ))

)   ( x 1  p ( v ) ( v ) ) 

p 1

( e  a1 ( v  v 0 ) ( x  (  (

 (v ) e a (v v0 )

 (v )



p 1

)

)  

  (v )( p  1)(e a (v v0 )) p1

(v )

(v ))

p 1

 (  ( v ) e  a1 ( v  v 0 ) )  





(v )

  

(10)



(v ))

p 1



( (v ) 1 (v )  e a1 (v v0 )  (v ))

p 1

p 1

1 p11

(13)

(v ) From the inequality (see [5])

 (v )

1 E 1 1 1 1 1 [G (  1)  ]E n  G n  (G  E ) n  n n G  0 E  0 n  1

)  x 1 p ( ( v ) ) x 1  p ( v )

 (v )(( x  (v ))

  

  (v )e a1 (v v0 )  (v )

e  a1 ( v  v 0 ) )   (x

e  a1 ( v  v 0 ) ) 

define

 (  ( v ) e  a1 ( v  v 0 ) )   ( v )

G  ( 1 (v) (v)) ,

From Lemma 2, we have

( x p 1 (v)) 

E  (ea1 (v v0 )  (v)) 

we have

1

  [ x(v)(1  h)  x (v)h] p  2 dh  x  (v)( p  1) (11)

  

0

 x p  2 (v)( p  1) x  (v )

( 1 (v) (v)  ea1 (v v0 )  (v)) 



1 p11

( (v))

1 p11

1 p11

p p 1

( (v)) 1 p11

Then from (6) and (9)–(11),



  (v )

 ( e (( v(vv))) ) p1 ( p  1) 1 1 ( (v)) p

( (v))(ea (v v0 ))

( p  1) 1

(14)

1

a

  (v ) r (v ) x p 1 ( ( v ))   (v )    (v )  ( v ) e  a ( v  v 0 ) x p 1 ( v ) 





p 1

 x ( ( v ))( p  1) x  1 ( v )  ( (ex a ((vv)) v0 ) )  

   

x ( v ) ( v ) ( 

p 1

1





 x 1 p ( v ) ( ( x

  (v )  r (v )  ( v )   (v )    (v) e a (v v0 )

1   (v )  1 v 1 1 p  v (e a ( s v0 )) s ( v (e a ( s v0 )) p1 s )  2  2 

)  (v )

 x 1 p ( ( v ) ) ( ( x e a (v  v0 ))

(v ))



e a (v v0 )

1



( x  ( v )) p 1 e a ( v  v0 )

)

     

1 p 1

x

p

   

1

0

From (13), (14),

  (v )

1 p 1

1

  ( (v))C (v)  (v)  (e a (v v0 )) p1

( ( v ))



 ( e  a1 ( v  v 0 )  ( v ))   ( t )

1 p11

p

( (v)) (v)( p  1) p1 ( (v ))

Define

(12)

1

1

G   p1 (v)( p  1) p1  1 ( (v))

From (6)

 (e (v v )) p1 0 x  ( (v))   a  (e (v v )) p11  a 0 x(v)  x( (v)). 1

1

(e a (v v0 )) p  ( (v ))

   x  (v),  

and

E  C p 1 (v ) ( (v))(

Then, from (8), (9) and (12),

1

p1 p

(v ) p1

p (e a (v v0 )) p ( p  1) p ) 1  Applying variation of the Young inequality

3

(15)

MATEC Web of Conferences 228, 01004 (2018) CAS 2018 p

https://doi.org/10.1051/matecconf/201822801004

p

1



E p1 ( p  1) 1  G p1  G ( p  1) 1 pE p1 

v

  ( s ) H (v s )s   (v2 ) H (v v2 )

v2

v

we have 1

 (e a (v v0 )) p1 



1 p11

1 p11

   ( ( s )) H  s (v s)s.

( (v))( p  1)

( (v)) (v)   ( (v))C (v )

From (17), (19) and (20), we have (16)

t

  ( s ) H ( v s )  s

  p ( (v))( p p e a (v v0 ) p 1 (v)) 1 C p (v)

t2

v

  C ( s) H (v s)) ( ( s )) v2

Using (16) in (15), we conclude that



t

t2

[ ( s ) 

 H s (v s )s   (v2 ) H (v v2 )

C (v) ( ( s )) ]s   (v2 ) p p p 1 ( s )e a ( s t0 ) p

p

v



( v s )

( ( s))

(21)

( ( s )) H (v s )s

  h (v s ) H

p 1 p

v2

(v s ) 1 ( ( s )) ( ( s ))s

  (v2 ) H (v v2 ) v

1

  ( p  1)(e a ( s v0 )) p1  ( s )

1 p11

v2



1 p11

( ( s ))

( ( s )) H (v s ) s

Define

G  

1 p11

1

( ( s))( p  1)(e a ( s v0 )) p1

 ( s) H (v s)

(17)

 C ( s ) H (v s )  H  s (v s ) H

1 p11

v

satisfying

 ( ( s )) h(v s ) H

1 p11

v2

D  {(v s )  v0  s  v v s  V } ) belongs to a class H if (i) H (v v )  0 , v  v0 , H (v s )  0 , v  s  v0 ; (ii) H has a rd-continuous  -partial derivative s H (v s )  0 on D0 with the second variable and p 1 p

1

  (e a ( s v0 )) p1  ( s )( p  1)

This contradiction completes the proof. Further, a function H  Crd ( D [0  )) (where

1

(20)

v2

p1 p

(v s ) h (v  s )  ( ( s))  u   ( (v))

E 

Theorem 2. If (H1), (H2), and (3) hold, and for  H , v2  V , 1 lim sup H (v v )  [ ( s) H (v s)

H

.

t

v 

( p  p

here

p 1

2

1

Using the inequality

t0

( s )e a ( s v )) h (v s )]s   p 

 1

(18)

Ev  Gv   (  1)  1  

h (v s )  max{h(v s ), 0} . Then (1) is

we have

oscillatory. Proof. Let

x is a non-oscillatory solution of (1). x(v)  0 , x( (v))  0 , v  [v1  )T , v1  v0 . Define  (v ) as (9), then by Theorem 1, we obtain (15).

t

 h (v s) t2

v

 

Then

v2



v

v2

v

1

) ( ( s ))

v

v

    ( s ) H (v s )s

( ( s )) H

p 1 p

(v s ) ( ( s)) s

1

( ( s))(e a ( s t0 )) p1 ( p  1) ( s ) H (v s) v

hp (v s ) s p p p 1 ( s )e a ( s v)

1 p11

 1  hp (v s )  e ( s  v )   ( s ) H (v s )  s p p 1 v2   a p  ( s)  v

.

(19)

v2

which contradicts (18). The proof is complete.

Then

4

( ( s))s

(22)

 (v2 ) H (v v2 ) 

 ( ( s)) H (v s) s

   ( ( s )) H (v s)C ( s)s

1

By (22), (21),

p p1

v2

1 p11

v2

1 p11 1

v2





 ( s ) H ( v s )  s 

  (e a ( s v0 )) p1  ( s )( p  1)(

E  1  G  0 G

MATEC Web of Conferences 228, 01004 (2018) CAS 2018

https://doi.org/10.1051/matecconf/201822801004

3 Conclusion

2.

This paper is concerned with a delay dynamic equation of Euler type with p-Laplacin like operators, by using new transformation and inequality technique with specific analytical skills, new criteria for the oscillation of the equations are established. under condition (3) holds. It is a interesting problem for future research of develop a different method to study equation (1) under the condition that integral formulae of (3) bounded.

3. 4. 5.

Acknowledgements This study was funded by Shandong Provincial Science Foundation (ZR2016AM17, ZR2017MA043), and the Natural Foundation of China (617043180).

6.

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