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Asymptotic properties of solutions of third-order nonlinear neutral dynamic equations

Abstract

The purpose of this article is to give oscillation criteria for the third-order neutral dynamic equation ( r 2 ( t ) [ ( r 1 ( t ) [ y ( t ) + p ( t ) y ( τ ( t ) ) ] ) ] γ ) +f(t,y(δ(t)))=0, where γ1 is a ratio of odd positive integers with r 1 (t), r 2 (t), and p(t) are positive real-valued rd-continuous functions defined on T. We give new results for the third-order neutral dynamic equations and an example to illustrate the importance of our results.

1 Introduction

In the present article, we are concerned with oscillations of the third-order nonlinear neutral dynamic equation

( r 2 ( t ) [ ( r 1 ( t ) [ y ( t ) + p ( t ) y ( τ ( t ) ) ] ) ] γ ) +f ( t , y ( δ ( t ) ) ) =0
(1)

on a time scale T. Throughout this paper it is assumed that γ1 is a ratio of odd positive integers, τ(t):TT and δ(t):TR are rd-continuous functions such that τ(t)t, δ(t)t, lim t δ(t)= lim t τ(t)= and δ (t)>0 is rd-continuous, r 1 (t), r 2 (t) and p(t) are positive real valued rd-continuous functions defined on T, 0p(t)p<1 is increasing. We define the time scale interval [ t 0 , ) T by [ t 0 , ) T =[ t 0 ,)T. Furthermore, f:T×RR is a continuous function such that uf(t,u)>0 for all u0 and there exists a rd-continuous positive function q(t) defined on T such that |f(t,u)|q(t)| u γ |.

We use throughout this paper the following notations for convenience and for shortening the equations:

x ( t ) = y ( t ) + p ( t ) y ( τ ( t ) ) , x [ 1 ] = ( r 1 x ) , x [ 2 ] = r 2 ( x [ 1 ] ) γ , x [ 3 ] = ( x [ 2 ] ) .
(2)

A nontrivial function y(t) is said to be a solution of (1) if x C rd 1 [ t y ,), r 1 x C rd 1 [ t y ,) and x [ 2 ] C rd 1 [ t y ,) for t y t 0 and y(t) satisfies equation (1) for t y t 0 . A solution of (1) which is nontrivial for all large t is said to be oscillatory if it is neither eventually positive nor eventually negative. Otherwise, it is called nonoscillatory.

Recently, there has been many important research activity on the oscillatory behavior of dynamic equations. For example, on second-order dynamic equations, Saker [1], and Agarwal et al. [2], Saker [3], Hassan [4] and Candan [5, 6] considered the following nonlinear dynamic equations:

( r ( t ) ( ( y ( t ) + p ( t ) y ( t τ ) ) ) γ ) + f ( t , y ( t δ ) ) = 0 , ( r ( t ) ( ( y ( t ) + p ( t ) y ( τ ( t ) ) ) ) γ ) + f ( t , y ( δ ( t ) ) ) = 0 , ( r ( t ) ( x ( t ) ) γ ) + p ( t ) x γ ( t ) = 0 ,

and

( r ( t ) ( ( y ( t ) + p ( t ) y ( τ ( t ) ) ) ) γ ) + c d f ( t , y ( θ ( t , ξ ) ) ) ξ=0,

respectively, and they gave sufficient conditions which guarantee that every solution of the equation oscillates. Moreover, there are also some papers on third-order dynamic equations. For instance, Erbe et al. [7] considered the third-order nonlinear dynamic equation

( c ( t ) ( a ( t ) x ( t ) ) ) +q(t)f ( x ( t ) ) =0.

Later, Erbe et al. [8] considered the third-order nonlinear dynamic equation

x (t)+p(t)x(t)=0

by giving Hille and Nehari type criteria. Then, Hassan [9] studied the third-order nonlinear dynamic equation

( a ( t ) ( ( r ( t ) x ( t ) ) ) γ ) +f ( t , x ( τ ( t ) ) ) =0.

Lastly, Wang and Xu [10] studied asymptotic properties of a certain third-order dynamic equation,

( r 2 ( t ) ( ( r 1 ( t ) x ( t ) ) ) γ ) +q(t)f ( x ( t ) ) =0.

As we see from all the above, our equation, a neutral dynamic equation, is more general than other third-order dynamic equations and therefore it is very important. For some other important articles on oscillations of second-order nonlinear neutral delay dynamic equation on time scales and oscillations of third-order neutral differential equations, we refer the reader to the papers [11, 12], and [13], respectively. We give [14, 15] as references for books on the time scale calculus.

2 Main results

Lemma 1 Assume that y is an eventually positive solution of (1) and

t 0 t r 1 ( t ) =, t 0 ( 1 r 2 ( t ) ) 1 γ t=.
(3)

Then, there is a t 1 [ t 0 , ) T such that either

(i)x(t)>0, x (t)>0, x [ 1 ] (t)>0,t [ t 1 , ) T ,

or

(ii)x(t)>0, x (t)<0, x [ 1 ] (t)>0,t [ t 1 , ) T .

Proof Assume that y(t)>0 for t t 0 and therefore y(τ(t))>0 and y(δ(t))>0 for t t 1 > t 0 . Consequently, x(t)>0, eventually. Using (2) in (1) and the fact that |f(t,u)|q(t)| u γ |, we obtain

x [ 3 ] (t)+q(t) ( y ( δ ( t ) ) ) γ 0,t [ t 1 , ) T .
(4)

Hence, we conclude that x [ 2 ] (t) is a strictly decreasing function on [ t 1 , ) T . We claim that x [ 2 ] (t)>0 on [ t 1 , ) T . If not, then there exists a t 2 [ t 1 , ) T such that x [ 2 ] (t)<0 on [ t 2 , ) T . Then, there exist a negative constant c and t 3 [ t 2 , ) T such that

x [ 2 ] (t)c<0,t [ t 3 , ) T

and it follows that

x [ 1 ] (t) ( c r 2 ( t ) ) 1 γ .
(5)

Integrating (5) from t 3 to t and using (3), we obtain

r 1 (t) x (t) r 1 ( t 3 ) x ( t 3 )+ c 1 γ t 3 t ( 1 r 2 ( s ) ) 1 γ s,

which implies that r 1 (t) x (t) as t. Therefore, there exists a t 4 [ t 3 , ) T such that

r 1 (t) x (t) r 1 ( t 4 ) x ( t 4 )<0,t [ t 4 , ) T .
(6)

Dividing both sides of (6) by r 1 (t) and integrating from t 4 to t, we obtain

x(t)x( t 4 ) r 1 ( t 4 ) x ( t 4 ) t 4 t ( 1 r 1 ( s ) ) s.

Hence, we see from (3) that x(t) as t, which contradicts the fact that x(t)>0, and therefore x [ 2 ] (t)>0 for t [ t 1 , ) T . As a result of x [ 1 ] (t)>0 for t [ t 1 , ) T it follows that r 1 (t) x (t)<0 on [ t 1 , ) T or r 1 (t) x (t)>0 on [ t 1 , ) T , which completes the proof. □

Lemma 2 Let y be an eventually positive solution of (1). Assume that Case (i) of Lemma  1 holds. Then, there exists a t 1 [ t 0 , ) T such that

x (t) r 2 ( t , t 1 ) r 1 ( t ) [ x [ 2 ] ( t ) ] 1 γ ,t [ t 1 , ) T ,
(7)

where r 2 (t, t 1 )= t 1 t s ( r 2 ( s ) ) 1 γ and

x(t) r 1 (t, t 1 ) [ x [ 2 ] ( t ) ] 1 γ ,t [ t 1 , ) T ,

where r 1 (t, t 1 )= t 1 t r 2 ( s , t 1 ) r 1 ( s ) s.

Proof Since x [ 2 ] (t) is strictly decreasing on [ t 1 , ) T , we have

r 1 ( t ) x ( t ) r 1 ( t ) x ( t ) r 1 ( t 1 ) x ( t 1 ) = t 1 t [ x [ 2 ] ( s ) ] 1 γ ( r 2 ( s ) ) 1 γ s ,

it follows that

x (t) [ x [ 2 ] ( t ) ] 1 γ r 1 ( t ) t 1 t s ( r 2 ( s ) ) 1 γ

or

x (t) r 2 ( t , t 1 ) r 1 ( t ) [ x [ 2 ] ( t ) ] 1 γ ,t [ t 1 , ) T .
(8)

Similarly, integrating (8) from t 1 to t, we obtain

x(t) [ x [ 2 ] ( t ) ] 1 γ t 1 t r 2 ( s , t 1 ) r 1 ( s ) s

or

x(t) r 1 (t, t 1 ) [ x [ 2 ] ( t ) ] 1 γ ,t [ t 1 , ) T .

This completes the proof. □

Lemma 3 Let y be an eventually positive solution of (1). Assume that Case (ii) of Lemma  1 holds. If

t 0 1 r 1 ( t ) t [ 1 r 2 ( s ) s q ( u ) u ] 1 / γ st=,
(9)

then lim t y(t)=0.

Proof Since Case (ii) of Lemma 1 is satisfied,

lim t x(t)=l0.

We claim that lim t x(t)=0. Assume that l>0. Then for any ϵ>0, we have l<x(t)<l+ϵ for sufficiently large t t 1 . Choose 0<ϵ< l ( 1 p ) p . On the other hand, since

x(t)=y(t)+p(t)y ( τ ( t ) ) ,

we have

y ( t ) x ( t ) p x ( τ ( t ) ) > l p ( l + ϵ ) = k ( l + ϵ ) > k x ( t ) , t t 2 t 1 ,

where k= l p ( l + ϵ ) l + ϵ >0. Then,

( y ( δ ( t ) ) ) γ k γ ( x ( δ ( t ) ) ) γ ,t t 3 t 2 .
(10)

Substituting (10) into (4), we obtain

x [ 3 ] (t)q(t) k γ ( x ( δ ( t ) ) ) γ ,t t 3 .
(11)

Integrating (11) from t to ∞, we get

x [ 2 ] (t) k γ t q(s) ( x ( δ ( s ) ) ) γ s,t t 3

or using x(δ(t))>l,

x [ 1 ] (t)kl [ 1 r 2 ( t ) t q ( s ) s ] 1 / γ ,t t 3 .
(12)

Integrating (12) from t to ∞ and dividing both sides by r 1 (t), we have

x (t) k l r 1 ( t ) t [ 1 r 2 ( u ) u q ( s ) s ] 1 / γ u,t t 3 .
(13)

Integrating (13) from t 3 to ∞, we obtain

x( t 3 )kl t 3 1 r 1 ( t ) t [ 1 r 2 ( s ) s q ( u ) u ] 1 / γ st,

which contradicts (9) and therefore l=0. By making use of 0y(t)x(t), we conclude that lim t y(t)=0. □

Theorem 2.1 Assume that δ(σ(t))=σ(δ(t)). Furthermore, suppose that (3), (9), and

t 0 Q(s)s=,
(14)

where Q(s)=q(s) ( 1 p ) γ , hold. Then, every solution y(t) of (1) is either oscillatory on [ t 0 , ) T or lim t y(t)=0.

Proof Assume that (1) has a nonoscillatory solution; without loss of generality we may suppose that y(t)>0 for t t 0 and therefore y(τ(t))>0 and y(δ(t))>0 for t t 1 > t 0 . In the case when y(t) is negative the proof is similar. As we see from Lemma 1 we have two cases to consider. First we assume that x(t) satisfies Case (i) in Lemma 1. Then, by using (2) we see that

y(t)x(t)p(t)x ( τ ( t ) ) (1p)x(t),t t 2 t 1

or

( y ( δ ( t ) ) ) γ ( 1 p ) γ ( x ( δ ( t ) ) ) γ ,t t 3 t 2 .
(15)

Substituting (15) into (4), we obtain

x [ 3 ] (t)q(t) ( 1 p ) γ ( x ( δ ( t ) ) ) γ =Q(t) ( x ( δ ( t ) ) ) γ ,t t 3 .
(16)

Furthermore, using Pötzche’s chain rule, we find

( ( x ( δ ( t ) ) ) γ ) = γ 0 1 [ h ( x ( δ ( t ) ) ) σ + ( 1 h ) x ( δ ( t ) ) ] γ 1 ( x ( δ ( t ) ) ) d h γ 0 1 [ h x ( δ ( t ) ) + ( 1 h ) x ( δ ( t ) ) ] γ 1 ( x ( δ ( t ) ) ) d h = γ ( x ( δ ( t ) ) ) γ 1 x ( δ ( t ) ) δ ( t ) > 0 .
(17)

Define the function

z(t)= x [ 2 ] ( t ) ( x ( δ ( t ) ) ) γ ,t t 3 .
(18)

It is easy to see that z(t)>0. Taking the derivative of z(t), we see that

z ( t ) = x [ 3 ] ( t ) ( x ( δ ( t ) ) ) γ + ( x [ 2 ] ( t ) ) σ ( 1 ( x ( δ ( t ) ) ) γ ) = x [ 3 ] ( t ) ( x ( δ ( t ) ) ) γ ( x [ 2 ] ( t ) ) σ ( ( x ( δ ( t ) ) ) γ ) ( x ( δ σ ( t ) ) ) γ ( x ( δ ( t ) ) ) γ .
(19)

Substituting (16) into (19) and using (17), respectively, we have

z ( t ) Q ( t ) ( x [ 2 ] ( t ) ) σ ( ( x ( δ ( t ) ) ) γ ) ( x ( δ σ ( t ) ) ) γ ( x ( δ ( t ) ) ) γ Q ( t ) ( x [ 2 ] ( t ) ) σ γ ( x ( δ ( t ) ) ) γ 1 x ( δ ( t ) ) δ ( t ) ( x ( δ σ ( t ) ) ) γ ( x ( δ ( t ) ) ) γ = Q ( t ) γ ( x [ 2 ] ( t ) ) σ x ( δ ( t ) ) δ ( t ) ( x ( δ σ ( t ) ) ) γ x ( δ ( t ) ) Q ( t ) γ ( x [ 2 ] ( t ) ) σ x ( δ ( t ) ) δ ( t ) ( x ( δ σ ( t ) ) ) γ + 1 .
(20)

Using (7) in (20) and the fact that x [ 2 ] (t) is strictly decreasing, we obtain from (18)

z ( t ) Q ( t ) γ δ ( t ) r 2 ( δ ( t ) , t 1 ) r 1 ( δ ( t ) ) ( x [ 2 ] ( δ ( t ) ) ) 1 γ x ( δ σ ( t ) ) ( x [ 2 ] ( t ) ) σ ( x ( δ σ ( t ) ) ) γ Q ( t ) γ δ ( t ) r 2 ( δ ( t ) , t 1 ) r 1 ( δ ( t ) ) ( z σ ( t ) ) γ + 1 γ .
(21)

Finally, integrating (21) from t 3 to t, we get

z(t)z( t 3 ) t 3 t [ Q ( s ) γ r 2 ( δ ( s ) , t 1 ) δ ( s ) r 1 ( δ ( s ) ) ( z σ ( s ) ) γ + 1 γ ] s t 3 t Q(s)s

and consequently

t 3 t Q(s)sz( t 3 ),

which contradicts (14). When Case (ii) holds, we can conclude from Lemma 3 that lim t y(t)=0. □

Theorem 2.2 Suppose that (3), (9) hold and δ(σ(t))=σ(δ(t)). Furthermore, assume that there exists a positive rd-continuous -differentiable function α(t) such that

lim sup t t 0 t [ α ( s ) Q ( s ) ( ( α ( s ) ) + γ + 1 ) γ + 1 ( r 1 ( δ ( s ) ) α ( s ) r 2 ( δ ( s ) , t 1 ) δ ( s ) ) γ ] s=,
(22)

where ( α ( s ) ) + =max{0, α (s)} and Q(s)=q(s) ( 1 p ) γ . Then, every solution y(t) of (1) is either oscillatory on [ t 0 , ) T or lim t y(t)=0.

Proof Suppose to the contrary that y(t) is nonoscillatory solution of (1). We assume that y(t)>0 for t t 0 , then y(τ(t))>0 and y(δ(t))>0 for t t 1 > t 0 . We first consider that x(t) satisfies Case (i) in Lemma 1. We proceed as in the proof of Theorem 2.1, and we obtain (16). Let us define the function

z(t)=α(t) x [ 2 ] ( t ) ( x ( δ ( t ) ) ) γ ,t t 3 .
(23)

It is clear that z(t)>0. Taking the derivative of z(t), we see that

z ( t ) = ( x [ 2 ] ( t ) ) σ ( α ( t ) ( x ( δ ( t ) ) ) γ ) + α ( t ) ( x ( δ ( t ) ) ) γ x [ 3 ] ( t ) = α ( t ) x [ 3 ] ( t ) ( x ( δ ( t ) ) ) γ + ( x [ 2 ] ( t ) ) σ ( ( x ( δ ( t ) ) ) γ α ( t ) α ( t ) ( ( x ( δ ( t ) ) ) γ ) ( x ( δ ( t ) ) ) γ ( x ( δ σ ( t ) ) ) γ ) .
(24)

Now using (16) in (24), we obtain

z (t)α(t)Q(t)+ α ( t ) z σ ( t ) α σ ( t ) α ( t ) ( x [ 2 ] ( t ) ) σ ( ( x ( δ ( t ) ) ) γ ) ( x ( δ ( t ) ) ) γ ( x ( δ σ ( t ) ) ) γ .
(25)

Substituting (17) into (25), we obtain

z (t)α(t)Q(t)+ α ( t ) z σ ( t ) α σ ( t ) γ α ( t ) ( x [ 2 ] ( t ) ) σ x ( δ ( t ) ) δ ( t ) x ( δ ( t ) ) ( x ( δ σ ( t ) ) ) γ .
(26)

By using (7) into (26), we obtain

z (t)α(t)Q(t)+ ( α ( t ) ) + z σ ( t ) α σ ( t ) γ α ( t ) r 2 ( δ ( t ) , t 1 ) δ ( t ) r 1 ( δ ( t ) ) ( z σ ( t ) α σ ( t ) ) λ ,
(27)

where λ= γ + 1 γ . Let

A λ =γ α ( t ) r 2 ( δ ( t ) , t 1 ) δ ( t ) r 1 ( δ ( t ) ) ( z σ ( t ) α σ ( t ) ) λ

and

B λ 1 = ( α ( t ) ) + λ ( r 1 ( δ ( t ) ) γ α ( t ) r 2 ( δ ( t ) , t 1 ) δ ( t ) ) 1 / λ .

By making use of the inequality

λA B λ 1 A λ (λ1) B λ ,λ>1,A,B0
(28)

in (27), we have

z (t)α(t)Q(t)+ ( ( α ( t ) ) + γ + 1 ) γ + 1 ( r 1 ( δ ( t ) ) α ( t ) r 2 ( δ ( t ) , t 1 ) δ ( t ) ) γ .
(29)

Integrating both sides of (29) from t 3 to t then yields

t 3 t ( α ( s ) Q ( s ) ( ( α ( s ) ) + γ + 1 ) γ + 1 ( r 1 ( δ ( s ) ) α ( s ) r 2 ( δ ( s ) , t 1 ) δ ( s ) ) γ ) sz( t 3 )z(t)z( t 3 ),

which contradicts (22).

When Case (ii) holds, we can conclude from Lemma 3 that lim t y(t)=0. □

Let D 0 {(t,s) T 2 :t>s t 0 } and D{(t,s) T 2 :ts t 0 }. The function H C rd (D,R) is said to belong to class if H(t,t)=0, t t 0 , H(t,s)>0 on D 0 and H has a continuous -partial derivative H s (t,s) on D 0 with respect to the second variable.

Theorem 2.3 Assume that (3) and (9) hold and δ(σ(t))=σ(δ(t)). Furthermore, α(t) is defined as in Theorem  2.2 and H such that

lim sup t 1 H ( t , t 0 ) t 0 t [ H ( t , s ) α ( s ) Q ( s ) ( α σ ( s ) C ( t , s ) γ + 1 ) γ + 1 ( r 1 ( δ ( s ) ) H ( t , s ) α ( s ) r 2 ( δ ( s ) , t 1 ) δ ( s ) ) γ ] s = ,
(30)

where C(t,s)=max{0, H s (t,s)+ H ( t , s ) ( α ( s ) ) + α σ ( s ) }. Then every solution y(t) of (1) is either oscillatory on [ t 0 , ) T or lim t y(t)=0.

Proof Assume that y(t) is a nonoscillatory solution of (1). Define z(t) as in (23). We proceed as in the proof of Theorem 2.2 to obtain (27). Multiplying both sides of (27) by H(t,s), integrating with respect to s from t 3 to t, we get

t 3 t H ( t , s ) α ( s ) Q ( s ) s t 3 t H ( t , s ) z ( s ) s + t 3 t H ( t , s ) ( α ( s ) ) + z σ ( s ) α σ ( s ) s t 3 t γ H ( t , s ) α ( s ) r 2 ( δ ( s ) , t 1 ) δ ( s ) r 1 ( δ ( s ) ) ( z σ ( s ) α σ ( s ) ) λ s ,
(31)

where λ= γ + 1 γ . Integrating by parts yields by (31)

t 3 t H ( t , s ) α ( s ) Q ( s ) s H ( t , t 3 ) z ( t 3 ) + t 3 t H s ( t , s ) z σ ( s ) s + t 3 t H ( t , s ) ( α ( s ) ) + z σ ( s ) α σ ( s ) s t 3 t γ H ( t , s ) α ( s ) r 2 ( δ ( s ) , t 1 ) δ ( s ) r 1 ( δ ( s ) ) ( z σ ( s ) α σ ( s ) ) λ s H ( t , t 3 ) z ( t 3 ) + t 3 t C ( t , s ) z σ ( s ) s t 3 t γ H ( t , s ) α ( s ) r 2 ( δ ( s ) , t 1 ) δ ( s ) r 1 ( δ ( s ) ) ( z σ ( s ) α σ ( s ) ) λ s .
(32)

Let

A λ = γ H ( t , s ) α ( s ) r 2 ( δ ( s ) , t 1 ) δ ( s ) r 1 ( δ ( s ) ) ( z σ ( s ) α σ ( s ) ) λ

and

B λ 1 = C ( t , s ) α σ ( s ) λ ( r 1 ( δ ( t ) ) γ H ( t , s ) α ( t ) r 2 ( δ ( t ) , t 1 ) δ ( t ) ) 1 / λ .

Then, using the inequality (28) in (32), we have

t 3 t H ( t , s ) α ( s ) Q ( s ) s H ( t , t 3 ) z ( t 3 ) + t 3 t ( α σ ( s ) C ( t , s ) γ + 1 ) γ + 1 ( r 1 ( δ ( s ) ) H ( t , s ) α ( s ) r 2 ( δ ( s ) , t 1 ) δ ( s ) ) γ s

or

1 H ( t , t 3 ) t 3 t [ H ( t , s ) α ( s ) Q ( s ) ( α σ ( s ) C ( t , s ) γ + 1 ) γ + 1 ( r 1 ( δ ( s ) ) H ( t , s ) α ( s ) r 2 ( δ ( s ) , t 1 ) δ ( s ) ) γ ] s z ( t 3 ) ,

which contradicts (30) and completes the proof.

When Case (ii) holds, we can conclude from Lemma 3 that lim t y(t)=0. □

Example 2.4 Consider the following third-order neutral nonlinear dynamic equation:

( t 3 [ ( t [ y ( t ) + 1 2 y ( t 2 ) ] ) ] 3 ) + 3 t y 3 ( t 2 ) =0,t [ t 0 , ) T , t 0 >0,
(33)

where γ=3, r 1 (t)=t, r 1 (t)= r 2 (t)= t 3 p(t)= 1 2 , τ(t)=δ(t)= t 2 , and q(t)=3 t 1 . We can verify that all conditions of Theorem 2.1 are satisfied, therefore every solution of (33) is oscillatory or lim t y(t)=0. In fact, y(t)= t 1 is a solution of (33).

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Candan, T. Asymptotic properties of solutions of third-order nonlinear neutral dynamic equations. Adv Differ Equ 2014, 35 (2014). https://doi.org/10.1186/1687-1847-2014-35

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Keywords

  • oscillation
  • dynamic equations
  • time scales
  • neutral equations