Skip to content

Advertisement

  • Research
  • Open Access

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

Advances in Difference Equations20142014:35

https://doi.org/10.1186/1687-1847-2014-35

  • Received: 11 September 2013
  • Accepted: 6 January 2014
  • Published:

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 γ 1 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.

Keywords

  • oscillation
  • dynamic equations
  • time scales
  • neutral equations

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 γ 1 is a ratio of odd positive integers, τ ( t ) : T T and δ ( t ) : T R 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 , 0 p ( t ) p < 1 is increasing. We define the time scale interval [ t 0 , ) T by [ t 0 , ) T = [ t 0 , ) T . Furthermore, f : T × R R is a continuous function such that u f ( t , u ) > 0 for all u 0 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 s t = ,
(9)

then lim t y ( t ) = 0 .

Proof Since Case (ii) of Lemma 1 is satisfied,
lim t x ( t ) = l 0 .
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 ) k l [ 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 ) k l t 3 1 r 1 ( t ) t [ 1 r 2 ( s ) s q ( u ) u ] 1 / γ s t ,

which contradicts (9) and therefore l = 0 . By making use of 0 y ( 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 ) ) ( 1 p ) 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 ) s z ( 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
λ A B λ 1 A λ ( λ 1 ) B λ , λ > 1 , A , B 0
(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 ) s z ( 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 : t s 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).

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Art and Science, Niğde University, Niğde, 51200, Turkey

References

  1. Saker SH: Oscillation of second-order nonlinear neutral delay dynamic equations on time scales. J. Comput. Appl. Math. 2006, 187: 123-141. 10.1016/j.cam.2005.03.039MATHMathSciNetView ArticleGoogle Scholar
  2. Agarwal RP, O’Regan D, Saker SH: Oscillation criteria for second-order nonlinear neutral delay dynamic equations. J. Math. Anal. Appl. 2004, 300: 203-217. 10.1016/j.jmaa.2004.06.041MATHMathSciNetView ArticleGoogle Scholar
  3. Saker SH: Oscillation criteria for a second-order quasilinear neutral functional dynamic equation on time scales. Nonlinear Oscil. 2011, 13: 407-428. 10.1007/s11072-011-0122-8MathSciNetView ArticleGoogle Scholar
  4. Hassan TS: Oscillation criteria for half-linear dynamic equations on time scales. J. Math. Anal. Appl. 2008, 345: 176-185. 10.1016/j.jmaa.2008.04.019MATHMathSciNetView ArticleGoogle Scholar
  5. Candan T: Oscillation of second-order nonlinear neutral dynamic equations on time scales with distributed deviating arguments. Comput. Math. Appl. 2011, 62: 4118-4125. 10.1016/j.camwa.2011.09.062MATHMathSciNetView ArticleGoogle Scholar
  6. Candan T: Oscillation criteria for second-order nonlinear neutral dynamic equations with distributed deviating arguments on time scales. Adv. Differ. Equ. 2013., 2013: Article ID 112Google Scholar
  7. Erbe L, Peterson A, Saker SH: Asymptotic behavior of solutions of a third-order nonlinear dynamic equation on time scales. J. Comput. Appl. Math. 2005, 181: 92-102. 10.1016/j.cam.2004.11.021MATHMathSciNetView ArticleGoogle Scholar
  8. Erbe L, Peterson A, Saker SH: Hille and Nehari type criteria for third-order dynamic equations. J. Math. Anal. Appl. 2007, 329: 112-131. 10.1016/j.jmaa.2006.06.033MATHMathSciNetView ArticleGoogle Scholar
  9. Hassan TS: Oscillation of third order nonlinear delay dynamic equations on time scales. Math. Comput. Model. 2009, 49: 1573-1586. 10.1016/j.mcm.2008.12.011MATHView ArticleGoogle Scholar
  10. Wang Y, Xu Z: Asymptotic properties of solutions of certain third-order dynamic equations. J. Comput. Appl. Math. 2012, 236: 2354-2366. 10.1016/j.cam.2011.11.021MATHMathSciNetView ArticleGoogle Scholar
  11. Saker SH, O’Regan D, Agarwal RP: Oscillation theorems for second-order nonlinear neutral delay dynamic equations on time scales. Acta Math. Sin. Engl. Ser. 2008, 24: 1409-1432. 10.1007/s10114-008-7090-7MATHMathSciNetView ArticleGoogle Scholar
  12. Saker SH: Oscillation of superlinear and sublinear neutral delay dynamic equations. Commun. Appl. Anal. 2008, 12(2):173-187.MATHMathSciNetGoogle Scholar
  13. Baculíková B, Dz̆urina J: Oscillation of third-order neutral differential equations. Math. Comput. Model. 2010, 52: 215-226. 10.1016/j.mcm.2010.02.011MATHView ArticleGoogle Scholar
  14. Bohner M, Peterson A: Dynamic Equations on Time Scales: An Introduction with Applications. Birkhäuser, Boston; 2001.View ArticleGoogle Scholar
  15. Bohner M, Peterson A: Advances in Dynamic Equations on Time Scales. Birkhäuser, Boston; 2003.MATHView ArticleGoogle Scholar

Copyright

© Candan; licensee Springer. 2014

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Advertisement