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Oscillation criteria for second-order nonlinear neutral dynamic equations with distributed deviating arguments on time scales

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Abstract

In this article, we establish some new oscillation criteria and give sufficient conditions to ensure that all solutions of nonlinear neutral dynamic equation of the form

( r ( t ) ( ( y ( t ) + p ( t ) y ( τ ( t ) ) ) ) γ ) + a b f ( t , y ( δ ( t , ξ ) ) ) ξ=0

are oscillatory on a time scale T, where γ1 is a quotient of odd positive integers.

1 Introduction

The aim of this article is to develop some oscillation theorems for a second-order nonlinear neutral dynamic equation

( r ( t ) ( ( y ( t ) + p ( t ) y ( τ ( t ) ) ) ) γ ) + a b f ( t , y ( δ ( t , ξ ) ) ) ξ=0
(1)

on a time scale T. Throughout this paper, it is assumed that γ1 is a quotient of odd positive integers, 0<a<b, τ(t):TT, is rd-continuous function such that τ(t)t and τ(t) as t, δ(t,ξ):T×[a,b]T is rd-continuous function such that decreasing with respect to ξ, δ(t,ξ)t for ξ[a,b], δ(t,ξ) as t, r(t)>0 and 0p(t)<1 are real valued rd-continuous functions defined on T, p(t) is increasing and

(H1) t 0 ( 1 r ( t ) ) 1 γ t=,

(H2) f:T×RR is a continuous function such that uf(t,u)>0 for all u0 and there exists a positive function q(t) defined on T such that |f(t,u)|q(t)| u γ |.

A nontrivial function y(t) is said to be a solution of (1) if y(t)+p(t)y(τ(t)) C r d 1 [ t y ,] and r(t) ( ( y ( t ) + p ( t ) y ( τ ( t ) ) ) ) γ C r d 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 called oscillatory if it has no last zero. Otherwise, a solution is called nonoscillatory.

We note that if T=R, we have σ(t)=t, μ(t)=0, y (t)= y (t) and, therefore, (1) becomes a second-order neutral differential equation with distributed deviating arguments

( r ( t ) ( ( y ( t ) + p ( t ) y ( τ ( t ) ) ) ) γ ) + a b f ( t , y ( δ ( t , ξ ) ) ) dξ=0.

If T=N, we have σ(t)=t+1, μ(t)=1, y (t)=y(t)=y(t+1)y(t) and therefore (1) becomes a second-order neutral difference equation with distributed deviating arguments

( r ( t ) ( ( y ( t ) + p ( t ) y ( τ ( t ) ) ) ) γ ) + ξ = a b 1 f ( t , y ( δ ( t , ξ ) ) ) =0

and if T=hN, h>0, we have σ(t)=t+h, μ(t)=h, y (t)= h y(t)= y ( t + h ) y ( t ) h and, therefore, (1) becomes a second-order neutral difference equation with distributed deviating arguments

h ( r ( t ) ( h ( y ( t ) + p ( t ) y ( τ ( t ) ) ) ) γ ) + k = a h b h 1 f ( t , y ( δ ( t , k h ) ) ) h=0.

In recent years, there has been important research activity about the oscillatory behavior of second-order neutral differential, difference and dynamic equations. For example, Grace and Lalli [1] considered the following second-order neutral delay equation

( a ( t ) ( x ( t ) + p ( t ) x ( t τ ) ) ) +q(t)f ( x ( t τ ) ) =0,t t 0

and Graef et al. [2] considered the nonlinear second-order neutral delay equation

( y ( t ) + p ( t ) y ( τ ( t ) ) ) +q(t)f ( y ( t δ ) ) =0,t t 0 .

Recently, Agarwal et al. [3] considered second-order nonlinear neutral delay dynamic equation

( r ( t ) ( ( y ( t ) + p ( t ) y ( τ ( t ) ) ) ) γ ) +f ( t , y ( t δ ) ) =0.
(2)

Later, Saker [4] considered (2) but he used different technique to prove his results. In [5] and [6], the authors considered the second order neutral functional dynamic equation of the form

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

which is more general than (2). For more papers related to oscillation of second-order nonlinear neutral delay dynamic equation on time scales, we refer the reader to [710]. For neutral equations with distributed deviating arguments, we refer the reader to the paper by Candan [11]. To the best of our knowledge, [12] is the only paper regarding to the distributed deviating arguments on time scales. The books [13, 14] gives time scale calculus and some applications.

2 Main results

Throughout the paper, we use the following notations for simplicity:

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

and θ 1 (t)=δ(t,a) and θ 2 (t)=δ(t,b).

Theorem 2.1 Assume that (H1) and (H2) hold. In addition, assume that r (t)0. Then every solution of (1) oscillates if the inequality

x [ 2 ] (t)+A(t) x [ 1 ] ( θ 1 ( t ) ) 0,
(4)

where

A(t)= ( b a ) q ( t ) ( 1 p ( θ 1 ( t ) ) ) γ r ( θ 1 ( t ) ) ( θ 2 ( t ) 2 ) γ

has no eventually positive solution.

Proof Let y(t) be a nonoscillatory solution of (1), without loss of generality, we assume that y(t)>0 for t t 0 , then y(τ(t))>0 and y(δ(t,ξ))>0 for t t 1 > t 0 and bξa. In the case when y(t) is negative, the proof is similar. In view of (1), (H2) and (3)

x [ 2 ] (t)+ a b q(t) y γ ( δ ( t , ξ ) ) ξ0
(5)

for all t t 1 , and we see that x [ 1 ] (t) is an eventually decreasing function. We claim that x [ 1 ] (t)>0 eventually. Assume not then there exists a t 2 t 1 such that x [ 1 ] ( t 2 )=c<0, then we have x [ 1 ] (t)c for t t 2 and it follows that

x (t) ( c r ( t ) ) 1 / γ .
(6)

Now integrating (6) from t 2 to t and using (H1), we obtain

x(t)x( t 2 )+ c 1 / γ t 2 t ( 1 r ( s ) ) 1 / γ sas t

which contradicts the fact that x(t)>0 for all t t 0 . Hence, x [ 1 ] (t) is positive. Therefore, one sees that there is a t 2 t 1 such that

x(t)>0, x (t)>0, x [ 1 ] (t)>0, x [ 2 ] (t)<0,t t 2 .
(7)

For t t 3 t 2 , this implies that

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

then we conclude that

y γ ( δ ( t , ξ ) ) ( 1 p ( δ ( t , ξ ) ) ) γ x γ ( δ ( t , ξ ) ) ,t t 4 t 3 ,ξ[a,b].
(8)

Multiplying (8) by q(t) and integrating both sides from a to b, we have

a b q(t) y γ ( δ ( t , ξ ) ) ξ a b q(t) ( 1 p ( δ ( t , ξ ) ) ) γ x γ ( δ ( t , ξ ) ) ξ.
(9)

Substituting (9) into (5), we obtain

x [ 2 ] (t)+ a b q(t) ( 1 p ( δ ( t , ξ ) ) ) γ x γ ( δ ( t , ξ ) ) ξ0.
(10)

On the other hand, we can verify that x (t)0 for t t 4 and, therefore, we obtain

x(t)=x( t 4 )+ t 4 t x (s)s(t t 4 ) x (t) t 2 x (t),t t 5 2 t 4 .

From the last inequality, it can be easily seen that

x ( δ ( t , ξ ) ) ( δ ( t , ξ ) 2 ) x ( δ ( t , ξ ) ) ( θ 2 ( t ) 2 ) x ( δ ( t , ξ ) ) ,t t 6 t 5 ,ξ[a,b].

Substituting the last inequality into (10), we have

x [ 2 ] (t)+ a b q(t) ( 1 p ( δ ( t , ξ ) ) ) γ ( θ 2 ( t ) 2 ) γ ( x ( δ ( t , ξ ) ) ) γ ξ0

and it can be found

x [ 2 ] (t)+(ba)q(t) ( 1 p ( θ 1 ( t ) ) ) γ ( θ 2 ( t ) 2 ) γ ( x ( θ 1 ( t ) ) ) γ 0,

or

x [ 2 ] (t)+ ( b a ) q ( t ) ( 1 p ( θ 1 ( t ) ) ) γ r ( θ 1 ( t ) ) ( θ 2 ( t ) 2 ) γ x [ 1 ] ( θ 1 ( t ) ) 0,

which is the inequality (4). As a consequence of this, we have a contradiction and therefore every solution of (1) oscillates. □

Theorem 2.2 Assume that (H1) and (H2) hold. In addition, assume that r (t)0, δ(t,ξ) is increasing with respect to t and that the inequality

lim sup t θ 1 ( t ) t A(s)s>1
(11)

holds. Then every solution of (1) oscillates.

Proof Let y(t) be a nonoscillatory solution of (1). We can proceed as in the proof of Theorem 2.1 to get (4). Integrating (4) from θ 1 (t) to t for sufficiently large t, we have

0 θ 1 ( t ) t ( x [ 2 ] ( s ) + A ( s ) x [ 1 ] ( θ 1 ( s ) ) ) s = x [ 1 ] ( t ) x [ 1 ] ( θ 1 ( t ) ) + θ 1 ( t ) t A ( s ) x [ 1 ] ( θ 1 ( s ) ) s x [ 1 ] ( t ) x [ 1 ] ( θ 1 ( t ) ) + x [ 1 ] ( θ 1 ( t ) ) θ 1 ( t ) t A ( s ) s = x [ 1 ] ( t ) + x [ 1 ] ( θ 1 ( t ) ) ( θ 1 ( t ) t A ( s ) s 1 ) > 0 .

By making use of (11), we reach to a contradiction therefore the proof is complete. □

Theorem 2.3 Assume that (H1) and (H2) hold. In addition, assume that r (t)0, δ(t,ξ) is increasing with respect to t and there exists a positive rd-continuous -differentiable function α(t) such that

lim sup t t 0 t ( α ( s ) Q ( s ) ( ( α ( s ) ) + ) 2 r ( θ 2 ( s ) ) 4 γ ( θ 2 ( s ) 2 ) γ 1 α ( s ) ) s=,
(12)

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

Proof Suppose to the contrary that y(t) is nonoscillatory solution of (1). We may assume without loss of generality that y(t)>0 for t t 0 , then y(τ(t))>0 and y(δ(t,ξ))>0 for t t 1 > t 0 and bξa. Proceeding as in the proof of Theorem 2.1, we obtain (7) and the inequality (10). Using (7) and Pötzsche’s chain rule [[15], Theorem 1], we obtain

( x γ ( t ) ) = γ 0 1 [ x ( t ) + h μ ( t ) x ( t ) ] γ 1 d h x ( t ) γ 0 1 ( x ( t ) ) γ 1 d h x ( t ) = γ ( x ( t ) ) γ 1 x ( t ) > 0 .
(13)

From (10) and (13), we obtain

x [ 2 ] (t)(ba)q(t) ( 1 p ( θ 1 ( t ) ) ) γ x γ ( θ 2 ( t ) ) =Q(t) x γ ( θ 2 ( t ) ) ,t t 4 .
(14)

Define the function

z(t)=α(t) x [ 1 ] ( t ) x γ ( θ 2 ( t ) ) ,t t 4 .
(15)

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

z ( t ) = ( x [ 1 ] ) σ ( t ) ( α ( t ) x γ ( θ 2 ( t ) ) ) + α ( t ) x γ ( θ 2 ( t ) ) x [ 2 ] ( t ) = α ( t ) x [ 2 ] ( t ) x γ ( θ 2 ( t ) ) + ( x [ 1 ] ) σ ( t ) ( x γ ( θ 2 ( t ) ) α ( t ) α ( t ) ( x γ ( θ 2 ( t ) ) ) x γ ( θ 2 ( t ) ) ( x σ ( θ 2 ( t ) ) ) γ ) .
(16)

Now using (14) in (16), we obtain

z (t)α(t)Q(t)+ α ( t ) z σ ( t ) α σ ( t ) α ( t ) ( x [ 1 ] ) σ ( t ) ( x γ ( θ 2 ( t ) ) ) x γ ( θ 2 ( t ) ) ( x σ ( θ 2 ( t ) ) ) γ .
(17)

On the other hand, as in the proof of Theorem 2.1, it can be shown that for sufficiently large t t 5

x(t) ( t 2 ) x (t),t t 5 2 t 4

and then

γ x γ 1 (t)γ ( t 2 ) γ 1 ( x ( t ) ) γ 1

or

γ x γ 1 ( θ 2 ( t ) ) γ ( θ 2 ( t ) 2 ) γ 1 ( x ( θ 2 ( t ) ) ) γ 1 ,t t 6 t 5 .
(18)

Since x [ 2 ] (t)<0, we have

x [ 1 ] (t)> x [ 1 ] ( σ ( t ) ) .
(19)

Multiplying (18) by x ( θ 2 (t)) and using (19), it follows that

γ x γ 1 ( θ 2 ( t ) ) x ( θ 2 ( t ) ) γ ( θ 2 ( t ) 2 ) γ 1 ( x ( θ 2 ( t ) ) ) γ γ ( θ 2 ( t ) 2 ) γ 1 r ( θ 2 ( σ ( t ) ) ) r ( θ 2 ( t ) ) ( x ( θ 2 ( σ ( t ) ) ) ) γ γ ( θ 2 ( t ) 2 ) γ 1 ( x [ 1 ] ) σ ( θ 2 ( t ) ) r ( θ 2 ( t ) ) .
(20)

From (13), for sufficiently large t t 7 t 6 , we have

( x γ ( θ 2 ( t ) ) ) γ x γ 1 ( θ 2 ( t ) ) x ( θ 2 ( t ) ) .
(21)

From (20) and (21), it follows that

( x γ ( θ 2 ( t ) ) ) γ ( θ 2 ( t ) 2 ) γ 1 ( x [ 1 ] ) σ ( θ 2 ( t ) ) r ( θ 2 ( t ) ) .
(22)

Substituting (22) into (17), we obtain

z (t)α(t)Q(t)+ α ( t ) z σ ( t ) α σ ( t ) γ ( θ 2 ( t ) 2 ) γ 1 α ( t ) ( α σ ( t ) ) 2 r ( θ 2 ( t ) ) ( z σ ( t ) ) 2 .

Using the fact um u 2 1 4 m , m>0, we have

z ( t ) α ( t ) Q ( t ) + ( α ( t ) ) + α σ ( t ) ( z σ ( t ) γ ( θ 2 ( t ) 2 ) γ 1 α ( t ) ( ( α ( t ) ) + ) α σ ( t ) r ( θ 2 ( t ) ) ( z σ ( t ) ) 2 ) ( α ( t ) Q ( t ) ( ( α ( t ) ) + ) 2 r ( θ 2 ( t ) ) 4 γ ( θ 2 ( t ) 2 ) γ 1 α ( t ) ) .

Integrating the last inequality from t 7 to t, we obtain

z( t 7 )<z(t)z( t 7 ) t 7 t ( α ( s ) Q ( s ) ( ( α ( s ) ) + ) 2 r ( θ 2 ( s ) ) 4 γ ( θ 2 ( s ) 2 ) γ 1 α ( s ) ) s

or

z( t 7 )> t 7 t ( α ( s ) Q ( s ) ( ( α ( s ) ) + ) 2 r ( θ 2 ( s ) ) 4 γ ( θ 2 ( s ) 2 ) γ 1 α ( s ) ) s

which contradicts (12). Therefore, the proof is complete. □

Theorem 2.4 Assume that (H1) and (H2) hold and σ(t)t for each tT. Let α(t), δ(t,ξ), and Q(s) be as defined in Theorem  2.3. If

lim sup t t 0 t ( α ( s ) Q ( s ) ( ( α ( s ) ) + ) 2 r ( θ 2 ( s ) ) 2 3 γ ( μ ( θ 2 ( s ) ) ) γ 1 α ( s ) ) s=,

then every solution of (1) is oscillatory on [ t 0 ,).

Proof Following the same lines as in the proof of Theorem 2.1, we get (7) and (10). Using the inequality,

x γ y γ 2 1 γ ( x y ) γ ,γ1,

we have

( x γ ( t ) ) = x γ ( σ ( t ) ) x γ ( t ) μ ( t ) 2 1 γ ( x ( σ ( t ) ) x ( t ) ) γ μ ( t ) = 2 1 γ ( μ ( t ) ) γ 1 ( x ( σ ( t ) ) x ( t ) μ ( t ) ) γ = 2 1 γ ( μ ( t ) ) γ 1 ( x ( t ) ) γ .
(23)

Now setting z(t) by (15), using (17) and (23) we see that

z (t)α(t)Q(t)+ ( α ( t ) ) + z σ ( t ) α σ ( t ) 2 1 γ ( μ ( θ 2 ( t ) ) ) γ 1 α ( t ) ( α σ ( t ) ) 2 r ( θ 2 ( t ) ) ( z σ ( t ) ) 2 .

The remaining part of the proof is similar to that of Theorem 2.3, hence it is omitted. □

Example 2.5

Consider the following second-order neutral nonlinear dynamic equation

( ( ( y ( t ) + ( t + a 1 t + a ) y ( τ ( t ) ) ) ) 5 / 3 ) + a b t 1 / 3 y(tξ)ξ=0,tT

where γ= 5 3 , r(t)=1, p(t)=( t + a 1 t + a ), q(t)= t 1 / 3 . One can verify that the conditions of Theorem 2.3 are satisfied. Note that taking α(s)=s, we see that

lim sup t t 0 t ( α ( s ) Q ( s ) ( ( α ( s ) ) + ) 2 r ( θ 2 ( s ) ) 4 γ ( θ 2 ( s ) 2 ) γ 1 α ( s ) ) s = lim sup t t 0 t ( ( b a ) s 1 1 20 3 ( s b 2 ) 2 / 3 s ) s = .

Therefore, (1) is oscillatory.

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Correspondence to Tuncay Candan.

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Keywords

  • oscillation
  • dynamic equations
  • time scales
  • distributed deviating arguments