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

## 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 $γ⩾1$ is a quotient of odd positive integers.

## 1 Introduction

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

on a time scale $T$. Throughout this paper, it is assumed that $γ⩾1$ is a quotient of odd positive integers, $0, $τ(t):T→T$, 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 $0⩽p(t)<1$ are real valued rd-continuous functions defined on $T$, $p(t)$ is increasing and

(H1) $∫ t 0 ∞ ( 1 r ( t ) ) 1 γ △t=∞$,

(H2) $f:T×R→R$ is a continuous function such that $uf(t,u)>0$ for all $u≠0$ 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  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.  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.  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  considered (2) but he used different technique to prove his results. In  and , 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 . For neutral equations with distributed deviating arguments, we refer the reader to the paper by Candan . To the best of our knowledge,  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 $b⩾ξ⩾a$. In the case when $y(t)$ is negative, the proof is similar. In view of (1), (H2) and (3)

$x [ 2 ] (t)+ ∫ a b q(t) y γ ( δ ( t , ξ ) ) △ξ⩽0$
(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

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

$x [ 2 ] (t)+ ∫ a b q(t) ( 1 − p ( δ ( t , ξ ) ) ) γ x γ ( δ ( t , ξ ) ) △ξ⩽0.$
(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

$x [ 2 ] (t)+ ∫ a b q(t) ( 1 − p ( δ ( t , ξ ) ) ) γ ( θ 2 ( t ) 2 ) γ ( x △ ( δ ( t , ξ ) ) ) γ △ξ⩽0$

and it can be found

$x [ 2 ] (t)+(b−a)q(t) ( 1 − p ( θ 1 ( t ) ) ) γ ( θ 2 ( t ) 2 ) γ ( x △ ( θ 1 ( t ) ) ) γ ⩽0,$

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 $Q(s)=(b−a)q(s) ( 1 − p ( θ 1 ( s ) ) ) γ$. 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 $b⩾ξ⩾a$. Proceeding as in the proof of Theorem 2.1, we obtain (7) and the inequality (10). Using (7) and Pötzsche’s chain rule [, 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

$x [ 2 ] (t)⩽−(b−a)q(t) ( 1 − p ( θ 1 ( t ) ) ) γ x γ ( θ 2 ( t ) ) =−Q(t) x γ ( θ 2 ( t ) ) ,t⩾ t 4 .$
(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 $u−m u 2 ⩽ 1 4 m$, $m>0$, 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 )

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 $t∈T$. 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,

$x γ − y γ ⩾ 2 1 − γ ( x − y ) γ ,γ⩾1,$

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

$( ( ( y ( t ) + ( t + a − 1 t + a ) y ( τ ( t ) ) ) △ ) 5 / 3 ) △ + ∫ a b t − 1 / 3 y(t−ξ)△ξ=0,t∈T$

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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Candan, T. Oscillation criteria for second-order nonlinear neutral dynamic equations with distributed deviating arguments on time scales. Adv Differ Equ 2013, 112 (2013). https://doi.org/10.1186/1687-1847-2013-112 