# Oscillation criteria for second-order nonlinear neutral difference equations of mixed type

## Abstract

Some oscillation criteria are established for the second order nonlinear neutral difference equations of mixed type.

$Δ 2 ( x n + a x n - τ 1 ± b x n + τ 2 ) α = q n x n - σ 1 β + p n x x + σ 2 β , n ≥ n 0$

where α and β are ratio of odd positive integers with β ≥ 1. Results obtained here generalize some of the results given in the literature. Examples are provided to illustrate the main results.

2010 Mathematics Subject classification: 39A10.

## 1 Introduction

In this article, we study the oscillation behavior of solutions of mixed type neutral difference equation of the form,

$Δ 2 ( x n + a x n - τ 1 ± b x n + τ 2 ) α = q n x n - σ 1 β + p n x x + σ 2 β$

where n (n0) = {n0, n0 + 1, ...}, n0 is a nonnegative integer, a, b are real nonnegative constants, τ1, τ2, σ1, and σ2 are positive integers, {q n } and {p n } are positive real sequences and α, β are ratio of odd positive integers with β ≥ 1.

Let θ = max {τ1, σ1}. By a solution of Equation (E±) we mean a real sequence {x n } which is defined for nn0 - θ and satisfies Equation (E±) for all n (n0). A nontrivial solution of Equation (E±) is said to be oscillatory if it is neither eventually positive nor eventually negative. Otherwise it is known as nonoscillatory.

Equations of this type arise in a number of important applications such as problems in population dynamics when maturation and gestation are included, in cobweb models, in economics where demand depends on the price at an earlier time and in electric networks containing lossless transmission lines. Hence it is important and useful to study the oscillation behavior of solutions of neutral type difference Equation (E±).

The oscillation, nonoscillation and asymptotic behavior of solutions of Equation (E±), when b = 0 and p n ≡ 0 or a = 0 and p n ≡ 0 or b = 0 and q n ≡ 0 have been considered by many authors, see for example  and the reference cited therein. However, there are few results available in the literature regarding the oscillatory properties of neutral difference equations of mixed type, see for example . Motivated by the above observation, in this article we establish some new oscillation criteria for the Equation (E±) which generalize some of the results obtained in [13, 57].

In Section 2, we present conditions for the oscillation of all solutions of equation (E±). Examples are provided in Section 3 to illustrate the results.

## 2 Oscillation results

In this section, we obtain sufficient conditions for the oscillation of all solutions of Equation (E±). First we consider the Equation (E-), viz,

$Δ 2 ( x n + a x n - τ 1 - b x n + τ 2 ) α = q n x n - σ 1 β + p n x x + σ 2 β , n ∈ ℕ ( n 0 ) .$

To prove our main results we need the following lemma, which can be found in .

Lemma 2.1. Let A0, B0 and γ1. Then

$A γ + B γ ≥ 1 2 γ - 1 ( A + B ) γ ,$
(2.1)

and

$A γ - B γ ≥ ( A - B ) γ , i f A ≥ B .$
(2.2)

Theorem 2.2. Let σ1 > τ1, σ2 > τ2 and {q n } and {p n } are positive real nonincreasing sequences. Assume that the difference inequalities

i)

$Δ 2 y n - p n 2 β - 1 ( 1 + a β ) β / α y n + σ 2 β / α ≥ 0$
(2.3)

has no eventually positive increasing solution,

ii)

$Δ 2 y n - p n 2 β - 1 ( 1 + a β ) β / α y n - σ 2 + τ 1 β / α ≥ 0$
(2.4)

has no eventually positive decreasing solution,

iii)

$Δ 2 y n + q n b β y n - σ 1 - τ 2 β / α + p n b β y n + σ 2 - τ 2 β / α ≤ 0$
(2.5)

has no eventually positive solution.

Then every solution of Equation (E-) is oscillatory.

Proof. Let {x n } be a nonoscillatory solution of Equation (E-). Without loss of generality, we may assume that there exists n1 (n0) such that xn-θ> 0 for all nn1. Set $z n = ( x n + a x n - τ 1 - b x n + τ 2 ) α$.

Then

$Δ 2 z n = q n x n - σ 1 β + p n x n + σ 2 β > 0 , n ≥ n 1 ,$

which implies that {z n } and {Δz n } are of one sign for all n2n1. We claim that z n > 0 eventually. To prove it assume that z n < 0. Then we let

$0 < u n = - z n = ( b x n + τ 2 - a x n - τ 1 - x n ) α ≤ b α x n + τ 2 α .$

Thus

$x n β ≥ 1 b β u n - τ 2 β / α , n ≥ n 2 .$

From Equation (E-), we get

$0 = Δ 2 u n + q n x n - σ 1 β + p x x n + σ 2 β ≥ Δ 2 u n + q n b β u n - σ 1 - τ 2 β / α + p n b β u n + σ 2 - τ 2 β / α .$

Hence {u n } is a positive solution of inequality (2.5), a contradiction.

Therefore z n ≥ 0. We define

$y n = z n + a β z n - τ 1 - b β 2 β - 1 z n + τ 2 .$
(2.6)

Then, we have

$Δ 2 y n = Δ 2 z n + a β Δ 2 z n - τ 1 - b β 2 β - 1 Δ 2 z n + τ 2 = q n x n - σ 1 β + p n x n + σ 2 β + a β q n - τ 1 x n - σ 1 - τ 1 β + a β p n - τ 1 x n + σ 2 - τ 1 β - b β 2 β - 1 q n + τ 2 x n - σ 1 + τ 2 β - b β 2 β - 1 p n + τ 2 x n + σ 2 + τ 2 β .$
(2.7)

Using the inequality (2.1) in (2.7), we obtain

$Δ 2 y n ≥ q n 2 β - 1 ( x n - σ 1 + a x n - σ 1 - τ 1 ) β - b β 2 β - 1 q n + τ 2 x n - σ 1 + τ 2 β + p n 2 β - 1 ( x n + σ 2 + a x n + σ 2 - τ 1 ) β - b β 2 β - 1 p n + τ 2 x n + σ 2 + τ 2 β .$

Now using the inequality (2.2), we obtain

$Δ 2 y n ≥ q n 2 β - 1 z n - σ 1 β / α + p n 2 β - 1 z n + σ 2 β / α > 0 .$
(2.8)

Consequently {y n } and {Δy n } are of one sign, eventually. Now we shall prove that y n > 0. If not, then let

$0 < v n = - y n = b β 2 β - 1 z n + τ 2 - a β z n - τ 1 - z n ≤ b β 2 β - 1 z n + τ 2 .$

Hence

$z n ≥ 2 β - 1 b β v n - τ 2 ,$

and (2.8) implies

$0 ≥ Δ 2 v n + q n b β v n - σ 1 - τ 2 β / α + p n b β v n + σ 2 - τ 2 β / α .$

We obtain that {v n } is a positive solution of inequality (2.5), a contradiction. Next we consider the following two cases:

Case 1: Let Δz n < 0 for nn3n2. We claim that Δy n < 0 for nn3. If not, then we have y n > 0, Δy n > 0 and Δ2y n ≥ 0 which implies that $lim n → ∞ y n =∞$. On the other hand, z n > 0, Δz n < 0 implies that $lim n → ∞ z n =c<∞$. Then applying limits on both sides of (2.6) we obtain a contradiction. Thus Δy n < 0 for nn3. Using the monotonicity of {z n }, we now get

$y n - σ 1 = z n - σ 1 + a β z n - σ 1 - τ 1 - b β 2 β - 1 z n - σ 1 + τ 2 ≤ ( 1 + a β ) z n - σ 1 - τ 1 .$

This together with (2.8) implies

$Δ 2 y n ≥ q n 2 β - 1 ( 1 + a β ) β / α y n - σ 1 + τ 1 β / α .$

Thus {y n } is a positive decreasing solution of inequality (2.4), a contradiction. Case 2: Let Δz n > 0 for nn3. Now we consider the following two cases. Case (i): Assume that Δy n < 0 for nn3. Proceeding similarly as above and using the monotonicity of {z n } we obtain

$y n - σ 1 ≤ ( 1 + a β ) z n - σ 1 .$

Then using this in (2.8) we obtain

$Δ 2 y n ≥ q n 2 β - 1 z n - σ 1 β / α ≥ q n 2 β - 1 ( 1 + a β ) β / α y n - σ 1 β / α ≥ q n 2 β - 1 ( 1 + a β ) β / α y n - σ 1 + τ 1 β / α ,$

and again {y n } is a positive decreasing solution of inequality (2.4), a contradiction.

Case (ii): Assume that Δy n > 0 for nn3. Then $y n + σ 2 ≤ ( 1 + a β ) z n + σ 2$ which in view of (2.8) implies

$Δ 2 y n ≥ p n 2 β - 1 z n + σ 2 β / α ≥ p n 2 β - 1 ( 1 + a β ) β / α y n + σ 2 β / α ,$

that is, (2.3) has a positive increasing solution, a contradiction. The proof is complete.

Remark 2.1. Theorem 2.2 permits us to obtain various oscillation criteria for Equation (E-). Moreover we are able to study the asymptotic properties of solutions of Equation (E-) even if not all assumptions of Theorem 2.2 are satisfied. If the difference inequality (2.3) has an eventually positive increasing solution then the conclusion of Theorem 2.2 is replaced by "Every solution of Equation (E-) is either oscillatory or |x n | → ∞ as n → ∞".

Remark 2.2. In [2, Theorem 7.6.26], the author considered the Equation (E-) with α = β = 1, p n p, and q n q and obtain oscillation results with (1 + a - b) > 0. Hence Theorem 2.2 generalize and improve the results of [2, Theorem 7.6.26].

Remark 2.3. Applying existing conditions sufficient for the inequalities (2.3), (2.4), and (2.5) to have no above mentioned solutions, we immediately obtain various oscillation criteria for Equation (E-).

Theorem 2.3. Let σ1 > τ1, σ2 ≥ 2, and β = α. Assume that

(2.9)

and

(2.10)

and that the difference inequality (2.5) has no eventually positive solution. Then every solution of Equation (E-) is oscillatory.

Proof. Conditions (2.9) and (2.10) are sufficient for the inequality (2.3) to have no increasing positive solution and for (2.4) to have no decreasing positive solution, respectively (see e.g., [2, Lemma 7.6.15]). The proof then follows from Theorem 2.2.

Remark 2.4. Taking into account the result of , we see that the absence of positive solution of (2.5) can be replaced by the assumption that for the corresponding equation

$Δ 2 y n + q n b β y n - σ 1 - τ 2 β / α + p n b β y n + σ 2 - τ 2 β / α = 0$

every solution of this equation are oscillatory.

Next we consider the difference Equation (E+)

$Δ 2 ( x n + a x n - τ 1 + b x n + τ 2 ) α = q n x n - σ 1 β + p n x n + σ 2 β$

and present conditions for the oscillation of all solutions of Equation (E+).

Theorem 2.4. Assume that σ1τ1, σ2τ2 + 2, $q n * =min { q n - σ 1 , q n , q n + τ 2 }$ and $p n * =min { p n - σ 1 , p n , p n + τ 2 }$. If

$Δ 2 y n - p n * 4 β - 1 1 + a β + b β 2 β - 1 β / α y n - τ 2 + σ 2 β / α ≥ 0 . ,$
(2.11)

has no eventually positive increasing solution, and

$Δ 2 y n - p n * 4 β - 1 1 + a β + b β 2 β - 1 β / α y n - σ 2 + τ 2 β / α ≥ 0 ,$
(2.12)

has no eventually positive decreasing solution, then every solution of (E+) is oscillatory.

Proof. Let {x n } be a nonoscillatory solution of (E+). Without loss of generality, we assume that there exists an integer n1 (n0) such that xn-θ> 0 for all nn1. Setting

$z n = ( x n + a x n - τ 1 + b x n + τ 2 ) α$

and

$y n = z n + a β z n - τ 1 + b β 2 β - 1 z n + τ 2 .$
(2.13)

Then z n > 0, y n > 0 and

$Δ 2 z n = q n x n - σ 1 β + p n x n + σ 2 β ≥ 0 .$
(2.14)

Then {Δz n } is of one sign, eventually. On the other hand

$Δ 2 y n = q n x n - σ 1 β + p n x n + σ 2 β + a β q n - τ 1 x n - σ 1 - τ 1 β + a β p n - τ 1 x n + σ 2 - τ 1 β + b β 2 β - 1 q n + τ 2 x n - σ 1 + τ 2 β + b β 2 β - 1 p n + τ 2 x n + σ 2 + τ 2 β .$
(2.15)

Using (2.1) in (2.15) we obtain

$Δ 2 y n ≥ q n * 2 β - 1 ( x n - σ 1 + a x n - σ 1 - τ 1 ) β + q n + τ 2 2 β - 1 b β x n - σ 1 + τ 2 β + p n * 2 β - 1 ( x n + σ 2 + a x n + σ 2 - τ 1 ) β + p n + τ 2 2 β - 1 b β x n + σ 2 + τ 2 β$

or

$Δ 2 y n ≥ 1 4 β - 1 q n * z n - σ 1 β / α + p n * z n + σ 2 β / α , n ≥ n 1 .$
(2.16)

Next we consider the following two cases:

Case 1: Assume that Δz n > 0. Then Δy n > 0. In view of (2.16), we have

$Δ 2 y n + τ 2 ≥ 1 4 β - 1 p n + τ 2 * z n + σ 2 + τ 2 β / α .$
(2.17)

Applying the monotonicity of z n , we find

$y n + σ 2 = z n + σ 2 + a β z n - τ 1 + σ 2 + b β 2 β - 1 z n + τ 2 + σ 2 ≤ 1 + a β + b β 2 β - 1 z n + τ 2 + σ 2 .$
(2.18)

Combining (2.17) and (2.18) we have

$Δ 2 y n + τ 2 ≥ p n + τ 2 * 4 β - 1 1 + a β + b β 2 β - 1 β / α y n + σ 2 β / α .$
(2.19)

Thus

$Δ 2 y n - p n * 4 β - 1 1 + α β + b β 2 β - 1 β / α y n - τ 2 + σ 2 β / α ≥ 0 .$

Therefore {y n } is a positive increasing solution of the difference inequality (2.11), a contradiction.

Case 2: Assume that Δz n < 0. Then Δy n < 0. In view of (2.16) we see that

$Δ 2 y n - τ 1 ≥ 1 4 β - 1 q n - τ 1 * z n - τ 1 - σ 1 β / α .$

From the monotonicity of {z n } we find

$y n - σ 1 ≤ 1 + a β + b β 2 β - 1 z n - τ 1 - σ 1 .$

Combining the last two inequalities, we obtain

$Δ 2 y n - τ 1 ≥ q n - τ 1 * 4 β - 1 1 + a β + b β 2 β - 1 β / α y n - σ 1 . β / α$
(2.20)
$Δ 2 y n - q n * 4 β - 1 1 + a β + b β 2 β - 1 β / α y n - σ 1 + τ 1 β / α ≥ 0 .$

Therefore {y n } is a positive decreasing solution of the difference inequality (2.12), a contradiction. This completes the proof.

Theorem 2.5. Assume that σ1τ1, σ2τ2+2, β = α, $q n * =min { q n - σ 1 , q n , q n + r 2 }$ and $p n * = min { p n - σ 1 , p n , p m + r 2 }$. If

(2.21)

has no eventually positive increasing solution, and

(2.22)

has no eventually positive decreasing solution, then every solution of (E+) is oscillatory.

Proof. Conditions (2.21) and (2.22) are sufficient for the inequality (2.11) to have no increasing positive solution and for (2.12) to have no decreasing positive solution, respectively (see e.g., [2, Lemma 7.6.15]). The proof then follows from Theorem 2.4.

Remark 2.5. When a = β = 1, Theorem 2.5 involves result of Theorem 7.6.6 of .

Theorem 2.6. Let $β=α, δ 1 = σ 1 - τ 1 2 >0$, and $δ 2 = σ 2 - τ 2 2 >0$. Suppose that there exist two positive real sequence {ϕ n } and {ψ n } with Δϕ n ≥ 0 and Δψ n ≤ 0, such that

$q n * ≥ 4 α - 1 1 + a α + b a 2 α - 1 ϕ n + 1 ϕ n - δ 1 ,$
(2.23)

and

$p n * ≥ 4 α - 1 1 + a α + b α 2 α - 1 ψ n + 1 ψ n + δ 2 ,$
(2.24)

where $p n *$, $q n *$ are as in Theorem 2.4. If the difference inequality

$Δ v n + ϕ n - δ 1 v n - δ 1 ≥ 0$
(2.25)

has no eventually negative solution, and

$Δ v n + ψ n - δ 2 v n - δ 2 ≥ 0$
(2.26)

has no eventually positive solution, then every solution of Equation (E+) is oscillatory.

Proof. Let {x n } be a nonoscillatory solution of Equation (E+). Without loss of generality, we assume that there exists an integer n1 (n0) such that xn-θ> 0 for all nn1. Define z n and y n as in Theorem 2.4. Proceeding as in the proof of Theorem 2.4, we obtain (2.16). Next we consider the following two cases.

Case 1: Assume Δz n > 0. Clearly Δy n > 0. Then as in case 1 of Theorem 2.4, we find that {y n } is a positive increasing solution of inequality (2.19). Let $B n - Δ y n + ψ n y n + δ 2$. Then B n > 0. Using (2.24), we have

$Δ B n - Δ ψ n ψ n B n - ψ n + 1 B n + δ 2 = Δ 2 y n - Δ ψ n ψ n Δ y n - ψ n + 1 ψ n + δ 2 y n + 2 δ 2 ≥ Δ 2 y n - ψ n + 1 ψ n + δ 2 y n + 2 δ 2 ≥ Δ 2 y n - p n * 4 α - 1 1 + a α + b α 2 α - 1 y n - τ 2 + σ 2 ≥ 0 .$

Define B n = ψ n v n . Then {v n } is a positive solution of (2.26), a contradiction. Case 2: Assume that Δz n < 0. Clearly Δy n < 0. Then as in case 2 of Theorem 2.4, we find that {y n } is a positive decreasing solution of inequality (2.20). Let $A n - Δ y n + ϕ n y n - δ 1$. Then A n < 0. Using (2.23), we have

$Δ A n - Δ ϕ n ϕ n - ϕ n + 1 A n - δ 1 = Δ 2 y n - Δ ϕ n ϕ n Δ y n - ϕ n + 1 ϕ n - δ 1 y n - 2 δ 1 ≥ Δ 2 y n - ϕ n + 1 ϕ n + δ 1 y n - 2 δ 1 ≥ Δ 2 y n - q n * 4 α - 1 1 + a α + b α 2 α - 1 y n - σ 1 + τ 1 ≥ 0 .$

Define A n = ϕ n v n . Then {v n } is a negative solution of inequality (2.25), a contradiction. This completes the proof.

From Theorem 2.6 and the results given in  we have the following oscillation criteria for Equation (E+).

Corollary 2.7. Let $β=α, δ 1 = σ 1 - τ 1 2 >0$, and $δ 2 = σ 2 - τ 2 2 >0$. Suppose that there exist two positive real sequence {ϕ n } and {ψ n } with Δϕ n ≥ 0 and Δψ n ≤ 0 such that (2.23) and (2.24) holds. If

(2.27)

and

(2.28)

then every solution of Equation (E+) is oscillatory.

Proof. It is known (see ) that condition (2.27) is sufficient for inequality (2.25) to have no eventually negative solution. On the other hand, condition (2.28) is sufficient for inequality (2.26) to have no eventually positive solution.

Remark 2.6. From the results presented in this section, we observe that when the coefficient p n = 0 or the condition on {p n } is violated the conclusion of the theorem may be replaced by "Every solution {x n } of equation (E±) is oscillatory or x n → ∞ as n → ∞".

Once again from the proofs, we see that if q n = 0 or condition on {q n } is violated then the conclusion of the theorems may be replaced by "Every solution {x n } of Equation (E±) is oscillatory or x n → 0 as n → ∞".

## 3 Examples

In this section, we provide some examples to illustrate the main results.

Example 1. Consider the difference equation

$Δ 2 x n + 2 τ 1 x n - τ 1 - 2 - τ 2 x n + τ 2 = 1 2 2 σ 1 x n - σ 1 + 2 - σ 2 x n + σ 2 .$
(3.1)

Here $α = β = 1 , a = 2 τ 1 , b = 2 - τ 2 , q n = 2 σ 1 - 1 , p n = 2 - σ 2 - 1 .$.

It is easy to see that condition (2.9) of Theorem 2.3 is not satisfied and hence Equation ( 3 .1) has a nonoscillatory solution {x n } = {2n} → ∞ as n → ∞.

Example 2. Consider the difference equation

$Δ 2 x n + 2 - τ 1 x n - τ 1 - 2 τ 2 x n + τ 2 = 1 8 2 - σ 1 x n - σ 1 + 2 σ 2 x n + σ 2 .$
(3.2)

Here $α = β = 1 , a = 2 - τ 1 , b = 2 τ 2 , q n = 2 - σ 1 - 3 , p n = 2 σ 2 - 3 , σ 1 > τ 1 , σ 2 > τ 2 .$ . It is easy to see that condition (2.10) of Theorem 2.3 is not satisfied and hence Equation ( 3 .2) has a nonoscillatory solution ${ x n } = 1 2 n → 0$ as

n → ∞.

Example 3. Consider the difference equation

$Δ 2 x n + a x n - τ 1 + b x n - τ 2 α = q x n - σ 1 α + p x n + σ 2 α , n ≥ n 0 .$
(3.3)

where p > 0, q > 0 are constants and σ1 > τ1 and σ 2 > τ2. It is easy to see that $q n * =q$ and $p n * =p$. Assume that ϵ > 0. Let $ϕ n = 2 + ε 2 δ 1 δ 1 ( δ 1 + 1 ) δ 1 + 1$, $ψ n = 2 + ε 2 δ 2 δ 2 ( δ 2 + 1 ) δ 2 + 1$, where $δ i = σ i - τ i 2$, i = 1, 2.

Clearly {ϕ n } and {ψ n } satisfies the condition of Corollary 2.6. If

$q > 4 α - 1 2 + ε 2 2 δ 1 δ 1 ( δ 1 + 1 ) δ 1 + 1 2 1 + a α + b α 2 α - 1$

and

$p > 4 α - 1 2 + ε 2 2 δ 2 δ 2 ( δ 2 + 1 ) δ 2 + 1 2 1 + a α + b α 2 α - 1$

then condition (2.23) and (2.24) holds. Further we see that

$lim inf n → ∞ ∑ s = n - δ 1 n - 1 ϕ s - δ 1 = 2 + ε 2 δ 1 δ 1 ( δ 1 + 1 ) δ 1 + 1 δ 1 > δ 1 δ 1 + 1 δ 1 + 1$

and

By Corollary 2.6, we see that all solutions of Equation ( 3 .3) are oscillatory.

Remark 3.1. It would be interesting to extend the results of this article to the equation

$Δ a n Δ x n + b x n - τ 1 ± c x n + τ 2 α = q n x n - σ 1 β + p n x n + σ 2 γ$

where α, β, and γ are ratio of odd positive integers.

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Correspondence to Ethiraju Thandapani.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

ET framed the problem and NK solved the problem. SP modified and made necessary changes in the proof of the theorems. All authors read and approved the manuscript.

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Thandapani, E., Kavitha, N. & Pinelas, S. Oscillation criteria for second-order nonlinear neutral difference equations of mixed type. Adv Differ Equ 2012, 4 (2012). https://doi.org/10.1186/1687-1847-2012-4 