Oscillation and asymptotic properties of a class of second-order Emden–Fowler neutral differential equations

We consider a class of second-order Emden–Fowler equations with positive and nonpositve neutral coefficients. By using the Riccati transformation and inequalities, several oscillation and asymptotic results are established. Some examples are given to illustrate the main results.

However, the results of Liu et al. (2012) and Shi et al. (2016) cannot be applied to Eq. (1) due to −1 < p(t) ≤ 0 in (1), but, in Liu et al. (2012), Shi et al. (2016) the assumption is 0 ≤ p(t) ≤ 1 or p(t) > 1. Similarly, the results in Erbe et al. (2009) and Li et al. (2015) cannot be applied to Eq. (1) because there is another parameter β and the condition on function f in Li et al. (2015), Erbe et al. (2009) does not satisfy the hypothesis (H 5 ). In this paper, we will extend the results of Liu et al. (2012), Shi et al. (2016) to the case of −1 < p(t) ≤ 0 and improve the results of Erbe et al. (2009. By employing Riccati transformation, several new oscillation and asymptotic criteria are obtained under the assumptions that (H 1 ) − (H 5 ). Throughout this paper, we suppose that all inequalities hold for sufficiently large t. Without loss of generality, we only consider the positive solutions of (1).
In what follows, let We say a function H = H (t, s) belongs to a function class P, if it satisfies (ii) H has partial derivatives ∂H /∂t and ∂H /∂s on D 0 , such that and where h 1 and h 2 are nonnegative continuous functions on D 0 .
Proof Suppose x is a nonoscillatory solution of (1). Without loss of generality, there exists a t 1 ≥ t 0 , such that x(t) > 0, x(τ (t)) > 0, and x(σ (t)) > 0, for all t ≥ t 1 . From (1) and the hypothesis (H 5 ), we get Therefore, r|z ′ | α−1 z ′ is nonincreasing. We claim that z ′ > 0. Otherwise, if z ′ < 0, using the fact that r|z ′ | α−1 z ′ is nonincreasing, there exists a positive k > 0, such that That is, Integrating the above inequality from t 1 to t, we get It follows from (H 2 ) that We consider the following two cases.
Then we get which contradicts (7). Case 2 If x is bounded, from the definition of z and −1 < p(t) ≤ 0, z is also bounded, which also contradicts (7).
Hence, it is clear from the above discussion that z ′ > 0, and then z > 0 or z < 0. We consider each of two cases separately.
Suppose first that z > 0. Considering the definition of z and −1 < p(t) ≤ 0, we get From σ (t) ≤ t and the fact that r(z ′ ) α is nonincreasing, we obtain and there exist a positive constant K and a t 2 ≥ t 1 , such that From the fact that r(z ′ ) α is nonincreasing, we get .
where l is a constant. That is, for all sufficiently large t, z is bounded. We can easily prove that x is also bounded. From the fact that x is bounded, we get where a is a constant. We claim that a = 0. Otherwise, if a > 0, there exists a sequence {t n }, such that lim n→∞ t n = ∞ and lim n→∞ x(t n ) = a. Letting for large enough n, we obtain Then, from the definition of z(t) and p(t) ≥ −p 0 , we have which contradicts z < 0. Thus lim sup t→∞ x(t) = 0. By x > 0, we get The proof is complete. From Theorem 1, letting ρ = 1, we get the following corollary. Theorem 2 Assume that (H 2 ) − (H 5 ) hold. If there exist two functions H ∈ P and ρ ∈ C 1 ([t 0 , ∞), (0, +∞)), such that where is as in Theorem 1, then the conclusion of Theorem 1 remains intact.
Proof Suppose that x is a nonoscillatory solution of (1). Proceeding as in the proof of Theorem 1, we get z > 0 or z < 0.
Firstly, we consider z > 0. As the proof above (13) holds. That is

Multiplying this inequality by H(t, s) and integrating it from t 2 to t, we have
By the property of ∂H (t, s)/∂s = −h 2 (t, s) √ H (t, s) < 0, we conclude that ds to the latter inequality and multiplying this inequality by 1/H (t, t 0 ), we get Letting t → ∞ in (16), we can get a contradict to (14). If z < 0, repeating the proof of Theorem 1, we have lim t→∞ x(t) = 0. This completes the proof.
Corollary 2 Assume that (H 2 ) − (H 5 ) hold. If then the conclusion of Theorem 1 remains intact.
and for all T ≥ t 0 and for some constant θ > 1, where A + (t) = max{A(t), 0}, is as in Theorem 1, and H satisfies then every solution of Eq. (1) is oscillatory or tends to zero as t → ∞.
Proof Suppose that x is a nonoscillatory solution of Eq. (1). Then, as in the proof of Theorem 1, z > 0 or z < 0.

Thus,
Taking into account (19) and (21), we deduce that then and Now we will prove that 1 Proof See "Appendix".

Examples
In this section, we will present two examples to illustrate the main results.