Classifications of solutions of second-order nonlinear neutral differential equations of mixed type

Abstract

In this paper the authors classified all solutions of the second-order nonlinear neutral differential equations of mixed type,into four classes and obtained conditions for the existence/non-existence of solutions in these classes. Examples are provided to illustrate the main results. MSC: 34C15

On the oscillation of second-order half-linear functional differential equations with mixed neutral term

“…However, oscillation results for mixed neutral differential and dynamic equations are relatively scarce in the literature; some results can be found, for example, in [20][21][22][23][24][25][26][27][28][29][30][31][32], and the references cited therein. We would like to point out that the results obtained in [20][21][22][23][24][25][26][27][28][29][30][31][32] require both of p 1 and p 2 to be constants or bounded functions, and hence, the results established in these papers cannot be applied to the cases where lim t→∞ p 1 (t) = ∞ and /or lim t→∞ p 2 (t) = ∞. In view of the observations above, we wish to develop new sufficient conditions which can be applied to the cases where lim t→∞ p 1 (t) = ∞ and /or lim t→∞ p 2 (t) = ∞.…”

Global Non-monotonicity of Solutions to Nonlinear Second-Order Differential Equations

“…were reported by Thandapani et al [37]. Nonexistence of weakly oscillatory solutions for a third-order nonlinear functional differential equation…”

Classification of Solutions of Second Order Nonlinear Neutral Delay Dynamic Equations with Positive and Negative Coefficients

“…x(n) 2x(n 1)) n () 2 2 3 2 2 2 8(n 1) 1 x(n)(x (n) 1) (n n )(n 1) (n 1) 2 3 25 5n 10n 6 x (n 1) 0 (n n)(n 1) , (3.5) n2 . For this difference equation, all assumptions of Theorem (3.1) holds, but 6 (H ) The equation Hence, z (t) 0 Now proceeding as in the proof of Theorem 3.1, we obtain t r(t)z (t) lim f (x( (t))) due to 9 (H ) , which contradicts the assumption z (t) 0 for large 1…”

“…f (x( (t))) (q(t) Mh(t)) for 1 t [t , ) T , due to 2 (H ) - 5 (H ) Integrating the last inequality from 1 t to t, we obtain 1 t 11 t 1 r(t )z (t ) r(t)z (t) (q(s) Mh(s)) s f (x( (t))) f (x( (t ))) From 6 (H ) , we obtain t r(t)z (t) inf , f (x( (t))) lim Which contradicts the assumption z (t) 0 for large t. Thus, the theorem is proved.…”