Property A of differential equations with positive and negative term

In the paper, we elaborate new technique for the investigation of the asymptotic properties for third order differential equations with positive and negative term ( b(t) ( a(t)x′(t) )′)′ + p(t) f (x(τ(t)))− q(t)h(x(σ(t))) = 0. We offer new easily verifiable criteria for property A. We support our results with illustrative examples.

The investigation of the higher order differential equations (see [1][2][3][4][5][6][7]) essentially makes use of some generalization of a lemma of Kiguradze [5,6].In the lemma, from the fixed sign of the highest derivative, we deduce the structure of possible nonoscillatory solutions.Since the positive and negative term are included in (E), we are not able to fix the sign of the third order quasi-derivative for an eventually positive solution.So the authors mainly study properties of (E) in the partial case when either p(t) ≡ 0 or q(t) ≡ 0.
In what follows we shall assume that It will be shown that this condition reduces the influence of the negative term and this permit us to study property A. By property A we mean the situation when every nonoscillatory solution of (E) tends to zero at infinity.

Main results
In this paper we provide easily verifiable conditions for property A of studied equation.To simplify our notation, we denote ds ds, and du ds, where t ≥ t 1 ≥ t 0 and t 1 is large enough.
Theorem 2.1.Let for all t 1 large enough Then (E) has property A.
Proof.Assume that (E) possesses an eventually positive solution x(t) on (T x , ∞), T x ≥ t 0 .We introduce the auxiliary function z(t) associated to x(t) by (2.4) It follows from (H6) and the boundedness of h(u) that definition of function z(t) is correct and z(t) exists for all t ≥ T x .It is useful to notice that z(t) > x(t) > 0, z (t) < x (t) and Therefore, Kiguradze's lemma implies that either eventually, let us say for t ≥ t 1 ≥ T x .At first, assume that z(t) ∈ N 2 .Using the fact that b(t) a(t)z (t) is decreasing, we have In view of (2.6), we see that Setting the last estimate into (2.5),we see that that y(t) = a(t)z (t) is a positive solution of the differential inequality and, what is more, y(t)/B(t) is decreasing and b(t)y (t) > 0.
An integration from t 1 to ∞ yields b(t 1 )y (t 1 ) ≥ f ( ) which contradicts with (2.2).Now, taking lim sup on the both sides of (2.9), one gets a contradiction with (2.3).Now, we assume that z(t) ∈ N 0 .Since z(t) is positive and decreasing, there exists lim t→∞ z(t) = 2 ≥ 0. It follows from (2.4) that lim t→∞ x(t) = 2 .If we admit that > 0, then x(τ(t)) ≥ > 0, eventually.An integration of (2.5) yields Integrating from t to ∞ and consequently from t 1 to ∞ one gets For partial case of (E) we have the following easily verifiable criterion.
Our results are applicable also for the case when τ(t) ≡ t.
du ds dv which contradicts with (2.1) and the proof is complete.