Asymptotic Behaviour of Neutral Differential Equations of Third-order with Negative Term

We derive new comparison theorems and oscillation criteria for neutral differential equations of third order with negative term. We show that one can deduce oscillation criteria for the equation with negative term from those for the equation with positive term. We give some examples and show applications to equation with symmetric operator.


Introduction
Functional and neutral differential equations play an important role in many applications and have a long and rich history with a substantial contribution of Hungarian mathematicians, among them Tibor Krisztin who focused among others on asymptotic properties of delay and neutral functional differential equations of first order and applications, see e.g.[11] and [12].
Recently, much attention has been devoted to the oscillation of the neutral differential equation with a positive term 1 p(t) 1 r(t) x(t) + a(t)x γ(t) + q(t) f x δ(t) = 0, (E+) see e.g.[1,2,[7][8][9]15] and references given there.The aim of this paper is to consider the third-order neutral differential equation with negative term 1 r(t) where t ≥ t 0 .Moreover, we will consider the linear version of this equation 1 r(t) 1 p(t) z(t) + a(t)z γ(t) − q(t)z δ(t) = 0. (L−) Observe that differential operators in both equations, i.e. in equations (E+) and (E−), are mutually adjoint and therefore functions in operators are interchanged.
are called quasiderivatives of v.
A solution z of (E−) is said to be proper if it exists on the interval [t 0 , ∞) and satisfies the condition sup{|z(s)| : t ≤ s < ∞} > 0 for any t ≥ t 0 .
A proper solution is called oscillatory or nonoscillatory according to whether it does or does not have arbitrarily large zeros.
Later we show that (1.2) is equivalent to lim t→∞ v [2] (t) = ∞ (see Lemma 3.6).Hence, in case a(t) ≡ 0 this yields the original definition of property B introduced by I. Kiguradze for ordinary differential equations (see [10]).
To simplify notation, we set

Preliminaries: linear ODE and FDE
First, consider the special case of (L−), where a(t) ≡ 0 and δ(t) = t, i.e. the third-order linear differential equation 1 r(t) For completeness, we summarize basic results concerning the oscillatory behaviour of (2.1), which we will need in our later consideration.
It is a well-known fact (see for instance [10]) that all nonoscillatory solutions x of (2.1) can be divided into the two classes: M 1 = z solution of (2.1), ∃ T z : z(t)z [1] (t) > 0, z(t)z [2] (t) < 0 for t ≥ T z M 3 = z solution of (2.1), ∃ T z : z(t)z [1] (t) > 0, z(t)z [2] (t) > 0 for t ≥ T z .Definition 2.1.Equation (2.1) is said to have property B if any proper solution z of (2.1) is either oscillatory or satisfies The classification of nonoscillatory solutions of (2.4) and definitions of property B and weak property B are the same as for equation (2.1).One of our main tools will be the comparison method for third-order linear functional differential equations where q 1 (t) ≥ q 2 (t) > 0 and lim t→∞ δ a) If there exists a solution y ∈ M 1 of (2.5), then there exists a solution z ∈ M 1 of (2.6).
The following theorem extends Proposition 2.4 to functional differential equations.This result is interesting in the light of results from the book [17], where various criteria for weak property B are given."⇐=": Assume by contradiction that there exists a solution z ∈ M 3 of (2.4) such that lim t→∞ z [2] (t) = c > 0, i.e. there exists ε > 0 such that c − ε ≤ z [2] (t) ≤ c.
Integrating this inequality twice from t 0 to t we obtain (2.9) Consider the linear equation Then y = z is a solution of (2.10).By Theorem C, we get In view of (2.9), we get By Theorem A, the linear equation is nonoscillatory and it has a solution x ∈ M 1 by Proposition 2.2.Consider the delayed equation Then z = x is a solution of (2.11).Since x is increasing and δ(t) ≤ t, we have ≥ 1 for large t.
By the comparison theorem for the functional differential equations (Proposition 2.5), equation (2.4) has a solution x ∈ M 1 , a contradiction.
Example 2.8.Consider the equation where 0 < k < 1.A quick computation shows that condition (2.8) holds and therefore (2.12) has property B if and only if it has weak property B.
We finish this part by recalling a result concerning equivalence between property B and property A. Consider the equation (2.13) The equation (2.13) is said to have property A if any proper solution x of (2.13) is either oscillatory or satisfies Theorem D ([4, Theorem 1]).Equation (2.13) has property A if and only if equation (2.1) has property B.

Basic properties of neutral equations
In this section we establish some auxiliary results concerning the properties of solutions of neutral equation (E−).
Lemma 3.1.Let z be a nonoscillatory solution of (E−) and let v be defined by (1.1).Then v, v [1] , v [2]  are monotone for large t.
Proof.The proof is similar to the proof of Lemma 1 in [8] and therefore it is omitted.
Assume that v [2] (t) > 0 for t ≥ t 2 .Then there exists an M > 0 such that Integrating from t 2 to t we get Letting t → ∞ and using the fact that ∞ t 0 r(t) dt = ∞, we get v [1] (t) → ∞, i.e. v [1] (t) > 0 eventually, i.e. z is from the class M 3 .Now assume that v [2] (t) < 0 for t ≥ t 2 .Therefore v [1] is decreasing and there exists t 3 ≥ t 2 such that there are two possibilities, either v [1] (t) > 0 or v [1] (t) < 0 for t ≥ t 3 .Assume that v [1] (t) < 0. Then there exists a constant N > 0 such that Integrating this inequality from t 3 to t we have Letting t → ∞ and using the fact that ∞ t 0 p(t) dt = ∞ we get v(t) → −∞, i.e. v(t) < 0 eventually, a contradiction.Hence v [1] (t) > 0 and z is from the class M 1 .
for t ≥ T, where v is defined by (1.1).
Proof.The proof is similar to the proof of Lemma 2 in [8] and therefore it is omitted.
First, we prove a lemma which helps with characterizing solutions from the class M 1 .
Lemma 3.4.Assume that z is a solution of (E−) from the class M 1 .Then Without loss of generality we may assume that z is eventually positive, i.e. there exists T ≥ t 0 such that v(t) > 0, v [1] (t) > 0, v [2] (t) < 0 for t ≥ T.
Then we have v [1] (t) ≤ −lr(t) and by integrating from T to t we get Letting t → ∞ we get a contradiction.
The following lemmas summarize results concerning solutions from the class M 3 .
Lemma 3.5.Let z be an eventually positive solution of (E−) from the class M 3 , then Proof.Set y = v [1] and x = v [2] .Then z is a solution of (E−) if and only if v, y, x is a solution of the system v (t) = p(t)y(t) x (t) = q(t) f z δ(t) . (3.3) Let z ∈ M 3 .Then the vector v, y, x is a solution of system (3.3)such that sgn z(t) = sgn v(t) = sgn y(t) = sgn x(t) for large t.
We show that lim There exists T ≥ t 0 such that v(t) > 0, y(t) > 0, x(t) > 0 for t ≥ T. As y is eventually increasing, there exists T 1 ≥ T and K > 0 such that Integrating in [T 1 , t] we get Using the assumption and integrating in [T 2 , t] we obtain Using the assumption ∞ t 0 r(t) dt = ∞ we get lim t→∞ y(t) = ∞, which completes the proof of the first part of the assertion.Now integrating the first equation of (3.3) and using (3.4) we obtain From here and (3.1) Using the third equation of (3.3) and (3.5), there exists T 2 ≥ T 1 such that Using the fact that f (uv) ≥ f (u) f (v) and integrating the last inequality from T 2 to t we have Proof.Without loss of generality, we may assume that there exists t 1 ≥ t 0 such that z(t) > 0, v [1] (t) > 0 and v [2] (t) > 0 for t ≥ t 1 ."⇒": Observe that since t t 0 r(s) ds is increasing function, there exists t which proves the first part of the assertion."⇐": Suppose there exists a positive constant L such that v [2] (∞) < L, i.e. as v [2] is increasing Integrating this inequality twice from t 1 to t we obtain v(t) < v(t 1 ) + v [1] (t Since v(t) ≥ z(t), we get a contradiction with (3.6).
For equation (L−) we have the following characterization of solutions from the class M 3 .

Comparison theorems for the superlinear case
We state separately comparison theorems for neutral linear equation (L−) with the advanced argument δ(t) ≥ t and with delay argument δ(t) ≤ t.Similarly, we state comparison theorems for neutral nonlinear equation (E−).In this section we assume that lim sup In particular, if f (u) = u λ sgn u, then (4.1) is satisfied for λ ≥ 1.
Theorem 4.1.Assume that δ(t) ≥ t.If the linear ordinary differential equation has property B, then equation (L−) has also property B.
We extend the previous theorem for nonlinear equation (E−).From here and (4.3) Now we proceed similarly to the proof of the previous theorem.Consider the linear equation Taking K ≥ c 1 (1 − a 0 ), we get that equation (4.7) is a majorant of (4.6) for this choice.Now assume by contradiction, that (E−) has a solution z ∈ M 1 .Therefore, equation (4.7) has a solution y = v from the class M 1 .Using Proposition 2.5 a) we get that there exists a solution z ∈ M 1 of (4.6), a contradiction.Now assume by contradiction that equation (E−) has a solution z from the class M 3 such that lim t→∞ v [2] (t) < ∞.Then equation (4.7) has a solution y = v from the class M 3 such that lim t→∞ y [2] (t) < ∞.Using Proposition 2.5 b) we get a contradiction.Now we prove similar theorems for equations with delay, the main difference is in the fact that now we compare equations (L−) and (E−) with delay differential equations.Assume by contradiction that z ∈ M 1 and consider the delay equation This equation has a solution y = v satisfying y(t) > 0, y [1] (t) > 0, y [2] (t) < 0 for large t, i.e. y is a solution of (4.10) from the class M 1 .Since (4.9) holds, equation (4.10) is a majorant of (4.8) and by Proposition 2.5 a), M 1 = ∅ for (4.8), a contradiction.Now assume that z ∈ M 3 and assume by contradiction that lim t→∞ v [2] (t) < ∞.Consider the equation This equation has a solution y = v satisfying y(t) > 0, y [1] (t) > 0, y [2] (t) > 0 for large t, i.e. y is a solution of (4.11) from the class M 3 and moreover lim t→∞ y [2] (t) < ∞.Since (4.9) holds, equation (4.11) is a majorant of (4.8) and by Proposition 2.5 b) there exists a solution y ∈ M 3 of (4.8) such that z [2] (t) < ∞, a contradiction.From here and (4.9) we obtain Now we proceed similarly to the proof of the previous theorem.Consider the linear delay equation

.13)
Taking K ≥ c 1 (1 − a 0 ), we get that equation (4.13) is a majorant of (4.12) for this choice.Now assume by contradiction, that (E−) has a solution z ∈ M 1 .Therefore, equation (4.13) has a solution y = v from the class M 1 .Using Proposition 2.5 a) we get that there exists a solution z ∈ M 1 of (4.12), a contradiction.Now assume by contradiction that equation (E−) has a solution z from the class M 3 such that lim t→∞ v [2] (t) < ∞.Then equation (4.13) has a solution y = v from the class M 3 such that lim t→∞ y [2] (t) < ∞.Using Proposition 2.5 b) we get a contradiction.

Remark 4.5.
There exists various criteria for equation (2.13) to have property A. Using Theorem D and our comparison theorems for neutral equations we can derive new oscillation criteria for equations (E−) and (L−), moreover we can derive new criteria even in the case where g(t) = t.

u T 1 rLemma 3 . 6 .
(w) dw du ds and taking into account that (3.2) holds, we get lim t→∞ x(t) = ∞, which completes the proof.Let z be a nonoscillatory solution of (E−) from the class M 3 .Then 1) is said to have weak property B if any proper solution x of (2.1) is either oscillatory or belongs to M 3 .
2≥ t 1 such that for T ≥ t 2 we have s t 0 r(u) du ds Let (4.6)have property B for every K > 0 and let v(t) be defined by(1.1).Without loss of generality, we may assume that there exists t 1 ≥ t 0 such that z is a solution of (E−) and z δ(t) > 0 for t ≥ t 1 .Observe that if 0 < z(t) < ∞, then f being continuous, we can assume that there exists

Theorem 4.3. Assume
that δ(t) ≤ t.If the linear delay equation Theorem 4.4.Assume that (4.1) holds and δ(t) ≤ t.If for every K > 0 the linear delay equation Let (4.12) have property B for every K > 0 and let v(t) be defined by (1.1).Without loss of generality we may assume that there exists t 1 ≥ t 0 such that z is a solution of (E−) such that z δ(t) > 0 for t ≥ t 1 .. We proceed similarly to proof of Theorem 4.2.If 0 < z(t) < ∞, then f being continuous, we can assume that there exists c > 0 such that [18,lustrate this see Examples 6.1 and 6.2 below.Using the result of[8, Example 6.1] we get that equationy (t) + (1 − a 0 ) ∈ [0, 1).We show that this equation has property B for every k > 0. Consider the corresponding linear equationty (t) − k (1 − a 0 )t 3 y(t) = 0. − a 0 )t 3 x = 0 (6.4)we obtain that (6.4) has property A for every k > 0 and using Theorem D we get that (6.3) has property A for every k > 0. Now Theorem 4.1 gives the assertion.thususing[18,Theorem3.3]we have that (6.5) has property B for every K > 0. Now Theorem 4.4 yields the assertion.