OSCILLATORY AND ASYMPTOTIC PROPERTIES OF NONLINEAR FIRST ORDER DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT OF GENERALIZED TYPE

In the paper, we consider differential equations of neutral type with piecewise constant argument of generalized type, i.e., the argument is a general step function. Sufficient conditions for oscillation of all solutions of this type equations are obtained. The asymptotic behavior of the nonoscillating solutions is studied also. Appropriate examples are given to illustrate the recieved results.


Introduction
In his famous survey, A. D. Myshkis [17]among the functional differential equations first give several reasonable occasions to study differential equations with piecewise argument (in short DEPCA).Generally speaking they are equations with hybrid properties, which combine partial the properties of continuous systems, also part of discrete equations properties.Later with the works of J.
Akhmet in [3] first introduce the function "gama" in his consideration of only delayed situation.This function is widely used in several works of Chiu and Pinto [6], [7] and [8].In [19] and [20] Pinto has cleared the importance of the advanced and delayed intervals.Some specific results are given in [23], [10] and [7].Note that to our knowledge almost all investigations are devoted for equations of retarded type.The aim of this work is to extend part of the results investigated by not many autors [15] and [18] for the neutral case.
In the present paper the equation is investigated, where p ∈ R \ 0, τ > 0, q < 0 and γ(t) ].The full definition of γ(t) will be given below.The alternative case q > 0 is considered in Oscillatory and Asymptotic properties of Nonlinear First Order Differential Equations with Piecewise Constant Argument of Generalized Type part 2 to be appeared.

Preliminaries
In the present paper the equation is investigated, where p ∈ R \ {0} and q < 0(the case q > 0 is considered in the second part of this paper, to be appeared).Introduce two real sequences {t n } , {γ n } n = 0, 1, 2... such that t 0 = −τ, t n+1 = t n + 1 and n < γ n ≤ t n+1 , γ n+1 = γ n + 1 where γ(t) is a step function given by The function γ(t) can be rewrited in the form 2 ), n = 1, 2, 3....Note that the equation ( 1) is of advanced type in ).This decomposition will be present in all our results.
As usual with C we denote the Banach space of initial functions C = | and introduce the initial condition: Definition 1.The function y(t)is said to be a solution of initial value problem (IV P ) (1), (2) if:  (c) y(t) satisfies equation (1) in each interval [n − τ, n − τ + 1) n = 0, 1, 2....It is not difficult to chek that for arbitrary initial function the IV P (1), (2) has a unique solution (see Chapter 3 [16], Chapters 2 and 3 [14]).Definition 2. The solution of equation ( 1) is said to oscillate if the set of its zeros is unbounded above, otherwise it is called nonoscillating.Definition 3. The function f is said to finally enjoy the property K if there exists t 0 such that for t ≥ t 0 , f enjoys the property K.
Lemma 1.Let conditions (A) hold and let y(t) be a nonoscillating solution of equation (1).Then the following assertions are valid: Integrate the above equation from n − τ to t and obtain Since is a continious function.By passing to the limit as t → n we obtain Since there exists lim Therefore we have that lim In the same way assertion (b) is proved in the case when the nonoscillating solution y(t) is negative.
2 and obtain From equation ( 6) and the continuity of y(t), passing to the limit as t → n + 1 2 the equality follows.
We shall consider separatelly the cases p = 1 and p = 1.Let p = 1.Then from equation (7) it follows that if y(t) is solution of equation ( 1) the sequence {c n } ∞ n=1 satisfies the difference equations: . As in the above, we obtain the equality On the other hand integrating from n + 1 to t we obtain For t → n + 1 2 , we obtain Substituting y(n + 1 2 ) and y(n + 3 2 ) from equation ( 7) and equation ( 9) it follows Hence for p = 1 if equation ( 1) has a solution y(t), then the sequence {c n } ∞ n=1 satisfies the difference equation equation (11).

Main Results
We shall consider the asympthotic behavior of the nonoscillating solution of equation (1).We shall first consider the case when q < 0. The case q > 0 is considered in the second part of this paper which will be published later.

Choose a sequance {n
The contradiction obtained shows that lim sup t→∞ y(t) = 0 and lim t→∞ y(t) = 0.
This fact implies imediately that lim t→∞ z(t) = 0. Analogously it is proved the case when y(t) < 0 finally.
Remark 1. Theorem 2 naturally sets the question:What can we claim about the unbounded solutions of equation ( 1) when p ∈ (−1, 0) and whether it is true that, for each unbounded solution y(t) of equation ( 1), lim The answer to this question is negative.We shall show this by means of the following example: with γ 0 = 1 2 and with initial function where λ is the root of the equation F (λ) = 4λ 3 − 22λ 2 + 31 2 λ + 1 = 0 which is in the interval (0, 1).There exists such a root because F (0) = 1  2 > 0 and where We shall prove that y(t) is a solution of equation ( 12) with initial function ϕ(t) and that y(t) > 0 in [− 1 2 , ∞).From the definition of the function y(t) it follows that We shall prove that . From the definition of z(t) it follows that We shall prove by induction that y(n )) or Taking into account λ, we immediatelly verify that Then y(n+1) = λ n+1 and the inductive step is complete.From the equality proved, y(n) = λ n , and from definition of z(t) it immediately follows that y(t) satisfies equation (12) , k = 0, 1, 2, ..., then in order to prove that y(t) > 0 it suffices to prove that y(− 1 2 + k 4 ) > 0 for k = 0, 1, 2, ...
depends on k and is in one of the forms: y(n), y(n + 1 2 ) and y(− 1 4 + n 2 ).Obviously, y(n) > 0, n = 0, 1, 2, ... From equation ( 7) and (for f (t) = t) it follows that y(n 4 is considered.Thus we have established that the function y(t) defined above is a strictly positive(i.e., nonoscillating)solution of equation (12).Since y(n) = λ n and λ < 1, then lim inf t→∞ y(t) = 0. We shall prove that lim sup Finaly, the solution y(t) of equation ( 12) constructed above is nonoscillating, lim inf t→∞ y(t) = 0 and lim sup t→∞ Example 1 shows that the behavior of the solutions of equation ( 1) may be considerably more complex than that of the equations without a piecewise constant argument.For instance, for the neutral equation analogous to equation (12), It is known that its nonoscillating solutions either tend to 0 or diverge to ±∞.Remark 2. In Theorem 3 there is no claim concerning the bounded solutions of equation ( 1).The reason is the following:From Lemma 1 it follows that if y(t) is a bounded, nonoscillating solution of equation ( 1), then lim inf t→∞ y(t) = 0.It turns out, however, that equation (1) has solutions y(t) for which lim sup t→∞ y(t) > 0, i.e., for which the limit lim t→∞ y(t) does not exist.
In this case (as in Example 1 the behaviour of the nonoscillating solutions of equation ( 1) is more complex than that of the corresponding equations without a piecewise constant argument.We shall illustrate this fact by the following example.