ASYMPTOTIC AND OSCILLATORY BEHAVIOR OF SECOND ORDER NEUTRAL QUANTUM EQUATIONS WITH MAXIMA

In this study, the behavior of solutions to certain second order quantum (q-difference) equations with maxima are considered. In particular, the asymptotic behavior of non-oscillatory solutions is described, and sufficient conditions for os- cillation of all solutions are obtained.


introduction
Quantum calculus has been utilized since at least the time of Pierre de Fermat [8,Chapter B.5] to augment mathematical understanding gained from the more traditional continuous calculus and other branches of the discipline; see Kac and Cheung [4], for example.In this study we will analyze a second order neutral quantum (qdifference) equation D 2  q x(t) + p(t)x(q −k t) + r(t) max s∈{0,••• , } x(q −s t) = 0, (1.1) where the real scalar q > 1 and the q-derivatives are given, respectively, by the difference quotient D q y(t) = y(qt) − y(t) qt − t , and D 2 q y(t) = D q (D q y(t)) .
Equation (1.1) is a quantum version of studied by Luo and Bainov [5]; there the usual forward difference operator ∆y n := y n+1 − y n was used.For more results on differential and difference equations related to (1.1) and (1.2), please see the work by Bainov, Petrov, and Proytcheva [1, 2, 3], Luo and Bainov [5], Luo and Petrov [6], and Petrov [7].The particular appeal of (1.1) is that it is still a discrete problem, but with non-constant step size between domain points.

preliminary results
For q > 1, define the quantum half line by Let k, be non-negative integers, r : (0, ∞) q → [0, ∞), p : (0, ∞) q → R, and consider the second order neutral quantum (q-difference) equation where we assume Definition 2.1.A function f : (0, ∞) q → R eventually enjoys property P if and only if there exists t * ∈ (0, ∞) q such that for t ∈ [t * , ∞) q the function f enjoys property P.
A solution x of (2.1) is non-oscillatory if and only if x(t) < 0 or x(t) > 0 eventually; otherwise x is oscillatory.
Proof.We will prove (a); the proof of (b) is similar and thus omitted.Since x(t) > 0 eventually and r(t) ≥ 0, it follows from (2.4) that D 2 q z(t) ≤ 0 eventually and D q z is an eventually nonincreasing function.Then either there exists an (2.7) and (2.8) hold.So, let L := lim t→∞ D q z(t) ∈ R; then one of the following three cases holds: holds.Thus lim t→∞ x(t) = ∞.From (2.2) and (2.5) we see that lim t→∞ D q z(t) = −∞, a contradiction.
(ii) If L > 0, we arrive at a contradiction analogous to (i).
(iii) Assume L = 0. Since D q z is an eventually decreasing function, D q z(t) > 0 eventually and z is an eventually increasing function.Thus either lim t→∞ z(t) = M ∈ R, or lim t→∞ z(t) = ∞.If M > 0, then x(t) > z(t) > M/2 for large t ∈ (0, ∞) q , and from assumption (2.2) and equation (2.5) it follows that lim t→∞ D q z(t) = −∞, a contradiction.Using a similar argument we reach a contradiction if lim t→∞ z(t) = ∞.Therefore we assume there exists a finite limit, lim t→∞ z(t Thus for large t we have M/p < x(q −k t), and again from assumption (2.2) and equation (2.5) we have that lim t→∞ D q z(t) = −∞ = L, a contradiction of L = 0. Consequently lim t→∞ z(t) = 0, and since z is an eventually increasing function, z(t) < 0 eventually and (2.9) and ( 2 (2.14) Then the following assertions are valid.
Proof.We will prove (a); the proof of (b) is similar and thus omitted.From (2.4) it follows that D 2 q z(t) ≥ 0 eventually, and D q z is an eventually nondecreasing function.Assumption (2.2) implies that r(t) = 0 eventually, and thus either D q z(t) > 0 eventually or D q z(t) < 0. Suppose that D q z(t) > 0. Since D q z is a nondecreasing function, there exists a constant c > 0 such that D q z(t) ≥ c eventually.Then lim t→∞ D q z(t) = ∞.From (2.3) we obtain the inequality and therefore lim t→∞ x(t) = −∞.On the other hand, from (2.3) again and from the inequality z(t) > 0 there follows the estimate x(t) > −p(t)x(q −k t) ≥ x(q −k t).
The inequalities x(t) < 0 and x(t) > x(q −k t) eventually imply that x is a bounded function, a contradiction of the condition lim t→∞ x(t) = −∞ proved above.Thus D q z(t) < 0, and z is an eventually decreasing function.Let L = lim t→∞ D q z(t).Then lim t→∞ z(t) = −∞.From the inequality x(t) < z(t) it follows that lim t→∞ x(t) = −∞, and then (2.5) implies the relation lim t→∞ D q z(t) = ∞.The contradiction obtained shows that L = 0, that is lim t→∞ D q z(t) = 0. Suppose that z(t) < 0 eventually.Since z is a decreasing function, there exists a constant c < 0 such that z(t) ≤ c eventually.The inequality z(t) > x(t) implies that x(t) ≤ c eventually.From (2.5) it follows that lim t→∞ D q z(t) = ∞.The contradiction obtained shows that z(t) > 0, and since z is an eventually decreasing function, then there exists a finite limit M = lim t→∞ z(t).
that is x(q −k t) < −M.From (2.5) we obtain that lim t→∞ D q z(t) = ∞, a contradiction.Hence, M = 0, in other words lim t→∞ z(t) = 0. Since z is a decreasing function, z(t) > 0 eventually, and we have shown that if x is an eventually negative solution of (2.
Lemma 2.4 is readily verified.

main results
In this section we present the main results on the oscillatory and asymptotic behavior of solutions to (2.1).Theorem 3.1.Assume r satisfies (2.2), and If x is a nonoscillatory solution of (2.1), then lim t→∞ x(t) = 0.
Proof.Let x(t) > 0 eventually.Then Lemma 2.3 implies that z(t) < 0 eventually and lim t→∞ z(t) = 0. From (3.1) we have that so that x is bounded.Let c = lim sup t→∞ x(t), and suppose that c > 0. Choose an increasing quantum sequence of points {t i } from (0, ∞) q such that lim i→∞ t i = ∞ and lim i→∞ x(t i ) = c.Set d = lim sup i→∞ x(q −k t i ), and note that d ≤ c.Choose a subsequence of points {t j } ⊂ {t i } such that d = lim j→∞ x(q −k t j ), and pass to the limit in the inequality z(t j ) ≥ x(t j ) + px(q −k t j ) as j → ∞.We then see that a contradiction.Thus lim sup t→∞ x(t) = 0 and lim t→∞ x(t) = 0.The case where x(t) < 0 eventually is similar and is omitted.
Proof.Let x(t) > 0 eventually; the case where x(t) < 0 eventually is similar and is omitted.Since x is bounded, it follows from (2.3) that z is also bounded.Since (2.6) holds, Lemma 2.2 implies that z(t) < 0 eventually and lim t→∞ z(t) = 0.As in the proof of Theorem 3.1, let c = lim sup t→∞ x(t), and suppose that c > 0. Choose an increasing quantum sequence of points {t i } from (0, ∞) q such that lim i→∞ t i = ∞ and lim i→∞ x(t i ) = c.Set d = lim sup i→∞ x(q −k t i ), and note that d ≤ c.Choose a EJQTDE, 2009 No. 16, p. 5 subsequence of points {t j } ⊂ {t i } such that d = lim j→∞ x(q −k t j ), and pass to the limit in the inequality z(t j ) ≤ x(t j ) + P x(q −k t j ) as j → ∞.We then see a contradiction, so that lim sup t→∞ x(t) = 0 and lim t→∞ x(t) = 0.
Proof.Lemma 2.2 implies that either lim t→∞ z(t) = −∞ or lim t→∞ z(t) = 0. First, we consider lim t→∞ z(t) = −∞.Then z(t) > p(t)x(q −k t) ≥ px(q −k t), so that lim t→∞ x(t) = ∞.Next, we consider lim t→∞ z(t) = 0.In this case Lemma 2.2 implies that z is an eventually negative increasing function.If the solution x does not vanish at infinity, then there exist a constant c > 0 and an increasing quantum sequence of points {t i } from (t 0 , ∞) q such that t i+1 > q t i and x(t i ) > c/2 for each i ∈ N.Then, we have max s∈{0,••• , } x q −s t > c/2, t ∈ [t i , t i+ ] q .