Necessary and Sufficient Conditions for Nonoscillatory Solutions of Impulsive Delay Differential Equations ∗†

Monotonicity of solutions is an important property in the investigation of oscillatory behaviors of differential equations. A number of papers provide some existence criteria for eventually positive increasing solutions. However, relatively little attention is paid to eventually positive solutions that are also eventually decreasing solutions. For this reason, we establish several new and sharp oscillatory criteria for impulsive functional differential equations from this viewpoint.


Introduction
Let Υ = {t 0 , t 1 , ...} where 0 = t 0 < t 1 < t 2 < • • • and let R and N be the sets of real numbers and positive integers respectively.In this paper, we intend to establish necessary and sufficient conditions of existence of nonoscillatory solutions of impulsive differential equation (r(t)x ′ (t)) ′ + p(t)f (x(g(t))) = 0, t ∈ [0, ∞)\Υ, x under some or all of the following conditions (A1) lim k→∞ t k = +∞; (A2) p is a function on [0, ∞), and r is a positive and differentiable function on [0, ∞); (A3) g is a continuous function on [0, ∞) such that g(t) ≤ t for t ≥ 0 and lim t→∞ g(t) = ∞; (A4) f is a continuous function on R with uf (u) > 0 for u = 0 and inf |u|≥T {|f (u)|} > 0 for any T > 0; (A5) a k > 0 and b k > 0 for k ∈ N; (A6) there exists m > 0 such that A(0, t) ≥ m for t ≥ 0; and (A7) there exists M > 0 such that A(0, t) ≤ M for t ≥ 0, where "Monotonicity" of solutions is an important property for investigating oscillatory behaviors of functional differential equations.There are many papers (e.g.[2, 5, 7, 10-12, 13, 15]) which provide sufficient conditions to guarantee the solutions are increasing.For instance, if where by Lemma 2 in [12], then an eventually positive solution x of system (1)-(3) will satisfy x ′ (t) ≥ 0 eventually.However, the case where a solution x satisfies the condition x(t)x ′ (t) < 0 eventually has rarely been touched upon.To fill this gap, we will establish new and sharp oscillatory criteria from this viewpoint.Our technique is based on comparing our systems with their linearized systems (cf.[4][5][6][7][8][9]).However, the important point is that we are able to establish necessary and sufficient conditions.
In the subsequent discussions, we let g η = min t≥η g(t) for any η ≥ 0.
Note that if (A3) is assumed, then g η exists.
Definition 1.1 Let Λ be an interval in [0, ∞) and σ = inf Λ.For any φ ∈ P C([g σ , σ], R), a function x defined on [g σ , σ] ∪ Λ is said to be a solution of system (1)-(3) on Λ satisfying the initial value condition A partially ordered set is called a complete lattice if all subsets admit a supremum and an infimum.A complete lattice is recalled here because we will employ the well known Knaster-Tarski fixed point theorem.
Theorem 1.1 (Knaster-Tarski fixed point theorem) Let X be a set and f a function on X such that f (X) ⊆ X. Assume that (X, ≤) is a complete lattice and f (x We define the functions

Main results
We first provide a criterion to illustrate that the nonoscillatory solution x with x(t)x ′ (t) < 0 eventually indeed exists.
Proof.For the sake of convenience, we may assume that τ = 0. We first assume that the solution x is eventually positive, say that x(t) > 0 for t ≥ g 0 .Let c = inf t≥g0 {x(t)} and m f = inf u≥c {f (u)}.We note that r(t for t ≥ t kn ≥ • • • ≥ t k1 ≥ s > 0 where t k1 , . . ., t kn ∈ Υ. There are now two cases.First assume that there exists T > 0 such that x ′ (T ) < 0. In view of (6), Since r(t) > 0 for t ≥ 0, we may further see that x ′ (t) < 0 for t ≥ T .Next, if x ′ (t) ≥ 0 for all t ≥ g 0 , then x(t) is increasing on each interval (t k−1 , t k ] for k ∈ N. Similarly, we can verify that x(t)/A(0, t) is continuous and increasing for t ≥ 0. It follows from (A6) that So c ≥ min mx(0), min t∈[g0,0] x(t) > 0 and x(g(t)) ≥ c for t ≥ 0. In view of (A4), we see that f (x(g(t))) ≥ m f > 0 for t ≥ 0. We now divide (1) by B(0, t), and then integrate from 0 to t.By continuity of r(t)x ′ (t)/B(0, t), we have for t ≥ 0. By ( 5) and (A2), we may see that x ′ (t) < 0 eventually.This is a contradiction.Therefore, in both cases, x ′ (t) < 0 eventually.Second, we assume that the solution x is eventually negative.Let y(t) = −x(t) for sufficiently large t.Then y is an eventually positive solution of (r(t)y ′ (t)) where F (u) = −f (−u) for u ∈ R. We may observe that F is a continuous function on R with uF (u) > 0 for u = 0, and inf |u|≥T {|F (u)|} > 0 for any T > 0. In view of the above discussions, y ′ (t) < 0 eventually, which implies x ′ (t) > 0 eventually.The proof is complete.
Proof.Assume that the system (1)-(3) has a nonoscillatory solution x.By Lemma 2.1, we may see that x ′ (t) < 0 eventually.By (4) and Lemma 2 in [12], we may further see that x ′ (t) ≥ 0 eventually.This is a contradiction.The proof is complete.
We can give two examples to illustrate the condition (23).In the first example, the function f is concave on [0, ∞).Indeed, we note that for 0 < u 1 < u 2 .So f satisfies the condition (23).In particular, f (u) = u.In the second example, f is concave on [0, d] and decreasing on [d, ∞).Similarly, we may verify that f satisfies the condition (23).
Proof.In view of (A6) and (A7), we may see that Clearly, δ > 0. Assume that (35) holds.There exists T ∈ Υ such that T > τ and Clearly, X is a complete lattice and X is nonempty because of the fact that d 1 ∈ X.We define an operator S in X by ))dvds for t ≥ T and y ∈ X, where Given y 1 , y 2 ∈ X with y 1 ≤ y 2 .Then for t ≥ T .By the monotonicity of f , we may see that It follows that S(y 1 )(t) ≤ S(y 2 )(t) for t ≥ T .In view of (36), for y ∈ X.So S(X) ⊆ X and S is increasing in X.By the Knaster-Tarski fixed point Theorem, there exists x ∈ X such that S(x) = x.Clearly, x(t) ≥ d 1 > 0 for t ≥ T. Let T 1 > T such that g T1 > T .We note that x(g(t)) = w(x)(g(t)), x is an eventually positive solution of system ( 1)-( 3) with x ′ (t) ≤ 0 eventually.If t ≥ d : p(t) = 0 has measure zero for any d ≥ 0, then by (37), x ′ (t) < 0 eventually.
To see the converse, we first assume that system (1)-( 3) has an eventually positive solution x with x(t) ≥ d 1 and x ′ (t) ≤ 0 eventually.Without loss of generality, we assume that x(t) ≥ d 1 and x ′ (t) ≤ 0 for t ≥ g 0 .We divide (1) by B(0, t), and then integrate from 0 to t.We have We further divide (38) by A(0, t), and then integrate from 0 to t.We have In view of (38), x(t) is decreasing on each interval (t k−1 , t k ] for k ∈ N. By continuity of x(t)/A(0, t), we note that x(t) ≤ x(0)A(0, t) ≤ M x(0) for t ≥ 0. By (A4) and continuity of f , there exists δ > 0 such that f (u) ≥ δ for M x(0) ≥ u ≥ d 1 .By (39), it follows that for t ≥ 0. Since x(t) > 0 for t ≥ 0, we may further see that (35) holds.Second, we assume that system (1)-( 3) has an eventually negative solution x(t) with x(t) ≤ −d 1 and x ′ (t) ≥ 0 eventually.Then the system (8)-( 10) has an eventually positive solution y(t) with y(t) ≥ d 1 and y ′ (t) ≤ 0 eventually.By the above discussion, we may verify that (35) holds.The proof is complete.
Proof.Assume that the system (47)-( 49) has a nonoscillatory solution x.We may assume that x is eventually positive.The case that x is eventually negative is similar so we ignore it.We note that By Lemmas 2.1 and 2.2, we may assume that x(t) > 0 and x ′ (t) < 0 for t ≥ g 0 .Furthermore, lim t→∞ x(t) = 0.In view of f ′′ (0) = 0, we can see that there exists θ > δ > 0 such that f ′′ (u) ≥ 0 for δ ≥ u ≥ 0, or f ′′ (u) ≤ 0 for δ ≥ u ≥ 0. In the former case, f is convex on [0, δ].So f (u) ≥ f ′ (0)u for δ ≥ u ≥ 0. Since 0 ≤ x(t) ≤ δ eventually, we may see that x(t) is an eventually positive solution of By Theorem 2.2, we may further see that the equation has an eventually positive solution, which implies that its characteristic equation has a real root.But this is impossible because λ 2 + pf ′ (0)e −τ λ > 0 for λ ∈ R. In the later case, f is concave on [0, δ].By Corollary 2.4, we can see that has an eventually positive solution.By Theorem 2.2, we can further see that the equation ( 50) has an eventually positive solution.By the above discussion, this is also impossible.The proof is complete.

Examples
We illustrate our results by two examples.

Figure 3 We
Figure 3We have the following conclusions:(i) Assume that a k = b k for k ∈ N, and that r(t) = exp(−t) and p(t) = t for t ≥ 0. It is easy to check that (4) and (5) hold.By Corollary 2.1, the system (51)-(53) is oscillatory.(ii)Assume that Υ = N, b k = 1/e, r(t) = e −0.5t and p(t) = e 2t for t ≥ 0 and k ∈ N. We note that e s−t−1 ≤ B(s, t) ≤ e s−t+1 for t ≥ 1 and t ≥ s ≥ 0.(54)