Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients

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Datum:	2012
Hauptverfasser:	Tripathy, A.K., Panigrahi, S., Basu, R.
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Sprache:	English
Veröffentlicht:	Інститут математики НАН України 2012
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spelling	irk-123456789-1760212021-02-04T01:31:06Z Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients Tripathy, A.K. Panigrahi, S. Basu, R. 2012 Article Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients / A.K. Tripathy, S. Panigrahi, R. Basu // Нелінійні коливання. — 2012. — Т. 15, № 4. — С. 539-555. — Бібліогр.: 10 назв. — англ. 1562-3076 http://dspace.nbuv.gov.ua/handle/123456789/176021 517.9 en Нелінійні коливання Інститут математики НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
format	Article
author	Tripathy, A.K. Panigrahi, S. Basu, R.
spellingShingle	Tripathy, A.K. Panigrahi, S. Basu, R. Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients Нелінійні коливання
author_facet	Tripathy, A.K. Panigrahi, S. Basu, R.
author_sort	Tripathy, A.K.
title	Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients
title_short	Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients
title_full	Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients
title_fullStr	Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients
title_full_unstemmed	Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients
title_sort	oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients
publisher	Інститут математики НАН України
publishDate	2012
url	http://dspace.nbuv.gov.ua/handle/123456789/176021
citation_txt	Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients / A.K. Tripathy, S. Panigrahi, R. Basu // Нелінійні коливання. — 2012. — Т. 15, № 4. — С. 539-555. — Бібліогр.: 10 назв. — англ.
series	Нелінійні коливання
work_keys_str_mv	AT tripathyak oscillationresultsforfourthordernonlinearneutraldifferentialequationswithpositiveandnegativecoefficients AT panigrahis oscillationresultsforfourthordernonlinearneutraldifferentialequationswithpositiveandnegativecoefficients AT basur oscillationresultsforfourthordernonlinearneutraldifferentialequationswithpositiveandnegativecoefficients
first_indexed	2025-07-15T13:38:14Z
last_indexed	2025-07-15T13:38:14Z
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fulltext	UDC 517.9 OSCILLATION RESULTS FOR FOURTH ORDER NONLINEAR NEUTRAL DIFFERENTIAL EQUATIONS WITH POSITIVE AND NEGATIVE COEFFICIENTS ПРО ОСЦИЛЯЦIЮ ДЛЯ НЕЛIНIЙНОГО ДИФЕРЕНЦIАЛЬНОГО РIВНЯННЯ ЧЕТВЕРТОГО ПОРЯДКУ НЕЙТРАЛЬНОГО ТИПУ З ДОДАТНИМИ ТА ВIД’ЄМНИМИ КОЕФIЦIЄНТАМИ A. K. Tripathy Sambalpur Univ. Sambalpur – 768 019, India e-mail: arun_tripathy70@rediffmail.com S. Panigrahi, R. Basu* Univ. Hyderabad Hyderabad – 500 046, India e-mail: spsm@uohyd.ernet.in rakheebasu1983@gmail.com Unbounded oscillation and asymptotic behaviour of a class of nonlinear fourth order neutral differential equations with positive and negative coefficients of the form (r(t)(y(t) + p(t)y(t− τ))′′)′′ + q(t)G(y(t− α))− h(t)H(y(t− β)) = 0 and (r(t)(y(t) + p(t)y(t− τ))′′)′′ + q(t)G(y(t− α))− h(t)H(y(t− β)) = f(t) (E) are investigated under the assumption ∞∫ 0 t r(t) dt < ∞ for various ranges of p(t). Sufficient conditions are obtained for the existence of positive bounded soluti- ons of (E). Дослiджено необмежену осциляцiю та асимптотичну поведiнку класу нелiнiйних диференцiаль- них рiвнянь четвертого порядку нейтрального типу у виглядi (r(t)(y(t) + p(t)y(t− τ))′′)′′ + q(t)G(y(t− α))− h(t)H(y(t− β)) = 0 (1) та (r(t)(y(t) + p(t)y(t− τ))′′)′′ + q(t)G(y(t− α))− h(t)H(y(t− β)) = f(t) (E) ∗ Research supported by Department of Science and Technology (DST), New Delhi, India through Letter No SR/S4/MS: 541/08 dated September 30, 2008. ∗∗ Research supported by CSIR-New Delhi, India through the Letter No 09/414 (0876)/2009-EMR-I dated October 20, 2009. c© A. K. Tripathy, S. Panigrahi, R. Basu, 2012 ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 539 540 A. K. TRIPATHY, S. PANIGRAHI, R. BASU з додатними та вiд’ємними коефiцiєнтами за умови ∞∫ 0 t r(t) dt < ∞ для рiзних областей значень p(t). Отримано достатнi умови iснування додатних обмежених розв’язкiв рiвняння (E). 1. Introduction. In the last few years, there has been an increasing interest in the study of osci- llatory behaviour of solutions of neutral delay differential equations with positive and negative coefficients of first and second order, see, for example, [1, 4 – 8, 10]. However, very little work [10] is available on the study of oscillatory and asymptotic behaviour of solutions of fourth order equations which is due to the technical difficulties arising in its analysis. In this paper, we consider a class of nonlinear fourth order neutral delay differential equati- ons of the form (r(t)(y(t) + p(t)y(t− τ))′′)′′ + q(t)G(y(t− α))− h(t)H(y(t− β)) = 0 (1.1) and (r(t)(y(t) + p(t)y(t− τ))′′)′′ + q(t)G(y(t− α))− h(t)H(y(t− β)) = f(t), (1.2) where r, q and h are continuous and positive on [0,∞), p ∈ C([0,∞),R), f ∈ C([0,∞),R), G, H ∈ C(R,R) with uG(u) > 0, vH(v) > 0, for u, v 6= 0, H is bounded, G is nondecreasing and τ, α, β > 0 are constants. The main objective of this work is to study the oscillatory and asymptotic behaviour of solutions of (1.1) and (1.2), under the assumption ∞∫ 0 t r(t) dt < ∞. (A1) If h(t) ≡ 0, then (1.1) and (1.2) reduce to (r(t)(y(t) + p(t)y(t− τ))′′)′′ + q(t)G(y(t− α)) = 0 (1.3) and (r(t)(y(t) + p(t)y(t− τ))′′)′′ + q(t)G(y(t− α)) = f(t) (1.4) respectively. In [9], Parhi and Tripathy have studied (1.3) and (1.4), under the assumption (A1). If h(t) 6≡ 0, then nothing is known about the behaviour of solutions of (1.1)/(1.2). Therefore, an attempt is made here to study (1.1) and (1.2) under the same assumption in addition to ∞∫ 0 s r(s) ∞∫ s th(t) dt ds < ∞. (A2) ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 OSCILLATION RESULTS FOR FOURTH ORDER NONLINEAR NEUTRAL . . . 541 Because (1.1)/(1.2) is more general than (1.3)/(1.4), it is worth studying. Not only the present work is more illustrative than [9], but also some of the results are generalized and improved. By a solution of (1.1)/(1.2) we understand a function y ∈ C([−ρ,∞),R) such that (y(t) + +p(t)y(t − τ)) is twice continuously differentiable, (r(t)(y(t) + p(t)y(t − τ))′′) is twice conti- nuously differentiable and (1.1)/(1.2) is satisfied for t ≥ 0, where ρ = max{τ, α, β} and sup{\|y(t)\| : t ≥ t0} > 0 for every t ≥ t0. A solution y(t) of (1.1)/(1.2) is said to be oscillatory if it has arbitrarily large zeros; otherwise, it is called nonoscillatory. 2. Some preparatory results. To study the nonlinear functional differential equations of the type (1.1)/(1.2), we need the following results for our use in the sequel. Lemma 2.1 [9]. Let (A1) hold. If u(t) is an eventually positive twice continuously differenti- able function such that r(t)u′′(t) is twice continuously differentiable and (r(t)u′′(t))′′ ≤ 0, 6≡ 0 for large t, where r ∈ C([0,∞), (0,∞)), then one of the following cases holds for large t : (a) u′(t) > 0, u′′(t) > 0 and (r(t)u′′(t))′ > 0, (b) u′(t) > 0, u′′(t) < 0 and (r(t)u′′(t))′ > 0, (c) u′(t) > 0, u′′(t) < 0 and (r(t)u′′(t))′ < 0, (d) u′(t) < 0, u′′(t) > 0 and (r(t)u′′(t))′ > 0. Lemma 2.2 [9]. Suppose that the conditions of Lemma 2.1 hold. Then (i) the following inequalities hold for large t in the case (c) of Lemma 2.1: u′(t) ≥ −(r(t)u′′(t))′R(t), u′(t) ≥ −r(t)u′′(t) ∞∫ t ds r(s) , u(t) ≥ ktu′(t) and u(t) ≥ −k(r(t)u′′(t))′tR(t), where k > 0 is a constant and R(t) = ∫ ∞ t s− t r(s) ds and (ii) u(t) ≥ r(t)u′′(t)R(t) for large t in case (d) of Lemma 2.1. Lemma 2.3 [9]. If the conditions of Lemma 2.1 hold, then there exist constants k1 > 0 and k2 > 0 such that k1R(t) ≤ u(t) ≤ k2t for large t. Lemma 2.4 [9]. Let (A1) hold. Suppose that z(t) be a real valued twice continuously differenti- able function on [0,∞), such that r(t)z′′(t) is twice continuously differentiable with (r(t)z′′(t))′′ ≤ ≤ 0, 6≡ 0 for large t. If z(t) > 0 eventually, then one of the following cases holds for large t : (a) z′(t) > 0, z′′(t) > 0 and (r(t)z′′(t))′ > 0, (b) z′(t) > 0, z′′(t) < 0 and (r(t)z′′(t))′ > 0, (c) z′(t) > 0, z′′(t) < 0 and (r(t)z′′(t))′ < 0, (d) z′(t) < 0, z′′(t) > 0 and (r(t)z′′(t))′ > 0. If z(t) < 0 for large t, then either one of the cases (b) – (d) holds or one of the following cases holds for large t : (e) z′(t) < 0, z′′(t) < 0 and (r(t)z′′(t))′ > 0, (f) z′(t) < 0, z′′(t) < 0 and (r(t)z′′(t))′ < 0. ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 542 A. K. TRIPATHY, S. PANIGRAHI, R. BASU Lemma 2.5 [3]. Let p, y, z ∈ C([0,∞),R) be such that z(t) = y(t) + p(t)y(t − τ), for t ≥ τ ≥ 0, y(t) > 0 for t ≥ t1 > τ, lim inft→∞ y(t) = 0 and limt→∞ z(t) = L exists. Let p(t) satisfy one of the following conditions: (i) 0 ≤ p(t) ≤ p1 < 1, (ii) 1 < p2 ≤ p(t) ≤ p3, (iii) p4 ≤ p(t) ≤ 0, where pi is a constant, 1 ≤ i ≤ 4. Then L = 0. 3. Oscillation results for (1.1). In this section, sufficient conditions are established for unboun- ded oscillation and asymptotic behaviour of solutions of (1.1) under the assumption (A1). For our purpose, we need the following assumptions: (A3) there exists λ > 0 such that G(u) +G(v) ≥ λG(u+ v) for u, v > 0, u, v ∈ R, (A4) G(uv) = G(u)G(v), u, v ∈ R, (A5) G(−u) = −G(u), H(−u) = −H(u), u ∈ R, (A6) ∫ ∞ τ Q(t) d(t) = ∞, Q(t) = min{q(t), q(t− τ)}, t ≥ τ, (A7) ∫ ∞ t0 b(t)Q(t)G(R(t − α)) dt = ∞, where b(t) = min{Rγ(t), Rγ(t − τ)}, γ > 1, t0 ≥ ≥ ρ > 0, (A8) ∫ ∞ t0 Rγ(t)G(R(t− α))q(t) dt = ∞, γ > 1, t0 ≥ ρ > 0. Remark 3.1. Since R(t) < ∫ ∞ t s r(s) ds, we have that R(t) → 0 as t → ∞ in view of (A1). Remark 3.2. (A4) implies that G(−u) = −G(u). Indeed, G(1)G(1) = G(1) and G(1) > 0 imply that G(1) = 1. Further, G(−1)G(−1) = G(1) = 1 implies that (G(−1))2 = 1. Since G(−1) < 0 it follows that G(−1) = −1. Hence G(−u) = G(−1)G(−u) = −G(u). On the other hand, G(uv) = G(u)G(v) for u > 0, v > 0, and G(−u) = −G(u) imply that G(xy) = = G(x)G(y) for every x, y ∈ R. Remark 3.3. The prototype of G satisfying (A3), (A4) and (A5) is G(u) = (a+ b\|u\|γ)\|u\|µSgnu, where a ≥ 0, b > 0, γ ≥ 0 and µ ≥ 0 such that a+ b = 1. Theorem 3.1. Let 0 ≤ p(t) ≤ a < 1 or 1 < p(t) ≤ a < ∞. Suppose that (A1) – (A7) hold. Then every solution of (1.1) either oscillates or tends to zero as t → ∞. Proof. Due to Remark 3.1, b(t) → 0 as t → ∞. Hence (A7) implies that ∞∫ t0 Q(t)G(R(t− α)) dt = ∞. (3.1) Assume that y(t) is a nonoscillatory solution of (1.1). Then y(t) > 0 or y(t) < 0 for t ≥ t0 > ρ. Let y(t) > 0 for t ≥ t0. Setting z(t) = y(t) + p(t)y(t− τ), (3.2) ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 OSCILLATION RESULTS FOR FOURTH ORDER NONLINEAR NEUTRAL . . . 543 K(t) = ∞∫ t s− t r(s) ∞∫ s (θ − s)h(θ)H(y(θ − β)) dθ ds, (3.3) and w(t) = z(t)−K(t) = y(t) + p(t)y(t− τ)−K(t), (3.4) we obtain (r(t)w′′(t))′′ = −q(t)G(y(t− α)) ≤ 0, 6≡ 0 (3.5) for t ≥ t0 + α. Consequently, w(t), w′(t), (r(t)w′′(t)), (r(t)w′′(t))′ are monotonic on [t1,∞), t1 ≥ t0 + α. In what follows, we have two cases, viz. w(t) > 0 or < 0 for t ≥ t1. Suppose the former holds. By the Lemma 2.1, any one of the cases (a), (b), (c) and (d) holds. Suppose that any one of the cases (a), (b) and (d) holds. Upon using (A3), (A4) and (A6), Eq. (1.1) can be viewed as 0 = (r(t)w′′(t))′′ + q(t)G(y(t− α)) +G(a)(r(t− τ)w′′(t− τ))′′+ +G(a)q(t− τ)G(y(t− τ − α)) ≥ (r(t)w′′(t))′′ +G(a)(r(t− τ)w′′(t− τ))′′+ + λQ(t)G(y(t− α) + ay(t− α− τ)) ≥ (r(t)w′′(t))′′+ +G(a)(r(t− τ)w′′(t− τ))′′ + λQ(t)G(z(t− α)) for t ≥ t2 > t1, where we have used the fact that z(t) ≤ y(t) + ay(t− τ). From (3.3), it follows that K(t) > 0, K ′(t) < 0, and hence limt→∞K(t) exists due to (A2). Further, w(t) > 0 for t ≥ t1 implies that w(t) < z(t) for t ≥ t2 and thus the last inequality yields (r(t)w′′(t))′′ +G(a)(r(t− τ)w′′(t− τ))′′ + λQ(t)G(w(t− α)) ≤ 0, for t ≥ t2, that is, (r(t)w′′(t))′′ +G(a)(r(t− τ)w′′(t− τ))′′ + λG(k1)Q(t)G(R(t− α)) ≤ 0 due to (A4) and Lemma 2.3, for t ≥ t3 > t2. Integrating the above inequality from t3 to∞, we get λG(k1) ∞∫ t3 Q(t)G(R(t− α)) dt < ∞, a contradiction to (3.1). Next, we suppose that the case (c) holds. Upon using Lemmas 2.2 and 2.3, we have k(−r(t)w′′(t))′tR(t) ≤ w(t) ≤ k2t for t ≥ t4 > t3. Hence −[((−r(t)w′′(t))′)1−γ ]′ = (γ − 1)((−r(t)w′′(t))′)−γ(−r(t)w′′(t))′′ ≥ ≥ (γ − 1)LγRγ(t)q(t)G(y(t− α)), (3.6) ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 544 A. K. TRIPATHY, S. PANIGRAHI, R. BASU where L = k k2 > 0. Therefore, −[((−r(t)w′′(t))′)1−γ ]′ −G(a)[((−r(t− τ)w′′(t− τ))′)1−γ ]′ ≥ ≥ (γ − 1)Lγ [Rγ(t)q(t)G(y(t− α)) +G(a)Rγ(t− τ)q(t− τ)G(y(t− τ − α))] ≥ ≥ λ(γ − 1)Lγb(t)Q(t)G(z(t− α)) ≥ λ(γ − 1)Lγb(t)Q(t)G(w(t− α)) ≥ ≥ λ(γ − 1)LγG(k1)b(t)Q(t)G(R(t− α)) implies that λ(γ − 1)LγG(k1) ∞∫ t4 b(t)Q(t)G(R(t− α)) dt < ∞, which contradicts (A7).Hence the latter holds. Consequently, z(t) < K(t) andK(t) is bounded will imply that y(t) is bounded. From Lemma 2.4, it follows that any one of the cases (b) – (f) holds for t ≥ t2 > t1. In the cases (e) and (f) of Lemma 2.4, limt→∞w(t) = −∞ which contradicts the fact that y(t) is bounded and limt→∞w(t) exists. Consider the case (b) or (c), where −∞ < limt→∞w(t) ≤ 0. Consequently, 0 ≥ lim t→∞ w(t) = lim sup t→∞ [z(t)−K(t)] ≥ lim sup t→∞ [y(t)−K(t)] ≥ ≥ lim sup t→∞ y(t)− lim t→∞ K(t) = lim sup t→∞ y(t) implies that limt→∞ y(t) = 0. We may note that limt→∞K(t) = 0. Lastly, let the case (d) of Lemma 2.4 hold. Then limt→∞(r(t)w′′(t))′ exists. Hence integrating (3.5) from t2 to∞, we obtain ∞∫ t2 q(t)G(y(t− α)) dt < ∞, that is, ∞∫ t2 Q(t)G(y(t− α)) dt < ∞. (3.7) If lim inft→∞ y(t) > 0, then (3.7) yields, ∞∫ t2 Q(t) dt < ∞, which contradicts (A6) due to Remark 3.1. Hence lim inft→∞ y(t) = 0. Because limt→∞w(t) exists, using Lemma 2.5, limt→∞w(t) = 0 = limt→∞ z(t). Moreover, z(t) ≥ y(t) implies that limt→∞ y(t) = 0. ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 OSCILLATION RESULTS FOR FOURTH ORDER NONLINEAR NEUTRAL . . . 545 If y(t) < 0 for t ≥ t0, then we set x(t) = −y(t) for t ≥ t0 and (r(t)(x(t) + p(t)x(t− τ))′′)′′ + q(t)G(x(t− α))− h(t)H(x(t− β)) = 0. Proceeding as above, we obtain similar conclusion. Theorem 3.1 is proved. Remark 3.4. From Theorem 3.1, it revels that y(t) is bounded in the casew(t) < 0 for t ≥ t1, which further converges to zero as t → ∞. However, this fact is not required in the other case. Hence we have proved the following theorem. Theorem 3.2. Let 0 ≤ p(t) ≤ a < ∞. Suppose that (A1) – (A7) hold, then every unbounded solution of (1.1) oscillates. Theorem 3.3. Let 0 ≤ p(t) ≤ a < 1. If (A1), (A2), (A4), (A5) and (A8) hold, then every unbounded solution of (1.1) oscillates. Proof. Since R(t) → 0 as t → ∞, (A8) implies that ∞∫ t0 G(R(t− α))q(t) dt = ∞ (3.8) and hence ∞∫ t0 q(t) dt = ∞. (3.9) Let y(t) be a nonoscillatory solution of (1.1) such that y(t) is unbounded and y(t) > 0 for t ≥ t0 > 0. The case y(t) < 0 for t ≥ t0 > 0 is similar. We set z(t), K(t) and w(t) as in (3.2), (3.3) and (3.4) respectively to obtain (3.5) for t ≥ t0 + α. Consequently, each of w(t), w′(t), (r(t)w′′(t)) and (r(t)w′′(t))′ is of constant sign on [t1,∞), t1 ≥ t0 + α. Assume that w(t) > 0 for t ≥ t1. Then Lemma 2.1 holds. If any one of the cases (a) or (b) holds, then 0 < w′(t) = z′(t) − K ′(t) implies that z′(t) > 0 or < 0 for t ≥ t1. We note that z(t) is unbounded due to unbounded y(t). Hence z′(t) < 0 doesn’t arise. Ultimately, z′(t) > 0 and (1− p(t))z(t) < z(t)− p(t)z(t− τ) = y(t)− p(t)p(t− τ)y(t− 2τ) < y(t), that is, y(t) > (1− a)z(t) > (1− a)w(t) for t ≥ t2 > t1. Thus (3.5) yields G((1− a)w(t− α))q(t) ≤ −(r(t)w′′(t))′′, that is, G(k1(1− a))G(R(t− α))q(t) ≤ −(r(t)w′′(t))′′ (3.10) ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 546 A. K. TRIPATHY, S. PANIGRAHI, R. BASU due to Lemma 2.3 and (A4). Integrating (3.10) from t2 to∞, it follows that ∞∫ t2 q(t)G(R(t− α)) dt < ∞, a contradiction to (3.8). For the case (c) of Lemma 2.1, we proceed as in the proof of Theorem 3.1 to obtain (3.6). Using the same type of reasoning as above, (3.6) yields that −[((−r(t)w′′(t))′)1−γ ]′ ≥ (γ − 1)LγG((1− a)k1)R γ(t)q(t)G(R(t− α)) for t ≥ t2. Integrating the last inequality from t2 to∞, we obtain, ∞∫ t2 q(t)Rγ(t)G(R(t− α)) dt < ∞, a contradiction to (A8). In the case (d) of Lemma 2.1, limt→∞w(t) exists, that is, limt→∞ z(t) exists, a contradiction to our hypothesis. Due to Remark 3.4, the case w(t) < 0 doesn’t arise. Theorem 3.3 is proved. Theorem 3.4. Let −1 < a ≤ p(t) ≤ 0. If (A1), (A2), (A5) and (A8) hold, then every solution of (1.1) either oscillatory or tends to zero as t → ∞. Proof. Let y(t) be a nonoscillatory solution of (1.1) such that y(t) > 0 for t ≥ t0 > 0. Setting z(t), K(t) and w(t) as in (3.2), (3.3) and (3.4) we obtain (3.5) for t ≥ t0 + α and hence w(t) is monotone on [t1,∞), t1 ≥ t0 +α. Let w(t) > 0 for t ≥ t1. Suppose that one of the cases (a), (b) and (d) of Lemma 2.1 holds for t ≥ t1. From Lemma 2.3, we have y(t) ≥ w(t) ≥ k1R(t) for t ≥ t2 > t1 and hence (3.5) yields ∞∫ t3 q(t)G(R(t− α)) dt < ∞, t3 > t2 + α, a contradiction to (3.8). Next we consider the case (c). Proceeding as in the proof of Theorem 3.1, we obtain (3.6). Further, y(t) ≥ w(t) ≥ k1R(t) for t ≥ t2 by Lemma 2.3. Consequently, for t ≥ t3 > t2 + α, −[((−r(t)w′′(t))′)1−γ ]′ ≥ (γ − 1)LγG(k1)R γ(t)q(t)G(R(t− α)). Integrating the above inequality from t3 to∞, we get ∞∫ t3 q(t)Rγ(t)G(R(t− α)) dt < ∞, a contradiction to (A8). ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 OSCILLATION RESULTS FOR FOURTH ORDER NONLINEAR NEUTRAL . . . 547 If w(t) < 0 for t ≥ t1, then y(t) is bounded ultimately. Hence z(t) is bounded and so also w(t). In what follows, none of the cases (e) and (f) of Lemma 2.4 arises. In the case (b) or (c), −∞ < limt→∞w(t) ≤ 0. Using the fact that limt→∞K(t) = 0, we have limt→∞w(t) = = limt→∞ z(t). Hence 0 ≥ lim t→∞ w(t) = lim t→∞ z(t) = lim sup t→∞ [y(t) + p(t)y(t− τ)] ≥ lim sup t→∞ y(t) + lim inf t→∞ (ay(t− τ)) = = lim sup t→∞ y(t) + a lim sup t→∞ y(t− τ) = (1 + a) lim sup t→∞ y(t) implies that lim supt→∞ y(t) = 0, that is, limt→∞ y(t) = 0. Let the case (d) hold. Since lim t→∞ (r(t)w′′(t))′ exists, (3.5) yields that ∞∫ t2 q(t)G(y(t− α)) dt < ∞. (3.11) If lim inft→∞ y(t) > 0, then it follows from (3.11) that ∞∫ t2 q(t) dt < ∞, which contradicts (3.9). Hence lim inft→∞ y(t) = 0. Using Lemma 2.5, we assert that lim t→∞ w(t) = 0 = lim t→∞ z(t). Proceeding as above, we may show that lim supt→∞ y(t) = 0 and hence limt→∞ y(t) = 0. If y(t) < 0 for t ≥ t0, then one may proceed as above to obtain lim inft→∞ y(t) = 0, that is limt→∞ y(t) = 0. Thorem 3.4 is proved. Theorem 3.5. Let −∞ < p(t) ≤ 0. If (A1), (A2), (A5) and (A8) hold, then every unbounded solution of (1.1) is oscillatory. The proof of the theorem follows from the proof of Theorem 3.4. Hence the details are omitted. 4. Oscillation results for (1.2). This section is concerned with the oscillation and asymptotic behaviour of solutions of (1.2) with suitable forcing functions. We restrict our forcing functions which are allowed to change the sign eventually. Let the following hypotheses hold concerning the forcing function f(t) of (1.2): (A9) There existsF ∈C2([0,∞),R) such thatF (t) changes sign with−∞< lim inft→∞ F (t)< < 0 < lim supt→∞ F (t) < ∞, rF ′′ ∈ C2([0,∞),R) and (rF ′′)′′ = f. ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 548 A. K. TRIPATHY, S. PANIGRAHI, R. BASU (A10) There exists F ∈ C2([0,∞),R) such that F (t) changes sign with lim inft→∞ F (t) = = −∞, lim supt→∞ F (t) = +∞, rF ′′ ∈ C2([0,∞),R) and (rF ′′)′′ = f. Theorem 4.1. Let 0 ≤ p(t) ≤ a < ∞. Assume that (A1), (A2), (A3), (A4), (A5) and (A10) hold. If ∞∫ α b(t)Q(t)G(F+(t− α)) dt = ∞ = ∞∫ α b(t)Q(t)G(F−(t− α))dt, (A11) where F+(t) = max{0, F (t)} and F−(t) = max{−F (t), 0}, then every solution of (1.2) oscil- lates. Proof. Suppose on the contrary that y(t) is a nonoscillatory solution of (1.2) such that y(t) > > 0 for t ≥ t0 > ρ. Setting as in (3.2), (3.3) and (3.4), let V (t) = w(t)− F (t) = z(t)−K(t)− F (t). (4.1) Hence for t ≥ t0 + α, Eq. (1.2) becomes (r(t)V ′′(t))′′ = −q(t)G(y(t− α)) ≤ 0, 6≡ 0. (4.2) Consequently, V (t) is monotone on [t1,∞), t1 > t0 + α. Let V (t) > 0 for t ≥ t1. Then z(t) − K(t) > F (t) implies that z(t) − K(t) > 0 due to (A10) and hence z(t) − K(t) > > max{0, F (t)} = F+(t) for t ≥ t1, that is z(t) > K(t) + F+(t) > F+(t). (4.3) In view of Eq. (1.2), it is easy to verify that 0 = (r(t)V ′′(t))′′ + q(t)G(y(t− α)) +G(a)(r(t− τ)V ′′(t− τ))′′+ +G(a)q(t− τ)G(y(t− α− τ)) ≥ (r(t)V ′′(t))′′+ +G(a)(r(t− τ)V ′′(t− τ))′′ + λQ(t)G(z(t− α)) due to (A3) and (A4). Using (4.3), the last inequality yields 0 ≥ (r(t)V ′′(t))′′ +G(a)(r(t− τ)V ′′(t− τ))′′ + λQ(t)G(F+(t− α)), (4.4) for t ≥ t2 > t1. Assume that one of the cases (a), (b) and (d) of Lemma 2.1 holds. Then integrating (4.4) from t2 + α to∞, we obtain ∞∫ t2+α Q(t)G(F+(t− α))dt < ∞, a contradiction to (A11). We may note that b(t) → 0 as t → ∞ due to Remark 3.1. Consider the case (c) of Lemma 2.1. From Lemma 2.3 it follows that k(−r(t)V ′′(t))′tR(t) ≤ V (t) ≤ k2t, t ≥ t3 > t2. ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 OSCILLATION RESULTS FOR FOURTH ORDER NONLINEAR NEUTRAL . . . 549 Hence in view of (3.6), we have −[((−r(t)V ′′(t))′)1−γ ]′ ≥ (γ − 1)LγRγ(t)q(t)G(y(t− α)), for t ≥ t3. Proceeding as in Theorem 3.1, we obtain λ(γ − 1)LγG(k1) ∞∫ t4 b(t)q(t)G(F+(t− α)) dt < ∞, t4 > t3, which contradicts (A11). Consequently, V (t) < 0 for t ≥ t1. Thus any one of the cases (b) – (f) of Lemma 2.4 holds. If V (t) < 0, z(t)−K(t) < 0 ultimately due to (A10). In what follows, z(t) is bounded and so also y(t). Therefore, limt→∞ V (t) exists. Since z(t) = V (t) +K(t) +F (t), we have 0 ≤ lim inf t→∞ z(t) = lim inf t→∞ (V (t) +K(t) + F (t)) ≤ ≤ lim sup t→∞ V (t) + lim inf t→∞ (K(t) + F (t)) ≤ ≤ lim t→∞ V (t) + lim sup t→∞ K(t) + lim inf t→∞ F (t) = = lim t→∞ V (t) + lim t→∞ K(t) + lim inf t→∞ F (t) → −∞, which is absurd. If y(t) < 0 for t ≥ t0, we set x(t) = −y(t) to obtain x(t) > 0 for t ≥ t0 and (r(t)(x(t) + p(t)x(t− τ))′′)′′ + q(t)G(x(t− α))− h(t)H(x(t− β)) = f̃(t) due to (A5) where f̃(t) = −f(t). If we set F̃ (t) = −F (t), then F̃ (t) changes sign. Further, F̃+(t) = F−(t) and (r(t)F̃ ′′(t))′′ = f̃(t). Proceeding as above we obtain a contradiction. Theorem 4.1 is proved. Remark 4.1. In Theorem 4.1, V (t) < 0 implies that z(t) and y(t) are bounded simultaneously. This fact is unlikely true due to our assumption (A10). If (A9) is replaced by (A10), then bounded y(t) doesn’t provide any conclusion about the oscillatory behaviour of the solutions of (1.2). Hence with unbounded y(t), we have proved the following theorem: Theorem 4.2. Let 0 ≤ p(t) ≤ a < ∞. If (A1) – (A5), (A9) and (A11) hold, then every unbounded solution of (1.2) is oscillatory. Theorem 4.3. Let −1 < p(t) ≤ 0. Suppose that (A1), (A2), (A5) and (A10) hold. If ∞∫ α Rγ(t)q(t)G(F+(t− α)) dt = ∞ = ∞∫ α Rγ(t)q(t)G(F−(t− α)) dt γ > 1, (A12) then (1.2) is oscillatory. ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 550 A. K. TRIPATHY, S. PANIGRAHI, R. BASU Proof. For the sake of contradiction, let y(t) be a nonoscillatory solution of (1.2) such that y(t) > 0 for t ≥ t0 > ρ. The case y(t) < 0 can be similarly dealt with. For t ≥ t1 > t0, y(α(t)) > 0 and y(β(t)) > 0. Let’s set V (t) as in (4.1), so that we get (4.2). Consequently, V (t) is monotone on [t1,∞). Let V (t) > 0 for t ≥ t1. Then one of the cases (a) – (d) of Lemma 2.1 holds. Indeed, V (t) > 0, that is z(t)−K(t) > F (t) implies that z(t)−K(t) > 0 due to (A10). Hence (4.3) holds. Further, z(t)−K(t) > 0 yields that z(t) > K(t) > 0. Thus y(t) > z(t) > K(t) + F+(t) > F+(t) (4.5) for t ≥ t2 > t1. If any one of the cases (a), (b) and (d) holds, then using (4.5) in (4.2), we obtain ∞∫ t3 q(t)G(F+(t− α)) dt < ∞, t3 > t2 + α, a contradiction to (A12). Assume that case (c) holds. Proceeding as in Theorem 3.4 and upon using (4.5) in (3.6), we get −[((−r(t)V ′′(t))′)1−γ ]′ ≥ (γ − 1)LγG(k1)R γ(t)q(t)G(F+(t− α)), (4.6) for t ≥ t2 > t1. Integrating (4.6) from t2 to∞, we obtain a contradiction to (A12). Next, we suppose that V (t) < 0 for t ≥ t1. Then one of the cases (b) – (f) of Lemma 2.4 holds. Indeed, z(t) −K(t) < F (t) implies that z(t) −K(t) < 0 ultimately, due to (A10). Thus z(t) is bounded. Since V (t) is monotone, limt→∞ V (t) exists. Therefore, z(t) < K(t) + F (t) implies that lim inf t→∞ z(t) < lim inf t→∞ (K(t) + F (t)) ≤ lim sup t→∞ K(t) + lim inf t→∞ F (t) = = lim t→∞ K(t) + lim inf t→∞ F (t) → −∞, which is absurd. Theorem 4.3 is proved. Theorem 4.4. Let −1 < b ≤ p(t) ≤ 0. If (A1), (A2), (A5), (A9), and (A12) hold, then every unbounded solution of (1.2) oscillates. Proof. Let y(t) be an unbounded nonoscillatory solution of (1.2) such that y(t) > 0 for t ≥ t0 > ρ. The case when y(t) < 0 for t ≥ t0 > ρ is similar. Proceeding as in the proof of Theorem 4.3, we have the required contradiction when V (t) > 0 for t ≥ t1. Next, we suppose that V (t) < 0 for t ≥ t1. As a result, z(t) − K(t) < 0 due to (A9). Ultimately, we have two cases on z(t), viz. z(t) > 0 or z(t) < 0. If the former holds, and since y(t) is unbounded, then there exists {ηn}∞n=1 such that ηn → ∞, y(ηn) → ∞ as n → ∞ and y(ηn) = max{y(t) : t1 ≤ t ≤ ηn}. We may choose n large enough such that ηn − τ > t1. Hence it happens that z(ηn) ≥ (1 + b)y(ηn). ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 OSCILLATION RESULTS FOR FOURTH ORDER NONLINEAR NEUTRAL . . . 551 By Lemma 2.4, any one of the cases (b) – (f) holds. Assume that either case (b) or (c) holds true. Then limt→∞ \|V (t)\| < ∞ and z(t) = V (t) +K(t) + F (t) implies that ∞ = (1 + b) lim n→∞ y(ηn) ≤ lim n→∞ [\|V (ηn)\|+K(ηn) + \|F (ηn)\|] < ∞, (4.7) which is absurd. Suppose that any of the cases (d), (e) and (f) holds. For each of the cases V (t) is nonincreasing. Let limt→∞ V (t) = µ, µ ∈ [−∞, 0). If −∞ < µ < 0, then the conclusion follows from (4.7). It happens from (4.7) that∞ ≤ −∞ if µ = −∞. Hence the latter holds. As a result, y(t) < y(t − τ) for t ≥ t1, that is, y(t) is bounded for t ≥ t1 which contradicts to our hypothesis. Theorem 4.4 is proved. Theorem 4.5. Let −∞ < p(t) ≤ −1. If all the conditions of Theorem 4.4 hold, then every bounded solution of (1.2) oscillates. The proof of theorem follows from the proof of Theorem 4.4. Hence the details are omitted. Theorem 4.6. Let 1 < b1 ≤ p(t) ≤ b2 < 1 2 b21 and (A2) hold. Suppose that (A9) holds with −b1 − 1 16b2 ≤ F (t) ≤ b1 − 1 8b2 . If ∞∫ 0 s r(s) ∞∫ s tq(t) dt ds < ∞, then (1.2) admits a positive bounded solution. Proof. It is possible to choose T0 large enough such that ∞∫ T0 s r(s) ∞∫ s tq(t) dt ds < b1 − 1 16b2G(1) and ∞∫ T0 s r(s) ∞∫ s th(t) dt ds < b1 − 1 4b1H(1) . Let X = BC([T0,∞),R). Then X is a Banach space with respect to supremum norm defined by ‖x‖ = sup t≥T0 {\|x(t)\|}. Let S = { x ∈ X : b1 − 1 8b1b2 ≤ x(t) ≤ 1, t ≥ T0 } . ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 552 A. K. TRIPATHY, S. PANIGRAHI, R. BASU Hence S is a closed bounded convex subset of X. Define two maps Ω1 and Ω2 on S as follows; (Ω1y)(t) =  (Ω1y)(T1), T0 ≤ t ≤ T1, −y(t+ τ) p(t+ τ) + 2b21 + b1 − 1 4b1p(t+ τ) , t ≥ T1, and (Ω2y)(t) =  (Ω2y)(T1), T0 ≤ t ≤ T1, F (t+ τ) p(t+ τ) + K(t+ τ) p(t+ τ) − − 1 p(t+ τ) ∫ ∞ t+τ ( s− (t+ τ) r(s) ∫ ∞ s (u− s)q(u)G(y(u− α)) du ) ds, t ≥ T1, where K(t) is defined in (3.3). Indeed, K(t) = ∞∫ t s− t r(s) ∞∫ s (u− s)h(u)H(y(u− β)) du ds ≤ H(1) ∞∫ t s r(s) ∞∫ s uh(u) du ds < b1 − 1 4b1 implies that (Ω1y)(t) + (Ω2y)(t) ≤ 2b21 + b1 − 1 4b21 + b1 − 1 8b1b2 + b1 − 1 4b1 2 = b1 2 + b1 − 1 2b1 2 + b1 − 1 8b1b2 ≤ ≤ b1 2 + b1 − 1 2b1 2 + b1 − 1 8b1 2 = 4b1 2 + 5b1 − 5 8b1 2 < 1 and (Ω1y)(t) + (Ω2y)(t) ≥ − 1 b1 + 2b1 2 + b1 − 1 4b1b2 − b1 − 1 16b1b2 − b1 − 1 16b1b2 = = − 1 b1 + 2b1 2 + b1 − 1 4b1b2 − b1 − 1 8b1b2 = = − 1 b1 + 4b1 2 + b1 − 1 8b1b2 ≥ b1 − 1 8b1b2 , that is, Ω1y + Ω2y ∈ S. It is easy to verify that Ω1 is a contraction mapping. Next, we show that Ω2 is continuous. Let {yj(t)} be the sequence of continuous functions defined on S such that ‖yj−y‖ = 0 for all j → ∞. Because S is closed and bounded, (yj−y) ∈ ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 OSCILLATION RESULTS FOR FOURTH ORDER NONLINEAR NEUTRAL . . . 553 ∈ S and \|(Ω2yj)(t)− (Ω2y)(t)\| ≤ ≤ 1 p(t+ τ) ∞∫ t+τ s− t− τ r(s) ∞∫ s (u− s)h(u)\|H(yj(u− β))−H(y(u− β))\| du ds+ + 1 p(t+ τ) ∞∫ t+τ s− t− τ r(s) ∞∫ s (u− s)q(u)\|G(yj(u− α))−G(y(u− α))\| du ds. Because G and H are continuous functions, then it follows that ‖Ω2yj − Ω2y‖ = 0 as j → ∞. We know that Ω2 is uniformly bounded, there exist t1, t2 > 0 such that for t1 > t2 ≥ T1 and for all y(t) ∈ S, \|Ω2y(t1)− Ω2y(t2)\| ≤ ∣∣∣∣F (t1 + τ) p(t1 + τ) ∣∣∣∣+ ∣∣∣∣F (t2 + τ) p(t2 + τ) ∣∣∣∣+ ∣∣∣∣K(t1 + τ) p(t1 + τ) ∣∣∣∣+ ∣∣∣∣K(t2 + τ) p(t2 + τ) ∣∣∣∣+ + ∣∣∣∣∣∣ 1 p(t1 + τ) ∞∫ t1+τ s− t1 − τ r(s) ∞∫ s (u− s)q(u)G(y(u− α)) du ds ∣∣∣∣∣∣+ + ∣∣∣∣∣∣ 1 p(t2 + τ) ∞∫ t2+τ s− t2 − τ r(s) ∞∫ s (u− s)q(u)G(y(u− α)) du ds ∣∣∣∣∣∣ ≤ ≤ 2 ( b1 − 1 8b1b2 ) + 2 ( b1 − 1 4b1 2 ) + 2 ( b1 − 1 16b1b2 ) ≤ 7(b1 − 1) 8b1 2 implies that Ω2 is precompact. Hence verifying all the required conditions of Krasnosel’skii’s fixed point theorem it yields that Ω1 + Ω2 has a fixed point in S, that is, y(t) = −y(t+ τ) p(t+ τ) + 2b1 2 + b1 − 1 4b1p(t+ τ) + K(t+ τ) p(t+ τ) − − 1 p(t+ τ) ∞∫ t+τ s− (t+ τ) r(s) ∞∫ s (u− s)q(u)G(y(u− α))du  ds+ F (t+ τ) p(t+ τ) . Clearly, y(t) is a solution of (1.2) on [ b1 − 1 8b1b2 , 1 ] . Theorem 4.6 is proved. Remark 4.2. Theorems similar to Theorem 4.6 can be proved in the other ranges of p(t). ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 554 A. K. TRIPATHY, S. PANIGRAHI, R. BASU 5. Examples and discussions. Example 5.1. Consider (et(y(t) + e−4ty(t− π))′′)′′ + 8et+2πy(t− 2π)− − 50e−3t+ π 2 ( 1 + e2t−3π cos2 t ) y ( t− 3π 2 ) 1 + y2 ( t− 3π 2 ) = 6e2t cos t. (5.1) Indeed, if we choose F (t) = ( et 25 ) (9 sin t−12 cos t), then it is easy to verify that (r(t)F ′′(t))′′ = = f(t) = 6e2t cos t. Since F+(t− 2π) =  0, t ∈ [(2n+ 3)π + θ1, (2n+ 4)π + θ1], 3 5 et−2π sin(t− 2π − θ1), t ∈ [2(n+ 1)π + θ1, (2n+ 3)π + θ1], and F−(t− 2π) =  −3 5 et−2π sin(t− 2π − θ1), t ∈ [(2n+ 3)π + θ1, (2n+ 4)π + θ1], 0, t ∈ [2(n+ 1)π + θ1, (2n+ 3)π + θ1], for n = 0, 1, 2, . . . , then ∞∫ 2π b(t)Q(t)F+(t− 2π) dt = 24 5 e−π eθ1/2 ∞∑ n=0 (2n+3)π∫ 2(n+1)π e z 2 sin z dz = = 48 25 e−π eθ1/2 ∞∑ n=0 ( 2e (2n+3)π 2 + 2e(n+1)π ) = ∞, where F (t) = 3 5 et sin (t− θ1), θ1 = tan−1 ( 4 3 ) and z = t − θ1. Clearly, (A1) – (A5) and (A10) is satisfied. Hence by Theorem 4.1, every solution of (5.1) is oscillatory. In particular, y(t) = et sin t is such an oscillatory solution of (5.1). Example 5.2. Consider (et(y(t) + (1 + e−t)y(t− 2π))′′)′′ + e3ty(t− 4π)− e−t y(t− 6π) 1 + y2(t− 6π) = 0. (5.2) Clearly, (A1) – (A7) are satisfied. Hence by Theorem 3.1 every solution of (5.2) oscillates or tends to zero. It is learnt that the solution space of (1.1)/(1.2) is divided for bounded and unbounded soluti- ons. Due to the method incorporated here, we could not stop the bounded solutions of (1.1) as converging to zero. However, in case of unbounded solution, it oscillates. ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4 OSCILLATION RESULTS FOR FOURTH ORDER NONLINEAR NEUTRAL . . . 555 It is interesting to notice the solution space of forced equation (1.2) pertaining to (A9) or (A10). Emphasis will be given to forcing function as compared to the results concerning (1.1). It reveals that every unbounded solutions of (1.2) oscillates if (A9) holds except p(t) ≤ 1. 1. Chuanxi Q., Ladas G. Oscillation in differential equations with positive and negative coefficients // Can. Math. Bull. — 1990. — 33. — P. 442 – 450. 2. Edwards R. E. Functional analysis. — New York: Holt, Rinehart and Winston Inc., 1965. 3. Gyori I., Ladas G. Oscillation theory of delay differential equation with application. — Oxford: Claredon Press, 1991. 4. Li W. T., Quan H. S. Oscillation of higher order neutral differential equations with positive and negative coefficients // Ann. Different. Equat. — 1995. — 2. — P. 70 – 76. 5. Li W. T., Yan J. Oscillation of first order neutral differential equations with positive and negative coefficients // Collect. math. — 1999. — 50. — P. 199 – 209. 6. O’calan O. Oscillation of forced neutral differential equations with positive and negative coefficients // Comput. Math. and Appl. — 2007. — 54. — P. 1411 – 1421. 7. O’calan O. Oscillation of neutral differential equation with positive and negative coefficients // J. Math. Anal. and Appl. — 2007. — 331. — P. 644 – 654. 8. Parhi N., Chand S. On forced first order neutral differential equations with positive and negative coefficients // Math. Slovaca. — 2000. — 50. — P. 183 – 202. 9. Parhi N., Tripathy A. K. On oscillatory fourth order nonlinear neutral differential equations - I // Math. Slovaca. — 2004. — 54. — P. 389 – 410. 10. Tripathy A. K. Oscillation properties of a class of neutral differential equations with positive and negative coefficients // Fasc. Math. — 2010. — № 45. — P. 133 – 155. Received 20.04.11, after revision — 04.02.12 ISSN 1562-3076. Нелiнiйнi коливання, 2012, т . 15, N◦ 4

Oscillation results for fourth order nonlinear neutral differential equations with positive and negative coefficients

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