Oscillatory behavior of first order functional differential equations 9786010442382

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Oscillatory behavior of first order functional differential equations
 9786010442382

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

OSCILLATORY BEHAVIOR OF FIRST ORDER FUNCTIONAL DIFFERENTIAL EQUATIONS Ioannis P. Stavroulakis1,2 and Zhanat Kh. Zhunussova2 1

Department of Mathematics, University of Ioannina 451 10 Ioannina, HELLAS (Greece) 2 Faculty of Mechanics and Mathematics, Al-Farabi Kazakh National University, Almaty, 050040 KAZAKHSTAN

Almaty "Qazaq University" 2019

UDC 51(075.8) LBC 22.1я73 S 81 Recommended for publication by the decision of the Academic Council of the Faculty of Mechanics and Mathematics, Editorial and Publishing Council of Al-Farabi Kazakh National University (Protocol №5 dated 27.06.2019) Reviewers: Hassan El-Morshedy, Professor, Department of Mathematics, Faculty of Science, Damietta University, New Damietta 34517, Egypt Tamasha Aldibekov, Professor Department of Differential equations and control theory, Faculty of Mechanics and Mathematics, Al-Farabi kazakh National University, Almaty, Kazakhstan

Stavroulakis I.P., Zhunussova Zh.Kh. Oscillatory behavior of first order functional differential S 81 equations / I.P. Stavroulakis, Zh.Kh. Zhunussova. – Almaty: Qazaq University, 2019. – 192 p. ISBN 978-601-04-4238-2 Some theoretical foundations of oscillations of delay equations are expounded in the textbook: oscillations of delay equations, oscillations of difference equations. The tasks for independent work with solving of concrete examples, brief theory and solution algorithms of the problems, term tasks on sections of oscillations of delay and difference equations are presented. It is intended as a textbook for students of the high schools training on specialties "mathematics", "mechanics", "automation and control", "mathematical and computer modeling" and "informatics". It will be useful for the post-graduate students and scientific workers of the mathematical, economic and naturallytechnical specialties.

UDC 51(075.8) LBC 22.1я73 ISBN 978-601-04-4238-2

© Stavroulakis I.P., Zhunussova Zh.Kh., 2019 © Al-Farabi KazNU, 2019

CONTENTS 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Part I OSCILLATIONS OF DELAY EQUATIONS . . . . . . . . . .8 2. Oscillation Criteria for delay equation . . . . . . . . . . . . . . . . . . . . 8 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3. Oscillations in Critical State . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Part II OSCILLATIONS OF DIFFERENCE EQUATIONS . 71 4. Oscillation Criteria for difference equation . . . . . . . . . . . . . . .71 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 5. Oscillation Criteria for the first order linear difference equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .178

3

Preface Ordinary differential equations appear frequently in mathematical models that attempt to describe real-life situations in which the rate of change of the system depends only on its present stage. A basic limitation behind such a model is the assumption that all interactions in the system occur instantaneously. Mathematically an ordinary differential equation (of first order) is an equation of the form x (t) = f (t, x(t)), where f is a known function and x is the unknown function. Note that both x and x are evaluated at the same instant t. However, in many cases the past state of the system has to be taken into consideration. Delay differential equations or differential equations with retarded argument or hystero-defferential equations provide more realistic mathematical models for such systems in which the rate of change depends not only on their present stage but also on their past history. The theory of such systems has been developed in the beginning of this century. In his research on prey-predator population models and viscoelasticity, Volterra [95] formulated some rather general differential equations incorporating the past states of the system. A delay differential equation is an equation of the form x (t) = F (t, x(t), x(t − τ )), where F is a known function, τ > 0 is the delay or time lag or hysteresis and x is the unknown function. Minorsky [64] was 4

one of the first investigators to study delay differential equations and their effects on simple feedback control systems in which the communication time τ can not be neglected. An automatic feedback control system does not respond instantaneously to input signals but requires some time τ to process the information and react. Delay differential equations arise naturally in many other systems involving time lags such as population models, models for epidemics, economic models, nuclear reactrors, collision problems in electrodynamics, etc. The oscillation theory of differential equations is not a new area. It was originated by Sturm in 1836 [78]. Since then hundreds of papers have been published studying the oscillation theory of ordinary differential equations. The reader is referred to Swanson [79] and the bibliography cited therein. In the decade of 1970 a great number of papers were written extending known results from ordinary differential equations to delay differential equations. Of particular importance, however, has been the study of oscillations which are caused by the delay and which do not appear in the corresponding ordinary differential equation. In recent years there has been a great deal of interest in the study of oscillatory behaviour of the solutions to delay differential equations and also the discrete analogue delay difference equations. See, for example, the bibliography at the end of this manuscript and the references cited therein. This monograph presents the latest developments in the oscillation theory of first order delay differential equations and difference equations with variable argument. Numerous results presented here have been derived during collaborative work with colleagues mentioned in the references whom we wish to thank them all. 5

1

Introduction

The problem of establishing sufficient conditions for the oscillation of all solutions to the differential equation x (t) + p(t)x(τ (t)) = 0,

t ≥ t0 ,

(1)

where the functions p, τ ∈ C([t0, ∞), R+ ) (here R+ = [0, ∞)), τ (t) is nondecreasing, τ (t) < t for t ≥ t0 and limt→∞ τ (t) = ∞, has been the subject of many investigations. It is noteworthy to observe that a first-order linear differential equation of the form (1) without delay (τ (t) ≡ t) does not possess oscillatory solutions. Therefore the investigation of oscillatory solutions is of interest for equations of the form (1). The reader is referred to [2, 13, 17, 19, 23–29, 31–34, 36, 38, 39, 41-48, 50, 53-58, 60, 61, 65, 66, 70, 71, 76, 77, 81, 87-94, 96, 98-100, 106] and the references cited therein. By a solution of Eq.(1) we understand a continuously differentiable function defined on [τ (T0 ), ∞) for some T0 ≥ t0 and such that Eq.(1) is satisfied for t ≥ T0 . Such a solution is called oscillatory if it has arbitrarily large zeros, and otherwise it is called nonoscillatory. The oscillation theory of the (discrete analogue) delay difference equations Δx(n) + p(n)x(τ (n)) = 0, n = 0, 1, 2, ...,

(1)

Δx(n) + p(n)x(n − k) = 0, n = 0, 1, 2, ....

(1)

and where Δx(n) = x(n + 1) − x(n), k is a positive integer, p(n) is a sequence of nonnegative real numbers and τ (n) is a nondecreasing sequence of integers such that τ (n) ≤ n − 1 for all n ≥ 0 and 6

limn→∞ τ (n) = ∞, has also attracted growing attention in the last decades. The reader is referred to [1, 3–12, 14-18, 22-22, 30, 37, 49, 51, 52, 59, 62, 63, 67-69, 72-77, 82-86, 97, 101-105] and the references cited therein. By a solution of Eq.(1) we mean a sequence x(n) which is defined for n ≥ −k and which satisfies (1) for n ≥ 0. A solution x(n) of Eq.(1) is said to be oscillatory if the terms x(n) of the sequence are neither eventually positive nor eventually negative, and otherwise the solution is said to be nonoscillatory. (Analogously for Eq.(1) .) In this paper our main purpose is to present the state of the art on the oscillation of all solutions to Eq.(1) especially in the case where  t  t 1 p(s)ds ≤ p(s)ds < 1, and lim sup 0 < lim inf t→∞ e t→∞ τ (t) τ (t) and to (the discrete analogues) Eq.(1) when 0 < lim inf n→∞

n−1 

 p(i) ≤

i=n−k

k k+1

k+1 and lim sup n→∞

n 

p(i) < 1

i=n−k

and Eq.(1) when 0 < lim inf n→∞

n−1  i=τ (n)

n  1 p(i) ≤ p(i) < 1. and lim sup e n→∞ i=τ (n)

7

Part I

OSCILLATIONS OF DELAY EQUATIONS 2

Oscillation Criteria for Eq. (1)

In this section we study the delay equation x (t) + p(t)x(τ (t)) = 0,

t ≥ t0 .

(1)

where the functions p, τ ∈ C([t0, ∞), R+ ), τ (t) is nondecreasing, τ (t) < t for t ≥ t0 and limt→∞ τ (t) = ∞. The first systematic study for the oscillation of all solutions to Eq.(1) was made by Myshkis. In 1950 [65] he proved that every solution of Eq.(1) oscillates if 1 lim inf [t − τ (t)] lim inf p(t) > . t→∞ t→∞ e t→∞ (C1 ) In 1972, Ladas, Lakshmikantham and Papadakis [50] proved that the same conclusion holds if  t p(s)ds > 1. (C2 ) A := lim sup lim sup[t − τ (t)] < ∞ and

t→∞

τ (t)

In 1979, Ladas [48] established integral conditions for the oscillation of Eq.(1) with constant delay. Tomaras [92-94] extended this result to Eq.(1) with variable delay. For related 8

results see Ladde [55-57]. The following most general result is due to Koplatadze and Canturija [42]. If  t 1 a := lim inf p(s)ds > , (C3 ) t→∞ e τ (t) then all solutions of Eq.(1) oscillate; If  t 1 lim sup p(s)ds < , e t→∞ τ (t)

(N1 )

then Eq.(1) has a nonoscillatory solution. It is obvious that there is a gap between the conditions (C2 ) t and (C3 ) when the limit lim τ (t) p(s)ds does not exist. How t→∞ to fill this gap is an interesting problem which has been recently investigated by several authors. In 1988, Erbe and Zhang [29] developed new oscillation criteria by employing the upper bound of the ratio x(τ (t))/x(t) for possible nonoscillatory solutions x(t) of Eq.(1). Their result says that all the solutions of Eq.(1) are oscillatory, if 1 01−

A>1−

a2 , 2(1 − a) 9

(C5 )

while in 1992, Yu and Wang [99] and Yu, Wang, Zhang and Qian [100] obtained the condition √ 1 − a − 1 − 2a − a2 A>1− . (C6 ) 2 In 1990, Elbert and Stavroulakis [26] improved (C4 ) to the condition 1 (C7 ) A > 1 − (1 − √ )2 λ1 where λ1 is the smaller real root of the equation λ = eaλ . In more detail the main results in [26] are as follows: First define a function f = f (Δ) for 0 ≤ Δ ≤ 1e by f e−Δf = 1, 1 ≤ f ≤ e.

(2.1)

Observe that the function ue−Δu is strictly increasing in [1, e] from e−Δ to e1−Δe and therefore the function f (Δ) is uniquely defined. Since, by definition, f = eΔf , it is clear that f > 1 for Δ > 0 and so f = eΔ . Also f (0) = 1, f ( 1e ) = e and f is an increasing function of Δ. We also define a sequence {fi (Δ)}, 0 ≤ Δ ≤ 1 by f1 (Δ) = 1, fi+1 (Δ) = eΔfi (Δ) , i = 1, 2, ... . It is easily seen that for Δ > 0 fi+1 (Δ) > fi (Δ), i = 1, 2, ... . Observe that when 0 ≤ Δ ≤

1 e

then 1

f1 (Δ) = 1 < e, f2 (Δ) = eΔf1 < e e e = e, and in general 10

(2.2)

fi (Δ) < e, i = 1, 2, ... . That is, the sequence {fi (Δ)} is increasing and bounded from above. Thus, there exists a function f (Δ) such that lim fi (Δ) = f (Δ), 1 ≤ f (Δ) ≤ e

i→∞

and

f (Δ) = eΔf (Δ) .

By (2.1), it is clear that f (Δ) = f (Δ), hence lim fi (Δ) = f (Δ)

i→∞

and

1 fi (Δ) < f (Δ) when 0 < Δ ≤ . e 1 However, when Δ > e lim fi (Δ) = ∞,

i→∞

because otherwise the sequence {fi (Δ)} would have a finite limit f such that f = eΔf . Using the known inequality ex ≥ ex, . we have

f = eΔf > eΔf > f

which is a contradiction.

11

Lemma 2.1 ([26]) Assume that x(t) is a solution of Eq.(1) such that for some t x(t) > 0 on [τ3 (t), t], p(t) ≥ 0 on [τ2 (t), t] and  t τ (t)

p(s)ds ≥ Δ > 0 for τ (t) ≤ t ≤ t.

(2.3)

Let M, N be defined by M=

x(τ (t)) x(τ (t)) , N = min . τ2 (t)≤t≤τ (t) x(t) τ (t)≤t≤t x(t) min

Then N≥

⎧ ΔM ⎨e ⎩

when Δ >

1 e

min{f, eΔM } when Δ ≤

1 e

(2.4)

where f is defined by (2.1). Proof. From Eq.(1), we have x(τ (t)) x (t) = p(t) . − x(t) x(t) Integrating over [τ (t), t] for τ (t) ≤ t ≤ t, and using (2.3) and (2.4), we obtain  t   t x(τ (s)) x (s) x(τ (t)) − p(s) ds = log = ds ≥ x(t) x(s) τ (t) x(s) τ (t)  t p(s)ds ≥ Δ min{M, N }, ≥ min{M, N } τ (t)

and therefore

Hence

x(τ (t)) ≥ eΔ min{M,N } . x(t) N ≥ eΔ min{M,N } . 12

In the case where Δ > 1e we have min{M, N } = M, that is N ≥ eΔM , because otherwise N ≥ eΔN ≥ eΔN > N which is a contradiction. Also in the case where 0 < Δ ≤ have to consider the possibility min{M, N } = N. Then

1 e

we

N ≥ eΔN > 1 and N e−ΔN ≥ 1 and comparing the latter inequality with the definition of the function f (Δ), we see that f (Δ) ≤ N and therefore N ≥ min{f, eΔM },which was to be shown. The proof of the lemma is complete. Corollary 2.1 ([26]) If in Lemma 2.1 M ≥ f (Δ) then N ≥ f (Δ). Lemma 2.2 ([26]) Let the sequence of points {ti }ki=0 be such that ti = τ (ti+1 ) for i = 0, 1, ..., k − 1. Assume that x(t) is a solution of Eq.(1) such that x(t) > 0 on [τ3 (t0 ), tk ], p(t) ≥ 0 on [τ2 (t0 ), tk ] and  t p(s)ds ≥ Δ > 0 for t0 ≤ t ≤ tk . τ (t)

Let the sequence {Ni } be defined by Ni+1 =

x(τ (t)) , i = 0, 1, ...k. τ (ti )≤t≤ti x(t) min

Then Ni ≥ fi (Δ), i = 1, 2, ..., k + 1, where fi (Δ) is defined by (2.2). 13

(2.5)

Proof. Since x(t) > 0 on [τ3 (t0 ), tk ], it follows that x (t) ≤ 0 on [τ2 (t0 ), tk ] and by the definition of N1 , we have N1 ≥ 1 = f1 (Δ). When Δ > 1e we can apply Lemma 2.1 with t = t1 and we obtain M = N1 , N = N2 and N2 ≥ eΔN1 ≥ eΔf1 (Δ) = f2 (Δ) and, by induction, Ni ≥ fi (Δ), i = 1, 2, ..., k + 1. Similarly, when Δ ≤ f (Δ), we find

1 e,

taking into account that fi (Δ)
0 , A0 < 1.

Then there exists t∗ on (τ (

t),

t) such that √ 2  x(τ (t∗ )) 1 + 1 − A0 . < x(t∗ ) A0

14

(2.6)

Proof. Let



1 − A0 . A0 It is clear that 0 < λ < 1. Let t∗ be chosen so that  t p(s)ds = λA0 . λ=

1−

(2.7)

t∗

 t Such a t∗ exists because the function F (θ) = θ p(s)ds is continuous, F (

t) = 0 and F (τ (

t)) = A0 > λA0 . Hence t∗ ∈ t] and then (τ (

t),

t). Integrating Eq.(1) first over the interval [t∗ ,

∗ over [τ (

t), t ] and, taking into account (2.7) and the fact that t),

t], we obtain x(t) is nonincreasing on [τ2 (

 t  t ∗ x(t )−x(

t) = p(s)x(τ (s))ds ≥ x(τ (

t)) p(s)ds = λA0 x(τ (

t)), t∗

x(τ (

t)) − x(t∗ ) = ∗

≥ x(τ (t ))



t∗ τ (

t)

t∗



t∗ τ (

t)

p(s)x(τ (s))ds ≥

p(s)ds = (1 − λ)A0 x(τ (t∗ )).

Combining the last two inequalities, we find x(t∗ ) >

λ(1 − λ) 2 A x(τ (t∗ )) 1 − λA0 0

from which we obtain (2.6). The proof is complete. Since the function of A0 on the right hand side of (2.6) is strictly decreasing, we can extend Lemma 2.3 for the cases where  t p(s)ds > A0 , 0 < A0 < 1. τ (

t)

15

Thus we derive the following. Corollary 2.2 ([26]) Assume that Eq.(1) has a positive sot),

t], p(t) ≥ 0 on [τ2 (

t),

t], lution x(t) on [τ3 (

 t p(s)ds ≥ A0 > 0 , A0 < 1, τ (

t)

and

x(τ (t)) . τ (t)≤t≤t x(t)

N = min Then

 N
1e is the same, while the case 0 < Δ ≤ 1e is a new result. As far as the upper bound is concerned, in [42] it is given by  2 x(τ (t)) 2 , ≤ x(t) A0 while the upper bound in (2.6) is smaller. Under the assumptions of Lemma 2.3, we have √ 2  x(τ (t∗ )) 1 + 1 − A0 1< < x(t∗ ) A0 and the upper bound can not be essentially improved at least in the case when A0 is near to 1, because for A0 = 1 is is equal to 1. Of course when A0 is very small a further improvement may be possible. 16

Theorem 2.1 ([26]) Assume that when 1 0 1,

(2.9)

where f (a) is the unique root of the equation f e−af = 1 in the interval [1, e] for a ∈ [0, 1e ]. Then all solutions of Eq.(1) oscillate. Proof. Suppose the contrary. Then, we may assume, without loss of generality, that the solution x(t) is eventually positive. By (2.9) and the continuity of the function f (a) we can choose Δ and A0 so that the inequalities 0 < Δ < a, 0 < A0 < A and f (Δ)(1 − 1 − A0 )2 > 1 (2.10) hold. By the definition of a, for sufficiently large TΔ , we have  t p(s)ds ≥ Δ, t > TΔ . (2.11) τ (t)

Moreover we may suppose x(t) > 0 on [τ3 (TΔ ), ∞) and p(t) ≥ 0 on [τ2 (TΔ ), ∞). Also, by the definition of A, there are infinitely many {t j }∞ j=1 , tj > TΔ , tj → ∞ as j → ∞ such that  t j

τ (t j )

p(s)ds ≥ A0 .

(2.12)

Let’s choose t j sufficiently large and define k = k(j) by τk (t j ) > TΔ and τk+1 (t j ) ≤ TΔ . 17

Observe that k → ∞ as t j → ∞ since τk (t) → ∞ as t → ∞. Now, let t0 , t1 , ..., tk be defined by ti = τk−1 (t j ), i = 0, 1, ..., k, so we have tk = t j and ti = τ (ti+1 ), i = 0, 1, ..., k − 1. In view of (2.11), applying Lemma 2.2, we obtain Nk+1 =

min

τ ( t j )≤t≤ t j

x(τ (t)) ≥ fk+1 (Δ) x(t)

and, in view of (2.12), by Corollary 2.2, we have √ 2  1 + 1 − A0 . Nk+1 < A0 The last two inequalities yield fk+1 (Δ)(1 − 1 − A0 )2 < 1. Recall that when Δ ≤ tj → ∞, we obtain

1 e

f (Δ)(1 −

limk→∞ fk (Δ) = f (Δ) and letting

1 − A0 )2 ≤ 1,

which contradicts (2.10). The proof is complete. Example 2.1 ([26]) Consider the differential equation x (t) +

0.85 π √ (2a + cos t)x(t − ) = 0, 2 aπ + 2 18

with a = 1.137. We have p(t) = and



t

p(s)ds = t−π/2

Thus

0.85 √ (2a + cos t) aπ + 2 √ 0.85 π √ (aπ + 2 cos(t − )). 2 aπ + 2

√ 1 aπ − 2 √ = 0.367837 < a = lim inf p(s)ds = 0.85 t→∞ e aπ + 2 t−π/2 

t

and

 A = lim sup t→∞

t

p(s)ds = 0.85. t−π/2

We find f (a) = 2.6775 and √ f (a)(1 − 1 − A)2 = 1.0051... > 1. Hence, the conditions of Theorem 2.1 are satisfied and therefore all solutions of the above equation oscillate. Observe, however, that none of the conditions (C1 ) − (C6 ) is satisfied. In 1991 Kwong [47], using different techniques, improved (C4 ) to the condition ln λ1 + 1 A> , (C8 ) λ1 where λ1 is the smaller real root of the equation λ = eaλ . In 1994, Koplatadze and Kvinikadze [43] improved (C6 ), while in 1998, Philos and Sficas [70] and in 1999, Zhou and Yu [106] and Jaroˇs and Stavroulakis [38] derived the conditions A>1−

a2 a2 − λ1, 2(1 − a) 2 19

(C9 )

and



1 1 − 2a − a2 − (1 − √ )2 , 2 λ1 √ ln λ1 + 1 1 − a − 1 − 2a − a2 A> − , λ1 2

A>1−

1−a−

(C10 )

(C11 )

respectively. More precisely the main results in [38] are as follows: Set x(τ (t)) w(t) = x(t) We begin with the preliminary analysis of asymptotic behavior of the function ω(t) for a possible nonoscillatory solution x(t) of Eq.(1) in the case that a ≤1/e. For this purpose, assume that Eq.(1) has a solution x(t) which is positive for all large t. Dividing first Eq.(1) by x(t) and then integrating it from τ (t) to t leads to the integral equality  t p(s)w(s)ds, (2.13) w(t) = exp τ (t)

which holds for all sufficiently large t, say for t ≥ T1 , where both x(t) and x(τ (t)) are positive on [T1 , ∞). Lemma 2.4 ([38]). Suppose that a > 0 and Equation (1) has an eventually positive solution x(t). Then a ≤ 1/e and λ1 ≤ lim inf w(t) ≤ λ2 , t→∞

where λ1 and λ2 are the roots of the equation λ = eaλ . Proof. Let α = lim inf t→∞ w(t). From (2.13), we have  t w(t) = exp p(s)w(s)ds τ (t)

20

for sufficiently large t. This obviously implies that α ≥ exp aα which is impossible if a > 1/e, since a simple calculus argument shows that, in the case, λ < eaλ for all λ. This implies that Eq.(1) has no eventually positive solution if a > 1/e. On the other hand, if 0 < a ≤ 1/e, then λ = eaλ has roots λ1 ≤ λ2 , with equality λ1 = λ2 = e if and only if a = 1/e, and α ≥ eaα if and only if λ1 ≤ α ≤ λ2 . The next lemma is taken from [100] and it gives an upper bound for the function w(t) as t → ∞. Lemma 2.5 ([100]) Let 0 < a ≤ 1/e and x(t) be an eventually positive solution of Equation (1). Then lim sup w(t) ≤ t→∞

1−a−



2 . 1 − 2a − a2

Lemma 2.6 ([38]) Let 0 < a ≤ 1/e and let x(t) be an eventually positive solution of Equation (1). Then A≤

1 + ln λ1 − M, λ1

where λ1 is the smaller root of the equation λ = eaλ and M= lim inf t→∞

x(t) . x(τ (t))

Proof. Let θ be any number in (1/λ1 , 1). From Lemma 2.4 and the definition of M, there is a T1 > T such that x(τ (t)) > θλ1, t ≥ T1, x(t) 21

(2.14)

and

x(t) > θM, x(τ (t))

t ≥ T1 .

(2.15)

Now let t ≥ T1 . Since the function g(s) = x(τ (t))/x(s) is continuous, g(τ (t)) = 1 < θλ1 and g(t) > θλ1 , there is a t∗ (t) ∈ (τ (t), t) such that x(τ (t)) = θλ1 . x(t∗ (t)) Deviding (1) by x(t), integrating from τ (t) to t∗ (t), and taking into account (2.14), yields  t∗ (t)   t∗ (t) ln(θλ1 ) 1 x (s) ds = p(s)ds ≤ − . (2.16) θλ1 τ (t) x(s) θλ1 τ (t) Integrating (1) over [t∗ (t), t] and using (2.15) and the fact that x(τ (s) ≥ x(τ (t)) if s ≤ t, yields  t x(t) 1 1 x(t∗ (t)) x(t) − = ≤ p(s)ds ≤ − −θM x(τ (t)) x(τ (t)) θλ1 x(τ (t)) θλ1 t∗ (t) (2.17) Adding (2.16) and (2.17) yields  t 1 + ln(θλ1 ) p(s)ds ≤ − θM. θλ1 τ (t) Letting t → ∞ yields A≤

1 + ln(θλ1 ) − θM. θλ1

Letting θ → 1 completes the proof. The last two lemmas imply the following 22

Theorem 2.2 ([38]). Consider the differential equation (1) and assume that when A < 1 and 0 < a ≤ 1/e the following condition holds √ ln λ1 + 1 1 − a − 1 − 2a − a2 A> − , (C9 ) λ1 2 where λ1 is the smaller root of the equation λ = e aλ . Then all solutions of Equation (1) oscillate. Example 2.2 ([38]). Consider the delay differential equation x (t) +

0.6 π √ (2b + cos t)x(t − ) = 0, 2 bπ + 2

√ where b = ( 2(0.6e + 1))/(π(0.6e − 1)). Then  t √ 1 0.6(2b + cos u)/(bπ + 2)du = lim inf t→∞ e t−π/2 and



t

lim sup t→∞

0.6(2b + cos u)/(bπ +



2)du = 0.6.

t−π/2

Thus, according to the Theorem 2.2, all solutions are oscillatory. We remark that none of the conditions (C1 ) − (C10 ) can be applied to this equation. Furthermore, consider Eq.(1) and assume that τ (t) is continuously differentiable and that there exists θ > 0 such that p(τ (t))τ  (t) ≥ θp(t) eventually for all t. Under this additional 23

condition, in 2000, Kon, Sficas and Stavroulakis [41], established the following. Lemma 2.7 ([41]) Let 0 < a ≤ 1/e and let x(t) be an eventually positive solution of Eq.(1). Assume that there exists ω > 0 such that  t  τ (t) p(s)ds ≥ ω p(s)ds for all τ (t)≤u≤t. (2.18) τ (u)

u

Then lim supw(t) ≤ t→∞

1−a−



2 (1 − a)2 − 4A

,

(2.19)

where A is given by A=

eλ1 ωa − λ1 ωa − 1 . (λ1 ω)2

(2.20)

Proof. Let t > t0 be large enough so that τ (t) > t0 and T1 ≡ T1 (t) > t, t1 ≡ t1 (t) > t be such that τ (T1 ) = t ,  t1 p(s)ds = δ, t

where δ : 0 < δ < a is arbitrarily close to a. Integrating (1) from t to t1 , we obtain  t1 x(t) = x(t1 ) + p(s) x (τ (s))ds, t

while integrating from τ (s) to t for s < t1 , we have  t p(u) x (τ (u)) du. x(τ (s)) = x(t) + τ (s)

24

Combining the last two equalities, we obtain

  t1  t x(t) = x(t1 ) + p(s) x(t) + p(u) x (τ (u)) du ds. τ (s)

t

(2.21) Let 0 < λ < λ1 . Then the function ϕ(t) = x(t) e

λ

t t0

p(s)ds

, t≥a

is decreasing for appropriate a≥t0 since x(t) is also decreasing. Indeed, by Lemma 2.4, x(τ (t)) > λ x(t) for all sufficiently large t, and consequently 



0 = x (t) + p(t)x(τ (t)) ≥ x (t) + λ p(t)x(t) 

which implies ϕ (t) ≤ 0 for sufficiently large t. Substituting into (2.21), we derive for sufficiently large t x(t)≥   t1 ≥x(t1 )+δx(t)+ϕ(τ (t)) p(s) t

= x(t1 ) + δx(t)+ +ϕ(τ (t)) e therefore

−λ

 τ (t) t0

p(s) ds

 t1 t

p(s) 



p(u) e

−λ

 τ (u) t0

 p(ξ) dξ

du ds

τ (s)

t τ (s)

p(u) e



t1

x(t)≥x(t1 )+δx(t)+x(τ (t))

t

λ

 τ (t) τ (u)

p(ξ)dξ

t

p(s)

p(u) e

λ

 du ds and

 τ (t) τ (u)

 p(ξ)dξ

du ds.

τ (s)

t

(2.22) 25

In view of (2.18), we obtain  t   τ (t) λ τ (u) p(ξ)dξ p(u) e du ≥ τ (s)

= Thus





t1

t

p(s) t

p(u) e τ (s)

δ 1 ≥ − + λω λω δ 1 = − + λω λω



t

t u

p(ξ)dξ

du

τ (s)

 1  λω  tτ (s) p(ξ)dξ −1 . e λω λ

 τ (t) τ (u)

t1

p(s)e 

p(u) eλω

 p(ξ)dξ

λω

t t1

p(s)e

λω

du ds ≥

t τ (s)

p(ξ)dξ

s τ (s)

ds =

p(ξ)dξ−λω

s t

p(ξ)dξ

ds≥

t

 s δ 1 λωδ t1 ≥− p(s)e−λω t p(ξ)dξ ds = + e λω λω t  λωδ  t e δ −λω t1 p(ξ)dξ + =− 1 − e + λω (λω)2  eλωδ  δ −λωδ 1 − e = + =− λω (λω)2 δ 1 =− (eλωδ − 1), + λω (λω)2 and (2.22) yields x(t) ≥ x(t1 ) + δx(t) + A∗ x (τ (t)), where A∗ =

eλωδ − λωδ − 1 . (λω)2 26

(2.23)

From (2.23), we have x(t) ≥ d1 x (τ (t)), where we have set d1 =

A∗ . 1−δ

Observe that x(t1 ) ≥ d1 x (τ (t1 )) ≥ d1 x(t), since x(t) is decreasing and therefore (2.23) yields x(t) ≥ d2 x (τ (t)), where

A∗ d2 = . 1 − d1 − δ Following this iterative procedure (cf. [83], [84]), we obtain x(t) ≥ dn+1 x (τ (t)),

where

A∗ , n = 1, 2, .... 1 − dn − δ It is easy to see that the sequence {dn } is strictly increasing and bounded. dn+1 =

Therefore limn→∞ dn = d exists and satisfies d2 − (1 − δ)d + A∗ = 0. Since {dn } is strictly increasing it follows that 1 − δ − (1 − δ)2 − 4A∗ . d= 2 27

Observe that for all large t (1 − δ)2 − 4A∗ 2

1−δ− x(t) ≥ x(τ (t))

and since 0 < δ < a is arbitrarily close to a by letting λ→λ1 the last inequality leads to (2.19). The proof is complete. Remark 2.2 ([41]) Assume that τ (t) is continuously differentiable and that there exists ω > 0 such that 

p(τ (t)) τ (t) ≥ ω p(t)

(2.24)

eventually for all t. Then it is easy to see that (2.24) implies (2.18). Indeed, the function  t  τ (t) p(s)ds − ω p(s)ds, τ (t)≤u≤t, v(u) = τ (u)

u

satisfies the conditions v(t) = 0, and





v (u) = −p(τ (u))τ (u) + ωp(u) ≤ 0. If p(t) > 0 eventually for all t and 

p(τ (t)) τ (t) liminft→∞ = ω0 > 0, p(t) then ω can be any number satisfying 0 < ω < ω0 . Besides the case p(t) ≡ p > 0, τ (t) = t − τ or the case τ (t) = t − τ and p(t) is τ − periodic, there exists a class of functions which satisfy (2.24). 28

Theorem 2.3 ([41]) Consider the differential equation (1) and let A < 1, 0 < a ≤ 1e and there exist ω > 0 such that (2.18) be satisfied. Assume that 1−a−

nλ1 + 1 A > − λ1



(1 − a)2 − 4A , 2

(2.25)

where λ1 is the smaller root of the equation λ = eaλ and A is given by (2.20). Then all solutions of Eq.(1) oscillate. Proof. Assume, for the sake of contradiction, that x(t) is an eventually positive solution of Eq. (1). Then, as in Lemma 2.6, we obtain

nλ1 + 1 A≤ − M, λ1 where x(t) . M = liminft→∞ x(τ (t)) The last inequality, in view of Lemma 2.7, contradicts (2.25). The proof is complete. Remark 2.3 ([41]) Observe that when ω = 1, then A =

λ1 − λ1 a − 1 λ21

and (2.25) reduces to A > 2a +

29

2 − 1. λ1

(C12 )

In the case that a = 1e , then λ1 = e and (C12 ) leads to A >

4 − 1 ≈ 0.471517764. e

Example 2.3 ([41]) Consider the delay differential equation   √ 1  2 = 0, x (t) + p x t − qsin t − pe where p > 0, q > 0 and pq = 12 − 1e . Then    t √ 1 1 2 = a = liminft→∞ pds = liminft→∞ p a sin t + pe e τ (t) and A = limsupt→∞





t τ (t)

pds = limsupt→∞ p qsin

2



1 t+ pe

 =

1 1 = . e 2 Thus, according to Remark 2.3, all solutions of this equation oscillate. Observe, however, that none of the conditions (C1 ) − (C11 ) can be applied to this equation. = pq +

In 2003, Sficas and Stavroulakis [71] established the following: Lemma 2.8 ([71]) Let 0 < a ≤ 1e and let x(t) be an eventually positive solution of Eq.(1). Assume that condition (2.24) is satisfied. Then √ ln λ1 −1 + 1 + 2ω − 2ωλ1 M A ≤ + , (2.26) λ1 ωλ1 30

where λ1 is the smaller root of the equation λ = eaλ and M = x(t) lim inf x(τ (t)) . t→∞

Proof. Let θ be any number in (1/λ1, 1). From Lemma 2.4 and the definition of M, there is a T1 > T such that (see Lemma 2.6) x(τ (t)) (2.14) > θλ1 , t ≥ T1 , x(t) and x(t) (2.15) > θM, t ≥ T1 . x(τ (t)) Now let t ≥ T1 . Since the function g(s) = x(τ (t))/x(s) is continuous, g(τ (t)) = 1 < θλ1 , and g(t) > θλ1 , there is t∗ ≡ t∗ (t) ∈ (τ (t), t) such that x(τ (t)) = θλ1 . x(t∗ )

(2.27)

Dividing Eq.(1) by x(t), integrating from τ (t) to t∗ and taking into account (2.14), yields  t∗   t∗ 1 x (s) ln(θλ1 ) p(s)ds ≤ − (2.28) ds = θλ1 τ (t) x(s) θλ1 τ (t) Next we try to find an analogous inequality for  t Λ := p(s)ds. t∗

Integrating (1) from τ (s) to τ (t), we have  τ (t) p(u)x(τ (u))du, x(τ (s)) − x(τ (t)) = τ (s)

31

t∗ ≤ s ≤ t.

Thus, integrating (1) from t∗ to t and using (2.18), we obtain  t ∗ p(s)x(τ (s))ds = x(t ) − x(t) = t∗





t

τ (t)

p(s)[x(τ (t)) +

= t∗

 ≥ x(τ (t))

t

2



t∗



t

p(s)ds + x(τ (t))[ 

≥ x(τ (t))

p(u)x(τ (u))du]ds τ (s)

t∗ t

2





p(u)du)ds] τ (s)



t

p(s)ds + ωx(τ (t)) t∗

τ (t)

p(s)( t

p(s)( t∗

p(u)du)ds s

 t ω 2 = x(τ (t)) p(s)ds + x(τ (t))( p(s)ds)2 2 t∗ t∗ ω 2 = Λx(τ (t)) + Λ x(τ 2 (t)), 2 2 where τ (t) ≡ τ (τ (t)). Therefore t

x(t∗ ) x(t) Λ2 x(τ 2 (t)) ≤ − Λ+ ω 2 x(τ (t)) x(τ (t)) x(τ (t)) and taking into account (2.27) and (2.15), this is less than or equal to 1 − θM. θλ1 Since, by (2.14), x(τ 2 (t)) > θλ1 , x(τ (t)) we obtain 1 Λ2 − θM Λ + ωθλ1 ≤ 2 θλ1 32

or Λ2

ωθλ1 1 ) ≤ 0, + Λ + (θM − 2 θλ1

which leads to  −1 + 1 − 2ωθλ1 (θM − Λ≤ ωθλ1

1 θλ1 )

=

−1 +

√ 1 + 2ω − 2ωθ2 λ1 M , ωθλ1

since the other root is negative. Adding (2.28) and the last inequality, we obtain √  t ln(θλ1 ) −1 + 1 + 2ω − 2ωθ2 λ1 M p(s)ds ≤ + . θλ1 ωθλ1 τ (t) By letting θ → 1 we complete the proof. Theorem 2.4 ([71]) Consider the differential equation (1) and let A < 1, 0 < a ≤ 1e and suppose there exists ω > 0 such that (2.24) is satisfied. Assume that √ ln λ1 −1 + 1 + 2ω − 2ωλ1 B A > + , (2.29) λ1 ωλ1 where λ1 is the smaller root of the equation λ = eaλ and 1 − a − (1 − a)2 − 4A B= 2 where A is given by (2.20). Then all solutions of Eq.(1) oscillate. Proof. Assume, for the sake of contradiction, that x(t) is an eventually positive solution of Eq.(1). Then, by Lemma 2.8, we obtain (2.26) which, in view of Lemma 2.7, contradicts (2.29). The proof is complete. 33

Remark 2.4 ([71]) It is clear that in the above theorem ω can be replaced by ω0 , where ω0 is given in Remark 2.2. Remark 2.5 ([71]) Observe that when ω = 1, then (2.29) reduces to √ ln λ1 − 1 + 5 − 2λ1 + 2aλ1 . (C13 ) A> λ1 since from [41] it follows that 1 B =1−a− . λ1 In the case that a = 1e , then λ1 = e, and (C13 ) leads to √ 7 − 2e A> ≈ 0.459987065. e Remark 2.6 ([71]) It is to be noted that as a → 0, then all the previous conditions (C4 )-(C12 ) reduce to the condition (C2 ), i.e. A > 1. However condition (C13 ) leads to √ A > 3 − 1 ≈ 0.732 which is an essential improvement. Moreover (C13 ) improves all the above conditions when 0 < a ≤ 1e as well. Note that the value of the lower bound on A can not be less than 1 ≈ 0.367879441. e Thus the aim is to establish a condition which leads to a value as close as possible to 1e . For illustrative purpose, we give the values of the lower bound on A under these conditions when a = 1e : 34

(C4 ): (C5 ): (C6 ): (C7 ): (C8 ): (C9 ): (C10 ): (C11 ): (C12 ):

0.966166179 0.892951367 0.863457014 0.845181878 0.735758882 0.709011646 0.599215896 0.471517764 0.459987065

We see that condition (C13 ) essentially improves all the previous conditions (C4 ) − (C12 ). Example 2.4 ([71]) Consider the delay differential equation   √ 1  2 = 0, x (t) + p x t − q sin t − pe where p > 0, q > 0 and pq = 0.46 − 1e . Then    t √ 1 1 2 = a = liminft→∞ pds = liminft→∞ p a sin t + pe e τ (t) and A = limsupt→∞





t τ (t)

pds = limsupt→∞ p q sin

2



1 t+ pe

 =

1 = 0.46. e Thus, according to Remark 2.5, all solutions of this equation oscillate. = pq +

35

Observe, however, that none of the conditions (C1 ) − (C12 ) can be applied to this equation. Following this historical (and chronological) review we also mention that in the special case (called critical state) where  t  t 1 1 and lim p(s)ds ≥ p(s)ds = t→∞ τ (t) e e τ (t) this problem has been studied by several authors. See the next section.

36

EXERCISES 1. Show that the conditions (C1 ) − (C6 ) are not satisfied for the equation in Example 2.1. 2. Explain why none of the conditions (C1 ) − (C1 0) can be applied to the equation in Example 2.2. 3. Show that the inequality (2.25) reduces to (C1 2) in Remark 2.3. 4. Explain why none of the conditions (C1 ) − (C1 1) can be applied to the equation in Example 2.3. 5. Show that the inequality (2.29) reduces to (C1 3) in Remark 2.5. 6. Show the validity of the inequalities in Remark 2.6. 7. Explain why none of the conditions (C1 ) − (C1 2) can be applied to the equation in Example 2.4.

37

3

Oscillations in Critical State

In this section, we study the oscillation of all solutions to equation (1) x (t) + p(t)x(τ (t)) = 0, t ≥ t0 , or to the equation with constant delay of the form x (t) + p(t)x(t − τ ) = 0,

t ≥ t0 ,

(2)

where τ ∈ (0, ∞), the functions p(t), τ (t) ∈ C([t0, ∞), R+ ) (here R+ = [0, ∞)), τ (t) is nondecreasing, τ (t) < t for t ≥ t0 and limt→∞ τ (t) = ∞, in the critical case when  t  t 1 p(s)ds ≥ p(s)ds = and lim t→∞ τ (t) e τ (t)  t 1 1 p(s)ds = = or lim inf (3.1) t→∞ e e τ (t) or 1 1 lim p(t) = or lim inf p(t) = , (3.2) t→∞ t→∞ τe τe In 1986, Domshlak [15] first observed the following special critical situation: Among the equations of the form (2) with 1 , (3.3) t→∞ τe there exist equations such that their solutions are oscillatory in spite of the fact that the corresponding “limiting equation lim p(t) =

1 τ e x(t − τ ) = 0, t ≥ t0 , admits a non-oscillatory solution, namely x(t) = e−t/τ . x (t) +

38

(3.4)

Later, Elbert and Stavroulakis [27], Kozakiewicz [45], Li [60, 61], Tang and Yu [87], Yu and Tang [98], Yu and Wang [91] further investigated the oscillation of Eq.(1) or Eq.(2) in the critical case. In 1995, Elbert and Stavroulakis [27] investigated Eq.(1) in the case when we have 



t

1 p(s)ds ≥ e τ (t)

and lim

t→∞

t

1 p(s)ds = . e τ (t)

(3.5)

In connection with the delay function τ (t) in Eq.(1) we suppose that τ (t) is strictly increasing on [t0 , ∞), limt→∞ τ (t) = ∞, and its inverse is τ−1 (t) (τ−1 (t) > t). Let τ−k (t) be defined on [t0 , ∞) by τ−k−1 (t) = τ−1 (τ−k (t)) for k = 1, 2, ..., and let tk = τ−k (t0 ),

k = 1, 2, ... .

Clearly tk → ∞ as k → ∞. The coefficient p(t) is assumed to be a piecewise continuous function and satisfies the relation  t 1 p(s)ds ≥ . e τ (t) Let ϕ(t) be a continuous function on [τ (t0 ), t0 ]. A function x(t) is a solution of Eq.(1), associated with the initial function ϕ(t), if x(t) = ϕ(t) on [τ (t0 ), t0 ], x(t) is continuous on 39

[τ (t0 ), ∞), is differentiable almost every-where on (t0 , ∞), and satisfies Eq.(1). Among the functions p(t) we define a set Aλ for 0 < λ ≤ 1 as follows. Definition. The piece-wise continuous function p(t) : [t0 , ∞) → [0.∞) belongs to Aλ if  t 1 p(s)ds ≥ , t ≥ t1 , e τ (t) 

t

1 p(s)ds ≥ +λk e τ (t)



tk+1 tk

 1 , p(s)ds − e

tk < t ≤ tk+1 , k = 1, 2, ...,

(3.6)

for some λk ≥ 0, and lim inf λk = λ > 0. k→∞

t

We remark that if 1 e,

t

τ (t) p(s)ds

is a nonincreasing function and

> then p(t) ∈ A1 , because we may have λk = 1 in (3.6). However, the monotonicity is not a necessary condition ; e.g., in the case τ (t) = t − 1 the function τ (t) p(s)ds

1 (3.7) + (K sin2 πt/ta ), K > 0 and 0 ≤ a ≤ 2, e t belongs to A1 because t−1 (sin2 πs/sa )ds is a nonicreasing function. p(t) =

The following two lemmas have origin in [42] (see also [26]). 40

Lemma 3.1 ([27]) Assume that x(t) is a positive solution of Eq.(1) on [tk−2 , tk+1 ] for some k ≥ 2. Let N be defined by N=

x(τ (t)) tk ≤t≤tk+1 x(t) min

Then N < (2e)2 . Proof. Let L be the integral  tk+1 1 p(s)ds ≥ . L= e tk √ By Lemma 2.3, we obtain N < ((1 + 1 − L)/L)2 . Since the right-hand side is a decreasing function of L, we get N < ((1 + 1 − (1/e))/L)2 < (2e)2 . Lemma 3.2 ([27]) Assume that x(t) is a positive solution of Eq.(1) on [tk−3 , tk+1 ] for some k ≥ 3. and p(t) ∈ Aλ . Let M, N be defined by M= Then

x(τ (t)) , tk−1 ≤t≤tk x(t)

N=





min

1 + λk M > 1 and N ≥ exp M e

x(τ (t)) . tk ≤t≤tk+1 x(t) min

tk+1 tk

1 p(s)ds − e

 ≥ M.

Proof. Following the lines of the proof of Lemma 2.1, we have min{M, N } = M and, by (3.6), for tk ≤ t ≤ tk+1  t x(τ (t)) p(s)ds) ≥ ≥ exp(M x(t) τ (t) 41



  tk+1 1 1 ≥ exp M p(s)ds − + λk , e e tk which implies the inequality concerning N. On the other hand the solution x(t) is a strictly decreasing function on [tk−2 , tk+1 ] Hence x(τ (t))/x(t) > 1 on [tk−1 , tk ], and therefore M > 1. The proof of the lemma is complete. The next lemma deals with some properties of the following sequence. Let the sequence {ri }∞ i=0 be defined by the recurrence relation r0 = 1,

ri+1 = er1 /e for i = 0, 1, 2, ... .

(3.8)

Lemma 3.3 ([27]) For the sequence {ri }∞ i=0 in (3.8) the following relations hold: (a) ri < ri+1; (b) ri < e; (c) limi→∞ ri = e; (d) ri > e − 2e/(i + 2); Proof. The first two relations can be proved by induction. As a consequence of (a) and (b) the limi→∞ ri = r exists and it is finite. Then, by (3.8), we have r = er/e . It is easy to check that ex/e > x for x = e. This inequality implies that the limit r equals e. 42

(3.9)

Now we give the proof of (d). For i = 0 and i = 1 it is immediate. For i ≥ 1 the proof goes by induction, so we have ri+1 = eri /e > e1−2/(i+2) and it is sufficient to show e1−2/(i+2) > e − or f (x) = e−2/x +

2e , i+3

2 > 1 for x = i + 2. x+1

Since f  (x) =

2 −1/x x x −1/x (e + − )(e ) x2 x+1 x+1

and

1 x+1 = , x x we have f  (x) < 0 and f (x) > limx→∞ f (x) = 1, which was to be shown. The proof of the lemma is complete. e1/x > 1 +

Theorem 3.1 ([27]) Assume that the function p(t) in Eq.(1) belongs to Aλ for some λ ∈ (0, 1] and  ∞  ti  1 p(s)ds − ) = +∞, (3.10) e t i−1 i=1 Then every solution of Eq.(1) oscillates. Proof. Suppose the contrary. Then we may assume, without loss of generality, that there exists a solution x(t) such that

43

x(t) > 0 for t ≥ tk−3 for some k ≥ 3. Let the sequence {Ni }∞ i=0 be defined by x(τ (t)) min Ni = . (3.11) tk+i−1 ≤t≤tk+i x(t) By Lemma (3.2) we have N0 > 1 and    tk+i+1   1 Ni exp Ni λk+i ≥ Ni ; Ni+1 ≥ exp p(s)ds − e e tk+i (3.12) is nondecreasing. On the other therefore the sequence {Ni }∞ i=0 hand, by Lemma 3.1, it is bounded. Consequently the sequence converges. Let lim Ni = N. i→∞

Then (3.12) implies N ≥ exp(N/e). Hence by (3.9) we have N = e and 1 < N0 < N1 < ... < e.

(3.13)

From (3.12), in the view of (3.9), we obtain    tk+i+1 1 . Ni+1 ≥ Ni 1 + Ni λk+i p(s)ds − e tk+i Thus Ni+1 − Ni >

Ni2 λk+i



tk+i+1 tk+i

 1 . p(s)ds − e

(3.14)

From the definition of Aλ we know that λ = lim inf k→∞ λk > 0, so for any sufficiently small ε > 0 there exists a value κε such that λk+i > λ − ε for k + i > κε . 44

Thus, for such i’s from (3.14) and (3.13), we have 

 1 − ε) p(s)ds − Ni+1 − Ni > , e tk+i   tk+i+2 1 2 p(s)ds − Ni+2 − Ni+1 > Ni+1 (λ − ε) e tk+i+1   tk+i+2 1 . ≥ Ni2 (λ − ε) p(s)ds − e tk+i+1 .. . Ni2 (λ

tk+i+1

Summing up the inequalities above, we obtain   ∞ tk+j+1  1 e − Ni > Ni2 (λ − ε) for k + i ≥ κε . p(s)ds − e t k+j j=1 (3.15) The last inequality contradicts assumption (3.10). The proof is complete. In the next theorem we consider the case where the sum in (3.10) is convergent. Theorem 3.2 ([27]) Assume that p(t) ∈ Aλ , for some 0 < λ ≤ 1 and either  ∞  ti  2 1 > , p(s)ds − λ lim sup k (3.16) e e k→∞ ti=1 i=k

or λ lim inf k k→∞

∞   i=k

ti

1 p(s)ds − e ti=1

 >

Then every solution of Eq.(1) oscillates. 45

1 . 2e

(3.17).

t Note. If the function τ (t) p(s)ds is monotone, then the value of λ in conditions (3.16) and (3.17) of Theorem 3.2 is equal to one. Proof. Suppose the contrary. Then, as in the proof of Theorem 3.1, we have the sequence {Ni }∞ i=0 such that inequalities (3.12)-(3.15) hold. In particular, from (3.12) we have Ni+1 ≥ exp(Ni /e). Comparing the last inequality with (3.8), we obtain by induction N0 > r0 = 1,

Ni > r i

for i = 1, 2, ... .

Then, by Lemma 3.3(d), we have e − Ni < e − ri < 2e/(i + 2).

(3.18)

Multiplying (3.15) by k + i we obtain   ∞ tj+1  2e 1 (k + i) p(s)ds − > Ni2 (λ − ε)(k + i) i+2 e tj i=k+i

for k+i≥ κε . Taking the limit as i → ∞ we get   ∞ tj+1  1 , 2e ≥ e2 λ lim sup k p(s)ds − e k→∞ tj i=k

which contradicts (3.16). Now let A be defined by A = lim inf k k→∞

 ∞ 

tj+1 tj

i=k

46

p(s)ds −

1 e

 .

If A = ∞, then, by (3.16), every solution oscillates. Therefore we consider the case 0 < A < ∞. So for any sufficiently small ˆ = λ − ε > 0 and ε > 0 there exists a value κ ˆ ε such that for λ Aˆ = A − ε > 0.   ∞ tj+1  1 Aˆ ˆ and > p(s)ds − λk > λ for k ≥ κ ˆ ε . (3.19) e k tj j=k

If we use the inequality   x 1 ξ x 2 exp > x + exp 1− e 2 e e

for ξ < x < e.

In (3.12) we obtain for Ni > ξ and k + i > κ ˆε    tk+i+1   1 Ni ˆ exp Ni λ Ni+1 ≥ exp p(s)ds − e e tk+i  2    Ni 1 ξ 1− × > Ni + exp 2 e e    tk+i+1 1 ˆ × 1 + Ni λ p(s)ds − . e tk+i Consequently  tk+i+1   2  ξ 1 1 Ni 2ˆ Ni+1 − Ni > exp +ξ λ p(s)ds − 1− 2 e e e tk+i and summing up,   2   ∞  ∞ tj+1  Ni 1 1 ξ ˆ 1− e−Ni > exp +ξ 2 λ p(s)ds − 2 e j=1 e e tj j=k+i

47

or   2 ∞  ˆ Aˆ Ni ξ 1 ξ 2λ 1− . e − Ni > exp + 2 e j=1 e k+i

(3.20)

In particular the last inequality yields e − Ni > U0 /(k + i),

ˆ A. ˆ U0 = ξ 2 λ

By iteration we can improve this inequality to e − Ni >

Un , k+i

n = 0, 1, 2, ... .

(3.21)

Namely, by (3.20), we have 2   ∞  ˆ Aˆ Un 1 ξ 2λ ξ + exp e − Ni > 2 e j=1 e(k + j) k+i   ˆ Aˆ 1 Un2 ξ ξ 2λ Un+1 > exp + = , 2 2e e k+i k+i k+i where Un+1

  Un2 ξ ˆ A, ˆ n = 0, 1, 2, .... + ξ 2λ = 2 exp 2e e

(3.22)

From this it is clear that the sequence {Un }∞ n=0 is increasing. Moreover, comparing inequalities (3.18) and (3.21) we see that Un ≤ 2e. Therefore the sequence has a limit, say U, which satisfies the equation   U2 ξ ˆ A. ˆ + ξ 2λ U = 2 exp 2e e 48

This is a quadratic equation with real roots and therefore the descriminant is not negative; i.e., ˆ Aˆ ≥ 0. 1 − 2eξ/e−2 ξ 2 λ Let ε → 0 and ξ → ε. Then the last inequality becomes 1 − 2eλA ≥ 0, which contradicts (3.17). The proof of the theorem is complete. In the following theorem we give a criterion for nonoscillation. Theorem 3.3 ([27]) Let τ (t) = t − 1, p(t) = 1e + a(t), and t0 = 1 in Eq.(1); i.e., it has the form

 1  x (t) + + a(t) x(t − 1) = 0, t ≥ 1. (3.23) e Assume that

a(t) ≤ 1/8et2 . √ Then (3.23) has a solution x(t) ≥ te−t .

Proof. The proof is based on known comparison theorems (see Myshkis [66] or Elbert [25]). Let the functions A(t), B(t), C(t) be defined as 1 A(t) = + a(t), e 1 1 B(t) = , + e 8et2 1 1− 1 C(t) =  2t , t > 1. e 1− 1 t

49

By the assumption we have A(t) ≤ B(t). We are going to show that the inequality B(t) < C(t) also holds. Namely, for θ = 2t1 ∈ (0, 12 ), we have θ3 ( 12 θ2 − 14 θ + 2)

√ C(t) − B(t) = √ > 0. e 1 − 2θ[1 − θ + (1 + 12 θ2 ) 1 − 2θ] Now we will compare the differential equations x (t) + A(t)x(t − 1) = 0, z  (t) + B(t)z(t − 1) = 0, u (t) + C(t)u(t − 1) = 0. √ Let us observe that the function u(t) = te−t is a solution of the last differential equation. Let the initial function ϕ(t) be the √ −t function te on [0, 1], and let x(t) and z(t) be the solutions of the first and the second differential equations respectively, associated with this initial function ϕ(t). Then by the comparison theorems mentioned above we have √ x(t) ≥ z(t) > u(t) = te−t for t > 1. which was to be shown. The proof is complete. Note. It is to be emphasized that in Theorem 3.3 we require neither  t 1 p(t) ≥ 0 nor p(s)ds ≥ . e τ (t) Remark 3.1 ([27]) For (3.23) we have tk = k + 1 and   ∞ ∞  ti  1 1 = lim sup k lim sup k p(s)ds − a(t)dt ≤ . e 8e k→∞ k→∞ k ti−1 i=k

50

Now the question arises naturally whether or not the bounds in conditions (3.16) and (3.17) of Theorem 3.2 can be replaced by smaller ones. Remark 3.2 ([27]) Applying Theorems 3.1 ,3.2 we see that, under (3.7), Eq.(1) oscillates for any K > 0 if 0 ≤ a < 2 and for K > 1e if a = 2. On the other hand it has a nonoscillatory 1 if a = 2. solution for K < 8e The question arised, by Elbert and Stavroulakis in Remark 3.1, stimulated the following work. In 2000 Tang and Yu [87] proved the following theorem. Theorem 3.4 ([87]) Assume that τ (t) is strictly increasing on [t0 , ∞), (3.6) holds, and that   t  ∞ 1 1 dt > 2 . lim sup k p(t) p(s)ds − (3.24) e e k→∞ tk τ (t) Then all solutions of Eq.(1) oscillate. Remark 3.3 reduces to

It p(t) ∈ Aλ for some λ ∈ (0, 1], then (3.24)

λ lim sup k k→∞

∞   i=k

ti

1 p(s)ds − e ti−1



1 > , e

(3.25)

which shows that the right-hand side of (3.16) can be replaced by 1/e which is less than the original 2/e. In 2002 Yu and Tang [98] proved the following theorems. 51

Theorem 3.5 ([98]) (i) Assume that   2    t t 1 1 lim inf p(s)ds − p(s)ds > 3. t→∞ e 8e τ (t) t0 Then all solutions of Eq.(1) oscillate. (ii) Assume that (3.6) holds and    t 2  t 1 1 < 3. p(s)ds − p(s)ds lim sup e 8e t→∞ τ (t) t0

(3.26)

(3.27)

Then Eq.(1) has an eventually positive solution. Theorem 3.6 ([98]) Assume that (3.6) holds and p(t) ≡ 0 on any subinterval of [t0 , ∞) and    s   2  ∞ t 1 1 ds > 3 . p(s)ds p(s) p(ξ)dξ − lim inf t→∞ e 8e t0 t τ (s) (3.28) Then all solutions of Eq.(1) oscillate. Theorem 3.7 ([98]) Assume that (3.28) holds and  t 1 p(s)ds > for sufficiently large t. (3.29) e τ (t) Then all solutions of Eq.(1) oscillate. Corollary 3.1([98]) Assume that (3.29) holds and τ (t) is strictly increasing on [t0 , ∞). If   s  ∞ 1 1 ds > 2 , lim inf k p(s) p(ξ)dξ − (3.30) k→∞ e 8e tk τ (s) then all solutions of Eq.(1) oscillate. 52

Corollary 3.2 ([98]) Assume that τ (t) is strictly increasing on [t0 , ∞) and p(t) ∈ Aλ for some λ ∈ (0, 1], and that  ∞  ti  1 1 > . λ lim inf k p(s)ds − (3.31) k→∞ e 8e ti−1 i=k

Then all solutions of Eq.(1) oscillate. Remark 3.4 ([98]) The following example shows that 1/8e in (3.31) is the best possible. Thus, Theorem 3.4 and 3.7 or Corollary 3.1 or 3.2 answer the question raised in Remark 3.1 (See also Remark 3.2) Example 3.1 ([98]) Consider the delay differential equation     1 1 C t  = 0, t ≥ e, (3.32) x + x (t) + e ln 2 t t(ln t)1+α 2   where C, α > 0. Here τ (t) = 2t , p(t) = e ln1 2 1t + t(ln Ct)1+α and 



1 1 C 1 , p(s)ds = + − e αe ln 2 (ln t − ln 2)α (ln t)α τ (t) t

(3.33)

  t 2 1 p(s)ds − p(s)ds e τ (t) e

 1 C 1 × = − α(e ln 2)3 (ln t − ln 2)α (ln t)α  2

1 C 1− × ln t − 1 + , α (ln t)α ⎧   t  t 2 ⎨ C/e3 (ln 2)2 , α = 1, 1 p(s)ds − p(s)ds = 0, α > 1, lim t→∞ ⎩ e τ (t) e ∞, α < 1. 

t

53

By Theorem 3.5, every solution of Eq.(3.32) oscillates when α < 1 or α = 1 and C > (ln 2)2 /8 ,but Eq.(3.32) has an eventually positive solution when α > 1 or α = 1 and C < (ln 2)2 /8. Remark 3.5 ([98]) When α = 1, condition (3.33) implies that p(t) ∈ Aλ for λ = 1. Let t1 = e, tn = 2tn−1 = 2n e, n = 1, 2, · · · . Then  ∞  ti  1 = p(s)ds − λ lim k k→∞ e ti−1 =

c lim k e k→∞

∞ 

i=k

1 (i ln 2 + 1)(i ln 2 + 1 − ln 2) i=k  c > 1/8e, c > (ln 2)2 /8, = e(ln 2)2 < 1/8e, c < (ln 2)2 /8.

This shows that

1 8e

in condition (3.31) is the best possible.

Since the method which developed in [15] (Sturmian Comparison Method) is very effective in this critical case it was continued in [18] as well. In [24] the said method is followed which enables us to establish sufficient (and unimprovable in some sense) conditions for the oscillation of all solutions of Eq.(2) and also to obtain important information about: (a) The estimates of the length of the intervals between successive zeros. (b) The sets G = ∪(ai , bi ) ⊂ (t0 , ∞) such that the oscillatory properties are determined by the behavior of a(t) on G only and, therefore outside of G the coefficient a(t) may be arbitrary, i.e. in particular may be oscillatory. 54

The purpose in [24] is to describe a class (or classes) of equations of the form (2) with oscillatory solutions only under the condition 1 lim inf p(t) = , (3.2) k→∞ τe and also to investigate the above problems (a) and (b) for those equations. The following theorem has been given in [15] in a more general form. We state it here in a particular case with its proof so that our principal idea (Comparison Theorem) will become clear. Theorem 3.8 ([24]) Let ϕ(t) > 0 be a continuous function on (p − τ, q + τ ), q − p > τ such that  t  q π ϕ(t)dt = π and ϕ(s)ds < for t ∈ (p, q +τ ), (3.34) 2 p t−τ and on (p, q + τ ) p(t+τ ) ≥

ϕ(t)

sin

 t+τ t

ϕ(s)ds





t+τ

exp(−

s+τ

ϕ(s)(cot t

ϕ(ξ)dξ)ds) ≡

s

≡ b(t + τ ).

(3.35)

Then every solution of Eq.(2) has at least one zero on the interval (p − τ, q). Proof. Let p(t), b(t), x(t), y(t) be arbitrary continuous functions and x(t), y(t) be differentiable on (p, q), q − p > τ. Let

[x] ≡ x (t) + p(t)x(t − τ ), ˜ ≡ −y  (t) + b(t + τ )y(t + τ ).

[y] 55

It is easy to see that the following equality holds  q  q ˜ x(t) [y]dt + [p(t) − b(t)]x(t − τ )y(t)dt + p p+τ  q+τ  p+τ p(t)x(t − τ )y(t)dt − b(t)x(t − τ )y(t)dt + p q  q y(t) [x]dt − x(q)y(q) + x(p)y(p). (3.36) = p

Now define the function y(t) as follows  s+τ    t  t ϕ(s) cot ϕ(ξ)dξ)ds sin ϕ(s)ds. y(t) = exp − p

s

p

Then, in the view of (3.34), y(p) = y(q) = 0, y(t) > 0 on ˜ = 0. Now to prove the (p, q), y(t) < 0 on (q, q + τ ) and [y] theorem it suffices to show that Eq.(2) does not have a positive solution on (p−τ, q). To this end suppose for contradiction, that

[x] = 0 has a positive solution x(t) > 0 on (p − τ, q). Then the right-hand side of (3.36) is equal to zero while the left-hand side is strictly positive. The proof is complete. From Theorem 3.8 the following corollaries are immediate. Corollary 3.3 ([24]) Let {(pn , qn )}, pn → ∞, qn − pn > τ be an arbitrary sequence of intervals and ϕ(t) > 0 be a continuous function defined on G := ∪∞ n=1 (pn − τ, qn + τ ) such that on each (pn , qn ) all conditions of Theorem 3.8 are satisfied. Then every solution of Eq.(2) is oscillatory and has at least one zero on each interval (pn − τ, qn ) for n sufficiently large. 56

∞ Corollary 3.4 ([24]) If ϕ(t) > 0 on (t0 , ∞), ϕ(t)dt = ∞ and the condition (3.35) holds on the whole axis, then every solution of Eq.(2) is oscillatory  q and has at least one zero on each interval (p − τ, q) for which p ϕ(t)dt ≥ π and p is sufficiently large. The following result has been established in [18], Corollary 4.5. Theorem 3.9 ([24]) Let {pn }∞ 1 , limn→∞ pn = ∞ be an π arbitrary sequence ν ∈ (0, 2 ) be an arbitrary number, G := π ∪∞ n=1 (pn − τ, pn exp v ) and 1 lim inf p(t) = , t∈G,t→∞ eτ 1 2 τ )t } = D > . (3.37) t∈G,t→∞ eτ 8e Then every solution of Eq.(2) is oscillatory and has at least one zero on each interval (pn − τ, pn exp πν ) for ν 2 < 14 ( 8eD τ − 1) and n sufficiently large. lim inf {(p(t) −

Note that for the equation

 1  x (t) + + a(t) x(t − 1) = 0, t > 1 e

(3.23)

1 it has been proved in Theorem 3.3 that if the equality D = 8e holds in (3.36), then there exists a non-oscillatory solution. In that sense the strict inequality in (3.36) can not be replaced by the equality. However, both results may be improved. For example, one can obtain 1 1 (1 + 2 ), a(t) ≤ 2 8et ln t

57

√ √ by taking x(t) = t ln te−t in place of x(t) = te−t in Theorem 3.3. We can also improve the above Theorem 3.9 if we take ϕ(t) =

ν t ln t

in place of ϕ(t) = νt in the proof of Corollary 4.5 ([16]). This is done in the next theorem. Theorem 3.10 ([24]) Assume that  

 1 τ 1 t2 = p(t) − lim inf p(t) = , lim inf t→∞ t→∞ τe τe 8e and

  lim inf t→∞

   1 τ τ 2 2 t − ln t = C > . p(t) − τe 8e 8e

(3.38)

Then all solutions of Eq.(1) oscillate. Proof. Take ϕ(t) =

ν , t ln t

0 , (3.38) p(t + τ ) − lim inf t→∞ τe 8e 8e 61

and combining (3.39) and (3.38) we see that the condition (3.35) of Theorem 3.8 is satisfied on (t0 , ∞). By the virtue of Corollary 3.4, the proof is complete. Note that the number ν is not involved in (3.37). On the other hand, the condition  q ϕ(t)dt = π p

is equivalent to



q p

π νds = π ⇔ q = pexp ν . s ln s

Therefore the next statement is true. Theorem 3.11 ([24]) Assume that the conditions of Theorem 3.10 are satisfied. Then every solution of Eq.(2) is oscillaπ tory and has at least one zero on each interval (p − τ, pexp ν ) for ν 2 < 41 ( 8eC τ − 1) and p sufficiently large. Here we have a very important Remark 3.6 ([168). It is not necessary to impose conditions on p(t) for all t ≥ t0 . Indeed, Theorem 3.8 has a local character. Thus, based on Corollary 3.3, we see that the behavior of p(t) outside of G has no influence on the oscillatory properties of Eq.(2). We can, therefore, impose conditions on p(t) over the set G only changing lim inf t→∞ to lim inf t∈G,t→∞ . 

Theorem 3.11 ([24]) Let {pn } and ν be as in Theorem 3.9 exp πν ¯ = ∪∞ (p − τ, p ). Assume and G n n n=1  

 1 τ 1 t2 = , p(t) − lim inf p(t) = , lim inf t∈G,t→∞ t∈G,t→∞ eτ eτ 8e 62

and

  lim inf

t∈G,t→∞

1 p(t) − eτ



  τ τ 2 t − ln t = C > . 8e 8e 2

(3.37)

Then every solution of Eq.(2) is oscillatory and has at least exp π one zero on each interval (pn − τ, pn ν ) for ν 2 < 14 ( 8eC τ − 1) and n sufficiently large. Example 3.2 ([24]) Consider the equation (cf. Theorem 3.3) x (t) + a(t)x(t − 1) = 0, t ≥ 1, (3.40) where a(t) =



(2t − 1) ln t − 1

t(t − 1) ln t ln(t − 1) √ It is easy to see that x(t) = t ln te−t is a (non-oscillary) solution of (3.40). In this case one can check that      1 2 1 1 2 t − ln t = a(t) − lim inf t→∞ e 8e 8e 2e

and therefore condition (3.38) of Theorem 3.10 (as expected) τ in (3.38) can not is not satisfied. Thus the inequality C > 8e be replaced by the corresponding equality. Therefore condition (3.38) is unimprovable in this sense (cf. Remarks 3.4, 3.5). Remark 3.7 ([24]) Observe that Theorem 3.9 is an analogue of the classical Kneser’s Theorem [40]. If 1 lim inf t2 a(t) > t→∞ 4 63

then all solutions of the second order ordinary differential equation x (t) + a(t)x(t) = 0, (3.41) oscillate. Furthermore, Theorem 3.10 is an analogue of the following improvement of Kneser’s Theorem due to Hille [35]: If 

1 1 2 ln2 t > lim inf t a(t) − t→∞ 4 4 then all solutions of (3.41) oscillate. Observe that in the above theorems the conditions are formulated in terms of the values of the  t function p(t) itself (type A) and not in terms of the average t−τ p(s)ds as in [26, 27, 42, 45, 48, 54, 60] (type B). Sometimes conditions of type A (see [18]) yield stronger results. In general, however, conditions of type B are preferable. In the next theorem we derive conditions of type B. Theorem 3.12 ([24]) Let there exists a function p0 (t) such that 0 ≤ p0 (t) ≤ p(t),  t+τ  ∞ p0 (s) p0 (s)ds is bounded and ds = ∞. s t Assume that lim inf t→∞ Φ(t) = 1 and D := lim inf {[Φ(t) − 1]t2 } > 0,

(3.42)

t→∞

where



t+τ

Φ(t) := tψ(t) exp t





p0 (s)ds , ψ(t) := sψ(s) 64



t+τ t

p0 (s) ds. s

Then all solutions of Eq.(2) oscillate. Proof. Observe that condition (3.35) is equivalent to   t   s+τ   ϕ(t − τ ) p(t) ≥ exp − ϕ(s) cot ϕ(ξ)dξ ds , t sin t−τ ϕ(s)ds t−τ s (3.35) and setting νp0 (t + τ ) ϕ(t) := (t + τ ) 

in (3.35) , we obtain

  t+τ  νp0 (t) νp0 (s) p(t) ≥ exp − cot(νψ(s))ds . (3.43) t sin(νψ(t)) s t We will show that (3.42) implies the following condition  t+τ νp0 (s) t cot(νψ(s))ds ≥ 1. F (t, ν) := sin(vψ(t)) exp ν s t It is obvious that  t+τ  t+τ t p0 (s)ds ≤ tψ(t) ≤ p0 (s)ds t+τ t t

and therefore there exists t1 > t0 such that tψ(t) ≤ N1 for t > t1 . Furthermore

t sin(vψ(t)) = tψ(t) 1 − ν

= tψ(t) 1 −

= tψ(t) 1 −

 ν 2 ψ 2 (t) cos(θνψ(t)) 6  ν 2 t2 ψ 2 (t) cos(θνψ(t)) t2 6 

 ν2 ν2 A(t) ≥ tψ(t) 1 − 2 M t2 t

65

on (t0 , ∞), since A(t) is bounded. Now define π 1 , (1 − z cot z) f or z ∈ (0, z ), z < 0 0 z2 2 and observe that this function is bounded on (0, z 0] since limz→0 g(z) = 13 . Therefore 1 0 < 2 (1 − z cot z) ≤ N2 z which implies that 1 cot z ≥ (1 − N2 z 2 ) on (0, z0 ]. z Setting z = νψ(s) for t < s < t + τ and t > t2 , where t2 is sufficiently large, we obtain ν ν cot(νψ(s)) ≥ [1 − N22 νψ 2 (s)] s svψ(s)

 1 ν 2 s2 ψ 2 (s) = 1 − N2 sψ(s) s2  

1 1 ν 2 N2 N12 ν 2 N2 N12 ≥ ≥ , 1− 1− sψ(s) s2 sψ(s) t2 and 

 t+τ  t+τ ν 2 N2 N12 νa0 (s) p0 (s)ds exp cot(νψ(s))ds ≥ 1 − . exp 2 s t sψ(s) t t g(z) :=

Finally,

   t+τ ν 2M ν 2 N2 N12 p0 (s)ds F (t, v) ≥ tψ(t) 1 − 2 1− exp 2 t t sψ(s) t  

2 2 2 ν N2 N1 ν 2 N3 ν M [1 − ] ≥ Φ(t)(1 − ), = Φ(t) [1 − 2 t t2 t2 t > t3 > t2, 66

where N3 > M + N2 N12 . Now, from (3.42), we obtain Φ(t) ≥ 1 +

D−e t2

and therefore    ν 2 N3 D − ε − ν 2 N3 D−e 1 − = 1+ +o(t−2 ). F (t, ν) ≥ 1 + t2 t2 t2 which implies (3.44) by choosing ν 2 < ND3 . It is easy to see that if (3.44) holds then, because 0 ≤ p0 (t) ≤ p(t), inequality (3.43) holds, and in view of Corollary 3.4, the proof is complete. Example 3.3 ([24]) Consider the differential equation x (t) + a(t)x(t − 1) = 0, where

t ≥ 1,

  C(sin2 πt − γ) 1 1+ a(t) = , C > 0. e t2

(3.45)

(3.46)

It is easy to see that the following asymptotic formulae hold.

  t+1 1 a(s)ds 1 2 + 3C − 6γC 1− + + o(t−2 ) t = 2 s e 2t 6t t a(t)

t

 t+1 a(s)ds t

s



t+1 t

1 12C sin2 πt − 6C − 1 =1+ + + o(t−2 ) 2 2t 12t a(s)ds 1 1 −2  s+1 a(ξ)dξ = 1 + − 2 + o(t ) 2t 3t s s

ξ

67

 exp t

t+1



5 1 a(s)ds + o(t−2 ).  s+1 a(ξ)dξ = e 1 + − 2 2t 24t s s

ξ

Thus

   2 + 3C − 6γC 5 1 1 + ... 1+ − + ... Φ(t) = 1 − + 2t 6t2 2t 24t2 4C − 1 − 8γC + o(t−2 ) = 1+ 8t2 and if 4C − 1 − 8γC > 0 then condition 3.42 is satisfied. There1 all solutions of Eq.(3.45) oscillate. fore for γ < 12 and C > 4(1−2γ) Remark 3.8 ([24]). In the particular case that γ = 0 (3.46) becomes C 1 K sin2 πt , where K = . a(t) = + 2 e t e In this case it is known (see Remark 3.3) that Eq.(3.45) oscillates for K > 1e . On the other hand it has a non-oscillatory 1 1 solution for K ≤ 8e . Obviously there is a gap between 8e and 1 . Regarding this see also Remarks 3.1, 3.4 and 3.5. Here we e show that the condition (3.42) of Theorem 3.12 is satisfied for 1 K > 4e . Therefore, we essentially improve the said gap. Also note that for 0 < γ < 12 the function a(t) − 1e defined by (3.46) may not be always positive. In 1998, Diblik [13], generalized Theorem 3.10 as follows: Set ln1 t = ln t, lnk+1 t = ln(lnk t), k = 1, 2, · · ·

68

Theorem 3.13 ([11]) (i) Assume that for an integer k ≥ 2 and a constant θ > 1  1 τ p(t) ≥ 1 + (ln1 t)−2 + (ln1 t ln2 t)−2 + · · · (3.47) + 2 τ e 8et  +(ln1 t ln2 t · · · lnk−1 t)−2 + θ(ln1 t ln2 t · · · lnk t)−2 , as t → ∞ Then all solutions of Eq.(1.2) oscillate. (ii) Assume that for a positive integer kp(t) ≤

1  1 1 + (ln1 t)−2 + (ln1 t ln2 t)−2 + · · · + + τ e 8et2 (ln1 t ln2 t · · · lnk t)−2 ,

(3.48)

as t → ∞. Then there exists a positive solution x = x(t) of Eq.(2). Moreover −t x(t) < e τ t ln t ln2 t... lnk t, as t → ∞. In 1999, Tang, Yu and Wang [91] established the following comparison theorem. Theorem 3.14 ([91]) Assume that for sufficiently large t p(t) ≥

1 . τe

(3.49)

Then all solutions of Eq.(2) oscillate if and only if all solutions of the following second order ordinary differential equation

69

  2e 1 y (t) + p(t) − y(t) = 0, τ τe oscillate. 

t ≥ t0

(3.50)

Employing this comparison theorem and a wealth of results on oscillation of Eq.(3.50), many interesting oscillation and nonoscillation criteria can be obtained. In 2000 Tang and Yu [88] established the following more general comparison theorem in the case when p(t) − τ1e is oscillatory t and lim inf t→∞ t−τ p(s)ds = 1e . Theorem 3.15 ([88]) Assume that r(t) ∈ C([t0 , ∞), [0, ∞)) is a τ -periodic function and satisfies the following hypothesis  t 1 r(s)ds ≡ . (3.51) e t−τ Suppose that p(t) − r(t) ≥ 0 for sufficiently large t.

(3.52)

Then all solutions of Eq.(2) oscillate if and only if the Riccati inequality ω  (t) + r(t)ω 2 (t) + 2e2 [p(t) − r(t)] ≤ 0,

t ≥ t0 ,

(3.53)

has no eventually positive solution. As an application of Theorem 3.15, the following theorem is also given in [88]. Theorem 3.16 ([88]) Assume that there is a τ -periodic function r(t) ∈ C([t0 , ∞), [0, ∞)) such that (3.51) and (3.52) hold. Then the following statements are valid. 70

(i) If





r(s)ds

lim inf t→∞



t t0

 (p(s) − r(s))ds >

t

1 , 8e2

then all solutions of Eq.(2) oscillate; (ii) If there exist a T ≥ t0 such that for t ≥ T  t  ∞ 1 r(s)ds (p(s) − r(s))ds ≤ 2 , 8e T t

(3.54)

(3.55)

then Eq.(2) has an eventually positive solution. Example 3.4 ([88]) Applying Theorem 3.16 to the following delay equation 

π 1  −β x (t) + x(t − π) = 0, t ≥ , (3.56) (1 + sin 2t) + Ct πe 4 where C > 0 and β ∈ R. We see that all solutions of Eq.(3.56) oscillate if and only if β < 2 or β = 2 and C > π/8e. In 1995, Li [60], established the following. Theorem 3.17 ([60]) Assume that there exists a T0 ≥ t0 + τ such that  t 1 p(s)ds ≥ (3.57) for t ≥ T0 e t−τ and  

 t  ∞ 1 − 1 dt = ∞. (3.58) p(t) exp p(s)ds − e T0 t−τ Then all solutions of Eq.(2) oscillate.

71

EXERCISES 1. Explain Remark 3.2. 2. Explain the last sentence in Example 3.1. 3. Show the equality in Remark 3.5. 4. Show that the function x(t) is a solution of the equation (3.40) in Example 3.2. 5. Show the asymptotic formulae in Example 3.3. 6. Explain Remark 3.8. 7. Explain Example 3.4.

72

Part II

OSCILLATIONS OF DIFFERENCE EQUATIONS 4

Oscillation Criteria for Eq. (1)

In this section we study the difference equation Δx(n) + p(n)x(n − k) = 0, n = 0, 1, 2, ....

(1)

where Δx(n) = x(n + 1) − x(n), the usual forward difference operator, p(n) is a sequence of nonnegative real numbers and k is a positive integer. In 1981, Domshlak [14] was the first who studied this problem in the case where k = 1. Then, in 1989, Erbe and Zhang [30], established the following oscillation criteria for Eq.(1) . Theorem 4.1 ([30]) Assume that β := lim inf p(n) > 0 and lim sup p(n) > 1 − β. n→∞

(4.1)

n→∞

Then all solutions of Eq.(1) oscillate. Theorem 4.2 ([30]) Assume that lim inf p(n) > n→∞

kk (k + 1)k+1

Then all solutions of Eq.(1) oscillate. 73

(4.2)

Theorem 4.3 ([30]) Assume that n 

lim sup n→∞

p(i) > 1

(C2 )

i=n−k

Then all solutions of Eq.(1) oscillate. In the same year 1989 Ladas, Philos and Sficas [52] proved the following theorem. Theorem 4.4 ([52]) Assume that n−1 kk 1  lim inf p(i) > . n→∞ k (k + 1)k+1

(C3 )

i=n−k

Then all solutions of Eq.(1) oscillate. Therefore they improved the condition (4.2) by replacing the p(n) of (4.2) by the arithmetic mean of the terms p(n − k), ..., p(n − 1) in (C3 ) . a substantial improvement of this oscillation criterion has been presented, in 2004, by Philos, Purnaras and Stavroulakis [69]. kk  it should Concerning the constant (k+1) k+1 in (4.2) and (C3 ) be empasized that, as it is shown in [29], if sup p(n)
1− 4

(F1 )

and n 

p(i) ≥ d > 0 for large n and

i=n−k

lim sup n→∞

n  i=n−k

d4 > 1− 8



d3 1− + 4



d3 1− 2

−1 (F2 )

respectively. Unfortunately, the above conditions (F1 ) and (F2 ) are not correct. This is due to the fact that they are based on the following (false) discrete version of Koplatadze-Chanturia Lemma. (See [12] and [20]). Lemma A (False). Assume that x(n) is an eventually positive solution of Eq. (1) and that 75

n 

p(i) ≥ M > 0 for large n.

(4.3)

i=n−k

Then M2 x(n) > x(n − k) for large n. 4 As one can see, the erroneous proof of Lemma A is based on the following (false) statement. (See [12] and [20]). Statement A (False). If (4.3) holds, then for any large N, there exists a positive integer n such that n − k ≤ N ≤ n and N 

p(i) ≥

i=n−k

n  M M p(i) ≥ . , 2 2 i=N

It is obvious that all the oscillation results which have made use of the above Lemma A or Statement A are not correct. For details on this problem see the paper by Cheng and Zhang [12]. Here it should be pointed out that the following statement (see [52], [74]) is correct and it should not be confused with the Statement A. Statement 2.1 ([52], [74]) If n−1 

p(i) ≥ M > 0 for large n,

(4.3)

i=n−k

then for any large n, there exists a positive integer n∗ with n − k ≤ n∗ ≤ n − 1 such that n∗ n−1   M M p(i) ≥ p(i) ≥ . , 2 2 i=n∗ i=n−k

76

In 1995, Stavroulakis [74], on the basis of this correct Statement 2.1, established the following: Lemma 4.1 ([74]) Assume that p(n) is a sequence of nonnegative real numbers and that there exists M > 0 such that lim inf n→∞

n−1 

p(i) > M.

(4.4)

i=n−k

If x(n) is an eventually positive solution of Eq.(1) then, for every n sufficiently large, there exists an integer n∗ , with n−k ≤ n∗ ≤ n − 1, such that  2 x(n∗ − k) 2 . (4.5) ≤ x(n∗ ) M Proof. Let x(n) be an eventually positive solution of Eq.(1) . By (4.4), for n sufficiently large, say for n ≥ n0 , n−1 

p(i) > M.

i=n−k

Thus, for n ≥ n0 + k, we can find an integer n∗ with n − k ≤ n ≤ n − 1, such that (see [42, 52]) ∗



n  i=n−k

n−1  M M p(i) ≥ p(i) ≥ . and 2 2 i=n∗

From Eq.(1) , taking into account the last two inequalities and the fact that the sequence x(n) is decreasing, we have 77



n 

x(n − k) − x(n∗ + 1) =

(x(i) − x(i + 1)) =

i=n−k ∗

n 

 ≥





n 

p(i)x(i − k) ≥

i=n−k

p(i) x(n∗ − k) ≥

i=n−k

M x(n∗ − k). 2

Similarly, ∗

x(n ) − x(n + 1) =  ≥

n 

(x(i) − x(i + 1)) =

i=n∗ n 



p(i) x(n − k) ≥

n 

p(i)x(i − k) ≥

∗  n−1i=n 



p(i) x(n − k) ≥

i=n∗

i=n∗



M x(n − k). 2

Combining the last two inequalities, we obtain  2 M M MM ∗ ∗ x(n ) ≥ x(n − k) ≥ x(n∗ − k), x(n − k) = 2 2 2 2 that is, inequality (4.5). The proof is complete. Theorem 4.5 ([74]) Assume that a0 := lim inf n→∞

n−1 

 p(i) ≤

i=n−k

78

k k+1

k+1 ,

(4.4)

and

a20 . 4 n→∞ Then, all solutions of Eq.(1) oscillate. lim sup p(n) > 1 −

(4.5)

Proof. Assume, for the sake of contradiction, that x(n) is an eventually positive solution of Eq.(1) . Then eventually Δx(n) = x(n + 1) − x(n) ≤ −p(n)x(n − k) ≤ 0, and so x(n) is an eventually nonicreasing sequence of positive numbers. Summing up Eq.(1) from n − k to n − 1, we have n−1 

x(n) − x(n − k) +

p(i)x(i − k) = 0,

i=n−k

and, because x(n) is eventually nonicreasing, it follows that for n sufficiently large   n−1  p(i) x(n − k) ≤ 0 x(n) − x(n − k) + i=n−k



or x(n − k)

n−1 

p(i) +

i=n−k

x(n) −1 x(n − k)

 ≤ 0,

and using Lemma 4.1, for N sufficiently large there exists an integer n∗ with N − k ≤ n∗ ≤ N − 1 such that  n∗ −1  2  a p(i) + 0 − 1 ≤ 0. x(n∗ − k) 4 ∗ i=n −k

79

Now let λ(n) be a sequence such that p(λ(n)) → μ. For N = λ(n) + k + 1, n∗ satisfies λ(n) + 1 ≤ n∗ ≤ λ(n) + k or

n∗ − k ≤ λ(n) ≤ n∗ − 1. Thus,





x(n − k) p(λ(n)) +  ≤ x(n∗ − k)

∗ n −1

 a 2

p(i) +

i=n∗ −k

0

2

 −1

 a 2 0

2

≤ 

−1

≤ 0,

which in view of (4.5), leads to a contradiction. The proof is complete. In 1999, Domshlak [20] and in 2000 Cheng and Zhang [12] established the following lemmas, respectively, which may be looked upon as (exact) discrete versions of Koplatadze-Chanturia Lemma. Lemma 4.2 ([20]) Assume that x(n) is an eventually positive solution of Eq.(1) and that n−1 

p(i) ≥ M > 0 for large n.

(4.3)

M2 x(n) > x(n − k) for large n. 4

(4.6)

i=n−k

Then

80

Lemma 4.3 ([12]) Assume that x(n) is an eventually positive solution of Eq.(1) and that n−1 

p(i) ≥ M > 0 for large n.

(4.3)

x(n) > M k x(n − k) for large n.

(4.7).

i=n−k

Then In 2004, Stavroulakis [75], based on these lemmas improved the condition (4.5) as follows: Theorem 4.6 ([75]) Assume that k+1  k . 0 < α0 ≤ k+1 Then either one of the conditions lim sup n→∞

or lim sup n→∞

n−1 

p(i) > 1 −

i=n−k n−1 

α02 , 4

p(i) > 1 − α0k ,

(C4 )

(C˜4 )

i=n−k

implies that all solutions of Eq.(1) oscillate. Proof. Assume, for the sake of contadiction, that x(n) is an eventually positive solution of Eq.(1) . Then eventually Δx(n) = x(n + 1) − x(n) ≤ −p(n)x(n − k) ≤ 0, and so x(n) is an eventually nonincreasing sequence of positive numbers. 81

Summing up Eq.(1) from n − k to n − 1, we have n−1 

x(n) − x(n − k) +

p(i)x(i − k) = 0,

i=n−k

and, because x(n) is eventually nonincreasing, it follows that for all sufficiently large n   n−1  p(i) x(n − k) ≤ 0, x(n) − x(n − k) + i=n−k



or x(n − k)

n−1  i=n−k

x(n) p(i) + −1 x(n − k)

 ≤ 0.

Now, using Lemma 4.2, for all sufficiently large n, we have  n−1   α02 p(i) + x(n − k) −1 ≤0 4 i=n−k

or, using Lemma 4.3, for all sufficiently large n, we have   n−1  p(i) + α0k − 1 ≤ 0 x(n − k) i=n−k

respectively, which, in view of (C4 ) and (C˜4 ) , lead to a contradiction. The proof is complete. Remark 4.1 ([75]) From the above theorem it is now clear that k+1  n−1  k p(i) ≤ 0 < α0 := lim inf n→∞ k+1 i=n−k

82

and lim sup n→∞

n−1 

p(i) > 1 −

i=n−k

α02 4

is the correct oscillation condition by which the (false) condition (F1 ) should be replaced. Remark 4.2 ([75]) Observe the following: (i) When k = 1, 2

α02 , 4 (since, from the above mentioned conditions, it makes sense to  k k+1 investigate the case when α0 < k+1 ) and therefore condi  ˜ tion (C4 ) implies (C4 ) . α0k >

(ii) When k = 3, α03

α02 1 > when α0 > 4 4

while

α02 1 when α0 < . 4 4  So in this case the conditions (C4 ) and (C˜4 ) are independent. (iii) When k ≥ 4 α2 α0k < 0 , 4  ˜ and therefore condition (C4 ) implies (C4 ) . (iv) When k < 12 condition (C4 ) or (C˜4 ) implies (C2 ) . (v) When k ≥ 12 condition (C4 ) may hold but condition (C2 ) may not hold. α03
1 − α03 . 10 95

Thus condition (C˜4 ) is satisfied and therefore all solutions oscillate. Observe, however, that condition (C4 ) is not satisfied. If, on the other hand, in the above equation p(2n) =

nπ 8 746 8 , p(2n + 1) = + sin2 , 100 100 1000 2

n = 0, 1, 2, ...,

then it is easy to see that n−1 

24 p(i) = α0 = lim inf < n→∞ 100 i=n−3 and

n−1 

 4 3 4

24 746 α02 p(i) = lim sup + >1− . 100 1000 4 n→∞ i=n−3 84

In this case condition (C4 ) is satisfied and therefore all solutions oscillate. Observe, however, that condition (C˜4 ) is not satisfied. Example 4.2 ([75]) Consider the equation x(n + 1) − x(n) + p(n)x(n − 16) = 0,

n = 0, 1, 2, ...,

where p(17n) = p(17n + 1) = ... = p(17n + 15) =

2 , p(17n + 16) = 100

655 2 + , n = 0, 1, 2, .... 100 1000 Here k = 16 and it is easy to see that  17 n−1  32 16 p(i) = < α0 = lim inf n→∞ 100 17 i=n−16 and n−1 

32 655 α02 p(i) = lim sup + = 0.975 > 1 − . 100 1000 4 n→∞ i=n−16 We see that condition (C4 ) is satisfied and therefore all solutions oscillate. Observe, however, that lim sup n→∞

n  i=n−16

p(i) =

34 655 + = 0.995 < 1 100 1000

that is, condition (C2 ) is not satisfied. In 2006, Chatzarakis and Stavroulakis [7], improved the conditions (4.6) and (4.7) to the condition 85

x(n) >

M2 x(n − k) for large n. 2(2 − M )

Then using this (improved) upper bound for the condition lim sup n→∞

n−1 

p(i) > 1 −

i=n−k

x(n−k) x(n)

(4.8) derived

α02 , 2(2 − α0 )

(4.9)

4 ) . which essentially improves the conditions (C4 ) and (C Lemma 4.4 ([7]) Assume that x(n) is an eventually positive solution of Eq.(1) and that n−1 

p(i) ≥ M > 0 for large n.

(4.3)

i=n−k

Then x(n) >

M2 x(n − k) for large n. 2(2 − M )

(4.10)

Proof. Since x(n) is an eventually positive solution of Eq.(1) , then eventually Δx(n) = x(n + 1) − x(n) = −p(n)x(n − k) ≤ 0, and so x(n) is an eventually nonincreasing sequence of positive numbers. For all n consider now the following two possible cases: (i) p(n) ≥ M2 , and (ii) p(n) < M2 . In the case (i), from Eq.(1) , it is clear that x(n) = x(n + 1) + p(n)x(n − k) ≥ x(n + 1) + 86

M x(n − k). (4.11) 2

Also, summing up Eq.(1) from n − k to n − 1, and using the fact that the sequence x(n) is nonincreasing, we have x(n − k) − x(n) = 

p(i) x(n − k − 1) >

i=n−k

or

p(i)x(i − k) ≥

i=n−k



n−1 

n−1 

M x(n − k) 2

M x(n − k). (4.12) 2 Combining the inequalities (4.11) and (4.12), we obtain   M M x(n) + x(n − k) x(n) > x(n + 1) + 2 2 x(n − k) > x(n) +

which implies that M2 x(n − k). x(n) > 2(2 − M ) In the case (ii), there exists n∗ , n + 1 ≤ n∗ < n + k, such that ∗ n −1 n∗  M M p(i) < p(i) ≥ . and 2 2 i=n i=n Therefore n−1  i=n∗ −k

p(i) =

∗ n −1

i=n∗ −k

p(i) −

∗ n −1

i=n

p(i) ≥ M −

M M = . 2 2

Now summing up Eq.(1) first from n to n∗ , and then from n∗ − k to n − 1, and using the fact that the sequence x(n) is 87

nonincreasing, we have ∗



x(n) − x(n + 1) =  ≥



n 

n 

p(i)x(i − k) ≥

i=n



p(i) x(n∗ − k) ≥

i=n

or x(n) ≥ x(n∗ + 1) +

M x(n∗ − k) 2

M x(n∗ − k), 2

(4.13)

and then n−1 



x(n − k) − x(n) =  ≥

n−1 



p(i) x(n − k − 1) ≥

i=n∗ −k

or

p(i)x(i − k) ≥

i=n∗ −k

M x(n − k) 2

M x(n − k). (4.14) 2 Combining the inequalities (4.13), and (4.14) we obtain   M M ∗ x(n) + x(n − k) x(n) ≥ x(n + 1) + 2 2 x(n∗ − k) ≥ x(n) +

that is, x(n) >

M2 x(n − k). 2(2 − M )

The proof is complete.

88

Theorem 4.7 ([7]) Assume that  k+1 k 0 < α0 ≤ k+1 and lim sup n→∞

n−1 

p(i) > 1 −

i=n−k

α02 . 2(2 − α0 )

(4.9)

Then all solutions of Eq.(1) oscillate. Proof. Assume, for the sake of contadiction, that x(n) is an eventually positive solution of Eq.(1) . Then eventually Δx(n) = x(n + 1) − x(n) = −p(n)x(n − k) ≤ 0, and so x(n) is an eventually nonincreasing sequence of positive numbers. Summing up Eq.(1) from n − k to n − 1, we have x(n) − x(n − k) +

n−1 

p(i)x(i − k) = 0,

i=n−k

and, because x(n) is eventually nonincreasing, it follows that for all sufficiently large n  n−1   p(i) x(n − k) ≤ 0, x(n) − x(n − k) + i=n−k



or x(n − k)

n−1  i=n−k

x(n) p(i) + −1 x(n − k)

 ≤ 0.

Now, using Lemma 4.4, for all sufficiently large n, we have  n−1   α02 p(i) + x(n − k) − 1 ≤ 0, 2(2 − α0 ) i=n−k

89

which, in view of (4.9), leads to a contradiction. The proof is complete. Remark 4.3 ([7]) Observe the following: (i) When α0 → 0, then it is clear that the conditions (C4 ) , (C˜4 ) , and (4.9) reduce to lim sup n→∞

n−1 

p(i) > 1,

i=n−k 

which obviously implies (C2 ) . (ii) It always holds α02 α2 > 0, 2(2 − α0 ) 4 since α0 > 0 and therefore condition (C4 ) always implies (4.9). (iii) When k = 1, 2, α02 < α0k , 2(2 − α0 ) (since, from the above mentioned conditions, it makes sense to  k k+1 investigate the case when α0 ≤ k+1 ) and therefore condition  (4.9) implies (C˜4 ) . (iv) When k = 3,

√ α02 2 3 > α0 if 0 < α0 < 1 − , 2(2 − α0 ) 2

while

√  4 α02 2 3 3 . < α0 if 1 − < α0  2(2 − α0 ) 2 4 90

Therefore in this case the conditions (C˜4 ) and (4.9) are independent. (v) When k ≥ 4 α02 > α0k , 2(2 − α0 ) and therefore condition (C˜4 ) implies (4.9). (vi) When k ≥ 10 condition (4.9) may hold but condition (C2 ) may not hold. (vii) When k is large then α0 → 1e and in this case both conditions (C4 ) and (C˜4 ) imply (4.9). For illustrative purposes, we give the values of the lower bound of n−1  p(i) lim sup n→∞

i=n−k

under these conditions when k = 100 (α ≈ 0.366) : (C˜4 ) : (C4 ) : (4.9) :

0.999999 0.966511 0.959009

We see that condition (4.9) essentially improves the conditions (C˜4 ) , (C4 ) and also (C2 ) . We illustrate by the following examples. Example 4.3 ([7]) Consider the equation x(n + 1) − x(n) + p(n)x(n − 3) = 0, 91

n = 0, 1, 2, ...,

where 1 nπ 6731 1 , p(2n + 1) = + sin2 , n = 0, 1, 2, .... 10 10 10000 2 Here k = 3 and it is easy to see that  4 n−1  3 3 α0 = lim inf p(i) = ≈ 0.3164 < n→∞ 10 4 i=n−3 and p(2n) =

n−1 

lim sup n→∞

p(i) =

i=n−3

3 6731 + > 1 − α03 = 0.973. 10 10000

Thus, condition (C˜4 ) is satisfied and therefore all solutions oscillate. Observe, however, that condition (4.9) is not satisfied. If, on the other hand, in the above equation nπ 8 744 8 , p(2n + 1) = + sin2 , n = 0, 1, 2, .... 100 100 1000 2 it is easy to see that  4 n−1  24 3 p(i) = ≈ 0.3164 < α0 = lim inf n→∞ 100 4 i=n−3 and p(2n) =

lim sup n→∞

n−1  i=n−3

p(i) =

24 744 α02 + >1− ≈ 0.9836. 100 1000 2(2 − α0 )

In this case condition (4.9) is satisfied and therefore all solutions oscillate. Observe, however, that 0.984 < 1 − α03 ≈ 0.9861. and therefore condition (C˜4 ) is not satisfied. 92

Example 4.4 ([7]) Consider the equation x(n + 1) − x(n) + p(n)x(10) = 0,

n = 0, 1, 2, ...,

where p(11n + 1) = ... = p(11n + 10) =

35 , p(11n + 11) = 1000

613 35 + , n = 0, 1, 2, .... 1000 1000 Here k = 10 and it is easy to see that  9 n−1  35 10 p(i) = ≈ 0.3504 α0 = lim inf < n→∞ 100 11 i=n−10 =

and n−1 

35 613 α02 p(i) = lim sup + = 0.963 > 1− ≈ 0.9628. 100 1000 2(2 − α0 ) n→∞ i=n−10 We see that condition (4.9) is satisfied and therefore all solutions oscillate. Observe, however, that 0.963 < 1 − α010 ≈ 0.9999 α02 0.963 < 1 − ≈ 0.9693 4 and lim sup n→∞

n  i=n−10

p(i) =

35 963 + = 0.998 < 1. 1000 1000

Therefore none of the conditions (C˜4 ) , (C4 ) and (C2 ) is satisfied. 93

Also, Chen and Yu [8], derived the following oscillation cindition n  1 − α0 − 1 − 2α0 − α02 lim sup p(i) > 1 − . (C6 ) 2 n→∞ i=n−k

In 1998, Domshlak [19], studied the oscillation of all solutions and the existence of nonoscillatory solution of Eq.(1) with r periodic positive coefficients p(n), p(n + r) = p(n). It is very important that in the following cases where {r = k}, {r = k + 1}, {r = 2}, {k = 1, r = 3} and {k = 1, r = 4} the results obtained are stated in terms of necessary and sufficient conditions and it is very easy to check them. Reamrk 4.4. Observe that the conditions (C2 ) , (C3 ) , (C4 ) and (C6 ) for Eq.(1) are the discrete analogues of the condition (C2 ), (C3 ), (C4 ) and (C6 ) respectively for Eq.(1). In 2000, Shen and Stavroulakis [73], using new techniques, improved the previous results. Lemma 4.5 ([73]) Let the number h ≥ 0 be such that 1 pn−i ≥ h f or large n. k i=1 k

(2.1)

Assume that Eq.(1) has an eventually positive solution x(n). Then h ≤ k k /(k + 1)k+1 and lim sup n→∞

x(n) ≤ [d(h)]k , x(n − k) 94

(2.2)

where d(h) is the greater real root of the algebraic equation dk+1 − dk + h = 0, on the interval [0, 1].

(2.3)

We describe the number d(h) by the following proposition and remark. Proposition 4.1 ([73]) For the algebraic equation dk+1 − dk + h = 0, on [0, 1] the following statements hold true: (i) if h = 0, then it has exactly two different real roots d1 = 0 and d2 = 1. (ii) if 0 < h < k k /(k +1)k+1 , then it has exactly two different real roots d1 and d2 such that d1 ∈ (0, k/(k + 1)), d2 ∈ (k/(k + 1), 1). (iii) if h = k k /(k + 1)k+1 , then it has a unique real root d = k/(k + 1). Remark 4.5 ([73]) From Proposition 4.1, we see that the number d(h) in Lemma 4.5 satisfies ⎧ h=0 ⎨ = 1, d(h) is ∈ (k/(k + 1), 1), 0 < h < k k /(k + 1)k+1 ⎩ = k/(k + 1), h = k k /(k + 1)k+1 . Lemma 4.5 ([73]) Let the number M ≥ 0 be such that k 

p(n − i) ≥ M f or large n.

i=1

95

Assume that Eq.(1) has an eventually positive solution x(n). Then M ≤ k k+1 /(k + 1)k+1 and x(n − k)   lim sup p(n − i + j) ≤ [d(M )]k , x(n) i=1 j=1 n→∞ k

k

where d(M ) is the greater real root of the algebraic equation dk+1 − dk + M k = 0, on [0, 1]. Note that from this lemma we obtain a better and perhaps optimal bound which essentially improves (4.7). Proposition 4.2 ([73]) For the algebraic equation dk+1 − dk + M k = 0, on [0, 1]. the following statements hold true: (i) if M = 0, then it has exactly two different real roots d1 = 0 and d2 = 1. (ii) if k = 1 and 0 < M ≤ k k+1 /(k+1)k+1 , then it has exactly two different real roots d1 and d2 which satisfy d1 ∈ (0, k/(k + 1)), d2 ∈ (k/(k + 1), 1). (iii) if k = 1, then it has two real roots of the form √ √ 1 − 1 − 4M 1 + 1 − 4M and d2 = . d1 = 2 2

Remark 4.6 ([73]) The number d(M ) in Lemma 4.6 satisfies d(M ) is 96

⎧ ⎨ = 1, ∈ (k/(k√+ 1), 1), ⎩ = (1 + 1 − 4M )/2,

M =0 k = 1, 0 < M ≤ k k+1 /(k + 1)k+1 k = 1.

This implies that d(M ) ≤ 1 and the equality holds if and only if M = 0. Observe that ki=1 p(n − i) ≥ M implies k  k 

pn−i+j ≥ M k .

i=1 j=1

Thus, the last inequality in Lemma 4.6, leads to lim inf n→∞

xn ≥ [d(M )]−k M k . xn−k

Theorem 4.8 ([73]) Assume that 0 ≤ α0 ≤ k k+1 /(k + 1)k+1 and that there exists an integer l ≥ 1 such that ! # "k k k −k i=1 p(n − i) + [d(α0 )] i=1 j=1 p(n − i + j)+ lim sup l−1 k "m+1 −(m+1)k + m=0 [d(α0 /k)] n→∞ i=1 j=0 p(n − jk − i) > 1,

(4.15)

where d(α0 ) and d(α0 /k) are the greater real roots of the equations dk+1 − dk + α0k = 0 and dk+1 − dk + α0 /k = 0, respectively. Then all solutions of Eq.(1) oscillate.

97

√ Notice that when k = 1, d(μ0 ) = d(μ0 ) = (1 + 1 − 4μ0 )/2 (see [73]), and so condition (C10 ) reduces to ! # l−1 m+1   lim sup Cp(n) + p(n − 1) + C m+1 p(n − j − 1) > 1, n→∞

m=0

j=0

(4.16) where C = 2/(1 + 1 − 4μ0 ), μ0 = lim inf n→∞ p(n). Therefore, from Theorem 4.8, we have the following corollary. √

Corollary 4.1 ([73]) Assume that 0 ≤ μ0 ≤ 1/4 and that (4.16) holds. Then all solutions of the equation x(n + 1) − x(n) + p(n)x(n − 1) = 0

(4.17)

oscillate. A condition derived from (4.16) and which can be easier verified, is given in the next corollary. Corollary 4.2 ([73]) Assume that 0 ≤ μ0 ≤ 1/4 and that  2 √ 1 + 1 − 4μ0 . (C12 ) lim sup pn > 2 n→∞ Then all solutions of (4.17) oscillate. Remark 4.7 ([73]) Observe that when μ0 = 1/4, condition (4.18) reduces to lim sup p(n) > 1/4 n→∞

which can not be improved in the sense that the lower bound 1/4 can not be replaced by a smaller number. Indeed, by condition (N1 ) (Theorem 2.3 in [30]), we see that (4.17) has a nonoscillatory solution if sup p(n) < 1/4. 98

Note, however, that even in the critical state where lim p(n) = 1/4,

n→∞

(4.17) can be either oscillatory or nonoscillatory. For example, if p(n) = 14 + nc2 then (4.17) will be oscillatory in case c > 1/4 and nonoscillatory in case c < 1/4 (the Kneser-like theorem, [40]). Example 4.5 ([73]) Consider the equation   1 4 nπ x(n − 1) = 0, x(n + 1) − x(n) + + b sin 4 8 where b > 0 is a constant. It is easy to see that   1 nπ 1 lim inf p(n) = lim inf = , + b sin4 n→∞ n→∞ 4 8 4   1 1 4 nπ = + b. + b sin lim sup p(n) = lim sup 4 8 4 n→∞ n→∞ Therefore, by Corollary 4.2, all solutions oscillate. Observe, however, that none of the conditions 4.1, 4.2, (C3 ) , 4.4, (C˜4 ) , (C6 ) is satisfied. The following corollary concerns the case when k > 1. Corollary 4.3 ([73]) Assume that 0 ≤ α0 ≤ k k+1 /(k + 1)k+1 and that lim sup n→∞

n−1 

p(i) > 1 − [d(α0 )]−k α0k −

i=n−k

99

k[d(α0 /k)]−k β 2 , (C13 ) 1 − [d(α0 /k)]−k β

where d(α0 ), d(α0 /k) are as in Theorem 4.8. Then all solutions of Eq.(1) oscillate. In 2000, Shen and Luo [72] proved the following theorems. Theorem 4.9 ([72]) Assume that there exists some positive integer l such that ! k k  k   p(n − i) + p(n − i + j)+ lim sup n→∞

i=0

+

l−1  k m+1  

i=0 j=1

p(n − jk − i) > 1.

(4.19)

m=0 i=1 j=0

Then all solutions of Eq.(1) oscillate. Theorem 4.10 ([72]) Assume that there exists some positive integer l such that ! k k  k   p(n − i) + p(n − i + j)+ lim sup n→∞

i=1

+

l−1  k m+1  

i=1 j=1

p(n − jk − i) > 1.

(4.20)

m=0 i=1 j=0

Then all solutions of Eq.(1) oscillate. From Theorem 4.9 and Theorem 4.10 the following corollaries are derived,in which the numbers β and α0 are as in (4.1) and (4.4) respectively. Corollary 4.4 ([72]) Assume that n  kβ 2 p(i) > 1 − α0k+1 − . lim sup 1−β n→∞ i=n−k

100

(4.21)

Then all solutions of Eq.(1) oscillate. Corollary 4.5 ([73]) Assume that lim sup n→∞

n−1 

p(i) > 1 −

i=n−k

α0k

kβ 2 − . 1−β

(4.22)

Then all solutions of Eq.(1) oscillate. Following this historical (and chronological) review for the difference equation (1) we also mention that in the case where n−1  i=n−k

n−1  k k+1 k k+1 p(i) ≥ and lim p(i) = , n→∞ (k + 1)k+1 (k + 1)k+1 i=n−k

(cf. (3.5)) the oscillation of Eq.(1) has been studied in 1994 by Domshlak [17] and in 1998 by Tang [80] (see also Tang and Yu [83]). In a case when p(n) is asymptotically close to one of the periodic critical states, unimprovable results about oscillation preperties of the equation x(n + 1) − x(n) + p(n)x(n − 1) = 0 were obtained by Domshlak in 1999 [21] and in 2000 [22].

101

EXERCISES 1. Show the validity of the five cases in Remark 4. 2. 2. Check the values of liminf and limsup in Example 4.1. 3. The same in Example 4.2. 4. Show the validity of the seven cases in Remark 4. 3. 5. Check the values of liminf and limsup in Example 4.3. 6. Check the values of liminf and limsup in Example 4.4. 7. Show that when k=1 the condition (4.15) reduces to (4.16) in Theorem 4.8. 8. Explain Example 4.5.

102

5

Oscillation Criteria for Eq. (1)

In this section we study the first order linear difference equation with variable delay argument Δx(n) + p(n) x(τ (n)) = 0,

n ∈ N,

(1)

where Δx(n) = x(n + 1) − x(n), p : N → R+ , τ : N → N and lim τ (n) = +∞. n→+∞

By a proper solution of Eq.(1) we mean a function x : Nk0 →

R, k0 = min{τ (n) : n ∈ Nk }, Nk = {k, k + 1, ...} which satisfies Eq.(1) on Nk and sup{|x(i)| : i ≥ n} > 0 for n ∈ Nk0 . A proper solution x : N → R of Eq.(1) is said to be oscillatory (around zero) if for every positive integer n there exist n1 , n2 ∈ Nn , such that u(n1 )u(n2 ) ≤ 0. Otherwise, the solution is said to be non-oscillatory. In other words, a proper solution x is oscillatory if it is neither eventually positive nor eventually negative. In 1991, Philos [67] extended the oscillation criterion (C3 ) to the general case of the equation Eq.(1) , by establishing that, if the sequence τ (n) is increasing, then the condition ⎡ ⎤ n−1  1 (n − τ (n))n−τ (n) p(j)⎦ > lim sup lim inf ⎣ n−τ (n)+1 n→∞ n − τ (n) n→∞ (n − τ (n) + 1) j=τ (n)

3 ) (C suffices for the oscillation of all solutions of Eq.(1) . This oscillation result has recently improved substantially by Philos and 103

Purnaras [68] (the results in [68] extend the ones given in [69] concerning the special case of equation (1) . In 2008, Chatzarakis, Koplatadze and Stavroulakis [3, 4] established the following. Theorem 5.1 ([3]) Assume that τ (n) ≤ n − 1 and ( ) σ(n) = max τ (s) : 1 ≤ s ≤ n, s ∈ N , If

n 

lim sup n→+∞

p(i) > 1,

(5.1) (5.2)

i=σ(n)

then all proper solutions of Eq.(1) oscillate. Proof. Assume, for the sake of contradiction, that x0 : Nk0 → (0, +∞) is a positive proper solution of Eq.(1) . Since the function x0 is nonincreasing, for sufficiently large n ∈ Nk0 , it satisfies the following inequality Δx0 (n) + p(n) x0 (σ(n)) ≤ 0. Summing up the last inequality from σ(n) to n, and using the fact that the function x0 is nonincreasing and the function σ is nondecreasing, we have    n x0 (σ(n)) p(i) − 1 ≤ 0. i=σ(n)

Therefore for sufficiently large n n 

p(i) ≤ 1,

i=σ(n)

which contradicts (5.2). The proof is complete. 104

Remark 5.1 ([3]) If the sequence τ (n) is assumed to be nondecreasing then (5.2) leads to the condition lim sup n→+∞

n 

(C2 )

p(i) > 1,

i=τ (n)

which is the discrete analogue of the condition (C2 ) presented in Section 2. While in the special case of Eq.(1) the above condition (5.2) leads to the condition (C2 ) presented in Section 4. Next consider the difference inequality Δx(n) + q(n) x(σ(n)) ≤ 0,

n ∈ N,

(5.3)

lim σ(n) = +∞.

(5.4)

where q : N → R+ , σ : N → N and

n→+∞

Lemma 5.1 ([3]) Let n−1 

lim inf n→+∞

p(i) = α > 0,

(5.5)

i=τ (n)

σ(n) ≤ τ (n) ≤ n − 1,

p(n) ≤ q(n) for n ∈ N,

(5.6)

and x : Nk0 → (0, +∞) be a positive proper solution of (5.3) for a certain n0 ∈ N . Then Eq.(1) has a proper solution x∗ : Nn1 → (0, +∞) such that 0 < x∗ (n) ≤ x(n) f or n ∈ Nn1 , where n1 > k0 is a sufficiently large natural number. 105

(5.7)

Proof. Let x : Nk0 → (0, +∞) be a positive proper solution of (5.3). By (5.4) and (5.5), it is clear that there exists n1 ∈ Nk0 such that

x(σ(n)) > 0 and

n−1 

p(i) > 0 f or n ∈ Nn1 ,

(5.8)

i=τ (n)

From (5.3), we have x(n) ≥

+∞ 

q(i) x(σ(i)) for n ∈ Nn1 .

(5.9)

i=n

Assume that n∗ = min{τ (n) : n ∈ Nn1 } and consider the sequence of functions xi : Nn∗ → R (i = 1, 2, . . . } defined as follows x1 (n) = x(n) for n ∈ Nn∗ , ⎧ +∞ ⎪  ⎪ ⎨ p(i) xj−1 (τ (i)) for n ∈ Nn1 , xj (n) = i=k ⎪ ⎪ ⎩x(n) for n ∈ [n∗ , n1 ) (j = 2, 3, . . . ). (5.10) By (5.6), (5.9) and using the fact that the function x is nonincreasing, we have x2 (n) =

+∞  i=n

p(i) x1 (τ (i)) ≤

+∞ 

q(i) x(σ(i)) ≤ x(n) = x1 (n)

i=n

for n ≥ n1 . 106

Thus xj (n) ≤ xj−1 (n) for n ∈ Nn1 (j = 2, 3, . . . ).

(5.11)

Denote lim xj (n) = x∗ (n) (according to (5.11) this limit j→+∞

exists). Therefore, from (5.10), we get x∗ (n) =

+∞ 

p(i) x∗ (τ (i)) for n ∈ Nn1 ,

(5.12)

i=n

Now, we will show that x∗ (n) > 0 for n ≥ n1 . Assume, for the sake of contradiction, that there exists n2 ≥ n1 such that x∗ (n) = 0 for n ∈ Nn2 and x∗ (n) > 0 for n ∈ [n∗ , n2 ). Denote by N ∗ the set of natural numbers n for which τ (n) = n2 and n∗ = min N ∗ . By (5.12) and (5.6) it is clear that n∗ ≥ n2 . Therefore, if c = min{x∗ (τ (i)) : τ (n∗ ) ≤ i ≤ n∗ − 1} > 0, by (5.6) and (5.8), we have x∗ (n2 ) =

+∞  i=n2

p(i) x∗ (τ (i)) ≥

∗ n −1

i=τ (n∗ )

p(i) x∗ (τ (i)) ≥ c

∗ n −1

p(i) > 0,

i=τ (n∗ )

which, in view of x∗ (n2 ) = 0, leads to a contradiction. Therefore, x∗ (n) > 0 for n ≥ n1 . Hence Eq.(1) has a proper solution x∗ satisfying 0 < x∗ (n) ≤ x(n) for n ∈ Nn1 . The proof is complete The following results have been established in [4]. Let n0 ∈ N . Denote by Un0 the set of all proper solutions of Eq.(1) satisfying the condition x(n) > 0 for n ≥ n0 . Remark 5.2 ([4]) We will suppose that Un0 = ∅, if Eq.(1) has no solutions satisfying the condition x(n) > 0 for n ≥ n0 . 107

Lemma 5.1 ([4]) Assume that n0 ∈ N , Un0 = ∅, x ∈ Un0 , τ (n) ≤ n − 1, τ is a nondecreasing function and n−1 

lim inf n→+∞

p(i) = c > 0.

(5.13)

i=τ (n)

Then lim sup n→+∞

x(τ (n)) 4 ≤ 2. x(n + 1) c

(5.14)

Proof. By (5.13), for any ε ∈ (0, c), it is clear that n−1 

p(i) ≥ c − ε for n ∈ Nn0 .

(5.15)

i=τ (n)

Since x is a positive proper solution of Eq.(1) , then there exists n1 ∈ Nn0 such that x(τ (n)) > 0 for n ∈ Nn1 . Thus, from Eq.(1) we have x(n + 1) − x(n) = −p(n)x(τ (n)) ≤ 0 and so x is an eventually nonincreasing function of positive numbers. Now from inequality (5.15) it is clear that, there exists n∗ ≥ n such that ∗ n −1

i=n

c−ε p(i) < 2



and

n  i=n

108

p(i) ≥

c−ε . 2

(5.15)

c−ε , it is clear 2 ∗  that there exists n > n such that (5.15) is satisfied, while in c−ε the case where p(n) ≥ , then n∗ = n, and therefore 2 This is because in the case where p(n)
n − 1. Hence, n ≤ τ (n∗ ∗ ∗ )≤ n∗ − 1 and then ni=τ−1(n∗ ) p(i) ≤ ni=n−1 p(i) < c−ε 2 . This, in view of (5.15), leads to a contradiction. Thus, in both cases, we have τ (n∗ ) ≤ n − 1. Therefore, it is clear that n−1  i=τ (n∗ )

p(i) =

∗ n −1

i=τ (n∗ )

p(i) −

∗ n −1

p(i) ≥ (c − ε) −

i=n

c−ε c−ε = . 2 2

(5.16) Now, summing up Eq.(1) first from n to n and then from τ (n∗ ) to n − 1, and using the fact that the function u is nonin



109

creasing and the function τ is nondecreasing, we have ∗

x(n) − x(n∗ + 1) =  ≥



n 

n 

p(i)x(τ (i)) ≥

i=n



p(i) x(τ (n∗ )) ≥

i=n

or x(n) ≥

c−ε x(τ (n∗ )) 2

c−ε x(τ (n∗ )), 2

(5.17)

and then ∗

x(τ (n )) − x(n) = ⎛ ≥⎝

p(i)⎠ x(τ (n − 1)) ≥

i=τ (n∗ )

or

p(i)x(τ (i)) ≥

i=τ (n∗ )



n−1 

n−1 

c−ε x(τ (n − 1)) 2

c−ε x(τ (n − 1)). 2 Combining inequalities (5.17) and (5.18), we obtain x(τ (n∗ )) ≥

(5.18)

4 x(τ (n − 1)) ≤ x(n) (c − ε)2 and, for large n, we have x(τ (n)) 4 . ≤ x(n + 1) (c − ε)2 Hence, lim sup k→+∞

x(τ (n)) 4 , ≤ x(n + 1) (c − ε)2 110

(5.19)

which, for arbitrarily small values of ε, implies (5.14). The proof is complete. Lemma 5.3 ([4]) Assume that n0 ∈ N , Un0 = ∅, u ∈ Un0 , τ (n) ≤ n − 1, τ is a nondecreasing function and the condition (5.13) is satisfied. Then  n−1   4 p(i) = +∞ for any λ > 2 . (5.20) lim x(n) exp λ n→+∞ c i=1 Proof. Since all the conditions of Lemma 5.2 are satisfied, for any γ > c42 , there exists n1 ∈ Nn0 such that u(τ (n)) ≤γ u(n + 1)

for n ∈ Nn1 ,

Also, for any k ∈ Nn1 k k    Δx(n) x(n)  1− = = x(n + 1) x(n + 1) n=n n=n 1

1

(k − n1 ) −

k  n=n1

≤ (k − n1 ) −



x(n)  exp ln ≤ x(n + 1)

k  

1 + ln

n=n1

=−

k  n=n1

ln

x(n)  = x(n + 1)

x(n) x(k + 1) = ln , x(n + 1) x(n1 ) 111

(5.21)

or

k  Δx(n) x(k + 1) ≤ ln . x(n + 1) x(n ) 1 n=n 1

Moreover, from Eq.(1) , we have k k   Δx(n) x(τ (n)) p(n) =− . x(n + 1) x(n + 1) n=n n=n 1

1

Combining (5.21) with the last two relations, we obtain 

x(k + 1) ≥ x(n1 ) exp − γ

k 

 p(n) .

n=n1

Now, by (5.13), it is obvious that for λ > c42 , the last inequality yields  lim x(k + 1) exp λ

k→+∞

or

+∞

k 

p(i) = +∞. Therefore 

p(n) = +∞,

n=n1

n−1    p(i) = +∞, lim x(n) exp λ

n→+∞

which implies (5.20), since

i=n1 n−1

p(i) ≥

i=1

n−1 i=k1

p(i). The proof is

complete. Now we formulate Lemma 5.1 in terms of the set Un0 as follows. Lemma 5.1 ([4]) Assume that (5.13) is satisfied, and for sufficiently large n 112

σ(n) ≤ τ (n) ≤ n − 1,

p(n) ≤ q(n),

(5.6)

and x : Nn0 → (0, +∞) is a positive proper solution of (5.3). Then, there exists n1 ∈ Nn0 such that Un1 = ∅ and u∗ ∈ Un1 is the solution of Eq.(1) , which satisfies the condition 0 < x∗ (n) ≤ x(n) for n ∈ Nn1 .

(5.7)

By virtue of Lemma 5.1 , we can formulate Lemma 5.3 in the following more general form, where the function τ is not required to be nondecreasing. Lemma 5.4 ([4]) Assume that n0 ∈ N , Un0 = ∅, x ∈ Un0 , τ (n) ≤ n − 1 and the condition (5.13) is satisfied. Then, for any λ > c42 , the condition (5.20) holds. Proof. Since x : Nn0 → (0, +∞) is a solution of Eq.(1) , it is clear that x is a solution of the inequality Δx(n) + p(n) x(σ(n)) ≤ 0 for n ∈ Nn1 , where σ(n) = max{τ (i) : 1 ≤ s ≤ n, s ∈ N } and n1 > n0 is a sufficiently large number. First we will show that lim inf n→+∞

n−1 

p(i) = c.

(5.22)

i=σ(n)

Assume that (5.22) is not satisfied. Then there exists a se( ) +∞ quence ni i=1 of natural numbers such that σ(ni ) = τ (ni ) (i = 1, 2, . . . ) and 113

nj −1



lim inf j→+∞

p(i) = c1 < c.

(5.23)

i=σ(nj )

Also, from the definition of the function, σ and in view of σ(ni ) = τ (ni ), for any ni , there exists ni < ni such that σ(n) = σ(ni ) for ni ≤ n ≤ ni , lim ni = +∞ and σ(ni ) = τ (ni ). Thus i→+∞

ni −1



ni −1

p(j) =

j=τ (ni )

ni −1



p(j) =

j=σ(ni )



n i −1 

p(j) ≤

j=σ(ni )

p(j) (i = 1, 2, . . . ),

j=σ(ni )

and, by the virtue of (5.23), we have ni −1

lim inf i→+∞



j=τ (ni )

n i −1 

p(j) ≤ lim inf i→+∞

p(j) = c1 < c

j=σ(ni )

In view of (5.13), the last inequality leads to a contradiction. Therefore, (5.22) holds. Now, by Lemma 5.3, we conclude that the equation Δx(n) + p(n) x(σ(n)) = 0 has a solution x∗ which satisfies the condition 0 < x∗ (n) ≤ x(n) for n ∈ Nn1 ,

(5.7)

where n1 > n0 is a sufficiently large number. Hence, taking into account that the function σ is nondecreasing, in view of Lemma 5.3, we have n−1    p(i) = +∞, lim x∗ (n) exp λ n→+∞

where λ >

4 c2 .

i=1

Therefore, by (5.7), we get 114

n−1    4 lim x(n) exp λ p(i) = +∞ for any λ > 2 . n→+∞ c i=1

The proof is complete. ( )+∞ Lemma 5.5 ([4]) (Abel transformation) Let ai i=1 and ( )+∞ bi i=1 be sequences of nonnegative numbers and +∞ 

ai < +∞.

(5.24)

i=1

Then n 

ai bi = A1 b1 − An+1 bn+1 −

i=1

where Ai =

n 

Ai+1 (bi − bi+1 ),

i=1 +∞ j=i

aj .

Proof. Since (5.24) is satisfied, we have n  i=1

Ai+1 (bi − bi+1 ) =

n 

Ai+1 bi −

i=1

n+1 

Ai bi =

i=2

= A2 b1 − An+1 bn+1 + = A2 b1 − An+1 bn+1 − = A1 b1 − An+1 bn+1 −

n 

i=2 n  i=1

115

(Ai+1 − Ai )bi =

i=2 n 

ai bi = ai bi

or n 

ai bi = A1 b1 − An+1 bn+1 −

i=1

n 

Ai+1 (bi − bi+1 ).

i=1

The proof is complete. In 2002, Koplatadze, Kvinikadze and Stavroulakis [44] established the following Lemma. Lemma 5.6 ([44]) Let φ; ψ : N → (0, +∞), ψ be nonincreasing and (5.25) lim φ(n) = +∞, n→+∞

lim inf ψ(n) φ(n) = 0, n→+∞

(5.26)

where φ(n) = inf{φ(s) : s ≥ n, s ∈ N }. ( Then )+∞ there exists an increasing sequence of natural numbers ni i=1 such that lim ni = +∞,

i→+∞

i ) = φ(ni ), φ(n

i) ψ(n) φ(n) ≥ ψ(ni ) φ(n

(n = 1, 2, . . . , ni ; i = 1, 2, . . . ) Proof. Define the sets Ei (i = 1, 2) as follows n ∈ E1 ⇔ φ(n) = φ(n), ψ(s) ≥ φ(n) ψ(n) for s ∈ {1, . . . , n} n ∈ E2 ⇔ φ(s) According to (5.25) and (5.26), it is obvious that sup Ei = +∞ (i = 1, 2). Show that sup E1 ∩ E2 = +∞. 116

(5.27)

Let n0 ∈ E2 be such that n0 ∈ / E1 . By (5.26) there is n1 > n0 1 ) for n = n0 , n0 + 1, . . . , n1 and φ(n 1) = such that φ(n) = φ(n φ(n1 ). Since ψ is nonincreasing, we have ψ(n) ≥ φ(n 1 ) ψ(n1 ) for n = 1, . . . , n1 . φ(n) Therefore n1 ∈ E1 ∩ E2 . The above argument together with the fact that sup Ei = +∞ (i = 1, 2) imply that (5.27) holds. The proof is complete. Theorem 5.2 ([4]) Assume that n0 ∈ N , Un0 = ∅, τ (n) ≤ n−1 n−1  p(i) = c > 0, (5.13) lim inf n→+∞

i=τ (n)

and

n−1 

lim sup n→+∞

p(i) < +∞.

(5.28)

i=τ (n)

  Then there exists λ ∈ 1, c42 such that 



lim sup lim inf exp (λ + ε) ε→0+

×

+∞ 

n→+∞

n−1 



p(i) ×

i=1





τ (i)−1

p(i) · exp − (λ + ε)

i=1



p(l)

≤ 1.

(5.29)

l=1

Proof. Since Uk0 = ∅, Eq.(1) has a positive solution u : Nk0 → (0, +∞). First show that lim sup x(n) exp n→+∞

n−1  i=1

117

 p(i) < +∞,

(5.30)

Indeed, if n1 ∈ Nn0 , we have n  Δx(i) i=n1

x(i)

n  x(i + 1)

− (n − n1 ) = x(i) i=n1   n  x(i + 1) = exp ln − (n − n1 ) ≥ x(i) i=n =

1



n  

1 + ln

i=n1

or

x(i + 1)  x(n + 1) − (n − n1 ) = ln , x(i) x(n1 )

n  Δx(i) i=n1

x(i)

≥ ln

x(n + 1) . x(n1 )

Furthermore, by Eq.(1) , and taking into account that the function x is nonincreasing, we have n  Δx(i) i=n1

x(i)

=−

n  i=n1

p(i)

n  x(τ (i)) p(i). ≤− x(i) i=n 1

Combining the last two inequalities, we obtain x(n + 1) exp

n 

 p(i) ≤ x(n1 ),

i=n1

that is, (5.31) is fulfilled. On the other hand, since all the conditions of Lemma 5.4 are satisfied, we conclude that condition (5.20) holds for any λ > c42 . Denote by Λ the set of all λ for which  τ (n)−1   lim x(τ (n)) exp λ p(i) = +∞, (5.31) n→+∞

i=1

118

and λ0 = inf  Λ. In view of (5.20) and (5.30), it is obvious that λ0 ∈ 1, c42 . Thus, it suffices to show, that for λ = λ0 the inequality (5.29) holds. First, we will show that for any ε > 0 

τ (n)−1

lim x(τ (n)) exp (λ0 + ε)

n→+∞



 p(i) = +∞.

(5.32)

i=1

Indeed, if λ0 ∈ Λ then, by (5.31) it is obvious that condition (5.32) is fulfilled. If λ0 ∈ Λ then, according to the definition of λ0 , it is clear that there exists λn > λ0 such that λn → λ0 when n → +∞ and λn ∈ Λ, n = 1, 2, . . . . Thus, condition (5.31) holds for any λ = λn . However, for any ε > 0, there exists λn = λn (ε) such that λ0 < λn ≤ λ0 + ε. This insures the validity of (5.31) and (5.32) for any ε > 0. Similarly, we show that for any ε > 0, 



τ (n)−1

lim inf x(τ (n)) exp (λ0 − ε)



n→+∞

p(i) = 0.

(5.33)

i=1

Hence, by virtue of (5.32) and (5.33), it is clear that for any ε > 0, the functions τ (n)−1    p(i) , φ(n) = x(τ (n)) exp (λ0 + ε)

(5.34)

i=1

and



ψ(n) = exp − 2ε

n−1 

 p(i)

i=1

satisfy the conditions of Lemma 5.6 for sufficiently ( )+∞ large n. Hence, there exists an increasing sequence ni i=1 of natural 119

numbers, such that lim ni = +∞, and i→+∞

ψ(ni ) φ(ni ) ≤ ψ(n) φ(n) for n∗ ≤ n ≤ ni ,

(5.35)

φ(ni ) = φ(ni ) (i = 1, 2, . . . ),

(5.36)

where n∗ is a sufficiently large number. Now, given that 



τ (i)−1

x(τ (i)) exp (λ0 + ε)



p(l) ≥

l=1



τ (s)−1

≥ inf x(τ (s)) exp(λ0 + ε)



  p(l) : s ≥ i, s ∈ N = φ(i),

l=1 

Eq.(1) implies x(τ (nj )) ≥

+∞ 

p(i) x(τ (i)) ≥

i=τ (nj ) +∞ 





p(i) φ(i) · exp − (λ0 + ε)

i=τ (n)



τ (i)−1



p(l)

(j = 1, 2, . . . ),

l=1

i.e., i−1 i−1       p(i) φ(i)·exp −2ε p(l) ·exp 2ε p(l) ×

nj −1

x(τ (nj )) ≥



i=τ (nj )

l=1



τ (i)−1

× exp − (λ0 + ε)

 l=1

120

l=1

 p(l) +

+

+∞ 

τ (i)−1    p(i) φ(i) · exp − (λ0 + ε) p(l)

i=nj

(j = 1, 2, . . . ).

l=1

Thus, by (5.35), and using the fact that the function φ is nondecreasing, the last inequality yields nj −1    x(τ (nj )) ≥ φ(nj ) · exp − 2ε p(l) × l=1 τ (i)−1 i−1       p(i) · exp 2ε p(l) · exp − (λ0 + ε) p(l) +

nj −1

×



i=τ (nj )

+ φ(nj )

l=1

+∞ 

l=1





τ (i)−1

p(i) · exp − (λ0 + ε)

i=nj



p(l)

(j = 1, 2, . . . ).

l=1

(5.37) Also, in view of Lemma 5.5, we have 

nj −1



I(nj , ε) =

p(i) · exp 2ε

i=τ (nj )



i−1 

 p(l) ·

l=1

τ (i)−1

· exp − (λ0 + ε)



 p(l) =

l=1

 = exp 2ε

τ (nj )−1

 i=1

p(i)

+∞  



p(i) · exp − (λ0 + ε)

i=τ (nj )

τ (i)−1

 l=1

121



p(l) −

τ (i)−1 j −1 +∞  n     − exp 2ε p(i) p(i) · exp − (λ0 + ε) p(l) + i=1 nj −1 

+



i=nj

i i−1       exp 2ε p(l) − exp 2ε p(l) ×

i=nj

×

+∞ 

l=1

l=1

l=1





τ (i)−1

p(i) · exp − (λ0 + ε)

i=1



p(l)

(j = 1, 2, . . . ), (5.38)

l=1

Given that  exp 2ε

i 



i−1    p(l) − exp 2ε p(l) ≥ 0,

l=1

l=1

inequality (5.38) becomes 

I(nj , ε) ≥ exp 2ε

τ (nj )−1



p(i)

+∞  



τ (i)−1

exp − (λ0 + ε) 

− exp 2ε

p(i) ·

i=τ (nj )

i=1



 p(l) −

l=1 nj −1



p(i)

i=1

+∞ 

p(i) ·

i=nj

   p(l) . exp − (λ0 + ε) τ (i)−1

l=1

122

Therefore, by (5.37), we take nj −1 j )−1    τ (n    p(l) · exp 2ε p(l) × u(τ (nj )) ≥ φ(nj ) · exp − 2ε l=1 +∞ 

×

l=1





τ (i)−1

p(i) · exp − (λ0 + ε)



i=τ (nj )

p(l) .

l=1

Thus, (5.36) and (5.34) imply 

τ (nj )−1



exp (λ0 + ε)

p(i)

+∞  



τ (i)−1

p(i) · exp − (λ0 + ε)

i=τ (nj )

i=1





p(l) ≤

l=1

nj −1    p(i) . ≤ exp 2ε i=τ (nj )

From the last inequality, and taking into account that (5.28) is satisfied, we have 

τ (nj )−1

lim sup exp (λ0 + ε) j→+∞

 i=1



τ (i)−1

· exp − (λ0 + ε)



p(i)

+∞  

p(i)·

i=τ (nj )



p(l) ≤ exp(2εM ),

(5.39)

l=1

where M = lim sup

n−1

p(i). Hence, for any ε > 0, (5.39) gives

n→+∞ i=τ (n) n−1 +∞    p(i) p(i)· lim inf exp (λ0 + ε) n→+∞

i=1

123

i=n



τ (i)−1

· exp − (λ0 + ε)



 p(l) ≤ exp(2εM ),

l=1

which implies 

n−1 +∞    p(i) p(i)· lim sup lim inf exp (λ0 + ε) ε→0+

n→+∞

i=1



τ (i)−1



· exp − (λ0 + ε)

i=n

 p(l) ≤ 1.

l=1

The proof is complete. Remark 5.3 ([4]) Condition (5.28) is not a limitation since, by Remark 5.1 if n  lim sup p(i) > 1 n→+∞

i=τ (n)

then Un0 = ∅, for any n0 ∈ N . Remark 5.4 ([4]) In (5.13), without loss of generality, we may assume that c ≤ 1. Otherwise, for any n0 ∈ N, we have Un0 = ∅ [3]. Theorem 5.3 ([4]) Assume that all the conditions of Theorem 5.2 are satisfied. Then lim inf n→+∞

n−1 

p(i) ≤

i=τ (n)

1 ., e

(5.40)

Proof. Since all the conditions of Theorem 5.2 are satisfied,  4 there exists λ = λ0 ∈ 1, c2 such that the inequality (5.29) holds. 124

Assume that the condition (5.40) does not hold. Then, there exists n1 ∈ N and ε0 > 0 such that n−1 

p(i) ≥

i=τ (n)

1 + ε0 e

for n ∈ Nn1 .

Therefore, for any ε > 0  I(n, ε) = exp (λ0 + ε)

n−1 

p(i)

i=1



τ (i)−1

· exp − (λ0 + ε)



+∞ 

p(i) ·

i=n

 p(l) ≥

l=1 n−1   (λ + ε)(1 + ε )    0 0 p(i) × ≥ exp · exp (λ0 + ε) e i=1

×

+∞ 



p(i) · exp − (λ0 + ε)

i=n

Defining

i−1 

 p(l)

for n ∈ Nn1 .

(5.41)

l=1 i−1

p(l) = ai−1 , we will show that

l=1 +∞     p(i) · exp − (λ0 + ε)ai−1 ≥ lim inf exp (λ0 + ε)an−1 n→+∞

i=n

Indeed, since lim inf +∞ i=1

n−1

n→+∞ i=τ (n)

1 . λ0 + ε

p(i) = c > 0, it is obvious that

p(i) = +∞, i.e. lim ai = +∞. i→+∞

125

Therefore +∞     exp (λ0 + ε)an−1 p(i) · exp − (λ0 + ε)ai−1 = i=n

  = exp (λ0 + ε)an−1 × ×

+∞ 

  (ai − ai−1 ) · exp − (λ0 + ε)ai−1 =

i=n +∞   = exp (λ0 + ε)an−1 exp(−(λ0 + ε)ai−1 )

 ≥ exp (λ0 + ε)ai−1  = exp (λ0 + ε)ai−1

i=n  +∞   ai



ai

ds ≥

ai−1

exp(−(λ0 + ε)s)ds =

i=n ai−1  +∞ 

exp(−(λ0 + ε)s)ds =

ai−1

1 . λ0 + ε

Hence, by (5.41), we obtain  λ (1 + ε )  1 0 0 ≥ 1 + ε0 . · exp lim sup lim inf I(n, ε) ≥ n→+∞ λ0 e ε→0+ 



This contradicts (5.29) for λ = λ0 . The proof is complete. Theorem 5.4 ([4]) Assume that τ (n) ≤ n − 1, the conditions (5.13), (5.28) are satisfied and for any λ ∈ 1, c42 



lim sup lim inf ε→0+

n→+∞

 exp (λ+ ε)

n−1  i=1

126



p(i) ×

×

+∞ 





τ (i)−1



p(i) exp − (λ + ε)

i=n

p(l)

> 1.

(5.42)

l=1

Then all proper solutions of Eq.(1) oscillate. Proof. Assume that x : Nn0 → (0, +∞) is a positive proper solution of Eq.(1) . Then n 0 = ∅. Thus, in view of Theorem  U 4 5.2, there exists λ0 ∈ 1, c2 such that the condition (5.29) is satisfied for λ = λ0 . But this contradicts (5.42) and the proof is complete. Using Theorem 5.3, we can similarly prove the following theorem. Theorem 5.5 ([4]) Assume that τ (n) ≤ n − 1, n−1 

lim sup n→+∞

p(i) < +∞

(5.28)

1 p(i) > . e

(C3 )

i=τ (n)

and lim inf n→+∞

n−1  i=τ (n)

Then all proper solutions of Eq.(1) oscillate. Remark 5.5 ([4]) It is to be pointed out that condition (C3 ) is the discrete analogue of the condition (C3 ) for the difference equation (1) in the case of a general delay argument τ (n). Remark 5.6 ([4]) The condition (C3 ) is optimal for Eq.(1) under the assumption that lim (n − τ (n)) = +∞, since in this n→+∞

127

case the set of natural numbers increases infinitely in the interval [τ (n), n − 1] for n → +∞. Now, we are going to present two examples to show that the condition (C3 ) is optimal, in the sense that it cannot be replaced by the non-strong inequality. Example 5.6 ([4]) Consider Eq.(1) , where τ (n) = [an], p(n) = (n−λ − (n + 1)−λ )([an])λ ,

(5.43)

a ∈ (0, 1), λ = − ln−1 a, and [an] denotes the integer part of an. It is obvious that   n1+λ n−λ − (n + 1)−λ → λ for n → +∞. Therefore   λ n n−λ − (n + 1)−λ [an]λ → for n → +∞. e Hence, in view of (5.43) and (5.44), we have lim inf n→+∞

n−1  i=τ (n)

n−1   λ 1 e  −λ p(i) = lim inf i i − (i + 1)−λ [ai]λ · e k→+∞ λ i i=[an]

n−1  λ 1 λ 1 1 = lim inf = ln = e n→+∞ i e a e i=[an]

or lim inf n→+∞

n−1  i=τ (n)

1 p(i) = . e

Observe that all the conditions of Theorem 5.5 are satisfied except the condition (C3 ) . In this case, it is not guaranteed 128

that all solutions of Eq.(1) oscillate. Indeed, it is easy to see that the function u = n−λ is a positive solution of Eq.(1) . Example 5.2 ([4]) Consider Eq.(1) where τ (n) = [nα ], p(n) = (ln−λ n − ln−λ (n + 1)) lnλ [nα ],

(5.45)

a ∈ (0, 1), λ = − ln a, and [na ] denotes the integer part of na . It is obvious that   n ln1+λ n ln−λ n − ln−λ (n + 1) → λ f or

n → +∞.

Therefore   λ n ln n lnλ [na ] ln−λ n − ln−λ (n + 1) → e

for n → +∞. (5.46)

On the other hand n−1  i=[na ]

 n n−1  i+1  1 ds ds ln n ≥ = = ln i ln i s ln s ln[na ] i [na ] s ln s a i=[n ]

which tends to ln(1/a) as n → ∞, and  n−1 n−1 n−1  i   1 ds ds ln(n − 1) ≤ = == ln i ln i ln[na ] − 1 i−1 s ln s [na ]−1 s ln s a a i=[n ]

i=[n ]

which also tends to ln(1/α) as n → +∞. Together these two bounds imply n−1  1 1 = ln . lim n→+∞ i ln i a a i=[n ]

129

Hence, in view of (5.45) and (5.46), we obtain lim inf n→+∞

n−1 

n−1 

p(i) = lim inf n→+∞

i=[na ]

  lnλ [ia ] ln−λ i − ln−λ (i + 1) =

i=[na ]

n−1    1 λ λ e = lim inf i ln i lnλ [ia ] ln−λ i − ln−λ (i + 1) e n→+∞ λ i ln i e a i=[n ]

lim inf n→+∞

n−1  i=[na ]

1 λ 1 1 = ln = . i ln i e a e

We again observe that all the conditions of Theorem 5.5 are satisfied except the condition (C3 ) . In this case, it is not guaranteed that all solutions of Eq.(1) oscillate. Indeed, it is easy to see that the function x = ln−λ n is a positive solution of Eq.(1) . It is interesting to establish sufficient conditions for the oscillation of all solutions of the delay difference equation (1) in the case where (C2 ) nor (C˜3 ) or (C3 ) is fulfilled.This question has been investigated by several authors in the special case of equation (1) when neither (C2 ) nor (C3 ) is satisfied. (See Section 4). In the case of Eq.(1) with a general delay argument, this question was investigated for the first time in 2008, by Chatzarakis, Koplatadze and Stavroulakis [3]. In particular, the following results have been established in [3]. Lemma 5.7 ([3]) Assume that x is a positive proper solution of Eq.(1) , where p : N → R+ , τ : N → N is nondecreasing function, (5.47) 130

τ (n) ≤ n − 1, for n ∈ N, and lim inf n→+∞

n−1 

lim τ (n) = +∞

n→∞

p(i) = bsα ∈ (0, 1],

(5.48)

i=τ (n)

√ 2  x(τ (n)) 1 + 1 − bsα . ≤ lim sup bsα n→+∞ x(n + 1) √ If, additionally, p(n) ≥ 1 − 1 − α for large n, then √ x(τ (n)) 1−α+ 1−α ≤ lim sup . α2 n→+∞ x(n + 1) Then

(5.49)

(5.50)

Proof. By (5.48) it is clear that, for any ε ∈ (0, α) there exists k0 = k0 (ε) ∈ N such that n−1 

p(i) ≥ α − ε for n ∈ Nk0 ,

(5.51)

i=τ (n)

Since x is a positive proper solution of Eq.(1) , then there exists k1 ∈ Nk0 such that x(τ (n)) > 0 for n ∈ Nk1 . Thus, from Eq.(1) , we have x(n + 1) − x(n) = −p(n) x(τ (n)) ≤ 0 and so x is an eventually nonincreasing function of positive numbers.

131

From (5.51), it is clear that, if ω ∈ (0, α − ε), there exists n∗ ≥ n such that ∗ n −1



p(i) < ω and

i=n

n 

p(i) ≥ ω,

(5.52)

i=n

This is because in the case where p(n) < ω, it is clear that, there exists n∗ > n such that (5.52) is satisfied, while in the case where p(n) ≥ ω, then n∗ = n, and therefore ∗ n −1

p(i) =

i=n

n−1 

p(i) (by which we mean) = 0 < ω

i=n

and ∗

n 

p(i) =

i=n

n 

p(i) = p(n) ≥ ω.

i=n

That is, in both cases (5.52) is satisfied. Thus n−1  i=τ (n∗ )

p(i) =

∗ n −1

p(i) −

∗ n −1

i=τ (n∗ )

p(i) ≥ (α − ε) − ω.

i=n

Now, summing up Eq.(1) first from n to n∗ and then from τ (n∗ ) to n − 1, and using the fact that the function x is nonincreasing and the function τ is nondecreasing, we have  n∗  n∗   x(n) − x(n∗ + 1) = p(i) x(τ (i)) ≥ p(i) i=n

i=n

x(τ (n∗ )) ≥ ω x(τ (n∗ )) 132

or

x(n) ≥ x(n∗ + 1) + ωx(τ (n∗ )),

(5.53)

and then n−1 



x(τ (n )) − x(n) =

p(i) x(τ (i)) ≥

i=τ (n∗ )

⎛ ≥ ⎝

n−1 

⎞ p(i) ⎠ x(τ (n − 1)) ≥

i=τ (n∗ )

≥ [(α − ε) − ω]x(τ (n − 1)) or

x(τ (n∗ )) ≥ x(n) + [(α − ε) − ω]x(τ (n − 1)).

(5.54)

Combining inequalities (5.53) and (5.54), we obtain x(n) ≥ x(n∗ + 1) + ω[x(n) + ((α − ε) − ω)x(τ (n − 1))] or

ω[(α − ε) − ω] x(τ (n − 1)). (5.55) 1−ω Observe that the function f : (0, α − ε) → (0, 1) defined as x(n) ≥

ω[(α − ε) − ω] . (5.56) 1−ω attains its maximum at ω = 1 − 1 − (α − ε), which is equal to  2 fmax = 1 − 1 − (α − ε) . Thus, for ω = 1 − 1 − (α − ε) ∈ (0, α − ε) inequality (5.55) becomes  2 x(n) ≥ 1 − 1 − (α − ε) x(τ (n − 1)) f (ω) : =

133

or x(τ (n − 1)) ≤ x(n) and, for large n, we have x(τ (n)) ≤ x(n + 1)





2 1 + 1 − (α − ε) , α−ε 2 1 − (α − ε) . α−ε

1+

Hence, x(τ (n)) lim sup ≤ n→+∞ x(n + 1)



(5.57)

1+

2 1 − (α − ε) , α−ε

which, for arbitrarily small values of ε, implies (5.49). √ Next we consider the particular case where p(n) ≥ 1− 1 − α. In this case, from Eq.(1) we have √ x(n) = x(n + 1) + p(n) x(τ (n)) ≥ (1 − 1 − α) x(τ (n)). (5.58) Now, summing up Eq.(1) from τ (n) to n − 1, and using the fact that the function x is nonincreasing and the function τ is nondecreasing, we have x(τ (n)) − x(n) =

n−1 

p(i) x(τ (i)) ≥

i=τ (n)

⎛ ≥ ⎝

n−1 

⎞ p(i)⎠ x(τ (n − 1)) ≥

i=τ (n)

≥ (α − ε)x(τ (n − 1)) or x(τ (n)) ≥ x(n) + (α − ε) x(τ (n − 1)). 134

(5.59)

Combining inequalities (5.58) and (5.59), we obtain √ x(n) ≥ (1 − 1 − α)[x(n) + (α − ε)x(τ (n − 1))] √ x(τ (n − 1)) 1−α+ 1−α ≤ , x(n) α(α − ε) and, for large n, √ x(τ (n)) 1−α+ 1−α ≤ . x(n + 1) α(α − ε)

or

(5.60)

√ x(τ (n)) 1−α+ 1−α ≤ . lim sup α(α − ε) n→+∞ x(n + 1) The last inequality, for arbitrarily small values of ε, implies (5.50). The proof is complete. Hence

Theorem 5.6 ([3]) Assume that τ (n) ≤ n − 1, n−1 

p(i) = α ∈ (0, 1],

(5.48)

√  2 p(i) > 1 − 1 − 1 − α ,

(5.61)

( ) σ(n) = max τ (s) : 1 ≤ s ≤ n, s ∈ N ,

(5.1)

lim inf n→+∞

and lim sup n→+∞

where

i=τ (n)

n  i=σ(n)

Then all proper solutions of Eq.(1) oscillate.

135

If, additionally, p(n) ≥ 1 −



1 − α for large n, and √ n  1− 1−α , lim sup p(i) > 1 − α √ 1 − α n→+∞ i=σ(n)

(5.62)

then all proper solutions of Eq.(1) oscillate. Proof. We will first show that lim inf n→+∞

n−1 

p(i) = α.

(5.63)

i=σ(n)

Indeed, since τ (n) ≤ σ(n), then by (5.48), it is obvious that n−1 

lim inf n→+∞

n−1 

p(i) ≤ lim inf n→+∞

i=σ(n)

p(i) = α.

(5.64)

i=τ (n)

Thus, there exists a subsequence {ni }+∞ i=1 of natural numbers such that ni ↑ +∞ for i → +∞ and n−1 

lim inf n→+∞

p(i) = lim

n→+∞

i=σ(n)

n i −1 

p(j).

(5.65)

j=σ(ni )

On the other hand, from the definition of the function σ and taking into account that lim τ (n) = +∞ for any ni (i = n→+∞

1, 2, . . . ) there exists ni ≤ ni such that σ(n) = σ(ni ) since ni ≤ n ≤ ni , lim ni = +∞, and σ(ni ) = τ (ni ) (i = 1, 2, . . . ). Thus, n i −1  j=σ(ni )

i→+∞

ni −1

p(j) ≥



j=σ(ni )

ni −1

p(j) =



j=σ(ni )

136

ni −1

p(j) =



j=τ (ni )

p(j).

(5.66)

Combining inequalities (5.65) and (5.66), we obtain lim inf n→+∞

n−1  i=σ(n)

p(i) ≥ lim inf n→+∞

n−1 

p(i) = α.

(5.67)

i=τ (n)

The last inequality together with (5.64) imply (5.63). Now assume, for the sake of contradiction, that x is a positive proper solution of Eq.(1) . Then, for sufficiently large n, the function x is a positive proper solution of Δx(n) + p(n) x(σ(n)) ≤ 0. By Lemma 5.1, the equation Δx(n) + p(n) x(σ(n)) = 0

(5.68)

has a positive proper solution x∗ : Nk0 → (0, +∞), where k0 ∈ N is sufficiently large. Since (5.63) is satisfied, inequality (5.49) becomes √  2 1+ 1−α x∗ (σ(n)) lim sup , (5.69) ≤ α n→+∞ x∗ (n + 1) √ or, if p(n) ≥ 1 − 1 − α for sufficiently large n, then inequality (5.50) becomes √ x∗ (σ(n)) 1−α+ 1−α . (5.70) ≤ lim sup α2 n→+∞ x∗ (n + 1) √ In the case that (5.69) holds, for any ε ∈ (0, (1 − 1 − α)2 ) and for sufficiently large n, we have √   (5.71) x∗ (n + 1) ≥ (1 − 1 − α)2 − ε x∗ (σ(n)). 137

Now, summing up Eq.(5.68) from σ(n) to n, and using the fact that the function x∗ is nonincreasing and the function σ is nondecreasing, we have ⎛ ⎞ n  x∗ (σ(n)) ≥ x∗ (n + 1) + ⎝ p(i)⎠ x∗ (σ(n)). (5.72) i=σ(n)

Combining inequalities (5.71) and (5.72), we obtain   n  √ 2  x∗ (σ(n)) ≥ 1 − 1 − α − ε + p(i) x∗ (σ(n)). i=σ(n)

Hence n 

lim sup n→+∞

√  2 p(i) ≤ 1 − 1 − 1 − α + ε,

i=σ(n)

which, for arbitrarily small values of ε, becomes lim sup n→+∞

n 

√  2 p(i) ≤ 1 − 1 − 1 − α .

i=σ(n)

This, contradicts (5.61). In the case that (5.70) holds, following a similar procedure, we are led to the inequality √ n  1− 1−α p(i) ≤ 1 − α √ lim sup 1−α n→+∞ i=σ(n) which contradicts (5.62). The proof is complete. 138

Remark 5.7 ([3]) If α > 1, by (5.5), it is obvious, that the conditions of Theorem 5.1 are satisfied and therefore all proper solutions of Eq.(1) oscillate. Corollary 5.1 ([3]) Assume that 0 < α0 : = lim inf n→+∞

and lim sup n→+∞

n 

n−1 

 p(i) ≤

i=n−k

k k+1

k+1

2  √ p(i) > 1 − 1 − 1 − αo .

(5.61)

i=n−k

Then all proper solutions of Eq.(1) oscillate. If, additionally, for sufficiently large n, p(n) ≥ 1 − and √ n  1 − 1 − α0 p(i) > 1 − α0 √ , lim sup 1 − α0 n→+∞ i=n−k



1 − α0 , (5.62)

then all proper solutions of Eq.(1) oscillate. Now we present an example in which the condition (5.61 ) of the above Corollary is satisfied, while none of the conditions (C2 ) , (C3 ) , (4.5), (C4 ) , (C˜4 ) and (4.9), is satisfied. Example 5.3 ([3]) Consider the equation x(n + 1) − x(n) + p(n)x(n − 12) = 0,

n = 0, 1, 2, ...,

where p(13n + 1) = ... = p(13n + 12) = 139

35 , 1200

6 35 + , n = 0, 1, 2, .... 1200 10 Here n = 12 and it is easy to see that  13 n−1  35 12 p(i) = α0 = lim inf  0.3532 < n→∞ 100 13 i=n−12 p(13n + 13) =

n−1 

lim sup n→∞

and

i=n−12

n 

35 6 + = 0.950 100 10

35 950 + = 0.9791 > 1200 1000 i=n−12 2  √ > 1 − 1 − 1 − α0  0.9624.

lim sup n→∞

p(i) =

p(i) =

We see that the condition (5.61) of Corollary 5.1 is satisfied and therefore all solutions oscillate. Observe, however, that 0.9791 < 1,  13 12  0.3532, α0 = 0.35 < 13 α2  0.9693 4 α2 0.950 < 1 −  0.9693 4 0.950 < 1 − α12  0.9999, 0.692 < 1 −

and

α2 0.950 < 1 −  0.9628. 2(2 − α) 140

Therefore none of the conditions (C2 ) , (C3 ) , (4.5), (C4 ) , (C˜4 ) and (4.9), is satisfied. Theorem 5.7 ([3]) Assume that α ∈ (0, 1] and there exist k0 ∈ N and a function p ∈ Lloc (R+ , R+ ) such that t2 p (t) is nondecreasing function, p (i) ≤ p(i) for i ∈ Nk0 , (5.73) and  n−1 (5.74) lim inf p (s)ds ≥ α. n→+∞

τ (n)−1

Then condition (5.61) (or, if for sufficiently large n, p(n) ≥ √ 1− 1 − α, condition (5.62)) is sufficient for all proper solutions of Eq.(1) to oscillate. Proof. In view of Lemma 5.1 and Theorem 5.6, to prove Theorem 5.7, it suffices to show that lim inf n→+∞

n−1 

p(i) ≥ α.

(5.75)

i=τ (n)

By (5.73) and (5.74), we have n−1  i=τ (n)

  i n−1 n−1   (i − 1)i2 ds i−1 i p(i) ≥ ≥ p (s) ds ≥ p (i) 2 i i i−1 s i−1 i=τ (n)

i=τ (n)

 n−1  i τ (n) − 1  τ (n) − 1 n−1 ≥ p (s) ds = p (s) ds. τ (n) τ (n) i−1 τ (n)−1 i=τ (n)

Since τ (n) → +∞ for n → +∞, the last inequality, in view of (5.74), implies (5.75). The proof is complete 141

Corollary 5.2 ([3]) Consider Eq.(1) and let c ∈ (0, +∞), β ∈ (0, 1), c ln β ≥ −1 and for large n p(n) ≥ and lim sup n→+∞

n 

c , τ (n) ≤ [βn], n

 2 p(i) > 1 − 1 − 1 − γ ,

i=[βn]

where γ = ln β −c and [βn] denotes the integer part of βn. Then all proper solutions of Eq.(1) oscillate. Proof. Take p (t) = ct and α = ln β −c . Then it is easily shown that the conditions of Theorem 5.7 are satisfied. Analogously, if we take p (t) =

c t ln t ,

we have the following

Corollary 5.3 ([3]) Consider Eq.(1) and let c ∈ (0, +∞), β ∈ (0, 1), c ln β ≥ −1 and for large n p(n) ≥ and lim sup n→+∞

n 

c , τ (n) ≤ [nβ ]. n ln n  2 p(i) > 1 − 1 − 1 − γ ,

i=[nβ ]

where γ = ln β −c and [nβ ] denotes the integer part of nβ . Then all proper solutions of Eq.(1) oscillate. Recently new oscillation criteria for the solutions of Eq.(1) have been established by Chatzarakis, Philos and Stavroulakis [5,6] which essentially improve the corresponding criteria in [3], 142

as well as all the known corresponding criteria concerning the special case of Eq.(1) . In particular, the following results have been established in [5]. Consider the delay difference equation Δx(n) + p(n)x(τ (n)) = 0,

(1)

where (p(n))n≥0 is a sequence of nonnegative real numbers, and (τ (n))n≥0 is a sequence of integers such that τ (n) ≤ n − 1 for all n ≥ 0,

and

lim τ (n) = ∞

n→∞

Set k = − min τ (n). n≥0

(Clearly, k is a positive integer.) By a solution of the delay difference equation (1) , we mean a sequence of real numbers (x(n))n≥−k which satisfies Eq.(1) for all n ≥ 0. It is clear that, for each choice of real numbers c−k , c−k+1 , ..., c−1 , c0 , there exists a unique solution (x(n))n≥−k of Eq.(1) which satisfies the initial conditions x(−k) = c−k , x(−k + 1) = c−k+1 , ..., x(−1) = c−1 , x(0) = c0 . As usual, a solution (x(n))n≥−k of the delay difference equation (1) is called oscillatory if the terms x(n) of the sequence are neither eventually positive nor eventually negative, and otherwise the solution is said to be nonoscillatory. Lemma 5.8 ([5]) Assume that the sequence τ (n) is increasing, τ (n) ≤ n − 1 and set 143

α = lim inf n→∞

n−1 

p (j) .

(5.76)

j=τ (n)

Let (x(n))n≥−k be a nonoscillatory solution of the delay difference equation (1) . Then we have: (i) If 0 < α ≤ 12 , then √  x(n + 1) 1 1 − α − 1 − 2α . ≥ n→∞ x(τ (n)) 2 √ (ii) If 0 < α ≤ 6 − 4 2 and, in addition, lim inf

p(n) ≥

α for all large n, 2

(5.77)

(5.78)

then  x(n + 1) 1 2 lim inf ≥ 2 − 3α − 4 − 12α + α . n→∞ x(τ (n)) 4

(5.79)

Note. If 0 < α ≤ 12 , then 1 − 2α ≥ 0 and √  1 1 1 − α − 1 − 2α < . 2 2 √ √ Also, when 0 < α ≤ 6 − 4 2 (clearly, 6 − 4 2 < 12 ), we have 4 − 12α + α2 ≥ 0 and  1 1 0< 2 − 3α − 4 − 12α + α2 < . 4 2 0
1 − α − 1 − 2α . (5.80) 4 2 √ Therefore, in the case where 0 < α ≤ 6 − 4 2 and (5.78) holds, inequality (5.80) guarantees that (5.79) is an improvement of (5.77). Proof. Since the solution (x(n))n≥−k of the delay difference equation (1) is nonoscillatory, it is either eventually positive or eventually negative. As (−x(n))n≥−k is also a solution of Eq.(1) , we may (and do) restrict ourselves only to the case where x(n) > 0 for all large n. Let ρ ≥ −k be an integer such that x(n) > 0 for all n ≥ ρ, and consider an integer r ≥ 0 so that τ (n) ≥ ρ for n ≥ r (clearly, r > ρ). Then it follows immediately from Eq.(1) that Δx(n) ≤ 0 for every n ≥ r, which means that the sequence (x(n))n≥r is decreasing. Assume that 0 < α ≤ 12 , where α is defined by (5.76). Consider an arbitrary real number  with 0 <  < α. Then we can choose an integer n0 > r such that τ (n) ≥ r for n ≥ n0 , and n−1 

p(j) ≥ α −  for all n ≥ n0 .

(5.81)

j=τ (n)

Furthermore, let us consider an arbitrary real number ω with 0 < ω < α − . We will establish the following claim. Claim. For each n ≥ n0 , there exists an integer n∗ ≥ n such that τ (n∗ ) ≤ n − 1, and ∗

n 

p(j) ≥ ω,

j=n

145

(5.82)

and

n−1 

p(j) > (α − ) − ω.

(5.83)

j=τ (n∗ )

To prove this claim, let us consider an arbitrary integer n ≥ n0 . Assume, first, that p(n) ≥ ω, and choose n∗ = n. Then τ (n∗ ) = τ (n) ≤ n − 1. Moreover, we have ∗

n 

p(j) =

n 

p(j) = p(n) ≥ ω

j=n

j=n

and, by (5.81), n−1  j=τ (n∗ )

p(j) =

n−1 

p(j) ≥ a −  > (a − ) − ω.

j=τ (n)

So, (5.82) and (5.83) are fulfilled. Next, we suppose that p(n) < ω. It is not difficult to see that (5.81) guarantees that ∞ 

p(j) = ∞.

j=0

In particular, it holds ∞ 

p(j) = ∞.

j=n

Thus, as p(n) < ω, there always exists an integer n∗ > n so that ∗ n −1 p(j) < ω, (5.84) j=n

146

and (5.82) holds. We assert that τ (n∗ ) ≤ n − 1. Otherwise, τ (n∗ ) ≥ n. We also have τ (n∗ ) ≤ n∗ − 1. Hence, in view of (5.84), we get ∗ n −1

p(j) ≤

j=τ (n∗ )

∗ n −1

p(j) < ω.

j=n

On the other hand, (5.81) gives ∗ n −1

p(j) ≥ α −  > ω.

j=τ (n∗ )

We have arrived at a contradiction, which shows our assertion. Furthermore, by using (5.81) (for the integer n∗ ) as well as (5.84), we obtain n−1 

p(j) =

j=τ (n∗ )

∗ n −1

p(j) −

j=τ (n∗ )

∗ n −1

p(j) > (a − ) − ω

j=n

and consequently (5.83) holds true. Our claim has been proved. Next, we choose an integer N > n0 such that τ (n) ≥ n0 for n ≥ N . Let us consider an arbitrary integer n ≥ N . By our claim, there exists an integer n∗ ≥ n such that τ (n∗ ) ≤ n−1, and (5.82) and (5.83) hold. By taking into account the facts that the sequence (τ (s))s≥0 is increasing and that the sequence (x(t))t≥r is decreasing and by using (5.82), from Eq.(1) , we obtain ∗



x(n) − x(n + 1) = 



n 



n 

p(j)x(τ (j)) ≥

j=n

p(j) x(τ (n∗ )) ≥ ωx(τ (n∗ ))

j=n

147

and consequently x(n) ≥ x(n∗ + 1) + ωx(τ (n∗ )).

(5.85)

Furthermore, by taking again into account the facts that (τ (s))s≥0 is increasing and that (x(t))t≥r is decreasing and by using (5.83), from Eq.(1) , we derive  n∗  n−1   x(τ (n∗ )) − x(n) = p(j)x(τ (j)) ≥ p(j) x(τ (n − 1)) j=τ (n∗ )

j=n

> [(α − ) − ω] x(τ (n − 1)) and so x(τ (n∗ )) > x(n) + [(α − ) − ω]x(τ (n − 1)).

(5.86)

By (5.85) and (5.86), we get x(n) ≥ x(n∗ + 1) + ωx(τ (n∗ )) > ωx(τ (n∗ )) > > ω {x(n) + [(α − ) − ω]x(τ (n − 1))} and hence

(α − ) − ω x(τ (n − 1)). 1−ω We have thus proved that x(n) > ω

x(n) > ωλ1 x(τ (n − 1)) for all n ≥ N , where

(5.87)

(α − ) − ω . 1−ω Now, let n be an arbitrary integer with n ≥ N . By using our claim, we conclude that there exists an integer n∗ ≥ n such that λ1 =

148

τ (n∗ ) ≤ n − 1, and (5.82) and (5.83) are satisfied. Then (5.85) and (5.86) are also fulfilled. Moreover, in view of (5.87) (for the integer n∗ + 1), we have x(n∗ + 1) > ωλ1 x(τ (n∗ )).

(5.88)

By the use of (5.85), (5.88) and (5.86), we obtain x(n) ≥ x(n∗ + 1) + ωx(τ (n∗ )) > > ωλ1 x(τ (n∗ )) + ωx(τ (n∗ )) = ω(λ1 + 1)x(τ (n∗ )) > > ω(λ1 + 1) {x(n) + [(α − ) − ω]x(τ (n − 1))} , which gives [1 − ω(λ1 + 1)]x(n) > ω(λ1 + 1)[(α − ) − ω]x(τ (n − 1)). This implies, in particular, that 1 − ω(λ1 + 1) > 0. Consequently, x(n) > ω

(λ1 + 1)[(α − ) − ω] x(τ (n − 1)). 1 − ω(λ1 + 1)

Thus, it has been shown that x(n) > ωλ2 x(τ (n − 1)) for all n ≥ N , 149

where

(λ1 + 1)[(α − ) − ω] . 1 − ω(λ1 + 1) Following the above procedure, we can inductively construct a sequence of positive real numbers (λν )ν≥1 with λ2 =

1 − ω(λν + 1) > 0 (ν = 1, 2, ...) and λν+1 =

(λν + 1)[(α − ) − ω] 1 − ω(λν + 1)

(ν = 1, 2, ...)

such that x(n) > ωλν x(τ (n−1)) for all n ≥ N

(ν = 1, 2, ...). (5.89)

As λ1 > 0, we obtain λ2 =

(λ1 + 1)[(α − ) − ω] (α − ) − ω > = λ1 , 1 − ω(λ1 + 1) 1−ω

i.e., λ2 > λ1 . By an easy induction, one can see that the sequence (λν )ν≥1 is strictly increasing. Furthermore, by taking into account the fact that the sequence (x(t))t≥r is decreasing and by using (5.89) (for n = N ), we get x(τ (N − 1)) ≥ x(N ) > ωλν x(τ (N − 1)) (ν = 1, 2, ...). Therefore, for each ν ≥ 1, we have ωλν < 1, i.e., λν < This ensures that the sequence (λν )ν≥1 is bounded. Since (λν )ν≥1 is a strictly increasing and bounded sequence of positive real numbers, it follows that limν→∞ λν exists as a positive real number. Set Λ = lim λν . 1 ω.

ν→∞

150

Then (5.89) gives x(n) ≥ ωΛx(τ (n − 1)) for all n ≥ N .

(5.90)

By the definition of (λν )ν≥1 , we have Λ= i.e.,

(Λ + 1)[(α − ) − ω] , 1 − ω(Λ + 1)

ωΛ2 − [1 − (α − )]Λ + [(α − ) − ω] = 0.

Hence, either 0 1 / Λ= 1 − (α − ) − 1 − 2(α − ) + [(α − ) − 2ω]2 2ω or Λ=

0 1 / 1 − (α − ) + 1 − 2(α − ) + [(α − ) − 2ω]2 . 2ω

In both cases, it holds 0 1 / Λ≥ 1 − (α − ) − 1 − 2(α − ) + [(α − ) − 2ω]2 . 2ω Thus, from (5.90), it follows that 1/ x(n) ≥ 1 − (α − ) − 1 − 2(α − )+ 2 +[(α − ) − 2ω]2 x(τ (n − 1)),

(5.91)

for all n ≥ N . But, we can immediately see that the function 0 1/ 1 − (α − ) − 1 − 2(α − ) + [(α − ) − 2ω]2 f (ω) = 2 for 0 < ω < α −  151

attains its maximum at the point ω = α− 2 . So, by choosing ω = α− , from (5.91) we obtain 2  1 x(n) ≥ 1 − (α − ) − 1 − 2(α − ) x(τ (n − 1)), 2 for all n ≥ N.

(5.92)

Inequality (5.92) gives  1 x(n + 1) ≥ 1 − (α − ) − 1 − 2(α − ) x(τ (n)) 2 for every n ≥ N − 1 or  x(n + 1) 1 ≥ 1 − (α − ) − 1 − 2(α − ) for all n ≥ N − 1. x(τ (n)) 2 Consequently, lim inf n→∞

 x(n + 1) 1 ≥ 1 − (α − ) − 1 − 2(α − ) . x(τ (n)) 2

The last inequality holds true for all real numbers  with 0 <  < α. Hence, we can obtain (5.77). The proof of Part (i) of the lemma has been completed. In the remainder√ of the proof of the lemma, it will be assumed that 0 < α ≤ 6 − 4 2 ( which implies that 0 < α < 12 ) and, in addition, that (5.78) holds. Because of (5.78), we can consider an integer L ≥ N such that p(n) ≥ α2 for every n ≥ L. Then p(n) >

α− 2

for all n ≥ L., 152

(5.93)

By (5.92), we have x(n) ≥ θ1 x(τ (n − 1)) for all n ≥ L,

(5.94)

where

 1 1 − (α − ) − 1 − 2(α − ) . 2 Let us consider an arbitrary integer n ≥ L. By using (5.93) as well as (5.94) (for the integer n + 1), from Eq.(1) we obtain θ1 =

x(n) = x(n + 1) + p(n)x(τ (n)) > x(n + 1) + ≥ θ1 x(τ (n)) + and consequently

α− x(τ (n)) 2

α− x(τ (n)) 2

 α− x(τ (n))., θ1 + 2



x(n) >

(5.95)

Furthermore, by taking into account the facts that (τ (s))s≥0 is increasing and that (x(t))t≥r is decreasing and by using (5.81), from Eq.(1) , we derive x(τ (n)) − x(n) =

n−1 

p(j)x(τ (j)) ≥

j=τ (n)

⎛ ≥ ⎝

n−1 

⎞ p(j)⎠ x(τ (n − 1)) ≥

j=τ (n)

≥ (α − )x(τ (n − 1)) and hence x(τ (n)) ≥ x(n) + (α − )x(τ (n − 1)), 153

(5.96)

A combination of (5.95) and (5.96) gives   α− x(n) > θ1 + [x(n) + (α − )x(τ (n − 1))] , 2 i.e.,   

 α− α− x(n) > θ1 + (α − )x(τ (n − 1)). 1 − θ1 + 2 2 This guarantees, in particular, that   α− 1 − θ1 + > 0. 2 So,



 θ1 + α− (α − )  x(τ (n − 1)). 2 x(n) > 1 − θ1 + α− 2

We have thus proved that x(n) > θ2 x(τ (n − 1)) for all n ≥ L,  (α − ) θ1 + α− 2  .  θ2 = 1 − θ1 + α− 2 

where

By the arguments applied previously, a sequence of positive real numbers (θν )ν≥1 can inductively constructed, which satisfies   α− > 0 (ν = 1, 2, ...) 1 − θν + 2 

and θν+1

 θν + α− (α − ) 2   = 1 − θν + α− 2 154

(ν = 1, 2, ...);

this sequence is such that (5.94) holds, and x(n) > θν x(τ (n − 1)) for all n ≥ L

(ν = 2, 3, ...)., (5.97)

By the use of the definitions of θ1 and θ2 , it is a matter of elementary calculations to find θ2 = 1 − (α − ) − 1 − 2(α − ), i.e., θ2 = 2θ1 . So, θ2 > θ1 . By induction, we can easily verify that the sequence (θν )ν≥1 is strictly increasing. Furthermore, by taking into account the fact that (x(t))t≥r is decreasing and by using (for n = L) inequality (5.97), we obtain x(τ (L − 1)) ≥ x(L) > θν x(τ (L − 1)) (ν = 2, 3, ...). Hence, θν < 1 for every ν ≥ 2, which guarantees the boundedness of the sequence (θν )ν≥1 . Thus, limν→∞ θν exists as a positive real number. Define Θ = lim θν . ν→∞

Then it follows from (5.97) that x(n) ≥ Θx(τ (n − 1)) for all n ≥ L.

(5.98)

In view of the definition of (θν )ν≥1 , the number Θ satisfies   (α − ) Θ + α−  2 Θ= 1 − Θ + α− 2 or, equivalently, 2Θ2 − [2 − 3(α − )]Θ + (α − )2 = 0. 155

So, either Θ= or

 1 2 − 3(α − ) − 4 − 12(α − ) + (α − )2 4

 1 2 Θ= 2 − 3(α − ) + 4 − 12(α − ) + (α − ) . 4 √ Note that, because of 0 < α −  < 6 − 4 2, it holds 4 − 12(α − ) + (α − )2 > 0.

We always have  1 2 2 − 3(α − ) − 4 − 12(α − ) + (α − ) Θ≥ 4 and consequently (5.98) gives  1 x(n) ≥ 2 − 3(α − ) − 4 − 12(α − ) + (α − )2 4 x(τ (n − 1)) for all n ≥ L. Finally, we see that the last inequality can equivalently be written as follows  1 x(n+1) ≥ 2 − 3(α − ) − 4 − 12(α − ) + (α − )2 x(τ (n)) 4 for n ≥ L − 1, i.e.,  x(n + 1) 1 ≥ 2 − 3(α − ) − 4 − 12(α − ) + (α − )2 x(τ (n)) 4 for all n ≥ L − 1. Therefore,  x(n + 1) 1 2 lim inf ≥ 2 − 3(α − ) − 4 − 12(α − ) + (α − ) . n→∞ x(τ (n)) 4 156

As this inequality is satisfied for all real numbers  with 0 <  < α, we can conclude that (5.79) holds true. So, Part (ii) of the lemma has been proved. The proof of the lemma is complete. Theorem 5.8 ([5]) Assume that the sequence (τ (n))n≥0 is increasing, and n−1  α = lim inf p (j) . (5.76) n→∞

j=τ (n)

Then we have: (I) If 0 < α ≤ 12 , then the condition n 

lim sup n→∞

p (j) > 1 −

j=τ (n)

√  1 1 − α − 1 − 2α , 2

(5.99)

is sufficient for all solutions of the delay difference equation (1) to be oscillatory. √ (II) If 0 < α ≤ 6 − 4 2 and, in addition, p(n) ≥

α for all large n, 2

(5.78)

then the condition lim sup n→∞

n  j=τ (n)

 1 2 p (j) > 1− 2 − 3α − 4 − 12α + α , (5.100) 4

is sufficient for all solutions of Eq.(1) to be oscillatory. Proof. Suppoce, for the sake of contradiction, that the delay difference equation (1) admits a nonoscillatory solution 157

(x(n))n≥−k . Since (−x(n))n≥−k is also a solution of Eq.(1) , we can confine our discussion only to the case where the solution (x(n))n≥−k is eventually positive. Consider an integer ρ ≥ −k so that x(n) > 0 for every n ≥ ρ, and let r ≥ 0 be an integer such that τ (n) ≥ ρ for n ≥ r (clearly, r > ρ). Then from Eq.(1) we immediately obtain Δx(n) ≤ 0 for all n ≥ r, and consequently the sequence (x(n))n≥r is decreasing. Now, we consider an integer n0 > r such that τ (n) ≥ r for n ≥ n0 . Furthermore, we choose an integer N > n0 so that τ (n) ≥ n0 for n ≥ N . Then, by taking into account the facts that the sequence (τ (s))s≥0 is increasing and that the sequence (x(t))t≥r is decreasing, from Eq.(1) we obtain, for every n ≥ N , ⎛ ⎞ n n   x(τ (n)) − x(n + 1) = p(j)x(τ (j)) ≥ ⎝ p(j)⎠ x(τ (n)). j=τ (n)

j=τ (n)

Consequently, n 

p(j) ≤ 1 −

j=τ (n)

x(n + 1) x(τ (n))

for all n ≥ N ,

which gives n 

lim sup n→∞

p(j) ≤ 1 − lim inf n→∞

j=τ (n)

x(n + 1) . x(τ (n))

(5.101)

Assume, first, that 0 < α ≤ 12 . Then, by Lemma 5.8, inequality (5.77) is fulfilled, and so (5.101) leads to lim sup n→∞

n  j=τ (n)

p(j) ≤ 1 −

√  1 1 − α − 1 − 2α , 2

which contradicts condition (5.99). 158

√ Next, let us suppose that 0 < α ≤ 6 − 4 2 and that (5.78) holds. Then Lemma 5.8 ensures that (5.79) is satisfied. Thus, from (5.101), it follows that lim sup n→∞

n  j=τ (n)

 1 2 2 − 3α − 4 − 12α + α , p(j) ≤ 1 − 4

which contradicts condition (5.100). The proof of the theorem is complete. As it has already been mentioned, Theorem 5.6 is presented in [3] in a more general form in which it is not assumed that the sequence (τ (n))n≥0 is increasing. Instead, the sequence of integers (σ(n))n≥0 defined by σ(n) = max τ (s) for n ≥ 0, 0≤s≤n

is used which is clearly increasing. Moreover, as it has been shown in [3], it holds lim inf n→∞

n−1 

p(j) = lim inf n→∞

j=σ(n)

n−1 

p(j).

j=τ (n)

Following [3], one can use the last equality and apply Lemma 5.1 to establish the following generalization of Theorem 5.8. 

Theorem 5.8 ([5]) Let the sequence σ(n) = max0≤s≤n τ (s), n ≥ 0 and n−1  p (j) . (5.76) α = lim inf n→∞

j=τ (n)

159

Then we have:  (I) If 0 < α ≤ 12 , then the condition n 

lim sup n→∞

p (j) > 1 −

j=σ(n)

√  1 1 − α − 1 − 2α 2

is sufficient for all solutions of the delay difference equation (1) to be oscillatory. √  (II) If 0 < α ≤ 6 − 4 2 and, in addition, p(n) ≥

α for all large n, 2

(5.78)

then the condition lim sup n→∞

n  j=σ(n)

p (j) > 1 −

 1 2 − 3α − 4 − 12α + α2 4

is sufficient for all solutions of Eq.(1) to be oscillatory. Remark 5.8 ([3]) Observe the following: (i) When 0 < α ≤ 12 , it is easy to verify that √ √   2 1 1 − α − 1 − 2α > 1 − 1 − α , 2 and therefore the condition (5.99) is weaker than the condition (5.61). √ (ii) When 0 < α ≤ 6 − 4 2, we see that √ α 1− 1−α> , 2 160

√  1 − 1−α 1 2 − 3α − 4 − 12α + α2 > α √ 4 1−α and so the condition (5.100) is weaker than than the condition (5.62). Therefore Theorem 5.8 essentially improves Theorem 5.6. Remark 5.9 ([5]) In the special case of Eq.(1) we have seen that if  k+1 n−1  k α0 = lim inf p(j) ≤ , n→∞ k+1

and

j=n−k

the condition (Chen and Yu [8])   n   1 1 − α0 − 1 − 2α0 − α02 , (C6 ) lim sup p (j) > 1 − 2 n→∞ j=n−k

implies that all solutions of Eq.(1) oscillate. Observe that     √ 1 1 1 − α0 − 1 − 2α0 + α02 > 1 − α0 − 1 − 2α0 2 2 and therefore condition (C6 ) is weaker than the condition (5.99) in the special case of Eq.(1) . √ Observe, however, that when 0 < α0 ≤ 6 − 4 2, it is easy to show that       1 1 2 − 3α0 − 4 − 12α0 + α02 > 1 − α0 − 1 − 2α0 − α02 , 4 2 and therefore, in this case and when p(n) ≥ α2 for all large n, the condition (5.100) in Theorem 5.8 is weaker√than the above condition (C6 ) and especially, when α0 = 6−4 2  0.3431457, the lower bound in (C6 ) is 0.8929094, while in (5.100) is 0.7573593. 161

That is, an essential improvement. We illustrate the significance of these results by the following example. Example 5.4 ([5]) Consider the equation Δx(n) + p(n)x(n − 2) = 0, where

1474 1488 , p(3n + 1) = , 10000 10000 6715 p(3n + 2) = , n = 0, 1, 2, .... 10000 Here k = 2 and it is easy to see that p(3n) =

α0 = lim inf n→∞

n−1 

p(j) =

j=n−2

1474 1488 + = 10000 10000

 3 2  0.2962963, = 0.2962 < 3 and lim sup n→∞

n 

p(j) =

j=n−2

1474 1488 6715 + + = 0.9677. 10000 10000 10000

Observe that 0.9677 > 1 −

 √ 1 1 − α0 − 1 − 2α0  0.967317794, 2

162

that is, condition (5.99) of Theorem 5.8 is satisfied and therefore all solutions oscillate. Also, condition (C6 ) is satisfied. Observe, however, that 0.9677 < 1,  3 2 α0 = 0.2962 <  0.2962963, 3  2 √ 0.9677 < 1 − 1 − 1 − α0  0.974055774, and therefore none of the conditions (C2 ) , (C3 ) and (5.61) is satisfied. If, on the other hand, in the above equation p(3n) = p(3n+1) =

1481 6138 , p(3n+2) = , 10000 10000

n = 0, 1, 2, ...,

it is easy to see that n−1 

α0 = lim inf n→∞

p(j) =

j=n−2

1481 1481 + = 0.2962 < 10000 10000

 3 2  0.2962963, < 3 and lim sup n→∞

n 

p(j) =

j=n−2

1481 1481 6138 + + = 0.91. 10000 10000 10000

Furthermore, it is clear that p(n) ≥

α0 for all large n. 2 163

In this case

   1 2 0.91 > 1 − 2 − 3α0 − 4 − 12α0 + α0  0.904724375, 4 that is, condition (5.100) of Theorem 5.8 is satisfied and therefore all solutions oscillate. Observe, however, that 0.91 < 1,  3 2  0.2962963, α0 = 0.2962 < 3  2 √ 0.91 < 1 − 1 − 1 − α0  0.974055774,    1 2 1 − α0 − 1 − 2α0 − α0  0.930883291, 0.91 < 1 − 2 and therefore none of the conditions (C2 ) , (C3 ) , (5.61) and (C6 ) is satisfied. Very recently, the following results have been established in [6]. Lemma 5.9 ([6]). Assume that the sequence√(τ (n))n≥0 is increasing. Moreover, assume that 0 < α ≤ −1 + 2, where α = lim inf n→∞

n−1 

p (j) .

(5.76)

j=τ (n)

Then every nonoscillatory solution (x(n))n≥−k of the delay difference equation (1) satisfies  x(n + 1) 1 2 (5.102) ≥ 1 − α − 1 − 2α − α . lim inf n→∞ x(τ (n)) 2 164

Proof. Define q(t) = p(n) for n ≤ t < n + 1

(n = 0, 1, ...).

Clearly, q is a nonnegative real-valued function on the interval [0, ∞), which is continuous on each one of the intervals (n, n+1) (n = 0, 1, ...). Note that q(n) = p(n) for every integer n ≥ 0. Furthermore, consider the real-valued function σ defined on the interval [0, ∞) by σ(t) = τ (n) for n ≤ t < n + 1

(n = 0, 1, ...).

It is obvious that, for each n = 0, 1, ..., the function σ is continuous on (n, n + 1). We notice that σ(n) = τ (n) for all integers n ≥ 0. We can immediately see that σ(t) < t for all t ≥ 0,

and

lim σ(t) = ∞.

t→∞

Also, as the sequence (τ (n))n≥0 is assumed to be increasing, we observe that the function σ is increasing on [0, ∞). Let (x(n))n≥−k be a solution of the delay difference equation (1) . We define y(t) = x(n) + (Δx(n)) (t − n) for n ≤ t < n + 1

(n = −k, −k + 1, ...).

It is cleat that y(n) = x(n) for all integers n ≥ −k. Moreover,itiseasytoverifythatthereal-valuedfunctionyis continuousontheinterval[−k,∞).Also,weseethatyis 165

continuously differentiable on each one of the intervals (n, n + 1) (n = −k, −k + 1, ...) with y  (t) = Δx(n) for n < t < n + 1

(n = −k, −k + 1, ...).

Furthermore, as (x(n))n≥−k satisfies Eq.(1) for all integers n ≥ 0, we can easily conclude that the function y satisfies y  (t) + q(t)y (σ(t)) = 0 for n < t < n + 1

(n = 0, 1, ...). (5.103)  Next, assume that the solution (x(n))n≥−k of Eq.(1) is nonoscillatory. Then it is either eventually positive or eventually negative. As (−x(n))n≥−k is also a solution of Eq.(1) , we may (and do) restrict ourselves only to the case where x(n) > 0 for all large n. Let ρ ≥ −k be an integer such that x(n) > 0 for all n ≥ ρ, and consider an integer r ≥ 0 so that τ (n) ≥ ρ for n ≥ r (clearly, r > ρ). Then it follows immediately from Eq.(1) that Δx(n) ≤ 0 for every n ≥ r, which means that the sequence (x(n))n≥r is decreasing. Furthermore, it is not difficult to conclude that the function y is positive on the interval [ρ, ∞) and that y is decreasing on [r, ∞). Consider an arbitrary real number  with 0 <  < α. Then we can choose an integer n0 > r such that τ (n) ≥ r for n ≥ n0 , and n−1  p (j) > α −  for every n ≥ n0 . j=τ (n)

For any point t ≥ n0 , there exists an integer n ≥ n0 such that n ≤ t < n + 1, and consequently  t  n  t n−1  q(s)ds = q(s)ds ≥ q(s)ds = p(j) > α − . σ(t)

τ (n)

τ (n)

166

j=τ (n)

So, we have 

t

q(s)ds > α −  for all t ≥ n0 .

(5.104)

σ(t)

Furthermore, we will establish the following claim. Claim. For each point t ≥ n0 , there exists a t∗ > t such that σ(t∗ ) < t, and  t∗ q(s)ds = α − . (5.105) t

To prove this claim, let us consider an arbitrary point t ≥ n0 . Set  ν q(s)ds for ν ≥ t. f (ν) = t

We see that f (t) = 0. Moreover, it is not difficult to show that (5.104) guarantees that  ∞ q(s)ds = ∞ 0

and, in particular,





q(s)ds = ∞,

t

i.e., limν→∞ f (ν) = ∞. Thus, as the function f is continuous on the interval [t, ∞), there always exists a point t∗ > t so that f (t∗ ) = α−, i.e., such that (5.105) is satisfied. By using (5.104) (for the point t∗ ) as well as (5.105), we obtain  t∗  t∗  t q(s)ds = q(s)ds − q(s)ds > (α − ) − (α − ) = 0 σ(t∗ )

σ(t∗ )

t

and consequently we necessarily have σ(t∗ ) < t. Our claim has been proved. 167

Now, we choose an integer N > n0 such that τ (n) ≥ n0 for every n ≥ N . Let us consider an arbitrary point t ≥ N . By our claim, there exists a t∗ > t such that σ(t∗ ) < t, and (5.105) holds. From (5.103) it follows that  t∗ ∗ q(s)y (σ(s)) ds. (5.106) y(t) = y(t ) + t

Let s be any point with t ≤ s ≤ t∗ . As the function σ is increasing on [0, ∞), we have n0 ≤ σ(t) ≤ σ(s) ≤ σ(t∗ ) < t, and r ≤ σ(u) ≤ σ(t) for every u with σ(s) ≤ u ≤ t. Thus, by taking into account the fact that the function y is decreasing on [r, ∞), from (5.103) we obtain  t y(σ(s)) = y(t) + q(u)y (σ(u)) du σ(s) 

 t q(u)du y (σ(t)) ≥ y(t) + σ(s)

 s   s = y(t) + q(u)du − q(u)du y (σ(t)) . σ(s)

t

So, by applying (5.104) (for the point s), we get 

 s q(u)du y (σ(t)) . y(σ(s)) > y(t) + (α − ) − t

As this inequality holds true for all s with t ≤ s ≤ t∗ , it follows from (5.106) that 

   t∗  s ∗ y(t) ≥ y(t )+ q(s) y(t) + (α − ) − q(u)du y (σ(t)) ds t

t

168

= y(t∗ ) + 



q(s)ds y(t)+



t∗

q(s)ds −

t



t∗ t



t∗

+ (α − )



s

q(s)

  q(u)du ds y (σ(t))

t

t

and consequently, in view of (5.105), y(t) ≥ y(t∗ ) + (α − )y(t)+

 s     t∗ 2 q(s) q(u)du ds y (σ(t)) . + (α − ) − Noting the known formula

 s    t∗ q(s) q(u)du ds = t

t

or





t∗

q(s) we have  t∗







t∗

q(s) t

 q(s)ds du

u

t∗

q(u)du ds =

s

t∗

q(u)



t

t



t∗ t

s

(5.107)

t

t

 q(u)du ds,

s



q(s) q(u)du ds = t

 s 

 t∗    t ∗  t∗ 1 q(s) q(u)du ds + q(s) q(u)du ds = = 2 t t t s

 s   ∗  t∗ 1 t = q(s) q(u)du + q(u)du ds = 2 t t s

 t∗

 t∗  2  ∗ 1 t 1 = q(s) q(u)du ds = q(s)ds 2 t 2 t t t

169

and therefore, by (5.105),

 s   t∗ 1 q(s) q(u)du ds = (α − )2 . 2 t t Hence, (5.107) is written as 1 y(t) ≥ y(t∗ ) + (α − )y(t) + (α − )2 y (σ(t)) . 2 This gives

(5.108)

1 y(t) > (α − )y(t) + (α − )2 y (σ(t)) , 2 i.e.,

(α − )2 y(t) > y (σ(t)) . 2 [1 − (α − )] √ (Note that 0 < a −  < a ≤ −1 + 2 < 1.) We have thus proved that y(t) > λ1 y (σ(t))

for all t ≥ N ,

(5.109)

where

(α − )2 . λ1 = 2 [1 − (α − )] Let us again consider an arbitrary point t ≥ N . By using our claim, we conclude that there exists a t∗ > t such that σ(t∗ ) < t, and (5.105) holds. Then (5.108) is also fulfilled. Moreover, in view of (5.109) (for the point t∗ ), we have y(t∗ ) > λ1 y (σ(t∗ )) . But, since σ(t∗ ) < t and the function y is decreasing on [r, ∞), we also have y (σ(t∗ )) ≥ y(t). 170

Combining the last two inequalities, we obtain y(t∗ ) > λ1 y (t) and hence (5.108) yields 1 y(t) > λ1 y (t) + (α − )y(t) + (α − )2 y (σ(t)) 2 or

1 [1 − (α − ) − λ1 ] y(t) > (α − )2 y (σ(t)) . 2 This implies, in particular, that 1 − (α − ) − λ1 > 0. Consequently, y(t) >

(α − )2 y (σ(t)) . 2 [1 − (α − ) − λ1 ]

Thus, it has been shown that y(t) > λ2 y (σ(t))

for all t ≥ N ,

where

(α − )2 . 2 [1 − (α − ) − λ1 ] Following the above procedure, we can inductively construct a sequence of positive real numbers (λν )ν≥1 with λ2 =

1 − (α − ) − λν > 0 (ν = 1, 2, ...) and λν+1 =

(α − )2 2 [1 − (α − ) − λν ] 171

(ν = 1, 2, ...)

such that y(t) > λν y (σ(t))

for all t ≥ N

(ν = 1, 2, ...) .

(5.110)

As λ1 > 0, we obtain λ2 =

(α − )2 (α − )2 > = λ1 , 2 [1 − (α − ) − λ1 ] 2 [1 − (α − )]

i.e., λ2 > λ1 . By an easy induction, one can immediately see that the sequence (λν )ν≥1 is strictly increasing. Furthermore, by taking into account the fact that the function y is decreasing on [r, ∞) and using (5.110) (for t = N ), we get y(N ) > λν y (σ(N )) ≥ λν y(N )

(ν = 1, 2,...) .

Therefore, for each integer ν ≥ 1, we have λν < 1. This ensures that the sequence (λν )ν≥1 is bounded. Since (λν )ν≥1 is a strictly increasing and bounded sequence of positive real numbers, it follows that limν→∞ λν exists as a positive real number. Set Λ = lim λν . ν→∞

Then (5.110) gives y(t) ≥ Λy (σ(t))

for all t ≥ N .

Because of the definition of (λν )ν≥1 , it holds Λ=

(α − )2 , 2 [1 − (α − ) − Λ]

i.e., 1 Λ2 − [1 − (α − )] Λ + (α − )2 = 0. 2 172

(5.111)

Hence, either  1 Λ= 1 − (α − ) − 1 − 2(α − ) − (α − )2 2 or

 1 2 Λ= 1 − (α − ) + 1 − 2(α − ) − (α − ) . 2 In both cases, we have  1 2 1 − (α − ) − 1 − 2(α − ) − (α − ) Λ≥ 2

and consequently (5.111) yields  1 2 y(t) ≥ 1 − (α − ) − 1 − 2(α − ) − (α − ) y (σ(t)) 2 for all t ≥ N .

(5.112)

Let n be an arbitrary integer with n ≥ N . Then, by (5.112),  1 2 y(t) ≥ 1 − (α − ) − 1 − 2(α − ) − (α − ) y (σ(t)) 2 for n ≤ t < n + 1. But, y(σ(t)) = y(τ (n)) = x(τ (n)) for n ≤ t < n + 1. So,  1 y(t) ≥ 1 − (α − ) − 1 − 2(α − ) − (α − )2 x(τ (n)) 2 for n ≤ t < n + 1 and therefore lim t→(n+1)−0

y(t) ≥

 1 1 − (α − ) − 1 − 2(α − ) − (α − )2 x(τ (n)). 2

173

Note that limt→(n+1)−0 y(t) = y(n + 1) = x(n + 1). We have thus proved that  1 2 x(n + 1) ≥ 1 − (α − ) − 1 − 2(α − ) − (α − ) x(τ (n)) 2 for all n ≥ N .

(5.113)

Finally, we see that (5.113) is written as  x(n + 1) 1 ≥ 1 − (α − ) − 1 − 2(α − ) − (α − )2 x(τ (n)) 2 for every n ≥ N and consequently  x(n + 1) 1 2 lim inf ≥ 1 − (α − ) − 1 − 2(α − ) − (α − ) . n→∞ x(τ (n)) 2 The last inequality holds true for all real numbers  with 0 <  < α. Hence, we can obtain (5.102). The proof of our lemma is now complete. Theorem 5.9 ([6]) Assume that the sequence (τ (n))n≥0 is √ increasing, 0 ≤ α ≤ −1 + 2 where α = lim inf n→∞

n−1 

p (j) .

(5.76)

j=τ (n)

and lim sup n→∞

n  j=τ (n)

p (j) > 1 −

 1 1 − α − 1 − 2α − α2 , 2 174

(C6 )

Then all solutions of the delay difference equation (1) are oscillatory. Proof. Assume, for the sake of contradiction, that there exists a nonoscillatory solution (x(n))n≥−k of the delay difference equation (1) . Since (−x(n))n≥−k is also a solution of Eq.(1) , we can confine our discussion only to the case where the solution (x(n))n≥−k is eventually positive. Consider an integer ρ ≥ −k so that x(n) > 0 for every n ≥ ρ, and let r ≥ 0 be an integer such that τ (n) ≥ ρ for n ≥ r (clearly, r > ρ). Then from Eq.(1) we immediately obtain Δx(n) ≤ 0 for all n ≥ r, and consequently the sequence (x(n))n≥r is decreasing. Now, we choose an integer n0 > r such that τ (n) ≥ r for n ≥ n0 . Furthermore, we consider an integer N > n0 so that τ (n) ≥ n0 for n ≥ N . Then, as the sequence (τ (n))n≥0 is increasing and the sequence (x(n))n≥r is decreasing, it follows from Eq.(1) that, for every n ≥ N , ⎡ ⎤ n n   p(j)x(τ (j)) ≥ ⎣ p(j)⎦ x(τ (n)). x(τ (n)) − x(n + 1) = j=τ (n)

j=τ (n)

This gives n 

p(j) ≤ 1 −

j=τ (n)

x(n + 1) x(τ (n))

for all n ≥ N .

Hence, lim sup n→∞

n 

p(j) ≤ 1 − lim inf n→∞

j=τ (n)

175

x(n + 1) . x(τ (n))

But, in view of Lemma 5.9, inequality (5.102) holds. So, we obtain n   1 lim sup p(j) ≤ 1 − 1 − α − 1 − 2α − α2 , 2 n→∞ j=τ (n)

which contradicts condition (C6 ) . The proof of the theorem is complete. Following [3], and as in Theorem 5.8, we can establish the following generalization of Theorem 5.9. 

the sequence σ(n) = max0≤s≤n τ (s), Theorem 5.9 . Let √ n ≥ 0 and 0 < α ≤ −1 + 2, where α = lim inf n→∞

n−1 

p (j) .

(5.76)

j=τ (n)

Then the condition n   1 1 − α − 1 − 2α − α2 lim sup p (j) > 1 − 2 n→∞

(C˜6 )

j=σ(n)

is sufficient for all solutions of the delay difference equation (1) to be oscillatory. Remark 5.9 Observe that if α > 1e (Theorem 5.5), then all solutions of Eq.(1) oscillate. Therefore, in (5.61), (5.62) and (5.99) the conditions 0 < α ≤ 1, 0 < α < 1, 0 < α ≤ 12 and √ 0 < α ≤ −1 + 2 respectively, can be replaced by the following condition n−1  1 p (j) ≤ . 0 < α = lim inf n→∞ e j=τ (n)

176

Remark 5.10 Observe that, when 0 < α ≤ 1e , it is easy to see that √ √ 1 − α − 1 − 2α − α2 1− 1−α > >α √ 2 1−α √ √ 1 − α − 1 − 2α > (1 − 1 − α)2 > 2 and therefore the condition (C6 ) is weaker than the conditions (5.62), (5.99) and (5.61). Remark 5.11 ([6]) We note that, in the special case of the delay difference equation (1) , Lemma 5.9 and Theorem 5.9 have been presented by Chen and Yu [8]. We also notice that Lemma 5.9 and Theorem 5.9 should be looked upon as discrete analogues of corresponding results due to Yu, Wang, Zhang and Qian [100] concerning the solutions of the delay differential equation (1) (and, especially, of the delay differential equation Eq.(2)). We illustrate the significance of our results by the following example in which the delay difference equation (1) is considered with a variable delay argument. Example 5.5 ([6]) Let α be a real number with 0 < α ≤ 1/e,and define √  2 A1 = 1 − 1 − 1 − α , √  1 1 − α − 1 − 2α , 2 √ 1− 1−α A3 = 1 − α √ , 1−α

A2 = 1 −

177

and

 1 2 A4 = 1 − 1 − α − 1 − 2α − α . 2 Note that (cf. Remark 5.10) A1 > A2 > A3 > A4 . Next, we consider a positive real number d such that A4 − α < d < A3 − α (we notice that A4 > α). So, we have A1 > A2 > A3 > α + d > A4 . Furthermore, let β bea real   number with 0 < β < 1, and set α 1 c = − ln β and r = 2 + β ( β1 denotes the greatest integer less than or equal to β1 ). Consider now the delay difference equation (1) with ( 2 )  c , if n ∈ {1, 2, ...} \ ) r, r , ... ( p(n) = n d, if n ∈ 0, r, r2 , ...

and τ (0) = −1,

and

τ (n) = [βn] (n = 1, 2, ...).

Here (p(n))n≥0 is a sequence of positive real numbers, and (τ (n))n≥0 is a sequence of integers such that τ (n) ≤ n−1 for all n ≥ 0, and lim τ (n) = ∞. Moreover, we note that the sequence (τ (n))n≥0 n→∞ is increasing. We will first show that n−1  c lim = α. (5.114) n→∞ j j=[βn]

To this end, we obtain, for n sufficiently large,  n n−1 n−1  j+1   c ds ds n ≥c =c = c ln j s [βn] j [βn] s j=[βn]

j=[βn]

178

and  n−1 n−1 n−1  j   n−1 c ds ds ≤c =c = c ln . j [βn] − 1 j−1 s [βn]−1 s

j=[βn]

j=[βn]

But, it is easy to see that     1 n n−1 = lim c ln = c ln = α. lim c ln n→∞ n→∞ [βn] [βn] − 1 β From the above it is clear that (5.114) holds true. In particular, it follows from (5.114) that lim

n→∞

n r −1

j=[βrn ]

c = α. j

(5.115)

Observe that rn−1 < [βrn ] ≤ rn − 1 for large n.

(5.116)

Indeed, for any integer n ≥ 0, we have [βrn ] ≤ βrn < rn , and consequently [βrn ] ≤ rn − 1. On the other hand, for n ≥ 0, n n−1 n−1 − we obtain [βrn ] − rn−1  − 1) − r = (βr − 1) r  >(βr 1 1 1. But βr − 1 = β 2 + β − 1 > β 1 + β − 1 = β > 0 and so limn→∞ ((βr − 1) rn−1 − 1) = ∞, which quarantees that limn→∞ ([βrn ] − rn−1 ) = ∞. Hence, [βrn ] − rn−1 > 0, for all large n. Therefore, (5.116) has been proved. Now, in view of (5.116), we get n r −1

j=[βrn ]

p(j) =

n r −1

j=[βrn ]

c j

179

for all large n

and consequently, because of (5.115), n r −1

lim

n→∞

Furthermore, since d ≥ n−1  j=[βn]

p(j) = α.

(5.117)

j=[βrn ] c j

for all large j, we obtain

n−1  c p(j) ≥ j

for all large n,

j=[βn]

which, by virtue of (5.114), gives n−1 

lim inf n→∞

p(j) ≥ α.

(5.118)

j=[βn]

From (5.117) and (5.118) it follows that n−1 

lim inf n→∞

p(j) = α.

j=[βn]

Next, we shall prove that n  lim sup p(j) = α + d. n→∞

(5.119)

(5.120)

j=[βn]

Observe that n rn r −1  p(j) = p(j) + d for all large n, j=[βrn ]

j=[βrn ]

and so, because of (5.117), n

lim

n→∞

r 

p(j) = α + d.

j=[βrn ]

180

(5.121)

Furthermore, we see that     n ln [βn] ln β1 ln n ln[βn] = lim − = lim < 1, n→∞ n→∞ ln r ln r ln r ln r which implies that ln n ln[βn] − < 1 for sufficiently large n. ln r ln r Hence, for each large n, there exists at most one integer n∗ so that ln[βn] ln n ≤ n∗ ≤ ln r ln r

ln[βn] ≤ n∗ ln r ≤ ln n,

or

i.e., such that



[βn] ≤ rn ≤ n. By taking into account this fact, we obtain n  j=[βn]

p(j) ≤

n n−1   c c c +d= + +d j j n

j=[βn]

j=[βn]

for all large n. Thus, by using (5.114), we derive lim sup n→∞

n 

p(j) ≤ α + d.

(5.122)

j=[βn]

From (5.122) and (5.121) we conclude that (5.120) is always valid. Here, we observe that (5.119) coincides with (5.76). Moreover, as A4 < α + d < A 3 < A 2 < A 1 , 181

it follows from (5.120) that condition (C6 ) of Theorem 2.1 is satisfied and therefore all solutions of Eq.(1) are oscillatory. Observe, however, that none of the conditions (5.62), (5.99) and (5.61) is satisfied. In addition, we immediately see that conditions (C2 ) , (C˜3 ) and (C3 ) are also not satisfied.

182

EXERCISES 1. Check the validity of the relations in Example 5.1. 2. Similarly for Example 5.2. 3. Check the values of liminf and limsup in Example 5.3. 4. Check Corollary 5.2. 5. Check Corollary 5.3. 6. Check the validity of the inequalities stated in the Note (page 112). 7. Verify the inequalities in Remark 5.8. 8. Show the last inequality in Remark 5.9. 9. Verify all the relations in Example 5.4. 10. Show the validity of the inequalities in Remark 5.10. 11. Verify all the relations in Example 5.5.

183

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Stavroulakis Ioannis Petros Zhunussova Zhanat Khaphizovna OSCILLATORY BEHAVIOR OF FIRST ORDER FUNCTIONAL DIFFERENTIAL EQUATIONS Educational manual Typesetting Zh. Zhunussova Cover design Y. Gorbunov Cover design used photos from sites www.art-2026066_1280.com

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