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Mathematical and Computer Modelling 49 (2009) 1573–1586

Contents lists available at ScienceDirect

Mathematical and Computer Modelling

journal homepage: www.elsevier.com/locate/mcm

Oscillation of third order nonlinear delay dynamic equations on

time scales

Taher S. Hassan

Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, 35516, Egypt

a r t i c l e i n f o

Article history:

Received 11 April 2008

Received in revised form 17 December 2008

Accepted 31 December 2008

Keywords:

Oscillation

Delay nonlinear dynamic equations

Time scales

a b s t r a c t

It is the purpose of this paper to give oscillation criteria for the third order nonlinear delay

dynamic equation

?

on a time scale T, whereγ ≥ 1 is the quotient of odd positive integers, a and r are positive

rd-continuousfunctionsonT,andtheso-calleddelayfunctionτ : T → Tsatisfiesτ(t) ≤ t

for t ∈ T and limt→∞τ(t) = ∞ and f ∈ C (T × R,R). Our results are new for third

orderdelaydynamicequationsandextendmanyknownresultsforoscillationofthirdorder

dynamic equation. These results in the special cases when T = R and T = N involve and

improvesomeoscillationresultsforthirdorderdelaydifferentialanddifferenceequations;

when T = hN, T = qN0and T = N2our oscillation results are essentially new. Some

examples are given to illustrate the main results.

a(t)

??

r(t)x?(t)???γ??

+ f(t,x(τ (t))) = 0,

© 2009 Elsevier Ltd. All rights reserved.

1. Introduction

We are concerned with the oscillatory behavior of the third order nonlinear delay dynamic equation

?

on an arbitrary time scale T, whereγ ≥ 1 is a quotient of odd positive integers, a and r are positive rd-continuous functions

on T, and the so-called delay function τ : T → T satisfies τ(t) ≤ t, τ?(t) ≥ 0, for t ∈ T and limt→∞τ(t) = ∞.

The function f ∈ C (T × R,R) is assumed to satisfy uf (t,u) > 0 and there exists a positive rd-continuous function p(t)

on T such that

uγ

≥ p(t), for u ?= 0. Since we are interested in the oscillatory behavior of solutions near infinity,

we assume that supT = ∞, and define the time scale interval [t0,∞)Tby [t0,∞)T := [t0,∞) ∩ T. By a solution of

(1.1) we mean a nontrivial real-valued function x ∈ C1

a

∈ C1

vanishing in some neighborhood of infinity will be excluded from our consideration. A solution x of (1.1) is said to be

oscillatory if it is neither eventually positive nor eventually negative, otherwise it is called nonoscillatory.

The theory of time scale, which has recently received a lot of attention, was introduced by Hilger’s landmark paper [1],

a rapidly expanding body of literature has sought to unify, extend, and generalize ideas from discrete calculus, quantum

calculus, and continuous calculus to arbitrary time scale calculus, where a time scale T is an nonempty closed subset of the

reals, and the cases when this time scale is equal to the reals or to the integers represent the classical theories of differential

a(t)

??

r(t)x?(t)???γ??

+ f (t,x(τ (t))) = 0,

(1.1)

f(t,u)

rd[Tx,∞),Tx ≥ t0which has the property that rx?∈ C1

rd[Tx,∞),

??

rx????γ

rd[Tx,∞) and satisfies Eq. (1.1) on[Tx,∞), where Crdis the space of rd-continuous functions. The solutions

E-mail address: tshassan@mans.edu.eg.

0895-7177/$ – see front matter © 2009 Elsevier Ltd. All rights reserved.

doi:10.1016/j.mcm.2008.12.011

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

and of difference equations. Many other interesting time scales exist, and they give rise to many applications (see [2]).

Not only the new theory of the so-called ‘‘dynamic equations’’ unify the theories of differential equations and difference

equations, but also extends these classical cases to cases ‘‘in between’’, e.g., to the so-called q-difference equations when

T = qN0(which has important applications in quantum theory (see [3])) and can be applied on different types of time scales

likeT = hZ,T = N2

and time scale notation is assumed; for an excellent introduction to the calculus on time scales, see Bohner and Peterson [2,

4].Inthelastfewyears,therehasbeenincreasinginterestinobtainingsufficientconditionsfortheoscillation/nonoscillation

of solutions of different classes of dynamic equations on time scales, we refer the reader to the papers [5–10,17], and the

references cited therein. To the best of our knowledge, there is very little knownabout the oscillatory behavior of third order

dynamic equations. Recently, Erbe et al. [11–13] considered the third order dynamic equations

?

x???(t) + p(t)x(t) = 0,

and

?

respectively and established some sufficient conditions for oscillation. Therefore it will be of great interest to extend the

above results to establish oscillation criteria for (1.1) by using the generalized Riccati transformation technique. Our results

will improve and extend the results that have been established in [11–13].

0andT = Hnthespaceofharmonicnumbers.Inthisworkaknowledgeandunderstandingoftimescales

a(t)?

r(t)x?(t)????

+ p(t)f(x(t)) = 0,

a(t)

??

r(t)x?(t)???γ??

+ f(t,x(t)) = 0,

2. Main Results

Throughout this paper, we assume τoσ = σoτ and, for any given function φ(t) such that a(t)φ(t) is differentiable

function and a positive differentiable function Φ(t), for all sufficiently large T1∈ [t0,∞)T, and for any T > T1such that

τ (T) > T1, we let, for t ∈ [T,∞)T,

η1(t,T1) := Φ?(t) +

γ + 1

r (τ (t))τ?(t)δ1(τ(t),T1)Φ(t)((a(t)φ(t))σ)

?

and

?

where

?t

and

δ(t,T1) := δ1(t,T1)?δσ

results.

2γ

r (τ (t))τ?(t)δ(τ(t),T1)Φ(t)(a(t)φ(t))σ;

η2(t,T1) := Φ?(t) +

1

γ;

ψ1(t,T1) := Φ(t)

p(t) − (a(t)φ(t))?+γτ?(t)δ(τ(t),T1)

r (τ (t))

((a(t)φ(t))σ)2

?

;

ψ2(t,T1) := Φ(t)

p(t) − (a(t)φ(t))?+τ?(t)δ1(τ(t),T1)

r (τ (t))

((a(t)φ(t))σ)1+1

γ

?

,

δ1(t,T1) :=

T1

?s

1

γ(s)

a

,δ2(t,T1) :=

?t

T1

δ1(s,T1)

r(s)

?s,

2(t,T1)?γ−1,η+:= max{0,η},η−:= max{0,−η}.

Before stating our main results, we begin with the following lemmas which will play an important role in the proof of main

Lemma 2.1. Assume that

?∞

and

?∞

Furthermore, suppose that (1.1) has a positive solution x on [t0,∞)T. Then there exists a T ∈ [t0,∞)T, sufficiently large, so that

?

and either x?(t) > 0 on [T,∞)Tor limt→∞x(t) = 0.

t0

?t

1

γ(t)

a

= ∞,

?∞

t0

?t

r(t)= ∞,

(2.1)

t0

1

r(t)

?∞

t

?

1

a(s)

?∞

s

p(u)?u

?1

γ

?s?t = ∞.

(2.2)

a(t)

??

r(t)x?(t)???γ??

< 0,(r(t)x?(t))?> 0, on [T,∞)T,

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

1575

Proof. Pick t1∈ [t0,∞)T, so that x(τ(t)) > 0 on [t1,∞)T. Since x is a positive solution of (1.1), we have

?

on[t1,∞)T.Thena(t)

claim?

r(t)x?(t) ≤ r(t2)x?(t2) + a

a(t)

??

r(t)x?(t)???γ??

??

= −f(t,x(τ (t))) ≤ −p(t)xγ(τ (t)) < 0,

r(t)x?(t)???γ

(2.3)

isstrictlydecreasingon[t1,∞)Tandthus?

r(t)x?(t)??iseventuallyofonesign.We

r(t)x?(t)??> 0 on [t1,∞)T. If not, assume there exists t2≥ t1, sufficiently large, such that?

1

r(t)x?(t)??< 0 on

[t2,∞)Twhich, together with (2.3), implies that

γ(t2)?

r(t2)x?(t2)???t

t2

?s

1

γ(s)

a

.

Hence by (2.1), we have limt→∞r(t)x?(t) = −∞. Thus, there is t3∈ [t2,∞) such that r(t)x?(t) < 0 on [t3,∞)Twhich,

together with r(t)x?(t) is strictly decreasing on [t3,∞)T, implies that

?t

By (2.1), we have limt→∞x(t) = −∞, which contradicts the fact that x is a positive solution of (1.1). Then?

[t4,∞)Tor x?(t) < 0 on [t1,∞)T. If x?(t) < 0 on [t1,∞)T, we get limt→∞x(t) = l1≥ 0 and limt→∞r(t)x?(t) = l2≤ 0.

If we assume l1> 0, then x(τ(t)) ≥ l1, for t ≥ t5≥ t1. Integrating Eq. (1.1) from t to ∞, we get

−a(t)

t

Using the fact that f(t,x(τ (t))) ≥ p(t)xγ(τ(t)) ≥ lγ

??

and hence,

?

Again, integrating this inequality from t to ∞, we get

?∞

?∞

Finality, integrating the last inequality from t5to t, we get

?t

Hence by (2.2), we have limt→∞x(t) = −∞, which contradicts the fact that x is a positive solution of (1.1). Thus, we get

limt→∞x(t) = 0 and the proof is complete.

Lemma 2.2. Assume that x is a positive solution of Eq. (1.1) such that

?

x?(t) ≥δ1(t,t∗)

r(t)

and

γ(t)?

x(t) ≤ x(t3) + r(t3)x?(t3)

t3

?s

r(s).

r(t)x?(t)??> 0

on [t1,∞)Tand thus x?(t) is eventually of one sign. Then, we get either then there exists t4≥ t1such that x?(t) > 0 on

??

r(t)x?(t)???γ

≤ −

?∞

f(s,x(τ (s)))?s.

1p(t), we find that

−a(t)

r(t)x?(t)???γ

≤ −lγ

1

?∞

t

p(s)?s,

−?

r(t)x?(t)??≤ −l1

1

a(t)

?∞

t

p(s)?s

?1

γ

.

r(t)x?(t) ≤ −l1

t

?

?

1

a(s)

?∞

?∞

s

p(u)?u

?1

?1

γ

?s + l2

≤ −l1

t

1

a(s)

s

p(u)?u

γ

?s.

x(t) − x(t5) ≤ −l1

t5

1

r(s)

?∞

s

?

1

a(u)

?∞

u

p(v)?v

?1

γ

?u?s.

?

r(t)x?(t)??> 0,

x?(t) > 0,

on [t∗,∞)T, t∗≥ t0. Then

a

1

γ(t)?

r(t)x?(t)??,

(2.4)

x(t) ≥ δ2(t,t∗)a

on [t∗,∞)T.

1

r(t)x?(t)??,

(2.5)

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

r(t)x?(t)???γ

r(t)x?(t) ≥ r(t)x?(t) − r(t∗)x?(t∗)

?t

?

and, hence

?

Similarly, using (2.6) we see that

?

This completes the proof.

?

Now, we state and prove our main results.

Proof. Since a(t)

??

is strictly decreasing on [t∗,∞)T, we get, for t ∈ [t∗,∞)T

=

t∗

?

a(s)

??

r(s)x?(s)???γ?1

a

r(t)x?(t)???γ?1

γ

1

γ(s)

?s

≥

a(t)

??

γ?t

t∗

?s

1

γ(s)

a

,

(2.6)

x?(t) ≥δ1(t,t∗)

r(t)

a(t)

??

r(t)x?(t)???γ?1

γ,

for t ∈ [t∗,∞)T.

x(t) ≥ δ2(t,t∗)

a(t)

??

r(t)x?(t)???γ?1

γ,

for t ∈ [t∗,∞)T.

Theorem 2.1. Assume that (2.1) and (2.2) hold and that, for all sufficiently large T1 ∈ [t0,∞)T, there is a T > T1such that

τ (T) > T1and

?t

where the function φ(t) is a nonnegative function. Then every solution of Eq. (1.1) is either oscillatory or tends to zero.

Proof. Assume (1.1) has a nonoscillatory solution x on [t0,∞)T. Then, without loss of generality, there is a t1∈ [t0,∞)T,

sufficiently large, such that x(t) > 0 and x(τ(t)) > 0 on [t1,∞)T. Therefore from Lemma 2.1, we get

?

and either x?(t) > 0 for t ≥ t2 ≥ t1or limt→∞x(t) = 0. Let x?(t) > 0 on [t2,∞)T. Consider the generalized Riccati

substitution

???

By the product rule and then the quotient rule

?

+Φ(t)(a(t)φ(t))?+ Φ?(t)(a(t)φ(t))σ

?

xγ(τ(t))

?

From (1.1) and the definition of w(t), we have

limsup

t→∞

T

?

ψ1(s,T1) −

r(τ(s))η2

4γΦ(s)τ?(s)δ(τ(s),T1)

1(s,T1)

?

?s = ∞,

(2.7)

a(t)

??

r(t)x?(t)???γ??

< 0,

?

r(t)x?(t)??> 0,for t ∈ [t1,∞)T,

w(t) = Φ(t)a(t)

r(t)x?(t)??

x(τ(t))

?γ

+ φ(t)

?

.

w?(t) =

Φ(t)

xγ(τ(t))

??

??

a(t)

??

r(t)x?(t)???γ??

r(t)x?(t)???γ??

Φ(t)(xγ(τ(t)))?

xγ(τ(t))xγσ(τ(t))

+

?

Φ(t)

xγ(τ(t))

???

a(t)

??

r(t)x?(t)???γ?σ

= Φ(t)

a(t)

+ Φ(t)(a(t)φ(t))?+ Φ?(t)(a(t)φ(t))σ

??

+

Φ?(t)

xγσ(τ(t))−

a(t)

??

r(t)x?(t)???γ?σ

.

w?(t) = −Φ(t)f (t,x(τ (t)))

xγ(τ(t))

+ Φ(t)(a(t)φ(t))?+Φ?(t)

Φσ(t)wσ(t) − Φ(t)(xγ(τ(t)))?

xγ(τ(t))

?

a(t)

??

r(t)x?(t)??

x(τ(t))

?γ?σ

.

Using the fact that f (t,x(τ (t))) ≥ p(t)xγ(τ (t)), we get that

w?(t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))?+Φ?(t)

Φσ(t)wσ(t) − Φ(t)(xγ(τ(t)))?

xγ(τ(t))

?

a(t)

??

r(t)x?(t)??

x(τ(t))

?γ?σ

.

(2.8)

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

1577

Since x and τ are differentiable functions and τoσ = σoτ, we have xoτ is a differentiable function and (x(τ(t)))?=

x?(τ(t))τ?(t). Then, by using Chain Rule [2, Theorem 1.93], we obtain

(xγ(τ(t)))?≥ γ(xγ−1(τ (t)))x?(τ (t))τ?(t),

which implies

w?(t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))?+Φ?(t)

Φσ(t)wσ(t) − γΦ(t)τ?(t)x?(τ(t))

x(τ(t))

?

a(t)

??

??

r(t)x?(t)??

r(t)x?(t)???γ

x(τ(t))

?γ?σ

. (2.9)

We choose t3≥ t2such that τ(t) > t2, for t ≥ t3. Then, from (2.4) and using the fact that a(t)

decreasing on [t3,∞)T, we get that

x?(τ(t))

xσ(τ(t))≥δ1(τ(t),t2)

r (τ(t))

xσ(τ(t))

?

xσ(τ(t))

x(τ(t))

is strictly

a

1

γ(τ(t))?

a

r(τ(t))x?(τ(t))??

r(t)x?(t)???σ

??

≥δ1(τ(t),t2)

r (τ(t))

1

γ(t)?

=δ1(τ(t),t2)

r (τ(t))

a(t)

r (t)x?(t)??

?γ

r(t)x?(t)??

x(τ(t))

r(t)x?(t)??

a

1

γ(t)?

γ−1

σ

.

Therefore, from (2.5), we have

xσ(τ(t))≥δ1(τ(t),t2)?δσ

Using (2.10) in (2.9), we have

x?(τ(t))

2(τ(t),t2)?γ−1

r(τ(t))

?

a(t)

??

x(τ(t))

?γ?σ

,

for t ∈ [t3,∞)T.

(2.10)

w?(t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))?+Φ?(t)

Φσ(t)wσ(t)

??

−γΦ(t)τ?(t)δ(τ(t),t2)

r(τ(t))

??

a(t)

r(t)x?(t)??

x(τ(t))

?γ?σ?2

xσ(τ(t))

x(τ(t)),

for t ∈ [t3,∞)T.

Since x(t) is a strictly increasing function on [t3,∞)T, we have

w?(t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))?+Φ?(t)

Φσ(t)wσ(t)

??

−γΦ(t)τ?(t)δ(τ(t),t2)

r(τ(t))

??

a(t)

r(t)x?(t)??

x(τ(t))

?γ?σ?2

.

(2.11)

From the definition of w(t), we obtain

??

a(t)

??

r(t)x?(t)??

x(τ(t))

?γ?σ?2

=

?wσ(t)

=(wσ(t))2

Φσ(t)− (a(t)φ(t))σ

?2

(Φσ(t))2−

2(a(t)φ(t))σwσ(t)

Φσ(t)

+ ((a(t)φ(t))σ)2.

(2.12)

From (2.11), (2.12) and using the definitions of ψ1(t,t2) and η1(t,t2), we get

w?(t) ≤ −ψ1(t,t2) +η1(t,t2)

Φσ(t)wσ(t) −γΦ(t)τ?(t)δ(τ(t),t2)

r (τ(t))(Φσ(t))2

(wσ(t))2,

for t ∈ [t3,∞)T.

(2.13)

It is easy to check that

w?(t) ≤ −ψ1(t,t2) +

r(τ(t))η2

4γΦ(t)τ?(t)δ (τ(t),t2)−γΦ(t)τ?(t)δ (τ(t),t2)

1(t,t2)

r(τ(t))(Φσ(t))2

?

wσ(t) −

r(τ(t))Φσ(t)η1(t,t2)

2γΦ(t)τ?(t)δ (τ(t),t2)

?2

.

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

Consequently,

w?(t) ≤ −ψ1(t,t2) +

r(τ(t))η2

4γΦ(t)τ?(t)δ (τ(t),t2).

1(t,t2)

Integrating both sides from t3to t, we get

?t

which leads to a contradiction to (2.7).

t3

?

ψ1(s,t2) −

r(τ(s))η2

4γΦ(s)τ?(s)δ (τ(t),t2)

1(s,t2)

?

?s ≤ w(t3) − w(t) ≤ w(t3),

?

Theorem 2.2. Assume that (2.1) and (2.2) hold. If, for all sufficiently large T1∈ [t0,∞)T, there is a T > T1such that τ (T) > T1

and

?t

where the function φ(t) is a nonnegative function. then every solution of Eq. (1.1) is either oscillatory or tends to zero.

limsup

t→∞

T

?

ψ2(s,T1) −

rγ(τ(s))?(η2(s,T1))+

?γ+1

(γ + 1)γ+1(Φ(s)τ?(s)δ1(τ(s),T1))γ

?

?s = ∞,

(2.14)

Proof. Assume (1.1) has a nonoscillatory solution x on [t0,∞)T. Then, without loss of generality, there is a t1∈ [t0,∞)T,

sufficiently large, such that x(t) > 0 and x(τ(t)) > 0 on [t1,∞)T. Therefore from Lemma 2.1, we get

?

and either x?(t) > 0 for t ≥ t2≥ t1or limt→∞x(t) = 0. Let x?(t) > 0 on [t2,∞)T. We choose t3≥ t2such that τ(t) > t2,

for t ≥ t3. Using the fact that a(t)((r(t)x?(t))?)γis strictly decreasing on [t3,∞)Tand (2.4), we get

x?(τ(t)) ≥δ1(τ(t),t2)

r(τ(t))

≥δ1(τ(t),t2)

r(τ(t))

a(t)

??

r(t)x?(t)???γ??

< 0,

?

r(t)x?(t)??> 0,

for t ∈ [t1,∞)T,

?

?

a(τ(t))

??

r(t)x?(t)???σ

r(τ(t))x?(τ(t))???γ?1

,

γ

a

1

γ(t)?

for t ∈ [t3,∞)T.

(2.15)

We define w(t) as in the proof of Theorem 2.1. Using (2.15) in (2.9), we get, for t ∈ [t3,∞)T

w?(t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))?+Φ?(t)

Φσ(t)wσ(t)

??

−γΦ(t)τ?(t)δ1(τ(t),t2)

r(τ(t))

a1+1

γ(t)

r(t)x?(t)??

x(τ(t))

?γ+1

σ

xσ(τ(t))

x(τ(t)).

Since x(t) is strictly increasing function on [t3,∞)T, we have

w?(t) ≤ −Φ(t)p(t) + Φ(t)(a(t)φ(t))?+Φ?(t)

Φσ(t)wσ(t)

??

−γΦ(t)τ?(t)δ1(τ(t),t2)

r(τ(t))

a1+1

γ(t)

r(t)x?(t)??

x(τ(t))

?γ+1

σ

.

(2.16)

By using the inequality, see [14],

(u − v)1+1

γ≥ u1+1

γ+

1

γv1+1

γ−

?

1 +

1

γ

?

v

1

γu,

where u, v are constants and γ ≥ 1 is a quotient of odd positive integers. From the definition of w(t), we obtain

x(τ(t))

a1+1

≥(wσ(t))1+1

(Φσ(t))1+1

γ(t)

??

r(t)x?(t)??

?γ+1

σ

=

?wσ(t)

Φσ(t)− (a(t)φ(t))σ

?1+1

γ

γ

γ

+

1

γ((a(t)φ(t))σ)1+1

γ−

?

1 +

1

γ

?((a(t)φ(t))σ)

1

γwσ(t)

Φσ(t)

.

(2.17)

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1579

From (2.16), (2.17) and using the definitions of ψ2(t,t2) and η2(t,t2), we get, for t ∈ [t3,∞)T

w?(t) ≤ −ψ2(t,t2) +(η2(t,t2))+

Φσ(t)

wσ(t) −γΦ(t)τ?(t)δ1(τ(t),t2)

r(τ(t))(Φσ(t))λ

(wσ(t))λ,

(2.18)

where λ :=γ+1

γ. Define A and B by

Aλ:=γΦ(t)τ?(t)δ1(τ(t),t2)

r(τ(t))(Φσ(t))λ

(wσ(t))λ,

Bλ−1:=

r

1

λ(τ(t))(η2(t,t2))+

λ?γΦ(t)τ?(t)δ1(τ(t),t2)?1

λ

.

(2.19)

In view of (2.19), we have A and B are nonnegative numbers. Then, using the inequality (see [15])

λABλ−1− Aλ≤ (λ − 1)Bλ,

we get that

(2.20)

(η2(t,t2))+

Φσ(t)

wσ(t) −γΦ(t)τ?(t)δ1(τ(t),t2)

r(τ(t))(Φσ(t))λ

(wσ(t))λ≤

rγ(τ(t))?(η2(t,t2))+

?γ+1

(γ + 1)γ+1?Φ(t)τ?(t)δ1(τ(t),t2)?γ.

From this last inequality and (2.18), we get

w?(t) ≤

rγ(τ(t))?(η2(t,t2))+

?γ+1

(γ + 1)γ+1?Φ(t)τ?(t)δ1(τ(t),t2)?γ− ψ2(t,t2).

Integrating both sides from t3to t, we get

?t

which leads to a contradiction with (2.14).

t3

?

ψ2(s,t2) −

rγ(τ(s))?(η2(s,t2))+

?γ+1

(γ + 1)γ+1?Φ(s)τ?(s)δ1(τ(s),t2)?γ

?

?s ≤ w(t3) − w(t) ≤ w(t3),

?

Remark. Note that the results of Theorem 2.2 can be extended to Eq. (1.1) when 0 < γ ≤ 1 by putting φ = 0. We leave

the details to the interested reader.

By Theorems 2.1 and 2.2 by choosing φ(t) = 0 and Φ(t) = 1, we have the following other oscillation criteria:

Corollary 2.1. Assume that (2.1) and (2.2) hold. If

?∞

then every solution of Eq. (1.1) is either oscillatory or tends to zero.

t0

p(t)?t = ∞,

Similarly letting φ(t) = 0 in Theorems 2.1 and 2.2, we get the following results.

Corollary 2.2. Assume that (2.1) and (2.2) hold. If, for all sufficiently large T1∈ [t0,∞)T, there is a T > T1such that τ (T) > T1

and

?t

then every solution of Eq. (1.1) is either oscillatory or tends to zero.

limsup

t→∞

T

?

p(s)Φ(s) −

r(τ(s))(Φ?(s))2

4γΦ(s)τ?(s)δ (τ (s),T1)

?

?s = ∞,

(2.21)

Corollary 2.3. Assume that (2.1) and (2.2) hold. If, for all sufficiently large T1∈ [t0,∞)T, there is a T > T1such that τ (T) > T1

and

?t

then every solution of Eq. (1.1) is either oscillatory or tends to zero.

limsup

t→∞

T

?

p(s)Φ(s) −

rγ(τ(s))((Φ?(s))+)γ+1

(γ + 1)γ+1?Φ(s)τ?(s)δ1(τ (s),T1)?γ

?

?s = ∞,

(2.22)

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

Example 2.1. Consider the third order nonlinear delay dynamic equation

(tγ(x??(t))γ)?+

where β is a positive constant. We have

β

tγ+1xγ(τ(t)) = 0,

for t ∈ [t0,∞)T,

(2.23)

a(t) = tγ,

r (t) = 1,

p(t) =

β

tγ+1,

for t ∈ [t0,∞)T.

It is clear that condition (2.1) holds, since

?∞

and

?∞

byExample5.60in[2],then,forallsufficientlylargeT1,andforτ (T) > T1,wecanfindT∗> τ (T)suchthatδ1(τ(t),T1) ≥ 1,

for t ≥ T∗. Also,

?∞

?∞

≥

γ

t

s

and

?∞

so that condition (2.2) holds. For a time scaleT, we can chooseτ(t) such thatτ?(t) ≥ 1 andτ(σ(t)) = σ(τ(t)), for example

when T = R or T = Z, we choose τ(t) = t − τ, τ ≥ 0, when T = {t : t = qk,k ∈ N0,q > 1}, we choose τ(t) =

us take Φ(t) = tγ, then, by the Pötzsche chain rule

?1

= γ

0

≤ γσγ−1(t).

Thus, we assume T is a time scale satisfying σ(t) ≤ kt, for some k > 0,t ≥ Tk> T∗. Therefore

?t

?t

?

γ + 1

?

zero β >

γ+1

t0

?t

r(t)=

?∞

t0

?t = ∞,

(2.24)

t0

?t

1

γ(t)

a

=

?∞

t0

?t

t

= ∞,

(2.25)

s

p(u)?u = β

?

?∞

s

?u

uγ+1≥β

?1

γ

?∞

?β

?β

s

?−1

?1

?1

uγ

γ?∞

γ?∞

??

?u =β

γ

1

sγ,

t

1

a(s)

?∞

s

p(u)?u

γ

?s ≥

γ

t

?s

s2

?−1

??

?s =

?β

γ

?1

γ1

t,

t0

1

r(t)

?∞

t

?

1

a(s)

?∞

s

p(u)?u

?1

γ

?s?t ≥

?β

γ

?1

γ?∞

t0

?t

t

= ∞,

t

q, ect. Let

Φ?(t) = (tγ)?= γ

0

(t + hµ(t))γ−1dh

?1

((1 − h)t + hσ(t))γ−1dh

limsup

t→∞

Tk

?

p(s)Φ(s) −

?

?

?γ+1

γ

kγ2−1.

rγ(τ(s))((Φ?(s))+)γ+1

(γ + 1)γ+1(Φ(s)τ?(s)δ1(τ (s),T1))γ

γ

γ + 1

?

?

?s

≥ limsup

t→∞

Tk

β

s−

?γ+1

??γ+1kγ2−1

s

?

?s

≥

β −

γ

kγ2−1

limsup

t→∞

?t

Tk

?s

s

= ∞,

if β >

time scale where σ(t) ≤ kt, for some k > 0,t ≥ Tk, then, by Corollary 2.3, every solution of (2.23) is oscillatory or tends to

?

γ

γ+1

kγ2−1. Therefore, condition (2.22) is satisfied if β >

?

γ

γ+1

?γ+1

kγ2−1. We conclude that if [t0,∞)Tis a

?γ+1

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1581

Remark 2.1. If the assumption (2.2) is not satisfied, we have some sufficient conditions which ensure that every solution

x(t) of (1.1) oscillates or limt→∞x(t) exists (finite).

Example 2.2. Consider the third order nonlinear delay dynamic equation

(tγ−1((t

1

γx?(t))?)γ)?+

1

t3−2γxγ(τ(t)) = 0,

for t ∈ [t0,∞)T.

(2.26)

We have

a(t) = tγ−1,

Note that

?∞

?∞

and

?∞

by Example 5.60 in [4]. Then by Corollary 2.1 and Remark 2.1, every solution of (2.26) is oscillatory or limt→∞x(t) exists

(finite).

r (t) = t

1

γ,

p(t) = t2γ−3,

for t ∈ [t0,∞)T.

t0

?t

r(t)=

?t

1

γ(t)

?∞

=

t0

?∞

?t

t

1

γ

= ∞,

?t

t1−1

t0

a

t0

γ

= ∞,

t0

p(t)?t =

?∞

t0

1

t3−2γ?t = ∞

We are now ready to state and prove a Philos-type oscillation criteria for Eq. (1.1).

Theorem 2.3. Assume that (2.1) and (2.2) hold. Furthermore, suppose that there exist functions G,g ∈ Crd(D,R), where

D ≡ {(t,s) : t ≥ s ≥ t0} such that

G(t,t) = 0,

and G has a non positive continuous ?-partial derivative G?s(t,s) with respect to the second variable and satisfies, for all

sufficiently large T1∈ [t0,∞)T, there is a T > T1such that τ (T) > T1,

− G?s(t,s) −η1(s,T1)

Φσ(s)

Φσ(s)

t ≥ t0, G(t,s) > 0,

t > s ≥ t0,

(2.27)

G(t,s) =

g (t,s)

?

G(t,s),

(2.28)

and

limsup

t→∞

1

G(t,T)

?t

T

?

ψ1(s,T1)G(t,s) −

(g−(t,s))2r(τ(s))

4γΦ(s)τ?(s)δ(τ(s),T1)

?

?s = ∞.

(2.29)

Then every solution of Eq. (1.1) is oscillatory or tends to zero.

Proof. Assume (1.1) has a nonoscillatory solution x on [t0,∞)T. Then, without loss of generality, there is a t1∈ [t0,∞)T,

sufficiently large, such that x(t) > 0 and x(τ(t)) > 0 on [t1,∞)T. Therefore from Lemma 2.1, we get

?

and either x?(t) > 0 for t ≥ t2≥ t1or limt→∞x(t) = 0. Let x?(t) > 0 on [t2,∞)T. We choose t3≥ t2such that τ(t) > t2,

for t ≥ t3. We define w(t) also, as in the proof of Theorem 2.1. From (2.13), we have

ψ1(t,t2) ≤ −w?(t) +η1(t,t2)

r(τ(t))(Φσ(t))2

a(t)

??

r(t)x?(t)???γ??

< 0,

?

r(t)x?(t)??> 0, for t ∈ [t1,∞)T,

Φσ(t)wσ(t) −γΦ(t)τ?(t)δ(τ(t),t2)

(wσ(t))2,

for t ∈ [t3,∞)T.

(2.30)

In (2.30), replace t by s and multiply both sides by G(t,s), integrate with respect to s from t3to t, t ≥ t3,

?t

−

t3

t3

G(t,s)ψ1(s,t2)?s ≤ −

?t

?t

t3

G(t,s)w?(s)?s +

γG(t,s)Φ(s)τ?(s)δ(τ(s),t2)

r(τ(s))(Φσ(s))2

?t

t3

G(t,s)η1(s,t2)

Φσ(s)

wσ(s)?s

(wσ(s))2?s.

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

Integrating by parts and using (2.27) and then (2.28), we obtain

?t

+

t3

Φσ(s)

?t

and so

?t

t3

G(t,s)ψ1(s,t2)?s ≤ G(t,t3)w(t3) +

?t

≤ G(t,t3)w(t3) +

?t

γG(t,s)Φ(s)τ?(s)δ(τ(s),t2)

r(τ(s))(Φσ(s))2

wσ(s) −γG(t,s)Φ(s)τ?(s)δ(τ(s),t2)

r(τ(s))(Φσ(s))2

t3

G?s(t,s)wσ(s)?s

G(t,s)η1(s,t2)

wσ(s)?s −

?

?t

t3

(wσ(s))2?s

t3

−g (t,s)√G(t,s)

Φσ(s)

(wσ(s))2

?

?s,

t3

G(t,s)ψ1(s,t2)?s ≤ G(t,t3)w(t3) +

?t

t3

?g−(t,s)√G(t,s)

Φσ(s)

wσ(s)

−γG(t,s)Φ(s)τ?(s)δ(τ(s),t2)

r(τ(s))(Φσ(s))2

(wσ(s))2

?

?s.

(2.31)

It is easy to check that

g−(t,s)√G(t,s)

Φσ(s)

4γΦ(s)τ?(s)δ(τ(s),t2)−γG(t,s)Φ(s)τ?(s)δ(τ(s),t2)

which implies

g−(t,s)√G(t,s)

Φσ(s)

Using (2.32) in (2.31), we get

?t

which contradicts assumption (2.29). The proof is complete.

wσ(s) −γG(t,s)Φ(s)τ?(s)δ(τ(s),t2)

r(τ(s))(Φσ(s))2

−(t,s)r(τ(s))

(wσ(s))2

=

g2

r(τ(s))(Φσ(s))2

?

wσ(t) −

g−(t,s)r(τ(s))Φσ(s)

2γΦ(s)√G(t,s)τ?(s)δ(τ(s),t2)

?2

,

wσ(s) −γG(t,s)Φ(s)τ?(s)δ(τ(s),t2)

r(τ(s))(Φσ(s))2

(wσ(s))2≤

g2

−(t,s)r(τ(s))

4γΦ(s)τ?(s)δ(τ(s),t2).

(2.32)

1

G(t,t3)

t3

?

ψ1(s,t2)G(t,s) −

g2

−(t,s)r(τ(s))

4γΦ(s)τ?(s)δ(τ(s),t2)

?

?

?s ≤ |w(t3)|,

Theorem 2.4. Assume that (2.1) and (2.2) hold. Furthermore, suppose that there exist functions H,h ∈ Crd(D,R), where

D ≡ {(t,s) : t ≥ s ≥ t0} such that

H (t,t) = 0,

and H has a non positive continuous ?-partial derivative H?s(t,s) with respect to the second variable and satisfies, for all

sufficiently large T1∈ [t0,∞)T, there is a T > T1such that τ (T) > T1,

− H?s(t,s) −η2(s,T1)

Φσ(s)

and

?t

Then every solution of Eq. (1.1) is oscillatory or tends to zero.

t ≥ t0,

H (t,s) > 0,

t > s ≥ t0,

(2.33)

H (t,s) =

h(t,s)

Φσ(s)(H (t,s))

γ

γ+1,

(2.34)

limsup

t→∞

1

H (t,T)

T

?

ψ2(s,T1)H (t,s) −

(h−(t,s))γ+1rγ(τ(s))

(γ + 1)γ+1(Φ(s)τ?(s)δ1(τ(s),T1))γ

?

?s = ∞.

(2.35)

Proof. Assume (1.1) has a nonoscillatory solution x on [t0,∞)T. Then, without loss of generality, there is a t1∈ [t0,∞)T,

sufficiently large, such that x(t) > 0 and x(τ(t)) > 0 on [t1,∞)T. Therefore from Lemma 2.1, we get

?

and either x?(t) > 0 for t ≥ t2≥ t1or limt→∞x(t) = 0. Let x?(t) > 0 on [t2,∞)T. We choose t3≥ t2such that τ(t) > t2,

for t ≥ t3. We define w(t) also, as in the proof of Theorem 2.2. From (2.18) with (η2(t,t2))+replaced by η2(t,t2), we have

ψ2(t,t2) ≤ −w?(t) +η2(t,t2)

r(τ(t))(Φσ(t))λ

a(t)

??

r(t)x?(t)???γ??

< 0,

?

r(t)x?(t)??> 0, for t ∈ [t1,∞)T,

Φσ(t)wσ(t) −γΦ(t)τ?(t)δ1(τ(t),t2)

(wσ(t))λ,

for t ∈ [t3,∞)T.

(2.36)

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1583

In (2.36), replace t by s and multiply both sides by H (t,s), integrate with respect to s from t3to t, t ≥ t3,

?t

−

t3

Integrating by parts and using (2.33) and then (2.34), we obtain

?t

+

t3

?t

and so

?t

− H (t,s)γΦ(s)τ?(s)δ1(τ(s),t2)

r(τ(s))(Φσ(s))λ

t3

H (t,s)ψ2(s,t2)?s ≤ −

?t

?t

t3

H (t,s)w?(s)?s +

H (t,s)γΦ(s)τ?(s)δ1(τ(s),t2)

r(τ(s))(Φσ(s))λ

?t

t3

H (t,s)η2(s,t2)

Φσ(s)wσ(s)?s

(wσ(s))λ?s.

t3

H (t,s)ψ2(s,t2)?s ≤ H (t,t3)w(t3) +

?t

?t

H (t,s)γΦ(s)τ?(s)δ1(τ(s),t2)

r(τ(s))(Φσ(s))λ

t3

H?s(t,s)wσ(s)?s

H (t,s)η2(s,t2)

Φσ(s)wσ(s)?s −

?

?t

t3

(wσ(s))λ?s

≤ H (t,t3)w(t3) +

t3

−h(t,s)(H (t,s))

Φσ(s)

1

λ

wσ(s) − H (t,s)γΦ(s)τ?(s)δ1(τ(s),t2)

r(τ(s))(Φσ(s))λ

(wσ(s))λ

?

?s,

t3

H (t,s)ψ2(s,t2)?s ≤ H (t,t3)w(t3) +

?t

t3

?

h−(t,s)(H (t,s))

Φσ(s)

1

λ

wσ(s)

(wσ(s))λ

?

?s.

(2.37)

Again, define A ≥ 0 and B ≥ 0 by

Aλ:=γH (t,s)Φ(s)τ?(s)δ1(τ(s),t2)(wσ(s))λ

r(τ(s))(Φσ(s))λ

,

Bλ−1:=

h−(t,s)r

1

λ(τ(s))

λ?γΦ(s)τ?(s)δ1(τ(s),t2)?1

λ

,

and using the inequality (2.20), we get

h−(t,s)(H (t,s))

Φσ(s)

From this last inequality and (2.37), we get

?t

and this implies that

?t

which contradicts assumption (2.35). This completes the proof.

1

λ

wσ(s) −γH (t,s)Φ(s)τ?(s)δ1(τ(s),t2)(wσ(s))λ

r(τ(s))(Φσ(s))λ

≤

(h−(t,s))γ+1rγ(τ(s))

(γ + 1)γ+1(Φ(s)τ?(s)δ1(τ(s),t2))γ.

t3

?

H (t,s)ψ2(s,t2) −

(h−(t,s))γ+1rγ(τ(s))

(γ + 1)γ+1(Φ(s)τ?(s)δ1(τ(s),t2))γ

?

?s ≤ H (t,t3)w(t3),

1

H (t,t3)

t3

?

H (t,s)ψ2(s,t2) −

(h−(t,s))γ+1rγ(τ(s))

(γ + 1)γ+1(Φ(s)τ?(s)δ1(τ(s),t2))γ

?

?s ≤ |w(t3)|,

?

Remark 2.2. The results are in a form with a high degree of generality, thus with an appropriate choice of H (t,s) and

G(t,s) in Theorems 2.3 and 2.4, we can get several sufficient conditions for oscillation of Eq. (1.1). For instance, we can

choose H (t,s) = (t − s)α, α > 1. In this case

h(t,s) ≥

(t − s)

where λ =γ+1

1

α

λ

?αΦσ(s)(t − σ(s))α−1− η2(s,T1)(t − s)α?,

γ. We leave the details to the reader.

3. Applications

In this section, we apply the oscillation criteria to different types of time scale, for example if T = R then σ(t) =

t,µ(t) = 0,f?(t) = f?(t),?b

a(t)

+ f (t,x(τ (t))) = 0,

af(t)?t =?b

af(t)dt, and (1.1) becomes the third-order nonlinear delay differential equation

? ??

r(t)x?(t)???γ??

(3.1)

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

then we have from Theorems 2.1–2.4 and Corollaries 2.1–2.3 the following oscillation criteria for Eq. (3.1).

Theorem 3.1. Assume that

?∞

and

?∞

hold. If, for sufficiently large T1∈ [t0,∞), there is a T > T1such that τ (T) > T1and

?t

where the function φ(t) is a nonnegative function, then every solution of Eq. (3.1) is either oscillatory or tends to zero.

t0

dt

1

γ(t)

a

= ∞,

?∞

t0

dt

r(t)= ∞,

(3.2)

t0

1

r(t)

?∞

t

?

1

a(s)

?∞

s

p(u)du

?1

γ

dsdt = ∞,

(3.3)

limsup

t→∞

T

?

ψ1(s,T1) −

r(τ(s))η2

4γΦ(s)τ?(s)δ(τ(s),T1)

1(s,T1)

?

ds = ∞,

Theorem 3.2. Assume that (3.2) and (3.3) hold. If, for sufficiently large T1∈ [t0,∞), there is a T > T1such that τ (T) > T1and

?t

where the function φ(t) is a nonnegative function, then every solution of Eq. (3.1) is either oscillatory or tends to zero.

limsup

t→∞

T

?

ψ2(s,T1) −

rγ(τ(s))?(η2(s,T1))+

?γ+1

(γ + 1)γ+1(Φ(s)τ?(s)δ1(τ(s),T1))γ

?

ds = ∞,

Corollary 3.1. Assume that (3.2) and (3.3) hold. If

?∞

then every solution of Eq. (3.1) is either oscillatory or tends to zero.

t0

p(t)dt = ∞,

Corollary 3.2. Assume that (3.2) and (3.3) hold. If, for sufficiently large T1∈ [t0,∞), there is a T > T1such that τ (T) > T1and

?t

then every solution of Eq. (3.1) is either oscillatory or tends to zero.

limsup

t→∞

T

?

p(s)Φ(s) −

r(τ(s))(Φ?(s))2

4γΦ(s)τ?(s)δ(τ (s),T1)

?

ds = ∞,

Corollary 3.3. Assume that (3.2) and (3.3) hold. If, for sufficiently large T1∈ [t0,∞), there is a T > T1such that τ (T) > T1and

?t

then every solution of Eq. (3.1) is either oscillatory or tends to zero.

limsup

t→∞

T

?

p(s)Φ(s) −

rγ(τ(s))((Φ?(s))+)γ+1

(γ + 1)γ+1(Φ(s)τ?(s)δ1(τ (s),T1))γ

?

ds = ∞,

Theorem 3.3. Assume that (3.2) and (3.3) hold. Furthermore, suppose that there exist functions G,g ∈ C (D,R), where

D ≡ {(t,s) : t ≥ s ≥ t0} such that

G(t,t) = 0,

and G has a non positive continuous partial derivative∂G(t,s)

∂s

large T1∈ [t0,∞), there is a T > T1such that τ (T) > T1,

−∂G(t,s)

∂s

Φσ(s)

Φσ(s)

t ≥ t0,

G(t,s) > 0,

t > s ≥ t0,

with respect to the second variable and satisfies, for all sufficiently

−η1(s,T1)

G(t,s) =

g (t,s)

?

G(t,s),

and

limsup

t→∞

1

G(t,T)

?t

T

?

ψ1(s,T1)G(t,s) −

(g−(t,s))2r(τ(s))

4γΦ(s)τ?(s)δ(τ(s),T1)

?

ds = ∞.

Then every solution of Eq. (3.1) is oscillatory or tends to zero.

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

1585

Theorem 3.4. Assume that (3.2) and (3.3) hold. Furthermore, suppose that there exist functions H,h ∈ C (D,R), where

D ≡ {(t,s) : t ≥ s ≥ t0} such that

H (t,t) = 0,

and H has a non positive continuous partial derivative∂H(t,s)

∂s

large T1∈ [t0,∞), there is a T > T1such that τ (T) > T1,

−∂H (t,s)

∂s

Φσ(s)

and

?t

Then every solution of Eq. (3.1) is oscillatory or tends to zero.

t ≥ t0,

H (t,s) > 0,

t > s ≥ t0,

with respect to the second variable and satisfies, for all sufficiently

−η2(s,T1)

H (t,s) =

h(t,s)

Φσ(s)(H (t,s))

γ

γ+1,

limsup

t→∞

1

H (t,T)

T

?

ψ2(s,T1)H (t,s) −

(h−(t,s))γ+1rγ(τ(s))

(γ + 1)γ+1(Φ(s)τ?(s)δ1(τ(s),T1))γ

?

ds = ∞.

We note that above results made be viewed as an extension of the oscillation criteria that has been established by

Baculíková et al. [16].

If T = Z, then σ(t) = t + 1,µ(t) = 1,f?(t) = ?f(t),?b

?(a(t){?[r(t)?x(t)]}γ) + f (t,x(τ (t))) = 0,

then we have from Theorems 2.1–2.4 and Corollaries 2.1–2.3 the following oscillation criteria for Eq. (3.4).

af(t)?t =?b−1

t=af(t), and (1.1) becomes the third-order

nonlinear delay difference equation

(3.4)

Theorem 3.5. Assume that

∞

?

t=t0

1

a

1

γ(t)

= ∞,

∞

?

t=t0

1

r(t)= ∞,

(3.5)

and

∞

?

t=t0

1

r(t)

∞

?

s=t

?

1

a(s)

∞

?

u=s

p(u)

?1

γ

= ∞,

(3.6)

hold. If, for all sufficiently large N1∈ N, there is a N > N1such that τ (N) > N1and

t−1

?

where the function φ(t) is a nonnegative sequence, then every solution of Eq. (3.4) is either oscillatory or tends to zero.

limsup

t→∞

s=N

?

ψ1(s,N1) −

r(τ(s))η2

4γΦ(s)?τ(s)δ(τ(s),N1)

1(s,N1)

?

= ∞,

Theorem 3.6. Assume that (3.5) and (3.6) hold. If, for all sufficiently large N1∈ N, there is a N > N1such that τ (N) > N1and

t−1

?

where the function φ(t) is a nonnegative sequence, then every solution of Eq. (3.4) is either oscillatory or tends to zero.

limsup

t→∞

s=N

?

ψ2(s,N1) −

rγ(τ(s))?(η2(s,N1))+

?γ+1

(γ + 1)γ+1(Φ(s)?τ(s)δ1(τ(s),N1))γ

?

= ∞,

Corollary 3.4. Assume that (3.5) and (3.6) hold. If

?

then every solution of Eq. (3.4) is either oscillatory or tends to zero.

∞

t=t0

p(t) = ∞,

Corollary 3.5. Assume that (3.5) and (3.6) hold. If, for all sufficiently large N1∈ N, there is a N > N1such that τ (N) > N1and

t−1

?

then every solution of Eq. (3.4) is either oscillatory or tends to zero.

limsup

t→∞

s=N

?

p(s)Φ(s) −

r(τ(s))(?Φ(s))2

4γΦ(s)?τ(s)δ(τ (s),N1)

?

= ∞,

Page 14

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T.S. Hassan / Mathematical and Computer Modelling 49 (2009) 1573–1586

Corollary 3.6. Assume that (3.5) and (3.6) hold. If, for all sufficiently large N1∈ N, there is a N > N1such that τ (N) > N1and

t−1

?

then every solution of Eq. (3.4) is either oscillatory or tends to zero.

limsup

t→∞

s=N

?

p(s)Φ(s) −

rγ(τ(s))((?Φ(s))+)γ+1

(γ + 1)γ+1(Φ(s)?τ(s)δ1(τ (s),N1))γ

?

= ∞,

Theorem 3.7. Assume that (3.5) and (3.6) hold. Furthermore, suppose that there exist doubles sequences G,g on D, where

D ≡ {(t,s) : t ≥ s ≥ t0} such that, for all sufficiently large N1∈ N, there is a N > N1such that τ (N) > N1,

G(t,t) = 0,

−?sG(t,s) −η1(s,N1)

Φσ(s)

Φσ(s)

and

?

Then every solution of Eq. (3.4) is oscillatory or tends to zero.

t ≥ t0,

G(t,s) > 0,

t > s ≥ t0,

?

G(t,s) =

g (t,s)

G(t,s),

limsup

t→∞

1

G(t,N)

t−1

s=N

?

ψ1(s,N1)G(t,s) −

(g−(t,s))2r(τ(s))

4γΦ(s)?τ(s)δ(τ(s),N1)

?

= ∞.

Theorem 3.8. Assume that (3.5) and (3.6) hold. Furthermore, suppose that there exist doubles sequences H,h on D, where

D ≡ {(t,s) : t ≥ s ≥ t0} such that, for all sufficiently large N1∈ N, there is a N > N1such that τ (N) > N1,

H (t,t) = 0,

−?sH (t,s) −η2(s,N1)

Φσ(s)

and

?

Then every solution of Eq. (3.4) is oscillatory or tends to zero.

Similarly,wecanstateoscillationcriteriaformanyothertimescales,e.g.,T = hZ,h > 0,T =?

H0= 0,Hn=?n

Acknowledgement

t ≥ t0,

H (t,s) > 0,

t > s ≥ t0,

H (t,s) =

h(t,s)

Φσ(s)(H (t,s))

γ

γ+1,

limsup

t→∞

1

H (t,N)

t−1

s=N

?

ψ2(s,N1)H (t,s) −

(h−(t,s))γ+1rγ(τ(s))

(γ + 1)γ+1(Φ(s)?τ(s)δ1(τ(s),N1))γ

?

= ∞.

t : t = qk,k ∈ N0,q > 1?

,

T = N2

0:= {n2: n ∈ N0}, or T = {Hn : n ∈ N} where Hnis the so-called n-th harmonic number defined by

k=1

1

k,n ∈ N0.

The author would like to thank an anonymous referee for his/her careful reading of the entire manuscript, which helped

to improve the quality of this paper.

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