# Oscillation for Higher Order Dynamic Equations on Time Scales

Authors: Taixiang Sun; Qiuli He; Hongjian Xi; Weiyong Yu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We investigate the oscillation of the following higher order dynamic equation:{an(t)[(an-1(t)(⋯(a1(t)xΔ(t))Δ⋯)Δ)Δ]α}Δ+p(t)xβ(t)=0, on some time scale T, where n≥2, ak(t)  (1≤k≤n) and p(t) are positive rd-continuous functions on T and α,β are the quotient of odd positive integers. We give sufficient conditions under which every solution of this equation is either oscillatory or tends to zero.


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## Body

## 1. Introduction

In this paper, we investigate the oscillation of the following higher order dynamic equation:(E){an(t)[(an-1(t)(⋯(a1(t)xΔ(t))Δ⋯)Δ)Δ]α}Δ+p(t)xβ(t)=0,
on some time scale T, where n≥2, ak(t)(1≤k≤n) and p(t) are positive rd-continuous functions on T and α,β are the quotient of odd positive integers. Write
(1)Sk(t,x(t))={x(t),ifk=0,ak(t)Sk-1Δ(t,x(t)),if1≤k≤n-1,an(t)[Sn-1Δ(t,x(t))]α,ifk=n,
then (E) reduces to the following equation:
(2)SnΔ(t,x(t))+p(t)xβ(t)=0.Since we are interested in the oscillatory behavior of solutions near infinity, we assume that supT=∞ and t0∈T is a constant. For any a∈T, we define the time scale interval [a,∞)T={t∈T:t≥a}. By a solution of (2), we mean a nontrivial real-valued function x(t)∈Crd1[Tx,∞),Tx≥t0, which has the property that Sk(t,x(t))∈Crd1[Tx,∞) for 0≤k≤n and satisfies (2) on [Tx,∞), where Crd1 is the space of differentiable functions whose derivative is rd-continuous. The solutions vanishing in some neighborhood of infinity will be excluded from our consideration. A solution x(t) of (2) 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 the literature that 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 is an nonempty closed subset of the real numbers, and the cases when this time scale is equal to the real numbers or to the integers represent the classical theories of differential or of difference equations. Many other interesting time scales exist, and they give rise to many applications (see [2]). The new theory of the so-called “dynamic equations” not only unifies the theories of differential equations and difference equations, but also extends these classical cases to cases “in between,” for example, to the so-called q-difference equations when T={1,q,q2,…,qk,…}, which has important applications in quantum theory (see [3]). In this work, knowledge and understanding of time scales and time scale notation are assumed; for an excellent introduction to the calculus on time scales, see Bohner and Peterson [2, 4]. In recent years, there has been much research activity concerning the oscillation and nonoscillation of solutions of various equations on time scales, and we refer the reader to the papers [5–20].Recently, Erbe et al. in [21–23] considered the third-order dynamic equations
(3)(a(t)[r(t)xΔ(t)]Δ)Δ+p(t)f(x(t))=0,xΔΔΔ(t)+p(t)x(t)=0,(a(t){[r(t)xΔ(t)]Δ}γ)Δ+f(t,x(t))=0,
respectively, and established some sufficient conditions for oscillation.Hassan [24] studied the third-order dynamic equations
(4)(a(t){[r(t)xΔ(t)]Δ}γ)Δ+f(t,x(τ(t)))=0
and obtained some oscillation criteria, which improved and extended the results that have been established in [21–23].


## 2. Main Results

In this section, we investigate the oscillation of (2). To do this, we need the following lemmas.Lemma 1 (see [25]).
Assume that(5)∫t0∞[1an(s)]1/αΔs=∫t0∞Δsai(s)=∞∀1≤i≤n-1,
and 1≤m≤n. Then,(1)
liminft→∞Sm(t,x(t))>0 implies limt→∞Si(t,x(t))=∞ for 0≤i≤m-1;(2)
limsupt→∞Sm(t,x(t))<0 implies limt→∞Si(t,x(t))=-∞ for 0≤i≤m-1.Lemma 2 (see [25]).
Assume that (5) holds. If SnΔ(t,x(t))<0 and x(t)>0 for t≥t0, then there exists an integer 0≤m≤n with m+n even such that(1)
(-1)m+iSi(t,x(t))>0 for t≥t0 and m≤i≤n;(2)
ifm>1, then there exists T≥t0 such that Si(t,x(t))>0 for 1≤i≤m-1 and t≥T.Remark 3.
Letan(t)=⋯=a1(t)=1, and let T be the set of integers. Then, Lemmas 1 and 2 are Lemma 1.8.10 and Theorem 1.8.11 of [26], respectively.Lemma 4.
Assume that (5) holds. Furthermore, suppose that
(6)∫t0∞1an-1(u){∫u∞[1an(s)∫s∞‍p(v)Δv]1/αΔs}Δu=∞.
If x(t) is an eventually positive solution of (2), then there exists T≥t0 sufficiently large such that(1)
SnΔ(t,x(t))<0 for t≥T;(2)
eitherSi(t,x(t))>0 for t≥T and 0≤i≤n or limt→∞x(t)=0.Proof.
Pickt1≥t0 so that x(t)>0 on [t1,∞)T. It follows from (2) that
(7)SnΔ(t,x(t))=-p(t)xβ(t)<0fort≥t1.
By Lemma 2, we see that there exists an integer 0≤m≤n with m+n even such that (-1)m+iSi(t,x(t))>0 for t≥t1 and m≤i≤n, and x(t) is eventually monotone.
We claim thatlimt→∞x(t)≠0 implies m=n. If not, then Sn-1(t,x(t))<0(t≥t1) and Sn-2(t,x(t))>0(t≥t1), and there exist t2≥t1 and a constant c>0 such that x(t)≥c on [t2,∞)T. Integrating (2) from t into ∞, we get that for t≥t2(8)-an(t)[Sn-1Δ(t,x(t))]α=-Sn(t,x(t))≤-cβ∫t∞p(v)Δv.
Thus,
(9)Sn-1(t,x(t))≤-cβ/α×∫t∞[1an(s)∫s∞‍p(v)Δv]1/αΔsfort≥t2.
Again, integrating the above inequality from t2 into t, we obtain that for t≥t2(10)Sn-2(t,x(t))≤Sn-2(t2,x(t2))-cβ/α∫t2t1an-1(u){[1an(s)∫s∞‍p(v)Δv]1/α∫u∞[∫s∞‍1an(s)hhhhhhhhhhhhhhhhhhhhhhhh×∫s∞p(v)Δv]1/αΔs}Δu.
It follows from (6) that limt→∞Sn-2(t,x(t))=-∞, which is a contradiction to Sn-2(t,x(t))>0(t≥t1). The proof is completed.Lemma 5.
Assume thatx(t) is an eventually positive solution of (2) such that SnΔ(t,x(t))<0 for t≥T≥t0 and Si(t,x(t))>0 for t≥T and 0≤i≤n. Then,
(11)Si(t,x(t))≥Sn1/α(t,x(t))Bi+1(t,T)for0≤i≤n-1,t≥T,
and there exist T1>T and a constant c>0 such that
(12)x(t)≤cB1(t,T)fort≥T1,
where
(13)Bi(t,T)={∫Tt[1an(s)]1/αΔs,ifi=n,∫TtBi+1(s,T)ai(s)Δs,if1≤i≤n-1.Proof.
SinceSnΔ(t,x(t))<0 (t≥T), it follows that Sn(t,x(t)) is strictly decreasing on [T,∞)T. Then, for t≥T,
(14)Sn-1(t,x(t))≥Sn-1(t,x(t))-Sn-1(T,x(T))=∫Tt[Sn(s,x(s))an(s)]1/αΔs≥Sn1/α(t,x(t))Bn(t,T)Sn-2(t,x(t))≥Sn-2(t,x(t))-Sn-2(T,x(T))=∫TtSn-1(s,x(s))an-1(s)Δs≥Sn1/α(t,x(t))Bn-1(t,T)⋮S1(t,x(t))≥S1(t,x(t))-S1(T,x(T))=∫TtS2(s,x(s))a2(s)Δs≥Sn1/α(t,x(t))B2(t,T)S0(t,x(t))≥x(t)-x(T)=∫TtS1(s,x(s))a1(s)Δs≥Sn1/α(t,x(t))B1(t,T).
On the other hand, we have that for t≥T,
(15)Sn-1(t,x(t))=∫Tt[Sn(s,x(s))an(s)]1/αΔs+Sn-1(T,x(T))≤Sn-1(T,x(T))+Sn1/α(T,x(T))Bn(t,T).
Thus, there exist T1>T and a constant b1>0 such that
(16)Sn-1(t,x(t))≤b1Bn(t,T)fort≥T1.
Again,
(17)Sn-2(t,x(t))=Sn-2(T1,x(T1))+∫T1tSn-1(s,x(s))an-1(s)Δs≤Sn-2(T1,x(T1))+b1∫TtBn(s,T)an-1(s)Δs.
Thus, there exists a constant b2>0 such that
(18)Sn-2(t,x(t))≤b2∫TtBn(s,T)an-1(s)Δs=b2Bn-1(t,T)fort≥T1.
Again,
(19)Sn-3(t,x(t))=Sn-3(T1,x(T1))+∫T1tSn-2(s,x(s))an-2(s)Δs≤Sn-3(T1,x(T1))+b2∫TtBn-1(s,T)an-2(s)Δs.
Thus, there exists a constant b3>0 such that
(20)Sn-3(t,x(t))≤b3∫TtBn-1(s,T)an-2(s)Δs=b3Bn-2(t,T)fort≥T1.
The rest of the proof is by induction. The proof is completed.Lemma 6 (see [2]).
Letf:R→R be continuously differentiable and suppose that g:T→R is delta differentiable. Then, f∘g is delta differentiable and the formula
(21)(f∘g)Δ(t)=gΔ(t)∫01f′(hg(t)+(1-h)gσ(t))dh.Lemma 7 (see [27]).
IfA,B are nonnegative and λ>1, then
(22)Aλ-λABλ-1+(λ-1)Bλ≥0.Now, we state and prove our main results.Theorem 8.
Suppose that (5) and (6) hold. If there exists a positive nondecreasing delta differentiable function θ such that for all sufficiently large T∈[t0,∞)T and for any positive constants c1,c2, there is a T1>T such that
(23)limsupt→∞∫T1t[θ(s)p(s)-θΔ(s)B1α(s,T)δ1(t,T,c1,c2)]Δs=∞,
where
(24)δ1(t,T,c1,c2)={c1,ifα<β,1,ifα=β,c2B1α-β(t,T),ifα>β,
and B1(t,T) is as in Lemma 5. Then, every solution of (2) is either oscillatory or tends to zero.Proof.
Assume that (2) has a nonoscillatory solution x(t) on [t0,∞)T. Then, without loss of generality, there is a t1≥t0, sufficiently large, such that x(t)>0 for t≥t1. Therefore, we get from Lemma 4 that there exists t2≥t1 such that(i)
SnΔ(t,x(t))<0 for t≥t2;(ii)
eitherSi(t,x(t))>0 for t≥t2 and 0≤i≤n or limt→∞x(t)=0.Let Si(t,x(t))>0 for t≥t2 and 0≤i≤n. Consider
(25)w(t)=θ(t)Sn(t,x(t))xβ(t)fort≥t2.
It follows from Lemma 6 that
(26)(xβ)Δ(t)=βxΔ(t)∫01(hx(t)+(1-h)x(t)σ)β-1dh>0kkkkkkkkkkkkkkkkkkkkkkkkkkkkkfort≥t2.
Then,
(27)wΔ=[θxβ]ΔSnσ(·,x)+θxβSnΔ(·,x)=[θΔ(xβ)σ-θ(xβ)Δxβ(xβ)σ]Snσ(·,x)-θp≤θΔxβSn(·,x)-θp.
From (11) and (27), we get
(28)wΔ(t)≤θΔ(t)B1α(t,t2)xα-β(t)-θ(t)p(t)fort≥t2.
Now, we consider the following three cases.
Case 1. If α=β, then
(29)xα-β(t)=1fort≥t2.Case 2. If α>β, then it follows from (12) that there exist t3>t2 and a constant c2>0 such that
(30)xα-β(t)≤c2B1α-β(t,t2)fort≥t3.Case 3. If α<β, then
(31)x(t)≥x(t2)fort≥t2.
Thus,
(32)xα-β(t)≤c1=xα-β(t2)fort≥t2.
From (27)–(32), we obtain
(33)wΔ(t)≤θΔ(t)B1α(t,t2)δ1(t,t2,c1,c2)-θ(t)p(t)fort≥t3.
Integrating the above inequality from t3 into t, we have
(34)∫t3t[θ(s)p(s)-θΔ(s)B1α(s,t2)δ1(s,t2,c1,c2)]Δs≤w(t3)<∞,
which gives a contradiction to (23). The proof is completed.Theorem 9.
Suppose that (5) and (6) hold. If there exists a positive nondecreasing delta differentiable function θ such that for all sufficiently large T∈[t0,∞)T and for any positive constants c1,c2, there is a T1>T such that
(35)limsupt→∞∫T1t[(α/β)α(θΔ(s))α+1a1α(s)(α+1)α+1(B2(s,T)θ(s)δ2(s,T,c1,c2))αθ(s)p(s)kkkkkkkkkk-(α/β)α(θΔ(s))α+1a1α(s)(α+1)α+1(B2(s,T)θ(s)δ2(s,T,c1,c2))α]Δs=∞,
where
(36)δ2(t,T,c1,c2)={c1ifα<β,1,ifα=β,c2B1(β/α)-1(σ(t),T),ifα>β,
and B1(t,T),B2(t,T) are as in Lemma 5. Then, every solution of (2) is either oscillatory or tends to zero.Proof.
Assume that (2) has a nonoscillatory solution x(t) on [t0,∞)T. Then, without loss of generality, there is a t1≥t0, sufficiently large, such that x(t)>0 for t≥t1. Therefore, we get from Lemma 4 that there exists t2≥t1 such that(i)
SnΔ(t,x(t))<0 for t≥t2;(ii)
eitherSi(t,x(t))>0 for t≥t2 and 0≤i≤n or limt→∞x(t)=0.Let Si(t,x(t))>0 for t≥t2 and 0≤i≤n. Note that
(37)(xβ)Δ=βxΔ∫01(hx+(1-h)xσ)β-1dh=βxΔ∫01(hx+(1-h)xσ)βhx+(1-h)xσdh≥βxΔxβxσ.
From (11), we have
(38)(xβ)Δxβ≥βxΔxσ≥βSn1/α(·,x)B2(·,t2)a1xσ≥β(Snσ(·,x))1/αB2(·,t2)a1xσ=β(wσ)1/αa1(θσ)1/α(xσ)(β/α)-1B2(·,t2).
Then it follows from (27) that for t≥t2,
(39)wΔ=[θxβ]ΔSnσ(·,x)+θxβSnΔ(·,x)=[θΔ(xβ)σ-θ(xβ)Δxβ(xβ)σ]Snσ(·,x)-θp≤θΔwσθσ-βB2(·,t2)θa1(wσ)1+(1/α)(θσ)1+(1/α)(xσ)(β/α)-1-θp.
Now, we consider the following three cases.
Case 1. If α=β, then
(40)(xσ)(β/α)-1(t)=1fort≥t2.Case 2. If α>β, then it follows from (12) that there exist t3>t2 and a constant c such that
(41)x(t)≤cB1(t,t2)fort≥t3.
Thus,
(42)(xσ)(β/α)-1(t)≥c2B1(β/α)-1(σ(t),t2),
with c2=c(β/α)-1.
Case 3. If α<β, then
(43)x(t)≥x(t2)fort≥t2.
Thus,
(44)(xσ)(β/α)-1(t)≥c1=x(β/α)-1(t2).
From (39)–(44), we obtain that for t≥t3,
(45)wΔ≤wσθσθΔ-βB2(·,t2)θδ2(·,t2,c1,c2)a1(wσ)1+(1/α)(θσ)1+(1/α)-θp=-βB2(·,t2)θδ2(·,t2,c1,c2)a1×{(wσ)1+(1/α)(θσ)1+(1/α)-wσθσa1θΔβB2(·,t2)θδ2(·,t2,c1,c2)}-θp.
Let
(46)A=wσθσ,B=[αa1θΔ(α+1)βB2(·,t2)θδ2(·,t2,c1,c2)]α,
with λ=1+1/α. By Lemma 7, we have
(47)wΔ≤(α/β)α(θΔ)α+1a1α(α+1)α+1(B2(·,t2)θδ2(·,t2,c1,c2))α-θp.
Integrating the above inequality from t3 into t, it follows that
(48)∫t3t[(α/β)α(θΔ(s))α+1a1α(s)(α+1)α+1(B2(s,t2)θ(s)δ2(s,t2,c1,c2))αθ(s)p(s)kkkk-(α/β)α(θΔ(s))α+1a1α(s)(α+1)α+1(B2(s,t2)θ(s)δ2(s,t2,c1,c2))α]Δs≤w(t3)<∞,
which gives a contradiction to (35). The proof is completed.Remark 10.
The trick used in the proofs of Theorems8 and 9 is from [16].Theorem 11.
Suppose that (5) and (6) hold. If for all sufficiently large T∈[t0,∞)T,
(49)∫T∞p(u)[∫Tu‍Δsa1(s)]βΔu=∞,
then every solution of (2) is either oscillatory or tends to zero.Proof.
Assume that (2) has a nonoscillatory solution x(t) on [t0,∞)T. Then, without loss of generality, there is a t1≥t0, sufficiently large, such that x(t)>0 for t≥t1. Therefore, we get from Lemma 4 that there exists t2≥t1 such that(i)
SnΔ(t,x(t))<0 for t≥t2;(ii)
eitherSi(t,x(t))>0 for t≥t2 and 0≤i≤n or limt→∞x(t)=0.Let Si(t,x(t))>0 for t≥t2 and 0≤i≤n. Then, for t≥t2,
(50)x(t)=x(t2)+∫t2tS1(s,x(s))a1(s)Δs≥S1(t2,x(t2))∫t2tΔsa1(s).
It follows from (2) that
(51)-SnΔ(t,x(t))≥p(t)[S1(t2,x(t2))∫t2t‍Δsa1(s)]β.
Integrating the above inequality from t2 into ∞, we have
(52)Sn(t2,x(t2))≥S1β(t2,x(t2))∫t2∞‍p(u)[∫t2u‍Δsa1(s)]βΔu,
which gives a contradiction to (49). The proof is completed.Theorem 12.
Suppose that (5) and (6) hold. If for all sufficiently large T∈[t0,∞)T,
(53)limsupt→∞B1α(t,T)δ3(t,T,c1,c2)∫t∞p(s)Δs>1,
where
(54)δ3(t,T,c1,c2)={c1,c1isanypositiveconstant,ifα<β,1,ifα=β,c2B1β-α(t,T),c2isanypositiveconstant,ifα>β,
and B1(t,T) is as in Lemma 5, then every solution of (2) is either oscillatory or tends to zero.Proof.
Assume that (2) has a nonoscillatory solution x(t) on [t0,∞). Then, without loss of generality, there is a t1≥t0, sufficiently large, such that x(t)>0 for t≥t1. Therefore, we get from Lemma 4 that there exists t2≥t1 such that(i)
SnΔ(t,x(t))<0 for t≥t2;(ii)
eitherSi(t,x(t))>0 for t≥t2 and 0≤i≤n or limt→∞x(t)=0.Let Si(t,x(t))>0 for t≥t2 and 0≤i≤n. Then, it follows from (2) and (11) that for t≥t2,
(55)∫t∞p(s)xβ(s)Δs≤Sn(t,x(t))≤[x(t)B1(t,t2)]α.
Using the fact that x(t) is strictly increasing on [t2,∞)T, we obtain
(56)xβ(t)∫t∞p(s)Δs≤[x(t)B1(t,t2)]α.
Thus,
(57)B1α(t,t2)xβ-α(t)∫t∞p(s)Δs≤1.
Now, we consider the following three cases.
Case 1. If α=β, then
(58)xβ-α(t)=1fort≥t2.Case 2. If α>β, then it follows from (12) that there exist t3>t2 and a constant c such that
(59)x(t)≤cB1(t,t2)fort≥t3.
Thus,
(60)xβ-α(t)≥c2B1β-α(t,t2),
with c2=cβ-α.
Case 3. If α<β, then
(61)x(t)≥x(t2)fort≥t2.
Thus,
(62)xβ-α(t)≥c1=xβ-α(t2).
From (57)–(62), we obtain that for t≥t3,
(63)B1α(t,t2)δ3(t,t2,c1,c2)∫t∞p(s)Δs≤1,
which gives a contradiction to (53). The proof is completed.


## 3. Examples

In this section, we give some examples to illustrate our main results.Example 1.
Consider the following higher order dynamic equation:(64)SnΔ(t,x(t))+tγxβ(t)=0,
on an arbitrary time scale T with supT=∞, where n≥2, α,β and Sk(t,x(t))(0≤k≤n) are as in (2) with an(t)=tα-1,an-1(t)=⋯=a1(t)=t, and γ>-1. Then, every solution of (64) is either oscillatory or tends to zero.Proof.
Note that(65)∫t0∞[1an(s)]1/αΔs=∫t0∞[1sα-1]1/αΔs=∞,∫t0∞‍Δsai(s)=∫t0∞‍Δss=∞for1≤i≤n-1,∫t0∞p(s)Δs=∫t0∞sγΔs=∞,
by Example 5.60 in [4]. Pick t1>t0 such that
(66)∫t0t11an-1(u){∫ut1[1an(s)]1/αΔs}Δu>0.
Then,
(67)∫t0∞1an-1(u){∫u∞[1an(s)∫s∞p(v)Δv]1/αΔs}Δu≥[∫t1∞‍p(v)Δv]1/α×∫t0t11an-1(u)(∫ut1‍[1an(s)]1/αΔs)Δu=∞.
Let T∈[t0,∞)T, sufficiently large, and u1>T such that ∫Tu1‍(1/a1(s))Δs>1, then
(68)∫T∞p(u)[∫Tu‍1a1(s)Δs]βΔu≥∫u1∞p(u)[∫Tu‍1a1(s)Δs]βΔu≥∫u1∞‍p(u)Δu=∞.
Thus, conditions (5), (6), and (49) are satisfied. By Theorem 11, every solution of (64) is either oscillatory or tends to zero.Example 2.
Consider the following higher order dynamic equation:(69)SnΔ(t,x(t))+1t1+γxβ(t)=0,
on an arbitrary time scale T with supT=∞, where n≥2, Sk(t,x(t))(0≤k≤n) are as in (2) with an(t)=1,an-1(t)=t1/α,an-2(t)=⋯=a1(t)=t, 0<γ<min{1,β}, and α,β are the quotient of odd positive integers with α≥1. Then, every solution of (69) is either oscillatory or tends to zero.Proof.
Note that(70)∫t0∞[1an(s)]1/αΔs=∫t0∞Δs=∞,∫t0∞1an-1(s)Δs=∫t0∞1s1/αΔs=∞,∫t0∞‍1ai(s)Δs=∫t0∞‍1sΔs=∞for1≤i≤n-2.
Pick t1>t0 such that ∫t0t1‍(Δu/u1/α)>0, then
(71)∫t0∞1an-1(u){∫u∞‍[1an(s)∫s∞p(v)Δv]1/αΔs}Δu=∫t0∞1u1/α{∫u∞[∫s∞1vγ+1Δv]1/αΔs}Δu≥1γ∫t0∞‍1u1/α{∫u∞[∫s∞(vγ)Δvγ(vγ)σΔv]1/αΔs}Δu=1γ∫t0∞1u1/α[∫u∞(1sγ)1/αΔs]Δu≥1γ∫t0t11u1/α[∫t1∞(1sγ)1/αΔs]Δu=1γ[∫t1∞(1sγ)1/αΔs]∫t0t1Δuu1/α=∞.
LetM=max{c1,1,c2} with c1,c2 being positive constants, ρ=min{α,β}, and γ<τ<min{1,β}. Pick T1>T>0 such that
(72)1tγ≥2tτ≥2M[(1/2)n+(1/α)(t-2n-1T)]ρfort≥T1.
Let θ(t)=t, then
(73)B1(t,T)=∫Tt‍1a1(u1)×[∫Tu11a2(u2)kkkkkkkkk×[⋯[∫Tun-21an-1(un-1)[∫Tun-1‍Δun]1/αkkkkkkkkkkkkkkkk×Δun-1∫Tu11a2(u2)]⋯]Δu2[∫Tun-1‍Δun]1/α]Δu1=∫2n-1Tt1u1kkkkkkkk×[∫2n-2Tu11u2[⋯[∫2Tun-2[1un-1∫Tun-1‍Δun]1/αkkkkkkkkkkkkkkkkkkkkkkkk×Δun-1∫Tun-1‍]⋯∫Tun-1‍]Δu2∫Tun-1‍]Δu1≥(12)n+(1/α)(t-2n-1T),∫T1t[θ(s)p(s)-θΔ(s)B1α(s,T)δ1(s,T)]Δs=∫T1t[1sγ-1B1α(s,T)δ1(s,T,c1,c2)]Δs≥∫T1t‍[2tτ-M[(1/2)n+(1/α)(t-2n-1T)]ρ]Δs≥∫T1t1tτΔs.
Thus,
(74)limsupt→∞∫T1t[θ(s)p(s)-θΔ(s)B1α(s,T)δ1(s,T,c1,c2)]Δs=∞.
So conditions (5), (6), and (23) are satisfied. Then, by Theorem 8, every solution of (69) is either oscillatory or tends to zero.



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*Source: 268721-2013-10-02.xml*