# Nonoscillatory Solutions of Second-Order Superlinear Dynamic Equations with Integrable Coefficients

Authors: Quanwen Lin; Baoguo Jia
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The asymptotic behavior of nonoscillatory solutions of the
superlinear dynamic equation on time scales(r(t)xΔ(t))Δ+p(t)|x(σ(t))|γsgnx(σ(t))=0, γ>1, is discussed under the condition that P(t)=limτ→∞∫tτp(s)Δs exists
and P(t)≥0 for large t.


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## Body

## 1. Introduction

Consider the second-order superlinear dynamic equation(1.1)(r(t)xΔ(t))Δ+p(t)|x(σ(t))|γsgn⁡x(σ(t))=0,γ>1,
where(1.2)P(t)=lim⁡τ→∞∫tτp(s)Δs
exists and is finite. P(t)≥0 for large t.When𝕋=ℝ, r(t)=1, (1.1) is the second-order superlinear differential equation(1.3)x′′(t)+p(t)|x(t)|γsgn⁡x(t)=0,γ>1.When𝕋=ℕ, r(n)=1, (1.1) is the second-order superlinear difference equation(1.4)Δ2x(n)+p(n)|x(n+1)|γsgn⁡x(n+1)=0,γ>1.The following condition is introduced in [1].Condition (H).
We say that𝕋 satisfies Condition (H) provided one of the following holds.(1)
There exists a strictly increasing sequence{tn}n=0∞⊂𝕋 with limn→∞tn=∞ and for each n≥0 either σ(tn)=tn+1 or the real interval [tn,tn+1]⊂𝕋;(2)
𝕋∩ℝ=[T′,∞) for some T′∈𝕋.We note that time scales which satisfy Condition (H) include most of the important time scales, such asℝ, ℤ, and qℕ0, where q>1 and ℕ0 is the nonnegative integers and harmonic numbers {∑k=1n1/k:n∈ℕ} [2, Example 1.45].In [3], Naito proved the following result.Theorem 1.1.
IfP(t)=∫t∞p(s)ds≥0 for large t, then a nonoscillatory solution x(t) of (1.3) satisfies exactly one of the following three asymptotic properties:
(1.5)lim⁡t→∞x(t)=c≠0,lim⁡t→∞x(t)t=0,lim⁡t→∞x(t)=±∞,lim⁡t→∞x(t)t=c≠0.In this paper, we extend Theorem1.1 to superlinear dynamic equation (1.1) on time scale. As an application, we get the asymptotic behavior of each nonoscillation solution of the difference equation(1.6)Δ2x(n)+(an1+c+(-1)nbnc)|x(n+1)|γsgn⁡x(n+1)=0,γ>1,
where b>0, c>1, and a/c>b/2.


## 2. Main Theorems

Consider the second-order nonlinear dynamic equation(2.1)(r(t)xΔ(t))Δ+p(t)|x(σ(t))|γsgn⁡x(σ(t))=0,γ>0,
where r(t),p(t)∈C(𝕋,R), r(t)>0, t0∈𝕋, and ∫t0∞[r(t)]-1dt=∞. limt→∞∫t0tp(s)Δs exists and is finite.Lemma 2.1.
Suppose that𝕋 satisfies Condition (H). If x(t) is a positive solution of (1.1) on [T,∞), then the integral equation
(2.2)r(t)xΔ(t)xγ(t)=α+P(t)+∫t∞r(s)γ∫01[x(s)+hμ(s)xΔ(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs
is satisfied for t≥T, where α is a nonnegative constant.Proof.
Suppose thatx(t) is a positive solution of (1.1) on [T,∞).
In the first place, we will prove(2.3)∫T∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs<∞,
where xh(t)=x(t)+hμ(t)xΔ(t)=(1-h)x(t)+hx(σ(t))>0.
Multiplying both sides of (1.1) by 1/xγ(σ(t)), we get that
(2.4)(r(t)xΔ(t)xγ(t))Δ=-p(t)-r(t)γ∫01[xh(t)]γ-1dh[xΔ(t)]2xγ(t)xγ(σ(t)).
Integrating from T to t,
(2.5)r(t)xΔ(t)xγ(t)-r(T)xΔ(T)xγ(T)=-∫Ttp(s)Δs-∫Ttr(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs.
If (2.3) fails to hold, that is,
(2.6)∫T∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs=∞,
from (2.5), we have
(2.7)lim⁡t→∞r(t)xΔ(t)xγ(t)=-∞.
Without loss of generality, we can assume that for t≥T(2.8)r(T)xΔ(T)xγ(T)-∫Ttp(s)Δs<-1.
Otherwise, let L=maxt≥T|∫Ttp(s)Δs|. By (2.7), we can take a large T1>T such that r(T1)xΔ(T1)/xγ(T1)<-(2L+1). So we have
(2.9)r(T1)xΔ(T1)xγ(T1)-∫T1tp(s)Δs<-(2L+1)-[∫Ttp(s)Δs-∫TT1p(s)Δs]≤-(2L+1)-[-2L]=-1.
So we can replace T by T1>T such that (2.8) still holds.
From (2.5) and (2.8), we get for t≥T(2.10)r(t)xΔ(t)xγ(t)+∫Ttr(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs<-1.
In particular, we have
(2.11)xΔ(t)<0,fort≥T.
Therefore, x(t) is strictly decreasing.
Assume thatt=ti-1<ti=σ(t). Then, x(σ(t))<x(t), and so
(2.12)γ∫01[xh(s)]γ-1dh=γ∫01[(1-h)x(s)+hx(σ(s))]γ-1dh=[(1-h)x(s)+hx(σ(s))]γ|01x(σ(s))-x(s)=xγ(σ(s))-xγ(s)x(σ(s))-x(s).
If the real interval [ti-1,ti]⊂𝕋, then, for s∈[ti-1,ti], we have
(2.13)γ∫01[xh(s)]γ-1dh=γxγ-1(s).
Let
(2.14)y(t):=1+∫Ttr(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ-1(σ(s))Δs.
Hence, from (2.10), we get that
(2.15)-r(t)xΔ(t)xγ(t)>y(t).
From (2.14) and (2.15), we get that
(2.16)yΔ(t)=r(t)γ∫01[xh(t)]γ-1dh[xΔ(t)]2xγ(t)xγ(σ(t))>y(t)γ∫01[xh(t)]γ-1dh[-xΔ(t)]xγ(σ(t)).
Assume that t=ti-1<ti=σ(t). From (2.16) and (2.12), we get that
(2.17)y(σ(t))-y(t)y(t)(σ(t)-t)>xγ(σ(t))-xγ(t)x(σ(t))-x(t)x(t)-x(σ(t))xγ(σ(t))[σ(t)-t].
So,
(2.18)y(σ(t))y(t)>xγ(t)xγ(σ(t)),
that is,
(2.19)y(ti)y(ti-1)>xγ(ti-1)xγ(ti).
If the real interval [ti-1,ti]⊂𝕋, then, for t∈(ti-1,ti], it follows from (2.16) and (2.13) that
(2.20)y′(t)y(t)>γxγ-1(t)[-x′(t)]xγ(t),
that is,
(2.21)(ln⁡y(t))′>-(ln⁡xγ(t))′.
Integrating from ti-1 to t, we get that
(2.22)y(t)y(ti-1)>xγ(ti-1)xγ(t),t∈(ti-1,ti].
Let T=tn0, and let t∈(T,∞)𝕋. Then, there is an n>n0 such that t∈(tn-1,tn]𝕋. From (2.22) and (2.19), we get that
(2.23)y(t)y(tn-1)>xγ(tn-1)xγ(t),y(tn-1)y(tn-2)>xγ(tn-2)xγ(tn-1),…,y(tn0+1)y(tn0)>xγ(tn0)xγ(tn0+1).
Multiplying, we get that
(2.24)y(t)y(tn0)>xγ(tn0)xγ(t).
Using (2.15) again, we get
(2.25)-r(t)xΔ(t)xγ(t)>y(t)>y(tn0)xγ(tn0)xγ(t).
If we set L:=y(tn0)xγ(tn0), we get
(2.26)xΔ(t)<-Lr(t).
Integrating from T to t, we get that
(2.27)x(t)-x(T)<-∫TtLr(s)Δs⟶-∞,ast⟶∞,
which contradicts x(t)>0.
In (2.5), letting t→∞, replacing T by τ, and denoting α=limt→∞r(t)xΔ(t)/xγ(t), we get that
(2.28)α+∫τ∞p(s)Δs+∫τ∞r(s)γ∫01(xh(s))γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs=r(τ)xΔ(τ)xγ(τ).
We need to show thatα≥0.
Suppose thatα<0. Then, there exists a large T1 such that, for t>T1, we have
(2.29)r(t)xΔ(t)xγ(t)≤α2.
So,
(2.30)xΔ(t)≤αxγ(t)2r(t).
Thus,(2.31)M(T1)=:∫T1∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs≥-α2∫T1∞γ∫01[xh(s)]γ-1dh[-xΔ(s)]xγ(σ(s))Δs.
Assume that t=ti-1<ti=σ(t). From (2.12) and xΔ(t)<0, we have
(2.32)∫tσ(t)γ∫01[xh(s)]γ-1dh[-xΔ(s)]xγ(σ(s))Δs=γ∫01[xh(t)]γ-1dh[-xΔ(t)](σ(t)-t)xγ(σ(t))=xγ(t)-xγ(σ(t))xγ(σ(t))≥∫xγ(σ(t))xγ(t)1vdv=ln⁡xγ(t)xγ(σ(t))=ln⁡xγ(ti-1)xγ(ti).
If the real interval [ti-1,ti]⊂𝕋, from (2.13) we have, for t∈(ti-1,ti],
(2.33)∫ti-1tγ∫01[xh(s)]γ-1dh[-xΔ(s)]xγ(σ(s))Δs=∫ti-1tγxγ-1(s)[-x′(s)]xγ(s)ds=ln⁡xγ(ti-1)xγ(t).
From (2.31), (2.32), (2.33) and the additivity of the integral, it is easy to get
(2.34)M(T1)≥-α2lim⁡u→∞ln⁡xγ(T1)xγ(u).
So, for large u, we have
(2.35)ln⁡xγ(T1)xγ(u)≤-2M(T1)α+1.
Thus,
(2.36)xγ(u)≥xγ(T1)exp⁡(2M(T1)α-1).
By (2.30) and noticing that α<0, we get that
(2.37)xΔ(u)≤αxγ(T1)2r(u)exp⁡(2M(T1)α-1).
Integrating (2.37), we get that x(u)→-∞, which is a contradiction.
This completes the proof of the lemma.Consider the second-order superlinear dynamic equation(2.38)[r(t)xΔ(t)]Δ+p(t)|x(σ(t))|γsgn⁡x(σ(t))=0,γ>1,
where r(t)>0, ∫T∞(1/r(s))Δs=∞,(2.39)P(t)=lim⁡τ→∞∫tτp(s)Δs
exists and is finite, and P(t)≥0 for t≥T.Theorem 2.2.
Suppose that𝕋 satisfies Condition (H) and P(t)≥0 for t≥T. Then each nonoscillatory solution x(t) of (2.38) satisfies exactly one of the following three asymptotic properties:
(2.40)lim⁡t→∞x(t)=c≠0,(2.41)lim⁡t→∞x(t)∫TtΔs/r(s)=0,lim⁡t→∞x(t)=±∞,(2.42)lim⁡t→∞x(t)∫TtΔs/r(s)=c≠0.Proof.
Letx(t) be a nonoscillatory solution of (2.38), say, x(t)>0 for t≥T>0. From Lemma 2.1, it is known that x(t) satisfies the equality
(2.43)r(t)xΔ(t)=αxγ(t)+P(t)xγ(t)+γxγ(t)∫t∞r(s)∫01[x(s)+hμ(s)xΔ(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs
for t≥T. Therefore, we have
(2.44)r(t)xΔ(t)≥P(t)xγ(t),
for t≥T. Since P(t)≥0 for t≥T, it follows that xΔ(t)≥0 for t≥T. An integration by parts of (2.38) gives
(2.45)r(u)xΔ(u)-P(u)xγ(u)+γ∫tuP(s)xΔ(s)∫01[x(s)+hμ(s)xΔ(s)]γ-1dhΔs=r(t)xΔ(t)-P(t)xγ(t),
where u≥t≥T. Let t be fixed. Since P(s)xΔ(s)∫01[x(s)+hμ(s)xΔ(s)]γ-1dh is nonnegative, the integral term in (2.45) has a finite limit or diverges to ∞ as t→∞. If the latter case occurs, then r(u)xΔ(u)-P(u)xγ(u)→-∞ as u→∞, which is a contradiction to (2.44). Thus, the former case occurs, that is,
(2.46)∫t∞P(s)xΔ(s)∫01[x(s)+hμ(s)xΔ(s)]γ-1dhΔs<∞.
Define the function k1(t) as
(2.47)k1(t)=∫t∞P(s)xΔ(s)∫01[x(s)+hμ(s)xΔ(s)]γ-1dhΔs<∞
and the finite constant β=limu→∞[r(u)xΔ(u)-P(u)xγ(u)]. Then, equality (2.45) yields
(2.48)r(t)xΔ(t)=β+P(t)xγ(t)+γk1(t),
for t≥T. Observe by (2.44) that β≥0. From (2.44) and (2.46), it follows that
(2.49)∫t∞P2(s)xγ(s)r(s)∫01[x(s)+hμ(s)xΔ(s)]γ-1dhΔs≤k1(t)<∞.
Next, we define the function k2(t) by (noticing that γ>1)
(2.50)k2(t)=∫t∞P2(s)xγ(s)r(s)∫01[x(s)+hμ(s)xΔ(s)]γ-1dhΔs=∫t∞P2(s)xγ(s)r(s)∫01[(1-h)x(s)+hx(σ(s))]γ-1dhΔs≥∫t∞P2(s)xγ(s)r(s)∫01[(1-h)x(s)+hx(s)]γ-1dhΔs=∫t∞P2(s)x2γ-1(s)r(s)Δs
for t≥T. Dividing (2.48) by r(t) and integrating from T to t, we get
(2.51)x(t)=x(T)+∫Ttβr(s)Δs+∫TtP(s)xγ(s)r(s)Δs+γ∫Ttk1(s)r(s)Δs.
By Schwartz’s inequality and the fact that xΔ(t)≥0 for t≥T, the second integral term in (2.51) can be estimated as follows:
(2.52)∫TtP(s)xγ(s)r(s)Δs≤(∫TtP2(s)x2γ-1(s)r(s)Δs)1/2(∫Ttx(s)r(s)Δs)1/2≤k21/2(T)(∫TtΔsr(s))1/2x1/2(t),
for t≥T. From (2.51) and (2.52) and noticing that k1(t) is decreasing, we get that
(2.53)x(t)≤x(T)+∫Ttβr(s)Δs+k21/2(T)(∫TtΔsr(s))1/2x1/2(t)+γk1(T)∫TtΔsr(s).
The above inequality may be regarded as a quadratic inequality in x1/2(t). Then, we have
(2.54)x1/2(t)≤12k21/2(T)(∫TtΔsr(s))1/2+12D1/2(t)
for t≥T, where
(2.55)D(t)=k2(T)∫TtΔsr(s)+4[x(T)+(β+γk1(T))∫TtΔsr(s)].
It is obvious that D(t)=O(∫TtΔs/r(s)) as t→∞, and, consequently, there exists a positive constant m such that
(2.56)x(t)≤m∫TtΔsr(s)
for t≥T. Let T1(≥T) be an arbitrary number. It is clear that
(2.57)0≤∫TtP(s)xγ(s)r(s)Δs=∫TT1P(s)xγ(s)r(s)Δs+∫T1tP(s)xγ(s)r(s)Δs
for t≥T1. Arguing as in (2.52), we find
(2.58)∫T1tP(s)xγ(s)r(s)Δs≤[k2(T1)x(t)∫T1tΔsr(s)]1/2
for t≥T1, which when combined with (2.56) yields
(2.59)∫T1tP(s)xγ(s)r(s)Δs≤[mk2(T1)∫TtΔsr(s)∫T1tΔsr(s)]1/2≤[mk2(T1)]1/2∫TtΔsr(s),
for t≥T1≥T. Using (2.57) and (2.59) and noticing that ∫T∞Δs/r(s)=∞, we obtain
(2.60)0≤limsup⁡t→∞1∫TtΔs/r(s)∫TtP(s)xγ(s)r(s)Δs≤[mk2(T1)]1/2.
Since T1 is arbitary and k2(T1)→0 as T1→∞, letting T1→∞ in (2.60), we get
(2.61)0≤lim⁡t→∞1∫TtΔs/r(s)∫TtP(s)xγ(s)r(s)Δs=0.
Using L’Hospital’s rule of time scale (see Theorem 1.119 of [2]), we have
(2.62)lim⁡t→∞∫Tt(k1(s)/r(s))Δs∫TtΔs/r(s)=lim⁡t→∞k1(t)=0.
In view of (2.51), (2.61), and (2.62), we find limt→∞x(t)/∫TtΔs/r(s)=β.
Recall thatx(t) is nondecreasing for t≥T. Now, there are three cases to consider:(i)
β=0 and x(t) is bounded above,(ii)
β=0 and x(t) is unbounded,(iii)
β>0 (and hence x(t) is unbounded).
Case (i) implies (2.40) with c=limt→∞x(t)>0, while case (iii) implies (2.42) with c=β>0. It is also clear that case (ii) implies (2.41). This completes the proof.The following lemma is from [1].Lemma 2.3.
Suppose that𝕋 satisfies Condition (H). x(t)>0 is a solution of (1.1). Then, one has
(2.63)∫TtxΔ(s)xγ(σ(s))Δs≤x-γ+1(T)-x-γ+1(t)γ-1≤x-γ+1(T)γ-1.Using Lemma2.1, we can prove the following corollary.Corollary 2.4.
Under the assumptions of Lemma2.1, if x(t) is a positive solution of (1.1) on [T,∞), then the integral equation
(2.64)r(t)xΔ(t)xγ(σ(t))=α+P(t)+∫σ(t)∞r(s)[∫01γ(xh(s))γ-1dh][xΔ(s)]2xγ(s)xγ(σ(s))Δs
is satisfied for t≥T, where P(t)=∫t∞p(s)Δs, xh(s)=x(s)+hμ(s)xΔ(s).Proof.
In the left side of (2.2), using
(2.65)1xγ(t)=1xγ(σ(t))-(1xγ(t))Δμ(t)=1xγ(σ(t))+∫01γ(xh(t))γ-1dhxΔ(t)xγ(t)xγ(σ(t))μ(t)
and using (2.2), (2.65), the additivity of the integral, and [2, Theorem 1.75], we have that
(2.66)r(t)xΔ(t)xγ(t)=r(t)xΔ(t)xγ(σ(t))+r(t)∫01γ(xh(t))γ-1dh[xΔ(t)]2xγ(t)xγ(σ(t))μ(t)=α+P(t)+∫σ(t)∞r(s)[∫01γ(xh(s))γ-1dh][xΔ(s)]2xγ(s)xγ(σ(s))Δs+∫tσ(t)r(s)[∫01γ(xh(s))γ-1dh][xΔ(s)]2xγ(s)xγ(σ(s))Δs=α+P(t)+∫σ(t)∞r(s)[∫01γ(xh(s))γ-1dh][xΔ(s)]2xγ(s)xγ(σ(s))Δs+r(t)[∫01γ(xh(t))γ-1dh][xΔ(t)]2xγ(t)xγ(σ(t))μ(t).
From (2.66), we get (2.64).The following theorem can be regarded as a time scale version of [4, Theorem 1].Theorem 2.5.
Suppose that𝕋 satisfies Condition (H), r(t)>0 with ∫T∞[r(t)]-1Δt=∞, and suppose that limt→∞∫Ttp(s)Δs exists and is finite. Let P(t)=∫t∞p(s)Δs. Then, the superlinear dynamic equation (1.1) is oscillatory if
(2.67)limsup⁡t→∞∫TtP(s)r(s)Δs=∞.Proof.
Suppose thatx(t) is a nonoscillatory solution of (1.1) on [T,∞). Without loss of generality, assume that x(t) is positive for t∈[T,∞). From Corollary 2.4, x(t) satisfies the integral equation (2.64). Dropping the last integral term in (2.64), we have the inequality
(2.68)r(t)xΔ(t)xγ(σ(t))≥P(t).
Dividing (2.68) by r(t), integrating from T to t, and using Lemma 2.3, we find
(2.69)x-γ+1(T)γ-1≥∫TtxΔ(s)xγ(σ(s))Δs≥∫TtP(s)r(s)Δs.
This contradicts (2.67), and so (1.1) is oscillatory.Consider the second-order superlinear dynamic equation with forced term(2.70)[r(t)xΔ(t)]Δ+p(t)|x(σ(t))|γsgn⁡x(σ(t))=h(t),γ>1,
where r(t)>0, ∫T∞(1/r(s))Δs=∞, and(2.71)P(t)=lim⁡τ→∞∫tτp(s)Δs
exists and is finite.Lemma 2.6.
Suppose that(2.72)∫T∞|h(s)|Δs<+∞.
If x(t) is a positive solution of (2.70) and liminft→∞x(t)>0, then
(2.73)∫T∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs<∞,(2.74)r(t)xΔ(t)xγ(σ(t))=α+∫t∞[p(s)-h(s)xγ(σ(s))]Δs+∫σ(t)∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs
are satisfied for sufficiently large t, where xh(s)=x(s)+hμ(s)xΔ(s), α is a nonnegative constant.Proof.
The fact thatliminft→∞x(t)>0 implies the existence of t1≥T and m>0 such that x(t)≥m for t≥t1. Then, using (2.72), we find
(2.75)|∫t1th(s)xγ(σ(s))Δs|≤∫t1t|h(s)xγ(σ(s))|Δs≤1mγ∫t1t|h(s)|Δs≤M,t≥t1,
where M is some finite positive constant.
So,limτ→∞∫tτ[p(s)-h(s)/xγ(σ(s))]Δs exists and is finite.
Similar to the proof of Lemma2.1 and Corollary 2.4, it is easy to know that (2.73) and (2.74) hold.For subsequent results, we define(2.76)Φ0(t)=∫t∞[p(s)-k|h(s)|]Δs,t≥T,
where k is a positive constant. It is noted that, if (2.71) and (2.72) hold, then Φ0(t) is finite for any k. Assume that Φ0(t)>0 for sufficiently large t. Define, for a positive integer n and a positive constant ρ, the following functions:(2.77)Φ1(t)=∫σ(t)∞[Φ0(s)]2r(s)Δs,Φn+1(t)=∫σ(t)∞[Φ0(s)+ρΦn(s)]2r(s)Δs.
We introduce the following condition.Condition (A).
For everyρ>0, there exists a positive integer N such that Φn(t) is finite for n=1,2,…,N-1 and ΦN(t) is infinite.Theorem 2.7.
Suppose that (2.71), (2.72), and Condition (A) hold. Then, every solution x(t) of (2.70) is either oscillatory or satisfies
(2.78)liminf⁡t→∞x(t)=0.Proof.
Suppose on the contrary thatx(t) is a nonoscillatory solution of (2.70) and liminft→∞|x(t)|>0. Without loss of generality, let x(t) be eventually positive. By Lemma 2.6, x(t) satisfies (2.73) and (2.74). Further, there exist t1≥T and m>0 such that x(t)≥m for t≥t1. Let
(2.79)Φ0(t)=∫t∞[p(s)-|h(s)|mγ]Δs.
Then, from (2.74) we find
(2.80)r(t)xΔ(t)xγ(σ(t))≥Φ0(t)+∫σ(t)∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs≥Φ0(t)>0,
for t≥t1. From (2.80), we get
(2.81)xΔ(t)≥Φ0(t)xγ(σ(t))r(t)>0,t≥t1.
Applying (2.81) and noticing that γ>1, we find for t≥t1(2.82)∫σ(t)∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs≥∫σ(t)∞γ[x(s)]γ-1[Φ0(s)xγ(σ(s))]2r(s)xγ(s)xγ(σ(s))Δs≥γmγ-1∫σ(t)∞[Φ0(s)]2r(s)Δs=γmγ-1Φ1(t).
If N=1 in Condition (A), then the right side of (2.82) is infinite. This is a contradiction to (2.73).
Next, it follows from (2.80) and (2.82) that
(2.83)r(t)xΔ(t)xγ(σ(t))≥Φ0(t)+γmγ-1Φ1(t).
Using a similar technique and relations (2.83), we get
(2.84)∫σ(t)∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs≥γmγ-1∫σ(t)∞[Φ0(s)+γmγ-1Φ1(s)]2r(s)Δs=γmγ-1Φ2(t),t≥t1.
If N=2 in Condition (A), then the right side of (2.84) is infinite. This again contradicts (2.73).
A similar argument yields a contradiction for any integerN>2. This completes the proof of the theorem.Example 2.8.
We have(2.85)Δ2x(n)+(an1+c+b(-1)nnc)|x(n+1)|γsgn⁡x(n+1)=0,γ>1,
where b>0, c>1, and a/c>b/2. It is easy to see that
(2.86)∑k=n∞1k1+c≥∫n∞1t1+cdt=1cnc.
By [5], we have ∑k=n∞(-1)n/nc~(-1)n/2nc. So,
(2.87)∑k=n∞(-1)kkc=[1+o(1)](-1)n2nc.
Using (2.86) and (2.87), we get that, for large n,
(2.88)P(n)=∑k=n∞(ak1+c+b(-1)kkc)≥(a/c)+(b/2)[1+o(1)](-1)nnc>0.
By Theorem 2.2, each nonoscillatory solution x(t) of (2.85) satisfies exactly one of the following three asymptotic properties:
(2.89)lim⁡n→∞x(n)=c≠0,lim⁡n→∞x(n)n=0,lim⁡n→∞x(n)=±∞,lim⁡n→∞x(n)n=c≠0.



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*Source: 812165-2012-02-06.xml*