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.

1. Introduction

Consider the second-order superlinear dynamic equation(r(t)xΔ(t))Δ+p(t)|x(σ(t))|γsgnx(σ(t))=0,γ>1,
whereP(t)=limτ→∞∫tτp(s)Δs
exists and is finite. P(t)≥0 for large t.

When 𝕋=ℝ, r(t)=1, (1.1) is the second-order superlinear differential equationx′′(t)+p(t)|x(t)|γsgnx(t)=0,γ>1.

When 𝕋=ℕ, r(n)=1, (1.1) is the second-order superlinear difference equationΔ2x(n)+p(n)|x(n+1)|γsgnx(n+1)=0,γ>1.

The following condition is introduced in [1].

Condition (H).

We say that 𝕋 satisfies Condition (H) provided one of the following holds.

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;

𝕋∩ℝ=[T′,∞) for some T′∈𝕋.

We note that time scales which satisfy Condition (H) include most of the important time scales, such as ℝ, 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.

If P(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:
limt→∞x(t)=c≠0,limt→∞x(t)t=0,limt→∞x(t)=±∞,limt→∞x(t)t=c≠0.

In this paper, we extend Theorem 1.1 to superlinear dynamic equation (1.1) on time scale. As an application, we get the asymptotic behavior of each nonoscillation solution of the difference equationΔ2x(n)+(an1+c+(-1)nbnc)|x(n+1)|γsgnx(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(r(t)xΔ(t))Δ+p(t)|x(σ(t))|γsgnx(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 that 𝕋 satisfies Condition (H). If x(t) is a positive solution of (1.1) on [T,∞), then the integral equation
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 that x(t) is a positive solution of (1.1) on [T,∞).

In the first place, we will prove
∫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
(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,
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,
∫T∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs=∞,
from (2.5), we have
limt→∞r(t)xΔ(t)xγ(t)=-∞.
Without loss of generality, we can assume that for t≥Tr(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
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≥Tr(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
xΔ(t)<0,fort≥T.
Therefore, x(t) is strictly decreasing.

Assume that t=ti-1<ti=σ(t). Then, x(σ(t))<x(t), and so
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
γ∫01[xh(s)]γ-1dh=γxγ-1(s).
Let
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
-r(t)xΔ(t)xγ(t)>y(t).
From (2.14) and (2.15), we get that
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
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,
y(σ(t))y(t)>xγ(t)xγ(σ(t)),
that is,
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
y′(t)y(t)>γxγ-1(t)[-x′(t)]xγ(t),
that is,
(lny(t))′>-(lnxγ(t))′.
Integrating from ti-1 to t, we get that
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
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
y(t)y(tn0)>xγ(tn0)xγ(t).
Using (2.15) again, we get
-r(t)xΔ(t)xγ(t)>y(t)>y(tn0)xγ(tn0)xγ(t).
If we set L:=y(tn0)xγ(tn0), we get
xΔ(t)<-Lr(t).
Integrating from T to t, we get that
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
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 that α≥0.

Suppose that α<0. Then, there exists a large T1 such that, for t>T1, we have
r(t)xΔ(t)xγ(t)≤α2.
So,
xΔ(t)≤αxγ(t)2r(t).

Thus,
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
∫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=lnxγ(t)xγ(σ(t))=lnxγ(ti-1)xγ(ti).
If the real interval [ti-1,ti]⊂𝕋, from (2.13) we have, for t∈(ti-1,ti],
∫ti-1tγ∫01[xh(s)]γ-1dh[-xΔ(s)]xγ(σ(s))Δs=∫ti-1tγxγ-1(s)[-x′(s)]xγ(s)ds=lnxγ(ti-1)xγ(t).
From (2.31), (2.32), (2.33) and the additivity of the integral, it is easy to get
M(T1)≥-α2limu→∞lnxγ(T1)xγ(u).
So, for large u, we have
lnxγ(T1)xγ(u)≤-2M(T1)α+1.
Thus,
xγ(u)≥xγ(T1)exp(2M(T1)α-1).
By (2.30) and noticing that α<0, we get that
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[r(t)xΔ(t)]Δ+p(t)|x(σ(t))|γsgnx(σ(t))=0,γ>1,
where r(t)>0, ∫T∞(1/r(s))Δs=∞,P(t)=limτ→∞∫tτp(s)Δs
exists and is finite, and P(t)≥0 for t≥T.

Theorem 2.2.

Suppose that 𝕋 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:
limt→∞x(t)=c≠0,limt→∞x(t)∫TtΔs/r(s)=0,limt→∞x(t)=±∞,limt→∞x(t)∫TtΔs/r(s)=c≠0.

Proof.

Let x(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
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
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
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,
∫t∞P(s)xΔ(s)∫01[x(s)+hμ(s)xΔ(s)]γ-1dhΔs<∞.
Define the function k1(t) as
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
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
∫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)
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
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:
∫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
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
x1/2(t)≤12k21/2(T)(∫TtΔsr(s))1/2+12D1/2(t)
for t≥T, where
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
x(t)≤m∫TtΔsr(s)
for t≥T. Let T1(≥T) be an arbitrary number. It is clear that
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
∫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
∫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
0≤limsupt→∞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
0≤limt→∞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
limt→∞∫Tt(k1(s)/r(s))Δs∫TtΔs/r(s)=limt→∞k1(t)=0.
In view of (2.51), (2.61), and (2.62), we find limt→∞x(t)/∫TtΔs/r(s)=β.

Recall that x(t) is nondecreasing for t≥T. Now, there are three cases to consider:

β=0 and x(t) is bounded above,

β=0 and x(t) is unbounded,

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 that 𝕋 satisfies Condition (H). x(t)>0 is a solution of (1.1). Then, one has
∫TtxΔ(s)xγ(σ(s))Δs≤x-γ+1(T)-x-γ+1(t)γ-1≤x-γ+1(T)γ-1.

Using Lemma 2.1, we can prove the following corollary.

Corollary 2.4.

Under the assumptions of Lemma 2.1, if x(t) is a positive solution of (1.1) on [T,∞), then the integral equation
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
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
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 that 𝕋 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
limsupt→∞∫TtP(s)r(s)Δs=∞.

Proof.

Suppose that x(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
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
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[r(t)xΔ(t)]Δ+p(t)|x(σ(t))|γsgnx(σ(t))=h(t),γ>1,
where r(t)>0, ∫T∞(1/r(s))Δs=∞, andP(t)=limτ→∞∫tτp(s)Δs
exists and is finite.

Lemma 2.6.

Suppose that
∫T∞|h(s)|Δs<+∞.
If x(t) is a positive solution of (2.70) and liminft→∞x(t)>0, then
∫T∞r(s)γ∫01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs<∞,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 that liminft→∞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
|∫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 Lemma 2.1 and Corollary 2.4, it is easy to know that (2.73) and (2.74) hold.

For subsequent results, we defineΦ0(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: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 every ρ>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
liminft→∞x(t)=0.

Proof.

Suppose on the contrary that x(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
Φ0(t)=∫t∞[p(s)-|h(s)|mγ]Δs.
Then, from (2.74) we find
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
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∫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
r(t)xΔ(t)xγ(σ(t))≥Φ0(t)+γmγ-1Φ1(t).
Using a similar technique and relations (2.83), we get
∫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 integer N>2. This completes the proof of the theorem.

Example 2.8.

We have
Δ2x(n)+(an1+c+b(-1)nnc)|x(n+1)|γsgnx(n+1)=0,γ>1,
where b>0, c>1, and a/c>b/2. It is easy to see that
∑k=n∞1k1+c≥∫n∞1t1+cdt=1cnc.
By [5], we have ∑k=n∞(-1)n/nc~(-1)n/2nc. So,
∑k=n∞(-1)kkc=[1+o(1)](-1)n2nc.
Using (2.86) and (2.87), we get that, for large n,
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:
limn→∞x(n)=c≠0,limn→∞x(n)n=0,limn→∞x(n)=±∞,limn→∞x(n)n=c≠0.

Acknowledgment

The paper is supported by the National Natural Science Foundation of China (no. 10971232).

BaoguoB.ErbeL.PetersonA.Kiguradze-type oscillation theorems for second order superlinear dynamic equations on time scalesBohnerM.PetersonA.NaitoM.Nonoscillatory solutions of second order differential equations with integrable coefficientsWongJ. S. W.Oscillation criteria for second order nonlinear differential equations with integrable coefficientsErbeL.BaoguoJ.PetersonA.Nonoscillation for second order sublinear dynamic equations on time scales