AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation81216510.1155/2012/812165812165Research ArticleNonoscillatory Solutions of Second-Order Superlinear Dynamic Equations with Integrable CoefficientsLinQuanwen1JiaBaoguo2AndersonD.1Department of MathematicsMaoming UniversityMaoming 525000Chinammc.edu.cn2School of Mathematics and Computational ScienceSun Yat-Sen UniversityGuangzhou 510275Chinasysu.edu.cn201262201220120909201115112011231120112012Copyright © 2012 Quanwen Lin and Baoguo Jia.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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)=limτtτp(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)=limτtτp(s)Δs exists and is finite. P(t)0 for large t.

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

When 𝕋=, 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 .

Condition (H).

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

There exists a strictly increasing sequence {tn}n=0𝕋 with limntn= and for each n0 either σ(tn)=tn+1 or the real interval [tn,tn+1]𝕋;

𝕋=[T,) for some T𝕋.

We note that time scales which satisfy Condition (H) include most of the important time scales, such as , , and q0, where q>1 and 0 is the nonnegative integers and harmonic numbers {k=1n1/k:n} [2, Example 1.45].

In , Naito proved the following result.

Theorem 1.1.

If P(t)=tp(s)ds0 for large t, then a nonoscillatory solution x(t) of (1.3) satisfies exactly one of the following three asymptotic properties: limtx(t)=c0,limtx(t)t=0,  limtx(t)=±,limtx(t)t=c0.

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,   t0𝕋, and t0[r(t)]-1dt=. limtt0tp(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)+tr(s)γ01[x(s)+hμ(s)xΔ(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs is satisfied for tT, 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 Tr(s)γ01[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)γ01[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)γ01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs. If (2.3) fails to hold, that is, Tr(s)γ01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs=, from (2.5), we have limtr(t)xΔ(t)xγ(t)=-. Without loss of generality, we can assume that for tTr(T)xΔ(T)xγ(T)-Ttp(s)Δs<-1. Otherwise, let L=maxtT|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 tTr(t)xΔ(t)xγ(t)+Ttr(s)γ01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs<-1. In particular, we have xΔ(t)<0,for  tT. Therefore, x(t) is strictly decreasing.

Assume that t=ti-1<ti=σ(t). Then, x(σ(t))<x(t), and so γ01[xh(s)]γ-1dh=γ01[(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 γ01[xh(s)]γ-1dh=γxγ-1(s). Let y(t):=1+Ttr(s)γ01[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)γ01[xh(t)]γ-1dh[xΔ(t)]2xγ(t)xγ(σ(t))>y(t)γ01[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-,  as  t, which contradicts x(t)>0.

In (2.5), letting t, replacing T by τ, and denoting α=limtr(t)xΔ(t)/xγ(t), we get that α+τp(s)Δs+τr(s)γ01(xh(s))γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs=r(τ)xΔ(τ)xγ(τ).

We need to show that α0.

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)=:T1r(s)γ01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs-α2T1γ01[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)γ01[xh(s)]γ-1dh[-xΔ(s)]xγ(σ(s))Δs=γ01[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-1tγ01[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)-α2limulnxγ(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)=limτtτp(s)Δs exists and is finite, and P(t)0 for tT.

Theorem 2.2.

Suppose that 𝕋 satisfies Condition (H) and P(t)0 for tT. Then each nonoscillatory solution x(t) of (2.38) satisfies exactly one of the following three asymptotic properties: limtx(t)=c0,limtx(t)TtΔs/r(s)=0,limtx(t)=±,limtx(t)TtΔs/r(s)=c0.

Proof.

Let x(t) be a nonoscillatory solution of (2.38), say, x(t)>0 for tT>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)tr(s)01[x(s)+hμ(s)xΔ(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δs for tT. Therefore, we have r(t)xΔ(t)P(t)xγ(t), for tT. Since P(t)0 for tT, it follows that xΔ(t)0 for tT. An integration by parts of (2.38) gives r(u)xΔ(u)-P(u)xγ(u)+γtuP(s)xΔ(s)01[x(s)+hμ(s)xΔ(s)]γ-1dhΔs=r(t)xΔ(t)-P(t)xγ(t), where utT. 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, tP(s)xΔ(s)01[x(s)+hμ(s)xΔ(s)]γ-1dhΔs<. Define the function k1(t) as k1(t)=tP(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 tT. Observe by (2.44) that β0. From (2.44) and (2.46), it follows that tP2(s)xγ(s)r(s)01[x(s)+hμ(s)xΔ(s)]γ-1dhΔsk1(t)<. Next, we define the function k2(t) by (noticing that γ>1) k2(t)=tP2(s)xγ(s)r(s)01[x(s)+hμ(s)xΔ(s)]γ-1dhΔs=tP2(s)xγ(s)r(s)01[(1-h)x(s)+hx(σ(s))]γ-1dhΔstP2(s)xγ(s)r(s)01[(1-h)x(s)+hx(s)]γ-1dhΔs=tP2(s)x2γ-1(s)r(s)Δs for tT. 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+γTtk1(s)r(s)Δs. By Schwartz’s inequality and the fact that xΔ(t)0 for tT, 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)1/2k21/2(T)(TtΔsr(s))1/2x1/2(t), for tT. 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 tT, 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)mTtΔsr(s) for tT. Let T1(T) be an arbitrary number. It is clear that 0TtP(s)xγ(s)r(s)Δs=TT1P(s)xγ(s)r(s)Δs+T1tP(s)xγ(s)r(s)Δs for tT1. Arguing as in (2.52), we find T1tP(s)xγ(s)r(s)Δs[k2(T1)x(t)T1tΔsr(s)]1/2 for tT1, which when combined with (2.56) yields T1tP(s)xγ(s)r(s)Δs[mk2(T1)TtΔsr(s)T1tΔsr(s)]1/2[mk2(T1)]1/2TtΔsr(s), for tT1T. Using (2.57) and (2.59) and noticing that TΔs/r(s)=, we obtain 0limsupt1TtΔs/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 0limt1TtΔs/r(s)TtP(s)xγ(s)r(s)Δs=0. Using L’Hospital’s rule of time scale (see Theorem 1.119 of ), we have limtTt(k1(s)/r(s))ΔsTtΔs/r(s)=limtk1(t)=0. In view of (2.51), (2.61), and (2.62), we find limtx(t)/TtΔs/r(s)=β.

Recall that x(t) is nondecreasing for tT. 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=limtx(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 .

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))Δsx-γ+1(T)-x-γ+1(t)γ-1x-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 tT, where P(t)=tp(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 limtTtp(s)Δs exists and is finite. Let P(t)=tp(s)Δs. Then, the superlinear dynamic equation (1.1) is oscillatory if limsuptTtP(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-1TtxΔ(s)xγ(σ(s))ΔsTtP(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)=limτtτp(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 liminftx(t)>0, then Tr(s)γ01[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)γ01[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 liminftx(t)>0 implies the existence of t1T and m>0 such that x(t)m for tt1. Then, using (2.72), we find |t1th(s)xγ(σ(s))Δs|t1t|h(s)xγ(σ(s))|Δs1mγt1t|h(s)|ΔsM,  tt1, where M is some finite positive constant.

So, limτtτ[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,  tT, 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 liminftx(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 t1T and m>0 such that x(t)m for tt1. 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)γ01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))ΔsΦ0(t)>0, for tt1. From (2.80), we get xΔ(t)Φ0(t)xγ(σ(t))r(t)>0,  tt1. Applying (2.81) and noticing that γ>1, we find for tt1σ(t)r(s)γ01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δsσ(t)γ[x(s)]γ-1[Φ0(s)xγ(σ(s))]2r(s)xγ(s)xγ(σ(s))Δsγmγ-1σ(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)γ01[xh(s)]γ-1dh[xΔ(s)]2xγ(s)xγ(σ(s))Δsγmγ-1σ(t)[Φ0(s)+γmγ-1Φ1(s)]2r(s)Δs=γmγ-1Φ2(t),  tt1. 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=n1k1+cn1t1+cdt=1cnc. By , 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: limnx(n)=c0,limnx(n)n=0,limnx(n)=±,limnx(n)n=c0.

Acknowledgment

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

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