# Asymptotic Dichotomy in a Class of Odd-Order Nonlinear Differential Equations with Impulses

Authors: Kunwen Wen; Lijun Pan
Journal: Journal of Industrial Mathematics
 (2013)

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## Abstract

We investigate the oscillatory and asymptotic behavior of a class of odd-order nonlinear differential equations with impulses. We obtain criteria that ensure every solution is either oscillatory or (nonoscillatory and) zero convergent. We provide several examples to show that impulses play an important role in the asymptotic behaviors of these equations.


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## Body

## 1. Introduction

Impulsive effect, likewise, exists in a wide variety of evolutionary processes in which states are changed abruptly at certain moments in time, involving such fields as medicine and biology, economics, mechanics, electronics, telecommunications, and so forth. It has been observed that the solutions of quite a few first-or second-order impulsive differential equations are either oscillatory or (nonoscillatory and) zero convergent (see, e.g., [1–10]). For example, Bainov et al. studied the oscillation properties of first-order impulsive differential equations with deviating arguments [3]. Especially in [4], Chen investigated oscillations of second-order nonlinear differential with impulses, and he promposed that the impulses may change the oscillatory behavior of an equation. Based on [4], the authors were devoted to oscillations of impulsive differential equations (see, e.g., [5–10]). Such a dichotomy may yield useful information in real problems. The implications of this dichotomy are applied to the deflection of an elastic beam [11]. Thus, it is of interest to see whether similar dichotomies occur in different types of impulsive differential equations.One such type consists of impulsive differential equations which are important in the simulation of processes with jump conditions. But papers devoted to the study of asymptotic behaviors of higher order equations with impulses are quite rare. For this reason, Wen et al. studied in [12] the dichotomous properties of the following third-order nonlinear differential equation with impulses:
(1)(r(t)x′′(t))′+f(t,x)=0,t≥t0,t≠tk,x(i)(tk+)=gk[i](x(i)(tk)),i=0,1,2;k=1,2,…,x(i)(t0+)=x0(i),i=0,1,2,
where x(0)(t)=x(t), 0≤t0<t1<⋯<tk<⋯ such that limk→∞tk=∞. On the other hand, in [13], Chen and Wen investigated the oscillatory and asymptotic behaviors for odd-order nonlinear differential equations with impulses of the form
(2)x(2n+1)(t)+p(t)x(t)=0,t≥t0,t≠tk,x(i)(tk+)=ak[i]x(i)(tk),i=0,1,…,2n;k=1,2,…,x(i)(t0+)=x0(i),i=0,1,…,2n,
where n is a positive integer and 0≤t0<t1<⋯<tk<⋯ such that limk→+∞tk=+∞. They obtained some interesting results for assuring that every bounded solution of (2) is either oscillatory or nonoscillatory and zero convergent.In this paper, we will study a class of odd-order nonlinear differential equations with impulses of the form(3)(r(t)x(2n)(t))′+f(t,x)=0,t≥t0,t≠tk,x(i)(tk+)=gk[i](x(i)(tk)),i=0,1,…,2n;k=1,2,…,x(i)(t0+)=x0[i],i=0,1,…,2n,
where x(0)(t)=x(t),n is a positive integer, and 0≤t0<t1<⋯<tk<⋯ such that limk→+∞tk=+∞,
(4)x(i)(tk)=limh→0-x(i-1)(tk+h)-x(i-1)(tk)h,x(i)(tk+)=limh→0+x(i-1)(tk+h)-x(i-1)(tk+)h.
By a solution of (3), we mean a real function x=x(t) defined on [t0,+∞) such that(I)
x(i)(t0+)=x0[i] for i=0,1,…,2n;(II)
x(i)(t),i=0,1,…,2n, and (r(t)x(2n)(t))′ are continuous on [t0,+∞)∖{tk}; for i=0,1,…,2n,x(i)(tk+) and x(i)(tk-) exist, x(i)(tk-)=x(i)(tk) and x(i)(tk+)=gk[i](x(i)(tk)) for any tk;(III)
x(t) satisfies (r(t)x(2n)(t))′+f(t,x(t))=0 at each point t∈[t0,+∞)∖{tk}.A solution of (3) is said to be nonoscillatory if it is eventually positive or eventually negative. Otherwise, it is said to be oscillatory.We will establish oscillatory and asymptotic results of (3) based on combinations of the following conditions.(i)
r(t)>0 and r(t) is continuous on [t0,+∞);f(t,x) is continuous on [t0,+∞)×(-∞,+∞),xf(t,x)>0 for x≠0, and f(t,x)/φ(x)⩾p(t), where p(t) is positive and continuous on [t0,+∞) and φ is differentiable in (-∞,+∞) such that φ′(x)≥0 for x∈(-∞,+∞).(ii)
Fork=1,2,…, gk[i](x) are continuous in (-∞,+∞) and there exist positive numbers ak[i],bk[i] such that
(5)ak[i]≤gk[i](x)/x≤bk[i],i=0,1,2,…,2n.(iii)
(6)∫t0+∞‍∏t0<tk<s‍ak[i]bk[i-1]ds=+∞;i=1,2,…,2n-1,∫t0+∞1r(s)∏t0<tk<sak[2n]bk[2n-1]ds=+∞.‍Our plan is the following. We first obtain three theorems (Theorems1–3) to ensure every solution of (3) is either oscillatory or (nonoscillatory and) zero convergent. We will also illustrate our results with several examples. As applications of our results, we state three corollaries (Corollaries 5–7). These corollaries are new even for the special case (2).


## 2. Main Results

The main results of the paper are as follows.Theorem 1.
Assume that the conditions (i)–(iii) hold. Suppose further that there exists a positive integerk0 such that for k≥k0, ak[0]≥1,
(7)∑k=1+∞‍(bk[0]-1)<+∞,∫t0+∞∏t0<tk<s1bk[2n]p(s)ds=+∞.
Then every solution of (3) is either oscillatory or (nonoscillatory and) zero convergent.Theorem 2.
Assume that the conditions (i)–(iii) hold and thatφ(ab)≥φ(a)φ(b) for any ab>0. Suppose further that there exists a positive integer k0 such that for k≥k0,
(8)bk[0]≤1,bk[2n]≤1,bk[2n]≤φ(ak[0]),∫t0+∞‍p(s)ds=+∞.
Then every solution of (3) is either oscillatory or (nonoscillatory and) zero convergent.Theorem 3.
Assume that the conditions (i)–(iii) hold and thatφ(ab)≥φ(a)φ(b) for any ab>0. Suppose further that bk[2n]≤ak[0],{∏k=1n‍bk[0]} is bounded,
(9)∑k=1+∞‍max{|ak[0]-1|,|bk[0]-1|}<+∞,∑k=1+∞‍|bk[2n]-1|<+∞,∫t0+∞‍p(s)ds=+∞.
Then every solution of (3) is either oscillatory or (nonoscillatory and) zero convergent.Remark 4.
Whenn=1, (3) reduces to (1). Our Theorems 1–3 are Theorem 2.1, Theorem 2.3, Theorem 2.4 in [12], respectively. So our results generalize and contain results in [12].
Next, for (2), we will also be able to obtain some new results. It is easy to see that (2) has the form of (3) by setting r(t)=1,f(t,x(t))=p(t)x(t) and gk[i](x(i)(tk))=ak[i]x(i)(tk) for i=0,1,…,2n and k=1,2,…. Let
(10)ak[i]=bk[i],i=0,1,…,2n;k=1,2,…,φ(x)=x.
Our Theorems 1–3 can directly lead us to the following corollaries for (2).Corollary 5.
Assume that the conditions (i)–(iii) hold. Suppose further that there exists a positive integerk0 such that for k≥k0,ak[0]≥1,
(11)∑k=1+∞‍(ak[0]-1)<+∞,∫t0+∞‍∏t0<tk<s‍1ak[2n]p(s)ds=+∞.
Then every solution of (2) is either oscillatory or (nonoscillatory and) zero convergent.Corollary 6.
Assume that the conditions (i)–(iii) hold. Suppose further that there exists a positive integerk0 such that for k≥k0,
(12)ak[0]≤1,ak[2n]≤1,ak[2n]≤ak[0],∫t0+∞‍p(s)ds=+∞.
Then every solution of (2) is either oscillatory or (nonoscillatory and) zero convergent.Corollary 7.
Assume that the conditions (i)–(iii) hold. Suppose further thatak[2n]≤ak[0], {∏k=1nak[0]} is bounded,
(13)∑k=1+∞‍|ak[0]-1|<+∞,∑k=1+∞‍|ak[2n]-1|<+∞,∫t0+∞‍p(s)ds=+∞.
Then every solution of (2) is either oscillatory or (nonoscillatory and) zero convergent.Remark 8.
We note that the above corollaries for ensuring every solution of (2) to be either oscillatory or tend to zero with fixed sign eventually is also new for (2).Example 9.
Consider the equation(14)x′′′(t)+38tx3(t)=0,t≥12,t≠k,x(i)(k+)=(1+1k2)x(i)(k),i=0,1,2;k=1,2,…,x(12)=x0,x′(12)=x0[1],x′′(12)=x0[2],
where ak[i]=bk[i]=(1+(1/k2))≥1 for i=0,1,2;r(t)=1,p(t)=(3/8)t,tk=k, and φ(x)=x3. It is not difficult to see that conditions (i)–(iii) are satisfied. Furthermore,
(15)∑k=1+∞‍max{|ak[0]-1|,|bk[0]-1|}=∑k=1+∞‍1k2<+∞,∫t0+∞‍∏t0<tk<s‍1bk[2]p(s)ds=∫1/2+∞‍38∏1/2<tk<s‍k2k2+1sds=+∞.
Thus by Theorem 1, every solution of (14) is either oscillatory or (nonoscillatory and) zero convergent.Example 10.
Consider the equation(16)((2-sint)g(t)x(4)(t))′+tx3(t)=0,t≥12,t≠k,x(i)(k+)=(1+1k2)x(i)(k),x(4)(k+)=(1-1k2)x(4)(k),i=0,1,2,3;k=1,2,…,x(j)(12)=x0[j],j=0,1,2,3,4,
where ak[i]=bk[i]=(1+(1/k2)) for i=0,1,2,3,ak[4]=bk[4]=(1-(1/k2));p(t)=t,tk=k,φ(x)=x3, and
(17)r(t)=(2-sint)g(t),
where
(18)g(t)=|t-k-12|+1,t∈[k,k+1),k=1,2,….
Here, we do not assume that r(t) is bounded, monotonic, or differentiable. It is not difficult to see that conditions (i)–(iii) are satisfied. Furthermore,
(19)bk[4]=(1-1k2)≤(1+1k2)=ak[0],∑k=1+∞‍max{|ak[0]-1|,|bk[0]-1|}=∑k=1+∞‍1k2<+∞,∑k=1+∞‍|bk[4]-1|=∑k=1+∞‍1k2<+∞∫t0+∞‍p(s)ds=∫1/2+∞‍1sds=+∞.
Thus by Theorem 3, every solution of (16) is either oscillatory or (nonoscillatory and) zero convergent.Example 11.
Consider the equation(20)(t-1x(4)(t))′+(t-1-t-2)x(t)=0,t≥12,t≠k,x(k+)=x(k),x′(k+)=x′(k),x(i)(k+)=kk+1x(i)(k),i=2,3,4;k=1,2,…,x(j)(12)=x0[j],j=0,1,2,3,4,
where ak[0]=bk[0]=ak[1]=bk[1]=1,ak[i]=bk[i]=k/(k+1) for i=2,3,4;p(t)=t-1-t-2,r(t)=t-1,tk=k;φ(x)=x. It is not difficult to see that conditions (i)–(iii) are satisfied. Furthermore,
(21)bk[2n]=kk+1≤1=φ(ak[0]),∫t0+∞‍p(s)ds=∫1/2+∞‍(1s-1s2)ds=+∞.
Thus, by Theorem 2, every solution of (20) is either oscillatory or (nonoscillatory and) zero convergent. But the ordinary differential equation
(22)(t-1x(4)(t))′+(t-1-t-2)x(t)=0,
has a solution x(t)=-et which tends to -∞ as t→+∞. This example shows that impulses play an important role in oscillatory and asymptotic behaviors of equations.


## 3. Proofs

To prove our Theorems, we need the following Lemmas.Lemma 12 (Lakshmikantham et al. [1]).
Assume that(H0)
m∈PC′(R+,R) and m(t) is left-continuous at tk for k=1,2,…(H1)
fortk,k=1,2,… and t≥t0,
(23)m′(t)≤p(t)m(t)+q(t),t≠tk,m(tk+)≤dkm(tk)+bk,where p,q∈PC(R+,R),dk≥0, and bk are real constants. Then for t≥t0,
(24)m(t)≤m(t0)∏t0<tk<t‍dkexp(∫t0t‍p(s)ds)+∑t0<tk<t‍(∏tk<tj<t‍djexp(∫tkt‍p(s)ds))bk+∫t0t‍(∏s<tk<t‍dk)exp(∫st‍p(σ)dσ)q(s)ds.Remark 13.
If the inequalities in (23) are reversed, then the inequality in (24) should be reversed as well.Lemma 14.
Suppose that conditions (i)–(iii) hold andx(t) is a solution of (3). One has the following statements:(a)
if there exists someT≥t0 such that x(2n)(t)>0 and (r(t)x(2n)(t))′≥0 for t≥T, then there exists some T1≥T such that x(2n-1)(t)>0 for t≥T1;(b)
if there existsi∈{1,2,…,2n-1} and some T≥t0 such that x(i)(t)>0 and x(i+1)(t)≥0 for t≥T, then there exists some T1≥T such that x(i-1)(t)>0 for t≥T1.Proof.
First of all, we will prove the result of (a) to be true. Without loss of generality, we may assume thatx(2n)(t)>0 and (r(t)x(2n)(t))′≥0 for t≥t0. We assert that there exists some j such that x(2n-1)(tj)>0 for tj≥t0. If this is not true, then for any tk>t0, we have x(2n-1)(tk)≤0. Since x(2n-1)(t) is increasing on intervals of the form (tk,tk+1], we see that x(2n-1)(t)≤0 for t≥t0. Since r(t)x(2n)(t) is increasing on intervals of the form (tk,tk+1], we see that for t∈(t1,t2],
(25)r(t)x(2n)(t)≥r(t1)x(2n)(t1+);
that is,
(26)x(2n)(t)≥r(t1)r(t)x(2n)(t1+).
In particular,
(27)x(2n)(t2)≥r(t1)r(t2)x(2n)(t1+).
Similarly, for t∈(t2,t3], we have
(28)x(2n)(t)≥r(t2)r(t)x(2n)(t2+)≥r(t2)r(t)a2(2n)x(2n)(t2)≥r(t1)r(t)a2(2n)x(2n)(t1+).
By induction, we know that
(29)x(2n)(t)≥r(t1)r(t)∏t1<tk<t‍ak(2n)x(2n)(t1+),t≠tk.
From condition (ii), we have
(30)x(2n-1)(tk+)≥bk[2n-1]x(2n-1)(tk),t>t1,k=2,3,….
It follows from (29), (30), and Lemma 12, that for t≥t1,
(31)x(2n-1)(t)≥∏t1<tk<t‍bk[2n-1]{∏t1<tk<sak[2n]bk[2n-1]dsx(2n-1)(t1+)+x(2n)(t1+)r(t1)×∫t1t1r(s)∏t1<tk<sak[2n]bk[2n-1]ds}.
Note that ak[i]>0, bk[i]>0, and the second equality of condition (iii) hold. Thus, we get x(2n-1)(t)>0 for all sufficiently large t, which is contrary to x(2n-1)(t)≤0 for t≥t0. So there exists some j such that tj≥t0 and x(2n-1)(tj)>0. Since x(2n-1)(t) is increasing on intervals of the form (tj+λ,tj+λ+1] for each λ=0,1,2,…, then for t∈(tj+λ,tj+λ+1], we have
(32)x(2n-1)(t)≥∏tj≤tk<t‍ak[2n-1]x(2n-1)(tj)>0.
The assertion (a) is thus proved.
Next, we will prove the result of (b) to be true. Without loss of generality, we may assume thatx(i)(t)>0 and x(i+1)(t)≥0 for t≥t0. We assert there exists some j such that x(i-1)(tj)>0 for tj≥t0. If this is not true, then for any tk>t0, we have x(i-1)(tk)≤0. Since x(i-1)(t) is increasing on intervals of the form (tk,tk+1], we see that x(i-1)(t)≤0 for t≥t0. By x(i)(t)>0,x(i+1)(t)≥0, and t∈(tk,tk+1], we have that x(i)(t) is nondecreasing on (tk,tk+1]. For t∈(t1,t2], we have
(33)x(i)(t)≥x(i)(t1+).
In particular,
(34)x(i)(t2)≥x(i)(t1+).
Similarly, for t∈(t2,t3], we have
(35)x(i)(t)≥x(i)(t2+)≥a2(i)x(i)(t2)≥a2(i)x(i)(t1+).
By induction, we know that
(36)x(i)(t)≥∏t1<tk<t‍ak(i)x(i)(t1+),t>t1,t≠tk.
From condition (ii), we have
(37)x(i-1)(tk+)≥bk(i-1)x(i-1)(tk),k=2,3,….
Set u(t)=-x(i-1)(t). Then from (36) and (37), we see that
(38)u′(t)≤-∏t1<tk<t‍ak(i)x(i)(t1+),t>t1,t≠tk,u(tk+)≤bk(i-1)u(tk),k=2,3,….
It follows from Lemma 12 that
(39)u(t)≤u(t1+)∏t1<tk<t‍bk(i-1)-x(i)(t1+)∫t1t‍∏s<tk<t‍bk(i-1)∏t1<tk<t‍ak(i)ds=∏t1<tk<t‍bk(i-1){u(t1+)-x(i)(t1+)∫t1t‍∏t1<tk<s‍ak(i)bk(i-1)ds};
that is,
(40)x(i-1)(t)≥∏t1<tk<t‍bk(i-1),×{x(i-1)(t1+)+x(i)(t1+)∫t1t‍∏t1<tk<s‍ak(i)bk(i-1)ds}.
Note that ak(i)>0,bk(i)>0, and the first equality of condition (iii) holds. Thus, we get x(i-1)(t)>0 for all sufficiently large t. The relation x(i-1)(t)≤0 leads to a contradiction. So there exists some j such that tj>t0 and x(i-1)(tj)>0. Then
(41)x(i-1)(tj+)≥aj(i-1)x(i-1)(tj)>0.
Since x(i)(t)>0, then for every positive integer m, we have that x(i-1)(t) is increasing on (tj+m-1,tj+m]. For (tj,tj+1], we have
(42)x(i-1)(t)≥x(i-1)(tj+)>0.
In particular,
(43)x(i-1)(tj+1)≥x(i-1)(tj+)>0.
Similarly, for t∈(tj+1,tj+2], we have
(44)x(i-1)(t)≥x(i-1)(tj+1+)≥aj+1(i-1)x(i-1)(tj+1)>0.
By induction, for (tj+m-1,tj+m], we have
(45)x(i-1)(t)>0.
Summing up the above discussion, we know that there exists some T1≥T such that
(46)x(i-1)(t)>0,t≥T1.
The proof of Lemma 14 is complete.Remark 15.
We may prove in similar manners the following statements.(a′)
If we replace the condition (a) in Lemma14 “x(2n)(t)>0 and (r(t)x(2n)(t))′≥0 for t≥T” with “x(2n)(t)<0 and (r(t)x(2n)(t))′≤0 for t≥T,” then under conditions (i)–(iii), there exists some T1≥T such that x(2n-1)(t)<0 for t≥T1.(b′)
If we replace the condition (b) in Lemma14 “x(i)(t)>0 and x(i+1)(t)≥0 for t≥T” with “x(i)(t)<0 and x(i+1)(t)≤0 for t≥T,” then under conditions (i)–(iii), there exists some T1≥T such that x(i-1)(t)<0 for t≥T1.Lemma 16.
Suppose conditions (i)–(iii) hold. Letx=x(t) be a solution of (3).(a)
If there exists someT≥t0 such that x(t)>0 and (r(t)x(2n)(t))′<0 for t≥T, then x(2n)(t)>0 for all sufficiently large t.(b)
If there existsi∈{1,2,…,2n} and some T≥t0 such that x(t)>0 and x(i)(t)<0 for t≥T, then x(i-1)(t)>0 for all sufficiently large t.Proof.
First of all, we will prove the result of (a) to be true. We assert thatx(2n)(t)>0 for any tk≥T. If this is not true, then there exists some tj≥T such that x(2n)(tj)≤0. Since r(t)>0 and r(t)x(2n)(t) are strictly decreasing on (tj+m,tj+m+1] for m=0,1,2,… and for t∈(tj,tj+1], we have
(47)r(t)x(2n)(t)<r(tj)x(2n)(tj+)≤aj(2n)r(tj)x(2n)(tj)≤0.
Similarly, for t∈(tj+1,tj+2], we have
(48)r(t)x(2n)(t)<r(tj+1)x(2n)(tj+1+)≤aj(2n)aj+1(2n)r(tj)x(2n)(tj)≤0.
We can easily prove that, for any positive integer n≥2 and t∈(tj+n,tj+n+1], we have
(49)r(t)x(2n)(t)<aj(2n)aj+1(2n)…aj+n(2n)r(tj)x(2n)(tj)≤0.
Hence, x(2n)(t)<0 for t≥tj+1. By the result (a) of Remark 13, for sufficiently large t, we have x(2n-1)(t)<0. Using the result (b) of Remark 13 repeatedly, for all sufficiently large t, we get x(t)<0, which is contrary to x(t)>0 for t≥T. Hence we have x(2n)(tk)>0 for any tk≥T. So we get x(2n)(t)>0 for all sufficiently large t.
Next, we will prove the result of (b) to be true. We assert thatx(i-1)(tk)>0 for any tk≥T. If this is not true, then there exists some tj≥T such that x(i-1)(tj)<0. Since x(i-1)(t) is strictly monotony decreasing on (tj+n,tj+n+1] for n=0,1,2,… and for t∈(tj,tj+1], we have
(50)x(i-1)(t)<x(i-1)(tj+)≤aj(i-1)x(i-1)(tj)≤0.
Similarly, for t∈(tj+1,tj+2], we have
(51)x(i-1)(t)<x(i-1)(tj+1+)≤aj(i-1)aj+1(i-1)x(i-1)(tj)≤0.
We can easily prove that for any positive integer n≥2 and t∈(tj+n,tj+n+1], we have
(52)x(i-1)(t)<aj(i-1)…aj+1(i-1)j+n(i-1)x(i-1)(tj)≤0.
Hence, x(i-1)(t)<0 for t≥tj+1. By the result (b′) of Remark 8, for sufficiently large t, we have x(i-2)(t)<0. Similarly, by using the result (b′) of Remark 8 again, we can conclude that for all sufficiently large t, x(t)<0. That is contrary with x(t)>0 for t≥T. Hence, we have x(i-1)(tk)>0 for any tk≥T. So we get x(i-1)(t)>0 for all sufficiently large t.
The proof of Lemma16 is complete.Lemma 17.
Suppose conditions (i)–(iii) hold. Letx=x(t) be a solution of (3). Suppose that T≥t0 and x(t)>0 for t≥T. Then there exists some T′≥T and l∈{0,2,4,…,2n} such that for t≥T′,
(53)x(i)(t)>0,i=0,1,…,l;(-1)(i-1)x(i)(t)>0,i=l+1,…,2n;(r(t)x(2n)(t))′<0.Proof.
Letx(t)>0 for t≥T≥t0. By (3) and condition (i), we have
(54)(r(t)x(2n)(t))′=-f(t,x)≤-p(t)φ(x)<0.
By the result (a) of Lemma 16, there exists some T1≥T such that x(2n)(t)>0 for t≥T1. Without loss of generality, we may let x(2n)(t)>0 for t≥t0. Then x(2n-1)(t) is strictly increasing on (tk,tk+1]. If for any tk,x(2n-1)(tk)<0, then x(2n-1)(t)<0 for t≥T1. If there exists some tj such that x(2n-1)(tj)≥0, since x(2n-1)(t) is strictly monotony increasing and ak[2n-1]>0, then x(2n-1)(t)>0 for t>tj. Thus, there exists some T2≥T1 such that x(2n-1)(t)>0 for t≥T2. So one of the following statements holds:
(55)(A1)x(2n)(t)>0,x(2n-1)(t)>0,t≥T2;(56)(B1)x(2n)(t)>0,x(2n-1)(t)<0,t≥T2.
If (55) holds, then by the result (b) of Lemma 14, x(2n-2)(t)>0 for all sufficiently large t. Using the result (b) of Lemma 14 repeatedly, for all sufficiently large t, we can conclude that
(57)x(2n)(t)>0,x(2n-1)(t)>0,…,x′(t)>0,x(t)>0.
If (56) holds, by Lemma 16, we have x(2n-3)(t)<0 for all sufficiently large t. Similarly, there exists some T3≥T2 such that one of the following statements holds:
(58)(A2)x(2n-2)(t)>0,x(2n-3)(t)>0,t≥T3;(B2)x(2n-2)(t)>0,x(2n-3)(t)<0,t≥T3.
Repeating the discussion above, we can eventually get that there exist some T′≥T and l∈{0,2,4,…,2n} such that for t≥T′,
(59)x(i)(t)>0,i=0,1,…,l;(-1)(i-1)x(i)(t)>0,i=l+1,…,2n;(r(t)x(2n)(t))′<0.
The proof of Lemma 17 is complete.Lemma 18 (see [13]).
Suppose fort>0,x=x(t) is continuous at t≠tk, left-continuous at t=tk, and limt→tk+x(t) exists for k=1,2,…. Further, assume that(H2)
there existst¯∈R+, such that x(t)>0(<0) for t≥t¯;(H3)
x(t) is nonincreasing (resp., nondecreasing) on (tk,tk+1] for k=1,2,…;(H4)
∑k=1+∞‍[x(tk+)-x(tk)] is convergent.Then limt→∞x(t)=r exists and r≥0 (resp. ≤0).
We now turn to the proof of Theorem1.  Without loss of generality, we may assume that k0=1. If (3) has a nonoscillatory solution x=x(t), we may assume that x(t)>0 for t≥t0. By Lemma 17, there exists a T≥t0 and l∈{0,2,4,…,2n} such that for t≥T, (53) holds. Next, we rewrite (53) as
(60)x′(t)<0,…,x(2n)(t)>0,(r(t)x(2n)(t))′<0forl=0;(61)x′(t)>0,…,x(l)(t)>0,(-1)(i)x(i)(t)>0,(r(t)x(2n)(t))′<0forl≥2.
If (60) holds, then we see that the conditions (H2) and (H3) of Lemma 18 are satisfied. Furthermore, note that ∑k=1+∞‍(bk[0]-1)<+∞ and bk[0]≥ak[0]≥1, then one has
(62)∏k=1+∞‍bk[0]<+∞.
Sincex′(t)<0 and  t≥T, one obtains for any tk>T,
(63)x(tk)≤∏T<tj<tk‍bj[0]x(T+).
By (62) and (63), we know {x(tk)} is bounded. Thus, there exists M>0 such that |x(tk)|≤M. It follows from condition (ii) that
(64)|x(tk+)-x(tk)|≤|bk[0]-1||x(tk)|≤M(bk[0]-1).
From (64) and note that ∑k=1+∞(bk[0]-1) is convergent, we know that ∑k=1+∞[x(tk+)-x(tk)] is convergent. Therefore, the condition (H4) of Lemma 18 is also satisfied. By Lemma 18, we have limt→+∞x(t)=r≥0. We assert that r=0. If r>0, then there exists T1≥T, such that for any t≥T1,x(t)>r/2>0. Note that φ′(x)≥0; hence we may see that φ(x(t))≥φ(r/2) for t≥T1. Let Φ(t)=r(t)x(2n)(t). Then Φ(t)>0 for t≥T1. By conditions (i), (ii), and (3), one has for t≥T1,
(65)Φ′(t)=-f(t,x)≤-p(t)φ(x(t))≤-φ(r2)p(t),t≠tk,Φ(tk+)=r(tk+)x(2n)(tk+)≤bk[2n]r(tk)x(2n)(tk)≤bk[2n]Φ(tk),tk≥T1.
From (65) and Lemma 12, one gets for t≥T1,
(66)Φ(t)≤Φ(T1+)∏T1<tk<t‍bk[2n]-φ(r2)∫T1t‍∏s<tk<t‍bk[2n]p(s)ds=∏T1<tk<t‍bk[2n](Φ(T1+)-φ(r2)∫T1t‍∏T1<tk<s‍1bk[2n]p(s)ds).
It is easy to see from (7) and (66) that Φ(t)<0 for sufficiently large t. This is contrary to Φ(t)>0 for t≥T1. Thus, r=0; that is, limt→+∞x(t)=0.
If (61) holds, then let Ψ(t)=r(t)x(2n)(t)/φ(x(t)). Then Ψ(t)>0 for t≥T. By (3) and condition (i), one has for t≥T,
(67)Ψ′(t)=-f(t,x(t))φ(x(t))-r(t)x(2n)(t)φ′(x(t))x′(t)φ2(x(t))≤-f(t,x(t))φ(x(t))≤-p(t),t≠tk.
From the conditions (i) and (ii) and note that ak[0]≥1, we know that
(68)Ψ(tk+)=r(tk+)x(2n)(tk+)φ(x(tk+))≤r(tk)bk[2n]x(2n)(tk)φ(ak[0]x(tk))≤bk[2n]r(tk)x(2n)(tk)φ(x(tk))=bk[2n]Ψ(tk).
From (67), (68), and Lemma 12, one gets for t≥T,
(69)Ψ(t)≤Ψ(T+)∏T<tk<t‍bk[2n]-∫Tt‍∏u<tk<t‍bk[2n]p(u)du=∏T<tk<t‍bk[2n]{Ψ(T+)-∫Tt‍∏T<tk<u‍1bk[2n]p(u)du}.
It is easy to see from (7) and (69) that Ψ(t)<0 for sufficiently large t. This is contrary to Ψ(t)>0 for t≥T. Thus, every solution of (3) is oscillatory. The proof of Theorem 1 is complete.
We now give the proof of Theorem2.  Without loss of generality, we may assume that k0=1. If (3) has a nonoscillatory solution x=x(t), we may assume that x(t)>0 for t≥t0. By Lemma 17, there exists a T≥t0 and l∈{0,2,4,…,2n} such that for t≥T, (53) holds. Next, we may rewrite (53) as (60) and (61).
If (60) holds, then note that bk[0]≤1. Since x(t) is decreasing on (tj+l,tj+l+1] for tj≥T and each l=0,1,2,…, one has for t∈(tj+l,tj+l+1],
(70)x(t)<x(tj+l+)≤⋯≤x(tj+1)<x(tj+)≤bj[0]x(tj)≤x(tj).
So, since x(t) is decreasing and bounded on (tj,+∞), we know that x(t) is convergent as t→+∞. Let limt→+∞x(t)=r. Then r≥0. We assert that r=0. If r>0, then there exists T1≥T such that for t≥T1,x(t)>r/2>0. Since φ′(x)≥0, one has φ(x(t))≥φ(r/2). By (3) and condition (i), one has for t≥T,
(71)(r(t)x(2n)(t))′=-f(t,x)≤-p(t)φ(x(t))≤-φ(r2)p(t)<0,t≠tk.
From condition (ii) and noting that bk[2n]≤1, one has
(72)r(tk+)x(2n)(tk+)≤bk[2n]r(tk)x(2n)(tk)≤r(tk)x(2n)(tk),tk≥T1.
By (71), (72), and Lemma 12, one has for t≥T1,
(73)r(t)x(2n)(t)≤r(T1)x(2n)(T1+)-φ(r2)∫T1t‍p(s)ds.
It is easy to see from (8) and (73) that x(2n)(t)<0 for sufficiently large t. This is contrary to x(2n)(t)>0 for t≥T. Thus r=0, that is, limt→+∞x(t)=0;
If (61) holds, then let Ψ(t)=r(t)x(2n)(t)/φ(x(t)). We see that Ψ(t)>0 for t≥T. By (3) and the condition (i), one gets for t≥T(74)Ψ′(t)=-f(t,x(t))φ(x(t))-r(t)x(2n)(t)φ′(x(t))x′(t)φ2(x(t))≤-f(t,x(t))φ(x(t))≤-p(t),t≠tk.
From the conditions (i) and (ii), we know that
(75)Ψ(tk+)=r(tk)x(2n)(tk+)φ(x(tk+))≤r(tk)bk[2n]x(2n)(tk)φ(ak[0]x(tk))≤bk[2n]φ(ak[0])r(tk)x(2n)(tk)φ(x(tk))≤Ψ(tk).
From (74), (75), and Lemma 12, one gets for t≥T,
(76)Ψ(t)≤Ψ(T+)-∫Tt‍p(s)ds.
It is easy to see from (8) and (76) that Ψ(t)<0 for sufficiently large t. This is contrary to Ψ(t)>0 for t≥T. Thus, every solution of (3) is oscillatory. The proof of Theorem 2 is complete.
Finally, we give the proof of Theorem3. Without loss of generality, we may assume that k0=1. If (3) has a nonoscillatory solution x=x(t), we may assume that x(t)>0 for t≥t0. By Lemma 17, there exists a T≥t0 and l∈{0,2,4,…,2n} such that for t≥T, (53) hold. Next, we rewrite (53) as (60) and (61).
If (60) holds, then we see that the conditions (H2) and (H3) of Lemma 18 are satisfied. Furthermore, since x′(t)<0 for t≥T, we see that there exists some ti≥T such that for any tk>ti,
(77)x(tk)≤∏ti<tj<tk‍bj[0]x(ti+).
Since {∏k=1n‍bk[0]} is bounded, in view of (77), we know that {x(tk)} is bounded. Thus, there exists M1>0, such that |x(tk)|≤M1. It follows from the condition (ii) that
(78)|x(tk+)-x(tk)|≤max{|ak[0]-1|,|bk[0]-1|}|x(tk)|≤M1max{|ak[0]-1|,|bk[0]-1|}.
By (78), we know that ∑k=1+∞‍[x(tk+)-x(tk)] is convergent. Therefore, the condition (H4) of Lemma 18 is also satisfied. By Lemma 18, we know that limt→+∞x(t)=r≥0. We assert that r=0. If r>0, then there exists T1≥T, such that for t≥T1,x(t)>r/2>0. Since φ′(x)≥0, we see that φ(x(t))≥φ(r/2) for t≥T1. By bk[2n]≤ak[0], the boundedness of {∏k=1n‍bk[0]} and the condition (ii), we know that {∏k=1n‍bk[2n]} is bounded. Note that (r(t)x(2n)(t))′<0 for t≥T1. So we may also see that there exists M2>0 such that |r(tk)x(2n)(tk)|≤M2. Therefore, one has
(79)|(bk[2n]-1)r(tk)x(2n)(tk)|≤M2|bk[2n]-1|.
By (3) and the condition (i), one has
(80)(r(t)x(2n)(t))′=-f(t,x)≤-p(t)φ(x(t))<0,t≠tk.
Integrating (80) from T1 to t, it follows from (79) and the condition (ii) that
(81)r(t)x(2n)(t)-r(T1)x(2n)(T1+)≤∑T1<tk<t‍r(tk)[x(2n)(tk+)-x(2n)(tk)]-φ(r2)∫T1t‍p(s)ds≤∑T1<tk<t‍|(bk[2n]-1)r(tk)x(2n)(tk)|-φ(r2)∫T1t‍p(s)ds≤∑T1<tk<t‍M2|bk[2n]-1|-φ(r2)∫T1t‍p(s)ds.
Since∑k=1+∞|bk[2n]-1| is convergent, it is easy to see from (9) and (81) that x(2n)(t)<0 for sufficiently large t. This is contrary to x(2n)(t)>0 for t≥T. Thus r=0; that is, limt→+∞x(t)=0.
If (61) holds, then let Ψ(t)=r(t)x(2n)(t)/φ(x(t)). We see that Ψ(t)>0 for t≥T. Similar to the proof of (76), we also obtain
(82)Ψ(t)≤Ψ(T+)-∫Tt‍p(s)ds.
It is easy to see from (9) and (82) that Ψ(t)<0 for sufficiently large t. This is contrary to Ψ(t)>0 for t≥T. Thus, every solution of (3) is oscillatory. The proof of Theorem 3 is complete.



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*Source: 421231-2013-12-22.xml*