# Asymptotic Dichotomyin a Class of Third-Order Nonlinear Differential Equations with Impulses

Authors: Kun-Wen Wen; Gen-Qiang Wang; Sui Sun Cheng
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

Solutions of quite a few higher-order delay functional differential equations oscillate or converge to zero. In this paper, we obtain several such dichotomous criteria for a class of third-order
nonlinear differential equation with impulses.


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## Body

## 1. Introduction

It has been observed that the solutions of quite a few higher-order delay functional differential equations oscillate or converge to zero (see, e.g., the recent paper [1] in which a third order nonlinear delay differential equation with damping is considered). Such a dichotomy may yield useful information in real problems (see, e.g., [2] in which implications of this dichotomy are applied to the deflection of an elastic beam). Thus it is of interest to see whether similar dichotomies occur in different types of functional differential equations.One such type consists of impulsive differential equations which are important in simulation of processes with jump conditions (see, e.g., [3–22]). But papers devoted to the study of asymptotic behaviors of third-order equations with impulses are quite rare. For this reason, we study here the third-order nonlinear differential equation with impulses of the form
(1.1)(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=+∞,(1.2)x(i)(tk+)=limt→tk+x(i)(t),x(i)(tk-)=limt→tk-x(i)(t)
for i=0,1,2. Here gk[i],i=0,1,2 and k=1,2,…, are real functions and x0[i],i=0,1,2, are real numbers.By a solution of (1.1), we mean a real function x=x(t) defined on [t0,+∞) such that(i)
x(i)(t0+)=x0[i] for i=0,1,2;(ii)
x(i)(t),i=0,1,2, and (r(t)x′′(t))′are continuous on [t0,+∞)∖{tk}; for i=0,1,2,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′′(t))′+f(t,x)=0 at each point t∈[t0,+∞)∖{tk}.A solution of (1.1) is said to be nonoscillatory if it is eventually positive or eventually negative. Otherwise, it is said to be oscillatory.We will establish dichotomous criteria that guarantee solutions of (1.1) that are either oscillatory or zero convergent based on combinations of the following conditions.(A)
r(t) is positive and continuous on [t0,∞),f(t,x) is continuous on [t0,∞)×R,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 R such that φ′(x)≥0 for x∈R.(B)
For eachk=1,2,…,gk[i](x) is continuous in R and there exist positive numbers ak[i],bk[i] such that ak[i]≤gk[i](x)/x≤bk[i] for x≠0 and i=0,1,2.(C)
One has(1.3)∫t0∞∏t0<tk<s(ak[1]bk[0])ds=+∞,∫t0∞1r(s)∏t0<tk<s(ak[2]bk[1])ds=+∞.In the next section, we state four theorems to ensure that every solution of (1.1) either oscillates or tends to zero. Examples will also be given. Then in Section 3, we prove several preparatory lemmas. In the final section, proofs of our main theorems will be given.


## 2. Main Results

The main results of the paper are as follows.Theorem 2.1.
Assume that the conditions (A)–(C) hold. Suppose further that there exists a positive integerk0 such that for k≥k0,ak[0]≥1,(2.1)∑k=1+∞(bk[0]-1)<+∞,(2.2)∫t0+∞∏t0<tk<s(1bk[2])p(s)ds=+∞.
Then every solution of (1.1) either oscillates or tends to zero.Theorem 2.2.
Assume that the conditions (A)–(C) hold. Suppose further that there exists a positive integerk0 such that for k≥k0, bk[0]≤1,ak[1]≥1,(2.3)∏t0≤tk<+∞ak[0]≥σ>0,(2.4)∫t0+∞1r(s)(∫s+∞∏s<tk<u1bk[2]p(u)du)ds=+∞.
Then every solution of (1.1) either oscillates or tends to zero.Theorem 2.3.
Assume that the conditions (A)–(C) hold and thatφ(ab)≥φ(a)φ(b) for any ab>0. Suppose further that there exists a positive integer k0 such that for
(2.5)k≥k0,bk[0]≤1,bk[2]≤1,bk[2]≤φ(ak[0]),(2.6)∫t0+∞p(s)ds=+∞.
Then every solution of (1.1) either oscillates or tends to zero.Theorem 2.4.
Assume that the conditions (A)–(C) hold and thatφ(ab)≥φ(a)φ(b) for any ab>0. Suppose further that bk[2]≤ak[0], {∏k=1nbk[0]} is bounded, that
(2.7)∑k=1+∞max{|ak[0]-1|,|bk[0]-1|}<+∞,∑k=1+∞|bk[2]-1|<+∞,(2.8)∫t0+∞p(s)ds=+∞.
Then every solution of (1.1) either oscillates or tends to zero.Before giving proofs, we first illustrate our theorems by several examples.Example 2.5.
Consider the equation(2.9)(tx′′(t))′+etx(t)=0,t≥12,t≠k,x(i)(k+)=(1+1k2)x(i)(k),i=0,1,2;k=1,2,…,x(12)=x0[0],x′(12)=x0[1],x′′(12)=x0[2],
where ak[i]=bk[i]=(1+(1/k2))≥1 for i=0,1,2; p(t)=et, r(t)=t, tk=k, φ(x)=x. It is not difficult to see that conditions (A)–(C) are satisfied. Furthermore,
(2.10)∑k=1+∞(bk[0]-1)=∑k=1+∞1k2<+∞,∫t0+∞∏t0<tk<s1bk[2]p(s)ds=∫1/2+∞∏1/2<tk<sk2k2+1esds=+∞.
Thus by Theorem 2.1, every solution of (2.9) either oscillates or tends to zero.Example 2.6.
Consider the equation(2.11)(t(2-sint)g(t)x′′(t))′+t-3/2x3(t)=0,t≥12,t≠k,x(k+)=kk+1x(k),x(i)(k+)=x(i)(k),i=1,2;k=1,2,…,x(12)=x0[0],x′(12)=x0[1],x′′(12)=x0[2],
where ak[0]=bk[0]=k/(k+1), ak[i]=bk[i]=1 for i=1,2;  p(t)=t-3/2, tk=k, φ(x)=x3, and
(2.12)r(t)=t(2-sint)g(t),hereg(t)=|t-k-12|+1,t∈[k,k+1),k=1,2,….
Here, we do not assume that r(t) is bounded, monotonic, or differential. It is not difficult to see that conditions (A)–(C) are satisfied. Furthermore,
(2.13)∫t0+∞1r(s)(∫s+∞∏s<tk<u1bk(2)p(u)du)ds=∫1/2+∞1s(2-sins)g(s)(∫s+∞u-3/2du)ds≥∫1/2+∞13sg(s)(∫s+∞u-3/2du)ds≥∫1/2+∞29s(∫s+∞u-3/2du)ds=∫1/2+∞49sds=+∞.
Thus by Theorem 2.2, every solution of (2.11) either oscillates or tends to zero.Example 2.7.
Consider the equation(2.14)(e-2tx′′(t))′+e-2tx(t)=0,t≥12,t≠k,x(k+)=x(k),x′(k+)=x′(k),x′′(k+)=kk+1x′′(k),k=1,2,…,x(12)=x0[0],x′(12)=x0[1],x′′(12)=x0[2],
where ak[i]=bk[i]=1 for i=0,1,ak[2]=bk[2]=k/(k+1); p(t)=e-2t, r(t)=e-2t, tk=k; φ(x)=x. It is not difficult to see that conditions (A)–(C) are satisfied. Furthermore,
(2.15)∫t0+∞1r(s)(∫s+∞∏s<tk<u1bk(2)p(u)du)ds=∫1/2+∞e2s(∫s+∞∏s<tk<uk+1ke-2udu)ds≥∫1/2+∞e2s(∫s+∞e-2udu)ds=∫1/2+∞12ds=+∞.
Thus, by Theorem 2.2, every solution of (2.14) either oscillates or tends to zero.Note that the ordinary differential equation(2.16)(e-2tx′′(t))′+e-2tx(t)=0has a nonnegative solutionx(t)=et→+∞ as t→+∞. This example shows that impulses play an important role in oscillatory and asymptotic behaviors of equations under perturbing impulses.


## 3. Preparatory Lemmas

To prove our theorems, we need the following lemmas.Lemma 3.1 (Lakshmikantham et al. [3]).
Assume the following.(H0)
m∈PC′(R+,R) and m(t) is left-continuous at tk,k=1,2,….(H1)
Fortk,k=1,2,… and t≥t0,(3.1)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,
(3.2)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.Lemma 3.2.
Suppose that conditions (A)–(C) hold andx(t) is a solution of (1.1). One has the following statements.(a)
If there exists someT≥t0 such that x′′(t)>0 and (r(t)x′′(t))′≥0 for t≥T, then there exists some T1≥T such that x′(t)>0 for t≥T1.(b)
If there exists someT≥t0 such that x′(t)>0 and x′′(t)≥0 for t≥T, then there exists some T1≥T such that x(t)>0 for t≥T1.Proof.
First of all, we will prove that (a) is true. Without loss of generality, we may assume thatx′′(t)>0 and (r(t)x′′(t))′≥0 for t≥t0. We assert that there exists some j such that x′(tj)>0 for tj≥t0. If this is not true, then for any tk≥t0, we have x′(tk)≤0. Since x′(t) is increasing on intervals of the form (tk,tk+1], we see that x′(t)≤0 for t≥t0. Since r(t)x′′(t) is increasing on intervals of the form (tk,tk+1], we see that for (t1,t2],(3.3)r(t)x′′(t)≥r(t1)x′′(t1+),
that is,
(3.4)x′′(t)≥r(t1)r(t)x′′(t1+).
In particular,
(3.5)x′′(t2)≥r(t1)r(t2)x′′(t1+).
Similarly, for (t2,t3], we have
(3.6)x′′(t)≥r(t2)r(t)x′′(t2+)≥r(t2)r(t)a2[2]x′′(t2)≥r(t1)r(t)a2[2]x′′(t1+).
By induction, we know that for t>t1,(3.7)x′′(t)≥r(t1)r(t)∏t1<tk<tak[2]x′′(t1+),t≠tk.
From condition (B), we have
(3.8)x′(tk+)≥bk[1]x′(tk),k=2,3,….
Set m(t)=-x′(t). Then from (3.7) and (3.8), we see that for t>t1,(3.9)m′(t)≤-r(t1)r(t)∏t1<tk<tak[2]x′′(t1+),t≠tkm(tk+)≤bk[1]m(tk),k=2,3,….
It follows from Lemma 3.1 that
(3.10)m(t)≤m(t1+)∏t1<tk<tbk[1]-x′′(t1+)r(t1)∫t1t1r(s)∏s<tk<tbk[1]∏t1<tk<sak[2]ds=∏t1<tk<tbk[1]{m(t1+)-x′′(t1+)r(t1)∫t1t1r(s)∏t1<tk<sak[2]bk[1]ds}.
That is,
(3.11)′(t)≥∏t1<tk<tbk[1]{x′(t1+)+x′′(t1+)r(t1)∫t1t1r(s)∏t1<tk<sak[2]bk[1]ds}.
Note that ak[i]>0, bk[i]>0, and the second equality of condition (C) holds. Thus we get x′(t)>0 for all sufficiently large t. The relation x′(t)≤0 leads to a contradiction. Thus, there exists some j such that tj≥t0 and x′(tj)>0. Since x′(t) is increasing on intervals of the form (tj+λ,tj+λ+1] for λ=0,1,2,…, thus for t∈(tj,tj+1], we have
(3.12)x′(t)≥x′(tj+)≥aj[1]x′(tj)>0.
Similarly, for t∈(tj+1,tj+2],
(3.13)x′(t)≥x′(tj+1+)≥aj+1[1]x′(tj+1)≥aj[1]aj+1[1]x′(tj)>0.
We can easily prove that, for any positive integer λ≥2 and t∈(tj+λ,tj+λ+1],(3.14)x′(t)≥aj[1]aj+1[1]⋯aj+λ[1]x′(tj)>0.
Therefore, x′(t)>0 for t≥tj. Thus, (a) is true.
Next, we will prove that (b) is true. Without loss of generality, we may assume thatx′(t)>0 and x′′(t)≥0 for t≥t0. We assert that there exists some j such that x(tj)>0 for tj≥t0. If this is not true, then for any tk≥t0, we have x(tk)≤0. Since x(t) is increasing on intervals of the form (tk,tk+1], we see that x(t)≤0 for t≥t0. By x′(t)>0, x′′(t)≥0,t∈(tk,tk+1], we have that x′(t) is nondecreasing on (tk,tk+1]. For t∈(t1,t2], we have
(3.15)x′(t)≥x′(t1+).
In particular,
(3.16)x′(t2)≥x′(t1+).
Similarly, for t∈(t2,t3], we have
(3.17)x′(t)≥x′(t2+)≥a2[1]x′(t2)≥a2[1]x′(t1+).
By induction, we know that for t>t1,(3.18)x′(t)≥∏t1<tk<tak[1]x′(t1+),t≠tk.
From condition (B), we have
(3.19)x(tk+)≥bk[0]x(tk),k=2,3,….
Set u(t)=-x(t). Then from (3.18) and (3.19), we see that for t>t1,(3.20)u′(t)≤-∏t1<tk<tak[1]x′(t1+),t≠tk,u(tk+)≤bk[0]u(tk),k=2,3,….
It follows from Lemma 3.1 that
(3.21)u(t)≤u(t1+)∏t1<tk<tbk[0]-x′(t1+)∫t1t∏s<tk<tbk[0]∏t1<tk<sak[1]ds=∏t1<tk<tbk[0]{u(t1+)-x′(t1+)∫t1t∏t1<tk<sak[1]bk[0]ds}.
That is,
(3.22)x(t)≥∏t1<tk<tbk[0]{x(t1+)+x′(t1+)∫t1t∏t1<tk<sak[1]bk[0]ds}.
Note that ak[i]>0, bk[i]>0, and the first equality of condition (C) holds. Thus we get x(t)>0 for all sufficiently large t. The relation x(t)≤0 leads to a contradiction. So there exists some j such that tj≥t0 and x(tj)>0. Then
(3.23)x(tj+)≥aj[0]x(tj)>0.
Since x′(t)>0, we see that x(t) is strictly monotonically increasing on (tj+m,tj+m+1] for m=0,1,2,…. For t∈(tj,tj+1], we have
(3.24)x(t)≥x(tj+)>0.
In particular,
(3.25)x(tj+1)≥x(tj+)>0.
Similarly, for t∈(tj+1,tj+2], we have
(3.26)x(t)≥x(tj+1+)≥aj+1[0]x(tj+1)>0.
By induction, we have x(t)>0 for t∈(tj+m,tj+m+1]. Thus, we know that x(t)>0, for t≥tj. The proof of Lemma 3.2 is complete.Remark 3.3.
We may prove in similar manners the following statements.(a′)
If we replace the condition (a) in Lemma3.2 “x′′(t)>0 and (r(t)x′′(t))′≥0 for t≥T” with “x′′(t)<0 and (r(t)x′′(t))′≤0 for t≥T”, then there exists some T1≥T such that x′(t)<0 for t≥T1.(b′)
If we replace the condition (b) in Lemma3.2 “x′(t)>0 and x′′(t)≥0 for t≥T” with “x′(t)<0 and x′′(t)≤0 for t≥T”, then there exists some T1≥T such that x(t)<0 for t≥T1.Lemma 3.4.
Suppose that conditions (A)–(C) hold andx(t) is a solution of (1.1) such that x(t)>0 for t≥T, where T≥t0. Then there exists T′≥T such that either (a) x′′(t)>0, x′(t)<0 for t≥T′ or (b) x′′(t)>0, x′(t)>0 for t≥T′.Proof.
Without loss of generality, we may assume thatx(t)>0 for t≥t0. By (1.1) and condition (A), we have for t≥t0.(3.27)(r(t)x′′(t))′=-f(t,x)≤-p(t)φ(x)<0.
We assert that for any tk≥t0,x′′(tk)>0. If this is not true, then there exists some j such that x′′(tj)≤0, so x′′(tj+)≤aj[2]x′′(tj)≤0. Since r(t)x′′(t) is decreasing on (tj+k-1,tj+k] for k=1,2,…, we see that for t∈(tj,tj+1],(3.28)x′′(t)<r(tj)r(t)x′′(tj+)≤0.
In particular,
(3.29)x′′(tj+1)<r(tj)r(tj+1)x′′(tj+)≤0.
Similarly, for t∈(tj+1,tj+2], we have
(3.30)x′′(t)<r(tj+1)r(t)x′′(tj+1+)≤r(tj+1)r(t)aj+1[2]x′′(tj+1)≤r(tj)r(t)aj+1[2]x′′(tj+)≤0.
In particular,
(3.31)′′(tj+2)<r(tj)r(tj+2)aj+1[2]x′′(tj+)≤0.
By induction, for any t∈(tj+n-1,tj+n] for n=2,3,…, we have
(3.32)x′′(t)<r(tj)r(t)∏k=1n-1aj+k[2]x′′(tj+)≤0.
Hence, x′′(t)<0 for t≥tj. By Remark 3.3(a′), there exists T1≥tj such that x′(t)<0 for t≥T1; by Remark 3.3(b′), we get x(t)<0 for t≥T1, which is contrary to x(t)>0 for t≥t0. Hence, for any tk≥t0,x′′(tk)>0, since r(t)x′′(t) is decreasing on (tj+k-1,tj+k] for k=1,2,…, therefore x′′(t)>0 for t≥t0. It follows that x′(t) is strictly increasing on (tk,tk+1] for k=1,2,…. Furthermore, note that ak[1]>0,  k=1,2,…. we see that if for any tk,x′(tk)<0, then x′(t)<0 for t≥t0. If there exists some tj such that x′(tj)≥0, then x′(t)>0 for t>tj. The proof of Lemma 3.4 is complete.Lemma 3.5 (see [12]).
Suppose thatx(t) is continuous at t>0 and t≠tk, it is 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).


## 4. Proofs of Main Theorems

We now turn to the proof of Theorem2.1. Without loss of generality, we may assume that k0=1. If (1.1) has a nonoscillatory solution x=x(t), we first assume that x(t)>0 for t≥t0. By (1.1) and the condition (A), for t≥T≥t0, we get(4.1)(r(t)x′′(t))′=-f(t,x(t))≤-p(t)φ(x(t)),t≠tk.
From the condition (B), we know that(4.2)r(tk+)x′′(tk+)≤bk[2]r(tk)x′′(tk).
By Lemma 3.4, there exists a T≥t0 such that either (a) x′′(t)>0, x′(t)<0 for t≥T or (b) x′′(t)>0, x′(t)>0 for t≥T.Suppose that (a) holds. Then we see that the conditions (H2) and (H3) of Lemma 3.5 are satisfied. Furthermore, note that ∑k=1+∞(bk[0]-1)<+∞ and bk[0]≥ak[0]≥1. Then we have(4.3)∏k=1+∞bk[0]<+∞.
Since x′(t)<0,t≥T, we obtain for any tk>T,(4.4)x(tk)≤∏T<tj<tkbj[0]x(T+).
By (4.3) and (4.4), we know that the sequence {x(tk)} is bounded. Thus there exists M>0 such that |x(tk)|≤M. It follows from the condition (B) that(4.5)|x(tk+)-x(tk)|≤|bk[0]-1||x(tk)|≤M(bk[0]-1).
From (4.5) and the fact 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 3.5 is also satisfied. By Lemma 3.5, we know that 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 further that φ′(x)≥0; so we obtain φ(x(t))≥φ(r/2) for t≥T1. Let m(t)=r(t)x′′(t) for t≥T1. By (4.1) and (4.2), we have(4.6)m′(t)≤q(t),t≥T1,t≠tk,(4.7)m(tk+)≤bk[2]m(tk),tk≥T1,
where q(t)=-φ(r/2)p(t).  From (4.6), (4.7), and Lemma 3.1, we get for t≥T1,(4.8)m(t)≤m(T1+)∏T1<tk<tbk[2]+∫T1t(∏s<tk<tbk[2])q(s)ds=∏T1<tk<tbk[2]{m(T1+)-φ(r2)∫T1t(∏T1<tk<s1bk[2])p(s)ds}.
It is easy to see from (2.2) and (4.8) that m(t)<0 for sufficiently large t. This is contrary to m(t)>0 for t≥T1. Thus r=0, that is, limt→+∞x(t)=0.Suppose that (b) holds. LetΨ(t)=(r(t)x′′(t)/φ(x(t))) for t≥T. Then Ψ(t)>0 for t≥T. By (1.1) and the condition (A), we get, for t≥T,(4.9)Ψ′(t)=-f(t,x(t))φ(x(t))-r(t)x′′(t)φ′(x(t))x′(t)φ2(x(t))≤-f(t,x(t))φ(x(t))≤-p(t),t≠tk.
From the conditions (A), (B) and ak[0]≥1, we know that(4.10)Ψ(tk+)=r(tk)x′′(tk+)φ(x(tk+))≤r(tk)bk[2]x′′(tk)φ(ak[0]x(tk))≤bk[2]r(tk)x′′(tk)φ(x(tk))≤bk[2]Ψ(tk),tk≥T.
From (4.9), (4.10), and Lemma 3.1, we get, for t≥T,(4.11)Ψ(t)≤Ψ(T+)∏T<tk<tbk[2]-∫Tt(∏s<tk<tbk[2])p(s)ds=∏T<tk<tbk[2]{Ψ(T+)-∫Tt(∏T<tk<s1bk[2])p(s)ds}.
It is easy to see from (2.2) and (4.11) that Ψ(t)<0 for sufficiently large t. This is contrary to Ψ(t)>0 for t≥T, and hence we obtain a contradiction. Thus in case (b) x(t) must be oscillatory. The proof of Theorem 2.1 is complete.Next, we give the proof of Theorem2.2. Without loss of generality, we may assume that k0=1. If (1.1) has an eventually positive solution x=x(t) for t≥t0. By (1.1) and conditions (A) and (B), we have that (4.1) and (4.2) hold. By Lemma 3.4, there exists a T≥t0 such that either (a) x′′(t)>0, x′(t)<0 for t≥T or (b) x′′(t)>0, x′(t)>0 for t≥T.Suppose that (a) holds. Note thatbk[0]≤1 and for tj≥T and each l=0,1,2,…, x(t) is decreasing on (tj+l,tj+l+1]; we have for t∈(tj,tj+1](4.12)x(t)<x(tj+)≤bj[0]x(tj)≤x(tj).Similarly, fort∈(tj+1,tj+2], we have(4.13)x(t)<x(tj+1+)≤bj+1[0]x(tj+1)≤x(tj+1)≤x(tj).By induction, for eachl=0,1,2,…, we have(4.14)x(t)<x(tj+l)≤⋯≤x(tj+1)≤x(tj),t∈(tj+l,tj+l+1]so thatx(t) is decreasing 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≥t0, such that for t≥T1,x(t)>r/2>0. Since φ′(x)≥0, then φ(x(t))≥φ(r/2). Let m(t)=r(t)x′′(t) for t≥T1. Then By (4.1) and (4.2), we have that (4.6) and (4.7) hold. From (4.6), (4.7), and Lemma 3.1, we get for t≥T1,(4.15)m(+∞)≤m(t)∏t<tk<+∞bk[2]-φ(r2)∫t+∞∏s<tk<∞bk[2]p(s)ds.That is,(4.16)0≤limt→+∞r(t)x′′(t)≤r(t)x′′(t)∏t<tk<+∞bk[2]-φ(r2)∫t+∞∏s<tk<∞bk[2]p(s)ds.
It is easy to see from (4.16) that the following inequality holds:(4.17)x′′(t)≥φ(r/2)r(t)∫t+∞∏t<tk<s1bk[2]p(s)ds,t≥T1.
Note that ak[1]≥1; it follows from integrating (4.17) from t0 to t and by using the condition (B) that(4.18)x′(t)-x′(t0+)≥x′(t)-x′(t0+)+∑t0<tk<t(ak[1]-1)x′(tk)≥x′(t)-x′(t0+)+∑t0<tk<t[x′(tk+)-x′(tk)]≥φ(r2)∫t0t1r(s)(∫s+∞∏s<tk<u1bk[2]p(u)du)ds.
It is easy to see from (2.4) and (4.18) that x′(t)>0 for sufficiently large t. This is contrary to x′(t)<0 for t≥T1. Thus r=0, that is, limt→+∞x(t)=0.Suppose (b) holds. Without loss of generality, we may assume thatT=t0. Then we see that x′(t)>0,t≥t0. Since x(t) is nondecreasing on (tk,tk+1], for t∈(t0,t1], we have(4.19)x(t)≥x(t0+).In particular,(4.20)x(t1)≥x(t0+).Similarly, fort∈(t1,t2], we have(4.21)x(t)≥x(t1+)≥a1[0]x(t1)≥a1[0]x(t0+).By induction, we know that(4.22)x(t)≥∏t0<tk<tak[0]x(t0+),t>t0.
That is, x(t)≥∏t0<tk<tak[0]x(t0+) for t>t0. Note that bk[0]≤1 and ∏t0≤tk<+∞ak[0]≥σ>0. From the condition (B), we have x(t)≥σx(t0+). Since φ′(x)≥0, we have φ(x(t))≥φ(σx(t0+)). Let m(t)=r(t)x′′(t); by (4.1) and (4.2), we have, for t≥t0, that(4.23)m′(t)≤-φ(σx(t0+))p(t),t≠tk,m(tk+)≤bk[2]m(tk),tk>t0.
Similar to the proof of (4.17), we obtain(4.24)x′′(t)≥φ(σx(t0+))r(t)∫t+∞∏t<tk<s1bk[2]p(s)ds,t≥t0.
Let s(t)=-x′(t) for t≥t0. Then s(t)≤0. By (4.24) and the condition (B), and noting that ak[1]≥1, we have for t≥t0,(4.25)s′(t)≤-φ(σx(t0+))r(t)∫t+∞∏t<tk<s1bk[2]p(s)ds,t≠tk,s(tk+)≤ak[1]s(tk)≤s(tk),tk≥t0.By Lemma3.1, we get(4.26)0≤s(+∞)≤s(t)-φ(σx(t0+))∫t+∞1r(s)(∫s+∞∏s<tk<u1bk[2]p(u)du)ds.It follows that(4.27)0⩾x′(t)+φ(σx(t0+))∫t+∞1r(s)(∫s+∞∏s<tk<u1bk[2]p(u)du)ds.
In view of (4.27), we have, for t≥t0,(4.28)x′(t)≤-φ(σx(t0+))∫t+∞1r(s)(∫s+∞∏s<tk<u1bk[2]p(u)du)ds.
It is easy to see from (2.4) and (4.28) that x′(t)<0. This is contrary to x′(t)>0 for t≥t0. Thus in case (b) x(t) must be oscillatory. The proof of Theorem 2.2 is complete.We now give the proof of Theorem2.3. Without loss of generality, we may assume that k0=1. If (1.1) has an eventually positive solution, x=x(t) for t≥t0. By Lemma 3.4, there exists a T≥t0 such that either (a) x′′(t)>0, x′(t)<0, t≥T or (b) x′′(t)>0, x′(t)>0, t≥T holds.Suppose that (a) holds. Note thatbk[0]≤1, since for tj≥T and each l=0,1,2,…,x(t) is decreasing on (tj+l,tj+l+1]; then for t∈(tj,tj+1], we have(4.29)x(t)<x(tj+)≤bj[0]x(tj)≤x(tj).Similarly, fort∈(tj,tj+1], we have(4.30)x(t)<x(tj+1+)≤bj+1[0]x(tj+1)≤x(tj+1)≤x(tj).By induction, for anyt∈(tj+l,tj+l+1] for l=0,1,2,…, we have(4.31)x(t)<x(tj+l)≤⋯≤x(tj+1)≤x(tj).Sox(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, then φ(x(t))≥φ(r/2). By (1.1) and condition (A), we have for t≥T1(4.32)(r(t)x′′(t))′=-f(t,x)≤-p(t)φ(x(t))≤-φ(r2)p(t)<0,t≠tk,
From condition (B), and noting that bk[2]≤1, we have(4.33)r(tk+)x′′(tk+)≤bk[2]r(tk)x′′(tk)≤r(tk)x′′(tk),tk≥T1.
Let Φ(t)=r(t)x′′(t). Then Φ(t)>0 for t≥T1. By (4.32) and (4.33), we have for t≥T1, that(4.34)Φ′(t)≤-φ(r2)p(t),t≠tk,(4.35)Φ(tk+)≤Φ(tk),tk≥T1.
From (4.34), (4.35), and Lemma 3.1, we get, for t≥T1, that(4.36)Φ(t)≤Φ(T1+)-φ(r2)∫T1tp(s)ds,
It is easy to see from (2.6) and (4.36) 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 (b) holds, letΨ(t)=(r(t)x′′(t)/φ(x(t))) for t≥T. We see that Ψ(t)>0 for t≥T. By (1.1) and the condition (A), we get for t≥T(4.37)Ψ′(t)=-f(t,x(t))φ(x(t))-r(t)x′′(t)φ′(x(t))x′(t)φ2(x(t))≤-f(t,x(t))φ(x(t))≤-p(t),t≠tk.
From the conditions (A) and (B), we know that(4.38)Ψ(tk+)=r(tk)x′′(tk+)φ(x(tk+))≤r(tk)bk[2]x′′(tk)φ(ak[0]x(tk))≤bk[2]φ(ak[0])r(tk)x′′(tk)φ(x(tk))≤Ψ(tk),tk≥T.
From (4.37), (4.38), and Lemma 3.1, we get for t≥T(4.39)Ψ(t)≤Ψ(T+)-∫Ttp(s)ds.
It is easy to see from (2.6) and (4.39) that Ψ(t)<0 for sufficiently large t. This is contrary to Ψ(t)>0 for t≥T. Thus in case (b) x(t) must be oscillatory. The proof of Theorem 2.3 is complete.Finally, we give the proof of Theorem2.4. Without loss of generality, we may assume that k0=1. If (1.1) has an eventually positive solution, x=x(t) for t≥t0. By Lemma 3.4, there exists a T≥t0 such that either (a) x′′(t)>0, x′(t)<0, t≥T or (b) x′′(t)>0, x′(t)>0, t≥T holds.Suppose that (a) holds. We may easily see that the conditions (H2), (H3) of Lemma 3.5 are satisfied. Furthermore, since x′(t)<0, t≥T, then there exists some ti≥T, such that for t∈(ti,ti+1](4.40)x(t)≤x(ti+).In particular,(4.41)x(ti+1)≤x(ti+).Similarly, we have fort∈(ti+1,ti+2](4.42)x(t)≤x(ti+1+)≤bi+1[0]x(ti+1)≤bi+1[0]x(ti+).In particular,(4.43)x(ti+2)≤bi+1[0]x(ti+).By induction, we obtain for anytk>ti(4.44)x(tk)≤∏ti<tj<tkbj[0]x(ti+).
Since {∏k=1nbk[0]} is bounded and (4.44) holds, we know that {x(tk)} is bounded. Thus there exists M1>0, such that |x(tk)|≤M1. It follows from the condition (B) that(4.45)|x(tk+)-x(tk)|≤max{|ak[0]-1|,|bk[0]-1|}|x(tk)|≤M1max{|ak[0]-1|,|bk[0]-1|}.
By (4.45), we know that ∑k=1+∞[x(tk+)-x(tk)] is convergent. Therefore, the condition (H4) of Lemma 3.5 is also satisfied. By Lemma 3.5, 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 have φ(x(t))≥φ(r/2). Since (r(t)x′′(t))′<0, t≥T1, there exists some ti≥T1 such that for t∈(ti,ti+1](4.46)r(t)x′′(t)≤r(ti)x′′(ti+).In particular,(4.47)r(ti+1)x′′(ti+1)≤r(ti)x′′(ti+).Similarly, we have fort∈(ti+1,ti+2](4.48)r(t)x′′(t)≤r(ti+1)x′′(ti+1+)≤bi+1[2]r(ti+1)x′′(ti+1)≤bi+1[2]r(ti)x′′(ti+).In particular,(4.49)r(ti+2)x′′(ti+2)≤bi+1[2]r(ti)x′′(ti+).By induction, we obtain for anytk>ti(4.50)r(tk)x′′(tk)≤∏ti<tj<tkbj[2]r(ti)x′′(ti+).
By bk[2]≤ak[0] and the condition (B), we know that {∏k=1nbk[2]} is bounded, and from (4.50), we see that {r(tk)x′′(tk)} is bounded. There then exists M2>0 such that |r(tk)x′′(tk)|≤M2. Therefore, we have(4.51)|(bk[2]-1)r(tk)x′′(tk)|≤M2|bk[2]-1|.
By (1.1) and the condition (A), we have that (4.1) holds. Integrating (4.1) from T1 to t, it follows from (4.51) and φ(x(t))≥φ(r/2) for t≥T1 that(4.52)r(t)x′′(t)-r(T1)x′′(T1+)≤∑T1<tk<tr(tk)[x′′(tk+)-x′′(tk)]-∫T1tp(s)φ(x(s))ds≤∑T1<tk<tr(tk)[x′′(tk+)-x′′(tk)]-φ(r2)∫T1tp(s)ds≤∑T1<tk<t(bk[2]-1)r(tk)x′′(tk)-φ(r2)∫T1tp(s)ds≤∑T1<tk<t|(bk[2]-1)r(tk)x′′(tk)|-φ(r2)∫T1tp(s)ds≤∑T1<tk<tM2|bk[2]-1|-φ(r2)∫T1tp(s)ds.
Note that ∑k=1+∞|bk[2]-1| is convergent. Thus it is easy to see from (2.8) and (4.52) that x′′(t)<0 for sufficiently large t. This is contrary to x′′(t)>0 for t≥T. Thus r=0, that is, limt→+∞x(t)=0.Suppose that (b) holds. LetΨ(t)=(r(t)x′′(t)/φ(x(t))) for t≥T. We see that Ψ(t)>0 for t≥T. Similar to the proof of (4.39), we also obtain(4.53)Ψ(t)≤Ψ(T+)-∫Ttp(s)ds.
It is easy to see from (2.8) and (4.53) that Ψ(t)<0 for sufficiently large t. This is contrary to Ψ(t)>0 for t≥T. Thus in case (b) x(t) must be oscillatory. The proof of Theorem 2.4 is complete.



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*Source: 562634-2010-03-23.xml*