# Oscillatory Properties for Second-Order Impulsive Neutral Dynamic Equations with Positive and Negative Coefficients on Time Scales

Authors: Shuang Zhang; Qiaoluan Li
Journal: Journal of Mathematics
 (2021)

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## Abstract

We study oscillatory properties for second-order impulsive neutral dynamic equations with positive and negative coefficients on time scales. By using variable substitution, we obtain sufficient conditions for several dynamic equations to be oscillatory.


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## Body

## 1. Introduction

In this paper, we are concerned with the oscillation of the following second-order impulsive neutral dynamic equations with positive and negative coefficients:(1)atzΔtΔ+ptfxτt−qtfxζt=0,t∈JT\tk,ztk+=Mkztk,k∈N,zΔtk+=NkzΔtk,k∈N,where zt=xt+btxθt and JT:=t0,∞∩T,T is unbounded above time scale with tk∈T, 0≤t0<t1<t2<⋯<tk<⋯limk⟶∞tk=∞, and xtk+=limh⟶0+xtk+h and xΔtk+=limh⟶0+xΔtk+h, which represent right limits of xt and xΔt at t=tk in the sense of time scales. Furthermore, if tk is right-scattered, then xtk+=xtk and xΔtk+=xΔtk. xtk− and xΔtk− can be defined similarly. Throughout this paper, we assume that f is continuous on T, all the impulsive points tk are right-dense, and at, pt, and qt∈CrdT,ℝ+, where Crd denotes the set of rd-continuous functions. There exist positive constants αk, βk, ck, and dk such that αk≤Mkx/x≤βk and ck≤Nkx/x≤dk for x≠0 and k∈ℕ.During the past decades, the oscillation of impulsive differential equations and impulsive difference equations has been investigated by many authors [1–3]. In recent years, many researchers focus their attention on the oscillation of dynamic equations on time scales [4–9]. The theory of impulsive dynamic equations has received considerable attention [10]. Moreover, dynamic equations with positive and negative coefficients have been of great interest. Many results on the oscillatory properties for dynamic equations with positive and negative coefficients have been obtained [11, 12]. However, fewer papers are on the oscillation of impulsive neutral dynamic equations with positive and negative coefficients.For example, Huang and Feng [13] have considered the following equation:(2)xΔΔt+ft,xσt=0,JT≔t0,∞∩T,t≠tk,k=1,2,…,xtk+=gkxtk,k=1,2,…,xΔtk+=hkxΔtk,k=1,2,…,xt0+=xt0,k=1,2,…,xΔt0+=xΔt0,k=1,2,….By using Riccati transformation techniques, they obtained sufficient conditions for oscillation of all solutions.The rest of this paper is organized as follows: In Section2, we present some basic definitions and preliminary results. In Section 3, we state and prove several oscillatory results. Finally, two examples are given to illustrate our obtained results.


## 2. Preliminaries

In this section, we introduce definitions and preliminary results which are used in this paper. Throughout the paper, we define the intervala,b in T by(3)a,b≔t∈T:a≤t≤b.Open intervals and half-open intervals etc. are defined accordingly.Definition 1.
Fort∈T, we define the forward jump operator σ: T⟶T by σt:=ins∈T:s>t. If σt>t, then t is called right-scattered. Also, if t<supT and σt=t, then t is called right-dense.Definition 2.
A functionf: T⟶ℝ is rd-continuous if it is continuous at all right-dense points and its left-sided limit exists (and is finite) at a left-dense point. We denote the set of rd-continuous functions by CrdT,ℝ.Definition 3.
Assumef: T⟶ℝ is a function and let t∈Tk. Then, we define fΔt to be the number (provided it exists) with the property that given any ε>0, there is a neighborhood U of t (i.e., U=t−δ,t+δ∩T for some δ>0) such that(4)fσt−fs−fΔtσt−s≤εσt−s,for all s∈U. We call fΔt the delta derivative of f at t.Definition 4.
A functionx is a solution of (1), if it satisfies atzΔtΔ+ptfxτt−qtfxζt=0 a.e. on JT\tk, k=1,2,…, and for each k=1,2,…, x satisfies the impulsive conditions ztk+=Mkztk and zΔtk+=NkzΔtk, where zt=xt+btxθt.Definition 5.
A solutionx of (1) is oscillatory if it is neither eventually positive nor eventually negative; otherwise, it is called nonoscillatory. Equation (1) is called oscillatory if all solutions are oscillatory.Lemma 1.
(see [14]). Let the function m∈PC1T,ℝ satisfy the inequalities(5)mΔt≤ptmt+qt,t≠tk,mtk+≤dkmtk+bk,k=1,2,…,where p and q∈PC1T,ℝ and dk>0 and bk are constants; then(6)mt≤mt0∏t0<tk<tdkexp∫t0tpsΔs+∑t0<tk<t∏tk<tj<tdjexp∫tktpsΔsbk+∫t0t∏s<tk<tdkexp∫stpσΔσqsΔs,t≥t0,where PC=x:JT⟶ℝwhich  is  rd−continuous  except  attk,k=1,2,…,for  whichxtk−,xtk+,xΔtk−,andxΔtk+exist  withxtk−=xtkandxΔtk−=xΔtk.


## 3. Main Results

In this section, we first consider the dynamic equation(7)atzΔtΔ+ptfxτt−qtfxζt=0,ztk+=Mkztk,zΔtk+=ηkzΔtk,that is, the case ck=dk=ηk of equation (1).We assume the following conditions hold:(H1)
0≤bt<1(H2)
∑k=1∞∫tktk+11/atΔt∏i=1kηi=∞(H3)
f is nondecreasing, fu/u>0 for u≠0, and fuv≥fufv for uv>0(H4)
θt, τt, and ζt∈CrdT,T, θt≤t, limt⟶∞θt=∞, limt⟶∞τt=∞, and limt⟶∞ζt=∞(H5)
αk≥1, ∑k=1∞βk−1<∞, and 0<ηk≤1(H6)
θt≠tk, τt≠tk, and ζt≠tk for t≠tkTheorem 1.
Assume (H1)–(H6) are satisfied, if the inequality(8)yΔt+ptfyτtfSτt≤0,ytk+≤ηkytk,has no eventually positive solution, where(9)St=1−bt∫T1t1atΔt.Then, every bounded solution of (7) is oscillatory.Proof.
Assume thatxt is an eventually bounded positive solution of (7); then, there exists T∈T large enough and T≥t0 such that xt>0, xθt>0, xτt>0, and xζt>0 for all t∈T,∞. Let Ft=zt+∫t∞1/as∫TsqufxζuΔuΔs,t∈tk,tk+1; we find atFΔtΔ+ptfxτt=0, which implies that(10)atFΔtΔ=−ptfxτt<0.
Then,atFΔt is decreasing for all t∈tk,tk+1. We can claim that FΔt is ultimately greater than zero. If not, there must exist a m∈tj,tj+1, such that FΔm≤0.
For∀t∈m,tj+1, we have(11)atj+1FΔtj+1≤atFΔt≤amFΔm≤0,∫mtj+1FΔtΔt≤amFΔm∫mtj+11atΔt.
Making similar analysis on the intervaltk,tk+1, k=j+1,j+2, …, we obtain(12)∫tktk+1FΔtΔt≤ηk⋯ηj+1amFΔm∫tktk+11atΔt,t∈tk,tk+1.
Hence,(13)∫mtj+1FΔtΔt+∑k=j+1∞∫tktk+1FΔtΔt≤amFΔm∫mtj+11atΔt+∑k=j+1∞∫tktk+11atΔt∏i=j+1kηi.
Furthermore, we obtain(14)F∞≤amFΔm∫mtj+11atΔt+∑k=j+1∞∫tktk+11atΔt∏i=j+1kηi+Fm+∑k=j+1∞βk−1ztk.
By (H2) and (H5), we have Ft⟶−∞ as t⟶∞, which contradicts Ft>0. Therefore, FΔt is ultimately greater than zero. There exists T1≥T such that zΔt>FΔt>0, where t≥T1. By ztk+≥αkztk≥ztk, we have(15)zt≥∫T1tzΔsΔs>∫T1tFΔsΔs=∫T1tasFΔsasΔs≥atFΔt∫T1t1asΔs.
Letφt=atFΔt; then, xt≥zt1−bt≥φtSt. We see φt is a positive solution of (8), which contradicts the oscillation of (8). We complete the proof.We assume that the following conditions are satisfied:(H7)
bt is bounded(H8)
ζt∈CrdT,T is bijective, τt∈CrdT,T, τt≤ζt, ζ−1τt∈Crd1JT,T, limt⟶∞θt=∞, and limt⟶∞τt=∞(H9)
∫t0∞1/aη∫ζ−1τηηqsΔsΔη<∞(H10)
There existsL>0 and fu/u≥L for u≠0(H11)
pt−qζ−1τtζ−1τtΔ≥C>0Theorem 2.
Assume (H2) and (H5)–(H11) are satisfied; then, every bounded solution of (7) is oscillatory.Proof.
Assume thatxt is an eventually bounded positive solution of equation (7); then, there exists T∈T large enough and T≥t0 such that xt>0, xθt>0, xτt>0, and xζt>0 for all t∈T,∞. Let ut=zt−∫Tt1/aη∫ζ−1τηηqsfxζsΔsΔη,t∈tk,tk+1; we obtain that(16)atuΔtΔ=−pt−qζ−1τtζ−1τtΔfxτt≤−CLxτt<0.
Then,atuΔt is decreasing in tk,tk+1. We can obtain that uΔt is ultimately greater than zero. There exists T2≥T such that zΔt>uΔt>0 for all t∈T2,∞.
For∀t∈tk,tk+1 and t∈T2,∞, we have(17)∫T2tasuΔsΔΔs≤−CL∫T2txτsΔs.
Furthermore, we obtain(18)CL∫T2txτsΔs≤aT2uΔT2−atuΔt+∑T2<tk<tηk−1atkuΔtk≤aT2uΔT2<∞.
Therefore,xt and zt are integrable. However, we know zt is increasing in T2,∞ and nonintegrable from zΔt>0 and ztk+≥ztk, which contradicts the integrability of zt. We complete the proof.Next, we study the equation(19)atzΔtΔ+ptfxσt=0,ztk+=Mkztk,zΔtk+=ηkzΔtk,i.e., the case qt≡0, τt=σt, and ck=dk=ηk of (1).We assume the following conditions hold:(H12)
0≤bt<HtBk/Hθt, whereHt=∫t∞1/asΔs,Bk=1,tk∉θt,t,∏θt<tk<tαk,other(H13)
0<αk≤βk≤1, ηk≤1, and θt≠tk for t≠tk(H14)
∫t0t11/asΔs+η1∫t1t21/asΔs+η2η1∫t2t31/asΔs+⋯=∞(H15)
θt≤t and θt∈CrdT,T is increasing, limt⟶∞θt=∞, and σθt=θσt(H16)
There existsL>0 and fu/u≥L for u≠0; f is nondecreasing and fuv≥fufv for uv>0Theorem 3.
Assume (H12)–(H16) are satisfied. If(20)∫t0∞∏t0<tk<sαkηkpsf1−bσsHθσsHσsBkΔs=∞,then (19) is oscillatory.Proof.
Assume that (19) has a nonoscillatory solution xt, which is eventually positive; then, there exists T∈T large enough and T≥t0 such that xt>0 and xθt>0 for all t∈T,∞. We obtain that atzΔtΔ=−ptfxσt<0, where t∈tk,tk+1. Therefore, atzΔt is decreasing in tk,tk+1. We denote Iu=maxi:t0<ti<u.
Fortk<t≤s≤tk+1, we have(21)atk+1zΔtk+1≤aszΔs≤atzΔt.
Hence,(22)∫ttk+1zΔsΔs≤atzΔt∫ttk+11asΔs,s∈t,tk+1.
We can make similar analysis on the intervalstj,tj+1 and tIu,u, such that j=k+1,…,Iu−1. Thus,(23)zu−zt≤∫ttk+1zΔsΔs+∑j=k+1Iu−1∫tjtj+1zΔsΔs+∫tIuuzΔsΔs≤atzΔt∫ttk+11asΔs+ηk+1∫tk+1tk+21asΔs+⋯ηIu⋯ηk+1∫tIuu1asΔs.
By (H14), we obtain zΔt is ultimately greater than zero. Then, there exists T3≥T such that zΔt>0 for all t∈T3,∞, and(24)zt≥−atzΔt∫tu1asΔs.
Lettingu⟶∞, we have(25)zt≥−atzΔt∫t∞1asΔs=−atzΔtHt.
Therefore, we can claim thatzt/Ht is increasing in tk,tk+1.
Iftk∉θt,t, we obtain(26)ztHt≥zθtHθt.
Iftk∈θt,t, we obtain(27)ztHt≥∏θt<tk<tαkzθtHθt.
So(28)ztHt≥BkzθtHθt.
Therefore,(29)xt=zt−btxθt≥zt−btzθt≥zt1−btHθtHt1Bk.
Letωt=atzΔt/zt, t∈tk,tk+1; we have(30)ωΔt≤−ptfxσtzσt≤−ptLf1−bσtHθσtHσt1Bk,ωtk+≤ηkαkωtk.
By using Lemma1, we obtain for t>t0,(31)ωt≤ωt0∏t0<tk<tηkαk−L∫t0t∏s<tk<tηkαkpsf1−bσsHθσsHσs1BkΔs=∏t0<tk<tηkαkωt0−L∫t0t∏t0<tk<sαkηkpsf1−bσsHθσsHσs1BkΔs.
We get a contradiction ast⟶∞. We complete the proof.Forqt≡0 andαk=βk=1 in (1), it can be written as(32)atzΔtΔ+ptfxτt=0,ztk+=ztk,zΔtk+=NkzΔtk.We assume that the following conditions hold for (32):(H17)
θt, τt∈CrdT,T, θt≤t, τt≤t, limt⟶∞θt=∞, and limt⟶∞τt=∞(H18)
θt≠tk and τt≠tk for t≠tk(H19)
∫t0∞∏t0≤tk<sck1/asΔs=∞Lemma 2.
Assume (H19) is satisfied and xt is an eventually positive solution of equation (32); then, zΔt≥0 and zΔtk≥0, where t∈tk,tk+1.Proof.
Becausext is an eventually positive solution of equation (32), we obtain(33)atzΔtΔ=−ptfxτt<0,t∈tk,tk+1.
Then,atzΔt is decreasing in tk,tk+1. If there exists tj such that zΔtj=−l<0 for l>0 and we obtain for ∀t∈tj+n,tj+n+1, atzΔt≤−atjcjcj+1,…,cj+nl, then(34)zΔt≤−atjatl∏tj≤tk<tck,ztk+=ztk.
By Lemma1, we obtain for t>tj,(35)zt≤ztj−l∫tjtatjas∏tj≤tk<sckΔs.
We get a contradiction ast⟶∞. Then, zΔtk≥0.
For∀t∈tk−1,tk,(36)zΔt≥atkatzΔtk≥0.
We complete the proof.Theorem 4.
Assume (H1) and (H16)–(H19) are satisfied. If dk≤1 and(37)∫t0∞∏t0<tk<t1dkptf1−bτtKtΔt=∞,where(38)Kt=QτtQσt,Qt=∫Tt1asΔs,then (32) is oscillatory.Proof.
Suppose to the contrary thatxt is an eventually positive solution of (32); then, there exists T∈T large enough and T≥t0 such that xt>0, xθt>0, and xτt>0 for all t∈T,∞. Then, atzΔt is decreasing in tk,tk+1. According to atk+zΔtk+≤atkzΔtk, we obtain atzΔt is decreasing in T,∞.
For∀t∈tk,tk+1 and t≥τt≥T, we have(39)zσtzτt≤1Kt.It can be proved similar to Lemma 1 of [11] and so its proof is omitted here.
From Lemma2, we have zΔt≥0 and zΔtk≥0, where t≥T and t∈T. Let ωt=atzΔt/zt, t∈tk,tk+1. We have(40)ωΔt≤−ptfxτtzσt≤−ptLf1−bτtKt,ωtk+≤dkωtk.
Applying Lemma1, we obtain for t>t0,(41)ωt≤ωt0∏t0<tk<tdk−L∫t0t∏s<tk<tdkpsf1−bτsKsΔs=∏t0<tk<tdkωt0−L∫t0t∏t0<tk<s1dkpsf1−bτsKsΔs.
We get a contradiction ast⟶∞. This completes the proof of Theorem 4.


## 4. Examples

Example 1.
Consider the equation(42)1etzΔtΔ+ptxt−2−qtxt+2=0,t∈JT\2k,z2k+=Mkz2k,k∈N,zΔ2k+=kk+1zΔ2k,k∈N,where zt=xt+1/txt−2 and JT:=2,∞. Here, αk=1+1/k2+1, βk=1+1/k2, ηk=k/k+1, and ptSt−2=1. Obviously, (H1) and (H3)–(H6) are satisfied:(43)∑k=1∞∫tktk+11atΔt∏i=1kηi=∑k=1∞∫2k2k+2etΔt∏i=1kii+1=12∫24etΔt+13∫46etΔt+⋯>∫2∞Δt>∞.
So (H2) is satisfied. Taking account of Theorem 1 of [1], we know that(44)y′t+yt−2≤0,ytk+≤kk+1ytk,has no eventually positive solution. By Theorem 1, it is clear that every bounded solution of (42) is oscillatory.Example 2.
Consider the following equation:(45)t2etzΔtΔ+etxt+2−e−txt+4=0,t∈JT\2k,z2k+=Mkz2k,k∈N,zΔ2k+=kk+1zΔ2k,k∈N,where zt=xt+1/txt−2 and JT:=∪k=2∞k,k+1/2. Here, αk=1+1/k2+1, βk=1+1/k2, and ηk=k/k+1. Obviously, (H2), (H5)–(H8), and (H10) are satisfied:(46)∫t0∞1aη∫ζ−1τηηqsΔsΔη=∫2∞eηη2∫η−2ηe−sΔsΔη<∞,pt−qζ−1τtζ−1τtΔ=et−e2−t≥e2−1>0.
So (H9) and (H11) are satisfied. By Theorem 2, we see that every bounded solution of (45) is oscillatory.



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