# Comparison Criteria for Nonlinear Functional Dynamic Equations of Higher Order

Authors: Taher S. Hassan; Said R. Grace
Journal: Discrete Dynamics in Nature and Society
 (2016)

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## Abstract

We will consider the higher order functional dynamic equations with mixed nonlinearities of the formxnt+∑j=0Npjtϕγjxφjt=0, on an above-unbounded time scale T, where n≥2, xi(t)≔ri(t)ϕαixi-1Δ(t),i=1,…,n-1,   with  x0=x,ϕβ(u)≔uβsgn⁡u, and α[i,j]≔αi⋯αj. The function φi:T→T is a rd-continuous function such that limt→∞φi(t)=∞ for j=0,1,…,N. The results extend and improve some known results in the literature on higher order nonlinear dynamic equations.


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## Body

## 1. Introduction

In this paper, we consider comparison criteria for higher order nonlinear dynamic equation with mixed nonlinearities of the form(1)xnt+∑j=0Npjtϕγjxφjt=0,on an above-unbounded time scale T, where(i)
n≥2 is an integer, and xi(t)≔ri(t)ϕαixi-1Δ(t),i=1,2,…,n-1, t∈T, with rn=1, αn=1, and x0=x;(ii)
ϕβ(u)≔uβsgn⁡u for β>0.Without loss of generality we assume t0∈T. For A⊂T and B⊂R, we denote by Crd(A,B) the space of right-dense continuous functions from A to B and by Crd1(A,B) the set of functions in Crd(A,B) with right-dense continuous Δ-derivatives. Throughout this paper we make the following assumptions:(iii)
αi,γj>0,i=1,2,…,n-1 and j=0,1,…,N, are constants and ri∈Crdt0,∞T,(0,∞) for i=1,2,…,n-1, such that(2)∫t0∞ri-1/αisΔs=∞,i=1,2,…,n-1;(3)γj>γ0,j=1,2,…,l;γj<γ0,j=l+1,l+2,…,N.(iv)
pj∈Crdt0,∞T,R+ such that pj≢0, j=0,1,…,N on t0,∞T.(v)
φj∈Crd(T,T) rd-continuous function such that limt→∞φj(t)=∞, j=0,1,…,N, and we let φ(t)≔inf⁡φ0(t),φ1(t),…,φN(t) be a nondecreasing function on [t0,∞)T.Recall that the knowledge and understanding of time scales and time scale notation are assumed. For an excellent introduction to the calculus on time scales, see [1–3]. By a solution of (1) we mean a nontrivial real-valued function x∈Crd1[Tx,∞)T for some Tx≥t0 such that x[i]∈Crd1[Tx,∞)T,i=1,2,…,n-1, and x(t) satisfies (1) on [Tx,∞)T, where Crd is the space of right-dense continuous functions. An extendable solution x of (1) is said to be oscillatory if it is neither eventually positive nor eventually negative. Otherwise it is said to be nonoscillatory.In the last few years, there has been an increasing interest in obtaining sufficient conditions for the oscillation/nonoscillation of solutions of different classes of dynamic equations; we refer the reader to [4–13] and the references cited therein. Special cases of (1) have been studied by many authors. When αi=ri=N=1 and φit=t for i=1,2,…,n,pt=q0t=0, and q1t≥0, Grace et al. [14] established some oscillation criteria for higher order nonlinear dynamic equation of the form (4)xΔnt+ptxσtγ=0,where γ is the ratio of positive odd integers. In paper by Grace [15], some new criteria for the oscillation of the even order dynamic equation (5)rtxΔn-1tαΔ+ptxσtγ=0,where α and γ are the ratios of positive odd integers, were given. Recently, Hassan and Kong [16] obtained asymptotics and oscillation criteria for the nth-order half-linear dynamic equation with deviating argument (6)xn-1Δt+ptϕα1,n-1xgt=0,and Grace and Hassan [17] establish oscillation criteria for more general higher order dynamic equation (7)xnt+ptϕγxσφt=0.The purpose of this paper is to derive comparison criteria for higher order nonlinear dynamic equation with mixed nonlinearities (1).


## 2. Preliminaries

We will employ the following lemmas. Consider the inequality(8)rtϕαxΔtΔ+Qtϕγxφt≤0,where r and Q are positive real-valued, rd-continuous functions on T and r satisfies condition (2), φ:T→T is a rd-continuous function and φ(t)→∞ as t→∞, and α and γ are positive real numbers.Now, we present the following lemma.Lemma 1.
If inequality (8) has an eventually positive solution, then the equation(9)rtϕαxΔtΔ+Qtϕγxφt=0has also an eventually positive solution.Proof.
Letx(t) be an eventually positive solution of inequality (8). It is easy to see that xΔ>0 eventually. Let t0 be sufficiently large so that x(t)>0, xφ(t)>0, and y(t)≔r(t)ϕαxΔ(t) for t∈t0,∞T. Then, in view of(10)xt=xt0+∫t0tϕα-1ysrsΔs, there is t1≥t0 such that φ(t)≥t0, for t≥t1. Inequality (8) becomes(11)yΔt+Qtϕγxt0+∫t0φtϕα-1ysrsΔs≤0.Integrating (11) from t to v≥t≥t1 and letting v→∞, we have (12)yt≥Gt,yt,fort∈t1,∞T,where (13)Gt,yt≔∫t∞Qvϕγxt0+∫t0φvϕα-1ysrsΔsΔv.Now, we define a sequence of successive approximations wj(t) as follows:(14)w0t≔yt,wj+1t≔Gt,wjt,j=0,1,2,….It is easy to show that (15)0<wjt≤yt,wj+1t≤wjt,j=0,1,2,….Then, the sequence wj(t) is nonincreasing and bounded for each t≥t1. This means that we may define w(t)≔limj→∞wj(t)≥0. Since (16)0≤wt≤wjt≤yt,∀j≥0,we find that (17)∫t1twjsΔs≤∫t1tysΔs.By Lebesgue’s dominated convergence theorem on time scale, one can easily find (18)wt=Gt,wt.Therefore(19)wΔt=-Qtϕγxt0+∫t0φtϕα-1wsrsΔs=-Qtϕγmφt,where (20)mt≔xt0+∫t0tϕα-1wsrsΔs.Then (21)mt>0,rtϕαmΔt=wt,fort≥t1. Equation (19) then gives (22)rtϕαmΔtΔ+Qtϕγmφt=0.Hence (9) has a positive solution m(t). This completes the proof.The second one is cited from [18, 19].Lemma 2.
Assume that (3) holds. Then there exists an N-tuple (η1,η2,…,ηN) with ηj>0 satisfying(23)∑j=1Nγjηj=γ0,∑j=1Nηj=1.The next lemma is cited from [17] and improves the well-known lemma of Kiguradze.Lemma 3.
Assume that (2) holds. If (1) has an eventually positive solution x, then there exists an integer m∈0,n with m+n being odd such that(24)m≥1impliesxk>0fork=0,1,…,m-1,eventually, and(25)m≤nimplies-1m+kxk>0fork=m,…,n,eventually.


## 3. Main Results

In the following main theorem, we will use the following notations:α[h,k]≔αh⋯αk for 1≤h≤k≤n-1 and α[h,k]=1 for h>k and, for any u,v∈T, define Rjv,u, Pj(t), and P¯j(t), j=0,…,n-1, by the following recurrence formulas:(26) R j v , u ≔ ∫ u v R j - 1 s , u r m - j + 1 s Δ s 1 / α m - j + 1 , j = 1 , … , m , 1 , j = 0 ; P j t ≔ 1 r n - j t  ∫ t ∞ P j - 1 s Δ s 1 / α n - j , j = 1 , … , n - 1 , p t , j = 0 ; P ¯ j t ≔ 1 r n - j t  ∫ t ∞ P ¯ j - 1 s Δ s 1 / α n - j , j = 1 , … , n - 1 , p ¯ t , j = 0 ,with pt≔p0t+∏j=1Npj(t)/ηjηj and p¯t≔∑j=0Npj(t), provided the improper integrals involved are convergent.Theorem 4.
Assume that for sufficiently largeT∈[t0,∞)T the first-order dynamic equation(27)zΔt+Kmtϕγ0/α1,nzφt=0,forφt∈T,∞T,is oscillatory, where (28)Kmt≔Pn-m-1tRmφt,Tγ0/αm+1,n,for every number m∈{1,…,n-1} with m+n being odd. Then(1)
ifn is even, every solution of (1) is oscillatory,(2)
ifn is odd and(29)∫T∞P¯n-1tΔt=∞,then every solution of (1) either is oscillatory or tends to zero eventually.Proof.
Assume that (1) has a nonoscillatory solution x on [t0,∞)T. Then, without loss of generality, x(t)>0 and xφjt>0, for t∈[t0,∞)T and j=0,1,…,N. By Lemma 3, there exists an integer m, 0≤m<n, with n+m being odd such that (24) and (25) hold for t≥T1∈[t0,∞)T.
(I) Whenm≥1, from (1), we get(30)-xnt=∑j=0Npjtϕγjxφjt≥∑j=0Npjtϕγjxφt=ϕγ0xφt∑j=0Npjtxφtγj-γ0.From (23), we have (31)∑j=1Nγjηj-γ0∑j=1Nηj=0.Using the arithmetic-geometric mean inequality (see [20, Page 17]), we have (32)∑j=1Nηjvj≥∏j=1Nvjηj,foranyvj≥0,j=1,…,N. Then for t≥T1, (33)∑j=0Npjtxφtγj-γ0=p0t+∑j=1Nηjpjtηjxφtγj-γ0≥p0t+∏j=1Npjtηjηjxφtηjγj-γ0=p0t+∏j=1Npjtηjηj=pt.This together with (30) shows that (34)-xnt≥ptϕγ0xφt,fort≥T1.Integrating above inequality from t≥T1 to v∈[t,∞)T and then using the fact that x is strictly increasing and φ is nondecreasing, we get (35)-xn-1v+xn-1t≥∫tvpsϕγ0xφsΔs≥ϕγ0xφt∫tvpsΔs,and by (25) we see that x[n-1](v)>0. Hence by taking limits as v→∞ we have (36)xn-1t≥ϕγ0xφt∫t∞psΔs,which implies(37)xn-2tΔ≥ϕαn-1-1ϕγ0xφt1rn-1t∫t∞psΔs1/αn-1=ϕαn-1,n-1ϕγ0xφtP1t.Integrating above inequality (37) from t≥T1 to v∈[t,∞)T and letting v→∞, we get (38)-xn-2t≥ϕαn-1,n-1ϕγ0xφt∫t∞P1sΔs,-xn-3tΔ≥ϕαn-2,n-1ϕγ0xφt1rn-2t∫t∞P1sΔs1/αn-2=ϕαn-2,n-1ϕγ0xφtP2t. Continuing this process (n-m-3) times, we find(39)-xmtΔ≥ϕαm+1,n-1ϕγ0xφtPn-m-1t,fort≥T1.Also, from (24) and (25), we get (40)xm-1t=xm-1T1+∫T1tϕαm-1xms1rms1/αmΔs≥ϕαm-1xmt∫T1t1rms1/αmΔs=ϕαm-1xmtR1t,T1. It follows that (41)xm-2tΔ≥ϕαm-1,m-1xmtR1t,T1rm-1t1/αm-1. Then for t∈[T1,∞)T,(42)xm-2t≥xm-2t-xm-2T1≥∫T1tϕαm-1,m-1xmsR1s,T1rm-1s1/αm-1Δs≥ϕαm-1,m-1xmt∫T1tR1s,T1rm-1s1/αm-1Δs=ϕαm-1,m-1xmtR2t,T1.Analogously, we have(43)xt≥ϕα1,m-1xmtRmt,T1.Then for φ(t)∈[T1,∞)T(44)xφt≥ϕα1,m-1xmφtRmφt,T1.From (39) and (44), we get (45)-xmtΔ≥Pn-m-1tRmφt,T1γ0/αm+1,nϕγ0/α1,nxmφt=Kmtϕγ0/α1,nxmφt.Let zt≔xmt>0; we get (46)-zΔt≥Kmtϕγ0/α1,nzφt,or (47)zΔt+Kmtϕγ0/α1,nzφt≤0.In view of Corollary 2.3.5 in [21], there exists a positive solution of (27) which contradicts the assumption of the theorem.
(II) Whenm=0 (in this case n is odd), therefore(48)-1kxk>0fork=0,…,n.Since xΔ<0 eventually, then limt→∞x(t)=l≥0. Assume that l>0. Then for sufficiently large T2∈[T1,∞)T, we have xφjt≥l for t≥T2 and j=0,1,…,N. It follows that (49)ϕγjxφjt≥lγj≥Lfort∈T2,∞T,where L≔inf0≤j≤Nlγj>0. Then from (1), we have (50)-xnt=∑j=0Npjtϕγjxφjt≥L∑j=0Npjt=Lp¯t.Integrating from t to v∈[t,∞)T, we get(51)-xn-1v+xn-1t=L∫tvp¯sΔs.And by (48) we see that x[n-1](v)>0. Hence by taking limits as v→∞ we have(52)xn-1t≥L∫t∞p¯sΔs,which implies (53)xn-2tΔ≥L1/αn-11rn-1t∫t∞p¯sΔs1/αn-1=L1/αn-1P¯1t.Again integrating above inequality from t to v∈[t,∞)T and then taking v→∞, we get (54)-xn-2t≥L1/αn-1∫t∞P¯1sΔs,which implies (55)-xn-3tΔ≥L1/αn-2,n-11rn-2t∫t∞P¯1sΔs1/αn-2=L1/αn-2,n-1P¯2t.Continuing this process (n-3) times, we find (56)-xtΔ≥L1/α1,n-1P¯n-1tfort∈T2,∞T.Integrating above inequality T2 to t∈[T2,∞)T, we get (57)-xt+xT2≥L1/α1,n-1∫T2tP¯n-1sΔs.Hence by (29), we have limt→∞x(t)=-∞, which contradicts the fact that x>0 eventually. This shows that limt→∞x(t)=0. This completes the proof.Theorem 5.
Assume that for sufficiently largeT∈[t0,∞)T the second-order dynamic equation(58)rmtϕαmzΔtΔ+Qmtϕαmγ0/α1,nzφt=0,forφt∈T,∞T,is oscillatory, where(59)Qmt≔Pn-m-1tRmφt,T1γ0/αm+1,nR1φt,T1αmγ0/α1,n,for every integer number m∈{1,…,n-1} with m+n being odd. Then(1)
ifn is even, then every solution of (1) is oscillatory,(2)
ifn is odd and (29) holds, then every solution of (1) either is oscillatory or tends to zero eventually.Proof.
Assume that (1) has a nonoscillatory solution x on [t0,∞)T. Then, without loss of generality, x(t)>0 and xφjt>0,j=0,1,2,…,N on [t0,∞)T. By Lemma 3, there exists an integer m, 0≤m<n, with n+m being odd such that (24) and (25) hold for t≥T1∈[t0,∞)T.
(I) Whenm≥1, as seen in the proof of Theorem 4, we obtain, for t∈[T1,∞)T,(60)-xmtΔ≥ϕαm+1,n-1ϕγ0xφtPn-m-1t,(61)xm-1t≥ϕαm-1xmtR1t,T1.Hence, we have(62)xm-1tR1t,T1Δ≤0,fort∈T1,∞T.Since for t∈(T1,∞)T, (63)xm-1t=xm-1tR1t,T1R1t,T1,we have(64)xm-2tΔ=ϕαm-1-1xm-1tR1t,T1R1t,T1rm-1t1/αm-1.It follows from (62) that we have, for t∈T1,∞T,(65)xm-2t≥xm-2t-xm-2T1=∫T1tϕαm-1-1xm-1sR1s,T1R1s,T1rm-1s1/αm-1Δs≥ϕαm-1-1xm-1tR1t,T1∫T1tR1s,T1rm-1s1/αm-1Δs=ϕαm-1-1xm-1tR1t,T1R2t,T1.Continuing this process, we have (66)xt≥ϕα1,m-1-1xm-1tR1t,T1Rmt,T1.Then for φ(t)∈(T1,∞)T(67)xφt≥ϕα1,m-1-1xm-1φtR1φt,T1Rmφt,T1.From (60) and (67), we get(68)-xmtΔ≥ϕαmγ0/α1,nxm-1φtR1φt,T1Pn-m-1tRmφt,T1γ0/αm+1,n=Qmtϕαmγ0/α1,nxm-1φt.Set zt≔xm-1t>0; we have, for t≥T1, (69)-rmtϕαmzΔtΔ≥Qmtϕαmγ0/α1,nzφt,or(70)rmtϕαmzΔtΔ+Qmtϕαmγ0/α1,nzφt≤0.In view of Lemma 1, there exists a positive solution of (58) which contradicts the assumption of the theorem.
(II) Whenm=0, as shown in the proof of Theorem 4, we show that if (29) holds, then limt→∞x(t)=0. This completes the proof.Remark 6.
The conclusion of Theorems4 and 5 remains intact if assumption (29) is replaced by one of the following conditions:(71)∫T∞P¯0tΔt=∞,∫T∞P¯1tΔt=∞,⋮or∫T∞P¯n-2tΔt=∞.Theorem 7.
Assume that(72)∫t0∞P¯2tΔt=∞.And for sufficiently large T∈[t0,∞)T, the first-order dynamic equation (73)zΔt+Kn-1tϕγ0/α1,nzφt=0,forφt∈T,∞T,is oscillatory, where(74)Kn-1t≔ptRn-1φt,Tγ0.Then(1)
ifn is even, every solution of (1) is oscillatory,(2)
ifn is odd, then every solution of (1) either is oscillatory or tends to zero eventually.Theorem 8.
Assume that (72) holds and, for sufficiently large T∈[t0,∞)T, the second-order dynamic equation (75)rn-1tϕαn-1zΔtΔ+Qn-1tϕγ0/α1,n-2zφt=0,forφt∈T,∞T,is oscillatory, where(76)Qn-1t≔ptRn-1φt,T1γ0R1φt,T1γ0/α1,n-2.
Then(i)
ifn is even, every solution of (1) is oscillatory,(ii)
ifn is odd, every solution of (1) either is oscillatory or tends to zero eventually.Proof of Theorems7 and 8.
Assume that (1) has a nonoscillatory solution x on [t0,∞)T. Then, without loss of generality, it is sufficiently large, such that x(t)>0 and xφjt>0, for t∈[t0,∞)T and j=0,1,…,N. By Lemma 3, there exists an integer m, 0≤m<n, with n+m being odd such that (24) and (25) hold for t≥T1∈[t0,∞)T.
(I) Whenm≥1, we claim that (72) implies that m=n-1. In fact, if 1≤m≤n-3, then for t≥T1(77)xnt<0,xn-1t>0,xn-2t<0,xn-3t>0.Since xΔ(t)>0 on [T1,∞)T, then xt>x(T1)≔c1>0 for t≥T1. Then there exists T2∈[T1,∞)T such that xφjt≥l for t≥T2 and j=0,1,…,N. It follows that (78)ϕγjxφjt≥lγj≥L,fort∈T2,∞T,where L≔inf0≤j≤Nlγj>0. Then from (1), we have (79)-xnt=∑j=0Npjtϕγjxφjt≥L∑j=0Npjt=Lp¯t.Integrating from t to v∈[t,∞)T, we get (80)-xn-1v+xn-1t=L∫tvp¯sΔs.And by (48) we see that x[n-1](v)>0. Hence by taking limits as v→∞ we have (81)xn-1t≥L∫t∞p¯sΔs,which implies(82)xn-2tΔ≥L1/αn-11rn-1t∫t∞p¯sΔs1/αn-1=L1/αn-1P¯1t.Integrating above inequality from t to v∈[t,∞)T and then taking v→∞, we get (83)-xn-2t≥L1/αn-1∫t∞P¯1sΔs,which implies (84)-xn-3tΔ≥L1/αn-2,n-11rn-2t∫t∞P¯1sΔs1/αn-2=L1/αn-2,n-1P¯2t.Again, integrating above inequality from T2 to t∈[T2,∞)T and noting that x[n-3]>0 eventually, we get (85)xn-3T2-xn-3t≥L1/αn-2,n-1∫T2tP¯2sΔs.Then by (72), we have limt→∞x[n-3](t)=-∞, which contradicts the fact that x[n-3]>0 on [T2,∞)T. This shows that if (72) holds, then m=n-1. The rest of proof of (I) is similar to proof (I) of Theorems 4 and 5, respectively, with m=n-1 and hence can be omitted.
(II) Whenm=0 (in this case n is odd), therefore (86)-1kxk>0fork=0,…,n.Since xΔ<0 eventually, then limt→∞x(t)=l≥0. Assume that l>0. Then for sufficiently large T2∈[T1,∞)T, we have xφjt≥l for t≥T2 and j=1,…,N. It follows that (87)ϕγjxφjt≥lγj≥Lfort∈T2,∞T,where L≔inf0≤j≤Nlγj>0. Then from (1), we have (88)-xnt=∑j=0Npjtϕγjxφjt≥L∑j=0Npjt=Lp¯t.Integrating from t to v∈[t,∞)T, we get (89)-xn-1v+xn-1t=L∫tvp¯sΔs.And using (48) we see that x[n-1](v)>0. Hence by taking limits as v→∞ we have (90)xn-1t≥L∫t∞p¯sΔs,which implies (91)xn-2tΔ≥L1/αn-11rn-1t∫t∞p¯sΔs1/αn-1=L1/αn-1P¯1t.Integrating above inequality from t to v∈[t,∞)T and then taking v→∞, we get (92)-xn-2t≥L1/αn-1∫t∞P¯1sΔs,which implies(93)-xn-3tΔ≥L1/αn-2,n-11rn-2t∫t∞P¯1sΔs1/αn-2=L1/αn-2,n-1P¯2t.Again, integrating above inequality from T2 to t∈[T2,∞)T and noting that x[n-3]>0 eventually, we get(94)xn-3T2-xn-3t≥L1/αn-2,n-1∫T2tP¯2sΔs.Then by (72), we have limt→∞x[n-3](t)=-∞, which contradicts the fact that x[n-3]>0 on [T2,∞)T. This shows that limt→∞x(t)=0. This completes the proof.Remark 9.
The conclusion of Theorems7 and 8 remains intact if assumption (72) is replaced by one of the following conditions:(95)∫T∞P¯0tΔt=∞or∫T∞P¯1tΔt=∞.Theorem 10.
Assume that(96)∫T∞P¯0tΔt=∞.Then(1)
ifn is even, every solution of (1) is oscillatory,(2)
ifn is odd, then every solution of (1) either is oscillatory or tends to zero eventually.Proof.
Assume that (1) has a nonoscillatory solution x on [t0,∞)T. Then, without loss of generality, it is sufficiently large, such that x(t)>0 and xφjt>0, for t∈[t0,∞)T and j=0,1,…,N. By Lemma 3, there exists an integer m, 0≤m<n, with n+m being odd such that (24) and (25) hold for t≥T1∈[t0,∞)T.
(I) Whenm≥1, this implies that x(t) is strictly increasing on [t1,∞)T. Then for sufficiently large T2∈[T1,∞)T, we have x(φj(t))≥l for t≥T2. It follows that (97)ϕγjxφjt≥lγj≥Lfort∈T2,∞T,where L≔inf0≤j≤Nlγj>0. Equation (1) becomes(98)-xn-1tΔ=∑j=0Npjtϕγjxφjt≥L∑j=0Npjt=Lp¯tfort∈t2,∞T.Replacing t by s in (98), integrating from t2 to t∈[t2,∞)T, we obtain (99)-xn-1t+xn-1t2≥L∫t2tp¯sΔs.Hence by (96), we have limt→∞xn-1t=-∞, which contradicts the fact that xn-1t>0 eventually.
(II) Whenm=0 (in this case n is odd), therefore (100)-1kxk>0fork=0,…,n.This implies that x(t) is strictly decreasing on [t1,∞)T. Then limt→∞x(t)=l≥0. Assume that l>0. Then for sufficiently large t2∈[t1,∞)T, we have x(φj(t))≥l for t≥t2. It follows that(101)ϕγjxφjt≥lγj≥Lfort∈t2,∞T,where L≔inf0≤j≤Nlγj>0. As is case (I), we get a contradiction with the fact that xn-1t>0 eventually. This shows that limt→∞x(t)=0. This completes the proof.



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