Research article Special Issues

Oscillatory and asymptotic properties of higher-order quasilinear neutral differential equations

  • The objective of this paper is to study the oscillation criteria for odd-order neutral differential equations with several delays. We establish new oscillation criteria by using Riccati transformation. Our new criteria are interested in complementing and extending some results in the literature. An example is considered to illustrate our results.

    Citation: Clemente Cesarano, Osama Moaaz, Belgees Qaraad, Ali Muhib. Oscillatory and asymptotic properties of higher-order quasilinear neutral differential equations[J]. AIMS Mathematics, 2021, 6(10): 11124-11138. doi: 10.3934/math.2021646

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  • The objective of this paper is to study the oscillation criteria for odd-order neutral differential equations with several delays. We establish new oscillation criteria by using Riccati transformation. Our new criteria are interested in complementing and extending some results in the literature. An example is considered to illustrate our results.



    In this work, we consider the odd-order quasi-linear neutral differential equations of the form

    (r(t)[(x(t)+p(t)x(τ(t)))(n1)]α)+mκ=1qκ(t)xα(σκ(t))=0,  for   tt0, (1.1)

    where n3 is an odd integer, α is a ratio of positive odd integers and m is a positive integer. Throughout this work, we assume the following:

    (H1) r C1[t0,) and r(t)0, where

    t0r1/α(t)dt=;

    (H2) p, qκ C[t0,), p(t)[0,p0] such that p0 is a constant and qκ(t)>0;

    (H3) τ, σκ C[t0,), τ(t)t, σκ(t)t, limtτ(t)= and limtσκ(t)= for all κ=1,2,...,m.

    By a solution of (1.1), we mean xC([Tx,),R) with Txt0, which satisfies the properties

    (x+pxτ)C(n1)([Tx,),R)

    and

    r((x+pxτ)(n1))αC1([Tx,),R)

    and moreover satisfies (1.1) on [Tx,). We consider the nontrivial solutions of (1.1) existing on some half-line [Tx,) and satisfying the condition sup{|x(t)|:tt}>0 for any tTx. If there exists a t1t0 such that either x(t)>0 or x(t)<0 for all tt1, then x is said to be a nonoscillatory solution; otherwise, it is said to be an oscillatory solution.

    Delay differential equations as a subclass of functional differential equations take into account the dependence on the systems past history where the theory of delay differential equations has enhanced our understanding of the qualitative behavior of their solutions and it has benefited significantly and wide from it, where many applications showed in various fields as mathematical biology and epidemiology (for instance, transport phenomena, distributed networks, interaction of species) and other related fields, etc., see [1,2,3].

    Neutral delay differential equations are differential equations with delays, where the delays can appear in both the state variables and their time derivatives. There is considerable interest in studying of this type of equation because they are deemed to be adequate prescribing tool in modelling of the countless processes in all areas including problems concerning electric networks containing lossless transmission lines (as in high speed computers where such lines are used to interconnect switching circuits), in the study of vibrating masses attached to an elastic bar or in the solution of variational problems with time delays, or in the theory of automatic control and in neuro-mechanical systems in which inertia plays a major role, and in many areas of science as physical, biological and chemical, etc., see [4,5]. In addition, systems of delay differential equations were used to study stability properties of electrical power systems also, properties of delay differential equations were used in the study of singular fractional order differential equations, see [6,7] and the references cited therein.

    As a matter of fact, quasilinear (i.e., half-linear) (neutral) differential equations with deviating arguments (delayed or advanced arguments or mixed arguments) have numerous applications in physics and engineering (e.g., quasilinear (i.e., half-linear) differential equations arise in a variety of real world problems such as in the study of p-Laplace equations, porous medium problems, chemotaxis models, and so forth), see [8,9,10,11,12].

    For several years, an increasing interest in obtaining sufficient conditions for oscillatory and nonoscillatory behavior of different classes of differential equations has been observed, see [13,14,15,16,17,18] for second-order equations. While the development of the study of the second-order equations was in turn reflected on the even-order equations in the works [19,20,21,22,23,24,25,26,27,28]. The development of the study of the odd-order equations can also be traced through works [29,30,31,32,33,34], some of which are special cases of the studied equation.

    Many authors as Ladde and Zhang in [22,28] established a criterion for oscillatory behavior of the higher-order differential equation

    ((x(t)(n1))α)+q(t)xβ(τ(t))=0. (1.2)

    Grace [20] extended some new results to the equation

    (r(t)(x(t)(n1))α)+q(t)xβ(τ(t))=0, (1.3)

    under the assumptions that α is even,

    t0r1/α(t)dt= and  r(t)0.

    Agarwal et al. [19] studied Eq (1.3) under conditions

    t0r1/α(t)dt< and  t0q(t)dt=.

    Karpuz et al. [21] investigated the oscillatory behavior of linear neutral differential equations

    (x(t)+p(t)x(τ(t)))(n)+q(t)x(τ(t))=0,

    where n is an odd integer and 0p(t)<1.

    Li and Thandapani [31] established some oscillation criteria for certain higher-order neutral differential equation

    (x(t)+p(t)x(a+bt))(n)+q(t)x(c+dt)=0,

    with 0p(t)p0<.

    Yildiz et al. [27] examined the oscillation of odd-order neutral differential equation

    (x(t)+p(t)x(τ(t)))(n)+q(t)xα(τ(t))=0,

    where 0p(t)p1<1.

    In the present paper, we aim to improve the results in previous studies and present some new sufficient conditions which ensure that every solution of (1.1) oscillates or tends to zero.

    Here are some lemmas that we need during the next results.

    Lemma 2.1. [18, Lemma (2.3)] Let g(v)=CvDvα+1/α where C,D>0. Then g attains its maximum value on R at v=(αC/(α+1)D)α and

    maxvRg(v)=g(v)=αα(α+1)α+1Cα+1Dα. (2.1)

    Lemma 2.2. [34] Assume that c1,c2[0,) and γ>0. Then

    (c1+c2)γμ(cγ1+cγ2), (2.2)

    where

    μ:={1if γ12γ1if γ>1.

    Lemma 2.3. [35] Let fCn([t0,),(0,)). Assume that f(n)(t) is of fixed sign and not identically zero on [t0,) and that there exists a t1t0 such that f(n1)(t)f(n)(t)0 for all tt1. If limtf(t)0, then for every μ(0,1) there exists tμt1 such that

    f(t)μ(n1)!tn1|f(n1)(t)|for  ttμ.

    Through the rest of this paper, we will use the following definitions:

    z:=x+pxτ,
    η(t):=tt0r1/α(s)ds

    and

    σ(t)=min{σκ(t):κ=1,2,...,m}. (3.1)

    Lemma 3.1. Let x be a positive solution of (1.1). Then z(t)>0, (r((z)(n1))α)0 and there are two possible cases for derivatives of z:

    (I)z(t)>0,z(t)>0,z(n1)(t)>0,zn(t)0;(II)z(t)<0,z(t)>0,z(n1)(t)>0,zn(t)0.

    Proof. Assume that x is a positive solution of (1.1) on [t0,). Then, there exists t1t0 such that x(t)>0, x(σκ(t))>0 and x(τ(t))>0, for tt1. By the definition of z, it is easy to see that z(t)x(t)>0. Furthermore, from (1.1), we have (r((z)(n1))α)0. The rest of the proof is similar to proof of Lemma in [29]. Thus, the proof is complete.

    Lemma 3.2. Let x(t) be a positive solution of (1.1) and z(t) satisfy (II). If

    t0˜η(s)sn2ds=, (3.2)

    then limtx(t)=limtz(t)=0, where

    ˜η(t)=(1r(t)tmκ=1qκ(s)ds)1α.

    Proof. Let x be a positive solution of (1.1) on [t0,). Then, there exists t1t0 such that x(t)>0, x(σκ(t))>0 and x(τ(t))>0, for tt1. Since the corresponding function z(t)>0 and z(t)<0, then there exists a finite limit limtz(t)=c0. Let c>0. Then for any ϵ>0, we have ϵ+c >z(t)>c, eventually. It is easy to see that

    x(t)=z(t)p(t)x(τ(t))z(t)p(t)z(τ(t)),

    thus,

    x(t)cp0(ϵ+c)=cp0(ϵ+c)ϵ+c(ϵ+c).

    This implies that

    x(t)ϱz(t), (3.3)

    where ϱ=cp0(ϵ+c)/ϵ+c>0, that is,

    xα(σκ(t))ϱαzα(σκ(t)).

    Using (3.3) in (1.1), we obtain

    (r(t)((z(t))(n1))α)+mκ=1qκ(t)ϱαzα(σκ(t))0.

    By (3.1) and σ(t)<t, we see that

    (r(t)((z(t))(n1))α)+ϱαzα(σ(t))mκ=1qκ(t)0.

    Integrating last inequality from t to , we get

    r(t)((z(t))(n1))αϱαtzα(σ(s))mκ=1qκ(s)ds.

    By limtz(σ(t))>c, it follows that

    z(n1)(t)ϱc˜η(t). (3.4)

    Integrating (3.4) twice from t to , we have

    z(n3)(t)ϱctu˜η(s)dsdu=ϱct˜η(s)(st)ds.

    Repeating this procedure, we arrive at

    z(t)ϱc(n3)!t˜η(s)(st)n3ds.

    Now, integrating from t1 to , we see that

    z(t1)ϱc(n2)!t1˜η(s)(st1)n2dsϱc2n2(n2)!2t1˜η(s)sn2ds.

    This contradicts (3.2). Then we have limtz(t)=0.

    In the following lemma, we will use the notions

    ˜qκ1(t):=min{qκ(t),qκ(τ(t))},˜qκ2(t):=min{qκ(σ1(t)),qκ(σ1(τ(t)))}

    and

    ττ0>0; (3.5)
    (σ1(t))σ0>0. (3.6)

    Lemma 3.3. If x(t) is a positive solution of (1.1) and z(t) satisfy (I), (3.5) and σκτ=τσκ hold, then

    (r(t)(z(n1)(t))α+pα0τ0r(τ(t))(z(n1)(τ(t)))α)+zα(σ(t))μmκ=1˜qκ1(t)0. (3.7)

    Moreover, if (3.6) and (σκσ1)τ=τ(σκσ1) hold, then

    0(r(σ1(t))(z(n1)(σ1(t)))α)σ0+pα0(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)σ0τ0+zα(t)μmκ=1˜qκ2(t). (3.8)

    Proof. Let x be a positive solution of (1.1). Then, there exists t1t0 such that x(t)>0, x(σκ(t))>0 and x(τ(t))>0 for tt1. By Lemma 2, we see that

    zα(σ(t))μ(xα(σ(t))+pα0xα(τ(σ(t)))). (3.9)

    From (1.1), (3.5) and property  σκτ=τσκ, we get

    0=pα0τ(t)(r(τ(t))(z(n1)(τ(t)))α)+pα0mκ=1qκ(τ(t))xα(σκ(τ(t)))pα0τ0(r(τ(t))(z(n1)(τ(t)))α)+pα0mκ=1qκ(τ(t))xα(τ(σκ(t))).

    Using (1.1) with above inequality and taking (3.9) into account, we have

    0(r(t)(z(n1)(t))α)+pα0τ0(r(τ(t))(z(n1)(τ(t)))α)+mκ=1qκ(t)xα(σκ(t))+pα0mκ=1qκ(τ(t))xα(τ(σκ(t)))(r(t)(z(n1)(t))α)+pα0τ0(r(τ(t))(z(n1)(τ(t)))α)+1μmκ=1˜qκ1(t)zα(σκ(t))=(r(t)(z(n1)(t))α+pα0τ0(r(τ(t))(z(n1)(τ(t)))α))+1μmκ=1˜qκ1(t)zα(σκ(t)).

    By (3.1), we see that

    0(r(t)(z(n1)(t))α+pα0τ0(r(τ(t))(z(n1)(τ(t)))α))+1μzα(σ(t))mκ=1˜qκ1(t).

    Using (3.1) and (3.6) in (1.1), we are led to

    0=1(σ1(t))(r(σ1(t))(z(n1)(σ1(t)))α)+mκ=1qκ(σ1(t))xα(σκ(σ1(t)))1σ0(r(σ1(t))(z(n1)(σ1(t)))α)+mκ=1qκ(σ1(t))xα(σκ(σ1(t))). (3.10)

    Also, using (3.1) and (3.5) in (1.1), we obtain

    0=pα0(σ1(τ(t)))(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)+pα0mκ=1qκ(σ1(τ(t)))xα(σκ(σ1(τ(t))))pα0σ0τ0(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)+pα0mκ=1qκ(σ1(τ(t)))xα(τ(σκ(σ1(t)))). (3.11)

    Combining (3.10) with (3.11) and taking into account (3.9), one can see that

    01σ0(r(σ1(t))(z(n1)(σ1(t)))α)+pα0σ0τ0(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)+1μmκ=1˜qκ2(t)(x(σκ(σ1(t)))+x(τ(σκ(σ1(t)))))α.

    That is,

    01σ0(r(σ1(t))(z(n1)(σ1(t)))α)+pα0σ0τ0(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)+mκ=11μ˜qκ2(t)zα(σκ(σ1(t))).

    By the fact z>0, we note that z(σκ(σ1(t)))>z(t) which implies that

    01σ0(r(σ1(t))(z(n1)(σ1(t)))α)+pα0σ0τ0(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)+zα(t)mκ=11μ˜qκ2(t).

    The proof of lemma is complete.

    Theorem 3.1. Assume that (3.2), (3.5), σ(t)τ(t), σ(t)>0 and σκτ=τσκ hold. If there exists a function δC1([t0,),(0,)), such that

    lim supttt2[δ(s)μmκ=1˜qκ1(s)((n2)!)αμα(α+1)α+1(1+pα0τ0)r(s)(δ(s))α+1(δ(s)σn2   (s)σ(s))α]ds=, (3.12)

    then every solution of (1.1) is oscillatory or tends to zero.

    Proof. Let x be a positive solution of (1.1). Then, there exists t1t0 such that x(t)>0, x(σκ(t))>0 and x(τ(t))>0 for tt1. Let z  satisfying case (I). Define the positive function ω(t) by

    ω(t)=δ(t)r(t)(z(n1)   (t)) αzα(σ(t)). (3.13)

    Hence, by differentiating (3.13), we get

    ω(t)=δ(t)r((z)(n1))αzα(σ(t))+δ(t)(r(t)(z(n1)(t))α)zα(σ(t))αδ(t)r(t)(z(n1)(t))αzα1(σ(t))z(σ(t))σ(t)z2α(σ(t)). (3.14)

    Since z>0, z>0,we see that limtz0, using Lemma 2.3 with f=z, we see that

    z(t)μ(n2)!tn2z(n1)(t),

    for every μ(0,1). By zn(t)0, we get

    z(σ(t))μ(n2)!(σ(t))n2z(n1)(σ(t))μ(n2)!(σ(t))n2z(n1)(t). (3.15)

    Substituting (3.13) and (3.15) into (3.14) implies

    ω(t)δ(t)r(t)(z(n1)(t))αzα(σ(t))+δ(t)(r(t)(z(n1)(t))α)zα(σ(t))(z(n1)(t)z(σ(t)))α+1αδ(t)r(t)μσn2(t)σ(t)(n2)!δ(t)(r(t)(z(n1)(t))α)zα(σ(t))+δ(t)δ(t)ω(t)αδ(t)r(t)μσn2(t)σ(t)(n2)!(ω(t)δ(t)r(t))α+1α,

    that is,

    ω(t)δ(t)(r(t)(z(n1)  (t))α)zα(σ(t))+δ(t)δ(t)ω(t)αμσn2  (t)σ(t)(n2)!δ1/α  (t)r1/α  (t)ω(α+1)/α(t). (3.16)

    Now, define another positive function v(t) by

    v(t)=δ(t)r(τ(t))(z(n1)(τ(t)))αzα(σ(t)). (3.17)

    By differentiating (3.17), we get

    v(t)=δ(t)r(τ(t))(z(n1)(τ(t)))αzα(σ(t))+δ(t)(r(τ(t))(z(n1)(τ(t)))α)zα(σ(t))αδ(t)r(τ(t))(z(n1)(τ(t)))αzα1(σ(t))z(σ(t))σ(t)z2α(σ(t)). (3.18)

    From (3.15), σ(t)τ(t) and zn(t)0, we have

    z(σ(t))μ(n2)!(σ(t))n2z(n1)(σ(t))μ(n2)!(σ(t))n2z(n1)(τ(t)). (3.19)

    Substituting (3.19) and (3.17) into (3.18), implies

    v(t)δ(t)r(τ(t))(z(n1)(τ(t)))αzα(σ(t))+δ(t)(r(τ(t))(z(n1)(τ(t)))α)zα(σ(t))(z(n1)(τ(t))z(σ(t)))α+1αδ(t)r(τ(t))μσn2(t)σ(t)(n2)!δ(t)(r(τ(t))(z(n1)(τ(t)))α)zα(σ(t))+δ(t)δ(t)v(t)αδ(t)r(τ(t))μσn2(t)σ(t)(n2)!(v(t)δ(t)r(τ(t)))α+1α.

    By r(t)>0, we get

    v(t)δ(t)(r(τ(t))(z(n1)(τ(t)))α)zα(σ(t))+δ(t)δ(t)v(t)αμσn2(t)σ(t)(n2)!δ1/α(t)r1/α(t)v(α+1)/α(t). (3.20)

    Now, using inequalities (3.16) and (3.20), we get

    ω(t)+pα0τ0v(t)δ(t)(r(t)(z(n1)(t))α)+pα0τ0(r(τ(t))(z(n1)(τ(t)))α)zα(σ(t))+δ(t)δ(t)ω(t)αμσn2(t)σ(t)(n2)!δ1/α(t)r1/α(t)ω(α+1)/α(t)+pα0τ0(δ(t)δ(t)v(t)αμσn2(t)σ(t)(n2)!δ1/α(t)r1/α(t)v(α+1)/α(t)). (3.21)

    By (3.7), we obtain

    ω(t)+pα0τ0v(t)δ(t)mκ=1˜qκ1(t)μ+δ(t)δ(t)ω(t)αμσn2(t)σ(t)(n2)!δ1/α(t)r1/α(t)ω(α+1)/α(t)+pα0τ0(δ(t)δ(t)v(t)αμσn2(t)σ(t)(n2)!δ1/α(t)r1/α(t)v(α+1)/α(t)).

    Applying the following inequality inequality (2.1) with

    A=αμσn2  (t)σ(t)(n2)  !δ1/α  (t)r1/α  (t)   and  B=δ(t)δ(t),

    we get

    ω(t)+pα0τ0v(t)δ(t)μmκ=1˜qκ1(t)+((n2)!)αμα(α+1)α+1r(t)(δ(t))α+1(δ(t)σn2(t)σ(t))α+pα0((n2)!)ατ0μα(α+1)α+1r(t)(δ(t))α+1(δ(t)σn2(t)σ(t))α.

    Integrating the last inequality from t2 to t, we obtain

    tt2(δ(s)μmκ=1˜qκ1(s)((n2)!)αμα(α+1)α+1(1+pα0τ0)r(s)(δ(s))α+1(δ(s)σn2  (s)σ(s))α)dsω(t2)+pα0τ0v(t2).

    The proof is complete.

    Theorem 3.2. Assume that (3.2), (3.5), (3.6), σ(t)τ(t) and σκσ1τ=τσκσ1 hold. If there exists a function δC1([t0,),(0,)), such that

    lim supttt2[δ(s)μmκ=1˜qκ2(s)((n2)!)αμασ0(α+1)α+1(1+pα0τ0)r(σ1(s))(δ(s))α+1  (δ(s)sn2  )  α]ds=, (3.22)

    then every solution of (1.1) is oscillatory or tends to zero.

    Proof. Let x be a positive solution of (1.1). Then, there exist t1t0 such that x(t)>0, x(σκ(t))>0 and x(τ(t))>0 for tt1. Let z  satisfying case (I). Define the positive function by

    ω(t)=δ(t)r(σ1(t))(z(n1)  (σ1  (t)))αzα(t). (3.23)

    Hence, by differentiating (3.13), we get

    ω(t)=δ(t)r(σ1(t))(z(n1)(σ1(t)))αzα(t)+δ(t)(r(σ1(t))(z(n1)(σ1(t)))α)zα(t)αδ(t)r(σ1(t))(z(n1)(σ1(t)))αzα1(t)z(t)z2α(t). (3.24)

    Since z>0, z>0,we see that limtz0, using Lemma 2.3 with f=z, we obtain

    z(t)μ(n2)!tn2z(n1)(t), (3.25)

    for every μ(0,1). Thus, by σ1(t)>t and zn(t)0, we get

    z(t)μ(n2)!tn2z(n1)(t)μ(n2)!tn2z(n1)(σ1(t)). (3.26)

    Substituting (3.23) and (3.26) into (3.24) implies

    ω(t)δ(t)r(σ1(t))(z(n1)(σ1(t)))αzα(t)+δ(t)(r(σ1(t))(z(n1)(σ1(t)))α)zα(t)(z(n1)(σ1(t))z(t))α+1αδ(t)r(σ1(t))μtn2(n2)!δ(t)(r(σ1(t))(z(n1)(σ1(t)))α)zα(t)+δ(t)δ(t)ω(t)αδ(t)r(σ1(t))μtn2(n2)!(ω(t)δ(t)r(σ1(t)))α+1α,

    that is,

    ω(t)δ(t)(r(σ1  (t))(z(n1)  (σ1  (t)))α)zα(t)+δ(t)δ(t)ω(t)αμtn2(n2)!δ1/α  (t)r1/α  (σ1(t))ω(α+1)/α(t). (3.27)

    Now, define another positive function v(t) by

    v(t)=δ(t)r(σ1  (τ(t)))(z(n1)  (σ1  (τ(t))))αzα(t). (3.28)

    By differentiating (3.28), we get

    v(t)=δ(t)r(σ1(τ(t)))(z(n1)(σ1(τ(t))))αzα(t)+δ(t)(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)zα(t)αδ(t)r(σ1(τ(t)))(z(n1)(σ1(τ(t))))αzα1(t)z(t)z2α(t). (3.29)

    From (3.25), σ1(τ(t))t and zn(t)0, we have

    z(t)μ(n2)!tn2z(n1)(t)μ(n2)!tn2z(n1)(σ1(τ(t))). (3.30)

    Substituting (3.30) and (3.28) into (3.29), implies

    v(t)δ(t)r(σ1(τ(t)))(z(n1)(σ1(τ(t))))αzα(t)+δ(t)(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)zα(t)(z(n1)(σ1(τ(t)))z(t))α+1αδ(t)r(σ1(τ(t)))μtn2(n2)!δ(t)(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)zα(t)+δ(t)δ(t)v(t)αδ(t)r(σ1(τ(t)))μtn2(n2)!(v(t)δ(t)r(σ1(τ(t))))α+1α,

    By r(t)>0, we get

    v(t)δ(t)(r(σ1  (τ(t)))(z(n1)  (σ1  (τ(t))))α)zα(t)+δ(t)δ(t)v(t)αμtn2(n2)!δ1/α  (t)r1/α   (σ1  (t))v(α+1)/α(t). (3.31)

    Now, using inequalities (3.27) and (3.31), we get

    1σ0ω(t)+pα0σ0τ0v(t)δ(t)1σ0(r(σ1(t))(z(n1)(σ1(t)))α)zα(t)+δ(t)pα0σ0τ0(r(σ1(τ(t)))(z(n1)(σ1(τ(t))))α)zα(t)+δ(t)σ0δ(t)ω(t)αμtn2σ0(n2)!δ1/α(t)r1/α(σ1(t))ω(α+1)/α(t)+pα0σ0τ0(δ(t)δ(t)v(t)αμtn2(n2)!δ1/α(t)r1/α(σ1(t))v(α+1)/α(t)).

    By (3.9), we obtain

    1σ0ω(t)+pα0σ0τ0v(t)δ(t)μmκ=1˜qκ2(t)+δ(t)σ0δ(t)ω(t)αμtn2σ0(n2)!δ1/α(t)r1/α(σ1(t))ω(α+1)/α(t)+pα0τ0(δ(t)σ0δ(t)v(t)αμtn2σ0(n2)!δ1/α(t)r1/α(σ1(τ(t)))v(α+1)/α(t)).

    Applying the following inequality (2.1) with

    A=αμtn2σ0(n2)!δ1/α  (t)r1/α  (σ1  (t))   and  B=δ(t)σ0δ(t),

    we get

    1σ0ω(t)+pα0σ0τ0v(t)δ(t)μmκ=1˜qκ2(t)+((n2)!)αμασ0(α+1)α+1r(σ1(t))(δ(t))α+1(δ(t)tn2)α+pα0((n2)!)ατ0σ0μα(α+1)α+1r(σ1(t))(δ(t))α+1(δ(t)tn2)α.

    Integrating last the inequality from t2 to t, we obtain

    tt2[δ(s)μmκ=1˜qκ2(s)((n2)!)αμασ0(α+1)α+1(1+pα0τ0)r(σ1(s))(δ(s))α+1(δ(s)sn2)α]ds1σ0ω(t2)+pα0σ0τ0v(t2).

    The proof is complete.

    Example 3.1. Consider the odd order neutral delay differential equation

    (x(t)+1718x(tb))(n)+mk=1q0tnx(tb2κ)=0,n3,t1, (3.32)

    we note that

    μ=α=r(t)=1,b=b1>1,˜qκ2(s)=q0b2ntn,σ(t)=tb2,τ(t)=tb  and  set  δ(t)=tn1.

    It is easy to see that the conditions (3.5), (3.6) and (3.2) hold.

    Applying Theorem 3.2, we have that every solution of (3.32) is oscillatory or tends to zero as t when

    q0>(n2)!(n1)2b2n24m(1+1718b).

    Remark 3.1. If we consider the special case (x(t)+1718x(t/2))(3)+q0t3x(t/22)=0, then every solution is oscillatory or tends to zero if q0>46.22, while by using the result in [21], we have that every solution is oscillatory or tends to zero if q0>144. Consequently, our results apply to the equation (x(t)+1718x(t/2))(3)+70t3x(t/22)=0, while the other results fail to study this equation.

    In this study, oscillatory properties of a class of odd-order quasi-linear neutral differential equations are established. By introducing some Riccati substitution, we obtained new conditions that guarantee that all nonoscillatory solutions of (1.1) converge to zero. Our results extend and complement the previous results in the literature. An interesting issue is obtaining new criteria that ensure that all solutions of (1.1) oscillate.

    The authors is grateful to the editors and two anonymous referees for a very thorough reading of the manuscript and for some valuable notes that improved the manuscript.

    There are no competing interests



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