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Research article

On the upper bounds for the distance between zeros of solutions of a first-order linear neutral differential equation with several delays

  • Received: 29 April 2024 Revised: 27 June 2024 Accepted: 01 July 2024 Published: 07 August 2024
  • MSC : 34K11, 39A10, 39A99

  • This work is devoted to studying the distribution of zeros of a first-order neutral differential equation with several delays

    [y(t)+a(t)y(tσ)]+nj=1bj(t)y(tμj)=0,tt0.

    New estimations for the upper bounds of the distance between successive zeros are obtained. The properties of a positive solution of a first-order differential inequality with several delays in a closed interval are studied, and many results are established. We apply these results to a first-order neutral differential equation with several delays and also to a first-order differential equation with several delays. Our results for the differential equation with several delays not only provide new estimations but also improve many previous ones. Also, the results are formulated in a general way such that they can be applied to any functional differential equation for which studying the distance between zeros is equivalent to studying this property for a first-order differential inequality with several delays. Further, new estimations of the upper bounds for certain equations are given. Finally, a comparison with all previous results is shown at the end of this paper.

    Citation: Emad R. Attia. On the upper bounds for the distance between zeros of solutions of a first-order linear neutral differential equation with several delays[J]. AIMS Mathematics, 2024, 9(9): 23564-23583. doi: 10.3934/math.20241145

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  • This work is devoted to studying the distribution of zeros of a first-order neutral differential equation with several delays

    [y(t)+a(t)y(tσ)]+nj=1bj(t)y(tμj)=0,tt0.

    New estimations for the upper bounds of the distance between successive zeros are obtained. The properties of a positive solution of a first-order differential inequality with several delays in a closed interval are studied, and many results are established. We apply these results to a first-order neutral differential equation with several delays and also to a first-order differential equation with several delays. Our results for the differential equation with several delays not only provide new estimations but also improve many previous ones. Also, the results are formulated in a general way such that they can be applied to any functional differential equation for which studying the distance between zeros is equivalent to studying this property for a first-order differential inequality with several delays. Further, new estimations of the upper bounds for certain equations are given. Finally, a comparison with all previous results is shown at the end of this paper.



    Consider the first-order neutral differential equation with several delays

    [y(t)+a(t)y(tσ)]+nj=1bj(t)y(tμj)=0,tt0,(E1)

    where a,bjC([t0,),[0,)), 0<σ<μ1μ2μn, and the first-order differential equation with several delays

    y(t)+nj=1bj(t)y(tμj)=0,tt0.(E2)

    By a solution of Eq (E1) on [t+μn,), tt0, we mean a function yC([t,),R) such that y(t)+a(t)y(tσ)C1([t+μnσ,),R) and satisfies (E1) for all tt+μn. The existence and uniqueness of a solution y(t) of Eq (E1) with an initial function ϕC([t0μn,t0],R) can be proved using the method of steps as in [24, Thereoem 1.1.2]. Also, the existence of a positive solution for some neutral differential equations can be found in [25].

    A solution is said to be oscillatory if it is neither eventually negative nor eventually positive; otherwise, it is called non-oscillatory. If all solutions of a differential equation are oscillatory, then it is called oscillatory; otherwise, it is called non-oscillatory.

    Recently, neutral differential equations have arisen in many applications; see [1,5,23,24,27,28,29,34,35,36]. In this type of equation, the delays appear in both the unknown function and its derivatives. The qualitative properties of neutral differential equations have received a great deal of attention from many mathematicians; see [1,2,5,6,7,9,12,14,15,16,19,20,21,22,23,24,26,27,28,29,34,35,36,37,38,40,41,42]. In dynamical models, delay and oscillation effects are often formulated by means of external sources and/or nonlinear diffusion, perturbing the natural evolution of related systems; see, e.g., [30,31]. The oscillation of neutral and delay differential equations has been extensively developed and studied in many works; see [2,5,6,7,9,12,13,14,15,16,19,20,21,22,24,26,37,38,40,41]. On the other hand, the oscillation of first-order differential equations has numerous applications in the study of the oscillatory behavior of higher-order neutral differential equations; see [32,33]. However, only a few works have been interested in studying the distance between zeros of all solutions for first-order neutral and delay differential equations; see [2,3,4,6,7,8,10,11,15,17,18,37,38,39,40,41]. In this type of study, the interest is not only in the existence of zeros of solutions (i.e., proving the oscillation) but also in determining their locations. Therefore, studying the distance between zeros for first-order linear neutral and delay differential equations can give a deeper understanding of the dynamics of nonlinear systems of neutral differential equations that are used to model many real-life phenomena. This motivates us to study the distance between successive zeros for all solutions of Eqs (E1) and (E2).

    In the following, we display some results for the distance between zeros for Eqs (E1) and (E2):

    The distribution of zeros in first-order differential equations with one delay or several delays was studied by [3,4,8,10,11,17,18,39]. Further, many estimations for the upper bounds of the distance between zeros for first-order neutral differential equations were established by [2,6,7,15,37,38,40,41]. However, there are no results dealing with the distribution of zeros in Eq (E1).

    In this work, we obtain sufficient conditions ensuring that any solution of the first-order differential inequality with several delays

    Y(t)+nj=1Bj(t)Y(tηj)0,t[L+α,L0], (1.1)

    where L0L+α, Lt0, α0 such that 0<η1η2ηn, and BjC([L+α,),[0,)), j=1,2,...,n, cannot be positive on certain intervals. By using these results, many new estimations for the upper bounds of the distance between zeros for both Eqs (E1) and (E2) are obtained. A new approximation for the distance between successive zeros of a certain differential equation of the form (E2) is given, while all previous results cannot give this approximation. Since the distribution of zeros in all solutions of Eq (E1) has never been examined before, our results for Eq (E1) are obviously new. An illustrative example is given to show the applicability of our results to Eq (E1).

    In this section, we study the properties of a positive solution to the first-order differential inequality with several delays (1.1).

    Let Y(t) be a solution of inequality (1.1) on [L+α,L0], L0L+α such that

    Y(t)>0 for t[L+γ,L+α] and Y(t)0 for t[L+δ,L+β], (2.1)

    where γ,δ0 and β=max{α,ηn+γ}.

    Next, we prove some lemmas that play an important role in establishing the main results of this work.

    Assume that s{1,2,...,n} and the sequence of nonnegative real numbers {Msl}l1 is defined by

    Ms1=0,Ms0=1,ettηsnj=1jsMjl1Bj(v)dv1ttηsBs(v)dvMsl,l=1,2, . (2.2)

    Lemma 2.1. Assume that lN0, Y(t) is a solution of inequality (1.1) on [L+α,L0] such that (2.1) is satisfied with L0L+max{δ,α}+(l+1)ηn and Y(t)>0 on [L+α,L0]. Then

    Y(tηs)Y(t)Mslfort[L+max{δ,α}+(l+1)ηn,L0] (2.3)

    for s=1,2,...,n.

    Proof. In view of Y(t)0 for t[L+δ,L+β] and Y(t)>0 for t[L+γ,L0], it follows from (1.1) that

    Y(t)0 for t[L+δ,L0]. (2.4)

    Therefore,

    Y(tηs)Y(t)1=Ms0 for t[L+δ+ηs,L0][L+max{δ,α}+ηn,L0]. (2.5)

    Integrating (1.1) from tηs to t, it follows that

    Y(t)Y(tηs)+ttηsnj=1Bj(v)Y(vηj)dv0 for t[L+α+ηs,L0].

    Therefore,

    Y(t)Y(tηs)+ttηsBs(v)Y(vηs)dv+nj=1jsttηsBj(v)Y(v)Y(vηj)Y(v)dv0 (2.6)

    for t[L+α+ηn,L0]. Dividing (1.1) by Y(t), and integrating the resulting inequality from ζ to μ, L+α+ηnζμL0, we obtain

    Y(ζ)Y(μ)eμζnj1=1Bj1(v)Y(vηj1)Y(v)dv. (2.7)

    Substituting into (2.6), we obtain

    Y(t)Y(tηs)+ttηsBs(v)Y(vηs)dv+Y(t)nj=1jsttηsBj(v)Y(vηj)Y(v)etvnj1=1Bj1(v1)Y(v1ηj1)Y(v1)dv1dv0

    for t[L+α+ηn+ηs,L0].

    By (2.4), we have

    Y(t)Y(tηs)+Y(tηs)ttηsBs(v)dv+Y(t)ttηsnj=1jsBj(v)Y(vηj)Y(v)etvnj1=1j1sBj1(v1)Y(v1ηj1)Y(v1)dv1dv0

    for t[L+max{δ+ηs,α+ηn}+ηs,L0]. This leads to

    Y(t)Y(tηs)+Y(tηs)ttηsBs(v)dv+Y(t)(ettηsnj=1jsBj(v)Y(vηj)Y(v)dv1)0

    for t[L+max{δ+ηs,α+ηn}+ηs,L0]. That is,

    Y(tηs)(1ttηsBs(v)dv)Y(t)ettηsnj=1jsBj(v)Y(vηj)Y(v)dv

    for t[L+max{δ+ηs,α+ηn}+ηs,L0]. Therefore

    Y(tηs)Y(t)ettηsnj=1jsBj(v)Y(vηj)Y(v)dv1ttηsBs(v)dv>0 for t[L+max{δ+ηs,α+ηn}+ηs,L0]. (2.8)

    In view of (2.5), we obtain

    Y(tηs)Y(t)ettηsnj=1jsMj0Bj(v)dv1ttηsBs(v)dvMs1

    for t[L+max{δ+ηs,α+ηn}+ηn,L0][L+max{δ,α}+2ηn,L0]. Therefor,

    Y(vηj)Y(v)Mj1,for t[L+max{δ+ηj,α+ηn}+ηn,L0],1jn.

    Substituting into (2.8), we have

    Y(tηs)Y(t)ettηsnj=1jsMj1Bj(v)dv1ttηsBs(v)dvMs2

    for t[L+max{δ,α}+3ηn,L0]. By repeating this procedure, we obtain (2.3). The proof is complete.

    Lemma 2.2. Assume that lN0, 0<ϵ1, and Y(t) is a solution of inequality (1.1) on [L+α,L0] such that (2.1) is satisfied with L0L+max{δ,α}+(l+1+ϵ)ηn. If,

    ns=1(nj=1jsttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dv)1n11ttϵηsBs(v)etϵηsvηsnj=1Bj(v1)Mjl1dv1dv1(n1)nfortL+2ηn, (2.9)

    then Y(t) cannot be positive on [L+α,L0].

    Proof. Assume the contrary, and let Y(t)>0 on [L+α,L0]. This together with (2.1) implies that Y(t)>0 on [L+γ,L0]. Form (2.4), we have

    Y(t)0 for t[L+δ,L0]. (2.10)

    Integrating (1.1) from t to tϵηs, s=1,2,,n, we obtain

    Y(t)Y(tϵηs)+ttϵηsnj=1Bj(v)Y(vηj)dv0 for t[L+α+ϵηs,L0].

    Therefore,

    Y(t)Y(tϵηs)+ttϵηsBs(v)Y(vηs)dv+ttϵηsnj=1jsBj(v)Y(vηj)dv0 (2.11)

    for t[L+α+ϵηs,L0]. By (2.10), we obtain

    Y(t)Y(tϵηs)+Y(tηs)ttϵηsBs(v)dv+nj=1jsY(tηj)ttϵηsBj(v)dv0

    for t[L+max{δ,α}+ϵηs+ηn,L0]. That is,

    Y(tϵηs)Y(t)+Y(tϵηs)ttϵηsBs(v)etϵηsvηsnj=1Bj(v1)Mj1dv1dv+nj=1jsY(tϵηj)ttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj11dv1dv (2.12)

    for t[L+max{δ,α}+ϵηs+ηn,L0].

    From (2.7) and (2.11), we have

    Y(tϵηs)Y(t)+Y(tϵηs)ttϵηsBs(v)etϵηsvηsnj1=1Bj1(v1)Y(v1ηj1)Y(v1)dv1dv+nj=1jsY(tϵηj)ttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Y(v1ηj1)Y(v1)dv1dv

    for t[L+α+ϵηs+2ηn,L0]. This, together with (2.3) leads to

    Y(tϵηs)Y(t)+Y(tϵηs)ttϵηsBs(v)etϵηsvηsnj=1Bj(v1)Mjl1dv1dv+nj=1jsY(tϵηj)ttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dvl=1,2,

    for t[L+max{δ,α}+(l+1)ηn+ϵηs,L0]. From this and (2.12), we have

    Y(tϵηs)(1ttϵηsBs(v)etϵηsvηsnj=1Bj(v1)Mjl1dv1dv)>nj=1jsY(tϵηj)ttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dv,l=0,1,2,

    for t[L+max{δ,α}+(l+1)ηn+ϵηs,L0]. Using the arithmetic-geometric mean, we obtain

    Y(tϵηs)(1ttϵηsBs(v)etϵηsvηsnj=1Bj(v1)Mjl1dv1dv)>(n1)(nj=1jsY(tϵηj))1n1(nj=1jsttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dv)1n1

    for t[L+max{δ,α}+(l+1)ηn+ϵηs,L0]. Taking the product on both sides,

    ns=1Y(tϵηs)ns=1(1ttϵηsBs(v)etϵηsvηsnj=1Bj(v1)Mjl1dv1dv)>(n1)nns=1(nj=1jsY(tϵηj))1n1ns=1(nj=1jsttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dv)1n1

    for t[L+max{δ,α}+(l+1+ϵ)ηn,L0]. Therefore,

    ns=1Y(tϵηs)ns=1(1ttϵηsBs(v)etϵηsvηsnj=1Bj(v1)Mjl1dv1dv)>(n1)nns=1Y(tϵηs)ns=1(nj=1jsttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dv)1n1

    for t[L+max{δ,α}+(l+1+ϵ)ηn,L0]. Then

    ns=1(nj=1jsttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dv)1n11ttϵηsBs(v)etϵηsvηsnj=1Bj(v1)Mjl1dv1dv<1(n1)n

    for t[L+max{δ,α}+(l+1+ϵ)ηn,L0]. This contradicts with (2.9). The proof is complete.

    Lemma 2.3. Assume that lN0, 1k1k2n, 0<ϵ1, and Y(t) is a solution of inequality (1.1) on [L+α,L0] such that (2.1) is satisfied with L0L+max{δ,α}+(l+1)ηn+ϵηk2. If,

    k2s=k1ttϵηsBs(v)etϵηsvηsnj=1Bj(v1)Mjl1dv1dv1nj=s+1ttϵηsBj(v)etϵηsvηjnj1=1Bj1(v1)Mj1l1dv1dv1fortT+α+ϵηk2+ηn, (2.13)

    then Y(t) cannot be positive on [L+α,L0].

    Proof. As before, assume that Y(t)>0 on [L+α,L0]. Integrating (1.1) from t to tϵηs, s=1,2,,k2, it follows that

    Y(t)Y(tϵηs)+ttϵηsnj=1Bj(v)Y(vηj)dv0 for t[L+α+ϵηs,L0].

    Therefore,

    Y(t)Y(tϵηs)+ttϵηssj=1Bj(v)Y(vηj)dv+ttϵηsnj=s+1Bj(v)Y(vηj)dv0

    for t[L+α+ϵηs,L0]. The same reasoning as in Lemma 2.2 leads to

    Y(tϵηs)Y(t)+sj=1Y(tϵηj)ttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dv+Y(tϵηs)nj=s+1ttϵηsBj(v)etϵηsvηjnj1=1Bj1(v1)Mj1l1dv1dv

    for t[L+max{δ,α}+(l+1)ηn+ϵηs,L0]. Consequently,

    Y(tϵηs)(1nj=s+1ttϵηsBj(v)etϵηsvηjnj1=1Bj1(v1)Mj1l1dv1dv)>sj=1Y(tϵηj)ttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dv

    for t[L+max{δ,α}+(l+1)ηn+ϵηs,L0]. Taking the product on both sides,

    k2s=k1Y(tϵηs)k2s=k1(1nj=s+1ttϵηsBj(v)etϵηsvηjnj1=1Bj1(v1)Mj1l1dv1dv)>k2s=k1(sj=1Y(tηj)ttϵηsBj(v)etϵηjvηjnj1=1Bj1(v1)Mj1l1dv1dv)

    for t[L+max{δ,α}+(l+1)ηn+ϵηk2,L0]. That is,

    k2s=k1Y(tϵηs)k2s=k1(1nj=s+1ttϵηsBj(v)etϵηsvηjnj1=1Bj1(v1)Mj1l1dv1dv)>k2s=k1Y(tηs)k2s=k1(ttϵηsBs(v)etϵηsvηsnj1=1Bj1(v1)Mj1l1dv1dv)

    for t[L+max{δ,α}+(l+1)ηn+ϵηk2,L0]. Therefore,

    k2s=k1ttϵηsBs(v)etϵηsvηsnj1=1Bj1(v1)Mj1l1dv1dv1nj=s+1ttϵηsBj(v)etϵηsvηjnj1=1Bj1(v1)Mj1l1dv1dv<1

    for t[L+max{δ,α}+(l+1)ηn+ϵηk2,L0], which contradicts (2.13). The proof is complete.

    Assume that Dtt1(E), t1t0 is the upper bound of the distance between successive zeros of all solutions of a differential equation E on [t1,).

    Below, we obtain new approximations for the upper bound of the distance between successive zeros of all solutions of Eq (E2).

    Theorem 3.1. Let

    Bj(t)=bj(t),ηj=μj,j=1,2,,n.

    Assume that lN0, 0<ϵ1. If,

    ns=1(nj=1jsttϵμsbj(v)etϵμjvμjnj1=1bj1(v1)Mj1l1dv1dv)1n1(1ttϵμsbs(v)etϵμsvμsnj=1bj(v1)Mjl1dv1dv)1(n1)nfortt0+2μn,

    then Dt1(E2)(l+2+ϵ)μn and every solution of Eq (E2) is oscillatory.

    Proof. Assume the contrary, and let y(t) be a solution of Eq (E2) such that y(t)>0 on [L,L0], Lt0, L0L+(l+2+ϵ)μn. Therefore, y(t)0 for t[L+μn,L0]. Then, all assumptions of Lemma 2.2 are satisfied with α=γ=0, δ=β=μn, Bj(t)=bj(t), and ηj=μj, j=1,2,,n. Therefore, y(t) cannot be positive on [L,L+(l+2+ϵ)μn][L,L0]. This contradiction completes the proof.

    Using Lemma 2.3 instead of Lemma 2.2 in the proof of the preceding theorem, one can prove the following result, and hence the proof is omitted.

    Theorem 3.2. Let

    Bj(t)=bj(t),ηj=μj,j=1,2,,n.

    Assume that lN0, 1k1k2n, 0<ϵ1. If,

    k2s=k1ttϵμsbs(v)etϵμsvμsnj1=1bj1(v1)Mj1l1dv1dv(1nj=s+1ttϵμsbj(v)etϵμsvμjnj1=1bj1(v1)Mj1l1dv1dv)1fortt0+α+ϵμk2+μn, (3.1)

    then Dt1(E2)(l+2)μn+ϵμk2 and every solution of Eq (E2) is oscillatory.

    We obtain many upper bounds for the distance between zeros of all solutions of Eq (E1) with the following assumptions:

    (H1) Let RC1[[t+σ,),[0,)], tt0, bj(t)a(tμj)bj(tσ)a(tσ), j=1,2,,n, and R(t)a(tσ), tt1+σ for some t1t.

    (H2) Let NC1[[t+μn,),[0,)], tt0, bj(t)>0, j=1,2,,n, N(t)nj=1bj(t)a(tμj)bj(tσ), tt1+μn for some t1t.

    Lemma 3.1. Assume that (H1) holds, R(t)0 for tt1+σ. Let y(t) be a solution of Eq (E1) such that y(t)>0 on [L,L0], LL0+2μn, Lt1. Then, there exists a function z(t) that satisfies z(t)>0 on [L+2σ,L0] and z(t)0 on [L+σ+μn,L0], and

    z(t)+nj=1bj(t)1+R(t)z(t+σμj)0fort[L+2μnL0]. (3.2)

    Proof. Letting u(t)=y(t)+a(t)y(tσ), so u(t)>0 for [L+σ,L0] and

    u(t)=nj=1bj(t)y(tμj)0fort[L+μn,L0].

    Therefore u(t)0 for t[L+μn,L0]. Note that

    y(tμj)=u(tμj)a(tμj)y(tσμj).

    Then

    u(t)=nj=1bj(t)u(tμj)+nj=1bj(t)a(tμj)y(tσμj)fort[L+μn,L0]. (3.3)

    By (H1), we obtain

    u(t)nj=1bj(t)u(tμj)+nj=1bj(tσ)a(tσ)y(tσμj)fort[L+μn+σ,L0].

    Clearly,

    a(tσ)u(tσ)=nj=1bj(tσ)a(tσ)y(tμjσ).

    Therefore,

    u(t)nj=1bj(t)u(tμj)a(tσ)u(tσ)fort[L+μn+σ,L0].

    Consequently,

    u(t)+R(t)u(tσ)+nj=1bj(t)u(tμj)0fort[L+μn+σ,L0].

    Assume that v(t)=u(t)+R(t)u(tσ), so v(t)>0 on [L+2σ,L0]. Then

    v(t)=u(t)+R(t)u(tσ)+R(t)u(tσ)nj=1bj(t)u(tμj)+R(t)u(tσ)0 (3.4)

    for t[L+μn+σ,L0], and so v(t)0 on [L+μn+σ,L0]. Consequently,

    v(t)R(t)u(tσ)+nj=1bj(t)u(tμj)0fort[L+μn+σ,L0]. (3.5)

    In view of u(t)0 on [L+μn,L0], it follows that

    v(t)=u(t)+R(t)u(tσ)(1+R(t))u(tσ)fort[L+μn+σ,L0].

    Therefore,

    u(tσ)v(t)1+R(t)fort[L+μn+σ,L0] (3.6)

    and

    u(tμj)v(t+σμj)1+R(t+σμj)fort[L+μn+μj,L0σ+μj], j=1,2,,n. (3.7)

    Substituting into (3.5), we obtain

    v(t)R(t)1+R(t)v(t)+nj=1bj(t)1+R(t+σμj)v(t+σμj)0fort[L+2μn,L0].

    Let z(t)=v(t)1+R(t). Then z(t)>0 on [L+2σ,L0] and

    z(t)+nj=1bj(t)1+R(t)z(t+σμj)0fort[L+2μn,L0].

    Also, by using (3.4) and (3.6), we obtain

    z(t)=v(t)(1+R(t))v(t)R(t)(1+R(t))2nj=1bj(t)v(tμj)1+R(t)0

    for t[L+σ+μn,L0]. The proof is complete.

    Lemma 3.2. Assume that (H2) holds, and N(t)0 for tt1+σ. Let y(t) be a solution of Eq (E1) such that y(t)>0 on [L,L0], LL0+2μn, Lt1. Then, there exists a function z(t) that satisfies z(t)>0 on [L+2σ,L0] and z(t)0 on [L+σ+μn,L0], and

    z(t)+nj=1bj(t)1+N(t)z(t+σμj)0fort[L+2μn,L0].

    Proof. By the same method as in the proof of Lemma 2.1 we obtain (see (3.3))

    u(t)=nj=1bj(t)u(tμj)+nj=1bj(t)a(tμj)y(tσμj)fort[L+μn,L0], (3.8)

    where u(t)=y(t)+a(t)y(tσ), u(t)>0 for [L+σ,L0] and u(t)0 for t[L+μn,L0]. Note that

    u(t)=nj=1bj(t)y(tμj)bj(t)y(tμj)fort[L+μn,L0].

    Therefore,

    y(tσμj)1bj(tσ)u(tσ)fort[L+μn+σ,L0].

    This together with (3.8) implies that

    u(t)nj=1bj(t)u(tμj)u(tσ)nj=1bj(t)a(tμj)bj(tσ)fort[L+μn+σ,L0].

    In view of (H2), it follows that

    u(t)+nj=1bj(t)u(tμj)+N(t)u(tσ)0fort[L+μn+σ,L0]. (3.9)

    By using the same method as in Lemma 2.1, we obtain

    z(t)+nj=1bj(t)1+N(t)z(t+σμj)0fort[L+2μn,L0],

    where z(t)>0 on [L+2σ,L0], z(t)0 on t[L+μn+σ,L0], z(t)=v(t)1+N(t) and v(t)=u(t)+N(t)u(tσ), u(t)=y(t)+a(t)y(tσ). The proof is complete.

    Lemma 3.3. Assume that (H1) holds and br(t)|R(t)|, r{1,2,,n}. Let y(t) be a solution of Eq (E1) such that y(t)>0 for t[L,L0], LL0+μn+μr, Lt1. Then, there exists a function v(t) that satisfies v(t)>0 for t[L+2σ,L0] and v(t)0 for t[L+μn+μr,L0], and

    v(t)+br(t)|R(t)|1+R(t+σμr)v(t+σμr)+nj=1jrbj(t)1+R(t+σμj)v(t+σμj)0 (3.10)

    for t[L+μn+μr,L0].

    Proof. Letting u(t)=y(t)+a(t)y(tσ) and v(t)=u(t)+R(t)u(tσ), so u(t)0 for t[L+μn,L0] and v(t)>0 for [L+2σ,L0]. By (3.5), we have

    v(t)R(t)u(tσ)nj=1bj(t)u(tμj) for t[L+μn+σ,L0].

    Since u(t)0 for t[L+μn,L0], then

    v(t)R(t)u(tσ)nj=1bj(t)u(tμj)|R(t)|u(tσ)nj=1bj(t)u(tμj)|R(t)|u(tμr)nj=1bj(t)u(tμj)0

    for t[L+μn+μr,L0]. Therefore, v(t)0 for t[L+μn+μr,L0], and

    v(t)+(br(t)|R(t)|)u(tμr)+nj=1jrbj(t)u(tμj)0 for t[L+μn+μr,L0].

    Using (3.7), we obtain

    v(t)+br(t)|R(t)|1+R(t+σμr)v(t+σμr)+nj=1jrbj(t)1+R(t+σμj)v(t+σμj)0

    for t[L+μn+μr,L0], where v(t)>0 for [L+2σ,L0] and v(t)0 for t[L+μn+μr,L0]. The proof is complete.

    Lemma 3.4. Assume that (H2) holds and br(t)|N(t)|, r{1,2,,n}. Let y(t) be a solution of Eq (E1) such that y(t)>0 for t[L,L0], L0L+μn+μr, Lt1. Then, there is a function v(t) that satisfies v(t)>0 for t[L+2σ,L0] and v(t)0 for t[L+μn+μr,L0], and

    v(t)+br(t)|N(t)|1+N(t+σμr)v(t+σμr)+nj=1jrbj(t)1+N(t+σμj)v(t+σμj)0

    for t[L+μn+μr,L0].

    Proof. Using the same method as in the proof of Lemma 3.2, we obtain (see (3.9))

    u(t)+nj=1bj(t)u(tμj)+N(t)u(tσ)0 for t[L+μn+σ,L0],

    where u(t)=y(t)+a(t)y(tσ), u(t)>0 for t[L+σ,L0] and u(t)0 for t[L+μn,L0]. Letting v(t)=u(t)+N(t)u(tσ), then

    v(t)=u(t)+N(t)u(tσ)+N(t)u(tσ)nj=1bj(t)u(tμj)+N(t)u(tσ)

    for t[L+μn+σ,L0]. Therefore,

    v(t)nj=1bj(t)u(tμj)+N(t)u(tσ)nj=1bj(t)u(tμj)+|N(t)|u(tσ)

    for t[L+μn+σ,L0]. That is,

    v(t)nj=1bj(t)u(tμj)+|N(t)|u(tσ)0for t[L+μn+σ,L0].

    By using u(t)0 for t[L+μn,L0], we obtain

    v(t)(br(t)|N(t)|)u(tμr)nj=1jrbj(t)u(tμj)0for t[L+μn+μr,L0], (3.11)

    and

    v(t)=u(t)+N(t)u(tσ)(1+N(t))u(tσ) for t[L+μn+σ,L0].

    Therefore,

    u(tμj)v(t+σμj)1+N(t+σμj) for t[L+μn+μj,L0+μjσ].

    From this and (3.11), we have

    v(t)+br(t)|N(t)|1+N(t+σμr)v(t+σμr)+nj=1jrbj(t)1+N(t+σμj)v(t+σμj)0

    for t[L+μn+μr,L0], where v(t)>0 for [L+2σ,L0] and v(t)0 for t[L+μn+μr,L0]. The proof is complete.

    Theorem 3.3. Let

    Bj(t)=bj(t)1+R(t),ηj=μjσ,j=1,2,,n. (3.12)

    Assume that (H1) holds, lN0, 0<ϵ1, R(t)0 for tt1+σ. If condition (2.9) is satisfied, then Dt1(E1)(l+3+ϵ)μn(l+1+ϵ)σ and every solution of Eq (E2) is oscillatory.

    Proof. Assume the contrary, and let y(t) be a solution of Eq (E1) such that y(t)>0 on [L,L0], Lt0, L0L+(l+3+ϵ)μn(l+1+ϵ)σ. By using Lemma 3.1, then there exists a solution z(t) of (3.2) such that z(t)>0 on [L+2σ,L0] and z(t)0 on [L+σ+μn,L0]. Clearly

    max{δ,α}+(l+1+ϵ)ηn=(l+3+ϵ)μn(l+1+ϵ)σ,

    where α=2μn, δ=σ+μn, and ηn=μnσ. Applying Lemma 2.2 with γ=2σ, α=β=2μn, δ=σ+μn, so z(t) cannot be positive on [L+2μn,L0].

    Theorem 3.4. Assume that (H1) holds, lN0, 1k1k2n, 0<ϵ1, R(t)0 for tt1+σ. If condition (2.13) is satisfied such that Bj(t), ηj, j=1,2,,n, are defined by (3.12), then Dt1(E1)(l+3)μn+ϵμk2(l+1+ϵ)σ and every solution of Eq (E2) is oscillatory.

    Proof. Assume y(t) is a solution of Eq (E1) such that y(t)>0 on [L,L0], Lt0, L0L+(l+3)μn+ϵμk2(l+1+ϵ)σ. By using Lemma 3.1, there exists a solution z(t) of (3.2) such that z(t)>0 on [L+2σ,L0] and z(t)0 on [L+σ+μn,L0]. It is clear that

    max{δ,α}+(l+1)ηn+ϵηk2=(l+3)μn+ϵμk2(l+1+ϵ)σ,

    where α=2μn, δ=σ+μn, and ηj=μjσ. Applying Lemma 2.3 with γ=2σ, α=2μn, δ=σ+μn, so z(t) cannot be positive on [L+2μn,L0].

    The following two theorems can be proven using Lemma 3.2 instead of Lemma 3.1 in the proofs of Theorems 3.3 and 3.4, respectively.

    Theorem 3.5. Let

    Bj(t)=bj(t)1+N(t),ηj=μjσ,j=1,2,,n. (3.13)

    Assume that (H2) holds, lN0, 0<ϵ1, N(t)0 for tt1+σ. If condition (2.9) is satisfied, then Dt1(E1)(l+3+ϵ)μn(l+1+ϵ)σ and every solution of Eq (E2) is oscillatory.

    Theorem 3.6. Assume that (H2) holds, lN0, 1k1k2n, 0<ϵ1, N(t)0 for tt1+σ. If condition (2.13) is satisfied such that Bj(t), ηj, j=1,2,,n, are defined by (3.13), then Dt1(E1)(l+3)μn+ϵμk2(l+1+ϵ)σ and every solution of Eq (E2) is oscillatory.

    Theorem 3.7. Let r{1,2,,n},

    Bj(t)={bj(t)|R(t)|1+R(t+σμj)ifj=rbj(t)1+R(t+σμj)otherwise,ηj=μjσ,j=1,2,,n. (3.14)

    Assume that (H1) holds, lN0, 0<ϵ1, br(t)|R(t)| for tt1+σ. If condition (2.9) is satisfied, then Dt1(E1)(l+2+ϵ)μn+μr(l+1+ϵ)σ and every solution of Eq (E2) is oscillatory.

    Proof. Assume the contrary, and let y(t) be a solution of Eq (E1) such that y(t)>0 on [L,L0], Lt0, L0L+(l+2+ϵ)μn+μr(l+1+ϵ)σ. By using Lemma 3.3, there exists a solution v(t) of (3.10) on [L0+μn+μr,L] such that v(t)>0 on [L+2σ,L0] and v(t)0 on [L+μn+μr,L0]. Clearly

    max{δ,α}+(l+1+ϵ)ηn=(l+2+ϵ)μn+μr(l+1+ϵ)σ,

    where α=δ=μn+μr, and ηn=μnσ. Applying Lemma 2.2 with γ=2σ, α=δ=βμn+μr, so v(t) cannot be positive on [L+μn+μr,L0].

    Theorem 3.8. Assume that (H1) holds, lN0, 1k1k2n, 0<ϵ1, br(t)|R(t)| for tt1+σ, r{1,2,,n}. If condition (2.13) is satisfied with Bj(t) and ηj, j=1,2,,n, are defined by (3.14), then Dt1(E1)(l+2)μn+μr+ϵμk2(l+1+ϵ)σ and every solution of Eq (E2) is oscillatory.

    Proof. Assume the contrary, and let y(t) be a solution of Eq (E1) such that y(t)>0 on [L,L0], Lt0, L0L+(l+2)μn+μr+ϵμk2(l+1+ϵ)σ. By using Lemma 3.3, then there exists a solution v(t) of (3.10) on [L0+μn+μr,L] such that v(t)>0 on [L+2σ,L0] and v(t)0 on [L+μn+μr,L0]. Clearly

    max{δ,α}+(l+1)ηn+ϵηk2=(l+2)μn+μr+ϵμk2(l+1+ϵ)σ,

    where α=δ=μn+μr, and ηj=μjσ, j=1,2,,n. Applying Lemma 2.3 with γ=2σ, α=δ=β=μn+μr, so v(t) cannot be positive on [L+μn+μr,L0].

    Using Lemma 3.4 instead of Lemma 3.3 in the proofs of the preceding two theorems, we obtain the following results:

    Theorem 3.9. Let r{1,2,,n},

    Bj(t)={bj(t)|N(t)|1+N(t+σμj)ifj=rbj(t)1+N(t+σμj)otherwise,ηj=μjσ,j=1,2,,n. (3.15)

    Assume that (H2) holds, lN0, 0<ϵ1, br(t)|N(t)| for tt1+σ. If condition (2.9) is satisfied, then Dt1(E1)(l+2+ϵ)μn+μr(l+1+ϵ)σ and every solution of Eq (E2) is oscillatory.

    Theorem 3.10. Assume that (H2) holds, lN0, 1k1k2n, 0<ϵ1, br(t)|N(t)| for tt1+σ, r{1,2,,n}. If condition (2.13) is satisfied with Bj(t) and ηj=μjσ, j=1,2,,n, are defined by (3.15), then Dt1(E1)(l+2)μn+μr+ϵηk2(l+1+ϵ)σ and every solution of Eq (E2) is oscillatory.

    Remark 3.1. Following the same techniques used in the proof of our results, several sufficient criteria for the oscillation of both Eqs (E1) and (E2) can be obtained. For example, the conditions

    lim supt ns=1(nj=1jsttϵ(μsσ)bj(v)1+R(v)etϵ(μsσ)v(μsσ)nj1=1bj1(v1)1+R(v1)Mj1l1dv1dv)1n1(1ttϵ(μsσ)bs(v)1+R(v)etϵ(μsσ)v(μsσ)nj=1bj(v1)1+R(v1)Mjl1dv1dv)>1(n1)n,

    where the sequence {Msl}l0 is defined by (2.2) with Bj(t)=bj(t)1+R(t) and ηj=μjσ, j=1,2,,n, and

    lim supt k2s=k1ttϵμsbs(v)etϵμsvμsnj1=1bj1(v1)Mj1l1dv1dv(1nj=s+1ttϵμsbj(v)etϵμsvμjnj1=1bj1(v1)Mj1l1dv1dv)>1,

    where the sequence {Msl}l0 is defined by (2.2) with Bj(t)=q(t) and ηj=μj, j=1,2,,n, can be proved using the same proofs of Theorems 3.2 and 3.3, respectively.

    Example 4.1. Consider the first-order differential equation with several delays

    y(t)+b1(t)y(tμ1)+b2(t)y(tμ2)=0, (4.1)

    where b1(t)=185 and b2(t)=110, and μ1=δ, 0<δ<536, and μ2=1. Clearly,

    ttμ2b2(v)etμ2vμ22j1=1bj1(v1)dv1dv>1.

    Then, condition (3.1) is satisfied with l=1, k2=k3=3, and ϵ=1, and hence Dt1(4.1)4μ2=4, and every solution of Eq (4.1) is oscillatory for 0<δ<536. However, min1j2μj=μ1=δ. Then, δ can be chosen small enough such that all the results of [18] and [4,Theorem 2.2] fail to apply. Also, note that

    ttμ2b2(w)dw+1+ttμ2b1(w)dw1ttμ2b2(w)dwttμ2b2(w)wtμ2b2(w1)dw1dw<13100,

    and

    ttμ1b1(w)dw+1+ttμ1b2(w)dw1ttμ1b1(w)dwttμ1b1(w)wtμ1b1(w1)dw1dw=925δ(9δ290δ+50)518δ<1

    for sufficiently small δ. Therefore, [4,Theorem 2.1] cannot give an approximation better than 4μ2.

    Example 4.2. Consider the first-order neutral differential equation with several delays

    [y(t)+a(t)y(tσ)]+b1(t)y(tμ1)+b2(t)y(tμ2)=0,tπ, (4.2)

    where a(t)=a=max{2π99100,bπ299100}, b1(t)=b>0, and b2(t)=2+sin(2t), σ=π, μ1=2π3, and μ2=2π. It is clear that condition (H1) is satisfied with R(t)=a. Note that

    (ttπ22+sin(2v)1+adv)(ttπb1+adv)(1ttπ2b1+adv)(1ttπ2+sin(2v)1+adv)=2bπ(1+π2cos2(t))(1+a2π)(2+2abπ)2bπ(π1)(1+a2π)(2+2abπ)>1

    for b>150200π+1π(200π199). Then, condition (2.9) is satisfied with B1(t)=b1+a, B2(t)=2+sin(2t)1+a, η1=μ1σ=π2, η2=μ2σ=π, l=0 and ϵ=1. Therefore, all requirements of Theorem 3.3 with l=0 are satisfied, and hence Dπ(4.2)4μ22σ=6π.

    In this work, we obtained many new estimates for the upper bounds of the distance between adjacent zeros of Eqs (E1) and (E2). Our results are established in a general form (for inequality (1.1)), and hence they can be applied to any differential equation that can be transformed into an inequality of the form (1.1). Many new oscillation results for neutral differential equations with several not necessarily monotone delays can be obtained using the methods proposed in this work.

    The author declares that he has not used Artificial Intelligence (AI) tools in the creation of this article.

    The author extends their appreciation to Prince Sattam bin Abdulaziz University for funding this research work through the project number (PSAU/2024/01/29461).

    The authors declare no conflict of interest.



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