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Research article

Study on the oscillation of solution to second-order impulsive systems

  • Received: 02 September 2022 Revised: 29 December 2022 Accepted: 09 January 2023 Published: 13 July 2023
  • MSC : 34C10, 34C15, 34K11

  • In the present article, we set the if and only if conditions for the solutions of the class of neutral impulsive delay second-order differential equations. We consider two cases when it is non-increasing and non-decreasing for quotient of two positive odd integers. Our main tool is the Lebesgue's dominated convergence theorem. Examples illustrating the applicability of the results are also given, and state an open problem.

    Citation: Shyam Sundar Santra, Palash Mondal, Mohammad Esmael Samei, Hammad Alotaibi, Mohamed Altanji, Thongchai Botmart. Study on the oscillation of solution to second-order impulsive systems[J]. AIMS Mathematics, 2023, 8(9): 22237-22255. doi: 10.3934/math.20231134

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  • In the present article, we set the if and only if conditions for the solutions of the class of neutral impulsive delay second-order differential equations. We consider two cases when it is non-increasing and non-decreasing for quotient of two positive odd integers. Our main tool is the Lebesgue's dominated convergence theorem. Examples illustrating the applicability of the results are also given, and state an open problem.



    In modern era, delay differential equations (DEs) have become almost the center of interest. Many things in the world are directed by differential systems. We assume that the systems are independent of past state and future state depends on present state neutral delays. DEs are natural extensions of the delay DE which involve derivatives of the unknown at the delayed argument. Mathematical modeling with delay DEs is widely used for analysis and predictions in various areas of life science, for example, population dynamics, epidemiology, immunology, neutral networks, chemistry, physics, engineering, etc. The literature connected to impulsive delay differential system is vast.

    Below, we are going to provide some background of oscillation theory of impulsive DEs. The authors in [1] are concerned with the asymptotic behavior of a class of higher-order sublinear Emden-Fowler delay differential equations

    (q2(ι)ϱ(n1)(ι))+q1(ι)ϱν(τ(ι))=0,for ιι0, (1.1)

    where 0<ν<1 is a ratio of odd natural numbers, q2C1[ι0,), q2>0, q20, q1,τC[ι0,), τ(ι)<ι, limιτ(ι)=, q1(ι)0 and q2(ι) is not identically zero for large ι (for instanse, consider [2,3,4,5]). Shen et al. have taken the impulsive system (IS)

    {ϱ(ι)+q(ι)ϱ(ιμ1)=0,ιιk,ϱ(ι+k)ϱ(ιk)=Ik(ϱ(ιk)),kN, (1.2)

    where q,IkC(R,R), and they established the sufficient conditions for the oscillatory and asymptotic behavior of (1.2) [6]. Graef et al. in [7] considered the IS

    {(ϱ(ι)q2(ι)ϱ(ιμ2))+q1(ι)|ϱ(ιμ1)|ιsgnϱ(ιμ1)=0,ιι0,ϱ(ι+k)=ηkϱ(ιk),kN, (1.3)

    and considering q2(ι)PC([ι0,),R+) (q2(ι) piece wisely continuous in [ι0,)) set up results on sufficient conditions for oscillation (1.3). Shen et al. have established new sufficient conditions for oscillation of the IS

    {(ϱ(ι)q3(ι)ϱ(ιμ3))+q2(ι)ϱ(ιμ2)q1(ι)ϱ(ιμ1)=0,μ2μ1>0,ϱ(ι+k)=ηk(ϱ(ιk)),kN, (1.4)

    and established some new conditions for the oscillation of (1.4) when q3(ι)PC([ι0,),R+) and ηiJi(ϱ)ϱ1 [8]. Karpuz et al. [9] studied on advanced case, that is, taking a non homogeneous system and established the results for sufficient conditions for oscillation of (1.4). Tripathy et al. [10] have taken the following equations to establish the oscillatory and non-oscillatory character of a second order neutral impulsive differential system (IDS)

    {(ϱ(ι)η2ϱ(ιμ2))+η1ϱ(ιμ1)=0,ιιk,Δ(ϱ(ιk)η2ϱ(ιkμ2))+η1ϱ(ιkμ1)=0, (1.5)

    here kN, all coefficients and delays are constants. In [11] new result established for second-order neutral delay DS

    {(q3(ι)(ϱ(ι)+q2(ι)ϱ(ιμ2)))+q1(ι)(ϱ(ιμ1))=0,ιιk,Δ(q3(ιk)(ϱ(ιk)+q2(ιk)ϱ(ιkμ2)))+q4(ιk)(ϱ(ιkμ1))=0, (1.6)

    where kN. Santra et al. [12] observed the characteristic of solutions for first-order neutral delay IS of the form

    {(ϱ(ι)q2(ι)ϱ(ιμ2))+q1(ι)(ϱ(ιμ1))=0,ϱ(ι+k)=Jk(ϱ(ιk)),ϱ(ι+kμ3)=Jk(ϱ(ι+kμ3)), (1.7)

    here kN, taking varying values of the neutral coefficient q2. Also, Santra et al. in [13] established the necessary and sufficient results for oscillation of the solutions of the below systems with impulses applying Lebesgue's Dominated convergent theorem,

    {(ˊq(ι)(w(ι))α)+mi=1qi(ι)i(ϱ(σ(ι)))=0,Δ(ˊq(ιk)(w(ι))α)+mi=1qi(ιk)i(ϱ(σ(ιk)))=0, (1.8)

    where w(ι)=ϱ(ι)+ˊq(ι)ϱ(ˊσ(ι)), with

    Δϱ(η)=limsη+ϱ(s)limsηϱ(s),

    and 1ˊq(ι)0. In 2020, Li et al. studied the dynamic behavior of a computer worm system under a discontinuous control strategy and some conditions for globally asymptotically stable solutions of the discontinuous system were obtained by using the Bendixson–Dulac theorem, Green's formula and the Lyapunov function [14]. Also, they investigated the global dynamics of a controlled discontinuous diffusive SIR epidemic system under Neumann boundary conditions [15].

    The authors observed oscillatory and non-oscillatory both conditions for the solutions of the non linear neutral DE of the form

    {(q3(ι)(ϱ(ι)+q2(ι)ϱ(ιμ2)))+q1(ι)(ϱ(ιμ1))=(ι),Δ(q3(ιk)(ϱ(ιk)+q2(ιk)ϱ(ιkμ2)))+q1(ιk)(ϱ(ιkμ1))=(ιk),kN. (1.9)

    At last we observed some modern results in [16] where Tripathy and Santra improved oscillatory results of non-linear neutral IS of the form

    {(ˊq(ι)(w(ι))α)+mi=1qi(ι)ϱβi(σi(ι))=0,ιι0,ιιk,Δ(ˊq(ιk)(w(ιk))α)+mi=1qi(ι)ϱβi(σi(ιk))=0,kN,

    where w(ι)=ϱ(ι)+ˉq(ι)ϱ(ˊσ(ι)) with 1ˉq(ι)0 [17].

    Motivated by the above works, in this paper, we consider the IS

    {(ˊq(ι)(ϱ(ι))α)+mi=1qi(ι)i(ϱ(σi(ι)))=0,ιι0,Δ(ˊq(ιk)(ϱ(ιk))α)+mi=1qi(ιk)i(ϱ(σi(ιk)))=0,ιιk (1.10)

    where i,qi,ˊq,σi are continuous and α be the quotient of two positive odd integers which satisfy the given following postulate as

    (B1) σiC([0,),R), σi(ι)<ι, limισi(ι)=;

    (B2) ˊqC1([0,),R), qiC([0,),R); ˊq(ι)>0, qi(ι)0, for each ι0 & i=1,2,,m, qi(ι)0 in any [τ,);

    (B3) ϱiC(R,R) is non-decreasing and i(ϱ)ϱ>0 for ϱ0(i=1,2,,m);

    (B4) limι˚R(ι)= where

    ˚R(ι)=ι0(ˊq(η))1/αdη; (1.11)

    (B5) α be the quotient of two positive odd integers and the sequence ιk satisfies ι1<ι2<<ιk, as k.

    The main objective of this paper is to find out both necessary and sufficient conditions for the oscillation of all solutions to IS (1.10). In this direction, we refer [18,19,20,21,22,23,24,25,26,27,28,29,30,31,32] to the readers for more details on this study. All functional inequalities assumed here should be held eventually i.e., for all large ι that also satisfy whereas the domain is not clearly given.

    In Section 2, we recall some essential definition and necessary lemmas. Section 3 contains our main results in this work, while an example is presented to support the validity of our obtained results in Section 4. In Section 5, conclusion are presented.

    First we start and prove the following key lemma.

    Lemma 2.1. Consider postulates (B1)(B4), and that ϱ is converges to zero for the Eq (1.10). So there exist ι1ι0 and δ>0, so we have

    0<ϱ(ι)δ˚R(ι), (2.1)
    (˚R(ι)˚R(ι1))[ιmi=1qi(ι)i(ϱ(σi(ι)))dι+ιkιmi=1qi(ιk)i(ϱ(σi(ιk)))]1/αϱ(ι),ιι1. (2.2)

    Proof. Suppose ϱ is converge to zero. From (B1) There exists ι so that ϱ(ι)>0 and ϱ(σi(ι))>0, for all ιι and i=1,2,,m. Then by (1.10) we get

    {(ˊq(ι)(ϱ(ι))α)=mi=1qi(ι)i(ϱ(σi(ι)))0,Δ(ˊq(ιk)(ϱ(ιk))α)=mi=1qi(ιk)i(ϱ(σi(ιk)))0. (2.3)

    So, ˊq(ι)(ϱ(ι))α is non-increasing for ιι. Then ˊq(ι)(ϱ(ι))α>0. For contradiction let us consider

    ˊq(ι)(ϱ(ι))α0,

    at a certain time ιι. Applying qi0 in [τ,), and that (ϱ)>0 for ϱ>0, by (2.3), there exist ι2ι we get

    ˊq(ι)(ϱ(ι))αˊq(ι2)(ϱ(ι2))α<0,ιι2.

    From (B5), we get

    ϱ(ι)(ˊq(ι2)ˊq(ι))1/αϱ(ι2),ιι2.

    Taking integration from ι2 to ι, we get

    ϱ(ι)ϱ(ι2)+(ˊq(ι2))1/αϱ(ι2)(˚R(ι)˚R(ι2)). (2.4)

    Applying (B4), in the right part goes to ; so limιϱ(ι)=. That contradict to ϱ(ι)>0. Consequently

    ˊq(ι)(ϱ(ι))α>0,ιι.

    By ˊq(ι)(ϱ(ι))α is non-increasing, so we get

    ϱ(ι)(ˊq(ι1)ˊq(ι))1/αϱ(ι1),ιι1.

    Now we integrate the above inequality ι1 to ι and applying ϱ is continuous we have

    ϱ(ι)ϱ(ι1)+(ˊq(ι1))1/αϱ(ι1)(˚R(ι)˚R(ι1)).

    As limι˚R(ι)=, then δ>0 so that (2.1) satisfies. As

    ˊq(ι)(ϱ(ι))α>0,

    and non-increasing, so the limit of

    limιˊq(ι)(ϱ(ι))α,

    non negatively exists. Taking integration (1.10) from ι to τ, we have

    ˊq(τ)(ϱ(τ))αˊq(ι)(ϱ(ι))α+ιmi=1qi(η)i(ϱ(σi(η)))dη+ιkιmi=1qi(ιk)i(ϱ(σi(ιk)))=0.

    Calculating the limit when τ,

    ˊq(τ)(ϱ(τ))αιmi=1qi(ζ)i(ϱ(σi(η)))dη+ιιkmi=1qi(ιk)i(ϱ(σi(ιk))). (2.5)

    Therefore

    ϱ(ι)[1ˊq(ι)[ιmi=1qi(ζ)i(ϱ(σi(η)))dη+ιιkmi=1qi(ιk)i(ϱ(σi(ιk)))]]1/α.

    As ϱ(ι1)>0, integrating this inequality derives

    ϱ(ι)ζι1[1ˊq(ζ)ηmi=1qi(η)i(ϱ(σi(η)))dη+ιιkmi=1qi(ιk)i(ϱ(σi(ιk)))]1/αdζ.

    As the integrand is positive, increasing the lower limit from η to ι, and after that using the definition of ˚R(ι), we have

    ϱ(ι)(˚R(ι)˚R(ι1))[ιmi=1qi(ζ)i(ϱ(σi(ζ)))dζ+ιιkmi=1qi(ιk)i(ϱ(σi(ιk)))]1/α,

    this yields (2.2).

    Now for next result we consider the a constant γ, which satisfy (B5) with γ<α, so that

    i(z)zγ, (3.1)

    is non-increasing for z>0(i=1,2,,m).

    Example 3.1. An instant i(z)=|z|βsgn(z), with 0<β<γ holds this condition.

    Theorem 3.2. Letting (B1)(B5) and (3.1), each solution of (1.10) is oscillatory iff

    [0mi=1qi(η)i(δ˚R(σi(η)))dη+i=1mi=1qi(ιk)i(δ˚R(σi(ιk)))]=,δ>0.

    Proof. We prove the sufficient part by contradiction. For the purpose of sufficient part prove, at the beginning let ϱ is eventually positive solution. As, Lemma 2.1 satisfies, and so that there exists ι1ι0, we have

    ϱ(ι)(˚R(ι)˚R(ι1))w1/α(ι)0,ιι1, (3.2)

    where

    w(ι)=ιmi=1qi(η)i(ϱ(σi(η)))dη+ιkιmi=1qi(ιk)i(ϱ(σi(ιk)))0.

    As limι˚R(ι)=, then there exists ι2ι1, so that ˚R(ι)˚R(ι1)12˚R(ι), for ιι2. Then

    ϱ(ι)12˚R(ι)w1/α(ι). (3.3)

    Therefore

    w(ι)=mi=1qi(ι)i(ϱ(σi(ι))),Δw(ιk)=mi=1qi(ιki(ϱ(σi(ιk)))0.

    Therefore from the above we see that w0 and decreasing. As ϱ is positive, from (B3), i(ϱ(σi(ι))) is also positive, and by (B2), it gives us

    mi=1qi(ι)i(ϱ(σi(ι)))0,

    in any [τ,); thus w0 and w never be a constant in any interval [τ,). Thus w(ι) be also positive for ιι1. Calculating derivative,

    (w1γ/α(ι))=(1γα)wγ/α(ι)w(ι). (3.4)

    Integrating (3.4) from ι2 to ι, and applying w>0, we get

    w1γ/α(ι2)(1γα)[ιι2wγ/α(η)w(η)dηι2ιkwγ/α(ιk)Δw(ιk)]=(1γα)[ιι2wγ/α(η)(mi=1qi(η)i(ϱ(σi(η))))dη+ιkιwγ/α(ιk)mi=1qi(ιk)i(ϱ(σi(ιk)))]. (3.5)

    Next we search a lower bound for the right part of (3.5), which is not dependent of the solution ϱ. From (B3), (2.1), (3.1) and (3.3), we get

    i(ϱ(ι))=i(ϱ(ι))ϱγ(ι)ϱα(ι)i(δ˚R(ι))(δ˚R(ι))γϱγ(ι)i(δ˚R(ι))(δ˚R(ι))γ(˚R(ι)w1/α(ι)2)γ=i(δ˚R(ι))(2δ)γwγ/α(ι),ιι2.

    As w is non-increasing, γα>0, and σi(η)<η, it ensure us that

    i(ϱ(σi(η)))i(δ˚R(σi(η)))(2δ)γwγ/α(σi(η))i(δ˚R(σi(η)))(2δ)γwγ/α(η). (3.6)

    Returning to (3.5), we get

    w1γ/α(ι2)1γα(2δ)γ[ιι2mi=1qi(η)i(δ˚R(σi(η)))dη+ιkιmi=1qi(ιk)i(δ˚R(σi(ιk)))]. (3.7)

    As 1γα is positive, from (3.2) the right part goes to infinity as ι. It is a contradiction (3.7) and completes the sufficient part of the eventually positive solutions. Now we find solution for negative ϱ, for that we set the variables ˊϱ=ϱ and

    ˊi(ˊϱ)=i(ˊϱ).

    Thus (1.10) converted to positive solution of ˊϱ and ˊi in exchange with i. Write after ˊi satisfies (B3) and (3.1) then using the method for the solution ˊϱ from the above. In the subsequent part we prove the necessary condition by contrapositive thought. Whenever (3.2) does not satisfy we search an eventually positive solution which diverge to zero. Then for positive δ and for each positive ϵ there exists ι1ι0 if (3.2) does not satisfy so that

    ηmi=1qi(η)i(δ˚R(σi(η)))dη+ιkSmi=1qi(ιk)i(δ˚R(σi(ιk)))ϵ, (3.8)

    for all ηι1. Here ι1 rely on δ. Now here we are assume there exist set of continuous function

    Υ={ϱC([0,)):(ϵ2)1/α(˚R(ι)˚R(ι1))ϱ(ι)ϵ1/α(˚R(ι)˚R(ι1)),ιι1}.

    Next, we define an operator O on Υ by

    (Oϱ)(ι)={0,ιι1,ιι1[1ˊq(ζ)[ϵ2+ζmi=1qi(η)i(ϱ(σi(η)))dη+ιkSmi=1qi(ιk)i(ϱ(δ˚R(σi(ιk))))]]1/αdζ,ι>ι1.

    Here we see that when ϱ is continuous, Oϱ is also continuous on [0,). If Oϱ=ϱ, i.e., ϱ is a fixed point of O(ϱ) is a solution of (1.10). Initially we calculate (Oϱ)(ι) from below. Since ϱΥ, we get

    0ϵ1/α(˚R(ι)˚R(ι1))ϱ(ι).

    By (B3), we get 0i(ϱ(σi(η))) and by (B2) we get

    (Oϱ)(ι)0+ιι1[1ˊq(ζ)[ϵ2+0+0]]1/αdζ=(ϵ2)1/α(˚R(ι)˚R(ι1)).

    Then we calculate (Oϱ)(ι) from above. For ϱ in Υ, from (B2) and (B3), we get

    i(ϱ(σi(η)))i(δ˚R(σi(η))).

    From (3.8),

    (Oϱ)(ι)ιι1[1ˊq(η)[ϵ2+ηmi=1qi(ζ)i(δ˚R(σi(ζ)))dζ+ιkSmi=1qi(ιk)i(δ˚R(σi(ιk)))]]1/αdηϵ1/α(˚R(ι)˚R(ι1)).

    Thus, O maps Υ to Υ. Later on we will look for O in Υ. We are going to explain a sequence of function Υ by the iterative formula

    z0(ι)=ˉ0,ιι0,z1(ι)=(Oz0)(ι)={ˉ0,ι<ι1,ϵ1/α(˚R(ι)˚R(ι1)),ιι1,zn+1(ι)=(Ozn)(ι),n1,ιι1.

    Now when we fixed ι, we can get z1(ι)z0(ι). Applying that is non-decreasing and also using induction formula of mathematics, we can formulate that zn+1(ι)zn(ι). Thus, {zn} convergent sequence which converges to z pointwise. Here we find the fixed point z for the operator O in Υ applying dominated convergence theorem of Lebesgue. From consideration (3.8) shows that the solution is eventually positive i.e., does not converge to zero. Hence the proof of the theorem is complete.

    For subsequent theorem, let us consider there exists a continuously differentiable function σ0 satisfying

    0<σ0(ι)σi(ι),γ>0:γσ0(ι)(ιι0,i=1,2,,m). (3.9)

    Also, we suppose a constant γ, satisfy first part of (B5), and α<γ, such that

    i(z)zγ, (3.10)

    is non-decreasing for z>0(i=1,2,,m). The Example 3.1, i(z)=|z|βsgn(z) with γ<β holds this condition.

    Theorem 3.3. Under assumptions (B1)(B4), (3.9), (3.10), and ˊq(ι) is non-decreasing, every solution of (1.10) is converges to zero iff

    ι1[1ˊq(ζ)[ζmi=1qi(η)dη+ιkιmi=1qi(ιk)]]1/αdζ=. (3.11)

    Proof. Our aim to prove sufficient part by contradiction method. First we consider that the solution ϱ does not converges to zero. Applying similar logic same as in Lemma 2.1, we get ι1ι0 and ϱ(σi(ι)) is positive and

    ˊq(ι)(ϱ(ι))α>0,

    and non-increasing. Since ˊq(ι)>0 so ϱ(ι) is increasing for ιι1. From (B3), ϱ(ι)ϱ(ι1) and (3.10), we have

    i(ϱ(ι))i(ϱ(ι))ϱγ(ι)ϱγ(ι)i(ϱ(ι1))ϱγ(ι1)ϱγ(ι). (3.12)

    From (B1) we can find ι2ι1 and also σi(ι)ι1 when ιι2. Therefore

    i(ϱ(σi(ι)))i(ϱ(ι1))ϱγ(ι1)ϱγ(σi(ι)),ιι2. (3.13)

    Using this inequality, (2.5), we have σi(ι)σ0(ι) which shows that σ is increasing, and ϱ is also so, thus

    ˊq(ι)(ϱ(ι))αϱγ(σ0(ι))ϱγ(ι1)[ιmi=1qi(η)i(ϱ(ι1))dη+ιkιmi=1qi(ιk)i(ϱ(ι1))],

    for ιι2. From ˊq(ι)(ϱ(ι))α being non-increasing and σ0(ι)ι, we get

    ˊq(σ0(ι))(ϱ(σ0(ι)))αˊq(ι)(ϱ(ι))α.

    We apply this in the left part of the above inequality. Additionally, dividing by ˊq(σ0(ι))>0, uplift right and left part to 1α index, and divided by ϱβ/γ(σ0(ι))>0, we get

    ϱ(σ0(ι))ϱγ/α(σ0(ι))[1ˊq(σ0(ι))ϱγ(ι1)[ιmi=1qi(η)i(ϱ(ι1))dη+ιkιmi=1qi(ιk)i(ϱ(ι1))]]1/α,

    for ιι2. Multiply by σ0(ι)/β1 left part, and taking integration from ι1 to ι,

    1βιι1ϱ(σ0(η))σ0(η)ϱγ/α(σ0(η))dη1ϱγ/α(ι1)[ιι1[1ˊq(σ0(η))ηmi=1qi(ζ)i(ϱ(ι1))dζ+sιkmi=1qi(ιk)i(ϱ(ι1))]]1/αdη. (3.14)

    As α<γ, taking integration left part of above inequality, we finally reach

    1β(1γ/α)[ϱ1γ/α(σ0(η))]ιs=ι21γ(γ/α1)ϱ1γ/α(σ0(ι2)).

    Our main task is to show that (3.11) right part going to infinity as ι tends to infinity for that here apply

    min1imi(ϱ(ι1))>0,

    and ˊq(σ0(s))ˊq(s), (3.14) right part. For eventually negative solutions, we use the same change of variables as in Theorem 3.2, and proceed as above. To prove the necessary part we assume that (3.11) does not hold, and obtain an eventually positive solution that does not converge to zero. If (3.11) does not hold, then for each ϵ>0 there exists ι1ι0 such that

    ι1[1ˊq(η)ηmi=1qi(ζ)dζ+ιkιmi=1qi(ιk)]1/αdη<ϵ2(i(ϵ))1/α,ιι1. (3.15)

    Construct the continuous functions

    Υ={ϱC([0,)):ϵ2ϱ(ι)ϵ when ιι1}. (3.16)

    Now we define the operator O,

    (Oϱ)(ι)={0,ιι1,ϵ2+[ιι11ˊq(ζ)[ζmi=1qi(η)i(ϱ(σi(η)))dη+ιkSmi=1qi(ιk)i(ϱ(σi(ιk)))]]1/αdζ,ι>ι1.

    Note that if ϱ is continuous, for ι=ι1, O(ϱ) is a continuous function. Also as ϱ is a fixed point i.e., Oϱ=ϱ it give us that ϱ is a solution of (1.10). Our main criteria to calculate (Oϱ)(ι) from both equations for first part let ϱΥ. By 0<ϵ2ϱ, we have

    (Oϱ)(ι)ϵ2+0+0,

    on [ι1,). For the next part let ϱΥ. Then ϱϵ and from (3.15), we have

    (Oϱ)(ι)ϵ2+(i(ϵ))1/αιι1[1ˊq(η)ηmi=1qi(ζ)dζ+ιkSmi=1qi(ιk)]1/αdηϵ2+ϵ2=ϵ.

    Hence O is a rules from Υ to Υ. For finding a fixed point of O we can construct a sequence of function by recursive rules

    z0(ι)=ˉ0,ιι0,z1(ι)=(Oz0)(ι)=1,ιι1,zn+1(ι)=(Ozn)(ι),ιι1,n1.

    Now when we fixed ι, thus z1(ι)z0(ι). Applying is non-decreasing and also induction formula of mathematics, we can establish zn+1(ι)zn(ι) so that {zn} convergent sequence which converges to z in Υ pointwise. Hence, z be a positive solution of (1.10). This completes the proof.

    In this section, we are going to conclude the paper by providing two examples to show the effectiveness and feasibility of the main results.

    Example 4.1. Consider the IS

    {(eι(ϱ(ι))113)+1ι+1(ϱ(ι2))13+1ι+2(ϱ(ι1))53=0,(ek(ϱ(k))113)+1ι+4(ϱ(k2))13+1ι+5(ϱ(k1))53=0. (4.1)

    Comparing with said systems we get α=113, ˊq(ι)=eι, σ1(ι)=ι2, σ2(ι)=ι1, from (1.11)

    ˚R(ι)=ι0(ˊq(η))1αdη=ι0e3η11dη=113(e3ι111), (4.2)

    1(ϱ)=ϱ13 and 2(ϱ)=ϱ53. For β=73, we have

    0<max{γ1,γ2}=max{13,53}=53<73=β<113=α,

    and

    1(ϱ)ϱβ=ϱ13ϱ73=ϱ2,2(ϱ)ϱβ=ϱ53ϱ73=ϱ23,

    which both are non increasing. To verify (3.2), by employing (4.2), we have

    [0mi=1qi(η)i(δ˚R(σi(η)))dη+k=1mi=1qi(ιk)i(δ˚R(σi(ιk)))]0mi=1qi(η)i(δ˚R(σi(η)))dη0q1(η)1(δ˚R(σ1(η)))dη=01η+1(δ113(1e3(η2)11))13dη=,δ>0, (4.3)

    as integrand goes to + since η to positive infinity.

    One can see these results in Tables 1 and 2. We can see graphical representation of the inequality (4.3) for η[0,0.6] and η[0,1.75] in Figure 1 (a) and (b), respectively, for η=1.5 and δ{0.5,2.5} in Figure 2. Algorithmes 1 and 2 can be used for this purpose.

    Table 1.  Numerical results of the integral inequality (4.3) of IS for η{0.6,1.75} in Example 4.1.
    η∈[0, 0.6] η∈[0, 1.75]
    n η ˚R IS (4.3) η ˚R IS (4.3)
    1 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
    2 0.0545 0.2015 0.0784 0.1591 0.5962 0.2188
    3 0.1091 0.4060 0.1531 0.3182 1.2188 0.4122
    4 0.1636 0.6136 0.2247 0.4773 1.8690 0.5861
    5 0.2182 0.8243 0.2933 0.6364 2.5481 0.7444
    6 0.2727 1.0381 0.3592 0.7955 3.2572 0.8901
    7 0.3273 1.2552 0.4226 0.9545 3.9978 1.0254
    8 0.3818 1.4755 0.4839 1.1136 4.7713 1.1519
    9 0.4364 1.6991 0.5430 1.2727 5.5790 1.2709
    10 0.4909 1.9261 0.6002 1.4318 6.4226 1.3835
    11 0.5455 2.1564 0.6556 1.5909 7.3036 1.4905
    12 0.6000 2.3902 0.7094 1.7500 8.2236 1.5926

     | Show Table
    DownLoad: CSV
    Table 2.  Numerical results of the integral inequality (4.3) of IS for η[0,1.5] and δ{0.5,2.5} in Example 4.1.
    δ=0.5δ=0.5 δ=2.5δ=2.5
    η ˚R IS (4.3) ˚R IS (4.3)
    0.0000 0.0000 0.0000 0.0000 0.0000
    0.1364 0.5094 0.1312 0.5094 0.2244
    0.2727 1.0381 0.2490 1.0381 0.4259
    0.4091 1.5869 0.3562 1.5869 0.6090
    0.5455 2.1564 0.4546 2.1564 0.7773
    0.6818 2.7475 0.5458 2.7475 0.9333
    0.8182 3.3611 0.6310 3.3611 1.0790
    0.9545 3.9978 0.7110 3.9978 1.2157
    1.0909 4.6587 0.7865 4.6587 1.3449
    1.2273 5.3447 0.8581 5.3447 1.4674
    1.3636 6.0566 0.9263 6.0566 1.5840
    1.5000 6.7955 0.9915 6.7955 1.6955

     | Show Table
    DownLoad: CSV
    Figure 1.  2D plot numerical results of ˚R and the integral inequality of IS (4.3) for η{0.6,1.75} in Example 4.1.
    Figure 2.  Graphical representation of of the integral inequality of IS (4.3) for η=1.5 and δ{0.5,2.5} in Example 4.1.

    Therefore, all the postulates of Theorem 3.2 hold true. Hence, by Theorem 3.2 all solution of (4.1) is oscillatory.

    Example 4.2. Let us assume nonlinear IS

    {((ϱ(ι))13)+ι(ϱ(ι2))73+(ι+1)(ϱ(ι1))113=0((ϱ(3k))13)+(ι+3)ϱ(3k2)73+(ι+4)(ϱ(3k1))113=0. (4.4)

    Now comparing with given system we have α=13, ˊq(ι)=1, σ1(ι)=ι2, σ2(ι)=ι1, from (1.11)

    ˚R(ι)=ι0(ˊq(η))1αdη=ι0dη=ι, (4.5)

    1(ϱ)=ϱ73 and 2(ϱ)=ϱ113. For {\rm{ \mathsf{ γ}}} = \frac{5}{3} , thus

    \min \Big\{ {\rm{ \mathsf{ γ}}}_1, {\rm{ \mathsf{ γ}}}_2\Big\} = \Big\{ \frac{7}{3} , \frac{11}{3}\Big\} = \frac{7}{3} > \frac{5}{3} = {\rm{ \mathsf{ γ}}} > \frac{1}{3} = {\rm{ \mathsf{ α}}},

    also

    \begin{align*} \frac{\wp_1( \varrho)}{\varrho^{\rm{ \mathsf{ γ}}}} & = \frac{\varrho^{{{}^{7}\!\!\diagup\!\!{}_{3}\;}} }{ \varrho^{{{}^{5}\!\!\diagup\!\!{}_{3}\;}}} = \varrho^{2/3}, & \frac{\wp_2(\varrho )}{\varrho^{\rm{ \mathsf{ γ}}}} & = \frac{\varrho^{{{}^{11}\!\!\diagup\!\!{}_{3}\;}}}{\varrho^{{{}^{5}\!\!\diagup\!\!{}_{3}\;}}} = \varrho^{2}, \end{align*}

    two functions are increasing functions. To verify (3.11) we get

    \begin{align*} \int_{\iota_0}^\infty \bigg[ \frac{1}{\acute{\mathrm{q}}(\zeta)} & \int_S^\infty \sum\limits_{i = 1}^m \mathrm{q}_i(\eta) \, {\mathrm d}\eta + \sum\limits_{\iota_k\geq S} \sum\limits_{i = 1}^m \mathrm{q}_i( \iota_k) \bigg]^{1/{\rm{ \mathsf{ α}}}}\, {\mathrm d}\zeta \\ & \geq \int_{\iota_0}^\infty \bigg[ \frac{1}{\acute{\mathrm{q}}( \zeta)} \int_\zeta^\infty \sum\limits_{i = 1}^m \mathrm{q}_i(\eta) \, {\mathrm d} \eta \bigg]^{1/{\rm{ \mathsf{ α}}}} \, {\mathrm d} \zeta \\ & \geq \int_{\iota_0}^\infty \bigg[ \frac{1}{\acute{\mathrm{q}} (\zeta)} \int_\zeta^\infty \mathrm{q}_1(\eta)\, {\mathrm d}\eta \bigg]^{1/{\rm{ \mathsf{ α}}}} \, {\mathrm d}\zeta \\ &\geq \int_2^{\infty} \bigg[ \int_\zeta^\infty \eta\, {\mathrm d} \eta \bigg]^3 \, {\mathrm d} \zeta = \infty. \end{align*}

    Therefore, all postulate of Theorem 3.3 hold true. Hence, by Theorem 3.3, all solution of (4.4) is oscillatory or converges to zero.

    After concluding the paper and introducing [16,17,22,23,25,29,30,33,34,35], we have an open question that "Can we find the necessary and sufficient conditions for the oscillatory solution of the second order neutral impulsive delay differential system with several delays and arguments"?

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors are very grateful to the reviewers for their careful reading of the manuscript. M. Altanji's work has been supported by the Deanship of Scientific Research at King Khalid University through the Research Group program with Grant Number R.G.P2/91/44.

    The authors declare that they have no competing interests.

    Algorithm 1: MATLAB lines for calculation all variables in Example 4.1 when \eta changes and the \delta is constant.

    Algorithm 2: MATLAB lines for calculation all variables in Example 4.1 whenever \eta is constant and when the \delta changes.



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