# Oscillatory Behaviour of a First-Order Neutral Differential Equation in relation to an Old Open Problem

Authors: K. C. Panda; R. N. Rath; S. K. Rath
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

In this paper, we obtain sufficient conditions for oscillation and nonoscillation of the solutions of the neutral delay differential equationyt−∑j=1kpjtyrjt′+qtGygt−utHyht=ft, where pjandrj for each j and q,u,G,H,g,h,andf are all continuous functions and q≥0,u≥0,ht<t,gt<t,andrjt<t for each j. Further, each rjt, gt, and ht⟶∞ as t⟶∞. This paper improves and generalizes some known results.


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## Body

## 1. Introduction

Consider the neutral delay differential equation (NDDE in short) of the first order:(1)yt−∑j=1kpjtyrjt′+qtGygt−utHyht=ft,where pjandrj for each j and q,u,g,h,G,H,andf are in Cℝ,ℝ and q≥0,u≥0,gt<t,ht<t,andrjt<t for each j. Further, each rjt, gt, and ht⟶∞ as t⟶∞. We study the behavior of oscillation and nonoscillation of solutions of neutral differential equation (1) under the following assumptions:(2)xGx>0,forx≠0.There exists a real-valued bounded functionFt such that(3)F′t=ft.The functionFt in (3) satisfies(4)limt⟶∞Ft=0,(5)qt>0,∫t0∞qsds=∞,(6)∫t0∞utdt<∞.H is bounded and(7)uHu>0,foru≠0.Study of the oscillatory and asymptotic behaviour of neutral delay differential equations are undertaken by numerous authors due to its various applications (see [1]) in the fields of science and technology. Results concerned with oscillation and nonoscillation of the NDDE(8)yt−ptyt−τ′+qtGyσt=ft,are obtained in [2–5]. Further, the authors in [6–9] obtained oscillation results for the NDDE:(9)yt−ptyt−τ′+qtGyt−σ−utHyt−α=ft.The research papers [2–5, 7, 8] and many others while studying nonlinear NDDEs assume the condition “G is nondecreasing.” It is found that the authors in [3–5, 7, 8, 10] use the lemmas [1] (Lemma 1.5.1 and 1.5.2) as the main tool to study NDDEs (8) or (9). If liminft⟶∞yt=0 and limt⟶∞yt−ptyt−τ exists finitely then, these lemmas help us to evaluate limt⟶∞yt. Hence, the lemmas could only be applied to study (8) or (9) where there is only one functional delay term under the derivative but could not be applied to study (1) because of the presence of more than one functional delay term under the derivative. Further, “the note [1] (notes 1.8, page 31) suggests to extend these lemmas meant for the function with one delay to multiple delays and apply it to the study of NDDE (1) with several delays.” But, it seems difficult to extend these lemmas for the said purpose. The motivation behind this work is that no result in the literature appears to have an answer to the qualitative behaviour of solutions to the NDDE:(10)yt−e−t+e−4yt−4−e−t+e−5yt−5′+2et+1+2et+2+e2t−3y3t−1=4e−2t+4+4e−2t+5+2e−t.In this paper, we remove the condition “G is nondecreasing” that is assumed in [2–5, 7, 8] and study the oscillatory behaviour of solutions of the NDDE (1) and then apply the results to study the NDDE:(11)yt−∑j=1kpjtyrjt′+vtGygt=ft,where vt changes sign. The oscillatory behaviour of solutions of the discrete analogue of (1) with u = 0 is obtained in [11] by using two lemmas which are the discrete versions of Lemmas 1 and 2. We followed the methods of Rath et al. in some of our results and applied Lemmas 1 and 2 to answer the open problem [1] (Problem 2.8.3, p. 57). The oscillatory behaviour of the solution of the NDDE (1) in some results is proved to be different from that in [11].“Lett1 be a fixed positive real number, and(12)t0=mininft≥t1r1t,r2t,…,rkt,inft≥t1gt,inft≥t1ht.By a solution of (1), we mean a function y∈Ct0,∞,ℝ such that yt−∑i=1kpityrit is differentiable on t0,∞ and the neutral equation (1) is satisfied by yt for all t≥t1. It is known that (1) has a unique solution provided that an initial function ϕ∈Ct0,t1,ℝ is given to satisfy yt=ϕt for all t∈t0,t1. Such a solution is said to be nonoscillatory if it is eventually positive or eventually negative, otherwise it is called oscillatory.”In this work, we assume the existence of solutions of (1) and study only their qualitative behaviour. In the sequel, unless otherwise specified, when we write a functional inequality, it will be assumed to hold for all sufficiently large values of t.


## 2. Oscillation Results

In this section, we present some results which prove that (5) is sufficient for any solution of (1) to be oscillatory or tending to zero as t⟶∞. We need the following lemmas for our work.Lemma 1 (see [12]).
“Letut and vt be two real-valued continuous functions defined for t≥t0≥0. Then,(13)liminft⟶∞ut+liminft⟶∞vt≤liminft⟶∞ut+vt≤limsupt⟶∞ut+liminft⟶∞vtor liminft⟶∞ut+limsupt⟶∞vt≤limsupt⟶∞ut+vt≤limsupt⟶∞ut+limsupt⟶∞vt,provided that no sum is of the form ∞−∞.”Lemma 2 (see [12]).
“Letut and vt be two nonnegative real-valued continuous functions defined for t≥t0. Then,(14)liminft⟶∞ut×liminft⟶∞vt≤liminft⟶∞ut×vt≤limsupt⟶∞ut×liminft⟶∞vtor liminft⟶∞ut×limsupt⟶∞vt≤limsupt⟶∞ut×vt≤limsupt⟶∞ut×limsupt⟶∞vt,provided that no product is of the form 0×∞.”Theorem 1.
Suppose that (2)–(7) hold, assume that there exists a positive constant p such that pjt for j=1,2,….,k satisfies the following condition:(15)pjt≥0,for everyj=1,2,…k,∑j=1klim suppjt<p<1.Then, every solution of (1) oscillates or tends to zero as t⟶∞.Proof.
Letyt be any solution of (1) for t≥t0, where t0 is a positive real number. If it oscillates, then there is nothing to prove; otherwise, it leads to two distinct possibilities, either yt>0 or yt<0 for t≥t1>t0. Consider the first one i.e., yt>0 eventually. There exists positive real t2≥t1 such that for t≥t2, we have yt>0,ygt>0,yht>0, and yrjt>0 for each j. Let us define for t≥t1:(16)ct=∫t∞usHyhsds.
Clearly, due to the assumptions (6) and (7), ct is well defined and the improper integral ct1 is convergent at ∞, say to α. Then, ct1=at+ct, where at=∫t1tusHyhsds. Then, limt⟶∞at=α. Hence, limt⟶∞ct=limt⟶∞ct1−at. Thus, it follows that(17)ct⟶0,ast⟶∞.
Further,(18)c′t=−utHyht.
Fort≥t2, define(19)zt=yt−∑j=1kpjtyrjt,(20)wt=zt+ct−Ft.
From (1), (19), and (20), it follows due to (2) that(21)w′t=−qtGygt≤0.
Then, there existst3≥t2 such that wt is monotonic and is of a constant sign for t≥t3. For the sake of a contradiction, assume that yt is not bounded. Then, there exists a sequence yan such that(22)an⟶∞,yan⟶∞asn⟶∞,(23)yan=maxyt:t3≤t≤an.
From (4) and (17), it follows that for ϵ>0, we can find a positive real t4>t3 such that, for t≥t4, it implies Ft<ϵ and ct<ϵ. Since each rjt⟶∞ as t⟶∞, we may choose n large enough so that rjan≥t4 for each j. From (15) and using (19), (20), and (23), we obtain(24)wan=yan−∑j=1kpjanyrjan+can−Fan≥1−∑j=1kpjanyan−2ϵ>1−pyan−2ϵ.
Takingn⟶∞, we find limt⟶∞wt=∞, a contradiction as wt is monotonic decreasing. Hence, yt is bounded which implies wt and zt are bounded and limwt exists. Further, it follows that liminft⟶∞yt and limsupt⟶∞yt exists. We claim liminft⟶∞yt=0. Otherwise, let yt≥α>0. Next, since yt is bounded above, there exists β>0 such that yt≤β. Hence, we have 0<α≤yt≤β, which will be used for bounding the G term in (1) from below.
From the continuity ofG and the assumption (2), it follows that there exists a positive lower bound m for G on α,β. Hence, there exists t5 such that Gygt>m>0 for t>t5. Then, integrating (21) from t=t5 to s, we obtain(25)wt5−ws=∫t5sqtGygtdt≥m∫t5sqtdt.
In the above inequality, the left hand side is bounded, while the right hand side approaches+∞, as s⟶∞. Thus, we have a contradiction. This yields liminft⟶∞yt=0. From (4), monotonic nature of wt, and (20), it follows that limt⟶∞zt exists finitely. Let limt⟶∞zt=δ.
Then,liminft⟶∞zt=limsupt⟶∞zt=δ. Since pjt≥0, zt≤yt, and liminft⟶∞zt≤liminft⟶∞yt, it implies δ≤0. Then, using Lemmas 1 and 2 we obtain(26)0≥δ=lim supt⟶∞zt=lim supt⟶∞yt−∑j=1kpjtyrjt≥lim supt⟶∞yt+lim inft⟶∞−∑j=1kpjtyrjt≥lim supt⟶∞yt−lim supt⟶∞∑j=1kpjtyrjt≥limsupt⟶∞yt−∑j=1klimsupt⟶∞pjtyrjt≥limsupt⟶∞yt−∑j=1klimsupt⟶∞pjtlimsupt⟶∞yrjt≥limsupt⟶∞yt1−∑j=1klimsupt⟶∞pjt≥limsupt⟶∞yt1−p≥0.
Hence,limsupt⟶∞yt≤0, by (15), which implies that limt⟶∞yt=0. If yt<0 for t>t1, then we set xt=−yt to obtain xt>0 and then (1) reduces to(27)xt−∑j=1kpjtxrjt′+qtG˜xgt−utH˜xht=f˜t,where(28)f˜t=−ft,G˜v=−G−vH˜v=−H−v.
Further,(29)F˜t=−Ft,impliesF˜′t=f˜t.
In view of the above facts, it can be easily verified thatG˜, H˜, and F˜ satisfy the corresponding conditions satisfied by the functions G, H, and F in the theorem. Proceeding as in the proof for the case yt>0, we may complete the proof of the theorem.Theorem 2.
Suppose that (2)–(7) hold. Assume that there exists a positive constant p such that the functions pjt for j=1,2,….,k satisfies the condition:(30)pjt≤0,for everyj=1,2,…k,∑j=1klim infpjt>−p>−1.Then, every solution of (1) oscillates or tends to zero as t⟶∞.Proof.
Proceeding as in the proof of Theorem1 and setting ct, zt, and wt as in (16), (19), and (20), respectively, we obtain (21) and further prove yt is bounded with liminft⟶∞yt=0. From (4) and that wt is monotonic, it follows that limt⟶∞wt=limt⟶∞zt=δ∈R. As zt≥0 by (30), δ≥0. We claim δ=0, and if not, then δ>0, and this implies(31)δ=liminft⟶∞zt=liminft⟶∞yt−∑j=1kpjtyrjt≤liminft⟶∞yt+limsupt⟶∞−∑j=1kpjtyrjt≤∑j=1klimsupt⟶∞−pjtlimsupt⟶∞yrjt=∑j=1k−liminft⟶∞pjtlimsupt⟶∞yrjt≤plimsupt⟶∞yt≤pα.
Hence, we get(32)α≥δp>δ.
Again(33)δ=limsupt⟶∞zt=limsupt⟶∞yt−∑j=1kpjtyrjt≥limsupt⟶∞yt+liminft⟶∞−∑j=1kpjtyrjt=limsupt⟶∞yt+liminft⟶∞∑j=1k−pjtyrjt≥limsupt⟶∞yt+∑j=1kliminft⟶∞−pjtyrjt≥limsupt⟶∞yt+∑j=1kliminft⟶∞−pjtliminft⟶∞yrjt=limsupt⟶∞yt=α,a contradiction due to the inequality (32). Hence, we conclude δ=0 and from zt>yt, it follows that limt⟶∞yt≤0. Hence, limt⟶∞yt=0.
The proof for the caseyt≤0 for large t is similar. Hence, the theorem is proved.Theorem 3.
Suppose that (2)–(7) hold. Assume that the functions pjt for j=1,2,….,k satisfies the condition:(34)pjt<0,for everyj=1,2,…kand there exists,i∈1,2,3,…,ksuch thatlimsuppit−∑j≠iliminfpjt<−1.Then, every solution of (1) oscillates or tends to zero as t⟶∞.Proof.
Proceeding as in the proof of Theorem1 and setting ct, zt, and wt as in (16), (19), and (20), respectively, we obtain (21) and further prove yt is bounded with liminft⟶∞yt=0. From (4) and that wt is monotonic, it follows that limt⟶∞wt=limt⟶∞zt=δ∈R. As zt≥0, δ≥0. We claim δ=0. If not, then δ>0, and this implies(35)δ=liminft⟶∞zt=liminft⟶∞yt−∑j=1kpjtyrjt≤limsupt⟶∞yt+∑j≠i−pjtyrjt+liminft⟶∞−pityrit≤limsupt⟶∞yt+limsupt⟶∞∑j≠i−pjtyrjt+limsupt⟶∞−pitliminft⟶∞yrit≤limsupt⟶∞yt+∑j≠ilimsupt⟶∞−pjtlimsupt⟶∞yrit≤limsupt⟶∞yt1−∑j≠iliminft⟶∞pjt.
Again, we have(36)δ=limsupt⟶∞zt=limsupt⟶∞yt−∑j=1kpjtyrjt≥liminft⟶∞yt+limsupt⟶∞−∑j=1kpjtyrjt=limsupt⟶∞−pityrit+liminft⟶∞∑j≠i−pjtyrjt≥limsupt⟶∞yritliminft⟶∞−pit+∑j≠iliminft⟶∞−pjtyrjt≥limsupt⟶∞yt−limsupt⟶∞pit+∑j≠iliminft⟶∞−pjtliminft⟶∞yrjt≥limsupt⟶∞yt−limsupt⟶∞pit.
From (35) and (36), it follows that(37)limsupt⟶∞yt∑j≠iliminfpjt−1−limsuppit≤0.
Using (34), we obtain limt⟶∞yt=0. Thus, the theorem is proved.For the next two results we assume(38)−bj≤pjt≤0,j=1,2,…,k.Note that (38) is less restrictive than (34).Theorem 4.
Suppose that (2), (3), (6), and (7) hold. Assume that (38) holds. If(39)liminft⟶∞Ft=−∞,limsupt⟶∞Ft=+∞,then every solution of (1) oscillates.Proof.
Letyt be a nonoscillatory solution of (1) for t≥t1. If yt>0, then set ct, zt, and wt as in (16), (19), and (20), respectively, to obtain (21) for t≥t2>t1. As yt>0 and each pjt≤0, then zt≥0, and by (7), we have ct≥0. From (21), it follows wt≥0 or wt≤0 for t≥t3>t2. If wt≤0, then zt+ct−Ft≤0. This implies Ft≥zt+ct≥0. Consequently, liminft⟶∞Ft≥0, a contradiction. If wt≥0 for t≥t3, then limt⟶∞wt=α∈ℝ. Clearly,(40)zt+ct=wt+Ft.
We notice that(41)0≤liminfzt+ct≤limsupt⟶∞wt+liminft⟶∞Ft=limt⟶∞wt+liminft⟶∞Ft=−∞,a contradiction. Hence, yt<0 for t≥t1. Then, again setting ct, zt, and wt as in (16), (19), and (20), respectively, we obtain(42)w′t=−qtGygt≥0,fort≥t2,
Asyt<0 and each pjt≤0, then zt≤0, and by (7), we have ct≤0. From (42), it follows wt≥0 or wt≤0 for t≥t3>t2. If wt≥0, then zt+ct−Ft≥0. This implies Ft≤zt+ct≤0. Consequently, limsupt⟶∞Ft≤0, a contradiction.
Ifwt≤0 for t≥t3, then limt⟶∞wt=β∈ℝ. From (40), we find that(43)0≥limsupzt+ct≥liminft⟶∞wt+limsupt⟶∞Ft=limt⟶∞wt+limsupt⟶∞Ft=∞,a contradiction. Hence, yt is oscillatory. Thus, the proof is complete.Note that the above theorem is independent of any assumption onqt. Thus, it holds even for conditions weaker than (5). Further, the above theorem does not hold for the homogeneous NDDE associated with (1). Next, we present the following result which holds for the homogeneous NDDE associated with (1).Theorem 5.
Assume conditions (2)–(4), (6), (7), and (38) to hold. Further, assume grit=rigt for i=1,2,…,k. Suppose that there exists a real α>0 such that ri′t≥1/α. Further, suppose that there exists a real μ>0 such that for xi>0,i=1,2,…,k+1 and u>0,(44)∑i=1k+1Gxi≥μG∑i=1k+1xi,GuGxi≥Guxi.Let(45)G−u=−Gu,(46)∫t1∞q∗t=∞,whereq∗t=minqt,qr1t,qr2t,…,qrkt.Then, every solution of (1) oscillates or tends to zero as t⟶∞.Proof.
Letyt be a nonoscillatory solution of (1) for t≥t1. Then, yt>0 or yt<0 for t>t2≥t1. Let yt>0 for t>t2. Setting ct, zt, and wt as in (16), (19), and (20), respectively, we obtain (21). Note that (18) and (17) also hold. As yt>0 and each pjt≤0, then zt≥0. From (21), (4), and (17), it follows limt⟶∞wt=limt⟶∞zt=δ≥0. If δ=0, then limt⟶∞yt=0 from the fact yt≤zt. For the sake of contradiction, let δ>0. We choose ϵ>0 such that δ>ϵ, and we have δ−ϵ≤zt≤δ+ϵ for t≥t3>t2. As G is continuous and (2) holds, we have(47)a≤Gzt≤bfort≥t4.
From (21), we have(48)w′t+qtGygt=0.
Replacet by rit in (48), then we obtain, for i=1,2,…,k,(49)w′rit+qritGygrit=0.
On multiplying (49) by G−pigt and then adding the corresponding k equations for i=1,2,…,k to (48), we obtain(50)w′t+qtGygt+∑i=1kG−pigtw′rit+∑i=1kG−pigtqritGygrit=0.
From (38), we find bj>0 such that 0≤−pjt≤bj for each j. Then, this implies due to continuity of G and (2) that there exists a real d>0 such that G−pjt<d for each j. Note that w′<0, and then using rigt=grit, (44), (46), and (47) in (50), we obtain(51)0≥w′t+d∑i=1kw′rit+q∗tGygt+∑i=1kG−pigtGygrit≥w′t+d∑i=1kw′rit+q∗tGygt+∑i=1kG−pigtygrit≥w′t+d∑i=1kw′rit+μq∗tGygt+∑i=1k−pigtygrit≥w′t+d∑i=1kw′rit+μq∗tGygt+∑i=1k−pigtygrit≥w′t+dα∑i=1kw′ritri′t+μq∗tGzgt≥w′t+dα∑i=1kw′ritri′t+aμq∗t.
Integrating the above inequality fromt5>t4 to s, we obtain(52)wt5′−aμ1+kdα∫t5sq∗tdt≥ws,where t5′=minrit5:i=1,2,…,k. Then, taking limit s⟶∞, we obtain ws⟶−∞, a contradiction, due to (46). Thus, limt⟶∞yt=0. The proof for the case when yt is eventually negative is similar.
Note that the above result holds forft≡0. Further note that condition (46) implies (5), but the converse is not necessarily true. However, if qt is monotonic, then both the conditions are equivalent.
Next, we intend to present a result wherepjt, j=1,2,3,…,k, satisfies the following condition:(53)pjt>0,for everyj=1,2,…kand there exists,i∈1,2,3,…,ksuch thatliminfpit−∑j≠ilimsuppjt>1.
For that purpose, we give an example which would lead us to our next result.Example 1.
Consider the first-order NDDE with variable several delay:(54)yt−e−t+eyt−1−e−t+e2yt−2′+eyt−3/2=0.
Note that, in the above NDDE with the several delay term under a derivative sign,p1t=e−t+e and p2t=e−t+e2 satisfy (53). This NDDE has an unbounded solution yt=et tending to ∞ as t⟶∞, unlike other results presented so far.
The above example is the motivating point to the statement of our next result.Theorem 6.
Suppose that (2)–(7) hold. Assume that the functions pjt for j=1,2,….,k satisfy condition (53). Then, every bounded solution of (1) oscillates or tends to zero as t⟶∞.Proof.
Ifyt is any positive solution of (1), then proceeding as in the proof of Theorem 3, with the application of Lemmas 1 and 2 we obtain(55)δ=liminft⟶∞zt≤α1−liminft⟶∞pit,δ=limsupt⟶∞zt≥−α∑j≠ilimsupt⟶∞pjt,where α=limsupt⟶∞yt and δ=limt⟶∞zt. From these two inequalities and (53), we obtain δ=α=0. This completes the proof of the theorem.Theorem 7.
Suppose that (2) and (5)–(7) hold. Let pjt,j=1,2,…,k, satisfy (53). Consider the homogeneous NDDE:(56)yt−∑j=1kpjtyrjt′+qtGygt=ft.Then, every bounded solutionyt of (56) oscillates.Proof.
Letyt be an eventually positive solution of (56). Then, setting zt as in (19) we obtain(57)z′t=−qtGygt≤0,and it follows that zt is monotonic and of constant sign on some interval t1,∞. By this, we have two distinct possibilities, i.e., zt>0 or zt<0. As pjt,j=1,2,…,k, satisfies (53), then let i=1. Suppose that yt is bounded. This implies zt is bounded. As zt is monotonic also then λ:=limt⟶∞zt exists as a finite number. If zt>0 or zt<0, then limt⟶∞zt exists finitely. Integrating (57) from t = t0 to s and taking limit s⟶∞, we obtain(58)∫t0∞qtGygtdt<∞.
Now, we claim thatliminfyt=0. Taking integration on (57),(59)zt−λ=∫t∞qsGygsds.
Sincezt is bounded, the above integral is convergent. This in turn, by (5), it implies that  liminfs⟶∞Gygs=0. As Gx≠0 for x≠0, liminfs⟶∞ygs=0 and because limt⟶∞gt=∞, liminft⟶∞yt=0. Following the line of proof as in Theorem 6, we find that limt⟶∞zt=δ=0, which implies that zt>0 because z is decreasing. Since pit≥1, then yt>∑j=1kpjtyrjt≥yrit. Consequently, liminft⟶∞yt≠0, a contradiction. Hence, the bounded solution yt cannot be eventually positive. The proof for the case when yt is bounded and eventually negative solution of (56) is similar. Thus, every bounded solution yt oscillates, and the theorem is proved.Example 2.
Consider the first-order NDDE (10) with several delays under the derivative. It satisfies all the conditions of Theorem 1. As such the equation has a solution yt=e−t which tends to zero as t⟶∞. However, no result in the literature appears to have an answer to the qualitative behaviour of solutions to this NDDE.Remark 1.
In this section, we have not assumed the condition thatG is nondecreasing unlike the authors in [3, 4, 10].


## 3. Nonoscillation Results

In this section, we show that (5) is necessary for every solution of (1) to be oscillatory or tending to zero as t⟶∞. Or equivalently ∫t0∞qtdt<∞ is sufficient for (1) to have a bounded positive solution which does not tend to zero as t⟶∞, even if the limit exists. For this, we need the following lemma.Lemma 3 (Krasnoselskii’s fixed-point theorem [13]).
LetX be a Banach space and S be a bounded closed convex subset of X. Let AandB be operators from S to X such that Ax+By∈S for every pair of x,y∈S. If A is a contraction and B is completely continuous, then the equation(60)Ax+Bx=x,has a solution in S.Theorem 8.
Assume that (3) and (6) hold. Further, assume that one of the conditions of (15) or (30) hold. Then, (5) is a necessary condition for all solution of (1) to be oscillatory or tending to zero as t⟶∞.Proof.
Suppose condition (15) holds. The proof for the case when (30) holds, would follow on similar lines. Assume for the sake of contradiction, that (5) does not hold. This implies that there exists real t0 such that(61)∫t0∞qtdt<∞.
Thus, all we need to show is the existence of a bounded solutionyt of (1) with liminft⟶∞yt>0. From (3), we find a positive constant c and a positive real t1>t0>0 such that(62)Ft<c,fort≥t1.
Choose a positive constantL such that L≥5c/1−p. Since G∈Cℝ,ℝ, then let(63)β=maxGx:c≤x≤L,γ=maxHx|:c≤x≤L.
Let(64)μ=maxβ,γ.
From (6), we find t2>t1 such that t>t2 implies(65)μ∫t∞usds<c.
Then by using (61), one can fix t3>t2 such that for t≥t3 it follows that(66)μ∫t∞qsds<c.
ChooseT1>t3 such that(67)T0=mingT1,hT1,r1T1,r2T1,…,rkT1.
LetX=BCT0,∞,R, a Banach space of real-bounded, continuous functions with norm x=supt≥T0|xt|<∞. Define(68)S=y∈X:c≤yt≤L,t≥T0.
Clearly S is a bounded closed and convex subset ofX. Define two operators A and B: S⟶X as follows. For y∈S, define(69)Ayt=AyT1,T0≤t≤T1,∑j=1kpjtyrjt+Ft+3c,t≥T1,Byt=ByT1,T0≤t≤T1,∫t∞qsGygsds−∫t∞usHyhsds,t≥T1.
First, we show that ifx,y∈S, then Ax+By∈S. In fact, x=xt and y=yt∈S and for t≥T1, we obtain(70)Axt+Byt≤∑j=1kpjtxrjt+3c+∫t∞qsGygsds+Ft≤pL+5c≤L.
On the other hand, fort≥T1,(71)Axt+Byt≥3c−∫t∞usHyhsds−Ft≥3c−c−c≥c.
Hence,(72)c≤Axt+Byt≤L,fort≥T1.
Thus, we proved thatAx+By∈S for any x,y∈S. Next, we show that A is a contraction on S. In fact for x,y∈S and t≥T1 we have(73)Axt−Ayt≤supt≥T0∑j=1kpjtxrjt−yrjt≤∑j=1ksupt≥T0pjtxrjt−yrjt≤∑j=1ksupt≥T0pjtsupt≥T0xrjt−yrjt≤xt−yt∑j=1ksupt≥T0pjt≤px−y.
This implies A is a contraction because0<p<1. Next, by following the arguments given in [14] (Theorem 2.2), we show that B is completely continuous. Then, by Lemma 3, we can find x0 in S such that Ax0+Bx0=x0. Clearly, x0t is a bounded, positive solution of (1) with limit infimum greater than or equal to c>0. Thus, the theorem is proved.Remark 2.
Theorems for the casepjt satisfying (34) or (53) may be proved as in the proof of the above theorem with the following definition of the operators A and B. For x∈S, define(74)Axt=AxT1,ift∈T0,T1,xri−1tpiri−1t−∑j≠ipjri−1txrjri−1tpiri−1t+bλpiri−1t−Fri−1tpiri−1t,ift≥T1,Bxt=BxT1,ift∈T0,T,−1piri−1t∫ri−1t∞qsGxgsds+1piri−1t∫ri−1t∞usHxhsds,ift≥T1.
The functionr−1 used in the definition of the operators A and B is the inverse function of rt, which exists if rt is assumed to be monotonic. Further more, note that r−1rt=t.Remark 3.
For the results in this section, we assume none of the conditions (4), G is nondecreasing and xGx>0 unlike the corresponding results in [3, 4].
Taking into consideration of all the results i.e., Theorems1–3, 6, and 8 and Remark 2, we obtain the following theorem.Theorem 9.
Suppose that (2)–(4), (6), and (7) hold. Assume pjt in (1) to satisfy one of the four conditions (15), (30), (34), and (53). Then, (5) is both necessary and sufficient condition for every solution of (1) to be oscillatory or tending to zero as t⟶∞.Remark 4.
Ifpit=pt and p1t=p2t=⋯=pi−1t=pi+1t=⋯=pkt=0, then conditions (15), (30), (34), and (53) due to the boundedness of pjt reduce to the following conditions (i) 0≤pt≤p<1, (ii) −1<−p≤pt≤0, (iii) −p1≤pt≤−p<−1, and (iv) p1≥pt≥p>1, respectively. These conditions are assumed in [1, 3–5].Remark 5.
Taking the Remarks1, 3, and 4 into consideration, Theorem 9 generalizes and improves the results in [3, 4].


## 4. Application to Neutral Equations with Oscillating Coefficients

In this section, we find sufficient conditions so that every solution of the first-order neutral differential equation (11) oscillates or tends to zero as t⟶∞, where vt is allowed to change sign. Let v+t=maxvt,0 and v−t=max−vt,0. Then, vt=v+t−v−t, and equation (11) can be written as(75)yt−∑j=1kpjtyrjt′+v+tGygt−v−tGygt=ft.Now, we proceed as in the previous section by settingqt=v+t, ut=v−t, and Hx=Gx. Assumptions (5)–(7) would become(76)∫t0∞v+tdt=∞,(77)∫t0∞v−tdt=∞,(78)Gis bounded,respectively.Therefore, the study of (11) reduces to the study of (1) in results Theorems 1–3, 6, 8, 9, and Remark 2. From Theorem 9, we have the following theorem for (11) where vt changes sign.Theorem 10.
Suppose that (2)–(4), (77), and (78) hold. Assume pjt in (11) to satisfy one of the four conditions (15), (30), (34), or (53). Then, (76) is both necessary and sufficient condition for every solution of (11) (with vt changing sign) to be oscillatory or tending to zero as t⟶∞.From Theorems4 and 5, we have the following results for NDDE (11) with vt changing sign.Theorem 11.
Suppose that (2), (3), (77), and (78) hold. Assume that (38) and (39) hold, then every solution of (11) (with vt changing sign) oscillates.Theorem 12.
Assume conditions (2)–(4), (77), (78), and (38) to hold. Further, assume grit=rigt for i=1,2,…,k. Let there exist a real μ>0 such that for xi>0,i=1,2,…,k+1, and u>0, (44) and (45) hold. Suppose that (46) holds where qt=v+t. Further, suppose that there exists a real α>0 such that ri′t≥1/α. Then, every solution of (11) (with vt changing sign) oscillates or tends to zero as t⟶∞.For the results in this section, we need G to be bounded, continuous, and to satisfy (2). The prototype of such a functionG(y) is y2n sgn(y)/(1 + y2n). The following examples illustrate Theorem 10.Example 3.
Consider the higher-order NDDE:(79)yt−1/2eyt−1−1/2e2yt−2′+qtyt−3=ft,where(80)qt=sint,2kπ≤t≤2k+1π,k=0,1,2,…sintt4,2k+1π≤t≤2k+2π,k=0,1,2,…,(81)ft=sinte−t+3,2kπ≤t≤2k+1π,k=0,1,2,…,sinte−t+3t4,2k+1π≤t≤2k+2π,k=0,1,2,….
Clearly,(82)q+t=sint,2kπ≤t≤2k+1π,k=0,1,2,…0,2k+1π≤t≤2k+2π,k=0,1,2,…,(83)q−t=0,2kπ≤t≤2k+1π,k=0,1,2,…sintt4,2k+1π≤t≤2k+2π,k=0,1,2,…
It may be verified that the NDDE (79) satisfies all the conditions of Theorem 10. Hence, every solution of (79) oscillates or tends to zero as t⟶∞. As such, it admits a positive solution yt=e−t which tends to zero as t⟶∞.Example 4.
Consider the following higher-order NDDE:(84)yt−1/2eyt−1−1/2e2yt−2′+qtGyt−3=ft,where Gu=u2sgnu/1+u2 and qt as in (80) and(85)ft=e6e2t+e6−1sint,2kπ≤t≤2k+1π,k=0,1,2,…e6sinte2t+e6t4,2k+1π≤t≤2k+2π,k=0,1,2,…,satisfies all the conditions of Theorem 10. Hence, every solution of (84) oscillates or tends to zero as t⟶∞. As such, it admits a positive solution yt=e−t which tends to zero as t⟶∞.Remark 6.
The results of this article seem to be significant as no result in literature can be applied to the NDDEs (79) and (84).


## 5. Conclusion

The open problem [1] (Problem 2.8.3, p. 57) says the following:Extend the following result to equations with oscillating coefficients.Theorem 2.3.1 in [1]: under the assumptions that vt≥0 and(86)liminft⟶∞∫t−τtvsds>e−1,every solution of(87)y′t+vtyt−τ=0,t≥t0,oscillates.If we putft=0,Gy=y, and k=0 in (11), then the following corollary follows from Theorem 10.Corollary 1.
Suppose that (76) and (77) hold. Then, every bounded solution of(88)y′t+vtygt=0,(with vt changing sign) oscillates or tends to zero as t⟶∞.Again, if we putft=0,k=0 in (11), then the following corollary follows from Theorem 10.Corollary 2.
Suppose that (2), (76), and (77) hold. Then, every solution of(89)y′t+vtGygt=0,(with vt changing sign) oscillates or tends to zero as t⟶∞. Or, equivalently every unbounded solution of (89) oscillates.Note that (86) implies(90)∫t0∞vtdt=∞.Further, (76) is equivalent to (90) under the assumption (77). Thus, Corollaries 1 and 2 answer the open problem [1] (Problem 2.8.3, p.57) partially. Further, as the condition “G is nondecreasing” is not assumed, and vt in NDDE (11) has no fixed sign; therefore, by Remark 4, Theorem 10 of this article improves and generalizes the results in [7] (Theorem 2.2, Theorem 2.4, and Theorem 2.5) because of their additional assumptions (i) σ>α or σ<α, (ii) qt>ut−σ+α, and (iii) another sublinear condition on G, while studying the NDDE (9). Further, in this paper, Theorem 10 improves and generalizes the results in [4] (Theorems 2.2, 2.3, 2.5, 2.6, and Corollary 2.7) for the NDDE (8).



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