Oscillation of first order linear differential equations with several non-monotone delays

E.R. Attia, V. Benekas, H.A. El-Morshedy and I.P. Stavroulakis
From the journal Open Mathematics

Abstract

Consider the first-order linear differential equation with several retarded arguments

x(t)+k=1npk(t)x(τk(t))=0,tt0,

where the functions pk, τkC([t0, ∞), ℝ+), τk(t) < t for tt0 and limt→∞τk(t) = ∞, for every k = 1, 2, …, n. Oscillation conditions which essentially improve known results in the literature are established. An example illustrating the results is given.

MSC 2010: 34K11; 34K06

1 Introduction

This paper is concerned with the oscillation of the first order differential equation with several delays of the form

x(t)+k=1npk(t)x(τk(t))=0,tt0,(1)

where pk, τkC([t0, ∞), [0, ∞)), such that τk(t)<tandlimtτk(t)=,k = 1, 2, …, n.

Let T0 ∈ [t0, ∞), τ (t) = min{τk(t) : k = 1, …, n} and τ(−1)(t) = inf{τ (s) : st}. By a solution of Eq. (1) we understand a function xC([t0, ∞), ℝ) continuously differentiable on [τ(−1)(t0), ∞) which satisfies (1) for tτ(−1)(t0). Such a solution is called oscillatory if it has arbitrarily large zeros, and otherwise it is called nonoscillatory.

We assume throughout this work that there exist t1t0, a family of nondecreasing continuous functions {gk(t)}k=1n and a nondecreasing continuous function g(t) such that

τk(t)gk(t)g(t)t,tt1,k=1,2,...,n.

To simplify the notations, we denote by λ (ξ) the smaller root of the equation eξλ = λ, ξ ≥ 0 and

c(ξ)=1ξ12ξξ22,0ξ1e.

Next, we mention some known oscillation criteria for Eq. (1).

In the case that the arguments τk(t) are monotone the following sufficient oscillation conditions have been established. In 1978, Ladde [1] and in 1982 Ladas and Stavroulakis [2] obtained the sufficient oscillatory criterion

lim inftτmax(t)tk=1npk(s)ds>1e,

where τmax(t)=max1kn{τk(t)}.

In 1984, Hunt and Yorke [3] established the condition

lim inftk=1npk(t)(tτk(t))>1e,

where (tτk(t)) ≤ τ0, for some τ0 > 0, 1 ≤ kn.

Also, in 1984, Fukagai and Kusano [4] established the following result.

Assume that there is a continuous nondecreasing functionτ (t)such thatτk(t) ≤ τ (t) ≤ tfortt0, 1 ≤ kn. If

lim inftτ(t)tk=1npk(s)ds>1e,(2)

then all solutions of Eq. (1)oscillate. If, on the other hand, there exists a continuous nondecreasing functionτ (t) such thatτ (t) ≤ τk(t) fortt0, 1 ≤ kn, limt→∞τ (t) = ∞ andfor all sufficiently larget,

τ(t)tk=1npk(s)ds1e,

thenEq.(1)has a non-oscillatory solution.

In 2003, Grammatikopoulos, Koplatadze and Stavroulakis [5] improved the above results as follows:

Assume that the functionsτkare nondecreasing for allk ∈ {1, …, n,

0pi(t)pj(t)dt<+,i,j=1,...,n,

and

lim inftτk(t)tpk(s)ds>0,k=1,...,n.

If

k=1nlim inftτk(t)tpk(s)ds>1e,

then all solutions of Eq. (1)oscillate.

In the case of non-monotone arguments we mention the following known oscillation results. In 2015 Infante, Koplatadze and Stavroulakis [6] obtained the following sufficient oscillation conditions:

lim suptj=1ni=1ngj(t)tpi(s)exp(τi(s)gi(t)k=1npk(u)exp(τk(u)u=1np(v)dv)du)ds1n>1nn,

and

lim supϵ0+lim suptj=1ni=1ngj(t)tpi(s)exp(τi(s)gi(t)k=1n(λ(qk)ϵ)pk(u)du)ds1n>1nn,(3)

where

qk=lim inftτk(t)tpk(s)ds>0,k=1,2,...,n.(4)

Also, in 2015 Koplatadze [7] derived the following three conditions. The first one takes the form

lim suptj=1ni=1ngj(t)tpi(s)exp(nτi(s)gi(t)=1np(ξ)1nψk(ξ)dξ)ds1n>1nn1i=1nc(βi),

where k ∈ ℕ and

ψ1(t)=0,ψi(t)=expk=1nτk(t)t=1np(s)1nψi1(s)ds,i=2,3,...,

and

βi=lim inftgi(t)tpi(s)ds,i=1,2,...,n.(5)

The second condition is

lim suptj=1ni=1ngj(t)tpi(s)exp(n(λ(p¯)ϵ)τi(s)gi(t)=1np(ξ)1ndξ)ds1n>1nn1=1nc(β),

where ϵ ∈ (0, λ ()), and

0<p¯:=lim infti=1nτi(t)t=1np(s)1nds1e.

The third condition is

lim suptj=1ni=1ngj(t)tpi(s)τi(s)gi(t)=1np(ξ)1ndξds1n>0,(6)

and p¯>1e.

In 2016, Braverman, Chatzarakis and Stavroulakis [8] obtained the sufficient condition

lim supth(t)ti=1npi(u)ar(h(t),τi(u))du>1,

where

h(t)=max1inhi(t),and hi(t)=supt0stτi(s),i=1,2,...,n,(7)

and

a1(t,s)=expsti=1npi(u)du,ar+1(t,s)=expsti=1npi(u)ar(u,τi(u))du,rN.

Also, in 2016 Akca, Chatzarakis and Stavroulakis [9] obtained the sufficient condition

lim supth(t)ti=1npi(u)ar(h(u),τi(u))du>1+ln(λ(α))λ(α),(8)

where

0<α:=lim inftτmax(t)ti=1npi(s)ds1e.

2 Main results

To obtain our main results we need the following lemmas:

Lemma 2.1

([6]). Let x(t) be an eventually positive solution of Eq. (1). Then

lim inftx(τk(t))x(t)λ(qk),k=1,2,...,n,

whereqkis defined by(4).

Lemma 2.2

([10]). Assume that

lim inftg(t)tP(s)ds=α,

andx(t) is an eventually positive solution of the first order delay differential inequality

x(t)+P(t)x(g(t))0,tt1,

wherePC([t1, ∞), [0, ∞)). If 0 ≤ α*1e, then

lim inftx(t)x(g(t))c(α).

Theorem 2.3

Assume that

ρ:=lim inftg(t)tk=1npk(s)ds,0<ρ1e,

and

lim supt(g(t)tQ(v)dv+c(ρ)eg(t)ti=1npi(s)ds)>1,(9)

where

Q(t)=k=1ni=1npi(t)τi(t)tpk(s)egk(t)tj=1npj(v)dv+λ(ρ)ϵτk(s)gk(t)=1np(u)duds,

andϵ ∈ (0, λ (ρ)). Then all solutions of Eq.(1)are oscillatory.

Proof

Assume the contrary, i.e., there exists a nonoscillatory solution x(t) of (1). Because of the linearity of (1), we assume that x(t) is eventually positive. Therefore, there exists a sufficiently large t2t1 such that x(τk(t)) > 0, for all tt2, k = 1, 2, …, n. Thus, equation (1) implies that x(t) is nonincreasing for all tt2. Integrating (1) from τi(t) to t, we obtain

x(t)x(τi(t))+k=1nτi(t)tpk(s)x(τk(s))ds=0.(10)

Also, dividing (1) by x(t) and integrating the resulting equation from τk(s) to gk(t), we get

x(τk(s))=x(gk(t))eτk(s)gk(t)=1np(u)x(τ(u))x(u)du.

Substituting this into (10),

x(t)x(τi(t))+k=1nx(gk(t))τi(t)tpk(s)eτk(s)gk(t)=1np(u)x(τ(u))x(u)duds=0.

Multiplying the above equation by pi(t), and taking the sum over i, it follows that

x(t)+x(t)i=1npi(t)+k=1nx(gk(t))i=1npi(t)τi(t)tpk(s)eτk(s)gk(t)=1np(u)x(τ(u))x(u)duds=0.

The substitution y(t)=et0t=1np(s)dsx(t), reduces this equation to

y(t)+k=1ny(gk(t))egk(t)t=1np(s)dsi=1npi(t)τi(t)tpk(s)eτk(s)gk(t)=1np(u)x(τ(u))x(u)duds=0,(11)

which in turn, by integrating from g(t) to t, leads to

y(t)y(g(t))+g(t)tk=1ny(gk(v))egk(v)v=1np(s)dsi=1npi(v)τi(v)vpk(s)eτk(s)gk(v)=1np(u)x(τ(u))x(u)dudsdv=0.

Hence, the nonincreasing nature of y(t) implies that

y(t)y(g(t))+y(g(t))g(t)tk=1negk(v)v=1np(s)dsi=1npi(v)τi(v)vpk(s)eτk(s)gk(v)=1np(u)x(τ(u))x(u)dudsdv0,(12)

for all tt3 and some t3t2.

On the other hand, using the nonincreasing nature of x(t), equation (1) implies that

x(t)+x(g(t))k=1npk(t)0, for all tt3.(13)

Therefore, from [11, Lemma 2.1.2], we obtain lim inftx(g(t))x(t)λ(ρ). Thus, for sufficiently small ϵ > 0, we have

x(τ(t))x(t)x(g(t))x(t)>λ(ρ)ϵ, for all tt4,1n,(14)

for some t4t3. This together with (12) implies that

y(t)y(g(t))+y(g(t))g(t)tk=1negk(v)v=1np(s)dsi=1npi(v)τi(v)vpk(s)eλ(ρ)ϵτk(s)gk(v)=1np(u)dudsdv0,

for all tt5, where t5t4. That is

g(t)tk=1negk(v)v=1np(s)dsi=1npi(v)τi(v)vpk(s)eλ(ρ)ϵτk(s)gk(v)=1np(u)dudsdv1y(t)y(g(t)),(15)

for all tt5. Also, in view of Lemma 2.2, for sufficiently small ϵ > 0 and some t6t5, inequality (13) leads to

x(t)x(g(t))>c(ρ)ϵ, for all tt6.

Therefore,

y(t)y(g(t))=x(t)x(g(t))eg(t)ti=1npi(s)ds>c(ρ)ϵeg(t)ti=1npi(s)ds, for all tt6.

Combining this inequality with (15), it follows that

g(t)tQ(v)dv1c(ρ)ϵeg(t)ti=1npi(s)ds, for all tt6.

Then

lim suptg(t)tQ(v)dv+c(ρ)eg(t)ti=1npi(s)ds1+ϵlim supteg(t)ti=1npi(s)ds.(16)

Notice that, by integrating (13) from g(t) to t and using the nonincreasing nature of x(t), we obtain

g(t)tk=1npk(s)ds1x(t)x(g(t))1, for all tt3.

Now, letting ϵ → 0 in (16), we arrive at a contradiction with (9). The proof is complete. □

Remark 2.4

Theorem 2.3 is proved using the core idea of the proof of [12, Theorem 2.1] which is given for Equation (1) when n = 1. However, Theorem 2.3 produces a new oscillation criterion even for equations with only one non-monotone delay.

Using Lemma 2.1 instead of [11, Lemma 2.1.2], similar reasoning as in the proof of Theorem 2.3 implies the following result:

Theorem 2.5

Assume that

lim supt(g(t)tQ1(v)dv+c(ρ)eg(t)ti=1npi(s)ds)>1,

where

Q1(t)=k=1ni=1npi(t)τi(t)tpk(s)egk(t)tj=1npj(v)dv+τk(s)gk(t)=1nλ(q)ϵp(u)duds,

q is defined by (4), ρ is defined as in Theorem 2.3 and ϵ ∈ (0, λ (q)). Then all solutions of Equation (1) oscillate.

Theorem 2.6

Assume that

lim suptj=1nk=1ngj(t)tRk(s)ds1n+k=1nc(βk)nnek=1ngk(t)t=1np(s)ds>1nn,

where βk is defined by (5) with 0 < βk1eand

Rk(s)=egk(s)sj=1npj(u)dui=1npi(s)τi(s)spk(u)e(λ(ρ)ϵ)τk(u)gk(s)=1np(v)dvdu,

ρ is defined as in Theorem 2.3 and ϵ ∈ (0, λ(ρ )). Then all solutions of Equation (1) oscillate.

Proof

Let x(t) be a nonoscillatory solution of (1). As usual, we assume that x(t) is an eventually positive solution. Substituting (14) into (11), we obtain

y(t)+k=1ny(gk(t))egk(t)tj=1npj(s)dsi=1npi(t)τi(t)tpk(s)e(λ(ρ)ϵ)τk(s)gk(t)=1np(u)duds0,

for all tt2 and some t2t1 where ϵ ∈ (0, λ (ρ)) and y(t)=et0t=1np(s)dsx(t) Integrating from gj(t) to t and using the nonincreasing nature of y(t), we obtain

y(t)y(gj(t))+k=1ny(gk(t))gj(t)tRk(s)ds0, for all tt3,

and some t3t2. Using the relation between arithmetic and geometric mean, it follows that

y(gj(t))nk=1ny(gk(t))1nk=1ngj(t)tRk(s)ds1n+y(t), for all tt3.

Taking the product on both sides,

j=1ny(gj(t))nnk=1ny(gk(t))j=1nk=1ngj(t)tRk(s)ds1n+y(t)n, for all tt3.

Therefore,

j=1nk=1ngj(t)tRk(s)ds1n+y(t)nnnk=1ny(gk(t))1nn, for all tt3.(17)

Since y(t)=et0t=1np(s)dsx(t), then

y(t)nk=1ny(gk(t))=ek=1ngk(t)t=1np(s)dsx(t)nk=1nx(gk(t)).

Substituting in (17),

j=1nk=1ngj(t)tRk(s)ds1n+ek=1ngk(t)t=1np(s)dsx(t)nnnk=1nx(gk(t))1nn, for all tt3.(18)

On the other hand, the nonincreasing nature of x(t) and (1) imply that

x(t)+pk(t)x(gk(t))0, for all tt3,k=1,2,...,n,(19)

and hence, by Lemma 2.2, it follows that

x(t)nk=1nx(gk(t))>k=1nc(βk)ϵ, for all tt4,

for some t4t3 and sufficiently small ϵ > 0. This together with (18) leads to

j=1nk=1ngj(t)tRk(s)ds1n+k=1nc(βk)nnek=1ngk(t)t=1np(s)ds1nn+ϵnnek=1ngk(t)t=1np(s)ds,

for all tt4. Consequently,

lim suptj=1nk=1ngj(t)tRk(s)ds1n+k=1nc(βk)nnek=1ngk(t)t=1np(s)ds1nn+ϵnnlim suptek=1ngk(t)t=1np(s)ds.(20)

On the other hand, integrating (1) from gk(t) to t, we get

x(t)x(gk(t))+gk(t)t=1nx(τ(s))p(s)ds=0,k=1,2,...,n.

Then, using the nonincreasing nature of x(t), we obtain

gk(t)t=1np(s)dsx(gk(t))x(t)1, for all tt3,k=1,2,...,n.

But applying Lemma 2.2 to (19), we obtain

lim suptx(gk(t))x(t)<+,k=1,2,...,n.

Therefore, ek=1ngk(t)t=1np(s)ds is bounded. Now, allowing ϵ → 0 in (20), it follows that

lim suptj=1nk=1ngj(t)tRk(s)ds1n+k=1nc(βk)nnek=1ngk(t)t=1np(s)ds1nn.

This contradiction completes the proof. □

The following example shows that Theorem 2.3 can be applied but conditions (2), (3) and (6) fail to apply as well as (8) when r = 1.

Example 2.7

Consider the first order delay differential equation

x(t)+a(b+sin(t))xtπ2+a(b+cos(t))xτ¯(t)=0,t0,(21)

wherea=0.41.137π+2,b=1.784and

τ¯(t)=tπ2σsin2(300t),σ=1150.

Clearly

tπ2στ¯(t)tπ2.

Fig. 1

Equation (21) has the form (1) with p1(t) = a(b + sin (t)), p2(t) = a(b + cos (t)), τ1(t) = tπ2and τ2(t) = tπ2σ sin2(300t). Therefore, we can choose g1(t) = g2(t) = g(t) = tπ2and ϵ = 0.001. Then

g(t)tk=12pk(s)ds=abπ+2asin(t).

Consequently,

ρ=lim inftg(t)tk=12pk(s)ds=abπ2a0.2891659465,c(ρ)=1ρ12ρρ220.06470619,andλ(ρ)ϵ1.577422807.

LetI(t)=g(t)tQ(v)dv+c(ρ)eg(t)ti=12pi(s)ds.Then, the property that ex ≥ ex for x ≥ 0 leads to

I(t)eg(t)ti=12pi(v)g(v)vk=12pk(s)(g(v)v=12p(s)ds+λ(ρ)ϵg(s)g(v)l=12pl(u)du)dsdv+c(ρ)eg(t)ti=12pi(s)ds.

Now, using Maple, we get

I(t)0.49727+0.0029516sin(t)cos2(t)+0.27514sin(t)0.23059cos(t)0.015664cos2(t)0.083083cos(t)sin(t)+0.0037423cos3(t)+0.10144e0.16044sin(t).

ChooseTk=3π4+2πk,then

I(Tk)1.0019>1,forallkN,

which means that

lim supt(g(t)tQ(v)dv+c(ρ)eg(t)ti=12pi(s)ds)>1.

That is, condition (9) of Theorem 2.3 is satisfied and therefore all solutions of Eq. (21) oscillate.

We will show, however, that none of the conditions (2), (3), (6), and (8) (with r = 1) is satisfied. Indeed, notice that

τ(t)tk=12pk(s)dstπ2tk=12pk(s)ds=abπ+2asin(t).

Hence,

lim inftτ(t)tk=12pk(s)dsabπ2a0.2891659465<1e.

On the other hand, since h(t) = g(t), pi(t) ≤ a(b+1) for i = 1, 2, where h(t) is defined by (7), we have

a1(h(s),τi(s))=expτi(s)g(s)i=12pi(u)duexpsπ2σsπ2i=12pi(u)duexpsπ2σsπ22a(b+1)du=exp(2a(b+1)σ).

Therefore,

h(t)ti=12pi(s)a1(h(s),τi(s))dstπ2ta(b+sin(s)+b+cos(s))exp2a(b+1)σds=tπ2t2abds+tπ2tasin(s)ds+tπ2tacos(s)dsexp2a(b+1)σ=[abπ+2asin(t)]exp2a(b+1)σabπ+2aexp2a(b+1)σ<0.61188.

Since λ (ρ) ≈ 1.578422807, then

1+ln(λ(ρ))λ(ρ)>0.92.

Therefore, none of conditions (2) and (8) (with r = 1) is satisfied.

Also, since

g(t)ta(b+sin(s))ds=abπ2+a(sin(t)cos(t)),

and

τ¯(t)ta(b+cos(s))dstπ2σta(b+cos(s))dsa(b+1)tπ2σtπ2ds+tπ2ta(b+cos(s))ds=aσ(b+1)+abπ2+a(cos(t)+sin(t)).

Then, q1=abπ2a2andq2abπ2a2+aσ(b+1),where qi is defined by (4), i = 1, 2. Therefore, λ (q1) ≈ 1.134680932 and λ (q2) ≈ 1.136881841. Let

I1(t)=j=12i=12gj(t)tpi(s)exp(τi(s)gi(t)k=12(λ(qk)ϵ)pk(u)du)ds12.

Then

I1(t)<j=12i=12gj(t)tpi(s)exp(λ(q2)τi(s)gi(t)l=12pl(u)du)ds12j=12i=12gj(t)tpi(s)exp(2aλ(q2)(1+b)(ts+σ))ds12i=12tπ2σtpi(s)exp(2aλ(q2)(1+b)(ts+σ))dsi=12[tπ2σtπ2a(b+1)exp(2aλ(q2)(1+b)(ts+σ))ds+tπ2tpi(s)exp(2aλ(q2)(1+b)(ts+σ))ds]0.1363159006+0.08561813338sin(t)0.00925235722cos(t)0.02983977134cos2(t)0.00652549885sin(t)cos(t)<0.23771189<14.

Therefore, lim supϵ0+lim suptI1(t)<14,which implies that condition (3) is not satisfied.

Finally, since

i=12τi(t)tl=12pl(s)12dsi=1212τi(t)tl=12pl(s)ds12tπ2tl=12pl(s)ds+12tπ2σtl=12pl(s)ds=12tπ2tl=12pl(s)ds+12tπ2σtπ2l=12pl(s)ds+tπ2tl=12pl(s)dstπ2tl=12pl(s)ds+tπ2σtπ2a(b+1)dstπ2tl=12pl(s)ds+aσ(b+1),

then

p¯=lim infti=12τi(t)tl=12pl(s)12dsabπ2a+aσ(b+1)<0.2906549<1e.

Therefore, condition (6) fails to apply.

Acknowledgement

The authors would like to thank the referees for their valuable comments and suggestions.

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