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formerly Central European Journal of Mathematics

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

E.R. Attia
/ V. Benekas
/ H.A. El-Morshedy
/ I.P. Stavroulakis
• Corresponding author
• Department of Mathematics, University of Ioannina, 451 10, Ioannina, Greece
• Department of Mathematical Sciences, University of South Africa, 0003, Pretoria, South Africa
• Email
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Published Online: 2018-02-23 | DOI: https://doi.org/10.1515/math-2018-0010

Abstract

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

$x′(t)+∑k=1npk(t)x(τk(t))=0,t≥t0,$

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

$x′(t)+∑k=1npk(t)x(τk(t))=0,t≥t0,$(1)

where pk, τkC([t0, ∞), [0, ∞)), such that $\begin{array}{}{\tau }_{k}\left(t\right) 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 $\begin{array}{}\left\{{g}_{k}\left(t\right){\right\}}_{k=1}^{n}\end{array}$ and a nondecreasing continuous function g(t) such that

$τk(t)≤gk(t)≤g(t)≤t,t≥t1,k=1,2,...,n.$

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

$c(ξ)=1−ξ−1−2ξ−ξ22,0≤ξ≤1e.$

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

$lim inft→∞∫τmax(t)t∑k=1npk(s)ds>1e,$

where $\begin{array}{}{\tau }_{max}\left(t\right)=\underset{1\le k\le n}{max}\left\{{\tau }_{k}\left(t\right)\right\}.\end{array}$

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

$lim inft→∞∑k=1npk(t)(t−τk(t))>1e,$

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

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

Assume that there is a continuous nondecreasing function τ (t)such that τk(t) ≤ τ (t) ≤ t for tt0, 1 ≤ kn. If

$lim inft→∞∫τ∗(t)t∑k=1npk(s)ds>1e,$(2)

then all solutions of Eq. (1) oscillate. If, on the other hand, there exists a continuous nondecreasing function τ (t) such that τ (t) ≤ τk(t) for tt0, 1 ≤ kn, limt→∞ τ (t) = ∞ and for all sufficiently large t,

$∫τ∗(t)t∑k=1npk(s)ds≤1e,$

then Eq.(1) has a non-oscillatory solution.

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

Assume that the functions τk are nondecreasing for all k ∈ {1, …, n,

$∫0∞pi(t)−pj(t)dt<+∞,i,j=1,...,n,$

and

$lim inft→∞∫τk(t)tpk(s)ds>0,k=1,...,n.$

If

$∑k=1nlim inft→∞∫τk(t)tpk(s)ds>1e,$

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:

$lim supt→∞∏j=1n∏i=1n∫gj(t)tpi(s)exp⁡(∫τi(s)gi(t)∑k=1npk(u)exp⁡(∫τk(u)u∑ℓ=1npℓ(v)dv)du)ds1n>1nn,$

and

$lim supϵ→0+lim supt→∞∏j=1n∏i=1n∫gj(t)tpi(s)exp⁡(∫τi(s)gi(t)∑k=1n(λ(qk)−ϵ)pk(u)du)ds1n>1nn,$(3)

where

$qk=lim inft→∞∫τk(t)tpk(s)ds>0,k=1,2,...,n.$(4)

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

$lim supt→∞∏j=1n∏i=1n∫gj(t)tpi(s)exp⁡(n∫τi(s)gi(t)∏ℓ=1npℓ(ξ)1nψk(ξ)dξ)ds1n>1nn1−∏i=1nc(βi),$

where k ∈ ℕ and

$ψ1(t)=0,ψi(t)=exp∑k=1n∫τk(t)t∏ℓ=1npℓ(s)1nψi−1(s)ds,i=2,3,...,$

and

$βi=lim inft→∞∫gi(t)tpi(s)ds,i=1,2,...,n.$(5)

The second condition is

$lim supt→∞∏j=1n∏i=1n∫gj(t)tpi(s)exp⁡(n(λ(p∗¯)−ϵ)∫τi(s)gi(t)∏ℓ=1npℓ(ξ)1ndξ)ds1n>1nn1−∏ℓ=1nc(βℓ),$

where ϵ ∈ (0, λ ()), and

$0

The third condition is

$lim supt→∞∏j=1n∏i=1n∫gj(t)tpi(s)∫τi(s)gi(t)∏ℓ=1npℓ(ξ)1ndξds1n>0,$(6)

and $\begin{array}{}\overline{{p}_{\ast }}>\frac{1}{\mathrm{e}}.\end{array}$

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

$lim supt→∞∫h(t)t∑i=1npi(u)ar(h(t),τi(u))du>1,$

where

$h(t)=max1≤i≤nhi(t),and hi(t)=supt0≤s≤tτi(s),i=1,2,...,n,$(7)

and

$a1(t,s)=exp∫st∑i=1npi(u)du,ar+1(t,s)=exp∫st∑i=1npi(u)ar(u,τi(u))du,r∈N.$

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

$lim supt→∞∫h(t)t∑i=1npi(u)ar(h(u),τi(u))du>1+ln⁡(λ(α))λ(α),$(8)

where

$0<α:=lim inft→∞∫τmax(t)t∑i=1npi(s)ds≤1e.$

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

$lim inft→∞x(τk(t))x(t)≥λ(qk),k=1,2,...,n,$

where qk is defined by (4).

Lemma 2.2

([10]). Assume that

$lim inft→∞∫g(t)tP(s)ds=α∗,$

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

$x′(t)+P(t)x(g(t))≤0,t≥t1,$

where PC([t1, ∞), [0, ∞)). If 0 ≤ α*$\begin{array}{}\frac{1}{\mathrm{e}}\end{array}$, then

$lim inft→∞x(t)x(g(t))≥c(α∗).$

Theorem 2.3

Assume that

$ρ:=lim inft→∞∫g(t)t∑k=1npk(s)ds,0<ρ≤1e,$

and

$lim supt→∞(∫g(t)tQ(v)dv+c(ρ)e∫g(t)t∑i=1npi(s)ds)>1,$(9)

where

$Q(t)=∑k=1n∑i=1npi(t)∫τi(t)tpk(s)e∫gk(t)t∑j=1npj(v)dv+λ(ρ)−ϵ∫τk(s)gk(t)∑ℓ=1npℓ(u)duds,$

and ϵ ∈ (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

$x(t)−x(τi(t))+∑k=1n∫τi(t)tpk(s)x(τk(s))ds=0.$(10)

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

$x(τk(s))=x(gk(t))e∫τk(s)gk(t)∑ℓ=1npℓ(u)x(τℓ(u))x(u)du.$

Substituting this into (10),

$x(t)−x(τi(t))+∑k=1nx(gk(t))∫τi(t)tpk(s)e∫τk(s)gk(t)∑ℓ=1npℓ(u)x(τℓ(u))x(u)duds=0.$

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

$x′(t)+x(t)∑i=1npi(t)+∑k=1nx(gk(t))∑i=1npi(t)∫τi(t)tpk(s)e∫τk(s)gk(t)∑ℓ=1npℓ(u)x(τℓ(u))x(u)duds=0.$

The substitution $\begin{array}{}y\left(t\right)={\mathrm{e}}^{{\int }_{{t}_{0}}^{t}\sum _{\ell =1}^{n}{p}_{\ell }\left(s\right)ds}x\left(t\right),\end{array}$ reduces this equation to

$y′(t)+∑k=1ny(gk(t))e∫gk(t)t∑ℓ=1npℓ(s)ds∑i=1npi(t)∫τi(t)tpk(s)e∫τk(s)gk(t)∑ℓ=1npℓ(u)x(τℓ(u))x(u)duds=0,$(11)

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

$y(t)−y(g(t))+∫g(t)t∑k=1ny(gk(v))e∫gk(v)v∑ℓ=1npℓ(s)ds∑i=1npi(v)∫τi(v)vpk(s)e∫τk(s)gk(v)∑ℓ=1npℓ(u)x(τℓ(u))x(u)dudsdv=0.$

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

$y(t)−y(g(t))+y(g(t))∫g(t)t∑k=1ne∫gk(v)v∑ℓ=1npℓ(s)ds∑i=1npi(v)∫τi(v)vpk(s)e∫τk(s)gk(v)∑ℓ=1npℓ(u)x(τℓ(u))x(u)dudsdv≤0,$(12)

for all tt3 and some t3t2.

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

$x′(t)+x(g(t))∑k=1npk(t)≤0, for all t≥t3.$(13)

Therefore, from [11, Lemma 2.1.2], we obtain $\begin{array}{}\underset{t\to \mathrm{\infty }}{lim inf}\frac{x\left(g\left(t\right)\right)}{x\left(t\right)}\ge \lambda \left(\rho \right).\end{array}$ Thus, for sufficiently small ϵ > 0, we have

$x(τℓ(t))x(t)≥x(g(t))x(t)>λ(ρ)−ϵ, for all t≥t4,1≤ℓ≤n,$(14)

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

$y(t)−y(g(t))+y(g(t))∫g(t)t∑k=1ne∫gk(v)v∑ℓ=1npℓ(s)ds∑i=1npi(v)∫τi(v)vpk(s)eλ(ρ)−ϵ∫τk(s)gk(v)∑ℓ=1npℓ(u)dudsdv≤0,$

for all tt5, where t5t4. That is

$∫g(t)t∑k=1ne∫gk(v)v∑ℓ=1npℓ(s)ds∑i=1npi(v)∫τi(v)vpk(s)eλ(ρ)−ϵ∫τk(s)gk(v)∑ℓ=1npℓ(u)dudsdv≤1−y(t)y(g(t)),$(15)

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

$x(t)x(g(t))>c(ρ)−ϵ∗, for all t≥t6.$

Therefore,

$y(t)y(g(t))=x(t)x(g(t))e∫g(t)t∑i=1npi(s)ds>c(ρ)−ϵ∗e∫g(t)t∑i=1npi(s)ds, for all t≥t6.$

Combining this inequality with (15), it follows that

$∫g(t)tQ(v)dv≤1−c(ρ)−ϵ∗e∫g(t)t∑i=1npi(s)ds, for all t≥t6.$

Then

$lim supt→∞∫g(t)tQ(v)dv+c(ρ)e∫g(t)t∑i=1npi(s)ds≤1+ϵ∗lim supt→∞e∫g(t)t∑i=1npi(s)ds.$(16)

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

$∫g(t)t∑k=1npk(s)ds≤1−x(t)x(g(t))≤1, for all t≥t3.$

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 $lim supt→∞(∫g(t)tQ1(v)dv+c(ρ)e∫g(t)t∑i=1npi(s)ds)>1,$

where $Q1(t)=∑k=1n∑i=1npi(t)∫τi(t)tpk(s)e∫gk(t)t∑j=1npj(v)dv+∫τk(s)gk(t)∑ℓ=1nλ(qℓ)−ϵℓpℓ(u)duds,$

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 $lim supt→∞∏j=1n∏k=1n∫gj(t)tRk(s)ds1n+∏k=1nc(βk)nne∑k=1n∫gk(t)t∑ℓ=1npℓ(s)ds>1nn,$

where βk is defined by (5) with 0 < βk$\begin{array}{}\frac{1}{\text{e}}\end{array}$ and $Rk(s)=e∫gk(s)s∑j=1npj(u)du∑i=1npi(s)∫τi(s)spk(u)e(λ(ρ)−ϵ)∫τk(u)gk(s)∑ℓ=1npℓ(v)dvdu,$

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 $y′(t)+∑k=1ny(gk(t))e∫gk(t)t∑j=1npj(s)ds∑i=1npi(t)∫τi(t)tpk(s)e(λ(ρ)−ϵ)∫τk(s)gk(t)∑ℓ=1npℓ(u)duds≤0,$

for all tt2 and some t2t1 where ϵ ∈ (0, λ (ρ)) and $\begin{array}{}y\left(t\right)={\text{e}}^{{\int }_{{t}_{0}}^{t}\sum _{\ell =1}^{n}{p}_{\ell }\left(s\right)ds}x\left(t\right)\end{array}$ Integrating from gj(t) to t and using the nonincreasing nature of y(t), we obtain $y(t)−y(gj(t))+∑k=1ny(gk(t))∫gj(t)tRk(s)ds≤0, for all t≥t3,$

and some t3t2. Using the relation between arithmetic and geometric mean, it follows that $y(gj(t))≥n∏k=1ny(gk(t))1n∏k=1n∫gj(t)tRk(s)ds1n+y(t), for all t≥t3.$

Taking the product on both sides, $∏j=1ny(gj(t))≥nn∏k=1ny(gk(t))∏j=1n∏k=1n∫gj(t)tRk(s)ds1n+y(t)n, for all t≥t3.$

Therefore, $∏j=1n∏k=1n∫gj(t)tRk(s)ds1n+y(t)nnn∏k=1ny(gk(t))≤1nn, for all t≥t3.$(17)

Since $\begin{array}{}y\left(t\right)={\text{e}}^{{\int }_{{t}_{0}}^{t}\sum _{\ell =1}^{n}{p}_{\ell }\left(s\right)ds}x\left(t\right)\end{array}$, then $y(t)n∏k=1ny(gk(t))=e∑k=1n∫gk(t)t∑ℓ=1npℓ(s)dsx(t)n∏k=1nx(gk(t)).$

Substituting in (17), $∏j=1n∏k=1n∫gj(t)tRk(s)ds1n+e∑k=1n∫gk(t)t∑ℓ=1npℓ(s)dsx(t)nnn∏k=1nx(gk(t))≤1nn, for all t≥t3.$(18)

On the other hand, the nonincreasing nature of x(t) and (1) imply that $x′(t)+pk(t)x(gk(t))≤0, for all t≥t3,k=1,2,...,n,$(19)

and hence, by Lemma 2.2, it follows that $x(t)n∏k=1nx(gk(t))>∏k=1nc(βk)−ϵ∗, for all t≥t4,$

for some t4t3 and sufficiently small ϵ > 0. This together with (18) leads to $∏j=1n∏k=1n∫gj(t)tRk(s)ds1n+∏k=1nc(βk)nne∑k=1n∫gk(t)t∑ℓ=1npℓ(s)ds≤1nn+ϵ∗nne∑k=1n∫gk(t)t∑ℓ=1npℓ(s)ds,$

for all tt4. Consequently, $lim supt→∞∏j=1n∏k=1n∫gj(t)tRk(s)ds1n+∏k=1nc(βk)nne∑k=1n∫gk(t)t∑ℓ=1npℓ(s)ds≤1nn+ϵ∗nnlim supt→∞e∑k=1n∫gk(t)t∑ℓ=1npℓ(s)ds.$(20)

On the other hand, integrating (1) from gk(t) to t, we get $x(t)−x(gk(t))+∫gk(t)t∑ℓ=1nx(τℓ(s))pℓ(s)ds=0,k=1,2,...,n.$

Then, using the nonincreasing nature of x(t), we obtain $∫gk(t)t∑ℓ=1npℓ(s)ds≤x(gk(t))x(t)−1, for all t≥t3,k=1,2,...,n.$

But applying Lemma 2.2 to (19), we obtain $lim supt→∞x(gk(t))x(t)<+∞,k=1,2,...,n.$

Therefore, $\begin{array}{}{\text{e}}^{\sum _{k=1}^{n}{\int }_{{g}_{k}\left(t\right)}^{t}\sum _{\ell =1}^{n}{p}_{\ell }\left(s\right)ds}\end{array}$ is bounded. Now, allowing ϵ → 0 in (20), it follows that $lim supt→∞∏j=1n∏k=1n∫gj(t)tRk(s)ds1n+∏k=1nc(βk)nne∑k=1n∫gk(t)t∑ℓ=1npℓ(s)ds≤1nn.$

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 $x′(t)+a(b+sin⁡(t))xt−π2+a(b+cos⁡(t))xτ¯(t)=0,t≥0,$(21)

where $\begin{array}{}a=\frac{0.4}{1.137\pi +\sqrt{2}},b=1.784\end{array}$ and $τ¯(t)=t−π2−σsin2⁡(300t),σ=1150.$

Clearly $t−π2−σ≤τ¯(t)≤t−π2.$

Equation (21) has the form (1) with p1(t) = a(b + sin (t)), p2(t) = a(b + cos (t)), τ1(t) = t$\begin{array}{}\frac{\pi }{2}\end{array}$ and τ2(t) = t$\begin{array}{}\frac{\pi }{2}\end{array}$σ sin2(300t). Therefore, we can choose g1(t) = g2(t) = g(t) = t$\begin{array}{}\frac{\pi }{2}\end{array}$ and ϵ = 0.001. Then $∫g(t)t∑k=12pk(s)ds=abπ+2asin⁡(t).$

Consequently, $ρ=lim inft→∞∫g(t)t∑k=12pk(s)ds=abπ−2a≈0.2891659465,c(ρ)=1−ρ−1−2ρ−ρ22≈0.06470619,andλ(ρ)−ϵ≈1.577422807.$

Let $\begin{array}{}I\left(t\right)={\int }_{g\left(t\right)}^{t}Q\left(v\right)dv+c\left(\rho \right){\text{e}}^{{\int }_{g\left(t\right)}^{t}\sum _{i=1}^{2}{p}_{i}\left(s\right)ds}.\end{array}$ Then, the property that ex ≥ ex for x ≥ 0 leads to $I(t)≥e∫g(t)t∑i=12pi(v)∫g(v)v∑k=12pk(s)(∫g(v)v∑ℓ=12pℓ(s)ds+λ(ρ)−ϵ∫g(s)g(v)∑l=12pl(u)du)dsdv+c(ρ)e∫g(t)t∑i=12pi(s)ds.$

Now, using Maple, we get $I(t)≥0.49727+0.0029516sin⁡(t)cos2⁡(t)+0.27514sin⁡(t)−0.23059cos⁡(t)−0.015664cos2⁡(t)−0.083083cos⁡(t)sin⁡(t)+0.0037423cos3⁡(t)+0.10144e0.16044sin⁡(t).$

Choose $\begin{array}{}{T}_{k}=\frac{3\pi }{4}+2\pi k,\end{array}$ then $I(Tk)≈1.0019>1,forallk∈N,$

which means that $lim supt→∞(∫g(t)tQ(v)dv+c(ρ)e∫g(t)t∑i=12pi(s)ds)>1.$

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 $∫τ∗(t)t∑k=12pk(s)ds≤∫t−π2t∑k=12pk(s)ds=abπ+2asin⁡(t).$

Hence, $lim inft→∞∫τ∗(t)t∑k=12pk(s)ds≤abπ−2a≈0.2891659465<1e.$

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 $a1(h(s),τi(s))=exp∫τi(s)g(s)∑i=12pi(u)du≤exp∫s−π2−σs−π2∑i=12pi(u)du≤exp∫s−π2−σs−π22a(b+1)du=exp⁡(2a(b+1)σ).$

Therefore, $∫h(t)t∑i=12pi(s)a1(h(s),τi(s))ds≤∫t−π2ta(b+sin⁡(s)+b+cos⁡(s))exp2a(b+1)σds=∫t−π2t2abds+∫t−π2tasin⁡(s)ds+∫t−π2tacos⁡(s)dsexp2a(b+1)σ=[abπ+2asin⁡(t)]exp2a(b+1)σ≤abπ+2aexp2a(b+1)σ<0.61188.$

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 $∫τ¯(t)ta(b+cos⁡(s))ds≤∫t−π2−σta(b+cos⁡(s))ds≤a(b+1)∫t−π2−σt−π2ds+∫t−π2ta(b+cos⁡(s))ds=aσ(b+1)+abπ2+a(cos⁡(t)+sin⁡(t)).$

Then, $\begin{array}{}{q}_{1}=\frac{ab\pi }{2}-a\sqrt{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{q}_{2}\le \frac{ab\pi }{2}-a\sqrt{2}+a\sigma \left(b+1\right),\end{array}$ where qi is defined by (4), i = 1, 2. Therefore, λ (q1) ≈ 1.134680932 and λ (q2) ≈ 1.136881841. Let $I1(t)=∏j=12∏i=12∫gj(t)tpi(s)exp⁡(∫τi(s)gi(t)∑k=12(λ(qk)−ϵ)pk(u)du)ds12.$

Then $I1(t)<∏j=12∏i=12∫gj(t)tpi(s)exp⁡(λ(q2)∫τi(s)gi(t)∑l=12pl(u)du)ds12≤∏j=12∏i=12∫gj(t)tpi(s)exp⁡(2aλ(q2)(1+b)(t−s+σ))ds12≤∏i=12∫t−π2−σtpi(s)exp⁡(2aλ(q2)(1+b)(t−s+σ))ds≤∏i=12[∫t−π2−σt−π2a(b+1)exp⁡(2aλ(q2)(1+b)(t−s+σ))ds+∫t−π2tpi(s)exp⁡(2aλ(q2)(1+b)(t−s+σ))ds]≈0.1363159006+0.08561813338sin⁡(t)−0.00925235722cos⁡(t)−0.02983977134cos2⁡(t)−0.00652549885sin⁡(t)cos⁡(t)<0.23771189<14.$

Therefore, $\begin{array}{}\underset{ϵ\to {0}^{+}}{lim sup}\left(\underset{t\to \mathrm{\infty }}{lim sup}{I}_{1}\left(t\right)\right)<\frac{1}{4},\end{array}$ which implies that condition (3) is not satisfied.

Finally, since $∑i=12∫τi(t)t∏l=12pl(s)12ds≤∑i=1212∫τi(t)t∑l=12pl(s)ds≤12∫t−π2t∑l=12pl(s)ds+12∫t−π2−σt∑l=12pl(s)ds=12∫t−π2t∑l=12pl(s)ds+12∫t−π2−σt−π2∑l=12pl(s)ds+∫t−π2t∑l=12pl(s)ds≤∫t−π2t∑l=12pl(s)ds+∫t−π2−σt−π2a(b+1)ds≤∫t−π2t∑l=12pl(s)ds+aσ(b+1),$

then $p∗¯=lim inft→∞∑i=12∫τi(t)t∏l=12pl(s)12ds≤abπ−2a+aσ(b+1)<0.2906549<1e.$

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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Accepted: 2018-01-17

Published Online: 2018-02-23

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 83–94, ISSN (Online) 2391-5455,

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