R E S E A R C H
Open Access
Oscillatory behavior of third-order neutral
delay differential equations with distributed
deviating arguments
Yibing Sun
1*and Yige Zhao
1*Correspondence:
1School of Mathematical Sciences, University of Jinan, Jinan, P.R. China
Abstract
The main contribution of this paper is to establish some new criteria, which ensure that every solution of third-order neutral delay differential equations with distributed deviating arguments is either oscillatory or tends to zero. The obtained theorems extend and improve several known results in the literature. Two examples are provided to illustrate the main results.
MSC: 34K11
Keywords: Oscillation; Asymptotic property; Distributed deviating arguments;
Riccati transformation
1 Introduction
Over the last several years, an increasing attention has been given to the oscillation theory and asymptotic behavior of various classes of second-order and high-order differential equations and dynamic equations on time scales [1–17]. So far, much research activity concerns the oscillation problem of the third-order (TO) neutral differential and dynamic equations [18–26]. Recently, the research focus has been shifted to the study of the TO differential equations (DE) with distributed deviating arguments (DDA), and some results can be found in [27–35].
Li et al. [3] investigated
r(t)z(t)α–1z(t)+
b a
q(t,ξ)xg(t,ξ)α–1xg(t,ξ)dσ(ξ) = 0,
wherez(t) =x(t) +p(t)x(τ(t)), 0≤p(t)≤p0<∞andα≥1 is a constant. Under the meth-ods proposed by Li et al. [3,20], Jiang and Li [30] studied the following equation with DDA:
r(t)z(t)α–1z(t)+
b a
q(t,ξ)xg(t,ξ)α–1xg(t,ξ)dσ(ξ) = 0, (1.1)
whereα> 0 is a constant, and obtained several theorems for (1.1) whenever
∞ t0
r–α1(t)dt=∞, or
∞ t0
r–1α(t)dt<∞.
Furthermore, Elabbasy and Moaaz [31], and Wang et al. [32] examined a TODE with DDA under the assumption 0≤p(t)≤P< 1. However, the obtained oscillation theorems cannot be applied when p(t)≥1. Then Tunç [33] utilized a new technique, different from the existing methods, to give some criteria for a TODE with DDA, whenp(t)≥1.
The main objective here is to establish several oscillation criteria for the TO neutral delay (ND) DE with DDA
E2(t) +
b a
q(t,ξ)xg(t,ξ)α3–1
xg(t,ξ)dσ(ξ) = 0, (1.2)
wheret≥t0> 0,
E2(t) =r2(t)E1(t)α2–1
E1(t),
E1(t) =r1(t)z(t)α1–1
z(t), (1.3)
z(t) =x(t) +p(t)xτ(t),
andα1,α2, andα3are positive constants. We assume that:
(A1) ri(t)∈C([t0,∞), (0,∞)),
∞ t0 r
–α1
i
i (t)dt=∞,i= 1, 2;
(A2) p(t)∈C([t0,∞), [1,∞))withp(t)≡1, andq(t,ξ)∈C([t0,∞)×[a,b], [0,∞))with q(t,ξ)≡0eventually;
(A3) τ(t)∈C1([t0,∞),R),τ(t)≤t,τ(t) > 0andlim
t→∞τ(t) =∞;
(A4) g(t,ξ)∈C([t0,∞)×[a,b],R)is a nondecreasing function forξ, which satisfies
lim inft→∞g(t,ξ) =∞forξ∈[a,b];
(A5) σ(ξ)∈C([a,b],R)is nondecreasing and the integral of (1.2) is taken in the Riemann–Stieltjies sense.
This article is organized in the following manner. Section2presents three lemmas to prove our results. Section3establishes some new oscillation criteria for (1.2). Two exam-ples finalize this article.
2 Some lemmas
For simplicity, we use some notations for sufficiently larget1witht1≥t0as below:
g1(t) =g(t,a), g2(t) =g(t,b), d+(t) =max 0,d(t),
δ1(t,t1) =
t t1
1 r2(s)
1
α2
ds, δ2(t,t1) =
δ1(t,t1) r1(t)
1
α1
,
δ3(t,t1) =
t t1
δ2(s,t1)ds, t≥t1.
Furthermore, assume that
p1(t) = 1 p(τ–1(t))
1 – 1
p(τ–1(τ–1(t)))
> 0, (2.1)
p2(t) = 1 p(τ–1(t))
1 – δ3(τ
–1(τ–1(t)),t1)
p(τ–1(τ–1(t)))δ3(τ–1(t),t1)
q1(t) =
b a
q(t,ξ)pα3
1
g(t,ξ)dσ(ξ),
q2(t) =
b a
q(t,ξ)pα3
2
g(t,ξ)dσ(ξ).
Lemma 2.1 Assume that(A1)–(A5)hold.Furthermore,let x(t)be an eventually positive
solution of (1.2).Then z(t)satisfies either
(I) z(t) > 0, z(t) > 0, E1(t) > 0, E2(t)≤0,
or
(II) z(t) > 0, z(t) < 0, E1(t) > 0, E2(t)≤0.
Proof From the condition of Lemma2.1, there exists at1≥t0such that
x(t) > 0, xτ(t)> 0 and xg(t,ξ)> 0, ξ∈[a,b], (2.3)
fort≥t1. Then (1.2) implies that
E2(t) = –
b a
q(t,ξ)xg(t,ξ)α3–1xg(t,ξ)dσ(ξ)≤0, (2.4)
which means thatE2(t) is nonincreasing and of constant sign, andE1(t) is also of constant sign. We claim thatE1(t) > 0. To prove this, assume on the contrary thatE1(t) < 0 and there existsM1, s.t. fort≥t2≥t1,
E2(t)≤–M1< 0.
Then
E1(t)≥
M1 r2(t)
1
α2
> 0. (2.5)
We integrate (2.5) to get
E1(t)≤E1(t2) –M
1
α2
1
t t2
1 r2(s)
1
α2
ds.
Lettingt→ ∞and from (A1), we obtainlimt→∞E1(t) = –∞. Then there exist constants
M2andt3≥t2such that
E1(t)≤–M2< 0, t≥t3,
which yields that
z(t)≥
M2 r1(t)
1
α1
Integrating the latter inequality fromt3tot, we have
z(t)≤z(t3) –M
1
α1
2
t t3
1 r1(s)
1
α1
ds.
We use (A1) again to havelimt→∞z(t) = –∞, which contradictsz(t) > 0. Hence,E1(t) > 0
fort≥t1, andz(t) has properties (I) and (II).
Lemma 2.2 Assume that(A1)–(A5)and(2.1)hold.Furthermore,suppose that x(t)is an
eventually positive solution of (1.2)and z(t)has property(II)of Lemma2.1.If
∞ t0
1 r1(u)
∞ u
1 r2(v)
∞ v
q1(s)ds
1
α2
dv
1
α1
du=∞, (2.6)
thenlimt→∞x(t) = 0.
Proof One can see that (1.3) yields (see [33])
zτ–1(t)=xτ–1(t)+pτ–1(t)x(t), which can be rewritten as
x(t)≥ z(τ –1(t))
p(τ–1(t))–
z(τ–1(τ–1(t)))
p(τ–1(t))p(τ–1(τ–1(t))) (2.7)
≥p1(t)zτ–1(t). (2.8)
Combining (2.4) and (2.8), we get
E2(t)≤–
b a
q(t,ξ)pα3
1
g(t,ξ)zα3τ–1g(t,ξ)dσ(ξ)
≤–q1(t)zα3
τ–1g2(t)
, (2.9)
based on the fact thatz(t) < 0 fort≥t1. Sincez(t) has property (II) of Lemma2.1, one gets
limt→∞z(t) =l≥0. We claim thatl= 0. Indeed, if we assume on the contrary thatl> 0,
then there existst2≥t1s.t.τ–1(g2(t))≥t2andz(τ–1(g2(t)))≥l,t≥t2. Inequality (2.9) then yields
E2(t)≤–lα3q1(t). (2.10)
We integrate (2.10) to get
r2(t)E1(t)α2≥lα3
∞ t
q1(s)ds,
which indicates that
E1(t)≥
lα3
r2(t)
∞ t
q1(s)ds
1
α2
Integrating (2.11) fromtto∞, we have
–E1(t)≥l
α3
α2
∞ t
1 r2(v)
∞ v
q1(s)ds
1
α2
dv,
and then
z(t)≥l
α3
α1α2
1 r1(t)
∞ t
1 r2(v)
∞ v
q1(s)ds
1
α2
dv
1
α1
,
sincez(t) < 0 fort≥t2. Integrating the latter inequality fromt2to∞, we get
z(t2) >l
α3
α1α2
∞ t2
1 r1(u)
∞ u
1 r2(v)
∞ v
q1(s)ds
1
α2
dv
1
α1
du,
which contradicts (2.6). Thus, we obtainl= 0 andlimt→∞x(t) = 0, since 0 <x(t)≤z(t).
Lemma 2.3 Assume that(A1)–(A5)and(2.2)hold.Furthermore,suppose that x(t)is an
eventually positive solution of (1.2)and z(t)has property(I)of Lemma2.1.Then for t≥ t1≥t0,
r2(t)r1(t)z(t)α1α2+q2(t)zα3τ–1g1(t)≤0. (2.12)
Proof Property (I) ofz(t) yields
E2(t) =r2(t)
r1(t)
z(t)α1α2
> 0.
Applying the monotonicity ofE
1
α2
2 (t) gives
r1(t)z(t)α1=r1(t1)z(t1)α1+
t t1
r
1
α2
2 (s)(r1(s)(z(s))α1)
r
1
α2
2 (s)
ds
≥δ1(t,t1)r
1
α2
2 (t)
r1(t)z(t)α1
. (2.13)
It can be seen from (2.13) that
r1(t)(z(t))α1 δ1(t,t1)
≤0,
which, together withr1(t)(z(t))α1> 0, yields thatz(t)/δ2(t,t1) is nonincreasing fort≥t1.
Therefore,
z(t) =z(t1) +
t t1
z(s)
δ2(s,t1)δ2(s,t1)ds≥ δ3(t,t1) δ2(t,t1)z
(t), (2.14)
which leads to
z(t) δ3(t,t1)
Then fort≥t2≥t1,
zτ–1τ–1(t)≤δ3(τ
–1(τ–1(t)),t1)z(τ–1(t))
δ3(τ–1(t),t1) , (2.16)
due toτ–1(t)≤τ–1(τ–1(t)). Substituting (2.16) into (2.7), one has
x(t)≥p2(t)zτ–1(t), which leads to
xg(t,ξ)≥p2
g(t,ξ)zτ–1g1(t). (2.17)
Combining (1.2) and (2.17), we obtain (2.12).
3 Main results
We respectively consider two casesg1(t)≤τ(t) andg1(t)≥τ(t) fort≥t0. Now, we begin with the first case.
Theorem 3.1 Assume that(A1)–(A5), (2.1), (2.2),and(2.6)hold,and g1(t)≤τ(t).If there
existsρ(t)∈C1([t0,∞), (0,∞))s.t.
lim sup t→∞
t t∗
ρ(s)q2(s)
δ3(τ–1(g1(s)),t1) δ3(s,t1)
α3
– λ(ρ
+(s))α1α2+1 (ρ(s)γ(s)δ2(s,t1))α1α2
ds=∞, (3.1)
for t1and t∗with t∗≥t1≥t0,where
λ=
α1α2 α3
α1α2 1
α1α2+ 1
α1α2+1
,
γ(t) =
⎧ ⎨ ⎩
m1(δ3(t,t1))
α3
α1α2–1, m1is any positive constant, ifα1α2>α3,
m2, m2is any positive constant, ifα1α2≤α3,
then every solution of (1.2)is either oscillatory or tends to zero.
Proof Suppose that (1.2) has a nonoscillatory solutionx(t). We may assume that (2.3) holds fort≥t1≥t0. So we have thatz(t) is positive and satisfies the two properties fort≥t1.
We first consider property (I). Defineω(t) by
ω(t) =ρ(t)r2(t)((r1(t)(z
(t))α1))α2
zα3(t) , t≥t1. (3.2)
Thenω(t) > 0 and
ω(t) =ρ(t)r2(t)((r1(t)(z
(t))α1))α2
zα3(t) +ρ(t)
[r2(t)((r1(t)(z(t))α1))α2] zα3(t)
–α3r2(t)((r1(t)(z
(t))α1))α2z(t)
zα3+1(t)
=ρ
(t)
ρ(t)ω(t) +ρ(t)
[r2(t)((r1(t)(z(t))α1))α2]
zα3(t)
–α3ρ(t)
r2(t)((r1(t)(z(t))α1))α2z(t)
zα3+1(t) . (3.4)
Based on (2.12), we have
[r2(t)((r1(t)(z(t))α1))α2]
zα3(t) ≤–q2(t)
z(τ–1(g1(t))) z(t)
α3
. (3.5)
Sinceg1(t)≤τ(t), we getτ–1(g1(t))≤t. Applying (2.15), we obtain
z(τ–1(g1(t))) z(t) ≥
δ3(τ–1(g1(t)),t1)
δ3(t,t1) . (3.6)
From (2.13), we have
z(t)≥δ2(t,t1)r2(t)r1(t)z(t)α1α2α11α2. (3.7)
We combine (3.4)–(3.7) to conclude that
ω(t)≤ ρ
+(t)
ρ(t)ω(t) –ρ(t)q2(t)
δ3(τ–1(g1(t)),t1) δ3(t,t1)
α3
–α3δ2(t,t1) ρα11α2(t)
z
α3
α1α2–1(t)ω 1
α1α2+1(t). (3.8)
Next, we will computez
α3
α1α2–1(t) and consider the following two cases:
Case (i). Assume thatα1α2>α3. Sincez(t)/δ3(t,t1) is nonincreasing, due to (2.15), there exist constantsh1> 0 andt2≥t1such that
z(t) δ3(t,t1)≤
z(t2)
δ3(t2,t1)=h1, t≥t2,
and
z
α3
α1α2–1(t)≥m1δ3(t,t1)
α3
α1α2–1, (3.9)
wherem1=h
α3
α1α2–1
1 .
Case (ii). Assume thatα1α2≤α3. Sincez(t) > 0, there existsh2> 0 such that
z(t)≥z(t1) =h2, t≥t1,
and
z
α3
wherem2=h
α3
α1α2–1
2 . We combine (3.8) with (3.9) and (3.10) to have
ω(t)≤ ρ
+(t)
ρ(t)ω(t) –ρ(t)q2(t)
δ3(τ–1(g1(t)),t1) δ3(t,t1)
α3
–α3γ(t)δ2(t,t1) ρα11α2(t)
ωα11α2+1(t). (3.11)
By using the inequality (see [18])
Bu–Au1α+1≤ α α
(α+ 1)α+1 Bα+1
Aα , A> 0, (3.12)
where
B=ρ
+(t)
ρ(t), A=
α3γ(t)δ2(t,t1)
ρα11α2(t)
, α=α1α2,u=ω(t),
we obtain
ρ+(t) ρ(t)ω(t) –
α3γ(t)δ2(t,t1)
ρα11α2(t)
ωα11α2+1(t)
≤
α1α2 α3ρ(t)γ(t)δ2(t,t1)
α1α2 ρ
+(t) α1α2+ 1
α1α2+1
= λ(ρ
+(t))α1α2+1
(ρ(t)γ(t)δ2(t,t1))α1α2. (3.13)
We combine (3.11) and (3.13) to conclude that
ω(t)≤–ρ(t)q2(t)
δ3(τ–1(g1(t)),t1) δ3(t,t1)
α3
+ λ(ρ
+(t))α1α2+1 (ρ(t)γ(t)δ2(t,t1))α1α2.
We integrate the latter inequality to make
t t2
ρ(s)q2(s)
δ3(τ–1(g1(s)),t1) δ3(s,t1)
α3
– λ(ρ
+(s))α1α2+1 (ρ(s)γ(s)δ2(s,t1))α1α2
ds
≤ω(t2) –ω(t) <ω(t2),
which contradicts (3.1).
Secondly, we investigate property (II) and deducelimt→∞x(t) = 0.
Theorem 3.2 Assume that(A1)–(A5), (2.1), (2.2),and(2.6)hold.Furthermore,suppose
that g1(t)≤τ(t)andα1α2=α3.If there existsρ(t)s.t.
lim sup t→∞
t t∗
ρ(s)q2(s)
δ3(τ–1(g1(s)),t1) δ3(s,t1)
α3
– ρ
+(s) (δ3(s,t1))α3
ds=∞, (3.14)
Proof Suppose that (2.3) holds fort≥t1≥t0. Thenz(t) satisfies the two properties. We first consider property (I). From (2.13) and (2.14), we have
r1(t)zα1(t)≥
δ3(t,t1) δ2(t,t1)
α1
r1(t)
z(t)α1
≥
δ3(t,t1) δ2(t,t1)
α1
δ1(t,t1)r
1
α2
2 (t)
r1(t)z(t)α1,
and
zα3(t)≥δ3(t,t1)α3r2(t)r1(t)z(t)α1α2. (3.15)
Define ω(t) by (3.2). As in the proof of Theorem3.1, we get (3.3). SinceE2(t) > 0, (3.3) indicates
ω(t)≤ρ+(t)r2(t)((r1(t)(z
(t))α1))α2
zα3(t) +ρ(t)
[r2(t)((r1(t)(z(t))α1))α2]
zα3(t) .
Combining the latter inequality with (3.5), (3.6), and (3.15), we see that
ω(t)≤ ρ
+(t)
(δ3(t,t1))α3 –ρ(t)q2(t)
δ3(τ–1(g1(t)),t1) δ3(t,t1)
α3
. (3.16)
An integration of (3.16) fromt2(t2≥t1) totleads to
t t2
ρ(s)q2(s)
δ3(τ–1(g1(s)),t1) δ3(s,t1)
α3
– ρ
+(s) (δ3(s,t1))α3
ds≤ω(t2) –ω(t) <ω(t2),
for all sufficiently larget, which contradicts (3.14).
Secondly, if property (II) holds, thenlimt→∞x(t) = 0.
Theorem 3.3 Assume that(A1)–(A5), (2.1), (2.2),and(2.6)hold.Furthermore,suppose
that g1(t)≤τ(t)andα1α2=α3≥1.If there existsρ(t)s.t.
lim sup t→∞
t t∗
ρ(s)q2(s)
δ3(τ–1(g1(s)),t1) δ3(s,t1)
α3
– (ρ
+(s))2
4α3ρ(s)γ(s)δ2(s,t1)(δ3(s,t1))α3–1
ds=∞, (3.17)
for t1 and t∗ with t∗≥t1≥t0,whereγ(t)is given in Theorem3.1,then we get the same conclusion as in Theorem3.1.
Proof Suppose that (2.3) holds fort≥t1≥t0. Thenz(t) satisfies the two properties. We first consider property (I). Proceeding as in the proof of Theorem3.1, we get (3.11), and
ω(t)≤ ρ
+(t)
ρ(t)ω(t) –ρ(t)q2(t)
δ3(τ–1(g1(t)),t1) δ3(t,t1)
α3
–α3γ(t)δ2(t,t1)ω 2(t)
ρα11α2(t)
ω
1
From (3.2) and (3.15), one has
ωα11α2–1(t) =ρ 1
α1α2–1(t)
r2(t)((r1(t)(z(t))α1))α2
zα3(t)
1
α1α2–1
≥ρ
1
α1α2–1(t)δ3(t,t1)α3–1
. (3.19)
We substitute (3.19) into (3.18) to see that
ω(t)≤ ρ
+(t)
ρ(t)ω(t) –ρ(t)q2(t)
δ3(τ–1(g1(t)),t1) δ3(t,t1)
α3
–α3γ(t)δ2(t,t1)(δ3(t,t1))
α3–1
ρ(t) ω
2(t),
from which one gets
ω(t)≤–ρ(t)q2(t)
δ3(τ–1(g1(t)),t1) δ3(t,t1)
α3
+ (ρ
+(t))2
4α3ρ(t)γ(t)δ2(t,t1)(δ3(t,t1))α3–1 ,
by completing the square with respect toω(t). We integrate the latter inequality fromt2 (t2≥t1) totto obtain
t t2
ρ(s)q2(s)
δ3(τ–1(g1(s)),t1) δ3(s,t1)
α3
– (ρ
+(s))2
4α3ρ(s)γ(s)δ2(s,t1)(δ3(s,t1))α3–1
ds≤ω(t2),
for all sufficiently larget, which contradicts (3.17).
Secondly, if property (II) holds, thenlimt→∞x(t) = 0.
Next, we considerg1(t)≥τ(t) fort≥t0.
Theorem 3.4 Assume that conditions(A1)–(A5), (2.1), (2.2),and(2.6)hold,and g1(t)≥
τ(t).If there existsρ(t)s.t.
lim sup t→∞
t t∗
ρ(s)q2(s) – λ(ρ
+(s))α1α2+1 (ρ(s)γ(τ(s))δ2(τ(s),t1))α1α2
ds=∞, (3.20)
for t1and t∗with t∗≥t1≥t0,then we get the same conclusion as in Theorem3.1.
Proof Suppose that (2.3) holds fort≥t1≥t0. Thenz(t) satisfies the two properties. We first consider property (I). Defineν(t) by
ν(t) =ρ(t)r2(t)((r1(t)(z
(t))α1))α2
zα3(τ(t)) , t≥t1. (3.21)
Thenν(t) > 0 and
ν(t) =ρ(t)r2(t)((r1(t)(z
(t))α1))α2
zα3(τ(t)) +ρ(t)
[r2(t)((r1(t)(z(t))α1))α2] zα3(τ(t))
–α3r2(t)((r1(t)(z
(t))α1))α2(z(τ(t)))
zα3+1(τ(t))
=ρ
(t)
ρ(t)ν(t) +ρ(t)
[r2(t)((r1(t)(z(t))α1))α2]
zα3(τ(t))
–α3ρ(t)
r2(t)((r1(t)(z(t))α1))α2(z(τ(t)))
zα3+1(τ(t)) . (3.23)
Sinceτ–1(g1(t))≥t≥τ(t) andz(t) > 0, we have
z(τ–1(g1(t)))
z(τ(t)) ≥1, t≥t1,
which indicates that
[r2(t)((r1(t)(z(t))α1))α2]
zα3(τ(t)) ≤–q2(t), (3.24)
due to (2.12). Based on (3.7),E2(t)≤0 andτ(t)≤t, so one hasτ(t)≥t1and
zτ(t)≥δ2
τ(t),t1
r2(t)r1(t)z(t)α1α2
1
α1α2, (3.25)
fort≥t2>t1. Combining (3.9), (3.10), (3.23)–(3.25), we conclude that
ν(t)≤ρ
+(t)
ρ(t)ν(t) –ρ(t)q2(t)
–α3γ(τ(t))δ2(τ(t),t1) ρα11α2(t)
να11α2+1(t). (3.26)
Using (3.12) and (3.26) with
B=ρ
+(t)
ρ(t), A=
α3γ(τ(t))δ2(τ(t),t1)
ρα11α2(t)
,
one gets
ν(t)≤–ρ(t)q2(t) + λ(ρ
+(t))α1α2+1 (ρ(t)γ(τ(t))δ2(τ(t),t1))α1α2
.
Integrating the latter inequality fromt2tot, we have
t t2
ρ(s)q2(s) – λ(ρ
+(s))α1α2+1 (ρ(s)γ(τ(s))δ2(τ(s),t1))α1α2
ds≤ω(t2),
for all sufficiently larget, which contradicts (3.20).
Secondly, if property (II) holds, thenlimt→∞x(t) = 0.
Theorem 3.5 Assume that(A1)–(A5), (2.1), (2.2),and(2.6)hold.Furthermore,suppose
that g1(t)≥τ(t)andα1α2=α3.If there existsρ(t)s.t.
lim sup t→∞
t t∗
ρ(s)q2(s) –
ρ+(s) (δ3(τ(s),t1))α3
ds=∞, (3.27)
Proof Suppose that (2.3) holds fort≥t1≥t0. Thenz(t) satisfies the two properties. We first consider property (I). Proceeding as in the proof of Theorem3.4, we get (3.22), which implies
ν(t)≤ρ+(t)r2(t)((r1(t)(z
(t))α1))α2
zα3(τ(t)) +ρ(t)
[r2(t)((r1(t)(z(t))α1))α2]
zα3(τ(t)) . (3.28)
Applying (3.15), the monotonicity ofE2(t) and the fact thatτ(t)≤t, one hasτ(t)≥t1and
zα3τ(t)≥δ
3
τ(t),t1
α3
r2(t)r1(t)z(t)α1α2, (3.29) fort≥t2>t1. Combining (3.24), (3.28), and (3.29), one gets
ν(t)≤ ρ
+(t)
(δ3(τ(t),t1))α3 –ρ(t)q2(t). (3.30)
Upon integrating (3.30) fromt2tot, we obtain a contradiction to (3.27).
Secondly, if property (II) holds, thenlimt→∞x(t) = 0.
Theorem 3.6 Assume that(A1)–(A5), (2.1), (2.2),and(2.6)hold.Furthermore,suppose
that g1(t)≥τ(t)andα1α2=α3≥1.If there existsρ(t)s.t.
lim sup t→∞
t t∗
ρ(s)q2(s) –
(ρ+(s))2
4α3ρ(s)γ(τ(s))δ2(τ(s),t1)(δ3(τ(s),t1))α3–1
ds=∞,
for t1and t∗with t∗≥t1≥t0,then we get the same conclusion as in Theorem3.1.
We omit the proof of Theorem3.6here, since it is similar to that of Theorem3.3.
4 Examples
The following examples are given to show the applications of Theorems3.1and3.5.
Example4.1 Fort>k1≥1, consider a TONDDE with DDA
E2(t) +
k1+1
k1
10(t+ξ)x
t–k1– 1 ξ
4 3
x
t–k1– 1 ξ
dξ= 0, (4.1)
where
E2(t) =E1(t)4E1(t), E1(t) = (t–k1)z(t)–
2 3z(t),
z(t) =x(t) +4t+ 5
t+ 1x(t–k1).
Letα1= 1/3,α2= 5,α3= 7/3,a=k1,b=k1+ 1,r1(t) =t–k1,r2(t) = 1,τ(t) =t–k1,g(t,ξ) = t–k1– 1/ξ,σ(ξ) =ξ,p(t) = (4t+ 5)/(t+ 1),q(t,ξ) = 10(t+ξ). Chooset0=t1=k1. Then we obtainα1α2<α3, 4≤p(t) < 5,
δ1(t,t1) =δ1(t,k1) =t–k1,
δ2(t,t1) =
δ1(t,k1) t–k1
3 = 1,
δ3(t,t1) =δ3(t,k1) =t–k1,
δ3
τ–1(t),t1
=δ3(t+k1,k1) =t,
δ3
τ–1τ–1(t),t1
=δ3(t+ 2k1,k1) =t+k1,
δ3
τ–1g1(t),t1
=δ3
t– 1 k1
,k1
=t–k1– 1 k1 .
Furthermore, we deduce that
p1(t) >1 5
1 –1 4
= 3 20> 0,
p2(t) >1 5
1 –1 4·
t+k1 t
> 1 10> 0,
q1(t) >
k1+1
k1
3
20·10(t+ξ)dξ= 3 2
t+k1+ 1 2
,
q2(t) >
k1+1
k1
1
10·10(t+ξ)dξ=t+k1+ 1 2.
It is easy to verify that
∞ t0
1 r1(u)
∞ u
1 r2(v)
∞ v
q1(s)ds
1 α2 dv 1 α1 du > ∞ k1 1 u–k1
∞ u ∞ v 3 2
s+k1+ 1 2 ds 1 5 dv 3 du =∞.
Therefore, conditions (A1)–(A5), (2.1), (2.2), and (2.6) hold, andg1(t)≤τ(t). We choose ρ(t) =tandt∗=k1+ 2. Applying Theorem3.1, it remains to check (3.1), where
λ=
5 7
5 33
8
8 3
.
Then we get
t t∗
ρ(s)q2(s)
δ3(τ–1(g1(s)),t1) δ3(s,t1)
α3
– λ(ρ
+(s))α1α2+1 (ρ(s)γ(s)δ2(s,t1))α1α2
ds
>
t k1+2
s
s+k1+ 1 2
s–k
1–k11 s–k1
7 3 –( 5 7) 5 3(3
8)
8 3
(m2s)
5 3
ds→ ∞,
ast→ ∞, sincekt
1+2s
Example4.2 Fort>k1≥1, consider a TONDDE with DDA
E2(t) +
k1+l
k1 40ξ t x
t+ξ 2
2x
t+ξ 2
dξ= 0, (4.2)
wherelis a positive integer,
E2(t) =E1(t) –23
E1(t),
E1(t) = (t–k1)z(t)8z(t), z(t) =x(t) +5t+ 4k1
t+k1 x t 2 .
Letα1= 9,α2= 1/3,α3= 3,a=k1,b=k1+l,r1(t) =t–k1,r2(t) = 1,σ(ξ) =ξ,
τ(t) = t
2, g(t,ξ) = t+ξ
2 , p(t) =
5t+ 4k1 t+k1
, q(t,ξ) =40ξ t .
Chooset0=t1=k1. Then we haveα1α2=α3, 4 <p(t) < 5,
g1(t) =g(t,k1) =t+k1 2 ,
δ3
τ–1(t),t1
=δ3(2t,k1) = 2t–k1,
δ3
τ–1τ–1(t),t1
=δ3(4t,k1) = 4t–k1,
δ3
τ(t),t1
=δ3
t 2,k1
= t 2–k1>
t 4,
wheret≥t2> 4k1,δ1(t,t1),δ2(t,t1), andδ3(t,t1) are the same as in Example4.1. Further-more, we deduce that
p1(t) >1 5
1 –1 4
= 3 20> 0,
p2(t) > 1 5
1 –1 4·
4t–k1 2t–k1
> 1 20> 0,
q1(t) >
k1+1
k1
3 20·
40ξ
t dξ=
6k1l+ 3l2
t ,
q2(t) >
k1+1
k1
1 20·
40ξ
t dξ=
2k1l+l2
t .
Clearly, (2.6) holds. Choosingρ(t) =t2andt
∗=t2, one has
t t∗
ρ(s)q2(s) – ρ
+(s) (δ3(τ(s),t1))α3
ds > t t2
s2·2k1l+l 2
s –
2s (4s)3
ds→ ∞,
Funding
This research was supported by the National Natural Science Foundation of China (Grant No. 61803176, 61703180) and A Project of Shandong Province Higher Educational Science and Technology Program (Grant No. J18KA230).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to this work. They all read and approved the final version of the manuscript.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Received: 15 February 2019 Accepted: 17 July 2019 References
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