Infinitely many nonoscillatory solutions for second order nonlinear neutral delay differential equations

Nonlinear Analysis 70 (2009) 4274–4293

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Nonlinear Analysis

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Infinitely many nonoscillatory solutions for second order nonlinearneutral delay differential equationsZeqing Liu a,∗, Shin Min Kang ba Department of Mathematics, Liaoning Normal University, P. O. Box 200, Dalian, Liaoning 116029, People’s Republic of Chinab Department of Mathematics and The Research Institute of Natural Science, Gyeongsang National University, Chinju 660-701, Republic of Korea

a r t i c l e i n f o

Article history:Received 29 August 2006Accepted 29 September 2008

MSC:34K1534C10

Keywords:Second order nonlinear neutral delaydifferential equationInfinitely many nonoscillatory solutionsContraction mappingMann iterative sequence with errors

a b s t r a c t

In this paper we consider the second order nonlinear neutral delay differential equation[a(t) (x(t)+ b(t)x(t − τ))′

]′+ [h(t, x(h1(t)), x(h2(t)), . . . , x(hk(t)))]′

+ f (t, x(f1(t)), x(f2(t)), . . . , x(fk(t))) = g(t), t ≥ t0,

where τ > 0, a, b, g ∈ C([t0,+∞),R)with a(t) > 0 for t ≥ t0, h ∈ C1([t0,+∞)×Rk,R),f ∈ C([t0,+∞)× Rk,R), hl ∈ C1([t0,+∞),R) and fl ∈ C([t0,+∞),R)with

limt→+∞

hl(t) = limt→+∞

fl(t) = +∞, l = 1, . . . , k.

Under suitable conditions, by making use of the Banach fixed point theorem, we showthe existence of infinitely many nonoscillatory solutions, which are uncountable, forthe above equation, suggest several Mann type iterative approximation sequences witherrors for these nonoscillatory solutions and establish some error estimates between theapproximate solutions and the nonoscillatory solutions. Five nontrivial examples are givento illustrate the advantages of our results.

© 2008 Elsevier Ltd. All rights reserved.

1. Introduction and preliminaries

In recent years there has been much research activity concerning the oscillation and nonoscillation of solutions forvarious kinds of second order neutral delay differential equations, for example, see [1–10,12–14]. Huang [9] and Elbert [5]established a few oscillation and nonoscillation criteria for the second order linear differential equation

x′′(t)+ q(t)x(t) = 0, t ≥ 0, (1.1)

where q ∈ C([0,+∞),R+). Tang and Liu [12] studied the existence of bounded oscillation for the second order linear delaydifferential equation of unstable type

x′′(t) = p(t)x(t − τ), t ≥ t0, (1.2)

where τ > 0, p ∈ C([t0,+∞),R+) and p(t) 6≡ 0 on any interval of length τ . Agarwal, O’Regan and Saker [3] investigatedsome oscillation criteria for the second order nonlinear neutral delay dynamic equation[

r(t)((x(t)+ P(t)x(t − τ))4)γ]4+ f (t, x(t − δ)) = 0, t ≥ t0 (1.3)

∗ Corresponding author.E-mail addresses: [email protected] (Z. Liu), [email protected] (S.M. Kang).

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Z. Liu, S.M. Kang / Nonlinear Analysis 70 (2009) 4274–4293 4275

on a time scale T, where γ is a quotient of odd positive integers with r(t) and p(t) real-valued positive functions defined onT. In 1998, Kulenović and Hadžiomerspahić [10] discussed the second order linear neutral delay differential equation withpositive and negative coefficients

d2

dt2[x(t)+ cx(t − τ)]+ Q1(t)x(t − σ1)− Q2(t)x(t − σ2) = 0, t ≥ t0, (1.4)

where c ∈ R, τ > 0, σ1, σ2 ∈ [0,+∞),Q1,Q2 ∈ C([t0,+∞),R+), and under the conditions c 6= ±1, aQ1(t) ≥ Q2(t), t ≥t0 andother conditions, they gave some sufficient conditions for the existence of a nonoscillatory solution of Eq. (1.4). In 2004,Cheng and Annie [4] continued to investigate a few sufficient conditions, which guarantee the existence of a nonoscillatorysolution for Eq. (1.4) by omitting the conditions c 6= 1 and aQ1(t) ≥ Q2(t), t ≥ t0, which were used by Kulenović andHadžiomerspahić [10]. In 2005, Yu and Wang [13] studied the existence of a nonoscillatory solution for the second ordernonlinear neutral delay differential equations with positive and negative coefficients[

r(t) (x(t)+ P(t)x(t − τ))′]′+ Q1(t)f (x(t − σ1))− Q2(t)g(x(t − σ2)) = 0, t ≥ t0, (1.5)

where τ > 0, σ1, σ2 ∈ [0,+∞), P,Q1,Q2, r ∈ C([t0,+∞),R), f , g ∈ C(R,R). However, the results in [4,10,13] only dealtwith the existence of a nonoscillatory solution of Eqs. (1.4) and (1.5), respectively, and did not suggest the iterative approx-imations of the nonoscillatory solution and the existence of infinitely many nonoscillatory solutions of Eqs. (1.4) and (1.5).Our aim in this paper is to investigate the following second order nonlinear neutral delay differential equation[

a(t) (x(t)+ b(t)x(t − τ))′]′+ [h(t, x(h1(t)), x(h2(t)), . . . , x(hk(t)))]′

+ f (t, x(f1(t)), x(f2(t)), . . . , x(fk(t))) = g(t), t ≥ t0, (1.6)

where τ > 0, a, b, g ∈ C([t0,+∞),R) with a(t) > 0 for t ≥ t0, h ∈ C1([t0,+∞) × Rk,R), f ∈ C([t0,+∞) × Rk,R),hl ∈ C1([t0,+∞),R) and fl ∈ C([t0,+∞),R)with

limt→+∞

hl(t) = limt→+∞

fl(t) = +∞, l = 1, . . . , k.

Utilizing the contraction mapping principle, we give several existence results of infinitely many nonoscillatory solutions,which are uncountable, for Eq. (1.6), construct a few Mann type iterative approximation schemes with errors for thesenonoscillatory solutions and discuss error estimates between the approximate solutions and the nonoscillatory solutions.These results presented in this paper extend, improve and unify all results due to Cheng and Annie [4], Kulenović andHadžiomerspahić [10] and Yu and Wang [13]. Five nontrivial examples are considered to illustrate our results.By a solution of Eq. (1.6), we mean a function x ∈ C([t1 − τ ,+∞),R) for some t1 ≥ t0, such that x(t) + b(t)x(t − τ)

and a(t) (x(t)+ b(t)x(t − τ))′ are continuously differentiable in [t1,+∞) and such that Eq. (1.6) is satisfied for t ≥ t1. Asis customary, a solution of Eq. (1.6) is said to be oscillatory if it has arbitrarily large zeros and nonoscillatory otherwise. It isassumed throughout this paper that:(H) For any given constants M and N with M > N > 0, there exist four functions PM,N ,QM,N , RM,N ,WM,N ∈

C([t0,+∞),R+) satisfying

|f (t, u1, u2, . . . , uk)− f (t, u1, u2, . . . , uk)| ≤ PM,N(t)max{|ul − ul| : 1 ≤ l ≤ k},|h(t, u1, u2, . . . , uk)− h(t, u1, u2, . . . , uk)| ≤ RM,N(t)max{|ul − ul| : 1 ≤ l ≤ k}for t ∈ [t0,+∞), ul, ul ∈ [N,M] and ≤ l ≤ k;

(1.7)

|f (t, u1, u2, . . . , uk)| ≤ QM,N(t) and |h(t, u1, u2, . . . , uk)| ≤ WM,N(t)

for t ∈ [t0,+∞), ul ∈ [N,M] and 1 ≤ l ≤ k; (1.8)∫+∞

t0A(s)max{PM,N(s),QM,N(s), |g(s)|}ds < +∞ (1.9)

and ∫+∞

t0max

{RM,N(s)a(s)

,WM,N(s)a(s)

}ds < +∞, (1.10)

where

A(t) =∫ t

t0

1a(s)ds, t ∈ [t0,+∞).

Note that the function A is positive and increasing in (t0,+∞). Let X denote the Banach space of all continuous and boundedfunctions on [t0,+∞)with norm ‖x‖ = supt≥t0 |x(t)|, and

X(N,M) = {x ∈ X : N ≤ x(t) ≤ M, t ≥ t0} forM > N > 0.

4276 Z. Liu, S.M. Kang / Nonlinear Analysis 70 (2009) 4274–4293

It is easy to see that X(N,M) is a bounded closed and convex subset of X . For each function b : [t0,+∞)→ R, put

b+∞= lim inft→+∞

b(t) and b+∞ = lim supt→+∞

b(t).

The following lemma plays an important role in this paper.

Lemma 1.1 ([11]). Let {αn}n≥0, {βn}n≥0, {γn}n≥0 and {tn}n≥0 be four nonnegative real sequences satisfying the inequality

αn+1 ≤ (1− tn)αn + tnβn + γn, ∀n ≥ 0,

where {tn}n≥0 ⊂ [0, 1],∑∞

n=0 tn = +∞, limn→∞ βn = 0 and∑∞

n=0 γn < +∞. Then limn→∞ αn = 0.

2. Existence of infinitely many nonoscillatory solutions

Now we study those conditions under which Eq. (1.6) possesses infinitely many nonoscillatory solutions. Moreover,we establish several Mann iterative approximation schemes with errors for these nonoscillatory solutions. Under certainconditions, several error estimates between the nonoscillatory solutions and the approximate solutions are discussed.

Theorem 2.1. Assume that (H) holds and there exists a constant b0 satisfying

|b(t)| ≤ b0 <12, eventually. (2.1)

Let M be an arbitrary positive constant and N = ( 12 − b0)M. Then:(a) For any L ∈ ( 12M, (1− b0)M), there exist θ ∈ (0, 1) and T > t0+ τ such that for each x0 ∈ X(N,M), the Mann iterative

sequence with errors {xm}m≥0 generated by the following scheme:

xm+1(t) =

(1− αm − βm)xm(t)+ αm

{L− b(t)xm(t − τ)

+

∫+∞

t

1a(s)h(s, xm(h1(s)), xm(h2(s)), . . . , xm(hk(s)))ds

+ A(t)∫+∞

tf (s, xm(f1(s)), xm(f2(s)), . . . , xm(fk(s)))ds

+

∫ t

TA(s)f (s, xm(f1(s)), xm(f2(s)), . . . , xm(fk(s)))ds

− A(t)∫+∞

tg(s)ds−

∫ t

TA(s)g(s)ds

}+ βmγm(t), t ≥ T ,m ≥ 0,

(1− αm − βm)xm(T )+ αm

{L− b(T )xm(T − τ)

+

∫+∞

T


+ A(T )∫+∞

Tf (s, xm(f1(s)), xm(f2(s)), . . . , xm(fk(s)))ds

− A(T )∫+∞

Tg(s)ds

}+ βmγm(T ), t0 ≤ t < T ,m ≥ 0

(2.2)

converges to a nonoscillatory solution x ∈ X(N,M) of Eq. (1.6) and has the following error estimate:

‖xm+1 − x‖ ≤ (1− (1− θ)αm)‖xm − x‖ + 2Mβm, m ≥ 0, (2.3)

where {γm}m≥0 is an arbitrary sequence in X(N,M), {αm}m≥0 and {βm}m≥0 are any sequences in [0, 1] such that

∞∑m=0

αm = +∞ (2.4)

and∞∑m=0

βm < +∞ or there exists a sequence {ξm}m≥0 ⊆ [0,+∞) satisfying

βm = ξmαm,m ≥ 0 and limm→∞

ξm = 0; (2.5)

(b) Eq. (1.6) has infinitely many nonoscillatory solutions, which are uncountable.


Proof. First of all we show that (a) holds. Let L ∈ ( 12M, (1 − b0)M). It follows from (1.9), (1.10) and (2.1) that there existθ ∈ (0, 1) and T > t0 + τ satisfying

|b(t)| ≤ b0, t ≥ T , (2.6)

θ = b0 +∫+∞

T

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds, (2.7)∫

+∞

T

WN,M(s)a(s)

ds+∫+∞

TA(s)

(QN,M(s)+ |g(s)|

)ds ≤ min

{(1− b0)M − L, L−

12M}. (2.8)

Define a mapping SL : X(N,M)→ X by

SLx(t) =

L− b(t)x(t − τ)+∫+∞

t

1a(s)h(s, x(h1(s)), x(h2(s)), . . . , x(hk(s)))ds

+ A(t)∫+∞

tf (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))ds+

∫ t

TA(s)f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))ds

− A(t)∫+∞

tg(s)ds−

∫ t

TA(s)g(s)ds, t ≥ T , x ∈ X(N,M),

SLx(T ), t0 ≤ t < T , x ∈ X(N,M).

(2.9)

Clearly SLx is continuous for each x ∈ X(N,M). In view of (1.7), (2.6), (2.7) and (2.9), we conclude that for x, y ∈ X(N,M)and t ≥ T

|SLx(t)− SLy(t)| ≤ |b(t)||x(t − τ)− y(t − τ)| +∫+∞

t

1a(s)|h(s, x(h1(s)), x(h2(s)), . . . , x(hk(s)))

− h(s, y(h1(s)), y(h2(s)), . . . , y(hk(s)))|ds+ A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))

− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds+∫ t

TA(s)|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))

− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds

≤ b0‖x− y‖ + ‖x− y‖∫+∞

t

RN,M(s)a(s)

ds

+‖x− y‖A(t)∫+∞

tPN,M(s)ds+ ‖x− y‖

∫ t

TA(s)PN,M(s)ds

≤

(b0 +

∫+∞

t

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

)‖x− y‖

≤ θ‖x− y‖,

which gives that

‖SLx− SLy‖ ≤ θ‖x− y‖, x, y ∈ X(N,M). (2.10)

In light of (1.8), (2.6), (2.8) and (2.9), we get that for any x ∈ X(N,M) and t ≥ T

SLx(t) ≤ L+ |b(t)|x(t − τ)+∫+∞

t

1a(s)|h(s, x(h1(s)), x(h2(s)), . . . , x(hk(s)))|ds

+ A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds

+

∫ t

TA(s)|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds+ A(t)

∫+∞

t|g(s)|ds+

∫ t

TA(s)|g(s)|ds

≤ L+ b0M +∫+∞

t

WN,M(s)a(s)

ds+ A(t)∫+∞

tQN,M(s)ds

+

∫ t

TA(s)QN,M(s)ds+

∫+∞

tA(s)|g(s)|ds+

∫ t

TA(s)|g(s)|ds

≤ L+ b0M +∫+∞

t

WN,M(s)a(s)

ds+∫+∞

TA(s)QN,M(s)ds+

∫+∞

TA(s)|g(s)|ds


≤ L+ b0M +min{(1− b0)M − L, L−

12M}

≤ L+ b0M + (1− b0)M − L= M

and

SLx(t) ≥ L− |b(t)|x(t − τ)−∫+∞

t


− A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds−

∫ t

TA(s)|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds

− A(t)∫+∞

t|g(s)|ds−

∫ t

TA(s)|g(s)|ds

≥ L− b0M −∫+∞

t

WN,M(s)a(s)

ds− A(t)∫+∞

tQN,M(s)ds

−

∫ t

TA(s)QN,M(s)ds−

∫+∞

tA(s)|g(s)|ds−

∫ t

TA(s)|g(s)|ds

≥ L− b0M −∫+∞

t

WN,M(s)a(s)

ds−∫+∞

tA(s)QN,M(s)ds−

∫+∞

TA(s)|g(s)|ds

≥ L− b0M −min{(1− b0)M − L, L−

12M}

≥ L− b0M − L+12M

= N,

which imply that SL(X(N,M)) ⊂ X(N,M). It follows from (2.10) that SL is a contraction mapping and it has a unique fixedpoint x ∈ X(N,M), which is a nonoscillatory solution of Eq. (1.6). By means of (2.2), (2.9) and (2.10), we know that for anym ≥ 0 and t ≥ T ,

|xm+1(t)− x(t)| =∣∣∣∣(1− αm − βm)xm(t)+ αm {L− b(t)xm(t − τ)+

∫+∞

t


+ A(t)∫+∞

tf (s, xm(f1(s)), xm(f2(s)), . . . , xm(fk(s)))ds

+

∫ t


− A(t)∫+∞

tg(s)ds−

∫ t

TA(s)g(s)ds

}+ βmγm(t)− x(t)

∣∣∣∣≤ (1− αm − βm)|xm(t)− x(t)| + αm|SLxm(t)− SLx(t)| + βm|γm(t)− x(t)|≤ (1− αm − βm)|xm(t)− x(t)| + αmθ |xm(t)− x(t)| + 2Mβm≤ (1− (1− θ)αm)|xm(t)− x(t)| + 2Mβm,

which yields that

‖xm+1 − x‖ ≤ (1− (1− θ)αm)‖xm − x‖ + 2Mβm, m ≥ 0.

That is, (2.3) holds. Thus Lemma 1.1 and (2.3)–(2.5) ensure that limm→∞ xm = x.Now we show that (b) holds. By (1.9), (1.10) and (2.1), we know that for each L, K ∈ ( 12M, (1− b0)M)with L 6= K , there

exists T > t0 + τ satisfying (2.6) and (2.8) and∫+∞

T

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds < min

{|L− K |2M

,1− b02

}. (2.11)

Let the mapping SL be defined by (2.9), and define a mapping SK : X(N,M) → X by (2.9) with L replaced by K . As in theproof of (a), we conclude easily that the mappings SL and SK have the unique fixed points x, y ∈ X(N,M), respectively. That


is, x and y are solutions of Eq. (1.6) in X(N,M). In order to prove (b), we need only to show that x 6= y. In fact, (2.9) meansthat for t ≥ T

x(t) = SLx(t)

= L− b(t)x(t − τ)+∫+∞

t


+ A(t)∫+∞

tf (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))ds+

∫ t


− A(t)∫+∞

tg(s)ds−

∫ t

TA(s)g(s)ds (2.12)

and

y(t) = SKy(t)

= K − b(t)y(t − τ)+∫+∞

t

1a(s)h(s, y(h1(s)), y(h2(s)), . . . , y(hk(s)))ds

+ A(t)∫+∞

tf (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))ds

+

∫ t

TA(s)f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))ds− A(t)

∫+∞

tg(s)ds−

∫ t

TA(s)g(s)ds. (2.13)

It follows from (1.7) and (2.11)–(2.13) that for t ≥ T

|x(t)+ b(t)x(t − τ)− y(t)− b(t)y(t − τ)| =∣∣∣∣L− K + ∫ +∞

t

1a(s)

[h(s, x(h1(s)), x(h2(s)), . . . , x(hk(s)))

− h(s, y(h1(s)), y(h2(s)), . . . , y(hk(s)))] ds+ A(t)∫+∞

t[f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))

− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))] ds+∫ t

TA(s) [f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))

− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))] ds∣∣∣∣

≥ |L− K | −∫+∞

t

1a(s)|h(s, x(h1(s)), x(h2(s)), . . . , x(hk(s)))− h(s, y(h1(s)), y(h2(s)), . . . , y(hk(s)))|ds

− A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds

−

∫ t

TA(s)|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds

≥ |L− K | − ‖x− y‖(∫

+∞

t

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

)≥ |L− K | − 2M

(∫+∞

t

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

)> 0,

which implies that

x(t)− y(t)+ b(t)(x(t − τ)− y(t − τ)) 6= 0, t ≥ T ,

that is,

x(t) 6= y(t) or x(t − τ) 6= y(t − τ), t ≥ T ,

which yields that x 6= y. This completes the proof. �

Theorem 2.2. Assume that (H) holds and

b(t) ≥ 0, eventually, and b+∞ < 1. (2.14)


Let M be an arbitrary positive constant and N = 1−b+∞4 M. Then:

(a) For any L ∈ ( 3+b+∞4 M,M), there exist θ ∈ (0, 1) and T > t0 + τ such that for any x0 ∈ X(N,M), the Mann iterativesequence with errors {xm}m≥0 generated by (2.2) converges to a nonoscillatory solution x ∈ X(N,M) of Eq. (1.6) and has the errorestimate (2.3), where {γm}m≥0 is an arbitrary sequence in X(N,M), {αm}m≥0 and {βm}m≥0 are any sequences in [0, 1] satisfying(2.4) and (2.5);(b) Eq. (1.6) has infinitely many nonoscillatory solutions, which are uncountable.

Proof. Let L ∈ ( 3+b+∞4 M,M). (1.9), (1.10) and (2.14) guarantee that there exist θ ∈ (0, 1) and T > t0 + τ satisfying

0 ≤ b(t) ≤1+ b+∞2

, t ≥ T , (2.15)

θ =1+ b+∞2

+

∫+∞

T

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds, (2.16)

∫+∞

T

WN,M(s)a(s)

ds+∫+∞

TA(s)

(QN,M(s)+ |g(s)|

)ds ≤ min

{M − L, L+

1+ b+∞2

M

}. (2.17)

Define a mapping SL : X(N,M)→ X by (2.9). Clearly SLx is continuous for each x ∈ X(N,M). Using (1.7), (2.15) and (2.16),we deduce that for x, y ∈ X(N,M) and t ≥ T

|SLx(t)− SLy(t)| ≤ b(t)|x(t − τ)− y(t − τ)|

+

∫+∞

t


+ A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds

+

∫ t


≤1+ b+∞2‖x− y‖ + ‖x− y‖

∫+∞

t

RN,M(s)a(s)

ds

+‖x− y‖A(t)∫+∞


∫ t

TA(s)PN,M(s)ds

≤

(1+ b+∞2

+

∫+∞

t

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

)‖x− y‖

≤ θ‖x− y‖,

which means that (2.10) holds. In terms of (1.8), (2.9), (2.15) and (2.17), we know that for any x ∈ X(N,M) and t ≥ T

SLx(t) ≤ L+∫+∞

t


+ A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds+

∫ t


+ A(t)∫+∞

t|g(s)|ds+

∫ t

TA(s)|g(s)|ds

≤ L+∫+∞

t

WN,M(s)a(s)

ds+ A(t)∫+∞

tQN,M(s)ds+

∫ t

TA(s)QN,M(s)ds+ A(t)

∫+∞

t|g(s)|ds+

∫ t

TA(s)|g(s)|ds

≤ L+∫+∞

t

WN,M(s)a(s)

ds+∫+∞

TA(s)QN,M(s)ds+

∫+∞

TA(s)|g(s)|ds

≤ L+min

{M − L, L−

3+ b+∞4

M

}≤ M

and

SLx(t) ≥ L− |b(t)|x(t − τ)−∫+∞

t



− A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds−

∫ t


− A(t)∫+∞

t|g(s)|ds−

∫ t

TA(s)|g(s)|ds

≥ L−1+ b+∞2

M −∫+∞

t

WN,M(s)a(s)

ds− A(t)∫+∞

tQN,M(s)ds

−

∫ t

TA(s)QN,M(s)ds−

∫+∞

tA(s)|g(s)|ds−

∫ t

TA(s)|g(s)|ds

≥ L−1+ b+∞2

M −∫+∞

t

WN,M(s)a(s)

ds−∫+∞

TA(s)QN,M(s)ds−

∫+∞

TA(s)|g(s)|ds

≥ L−1+ b+∞2

M −min

{M − L, L−

3+ b+∞4

M

}

≥ L−1+ b+∞2

M − L+3+ b+∞4

M

= N,

which yields that SL(X(N,M)) ⊂ X(N,M). (2.10) guarantees that SL is a contraction mapping and therefore it possessesa unique fixed point x ∈ X(N,M), which is a nonoscillatory solution of Eq. (1.6). The rest of the proof is similar to that ofTheorem 2.1, and is omitted. This completes the proof. �


b(t) ≤ 0, eventually, and −1 < b+∞

< 0. (2.18)

Let M be an arbitrary positive constant and N = 1+b+∞4 M. Then:

(a) For each L ∈ ( 1+b+∞4 M, 1+b+∞2 M), there exist θ ∈ (0, 1) and T > t0 + τ such that for any x0 ∈ X(N,M), the Manniterative sequence with errors {xm}m≥0 generated by (2.2) converges to a nonoscillatory solution x ∈ X(N,M) of Eq. (1.6) and hasthe error estimate (2.3), where {γm}m≥0 is an arbitrary sequence in X(N,M), {αm}m≥0 and {βm}m≥0 are any sequences in [0, 1]satisfying (2.4) and (2.5);(b) Eq. (1.6) has infinitely many nonoscillatory solutions, which are uncountable.

Proof. Let L ∈ ( 1+b+∞4 M, 1+b+∞2 M). It follows from (1.9), (1.10) and (2.18) that there exist θ ∈ (0, 1) and T > t0 + τsatisfying

−1 <b+∞− 12

≤ b(t) ≤ 0, t ≥ T , (2.19)

θ =1− b

+∞

2+

∫+∞

T

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds, (2.20)∫

+∞

T

WN,M(s)a(s)

ds+∫+∞

TA(s)

(QN,M(s)+ |g(s)|

)ds ≤ min

{L−

1+ b+∞

4M,1+ b

+∞

2M − L

}. (2.21)

Define a mapping SL : X(N,M) → X by (2.9). Obviously, SLx is continuous for each x ∈ X(N,M). In view of (1.7), (2.9),(2.19) and (2.20), we infer that for x, y ∈ X(N,M) and t ≥ T

|SLx(t)− SLy(t)| ≤ |b(t)||x(t − τ)− y(t − τ)|

+

∫+∞

t


+ A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))

− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds+∫ t


− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds

≤1− b

+∞

2‖x− y‖ + ‖x− y‖

∫+∞

t

RN,M(s)a(s)

ds


+‖x− y‖A(t)∫+∞


∫ t

TA(s)PN,M(s)ds

≤

(1− b

+∞

2+

∫+∞

t

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

)‖x− y‖

≤ θ‖x− y‖,

which yields that (2.10) holds. In view of (1.8), (2.9), (2.19) and (2.21), we deduce that for any x ∈ X(N,M) and t ≥ T

SLx(t) ≤ L+ |b(t)|x(t − τ)+∫+∞

t


+ A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds+

∫ t


+ A(t)∫+∞

t|g(s)|ds+

∫ t

TA(s)|g(s)|ds

≤ L+1− b

+∞

2M +

∫+∞

t

WN,M(s)a(s)

ds+ A(t)∫+∞

tQN,M(s)ds

+

∫ t

TA(s)QN,M(s)ds+ A(t)

∫+∞

t|g(s)|ds+

∫ t

TA(s)|g(s)|ds

≤ L+1− b

+∞

2M +

∫+∞

t

WN,M(s)a(s)

ds+∫+∞

TA(s)QN,M(s)ds+

∫+∞

TA(s)|g(s)|ds

≤ L+1− b

+∞

2M +min

{L−

1+ b+∞

4M,1+ b

+∞

2M − L

}≤ L+

1− b+∞

2M +

1+ b+∞

2M − L

= M

and

SLx(t) ≥ L−∫+∞

t


− A(t)∫+∞

t|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds

−

∫ t

TA(s)|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds− A(t)

∫+∞

t|g(s)|ds−

∫ t

TA(s)|g(s)|ds

≥ L−∫+∞

t

WN,M(s)a(s)

ds− A(t)∫+∞

tQN,M(s)ds−

∫ t

TA(s)QN,M(s)ds− A(t)

∫+∞

t|g(s)|ds−

∫ t

TA(s)|g(s)|ds

≥ L−∫+∞

t

WN,M(s)a(s)

ds−∫+∞

TA(s)QN,M(s)ds−

∫+∞

TA(s)|g(s)|ds

≥ L−min{L−

1+ b+∞

4M,1+ b

+∞

2M − L

}≥ L− L+

1+ b+∞

4M

= N,

which imply that SL(X(N,M)) ⊂ X(N,M). It follows from (2.10) that SL is a contraction mapping and therefore it has aunique fixed point x ∈ X(N,M), which is a nonoscillatory solution of Eq. (1.6). The rest of the proof of Theorem 2.3 is exactlythe same as that of Theorem 2.1, and hence is omitted. This completes the proof. �


b(t) > 1, eventually, and 1 < b+∞≤ b+∞ < b2+∞ < +∞. (2.22)

Let M be an arbitrary positive constant and

N =M(b+∞ + δ2)[(b+∞ − δ)

2− b+∞ − δ2]

2(b+∞− δ)[(b+∞ + δ2)2 − b+∞ + δ]

,


where

δ = min

{1,b+∞− 12

,b2+∞− b+∞4b+∞

}.

Then:(a) For any L ∈

((b+∞ + δ2)

(M

b+∞−δ+ N

),(M + N

b+∞+δ2

)(b+∞− δ)

), there exist θ ∈ (0, 1) and T > t0 + τ such that

for each x0 ∈ X(N,M), the Mann iterative sequence with errors {xm}m≥0 generated by the following scheme:

xm+1(t) =


{L

b(t + τ)−xm(t + τ)b(t + τ)

+1

b(t + τ)

∫+∞

t+τ


+A(t + τ)b(t + τ)

∫+∞

t+τf (s, xm(f1(s)), xm(f2(s)), . . . , xm(fk(s)))ds

+1

b(t + τ)

∫ t+τ


−A(t + τ)b(t + τ)

∫+∞

t+τg(s)ds−

1b(t + τ)

∫ t+τ

TA(s)g(s)ds

}+ βmγm(t), t ≥ T ,m ≥ 0,


{L

b(T + τ)−xm(T + τ)b(T + τ)

+1

b(T + τ)

∫+∞

T+τ


+A(T + τ)b(T + τ)

∫+∞

T+τf (s, xm(f1(s)), xm(f2(s)), . . . , xm(fk(s)))ds

+1

b(T + τ)

∫ T+τ


−A(T + τ)b(T + τ)

∫+∞

T+τg(s)ds−

1b(t + τ)

∫ T+τ

TA(s)g(s)ds

}+ βmγm(T ), t0 ≤ t < T ,m ≥ 0

(2.23)

converges to a nonoscillatory solution x ∈ X(N,M) of Eq. (1.6) and has the error estimate (2.3), where {γm}m≥0 is an arbitrarysequence in X(N,M), {αm}m≥0 and {βm}m≥0 are any sequences in [0, 1] satisfying (2.4) and (2.5);(b) Eq. (1.6) has infinitely many nonoscillatory solutions, which are uncountable.

Proof. Let L ∈((b+∞ + δ2)

(M

b+∞−δ+ N

),(M + N

b+∞+δ2

)(b+∞− δ)

). It follows from (1.9), (1.10) and (2.22) that there

exist θ ∈ (0, 1) and T > t0 + τ satisfying

1 < b+∞− δ ≤ b(t) ≤ b+∞ + δ2 < (b

+∞− δ)2, t ≥ T , (2.24)

θ =1

b+∞− δ

(1+

∫+∞

T

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

), (2.25)∫

+∞

T

WN,M(s)a(s)

ds+∫+∞

TA(s)

(QN,M(s)+ |g(s)|

)ds

≤ min{(M +

N

b+∞ + δ2

)(b+∞− δ)− L, (b

+∞− δ)

(L

b+∞ + δ2− N

)−M

}. (2.26)


SLx(t) =

Lb(t + τ)

−x(t + τ)b(t + τ)

+1

b(t + τ)

∫+∞

t+τ


+A(t + τ)b(t + τ)

∫+∞

t+τf (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))ds

+1

b(t + τ)

∫ t+τ

TA(s)f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))ds−

A(t + τ)b(t + τ)

∫+∞

t+τg(s)ds

−1

b(t + τ)

∫ t+τ


SLx(T ), t0 ≤ t < T , x ∈ X(N,M).

(2.27)


Clearly SLx is continuous for each x ∈ X(N,M). By virtue of (1.7), (2.24), (2.26) and (2.27), we infer that for x, y ∈ X(N,M)and t ≥ T

|SLx(t)− SLy(t)| ≤1

b(t + τ)|x(t + τ)− y(t + τ)|

+1

b(t + τ)

∫+∞

t+τ

1a(s)|h(s, x(h1(s)), x(h2(s)), . . . , x(hk(s)))

− h(s, y(h1(s)), y(h2(s)), . . . , y(hk(s)))|ds

+A(t + τ)b(t + τ)

∫+∞

t+τ|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))

− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds+1

b(t + τ)

×

∫ t+τ


≤‖x− y‖b+∞− δ

(1+

∫+∞

t+τ

RN,M(s)a(s)

ds+ A(t + τ)∫+∞

t+τPN,M(s)ds+

∫ t+τ

TA(s)PN,M(s)ds

)≤‖x− y‖b+∞− δ

(1+

∫+∞

t+τ

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

)≤ θ‖x− y‖,

that is, (2.10) holds. It follows from (1.8), (2.24) and (2.26) that for any x ∈ X(N,M) and t ≥ T

SLx(t) ≤L

b(t + τ)−x(t + τ)b(t + τ)

+1

b(t + τ)

∫+∞

t+τ


+A(t + τ)b(t + τ)

∫+∞

t+τ|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))|ds

+1

b(t + τ)

∫ t+τ


+A(t + τ)b(t + τ)

∫+∞

t+τ|g(s)|ds+

1b(t + τ)

∫ t+τ

TA(s)|g(s)|ds

≤L

b+∞− δ−

N

b+∞ + δ2+

1b+∞− δ

(∫+∞

t+τ

WN,M(s)a(s)

ds+ A(t + τ)∫+∞

t+τQN,M(s)ds

+

∫ t+τ

TA(s)QN,M(s)ds+ A(t + τ)

∫+∞

t+τ|g(s)|ds+

∫ t+τ

TA(s)|g(s)|ds

)≤

Lb+∞− δ−

N

b+∞ + δ2+

1b+∞− δ

(∫+∞

t+τ

WN,M(s)a(s)

ds+∫+∞

TA(s)QN,M(s)ds+

∫+∞

TA(s)|g(s)|ds

)≤

Lb+∞− δ−

N

b+∞ + δ2+

1b+∞− δ

min{(M +

N

b+∞ + δ2

)(b+∞− δ)− L,

(b+∞− δ)

(L

b+∞ + δ2− N

)−M

}≤

Lb+∞− δ−

N

b+∞ + δ2+

1b+∞− δ

[(M +

N

b+∞ + δ2

)(b+∞− δ)− L

]= M

and

SLx(t) ≥L

b(t + τ)−x(t + τ)b(t + τ)

−1

b(t + τ)

∫+∞

t+τ


−A(t + τ)b(t + τ)

∫+∞


−1

b(t + τ)

∫ t+τ



−A(t + τ)b(t + τ)

∫+∞

t+τ|g(s)|ds−

1b(t + τ)

∫ t+τ

TA(s)|g(s)|ds

≥L

b+∞ + δ2−

Mb+∞− δ−

1b+∞− δ

(∫+∞

t+τ

WN,M(s)a(s)

ds

+ A(t + τ)∫+∞

t+τQN,M(s)ds+

∫ t+τ


∫+∞

t+τ|g(s)|ds+

∫ t+τ

TA(s)|g(s)|ds

)≥

L

b+∞ + δ2−

Mb+∞− δ−

1b+∞− δ

(∫+∞

t+τ

WN,M(s)a(s)

ds+∫+∞

TA(s)QN,M(s)ds+

∫+∞

TA(s)|g(s)|ds

)≥

L

b+∞ + δ2−

Mb+∞− δ−

1b+∞− δ

min{(M +

N

b+∞ + δ2

)(b+∞− δ)− L,

(b+∞− δ)

(L

b+∞ + δ2− N

)−M

}≥

L

b+∞ + δ2−

Mb+∞− δ−

1b+∞− δ

[(b+∞− δ)

(L

b+∞ + δ2− N

)−M

]= N,

which imply that SL(X(N,M)) ⊂ X(N,M). (2.10) ensures that SL is a contraction mapping and hence it has a unique fixedpoint x ∈ X(N,M), which is a nonoscillatory solution of Eq. (1.6). Notice that (2.10), (2.23) and (2.27) guarantee that for anym ≥ 0 and t ≥ T

|xm+1(t)− x(t)| =∣∣∣∣(1− αm − βm)xm(t)+ αm { L

b(t + τ)−xm(t + τ)b(t + τ)

+1

b(t + τ)

∫+∞

t+τ


+A(t + τ)b(t + τ)

∫+∞


+1

b(t + τ)

∫ t+τ


−A(t + τ)b(t + τ)

∫+∞

t+τg(s)ds−

1b(t + τ)

∫ t+τ

TA(s)g(s)ds

}+ βmγm(t)− x(t)

∣∣∣∣≤ (1− αm − βm)|xm(t)− x(t)| + αm|SLxm(t)− SLx(t)| + βm|γm(t)− x(t)|≤ (1− αm − βm)|xm(t)− x(t)| + αmθ |xm(t)− x(t)| + 2Mβm≤ (1− (1− θ)αm)|xm(t)− x(t)| + 2Mβm,

which gives (2.3). Thus Lemma 1.1 and (2.3)–(2.5) ensure that limm→∞ xm = x.Next we show that (b) holds. It follows from (1.9), (1.10) and (2.22) that for any different constants p, q ∈(

(b+∞ + δ2)(

Mb+∞−δ

+ N),(M + N

b+∞+δ2

)(b+∞− δ)

), there exists T > t0 + τ satisfying (2.24) and (2.26) and

∫+∞

T

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds < min

{|p− q|(b

+∞− δ)

2M(b+∞ + δ2),b+∞− δ − 12

}. (2.28)

Let the mappings Sp and Sq be defined by (2.8), but with p, q instead of L, respectively. It follows from (a) that Eq. (1.6) hassolutions x, y ∈ X(N,M) satisfying

x(t) = Spx(t)

=p

b(t + τ)−x(t + τ)b(t + τ)

+1

b(t + τ)

∫+∞

t+τ


+A(t + τ)b(t + τ)

∫+∞


+1

b(t + τ)

∫ t+τ



−A(t + τ)b(t + τ)

∫+∞

t+τg(s)ds−

1b(t + τ)

∫ t+τ

TA(s)g(s)ds, t ≥ T (2.29)

and

y(t) = Sqy(t)

=q

b(t + τ)−y(t + τ)b(t + τ)

+1

b(t + τ)

∫+∞

t+τ

1a(s)h(s, y(h1(s)), y(h2(s)), . . . , y(hk(s)))ds

+A(t + τ)b(t + τ)

∫+∞

t+τf (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))ds

+1

b(t + τ)

∫ t+τ

TA(s)f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))ds

−A(t + τ)b(t + τ)

∫+∞

t+τg(s)ds−

1b(t + τ)

∫ t+τ

TA(s)g(s)ds, t ≥ T . (2.30)

Next we assert that x 6= y. Note that (2.28)–(2.30) lead to∣∣∣∣x(t)+ x(t + τ)b(t + τ)− y(t)−

y(t + τ)b(t + τ)

∣∣∣∣ = ∣∣∣∣ pb(t + τ)

−q

b(t + τ)+

1b(t + τ)

×

∫+∞

t+τ

1a(s)

[h(s, x(h1(s)), x(h2(s)), . . . , x(hk(s)))ds− h(s, y(h1(s)), y(h2(s)), . . . , y(hk(s)))] ds

+A(t + τ)b(t + τ)

∫+∞

t+τ[f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))ds− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))] ds

+1

b(t + τ)

∫ t+τ

TA(s) [f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))ds− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))] ds

∣∣∣∣≥|p− q|

b+∞ + δ2−

1b+∞− δ

[∫+∞

t+τ

1a(s)|h(s, x(h1(s)), x(h2(s)), . . . , x(hk(s)))

− h(s, y(h1(s)), y(h2(s)), . . . , y(hk(s)))|ds+ A(t)∫+∞

t+τ|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))

− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))| ds

+

∫ t+τ


]≥|p− q|

b+∞ + δ2−‖x− y‖b+∞− δ

(∫+∞

t+τ

RN,M(s)a(s)

ds+ A(t + τ)∫+∞

t+τPN,M(s)ds+

∫ t+τ

TA(s)PN,M(s)ds

)≥|p− q|

b+∞ + δ2−

2Mb+∞− δ

(∫+∞

t+τ

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

)> 0, t ≥ T ,

which yields that

x(t)− y(t)+x(t + τ)b(t + τ)

−y(t + τ)b(t + τ)

6= 0, t ≥ T ,

that is,

x(t) 6= y(t) or x(t + τ) 6= y(t + τ), t ≥ T .

Therefore, x 6= y. This completes the proof. �


b(t) < −1, eventually, and −∞ < b+∞≤ b+∞ < −1. (2.31)

Let M be an arbitrary positive constant and N = M(1+b+∞)2(3+2b+∞+b+∞)

. Then:

(a) For any L ∈(−14M(1+ b+∞),−

12M(1+ b+∞)

), there exist θ ∈ (0, 1) and T > t0+τ such that for each x0 ∈ X(N,M),

the Mann iterative sequence with errors {xm}m≥0 generated by the following scheme:


xm+1(t) =


{−L

b(t + τ)−xm(t + τ)b(t + τ)

+1

b(t + τ)

∫+∞

t+τ


+A(t + τ)b(t + τ)

∫+∞


+1

b(t + τ)

∫ t+τ


−A(t + τ)b(t + τ)

∫+∞

t+τg(s)ds−

1b(t + τ)

∫ t+τ

TA(s)g(s)ds

}+ βmγm(t), t ≥ T ,m ≥ 0,


{−L

b(T + τ)−xm(T + τ)b(T + τ)

+1

b(T + τ)

∫+∞

T+τ


+A(T + τ)b(T + τ)

∫+∞

T+τf (s, xm(f1(s)), xm(f2(s)), . . . , xm(fk(s)))ds

+1

b(T + τ)

∫ T+τ


−A(T + τ)b(T + τ)

∫+∞

T+τg(s)ds−

1b(t + τ)

∫ T+τ

TA(s)g(s)ds

}+ βmγm(T ), t0 ≤ t < T ,m ≥ 0

(2.32)

converges to a nonoscillatory solution x ∈ X(N,M) of Eq. (1.6) and has the error estimate (2.3), where {γm}m≥0 is an arbitrarysequence in X(N,M), {αm}m≥0 and {βm}m≥0 are any sequences in [0, 1] satisfying (2.4) and (2.5);(b) Eq. (1.6) has infinitely many nonoscillatory solutions, which are uncountable.

Proof. Let L ∈(−14M(1+ b+∞),−

12M(1+ b+∞)

). It follows from (1.9), (1.10) and (2.31) that there exist θ ∈ (0, 1) and

T > t0 + τ satisfying

12(1+ 2b

+∞+ b+∞) < b(t) <

12(b+∞ − 1) < −1, t ≥ T , (2.33)

θ =2

1− b+∞

(1+

∫+∞

T

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

), (2.34)∫

+∞

T

WN,M(s)a(s)

ds+∫+∞

TA(s)

(QN,M(s)+ |g(s)|

)ds

≤ min

{−L−

M(1+ b+∞)2

,b+∞ − 1

1+ 2b+∞+ b+∞

[L+

14M(1+ b+∞)

]}. (2.35)


SLx(t) =

−Lb(t + τ)

−x(t + τ)b(t + τ)

+1

b(t + τ)

∫+∞

t+τ


+A(t + τ)b(t + τ)

∫+∞


+1

b(t + τ)

∫ t+τ


−A(t + τ)b(t + τ)

∫+∞

t+τg(s)ds−

1b(t + τ)

∫ t+τ


SLx(T ), t0 ≤ t < T , x ∈ X(N,M).

(2.36)

It is clear that SLx is continuous for each x ∈ X(N,M). By virtue of (1.7), (2.33), (2.34) and (2.36), we conclude immediatelythat for x, y ∈ X(N,M) and t ≥ T

|SLx(t)− SLy(t)| ≤1

|b(t + τ)||x(t + τ)− y(t + τ)| +

1|b(t + τ)|

∫+∞

t+τ

1a(s)|h(s, x(h1(s)), x(h2(s)), . . . , x(hk(s)))

− h(s, y(h1(s)), y(h2(s)), . . . , y(hk(s)))|ds

+A(t + τ)|b(t + τ)|

∫+∞

t+τ|f (s, x(f1(s)), x(f2(s)), . . . , x(fk(s)))− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds


+1

|b(t + τ)|

∫ t+τ


− f (s, y(f1(s)), y(f2(s)), . . . , y(fk(s)))|ds

≤2‖x− y‖

1− b+∞

(1+

∫+∞

t+τ

RN,M(s)a(s)

ds+ A(t + τ)∫+∞

t+τPN,M(s)ds+

∫ t+τ

TA(s)PN,M(s)ds

)≤2‖x− y‖

1− b+∞

(1+

∫+∞

t+τ

RN,M(s)a(s)

ds+∫+∞

TA(s)PN,M(s)ds

)≤ θ‖x− y‖,

which yields (2.10). Using (1.8), (2.35) and (2.36), we conclude that for any x ∈ X(N,M) and t ≥ T

SLx(t) ≤L

|b(t + τ)|+x(t + τ)|b(t + τ)|

+1

|b(t + τ)|

∫+∞

t+τ


+A(t + τ)|b(t + τ)|

∫+∞


+1

|b(t + τ)|

∫ t+τ


+A(t + τ)|b(t + τ)|

∫+∞

t+τ|g(s)|ds+

1|b(t + τ)|

∫ t+τ

TA(s)|g(s)|ds

≤2L

1− b+∞+

2M

1− b+∞+

2

1− b+∞

(∫+∞

t+τ

WN,M(s)a(s)

ds

+ A(t + τ)∫+∞

t+τQN,M(s)ds+

∫ t+τ


∫+∞

t+τ|g(s)|ds+

∫ t+τ

TA(s)|g(s)|ds

)≤2L+ 2M

1− b+∞+

2

1− b+∞

(∫+∞

t+τ

WN,M(s)a(s)

ds+∫+∞

TA(s)QN,M(s)ds+

∫+∞

TA(s)|g(s)|ds

)≤2L+ 2M

1− b+∞+

2

1− b+∞min

{−L−

M(1+ b+∞)2

,b+∞ − 1

1+ 2b+∞+ b+∞

(L+

M(1+ b+∞)4

)}

≤2L+ 2M

1− b+∞+

2

1− b+∞

(−L−

M(1+ b+∞)2

)= M

and

SLx(t) ≥L

|b(t + τ)|−x(t + τ)|b(t + τ)|

−1

|b(t + τ)|

∫+∞

t+τ


−A(t + τ)|b(t + τ)|

∫+∞


−1

|b(t + τ)|

∫ t+τ


−A(t + τ)|b(t + τ)|

∫+∞

t+τ|g(s)|ds−

1|b(t + τ)|

∫ t+τ

TA(s)|g(s)|ds

≥−2L

1+ 2b+∞+ b+∞

−2N

1+ 2b+∞+ b+∞

+2

b+∞ − 1

(∫+∞

t+τ

WN,M(s)a(s)

ds

+ A(t + τ)∫+∞

t+τQN,M(s)ds+

∫ t+τ


∫+∞

t+τ|g(s)|ds+

∫ t+τ

TA(s)|g(s)|ds

)≥

−2L− 2N

1+ 2b+∞+ b+∞

+2

b+∞ − 1

(∫+∞

t+τ

WN,M(s)a(s)

ds+∫+∞

TA(s)QN,M(s)ds+

∫+∞

TA(s)|g(s)|ds

)

≥−2L− 2N

1+ 2b+∞+ b+∞

+2

b+∞ − 1min

{−L−

M(1+ b+∞)2

,b+∞ − 1

1+ 2b+∞+ b+∞

(L+

M(1+ b+∞)4

)}


≥−2L− 2N

1+ 2b+∞+ b+∞

+2

b+∞ − 1

[b+∞ − 1

1+ 2b+∞+ b+∞

(L+

M(1+ b+∞)4

)]

≥

12M(1+ b+∞)− 2N

1+ 2b+∞+ b

≥ N,

which imply that SL(X(N,M)) ⊂ X(N,M). (2.10) ensures that SL is a contraction mapping and therefore it has a uniquefixed point x ∈ X(N,M), which is a nonoscillatory solution of Eq. (1.6). The remainder of the proof is completely the sameas that of Theorem 2.4, and hence is omitted. This completes the proof. �

Remark 2.1. Theorems 2.1–2.5 extend, improve and unify all results [4,10,13].

3. Examples

Finally we give five examples of applications of the results presented in Section 2.

Example 3.1. Consider the following second order linear neutral delay differential equation with positive and negativecoefficients

d2

dt2

[x(t)+

13x(t − τ)

]+

x(t − σ1)(1+ t)(1+ t3)

−x(t − σ2)1+ t3

= 0, t ≥ t0, (3.1)

where τ > 0, σ1 ≥ 0, σ2 ≥ 0 and t0 > 0 are constants. On the one hand, because condition (4) in [10]

0 ≤ aQ1(t)− Q2(t) =a

(1+ t)(1+ t3)−

11+ t3

=a− 1− t1+ t3

for every t ≥ T1 and a > 0

does not hold, where T1 is large enough, thus the theorem in [10] is inapplicable for Eq. (3.1). On the other hand, let k = 2and

a(t) = 1, A(t) = t − t0, b(t) =13, g(t) = h(t, u, v) = 0,

f (t, u, v) =u

(1+ t)(1+ t3)−

v

1+ t3, (t, u, v) ∈ [t0,+∞)× R2.

Clearly, b0 = 13 . For any givenM > N > 0, take

PM,N(t) =2+ t

(1+ t)(1+ t3), QM,N(t) =

M(2+ t)(1+ t)(1+ t3)

, t ∈ [t0,+∞).

It is easy to verify that the conditions (H) and (2.1) are satisfied. LetM > 0 and N = 16M . It follows from Theorem 2.1 that

Eq. (3.1) possesses infinitelymany nonoscillatory solutions in X(N,M), and for any L ∈ ( 12M,23M), there exist θ ∈ (0, 1) and

T > t0+τ such that theMann iterative sequencewith error {xm}m≥0 generated by (2.2) converges to a nonoscillatory solutionx ∈ X(N,M) of Eq. (3.1) and has the error estimate (2.3), where {γm}m≥0 is an arbitrary sequence in X(N,M), {αm}m≥0 and{βm}m≥0 are any sequences in [0, 1] satisfying (2.4) and (2.5).

Example 3.2. Consider the secondorder nonlinear neutral delay differential equationwith positive andnegative coefficients[t2(x(t)+

(5t1+ t2

sin(1−√t)+

t2 − 2t −√1+ t

1+ 3t2

(1+

1t

)t)x(t − τ)

)′]′

+x2(t − σ1)

1+ t32−x(t − σ2)(1− x(t − σ2))

1+ t32

= 0, t ≥ t0, (3.2)

where τ > 0, σ1 ≥ 0, σ2 ≥ 0 and t0 > 0 are constants. Let k = 2 and

a(t) = t2, A(t) =1t0−1t,

b(t) =5t1+ t2

sin(1−√t)+

t2 − 2t −√1+ t

1+ 3t2

(1+

1t

)t,

f (t, u, v) =u2 − v(1− v)

1+ t32

, (t, u, v) ∈ [t0,+∞)× R2.


Obviously, b+∞ = e3 . For anyM > N > 0, choose

PM,N(t) =1+ 4M

1+ t32, QM,N(t) =

M(1+ 2M)

1+ t32

, t ∈ [t0,+∞).

It is easy to see that the conditions (H) and (2.14) are fulfilled. Let M > 0 and N = 3−e12 M . It follows from Theorem 2.2

that Eq. (3.2) possesses infinitely many nonoscillatory solutions in X(N,M). Moreover, for any L ∈ ( 9+e12 M,M), there existθ ∈ (0, 1) and T > t0 + τ such that the Mann iterative sequence with error {xm}m≥0 generated by (2.2) converges to anonoscillatory solution x ∈ X(N,M) of Eq. (3.2) and has the error estimate (2.3), where {γm}m≥0 is an arbitrary sequence inX(N,M), {αm}m≥0 and {βm}m≥0 are any sequences in [0, 1] satisfying (2.4) and (2.5). However, we cannot use the theoremin [4] and Theorem 2 in [13] to prove the existence of nonoscillatory solutions for Eq. (3.2) because each of the conditions(C1) and (C3) in [13]

xf (x) = x3 > 0, xg(x) = x2(1− x) > 0 for x 6= 0

and

aQ1(t)− Q2(t) =a− 1

1+ t32is eventually nonnegative for every a > 0

does not hold.

Example 3.3. Consider the second order nonlinear neutral delay differential equation[(1+ t2)

(x(t)+

11− 3t2

(t + t2 −

1t

)(1+ sin2

(t −1t

))x(t − τ)

)′]′+

[t2 + 1

t

1+ t2√1+ x2(t − 1)x2(t − 2)+

x(t − 1) sin3(t − 1t +√1+ t2)

1+ x2(t − 3)

]′

+1

1+ t2

x(t2)x2(t ln(1+ t2))− tx2(t2)

(1+ t)(2+ sin

( √t

1+t2x3(2t − 1)

))

=1+√t

1+ t2

[(1+ cos(t + t2)) ln(1+ 2|t|)− (1+ sin(1− t2)) ln(1+

√1+ |t|)

], t ≥ t0, (3.3)

where τ > 0 and t0 > 0 are constants. Let k = 3 and

a(t) = 1+ t2, A(t) = arctan t − arctan t0,

b(t) =

(t + t2 − 1

t

) (1+ sin2

(t − 1

t

))1− 3t2

,

g(t) =1+√t

1+ t2

[(1+ cos(t + t2)) ln(1+ 2|t|)− (1+ sin(1− t2)) ln(1+

√1+ |t|)

],

h1(t) = t − 1, h2(t) = t − 2, h3(t) = t − 3,f1(t) = t2, f2(t) = t ln(1+ t2), f3(t) = 2t − 1,

h(t, u, v, w) =t2 + 1

t

1+ t2√1+ u2v2 +

u1+ w2

sin3(t −1t+

√1+ t2

),

f (t, u, v, w) =1

1+ t2

uv2 − tu2

(1+ t)(2+ sin

(w3√t

1+t2

))

for (t, u, v, w) ∈ [t0,+∞)× R3. Clearly, b+∞= −

23 . For anyM > N > 0, select

PM,N(t) =M1+ t2

(2+ 3M +

3M√t

1+ t2

),

QM,N(t) =M2

1+ t2

(M +

t1+ t

),

RM,N(t) =2M3√1+ N4

+2M2

(1+ N2)2+

11+ N2

,


WM,N(t) =√1+M4 +

M1+ N2

for t ∈ [t0,+∞). Consequently, we easily verify that the conditions (H) and (2.18) hold. Let M > 0 and N = M12 . It

follows from Theorem 2.3 that Eq. (3.3) has infinitely many nonoscillatory solutions in X(N,M). On the other hand, forany L ∈ ( 112M,

16M), there exist θ ∈ (0, 1) and T > t0 + τ such that the Mann iterative sequence with error {xm}m≥0

generated by (2.2) converges to a nonoscillatory solution x ∈ X(N,M) of Eq. (3.3) and has the error estimate (2.3), where{γm}m≥0 is an arbitrary sequence in X(N,M), {αm}m≥0 and {βm}m≥0 are any sequences in [0, 1] satisfying (2.4) and (2.5).However, the results in [4,10,13] are inapplicable for Eq. (3.3).

Example 3.4. Consider the second order nonlinear neutral delay differential equation[t3(x(t)+

11+ t2 + t4

(1− t2 + t3 + 4t4 +

1t

)(2+ cos(t + t2 − 1))x(t − τ)

)′]′+

[t2x2(t − 1)

(1+ t)(1+ x2(t − (2+√t) sin

√1+ t2))

]′+

(t + 1t )x

2(t2 − t)x(t ln(1+ t)+√1+ t cos(1− t2))

(1+ t3)[2+ sin(x(t2 − t)x(t ln(1+ t)+√1+ t cos(1− t2)))]

=

(1+ t sin(2t−t2)

1+t2

)ln(1+

√1+ t + t2)

t32 +

√1+ 1

2 sin2 t

, t ≥ t0, (3.4)


a(t) = t3, A(t) =12

(1t20−1t2

),

b(t) =

(1− t2 + t3 + 4t4 + 1

t

)(2+ cos(t + t2 − 1))

1+ t2 + t4,

g(t) =

(1+ t sin(2t−t2)

1+t2

)ln(1+

√1+ t + t2)

t32 +

√1+ 1

2 sin2 t

,

h1(t) = t − 1, h2(t) = t − (2+√t) sin

√1+ t2,

f1(t) = t2 − t, f2(t) = t ln(1+ t)+√1+ t cos(1− t2),

h(t, u, v) =t2u2

(1+ t)(1+ v2),

f (t, u, v) =(t + 1

t )u2v

(1+ t3)(2+ sin(uv))

for (t, u, v) ∈ [t0,+∞)× R2. Clearly, b+∞= 4 < b+∞ = 12 < b2+∞ = 16. For anyM > N > 0, select

PM,N(t) =M2(3+ 2M2)

(t + 1

t

)1+ t3

,

QM,N(t) =M3

(t + 1

t

)1+ t3

,

RM,N(t) =2Mt2

(1+ N2)(1+ t)

(1+

M2

1+ N2

),

WM,N(t) =M2t2

(1+ N2)(1+ t)

for t ∈ [t0,+∞). It is easy to show that the conditions (H) and (2.22) are satisfied. Let M > 0 and N = 12352544335M .

It follows from Theorem 2.4 that Eq. (3.4) has infinitely many nonoscillatory solutions in X(N,M). Furthermore, for anyL ∈ ( 75997612177340M,

10525428728015108 M), there exist θ ∈ (0, 1) and T > t0 + τ such that the Mann iterative sequence with error {xm}m≥0

generated by (2.23) converges to a nonoscillatory solution x ∈ X(N,M) of Eq. (3.4) and has the error estimate (2.3), where{γm}m≥0 is an arbitrary sequence in X(N,M), {αm}m≥0 and {βm}m≥0 are any sequences in [0, 1] satisfying (2.4) and (2.5).However, the results in [4,10,13] are inapplicable for Eq. (3.4).


Example 3.5. Consider the second order nonlinear neutral delay differential equation[(1+ t)

(x(t)+

(−√2t2 − (t3 +

√1+ t + 1) sin 1t

2+ t2+ cos

(t +

1t2

))x(t − τ)

)′]′

+

[(1t+1t2

)x3(t3 + 2t2 −

√1+ t2 − 1)

(1+ sin(x(t2 − t − cos(1− t2)))

)]′

+

x2(t2 + 5

√1+ t2 sin2(t − 1

t + 1)− 2)

(1+ t2)[1+

√1+ x2

(t2 ln(1+ t)−

√1+ t√1+ t2

)]=

1

1+ t32sin(t −

√t + 1)

[1+ cos(1+

√1− t + 4t5)

], t ≥ t0, (3.5)


a(t) = 1+ t, A(t) = ln(1+ t)− ln(1+ t0),

b(t) =−√2t2 − (t3 +

√1+ t + 1) sin 1t

2+ t2+ cos

(t +

1t2

),

g(t) =1

1+ t32sin(t −

√t + 1)[1+ cos(1+

√1− t + 4t5)],

h1(t) = t3 + 2t2 −√1+ t2 − 1, h2(t) = t2 − t − cos(1− t2),

f1(t) = t2 + 5

√1+ t2 sin2

(t −1t+ 1

)− 2,

f2(t) = t2 ln(1+ t)−√1+ t

√1+ t2,

h(t, u, v) =(1t+1t2

)u3(1+ sin v),

f (t, u, v) =u2

(1+ t2)(1+√1+ v2)

for (t, u, v) ∈ [t0,+∞)× R2. Clearly, b+∞= −2−

√2 < −

√2 = b+∞ < −1. For anyM > N > 0, select

PM,N(t) =M

(1+ t2)(1+√1+ N2)

(2++

M2√1+ N2(1+

√1+ N2)

),

QM,N(t) =M2

(1+√1+ N2)(1+ t2)

,

RM,N(t) = M2(6+M)(1t+1t2

),

WM,N(t) = 2M3(1t+1t2

)for t ∈ [t0,+∞). Consequently, we easily verify that the conditions (H) and (2.31) hold. Let M > 0 and N =

√2−1

6√2+2M . It

follows from Theorem 2.5 that Eq. (3.5) has infinitely many nonoscillatory solutions in X(N,M). On the other hand, for anyL ∈ (

√2−14 M,

√2−12 M), there exist θ ∈ (0, 1) and T > t0 + τ such that the Mann iterative sequence with error {xm}m≥0

generated by (2.32) converges to a nonoscillatory solution x ∈ X(N,M) of Eq. (3.5) and has the error estimate (2.3), where{γm}m≥0 is an arbitrary sequence in X(N,M), {αm}m≥0 and {βm}m≥0 are any sequences in [0, 1] satisfying (2.4) and (2.5).However, we do not invoke the results in [4,10,13] to show the existence of nonoscillatory solutions of Eq. (3.5).

Acknowledgements

The authors are grateful to the reviewers for their valuable comments and suggestions. This work was supported by theScience Research Foundation of Educational Department of Liaoning Province (2008352).


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Infinitely many nonoscillatory solutions for second order nonlinear neutral delay differential equations