We give some new criteria for the existence of non-oscillatory solutions of (1.1)

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ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

EXISTENCE OF NON-OSCILLATORY SOLUTIONS TO FIRST-ORDER NEUTRAL DIFFERENTIAL EQUATIONS

TUNCAY CANDAN

Abstract. This article presents sufficient conditions for the existence of non- oscillatory solutions to first-order differential equations having both delay and advance terms, known as mixed equations. Our main tool is the Banach contraction principle.

1. Introduction

In this article, we consider a first-order neutral differential equation d

dt[x(t) +P1(t)x(t−τ1) +P2(t)x(t+τ2)]

+Q1(t)x(t−σ1)−Q2(t)x(t+σ2) = 0,

(1.1) where P_i ∈C([t₀,∞),R), Q_i ∈ C([t₀,∞),[0,∞)),τ_i > 0 and σ_i ≥0 for i= 1,2.

We give some new criteria for the existence of non-oscillatory solutions of (1.1).

Recently, the existence of non-oscillatory solutions of first-order neutral functional differential equations has been investigated by many authors. Yu and Wang [16] showed that the equation

d

dt[x(t) +px(t−c)] +Q(t)x(t−σ) = 0, t≥t0

has a non-oscillatory solution forp≥0. Later, in 1993, Chen et al [9] studied the same equation and they extended the results to the casep∈R\{−1}. Zhang et al [17] investigated the existence of non-oscillatory solutions of the first-order neutral delay differential equation with variable coefficients

d

dt[x(t) +P(t)x(t−τ)] +Q1(t)x(t−σ1)−Q2(t)x(t−σ2) = 0, t≥t0. They obtained sufficient conditions for the existence of non-oscillatory solutions depending on the four different ranges ofP(t). In [10], existence of non-oscillatory solutions of first-order neutral differential equations

d

dt[x(t)−a(t)x(t−τ)] =p(t)f(x(t−σ)) was studied.

2010Mathematics Subject Classification. 34K11, 34C10.

Key words and phrases. Neutral equations; fixed point; non-oscillatory solution.

c

2016 Texas State University.

Submitted October 14, 2015. Published January 27, 2016.

1

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On the other hand, there has been research activities about the oscillatory be- havior of first and higher order neutral differential equations with advanced terms.

For instance, in [1] and [5], n-th order neutral differential equations with advanced term of the form

[x(t) +ax(t−τ) +bx(t+τ)]⁽ⁿ⁾+δ(q(t)x(t−g) +p(t)x(t+h)) = 0 and

[x(t)+λax(t−τ)+µbx(t+τ)]⁽ⁿ⁾+δZ d c

q(t, ξ)x(t−ξ)dξ+ Z d

c

p(t, ξ)x(t+ξ)dξ

= 0, were studied, respectively.

This article was motivated by the above studies. To the best of our knowledge, this current paper is the only paper regarding to the existence of non-oscillatory solutions of neutral differential equation with advanced term. Some other papers for the existence of non-oscillatory solutions of first, second and higher order neutral functional differential and difference equations; see [13, 18, 6, 7, 8, 15] and the references contained therein. We refer the reader to the books [14, 12, 4, 11, 2, 3]

on the subject of neutral differential equations.

Let m= max{τ1, σ1}. By a solution of (1.1) we mean a function x∈ C([t1− m,∞),R), for some t1 ≥ t0, such that x(t) +P1(t)x(t−τ1) +P2(t)x(t+τ2) is continuously differentiable on [t1,∞) and (1.1) is satisfied fort≥t1.

As it is customary, a solution of (1.1) is said to be oscillatory if it has arbitrarily large zeros. Otherwise the solution is called non-oscillatory.

The following theorem will be used to prove the theorems.

Theorem 1.1 (Banach’s Contraction Mapping Principle). A contraction mapping on a complete metric space has exactly one fixed point.

2. Main Results

To show that an operatorS satisfies the conditions for the contraction mapping principle, we consider different cases for the ranges of the coefficients P1(t) and P2(t).

Theorem 2.1. Assume that 0≤P₁(t)≤p₁<1,0≤P₂(t)≤p₂<1−p₁ and Z ∞

t₀

Q1(s)ds <∞, Z ∞

t₀

Q2(s)ds <∞, (2.1) then (1.1)has a bounded non-oscillatory solution.

Proof. Because of (2.1), we can choose at1> t0,

t1≥t0+ max{τ1, σ1} (2.2)

sufficiently large such that Z ∞

t

Q1(s)ds≤ M₂−α M2

, t≥t1, (2.3)

Z ∞ t

Q₂(s)ds≤ α−(p₁+p₂)M₂−M₁ M2

, t≥t₁, (2.4) whereM1andM2 are positive constants such that

(p₁+p₂)M₂+M₁< M₂ and α∈ (p₁+p₂)M₂+M₁, M₂ .

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Let Λ be the set of all continuous and bounded functions on [t0,∞) with the supremum norm. Set

Ω ={x∈Λ :M1≤x(t)≤M2, t≥t0}.

It is clear that Ω is a bounded, closed and convex subset of Λ. Define an operator S: Ω→Λ as follows:

(Sx)(t) =











α−P1(t)x(t−τ1)−P2(t)x(t+τ2) +R∞

t [Q1(s)x(s−σ1)−Q2(s)x(s+σ2)]ds, t≥t1,

(Sx)(t1), t0≤t≤t1.

Obviously, Sxis continuous. For t ≥t1 and x∈Ω, from (2.3) and (2.4), respectively, it follows that

(Sx)(t)≤α+ Z ∞

t

Q1(s)x(s−σ1)ds≤α+M2

Z ∞ t

Q1(s)ds≤M2

and

(Sx)(t)≥α−P1(t)x(t−τ1)−P2(t)x(t+τ2)− Z ∞

t

Q2(s)x(s+σ2)ds

≥α−p1M2−p2M2−M2

Z ∞ t

Q2(s)ds≥M1.

This means that SΩ ⊂ Ω. To apply the contraction mapping principle, the re- maining is to show that S is a contraction mapping on Ω. Thus, ifx1, x2∈Ω and t≥t1,

|(Sx1)(t)−(Sx₂)(t)|

≤P1(t)|x1(t−τ1)−x2(t−τ1)|+P2(t)|x1(t+τ2)−x2(t+τ2)|

+ Z ∞

t

(Q1(s)|x1(s−σ1)−x2(s−σ1)|+Q2(s)|x1(s+σ2)−x2(s+σ2)|)ds or

|(Sx1)(t)−(Sx2)(t)|

≤ kx1−x2k

p1+p2+ Z ∞

t

(Q1(s) +Q2(s))ds

≤

p1+p2+M2−α

M₂ +α−(p1+p2)M2−M1

M₂

kx1−x2k

=λ1kx1−x2k, whereλ₁= (1−^M_M¹

2). This implies that

kSx₁−Sx₂k ≤λ₁kx₁−x₂k,

where the supremum norm is used. Sinceλ1<1,Sis a contraction mapping on Ω.

ThusShas a unique fixed point which is a positive and bounded solution of (1.1).

This completes the proof.

Theorem 2.2. Assume that 0 ≤P1(t)≤ p1 < 1, p1−1 < p2 ≤ P2(t) ≤ 0 and (2.1)hold, then (1.1)has a bounded non-oscillatory solution.

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Proof. Because of (2.1), we can choose a t1 > t0 sufficiently large satisfying (2.2) such that

Z ∞ t

Q1(s)ds≤(1 +p2)N2−α N2

, t≥t1, (2.5)

Z ∞ t

Q₂(s)ds≤ α−p₁N₂−N₁ N2

, t≥t₁, (2.6)

whereN₁ andN₂ are positive constants such that

N₁+p₁N₂<(1 +p₂)N₂ and α∈(N₁+p₁N₂,(1 +p₂)N₂).

Ω ={x∈Λ :N1≤x(t)≤N2, t≥t0}.

(Sx)(t) =











α−P1(t)x(t−τ1)−P2(t)x(t+τ2) +R∞

(Sx)(t1), t0≤t≤t1.

Obviously, Sxis continuous. For t ≥t1 and x∈Ω, from (2.5) and (2.6), respectively, it follows that

(Sx)(t)≤α−p2N2+N2

Z ∞ t

Q1(s)ds≤N2, (Sx)(t)≥α−p1N2−N2

Z ∞ t

Q2(s)ds≥N1.

This proves thatSΩ⊂Ω. To apply the contraction mapping principle, it remains to show thatS is a contraction mapping on Ω. Thus, ifx₁, x₂∈Ω andt≥t₁,

|(Sx1)(t)−(Sx2)(t)| ≤ kx1−x2k

p1−p2+ Z ∞

t

(Q1(s) +Q2(s))ds

≤λ2kx1−x2k, whereλ2= (1−^N_N¹

2). This implies

kSx1−Sx2k ≤λ2kx1−x2k,

Theorem 2.3. Assume that 1< p1≤P1(t)≤p1₀ <∞,0 ≤P2(t)≤p2< p1−1 and (2.1)hold, then (1.1)has a bounded non-oscillatory solution.

Proof. In view of (2.1), we can choose at1> t0,

t1+τ1≥t0+σ1, (2.7)

sufficiently large such that Z ∞

t

Q1(s)ds≤ p1M4−α M4

, t≥t1, (2.8)

Z ∞ t

Q2(s)ds≤α−p1₀M3−(1 +p2)M4

M4

, t≥t1, (2.9)

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whereM3andM4 are positive constants such that

p1₀M3+ (1 +p2)M4< p1M4 and α∈ p1₀M3+ (1 +p2)M4, p1M4 . Let Λ be the set of all continuous and bounded functions on [t0,∞) with the supremum norm. Set

Ω ={x∈Λ :M3≤x(t)≤M4, t≥t0}.

It is clear that Ω is a bounded, closed and convex subset of Λ. Define a mapping S: Ω→Λ as follows:

(Sx)(t) =











1

P₁(t+τ₁){α−x(t+τ1)−P2(t+τ1)x(t+τ1+τ2) +R∞

t+τ₁[Q1(s)x(s−σ1)−Q2(s)x(s+σ2)]ds}, t≥t1,

(Sx)(t₁), t₀≤t≤t₁.

Clearly,Sxis continuous. For t≥t1 andx∈Ω, from (2.8) and (2.9), respectively, it follows that

(Sx)(t)≤ 1 P₁(t+τ₁)

α+M₄

Z ∞ t

Q₁(s)ds

≤ 1 p₁

α+M₄

Z ∞ t

Q₁(s)ds

≤M₄ and

(Sx)(t)≥ 1 P1(t+τ1)

α−(1 +p2)M4−M4

Z ∞ t

Q2(s)ds

≥ 1 p₁₀

α−(1 +p2)M4−M4

Z ∞ t

Q2(s)ds

≥M3.

This means that SΩ⊂Ω. To apply the contraction mapping principle it remains to show thatS is a contraction mapping on Ω. Thus, ifx₁, x₂∈Ω andt≥t₁,

|(Sx1)(t)−(Sx₂)(t)| ≤ 1 p1

kx1−x₂k

1 +p₂+ Z ∞

t

(Q₁(s) +Q₂(s))ds

≤λ₃kx₁−x₂k, whereλ3= (1−^p_p¹⁰^M³

1M₄ ). This implies

kSx₁−Sx₂k ≤λ₃kx₁−x₂k,

Theorem 2.4. Assume that 1< p1≤P1(t)≤p1₀ <∞,1−p1 < p2 ≤P2(t)≤0 and (2.1)hold, then (1.1)has a bounded non-oscillatory solution.

Proof. In view of (2.1), we can choose a t1 > t0 sufficiently large satisfying (2.7) such that

Z ∞ t

Q₁(s)ds≤ (p1+p2)N4−α

N₄ , t≥t₁, (2.10)

Z ∞ t

Q2(s)ds≤ α−p1₀N3−N4

N₄ , t≥t1, (2.11)

whereN3 andN4 are positive constants such that

p₁₀N₃+N₄<(p₁+p₂)N₄ and α∈ p₁₀N₃+N₄,(p₁+p₂)N₄ .

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Ω ={x∈Λ :N3≤x(t)≤N4, t≥t0}.

(Sx)(t) =











1

P₁(t+τ₁){α−x(t+τ₁)−P₂(t+τ₁)x(t+τ₁+τ₂) +R∞

(Sx)(t1), t0≤t≤t1.

Clearly,Sxis continuous. Fort≥t1andx∈Ω, from (2.10) and (2.11), respectively, it follows that

(Sx)(t)≤ 1 P₁(t+τ₁)

α−p2N4+N4

Z ∞ t

Q1(s)ds

≤ 1 p₁

α−p2N4+N4

Z ∞ t

Q1(s)ds

≤N4

and

(Sx)(t)≥ 1 P₁(t+τ₁)

α−N4−N4

Z ∞ t

Q2(s)ds

≥ 1 p₁₀

α−N₄−N₄ Z ∞

t

Q₂(s)ds

≥N₃.

This proves thatSΩ⊂Ω. To apply the contraction mapping principle it remains to show thatS is a contraction mapping on Ω. Thus, ifx1, x2∈Ω andt≥t1,

|(Sx₁)(t)−(Sx₂)(t)| ≤ 1 p1

kx₁−x₂k

1−p₂+ Z ∞

t

(Q₁(s) +Q₂(s))ds

≤λ4kx1−x2k, whereλ4= (1−^p_p¹⁰^N³

1N₄ ). This implies

Theorem 2.5. Assume that −1 < p₁ ≤P₁(t)≤0, 0≤P₂(t)≤p₂ <1 +p₁ and (2.1)hold, then (1.1)has a bounded non-oscillatory solution.

Z ∞ t

Q1(s)ds≤(1 +p1)M6−α M6

, t≥t1, (2.12)

and

Z ∞ t

Q₂(s)ds≤ α−p₂M₆−M₅ M6

, t≥t₁, (2.13)

whereM5andM6 are positive constants such that

M₅+p₂M₆<(1 +p₁)M₆ and α∈(M₅+p₂M₆,(1 +p₁)M₆).

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Ω ={x∈Λ :M₅≤x(t)≤M₆, t≥t₀}.

(Sx)(t) =











α−P₁(t)x(t−τ₁)−P₂(t)x(t+τ₂) +R∞

(Sx)(t1), t0≤t≤t1.

Obviously,Sxis continuous. Fort≥t₁ andx∈Ω, from (2.12) and (2.13), respectively, it follows that

(Sx)(t)≤α−p1M6+M6

Z ∞ t

Q1(s)ds≤M6, (Sx)(t)≥α−p2M6−M6

Z ∞ t

Q2(s)ds≥M5.

This proves thatSΩ⊂Ω. To apply the contraction mapping principle it remains to show thatS is a contraction mapping on Ω. Thus, ifx₁, x₂∈Ω,t≥t₁,

|(Sx1)(t)−(Sx2)(t)| ≤ kx1−x2k

−p1+p2+ Z ∞

t

(Q1(s) +Q2(s))ds

≤λ5kx1−x2k, whereλ5= (1−^M_M⁵

6). This implies

kSx₁−Sx₂k ≤λ₅kx₁−x₂k,

where the supremum norm is used. Sinceλ₅<1,Sis a contraction mapping on Ω.

Theorem 2.6. Assume that−1< p1 ≤P1(t)≤0,−1−p1 < p2≤P2(t)≤0 and (2.1)hold, then (1.1)has a bounded non-oscillatory solution.

Z ∞ t

Q₁(s)ds≤(1 +p1+p2)N6−α

N₆ , t≥t₁, (2.14)

and

Z ∞ t

Q2(s)ds≤α−N5

N₆ , t≥t1, (2.15)

whereN5 andN6 are positive constants such that

N5<(1 +p1+p2)N6 and α∈(N5,(1 +p1+p2)N6).

Let Λ be the set of continuous and bounded functions on [t0,∞) with the supremum norm. Set

Ω ={x∈Λ :N₅≤x(t)≤N₆, t≥t₀}.

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(Sx)(t) =











α−P1(t)x(t−τ1)−P2(t)x(t+τ2) +R∞

t [Q₁(s)x(s−σ₁)−Q₂(s)x(s+σ₂)]ds, t≥t₁,

(Sx)(t1), t0≤t≤t1.

Obviously,Sxis continuous. Fort≥t1 andx∈Ω, from (2.14) and (2.15), respectively, it follows that

(Sx)(t)≤α−p1N6−p2N6+N6

Z ∞ t

Q1(s)ds≤N6, (Sx)(t)≥α−N₆

Z ∞ t

Q₂(s)ds≥N₅.

This proves thatSΩ⊂Ω. To apply the contraction mapping principle it remains to show thatS is a contraction mapping on Ω. Thus, ifx1, x2∈Ω andt≥t1,

|(Sx1)(t)−(Sx2)(t)| ≤ kx1−x2k

−p1−p2+ Z ∞

t

(Q1(s) +Q2(s))ds

≤λ₆kx1−x₂k, whereλ6= (1−^N_N⁵

6). This implies

where the supremum norm is used. Sinceλ₆<1,Sis a contraction mapping on Ω.

Theorem 2.7. Assume that −∞< p1₀ ≤ P1(t)≤ p1 < −1, 0 ≤ P2(t) ≤ p2 <

−p1−1 and (2.1)hold, then (1.1)has a bounded non-oscillatory solution.

Z ∞ t

Q1(s)ds≤p1₀M7+α

M₈ , t≥t1, (2.16)

and

Z ∞ t

Q₂(s)ds≤(−p₁−1−p₂)M₈−α M8

, t≥t₁, (2.17) whereM7andM8 are positive constants such that

−p1₀M7<(−p1−1−p2)M8 and α∈(−p1₀M7,(−p1−1−p2)M8). Let Λ be the set of all continuous and bounded functions on [t₀,∞) with the supremum norm. Set

Ω ={x∈Λ :M7≤x(t)≤M8, t≥t0}.

(Sx)(t) =











−1

P1(t+τ1){α+x(t+τ1) +P2(t+τ1)x(t+τ1+τ2)

−R∞

t+τ1[Q₁(s)x(s−σ₁)−Q₂(s)x(s+σ₂)]ds}, t≥t₁

(Sx)(t1), t0≤t≤t1.

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(Sx)(t)≤ −1 p1

α+M8+p2M8+M8

Z ∞ t

Q2(s)ds

≤M8

and

(Sx)(t)≥ −1 p₁₀

α−M8

Z ∞ t

Q1(s)ds

≥M7.

This implies thatSΩ⊂Ω. To apply the contraction mapping principle it remains to show thatS is a contraction mapping on Ω. Thus, ifx1, x2∈Ω andt≥t1,

|(Sx1)(t)−(Sx2)(t)| ≤ −1

p₁kx1−x2k

1 +p2+ Z ∞

t

(Q1(s) +Q2(s))ds

≤λ7kx1−x2k, whereλ7= (1−^p_p¹⁰^M⁷

1M₈ ). This implies

kSx1−Sx₂k ≤λ₇kx1−x₂k,

Theorem 2.8. Assume that−∞< p1₀≤P1(t)≤p1<−1,p1+1< p2≤P2(t)≤0 and (2.1)hold, then (1.1)has a bounded non-oscillatory solution.

Z ∞ t

Q1(s)ds≤ p1₀N7+p2N8+α N8

, t≥t1, (2.18)

and

Z ∞ t

Q₂(s)ds≤ (−p1−1)N8−α

N₈ , t≥t₁, (2.19)

whereN₇ andN₈ are positive constants such that

−p1₀N7−p2N8<(−p1−1)N8 and α∈(−p1₀N7−p2N8,(−p1−1)N8).

Let Λ be the set of continuous and bounded functions on [t0,∞) with the supremum norm. Set

Ω ={x∈Λ :N₇≤x(t)≤N₈, t≥t₀}.

(Sx)(t) =







−1

P₁(t+τ₁){α+x(t+τ1) +P2(t+τ1)x(t+τ1+τ2)

−R∞

(Sx)(t1), t0≤t≤t1.

(Sx)(t)≤ −1 p₁

α+N8+N8

Z ∞ t

Q2(s)ds

≤N8

and

(Sx)(t)≥ −1 p1₀

α+p₂N₈−N₈ Z ∞

t

Q₁(s)ds

≥N₇.

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These prove that SΩ⊂Ω. To apply the contraction mapping principle it remains to show thatS is a contraction mapping on Ω. Thus, ifx1, x2∈Ω,t≥t1,

|(Sx1)(t)−(Sx2)(t)| ≤ −1 p1

kx1−x2k

1−p2+ Z ∞

t

(Q1(s) +Q2(s))ds

≤λ₈kx1−x₂k, whereλ₈= (1−^p_p¹⁰^N⁷

1N₈ ). This implies

Example 2.9. Consider the equation h

x(t)−1

2x(t−2π) +1

2 −exp(−t 2)

x(t+ 5π)i0

+1

2exp(−t

2)x(t−4π)−exp(−t

2)x(t+5π

2 ) = 0, t >−2 ln(1/2)

(2.20)

and note that P₁(t) =−1

2, P₂(t) = 1

2−exp(−t

2), Q₁(t) =1

2exp(−t

2), Q₂(t) = exp(−t 2).

A straightforward verification yields that the conditions of Theorem 2.5 are valid.

We note thatx(t) = 2 + sint is a non-oscillatory solution of (2.20).

Example 2.10. Consider the equation h

x(t)− 1 exp(1)

3

4−exp(−t)

x(t−1)−exp(1/4)1

4+ exp(−t) x(t+1

4)i0

+ exp(−t−1)x(t−1)−exp(−t+1

4)x(t+1

4) = 0, t≥3 2

(2.21)

and note that

P1(t) =− 1 exp(1)

3

4−exp(−t)

, P2(t) =−exp(1 4)1

4 + exp(−t) , Q₁(t) = exp(−t−1), Q₂(t) = exp(−t+1

4).

It is easy to verify that the conditions of Theorem 2.6 are valid. We note that x(t) = 1 + exp(−t) is a non-oscillatory solution of (2.21).

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Tuncay Candan

Department of Mathematics, Faculty of Arts and Sciences, Ni˘gde University, Ni˘gde 51200, Turkey

E-mail address:tcandan@nigde.edu.tr