EXISTENCE OF NONOSCILLATORY SOLUTIONS FOR SECOND-ORDER NONLINEAR NEUTRAL DIFFERENTIAL EQUATIONS WITH VARIABLE DELAYS

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EXISTENCE OF NONOSCILLATORY SOLUTIONS FOR SECOND-ORDER NONLINEAR NEUTRAL DIFFERENTIAL

EQUATIONS WITH VARIABLE DELAYS

SHYAM SUNDAR SANTRA1_,_§

Abstract. In this work, an attempt is made to discuss the existence of nonoscillatory solutions of second order nonlinear neutral differential equations with variable delays. The main tools are Lebesgue’s dominated convergence theorem and Banach contraction principle to obtain new sufficient conditions for the existence of nonoscillatory solutions. This problem is considered in various ranges of the neutral coefficient. Further, two illustrative examples showing applicability of the new results are included.

Keywords: Existence of positive solutions, neutral, delay, non-linear, Lebesgue’s domi-nated convergence theorem, Banach’s fixed point theorem.

AMS Subject Classification: 34C10, 34C15, 34K40.

1. Introduction

This article is concerned with sufficient conditions for existence of positive solutions of a nonlinear neutral second-order delay differential equation

d dt

r(t) d

dt

x(t) +p(t)x(τ(t))

+q(t)G x(σ(t))

= 0. (1)

Suppose that the following assumptions hold.

(A1) r∈C([t0,∞),(0,∞)), p∈C([t0,∞),R) andq ∈C([t0,∞),(0,∞)), where q is not identically zero;

(A2) G∈C(R,R) is nondecreasing such thatxG(x)>0 for x6= 0;

(A3) τ, σ ∈ C([t0,∞),(0,∞)) such that σ(t), τ(t) ≤ t for t ≥ t0, σ(t), τ(t) → ∞ as

t→ ∞with invertible τ when necessary.

Culakova et al. [1] considered (1) and studied existence of nonoscillatory solutions whenp∈C([t0,∞),(−∞,0)). In [7], under various ranges ofp, Santra studied oscillatory behaviour of the solutions of the following neutral differential equations

d dt

x(t) +p(t)x(t−τ)

+q(t)G x(t−σ) = 0

1 _{Department of Mathematics, Sambalpur University, Sambalpur - 768019, India.} e-mail: [email protected]; ORCID: https://orcid.org/0000-0001-9740-3081.

§ Manuscript received: July 21, 2017; accepted: August 24, 2017.

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and

d dt

x(t) +p(t)x(t−τ)+q(t)G x(t−σ)=f(t) (2)

Also, sufficient conditions are obtained for existence of bounded nonoscillatory solutions of (2). The motivation of the present work come from the above studies. Hence, the objective of this work is to study existence of positive solutions of (1) for any|p(t)|<+∞.

Oscillation and nonoscillation of functional differential equations have been studied in recent years. For further work to this type of equations, one may follow [1, 2, 11, 12] and the references cited therein. The existence of nonoscillatory solution of functional differential equations received much less attention, which is mainly due to the technical difficulties arising in its analysis.

By a solution of (1) we mean a continuously differentiable functionx(t) which is defined fort≥T∗ = min{τ(t0), σ(t0)} such that x(t) satisfies (1) for allt ≥t0. In the sequel, it will always be assumed that the solutions of (1) exist on some half line [t1,∞), t1 ≥ t0. A solution of (1) is said to be oscillatory, if it has arbitrarily large zeros; otherwise, it is called nonoscillatory. Equation (1) is called oscillatory, if all its solutions are oscillatory.

2. Main Results

Theorem 2.1. Assume that(A1)–(A3)hold and p ∈C(R+,[0,1)). Furthermore assume

thatG is Lipschitzian on the intervals of the form[a, b], 0< a < b <∞. If

(A4) R₀∞_r(η)1

R∞ η q(ζ)dζ

dη <+∞,

then (1) has a bounded nonoscillatory solution.

Proof. Let 0≤p(t)≤p <1,t∈R+ andp >0. Due to (A4), it is possible to find T > T∗ such that

Z ∞ T

1

r(η) Z ∞

η

q(ζ)dζ

dη < 1−p

5L ,

whereL= max{L1, G(1)},L1 is the Lipschitz constant ofGon

7(1−p) 10 ,1

fort≥t0. Let

Y = BC([t0,∞),R) be the space of real valued continuous functions on [t0,∞]. Indeed,

Y is a Banach space with respect to supremum norm defined by

kxk= sup{|x(t)|:t≥t0}. Define

S=

v∈Y : 7(1−p)

10 ≤v(t)≤1, t≥t0

.

Note thatS is a closed and convex subspace of Y. Let Φ :S→S be such that

(Φx)(t) =   

(Φx)(T), t∈[t0, T] −p(t)x(τ(t)) + 9+p₁₀ −R∞

t 1 r(η)

R∞

η q(ζ)G(x(σ(ζ)))dζ

dη, t≥T.

For everyx∈S, (Φx)(t)≤ 9+p₁₀ <1 and (Φx)(t)≥ −p+9 +p

10 − 1−p

5 = 7

(3)

implies that Φx∈S. Forx1, x2 ∈S, it follows that |(Φx1)(t)−(Φx2)(t)| ≤p|x1(τ(t))−x2(τ(t))|

+ Z ∞

t 1

r(η) Z ∞

η

q(ζ)|G x1(σ(ζ))

−G x2(σ(ζ))

|dζ

dη,

≤pkx1−x2k+kx1−x2kL1 Z ∞

t 1

r(η) Z ∞

η

q(ζ)dζ

dη

≤

p+1−p 5

kx1−x2k = 4p+ 1

5 kx1−x2k.

Therefore,kΦx1−Φx2k ≤ 4p+1₅ kx1−x2kimplies that Φ is a contraction. By using Banach’s fixed point theorem, it follows that Φ has a unique fixed point x(t) in

7(1−p)

10 ,1

. This

completes the proof of the theorem.

Theorem 2.2. Assume that (A1)–(A4) hold and 1< p1 ≤p(t)≤p2<∞, where p21 ≥p2

fort∈R+. Furthermore assume thatGbe Lipschitzian on the intervals of the form[α, β], 0< α < β <∞. Then (1)has a bounded nonoscillatory solution.

Proof. Due to (A4), it is possible to find T > T∗ such that

Z ∞ T

1

r(η) Z ∞

η

q(ζ)dζ

dη < p1−1

3L ,

whereL= max{L1, L2},L1 is the Lipschitz constant ofG on [α, β],L2 =G(β) with

α= 3λ(p 2

1−p2)−p2(p1−1) 3p12p2

β= p1−1 + 3λ 3p1

, λ > p2(p1−1)

3(p12−p2)

>0.

LetY =BC([t0,∞),R) be the space of real valued functions defined on [t0,∞). Indeed,

Y is a Banach space with respect to supremum norm defined by

||x||= sup{|x(t)|:t≥t0}. Define

S=

u∈Y : α≤u(t)≤β, t≥t0

.

Let Φ :S →S be such that

(Φx)(t) =   

(Φx)(T), t∈[t0, T]

−_p(τx(τ−−11(t))_(t)) +_p(τ−λ1_(t))+_p(τ−11_(t))

Rτ−1(t) T

1 r(η)

R∞

η q(ζ)G x(σ(ζ))

dζ

dη, t≥T.

For everyx∈S,

(Φx)(t)≤ G(β)

p(τ−1₍_t₎₎

Z τ−1(t) T

1

r(η) Z ∞

η

q(ζ)dζ

dη+ λ

p(τ−1₍_t₎₎ ≤ G(β)

p(τ−1₍_t₎₎ Z ∞

T 1

r(η) Z ∞

η

q(ζ)dζ

dη+ λ

p(τ−1₍_t₎₎ ≤ 1

p1

p1−1 3 +λ

(4)

and

(Φx)(t)≥ −x(τ

−1₍_t₎₎

p(τ−1₍_t₎₎ +

λ p(τ−1₍_t₎₎−

β p1

+ λ

p2 =α

implies that Φx∈S. Forx1, x2 ∈S, it follows that |(Φx1)(t)−(Φx2)(t)| ≤

1

|p(τ−1₍_t_))||x1(τ

−1₍_t₎₎₋_x

2(τ−1(t))| + L1

|p(τ−1₍_t_))|

Z τ−1(t) T

1

r(η) Z ∞

η

q(ζ)|x1(σ(ζ))−x2(σ(ζ))|dζ

dη,

≤ 1

p1

||x1−x2||+

L1

p1

||x1−x2||

Z τ−1(t) T

1

r(η) Z ∞

η

q(ζ)dζ

dη

< 1 p1

||x1−x2||

1 +p1−1 3

,

that is,

||Φx1−Φx2|| ≤

1

p1

+p1−1 3p1

||x1−x2||.

Since

1 p1 +

p1−1 3p1

<1, then Φ is a contraction mapping of S into S. Note that S is a

closed convex subset ofY and hence we apply Banach’s fixed point toS. So, we conclude that Φ has a unique fixed point on [α, β]. It is easy to verify that

x(t) =   

(Φx)(T), t∈[t0, T]

−x(τ_p(τ−−11(t))_(t))+_p(τ−λ1_(t))+ _p(τ−11_(t))

Rτ−1(t) T

1 r(η)

R∞

η q(ζ)G x(σ(ζ))

dζ

dη, t≥T.

is a positive bounded solution of (1) on [α, β]. This completes the proof of the theorem.

Theorem 2.3. Assume that (A1)–(A3) hold and −1 < −p ≤ p(t) ≤ 0, where p > 0,

t∈R+. Furthermore assume that (A5) R(t) =Rt

0 dη

r(η) and limt→∞R(t) = +∞ (A6) R∞

0 q(η)G(εR(σ(η)))dη <+∞ for every ε >0

hold, then (1) has a unbounded nonoscillatory solution.

Proof. Due to (A6), it is possible to find ε >0 such that

Z ∞ T

q(η)G(εR(σ(η)))dη≤ ε

3, T ≥T

∗_.

Let’s consider

M =

x:x∈C([t0,+∞),R), x(t) = 0f or t∈[t0, T]and

ε

3[R(t)−R(T)]≤x(t)≤ε[R(t)−R(T)]

and define Φ :M →C([t0,+∞),R) such that

(Φx)(t) =   

0, t∈[t0, T) −p(t)x(τ(t)) +Rt

T 1 r(η)

ε 3 +

R∞

η q(ζ)G x(σ(ζ)))dζ

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For everyx∈M,

(Φx)(t)≥ Z t

T 1

r(η)

ε

3+ Z ∞

η

q(ζ)G x(σ(ζ))dζ

dη

≥ ε 3

Z t T

dη r(η) =

ε

3[R(t)−R(T)]

andx(t)≤εR(t) implies that

(Φx)(t)≤ −p(t)x(τ(t)) + 2ε 3

Z t T

dη r(η)

≤pε[R(τ(t))−R(T)] + 2ε

3 [R(t)−R(T)]

≤pε[R(t)−R(T)] + 2ε

3[R(t)−R(T)]

=

p+2 3

ε[R(t)−R(T)]

≤ε[R(t)−R(T)]

implies that (Φx)(t)∈M. Definevn: [T−ρ,+∞)→R by the recursive formula

vn(t) = (Φvn−1)(t), n≥1,

with the initial condition

v0(t) = (

0, t∈[t0, T) ε

3[R(t)−R(T)], t≥T. Inductively, it is easy to verify that

ε

3[R(t)−R(T)]≤vn−1(t)≤vn(t)≤ε[R(t)−R(T)].

fort≥T. Therefore, for t≥t0, limn→∞vn(t) exists. Let limn→∞vn(t) =v(t) fort≥t0. By the Lebesgue’s dominated convergence theoremv∈M and (Φv)(t) =v(t), wherev(t) is a solution of (1) on [t0,∞) such that v(t) > 0. Note that limt→∞_R(t)z(t) = ε₃, where z(t) =x(t) +p(t)x(τ(t)). This completes the proof.

Theorem 2.4. Assume that (A1)−(A4)hold and p∈C(R+,(−1,0]). Then (1) admits

a bounded nonoscillatory solutions.

Proof. Let−1<−p≤p(t)≤0,p >0 for t∈R+. Due to (A4),

G(ε) Z ∞

T 1

r(η) Z ∞

η

q(ζ)dζ

dη≤ ε

3, T ≥T

∗_,

whereε >0 is a constant. Consider

M =

x∈C([t0,+∞),R) :x(t) = ε

3, t∈[t0, T];

ε

3 ≤x(t)≤ε f or t≥T

and let Φ :M →M be defined by

(Φx)(t) =    ε

3, [t0, T]

−p(t)x(τ(t)) + ε₃ +Rt T

1 r(η)

R∞

η q(ζ)G x(σ(ζ))

dζ

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For everyx∈M, (Φx)(t)≥ ε 3 and (Φx)(t)≤pε+ ε

3 +G(ε) Z t

T 1

r(η) Z ∞

η

q(ζ)dζ

dη

≤pε+ ε 3 +

ε

3 =

p+2 3

ε≤ε

implies that Φx∈M. The rest of the proof follows from Theorem 2.3.

Theorem 2.5. Assume that(A1)−(A4)hold and−∞<−p1 ≤p(t)≤ −p2<−1, where

p1, p2 > 0 such that 3p2 > p1 for t ∈ R+. Furthermore assume that G is Lipschitzian

on the interval of the form [a, b], 0 < a < b < ∞. Then equation (1) has a bounded nonoscillatory solution.

Proof. Due to (A4), it is possible to find T > T∗ such that

Z ∞ T

1

r(η) Z ∞

η

q(ζ)dζ

dη <p2−1

3L ,

whereL= max{L1, G(1)},L1 is the Lipschitz constant ofG on (α,1), α= (p2−1)(3p2_3p1p2−p1). LetY =BC([t0,∞),R) be the space of real valued continuous functions defined on [t0,∞). Indeed,Y is a Banach space with the supremum norm defined by

||x||= sup{|x(t)|:t≥t0}. Define

S=

v∈Y : α ≤v(t)≤1, t≥t0

.

and note thatS is a closed and convex subspace ofY. Let Ψ :S →S be such that

(Ψx)(t) =   

(Ψx)(T), t∈[t0, T] −x(τ_p(τ−−11_(t))(t)) −

p2−1

p(τ−1_(t))+_p(τ−11_(t))

Rτ−1(t) T

1 r(η)

R∞

η q(ζ)G x(σ(ζ))

dζ

dη, t≥T.

For everyx∈S,

(Ψx)(t)≤ −x(τ

−1₍_t₎₎

p(τ−1₍_t₎₎ −

p2−1

p(τ−1₍_t₎₎ ≤ 1

p2

+p2−1

p2 = 1

and

(Ψx)(t)≥ − p2−1

p(τ−1₍_t₎₎+ 1

p(τ−1₍_t₎₎

Z τ−1(t) T

1

r(η) Z ∞

η

q(ζ)G x(σ(ζ))

dζ

dη

≥ p2−1

p1

+ G(1)

p(τ−1₍_t₎₎

Z τ−1(t) T

1

r(η) Z ∞

η

q(ζ)dζ

dη

≥ p2−1

p1

− G(1)

p2 Z ∞

T 1

r(η) Z ∞

η

q(ζ)dζ

dη

≥ p2−1

p1

− p2−1 3p2

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implies that Ψx∈S. Forx1, x2 ∈S, it follows that |(Ψx1)(t)−(Ψx2)(t)| ≤

1

|p(τ−1₍_t_))||x1(τ

−1₍_t₎₎₋_x

2(τ−1(t))| + L1

|p(τ−1₍_t_))|

Z τ−1(t) T

1

r(η) Z ∞

η

q(ζ)|x1(σ(ζ))−x2(σ(ζ))|dζ

dη,

≤ 1

p2

||x1−x2||+

p2−1 3p2

||x1−x2|| =µ||x1−x2||,

where µ= _p21

1 + p2₃−1

<1. Therefore, Ψ is a contraction. Hence, by Banach’s fixed

point theorem Ψ has a unique fixed pointx ∈ S. It is easy to see that limt→∞x(t) 6= 0.

This completes the proof of the theorem.

3. Examples and Open Problem Example 3.1. Consider the equation

d dt

etd dt

x(t)−e−(t+1)x(t−1)

+2

ee −2

3t x(t−3) 1

3 _{= 0}_{, t}_≥₃_, ₍₃₎

and note that r(t) = et, p(t) = −e−(t+1), q(t) = _e2e−23t, τ(t) = t−1, σ(t) = t−3 and

G(x) = x13. A straightforward verification yields that the conditions of Theorem 2.4 are

satisfied. We note thatx(t) =e−t is a nonoscillatory solution of (3).

Example 3.2. Consider the equation

d dt

e−3td dt

x(t)−e−7tx(t−2)

+ 28e−10t x(t−2)3 = 0, t≥2, (4)

and note that r(t) = e−3t, p(t) = −e−7t, q(t) = 28e−10t, τ(t) = t−2, σ(t) = t−2 and

G(x) =x3. Clearly all conditions of Theorem 2.3 are satisfied. In particular, x(t) =e3t

is a positive unbounded solution of (4).

Remark 3.1. It is interesting to study the necessary and sufficient conditions for the oscillation of (1) for any |p(t)|<+∞ .

Acknowledgement: This work is supported by the Department of Science and Tech-nology (DST), New Delhi, India, through the bank instruction order No. DST/INSPIRE Fellowship/2014/140, dated Sept. 15, 2014.

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