# Positive periodic solutions for neutral functional differential equations

## Positive periodic solutions for neutral functional differential equations

Journal Pre-proof Positive periodic solutions for neutral functional differential equations Weibing Wang, Jianhua Shen PII: DOI: Reference: S0893-96...

Journal Pre-proof Positive periodic solutions for neutral functional differential equations Weibing Wang, Jianhua Shen

PII: DOI: Reference:

S0893-9659(19)30480-X https://doi.org/10.1016/j.aml.2019.106154 AML 106154

To appear in:

Applied Mathematics Letters

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Positive periodic solutions for neutral functional differential equations Weibing Wang1 1 Department

Jianhua Shen2

of Mathematics, Hunan University of Science and Technology Xiangtan, Hunan 411201, P.R. China

2 Department

of Mathematics, Hangzhou Normal University

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Hangzhou, Zhejiang 310036, P.R. China Abstract In this paper, we investigate the existence of positive periodic solution for the first order neutral differential equation. The results are established using a fixed point theorem in cones of Banach space. Keywords: Neutral differential equation, Positive periodic solution, Fixed point

2010 Mathematical Subject Classification: 34K13, 39A10.

1

Introduction

In this paper we consider the existence of positive periodic solutions for the following neutral differential equation(NFDE) (x(t) − cx(t − τ ))0 = a(t)g(x(t − δ(t))) − p(t),

(1.1)

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where c, τ are constants and the functions a, δ, p, g satisfy the following assumptions: (H1) a, p, δ are ω-periodic continuous functions, ˜ 6= 0, p˜ 6= 0, Rω R ω ω > 0 is a constant, a p˜a(t) ≥ 0 for all t ∈ R, here a ˜ = 0 a(s)ds, p˜ = 0 p(s)ds, (H2) g : (0, +∞) → (0, +∞) is continuous. NFDE provides good models in many fields such as physics, biology and mechanics [1]. In recent years, there has been considerable interest in the existence of positive periodic solutions for NFDE, see [2, 3, 4] and references therein. In [5, 6], by using Krasnoselskii fixed point theorem, author(s) discussed the existence of positive periodic solution for the following equation (x(t) − cx(t − h(t)))0 = −a(t)x(t) + f (t, x(t − h(t))), (1.2) (x(t) − c(t)x(t − τ ))0 = −a(t)x(t) + f (t, x(t − τ )),

(1.3)

respectively. In order to apply the fixed point theorem, author(s) constructed two appropriate operators from the linear term a(t)x(t) : one is completely continuous and the other is contractive. However, this technique is invalid for (1.1) even though g is a linear function. The aim of this paper is, by applying the fixed point theorem in cone and some new techniques, to establish some sufficient conditions which guarantee the existence of positive periodic solutions for (1.1). 1

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2

Preliminaries

Let Eω = {u ∈ C(R, R) : u(t + ω) = u(t), t ∈ R} with the norm kuk = max0≤t≤ω |u(t)|. It is clear that Eω is a Banach space. Let h, k ∈ Eω and consider the differential equation x0 (t) = −h(t)x(t) + k(t).

(2.1)

˜ 6= 0, then (2.1) has a unique ω-periodic solution Lemma 2.1. [7] Assume that h Rs Z t+ω exp t h(r)dr Rω x(t) = k(s)ds. exp 0 h(r)dr − 1 t

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Lemma 2.2. [8] Let X be a Banach space and K be a cone in X. Suppose that Ω1 and ¯ 1 ⊂ Ω2 and Φ : K ∩ (Ω ¯ 2 \Ω1 ) → K is a Ω2 are open subsets of X such that 0 ∈ Ω1 ⊂ Ω completely continuous operator such that (i) inf kΦuk > 0, u 6= λΦu for u ∈ K ∩ ∂Ω1 and λ ≥ 1, and u 6= λΦu for u ∈ K ∩ ∂Ω2 and 0 < λ ≤ 1, or (ii) inf kΦuk > 0, u 6= λΦu for u ∈ K ∩ ∂Ω2 and λ ≥ 1, and u 6= λΦu for u ∈ K ∩ ∂Ω1 and 0 < λ ≤ 1. ¯ 2 \Ω1 ). Then Φ has a fixed point in K ∩ (Ω Define A : Eω → Eω by

(Ax)(t) = x(t) − cx(t − τ ), x ∈ Eω .

Lemma 2.3. [9] If |c| 6= 1 then A has a bounded inverse A−1 on Eω and for all x ∈ Eω  X  cj x(t − τ j), |c| < 1,    j≥0 (A−1 x)(t) = X   − c−j x(t + τ j), |c| > 1.   j≥1

Consider the equation

¡ ¢ u0 (t) = a(t)g (A−1 u)(t − δ(t)) − p(t).

(2.2)

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From the definition of A, A−1 , one easily check that u is an ω-periodic solution of (2.2) if and only if A−1 u is an ω-periodic solution of (1.1). Now, suppose that u is an ω-periodic solution of (2.2) and let v = bu , here 0 < b < 1 is a constant,then v ∈ Eω satisfies ¡ ¢ v 0 (t) + p(t)v ln b = a(t)v(t)g (A−1 logb v)(t − δ(t)) ln b. (2.3) Moreover,

Rs ¡ ¢ exp t p(r)(ln b)dr Rω v(t) = a(s)v(s)g (A−1 logb v)(s − δ(s)) (ln b)ds exp 0 p(r)(ln b)dr − 1 t Z t+ω ¡ ¢ Gb (t, s)v(s)g (A−1 logb v)(s − δ(s)) |a(s) ln b|ds, = Z

t+ω

t

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where

Rs exp t p(r)(ln b)dr Rω Gb (t, s) = . | exp 0 p(r)(ln b)dr − 1|

Let Mb = max{Gb (t, s) : t ≤ s ≤ t + ω}, mb = min{Gb (t, s) : t ≤ s ≤ t + ω}, then 0 < mb < Mb < +∞. Define the operator T and the cone K on Eω by Z t+ω ¡ ¢ (T v)(t) = Gb (t, s)v(s)g (A−1 logb v)(s − δ(s)) |a(s) ln b|ds, t

mb . Mb One easily show that A−1 logb v is an ω-periodic solution of (1.1) if T v = v ∈ Eω and v > 0. K = {v ∈ Eω : v(t) ≥ kb kvk}, kb =

|c|

Lemma 2.4. (1) If c ∈ (−1, 0], u ∈ K with 0 < kuk = r < kb1+c , then 1 1 logb (kbc r1+c ) ≤ A−1 logb u ≤ logb (kb r1+c ). 2 1−c 1 − c2

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0<

(2) If c ∈ [0, 1), u ∈ K with 0 < kuk = r < 1, then 0<

1 1 logb r ≤ A−1 logb u ≤ logb (kb r). 1−c 1−c

|c|

Proof. We only consider case (1). The condition u ∈ K with kuk = r ≤ kb1+c follows that kb r ≤ u ≤ r < 1 and 0 < logb r ≤ logb u ≤ logb (kb r) for all t ∈ R. By Lemma 2.3, noticing that |c| = −c, we have X X (A−1 logb u)(t) = c2j logb u(t − 2τ j) − |c|2j−1 logb u(t − τ (2j − 1)) j≥0

j≥1

|c| 1 1 logb r − logb (kb r) = logb (kbc r1+c ), 2 2 1−c 1−c 1 − c2 1 |c| 1 (A−1 logb u)(t) ≤ logb (kb r) − logb r = logb (kb r1+c ). 2 2 2 1−c 1−c 1−c ≥

Set

Main results

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logb (kbc s1+c ) logb (kb s) logb (kb s1+c ) logb s , ν (s) = . , ψ (s) = , µ (s) = b b b 1 − c2 1 − c2 1−c 1−c

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ϕb (s) =

Theorem 3.1. Let (H1), (H2) hold and c ∈ (−1, 0]. Assume that there exist constants 0 < |c|

b < 1, 0 < r < R < kb1+c such that

sup{g(u) : ϕb (α) ≤ u ≤ ψb (α)} <

p˜ < inf{g(u) : ϕb (β) ≤ u ≤ ψb (β)} a ˜

where {α, β} = {r, R}. Then (1.1) has at least one positive ω-periodic solution. 3

(3.1)

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Proof. Without loss of generality, we assume that α = r, β = R. Let Ωr = {v ∈ Eω : kvk < ¯ R \ Ωr ) → K. Notice that r}, ΩR = {v ∈ Eω : kvk < R}. At first, we show that T : K ∩ (Ω ¯ r ≤ γ ≤ R if v ∈ K ∩ (ΩR \ Ωr ) with ||v|| = γ. From Lemma 2.4, we have (A−1 logb v)(t) ≥

1 1 logb (kbc γ 1+c ) ≥ logb (kbc R1+c ) > 0 2 2 1−c 1−c

(3.2)

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¯ R \ Ωr ) with ||v|| = γ. Hence, g((A−1 logb v)(t − δ(t))) is well-defined for for any v ∈ K ∩ (Ω ¯ R \ Ωr ). From 0 < mb ≤ Gb (t, s) ≤ Mb , we obtain that for v ∈ K ∩ (Ω ¯ R \ Ωr ), any v ∈ K ∩ (Ω Z t+ω ¡ ¢ 0 ≤ (T v)(t) ≤ Mb v(s)g (A−1 logb v)(s − δ(s)) |a(s) ln b|ds Zt ω ¡ ¢ = Mb v(s)g (A−1 logb v)(s − δ(s)) |a(s) ln b|ds < +∞, Z 0t+ω ¡ ¢ (T v)(t) ≥ mb v(s)g (A−1 logb v)(s − δ(s)) |a(s) ln b|ds Zt ω ¡ ¢ = mb v(s)g (A−1 logb v)(s − δ(s)) |a(s) ln b|ds ≥ kb kT vk. 0

On the other hand, Rs Z t+2ω ¡ ¢ exp t+w p(r)(ln b)dr Rω (T v)(t + ω) = a(s)v(s)g (A−1 logb v)(s − δ(s)) (ln b)ds exp 0 p(r)(ln b)dr − 1 t+ω R Z t+ω ξ+ω ¡ ¢ exp t+w p(r)(ln b)dr Rω = a(ξ + ω)v(ξ + ω)g (A−1 logb v)(ξ + ω − δ(ξ + ω)) (ln b)dξ exp 0 p(r)(ln b)dr − 1 t Rξ Z t+ω ¡ ¢ exp t p(r)(ln b)dr Rω = a(ξ)v(ξ)g (A−1 logb v)(ξ − δ(ξ)) (ln b)dξ = (T v)(t). exp 0 p(r)(ln b)dr − 1 t

¯ R \Ωr ) → K. In addition, one easily checks that T is completely continuous. Hence, T : K∩(Ω Next, we show that v 6= λT v for v ∈ K ∩ ∂Ωr and 0 < λ ≤ 1. If it is not true, there exist v ∈ K ∩ ∂Ωr and 0 < λ ≤ 1 with v = λT v. Hence, ¡ ¢ v 0 (t) + p(t)v ln b = λa(t)v(t)g (A−1 logb v)(t − δ(t)) ln b. (3.3) Since v(t) ≥ kb r > 0, we rewrite (3.3) as

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¡ ¢ (ln v(t))0 + p(t) ln b = λa(t)g (A−1 logb v)(t − δ(t)) ln b.

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Integrating (3.4) from 0 to ω, we obtain that Z ω ¡ ¢ p˜ = λ a(s)g (A−1 logb v)(s − δ(s)) ds, 0

which implies that

p˜ ≤ a ˜ sup{g(v) : ϕb (r) ≤ v ≤ ψb (r)} if a 0, p˜ ≥ a ˜ sup{g(v) : ϕb (r) ≤ v ≤ ψb (r)} if a  0. 4

(3.4)

(3.5)

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Hence, sup{g(v) : ϕb (r) ≤ v ≤ ψb (r)} ≥ p˜/˜ a. This is a contradiction. Finally, we show that inf kT vk > 0, T v 6= λAv for v ∈ K ∩ ∂ΩR , λ ≥ 1. Suppose that inf v∈K∩∂ΩR kT vk = 0. There exists {vn } ⊆ K ∩ ∂ΩR such that kT vn k → 0 as n → ∞. Noting that vn (t) ≥ kb R > 0, a 6≡ 0 and Z ω ¢ ¡ 0 ≤ mb vn (s)g (A−1 logb vn )(s − δ(s)) |a(s) ln b|ds ≤ kT vn k → 0, 0

¡

¢ we obtain that g (A−1 logb vn )(s − δ(s)) → 0 as n → ∞, which is impossible since ϕb (R) ≤ A−1 logb vn ≤ ψb (R), inf{g(u) : ϕb (R) ≤ u ≤ ψb (R)} >

p˜ > 0. a ˜

Suppose that there exist u ∈ K ∩ ∂ΩR and λ ≥ 1 with u = λT u. Then ¡ ¢ (ln u(t))0 + p(t) ln b = λa(t)g (A−1 logb u)(t − δ(t)) ln b.

(3.6)

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Integrating (3.6) from 0 to ω, we obtain Z ω ¡ ¢ p˜ = λ a(s)g (A−1 logb u)(s − δ(s)) ds, 0

p˜ ≥ a ˜ inf{g(u) : ϕb (R) ≤ u ≤ ψb (R)} if a 0, p˜ ≤ a ˜ inf{g(u) : ϕb (R) ≤ u ≤ ψb (R)} if a  0,

¯ R \Ωr ) with T v = v, which is which contradicts (3.1). By Lemma 2.2, there exists v ∈ K ∩ (Ω −1 the positive solution of (2.3) and A logb v is an ω-periodic solution of (1.1). The fact (3.2) implies that A−1 logb v > 0. The proof is completed. Similarly, if c ∈ [0, 1), we have the following result.

Theorem 3.2. Let (H1) − (H2) hold and c ∈ [0, 1). Assume that there exist constants 0 < b < 1, 0 < r < R < 1 such that sup{g(u) : µb (α) ≤ u ≤ νb (α)} <

p˜ < inf{g(u) : µb (β) ≤ u ≤ νb (β)}, a ˜

(3.7)

where {α, β} = {r, R}. Then (1.1) has at least one positive ω-periodic solution. Consider the equation

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(x(t) − cx(t − τ ))0 = a(t)xα (t − δ(t)) − p(t),

(3.8)

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where c, τ are constants, a, p, δ are ω-periodic continuous functions, a, p are of constant sign and a ˜ · p˜ > 0.

Corollary 3.1. (3.8) has at least a positive ω-periodic solution if one of the following conditions is satisfied: (C1) c ∈ (−1, 0], α > 0, a ˜ · |˜ p|α · p˜−1 < (1 + c)α , (C2) c ∈ [0, 1), α > 0, a ˜ · |˜ p|α · p˜−1 < (1 − c)α , (C3) c ∈ (−1, 0], α < 0, a ˜ · |˜ p|α · p˜−1 > (1 + c)α , α (C4) c ∈ [0, 1), α < 0, a ˜ · |˜ p| · p˜−1 > (1 − c)α . 5

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Proof. We only consider the case that (C1) is satisfied. In fact, g(u) = uα , kb = e|˜p| ln b . Let |c|

ε > 0 be sufficiently small, rε = ε, Rε = (1 − ε)kb1+c , then as ε → 0+ , µ ¶α µ ¶α 1−c 1+c 1 sup{g(u) : ϕb (Rε ) ≤ u ≤ ψb (Rε )} = |˜ p| + logb (1 − ε) → |˜ p| , 1 − c2 1 − c2 1+c µ ¶α 1 c 1+c inf{g(u) : ϕb (rε ) ≤ u ≤ ψb (rε )} = logb kb ε → +∞. 1 − c2 |c|

Thus there exists ε0 > 0 such that 0 < rε0 < Rε0 < kb1+c and sup{g(u) : ϕb (Rε0 ) ≤ u ≤ ψb (Rε0 )} <

p˜ < inf{g(u) : ϕb (rε0 ) ≤ u ≤ ψb (rε0 )}. a ˜

By Theorem 3.1,(3.8) has at least a positive ω-periodic solution.

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Remark 3.1. If |˜ a| < 1, the equation

x0 (t) = a(t)x(t − δ(t)) − p(t),

(3.9)

has at least a positive ω-periodic solution, where a, δ, p are ω-periodic continuous functions, a, p are of constant sign and a ˜ · p˜ > 0.

Acknowledgments

The authors wish to express their thanks to the referee for his/her very valuable suggestions and careful corrections. The work is supported by Hunan Provincial Natural Science Foundation of China.

References

[1] J.K. Hale, Theory of Functional Differential Equations, Springer-Verlag, New York, 1977.

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[2] W. Han, J. Ren, Some results on second-order neutral functional differential equations with infinite distributed delay, Nonlinear Anal. 70 (2009) 1393-1406. [3] Y. Li, Positive periodic solutions of periodic neutral Lotka-Volterra system with state dependent delays, J. Math. Anal. Appl. 330 (2007) 1347-1362.

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[4] Z. Yang, Z. Zhou, Periodic solutions of a class of neutral differential models with feedback control, Appl. Math. Comput. 189 (2007) 996-1009. [5] Y. Luo, W. Wang, J. Shen, Existence of positive periodic solutions for two kinds of neutral functional differential equations, Appl. Math. Lett. 21 (2008) 581-587. [6] T. Candan, Existence of positive periodic solutions of first order neutral diferential equations with variable coeffcients, Appl.Math.Lett. 52(2016)142-148. 6

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[7] W. Wang,Y. Luo, Periodic solutions for a class of second order differential equations, Electron. J. Qual. Theo. 41(2013)1-10. [8] D. Guo,V.Lakshmikantham, Nonlinear Problems in Abstract Cones. Academic Press, Orlando, FL(1988)

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[9] S. Lu, W. Ge, On the existence of periodic solutions for neutral functional differential equation, Nonlinear Anal. TMA. 54 (2003) 1285-1306.

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CRediT author statement Weibing

Wang:

Conceptualization,

Writing-Original

preparation, Formal analysis.

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Jianhua Shen: Writing- Reviewing and Editing

draft