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Positive periodic solutions for neutral multi-delay logarithmic population model Journal of Inequalities and Applications 2012, 2012:10 doi:10.1186/1029-242X-2012-10 Mei-Lan Tang csutmla

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Positive periodic solutions for neutral multi-delay logarithmic population model

Journal of Inequalities and Applications 2012, 2012:10 doi:10.1186/1029-242X-2012-10

Mei-Lan Tang (csutmlang@163.com) Xian-Hua Tang (tangxh@mail.csu.edu.cn)

ISSN 1029-242X

Article type Research

Publication date 16 January 2012

Article URL http://www.journalofinequalitiesandapplications.com/content/2012/1/10

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Journal of Inequalities and

Applications

Positive periodic solutions for neutral multi-delay

logarithmic population model

Mei-Lan Tang∗ and Xian-Hua Tang

School of Mathematical Science and Computing Technology,

Central South University, Changsha, 410083, China

∗Corresponding author: csutmlang@163.com

Email address:

XHT: tangxh@mail.csu.edu.cn

Abstract

Based on an abstract continuous theorem of k -set contractive operator and some analysis

skill, a new result is obtained for the existence of positive periodic solutions to a neutral multi-delay logarithmic population model Some sufficient conditions obtained in this article for the existence of positive periodic solutions to a neutral multi-delay logarithmic population model are easy to check Furthermore, our main result also weakens the condition in the existing results An example is used to illustrate the applicability of the main result.

Keywords: positive periodic solution; existence; k -set contractive operator; logarithmic

population model.

MSC 2010: 34C25; 34D40.

1 Introduction

In recent years, there has been considerable interest in the existence of periodic solutions

of functional differential equations (see, for example, [1–7]) It is well known that the environments of most natural populations change with time and that such changes induce variation in the growth characteristics of populations Among many population models, the neutral logarithmic population model has recently attracted the attention of many mathematicians and biologists

Let ω > 0 be a constant, C ω = {x : x ∈ C(R, R), x(t + ω) = x(t)}, with the norm defined by |x|0 = maxt∈[0,ω] |x(t)|, and C1

ω = {x : x ∈ C1(R, R), x(t + ω) = x(t)}, with the norm defined by kxk0 = max{|x|0, |x 0 |0}, then C ω , C1

¯h = 1

ω

Rω

Lu and Ge [8] studied the existence of positive periodic solutions for neutral logarithmic

population model with multiple delays Based on an abstract continuous theorem of k -set

contractive operator, Luo and Luo [9] investigate the following periodic neutral multi-delay logarithmic population model:

dN



r(t) −

n

X

j=1

a j (t) ln N(t − σ j (t)) −

m

X

i=1

b i (t) d

dt ln N(t − τ i (t))



where r(t), a j (t), b i (t), σ j (t), τ i (t) are all in C ω with ¯r > 0, σ j (t) > 0 and τ i (t) > 0,∀t ∈

[0, ω], ∀j ∈ {1, 2, , n}, ∀i ∈ {1, 2, , m} Furthermore, b i (t) ∈ C1(R, R), σ j (t) ∈

C1(R, R), τ i (t) ∈ C2(R, R) and σ 0

j (t) < 1, τ 0

i (t) < 1, ∀j ∈ {1, 2, , n}, ∀i ∈ {1, 2, , m} Since σ 0

j (t) < 1, ∀t ∈ [0, ω], t − σ j (t) has a unique inverse Let µ j (t) be the inverse of

For convenience, denote Γ(t) = Pn

j=1

a j (µ j (t)) 1−σ 0

j (µ j (t)) − Pm

i=1

b 0

i (γ i (t)) 1−τ 0

i (γ i (t)) Luo and Luo [9] obtain the following sufficient condition on the existence of positive periodic solutions for neutral logarithmic population model with multiple delays

Theorem A Assume the following conditions hold:

(H1 0 ) There exists a constant θ > 0 such that |Γ(t)| > θ, ∀t ∈ [0, ω].

(H2 0) Pn

j=1 |a j |0ω +Pm

i=1 |b i |0|1 − τ 0

i | 1/20 < 1 andPm

i=1 |b i |0|1 − τ 0

i |0 < 1.

Then Equation (1) has at least an ω-positive periodic solution.

The purpose of this article is to further consider the existence of positive periodic solutions to a neutral multi-delay logarithmic population model (1) We will present some new sufficient conditions for the existence of positive periodic solutions to a neutral

multi-delay logarithmic population model In this article, we will replace the assumption (H1 0):

|Γ(t)| > θ in Theorem A by different assumption Γ(t) > 0, ∀t ∈ [0, ω] (or Γ(t) < 0, ∀t ∈

[0, ω]) Obviously, it is more easy to check Γ(t) > 0, ∀t ∈ [0, ω], than to find a constant

A will be greatly weakened Pn

j=1 |a j |0ω +Pm

i=1 |b i |0|1 − τ 0

i | 1/20 < 1 in Theorem A is replaced

by 1

2

Pn

j=1 |a j |0ω +Pm

i=1 |b i |0|1 − τ 0

i | 1/20 < 1 in this article.

2 Main lemmas

Under the transformation N(t) = e x(t), then Equation (1) can be rewritten in the following form:

x 0 (t) = r(t) −

n

X

j=1

a j (t)x(t − σ j (t)) −

m

X

i=1

where c i (t) = b i (t)(1 − τ 0

i (t)), i = 1, 2, , m.

It is easy to see that in this case the existence of positive periodic solution of Equation (1) is equivalent to the existence of periodic solution of Equation (2) In order to investigate the existence of periodic solution of Equation (2), we need some definitions and lemmas

Definition 1 Let E be a Banach space, S ⊂ E be a bounded subset, denote

α E (S) = inf{δ > 0| there is a f inite number of subsets S i ⊂ S such that S =[

i

Definition 2 Let E1 and E2 be Banach spaces, D ⊂ E1, A : D → E2 be a continuous and

bounded set S ⊂ D, then A is called k-set contractive operator on D.

Definition 3 Let X, Y be normed vector spaces, L : DomL ⊂ X → Y be a linear mapping.

This mapping L will be called a Fredholm mapping of index 0 if dimKerL = codimImL <

∞ and ImL is closed in Y [3].

Assume that L : DomL ⊂ X → Y is a Fredholm operator with index 0, from [3], we know that sup{δ > 0|δα X (B) ≤ α Y (L(B))} exists for any bounded set B ⊂ DomL, so we

can define

Now let L : X → Y be a Fredholm operator with index 0, X and Y be Banach spaces,

Ω ⊂ X be an open and bounded set, and let N : ¯ Ω → Y be a k-set contractive operator with k < l(L) By using the homotopy invariance of k-set contractive operator’s topological degree D[(L, N ), Ω], Petryshyn and Yu [10] proved the following result.

Lemma 1 [10] Assume that L : X → Y is a Fredholm operator with index 0, r ∈ Y is

bounded, open, and symmetric about 0 ∈ Ω Furthermore, we also assume that

where[·, ·] is a bilinear form on Y × X, and Q is the projection of Y onto Coker, where

In the rest of this article, we set Y = C ω , X = C1

ω

Nx = −

n

X

j=1

a j (t)x(t − σ j (t)) −

m

X

i=1

then Equation (2) is equivalent to the equation

where r = r(t) Clearly, Equation (2) has an ω-periodic solution if and only if Equation (5) has a solution x ∈ C1

ω

Lemma 2 [7] The differential operator L is a Fredholm operator with index 0, and satisfies

l(L) ≥ 1.

Lemma 4 [8] Suppose τ ∈ C1

inverse µ(t) satisfying µ ∈ C(R, R)with µ(a + ω) = µ(a) + ω.

Lemma 5 [11] Let x(t) be continuous differentiable T -periodic function (T > 0) Then

max

t∈[t ∗ ,t ∗ +T ] |x(t)| ≤ |x(t ∗ )| + 1

2

T

Z

0

|x 0 (s)|ds.

3 Main results

Let µ j (t) be the inverse of t−σ j (t), γ j (t) be the inverse of t−τ i (t) and Γ(t) =Pn

j=1

a j (µ j (t)) 1−σ 0

j (µ j (t)) −

Pm

i=1

b 0

i (γ i (t))

1−τ 0

i (γ i (t))

Theorem 1 Assume the following conditions hold:

(H1) If Γ(t) > 0, ∀t ∈ [0, ω] (or Γ(t) < 0, ∀t ∈ [0, ω]);

2

Pn

j=1 |a j |0ω +Pm

i=1 |b i |0|1 − τ 0

i | 1/20 < 1 and Pm

i=1 |b i |0|1 − τ 0

i |0 < 1.

Then Equation (1) has at least an ω-positive periodic solution.

Proof Suppose that x(t) is an arbitrary ω -periodic solution of the following operator

equation

where L and N are defined by Equations (3) and (4), respectively Then x(t) satisfies

x 0 (t) = λ



r(t) −

n

X

j=1

a j (t)x(t − σ j (t)) −

m

X

i=1

c i (t)x 0 (t − τ i (t))



Integrating both sides of Equation (7) over [0, ω] gives

ω

Z

0



r(t) −

n

X

j=1

a j (t)x(t − σ j (t)) +

m

X

i=1

b 0

i (t)x(t − τ i (t))



i.e.,

ω

Z

0



Xn

j=1

a j (t)x(t − σ j (t)) −

m

X

i=1

b 0 i (t)x(t − τ i (t))



Let t − σ j (t) = s, i.e., t = µ j (s) Lemma 4 implies that

a j (µ j (s))

1 − σ 0

j (µ j (s)) ∈ C ω ,

a j (µ j (s))

1 − σ 0

j (µ j (s)) x(s) ∈ C ω .

Lemma 4 implies µ j (0 + ω) = µ j (0) + ω, γ i (0 + ω) = γ i (0) + ω, ∀j ∈ {1, , n}, i ∈

{1, , m}.

Noting that σ j (0) = σ j (ω), τ i (0) = τ i (ω), then

ω

Z

0

a j (µ j (s))

1 − σ 0

j (µ j (s)) ds =

ω−σZj (ω)

−σ j(0)

a j (µ j (s))

1 − σ 0

j (µ j (s)) ds =

ω

Z

0

a j (t)dt = ω¯a j , j = 1, , n, (10)

ω

Z

0

b 0

i (γ i (s))

1 − τ 0

i (γ i (s)) ds =

ω−τZi (ω)

−τ i(0)

b 0

i (γ i (s))

1 − τ 0

i (γ i (s)) ds =

ω

Z

0

b 0

i (t)dt = 0, i = 1, , m. (11)

Noting that Γ(t) > 0, we have

¯

ω

ω

Z

0

Γ(t)dt = 1

ω

ω

Z

0



Xn

j=1

a j (µ j (t))

1 − σ 0

j (µ j (t)) −

m

X

i=1

b 0

i (γ i (t))

1 − τ 0

i (γ i (t))



dt =

n

X

j=1

Furthermore

ω

Z

0

a j (t)x(t − σ j (t))dt =

ω−σZj (ω)

−σ j(0)

a j (µ j (s))

1 − σ 0

j (µ j (s)) x(s)ds

=

ω

Z

0

a j (µ j (s))

1 − σ 0

ω

Z

0

b 0

i (t)x(t − τ i (t))dt =

ω−τZi (ω)

−τ i(0)

b 0

i (γ i (s))

1 − τ 0

i (γ i (s)) x(s)ds

=

ω

Z

0

b 0

i (γ i (s))

1 − τ 0

Combining (13) and (14) with (9) yields

ω

Z

0

Since Γ(t) > 0, it follows from the extended integral mean value theorem that there exists η ∈ [0, ω] satisfying

x(η)

ω

Z

0

i.e.,

x(η) = r¯¯

By Lemma 5, we obtain

2

ω

Z

0

|x 0 (t)|dt.

So

|x|0 ≤ | r¯¯

Γ| +

1 2

ω

Z

0

Multiplying both sides of Equation (7) by x 0 (t) and integrating them over [0, ω], we

have

ω

Z

0

|x 0 (t)|2dt

=

ω

Z

0

x 0 (t)2dt

=

¯

¯

¯

ω

Z

0

x 0 (t)2dt

¯

¯

¯

= λ

¯

¯

¯

ω

Z

0

r(t)x 0 (t)dt −

ω

Z

0

n

X

j=1

a j (t)x(t − σ j (t))x 0 (t)dt −

ω

Z

0

m

X

i=1

c i (t)x 0 (t − τ i (t))x 0 (t)dt

¯

¯

¯

ω

Z

0

|x 0 (t)|dt +

n

X

j=1

|a j |0|x|0

ω

Z

0

|x 0 (t)|dt +

m

X

i=1

ω

Z

0

|c i (t)x 0 (t − τ i (t))x 0 (t)|dt.

(19)

Cauchy–Schwarz inequality implies

ω

Z

0

¯

¯

¯

ω

Z

0

r(t)x 0 (t)dt −

ω

Z

0

n

X

j=1

a j (t)x(t − σ j (t))x 0 (t)dt

−

ω

Z

0

m

X

i=1

c i (t)x 0 (t − τ i (t))x 0 (t)dt

¯

¯

¯

≤



|r|0+

n

X

j=1

|a j |0|x|0









ω

Z

0

|x 0 (t)|2dt





1/2

m

X

i=1





ω

Z

0

|c i (t)x 0 (t − τ i (t))|2dt





1/2



ω

Z

0

|x 0 (t)|2dt





1/2

.





ω

Z

0

|c i (t)x 0 (t − τ i (t))|2dt





1/2

=





ω

Z

0

1

1 − τ 0

i (γ i (t)) |c i (γ i (t))x

0 (t)|2dt





1/2

=





ω

Z

0

(1 − τ 0

i (γ i (t)))|b i (γ i (t))x 0 (t)|2dt





1/2

(21)

i | 1/20 |b i |0





ω

Z

0

|x 0 (t)|2dt





1/2

.

Substituting Equations (18)and (21) into (20) gives

ω

Z

0

|x 0 (t)|2dt ≤



|r|0+

n

X

j=1

|a j |0|x|0









ω

Z

0

|x 0 (t)|2dt





1/2

ω 1/2

+

m

X

i=1

|1 − τ 0

i | 1/20 |b i |0





ω

Z

0

|x 0 (t)|2dt





≤



|r|0+

n

X

j=1

|a j |0

¯

r

¯ Γ









ω

Z

0

|x 0 (t)|2dt





1/2



1

2ω

n

X

j=1

|a j |0+

m

X

i=1

|1 − τ i 0 | 1/20 |b i |0









ω

Z

0

|x 0 (t)|2dt



.

2ωPn j=1 |a j |0 +Pm

i=1 |1 − τ 0

i | 1/20 |b i |0 < 1, it follows from

Equa-tion (22) that there exists constant M > 0 such that





ω

Z

0

|x 0 (t)|2dt





1/2

Then

|x|0 ≤

¯

¯r¯

¯ Γ

¯

¯+1 2

ω

Z

0

|x 0 (t)|dt ≤

¯

¯r¯

¯ Γ

¯

¯+1

1/2 := M1. (24)

Again from (7), we get

|x 0 |0 ≤ |r|0+

n

X

j=1

|a j |0|x|0+

m

X

i=1

Condition Pm

i=1 |c i |0 ≤Pm

i=1 |b i |0|1 − τ 0

i |0 < 1 implies that

|x 0 |0 ≤ |r|0+

Pn

j=1 |a j |0M1

1 −Pm i=1 |c i |0 := M2. (26)

Let M3 > max{M1, M2, |¯ r/Pn

Pm

except (R2) hold Next, we prove that the condition (R2) of Lemma 1 is also satisfied We

define a bounded bilinear form [·, ·] on C ω × C1

ω as follows:

[y, x] =

ω

Z

0

Define Q : Y → CokerL by

ω

ω

Z

0

y(t)dt.

Obviously,

n

Without loss of generality, we may assume that x = M3 Thus

[QN(x) + Qr, x][QN(−x) + Qr, x]

= M32





ω

Z

0

n

X

j=1

ω

Z

0

a j (t)dt









ω

Z

0

n

X

j=1

ω

Z

0

a j (t)dt



= ω2M32



r − M¯ 3

n

X

j=1

¯a j







r + M¯ 3

n

X

j=1

¯a j





< 0.

Therefore, by Lemma 1, Equation (1) has at least an ω-positive periodic solution.

Since |1−τ 0

i |0 < 1, then |1−τ 0

i |0 < |1−τ 0

i | 1/20 SoPm i=1 |b i |0|1−τ 0

i |0 <Pm i=1 |b i |0|1−τ 0

i | 1/20 From Theorem 1, we have

Corollary 1 Assume that the following conditions hold

(H1 0 ) If Γ(t) > 0, ∀t ∈ [0, ω] (or Γ(t) < 0, ∀t ∈ [0, ω]).

(H2 0) 1

2

Pn

j=1 |a j |0ω +Pm

i=1 |b i |0|1 − τ 0

i | 1/20 < 1 and |1 − τ 0

i |0 < 1, i = 1, , m.

Then Equation (1) has at least an ω-positive periodic solution.

4 Example

Example 1 is given to illustrate the effectiveness of our new sufficient conditions, also to demonstrate the difference between the proposed result in this paper and the result in [9]

Example 1 Consider the following equation:

dN

"

8(cos

64(3 − cos t)

d

dt ln N(t − π)

#

(29)

32(cos2t + 1) sin t − 1

64(3 − cos t) cos t.

8(cos2t + 1),

b1(t) = 1

64(3 − cos t), σ1(t) = τ1(t) = π So ¯ r = 1

128 > 0, σ 0

1(t) = τ 0

1(t) = 0, µ1(t) = γ1(t) =

π + t Thus

Γ(t) = a1(µ1(t)) − b 0

1(γ1(t)) = 1

8(cos

64sin t > 0,

1

2|a1|0ω + |b1|0|1 − τ

0

i | 1/20 = 4π + 1

The conditions in Theorem 1 in this article are satisfied Hence Equation (29) has at least

2) in Theorem A(Theorem 3.1

in [9]) is not satisfied Since

|a1|0ω + |b1|0|1 − τ 0

i | 1/20 = 8π + 1

(H 0

1) in Theorem A (Theorem 3.1 in [9]) is satisfied, it is more complex to check the

condition |Γ(t)| > θ, ∀t ∈ [0, ω] in Theorem A than to test Γ(t) > 0, ∀t ∈ [0, ω] This

example illustrates the advantages of the proposed results in this paper over the existing ones.

Competing interests

The authors declare that they have no competing interests

Authors’ contributions

All authors contributed equally to the manuscript and read and approved the final manuscript

Acknowledgments

The authors are grateful to the referees for their valuable comments which have led to improvement of the presentation This study was partly supported by the Zhong Nan

Da Xue Qian Yan Yan Jiu Ji Hua under grant No 2010QZZD015, Hunan Scientific Plan under grant No 2011FJ6037, NSFC under grant No 61070190 and NFSS under grant 10BJL020

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