Existence of positive periodic solutions for a neutral population model with delays and impulse

Existence of positive periodic solutions for a neutral population model with delays and impulse

Nonlinear Analysis 69 (2008) 3919–3930 www.elsevier.com/locate/na Existence of positive periodic solutions for a neutral population model with delays...

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Nonlinear Analysis 69 (2008) 3919–3930 www.elsevier.com/locate/na

Existence of positive periodic solutions for a neutral population model with delays and impulseI Qi Wang ∗ , Binxiang Dai School of Mathematical Science and Computing Technology, Central South University, Changsha, 410075, PR China Received 19 January 2007; accepted 18 October 2007

Abstract In this paper, by using the theory of abstract continuous theorem of k-set contractive operator and some analysis techniques, we obtain some sufficient conditions for the existence of positive periodic solutions for a neutral population model with delays and impulse. c 2007 Elsevier Ltd. All rights reserved.

MSC: 34C25; 34A39 Keywords: Neutral delay population model; Impulse; k-set contractive operator; Positive periodic solution

1. Introduction In 1993, Kuang in [1] proposed an open problem (Open problem 9.2) to obtain sufficient conditions for the existence of a positive periodic solution of the following equation dN = N (t)[a(t) − β(t)N (t) − b(t)N (t − τ (t)) − c(t)N 0 (t − τ (t))], dt

(1.1)

where a(t), β(t), b(t), c(t), τ (t) are all in Cω with τ (t) ≥ 0, a(t) ≥ 0, t ∈ [0, ω]. In [2], Fang and Li studied (1.1) and gave an answer to the open problem 9.2 of [1]. But paper [2] required that c(t) β(t) ≥ 0, b(t) ≥ 0, c(t) ≥ 0, t ∈ [0, ω] and c00 (t) < b(t) or c00 (t) ≤ b(t), β(t) > 0, t ∈ [0, ω], where c0 (t) = 1−τ 0 (t) . By using the continuation theorem, Yang and Cao [3] studied the following equation " # n n X X dN 0 = N (t) a(t) − β(t)N (t) − bi (t)N (t − τi (t)) − ci (t)N (t − γi (t)) , (1.2) dt i=1 i=1

I Research supported by the National Natural Science Foundation of China (10471153) and the Hunan Provincial Natural Science Foundation of China (06JJ20006). ∗ Corresponding author. E-mail addresses: [email protected] (Q. Wang), [email protected] (B. Dai).

c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.10.033

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where a(t), β(t), bi (t), σi (t), τi (t), γi (t), i ∈ {1, 2, . . . , n} are all T -periodic functions. But the verification of the assumption that the operator N : Ω¯ → Ω is L-compact is difficult and incomplete, for the concrete example the reader can refer to [4]. In [4], Lu and Ge used a different method from that followed by the above authors and considered the following equation with finite delays " # n m X X dN = N (t) a(t) − β(t)N (t) − b j (t)N (t − σ j (t)) − ci (t)N 0 (t − τi (t)) . (1.3) dt j=1 i=1 They established some criteria to guarantee the existence of positive periodic solutions of Eq. (1.3). On the other side, birth of many species is an annual birth pulse or harvesting. To have a more accurate description of many mathematical ecology systems, we need to consider the use of impulsive differential equations (see for example, the monographs of Samoilenko and Perestyuk [5], Laksmikantham et al. [6], Zavalishchin [7], Liu [8], Nieto et al. [9,10]). Some qualitative properties such as oscillation, periodicity, asymptotic behavior and stability properties have been investigated extensively by many authors over the past few years (see for example, the monographs of Nieto et al. [11–13] and others [14–27]). However there are few published papers discussing the impulsive neutral differential system. Our method is different from that in [28,29] In this paper, we investigate the following periodic neutral population model with delays and impulse " #  n m   dN = N (t) a(t) − e(t)N (t) − X b (t)N (t − σ (t)) − X c (t)N 0 (t − τ (t)) , t 6= t , j j i i k (1.4) dt j=1 i=1   + − N (tk ) = (1 + θk )N (tk ), t = tk with initial conditions: N (t) = ϕ(t),

N 0 (t) = ϕ 0 (t), −σ ≤ t ≤ 0, \ ϕ ∈ C([−σ, 0], [0, ∞)) C 1 ([−σ, 0], [0, ∞)),

ϕ(0) > 0,

(1.5)

where a(t), e(t), b j (t), ci (t), σ j (t), τi (t) are all in PCω with σ j (t) ≥ 0, τi (t) ≥ 0, t ∈ [0, ω], 1 + θk > 0, σ = max0≤t≤ω {max{σ j (t), τi (t)}}, i = 1, 2, . . . , n, j = 1, 2, . . . , m. Furthermore, σ j0 (t) < 1, t ∈ [0, ω], ci ∈ C 1 (R, R) and τi (t) ∈ C 2 (R, R) with τi (t) < 1, t ∈ [0, ω], i ∈ {1, 2, . . . , m}, j ∈ {1, 2, . . . , n}. For the ecological justification of (1.4) and the similar types refer to [1,30–32]. Our aim is, by using the abstract continuation theorem for the k-set contract operator, to establish some criteria to guarantee the existence of positive periodic solutions of system (1.4) and (1.5). Our results show that under appropriate periodic impulse perturbations, the impulsive system neutral delay population model (1.4) and (1.5) preserves the original periodicity of the neutral system without impulse. On the other hand, by using the theory of the abstract continuous theorem of the k-set contractive operator, we can avoid the difficulty in verifying the conditions of the continuation theorem [3,33]. We also extend the known results in [4]. Throughout this paper, we make the following notation and assumptions: Let ω > 0 be a constant and Cω = {x | x ∈ C(R, R), x(t + ω) ≡ x(t)} Cω1

with the norm defined by |x|0 = max |x(t)|; t∈[0,ω]

= {x | x ∈ C (R, R), x(t + ω) ≡ x(t)} 1

with the norm defined by kxk = max {|x|0 , |x 0 |0 }; t∈[0,ω]

PC = {x | x : R → R + , lim x(s) = x(t), if t 6= tk ; lim x(t) = x(tk ), lim x(t)exists , k ∈ Z + }; s→t

t→tk−

t→tk+

PC 1 = {x | x : R → R+ , x 0 ∈ PC}; PCω = {x | x ∈ PC, x(t + ω) ≡ x(t)} PCω1

= {x | x ∈ PC , x(t + ω) ≡ x(t)} 1

with the norm defined by |x|0 = max {|x(t)|}; t∈[0,ω]

with the norm defined by kxk = max {|x|0 , |x 0 |0 }. t∈[0,ω]

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Then those spaces are all Banach spaces. We also denote Z ω Z ω ¯h = 1 h(s)ds, |h|1 = |h(s)|ds, for any h ∈ PCω , 4k = 1 + θk , k = 1, 2, . . . , ω 0 0 and make the following assumptions: (H1) (H2) (H3) (H4)

0 < t1 < t2 < · · · , are fixed points with limk→∞ tk = +∞; {θk } is a real sequence with 4k = 1 + θk > 0; Π0
Definition 1.1. A function N (t) ∈ C([−σ, 0], [0, ∞)) is said to be a solution of the initial value problem (1.4) and (1.5) on [−σ, ∞) if (a) N (t) is absolutely continuous on each interval (tk−1 , tk ], k = 1, 2, . . . ; (b) for any tk , k = 1, 2, . . . , N (tk+ ) and N (tk− ) exist and N (tk− ) = N (tk ); (c) N (t) satisfies (1.4) and (1.5) for almost everywhere (for short a.e.) in [0, ∞]\{tk } and satisfies N (tk+ ) = (1 + θk )N (tk ) = 4k N (tk ) for t = tk , k ∈ Z + = {1, 2, . . .}. Under the assumptions (H1)–(H4), we consider the following system: " # n m X X dy 0 = y(t) a(t) − E(t)y(t) − B j (t)y(t − σ j (t)) − Ci (t)y (t − τi (t)) dt j=1 i=1

(1.6)

with initial conditions: y(t) = ϕ(t),

y 0 (t) = ϕ 0 (t), −σ ≤ t ≤ 0, ϕ(0) > 0, \ ϕ ∈ C([−σ, 0], [0, ∞)) C 1 ([−σ, 0], [0, ∞)),

(1.7)

where Y

B j (t) = b j (t)

4k ,

E(t) = β(t)

Y

4k ,

Ci (t) = ci (t)

0
0
Y

4k .

(1.8)

0
By a solution of (1.6) and (1.7), it means an absolutely continuous function y(t) defined on [−σ, ∞) that satisfies (1.6) a.e., for t ≥ 0, and y(t) = ϕ(t), y 0 (t) = ϕ 0 (t) on [−σ, 0]. The proof of Lemma 1.1 is similar to that of Theorem 1 in [25]. Lemma 1.1. Assume that that (H1)–(H4) hold, then Q y(t) = y(t, 0, ϕ) is a solution of (1.6) and (1.7) on [−σ, ∞), then N (t) = 0
j=1

i=1

 =

Y 0


m X i=1

Y

4k y 0 (t) −

4k y(t) a(t) − e(t)

0
0
 ci (t)

Y 0
Y

4k y 0 (t − τi (t))

4k y(t) −

n X j=1

b j (t)

Y 0
4k y(t − σ j (t))

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( Y

=

" y (t) − y(t) a(t) − E(t)y(t) − 0

4k

0
B j (t)y(t − σ j (t)) −

j=1

On the other hand, for any t = tk , k = 1, 2, . . . , Y Y 4k y(t) = 4k y(tk ), N (tk+ ) = lim t→tk+

n X

0≤t j
m X

#) Ci (t)y (t − τi (t)) 0

= 0. (1.9)

i=1

and

N (tk ) =

Y

4k y(tk ).

(1.10)

0≤t j
0≤t j ≤tk

Thus, N (tk+ ) = 4k N (tk ),

k = 1, 2, . . . .

(1.11)

It follows from (1.9)–(1.11) that N (t) is a solution of (1.4) and (1.5). (2) Since N (t) is absolutely continuous on each interval (tk , tk+1 ], k = 0, 1, 2, . . . , and in view of (1.11), it follows that for any k = 1, 2, . . . , Y Y + 4−1 N (t ) = 4−1 and y(tk+ ) = j k j N (tk ) = y(tk ), 0≤t j
0≤t j ≤tk

y(tk− )

=

Y 0≤t j
− 4−1 j N (tk )

Y

=

− 4−1 j N (tk ) = y(tk )

(1.12)

0≤t j ≤tk−1

which implies that y(t) is continuous on [−σ, ∞). It Q is easy to prove that y(t) is absolutely continuous on [−σ, ∞). Similar to the proof of (1), we can check that y(t) = 0
n X

B j (t)ex(t−σ j (t)) −

j=1

m X

Ci (t)x 0 (t − τi (t))ex(t−τi (t)) .

(1.13)

i=1

In the following section, we only discuss the existence of a periodic solution for (1.13). We organize this paper as follows: in Section 2, we present the main lemmas; in Section 3, we present the main theorem; in Section 4, we present an example. 2. Main lemmas In order to study Eq. (1.13), we make some preparations. Let E be a Banach space. For a bounded subset A ⊂ E, ( ) [ α E (A) = inf δ > 0 | There is a finite number of subsets Ai ⊂ A such that A = Ai and diam(Ai ) ≤ δ i

denotes the Kuratoskii measure of non-compactness, where diam(Ai ) denotes the diameter of set Ai . Let X, Y be two Banach spaces and Ω be a bounded open subset of X . A continuous and bounded map N : Ω¯ → Y is called k-set contractive if for any bounded set A ⊂ Ω , we have αY (N (A)) ≤ kα X (A), where k is a non-negative constant. For a Fredholm operator L : X → Y with index zero, according to [33,34], we define l(L) = sup{r ≥ 0|r α X (A) ≤ αY (L(A)), for all bounded subset A ⊂ X }. Lemma 2.1 ([33,34]). Let L : X → Y be a Fredholm operator with zero index, and a ∈ Y be a fixed point. Suppose that N : Ω → Y is called a k-set contractive with k < l(L), where Ω ⊂ X is bounded, open and symmetric about 0 ∈ Ω . Further, we also assume that (1) L x 6= λN x + λa, for x ∈ ∂Ω , λ ∈ (0, 1), and

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(2) [Q N (x) + Qa, x] × [Q N (−x) + Qa, x] < 0, for x ∈ ker L

T

∂Ω ;

where [·, ·] is a bilinear form on Y × X and Q is the projection of Y onto Coker (L), where Coker (L) is the cokernel of the operator L. Then there is an x ∈ Ω¯ such that L x − N x = a. In order to use Lemma 2.1 to study Eq. (1.13), we set Y = Cω , X = Cω1 , L : Cω1 → Cω ,

Lx =

dx , dt

(2.1)

and N : Cω1 → Cω defined by N x = −E(t)ex(t) −

n X

B j (t)ex(t−σ j (t)) −

j=1

m X

Ci (t)x 0 (t − τi (t))ex(t−τi (t)) .

(2.2)

i=1

It is easy to see from [35] that L is a Fredholm operator with index zero. Thus Eq. (1.13) has a positive ω-periodic solution if and only if L x = N x + a for some x ∈ Cω1 , where a = a(t). Lemma 2.2 ([36]). The differential operator L is a Fredholm operator with index zero, and satisfies l(L) ≥ 1. Pm Lemma 2.3. Let r0 , r1 be two positive constants, and Ω = {x|x ∈ Cω1 , |x|0 < r0 , |x 0 |0 < r1 }. If k = ( i=1 |ci |0 )er0 , then N : Ω → Cω is a k-set contractive map. Proof. Let A ⊂ Ω¯ S be a bounded subset and let η = αCω (A). Then for any ε > 0, there is a finite family of subsets of Ai satisfying A = Ai with dim(Ai ) ≤ η + ε. Now let g(t, x(t), y1 , . . . , yn , z 1 , . . . , z m , w1 , . . . , wm ) = E(t)ex(t) +

n X

B j (t)e y j +

j=1

m X

Ci (t)wi ezi .

i=1

Since g(t, x(t), y1 , . . . , yn , z 1 , . . . , z m , w1 , . . . , wm ) is uniformly continuous on any compact subset of R×SR 2m+n+1 , A and Ai are pre-compact in Cω , it follows that there is a finite family of subsets Ai j of Ai such that Ai = Ai j with |g(t, x(t), x(t − σ1 (t)), . . . , x(t − σn (t)), x(t − τ1 (t)), . . . , x(t − τm (t)), u 0 (t − τ1 (t)), . . . , u 0 (t − τm (t))) − g(t, u(t), u(t − σ1 (t)), . . . , u(t − σn (t)), u(t − τ1 (t)), . . . , u(t − τm (t)), u 0 (t − τ1 (t)), . . . , u 0 (t − τm (t)))| ≤ ε, for any x, u ∈ Ai j . Therefore, we have |N x − N u|0 = sup |g(t, x(t), x(t − σ1 (t)), . . . , x(t − σn (t)), x(t − τ1 (t)), . . . , x(t − τm (t)), t∈[0,ω] 0

x (t − τ1 (t)), . . . , x 0 (t − τm (t))) − g(t, u(t), u(t − σ1 (t)), . . . , u(t − σn (t)),

u(t − τ1 (t)), . . . , u(t − τm (t)), u 0 (t − τ1 (t)), . . . , u 0 (t − τm (t)))| ≤ sup |g(t, x(t), x(t − σ1 (t)), . . . , x(t − σn (t)), x(t − τ1 (t)), . . . , x(t − τm (t)), t∈[0,ω] 0

x (t − τ1 (t)), . . . , x 0 (t − τm (t))) − g(t, x(t), x(t − σ1 (t)), . . . , x(t − σn (t)),

x(t − τ1 (t)), . . . , x(t − τm (t)), u 0 (t − τ1 (t)), . . . , u 0 (t − τm (t)))| + sup |g(t, x(t), x(t − σ1 (t)), . . . , x(t − σn (t)), x(t − τ1 (t)), . . . , t∈[0,ω]

x(t − τm (t)), u 0 (t − τ1 (t)), . . . , u 0 (t − τm (t))) − g(t, u(t), u(t − σ1 (t)), . . . , u(t − σn (t)), u(t − τ1 (t)), . . . , u(t − τm (t)), u 0 (t − τ1 (t)), . . . , u 0 (t − τm (t)))| m X ≤ |ci |0 |x 0 (t − τi (t)) − u 0 (t − τi (t))|er0 + ε i=1

≤ k(η + ε) + ε.

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As ε is arbitrarily small, it is easy to obtain αCω (N (A)) ≤ kαCω0 (A), 

thus we prove this lemma.

Lemma 2.4. Suppose τ ∈ Cω1 and τ 0 (t) < 1, t ∈ [0, ω]. Then the function t −τ (t) has a unique inverse µ(t) satisfying µ ∈ C(R, R) with µ(a + ω) = µ(a) + ω, ∀a ∈ R and if g ∈ PCω , τ 0 (t) < 1, t ∈ [0, ω], then g(µ(t)) ∈ PCω . Proof. Since τ 0 (t) < 1, t ∈ [0, ω] and t − τ (t) is continuous on R, it follows that t − τ (t) has a unique inverse function µ(t) ∈ C(R, R) on R. For any a ∈ R, by the condition τ 0 (t) < 1, one can find that there is a unique solution t¯ for the equation t − τ (t) = a; also there is a unique solution t for the equation t − τ (t) = a + ω, thus µ(a) = t¯ = a + τ (t¯ ),

µ(a + ω) = t.

As ω + a + τ (t¯ ) − τ (ω + a + τ (t¯ )) = ω + a + τ (t¯ ) − τ (a + τ (t¯ )) = ω + a + τ (t¯ ) − τ (t¯ ) = ω + a. It follows that t = ω + a + τ (t¯ ). Since µ(a + ω) = t, we have µ(a + ω) = t = ω + a + τ (t¯ ) and µ(a + ω) = t = µ(a) + ω. We can easily obtain that if g ∈ PCω , τ 0 (t) < 1, t ∈ [0, ω], then g(µ(t + ω)) = g(µ(t) + ω) = g(µ(t)), t ∈ R, where µ(t) is the unique inverse function of t − τ (t), which together with µ(t) ∈ C(R, R) implies that g(µ(t)) ∈ PCω .  Lemma 2.5 ([4]). Let 0 ≤ α ≤ ω be a constant, s ∈ Cω such that maxt∈[0,ω] |s(t)| ≤ α. Then for ∀x ∈ Cω1 , we have Z ω Z ω |x(t) − x(t − s(t))|2 dt ≤ 2α 2 |x 0 (t)|2 dt. (2.3) 0

0

And further, suppose s ∈ Cω1 and s 0 (t) < 1, t ∈ R. Then for x ∈ Cω1 , we have the following conclusions: (1) There R ω exists a unique integer m such R ω that l = |s(t) − mω|0 < ω; (2) 0 |x(t) − x(t − s(t))|2 dt ≤ 2l 2 0 |x 0 (t)|2 dt. As σ j0 (t) < 1, t ∈ [0, ω], from Lemma 2.5, we can choose an integer m j (σ j (t)), j ∈ {1, 2, . . . , n} such that l j = |σ j − m j ω|0 ,

(2.4)

and ω

Z

|x(t) − x(t − σ j (t))| dt ≤ 2

0

2l 2j

Z 0

ω

|x 0 (t)|2 dt,

x ∈ Cω1 .

(2.5)

3. Main theorem Since τi0 (t) < 1, σ j0 (t) < 1, t ∈ [0, ω], we see thatτi (t), σ j (t) all have their inverse functions. Throughout the following part, we set γi (t), µ j (t) to represent the inverse function of t − τi (t), t − σ j (t) respectively. We also denote Γ (t) = E(t) +

n X j=1

where C0,i (t) =

Ci (t) ,i 1−τi0 (t)

0 (γ (t)) m X C0,i B j (µ j (t)) i − , 0 0 1 − σ j (µ j (t)) i=1 1 − τi (γi (t))

(3.1)

= 1, 2, . . . , m. The proof of the following theorem is similar to Theorem 3.1 of [4].

Theorem 3.1. Suppose that the following conditions hold. P (L1 ) a¯ > 0, Γ (t) ≥ 0, t ∈ [0, ω] and E¯ + nj=1 B¯ j > 0, where E¯ = defined by (3.1).

R 1 ω ω 0

E(t)dt, B¯ j =

R 1 ω ω 0

B j (t)dt, Γ (t) is

Q. Wang, B. Dai / Nonlinear Analysis 69 (2008) 3919–3930

3925

(L2 ) There exists a constant r0 > 0 such that ( ) n m m X X √ X |Ci |0 , 2 max |B j |0l j + |C0,i |0 er0 < 1 i=1

j=1

i=1

and "

   r0 > H + ω   

r0 |a 0 |1

+ er0

|E 0 |

1

j=1

1 2

¯ E+

Pa¯n

j=1

 21

# |B 0j |1

n m √ P P 2 |B j |0l j + |C0,i |0

1 − er0

j=1

where H = | ln

+

n P

  !  , 

i=1

|. Then the Eq. (1.4) has at least one positive ω-periodic solution, where l j is defined

B¯ j

by (2.4). Proof. Let u(t) be an arbitrary ω-periodic solution of the operator equation as follows Lu = λN u + λa,

λ ∈ (0, 1),

where N , L are defined by (2.1) and (2.2) respectively. Then " # n m X X u(t−τi (t)) 0 u(t) u(t−σ j (t)) 0 u (t) = λ a(t) − E(t)e − B j (t)e − Ci (t)u (t − τi (t))e . j=1

Integrating both sides of (3.2) over [0, ω], we have # Z ω" n m X X u(t) u(t−σ j (t)) 0 u(t−τi (t)) aω ¯ = E(t)e + B j (t)e + Ci (t)u (t − τi (t))e dt. 0

j=1

−σ j (0)

By Lemma 2.4, we have Z Z ω B j (t)eu(t−σ j (t)) dt = 0

ω

0

Similarly, we have Z ω Z u(t−τi (t)) 0 C0,i (t)e dt =

0

0

(3.3)

i=1

Let t − σ j (t) = s, then t = µ j (s) and Z ω Z ω−σ j (ω) B j (t)eu(t−σ j (t)) dt = 0

(3.2)

i=1

B j (µ j (s)) u(s) e ds. 1 − σ j0 (µ j (s))

B j (µ j (s)) u(s) e ds. 1 − σ j0 (µ j (s)) 0 (γ (s)) C0,i i

ω

1 − τi0 (γi (s))

eu(s) ds.

So from (3.3), we obtain Z ω Γ (s)eu(s) ds = aω. ¯

(3.4)

0

Considering assumption (L1 ), we have Γ (t) ≥ 0, t ∈ [0, ω], then it follows from the integro mean value theorem that there is a ζ ∈ [0, ω] such that Z ω eu(ζ ) Γ (s)ds = aω, ¯ 0

i.e.

eu(ζ )

=

a¯ Γ¯

thus u(ζ ) = ln ω

Z |u|0 ≤ H + 0

a¯ Γ¯

|u 0 (s)|ds.

= ln

¯ E+

Pa¯n

j=1

B¯ j

. Let H = | ln

¯ E+

Pa¯n

j=1

B¯ j

|, then (3.5)

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Q. Wang, B. Dai / Nonlinear Analysis 69 (2008) 3919–3930

1 , |x| < r , |x 0 | < r }, where r is defined by (L ) and r is a constant satisfying the condition Let Ω = {x|x ∈ C 0 0 0 1 0 2 1 Pω

r1 >

|a|0 +|E|0 er0 + nj=1 |B j |0 er0 Pm . 1− i=1 |Ci |0 er0

Lu 6= λN u + λa,

In what follows, we will prove that

λ ∈ (0, 1), and for all u ∈ ∂Ω .

Consider the situation contrary to the above one. Then there must exist a λ ∈ (0, 1) and a u ∈ ∂Ω such that Lu = λN u + λa, thus " # n m X X u(t−τi (t)) 0 u(t) u(t−σ j (t)) 0 u (t) = λ a(t) − E(t)e − B j (t)e − Ci (t)u (t − τi (t))e . (3.6) j=1

i=1

In view of u ∈ ∂Ω , we see either |u|0 = r0 and ≤ r1 , or |u 0 |0 = r1 and |u|0 ≤ r0 . Case 1. If |u|0 = r0 , then multiplying both sides of (3.6) by u 0 (t) and integrating over t ∈ [0, ω], we get Z " Z ω Z ω n Z ω ω X 0 2 a(t)u 0 (t)dt − E(t)eu(t) u 0 (t)dt − B j (t)eu(t−σ j (t)) u 0 (t)dt |u (t)| dt = λ 0 0 0 j=1 0 # Z m X ω − Ci (t)u 0 (t − τi (t))eu(t−τi (t)) u 0 (t)dt 0 i=1 Z " Z ω n Z ω ω X a(t)u 0 (t)dt − E(t)eu(t) u 0 (t)dt − B j (t)eu(t−σ j (t)) u 0 (t)dt ≤ 0 0 0 j=1 # Z m ω X 0 u(t−τi (t)) 0 − Ci (t)u (t − τi (t))e u (t)dt 0 i=1 Z " Z Z n ω ω ω X = −a 0 (t)u(t)dt + E 0 (t)eu(t) dt + B 0j (t)eu(t) dt 0 0 0 j=1 |u 0 |0

+

ω

n Z X

m Z X

B j (t)(eu(t) − eu(t−σ j (t)) )u 0 (t)dt −

j=1 0

≤ |u|0 |a 0 |1 + e|u|0 |E 0 |1 +

n X

# |B 0j |1 +

j=1

+

m Z X i=1

ω

0

i=1

"

ω

n Z X

ω

# Ci (t)u 0 (t − τi (t))eu(t−τi (t)) u 0 (t)dt

|B j (t)(eu(t) − eu(t−σ j (t)) )u 0 (t)|dt

j=1 0

|Ci (t)u 0 (t − τi (t))eu(t−τi (t)) u 0 (t)|dt.

(3.7)

0

As |e

u(t)

u(t−σ j (t))

−e

Z u(t) s |= e ds = eξ |u(t) − u(t − σ j (t))|, u(t−σ j (t))

where ξ is between u(t − σ j (t)) and u(t), it follows that Z u(t) u(t) u(t−σ j (t)) s |e −e |= e ds ≤ e|u|0 |u(t) − u(t − σ j (t))|, u(t−σ j (t)) which shows that Z n Z ω n X X u(t) u(t−σ j (t)) 0 |u|0 |B j (t)(e −e )u (t)|dt ≤ e |B j |0 j=1 0 |u|0

≤e

ω

j=1

0

n X

Z

j=1

|(u(t) − u(t − σ j (t)))u 0 (t)|dt ω

|B j |0 0

|u (t)| dt 0

2

 1 Z 2

0

ω

|u(t) − u(t − σ j (t))|

2

1 2

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Q. Wang, B. Dai / Nonlinear Analysis 69 (2008) 3919–3930

so by (2.5), we have n Z X

ω

|B j (t)(eu(t) − eu(t−σ j (t)) )u 0 (t)|dt ≤



2e|u|0

n X

j=1 0

Similarly, we get m Z m Z ω X X 0 u(t−τi (t)) 0 |u|0 |Ci (t)u (t − τi (t))e u (t)|dt ≤ e i=1

≤e

ω

m Z X

m X

|Ci (t)u 0 (t − τi (t))u 0 (t)|dt ω

|u (t)| dt 2

0

1 2

ω

Z

0

i=1

≤ e|u|0

0 ω

Z |C0,i |0

j=1

+e

m X

ω

Z

j=1

|u 0 (t)|2 dt.

|C0,i |0

√ P Pm In view of e|u|0 ( 2 nj=1 |B j |0l j + i=1 |C0,i |0 ) < 1, one can get from (3.10) that ! n P 0 0 r |B j |1 r0 |a|1 + e 0 |E |1 + Z ω j=1 0 2 !. |u (t)| dt ≤ m n √ P 0 P |C0,i |0 |B j |0l j + 1 − er0 2 i=1

j=1

Thus by (3.5), it follows that 1

r0 = |u|0 ≤ H + ω 2

ω

Z

|u 0 (t)|2 dt

1

2

0

   ≤ H +ω   

+ er0

r0 |a 0 |1

|E 0 |1

+

n P j=1

1 2

1 − er0

! |B 0j |1

n m √ P P 2 |B j |0l j + |C0,i |0

 21   !  

i=1

j=1

< r0 , which is a contradiction. Case 2. If |u 0 |0 = r1 and |u|0 ≤ r0 , then by (3.6) we have |u 0 |0 ≤ |a|0 + |E|0 e|u|0 +

n X

|B j |0 e|u|0 +

j=1

≤ |a|0 + |E|0 er0 +

n X j=1

|B j |0 er0 +

m X

|Ci |0 |u 0 |0 e|u|0

i=1 m X i=1

i

(3.9)

ω

|u 0 (t)|2 dt

0

(3.10)

0

i=1

i

|u 0 (t)|2 dt.

Substituting (3.8) and (3.9) into (3.7) and considering |u|0 = r0 , then we have ! Z ω Z n n X √ |u| X 0 2 0 |u|0 0 0 0 |u (t)| dt ≤ |u|0 |a |1 + e |E |1 + |B j |1 + 2e |B j |0l j |u|0

! 12 Ci (γi (s)) 0 2 1 − τ 0 (γ (s)) u (t) dt

0

i=1

0

(3.8)

0

i=1 |u|0

|u 0 (t)|2 dt.

0

j=1

0

ω

Z |B j |0l j

|Ci |0 |u 0 |0 er0 .

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Q. Wang, B. Dai / Nonlinear Analysis 69 (2008) 3919–3930

Considering the assumption ) ( m m n X X √ X |C0,i |0 er0 < 1, |Ci |0 , 2 |B j |0l j + max i=1

we get

i=1

j=1

Pm

i=1 |Ci |0

er0

< 1. It follows that n P

|a|0 + |E|0 er0 +

|B j |0 er0

j=1

r1 = |u 0 |0 ≤ 1−

m P

< r1 ,

|Ci |0 er0

i=1

which is a contradiction. So Lu 6= λN u + λa,

λ ∈ (0, 1), and ∀u ∈ ∂Ω .

Next, we will define a bounded bilinear form [·, ·] on Cω × Cω1 by [y, x] = Rω Coker(L) by y 7→ ω1 0 y(t)dt. Obviously, n o \ u|u ∈ Ker L ∂Ω = {u|u ≡ r0 or − r0 }.

Rω 0

y(t)x(t)dt. We also define Q : y →

Without loss of generality, we assume u ≡ r0 , thus [Q N (u) + Q(a), u] · [Q N (−u) + Q(a), u] "Z Z ω n Z ω X 2 r0 = r0 a(t)dt − e E(t)dt + 0

"Z

0

ω

E(t)dt +

0

0

" =

ω

a(t)dt − e

×

r02

r0

a¯ − e

E¯ +

n X

!# B j (t)dt

j=1 0

Z

−r0

ω

n Z X

ω

B j (t)dt

j=1 0

!# " B¯ j

!#

−r0

E¯ +

a¯ − e

j=1

n X

!# B¯ j

.

j=1

Pa¯n As r0 > | ln E+ ¯ ¯ |, we see j=1 B j a¯ a¯ r0 > ln , ≥ ln n n P P ¯ E+ B¯ j E¯ + B¯ j j=1 j=1

and a¯ a¯ −r0 < − ln . ≤ ln n n P P ¯ E+ B¯ j E¯ + B¯ j j=1 j=1 Then we have er0 > E¯ +

a¯ n P

,

e−r0 <

B¯ j

E¯ +

j=1

a¯ n P

, B¯ j

j=1

i.e. r0

a¯ − e

E¯ +

n X j=1

! B¯ j

< 0,

−r0

a¯ − e

E¯ +

n X j=1

! B¯ j

> 0.

(3.11)

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Q. Wang, B. Dai / Nonlinear Analysis 69 (2008) 3919–3930

It follows from (3.11) that [Q N (u) + Q(a), u] · [Q N (−u) + Q(a), u] < 0. Therefore by Lemma 2.1, we get that the Eq. (1.13) has at least an ω-periodic solution and so system (1.4) and (1.5) has at least a positive ω-periodic solution. The proof is completed.  Consider the following equation ! m n X X dN 0 ci (t)N (t − τi (t)) , b j (t)N (t − σ j (t)) − = N (t) a(t) − β(t)N (t) − dt i=1 j=1

(3.12)

which is a special case of system (1.4) and (1.5) without impulse. Corollary 3.2. Suppose that the following conditions hold. P (L01 ) a¯ > 0, Γ (t) ≥ 0, t ∈ [0, ω] and β¯ + nj=1 b¯ j > 0, where Γ (t) is defined as (3.1). (L02 ) There exists a constant r0 > 0 such that ( ) m n m X X √ X max |ci |0 , 2 |b j |0l j + |c0,i |0 er0 < 1 i=1

j=1

i=1

and "

   r0 > H + ω   

r0 |a 0 |1

+ er0

|β 0 |

1

+

j=1

1 2

1 − er0 ¯

j=1 b j

|b0j |1

n m √ P P 2 |b j |0l j + |c0,i |0 j=1

Pa¯n where H = | ln β+ ¯

 12

#

n P

  !  , 

i=1

|. Then the Eq. (3.12) has at least one positive ω-periodic solution.

4. An example Consider the following equation        sin t sin t 1 1 0 0 N (t) = N (t) 2 − 4N (t) + 3N t − 2π − − 1 + sin t N t − 2 − cos t , 20 10 2 2 t 1 sin t 0 where a(t) = 2, β(t) = 4, b1 (t) = −3, c1 (t) = sin 10 (1 + 2 sin t), σ1 (t) = 2π + 20 , σ1 (t) = c1 (t) 1 1 sin t 0 2 + 2 cos t, τ1 (t) = − 2 sin t < 1, c0,1 (t) = 1−τ 0 (t) = 10 , l1 = |σ1 − 2π| ≤ 0.05, then

cos t 20

< 1, τ1 (t) =

1

max

( m X

√ |ci |0 , 2

i=1

n X j=1

|b j |0l j +

m X

)

n o | ln a¯ | √ ¯ b¯ 1 |c0,i |0 er0 = max |c1 |0 , 2|b1 |0l1 + |c0,1 |0 e β+

i=1

 √ 2 3 √ = max 0.15, 2 + 0.1 e| 4−3+1 | = 0.1 + 0.15 2 < 1, 20 

0 (γ (t)) c0,1 1 b1 (µ1 (t)) − 0 0 1 − σ1 (µ1 (t)) 1 − τ1 (γ1 (t)) −3 cos(γ1 (t))   > 0, − = 4+ 1 1 − 20 cos t 10 1 + 21 sin(γ1 (t))

Γ (t) = β(t) +

t ∈ [0, 2π ].

By using Corollary 3.2, we see that this equation has at least a positive 2π -periodic solution.

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Q. Wang, B. Dai / Nonlinear Analysis 69 (2008) 3919–3930

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