Global attractivity and positive almost periodic solution for delay logistic differential equation

Nonlinear Analysis 68 (2008) 54–72 www.elsevier.com/locate/na

Global attractivity and positive almost periodic solution for delay logistic differential equationI Xitao Yang a,b,∗ , Rong Yuan b a Department of Mathematics, Hunan University of Science and Technology, Xiangtan 411201, Hunan, People’s Republic of China b Department of Mathematics, Beijing Normal University, Beijing 100875, People’s Republic of China

Received 25 February 2006; accepted 23 October 2006

Abstract In this paper, we establish some sufficient conditions for the global attractivity of positive solutions and the unique existence of positive almost periodic solutions for delay logistic differential equations of the form x 0 (t) = r (t)x(t)(a(t) − b(t)x(t) − L(xt )). Our results extend and improve corresponding ones in the literature. c 2006 Elsevier Ltd. All rights reserved.

Keywords: Global attractivity; Almost periodicity; Logistic differential equation

1. Introduction Consider the scalar delay differential equation dx(t) = r (t)x(t)(a(t) − b(t)x(t) − L(xt )), (1.1) dt where xt (θ ) = x(t + θ ) for θ ∈ [−1, 0], L is a bounded linear functional from C = C([−1, 0], R) to R with L(1) = 1 and L(ϕ) ≥ 0 for ϕ ≥ 0, a(t), b(t) and r (t) are positive continuous functions. It follows from [5] that for any ϕ ∈ C + = {ϕ ∈ C : ϕ(0) > 0}, the solution x(t, 0, ϕ) of (1.1) with x0 = ϕ exists on [0, ∞) and x(t, 0, ϕ) > 0 for all t ≥ 0. The equation may be regarded as a prototypical model for the variation of the population of an organism when there is density-dependent growth which depends on population in the interval [t − 1, t]. For the special case of (1.1), equation dx(t) = x(t)(a(t) − b(t)x(t) − L(xt )) dt

(1.2)

I The research was partly supported by NSFC(10671021). ∗ Corresponding author at: Department of Mathematics, Hunan University of Science and Technology, Xiangtan 411201, Hunan, People’s Republic of China. E-mail addresses: [email protected] (X. Yang), [email protected] (R. Yuan).

c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.10.031

X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

55

was extensively studied (see [2–4,6–12,15,16]). When a(t) = a, b(t) = b, here a and b are positive constants, from a Tang’s recent results in [11], we see that the positive solution 1+b to (1.1) is globally attractive if ∞

Z

r (s)ds = ∞

(1.3)

0

and one of the following three conditions holds: (1) b > 1; (2) b = 1 and there is a constant A > 0 such that for large t, Z t r (s)ds ≤ A;

(1.4)

t−1

(3) 0 < b < 1 and for large t, Z t 1 ln(1 − b) b a − . r (s)ds ≤ − 1 + b t−1 2 b 6

(1.5)

Now, a natural question is whether there are conditions or not, similar to the above conditions (1), (2) and (3), which can guarantee the global attractivity of positive solutions to Eq. (1.1) in the case that a(t) and b(t) are not constants. In present paper, we will give a positive answer to the above question. Moreover, when Eq. (1.1) is almost periodic, we also prove that similar conditions can guarantee the existence and uniqueness of an almost periodic solution that is globally attractive. Our results improve those in the literature. Let kϕk = supθ∈[−1,0] |ϕ(θ )| and Sα+ = {ϕ ∈ C + : kϕk ≤ α} for α > 0. Denote e∗ = lim inft→∞ e(t) and ∗ e = lim supt→∞ e(t) for any bounded function e(t). Throughout this paper, we assume that (1.3) holds. We begin with some definitions. Definition 1 ([1]). A function f (t) is said to be almost periodic, if for any  > 0, there is a constant l() > 0 such that in any interval of length l() there exists τ such that the inequality | f (t + τ ) − f (t)| <  is satisfied for all t ∈ R. The number τ is called the -translation number of f (t). It follows from [1] that a( f, λ) = RT limT →∞ T1 0 f (t)e−iλt dt exists for each λ ∈ R and the set Λ f = {λ ∈ R : a( f, λ) 6= 0} is countable. The P sets Λ f and mod( f ) = { Nj=1 n j λ j : n j ∈ Z, N ∈ N, λ j ∈ Λ f } are called the spectrum and module of f (t), respectively. Denote by H ( f (t)) = {g(t) : there is {tn } ⊂ R such that limn→∞ f (t + tn ) = g(t) for all t ∈ R} the hull of f (t). It’s easy to see that if f (t) is almost periodic, then for all f 1 (t) ∈ H ( f (t)), lim inf f (t) = inf f (t) = inf f 1 (t), t→∞

t∈R

t∈R

lim sup f (t) = sup f (t) = sup f 1 (t). t→∞

t∈R

t∈R

Definition 2. The positive solution N (t) to (1.1), in the sense that N (t) > 0 for all t ≥ 0, is said to be globally attractive, if for any ϕ ∈ C + , lim (x(t, 0, ϕ) − N (t)) = 0.

t→∞

The remainder of this paper is organized as follows. The main results are established in Section 2 while the proofs are left to Section 4. In Section 3, we will give some lemmas which are useful for the proofs of the main results. For simplicity and convenience, the proof of Lemma 6 is put in the Appendix.

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X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

2. Main results Theorem 1. Assume that a∗ > 0, b∗ > 0 and Z t M = lim sup r (s)ds < ∞. t→∞

(2.1)

t−1

Then for any ϕ ∈ C + , x ≤ lim inf x(t, 0, ϕ) ≤ lim sup x(t, 0, ϕ) ≤ x, t→∞

(2.2)

t→∞

where x(t, 0, ϕ) is a solution to (1.1), x =

a∗ b∗ +exp(−a ∗ M)

a∗ and x = max( a∗b−x ∗ , b∗ +exp(((b∗ +1)x−a )M) ). ∗

The following corollary can be obtained from Theorem 1 and Theorem 1 in [13]. Corollary 1. Assume that the conditions of Theorem 1 are satisfied, and a(t), b(t) and r (t) are ω-periodic with ω > 0. Then (1.1) admits a positive ω-periodic solution. Theorem 2. Assume that N (t) is a positive solution to (1.1) with 0 < N∗ ≤ N ∗ < ∞ and b∗ > globally attractive.

N∗ N∗ .

Then N (t) is

Theorem 3. Assume that the conditions of Theorem 1 hold, N (t) is a positive solution to (1.1) and b∗ = N (t) is globally attractive.

N∗ N∗ .

Then

a If a(t) = a and b(t) = b are positive constants, Eq. (1.1) admits a positive solution 1+b . If we consider the global ∗ N a attractivity of the positive solution N (t) ≡ 1+b , then N∗ = 1 and b∗ = b. Thus we can see that Theorems 2 and 3 will reduce to the results obtained by Theorems 3.1 and 3.2 in [11] in this case.

Theorem 4. Assume that the conditions of Theorem 1 hold, N (t) is a positive solution to (1.1), c0 := N ∗ δ0 M ≤

1 ln(1 − c0 ) c0 − − , 2 c0 6

δ0 =

b∗ N∗ N∗

N ∗ b∗ . N∗ b∗

< 1 and (2.3)

Then N (t) is globally attractive. Remark 1. When a(t) = a and b(t) = b are positive constants, the condition (2.3) is weaker than that of (1.5). To see this, we consider the equation   5   + 3 ln 2 t 2 + 10 8 1 dx(t) = x(t) 1 − x(t) − x(t − 1) . (2.4) dt 2 1 + t2 For the positive solution N (t) = 23 to Eq. (2.4), we can see that δ0 = 1, c0 = R t ( 5 +3 ln 2)s 2 +10 M = lim supt→∞ t−1 8 1+s 2 ds = 58 + 3 ln 2. From the fact that N ∗ δ0 M =

1 2,

R∞ 0

( 58 +3 ln 2)s 2 +10 ds 1+s 2

= ∞ and

5 1 ln(1 − c0 ) c0 + 2 ln 2 = − − , 12 2 c0 6

the conditions of Theorem 4 are satisfied. Thus the positive solution N (t) = 23 to (2.4) is globally attractive. However, since   5 2 Z 8 + 3 ln 2 s + 10 2 t 5 1 ln(1 − c0 ) c0 ds > + 2 ln 2 = − − , for t ∈ R, 2 3 t−1 12 2 c0 6 1+s we cannot conclude the global attractivity of the above solution from the results in [11].

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In the following, we would like to consider that Eq. (1.1) is almost periodic, i.e., assume that a(t), b(t) and r (t) are almost periodic, because of seasonal variation which need not be exactly periodic but, rather, almost periodic. In this case, the condition (2.1) always holds. Thus, if a∗ > 0 and b∗ > 0, Theorem 1 holds. Let x and x be the constants given in Theorem 1. Theorem 5. Assume that a(t), b(t) and r (t) are almost periodic functions, a∗ > 0 and b∗ > xx . Then (1.1) has a unique positive almost periodic solution p(t) with mod( p(t)) ⊆ mod(a(t), b(t), r (t)) and there are positive constants γ and B such that for each α > 0, there is a T (α) > 0 such that for any ϕ ∈ Sα+ , e   Z t r (s)ds , t ≥ T (α). (2.5) |x(t, 0, ϕ) − p(t)| ≤ B sup |x(T + θ, 0, ϕ) − p(T + θ )| exp −e γ T (α)

−1≤θ ≤0

Remark 2. Let N = b∗ N N

<

b∗ x x .

a∗ b∗

and N =

1 b∗ (a∗

−

a∗ b∗ ).

∗

When r (t) = 1, L(ϕ) = ϕ(−1), b∗ > max(1, ab∗ ), we see that

Thus the Theorem in [16] is improved to some extent.

Theorem 6. Assume that a(t), b(t) and r (t) are almost periodic functions, a∗ > 0 and b∗ = xx . Then (1.1) has a unique positive almost periodic solution p(t) with mod( p(t)) ⊆ mod(a(t), b(t), r (t)) and p(t) is globally attractive. Theorem 7. Assume that a(t), b(t) and r (t) are almost periodic functions, a∗ > 0, b∗ > 0, e c0 := c0 ) e c0 1 ln(1 − e − , xe δ0 M < − e c0 2 6

xb∗ e δ0 = . xb∗

b∗ x x

< 1 and (2.6)

Then (1.1) has a unique positive almost periodic solution p(t) with mod( p(t)) ⊆ mod(a(t), b(t), r (t)) and p(t) is globally attractive. 3. Some lemmas Lemma 1. Assume that τ1 = supt≥t0

Rt t−τ

D + u(t) ≤ r (t)(−α(t)u(t) + β(t)

r (s)ds < ∞ and sup

u(s)),

t−τ ≤s≤t

t ≥ t0 ,

where τ ≥ 0, α(t), β(t) and r (t) are continuous functions with r (t) > 0, α(t) ≥ α0 > 0 and 0 < β(t) ≤ qα(t) for all t ≥ t0 with 0 ≤ q < 1. Then   Z t r (s)ds , t ≥ t0 , (3.1) u(t) ≤ G exp −γ t0

where G = supt0 −τ ≤s≤t0 u(s) and γ = inft≥t0 {ω(t) : ω(t) − α(t) + β(t)eω(t)τ1 = 0} > 0. Proof. In the case τ > 0, similar to the argument of Lemma 2.1 in [14], we can see that for any t ≥ t0 there is a unique positive ω(t) such that ω(t) − α(t) + β(t)eω(t)τ1 = 0 and γ > 0. In the case that τ = 0, we see that τ1 = 0 and ω(t) = α(t) − β(t) ≥ (1 − q)α0 for all t ≥ t0 . This also implies that γ > 0. Now, we consider two cases. Case 1. G > 0. Suppose that (3.1) fails. Then there would be a k > 1 such that   Z t r (s)ds u(t) > kG exp −γ t0

Rt for some t ≥ t0 . Set v(t) = kG exp(−γ t0 r (s)ds) and w(t) = u(t) − v(t). Let ξ = inf{t ≥ t0 : w(t) ≥ 0}. Since w(t0 ) = u(t0 ) − kG < 0, one has ξ > t0 , w(ξ ) = 0, u(s) < v(s) for s ∈ [t0 − τ, ξ ) and D + w(ξ ) = lim sup h→0+

w(ξ + h) − w(ξ ) ≥ 0. h

(3.2)

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X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

Since G > 0, we have D + w(ξ ) = D + u(ξ ) − v 0 (ξ ) ≤ r (ξ )(−α(ξ )u(ξ ) + β(ξ ) 



sup

ξ −τ ≤s≤ξ

≤ r (ξ ) −α(ξ )kG exp −γ

Z

ξ

t0



Z

= r (ξ )kG exp −γ

ξ

< r (ξ ) exp −γ

Z

ξ

ξ

r (s)ds



t0

!



r (s)ds + β(ξ )

sup

ξ −τ ≤s≤ξ

r (s)ds (γ − α(ξ )) + β(ξ )



v(s) + kGγ r (ξ ) exp −γ

sup

Z

ξ

r (s)ds



t0



t0



u(s)) + kGγ r (ξ ) exp −γ

Z

ξ −τ ≤s≤ξ

ξ

 Z exp γ



r (v)dv + γ

s



r (s)ds kG(γ − α(ξ ) + β(ξ )eγ τ1 ).

t0

Let ω(ξ ) satisfy ω(ξ ) − α(ξ ) + β(ξ )eω(ξ )τ1 = 0. Then according to the definition of γ , we have γ − α(ξ ) + β(ξ )eγ τ1 = γ − α(ξ ) + β(ξ )eγ τ1 − (ω(ξ ) − α(ξ ) + β(ξ )eω(ξ )τ1 ) = −(ω(ξ ) − γ ) − β(ξ )(eω(ξ )τ1 − eγ τ1 ) ≤ 0. Thus we see that D + w(ξ ) < 0, which contradicts (3.2). Therefore (3.1) holds. Case 2. G = 0. For any  > 0, similar to the previous argument, we can prove that   Z t u(t) ≤  exp −γ r (s)ds , t ≥ t0 . t0

Setting  → 0, it follows that (3.1) also holds. This completes the proof.



Lemma 2 (Tang [11]). Let 0 < υ < 1. Then the system of inequalities i h   ln(1−υ) x − −1 exp (υ + x) −  υ 6    h  i x y≤    υ exp (υ + x) − ln(1−υ) − x6 + x υ i h  y ln(1−υ)  + 1 − exp (υ − y) −   υ 6   h  i y  x ≤ ln(1−υ) y − υ exp (υ − y) − υ + 6y has a unique solution x = y = 0 in the region D = {(x, y) : 0 ≤ x < 1, y ≥ 0}. Lemma 3. Assume that f (t, ϕ) : R × C → R, is an almost periodic function in t uniformly on C, and for each D > 0, there is H (D) > 0 such that | f (t, ϕ)| ≤ H (D) for all t ∈ R and ϕ ∈ C with kϕk ≤ D. If there is a solution x(t) (t ≥ 0) to the equation x 0 = f (t, xt ),

(3.3)

such that A ≤ lim inft→∞ x(t) ≤ lim supt→∞ x(t) ≤ B, then Eq. (3.3) has a solution p(t) which is defined on R such that A ≤ p(t) ≤ B,

(3.4)

t ∈ R.

Proof. Let D = supt≥0 |x(t)|. From the almost periodicity of f , we can see that there is a H (D) > 0 such that | f (t, ϕ)| ≤ H (D) for all t ∈ R, ϕ ∈ C with kϕk ≤ D. Let S = {ϕ ∈ C : kϕk ≤ D, |ϕ(s) − ϕ(t)| ≤ H (D)|s − t|}. There is {tn } ⊂ R such that tn → ∞ and f (t + tn , ϕ) converges uniformly on S. By the Ascoli theorem and the diagonal process, we can choose {tn j } ⊂ {tn } such that x(t + tn j ) → p(t) on any compact subset of R as j → ∞ and p(t) satisfies (3.3) on R. We want to prove that (3.4) holds. For each  > 0, there is a T > 0 such that A −  ≤ x(t) ≤ B + ,

t ≥ T.

(3.5)

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X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

For any fixed t ∈ R, there is J > 0 such that t + tn j > T for all j > J . It follows from (3.5) that A −  ≤ x(t + tn j ) ≤ B + ,

j ≥ J.

Setting j → ∞ and  → 0, we see that (3.4) holds. This completes the proof.



Lemma 4. Assume that f (t, ϕ) : R × C → R, is an almost periodic function in t uniformly on C, and for each D > 0, there is continuous function H (D) > 0 such that | f (t, ϕ)| ≤ H (D) for all t ∈ R and ϕ ∈ C with kϕk ≤ D. Let S be a compact set of R. If ∀g ∈ H ( f ), the equation dx(t) = g(t, xt ) dt admits a unique solution pg (t), defined on R, whose range is in S, then all these solutions pg (t) are almost periodic and mod( pg ) ⊆ mod( f ). The proof is similar to that of Theorem10.1 in [1]. We omit it here. Lemma 5. Assume that (1.3) holds and r (t) is almost periodic. Then for each r1 (t) ∈ H (r (t)), Z 0 Z ∞ r1 (s)ds = r1 (s)ds = ∞. −∞

(3.6)

0

Proof. First, we prove that Z 0 r (s)ds = ∞.

(3.7)

−∞

Otherwise, we would have Z 0 r (s)ds < ∞. −∞

Since r (t) is almost periodic, there exist {tn } ⊂ R such that tn → −∞ and r (t + tn ) → r (t) uniformly on R as n → ∞. For any fixed T > 0, there is N ∈ N such that T + tn < 0 for all n > N . It follows that for all n > N , Z T Z T +tn Z 0 r (s + tn )ds = r (s)ds ≤ r (s)ds. 0

tn

−∞

Setting n → ∞ and T → ∞, we obtain that Z ∞ Z 0 r (s)ds ≤ r (s)ds < ∞, 0

−∞

which contradicts (1.3). Thus (3.7) holds. For each r1 (t) ∈ H (r (t)), there exist {τn } ⊂ R such that r (t + τn ) → r1 (t) uniformly on R as n → ∞. Now we want to prove that Z 0 r1 (s)ds = ∞. (3.8) −∞

Otherwise, from the previous argument, we would know Z ∞ Z 0 r1 (s)ds ≤ r1 (s)ds < ∞. 0

−∞

Since r1 (t − τn ) → r (t) uniformly on R as n → ∞, for all T > 0, we have Z T Z T −τn Z ∞ r1 (s − τn )ds = r1 (s)ds ≤ r1 (s)ds. 0

−τn

−∞

(3.9)

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X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

Setting n → ∞ and T → ∞, we obtain that Z ∞ Z ∞ r1 (s)ds < ∞, r (s)ds ≤ 0

−∞

which contradicts (1.3). Similarly, we can prove that proof. 

R∞ 0

r1 (s)ds = ∞. Thus (3.6) holds. This completes the

Lemma 6. Let I be an open interval. Assume that the following conditions hold. (i) y(t) on I satisfies the equation dy(t) = −r (t)(1 + y(t))(b(t)N (t)y(t) + L(Nt yt )). (3.10) dt (ii) There are t0 , t, t ∈ I such that min( t, t ) > t0 + 2, [t0 − 1, max( t, t )] ⊂ I , y(t0 ) = 0, t and t are local maximum and minimum points of y(t) on I with y(t) > 0 and y(t) < 0, respectively. (iii) For t ∈ I , Z t 1 ln(1 − c) c − + η, (3.11) b1 < b(t) < b2 , N2 δ r (s)ds < − 2 c 6 t−1 N1 < N (t) < N2 ,

−v1 < y(t) < u 1 ,

where N (t) satisfies (1.1) on I , η ≥ 0, δ := v1 are positive with v1 < 1. Then

(3.12)

b2 N2 b1 N1 ,

c := b1 N1 /N2 with c < 1 and c 6= u 1 , N2 , N1 , b1 , b2 , u 1 and

1 + y(t) 1 v1 1 ln ≤ − ln(1 − c) − + g(v1 , c, η), v1 + c 1 − vc1 y(t) c 6

(3.13)

1 1 + y(t) 1 u1 ln ≤ − ln(1 − c) + + η, c − u 1 1 + uc1 y(t) c 6

(3.14)

where g(v1 , c, η) = max(η, 1c ·

exp((c+v1 )η)−1 v1 +c exp(−(c+v1 )A)

− ηc ) and A =

1 2

−

ln(1−c) c

− 6c .

We put the proof of Lemma 6 in the Appendix to be read conveniently. 4. The proofs of main results Proof of Theorem 1. For any ϕ ∈ C + , let x(t) = x(t, 0, ϕ). For any  ∈ (0, a∗ ), there exists T > 1 such that for t ≥ T, a∗ −  < a(t) ≤ a ∗ + , Z t r (s)ds ≤ M + .

b∗ −  < b(t) ≤ b∗ + ,

(4.1) (4.2)

t−1

It follows from (1.1) that x 0 (t) < r (t)(a ∗ + )x(t) for t ≥ T . This implies that for t ≥ T + 1 and θ ∈ [−1, 0], x(t) ≤ exp((a ∗ + )(M + ))x(t + θ ).

(4.3)

By (1.1) and (4.1)–(4.3), we obtain x 0 (t) ≤ r (t)x(t)(a ∗ +  − (b∗ − )x(t) − e−(a This yields that "

∗

∗ +)(M+)

1 b∗ −  + e−(a +)(M+) x(t) ≤ − x(T + 1) a∗ +  #−1 ∗ b∗ −  + e−(a +)(M+) + . a∗ + 

!

x(t)),

 Z exp −

t

t ≥ T + 1.

(a + )r (s)ds ∗



T +1

(4.4)

X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

61

Setting t → ∞ and  → 0, we get that lim sup x(t) ≤ x.

(4.5)

t→∞

For the above , there is a T1 > T such that x(t) < x + ,

t ≥ T1 .

(4.6)

It follows from (1.1) and (4.2) that x 0 (t) ≥ r (t)x(t)(a∗ −  − (b∗ + )(x + ) − x − ),

t ≥ T1 + 1.

Noting (4.2) and the fact that a1 () :≡ (b∗ + )(x + ) + x + 2 − a∗ > 0, we obtain that for t ≥ T1 + 2 and θ ∈ [−1, 0], x(t) ≥ exp(−a1 ()(M + ))x(t + θ ). This implies from (1.1) and (4.2) that x 0 (t) ≥ r (t)x(t)(a∗ −  − (b∗ + )x(t) − exp(a1 ()(M + ))x(t)),

t ≥ T1 + 2.

It follows that  x(t) ≥

  Z t  1 b∗ +  + exp(a1 ()(M + )) − exp − (a∗ − )r (s)ds x(T1 + 2) a∗ −  T1 +2 −1 ∗ b +  + exp(a1 ()(M + )) + , t ≥ T1 + 2. a∗ − 

(4.7)

Setting t → ∞ and  → 0, we obtain that lim inf x(t) ≥ t→∞

b∗

+ exp(((b∗

a∗ . + 1)x − a∗ )M)

(4.8)

If a∗ ≤ x, from (4.5) and (4.8), we see that (2.2) holds. In the case that a∗ > x, we can choose the  in (4.1), (4.2) and (4.6) such that a∗ −  > x + . From (1.1) and (4.1)–(4.6), we obtain that x 0 (t) ≥ r (t)x(t)(a∗ −  − (b∗ + )x(t) − x − ). This yields that for t ≥ T1 + 2,    Z t  −1 1 b∗ +  b∗ +  x(t) ≥ − exp − . (a∗ − x − 2)r (s)ds + x(T1 + 2) a∗ − x − 2 a∗ − x − 2 T1 +2

(4.9)

This implies that a∗ − x . b∗ Combining this with (4.5) and (4.8), we can also see that (2.2) holds. lim inf x(t) ≥ t→∞



Proof of Theorem 2. Take an 0 ∈ (0, min(b∗ , N∗ )) such that (b∗ − 0 )(N∗ − 0 ) > (N ∗ + 0 ).

(4.10)

For this 0 , we can choose T > 0 such that b∗ − 0 < b(t) < b∗ + ,

N∗ − 0 < N (t) < N ∗ + 0 . x(t) N (t)

(4.11)

For any ϕ ∈ C + , let x(t) = x(t, 0, ϕ) and y(t) = − 1. Then y(t) satisfies (3.10) on [0, ∞). Set v(t) = |y(t)| and v = lim supt→∞ v(t). First, we prove that v < ∞. If v = ∞, then there exists t > T + 1 such that v(t) ≤ v(t),

0 ≤ t ≤ t.

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X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

It follows from (1.1), (4.9) and (4.10) that D + v(t) = −sgn y(t)r (t)(1 + y(t))(b(t)N (t)y(t) + L(Nt yt )) < r (t)(1 + y(t))(−(b∗ − 0 )(N∗ − 0 )v(t) + (N ∗ + 0 )v(t)) < 0, which is a contradiction. Thus v < ∞. Next, we prove that v = 0. Otherwise, there would exist an 1 ∈ (0, min(v, b∗ , N∗ )) such that (b∗ − 1 )(N∗ − 1 )(v − 1 ) > (N ∗ + 1 )(v + 1 ). For this 1 , there exists a T1 > 0 such that for t ≥ T1 , v(t) < v + ,

N∗ −  < N (t) < N ∗ + 1 .

We claim that D + v(t) ≤ 0 for t ≥ T1 + 1. Otherwise, from the definition of v, it follows that there exists a t ∗ > T1 + 1 such that v(t ∗ ) ≥ v − 1 ,

D + v(t ∗ ) > 0.

On the other hand, by directly calculating the upper right derivative of v(t), we have D + v(t ∗ ) = −sgn y(t ∗ )r (t ∗ )(1 + y(t ∗ ))(b(t ∗ )N (t)y(t ∗ ) + L(Nt ∗ yt ∗ )) < r (t ∗ )(1 + y(t ∗ ))(−(b∗ − 1 )(N∗ − 1 )(v − 1 ) + (N ∗ + 1 )(v + 1 )) < 0. This is also a contradiction. Thus we have limt→∞ |y(t)| = v. We consider two cases. If y(t) is oscillating, then v = 0. If y(t) is not oscillating, we also would like to prove Claim. v = 0. Indeed, otherwise, we would have v > 0. Without loss of generality, we assume that y(t) > 0 eventually. This implies that limt→∞ y(t) = v. Take a T2 > 0 such that b(t) > b2∗ , y(t) > v2 , N (t) > N2∗ and y 0 (t) < 0 for t ≥ T2 . It follows that   Z t b∗ N∗ v N∗ v 1 + y(t) r (s)ds ≤− + ln . 1 + y(T2 + 1) 8 4 T2 +1 Noting that y(T2 + 1) ≥ v, we obtain that   Z ∞ N∗ v −1 1 + y(T2 + 1) b∗ N∗ v + < ∞, ln r (s)ds ≤ 8 4 1+v T2 +1 which contradicts (1.3). Thus the claim holds. The proof is completed.



Proof of Theorem 3. For any ϕ ∈ C + , setting y(t) = x(t,0,ϕ) N (t) − 1, then y(t) satisfies (3.10) on [0, ∞). Since x(t, 0, ϕ) > 0 for all t ≥ 0, we can see that y(t) > −1. It suffices to prove that limt→∞ y(t) = 0. If y(t) is not oscillating, similar to the argument in Claim of Theorem 2, we can see that limt→∞ y(t) = 0. Assume that y(t) is oscillating. Let u = lim supt→∞ y(t) and −v = lim inft→∞ y(t). From Theorem 1, we see that 0 ≤ u < ∞ and 0 ≤ v < 1. Choose two sequences {tn } and {τn } such that y 0 (tn ) = y 0 (τn ) = 0, y(tn ) → −v,

y(tn ) < 0 < y(τn ),

y(τn ) → u,

tn , τn → ∞,

n → ∞.

It follows from (3.10) that N (tn )b(tn )y(tn ) = −L(Ntn ytn ),

N (τn )b(τn )y(τn ) = −L(Nτn yτn ).

(4.12)

For each  ∈ (0, min(b∗ , N∗ , 1 − v)), there exist T > 0 such that for t ≥ T , N∗ −  < N (t) < N ∗ + , b∗ −  < b(t) < b∗ + ,

−v −  < y(t) < u + ε, Z

t t−1

r (s)ds < M + .

(4.13) (4.14)

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One can take a N ∈ N such that tn , τn > T + 1 for n > N . It follows from (4.12) that y(τn ) ≤

(v + )(N ∗ + ) , (b∗ − )(N∗ − )

−y(tn ) ≤

(u + )(N ∗ + ) . (b∗ − )(N∗ − )

Setting n → ∞ and  → 0, we obtain that u = v. It follows from (4.12) that there exist ξn ∈ [τn − 1, τn ] such that y(ξn ) = 0 and y(s) > 0 for s ∈ (ξn , τn ]. From (3.10), we obtain that for n > N and t ∈ (ξn , τn ], y 0 (t) = −r (t)(N (t)b(t)y(t) + L(Nt yt )) 1 + y(t) ≤ r (t)(−(b∗ − )(N∗ − )y(t) + (v + )(N ∗ + )).

(4.15)

Let y(ηn ) = maxξn ≤s≤τn y(s). Noting that y(ξn ) = 0 and y 0 (τn ) = 0, we can see that y 0 (ηn ) = 0. It follows that N (ηn )b(ηn )y(ηn ) = −L(Nηn yηn ). From (4.13) and (4.14), this implies that for t ∈ (ξn , τn ], (b∗ − )(N∗ − )y(t) < N (ηn )b(ηn )y(ηn ) = −L(Nηn yηn ) ≤ (v + )(N ∗ + ). By (4.15), we can see that (v

+ )(N ∗

(v + )(N ∗ + )(1 + y(τn )) 1 ln ≤ M + . + ) + (b∗ − )(N∗ − ) (v + )(N ∗ + ) − (b∗ − )(N∗ − )y(τn )

Note the facts that N ∗ = b∗ N∗ and 1 + v > 0. Setting n → ∞ and  → 0, we obtain that u≤

exp(N ∗ (v + 1)M) − 1 v. v + exp(N ∗ (v + 1)M)

Thus u = v = 0. This completes the proof.



Proof of Theorem 4. For any ϕ ∈ C + , setting y(t) = x(t,0,ϕ) N (t) − 1, then y(t) satisfies (3.10) on [0, ∞). Since x(t, 0, ϕ) > 0 for all t ≥ 0, we can see that y(t) > −1. We only need to prove that limt→∞ y(t) = 0. If y(t) is not oscillating, similar to the argument in Claim of Theorem 2, we can see that limt→∞ y(t) = 0. Assume that y(t) is oscillating. Let u = lim supt→∞ y(t) and −v = lim inft→∞ y(t). From Theorem 1, we see that 0 ≤ u < ∞ and 0 ≤ v < 1. For each η > 0, it follows from (2.3) that there is an 0 (η) ∈ (0, min(b∗ , N∗ , 1 − v)) ∗ −) such that c0 () :≡ (b∗ −)(N < 1, c0 () 6= u +  and N ∗ + (N ∗ + )δ0 ()M <

1 ln(1 − c0 ()) c0 () − − +η 2 c0 () 6

for any  ∈ (0, 0 (η)), where δ0 () = b∗ −  < b(t) < b∗ + , N∗ −  < N (t) < N ∗ + ,

(b∗ +)(N ∗ +) (b∗ −)(N∗ −) .

Thus we can take a T > 1 such that for t ≥ T , Z t 1 ln(1 − c0 ()) c0 () − + η, (N ∗ + )δ0 () r (s)ds < − 2 c0 () 6 t−1 −v −  < y(t) < u + .

There exist {sn } such that {sn } strictly increase, sn → ∞ as n → ∞, s2 − s1 > 2, sn > T + 1 and y(sn ) = 0 for all n ∈ N. Choose two sequences {tk } and {τk } in [s2 , ∞) such that tk and τk are local maximum and minimum points of y(t) with y(tk ) > 0, y(tk ) → u and y(τk ) < 0, y(τk ) → −v, respectively. Let I = (T, ∞) and t0 = s1 . Applying Lemma 6, we have 1 1 + y(tk ) 1 v1 ln ≤− ln(1 − c0 ()) − + g(v1 , c0 (), η) c () 0 v1 + c0 () 1 − c0 () 6 y(tk ) v1

1 1 + y(τk ) 1 u1 ln ≤− ln(1 − c0 ()) + + η, c () 0 c0 () − u 1 1 + c0 () 6 u 1 y(τk )

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where u 1 = u +  and v1 = v + . This implies that h  i  ln(1−c0 ()) v1 exp (c () + v ) − − + g(v , c (), η) −1  0 1 1 0 c0 () 6    h  i y(tk ) ≤ v1   v1 0 ())  c exp (c0 () + v1 ) − ln(1−c − + g(v , c (), η) + v 1 0 1 c0 () 6 h i  ln(1−c0 ())  1 − exp (c0 () − u 1 ) − c0 () + u61 + η     i u 1 . h  −y(τk ) ≤  u1 0 ()) + + η u 1 − c0 () exp (c0 () − u 1 ) − ln(1−c c0 () 6 Setting k → ∞,  → 0 and η → 0, we can obtain the system of inequalities in Lemma 2. Thus u = v = 0. The proof is completed.  Proof of Theorem 5. Since the uniqueness of the almost periodic solution implies the module containment, we only need to prove existence, uniqueness and the relation (2.5). Existence. From Lemma 4, it suffices to prove that for each (a1 (t), b1 (t), r1 (t)) ∈ H (a(t), b(t), r (t)), the following equation x 0 (t) = r1 (t)x(t)(a1 (t) − b1 (t)x(t) − L(xt ))

(4.16)

has a unique solution q(t), defined on R, whose range is in [ x, x ]. By Lemma 5 and the facts that a(t), b(t) and r (t) are almost periodic, we can see that the relation (2.2) also holds for any positive solution x(t, 0, ϕ) to (4.16) and the conditions of Theorem 5 are satisfied if r (t), a(t) and b(t) are replaced by r1 (t), a1 (t) and b1 (t), respectively. Thus, without loss of generality, we only need to prove that Eq. (1.1) has a unique solution q(t), defined on R, whose range is in [ x, x ]. Since r (t) is almost periodic, we see that (2.1) holds. From Theorem 1 and Lemma 3, we know that Eq. (1.1) admits a solution which possesses all the above properties. Suppose that (1.1) has two such solutions q (1) (t) and (1) q (2) (t) defined on R. Let v(t) = | ln qq (2) (t) |. Then (t) 1 1 (1) |q (t) − q (2) (t)| ≤ v(t) ≤ |q (1) (t) − q (2) (t)|. x x By directly calculating the upper right derivative of v(t), we have (1)

D + v(t) = −sgn (q (1) (t) − q (2) (t))r (t)(−b(t)(q (1) (t) − q (2) (t)) + L(qt

(2)

− qt )) !

≤ r (t) −b∗ |q (1) (t) − q (2) (t)| + sup |q (1) (t + θ ) − q (2) (t + θ )| −1≤θ ≤0

! ≤ r (t) −b∗ xv(t) + x

sup |v(t + θ )| . −1≤θ≤0

It follows from Lemma 1 that there is a constant γ1 > 0 such that   Z t x v(t) ≤ ln exp −γ1 r (s)ds , t ≥ t0 , t, t0 ∈ R. x t0 By (1.3) and Lemma 5, for each t ∈ R, setting t0 → −∞, we see that v(t) = 0. Thus q (1) (t) = q (2) (t) for all t ∈ R. According to Lemma 4, Eq. (1.1) has a unique almost periodic solution p(t) whose range is in [ x, x ] and mod( p(t)) ⊂ mod(a(t), b(t), r (t)). Uniqueness. Suppose that p1 (t) is another positive almost periodic solution. From Theorem 1 and the almost periodicity of p1 (t), we see that x ≤ p1 (t) ≤ x for all t ∈ R. Thus the uniqueness follows. Stability. For any ϕ ∈ C + , let x(t) = x(t, 0, ϕ) and w(t) = | ln x(t) p(t) | (t ≥ 0). There is an η0 > 0 such that b∗ (x(η0 ) − η0 ) > x + η0 , a∗ 0 , b∗ +exp(((b∗ +1)(x+η where x(η0 ) = max( a∗ −x−η ). In terms of the almost periodicity of a(t), b(t) and b∗ 0 )−a∗ )M) Rt t−1 r (s)ds, we can see that for t ≥ 0,

a∗ ≤ a(t) ≤ a ∗ ,

b∗ ≤ b(t) ≤ b∗ ,

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X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

Z

t

r (s)ds ≤ M.

t−1

From the proof of (4.4) in Theorem 1, we know that (4.4) holds for T = 0 and  = 0. It’s easy to see that the closure of {x1 (·, 0, ϕ) : ϕ ∈ Sα+ } is a compact subset of C for each α > 0. This implies that there is a T1 (α) > 1 such that x(t, 0, ϕ) < x + η0 ,

ϕ ∈ Sα+ , t ≥ T1 (α).

Similar to the argument of (4.7) and (4.9) in Theorem 1, we have for t ≥ T1 (α) + 2,    Z t  1 b∗ + exp(((b∗ + 1)(x + η0 ) − a∗ )M) x(t) ≥ − exp − a∗r (s)ds x(T1 (α) + 2) a∗ T1 (α)+2  b∗ + exp(((b∗ + 1)(x + η0 ) − a∗ )M) −1 + a∗ and  x(t) ≥

1 b∗ − x(T1 (α) + 2) a∗ − x − η0



 Z exp −

t

b∗ (a∗ − x − η0 )r (s)ds + a ∗ − x − η0 T1 (α)+2 

−1

in the case a∗ > x + η0 . This implies that there is a T (α) > T1 (α) + 3 such that x(t) ≥ x(η0 ) − η0 ,

t ≥ T (α) − 1.

Similar to the previous argument, we can obtain that 1 w(t) ≤ x(η0 ) − η0

 Z sup |x(T (α) + θ, 0, ϕ) − p(T (α) + θ )| exp −e γ −1≤θ ≤0

t T (α)



r (s)ds ,

where e γ is the unique positive root of λ − b∗ (x(η0 ) − η0 ) + (x + η0 )eλM = 0. Setting B = we can see that (2.5) holds. This completes the proof. 

t ≥ T (α),

1 x(η0 )−η0

and T = T (α),

∗

Proof of Theorem 6. It follows from Theorem 1 that xx ≥ pp∗ for any positive solution p(t) to Eq. (1.1). Thus the global attractivity of positive almost periodic solution follows from Theorems 2 and 3. It’s easy to see that the uniqueness of an almost periodic solution to Eq. (1.1) follows from the global attractivity. Thus we only need to prove the existence of an almost periodic solution to Eq. (1.1). By the argument of Theorem 5, it suffices to prove that the solution to Eq. (1.1) defined on R, whose range is in [ x, x ], is unique. Suppose that x(t) and N (t) are two solutions to (1.1) which possess these properties. Let y(t) = Nx(t) (t) − 1. Then y(t) satisfies (3.10) on R. Now we want to prove that y(t) = 0 for all R. Otherwise, there would exist a τ0 ∈ R such that y(τ0 ) 6= 0. We consider the following two cases. Case 1. There is T < τ0 such that y(t) ≥ 0 or y(t) ≤ 0 for all t ≤ T . Without loss of generality, we assume that y(t) ≥ 0 for all t ≤ T . It follows from (3.10) that y 0 (t) ≤ 0 for t ≤ T . By the fact that y(t) is bounded on R, we can R0 see that limt→−∞ y(t) exists, denoted by l. From (1.3) and Lemma 5, we can see that −∞ r (s)ds = ∞. Similar to the argument in Claim of Theorem 2, we can obtain that l = 0. Thus y(t) = 0 for t ≤ T . Using the uniqueness of solution for (3.10), we see that y(t) = 0 for all R, a contradiction. Case 2. There exist {tn } such that {tn } strictly decrease, tn → −∞ as n → ∞, y(tn ) = 0 for all n ∈ N and u = supt≤t1 y(t) > 0, v = − inft≤t1 y(t) > 0. It’s easy to see that v < 1. For any  ∈ (0, min(u, 1 − v)), there is a t < t1 such that y(t) > u − . Suppose t ∈ [t j+1 , t j ]. Let y(τ ) = maxt j+1 ≤t≤t j y(t). Then y(τ ) ≥ y(t) > 0 and y˙ (τ ) = 0. This yields b(τ )N (τ )y(τ ) + L(Nτ yτ ) = 0.

(4.17)

This implies that b∗ x y(t) ≤ b(τ )N (τ )y(τ ) = −L(Nτ yτ ) ≤ xv. Thus we have u ≤ v. Similarly, we can obtain that v ≤ u. Thus u = v. To complete the proof, we would like to prove the following two claims. Claim I. (v + )x − x y(t) > 0 for all t ≤ t1 and  > 0.

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Otherwise, there would exist a e t ∈ (−∞, t1 ] and  > 0 such that t) ≤ 0. (v + )x − x y(e This implies that y(e t) > 0. Assume that e t ∈ (t j0 +1 , t j0 ) for some j0 ∈ N. Let y(e τ ) = max j0 +1≤t≤ j0 y(t). Then y(e τ ) ≥ y(e t) > 0 and y 0 (e τ ) = 0. It follows that b(e τ )N (e τ )y(e τ ) + L(Ne τ ye τ ) = 0. Using the almost periodicity of b(t), we have b∗ N∗ y(e t) ≤ b(e τ )N (e τ )y(e τ ) = −L(Ne τ ye τ ) < (v + )x, which is a contradiction. Thus Claim I holds. Claim II. −b(t)N (t)y(t) − L(Nt yt ) ≤ e δ0 ((v + )x − b∗ x y(t)) for all t ≤ t1 , where e δ0 = In fact, for each t ∈ (−∞, t1 ], if y(t) ≥ 0, then

b∗ x b∗ x

≥ 1.

−b(t)N (t)y(t) − L(Nt yt ) ≤ (v + )x − b∗ x y(t) ≤ e δ0 ((v + )x − b∗ x y(t)). Thus, in this case, Claim II holds. If y(t) < 0, then δ0 ((v + )x − b∗ x y(t)). −b(t)N (t)y(t) − L(Nt yt ) ≤ e δ0 (v + )x − b∗ x y(t) = e Therefore, Claim II holds. From the above Claims I and II, we see that y 0 (t) b(t)N (t)y(t) + L(Nt yt ) e = −r (t) ≤ δ0r (t), x(1 + y(t))(v +  − y(t)) (v + )x − b∗ x y(t) which yields  0 1 + y(t) ln ≤ (1 + v + )xe δ0r (t), v +  − y(t)

t ≤ t1 ,

t ≤ t1 .

(4.18)

By (4.17), there is a s0 ∈ [τ − 1, τ ] such that y(s0 ) = 0 and y(s) > 0 for s ∈ (s0 , τ ]. Integrating (4.18) from s0 to τ , we obtain that Z τ 1 + y(τ ) e ln ≤ (v +  + 1)x δ0 δ0 M. r (s)ds ≤ (v +  + 1)xe 1 y(τ ) 1 − v+ s0 It follows that y(t) ≤ y(τ ) ≤

e(v++1)x δ0 M − 1 e

1+

1 (v++1)xe δ0 M v+ e

.

Thus we have e(v+1)x δ0 M − 1 e

u≤

v + e(v+1)xeδ0 M

v < v,

contradicting the fact that u = v. This completes the proof.



Proof of Theorem 7. It is easily shown that for any positive solution p(t) to (1.1), c0 = p∗ p∗ ,

b∗ p∗ p∗

≥ e c0 . If c0 ≥ 1,

from Theorems 2 and 3, we can see that the positive solution p(t) to (1.1) is globally attractive. If i.e., b∗ ≥ c0 < 1, by Theorem 4 and the fact that 1 ln(1 − e c0 ) e c0 1 ln(1 − c0 ) c0 p ∗ δ0 M ≤ xe δ0 M ≤ − − ≤ − − , e 2 c0 6 2 c0 6 we can also see that the positive solution p(t) to (1.1) is globally attractive. Thus we only need to prove the existence of almost periodic solution for Eq. (1.1). By the argument of Theorem 5, it suffices to prove that the solution to Eq. (1.1) defined on R, whose range is in [ x, x ], is unique. Suppose that x(t) and N (t) are two solutions to (1.1)

X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

67

which possess these properties. Let y(t) = Nx(t) (t) − 1. Then y(t) satisfies Eq. (3.10) on R. We want to prove that y(t) = 0 for all t ∈ R. Otherwise, there would exist a τ0 ∈ R such that y(τ0 ) 6= 0. There are also the following two cases. Case 1. There is a T < τ0 such that y(t) ≥ 0 or y(t) ≤ 0 for all t ≤ T . Similar to the argument of Case 1 in Theorem 6, we can obtain a contradiction. Case 2. There exist {sn } such that {sn } strictly decrease, sn → −∞ as n → ∞, y(sn ) = 0 for all n ∈ N and u = supt≤s1 y(t) > 0, v = − inft≤t1 y(t) > 0. It’s easy to see that 0 ≤ u < ∞ and 0 ≤ v < 1. For any η ∈ (0, min(u, 1 − v)), there is t, t < t1 such that y(t) > u − η and y(t) < −v + η. Suppose t ∈ [s j+1 , s j ] and t ∈ [si+1 , si ]. Let y(τ ) = maxs j+1 ≤t≤s j y(t) and y(ξ ) = minsi+1 ≤t≤si y(t). Then y(τ ) ≥ y(t) > 0, y(ξ ) ≤ y(t) < 0, Rt y˙ (τ ) = 0 and y˙ (ξ ) = 0. It’s easy to see that t−1 r (s)ds is almost periodic. For each η > 0, by (2.3), we can find an 0 (η) ∈ (0, min(b∗ , N∗ , u, 1 − v)) such that e c0 () :≡ δ0 ()M < (x + )e

(b∗ −)(x−) x+

< 1, e c0 () 6= u +  and

c0 () 1 ln(1 − e c0 ()) e − − +η e 2 c0 () 6

for any  ∈ (0, 0 (η)), where e δ0 () = that for t ∈ I = (−∞, s1 ], b∗ −  < b(t) < b + , ∗

(b∗ +)(x+) (b∗ −)(x−) .

(x + )e δ0 ()

Z

In terms of the almost periodicity of a(t), b(t) and r (t), we see t

r (s)ds ≤

t−1

x −  < N (t) < x + ,

−v −  < y(t) < u + .

c0 () c0 ()) e 1 ln(1 − e − − + η, e c0 () 2 6

There is s N such that s N < min(τ, ξ ) − 2. Let t0 = s N and I = (−∞, s1 ). Applying Lemma 6, we obtain that 1 1 + y(τ ) 1 v1 ln + g(v1 ,e c0 (), η) ≤− ln(1 − e c0 ()) − e c () 0 e c0 () v1 + e c0 () 1 − 6 y(τ ) v1

1 1 + y(ξ ) 1 u1 ln ≤− ln(1 − e c0 ()) + + η, e c () 0 e e c0 () − u 1 1 + c0 () 6 u 1 y(ξ ) where v1 = v +  and u 1 = u + . Similar to the argument in Theorem 4, we can see that the system of inequalities in Lemma 2 holds for x = v and y = u. This implies that u = v = 0, which is a contradiction. This completes the proof.  Acknowledgments The authors are grateful to the referee for carefully reading the paper and for many valuable comments which led to a significant improvement of the original manuscript. Appendix. Proof Lemma 6 We divide this proof into two parts: The proof of (3.13) and (3.14), respectively. The proof of (3.13). To prove that (3.13) holds, we first prove the following two claims. Claim 1. v1 N2 − b1 N1 y(t) > 0 for all t ∈ [t0 , t]. Otherwise, there would exist s0 ∈ [t0 , t] such that v1 N2 − b∗ N1 y(s0 ) ≤ 0. This implies that y(s0 ) > 0. Let y(τ ) = maxt0 ≤t≤t y(t). Then y˙ (τ ) = 0 and y(τ ) ≥ y(s0 ). This implies that 0 = b(τ )N (τ )y(τ ) + L(Nτ yτ ). It follows that b1 N1 y(s0 ) < b(τ )N (τ )y(τ ) = −L(Nτ yτ ) < v1 N2 , contradicting (5.1). Thus Claim 1 holds.

(5.1)

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X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

Claim 2. −b(t)N (t)y(t) − L(Nt yt ) ≤ δ(v1 N2 − b1 N1 y(t)) for all t ∈ [t0 , t]. In fact, for each t ∈ [t0 , t], if y(t) ≥ 0, we can see that Claim 2 holds. If y(t) < 0, then −b(t)N (t)y(t) − L(Nt yt ) ≤ −b2 N2 y(t) + v1 N2 < δ(v1 N2 − b1 N1 y(t)). Hence Claim 2 holds. Since y˙ (t) = 0, we have b(t)N (t)y(t) + L(Nt yt ) = 0. From the facts that y(t) > 0 and N (t) > 0 for all t ≥ 0, we see that there exists ξ ∈ [t − 1, t] such that y(ξ ) = 0 and y(t) > 0 for t ∈ (ξ, t]. Using Claims 1, 2 and (3.10), we have   1 + y(t) 0 1 ln ≤ N2 δr (t), t ∈ [t0 , t ]. (5.2) c + v1 v1 − cy(t) Integrating (5.2) from s to ξ , we obtain that for s ∈ [t0 , ξ ],   Rξ exp −(c + v1 )N2 δ s r (θ )dθ − 1  . y(s) ≥ Rξ 1 + vc1 exp −(c + v1 )N2 δ s r (θ )dθ Substituting this into (3.10), we have y 0 (t) ≤ r (t)(−b1 N1 y(t) − N2 R(t)), 1 + y(t)

ξ ≤t ≤t

where   Rξ exp −(c + v1 )N2 δ t−1 r (θ )dθ − 1  . R(t) = Rξ 1 + vc1 exp −(c + v1 )N2 δ t−1 r (θ )dθ It follows that for ξ ≤ t ≤ t,   1 + y(t) 0 −b1 N1 y(t) − N2 R(t) 1 ln ≤ r (t) N2 (c + v1 ) v1 − cy(t) (v1 − cy(t))N2 ≤ r (t)

−R(t) , v1

R(t) ≥ −v1 .

Combining (5.2) with (5.3), we obtain that     1 1 + y(t) 0 −R(t) ln ≤ N2 δr (t) min 1, , c + v1 v1 − cy(t) v1

ξ ≤ t ≤ t.

(5.3)

(5.4)

There are two cases R t that need to be considered. 1 Case 1. N2 δ ξ r (s)ds ≤ − v11+c ln 1−v 1+c := d. In this case, we continue to consider two cases. Case 1.1. v1 ≤ c. Integrating (5.4) from ξ to t and using Lemma 2.1 in [11], we obtain Z t 1 1 + y(t) ln ≤ N δ r (s)ds 2 v1 + c 1 − vc1 y(t) ξ ≤− Case 1.2. v1 > c. Let A =

1 − v1 1 v1 1 ln ≤ − ln(1 − c) − . v1 + c 1+c c 6 ln(1−c) c

− 6c . Integrating (5.4) from ξ to t and using (3.11), we obtain that   Rξ Z 1 − exp −(c + v1 )N2 δ s−1 r (θ )dθ 1 1 + y(t) N2 δ t   ds ln ≤ r (s) Rξ v1 + c 1 − vc1 y(t) v1 ξ 1 + vc1 exp −(c + v1 )N2 δ s−1 r (θ )dθ 1 2

−

(5.5)

X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

   Rs Z t 1 − exp −(c + v ) A + η − N δ r (θ )dθ 1 2 ξ N2 δ    ds ≤ r (s) Rs c v1 ξ 1 + v1 exp −(c + v1 ) A + η − N2 δ ξ r (θ )dθ    Rt Z 1 + vc1 exp −(c + v1 ) A + η − N2 δ ξ r (s)ds 1 N2 δ t r (s)ds − ln . = v1 ξ v1 c 1 + vc1 exp(−(c + v1 )(A + η))

69

(5.6)

If d ≤ A + η, from (5.6) and Lemma 2.2 in [11], we obtain that 1 + vc1 exp(−(c + v1 )(A + η − d)) 1 1 1 + y(t) d − ln ≤ ln v1 + c 1 − vc1 y(t) v1 v1 c 1 + vc1 exp(−(c + v1 )(A + η))   1 (1 + c)v1 + c(1 − v1 ) exp(−(c + v1 )(A + η − d)) d + ln − ln(1 + c) = v1 v1 c v1 + c exp(−(c + v1 )(A + η − d)) 1 d + [− ln(1 + c) + v1 c(A + η − d)] ≤ v1 v1 c (1 − v1 ) ln(1 − v1 ) 1+c ln(1 + c) − = A+η− c(v1 + c) v1 (v1 + c) ! ! 2 c1 v12 1 c 1 v1 1− − + 1− − ≤ A+η− v1 + c 2 6 v1 + c 2 6 =−

ln(1 − c) v1 − + η. c 6

(5.7)

ln(1−c) 1 1 Note that f (x) = − x+c ln 1−x + 6c increase on [0, 1) and f ( 1+c 1+c − 2 + c 2 ) < 0 for c ∈ (0, 1). If d > A + η, Rt then v1 > (1 + c)/2. Since N2 δ ξ r (s)ds < A + η, from (5.6) and Lemma 2.4 in [11], it follows that

1 1 + y(t) 1 1 ln ≤ (A + η) − ln v1 + c 1 − vc1 y(t) v1 v1 c 1 + =

1 A − ln v1 v1 c 1 +

≤−

c v1 exp(−(c 1 + vc1

c v1

c v1

1+

+ v1 )(A + η))

exp(−(c + v1 )A)

+ q(v1 , c, η)

ln(1 − c) v1 − + q(v1 , c, η), c 6

where 1 + v1 exp(−(c + v1 )(A + η)) 1 η ln + . v1 v1 c 1 + vc1 exp(−(c + v1 )A) c

q(v1 , c, η) = Since

1 v1 exp((c + v1 )η) + c exp(−(c + v1 )A) η + ln c v1 c v1 + c exp(−(c + v1 )A) 1 exp((c + v1 )η) − 1 η ≤ · − , c v1 + c exp(−(c + v1 )A) c

q(v1 , c, η) = −

we see that 1 1 + y(t) ln(1 − c) v1 1 exp((c + v1 )η) − 1 η ln ≤− − + · − . v1 + c 1 − vc1 y(t) c 6 c v1 + c exp(−(c + v1 )A) c

(5.8)

70

X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

Rt Rt Case 2. N2 δ ξ r (s)ds > d. In this case, there exists s0 ∈ (ξ, t] such that N2 δ s0 r (s)ds = d. Integrating (5.4) and using (3.11), we obtain that   Rξ Z s0 Z t 1 − exp −(c + v )N δ r (θ )dθ 1 2 s−1 N2 δ 1 1 + y(t)   ds r (s)ds + ln ≤ N2 δ r (s) Rξ c v1 + c 1 − vc1 y(t) v 1 ξ s0 1 + v1 exp −(c + v1 )N2 δ s−1 r (θ )dθ    Rs Z t Z s0 1 − exp −(c + v ) A + η − N δ r (θ )dθ 1 2 ξ N2 δ    ds ≤ N2 δ r (s)ds + r (s) Rs c v 1 ξ s0 1 + exp −(c + v ) A + η − N δ r (θ )dθ v1

1

2

ξ

N2 δ r (s)ds v1 s0 ξ    Rt 1 + vc1 exp −(c + v1 ) A + η − N2 δ ξ r (s)ds 1    − ln v1 c 1 + c exp −(c + v ) A + η − N δ R s0 r (s)ds 1 2 ξ v1

= N2 δ

Z

s0

≤ A+η+

r (s)ds +

Z

t

1 − v1 1 d− ln v1 v1 c 1 +

1+ c v1

c v1

exp(−(c + v1 )d)

1 − v1 1 1 − v1 ln − ln(1 + c) + η v1 (c + v1 ) 1+c v1 c ln(1 − c) v1 ≤− − + η. c 6 From (5.5) and (5.7)–(5.9), we see that (3.13) holds.  = A−

The proof of (3.14). Similar to the argument of Claims 1 and 2, we can prove the following two claims. Claim 3. u 1 N2 + b1 N1 y(t) > 0 for all t ∈ [t0 , t]. Claim 4. b(t)N (t)y(t) + L(Nt yt ) ≤ δ(u 1 N2 + b1 N1 y(t)) for all t ∈ [t0 , t]. It follows that   1 + y(t) 0 −1 ln ≤ N2 δr (t), t ∈ [t0 , t]. u1 − c u 1 + cy(t)

(5.9)

(5.10)

Since y˙ (t) = 0, there is a τ ∈ [t − 1, t] such that y(τ ) = 0 and y(t) < 0 for t ∈ (τ, t]. We now consider the following two cases. Rt c 1+x 1 Case 1. N2 δ τ r (s)ds ≤ A + η − u 11−c ln 1+u 1+c . Since the function x−c ln 1+c decreases on x ∈ [0, ∞), we have c 1 + u1 ln ≤ ln(1 + c). u1 − c 1+c It follows that 1 ln(1 + u 1 ) ln(1 + c) − ≥ 0. c(u 1 − c) u 1 (u 1 − c)

(5.11)

Integrating (5.10) from τ to t and using (3.11) and Lemma 2.5 in [11], we obtain that Z t 1 1 + y(t) ln ≤ N δ r (s)ds 2 u 1 − c 1 + uc1 y(t) τ 1 1 + u1 ln u1 − c 1+c 1+c (1 + u 1 ) ln(1 + u 1 ) ≤ A+ ln(1 + c) − +η c(u 1 − c) u 1 (u 1 − c) ln(1 − c) u 1 + + η. ≤− c 6 ≤ A+η−

(5.12)

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X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

Rt 1 Case 2. N2 δ τ r (s)ds > A + η − u 11−c ln 1+u 1+c . Noting the fact that A ≥ h ∈ [τ, t ] such that Z h 1 1 + u1 r (s)ds = A + η − N2 δ ln . u − c 1+c 1 τ

1 x−c

ln 1+x 1+c for all x ≥ 0, we can choose

(5.13)

For t ∈ [h, t] and s ∈ [t − 1, τ ], we have Z τ Z τ r (s)ds r (s)ds ≤ N2 δ N2 δ t−1

s

≤ A + η − N2 δ We can easily shown that if x ≤ c exp((u 1 − c)x) < 1, u1 c exp((u 1 − c)x) > 1, u1

1 u 1 −c

h

Z τ

r (s)ds =

1 1 + u1 ln . u1 − c 1+c

1 ln 1+u 1+c , then

u 1 > c,

(5.14)

u 1 < c.

(5.15)

Integrating (5.10) from s to τ , from (5.14) and (5.15), we obtain that Rτ
exp (u 1 − c)N2 δ s r (θ )dθ − 1 Rτ
. y(s) ≤ 1 − uc1 exp (u 1 − c)N2 δ s r (θ )dθ Substituting this into (5.10), we obtain that for t ∈ [τ, t ], −

y 0 (t) ≤ r (t)(b1 N1 y(t) + N2 Q(t)), 1 + y(t)

(5.16)

where Q(t) =

exp (u 1 − c)N2 δ 1−

c u1

exp

Rτ

t−1 r (s)ds − 1 Rτ
. (u 1 − c)N2 δ t−1 r (s)ds




Note that y(t) < 0 for t ∈ (τ, t]. If Q(t) ≤ u 1 , from Claim 3, we have Q(t) b1 N1 y(t) + N2 Q(t) ≤ , b1 N1 y(t) + N2 u 1 u1

τ ≤ t ≤ t.

This implies from (5.16) that for τ ≤ t ≤ t and Q(t) ≤ u 1 ,   −1 1 + y(t) 0 Q(t) ln ≤ N2 δr (t) . u1 − c u 1 + cy(t) u1 Combining (5.10) with (5.17), we obtain that     −1 1 + y(t) 0 Q(t) ln ≤ N2 δr (t) min 1, , u1 − c u 1 + cy(t) u1 Note that f (x) =

exp((u 1 −c)x)−1 1− uc exp((u 1 −c)x)

(5.17)

τ ≤ t ≤ t.

(5.18)

increases on R. Integrating (5.18) from τ to t and using (3.11), we obtain that

1

1 1 + y(t) ln ≤ N2 δ c − u 1 1 + uc1 y(t)

Z

≤ N2 δ

Z

h τ h τ

N2 δ r (s)ds + u1

Z

N2 δ r (s)ds + u1

Z

t

r (s)

h t h

r (s)

exp (u 1 − c)N2 δ 1−

c u1

exp

Rτ

s−1 r (θ )dθ − 1 Rτ
ds (u 1 − c)N2 δ s−1 r (θ )dθ

exp (u 1 − c) A + η − N2 δ 1−

c u1




Rs



r (θ )dθ − 1 Rs

ds exp (u 1 − c) A + η − N2 δ τ r (θ )dθ τ

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X. Yang, R. Yuan / Nonlinear Analysis 68 (2008) 54–72

Z N2 δ t r (s)ds u1 h τ    Rt c exp −(c − u ) A + η − N δ 1 − r (s)ds 1 2 τ u1 1    . ln + u 1 c 1 − c exp −(c − u ) A + η − N δ R h r (s)ds 1 2 τ u1

= N2 δ

Z

h

r (s)ds −

By (5.14) and (5.15), we see that g(x) = − ux1 + u11 c ln(1 − uc1 exp(A + η − x)) increases for A + η − x ≤ A + η. By using (5.13) and Lemma 2.5 in [11], we have Z h Z N2 δ t 1 1 + y(t) 1 + u1 r (s)ds − ln N δ = r (s)ds 2 c − u 1 1 + uc1 y(t) u1 u1 τ τ    Rt c r (s)ds 1 − exp −(c − u ) A + η − N δ 1 2 τ u1 1 ln + 1+u 1 c u1c 1 − u 1 · 1+c Z h A + η ln(1 + c) 1 + u1 r (s)ds − N2 δ + ≤ u1 u1 cu 1 τ 1 + u1 1+c ln(1 + c) − ln(1 + u 1 ) = A+η+ c(u 1 − c) u 1 (u 1 − c) ln(1 − c) u 1 ≤− + + η. c 6 From (5.12) and (5.19), we can see that (3.14) holds. This completes the proof. 

1 u 1 −c

1 ln 1+u 1+c ≤

(5.19)

References [1] A.M. Fink, Almost Periodic Differential Equations, in: Lecture Notes in Mathematics, vol. 377, Springer-Verlag, 1974. [2] K. Gopalsamy, Stability and Oscilations in Delay Differential Equation of Population Dynamics, Kluwer Academic Publishers, Dordredit, 1992. [3] K. Gopalsamy, Global asymtotic stability in an almost periodic Lotka–Volterra system, J. Aust. Math. Soc. Ser.B (27) (1986) 346–360. [4] K. Gopalsamy, X.Z. He, L.Z. Wen, Global attractivity and oscilation in an almost periodic delay logistic equation, Nonlinear Times Dig. 1 (1994) 9–23. [5] J.K. Hale, Theory of Functional Differential Equations, Spring-Verlag, New York, 1977. [6] Z. Hou, J.S. Cassell, Global attractivity of a single species population model, NoDEA Nonlinear Differential Equations Appl. 5 (1998) 167–180. [7] Y. Kuang, Delay Differential Equations with Application in Population Dynamics, Academic Press, Boston, 1993. [8] S.M. Lenhart, C.C. Travis, Global stability of a boilogical model with time delay, Proc. Amer. Math. Soc. 96 (1986) 75–78. [9] S. Murakami, Almost periodic solution of a system of integrodifferential equations, Tohoku Math. J. 39 (1987) 71–79. [10] G. Seifert, On a delay-differential eqution single specie population variations, Nonlinear Anal. TMA 11 (1987) 1051–1059. [11] X.H. Tang, Global attractivity for a delay logistic equation with instantaneous term, Nonlinear Anal. 59 (2004) 211–233. [12] X.H. Tang, X. Zou, 3/2-type criteria for global attractivity of Lotka–Volterra competition system without instantaneous negative feedback, J. Differential Equations 186 (2002) 420–439. [13] Z. Teng, L. Chen, The positive periodic Kolmogorov type system with delays, Acta Math. Appl. 22 (1999) 446–456 (in Chinese). [14] H. Tian, The exponential asymptotic stability of singularly perturbed delay differential equations with a bounded lag, J. Math. Anal. Appl. 270 (2002) 143–149. [15] J.S. Yu, Global attractivity of the zero solution of a class of functional differential equation and its applications, Sci. China 39A (1996) 225–237. [16] R. Yuan, J. Hong, The existence of almost periodic solution of a population equation with delay, Appl. Anal. 61 (1996) 45–52.