Some results on second-order neutral functional differential equations with infinite distributed delay

Nonlinear Analysis 70 (2009) 1393–1406 www.elsevier.com/locate/na

Some results on second-order neutral functional differential equations with infinite distributed delayI Weiwei Han, Jingli Ren ∗ Department of Mathematics, Zhengzhou University, Zhengzhou 450001, PR China Received 21 January 2008; accepted 7 February 2008

Abstract In this paper, we consider two types of second-order neutral functional differential equations with infinite distributed delay. By choosing available operators and applying Krasnoselskii’s fixed-point theorem, we obtain sufficient conditions for the existence of periodic solutions to such equations. c 2008 Published by Elsevier Ltd

MSC: 34K40; 34K13 Keywords: Neutral functional differential equation; Positive periodic solutions; Fixed point; Distributed delay; Second order; Operator

1. Introduction Consider the following two types of second-order neutral functional differential equations with infinite distributed delay !00 Z Z 0

x(t) − c

K (r )x(t + r )dr

0

= a(t)x(t) − λb(t)

−∞

K (r ) f (x(t + r ))dr,

(1.1)

−∞

and Z

0

x(t) − c

!00 K (r )x(t + r )dr

= −a(t)x(t) + λb(t)

−∞

Z

0

K (r ) f (x(t + r ))dr,

(1.2)

−∞

here λ is a positive parameter; c is a constant with |c| < 1; a(t) ∈ C(R, (0, ∞)), b(t) ∈ C(R, (0, ∞)), a(t) and b(t) R0 are ω-periodic functions; f (x) ∈ C(R, (0, ∞)); K (r ) ∈ C((−∞, 0], [0, ∞)) and −∞ K (r )dr = 1. Neutral functional differential equations are not only an extension of ordinary delay differential equations but also provide good models in many fields including Biology, Mechanics and Economics [1,2,4,15]. For example, in I Research was supported by the National Natural Science Foundation of China (No. 60504037) and the Outstanding Youth Foundation of Henan Province of China (No. 0612000200). ∗ Corresponding author. E-mail address: [email protected] (J. Ren).

c 2008 Published by Elsevier Ltd 0362-546X/$ - see front matter doi:10.1016/j.na.2008.02.018

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population dynamics, since a growing population consumes more (or less) food than a matured one, depending on individual species, this leads to neutral functional equations [4]. The study on neutral functional differential equations is more intricate than ordinary delay differential equations, that is why comparing plenty of results on the existence of positive periodic solutions for various types of first-order or second-order ordinary delay differential equations, studies on positive periodic solutions for neutral differential equations are relatively less, and most of them are confined to first-order neutral differential equations, see [3,5–8,13]. Very recently, in [11], Wu and Wang discussed the secondorder neutral delay differential equation (x(t) − cx(t − δ))00 + a(t)x(t) = λb(t) f (x(t − τ (t))),

(1.3)

where λ is a positive parameter, δ and c are constants with |c| 6= 1, a(t), b(t) ∈ C(R, (0, ∞)), f ∈ C([0, ∞), [0, ∞)), and a(t), b(t), τ (t) are ω-periodic functions. The key step in [11] is the application of a theorem of Zhang in [14] for the neutral operator (Ax)(t) = x(t) − cx(t − δ), and the fixed-point index theorem, to obtain the existence of positive periodic solutions for (1.3) with c < 0. In the paper we continue on the research of second-order neutral delay differential equation. To be concrete, we consider the equations with infinite distributed delay, i.e. (1.1) and (1.2). The delay arises from the models by employing a distributed time lag approach in which the contributions of time delay are expressed as a weighted response over a finite interval of past time through appropriately chosen memory kernels [9,10]. For (1.1) and (1.2), the said theorem of Zhang for neutral operator in [14] does not apply. To get around this, in this paper, we obtain various sufficient conditions for the existence of positive periodic solutions of (1.1) and (1.2) by employing two available operators and applying Krasnoselskii’s fixed-point theorem. The techniques used are quite different from that in [11] and our results are more general than those in [11] in two aspects. First, when c < 0, our result enlarges the range of c in [11]. Second, we also establish results for the existence of positive solutions for (1.1) and (1.2) when c > 0, the case for which [11] has not been discussed. Besides, as far as we know, up to this point there has been no result on (1.1) even for the simple case c = 0, and in this paper we present the Green function and integrated form of (1.1) for the first time. This should be helpful for further studies in this type of equations. 2. Some lemmas Let X = {x(t) ∈ C(R, R) : x(t + ω) = x(t), t ∈ R} with norm kxk = supt∈[0,ω] |x(t)|. Clearly, (X, k · k) is a Banach space. Define Cω+ = {x(t) ∈ C(R, (0, +∞)) : x(t + ω) = x(t)},

Cω− = {x(t) ∈ C(R, (−∞, 0)) : x(t + ω) = x(t)}.

Denote M = max{a(t) : t ∈ [0, ω]}, A1 = A2 =

exp(− βw 2 ) β(1 − exp(−βw)) cos βω 2 2β sin

βω 2

,

,

B2 =

m = min{a(t) : t ∈ [0, ω]}, B1 =

β=

√ M,

1 + exp(−βw) , 2β(1 − exp(−βw))

1 2β sin βω 2

,

and F(x) = b(t) f (x(t + r )) − ca(t)x(t + r ). The following is Krasnoselskii’s fixed-point theorem which our results will be based on. Theorem A ([12]). Let X be a Banach space. Assume K is a bounded closed convex subset of X . If Q, S: K → X satisfy (i) Qx + Sy ∈ K , ∀x, y ∈ K , (ii) S is a contractive operator, and (iii) Q is a completely continuous operator in K , then Q + S has a fixed point in K .

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In order to obtain our results, we first give some lemmas. Lemma 2.1. The equation y 00 (t) − M y(t) = h(t),

h ∈ Cω− ,

(2.1)

has a unique ω-periodic solution Z t+ω y(t) = G 1 (t, s)(−h(s))ds,

(2.2)

t

where G 1 (t, s) =

exp(−β(s − t)) + exp(β(s − t − w)) , 2β(1 − exp(−βw))

s ∈ [t, t + ω].

(2.3)

Proof. First it is easy to see that the associate homogeneous equation of (2.1) has solutions y(t) = c1 exp(βt) + c2 exp(−βt). Applying the method of variation of parameters, we get c10 (t) =

h(t) exp(−βt) , 2β

c20 (t) =

h(t) exp(βt) . −2β

Since y(t), y 0 (t) are periodic functions, we have Z t+w h(s) exp(−β(s − w)) c1 (t) = ds, 2β(1 − exp(βw)) t

c2 (t) =

Z

t+w

t

h(s) exp(β(s − w)) ds. 2β(exp(−βw) − 1)

Therefore Z y(t) =

t+ω

G 1 (t, s)(−h(s))ds,

t

where G 1 (t, s) is as given in (2.3).  R t+ω Lemma 2.2. t G 1 (t, s)ds = M1 and 0 < A1 ≤ G 1 (t, s) ≤ B1 for all t ∈ [0, ω] and s ∈ [t, t + ω]. Proof. By the definition of G 1 (t, s), it is easy to see that

∂G 1 (t,s) ∂s

= 0 only if s = t + ω2 . Then from the fact that

exp( −βω ω 2 ) )= = A1 , 2 β(1 − exp(−βω)) 1 + exp(−βω) G 1 (t, t) = G 1 (t, t + ω) = = B1 , 2β(1 − exp(−βω)) G 1 (t, t +

we know 0 < A1 ≤ G 1 ≤ B1 . Besides, Z t+w Z t+w exp(−β(s − t)) + exp(β(s − t − w)) G 1 (t, s)ds = ds 2β(1 − exp(−βw)) t t   t+w 1 1 1 = − exp(−β(s − t)) + exp(β(s − t − w)) 2β(1 − exp(−βw)) β β   t 1 1 1 = − (exp(−βw) − 1) + (1 − exp(−βw)) 2β(1 − exp(−βw)) β β 1 2 = · (1 − exp(−βw)) 2β(1 − exp(−βw)) β 1 1 = 2 = . M β The proof is completed.



(2.4)

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Now we will study the following generalization of (2.1), y 00 (t) − a(t)y(t) = h(t),

h ∈ Cω− .

(2.5)

Define T1 , B1 : X → X by Z t+ω (T1 h)(t) = G 1 (t, s)(−h(s))ds,

(B1 y)(t) = [−M + a(t)]y(t).

(2.6)

t

Clearly T1 , B1 are completely continuous, (T1 h)(t) > 0 for h(t) < 0 and kB1 k ≤ M − m. By Lemma 2.1, the solution of (2.5) can be written in the form y(t) = (T1 h)(t) + (T1 B1 y)(t).

(2.7)

Since kT1 B1 k ≤ kT1 kkB1 k ≤

1 m · (M − m) = 1 − < 1, M M

(2.8)

then y(t) = (I − T1 B1 )−1 (T1 h)(t).

(2.9)

If we define P1 : X → X by (P1 h)(t) = (I − T1 B1 )−1 (T1 h)(t),

(2.10)

it is obvious that y(t) = (P1 h)(t) is the unique ω-periodic solution of (2.5) for any h ∈ Cω− . Moreover, we have Lemma 2.3. P1 is completely continuous and satisfies (T1 h)(t) ≤ (P1 h)(t) ≤

M kT1 hk, m

h ∈ Cω− .

(2.11)

Proof. By Neumann expansions of P1 , we have P1 = (I − T1 B1 )−1 T1 = (I + T1 B1 + (T1 B1 )2 + · · · + (T1 B1 )n + · · ·)T1

(2.12)

= T1 + T1 B1 T1 + (T1 B1 ) T1 + · · · + (T1 B1 ) T1 + · · · . 2

n

Since T1 and B1 are completely continuous, so is P1 . Moreover, by (2.12), and recalling that kT1 B1 k ≤ 1 − (T1 h)(t) > 0 for h(t) < 0, we get (T1 h)(t) ≤ (P1 h)(t) ≤

M k(T1 h)(t)k, m

m M,

and

h ∈ Cω− . 

Similar to the above discussion, we have Lemma 2.4. The equation y 00 (t) + M y(t) = h(t),

h ∈ Cω+ ,

has a unique ω-periodic solution Z t+ω y(t) = G 2 (t, s)h(s)ds,

(2.13)

(2.14)

t

where G 2 (t, s) =

cos β( ω2 + t − s) 2β sin βω 2

,

s ∈ [t, t + ω].

(2.15)

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Proof. First it is easy to see that the associate homogeneous equation of (2.13) has solution y(t) = c1 cos βt + c2 sin βt. Applying the method of variation of parameters, we get c10 (t) =

− sin βt h(t), 2β

c20 (t) =

cos βt h(t). 2β

Noticing that y(t), y 0 (t) are periodic functions, we have Z t+ω Z t+ω h(s) cos(s − ω2 ) h(s) sin(s − ω2 ) c1 (t) = ds, c (t) = ds. 2 2β sin βω 2β sin βω t t 2 2

(2.16)

Therefore y(t) = c1 (t) cos βt + c2 (t) sin βt Z t+ω = G 2 (t, s)h(s)ds, t

where G 2 (t, s) is as defined in (2.15).  R t+ω Lemma 2.5. t G 2 (t, s)ds = M1 . Furthermore, if max{a(t) : t ∈ [0, ω]} < ( πω )2 , then 0 < A2 ≤ G 2 (t, s) ≤ B2 for all t ∈ [0, ω] and s ∈ [t, t + ω]. Proof. By the definition of G 2 (t, s), we have Z t+ω Z t+ω cos β( ω2 + t − s) G 2 (t, s)ds = ds 2β sin βω t t 2 t+ω − sin β( ω2 + t − s) = 2β 2 sin βω 2 t 1 1 = + 2 2β 2 2β 1 = . M On the other hand, it is easy to see that G 2 (t, t) = G 2 (t, t + ω) = G 2 (t, t +

cos βω 2

∂G 2 (t,s) ∂s

2β sin βω 2

= 0 only if s = t + ω2 . Then from the fact that

= A2 ,

ω 1 )= = B2 , 2 2β sin ω2

since max{a(t) : t ∈ [0, ω]} < ( πω )2 , we have 0 < βω 2 < 0 < A2 ≤ G 2 (t, s) ≤ B2 for all t ∈ [0, ω] and s ∈ [t, t + ω].

π 2

βω and then 2β sin βω 2 > 0, 0 < cos 2 < 1, hence 

Now we study the following equation corresponding to (2.13), y 00 (t) + a(t)y(t) = h(t),

h ∈ Cω+ .

Define T2 , B2 : X → X by Z t+ω (T2 h)(t) = G 2 (t, s)h(s)ds,

(2.17)

(B2 y)(t) = [M − a(t)]y(t).

(2.18)

t

Clearly T2 , B2 are completely continuous, (T2 h)(t) > 0 for h(t) > 0 and kB2 k ≤ M − m. By Lemma 2.4, the solution of (2.17) can be written in the form y(t) = (T2 h)(t) + (T2 B2 y)(t).

(2.19)

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W. Han, J. Ren / Nonlinear Analysis 70 (2009) 1393–1406 1 M

Since kT2 B2 k ≤ kT2 kkB2 k ≤

· (M − m) = 1 −

m M

< 1, we have

y(t) = (I − T2 B2 )−1 (T2 h)(t).

(2.20)

Define P2 : X → X by (P2 h)(t) = (I − T2 B2 )−1 (T2 h)(t). Obviously, for any h ∈ solution of (2.17).

(2.21) ( πω )2 ,

if max{a(t) : t ∈ [0, ω]} <

Cω+ ,

y(t) = (P2 h)(t) is the unique positive ω-periodic

Lemma 2.6. P2 is completely continuous. Furthermore, if max{a(t) : t ∈ [0, ω]} < ( πω )2 , then M k(T2 h)(t)k, m

(T2 h)(t) ≤ (P2 h)(t) ≤

for all h ∈ Cω+ .

(2.22)

Now we consider (1.1). It can be rewritten as !00 Z 0 x(t) − c K (r )x(t + r )dr −∞

!

0

Z

K (r )x(t + r )dr

= a(t) x(t) − c −∞

! K (r )x(t + r )dr

= a(t) x(t) − c

Z

R0 −∞

y 00 (t) − a(t)y(t) =

Z

0

Z

0

K (r ) f (x(t + r ))dr.

−∞

K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr.

+

−∞

Taking y(t) = x(t) − c

K (r )x(t + r )dr − λb(t)

−∞

0

Z

0

Z + ca(t)

(2.23)

−∞

K (r )x(t + r )dr , then (2.23) is transformed into 0

K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr.

(2.24)

−∞

Define operators Q 1 , S : X → X by Z

(Q 1 x)(t) = P1

0

! K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr ,

−∞

(Sx)(t) = c

Z

0

(2.25)

K (r )x(t + r )dr.

−∞

Comparing (2.24) to (2.5), it is easy to see that the existence of periodic solutions for (1.1) is equivalent to the existence of solutions for the operator equation (2.26)

Q 1 x + Sx = x in X . Similarly, we consider (1.2). It can be rewritten as !00 Z 0

x(t) − c

K (r )x(t + r )dr

−∞

Z

0

= −a(t) x(t) − c

! K (r )x(t + r )dr

+ λb(t)

−∞

Z

0

= −a(t) x(t) − c −∞

Z

0 −∞

! K (r )x(t + r )dr

Z

0

+ −∞

K (r ) f (x(t + r ))dr − ca(t)

Z

0

K (r )x(t + r )dr.

−∞

K (r )[λb(t) f (x(t + r )) − ca(t)x(t + r )]dr.

(2.27)

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W. Han, J. Ren / Nonlinear Analysis 70 (2009) 1393–1406

Take y(t) = x(t) − c

R0 −∞

y 00 (t) + a(t)y(t) =

K (r )x(t + r )dr , then (2.27) is transformed into

Z

0

K (r )[λb(t) f (x(t + r )) − ca(t)x(t + r )]dr.

(2.28)

−∞

Define operators Q 2 , S : X → X by (Q 2 x)(t) = P2

Z

0

! K (r )[λb(t) f (x(t + r )) − ca(t)x(t + r )]dr ,

−∞

(Sx)(t) = c

Z

0

(2.29)

K (r )x(t + r )dr.

−∞

Contrast (2.28) to (2.17), the existence of periodic solution for (1.2) is equivalent to the existence of solution for operator equation (2.30)

Q 2 x + Sx = x in X . Moreover, by the complete continuity of Pi (i = 1, 2), it is easy to see that Lemma 2.7. Q i (i = 1, 2) is completely continuous in X . Besides, we have Lemma 2.8. If |c| < 1, S is a contractive operator. R0 Proof. For any x1 , x2 ∈ X , we have from −∞ K (r )dr = 1 that Z Z 0 0 K (r )x1 (t + r )dr − c K (r )x2 (t + r )dr |(Sx)1 (t) − (Sx)2 (t)| = c −∞ −∞ Z 0 K (r )[x1 (t + r ) − x2 (t + r )]dr = c −∞ Z 0 K (r )dr ≤ |c|kx1 − x2 k −∞ = |c|kx1 − x2 k which implies that k(Sx)1 (t) − (Sx)2 (t)k ≤ |c|kx1 − x2 k. Obviously, S is a contractive operator if |c| < 1.



3. Main results In this part, we first consider a special case when λ = 1, next we consider the general one. Case 1: λ = 1. Theorem 3.1. If c ∈ (0, 1) and (1 − c)c ≤ F(x) ≤ 1 − c

for all t ∈ [0, ω] and x ∈



 c 1 , , M m

then (1.1) has at least one positive ω-periodic solution x(t) with 0 <

c M

(3.1) ≤ x(t) ≤

1 m.

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Proof. Let K 1 = {x ∈ X : x ∈ [ Mc , m1 ]}. It is obvious that K 1 is a bounded closed convex set in X . Moreover, for any x ∈ K 1 , it is easy to verify that Q 1 , S which defined as (2.25) are continuous and (Q 1 x)(t + ω) = (Q 1 x)(t), (Sx)(t + ω) = (Sx)(t), that is, Q 1 (K 1 ) ⊂ X, S(K 1 ) ⊂ X . Next, we claim that Q 1 x + Sy ∈ K 1 for all x, y ∈ K 1 . Since F(x) ≥ (1 − c)c > 0 which implies R0 −∞ K (r )[−b(t) f (x(t + r )) + ca(t)x(t + r )]dr < 0, then for any x, y ∈ K 1 , by Lemmas 2.2 and 2.3, we have (Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−b(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

0

K (r )y(t + r )dr

−∞

! Z 0 Z 0



K (r )y(t + r )dr K (r )[−b(t) f (x(t + r )) + ca(t)x(t + r )]dr + c

T1

−∞ −∞ Z t+ω Z 0 M ≤ G 1 (t, s) K (r ) [b(s) f (x(s + r )) − ca(s)x(s + r )] dr ds max m t∈[0,ω] t −∞ Z 0 +c K (r )y(t + r )dr M ≤ m

−∞

≤ = = =

Z Z 0 Z M t+ω c 0 G 1 (t, s)(1 − c) K (r )dr ds + K (r )dr m t m −∞ −∞ Z M t+ω c G 1 (t, s)(1 − c)ds + m t m 1 c M (1 − c) + m M m 1 . m

(3.2)

On the other hand, by Lemmas 2.2 and 2.3, (Q 1 x)(t) + (Sy)(t) ! Z Z 0 K (r )[−b(t) f (x(t + r )) + ca(t)x(t + r )]dr + c = P1 t+ω

Z

G 1 (t, s)

t t+ω

Z

Z

0

G 1 (t, s)(1 − c)c

t

Z

Z

0

G 1 (t, s)(1 − c)cds +

t

= (1 − c)c · c . M

Z

0

K (r )y(t + r )dr

−∞

K (r )dr ds +

−∞ t+ω

=

=

K (r ) [b(s) f (x(s + r )) − ca(s)x(s + r )] dr ds + c

−∞

≥

K (r )y(t + r )dr

−∞

−∞

≥

0

c2 M

c2 M

1 c2 + M M (3.3)

Combining (3.2) and (3.3), we get Q 1 x + Sy ∈ K 1 for all x, y ∈ K 1 . Moreover, from Lemmas 2.7 and 2.8, Q 1 is completely continuous and S is a contractive operator in X . Hence by Theorem A, Q 1 + S has a fixed point x ∈ K 1 , that is to say, (1.1) has a positive ω-periodic solution x(t) with 0 < Mc ≤ x(t) ≤ m1 .  Theorem 3.2. If c = 0 and 0 < F(x) ≤ 1, then (1.1) has at least one positive ω-periodic solution x(t) with 0 < x(t) ≤ m1 .

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Proof. In this degenerate case, S is the zero operator. Let K 1 = {x ∈ X : x ∈ [0, m1 ]} as in Theorem 3.1. Following the steps in the proof of Theorem 3.1, we easily get that the Eq. (1.1) has at least a nonnegative ω-periodic solution x(t) with 0 ≤ x(t) ≤ m1 . Since F(x) > 0, it is easy to see from (2.26) and (3.3) that x(t) > 0, i.e., (1.1) has at least a positive ω-periodic solution x(t) with 0 < x(t) ≤ m1 .  m Theorem 3.3. Assume that c ∈ (− M , 0) and if

− cM < F(x) ≤ m,

for all t ∈ [0, ω] and x ∈ [0, 1],

(3.4)

then (1.1) has at least one positive ω-periodic solution x(t) with 0 < x(t) ≤ 1. Proof. Let K 2 = {x ∈ X : x ∈ [0, 1]}. Obviously, K 2 is a bounded closed convex set and Q 1 (K 2 ) ⊂ X, S(K 2 ) ⊂ X . Lemmas 2.7 and 2.8 still hold, and we only need to prove Q 1 x + Sy ∈ K 2 for all x, y ∈ K 2 . In fact, for any x, y ∈ K 2 , we have from Lemmas 2.2 and 2.3 that (Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−b(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

0

K (r )y(t + r )dr

−∞

! Z 0 Z 0



K (r )y(t + r )dr K (r )[−b(t) f (x(t + r )) + ca(t)x(t + r )]dr + c

T1

−∞ −∞ Z t+ω Z 0 M max ≤ G 1 (t, s) K (r ) [b(s) f (x(s + r )) − ca(s)x(s + r )] dr ds m t∈[0,ω] t −∞ Z 0 +c K (r )y(t + r )dr M ≤ m

−∞

Z Z 0 M t+ω ≤ G 1 (t, s)m K (r )dr ds m t −∞ M 1 = ·m· m M = 1.

(3.5)

On the other hand, by Lemmas 2.2 and 2.3, (Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−b(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞ t+ω

Z ≥

G 1 (t, s)

t

>

Z

t+ω

G 1 (t, s)

t

= −cM ·

0

K (r )y(t + r )dr

−∞

Z

0

−∞ Z 0

K (r ) [b(s) f (x(s + r )) − ca(s)x(s + r )] dr ds + c

Z

0

K (r )y(t + r )dr

−∞

K (r )(−cM)dr ds + c

−∞

1 +c M

= 0.

(3.6)

So from (3.5) and (3.6), we get Q 1 x + Sy ∈ K 2 for all x, y ∈ K 2 . Since F(x) > −cM, from (2.26) and (3.6), it is clear that x(t) > 0, hence 0 < x(t) ≤ 1, that is to say the Eq. (1.1) has at least a positive ω-periodic solution x(t) with 0 < x(t) ≤ 1.  Considering the operator Eq. (2.30) in a similar way as above, we can get the following results:

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Theorem 3.4. If max{a(t) : t ∈ [0, ω]} < ( πω )2 , c ∈ (0, 1), and   c 1 (1 − c)c ≤ F(x) ≤ 1 − c for ∀t ∈ [0, ω] and x ∈ , , M m then (1.2) has at least one positive ω-periodic solution x(t) with 0 <

(3.7) c M

≤ x(t) ≤

1 m.

Theorem 3.5. If max{a(t) : t ∈ [0, ω]} < ( πω )2 , c = 0, and 0 < F(x) ≤ 1, then (1.2) has at least one positive ω-periodic solution x(t) with 0 < x(t) ≤ m1 . m , 0), and Theorem 3.6. If max{a(t) : t ∈ [0, ω]} < ( πω )2 , c ∈ (− M

− cM < F(x) ≤ m,

for ∀t ∈ [0, ω], x ∈ [0, 1],

(3.8)

then (1.2) has at least one positive ω-periodic solution x(t) with 0 < x(t) ≤ 1. Case 2: λ > 0. In this case, furthermore, we denote L(r ) = max f (x),

l(r ) = min f (x), 0≤x≤r

0≤x≤r

Let f 0 = limx→0+

f (x) x

and f ∞ = limx→∞

f (x) x .

i 0 = number of zeros in the set { f 0 , f ∞ },

b0 = min b(t). t∈[0,w]

Note i ∞ = number of infinities in the set { f 0 , f ∞ }.

From the definition of i 0 , i ∞ , it is clear that i 0 , i ∞ = 1, 2. Now we will show that (1.1) and (1.2) have i 0 or i ∞ positive w-periodic solution(s) for sufficiently large or small λ, respectively. Theorem 3.7. Assume λb(t) f (x(t + r )) > ca(t)x(t + r ) holds, c ∈ [0, 1). If f 0 = 0, then (1.1) has a positive R w 1ω ω-periodic solution for λ ≥ l(r )cMr . b(s)ds 1

0

Proof. IfR f 0 = 0, we can choose 0 < r1 < 1 so that f (x) ≤ εx for 0 ≤ x ≤ r1 , where ε > 0 satisfies w λM m B1 ε 0 b(s)ds + c ≤ 1. Let K 1 = {x ∈ X : 0 ≤ x(t) ≤ r1 }, it is easy to see that K 1 is a bounded closed convex set of X , and Q 1 (K 1 ) ⊂ X , S(K 1 ) ⊂ X . Next, we will show that Q 1 x + Sy ∈ K 1 for all x, y ∈ K 1 . Since for any x, y ∈ K 1 , (Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

0

K (r )y(t + r )dr

−∞

! Z 0 Z 0



K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c K (r )y(t + r )dr

T1

−∞ −∞ Z Z 0 t+w M ≤ max G 1 (t, s) K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds m t∈[0,w] t −∞ Z 0 +c K (r )y(t + r )dr M ≤ m

M ≤ m

Z

−∞ t+w t

G 1 (t, s)

Z

0

−∞ Z 0

K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds + c

Z

0

K (r )y(t + r )dr

−∞

Z Z 0 M t+w G 1 (t, s) K (r )λb(s) f (x(s + r ))dr ds + c K (r )y(t + r )dr m t −∞ −∞ Z w M ≤ λB1r1 ε b(s)ds + cr1 m 0 ≤ r1 . ≤

(3.9)

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W. Han, J. Ren / Nonlinear Analysis 70 (2009) 1393–1406

On the other hand, by f (x) ≥ l(r1 ) for x ∈ K 1 , we have for any x, y ∈ K 1 that (Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr

≥ T1

Z

t+w

≥ Z

t+w

K (r )y(t + r )dr

−∞

G 1 (t, s)

t

≥

0

+c

−∞

Z

K (r )y(t + r )dr

−∞

!

0

Z

0

G 1 (t, s)

t

Z

0

−∞ Z 0

K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds + c

Z

0

K (r )y(t + r )dr

−∞

K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds

−∞

≥ λA1l(r1 )

Z

w

b(s)ds − c A1 Mr1 ω ≥ 0,

(3.10)

0

if λ ≥

cMr R w 1ω . l(r1 ) 0 b(s)ds

Then from (3.9) and (3.10), we get Q 1 x + Sy ∈ K 1 for all x, y ∈ K 1 . From Lemmas 2.7 and 2.8, Q 1 is a completely continuous operator, S is a contractive operator. Therefore, Q 1 + S has fixed point x ∈ K 1 by Theorem A. Since λb(t) f (x(t + r )) > ca(t)x(t + r ), from (2.26) and (3.10), we can get 0 < x(t) ≤ r1 , which means x(t) is a positive w-periodic solution of (1.1).  Theorem 3.8. Assume λb(t) f (x(t + r )) > ca(t)x(t + r ) holds, c ∈ [0, 1). If f ∞ = 0, then (1.1) has a positive 1 +c R wA1 Mr2 ω . ω-periodic solution for λ ≥ (1−c)H A l(r ) b(s)ds 1

2

0

Proof. If f ∞ = 0, there is an H1 > 1 such that f (x) ≤ εx for x ≥ H1 . Let r2 = max{2, B1AH1 1 }, where ε > 0 satisfies Rw 2m λr2 ε 0 b(s)ds − cm H1 ω ≥ (1−c)r B1 M . Let K 2 = {x ∈ X : H1 ≤ x(t) ≤ r2 }, it is easy to see that K 2 is a bounded closed convex set of X . Next, we will show that Q 1 x + Sy ∈ K 2 for all x, y ∈ K 2 . Since for any x, y ∈ K 2 , (Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

0

K (r )y(t + r )dr

−∞

! Z 0 Z 0



K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c K (r )y(t + r )dr

T1

−∞ −∞ Z Z 0 Z 0 M t+w ≤ G 1 (t, s) K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds + c K (r )y(t + r )dr m t −∞ −∞   Z w M ≤ B1 λr2 ε b(s)ds − cm H1 ω + cr2 m 0 ≤ r2 . M ≤ m

On the other hand, by f (x) ≥ l(r2 ) for x ∈ [H1 , r2 ], we have (Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

Z

0

≥ T1 −∞

0

K (r )y(t + r )dr

−∞

! K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr

Z

0

+c −∞

K (r )y(t + r )dr

(3.11)

1404

W. Han, J. Ren / Nonlinear Analysis 70 (2009) 1393–1406

Z

t+w

≥

G 1 (t, s)

t

Z

0

K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds + c

−∞

Z

t+w

Z

0

≥ A1 t

Z

Z

0

K (r )y(t + r )dr

−∞

K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds + c

Z

−∞

0

K (r )y(t + r )dr

−∞

t+w

(λb(s)l(r2 ) − cMr2 )ds + cH1 Z w ≥ λA1l(r2 ) b(s)ds − c A1 Mr2 ω + cH1 , ≥ A1

t

0

which means (Q 1 x)(t) + (Sy)(t) ≥ H1 ,

(3.12)

(1−c)H1 +c R wA1 Mr2 ω . A1 l(r2 ) 0 b(s)ds

if λ ≥ From (3.11) and (3.12), we get Q 1 x + Sy ∈ K 2 for all x, y ∈ K 2 . Again by Lemmas 2.7 and 2.8, we know Q 1 + S has fixed point x ∈ K 2 by Theorem A. Furthermore, H1 ≤ x(t) ≤ r2 , which means x(t) is a positive w-periodic solution of (1.1).  Theorem 3.9. Assume λb(t) f (x(t + r )) > ca(t)x(t + r ) holds, c ∈ [0, 1). If f 0 = ∞, then (1.1) has a positive (1−c)mr Rw3 ω-periodic solution for 0 < λ ≤ M B L(r ) b(s)ds 1

3

0

Proof. If f 0 = ∞, we can choose 0 ≤ r3 ≤ 1 so that f (x) ≥ ηx for 0 ≤ x ≤ r3 , where the constants η > 0 satisfy λb0 η − cM ≥ 0. let K 3 = {x ∈ X : 0 ≤ x(t) ≤ r3 }, it is easy to see that K 3 is a bounded closed convex set of X . Next, we will show that Q 1 x + Sy ∈ K 3 for all x, y ∈ K 3 . Since for any x, y ∈ K 3 , (Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

Z

0

≥ T1 Z

t+w

! K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr

G 1 (t, s)

t

Z

t+w

≥

G 1 (t, s)

t

K (r )y(t + r )dr

−∞ 0

Z +c

K (r )y(t + r )dr

−∞

−∞

≥

0

Z

0

−∞ Z 0

K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds + c

Z

0

K (r )y(t + r )dr

−∞

K (r )[λb0 η − cM]x(s + r )dr ds

−∞

≥ 0.

(3.13)

On the other hand, by f (x) ≤ L(r3 ) for x ∈ [0, r3 ], thus if λ ≤

(1−c)mr Rw3 , M B1 L(r3 ) 0 b(s)ds

(Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

0

we have

K (r )y(t + r )dr

−∞

! Z 0 Z 0



K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c K (r )y(t + r )dr

T1

−∞ −∞ Z Z 0 Z 0 M t+w ≤ G 1 (t, s) K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds + c K (r )y(t + r )dr m t −∞ −∞ Z t+w M ≤ G 1 (t, s)λb(s)L(r3 )ds + cr3 m t Z w M ≤ λ B1 L(r3 ) b(s)ds + cr3 m 0 ≤ r3 . M ≤ m

(3.14)

1405

W. Han, J. Ren / Nonlinear Analysis 70 (2009) 1393–1406

By (3.13) and (3.14), we get Q 1 x + Sy ∈ K 3 for all x, y ∈ K 3 . Again by Lemmas 2.7 and 2.8, we know Q 1 + S has fixed point x ∈ K 3 by Theorem A. Since λb(t) f (x(t + r )) > ca(t)x(t + r ), from (2.26) and (3.13), we can get 0 < x(t) ≤ r3 , which means x(t) is a positive w-periodic solution of (1.1).  Theorem 3.10. Assume λb(t) f (x(t + r )) > ca(t)x(t + r ) holds, c ∈ [0, 1). If f ∞ = ∞, then (1.1) has a positive 4 +cm R wM B1 H2 ω . ω-periodic solution for 0 < λ ≤ (1−c)mr M B L(r ) b(s)ds 1

4

0

Proof. If f ∞ = ∞, we can choose an H2 > 1 so that f (x) ≥ ηx for x ≥ H2 , let r4 = max{2, B1AH1 2 }, where the Rw constants η > 0 satisfies λA1 ηH2 0 b(s)ds − c A1 Mr4 ω + cH2 ≥ H2 . Let K 4 = {x ∈ X : H2 ≤ x(t) ≤ r4 }, it is easy to see that K 4 is a bounded closed convex set of X . Next, we will show that Q 1 x + Sy ∈ K 4 for all x, y ∈ K 4 . Since for any x, y ∈ K 4 (Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr

≥ T1 −∞

Z

t+w

≥

Z

0

+c

t

Z

0

K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds + c

−∞

Z

t+w

K (r )y(t + r )dr

−∞

G 1 (t, s)

≥ A1

K (r )y(t + r )dr

−∞

!

0

Z

0

Z

0

K (r )y(t + r )dr

−∞

(λb(s)ηH2 − cMr4 )ds + cH2

t

≥ A1 ληH2

w

Z

b(s)ds − c A1 Mr4 ω + cH2 .

(3.15)

0

On the other hand, by f (x) ≤ L(r4 ) for x ∈ [H2 , r4 ], then if λ ≤

(1−c)mr4 +cm R wM B1 H2 ω , M B1 L(r4 ) 0 b(s)ds

(Q 1 x)(t) + (Sy)(t) ! Z 0 Z = P1 K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c −∞

0

we have

K (r )y(t + r )dr

−∞

! Z 0 Z 0



K (r )[−λb(t) f (x(t + r )) + ca(t)x(t + r )]dr + c K (r )y(t + r )dr

T1

−∞ −∞ Z Z 0 Z 0 M t+w ≤ G 1 (t, s) K (r )[λb(s) f (x(s + r )) − ca(s)x(s + r )]dr ds + c K (r )y(t + r )dr m t −∞ −∞ Z M t+w ≤ G 1 (t, s)(λb(s)L(r4 ) − cm H2 )ds + cr4 m t Z w M ≤ B1 (λL(r4 ) b(s)ds − cm H2 ω) + cr4 , m 0 ≤ r4 . M ≤ m

(3.16)

By (3.15) and (3.16), we get Q 1 x + Sy ∈ K 4 for all x, y ∈ K 4 . Again by Lemmas 2.7 and 2.8, we know Q 1 + S has fixed point x ∈ K 4 by Theorem A. Furthermore, H2 ≤ x(t) ≤ r4 , which means x(t) is a positive w-periodic solution of (1.1).  From Theorems 3.7 and 3.8, we can easily get Theorem 3.11. Assume λb(t) f (x(t + r )) > ca(t)x(t + r ) holds, c ∈ [0, 1). If f 0 = f ∞ = 0, then (1.1) has two 1 +c R w 1ω R wA1 Mr2 ω }. positive periodic for λ ≥ max{ l(r )cMr , (1−c)H b(s)ds A l(r ) b(s)ds 1

0

1

2

0

1406

W. Han, J. Ren / Nonlinear Analysis 70 (2009) 1393–1406

From Theorems 3.9 and 3.10, we know Theorem 3.12. Assume λb(t) f (x(t + r )) > ca(t)x(t + r ) holds, c ∈ [0, 1). If f 0 = f ∞ = ∞, then (1.1) has two (1−c)mr 4 +cm Rw3 R wM B1 H2 ω }. positive periodic for 0 < λ ≤ min{ M B L(r , (1−c)mr ) b(s)ds M B L(r ) b(s)ds 1

3

0

1

4

0

Remark 3.1. If adding the assumption max{a(t) : t ∈ [0, ω]} < ( πω )2 , then Theorems 3.7–3.12 also hold for the Eq. (1.2). References [1] H.I. Freedman, J. Wu, Periodic solutions of single-species models with periodic delay, SIAM Journal of Mathematical Analysis 23 (1992) 689–701. [2] J.K. Hale, Theory of Functional Differential Equations, Springer-Verlag, New York, 1977. [3] H.F. Huo, Existence of positive periodic solutions of a neutral delay Lotka–Volterra system with impulses, Computers and Mathematics with Applications 48 (2004) 1833–1846. [4] Y. Kuang, Delay Differential Equations with Applications in Population Dynamics, Academic Press, New Work, 1993. [5] Z. Li, X. Wang, Existence of positive periodic solutions for neutral functional differential equations, Electronic Journal of Differential Equations 34 (2006) 1–8. [6] Y.K. Li, Positive periodic solutions of periodic neutral Lotka–Volterra system with state dependent delays, Journal of Mathematical Analysis and Applications 330 (2007) 1347–1362. [7] G.R. Liu, J.R. Yan, F.Q. Zhang, Positive periodic solutions for a neutral delay Lotka–Volterra system, Nonlinear Analysis TMA 67 (2007) 2642–2653. [8] Y. Luo, W.B. Wang, J.H. Shen, Existence of positive periodic solutions for two kinds of neutral functional differential equations, Applied Mathematics Letters (2007), doi:10.1016/j.aml. [9] V.B. Lal, Karmeshu, S. Kaicker, Modeling innovation diffusion with distributed time lag, Technological Forecasting and Social Change 34 (1988) 103–113. [10] G.P. Rangaiah, P.R. Krishnaswamy, Gamma distribution model—A limiting form of gamma distributed time delay model, Chemical Engineering Science 43 (1988) 929–932. [11] J. Wu, Z.C. Wang, Two periodic solutions of second-order neutral functional differential equations, Journal of Mathematical Analysis and Applications 329 (2007) 677–689. [12] B.L. You, Ordinary Differential Equation Complementary Curriculum, People Education Press, Beijing, 1982. [13] Z.H. Yang, Z.H. Zhou, Periodic solutions of a class of neutral differential models with feedback control, Applied Mathematics and Computation 189 (2007) 996–1009. [14] M.R. Zhang, Nonuniform non-resonance at the first eigenvalue of the p-Laplacian, Nonlinear Analysis TMA 29 (1997) 41–51. [15] Z.X. Zheng, Theory of Functional Differential Equations, Anhui Educational Publishing House, Hefei, 1994.