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Positive periodic solutions of a periodic neutral delay logistic equation with impulses
Computers and Mathematics with Applications 56 (2008) 2189–2196
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Positive periodic solutions of a periodic neutral delay logistic equation with impulsesI Yongkun Li Department of Mathematics, Yunnan University, Kunming, Yunnan 650091, People’s Republic of China
article
info
a b s t r a c t
Article history: Received 4 April 2006 Received in revised form 1 March 2008 Accepted 25 March 2008
By using a fixed point theorem of strict-set-contraction, some criteria are established for the existence of positive periodic solutions for a periodic neutral functional differential equation with impulses of the form
dx(t )
Keywords: Positive periodic solution Neutral functional differential equation Impulse Strict-set-contraction
dt
" = x(t ) a(t ) −
m X
bi (t )x(t − τi (t )) −
i=1
x(tk ) − x(tk ) = −Ik (x(tk )),
k = 1, 2, . . . .
+
n X
# cj (t )x (t − σj (t )) , 0
j=1
© 2008 Elsevier Ltd. All rights reserved.
1. Introduction Recently, by using the continuation theory for k-set-contractions, Lu [1] studied the existence of positive periodic solutions of the following neutral logistic equation dN (t ) dt
" = N (t ) a(t ) − β(t )N (t ) −
m X
bi (t )N (t − τi (t )) −
n X
# cj (t )N (t − σj (t )) , 0
(1.1)
j=1
i=1
where a, β, bi , cj ∈ C (R, [0, +∞)), τi , σj ∈ C (R, [0, +∞)) (i = 1, 2, . . . , m, j = 1, 2, . . . , n) are ω-periodic functions. Also, Yang and Cao [2], by using Mawhin’s continuation theorem [3], investigated the existence of positive periodic solutions of (1.1) in the case of m = n. However, the main results obtained in [1,2] all required that cj ∈ C 1 , σj ∈ C 2 and σj0 < 1 (j = 1, 2, . . . , n). In this paper, we are concerned with the following neutral delay logistic equation with impulses
dx(t ) dt
" = x(t ) a(t ) −
m X i=1
bi (t )x(t − τi (t )) −
x(tk ) − x(tk ) = −Ik (x(tk )), +
k = 1, 2, . . . ,
n X j=1
# cj (t )x (t − σj (t )) , 0
(1.2)
where a, bi , cj ∈ C (R, R ) and τi , σj ∈ C (R, R), i = 1, 2, . . . , m, j = 1, 2, . . . , n are ω-periodic functions, Ik ∈ C (R+ , R+ ), there exists a positive integer p such that tk+p = tk + ω, Ik+p = Ik , k ∈ Z, [0, ω) ∩ {tk : k ∈ Z} = {t1 , t2 , . . . , tp }. For the ecological justification of (1.1), one can refer to [4–7]. The main purpose of this paper is to establish criteria to guarantee the existence of positive periodic solutions of (1.2) by using a fixed point theorem of strict-set-contraction [Theorems 3, 8]. +
I This work is supported by the National Natural Sciences Foundation of People’s Republic of China under Grant 10361006 and the Natural Sciences Foundation of Yunnan Province under Grant 2003A0001M. E-mail address: [email protected].
0898-1221/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2008.03.044
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Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
For convenience, we introduce the notation R − 0ω a(s)ds
Θ := e
,
ω
Z
∆=
" m X
0
Π=
ω
Z
" m X
0
bi (s) +
i=1
n X
Θ bi (s) −
i=1
n X
# cj (s) ds,
j=1
# cj (s) ds,
f M = max {f (t )},
j=1
t ∈[0,ω]
f m = min {f (t )}, t ∈[0,ω]
where f is a continuous ω-periodic function. Throughout this paper, we assume that: Rω
− 0 a(s)ds (H1 ) Θ < 1. P := e P n (H2 ) m Θ bi (t ) − j=1 cj (t ) ≥ 0. i=1 Pm Pn 2 (H3 ) (1 + am ) 1Θ−Θ∆ ≥ maxt ∈[0,ω] i=1 bi (t ) + j=1 cj (t ) . Pm Pn M (H4 ) ΠΘ((a1−−Θ1)) ≤ mint ∈[0,ω] i=1 Θ bi (t ) − j=1 cj (t ) . P (H5 ) 1Θ−2Θ∆ nj=1 cjM < 1.
2. Preliminaries The following lemma is fundamental to our discussion. Since the proof is similar to those in the literature [9,10], we omit the proof. Lemma 2.1. x(t ) is an ω-periodic solution of (1.2) is equivalent to x(t ) is an ω-periodic solution of the following: x(t ) =
Z
t +ω
G(t , s)x(s)
" m X
t
bi (s)x(s − τi (s)) +
i=1
n X
# cj (s)x (s − σj (s)) + 0
j=1
X
G(t , tk )Ik (x(tk )),
(2.1)
k:tk ∈[t ,t +ω)
where G(t , s) =
e−
Rs
t a(θ)dθ
1 − e−
Rω
0 a(θ)dθ
,
s ∈ [t , t + ω].
(2.2)
In order to obtain the existence of a periodic solution of system (2.1), we first make the following preparations: Let E be a Banach space and K be a cone in E. The semi-order induced by the cone K is denoted by ‘‘≤’’. That is, x ≤ y if and only if y − x ∈ K . In addition, for a bounded subset A ⊂ E, let αE (A) denote the (Kuratowski) measure of non-compactness defined by
( αE (A) = inf δ > 0 : there is a finite number of subsets Ai ⊂ A ) such that A =
[
Ai and diam(Ai ) ≤ δ
,
i
where diam(Ai ) denotes the diameter of the set Ai . Let E, F be two Banach spaces and D ⊂ E, a continuous and bounded map Φ : D → F is called k-set contractive if for any bounded set S ⊂ D we have
αF (Φ (S )) ≤ kαE (S ). Φ is called strict-set-contractive if it is k-set-contractive for some 0 ≤ k < 1. The following lemma cited from Ref. [8,11] is useful for the proof of our main results of this paper. Lemma 2.2 ([8,11]). Let K be a cone of the real Banach space X and Kr ,R = {x ∈ K : r ≤ kxk ≤ R} with R > r > 0. Suppose that Φ : Kr ,R → K is strict-set-contractive such that one of the following two conditions is satisfied: (i) Φ x ≤ 6 x, ∀x ∈ K , kxk = r and Φ x ≥ 6 x, ∀x ∈ K , kxk = R. (ii) Φ x ≥ 6 x, ∀x ∈ K , kxk = r and Φ x ≤ 6 x, ∀x ∈ K , kxk = R. Then Φ has at least one fixed point in Kr ,R .
Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
2191
Define PC (R) = x : R → R, x|(tk ,tk+1 ) ∈ C ((tk , tk+1 ), R), ∃x(tk− ) = x(tk ), x(tk+ ), k ∈ Z
and
PC 1 (R) = x : R → R, x|(tk ,tk+1 ) ∈ C 1 ((tk , tk+1 ), R), ∃x0 (tk− ) = x0 (tk ), x(tk+ ), k ∈ Z . Set
X = {x : x ∈ PC (R), x(t + ω) = x(t ), t ∈ R} with the norm defined by |x|0 = supt ∈[0,ω] {|x(t )|}, and
Y = x : x ∈ PC 1 (R), x(t + ω) = x(t ), t ∈ R
with the norm defined by |x|1 = max{|x|0 , |x0 |0 }. Then X and Y are all Banach spaces. Define the cone K in Y by K = {x ∈ Y, x(t ) ≥ Θ |x|1 , t ∈ [0, ω]}.
(2.3)
Let the map Φ be defined by
(Φ x)(t ) =
Z
t +ω
G(t , s)x(s)
" m X
t
n X
bi (s)x(s − τi (s)) +
# G(t , tk )Ik (x(tk )),
(2.4)
k:tk ∈[t ,t +ω)
j=1
i=1
X
cj (s)x (s − σj (s)) + 0
where x ∈ K , t ∈ R, and G(t , s) is defined by (2.2). Definition 2.1 ([12]). The set F is said to be quasi-equicontinuous in [0, ω] if for any > 0 there exists a δ > 0 such that if x ∈ F , k ∈ Z, t 0 , t 00 ∈ (tk−1 , tk ] ∩ [0, ω], and |t 0 − t 00 | < d, then |x(t 0 ) − x(t 00 )| < . Lemma 2.3 (Compactness criterion [12]). The set F ⊂ X is relatively compact if and only if: (a) F is bounded, that is, kxk ≤ M for each x ∈ F and some M > 0; (b) F is quasi-equicontinuous in [0, ω]. In what follows, we will give some lemmas concerning K and Φ defined by (2.3) and (2.4), respectively. Lemma 2.4. Assume that (H1 )–(H3 ) hold. (i) If aM ≤ 1, then Φ : K → K is well defined. (ii) If (H4 ) holds and aM > 1, then Φ : K → K is well defined. Proof. For any x ∈ K , it is clear that Φ x ∈ Y. In view of (H2 ), for x ∈ K , t ∈ [0, ω], we have m X
bi (t )x(t − τi (t )) +
n X
cj (t )x0 (t − σj (t )) ≥
bi (t )x(t − τi (t )) −
≥
m X
Θ bi (t )|x|1 −
i=1
=
n X
cj (t )|x0 (t − σj (t ))|
j=1
i=1
j=1
i=1
m X
n X
cj (t )|x|1
j=1
" m X
Θ bi (t ) −
i=1
n X
# cj (t ) |x|1 ≥ 0.
(2.5)
j=1
Therefore, for x ∈ K , t ∈ [0, ω], we find
|Φ x|0 ≤
1 1−Θ
ω
Z
x(s)
" m X
0
bi (s)x(s − τi (s)) +
i=1
#
n X
cj (s)x (s − σj (s)) ds +
1
0
j=1
p X
1 − Θ k=1
Ik (x(tk ))
and
Θ (Φ x)(t ) ≥ 1−Θ
Z
Θ = 1−Θ
Z
t +ω
x(s)
t
≥ Θ |Φ x|0 .
0
" m X
bi (s)x(s − τi (s)) +
n X
i=1
ω
x(s)
" m X i=1
# cj (s)x (s − σj (s)) ds + 0
j=1
#
n
bi (s)x(s − τi (s)) +
X j=1
cj (s)x (s − σj (s)) ds + 0
p Θ X Ik (x(tk )) 1 − Θ k=1
p Θ X Ik (x(tk )) 1 − Θ k=1
(2.6)
2192
Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
Now, we show that (Φ x)(t ) ≥ Θ |(Φ x)|1 , t ∈ [0, ω]. From (2.4), we have
(Φ x) (t ) = G(t , t + ω)x(t + ω) 0
" m X
bi (t + ω)x(t + ω − τi (t + ω)) +
i=1
− G(t , t )x(t )
" m X
n X
# cj (t + ω)x (t + ω − σj (t + ω)) 0
j=1
bi (t )x(t − τi (t )) +
n X
# cj (t )x (t − σj (t )) + a(t )(Φ x)(t ) 0
j=1
i=1
" m X
= a(t )(Φ x)(t ) − x(t )
bi (t )x(t − τi (t )) +
i=1
n X
# cj (t )x (t − σj (t )) . 0
(2.7)
j=1
It follows from (2.7) that if (Φ x)0 (t ) ≥ 0, then
(Φ x)0 (t ) ≤ a(t )(Φ x)(t ) ≤ aM (Φ x)(t ) ≤ (Φ x)(t ).
(2.8)
On the other hand, from (2.5), (2.7) and (H3 ), if (Φ x)0 (t ) < 0, then
−(Φ x) (t ) = x(t ) 0
" m X
ai (t )x(t − τi (t )) +
≤ |x|
" m X
# cj (t )x (t − σj (t )) − a(t )(Φ x)(t ) 0
j=1
i=1 2 1
n X
bi (t ) +
n X
# cj (t ) − am (Φ x)(t )
j=1
i=1
# Z ω "X n m X Θ2 2 cj (s) ds − am (Φ x)(t ) |x|1 Θ bi (s) − ≤ (1 + a ) 1−Θ 0 j=1 i=1 # " Z ω n m X X Θ m |x|1 cj (s) ds − am (Φ x)(t ) Θ |x|1 bi (s) − = (1 + a ) Θ |x|1 1−Θ 0 j=1 i=1 " # Z t +ω n m X X 0 m cj (s)|x (s − σj (s))| ds − am (Φ x)(t ) ≤ (1 + a ) bi (s)x(s − τi (s)) − G(t , s)x(s) m
t
≤ (1 + a ) m
Z
t +ω
G(t , s)x(s)
i=1
j=1
" m X
n X
t
bi (s)x(s − τi (s)) +
i=1
# cj (s)x (s − σj (s)) ds − am (Φ x)(t ) 0
j=1
= (1 + am )(Φ x)(t ) − am (Φ x)(t ) = (Φ x)(t ).
(2.9)
It follows from (2.8) and (2.9) that |(Φ x)0 |0 ≤ |Φ x|0 . So |Φ x|1 = |Φ x|0 . By (2.6) we have
(Φ x)(t ) ≥ Θ |Φ x|1 . Hence, Φ x ∈ K . The proof of (i) is complete. (ii) In view of the proof of (i), we only need to prove that (Φ x)0 (t ) ≥ 0 implies
(Φ x)0 (t ) ≤ (Φ x)(t ). From (2.5), (2.7), (H2 ) and (H4 ), we obtain
(Φ x) (t ) ≤ a(t )(Φ x)(t ) − Θ |x|1 0
" m X
bi (t )x(t − τi (t )) −
i=1
" ≤ a(t )(Φ x)(t ) − Θ |x|
2 1
Θ
m X
# cj (t )|x (t − σj (t ))| 0
j=1
bi (t ) −
i=1
aM − 1
n X
n X
# cj (t )
j=1
Z
ω
" m X
n X
#
cj (s) ds bi (s) + Θ (1 − Θ ) 0 j=1 i=1 " # Z t +ω m n X X 1 M M ≤ a (Φ x)(t ) − (a − 1) |x|1 bi (s)|x|1 + cj (s)|x|1 ds 1−Θ t i=1 j=1
≤ a (Φ x)(t ) − Θ |x| M
2 1
Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
≤ aM (Φ x)(t ) − (aM − 1)
Z
t +ω
G(t , s)x(s)
" m X
t
≤ a (Φ x)(t ) − (a − 1) M
M
Z
bi (s)x(s − τj (s)) +
i=1 t +ω
G(t , s)x(s)
" m X
t
n X
2193
# cj (s)|x0 (s − σj (s))| ds
j=1
bi (s)x(s − τj (s, x(s))) +
i=1
n X
# cj (s)x (s − σj (s)) ds 0
j=1
= aM (Φ x)(t ) − (aM − 1)(Φ x)(t ) = (Φ x)(t ). The proof of (ii) is complete.
Lemma 2.5. Assume that (H1 )–(H3 ) hold and R
Pn
j=1
cjM < 1.
¯ R → K is strict-set-contractive, (i) If aM ≤ 1, then Φ : K Ω T ¯ R → K is strict-set-contractive, (ii) If (H4 ) holds and aM > 1, then Φ : K Ω T
where ΩR = {x ∈ Y : |x|1 < R}. Proof. We only need to prove P (i), sincethe proof of (ii) is similar. It is easy to see that Φ is continuous and bounded. Now n ¯ R . Let η = αY (S ). Then, for any positive number we prove that αY (Φ (S )) ≤ R j=1 cjM αY (S ) for any bounded set S ⊂ Ω
ε< R
Pn
cjM η, there is a finite family of subsets {Si } satisfying S =
j=1
S
i
Si with diam(Si ) ≤ η + ε . Therefore
|x − y|1 ≤ η + ε for any x, y ∈ Si .
(2.10)
As S and Si are precompact in X, it follows that there is a finite family of subsets {Sij } of Si such that Si =
S
j
Sij and
|x − y|0 ≤ ε for any x, y ∈ Sij .
(2.11)
In addition, for any x ∈ S and t ∈ [0, ω], we have
|(Φ x)(t )| =
t +ω
Z
G(t , s)x(s)
t
≤
" m X
bi (s)x(s − τi (s)) +
1−Θ
Z 0
ω
" m X
# cj (s)x (s − σj (s)) ds 0
j=1
i=1
R2
n X
bi (s) +
i=1
n X
# cj (s) ds := H
j=1
and
# " n m X X 0 cj (t )x (t − σj (t )) bi (t )x(t − τi (t )) + |(Φ x) (t )| = a(t )(Φ x)(t ) − x(t ) j=1 j=1 ! n m X X ≤ aM H + R2 cjM . bM + i 0
i=1
j=1
Applying Lemma 2.3, we know that Φ (S ) is precompact in X. Then, there is a finite family of subsets {Sijk } of Sij such that S Sij = k Sijk and
|Φ x − Φ y|0 ≤ ε for any x, y ∈ Sijk . From (2.7), (2.10)–(2.12) and (H2 ), for any x, y ∈ Sijk , we obtain
( " # m n X X 0 |(Φ x) − (Φ y) |0 = max a(t )(Φ x)(t ) − a(t )(Φ y)(t ) − x(t ) bi (t )x(t − τi (t )) + cj (t )x (t − σj (t )) t ∈[0,ω] j=1 j=1 " # ) m n X X 0 + y(t ) bi (t )y(t − τi (t )) + cj (t )y (t − σj (t )) i=1 j=1 ( " # m n X X 0 ≤ max {|a(t )[(Φ x)(t ) − (Φ y)(t )]|} + max x(t ) bi (t )x(t − τi (t )) + cj (t )x (t − σj (t )) t ∈[0,ω] t ∈[0,ω] i=1 j=1 " # ) m n X X 0 − y(t ) bi (t )y(t − τj (t )) + cj (t )y (t − σj (t )) j=1 j=1 0
0
(2.12)
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Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
( " ! m n X X 0 ≤ a |(Φ x) − (Φ y)|0 + max x(t ) bi (t )x(t − τi (t )) + cj (t )x (t − σj (t )) t ∈[0,ω] j=1 j=1 !# ) n m X X 0 bi (t )y(t − τi (t )) + cj (t )y (t − σj (t )) − i=1 j=1 ) ( " # n m X X 0 + max y(t ) cj (t )y (t − σj (t )) [x(t ) − y(t )] bi (t )y(t − τi (t )) + t ∈[0,ω] j=1 i=1 ( ) q m X X M 0 0 ≤ a ε + R max bi (t )|x(t − τi (t )) − y(t − τi (t ))| + cj (t )|x (t − σj (t )) − y (t − σj (t ))| M
t ∈[0,ω]
( + ε max
t ∈[0,ω]
i=1
j=1
n
q
X
X
bi (t )y(t − τi (t , y(t ))) +
i=1
! + R(η + ε)
BM i
n X
i=1
= Rη
n X
cj (t )|y (t − σj (t ))|
j=1
m X
≤ aM ε + Rε
) 0
! cjM
m X
+ Rε
j=1
bM i +
i=1
n X
! cjM
j=1
! + Hˆ ε,
cjM
(2.13)
j=1
Pm
ˆ = aM + 2R i=1 bM where H i + 2R From (2.12) and (2.13) we have |Φ x − Φ y|1 ≤
R
n X
Pn
j=1
cjM .
! η + Hˆ ε for any x, y ∈ Sijk .
cjM
j=1
As ε is arbitrary small, it follows that
αY (Φ (S )) ≤ R
n X
! αY (S ).
cjM
j=1
Therefore, Φ is strict-set-contractive. The proof of Lemma 2.5 is complete.
3. Main result Our main result of this paper is as follows: Theorem 3.1. Assume that (H1 )–(H3 ), (H5 ) hold. (i) If aM ≤ 1, then system (1.2) has at least one positive ω-periodic solution. (ii) If (H4 ) holds and aM > 1, then system (1.2) has at least one positive ω-periodic solution. Θ (1−Θ )
Proof. We only need to prove (i), since the proof of (ii) is similar. Let R = 1Θ−2Θ and 0 < r < . Then we have 0 < r < R. Π ∆ From Lemmas 2.4 and 2.5, we know that Φ is strict-set-contractive on Kr ,R . In view of (2.7), we see that if there exists x∗ ∈ K such that Φ x∗ = x∗ , then x∗ is one positive ω-periodic solution of system (1.2). Now, we shall prove that condition (ii) of Lemma 2.2 holds. First, we prove that Φ x 6≥ x, ∀x ∈ K , |x|1 = r. Otherwise, there exists x ∈ K , |x|1 = r such that Φ x ≥ x. So |x| > 0 and Φ x − x ∈ K , which imply that
(Φ x)(t ) − x(t ) ≥ Θ |Φ x − x|1 ≥ 0 for any t ∈ [0, ω].
(3.1)
Moreover, for t ∈ [0, ω], we have
(Φ x)(t ) =
Z
t +ω
G(t , s)x(s)
t
≤
" m X
bi (s)x(s − τi (s)) +
j=1
i=1
1 1−Θ
Z r |x|0 0
ω
" m X i=1
n X
bi (s) +
n X j=1
# cj (s) ds
# cj (s)x (s − σj (s)) ds 0
Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
1
=
1−Θ
2195
Π |x|0
< Θ |x|0 .
(3.2)
In view of (3.1) and (3.2), we have
|x|0 ≤ |Φ x| < Θ |x|0 < |x|0 , which is a contradiction. Finally, we prove that Φ x 6≤ x, ∀x ∈ K , |x|1 = R also hold. For this case, we only need to prove that
Φ x 6< x,
x ∈ K , |x|1 = R.
Suppose, for the sake of contradiction, that there exist x ∈ K and |x|1 = R such that Φ x < x. Thus x − Φ x ∈ K \ {0}. Furthermore, for any t ∈ [0, ω], we have x(t ) − (Φ x)(t ) ≥ Θ |x − Φ x|1 > 0.
(3.3)
In addition, for any t ∈ [0, ω], we find
(Φ x)(t ) =
t +ω
Z
G(t , s)x(s)
" m X
t
bi (s)x(s − τi (s)) +
i=1
Θ ≥ Θ |x|21 1−Θ
Z
ω
" m X
0
#
n X
cj (s)x (s − σj (s)) ds 0
j=1
Θ bi (s) −
n X
# cj (s) ds
j=1
i=1
Θ2 1R2 1−Θ = R.
=
(3.4)
From (3.3) and (3.4), we obtain
|x| > |Φ x|0 ≥ R, which is a contradiction. Therefore, conditions (i) and (ii) of Lemma 2.2 hold. By Lemma 2.2, we see that Φ has at least one non-zero fixed point in K . Thus, system (2.1) has at least one positive ω-periodic solution. Therefore, it follows from Lemma 2.1 that system (1.2) has a positive ω-periodic solution. The proof of Theorem 3.1 is complete. Remark 3.1. Since we do not restrict σj to be non-negative, in fact, (1.2) is a neutral functional differential equation of mixed type. An example Consider the following non-impulsive equation x0 (t ) = x(t )
1 + cos t 4π
− (2 + sin t )x(t − τ1 (t )) −
1 − sin t 0 x (t − σ1 (t )) , 20
(3.5)
where τ , σ ∈ C (R, R) are 2π -periodic functions. Obviously, a(t ) =
1 + cos t 4π
,
b1 (t ) = 2 + sin t ,
c1 (t ) =
1 − sin t 20
.
Furthermore, we have 1
Θ = e− 2 , ∆=
Z
2π
max {Θ b1 (t ) − c1 (t )} > 3 > 0,
0≤t ≤2π
1
[Θ b1 (s) − c1 (s)]ds = 16π e− 2 −
0
(1 + aM )
∆=
Θ2 ∆ > 59 and 1−Θ
1−Θ
Θ 2∆
1 10
99 10
1
π + 6π e− 2 ,
max {b1 (t ) + c1 (t )} < 4,
0≤t ≤2π
π,
c1M < 1.68 × 10−3 < 1.
Hence, (H1 )–(H3 ) and (H5 ) hold and aM ≤ 1. According to Theorem 3.1, system (3.5) has at least one positive 2π -periodic solution. Remark 3.2. Since we do not require σ10 (t ) < 1, the results in [1,2] cannot be used to obtain the existence of positive 2π -periodic solutions of (3.5).
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Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
Acknowledgment The author thanks the anonymous referee for his/her valuable comments and suggestions to improve the content of this article. References [1] S. Lu, On the existence of positive periodic solutions for neutral functional differential equation with multiple deviating arguments, J. Math. Anal. Appl. 280 (2003) 321–333. [2] Z. Yang, J. Cao, Sufficient conditions for the existence of positive periodic solutions of a class of neutral delay models, Appl. Math. Comput. 142 (2003) 123–142. [3] R.E. Gaines, J.L. Mawhin, Coincidence Degree Theory and Nonlinear Differential Equations, Springer-Verlag, Berlin, 1977. [4] Y. Kuang, Delay Differential Equations with Applications in Population Dynamics, Academic Press, New York, 1993. [5] Y. Kuang, A. Feldstein, Boundedness of solutions of a nonlinear nonautonomous neutral delay equation, J. Math. Anal. Appl. 156 (1991) 293–304. [6] K. Gopalsamy, X. He, L. Wen, On a periodic neutral logistic equation, Glasgow Math. J. 33 (1991) 281–286. [7] K. Gopalsamy, B.G. Zhang, On a neutral delay logistic equation, Dyn. Stab. Syst. 2 (1988) 183–195. [8] N.P. Các, J.A. Gatica, Fixed point theorems for mappings in ordered Banach spaces, J. Math. Anal. Appl. 71 (1979) 547–557. [9] J.J. Nieto, Impulsive resonance periodic problems of first order, Appl. Math. Lett. 15 (2002) 489–493. [10] X. Li, X. Lin, D. Jiang, X. Zhang, Existence and multiplicity of positive periodic solutions to functional differential equations with impulse effects, Nonlinear Anal. 62 (2005) 683–701. [11] D. Guo, Positive solutions of nonlinear operator equations and its applications to nonlinear integral equations, Adv. Math. 13 (1984) 294–310 (in Chinese). [12] D. Bainov, P. Simeonov, Impulsive differential equations: Periodic solutions and applications, in: Pitman Monographs Surv. Pure Appl. Math., vol. 66, 1993.
Contents lists available at ScienceDirect
Computers and Mathematics with Applications journal homepage: www.elsevier.com/locate/camwa
Positive periodic solutions of a periodic neutral delay logistic equation with impulsesI Yongkun Li Department of Mathematics, Yunnan University, Kunming, Yunnan 650091, People’s Republic of China
article
info
a b s t r a c t
Article history: Received 4 April 2006 Received in revised form 1 March 2008 Accepted 25 March 2008
By using a fixed point theorem of strict-set-contraction, some criteria are established for the existence of positive periodic solutions for a periodic neutral functional differential equation with impulses of the form
dx(t )
Keywords: Positive periodic solution Neutral functional differential equation Impulse Strict-set-contraction
dt
" = x(t ) a(t ) −
m X
bi (t )x(t − τi (t )) −
i=1
x(tk ) − x(tk ) = −Ik (x(tk )),
k = 1, 2, . . . .
+
n X
# cj (t )x (t − σj (t )) , 0
j=1
© 2008 Elsevier Ltd. All rights reserved.
1. Introduction Recently, by using the continuation theory for k-set-contractions, Lu [1] studied the existence of positive periodic solutions of the following neutral logistic equation dN (t ) dt
" = N (t ) a(t ) − β(t )N (t ) −
m X
bi (t )N (t − τi (t )) −
n X
# cj (t )N (t − σj (t )) , 0
(1.1)
j=1
i=1
where a, β, bi , cj ∈ C (R, [0, +∞)), τi , σj ∈ C (R, [0, +∞)) (i = 1, 2, . . . , m, j = 1, 2, . . . , n) are ω-periodic functions. Also, Yang and Cao [2], by using Mawhin’s continuation theorem [3], investigated the existence of positive periodic solutions of (1.1) in the case of m = n. However, the main results obtained in [1,2] all required that cj ∈ C 1 , σj ∈ C 2 and σj0 < 1 (j = 1, 2, . . . , n). In this paper, we are concerned with the following neutral delay logistic equation with impulses
dx(t ) dt
" = x(t ) a(t ) −
m X i=1
bi (t )x(t − τi (t )) −
x(tk ) − x(tk ) = −Ik (x(tk )), +
k = 1, 2, . . . ,
n X j=1
# cj (t )x (t − σj (t )) , 0
(1.2)
where a, bi , cj ∈ C (R, R ) and τi , σj ∈ C (R, R), i = 1, 2, . . . , m, j = 1, 2, . . . , n are ω-periodic functions, Ik ∈ C (R+ , R+ ), there exists a positive integer p such that tk+p = tk + ω, Ik+p = Ik , k ∈ Z, [0, ω) ∩ {tk : k ∈ Z} = {t1 , t2 , . . . , tp }. For the ecological justification of (1.1), one can refer to [4–7]. The main purpose of this paper is to establish criteria to guarantee the existence of positive periodic solutions of (1.2) by using a fixed point theorem of strict-set-contraction [Theorems 3, 8]. +
I This work is supported by the National Natural Sciences Foundation of People’s Republic of China under Grant 10361006 and the Natural Sciences Foundation of Yunnan Province under Grant 2003A0001M. E-mail address: [email protected].
0898-1221/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2008.03.044
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Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
For convenience, we introduce the notation R − 0ω a(s)ds
Θ := e
,
ω
Z
∆=
" m X
0
Π=
ω
Z
" m X
0
bi (s) +
i=1
n X
Θ bi (s) −
i=1
n X
# cj (s) ds,
j=1
# cj (s) ds,
f M = max {f (t )},
j=1
t ∈[0,ω]
f m = min {f (t )}, t ∈[0,ω]
where f is a continuous ω-periodic function. Throughout this paper, we assume that: Rω
− 0 a(s)ds (H1 ) Θ < 1. P := e P n (H2 ) m Θ bi (t ) − j=1 cj (t ) ≥ 0. i=1 Pm Pn 2 (H3 ) (1 + am ) 1Θ−Θ∆ ≥ maxt ∈[0,ω] i=1 bi (t ) + j=1 cj (t ) . Pm Pn M (H4 ) ΠΘ((a1−−Θ1)) ≤ mint ∈[0,ω] i=1 Θ bi (t ) − j=1 cj (t ) . P (H5 ) 1Θ−2Θ∆ nj=1 cjM < 1.
2. Preliminaries The following lemma is fundamental to our discussion. Since the proof is similar to those in the literature [9,10], we omit the proof. Lemma 2.1. x(t ) is an ω-periodic solution of (1.2) is equivalent to x(t ) is an ω-periodic solution of the following: x(t ) =
Z
t +ω
G(t , s)x(s)
" m X
t
bi (s)x(s − τi (s)) +
i=1
n X
# cj (s)x (s − σj (s)) + 0
j=1
X
G(t , tk )Ik (x(tk )),
(2.1)
k:tk ∈[t ,t +ω)
where G(t , s) =
e−
Rs
t a(θ)dθ
1 − e−
Rω
0 a(θ)dθ
,
s ∈ [t , t + ω].
(2.2)
In order to obtain the existence of a periodic solution of system (2.1), we first make the following preparations: Let E be a Banach space and K be a cone in E. The semi-order induced by the cone K is denoted by ‘‘≤’’. That is, x ≤ y if and only if y − x ∈ K . In addition, for a bounded subset A ⊂ E, let αE (A) denote the (Kuratowski) measure of non-compactness defined by
( αE (A) = inf δ > 0 : there is a finite number of subsets Ai ⊂ A ) such that A =
[
Ai and diam(Ai ) ≤ δ
,
i
where diam(Ai ) denotes the diameter of the set Ai . Let E, F be two Banach spaces and D ⊂ E, a continuous and bounded map Φ : D → F is called k-set contractive if for any bounded set S ⊂ D we have
αF (Φ (S )) ≤ kαE (S ). Φ is called strict-set-contractive if it is k-set-contractive for some 0 ≤ k < 1. The following lemma cited from Ref. [8,11] is useful for the proof of our main results of this paper. Lemma 2.2 ([8,11]). Let K be a cone of the real Banach space X and Kr ,R = {x ∈ K : r ≤ kxk ≤ R} with R > r > 0. Suppose that Φ : Kr ,R → K is strict-set-contractive such that one of the following two conditions is satisfied: (i) Φ x ≤ 6 x, ∀x ∈ K , kxk = r and Φ x ≥ 6 x, ∀x ∈ K , kxk = R. (ii) Φ x ≥ 6 x, ∀x ∈ K , kxk = r and Φ x ≤ 6 x, ∀x ∈ K , kxk = R. Then Φ has at least one fixed point in Kr ,R .
Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
2191
Define PC (R) = x : R → R, x|(tk ,tk+1 ) ∈ C ((tk , tk+1 ), R), ∃x(tk− ) = x(tk ), x(tk+ ), k ∈ Z
and
PC 1 (R) = x : R → R, x|(tk ,tk+1 ) ∈ C 1 ((tk , tk+1 ), R), ∃x0 (tk− ) = x0 (tk ), x(tk+ ), k ∈ Z . Set
X = {x : x ∈ PC (R), x(t + ω) = x(t ), t ∈ R} with the norm defined by |x|0 = supt ∈[0,ω] {|x(t )|}, and
Y = x : x ∈ PC 1 (R), x(t + ω) = x(t ), t ∈ R
with the norm defined by |x|1 = max{|x|0 , |x0 |0 }. Then X and Y are all Banach spaces. Define the cone K in Y by K = {x ∈ Y, x(t ) ≥ Θ |x|1 , t ∈ [0, ω]}.
(2.3)
Let the map Φ be defined by
(Φ x)(t ) =
Z
t +ω
G(t , s)x(s)
" m X
t
n X
bi (s)x(s − τi (s)) +
# G(t , tk )Ik (x(tk )),
(2.4)
k:tk ∈[t ,t +ω)
j=1
i=1
X
cj (s)x (s − σj (s)) + 0
where x ∈ K , t ∈ R, and G(t , s) is defined by (2.2). Definition 2.1 ([12]). The set F is said to be quasi-equicontinuous in [0, ω] if for any > 0 there exists a δ > 0 such that if x ∈ F , k ∈ Z, t 0 , t 00 ∈ (tk−1 , tk ] ∩ [0, ω], and |t 0 − t 00 | < d, then |x(t 0 ) − x(t 00 )| < . Lemma 2.3 (Compactness criterion [12]). The set F ⊂ X is relatively compact if and only if: (a) F is bounded, that is, kxk ≤ M for each x ∈ F and some M > 0; (b) F is quasi-equicontinuous in [0, ω]. In what follows, we will give some lemmas concerning K and Φ defined by (2.3) and (2.4), respectively. Lemma 2.4. Assume that (H1 )–(H3 ) hold. (i) If aM ≤ 1, then Φ : K → K is well defined. (ii) If (H4 ) holds and aM > 1, then Φ : K → K is well defined. Proof. For any x ∈ K , it is clear that Φ x ∈ Y. In view of (H2 ), for x ∈ K , t ∈ [0, ω], we have m X
bi (t )x(t − τi (t )) +
n X
cj (t )x0 (t − σj (t )) ≥
bi (t )x(t − τi (t )) −
≥
m X
Θ bi (t )|x|1 −
i=1
=
n X
cj (t )|x0 (t − σj (t ))|
j=1
i=1
j=1
i=1
m X
n X
cj (t )|x|1
j=1
" m X
Θ bi (t ) −
i=1
n X
# cj (t ) |x|1 ≥ 0.
(2.5)
j=1
Therefore, for x ∈ K , t ∈ [0, ω], we find
|Φ x|0 ≤
1 1−Θ
ω
Z
x(s)
" m X
0
bi (s)x(s − τi (s)) +
i=1
#
n X
cj (s)x (s − σj (s)) ds +
1
0
j=1
p X
1 − Θ k=1
Ik (x(tk ))
and
Θ (Φ x)(t ) ≥ 1−Θ
Z
Θ = 1−Θ
Z
t +ω
x(s)
t
≥ Θ |Φ x|0 .
0
" m X
bi (s)x(s − τi (s)) +
n X
i=1
ω
x(s)
" m X i=1
# cj (s)x (s − σj (s)) ds + 0
j=1
#
n
bi (s)x(s − τi (s)) +
X j=1
cj (s)x (s − σj (s)) ds + 0
p Θ X Ik (x(tk )) 1 − Θ k=1
p Θ X Ik (x(tk )) 1 − Θ k=1
(2.6)
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Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
Now, we show that (Φ x)(t ) ≥ Θ |(Φ x)|1 , t ∈ [0, ω]. From (2.4), we have
(Φ x) (t ) = G(t , t + ω)x(t + ω) 0
" m X
bi (t + ω)x(t + ω − τi (t + ω)) +
i=1
− G(t , t )x(t )
" m X
n X
# cj (t + ω)x (t + ω − σj (t + ω)) 0
j=1
bi (t )x(t − τi (t )) +
n X
# cj (t )x (t − σj (t )) + a(t )(Φ x)(t ) 0
j=1
i=1
" m X
= a(t )(Φ x)(t ) − x(t )
bi (t )x(t − τi (t )) +
i=1
n X
# cj (t )x (t − σj (t )) . 0
(2.7)
j=1
It follows from (2.7) that if (Φ x)0 (t ) ≥ 0, then
(Φ x)0 (t ) ≤ a(t )(Φ x)(t ) ≤ aM (Φ x)(t ) ≤ (Φ x)(t ).
(2.8)
On the other hand, from (2.5), (2.7) and (H3 ), if (Φ x)0 (t ) < 0, then
−(Φ x) (t ) = x(t ) 0
" m X
ai (t )x(t − τi (t )) +
≤ |x|
" m X
# cj (t )x (t − σj (t )) − a(t )(Φ x)(t ) 0
j=1
i=1 2 1
n X
bi (t ) +
n X
# cj (t ) − am (Φ x)(t )
j=1
i=1
# Z ω "X n m X Θ2 2 cj (s) ds − am (Φ x)(t ) |x|1 Θ bi (s) − ≤ (1 + a ) 1−Θ 0 j=1 i=1 # " Z ω n m X X Θ m |x|1 cj (s) ds − am (Φ x)(t ) Θ |x|1 bi (s) − = (1 + a ) Θ |x|1 1−Θ 0 j=1 i=1 " # Z t +ω n m X X 0 m cj (s)|x (s − σj (s))| ds − am (Φ x)(t ) ≤ (1 + a ) bi (s)x(s − τi (s)) − G(t , s)x(s) m
t
≤ (1 + a ) m
Z
t +ω
G(t , s)x(s)
i=1
j=1
" m X
n X
t
bi (s)x(s − τi (s)) +
i=1
# cj (s)x (s − σj (s)) ds − am (Φ x)(t ) 0
j=1
= (1 + am )(Φ x)(t ) − am (Φ x)(t ) = (Φ x)(t ).
(2.9)
It follows from (2.8) and (2.9) that |(Φ x)0 |0 ≤ |Φ x|0 . So |Φ x|1 = |Φ x|0 . By (2.6) we have
(Φ x)(t ) ≥ Θ |Φ x|1 . Hence, Φ x ∈ K . The proof of (i) is complete. (ii) In view of the proof of (i), we only need to prove that (Φ x)0 (t ) ≥ 0 implies
(Φ x)0 (t ) ≤ (Φ x)(t ). From (2.5), (2.7), (H2 ) and (H4 ), we obtain
(Φ x) (t ) ≤ a(t )(Φ x)(t ) − Θ |x|1 0
" m X
bi (t )x(t − τi (t )) −
i=1
" ≤ a(t )(Φ x)(t ) − Θ |x|
2 1
Θ
m X
# cj (t )|x (t − σj (t ))| 0
j=1
bi (t ) −
i=1
aM − 1
n X
n X
# cj (t )
j=1
Z
ω
" m X
n X
#
cj (s) ds bi (s) + Θ (1 − Θ ) 0 j=1 i=1 " # Z t +ω m n X X 1 M M ≤ a (Φ x)(t ) − (a − 1) |x|1 bi (s)|x|1 + cj (s)|x|1 ds 1−Θ t i=1 j=1
≤ a (Φ x)(t ) − Θ |x| M
2 1
Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
≤ aM (Φ x)(t ) − (aM − 1)
Z
t +ω
G(t , s)x(s)
" m X
t
≤ a (Φ x)(t ) − (a − 1) M
M
Z
bi (s)x(s − τj (s)) +
i=1 t +ω
G(t , s)x(s)
" m X
t
n X
2193
# cj (s)|x0 (s − σj (s))| ds
j=1
bi (s)x(s − τj (s, x(s))) +
i=1
n X
# cj (s)x (s − σj (s)) ds 0
j=1
= aM (Φ x)(t ) − (aM − 1)(Φ x)(t ) = (Φ x)(t ). The proof of (ii) is complete.
Lemma 2.5. Assume that (H1 )–(H3 ) hold and R
Pn
j=1
cjM < 1.
¯ R → K is strict-set-contractive, (i) If aM ≤ 1, then Φ : K Ω T ¯ R → K is strict-set-contractive, (ii) If (H4 ) holds and aM > 1, then Φ : K Ω T
where ΩR = {x ∈ Y : |x|1 < R}. Proof. We only need to prove P (i), sincethe proof of (ii) is similar. It is easy to see that Φ is continuous and bounded. Now n ¯ R . Let η = αY (S ). Then, for any positive number we prove that αY (Φ (S )) ≤ R j=1 cjM αY (S ) for any bounded set S ⊂ Ω
ε< R
Pn
cjM η, there is a finite family of subsets {Si } satisfying S =
j=1
S
i
Si with diam(Si ) ≤ η + ε . Therefore
|x − y|1 ≤ η + ε for any x, y ∈ Si .
(2.10)
As S and Si are precompact in X, it follows that there is a finite family of subsets {Sij } of Si such that Si =
S
j
Sij and
|x − y|0 ≤ ε for any x, y ∈ Sij .
(2.11)
In addition, for any x ∈ S and t ∈ [0, ω], we have
|(Φ x)(t )| =
t +ω
Z
G(t , s)x(s)
t
≤
" m X
bi (s)x(s − τi (s)) +
1−Θ
Z 0
ω
" m X
# cj (s)x (s − σj (s)) ds 0
j=1
i=1
R2
n X
bi (s) +
i=1
n X
# cj (s) ds := H
j=1
and
# " n m X X 0 cj (t )x (t − σj (t )) bi (t )x(t − τi (t )) + |(Φ x) (t )| = a(t )(Φ x)(t ) − x(t ) j=1 j=1 ! n m X X ≤ aM H + R2 cjM . bM + i 0
i=1
j=1
Applying Lemma 2.3, we know that Φ (S ) is precompact in X. Then, there is a finite family of subsets {Sijk } of Sij such that S Sij = k Sijk and
|Φ x − Φ y|0 ≤ ε for any x, y ∈ Sijk . From (2.7), (2.10)–(2.12) and (H2 ), for any x, y ∈ Sijk , we obtain
( " # m n X X 0 |(Φ x) − (Φ y) |0 = max a(t )(Φ x)(t ) − a(t )(Φ y)(t ) − x(t ) bi (t )x(t − τi (t )) + cj (t )x (t − σj (t )) t ∈[0,ω] j=1 j=1 " # ) m n X X 0 + y(t ) bi (t )y(t − τi (t )) + cj (t )y (t − σj (t )) i=1 j=1 ( " # m n X X 0 ≤ max {|a(t )[(Φ x)(t ) − (Φ y)(t )]|} + max x(t ) bi (t )x(t − τi (t )) + cj (t )x (t − σj (t )) t ∈[0,ω] t ∈[0,ω] i=1 j=1 " # ) m n X X 0 − y(t ) bi (t )y(t − τj (t )) + cj (t )y (t − σj (t )) j=1 j=1 0
0
(2.12)
2194
Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
( " ! m n X X 0 ≤ a |(Φ x) − (Φ y)|0 + max x(t ) bi (t )x(t − τi (t )) + cj (t )x (t − σj (t )) t ∈[0,ω] j=1 j=1 !# ) n m X X 0 bi (t )y(t − τi (t )) + cj (t )y (t − σj (t )) − i=1 j=1 ) ( " # n m X X 0 + max y(t ) cj (t )y (t − σj (t )) [x(t ) − y(t )] bi (t )y(t − τi (t )) + t ∈[0,ω] j=1 i=1 ( ) q m X X M 0 0 ≤ a ε + R max bi (t )|x(t − τi (t )) − y(t − τi (t ))| + cj (t )|x (t − σj (t )) − y (t − σj (t ))| M
t ∈[0,ω]
( + ε max
t ∈[0,ω]
i=1
j=1
n
q
X
X
bi (t )y(t − τi (t , y(t ))) +
i=1
! + R(η + ε)
BM i
n X
i=1
= Rη
n X
cj (t )|y (t − σj (t ))|
j=1
m X
≤ aM ε + Rε
) 0
! cjM
m X
+ Rε
j=1
bM i +
i=1
n X
! cjM
j=1
! + Hˆ ε,
cjM
(2.13)
j=1
Pm
ˆ = aM + 2R i=1 bM where H i + 2R From (2.12) and (2.13) we have |Φ x − Φ y|1 ≤
R
n X
Pn
j=1
cjM .
! η + Hˆ ε for any x, y ∈ Sijk .
cjM
j=1
As ε is arbitrary small, it follows that
αY (Φ (S )) ≤ R
n X
! αY (S ).
cjM
j=1
Therefore, Φ is strict-set-contractive. The proof of Lemma 2.5 is complete.
3. Main result Our main result of this paper is as follows: Theorem 3.1. Assume that (H1 )–(H3 ), (H5 ) hold. (i) If aM ≤ 1, then system (1.2) has at least one positive ω-periodic solution. (ii) If (H4 ) holds and aM > 1, then system (1.2) has at least one positive ω-periodic solution. Θ (1−Θ )
Proof. We only need to prove (i), since the proof of (ii) is similar. Let R = 1Θ−2Θ and 0 < r < . Then we have 0 < r < R. Π ∆ From Lemmas 2.4 and 2.5, we know that Φ is strict-set-contractive on Kr ,R . In view of (2.7), we see that if there exists x∗ ∈ K such that Φ x∗ = x∗ , then x∗ is one positive ω-periodic solution of system (1.2). Now, we shall prove that condition (ii) of Lemma 2.2 holds. First, we prove that Φ x 6≥ x, ∀x ∈ K , |x|1 = r. Otherwise, there exists x ∈ K , |x|1 = r such that Φ x ≥ x. So |x| > 0 and Φ x − x ∈ K , which imply that
(Φ x)(t ) − x(t ) ≥ Θ |Φ x − x|1 ≥ 0 for any t ∈ [0, ω].
(3.1)
Moreover, for t ∈ [0, ω], we have
(Φ x)(t ) =
Z
t +ω
G(t , s)x(s)
t
≤
" m X
bi (s)x(s − τi (s)) +
j=1
i=1
1 1−Θ
Z r |x|0 0
ω
" m X i=1
n X
bi (s) +
n X j=1
# cj (s) ds
# cj (s)x (s − σj (s)) ds 0
Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
1
=
1−Θ
2195
Π |x|0
< Θ |x|0 .
(3.2)
In view of (3.1) and (3.2), we have
|x|0 ≤ |Φ x| < Θ |x|0 < |x|0 , which is a contradiction. Finally, we prove that Φ x 6≤ x, ∀x ∈ K , |x|1 = R also hold. For this case, we only need to prove that
Φ x 6< x,
x ∈ K , |x|1 = R.
Suppose, for the sake of contradiction, that there exist x ∈ K and |x|1 = R such that Φ x < x. Thus x − Φ x ∈ K \ {0}. Furthermore, for any t ∈ [0, ω], we have x(t ) − (Φ x)(t ) ≥ Θ |x − Φ x|1 > 0.
(3.3)
In addition, for any t ∈ [0, ω], we find
(Φ x)(t ) =
t +ω
Z
G(t , s)x(s)
" m X
t
bi (s)x(s − τi (s)) +
i=1
Θ ≥ Θ |x|21 1−Θ
Z
ω
" m X
0
#
n X
cj (s)x (s − σj (s)) ds 0
j=1
Θ bi (s) −
n X
# cj (s) ds
j=1
i=1
Θ2 1R2 1−Θ = R.
=
(3.4)
From (3.3) and (3.4), we obtain
|x| > |Φ x|0 ≥ R, which is a contradiction. Therefore, conditions (i) and (ii) of Lemma 2.2 hold. By Lemma 2.2, we see that Φ has at least one non-zero fixed point in K . Thus, system (2.1) has at least one positive ω-periodic solution. Therefore, it follows from Lemma 2.1 that system (1.2) has a positive ω-periodic solution. The proof of Theorem 3.1 is complete. Remark 3.1. Since we do not restrict σj to be non-negative, in fact, (1.2) is a neutral functional differential equation of mixed type. An example Consider the following non-impulsive equation x0 (t ) = x(t )
1 + cos t 4π
− (2 + sin t )x(t − τ1 (t )) −
1 − sin t 0 x (t − σ1 (t )) , 20
(3.5)
where τ , σ ∈ C (R, R) are 2π -periodic functions. Obviously, a(t ) =
1 + cos t 4π
,
b1 (t ) = 2 + sin t ,
c1 (t ) =
1 − sin t 20
.
Furthermore, we have 1
Θ = e− 2 , ∆=
Z
2π
max {Θ b1 (t ) − c1 (t )} > 3 > 0,
0≤t ≤2π
1
[Θ b1 (s) − c1 (s)]ds = 16π e− 2 −
0
(1 + aM )
∆=
Θ2 ∆ > 59 and 1−Θ
1−Θ
Θ 2∆
1 10
99 10
1
π + 6π e− 2 ,
max {b1 (t ) + c1 (t )} < 4,
0≤t ≤2π
π,
c1M < 1.68 × 10−3 < 1.
Hence, (H1 )–(H3 ) and (H5 ) hold and aM ≤ 1. According to Theorem 3.1, system (3.5) has at least one positive 2π -periodic solution. Remark 3.2. Since we do not require σ10 (t ) < 1, the results in [1,2] cannot be used to obtain the existence of positive 2π -periodic solutions of (3.5).
2196
Y. Li / Computers and Mathematics with Applications 56 (2008) 2189–2196
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