Electronic Journal of Differential Equations, Vol. 2007(2007), No. 110, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
POSITIVE PERIODIC SOLUTIONS OF NEUTRAL LOGISTIC
DIFFERENCE EQUATIONS WITH MULTIPLE DELAYS
YONGKUN LI, QIAN CHEN
Abstract. Using a fixed point theorem of strict-set-contraction, we some established the existence of positive periodic solutions for the neutral logistic
difference equation, with multiple delays,
p
q
h
i
X
X
∆x(n) = x(n) a(n) −
ai (n)x(n − τi (n)) −
cj (n)∆x(n − σj (n)) .
i=1
j=1
1. Introduction
Using the continuation theory for k-set-contractions and the Mawhin’s continuation theorem, [2], the existence of positive periodic solutions for the following
neutral logistic differential equation with multiple delays
n
n
h
i
X
X
dN (t)
= N (t) a(t)−β(t)N (t)−
bj (t)Nj (t−τj (t))−
cj (t)Nj0 (t−σj (t)) (1.1)
dt
j=1
j=1
are investigated in [3, 4, 11, 12, 13]. Where a(t), βj (t), bj (t), cj (t), τj (t), γj (t) are
continuous ω-periodic functions and a(t) ≥ 0, βj (t) ≥ 0, bj (t) ≥ 0, cj (t) ≥ 0 (j =
1, 2, . . . , n). For the ecological justification of (1.1), see for example [5, 6, 8, 9, 10].
Given a, b be integers and a < b, we employ intervals to denote discrete sets such
as Z[a, b] = {a, a + 1, . . . , b}, Z[a, b) = {a, . . . , b − 1}, Z[a, ∞) = {a, a + 1, . . . }, etc.
Let ω ∈ Z[1, ∞) be fixed. Throughout this work, we denote the product of y(n)
Qn=b
Qn=b
from n = a to n = b by n=a y(n) with the understanding that n=a y(n) = 1 for
all a > b.
The main purpose of this paper is to use a fixed point theorem of strict-setcontraction [1, Theorem 3] to establish the existence of positive periodic solutions
of the following neutral logistic difference equation, with multiple delays,
p
q
h
i
X
X
∆x(n) = x(n) a(n) −
ai (n)x(n − τi (n)) −
cj (n)∆x(n − σj (n)) ,
i=1
(1.2)
j=1
2000 Mathematics Subject Classification. 34K13, 34K40, 92B05.
Key words and phrases. Positive periodic solution; neutral functional difference equation;
strict-set-contraction.
c
2007
Texas State University - San Marcos.
Submitted May 22, 2007. Published August 14, 2007.
Supported by the National Natural Sciences Foundation of China.
1
2
Y. LI, Q. CHEN
EJDE-2007/110
where a, ai , cj ∈ (Z(−∞, ∞), R+ ) and τi , σj ∈ (Z(−∞, ∞), R), i = 1, 2, . . . , p,
j = 1, 2, . . . , q are ω-periodic functions. To the best of our knowledge, this is the
first paper to study the existence of periodic solutions of neutral logistic difference
equations delays.
For convenience, we introduce the notation
Θ :=
ω−1
Y
(1 + a(k)),
Γ=
p
ω−1
X hX
s=0
Θ−1
s=0
k=0
Π=
ω−1
Xh
i=1
ai (s) +
q
X
p
X
ai (s) −
i=1
i
cj (s) ,
fM =
min
{f (n)},
i
cj (s) ,
j=1
max
{f (n)},
n∈Z[0,ω−1]
j=1
fm =
q
X
n∈Z[0,ω−1]
where f is a continuous ω-periodic function. In this paper, we assume that
Qω−1
(H1) k=0
> 1.
Pp (1 + a(k)) P
q
(H2) Θ i=1 ai (n) − j=1 cj (n) ≥ 0.
nP
o
Pq
p
Γ
≥ maxn∈Z[0,ω−1]
(H3) (1 + am ) Θ(Θ−1)
i=1 ai (n) +
j=1 cj (n) .
n
o
M
Pp
Pq
−1)Θ
≤ minn∈Z[0,ω] Θ−1 i=1 ai (n) − j=1 cj (n) .
(H4) Π(aΘ−1
Pq
M
(H5) Θ−1
j=1 cj < 1.
ΘΓ
2. Preliminaries
To obtain the existence of a periodic solution of system (1.2), 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 = ∪i Ai and diam(Ai ) ≤ δ ,
where diam(·) denotes the diameter of the set.
Let E, F be two Banach spaces and D ⊂ E, a continuous and bounded map
Φ : Ω̄ → F is called k-set contractive if for any bounded set S ⊂ D we have
αF (Φ(S)) ≤ kαE (S).
A function Φ is called strict-set-contractive if it is k-set-contractive for some 0 ≤
k < 1.
The following lemma is useful for the proof of our main results of this paper.
Lemma 2.1 ([1, 7]). Let K be a cone of the real Banach space E and Kr,R = {x ∈
K : r ≤ kxk ≤ R} with R > r > 0. Suppose that Φ : Kr,R → K is strict-setcontractive such that one of the following two conditions is satisfied:
(i) Φx x, for all x ∈ K, kxk = r and Φx x, for all x ∈ K, kxk = R.
(ii) Φx x, for all x ∈ K, kxk = r and Φx x, for all x ∈ K, kxk = R.
Then Φ has at least one fixed point in Kr,R .
EJDE-2007/110
POSITIVE PERIODIC SOLUTIONS
3
Finding an ω-periodic solution of (1.1) is equivalent to finding an ω-periodic
solution of the equation
x(n) =
n+ω−1
X
p
q
hX
i
X
G(n, s)x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
s=n
i=1
j=1
where
Qn+ω−1
k=s+1 (1 + a(k))
,
G(n, s) = Qω−1
k=0 (1 + a(k)) − 1
n ≤ s ≤ n + ω − 1.
It is easy to see that, for s ∈ Z[k, k + ω − 1] we have
Qω−1
1
k=0 (1 + a(k))
≤ G(k, s) ≤ Qω−1
.
Qω−1
k=0 (1 + a(k)) − 1
k=0 (1 + a(k)) − 1
Let
X = Y = {x : Z(−∞, ∞) → R, x(n + ω) = x(n), n ∈ Z(−∞, ∞)},
X with the norm |x|0 = maxn∈Z[0,ω−1] {|x(n)|}, x ∈ X and Y with the norm |y|1 =
maxn∈Z[0,ω−1] {|y|0 , |∆y|0 }, y ∈ Y. Then X, Y are Banach spaces. Note that solving
(1.2) is equivalent to solving
x = Φx,
(2.1)
where
(Φx)(n) =
n+ω−1
X
s=n
p
q
hX
i
X
G(n, s)x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
i=1
j=1
for x ∈ Y.
To apply Lemma 2.1 to (1.2), we define the cone K in Y by
K = {x ∈ Y : x(n) ≥ 0 and x(n) ≥ Θ−1 |x|1 , n ∈ Z[0, ω − 1]}.
(2.2)
In what follows, we will give some lemmas concerning K and Φ defined by (2.1)
and (2.2), respectively.
Lemma 2.2. 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 each x ∈ K, it is clear that Φx ∈ Y. In view of (2.2), for n ∈ Z(−∞, ∞),
we obtain
(Φx)(n + ω)
=
n+2ω−1
X
p
q
hX
i
X
G(n + ω, s)x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
s=n+ω
=
n+ω−1
X
u=n
q
X
+
i=1
j=1
p
hX
G(n + ω, u + ω)x(u + ω)
ai (u + ω)x(u + ω − τi (u + ω))
i=1
i
cj (u + ω)∆x(u + ω − σj (u + ω))
j=1
= (Φx)(n).
4
Y. LI, Q. CHEN
EJDE-2007/110
That is, (Φx)(n + ω) = (Φx)(n), n ∈ Z(−∞, ∞). So Φx ∈ Y. In view of (H2), for
x ∈ K, n ∈ Z(−∞, ∞), we have
p
q
n+ω−1
hX
i
X
X
G(n, s)x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
s=n
≥
i=1
G(n, s)x(s)
s=n
≥
j=1
p
hX
n+ω−1
X
ai (s)x(s − τi (s)) −
i=1
q
X
i
cj (s)|∆x(s − σj (s))|
(2.3)
j=1
p
q
hX
i
X
G(n, s)x(s)
ai (s)Θ−1 |x|1 −
cj (s)|x|1 ≥ 0.
n+ω−1
X
s=n
i=1
j=1
Therefore, for x ∈ K, n ∈ Z[0, ω − 1]. We find that
q
p
n+ω−1
i
hX
X
X
Θ
cj (s)∆x(s − σj (s))
ai (s)x(s − τi (s)) +
x(s)
|Φx|0 ≤
Θ − 1 s=n
j=1
i=1
and
(Φx)(n) ≥
=
p
q
ω−1
i
hX
X
1 X
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
x(s)
Θ − 1 s=0
i=1
j=1
p
q
ω−1
hX
i (2.4)
X
1 X
x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
Θ − 1 s=0
i=1
j=1
≥ Θ−1 |Φx|0 .
Now, we show that (Φx)(n) ≥ Θ−1 |(Φx)|1 , n ∈ Z[0, ω − 1]. From (2.2), we have
∆(Φx)(n)
= ∆(Φx)(n + 1) − ∆(Φx)(n)
p
q
n+ω
hX
i
X
X
=
G(n + 1, s)x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
s=n+1
−
i=1
n+ω−1
X
p
hX
G(n, s)x(s)
s=n
j=1
ai (s)x(s − τi (s)) +
i=1
q
X
i
cj (s)∆x(s − σj (s))
j=1
p
hX
= G(n + 1, n + ω)x(n + ω)
ai (n + ω)x(n + ω − τi (n + ω))
i=1
+
q
X
i
cj (n + ω)∆x(n + ω − σj (n + ω))
j=1
+
n+ω−1
X
p
q
hX
i
X
G(n + 1, s)x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
s=n
i=1
j=1
p
q
hX
i
X
− G(n + 1, n)x(n)
ai (n)x(n − τi (n)) +
cj (n)∆x(n − σj (n))
i=1
−
n+ω−1
X
s=n
p
hX
G(n, s)x(s)
i=1
j=1
ai (s)x(s − τi (s)) +
q
X
j=1
i
cj (s)∆x(s − σj (s))
EJDE-2007/110
POSITIVE PERIODIC SOLUTIONS
5
p
hX
= [G(n + 1, n + ω) − G(n + 1, n)]x(n)
ai (n)x(n − τi (n))
i=1
+
q
X
i
cj (n)∆x(n − σj (n))
j=1
+
n+ω−1
X
p
hX
[G(n + 1, s) − G(n, s)]x(s)
ai (s)x(s − τi (s))
s=n
+
q
X
i=1
i
cj (s)∆x(s − σj (s))
j=1
= a(n)
n+ω−1
X
p
q
hX
i
X
G(n, s)x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
s=n
i=1
j=1
p
q
i
hX
X
ai (n)x(n − τi (n)) +
cj (n)∆x(n − σj (n))
− x(n)
i=1
j=1
p
q
hX
i
X
= a(n)(Φx)(n) − x(n)
ai (n)x(n − τi (n)) +
cj (n)∆x(n − σj (n)) .
i=1
(2.5)
j=1
It follows from the above inequality and (2.3) that if (Φx)(n) ≥ 0, then
(∆Φx)(n) ≤ a(n)(Φx)(n) ≤ aM (Φx)(n) ≤ (Φx)(n).
(2.6)
On the other hand, from (2.5) and (H3), if (∆Φx)(n) < 0, then
− ∆(Φx)(n)
p
q
hX
i
X
= x(n)
ai (n)x(n − τi (n)) +
cj (n)∆x(n − σj (n)) − a(n)(Φx)(n)
i=1
≤ |x|21
p
hX
j=1
ai (n) +
i=1
q
X
i
cj (n) − am (Φx)(n)
j=1
p
q
ω−1
i
Xh
X
X
1
2
−1
|x|1
Θ
ai (s) −
cj (s) − am (Φx)(n)
≤ (1 + a )
Θ(Θ − 1)
s=0
i=1
j=1
m
m
n+ω−1
X
m
n+ω−1
X
= (1 + a )
s=n
≤ (1 + a )
p
q
h
i
X
X
1
−1
−1
Θ |x|1 Θ |x|1
ai (s) − |x|1
cj (s) − am (Φx)(n)
Θ−1
i=1
j=1
q
h
i
X
G(n, s)x(s) a(s)x(s − τi (s)) −
cj (s)|∆x(s − σj (s))|
s=n
j=1
− am (Φx)(n)
= (1 + am )(Φx)(n) − am (Φx)(n)
= (Φx)(n).
(2.7)
It follows from (2.6) and (2.7) that |(∆Φx)|0 ≤ |Φx|0 . So |Φx|1 = |Φx|0 . By (2.4)
we have (Φx)(n) ≥ Θ−1 |Φx|1 . Hence, Φx ∈ K. The proof of part (i) is complete.
6
Y. LI, Q. CHEN
EJDE-2007/110
Part (ii): In view of the proof of (i), we only need to prove that (∆Φx)(n) ≥ 0
implies
(∆Φx)(n) ≤ (Φx)(n).
From (2.3), (2.5), (H2) and (H4), we obtain
(∆Φx)(n)
−1
≤ a(n)(Φx)(n) − Θ
|x|1
p
hX
ai (n)x(n − τi (n)) −
q
X
i=1
≤ a(n)(Φx)(n) − Θ−1 |x|21
j=1
p
hX
Θ−1 ai (n) −
i=1
≤ aM (Φx)(n) − |x|21
aM − 1
Θ−1
≤ aM (Φx)(n) − (aM − 1)
s=n
n+ω−1
X
+
i
cj (n)
n+ω−1
X
i=1
ai (s) +
q
X
i
cj (s)
j=1
p
q
hX
i
X
Θ
Θ−1 |x|1
ai (s)|x|1 +
cj (s)|x|1
Θ−1
i=1
j=1
p
hX
ai (s)x(s − τi (s))
G(n, s)x(s)
s=n
q
X
q
X
j=1
p
n+ω−1
X hX
s=n
≤ aM (Φx)(n) − (aM − 1)
i
cj (n)|∆x(n − σj (n))|
i=1
i
cj (s)|∆x(s − σj (s))|
j=1
≤ aM (Φx)(n) − (aM − 1)
n+ω−1
X
p
hX
G(n, s)x(s)
ai (s)x(s − τi (s))
s=n
+
q
X
i=1
i
cj (s)∆x(s − σj (s))
j=1
M
= a (Φx)(n) − (aM − 1)(Φx)(n)
= (Φx)(n).
The proof of (ii) is complete.
Lemma 2.3. Φ(S) is precompact in X for any bounded set S in Y.
Proof. Let d be a constant and S = {x ∈ Y : kxk1 < d} is a bounded set. We prove
that ΦS is compact. To do this, we must show that any sequence in ΦS contains a
convergent subsequence. Thus, let {xm , m ∈ Z[1, ∞)} be a sequence in S. Let us
show that {Φxm , m ∈ Z[1, ∞)} has a convergent subsequence. For convenience, we
set
p
q
hX
i
X
M (x(n)) = x(n)
ai (n)x(n − τi (n)) +
cj (n)∆x(n − σj (n)) ,
i=1
j=1
then
(Φx)(n) =
n+ω−1
X
s=n
G(n, s)x(s)M (x(s)).
EJDE-2007/110
POSITIVE PERIODIC SOLUTIONS
7
It is obvious that the sequence {M (xm (0))}, m ∈ Z[1, ∞), is bounded; so the sequence {M (xm (0))} contains a convergent subsequence. So let the sequence {xm,0 },
be a subsequence of {xm } such that {M (xm,0 (0))} is convergent.
Again, {M (xm,0 (1))}, m ∈ Z[1, ∞), contains a convergent subsequence. So, let
{xm,1 } be a subsequence of {xm,0 , such that {M (xm,1 (1))} is convergent. Observe
that {xm,1 } is a subsequence of {xm } and that {M (xm,1 (0))} and {M (xm,1 (1))}
are convergent.
Now, {M (xm,1 (−1)} contains a convergent subsequence. Let {x−1,m,1 } be a subsequence of {xm,1 } such that {M (x−1,m,1 (−1))} is convergent. Also, {x−1,m,1 } is
a subsequence of {xm } and {M (x−1,m,1 (−1))}, {M (x−1,m,1 (0))}, {M (x−1,m,1 (1))}
are convergent.
Continuing in this fashion we find, for each l ∈ Z(−∞, ∞), a subsequence
{x−(l+1),m,(l+1) }, m ∈ Z[1, ∞)} of {x−l,m,l , m ∈ Z[1, ∞)} such that
{M (x−(l+1),m,(l+1) (l + 1)), m ∈ Z[1, ∞)} and {M (x−(l+1),m,(l+1) (−(l + 1))), m ∈
Z[1, ∞)} is convergent. Observe that also the sequences {M (x−(l+1),m,l+1 (−l)), m ∈
Z[1, ∞)}, . . . , {M (x−(l+1),m,l+1 (0)), m ∈ Z[1, ∞)}, . . . , {M (x−(l+1),m,l+1 (l)), m ∈
Z[1, ∞)} are convergent.
Consider now the sequence {x−u,u,u , u ∈ Z[1, ∞)}. Observe that it is a subsequence of {xm , m ∈ Z[1, ∞)}, and also that {M (x−u,u,u (l)), u ∈ Z[1, ∞)} is
convergent for all l ∈ Z(−∞, ∞). Let us show that {Φx−u,u,u , u ∈ Z[1, ∞)} is a
Cauchy sequence.
Since {M (x−u,u,u (l)), u ∈ Z[1, ∞)} is convergent for all l ∈ Z(−∞, ∞), let > 0
be given, there exists u0 ∈ Z[1, ∞) such that e, g ∈ Z[1, ∞) with e, g ≥ u0 ,
sup
|M (x−e,e,e (n)) − M (x−g,g,g (n))| <
n∈Z[0,ω−1]
ε(Θ − 1)
.
Θ
Therefore, if e, g ∈ Z[1, ∞) with e, g ≥ u0 , for all n ∈ Z, we have
k(Φx−e,e,e )(n) − (Φx−g,g,g )(n)k ≤
n+ω−1
X
G(n, u)|M (x−e,e,e (u)) − M (x−g,g,g (u))|
u=n
≤
Θ
sup |M (x−e,e,e (n)) − M (x−g,g,g (n))|
Θ − 1 n∈[0,T −1]
< ε.
This proves that {Φx−u,u,u , u ∈ Z[1, ∞)} is a Cauchy sequence in X, and with this,
the proof of Lemma 2.3 is complete.
Pq
M
Lemma 2.4. Assume that (H1)–(H3) hold and R j=1 cj < 1.
T
(i) If aM ≤ 1, then Φ : K Ω̄R → K is strict-set-contractive,
T
(ii) If (H4) holds and aM > 1, then Φ : K Ω̄R → K is strict-set-contractive,
where ΩR = {x ∈ Y : |x|1 < R}.
Proof. We prove only (i), since the proof of (ii) is similar. It is easy
Pqto see that Φ
is continuous and bounded. Now we prove that αY (Φ(S)) ≤ R j=1 cM
j αY (S)
for any P
bounded set S ⊂ Ω̄R . Let η = αY (S). Then, for any positive number
S
q
ε < R j=1 cM
j η, there is a finite family of subsets {Si } satisfying S =
i Si
with diam(Si ) ≤ η + ε. Therefore
|x − y|1 ≤ η + ε
for all x, y ∈ Si .
(2.8)
8
Y. LI, Q. CHEN
EJDE-2007/110
Since S and Si are precompact
S in X, it follows that there is a finite family of subsets
{Sij } of Si such that Si = j Sij and
|x − y|0 ≤ ε
for all x, y ∈ Sij .
(2.9)
By Lemma 2.3, we know that Φ(S) is precompact
in X. Then, there is a finite
S
family of subsets {Sijk } of Sij such that Sij = k Sijk and
|Φx − Φy|0 ≤ ε
for all x, y ∈ Sijk .
(2.10)
From (2.3), (2.5) and (2.8)-(2.10) and (H2), for any x, y ∈ Sijk , we obtain
|(∆Φx) − (∆Φy)|0
n
= max
a(n)(Φx)(n) − a(n)(Φy)(n)
n∈Z[0,ω−1]
p
q
hX
i
X
− x(n)
ai (n)x(n − τi (n)) +
cj (n)∆x(n − σj (n))
i=1
j=1
p
q
io
hX
X
ai (n)y(n − τi (n)) +
cj (n)∆y(n − σj (n)) + y(n)
i=1
≤
j=1
{|a(n)[(Φx)(n) − (Φy)(n)]|}
max
n∈Z[0,ω−1]
+
max
n∈Z[0,ω−1]
p
q
i
hX
n
X
ai (n)x(n − τi (n)) +
cj (n)∆x(n − σj (n))
x(n)
i=1
j=1
p
q
hX
io
X
− y(n)
ai (n)y(n − τi (n)) +
cj (n)∆y(n − σj (n)) i=1
j=1
≤ aM |(Φx) − (Φy)|0
p
q
h X
n
X
ai (n)x(n − τi (n)) +
cj (n)∆x(n − σj (n))
+ max
x(n)
n∈Z[0,ω−1]
−
p
X
i=1
ai (n)x(n − τi (n)) +
i=1
+
+
io
cj (n)∆x(n − σj (n)) j=1
max
n∈Z[0,ω−1]
q
X
j=1
q
X
p
n
hX
ai (n)y(n − τi (n))
y(n)
i=1
o
i
cj (n)∆y(n − σj (n)) [x(n) − y(n)]
j=1
≤ aM ε + R
max
n∈Z[0,ω−1]
+
q
X
p
nX
ai (n)|x(n − τi (n)) − y(n − τi (n))|
i=1
o
cj (n)|∆x(n − σj (n)) − ∆y(n − σj (n))|
j=1
+ε
max
n∈Z[0,ω−1]
p
nX
i=1
ai (n)y(n − τi (n)) +
q
X
j=1
o
cj (n)|∆y(n − σj (n))|
EJDE-2007/110
POSITIVE PERIODIC SOLUTIONS
9
q
q
X
X
cM
+ Rε bM +
cM
≤ aM ε + Rε bM + R(η + ε)
j
j
j=1
= Rη
q
X
j=1
cM
+ Ĥε,
j
j=1
Pq
where Ĥ = aM + 2RbM + 2R j=1 cM
j . From the above inequality and (2.10), we
have
q
X
|Φx − Φy|1 ≤ R
cM
η + Ĥε for all x, y ∈ Sijk .
j
j=1
Since ε is arbitrary small, it follows that
q
X
αCω1 (Φ(S)) ≤ R
cM
αCω1 (S).
j
j=1
Therefore, Φ is strict-set-contractive. The proof of Lemma 2.4 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 ωperiodic solution.
Proof. We only need to prove (i), since the proof of (ii) is similar. Let R = Θ(Θ−1)
Γ
and 0 < r < (Θ−1)
.
Then
we
have
0
<
r
<
R.
From
Lemmas
2.2
and
2.4,
we
know
ΘΠ
that Φ is strict-set-contractive on Kr,R . In view of (2.5), we see that if there exists
x∗ ∈ K such that Φx∗ = x∗ , then x∗ is one positive ω-periodic solution of system
(1.2). Now, we shall prove that condition (ii) of Lemma 2.1 hold.
First, we prove that Φx x, for all x ∈ K, |x|1 = r. Otherwise, there exists
x ∈ K, |x|1 = r such that Φx ≥ x. So |x| > 0 and Φx − x ∈ K, which implies that
(Φx)(n) − x(n) ≥ Θ−1 |Φx − x|1 ≥ 0
for all t ∈ [0, ω].
(3.1)
Moreover, for t ∈ [0, ω], we have
(Φx)(n) =
n+ω−1
X
p
q
hX
X
G(n, s)x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
s=n
≤
i=1
p
ω−1
X hX
Θ
r|x|0
Θ−1
s=0
i=1
ai (s) +
j=1
q
X
i
cj (s)
j=1
r
=
Π|x|0
Θ−1
< Θ−1 |x|0 .
In view of the above inequality and (3.1), we have
|x|0 ≤ |Φx| < Θ−1 |x|0 < |x|0 ,
which is a contradiction.
10
Y. LI, Q. CHEN
EJDE-2007/110
Finally, we prove that Φx x, for all x ∈ K, |x|1 = R holds. For this case, we
need to prove that
Φx ≮ x x ∈ K, |x|1 = R.
Suppose, for the sake of contradiction, that there exists x ∈ K and |x|1 = R such
that Φx < x. Thus x − Φx ∈ K \ {0}. Furthermore, for any t ∈ [0, ω], we have
x(n) − (Φx)(n) ≥ Θ−1 |x − Φx|1 > 0.
(3.2)
In addition, for any t ∈ [0, ω], we find
p
q
n+ω−1
hX
X
X
(Φx)(n) =
G(n, s)x(s)
ai (s)x(s − τi (s)) +
cj (s)∆x(s − σj (s))
s=n
i=1
ω−1
Xh
j=1
p
X
q
X
≥
i
1
Θ−1
ai (s) −
cj (s)
Θ−1 |x|21
Θ−1
s=0
i=1
j=1
=
1
ΓR2 = R.
Θ(Θ − 1)
From this inequality and (3.2), we obtain
|x| > |Φx|0 ≥ R,
which is a contradiction. Therefore, conditions (i) and (ii) hold. By Lemma 2.1, we
see that Φ has at least one nonzero fixed point in K. Therefore, system (1.2) has at
least one positive ω-periodic solution. The proof of Theorem 3.1 is complete.
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Department of Mathematics, Yunnan University, Kunming, Yunnan 650091, China
E-mail address: yklie@ynu.edu.cn