Electronic Journal of Differential Equations, Vol. 2008(2008), No. 88, pp. 1–11.
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 FOR A PREDATOR-PREY
MODEL WITH TIME DELAYS AND IMPULSIVE EFFECT
SHAN GAO, YONGKUN LI
Abstract. In this article, a two-species predator-prey model with time delays
and impulsive effect is investigated. By using Mawhin’s continuation theorem
of coincidence degree theory, sufficient conditions are obtained for the existence
of positive periodic solutions.
1. Introduction
In the past few years, predator-prey models and with many kinds of functional
responses have been of great interest to both applied mathematicians and ecologists
see references in this article. Recently, by using Floquet theory of linear periodic
impulsive equation, Song and Li [15] considered the following T -periodic predatorprey model with modified Leslie-Gower and Holling-type II schemes and impulsive
effect

a1 (t)y(t) 

ẋ(t) = x(t) r1 (t) − b1 (t)x(t) −
x(t) + k1 (t) 
t 6= τk , k ∈ Z+ ,

a2 (t)y(t) 

ẏ(t) = y(t) r2 (t) −
x(t) + k2 (t)
)
+
x(τk ) = (1 + hk )x(τk )
t = τk , k ∈ Z+ .
y(τk+ ) = (1 + gk )y(τk )
where b1 (t), ri (t), ai (t), ki (t)(i = 1, 2) are continuous ω-periodic functions such that
b1 (t) > 0, ri (t) > 0, ai (t) > 0, ki (t) > 0(i = 1, 2) and Z+ = {1, 2, . . . }; hk , gk (k ∈
Z+ ) are constants and there exists an integer q > 0 such that hk+q = hk , gk+q =
gk , τk+q = τk + ω, and 1 + hk > 0, 1 + gk > 0 for all k ∈ Z+ . They obtain some
conditions for the linear stability of trivial periodic solution and semitrivial periodic
solutions.
However, as pointed out in [9], naturally, more realistic and interesting models
of single or multiple species growth should take into account both the seasonality
of the changing environment and the effects of time delays.
2000 Mathematics Subject Classification. 34K13, 34K45, 92D25.
Key words and phrases. Predator-prey model; impulse; positive periodic solution;
coincidence degree.
c
2008
Texas State University - San Marcos.
Submitted April 7, 2008. Published June 21, 2008.
Supported by the National Natural Sciences Foundation of China and the Natural Sciences
Foundation of Yunnan Province, China.
1
2
S. GAO, Y. LI
EJDE-2008/88
In this paper, we consider the following ω-periodic predator-prey system with
time delays and impulses:

a1 (t)y(t − σ1 (t)) 

ẋ(t) = x(t) r1 (t) − b1 (t)x(t − τ (t)) −
x(t − τ1 (t)) + k1 (t) 
t 6= tk , k ∈ Z+ ,

a2 (t)y(t − σ2 (t)) 

ẏ(t) = y(t) r2 (t) −
x(t − τ2 (t)) + k2 (t)
)
+
x(tk ) = Ik (x(tk )) + x(t−
k)
t = tk , k ∈ Z+ .
−
y(t+
k ) = Jk (y(tk )) + y(tk )
(1.1)
−
+
−
where x(t+
),
x(t
),
y(t
),
y(t
)
represent
the
right
and
the
left
limit
of
x(t
)
k and
k
k
k
k
y(tk ), respectively, in this paper, we assume that x, y are left continuous at tk ; b1 (t),
τ (t), ai (t), ri (t), ki (t), σi (t), τi (t) (i = 1, 2) are all positive periodic continuous
functions with period ω > 0 and Z+ = {1, 2, . . . }; Ik , Jk ∈ C(R+ , R) satisfy that
Ik (u) > −u, Jk (v) > −v, and there exists a positive integer p such that tk+p =
tk + ω, Ik+p = Ik , Jk+p = Jk , k ∈ Z+ . Without loss of generality, we also assume
that [0, ω) ∩ {tk : k ∈ Z+ } = {t1 , t2 , . . . , tp }.
Our purpose of this paper is by using continuation theorem of coincidence degree theory [6] to establish criteria to guarantee the existence of positive periodic
solutions of system (1.1).
2. Notation and preliminaries
To obtain our main result of this paper, we first need to make the following
preparations. For any non-negative integer q, let
C (q) [0, ω; t1 , t2 , . . . , tp ]
n
(q) −
= x : [0, ω] → R such that x(q) (t) exists for t 6= t1 , . . . , tp ; x(q) (t+
(tk )
k ), x
o
exists at t1 , . . . , tp ; and x(j) (tk ) = x(j) (t−
),
k
=
1,
2,
.
.
.
,
p,
j
=
0,
1,
2,
.
.
.
,
q
k
with the norm
kxkq = max{ sup |x(j) (t)|}qj=0 .
t∈[0,ω]
(q)
It is easy to see that C [0, ω; t1 , t2 , . . . , tp ] is a Banach space and the functions
in C[0, ω; t1 , t2 , . . . , tp ] are continuous with respect to t different from t1 , t2 , . . . , tp .
Let
P Cω = x ∈ C[0, ω; t1 , t2 , . . . , tp ] : x(0) = x(ω)
with the same norm as that of C[0, ω; t1 , t2 , . . . , tp ].
Let X, Y be normed vector spaces, L : Dom L ⊂ X → Y be a linear mapping,
and N : X → Y be a continuous mapping. The mapping L will be called a Fredholm
mapping of index zero if dim ker L = codim Im L < +∞ and Im L is closed in Y .
If L is a Fredholm mapping of index zero, and there exist continuous projectors:
P : X → X and Q : Y → Y such that Im P = ker L, ker Q = Im L = Im(I − Q).
It follows that mapping L|Dom L∩ker P : (I − P )X → Im L is invertible. We denote
the inverse of that mapping by KP . If Ω is an open bounded subset of X, the
mapping N will be called L-compact on Ω if QN (Ω) is bounded and KP (I − Q)N :
Ω → X is compact. Since Im Q is isomorphic to ker L, there exists an isomorphism
J : Im Q → ker L.
EJDE-2008/88
POSITIVE PERIODIC SOLUTIONS
3
Definition 2.1. The set F is said to be quasi-equicontinuous in [0, ω] if for any
> 0 there exists δ > 0 such that if x ∈ F , k ∈ Z+ , t1 , t2 ∈ (tk−1 , tk ) ∩ [0, ω],
|t1 − t2 | < δ, then |x(t1 ) − x(t2 )| < .
Lemma 2.2 ([2]). The set F ⊂ P Cω is relatively compact if and only if
(1) F is bounded, that is, kf kP Cω = kf k0 = supt∈[0,ω] |f (t)| ≤ M for each
f ∈ F and some M > 0;
(2) F is quasi-equicontinuous in Dom f .
Now, we introduce Mawhin’s continuation theorem.
Lemma 2.3 ([6]). Let Ω ⊂ X be an open bounded set and let N : X → Y be a
continuous operator which is L-compact on Ω. Assume
(a) for each λ ∈ (0, 1), x ∈ ∂Ω ∩ Dom L, Lx 6= λN x,
(b) for each x ∈ ∂Ω ∩ ker L, QN x 6= 0,
(c) deg(JQN, Ω ∩ ker L, 0) 6= 0.
Then Lx = N x has at least one solution in Ω ∩ Dom L.
Throughout this paper, we assume that there exist p1k , p2k , q1k , q2k ∈ R, k ∈ Z+
such that
Ik (u)
Ik (u)
≥ q1k > −1, sup
≤ p1k < +∞,
inf
u>0
u
u
u>0
Jk (v)
Jk (v)
inf
≥ q2k > −1, sup
≤ p2k < +∞.
v>0
v
v
v>0
For convenience, we introduce the notation
Z
1 ω
f=
f (t) dt, f M = max {f (t)},
ω 0
t∈[0, ω]
l1k = max | ln(1 + p1k )|, | ln(1 + q1k )| , l2k = max | ln(1 + p2k )|, | ln(1 + q2k )|
where f is a continuous ω-periodic function and k ∈ Z+ .
3. Main result
Let
Qp
r +
1
1
ω
ln(
r +
2
1
ω
ln(
k=1 (1
+ p1k )) ,
b1
p
p
Y
X
M1 = H1 + 2ωr1 + ln
(1 + p1k ) +
l1k ;
H1 = ln
H2 = ln
M2 = H2 − 2ωr2 − ln
H3 = ln
k=1
Qp
k=1 (1
a2
( k2 )
p
Y
k=1
+ q2k )) ,
p
X
(1 + p2k ) −
l2k ;
k=1
1
ω
(k M + eM1 )(r +
2
2
k=1
ln(
Qp
k=1 (1
+ p2k ))) a2
p
p
Y
X
M3 = H3 + 2ωr2 + ln
(1 + p2k ) +
l2k ;
k=1
k=1
,
4
S. GAO, Y. LI
H4 = ln
1
ω
r1 +
ln(
EJDE-2008/88
Qp
k=1 (1
+ q1k )) − eM3 ( ak11 ) ,
b1
p
p
Y
X
M4 = H4 − 2ωr1 − ln
(1 + p1k ) −
l1k ;
k=1
k=1
Our main result of this paper is as follows:
Theorem 3.1. If
r1 +
p
1 Y
ln
(1 + p1k ) > 0,
ω
k=1
r2 +
r1 +
1
ln
ω
1
ln
ω
p
Y
k=1
p
Y
(1 + q2k ) > 0,
k=1
a1
(1 + q1k ) − eM3 ( ) > 0,
k1
then (1.1) has at least one ω-periodic positive solution.
Proof. Let x(t) = eu(t) , y(t) = ev(t) then (1.1) is reformulated as

a1 (t) exp{v(t − σ1 (t))} 

exp{u(t − τ1 (t))} + k1 (t) 

a2 (t) exp{v(t − σ2 (t))}


v̇(t) = r2 (t) −
exp{u(t − τ2 (t))} + k2 (t)
for t 6= tk , k ∈ Z+ , and
)
+
u(tk ) = ek (u(tk )) + u(t−
k)
t = tk , k ∈ Z+ ,
−
v(t+
k ) = fk (v(tk )) + v(tk )
u̇(t) = r1 (t) − b1 (t) exp{u(t − τ (t))} −
(3.1)
where
I (exp{u(t )}) + exp{u(t )} k
k
k
,
exp{u(tk )}
J (exp{v(t )}) + exp{v(t )} k
k
k
fk (v(tk )) = ln
.
exp{v(tk )}
ek (u(tk )) = ln
It is easy to see that
ln(1 + q1k ) ≤ ek (u(tk )) ≤ ln(1 + p1k ),
ln(1 + q2k ) ≤ fk (v(tk )) ≤ ln(1 + p2k ).
If system (3.1) has an ω-periodic solution (u(t), v(t)), then
(eu(t) , ev(t) ) = (x∗ (t), y ∗ (t))
is a positive ω-periodic solution to system (1.1). So, in the following, we discuss
the existence of ω-periodic solution to system (3.1). Here, we denote
a1 (t) exp{v(t − σ1 (t))}
,
exp{u(t − τ1 (t))} + k1 (t)
a2 (t) exp{v(t − σ2 (t))}
B(t) = r2 (t) −
.
exp{u(t − τ2 (t))} + k2 (t)
A(t) = r1 (t) − b1 (t) exp{u(t − τ (t))} −
EJDE-2008/88
POSITIVE PERIODIC SOLUTIONS
5
To use the continuation theorem of coincidence degree theory to establish the
existence of an ω-periodic solution of (3.1), we take
X = P Cω × P Cω ,
Y = X × R2p .
Then X is a Banach space with the norm
kxkX = k(u, v)kX = kuk0 + kvk0 = sup |u(t)| + sup |v(t)|,
t∈[0,ω]
t∈[0,ω]
and Y is also a Banach space with the norm
kzkY = kxkX + kyk2 ,
x ∈ X, y ∈ R2p ,
where k · k2 in Rn is defined as
kξk2 = k(ξ1 , ξ2 , . . . , ξn )k2 =
n
X
|ξi |.
i=1
So if x = (u, v) ∈ X ∩ R2 , then kxkX = kxk2 . Let
n
o
Dom L = (u, v) : u, v ∈ C (1) [0, ω; t1 , t2 , . . . , tp ]; u(0) = u(ω), v(0) = v(ω) ,
L : Dom L ∩ X → Y
(u, v) → (u̇, v̇, ∆u(t1 ), . . . , ∆u(tp ), ∆v(t1 ), . . . , ∆v(tp )),
and let N : X → Y with
N (x) = N (u, v) = A(t), B(t), ∆u(t1 ), . . . , ∆u(tp ), ∆v(t1 ), . . . , ∆v(tp ) .
n
o
Obviously, ker L = (u, v) : u, v ∈ R, t ∈ [0, ω] = R2 ,
Z
n
Im L = z = (f, g, c1 , . . . , cp , d1 , . . . , dp ) ∈ Y :
ω
f (s) ds +
0
Z
ω
g(s) ds +
0
p
X
dk = 0
p
X
ck = 0,
k=1
o
k=1
and dim ker L = codim Im L = 2. So that, Im L is closed in Y , L is a Fredholm
mapping of index zero. Define the two projectors
Z ω
1
Px = ,
x(t) dt,
ω 0
Z
p
p
1Z ω
X
X
1 ω
Qz =
f (s) ds +
ck ,
g(s) ds +
dk , 0, . . . , 0 .
| {z }
ω 0
ω 0
k=1
k=1
2p
It is easy to show that P and Q are continuous and satisfy
Im P = ker L = R2 , Im L = ker Q = Im(I − Q).
Further, let LP = LDom L∩ker P and the generalized inverse KP = L−1
P is given by
Z Z
p
p
Z t
X
X
1 ω t
1X
ck −
f (s) ds dt −
ck +
tk ck ,
KP z =
f (s) ds +
ω 0 0
ω
0
t>tk
k=1
k=1
Z t
Z Z
p
p
X
X
1 ω t
1X
g(s) ds +
dk −
g(s) ds dt −
dk +
tk d k .
ω 0 0
ω
0
t>t
k
k=1
k=1
6
S. GAO, Y. LI
EJDE-2008/88
Thus, the expression of QN x is
1Z
ω
ω
A(s) ds +
0
p
X
k=1
1
ek (u(tk )) ,
ω
ω
Z
B(s) ds +
0
p
X
k=1
fk (v(tk )) , 0, . . . , 0 ,
| {z }
2p
and then
KP (I − Q)N x
Z Z
p
Z t
X
1X
1 ω t
A(s) ds dt +
tk ek (u(tk ))
=
A(s) ds +
ek (u(tk )) −
ω 0 0
ω
0
t>tk
k=1
Z ω
p
1
t
1
t X
+( − )
A(s) ds − ( + )
ek (u(tk )) ,
2 ω 0
2 ω
k=1
Z Z
Z t
p
X
1 ω t
1X
tk fk (v(tk ))
B(s) ds +
fk (v(tk )) −
B(s) ds dt +
ω 0 0
ω
0
t>tk
k=1
Z ω
p
1
t
1
t X
+( − )
B(s) ds − ( + )
fk (v(tk )) .
2 ω 0
2 ω
k=1
Hence, QN and KP (I − Q)N are both continuous. Using Lemma 2.2, it is easy to
show that KP (I − Q)N (Ω) is compact for any open bounded set Ω ⊂ X. Moreover,
QN (Ω) is bounded. Therefore, N is L-compact on Ω for any open bounded set
Ω ⊂ X.
Now, it needs to show that there exists a domain Ω that satisfies all the requirements given in Lemma 2.3. Corresponding to operator equation Lx = λN x,
λ ∈ (0, 1), x = (u, v), we get

h
a1 (t) exp{v(t − σ1 (t))} i

u̇(t) = λ r1 (t) − b1 (t) exp{u(t − τ (t))} −
exp{u(t − τ1 (t))} + k1 (t) 
h

a2 (t) exp{v(t − σ2 (t))} i


v̇(t) = λ r2 (t) −
exp{u(t − τ2 (t))} + k2 (t)
for t 6= tk , k ∈ Z+ , and
)
+
u(tk ) = λek (u(tk )) + u(t−
k)
t = tk , k ∈ Z+ .
−
v(t+
k ) = λfk (v(tk )) + v(tk )
(3.2)
Suppose x = (u, v) is an ω-periodic solution to system (3.2). By integrating (3.2)
over [0, ω] we obtain
Z
ω
r1 (t) dt +
0
p
X
ek (u(tk ))
k=1
Z
=
ω
h
b1 (t) exp{u(t − τ (t))} +
0
Z
ω
r2 (t) dt +
0
p
X
k=1
Z
fk (v(tk )) =
0
a1 (t) exp{v(t − σ1 (t))} i
dt,
exp{u(t − τ1 (t))} + k1 (t)
ω
a2 (t) exp{v(t − σ2 (t))}
dt.
exp{u(t − τ2 (t))} + k2 (t)
(3.3)
EJDE-2008/88
POSITIVE PERIODIC SOLUTIONS
From (3.2) and (3.3), we have
Z ω
Z
|u̇(t)| dt < 2
0
r1 (t) dt +
0
ω
Z
k=1
p
X
ω
Z
|v̇(t)| dt < 2
0
p
X
ω
r2 (t) dt +
0
7
ek (u(tk )),
(3.4)
fk (v(tk )).
(3.5)
k=1
Since u(t), v(t) ∈ P Cω , there exist ξ1 , ξ2 , η1 , η2 ∈ [0, ω] such that
u(ξ1 ) = min u(t),
u(η1 ) = max u(t),
v(ξ2 ) = min v(t),
v(η2 ) = max v(t).
t∈[0,ω]
t∈[0,ω]
t∈[0,ω]
(3.6)
t∈[0,ω]
Then by (3.3) and (3.6), we obtain
Z
p
1X
1 ω
ek (u(tk )) ≥
b1 (t)eu(ξ1 ) dt ,
ω
ω 0
k=1
Z
p
X
1
1 ω a2 (t)ev(η2 )
r2 +
fk (v(tk )) ≤
dt ;
ω
ω 0
k2 (t)
r1 +
k=1
that is,
u(ξ1 ) ≤ ln
r1 +
1
ω
Pp
k=1
ek (u(tk )) i
b1
≤ ln
r1 +
≥ ln
r2 +
1
ω
ln
Qp
k=1 (1
+ p1k ) =: H1
+ q2k ) =: H2 .
b1
and
v(η2 ) ≥ ln
r2 +
1
ω
Pp
k=1 fk (v(tk ))
a2
k2
1
ω
ln
Qp
k=1 (1
a2
k2
Hence
Z
ω
u(t) ≤ u(ξ1 ) +
|u̇(t)| dt +
0
p
X
|ek (u(tk ))|
k=1
≤ H1 + 2ωr1 + ln
p
Y
p
X
(1 + p1k ) +
l1k =: M1
k=1
k=1
and
Z
v(t) ≥ v(η2 ) −
ω
|v̇(t)| dt −
0
≥ H2 − 2ωr2 − ln
p
X
|fk (v(tk ))|
k=1
p
Y
k=1
p
X
(1 + p2k ) −
l2k =: M2 .
k=1
So we have
r2 +
Z
Z
p
1X
1 ω a2 (t)ev(ξ2 )
1 ω a2 (t)ev(ξ2 )
fk (v(tk )) ≥
dt
≥
dt ;
ω
ω 0 k2 (t) + eM1
ω 0 k2M + eM1
k=1
that is,
k M + eM1 r + 1 Pp f (v(t )) 2
k
2
k=1 k
ω
v(ξ2 ) ≤ ln
a2
8
S. GAO, Y. LI
≤ ln
k M + e M1 r +
2
2
EJDE-2008/88
1
ω
ln(
Qp
k=1 (1
+ p2k )) a2
=: H3 .
Thus
ω
Z
|v̇(t)| dt +
v(t) ≤ v(ξ2 ) +
0
p
X
|fk (v(tk ))|
k=1
p
Y
≤ H3 + 2ωr2 + ln
p
X
(1 + p2k ) +
l2k =: M3 .
k=1
k=1
Similarly, we have
r1 +
Z
Z
p
1 ω
1X
1 ω a1 (t)eM3
ek (u(tk )) ≤
b1 (t)eu(η1 ) dt +
dt;
ω
ω 0
ω 0
k1 (t)
k=1
that is,
u(η1 ) ≥ ln
≥ ln
h r1 +
1
ω
h r1 +
1
ω
Pp
k=1
ek (v(tk )) − eM3
a1
k1
i
b1
ln
Qp
k=1 (1
+ q1k ) − eM3
a1
k1
b1
i
=: H4 .
Then
Z
u(t) ≥ u(η1 ) −
ω
|u̇(t)| dt −
0
p
X
|ek (u(tk ))|
k=1
≥ H4 − 2ωr1 − ln
p
Y
p
X
(1 + p1k ) −
l1k =: M4 .
k=1
k=1
Now, we have
M4 ≤ u(t) ≤ M1 ,
M2 ≤ v(t) ≤ M3 .
Let D = |M1 | + |M2 | + |M3 | + |M4 |. We have
kxkX = kuk0 + kvk0 ≤ D.
Clearly, D is independent of λ. Denote M = D + D0 , where D0 is taken sufficiently
large such that each solution (u∗ , v ∗ ) of
Z ω
Z ωh
p
X
a1 (t)ev i
r1 (t) dt +
ek (u(tk )) =
b1 (t)eu + u
dt,
e + k1 (t)
0
0
k=1
(3.7)
Z ω
Z ω
p
X
a2 (t)ev
r2 (t) dt +
fk (v(tk )) =
dt
u
0
0 e + k2 (t)
k=1
∗
∗
satisfies k(u , v )kX < D0 , and we can obtain D0 by repeating the above arguments.
Then k(u∗ , v ∗ )kX < M .
Let Ω = x = (u, v) ∈ X, kxkX < M , which satisfies condition (a) of Lemma 2.3.
When x ∈ ∂Ω ∩ ker L = ∂Ω ∩ R2 , x is a constant vector in R2 with kxkX = M .
Then
p
i
1hZ ω X
a1 (t) exp{v} QN x =
r1 (t) − b1 (t) exp{u} −
dt +
ek (u(tk )) ,
ω 0
exp{u} + k1 (t)
k=1
EJDE-2008/88
POSITIVE PERIODIC SOLUTIONS
9
Z
p
i
X
a2 (t) exp{v} 1 h ω
r2 (t) −
dt +
fk (v(tk )) , 0, . . . , 0
| {z }
ω
exp{u} + k2 (t)
0
k=1
2p
6= 0,
which shows that condition (b) in Lemma 2.3 holds.
Finally, we prove that condition (c) in Lemma 2.3 is satisfied. The isomorphism
J of Im Q onto ker L can be defined by
J : Im Q → X,
(f, g, c1 , . . . , cp , d1 , . . . , dp ) → (f, g).
For x ∈ ker L ∩ Ω, we have
p
1hZ ω i
X
a1 (t) exp{v} r1 (t) − b1 (t) exp{u} −
dt +
JQN x =
ek (u(tk )) ,
ω 0
exp{u} + k1 (t)
k=1
Z
p
ω
h
i
X
1
a2 (t) exp{v}
r2 (t) −
dt +
fk (v(tk ))
ω
exp{u}
+ k2 (t)
0
k=1
Z
p
ev ω
a1 (t)
1X
u
dt
+
ek (u(tk )),
= r 1 − b1 e −
ω 0 eu + k1 (t)
ω
k=1
Z
p
ev ω
1X
a2 (t)
r2 −
dt
+
fk (v(tk )) .
u
ω 0 e + k2 (t)
ω
k=1
Denote ϕ : Dom L × [0, 1] → X as the form
Z
ev ω
a2 (t)
u
ϕ(u, v, µ) = r1 − b1 e , r2 −
dt
ω 0 eu + k2 (t)
p
p
ev Z ω
1X
1X
a1 (t)
dt
+
e
(u(t
)),
f
(v(t
))
,
+µ −
k
k
k
k
ω 0 eu + k1 (t)
ω
ω
k=1
k=1
where µ ∈ [0, 1] is a parameter. With the mapping ϕ, we have ϕ(u, v, µ) 6= 0 for
(u, v) ∈ ∂Ω∩ker L. Otherwise, there exists a constant vector (u, v) with k(u, v)kX =
M implies ϕ(u, v, µ) = 0; i.e.,
Z
p
µev ω
µX
a1 (t)
r1 − b1 e u −
dt
+
ek (u(tk )) = 0
ω 0 eu + k1 (t)
ω
k=1
and
r2 −
ev
ω
Z
0
ω
p
a2 (t)
µX
dt
+
fk (v(tk )) = 0.
eu + k2 (t)
ω
k=1
Similar to the above discussion, we know that k(u, v)kX < M , which contradicts
k(u, v)kX = M . From the property of coincidence degree theory, we can obtain
deg(JQN x, Ω ∩ ker L, 0) = deg(ϕ(u, v, 1), Ω ∩ ker L, 0)
= deg(ϕ(u, v, 0), Ω ∩ ker L, 0).
Obviously, the following algebraic equation has a unique solution (u∗ , v ∗ )
r1 − b1 eu = 0,
Z
ev ω
a2 (t)
r2 −
dt = 0.
ω 0 eu + k2 (t)
10
S. GAO, Y. LI
EJDE-2008/88
So
deg(JQN x, Ω ∩ ker L, 0) = deg(ϕ(u, v, 0), Ω ∩ ker L, 0) = 1 6= 0.
By Lemma 2.3, the system (1.1) has at least one ω-periodic solution in Ω. The
proof is complete.
4. An Example
Consider the system

a1 (1 + θ cos 2πt)y(t − σ1 (t)) 

ẋ(t) = x(t) r1 + sin 2πt − b1 x(t − τ (t)) −

x(t − τ1 (t)) + k1

a2 (1 + θ cos 2πt)y(t − σ2 (t))


ẏ(t) = y(t) r2 −
x(t − τ2 (t)) + k2
(4.1)
for t 6= tk , k ∈ Z+ , and
)
−
x(t+
k ) = (1 − h)x(tk )
t = tk , k ∈ Z+ ,
−
y(t+
k ) = (1 − g)y(tk )
where r1 = 1.1, r2 = 1.25, b1 = 0.6, a1 = 0.05, a2 = 0.7, k1 = k2 = 1, h = 0.5,
g = 0.7, θ = 0.5. Obviously, in this case, ω = 1, p = 1 and
H1 < −0.38,
M1 < 1.82,
H3 < −0.66,
M3 < 1.84,
0.40 < r1 + ln(1 − h) < 0.41, 0.04 < r2 + ln(1 − g) < 0.05,
a1
> 0.40 − 0.32 = 0.08 > 0.
r1 + ln(1 − h) − eM3
k1
So that, all conditions of Theorem 3.1 are satisfied. Therefore, (4.1) has at least
one ω-periodic positive solution.
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Shan Gao
Department of Mathematics, Yunnan University, Kunming, Yunnan, 650091, China
E-mail address: 2002711036@163.com
Yongkun Li
Department of Mathematics, Yunnan University, Kunming, Yunnan, 650091, China
E-mail address: yklie@ynu.edu.cn