Appl. Math. J. Chinese Univ. Ser. B
2006, 21(3): 302-312
POSITIVE PERIODIC SOLUTION FOR A NONAUTONOMOUS
LOGISTIC MODEL WITH LINEAR FEEDBACK REGULATION
Ding Xiaoquan Cheng Shuhan
Abstract. A nonautonomous delayed logistic model with linear feedback regulation is proposed in this paper. Sufficient conditions are derived for the existence, uniqueness and global
asymptotic stability of positive periodic solution of the model.
§1 Introduction
Gopalsamy and Weng[1,2] have considered the following autonomous delayed model with
linear feedback regulation:
¶
µ
a1 x1 (t) + a2 x1 (t − τ )
0
− cx2 (t) ,
x1 (t) = rx1 (t) 1 −
K
(1.1)
0
x2 (t) = −ax2 (t) + bx1 (t),
where x1 (t) is the density of the population, x2 (t) is the regulative variable. a, b, c and k are
positive constants, a1 , a2 and τ are nonnegative constants. Many authors[2−4] have studied the
global stability of the positive equilibrium of system (1.1).
However, in realistic world, the environment is always varying periodically with time. This
effect of changing environment motivates us to consider the following non-autonomous delayed
logistic model with linear feedback regulation:
x01 (t) = x1 (t)(r(t) − a1 (t)x1 (t) − a2 (t)x1 (t − τ1 ) − c(t)x2 (t − τ2 )),
x02 (t) = −a(t)x2 (t) + b(t)x1 (t − τ3 ),
(1.2)
with initial conditons
xi (θ) = ψi (θ), θ ∈ [−τ, 0], ψi (0) > 0, ψi ∈ C([−τ, 0], R+ ), i = 1, 2,
(1.3)
where a(t), b(t), c(t), r(t), a1 (t) and a2 (t) are continuous ω-periodic functions, a(t), b(t), c(t) and
r(t) are strictly positive, a1 (t) and a2 (t) are nonnegative. τ1 , τ2 and τ3 are nonnegative constants, τ = max {τ1 , τ2 , τ3 }.
Received:2005-04-12.Revised:2006-01-18
MR Subject Classification:34C11, 34K35, 92D25.
Keywords: logistic model, periodic solution, global asymptotic stability, linear feedback regulation.
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303
We adopt the following notations throughout this paper:
Z
1 b
f (t) dt, f l := min f (t), f u := max f (t)
f :=
ω a
t∈[0,ω]
t∈[0,ω]
where f is a continuous ω-periodic function.
The organization of this paper is as follows. In §2, sufficient conditions are established for
the persistence of system (1.2). In §3, by using the continuation theorem of coincidence degree
theory, we show the existence of positive ω-periodic solutions of system (1.2). In §4, a set
of sufficient conditions are derived for the uniqueness and global stability of positive periodic
solution of system (1.2).
§2 Uniform persistence
We first show the positivity and boundedness of solutions of system (1.2) with initial conditions (1.3).
Lemma 2.1. Solutions of system (1.2) with initial conditions (1.3) are positive for all t ≥ 0.
Proof. Let (x1 (t), x2 (t))T be a solution of system (1.2) with initial conditions (1.3). From
system (1.2) we obtain
Z t
x1 (t) = x1 (0) exp{ [r(s) − a1 (s)x1 (s) − a2 (s)x1 (s − τ1 ) − c(s)x2 (s − τ2 )]ds} > 0,
0
Z
x2 (t) = {x2 (0) +
t
for x1 (0) > 0,
Z s
Z t
[b(s)x1 (s − τ3 ) exp (
a(u)du)]ds} exp (−
a(s)ds) > 0, for x2 (0) > 0.
0
0
0
T
Lemma 2.2.Let (x1 (t), x2 (t)) be a solution of system (1.2) with initial conditions (1.3), then
there exists a T1 > 0 such that
0 < x1 (t) ≤ M1 ,
where
M1 =
al1
0 < x2 (t) ≤ M2 ,
ru
,
+ al2 e−ru τ1
for t ≤ T1 ,
(2.1)
bu
M1 .
al
(2.2)
M2 > M2∗ =
Proof. Suppose (x1 (t), x2 (t))T is a solution of system (1.2) which satisfies (1.3). According to
the first equation of system (1.2), it follows from the positivity of the solution that
x01 (t) ≤ x1 (t)(ru − al1 x1 (t) − al2 x1 (t − τ1 ))
(2.3)
A standard comparison argument (see, for example, the proof of Lemma 2.1. in [5]) shows that
there is a T11 > 0 such that
x1 (t) ≤
al1
ru
:= M1 ,
+ al2 e−ru τ1
for t ≥ T11 .
It follows from (2.4) and the second equation of system (1.2) that for t ≥ T11 + τ3 ,
x02 (t) ≤ −al x2 (t) + bu M1 .
(2.4)
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A standard comparison argument shows that
lim sup x2 (t) ≤
t→+∞
bu
M1 := M2∗ .
al
(2.5)
It follows from (2.5) that there exists an M2 > M2∗ and a T1 > T11 + τ3 such that
x2 (t) < M2
for t ≥ T1
Theorem 2.1. Suppose that system (1.2) satisfies the following:
(H1)
r l > c u M2 ,
where M2 is defined in (2.2). Then system (1.2) is uniformly persistent.
Proof. Suppose (x1 (t), x2 (t))T is a solution of system (1.2) which satisfies (1.3). According to
the first equation of system (1.2), we have for t > T1 ,
x01 (t) ≥ x1 (t)(rl − au1 x1 (t) − au2 x1 (t − τ1 ) − cu M2 ) ≥ x1 (t)(rl − (au1 + au2 )M1 − cu M2 ), (2.6)
which leads to
x1 (t − τ1 ) ≤ x1 (t)e−(r
l
u
u
−(au
1 +a2 )M1 −c M2 )τ1
, for t > T1 + τ1 .
(2.7)
It follows form (2.6) and (2.7) that
x01 (t) ≥ x1 (t)(rl − (au1 + au2 e−(r
l
u
u
−(au
1 +a2 )M1 −c M2 )τ1
)x1 (t) − cu M2 ).
A standard comparison argument shows that
lim inf ≥
t→+∞
au1
+
rl − cu M2
l −(au +au )M −cu M )τ
u
−(r
1
2 1
1
2
a2 e
:= m∗1 .
(2.8)
Therefore, there exists a T2 > T1 + τ1 and a positive constant m1 < m∗1 such that x1 (t) > m1
for t ≥ T2 .
In addition, from the second equation of system (1.2) we obtain
x02 (t) ≥ −au x2 (t) + bl m1 , for t > T2 + τ3 .
A standard comparison argument shows that
lim inf x2 (t) ≥
t→+∞
bl m1
:= m∗2 .
au
Therefore, there exists a T3 > T2 + τ3 and a positive constant m2 < m∗2 such that x2 (t) > m2
for t ≥ T3 .
We now let
D = {(x1 , x2 )|m1 ≤ x1 ≤ M1 , m2 ≤ x2 ≤ M2 },
(2.9)
then D is a bounded compact region in R2+ which has the positive distance from coordinate
planes. From what has been discussed above, we obtain that there exist T > T3 , if t ≥ T , then
every positive solution of systems (1.2) and (1.3) eventually enters and remains in the region
D. The proof is completed.
Ding Xiaoquan,et al.
POSITIVE PERIODIC SOLUTIONS FOR A NONAUTONOMOUS ...
305
§3 Existence of periodic solutions
In order to obtain the existence of positive periodic solutions of (1.2), for convenience, we
shall summarize in the following a few concepts and results from [6, p. 40] that will be basic
for this section.
Let X, Y be real Banach spaces, L : DomL ⊂ X → Y be a linear mapping, and N : X → Y
be a continuous mapping. The mapping L is called a Fredholm mapping of index zero if
dim KerL = codimImL < +∞ and ImL 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 ImP =
KerL, KerQ = ImL = Im(I−Q), then the restriction LP of L to DomL∩KerP : (I−P )X → ImL
is invertible. Denote the inverse of LP 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 : Q → X is compact.
Since ImQ is isomorphic to KerL, there exists isomorphisms J : ImQ → KerL.
Lemma 3.1. Let Ω ⊂ X be an open bounded set, L be a Fredholm mapping of index zero
and N be L-compact on Ω. Assume
(a) for each λ ∈ (0, 1), x ∈ ∂Ω ∩ DomL, Lx 6= λN x;
(b) for each x ∈ ∂Ω ∩ KerL, QN x 6= 0;
(c) deg{JQN, Ω ∩ KerL, 0} 6= 0.
Then Lx = N x has at least one solution in Ω ∩ DomL.
We are now in a position to state our result on the existence of periodic solutions of system
(1.2).
Theorem 3.1.Suppose that system (1.2) satisfies
µ the
¶u following:
b
(H2)
a1 + a2 >
ce2rω ,
a
Then system (1.2) with conditions (1.3) has at least one positive -periodic solution.
Proof. Let
y1 (t) = ln[x1 (t)],
y2 (t) = ln[x2 (t)].
(3.1)
On substituting (3.1) into (1.2), we rewrite (1.2) in the form
y10 (t) = r(t) − a1 (t)ey1 (t) − a2 (t)ey1 (t−τ1 ) − c(t)ey2 (t−τ2 ) ,
(3.2)
y20 (t) = −a(t) + b(t)ey1 (t−τ3 )−y2 (t) .
T
It is easy to see that if system (3.2) has one ω-periodic solution (y1∗ (t), y2∗ (t)) , then
T
T
x∗ = (x∗1 (t), x∗2 (t)) = (exp[y1∗ (t)], exp[y2∗ (t)]) is a positive ω-periodic solution of system (1.2).
Therefore, to complete the proof it suffices to show that system (3.2) has one ω-periodic solution.
Take
X = Y = {y = (y1 (t), y2 (t))T ∈ C(R, R2 ) : yi (t + ω) = yi (t), i = 1, 2}
and
T
kyk = k(y1 (t), y2 (t)) k = max (|y1 (t)| + |y2 (t)|),
t∈[0,ω]
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then X and Y are Banach spaces with the norm k · k. Set
L(y1 (t), y2 (t))T = (y10 (t), y20 (t))T ,
L : DomL ⊂ X → Y,
where DomL = {(y1 (t), y2 (t))T ∈ C 1 (R, R2 )} and
#
# "
"
r(t) − a1 (t)ey1 (t) − a2 (t)ey1 (t−τ1 ) − c(t)ey2 (t−τ2 )
y1 (t)
=
N : X → Y, N
−a(t) + b(t)ey1 (t−τ3 )−y2 (t)
y2 (t)
With these notations system (3.2) can be written in the form
Ly = N y,
y∈X
Rω
Obviously, KerL = R2 , ImL = {(y1 (t), y2 (t))T ∈ Y : 0 yi (t)dt = 0, i = 1, 2} is closed in Y ,
and dimKerL = codimImL = 2. Therefore L is a Fredholm mapping of index zero. Now define
two projectors P : X → X and Q : Y → Y as
"
#
"
# " # "
#
y1 (t)
y1 (t)
y1
y1 (t)
P
=Q
=
,
∈X=Y
y2 (t)
y2 (t)
y2
y2 (t)
then P and Q are continuous projectors such that
ImP = KerL, KerQ = ImL = Im(I − Q)
Furthermore, through an easy computation we find that the inverse KP of LP has the form
KP : ImL → DomL ∩ KerP,
Z t
Z Z
1 ω t
KP (y) =
y(s)ds −
y(s)dsdt.
ω 0 0
0
Then QN : X → Y and KP (I − Q)N : X → X read
" Rω
#
1
y1 (t)
y1 (t−τ1 )
y2 (t−τ2 )
[r(t)
−
a
(t)e
−
a
(t)e
−
c(t)e
]dt
1
2
R
QN y = ω 0
,
1 ω
y1 (t−τ3 )−y2 (t)
]dt
ω 0 [−a(t) + b(t)e
Z
KP (I − Q)N y =
t
N y(s)ds −
0
1
ω
Z
ω
Z
t
N y(s)dsdt − (
0
0
t
1
− )
ω 2
Z
ω
N y(s)ds.
0
Clearly, QN and KP (I −Q)N are continuous. By using Arzela-Ascoli theorem, it is not difficult
to prove 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 with any open bounded set Ω ∈ X.
In order to apply Lemma 3.1, we need to search for an appropriate open, bounded subset
Ω.
Corresponding to the operator equation Ly = λN y, λ ∈ (0, 1), we have
y10 (t) = λ[r(t) − a1 (t)ey1 (t) − a2 (t)ey1 (t−τ1 ) − c(t)ey2 (t−τ2 ) ],
y20 (t) = λ[−a(t) + b(t)ey1 (t−τ3 )−y2 (t) ].
(3.3)
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307
Suppose that (y1 (t), y2 (t))T ∈ X is a solution of (3.3) for a certain λ ∈ (0, 1). Integrating (3.3)
over the interval [0, ω] leads to
Z ω
Z ω
λ[r(t) − a1 (t)ey1 (t) − a2 (t)ey1 (t−τ1 ) − c(t)ey2 (t−τ2 ) ]dt =
y10 (t)dt = 0,
(3.4)
0
0
Z
ω
Z
λ[−a(t) + b(t)ey1 (t−τ3 )−y2 (t) ]dt =
0
That is
Z
ω
0
ω
y20 (t)dt = 0,
Z
[a1 (t)ey1 (t) + a2 (t)ey1 (t−τ1 ) + c(t)ey2 (t−τ2 ) ]dt =
0
Z
Z
ω
b(t)ey1 (t−τ3 )−y2 (t) dt =
ω
r(t)dt = rω,
(3.6)
0
ω
a(t)dt = aω,
0
(3.5)
(3.7)
0
It follows from (3.3)-(3.7) that
Z ω
Z ω
0
|y1 (t)|dt ≤
λ[r(t) + a1 (t)ey1 (t) + a2 (t)ey1 (t−τ1 ) + c(t)ey2 (t−τ2 ) ]dt <
0
0
Z ω
[r(t) + a1 (t)ey1 (t) + a2 (t)ey1 (t−τ1 ) + c(t)ey2 (t−τ2 ) ]dt = 2rω,
(3.8)
0
Z
0
ω
Z
|y20 (t)|dt ≤
ω
0
Z
λ[a(t) + b(t)ey1 (t−τ3 )−y2 (t) ]dt <
(3.9)
ω
[a(t) + b(t)e
y1 (t−τ3 )−y2 (t)
]dt = 2aω,
0
Since (y1 (t), y2 (t))T ∈ X, there exist ξi , ηi ∈ [0, ω](i = 1, 2) such that
yi (ξi ) = min yi (t),
yi (ηi ) = max yi (t),
t∈[0,ω]
r
Sa , H a
Denote Sa = a1 + a2 , Ra =
Z
ey1 (ξ1 ) Sa ω = ey1 (ξ1 )
t∈[0,ω]
i = 1, 2.
(3.10)
u
=
b
Sa −( a
) ce2rω
r,
Sa
Z
ω
[a1 (t) + a2 (t)]dt ≤
0
ω
from (3.6) we obtain
[a1 (t)ey1 (t) + a2 (t)ey1 (t−τ1 ) ]dt < rω,
0
which implies
y1 (ξ1 ) ≤ ln Ra .
This, together with (3.8), leads to
Z
y1 (t) ≤ y1 (ξ1 ) +
0
ω
|y10 (t)|dt < ln Ra + 2rω.
(3.11)
It follows from (3.3) and (3.10) that
0 = y20 (η2 ) = λ[−a(η2 ) + b(η2 )ey1 (η2 −τ3 )−y2 (η2 ) ].
This, together with (3.11), yields
ey2 (η2 ) =
b(η2 ) y1 (η2 −τ3 )
e
≤
a(η2 )
µ ¶u
µ ¶u
b
b
eln Ra +2rω =
Ra e2rω .
a
a
(3.12)
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On the other hand, from (3.6) we can derive
Z ω
Z ω
y1 (η1 )
y1 (η1 )
e
Sa ω =e
[a1 (t) + a2 (t)]dt ≥
[a1 (t)ey1 (t) + a2 (t)ey1 (t−τ1 ) ]dt =
0
0
Z ω
Z ω
y2 (t−τ2 )
y2 (η2 )
rω −
c(t)e
dt > rω − e
c(t)dt > Ha ω,
0
0
which implies
y1 (η1 ) ≥ ln Ha − ln Sa .
This, together with (3.8), leads to
Z
ω
y1 (t) ≥ y1 (η1 ) −
0
|y10 (t)|dt > ln Ha − ln Sa − 2rω.
(3.13)
It follows from (3.11) and (3.13) that
max |y1 (t)| < max {| ln Ra |, | ln Ha − ln Sa |} + 2rω := R1 .
t∈[0,ω]
(3.14)
Furthermore, from (3.12) we obtain
(¯ µ ¶ ¯ ¯ µ ¶ ¯)
u
l
¯
b ¯¯
b ¯¯ ¯¯
,
ln
|y2 (η2 )| ≤ max ¯¯ln
¯
¯ + R1 := R2
a ¯ ¯
a ¯
This, together with (3.9), leads to
Z
|y2 (t)| ≤ |y2 (η2 )| +
0
ω
|y20 (t)|dt < R2 + 2aω := R3 .
(3.15)
Clearly, R1 , R2 and R3 in (3.14) and (3.15) are independent of λ.Denote M = R1 + R2 +
R3 + R0 , here R0 is taken sufficiently large such that the unique solution (y1∗ , y2∗ )T of the system
of algebraic equation
r − a1 ey1 − a2 ey1 − cey2 = 0
(3.16)
− a + bey1 −y2 = 0
satisfies k(y1∗ , y2∗ )T k < M , where
y1∗ = ln
ra
,
aa1 + a2 + bc
y2∗ = ln
rb
.
aa1 + a2 + bc
We now take Ω = {(y1 (t), y2 (t))T ∈ X : k(y1 (t), y2 (t))T k < M }. This satisfies condition (a)
in Lemma 3.1. When (y1 (t), y2 (t))T ∈ ∂Ω ∩ KerL = ∂Ω ∩ R2 , (y1 (t), y2 (t))T is an constant
vector in R2 with |y1 | + |y2 | = M . Thus, we have
" # "
# " #
y1
r − a1 ey1 − a2 ey1 − cey2
0
QN
=
6=
−a + bey1 −y2
0
y2
This proves that condition (b) in Lemma 3.1 is satisfied.
Ding Xiaoquan,et al.
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309
Taking J = I : ImQ → KerL, (y1 , y2 )T → (y1 , y2 )T , a direct calculation shows that
deg(JQN (y1 , y2 )T , (r − a1 ey1 − a2 ey1 − cey2 , −a + bey1 −y2 )T , Ω ∩ KerL, (0, 0)T ) = 1.
By now we have proved that Ω satisfies all the requirements in Lemma 3.1. Hence, (3.2) has
at least one ω-periodic solution. Accordingly, system (1.2) has at least one positive ω-periodic
solution. This completes the proof.
§4 Uniqueness and global stability
In this section we formulate the uniqueness and global stability of ω-periodic solution z ∗ (t) =
(x∗1 (t), x∗2 (t))T in Theorem 3.1. It is immediate that if z ∗ (t) is globally asymptotically stable
then z ∗ (t) is in fact unique.
Theorem4.1.In addition to (H2), assume further that the following hold:
(H3)
lim inf t→+∞ Ai (t) > 0,
i = 1, 2,
where
A1 (t) = a1 (t) + a2 (t) − b(t + τ3 ) − [r(t) + (2a1 (t) + a2 (t))M1 + c(t)M2 ]
Z t+τ1
Z t+2τ1
a2 (s)ds − a2 (t + τ1 )M1
a2 (s)ds
t
t+τ1
t+τ1 +τ2
(4.1)
Z
A2 (t) = a(t) − c(t + τ2 ) − c(t + τ2 )M1
a2 (s)ds,
t+τ2
where M1 , M2 are defined in (2.2).Then system (1.2) with conditions (1.3) has a unique positive
ω-periodic solution z ∗ (t) = (x∗1 (t), x∗2 (t))T which is globally asymptotically stable.
Proof. Due to the conclusion of Theorem 3.1, we only need to show the global asymptotic stability of positive periodic solutions of system (1.2) with initial condition (1.3).Let (x∗1 (t), x∗2 (t))T
be a positive ω-periodic solution of system (1.2) with initial condition (1.3). Suppose that
(x1 (t), x2 (t))T is any positive solution of system (1.2) with initial condition (1.3).It follows
from Lemma 2.2 that there exist positive constants T > 0, Mi (defined by (2.2)) such that for
all t ≥ T ,
0 < xi (t) ≤ Mi ,
0 < x∗i (t) ≤ Mi ,
i = 1, 2.
(4.2)
Let
V11 = | ln x1 (t) − ln x∗1 (t)|.
(4.3)
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Calculating the upper right derivative of V11 (t) along solutions of system (1.2), it follows that
µ
¶
ẋ1 (t) ẋ∗1 (t)
+
D V11 (t) =
−
sgn(x1 (t) − x∗1 (t) =
x1 (t) x∗1 (t)
− sgn(x1 (t) − x∗1 (t))[(a1 (t) + a2 (t))(x1 (t) − x∗1 (t)) + c(t)(x2 (t − τ2 )−
Z t
∗
x2 (t − τ2 )) − a2 (t)
{[r(u) − a1 (u)x1 (u) − a2 (u)x1 (u − τ1 ) − c(u)−
t−τ1
x2 (u − τ2 )](x1 (u) − x∗1 (u)) − a1 (u)x∗1 (u)(x1 (u) − x∗1 (u)) − a2 (u)x∗1 (u)−
(x1 (u − τ1 ) − x∗1 (u − τ1 )) − c(u)x∗1 (u)(x2 (u − τ2 ) − x∗2 (u − τ2 ))}du] ≤
(4.4)
− (a1 (t) + a2 (t))|x1 (t) − x∗1 (t)| + c(t)|x2 (t − τ2 ) − x∗2 (t − τ2 )| + a2 (t)+
Z t
{[r(u) + a1 (u)x1 (u) + a2 (u)x1 (u − τ1 ) + c(u)x2 (u − τ2 )]+
t−τ1
|x1 (u) − x∗1 (u)| + a1 (u)x∗1 (u)|x1 (u) − x∗1 (u)| + a2 (u)x∗1 (u)+
|x1 (u − τ1 ) − x∗1 (u − τ1 )| + c(u)x∗1 (u)|x2 (u − τ2 ) − x∗2 (u − τ2 )|}du.
It follows from (4.2) and (4.4) that for t ≥ T + τ ,
D+ V11 (t) ≤ − (a1 (t) + a2 (t))|x1 (t) − x∗1 (t)| + c(t)|x2 (t − τ2 ) − x∗2 (t − τ2 )|+
Z t
a2 (t)
{[r(u) + (2a1 (u) + a2 (u))M1 + c(u)M2 ]|x1 (u) − x∗1 (u)|+
(4.5)
t−τ1
a2 (u)M1 |x1 (u − τ1 ) − x∗1 (u − τ1 )| + c(u)M1 |x2 (u − τ2 ) − x∗2 (u − τ2 )|}du.
Define
Z
t+τ1
Z
t
a2 (s){[r(u) + (2a1 (u) + a2 (u))M1 + c(u)M2 ]|x1 (u) − x∗1 (u)|+
V12 (t) =
t
s−τ1
a2 (u)M1 |x1 (u − τ1 ) −
x∗1 (u
− τ1 )| + c(u)M1 |x2 (u − τ2 ) −
x∗2 (u
(4.6)
− τ2 )|}duds.
It follows from (4.5) and (4.6) that for t ≥ T + τ ,
D+ V11 (t) + V̇12 ≤ − (a1 (t) + a2 (t))|x1 (t) − x∗1 (t)| + c(t)|x2 (t − τ2 ) − x∗2 (t − τ2 )|+
Z t+τ1
a2 (s)ds{r(t) + (2a1 (t) + a2 (t))M1 + c(t)M2 ]|x1 (t) − x∗1 (t)|+
(4.7)
t
a2 (t)M1 |x1 (t − τ1 ) − x∗1 (t − τ1 )| + c(t)M1 |x2 (t − τ2 ) − x∗2 (t − τ2 )|}
We now define
V1 (t) = V11 (t) + V12 (t) + V13 (t),
(4.8)
in which
Z
Z
t
l+τ1 +τ2
[c(l + τ2 ) +
V13 (t) =
t−τ2
Z
t
Z
l+τ2
M1 a2 (s)c(l + τ2 )ds|x2 (l) − x∗2 (l)|]dl+
l+2τ1
a2 (s)a2 (l + τ1 )ds|x1 (l) −
M1
t−τ1
l+τ1
(4.9)
x∗1 (l)|dl.
Ding Xiaoquan,et al.
POSITIVE PERIODIC SOLUTIONS FOR A NONAUTONOMOUS ...
311
Then it follows from (4.7), (4.8) and (4.9) that for t ≥ T + τ ,
Z
+
t+τ1
D V1 (t) ≤ − [a1 (t) + a2 (t) − (r(t) + (2a1 (t) + a2 (t))M1 + c(t)M2
Z
a2 (s)ds−
t
t+2τ1
a2 (t + τ1 )M1
t+τ1
a2 (s)ds]|x1 (t) − x∗1 (t)|+
Z
t+τ1 +τ2
[c(t + τ2 ) + c(t + τ2 )M1
t+τ2
(4.10)
a2 (s)ds]|x2 (t) − x∗2 (t)|.
Next, let
V21 (t) = |x2 (t) − x∗2 (t)|.
(4.11)
Calculating the upper right derivative of V11 (t) along solutions of system (1.2), it follows that
D+ V21 (t) =(x2 (t) − x∗2 (t))sgn(x2 (t) − x∗2 (t)) ≤
− a(t)|x2 (t) − x∗2 (t)| + b(t)|x1 (t − τ3 ) − x∗1 (t − τ3 )|.
Define
Z
t
V22 (t) =
t−τ3
a(s + τ3 )|x1 (s) − x∗1 (s)|ds.
(4.12)
(4.13)
It follows from (4.12) and (4.13) that
D+ V21 (t) + V̇22 (t) ≤ −a(t)|x2 (t) − x∗2 (t)| + b(t + τ3 )|x1 (t) − x∗1 (t)|.
(4.14)
We now define
V (t) = V1 (t) + V21 (t) + V22 (t),
(4.15)
then it follows from (4.10), (4.14) and (4.15) that for t ≥ T + τ ,
D+ V (t) ≤ −A1 (t)|x1 (t) − x∗1 (t)| − A2 (t)|x2 (t) − x∗2 (t)|,
(4.16)
where A1 (t) and A2 (t) are defined in (4.1).
By the assumption (H3), there exist constants α1 > 0, α2 > 0 and a T ∗ > T + τ such that
A1 (t) ≥ α1 > 0,
A2 (t) ≥ α2 > 0 for t ≥ T ∗
(4.17)
Integrating both sides of (4.16) over interval [T ∗ , t], we derive that for t ≥ T ∗ ,
Z t
Z t
A2 (s)|x2 (s) − x∗2 (s)|ds ≤ V (T ∗ ).
A1 (s)|x1 (s) − x∗1 (s)|ds +
V (t) +
(4.18)
It follows from (4.17) and (4.18) that for t ≥ T ∗ ,
Z t
Z
∗
V (t) + α1
|x1 (s) − x1 (s)|ds + α2
|x2 (s) − x∗2 (s)|ds ≤ V (T ∗ ).
(4.19)
|x2 (s) − x∗2 (s)|ds < +∞.
(4.20)
T∗
T∗
T∗
T∗
Therefore, V (t) is bounded on [T ∗ , +∞), and also
Z
Z +∞
|x1 (s) − x∗1 (s)|ds < +∞,
T∗
t
+∞
T∗
312
Appl. Math. J. Chinese Univ. Ser. B
Vol. 21, No. 3
By Lemma 2.2, |x1 (t) − x∗1 (t)| and |x2 (t) − x∗2 (t)| are bounded for t ≥ T ∗ . On the other
hand, it is easy to see that ẋ1 (t), ẋ2 (t), ẋ∗1 (t) and ẋ∗2 (t) are bounded for t ≥ T ∗ .Therefore,
|x1 (t) − x∗1 (t)| and |x2 (t) − x∗2 (t)| are uniformly continuous on [T ∗ , +∞). By Barbalat’s Lemma
(Lemma 1.2.2 and Lemma 1.2.3[1] ), one can conclude that
lim |x1 (t) − x∗1 (t)| = 0,
t→+∞
lim |x2 (t) − x∗2 (t)| = 0.
t→+∞
The proof is complete.
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Information Science and Engineering College , Shandong Agricultural University, Shandong Taian
271018, China.