POSITIVE PERIODIC SOLUTIONS OF
NONAUTONOMOUS FUNCTIONAL DIFFERENTIAL
EQUATIONS DEPENDING ON A PARAMETER
GUANG ZHANG AND SUI SUN CHENG
Received 5 November 2001
This article investigates the existence of positive periodic solutions for a firstorder functional differential equations of the form
y (t) = −a(t)y(t) + λh(t) f y t − τ(t) ,
(1)
where a = a(t), h = h(t), and τ = τ(t) are continuous T-periodic functions.
We will also assume that T > 0, λ > 0, f = f (t) as well as h = h(t) are positive,
T
0 a(t) dt > 0.
Functional differential equations with periodic delays appear in a number of
ecological models. In particular, our equation can be interpreted as the standard
Malthus population model y = −a(t)y subject to perturbation with periodical
delay. One important question is whether these equations can support positive
periodic solutions. Such questions have been studied extensively by a number
of authors (cf. [1, 2, 3, 4, 6, 7] and the references therein). In this paper, we are
concerned with the existence and nonexistence of periodic solutions when the
parameter λ varies. For this purpose, we call a continuously differentiable and
T-periodic function a periodic solution of (1) associated with λ∗ if it satisfies
(1) when λ = λ∗ . We show that there exists λ∗ > 0 such that (1) has at least one
positive T-periodic solution for λ ∈ (0, λ∗ ] and does not have any T-periodic
positive solutions for λ > λ∗ . Our technique is based on the well-known upper
and lower solutions method (cf. [5]).
We proceed from (1) and obtain
t
y(t) exp
a(s) ds
0
= λ exp
t
0
Copyright © 2002 Hindawi Publishing Corporation
Abstract and Applied Analysis 7:5 (2002) 279–286
2000 Mathematics Subject Classification: 34B15, 34K13
URL: http://dx.doi.org/10.1155/S1085337502000878
a(s) ds h(t) f y t − τ(t) .
(2)
280
Bifurcation in functional differential equations
After integration from t to t + T, we obtain
y(t) = λ
t+T
G(t, s)h(s) f y s − τ(s)
ds,
(3)
t
where
s
G(t, s) =
a(u) du
exp
T
exp
t
.
(4)
a(u) du − 1
0
Note that the denominator in G(t, s) is not zero since we have assumed that
0 a(t) dt > 0.
It is not difficult to check that any T-periodic function y(t) that satisfies (3)
is also a T-periodic solution of (1). Note further that
T
0 < N ≡ min G(t, s) ≤ G(t, s) ≤ max G(t, s) ≡ M,
0≤s,t ≤T
1≥
0≤t,s≤T
t ≤ s ≤ t + T,
min0≤s,t≤T G(t, s) N
G(t, s)
≥
=
> 0.
max0≤s,t≤T G(t, s) max0≤s,t≤T G(t, s) M
(5)
Now let X be the set of all real T-periodic continuous functions, endowed
with the usual linear structure as well as the norm
y = sup y(t).
(6)
0≤t ≤T
Then X is a Banach space with cones
Φ = y(t) ∈ X : y(t) ≥ 0 ,
(7)
Ω = y(t) : y(t) ≥ σ y , t ∈ R ,
where σ = N/M.
Define a mapping F : X → X by
(F y)(t) = λ
t+T
G(t, s)h(s) f y s − τ(s)
ds.
(8)
t
Then it is easily seen that F is completely continuous on bounded subsets of Ω
and for y ∈ Φ,
(F y)(t) ≤ λM
T
h(s) f y s − τ(s)
ds
(9)
ds ≥ σ F y .
(10)
0
so that
(F y)(t) ≥ λN
T
0
h(s) f y s − τ(s)
G. Zhang and S. S. Cheng 281
That is, FΦ is contained in Ω.
Lemma 1. The mapping F maps Φ into Ω.
Lemma 2. Suppose that
lim
u→+∞
f (u)
= +∞.
u
(11)
Let I be a compact subset of (0, +∞). Then there exists a constant bI > 0 such that
u < bI for all λ ∈ I and all possible T-periodic positive solutions u of (1) associated
with λ.
Proof. Suppose to the contrary that there is a sequence {un } of T-periodic positive solutions of (1) associated with {λn } such that λn ∈ I for all n and un →
+∞ as n → ∞. Since un ∈ Ω,
min un (t) ≥ σ un .
(12)
0≤t ≤T
By (11), we may choose R f > 0 such that f (u) ≥ ηu for all u ≥ R f , and there exists
n0 such that σ un0 ≥ R f , where η satisfies
T
σηNλn0
h(s) ds > 1.
(13)
0
Thus, we have
un ≥ un (t) = λn
0
0
0
≥ σηNλn0
T
0
t+T
t
G(t, s)h(s) f un0 s − τ(s)
ds
(14)
h(s)un0 ds > un0 .
This is a contradiction. The proof is complete.
Lemma 3. Suppose that
f is nondecreasing on [0, +∞) and f (0) > 0.
(15)
Let (1) have a T-periodic positive solution y(t) associated with λ > 0. Then (1) also
has a positive T-periodic solution associated with λ ∈ (0, λ).
Proof. In view of (3) and (15), we have
y(t) = λ
≥λ
t+T
t
t+T
t
G(t, s)h(s) f y s − τ(s)
G(t, s)h(s) f y s − τ(s)
ds
t+T
ds,
0<λ
(16)
G(t, s)h(s) f (0) ds.
t
282
Bifurcation in functional differential equations
Let y 0 (t) = y(t),
t+T
y k+1 (t) = λ
G(t, s)h(s) f y k s − τ(s)
t
ds,
k = 0, 1, 2, . . .,
(17)
k = 0, 1, 2, . . ..
(18)
y (t) = 0, and
0
y
k+1
(t) = λ
t+T
G(t, s)h(s) f y s − τ(s)
k
t
ds,
Clearly, we have
y 0 (t) ≥ y 1 (t) ≥ · · · ≥ y k (t) ≥ y (t) ≥ · · · ≥ y (t) ≥ y (t).
k
1
(19)
0
If we now let y(t) = limk→∞ y k (t), then y(t) satisfies (3). Clearly, we have
y(t) ≥ y (t) = λ
1
t+T
G(t, s)h(s) f (0) ds > 0.
(20)
t
This completes our proof.
Lemma 4. Suppose that (11) and (15) hold. Then there exists λ∗ > 0 such that (1)
has a T-periodic positive solution.
Proof. Let
β(t) =
t+T
G(t, s)h(s) ds,
t
M f = max f β t − τ(t) ,
0≤t ≤T
λ∗ =
1
.
Mf
(21)
We have
β(t) =
t+T
t
G(t, s)h(s) ds ≥ λ∗
t+T
t+T
0 < λ∗
G(t, s)h(s) f β s − τ(s)
ds,
t
(22)
G(t, s)h(s) f (0) ds.
t
Let y 0 (t) = β(t),
y k+1 (t) = λ∗
t+T
t
G(t, s)h(s) f y k s − τ(s)
ds,
k = 0, 1, 2, . . .,
(23)
k = 0, 1, 2, . . ..
(24)
y (t) = 0, and
0
y
k+1
(t) = λ∗
t+T
t
G(t, s)h(s) f y s − τ(s)
k
ds,
G. Zhang and S. S. Cheng 283
Clearly, we have
y 0 (t) ≥ y 1 (t) ≥ · · · ≥ y k (t) ≥ y (t) ≥ · · · ≥ y (t) ≥ y (t).
k
1
0
(25)
If we now let y(t) = limk→∞ y k (t), then y(t) satisfies (3). Clearly, we have
y(t) ≥ y (t) = λ∗
1
t+T
G(t, s)h(s) f (0) ds > 0.
(26)
t
The proof is complete.
∗
Theorem 5. Suppose that (11) and (15) hold. Then there exists λ > 0 such that
(1) has at least one positive T-periodic solution for λ ∈ (0, λ∗ ] and does not have
any T-periodic positive solutions for λ > λ∗ .
Proof. Suppose to the contrary that there is a sequence {un } of T-periodic positive solutions of (1) associated with {λn } such that limn→∞ λn = ∞. Then either
we have un j → +∞ as j → ∞ or there is M̃ > 0 such that un ≤ M̃. Assume the
former case holds. Note that un ∈ Ω and thus
min un (t) ≥ σ un .
0≤t ≤T
(27)
By (11), we may choose R f > 0 and η1 > 0 such that f (u) ≥ η1 u when σu ≥
R f . On the other hand, there exist {tn } ⊂ [0, T] such that un j (tn j ) = un j and
un j (tn j ) = 0 by the periodicity of {un j (t)}. In view of (1), we have
a tn j un j = a tn j u tn j = λn j h tn j f un j tn j − τ tn j
≥ λn j η1 σh tn j un j (28)
for all large j. That is, we have λn j ≤ a(tn j )/(η1 σh(tn j )). Note that a(t)/h(t) is
bounded. Thus, we obtain a contradiction.
Next, suppose that the latter case holds. In view of (15), there exists η2 > 0
such that f (0) ≥ η2 M̃. Then as above, we will obtain
a tn un = a tn u tn = λn h tn f un tn − τ tn
≥ λn η2 h tn M̃ ≥ λn η2 h tn un for all n. A contradiction will again be reached.
(29)
284
Bifurcation in functional differential equations
Thus, there exists λ∗ > 0 such that (1) has at least one positive T-periodic
solution for λ ∈ (0, λ∗ ) and no T-periodic positive solutions for λ > λ∗ .
Finally, we assert that (1) has at least one T-periodic positive solution for
λ = λ∗ . Indeed, let {λn } satisfy 0 < λ1 < · · · < λk < λ∗ and limk→∞ λk = λ∗ . Since
un (t) is T-periodic positive solution of (1) associated with λn and Lemma 2
implies that the set {un (t)} of solutions is uniformly bounded in Ω, the sequence {un (t)} has a subsequence converging to u(t) ∈ Ω. We can now apply
the Lebesgue convergence theorem to show that u(t) is a T-periodic positive
solution of (1) associated with λ = λ∗ . The proof is complete.
Example 6. Consider the equation
x (t) + a(t)x(t) = λh(t) xγ t − τ(t) + 1 ,
γ > 1,
(30)
where a, h, and τ satisfy the same assumptions stated for (1). In view of Theorem
5, there exists a λ∗ > 0 such that (30) has at least one T-periodic positive solution
for λ ∈ (0, λ∗ ] and no T-periodic positive solution for λ > λ∗ .
Example 7. Consider the equation
y (t) = −ay(t) + λb y 2 (t) + ε ,
(31)
where a, b, ε > 0. Note that the function f (x) = (x2 + ε) satisfies (11) and (15) in
Theorem 5. Therefore Theorem 5 may be applied. However, we may give a direct
√
proof that, for λ > a/(2b ε), this equation cannot have any positive 2π-periodic
solutions associated with λ. Indeed, assume to the contrary that y(t) is such a
solution. Then y (ξ) = 0 for some ξ ∈ [0, 2π]. Hence
−ay(ξ) + λby 2 (ξ) + λbε = 0.
(32)
However, since the discriminant of the quadratic equation
λbx2 − ax + λbε = 0
(33)
a2 − 4λ2 b2 ε < 0,
(34)
satisfies
a contradiction is obtained. We remark that when ε = 0, our equation reduces to
the well-known logistic equation.
Similarly, we can consider the equation
x (t) = a(t)x(t) − λh(t) f x t − τ(t) ,
(35)
G. Zhang and S. S. Cheng 285
where a = a(t), h = h(t), and f = f (t) satisfy the same assumptions stated for (1).
By (35), we have
x(t) =
t+T
H(t, s)h(s) f x s − τ(s)
ds,
(36)
t
where
s
exp − a(u) du
H(t, s) =
t+T
a(u) du
exp
t
s
T
=
T
1 − exp − a(u) du
exp
a(u) du − 1
0
(37)
0
which satisfies
M ≥ H(t, s) ≥ N,
t ≤ s ≤ t + T,
(38)
for some M and N > 0, and σ = N/M ≤ 1.
Theorem 8. Suppose that (11) and (15) hold. Then there exists λ∗ > 0 such that
(35) has at least one positive T-periodic solution for λ ∈ (0, λ∗ ] and no T-periodic
positive solution for λ > λ∗ .
Acknowledgment
This work was supported by the Natural Science Foundation of Shanxi Province
and Yanbei Normal College. Part of this work was done during the first author’s
visit to the Institute of Applied Mathematics, Academy of Mathematics and System Sciences, Chinese Academy of Sciences. The first author wishes to express
his thanks to Professor Daomin Cao for his kind invitation and nice hospitality.
We also thank the referee for his helpful criticisms.
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Guang Zhang: Department of Mathematics, Yanbei Normal College and Datong College, Datong, Shanxi 037000, China
E-mail address: dtgzhang@yahoo.com.cn
Sui Sun Cheng: Department of Mathematics, Tsing Hua University, Hsinchu,
Taiwan 30043, Taiwan
E-mail address: sscheng@math.nthu.edu.tw