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Periodic solutions of first-order nonlinear functional differential equations
Nonlinear Analysis 68 (2008) 845–861 www.elsevier.com/locate/na
Periodic solutions of first-order nonlinear functional differential equationsI X.H. Tang ∗ , Zhiyuan Jiang School of Mathematical Sciences and Computing Technology, Central South University, Changsha, Hunan 410083, PR China Received 25 September 2006; accepted 27 November 2006
Abstract In this paper, we obtain a set of “easily verifiable” sufficient conditions for the existence of the periodic solutions of the firstorder linear functional differential equations with periodic perturbation x 0 (t) = l(x)(t) + f (t, xt ), where l : C(R) → C(R) is a linear bounded operator. These conditions generalize and improve the known results given in the literature. c 2006 Elsevier Ltd. All rights reserved.
MSC: 34K13 Keywords: Functional differential equations; Periodic solution; Periodic perturbation
1. Introduction Let BC(R, R) be the Banach space of bounded continuous functions x : R → R with the sup norm kxk = sup{|x(t)| : t ∈ R} and let Cω (R) = {x ∈ BC(R, R) : x(t + ω) = x(t), t ∈ R}, where ω > 0. Define the norms as follows: kxk0 = max{|x(t)| : 0 ≤ t ≤ ω},
ω
Z
|x(t)|dt,
kxk1 = 0
∀x ∈ Cω (R).
Consider the first-order functional differential equations x 0 (t) = l(x)(t) + f (t, xt ), I This work was partially supported by the NNSF of China (No. 10471153). ∗ Corresponding author.
E-mail address: [email protected] (X.H. Tang). c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.11.041
(1.1)
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
and the particular case x 0 (t) = l(x)(t) + g(t),
(1.2)
where l : C(R) → C(R) is a linear bounded operator and l(Cω (R)) ⊆ Cω (R), g ∈ Cω (R), and f ∈ C(R × BC(R, R), R), for any t ∈ R, f (t, ·) mapping a bounded set in BC(R, R) into a bounded set in R and satisfying f (t + ω, ϕ) = f (t, ϕ) for (t, ϕ) ∈ R × BC(R, R). Everywhere in what follows, we will assume that the operator l is nontrivial and admits the representation l(x)(t) = −a(t)x(t) + l1 (x)(t) − l2 (x)(t),
(1.3)
where a ∈ Cω (R), l1 , l2 : Cω (R) → Cω (R) are linear and satisfy the condition l1 (x)(t) ≥ 0,
l2 (x)(t) ≥ 0,
∀t ∈ [0, ω] if x(t) ≥ 0, ∀t ∈ [0, ω].
In the representation (1.3), we list the instantaneous term a(t)x(t) as an independent one, because the instantaneous term always plays an important role in the existence and stability of the periodic solution for Eq. (1.1). The common particular case of Eq. (1.2) is the following linear equation with deviating arguments: x 0 (t) = p0 (t)x(t) +
n X
pi (t)x(t − τi (t)) + g(t),
(1.4)
i=1
where g ∈ Cω (R) and p0 , pi , τi ∈ Cω (R), i = 1, 2, . . . , n. For Eq. (1.4), we find that l1 (x)(t) =
n X
[ pi (t)]+ x(t − τi (t)),
l2 (x)(t) =
i=1
n X
[ pi (t)]− x(t − τi (t)),
i=1
where here and in the sequel, [x]+ = (|x| + x)/2, [x]− = (|x| − x)/2. It is known (see, e.g., [10]) that Eq. (1.2) has a unique ω-periodic solution if and only if the corresponding homogeneous equation x 0 (t) = l(x)(t),
(1.5)
has only a trivial ω-periodic solution. In view of this fact, [1,10,16] gave a set of conditions which guarantee that Eq. (1.2) has a unique ω-periodic solution. In the paper [11], Ma, Yu and Wang further proved that the homogeneous equation (1.5) having only a trivial ω-periodic solution implies that Eq. (1.1) has at least one ω-periodic solution under the additional restriction (H1) limkϕk→∞
| f (t,ϕ)| kϕk
= 0 uniformly in t ∈ R.
In this paper, our main purpose is to derive a set of “easily verifiable” sufficient conditions for the existence of the periodic solutions of Eq. (1.1). These conditions generalize and improve the known results given in the literature (see [1,4,6,7,10,16]). For example, consider Eq. (1.4) with p0 (t) ≥ 0 and pi (t) ≤ 0, i = 1, 2, . . . , n. In paper [1], it is shown that if n P
k p 0 k1 <
k pi k1
i=1
1+
n P
(1.6) k pi k1
i=1
and n X
k pi k1 < 4 (1 − k p0 k1 ) ,
(1.7)
i=1
or ω
n Z X i=1
0
| pi (t)| exp
ω
Z t
p0 (s)ds
< 4,
(1.8)
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
then Eq. (1.4) has a unique ω-periodic solution. But in view of Corollary 2.1 in Section 2 and Corollary 3.4 in Section 3, condition (1.7) can be improved to n X
k pi k1 < 2 + 2 (1 − k p0 k1 )1/2 ;
(1.9)
i=1
and conditions (1.6) and (1.8) can be replaced by the following better conditions (see Example 4.1): Z β+ω n Z β+ω X ek p 0 k 1 − 1 < p0 (s)ds dt | pi (t)| exp i=1
β
< 1 + 2e
t k p0 k1 /2
+e
k p 0 k1
,
β ∈ [0, ω].
(1.10)
Other related results can be found in [2,3,5,8,9,12–15]. For Eq. (1.1), we always assume that (H1) holds in what follows. 2. The case a(t) ≥ 0 Theorem 2.1. Assume that a(t) ≥ 0 and that for any α ∈ [0, ω], α+ω
Z α
α+ω
Z l1 (1)(t) exp − t
and that for any β ∈ [0, ω], Z Z β+ω l2 (1)(t) exp − β
β+ω
R R α+ω α+ω l2 (1)(t) exp − t a(s)ds dt 1 − e−kak1 + α R a(s)ds dt < , R α+ω α+ω 1+ α l2 (1)(t) exp − t a(s)ds dt
(
a(s)ds dt < e−kak1 + min
2(1 + c)e−kak1 p
(1 − c)2 + 4ce−kak1 + (1 − c) Z β+ω ) Z β+ω 1 . l1 (1)(t) exp − a(s)ds dt + 1− c β t 0≤c≤1
t
(2.1)
(2.2)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Proof. According to theorems in [7,8], it is sufficient to show that homogeneous equation (1.5) has only a trivial ω-periodic solution. Assume the contrary. Let x(t) be a nontrivial ω-periodic solution of Eq. (1.5). Then kxk0 > 0. Set x∗ = min{x(t) : 0 ≤ t ≤ ω},
x ∗ = max{x(t) : 0 ≤ t ≤ ω}.
(2.3)
Choose t∗ ∈ [0, ω] and t ∗ ∈ [t∗ , t∗ + ω] such that x(t∗ ) = x∗ ,
x(t ∗ ) = x ∗ .
(2.4)
Since l(x) is linear, we can let x ∗ = kxk0 . From (1.3) and (1.5), we have 0 Z t Z t x(t) exp = [l1 (x)(t) − l2 (x)(t)] exp a(s)ds . a(s)ds
(2.5)
0
0
There are two possible cases: Case 1. x∗ = x(t∗ ) ≥ 0. Integrating (2.5) from t∗ to t ∗ , we obtain ! Z t∗ Z t∗ Z t Z t∗ ∗ x exp a(s)ds − x∗ exp a(s)ds = [l1 (x)(t) − l2 (x)(t)] exp a(s)ds dt 0
0
0
t∗
≤ x
∗
Z
t∗ t ∗ −ω
l1 (1)(t) exp
Z
t
a(s)ds dt.
0
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
It follows that x ∗ ≤ x∗ + x ∗
Z
t∗ t ∗ −ω
l1 (1)(t) exp −
!
t∗
Z
a(s)ds dt.
(2.6)
t
On the other hand, the integration of (2.5) from t ∗ − ω to t ∗ yields Z t ∗ Z ω Z ∗ [l1 (x)(t) − l2 (x)(t)] exp − a(s)ds = x 1 − exp − t ∗ −ω
0
It follows that Z Z t∗ x∗ l2 (1)(t) exp − t ∗ −ω
a(s)ds dt. t
!
t∗
!
t∗
a(s)ds dt ≤ x
∗
"Z
t∗
l1 (1)(t) exp −
t ∗ −ω
t
!
t∗
Z
−kak1
a(s)ds dt − 1 − e
#
.
t
(2.7) If l2 (1)(t) ≡ 0, then Z t∗ Z l2 (1)(t) exp − t ∗ −ω
!
t∗
a(s)ds dt = 0, t
and it follows from (2.1) and (2.7) that ! Z t∗ Z t∗ 0≤ l1 (1)(t) exp − a(s)ds dt − 1 − e−kak1 < 0. t ∗ −ω
t
This is impossible. Here and in the sequel, we need the following equation: ! ! Z t∗ Z t ∗ +ω Z t ∗ +ω Z t∗ li (1)(t) exp − a(s)ds dt = li (1)(t) exp − a(s)ds dt, t ∗ −ω
t∗
t
i = 1, 2.
t
If l2 (1)(t) 6≡ 0, then in view of (2.1), (2.6) and (2.7), we successively get contradiction R ∗ R t∗ t −kak1 t ∗ −ω l1 (1)(t) exp − t a(s)ds dt − 1 − e R ∗ 1≤ R t∗ t l (1)(t) exp − a(s)ds dt ∗ 2 t −ω t ! Z t∗ Z t∗ + l1 (1)(t) exp − a(s)ds dt t ∗ −ω
t
< 1. Case 2. m = −x(t∗ ) > 0. Set x ∗ = M and c = m/M. Then 0 < c ≤ 1. If p Z t∗ (1 − c)2 + 4ce−kak1 + (1 − c) , a(s)ds ≤ ln 2e−kak1 t∗ then integrating (2.5) from t∗ to t ∗ , we have ! Z Z t∗ Z t∗ M exp a(s)ds + m exp a(s)ds = 0
0
t∗
(2.8)
[l1 (x)(t) − l2 (x)(t)] exp
t
Z
a(s)ds dt
0
t∗
Z
t∗
l1 (1)(t) exp
≤ M
Z
a(s)ds dt 0
t∗
Z
t
t∗
+m t∗
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
849
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
It follows that t∗
Z exp −
t∗
Z
t∗
≤
! # " Z t∗ Z t∗ M a(s)ds dt a(s)ds + 1− l1 (1)(t) exp − m t∗ t ! Z ∗ !
t
l2 (1)(t) exp −
t∗
a(s)ds dt.
(2.9)
t
On the other hand, integrating (2.5) from t ∗ − ω to t∗ , we obtain ! Z t∗ Z t∗ Z t Z t ∗ −ω a(s)ds + m exp a(s)ds = [l2 (x)(t) − l1 (x)(t)] exp a(s)ds dt M exp 0
t ∗ −ω
0
Z
0
t∗
≤ M t ∗ −ω Z t∗
l2 (1)(t) exp
Z
t
a(s)ds dt Z t l1 (1)(t) exp a(s)ds dt. 0
+m t ∗ −ω
0
It follows that e
" ! Z ! # Z t∗ Z t∗ t∗ m + exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt M t∗ t ∗ −ω t ! Z Z ∗
−kak1
t∗
≤ t ∗ −ω
t
l2 (1)(t) exp −
a(s)ds dt.
(2.10)
t
Combining (2.8)–(2.10), we have ! Z t∗ Z Z t∗ l2 (1)(t) exp − a(s)ds dt ≥ e−kak1 + exp − t ∗ −ω
t
!
t∗
a(s)ds t∗
" ! # Z t∗ Z t∗ M 1− l1 (1)(t) exp − + a(s)ds dt m t∗ t " ! Z ! # Z t∗ Z t∗ t∗ m exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt + M t∗ t ∗ −ω t " ! # ! Z t∗ Z t∗ Z t∗ 1 1− l1 (1)(t) exp − a(s)ds dt = e−kak1 + exp − a(s)ds + c t∗ t t∗ " ! # ! Z Z t∗ Z t∗ t∗ + c exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt t ∗ −ω
t∗
t
! # Z t∗ 1 a(s)ds dt 1− l1 (1)(t) exp − a(s)ds + ≥ e + (1 + c) exp − c t ∗ −ω t t∗ ! # " Z t∗ Z t∗ 2(1 + c)e−kak1 1 −kak1 ≥ e +p 1− l1 (1)(t) exp − a(s)ds dt + t t ∗ −ω (1 − c)2 + 4ce−kak1 + (1 − c) c ( 2(1 + c)e−kak1 −kak1 ≥ e + min p 0≤c≤1 (1 − c)2 + 4ce−kak1 + (1 − c) " ! #) Z t∗ Z t∗ 1 + 1− l1 (1)(t) exp − a(s)ds dt , c t ∗ −ω t −kak1
Z
t∗
!
which contradicts (2.2) with β = t ∗ − ω or β = t ∗ .
"
Z
t∗
(2.11)
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
If t∗
Z
t∗
p (1 − c)2 + 4ce−kak1 + (1 − c) a(s)ds > ln , 2e−kak1
(2.12)
then integrating (2.5) from t ∗ to t∗ + ω, we have ! Z Z t∗ +ω Z t∗ a(s)ds + m exp a(s)ds = M exp 0
t∗ +ω
t∗
0
[l2 (x)(t) − l1 (x)(t)] exp
t
Z
a(s)ds dt
0 t∗ +ω
Z
l2 (1)(t) exp
Z
t
a(s)ds dt Z t Z t∗ +ω +m a(s)ds dt. l1 (1)(t) exp
≤ M
t∗
0
t∗
It follows that Z exp −
t∗ +ω
0
Z t∗ +ω Z t∗ +ω m a(s)ds + 1− l1 (1)(t) exp − a(s)ds dt M t∗ t∗ t Z t∗ +ω Z t∗ +ω ≤ l2 (1)(t) exp − a(s)ds dt.
t∗
(2.13)
t
On the other hand, integrating (2.5) from t∗ to t ∗ , we have ! Z t∗ Z t∗ Z t Z t∗ a(s)ds + m exp a(s)ds = [l1 (x)(t) − l2 (x)(t)] exp a(s)ds dt M exp 0
0
0
t∗
Z
t∗
l1 (1)(t) exp
≤ M
Z
t
a(s)ds dt
0
t∗ t∗
Z +m
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
t∗
It follows that " Z t∗ +ω Z t∗ Z t∗ +ω # M e exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt + m t∗ t∗ t Z t∗ +ω Z t∗ ≤ l2 (1)(t) exp − a(s)ds dt. −kak1
t∗
t
Combining (2.12)–(2.14), we have Z Z t∗ +ω Z t∗ +ω a(s)ds dt ≥ e−kak1 + exp − l2 (1)(t) exp − t∗
t
Z
t∗ +ω
Z l1 (1)(t) exp −
t∗ +ω
t∗ +ω
a(s)ds
t∗
m 1− a(s)ds dt M t∗ t " Z t∗ Z t∗ +ω # Z t∗ +ω M + exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt m t∗ t∗ t Z t∗ +ω Z t∗ +ω Z t∗ +ω = e−kak1 + exp − a(s)ds + c 1 − l1 (1)(t) exp − a(s)ds dt t∗ t∗ t " Z t∗ +ω Z t∗ Z t∗ +ω # 1 + exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt c t∗ t∗ t Z t∗ +ω 1 1 −kak1 ≥ e + 1+ exp − a(s)ds + c − c c t∗ +
(2.14)
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
851
Z t∗ +ω Z t∗ +ω 1 a(s)ds dt + 1− l1 (1)(t) exp − c t∗ t ! Z t∗ 1 −kak1 1 −kak1 =e + 1+ e exp a(s)ds + c − c c t∗ Z t∗ +ω Z t∗ +ω 1 a(s)ds dt + 1− l1 (1)(t) exp − c t∗ t hp i (1 + c) (1 − c)2 + 4ce−kak1 + (1 − c) 1 − c2 ≥ e−kak1 + − 2c c Z t∗ +ω Z t∗ +ω 1 1− a(s)ds dt + l1 (1)(t) exp − c t∗ t Z t∗ +ω Z t∗ +ω 2(1 + c)e−kak1 1 −kak1 =e +p l1 (1)(t) exp − a(s)ds dt + 1− t∗ t (1 − c)2 + 4ce−kak1 + (1 − c) c ( 2(1 + c)e−kak1 −kak1 ≥ e + min p 0≤c≤1 (1 − c)2 + 4ce−kak1 + (1 − c) Z t∗ +ω ) Z t∗ +ω 1 + 1− l1 (1)(t) exp − , a(s)ds dt c t∗ t which contradicts (2.2). Cases 1 and 2 complete the proof.
If a(t) ≡ 0, then conditions (2.1) and (2.2) imply R α+ω Z α+ω l2 (1)(t)dt kl2 (1)k1 α , kl1 (1)k1 = l1 (1)(t)dt < = R α+ω 1 + kl2 (1)k1 1+ α l2 (1)(t)dt α and kl2 (1)k1 =
β+ω
Z β
l2 (1)(t)dt < 1 + min
0≤c≤1
Z β+ω 1 1− l1 (1)(t)dt 1+c+ c β
= 2 + 2 (1 − kl1 (1)k1 )1/2 , respectively. Thus, we have the following corollary. Corollary 2.1. Assume that a(t) ≡ 0 and that kl2 (1)k1 1 + kl2 (1)k1
(2.15)
kl2 (1)k1 < 2 + 2 (1 − kl1 (1)k1 )1/2 .
(2.16)
kl1 (1)k1 < and
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. If a(t) ≥ 0(6≡ 0) and l1 (1)(t) ≡ 0, then condition (2.1) holds naturally, and condition (2.2) implies ( ) Z β+ω Z β+ω 2(1 + c)e−kak1 1 −kak1 l2 (1)(t) exp − a(s)ds dt < e + min p + 0≤c≤1 β t (1 − c)2 + 4ce−kak1 + (1 − c) c = 1 + e−kak1 + 2e−kak1 /2 . Thus, we have the following corollary.
852
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Corollary 2.2. Assume that a(t) ≥ 0 and l1 (1)(t) ≡ 0, that either a(t) 6≡ 0 or l2 (1)(t) 6≡ 0, and that for any β ∈ [0, ω], Z β+ω Z β+ω a(s)ds dt < 1 + e−kak1 + 2e−kak1 /2 . (2.17) l2 (1)(t) exp − β
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 2.3. Assume that a(t) ≥ 0 and that for any α ∈ [0, ω], Z α+ω Z α+ω a(s)ds dt < 1 − e−kak1 , l1 (1)(t) exp − α
(2.18)
t
and that for any β ∈ [0, ω], β+ω
Z β
β+ω
Z [l2 (1)(t) + l1 (1)(t)] exp −
a(s)ds dt < 1 + 3e−kak1 .
(2.19)
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Proof. We only need to verify that conditions (2.18) and (2.19) imply (2.1) and (2.2). It is obvious that (2.18) implies (2.1). On the other hand, by (2.18), ( Z β+ω ) Z β+ω 1 2(1 + c)e−kak1 + l1 (1)(t) exp − a(s)ds dt 1− min p 0≤c≤1 β t (1 − c)2 + 4ce−kak1 + (1 − c) c Z β+ω Z β+ω 1 −kak1 ≥ min (1 + c)e l1 (1)(t) exp − a(s)ds dt + 1− 0≤c≤1 c β t Z β+ω Z β+ω = 2e−kak1 + 1 − l1 (1)(t) exp − a(s)ds dt. β
t
Thus, it follows from (2.19) that (2.2) holds also. The proof is complete.
Similarly, it follows from (2.1) that ( Z β+ω ) Z β+ω 2(1 + c)e−kak1 1 min p + 1− l1 (1)(t) exp − a(s)ds dt 0≤c≤1 β t (1 − c)2 + 4ce−kak1 + (1 − c) c Z β+ω Z β+ω 1 −kak1 l1 (1)(t) exp − a(s)ds dt 1− ≥ min (1 + c)e + 0≤c≤1 c β t Z β+ω 1/2 Z β+ω −kak1 −kak1 /2 l1 (1)(t) exp − a(s)ds dt . ≥ e + 2e 1− β
t
Thus, we have the following corollary. Corollary 2.4. Assume that a(t) ≥ 0 and that (2.1) holds for any α ∈ [0, ω]. Suppose that for any β ∈ [0, ω], Z β+ω Z β+ω l2 (1)(t) exp − a(s)ds dt β
< 2e
t −kak1
−kak1 /2
+ 2e
Z 1−
β+ω β
β+ω
Z l1 (1)(t) exp −
a(s)ds dt
1/2
.
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution.
(2.20)
853
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Theorem 2.2. Assume that a(t) ≥ 0 and that for any α ∈ [0, ω] R R α+ω α+ω Z α+ω Z α+ω l1 (1)(t) exp − t a(s)ds dt − 1 − e−kak1 α R a(s)ds dt < l2 (1)(t) exp − e−kak1 , R α+ω α+ω −kak t α 1 l1 (1)(t) exp − t a(s)ds dt + e α (2.21) and that for any β ∈ [0, ω] Z Z β+ω l1 (1)(t) exp − β
β+ω
(
a(s)ds dt < 1 + min
2(1 + c) p
(1 − c)2 + 4cekak1 + (1 − c) ) Z β+ω Z β+ω 1 −kak1 . e − a(s)ds dt l2 (1)(t) exp − + c t β 0≤c≤1
t
(2.22) Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Proof. According to theorems in [7,8], it is sufficient to show that homogeneous equation (1.5) has only a trivial ω-periodic solution. Assume the contrary. Let x(t) be a nontrivial ω-periodic solution of Eq. (1.5) and let M, m, c, x∗ , x ∗ , t∗ and t ∗ be as in the proof of Theorem 2.1. There are two possible cases: Case 1. x∗ = x(t∗ ) ≥ 0. Integrating (2.5) from t ∗ to t∗ + ω, we obtain ! Z t∗ +ω Z t∗ +ω Z t Z t∗ ∗ x exp a(s)ds − x∗ exp a(s)ds = a(s)ds dt [l2 (x)(t) − l1 (x)(t)] exp 0
t∗
0
≤ x∗
0
Z
t∗ +ω
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
t∗
It follows that x ∗ e−kak1 ≤ x∗ + x ∗
Z
t∗ +ω
t∗ +ω
Z l2 (1)(t) exp −
t∗
a(s)ds dt.
(2.23)
t
On the other hand, the integration of (2.5) from t∗ to t∗ + ω yields Z ω Z t∗ +ω Z x∗ 1 − exp − a(s)ds = [l1 (x)(t) − l2 (x)(t)] exp − 0
t∗
It follows that Z t∗ +ω Z x∗ l1 (1)(t) exp − t∗
≤x
∗
t∗ +ω
t∗ +ω
a(s)ds dt.
t
a(s)ds dt − 1 − e
−kak1
t t∗ +ω
Z
t∗ +ω
Z l2 (1)(t) exp −
t∗
a(s)ds dt.
t
Thus in view of (2.21), (2.23) and (2.24), we successively get the contradiction Z t∗ +ω Z t∗ +ω ∗ kak1 ∗ x ≤e x∗ + x l2 (1)(t) exp − a(s)ds dt t∗ t∗ +ω
t
Z t∗ +ω kak1 ∗ l2 (1)(t) exp − a(s)ds dt ≤e x t∗ t R R t∗ +ω t +ω l1 (1)(t) exp − t ∗ a(s)ds dt + e−kak1 t∗ R ×R t +ω t∗ +ω l1 (1)(t) exp − t ∗ a(s)ds dt − 1 − e−kak1 t∗ Z
< x ∗.
(2.24)
854
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Case 2. m = −x(t∗ ) > 0. If p Z t∗ (1 − c)2 + 4cekak1 + (1 − c) , a(s)ds ≤ ln 2 t∗
(2.25)
then integrating (2.5) from t∗ to t ∗ , we have ! Z t∗ Z Z t∗ a(s)ds + m exp M exp a(s)ds = 0
0
t∗
t
Z
[l1 (x)(t) − l2 (x)(t)] exp
a(s)ds dt
0
t∗ t∗
Z
l1 (1)(t) exp
≤ M
Z
t
a(s)ds dt
0
t∗ t∗
Z +m
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
t∗
It follows that " ! Z ∗ ! # Z t∗ Z t∗ t m exp − a(s)ds − l2 (1)(t) exp − a(s)ds dt 1+ M t∗ t∗ t ! Z ∗ Z ∗ t
t
l1 (1)(t) exp −
≤ t∗
a(s)ds dt.
(2.26)
t
On the other hand, integrating (2.5) from t ∗ − ω to t∗ , we obtain ! Z t∗ Z t∗ Z t Z t ∗ −ω M exp a(s)ds + m exp a(s)ds = [l2 (x)(t) − l1 (x)(t)] exp a(s)ds dt 0
0
t ∗ −ω
Z
0
t∗
≤ M t ∗ −ω Z t∗
l2 (1)(t) exp
+m t ∗ −ω
Z
t
0
l1 (1)(t) exp
Z
a(s)ds dt t a(s)ds dt.
0
It follows that t∗
Z exp −
t∗
Z
t∗
≤ t ∗ −ω
" ! # Z t∗ Z t∗ M −kak1 e − a(s)ds + l2 (1)(t) exp − a(s)ds dt m t ∗ −ω t ! Z ∗ !
t
l1 (1)(t) exp −
a(s)ds dt.
Combining (2.25)–(2.27), we have ! Z t∗ Z Z t∗ l1 (1)(t) exp − a(s)ds dt ≥ 1 + exp − t ∗ −ω
(2.27)
t
t
!
t∗
a(s)ds t∗
" ! Z ∗ ! # Z t∗ Z t∗ t m + a(s)ds dt exp − a(s)ds − l2 (1)(t) exp − M t∗ t∗ t " ! # Z t∗ Z t∗ M −kak1 e − + l2 (1)(t) exp − a(s)ds dt m t ∗ −ω t " ! # ! Z t∗ Z t∗ Z t∗ 1 −kak1 e − l2 (1)(t) exp − a(s)ds dt ≥ 1 + (1 + c) exp − a(s)ds + c t ∗ −ω t t∗ " ! # Z t∗ Z t∗ 1 −kak1 2(1 + c) + e ≥ 1+ p − l2 (1)(t) exp − a(s)ds dt t ∗ −ω t (1 − c)2 + 4cekak1 + (1 − c) c
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
( ≥ 1 + min
0≤c≤1
2(1 + c) p
(1 − c)2 + 4cekak1
"
1 −kak1 + e − + (1 − c) c
t∗
Z
l2 (1)(t) exp −
t ∗ −ω
!
t∗
Z
a(s)ds dt
,
t
which contradicts (2.22) with β = t ∗ − ω or β = t ∗ . If p Z t∗ (1 − c)2 + 4cekak1 + (1 − c) a(s)ds > ln , 2 t∗
(2.28)
then integrating (2.5) from t ∗ to t∗ + ω, we have ! Z t∗ +ω Z Z t∗ a(s)ds + m exp a(s)ds = M exp 0
#)
t∗ +ω
t∗
0
[l2 (x)(t) − l1 (x)(t)] exp
t
Z
a(s)ds dt
0 t∗ +ω
Z
l2 (1)(t) exp
Z
t
a(s)ds dt Z t Z t∗ +ω +m l1 (1)(t) exp a(s)ds dt.
≤ M
t∗
0
t∗
0
It follows that Z t∗ +ω Z t∗ +ω Z t∗ +ω M 1+ exp − a(s)ds − l2 (1)(t) exp − a(s)ds dt m t∗ t∗ t Z t∗ +ω Z t∗ +ω ≤ l1 (1)(t) exp − a(s)ds dt. t∗
(2.29)
t
On the other hand, integrating (2.5) from t∗ to t ∗ , we have ! Z t∗ Z t∗ Z t Z t∗ M exp a(s)ds + m exp a(s)ds = a(s)ds dt [l1 (x)(t) − l2 (x)(t)] exp 0
0
0
t∗
Z
t∗
l1 (1)(t) exp
≤ M
Z
t∗
+m
a(s)ds dt
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
t∗
It follows that Z exp −
0
t∗
Z
t
" Z t∗ +ω # Z t∗ m −kak1 a(s)ds + l2 (1)(t) exp − a(s)ds dt e − M t∗ t∗ t Z t∗ +ω Z t∗ ≤ l1 (1)(t) exp − a(s)ds dt. t∗ +ω
t∗
t
Combining (2.28)–(2.30), we have Z t∗ +ω Z Z t∗ +ω l1 (1)(t) exp − a(s)ds dt ≥ 1 + exp − t∗
t
t∗ +ω
a(s)ds
t∗
Z t∗ +ω Z t∗ +ω Z t∗ +ω M exp − a(s)ds − l2 (1)(t) exp − a(s)ds dt + m t∗ t∗ t " Z t∗ +ω # Z t∗ m + e−kak1 − l2 (1)(t) exp − a(s)ds dt M t∗ t Z t∗ +ω 1 −kak1 1 exp − a(s)ds + c − e ≥ 1+ 1+ c c t∗
(2.30)
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Z t∗ +ω Z t∗ +ω 1 −kak1 a(s)ds dt + e − l2 (1)(t) exp − c t∗ t ! Z t∗ 1 −kak1 = 1+ 1+ e exp a(s)ds c t∗ Z t∗ +ω Z t∗ +ω 1 −kak1 1 −kak1 + c− e + e − a(s)ds dt l2 (1)(t) exp − c c t∗ t i hp (1 + c) (1 − c)2 + 4cekak1 + (1 − c) e−kak1 1 − c2 −kak1 ≥ 1+ − e 2c c Z t∗ +ω Z t∗ +ω 1 −kak1 e − a(s)ds dt + l2 (1)(t) exp − c t∗ t Z t∗ +ω Z t∗ +ω 2(1 + c) 1 −kak1 l2 (1)(t) exp − a(s)ds dt = 1+ p − + e t∗ t (1 − c)2 + 4cekak1 + (1 − c) c ( 2(1 + c) ≥ 1 + min p 2 0≤c≤1 (1 − c) + 4cekak1 + (1 − c) Z t∗ +ω ) Z t∗ +ω 1 −kak1 + e − l2 (1)(t) exp − , a(s)ds dt c t∗ t which contradicts (2.22) with β = t∗ . Cases 1 and 2 complete the proof.
Like Corollaries 2.1, 2.2 and 2.4, we have the following corollaries. Corollary 2.5. Assume that a(t) ≡ 0 and that kl1 (1)k1 1 + kl1 (1)k1
(2.31)
kl1 (1)k1 < 2 + 2 (1 − kl2 (1)k1 )1/2 .
(2.32)
kl2 (1)k1 < and
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 2.6. Assume that a(t) ≥ 0 and l2 (1)(t) ≡ 0, and that for any β ∈ [0, ω], Z β+ω Z β+ω a(s)ds dt < 1 + e−kak1 + 2e−kak1 /2 . 1 − e−kak1 < l1 (1)(t) exp − β
(2.33)
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 2.7. Assume that a(t) ≥ 0 and that (2.21) holds for any α ∈ [0, ω]. Suppose that for any β ∈ [0, ω], Z β+ω Z β+ω l1 (1)(t) exp − a(s)ds dt β
t
Z < 1 + e−kak1 + 2e−kak1 /2 e−kak1 −
β
β+ω
β+ω
Z l2 (1)(t) exp −
1/2 a(s)ds dt .
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Remark 2.1. Corollaries 2.1 and 2.5 were obtained in [6].
(2.34)
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
3. The case a(t) ≤ 0 Define the operator (see [4]) ϑ : Cω (R) → Cω (R) by ϑ(x)(t) = x(ω − t)
for t ∈ R.
Now if x(t) is a solution of Eq. (1.5), then y(t) = ϑ(x)(t) is a solution of y 0 (t) = ϑ(a)(t)y(t) − ϑ(l1 (ϑ(y)))(t) + ϑ(l2 (ϑ(y)))(t),
(3.1)
and, vice versa, if y(t) is a solution of Eq. (3.1), then x(t) = ϑ(y)(t) is a solution of Eq. (1.5). Therefore, Eq. (1.5) has only the trivial ω-periodic solution if and only if Eq. (3.1) has only the trivial ω-periodic solution. Note that kak1 = kϑ(a)k1 , and for i = 1, 2, Z Z α+ω ϑ(li (ϑ(1)))(t) exp − α
α+ω
|ϑ(a)(s)|ds dt =
t
α+ω ˆ
Z αˆ
Z t li (1)(t) exp − |a(s)|ds dt, αˆ
where αˆ = ω − α. Therefore, if a(t) ≤ 0 for t ∈ R, then one can obtain the following theorems and corollaries by using the results obtained in Section 2 for Eq. (3.1). Theorem 3.1. Assume that a(t) ≤ 0 and that for any α ∈ [0, ω], α+ω
Z α
R R α+ω t Z t l1 (1)(t) exp − α |a(s)|ds dt 1 − e−kak1 + α R l2 (1)(t) exp − |a(s)|ds dt < , R α+ω t α 1+ α l1 (1)(t) exp − α |a(s)|ds dt
and that for any β ∈ [0, ω], ( Z t Z β+ω l1 (1)(t) exp − |a(s)|ds dt < e−kak1 + min p β
2(1 + c)e−kak1
(1 − c)2 + 4ce−kak1 + (1 − c) Z t ) Z β+ω 1 + . 1− l2 (1)(t) exp − |a(s)|ds dt c β β
β
(3.2)
0≤c≤1
(3.3)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 3.1. Assume that a(t) ≤ 0 and l2 (1)(t) ≡ 0, that either a(t) 6≡ 0 or l1 (1)(t) 6≡ 0, and that for any β ∈ [0, ω], Z t Z β+ω l1 (1)(t) exp − |a(s)|ds dt < 1 + e−kak1 + 2e−kak1 /2 . (3.4) β
β
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 3.2. Assume that a(t) ≤ 0 and that for any α ∈ [0, ω], Z t Z α+ω l2 (1)(t) exp − |a(s)|ds dt < 1 − e−kak1 ,
(3.5)
and that for any β ∈ [0, ω], Z t Z β+ω [l1 (1)(t) + l2 (1)(t)] exp − |a(s)|ds dt < 1 + 3e−kak1 .
(3.6)
α
β
α
β
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution.
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Corollary 3.3. Assume that a(t) ≤ 0 and that (3.2) holds for any α ∈ [0, ω]. Suppose that for any β ∈ [0, ω], Z t Z β+ω |a(s)|ds dt l1 (1)(t) exp − β
β
< 2e
−kak1
−kak1 /2
+ 2e
Z 1−
β+ω β
1/2 Z t |a(s)|ds dt . l2 (1)(t) exp −
(3.7)
β
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Theorem 3.2. Assume that a(t) ≤ 0 and that for any α ∈ [0, ω], R R α+ω t Z t Z α+ω l (1)(t) exp − a(s)ds dt − 1 − e−kak1 2 α α R |a(s)|ds dt < R l1 (1)(t) exp − e−kak1 , α+ω t −kak1 α α l (1)(t) exp − |a(s)|ds dt + e 2 α α and that for any β ∈ [0, ω], ( Z t Z β+ω l2 (1)(t) exp − |a(s)|ds dt < 1 + min p β
β
0≤c≤1
+
1 −kak1 e c
(3.8)
2(1 + c)
(1 − c)2 + 4cekak1 + (1 − c) Z t ) Z β+ω − . l1 (1)(t) exp − |a(s)|ds dt β
β
(3.9)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 3.4. Assume that a(t) ≤ 0 and l1 (1)(t) ≡ 0, and that for any β ∈ [0, ω], Z t Z β+ω −kak1 1−e < l2 (1)(t) exp − |a(s)|ds dt < 1 + e−kak1 + 2e−kak1 /2 . β
β
(3.10)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 3.5. Assume that a(t) ≤ 0 and that (3.8) holds for any α ∈ [0, ω]. Suppose that for any β ∈ [0, ω], Z t Z β+ω l2 (1)(t) exp − |a(s)|ds dt β
β
Z < 1 + e−kak1 + 2e−kak1 /2 e−kak1 −
β
β+ω
Z t 1/2 l1 (1)(t) exp − |a(s)|ds dt . β
(3.11)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. 4. Examples and remarks Example 4.1. Consider the first-order functional differential equation x 0 (t) = −ax(t) + b(1 + sin t)x(t − τ ) + f (t), where a, b, τ ∈ R and f ∈ C2π (R). For any α ∈ R, Z α+2π |b| |b|(|a| sin α + cos α) −|a|(t−α) −2π|a| |b|(1 + sin t)e dt = 1 − e + , |a| a2 + 1 α
(4.1)
(4.2)
and Z α
α+2π
−|a|(α+2π−t)
|b|(1 + sin t)e
|b| |b|(|a| sin α − cos α) −2π|a| + dt = 1 − e . |a| a2 + 1
(4.3)
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
859
Case (i). a > 0, b > 0. In this case, l1 (1)(t) = b(1 + sin t) and l2 (1)(t) ≡ 0. In view of Corollary 2.3, if for any α ∈ [0, 2π], Z α |b| |a| l1 (1)(t)e−|a|(α−t) dt ≤ 1 − e−2π|a| < 1 − e−2π|a| , 1+ √ |a| a2 + 1 α−2π or if |b| <
|a| 1 + √|a|
,
(4.4)
a 2 +1
then Eq. (4.1) has a unique 2π -periodic solution. In view of Corollary 2.6, if for any β ∈ [0, 2π], |a| −2π|a| −2π|a| |b| 1−e < 1−e 1− √ |a| a2 + 1 Z β+2π ≤ l1 (1)(t)e−|a|(β+2π−t) dt β |b| |a| 1+ √ ≤ 1 − e−2π|a| |a| a2 + 1 −2π|a| −π|a| < 1+e + 2e , or if |a| 1 − √|a|
a 2 +1
|a| 1 + e−π|a| , < |b| < |a| −π|a| √ 1+ 1−e 2
(4.5)
a +1
then Eq. (4.1) has also a unique 2π-periodic solution. But, by Corollaries 2.1 and 2.5, the conditions corresponding to (4.4) and (4.5) are, respectively, |b| <
|a| , 1 + 2π |a|
and
2π|a| < 2 + 2(1 − 2π |b|)1/2 ,
(4.6)
and |a| 2 + 2(1 − 2π |a|)1/2 < |b| < . 1 − 2π |a| 2π
(4.7)
By Corollary 1.1 in [1], the conditions corresponding to (4.4) and (4.5) are, respectively, 4|a| |a| |b| < min , , 1 + 2π |a| e2π|a| − 1 1 − |a|
(4.8)
a 2 +1
and |a| 4|a| < |b| < 1 − 2π |a| e2π|a| − 1 1 −
|a|
.
(4.9)
a 2 +1
It is not difficult to see that conditions (4.4) and (4.5) are weaker than (4.6) and (4.7), and weaker than (4.8) and (4.9), respectively. Case (ii). a > 0, b < 0. In this case, l1 (1)(t) ≡ 0 and l2 (1)(t) = |b|(1 + sin t). In view of Corollary 2.2, if for any β ∈ [0, 2π],
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861 β+2π
Z β
|b| |b|(|a| sin β − cos β) l2 (1)(t)e−|a|(β+2π−t) dt = 1 − e−2π|a| + |a| a2 + 1 |b| |a| ≤ 1 − e−2π|a| 1+ √ |a| a2 + 1 < 1 + e−2π|a| + 2e−π|a| ,
or if |a| 1 + e−π|a| , |b| < |a| −π|a| 1−e 1+ √
(4.10)
a 2 +1
then Eq. (4.1) has a unique 2π -periodic solution. But, by Corollary 2.1, the condition corresponding to (4.10) is |b| <
4 − 2π |a| . 2π
(4.11)
It is not difficult to see that condition (4.10) is weaker than (4.11). Case (iii). a < 0, b > 0. In this case, l1 (1)(t) = b(1 + sin t) and l2 (1)(t) ≡ 0. In view of Corollary 3.1, if (4.10) holds, then Eq. (4.1) has a unique 2π-periodic solution. But, by Corollary 2.5, the condition corresponding to (4.10) is (4.11). Case (iv). a < 0, b < 0. In this case, l1 (1)(t) ≡ 0 and l2 (1)(t) = |b|(1 + sin t). In view of Corollaries 3.2 and 3.4, If (4.4) or (4.5) holds, then Eq. (4.1) has a unique 2π -periodic solution. But, by Corollaries 2.1 and 2.5, the conditions corresponding to (4.4) and (4.5) are (4.6) and (4.7), respectively. By Corollary 1.1 in [1], the conditions corresponding to (4.4) and (4.5) are (4.8) and (4.9), respectively. Remark 4.1. Example 4.1 shows that the conditions in Corollaries 2.1 and 2.5 in which the instantaneous term −a(t)x(t) is treated as l1 (x)(t) or l2 (x)(t) are stronger than ones in the other theorems and corollaries in this paper. Remark 4.2. Example 4.1 shows that condition (1.10) is weaker than (1.6) and (1.8) in general. Acknowledgments The authors are grateful to the referees for their valuable comments and suggestions. References [1] R. Hakl, A. Lomtatidze, B. P˚uzˇ a, On periodic solutions of first order linear functional differential equations, Nonlinear Analysis 49 (2002) 929–945. [2] R. Hakl, A. Lomtatidze, B. P˚uzˇ a, On a boundary value problem for first-order scalar functional differential equations, Nonlinear Analysis 53 (2003) 391–405. ˇ [3] R. Hakl, A. Lomtatidze, J. Sremr, On a periodic-type boundary value problem for first order nonlinear functional differential equations, Nonlinear Analysis 51 (2002) 425–447. ˇ [4] R. Hakl, A. Lomtatidze, J. Sremr, A periodic type boundary-value problem for first order nonlinear functional differential equations, Nelinijni Kolyvannya 5 (3) (2002) 416–432 (in English). [5] R. Hakl, A. Lomtatidze, I.P. Stavroulakis, On a boundary value problem for scalar linear functional differential equations, Abstract and Applied Analysis 1 (2004) 45–67. [6] A.G. Lomtatidze, R. Hakl, B. P˚uzˇ a, On the periodic boundary value problem for first-order functional-differential equations, Differential Equations 39 (3) (2–3) 344–352 (English. Russian original). Translation from Differentsial’nye Uravneniya 39 (3) (2003) 320–327. [7] J.K. Hale, Periodic and almost periodic solutions of functional-differential equations, Archive for Rational Mechanics and Analysis 15 (1964) 289–304. [8] T. Jankowski, Existence of solutions of boundary value problems for differential equations with delayed arguments, Journal of Computational and Applied Mathematics 156 (2003) 239–252. [9] D. Jiang, J.J. Nieto, W. Zuo, On monotone method for first and second order periodic boundary value problems and periodic solutions of functional differential equations, Journal of Mathematical Analysis and Applications 289 (2004) 691–699. [10] I. Kiguradze, B. P˚uzˇ a, On periodic solutions of systems of linear functional differential equations, Archive der Mathematik 33 (3) (1997) 197–212.
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[11] S.W. Ma, J.S. Yu, Z.C. Wang, Periodic solutions of functional differential equations with periodic perturbation, Chinese Science Bulletin 43 (1998) 1386–1388 (in Chinese). [12] K.G. Mavridis, P.Ch. Tsamatos, Positive solutions for a Floquet functional boundary value problem, Journal of Mathematical Analysis and Applications 296 (2004) 165–182. [13] J.J. Nieto, R. Rodr´ıguez-L´opez, Periodic boundary value problem for non-Lipschitzian impulsive functional differential equations, Journal of Mathematical Analysis and Applications 318 (2006) 593–610. [14] J.J. Nieto, R. Rodr´ıguez-L´opez, Remarks on periodic boundary value problems for functional differential equations, Journal of Computational and Applied Mathematics 158 (2003) 339–353. [15] J.J. Nieto, Differential inequalities for functional perturbations of first order ordinary differential equations, Applied Mathematics Letters 15 (2002) 173–179. [16] B. P˚uzˇ a, On periodic solutions of linear differential systems with deviating arguments, Differentsial’nye Uravneniya 31 (11) (1995) 1937–1938 (in Russian).
Periodic solutions of first-order nonlinear functional differential equationsI X.H. Tang ∗ , Zhiyuan Jiang School of Mathematical Sciences and Computing Technology, Central South University, Changsha, Hunan 410083, PR China Received 25 September 2006; accepted 27 November 2006
Abstract In this paper, we obtain a set of “easily verifiable” sufficient conditions for the existence of the periodic solutions of the firstorder linear functional differential equations with periodic perturbation x 0 (t) = l(x)(t) + f (t, xt ), where l : C(R) → C(R) is a linear bounded operator. These conditions generalize and improve the known results given in the literature. c 2006 Elsevier Ltd. All rights reserved.
MSC: 34K13 Keywords: Functional differential equations; Periodic solution; Periodic perturbation
1. Introduction Let BC(R, R) be the Banach space of bounded continuous functions x : R → R with the sup norm kxk = sup{|x(t)| : t ∈ R} and let Cω (R) = {x ∈ BC(R, R) : x(t + ω) = x(t), t ∈ R}, where ω > 0. Define the norms as follows: kxk0 = max{|x(t)| : 0 ≤ t ≤ ω},
ω
Z
|x(t)|dt,
kxk1 = 0
∀x ∈ Cω (R).
Consider the first-order functional differential equations x 0 (t) = l(x)(t) + f (t, xt ), I This work was partially supported by the NNSF of China (No. 10471153). ∗ Corresponding author.
E-mail address: [email protected] (X.H. Tang). c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.11.041
(1.1)
846
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
and the particular case x 0 (t) = l(x)(t) + g(t),
(1.2)
where l : C(R) → C(R) is a linear bounded operator and l(Cω (R)) ⊆ Cω (R), g ∈ Cω (R), and f ∈ C(R × BC(R, R), R), for any t ∈ R, f (t, ·) mapping a bounded set in BC(R, R) into a bounded set in R and satisfying f (t + ω, ϕ) = f (t, ϕ) for (t, ϕ) ∈ R × BC(R, R). Everywhere in what follows, we will assume that the operator l is nontrivial and admits the representation l(x)(t) = −a(t)x(t) + l1 (x)(t) − l2 (x)(t),
(1.3)
where a ∈ Cω (R), l1 , l2 : Cω (R) → Cω (R) are linear and satisfy the condition l1 (x)(t) ≥ 0,
l2 (x)(t) ≥ 0,
∀t ∈ [0, ω] if x(t) ≥ 0, ∀t ∈ [0, ω].
In the representation (1.3), we list the instantaneous term a(t)x(t) as an independent one, because the instantaneous term always plays an important role in the existence and stability of the periodic solution for Eq. (1.1). The common particular case of Eq. (1.2) is the following linear equation with deviating arguments: x 0 (t) = p0 (t)x(t) +
n X
pi (t)x(t − τi (t)) + g(t),
(1.4)
i=1
where g ∈ Cω (R) and p0 , pi , τi ∈ Cω (R), i = 1, 2, . . . , n. For Eq. (1.4), we find that l1 (x)(t) =
n X
[ pi (t)]+ x(t − τi (t)),
l2 (x)(t) =
i=1
n X
[ pi (t)]− x(t − τi (t)),
i=1
where here and in the sequel, [x]+ = (|x| + x)/2, [x]− = (|x| − x)/2. It is known (see, e.g., [10]) that Eq. (1.2) has a unique ω-periodic solution if and only if the corresponding homogeneous equation x 0 (t) = l(x)(t),
(1.5)
has only a trivial ω-periodic solution. In view of this fact, [1,10,16] gave a set of conditions which guarantee that Eq. (1.2) has a unique ω-periodic solution. In the paper [11], Ma, Yu and Wang further proved that the homogeneous equation (1.5) having only a trivial ω-periodic solution implies that Eq. (1.1) has at least one ω-periodic solution under the additional restriction (H1) limkϕk→∞
| f (t,ϕ)| kϕk
= 0 uniformly in t ∈ R.
In this paper, our main purpose is to derive a set of “easily verifiable” sufficient conditions for the existence of the periodic solutions of Eq. (1.1). These conditions generalize and improve the known results given in the literature (see [1,4,6,7,10,16]). For example, consider Eq. (1.4) with p0 (t) ≥ 0 and pi (t) ≤ 0, i = 1, 2, . . . , n. In paper [1], it is shown that if n P
k p 0 k1 <
k pi k1
i=1
1+
n P
(1.6) k pi k1
i=1
and n X
k pi k1 < 4 (1 − k p0 k1 ) ,
(1.7)
i=1
or ω
n Z X i=1
0
| pi (t)| exp
ω
Z t
p0 (s)ds
< 4,
(1.8)
847
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
then Eq. (1.4) has a unique ω-periodic solution. But in view of Corollary 2.1 in Section 2 and Corollary 3.4 in Section 3, condition (1.7) can be improved to n X
k pi k1 < 2 + 2 (1 − k p0 k1 )1/2 ;
(1.9)
i=1
and conditions (1.6) and (1.8) can be replaced by the following better conditions (see Example 4.1): Z β+ω n Z β+ω X ek p 0 k 1 − 1 < p0 (s)ds dt | pi (t)| exp i=1
β
< 1 + 2e
t k p0 k1 /2
+e
k p 0 k1
,
β ∈ [0, ω].
(1.10)
Other related results can be found in [2,3,5,8,9,12–15]. For Eq. (1.1), we always assume that (H1) holds in what follows. 2. The case a(t) ≥ 0 Theorem 2.1. Assume that a(t) ≥ 0 and that for any α ∈ [0, ω], α+ω
Z α
α+ω
Z l1 (1)(t) exp − t
and that for any β ∈ [0, ω], Z Z β+ω l2 (1)(t) exp − β
β+ω
R R α+ω α+ω l2 (1)(t) exp − t a(s)ds dt 1 − e−kak1 + α R a(s)ds dt < , R α+ω α+ω 1+ α l2 (1)(t) exp − t a(s)ds dt
(
a(s)ds dt < e−kak1 + min
2(1 + c)e−kak1 p
(1 − c)2 + 4ce−kak1 + (1 − c) Z β+ω ) Z β+ω 1 . l1 (1)(t) exp − a(s)ds dt + 1− c β t 0≤c≤1
t
(2.1)
(2.2)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Proof. According to theorems in [7,8], it is sufficient to show that homogeneous equation (1.5) has only a trivial ω-periodic solution. Assume the contrary. Let x(t) be a nontrivial ω-periodic solution of Eq. (1.5). Then kxk0 > 0. Set x∗ = min{x(t) : 0 ≤ t ≤ ω},
x ∗ = max{x(t) : 0 ≤ t ≤ ω}.
(2.3)
Choose t∗ ∈ [0, ω] and t ∗ ∈ [t∗ , t∗ + ω] such that x(t∗ ) = x∗ ,
x(t ∗ ) = x ∗ .
(2.4)
Since l(x) is linear, we can let x ∗ = kxk0 . From (1.3) and (1.5), we have 0 Z t Z t x(t) exp = [l1 (x)(t) − l2 (x)(t)] exp a(s)ds . a(s)ds
(2.5)
0
0
There are two possible cases: Case 1. x∗ = x(t∗ ) ≥ 0. Integrating (2.5) from t∗ to t ∗ , we obtain ! Z t∗ Z t∗ Z t Z t∗ ∗ x exp a(s)ds − x∗ exp a(s)ds = [l1 (x)(t) − l2 (x)(t)] exp a(s)ds dt 0
0
0
t∗
≤ x
∗
Z
t∗ t ∗ −ω
l1 (1)(t) exp
Z
t
a(s)ds dt.
0
848
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
It follows that x ∗ ≤ x∗ + x ∗
Z
t∗ t ∗ −ω
l1 (1)(t) exp −
!
t∗
Z
a(s)ds dt.
(2.6)
t
On the other hand, the integration of (2.5) from t ∗ − ω to t ∗ yields Z t ∗ Z ω Z ∗ [l1 (x)(t) − l2 (x)(t)] exp − a(s)ds = x 1 − exp − t ∗ −ω
0
It follows that Z Z t∗ x∗ l2 (1)(t) exp − t ∗ −ω
a(s)ds dt. t
!
t∗
!
t∗
a(s)ds dt ≤ x
∗
"Z
t∗
l1 (1)(t) exp −
t ∗ −ω
t
!
t∗
Z
−kak1
a(s)ds dt − 1 − e
#
.
t
(2.7) If l2 (1)(t) ≡ 0, then Z t∗ Z l2 (1)(t) exp − t ∗ −ω
!
t∗
a(s)ds dt = 0, t
and it follows from (2.1) and (2.7) that ! Z t∗ Z t∗ 0≤ l1 (1)(t) exp − a(s)ds dt − 1 − e−kak1 < 0. t ∗ −ω
t
This is impossible. Here and in the sequel, we need the following equation: ! ! Z t∗ Z t ∗ +ω Z t ∗ +ω Z t∗ li (1)(t) exp − a(s)ds dt = li (1)(t) exp − a(s)ds dt, t ∗ −ω
t∗
t
i = 1, 2.
t
If l2 (1)(t) 6≡ 0, then in view of (2.1), (2.6) and (2.7), we successively get contradiction R ∗ R t∗ t −kak1 t ∗ −ω l1 (1)(t) exp − t a(s)ds dt − 1 − e R ∗ 1≤ R t∗ t l (1)(t) exp − a(s)ds dt ∗ 2 t −ω t ! Z t∗ Z t∗ + l1 (1)(t) exp − a(s)ds dt t ∗ −ω
t
< 1. Case 2. m = −x(t∗ ) > 0. Set x ∗ = M and c = m/M. Then 0 < c ≤ 1. If p Z t∗ (1 − c)2 + 4ce−kak1 + (1 − c) , a(s)ds ≤ ln 2e−kak1 t∗ then integrating (2.5) from t∗ to t ∗ , we have ! Z Z t∗ Z t∗ M exp a(s)ds + m exp a(s)ds = 0
0
t∗
(2.8)
[l1 (x)(t) − l2 (x)(t)] exp
t
Z
a(s)ds dt
0
t∗
Z
t∗
l1 (1)(t) exp
≤ M
Z
a(s)ds dt 0
t∗
Z
t
t∗
+m t∗
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
849
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
It follows that t∗
Z exp −
t∗
Z
t∗
≤
! # " Z t∗ Z t∗ M a(s)ds dt a(s)ds + 1− l1 (1)(t) exp − m t∗ t ! Z ∗ !
t
l2 (1)(t) exp −
t∗
a(s)ds dt.
(2.9)
t
On the other hand, integrating (2.5) from t ∗ − ω to t∗ , we obtain ! Z t∗ Z t∗ Z t Z t ∗ −ω a(s)ds + m exp a(s)ds = [l2 (x)(t) − l1 (x)(t)] exp a(s)ds dt M exp 0
t ∗ −ω
0
Z
0
t∗
≤ M t ∗ −ω Z t∗
l2 (1)(t) exp
Z
t
a(s)ds dt Z t l1 (1)(t) exp a(s)ds dt. 0
+m t ∗ −ω
0
It follows that e
" ! Z ! # Z t∗ Z t∗ t∗ m + exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt M t∗ t ∗ −ω t ! Z Z ∗
−kak1
t∗
≤ t ∗ −ω
t
l2 (1)(t) exp −
a(s)ds dt.
(2.10)
t
Combining (2.8)–(2.10), we have ! Z t∗ Z Z t∗ l2 (1)(t) exp − a(s)ds dt ≥ e−kak1 + exp − t ∗ −ω
t
!
t∗
a(s)ds t∗
" ! # Z t∗ Z t∗ M 1− l1 (1)(t) exp − + a(s)ds dt m t∗ t " ! Z ! # Z t∗ Z t∗ t∗ m exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt + M t∗ t ∗ −ω t " ! # ! Z t∗ Z t∗ Z t∗ 1 1− l1 (1)(t) exp − a(s)ds dt = e−kak1 + exp − a(s)ds + c t∗ t t∗ " ! # ! Z Z t∗ Z t∗ t∗ + c exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt t ∗ −ω
t∗
t
! # Z t∗ 1 a(s)ds dt 1− l1 (1)(t) exp − a(s)ds + ≥ e + (1 + c) exp − c t ∗ −ω t t∗ ! # " Z t∗ Z t∗ 2(1 + c)e−kak1 1 −kak1 ≥ e +p 1− l1 (1)(t) exp − a(s)ds dt + t t ∗ −ω (1 − c)2 + 4ce−kak1 + (1 − c) c ( 2(1 + c)e−kak1 −kak1 ≥ e + min p 0≤c≤1 (1 − c)2 + 4ce−kak1 + (1 − c) " ! #) Z t∗ Z t∗ 1 + 1− l1 (1)(t) exp − a(s)ds dt , c t ∗ −ω t −kak1
Z
t∗
!
which contradicts (2.2) with β = t ∗ − ω or β = t ∗ .
"
Z
t∗
(2.11)
850
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
If t∗
Z
t∗
p (1 − c)2 + 4ce−kak1 + (1 − c) a(s)ds > ln , 2e−kak1
(2.12)
then integrating (2.5) from t ∗ to t∗ + ω, we have ! Z Z t∗ +ω Z t∗ a(s)ds + m exp a(s)ds = M exp 0
t∗ +ω
t∗
0
[l2 (x)(t) − l1 (x)(t)] exp
t
Z
a(s)ds dt
0 t∗ +ω
Z
l2 (1)(t) exp
Z
t
a(s)ds dt Z t Z t∗ +ω +m a(s)ds dt. l1 (1)(t) exp
≤ M
t∗
0
t∗
It follows that Z exp −
t∗ +ω
0
Z t∗ +ω Z t∗ +ω m a(s)ds + 1− l1 (1)(t) exp − a(s)ds dt M t∗ t∗ t Z t∗ +ω Z t∗ +ω ≤ l2 (1)(t) exp − a(s)ds dt.
t∗
(2.13)
t
On the other hand, integrating (2.5) from t∗ to t ∗ , we have ! Z t∗ Z t∗ Z t Z t∗ a(s)ds + m exp a(s)ds = [l1 (x)(t) − l2 (x)(t)] exp a(s)ds dt M exp 0
0
0
t∗
Z
t∗
l1 (1)(t) exp
≤ M
Z
t
a(s)ds dt
0
t∗ t∗
Z +m
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
t∗
It follows that " Z t∗ +ω Z t∗ Z t∗ +ω # M e exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt + m t∗ t∗ t Z t∗ +ω Z t∗ ≤ l2 (1)(t) exp − a(s)ds dt. −kak1
t∗
t
Combining (2.12)–(2.14), we have Z Z t∗ +ω Z t∗ +ω a(s)ds dt ≥ e−kak1 + exp − l2 (1)(t) exp − t∗
t
Z
t∗ +ω
Z l1 (1)(t) exp −
t∗ +ω
t∗ +ω
a(s)ds
t∗
m 1− a(s)ds dt M t∗ t " Z t∗ Z t∗ +ω # Z t∗ +ω M + exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt m t∗ t∗ t Z t∗ +ω Z t∗ +ω Z t∗ +ω = e−kak1 + exp − a(s)ds + c 1 − l1 (1)(t) exp − a(s)ds dt t∗ t∗ t " Z t∗ +ω Z t∗ Z t∗ +ω # 1 + exp − a(s)ds − l1 (1)(t) exp − a(s)ds dt c t∗ t∗ t Z t∗ +ω 1 1 −kak1 ≥ e + 1+ exp − a(s)ds + c − c c t∗ +
(2.14)
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
851
Z t∗ +ω Z t∗ +ω 1 a(s)ds dt + 1− l1 (1)(t) exp − c t∗ t ! Z t∗ 1 −kak1 1 −kak1 =e + 1+ e exp a(s)ds + c − c c t∗ Z t∗ +ω Z t∗ +ω 1 a(s)ds dt + 1− l1 (1)(t) exp − c t∗ t hp i (1 + c) (1 − c)2 + 4ce−kak1 + (1 − c) 1 − c2 ≥ e−kak1 + − 2c c Z t∗ +ω Z t∗ +ω 1 1− a(s)ds dt + l1 (1)(t) exp − c t∗ t Z t∗ +ω Z t∗ +ω 2(1 + c)e−kak1 1 −kak1 =e +p l1 (1)(t) exp − a(s)ds dt + 1− t∗ t (1 − c)2 + 4ce−kak1 + (1 − c) c ( 2(1 + c)e−kak1 −kak1 ≥ e + min p 0≤c≤1 (1 − c)2 + 4ce−kak1 + (1 − c) Z t∗ +ω ) Z t∗ +ω 1 + 1− l1 (1)(t) exp − , a(s)ds dt c t∗ t which contradicts (2.2). Cases 1 and 2 complete the proof.
If a(t) ≡ 0, then conditions (2.1) and (2.2) imply R α+ω Z α+ω l2 (1)(t)dt kl2 (1)k1 α , kl1 (1)k1 = l1 (1)(t)dt < = R α+ω 1 + kl2 (1)k1 1+ α l2 (1)(t)dt α and kl2 (1)k1 =
β+ω
Z β
l2 (1)(t)dt < 1 + min
0≤c≤1
Z β+ω 1 1− l1 (1)(t)dt 1+c+ c β
= 2 + 2 (1 − kl1 (1)k1 )1/2 , respectively. Thus, we have the following corollary. Corollary 2.1. Assume that a(t) ≡ 0 and that kl2 (1)k1 1 + kl2 (1)k1
(2.15)
kl2 (1)k1 < 2 + 2 (1 − kl1 (1)k1 )1/2 .
(2.16)
kl1 (1)k1 < and
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. If a(t) ≥ 0(6≡ 0) and l1 (1)(t) ≡ 0, then condition (2.1) holds naturally, and condition (2.2) implies ( ) Z β+ω Z β+ω 2(1 + c)e−kak1 1 −kak1 l2 (1)(t) exp − a(s)ds dt < e + min p + 0≤c≤1 β t (1 − c)2 + 4ce−kak1 + (1 − c) c = 1 + e−kak1 + 2e−kak1 /2 . Thus, we have the following corollary.
852
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Corollary 2.2. Assume that a(t) ≥ 0 and l1 (1)(t) ≡ 0, that either a(t) 6≡ 0 or l2 (1)(t) 6≡ 0, and that for any β ∈ [0, ω], Z β+ω Z β+ω a(s)ds dt < 1 + e−kak1 + 2e−kak1 /2 . (2.17) l2 (1)(t) exp − β
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 2.3. Assume that a(t) ≥ 0 and that for any α ∈ [0, ω], Z α+ω Z α+ω a(s)ds dt < 1 − e−kak1 , l1 (1)(t) exp − α
(2.18)
t
and that for any β ∈ [0, ω], β+ω
Z β
β+ω
Z [l2 (1)(t) + l1 (1)(t)] exp −
a(s)ds dt < 1 + 3e−kak1 .
(2.19)
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Proof. We only need to verify that conditions (2.18) and (2.19) imply (2.1) and (2.2). It is obvious that (2.18) implies (2.1). On the other hand, by (2.18), ( Z β+ω ) Z β+ω 1 2(1 + c)e−kak1 + l1 (1)(t) exp − a(s)ds dt 1− min p 0≤c≤1 β t (1 − c)2 + 4ce−kak1 + (1 − c) c Z β+ω Z β+ω 1 −kak1 ≥ min (1 + c)e l1 (1)(t) exp − a(s)ds dt + 1− 0≤c≤1 c β t Z β+ω Z β+ω = 2e−kak1 + 1 − l1 (1)(t) exp − a(s)ds dt. β
t
Thus, it follows from (2.19) that (2.2) holds also. The proof is complete.
Similarly, it follows from (2.1) that ( Z β+ω ) Z β+ω 2(1 + c)e−kak1 1 min p + 1− l1 (1)(t) exp − a(s)ds dt 0≤c≤1 β t (1 − c)2 + 4ce−kak1 + (1 − c) c Z β+ω Z β+ω 1 −kak1 l1 (1)(t) exp − a(s)ds dt 1− ≥ min (1 + c)e + 0≤c≤1 c β t Z β+ω 1/2 Z β+ω −kak1 −kak1 /2 l1 (1)(t) exp − a(s)ds dt . ≥ e + 2e 1− β
t
Thus, we have the following corollary. Corollary 2.4. Assume that a(t) ≥ 0 and that (2.1) holds for any α ∈ [0, ω]. Suppose that for any β ∈ [0, ω], Z β+ω Z β+ω l2 (1)(t) exp − a(s)ds dt β
< 2e
t −kak1
−kak1 /2
+ 2e
Z 1−
β+ω β
β+ω
Z l1 (1)(t) exp −
a(s)ds dt
1/2
.
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution.
(2.20)
853
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Theorem 2.2. Assume that a(t) ≥ 0 and that for any α ∈ [0, ω] R R α+ω α+ω Z α+ω Z α+ω l1 (1)(t) exp − t a(s)ds dt − 1 − e−kak1 α R a(s)ds dt < l2 (1)(t) exp − e−kak1 , R α+ω α+ω −kak t α 1 l1 (1)(t) exp − t a(s)ds dt + e α (2.21) and that for any β ∈ [0, ω] Z Z β+ω l1 (1)(t) exp − β
β+ω
(
a(s)ds dt < 1 + min
2(1 + c) p
(1 − c)2 + 4cekak1 + (1 − c) ) Z β+ω Z β+ω 1 −kak1 . e − a(s)ds dt l2 (1)(t) exp − + c t β 0≤c≤1
t
(2.22) Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Proof. According to theorems in [7,8], it is sufficient to show that homogeneous equation (1.5) has only a trivial ω-periodic solution. Assume the contrary. Let x(t) be a nontrivial ω-periodic solution of Eq. (1.5) and let M, m, c, x∗ , x ∗ , t∗ and t ∗ be as in the proof of Theorem 2.1. There are two possible cases: Case 1. x∗ = x(t∗ ) ≥ 0. Integrating (2.5) from t ∗ to t∗ + ω, we obtain ! Z t∗ +ω Z t∗ +ω Z t Z t∗ ∗ x exp a(s)ds − x∗ exp a(s)ds = a(s)ds dt [l2 (x)(t) − l1 (x)(t)] exp 0
t∗
0
≤ x∗
0
Z
t∗ +ω
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
t∗
It follows that x ∗ e−kak1 ≤ x∗ + x ∗
Z
t∗ +ω
t∗ +ω
Z l2 (1)(t) exp −
t∗
a(s)ds dt.
(2.23)
t
On the other hand, the integration of (2.5) from t∗ to t∗ + ω yields Z ω Z t∗ +ω Z x∗ 1 − exp − a(s)ds = [l1 (x)(t) − l2 (x)(t)] exp − 0
t∗
It follows that Z t∗ +ω Z x∗ l1 (1)(t) exp − t∗
≤x
∗
t∗ +ω
t∗ +ω
a(s)ds dt.
t
a(s)ds dt − 1 − e
−kak1
t t∗ +ω
Z
t∗ +ω
Z l2 (1)(t) exp −
t∗
a(s)ds dt.
t
Thus in view of (2.21), (2.23) and (2.24), we successively get the contradiction Z t∗ +ω Z t∗ +ω ∗ kak1 ∗ x ≤e x∗ + x l2 (1)(t) exp − a(s)ds dt t∗ t∗ +ω
t
Z t∗ +ω kak1 ∗ l2 (1)(t) exp − a(s)ds dt ≤e x t∗ t R R t∗ +ω t +ω l1 (1)(t) exp − t ∗ a(s)ds dt + e−kak1 t∗ R ×R t +ω t∗ +ω l1 (1)(t) exp − t ∗ a(s)ds dt − 1 − e−kak1 t∗ Z
< x ∗.
(2.24)
854
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Case 2. m = −x(t∗ ) > 0. If p Z t∗ (1 − c)2 + 4cekak1 + (1 − c) , a(s)ds ≤ ln 2 t∗
(2.25)
then integrating (2.5) from t∗ to t ∗ , we have ! Z t∗ Z Z t∗ a(s)ds + m exp M exp a(s)ds = 0
0
t∗
t
Z
[l1 (x)(t) − l2 (x)(t)] exp
a(s)ds dt
0
t∗ t∗
Z
l1 (1)(t) exp
≤ M
Z
t
a(s)ds dt
0
t∗ t∗
Z +m
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
t∗
It follows that " ! Z ∗ ! # Z t∗ Z t∗ t m exp − a(s)ds − l2 (1)(t) exp − a(s)ds dt 1+ M t∗ t∗ t ! Z ∗ Z ∗ t
t
l1 (1)(t) exp −
≤ t∗
a(s)ds dt.
(2.26)
t
On the other hand, integrating (2.5) from t ∗ − ω to t∗ , we obtain ! Z t∗ Z t∗ Z t Z t ∗ −ω M exp a(s)ds + m exp a(s)ds = [l2 (x)(t) − l1 (x)(t)] exp a(s)ds dt 0
0
t ∗ −ω
Z
0
t∗
≤ M t ∗ −ω Z t∗
l2 (1)(t) exp
+m t ∗ −ω
Z
t
0
l1 (1)(t) exp
Z
a(s)ds dt t a(s)ds dt.
0
It follows that t∗
Z exp −
t∗
Z
t∗
≤ t ∗ −ω
" ! # Z t∗ Z t∗ M −kak1 e − a(s)ds + l2 (1)(t) exp − a(s)ds dt m t ∗ −ω t ! Z ∗ !
t
l1 (1)(t) exp −
a(s)ds dt.
Combining (2.25)–(2.27), we have ! Z t∗ Z Z t∗ l1 (1)(t) exp − a(s)ds dt ≥ 1 + exp − t ∗ −ω
(2.27)
t
t
!
t∗
a(s)ds t∗
" ! Z ∗ ! # Z t∗ Z t∗ t m + a(s)ds dt exp − a(s)ds − l2 (1)(t) exp − M t∗ t∗ t " ! # Z t∗ Z t∗ M −kak1 e − + l2 (1)(t) exp − a(s)ds dt m t ∗ −ω t " ! # ! Z t∗ Z t∗ Z t∗ 1 −kak1 e − l2 (1)(t) exp − a(s)ds dt ≥ 1 + (1 + c) exp − a(s)ds + c t ∗ −ω t t∗ " ! # Z t∗ Z t∗ 1 −kak1 2(1 + c) + e ≥ 1+ p − l2 (1)(t) exp − a(s)ds dt t ∗ −ω t (1 − c)2 + 4cekak1 + (1 − c) c
855
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
( ≥ 1 + min
0≤c≤1
2(1 + c) p
(1 − c)2 + 4cekak1
"
1 −kak1 + e − + (1 − c) c
t∗
Z
l2 (1)(t) exp −
t ∗ −ω
!
t∗
Z
a(s)ds dt
,
t
which contradicts (2.22) with β = t ∗ − ω or β = t ∗ . If p Z t∗ (1 − c)2 + 4cekak1 + (1 − c) a(s)ds > ln , 2 t∗
(2.28)
then integrating (2.5) from t ∗ to t∗ + ω, we have ! Z t∗ +ω Z Z t∗ a(s)ds + m exp a(s)ds = M exp 0
#)
t∗ +ω
t∗
0
[l2 (x)(t) − l1 (x)(t)] exp
t
Z
a(s)ds dt
0 t∗ +ω
Z
l2 (1)(t) exp
Z
t
a(s)ds dt Z t Z t∗ +ω +m l1 (1)(t) exp a(s)ds dt.
≤ M
t∗
0
t∗
0
It follows that Z t∗ +ω Z t∗ +ω Z t∗ +ω M 1+ exp − a(s)ds − l2 (1)(t) exp − a(s)ds dt m t∗ t∗ t Z t∗ +ω Z t∗ +ω ≤ l1 (1)(t) exp − a(s)ds dt. t∗
(2.29)
t
On the other hand, integrating (2.5) from t∗ to t ∗ , we have ! Z t∗ Z t∗ Z t Z t∗ M exp a(s)ds + m exp a(s)ds = a(s)ds dt [l1 (x)(t) − l2 (x)(t)] exp 0
0
0
t∗
Z
t∗
l1 (1)(t) exp
≤ M
Z
t∗
+m
a(s)ds dt
l2 (1)(t) exp
Z
t
a(s)ds dt.
0
t∗
It follows that Z exp −
0
t∗
Z
t
" Z t∗ +ω # Z t∗ m −kak1 a(s)ds + l2 (1)(t) exp − a(s)ds dt e − M t∗ t∗ t Z t∗ +ω Z t∗ ≤ l1 (1)(t) exp − a(s)ds dt. t∗ +ω
t∗
t
Combining (2.28)–(2.30), we have Z t∗ +ω Z Z t∗ +ω l1 (1)(t) exp − a(s)ds dt ≥ 1 + exp − t∗
t
t∗ +ω
a(s)ds
t∗
Z t∗ +ω Z t∗ +ω Z t∗ +ω M exp − a(s)ds − l2 (1)(t) exp − a(s)ds dt + m t∗ t∗ t " Z t∗ +ω # Z t∗ m + e−kak1 − l2 (1)(t) exp − a(s)ds dt M t∗ t Z t∗ +ω 1 −kak1 1 exp − a(s)ds + c − e ≥ 1+ 1+ c c t∗
(2.30)
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Z t∗ +ω Z t∗ +ω 1 −kak1 a(s)ds dt + e − l2 (1)(t) exp − c t∗ t ! Z t∗ 1 −kak1 = 1+ 1+ e exp a(s)ds c t∗ Z t∗ +ω Z t∗ +ω 1 −kak1 1 −kak1 + c− e + e − a(s)ds dt l2 (1)(t) exp − c c t∗ t i hp (1 + c) (1 − c)2 + 4cekak1 + (1 − c) e−kak1 1 − c2 −kak1 ≥ 1+ − e 2c c Z t∗ +ω Z t∗ +ω 1 −kak1 e − a(s)ds dt + l2 (1)(t) exp − c t∗ t Z t∗ +ω Z t∗ +ω 2(1 + c) 1 −kak1 l2 (1)(t) exp − a(s)ds dt = 1+ p − + e t∗ t (1 − c)2 + 4cekak1 + (1 − c) c ( 2(1 + c) ≥ 1 + min p 2 0≤c≤1 (1 − c) + 4cekak1 + (1 − c) Z t∗ +ω ) Z t∗ +ω 1 −kak1 + e − l2 (1)(t) exp − , a(s)ds dt c t∗ t which contradicts (2.22) with β = t∗ . Cases 1 and 2 complete the proof.
Like Corollaries 2.1, 2.2 and 2.4, we have the following corollaries. Corollary 2.5. Assume that a(t) ≡ 0 and that kl1 (1)k1 1 + kl1 (1)k1
(2.31)
kl1 (1)k1 < 2 + 2 (1 − kl2 (1)k1 )1/2 .
(2.32)
kl2 (1)k1 < and
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 2.6. Assume that a(t) ≥ 0 and l2 (1)(t) ≡ 0, and that for any β ∈ [0, ω], Z β+ω Z β+ω a(s)ds dt < 1 + e−kak1 + 2e−kak1 /2 . 1 − e−kak1 < l1 (1)(t) exp − β
(2.33)
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 2.7. Assume that a(t) ≥ 0 and that (2.21) holds for any α ∈ [0, ω]. Suppose that for any β ∈ [0, ω], Z β+ω Z β+ω l1 (1)(t) exp − a(s)ds dt β
t
Z < 1 + e−kak1 + 2e−kak1 /2 e−kak1 −
β
β+ω
β+ω
Z l2 (1)(t) exp −
1/2 a(s)ds dt .
t
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Remark 2.1. Corollaries 2.1 and 2.5 were obtained in [6].
(2.34)
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
3. The case a(t) ≤ 0 Define the operator (see [4]) ϑ : Cω (R) → Cω (R) by ϑ(x)(t) = x(ω − t)
for t ∈ R.
Now if x(t) is a solution of Eq. (1.5), then y(t) = ϑ(x)(t) is a solution of y 0 (t) = ϑ(a)(t)y(t) − ϑ(l1 (ϑ(y)))(t) + ϑ(l2 (ϑ(y)))(t),
(3.1)
and, vice versa, if y(t) is a solution of Eq. (3.1), then x(t) = ϑ(y)(t) is a solution of Eq. (1.5). Therefore, Eq. (1.5) has only the trivial ω-periodic solution if and only if Eq. (3.1) has only the trivial ω-periodic solution. Note that kak1 = kϑ(a)k1 , and for i = 1, 2, Z Z α+ω ϑ(li (ϑ(1)))(t) exp − α
α+ω
|ϑ(a)(s)|ds dt =
t
α+ω ˆ
Z αˆ
Z t li (1)(t) exp − |a(s)|ds dt, αˆ
where αˆ = ω − α. Therefore, if a(t) ≤ 0 for t ∈ R, then one can obtain the following theorems and corollaries by using the results obtained in Section 2 for Eq. (3.1). Theorem 3.1. Assume that a(t) ≤ 0 and that for any α ∈ [0, ω], α+ω
Z α
R R α+ω t Z t l1 (1)(t) exp − α |a(s)|ds dt 1 − e−kak1 + α R l2 (1)(t) exp − |a(s)|ds dt < , R α+ω t α 1+ α l1 (1)(t) exp − α |a(s)|ds dt
and that for any β ∈ [0, ω], ( Z t Z β+ω l1 (1)(t) exp − |a(s)|ds dt < e−kak1 + min p β
2(1 + c)e−kak1
(1 − c)2 + 4ce−kak1 + (1 − c) Z t ) Z β+ω 1 + . 1− l2 (1)(t) exp − |a(s)|ds dt c β β
β
(3.2)
0≤c≤1
(3.3)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 3.1. Assume that a(t) ≤ 0 and l2 (1)(t) ≡ 0, that either a(t) 6≡ 0 or l1 (1)(t) 6≡ 0, and that for any β ∈ [0, ω], Z t Z β+ω l1 (1)(t) exp − |a(s)|ds dt < 1 + e−kak1 + 2e−kak1 /2 . (3.4) β
β
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 3.2. Assume that a(t) ≤ 0 and that for any α ∈ [0, ω], Z t Z α+ω l2 (1)(t) exp − |a(s)|ds dt < 1 − e−kak1 ,
(3.5)
and that for any β ∈ [0, ω], Z t Z β+ω [l1 (1)(t) + l2 (1)(t)] exp − |a(s)|ds dt < 1 + 3e−kak1 .
(3.6)
α
β
α
β
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution.
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
Corollary 3.3. Assume that a(t) ≤ 0 and that (3.2) holds for any α ∈ [0, ω]. Suppose that for any β ∈ [0, ω], Z t Z β+ω |a(s)|ds dt l1 (1)(t) exp − β
β
< 2e
−kak1
−kak1 /2
+ 2e
Z 1−
β+ω β
1/2 Z t |a(s)|ds dt . l2 (1)(t) exp −
(3.7)
β
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Theorem 3.2. Assume that a(t) ≤ 0 and that for any α ∈ [0, ω], R R α+ω t Z t Z α+ω l (1)(t) exp − a(s)ds dt − 1 − e−kak1 2 α α R |a(s)|ds dt < R l1 (1)(t) exp − e−kak1 , α+ω t −kak1 α α l (1)(t) exp − |a(s)|ds dt + e 2 α α and that for any β ∈ [0, ω], ( Z t Z β+ω l2 (1)(t) exp − |a(s)|ds dt < 1 + min p β
β
0≤c≤1
+
1 −kak1 e c
(3.8)
2(1 + c)
(1 − c)2 + 4cekak1 + (1 − c) Z t ) Z β+ω − . l1 (1)(t) exp − |a(s)|ds dt β
β
(3.9)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 3.4. Assume that a(t) ≤ 0 and l1 (1)(t) ≡ 0, and that for any β ∈ [0, ω], Z t Z β+ω −kak1 1−e < l2 (1)(t) exp − |a(s)|ds dt < 1 + e−kak1 + 2e−kak1 /2 . β
β
(3.10)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. Corollary 3.5. Assume that a(t) ≤ 0 and that (3.8) holds for any α ∈ [0, ω]. Suppose that for any β ∈ [0, ω], Z t Z β+ω l2 (1)(t) exp − |a(s)|ds dt β
β
Z < 1 + e−kak1 + 2e−kak1 /2 e−kak1 −
β
β+ω
Z t 1/2 l1 (1)(t) exp − |a(s)|ds dt . β
(3.11)
Then Eq. (1.1) has at least one ω-periodic solution and Eq. (1.2) has a unique ω-periodic solution. 4. Examples and remarks Example 4.1. Consider the first-order functional differential equation x 0 (t) = −ax(t) + b(1 + sin t)x(t − τ ) + f (t), where a, b, τ ∈ R and f ∈ C2π (R). For any α ∈ R, Z α+2π |b| |b|(|a| sin α + cos α) −|a|(t−α) −2π|a| |b|(1 + sin t)e dt = 1 − e + , |a| a2 + 1 α
(4.1)
(4.2)
and Z α
α+2π
−|a|(α+2π−t)
|b|(1 + sin t)e
|b| |b|(|a| sin α − cos α) −2π|a| + dt = 1 − e . |a| a2 + 1
(4.3)
X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861
859
Case (i). a > 0, b > 0. In this case, l1 (1)(t) = b(1 + sin t) and l2 (1)(t) ≡ 0. In view of Corollary 2.3, if for any α ∈ [0, 2π], Z α |b| |a| l1 (1)(t)e−|a|(α−t) dt ≤ 1 − e−2π|a| < 1 − e−2π|a| , 1+ √ |a| a2 + 1 α−2π or if |b| <
|a| 1 + √|a|
,
(4.4)
a 2 +1
then Eq. (4.1) has a unique 2π -periodic solution. In view of Corollary 2.6, if for any β ∈ [0, 2π], |a| −2π|a| −2π|a| |b| 1−e < 1−e 1− √ |a| a2 + 1 Z β+2π ≤ l1 (1)(t)e−|a|(β+2π−t) dt β |b| |a| 1+ √ ≤ 1 − e−2π|a| |a| a2 + 1 −2π|a| −π|a| < 1+e + 2e , or if |a| 1 − √|a|
a 2 +1
|a| 1 + e−π|a| , < |b| < |a| −π|a| √ 1+ 1−e 2
(4.5)
a +1
then Eq. (4.1) has also a unique 2π-periodic solution. But, by Corollaries 2.1 and 2.5, the conditions corresponding to (4.4) and (4.5) are, respectively, |b| <
|a| , 1 + 2π |a|
and
2π|a| < 2 + 2(1 − 2π |b|)1/2 ,
(4.6)
and |a| 2 + 2(1 − 2π |a|)1/2 < |b| < . 1 − 2π |a| 2π
(4.7)
By Corollary 1.1 in [1], the conditions corresponding to (4.4) and (4.5) are, respectively, 4|a| |a| |b| < min , , 1 + 2π |a| e2π|a| − 1 1 − |a|
(4.8)
a 2 +1
and |a| 4|a| < |b| < 1 − 2π |a| e2π|a| − 1 1 −
|a|
.
(4.9)
a 2 +1
It is not difficult to see that conditions (4.4) and (4.5) are weaker than (4.6) and (4.7), and weaker than (4.8) and (4.9), respectively. Case (ii). a > 0, b < 0. In this case, l1 (1)(t) ≡ 0 and l2 (1)(t) = |b|(1 + sin t). In view of Corollary 2.2, if for any β ∈ [0, 2π],
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X.H. Tang, Z. Jiang / Nonlinear Analysis 68 (2008) 845–861 β+2π
Z β
|b| |b|(|a| sin β − cos β) l2 (1)(t)e−|a|(β+2π−t) dt = 1 − e−2π|a| + |a| a2 + 1 |b| |a| ≤ 1 − e−2π|a| 1+ √ |a| a2 + 1 < 1 + e−2π|a| + 2e−π|a| ,
or if |a| 1 + e−π|a| , |b| < |a| −π|a| 1−e 1+ √
(4.10)
a 2 +1
then Eq. (4.1) has a unique 2π -periodic solution. But, by Corollary 2.1, the condition corresponding to (4.10) is |b| <
4 − 2π |a| . 2π
(4.11)
It is not difficult to see that condition (4.10) is weaker than (4.11). Case (iii). a < 0, b > 0. In this case, l1 (1)(t) = b(1 + sin t) and l2 (1)(t) ≡ 0. In view of Corollary 3.1, if (4.10) holds, then Eq. (4.1) has a unique 2π-periodic solution. But, by Corollary 2.5, the condition corresponding to (4.10) is (4.11). Case (iv). a < 0, b < 0. In this case, l1 (1)(t) ≡ 0 and l2 (1)(t) = |b|(1 + sin t). In view of Corollaries 3.2 and 3.4, If (4.4) or (4.5) holds, then Eq. (4.1) has a unique 2π -periodic solution. But, by Corollaries 2.1 and 2.5, the conditions corresponding to (4.4) and (4.5) are (4.6) and (4.7), respectively. By Corollary 1.1 in [1], the conditions corresponding to (4.4) and (4.5) are (4.8) and (4.9), respectively. Remark 4.1. Example 4.1 shows that the conditions in Corollaries 2.1 and 2.5 in which the instantaneous term −a(t)x(t) is treated as l1 (x)(t) or l2 (x)(t) are stronger than ones in the other theorems and corollaries in this paper. Remark 4.2. Example 4.1 shows that condition (1.10) is weaker than (1.6) and (1.8) in general. Acknowledgments The authors are grateful to the referees for their valuable comments and suggestions. References [1] R. Hakl, A. Lomtatidze, B. P˚uzˇ a, On periodic solutions of first order linear functional differential equations, Nonlinear Analysis 49 (2002) 929–945. [2] R. Hakl, A. Lomtatidze, B. P˚uzˇ a, On a boundary value problem for first-order scalar functional differential equations, Nonlinear Analysis 53 (2003) 391–405. ˇ [3] R. Hakl, A. Lomtatidze, J. Sremr, On a periodic-type boundary value problem for first order nonlinear functional differential equations, Nonlinear Analysis 51 (2002) 425–447. ˇ [4] R. Hakl, A. Lomtatidze, J. Sremr, A periodic type boundary-value problem for first order nonlinear functional differential equations, Nelinijni Kolyvannya 5 (3) (2002) 416–432 (in English). [5] R. Hakl, A. Lomtatidze, I.P. Stavroulakis, On a boundary value problem for scalar linear functional differential equations, Abstract and Applied Analysis 1 (2004) 45–67. [6] A.G. Lomtatidze, R. Hakl, B. P˚uzˇ a, On the periodic boundary value problem for first-order functional-differential equations, Differential Equations 39 (3) (2–3) 344–352 (English. Russian original). Translation from Differentsial’nye Uravneniya 39 (3) (2003) 320–327. [7] J.K. Hale, Periodic and almost periodic solutions of functional-differential equations, Archive for Rational Mechanics and Analysis 15 (1964) 289–304. [8] T. Jankowski, Existence of solutions of boundary value problems for differential equations with delayed arguments, Journal of Computational and Applied Mathematics 156 (2003) 239–252. [9] D. Jiang, J.J. Nieto, W. Zuo, On monotone method for first and second order periodic boundary value problems and periodic solutions of functional differential equations, Journal of Mathematical Analysis and Applications 289 (2004) 691–699. [10] I. Kiguradze, B. P˚uzˇ a, On periodic solutions of systems of linear functional differential equations, Archive der Mathematik 33 (3) (1997) 197–212.
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