Hindawi Publishing Corporation
International Journal of Differential Equations
Volume 2011, Article ID 916279, 21 pages
doi:10.1155/2011/916279
Research Article
Periodic Solutions for a Class of n-th Order
Functional Differential Equations
Bing Song,1, 2 Lijun Pan,3 and Jinde Cao1
1
Department of Mathematics, Southeast University, Nanjing 210096, China
JiangSu Institute of Economic Trade Technology, Nanjing 211168, China
3
School of Mathematics, Jia Ying University, Meizhou Guangdong, 514015, China
2
Correspondence should be addressed to Jinde Cao, jdcao@seu.edu.cn
Received 10 May 2011; Accepted 14 July 2011
Academic Editor: Peiguang Wang
Copyright q 2011 Bing Song et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
We study the existence of periodic solutions for n-th order functional differential equations
k
i
xn t n−1
i0 bi x t fxt − τt pt. Some new results on the existence of periodic
solutions of the equations are obtained. Our approach is based on the coincidence degree theory
of Mawhin.
1. Introduction
In this paper, we are concerned with the existence of periodic solutions of the following n-th
order functional differential equations:
xn t n−1 k
bi xi t fxt − τt pt,
1.1
i0
where bi , i 0, 1, . . . , n − 1 are constants, k is a positive odd, f ∈ C1 R, R for ∀x ∈ R, p ∈
CR, R with pt T pt.
In recent years, there are many papers studying the existence of periodic solutions of
first-, second- or third-order differential equations 1–12. For example, in 5, Zhang and
Wang studied the following differential equations:
x t ax2k−1 t bx 2k−1 t cx2k−1 t g t, xt − τ1 , x t − τ2 pt.
1.2
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International Journal of Differential Equations
The authors established the existence of periodic solutions of 1.2 under some conditions on
a, b, c, and 2k − 1.
In 13–24, periodic solutions for n, 2n, and 2n 1 th order differential equations were
discussed. For example, in 22, 24, Pan et al. studied the existence of periodic solutions of
higher order differential equations of the form
xn t n−1
bi xi t ft, xt, xt − τ1 t, . . . , xt − τm t pt.
1.3
i1
The authors obtained the results based on the damping terms xi t and the delay τi t.
In present paper, by using Mawhin’s continuation theorem, we will establish some
theorems on the existence of periodic solutions of 1.1. The results are related to not only bi
and ft, x but also the positive odd k. In addition, we give an example to illustrate our new
results.
2. Some Lemmas
We investigate the theorems based on the following lemmas.
Lemma 2.1 see 17. Let n1 > 1, α ∈ 0, ∞ be constants, s ∈ CR, R with st T st, and
st ∈ −α, α, for all t ∈ 0, T . Then for ∀x ∈ C1 R, R with xt T xt, one has
T
|xt − xt − st|n1 dt ≤ 2αn1
0
T
n1
x t dt.
2.1
0
Lemma 2.2. Let k ≥ 1, α ∈ 0, ∞ be constants, s ∈ CR, R with st T st, and st ∈
−α, α, for all t ∈ 0, T . Then for ∀x ∈ C1 R, R with xt T xt, one has
T
T
T
k1/k
k1
k
k1
k
k1/k 1/k
x t
dt ≤ 2α
k
dt .
k − 1
|xt| dt x t − x t − st
0
0
0
2.2
Proof. Let Ft xk t. By Lemma 2.2, one has
T T
k1/k
k
k
dt |Ft − Ft − st|k1/k dt
x t − x t − st
0
0
≤ 2αk1/k
T
k1/k
F t
dt
0
T k1/k
k−1
2αk1/k
dt
kx tx t
0
2α
k1/k k1/k
T
k
0
k1/k
dt.
|xt|k−1k1/k x t
2.3
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3
By inequality
xy ≤
xp y q
,
p
q
x ≥ 0, y ≥ 0,
1 1
1,
p q
2.4
one has
k1/k k − 1|xt|k1 |x t|k1
.
≤
|xt|k−1k1/k x t
k
k
2.5
Thus we obtain
T T
T
k1/k
k1
k
k1
k
k1/k 1/k
x t dt .
dt ≤ 2α
k
k − 1
|xt| dt x t − x t − st
0
0
0
2.6
Lemma 2.3. If k ≥ 1 is an integer, x ∈ Cn R, R, and xt T xt, then
T
k
x t dt
1/k
0
≤T
T
1/k
k
x t dt
≤ ··· ≤ T
n−1
T
0
1/k
n k
x t dt
.
2.7
0
The proof of Lemma 2.3 is easy, here we omit it.
We first introduce Mawhin’s continuation theorem.
Let X and Y be Banach spaces, L : DL ⊂ X → Y are a Fredholm operator of index
zero, here DL denotes the domain of L. P : X → X, Q : Y → Y be projectors such that
Im P Ker L,
Ker Q Im L,
X Ker L ⊕ Ker P,
Y Im L ⊕ Im Q.
2.8
It follows that
L|DL∩Ker P : DL ∩ Ker P −→ Im L
2.9
is invertible, we denote the inverse of that map by Kp . Let Ω be an open bounded subset of
Ø, the map N : X → Y will be called L-compact in Ω, if QNΩ is bounded
X, DL ∩ Ω /
and Kp I − QN : Ω → X is compact.
Lemma 2.4 see 25. Let L be a Fredholm operator of index zero and let N be L-compact on Ω.
Assume that the following conditions are satisfied:
i Lx /
λNx, for all x ∈ ∂Ω ∩ DL, λ ∈ 0, 1;
ii QNx /
0, for all x ∈ ∂Ω ∩ Ker L;
iii deg{QNx, Ω ∩ Ker L, 0} /
0,
then the equation Lx Nx has at least one solution in Ω ∩ DL.
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International Journal of Differential Equations
Now, we define Y {x ∈ CR, R | xt T xt} with the norm |x|∞ maxt∈0,T {|xt|} and X {x ∈ Cn−1 R, R | xt T xt} with norm x max{|x|∞ , |x |∞ , . . . , |xn−1 |∞ }. It is easy to see that X, Y are two Banach spaces. We also define
the operators L and N as follows:
Lx xn , DL {x | x ∈ Cn R, R, xt T xt},
L : DL ⊂ X −→ Y,
N : X −→ Y,
Nx −
n−1 k
bi xi t − ft, xt − τt pt.
2.10
i1
It is easy to see that 1.1 can be converted to the abstract equation Lx Nx. Moreover, from
T
the definition of L, we see that ker L R, dimker L 1, Im L {y | y ∈ Y, 0 ysds 0} is
closed, and dimY \Im L 1, one has codimIm L dimker L. So L is a Fredholm operator
with index zero. Let
P : X −→ ker L,
P x x0,
Q : Y −→ Y \ Im L,
1
Qy T
T
ytdt,
2.11
0
and let
L|DL∩Ker P : DL ∩ Ker P −→ Im L.
2.12
Then L|DL∩Ker P has a unique continuous inverse Kp . One can easily find that N is L-compact
in Ω, where Ω is an open bounded subset of X.
3. Main Result
Theorem 3.1. Suppose n 2m 1, m > 0 an integer and the following conditions hold:
H1 The function f satisfies
ft, x ≤ γ,
lim x→∞
xk ft, x − f t, y ≤ βxk − yk ,
3.2
|b0 | > γ θ2 .
3.3
3.1
where γ ≥ 0.
H2 H3 There is a positive integer 0 < s ≤ m such that
b2s /
0,
0,
b2s /
b2si 0,
if s m,
i 1, 2, . . . , 2m − 2s, if 0 < s < m.
3.4
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5
H4 A2 2s, k θ1 T
2s−1k
k|b0 |T
θ1 T k
γ θ2
2s
γ θ2
A1 2s, k θ1 T 2s−1k
|b0 | − γ − θ2
k−1/k
A1 2s, k θ1 T 2s−1k
|b0 | − γ − θ2
< |b2s |,
if 1 < s ≤ m,
k−1/k
k
A1 2, k θ1 T k
2 A1 2, k θ1 T
k|b0 |T
< |b2 |,
|b0 | − γ − θ2
|b0 | − γ − θ2
3.5
if s 1,
s−ik
, θ1 2k/k1 β|τt|∞ k1/k1 , θ2 where A1 s, k si1 |bi |T s−ik , A2 s, k s−2
i1 |bi |T
k/k1
. Then 1.1 has at least one T -periodic solution.
2k/k1 β|τt|∞ k1/k1 k − 1
Proof. Consider the equation
Lx λNx,
λ ∈ 0, 1,
3.6
where L and N are defined by 2.10. Let
Ω1 DL
, Lx λNx
x∈
Ker L
for some λ ∈ 0, 1 .
3.7
For x ∈ Ω1 , one has
xn t λ
2s
k
bi xi t λft, xt − τt λpt,
λ ∈ 0, 1.
3.8
i0
Multiplying both sides of 3.8 by xt, and integrating them on 0, T , one has for λ ∈ 0, 1
T
x
n
txtdt λ
0
2s
T k
xi t xtdt
bi
0
i0
λ
T
ft, xt − τtxtdt λ
0
3.9
T
ptxtdt.
0
Since for any positive integer i,
T
0
x2i−1 txtdt 0,
3.10
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International Journal of Differential Equations
and in view of n 2m 1 and k is odd, it follows from 3.3 and 3.9 that
|b0 |
T
|xt|k1 dt
0
≤
2s
T
T T
i k
ft, xt − τt |xt|dt pt|xt|dt
|bi |
x t |xt|dt 0
i1
≤
2s
|bi |
T
0
T T
i k
x t |xt|dt ft, xt|xt|dt
0
i1
0
0
ft, x − ft, xt − τt|xt|dt T
0
pt|xt|dt.
0
3.11
By using Hölder inequality and Lemma 2.1, from 3.11, we obtain
|b0 |
T
|xt|k1 dt
0
≤
T
T
1/k1 ⎡
k/k1
2s
k1
k1
i
⎣ |bi |
|xt| dt
x t dt
0
0
i1
T
k/k1
ft, xtk1/k dt
0
T
k/k1
ft, x − ft, xt − τtk1/k dt
0
T
k/k1 ⎤
⎦
ptk1/k dt
0
≤
T
0
|xt|
k1
T
1/k1 ⎡
k/k1
2s
2s k1
2s−ik
⎣
dt
|bi |T
x t dt
0
i1
T
k/k1
ft, xtk1/k dt
0
T
k/k1
ft, x − ft, xt − τtk1/k dt
0
k/k1
.
pt ∞ T
3.12
International Journal of Differential Equations
7
So
T
|b0 |
k/k1
|xt|
k1
dt
0
≤ A1 2s, k
T
k/k1
2s k1
x t dt
0
T
T
k/k1
ft, xtk1/k dt
3.13
0
k/k1
ft, xt − ft, xt − τtk1/k dt
u1 ,
0
where u1 is a positive constant. Choosing a constant ε > 0 such that
3.14
γ ε θ2 < |b0 |,
γ ε θ2 A1 2s, k θ1 T 2s−1k
A2 2s, k θ1 T 2s−1k |b0 | − γ ε − θ2
k−1/k
2s−1k
2s A1 2s, k θ1 T
k|b0 |T
< |b2s |, if 1 < s ≤ m,
|b0 | − γ ε − θ2
θ1 T k
k−1/k
A1 2, k θ1 T k
A1 2, k θ1 T k
2
k|b0 |T
< |b2 |,
|b0 | − γ ε − θ2
|b0 | − γ ε − θ2
γ ε θ2
if s 1,
3.15
for the above constant ε > 0, we see from 3.1 that there is a constant δ > 0 such that
ft, xt < γ ε |xt|k ,
for |xt| > δ, t ∈ 0, T .
3.16
Δ2 {t ∈ 0, T : |xt| > δ}.
3.17
Denote
Δ1 {t ∈ 0, T : |xt| ≤ δ},
Since
T
0
ft, xtk1/k dt ≤
Δ1
ft, xtk1/k dt Δ2
k1/k
k1/k
T γ ε
≤ fδ
ft, xtk1/k dt
T
|xt|k1 dt
0
k1/k
k1/k
fδ
T γ ε
T
0
|xt|k1 dt,
3.18
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International Journal of Differential Equations
using inequality
a bl ≤ al bl
for a ≥ 0, b ≥ 0, 0 ≤ l ≤ 1,
3.19
it follows from 3.18 that
T
ft, xtk1/k dt
k/k1
≤ fδ T
k/k1
γ ε
T
0
k/k1
|xt|
k1
dt
.
3.20
0
From 3.2 and by Lemma 2.2, one has
T
k/k1
ft, xt − ft, xt − τtk1/k dt
0
T
k/k1
k1/k
k
k
≤β
dt
x t − x t − τt
0
≤2
k/k1
β|τt|∞ k
1/k1
k − 1
T
|xt|
≤2
β|τt|∞ k
dt T
0
1/k1 ⎣
k − 1
k/k1
k1
x t dt
0
⎡
k/k1
k1
k/k1
T
k/k1
k1
|xt|
dt
0
T
k/k1 ⎤
⎦
k1
x t dt
0
≤2
k/k1
2
β|τt|∞ k
k/k1
θ2
T
0
1/k1
k − 1
k/k1
T
1/k1 2s−1k
T
dt
dt
T
k/k1
2s k1
x t dt
0
k/k1
|xt|
|xt|
0
β|τt|∞ k
k1
k/k1
k1
θ1 T
2s−1k
T
k/k1
2s k1
x t dt
.
0
3.21
International Journal of Differential Equations
9
Substituting the above formula into 3.13, one has
T
k/k1
k1
|b0 | − γ ε − θ2
|xt| dt
0
3.22
k/k1
T 2s k1
2s−1k
≤ A1 2s, k θ1 T
u2 ,
x t dt
0
where u2 is a positive constant. That is
T
k/k1
|xt|
k1
dt
0
A1 2s, k θ1 T 2s−1k
≤
|b0 | − γ ε − θ2
T
k/k1
2s k1
x t dt
u3 ,
3.23
0
where u3 is a positive constant.
On the other hand, multiplying both sides of 3.8 by x2s t, and integrating on 0, T ,
one has
T
xn tx2s tdt
0
2s
i0
T T
T
k
i
2s
2s
x t x tdt bi
ft, xt − τtx tdt ptx2s tdt.
0
0
3.24
0
If 1 < s ≤ m, since
T
T x2m1 tx2s tdt 0,
0
k
x2s−1 t x2s tdt 0,
3.25
0
T
0
k 2s
xt x
tdt −k
T
0
xtk−1 x2s−1 tx tdt,
3.26
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International Journal of Differential Equations
by using Hölder inequality and Lemma 2.1, from 3.23, one has
|b2s |
T 2s k1
x t dt
0
T
2s−2 2s i k ≤ x t
|bi |x t ft, xt − τt pt dt
0
i1
k|b0 |
T
|xt|k−1 x2s−1 tx tdt
0
≤
T
2s k1
x t dt
T
1/k1 ⎡
k/k1
2s−2
2s k1
⎣ |bi |T 2s−ik
x t dt
0
0
i1
T
3.27
k/k1
ft, xtk1/k dt
0
T
k/k1
ft, xt − ft, xt − τk1/k dt
0
pt∞ T k/k1
k|b0 |x t∞
T
|xt|k−1 x2s−1 tdt.
0
Since x0 xT , there exists ξ ∈ 0, T such that x ξ 0. So for t ∈ 0, T x t x ξ t
x σdσ.
3.28
ξ
Using Hölder inequality and Lemma 2.1, one has
x t ≤
∞
T
x tdt ≤ T k/k1
0
T
1/k1
k1
x t dt
0
≤ T 2s−1−1/k1
T
3.29
1/k1
2s k1
x t dt
.
0
Using inequality
1
T
T
0
1/r
|xt|r ≤
1
T
T
1/l
l
|xt| 0
for 0 ≤ r ≤ l, ∀x ∈ R.
3.30
International Journal of Differential Equations
11
and applying Hölder inequality and by Lemma 2.1, we obtain
T
|xt|
x2s−1 tdt ≤
k−1 T
0
k
k−1/k T
|xt| dt
0
≤T
1/k
2s−1 k
t dt
x
0
1/k1
T
|xt|
k1
k−1/k1 T
dt
0
≤T
1 1/k1
1/k1
2s−1 k1
t dt
x
0
T
|xt|
k1
k−1/k1 T
dt
0
1/k1
2s k1
x t dt
.
0
3.31
Substituting the above formula, 3.20, 3.27, and 3.30 into 3.26, one has
T 2s k1
|b2s |
x t dt
0
⎛
⎞1/k1 ⎧
k/k1
T
⎨
k1
T 2s 2s k1
2s−1k
⎝
⎠
A 2s, k θ1 T
≤
x t dt
x t dt
⎩ 2
0
0
γ ε θ2
T
k/k1
k1
|xt|
dt
0
⎫
k/k1 ⎬
pt∞ fδ T
⎭
k|b0 |T
2s
T
2/k1 T
2s k1
x t dt
0
k−1/k1
|xt|k1 dt
.
0
3.32
Then, one has
|b2s | − A2 2s, k − θ1 T
2s−1k
⎞k/k1
⎛
k1
T
2s
⎝ x t dt⎠
0
≤ k|b0 |T
2s
T
1/k1 T
2s k1
x t dt
0
|xt|k1 dt
0
T
k/k1
2s k1
γ ε θ2
u4 ,
x t dt
0
k−1/k1
3.33
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International Journal of Differential Equations
where u4 is a positive constant. Using inequality
a bl ≤ al bl
3.34
for a ≥ 0, b ≥ 0, 0 ≤ l ≤ 1,
it follows from 3.23 that
T
k−1/k1
k1
|xt|
dt
0
≤
k−1/k T
2s−1k
A1 2s, k θ1 T
|b0 | − γ ε − θ2
3.35
k−1/k1
2s k1
x t dt
u5 ,
0
where u5 is a positive constant. Substituting the above formula and 3.23 into 3.33, one has
⎧
⎪
⎪
⎪
⎨
|b2s | − A2 2s, k − θ1 T
⎪
⎪
⎪
⎩
−k|b0 |T 2s
≤ u5 k|b0 |T
2s
2s−1k
−
A1 2s, k θ1 T 2s−1k
|b0 | − γ ε − θ2
γ ε θ2
⎫
⎪
k/k1
k−1/k ⎪
⎪
⎬ T A1 2s, k θ1 T 2s−1k
2s k1
x t dt
⎪
|b0 | − γ ε − θ2
⎪
0
⎪
⎭
T
3.36
1/k1
2s k1
x t dt
u6 ,
0
where u6 is a positive constant.
T
T
T
If s 1, since 0 x tk x tdt 0, 0 xtk x tdt −k 0 xtk−1 x t2 dt, from
3.24, one has
T
b2
k1
dt
x t
0
−kb0
T
xt
k−1 2
x t dt −
0
T
ft, xt − τx tdt 0
T
3.37
ptx tdt.
0
Applying the above method, one has
⎧
⎨
⎩
|b2 | − θ1 T k −
×
T
0
k−1/k ⎫
⎬
k
A1 2, k θ1 T k
A1 2, k θ1 T
− k|b0 |T 2
⎭
|b0 | − γ ε − θ2
|b0 | − γ ε − θ2
γ ε θ2
k1
x t dt
k/k1
≤ u7 k|b0 |T 2
T
0
k1
x t dt
1/k1
u8 ,
3.38
International Journal of Differential Equations
13
where u7 , u8 is a positive constant. Hence there is a constant M1 , M2 > 0 such that
T 2s k1
x t dt ≤ M1 ,
3.39
0
T
3.40
|xt|k1 dt ≤ M2 .
0
From 3.5, using Hölder inequality and Lemma 2.1, one has
T
T T
2s
n i k
|bi |
x tdt ≤
x t dt ft, xtdt
0
0
i0
T
0
ft, xt − ft, xt − τtdt T
0
ptdt
0
2s
≤
|bi |T 2s−ik1/k1 θ1 T 2s−1k1/k1
T
0
i1
k/k1
2s k1
x t dt
|b0 | γ ε θ2 T 1/k1
T
k1
|xt|
k/k1
dt
0
pt∞ fδ T
2s
2s−ik1/k1
2s−1k1/k1
θ1 T
≤
M1 k/k1
|bi |T
i1
|b0 | γ ε θ2 M2 k/k1 pt∞ fδ T M,
3.41
where M is a positive constant. We claim that
T
i n n−i−1
x t ≤ T
x tdt,
i 1, 2, . . . , n − 1.
3.42
0
In fact, noting that xn−2 0 xn−2 T , there must be a constant ξ1 ∈ 0, T such that
xn−1 ξ1 0, we obtain
T
t
T
n−1 n−1
n
x sds ≤ xn−1 ξ1 xn tdt xn tdt.
t x
ξ1 x
ξ1
0
0
3.43
Similarly, since xn−3 0 xn−3 T , there must be a constant ξ2 ∈ 0, T such that xn−2 ξ2 0, from 3.43 we get
T
t
T
n−2 n−2
n−1
x
t x
ξ2 sds ≤ xn−1 tdt ≤ T xn tdt.
x
ξ2
0
0
3.44
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International Journal of Differential Equations
By induction, we conclude that 3.42 holds. Furthermore, one has
T
i x t ≤ T n−i−1 xn tdt ≤ T n−i−1 M,
∞
3.45
i 1, 2, . . . , n − 1.
0
1/k1
It follows from 3.39 that there exists a ξ ∈ 0, T such that |xξ| ≤ M2
Lemma 2.1, we get
|xt|∞ ≤ xξ T
t
. Applying
x tdt ≤ M2
1/k1
ξ
k/k1
T
1/k1
k1
x t dt
0
1/k1
≤ M2
T 2s−1k/k1
T
1/k1
3.46
2s k1
x t dt
0
1/k1
M2
1/k1
T 2s−1k/k1 M1
.
It follows that there is a constant A > 0 such that x ≤ A. Thus Ω1 is bounded.
Let Ω2 {x ∈ Ker L, QNx 0}. Suppose x ∈ Ω2 , then xt d ∈ R and satisfies
QNx 1
T
T
−b0 dk − ft, d pt dt 0.
3.47
0
We will prove that there exists a constant B > 0 such that |d| ≤ B. If |d| ≤ δ, taking δ B, we
get |d| ≤ B. If |d| > δ, from 3.47, one has
1 T
−ft, d pt dt
|b0 ||d| T 0
k
1
≤
T
T
0
ft, ddt pt ≤ γ ε |d|k pt .
∞
∞
3.48
Thus
1/k
pt
∞
.
|d| ≤
|b0 | − γ ε
3.49
Taking |pt|∞ /|b0 | − γ ε1/k B, one has |d| ≤ B, which implies Ω2 is bounded. Let Ω
be a nonempty open bounded subset of X such that Ω ⊃ Ω1 ∪ Ω2 . We can easily see that L is a
Fredholm operator of index zero and N is L-compact on Ω. Then by the above argument, we
International Journal of Differential Equations
15
have
i Lx /
λNx, for all x ∈ ∂Ω ∩ DL, λ ∈ 0, 1,
ii QNx / 0, for all x ∈ ∂Ω ∩ Ker L.
At last we will prove that condition iii of Lemma 2.4 is satisfied. We take
H : Ω ∩ Ker L × 0, 1 −→ Ker L,
1 − μ T
H d, μ μd −b0 dk − ft, d pt dt.
T
0
3.50
From assumptions H1 and H2 , we can easily obtain Hd, μ / 0, for all d, μ ∈ ∂Ω ∩
Ker L × 0, 1, which results in
deg{QN, Ω ∩ Ker L, 0} deg{H·, 0, Ω ∩ Ker L, 0} deg{H·, 1, Ω ∩ Ker L, 0} / 0.
3.51
Hence, by using Lemma 2.2, we know that 1.1 has at least one T -periodic solution.
Theorem 3.2. Suppose n 4m 1, m > 0 an integer and conditions H1 , H2 hold. If
H5 there is a positive integer 0 < s ≤ m such that
b4s−3i 0,
b4s−3 /
0,
i 1, 2, . . . , 4m − 4s 3,
3.52
H6 A2 4s − 3, k θ1 T
k|b0 |T
4s−3
4s−4k
γ θ2
A1 4s − 3, k θ1 T 4s−4k
|b0 | − γ − θ2
A1 4s − 3, k θ1 T 4s−4
|b0 | − γ − θ2
k−1/k
< |b4s−3 |,
γ θ2 A1 1, k θ1 < |b1 |,
θ1 |b0 | − γ − θ2
if 1 < s ≤ m,
3.53
if s 1,
then 1.1 has at least one T -periodic solution.
Proof. From the proof of Theorem 3.1, one has
T
0
k/k1
k1
|xt|
dt
A1 4s − 3, k θ1 T 4s−4k
≤
|b0 | − γ ε − θ2
T
k/k1
4s−3 k1
t dt
x
u9 ,
0
3.54
16
International Journal of Differential Equations
where u9 is a positive constant. Multiplying both sides of 3.8 by x4s−3 t, and integrating
on 0, T , one has
T
x
n
tx
4s−3
tdt − λ
0
4s−3
T k
xi t x4s−3 tdt
bi
0
i0
−λ
T
ft, xt − τx4s−3 tdt λ
0
T
3.55
ptx4s−3 tdt.
0
Since
T
x
4m1
tx
4s−3
tdt −1
2m−2s2
0
T 2
x2m2s−1 t dt,
3.56
0
then it follows from 3.55 and 3.56 that
b4s−3
T 4s−4
k
T 4s−3 k1
xi t x4s−3 tdt
bi
t dt ≤ −
x
0
0
i0
−
T
ft, xt − τx4s−3 tdt 0
T
3.57
ptx4s−3 tdt.
0
By using the same way as in the proof of Theorem 3.1, the following theorems can be proved
in case 1 < s ≤ m or s 1.
Theorem 3.3. Suppose n 4m 1, m > 0 for a positive integer and conditions H1 , H2 hold. If
H7 there is a positive integer 0 < s ≤ m such that
b4s−1 /
0,
b4s−1i 0,
i 1, 2, . . . , 4m − 4s 1,
3.58
H8 A1 4s − 1, k θ1 T 4s−2k
A2 4s − 1, k θ1 T
|b0 | − γ − θ2
k−1/k
4s−2k
4s−1 A1 4s − 1, k θ1 T
k|b0 |T
< |b4s−1 |,
|b0 | − γ − θ2
4s−2k
γ θ2
3.59
then 1.1 has at least one T -periodic solution.
Theorem 3.4. Suppose n 4m 3, m ≥ 0 an integer and conditions H1 , H2 hold. If
H9 there is a positive integer 0 ≤ s ≤ m such that
b4s1 /
0,
b4s1i 0,
i 1, 2, . . . , 4m − 4s 1,
3.60
International Journal of Differential Equations
17
H10 A2 4s 1, k θ1 T
k|b0 |T
4s1
4sk
γ θ2
A1 4s 1, k θ1 T 4sk
|b0 | − γ − θ2
A1 4s 1, k θ1 T 4sk
|b0 | − γ − θ2
k−1/k
γ θ2 A1 1, k θ1 θ1 < |b1 |,
|b0 | − γ − θ2
if 0 < s ≤ m,
< |b4s1 |,
if s 0,
3.61
then 1.1 has at least one T -periodic solution.
Theorem 3.5. Suppose n 4m 3,m > 0 an integer and conditions H1 , H2 hold. If
H11 there is a positive integer 0 < s ≤ m such that
b4s−1 /
0,
b4s−1i 0,
i 1, 2, . . . , 4m − 4s 3,
3.62
H12 A1 4s − 1, k θ1 T 4s−2k
A2 4 − 1, k θ1 T
|b0 | − γ − θ2
k−1/k
4s−2k
4s−1 A1 4s − 1, k θ1 T
k|b0 |T
< |b4s−1 |,
|b0 | − γ − θ2
4s−2k
γ θ2
3.63
then 1.1 has at least one T -periodic solution.
Theorem 3.6. Suppose n 4m, m > 0 an integer and conditions H1 hold. If
H13 b0 > γ θ2 ,
3.64
H14 there is a positive integer 0 < s ≤ 2m such that
b2s−1 /
0,
0,
b2s−1 /
b2s−1i 0,
if s 2m,
i 1, 2, . . . , 4m − 2s, if 0 < s < 2m,
3.65
18
International Journal of Differential Equations
H15 A2 2s − 1, k θ1 T
kb0 T
2s−1
2s−2k
γ θ2
A1 2s − 1, k θ1 T 2s−2k
b0 − γ − θ2
A1 2s − 1, k θ1 T 2s−2k
b0 − γ − θ2
k−1/k
γ θ2 A1 1, k θ1 θ1 < |b1 |,
b0 − γ − θ2
< |b2s−1 |,
if 1 < s ≤ 2m,
if s 1,
3.66
then 1.1 has at least one T -periodic solution.
Theorem 3.7. Suppose n 4m 2, m > 0 an integer and conditions H1 hold. If
H16 −b0 > γ θ2 ,
3.67
H17 there is a positive integer 0 < s ≤ 2m 1 such that
b2s−1 /
0,
b2s−1i 0,
0,
b2s−1 /
if s 2m 1,
i 1, 2, . . . , 4m − 2s, if 0 < s < 2m 1,
3.68
H18 A2 2s − 1, k θ1 T
−kb0 T
2s−1
2s−2k
γ θ2
A1 2s − 1, k θ1 T 2s−2k
−b0 − γ − θ2
A1 2s − 1, k θ1 T 2s−2k
−b0 − γ − θ2
k−1/k
γ θ2 A1 1, k θ1 θ1 < |b1 |,
−b0 − γ − θ2
< |b2s−1 |,
if 1 < s ≤ 2m 1,
if s 1,
3.69
then 1.1 has at least one T -periodic solution.
Theorem 3.8. Suppose n 4m, m > 0 is an integer, and conditions H1 , H13 hold. If
H19 there is a positive integer 0 < s ≤ m such that
b4s−2 /
0,
b4s−2i 0, i 1, 2, . . . , 4m − 4s 1,
3.70
International Journal of Differential Equations
19
H20 4s−3k
A2 4s − 2, k θ1 T
kb0 T
γ θ2
4s−2
γ θ2
A1 4s − 2, k θ1 T 4s−3k
b0 − γ − θ2
A1 4s − 2, k θ1 T 4s−3k
b0 − γ − θ2
k−1/k
< |b4s−2 |,
if 1 < s ≤ m,
k−1/k
k
A1 2, k θ1 T k
2 A1 2, k θ1 T
kb0 T
< |b2 |,
b0 − γ − θ2
b0 − γ − θ2
3.71
if s 1,
then 1.1 has at least one T -periodic solution.
Theorem 3.9. Suppose n 4m, m > 1 an integer and conditions H1 , H13 hold. If
H21 there is a positive integer 1 < s ≤ m such that
b4s−4i 0,
b4s−4 /
0,
i 1, 2, . . . , 4m − 4s 3,
3.72
H22 A2 4s − 4, k θ1 T
kb0 T
4s−4
4s−5k
γ θ2
A1 4s − 4, k θ1 T 4s−5k
b0 − γ − θ2
k−1/k
A1 4s − 4, k θ1 T 4s−5k
b0 − γ − θ2
3.73
< |b4s−4 |,
then 1.1 has at least one T -periodic solution.
Theorem 3.10. Suppose n 4m 2, m ≥ 1 an integer and conditions H1 , H16 hold. If
H23 there is a positive integer 1 ≤ s ≤ m such that
b4si 0,
b4s /
0,
i 1, 2, . . . , 4m − 4s 1,
3.74
H24 A2 4s, k θ1 T 4s−1k −kb0 T 4s
γ θ2
A1 4s, k θ1 T 4s−1k
−b0 − γ − θ2
k−1/k
A1 4s, k θ1 T 4s−1k
−b0 − γ − θ2
then 1.1 has at least one T -periodic solution.
< |b4s |,
3.75
20
International Journal of Differential Equations
Theorem 3.11. Suppose n 4m 2, m ≥ 1 is an integer, and conditions H1 , H16 hold. If
H25 there is a positive integer 1 ≤ s ≤ m such that
b4s−2 / 0,
b4s−2i 0,
i 1, 2, . . . , 4m − 4s 3,
3.76
H26 A2 4s − 2, k θ1 T
−kb0 T
γ θ2
θ1 T k
4s−2
4s−3k
γ θ2
A1 4s − 2, k θ1 T 4s−3k
−b0 − γ − θ2
A1 4s − 2, k θ1 T 4s−3k
−b0 − γ − θ2
k−1/k
< |b4s−2 |,
if 1 < s ≤ m,
k−1/k
k
A1 2, k θ1 T k
2 A1 2, k θ1 T
− kb0 T
< |b2 |,
−b0 − γ − θ2
−b0 − γ − θ2
3.77
if s 1,
then 1.1 has at least one T -periodic solution.
The proofs of Theorem 3.3–3.11 are similar to that of Theorem 3.1.
Example 3.12. Consider the following equation:
" #
$%3
3 1 3
1
1
π
3
x t 300 x t cos t,
x t xt sin t x t −
50
100
300
10
5
3.78
where n 5, k 3, b4 b3 0, b2 300, b1 1/50, b0 1/100, ft, x 1/300sin tx3 , pt cos t, τt π/10. Thus, T 2π, γ 1/300, A1 2, k |b1 |2π3 |b2 | 1/50 × 2π3 200. Obviously assumptions H1 –H3 hold and
θ1 T k
γ θ2
k−1/k
k
A1 2, k θ1 T k
2 A1 2, k θ1 T
k|b0 |2π
< |b2 |.
|b0 | − γ − θ2
|b0 | − γ − θ2
3.79
By Theorem 3.1, we know that 3.78 has at least one 2π-periodic solution.
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