Volume 2011, Article ID 192156, 11 pagesdoi:10.1155/2011/192156 Research Article Existence and Uniqueness of Periodic Solution for Nonlinear Second-Order Ordinary Differential Equations
Trang 1Volume 2011, Article ID 192156, 11 pages
doi:10.1155/2011/192156
Research Article
Existence and Uniqueness of
Periodic Solution for Nonlinear Second-Order
Ordinary Differential Equations
Jian Zu
College of Mathematics, Jilin University, Changchun 130012, China
Correspondence should be addressed to Jian Zu,zujian1984@gmail.com
Received 22 May 2010; Accepted 6 March 2011
Academic Editor: Kanishka Perera
Copyrightq 2011 Jian Zu 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 periodic solutions for nonlinear second-order ordinary differential problem x
ft, x, x 0 By constructing upper and lower boundaries and using Leray-Schauder degree theory, we present a result about the existence and uniqueness of a periodic solution for second-order ordinary differential equations with some assumption
1 Introduction
The study on periodic solutions for ordinary differential equations is a very important branch
in the differential equation theory Many results about the existence of periodic solutions for second-order differential equations have been obtained by combining the classical method of lower and upper solutions and the method of alternative problemsThe Lyapunov-Schmidt method as discussed by many authors 1 10 In 11, the author gives a simple method to discuss the existence and uniqueness of nonlinear two-point boundary value problems In this paper, we will extend this method to the periodic problem
We consider the second-order ordinary differential equation
x ft, x, x 0. 1.1
Throughout this paper, we will study the existence of periodic solutions of 1.1 with the following assumptions:
H1 f, f x , and f xare continuous inR × R × R, and
f
t, x x
ft 2π, x, x
Trang 22 Boundary Value Problems
H2
N2< α − γ
2
4 ≤ β < N 12
,
sinπ
4α − γ2
4N <
1− γ2
4α if N > 0,
γ < 4N 1
π
1− β
N 12
,
1.3
where N is some positive integer,
α inf
R 3
f x
R 3
f x
R 3
f x. 1.4 The following is our main result
2 Basic Lemmas
The following results will be used later
x 0 xh 0, xt > 0 for t ∈ 0, h, 2.1
then
h
0
x txtdt ≤ h
4
h
0
and the constant h/4 is optimal.
xa xb 0, then
b
a
x2tdt ≤ b − a2
π2
b
a
x2tdt. 2.3
Consider the periodic boundary value problem
x ptx qtx 0,
x 0 x2π, x0 x2π. 2.4
Trang 3Lemma 2.3 Suppose that p, q are L2-integrable 2π-periodic function, where p, q satisfy the condition (H2), with
α inf
0,2π q t, β sup
0,2π q t, γ sup
0,2π
p t, 2.5
then2.4 has only the trivial 2π-periodic solution xt ≡ 0.
have
e
t
t0 psds
x
e t0 t psds
where t0∈ 0, 2π is undetermined.
Firstly, we prove that xt has at least one zero in 0, 2π If xt / 0, we may assume
xt0 2π Then,
0
t02π
t0
e
t t0 psds
x
dt −
t02π
t0
e
t t0 psds
q txdt < 0, 2.7
we could get a contradiction
Without loss of generality, we may assume that x0 x2π 0, x0 x2π
Secondly, we prove that xt has at least 2N 2 zeros on 0, 2π Considering the initial
value problem
ϕ− γϕ αϕ 0, ϕ 0 0, ϕ0 A. 2.8 Obviously,
ϕ t 2A
4α − γ2e γt/2sin
4α − γ2
is the solution of2.8 and
ϕt 2A
α
4α − γ2e γt/2sin
⎛
⎜
4α − γ2
⎞
⎟
Trang 44 Boundary Value Problems
where θ ∈ 0, π/2 with sin θ
4α − γ2/4α Since
N <
4α − γ2
holds under the assumptions ofH2, there is a t0∈ 0, π, such that
4α − γ2
2 t0 θ π, i.e., π
2 ≤
4α − γ2
Now, let N > 0 By the conditions H2, 2.11, and 2.12, we have
sin
4α − γ2
2 t0 sin θ
4α − γ2
π
4α − γ2
4N ,
2.13
π
2 < π
4α − γ2
2.14
Since sin t is decreasing in π/2, π , we have 0 < t0< π/2N Therefore,
ϕt > 0, ϕt > 0, for t ∈ 0, t0, ϕt0 0. 2.15
We also consider the initial value problem
ψ γψ αψ 0, ψ t0 ϕt0, ψt0 0. 2.16 Clearly,
ψ t 2
α
4α − γ2ϕ t0e −γt−t0/2sin
⎛
⎜
4α − γ2
2 t − t0 θ
⎞
is the solution of2.16, where θ is the same as the previous one, and
ψt − 2α
4α − γ2ϕ t0e −γt−t0/2sin
4α − γ2
2 t − t0. 2.18
Hence, there exists a t1∈ 0, 2π with t1− t0 ∈ 0, π, such that
4α − γ2
2 t1− t0 θ π. 2.19
Trang 5From2.12 and 2.19, it follows that
4α − γ2
4 t1 π − θ, i.e., π
2 ≤
4α − γ2
ByH2 and 2.21, we have
sin
4α − γ2
4 t1 sin θ
4α − γ2
4α − γ2
4N .
2.22
Since sin t is decreasing on π/2, π, we have 0 < t1< π/N, and
ψt < 0, ψt > 0, for t ∈ t0, t1. 2.23
We now prove that xt has a zero point in 0, t1 If on the contrary xt > 0 for t ∈
0, t1, then we would have the following inequalities:
x t ≤ ϕt, for t ∈ 0, t0, 2.24
x t ≤ ψt, for t ∈ t0, t1. 2.25
In fact, from2.4, 2.8, and 2.15, we have
ϕtxt − ϕtxt
ϕtxt ϕtxt − ϕtxt − ϕtxt
γϕt − αϕtx t − ϕt−ptxt − qtxt
γ p tϕtxt −ptϕtxt − ϕtxtq t − αϕ txt
≥−ptϕtxt − ϕtxt,
2.26
with t ∈ 0, t0 Setting y ϕtxt − ϕtxt, and since
we obtain
ye t0psds
≥ 0, t ∈ 0, t0. 2.28
Trang 66 Boundary Value Problems
Notice that ϕ0 x0 0, which implies
y 0 0, ye 0t psds ≥ 0, t ∈ 0, t0. 2.29
So, we have
ϕtxt − ϕtxt ≥ 0, t ∈ 0, t0, i.e., ϕ t
x t
≥ 0, t ∈ 0, t0. 2.30
Integrating from 0 to t ∈ 0, t0, we obtain
0≤ t
0
ϕ s
x s
x t− limt → 0
ϕ t
ϕ t
ϕ0
x0. 2.31 Therefore,
ϕ t
x t ≥ 1, t ∈ 0, t0, 2.32
which implies2.24 By a similar argument, we have 2.25 Therefore, 0 < xt1 ≤ ψt1 0,
a contradiction, which shows that xt has at least one zero in 0, t1, with t1 < π/N.
We let xt1 0, t1 ∈ 0, t1 If t1 t1 < 2π, then from a similar argument, there is a
t2∈ t1, t1 t1, such that xt2 0 and so on So, we obtain that xt has at least 2N 2 zeros
on0, 2π.
Thirdly, we prove that xt has at least 2N 3 zeros on 0, 2π If, on the contrary, we assume that xt only has 2N 2 zeros on 0, 2π, we write them as
Obviously,
x
t i
/
0, i 0, 1, , 2N 1. 2.34
Without loss of generality, we may assume that xt0 > 0 Since
x
t i
x
t i1
we obtain xt 2N1 < 0, which contradicts xt 2N1 xt0 > 0 Therefore, xt has at least 2N 3 zeros on 0, 2π.
Trang 7Finally, we prove Lemma2.3 Since xt has at least 2N 3 zeros on 0, 2π, there are two zeros ξ1and ξ2with 0 < ξ2− ξ1≤ π/N 1 By Lemmas2.1and2.2, we have
ξ2
ξ1
x2tdt −
ξ2
ξ1
x txtdt
ξ2
ξ1
p txtxtdt
ξ2
ξ1
q tx2tdt
≤
γ
4ξ2− ξ1 β
π2ξ2− ξ12
ξ2
ξ1
x2tdt.
2.36
FromH2, it follows that
γ
4ξ2− ξ1 β
4N 1
β
N 12 < 1. 2.37 Hence,
ξ2
ξ1
which implies xt 0 for t ∈ ξ1, ξ2 Also xξ1 0 Therefore, xt ≡ 0 for t ∈ 0, 2π, a
contradiction The proof is complete
Firstly, we prove the existence of the solution Consider the homotopy equation
x αx λ−ft, x, x
αx≡ λFt, x, x
where λ ∈ 0, 1 and α infR 3f x When λ 1, it holds 1.1 We assume that Φt is the fundamental solution matrix of x αx 0 with Φ0 I Equation 3.1 can be transformed into the integral equation
x
x
t Φt
x0
x0
t
0
Φ−1s
0
λF s, xs, xs
ds
. 3.2
FromH1, xt is a 2π-periodic solution of 3.2, then
I − Φ2π
x0
x0
Φ2π
2π
0
Φ−1s
0
λF s, xs, xs
ds. 3.3
ForI − Φ2π is invertible,
x0
x0
I − Φ2π−1Φ2π
2π
0
Φ−1s
0
λF s, xs, xs
ds. 3.4
Trang 88 Boundary Value Problems
We substitute3.4 into 3.2,
x
x
t ΦtI − Φ2π−1Φ2π
2π
0
Φ−1s
0
λF s, xs, xs
ds
Φt
t
0
Φ−1s
0
λF s, xs, xs
ds.
3.5
Define an operator
P λ : C10, 2π −→ C10, 2π, 3.6 such that
P λ
x
x
t ≡ ΦtI − Φ2π−1Φ2π
2π
0
Φ−1s
0
λF s, xs, xs
ds
Φt
t
0
Φ−1s
0
λF s, xs, xs
ds.
3.7
Clearly, P λ is a completely continuous operator in C10, 2π.
There exists B > 0, such that every possible periodic solution xt satisfies x ≤ B · denote the usual normal in C10, 2π If not, there exists λ k → λ0and the solution x k t with
x k → ∞ k → ∞.
We can rewrite3.1 in the following form:
xk αx k −λ k
1
0
f x
t, x k , θx k
dθxk − λ k
1
0
f x t, θx k , 0 dθx k − λ k f t, 0, 0 λ k αx k 3.8
Let y k x k /x k t ∈ R, obviously y k 1 k 1, 2, It satisfies the following
problem:
y k αy k −λ k
1
0f x
t, x k , θxk
dθyk − λ k
1
0f x t, θx k , 0 dθy k − λ k f t, 0, 0/x k λ k αy k ,
3.9
in which we have
f t, 0, 0
x k −→ 0 k −→ ∞. 3.10 Since{y k }, {y
k} are uniformly bounded and equicontinuous, there exists continuous function
1 , such that limk → ∞ y k t
0f x t, θx k , 0dθ}∞1 and
Trang 90f xt, x k , θxk dθ}∞1 are uniformly bounded By the Hahn-Banach theorem, there exists
L2-integrable function pt, qt, and a subsequence of {k}∞1 denote it again by {k}∞
1, such that
1
0
f x t, θx k , 0 dθ −→ qt, ω
1
0
f x
t, x k , θxk
dθ −→ pt, ω 3.11
where−→ denotes “weakly converges to” in L ω 20, 2π As a consequence, we have
ut αut −λ0p tut − λ0q tut λ0αu t, 3.12 that is,
ut λ0p tut λ0q t 1 − λ0αu t 0. 3.13 Denote thatpt λ0pt, qt λ0qt 1 − λ0α, then we get
pt λ0p t ≤ γ, λ0α 1 − λ0α ≤ qt ≤ λ0β 1 − λ0α, 3.14
which also satisfy the conditionH2 Notice that pt and qt are L2-integrable on0, 2π, so
Therefore, PC10, 2π is bounded.
Denote
Ω x ∈ C10, 2π, x < B 1,
h λ x x − P λ x.
3.15
Because 0 / ∈ h λ ∂Ω for λ ∈ 0, 1, by Leray-Schauder degree theory, we have
degx − Px, Ω, 0 degh1x, Ω, 0 degh0x, Ω, 0 / 0. 3.16
So, we conclude that P has at least one fixed point in Ω, that is, 1.1 has at least one solution Finally, we prove the uniqueness of the equation when the conditionH1 and H2
holds Let x1t and x2t be two 2π-periodic solutions of the problem Denote x0t x1t −
x2t, t ∈ 0, 2π, then x0t is a solution of the following problem:
x
1
0
f x
t, x2 x0, x2 θx
0
1
0
f x
t, x2 θx0, x2
dθx 0,
x 0 x2π, x0 x2π.
3.17
By Lemma2.3, we have x0t ≡ 0 for t ∈ 0, 2π.
Trang 1010 Boundary Value Problems Letxt 2kπ xt, t ∈ 0, 2π, k ∈ Z We have
xt 2kπ xt −ft, x, x
−ft, x, x
−ft 2kπ, x, x
, 3.18
with t ∈ 0, 2π, k ∈ Z Denote xt 2kπ t ∈ 0, 2π by xt t ∈ R So, xt is the solution
of the problem1.1 The proof is complete
4 An Example
Consider the system
x2
3sin tx 6x cos x pt, 4.1
where pt pt 2π is a continuous function Obviously,
α inf
R 3
f x
inf
R 36 − sin x 5,
β sup
R 3
f x
sup
R 3
6 − sin x 7,
γ sup
R 3
f x sup
R 3
23sin t
23
4.2
satisfy Theorem1.1, then there is a unique 2π-periodic solution in this system.
Acknowledgments
The author expresses sincere thanks to Professor Yong Li for useful discussion He would like
to thank the reviewers for helpful comments on an earlier draft of this paper
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