Volume 2011, Article ID 192156,11pages 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,[email protected] Received 22 May 2010; Accepted 6 March 2011
Academic Editor: Kanishka Perera
Copyrightq2011 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 methodas discussed by many authors1–10. In11, 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
xft, x, x 0. 1.1
Throughout this paper, we will study the existence of periodic solutions of 1.1 with the following assumptions:
H1f, fx, and fxare continuous inR×R×R, and f
t, x x f
t2π, x, x
, 1.2
H2
N2< α−γ2
4 ≤β <N12, sinπ
4α−γ2 4N <
1− γ2
4α ifN >0, γ < 4N1
π
1− β
N12
,
1.3
whereNis some positive integer, αinf
R3
fx
, βsup
R3
fx
, γsup
R3
fx. 1.4
The following is our main result.
Theorem 1.1. Assume thatH1andH2hold, then1.1has a unique 2π-periodic solution.
2. Basic Lemmas
The following results will be used later.
Lemma 2.1see12. Letx∈C10, h,R h >0with
x0 xh 0, xt>0 fort∈0, h, 2.1
then
h 0
xtxtdt≤ h 4
h 0
x2tdt, 2.2
and the constanth/4 is optimal.
Lemma 2.2see12. Letx∈C1a, b,R a, b∈R, a < bwith the boundary value conditions xa xb 0, then
b a
x2tdt≤ b−a2 π2
b a
x2tdt. 2.3
Consider the periodic boundary value problem
xptxqtx0,
x0 x2π, x0 x2π. 2.4
Lemma 2.3. Suppose that p, q are L2-integrable 2π-periodic function, where p, q satisfy the condition (H2), with
α inf
0,2πqt, β sup
0,2πqt, γ sup
0,2π
pt, 2.5
then2.4has only the trivial 2π-periodic solutionxt≡0.
Proof. If on the contrary,2.4has a nonzero 2π-periodic solutionxt, then using2.4, we have
e
t
t0psds
x
e
t
t0psdsqtx0, 2.6
wheret0∈0,2πis undetermined.
Firstly, we prove thatxthas at least one zero in0,2π. Ifxt/0, we may assume xt > 0. Since xt is a 2π-periodic solution, there exists at0 ∈ 0,2πwith xt0 0 xt02π. Then,
0 t02π
t0
e
t
t0psds
x
dt− t02π
t0
e
t
t0psdsqtxdt <0, 2.7
we could get a contradiction.
Without loss of generality, we may assume thatx0 x2π 0, x0 x2π A >0; then there exists a sufficiently smallδ >0 such thatxδ/2>0, x2π−δ/2<0. Since xtis a continuous function, there must exist at∈δ/2,2π−δ/2withxt 0.
Secondly, we prove thatxthas at least 2N2 zeros on0,2π. Considering the initial value problem
ϕ−γϕαϕ0, ϕ0 0, ϕ0 A. 2.8
Obviously,
ϕt 2A
4α−γ2
eγt/2sin
4α−γ2
2 t 2.9
is the solution of2.8and
ϕt 2A α
4α−γ2eγt/2sin
⎛
⎜⎝
4α−γ2 2 tθ
⎞
⎟⎠, 2.10
whereθ∈0, π/2with sinθ
4α−γ2/4α. Since
N <
4α−γ2
2 < N1 2.11
holds under the assumptions ofH2, there is at0∈0, π, such that
4α−γ2
2 t0θπ, i.e., π 2 ≤
4α−γ2
2 t0< π. 2.12
Now, letN >0. By the conditionsH2,2.11, and2.12, we have
sin
4α−γ2
2 t0sinθ
4α−γ2
4α >sinπ
4α−γ2
4N , 2.13
π 2 < π
4α−γ2
4N < π. 2.14
Since sintis decreasing inπ/2, π, we have 0< t0< π/2N. Therefore,
ϕt>0, ϕt>0, fort∈0, t0, ϕt0 0. 2.15
We also consider the initial value problem
ψγψαψ0, ψt0 ϕt0, ψt0 0. 2.16
Clearly,
ψt 2 α
4α−γ2ϕt0e−γt−t0/2sin
⎛
⎜⎝
4α−γ2
2 t−t0 θ
⎞
⎟⎠ 2.17
is the solution of2.16, whereθis the same as the previous one, and
ψt − 2α
4α−γ2
ϕt0e−γt−t0/2sin
4α−γ2
2 t−t0. 2.18
Hence, there exists at1∈0,2πwitht1−t0 ∈0, π, such that
4α−γ2
2 t1−t0 θπ. 2.19
Then,
ψt1 0. 2.20
From2.12and2.19, it follows that
4α−γ2
4 t1π−θ, i.e., π 2 ≤
4α−γ2
4 t1 < π. 2.21
ByH2and2.21, we have
sin
4α−γ2
4 t1sinθ
4α−γ2
4α >sinπ 4α−γ2
4N . 2.22
Since sintis decreasing onπ/2, π, we have 0< t1< π/N, and
ψt<0, ψt>0, fort∈t0, t1. 2.23 We now prove thatxthas a zero point in0, t1. If on the contraryxt > 0 fort ∈ 0, t1, then we would have the following inequalities:
xt≤ϕt, fort∈0, t0, 2.24
xt≤ψt, fort∈t0, t1. 2.25
In fact, from2.4,2.8, and2.15, we have ϕtxt−ϕtxt
ϕtxt ϕtxt−ϕtxt−ϕtxt
γϕt−αϕt
xt−ϕt
−ptxt−qtxt
γpt
ϕtxt
−pt
ϕtxt−ϕtxt
qt−α
ϕtxt
≥
−pt
ϕtxt−ϕtxt ,
2.26
witht∈0, t0. Settingyϕtxt−ϕtxt, and since
y≥ −pty, 2.27
we obtain
yet0psds
≥0, t∈0, t0. 2.28
Notice thatϕ0 x0 0, which implies
y0 0, ye0tpsds≥0, t∈0, t0. 2.29
So, we have
ϕtxt−ϕtxt≥0, t∈0, t0, i.e., ϕt
xt
≥0, t∈0, t0. 2.30
Integrating from 0 tot∈0, t0, we obtain
0≤ t
0
ϕs xs
ds ϕt xt− lim
t→0
ϕt
xt ϕt
xt−ϕ0
x0. 2.31
Therefore,
ϕt
xt ≥1, t∈0, t0, 2.32
which implies2.24. By a similar argument, we have2.25. Therefore, 0< xt1≤ψt1 0, a contradiction, which shows thatxthas at least one zero in0, t1, witht1 < π/N.
We letxt1 0, t1 ∈ 0, t1. If t1t1 < 2π, then from a similar argument, there is a t2∈t1, t1t1, such thatxt2 0 and so on. So, we obtain thatxthas at least 2N2 zeros on0,2π.
Thirdly, we prove thatxthas at least 2N3 zeros on0,2π. If, on the contrary, we assume thatxtonly has 2N2 zeros on0,2π, we write them as
0t0< t1<· · ·< t2N12π. 2.33
Obviously,
x ti
/0, i0,1, . . . ,2N1. 2.34
Without loss of generality, we may assume thatxt0>0. Since
x ti
x ti1
<0, i0,1, . . . ,2N, 2.35
we obtainxt2N1 <0, which contradictsxt2N1 xt0> 0. Therefore,xthas at least 2N3 zeros on0,2π.
Finally, we prove Lemma2.3. Sincexthas at least 2N3 zeros on0,2π, there are two zerosξ1andξ2with 0< ξ2−ξ1≤π/N1. By Lemmas2.1and2.2, we have
ξ2
ξ1
x2tdt− ξ2
ξ1
xtxtdt ξ2
ξ1
ptxtxtdt ξ2
ξ1
qtx2tdt
≤ γ
4ξ2−ξ1
β
π2ξ2−ξ12 ξ2
ξ1
x2tdt.
2.36
FromH2, it follows that γ
4ξ2−ξ1 β
π2ξ2−ξ12≤ πγ
4N1 β
N12 <1. 2.37 Hence,
ξ2
ξ1
x2tdt0, 2.38
which impliesxt 0 fort∈ξ1, ξ2. Alsoxξ1 0. Therefore,xt≡ 0 for t ∈0,2π, a contradiction. The proof is complete.
3. Proof of Theorem 1.1
Firstly, we prove the existence of the solution. Consider the homotopy equation xαxλ
−f t, x, x
αx
≡λF t, x, x
, 3.1
whereλ ∈ 0,1and α infR3fx. Whenλ 1, it holds1.1. We assume thatΦtis the fundamental solution matrix ofxαx0 withΦ0 I. Equation3.1can be transformed into the integral equation
x x
t Φt
x0 x0
t
0
Φ−1s
0
λFs, xs, xs
ds
. 3.2
FromH1,xtis a 2π-periodic solution of3.2, then
I−Φ2π x0
x0
Φ2π 2π
0
Φ−1s
0
λFs, xs, xs
ds. 3.3
ForI−Φ2πis invertible, x0
x0
I−Φ2π−1Φ2π 2π
0
Φ−1s
0
λFs, xs, xs
ds. 3.4
We substitute3.4into3.2, x
x
t ΦtI−Φ2π−1Φ2π 2π
0
Φ−1s
0
λFs, xs, xs
ds
Φt t
0
Φ−1s
0
λFs, xs, xs
ds.
3.5
Define an operator
Pλ:C10,2π−→C10,2π, 3.6
such that
Pλ
x x
t≡ΦtI−Φ2π−1Φ2π 2π
0
Φ−1s
0
λFs, xs, xs
ds
Φt t
0
Φ−1s
0
λFs, xs, xs
ds.
3.7
Clearly,Pλis a completely continuous operator inC10,2π.
There existsB >0, such that every possible periodic solutionxtsatisfiesx ≤B· denote the usual normal inC10,2π. If not, there existsλk → λ0and the solutionxktwith xk → ∞k → ∞.
We can rewrite3.1in the following form:
xkαxk−λk 1 0
fx
t, xk, θxk
dθxk−λk 1 0
fxt, θxk,0dθxk−λkft,0,0 λkαxk. 3.8
Letyk xk/xkt ∈ R, obviouslyyk 1k 1,2, . . .. It satisfies the following problem:
ykαyk−λk
1
0fx
t, xk, θxk
dθyk−λk
1
0fxt, θxk,0dθyk−λkft,0,0/xkλkαyk, 3.9
in which we have
ft,0,0
xk −→0 k−→ ∞. 3.10
Since{yk},{yk}are uniformly bounded and equicontinuous, there exists continuous function ut,vt and a subsequence of{k}∞1 denote it again by{k}∞1 , such that limk→ ∞ykt ut, limk→ ∞ykt vt uniformly in R. Using H1 and H2,{1
0fxt, θxk,0dθ}∞1 and
{1
0fxt, xk, θxkdθ}∞1 are uniformly bounded. By the Hahn-Banach theorem, there exists L2-integrable functionpt,qt, and a subsequence of{k}∞1 denote it again by{k}∞1, such that
1 0
fxt, θxk,0dθ−→ω qt, 1
0
fx
t, xk, θxk
dθ−→ω pt, 3.11
where−→ω denotes “weakly converges to” inL20,2π. As a consequence, we have
ut αut −λ0ptut−λ0qtut λ0αut, 3.12
that is,
ut λ0ptut
λ0qt 1−λ0α
ut 0. 3.13
Denote thatpt λ0pt,qt λ0qt 1−λ0α, then we get
ptλ0pt≤γ, λ0α 1−λ0α≤qt ≤λ0β 1−λ0α, 3.14 which also satisfy the conditionH2. Notice thatpt andqt areL2-integrable on0,2π, so utsatisfies Lemma2.3. Hence, we haveut≡0 fort∈0,2π, which contradictsu1.
Therefore, PC10,2πis bounded.
Denote
Ω
x∈C10,2π,x< B1 , hλx x−Pλx.
3.15
Because 0∈/hλ∂Ωforλ∈0,1, by Leray-Schauder degree theory, we have
degx−P x,Ω,0 degh1x,Ω,0 degh0x,Ω,0/0. 3.16
So, we conclude thatPhas at least one fixed point inΩ, that is,1.1has at least one solution.
Finally, we prove the uniqueness of the equation when the conditionH1andH2 holds. Letx1tandx2tbe two 2π-periodic solutions of the problem. Denotex0t x1t− x2t, t∈0,2π, thenx0tis a solution of the following problem:
x 1
0
fx
t, x2x0, x2θx0
dθx 1
0
fx
t, x2θx0, x2
dθx0,
x0 x2π, x0 x2π.
3.17
By Lemma2.3, we havex0t≡0 fort∈0,2π.
Letxt 2kπ xt, t∈0,2π, k∈Z. We have
xt2kπ xt −f t, x, x
−f t,x, x
−f
t2kπ,x, x
, 3.18
witht∈0,2π, k∈Z. Denotext 2kπ t∈0,2πbyxt t∈R. So,xtis the solution of the problem1.1. The proof is complete.
4. An Example
Consider the system
x2
3sintx6xcosxpt, 4.1
wherept pt2πis a continuous function. Obviously, αinf
R3
fx
inf
R36−sinx 5, βsup
R3
fx
sup
R3 6−sinx 7, γsup
R3
fxsup
R3
2 3sint
2 3
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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