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Hindawi Publishing Corporation Boundary Value Problems Volume 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 March 2011 Academic Editor: Kanishka Perera Copyright q 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 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 secondorder ordinary differential equations with some assumption 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 problems The 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 1.1 Throughout this paper, we will study the existence of periodic solutions of 1.1 with the following assumptions: H1 f, fx , and fx are continuous in R × R × R, and f t, x x f t 2π, x, x , 1.2 Boundary Value Problems H2 γ2 ≤ β < N 12 , π 4α − γ γ2 sin < 1− if N > 0, 4N 4α β 4N 1 , γ< 1− π N 12 N2 < α − 1.3 where N is some positive integer, α inf fx , R3 γ supfx β sup fx , R3 R3 1.4 The following is our main result Theorem 1.1 Assume that H1 and H2 hold, then 1.1 has a unique 2π-periodic solution Basic Lemmas The following results will be used later Lemma 2.1 see 12 Let x ∈ C1 0, h, R h > 0 with x0 xh 0, for t ∈ 0, h, 2.1 x tdt, 2.2 xt > then h xtx tdt ≤ h h and the constant h/4 is optimal Lemma 2.2 see 12 Let x ∈ C1 a, b, R a, b ∈ R, a < b with the boundary value conditions xa xb 0, then b x2 tdt ≤ a b − a2 π2 b x tdt 2.3 a Consider the periodic boundary value problem x ptx qtx 0, x0 x2π, x 0 x 2π 2.4 Boundary Value Problems Lemma 2.3 Suppose that p, q are L2 -integrable 2π-periodic function, where p, q satisfy the condition (H2 ), with α inf qt, 0,2π γ sup pt, β sup qt, 0,2π 0,2π 2.5 then 2.4 has only the trivial 2π-periodic solution xt ≡ Proof If on the contrary, 2.4 has a nonzero 2π-periodic solution xt, then using 2.4, we have t t psds psds e t0 x e t0 qtx 0, 2.6 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 xt > Since xt is a 2π-periodic solution, there exists a t0 ∈ 0, 2π with x t0 x t0 2π Then, 0 t0 2π e t psds x t0 t0 dt − t0 2π t e t0 psds qtxdt < 0, 2.7 t0 we could get a contradiction Without loss of generality, we may assume that x0 x2π 0, x 0 x 2π A > 0; then there exists a sufficiently small δ > such that xδ/2 > 0, x2π − δ/2 < Since xt is a continuous function, there must exist a t ∈ δ/2, 2π − δ/2 with xt Secondly, we prove that xt has at least 2N zeros on 0, 2π Considering the initial value problem ϕ − γϕ αϕ 0, ϕ 0 A ϕ0 0, 2.8 Obviously, 2A eγt/2 sin ϕt 4α − γ 4α − γ 2 t 2.9 is the solution of 2.8 and ϕ t 2A ⎞ ⎛ 4α − γ α ⎟ ⎜ t θ⎠, eγt/2 sin⎝ 4α − γ 2.10 Boundary Value Problems where θ ∈ 0, π/2 with sin θ 4α − γ /4α Since N< 4α − γ 2 2.11 By the conditions H2 , 2.11, and 2.12, we have sin 4α − γ 2 t0 sin θ π < π 4α − γ > sin 4α 4α − γ 4N π 4α − γ 4N 2.13 , 2.14 < π Since sin t is decreasing in π/2, π , we have < t0 < π/2N Therefore, ϕ t > 0, ϕt > 0, for t ∈ 0, t0 , ϕ t0 2.15 We also consider the initial value problem ψ γψ αψ 0, ψ t0 ψt0 ϕt0 , 2.16 Clearly, ψt ⎛ α ⎜ ϕt0 e−γt−t0 /2 sin⎝ 4α − γ 4α − γ 2 ⎞ ⎟ t − t0 θ⎠ 2.17 is the solution of 2.16, where θ is the same as the previous one, and 2α ϕt0 e−γt−t0 /2 sin ψ t − 4α − γ 4α − γ 2 t − t0 2.18 Hence, there exists a t1 ∈ 0, 2π with t1 − t0 ∈ 0, π, such that 4α − γ 2 t1 − t0 θ π 2.19 Boundary Value Problems Then, ψt1 2.20 From 2.12 and 2.19, it follows that 4α − γ t1 π − θ, i.e., π ≤ 4α − γ t1 < π 2.21 By H2 and 2.21, we have sin 4α − γ t1 sin θ π 4α − γ 4α − γ > sin 4α 4N 2.22 Since sin t is decreasing on π/2, π, we have < t1 < π/N, and ψ t < 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 > for t ∈ 0, t1 , then we would have the following inequalities: xt ≤ ϕt, for t ∈ 0, t0 , 2.24 xt ≤ ψt, for t ∈ t0 , t1 2.25 In fact, from2.4, 2.8, and 2.15, we have ϕ txt − ϕtx t ϕ txt ϕ tx t − ϕ tx t − ϕtx t γϕ t − αϕt xt − ϕt −ptx t − qtxt γ pt ϕ txt −pt ϕ txt − ϕtx t qt − α ϕtxt ≥ −pt ϕ txt − ϕtx t , 2.26 with t ∈ 0, t0 Setting y ϕ txt − ϕtx t, and since y ≥ −pty, 2.27 we obtain t ye psds ≥ 0, t ∈ 0, t0 2.28 Boundary Value Problems Notice that ϕ0 x0 0, which implies y0 0, t ye psds ≥ 0, 2.29 t ∈ 0, t0 So, we have ϕ txt − ϕtx t ≥ 0, t ∈ 0, t0 , i.e., ϕt xt ≥ 0, t ∈ 0, t0 2.30 Integrating from to t ∈ 0, t0 , we obtain 0≤ t ϕs xs ds ϕt ϕt ϕ 0 ϕt − lim − xt t → xt xt x 0 2.31 Therefore, ϕt ≥ 1, xt t ∈ 0, t0 , 2.32 which implies 2.24 By a similar argument, we have 2.25 Therefore, < 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 1 t ∈ t , t t1 , such that xt2 and so on So, we obtain that xt has at least 2N zeros on 0, 2π Thirdly, we prove that xt has at least 2N zeros on 0, 2π If, on the contrary, we assume that xt only has 2N zeros on 0, 2π, we write them as t0 < t1 < · · · < t2N1 2π 2.33 Obviously, 0, x ti / i 0, 1, , 2N 2.34 Without loss of generality, we may assume that x t0 > Since x ti x ti1 < 0, i 0, 1, , 2N, 2.35 we obtain x t2N1 < 0, which contradicts x t2N1 x t0 > Therefore, xt has at least 2N zeros on 0, 2π Boundary Value Problems Finally, we prove Lemma 2.3 Since xt has at least 2N zeros on 0, 2π, there are two zeros ξ1 and ξ2 with < ξ2 − ξ1 ≤ π/N 1 By Lemmas 2.1 and 2.2, we have ξ2 x tdt − ξ2 ξ1 xtx tdt ξ2 ξ1 ptxtx tdt ξ2 ξ1 qtx2 tdt ξ1 γ β ≤ ξ2 − ξ1 ξ2 − ξ1 2 π ξ2 2.36 2 x tdt ξ1 From H2 , it follows that γ β πγ β < ξ2 − ξ1 ξ2 − ξ1 2 ≤ 4N 1 N 12 π 2.37 Hence, ξ2 x tdt 0, 2.38 ξ1 which implies x t for t ∈ ξ1 , ξ2 Also xξ1 Therefore, xt ≡ for t ∈ 0, 2π, a contradiction The proof is complete Proof of Theorem 1.1 Firstly, we prove the existence of the solution Consider the homotopy equation x αx λ −f t, x, x αx ≡ λF t, x, x , 3.1 where λ ∈ 0, 1 and α infR3 fx When λ 1, it holds 1.1 We assume that Φt is the fundamental solution matrix of x αx with Φ0 I Equation 3.1 can be transformed into the integral equation x x t Φt x0 x 0 t Φ s −1 λFs, xs, x s ds 3.2 From H1 , xt is a 2π-periodic solution of 3.2, then I − Φ2π x0 x 0 Φ2π 2π −1 Φ s λFs, xs, x s ds 3.3 For I − Φ2π is invertible, x0 x 0 −1 I − Φ2π Φ2π 2π −1 Φ s λFs, xs, x s ds 3.4 Boundary Value Problems We substitute 3.4 into 3.2, x x 2π −1 t ΦtI − Φ2π Φ2π Φ s Φt Φ−1 s ds 3.5 ds λFs, xs, x s λFs, xs, x s t −1 Define an operator 3.6 Pλ : C1 0, 2π −→ C1 0, 2π, such that Pλ x x −1 t ≡ ΦtI − Φ2π Φ2π 2π −1 Φ s λFs, xs, x s Φt t −1 Φ s λFs, xs, x s ds 3.7 ds Clearly, Pλ is a completely continuous operator in C1 0, 2π There exists B > 0, such that every possible periodic solution xt satisfies x ≤ B · denote the usual normal in C1 0, 2π If not, there exists λk → λ0 and the solution xk t with xk → ∞ k → ∞ We can rewrite 3.1 in the following form: xk αxk −λk fx t, xk , θxk dθxk − λk fx t, θxk , 0dθxk − λk ft, 0, 0 λk αxk 3.8 Let yk xk /xk t ∈ R, obviously yk k 1, 2, It satisfies the following problem: yk αyk −λk 1 fx t, xk , θxk dθyk − λk fx t, θxk , 0dθyk − λk ft, 0, 0/xk λk αyk , 3.9 in which we have ft, 0, 0 −→ xk k −→ ∞ 3.10 Since {yk }, {yk } are uniformly bounded and equicontinuous, there exists continuous function ∞ ut, vt and a subsequence of {k}∞ denote it again by {k}1 , such that limk → ∞ yk t ∞ ut, limk → ∞ yk t vt uniformly in R Using H1 and H2 , { fx t, θxk , 0dθ}1 and Boundary Value Problems ∞ { fx t, xk , θxk dθ}1 are uniformly bounded By the Hahn-Banach theorem, there exists ∞ L2 -integrable function pt, qt, and a subsequence of {k}∞ denote it again by {k}1 , such that ω fx t, θxk , 0dθ −→ qt, ω fx t, xk , θx k dθ −→ pt, 3.11 ω where −→ denotes “weakly converges to” in L2 0, 2π As a consequence, we have u t αut −λ0 ptu t − λ0 qtut λ0 αut, 3.12 u t λ0 ptu t λ0 qt 1 − λ0 α ut 3.13 that is, Denote that pt λ0 pt, qt λ0 qt 1 − λ0 α, then we get pt λ0 pt ≤ γ, λ0 α 1 − λ0 α ≤ qt ≤ λ0 β 1 − λ0 α, 3.14 which also satisfy the condition H2 Notice that pt and qt are L2 -integrable on 0, 2π, so ut satisfies Lemma 2.3 Hence, we have ut ≡ for t ∈ 0, 2π, which contradicts u Therefore, PC1 0, 2π is bounded Denote Ω x ∈ C1 0, 2π, x < B , 3.15 hλ x x − Pλ x Because ∈ / hλ ∂Ω for λ ∈ 0, 1, by Leray-Schauder degree theory, we have degx − P x, Ω, 0 degh1 x, Ω, 0 degh0 x, Ω, 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 condition H1 and H2 holds Let x1 t and x2 t be two 2π-periodic solutions of the problem Denote x0 t x1 t − x2 t, t ∈ 0, 2π, then x0 t is a solution of the following problem: x fx t, x2 x0 , x2 θx dθx x0 x2π, By Lemma 2.3, we have x0 t ≡ for t ∈ 0, 2π fx t, x2 θx0 , x2 dθx 0, x 0 x 2π 3.17 10 Boundary Value Problems Let xt 2kπ xt, t ∈ 0, 2π, k ∈ Z We have x t 2kπ x t −f t, x, x −f t, x, x −f t 2kπ, x, x , 3.18 with t ∈ 0, 2π, k ∈ Z Denote xt 2kπ t ∈ 0, 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