Poincaré Map and Periodic Solutions of First-Order Impulsive Differential Equations on Moebius Stripe

and Applied Analysis 3 Lemma 7. Assume that conditions (H1), (H2) hold. Suppose that x(t, t0, x0) is a solution of (1) satisfying initial value x(t + 0 ) = x0. Then −x(t + T, t0, x0) is also a solution of (1), and −x (t + T, t0, x0) = x (t, t0, −x (t0 + T, t0, x0)) , t ∈ R. (5) Proof. Let φ(t) ≡ −x(t + T, t0, x0), ψ(t) ≡ x(t, t0, −x(t0 + T, t0, x0)). Then for t ̸ = τk, k ∈ Z, we have by (H2) that dφ (t) dt = − dx (t + T, t0, x0) dt = − f (t + T, x (t + T, t0, x0)) = f (t, −x (t + T, t0, x0)) = f (t, φ (t)) . (6) For t = τk, k ∈ Z , it follows from (H1), (H2) that τk + T = τk+q, k ∈ Z + and t=τk = − x (τk + T + , t0, x0) + x (τk + T, t0, x0) = − x (τ + k+q , t0, x0) + x (τk+q, t0, x0) = − Ik+q (x (τk+q, t0, x0)) = − Ik (x (τk+q, t0, x0)) = −Ik (x (τk + T, t0, x0)) = Ik (−x (τk + T, t0, x0)) = Ik (φ (τk)) . (7) Thus, we proved that φ(t) ≡ −x(t + T, t0, x0) is a solution of (1). On the other hand, it is obvious that φ (t) t=t0 = −x (t0 + T, t0, x0) = ψ (t) t=t0 . (8) Hence, by uniqueness theorem we have that φ(t) ≡ ψ(t), t ∈ R. This completes the proof. LetD denotes the stripe area on the plain {(t, x) | (t, x) ∈ R ×R} between two lines t = t0 and t = t0 + T; that is, D = {(t, x) | t0 ≤ t ≤ t0 + T, −∞ < x < +∞} . (9) Assume that x(t, t0, x0) exists for all t ∈ [t0, +∞). Define L0 = {(t, x(t, t0, x0)) | t0 ≤ t ≤ t0 + T}. In general, we denote Lk (k ≥ 1) by Lk = {(t, x (t, t0, −xk)) | t0 ≤ t ≤ t0 + T} , (10) where xk = x(t0 + T + , t0, −xk−1), t ≥ t0. It follows from Lemma 7 that Lk has the form Lk = {(t, (−1) k x (t + kT, t0, x0)) | t0 ≤ t ≤ t0 + T} . (11) We now introduce an equivalence relation ∼ on D such that for (t, x), (t, x) ∈ D (t, x) ∼ (t 󸀠 , x 󸀠 ) iff 󵄨󵄨󵄨󵄨t − t 󵄨󵄨󵄨󵄨 = T, x = −x 󸀠 . (12) Then we denote the corresponding quotient space by M2. From geometric point of view,M2 is obtained by considering O x