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## i.e. e −iαT − e

i.e. e −iαT − e +iαT = 0 ρ1 = ρ2 which is a contradiction. Therefore k1 = 0 = k2, showing the linear independence. Thus: For |K| �= 1 the solution u(t) to (1) can be written in the form with U (1,2) (t) being T-periodic. u(t) = c1e iαt U (1) (t) + c2e −iαt U (2) (t) 1. for |K| < 1 the solutions are quasi-periodic since α ∈ R 2. for |K| > 1 one solution grows exponentially, while the other decays exponentially since α ∈ iR. The growing solution renders the state u(t) = 0 linearly unstable. ρ i −1

Goal: determine α(δ, ǫ). For a response that is subharmonic with respect to the period of the forcing one has ρ = e iαπ = −1, i.e. α = 1. We therefore expand with Insert O(1): O(ǫ): α = 1 + ǫα1 + ǫ 2 α2 + . . . u = e iαt ψ(t) = e i(ǫα1+ǫ 2 α2t+...)t e it ψ(t) � �� � φ(t) φ(t) = φ0(t) + ǫφ1(t) + . . . δ = 1 + ǫδ1 + . . . ü = − � ǫα1 + ǫ 2 α2t + . . . � 2 e i(ǫα1+ǫ 2 α2t+...)t φ + 2i � ǫα1 + ǫ 2 α2t + . . . � e i(ǫα1+ǫ 2 α2t+...)t ˙ φ + +e i(ǫα1+ǫ 2 α2t+...)t ¨ φ ¨φ0 + φ0 = 0 ⇒ φ0(t) = c1 cost + c2 sin t ¨φ1 + φ1 = −2iα1 ˙ φ0 − δ1φ0 − cos 2t φ0 ≡ F1(t) F1(t) = −2iα1 (−c1 sin t + c2 cos t) − δ1 (c1 cost + c2 sin t) − cos 2t (c1 cos t + c2 sin t) � = cos t −2iα1c2 − δ1c1 − 1 2 c1 � � + sin t 2iα1c1 − δ1c2 + 1 2 c2 � − 1 2 c1 cos 3t − 1 2 c2 sin 3t To avoid secular terms we need −2iα1c2 − δ1c1 − 1 2 c1 = 0 2iα1c1 − δ1c2 + 1 2 c2 = 0 � � � � δ1 + 1 2iα1 2 2iα1 −δ1 + 1 � � � � = 0 α1 = ± 1 2 −δ 2 1 + 1 � Insert � δ1 + 1 � 2 2 4 + 4α2 1 = 0 δ 2 1 − 1 4 c1 ∓ = ±i1 2 � 1 4 − δ2 1 � 1 4 − δ2 1c2 = 0 17

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