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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

Consider the two

Consider the two linearly independent solutions � u1(0) = 1 u1(t) with ˙u1(0) = 0 u2(t) with � u2(0) = 0 ˙u2(0) = 1 Since u1,2 are linearly independent their Wronskian is non-zero � � W(u1(t), u2(t)) = � � u1(t) � u2(t) � � ˙u1(t) ˙u2(t) � = � � � � u1(0) � u2(0) � � ˙u1(0) ˙u2(0) � = 1 �= 0 Use u1,2 to express the general solution as u(t) = c1u1(t) + c2u2(t) where the c1,2 are given by the initial conditions c1 = u(0) c2 = ˙u(0) (2) Now consider u(t + T). For this discrete time translation u1,2(t + T) are also solutions of (1) and can be expanded in terms of u1,2(t) using (2). This gives uj(T + t) = uj(T)u1(t) + ˙uj(T)u2(t) j = 1, 2 u(t + T) = c1u1(t + T) + c2u2(t + T) The relation u(t + T) = ρu(t) yields then = c1 [u1(T)u1(t) + ˙u1(T)u2(t)] + c2 [u2(T)u1(t) + ˙u2(T)u2(t)] c1 [u1(T)u1(t) + ˙u1(T)u2(t)] + c2 [u2(T)u1(t) + ˙u2(T)u2(t)] = ρ [c1u1(t) + c2u2(t)] Since u1(t) and u2(t) are linearly independent we get c1u1(T) + c2u2(T) = ρc1 c1 ˙u1(T) + c2 ˙u2(T) = ρc2 requiring � ��� u1(T) − ρ u2(T) ˙u1(T) ˙u2(T) − ρ Note: ρ 2 − ρ (u1(T) + ˙u2(T)) � �� � ≡2K • ρ is called the Floquet multipler � � � � = 0 + u1(T) ˙u2(T) − ˙u1(T)u2(T) = 0 � �� � W=1 ρ 2 − 2Kρ + 1 = 0 ρ1,2 = K ± √ K 2 − 1 14

• it is also convenient to introduce α via ρ1 = e iαT which characterizes the Floquet exponent iαT. Because W = 1 one has ρ1ρ2 = 1 implying ρ2 = e −iαT . The evolution of u(t) during one period of Q(t) is therefore determined by K 1. K = 1: ρ1,2 = 1 if K = 1 because ρ = 1 one has a T-periodic solution. 2. K = −1: ρ1,2 = −1 if K = −1 because ρ = −1 one has a 2T-periodic solution u(t + 2T) = −u(t + T) = u(T) 3. |K| < 1: ρ1,2 are complex with |ρ1,2| = 1, α ∈ R 4. |K| > 1: |ρ1| > 1 and |ρ2| < 1 are real and α ∈ iR For |K| �= 1 one obtains two solution u (1,2) (t) satisfying Factoring out the Floquet multiplier e iαT via we get u (1) (t + T) = ρ1u (t) (t) = e iαT u (1) (t) u (2) (t + T) = ρ2u (2) (t) = e −iαT u (2) (t) u (1) (t) = e iαt U (1) (t) U (1) (t + T) = u (1) (t + T)e −iα(t+T) = e iαT u (1) (t)e −iα(t+T) = U (1) (t) i.e. U (1) (t) is T-periodic. Analogously U (1) (t) and U (2) (t) are T-periodic. u (2) (t) = e −iαt U (2) (t) The u (1,2) (t) are independent linear combinations of the two linearly independent solutions u1(t) and u2(t) ⇒ u (1,2) (t) are also linearly independent. More formally: assume there are k1 �= 0 �= k2 with k1u (1) (t) + k2u (2) (t) = 0 for all t k1e iαt U (1) (t) + k2e −iαt U (2) (t) = 0 k1e iα(t+T) U (1) (t + T) + k2e −iα(t+T) U (2) (t + T) = 0 Using U (1,2) (t + T) = U (1.2) (t) non-zero ki requires e iαt U (1) (t)e −iα(t+T) U (2) (t) − e −iαt U (2) (t)e iα(t+T) U (1) (t) = 0 15

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