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# Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

## O(ǫ 2 ): avoid secular

O(ǫ 2 ): avoid secular terms Thus Even solution: O(ǫ): leading to O(ǫ 2 ): leading to ü2 + 4u2 = −δ2 sin 2t − cos 2t 1 sin 4t � 24 = − sin 2t +δ2 + 1 � 1 − 2 24 1 1 sin 6t 2 24 δ2 = − 1 48 δ = 4 − 1 48 ǫ2 + . . . u = sin 2t + ǫ 1 sin 4t + . . . 24 ü1 + 4u1 = −δ1 cos 2t − cos 2 2t = −δ1 cos 2tδ1 − 1 1 − cos 4t 2 2 δ1 = 0 u1 = − 1 1 + cos 4t 8 24 � − 1 � 1 + cos 4t 24 ü2 + 4u2 = −δ2 cos 2t − cos 2t 8 � = cos 2t −δ2 + 1 � 1 − − 8 48 1 cos 6t 48 δ2 = 1 1 5 − = 8 48 48 Forcing Strength ε 1 0.5 0 -0.5 -1 n=0 � u = cos 2t + ǫ − 1 � 1 + cos 4t + . . . 8 24 n=1 n=2 0 1 2 3 4 5 Frequency δ Figure 1: Periodic solutions of the Mathieu equation. The growth and decay of solutions away from the lines is discussed later. 12

1.1.2 Floquet Theory In the discussion of the Mathieu equation we only obtained period solutions, which in turn required a specific combination of the forcing and the frequency, δ = δ(ǫ). We would like to have information about the solutions also away from these lines in parameter space. Consider with Q(t) a T-periodic function Notes: • the solutions to (1) need not be periodic: ü + Q(t)u = 0 (1) Q(t + T) = Q(t) – for Q = c > 0 the solutions would be growing and decaying ⇒ expect that this character can persiste even with Q time-dependent – for Q = c < 0 the solutions would be periodic with a period that need not be related to the period T of Q(t) ⇒ expect that the underlying period persists to some extent with Q time-dependent ⇒ expect that quasi-periodic solutions are possible with two incommensurate frequencies • for Q = const. (1) is invariant for translations in time by any amount: for any ∆t the function u(t + ∆t) is a solution if u(t) is a solution • for T-periodic Q(t) eq.(1) is invariant only under translations by an integer multiple of T ⇒ with u(t) also u(t + ∆t) is a solution if ∆t = nT , n integer • for constant coefficients the solution is given by a complex exponential, u ∝ e iαt , α ∈ C. For arbitrary ∆t one has then u(t + ∆t) = e iα∆t u(t) • does one have a similar relation if Q(t) is T-periodic and one considers only shifts ∆t = T? If one considers the discrete temporal evolution u(t = n∆t) with ∆t = T the coefficient is constant, Q(t + nT). We show now: The discrete time translation symmetry together with the linearity of the equation induces a simple relationship between u(t) and u(t + T) u(t + T) = ρu(t) ρ ∈ C The basic idea is to use the general solution u(t) = c1u1(t) + c2u2(t) to express the solution u(t + T) in terms of u1(t) and u2(t). 13

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