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Lecture Note Sketches Hermann Riecke - ESAM Home Page

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 **Note**s: • 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

- Page 1 and 2: Lecture Note Sketches Perturbation
- Page 3 and 4: 4 Fronts and Their Interaction 87 4
- Page 5 and 6: References [1] P. Coullet, C. Elphi
- Page 7 and 8: 1.1.1 The Mathieu Equation Consider
- Page 9 and 10: • case δ1 = + 1 2 � � 1 ü2
- Page 11: • we can assume from the start th
- Page 15 and 16: • it is also convenient to introd
- Page 17 and 18: Goal: determine α(δ, ǫ). For a r
- Page 19 and 20: Forcing Strength ε 1 0.5 0 -0.5 n=
- Page 21 and 22: 10 5 −10 0 0 −5 t 50 100 150 20
- Page 23 and 24: O(ǫ): with Note: y0(ψ, T) = R(T)
- Page 25 and 26: i.e. Now the nonlinear problem: Fix
- Page 27 and 28: • α < 0: Notes: - in-phase solut
- Page 29 and 30: Since L is singular we expect that
- Page 31 and 32: • using the other left 0-eigenvec
- Page 33 and 34: Thus, m = n + 1 or αmn = 0 Alterna
- Page 35 and 36: • a nonlinear saturating term nee
- Page 37 and 38: • Non-resonant forcing does not i
- Page 39 and 40: with O(1) forcing (cf. in Sec.1.2.1
- Page 41 and 42: Thus, if ∆ > 0 both solutions R2
- Page 43 and 44: Here we can get an integral express
- Page 45 and 46: which diverge at the upper limit fo
- Page 47 and 48: as in the case a > 0. But: now the
- Page 49 and 50: • if the integration by parts lea
- Page 51 and 52: First consider maximum at the lower
- Page 53 and 54: Is this series for I(x; ǫ) asympto
- Page 55 and 56: • since we kept only the first no
- Page 57 and 58: It looks like a case for Watson’s
- Page 59 and 60: Extend integration to (−∞, +∞
- Page 61 and 62: Bound the integral term ���
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π i to ensure decay of the exponen

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using � ∞ 0 e−u ln u du = γ

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using the series expansion from the

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Example: Behavior near the saddle p

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Limiting behavior of the contours v

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Write xρ = (X + iY ) ρ ≡ Φ + i

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3 Nonlinear Schrödinger Equation C

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3.1 Some Properties of the NLS Cons

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Notes: Thus: ˙x = 1 2 m ˙x2 + V (

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• the boost velocity c or the bac

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and Note: ∂φ0 � i 1 2 λ20 ψ0

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• with increasing amplitude the p

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4 Fronts and Their Interaction Cons

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a) • this equation can be read as

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• the coefficient of ψ is chosen

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Using that ψL,R satisfy the O(ǫ 0

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Analogously for x > xm: 0 = ǫ∂Tx

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- L = 0 corresponds to a pure gas p