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Lecture Note Sketches Hermann Riecke - ESAM Home Page

At O(ǫ) we get ∂ˆtu1 − v1 = −∂Tu0 ≡ I1 = −∂TAe iω0 ˆt + c.c ∂ˆtv1 + ω 2 0 u1 = − � ∂Tv0 + βv0 + αu 3 0 − f cosωˆt � ≡ I2 iω0ˆt = e The solvability condition is given by � −iω0∂TA − βiω0A − 3|A| 2 Aα + 1 2 feiΩT � 2π ω 0 0 � 1, − i ω0 � � −iω0ˆt I1 e I2 Only the terms ∝ e iω0 ˆt in Ii contribute to this condition Thus � dˆt = 0 0 = −2∂TA − βA + 3i α |A| ω0 2 A − i fe 2ω0 iΩT ∂TA = − 1 3α βA + i |A| 2 2ω0 2 A − i fe 4ω0 iΩT We can remove the explicit T-dependence in the equation by writing yielding � ∂T A = − 1 � β − iΩ 2 A = Ae iΩT � + [. . .] e 3iω0 ˆt + c.c. A + i 3α |A| 2ω0 2 A − i f (10) 4ω0 To compare with the previous result for the Duffing oscillator introduce amplitude and phase A(T) = 1 2 R(T)e−iφ(T) which results in ∂TR + (iΩ − i∂Tφ)R = − 1 3iα βR + R 2 8ω0 3 − and yields after splitting into real and imaginary parts in agreement with our previous result (7,8). **Note**s: ∂TR = − 1 f βR + sin φ 2 2ω0 ΩR − ∂TφR = 3 α R 8 ω0 3 − f cosφ 2ω0 30 if e 2ω0 iφ � �� � f sin φ− 2ω0 if cos φ 2ω0

• using the other left 0-eigenvector would result in an equivalent solvability condition leading to the complex conjugate of (10). The complex amplitude equation (10) suggests that in general the complex amplitude satisfies an equation of the form A ′ = µA − γ|A| 2 A + νf (11) where µ ≡ µr + iµi, γ ≡ γr + iγi, ν ≡ νr + iνi are complex coefficients. **Note**s: • Comparing (33) with (10) shows that the Duffing oscillator does not lead to the most general amplitude equation for a forced oscillator: – for the Duffing oscillator one has γr = 0: no nonlinear dissipation of the oscillation amplitude, only linear damping ⇒ the saturation of the oscillation amplitude occurs through a change of the natural frequency of the oscillator with increasing amplitude, which renders the forcing less effective with increasing oscillation amplitude • The fact that νr = 0 in (10) is of no significance: for arbitrary ν ≡ ¯νe iδ with ˆν ∈ R replacing A → Âeiδ leads to 1.3 Symmetries 1.3.1 No Forcing A ′ = µA − γ|A| 2 A + ˆνf (12) In the absence of forcing (4) is invariant under arbitrary time translations t → t + ∆t i.e. if y(t) is a solution to (4) so is y(t + ∆t). This invariance must be reflected in the resulting amplitude equation (12). However, A does not depend on t. How is it affected by translations in the fast time t? Consider a solution y(t) and the time-shifted solution y(t + ∆t) and their expansions in terms of complex amplitudes A and B, respectively, The expansion implies: y(t) = A(T)e iωt + c.c. + h.o.t. y(t + ∆t) = B(T)e iωt + c.c. + h.o.t. • if y(t) is a solution of the original equation then A(T) is a solution of the amplitude equation and vice versa 31

- 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 and 12: • we can assume from the start th
- Page 13 and 14: 1.1.2 Floquet Theory In the discuss
- 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: Since L is singular we expect that
- 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 ���
- Page 63 and 64: π i to ensure decay of the exponen
- Page 65 and 66: using � ∞ 0 e−u ln u du = γ
- Page 67 and 68: using the series expansion from the
- Page 69 and 70: Example: Behavior near the saddle p
- Page 71 and 72: Limiting behavior of the contours v
- Page 73 and 74: Write xρ = (X + iY ) ρ ≡ Φ + i
- Page 75 and 76: 3 Nonlinear Schrödinger Equation C
- Page 77 and 78: 3.1 Some Properties of the NLS Cons
- Page 79 and 80: 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