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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Turning points of R∞(Ω) correspond to dΩ dR∞ Amplitude f 12 10 8 6 4 2 f=50 beta=0.5 dΩ1,2 dR∞ 0 -50 0 Frequency 50 = 0 = 3 4 αR∞ ± −2f2 � f2 4 − β2 Figure 7: Dependence of the amplitude of the phase-locked solution on the forcing frequency with damping (f = 50, β 2 = 0.5). 1.2.2 Duffing Oscillator as a System of Equations For simplicity demonstrate the approach with the Duffing oscillator again near the 1:1resonance ¨y + ω 2 0 + ǫ ˙y + ǫαy 3 = ǫf cos ωt ω = ω0 + ǫΩ Rewrite in terms of a system of first-order equations with Thus i.e. � u L v R 2 ∞ u = y v = ˙y 1 R 3 ∞ ˙u − v = 0 ˙v + ω 2 0 u = −ǫ � βv + αu 3 − f cosωt � � � ∂t −1 ≡ Expanding as usually � u v ω 2 0 ∂t � � � u v � U = A(T) V � � = 0 −ǫ {βv + αu 3 − f cosωt} � e iω0t + O(ǫ) + c.c. We need to determine an evoluation equation for the complex amplitude A(T). We have from the terms proportional to eiω0t � � � U iω0 L0 ≡ V � � � −1 U = 0 V ⇒ ω 2 0 iω0 28 � U V � � � � 1 = iω0

Since L is singular we expect that at O(ǫ) secular terms will arise, which will imply a solvability condition. Previously, this condition simply amounted to setting the terms proportional to cosω0t and to sin ω0t to 0. Dealing with a system of equations it seems at first glance as if there are too many solvability conditions for this single complex amplitude: there are two complex equations for each Fourier mode. We need to invoke explicitly the Fredholm Alternative Theorem, for which we need the left zero-eigenvector (U + (t), V + (t)) of L. For that we need to define a scalar product. For yi(t) ≡ (ui(t), vi(t)) choose 〈y1(t),y2(t)〉 ≡ � 2π ω 0 The left zero-eigenvector (U + (t), V + (t)) is defined by for any � u(t) v(t) � . Use integration by parts � 2π ω 0 0 0 = = = Thus � U + **Note**: 0 � � � + + ∗ ∂t −1 U (t), V (t) ω2 0 ∂t � � (u1(t) ∗ , v1(t) ∗ � u2(t) ) v2(t) � � u(t) v(t) U +∗ (∂ˆtu − v) + V +∗ � ω 2 0 u + ∂ˆtv � dˆt � dt � = 0 −u∂ˆtU +∗ − U +∗ V + V +∗ ω 2 0 U − V ∂ˆtV +∗ dˆt � �� −∂ˆt ω 2 0 V + � = −1 −∂ˆt � U + 0 V + 0 � � U + V + � ∗� t � u(ˆt) v(ˆt) � e ±iω0ˆt � 1 = ± i ω0 � dˆt � ±iω0ˆt e • L is not symmetric, therefore the left and the right 0-eigenvectors differ from each other. • There are two left 0-eigenvectors, which are complex conjugate of each other because the original equation is real. Including the slow time T the expanded Duffing equations becomes ∂ˆtu − v = −ǫ∂Tu ∂ˆtv + ω 2 0u = −ǫ � ∂Tv + βv + αu 3 − f cosωˆt � 29

- 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: • α < 0: Notes: - in-phase solut
- 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 ���
- 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
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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