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and with **Note**: λ(T) = λ0(T) + ǫλ1(T) + . . ., ω(T) = ω0(T) + ǫω1(T) + . . . ω0(T) = 1 2 λ0(T) 2 • this is not a weakly nonlinear analysis: λ = O(1) • the phase φ evolves on the O(1) time scale, d φ = ω, but the frequency changes on the dt slow time scale as the solution evolves along the family of solutions • in the general case (q �= 0) one would have to introduce a spatial phase θ(x, t) as well Rewrite NLS in terms of ψ rather than Ψ Insert expansion of ψ O(ǫ 0 ) : confirms O(ǫ): Θ = λ(T)θ(x, t) iωψ − i 2 λ2 ∂ 2 Θ ψ − i|ψ|2 ψ = ǫ � −∂Tψ + e −iφ P(Ψ, ∂xΨ, . . . � i ω0 ���� 1 2 λ2 0 ψ0 − i 2 λ2 0 ∂2 Θ ψ0 − i|ψ0| 2 ψ0 = 0 ω0 = 1 2 λ2 0 Lψ1 ≡ iω0ψ1 − i 2 λ20∂ 2 Θψ1 − i � 2|ψ0| 2 ψ1 + ψ 2 0ψ ∗� 1 = −∂Tψ0 − iω1ψ0 + iλ0λ1∂ 2 Θψ0 + e −iφ P In general, the unperturbed soliton ψ0 is part of a four-parameter family of solutions, ψ0 = ψ0(x, t; x0, φ0, λ, q). Here we are keeping the parameter q = 0 fixed. The linear operator L is singular: Take derivatives of the O(1)-equation with respect to the parameters Θ0, φ0: � = i 1 ∂Θ0 � i 1 2 λ2 0 ψ0 − i 2 λ2 0 ∂2 Θ ψ0 − i|ψ0| 2 ψ0 i 1 2 λ2 0ψ0 − i 2 λ2 0∂ 2 Θψ0 − i|ψ0| 2 ψ0 = 0 2 λ20 ∂Θ0ψ0 − i 2 λ20 ∂2 Θ∂Θ0ψ0 − 2i|ψ0| 2 ∂Θ0ψ0 − iψ 2 0∂Θ0ψ ∗ 0 = L (∂θ0ψ0) = 0 82

and **Note**: ∂φ0 � i 1 2 λ20 ψ0 − i 2 λ20 ∂2 Θψ0 − i|ψ0| 2 � ψ0 = L (∂φ0ψ0) = Liψ0 = 0 • thus L is singular and has two eigenvectors with vanishing eigenvalue • these zero eigenvalues arise because the solution ψ0 around which the equations are linearized breaks two continuous symmetries of the original system: Θ → Θ + ∆θ translation symmetry in space φ → φ + ∆φ translation symmetry in time this is a general result: breaking a continuous translation symmetry leads to 0 eigenvalues. The eigenvectors (modes) are called translation modes or Goldstone modes. • breaking a discrete symmetry leads to a discrete family of solutions. This does not imply that the linearization around that solution has a vanishing eigenvalue since one cannot go continuously from one of the solutions in the family to another (no derivative with respect to a continuous parameter can be taken). • the existence of a continuous symmetry ϕ → ϕ+∆ϕ alone does not induce a vanishing eigenvalue: if the solution ψ0 does not break the continuous symmetry, ∂ϕψ0 = 0 and does not represent an eigenvector. • the full solution is part of a four-parameter familty because of the continuous variation of the amplitude and of the velocity of the solution → additional vanishing eigenvalues. it turns out that two of the 0-eigenvalues of iL are associated with proper eigenvectors, whereas the other two have generalized eigenvectors. Ψφ0 : iLiψ0 = 0 Ψx0 : iL∂Θψ0 = 0 Ψλ : iL (Θ∂Θψ0 + ψ0) = iλ 2 0ψ0 Ψq : iLiΘψ0 = −λ 2 0∂Θψ0 Since the linearized operator is singular the equation at O(ǫ) can only be solved if certain solvability conditions are satisfied (Fredholm alternative). To get the solvability conditions we need to project the O(ǫ)-equation onto the relevant left eigenvectors. Projections need a scalar product. For functions the scalar product typically involves some integral over the domain. Here we can make iL (not L, though) self-adjoint5 by a suitable choice of the scalar product. Choose � �� ∞ 〈ψ1, ψ2〉 = ℜ ψ −∞ ∗ 1ψ2dΘ 5**Note** because of the terms iω0 and 1 2iλ20 ∂2 Θ L is not self-adjoint, but suggest that iL may be self-adjoint. 83

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Lecture Note Sketches Perturbation

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4 Fronts and Their Interaction 87 4

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References [1] P. Coullet, C. Elphi

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1.1.1 The Mathieu Equation Consider

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• case δ1 = + 1 2 � � 1 ü2

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• we can assume from the start th

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1.1.2 Floquet Theory In the discuss

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• it is also convenient to introd

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Goal: determine α(δ, ǫ). For a r

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Forcing Strength ε 1 0.5 0 -0.5 n=

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10 5 −10 0 0 −5 t 50 100 150 20

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O(ǫ): with Note: y0(ψ, T) = R(T)

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i.e. Now the nonlinear problem: Fix

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• α < 0: Notes: - in-phase solut

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Since L is singular we expect that

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- 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
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- 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
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- 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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- Page 65 and 66: using � ∞ 0 e−u ln u du = γ
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- 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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- Page 95 and 96: Analogously for x > xm: 0 = ǫ∂Tx
- Page 97 and 98: - L = 0 corresponds to a pure gas p