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Lecture Note Sketches Hermann Riecke - ESAM Home Page

1 0.5 0 -0.5 -1 0 1 2 3 4 5 6 **Note**s: Forcing Subharmonic response (odd) Subharmonic response (even) • in both cases the solutions have a period that is twice as long as that of the forcing: subharmonic response. • δ − δ0 = ǫδ1 + ǫ 2 δ2 + . . . characterizes the detuning between the forcing frequency and the natural frequency of the unforced oscillator • for δ = + 1 the solution contains only sin nt to the order considered, i.e. it is odd, and the solution is even. 2 for δ = −1 2 • for the forcing the transformation ǫ → −ǫ is equivalent to t → t + π . Shifting the time 2 interchanges the even and the odd solution: sin t → cost and cost → − sin t⇒ Thus: by π 2 δeven(ǫ) = δodd(−ǫ) • for any n (or δ0) the Mathieu equation is even under time reversal, i.e. the equation does not change form under the replacement t → −t ≡ ¯t since only even derivatives arise and the coefficients are even in t. I.e. if u(t) is a solution, so is u(−t) d2 dt2u(−t) + (δ + ǫ cos (2t))u(−t) = d2 d(−t) 2u(−t) + (δ + ǫ cos (−2t))u(−t) = d2 d(¯t) 2u(¯t) + (δ + ǫ cos (2¯t)) u(¯t) = 0 • if we were to find a solution u(t) that is of mixed parity (not even nor odd) then – ue ≡ u(t) + u(−t) is even – uo ≡ u(t) − u(−t) is odd and since the Mathieu equation is linear ue and uo are also solutions 10

• we can assume from the start that u(t) is either even or odd and include only sinfunctions or cos-functions **Note**: • the Mathieu equation is linear, therefore the amplitude of the solution is undetermined (in the absence of initial conditions) – we do not need to keep the higher-order homogenous solution since they only change the overall amplitude of the solution. – fix the amplitude of the leading-order solution by a suitable normalization condition �� 2π u(t) costdt = π for δ = +1 2 u(t) sin t dt = π for δ = −1 2 iii) case n = 2, δ = 4 **Note**: �0 2π 0 • the oscillation period is the same as that of the forcing: harmonic response Use the normalization conditions O(1): u0(t) = �� 2π 0 u0(t) sin 2t dt = π � 2π 0 u0(t) cos 2t dt = π � 2π 0 uk≥1(t) sin 2t dt = 0 odd � 2π 0 uk≥1(t) cos2t dt = 0 even � sin 2t odd solution, out of phase with respect to the forcing cos 2t even solution, in phase with respect to the forcing with amplitudes fixed by the normalization Odd solution: O(ǫ): thus **Note**: ü1 + 4u1 = −δ1 sin 2t − cos 2t sin 2t = −δ1 sin 2t − 1 sin 4t 2 δ1 = 0 u1 = 1 sin 4t 24 • because of the normalization we do not keep the homogeneous solution 11

- 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: • case δ1 = + 1 2 � � 1 ü2
- 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 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 (−∞, +∞
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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