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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Since sin t = t − 1 6 t3 + 1 120 t5 . . . one might expect t − 1 6 t3 < sin t < t This can be shown to be true for all t > 0. Estimate now Thus, for t = x −α with α > 1 4 � � x sin 2 t − xt 2 � � = x |sin t + t| |sin t − t| < x 2t 1 6 t3 = 1 3 xt4 one has To leading order one then has also and also � x −α � �x sin 2 t − xt 2� � < 1 3 x1−4α → 0 x → ∞ 0 e −x sin2 t ∼ e −xt 2 e −xsin2 t dt ∼ � x −α 0 t ≤ x −α , x → ∞ e −xt2 dt x → ∞ What about the integrals from x−α to ǫ? They are both exponentially small compared to I(x): � ǫ x−α e −x sin2 t −xsin dt < e 2 x−α ǫ � ǫ x −α e −xt2 dt < e −x1−2α ǫ For the latter estimate to be meaningful we need to choose α < 1 2 . Example: Modified Bessel function In(x) = 1 π � π 0 excos t cosnt dt φ(t) has its maximum at t = 0. Try to approximate both cosines by lowest order terms and consider � ǫ e x 1dt = ǫe x (27) Why does this result depend on the artificial parameter ǫ? 0 The maximum at the boundary is at the same time a local maximum: need to keep the quadratic terms of φ(t) (cf. (24)). In(x; ǫ) = 1 π � ǫ 0 = 1 π ex1 2 e x−1 2 xt2 dt � 2π x Now there is no ǫ-dependence any more, as required. Example: I(x) = � ∞ 0 e−1t e−xtdt 56 = 1 √ 2πx e x

It looks like a case for Watson’s lemma with f(t) = e −1 t . However: e −1 t and all its derivatives vanish at t = 0 ⇒ Watson’s lemma would yield I(x) = 0 The dominant contribution to the integral comes from the maximum of the integrand. In this case the maximum depends on x, **Note**: tmax = 1 √ x • a maximum that depends on x is called a movable maximum. • to apply Laplace’s method to a movable maximum transform t into a coordinate in which the maximum is fixed, independent of x. We could introduce a shift of the coordinate s = t − tmax This would lead to complicated expressions in e −1 t . Instead, the maximum can be fixed by a rescaling s = √ xt smax = 1 I(x) = 1 � ∞ √ x 0 e −√ x( 1 s +s) ds Using (26) and noting that √ x plays the role of x in (26) we get and φ ′′ (1) = −2 **Note**: 3 I(x) ∼ 1 � √ x 2π 2 √ x e−2√ x = √ π x 3 4 e −2√ x • I(x) decreases faster than any power. Watson’s lemma can only generate results that depend on x as a power series. 3 Striling’s formula in Hinch p.33 is nice example for movable maximum and for important formula 57

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Lecture Note Sketches Perturbation

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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 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: • since we kept only the first no
- 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 (
- Page 81 and 82: • the boost velocity c or the bac
- Page 83 and 84: and Note: ∂φ0 � i 1 2 λ20 ψ0
- Page 85 and 86: • with increasing amplitude the p
- Page 87 and 88: 4 Fronts and Their Interaction Cons
- Page 89 and 90: a) • this equation can be read as
- Page 91 and 92: • the coefficient of ψ is chosen
- Page 93 and 94: Using that ψL,R satisfy the O(ǫ 0
- Page 95 and 96: Analogously for x > xm: 0 = ǫ∂Tx
- Page 97 and 98: - L = 0 corresponds to a pure gas p