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Lecture Note Sketches Hermann Riecke - ESAM Home Page

One can show by other methods that **Note**: I(x) ∼ ln x x x → ∞ • Integration by parts can only generate series in integer powers of x. Whenever the asymptotic series is not a power series integration by parts must fail. Therefore, no kind of integration by parts can work for the Stieltjes integral for large x. 2.3 Laplace Integrals Laplace integrals have the form � b I(x) = f(t)e xφ(t) dt a For large x the contributions to the integral for t near the maximal value(s) of φ(t) dominate exponentially. Therefore it is sufficient to restrict the integral to a vicinity of the maximum (maxima). If φ(t) has a maximum at c ∈ (a, b) consider the approximation � c+ǫ I(x; ǫ) = f(t)e xφ(t) dt If the maximum of φ(t) occurs at a consider the approximation and analogously if the maximum is at b. c−ǫ � a+ǫ I(x; ǫ) = f(t)e xφ(t) dt a To evaluate I(x; ǫ) it is then sufficient to use an approximation for f(t) that is valid in the vicinity of the maximum. For this to work one needs that • the final approximation does not depend on ǫ • the asymptotic expansions of I(x) and I(x; ǫ) are identical This is actually the case, because typically I(x) and I(x; ǫ) differ only by terms that are exponentially small in x. To show this estimate I(x) − I(x; ǫ) and I(x; ǫ). For simplicity assume the maximum occurs at c = 0, which could be inside the interval or at its boundary. 50

First consider maximum at the lower boundary, c = 0 = a, i.e. φ(t) is decreasing from a to b �� � b |I(x) − I(x; ǫ)| = � � f(t)e xφ(t) � �� � � � b � dt� � < exφ(ǫ) � � f(t)dt� � = exφ(c) e x(φ(ǫ)−φ(c)) �� � � b � � � f(t)dt� � Compare with ǫ � ǫ I(x; ǫ) = f(t)e xφ(t) dt 0 In the interval [0, ǫ] φ(t) can be approximated by φ(t) ≈ φ(c) − γtm , γ > 0, and f(t) by f(t) = f(c) + δtm′ . If φ does not have a local maximum at t = c one has m = 1 I(x; ǫ) ≈ e xφ(c) � ǫ 0 � f(c) + δt m′� e −xγt dt = e xφ(c) ǫ � f(c) � −γxǫ 1 − e γx � + δ 1 (−γ) m′ ǫ dm′ dxm′ � 1 � −γxǫ 1 − e γx ��� Independent of the value of m ′ , for large x the terms inside the curly braces amount to polynomial terms in x −1 with exponentially small corrections. Therefore one has I(x, ; ǫ) ≈ C x −ρ e xφ(c) for some ρ > 0 and some x-independent constant C. Thus |I(x) − I(x; ǫ)| |I(x; ǫ| ≈ � � � � b ǫ f(t)dt � � � C x ρ e x(φ(ǫ)−φ(c)) ≈ x → ∞ � � � � b ǫ f(t)dt � � � C x ρ e −ǫγx ���� → exponentially 0 (21) If the c ∈ (a, b) one has a local maximum and m ≥ 2. Then one has to estimate an integral of the form � ǫ � f(c) + δt m′� e −xγtm dt 0 With some more work one finds that they also decay with x less than exponentially and one obtains again an estimate like (21). Example 1: I(x) = � a 0 (1 − t)−1 e −xt dt Here φ(t) = −t. It is maximal at t = 0. Expand therefore f(t) around t = 0. � ǫ I(x; ǫ) ∼ 0 (1 + . . .) e −xt dt = 1 � −ǫx 1 − e x � The term containing ǫ is exponentially smaller than the other term (it is subdominant): Go to higher orders in the expansion � ǫ ∞� I(x; ǫ) ∼ 0 n=0 I(x) ∼ 1 x (−t) n e −xt dt = 51 x → ∞ ∞� � ǫ n=0 0 (−t) n e −xt dt

- 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=
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- 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: • if the integration by parts lea
- 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 (
- 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