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Lecture Note Sketches Hermann Riecke - ESAM Home Page

• A series � n fn (x − x0) αn is called asymptotic to f(x), if for any fixed N � � � �f(x) − � N� n f(x) ∼ � n fn (x − x0) αn fn (x − x0) αn , � � � αN � ≪ |x − x0| � for x → x0 • The series need not be an expansion in integer powers of x − x0. • If a series is asymptotic to f(x) it does not have to converge to f(x) for fixed x when N → ∞. In fact, the asymptotic series typically approaches f(x) as N is increased up to some intermediate value of N and then it diverges when N is increased further. Example 1 Consider small-x behavior of I(x) = � 1 0 The series � 1 converges uniformly to 1 sin xt t an+1 an for all x ∈ [0, 1]. = 0 1 sin xt dt t � 1 1 1 sin xt dt = t 0 t 1 (2n − 1)(2n − 2) x2 < Therefore we integrate term by term to get � 1 0 � xt − 1 3! (xt)3 � + . . . dt x − 1 3! x3 t 2 + 1 5! x5 t 4 + . . . 1 (2n − 1) (2n − 2) 1 1 1 sin xt dt ∼ x − t 3! 3 x3 + 1 1 5! 5 x5 . . . Example 2 I(x) = � ∞ x ta−1 e −t dt Determine the behavior of the incomplete Γ-function for small x � ∞ Γ(a, x) = x t a−1 e −t dt → 0 for n → ∞ We would like to expand the exponential and then integrate term by term. This would lead to terms � ∞ a−1 1 t n! (−t)n x 44

which diverge at the upper limit for large enough n. To treat the upper limit we need to keep the exponential. Depending on a we need to do this in different ways: i) a > 0 For a > 0 the lower limit x → 0 poses no difficulty. Use it to rewrite � ∞ x t a−1 e −t dt = � ∞ 0 = Γ(a) − t a−1 e −t dt − n=0 � x t a−1 0 ∞� (−t) n dt (19) n! n=0 ∞� (−1) n xn+a (n + a) n! since the second term can be integrated term by term ii) a ≡ −ā < 0 but not an integer Now we also have to worry about the divergence at x = 0 in the first term in (19). Divide the integral into two parts: • one in which the upper limit t → ∞ does not require to keep the exponential function include in this integral as many terms of the expansion of the exponential as possible need t −ā+N → 0 for t → ∞ i.e. N < ā • one in which the exponential has to be kept Choose N as the largest integer smaller than ā: Then � ∞ x t a−1 e −t dt = � ∞ t −ā−1 x N < ā < N + 1 N� (−t) n � ∞ dt + t n! −ā−1 n=0 x ∞� n=N+1 By construction the first integral converges at the upper limit t → ∞. � ∞ t −ā−1 x N� (−t) n dt = − n! n=0 N� (−1) n xn−ā (n − ā) n! n=0 (−t) n n! dt The second integral cannot be done term by term because of the upper limit. But it poses no problem at the lower limit t = x → 0: the term in the integrand with the lowest power in t is t −ā−1+N+1 with N − ā > −1 Exploit the integral over (0, ∞) and retain again the exponential function strategically � ∞ t −ā−1 ∞� (−t) n � ∞ dt = t n! −ā−1 � e −t N� (−t) − n � � x dt − t n! −ā−1 ∞� (−t) n n! dt. x n=N+1 0 45 n=0 0 n=N+1

- 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 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: Here we can get an integral express
- 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
- 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
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Analogously for x > xm: 0 = ǫ∂Tx

- Page 97 and 98:
- L = 0 corresponds to a pure gas p